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https://www.reddit.com/r/learnmath/comments/9k84dd/discrete_math_prove_ab_divisible_by_n_a_mod_n_b/
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[Discrete Math] Prove a-b divisible by n <- > a mod n = b mod n : r/learnmath
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•7 yr. ago
humbleharbinger
[Discrete Math] Prove a-b divisible by n <- > a mod n = b mod n
RESOLVED
I'm trying to understand this proof. I can prove that a mod n = b mod n - > a-b is divisible by n. But I can't prove the other way. My main question is how do we find out that "the remainder when a-b is divided by n is the same as the remainder when r1 - r2 is divided by n." and then how do we find out that the remainder of (r1 - r2) /n is 0?
The proof is as follows...
Proof Assume n divides a-b and prove a mod n = b mod n.
We know we can write a = q1 n + r1 and b = q2 n + r2, with remainders r1 and r2 both between 0 and n.
Then a-b = (q1 - q2) n + (r1 - r2).
Because n goes evenly into (q1 - q2) n, the remainder when a-b is divided by n is the same as the remainder when r1 - r2 is divided by n.
Since a-b is divisible by n, the remainder when r1 - r2 is divided by n must be 0. So r1 - r2 is a multiple of n.
But r1 and r2 are both numbers between 0 and n, so the only way r1 - r2 can be an even multiple of n is for it to equal 0 n = 0.
So r1 = r2 and a mod n = b mod n.
The proof is complete.
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What is the meaning of the Chinese character 天?
·
The Chinese character 天 (tiān) means "heaven" or "sky." It can also imply "day" in certain contexts. In traditional Chinese culture, 天 is associated with the cosmos and the natural order, often representing a higher power or divine force. The character is also used in various compound words and phrases, such as 天气 (tiānqì), which means "weather."
Huang Yi
Knows Mandarin Chinese
·
Author has 1.6K answers and 935.6K answer views
·
5y
天 means “sky” and “day”. And 天 was believed as the symbol of the rule of the world, the “heaven”.
For example, 天 in 天气 (weather) is about the sky; and in 天天 (everyday) it’s about the day. The emperors told people that they’re “天子”, the sons of the Heaven.
Qing
laugh everyday
·
2y
Originally Answered: What is the meaning of the Chinese character "天"?
·
1.sky:顶天立地
2.At the top:天窗，天桥
3.A period of 24 hours in a day and night：今天/天越来越长了。
4.Days：三天三夜/忙了一天了，早点儿休息吧。
5.A certain time of day：天儿还早呢。
6.Season：春天，夏天
7.Weather：天晴了。
8.Natural：天生的，天性
9.Nature：天灾
10.The surname
Related questions
What does "天 问 一 号" mean?
What does the Chinese character 祿 mean?
What is the stroke order of the Chinese character "天"?
Why is the character 天 written differently in Chinese and Japanese? (see details)
What do these three Chinese characters mean?
Sylvia Lee
Lives in China
·
5y
天 / tiān /
名词 noun
1 （天空） sky; heaven:
under the open sky; in the open;露天
sail the blue heavens在蓝天中翱翔
2 （一昼夜; 白天） day:
in two or three days;过两天
a few days ago;前几天
3 （一天里的某一段时间） a period of time in a day:
around four in the morning; at dawn;五更天
It's still early.天儿还早呢。
4 （季节） season:
summer;夏天
the rainy season黄梅天
5 （天气） weather:
overcast; cloudy day;阴天
wet weather;雨天
6 （自然） nature:
combat nature; conquer nature;战天斗地
Man will conquer nature.人定胜天。
7 （迷信者指自然界的主宰者; 造物） God; Heaven:
Good Heavens!天哪!
God knows!天知道!
8 （迷信者指神佛仙人的处所） Heaven:
go to Heaven; die上西天
9 （姓氏） a surname:
Tian Gao天高
形容词adjective
1 （位置在顶部的; 凌空架设的） overhead:
c
天 / tiān /
名词 noun
1 （天空） sky; heaven:
under the open sky; in the open;露天
sail the blue heavens在蓝天中翱翔
2 （一昼夜; 白天） day:
in two or three days;过两天
a few days ago;前几天
3 （一天里的某一段时间） a period of time in a day:
around four in the morning; at dawn;五更天
It's still early.天儿还早呢。
4 （季节） season:
summer;夏天
the rainy season黄梅天
5 （天气） weather:
overcast; cloudy day;阴天
wet weather;雨天
6 （自然） nature:
combat nature; conquer nature;战天斗地
Man will conquer nature.人定胜天。
7 （迷信者指自然界的主宰者; 造物） God; Heaven:
Good Heavens!天哪!
God knows!天知道!
8 （迷信者指神佛仙人的处所） Heaven:
go to Heaven; die上西天
9 （姓氏） a surname:
Tian Gao天高
形容词adjective
1 （位置在顶部的; 凌空架设的） overhead:
ceiling;天棚
skylight;天窗
2 （天然的; 天生的） natural; innate:
natural instincts;天性
natural calamity天灾
以上来源于：《新汉英大辞典》
reference to New Age Chinese-English Dictionary
MM8787v5
It is used in the industry
·
Author has 305 answers and 60.9K answer views
·
Updated 2y
Originally Answered: What is the meaning of the Chinese character "天"?
·
The Chinese character "天" is a noun. It simply means “sky”.
Another common meaning is ‘day’. One day is “一天", where “一" is “one”.
“天" can be used with the Chinese character, “下", which means “under”, to form the Chinese word “天下", which means the world (under the sky, literally).
It is also used to express surprise at times. An example is “我的天”, which means “Oh my god”.
If you would like to learn how to write this Chinese character, here is an interesting step-by-step guide:
The simple Traditional Chinese Character meaning "Sky"
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The Chairman's Bao
Creator of World Leading Chinese Language App
·
Author has 140 answers and 281.4K answer views
·
5y
On its own 天 means ”sky”, “heaven” or “day” and within these categories, you’ll find other related words.
天空 (sky), 天堂 (heaven), 天文 (astronomy), 天下 (everything under the sky/the whole world/ the whole of China), 天气 (weather),今天 (today), 昨天 (yesterday), 明天 (tomorrow) etc
It is also used in phrases such as 天啊 and 天知道 which can be translated to “God Knows!” or “Good Heavens!”
Related questions
What is the meaning of the Chinese character “th”?
What do these Chinese characters mean/say?
Can you provide the meaning of the Chinese character "雞"?
How do I write the Chinese characters 天 well?
Why do Chinese characters avoid using circular shapes, and what does this mean for characters like 中 and 天?
Alex Li
Bachelor from China University of Geosciences (Wuhan) (Graduated 2002)
·
5y
day. e.g. 一天 = one day
the sky. e.g. 蓝天 = blue sky
god (in Chinese sense). e.g. 我的天 = my god
these are the most frequently used meanings of this character.
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Hoong-Chien Lee
Knows Chinese
·
5y
The character means heaven, the physical one as well as the spiritual one. It also means day. 一天 （yi tian), one day; 今天 (jin tian), today; etc.
Stefan Wendeler
Knows Mandarin Chinese
·
5y
Instead of answering the question (I mean just Google it), I'll recommend downloading the “Pleco” dictionary app. You can hand-write characters, type pinyin or search for a chinese translation from English.
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Sean Xiong
Knows Chinese
·
8y
Related
How do I write the Chinese characters 天 well?
The correct sequence of writing 天 is：
The upper horizontal line (it is called 橫)
The lower horizontal line
The 丿 (it is called 撇)
The ㇏ (it is called 捺)
There are only four strokes in 天
For the two horizontal lines, they are written from left to right. They’re not absolutely horizontal. They have a little gradient and the right part is gradually slightly higher than the left part.
The lower part of 天 is not exactly 人. The left part of 人 is like a ノ. However the left part of 人 in 天 is like 丿，there is a vertical part on the top.
The only difference between 天 and 夭 is the first stroke. The first stroke o
The correct sequence of writing 天 is：
The upper horizontal line (it is called 橫)
The lower horizontal line
The 丿 (it is called 撇)
The ㇏ (it is called 捺)
There are only four strokes in 天
For the two horizontal lines, they are written from left to right. They’re not absolutely horizontal. They have a little gradient and the right part is gradually slightly higher than the left part.
The lower part of 天 is not exactly 人. The left part of 人 is like a ノ. However the left part of 人 in 天 is like 丿，there is a vertical part on the top.
The only difference between 天 and 夭 is the first stroke. The first stroke of 天 is quite straight and written from left to right. The first stroke of 夭 is a little bit curved and written from right to left.
The attached image may help you to understand what I said.
Peter Tan
Knows Chinese
·
Author has 2K answers and 830K answer views
·
May 10
Related
Why do Chinese characters avoid using circular shapes, and what does this mean for characters like 中 and 天?
Chinese characters are pictographs, they are picture-like and in the beginning they were pictures like these
These are 中天圆袁猿楚初員衣地。These “characters” are hand written so some that appear round may be square and some square may be round. Some are reduced to a horizontal line。Maybe some are there but could not be seen.
Why do Chinese characters avoid using circular shapes, and what does this mean for characters like 中 and 天?
Probably because of “yi”, it is for you 衣 which we wear for the coverings of those who are of the 蠃luo species the 袁 which clearly has 衣 in it。Notice the hand holding 衣 in袁 in t
Chinese characters are pictographs, they are picture-like and in the beginning they were pictures like these
These are 中天圆袁猿楚初員衣地。These “characters” are hand written so some that appear round may be square and some square may be round. Some are reduced to a horizontal line。Maybe some are there but could not be seen.
Why do Chinese characters avoid using circular shapes, and what does this mean for characters like 中 and 天?
Probably because of “yi”, it is for you 衣 which we wear for the coverings of those who are of the 蠃luo species the 袁 which clearly has 衣 in it。Notice the hand holding 衣 in袁 in the Oracle scripts and the circle in the 楚 script for 袁？
There are three possible explanations
The round circle is a pronunciation hint.
The circle is the head (view from top)
It is the the man who was holding the 衣。
so Why do Chinese characters avoid using circular shapes？
aVoid? Void is 空白 which is empty. To avoid this， what is there but is not seen is “written” as a horizontal line and what is a circle is written as a square. Of course this is just a “hypothesis”, maybe we could know when we see and know the word from its beginning.
Or maybe because a circle is a “character” that has no beginning and has no ending. But even the Great I’Am is the Alpha and Omega，the first and the last, the Aleph and Tav of which there is no equivalent in Chinese (I think)。
Kirby Cho
Knows Korean
·
Author has 2.2K answers and 15.9M answer views
·
Updated 2y
Related
What does the Chinese character "제" mean?
Well, there are 23 Hanja characters (with different meanings) read as “제” /jé/.
Typed version
[Disclaimer: I will be using the Korean semantics, so the definition may differ from the original Chinese definition.]
濟 (제) /jé/:
The semantic word is “건너다”/gŏn nŏ da/ which means [transitive] “to cross (a street, river, ridge, etc.)”, “to be moved from one side to another”
制 (제) /jé/:
The semantic word is “억제(抑制)하다” /ŏk jé ha da/ means “to suppress”, “to control”, “to repress”, “to restrain”
題 (제) /jé/:
The semantic words are “표제(標題/ 表題)” /pyo jé/ meaning “title”, “heading” or
“제목(題目)” /jé mok/ meaning “a ti
Well, there are 23 Hanja characters (with different meanings) read as “제” /jé/.
Typed version
[Disclaimer: I will be using the Korean semantics, so the definition may differ from the original Chinese definition.]
濟 (제) /jé/:
The semantic word is “건너다”/gŏn nŏ da/ which means [transitive] “to cross (a street, river, ridge, etc.)”, “to be moved from one side to another”
制 (제) /jé/:
The semantic word is “억제(抑制)하다” /ŏk jé ha da/ means “to suppress”, “to control”, “to repress”, “to restrain”
題 (제) /jé/:
The semantic words are “표제(標題/ 表題)” /pyo jé/ meaning “title”, “heading” or
“제목(題目)” /jé mok/ meaning “a title of a book or other work.”
製 (제) /jé/:
The semantic words are “만들다” /man dŭl da/ meaning “to make”, “to establish” or
“짓다” /jit da/ meaning [transitive] “to make something (food, a house, etc.)”, “to build”
第 (제) /jé/:
The semantic word is “차례(次例)” /cha ryé/ meaning “order”, “sequence”; “turn (in a game, line, etc.)”
際 (제) /jé/:
The semantic word is “즈음” /jŭ-ŭm/ meaning “the time”, “an occasion”
“때” /ttaé/ meaning “time”, “moment”
“기회(機會)” /gi hoé/ meaning “opportunity”
諸 (제) /jé/:
The semantic word is “모든” /mo dŭn/ meaning “all”, “every”, “each”
提 (제) /jé/:
The semantic word is “끌다” /kkŭl da/ meaning “to pull”, “to draw”, “to drag”
除 (제) /jé/:
The semantic word is “덜다” /dŏl da/ meaning “to subtract”, “to cut down”
“없애다” /ŏps aé da/ meaning “to eliminate”
“나누다” /na nu da/ meaning “to divide (into pieces)”, “to split up”, “to share”
帝 (제) /jé/:
The semantic word is “임금” meaning “monarch”
“황제(皇帝)” /hwang jé/ meaning “emperor”
堤 (제) /jé/:
The semantic word is “방죽” /bang juk/ meaning “bank”, “embankment”, “dike”
“둑” /duk/ meaning “bank”, “embankment”, “dike”
劑 (제) /jé/:
The semantic word is “약제(藥劑)” /yak jé/ meaning “medicine”, “drugs”, “chemicals”
祭 (제) /jé/:
The semantic word is “제사(祭祀)” /jé sa/ meaning “sacrificial rites in honor of an ancestor or god.”
弟 (제) /jé/:
The semantic word is “아우” /a u/ meaning “one’s younger brother”
齊 (제) /jé/:
The semantic word is “가지런하다” /ga ji rŏn ha da/ meaning “arrange evenly”, “arrange orderly”
悌 (제) /jé/:
The semantic word is “공경(恭敬)하다” /gong gyŏng ha da/ meaning “respect”, “honor”, “(formal) revere”
梯 (제) /jé/:
The semantic word is “사다리” /sa da ri/ meaning “ladder”, “climbing tool”
霽 (제) /jé/:
The semantic word is “비가 개다” /bi ga gaé da/ meaning “to clear up the rain”
“비가 그치다” /bi ga gŭ chi da/ meaning “to cease the rain”
蹄 (제) /jé/:
The semantic word is “발굽” /bal gup/ meaning “hoof (pl. hooves)”
“발길질하다” /bal gil jil ha da/ meaning “to (give a) kick”
啼 (제) /jé/:
The semantic word is “울다” /ul da/ meaning “to cry”, “to weep”
臍 (제) /jé/:
The semantic word is “배꼽” /baé kkop/ meaning “(anatomy) navel”, “belly button”
薺 (제) /jé/:
The semantic word is “냉이” /naéng-i/ meaning “shepherd's purse (Capsella bursa-pastoris)”
醍 (제) /jé/:
The semantic word is “제호(醍醐)” /jé ho/ meaning “the fifth and last flavor according to the Nirvana Sutra, sarpir-maṇḍa (in Sanskrit)”
“맑은 술” /malg-ŭn sul/ meaning “clear alcohol”
I hope that this helps?
Dachuan Yang
B.A. in Food Science, Shandong Agricultural University
·
8y
Related
How do I write the Chinese characters 天 well?
As long as the stroke of Chinese characters are complex and not related to pronunciation , it's a barrier for beginners.
In fact there was four fundamental strokes that you will be aware of the differences between 天and 夭 after you grasp them.It's Heng,Shu,Pie,Na.
For beginners who learn the Chinese handwriting by pen，These strokes can be simplified as four kinds of short segments which have their own specific direction .Heng and Shu are perpendicular and Pie and Na both have a angle with them,in different character the angles are different.And as a difference between them,Pie points to the lower
As long as the stroke of Chinese characters are complex and not related to pronunciation , it's a barrier for beginners.
In fact there was four fundamental strokes that you will be aware of the differences between 天and 夭 after you grasp them.It's Heng,Shu,Pie,Na.
For beginners who learn the Chinese handwriting by pen，These strokes can be simplified as four kinds of short segments which have their own specific direction .Heng and Shu are perpendicular and Pie and Na both have a angle with them,in different character the angles are different.And as a difference between them,Pie points to the lower left corner and Na points to the right one.
Then we back to 天,smart reader would found that the 天was composed of two Heng,a Pie and a Na.And the top of 夭 is a Pie with small angle.
At the end of the answer,I'd like to show my new year’s greetings to all of you by Chinese characters, happy new year!
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Polynomial
A polynomial is a function in one or more variables that consists of a sum of variables raised to nonnegative, integral powers and multiplied by coefficients from a predetermined set (usually the set of integers; rational, real or complex numbers; but in abstract algebra often an arbitrary field). Note that a constant is also a polynomial.
For example, these are polynomials:
, in the variable
, in the variables and
, in the variable
, in the variable
, in any variable
, in the variable
, in the variable
However,
are functions, but not polynomials, in the variable
Contents
1 Introductory Topics
1.1 A More Precise Definition
1.2 The Degree of a Polynomial
1.3 Finding Roots of Polynomials
1.3.1 What is a root?
1.3.2 The Fundamental Theorem of Algebra
1.3.3 Factoring
1.3.4 The Rational Root Theorem
1.3.5 Descarte's Law of Signs
1.4 Binomial Theorem
1.5 Special Values
2 Intermediate Topics
3 Olympiad Topics
4 See also
Introductory Topics
A More Precise Definition
A polynomial in one variable is a function . Here, is the th coefficient and . Often, the leading coefficient of a polynomial will be equal to 1. In this case, we say we have a monic polynomial.
The Degree of a Polynomial
The simplest piece of information that one can have about a polynomial of one variable is the highest power of the variable which appears in the polynomial. This number is known as the degree of the polynomial and is written . For instance, and . When a polynomial is written in the form with , the integer is the degree of the polynomial.
The degree, together with the coefficient of the largest term, provides a surprisingly large amount of information about the polynomial: how it behaves in the limit as the variable grows very large (either in the positive or negative direction) and how many roots it has.
Finding Roots of Polynomials
What is a root?
A root is a value for a variable that will make the polynomial equal zero. For an example, 2 is a root of because . For some polynomials, you can easily set the polynomial equal to zero and solve or otherwise find roots, but in some cases it is much more complicated.
The Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra states that any polynomial with complex coefficients can be written as
where is a constant, the are (not necessarily distinct) complex numbers and is the degree of the polynomial in exactly one way (not counting re-arrangements of the terms of the product). It's very easy to find the roots of a polynomial in this form because the roots will be . This also tells us that the degree of a given polynomial is at least as large as the number of distinct roots of that polynomial. In quadratics roots are more complex and can simply be the square root of a prime number.
Factoring
Different methods of factoring can help find roots of polynomials. Consider this polynomial:
This polynomial easily factors to:
Now, the roots of the polynomial are clearly -3, -2, and 2.
The Rational Root Theorem
We are often interested in finding the roots of polynomials with integral coefficients. Consider such a polynomial . The Rational Root Theorem states that if has a rational root and this fraction is fully reduced, then is a divisor of and is a divisor of . This is convenient because it means we must check only a small number of cases to find all rational roots of many polynomials. It is also especially convenient when dealing with monic polynomials.
Descarte's Law of Signs
By the Fundamental Theorem of Algebra, the maximum number of distinct factors (not necessarily real) of a polynomial of degree n is n. This tells us nothing about whether or not these roots are positive or negative. Descarte's Rule of Signs says that for a polynomial , the number of positive roots to the equation is equal to the number of sign changes in the coefficients of the polynomial, or is less than that number by a multiple of 2. The number of negative roots to the equation is the number of sign changes in the coefficients of , or is less than that by a multiple of 2.
Binomial Theorem
The Binomial Theorem can be very useful for factoring and expanding polynomials.
Special Values
Given the coefficients of a polynomial, it is very easy to figure out the value of the polynomial on different inputs. In some cases, the reverse is also true. The most obvious example is also the simplest: for any polynomial , so the value of a polynomial at 0 is also the constant coefficient.
Similarly, , so the value at 1 is equal to the sum of the coefficients.
In fact, the value at any point gives us a linear equation in the coefficients of the polynomial. We can solve this system and find a unique solution when we have as many equations as we do coefficients. Thus, given the value of a polynomial and different points, we can always find the coefficients of the polynomial.
Intermediate Topics
Complex numbers
Fundamental Theorem of Algebra
Roots of unity
Olympiad Topics
Vieta's formulas
Newton's identities
Newton's sums
See also
Algebra
Retrieved from "
Categories:
Algebra
Polynomials
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Operations Research
Maximizing a piecewise-linear convex function
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I am working on an optimization problem where some of the terms of the objective function to maximize are expressed as a piecewise linear function of variables:
z={c−x,c+x,x≤0x>0
as depicted below
When I have c−≥c+, I can solve the problem by adding a new variable x′, and two constraints:
x′≤c−x
x′≤c+x
But what do I do when c−<c+ ? I don't think there is a way to express the problem as a linear programming problem in that case, is there ?
I have heard about SOS constraints. Are they the canonical way to solve this kind of problem ? If my problem contains many such piecewise linear functions, is it reasonable to expect a solver from being able to solve such a problem with thousands of SOS1 constraints ?
Example
maximizesubject toamax(x,0)+bmin(x,0)+cmax(y,0)+dmin(y,0)x,y∈[−1,1]
solver
constraint
linearization
nonconvex-programming
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edited Mar 11, 2021 at 7:50
Rodrigo de Azevedo
30922 silver badges1111 bronze badges
asked Mar 11, 2021 at 7:39
lovasoalovasoa
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1
1
These things are difficult to predict. Best advice is: try it out. In general: if I am developing a large model and I did not try out many variants and formulations, I did not do my job.
– Erwin Kalvelagen
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Mar 12, 2021 at 17:13
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within CPLEX you can use piecewise linear function in the objective with all APIs.
Let me show how to do that in OPL. Let me use the example from Making Optimization Simple
Piecewise linear function
```
/
let s now deal with a new information : for a given bus size if we take more than 4 then we get a 20% discount.
This moves our function cost from linear to piecewise linear. Let's have a look at how to deal with this in OPL.
We would then write:
/
int nbKids=300;
float costBus40=500;
float costBus30=400;
// If we take more than 4 buses for a given size, 20% cheaper for additional buses
dvar int+ nbBus40;
dvar int+ nbBus30;
pwlFunction pricePerQuantity=piecewise{1->4;0.8};
assert forall(k in 1..4 ) pricePerQuantity(k)==k;
assert forall(k in 5..10) as:abs(pricePerQuantity(k)-4-(k-4)0.8)<=0.00001;
minimize
costBus40pricePerQuantity(nbBus40) +pricePerQuantity(nbBus30)costBus30;
subject to
{
40nbBus40+nbBus3030>=nbKids;
}
```
NB: I am an IBMer
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answered Mar 11, 2021 at 9:05
Alex FleischerAlex Fleischer
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In your example, the optimization domain is still convex, isn't it ? This is the equivalent of the case where c−≥c+ in my original question, which can be solved with traditional linear programming. Would your example work as well and give an optimal solution in a reasonable amount of time if taking more buses had an extra fare associated to it ?
– lovasoa
Commented
Mar 11, 2021 at 10:12
1
Hi, if not convex it will work. Could be slower sometimes.
– Alex Fleischer
Commented
Mar 11, 2021 at 10:55
Do you know what cplex uses to be able to solve this class of problems ? If there is a large number of places where the optimization domain is not convex, how can it find an optimal solution in a reasonable amount of time ?
– lovasoa
Commented
Mar 12, 2021 at 10:40
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Home > Lymph nodes & spleen, nonlymphoma > Anatomy & histology-lymph nodes
Lymph nodes & spleen, nonlymphoma
Lymph nodes-general
Anatomy & histology-lymph nodes
Authors:Nikhil Sangle, M.D., Nat Pernick, M.D.
Last author update: 1 November 2013
Last staff update: 24 July 2025 (update in progress)
Copyright: 2003-2025, PathologyOutlines.com, Inc.
PubMed Search: Lymph nodes [title] anatomy [title]
Page views in 2025 to date: 19,613
Table of Contents
Definition / general | Terminology | Embryology | Age related changes | Diagrams / tables | Gross description | Microscopic (histologic) description | Microscopic (histologic) images | Positive stains | Negative stains | Electron microscopy images
Cite this page: Sangle N, Pernick N. Anatomy & histology-lymph nodes. PathologyOutlines.com website. Accessed August 23rd, 2025.
Definition / general
A secondary lymphoid organ, where B and T cells proliferate in response to exogenous antigen; primary lymphoid organs are bone marrow and thymus
Other secondary lymphoid organs are spleen and Peyer patches
Tertiary lymphoid organs are tissues with few lymphocytes that recruit more when inflammation is present
Lymph nodes are organized to detect and inactivate foreign antigens present in lymph fluid that drains skin, GI tract and respiratory tract, the major organs in contact with the environment
Terminology
Afferent lymph vessels:
Penetrate capsule, enter marginal sinus, communicate with intranodal sinuses, then become efferent vessels, which lack an endothelial lining
Intranodal vessels contain littoral cells or histiocytes with phagocytic properties
Capsule:
Thin fibrous connective tissue covering of lymph node
May be thicker at hilus
Connected to fibrous trabeculae which penetrate the node
Capsule may contain smooth muscle cells
(Anat Rec 1975;183:517)
Cortex:
Subcapsular portion of node with largest number of follicles (primary or secondary)
Primary follicle:
Round aggregates of small, dark staining inactive (naïve) B lymphocytes, usually near the capsule, within a network of follicular dendritic cell processes
No germinal center present
Secondary follicle:
Arises from primary follicle that develops germinal centers (see below) due to antigenic stimulation of B cells and production of antibodies
Contains pale staining germinal center which may be polarized towards site of antigen entry
Surrounded by mantle zone and marginal zone lymphocytes
Germinal center:
Contains predominantly B lymphocytes (including centroblasts and centrocytes) and scattered follicular T helper cells and T regs
Also tingible body macrophages and follicular dendritic cells
Mantle zone:
Tightly packed small B lymphocytes of the primary follicles, pushed aside by the germinal centers
Marginal zone:
Less packed small B lymphocytes with more cytoplasm
Light zone on outer rim of mantle zone
Contains a mix of post-follicular memory B cells derived after stimulation of recirculating cells from T cell dependent antigen and naïve B cells
Often not well developed in lymph nodes
Paracortex:
Tissue between cortical follicles and medulla (see below)
Contains predominantly dark staining mature T cells, B immunoblasts, interdigitating dendritic cells, plasmacytoid dendritic cells, histiocytes and high endothelial venules (postcapillary venules lined by plump endothelial cells that express leukocyte adhesion molecules and contain intraluminal lymphocytes)
Expands during cell mediated immunological reactions
Has coarse network of reticulin fibers
Medulla:
Portion of node closest to hilum
Contains the medullary cords, sinuses and vessels but minimal number of follicles
Medullary cords:
Found in hilar region between the sinuses, composed mostly of small B and T lymphocytes, plasmacytoid lymphocytes, plasmablasts and plasma cells
Sinuses:
Carry lymph from afferent to efferent lymphatics
Subcapsular sinus is below capsule and partially lined by endothelium
Becomes "medullary" as it approaches the hilum and is lined by macrophages
Also contains mast cells and plasma cells
Vessels:
Blood enters and leaves lymph node at hilus
Embryology
Develop from lateral plate mesoderm (on either side of intermediate mesoderm)
First, lymphatic sacs arise from endothelial outgrowths of large central veins at week 5
Second, lymphatic plexus develops from lymphatic sacs
Third, plexuses are invaded by mesenchymal cells that proliferate and aggregate to form lymph nodes
Small collections of lymphoblasts are present by first trimester
By second trimester, cortex is distinguishable from medulla and primary follicles are present
References: Martini: Human Anatomy, 9th Edition, 2017
Age related changes
Germinal centers are more common in infants and children, decrease in young adults, often absent in elderly (Am J Pathol 1975;78:7)
Germinal centers are more common in mesenteric and cervical lymph nodes ( J Clin Pathol 1980;33:454)
Hyaline deposits increased with age
Peripheral lymph nodes, with little antigenic stimulation, often have replacement by fat, particularly in axillary, cubital and popliteal nodes
Lymphocyte depletion, fibrosis and hyaline deposits are associated with chronic disease, particularly cancer
Diagrams / tables
Images hosted on other servers:
Lymph node structures
Lymph node architecture
Gross description
Ovoid with gray-tan cut surface
Microscopic (histologic) description
At low power, lymph node structures are capsule, cortex and medulla, follicles, paracortex, sinuses
Germinal center: round / oval zone containing pale staining cells, surrounded by darker cells
Mantle zone: small unchallenged B cells surrounding pale staining germinal centers
Marginal zone: light zone surrounding follicles; contains postfollicular memory B cells derived after stimulation of recirculating cells from T cell dependent antigen; named "marginal cells" due to location at interface of lymphoid white pulp and nonlymphoid red pulp in the spleen; however, marginal zone is rarely seen except in mesenteric nodes (APMIS 2002;110:325)
Sinuses: direct the flow of lymph from the afferent lymphatics, to the subcapsular sinus, to the trabecular sinus, to the medullary sinus, to the efferent lymphatics (see diagrams) (Toxicol Pathol 2006;34:409)
Centroblasts:
Large noncleaved follicular center cells (B cells) with moderate amounts of basophilic cytoplasm, large round nuclei, open chromatin, multiple peripheral nucleoli
Frequent mitotic figures
Centrocytes:
Large and small cleaved follicular center cells (B cells) with scant cytoplasm and inconspicuous nucleoli
Immunoblasts:
Large B cells scattered throughout the paracortex
Intermediate between small B cell and a plasma cell
Prominent single nucleoli
Express B cell markers (CD20, CD79a, PAX5) and CD30
Macrophages:
Process antigens via phagocytosis
Related to circulatory monocytes
Are present throughout the lymph node
May contain thyroglobulin in lymph nodes draining thyroid tumors (J Clin Pathol 2001;54:314)
Abundant cytoplasm with medium to large nuclei with vesicular chromatin
Tingible body macrophages have clear cytoplasm and contain apoptotic bodies, which gives node a starry sky pattern
Mast cells:
Present in T cell areas (World J Surg Oncol 2003;1:25)
Difficult to detect
Distinct cytoplasmic boundaries, faintly granular cytoplasm, large pale nuclei
Some cells are elongated and resemble fibroblasts
NK cells:
Distinct group of non T, non B lymphocytes (5 - 10% of peripheral blood lymphocytes) with large granular lymphocyte morphology on Wright-Giemsa stains
NK cells derive from a common lymphoid progenitor with T cells
First line of defense against various infections, by recognizing and killing target cells and producing cytokines, particularly interferon-gamma, which enhance the innate immune response
Capable of lysing certain target cells (virally infected and tumor cells) without prior activation or major histocompatibility complex restriction (hence named "natural killers" that are part of "innate" immune system) (Wikipedia: Natural killer cell [Accessed 26 March 2021])
Do not rearrange their receptor genes, as B / T cells do but rely on a fixed number of NK cell receptors (inhibitory and activating) that recognize MHC class I and class I-like molecules and other ligands
Appear to have capability for memory-like responses (EMBO Rep 2009;10:1103)
Important for immunomodulation and regulation of hematopoiesis
Plasma cells:
Abundant basophilic cytoplasm (due to high content of rough endoplasmic reticulum) with paranuclear hof (highlighted by Giemsa stain, due to Golgi apparatus)
Have eccentrically placed nucleus with spoke wheel (clock face) chromatin due to small clumps of chromatin on nuclear membrane in an otherwise round and clear nucleus
May have Russell bodies (intracytoplasmic PAS+ globules)
Microscopic (histologic) images
Contributed by Nikhil Sangle, M.D.
Centroblasts
Centrocytes
AFIP images
Normal lymph node
Primary follicle
Lymph node
Secondary follicle
Secondary follicle
Paracortical T zone
Interdigitating dendritic cells
Smooth muscle proliferation in lymph node hilum
Sclerosis in an inguinal lymph node
Positive stains
B cells: CD19, CD20, CD22, CD79
Germinal center B cells: strong and dense bcl6 and CD10
B cells in primary follicules and mantle zones: IgD, IgM, CD21, CD23, occasionally CD5
Antigen stimulated B cells with the capacity to differentiate toward plasma cells express MUM1 / IRF4 and CD138
T cells: CD2, CD3, variable CD4 and CD8
Follicular helper T cells: C3, CD4, CD57, PD1 or CD279
T regs: CD4, CD25 and FOXP3
Premature B and T cells: TdT (terminal deoxynucleotidyl transferase)
Follicular dendritic cells: CD21, CD23, CD35
Macrophages: CD68, lysozyme
NK cells:
CD56, CD57, CD16 (> 90% are CD16+ / CD56+)
Also cytoplasmic (not surface) CD3, CD2, CD7, CD8 (up to 80%), perforin, granzyme B, TIA1
Germinal centers have strong dense BCL6 and CD10 expression
Negative stains
bcl2 (not expressed in germinal center B lymphocytes)
Electron microscopy images
Images hosted on other servers:
Plasma cells
Back to top
Home > Lymph nodes & spleen, nonlymphoma > Anatomy & histology-lymph nodes
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Using Transformations To Graph Quadratic Functions
Makayla T.
asked • 09/13/18
The parent function f(x)=x^2 is translated 2 units left and 3 units up. Write the quadratic function in vertex form.
The vertex form of a quadratic function is f(x)=a(x-h)^2+k.
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Nathan B.answered • 09/13/18
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f(x)=a(x - h)2 + k
Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.
The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):
x - h = 0
-2 - h = 0
-h = 2
h = -2
So the function ends up looking like:
f(x)=a(x - (-2))2 + 2
Subtracting a negative cancels the signs out to make a positive:
f(x)=a(x+ 2)2 + 2
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Mark M.
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a = 1 since the original function is f(x) = x 2.
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inequality - Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ - Mathematics Stack Exchange
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Prove 2 a b a+b≤a b−−√2 a b a+b≤a b
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a a and b b are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).
Here's my attempt:
2 a b a+b≤a b−−√2 a b a+b≤a b
2 a 2 b 2(a+b)2≤a b 2 a 2 b 2(a+b)2≤a b
2 a 2 b 2 a 2+2 a b+b 2−a b≤0 2 a 2 b 2 a 2+2 a b+b 2−a b≤0
2 a 2 b 2−a b(a 2+2 a b+b 2)a 2+2 a b+b 2≤0 2 a 2 b 2−a b(a 2+2 a b+b 2)a 2+2 a b+b 2≤0
2 a 2 b 2−a 3 b−2 a 2 b 2−a b 3 a 2+2 a b+b 2≤0 2 a 2 b 2−a 3 b−2 a 2 b 2−a b 3 a 2+2 a b+b 2≤0
−a 3 b−a b 3 a 2+2 a b+b 2≤0−a 3 b−a b 3 a 2+2 a b+b 2≤0
Because a a and b b are positive, this guarantees the left side is negative, making the inequality true.
inequality
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edited Feb 18, 2021 at 9:26
Martin Sleziak
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asked Jan 31, 2016 at 5:46
ChrisChris
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1 This cannon be correct since you have equality if a=b a=b.A.S. –A.S. 2016-01-31 05:48:55 +00:00 Commented Jan 31, 2016 at 5:48
It's mostly correct (except you should write it backwards and replace < with <=) but a little complicated. It is simpler to square each side, simplify by a b a b and get rid of denominators Ewan Delanoy –Ewan Delanoy 2016-01-31 05:49:28 +00:00 Commented Jan 31, 2016 at 5:49
I tried this approach and got 0 <=<=(a−b)2(a−b)2, which could be 0 0 if a a and b b are equal, or positive in any other case, which makes the inequality true. Hopefully I did the math right.Chris –Chris 2016-01-31 06:03:53 +00:00 Commented Jan 31, 2016 at 6:03
At the second step of your derivation, why didn't you square the 2 2?user3733558 –user3733558 2021-02-18 09:42:25 +00:00 Commented Feb 18, 2021 at 9:42
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This is just a rearrangement of the famous AM-GM inequality!
It states that a+b 2≥a b−−√a+b 2≥a b
for a,b≥0 a,b≥0.
The proof is clear from the picture!
Now multiply by a b−−√a b on both sides and rearrange with ease to get the desired: 2 a b a+b≤a b−−√2 a b a+b≤a b.
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answered Jan 31, 2016 at 6:11
MaffredMaffred
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Let a,b∈R a,b∈R with a,b>0 a,b>0.
2 a b a+b≤a b−−√⟺2 a+b≤a b−−√a b⟺2 a+b≤1 a b−−√⟺a b−−√≤a+b 2⟺2 a b−−√≤a+b⟺0≤a−2 a b−−√+b⟺0≤(a−−√−b√)2 2 a b a+b≤a b⟺2 a+b≤a b a b⟺2 a+b≤1 a b⟺a b≤a+b 2⟺2 a b≤a+b⟺0≤a−2 a b+b⟺0≤(a−b)2
where the last inequality is always true.
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answered Feb 18, 2021 at 9:51
C SquaredC Squared
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A much simpler way is to use the relation between the harmonic and geometric means:
2 1 a+1 b≤a b−−√2 1 a+1 b≤a b
adding the fraction in the denominator of the left hand side: we have:
2 b a b+a a b≤a b−−√2 b a b+a a b≤a b
And the relation you listed follows.
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answered Jan 31, 2016 at 6:49
MrXMrX
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0Prove that 2 x 2−2 x+1 2−−−−−−−√≥1 x+1 x 2 x 2−2 x+1 2≥1 x+1 x for 0<x<1.0<x<1.
4Proof: a<a b−−√<a+b 2<b a<a b<a+b 2<b
2Prove a+2 a 2+3 a 3<6 a 6 a+2 a 2+3 a 3<6 a 6 for a>1 a>1.
0Prove inequality: 1 2√+1(2√)2+⋅⋅⋅+1(2√)n≤1 2√−1,∀n∈N 1 2+1(2)2+⋅⋅⋅+1(2)n≤1 2−1,∀n∈N
7Prove 1 3>2 a b 2(a+a 2+b 2√)3 1 3>2 a b 2(a+a 2+b 2)3
1Using Cauchy-Schwarz to show 2(a+b)−−−−−−−√−a 2+b 2 a+b−−−−√≥2 a b a+b−−−√,a,b>0 2(a+b)−a 2+b 2 a+b≥2 a b a+b,a,b>0
2Show that we cant find three numbers among 2 a b a+b,a b−−√,a+b 2,a 2+b 2 2−−−−√2 a b a+b,a b,a+b 2,a 2+b 2 2 that are in arithmetic progression.
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https://www.youtube.com/watch?v=HCTAnKNRNa0
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Find Triangle Lengths & Angles Using Extended Ratios | Geometry | Eat Pi
Eat Pi
19200 subscribers
8 likes
Description
503 views
Posted: 21 Feb 2024
In this video, I teach you how to find the lengths and angles of a triangle using ratios, specifically an extended ratio. An extended ratio is simply a ratio with 3 numbers rather than 2.
0:00 - The lengths of the sides of triangle are in the extended ratio 6:7:9 and the perimeter of the triangle is 88 cm. What are the lengths of the sides?
2:18 - The lengths of the sides of triangle are in the extended ratio 4:3:2. What is the measure of the largest angle?
If you have any questions, please leave them in the comment section below! Also, if you find the videos helpful, please like, share, and subscribe!
2 comments
Transcript:
The lengths of the sides of triangle are in the extended ratio 6:7:9 and the perimeter of the triangle is 88 cm. What are the lengths of the sides? what's up you freaking Geniuses so in this video I'm going to teach you how to find the lengths and angles of triangles using extended ratios all right so let's jump into this first problem right here so it says the lengths of the sides of a triangle are in the extended ratio 679 the perimeter of the triangle is 88 CM what are the lengths of the sides all right so it's basically telling us that we have this random triangle right here oh that's pretty good all right and then it says that we can relate all three sides using this ratio and since we have three different numbers in our ratio that's why it's called an extended ratio rather than just a regular ratio okay so we can again relate them using six 7 and N so we can basically set this up as saying let's say this side is 6X this side we can say is 7x and then this bottom side let's call it 9x and the reason we're adding all these X's is because we don't know what the actual lengths of each of these sides are right all we know is that the ratio of all these three sides is 6 to 7 to 9 but the problem does tell us that the perimeter of the triangle is 88 CM right so if we add up all three sides they should add up to 88 all right so let's add these three up so let's say 6X + 7 x + 9x is equal to 8 88 all right so here 6 + 7 + 9 is = to 22 so we get 22x is equal to 88 we can divide both sides by 22 so here we get that X is equal to 4 right so now that we know what x is equal to now we can plug it in to each of these so let's plug it in here so first of all 6 4 is equal to 24 this side over here would be 7 4 which is equal to 28 and the bottom side over here 9 4 is equal to 36 and let's not forget our units so here it tells us uh we're working in centimeters right so this side right here is 24 cm this one's 28 cm and this one's 36 cmet all right so now this problem says The lengths of the sides of triangle are in the extended ratio 4:3:2. What is the measure of the largest angle? the measures of the angles of a triangle are in the extended ratio 432 what is the measure of the largest angle right so again we have just kind of a random triangle over here and now we're given the ratio of all the angles so again it's 4 32 all right so let's just say this one's uh four this one's three and this one's two and again we have to add our X's because we don't know what they actually are right so we're going to say 4X 3x and 2x okay so again in this case we want to solve for x and the way that we're going to do that is by well first of all we need to know the fact that when whenever you add up all the angles inside of any triangle they should add up to 180° right so if we add all these three up they should add up to 180 so here we can say 2x + 3x + 4x is equal to 180 all right so adding these up together 2 + 3 is 5 and 5 + 4 is 9 so we get 9x is equal to 180 all right so then we can divide both sides by 9 so then here we get that X is equal to 20 okay so now that we know what x is equal to what was the question so it says what is the measure of the largest angle all right well the largest angle uh would be this one that we labeled as 4X right so if we plug in 20 for X over here well 4 20 is going to be equal to 80 okay so this angle right here is equal to 80° boom so if you found the video helpful definitely leave a thumbs up down below and if you have any other questions or want to see any other examples just let me know in the comment section below
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https://proofwiki.org/wiki/Definition:Idempotent_Semigroup
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Definition:Idempotent Semigroup
From ProofWiki
Jump to navigation Jump to search
Contents
1 Definition
2 Also known as
3 Examples
3.1 Idempotent Semigroup with Relation induced by Inverse
4 Also see
5 Sources
Definition
An idempotent semigroup is a semigroup whose operation is idempotent.
That is, a semigroup $\struct {S, \circ}$ is idempotent if and only if:
: $\forall x \in S: x \circ x = x$
Also known as
There are extensive bodies of research covering idempotent semigroups.
In this literature, band is a common synonym for idempotent semigroup.
Examples
Idempotent Semigroup with Relation induced by Inverse
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
: $\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
Also see
Properties of Idempotent Semigroup
Definition:Semilattice
Results about idempotent semigroups can be found here.
Sources
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19$
Retrieved from "
Categories:
Definitions/Idempotent Semigroups
Definitions/Semigroups
Definitions/Idempotence
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https://openstax.org/books/university-physics-volume-1/pages/17-summary
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University Physics Volume 1
Summary
University Physics Volume 1Summary
Search for key terms or text.
Summary
## 17.1 Sound Waves
Sound is a disturbance of matter (a pressure wave) that is transmitted from its source outward. Hearing is the perception of sound.
Sound can be modeled in terms of pressure or in terms of displacement of molecules.
The human ear is sensitive to frequencies between 20 Hz and 20 kHz.
## 17.2 Speed of Sound
The speed of sound depends on the medium and the state of the medium.
In a fluid, because the absence of shear forces, sound waves are longitudinal. A solid can support both longitudinal and transverse sound waves.
In air, the speed of sound is related to air temperature T by
v is the same for all frequencies and wavelengths of sound in air.
## 17.3 Sound Intensity
Intensity is the same for a sound wave as was defined for all waves, where P is the power crossing area A. The SI unit for I is watts per meter squared. The intensity of a sound wave is also related to the pressure amplitude
where is the density of the medium in which the sound wave travels and is the speed of sound in the medium.
Sound intensity level in units of decibels (dB) is
where is the threshold intensity of hearing.
The perception of frequency is pitch. The perception of intensity is loudness and loudness has units of phons.
## 17.4 Normal Modes of a Standing Sound Wave
Unwanted sound can be reduced using destructive interference.
Sound has the same properties of interference and resonance as defined for all waves.
In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics.
## 17.5 Sources of Musical Sound
Some musical instruments can be modeled as pipes that have symmetrical boundary conditions: open at both ends or closed at both ends. Other musical instruments can be modeled as pipes that have anti-symmetrical boundary conditions: closed at one end and open at the other.
Some instruments, such as the pipe organ, have several tubes with different lengths. Instruments such as the flute vary the length of the tube by closing the holes along the tube. The trombone varies the length of the tube using a sliding bar.
String instruments produce sound using a vibrating string with nodes at each end. The air around the string oscillates at the frequency of the string. The relationship for the frequencies for the string is the same as for the symmetrical boundary conditions of the pipe, with the length of the pipe replaced by the length of the string and the velocity replaced by
## 17.6 Beats
When two sound waves that differ in frequency interfere, beats are created with a beat frequency that is equal to the absolute value of the difference in the frequencies.
## 17.7 The Doppler Effect
The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer.
The actual change in frequency is called the Doppler shift.
## 17.8 Shock Waves
The Mach number is the velocity of a source divided by the speed of sound,
When a sound source moves faster than the speed of sound, a shock wave is produced as the sound waves interfere.
A sonic boom is the intense sound that occurs as the shock wave moves along the ground.
The angle the shock wave produces can be found as
A bow wake is produced when an object moves faster than the speed of a mechanical wave in the medium, such as a boat moving through the water.
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Trichotillomania and its treatment: an updated review and recommendations
Megan C. DuBoisa Department of Psychological Sciences, Kent State University, Kent, OH, USA
,
Bridget M. Felera Department of Psychological Sciences, Kent State University, Kent, OH, USA
,
Christopher A. Flessnera Department of Psychological Sciences, Kent State University, Kent, OH, USA
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Martin E. Franklinb Department of Psychiatry, University of Pennsylvania School of Medicine, Philadelphia, PA, USA;c Rogers Behavioral Health, Philadelphia, PA, USACorrespondencemartin.franklin@rogersbh.org
Received 25 May 2025, Accepted 02 Sep 2025, Accepted author version posted online: 10 Sep 2025, Published online: 17 Sep 2025
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ABSTRACT
Introduction
Trichotillomania is an impulse control disorder in which individuals fail to resist urges to pull out their own hair and is associated with significant psychiatric comorbidity and functional impairment in affected children, adolescents, and adults. Onset in childhood or adolescence is typical, yet the literature on phenomenology, psychopathology, and treatment outcome involving pediatric samples remains particularly sparse. Efficacious treatments have been developed and found efficacious, most notably cognitive-behavioral interventions known collectively as habit reversal training, although relapse in adults appears to be somewhat common.
Areas covered
Herein, the authors give an overview of the latest developments in the treatment of trichotillomania and provide their own expert recommendations for the management of the condition. This article is based on searches using the PubMed and PsycINFO databases for peer-reviewed articles from 2011 through to April 2025.
Expert opinion
Recent developments in pharmacotherapies, both alone and in combination with cognitive behavioral treatments, hold promise, although further efforts are needed to examine their efficacy, effectiveness, and durability. Dissemination of information about trichotillomania and its treatment remains a critical next step in the field, since many affected individuals and their families experience difficulties with finding local treatment providers with sufficient knowledge to deliver interventions known to reduce hair pulling behaviors and associated symptoms.
KEYWORDS:
Trichotillomania
behavior therapy
habit reversal training
Impulse control disorder
psychotherapy
Article highlights
Trichotillomania (TTM) is a distressing and impairing psychiatric condition that involves repeatedly pulling one’s hair, resulting in visible hair-loss. TTM is associated with significant impairment in a variety of domains (e.g. social/family, work/school) and often comorbid with other disorders such as anxiety, depression, or obsessive-compulsive disorder.
TTM commonly emerges between ages 9 and 13, though hair-pulling behaviors in early childhood are not unheard of. Several important developmental considerations exist with respect to assessment, functional impairment, pulling style (i.e. focused versus automatic), and treatment outcomes.
Assessment tools are underdeveloped for pediatric populations. While several validated measures exist for adults, only two psychometrically sound tools (e.g. TSC, MIST-C) are currently available for children. The most commonly used measures were normed on adults, raising concerns about developmental appropriateness and accuracy in youth.
Behavior therapy (e.g. habit-reversal therapy, acceptance-enhanced behavior therapy) remains the most established treatment for TTMhile individual, one-to-one habit-reversal therapy (HRT) is well-supported for adults and shows promise for youth, stepped-care and web-based interventions offer potential to reduce treatment barriers. However, these approaches require additional research, particularly in pediatric populations where evidence remains scarce.
Pharmacological treatments for TTM have yielded inconsistent results. SSRIs show limited benefit, while olanzapine, an atypical antipsychotic, and clomipramine, a tricyclic antidepressant, both show greater efficacy but not without adverse side effects. Newer agents such as NAC or memantine show potential but require more rigorous and replicated evaluation, particularly in younger populations.
Practitioners are encouraged to seek out resources that may be beneficial in conceptualization and treatment of TTM. The TLC Foundation for BFRBs is one such resource.
Declaration of interest
The authors have no relevant affiliations or financial involvement with any organization or entity with a financial interest in or financial conflict with the subject matter or materials discussed in the manuscript. This includes employment, consultancies, honoraria, stock ownership or options, expert testimony, grants or patents received or pending, or royalties.
Reviewer disclosures
Peer reviewers on this manuscript have no relevant financial or other relationships to disclose.This manuscript was not funded.
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Efficient algorithm to determine the shape from the inputted four points
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I am trying to solve Determine the shape problem on Uva. From what I can get after reading the question it that it is an Ad Hoc geometry problem in which we have to use some geometry theorem to determine what shape the four points we take as input form on a 2D plane. After spending many hours I still cannot think of any efficient algorithm which can solve the problem efficiently in the given time limit.I tried using distance formula and slopes but was not of much help.Please suggest some good algorithm or theorem I can use to solve this problem.
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asked Jun 24, 2015 at 5:09
Alpa8Alpa8
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My first thought was the following:
Determine the lengths of each side. Use the formula sqrt((a - x)^2 + (b - y)^2).
Determine each of the angles.
Determine which shape:
If all angles 90
If all sides equal, return SQUARE
Else, return RECTANGLE
Else if angle 1 = 3 and angle 2 = 4 equal
If all sides equal, return RHOMBUS
Else, return Parallelogram
Else if angle 1 + 2 == 180, or angle 3 + 4 == 180, return TRAPEZIUM
Else, return ORDINARY QUADRILATERAL
But consider the logic followed here; they've used vector math, and simply calculate whether something is a right angle or whether two lines are parallel, rather than calculating all angles.
Prepare some functions to calculate whether two lines are parallel, and whether an angle is a right angle (using vector mathematics).
Sort the points so they're in the correct rotational order (the actual order is ambiguous for concave shapes, but it shouldn't matter, should still return ORDINARY QUAD).
Determine the lengths of each line.
Determine the shape, given the lengths and whether lines are parallel or angles right.
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edited Jun 24, 2015 at 5:35
answered Jun 24, 2015 at 5:16
LukeLuke
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Firstly, different shapes have the following relationship:
Ordinary Quadrilateral --(with one pair of parallel sides)--> Trapezium Trapezium --(with additional pair of parallel sides)--> Parallelogram Parallelogram --(with four equal straight lines)--> Rhombus--(with four 90 degree angles)-->square |--(with four 90 degree angles)--> Rectangle --(with four equal lines)--> square
Given four points, A, B, C, D, take random one(say A), and we need to calculate statistics(angle and length) of following four pairs of points(not all), including
(1) AB and CD, A1, L1 (2) AC and BD, A2, L2 (3) AB and AC, A3, L3 (4) AB and AD, A4, L4
And then I think the trick comes to how to organize the branches so that we could have minimal computation and code path. My proposal is listed as follows:
A3 = getAngle(AB, AC) A4 = getAngle(AB, AD) if A3 > A4 we know AD is the diagonal line, then use A, B, C to calculate else we know AC is the diagonal line, then use A, B, D to calculate # following suppose we use A, B, C to do the calculation, we could easily do the A, B, D thing if define new variables L3 = LengthEqual(AB, AC) A1 = getAngle(AB, AD) if A3 == 90 && A1 == 0 if L3 == True Square else Rectangle else A2 = getAngle(AC, BD) L2 = LengthEqual(AC, BD) if A2 == 0 && A1 == 0 if L2 == True Rhombus else Parallelogram else if A2 == 0 || A1 == 0 Trapezium else Ordinary Quadrilateral
In this means, we could achieve relatively less computation to wisely branch to the results we want. Hope this helps.
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edited Jun 24, 2015 at 7:11
answered Jun 24, 2015 at 6:53
lordofirelordofire
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https://projecteuclid.org/journals/statistical-science/volume-39/issue-1/Causal-Inference-Methods-for-Combining-Randomized-Trials-and-Observational-Studies/10.1214/23-STS889.full
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Causal Inference Methods for Combining Randomized Trials and Observational Studies: A Review
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Home>Journals>Statist. Sci.>Volume 39>Issue 1>Article
February 2024 Causal Inference Methods for Combining Randomized Trials and Observational Studies: A Review
Bénédicte Colnet, Imke Mayer, Guanhua Chen, Awa Dieng, Ruohong Li, Gaël Varoquaux, Jean-Philippe Vert, Julie Josse, Shu Yang
Author Affiliations +
Bénédicte Colnet,1 Imke Mayer,2 Guanhua Chen,3 Awa Dieng,4 Ruohong Li,5 Gaël Varoquaux,6 Jean-Philippe Vert,7 Julie Josse,8 Shu Yang 9
1 Bénédicte Colnet is Ph.D. candidate, Soda project-team, INRIA Saclay, Palaiseau, France
2 Imke Mayer is Research Scientist, Owkin, London, UK
3 Guanhua Chen is Associate Professor, Department of Biostatistics and Medical Informatics, University of Wisconsin-Madison, Madison, WI 53726, USA
4 Awa Dieng is Research Associate, Google DeepMind, Montreal, Canada
5 Ruohong Li is Senior Machine Learning Scientist, Microsoft, Kirkland, WA 98033 , USA
6 Gaël Varoquaux is Research Director, Soda project-team, INRIA Saclay, Paries, France
7 Jean-Philippe Vert is Chief R&D Officer, Owkin, Paris, France
8 Julie Josse is Head, Premedical project Inria team, University of Montpellier, France
9 Shu Yang is Associate Professor, Department of Statistics, North Carolina State University, 2311 Stinson Drive, Raleigh, NC 27695, USA
Statist. Sci. 39(1): 165-191 (February 2024). DOI: 10.1214/23-STS889
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Abstract
With increasing data availability, causal effects can be evaluated across different data sets, both randomized controlled trials (RCTs) and observational studies. RCTs isolate the effect of the treatment from that of unwanted (confounding) co-occurring effects but they may suffer from unrepresentativeness, and thus lack external validity. On the other hand, large observational samples are often more representative of the target population but can conflate confounding effects with the treatment of interest. In this paper, we review the growing literature on methods for causal inference on combined RCTs and observational studies, striving for the best of both worlds. We first discuss identification and estimation methods that improve generalizability of RCTs using the representativeness of observational data. Classical estimators include weighting, difference between conditional outcome models and doubly robust estimators. We then discuss methods that combine RCTs and observational data to either ensure unconfoundedness of the observational analysis or to improve (conditional) average treatment effect estimation. We also connect and contrast works developed in both the potential outcomes literature and the structural causal model literature. Finally, we compare the main methods using a simulation study and real world data to analyze the effect of tranexamic acid on the mortality rate in major trauma patients. A review of available codes and new implementations is also provided.
Funding Statement
SY is partially supported by NSF SES 2242776, NIH 1R01AG066883 and FDA 1U01FD007934.
Acknowledgments
This work was initiated by a SAMSI working group jointly led by JJ and SY in the 2020 causal inference program. We would like to acknowledge the helpful discussions during the SAMSI working group meetings. We also would like to acknowledge the discussions and insights from the Traumabase group and physicians, in particular, Drs. François-Xavier AGERON, Tobias GAUSS and Jean-Denis MOYER. In addition, none of the data analysis part could have been done without the help of Dr. Ian ROBERTS and the CRASH-3 group, who shared with us the clinical trial data. Part of this work was performed while JJ was a visiting researcher at Google Brain Paris. Finally, we would like to warmly thank Issa DAHABREH for his comments, suggestions of additional references and insightful discussions.
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Bénédicte Colnet.Imke Mayer.Guanhua Chen.Awa Dieng.Ruohong Li.Gaël Varoquaux.Jean-Philippe Vert.Julie Josse.Shu Yang."Causal Inference Methods for Combining Randomized Trials and Observational Studies: A Review."Statist. Sci.39(1)165 - 191,February 2024.
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Published: February 2024
First available in Project Euclid: 18 February 2024
MathSciNet: MR4718532
Digital Object Identifier: 10.1214/23-STS889
Keywords: Causal effect generalization , data integration , double robustness , heterogeneous data , S-admissibility , transportability
Rights: Copyright © 2024 Institute of Mathematical Statistics
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Bénédicte Colnet, Imke Mayer, Guanhua Chen, Awa Dieng, Ruohong Li, Gaël Varoquaux, Jean-Philippe Vert, Julie Josse, Shu Yang "Causal Inference Methods for Combining Randomized Trials and Observational Studies: A Review," Statistical Science, Statist. Sci. 39(1), 165-191, (February 2024)
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Zinc-Dependent Enzymes in Zinc-Depleted Rats; Intestinal Alkaline Phosphatase - ScienceDirect
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Abstract
References (49)
Cited by (28)
The American Journal of Clinical Nutrition
Volume 22, Issue 9, September 1969, Pages 1250-1263
Zinc-Dependent Enzymes in Zinc-Depleted Rats; Intestinal Alkaline Phosphatase12
Author links open overlay panel REINHOLD JOHN G.PH.D.(Professor of Biochemistry.), KFOURY GEORGE A.PHARM.D.(Senior Research Assistant.)
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SUMMARY
The effect of zinc depletion upon the behavior of the intestinal alkaline phosphatases of the rat has been examined by means of electrophoresis on polyacrylamide gels, chromatography on DEAE-cellulose, and kinetic and inhibition studies both of mucosal homogenates and of preparations partly purified by the method of Morton (30).
Electrophoretic separations gave less well resolved phosphatase isoenzyme patterns in the homogenates of depleted rats than in their pair-fed controls. In addition, a slowly migrating isoenzyme was absent or attenuated in electropherograms of the partly purified preparations.
Four isoenzymes were separated by chromatography of the partially purified phosphatase of the control rats. Their counterparts were detected in all eluates of similar preparations from zinc-depleted rats. The distribution of phosphatase activity was studied in three of the isoenzymes. This differed from that in the controls in each of three series of rats, but variability in the latter prevented evaluation of the significance of these differences.
Kinetic studies showed that the action of zinc in increasing reaction velocity was dependent upon pH, the effect being minor at pH 9.5 but marked at pH 10.5. Addition of zinc to the phosphatase measurement systems caused activity to increase to the same extent in zinc-depleted and control rats. Zinc depletion significantly decreased the affinity of phosphatase for zinc and for magnesium, but had no effect upon that for p-nitrophenylphosphate. A marked increase in the affinity of phosphatase for the latter was effected by zinc both in depleted and control rats.
An enhanced susceptibility to inhibition by l-phenylalanine was characteristic of partly purified enzyme preparations from one group of zinc-depleted rats.
It is concluded that zinc depletion in the rat is associated with alterations in enzyme behavior that probably stem from modifications in enzyme structure brought about by zinc depletion. These are not rectified by addition of zinc to enzyme assay systems. It is suggested that they lead to an increase in protein turnover.
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View more references
Cited by (28)
Generation of particle assemblies mimicking enzymatic activity by processing of herbal food: The case of rhizoma polygonati and other natural ingredients in traditional Chinese medicine
2021, Nanoscale Advances Show abstract Processed herbs have been widely used in eastern and western medicine; however, the mechanism of their medicinal effects has not yet been revealed. It is commonly believed that a central role is played by chemically active molecules produced by the herbs' metabolism. In this work, processed rhizoma polygonati (RP) and other herbal foods are shown to exhibit intrinsic phosphatase-like (PL) activity bounded with the formation of nano-size flower-shaped assembly. Via quantum mechanical calculations, an enzymatic mechanism is proposed. The enzymatic activity may be induced by the interaction between the sugar molecules distributed on the surface of the nanoassemblies and the phosphatase substrate via either a hydroxyl group or the deprotonated hydroxyl group. Meanwhile, the investigation was further extended by processing some fresh herbs and herbal food through a similar protocol, wherein other enzymatic activities (such as protease, and amylase) were observed. The PL activity exhibited by the processed natural herbs was found to be able to effectively inhibit cancer cell growth via phosphatase signaling, possibly by crosstalk with kinase signaling or DNA damage by either directly binding or unwinding of DNA, as evidenced by high-resolution atomic-force microscopy (HR-AFM). In this work, the neologism herbzyme (herb + enzyme) is proposed. This study represents the first case of scientific literature introducing this new term. Besides the well-known pharmacological properties of the natural molecules contained in herbs and herbal food, there exists an enzymatic/co-enzymatic activity attributed to the nanosized assemblies.
### Availability of supplemental amino acid-chelated trace elements in diets containing tricalcium phosphate and phytate to rainbow trout, Oncorhynchus mykiss
2003, Aquaculture Show abstract This study investigated the availability of amino acid-chelated trace elements to rainbow trout fed diets containing tricalcium phosphate (TCP) and phytate (PH) that are known mineral inhibitors. Six semi-purified diets were supplemented with trace elements either from the sulfate (SF, diets 1 and 2) or amino acid-chelate (AM, diets 3–6). Diets 1 (SF) and 3 (AM) contained neither TCP nor PH. Diet 4 (AMPO) included TCP while Diet 5 (AMPH) contained PH alone. Diets 2 (SF++) and 6 (AM++), on the other hand, contained both inhibitors. Rainbow trout weighing 1.6 g were fed the experimental diets three times daily to near satiation for 18 weeks. Growth, whole body mineral deposition, and enzyme activity/expression were measured after the trial. Growth of fish fed the SF++ diet was significantly lower than the rest of the groups. Whole body Cu content was significantly higher in AM groups than in SF groups with or without TCP and PH. Whole body and bone Zn contents were significantly higher in fish fed the diet containing AM alone compared to the rest. Alkaline phosphatase activity was significantly higher in fish fed the diet containing AM alone than with fish fed the SF diet with or without TCP and PH. CuZnSOD and DNA polymerase were highly expressed in the AM compared to the SF groups. The results show that availability of trace elements is significantly affected by their chemical form and that amino acid-chelates are more beneficial for rainbow trout even in the presence of TCP and PH compared to the inorganic sources tested.
### Bioavailability and tissue distribution of amino acid-chelated trace elements in rainbow trout Oncorhynchus mykiss
2003, Fisheries Science
### Bioavailability of amino acids chelated and glass embedded zinc to rainbow trout, Oncorhynchus mykiss, fingerlings
2001, Aquaculture Nutrition
### Zinc and insulin metabolism
1981, Biological Trace Element Research
### The effects of early zinc deficiency on DNA and protein synthesis in the rat
1970, British Journal of Nutrition
View all citing articles on Scopus
1
From the Department of Biochemistry, American University of Beirut, Beirut, Lebanon
2
Supported by National Institute of Arthritis and Metabolic Diseases Research Grant AM-09622.
View full text
Copyright © 1969 American Society for Nutrition. Published by Elsevier Inc. All rights reserved.
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Chicken McNugget Theorem
The Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) states that for any two relatively primepositive integers, the greatest integer that cannot be written in the form for nonnegative integers is .
A consequence of the theorem is that there are exactly positive integers which cannot be expressed in the form . The proof is based on the fact that in each pair of the form , exactly one element is expressible.
Contents
1 Origins
2 Proof Without Words
3 Proof 1
4 Proof 2
5 Corollary
6 Generalization
7 Problems
7.1 Introductory
7.2 Intermediate
7.3 Olympiad
7.4 See Also
Origins
There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.
Proof Without Words
Example using m= and n=
Proof 1
Definition. An integer will be called purchasable if there exist nonnegative integers such that .
We would like to prove that is the largest non-purchasable integer. We are required to show that:
(1) is non-purchasable
(2) Every is purchasable
Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.
Lemma. Let be the set of solutions to . Then for any .
Proof: By Bezout's Lemma, there exist integers such that . Then . Hence is nonempty. It is easy to check that for all . We now prove that there are no others. Suppose and are solutions to . Then implies . Since and are coprime and divides , divides and . Similarly . Let be integers such that and . Then implies We have the desired result.
Lemma. For any integer , there exists unique such that .
Proof: By the division algorithm, there exists one and only one such that .
Lemma. is purchasable if and only if .
Proof: If , then we may simply pick so is purchasable. If , then if and if , hence at least one coordinate of is negative for all . Thus is not purchasable.
Thus the set of non-purchasable integers is . We would like to find the maximum of this set. Since both are positive, the maximum is achieved when and so that .
Proof 2
We start with this statement taken from Proof 2 of Fermat's Little Theorem:
"Let . Then, we claim that the set , consisting of the product of the elements of with , taken modulo , is simply a permutation of . In other words,
Clearly none of the for are divisible by , so it suffices to show that all of the elements in are distinct. Suppose that for . Since , by the cancellation rule, that reduces to , which is a contradiction."
Because and are coprime, we know that multiplying the residues of by simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form , is and is the original residue. We now prove the following lemma.
Lemma: For any nonnegative integer , is the least purchasable number .
Proof: Any number that is less than and congruent to it can be represented in the form , where is a positive integer. If this is purchasable, we can say for some nonnegative integers . This can be rearranged into , which implies that is a multiple of (since ). We can say that for some positive integer , and substitute to get . Because , , and . We divide by to get . However, we defined to be a positive integer, and all positive integers are greater than or equal to . Therefore, we have a contradiction, and is the least purchasable number congruent to .
This means that because is purchasable, every number that is greater than and congruent to it is also purchasable (because these numbers are in the form where ). Another result of this Lemma is that is the greatest number that is not purchasable. , so , which shows that is the greatest number in the form . Any number greater than this and congruent to some is purchasable, because that number is greater than . All numbers are congruent to some , and thus all numbers greater than are purchasable.
Putting it all together, we can say that for any coprime and , is the greatest number not representable in the form for nonnegative integers .
Corollary
This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma.
Lemma: For any integer , exactly one of the integers , is not purchasable.
Proof: Because every number is congruent to some residue of permuted by , we can set for some . We can break this into two cases.
Case 1: . This implies that is not purchasable, and that . is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself that is purchasable. Therefore, and , so is purchasable.
Case 2: . This implies that is purchasable, and that . Again, because is the least number congruent to itself that is purchasable, and because and , is not purchasable.
We now limit the values of to all integers , which limits the values of to . Because and are coprime, only one of them can be a multiple of . Therefore, , showing that is not an integer and that and are integers. We can now set limits that are equivalent to the previous on the values of and so that they cover all integers form to without overlap: and . There are values of , and each is paired with a value of , so we can make different ordered pairs of the form . The coordinates of these ordered pairs cover all integers from to inclusive, and each contains exactly one not-purchasable integer, so that means that there are different not-purchasable integers from to . All integers greater than are purchasable, so that means there are a total of integers that are not purchasable.
In other words, for every pair of coprime integers , there are exactly nonnegative integers that cannot be represented in the form for nonnegative integers .
Generalization
If and are not relatively prime, then we can simply rearrange into the form and are relatively prime, so we apply Chicken McNugget to find a bound We can simply multiply back into the bound to get Therefore, all multiples of greater than are representable in the form for some non-negative integers .
Problems
Introductory
Marcy buys paint jars in containers of and . What's the largest number of paint jars that Marcy can't obtain?
Answer: containers
Bay Area Rapid food sells chicken nuggets. You can buy packages of or . What is the largest integer such that there is no way to buy exactly nuggets? Can you Generalize? (Source: The Art and Craft of Problem Solving)
Answer:
If a game of American Football has only scores of field goals ( points) and touchdowns with the extra point ( points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?
Answer: points
The town of Hamlet has people for each horse, sheep for each cow, and ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
(Source: AMC 10B 2015 Problem 15)
Answer:
In the state of Coinland, coins have values and cents. Suppose is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of (Source: 2023 AMC 12B Problems/Problem 16)
Answer:
Intermediate
Ninety-four bricks, each measuring are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes or or to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? (Source: AIME)
Find the sum of all positive integers such that, given an unlimited supply of stamps of denominations and cents, cents is the greatest postage that cannot be formed. (Source: AIME II 2019 Problem 14)
Olympiad
On the real number line, paint red all points that correspond to integers of the form , where and are positive integers. Paint the remaining integer points blue. Find a point on the line such that, for every integer point , the reflection of with respect to is an integer point of a different color than . (Source: India TST)
See Also
Theorem
Prime
Retrieved from "
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Mario Bunge La ciencia. Su método y su filosofía ¿Qué es la ciencia?
Mario Bunge La ciencia. Su método y su filosofía 6 1. Introducción Mientras los animales inferiores sólo están en el mundo, el hombre trata de entenderlo; y sobre la base de su inteligencia imperfecta pero perfectible, del mundo, el hombre intenta enseñorearse de él para hacerlo más confortable. En este proceso, construye un mundo artificial: ese creciente cuerpo de ideas llamado "ciencia", que puede caracterizarse como conocimiento racional, sistemático, exacto, verificable y por consiguiente falible. Por medio de la investigación científica, el hombre ha alcanzado una reconstrucción conceptual del mundo que es cada vez más amplia, profunda y exacta.
Un mundo le es dado al hombre; su gloria no es soportar o despreciar este mundo, sino enriquecerlo construyendo otros universos. Amasa y remoldea la naturaleza sometiéndola a sus propias necesidades animales y espirituales, así como a sus sueños: crea así el mundo de los artefactos y el mundo de la cultura. La ciencia como actividad —como investigación— pertenece a la vida social; en cuanto se la aplica al mejoramiento de nuestro medio natural y artificial, a la invención y manufactura de bienes materiales y culturales, la ciencia se convierte en tecnología. Sin embargo, la ciencia se nos aparece como la más deslumbrante y asombrosa de las estrellas de la cultura cuando la consideramos como un bien en sí mismo, esto es como una actividad productora de nuevas ideas (investigación científica). Tratemos de caracterizar el conocimiento y la investigación científicos tal como se los conoce en la actualidad.
2. Ciencia formal y ciencia fáctica No toda la investigación científica procura el conocimiento objetivo. Así, la lógica y la matemática —esto es, los diversos sistemas de lógica formal y los diferentes capítulos de la matemática pura— son racionales, sistemáticos y verificables, pero no son objetivos; no nos dan informaciones acerca de la realidad: simplemente, no se ocupan de los hechos. La lógica y la matemática tratan de entes ideales; estos entes, tanto los abstractos como los interpretados, sólo existen en la mente humana. A los lógicos y matemáticos no se les da objetos de estudio: ellos construyen sus propios objetos. Es verdad que a menudo lo hacen por abstracción de objetos reales (naturales y sociales); más aún, el trabajo del lógico o del matemático satisface a menudo las necesidades del naturalista, del sociólogo o del tecnólogo, y es por esto que la sociedad los tolera y, ahora, hasta los estimula. Pero la materia prima que emplean los lógicos y los matemáticos no es fáctica sino ideal.
Por ejemplo, el concepto de número abstracto nació, sin duda, de la coordinación Mario Bunge La ciencia. Su método y su filosofía 7 (correspondencia biunívoca) de conjuntos de objetos materiales, tales como dedos, por una parte, y guijarros, por la otra; pero no por esto aquel concepto se reduce a esta operación manual, ni a los signos que se emplean para representarlo. Los números no existen fuera de nuestros cerebros, y aun allí dentro existen al nivel conceptual, y no al nivel fisiológico. Los objetos materiales son numerables siempre que sean discontinuos; pero no son números; tampoco son números puros (abstractos) sus cualidades o relaciones. En el mundo real encontramos 3 libros, en el mundo de la ficción construimos 3 platos voladores. ¿Pero quién vio jamás un 3, un simple 3?
La lógica y la matemática, por ocuparse de inventar entes formales y de establecer relaciones entre ellos, se llaman a menudo ciencias formales, precisamente porque sus objetos no son cosas ni procesos, sino, para emplear el lenguaje pictórico, formas en las que se puede verter un surtido ilimitado de contenidos, tanto fácticos como empíricos. Esto es, podemos establecer correspondencias entre esas formas (u objetos formales), por una parte, y cosas y procesos pertenecientes a cualquier nivel de la realidad por la otra. Así es como la física, la química, la fisiología, la psicología, la economía, y las demás ciencias recurren a la matemática, empleándola como herramienta para realizar la más precisa reconstrucción de las complejas relaciones que se encuentran entre los hechos y entre los diversos aspectos de los hechos; dichas ciencias no identifican las formas ideales con los objetos concretos, sino que interpretan las primeras en términos de hechos y de experiencias (o, lo que es equivalente, formalizan enunciados fácticos).
Lo mismo vale para la lógica formal: algunas de sus partes —en particular, pero no exclusivamente, la lógica proposicional bivalente— pueden hacerse corresponder a aquellas entidades psíquicas que llamamos pensamientos. Semejante aplicación de las ciencias de la forma pura a la inteligencia del mundo de los hechos, se efectúa asignando diferentes interpretaciones a los objetos formales. Estas interpretaciones son, dentro de ciertos límites, arbitrarias; vale decir, se justifican por el éxito, la conveniencia o la ignorancia. En otras palabras el significado fáctico o empírico que se les asigna a los objetos formales no es una propiedad intrínseca de los mismos. De esta manera, las ciencias formales jamás entran en conflicto con la realidad. Esto explica la paradoja de que, siendo formales, se "aplican" a la realidad: en rigor no se aplican, sino que se emplean en la vida cotidiana y en las ciencias fácticas a condición de que se les superpongan reglas de correspondencia adecuada. En suma, la lógica y la matemática establecen contacto con la realidad a través del puente del lenguaje, tanto el ordinario como el científico.
Tenemos así una primera gran división de las ciencias, en formales (o ideales) y fácticas (o materiales). Esta ramificación preliminar tiene en cuenta el objeto o tema de las respectivas disciplinas; también da cuenta de la diferencia de especie entre los enunciados que se proponen establecer las ciencias formales y las fácticas: mientras los enunciados formales consisten en relaciones entre signos, los enunciados de las ciencias fácticas se refieren, en su mayoría, a entes extracientíficos: a sucesos y procesos. Nuestra división también tiene en Mario Bunge La ciencia. Su método y su filosofía 8 cuenta el método por el cual se ponen a prueba los enunciados verificables: mientras las ciencias formales se contentan con la lógica para demostrar rigurosamente sus teoremas (los que, sin embargo, pudieron haber sido adivinados por inducción común o de otras maneras), las ciencias fácticas necesitan más que la lógica formal: para confirmar sus conjeturas necesitan de la observación y/o experimento. En otras palabras, las ciencias fácticas tienen que mirar las cosas, y, siempre que les sea posible, deben procurar cambiarlas deliberadamente para intentar descubrir en qué medida sus hipótesis se adecuan a los hechos.
Cuando se demuestra un teorema lógico o matemático no se recurre a la experiencia: el conjunto de postulados, definiciones, reglas de formación de las expresiones dotadas de significado, y reglas de inferencia deductiva —en suma, la base de la teoría dada—, es necesaria y suficiente para ese propósito. La demostración de los teoremas no es sino una deducción: es una operación confinada a la esfera teórica, aun cuando a veces los teoremas mismos (no sus demostraciones) sean sugeridos en alguna esfera extramatemática y aun cuando su prueba (pero no su primer descubrimiento) pueda realizarse con ayuda de calculadoras electrónicas. Por ejemplo, cualquier demostración rigurosa del teorema de Pitágoras prescinde de las mediciones, y emplea figuras sólo como ayuda psicológica al proceso deductivo: que el teorema de Pitágoras haya sido el resultado de un largo proceso de inducción conectado a operaciones prácticas de mediciones de tierras, es objeto de la historia, la sociología y la psicología del conocimiento.
La matemática y la lógica son, en suma, ciencias deductivas. El proceso constructivo, en que la experiencia desempeña un gran papel de sugerencias, se limita a la formación de los puntos de partida (axiomas). En matemática la verdad consiste, por esto, en la coherencia del enunciado dado con un sistema de ideas admitido previamente: por esto, la verdad matemática no es absoluta sino relativa a ese sistema, en el sentido de que una proposición que es válida en una teoría puede dejar de ser lógicamente verdadera en otra teoría. (Por ejemplo, en el sistema de aritmética que empleamos para contar las horas del día, vale la proposición de 24 + 1 = 1.) Más aún las teorías matemáticas abstractas, esto es, que contienen términos no interpretados (signos a los que no se atribuye un significado fijo, y que por lo tanto pueden adquirir distintos significados) pueden desarrollarse sin poner atención al problema de la verdad.
Considérese el siguiente axioma de cierta teoría abstracta (no interpretada): "Existe por lo menos un x tal que es F". Se puede dar un número ilimitado de interpretaciones (modelos) de este axioma, dándose a x y F otros tantos significados. Si decimos que S designa punto, obtenemos un modelo geométrico dado: si adoptamos la convención de que L designa número, obtenemos un cierto modelo aritmético, y así sucesivamente. En cuanto "llenamos" la forma vacía con un contenido específico (pero todavía matemático), obtenemos un sistema de entes lógicos que tienen el privilegio de ser verdaderos o falsos dentro del sistema dado de proposiciones: a partir de ahí tenemos que habérnoslas con el problema de la verdad Mario Bunge La ciencia. Su método y su filosofía 9 matemática. Aún así tan sólo las conclusiones (teoremas) tendrán que ser verdaderas: los axiomas mismos pueden elegirse a voluntad. La batalla se habrá ganado si se respeta la coherencia lógica esto es, si no se violan las leyes del sistema de lógica que se ha convenido en usar.
En las ciencias fácticas, la situación es enteramente diferente. En primer lugar, ellas no emplean símbolos vacíos (variables lógicas) sino tan sólo símbolos interpretados; por ejemplo no involucran expresiones tales como 'x es F', que no son verdaderas ni falsas. En segundo lugar, la racionalidad —esto es, la coherencia con un sistema de ideas aceptado previamente— es necesaria pero no suficiente para los enunciados fácticos; en particular la sumisión a algún sistema de lógica es necesaria pero no es una garantía de que se obtenga la verdad. Además de la racionalidad, exigimos de los enunciados de las ciencias fácticas que sean verificables en la experiencia, sea indirectamente (en el caso de las hipótesis generales), sea directamente (en el caso de las consecuencias singulares de las hipótesis). Únicamente después que haya pasado las pruebas de la verificación empírica podrá considerarse que un enunciado es adecuado a su objeto, o sea que es verdadero, y aún así hasta nueva orden. Por eso es que el conocimiento fáctico verificable se llama a menudo ciencia empírica.
En resumidas cuentas, la coherencia es necesaria pero no suficiente en el campo de las ciencias de hechos: para anunciar que un enunciado es (probablemente) verdadero se requieren datos empíricos (proposiciones acerca de observaciones o experimentos). En última instancia, sólo la experiencia puede decirnos si una hipótesis relativa a cierto grupo de hechos materiales es adecuada o no. El mejor fundamento de esta regla metodológica que acabamos de enunciar es que la experiencia le ha enseñado a la humanidad que el conocimiento de hecho no es convencional, que si se busca la comprensión y el control de los hechos debe partirse de la experiencia. Pero la experiencia no garantizará que la hipótesis en cuestión sea la única verdadera: sólo nos dirá que es probablemente adecuada, sin excluir por ello la posibilidad de que un estudio ulterior pueda dar mejores aproximaciones en la reconstrucción conceptual del trozo de realidad escogido. El conocimiento fáctico, aunque racional, es esencialmente probable: dicho de otro modo: la inferencia científica es una red de inferencias deductivas (demostrativas) y probables (inconcluyentes).
Las ciencias formales demuestran o prueban: las ciencias fácticas verifican (confirman o disconfirman) hipótesis que en su mayoría son provisionales. La demostración es completa y final; la verificación es incompleta y por eso temporaria. La naturaleza misma del método científico impide la confirmación final de las hipótesis fácticas. En efecto los científicos no sólo procuran acumular elementos de prueba de sus suposiciones multiplicando el número de casos en que ellas se cumplen; también tratan de obtener casos desfavorables a sus hipótesis, fundándose en el principio lógico de que una sola conclusión que no concuerde con los hechos tiene más peso que mil confirmaciones. Por ello, mientras las teorías formales pueden ser llevadas a un estado de perfección (o estancamiento), los sistemas relativos a los hechos son esencialmente defectuosos: cumplen, pues, la condición necesaria para ser Mario Bunge La ciencia. Su método y su filosofía 10 perfectibles. En consecuencia si el estudio de las ciencias formales vigoriza el hábito del rigor, el estudio de las ciencias fáctiles puede inducirnos a considerar el mundo como inagotable, y al hombre como una empresa inconclusa e interminable.
Las diferencias de método, tipo de enunciados y referentes que separan las ciencias fácticas de las formales, impiden que se las examine conjuntamente más allá de cierto punto. Por ser una ficción seria, rigurosa y a menudo útil, pero ficción al cabo, la ciencia formal requiere un tratamiento especial. En lo que sigue nos concentraremos en la ciencia fáctica. Daremos un vistazo a las características peculiares de las ciencias de la naturaleza y de la cultura en su estado actual, con la esperanza de que la ciencia futura enriquezca sus cualidades o, al menos, de que las civilizaciones por venir hagan mejor uso del conocimiento científico.
Los rasgos esenciales del tipo de conocimiento que alcanzan las ciencias de la naturaleza y de la sociedad son la racionalidad y la objetividad. Por conocimiento racional se entiende: a) que está constituido por conceptos, juicios y raciocinios y no por sensaciones, imágenes, pautas de conducta, etc. Sin duda, el científico percibe, forma imágenes (por ejemplo, modelos visualizables) y hace operaciones; por tanto el punto de partida como el punto final de su trabajo son ideas; b) que esas ideas pueden combinarse de acuerdo con algún conjunto de reglas lógicas con el fin de producir nuevas ideas (inferencia deductiva). Estas no son enteramente nuevas desde un punto de vista estrictamente lógico, puesto que están implicadas por las premisas de la deducción; pero no gnoseológicamente nuevas en la medida en que expresan conocimientos de los que no se tenía conciencia antes de efectuarse la deducción; c) que esas ideas no se amontonan caóticamente o, simplemente, en forma cronológica, sino que se organizan en sistemas de ideas, esto es en conjuntos ordenados de proposiciones (teorías).
Que el conocimiento científico de la realidad es objetivo, significa: a) que concuerda aproximadamente con su objeto; vale decir que busca alcanzar la verdad fáctica; b) que verifica la adaptación de las ideas a los hechos recurriendo a un comercio peculiar con los hechos (observación y experimento), intercambio que es controlable y hasta cierto punto reproducible.
Ambos rasgos de la ciencia fáctica, la racionalidad y la objetividad, están íntimamente soldados. Así, por ejemplo, lo que usualmente se verifica por medio del experimento es alguna consecuencia —extraída por vía deductiva— de alguna hipótesis; otro ejemplo: el cálculo no sólo sigue a la observación sino que siempre es indispensable para planearla y registrarla. La racionalidad y objetividad del conocimiento científico pueden analizarse en un cúmulo de características a las que pasaremos revista en lo que sigue.
3. Inventario de las principales características de la ciencia fáctica Mario Bunge La ciencia. Su método y su filosofía 11 1) El conocimiento científico es fáctico: parte de los hechos, los respeta hasta cierto punto, y siempre vuelve a ellos. La ciencia intenta describir los hechos tal como son, independientemente de su valor emocional o comercial: la ciencia no poetiza los hechos ni los vende, si bien sus hazañas son una fuente de poesía y de negocios. En todos los campos, la ciencia comienza estableciendo los hechos; esto requiere curiosidad impersonal, desconfianza por la opinión prevaleciente, y sensibilidad a la novedad.
Los enunciados fácticos confirmados se llaman usualmente "datos empíricos"; se obtienen con ayuda de teorías (por esquemáticas que sean) y son a su vez la materia prima de la elaboración teórica. Una subclase de datos empíricos es de tipo cuantitativo; los datos numéricos y métricos se disponen a menudo en tablas, las más importantes de las cuales son las tablas de constantes (tales como las de los puntos de fusión de las diferentes sustancias).
Pero la recolección de datos y su ulterior disposición en tablas no es la finalidad principal de la investigación: la información de esta clase debe incorporarse a teorías si ha de convertirse en una herramienta para la inteligencia y la aplicación. ¿De qué sirve conocer el peso específico del hierro si carecemos de fórmulas mediante las cuales podemos relacionarlos con otras cantidades?
No siempre es posible, ni siquiera deseable, respetar enteramente los hechos cuando se los analiza, y no hay ciencia sin análisis, aun cuando el análisis no sea sino un medio para la reconstrucción final de los todos. El físico atómico perturba el átomo al que desea espiar; el biólogo modifica e incluso puede matar al ser vivo que analiza; el antropólogo empeñado en el estudio de campo de una comunidad provoca en ella ciertas modificaciones. Ninguno de ellos aprehende su objeto tal como es, sino tal como queda modificado por sus propias operaciones; sin embargo, en todos los casos tales cambios son objetivos, y se presume que pueden entenderse en términos de leyes: no son conjurados arbitrariamente por el experimentador. Más aún, en todos los casos el investigador intenta describir las características y el monto de la perturbación que produce en el acto del experimento; procura, en suma estimar la desviación o "error" producido por su intervención activa. Porque los científicos actúan haciendo tácitamente la suposición de que el mundo existiría aun en su ausencia, aunque desde luego, no exactamente de la misma manera.
2) El conocimiento científico trasciende los hechos: descarta los hechos, produce nuevos hechos, y los explica. El sentido común parte de los hechos y se atiene a ellos: a menudo se imita al hecho aislado, sin ir muy lejos en el trabajo de correlacionarlo con otros o de explicarlo. En cambio, la investigación científica no se limita a los hechos observados: los científicos exprimen la realidad a fin de ir más allá de las apariencias; rechazan el grueso de los hechos percibidos, por ser un montón de accidentes, seleccionan los que consideran que son relevantes, controlan hechos y, en lo posible, los reproducen. Incluso producen cosas nuevas desde instrumentos hasta partículas elementales; obtienen nuevos compuestos químicos, nuevas variedades vegetales y animales, y al menos en principio, crean nuevas Mario Bunge La ciencia. Su método y su filosofía 12 pautas de conducta individual y social.
Más aún, los científicos usualmente no aceptan nuevos hechos a menos que puedan certificar de alguna manera su autenticidad; y esto se hace, no tanto contrastándolos con otros hechos, cuanto mostrando que son compatibles con lo que se sabe. Los científicos descartan las imposturas y los trucos mágicos porque no encuadran en hipótesis muy generales y fidedignas, que han sido puestas a prueba en incontables ocasiones. Vale decir, los científicos no consideran su propia experiencia individual como un tribunal inapelable; se fundan, en cambio, en la experiencia colectiva y en la teoría.
Hay más: el conocimiento científico racionaliza la experiencia en lugar de limitarse a describirla; la ciencia da cuenta de los hechos no inventariándolos sino explicándolos por medio de hipótesis (en particular, enunciados de leyes) y sistemas de hipótesis (teorías). Los científicos conjeturan lo que hay tras los hechos observados, y de continuo inventan conceptos (tales como los del átomo, campo, masa, energía, adaptación, integración, selección, clase social, o tendencia histórica) que carecen de correlato empírico, esto es, que no corresponden a preceptos, aun cuando presumiblemente se refieren a cosas, cualidades o relaciones existentes objetivamente. No percibimos los campos eléctricos o las clases sociales: inferimos su existencia a partir de hechos experimentables y tales conceptos son significativos tan sólo en ciertos contextos teóricos.
Este trascender la experiencia inmediata, ese salto del nivel observacional al teórico, le permite a la ciencia mirar con desconfianza los enunciados sugeridos por meras coincidencias; le permite predecir la existencia real de las cosas y procesos ocultos a primera vista pero que instrumentos (materiales o conceptuales) más potentes pueden descubrir. Las discrepancias entre las previsiones teóricas y los hallazgos empíricos figuran entre los estímulos más fuertes para edificar teorías nuevas y diseñar nuevos experimentos. No son los hechos por sí mismos sino su elaboración teórica y la comparación de las consecuencias de las teorías con los datos observacionales, la principal fuente del descubrimiento de nuevos hechos.
3) La ciencia es analítica: la investigación científica aborda problemas circunscriptos, uno a uno, y trata de descomponerlo todo en elementos (no necesariamente últimos o siquiera reales). La investigación científica no se planta cuestiones tales como "¿Cómo es el universo en su conjunto?", o "¿Cómo es posible el conocimiento?" Trata, en cambio, de entender toda situación total en términos de sus componentes; intenta descubrir los elementos que explican su integración.
Los problemas de la ciencia son parciales y así son también, por consiguiente, sus soluciones; pero, más aún: al comienzo los problemas son estrechos o es preciso estrecharlos. Pero, a medida que la investigación avanza, su alcance se amplía. Los resultados de la ciencia son generales, tanto en el sentido de que se refieren a clases de objetos (por ejemplo, la lluvia), como en que están, o tienden a ser incorporados en síntesis conceptuales llamadas teorías. El análisis, tanto de los problemas como de las cosas, no es tanto un Mario Bunge La ciencia. Su método y su filosofía 13 objetivo como una herramienta para construir síntesis teóricas. La ciencia auténtica no es atomista ni totalista.
La investigación comienza descomponiendo sus objetos a fin de descubrir el "mecanismo" interno responsable de los fenómenos observados. Pero el desmontaje del mecanismo no se detiene cuando se ha investigado la naturaleza de sus partes; el próximo paso es el examen de la interdependencia de las partes, y la etapa final es la tentativa de reconstruir el todo en términos de sus partes interconectadas. El análisis no acarrea el descuido de la totalidad; lejos de disolver la integración, el análisis es la única manera conocida de descubrir cómo emergen, subsisten y se desintegran los todos. La ciencia no ignora la síntesis: lo que sí rechaza es la pretensión irracionalista de que las síntesis pueden ser aprehendidas por una intuición especial, sin previo análisis.
4) La investigación científica es especializada: una consecuencia del enfoque analítico de los problemas es la especialización. No obstante la unidad del método científico, su aplicación depende, en gran medida, del asunto; esto explica la multiplicidad de técnicas y la relativa independencia de los diversos sectores de la ciencia.
Sin embargo, es menester no exagerar la diversidad de las ciencias al punto de borrar su unidad metodológica. El viejo dualismo materia-espíritu había sugerido la división de las ciencias en Naturwissenschaften, o ciencias de la naturaleza, y Geisteswissenschaften, o ciencias del espíritu. Pero estos géneros difieren en cuanto al asunto, a las técnicas y al grado de desarrollo, no así en lo que respecta al objetivo, método y alcance. El dualismo razón-experiencia había sugerido, a su vez, la división de las ciencias fácticas en racionales y empíricas. Menos sostenible aún es la dicotomía ciencias deductivas—ciencias inductivas, ya que toda empresa científica —sin excluir el dominio de las ciencias formales— es tan inductiva como deductiva, sin hablar de otros tipos de inferencia.
La especialización no ha impedido la formación de campos interdisciplinarios tales como la biofísica, la bioquímica, la psicofisiología, la psicología social, la teoría de la información, la cibernética, o la investigación operacional. Con todo, la investigación tiende a estrechar la visión del científico individual; un único remedio ha resultado eficaz contra la unilateralidad profesional, y es una dosis de filosofía.
5) El conocimiento científico es claro y preciso: sus problemas son distintos, sus resultados son claros. El conocimiento ordinario, en cambio, usualmente es vago e inexacto; en la vida diaria nos preocupamos poco por definiciones precisas, descripciones exactas, o mediciones afinadas: si éstas nos preocuparan demasiado, no lograríamos marchar al paso de la vida. La ciencia torna impreciso lo que el sentido común conoce de manera nebulosa; pero, desde luego la ciencia es mucho más que sentido común organizado: aunque proviene del sentido común, la ciencia constituye una rebelión contra su vaguedad y superficialidad. El conocimiento científico procura la precisión; nunca está enteramente libre de vaguedades, Mario Bunge La ciencia. Su método y su filosofía 14 pero se las ingenia para mejorar la exactitud; nunca está del todo libre de error, pero posee una técnica única para encontrar errores y para sacar provecho de ellos.
La claridad y la precisión se obtienen en ciencia de las siguientes maneras: a) los problemas se formulan de manera clara; lo primero, y a menudo lo más difícil, es distinguir cuáles son los problemas; ni hay artillería analítica o experimental que pueda ser eficaz si no se ubica adecuadamente al enemigo; b) la ciencia parte de nociones que parecen claras al no iniciado; y las complica, purifica y eventualmente las rechaza; la transformación progresiva de las nociones corrientes se efectúa incluyéndolas en esquemas teóricos. Así, por ejemplo, "distancia" adquiere un sentido preciso al ser incluida en la geometría métrica y en la física; c) la ciencia define la mayoría de sus conceptos: algunos de ellos se definen en términos de conceptos no definidos o primitivos, otros de manera implícita, esto es, por la función que desempeñan en un sistema teórico (definición contextual). Las definiciones son convencionales, pero no se las elige caprichosamente: deben ser convenientes y fértiles. (¿De qué vale, por ejemplo, poner un nombre especial a las muchachas pecosas que estudian ingeniería y pesan más de 50 kg?) Una vez que se ha elegido una definición, el discurso restante debe guardarte fidelidad si se quiere evitar inconsecuencias; d) la ciencia crea lenguajes artificiales inventando símbolos (palabras, signos matemáticos, símbolos químicos, etc.; a estos signos se les atribuye significados determinados por medio de reglas de designación (tal como "en el presente contexto H designa el elemento de peso atómico unitario"). los símbolos básicos serán tan simples como sea posible, pero podrán combinarse conforme a reglas determinadas para formar configuraciones tan complejas como sea necesario (las leyes de combinación de los signos que intervienen en la producción de expresiones complejas se llaman reglas de formación); e) la ciencia procura siempre medir y registrar los fenómenos. Los números y las formas geométricas son de gran importancia en el registro, la descripción y la inteligencia de los sucesos y procesos. En lo posible, tales datos debieran disponerse en tablas o resumirse en fórmulas matemáticas. Sin embargo, la formulación matemática, deseable como es, no es una condición indispensable para que el conocimiento sea científico; lo que caracteriza el conocimiento científico es la exactitud en un sentido general antes que la exactitud numérica o métrica, la que es inútil si media la vaguedad conceptual. Más aún, la investigación científica emplea, en medida creciente, capítulos no numéricos y no métricos de la matemática, tales como la topología, la teoría de los grupos, o el álgebra de las clases, que no son ciencias del número y la figura, sino de la relación.
6) El conocimiento científico es comunicable: no es inefable sino expresable, no es privado sino público. El lenguaje científico comunica información a quienquiera haya sido adiestrado para entenderlo. Hay, ciertamente, sentimientos oscuros y nociones difusas, incluso en el Mario Bunge La ciencia. Su método y su filosofía 15 desarrollo de la ciencia (aunque no en la presentación final del trabajo científico); pero es preciso aclararlos antes de poder estimar su adecuación. Lo que es inefable puede ser propio de la poesía o de la música, no de la ciencia, cuyo lenguaje es informativo y no expresivo o imperativo. La inefabilidad misma es, en cambio, tema de investigación científica, sea psicológica o lingüística.
La comunicabilidad es posible gracias a la precisión; y es a su vez una condición necesaria para la verificación de los datos empíricos y de las hipótesis científicas. Aun cuando, por "razones" comerciales o políticas, se mantengan en secreto durante algún tiempo unos trozos del saber, deben ser comunicables en principio para que puedan ser considerados científicos.
La comunicación de los resultados y de las técnicas de la ciencia no sólo perfecciona la educación general sino que multiplica las posibilidades de su confirmación o refutación. La verificación independiente ofrece las máximas garantías técnicas y morales, y ahora es posible, en muchos campos, en escala internacional. Por esto, los científicos consideran el secreto en materia científica como enemigo del progreso de la ciencia; la política del secreto científico es, en efecto, el más eficaz originador de estancamiento en la cultura, en la tecnología y en la economía, así como una fuente de corrupción moral.
7) El conocimiento científico es verificable: debe aprobar el examen de la experiencia. A fin de explicar un conjunto de fenómenos, el científico inventa conjeturas fundadas de alguna manera en el saber adquirido. Sus suposiciones pueden ser cautas o audaces simples o complejas; en todo caso deben ser puestas a prueba. El test de las hipótesis fácticas es empírico, esto es, observacional o experimental. El haberse dado cuenta de esta verdad hoy tan trillada es la contribución inmortal de la ciencia helenística. En ese sentido, las ideas científicas (incluidos los enunciados de leyes) no son superiores a las herramientas o a los vestidos: si fracasan en la práctica, fracasan por entero.
La experimentación puede calar más profundamente que la observación, porque efectúa cambios en lugar de limitarse a registrar variaciones: aísla y controla las variables sensibles o pertinentes. Sin embargo los resultados experimentales son pocas veces interpretables de una sola manera. Más aún, no todas las ciencias pueden experimentar; y en ciertos capítulos de la astronomía y de la economía se alcanza una gran exactitud sin ayuda del experimento.
La ciencia fáctica es por esto empírica en el sentido de que la comprobación de sus hipótesis involucra la experiencia; pero no es necesariamente experimental y en particular no es agotada por las ciencias de laboratorio, tales como la física.
La prescripción de que las hipótesis científicas deben ser capaces de aprobar el examen de la experiencia es una de las reglas del método científico; la aplicación de esta regla depende del tipo de objeto, del tipo de la hipótesis en cuestión y de los medios disponibles. Por esto se necesita una multitud de técnicas de verificación empírica. La verificación de la fórmula de un compuesto químico se hace de manera muy diferente que la verificación de un cálculo astronómico o de una hipótesis concerniente al pasado de las rocas o de los hombres. Las Mario Bunge La ciencia. Su método y su filosofía 16 técnicas de verificación evolucionan en el curso del tiempo; sin embargo, siempre consisten en poner a prueba consecuencias particulares de hipótesis generales (entre ellas, enunciados de leyes). Siempre se reducen a mostrar que hay, o que no hay, algún fundamento para creer que las suposiciones en cuestión corresponden a los hechos observados o a los valores medidos. La verificabilidad hace a la esencia del conocimiento científico; si así no fuera, no podría decirse que los científicos procuran alcanzar conocimiento objetivo.
8) La investigación científica es metódica: no es errática sino planeada. Los investigadores no tantean en la oscuridad: saben lo que buscan y cómo encontrarlo. El planeamiento de la investigación no excluye el azar; sólo que, a hacer un lugar a los acontecimientos imprevistos es posible aprovechar la interferencia del azar y la novedad inesperada. Más aún a veces el investigador produce el azar deliberadamente. Por ejemplo, para asegurar la uniformidad de una muestra, y para impedir una preferencia inconsciente en la elección de sus miembros, a menudo se emplea la técnica de la casualización, en que la decisión acerca de los individuos que han de formar parte de ciertos grupos se deja librada a una moneda o a algún otro dispositivo. De esta manera, el investigador pone el azar al servicio de orden: en lo cual no hay paradoja, porque el acaso opera al nivel de los individuos, al par que el orden opera en el grupo con totalidad.
Todo trabajo de investigación se funda sobre el conocimiento anterior, y en particular sobre las conjeturas mejor confirmadas. (Uno de los muchos problemas de la metodología es, precisamente averiguar cuáles son los criterios para decidir si una hipótesis dada puede considerarse razonablemente confirmada, eso es, si el peso que le acuerdan los fundamentos inductivos y de otro orden basta para conservarla). Más aun, la investigación procede conforme a reglas y técnicas que han resultado eficaces en el pasado pero que son perfeccionadas continuamente, no sólo a la luz de nuevas experiencias, sino también de resultados del examen matemático y filosófico. Una de las reglas de procedimiento de la ciencia fáctica es la siguiente: las variables relevantes (o que se sospecha que son sensibles) debieran variarse una cada vez.
La ciencia fáctica emplea el método experimental concebido en un sentido amplio. Este método consiste en el test empírico de conclusiones particulares extraídas de hipótesis generales (tales como "los gases se dilatan cuando se los calienta" o "los hombres se rebelan cuando se los oprime"). Este tipo de verificación requiere la manipulación de la observación y el registro de fenómenos; requiere también el control de las variables o factores relevantes; siempre que fuera posible debiera incluir la producción artificial deliberada de los fenómenos en cuestión, y en todos los casos exige el análisis de los datos obtenidos en el curso de los procedimientos empíricos. Los datos aislados y crudos son inútiles y no son dignos de confianza; es preciso elaborarlos, organizarlos y confrontarlos con las conclusiones teóricas.
El método científico no provee recetas infalibles para encontrar la verdad: sólo contiene un conjunto de prescripciones falibles (perfectibles) para el planeamiento de observaciones y Mario Bunge La ciencia. Su método y su filosofía 17 experimentos, para la interpretación de sus resultados, y para el planteo mismo de los problemas. Es, en suma, la manera en que la ciencia inquiere en lo desconocido. Subordinadas a las reglas generales del método científico, y al mismo tiempo en apoyo de ellas, encontramos las diversas técnicas que se emplean en las ciencias especiales: las técnicas para pesar, para observar por el microscopio, para analizar compuestos químicos, para dibujar gráficos que resumen datos empíricos, para reunir informaciones acerca de costumbres, etc. La ciencia es pues, esclava de sus propios métodos y técnicas mientras éstos tienen éxito: pero es libre de multiplicar y de modificar en todo momento sus reglas, en aras de mayor racionalidad y objetividad.
9) El conocimiento científico es sistemático: una ciencia no es un agregado de informaciones inconexas, sino un sistema de ideas conectadas lógicamente entre sí. Todo sistema de ideas caracterizado por cierto conjunto básico (pero refutable) de hipótesis peculiares, y que procura adecuarse a una clase de hechos, es una teoría. Todo capítulo de una ciencia especial contiene teorías o sistemas de ideas que están relacionadas lógicamente entre sí, esto es, que están ordenadas mediante la relación "implica". Esta conexión entre las ideas puede calificarse de orgánica, en el sentido de que la sustitución de cualquiera de las hipótesis básicas produce un cambio radical en la teoría o grupo de teorías.
El fundamento de una teoría dada no es un conjunto de hechos sino, más bien, un conjunto de principios, o hipótesis de cierto grado de generalidad (y, por consiguiente, de cierta fertilidad lógica). Las conclusiones (o teoremas) pueden extraerse de los principios, sea en la forma natural, o con la ayuda de técnicas especiales que involucran operaciones matemáticas.
El carácter matemático del conocimiento científico —esto es, el hecho de que es fundado, ordenado y coherente— es lo que lo hace racional. La racionalidad permite que el progreso científico se efectúe no sólo por la acumulación gradual de resultados, sino también por revoluciones. Las revoluciones científicas no son descubrimientos de nuevos hechos aislados, ni son perfeccionamientos en la exactitud de las observaciones, sino que consisten en la sustitución de hipótesis de gran alcance (principios) por nuevos axiomas, y en el reemplazo de teorías enteras por otros sistemas teóricos. Sin embargo, semejantes revoluciones son a menudo provocadas por el descubrimiento de nuevos hechos de los que no dan cuenta las teorías anteriores, aunque a veces se encuentran en el proceso de comprobación de dichas teorías; y las nuevas teorías se tornan verificables en muchos casos, merced a la invención de nuevas técnicas de medición, de mayor precisión.
10) El conocimiento científico es general: ubica los hechos singulares en pautas generales, los enunciados particulares en esquemas amplios. El científico se ocupa del hecho singular en la medida en que éste es miembro de una clase o caso de una ley; más aún, presupone que todo hecho es clasificable y legal. No es que la ciencia ignore la cosa individual o el hecho Mario Bunge La ciencia. Su método y su filosofía 18 irrepetible; lo que ignora es el hecho aislado. Por esto la ciencia no se sirve de los datos empíricos —que siempre son singulares— como tales; éstos son mudos mientras no se los manipula y convierte en piezas de estructuras teóricas.
En efecto, uno de los principios ontológicos que subyacen a la investigación científica es que la variedad y aun la unicidad en algunos respectos son compatibles con la uniformidad y la generalidad en otros respectos. Al químico no le interesa ésta o aquella hoguera, sino el proceso de combustión en general: trata de descubrir lo que comparten todos los singulares.
El científico intenta exponer los universales que se esconden en el seno de los propios singulares; es decir, no considera los universales ante rem ni post rem sino in re: en la cosa, y no antes o después de ella. Los escolásticos medievales clasificarían al científico moderno como realista inmanentista, porque, al descartar los detalles al procurar descubrir los rasgos comunes a individuos que son únicos en otros respectos, al buscar las variables pertinentes (o cualidades esenciales) y las relaciones constantes entre ellas (las leyes), el científico intenta exponer la naturaleza esencial de las cosas naturales y humanas.
El lenguaje científico no contiene solamente términos que designan hechos singulares y experiencias individuales, sino también términos generales que se refieren a clases de hechos.
La generalidad del lenguaje de la ciencia no tiene, sin embargo, el propósito de alejar a la ciencia de la realidad concreta: por el contrario, la generalización es el único medio que se conoce para adentrarse en lo concreto, para apresar la esencia de las cosas (sus cualidades y leyes esenciales). Con esto, el científico evita en cierta medida las confusiones y los engaños provocados por el flujo deslumbrador de los fenómenos. Tampoco se asfixia la utilidad en la generalidad: por el contrario, los esquemas generales de la ciencia encuadran una cantidad ilimitada de casos específicos, proveen leyes de amplio alcance que incluyen y corrigen todas las recetas válidas de sentido común y de la técnica precientífica.
11) El conocimiento científico es legal: busca leyes (de la naturaleza y de la cultura) y las aplica. El conocimiento científico inserta los hechos singulares en pautas generales llamadas "leyes naturales" o "leyes sociales". Tras el desorden y la fluidez de las apariencias, la ciencia fáctica descubre las pautas regulares de la estructura y del proceso del ser y del devenir. En la medida en que la ciencia es legal, es esencialista: intenta legar a la raíz de las cosas.
Encuentra la esencia en las variables relevantes y en las relaciones invariantes entre ellas.
Hay leyes de hechos y leyes mediante las cuales se pueden explicar otras leyes. El principio de Arquímedes pertenece a la primera clase; pero a su vez puede deducirse de los principios generales de la mecánica; por consiguiente, ha dejado de ser un principio independiente, y ahora es un teorema deducible de hipótesis de nivel más elevado. Las leyes de la física proveen la base de las leyes de las combinaciones químicas; las leyes de la fisiología explican ciertos fenómenos psíquicos; y las leyes de la economía pertenecen a los fundamentos de la sociología. Es decir, los enunciados de las leyes se organizan en una estructura de niveles.
Ciertamente, los enunciados de las leyes son transitorios; pero ¿son inmutables las leyes Mario Bunge La ciencia. Su método y su filosofía 19 mismas? Si se considera a las leyes como las pautas mismas del ser y del devenir, entonces debieran cambiar junto con las cosas mismas; por lo menos, debe admitirse que, al emerger nuevos niveles, sus cualidades peculiares se relacionan entre sí mediante nuevas leyes. Por ejemplo, las leyes de la economía han emergido en el curso de la historia sobre la base de otras leyes (biológicas y psicológicas) y, más aún, algunas de ellas cambian con el tipo de organización social.
Por supuesto, no todos los hechos singulares conocidos han sido ya convertidos en casos particulares de leyes generales; en particular los sucesos y procesos de los niveles superiores han sido legalizados sólo en pequeña medida. Pero esto se debe en parte al antiguo prejuicio de que lo humano no es legal, así como a la antigua creencia pitagórica de que solamente las relaciones numéricas merecen llamarse "leyes científicas". Debiera emplearse el stock íntegro de las herramientas conceptuales en la búsqueda de las leyes de la mente y de la cultura; más aún, acaso el stock de que se dispone es insuficiente y sea preciso inventar herramientas radicalmente nuevas para tratar los fenómenos mentales y culturales, tal como el nacimiento de la mecánica moderna hubiera sido imposible sin la invención expresa del cálculo infinitesimal.
Pero el ulterior avance en el progreso de la legalización de los fenómenos no físicos requiere por sobre todo, una nueva actitud frente al concepto mismo de ley científica. En primer lugar, es preciso comprender que hay muchos tipos de leyes (aun dentro de una misma ciencia), ninguno de los cuales es necesariamente mejor que los tipos restantes. En segundo lugar, debiera tornarse un lugar común entre los científicos de la cultura el que las leyes no se encuentran por mera observación y el simple registro sino poniendo a prueba hipótesis: los enunciados de leyes no son, en efecto, sino hipótesis confirmadas. Y cómo habríamos de emprender la confección de hipótesis científicas si no presumiéramos que todo hecho singular es legal?
12) La ciencia es explicativa: intenta explicar los hechos en términos de leyes, y las leyes en términos de principios. Los científicos no se conforman con descripciones detalladas; además de inquirir cómo son las cosas, procuran responder al por qué: por qué ocurren los hechos como ocurren y no de otra manera. La ciencia deduce proposiciones relativas a hechos singulares a partir de leyes generales, y deduce las leyes a partir de enunciados nomológicos aún más generales (principios). Por ejemplo, las leyes de Kepler explicaban una colección de hechos observados del movimiento planetario; y Newton explicó esas leyes deduciéndolas de principios generales explicación que permitió a otros astrónomos dar cuenta de las irregularidades de las órbitas de los planetas que eran desconocidas para Kepler.
Solía creerse que explicar es señalar la causa, pero en la actualidad se reconoce que la explicación causal no es sino un tipo de explicación científica. La explicación científica se efectúa siempre en términos de leyes, y las leyes causales no son sino una subclase de las leyes científicas. Hay diversos tipos de leyes científicas y, por consiguiente, hay una Mario Bunge La ciencia. Su método y su filosofía 20 variedad de tipos de explicación científica: morfológicas, cinemáticas, dinámicas, de composición, de conservación, de asociación, de tendencias globales, dialécticas, teleológicas, etc.
La historia de la ciencia enseña que las explicaciones científicas se corrigen o descartan sin cesar. ¿Significa esto que son todas falsas? En las ciencias fácticas, la verdad y el error no son del todo ajenos entre sí: hay verdades parciales y errores parciales; hay aproximaciones buenas y otras malas. La ciencia no obra como Penélope, sino que emplea la tela tejida ayer.
Las explicaciones científicas no son finales pero son perfectibles.
13) El conocimiento científico es predictivo: Trasciende la masa de los hechos de experiencia, imaginando cómo puede haber sido el pasado y cómo podrá ser el futuro. La predicción es, en primer lugar, una manera eficaz de poner a prueba las hipótesis; pero también es la clave del control y aun de la modificación del curso de los acontecimientos. La predicción científica en contraste con la profecía se funda sobre leyes y sobre informaciones específicas fidedignas, relativas al estado de cosas actual o pasado. No es del tipo "ocurrirá E", sino más bien de este otro: "ocurrirá E1 siempre que suceda C1, pues siempre que sucede C es seguido por o está asociado con E". C y E designan clases de sucesos en tanto que C1 y E1 denotan los hechos específicos que se predicen sobre la base del o los enunciados que conectan a C con E en general.
La predicción científica se caracteriza por su perfectibilidad antes que por su certeza. Más aún, las predicciones que se hacen con la ayuda de reglas empíricas son a veces más exactas que las predicciones penosamente elaboradas con herramientas científicas (leyes, informaciones específicas y deducciones); tal es el caso con frecuencia de los pronósticos meteorológicos, de la prognosis médica y de la profecía política. Pero en tanto que la profecía no es perfectible y no puede usarse para poner a prueba hipótesis, la predicción es perfectible y, si falla, nos obliga a corregir nuestras suposiciones, alcanzando así una inteligencia más profunda. Por esto la profecía exitosa no es un aporte al conocimiento teórico, en tanto que la predicción científica fallida puede contribuir a él.
Puesto que la predicción científica depende de leyes y de ítems de información específica, puede fracasar por inexactitud de los enunciados de las leyes o por imprecisión de la información disponible. (También puede fallar, por supuesto, debido a errores cometidos en el proceso de inferencia lógica o matemática que conduce de las premisas (leyes e informaciones) a la conclusión (enunciado predictivo)). Una fuente importante de fallos en la predicción es el conjunto de suposiciones acerca de la naturaleza del objeto (sistema físico, organismo vivo, grupo social, etc.) cuyo comportamiento ha de predecirse. Por ejemplo, puede ocurrir que creamos que el sistema en cuestión está suficientemente aislado de las perturbaciones exteriores, cuando en rigor éstas cuentan a la larga; dado que la aislación es una condición necesaria de la descripción del sistema con ayuda de un puñado de enunciados de leyes, no debiera sorprender que fuera tan difícil predecir el comportamiento Mario Bunge La ciencia. Su método y su filosofía 21 de sistemas abiertos tales como el océano, la atmósfera, el ser vivo o el hombre.
Puesto que la predicción científica se funda en las leyes científicas, hay tantas clases de predicciones como clases de enunciado nomológicos. Algunas leyes nos permiten predecir resultados individuales, aunque no sin error si la predicción se refiere al valor de una cantidad. Otras leyes; incapaces de decirnos nada acerca del comportamiento de los individuos (átomos, personas, etc.) son en cambio la base para la predicción de algunas tendencias globales y propiedades colectivas de colecciones numerosas de elementos similares; son las leyes estadísticas. Las leyes de la historia son de este tipo; y por esto es casi imposible la predicción de los sucesos individuales en el campo de la historia, pudiendo preverse solamente el curso general de los acontecimientos.
14) La ciencia es abierta: no reconoce barreras a priori que limiten el conocimiento. Si un conocimiento fáctico no es refutable en principio, entonces no pertenece a la ciencia sino a algún otro campo. Las nociones acerca de nuestro medio, natural o social, o acerca del yo, no son finales: están todas en movimiento, todas son falibles. Siempre es concebible que pueda surgir una nueva situación (nuevas informaciones o nuevos trabajos teóricos) en que nuestras ideas, por firmemente establecidas que parezcan, resulten inadecuadas en algún sentido. La ciencia carece de axiomas evidentes: incluso los principios más generales y seguros son postulados que pueden ser corregidos o reemplazados. A consecuencia del carácter hipotético de los enunciados de leyes, y de la naturaleza perfectible de los datos empíricos la ciencia no es un sistema dogmático y cerrado sino controvertido y abierto. O, más bien, la ciencia es abierta como sistema porque es falible y por consiguiente capaz de progresar. En cambio, puede argüirse que la ciencia es metodológicamente cerrada no en el sentido de que las reglas del método científico sean finales sino en el sentido de que es autocorrectiva: el requisito de la verificabilidad de las hipótesis científicas basta para asegurar el progreso científico.
Tan pronto como ha sido establecida una teoría científica, corre el peligro de ser refutada o, al menos, de que se circunscriba su dominio. Un sistema cerrado de conocimiento fáctico que excluya toda ulterior investigación, puede llamarse sabiduría pero es en rigor un detritus de la ciencia. El sabio moderno, a diferencia del antiguo no es tanto un acumulador de conocimientos como un generador de problemas. Por consiguiente, prefiere los últimos números de las revistas especializadas a los manuales, aun cuando estos últimos sean depósitos de verdad más vastos y fidedignos que aquellas. El investigador moderno ama la verdad pero no se interesa por las teorías irrefutables. Una teoría puede haber permanecido intocada no tanto por su alto contenido de verdad cuanto porque nadie la ha usado. No se necesita emprender una investigación empírica para probar la tautología de que ni siquiera los científicos se casan con solteronas.
Los modernos sistemas de conocimiento científico son como organismos en crecimiento: mientras están vivos cambian sin pausa. Esta es una de las razones por las cuales la ciencia Mario Bunge La ciencia. Su método y su filosofía 22 es éticamente valiosa: porque nos recuerda que la corrección de errores es tan valiosa como el no cometerlos y que probar cosas nuevas e inciertas es preferible a rendir culto a las viejas y garantizadas. La ciencia, como los organismos, cambia a la vez internamente y debido a sus contactos con sus vecinos; esto es, resolviendo sus problemas específicos y siendo útil en otros campos.
15) La ciencia es útil: porque busca la verdad, la ciencia es eficaz en la provisión de herramientas para el bien y para el mal. El conocimiento ordinario se ocupa usualmente de lograr resultados capaces de ser aplicados en forma inmediata; con ello no es suficientemente verdadero, con lo cual no puede ser suficientemente eficaz. Cuando se dispone de un conocimiento adecuado de las cosas es posible manipularlas con éxito. La utilidad de la ciencia es una consecuencia de su objetividad; sin proponerse necesariamente alcanzar resultados aplicables, la investigación los provee a la corta o a la larga. La sociedad moderna paga la investigación porque ha aprendido que la investigación rinde. Por este motivo, es redundante exhortar a los científicos a que produzcan conocimientos aplicables: no pueden dejar de hacerlo. Es cosa de los técnicos emplear el conocimiento científico con fines prácticos, y los políticos son los responsables de que la ciencia y la tecnología se empleen en beneficio de la humanidad. Los científicos pueden, a lo sumo, aconsejar acerca de cómo puede hacerse uso racional, eficaz y bueno de la ciencia.
La técnica precientífica era primordialmente una colección de recetas pragmáticas no entendidas, muchas de las cuales desempeñaban la función de ritos mágicos. La técnica moderna es, en medida creciente —aunque no exclusivamente—, ciencia aplicada. La ingeniería es física y química aplicadas, la medicina es biología aplicada, la psiquiatría es psicología y neurología aplicadas; y debiera llegar el día en que la política se convierta en sociología aplicada.
Pero la tecnología es más que ciencia aplicada: en primer lugar porque tiene sus propios procedimientos de investigación, adaptados a circunstancias concretas que distan de los casos puros que estudia la ciencia. En segundo lugar, porque toda rama de la tecnología contiene un cúmulo de reglas empíricas descubiertas antes que los principios científicos en los que —si dichas reglas se confirman— terminan por ser absorbidas. La tecnología no es meramente el resultado de aplicar el conocimiento científico existente a los casos prácticos: la tecnología viva es esencialmente, el enfoque científico de los problemas prácticos, es decir, el tratamiento de estos problemas sobre un fondo de conocimiento científico y con ayuda del método científico. Por eso la tecnología, sea de las cosas nuevas o de los hombres, es fuente de conocimientos nuevos.
La conexión de la ciencia con la tecnología no es por consiguiente asimétrica. Todo avance tecnológico plantea problemas científicos cuya solución puede consistir en la invención de nuevas teorías o de nuevas técnicas de investigación que conduzcan a un conocimiento más adecuado y a un mejor dominio del asunto. La ciencia y la tecnología constituyen un ciclo de Mario Bunge La ciencia. Su método y su filosofía 23 sistemas interactuantes que se alimentan el uno al otro. El científico torna inteligible lo que hace el técnico y éste provee a la ciencia de instrumentos y de comprobaciones; y lo que es igualmente importante el técnico no cesa de formular preguntas al científico añadiendo así un motor externo al motor interno del progreso científico. La continuación de la vida sobre la Tierra depende del ciclo de carbono: los animales se alimentan de plantas, las que a su vez obtienen su carbono de lo que exhalan los animales. Análogamente la continuación de la civilización moderna depende, en gran medida del ciclo del conocimiento: la tecnología moderna come ciencia, y la ciencia moderna depende a su vez del equipo y del estímulo que le provee una industria altamente tecnificada.
Pero la ciencia es útil en más de una manera. Además de constituir el fundamento de la tecnología, la ciencia es útil en la medida en que se la emplea en la edificación de concepciones del mundo que concuerdan con los hechos, y en la medida en que crea el hábito de adoptar una actitud de libre y valiente examen, en que acostumbra a la gente a poner a prueba sus afirmaciones y a argumentar correctamente. No menor es la utilidad que presta la ciencia como fuente de apasionantes rompecabezas filosóficos, y como modelo de la investigación filosófica.
En resumen, la ciencia es valiosa como herramienta para domar la naturaleza y remodelar la sociedad; es valiosa en sí misma, como clave para la inteligencia del mundo y del yo; y es eficaz en el enriquecimiento, la disciplina y la liberación de nuestra mente.
Mario Bunge La ciencia. Su método y su filosofía 24 ¿Cuál es el método de la ciencia?
Mario Bunge La ciencia. Su método y su filosofía 25 Mario Bunge La ciencia. Su método y su filosofía 1 G. Bocaccio Vita di Dante, en Il comento alla Divina Commedia e gli altri scriti intorno a Dante (Bari, Laterza 1918), I, p. 37. Subrayado mío.
2 D. Hume, A Treatise of Human Nature (London, Everyman, 1911) I, p. 105. Subrayado mío.
26 "The lame in the path outstrips the swift who wander from it." F. Bacon 1. La ciencia, conocimiento verificable En su deliciosa biografía del Dante (ca. 1360), Boccaccio1 expuso su opinión —que no viene al caso— acerca del origen de la palabra "poesía" concluyendo con este comentario: "otros lo atribuyen a razones diferentes, acaso aceptables; pero ésta me gusta más". El novelista aplicaba, al conocimiento acerca de la poesía y de su nombre el mismo criterio que podría apreciarse para apreciar la poesía misma: el gusto. Confundía así valores situados en niveles diferentes: el estético, perteneciente a la esfera de la sensibilidad, y el gnoseológico, que no obstante estar enraizado en la sensibilidad está enriquecido con una cualidad emergente: la razón.
Semejante confusión no es exclusiva de poetas: incluso Hume, en una obra célebre por su crítica mortífera de varios dogmas tradicionales escogió el gusto como criterio de verdad. En su Treatise of Human Nature (1739) puede leerse2: "No es sólo en poesía y en música que debemos seguir nuestro gusto, sino también en la filosofía (que en aquella época incluía también a la ciencia). Cuando estoy convencido de algún principio, no es sino una idea que me golpea (strikes) con mayor fuerza. Cuando prefiero un conjunto de argumentos por sobre otros, no hago sino decidir, sobre la base de mi sentimiento, acerca de la superioridad de su influencia". El subjetivismo era así la playa en que desembarcaba la teoría psicologista de las "ideas" inaugurada por el empirismo de Locke.
El recurso al gusto no era, por supuesto, peor que el argumento de autoridad, criterio de verdad que ha mantenido enjaulado al pensamiento durante tanto tiempo y con tanta eficacia.
Desgraciadamente, la mayoría de la gente, y hasta la mayoría de los filósofos, aún creen —u obran como si creyeran— que la manera correcta de decir el valor de verdad de un enunciado es someterlo a la prueba de algún texto: es decir verificar si es compatible con (o deducible de) frases más o menos célebres tenidas por verdades eternas, o sea, principios infalibles de alguna escuela de pensamiento. En efecto, son demasiados los argumentos filosóficos que se ajustan al siguiente molde: "X está equivocado, porque lo que dice contradice lo que escribió el maestro Y", o bien "el X-ismo es falso porque sus tesis son incompatibles con las proposiciones fundamentales de Y-ismo". Los dogmáticos —antiguos y modernos fuera y Mario Bunge La ciencia. Su método y su filosofía 3 Aristóteles, Analíticos Posteriores, libro II, cap. XIX 110 b.
4 W. James, Pragmatism, (New York, Meridian Books, 1935), p. 134.
27 dentro de la profesión científica, maliciosos o no— obran de esta manera aun cuando no desean convalidar creencias que simplemente no pueden ser comprobadas, sea empíricamente, sea racionalmente. Porque "dogma" es, por definición, toda opinión no confirmada de la que no se exige verificación porque se la supone verdadera y, más aún, se la supone fuente de verdades ordinarias.
Otro criterio de verdad igualmente difundido ha sido la evidencia. Según esta opinión, verdadero es aquello que parece aceptable a primera vista, sin examen ulterior: aquello, en suma, que se intuye. Así, Aristóteles 3 afirmaba que la intuición "aprehende las premisas primarias" de todo discurso, y es por ello "la fuente que origina el conocimiento científico".
No sólo Bergson, Husserl y mucho otros intuicionistas e irracionalistas han compartido la opinión de que las esencias pueden cogerse sin más: también el racionalismo ingenuo, tal como el que sostenía Descartes, afirma que hay principios evidentes que, lejos de tener que someterse a prueba alguna, son la piedra de toque de toda otra proposición, sea formal o fáctica.
Finalmente, otros han favorecido las "verdades vitales" (o las "mentiras vitales"), esto es, las afirmaciones que se creen o no por conveniencia, independientemente de su fundamento racional y/o empírico. Es el caso de Nietzsche y los pragmatistas posteriores, todos los cuales han exagerado el indudable valor instrumental del conocimiento fáctico, al punto de afirmar que "la posesión de la verdad, lejos de ser (...) un fin en sí, es sólo un medio preliminar para alcanzar otras satisfacciones vitales"4, de donde "verdadero" es sinónimo de "útil".
Pregúntese a un científico si cree que tiene derecho a suscribir una afirmación en el campo de las ciencias tan sólo porque le guste, o porque la considere un dogma inexpugnable o porque a él le parezca evidente, o porque la encuentre conveniente. Probablemente conteste más o menos así: ninguno de esos presuntos criterios de verdad garantiza la objetividad, y el conocimiento objetivo es la finalidad de la investigación científica. Lo que se acepta sólo por gusto o por autoridad, o por parecer evidente (habitual) o por conveniencia, no es sino creencia u opinión, pero no es conocimiento científico. El conocimiento científico es a veces desagradable, a menudo contradice a los clásicos (sobre todo si es nuevo), en ocasiones tortura al sentido común y humilla a la intuición; por último, puede ser conveniente para algunos y no para otros. En cambio aquello que caracteriza al conocimiento científico es su verificabilidad: siempre es susceptible de ser verificado (confirmado o disconfirmado).
2. Veracidad y verificabilidad Obsérvese que no pretendemos que el conocimiento científico, por contraste con el ordinario, Mario Bunge La ciencia. Su método y su filosofía 28 el tecnológico o el filosófico, sea verdadero. Ciertamente lo es con frecuencia, y siempre intenta serlo más y más. Pero la veracidad, que es un objetivo, no caracteriza el conocimiento científico de manera tan inequívoca como el modo, medio o método por el cual la investigación científica plantea problemas y pone a prueba las soluciones propuestas.
En ocasiones, puede alcanzarse una verdad con sólo consultar un texto. Los propios científicos recurren a menudo a un argumento de autoridad atenuada: lo hacen siempre que emplean datos (empíricos o formales) obtenidos por otros investigadores —cosa que no pueden dejar de hacer, pues la ciencia moderna es, cada vez más, una empresa social—. Pero, por grande que sea la autoridad que se atribuye a una fuente, jamás se la considera infalible: si se aceptan sus datos, es sólo provisionalmente y porque se presume que han sido obtenidos con procedimientos que concuerdan con el método científico, de manera que son reproducibles por quienquiera que se disponga a aplicar tales procedimientos. En otras palabras: un dato será considerado verdadero hasta cierto punto, siempre que pueda ser confirmado de manera compatible con los cánones del método científico.
En consecuencia, para que un trozo de saber merezca ser llamado "científico", no basta —ni siquiera es necesario— que sea verdadero. Debemos saber, en cambio, cómo hemos llegado a saber, o a presumir, que el enunciado en cuestión es verdadero: debemos ser capaces de enumerar las operaciones (empíricas o racionales) por las cuales es verificable (confirmable o disconfirmable) de una manera objetiva al menos en principio. Esta no es sino una cuestión de nombres: quienes no deseen que se exija la verificabilidad del conocimiento deben abstraerse de llamar "científicas" a sus propias creencias, aun cuando lleven bonitos nombres con raíces griegas. Se las invita cortésmente a bautizarlas con nombres más impresionantes, tales como "reveladas, evidentes, absolutas, vitales, necesarias para la salud del Estado, indispensables para la victoria del partido", etc.
Ahora bien, para verificar un enunciado —porque las proposiciones, y no los hechos, son verdaderas y falsas y pueden, por consiguiente, ser verificadas— no basta la contemplación y ni siquiera el análisis. Comprobamos nuestras afirmaciones confrontándolas con otros enunciados. El enunciado confirmatorio (o disconfirmatorio), que puede llamarse el verificans, dependerá del conocimiento disponible y de la naturaleza de la proposición dada, la que puede llamarse verificandum. Los enunciados confirmatorios serán enunciados referentes a la experiencia si lo que se somete a prueba es una afirmación fáctica, esto es, un enunciado acerca de hechos, sean experimentados o no. Observemos, de pasada, que el científico tiene todo el derecho de especular acerca de hechos inexperienciales, esto es, hechos que en una etapa del desarrollo del conocimiento están más allá de alcance de la experiencia humana; pero entonces está obligado a señalar las experiencias que permiten inferir tales hechos inobservados o aun inobservables; vale decir tiene la obligación de anclar sus enunciados fácticos en experiencias conectadas de alguna manera con los hechos transempíricos que supone. Baste recordar la historia de unos pocos inobservables distinguidos: la otra cara de la Luna, las ondas luminosas, los átomos, la conciencia, la lucha de clases y la opinión Mario Bunge La ciencia. Su método y su filosofía 29 pública.
En cambio, si lo que se ha verificado no es una proposición referente al mundo exterior sino un enunciado respecto al comportamiento de signos (tal como por ej. 2 + 3 = 5), entonces los enunciados confirmatorios serán definiciones, axiomas, y reglas que se adoptan por una razón cualquiera (p. ej., porque son fecundas en la organización de los conceptos disponibles y en la elaboración de nuevos conceptos). En efecto, la verificación de afirmaciones pertenecientes al dominio de las formas (lógica y matemática) no requiere otro instrumento material que el cerebro; sólo la verdad fáctica —como en el caso de "la Tierra es redonda"— requiere la observación o el experimento.
Resumiendo: la verificación de enunciados formales sólo incluye operaciones racionales, en tanto que las proposiciones que comunican información acerca de la naturaleza o de la sociedad han de ponerse a prueba por ciertos procedimientos empíricos tales como el recuento o la medición. Pues, aunque el conocimiento de los hechos no provienen de la experiencia pura —por ser la teoría un componente indispensable de la recolección de informaciones fácticas— no hay otra manera de verificar nuestras sospechas que recurrir a la experiencia, tanto "pasiva" como activa.
3. Las proposiciones generales verificables: hipótesis científicas La descripción que antecede satisfará, probablemente, a cualquier científico contemporáneo que reflexione sobre su propia actividad. Pero no resolverá la cuestión para el metacientífico o epistemólogo, para quien los procedimientos, las normas y a veces hasta los resultados de la ciencia son otros tantos problemas. En efecto, el metacientífico no puede dejar de preguntarse cuáles son las afirmaciones verificables, cómo se llega a afirmarlas, cómo se las comprueba, y en qué condiciones puede decirse que han sido confirmadas. Tratemos de esbozar una respuesta a estas preguntas.
En primer lugar si hemos de tratar el problema de la verificación, debemos averiguar qué se puede verificar, ya que no toda afirmación —ni siquiera toda afirmación significativa— es verificable. Así, por ejemplo, las definiciones nominales —tales como "América es el continente situado al oeste de Europa"— se aceptan o rechazan sobre la base del gusto, de la conveniencia, etc., pero no pueden verificarse, y ello simplemente porque no son verdaderas ni falsas. Por ejemplo, si convenimos en llamar "norte-sur" a la dirección que toma normalmente la aguja de una brújula, semejante nombre puede gustarnos o no, pero es inverificable: no es sino un nombre, no se funda sobre elemento de prueba alguno y ninguna operación podría confirmarlo o disconfirmarlo. En cambio lo que puede confirmarse o disconfirmarse es una afirmación fáctica que contenga a ese término tal como "la 5ª Avenida corre de sur a norte". La verificación de esa afirmación es posible, y puede hacerse con la ayuda de una brújula.
No sólo las definiciones nominales sino también las afirmaciones acerca de fenómenos Mario Bunge La ciencia. Su método y su filosofía 30 sobrenaturales son inverificables, puesto que por definición trascienden todo cuanto está a nuestro alcance, y no se las puede poner a prueba con ayuda de la lógica ni de la matemática.
Las afirmaciones acerca de la sobrenaturaleza son inverificables no porque no se refieran a hechos —pues a veces pretenden hacerlo—, sino porque no se dispone de método alguno mediante el cual se podrá decidir cuál es su valor de verdad. En cambio, muchas de ellas son perfectamente significativas para quien se tome el trabajo de ubicarlas en su contexto sin pretender reducirlas, por ejemplo, a conceptos científicos. La verificación torna más exacto el significado, pero no produce significado alguno. Más bien al contrario, la posesión de un significado determinado es una condición necesaria para que una proposición sea verificable.
Pues, ¿cómo habríamos de disponernos a comprobar lo que no entendemos?
Ahora bien, los enunciados verificables son de muchas clases. Hay proposiciones singulares tales como "este trozo de hierro está caliente"; particulares o existenciales, tales como "algunos trozos de hierro están calientes" (que es verificablemente falsa). Hay, además, enunciados de leyes, tales como "todos los metales se dilatan con el calor" (o mejor, "para todo x, si x es un trozo de metal que se calienta, entonces x se dilata"). Las proposiciones singulares y particulares pueden verificarse a menudo de manera inmediata, con la sola ayuda de los sentidos o eventualmente, con el auxilio de instrumentos que amplíen su alcance; pero otras veces exigen operaciones complejas que implican enunciados de leyes y cálculos matemáticos, como es el caso de "la distancia media entre la Tierra y el Sol es de unos 1.500 millones de kilómetros".
Cuando un enunciado verificable posee un grado de generalidad suficiente, habitualmente se lo llama hipótesis científica. O, lo que es equivalente, cuando una proposición general (particular o universal) puede verificarse sólo de manera indirecta —esto es, por el examen de algunas de sus consecuencias— es conveniente llamarla "hipótesis científica". Por ejemplo, "todos los trozos de hierro se dilatan con el calor", y a fortiori, "todos los metales se dilatan con el calor", son hipótesis científicas: son puntos de partida de raciocinios y, por ser generales, sólo pueden ser confirmados poniendo a prueba sus consecuencias particulares, esto es, probando enunciados referentes a muestras específicas de metal.
Solía creerse que el discurso científico no incluye elementos hipotéticos sino tan sólo hechos, y, sobre todo, lo que en inglés se denominan hard facts. Ahora se comprende que el núcleo de toda teoría científica es un conjunto de hipótesis verificables. Las hipótesis científicas son, por una parte, remates de cadenas inferenciales no demostrativas (analógicas o inductivas) más o menos oscuras; por otra parte, son puntos de partida de cadenas deductivas cuyos últimos eslabones —los más próximos a los sentidos, en el caso de la ciencia fáctica—, deben pasar la prueba de la experiencia.
Más aún: habitualmente se concuerda en que debiera llamarse "hipótesis" no sólo a las conjeturas de ensayo, sino también a las suposiciones razonablemente confirmadas o establecidas, pues probablemente no hay enunciados fácticos generales perfectos. La experiencia ha sugerido adoptar este sentido de la palabra "hipótesis". Considérese, por Mario Bunge La ciencia. Su método y su filosofía 5 P. W. Bridgam, Reflections of a Physicist (N. York, Philosophical Library, 1955), p. 83.
31 ejemplo, la ley de Newton de la gravedad, que ha sido confirmada en casi todos los casos con una precisión asombrosa. Tenemos dos razones para llamarla hipótesis: la primera es que ha pasado la prueba sólo un número finito de veces; la segunda, es que hemos terminado por aprehender que incluso ese célebre enunciado de ley es tan sólo una primera aproximación de un enunciado más exacto incluido en la teoría general de la relatividad, que tampoco es probable que sea definitiva.
4. El método científico ¿ars inveniendi?
Hemos convenido en que un enunciado fáctico general susceptible de ser verificado puede llamarse hipótesis, lo que suena más respetable que corazonada, sospecha, conjetura, suposición o presunción, y es también más adecuado que estos términos, ya que la etimología de “hipótesis” es punto de partida, que ciertamente lo es una vez que se ha dado con ella. Abordemos ahora el segundo problema que nos propusimos, a saber: ¿existe una técnica infalible para inventar hipótesis científicas que sean probablemente verdaderas? En otras palabras: ¿existe un método, en el sentido cartesiano de conjunto de "reglas ciertas y fáciles" que nos conduzca a enunciar verdades fácticas de gran extensión?
Muchos hombres, en el curso de muchos siglos, han creído en la posibilidad de descubrir la técnica del descubrimiento, y de inventar la técnica de la invención. Fue fácil bautizar al niño no nacido, y se lo hizo con el nombre de ars inveniendi. Pero semejante arte jamás fue inventado. Lo que es más, podría argüirse que jamás se lo inventará, a menos que se modifique radicalmente la definición de "ciencia"; en efecto, el conocimiento científico por oposición a la sabiduría revelada, es esencialmente falible, esto es, susceptible de ser parcial o aun totalmente refutado. La falibilidad del conocimiento científico, y, por consiguiente, la imposibilidad de establecer reglas de oro que nos conduzcan derechamente a verdades finales, no es sino el complemento de aquella verificabilidad que habíamos encontrado en el núcleo de la ciencia.
Vale decir, no hay reglas infalibles que garanticen por anticipado el descubrimiento de nuevos hechos y la invención de nuevas teorías, asegurando así la fecundidad de la investigación científica: la certidumbre debe buscarse tan solo en las ciencias formales. ¿Significa esto que la investigación científica es errática e ilegal, y por consiguiente que los científicos lo esperan todo de la intuición o de la iluminación? Ta es la moraleja que algunos científicos y filósofos eminentes han extraído de la inexistencia de leyes que nos aseguren contra la infertilidad y el error. Por ejemplo, Bridgman —el expositor del operacionismo— ha negado la existencia del método científico, sosteniendo que "la ciencia es lo que hacen los científicos, y hay tantos métodos científicos como hombres de ciencia"5.
Mario Bunge La ciencia. Su método y su filosofía 6 C. Huyghens Traité de la lumière (París, Gauthier-Villars 1920), p. 5.
7 J. C. Maxwell, A Treatise of Electricity and Magnetism, 3ª ed. (Oxford, University Press 1937), II, pp.
434 y ss.
32 Es verdad que en ciencia no hay caminos reales; que la investigación se abre camino en la selva de los hechos, y que los científicos sobresalientes elaboran su propio estilo de pesquisa. Sin embargo esto no debe hacernos desesperar de la posibilidad de descubrir pautas, normalmente satisfactorias de plantear problemas y poner a prueba hipótesis. Los científicos que van en pos de la verdad no se comportan ni como soldados que cumplen obedientemente las reglas de la ordenanza (opiniones de Bacon y Descartes), ni como los caballeros de Mark Twain, que cabalgaban en cualquier dirección para llegar a Tierra Santa (opinión de Bridgman). No hay avenidas hechas en ciencia, pero hay en cambio una brújula mediante la cual a menudo es posible estimar si se está sobre una huella promisoria. Esta brújula es el método científico, que no produce automáticamente el saber, pero que nos evita perdernos en el caos aparente de los fenómenos, aunque sólo sea porque nos indica cómo no plantear los problemas y cómo no sucumbir al embrujo de nuestros prejuicios predilectos.
La investigación no es errática sino metódica; sólo que no hay una sola manera de sugerir hipótesis, sino muchas maneras: las hipótesis no se nos imponen por la fuerza de los hechos, sino que son inventadas para dar cuenta de los hechos. Es verdad que la invención no es ilegal, sino que sigue ciertas pautas; pero éstas son psicológicas antes que lógicas, son peculiares de los diversos tipos intelectuales, y, por añadidura, los conocemos poco, porque apenas se los investiga. Hay, ciertamente, reglas que facilitan la invención científica, y en especial la formulación de hipótesis; entre ellas figuran las siguientes: el sistemático reordenamiento de los datos, la supresión imaginaria de factores con el fin de descubrir las variables relevantes, el obstinado cambio de representación en busca de analogías fructíferas.
Sin embargo, las reglas que favorecen o entorpecen el trabajo científico no son de oro sino plásticas; más aún, el investigador rara vez tiene conciencia del camino que ha tomado para formular sus hipótesis. Por esto la investigación científica puede planearse a grandes líneas y no en detalle, y aún menos puede ser regimentada.
Algunas hipótesis se formulan por vía inductiva, esto es, como generalizaciones sobre la base de la observación de un puñado de casos particulares. Pero la inducción dista de ser la única o siquiera la principal de las vías que conducen a formular enunciados generales verificables.
Otras veces, el científico opera por analogía; por ejemplo la teoría ondulatoria de la luz le fue sugerida a Huyghens (1690) por una comparación con las olas6. En algunos casos el principio heurístico es una analogía matemática; así, por ejemplo, Maxwell (1873) predijo la existencia de ondas electromagnéticas sobre la base de una analogía formal entre sus ecuaciones del campo y la conocida ecuación de las ondas elásticas 7. Ocasionalmente, el investigador es guiado por consideraciones filosóficas; así fue como procedió Oersted (1820); buscó Mario Bunge La ciencia. Su método y su filosofía 8 Véase, p. ej. S. F. Mason, A History of the Sciences (London, Routledge & Kegan Paul, 1953). p. 386 9 D. Bohm, "A proposed Explanation of Quantum Theory in Terms of Hidden Variables at a Sub Quantum Mechanical Level", en Colston Papers (London, Butterworths Scientific Publications 1957) IX, p. 33.
33 deliberadamente una conexión entre la electricidad y el magnetismo, obrando sobre la base de la convicción a priori de que la estructura de todo cuanto existe es polar, y que todas las "fuerzas" de la naturaleza están conectadas orgánicamente entre sí 8. La convicción filosófica de que la complejidad de la naturaleza es ilimitada le llevó a Bohm a especular sobre un nivel subcuántico, fundándose en una analogía con el movimiento browniano clásico9. Ni siquiera la fantasía teológica ha dejado de contribuir, aunque por cierto en mínima medida; recuérdese el principio de la mínima acción de Maupertuis (1747), formulado en la creencia de que el Creador lo había dispuesto todo de la manera más económica posible.
A las hipótesis científicas se llega, en suma, de muchas maneras: hay muchos principios heurísticos, y el único invariante es el requisito de verificabilidad. La inducción, la analogía y la deducción de suposiciones extracientíficas (p. ej. filosóficas) proveen puntos de partida que deben ser elaborados y probados.
5. El método científico, técnica de planteo y comprobación Los especialistas científicos habitualmente no se interesan por el problema de la génesis de las hipótesis científicas; esta cuestión es de competencia de las diversas ciencias de la ciencia. El proceso que conduce a la enunciación de una hipótesis científica puede estudiarse en diversos niveles; el lógico, el psicológico y el sociológico. El lógico se interesará por la inferencia plausible como conexión inversa (no deductiva) entre proposiciones singulares y generales. El psicólogo investigará la etapa de la "iluminación" o relámpago en el proceso de resolución de los problemas, etapa en que se produce la síntesis de elementos anteriormente inconexos; también se propondrá estudiar fenómenos tales como los estímulos e inhibiciones que caracterizan al trabajo en equipo. El sociólogo inquirirá por qué determinada estructura social favorece ciertas clases de hipótesis mientras desalienta a otras.
El metodólogo, en cambio no se ocupará de la génesis de las hipótesis, sino del planteo de los problemas que las hipótesis intentan resolver y de su comprobación. El origen del nexo entre el planteo y la comprobación —esto es, el surgimiento de la hipótesis— se lo deja a otros especialistas. El motivo es, nuevamente, una cuestión de nombres: lo que hoy se llama "método científico" no es ya una lista de recetas para dar con las respuestas correctas a las preguntas científicas, sino el conjunto de procedimientos por los cuales: a) se plantean los problemas científicos y, b) se ponen a prueba las hipótesis científicas.
El estudio del método científico es, en una palabra, la teoría de la investigación. Esta teoría es descriptiva en la medida en que descubre pautas en la investigación científica (y aquí Mario Bunge La ciencia. Su método y su filosofía 34 interviene la historia de la ciencia, como proveedora de ejemplos). La metodología es normativa en la medida en que muestra cuáles son las reglas de procedimiento que pueden aumentar la probabilidad de que el trabajo sea fecundo. Pero las reglas discernibles en la práctica científica exitosa son perfectibles, no son cánones intocables, porque no garantizan la obtención de la verdad; pero, en cambio, facilitan la detección de errores.
Si la hipótesis que ha de ser puesta a prueba se refiere a objetos ideales (números, funciones, figuras, fórmulas lógicas, suposiciones filosóficas, etc.), su verificación consistirá en la prueba de su coherencia —o incoherencia— con enunciados (postulados, definiciones, etc.) previamente aceptados. En este caso, la confirmación puede ser una demostración definitiva.
En cambio, si el enunciado en cuestión se refiere (de manera significativa) a la naturaleza o a la sociedad, puede ocurrir, o bien que podamos averiguar su valor de verdad con la sola ayuda de la razón, o que debamos recurrir, además, a la experiencia.
El análisis lógico basta cuando el enunciado que se pone a prueba es de alguno de los siguientes tipos: a) una simple tautología, o sea, un enunciado verdadero en virtud de su sola forma, independientemente de su contenido (como el caso de "El agua moja o no moja"); b) una definición, o equivalencia entre dos grupos de términos (como en el caso de "Los seres vivos se alimentan, crecen y se reproducen); c) una consecuencia de enunciados fácticos que poseen una extensión o alcance mayor (como ocurre cuando se deduce el principio de la palanca de la ley de conservación de la energía). Vale decir, el análisis lógico y matemático comprobará la validez de los enunciados (hipótesis) que son analíticos en determinado contexto. Muchos enunciados no son intrínsecamente analíticos: su analiticidad es relativa o contextual, como lo demuestra el hecho de que esta propiedad puede perderse, si se estrecha o amplía el contexto, o si se reagrupan los enunciados de la teoría correspondiente, de manera tal que los antiguos teoremas se conviertan en postulados y viceversa.
Vale decir, la mera referencia a los hechos no basta para decidir qué herramienta, si el análisis o la experiencia, ha de emplearse. Para convalidar una proposición hay que empezar por determinar su status y estructura lógica. En consecuencia, el análisis lógico (tanto sintáctico como semántico) es la primera operación que debiera emprenderse al comprobar las hipótesis científicas, sean fácticas o no. Esta norma debiera considerarse como una regla del método científico.
Los enunciados fácticos no analíticos —esto es, las proposiciones referentes a hechos, pero indecidibles con la sola ayuda de la lógica— tendrán que concordar con los datos empíricos o adaptarse a ellos. Esta norma, que distaba de ser obvia antes del siglo XVIII, y que contradice tanto el apriorismo escolástico como el racionalismo cartesiano, es la segunda regla del método científico. Podemos enunciarla de la siguiente manera: el método científico, aplicado a la comprobación de afirmaciones informativas, se reduce al método experimental.
6. El método experimental Mario Bunge La ciencia. Su método y su filosofía 35 La experimentación involucra la modificación deliberada de algunos factores, es decir, la sujeción del objeto de experimentación a estímulos controlados. Pero lo que habitualmente se llama "método experimental" no envuelve necesariamente experimentos en el sentido estricto del término, y puede aplicarse fuera del laboratorio. Así, por ejemplo, la astronomía no experimenta con cuerpos celestes (por el momento) pero es una ciencia empírica porque aplica el método experimental. En lugar de elaborar una definición del término, veamos cómo funcionó en un caso famoso tan conocido que casi siempre se lo entiende mal.
Adams y Le Verrier descubrieron el planeta Neptuno procediendo de una manera que es típica de la ciencia moderna. Sin embargo, no ejecutaron un solo experimento; ni siquiera partieron de "hechos sólidos". En efecto, el problema que se plantearon fue el de explicar ciertas irregularidades halladas en el movimiento de los planetas exteriores (a la Tierra); pero esas irregularidades no eran fenómenos observables: consistían en discrepancias entre las órbitas observadas y las calculadas. El hecho que debía explicar no era un conjunto de datos de los sentidos, sino un conflicto entre datos empíricos y consecuencias deducidas de los principios de la mecánica celeste.
La hipótesis que propusieron para explicar la discrepancia fue que un planeta transuraniano inobservado perturbaba el movimiento de los planetas exteriores entonces conocidos.
También podrían haber imaginado que la ley de Newton de la gravitación falla a grandes distancias, pero esto era apenas concebible en una época en que la Weltanschauung prevaleciente entre los científicos incluía una fe dogmática en la física newtoniana. De esta hipótesis, unida a los principios aceptados de la mecánica celeste y ciertas suposiciones específicas (referentes, entre otras, al plano de la órbita), Adams y Le Verrier dedujeron consecuencias observables con la sola ayuda de la lógica y la matemática: predijeron el lugar en que se encontraba el "nuevo" planeta en tal y cual noche. La observación del cielo y el descubrimiento no fueron sino el último eslabón de un largo proceso por el cual se probaron conjuntamente varias hipótesis.
No es fácil decidir si una hipótesis concuerda con los hechos. En primer lugar, la verificación empírica rara vez puede determinar cuál de los componentes de una teoría dada ha sido confirmado o disconfirmado; habitualmente se prueban sistemas de proposiciones antes que enunciados aislados. Pero la principal dificultad proviene de la generalidad de las hipótesis científicas. La hipótesis de Adams y Le Verrier era general, aun cuando ello no es aparente a primera vista: tácitamente habían supuesto que el planeta existía en todo momento dentro de un largo lapso; y comprobaron la hipótesis tan sólo para unos pocos breves intervalos de tiempo. En cambio, las proposiciones fácticas singulares no son tan difíciles de probar. Así, por ejemplo, no es difícil comprobar si "El Sr. Pérez, que es obeso, es cardíaco"; bastan una balanza y un estetoscopio. Lo difícil de comprobar son las proposiciones fácticas generales, esto es, los enunciados referentes a clases de hechos y no a hechos singulares. La razón es sencilla: no hay hechos generales, sino tan sólo hechos singulares; por consiguiente, la frase "adecuación de las ideas a los hechos" está fuera de la cuestión en lo que respecta a las Mario Bunge La ciencia. Su método y su filosofía 36 hipótesis científicas.
Supongamos que se sugiere la hipótesis "los obesos son cardíacos", sea por la observación de cierto número de correlaciones entre la obesidad y las enfermedades del corazón (esto es, por inducción estadística, sea sobre la base del estudio de la función del corazón en la circulación (esto es, por deducción). El enunciado general "los obesos son cardíacos" no se refiere solamente a nuestros conocidos, sino a todos los gordos del mundo; por consiguiente, no podemos esperar verificarlo directamente (esto es, por el examen de un inexistente "gordo general") ni exhaustivamente (auscultando a todos los seres humanos presentes, pasados y futuros). La metodología nos dice cómo debemos proceder; en este caso, examinaremos sucesivamente los miembros de una muestra suficientemente numerosa de personas obesas.
Vale decir, probamos una consecuencia particular de nuestra suposición general. Esta es una tercera máxima del método científico: obsérvense singulares en busca de elementos de prueba universales.
Hasta aquí todo parece sencillo; pero los problemas relacionados con la prueba real distan de ser triviales, y algunos de ellos no han sido resueltos satisfactoriamente. Debemos recurrir a las técnicas del planteo de problemas de este tipo, es decir, a las técnicas de diseño de los procedimientos empíricos adecuados. Esta técnica nos aconseja comenzar por decidir lo que hemos de entender por "obeso" y por "cardíaco", lo que no es en modo alguno tarea sencilla, ya que el umbral de obesidad es en gran medida convencional. O sea, debemos empezar por determinar el exacto sentido de nuestra pregunta. Y ésta es una cuarta regla del método científico, a saber: formúlese preguntas precisas.
Luego procederemos a elegir la técnica experimental (clase de balanza, tipo de examen de corazón, etc.) y la manera de registrar datos y de ordenarlos. Además debemos decidir el tamaño de la muestra que habremos de observar y la técnica de escoger sus miembros, con el fin de asegurar que será una fiel representante de la población total. Sólo una vez realizadas estas operaciones preliminares podremos visitar al Sr. Pérez y a los demás miembros de la muestra, con el fin de reunir datos. Y aquí se nos muestra una quinta regla del método científico: la recolección y el análisis de datos deben hacerse conforme a las reglas de la estadística.
Después que los datos han sido reunidos, clasificados y analizados, el equipo que tiene a su cargo la investigación podrá realizar una inferencia estadística concluyendo que "el N % de los obesos son cardíacos". Más aún, habrá que estimar el error probable de esta afirmación.
Obsérvese que la hipótesis que había motivado nuestra investigación era un enunciado universal de la forma "para todo x, si x es F, entonces x es G". Por otro lado, el resultado de la investigación es un enunciado estadístico, a saber: "de la clase de las personas obesas, una subclase que llega a su N/100ava parte está compuesta por cardíacos". Esto es, nuestra hipótesis de trabajo ha sido corregida. ¿Debemos contentarnos con esta respuesta? Nos gustaría formular otras preguntas: deseamos entender la ley que hemos hallado, nos gustaría deducirla de las leyes de la fisiología humana. Y aquí se aplica una sexta regla del método Mario Bunge La ciencia. Su método y su filosofía 37 científico, a saber: no existen respuestas definitivas, y ello simplemente porque no existen preguntas finales.
7. Métodos teóricos Toda ciencia fáctica especial elabora sus propias técnicas de verificación; entre ellas, las técnicas de medición son típicas de la ciencia moderna. Pero en todos los casos estas técnicas, por diferentes que sean, no constituyen fines en sí mismos; todas ellas sirven para contrastar ciertas ideas con ciertos hechos por la vía de la experiencia. O, si se prefiere, el objetivo de las técnicas de verificación es probar enunciados referentes a hechos por vía del examen de proposiciones referentes a la experiencia (y en particular, al experimento). Este es el motivo por el cual los experimentadores no tienen por qué construir cada uno de sus aparatos e instrumentos, pero deben en cambio diseñarlos y/o usarlos a fin de poner a prueba ciertas afirmaciones. Las técnicas especiales, por importantes que sean, no son sino etapas de la aplicación del método experimental, que no es otra cosa que el método científico en relación con la ciencia fáctica, y la ciencia, por fáctica que sea, no es un montón de hechos sino un sistema de ideas.
En el párrafo anterior ejemplificamos el método experimental analizando el proceso de verificación que requeriría el enunciado "los obesos son cardíacos"; encontramos que esta hipótesis requería una precisión cuantitativa, y después de una investigación imaginaria adoptamos, en su lugar, cierta generalización empírica del tipo de los enunciados estadísticos.
Ahora bien: las generalizaciones empíricas tan caras a Aristóteles y a Bacon, y aun cuando se las formule en términos estadísticos, no son distintivas de la ciencia moderna. El tipo de hipótesis característico de la ciencia moderna no es el de los enunciados descriptivos aislados cuya función principal es resumir experiencias. Lo peculiar de la ciencia moderna es que consiste en su mayor parte en teorías explicativas, es decir, en sistemas de proposiciones que pueden clasificarse en: principios, leyes, definiciones, etc., y que están vinculadas entre sí mediante conectivas lógicas (tales como "y, o, si... entonces", etc.).
Las teorías dan cuenta de los hechos no sólo describiéndolos de manera más o menos exacta, sino también proveyendo modelos conceptuales de los hechos, en cuyos términos puede explicarse y predecirse, al menos en principio, cada uno de los hechos de una clase. Las posibilidades de una hipótesis científica no se advierten por entero antes de incorporarlas en una teoría; y es sólo entonces cuando puede encontrársele varios soportes. Al sumergirse en una teoría, el enunciado dado es apoyado —o aplastado— por toda la masa del saber disponible; permaneciendo aislado es difícil de confirmar y de refutar y, sobre todo, sigue sin ser entendido.
La conversión de las generalizaciones empíricas en leyes teóricas envuelve trascender la esfera de los fenómenos y el lenguaje observacional: ya no se trata de hacer afirmaciones acerca de hechos observables, sino de adivinar su "mecanismo" interno (el que, desde luego Mario Bunge La ciencia. Su método y su filosofía 38 no tiene por qué ser mecánico). Supóngase que un psicólogo desea estudiar las correlaciones entre cierto estímulo observable S y cierta conducta observable R, que —a modo de ensayo— considera como la respuesta al estímulo dado. Si, después de una sucesión de experimentos, llegara a confirmar su hipótesis de trabajo y deseara trascender las fronteras de la psicología fenomenista, intentaría elaborar, digamos, un modelo neurológico que explicara el nexo S-R en términos fisiológicos. No es tarea fácil: el psicólogo tiene que inventar diversas hipótesis acerca de otros tantos canales nerviosos posibles que conecten los hechos observables extremos, S y R. Análogamente, los físicos atómicos imaginan diversos mecanismos ocultos que conectan los fenómenos macroscópicos con su soporte microscópico.
Pero nuestro psicólogo no andará del todo a tientas: podrá probar si su conexión concuerda con algunos de los esquemas pavlovianos de los reflejos, o con cualquier otro mecanismo.
Cada una de sus hipótesis —sea que consistan en suponer que interviene un reflejo innato o condicionado— tendrá que especificar el aparato receptor, el nervio aferente, la estación central, el nervio eferente, el órgano receptor, etc. Más aún, sus varias hipótesis de trabajo tendrán que ser compatibles con el saber más firmemente establecido (aunque no inamovible) y tendrán que ser puestas a prueba mediante técnicas especiales (excitación o destrucción de nervios, registro de impulsos nerviosos, etc.) Vale la pena emprender esta difícil tarea: la eventual confirmación de una de las hipótesis puestas a prueba no sólo explicará el nexo S-R dado, sino que también lo ubicará en su contexto: además, apoyará la hipótesis misma de que tal nexo no es accidental. Pues, aunque suene a paradoja, un enunciado fáctico es tanto más fidedigno cuanto mejor está apoyado por consideraciones teóricas.
Es importante advertir, en efecto, que la experiencia dista de ser el único juez de las teorías fácticas, o siquiera el último. Las teorías se contrastan con los hechos y con otras teorías. Por ejemplo, una de las pruebas de la generalización de una teoría dada es averiguar si la nueva teoría se reduce a la vieja dentro de un cierto dominio, de modo tal que cubra por lo menos el mismo grupo de hechos. Más aún, el grado de sustentación o apoyo de las teorías no es idéntico a su grado de confirmación. Las teorías no se constituyen ex nihilo, sino sobre ciertas bases: éstas las sostienen antes y después de la prueba; la prueba misma, si tiene éxito, provee los apoyos restantes de la teoría y fija su grado de confirmación. Aun así el grado de confirmación de una teoría no basta para determinar la probabilidad de la misma.
8. En qué se apoya una hipótesis científica Una hipótesis de contenido fáctico no sólo es sostenida por la confirmación empírica de cierto número de sus consecuencias particulares (p. ej. predicciones). Las hipótesis científicas están incorporadas en teorías o tienden a incorporarse en ellas; y las teorías están relacionadas entre sí, constituyendo la totalidad de ellas la cultura intelectual. Por esto, no debiera sorprender que las hipótesis científicas tengan soportes no sólo científicos, sino Mario Bunge La ciencia. Su método y su filosofía 39 también extracientíficos: los primeros son empíricos y racionales, los últimos son psicológicos y culturales. Expliquémonos.
Cuanto más numerosos sean los hechos que confirman una hipótesis, cuanto mayor sea la precisión con que ella reconstruye los hechos, y cuanto más vastos sean los nuevos territorios que ayuda a explorar, tanto más firme será nuestra creencia en ella, esto es, tanto mayor será la probabilidad que le asignemos. Esto es, esquemáticamente dicho, lo que se entiende por el soporte empírico de las hipótesis fácticas. Pero la experiencia disponible no puede ser considerada como inapelable: en primer lugar, porque nuevas experiencias pueden mostrar la necesidad de un remiendo: en segundo término, porque la experiencia científica no es pura, sino interpretada, y toda interpretación se hace en términos de teorías, motivo por el cual la primera reacción de los científicos experimentados ante informaciones sobre hechos que parecerían trastornar teorías establecidas es de escepticismo.
Cuanto más estrecho sea el acuerdo de la hipótesis en cuestión con el conocimiento disponible de mismo orden, tanto más firme es nuestra creencia en ella; semejante concordancia es particularmente valiosa cuando consiste en una compatibilidad con enunciados de leyes. Esto es lo que hemos designado con el nombre de soporte racional de las hipótesis fácticas. Este es, dicho sea de paso, el motivo por el cual la mayoría de los científicos desconfían de los informes acerca de la llamada percepción extransensorial, porque los llamados fenómenos psi contradicen el cuerpo de hipótesis psicológicas y fisiológicas bien establecidas. En resumen, las teorías científicas deben adecuarse, sin duda, a los hechos, pero ningún hecho aislado es aceptado en la comunidad de los hechos controlados científicamente a menos que tenga cabida en alguna parte del edificio teórico establecido.
Desde luego, el soporte racional no es garantía de verdad; si lo fuera, las teorías fácticas serían invulnerabes a la experiencia. Los soportes empíricos y racionales de las hipótesis fácticas son interdependientes.
En cuanto a los soportes extracientíficos de las hipótesis científicas, uno de ellos es de carácter psicológico: influye sobre nuestra elección de las suposiciones y sobre el valor que le asignamos a su concordancia con los hechos. Por ejemplo, los sentimientos estéticos que provocan la simplicidad y la unidad lógica estimulan unas veces y otras obstaculizan la investigación sobre la validez de las teorías. Esto es lo que hemos denominado el soporte psicológico de las hipótesis fácticas; a menudo es oscuro, y no sólo está vinculado a características personales, sino también sociales.
Lo que hemos llamado soporte cultural de las hipótesis fácticas consiste en su compatibilidad con alguna concepción del mundo, y en particular, con la Zeitgeist prevaleciente. Es obvio que tendemos a asignar mayor peso a aquellas hipótesis que congenian con nuestro fondo cultural y, en particular, con nuestra visión del mundo, que aquellas hipótesis que lo contradicen. La función dual del soporte cultural de las conjeturas científicas se advierte con facilidad: por una parte, nos impulsa a poner atención en ciertas clases de hipótesis y hasta interviene en la sugerencia de las mismas; por otra parte, puede impedirnos apreciar otras Mario Bunge La ciencia. Su método y su filosofía 40 posibilidades, por lo cual puede constituir un factor de obstinación dogmática. La única manera de minimizar este peligro es cobrar conciencia del hecho de que las hipótesis científicas no crecen en un vacío cultural.
Los soportes empíricos y racionales son objetivos, en el sentido de que en principio son susceptibles de ser sopesados y controlados conforme a patrones precisos y formulables. En cambio, los soportes extracientíficos son, en gran medida, materia de preferencia individual, de grupo o de época; por consiguiente, no debieran ser decisivos en la etapa de la comprobación, por prominentes que sean en la etapa heurística. Es importante que los científicos sean personas cultas, aunque sólo sea para que adviertan la fuerte presión que ejercen los factores psicológicos y culturales sobre la formulación, elección, investigación y credibilidad de las hipótesis fácticas. La presión, para bien o para mal, es real y nos obliga a tomar partido por una u otra concepción del mundo; es mejor hacerlo conscientemente que inadvertidamente.
La enumeración anterior de los tipos de soportes de las hipótesis científicas no tenía otro propósito que mostrar que el método experimental no agota el proceso que conduce a la aceptación de una suposición fáctica. Este hecho podría invocarse en favor de la tesis de que la investigación científica es un arte.
9. La ciencia: técnica y arte La investigación científica es legal, pero sus leyes —las reglas del método científico— no son pocas, ni simples, ni infalibles, ni bien conocidas: son, por el contrario numerosas, complejas, más o menos eficaces, y en parte desconocidas. El arte de formular preguntas y de probar respuestas —esto es, el método científico— es cualquier cosa menos un conjunto de recetas; y menos técnica todavía es la teoría del método científico. La moraleja es inmediata: desconfíese de toda descripción de la vida de la ciencia —y en primer lugar de la presente— pero no se descuide ninguna. La investigación es una empresa multilateral que requiere el más intenso ejercicio de cada una de las facultades psíquicas, y que exige un concurso de circunstancias sociales favorables; por este motivo, todo testimonio personal, perteneciente a cualquier período, y por parcial que sea, puede echar alguna luz sobre algún aspecto de la investigación.
A menudo se sostiene que la medicina y otras ciencias aplicadas son artes antes que ciencias, en el sentido de que no pueden ser reducidas a la simple aplicación de un conjunto de reglas que pueden formularse todas explícitamente y que pueden elegirse sin que medie el juicio personal. Sin embargo, en este sentido la física y la matemática también son artes: ¿quién conoce recetas hechas y seguras para encontrar leyes de la naturaleza o para adivinar teoremas? Si "arte" significa una feliz conjunción de experiencia, destreza, imaginación, visión y habilidad para realizar inferencias del tipo no analítico, entonces no sólo son artes la medicina, la pesquisa criminal, la estrategia militar, la política y la publicidad, sino también Mario Bunge La ciencia. Su método y su filosofía 41 toda otra disciplina. Por consiguiente, no se trata de si un campo dado de la actividad humana es un arte, sino si, además, es científico.
La ciencia es ciertamente comunicable; si un cuerpo de conocimiento no es comunicable, entonces por definición no es científico. Pero esto se refiere a los resultados de la investigación antes que a las maneras en que éstos se obtienen; la comunicabilidad no implica que el método científico y las técnicas de las diversas ciencias especiales puedan aprenderse en los libros: los procedimientos de la investigación se dominan investigando, y los metacientíficos debieran por ello practicarlos antes de emprender su análisis. No se sabe de obra maestra alguna de la ciencia que haya sido engendrada por la aplicación consciente y escrupulosa de las reglas conocidas del método científico; la investigación científica es practicada en gran parte como un arte no tanto porque carezca de reglas cuanto porque algunas de ellas se dan por sabidas, y no tanto porque requiera una intuición innata cuanto porque exige una gran variedad de disposiciones intelectuales. Como toda otra experiencia, la investigación puede ser comprendida por otros pero no es íntegramente transferible; hay que pagar por ella el precio de un gran número de errores, y por cierto que al contado. Por consiguiente, los escritos sobre el método científico pueden iluminar el camino de la ciencia, pero no pueden exhibir toda su riqueza, y sobre todo, no son un sustituto de la investigación misma, del mismo modo que ninguna biblioteca sobre botánica puede reemplazar a la contemplación de la naturaleza, aunque hace posible que la contemplación sea más provechosa.
10. La pauta de la investigación científica La variedad de habilidades y de información que exige el tratamiento científico de los problemas ayuda a explicar la extremada división del trabajo prevaleciente en la ciencia contemporánea, en la que encuentra lugar toda capacidad natural y toda habilidad adquirida.
Es posible apreciar esta variedad exponiendo la pauta general de la investigación científica.
Creo que esa pauta —o sea, el método científico— es, a grandes líneas, la siguiente: 1 PLANTEO DEL PROBLEMA 1.1 Reconocimiento de los hechos: examen del grupo de hechos, clasificación preliminar y selección de los que probablemente sean relevantes en algún respecto.
1.2 Descubrimiento del problema: hallazgo de la laguna o de la incoherencia en el cuerpo del saber.
1.3 Formulación del problema: planteo de una pregunta que tiene probabilidad de ser la correcta; esto es, reducción del problema a su núcleo significativo, probablemente soluble y probablemente fructífero, con ayuda de conocimiento disponible.
Mario Bunge La ciencia. Su método y su filosofía 42 2 CONSTRUCCIÓN DE UN MODELO TEÓRICO 2.1 Selección de los factores pertinentes: invención de suposiciones plausibles relativas a las variables que probablemente son pertinentes.
2.2 Invención de las hipótesis centrales y de las suposiciones auxiliares: propuesta de un conjunto de suposiciones concernientes a los nexos entre las variables pertinentes; p. ej. formulación de enunciados de ley que se espera puedan amoldarse a los hechos observados.
2.3 Traducción matemática: cuando sea posible, traducción de las hipótesis, o de parte de ellas, a alguno de los lenguajes matemáticos.
3 DEDUCCIÓN DE CONSECUENCIAS PARTICULARES 3.1 Búsqueda de soportes racionales: deducción de consecuencias particulares que pueden haber sido verificadas en el mismo campo o en campos contiguos.
3.2 Búsqueda de soportes empíricos: elaboración de predicciones (o retrodicciones) sobre la base de modelo teórico y de datos empíricos, teniendo en vista técnicas de verificación disponibles o concebibles.
4 PRUEBA DE LAS HIPÓTESIS 4.1 Diseño de la prueba: planeamiento de los medios para poner a prueba las predicciones; diseño de observaciones, mediciones, experimentos y demás operaciones instrumentales.
4.2 Ejecución de la prueba: realización de las operaciones y recolección de datos. 4.3 Elaboración de los datos: clasificación, análisis, evaluación, reducción, etc., de los datos empíricos.
4.4 Inferencia de la conclusión: interpretación de los datos elaborados a la luz del modelo teórico.
5 INTRODUCCIÓN DE LAS CONCLUSIONES EN LA TEORÍA 5.1 Comparación de las conclusiones con las predicciones: contraste de los resultados de la prueba con las consecuencias del modelo teórico, precisando en qué medida éste puede considerarse confirmado o disconfirmado (inferencia probable).
5.2 Reajuste del modelo: eventual corrección o aun reemplazo del modelo.
5.3 Sugerencias acerca de trabajo ulterior: búsqueda de lagunas o errores en la teoría y/o los procedimientos empíricos, si el modelo ha sido disconfirmado; si ha sido confirmado, examen de posibles extensiones y de Mario Bunge La ciencia. Su método y su filosofía 43 posibles consecuencias en otros departamentos del saber.
11. Extensibilidad del método científico Para elaborar conocimiento fáctico no se conoce mejor camino que el de la ciencia. El método de la ciencia no es, por cierto, seguro; pero es intrínsecamente progresivo, porque es auto-correctivo: exige la continua comprobación de los puntos de partida, y requiere que todo resultado sea considerado como fuente de nuevas preguntas. Llamemos filosofía científica a la clase de concepciones filosóficas que aceptan el método de la ciencia como la manera que nos permite: a) plantear cuestiones fácticas "razonables" (esto es, preguntas que son significativas, no triviales, y que probablemente pueden se respondidas dentro de una teoría existente o concebible); y b) probar respuestas probables en todos los campos especiales del conocimiento.
No debe confundirse la filosofía científica con el cientificismo en cualquiera de sus dos versiones: el enciclopedismo científico y el reduccionismo naturalista. El enciclopedismo científico pretende que la única tarea de los filósofos es recoger los resultados más generales de la ciencia, elaborando una imagen unificada de los mismos, y preferiblemente formulándolos todos en un único lenguaje (p. ej., el de la física). En cambio, la filosofía, científica o no, analiza lo que se le presente y, a partir de este material, construye teorías de segundo nivel, es decir teorías de teorías; la filosofía será científica en la medida en que elabore de manera racional los materiales previamente elaborados por la ciencia. Así es como puede entenderse la extensión del método científico al trabajo filosófico.
En cuanto al cientificismo concebido como reduccionismo naturalista —y que a veces se superpone con el enciclopedismo científico como ocurre con el fisicalismo—, puede describírselo como una tentativa de resolver toda suerte de problemas con ayuda de las técnicas creadas por las ciencias naturales, desdeñando las cualidades específicas, irreductibles, de cada nivel de la realidad. El cientificismo radical de esta especie sostendría, por ejemplo, que la sociedad no es más que un sistema físico-químico (o, a lo sumo, biológico), de donde los fenómenos sociales debieran estudiarse exclusivamente mediante la ayuda de metros, relojes, balanzas y otros instrumentos de la misma clase. En cambio, la filosofía científica favorece la elaboración de técnicas específicas en cada campo, con la única condición de que estas técnicas cumplan las exigencias esenciales del método científico en lo que respecta a las preguntas y a las pruebas. De esta manera es como puede entenderse la extensión del método científico a todos los campos especiales del conocimiento.
Pero también debería emplearse el método de la ciencia en las ciencias aplicadas y, en general, en toda empresa humana en que la razón haya de casarse con la experiencia; vale decir, en todos los campos excepto en arte, religión y amor. Una adquisición reciente del método científico es la investigación operativa (operations research), esto es, el conjunto de procedimientos mediante los cuales los dirigentes de empresas pueden obtener un Mario Bunge La ciencia. Su método y su filosofía 10 Véase P. M. Morse y G. E. Kimball, Methods of Operations Research, ed. rev. (Cambridge, Mass., The Technology Press of Massachussets Institute of Technology; N. York, John Wiley & Sons, 1951).
44 fundamento cuantitativo para tomar decisiones, y los administradores pueden adquirir ideas para mejorar la eficiencia de la organización10. Pero, desde luego la extensión del método científico a las cosas humanas está aún en su infancia. Pídasele a un político que pruebe sus afirmaciones, no recurriendo a citas y discursos, sino confrontándolos con hechos certificables (tal como se recogen y elaboran, por ejemplo, con ayuda de las técnicas estadísticas). Si es honesto, cosa que puede suceder, o bien: a) admitirá que no entiende la pregunta, o b) concederá que todas sus creencias son, en el mejor de los casos, enunciados probables, ya que sólo pueden ser probados imperfectamente, o c) llegará a la conclusión de que muchas de sus hipótesis favoritas (principios, máximas, consignas) tienen necesidad urgente de reparación. En este último caso puede terminar por admitir que una de las virtudes del método de la ciencia es que facilita la regulación o readaptación de las ideas generales que guían (o justifican) nuestra conducta consciente, de manera tal que ésa pueda corregirse con el fin de mejorar los resultados.
Desgraciadamente, la cientifización de la política la haría más eficaz, pero no necesariamente mejor, porque el método puede dar la forma y no el contenido; y el contenido de la política está determinado por intereses que no son primordialmente culturales o éticos, sino materiales. Por esto, una política científica puede dirigirse a favor o en contra de cualquier grupo social: los objetivos de la estrategia política, así como los de la investigación científica aplicada, no son fijados por patrones científicos, sino por intereses sociales. Esto muestra a la vez el alcance y los límites del método científico: por una parte, puede producir saber, eficiencia y poder; por la otra, este saber, esta eficiencia y este poder pueden usarse para bien o para mal, para libertar o para esclavizar.
12. El método científico: ¿un dogma más?
¿Es dogmático favorecer la extensión del método científico a todos los campos del pensamiento y de la acción consciente? Planteamos la cuestión en términos de conducta. El dogmático vuelve sempiternamente a sus escrituras, sagradas o profanas, en búsqueda de la verdad; la realidad le quemaría los papeles en los que imagina que está enterrada la verdad: por esto elude el contacto con los hechos. En cambio, para el partidario de la filosofía científica todo es problemático: todo conocimiento fáctico es falible (pero perfectible), y aun las estructuras formales pueden reagruparse de maneras más económicas y racionales; más aún, el propio método de la ciencia será considerado por él como perfectible, como lo muestra la reciente incorporación de conceptos y técnicas estadísticas. Por consiguiente, el partidario del método científico no se apegará obstinadamente al saber, ni siquiera a los medios consagrados para adquirir conocimiento, sino que adoptará una actitud investigadora; se Mario Bunge La ciencia. Su método y su filosofía 45 esforzará por aumentar y renovar sus contactos con los hechos y el almacén de las ideas mediante las cuales los hechos pueden entenderse, controlarse y a veces reproducirse.
No se conoce otro remedio eficaz contra la fosilización del dogma —religioso, político, filosófico o científico— que el método científico, porque es el único procedimiento que no pretende dar resultados definitivos. El creyente busca la paz en la aquiescencia; el investigador, en cambio, no encuentra paz fuera de la investigación y la disensión: está en continuo conflicto consigo mismo, puesto que la exigencia de buscar conocimiento verificable implica un continuo inventar, probar y criticar hipótesis. Afirmar y asentir es más fácil que probar y disentir; por esto hay más creyentes que sabios, y por esto, aunque el método científico es opuesto al dogma, ningún científico y ningún filósofo científico debieran tener la plena seguridad de que han evitado todo dogma.
De acuerdo con la filosofía científica, el peso de los enunciados —y por consiguiente su credibilidad y su eventual eficacia práctica— depende de su grado de sustentación y de confirmación. Si, como estimaba Demócrito, una sola demostración vale más que el reino de los persas, puede calcularse el valor del método científico en los tiempos modernos. Quienes lo ignoran íntegramente no pueden llamarse modernos; y quienes lo desdeñan se exponen a no ser veraces ni eficaces.
Mario Bunge La ciencia. Su método y su filosofía 46 ¿Qué significa “ley científica”?
Mario Bunge La ciencia. Su método y su filosofía 47 Mario Bunge La ciencia. Su método y su filosofía 48 1. Cuatro significados del término "ley científica" Probablemente la mayoría de los científicos y metacientíficos concuerden en que la corriente central de la investigación científica consiste en la búsqueda, explicación y aplicación de las leyes científicas. Sin embargo, sólo unos pocos estudiosos de la ciencia concuerdan respecto de lo que designa el .término "ley" en el contexto de la ciencia. Así, por ejemplo, la expresión "ley de Newton del movimiento" se interpreta unas veces como cierta pauta objetiva del movimiento mecánico. Otras veces los mismos términos designan la fórmula de Newton "Fuerza = masa x aceleración", o cualquier otro enunciado que la incluye. Finalmente, "la ley de Newton del movimiento" se entiende a veces como una regla de procedimiento por medio de la cual se puede predecir o controlar las trayectorias de los cuerpos. En el primer caso se hace referencia a un trozo de la realidad física; en el segundo, el designado (designatum) es una pieza del conocimiento; en el tercero, es una regla de acción.
A cuál de los tres designados se refiere el científico cuando habla acerca de la "ley de Newton del movimiento, dependerá de las circunstancias o del contexto en que usa la expresión, así como de su filosofía explícita o tácita. Si concede que el mundo físico subsiste aun cuando no haya quien lo perciba o lo piense, entonces la expresión en cuestión podrá significar una conexión objetiva entre las cualidades fuerza, masa y aceleración, sea que se las mida o no.
En cambio, si el científico no asigna existencia autónoma a los objetos físicos, entonces entenderá por "ley científica" una relación invariante entre términos anclados de alguna manera a datos de los sentidos (los cuales funcionarán como términos últimos o "hechos atómicos y no como señales elementales de nuestro comercio con las cosas). Y si sólo accede a hablar acerca de operaciones posibles, entonces podrá significar por "ley científica" cierta pauta de la conducta humana (p. ej.. la predicción) en relación con cierta clase de datos empíricos (cuya totalidad llamará "sistema de cuerpos en movimiento", o algo parecido), y cierto tipo de objetivo. En particular nuestro científico podrá sostener que tan sólo las "ecuaciones de laboratorio merecen ser llamadas leyes naturales, pues ellas —y no los principios de los cuales eventualmente se derivan— son comprobables directamente en el laboratorio. Finalmente, cualquiera que sea la preferencia filosófica de nuestro científico, si ha oído hablar de la física teórica contemporánea podrá admitir que hay una clase especial de enunciados que se refieren a las leyes mismas, y que operan como principios reguladores, tales como: "las leyes naturales no dependen de los sistemas de referencia ni, en particular, del cuadro de referencia del observador".
En total debiéramos distinguir, pues, por lo menos cuatro significados del término "ley" en Mario Bunge La ciencia. Su método y su filosofía 49 el contexto de las ciencias fácticas.
2. Nomenclatura propuesta.
Nunca se señala semejante variedad semántica. Sin embargo, debiera ser de utilidad distinguir entre los diversos significados del término "ley" tal como se lo usa en las ciencias naturales y sociales, así como la consiguiente adopción de una nomenclatura uniforme. Puesto que los cuatro significados corresponden al mismo término. sería conveniente añadirles subíndices con el fin de eliminar la ambigüedad señalada. Permítaseme proponer las siguientes reglas de designación: (1) Ley1, o simplemente ley, denota toda pauta inmanente del ser o del devenir; esto es, toda relación constante y objetiva en la naturaleza, en la mente o en la sociedad.
(2) Ley2 o enunciado nomológico o enunciado de ley, designa toda hipótesis general que tiene como referente mediato una ley1, y que constituye una reconstrucción conceptual de ella. Todo enunciado de ley tiene, en realidad. dos referentes: uno es la pauta de cierta clase de hechos, al que se supone que se adecua (nunca perfectamente) el enunciado en cuestión, podemos llamarlo el referente mediato del enunciado de ley. El referente inmediato de un enunciado nomológico es, en cambio, el modelo teórico al que se aplica exactamente. Así, por ejemplo, la mecánica analítica se refiere en forma mediata a las partículas materiales, siendo su referente inmediato el concepto llamado "sistema de puntos materiales".
(3) Ley3, o enunciado nomopragmático, designa toda regla mediante la cual puede regularse (exitosamente o no) una conducta. Las leyes 3 son casi siempre consecuencias de leyes 2 en conjunción con ítems de información específica. Una clase conspicua de este tipo de ley es la de los enunciados nomológicos predictivos, esto es, las proposiciones mediante las cuales se hacen predicciones (o retrodicciones) de sucesos singulares.
(4) Ley4, o enunciado metanomologico, designa todo principio general acerca de la forma y/o alcance de los enunciados de ley pertenecientes a algún capítulo de la ciencia fáctica.
Las leyes1 son estructuras nómicas (pautas invariantes .) al nivel óntico. Las leyes2 son proposiciones (que a menudo toman la forma de ecuaciones) acerca de pautas objetivas: son pautas al nivel del conocimiento. Las leyes3, son relaciones invariantes al nivel pragmático: son guías para la acción fundada científicamente. Y las leyes 4 son prescripciones metodológicas y/o principios ontológicos (hipótesis acerca de rasgos conspicuos de la realidad).
3. Ejemplificación de las distinciones Consideremos nuevamente la ley del movimiento mecánico. Ésta puede considerarse como una pauta objetiva (ley1) que diversos enunciados de ley (leyes2) reconstruyen en diferentes aproximaciones. A saber: (a) la ley de Aristóteles "La fuerza es igual a la resistencia Mario Bunge La ciencia. Su método y su filosofía 50 multiplicada por la velocidad"; (b) y "La fuerza es igual a la masa multiplicada por la aceleración"; © la ley de Einstein "La fuerza es igual a la velocidad de variación del impulso"; (d) el teorema de Ehrenfest "La fuerza media es igual al valor medio de la velocidad de variación del impulso"; (e) el teorema de Broglie-Bohm "La fuerza exterior más la fuerza cuántica es igual a la velocidad de variación del impulso «ocultos»".
Al no haber "hechos generales", no es posible verificar directamente hipótesis generales como son los enunciados de leyes; ni es posible aplicarlos sin más. Sólo pueden comprobarse y usarse las consecuencias particulares de hipótesis científicas. Por consiguiente, ninguno de los enunciados de ley que acabamos de mencionar puede considerarse como una ley,, esto es, como una regla de acción. Pero ciertos teoremas deducidos de esas leyes serán leyes. Por ejemplo, toda solución de la ecuación de Newton con condiciones iniciales dadas (posición inicial x 0 y velocidad inicial v0) será una ley. Así, la ecuación de Galileo: x (t) = x 0 + v0 t + ½g t2 es una consecuencia verificable de la segunda ley del movimiento de Newton, F = ma, y se usa para predecir, por ejemplo, tiempos de caída (siempre que se especifiquen los valores de las variables y parámetros, esto es, a condición de que el enunciado universal se convierta en singular); por esto, la ley de Galileo es una típica "ecuación de laboratorio” que cumple nuestra definición de enunciado nomopragmático (ley3).
Obsérvese que, a diferencia de la correspondiente ley2 una ley3 puede incluir ítems de informaciones específicas, tales como la posición y velocidad iniciales de un cuerpo, o el contorno de una membrana vibrante. Más aún, las leyes3 no serán, en general, invariantes respecto de las mismas transformaciones que dejan invariantes a las correspondientes leyes2.
Vale decir, mientras que las leyes de los hechos no dependen de nuestro "punto de vista" (sistema de referencia, unidades de medición, y otras convenciones), las leyes 3, sí dependen de nuestro punto de vista. En otras palabras, la descripción de los fenómenos del presente, del pasado o del futuro depende esencialmente del operador, aun cuando los fenómenos mismos ocurran sin nuestra intervención. Las transformaciones que dejan invariante a la ley de Newton del movimiento (pero no a sus consecuencias) son las que constituyen el grupo de Galileo (X = x - vt). El principio de la relatividad del movimiento es el enunciado metanomológico (ley4) que corresponde a la ley de Newton del movimiento; en efecto, dicho principio se refiere a esta ley del movimiento (y, específicamente, a sus propiedades de invariancia respecto de cierto conjunto de cambios en la representación de los fenómenos).
4. Justificación de la distinción entre leyes y enunciados de leyes La distinción entre las leyes1 y sus reconstrucciones conceptuales (leyes 2) debiera ser obvia para todo no idealista, aunque sólo sea por el hecho de que suele suponerse que un referente Mario Bunge La ciencia. Su método y su filosofía 51 mediato único (una ley1) les corresponde a los diversos enunciados de leyes (leyes 2) de un cierto tipo, que se suceden históricamente. Semejante distinción está involucrada en la noción misma de perfectibilidad de la descripción científica de los hechos, que contrasta con la presunta constancia de las pautas de los hechos (presunción ésta que es indudablemente correcta en primera aproximación y en relación con cada uno de los niveles de la realidad, no así en relación con la totalidad de la realidad, puesto que la emergencia de nuevos niveles va acompañada de la emergencia de nuevas leyes).
Lo que habitualmente designa el término "ley de la realidad física o cultural" no depende de nuestro conocimiento, a menos que se trate de una ley del proceso cognoscitivo. Antes bien, nuestro conocimiento de las leyes1 (esto es, las leyes 2) presupone la existencia de pautas objetivas. Si no hacemos esta distinción, podemos caer en la visión mágica del mundo exterior, propuesta por Chesterton (según la cual "no hay leyes, sino tan sólo repeticiones misteriosas" [ weird]), o rendirnos a la conclusión no menos nihilista de Bridgman, de que "la naturaleza es intrínsecamente y en sus elementos incomprensible y no está sujeta a la ley".
Normalmente, los científicos no aceptan ninguna de estas versiones del contingentismo, sino que se inclinan más bien a admitir el principio leibniziano de la inagotabilidad de los actuales, el correspondiente principio de Waismann de la textura abierta de los conceptos empíricos, y la hipótesis de que en el mundo exterior no hay repeticiones sino tan sólo leyes (probablemente) constantes, siendo la repetición una ficción inventada por el hombre para arreglárselas con la variedad y la novedad.
Las leyes1 no son verdaderas ni falsas: simplemente son. Sólo las leyes2 pueden ser más o menos exactas. Las leyes1, aunque objetivas, no son objetos sensibles sino inteligibles: no percibimos las leyes, sino que las inferimos a partir de los fenómenos, tal como inferimos todo otro universal fáctico. Éste es el motivo por el cual los empiristas deben negar la existencia de las leyes, porque las leyes objetivas no son observables. Semejante inferencia dista de ser directa: no "aprehendemos” leyes, (a duras penas "aprehendemos los singulares”) en su pureza, sin distorsión. El proceso del descubrimiento científico es cualquier cosa menos un mero reflejo de los hechos sobre la conciencia, por la vía de la percepción y de la inducción; es, por el contrario, un arduo trabajo de ensayo de reconstrucción, por medio de conceptos teóricos más o menos elaborados. En particular, las construcciones conceptuales llamadas "leyes científicas" (nuestras leyes2) son las reconstrucciones cambiantes de las leyes objetivas en el nivel del pensamiento racional. O, si se prefiere, las construcciones conceptuales llamadas "leyes2" son la proyección deformada e incompleta de las leyes1 sobre el plano conceptual.
En suma, mientras las leyes de la naturaleza, del pensamiento y de la sociedad (leyes1) son la estructura de la realidad, los correspondientes enunciados nomológicos (leyes2) pertenecen a nuestros modelos ideales de la realidad, por lo cual se aplican —en el mejor de los casos— sólo aproximadamente, nunca con toda la exactitud deseada.
Mario Bunge La ciencia. Su método y su filosofía 52 5. Justificación de la necesidad de las distinciones restantes Ocupémonos ahora de esas pautas que empleamos en la descripción de fenómenos singulares en términos de esquemas generales, y que usamos cuando enriquecemos nuestra experiencia y la hacemos más exitosa. Los enunciados nomopragmáticos (leyes3) no se consideran habitualmente como proposiciones que pertenecen a una clase aparte, acaso porque rara vez son axiomas independientes. En efecto, casi siempre son aplicaciones de leyes2 a situaciones o a clases de situaciones especificas.
Los enunciados nomopragmáticos se deducen casi siempre de leyes 2 en conjunción con datos empíricos (esto es, proposiciones particulares que se refieren a miembros de esa subclase de hechos que llamamos "experiencia") . Esta peculiaridad se advierte claramente en el caso de los enunciados predictivos deducidos de los enunciados nomológicos y de las condiciones iniciales. Pero los datos empíricos no tienen por qué ser condiciones iniciales, valores de frontera, o trozos análogos de información específica. Considérese la ley (aproximadamente verdadera) de Cuvier, de la correlación morfológica; una consecuencia de esta ley2 es la conocida directiva para predicciones paleontológicas: "La reconstrucción del organismo entero sólo requiere el examen de una parte de sus restos", ejemplo típico de ley,.
Obsérvese, de paso, que en relación con el uso de las leyes2 ocurre la siguiente inversión de la relación hecho-ley: al establecer enunciados de leyes asignamos prioridad a los hechos, al menos en una etapa dada de la investigación; pero al aplicar las leyes2 razonamos como si las leyes planearan por encima de los hechos cuya estructura y tiempo son en realidad. Así, por ejemplo, con frecuencia pronunciamos frases de esta clase: "El hecho E es imposible porque su producción violaría la ley L". Con ello no queremos decir que los sucesos están sujetos a nuestros enunciados nomológicos; ni siquiera significamos que los hechos deben obedecer a las leyes1. Sólo estamos haciendo predicciones sobre la base de enunciados de leyes.
¿Toda ley3 no es sino consecuencia lógica de una ley2, en conjunción con informaciones especificas? Decididamente, esto no es así en la mecánica cuántica que —a diferencia de la física clásica— contiene postulados que se refieren explícitamente a resultados posibles de los experimentos, a diferencia de aquellos que se refieren a posibles aspectos de las cosas en sí. Sea, por ejemplo, el siguiente axioma de la mecánica cuántica en su formulación e interpretación habituales: "Los autovalores an de un operador A op, son los únicos resultados posibles de una medición exacta de la variable dinámica A representada por ese operador".
O, si no, este otro postulado: "La probabilidad de hallar el valor an al medir la variable A es igual a c2 n , donde “c” designa el n-ésimo coeficiente del desarrollo de la función de estado en funciones propias del operador que representa a A". Ambos postulados son típicas leyes 3 porque no se refieren a las cualidades A de las cosas en sí, sino, al contrario, a las cualidades tales como nos son conocidas en la experimentación (donde se manifiestan acopladas con las cualidades del dispositivo experimental).
Mario Bunge La ciencia. Su método y su filosofía 53 Si se arguyera que las que acabamos de mencionar no son leyes sino “meras" reglas semánticas que asignan un contenido empírico a ciertos símbolos (Aop, an y cn), podrá argüirse que las leyes, no son reglas convencionales de significado, que establecen vínculos arbitrarios entre signos y designados, sino que, por el contrario, se supone que expresan pautas constantes de la experiencia, tal como lo prueba el hecho de que nos permiten recoger nuevas informaciones empíricas, así como controlar (al menos estadísticamente) ciertos procesos físicos.
Otra objeción podría ser la siguiente. Los ejemplos aducidos han sido tomados de la ciencia física; ¿hay algún motivo para sostener la distinción propuesta en el campo de las ciencias sociales? La respuesta es ésta: precisamente en las ciencias del hombre es donde debiera ser de mayor utilidad la distinción entre enunciado nomológico y enunciado nomopragmático.
El sociólogo manipula leyes sociológicas (leyes 2 que pretenden dar cuenta de las leyes sociales, o leyes del nivel social; pero también manipula reglas, prescripciones propuestas, e ideales de política social. Si no distingue las dos clases de enunciados puede confundir proposiciones científicas con consignas (las que pueden apoyarse sobre consideraciones científicas, pero que no son enunciados de las ciencias sociales). Las leyes sociológicas (leyes2) no son ideales ni imperativos; tan sólo ciertos enunciados universales acerca de la práctica social (leyes3) pueden convertirse en ideales o normas para ciertos grupos sociales en ciertas circunstancias (y a condición de que se los reformule en un lenguaje normativo).
Desde luego, los ideales y las normas sociales, así como las propuestas de acción social, serán viables en la medida en que se funden sobre leyes 2 que encuadren con suficiente exactitud los hechos sociales. Pero esta relación de dependencia de los enunciados sociales nomopragmáticos respecto de las leyes sociológicas no implica que ambas clases de enunciados se recubren; los enunciados acerca de las pautas sociales pertenecen a las ciencias sociales, en tanto que los que se refieren a la política social pertenecen a la tecnología social.
Por último consideremos el cuarto significado de "ley científica". Probablemente fue en la física moderna donde se advirtió por primera vez la necesidad de disponer de enunciados explícitos de leyes acerca de las leyes. Las leyes4 no son requisitos lógicos o metodológicos conocidos de antiguo, tales como "Los enunciados nomológicos deben ser generales, significativos y verificables". Las leyes4 son reglas que guían la construcción de las teorías.
Miembros conspicuos de esta clase de leyes son los siguientes: (a) el principio de la covariancia general ("Las ecuaciones que expresan leyes físicas deben ser invariantes de forma respecto de transformaciones generales y continuadas de coordenadas”); (b) el principio de la mecánica cuántica conforme al cual "Las cantidades observables deben representarse por operadores lineales hermíticos". El que estos principios (o reglas) se conserven en el futuro, queda por verse. Lo que nos interesa en este momento es que estos enunciados forman una clase aparte. Podría argüirse que son metacientíficos, o epistemológicos, puesto que hablan acerca de entes y procedimientos científicos; pero esto Mario Bunge La ciencia. Su método y su filosofía 54 sólo mostraría que la metaciencia no está del todo por encima de la ciencia, sino que está en parte, ocluida en ella.
6. Aplicación de la distinción entre leyes1 y leyes2: ¿son necesarias las leyes científicas?
Con excepción de los empiristas estrictos y de ciertos idealistas objetivos, habitualmente se sostiene, o se implica, que las leyes científicas son necesarias en algún sentido. El análisis de esta proposición requiere un examen semántico previo de los términos que ella pone en relación, que son “ley científica" y "necesario".
El término "ley científica", en la proposición "Las leyes científicas son necesarias", designa por lo común, e indistintamente, pautas objetivas de la naturaleza, de la mente o de la sociedad (esto es, nuestras leyes 1), y enunciados nomológicos (leyes2) . Esta ambigüedad es una de las fuentes de la controversia acerca de la necesidad de las leyes. En cuanto al término “necesario”, se le asignan muchos más significados de los cuales los siguientes son pertinentes a nuestro propósito: (a) "necesario' es equivalente de relación constante y biunívoca (uno a uno) entre dos o más colecciones de objetos (p. ej., propiedades); (b) "necesario” es aquello que no podría ser de otra manera (lo opuesto de contingente); (c) "necesidad" significa conexidad lógica y, en particular, analiticidad (deducibilidad a partir de premisas admitidas anteriormente). Designemos las dos primeras acepciones con el término necesidad fáctica, y llamemos necesidad lógica a la conexidad lógica. Dejaremos de lado otros significados de "necesario”, sea porque no tienen sentido en el presente contexto (como ocurre con la equivalencia de necesidad y legalidad, ecuación que convierte en tautológico al enunciado que estamos examinando), sea porque pueden incluirse en la necesidad fáctica, o bien porque equivalen a la categoría seudopsicológica de inconcebibilidad.
Tenemos dos clases de objetos generales (leyes 1 y leyes2) y dos predicados "lógicamente necesario” (que simbolizaremos con L), y “fácticamente necesario" (que designaremos con F). Por consiguiente, a priori hay cuatro posibilidades: LF (necesidad lógica y fáctica), L¬F (necesidad lógica y contingencia fáctica), ¬LF (contingencia lógica y necesidad fáctica) y ¬(LF) (contingencia lógica fáctica). Examinémoslas.
(a) Leyes1. Debemos excluir las posibilidades LF y L¬F en relación con las pautas objetivas, pues la necesidad lógica es una propiedad de los enunciados y no de los objetos concretos; quedan dos posibilidades ¬LF y ¬(LF). Propondré un argumento en favor de la tesis de que las leyes1 son fácticamente necesarias y lógicamente contingentes.
Si las leyes1 fuesen aisladas, si no constituyeran sistemas, entonces podría pensarse que son fácticamente contingentes, esto es, que podrían no haber sido lo que son. Pero las leyes constituyen sistemas nómicos regionales (esto es, redes que caracterizan cada nivel de la realidad); por consiguiente, cada una de las leyes no es contingente. Sin embargo, podría objetarse que nada nos garantiza la constancia de las leyes: ellas podrían cambiar y, más aún, ciertamente lo hacen cada vez que emergen nuevos niveles de la realidad. La cuestión es Mario Bunge La ciencia. Su método y su filosofía 11 Para la elaboración de este punto, cf. el libro del autor Causality: The Place of the Causal Principie in Modern Science (Cambridge, Mass., Harvard University Press, 1959), cap. 12. Hay una versión española publicada por la Editorial Universitaria de Buenos Aires, 1961. (N. del E.).
55 averiguar si la variación de las leyes 1 —que es perfectamente concebible—, es a su vez contingente o necesaria. Habiendo admitido que las leyes1 constituyen sistemas, debiéramos concluir que, si cambian, entonces lo hacen de manera necesaria, y en particular de manera legal, de modo que presumiblemente existen leyes de la variación de las leyes. Concluimos que es verosímil que las leyes sean fácticamente necesarias, pero es seguro que son lógicamente contingentes.
(b) Leyes2. Puesto que éstas son construcciones conceptuales (constructs), ¬LF y ¬(LF) no son posibles: examinaremos entonces las posibilidades restantes, que son LF y L¬F. Argüiré que las leyes2 son fácticamente contingentes y lógicamente necesarias en cierto sentido.
Tomado aisladamente, todo enunciado nomológico es lógicamente contingente, puesto que un mismo grupo de fenómenos puede describirse por medio de un número ilimitado de hipótesis universales que merecen el nombre de "leyes". Esto, que es en esencia el argumento de Russell contra el principio de legalidad, vale para generalizaciones tales como las llamadas curvas empíricas, ya que por un número finito de puntos puede hacerse pasar infinitas curvas. Pero deja de valer cuando el enunciado nomológico en cuestión es incluido en una teoría, esto es, cuando se pone en contacto lógico con otros enunciados de leyes: en este caso, no sólo tiene el apoyo de sus casos favorables, sino que también gana el apoyo de hipótesis relacionadas con él, adquiriendo así, en cierta medida, el carácter de lógicamente necesario. Lo mismo se aplica a fortiori a aquellos enunciados nomológicos que son deducibles de axiomas o principios: son analíticos derechamente.
Parece, pues, que podemos concluir: (a) que las leyes1 (pautas objetivas) son fácticamente necesarias pero lógicamente contingentes; (b) que las leyes2 (enunciados nomológicos) son fácticamente contingentes y lógicamente necesarias, no en el sentido de ser impuestas por axiomas lógicos, o por principios inmutables de la razón, sino porque están o tienden a estar relacionadas lógicamente con otros enunciados de leyes.
7. Aplicación de la distinción entre leyes2 y leyes3: ¿es la causalidad una propiedad intrínseca de las leyes?
La distinción propuesta puede contribuir a eliminar varios malos entendidos en las filosofías de la ciencia corrientes. Por ejemplo, la distinción entre enunciados nomológicos (leyes2) y nomopragmáticos (leyes3) ayuda a aclarar la diferencia de especie que separa la explicación científica de la predicción científica, que se niega tan a menudo11. La diferencia es habitualmente borrada por los metacientíficos que restringen sus análisis a la estructura lógica de ambas operaciones, que en efecto es una. Lo que deseo explicar es el hecho de que Mario Bunge La ciencia. Su método y su filosofía 56 sobre la base de enunciados de leyes causales (o parcialmente causales) se pueden proponer explicaciones causales (o parcialmente causales), esto es, explicaciones en términos de causa, pero muy pocas veces predicciones que sean "causales" en la misma medida.
En efecto, la mayoría de las predicciones que se hacen sobre la base de enunciados de leyes —sean causales o no— tienen un componente estadístico que puede estar ausente de la correspondiente ley2. Así, por ejemplo, las predicciones astronómicas concernientes a las posiciones de los cuerpos celestes son siempre parcialmente estadísticas, en el sentido de que incluyen la estimación del error probable. En compensación, los enunciados de leyes2 con fines de verificación, predicción o acción —esto es, las leyes3— pueden tener un ingrediente causal ausente de la correspondiente ley2 (si es que tienen correspondencia en el nivel gnoseológico). Esto ocurre toda vez que podemos controlar algunas de las variables relacionadas por el enunciado nomológico en consideración. El conjunto de las variables bajo control experimental se llama a menudo “causa”, si, al cambiar sus valores de manera prescripta, se produce invariablemente cierto efecto de una manera unívoca, sin que a su vez influya apreciablemente sobre la "causa". Sin embargo, esto no basta para asegurar que la relación dada —esto es, la ley2— sea ella misma causal. Para asegurarlo tendríamos que probar que, eligiendo el conjunto complementario de variables como parámetros bajo control experimental (esto es, manipulando el efecto anterior como causa), la conexión variará. Pues si la conexión permanece invariable (si la relación es simétrica), entonces no puede llamarse propiamente causal, pues, por definición, la causación es una conexión asimétrica.
En conclusión, los ingredientes causal y estadístico de una ley natural o social dada no son siempre propiedades intrínsecas de ella, sino que varían según que se trate de un enunciado nomológico o nomopragmático. Por esto es útil, cuando se discute el problema de la legalidad y de la causalidad, aclarar si se hace referencia a leyes o a enunciados que se usan con fines predictivos o con otros propósitos vinculados con la experiencia. Adviértase, de pasada, que el mero hecho de que puede trazarse una distinción neta entre enunciados nomológicos y enunciados nomopragmáticos constituye un argumento en contra de la pretensión operacionista de que el significado de una proposición sintética consiste en la técnica de su verificación.
8. Los ideales de la ciencia en términos de los diversos niveles de significación de "ley” Sobre la base de las distinciones elaboradas y justificadas en lo que precede, podríamos comprimir los ideales de la investigación científica fundamental en las siguientes máximas: (1) Legalidad. Los hechos singulares (sucesos y procesos) tales como el lanzamiento del sputnik, la última pesadilla del lector, o la última huelga de la historia, deberán considerarse como casos particulares de leyes 1, (o, más exactamente, como secciones de haces de leyes1).
(2) Cognoscibilidad. Las leyes1 no son perceptibles pero son cognoscibles. Su conocimiento se corporiza en hipótesis generales (particulares o universales) que pueden llamarse "leyes2" Mario Bunge La ciencia. Su método y su filosofía 57 o "enunciados nomológicos".
(3) Limitación y perfectibilidad. Toda ley2 tiene un dominio de validez peculiar y es falible porque depende en parte de la experiencia; pero todo enunciado nomológico puede perfeccionarse tanto en extensión como en precisión.
(4) Generalidad del conocimiento fáctico. Los enunciados fácticos singulares son deducibles de enunciados fácticos generales (hipótesis llamadas “leyes2”). A esto se reduce, desde el punto de vista lógico, la explicación científica de los hechos.
(5) Sistematicidad. Las leyes2 constituyen sistemas lógicamente organizados o, al menos, organizables. La mayoría de las leyes2 son deducibles de hipótesis de tipo más elevado; las de máximo grado en un contexto dado se llaman "axiomas" o "principios". Esto es, la mayoría de los enunciados nomológicos son aplicables en términos de leyes 2 de un grado de generalidad aún mayor (p. ej., las ecuaciones de movimiento son deducibles de principios variacionales). En esto consiste la explicación científica de las leyes.
(6) Generalidad de los enunciados empíricos. Los enunciados empíricos singulares (los que se refieren a la subclase de hechos que llamamos "experiencia") son deducibles de hipótesis que pueden llamarse "leyes 3". Estas últimas proposiciones son las herramientas de la predicción; contienen variables (ligadas), tales como el tiempo y/o constantes descriptivas que resumen ítems de información específica (tal como los precios del trigo de un año dado).
(7) La legalidad de las leyes. Los enunciados nomológicos (leyes 2) encuadran en ciertos esquemas generales que pueden denominarse "leyes”. La exigencia (inadecuada) de que todas las leyes debieran ser expresables como ecuaciones diferenciales, y el principio (plausible) de covariancia pertenecen a esta clase de proposiciones (o, mejor, de propuestas).
Pueden considerarse como prescripciones metodológicas y/o como suposiciones ontológicas.
Dado que siete es un número célebre por sus propiedades, podemos terminar en este punto nuestra tentativa de caracterizar la ciencia en términos de los diversos significados de la palabra "ley", multiplicidad semántica que ha originado famosos embrollos.
Mario Bunge La ciencia. Su método y su filosofía 58 Filosofar científicamente y encarar la ciencia filosóficamente Mario Bunge La ciencia. Su método y su filosofía 59 Mario Bunge La ciencia. Su método y su filosofía 60 1. Lugar de la epistemología en la universidad argentina.
Es fácil advertir cuán modesto es el lugar que actualmente ocupa la filosofía de la ciencia en nuestras universidades. Si se exceptúan los pintorescos cursos de "epistemología de la ingeniería" de años recientes, la filosofía de la ciencia se ensena solamente en las facultades de filosofía, y en éstas no ocupa un lugar importante. ¿Qué importancia puede dársele a uno de los pocos cursos de filosofía sistemática que figuran en un plan de estudios que parece confeccionado a la medida de especialistas en filosofía grecorromana y medieval? ¿Qué importancia puede tener un único curso de filosofía de la ciencia, comparado con todos los cursos de filosofías y de lenguas muertas? Es una de tantas materias, acaso la más humilde de todas.
Tan poca importancia se le asigna a la filosofía de la ciencia en nuestra universidad, que el estudiante es lanzado a ella inerme. No se le dota, por ejemplo, de nociones científicas de nivel universitario; no se le equipa con las herramientas de la lógica moderna y del análisis lógico del lenguaje; ni siquiera se le exige un conocimiento suficiente del inglés, del alemán y del francés. Es claro que a menudo se hallaba consuelo en la circunstancia de que tampoco se exigían estos requisitos elementales a quienes ensenaban la materia o simulaban hacerlo.
La filosofía de la ciencia está arrinconada en el plan de estudios y, en general, en el panorama filosófico del país. Entre nosotros no se considera deseable que el filósofo se inspire en el modo de proceder del científico, quien comienza por los hechos, luego los describe y más tarde formula hipótesis y construye teorías para explicarlos; después deduce de ellas conclusiones particulares verificables, recurre eventualmente a nuevas observaciones o a nuevos cálculos, y contrasta sus conclusiones con estos resultados; y, finalmente, si lo halla necesario, corrige sus conjeturas sin compasión. Este severo carácter autocorrectivo de la investigación científica no suele estimarse superior al carácter oracular habitual en la filosofía tradicional, la que no siempre titubeaba en formular conjeturas sin fundamento y sin verificación.
Entre nosotros apenas se considera interesante la riquísima problemática filosófica que suscita la ciencia: para algunos, dicha problemática es demasiado estrecha, para otros demasiado árida, y para la mayoría de los filósofos y de los científicos ella apenas existe: se cree vulgarmente, en efecto, que la ciencia carece de problemas filosóficos y que no es más que una máquina de buscar datos. Entre nosotros suele encontrarse más cómodo adoptar una postura especulativa y de desprecio por los hechos y por la razón que adoptar una actitud crítica fundada en los hechos y que haga pleno uso de los instrumentos de la razón: es más Mario Bunge La ciencia. Su método y su filosofía 61 fácil proclamar la bancarrota de la razón y las limitaciones de la ciencia, anunciando que se está en posesión de fórmulas definitivas, o bien de una peculiar intuición que ahorraría el trabajoso camino de la investigación. Se busca la explicación última de todas las cosas sin atender a las explicaciones provisionales y perfectibles de la ciencia.
¿A qué se deben el descuido de la epistemología y el desdén por la actitud científica entre nosotros?
2. Algunos de los motivos del atraso de la epistemología en Latinoamérica La epistemología apenas se cultiva en Latinoamérica, y ni siquiera goza en ella de buena reputación. La reputación ambigua de la epistemología en estas tierras parece deberse, entre otros, a los siguientes motivos: a) En nuestro medio aún no se ha difundido la noticia de que la ciencia se está convirtiendo en el núcleo de la cultura moderna; ni suele estimarse que para filosofar con sentido, rigor y fruto en pleno siglo XX sea necesario estar al corriente de las grandes conquistas y de los grandes problemas de la ciencia, así como adoptar una actividad científica ante los problemas filosóficos.
b) Durante el último medio siglo han proliferado en Europa, y se han exportado a Latinoamérica, las corrientes irracionalistas. Al negarse la razón y exaltarse en su lugar la intuición, al rechazarse el dato fundado y abrazarse al mito, se niega la ciencia, que es un enfoque racional del mundo, y por consiguiente se niega la epistemología que es la teoría de ese enfoque racional de los hechos materiales y espirituales. En algunos países, el irracionalismo moderno puede interpretarse como síntoma de decadencia social; en nuestra América, tan necesitada de razón, esa mercancía importada goza de gran consumo porque es el complemento intelectual del analfabetismo y del atraso técnico y científico. El irracionalista europeo puede tolerar la ciencia a condición de que no conforme la visión del mundo: la Weltanschauung ha de seguir siendo mítica y no científica, pues quien conoce algo acerca del reloj del mundo puede pretender corregir su atraso. Entre nosotros, la prédica irracionalista es menos compleja: es el complemento filosófico de las pretensiones por retornar a la colonia, a la economía pastoril, a la cultura tradicional de corte predominantemente histórico-literario. No es dable esperar estímulos a la investigación epistemológica en un medio donde las corrientes oscurantistas gozan de mayor prestigio y poder que las iluministas, en un medio donde se habla más de la pretendida crisis de la ciencia que de sus éxitos.
c) El nivel científico de Latinoamérica es bajo, aunque sube rápidamente. Tenemos un notable déficit de científicos: necesitamos con angustiosa urgencia matemáticos, físicos, químicos, biólogos, psicólogos y sociólogos que contribuyan a la explotación racional de nuestras riquezas, a suplir nuestras deficiencias económicas y a superar la etapa de la cultura colonial.
¿Cómo asombrarse de que entre los escasos científicos latinoamericanos, recargados de Mario Bunge La ciencia. Su método y su filosofía 62 tareas de toda índole, no haya surgido un número ponderable de epistemólogos?
Presumiblemente, a lo sumo diez de cada cien científicos suelen tener inquietudes filosóficas, y de estos diez apenas uno se resuelve a encararlas de manera sistemática. En países cuyos científicos puros no llegan a mil, apenas puede esperarse que haya diez epistemólogos.
d) Los filósofos de tipo tradicional no son los únicos escépticos acerca de la utilidad de la epistemología: también la mayoría de los científicos suelen considerarla pasatiempo de profesores jubilados o de discutidores sin prisa por alcanzar resultados "positivos". Es un hecho que, hasta hace una veintena de años, casi todos los científicos que abordaban cuestiones filosóficas lo hacían al promediar su carrera o al terminarla. Este fenómeno no se debe solamente a la información unilateral que suele recibir el especialista: en parte se debe a que, para poder advertir la existencia de problemas filosóficos en el seno mismo de una especialidad científica, y para dedicarse a abordarlos, se necesita adquirir cierta experiencia y despojarse, así sea transitoriamente, de la prisa juvenil que reclama la obtención de resultados inmediatos aun a costa de la profundidad de su comprensión. Esta prisa es particularmente justificable entre nosotros: nuestros científicos, en su mayoría jóvenes, tienen aguda conciencia de que América latina no terminará de incorporarse al mundo culto mientras la aventura bélica, política y deportiva gocen en ella de mayor prestigio y protección que esa estupenda aventura intelectual que es la ciencia. Pero tarde o temprano nuestros investigadores advertirán —como les ha ocurrido a casi todos los científicos de primera línea— que quien encuentra grandes soluciones es quien enfoca los problemas con más amplitud, quien adopta una actitud filosófica ante la ciencia, es decir, quien sitúa el problema dado en su contexto más amplio y está dispuesto a revisar los fundamentos mismos de las teorías o de las técnicas. Así nació la ciencia moderna y así se renovó en el curso del último siglo.
Todas estas circunstancias contribuyen a crear un clima poco propicio para la investigación epistemológica. Afortunadamente, todas ellas son sólo aspectos de nuestra inmadurez económico-social y cultural; por lo tanto, es dado predecir que habrán de extinguirse a medida que nos desarrollemos.
Pero ya es hora de averiguar qué se entiende por "epistemología".
3. Filosofía y ciencia Cuando decimos "filosofía y ciencia”, el signo "y" puede significar la afirmación simultánea de ambos términos, o bien una relación cualquiera entre ellos. Si queremos ser más precisos, debemos recurrir, no ya a una conjunción, sino a las preposiciones, por figurar éstas entre los equivalentes lingüísticos de las relaciones lógicas. Juguemos, pues, un rato con las preposiciones, como una de las maneras de averiguar el nombre más correcto de nuestra disciplina.
Empecemos por "de". Si decimos "filosofía de la ciencia", damos a entender que se trata del Mario Bunge La ciencia. Su método y su filosofía 63 examen filosófico de la ciencia: de sus problemas, métodos, técnicas. estructura lógica, resultados generales, etc. Y así es: de todo esto se ocupa la epistemología; pero también de algo más. Probemos “en”. Por "filosofía en la ciencia” o, más exactamente, "filosofía de la filosofía en la ciencia" debiéramos entender, quizás, el estudio de las implicaciones filosóficas de la ciencia, el examen de las categorías e hipótesis que intervienen en la investigación científica, o que emergen en la síntesis de sus resultados. Por ejemplo, las categorías de materia, espacio, tiempo, transformación, conexión, ley y causación; e hipótesis tales como "La naturaleza es cognoscible o "Todos los sucesos son legales". De acuerdo: también de esto se ocupa la epistemología; y sin embargo no basta. ¿Qué nos dirá la expresión "filosofía desde la ciencia? Sugiere que se trata de una filosofía que hace pie en la ciencia, que ha sustituido la especulación sin freno por la investigación guiada por el método científico, exigiendo que todo enunciado tenga sentido y que la mayoría de las aseveraciones sean verificables.
Y ¿qué designa "filosofía con la ciencia"? Esta expresión sugiere —ambiguamente— que se trata de una filosofía que acompaña a la ciencia, que no se queda detrás de ella, que no especula sobre el ser y el tiempo al margen de las ciencias que se ocupan de los distintos tipos de ser y de acaecer: que es, en suma, una disciplina que no emplea conocimientos anacrónicos ni trata de forzar puertas ya abiertas. Examinemos, por último, la expresión "filosofía para la ciencia". Sugiere una filosofía que no se limita a nutrirse de la ciencia, sino que aspira a serle útil, al señalar, por ejemplo, las diferencias que existen entre la definición y el dato, o entre la verdad de hecho y la proposición que es verdadera o falsa independientemente de los hechos: será ésta una filosofía que no sólo escarbe los fundamentos de las ciencias para poner en descubierto las hipótesis filosóficas que ellas admiten en un momento dado, sino que además aclare la estructura y función de los sistemas científicos, señalando relaciones y posibilidades inexploradas.
Todo eso es, en efecto, la epistemología: filosofía de, en, desde, con y para la ciencia. Para ser equitativos con las cinco preposiciones, convengamos en no emplear ninguna de ellas, eligiendo en cambio un término único que posea todos esos significados. ¿Por qué no epistemología, que etimológicamente significa teoría de la ciencia? O ¿por qué no metaciencia, que significa ciencia de la ciencia? Cualquiera de estas denominaciones tiene la ventaja de que no reduce el ámbito de la disciplina en cuestión a un capítulo de la teoría del conocimiento, sino que permite abarcar todos los aspectos que pueden presentarse en el examen de la ciencia: el lógico, el gnoseológico, y eventualmente el ontológico.
Pero ¿no podríamos proseguir el juego con otras preposiciones, tales como "contra", "sobre" o “bajo"? Es verdad, éstas sirven para caracterizar otras tantas relaciones posibles entre la filosofía y la ciencia; pero veremos que no son adecuadas. En efecto, "filosofía contra la ciencia" es toda filosofía irracionalista o aquella que, sin serlo del todo, es enemiga del método científico.
Aunque escasas y escuetas, hay, sin embargo, filosofías de la ciencia que niegan extensión Mario Bunge La ciencia. Su método y su filosofía 64 y valor a la ciencia o la amputan radicalmente, y que además no encaran los problemas de la ciencia de manera científica o siquiera inteligible. Una epistemología que no sea parasitaria, sino que se esfuerce por ser útil a la ciencia, debe empezar por respetarla, aunque no necesariamente con servilismo, ya que la ciencia siempre puede aprender de la critica filosófica fundada. Quien filosofa contra la ciencia o aun al margen de ella, imita a los escolásticos que rehusaban mirar por el anteojo astronómico de Galileo.
En cuanto a las preposiciones "sobre" y "bajo”, al enlazar los términos 'filosofía" y “ciencia" sirven para designar concepciones muy estrechas del lugar y de la función de la epistemología. Si decimos "filosofía sobre la ciencia", significamos una ciencia superior en valor y poder a las ciencias particulares: una scientia rectrix con tales pretensiones de rectoría que los científicos se burlan de ella y con razón, pues la investigación científica no tolera úcases. Por su parte la expresión "filosofía bajo la ciencia" sugiere la posición inversa, de dependencia unilateral de la filosofía respecto de la ciencia: es éste un error que los epistemólogos no cometen en los hechos, aunque a veces lo proclaman como la más excelsa de las virtudes epistemológicas. La filosofía de la ciencia no sólo comporta el examen de los supuestos filosóficos de la investigación científica, sino que tiene derecho a una elaboración creadora en un nivel diferente del científico aunque reposa sobre él: el nivel metacientífico.
No hay pensador más entremetido que el epistemólogo: hoy señala una hipótesis filosófica oculta en un sistema teórico, mañana le discutirá al científico el derecho a usar cierta categoría en determinado contexto, y pasado mañana propondrá una teoría sobre determinada clase de conceptos o de operaciones de la ciencia. La epistemología no está por encima ni por debajo de la ciencia: está a la vez en la raíz, en los frutos y en el propio tronco del árbol de la ciencia.
Es necesario distinguir los problemas metacientíficos de los científicos, pero no hay por qué inventar un abismo que los separe: acaso no exista problema científico que no suscite problemas filosóficos, ni problema filosófico que pueda abordarse con esperanza de éxito si no es adoptando una actitud científica.
Algunos filósofos carentes de formación científica son culpables de las filosofías de la ciencia que son anticientíficas o por lo menos acientíficas, del mismo modo que los científicos sin formación filosófica suelen ser los creyentes más fervorosos en la existencia de la filosofía de la ciencia, que a menudo es aquella que han aprendido en el libro de epistemología con que se han cruzado. No existe la filosofía de la ciencia en cuanto teoría única: apenas hay intentos, si bien cada vez más serios, por "cientificizar” la epistemología y, en general, la filosofía. La situación imperante en este dominio recuerda a la reinante en la física antes de la síntesis newtoniana, o en la biología antes de la síntesis darwiniana: hay muchos resultados dispersos que rompen los moldes caducos de las distintas escuelas, resultados que será preciso ir integrando, cortando para ello las alambradas de púas tendidas entre las escuelas que han hecho contribuciones positivas a la filosofía científica de la ciencia. Quienes emprendan la labor de podar las ramas secas, desarrollar las verdes y coordinarlas en sistemas coherentes —pero transitorios—, cumplirán la misión del sinoptikós de Platón. Pero no lo harán ya al Mario Bunge La ciencia. Su método y su filosofía 65 margen de la ciencia, no lo harán ignorando el saber moderno, sino que se fundarán sobre él.
Toda época ha intentado integrar los conocimientos; nuestra época, la era de la ciencia, intenta integrar conocimientos más o menos verificados, pero no pretende elaborar síntesis cristalizadas.
4. Disciplinas contiguas a la epistemología Si uno de los cometidos del epistemólogo es analizar la estructura lógica de las teorías científicas, entonces la lógica es, una de sus herramientas de trabajo. Naturalmente, el epistemólogo se servirá de la lógica de su siglo, sin ser necesariamente un especialista en ella, del mismo modo que el biólogo emplea la física de su siglo sin ser él mismo físico. Y la lógica de nuestro tiempo —me refiero a la lógica científica— se compone, esencialmente, de la lógica simbólica, o logística, y de la lógica inductiva o de la inferencia, probable. El epistemólogo que ignore la lógica formal moderna podrá confundir expresiones del tipo "Sócrates es mortal" con las del tipo "Sócrates fue maestro de Platón". Y quien ignore la existencia de la lógica de la inferencia no demostrativa, no advertirá las diferencias existentes entre el proceso constructivo de una teoría científica y su posterior reordenamiento racional.
Algo similar puede decirse de la semiótica o ciencia de los signos —y en particular, de los lenguajes—, en la que caben la sintaxis o teoría de las relaciones entre los signos, la semántica o teoría de las relaciones entre los signos y aquello que designan, y la pragmática o teoría del uso de los signos. Dado que toda ciencia emplea signos, el epistemólogo hará bien en emplear los resultados de la semiótica al analizar el lenguaje de la ciencia. Pero no exageremos. Aunque hay quienes sostienen que la filosofía de la ciencia es sólo lógica de la ciencia o a lo sumo análisis sintáctico y semántico del lenguaje científico; y aunque los formalistas afirman que el epistemólogo sólo debe interesarse por la estructura lógica de las teorías acabadas, es un hecho que las ciencias de la realidad no sólo trabajan con conceptos, sino también con cosas, tanto naturales como artificiales. Siendo los actos del científico tan importantes como su pensamiento, la epistemología no debiera limitarse a la lógica y el lenguaje de la ciencia: no debiera ser sólo teoría de teorías, sino también teoría de actos, es decir, metodología y no sólo metateoría. Por consiguiente, la lógica y la teoría de los signos son herramientas importantes del epistemólogo, pero no las únicas.
Muchos epistemólogos hallan tan interesante y fructífero el estudio del proceso de descubrimiento e invención como el de la exposición y justificación de los resultados. Más aún, la historia de la ciencia, si en ella se incluye la más reciente, es nada menos que la proveedora de la materia prima de la epistemología. ¿Por qué ha de interesar la dinámica de la ciencia menos que su estática? Rara vez un interés profundo por las ideas y los actos no lleva a inquirir sobre sus orígenes y desarrollo. Todavía más: la filiación histórica de unas y otros ayuda a comprenderlos. Así como el estado actual de una especie biológica no se entiende adecuadamente si no es como etapa de un proceso, así tampoco se entiende Mario Bunge La ciencia. Su método y su filosofía 66 acabadamente el quehacer científico si sólo se pone atención en sus resultados. Muchos de los esfuerzos del científico del pasado parecen tontos, y milagrosos los éxitos del moderno, si no se los ubica en su contexto histórico. Quien sostiene que el epistemólogo sólo debe ocuparse de la estructura lógica y —de haberlo— del fundamento empírico de las teorías acabadas, adopta una actitud fijista que lleva a petrificar los resultados, a olvidar que todos ellos son aproximados y perfectibles. Si se desea estudiar en forma cabal una transformación —y la ciencia es cambiante en grado sumo— es menester adoptar una actitud transformista capaz de captar la dinámica de la averiguación científica.
Otro tanto puede decirse de la historia de la filosofía: a menudo se supone que el epistemólogo nada tiene que aprender de los filósofos del pasado, quienes no habrían hecho sino apilar error sobre error. Quien adopta esta actitud arrogante ante sus antecesores se expone a descubrir la pólvora en el mejor de los casos, y la piedra filosofal en el peor.
Además, desdeña una de las fuentes de la actividad científica y, a la vez, uno de sus principales resultados, a saber, ciertos principios filosóficos referentes a la realidad en su conjunto, al conocimiento en general, etc. Estos principios participan —habitualmente en forma implícita— de la investigación científica, aunque sólo sea porque intervienen en la visión del mundo del investigador. La adopción de una actitud científica en filosofía, y el tratamiento riguroso de problemas metacientíficos, no implica desdeñar la totalidad de la filosofía tradicional; implica, más bien, abordar íntegramente su problemática, pero ahora sobre la base de los conocimientos científicos actuales y de las técnicas filosóficas actuales.
Desde luego el epistemólogo científico desestimará ciertos problemas tradicionales por considerarlos meros enredos verbales, y concederá a otros problemas mucha menor importancia de la que tuvieron en el pasado. Pero, en compensación, abordará problemas acerca de cuyo solo enunciado no podían tener idea sus antecesores, tales como la estrategia de la experimentación, o las relaciones entre la probabilidad y la frecuencia, o la técnica de la construcción de teorías. El epistemólogo, en suma, no tiene por qué fingir que ha cortado todo vínculo con el pasado, ya que sobre el pasado se encarama, por radicales que sean las novedades que enuncia: si no quiere recaer en viejos errores, se esforzará por asimilar el pasado en lugar de desdeñarlo.
El epistemólogo que descuida o desdeña la historia de las ideas científicas y filosóficas adopta una postura tan altanera y cerrada como la del historiador de la filosofía que ignora la existencia de la filosofía de la ciencia o la confunde con el movimiento negador o retaceador de la ciencia. El fijista que ignora la historia de las ideas suele tomar por definitiva la teoría más reciente, rodeándola de un caparazón escolástico que más tarde podrá dificultar su desarrollo interno y su crítica epistemológica. Así ocurrió con la mecánica de Newton, así ocurre con la mecánica cuántica. Al proceder de esta manera, lejos de ser útil al progreso científico, el epistemólogo fijista podrá llegar a obstaculizarlo. Además, el fijista —que se priva nada menos que de contemplar la formación y el desarrollo de los conceptos— suele caer en la tentación de filosofar acerca de una ciencia atemporal, perfecta, inexistente, Mario Bunge La ciencia. Su método y su filosofía 67 imitando así al metafísico que inventa un "ser" inmutable e inaccesible, allende el acaecer ordinario. La epistemología, en suma, sin confundirse con la historia de las ideas y de las prácticas de la ciencia y de la filosofía, debe hacer uso de ellas, para poder ubicar su objeto en su contexto histórico.
Los empiristas tradicionales buscaban el significado de las ideas en sus raíces psicológicas: creyendo hacer filosofía hacían psicología del conocimiento. Los materialistas vulgares encontraban el significado de las ideas en su correlación con el medio natural y social en que ellas nacen y se desarrollan: creyendo hacer filosofía hacían sociología del conocimiento. La psicología y la sociología del conocimiento son o aspiran a ser ciencias particulares, no forman parte de la epistemología, aunque a menudo se las confunde con ésta, porque las tres hablan sobre la ciencia. Mientras la psicología de la ciencia estudia el correlato psíquico del concepto y del acto del científico; y mientras la sociología de la ciencia estudia la función social de la ciencia y eventualmente la responsabilidad social del científico, la filosofía de la ciencia, por su parte, se ocupa de los aspectos lógicos, gnoseológicos y ontológicos de la ciencia, y no del comportamiento individual o social del investigador científico. Sin embargo, sería miope el epistemólogo que no aprovechase las conclusiones que le brindan la psicología y la sociología del conocimiento, pues ellas le permiten ubicar y comprender más adecuadamente su objeto.
Las disciplinas que hemos mencionado —la epistemología, la lógica, la teoría del lenguaje, la historia de la ciencia y de la filosofía y la psicología y la sociología de la ciencia— se esfuerzan por saber qué es el saber. Por consiguiente, aunque difieren, distan de ser ajenas entre sí: cada una de ellas ilumina una faceta de un mismo objeto: el saber verificable.
5. Ciencias y humanidades Apenas se discute ya que la ciencia es lo que distingue la cultura contemporánea de las anteriores. No sólo es el fundamento de la tecnología que está dando una fisonomía inconfundible a nuestra cultura material, sino que de continuo absorbe disciplinas que otrora fueron artísticas y filosóficas: ayer, la antropología, la psicología y la economía; hoy, la sociología y la historia; mañana, quizá, la estética y la ética. Además, la concepción del mundo del hombre contemporáneo se funda, en medida creciente, sobre los resultados de la ciencia: el dato reemplaza al mito, la teoría a la fantasía, la predicción a la profecía. La cultura social y la personal se tornan, en suma, cada vez más científicas. Hace un siglo, quien ignoraba La Ilíada era tildado de ignorante. Hoy lo es, con igual justicia, quien ignora los rudimentos de la física, de la biología, de la economía y de las ciencias formales. Con razón, porque estas disciplinas nos ayudan mejor que Homero a desenvolvernos en la vida moderna; y no sólo son más útiles, sino que también son intelectualmente más ricas.
Semejante actitud no implica desdén para con las artes y las llamadas humanidades; no significa que sea digno de admiración el especialista que permanece insensible a la belleza o Mario Bunge La ciencia. Su método y su filosofía 68 que menosprecia la investigación filológica. Lo criticable es que, en el siglo de los mayores avances sociales y de la energía nuclear, se siga sosteniendo que la literatura y la crítica literaria deben seguir siendo el eje de la cultura o por lo menos la base de la formación cultural. Modernicemos el concepto de humanidades y equilibremos los diversos ingredientes de la educación, ofreciendo las posibilidades de una educación integral y actual. Si la vida no es ni debe ser puro goce, y si la cultura no es ni debe limitarse a ser comentario de textos, entonces es preciso que renovemos las ideas acerca del lugar que deben desempeñar las artes y las humanidades en la educación moderna. Sostener que el goce estético y la educación para refinarlo deben ocupar un lugar más importante que la búsqueda de la verdad, de la utilidad y del bien social, no es hoy signo de cultura refinada, sino de incultura, de egoísmo, de frivolidad propia de salones victorianos.
¿Cómo es posible seguir sosteniendo que la ciencia y la filosofía de la ciencia son áridas, inhumanas o deshumanizadas, siendo por ello preciso dulcificarlas y dignificarlas mediante una dosis de las llamadas humanidades? ¿Acaso las teorías científicas y metacientíficas se encuentran en la naturaleza, para que pueda tildárselas de inhumanas? ¿No son acaso creaciones humanas, que suelen costar un esfuerzo de imaginación y de concepción mayor que la mayoría de las obras literarias y de crítica literaria? ¿Acaso las obras científicas y metacientíficas no emplean, además de elementos sensibles y del lenguaje diario, almacenes de experiencias, instrumentales conceptuales y lenguajes enormemente más ricos que los que usa el escritor? Consúltese cualquier revista científica y se advertirá cuán ardorosa —aunque controlada—es la imaginación requerida para inventar una teoría, o para hacer un cálculo aproximado, o para diseñar un instrumento. Sólo cree que la ciencia es pobre en concepto y en imágenes, y que la investigación científica carece de poesía, quien tiene pobres informaciones acerca de la vida de la ciencia. Junto con la filosofía, ella constituye la más rica creación del espíritu. ¿Por qué, entonces, oponer las humanidades a las ciencias, como si éstas fuesen menos humanas que aquéllas, y como si no fuesen precisamente las ciencias las que alcanzan el conocimiento más profundo y adecuado del hombre? Dígase más bien que las ciencias y las llamadas humanidades no son antagónicas sino complementarias, aun reconociendo que en la época contemporánea el centro de la cultura se desplaza de las humanidades a las ciencias.
¿Cómo lograr eficazmente la integración de la ciencia y de las humanidades en la enseñanza universitaria? La solución que suele ofrecerse en algunos países consiste en agregar trabajos de laboratorio al plan de estudios de las humanidades, y literatura al plan de estudios de ciencia. No debe asombrar que esta solución sumista fracase: lo que se agrega se considera materia "blanda", que se tolera y estudia a desgano, sin que deje rastros. No se logra una reorientación de los estudios universitarios y de la mentalidad de los estudiantes con el mero agregado de cursos. Si lo que se busca es una síntesis, debe ensayarse una solución integradora y no aditiva. ¿Por qué no ensayar el cultivo de una actitud filosófica en las ciencias naturales y sociales, y de una actitud científica en la filosofía y en las llamadas Mario Bunge La ciencia. Su método y su filosofía 69 humanidades? No hay por qué buscar la ciencia fuera de las humanidades, cuando lo que se requiere es encararlas en forma científica; ni hay por qué buscar la filosofía fuera de la ciencia, cuando se sabe que ésta posee sustancia filosófica.
La epistemología es terreno particularmente adecuado para advertir la integración de la ciencia, de la filosofía y de las humanidades, y para promoverla. La epistemología se ocupa de los fundamentos y procedimientos de todas las ciencias, desde la geología hasta la lingüística; la epistemología muestra que la ciencia moderna es una actividad eminentemente espiritual. sirviéndose de la manualidad como de un medio. No es difícil mostrarle al estudiante de ciencia que el quehacer científico no es ajeno al espiritual, ya que se propone edificar sistemas de ideas; que, por añadidura, estos sistemas de ideas suponen hipótesis filosóficas y conducen al establecimiento de otras; y que toda ciencia plantea, a su vez, arduos problemas a la historia de las ideas, a la sociología y a otras disciplinas que suelen o solían considerarse humanísticas. No es necesario inyectarle humanidades al científico; basta mostrarle que su propia ciencia las incluye o está relacionada con ellas. Exíjasele precisión conceptual al estudiante de ciencias y terminará esforzándose por afilar su lógica y por pulir su expresión literaria; muéstresele el valor intrínseco y social de la ciencia y convénzaselo de que es conveniente la transparencia lógica de los edificios teóricos para saber cómo repararlos o ampliarlos: de esta manera aprenderá a reconocer en su ciencia bastante más que el estudio de una determinada clase de objetos.
No conseguiremos que el científico sea un hombre culto obligándole a estudiar temas que no le interesan. Estimulémosle, en cambio, a que advierta la raíz gnoseológica y la armazón lógica de su especialidad; habituémosle a que repare en las conexiones de su especialidad con las demás disciplinas; acostumbrémosle a la idea de que su materia tiene un pasado y una función social, de la que en gran parte depende su futuro. Para conseguir todo esto lo más eficaz son las oportunas acotaciones del propio instructor de ciencias; pero como en todas partes son contados los profesores de ciencias que poseen información filosófica e histórico-social, conviene ensayar cursos especiales de filosofía y de historia de la ciencia. A la luz de estas disciplinas, el especialista y el aprendiz de especialista comprenden que la filosofía y las llamadas humanidades no son del todo exteriores a su materia; y al advertirlo se esforzarán por profundizar en estas dimensiones extracientíficas de su especialidad. Así, insensiblemente, se convertirá en un especialista culto. En cambio, del especialista que niega resueltamente que su ciencia tenga relación con la filosofía; de quien se desinteresa totalmente de la estructura lógica, de la evolución histórica o de la función social de su propia especialidad, de éste no puede decirse que sea un hombre culto aun cuando lea novelas o visite exposiciones de pintura. Será tan inculto por desechar todo el saber acerca de lo que a él le interesa saber, que ignorará qué es su propia ciencia.
6. Los estudios epistemológicos en la formación del científico Mario Bunge La ciencia. Su método y su filosofía 70 Hay, sin embargo, quienes piensan que, aunque el científico cobre conciencia de las implicaciones y proyecciones no científicas de su propio trabajo, no por ello será será más eficaz en su especialidad: conceden que será más culto y que por consiguiente vivirá una vida más racional y más rica, pero arguyen que, en cambio, no descubrirá ni inventará más ni mejor, sino al contrario, pues se distraerá con las lecturas y meditaciones marginales a su especialidad. Esta difundida opinión refleja, sin duda, una preocupación responsable por ahorrar desvíos inútiles, pero no ha sido compartida por los grandes maestros del pensamiento científico, y es más bien típica de quienes toman los instrumentos por fines.
El estudiante de ciencias o el científico que alguna vez dedique una parte de su tiempo a estudios epistemológicos podrá obtener de éstos algunos de los siguientes beneficios: a) no será prisionero de una filosofía incoherente y adoptada inconscientemente; podrá entonces corregir, sistematizar y enriquecer las opiniones filosóficas que de todas maneras integran su visión del mundo; b) no confundirá lo que se postula con lo que se deduce, la convención con el dato empírico, la cosa con sus cualidades, el objeto con su conocimiento, la verdad con su criterio, y así sucesivamente. Esto le ahorrará buscar demostraciones de definiciones, le impedirá confundir prueba lógico-matemática con verificación empírico-lógica y le ayudará a sopesar el soporte empírico de las teorías; no confundirá materia con masa ni atribuirá masa a toda cantidad de energía; no tomará "precedencia" ni “predictibilidad” por “causalidad”, y no reducirá la explicación científica a su especie causal.
En general, se esforzará por entender los términos que emplea, tal como se esforzaron, antes que él, los científicos con mentalidad filosófica que construyeron la ciencia moderna; c) se habituará a explicar las suposiciones e hipótesis, lo que le permitirá saber qué es lo que hay que corregir cuando la teoría no concuerda satisfactoriamente con los hechos; d) se acostumbrará a ordenar sistemáticamente las ideas y a depurar el lenguaje; se habituará, en suma, a buscar la coherencia y la claridad; e) afilará su bisturí crítico: la meditación epistemológica, al habituar a exigir pruebas, es buen preventivo del dogmatismo; f) el científico con alguna formación epistemológica podrá mejorar la estrategia de la investigación, al proceder con mayor cuidado en el planeamiento de los experimentos o de los cálculos y en la formulación de las hipótesis, así como en la evaluación de las consecuencias de unos y otras. La epistemología ciertamente no ayuda a medir ni a resolver ecuaciones, pero en cambio ayuda a ubicar estas operaciones en el proceso de la investigación; g) su atención se desplazará del resultado al problema, de la receta a la explicación, de la ley empírica a la ley teórica. Ninguna teoría de contenido fáctico le satisfará en forma definitiva: siempre encontrará alguna objeción que hacerle. El estudio de la epistemología, al tornarlo protestón, podrá estimularlo a explorar nuevos territorios; h) la filosofía y la historia de la ciencia le acostumbrarán a considerar la marcha de la ciencia, Mario Bunge La ciencia. Su método y su filosofía 71 no como un desarrollo meramente aditivo, sino como un proceso en que cada solución plantea nuevos problemas, en que viejas hipótesis desechadas por un motivo pueden volver a cobrar interés por otro motivo, y en que cada problema tiene varias capas y, por lo tanto, varios niveles de solución. En cambio, para quien no enfoca la ciencia con una actitud filosófica e histórica, toda fórmula científica es trivial en cuanto a manejarla, y la teoría más reciente es la definitiva o por lo menos la penúltima. ¿No hay textos que califican de evidentes los principios de Newton, y no hay científicos que esperan con impaciencia la teoría futura?; i) se ampliará su horizonte, al enriquecerse el surtido de relaciones lógicas y de posibilidades de interpretación; j) obrará con cautela cuando tantee terreno nuevo: extremará las exigencias de la verificación, dudará del valor de los datos empíricos que encajen en teorías endebles —o al menos los pondrá en cuarentena— y no dejará que los detalles le oculten lo esencial. Pero no por ello perderá coraje: antes bien, sentirá respeto por las teorías consagradas, aunque no reverencia por ellas. Así como no hay grandes hombres para su valet, tampoco hay teoría intocable para el científico que adopta una actitud filosófica, pues ve a la ciencia, por así decirlo, en pantuflas.
Por todos estos motivos conviene al desarrollo de la ciencia que los instructores de ciencia llamen la atención sobre los problemas filosóficos y las raíces históricas de las cuestiones científicas; por los mismos motivos conviene incluir el estudio de la filosofía y de la historia de la ciencia en los planes de estudio de las diversas ciencias particulares. Con ello no se agregarán conocimientos específicos acerca del mundo, pero sí se facilitará la correcta comprensión, profundización, ordenación y evaluación de dichos conocimientos. El científico o estudiante de ciencias que dedique alguna atención a este género de estudios no se distraerá necesariamente, sino que recibirá estímulos para encarar su tarea con mayor profundidad y responsabilidad, y hasta con más amor: advertirá que su trabajo es más complejo, más importante y hasta más bello de lo que había creído.
Desde luego, existe el peligro de que alguno se pase al campo de la epistemología o al de la historia de la ciencia. ¡Enhorabuena si lo hace! ¿No protestamos acaso por la escasez de filósofos e historiadores de la ciencia que conocen el objeto de sus estudios?
7. El aprendizaje y la enseñanza de la epistemología Si no es difícil lograr que el estudiante de ciencias llegue a adoptar una actitud filosófica ante su propia especialidad, es de temer en cambio que, en las condiciones actuales, no sea fácil inducir a los estudiantes de filosofía a que adopten una actitud científica. En primer lugar, es la inmadurez de la propia epistemología, la que torna su estudio accidentado. En segundo lugar, porque nuestros estudiantes no han sido preparados para adoptar una actitud científica sino para lo contrario: salvo excepciones, se les ha inculcado indiferencia y a veces desprecio por la ciencia, y no se les ha dado la formación científica indispensable para abordar con Mario Bunge La ciencia. Su método y su filosofía 72 profundidad el estudio de la epistemología.
¿Es posible que existan estudiantes de preceptiva literaria que no sepan leer y escribir? No, porque se trata de una disciplina que versa sobre el lenguaje escrito y por lo tanto lo presupone. Y nadie haría caso de un analfabeto que pretendiera ensenarla. En cambio, entre nosotros no provoca asombro y escándalo el que se ensene filosofía de la ciencia a estudiantes que, en el mejor de los casos, sólo están equipados con los recuerdos de las nociones científicas elementales que aprendieron en la escuela secundaria. Y han sido contados los que, en nuestro medio, se han escandalizado de que hubiese audaces que simularan ensenar filosofía de la ciencia sin haber hecho jamás investigación científica, sin siquiera haber estudiado ciencias en el nivel universitario. Esto no ocurre, desde luego, en los centros culturales avanzados, donde la epistemología es ensenada por personas que investigan o han investigado en algún campo de la ciencia, a alumnos que poseen una preparación científica de nivel universitario. No se conoce otra vía para alcanzar un conocimiento adecuado del objeto mismo de la epistemología.
Ni siquiera basta tener nociones sobre la ciencia clásica si se quiere filosofar con provecho sobre la ciencia actual. Para hacer filosofía de la ciencia viva, para hacer epistemología útil a la ciencia, para poder detectar y abordar la problemática filosófica suscitada por la investigación científica que se está haciendo ante nuestra vista, es necesario —aunque ciertamente no es suficiente— tener un conocimiento de primera mano de esa misma ciencia actual. Y esto no le es dado, en toda su amplitud, a un solo individuo. Por esto, la epistemología como cualquier otra rama del saber y acaso más que otras, es una empresa colectiva, a la que contribuyen numerosos especialistas, filósofos de la lógica, de la matemática, de la física, de la biología, de las ciencias socio-históricas, etcétera.
La filosofía de la ciencia que no es ensenada por científicos a estudiantes que poseen una formación científica discreta, tiene mucho de farsa. Es hora de que el estudio de la epistemología cobre entre nosotros la seriedad que lo caracteriza en otras partes. Es hora de facilitar, a quienes deseen estudiarla con seriedad, los instrumentos lógicos, semióticos y científicos necesarios. Esta reforma es propugnada, entre nosotros, por la novísima Agrupación Rioplatense de Lógica y Filosofía Científica. Mientras ello no llegue, será conveniente que alumnos y profesores extrememos la modestia ante las ciencias que hayamos de examinar, tratando de entender sus rudimentos antes de criticarlas.
Para facilitar el aprendizaje científico previo a cualquier consideración epistemológica seria, se contará con la colaboración de científicos y estudiantes de ciencias, quienes estarán a disposición de los estudiantes de epistemología y, a su vez, tendrán oportunidad de informarse, por estos últimos, acerca de los problemas tradicionales de la filosofía, con muchos de los cuales entroncan los problemas filosóficos de la actualidad. En cuanto a las numerosas limitaciones del profesor, para subsanarlas aunque sea en parte, se solicitará el concurso de colegas y de especialistas en diversas ramas de la ciencia, para que expongan los problemas filosóficos que han encontrado en el curso de sus investigaciones. De esta Mario Bunge La ciencia. Su método y su filosofía 73 manera, cada uno de los participantes del curso aprenderá algo.
El curso de epistemología no tendrá una orientación filosófica definida: su objetivo será facilitar la adquisición de información amplia y objetiva, promover la discusión y, sobre todo, incitar a la meditación independiente. Desde luego, el profesor tendrá una orientación definida o la buscará, ya que el pensador sin brújula y sin norte suele carecer de ideas originales y coherentes, así como del entusiasmo necesario para proseguir la búsqueda y para contagiarlo a los demás. No hay filosofía viva sin diálogo y sin cierta parcialidad compatible con la objetividad; al mismo tiempo que se filosofa sobre un tema dado se dialoga con alguien y se teoriza contra alguien, aun cuando en la exposición final no se trasunten el diálogo ni la polémica.
Se tendrán en cuenta las principales orientaciones filosóficas, sin excluir las anticientíficas, aunque sólo sea para analizarlas científicamente. Pero no se tomarán por temas de estudio las escuelas y los autores, sino los problemas epistemológicos: ya es hora de abandonar el enfoque exclusiva y predominantemente escolástico e histórico de los problemas filosóficos; es hora de abordarlos sistemáticamente, como lo han hecho quienes han dicho algo nuevo.
La tarea informativa quedará, así, subordinada a la labor formativa o, mejor, autoformativa; los autores servirán de peldaños y no de cadenas. Se preferirá el diálogo vivo a la recitación de datos, y la discusión inacabada al oráculo definitivo. Se tendrá la pretensión de guardar fidelidad al lema que eligieron los discípulos de uno de los fundadores de la ciencia moderna: Provando e riprovando.
Se intentará, en suma, adoptar una actitud científica ante los problemas epistemológicos, con la esperanza de que produzca frutos que convenzan a los científicos de la conveniencia de encarar filosóficamente la ciencia, y que persuada a los filósofos de que la filosofía rigurosa y fecunda no es un género literario sino una ciencia.
Mario Bunge La ciencia. Su método y su filosofía 74 FUENTES Este volumen contiene cuatro ensayos tomados, con algunas modificaciones, del libro del autor Metascientific Queries (Springfield, Ill. Charles C. Thomas, 1959). Los cuatro fueron publicados aisladamente en castellano, pero son hoy difíciles de hallar: el primero por la Facultad de Ingeniería de la Universidad de Buenos Aires (1958), el segundo por la Facultad de Filosofía y Letras de la misma Universidad (1958), el tercero por la Universidad Nacional de México (1958) y el cuarto por la revista Ciencia e Investigación (13, 244, 1957).
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[FREE] In a resonance tube experiment, two consecutive resonances are observed when the lengths of the air columns - brainly.com
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In a resonance tube experiment, two consecutive resonances are observed when the lengths of the air columns are 16 cm and 49 cm. If the frequency of the tuning fork used is 500 Hz, the velocity of sound in air is:
A. 310 ms⁻¹
B. 320 ms⁻¹
C. 330 ms⁻¹
D. 340 ms⁻¹
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The velocity of sound in air is found to be 330 ms⁻¹ when two consecutive resonances in a resonance tube experiment are observed at 16 cm and 49 cm and the frequency used is 500 Hz. The correct option is C.
In a resonance tube experiment, we observe consecutive resonances at lengths of the air column. The distance between two consecutive resonances in a tube with one end closed is equal to half the wavelength ( rac{ \text{λ}}{2}). The frequency ( ) provided was 500 Hz, and the two consecutive resonances lengths were 16 cm and 49 cm, which means the half-wavelength difference is 49 cm - 16 cm = 33 cm or 0.33 m. Since this difference represents half a wavelength, the full wavelength (λ) is 0.33 m 2 = 0.66 m. The velocity of sound (v) can be calculated using the formula = λ. Substituting the given values gives us = 500 Hz and λ = 0.66 m:
\tf = 500 Hz 0.66 m = 330 m/s
Hence, the velocity of sound in air is 330 ms⁻¹ which corresponds to option C.
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The velocity of sound in air, calculated from the resonance tube experiment where resonances were observed at 16 cm and 49 cm, is 330 m/s. This is derived using the wavelength difference and the frequency of the tuning fork. The chosen option is C.
Explanation
In the resonance tube experiment, we are examining how sound waves behave in air columns of different lengths in a tube. The lengths of the air columns where consecutive resonances are observed are given as 16 cm (0.16 m) and 49 cm (0.49 m).
Finding Wavelength:
The difference in length between these two resonances corresponds to half the wavelength (λ/2) of the sound, because the first resonance is at the fundamental frequency, and the second is the first overtone.
Therefore:
2 λ=49 cm−16 cm=33 cm=0.33 m
Calculating Complete Wavelength:
To get the full wavelength (λ), we multiply by 2:
λ=2×0.33 m=0.66 m
Using the Wave Speed Formula:
The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by:
v=f×λ
Given that the frequency of the tuning fork used is 500 Hz, we substitute the values into the equation:
v=500 Hz×0.66 m=330 m/s
Thus, the velocity of sound in air is 330 m/s. This corresponds to option C.
Examples & Evidence
An example can be a similar experiment where different lengths of air columns are tested in a tube to observe how sound waves resonate at different frequencies. This helps to understand the concept of resonance and sound wave propagation in various mediums.
The formula for wave speed, v = f λ, is widely used in physics to relate frequency, wavelength, and speed of sound in various mediums, confirming the calculations are valid in this context.
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The velocity of sound in air is 330 m/s.
In the resonance tube experiment, two consecutive resonances are found at air column lengths of 16 cm and 49 cm. To find the velocity of sound in air using these measurements, we must recognize that the difference in length between the two resonances corresponds to the length of one wavelength (λ). This means that the second resonance occurs at a length that is one full wavelength longer than the length at the first resonance. So, the wavelength would be 49 cm - 16 cm = 33 cm = 0.33 meters.
Since the frequency (f) of the tuning fork is given as 500 Hz, we can use the formula for wave speed (v), which is v = f
λ.
Substituting the given values into the equation, we get:
v = 500 Hz 0.33 m = 165 m/s
However, since each resonance corresponds to a half-wavelength difference and not a full wavelength, we must multiply the calculated velocity by 2 to account for this:
v = 2 165 m/s = 330 m/s
Therefore, the velocity of sound in air is 330 m/s which corresponds to option C.
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Chimera (mythology)
In classic Greek mythology, the Chimera is depicted as a fearsome, fire-breathing creature composed of parts from three different animals: a lion's head with a thick mane, a goat's body, and a serpent's tail. Sometimes illustrated with a second fire-breathing head resembling a goat, this monstrous being is believed to have originated from the union of Typhon, a storm giant, and Echidna, a half-woman, half-dragon figure known for spawning various other legendary beasts. The Chimera terrorized the region around Lycia, attacking farms and consuming livestock and people, until the hero Bellerophon, aided by the winged horse Pegasus, was sent to confront it. Bellerophon ultimately defeated the beast using a clever strategy involving a lead weapon that blocked the Chimera's throat, leading to its fiery demise.
The myth of the Chimera has transcended its ancient roots to inspire various forms of modern speculative fiction, including video games, comics, and films. Additionally, the term "chimera" has been adopted in contemporary genetics and psychology, reflecting the phenomenon of chimerism, where an individual may possess cells from multiple organisms. This mythical creature serves as a metaphor for unattainable dreams or delusions, highlighting the potential emotional turmoil that can arise from such illusions. The enduring legacy of the Chimera illustrates its cultural significance and versatility, resonating across both artistic and scientific fields.
Published in: 2023
By: Dewey, Joseph, PhD
Go to EBSCOhost and sign in to access more content about this topic.
Chimera (mythology)
In classic Greek mythology, the Chimera was a monstrous fire-breathing creature made up of parts of three different animals. Tradition held that the three-headed creature had the head and rich, thick mane of a male lion, the middle section of a farm goat (the name comes from the Greek for "she-goat"), and the head and thick, coiling tail of a serpent. From the middle, the creature also sprouted a fire-breathing goat’s head. Some renderings portrayed the Chimera with leathery wings.
The dramatic death of the Chimera, recounted by the epic poet Homer and later by Hesiod, marks one of the most famous scenes in Greek mythology; indeed, the terrifying creature had become such an element of the Greek culture the idea of the Chimera survived, stories passed from generation to generation of the existence of creatures made up of elements of multiple animals haunting the countryside. The Chimera remains a staple character type in video games, anime, graphic novels, and science-fiction stories and films. In addition, the essence of the Chimera myth has found expression in two unlikely fields: genetics and psychology.
Background
According to Hesiod, the Chimera was the offspring of Typhon, the menacing storm giant responsible for massive tempests that destroyed entire armadas, and Echidna, a cave-dwelling half woman, half dragon responsible for generating a variety of illnesses as well as the process of corruption and decay. Typhon was cursed by Zeus for challenging his supremacy, and his mating with Echidna produced only monsters who each became a curse on humanity. In addition to the Chimera, the two produced Cerebus, the three-headed hound who guarded the passage into Hades; Hydra, a giant water snake whose nine heads could regenerate; and Orthus, a two-headed dog with a serpent’s tail. She produced the Sphinx, the half lion, half human who cursed Thebes, and the Nemeian Lion, which was killed by the Greek hero Heracles.
The Chimera is known largely for the story of its dramatic death, first recounted in book 6 of Homer’s Iliad. The Chimera had long terrorized the countryside around the town of Lycia in Anatolia (present-day southwestern Turkey). Unprovoked, the fire-breathing creature raided farms, consumed animals as well as people, and laid waste to farmhouses. The countryside was strewn with heaps of burned rubble and charred bodies. The king, Iobates, realized the enormity of the threat but had no strategy for destroying the creature. None of his soldiers were willing to challenge the creature. Then, unexpectedly, a visitor to court arrived from Greece—the young, courageous Prince Bellerophon came bearing a letter of introduction from Iobates’s son-in-law.
However, the letter also told the king that Bellerophon had made inappropriate advances toward his daughter (who secretly harbored desires for the handsome prince and had tried to seduce him). The king was asked to arrange the death of Bellerophon. But the king feared repercussions from the gods should he kill Bellerophon. Uncertain how to proceed, the king opted to send Bellerophon to slay the Chimera, which he believed was a suicide mission. However, Bellerophon enjoyed the favor of the gods, most particularly Athena, the goddess of courage and wisdom, who sent him a gift of Pegasus, the magnificent winged horse, according to the poet Pindar. Astride Pegasus, Bellerophon sought and killed the Chimera. Later accounts tell how he cleverly attached a lump of lead to his lance and plunged it into the mouth of the monster’s middle head. The intense heat from her fiery breath melted the lead and, in turn, blocked the monster’s throat. The fire, having nowhere to go, quickly incinerated the entire insides of the Chimera.
Impact
It is no surprise that the Chimera, like so many other fantastic creatures from the myths and legends of antiquity, has become a template for ferocious monsters in speculative fiction and onscreen. Indeed, the figure of a menacing fire-breathing hybrid creature, often winged, has become a familiar figure in heroic adventure narratives, classic science-fiction television series, Japanese anime features, role-playing board games, comic books, and video games. The concept of creating an original creature from parts of other creatures is the heart of the Chimera figure. In fact, Disney Studio animators working on the storyboards for their hugely successful 1991 animated feature Beauty and the Beast drew their inspiration from the Chimera model, creating the Beast out of elements of the buffalo, the gorilla, the lion, the bear, the boar, and the wolf.
Apart from its influence in pop culture, however, the Chimera concept has lent itself to two critical areas not associated with mythic figures or heroic adventure: genetics and psychology. After the advent of DNA testing, researchers discovered that animals, including humans, can have the DNA of multiple organisms in the same body; that is, a single human can have another person’s DNA in some cells and his or her own in others. The phenomenon, known as "chimerism," has long been recognized in the case of fraternal twins. Similarly, blood transfusions, transplants, fetal development, and possibly even breastfeeding or sexual intercourse may permanently implant a small amount of one person’s cells into another’s organs or bloodstream, a situation termed "microchimerism." Researchers are just beginning to explore the frequency of chimerism and microchimerism, their potential effects on health, disease, and behavior, and the consequences for medicine, law, and psychology. The ethics of creating chimeras, including nonhuman-human, has also arisen around organ transplantation and genetic engineering research.
More commonly, the figure of the Chimera appears in pop psychology. Given the fantastic nature of the creature and the reality that no such creatures actually exists save in the imagination, the term "chimera" has been used to describe any delusional dream or wish that, given a person’s reality, is simply impossible to attain. These illusions sustain unrealistic hopes and can cause significant emotional turmoil at the point in which that person comes at last to understand the illusionary nature of the dream. The concept, therefore, brings together two elements of the mythic creature: that it is not real and that it can, nevertheless, be quite dangerous.
Bibliography
Evans, Hestia, comp. Mythology: The Gods, Heroes, and Monsters. Cambridge: Candlewick, 2007. Print.
Evslin, Bernard. Heroes, Gods, and Monsters of the Greek Myths. New York: Random House, 1966. Print.
Fitzgerald, Robert, trans. and ed. The Iliad of Homer. New York: Farrar, 2004. Print.
Kaplan, Matt. Science of the Magical: From the Holy Grail to Love Potions to Superpowers. New York: Simon, 2015. Print.
Kean, Sam. "The You in Me." Psychology Today. Sussex, 9 June 2016. Web. 13 July 2016.
"Khimaira." Theoi. Aaron J. Atsma, n.d. Web. 13 July 2016.
Kuře, Josef. "Etymological Background and Further Clarifying Remarks Concerning Chimeras and Hybrids." Chimeras and Hybrids in Comparative European and International Research: Scientific, Ethical, Philosophical and Legal Aspects. Ed. Jochen Taupitz and Marion Weschka. Berlin: Springer, 2009. 7–20. Print.
López-Ruiz, Carolina. Gods, Heroes, and Monsters: A Sourcebook of Greek, Roman, and Near Eastern Myths in Translation. New York: Oxford UP, 2013. Print.
Robson, David. "Is Another Human Living Inside You?" BBC Future. BBC, 18 Sept. 2015. Web. 13 July 2016.
Related Topics
Homer
Hesiod
Anime
Zeus (deity)
Sphinx (mythological creature)
Heracles’s Twelve Labors
Iliad by Homer
Bellerophon
Athena (deity)
Bellerophon and the Battle with Chimera
Beauty and the Beast (film)
Chimera (genetics)
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X X X X X X X X: Xi Worksheet Inequalities, Modulus &complex Numbers | PDF | Complex Number | Mathematical Objects
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X X X X X X X X: Xi Worksheet Inequalities, Modulus &complex Numbers
This document contains 30 problems involving solving inequalities, modulus operations, and complex numbers. The problems include solving inequalities for variables, finding square roots and …
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XI WORKSHEET INEQUALITIES, MODULUS &COMPLEX NUMBERS
1.
Solve for
x
:
2 4 1
x x
2.
Solve for
x
:
2
5 6 0
x x
3.
Solve for
x
:
2
3 1 7 0
x x
4.
Solve for
x
:
1 3 2 2
x x
5.
Solve for
x
:
3 1 2
x x x
6.
Solve for
x
:
2 4 4
x x
7.
Solve for
x
:
1 2
x x x
8.
Solve for
x
: (i)
3 2 0 1
x x
(ii)
1 5 2 3 2 3 7 3
x x
9.
Solve for
x
:
2
2
x x x
10.
Solve for
x
:
2
2 2
x x x x
11.
Solve for
x
:
3 2 2 2 2 7 1 5 5 5
x x x
. 12.
Solve for
x
:
3 5 7 9 1 5 7 5 7
x x x x
. 13.
Express the following in the form of
a bi
(i)
4 5 1 1 1 5
i i
(ii)
2 2 2 3 7
i i
14.
Find square roots of
7 7 3 6
i
. 15.
Find the conjugate of (i)
1 2 5 3 9
i i i
(ii)
1 1 1 3 1 2 6
i i z
16.
Find the modulus of (i)
2 2 3
i
(ii)
2
2 4 1
i i i
17.
If
4 3
a b i i
, then evaluate
2 2
a b
. 18.
If
1 2 1 3 1 4 1 5 ,
i i i i A B i
then evaluate
2 2
A B
. 19.
Find the real numbers
&
a b
such that
2 3
a b i i
is the conjugate of
1 0 1 1
i
. 20.
If
1,1
z i
find
z
. 21.
If
1 2
3 2 , 1
z i z i
then find Re
1 2
1
z z
and Im
1 2
z z
22.
Four times the conjugate of complex number is subtracted from the complex number itself to get
1 5 1 5
i
. Write the modulus of the complex number. 23.
A complex number
5 2
i
is first reflected in imaginary axis and then in real axis, find the complex number is final position. 24.
Find
2,2
z z if z
is purely imaginary. 25.
Find the conjugate of
1 0 1 1 0 1
1 1 1 1 1
i i i i
. 26.
If
1 2 2 3,3 3
i x i i y i i i i
then find
&,
x y
when
,
x y
. 27.
If
1 2 1 2
2 1 2
z z z z
and
2
1
z
then find
1
z
. 28.
If
1 2 1 2
3 1 1 3
z z z z
and
2
1,
z
then find
1
z
. 29.
If
z x yi
and
3
,
z p qi
then prove that
2 2
2
x y p q p q
. 30.
If
1 2
&
z z
are different complex numbers and
1
| | 1,
z
then find
1 2 1 2
1
z z z z
.
adDownload to read ad-free
ANSWERS 1.
Ans.
5 3 3
x or x
2.
Ans.
2 3 3 2
x or x
3.
Ans.
1 2
x or x
4.
Ans.
2 2 3
x or x
5.
Ans.
5, 3 3, 2 1,
6.
Ans.
0 8
x or x
7.
Ans.
1 3
x or x
8.
Ans. (i)
3 2
x
(ii)
9 15 4 4
x
9.
Ans.
1 3
x
10.
Ans.
[0, ) 2
11.
Ans.
5, 5,7
12.
Ans.
11 9
13.
Ans. (i)
0 0
i
(ii)
0 2
i
14.
Ans.
2 9
i
15.
Ans. (i)
5 3 2 6 41
i
(ii)
1 1 1 3 1 2 6
i z i z
16.
Ans. (i) 4 (ii)
5 6
17.
Ans. 25 18.
Ans. 22100 19.
Ans.
1, 4
a b
20.
Ans.
1 2
i
21.
Ans.
1 1,26 2
22.
Ans.
34
23.
Ans.
5 2
i
24.
Ans. 2 25.
Ans.
1 2
i
26.
Ans.
3, 1
x y
27.
Ans. 2 28.
Ans.
1 3
29.
Ans. Prove 30.
Ans. 1
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Mathematical Treasure: Geometry of Jakob Steiner | Mathematical Association of America
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Home » MAA Publications » Periodicals » Convergence » Mathematical Treasure: Geometry of Jakob Steiner
Mathematical Treasure: Geometry of Jakob Steiner
Author(s):
Frank J. Swetz (The Pennsylvania State University)
Jakob Steiner (1796–1863) was a Swiss mathematician who specialized in geometry. Considered one of the greatest pure geometers who ever lived, he introduced the concept of geometric forms, perfected the theory of duality in geometry, and contributed much to the development of modern projective geometry. Shown above is the title page of his Lectures, originally published in 1867 after his death by his former student Heinrich Schröter (or Schröder). The copy shown above is from the second edition of 1876, to which Schröder added his own second part, “The theory of conic sections based on projective properties,” further demonstrating Steiner’s theories.
On page 134, Schröder employed the concept of Steiner points to examine the relationship of the lines in Figure 30.
TheSpecial Collectionsstaff at the Linderman Library of Lehigh University in Bethlehem, Pennsylvania, is pleased to cooperate with the Mathematical Association of America to exhibit this and other items from the Library’s holdings in “Mathematical Treasures.” In particular, Convergence would like to thank Lois Fischer Black, Curator, Special Collections, and Ilhan Citak, Archives and Special Collections Librarian,for their kind assistance in helping to make this display possible.You may use these images in your classroom; all other uses require permission from the Special Collections staff, Linderman Library, Lehigh University.
Index of Mathematical Treasures
Frank J. Swetz (The Pennsylvania State University), "Mathematical Treasure: Geometry of Jakob Steiner," Convergence (October 2013)
Convergence
Tags:
Geometry
History of Mathematics
Projective Geometry
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https://surgeryreference.aofoundation.org/orthopedic-trauma/adult-trauma/foot-phalanges/proximal-hallux-diaphyseal-and-extraarticular-fractures/k-wire-fixation
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K-wire fixation for Diaphyseal and extraarticular end segment fractures of the proximal hallux
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K-wire fixation
Diaphyseal and extraarticular end segment fractures of the proximal hallux
Diaphyseal and extraarticular end segment fractures of the proximal hallux
Select a chapter
1/7 – Principles
2/7 – Possible K-wire configurations
3/7 – Patient preparation and approach
4/7 – Reduction and preliminary fixation
5/7 – K-wire insertion (non-axial)
6/7 – K-wire insertion (axial)
7/7 – Aftercare
1. Principles
Introduction
K-wires can be applied to many different fracture patterns. Considerations include fracture obliquity, comminution, and soft tissue status.
As K-wires cannot compress the fracture, the fracture needs to be reduced before K-wire insertion. This is particularly true for articular fractures.
The preferred K-wire configuration would be extra-articular to prevent damage to the articular surface. Axially inserted wires that break may be difficult to retrieve.
Threaded K-wires are generally not preferred as they are more challenging to extract and potentially maintain fracture gaps.
Reduction of the proximal phalanx should be anatomic when possible.
The goal of an intramedullary wire would be to secure the fracture site in a soft-tissue-friendly manner while maintaining alignment.
K-wires should be sized appropriately to the bone. Two wires should be used in the hallux. Typically wires thicker than 2.0 mm are not needed. The surgeon should be careful to cool the K-wire during insertion to reduce the risk of thermal necrosis.
Timing of surgery
The timing of surgery is influenced by the soft tissue injury and the patient's physiologic status.
Dislocation or injuries associated with the skin at risk requires immediate intervention regardless of the amount of soft tissue swelling.
If possible, swelling should be significantly decreased before surgery, which can take up to two weeks in some instances.
Open fractures should be promptly irrigated and debrided, and treated with antibiotics. Definitive fracture fixation may not be possible during this setting.
Forefoot fractures do not contribute to physiologic instability. If there is no soft tissue at risk, urgent intervention is not required.
Toe fractures versus long bone fractures
Toe fractures are different from long bone fractures.
The bones are very small
Fracture gaps are small
Fixation devices don't need to counter large forces
Thus, following the AO principles is less critical than for long bones, and often isolated screws or K-wires alone will work satisfactorily.
Restoration of length, rotation, and angulation are important for cosmesis.
The hallux is particularly critical due to its importance for walking.
2. Possible K-wire configurations
Simple oblique fracture
Diverging wires
Depending on the size of the fragment, more than one K-wire may be necessary to provide stability to the fracture.
Both K-wires can be inserted from the medial side of the bone in a diverging pattern.
Divergent K-wires prevent postoperative distraction of the fracture.
Crossing wires
K-wires can also be inserted in a crossed fashion. The two K-wires should not cross at the fracture site as this would be a less stable construct.
Axial wiring
Axially inserted K-wires may be used similarly to an intramedullary nail.
Typically, the insertion starts at the tip of the toe and continues across the fracture site. Depending on the fracture's location, the wire may just be inserted into the base of the proximal phalanx.
However, the K-wire may also be inserted to cross the MTP joint.
Simple transverse fracture
Axial wiring
In most cases, the surgeon will prefer axial wiring. However, in some cases (if the surgeon wishes to avoid crossing the IP joint), medial wiring may also be possible.
Medial wiring
Multiple configurations are possible. Ideally, one should avoid trans articular wires to prevent damage to the articular surface.
Both K-wires can be inserted from the medial side of the bone in a diverging pattern.
Divergent K-wires prevent postoperative distraction of the fracture.
Crossing wiring
K-wires can also be inserted in a crossed fashion. The two K-wires should not cross at the fracture site as this would be a less stable construct.
Multifragmentary fractures
An intramedullary K-wire can be used but will not control rotation or maintenance of length. However, it will maintain alignment and provide sufficient stability for fracture healing.
Non-parallel wires may have an improved capacity to maintain length and rotation. But significant comminution may prevent their placement.
3. Patient preparation and approach
Patient preparation
This procedure is typically performed with the patient placed supine and the knee flexed 90°.
Approach
An open approach may be required to reduce the fracture and remove soft-tissue interposition.
Typically the medial approach to the halluxis used.
Consider inserting the K-wires outside the skin incision to prevent contamination.
4. Reduction and preliminary fixation
Throughout this treatment option, illustrations of a generic fracture pattern are shown in four different ways:
Unreduced fracture
Reduced fracture
Fracture reduced and fixed provisionally
Fracture fixed definitively
Indirect reduction
Grasp the digit's terminus and use a combination of axial and angular traction to exaggerate the deformity.
Realign the toe and release the traction.
A sturdy cylindrical object at the fracture's apex can be used as a fulcrum to facilitate reduction.
If closed reduction is not successful, open reduction may be required.
Direct reduction
Debridement
If an anatomic reduction is desired, it is crucial to debride the hematoma and invaginated periosteum from the fracture site.
Reduction and preliminary fixation using a reduction clamp
Restore anatomical axial rotation, length, and angulation using one or more of the below techniques:
A small, pointed reduction forceps can be used for larger fragments. Be careful not to apply excessive force as this can lead to fragmentation. If possible, apply the reduction clamp so that the forces created by the clamp are at right angles to the fracture line. This clamp placement helps in reducing the fracture and in applying compression.
Use a periosteal elevator as a lever to reduce the fracture.
Insert provisional K-wires to maintain the reduction (if not already inserted using the K-wire technique).
Reduction and preliminary fixation using K-wire as a joystick
K-wires can be inserted into the fragment and used as a joystick. Once reduction is accomplished, the wire may be advanced to secure the reduction.
Quality control
Use image intensification to confirm the reduction.
5. K-wire insertion (non-axial)
K-wires are inserted percutaneously according to the preoperative plan.
Caution should be made to prevent thermal injury when inserting a K-wire in dense bone.
Intraoperative fluoroscopy is particularly useful to aid the correct position of the K-wire.
The stability of the reduction is assessed, and if needed, additional K-wires are inserted.
When satisfactory stability is achieved, cut the K-wires leaving adequate K-wire protruding through the skin to facilitate removal.
Assess the skin pin interface and release any excessive skin traction.
6. K-wire insertion (axial)
Marking K-wire track
To avoid unnecessary radiation from image intensification, mark the K-wire's planned track with a skin marker on the distal phalanx in both the AP and the lateral aspects.
K-wire insertion into the distal phalanx
A 16 gauge hypodermic needle or a drill guide can be used to prevent the K-wire from slipping during insertion. Either instrument facilitates K-wire insertion along the longitudinal axis of the phalanx.
If a K-wire is mistakenly inserted at an angle to the phalanx axis, it is recommended to leave it in until a second K-wire has been inserted in the correct orientation. Leaving the first K-wire in place prevents the second wire from going unintentionally along the wrong track.
Optional reduction and fixation technique: Use K-wire as a joystick
Outside-in technique
Advance the K-wire up to the fracture line but not across it. Use the K-wire as a joystick to reduce the fracture.
Once reduction is satisfactory, the wire can be advanced across the fracture plane and into the phalanx base.
Inside-out technique
In open fractures, another option is to introduce a double-ended K-wire in a retrograde fashion with the inside-out technique.
Flex the distal fragment to gain an optimal view of the fracture surface. Insert the K-wire through a drill guide and advance it along the medullary canal and through the distal tip of the tuft, piercing the soft tissues until it exits the skin.
Leaving the drill guide in place for soft-tissue protection, pull on the K-wire's distal end until it is flush with the fracture surface.
Use the K-wire as a joystick to reduce the fracture. Then advance it through the fracture up to the base of the fractured phalanx.
Proximal or comminuted fractures
In proximal or comminuted, advance the K-wire across the joint and into the phalanx base (or metatarsal) proximal to the injured phalanx.
Two K-wires need to be used in the hallux.
Cut the K-wire
K-wire protrusion through the skin
Cut the K-wire so that it protrudes through the skin, about 1 cm from the tip of the toe.
With the wire placed in a finished position, the portion of the wire nearest the skin is clamped with the Kocher.
Using a sucker tip that goes over the tip of the K-wire, the K-wire is bent 170°.
Leaving the K-wire to protrude through the skin in this way has the advantage of its being easy to remove. The disadvantages are patient discomfort and the risk of pin-track infection.
7. Aftercare
An appropriate well-padded dressing should be applied to protect the surgical incision.
The skin pin interface should be similarly well-padded but with dressings that can be readily removed to inspect for pin site infection.
Immediate postoperative treatment is rest, ice, and elevation.
The patient should be encouraged to begin early weight-bearing as permitted by the stability of the fracture. In general, patients can be weight-bearing as tolerated.
A stiff-soled shoe can be used to protect the surgical site.
Patients must exercise their ankle and subtalar joints to prevent stiffness (eg, by stretching their Achilles).
X-ray the toe at six weeks to confirm satisfactory union and remove K-wires if present. Once the fracture is united, the orthosis may be gradually discontinued.
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LECTURE 14 Second Order Linear Equations, General Theory 1. Standard Form A second order linear differential equation is a differential equation of the form A(x)y′′ + B(x)y′ + C(x)y = D(x) .
(14.1) (Here A, B, C and D are certain prescribed functions of x.) As in the case of first order linear equations, in any interval where A(x) ̸= 0, we can replace such an equation by an equivalent one in standard form: y′′ + p(x)y′ + q(x)y = g(x) (14.2) where p(x) = B(x) A(x) q(x) = C(x) A(x) g(x) = D(x) A(x) (14.3) 2. Homogeneous vs. Non-homogeneous Linear Differential Equations In the development that follows it will be important to distinguish between the case when the right hand side of y′′ + p(x)y′ + q(x)y = g(x) (14.4) is zero or non-zero. We shall say that a second order linear ODE is homogeneous if it can be written in the form y′′ + p(x)y′ + q(x)y = 0 (14.5) otherwise (if g(x) ̸= 0) we shall say that it is non-homogeneous. Note that this terminology is completely unrelated to homogeneous equations of degree zero (the topic of the preceding lecture).
3. Differential Operator Notation Consider the general second order linear differential equation φ′′ + p(x)φ′ + q(x)φ = g(x) .
(14.6) We shall often write differential equations like this as L[φ] = g(x) (14.7) where L is the linear differential operator L = d2 dx2 + p(x) d dx + q(x) .
(14.8) 1 4. GENERAL THEOREMS 2 That is to say, L is the operator that acts on a function φ by L[φ] = d2 dx2 + p(x) d dx + q(x) φ = d2φ dx2 + p(x) dφ dx + q(x)φ .
(14.9) 4. General Theorems The following theorem tells us the conditions for the existence and uniqueness of solutions of a second order linear differential equation.
Theorem 14.1. If the functions p, q and g are continuous on an open interval I ⊂R containing the point xo, then in some interval about xo there exists a unique solution y = φ(x) to the differential equation y′′ + p(x)y′ + q(x)y = g(x) (14.10) satisfying the prescribed initial conditions y(xo) = yo y′(xo) = y′ o .
(14.11) Note how this theorem is analogous to the corresponding theorem for first order linear ODE’s. Note also that the conditions for existence and uniqueness are fairly lax - all we require is the continuity of the functions p, q, and g around a given initial point. Finally, we note that the form of the initial conditions involves the specification of both y(x) and its derivative y′(x) at an initial point xo.
I should also point out that the preceding theorem does not address the issue of how to construct a solution of a second order linear ODE. Indeed, the actual construction of solutions to second order linear ODE is sufficiently complicated to that we shall spend 90% of the remaining lectures on techniques of solution. The next two theorems at least tell us the basic ingredients for a general solution of a second order linear ODE.
Theorem 14.2. (The Superposition Principle) If y = y1(x) and y = y2(x) are two solutions of the differ-ential equation L[y] = d2y dx2 + p(x)dy dx + q(x)y = 0 (14.12) then any linear combination y = c1y1(x) + c2y2(x) (14.13) of y1(x) and y2(x), where c1 and c2 are constants, is also a solution of (14.12).
Proof.
L[c1y1 + c2y2] = d2 dx2 (c1y1 + c2y2) + p(x) d dx (c1y1 + c2y2) + q(x) (c1y1 + c2y2) = c1 d2y1 dx2 + p(x) dy1 dx + q(x)y1 + c2 d2y3 dx2 + p(x) dy2 dx + q(x)y2 = c1 · 0 + c2 · 0 = 0 (14.14) The fact that a linear combination of solutions of a linear, homogeneous differential equation is also a solution is extremely important. The theory of linear homogeneous equations, including differential equations involving higher derivatives depends strongly on the superposition principle.
4. GENERAL THEOREMS 3 Example 14.3.
y1(x) = cos(x) (14.15) and y2(x) = sin(x) (14.16) are both solutions of y′′ + y = 0.
(14.17) It is easy to check that any linear combination of y1 and y2 is also a solution.
Example 14.4.
y1(x) = 1 (14.18) and y2(x) = x1/2 (14.19) are both solutions of yy′′ + (y′)2 = 0.
(14.20) However, it is easy to check that y1 + y2 = 1 + √x is not a solution of (14.20). The reason for this lies in the fact that (14.20) is not linear.
Given two solutions y1 and y2 of a second order linear homogeneous differential equation L[y] = 0 , (14.21) we can construct an infinite number of other solutions y(x) = c1y1(x) + c2y2(x) (14.22) by letting c1 and c2 run through R. The following question then arises: are all the solutions of (14.21) capable of being expressed in form (14.22) for some choice of c1 and c2?
This will not always be the case; and so we shall say that two solutions y1 and y2 form a fundamental set of solutions to (14.21) if every solution of (14.21) can be expressed as a linear combination of y1 and y2.
Theorem 14.5. If p and q are continuous on an open interval I = (α,β) and if y1 and y2 are solutions of the differential equation L[y] = y′′ + p(x)y′ + q(x)y = 0 (14.23) satisfying W(y1,y2) = y1(x)y′ 2(x) −y′ 1(x)y2(x) ̸= 0 (14.24) at every point x ∈I, then any other solution of (14.23) on the interval I can be expressed uniquely as a linear combination of y1 and y2.
Proof.
Let y1 and y2 be two given solutions on an interval I and let Y be an any other solution on I. Choose a point xo ∈I. From our basic uniqueness and existence theorem (Theorem 3.2), we know that there is only solution y(x) of (14.23) such that y(xo) = Y (xo) y′(xo) = Y ′(xo) .
(14.25) namely, Y (x). Therefore if we can show that a solution of the form y(x) = c1y1(x) + c2y2(x) satisfies the initial conditions (14.25), then we must have Y (x) = c1y1(x) + c2y2(x) and so Y (x) is a linear combination of y1(x) and y2(x).
4. GENERAL THEOREMS 4 Thus, we now seek to define constants c1 and c2 so that these initial conditions can be matched. We thus set c1y1(xo) + c2y2(xo) = yo c1y′ 1(xo) + c2y′ 2(xo) = y′ o .
(14.26) This is just a series of two equations with two unknowns. Solving the first equation for c1 yields c1 = yo −c2y2(xo) y1(xo) .
(14.27) Plugging this into the second equation yields yo −c2y2(xo) y1(xo) y′ 1(xo) + c2y′ 2(xo) = y′ o (14.28) or yoy′ 1(xo) −c2y2(xo)y′ 1(xo) + c2y1(xo)y′ 2(xo) = y1(xo)y′ o (14.29) or c2 = y1(xo)y′ o −y′ 1(xo)yo y1(xo)y′ 2(xo) −y′ 1(xo)y2(xo) .
(14.30) Plugging this expression for c2 into (14.27) yields c1 = yoy′ 2(xo) −y2(xo)y′ o y1(xo)y′ 2(xo) −y′ 1(xo)y2(xo) .
(14.31) Thus, we can solve for c1 and c2 whenever the denominator W(y1,y2) = y1(xo)y′ 2(xo) −y′ 1(xo)y2(xo) (14.32) does not vanish. Thus, so long as y1 and y2 satisfy (14.23) we can always express any solution as a linear combination of y1 and y2.
Remark: The quantity W(y1,y2) = y1(xo)y′ 2(xo) −y′ 1(xo)y2(xo) (14.33) is called the Wronskian of y1 and y2.
Example 14.6. Show that y1(x) = cos(x) (14.34) and y2(x) = sin(x) (14.35) are form a set of fundamental solutions to the differential equation y′′ + y = 0 .
(14.36) We simply have to check that the Wronskian does not vanish: W(y1, y2) = y1(xo)y′ 2(xo) −y′ 1(xo)y2(xo) = cos(x) (cos(x)) −(−sin(x)) sin(x) = 1 ̸= 0 .
(14.37) Since the Wronskian does not vanish, y1 and y2 must be linearly independent.
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Precalculus 11-04 Lines and Planes in Space
Mr. Wright's Math Extravaganza
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Description
125 views
Posted: 14 Apr 2023
Mr. Wright's Precalculus
The corresponding textbook can be purchased at or
Write an equation for a line in three dimensions.
Write an equation for a plane.
Find the angle between two planes.
Graph a plane.
Transcript:
lesson 11 4 lines and planes in space [Music] let's talk about three-dimensional lines not as simple as just y equals MX plus b if you expand that with a z you have like ax was b y Plus c z equals something that is actually a plane not a line so here's imagine this Lionel goes through points p and Q so you can see in the picture Q is X Y Z p is X1 y1 Z1 V is a direction Vector for l so it's going the same direction as l so we can call that ABC if we start at p and move any distance in the direction V to get some point Q so we're starting at p and we're going some Direction and the direction as V it's they're parallel to each other we end up at Q therefore Vector p q is the same as T times vector v because they are parallels they're a multiple of each other so to find Vector PQ it's just Q x y z minus the point p x 1 y 1 Z1 and then the TV we just distribute the T through the a b and c this is known as the general form we can take that general form and change it into What's called the parametric equations of a line just take each component the X component the Y component and the Z component and then solve them for X Y and Z and you end up with that this is what we used when we did 3D systems of equations that got many solutions remember we're going to say oh let Z equal a and in this case we'll let Z equal t and what works out or we can solve them all for T and get the symmetric equations of a line with all equal to each other because they're all equal to T the most common one used is the parametric equations of a line so let's find a set of parametric equations a line that pass through those two points so first we need a Direction vector so that would be our vector v and just subtract the two points so we'll have four minus one zero minus three 1 minus negative two so V is 3 negative 3 3. that is our a our B and RC now um let's we have enough stuff plug in to the formula so as x equals a t plus X1 y equals a b t plus y1 and Z equals c t plus Z1 so a is 3 T and we just need to pick one of the points let's pick the first one as x one y one Z One it doesn't matter if you use the second Point you'll also get the equation of line it'll just look different if you pick different values of T then you get the other one B is negative 3 and y1 is 3. C is three and Z1 is negative two so that is our answer if you were to plug in different numbers of t such as 0 you get 1 3 negative 2. plugged in a one you'd get four zero one oh that's our other point so it seems to work if we had used our other point you could plug in values for T and get the first point so what this is this just gives you all the points on the line planes planes are like three-dimensional lines they're infinitely long in two Dimensions but infinitely thin and the third dimension if we look at our picture here says PQ Dot N equals zero because they are perpendicular well where's p and Q all right well let's take a point here and call it P and a point over here and we'll call it Q so we have a vector PQ Point p can be X1 y1 Z1 Point Q can be x y z all right they're perpendicular n is called the normal Vector to the plane because it's perpendicular in math the word normal means perpendicular and so if we dot product those we end up with standard form where it's the x times the X remember p q is going to be the difference in the X's difference in the Y's difference in the Z's so if I dot product it's multiply the X's plus multiply the Y's plus multiply the Z's and it equals zero if we knew the numbers for X1 y1 Z1 we could multiply it out and get the general form which is that let's find the general form of the equation of a plane passing through those three points first of all we need to find the normal vector n all right how we're going to do this is start by find two vectors in the plane and so what I'm going to do is I'm going to find Vector a b which would be 1 minus 3 5 minus 2 0 minus 2. or negative 2 3 negative 2. and I'm also going to find Vector b c which would be 1 minus one negative three minus five and one minus zero or zero negative eight one then we cross product because the cross product result is perpendicular to the two vectors so our n is going to be equal to a b cross product with b c so we have the i j k negative 2 3 negative two zero negative eight one copy the first two columns looks like I have three I zero J and 16k minus 0 K minus 16 I minus a negative 2 J so it looks like we have negative 13 eyes two J's and 16 K's or negative 13 2 16. that is our a our B and our c all right so now we fill in the standard form which is a x minus X1 plus b times y minus y1 plus c times Z minus Z1 equals zero if I need to pick a point any of the points on the plane will work so I'm just going to pick point B one five zero so a so I'm over here negative 13 x minus X1 which is 1 from our point B little B is 2 minus five X1 y1 Z1 and C is 16. z minus zero all right now we just have to simplify to our general form so we'd have just distribute so it'll be negative 13x plus 13. 2 y minus 10 16z equals 0. so it looks like I have negative 13x plus 2i Plus 16z plus 3 equals zero we can find the angle between two planes because using geometry we can show that the normal vectors the angle between the normal vectors is the same as the angle between the planes if you remember to find the angle between two vectors it's the dot product equals the magnitude times the magnitude times cosine of the angle we can also find the distance between a point and the plane if we have Point p which is what we're looking for for the distance from and point Q which is on our plane we just need to find the projection of that onto the normal vector and the magnitude of that will be the distance so if you remember the projection is simply or the magnitude of the projection will simply be the dot product of p q with n divided by the magnitude of n now let's go over graphing planes in space first thing to do is to find the intercepts then you plot the intercepts and then you draw a triangle to represent the plane so what you're going to do is you're actually going to draw the plane that's in one octane and it just remember it extends in all directions so let's sketch that all right so first of all we need to find the x-intercepts or all the intercepts find the x-intercept you let the other variables be zero and we'll get X is eight and I can put that on my graph and find the y-intercept the Y is y everything else is zero I think you get Y is 6. put that on the graph on the axis and then the Z intercept the X and Y are zeros and the Z is not and looks like you get Z is four so put that point there then we're going to connect these with lines and so that is the plane that goes to the first Octane and often it's kind of shaded I remember that just represents the slanted plane uh it it really truly extends in all directions
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Identifying perfect square form (video) | Khan Academy
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Skip to lesson content
Algebra 1
Course: Algebra 1>Unit 13
Lesson 8: Factoring quadratics with perfect squares
Perfect square factorization intro
Factoring quadratics: Perfect squares
Perfect squares intro
Factoring perfect squares
Identifying perfect square form
Factoring perfect squares: negative common factor
Factoring perfect squares: missing values
Factoring perfect squares: shared factors
Perfect squares
Math>
Algebra 1>
Quadratics: Multiplying & factoring>
Factoring quadratics with perfect squares
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Identifying perfect square form
VA.Math: A.EO.2.c, A2.EO.3.bVA.Math.2023: A.EO.2.c, A2.EO.3.b
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Sal shows how we can identify that a trinomial has the "perfect square" form.
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Vera Chu 8 years ago Posted 8 years ago. Direct link to Vera Chu's post “Can the answer also be ne...” more Can the answer also be negative?(-5x-2)?
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joshua a year ago Posted a year ago. Direct link to joshua's post “(Ax)² = (Ax) (Ax) Ax² =...” more (Ax)² = (Ax) (Ax)
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Fernando 8 years ago Posted 8 years ago. Direct link to Fernando's post “Is possible to factor thi...” more Is possible to factor this quadratic:
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Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “No it wouldn't work. Ye...” more No it wouldn't work.
Yes... your quadratic has perfect squares at both ends, but the middle term is incorrect.
The terms on the ends of your factors would need to be: (5x + 3)^2
If you multiply this out, you get: 25x^2 + 30x + 9
Hope this helps.
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Kenzzzzzzzzzzie 4 years ago Posted 4 years ago. Direct link to Kenzzzzzzzzzzie's post “This seems sort of compli...” more This seems sort of complicated... Why don't we just factor the binomial like we usually would? Like for example, when Sal had 25x^2 + 20x + 4, I did it normally and got the same answer that he did, (5x+2)^2.
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winter a year ago Posted a year ago. Direct link to winter's post “Hello, I'm studying for a...” more Hello, I'm studying for an exam and our workbook has this question:
The product of two numbers is 120, and the sum of their squares is 289. The sum of the number is __."
Does anyone know a course/video that talks about these types of problems? Or does anyone know how to solve this? I know what the answer is but I don't really understand how this works.
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Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “Use X = one number and Y ...” more Use X = one number and Y = 2nd number
Translate the first part of the 1st sentence and you get:
xy=120
Translate the second part of the 1st sentence and you get:
x^2+y^2=289
You now have a system of equations.
-- Solve the 1st equation for one of the 2 variables by dividing by sides: y=120/x
-- Substitute this into the 2nd equation in place of Y to get: x^2+(120/x)^2 = 289
-- Do the exponent: x^2+14400/x^2 = 289
-- Multiply both sides by x^2: x^4 + 14400 = 289x^2
-- Subtract 289x^2 from both sides: x^4-289x^2+14400 = 0
-- Factor: (x^2-225)(x^2-64) = 0
-- Factor more: (x-15)(x+15)(x-8)(x+8)=0
-- Split the factors apart and solve each.
x-15=0 creates x=15
x+15=0 creates x=-15
x-8=0 creates x=8
x+8=0 creates x=-8
To find y: use y=120/x
x=15 creates y=120/15 = 8
x=-15 creates y=120/(-15) = -8
x=8 creates y=120/8 = 15
x=-8 creates y=120/(-8) = -15
Basically, the two numbers are either 8 and 15, or -8 and -15.
Hope this helps.
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Le_Go-J 5 years ago Posted 5 years ago. Direct link to Le_Go-J's post “Why is it 2ABx and not AB...” more Why is it 2ABx and not ABx^2 @ 1:38 ?
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Bradley Reynolds 5 years ago Posted 5 years ago. Direct link to Bradley Reynolds's post “It is 2ABx because he is ...” more It is 2ABx because he is adding not multiplying. If Sal were to multiply ABx by ABx the result would be (ABx)^2.
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Milkdromeda2b a month ago Posted a month ago. Direct link to Milkdromeda2b's post “Me thinking in my head: "...” more Me thinking in my head: " What if you do it the other way round?"
Next second: He does it the other way round!
Also, is this the same as completing the square that I am always hearing about?
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Kim Seidel a month ago Posted a month ago. Direct link to Kim Seidel's post “This is a concept you nee...” more This is a concept you need to know to use the complete the square method of solving quadratic equations. Completing the square means we can take any quadratic polynomial and force it to into the form of a perfect square trinomial. This video teaches you how to recognize what is a perfect square trinomial and how to factor it quickly.
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Kenzzzzzzzzzzie 4 years ago Posted 4 years ago. Direct link to Kenzzzzzzzzzzie's post “I'm a bit confused about ...” more I'm a bit confused about the second term in the perfect square form.
In this video, Sal said that 20x (the center term of 25x^2 + 20x + 4) is 2.A.B.x, implying that A is the coefficient of 25^2 and x was just added on because that's how the formula works.
However, in the "Practice: Perfect Squares" exercise, a solution says that 14x (the center term of x^2 + 14x + 49) is twice the root of 1x^2 and 49, implying that A is the coefficiant AND the variable of 1x^2.
Can someone please explain which one is correct, if both are, or if I messed something up? I'm not sure if this matters much, but it confused me when I got to Completing the Square in the next unit.
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Kim Seidel 4 years ago Posted 4 years ago. Direct link to Kim Seidel's post “Consider what happens whe...” more Consider what happens when you multiply: (ax+b)(ax+b)
You get: (ax)^2+abx+abx+b^2 which simplifies to a^2x^2+2ab+b^2
The "a" refers to the square root of the coefficient of the x^2 term. The "b" refers to the square root of the constant term.
For 25x^2 + 20x + 4: a=5 and b=2. 2ab = 2(5)(2) = 20
For x^2 + 14x + 49: a=1 and b=7. 2ab = 2(1)(7) = 7
Hope this helps.
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April Turgutalp 2 years ago Posted 2 years ago. Direct link to April Turgutalp's post “If Ax Ax is Ax squared,...” more If Ax Ax is Ax squared, then why isn't ABx ABx not ABx squared? Why is it 2ABx?
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Hank 2 years ago Posted 2 years ago. Direct link to Hank's post “We're adding ABx and ABx ...” more We're adding ABx and ABx not multiplying
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tiffany.quan08 2 years ago Posted 2 years ago. Direct link to tiffany.quan08's post “why do you use (ax+b)^2 a...” more why do you use (ax+b)^2 and not use a^2+2ab+b^2
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David Severin 2 years ago Posted 2 years ago. Direct link to David Severin's post “because a could be a coef...” more because a could be a coefficient with the variable x.
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Video transcript
[Voiceover] We wanna figure out what AX plus B squared is, and I encourage you to pause the video and figure out what that is in terms of capital-A and capital-B. So let's work through it. This is the same thing as multiplying AX plus B times AX plus B. So let me fill that in. This is AX there, another AX there. I just wrote it in that order to make the color switching a little bit easier. So AX plus B times AX plus B. Well, what's that going to be equal to? Well, if you take this AX and you multiply it times that AX, you're going to get AX squared. AX, the entire thing squared. And then if you take, if you take this AX and then multiply it times this B, you're going to get ABX. Then if you take this B and you multiply it times this AX, you're going to get another ABX. ABX. And then last but not least, if you take this B and multiply it times the other B, it's going to be plus B squared. And so what are you left with? Well, you're going to be left with A, I'll write it like this, AX squared, we actually if we want, well, I'll write it in a different way in a second, and then you have plus, plus two, that's a slightly different color. I'm gonna do that other color. Plus two ABX, and then finally plus B squared. Plus B squared. Now, I said I could write it in a slightly different way, what I could do is just rewrite-out AX squared as being the same thing. This is the same thing as A-squared X-squared, and then I can write out everything else the same way. Plus two ABX, and then plus B squared. Now, why did I, what's interesting about doing this? Well, now we can see the pattern for the square of any binomial or binomial like this, so for example, if someone were to walk up to you and say "alright, I have a trinomial of the form," let's say they have a trinomial of the form 25X squared plus 20X plus 4, and they were to tell you to factor this, well, actually, let's just do that. Why don't you pause the video and see if you could factor this as the product of two binomials. Well, when you look at this, you'd say "well look, there's 25X squared, "that looks like a perfect square. "25X squared, that's the same thing as five-squared "x-squared," or you could write it as five X squared. This four here, that's a perfect square. That's the same thing as two squared. And let's see, 20, right over here, if we want it to fit this pattern, we would see that A is five and B is two, and so let's see, what would be two times AB? Well, five times two AB would be 10, and then two times that would be 20. So this right over here is, that is plus two times five. Two times five times two times two X. Times two X, I'll do it in this color. Times two X. So you see that this completely matches this pattern here where A is equal to five and B is equal to two. Once again, this is AX, the whole thing squared, then you have two times A times BX, you see that there, and then finally you have the B squared. So if you wanted to factor this, you could say "well, this is just going to the same thing "since we know what A and B are, "this is going to be five X plus two." Five X plus two. Five X plus two, whole thing squared. So the whole point of doing this is to start recognizing when we actually have perfect squares, especially perfect squares where the leading coefficient isn't one.
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Matchsticks Builds Hexagon
The document investigates building hexagons using matchsticks and finding patterns in the number of triangles and matchsticks. It shows that adding another hexagon increases the number of tr…
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MATCHSTICKS BUILD HEXAGONS
INTRODUCTION
Mathe matic s is a metho dical applic ation of matter. It makes our life orderly and preve nts cha os. It is imp orta nt to dev elo p pro ble m sol vin g act ivi tie s to bui ld new mat hem ati cal knowledge, applying knowledge, adapting a variety of appropriate problem solving strategies to daily life.This investigation aims to determine the number of triangles and matchsticks in a hexagon when another hexagon is added. It will also be interesting to discover formulas for calculating the number of triangles inside the hexagon and the number of matchsticks used to form it.
Situation
Suppose we have matchsticks and form it into a hexagon. Then connect each opposite vertices with a matchstick to form six triangles inside of it.Figure 1 Figure 2
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Figure 3 Figure 4
Investigate
The hexagon is defined as a polygon with six sides. In this investigation, the sides of the polygon and the triangles inside of it are formed by the matchsticks.As seen in the illustrations, every term, the number of triangles and matchsticks used in each illustration increases. In figure one, in a one hexagon, there were six triangles inside of it and twelve matchsticks were used to complete the situation. On the second figure, in two hexagons, there were 12 triangles and we added another 11 matchsticks to form two connected hexagons. The same procedure will follow on the third and so on.It can be seen that the number of triangles inside the hexagon is added by six as we put another hexagon. On the other hand, 12 matchsticks were used to form a hexagon and connect its opposite verti ces to form six triangles inside of it. Then, 11 matchsticks were added as we put another hexagon and connect its opposite vertices to form six triangles inside of it.The number of triangles inside the hexagon and the number of matchsticks used is shown in the table below.Number of Hexagons Process Number of triangles Number of matchsticks 1 6 1 2 2 1 2 2 3 3 1 8 3 4
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4 2 4 4 5
⋮ ⋮ ⋮ ⋮
The sequences for this situation w ere as follows:A. Triangles – 6, 12, 18, 24, …B. Matchsticks – 12, 23, 34, 45, …Also, the sums of these terms are as follows:A. Sum of the terms of the triangles:
S
1
=
6
S
2
=
6
+
12
=
18
S
3
=
6
+
12
+
18
=
36
S
4
=
6
+
12
+
18
+
24
=
60
B. Sum of the terms of the matchsticks:
S
1
=
12
S
2
=
12
+
23
=
35
S
3
=
12
+
23
+
34
=
69
S
4
=
12
+
23
+
34
+
45
=
114
STATEMENT OF THE PROBLEM
This mathematical investigation aims to determine the rule in finding the terms and the sum of the terms of the triangles inside the hexagon and the matchsticks used to form a hexagon with six triangles inside of it.Specifically, it sought to answer the following questions:1.Wha t is the rul e in fin din g th e
nth
term of the triangles inside a hexagon and the matchsticks used to form it?2.Wha t is t he ru le i n fi ndi ng th e sum of th e
nth
term of the triangles inside a hexagon and the matchsticks used to form it?
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DATA GATHERING AND CONJECTURES
Conjecture 1 Looking for patterns on the
nth
term in the table below,Number of Hexagons Process
nth
term(number of triangles)
nth
term(number of matchsticks)1 6 6(1)1 2 1 1(1)+1 2 1 2 6(2)2 3 1 1(2)+1 3 1 8 6(3)3 4 1 1(3)+1 4 2 4 6(4)4 5 1 1(4)+1
⋮ ⋮ ⋮ ⋮
n n
6
n n
11
n
+
1
In each term, the six triangles inside the hexagon is multiplied by the number of hexagons being added and 11 matchsticks were added to form another hexagon with six triangles inside of it. Based on the pattern, the
nth
term of the triangles inside the hexagon can be obtained using
T
n
=
6
n
, where n is a natural number and the
nth
term of the matchsticks can be obtained using
M
n
=
11
n
+
1
, where n is a natural number.
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圆内的均匀随机点
最新推荐文章于 2022-08-15 22:57:11 发布
thulium_于 2019-10-17 15:39:22 发布
阅读量2.7k收藏 13
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CC 4.0 BY-SA版权
分类专栏：算法
原文链接：
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圆内的均匀随机点
前言
最近遇到一个问题，需要在以一个坐标为中心的区域内生成一组均匀分布的随机点，首先想到的就是以圆作为区域。
圆内随机点
方法1:
根据x2+y2=R2x2+y2=R2,那么自让想到可以先随机生成[-R,R]间的横坐标x，然后生成[−R2−X2−−−−−−−√,R2−X2−−−−−−−√−R2−X2,R2−X2]范围内的随机数y，那么(x,y)自然也就是在圆内的随机点了。
写一段代码看一看:
cobol
def random_point_in_circle(point_num, radius): for i in range(2,point_num+1): x=random.uniform(-radius,radius) y_max=math.sqrt(radiusradius-xx) y=random.uniform(-y_max,y_max) plt.plot(x,y,'',color="blue") def main(): pi = np.pi theta = np.linspace(0, pi 2, 1000) R = 1 x = np.sin(theta) R y = np.cos(theta) R plt.figure(figsize=(6, 6)) plt.plot(x, y, label="cycle", color="green", linewidth=2) plt.title("random_points_in_circle") random_point_in_circle(4000, R) plt.legend() plt.show() if __name__=="__main__": main()
AI运行代码
看到这个图应该立刻就知道哪里出错了，当x越靠近圆的边缘的话，y的范围就会越小，所以两边边缘的点会非常密集，不能算"均匀分布"。
方法2:
然后就会想到能否利用面积这个概念呢？因为上一个方法出错在边缘处，即y的范围会随着x的范围的变化而发生变化，所以如果在一个矩形区域内生成随机点，就会是均匀分布的；然后如果在圆内就保留下来这个点:
cobol
def random_point_in_circle(point_num, radius): for i in range(2,point_num+1): while True: x=random.uniform(-radius,radius) y=random.uniform(-radius,radius) if(x2)+(y2)<(radius2): break plt.plot(x,y,'',color="blue") def main(): pi = np.pi theta = np.linspace(0, pi 2, 1000) R = 1 x = np.sin(theta) R y = np.cos(theta) R plt.figure(figsize=(6, 6)) plt.plot(x, y, label="cycle", color="green", linewidth=2) plt.title("random_points_in_circle") random_point_in_circle(4000, R) plt.legend() plt.show() if __name__=="__main__": main()
AI运行代码
效果很OK:
但是这种方法的缺点就是会有较大的开销，想想看我们是按矩形范围内产生的点，最后会在圆内的点的概率只有πR2(2R)2=π4πR2(2R)2=π4
方法3:
那么我们能否考虑用极坐标呢，可以消除y的范围对x的范围敏感的问题。利用x=R∗cos(θ)x=R∗cos(θ)与y=R∗sin(θ)y=R∗sin(θ)，先随机生成[0,2π0,2π]内的θθ，然后随机生成[0,R]内的r:
cobol
def random_point_in_circle(point_num, radius): for i in range(2,point_num+1): theta=random.random()2np.pi r=random.uniform(0,radius) x=rmath.cos(theta) y=rmath.sin(theta) plt.plot(x,y,'',color="blue") def main(): pi = np.pi theta = np.linspace(0, pi 2, 1000) R = 1 x = np.sin(theta) R y = np.cos(theta) R plt.figure(figsize=(6, 6)) plt.plot(x, y, label="cycle", color="green", linewidth=2) plt.title("random_points_in_circle") random_point_in_circle(4000, R) # 修改此处来显示不同算法的效果 plt.legend() plt.show() if __name__=="__main__": main()
AI运行代码
边缘的点会比较稀疏的原因是这样的，由于r是在[0,R]之间等概率产生的，所以可以认为同一个r的生成的随机点是相同的，但是圆的半径会变大，同样数量的点就会显得稀疏了。
方法4:
在这里我们先引入一条定理:令R=r2R=r2,R在[0,1]上是均匀分布，θθ在[0,2π0,2π]上是均匀分布，且R与θθ相互独立，令
x=r∗cos(θ)=R−−√∗cos(θ)x=r∗cos(θ)=R∗cos(θ)
y=r∗sin(θ)=R−−√∗sin(θ)y=r∗sin(θ)=R∗sin(θ)
那么我们有(x,y)是均匀分布。
如果要证明(x,y)是均匀分布，由对称性我们只需要证明(x,y)在第一象限为均匀分布即可，即需要证明(x,y)的联合概率密度f(x,y)=1S=1π/4f(x,y)=1S=1π/4
首先我们知道连续性随机向量变换的联合分布的一个定理:
设(X,Y)是联合概率密度为f(x,y)f(x,y)的连续性随机向量，g1(x,y),g2(x,y)g1(x,y),g2(x,y)ξ=g1(X,Y),η=g2(X,Y)ξ=g1(X,Y),η=g2(X,Y)。如果对任何非负连续的二元函数h(μ,υ)h(μ,υ)成立，则有:
∬h[g1(x,y),g2(x,y)]f(x,y)dxdy=∬h(μ,υ)p(μ,υ)dμdυ∬h[g1(x,y),g2(x,y)]f(x,y)dxdy=∬h(μ,υ)p(μ,υ)dμdυ
放上代码:
cobol
def random_point(car_num,radius): for i in range(1, car_num + 1): theta = random.random() 2 np.pi r = random.uniform(0, radius) x = math.cos(theta) (r 0.5) y = math.sin(theta) (r 0.5) plt.plot(x, y, '', color="blue") def main(): pi = np.pi theta = np.linspace(0, pi 2, 1000) R = 1 x = np.sin(theta) R y = np.cos(theta) R plt.figure(figsize=(6, 6)) plt.plot(x, y, label="cycle", color="green", linewidth=2) plt.title("random_points_in_circle") random_point(4000, R) # 修改此处来显示不同算法的效果 plt.legend() plt.show() if __name__=="__main__": main()
AI运行代码
结果如下:
应该是比较满意的了。
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【算法】均匀 的生成 圆 内的 随机 点
weixin_30702413的博客
04-15 607
算法 1 设半径为$R$。 $x = r\ast cos(\theta)$ $y = r \ast sin(\theta)$ 其中 $0\leqslant r \leqslant R$，$t$为0-1 均匀分布 产生的 随机 数，$r = sqrt(t) \ast R$，$\theta = 2\pi \ast t, t \sim U(0, 1)$ 证明：url 1 imp...
三角形或 圆 内 均匀分布 随机 点 的计算
纯，属虚构
04-15 6286
圆： 均匀 产生角度，均匀 产生半径。半径的概率应与其长度一致。 clear for i=1:1000; theta=rand(); x=rand(); r=sqrt(x); % 均匀 的半径 x1(i)=rcos(2pitheta); x2(i)=rsin(2pitheta); end
matlab：圆 内 均匀 随机 取 点
qq_32515081的博客
04-29 7975
文章目录1.舍选法2.反函数法3.代码 这里介绍两种 圆 内 均匀 取 点 的方法：舍选法和反函数法。 1.舍选法 顾名思义，舍选法是指在正方形（边长等于 圆 直径）的上面 均匀 撒 点，然后在正方形上画 圆，超出 圆 形的舍弃掉，只保留 圆 内的 点： 效果如下 2.反函数法 一般取某个概率密度函数下的 随机 数，会用到反函数法，这里的反函数指的是累积分布函数的反函数。如图所示，纵轴为概率，横轴为 随机 数的取值，可以看到在纵轴区间相等的间隔内，横轴不同的 随机 数被取到的概率是不同的，但是相同间隔内的概率是相等的，也
478. 在 圆 内 随机 生成 点
宫水三叶的刷题日记
06-18 415
题目描述 这是 LeetCode 上的 「478. 在 圆 内 随机 生成 点」 ，难度为 「中等」。 Tag : 「数学」、「随机 化」 给定 圆 的半径和 圆 心的位置，实现函数 randPoint，在 圆 中产生 均匀 随机 点。 实现 Solution 类: Solution(double radius, double x_center, double y_center)用 圆 的半径 radius 和 圆 心的位置 初始化对象randPoint()返回 圆 内的 一个 随机 点
在 圆 内 随机 生成 点
github_36510643的博客
08-15 1238
Solution(double radius, double x_center, double y_center) 用 圆 的半径 radius 和 圆 心的位置 (x_center, y_center) 初始化对象。输出: [null, [-0.02493, -0.38077], [0.82314, 0.38945], [0.36572, 0.17248]]randPoint() 返回 圆 内的 一个 随机 点。圆 周上的一 点 被认为在 圆 内。给定 圆 的半径和 圆 心的位置，实现函数 randPoint ，在 圆 中产生 均匀 随机 点。...
在 圆 形里 均匀 随机 一个 点
05-30
要在 圆 形中 均匀 随机 选择 一个 点，可以使用极坐标。具体步骤如下： 1. 生成 一个 随机 数r，范围在[0,1]之间，表示 点 到 圆 心的距离。 2. 生成 一个 随机 数theta，范围在[0,2π]之间，表示 点 与 圆 心的连线与x轴正方向的夹角。 3. 将极坐标转换为直角坐标，公式为x = r cos(theta)，y = r sin(theta)。 4. 将x和y乘以 圆 的半径r，得到 点 在 圆 上的坐标。 这样就可以在 圆 形中 均匀 随机 选择 一个 点 了。
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[FREE] Find the volume of the indicated region: the tetrahedron cut off from the first octant by the plane - brainly.com
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Find the volume of the indicated region: the tetrahedron cut off from the first octant by the plane 6 x+2 y+5 z=1.
A. 15
B. 30
C. 20
D. 10
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The volume is 10 cubic units. The volume of the tetrahedron cut off from the first octant by the plane x/6+y/2+z/5=1 is calculated using triple integrals in calculus. In this case, the x, y, and z intercepts are used to calculate the base area and height of the tetrahedron.
In mathematics, particularly in calculus, we can solve this problem by first rewriting the equation of the plane to the form z = f(x, y).
Here, after checking the boundaries when x, y, and z are zero, we find that the volume of the tetrahedron can be determined by the triple integral of dz dy dx. The volume V of a tetrahedron is given by: 1/6 base area height. We can use the x, y, and z intercepts to find these values.
The intercepts are x = 6, y = 2 and z = 5 from the given equation. Therefore, the volume V = 1/6 6 2 5 = 1/6 60 = 10 cubic units. So, the correct answer is (d) 10.
Learn more about the topic of Volume of Tetrahedron here:
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The volume of the tetrahedron defined by the plane 6 x+2 y+5 z=1 and cut off in the first octant is calculated to be 10 cubic units. This is derived using the base area of a right triangle and the height determined by the z-intercept. Therefore, the answer is option (D) 10.
Explanation
To find the volume of the tetrahedron cut off from the first octant by the plane 6 x+2 y+5 z=1, we start by determining the intercepts of the plane with the axes.
Finding Intercepts:
x-intercept: To find the x-intercept, set y=0 and z=0:
6 x=1⟹x=6
y-intercept: Set x=0 and z=0:
2 y=1⟹y=2
z-intercept: Set x=0 and y=0:
5 z=1⟹z=5
Thus, the intercepts are (6,0,0), (0,2,0), and (0,0,5).
Volume Formula: The formula for the volume V of a tetrahedron is given by:
V=6 1×Base Area×Height
In this case, the base is the triangle formed by the x-y plane with vertices at the intercept points.
Base Area Calculation: The base of the tetrahedron in the x-y plane is a right triangle with:
Base = 6 (x-intercept)
Height = 2 (y-intercept)
Area of the triangle A:
A=2 1×Base×Height=2 1×6×2=6
Height Calculation: The height of the tetrahedron corresponds to the z-intercept, which is 5.
Calculate Volume: Now substituting these values back into the volume formula gives us:
V=6 1×Base Area×Height=6 1×6×5=5
Therefore, the volume of the tetrahedron is 10 cubic units.
Thus, the correct answer is (D) 10.
Examples & Evidence
To further understand this concept, imagine a triangular pyramid where the base area is a triangle and the height rises perpendicularly from the base to the apex. Different base lengths and heights can result in different volumes, which can be calculated in the same manner by substituting the respective values into the volume formula for a tetrahedron.
The formula for the volume of a tetrahedron is widely recognized and can be found in many geometry textbooks and official mathematics resources. The process of finding intercepts from the plane equation is a standard procedure in solid geometry.
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Community Answer Find the volume of the indicated region. The tetrahedron bounded by the coordinate planes and the plane x/8 + y/10 + z/5 = 1
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Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points.
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Consider the following incomplete deposit slip: | Description | Amount ($) | | ---: | Cash, including coins | 150 | 75 | | Check 1 | 564 | 81 | | Check 2 | 2192 | 43 | | Check 3 | 4864 | 01 | | Subtotal | ???? | ?? | | Less cash received | ???? | ?? | | Total | 7050 | 50 | | | | | How much cash did the person who filled out this deposit slip receive?
Which sequence is generated by the function f(n+1)=f(n)−2 for f(1)=10? A. −2,8,18,28,38,… B. 10,8,6,4,2,… C. 8,18,28,38,48,… D. −10,−12,−14,−16,−18,…
i) Write down the expansion of (1+x)3. ii) Find the first four terms in the expansion (1−x)−4 in ascending powers of x. For what values is the expansion valid? iii) When the expansion is valid, find the values of a and b in the equation below. (1−x)4(1+x)3=1+7 x+a x 2+b x 3+……
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Solve the simultaneous equations: [ \begin{array}{r} 5x+y=21 \ x-3y=9 \end{array} ]
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5.2
Forms of Linear Equations
Written by:Andrew Gloag | Eve Rawley |
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Forms of Linear Equations
You’ve already encountered another useful form for writing linear equations: standard form. An equation in standard form is written , where , and are all integers and is positive. (Note that the in the standard form is different than the in the slope-intercept form.)
One useful thing about standard form is that it allows us to write equations for vertical lines, which we can’t do in slope-intercept form.
For example, let’s look at the line that passes through points (2, 6) and (2, 9). How would we find an equation for that line in slope-intercept form?
First we’d need to find the slope: . But that slope is undefined because we can’t divide by zero. And if we can’t find the slope, we can’t use point-slope form either.
If we just graph the line, we can see that equals 2 no matter what is. There’s no way to express that in slope-intercept or point-slope form, but in standard form we can just say that , or simply .
Converting to Standard Form
To convert an equation from another form to standard form, all you need to do is rewrite the equation so that all the variables are on one side of the equation and the coefficient of is not negative.
Rewrite the following equations in standard form:
We need to rewrite each equation so that all the variables are on one side and the coefficient of is not negative.
a)
Subtract from both sides to get .
Add 7 to both sides to get .
Flip the equation around to put it in standard form: .
b)
Distribute the –3 on the right-hand-side to get .
Add to both sides to get .
Add 2 to both sides to get . Flip that around to get .
c)
Find the common denominator for all terms in the equation – in this case that would be 6.
Multiply all terms in the equation by 6:
Subtract from both sides:
Subtract 3 from both sides:
The equation in standard form is .
Graphing Equations in Standard Form
When an equation is in slope-intercept form or point-slope form, you can tell right away what the slope is. How do you find the slope when an equation is in standard form?
Well, you could rewrite the equation in slope-intercept form and read off the slope. But there’s an even easier way. Let’s look at what happens when we rewrite an equation in standard form.
Starting with the equation , we would subtract from both sides to get . Then we would divide all terms by and end up with .
That means that the slope is and the intercept is . So next time we look at an equation in standard form, we don’t have to rewrite it to find the slope; we know the slope is just , where and are the coefficients of and in the equation.
Finding the Slope and Intercept
Find the slope and the intercept of the following equations written in standard form.
a)
, and , so the slope is , and the intercept is .
b)
, and , so the slope is , and the intercept is .
c)
, and , so the slope is , and the intercept is .
Once we’ve found the slope and intercept of an equation in standard form, we can graph it easily. But if we start with a graph, how do we find an equation of that line in standard form?
First, remember that we can also use the cover-up method to graph an equation in standard form, by finding the intercepts of the line. For example, let’s graph the line given by the equation .
To find the intercept, cover up the term (remember, the intercept is where ):
The intercept is (2, 0).
To find the intercept, cover up the term (remember, the intercept is where :
The intercept is (0, -3).
We plot the intercepts and draw a line through them that extends in both directions:
Now we want to apply this process in reverse—to start with the graph of the line and write the equation of the line in standard form.
Finding and Re-Writing the Equation of a Line
Find the equation of each line and write it in standard form.
a)
We see that the intercept is and the intercept is
We saw that in standard form : if we “cover up” the term, we get , and if we “cover up” the term, we get .
So we need to find values for and so that we can plug in 3 for and -4 for and get the same value for in both cases. This is like finding the least common multiple of the and intercepts.
In this case, we see that multiplying by 4 and multiplying by –3 gives the same result:
Therefore, and and the equation in standard form is.
b)
We see that the intercept is and the intercept is
The values of the intercept equations are already the same, so and . The equation in standard form is.
c)
We see that the intercept is and the intercept is
Let’s multiply the intercept equation by
Then we see we can multiply the intercept again by 4 and the intercept by 3, so we end up with and .
The equation in standard form is.
Examples
Find the slope and the intercept of the following equations written in standard form.
Example 1
, and , so the slope is , and the intercept is .
Example 2
, and , so the slope is , and the intercept is .
Review
For 1-6, rewrite the following equations in standard form.
For 7-12, find the slope and intercept of the following lines.
For 13-14, find the equation of each line and write it in standard form.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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Please be kind. Is using two different panel Colors on your windows a thing? Like using two blue panels on the outside and two sage in the middle?
Large windows.
Pulling together a downstairs rec room / living room and want peaceful coastal / romantic vibes.
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Tina Collette Russell
I have seen some really pretty treatments where they are using two different pannels like you are describing... and I have seen where they put a "curtain hook" about midway up the outside of the window frame, then take the inside panel colour and "swoo…
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Hello there!
Teaching Multiplication and Division Fact Families Saves Time in the Long Run
With so much to teach and so little time, teachers start looking for ways to save time. It’s natural to wonder if certain topics, like multiplication and division fact families, are truly essential or if we can simply skim over them and focus on memorizing facts. Hold that thought! Before dismissing fact families’ significance, let’s pause and consider their value. It’s definitely worth spending time on teaching multiplication and division fact families because this skill lays the foundation for building strong math skills making problem-solving easier.
Understanding the Relationship between Multiplication and Division
By teaching fact families, students gain a deep understanding of the relationship between multiplication and division. Fact families highlight the inverse relationship between these two operations. For example, in the fact family 4 × 5 = 20, 5 × 4 = 20, 20 ÷ 5 = 4, and 20 ÷ 4 = 5, students recognize that multiplication and division are interconnected. This builds number sense, and students can also learn to apply this knowledge to solve related problems efficiently.
Building Computational Fluency With Multiplication and Division Fact Families
Proficiency in multiplication and division fact families forms the basis for developing computational fluency. When students internalize the idea of fact families, they can leverage their knowledge of a few key facts to learn and apply many related facts. This makes students more efficient. By learning one fact, they now have four facts memorized.
Promoting Number Sense
Fact families help students develop a better understanding of numbers. By looking at different combinations of numbers in fact, families, students learn how numbers are connected to each other. They develop a natural sense of how big or small numbers are, how to estimate, and how operations affect quantities. Having this number sense is important for making smart math choices and using math skills in real-life situations.
Fostering Mathematical Confidence
Teaching fact families empowers students and boosts their mathematical confidence. As students grasp the connections between multiplication and division, they gain a sense of ownership and mastery over these operations. This confidence motivates them to tackle more challenging mathematical tasks and promotes a positive attitude toward learning mathematics.
Saving time in the long run
Teaching multiplication and division fact families is vital because it builds a solid foundation for mathematical fluency, problem-solving skills, number sense, and overall confidence in mathematics. By providing students with a deep understanding of the relationships within fact families, we equip them with essential tools to navigate mathematical challenges and succeed in their future mathematical endeavors.
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https://en.wikipedia.org/wiki/Rule_of_division_(combinatorics)
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Rule of division (combinatorics)
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From Wikipedia, the free encyclopedia
Counting principle
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| | This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Rule of division" combinatorics – news · newspapers · books · scholar · JSTOR (June 2020) (Learn how and when to remove this message) |
In combinatorics, the rule of division is a counting principle. It states that there are n/d ways to do a task if it can be done using a procedure that can be carried out in n ways, and for each way w, exactly d of the n ways correspond to the way w.
In a nutshell, the division rule is a common way to ignore "unimportant" differences when counting things.
Applied to Sets
[edit]
In the terms of a set: "If the finite set A is the union of n pairwise disjoint subsets each with d elements, then n = |A|/d."
As a function
[edit]
The rule of division formulated in terms of functions: "If f is a function from A to B where A and B are finite sets, and that for every value y ∈ B there are exactly d values x ∈ A such that f (x) = y (in which case, we say that f is d-to-one), then |B| = |A|/d."
Examples
[edit]
Example 1
How many different ways are there to seat four people around a circular table, where two seatings are considered the same when each person has the same left neighbor and the same right neighbor?
: To solve this exercise we must first pick a random seat, and assign it to person 1, the rest of seats will be labeled in numerical order, in clockwise rotation around the table. There are 4 seats to choose from when we pick the first seat, 3 for the second, 2 for the third and just 1 option left for the last one. Thus there are 4! = 24 possible ways to seat them. However, since we only consider a different arrangement when they don't have the same neighbours left and right, only 1 out of every 4 seat choices matter.
: Because there are 4 ways to choose for seat 1, by the division rule (n/d) there are 24/4 = 6 different seating arrangements for 4 people around the table.
Example 2
We have 6 coloured bricks in total, 4 of them are red and 2 are white, in how many ways can we arrange them?
: If all bricks had different colours, the total of ways to arrange them would be 6! = 720, but since they don't have different colours, we would calculate it as following:
: 4 red bricks have 4! = 24 arrangements
: 2 white bricks have 2! = 2 arrangements
: Total arrangements of 4 red and 2 white bricks = 6!/4!2! = 15.
See also
[edit]
Combinatorial principles
Notes
[edit]
^ Jump up to: a b c Rosen 2012, pp.385-386
References
[edit]
Rosen, Kenneth H (2012). Discrete Mathematics and Its Applications. McGraw-Hill Education. ISBN 978-0077418939.
Further reading
[edit]
Leman, Eric; Leighton, F Thompson; Meyer, Albert R; Mathematics for Computer Science, 2018.
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Statement-1:Potassium chromate solution in acetic acid precipitates only Ba^(2+) as BaCrO_4 in group V^(th) Statement-2:SrCrO_4 and CaCrO_4 are not precipitated with potassium chromate solution in acetic acid.
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HomeClass 12CHEMISTRYStatement-1:Potassium chromate solution ...
Statement-1:Potassium chromate solution in acetic acid precipitates only B a 2+ as B a C r O 4 in group V t h
Statement-2:S r C r O 4 and C a C r O 4 are not precipitated with potassium chromate solution in acetic acid.
A
Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False.
D
Statement-1 is False, Statement-2 is True.
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B
K s p of S r C r O 4 and C a C r O 4 are much higher than B a C r O 4. Addition of acetic acid lowers the concentration of C r O 4 2-.Hence B a C r O 4 is precipitated only because its K s p value is lower.
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The blank space in the statement “The equation x 2 + y 2 + D x + E y + F = 0 is called the _____ form of the equation of a circle .”
Views: 5,193 students
Updated on: Dec 15, 2023
Text SolutionText solutionverified iconVerified
Step 1. The equation x^2 + y^2 + Dx + Ey + F = 0 is called the general form of the equation of a circle.
In this form of the equation, the coefficients D, E, and F represent the values that determine the position and size of the circle.
Step 2. The general form of the equation of a circle is used to represent a circle on a coordinate plane.
Step 3. The equation represents a circle centered at coordinates (-D/2, -E/2) with a radius of r, where r is determined by the values of D, E, and F.
Step 4. By rearranging the terms, you can also determine the center and radius of the circle.
To find the center of the circle, use the formulas:
-x = D/2
-y = E/2
To find the radius of the circle, use the formula:
-r2 = F - (D2 + E2)/4
Step 5. By substituting the given values of D, E, and F into the formulas for the center and radius, you can determine the specific properties of the circle described by the equation.
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| Question Text | The blank space in the statement “The equation x 2 + y 2 + D x + E y + F = 0 is called the _____ form of the equation of a circle .” |
| Updated On | Dec 15, 2023 |
| Topic | All Topics |
| Subject | Algebra 2 |
| Class | High School |
| Answer Type | Text solution:1 |
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In two sample testing the following could be possible target parameters:
1) mu1-mu2 (mean difference; difference in averages)
2) sigma1^2/sigma2^2 (ratio of variances; difference in variability, or spread; compare variation)
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Terms in this set (20)
In two sample testing the following could be possible target parameters:
1) mu1-mu2 (mean difference; difference in averages)
2) sigma1^2/sigma2^2 (ratio of variances; difference in variability, or spread; compare variation)
(sigma of xbar1-xbar2) = √[(s1^2/n1 + s2^2/n2]
Equation for the sample standard deviation of two samples.
(mu1-mu2) = (xbar1-xbar2)± z(alpha/2)(√[s1^2/n1 + s2^2/n2])
Confidence interval equation for mu1-mu2
-5.83 ± 4.08
(because 0 is not in the CI, the difference in the sample means appears to indicate a real difference in retention, reject the null hypothesis)
(Ho: mu1-mu2 = 0; Ha: mu1-mu2≠0, where alpha = 0.05)
(z-score = xbar1-xbar2-0/sigma of xbar1-xbar2 = -5.83/2.08 = -2.799, which has a magnitude greater than 1.96, so we reject the null hypothesis)
Example problem: Two samples concerning retention rates for first-year students at private and public institutions were obtained from the Department of Education's data base to see if there was a significant difference in the two types of colleges. The data is shown in the diagram. What does a 95% confidence interval tell us about retention rates?
What is the z-score?
Image 30: Image: -5.83 ± 4.08
(because 0 is not in the CI, the difference in the sample means appears to indicate a real difference in retention, reject the null hypothesis)
(Ho: mu1-mu2 = 0; Ha: mu1-mu2≠0, where alpha = 0.05)
(z-score = xbar1-xbar2-0/sigma of xbar1-xbar2 = -5.83/2.08 = -2.799, which has a magnitude greater than 1.96, so we reject the null hypothesis)
(two samples are randomly selected; sample sizes are both greater than 30)
Name two conditions required for valid large-sample inferences about (mu1-mu2)
sp^2 = [(n1-1)s1^2 + (n2-1)s2^2]/(n1+n2-2)
What is the equation for the sample variance when using the t-distribution (pooled sample estimator for sigma^2)
(mu1-mu2) = (xbar1-xbar2) ± t(alpha/2)√(sp^2[1/n1+1/n2])
(degrees of freedom = n1+n2-2)
Confidence interval equation for mu1-mu2, when we use (t) score instead, due to small sample sizes.
t = [(xbar1-xbar2)-Do]/[√sp^2(1/n1+1/n2)]
(rejection region: t>talpha)
Test statistic for t-distribution of difference between means (mu1-mu2), where
Ho: (mu1-mu2) = Do
Ha: (,u1-mu2)>Do or < Do
(samples randomly selected; have approximately normal population distributions; population variances approximately equal)
Conditions that need to be satisfied to use t-test for two samples (small sample sizes)
-this is not a paired t-test
(t of alpha = 2.021, from table, so if the t-score calculated is greater than this value, we can reject the null hypothesis)
(sp^2 = 242.5, from calculation; t = -0.832, which has a magnitude less than 2.021, so we fail to reject the null hypothesis, and say there is no difference between these two times)
Does class time affect performance? The test performance of students in two sections of international trade, meeting at different times, were compared.
Image 31: Image: (t of alpha = 2.021, from table, so if the t-score calculated is greater than this value, we can reject the null hypothesis)
(sp^2 = 242.5, from calculation; t = -0.832, which has a magnitude less than 2.021, so we fail to reject the null hypothesis, and say there is no difference between these two times)
(If variances are unequal, we calculate degrees of freedom using the following: df = [s1^2/n1 + s2^2/n2]^2/[(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)] = 26.15 = 26, so t of alpha at df = 26 = 2.056)
(t = 72-86/√s1^2/n1+s2^2/n2 = -3.16, which has a magnitude greater than t of alpha, so the null hypothesis is now rejected, showing that there is a difference between the results of the two time points tested)
Several students in the 8:00 am class sleep, so we get rid of these samples, making the sample for the 8am class slightly different (this example is solved differently in a different problem). Find the new test statistic.
![Image 32: Image: (If variances are unequal, we calculate degrees of freedom using the following: df = [s1^2/n1 + s2^2/n2]^2/[(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)] = 26.15 = 26, so t of alpha at df = 26 = 2.056)
(t = 72-86/√s1^2/n1+s2^2/n2 = -3.16, which has a magnitude greater than t of alpha, so the null hypothesis is now rejected, showing that there is a difference between the results of the two time points tested)](
F
Since variances do not follow a normal distribution (due to the squaring involved) we use a different technique for inferences about variances. aka the _____ statistic
equal
The F random variable is a ratio with n1-1 numerator and n2-1 denominator degrees of freedom. The ratio tests that the variances are _____.
1; larger; right; 0; 1
(rejection region if F> F of alpha, then we reject the null)
If the ratio of variances in an F test are too far from ___ in value, in either direction the likelihood that both samples share a population variance drops. For convenience the _ sample variance is usually put in the numerator. Distribution is skewed to the , since s1^2/s2^2 cannot be less than _, but can increase without bound (the total area needs to be equal to ____)
(s1^2/s2^2)(1/F upper of alpha/2) < (sigma1^2/sigma2^2) < (s1^2/s2^2)(1/F lower of alpha/2)
Confidence interval equation for sigma1^2/sigma2^2. If 1 falls in the confidence interval, then we fail to reject the null hypothesis.
same (here if we acted like these were two samples, the assumption that each sample is randomly selected would be violated, but if we use a paired t-test this will not be violated)
(rejection region is if t>t of alpha, so we would reject the null)
In a paired t-test, we define a new variable known as the difference between two regions of the _____ sample.
t = dbar/(sd/√n)
What is the test statistic equation for a paired t-test.
xbar of d ± z(alpha/2)(sigma of d/√nd)
(xbar of d = sample mean difference, sd = sample standard deviation of the differences, nd = number of pairs observed)
Confidence interval for a paired difference (mu of d = mu1-mu2)
(xbar of d = 4, sd = 3)
mu1-mu2 = 4 ± (1.8333/√10) = 4±1.74
(since 0 is not in the interval, one program does seem to work more effectively; rejected null)
Sample Problem: Suppose ten pairs of puppies were housetrained using two different methods: one puppy from each pair was paper-trained, with the paper gradually moved outside, and the other was taken out every three hours and twenty minutes after each meal. The number of days until the puppies were considered housetrained (three days straight without an accident) were compared. Nine of the ten paper-trained dogs took longer than the other paired dog to complete training, with the average difference equal to 4 days, with a standard deviation of 3 days. What is the 90% confidence interval on the difference in successful training?
(z = 1.75/10.35/√150 = 2.07, where z of 0.05 = 1.96)
(because z(2.07) is greater than z of alpha, we reject the null hypothesis)
Sample Problem: Suppose 150 items were priced at two online stores, "cport" and "warriorwoman", where the mean difference is $1.75, and the standard deviation is $10.35. Test at the 95% level that the difference in the two stores is zero.
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https://www.youtube.com/watch?v=QIlDmDOWZ84
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Calculus Help: Differential Equations - Bernoulli's Equations - dy/dx (x^3 y^3+xy)=1
Calculus Physics Chem Accounting Tam Mai Thanh Cao
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Here is the technique to solve this question and how to find them in step-by-step
DifferentialEquations
Bernoulli
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Transcript:
hello everybody so today i will talk about calculus the topic today is about differential equations so now i will show you how to answer we need to divide the y over the x we need to open parenthesis next we need to divide both sides by s to the power of 3 next we can apply about bronulate equation so i need to put about v equals to 1 over s square later that we do first derivative for the both sides because we don't have negative number two so we need to divide both sides by negative number two [Applause] so 1 over s where we change by v this one will change by this one so i need to boot this one go to the left and y to the power of 3 we put on the right so we have the velocity number this one should be about negative negative y to the power of three next we multiply both sides by number two so we have v and first derivative of v that is about linear differential equations now we can use about interwriting factor so we we need to find about mu y equals to e to the power of the integration of q y q y that is about this one in here the integration of q y d y you got about e to the power of y square that means we need to multiply both sides by this one next we need to put the integration for the both sides now we can apply about the product rule of derivative so we have this one in here we know that first the reflective of y that is about first derivative of a multiplied by b plus first derivative of b multiplied by a so this is about first degree for t of a multiplied by b this one that is about first the refer to of b multiplied by i so the integration of this one we have y equals to a b so the integration of this one we have v multiply by e to the power of y square now i need to concentrate about this one only so y to the power of 3 we can change into y and y square next we need to use about integration by substitution so i need to put about u equals to y square later that we need to do first derivative for the both sides first derivative of y to the power of 2 we have 2 y d y so this one we change by u this one will change by u 2 y d y that is about d u now we can use about integration bypass so i need to put about t equals to u that we need to do first the refer to for the both sides dv that is about this one the interruption of this one you got about v equals to e to the power of u so we have t multiplied by v but we have the negative so we keep about a negative side in here minus we have dt multiplied by v so we have the integration e to the power of udu so the integration of this one we have e to the power of u and then we put about c about u that is about this one now we go back in here the integration of this one that is about this one later that we divide both sides by this one about v that is about this one and this is the final answer this is the end thank you for watching
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https://www.gymglish.com/es/conjugacion/espanol/verbo/salir
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Conjugación en español del verbo SALIR - Conjugador español
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yo sal go
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Pretérito perfecto
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Pretérito pluscuamperfecto
yo había sal ido
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él/ella/usted había sal ido
nosotros/as habíamos sal ido
vosotros/as habíais sal ido
ellos/ellas/ustedes habían sal ido
Pretérito indefinido
yo sal í
tú sal iste
él/ella/usted sal ió
nosotros/as sal imos
vosotros/as sal isteis
ellos/ellas/ustedes sal ieron
Futuro
yo sal dré
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él/ella/usted sal drá
nosotros/as sal dremos
vosotros/as sal dréis
ellos/ellas/ustedes sal drán
Futuro perfecto
yo habré sal ido
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él/ella/usted habrá sal ido
nosotros/as habremos sal ido
vosotros/as habréis sal ido
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Condicional
yo sal dría
tú sal drías
él/ella/usted sal dría
nosotros/as sal dríamos
vosotros/as sal dríais
ellos/ellas/ustedes sal drían
Condicional perfecto
yo habría sal ido
tú habrías sal ido
él/ella/usted habría sal ido
nosotros/as habríamos sal ido
vosotros/as habríais sal ido
ellos/ellas/ustedes habrían sal ido
Futuro próximo
yo voy a sal ir
tú vas a sal ir
él/ella/usted va a sal ir
nosotros/as vamos a sal ir
vosotros/as vais a sal ir
ellos/ellas/ustedes van a sal ir
Presente continuo
yo estoy sal iendo
tú estás sal iendo
él/ella/usted está sal iendo
nosotros/as estamos sal iendo
vosotros/as estáis sal iendo
ellos/ellas/ustedes están sal iendo
Pretérito anterior
yo hube sal ido
tú hubiste sal ido
él/ella/usted hubo sal ido
nosotros/as hubimos sal ido
vosotros/as hubisteis sal ido
ellos/ellas/ustedes hubieron sal ido
Subjuntivo
Presente
yo sal ga
tú sal gas
él/ella/usted sal ga
nosotros/as sal gamos
vosotros/as sal gáis
ellos/ellas/ustedes sal gan
Pretérito perfecto
yo haya sal ido
tú hayas sal ido
él/ella/usted haya sal ido
nosotros/as hayamos sal ido
vosotros/as hayáis sal ido
ellos/ellas/ustedes hayan sal ido
Pretérito imperfecto
yo sal iera/sal iese
tú sal ieras/sal ieses
él/ella/usted sal iera/sal iese
nosotros/as sal iéramos/sal iésemos
vosotros/as sal ierais/sal ieseis
ellos/ellas/ustedes sal ieran/sal iesen
Pretérito pluscuamperfecto
yo hubiera sal ido/hubiese sal ido
tú hubieras sal ido/hubieses sal ido
él/ella/usted hubiera sal ido/hubiese sal ido
nosotros/as hubiéramos sal ido/hubiésemos sal ido
vosotros/as hubierais sal ido/hubieseis sal ido
ellos/ellas/ustedes hubieran sal ido/hubiesen sal ido
Imperativo
Presente
(tú) sal
(usted) sal ga
(nosotros/as) sal gamos
(vosotros/as) sal id
(ustedes) sal gan
negativo
(tú) no sal gas
(usted) no sal ga
(nosotros/as) no sal gamos
(vosotros/as) no sal gáis
(ustedes) no sal gan
Gerundio
Simple
sal iendo
Compuesto
habiendo sal ido
Participo
Pasado
sal ido
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On Maximal 3-Planar Graphs Michael Hoffmann, Meghana M. Reddy, and Shengzhe Wang Department of Computer Science, ETH Zürich, Switzerland {hoffmann, meghana.mreddy}@inf.ethz.ch, wangshe@ethz.ch Abstract A graph is 3-planar if it admits a drawing in the plane such that every edge is crossed at most three times. A 3-planar graph is maximal 3-planar if addition of any edge results in a graph that is not 3-planar. A 3-planar graph on n vertices has at most 5.5n −11 edges, and a 3-planar graph that has exactly 5.5n −11 edges is an optimal 3-planar graph. In contrast to planar graphs where maximal and optimal graphs coincide, a maximal 3-planar graph may have fewer edges than an optimal 3-planar graph. In this paper, we study properties of maximal 3-planar graphs. First, we characterize the graphs on nine vertices that are (maximal) 3-planar. Second, we show that—in contrast to maximal 1– and 2-planar graphs—maximal 3-planar graphs may contain cut vertices.
Third, we give a first upper bound on the minimal edge density by constructing maximal 3-planar graphs on n vertices with only 2.375n + O(1) edges.
1 Introduction Planar graphs are graphs that can be drawn on the plane without crossings. If the addition of any edge to a planar graph makes it impossible for the resulting graph to admit a plane drawing, the planar graph is said to be a maximal planar graph. Maximal planar graphs are well-studied and have many interesting properties. For example, the number of edges in maximal planar graphs is solely dependent on the number of vertices. Specifically, every maximal planar graph on n ≥3 vertices has 3n −6 edges. It is natural to further explore the edge density for various families of beyond-planar graphs, which have been extensively studied over the past decade [5, 8].
In this paper, we focus on 3-planar graphs. A graph is k-planar if it admits a drawing where each edge has at most k crossings. Pach and Tóth gave upper bounds on the number of edges in a k-planar graph, which they used to improve the Crossing Lemma.
Unlike planar graphs, the class of k-planar graphs is not closed under edge contractions.
Thus, many useful properties of minor closed classes do not apply to k-planar graphs.
As k gets larger, the density of a k-planar graph on n vertices clearly increases, but the exact correlation is still unknown. Pach and Tóth showed a general upper bound of 4.108 √ kn edges for k-planar graphs. For small values of k, i.e., k = 1 and k = 2, they also gave tight upper bounds. The class of 1-planar graphs was introduced by Ringel in the context of planar graph colorings. A 1-planar graph has at most 4n −8 edges, and there are infinitely many optimal 1-planar graphs that achieve the bound . A 2-planar graph on n vertices has at most 5n −10 edges, and there are infinitely many optimal 2-planar graphs that achieve this bound .
A lot is left to be explored for maximal k-planar graphs, where a graph is maximal k-planar if there is no edge that can be added such that the resulting graph is still k-planar.
In contrast to planar graphs, maximal k-planar graphs are not necessarily optimal k-planar graphs. Indeed the gap between maximal and optimal can be very large for k-planar graphs.
Hudák et al. showed an infinite family of maximal 1-planar graphs with 8 3n+O(1) ≈2.667n edges, and a sparser construction by Brandenburg et al. only has 45 17n + O(1) ≈2.647n edges. Brandenburg et al. also proved that every maximal 1-planar graph has at least 40th European Workshop on Computational Geometry, Ioannina, Greece, March 13–15, 2024.
This is an extended abstract of a presentation given at EuroCG’24. It has been made public for the benefit of the community and should be considered a preprint rather than a formally reviewed paper. Thus, this work is expected to appear eventually in more final form at a conference with formal proceedings and/or in a journal.
31:2 On Maximal 3-Planar Graphs 28 13n −O(1) ≈2.153n edges. This lower bound was further improved by Barát and Tóth to 20 9 n ≈2.22n.
Interestingly, the density even decreases for maximal k-planar graphs when k increases from 1 to 2. Hoffmann and M. Reddy showed that every maximal 2-planar graph on n ≥5 vertices has at least 2n edges, and they also described an infinite family of maximal 2-planar graphs on n vertices with only 2n + O(1) edges. So, even though they allow more crossings, some maximal 2-planar graphs have a lower edge density than any maximal 1-planar graph.
Results.
First, in Section 3 we characterize the graphs on nine vertices that are (maximal) 3-planar. Next, in Section 4 we exhibit maximal 3-planar graphs that contain vertices for which all incident edges are crossed, in every simple 3-plane drawing of the graph. As a consequence, maximal 3-planar graphs are not necessarily 2-connected and may contain vertices of degree one. In contrast, all maximal 1-and 2-planar graphs are 2-connected. Finally, in Section 5, we construct maximal 3-planar graphs on n vertices with only 2.375n + O(1) edges.
2 Preliminaries A drawing is simple if any pair of edges has at most one common point, including endpoints.
To analyze k-plane drawings of a graph, one typical restriction is to consider a drawing that minimizes the total number of crossings among all k-plane drawings of the graph, which is called a crossing-minimal k-plane drawing. The benefit of such a restriction is that for k ≤3, a crossing-minimal k-plane drawing is always simple . Consequently, 3-planar graphs always admit a simple 3-plane drawing.
▶Lemma 1. If a 3-planar graph G is not maximal 3-planar, then there exists a simple 3-plane drawing of G that is not maximal 3-plane.
Proof. If G is not maximal 3-planar, then there exists a pair u, v of nonadjacent vertices such that G′ = G ∪e is 3-planar where e = (u, v). Take any simple 3-plane drawing of G′ and remove e to obtain a simple 3-plane drawing of G that is not maximal.
◀ In all figures of this paper, the edges are colored to indicate their number of crossings.
Uncrossed edges are shown green, singly crossed edges are shown purple, doubly crossed edges are shown orange, and triply crossed edges are shown blue. Edges with an undetermined number of crossings are shown black.
3 Characterization of (Maximal) 3-planar Graphs on 9 Vertices Angelini et al. showed that K8 is 3-planar, while K9 is 4-planar (but not 3-planar).
This motivated us to study the set of 3-planar graphs on nine vertices. With the help of a computer program, we can enumerate all possible simple 3-plane drawings of a given graph; see Section 6.
The basic idea is to check all graph structures on nine vertices in decreasing order based on the number of edges. Specifically, starting from K9 \ K2, which is generated by removing a single edge from a K9, we check if the given graph admits a 3-plane drawing. Further, if we want to remove two edges from a clique, we can either remove two independent edges or remove a path P3 of length two. If we restrict to a graph with nine vertices, that is equivalent to saying that any graph with nine vertices and thirty-four edges will be isomorphic to either K9 \ (K2 + K2) or K9 \ P3. We have a similar argument if we remove three edges.
M. Hoffmann, M. M. Reddy and S. Wang 31:3 x5 x1 x3 x7 x9 x0 x2 x4 x6 x8 x5 x1 x3 x7 x9 x0 x2 x4 x6 x8 x10 Figure 1 (Left) Drawing D1 of graph G1 by removing five independent edges from K10. (Right) Drawing D′ 1 of graph G′ 1 by inserting a new vertex x10 to G1 and adding five edges incident to x10.
Using our computer program, we were able to verify that all graphs on nine vertices that have at least 34 edges do not admit a simple 3-plane drawing, while all the remaining graphs on nine vertices are 3-planar. The result can be verified with the code in our repository .
▶Theorem 2. A graph on nine vertices is 3-planar if and only if it has at most 33 edges, and it is maximal 3-planar if and only if it has exactly 33 edges.
We therefore notice that maximal 3-planar graphs and maximum 3-planar graphs on n vertices coincide for n ≤9.
4 Uncrossed Edge in Maximal 3-planar Graphs Though we allow crossings in beyond-planar graphs, it does not necessarily mean that every edge will have a crossing. In a more general sense, crossings are not equally distributed over the edges. In maximal k-planar graphs where k ≤2, it has been shown that every vertex must be incident to an uncrossed edge in every crossing-minimal k-plane drawing . But when k = 3, the situation is different and we have the following ▶Lemma 3. There exist infinitely many maximal 3-planar graphs that each contains a vertex v such that all edges incident to v are crossed in every simple 3-plane drawing of the graph.
Proof. The proof of Lemma 3 is based on a graph G′ 1 that has the required properties. Then we can use G′ 1 to create larger graphs. To construct a graph G′ 1, we start from a base graph G1 that is isomorphic to K10 minus five independent edges. Specifically, we take the vertex set as {x0, x1, ..., x9}, and include all edges of the induced complete graph except the five edges {x0, x1}, {x2, x3}, {x4, x5}, {x6, x7} and {x8, x9}. We enumerated all simple 3-plane drawings of G1 using our program and found that this graph has exactly one simple 3-plane drawing up to automorphism. This drawing is illustrated in Fig. 1, and let this drawing be D1. We can observe that it is impossible to add an edge to D1 while maintaining 3-planarity, thus it proves G1 is a maximal 3-planar graph from Lemma 1.
We can further obtain a new graph G′ 1 by adding a new vertex x10 to G1 and connecting it to the five vertices x0, x2, x4, x6, x8. We claim that the drawing D′ 1 illustrated in Fig. 1(right) EuroCG’24 31:4 On Maximal 3-Planar Graphs x10 Figure 2 Graph G2 by gluing two copies of G1 on a merged vertex.
is the unique simple 3-plane drawing of G′ 1 (up to automorphisms). To see this, consider every face of D1. It turns out that there exists exactly one face where the vertex x10 can be placed such that the five edges to vertices x0, x2, x4, x6, x8 can be added while maintaining 3-planarity, and the resulting drawing is D′ 1. This concludes that the graph G′ 1 is maximal 3-planar. We further note that x10 is not incident to any uncrossed edge in any 3-plane drawing of G′ 1.
Further we can take two copies of the graph G′ 1 and merge the two vertices with degree five from each of the copies into a single vertex with degree ten to obtain a new graph G2. It can be interpreted as gluing two copies of G′ 1 together at the vertex x10. Refer to Fig. 2 for one possible 3-plane drawing of G2.
We claim that G2 is a maximal 3-planar graph. Consider an arbitrary simple 3-plane drawing of G2. The subdrawings corresponding to the two copies of G1 that are still simple drawings should be the same as D1. It can also be viewed as adding a copy of D1 into an existing drawing D1. Again, considering all the faces of D1 we can observe that the drawing shown in Fig. 2 is the only possible 3-plane drawing of G2 up to automorphism.
This replication can be repeated infinitely many times to obtain larger graphs, and in each such graph the vertex x10 is not incident to any uncrossed edge in every simple 3-plane drawing, concluding the proof.
◀ Clearly, x10 is a cut vertex in G2 and every graph obtained by following the replicating procedure. Thus, we have the following ▶Theorem 4. There exist infinitely many maximal 3-planar graphs that are not 2-connected.
5 Number of Edges in Sparse Maximal 3-planar Graphs We describe a construction of sparse maximal 3-planar graphs based on G1 in this section.
▶Theorem 5. There exist an infinite family of maximal 3-planar graphs on n vertices with at most 2.375n + O(1) edges.
Proof. We construct a nested graph with arbitrarily many layers where each layer is a variation of G1 as shown in Fig. 3. Specifically, in each layer, we add a hermit vertex M. Hoffmann, M. M. Reddy and S. Wang 31:5 x5 x3 x1 x7 x9 x0 x2 x4 x6 x8 x′ 5 x′′ 5 x′ 4 x′′ 4 Figure 3 A single layer in the nested graph structure. Some labels are omitted for simplicity.
connecting to two endpoints of each planar edge, and a triangle to each original vertex, where a hermit is a degree-two vertex.
Further, as shown in Fig. 3, consecutive layers are connected with each other with orange dashed lines, which represent half edges. Specifically, suppose vertices from layer i are indexed as xi j for 0 ≤j ≤9, and corresponding triangle vertices are labelled as xi ′ j and xi ′′ j .
To connect layers i and i + 1, an edge is added between xi ′ j and xi+1 ′ j+1 for even j. To close the innermost and outermost layers, we simply use two 5-stars respectively to complete the graph.
We argue that the constructed graph still admits few simple 3-plane drawings. We start from each layer based on the unique simple drawing D1. For those inserted triangles, since the drawing is symmetric considering the innermost face and the outermost face, it is enough to argue about the outer triangles. We note that outer vertices xi j of layer i for odd j should be reachable from a single face because all vertices from other layers have to be drawn in a single face of the drawing for layer i. Thus the only eligible face is the outermost face.
Observe that xi ′′ j connects to xi j and xi ′ j , so the drawing of triangles has to be the same as shown in Fig. 3. We then note that each planar edge is enclosed by blue curves. Thus EuroCG’24 31:6 On Maximal 3-Planar Graphs every hermit vertex can only be drawn in faces on two sides of the planar edge. For the closing 5-star, the argument is similar as stated for graph G′ 1. Then consider all possible simple 3-plane drawing of such a nested graph, it is impossible to add any edge in any of the drawing, and this concludes that the graph is maximal 3-planar.
For each layer, we have 40 vertices and 90 edges. And between two layers, there are 5 edges connecting them. We can charge these 5 edges to one layer, and extend to arbitrarily many layers. In total we will have 40k + 2 vertices and 95k + 5 edges for a graph with k layers. And it gives an edge density of 95k+5 40k+2 · n = 2.375n + O(1) where n is the number of vertices. This concludes the proof.
◀ It is possible that the construction for 2-planar graphs with roughly 2n edges can be extended to the 3-planar case; see Fig. 4. But new ideas are needed to prove maximality.
v1 0 v1 1 v1 3 v1 4 v1 5 v1 6 v1 7 v1 8 v1 9 v1 10 v1 11 v1 12 v1 13 v1 2 v2 11 v3 0 v2 12 v2 13 v2 8 v2 9 v2 10 v2 7 v2 6 v2 5 v2 4 v2 3 v2 2 v2 1 v2 0 v3 11 v3 12 v3 13 v3 1 v3 2 v3 3 v3 4 v3 5 v3 6 v3 7 v3 8 v3 9 v3 10 Figure 4 Nested C14 is a possible maximal 3-planar graph.
6 Enumeration of 3-plane drawings In this section, we sketch the program we used to enumerate all possible simple 3-plane drawings of small graphs. This program was used to show certain properties of 3-planar graphs, like maximality. The basic idea is similar to previous work [1, 6]. We adapted the code for 2-planar graphs from and extended it to 3-planar graphs. It is available in our repository .
To enumerate all simple k-plane drawings of a graph up to strong isomorphism (that is, up to a homeomorphism of the plane), we enumerate combinations of all possible drawings of each edge. We fix a labeling of the vertices and an ordering of the edges to then use M. Hoffmann, M. M. Reddy and S. Wang 31:7 depth first search to explore all possible simple 3-plane drawings. The restriction to simple drawings is without loss of generality by Lemma 1.
We add edges one by one to the current drawing, and try to complete the given graph.
Whenever we add an edge, we take every valid drawing of the edge into consideration. After we run out of different ways to draw the current edge, we backtrack and try to draw the previous edge differently. Whenever we successfully added all edges into the drawing, we found a simple 3-plane drawing of the given graph. We record the drawing and continue.
The time complexity of such a search is exponential in the number of edges, and thus it can be used for small graphs only. In practice, it takes roughly one hour to enumerate drawings for graphs on 9 vertices, and 30 hours for graphs on 10 vertices. For larger graphs it seems challenging to enumerate all simple 3-plane drawings within reasonable time limits.
References 1 Patrizio Angelini, Michael A. Bekos, Michael Kaufmann, and Thomas Schneck. Efficient generation of different topological representations of graphs beyond-planarity. In Proc.
Graph Drawing and Network Visualization (GD’19), pages 253–267, 2019.
2 Patrizio Angelini, Michael A. Bekos, Michael Kaufmann, and Thomas Schneck. Efficient generation of different topological representations of graphs beyond-planarity. J. Graph Algorithms Appl., 24(4):573–601, 2020.
URL: doi:10.7155/JGAA.00531.
3 János Barát and Géza Tóth. Improvements on the density of maximal 1-planar graphs, 2015. arXiv:1509.05548.
4 Franz J. Brandenburg, David Eppstein, Andreas Gleißner, Michael T. Goodrich, Kathrin Hanauer, and Josef Reislhuber. On the density of maximal 1-planar graphs. In Proc. Graph Drawing (GD’13), pages 327–338, 2013. doi:10.1007/978-3-642-36763-2_29.
5 Walter Didimo, Giuseppe Liotta, and Fabrizio Montecchiani. A survey on graph drawing beyond planarity. ACM Comput. Surv., 52(1), 2019. doi:10.1145/3301281.
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ReviewVolume 80, Issue 8p1050-1057 August 2005
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Medication-Induced Hyperprolactinemia
Mark E.Molitch, MD
Mark E.Molitch, MD
Correspondence
Address reprint requests and correspondence to Mark E. Molitch, MD, Division of Endocrinology, Metabolism and Molecular Medicine, Northwestern University Feinberg School of Medicine, 303 E Chicago Ave (Tarry 15-731), Chicago, IL 60611
molitch@northwestern.edu
Footnotes
1
Dr Molitch is currently receiving research support from Pfizer, Inc, Novartis Pharmaceuticals Corp, Sanofi-Aventis Pharmaceuticals, and Amgen Inc and is serving as a consultant to Abbott Laboratories.,
Affiliations
Division of Endocrinology, Metabolism and Molecular Medicine, Northwestern University Feinberg School of Medicine, Chicago, Ill
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1molitch@northwestern.edu
Affiliations & Notes Article Info
Division of Endocrinology, Metabolism and Molecular Medicine, Northwestern University Feinberg School of Medicine, Chicago, Ill
1
Dr Molitch is currently receiving research support from Pfizer, Inc, Novartis Pharmaceuticals Corp, Sanofi-Aventis Pharmaceuticals, and Amgen Inc and is serving as a consultant to Abbott Laboratories.,
DOI: 10.4065/80.8.1050 External LinkAlso available on ScienceDirect External Link
Copyright: © 2005 Mayo Foundation for Medical Education and Research.
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Abstract
Keywords
Medications Causing Hyperprolactinemia
Assessment of A Patient with Suspected Medication-Induced Hyperprolactinemia
Patient Treatment for Medication-Induced Hyperprolactinemia
Conclusions
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Abstract
Keywords
Medications Causing Hyperprolactinemia
Assessment of A Patient with Suspected Medication-Induced Hyperprolactinemia
Patient Treatment for Medication-Induced Hyperprolactinemia
Conclusions
REFERENCES
Article metrics
Related Articles
Abstract
Medication use is a common cause of hyperprolactinemia, and it is important to differentiate this cause from pathologic causes, such as prolactinomas. To ascertain the frequency of this clinical problem and to develop treatment guidelines, the medical literature was searched by using PubMed and the reference lists of other articles dealing with hyperprolactinemia due to specific types of medications. The medications that most commonly cause hyperprolactinemia are antipsychotic agents; however, some newer atypical antipsychotics do not cause this condition. Other classes of medications that cause hyperprolactinemia include antidepressants, antihypertensive agents, and drugs that increase bowel motility. Hyperprolactinemia caused by medications is commonly symptomatic, causing galactorrhea, menstrual disturbance, and impotence. It is important to ensure that hyperprolactinemia in an individual patient is due to medication and not to a structural lesion in the hypothalamic/pituitary area; this can be accomplished by (1) stopping the medication temporarily to determine whether prolactin levels return to normal, (2) switching to a medication that does not cause hyperprolactinemia (in consultation with the patient's psychiatrist for psychoactive medications), or (3) performing magnetic resonance imaging or computed tomography of the hypothalamic/pituitary area. If the patient's hyperprolactinemia is symptomatic, treatment strategies include switching to an alternative medication that does not cause hyperprolactinemia, using estrogen or testosterone replacement, or, rarely, cautiously adding a dopamine agonist.
Keywords
D 2 (dopamine D 2)
HIV (human immunodeficiency virus)
MRI (magnetic resonance imaging)
PRL (prolactin)
TRH (thyrotropin-releasing hormone)
Critical to an understanding of the ways in which medications affect prolactin (PRL) secretion is an explanation of the neuroendocrine regulation of PRL secretion. The hypothalamus predominantly influences PRL secretion through 1 or more PRL inhibitory factors that reach the pituitary via the hypothalamic/pituitary portal vessels. Disruption of the pituitary stalk leads to moderately increased PRL secretion and to decreased secretion of the other pituitary hormones. Dopamine is the predominant physiological inhibitory factor; blockade of endogenous dopamine receptors by various drugs, including antipsychotic agents, causes PRL secretion to increase. There are PRL-releasing factors as well. Although thyrotropin-releasing hormone (TRH) causes a rapid release of PRL, numerous different experimental approaches have failed to clarify the physiological role of TRH as a PRL-releasing factor. Vasoactive intestinal peptide has stimulatory effects that are selective for PRL; also, a part of the vasoactive intestinal peptide precursor, a similarly sized peptide known as peptide histidine methionine, exerts PRL-releasing properties in humans. The precise roles of these peptides and other PRL-releasing factors such as TRH are unclear.1
1.
Molitch, ME
Disorders of prolactin secretion
Endocrinol Metab Clin North Am. 2001; 30:585-610
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The clinical consequences of hyperprolactinemia include galactorrhea and hypogonadotropic hypogonadism, the latter manifesting as oligomenorrhea or amenorrhea in women, erectile dysfunction in men, and loss of libido and infertility in both sexes.1
1.
Molitch, ME
Disorders of prolactin secretion
Endocrinol Metab Clin North Am. 2001; 30:585-610
Full Text
Full Text (PDF)
Scopus (93)
PubMed
Google Scholar
The hypogonadism primarily is due to the hyperprolactinemia causing a decrease in the pulsatile secretion of gonadotropin-releasing hormone by the hypothalamus.1
1.
Molitch, ME
Disorders of prolactin secretion
Endocrinol Metab Clin North Am. 2001; 30:585-610
Full Text
Full Text (PDF)
Scopus (93)
PubMed
Google Scholar
The differential diagnosis of hyperprolactinemia includes medication-induced hyperprolactinemia. It is important to differentiate medication-induced hyperprolactinemia from pathological causes, such as PRL-producing tumors (prolactinomas), hypothalamic disease, hypothyroidism, and renal insufficiency.1
1.
Molitch, ME
Disorders of prolactin secretion
Endocrinol Metab Clin North Am. 2001; 30:585-610
Full Text
Full Text (PDF)
Scopus (93)
PubMed
Google Scholar
Many medications reportedly cause hyperprolactinemia (Table 1). The medical literature was searched by using PubMed and the reference lists of other articles dealing with hyperprolactinemia due to specific types of medications. In this review, I critically evaluate the literature regarding the frequencies at which such medications cause hyperprolactinemia and whether the degree of hyperprolactinemia is sufficient to cause symptoms. The assessment and treatment of patients with suspected medication-induced hyperprolactinemia are described as well.
Antipsychotics (neuroleptics)
Phenothiazines
Thioxanthenes
Butyrophenones
Atypical antipsychotics
Antidepressants
Tricyclic and tetracyclic antidepressants
Monoamine oxidase inhibitors
Selective serotonin reuptake inhibitors
Other
Opiates and cocaine
Antihypertensive medications
Verapamil
Methyldopa
Reserpine
Gastrointestinal medications
Metoclopramide
Domperidone
Histamine 2 receptor blockers?
Protease inhibitors?
Estrogens
TABLE 1
Medications That May Cause Hyperprolactinemia
Open table in a new tab
Medications Causing Hyperprolactinemia
Antipsychotic (Neuroleptic) Medications
The medications that most commonly cause hyperprolactinemia are antipsychotic (neuroleptic) agents (Table 2). These drugs are dopamine receptor blockers. Their antipsychotic effects are mediated by dopamine D 2 (D 2) and dopamine D 4 receptors in the mesolimbic area of the brain, and their extrapyramidal adverse effects are mediated through D 2 receptors in the striatal area.2,3
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| | Increase in prolactin† |
| Antipsychotics | |
| Typical | |
| Phenothiazines | +++ |
| Butyrophenones | +++ |
| Thioxanthenes | +++ |
| Atypical | |
| Risperidone | +++ |
| Molindone | ++ |
| Clozapine | 0 |
| Quetiapine | + |
| Ziprasidone | 0 |
| Aripiprazole | 0 |
| Olanzapine | + |
| Antidepressants | |
| Tricyclics | |
| Amitriptyline | + |
| Desipramine | + |
| Clomipramine | +++ |
| Nortriptyline | − |
| Imipramine | CR |
| Maprotiline | CR |
| Amoxapine | CR |
| Monoamine oxidase inhibitors | |
| Pargyline | +++ |
| Clorgyline | +++ |
| Tranylcypromine | ± |
| Selective serotonin reuptake inhibitors | |
| Fluoxetine | CR |
| Paroxetine | ± |
| Citalopram | ± |
| Fluvoxamine | ± |
| Other | |
| Nefazodone | 0 |
| Bupropion | 0 |
| Venlafaxine | 0 |
| Trazodone | 0 |
TABLE 2
Effects of Psychotropic Medications on Prolactin Levels
CR = isolated case reports of hyperprolactinemia but generally no increase in prolactin levels.
†
0 = no effect; ± = minimal increase but not to abnormal levels; + = increase to abnormal levels in small percentage of patients; ++ = increase to abnormal levels in 25% to 50% of patients; +++ = increase to abnormal levels in more than 50% of patients.
Open table in a new tab
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and aripiprazole30
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much less commonly elevate PRL levels. It is believed that the lack of effect of these atypical agents is due to their being only transiently and weakly bound to the D 2 receptor31
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Frequent sampling for PRL levels in patients taking clozapine or olanzapine long-term shows that PRL levels increase quickly 1.5-fold to 2.5-fold within 2 to 4 hours after the drugs are taken, only to decrease to baseline within 8 hours, thus supporting the hypothesis that these agents only transiently bind to the tuberoinfundibular D 2 receptor. In contrast, risperidone causes a similar doubling of PRL levels, but the effects persist for 24 hours.25
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Hyperprolactinemia caused by these drugs is accompanied usually by decreased libido, erectile dysfunction in men, and galactorrhea and amenorrhea in women.11,12,15,33–36
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and preliminary data suggest that there may be an increased risk of osteopenia.20,37
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Tricyclic antidepressants have caused mild hyperprolactinemia in some patients, but the data are scant (Table 2). In 1 series, amitriptyline reportedly caused a 100% increase in PRL levels in 2 (14%) of 14 patients.39
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it caused no change in PRL levels in another study of 24 patients.40
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reported that neither nortriptyline nor mianserin caused increases in PRL levels, but no specific data were given in that report. Clomipramine has reportedly caused hyperprolactinemia in 60% of men and 87.5% of women.41
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The mechanisms by which these drugs increased PRL levels are uncertain, and this class of drugs likely facilitates several possible stimulatory pathways.
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In clinical trials of fluoxetine in 5920 patients, the PRL level was not measured, but galactorrhea was reported in only 0.07%, breast enlargement in 0.08%, and breast pain in 0.25% of subjects.42
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In a second study of 8 patients treated with paroxetine, PRL levels increased similarly, from mean ± SD levels of 5.8+3.0 µg/L to 8.4+4.4 µg/L,48
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Two drugs commonly used to increase gastrointestinal motility and stomach emptying in patients with gastroparesis diabeticorum, metoclopramide and domperidone, are dopamine receptor blockers. These drugs cause hyperprolactinemia in more than 50% of patients and commonly cause symptoms of amenorrhea and galactorrhea in women and impotence in men.83–85
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10 of 18 patients who were infected (opportunistic) were found to be hyperprolactinemic, whereas only 2 of 28 noninfected patients were hyperprolactinemic. Also, of 20 patients taking protease inhibitors, 4 (20%) were hyperprolactinemic; of 26 patients not taking protease inhibitors, 6 (23%) were hyperprolactinemic.94
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effect on PRL levels of estrogen replacement therapy after oophorectomy or at menopause with varying dosages of up to 1.25 mg of conjugated estrogens or 50 µg of estradiol daily. Therefore, in a given patient who is found to be hyperprolactinemic while taking oral contraceptives, it is uncertain whether the estrogen plays a role.
Other Medications
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but no subsequent studies have confirmed this finding.
Assessment of A Patient with Suspected Medication-Induced Hyperprolactinemia
In patients taking medications known to cause hyperprolactinemia, it is critical to establish that the medication is the cause (Table 3). Minor PRL elevations in the range seen with most of these medications also may be seen in patients with PRL-secreting microadenomas and, more importantly, in those with large mass lesions that can cause PRL elevation due to hypothalamic/stalk dysfunction, resulting in a decrease in dopamine reaching the lactotrophs, ie, a disinhibition of PRL secretion.109–111
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Diagnosis
Confirm that the medication is the cause of hyperprolactinemia
Discontinue medication for 3 or 4 days
Image hypothalamic/pituitary area with use of magnetic resonance imaging or computed tomography
Treatment
Switch to an alternative medication that does not cause hyperprolactinemiaor
Treat the problem caused by the hyperprolactinemia
Estrogen or progesterone
Testosterone
Bisphosphonate or
Add dopamine agonist cautiously (rarely necessary)
TABLE 3
Treatment Strategies for the Patient With Symptomatic Medication-Induced Hyperprolactinemia
With psychoactive medications, this must be done cautiously in consultation with the patient's psychiatrist.
Open table in a new tab
A careful patient history may elicit symptoms or documentation of hyperprolactinemia coinciding in time with initiation of medication. However, usually such a history is not forthcoming, and in these circumstances, the simplest approach is to have the patient discontinue the medication because PRL levels generally return to normal within 3 to 4 days, at least with phenothiazines.9
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Meltzer, HY ∙ Fang, VS
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Some antidepressants may have more prolonged action, but there are no data showing when their PRL-increasing effects wear off. In all patients taking psychoactive agents, discontinuation of the medication must be done in careful consultation with the psychiatrist or other prescribing physician to avoid sudden exacerbation of any underlying psychiatric disorder.
If a patient cannot stop taking a given medication for even a short time, an alternative drug that does not cause hyperprolactinemia can be substituted for several days to determine whether PRL levels decrease (discussed subsequently). Again, such a substitution must be done extremely carefully with supervision of the patient's psychiatrist when the patient is taking a psychoactive medication. If such a substitution is not feasible, then the patient should undergo magnetic resonance imaging (MRI) of the hypothalamic/pituitary area or, if MRI is unavailable, computed tomography.112
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Naidich, MJ ∙ Russell, EJ
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It is much more important to exclude large mass lesions in such a patient rather than to establish the presence of a microadenoma.
Patient Treatment for Medication-Induced Hyperprolactinemia
The first step in treatment is to determine whether the patient has symptoms related to the hyperprolactinemia. In a woman with normal, regular menses, if nonbothersome galactorrhea is the only reason to perform the PRL measurement, simple reassurance may be all that is needed. In contrast, if a patient has highly symptomatic hyperprolactinemia that, along with amenorrhea, causes decreased libido, bothersome galactorrhea, impotence, or osteoporosis, then a more active treatment strategy is necessary.
Assuming that the patient needs to continue taking medication for an underlying disorder, switching to another drug in the same class that does not cause hyperprolactinemia is the easiest way to correct the problem. Thus, for a patient with antipsychotic-induced hyperprolactinemia, switching to drugs such as olanzapine, clozapine, or quetiapine may eliminate hyperprolactinemia.17,27,113
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Similarly, for a patient with antidepressant-induced hyperprolactinemia, switching to an alternative antidepressant may be successful. Again, all such medication changes must be done under the supervision of the patient's psychiatrist, and consideration must be given to other potential adverse effects of these alternative medications. There are many antihypertensive agents; therefore, switching a patient from verapamil to an alternative generally should not be a problem. However, for a patient with gastroparesis, no good alternatives to metoclopramide exist currently in the United States, although erythromycin has been used in some studies. Because of its association with cardiac dysrhythmias, cisapride is no longer available in the United States.
If a patient has symptomatic hyperprolactinemia and cannot be switched from his or her medication, other treatments could be considered. If the major concern is decreased estrogen or testosterone levels, then simple substitution with estrogen or testosterone can be done. If the concern is osteoporosis, a bisphosphonate could be used.
The most difficult treatment modality is to treat a patient with a dopamine agonist while continuing current medication. This modality has been used primarily in small numbers of patients with antipsychotic-induced hyperprolactinemia; there is a small risk of the dopamine agonist exacerbating the underlying psychosis, and the dopamine agonist is not always successful in normalizing PRL levels. In 1 series of 7 patients with hyperprolactinemia and galactorrhea from various antipsychotic medications who were treated with bromocriptine, 2 achieved normal PRL levels, 4 experienced considerable decreases in PRL levels, and galactorrhea improved in all.114
114.
Beumont, P ∙ Bruwer, J ∙ Pimstone, B ...
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In another series of 9 patients with hyperprolactinemia from thioridazine, 4 achieved normal PRL levels with bromocriptine with no worsening of psychiatric status.115
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In a third series of 6 women with hyperprolactinemia and amenorrhea or oligomenorrhea from various antipsychotics, 4 achieved normal PRL levels and experienced menstrual irregularity with bromocriptine, but in 1 of these 4, mental status worsened.116
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In a more recent series of 4 patients with risperidone-induced hyperprolactinemia, bromocriptine or cabergoline reduced PRL levels and alleviated hypogonadism in 3, with no worsening of the underlying psychosis.117
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In another recent series, 11 of 19 patients with symptomatic, risperidoneinduced hyperprolactinemia experienced remission of clinical signs and normalization of PRL levels without exacerbation of the underlying psychopathology when treated with cabergoline.118
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However, other case reports document the worsening of mental status with bromo-criptine,119
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so this complication must always be looked for carefully when adding a dopamine agonist to antipsychotic therapy.
Conclusions
Hyperprolactinemia caused by medications is commonly symptomatic, causing galactorrhea, menstrual disturbance, and impotence. It is important to ensure that hyperprolactinemia in an individual patient is due to medication and not to a structural lesion in the hypothalamic/pituitary area. A careful patient history may elicit symptoms or documentation of hyperprolactinemia coinciding in time with initiation of a medication; however, such a history usually is not forthcoming. In such circumstances, either the medication should be stopped temporarily to determine whether PRL levels return to normal or the drug should be switched to one that does not cause hyperprolactinemia. In cases of suspected psychoactive medication–induced hyperprolactinemia, changing or stopping medications must be done in consultation with the patient's psychiatrist. If this cannot be done, MRI or computed tomography will exclude a structural lesion. Treatment is needed only if the hyperprolactinemia is symptomatic. Treatment strategies include switching to an alternative medication that does not cause hyperprolactinemia, using estrogen or testosterone replacement, or cautiously adding a dopamine agonist.
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2022 USAJMO Problems/Problem 4
Problem
Let be a rhombus, and let and be points such that lies inside the rhombus, lies outside the rhombus, and . Prove that there exist points and on lines and such that is also a rhombus.
Solution
(Image of the solution is here )
Let's draw () perpendicular bisector of . Let be intersections of with and , respectively. is a kite. Let mid-point of . Let mid-point of (and also is mid-point of ). are on the line .
, , and so (side-side-side). By spiral similarity, . Hence, we get
Similarly, , , and so (side-side-side). From spiral similarity, . Thus,
If we can show that , then the kite will be a rhombus.
By spiral similarities, and . Then, .
. Then, . Also, in the right triangles and , . Therefore,
and we get .
(Lokman GÖKÇE)
See Also
2022 USAJMO (Problems • Resources)
Preceded by
Problem 3Followed by
Problem 5
1•2•3•4•5•6
All USAJMO Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
Retrieved from "
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https://www.ck12.org/flexi/cbse-math/higher-order-metric-conversions/convert-5-minutes-to-seconds/
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Convert 5 minutes to seconds. - Steps | CK-12 Foundation
All Subjects
CBSE Math
Higher Order Metric Conversions
Question
Convert 5 minutes to seconds.
Flexi Says:5 minutes is 300 seconds.
Time conversion involves understanding the relationship between hours and minutes. Since 1 minute is equivalent to 60 seconds, converting minutes to seconds is a straightforward multiplication process.
Steps to convert minutes to seconds
Step 1: Identify the given time: The time provided is 5 minutes.
Step 2: Convert the minutes to seconds: Multiply the number of minutes by 60 to get the equivalent seconds:
5 × 60 = 300 seconds.
Final Answer: 5 minutes is 300 seconds.
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Temperature-sensitive foaming agent developed for smart foam drainage technology - PMC
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RSC Adv
. 2022 Aug 17;12(36):23447–23453. doi: 10.1039/d2ra04034d
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Temperature-sensitive foaming agent developed for smart foam drainage technology
Wenfeng Jia
Wenfeng Jia
1 State Key Laboratory of Petroleum Resources and Prospecting, China University of Petroleum, Beijing 102249, China, Email: jiawf@cup.edu.cn, Email: xianchang@cup.ecu.cn
2 Unconventional Oil and Gas Institute, China University of Petroleum, Beijing, 102249, PR China
Find articles by Wenfeng Jia
1,2,✉, Chenggang Xian
Chenggang Xian
1 State Key Laboratory of Petroleum Resources and Prospecting, China University of Petroleum, Beijing 102249, China, Email: jiawf@cup.edu.cn, Email: xianchang@cup.ecu.cn
2 Unconventional Oil and Gas Institute, China University of Petroleum, Beijing, 102249, PR China
Find articles by Chenggang Xian
1,2,✉, Junwen Wu
Junwen Wu
3 Sinopec Research Institute of Petroleum Exploration and Development, Beijing 102206, PR China, Email: wujunwen@iccas.ac.cn
Find articles by Junwen Wu
3,✉
Author information
Article notes
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1 State Key Laboratory of Petroleum Resources and Prospecting, China University of Petroleum, Beijing 102249, China, Email: jiawf@cup.edu.cn, Email: xianchang@cup.ecu.cn
2 Unconventional Oil and Gas Institute, China University of Petroleum, Beijing, 102249, PR China
3 Sinopec Research Institute of Petroleum Exploration and Development, Beijing 102206, PR China, Email: wujunwen@iccas.ac.cn
✉
Corresponding author.
Received 2022 Jun 30; Accepted 2022 Aug 1; Collection date 2022 Aug 16.
This journal is © The Royal Society of Chemistry
PMC Copyright notice
PMCID: PMC9382362 PMID: 36090426
This article has been corrected. See RSC Adv. 2022 Sep 21;12(41):26657.
Abstract
The conventional foam drainage technology needs to be defoamed, which is not convenient for its popularization and application. In view of this problem, from the point of molecular design, a temperature-responsive surfactant was designed and synthesized. In the synthetic process, polyoxyethylene alkyl ether carboxylic acid, diethanolamine and sodium chloroacetate were used as raw materials. First, polyoxyethylene alkyl ether carboxylic acid reacted with diethanolamine to generate a tertiary amine with hydroxyl catalyzed by sulfoxide chloride, and the intermediate product then reacted with sodium chloroacetate by the quaternary amine reaction to afford the target temperature-responsive surfactant. The foaming agent can achieve conformational transformation in the temperature range of 20 °C to 120 °C, resulting in the structural change of the self-assembly and regulating the stability of the foam, which makes the formed foam burst rapidly at low temperatures and be super-stable at high temperatures. The indoor evaluations show that the foaming height of the foaming agent is basically unchanged at the same temperature after 4 temperature-changing cycles, and the temperature-controlled defoaming rate reaches 90%, indicating that it has the intelligent temperature response switching performance of “high-temperature defoaming, low-temperature defoaming”. Its preparation process is simple, low-cost, and environmentally friendly. It is expected to be popularized and applied in the field of gas fields, expand the application scope of foam drainage technology, reduce the cost of foam drainage, and help the efficient development of gas fields.
A temperature-sensitive surfactant with Gemini structure, possessing intelligent temperature response switching performance, was synthesized for smart foam drainage technology.
1. Introduction
In the later stage of natural gas exploitation, due to the low pressure of the gas well and the decrease in drainage capacity, the large amount of liquid produced by the gas well leads to a decrease in exploitation efficiency. Foam drainage technology is widely used in natural gas recovery because of its simple operation and excellent drainage effect.1 In this process, the foaming agent is added to the bottom of the well, so that the bottom hole fluid is stirred by natural gas and fully mixed with the foaming agent to form a large amount of foam, reducing the friction loss and gravity gradient in the tubing during the self-flowing stage to effectively reduce the bottom hole back pressure and make the liquid be continuously lifted.2 The regeneration capacity of the foam is very strong, and when its aqueous solution is carried to the ground pipelines and separation equipment, it is repeatedly agitated, resulting in foam accumulating in the separator. When the surfactant is excessive or the foam produced is too stable, this phenomenon is particularly serious. A large amount of foam will be brought to the gathering and transportation pipeline, causing blockage and resulting in an increase in the collection and transportation pressure.3 Therefore, it is necessary to inject a defoamer at the entrance of the separator to achieve defoaming, inhibit the regeneration of foam and facilitate the separation of gas and water.4 However, the addition of defoaming agents will limit the recycling of foaming agents, and the introduction of a defoaming device will increase the production cost.5
Foam drainage technology has entered the personalized application stage, endowing foam with intelligent responsiveness, making it foam at wellbore temperature and defoam at ground temperature, removing the defoaming process of conventional foam drainage technology and simplifying the foam drainage process. It is of great research value to reduce the cost and expand the application range of foam drainage gas production.
Temperature-sensitive foam is foam whose stability can be controlled by temperature.6,7 The foaming agent is the most important and basic component of the foam system, so the construction of a temperature-sensitive foaming agent is the key to building a temperature-sensitive foam system. In 2011, the Fameau group first carried out research on temperature-sensitive foam, and an ultra-stable temperature-responsive foam was first reported by the co-assembly of 12-hydroxystearic acid (12-HSA) and ethanolamine or hexanolamine salt, and reversible change could be achieved in the temperature range of 20 °C to 50 °C.8–11 However, the upper limit of the response temperature of this foam cannot meet the requirements of the gas well drainage temperature. Raghavan's group investigated the viscosity of the C 22 cationic surfactant, erucic acid dihydroxyethyl methyl ammonium chloride (EHAC), in saline solutions using steady-state and dynamic rheological methods.12,13 The surfactant (EHAC) self-assembles into large worm-like micelles upon heating (∼90 °C), generating an unusually strong viscoelasticity (viscosity > 10 Pa s). The upper limit of the response temperature is high, but its foaming and foaming stabilization ability has not been systematically studied.
Novel smart foam is currently the main research direction, and a variety of ideas have been proposed for the development of temperature-sensitive foam.14–16 Regarding the development of temperature-sensitive foaming agents, the basic principle is regulating the breaking and recombination of self-assemblies in the solution with temperature changes.17–19
From the perspective of molecular design, the two surfactants are connected by linking groups, which reduces the electrostatic repulsion between the head groups and enables them to align closely at the interface,20 making it easier to self-assemble micelles. Additionally, the temperature-sensitive ethoxy chain segment is introduced into Gemini surfactants;21 with the increase in temperature, the ethoxy chain dehydrates, the hydrophilic head group area decreases, and the hydrophobic chain segment length increases, promoting micelle growth. Based on this idea, a novel surfactant has been synthesized to achieve conformational transformation in the temperature range of 20 °C to 120 °C, which leads to changing the assembly structure and adjusting the stability of the foam. Temperature-sensitive foam bursts quickly at low temperatures, and has super stability at high temperatures, which is of great significance to simplifying the foam defoaming process and effectively improving the recovery of gas reservoirs.
2. Experimental section
2.1. Chemicals
Polyoxyethylene alkyl ether carboxylic acid, diethanolamine, acetone, sulfoxide chloride (SOCl 2), ethanol (95%), sodium chloroacetate, and ether were purchased from Beijing Chemical Reagent Plant, China; these chemicals were of analytical grade and were not purified further before use. The water used in the tests was deionized water.
All the glassware was cleaned by scouring powder and then rinsed with ultrapure water before the experiments.
2.2. Synthesis of the temperature-sensitive Gemini surfactant GACB
The detailed preparation method of the temperature-sensitive Gemini surfactant molecule is as follows:
In a typical procedure, polyoxyethylene alkyl ether carboxylic acid (30.0 g) and diethanolamine (6.2 g) were dissolved in acetone (100 mL) at 1 : 1.5 (mol mol−1). SOCl 2 (5 mL) as a catalyst was added dropwise at low temperature using a constant pressure drip funnel. After the completion of dropping, the mixture was heated to 60 °C for the reaction with circulating water cooling. The reaction time was 6 h, and the resulting liquid was distilled under reduced pressure to remove the solvent and unreacted diethanolamine to afford the intermediate.
The intermediate product (25 g) and sodium chloroacetate (7 g) were dissolved in a mixture of ethanol and water (5 : 1, v/v), and the reaction was carried out by heating the system to 80 °C in a round-bottom flask and cooled by circulating water. The reaction time was 12 h. After the end of the reaction, the solvent was removed by reduced pressure distillation, and the mixture was washed and filtered with ethanol. Finally, the remaining product was recrystallized with a mixed solution of ethanol and ether to afford the target product. The synthetic route of the temperature-sensitive Gemini surfactant GACB is shown in Fig. 1.
Fig. 1. Synthetic route of the temperature-sensitive Gemini surfactant GACB.
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2.3. Experimental instruments
The 1 H NMR spectra were recorded on a Bruker Advance III 400WB NMR. The surface tension of the surfactant solutions was studied using a KRÜSS K100 surface tensiometer. A Ross–Miles foaming device (Model 2151) and a high-temperature and high-pressure visualization foam instrument (Jiangsu Baobo Machinery Manufacturing Co., LTD) were used to investigate the foaming properties and temperature-sensitive characteristics. The glassware used in the experiment was washed with detergent and deionized water.
2.4. Molecular structure characterization
A small amount of the sample was dissolved in D 2 O, injected into a glass NMR tube with a diameter of 5 mm, and the 1 H NMR spectrum of the product at room temperature was recorded on a Bruker AV III HD (400 MHz, magnetic field strength: 9.4 T), and the map data were peaked and integrated with the MestReNova software to determine the product structure characteristics.
2.5. Equilibrium surface tension
The equilibrium surface tension was determined by the platinum ring method using a KRÜSS K100 surface tensiometer. The maximum uncertainty on γ values is ±0.05 mN m−1. All the measurements were performed at 25.0 ± 0.1 °C and repeated at least three times to obtain an average value.
2.6. Foam performance evaluation methods
The Ross–Miles test (suitable for evaluation at 20–90 °C) was applied to determine the foam properties, including the foaming and stability properties. The experimental procedures are as follows:
(1) Put the cylindrical glass container (50 mm internal diameter, 1100 mm height) in a water bath at a constant temperature of 80.0 ± 0.1 °C. (2) Wash the Ross–Miles foaming device with ultrapure water, and then wash it two to three times with the fluid to be tested. (3) Put 50 cm 3 of the solution to be tested into the Ross–Miles foaming device and heat to 80.0 ± 0.1 °C. (4) Inject 200 cm 3 of the preheated solution into the dropping funnel. (5) Put the dropping funnel onto the pipe rack and make it vertical to the cylindrical glass container. (6) Open the dropping funnel piston and keep the solution flowing into the center of the calibration tube. When the solution flows out of the dropping funnel, start the stopwatch immediately and record the maximum foaming height marked as H max, which is used to evaluate the foaming property, and the time when the foam height reduces to half of the original height, which is called half-life period. The stability of foam is evaluated by the half-life period, marked as t 1/2.
The high-temperature and high-pressure visualization test (suitable for evaluation at 20–160 °C) is as follows: (1) wash the high-temperature and high-pressure visualization device with ultrapure water, and then wash it two to three times with the fluid to be tested. (2) Set the test temperature of the high-temperature and high-pressure foam device and wait for the temperature to rise to the operating temperature. (3) Pour 100 mL of the test solution into a graduated cylinder and pump it into the testing kettle. (4) Turn on the stirring device at a constant speed of 1000 rpm for 3 min to generate foam. (5) After the end of the stirring, record the maximum height of the foam H max and the half-life period t 1/2.
The experiments were repeated at least three times, and the average value was collected.
3. Results and discussion
3.1. Molecular structure characterization of GACB
The 1 H NMR spectra of the temperature-sensitive Gemini surfactant GACB dissolved in D 2 O are shown in Fig. 2. The peaks at δ 0.83 ppm and 1.24 ppm came from the CH 3 and CH 2 groups at the tail of the surfactant [CH 3–(CH 2)n–], respectively. The chemical shift of methylene [–CH 2–CH 2–(O–CH 2–CH 2)n–] linked to the ethoxy group shifts to the left, with characteristic peaks occurring at δ 1.52 ppm and δ 3.17 to 3.20 ppm. Characteristic peaks of methylene (–O–CH 2–CH 2–O–CH 2–CH 2–CH 2–) adjacent to quaternary ammonium groups [–(CH 2)2–N–(CH 2–C O–O)2] appear at δ 3.61 to 3.64 ppm. At δ 3.81 to 3.82 ppm is the characteristic peak of the methylene group (–O C–CH 2–) connected to the carbonyl group. The peak at δ 3.81 to 3.82 ppm is caused by methylene (–C O–O–CH 2–) connected to the ester group. The integration area of each peak (a–g) and the ratio of different hydrogen atoms in the sample are 3 : 18 : 2 : 2 : 42 : 4 : 1. The results show that the structure of the product coincided with the molecular structure of the target compound.
Fig. 2. 1 H NMR spectra of GACB dissolved in D 2 O.
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Table 1 shows the weight of the raw materials and products and yield at each step in the above synthetic process of the temperature-sensitive Gemini surfactant GACB. The yield of the single-step reaction reached more than 80%, and the total yield of GACB was 79.12%.
Synthetic yield of the temperature-sensitive Gemini surfactant GACB.
| Reaction process | Raw material weight (g) | Product weight (g) | Yield (%) |
--- --- |
| | 30.00 | 28.75 | 95.83 |
| | 25.00 | 20.64 | 82.56 |
| Total yield | 79.12 |
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3.2. Performance evaluation of GACB
3.2.1. Surface tension
The surface tension of GACB solutions as a function of concentration was studied using a KRÜSS K100 surface tensiometer to determine the critical micellar concentrations (cmc). As shown in Fig. 3, when the concentration of GACB reached about 0.0018% (w/w), the variation amplitude of surface tension decreased. The minimum surface tension of the GACB solution is 33.84 mN m−1 in the present test temperature and concentration range. It can be seen clearly that the changes in surface tension of the GACB solution with different concentrations can be divided into two parts: the slope area and platform area. The intersection of these two areas is the critical micellar concentration (cmc) of the system. Micelles begin to form in solutions above this concentration, and so the changing amplitude of surface tension begins to decrease. At the test temperature of 25 °C, the cmc of GACB is about 0.0015%. Surfactant solutions with a concentration of 0.20% or more would be used in the subsequent tests.
Fig. 3. Surface tension of the GACB solution as a function of concentration (T = 25 °C).
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3.2.2. Foaming performance under different temperatures
A Ross–Miles foaming device was used to investigate the foam properties at 20, 40, 60, and 80 °C, including the foaming and stability properties. As shown in Fig. 4, the foam height of GACB at 60 °C reached 340 mm and the half-life period was over 6 min. The results show that the foam formed by GACB still processed good foaming properties and stability at higher temperatures, and the foaming property and foam stability can be enhanced with the increase in temperature.
Fig. 4. Half-life period t 1/2 and maximum foaming height H max of foams produced with 0.2% GACB solution at different temperatures.
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The high-temperature and high-pressure visualization test was used to determine the foam properties at 20–120 °C with 0.2% GACB solution. As shown in Fig. 5, when the temperature was above 100 °C, the foaming performance increased while the foam stability decreased. Compared with that at 40 °C, the maximum foam height of GACB increased by about 1 time at 120 °C, and the half-life of foam decreased from 40 min to 10–15 min. This is mainly due to the following changes when the temperature rises: (1) the surface viscosity of the liquid film decreases, and the liquid film drainage rate increases; (2) the molecular movement in the foam increases, the liquid film becomes thinner, and the “gas channeling” increases; (3) the liquid vapor pressure increases, and the rapid evaporation of the liquid film makes it thin. However, the foaming ability of the surfactant is still high at 120 °C, which proves that the foaming agent selected in this paper has a good temperature response and the liquid film has high strength and elasticity at high temperatures and can resist the dispersion caused by the intense Brownian motion of molecules at high temperatures and the foam coalescence and disproportionation caused by the thinning of the foam liquid film.
Fig. 5. The maximum height of the foam H max and the half-life period t 1/2 of 0.2% GACB solution under different temperatures determined by the high-temperature and high-pressure visualization test.
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As seen by the observation window of the high-temperature and high-pressure foam apparatus, it was found that the water phase started to accelerate evaporation or even boil under high temperature, as shown in Fig. 6. The evaporated gas can continuously stir up the solution to form new foam and the rapid loss of the liquid phase also leads to a decrease in the foam volume, which was the reason that the foam performance changed in the high-temperature environment.
Fig. 6. Images of the GACB foam morphology under different temperatures: (a) 20 °C; (b) 40 °C; (c) 60 °C; (d) 80 °C; (e) 100 °C; (f) 120 °C.
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3.3. Temperature response performance
In order to further verify the temperature response performance of the foaming agent, the high-temperature and high-pressure foam apparatus was used to investigate the same foaming agent at various temperatures from 25 to 120 °C. The heating rate was about 0.5 °C min−1, and the cooling rate was about 0.1 °C min−1. The change of the foam height after every heating or cooling was recorded, and the test results are shown in Fig. 7. During the heating process, the foaming height of the temperature-sensitive foaming agent increased significantly with the increase in temperature. During the cooling process, the foaming height decreased gradually. The foam at 120 °C even exceeded the imaging range of the foam apparatus window.
Fig. 7. Images of the foaming height of 0.2% GACB solution at a single heating procedure (from 25 to 120 °C) and cooling procedure (from 120 °C to 25 °C) at atmospheric pressure.
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The system was further heated and cooled, repeated many times, and the foaming height after the temperature change was recorded, as shown in Fig. 8. The foaming height of GACB at room temperature was less than 30 mm, and after heating, the foaming height of the foaming agent reached more than 40 mm. After four temperature-changing cycles, the foaming height at the same temperature was basically unchanged. Therefore, it is proved that GACB has temperature response performance and can realize multiple cycles of foaming height variation with temperature.
Fig. 8. Foaming height of 0.2% GACB solution after many heating (120 °C) and cooling (25 °C) procedures at atmospheric pressure.
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The temperature response mechanism was explained as follows: hydrogen bonds are formed in the low-temperature environment between the ethoxy groups and water molecules in GACB, which enhances its water solubility and shows the characteristics of hydrophilic groups. The ethoxy groups are dehydrated because the hydrogen bonds break with the increase in temperature, which results in the ethoxy groups showing the characteristics of hydrophobic groups. According to the theory of stacking parameters, this change can promote the growth of micelles and make the original spherical micelles gradually grow into rod micelles with higher viscosity. Gemini surfactants consist of two single-chain surfactants through the linking groups in the molecules, which greatly reduces the electrostatic repulsion between the head groups, so that molecules can be arranged more closely at the interface, lowering the surface tension and critical micellar concentration, and are more easily self-assembled to form micelles in the solution, thus counteracting the negative effects of high temperature on the foam system. The change of micellar morphology caused by ethoxy groups with temperature, or the micelle growth due to the Gemini structure, will make the liquid drainage rate of the foam system decrease significantly at high temperatures, so that the foam will be more stable at high temperature and burst at low temperature, giving the foam temperature response performance.
4. Conclusions
The current conventional foam drainage technology needs defoaming treatment, which is not convenient for its popularization and application. Based on ethoxy groups with temperature stimulus response, a temperature-responsive Gemini surfactant molecule GACB was designed and synthesized. Conformational transitions of the surfactant result in a change of the self-assembly structure in the temperature range of 20 to 120 °C, regulating the stability of the foam, rapidly breaking the foam at low temperature, and affording superb stability at high temperature. The indoor evaluations show that the foaming height of the foaming agent at the same temperature is basically unchanged after four temperature-changing cycles, indicating that it has the intelligent temperature response performance of “high-temperature foam stabilization and low-temperature foam elimination”. It is of great significance to simplify the foam drainage procedure, expand the application scope of foam drainage technology, reduce the foam drainage cost and effectively improve the recovery of gas reservoirs.
Data availability
The data that support this study will be shared upon reasonable request to the corresponding author.
Conflicts of interest
The authors declare no conflict of interest.
Supplementary Material
Acknowledgments
This study was supported by the Strategic Cooperation Technology Projects of CNPC and CUPB (ZLZX2020-01).
References
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Data Availability Statement
The data that support this study will be shared upon reasonable request to the corresponding author.
Articles from RSC Advances are provided here courtesy of Royal Society of Chemistry
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Abstract
1. Introduction
2. Experimental section
3. Results and discussion
4. Conclusions
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Text solutionVerified
Concepts:
Parabola, Vertex, Quadratic equations
Explanation:
To find the vertex of the parabola given by the equation y 2=4 a x, we can analyze the standard form of a parabola. The equation y 2=4 a x represents a parabola that opens to the right. In this form, the vertex is located at the point (0,0). This is because the equation is centered at the origin and does not have any additional terms that would shift it away from this point.
Step by Step Solution:
Step 1
Identify the standard form of the parabola, which is y 2=4 a x. This indicates that the parabola opens to the right.
Step 2
Recognize that the vertex of the parabola in this form is at the point (0,0).
Final Answer:
The vertex of the parabola y 2=4 a x is at the point (0, 0).
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https://math.libretexts.org/Courses/Fresno_City_College/Math_3A%3A_College_Algebra_-_Fresno_City_College/01%3A_Solving_Equations/1.02%3A_Solving_Absolute_Value_Equations
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≤
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Write the equivalent equations.|x|=8x=−8 or x=8x=±8
|y|=−6No solution
Write the equivalent equations.Since −0=0,|z|=0z=−0 or z=0z=0
±2
±11
x=4, x=−23
x=−1, x=52
2|x−7|+5=9
2|x−7|=4
|x−7|=2
x−7=−2
x−7=2
x=5
x=9
x=8, x=0
x=8, x=2
Isolate the absolute value term.An absolute value cannot be negative.|23x−4|=−8|23x−4|=−8No solution
Write the equivalent equations.Solve each equation.Check.We leave the check to you.5x−1=−(2x+3)5x−1=−2x−37x−1=−37x=−2x=−27|5x−1|=|2x+3|ororor5x−1=2x+33x−1=33x=4x=43x=43
x=−25, x=52
x=3,x=19
|5x−6|≤4
−4≤5x−6≤4
2≤5x≤10
25≤x≤2
[25,2]
|x|>4
x<−4
x>4
(−inf,−4)∪(4,inf)
|2x−3|≥5
2x−3≤−5
2x−3≥5
2x≤−2
2x≥8
x≤−1
x≥4
(−inf,−1]∪[4,inf)
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1.2: Solving Absolute Value Equations
Last updated
: Jul 24, 2023
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1.1: Solving Linear Equations
1.3: Solving Quadratic Equations
Page ID
: 134532
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
By the end of this section, you will be able to:
Solve absolute value equations
Solve absolute value inequalities with “less than”
Solve absolute value inequalities with “greater than”
Solve applications with absolute value
Before you get started, take this readiness quiz.
Evaluate: −|7|−|7|.
If you missed this problem, review [link].
Fill in <,>,<,>,<,>,<,>, or == for each of the following pairs of numbers.
ⓐ |−8|___−|−8||−8|___−|−8| ⓑ 12___−|−12|12___−|−12| ⓒ |−6|___−6|−6|___−6 ⓓ −(−15)___−|−15|−(−15)___−|−15|
If you missed this problem, review [link].
Simplify: 14−2|8−3(4−1)|14−2|8−3(4−1)|.
If you missed this problem, review [link].
Solve Absolute Value Equations
As we prepare to solve absolute value equations, we review our definition of absolute value.
ABSOLUTE VALUE
The absolute value of a number is its distance from zero on the number line.
The absolute value of a number n is written as |n||n| and |n|≥0|n|≥0 for all numbers.
Absolute values are always greater than or equal to zero.
We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example:
−5−5 is 5 units away from 0, so |−5|=5|−5|=5.
55 is 5 units away from 0, so |5|=5|5|=5.
Figure 1.2.11.2.1 illustrates this idea.
For the equation |x|=5,|x|=5, we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and −5−5 are five units from zero on the number line. They are the solutions to the equation.
If|x|=5thenx=−5 or x=5Ifthen|x|=5x=−5 or x=5
The solution can be simplified to a single statement by writing x=±5x=±5. This is read, “x is equal to positive or negative 5”.
We can generalize this to the following property for absolute value equations.
ABSOLUTE VALUE EQUATIONS
For any algebraic expression, u, and any positive real number, a,
if|u|=athenu=−a or u=a
ifthen|u|=au=−a or u=a
Remember that an absolute value cannot be a negative number.
Example 1.2.11.2.1
Solve:
|x|=8|x|=8
|y|=−6|y|=−6
|z|=0|z|=0
Solution a
: |x|=8Write the equivalent equations.x=−8 or x=8x=±8
Solution b
: |y|=−6No solution
Since an absolute value is always positive, there are no solutions to this equation.
Solution c
: |z|=0Write the equivalent equations.z=−0 or z=0Since −0=0,z=0
Both equations tell us that z=0z=0 and so there is only one solution.
EXERCISE 1.2.21.2.2
Solve:
|x|=2|x|=2
|y|=−4|y|=−4
|z|=0|z|=0
Answer a
: ±2
Answer b
: no solution
Answer c
: 0
EXERCISE 1.2.31.2.3
Solve:
|x|=11|x|=11
|y|=−5|y|=−5
|z|=0|z|=0
Answer a
: ±11
Answer b
: no solution
Answer c
: 0
To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.
How to Solve Absolute Value Equations
Example 1.2.41.2.4
Solve |5x−4|−3=8|5x−4|−3=8.
Solution
EXERCISE 1.2.51.2.5
Solve: |3x−5|−1=6|3x−5|−1=6.
Answer
: x=4, x=−23
EXERCISE 1.2.61.2.6
Solve: |4x−3|−5=2|4x−3|−5=2.
Answer
: x=−1, x=52
The steps for solving an absolute value equation are summarized here.
SOLVE ABSOLUTE VALUE EQUATIONS.
Isolate the absolute value expression.
Write the equivalent equations.
Solve each equation.
Check each solution.
Example 1.2.71.2.7
Solve 2|x−7|+5=92|x−7|+5=9.
Solution
: | | |
--- |
| | 2|x−7|+5=9 |
| Isolate the absolute value expression. | 2|x−7|=4 |
| | |x−7|=2 |
| Write the equivalent equations. | x−7=−2 or x−7=2 | |
| Solve each equation. | x=5 or x=9 | |
| Check: | |
Exercise 1.2.81.2.8
Solve: 3|x−4|−4=83|x−4|−4=8.
Answer
: x=8, x=0
Exercise 1.2.91.2.9
Solve: 2|x−5|+3=92|x−5|+3=9.
Answer
: x=8, x=2
Remember, an absolute value is always positive!
Example 1.2.101.2.10
Solve: |23x−4|+11=3|23x−4|+11=3.
Solution
: |23x−4|=−8Isolate the absolute value term.|23x−4|=−8An absolute value cannot be negative.No solution
Exercise 1.2.111.2.11
Solve: |34x−5|+9=4|34x−5|+9=4.
Answer
: No solution
Exercise 1.2.121.2.12
Solve: |56x+3|+8=6|56x+3|+8=6.
Answer
: No solution
Some of our absolute value equations could be of the form |u|=|v||u|=|v| where u and v are algebraic expressions. For example, |x−3|=|2x+1||x−3|=|2x+1|.
How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, u, and a positive real number, a, if |u|=a|u|=a, then u=−au=−a or u=au=a.
This tell us that
if|u|=|v|then|u|=vor|u|=−vand sou=v or u=−voru=−v or u=−(−v)ifthenand so|u|=|v||u|=vu=v or u=−voror|u|=−vu=−v or u=−(−v)
This leads us to the following property for equations with two absolute values.
EQUATIONS WITH TWO ABSOLUTE VALUES
For any algebraic expressions, u and v,
if|u|=|v|thenu=−v or u=v
ifthen|u|=|v|u=−v or u=v
When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.
Example 1.2.131.2.13
Solve: |5x−1|=|2x+3||5x−1|=|2x+3|.
Solution
: |5x−1|=|2x+3|Write the equivalent equations.5x−1=−(2x+3)or5x−1=2x+35x−1=−2x−3or3x−1=3Solve each equation.7x−1=−33x=47x=−2x=43x=−27orx=43Check.We leave the check to you.
Exercise 1.2.141.2.14
Solve: |7x−3|=|3x+7||7x−3|=|3x+7|.
Answer
: x=−25, x=52
Exercise 1.2.151.2.15
Solve: |6x−5|=|3x+4||6x−5|=|3x+4|.
Answer
: x=3,x=19
Solve Absolute Value Inequalities with “Less Than”
Let’s look now at what happens when we have an absolute value inequality. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work. Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation |x|=5|x|=5, we saw that both 5 and −5−5 are five units from zero on the number line. They are the solutions to the equation.
|x|=5x=−5orx=5
x=−5|x|=5orx=5
What about the inequality |x|≤5|x|≤5? Where are the numbers whose distance is less than or equal to 5? We know −5−5 and 5 are both five units from zero. All the numbers between −5−5 and 5 are less than five units from zero (Figure 1.2.21.2.2).
In a more general way, we can see that if |u|≤a|u|≤a, then −a≤u≤a−a≤u≤a (Figure 1.2.31.2.3).
Figure 1.2.31.2.3.
This result is summarized here.
ABSOLUTE VALUE INEQUALITIES WITH << OR ≤≤
For any algebraic expression, u, and any positive real number, a,
if|u|<a,then −a<u<aif|u|≤a,then −a≤u≤a
if|u|<a,then −a<u<aif|u|≤a,then −a≤u≤a
After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution.
Example 1.2.161.2.16
Solve |x|<7|x|<7. Graph the solution and write the solution in interval notation.
Solution
: | | |
--- |
| | |
| Write the equivalent inequality. | |
| Graph the solution. | |
| Write the solution using interval notation. | |
Check:
To verify, check a value in each section of the number line showing the solution. Choose numbers such as −8,−8, 1, and 9.
EXERCISE 1.2.171.2.17
Graph the solution and write the solution in interval notation: |x|<9|x|<9.
Answer
EXERCISE 1.2.181.2.18
Graph the solution and write the solution in interval notation: |x|<1|x|<1.
Answer
Example 1.2.191.2.19
Solve |5x−6|≤4|5x−6|≤4. Graph the solution and write the solution in interval notation.
Solution
: | | |
--- |
| Step 1. Isolate the absolute value expression. It is isolated. | |5x−6|≤4 |
| Step 2. Write the equivalent compound inequality. | −4≤5x−6≤4 |
| Step 3. Solve the compound inequality. | 2≤5x≤10 25≤x≤2 |
| Step 4. Graph the solution. | |
| Step 5. Write the solution using interval notation. | [25,2] |
| Check: The check is left to you. | |
EXERCISE 1.2.201.2.20
Solve |2x−1|≤5|2x−1|≤5. Graph the solution and write the solution in interval notation:
Answer
EXERCISE 1.2.211.2.21
Solve |4x−5|≤3|4x−5|≤3. Graph the solution and write the solution in interval notation:
Answer
SOLVE ABSOLUTE VALUE INEQUALITIES WITH << OR ≤≤
Isolate the absolute value expression.
Write the equivalent compound inequality.
|u|<ais equivalent to−a<u<a|u|≤ais equivalent to−a≤u≤a
|u|<a|u|≤ais equivalent tois equivalent to−a<u<a−a≤u≤a
3. Solve the compound inequality.
4. Graph the solution
5. Write the solution using interval notation.
Solve Absolute Value Inequalities with “Greater Than”
What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line.
We started with the inequality |x|≤5|x|≤5. We saw that the numbers whose distance is less than or equal to five from zero on the number line were −5−5 and 5 and all the numbers between −5−5 and 5 (Figure 1.2.41.2.4).
Figure 1.2.41.2.4.
Now we want to look at the inequality |x|≥5|x|≥5. Where are the numbers whose distance from zero is greater than or equal to five?
Again both −5−5 and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than −5−5 and greater than 5 on the number line (Figure 1.2.51.2.5).
Figure 1.2.51.2.5.
In a more general way, we can see that if |u|≥a|u|≥a, then u≤−au≤−a or u≤au≤a. See Figure.
Figure 1.2.61.2.6.
This result is summarized here.
ABSOLUTE VALUE INEQUALITIES WITH >> OR ≥≥
For any algebraic expression, u, and any positive real number, a,
if|u|>a,then u<−a or u>aif|u|≥a,then u≤−a or u≥a
ifif|u|>a,|u|≥a,then u<−a or u>athen u≤−a or u≥a
Example 1.2.221.2.22
Solve |x|>4|x|>4. Graph the solution and write the solution in interval notation.
Solution
: | | |
--- |
| | |x|>4 |
| Write the equivalent inequality. | x<−4 or x>4 |
| Graph the solution. | |
| Write the solution using interval notation. | (−inf,−4)∪(4,inf) |
| Check: | |
To verify, check a value in each section of the number line showing the solution. Choose numbers such as −6,−6, 0, and 7.
EXERCISE 1.2.231.2.23
Solve |x|>2|x|>2. Graph the solution and write the solution in interval notation.
Answer
EXERCISE 1.2.241.2.24
Solve |x|>1|x|>1. Graph the solution and write the solution in interval notation.
Answer
Example 1.2.251.2.25
Solve |2x−3|≥5|2x−3|≥5. Graph the solution and write the solution in interval notation.
Solution
: | | |
--- |
| | |2x−3|≥5 |
| Step 1. Isolate the absolute value expression. It is isolated. | |
| Step 2. Write the equivalent compound inequality. | 2x−3≤−5 or 2x−3≥5 |
| Step 3. Solve the compound inequality. | 2x≤−2 or 2x≥8 x≤−1 or x≥4 |
| Step 4. Graph the solution. | |
| Step 5. Write the solution using interval notation. | (−inf,−1]∪[4,inf) |
| Check: The check is left to you. | |
EXERCISE 1.2.261.2.26
Solve |4x−3|≥5|4x−3|≥5. Graph the solution and write the solution in interval notation.
Answer
EXERCISE 1.2.271.2.27
Solve |3x−4|≥2|3x−4|≥2. Graph the solution and write the solution in interval notation.
Answer
SOLVE ABSOLUTE VALUE INEQUALITIES WITH > OR ≥.
Isolate the absolute value expression.
Write the equivalent compound inequality.
|u|>ais equivalent tou<−a or u>a|u|≥ais equivalent tou≤−a or u≥a|u|>ais equivalent tou<−a or u>a|u|≥ais equivalent tou≤−a or u≥a
3. Solve the compound inequality.
4. Graph the solution
5. Write the solution using interval notation.
Solve Applications with Absolute Value
Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.
|actual-ideal|≤tolerance
Example 1.2.28
The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by 0.075 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?
Solution
: Let x= the actual measurementUse an absolute value inequality to express this situation.|actual-ideal|≤tolerance|x−60|≤0.075Rewrite as a compound inequality.−0.075≤x−60≤0.075Solve the inequality.59.925≤x≤60.075Answer the question.The diameter of the rod can be between59.925mm and 60.075mm.
ExERCISE 1.2.29
The ideal diameter of a rod needed for a machine is 80 mm. The actual diameter can vary from the ideal diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?
Answer
: The diameter of the rod can be between 79.991 and 80.009 mm.
ExERCISE 1.2.30
The ideal diameter of a rod needed for a machine is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?
Answer
: The diameter of the rod can be between 74.95 and 75.05 mm.
Access this online resource for additional instruction and practice with solving linear absolute value equations and inequalities.
Solving Linear Absolute Value Equations and Inequalities
Key Concepts
Absolute Value
The absolute value of a number is its distance from 0 on the number line.
The absolute value of a number n is written as |n| and |n|≥0 for all numbers.
Absolute values are always greater than or equal to zero.
Absolute Value Equations
For any algebraic expression, u, and any positive real number, a,
if|u|=athenu=−a or u=a
Remember that an absolute value cannot be a negative number.
How to Solve Absolute Value Equations
Isolate the absolute value expression.
Write the equivalent equations.
Solve each equation.
Check each solution.
Equations with Two Absolute Values
For any algebraic expressions, u and v,
if|u|=|v|thenu=−v or u=v
Absolute Value Inequalities with < or ≤
For any algebraic expression, u, and any positive real number, a,
if|u|=athen−a<u<aif|u|≤athen−a≤u≤a
How To Solve Absolute Value Inequalities with < or ≤
Isolate the absolute value expression.
Write the equivalent compound inequality.
|u|<ais equivalent to−a<u<a|u|≤ais equivalent to−a≤u≤a
Solve the compound inequality.
Graph the solution
Write the solution using interval notation
Absolute Value Inequalities with > or ≥
For any algebraic expression, u, and any positive real number, a,
if|u|>a,then u<−a or u>aif|u|≥a,then u≤−a or u≥a
How To Solve Absolute Value Inequalities with > or ≥
Isolate the absolute value expression.
Write the equivalent compound inequality.
|u|>ais equivalent tou<−a or u>a|u|≥ais equivalent tou≤−a or u≥a
Solve the compound inequality.
Graph the solution
Write the solution using interval notation
1.1: Solving Linear Equations
1.3: Solving Quadratic Equations
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344
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https://pmc.ncbi.nlm.nih.gov/articles/PMC3622115/
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ESTIMATING STRAIN-SPECIFIC AND OVERALL EFFICACY OF POLYVALENT VACCINES AGAINST RECURRENT PATHOGENS FROM A CROSS-SECTIONAL STUDY - PMC
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Biometrics
. Author manuscript; available in PMC: 2014 Mar 1.
Published in final edited form as: Biometrics. 2013 Feb 4;69(1):235–244. doi: 10.1111/j.1541-0420.2012.01826.x
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ESTIMATING STRAIN-SPECIFIC AND OVERALL EFFICACY OF POLYVALENT VACCINES AGAINST RECURRENT PATHOGENS FROM A CROSS-SECTIONAL STUDY
Kari Auranen
Kari Auranen
1 Department of Vaccination and Immune Protection, National Institute for Health and Welfare, Finland
Find articles by Kari Auranen
1,, Hanna Rinta-Kokko
Hanna Rinta-Kokko
2 Department of Vaccination and Immune Protection, National Institute for Health and Welfare, Finland
Find articles by Hanna Rinta-Kokko
2, M Elizabeth Halloran
M Elizabeth Halloran
3 Center for Statistics and Quantitative Infectious Diseases, Vaccine and Infectious Disease Division, Fred Hutchinson Cancer Research Center and Department of Biostatistics, University of Washington
Find articles by M Elizabeth Halloran
3
Author information
Article notes
Copyright and License information
1 Department of Vaccination and Immune Protection, National Institute for Health and Welfare, Finland
2 Department of Vaccination and Immune Protection, National Institute for Health and Welfare, Finland
3 Center for Statistics and Quantitative Infectious Diseases, Vaccine and Infectious Disease Division, Fred Hutchinson Cancer Research Center and Department of Biostatistics, University of Washington
Email: kari.auranen@thl.fi
Issue date 2013 Mar.
PMC Copyright notice
PMCID: PMC3622115 NIHMSID: NIHMS409126 PMID: 23379663
The publisher's version of this article is available at Biometrics
Summary
Evaluating vaccine efficacy for protection against colonisation with bacterial pathogens is an area of growing interest. Colonisation of the nasopharynx is an asymptomatic carrier state responsible for person-to-person transmission. It differs from most clinical outcomes in that it is common, recurrent and observed only in its prevalent state. To estimate rates of acquisition and clearance of colonisation requires repeated active sampling of the same individuals over time, an expensive and invasive undertaking. Motivated by feasibility constraints in efficacy trials with colonisation endpoints, investigators have been estimating vaccine efficacy from cross-sectional studies without principled methods. We present two examples of vaccine studies estimating vaccine efficacy from cross-sectional data on nasopharyngeal colonisation by Streptococcus pneumoniae (pneumococcus). This study presents a framework for defining and estimating strain-specific and overall vaccine efficacy for susceptibility to acquisition of colonisation (VE acq) when there is a large number of strains with mutual interactions and recurrent dynamics of colonisation. We develop estimators based on one observation of the current status per study subject, evaluate their robustness, and re-analyse the two vaccine trials. Methodologically, the proposed estimators are closely related to case-control studies with prevalent cases, with appropriate consideration of the at-risk time in choosing the controls.
Keywords: Case-control studies, Multiple colonisation, Pneumococcus, Polyvalent vaccine, Recurrent infection, Risk-set sampling, Strain interaction, Vaccine efficacy
1. Introduction
Vaccine efficacy (VE) is generally defined as a relative reduction in some measure of risk in the vaccinated group compared to the unvaccinated group (Halloran, Longini, and Struchiner, 2010). When a vaccine is polyvalent, that is, it contains antigens against two or more, but not necessarily all, related strains of a microorganism, assessing both the strain-specific efficacies and the overall (aggregate) efficacy against all strains is of interest. Some bacterial pathogens asymptomatically colonise the nasopharyngeal passages, called the carrier state. Colonisation with any strain may be cleared and recur repeatedly. Colonisation only rarely leads to manifest disease, such as bacteremia or meningitis, but it is a prerequisite for disease and the only source of transmission. Thus, the effect of vaccination on colonisation is of interest. Examples of such bacterial pathogens with polyvalent vaccines either available or under development include Streptococcus pneumoniae (pneumococcus), Neisseria meningitidis (Tan, Carlone, and Borrow, 2010), and group B streptococci (Edwards, 2008). Current formulations of pneumococcal conjugate vaccine contain antigens of 7, 10 or 13 of the known 93 serotypes (MMWR, 2010).
A straightforward approach to define and estimate vaccine efficacy for susceptibility based on the hazard rate of acquiring colonisation (VE acq) for individual strains would be to treat colonisation by each strain as separate and independent outcomes. To estimate rates of acquisition and clearance of colonisation requires repeated active sampling of the same individuals over time, an expensive and invasive undertaking. Thus, investigators have turned to using cross-sectional data to estimate vaccine efficacy without, however, explicitly defining the estimand of interest and its underlying assumptions (Rinta-Kokko et al. 2009).
We present two randomised controlled trials of polyvalent pneumococcal vaccines in which assessment of serotype-specific and overall vaccine efficacy is based on cross-sectional data (Table 1). In a vaccine study in South Africa (Mbelle et al., 1999), 500 infants were randomised to either a 9-valent pneumococcal conjugate vaccine or control vaccine and given 3 doses at 6, 10, and 14 weeks of age. Serotype-specific colonisation was measured once at the age of 9 months. Samples were available from 481 children. In a vaccine trial in Finland in 1995–1999 (Kilpi et al., 2001), 2497 children were randomised to a 7-valent pneumococcal conjugate vaccine or control and given 3 doses at 2, 4 and 6 months of age, and a booster at 12 months of age. Serotype-specific colonisation was measured before the booster at 12 months of age and approximately 6 months after the booster at 18 months of age. Here we analyse measurements made at least 6 months after the booster (2403 children).
Table 1.
Vaccine efficacy against colonisation in two vaccine studies. Upper part. Data from a randomised, double-blind vaccine study on the effect of an experimental 9-valent pneumococcal conjugate vaccine on colonisation in Soweto, South Africa (Mbelle et al., 1999). Altogether 500 infants were recruited; 481 completed the 9-month follow-up. Children received either pneumococcal vaccine (Wyeth Lederle Vaccines, n=242) or placebo (n=239). The vaccine was given at 6, 10 and 14 weeks of age and the estimates of vaccine efficacy are based on measurements of colonisation at 9 months of age. Estimates are shown individually for seven of the nine vaccine serotypes (serotypes 1 and 5 were not found at all in the study), for all vaccine serotypes and for all non-vaccine serotypes. Lower part: Data from a randomised, double-blind vaccine study on the effect of a 7-valent pneumococcal conjugate vaccine on colonisation in Finland (Kilpi et al., 2001). Children received either of two pneumococcal vaccines (Wyeth Lederle Vaccines, n=831, or Merck Sharp and Dohme, n=835), or HBV vaccine (Merck Sharp and Dohme, n=831) as a control. The vaccine was given at 2, 4, 6 and 12 months of age and the estimates of vaccine efficacy are based on measurements of colonisation at 18 months of age. Estimates are shown individually for the seven vaccine serotypes (in the South African dataset, there were no samples with two of the vaccine serotypes, 1 and 5), for all vaccine serotypes, and for all non-vaccine serotypes. For both studies, the estimates of vaccine efficacy are determined against all vaccine serotypes and non-vaccine serotypes using estimators (4) and (5) for VE W|0 and VE V̄|0, respectively. For comparison, the last column shows estimates of vaccine efficacy (HRK), based on simple odds ratios (Rinta-Kokko et al., 2009; see text)
| Serotype | Number of samples | Estimates of vaccine efficacy |
:---
| | Vaccinees | Controls | VE W|0 (90% CI†) | HRK (90% CI†) |
| South Africa |
| 4 | 3 | 5 | 0.54 (−0.54, 0.86) | 0.41 (−0.97, 0.82) |
| 6B | 8 | 28 | 0.78 (0.57, 0.89) | 0.74 (0.49, 0.87) |
| 9V | 1 | 2 | 0.62 (−1.89, 0.95) | 0.51 (−2.71, 0.93) |
| 14 | 3 | 6 | 0.62 (−0.24, 0.88) | 0.51 (−0.58, 0.85) |
| 18C | 2 | 0 | – | – |
| 19F | 19 | 32 | 0.55 (0.25, 0.73) | 0.45 (0.09, 0.67) |
| 23F | 7 | 14 | 0.62 (0.17, 0.83) | 0.52 (−0.04, 0.78) |
| |
| Vaccine types | 43 | 87 | 0.62 (0.46, 0.74) | 0.62 (0.46, 0.74) |
| Non-vaccine types | 87 | 58 | −0.26 (−0.81, 0.12) | −0.75 (−1.44, −0.26) |
| 0 | 112 | 94 | | |
| |
| Finland |
| 4 | 1 | 2 | 0.77 (−0.76, 0.97) | 0.75 (−0.87, 0.97) |
| 6B | 44 | 39 | 0.47 (0.24, 0.64) | 0.45 (0.20, 0.62) |
| 9V | 4 | 4 | 0.53 (−0.50, 0.85) | 0.50 (−0.60, 0.84) |
| 14 | 13 | 10 | 0.39 (−0.22, 0.70) | 0.35 (−0.30, 0.68) |
| 18C | 6 | 4 | 0.30 (−1.03, 0.76) | 0.25 (−1.17, 0.74) |
| 19F | 44 | 32 | 0.36 (0.05, 0.56) | 0.32 (0.00, 0.54) |
| 23F | 47 | 34 | 0.35 (0.06, 0.56) | 0.32 (0.01, 0.53) |
| |
| Vaccine types | 159 | 125 | 0.41 (0.27, 0.52) | 0.41 (0.27, 0.52) |
| Non-vaccine types | 268 | 111 | −0.16 (−0.42, 0.06) | −0.25 (−0.52, −0.02) |
| 0 | 1176 | 564 | | |
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For vaccine efficacy against non-vaccine strains, the reference state is 0 and estimator (5) was used.
†
CI = confidence interval.
Rinta-Kokko et al. (2009) proposed to estimate VE acq based on one cross-sectional study under stationary conditions using simple ratios of the odds of vaccination in those colonised with the types targeted by the vaccine versus those not colonised. However, that analysis was not adjusted for the possible difference in the times spent susceptible by the vaccinees and controls. Colonisation is usually common and recurrent even on the strain level. In addition, the time spent at-risk for acquiring a new colonisation is affected by between-strain interactions induced by intra-species competition (Lipsitch et al., 2000; Melegaro et al., 2007). Any reduction in colonisation within a vaccinated host by pneumococcal strains targeted by the vaccine is thus counterbalanced by an increase in non-vaccine type colonisation (e.g. Mbelle et al., 1999), a phenomenon we call within-host replacement. For these reasons, the vaccine efficacy estimands may have to be conditioned on the appropriate states of colonisation in which an individual is susceptible to colonisation by the strain(s) targeted by the vaccine. Without such adjustment, the biological protective efficacy on the causal pathway and the estimated (outcome) efficacy may be different, confounding the interpretation and comparability of estimates of vaccine efficacy across different epidemiologic settings (cf. Struchiner et al., 1994).
Here we present a framework for defining vaccine efficacy for multiple interacting strains which can be acquired and carried repeatedly. We develop two new estimands of VE acq in the multi-type setting. We propose estimators of VE acq for individual strains as well as for any collection of select (target) strains based solely on one measurement per study subject. We generalise the model to allow estimation of combined efficacy against acquisition and duration of colonisation as a more general estimand within a broader class of vaccine efficacy on colonisation. We present simulations and discuss robustness of estimation to departures from model assumptions.
2. Models of colonisation and vaccine effects
Figure 1A depicts two alternative models (A and B) to describe recurrent colonisation and competition among multiple strains of a pathogen within an individual (Lipsitch, 1997; Melegaro et al. 2007; Auranen et al., 2010).
Figure 1.
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1A. ModelAof colonisation. The model has two strains and four states of colonisation. The states are uncolonised (0), colonised with strain 1, colonised with strain 2, and colonised with both strains (1 and 2). The model is governed by eight transition hazards, which fulfil conditions (A1) and (A2) if q 1, 12/q 2, 12 = q 0, 2/q 0, 1. In the text, this model is considered for an arbitrary number of strains, separately for the vaccinated (T) and control (C) individuals. 1B. ModelBof colonisation. The model has two strains and three states of colonisation. The states are uncolonised (0), colonised with strain 1, and colonised with strain 2. The model is governed by six transition hazards, which fulfil conditions (B1) and (B2) if q 1, 2/q 2, 1 = q 0, 2/q 0, 1. In the aggregated model, states 0, 1 and 2 correspond to aggregated states , and (see Section 5).
In model A for two strains, an individual can be either susceptible (i.e., uncolonised in state 0) or colonised with either of the two strains (in states 1 or 2). In addition, the individual may be simultaneously colonised by both strains (co-colonised state 1&2). Figure 1A presents the model only for two strains, but we consider it for an arbitrary number of distinguishable strains. However, in this paper we do not allow simultaneous colonisation with more than two strains, which is supported by the rarity of such multiple colonisation by pneumococci (Kaltoft et al., 2008). For n strains, the set of n(n+1)/2+1 possible states of colonisation is denoted by = {0} ∪ S ∪ {(1, 2), (1, 3), …, (1, n), (2, 3), …, (n−1, n)} where S = {1, …, n} is the index set of the strains.
There are altogether 2 n 2 transition hazards between the directed pairs of states in . For clarity, we index the individual states explicitly by strain indices, i.e., q i,j denotes the hazard of transition from state i to state j, i, j ∈ S ∪ {0}, i ≠ j, and q i,ij and q ij,i denote the hazards of the two reverse transitions between states i and (i, j), i, j ∈ S, i ≠ j. Occasionally, it is more convenient to use an alternative naming convention in which the states are enumerated as {, …, [n(n + 1)/2 + 1]}. Then q[h][k] denotes the hazard of transition from state [h] to state [k] ([h], [k] ∈ , [h] ≠ [k]).
We assume the hazards of transition in the unvaccinated subjects (“C” for controls) under model A fulfil two conditions:
q j,i j C/q i,i j C=q 0,i C/q 0,j C for all i,j∈S,i≠j,(A1)
q i,0 C=q i j,j C≡μ for all i,j∈S,i≠j.(A2)
For any two strains, assumption (A1) states the ratio of their two hazards of acquisition to the doubly-colonised state (j to (i, j) vs. i to (i, j)) equals the ratio of the respective hazards of colonisation from state 0 (0 to i vs. 0 to j). Competition between strains is quantified in terms of relative hazards q j,i j C/q 0,i C, values < 1 indicating reduced acquisition of strain i due to the currently colonising strain j. Consequently, we refer to condition (A1) as symmetry in competition because it implies q j,i j C/q 0,i C=q i,i j C/q 0,j C for any given pair of strains i and j.
Conditions (A1) and (A2) still allow each strain i to have its own hazard of colonisation q 0,i. According to (A2), however, the hazards of clearance from the singly-colonised states to state 0 and the hazards of clearance from doubly-colonised states to singly-colonised states are equal for all strains and denoted by μ. In Sections 5 and 6 we relax this assumption.
Illustration
To illustrate these concepts, assume there were four strains, with the vaccine containing antigens of two strains, called the vaccine strains. The other two are non-vaccine strains. Denote the two vaccine strains 1 and 2, and the non-vaccine strains 3 and 4. Let the hazards of colonisation in uncolonised, unvaccinated individuals be q 0, 1 = 0.07, q 0, 2 = 0.06, q 0, 3 = 0.05, and q 0, 4 = 0.04 per time unit. Assume the hazard of clearance (μ) is 0.17 per time unit. Let the relative hazards for acquisition of double carriage (q j,ij/q 0,i) be 0.4 for all i, j, according to eq. (A1). The value 0.4 < 1 indicates considerable competition against double colonisation. For example, if a person is already colonised with strain 1, 3, or 4, the hazard of becoming doubly colonised with strain 2 would be 0.06 · 0.4 = 0.024.
Model B
Figure 1B presents model B for two strains. Again, an individual can be either susceptible or colonised with either of the two strains, but there is no co-colonisation. In a model for n strains, the conditions corresponding to (A1) and (A2) are:
q j,i C/q i,j C=q 0,i C/q 0,j C,i,j∈S,i≠j,(B1)
q i,0 C=μ,i∈S.(B2)
Under model B, competition between strains is described in terms of the ratios q j,i C/q 0,i C of the reverse transition hazards between the singly colonised states, values < 1 again indicating competition. Although many results derived in this paper are applicable for any number of strains also under model B, the main use of this model is with three states only, obtained by aggregating states of model A (see Section 5).
The vaccine model
We assume the direct biological effect of vaccination acts on susceptibility to acquisition of colonisation, the primary estimand of interest thus being VE acq. Under model A a strain-specific efficacy estimand VE acq,i is defined as the relative reduction in the hazard of acquisition of strain i if vaccinated (T) compared to if receiving control (C):
VE acq,i=1-q 0,i T/q 0,i C=1-q j,i j T/q j,i j C,i,j∈S,i≠j.(A3)
VE acq,i is defined separately for each state from which a transition is possible to state i or (i, j), i.e., for all states in which an individual is susceptible to acquisition of strain i. The definition assumes no interaction between the current state of colonisation and vaccine efficacy for strain i.
Illustration, cont'd
In the four strain example above, let VE acq, 1 = 0.7 and VE acq, 2 = 0.4. Under (A3), the efficacy against colonisation by strain 1 is the same whether an individual is uncolonised or colonised with strain 2, 3, or 4. The hazard of an uncolonised vaccinated person being colonised with strain 1 would be (1 − 0.7)0.07 = 0.21 (per time unit). However, due to the competition described above, the hazard of a colonised vaccinated person being colonised with strain 1 would be (1 − 0.7)0.4 · 0.07 = 0.064. The same arguments hold for strain 2. The hazards of the non-vaccine strains 3 and 4 are unaffected by vaccination.
Initially we assume that vaccination does not affect the hazard of clearance of colonisation:
q i,0 T=q i j,j T=μ,i,j∈S,i≠j.(A4)
Under model B, the analogous vaccine efficacy estimand is
VE acq,i=1-q 0,i T/q 0,i C=1-q j,i T/q j,i C,i,j∈S,i≠j,(B3)
and the assumption about the hazard of clearance is (B4) q i,0 T=μ,i∈S. It is straightforward to show that with the assumed vaccine models, (A3)–(A4) or (B3)–(B4), the respective conditions (A1)–(A2) or (B1)–(B2) hold for the vaccinated individuals as well. In the sequel, we drop “acq” in the notation for VE acq,i, if not required for clarity.
3. Overall vaccine efficacy for multiple strains
We denote the index sets of the non-vaccine and vaccine strains by V̄ and V, so S = V̄ ∪ V. We also define the set V̄ 0 = V̄ ∪ {0} as the index set of those (single) states against which the vaccine does not confer direct biological protection. In particular, when conditioning on the current state of colonisation, assuming no cross-reactivity of immunity between strains, the true direct biological vaccine efficacy against a non-vaccine strain is zero, i.e., VE i = 0 for i ∈ V̄. In contrast, the efficacy against any vaccine strain is a priori unknown. We denote the index set of these target strain(s) in a vaccine study by W ⊆ V. Typically W = {i} separately for each i ∈ V (for strain-specific estimates for vaccine efficacy) or W = V (for overall efficacy against all vaccine-strains). However, there may be interest in a proper subset of (vaccine) strains as the target, e.g., “new” strains in a vaccine with a larger valency.
Partition of the states of colonisation
To define the estimands of overall efficacy, we need to specify a partition of the states of colonisation into three disjoint sets (Figure 2). The sets correspond to (a) states in which none of the vaccine-strains is involved, denoted ; (b) states in which any one of the target strains of the vaccine is involved either alone or with any of the non-vaccine strains, denoted ; and (c) the rest = ( ∪ ). The sets and generally depend on the choice of target strains W against which vaccine efficacy is considered. For W = V, contains only the doubly-colonised states with two vaccine-strains (Figure 2A). These states are not included in because our definition of vaccine efficacy does not involve acquisition of secondary vaccine strains (see below). For strain-specific efficacy, also includes all states with non-target vaccine-strains (Figure 2B).
Figure 2.
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Partition of states for estimation of vaccine efficacy. The figure presents the seven states of colonisation in model A for three strains. Strains 1 and 2 are vaccine strains and strain 3 is a non-vaccine strain, i.e., S = {1, 2, 3} and V = {1, 2}. 2A. Overall vaccine efficacy. The figure shows the partition for estimation of overall vaccine efficacy against strains 1 and 2. Here = {1, 2, (1, 3), (2, 3)} and = {0, 3}. The complement = {(1, 2)} consists of the (only) doubly-colonised state with two vaccine strains. The aggregate model has a chain structure ↔ ↔ . 2B. Strain-specific vaccine efficacy. The figure shows the partition for vaccine efficacy against strain 1. Here = {1, (1, 3)}, = {0, 3}, and = {2, (1, 2), (2, 3)}. Generally, for an individual strain i ∈ S, define = {i} ∪ {(i, j); j ∈ V̄, j ≠ i} as the set of those states in which strain i is involved either alone or with any of the non-vaccine strains. Then = ∪i∈W and = (∪i∈V̄) ∪ {0}.
Overall vaccine efficacy
Generalising the strain-specific estimands in (A3) and (A4), we define overall (aggregate) vaccine efficacy against all strains in a target set W as the relative reduction in their overall hazard of acquisition. To make this definition precise in a multi-state setting, we specify a distribution for the time spent at risk to acquire the target strains among the non-vaccine states. We consider two estimands based on transitions either from state 0 only, or alternatively from any of the non-vaccine states V̄ 0.
The first estimand specifies the overall vaccine efficacy against a given collection of target strains as the relative reduction in their overall hazard of acquisition from state 0. The target states W then include all states to which there is a direct transition (“arrow”) from state 0 and the overall hazard is simply the sum of the hazards of colonisation q 0,i, i ∈ W. For a given set of vaccine-strains W (⊆ V), the overall vaccine efficacy estimand is thus defined as
VE W∣0:=1-∑i∈W q 0,i T/∑i∈W q 0,i C={∑i∈W q 0,i C·VE i}/∑i∈W q 0,i C.(1)
The rightmost equation expresses overall efficacy as a weighted average of strain-specific efficacies, the weights being the hazards of colonisation. In a strain-specific analysis, i.e., when W = {i}, it holds naturally that VE{i}|0 = VE i as in the left part of (A3) and (B3).
The second estimand considers vaccine efficacy based on transitions from set V̄ 0, i.e., from state 0 or any of the singly-colonised states with a non-vaccine strain. This seems a natural extension of the “susceptibility set” from state 0 to states against which the vaccine does not have a biological effect mechanism. The set of target states now includes all states to which there is a direct transition (“arrow”) from any of the states in V̄ 0. To define a unique, time-independent hazard of transition between the two collections of states (V̄ 0 and ), we need in addition to specify a mixing distribution for the time spent among the states in V̄ 0. It proves useful to apply the stationary distribution. For a given set W of target strains, the alternative overall vaccine efficacy estimand is then defined as:
VE W∣V¯0:=1-∑[h]∈V¯0 p¯[h]∣V¯0 T(∑[k]∈W q[h][k]T)/∑[h]∈V¯0 p¯[h]∣V¯0 C(∑[k]∈W q[h][k]C),(2)
where p̄[h]|V̄ 0 denotes the stationary probability of state [h], conditioned upon being in V̄ 0.
Expression (2) is also a weighted average of strain-specific efficacies (see Web Appendix A). The two definitions thus fully agree for any strain-specific efficacy as in (A3). Moreover, for the overall efficacy they will yield similar numerical values if between-strain competition is homogeneously symmetric across all strains or acquisition of double colonisation is not common (see Web Appendix A).
Analogous to (1), we also define vaccine efficacy against the non-vaccine strains as the relative reduction in the overall hazard of acquisition of any of the non-vaccine strains from the uncolonised state:
VE V¯∣0:=1-∑j∈V¯q 0,j T/∑j∈V¯q 0,j C.(3)
By the assumption VE i = 0 for any non-vaccine strain, the theoretical value of VE V̄|0 is 0. The estimand can naturally be restricted to any individual non-vaccine strain.
Illustration cont'd
In the four strain example, the overall vaccine efficacy against the two vaccine strains, based on definition (1), is VE{1, 2}|0 = (0.07·0.7+0.06·0.4)/(0.07+0.06) = 0.56. Based on definition (2), the overall vaccine efficacy VE{1, 2}|{0, 3, 4} is also 0.56, calculated as
1-p¯0∣V¯0 T((1-0.7)0.07+(1-0.4)0.06)+p¯{3,4}∣V¯0 T·0.4((1-0.7)0.07+(1-0.4)0.06)p¯0∣V¯0 C(0.07+0.06)+p¯{3,4}∣V¯0 C·0.4(0.07+0.06),
where (p 0∣V¯0 T,p{3,4}∣V¯0 T)=(0.253,0.134)/0.387 and (p 0∣V¯0 C,p{3,4}∣V¯0 C)=(0.197,0.103)/0.3 are the conditioned stationary probabilities of states 0 and {3, 4} in the vaccinees and controls. In this model, the competition parameter has the same value (0.4) for all pairs of strains and the overall efficacies based on (1) and (2) are thus equivalent (cf. Web Appendix A).
Throughout, we assume no interference among study subjects, and that the rate of exposure to colonisation does not change. This would be plausible if a small proportion of the population took part in the trial, such as a portion of infants. Under randomisation, expected exposure to colonisation would be the same in vaccinees and non-vaccinees. However, even under these circumstances, the hazards of acquiring non-vaccine strains would differ between vaccinees and non-vaccinees if not conditioned on current state of colonisation, because total time at-risk is distributed differently among the states of colonisation. The very motivation of this research is to elucidate the definition and estimation of vaccine efficacy in such situations.
4. Cross-sectional estimation of vaccine efficacy against colonisation
Here we present the main results of this paper by formulating expressions for VE acq in (1) and (2) in terms of the stationary distribution, the stationarity requirement being imposed separately on the unvaccinated and vaccinated groups. Estimation of vaccine efficacy can then be based on these expressions, relying solely on cross-sectional measurement of colonisation. The following results are valid under model A. Derivation of results, proofs, and applicability under model B are shown in Web Appendix A. Let p[k]C and p[k]T denote the stationary probabilities of state [k] ∈ in the non-vaccinated (C) and vaccinated groups (T).
Result 1
The overall vaccine efficacy against a collection of vaccine strains W (⊆ V), as defined by (1), can be expressed in terms of the stationary probabilities as
VE W∣0=1-p W T/p V¯0 T p W C/p V¯0 C≡1-(∑[k]∈W p[k]T)/(∑[k]∈V¯0 p[k]T)(∑[k]∈W p[k]C)/(∑[k]∈V¯0 p[k]C).(4)
Expression (4) for VE W|0 has the form `one minus an odds ratio’. The odds of being vaccinated (T vs. C) are compared between those colonised with the target strains and those colonised with the non-vaccine strains. The possible target states W include all singly-colonised states for any target strain. The reference states involve all states in which no vaccine-strain is involved.
Instead of using as the reference set, it is possible to restrict the comparisons to state 0 (uncolonised). With this choice and using the non-vaccine strains as the target set, we obtain the following expression for vaccine efficacy against the non-vaccine strains to estimate (3):
VE V¯∣0=1-{(∑[k]∈V¯p[k]T)/p 0 T}/{(∑[k]∈V¯p[k]C)/p 0 C}.(5)
If the model is correctly specified, the theoretical value of (5) is 0, so the estimate of VE V̄| 0 should be close to zero (see Web Appendix A).
Result 2
The overall vaccine efficacy against a collection of vaccine strains W (⊆ V), as defined by (2), can be expressed in terms of the stationary probabilities as
VE W∣V¯0=1-p W T/p V¯0 T p W C/p V¯0 C≡1-(∑[k]∈W p[k]T)/(∑[k]∈V¯0 p[k]T)(∑[k]∈W p[k]C)/(∑[k]∈V¯0 p[k]C).(6)
As with VE W|0 in (4), the reference set includes all those states in which no vaccine-strain is involved. Instead of the singly colonised states W, the target states now include all those states in which one of the target strains is involved either alone or together with a non-vaccine strain (Figure 2).
Expressions (4)–(6) provide estimators in terms of proportions of the subjects in different states of colonisation in the stationary phase. Thus, under stationary conditions both strain-specific efficacies and overall efficacy can be estimated without expensive and invasive longitudinal observation using standard methods for estimating odds ratios.
Illustration, cont’d
Table 2 demonstrates how to calculate estimates based on VE W|V̄ 0 in (6) from observed strain distributions.
Table 2.
Estimation of vaccine efficacy from cross-sectional data. The table presents one simulated data set for 1000 vaccinated and 1000 control (non-vaccinated) individuals at the stationary phase in model A for two vaccine strains (1 and 2) and two non-vaccine strains (3 and 4), with hazards of colonisation 0.07, 0.06, 0.05, and 0.04 per time unit. The hazard of clearance (μ) was 0.17 per time unit, and the relative hazards for acquisition of double carriage (q j, ij/q 0, i) was 0.4 for all i; j, see eq. (A1). For each strain, the numbers of observations are given for the singly and the three doubly-colonised states with that strain. The last column shows how to calculate vaccine efficacy against individual and all vaccine-strains (VS) by using estimator (6) for VE W|V̄0, and vaccine efficacy against individual and all non-vaccine strains (NVS) by using estimator (5). The true values of vaccine efficacy are given in the first column. The true value of the overall efficacy against the vaccine-strains is VE V| V̄0 = VE{1;2}|{0, 3, 4} = 0.56, based on definition (2) and equation (E1)
| True efficacy | States of colonisation | Numbers of samples in vaccinees | Numbers of samples in controls | Estimates of vaccine efficacy VE W|V̄0 |
:--- :---
| 0.70 (VE 1) | 1+1&2+1&3+1&4 | 61 + 4 + 7 + 6 | 162 + 29 + 18 + 9 | 1-(61+7+6)/(522+260+13)(162+18+9)/(403+207+9)=0.70 |
| 0.40 (VE 2) | 2+1&2+2&3+2&4 | 107 + 4 + 12 + 8 | 136 + 29 + 17 + 10 | 1-(107+12+8)/(522+260+13)(136+17+10)/(403+207+9)=0.39 |
| 0.00 (VE 3) | 3+1&3+2&3+3&4 | 131 + 7 + 12 + 13 | 111 + 18 + 17 + 9 | 1-131/522 111/403=0.09∗ |
| 0.00 (VE 4) | 4+1&4+2&4+3&4 | 129 + 6 + 8 + 13 | 96 + 9 + 10 + 9 | 1-129/522 96/403=-0.04∗ |
| |
| 0.56 (all VS) | VS+VS&NVS+VS&VS† | 168 + 33 + 4 | 298 + 54 + 29 | 1-(168+33)/(522+260+13)(298+54)/(403+207+9)=0.56 †† |
| 0.00 (all NVS) | NVS+VS & NVS+NVS& NVS† | 260 + 33 + 13 | 207 + 54 + 9 | 1-260/522 207/403=0.03∗ |
| 0 | – | 522 | 403 | – |
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For efficacy against non-vaccine strains, the reference state is 0 and estimator (5) was used.
†
VS&NVS is the number of subjects colonised with a vaccine-strain and a non-vaccine strain, VS&VS is the number of subjects colonised with two vaccine-strains, NVS&NVS is the number of subjects colonised with two non-vaccine strains.
††
For the vaccine-strains, the overall estimate is based on eq. (6) for VE V|V̄0 = VE{1, 2}|{0, 3, 4}. An alternative estimate given by eq. (4) is VE V|0 = VE{1, 2}|0 = 1− (168/(522+260+13))/(298/(403+207+9)) = 0.56.
5. The aggregated model and sufficient conditions
To analyse the robustness of the proposed estimators to departures from the assumptions in Section 2, we formulate a sufficient set of conditions that still allows valid estimation of VE acq from cross-sectional data. Consider the partition of the original set of colonisation states into three aggregated states (Figures 2A and 2B). The partition consists of the reference set , i.e., the colonisation states, including state 0, for which the vaccine has no direct effect on the hazard of entering these states, the target set , and the “rest” . We specify a new Markov model for the three aggregated states by defining the hazards of transition from one aggregate state to another in terms of the underlying transition hazards as follows:
q U,P=∑[h]∈U p¯[h]U(∑[k]∈P q[h][k]),U,P∈{V¯0,W,R},U≠P,(7)
where p̄[h]|U denotes the stationary probability of the state [h], conditioned upon the aggregate state U, and q[h][k] is the transition hazard from state [h] to state [k] (see Deng et al. (2009) for theoretical justification).
Under the aggregated model, it is natural to define another VE acq for susceptibility of acquisition of strains in set W as
VE W∣V¯0=1-q V¯0,W T/q V¯0,W C,(8)
i.e., the relative reduction in the hazard due to vaccination. This estimand is equivalent to (2) (see Web Appendix A for details). In addition, expression (6) for VE W|V̄ 0 can be interpreted as based on the aggregated model since the stationary distribution of the aggregated process is obtained by summing up the respective stationary probabilities under model A. It follows that, for any choice of the target strains, it is possible to formulate a smaller set of conditions directly in the context of the aggregated model that allows cross-sectional estimation of VE acq. In the following we summarise these conditions.
For W = V, i.e., when estimating overall efficacy against all vaccine strains, the aggregated model has a simple chain structure of three states: ↔ ↔ (Figure 2A). It is then easy to show by the reversibility of chain models that the only condition required for cross-sectional estimation is the similarity of clearance hazards of the aggregate states in the vaccinees and controls: (B5) q W,V¯0 C=q W,V¯0 T.
Likewise, if the reverse hazards and satisfy the symmetry condition ( / = / ), the essential model structure at stationarity is again a chain between three states: ↔ ↔ and (B5) is sufficient to estimate vaccine efficacy against any collection of vaccine strains (W ⊂ V). Only when W is a strict subset of V and the competition is not symmetric, the aggregated model has the triangle structure (Figure 1B), and the conditions are more restrictive. Specifically, they are (B1)–(B4) for model B with three states corresponding to , , and (see Web Appendix A for more details).
6. Simulations, sensitivity analyses, and data analyses
Simulated examples
We conducted a simulation study with four vaccine and five non-vaccine strains under model A to examine the performance of the proposed cross-sectional estimators. We chose the hazards of colonisation, q 0,i, i = 1, …, 9, and the hazard of clearance μ so the stationary strain distribution mimics that typical of serotypes of S. pneumoniae, with prevalence ranging from approximately 20% for the most common type to < 2% for rare types. Repetitions of samples in a cohort of 1000 vaccinees and 1000 unvaccinated controls were taken at the stationary phase of the process satisfying conditions (A1)–(A4).
Estimates of VE acq based on VE W|V̄ 0 or VE W|0 were similar (Table 3), with both producing estimates close to the true values. Coverage probabilities are almost equal to the nominal confidence interval (90%). The use of VE W|V̄ 0 produces somewhat more precise estimates because a larger portion of the data is used. The slightly negative bias in the point estimates is due to bias in odds-ratio estimates from finite samples (see Web Appendix C Table C1).
Table 3.
Performance of cross-sectional estimators of vaccine efficacy assuming perfect detection of the singly and doubly-colonised states. The results are based on 1000 simulated data sets, each with 1000 vaccinees and 1000 controls. The model consisted of 4 vaccine (shown in boldface) and 5 non-vaccine strains, with hazards of colonisation 0.0075, 0.003, 0.002, 0.001, 0.005, 0.003, 0.0009, 0.0005, and 0.0003 per time unit (day). The relative hazard for acquisition of double carriage (q j, ij/q 0, i) was 0.4 for all i, j. The hazard of clearance μ was 0.017 per day. The simulation was started from state 0 (uncolonised) for each individual and the sample for calculating the vaccine efficacy was taken at day 183. The true vaccine efficacy was 0.70 against strains 1 and 3, and 0.40 against strains 2 and 4. The true value of the overall efficacy is VE V|V̄0 = 0:61, based on definition (2) and equation (E1). The program code to analyse an example data with 9 strains is provided in a Web Appendix B.
| Strain | True efficacy | No. of singly and doubly colonised samples† (% of all samples) | Estimates of vaccine efficacy (SE log(OR); 90 % cov. prob.)†† |
:--- :--- |
| | | Vaccinees | Controls | VE W|V̄ 0 | VE W|0 |
| 1 | 0.7 | 66 + 19(8.5) | 166 + 61(22.7) | 0.70(0.14; 0.91) | 0.70(0.16; 0.91) |
| 2 | 0.4 | 53 + 16(6.9) | 66 + 31(9.7) | 0.39(0.18; 0.90) | 0.39(0.19; 0.90) |
| 3 | 0.7 | 18 + 6(2.4) | 44 + 22(6.6) | 0.70(0.27; 0.90) | 0.69(0.29; 0.91) |
| 4 | 0.4 | 17 + 6(2.3) | 22 + 11(3.3) | 0.38(0.30; 0.91) | 0.37(0.33; 0.92) |
| 5 | 0 | 146 + 34(18.0) | 110 + 46(15.6) | −0.01(0.14; 0.90) | −0.01(0.14; 0.90) |
| 6 | 0 | 88 + 24(11.2) | 66 + 31(9.7) | −0.01(0.18; 0.92) | −0.01(0.18; 0.92) |
| 7 | 0 | 26 + 9(3.5) | 20 + 10(3.0) | −0.06(0.31; 0.92) | −0.06(0.31; 0.92) |
| 8 | 0 | 15 + 5(2.0) | 11 + 6(1.7) | −0.13(0.42; 0.90) | −0.13(0.42; 0.90) |
| 9 | 0 | 9 + 3(1.2) | 6 + 4(1.0) | −0.18(0.56; 0.92) | −0.18(0.56; 0.92) |
| |
| Vaccine strains | 0.61 | 154 + 41(19.5) | 298 + 96(39.4) | 0.61(0.10; 0.90) | 0.61(0.11; 0.89) |
| Non-vaccine strains | 0 | 284 + 55(33.9) | 213 + 82(29.5) | 0.00(0.11; 0.91) | 0.00(0.11; 0.91) |
| 0 | | 501(50.1) | 378(37.8) | – | – |
Open in a new tab
For efficacy against non-vaccine strains, the reference state is 0 and estimator (5) was used.
†
The mean number of samples in the 1000 simulated data sets.
††
The average standard error of the log odds ratio, based on the 1000 simulated data sets. The coverage probability of a 90% confidence interval, based on the 1000 simulated data sets.
Table 4 presents estimates of VE acq in a similar scenario, except the observed data for any individual included at most one strain. This was achieved by randomly dropping one strain in any simulated sample with two simultaneous strains (approximately 20% of positive samples) and using estimator (4). As a general finding, this type of symmetry in missing data does not appear to affect the efficacy estimates. The estimates were not biased under heterogeneous clearance rates across individuals either (data not shown).
Table 4.
Performance of cross-sectional estimation of vaccine efficacy under imperfect detection of double colonisation. The simulated complete data were the same as used in Table 3. Consequently, all doubly-colonised states were recorded as singly-colonised, with 50% chance of either of the colonising strains being detected.
| Strain | True efficacy (VE i) | No. of colonised samples† (% of all samples) | Estimates of vaccine efficacy (SE log(OR); 90% cov. prob.)†† |
:--- :--- |
| | | Vaccinees | Controls | VE W|V̄ 0 | HRK |
| 1 | 0.7 | 75(7.5) | 196(19.6) | 0.70(0.15; 0.90) | 0.66(0.14; 0.78) |
| 2 | 0.4 | 61(6.1) | 82(8.2) | 0.42(0.18; 0.90) | 0.27(0.18; 0.72) |
| 3 | 0.7 | 21(2.1) | 55(5.5) | 0.71(0.27; 0.89) | 0.64(0.27; 0.84) |
| 4 | 0.4 | 20(2.0) | 28(2.8) | 0.40(0.30; 0.90) | 0.24(0.30; 0.83) |
| 5 | 0 | 163(16.3) | 133(13.3) | 0.07(0.14; 0.85) | −0.28(0.13; 0.40) |
| 6 | 0 | 100(10.0) | 81(8.1) | 0.07(0.16; 0.86) | −0.27(0.16; 0.59) |
| 7 | 0 | 31(3.1) | 25(2.5) | 0.05(0.28; 0.88) | −0.27(0.28; 0.82) |
| 8 | 0 | 17(1.7) | 14(1.4) | 0.00(0.38; 0.91) | −0.34(0.38; 0.86) |
| 9 | 0 | 10(1.0) | 8(0.8) | −.06(0.50; 0.91) | −0.42(0.49; 0.88) |
| |
| Vaccine strains | 0.61 | 177(17.7) | 361(36.1) | 0.62(0.11; 0.88) | 0.62(0.11; 0.88) |
| Non-vaccine strains | 0 | 321(32.1) | 261(26.1) | 0.08(0.11; 0.80) | −0.33(0.10; 0.11) |
| 0 | | 502(50.2) | 378(37.8) | | |
Open in a new tab
For efficacy against non-vaccine strains, the reference state is 0 and estimator (5) was used.
†
The mean number of samples in the 1000 simulated data sets.
††
The average standard error for the log odds ratio, based on the 1000 simulated data sets. The coverage probability of a 90% confidence interval, based on the 1000 simulated data sets.
The last column in Table 4 presents the estimates of VE acq proposed by Rinta-Kokko et al. (2009) (HRK), based on simple odds ratios obtained by splitting the observations into two complementary categories for each choice of the target strain(s). Such estimates are straightforward to calculate in a setting in which observations do not record double colonisation. However, this estimator does not take into account interactions between strains, thus systematically underestimates the strain-specific efficacies. In particular, estimates for the non-vaccine strains are grossly negative. Only the overall efficacy against all vaccine-strains agrees with the true value, which follows directly from the fact that expression (6) is equivalent to the simple odds ratio in this case.
Sensitivity to model assumptions
To study sensitivity of the cross-sectional estimates to departures from model assumptions, we performed a simulation study for 9 strains, similarly as in Table 3. To estimate overall vaccine efficacy, the only assumption required is (B5). When vaccine enhances clearance of vaccine strains, violating (B5), the mean of the cross-sectional estimator (6) was close to a combined efficacy 1 − (1 − VE W| V̄ 0)(1−VE D) where VE D is one minus the ratio of mean duration of colonisation in vaccinees to that in controls (Web Appendix C Table C2). The combined efficacy could not be estimated as well if vaccine decelerates clearance, which, however, is an implausible assumption. The estimate based on (6) can thus be interpreted more generally as a summary measure of the effect of being vaccinated on susceptibility to acquisition and on duration of colonisation. If individuals are infectious throughout colonisation, this quantity is related to the transmission potential of the strain in question (Rinta-Kokko et al. 2009; Becker and Starczak 1998). Similar results were obtained in extensive simulation studies for strain-specific efficacy against the transmission potential under symmetric competition between strains.
The theoretical results of Section 5 further mean the cross-sectional estimators for overall and strain-specific efficacy are also applicable with any difference in the clearance rates of the target strain and those in the class “rest”, at least under symmetric competition. We verified this by simulations in which the ratio of clearance rates for the target and the rest was as large (small) as 2 (0.5) (Web Appendix C Table C3). The same finding applies to the combined efficacy when the vaccine affects clearance (data not shown).
Analysis of two pneumococcal vaccine trials
We used the cross-sectional estimators of vaccine efficacy to analyse data from the two randomised, controlled trials of the effect of pneumococcal conjugate vaccination on colonisation (Table 1). In both studies, double colonisation had not been recorded and, consequently, we calculated estimates based on (4) (for the vaccine types) and (5) (for the non-vaccine types). These can be interpreted as estimates of VE acq in those individuals susceptible to acquisition. In both studies, vaccine efficacy against colonisation is estimated to be in the range 0.4–0.8 for most serotypes, although confidence intervals are wide. The overall efficacy against the non-vaccine types does not differ from 0 in either of the studies. The HRK efficacy estimates based on the simple odds ratios that do not adjust for within-host replacement in the last column are consistently smaller, demonstrating that the adjustment for time at-risk makes a difference.
7. Discussion
In this paper we consider estimation of VE acq, vaccine efficacy against susceptibility to asymptomatic infection by a microorganism with multiple interacting strains, from cross-sectional data. We propose two estimands of VE acq which differ in the way they condition on being at-risk to acquire the target strains. Specifically, VE W|0 is based on the relative reduction in the overall hazard of acquiring the target strains in an uncolonised individual whereas VE W|V̄ 0 is based on the corresponding hazard in all those susceptible to acquiring the target strains. The first has an individual level interpretation and would ideally be estimated from longitudinal data in the context of a (Markov) transition model. The second has a population level interpretation as the average efficacy under stationary conditions. Both estimands can be used for any set of strains and are coherent in the sense that the efficacy against all target strains is a weighted average of the respective strain-specific efficacies.
In practice, estimates of VE W|0 and VE W|V̄ 0 are similar when double colonisation is un-common. Moreover, we showed assuming symmetry in competition, the two estimands are equivalent in any strain-specific analysis. For better statistical efficiency, estimates of VE W|V̄ 0 seem therefore preferable on most occasions. Under quite general conditions, the estimators yield the combined efficacy against susceptibility to acquisition and duration of colonisation. In particular, no assumption is needed about the clearance rates of the target strains and the rest. If the overall clearance rate of the target strains is not affected by vaccination, both estimators collapse to VE acq. When competition between the target strain and the other vaccine-strains is not very asymmetric, the key assumption is stationarity.
Pneumococcal vaccine trials are most often conducted in infants, in whom a relatively stationary colonisation is achieved in unvaccinated individuals a few months after birth (e.g. Granat et al., 2007). In vaccinees, such data are more scarce, but it is plausible that a similar time period is sufficient to achieve stationarity (Nohynek et al., 2008). In our simulations, taking samples six months after vaccination was enough for successful cross-sectional estimation of vaccine efficacy. If using the proposed method in a specific epidemiologic setting, the appropriate sampling time should be examined by simulation.
To differentiate the effect of vaccination on acquisition and clearance rates requires longitudinal data. Vaccine-induced changes in the average clearance rate of vaccine strains through direct biological enhancement by vaccination on clearance is not supported by available data in the pneumococcal context (O’Brien et al., 1997; Dagan et al., 2002). Vaccination could also affect average clearance rates by changing the within-host mixture of vaccine-strains. This situation would require large differences in strain-specific efficacy for susceptibility and a wide distribution of clearance rates across different strains. Such differences are not evident between the pools of pneumococcal vaccine and non-vaccine serotypes although differences exist among individual serotypes (e.g. Lipsitch et al., 2012). Nevertheless, in these situations, we show the combined efficacy against susceptibility and duration is estimable from cross-sectional data if vaccination does not markedly lengthen duration of vaccine-strain colonisation.
The difference between estimates of VE W|V̄ 0 and the simple HRK estimates can be considerable for strain-specific efficacies (Tables 1 and 4). For pivotal trials with colonisation endpoints, VE W|V̄ 0 appears more justifiable as a measure of the biological effect of the vaccine on transmission potential or susceptibility to acquisition. This estimand is also more likely comparable across epidemiological settings with differing levels of competing exposure by non-vaccine strains. By contrast, the simpler HRK estimate with an anticipated negative efficacy for non-vaccine strains may be more informative about the extent of replacement colonisation. Furthermore, a difference between the HRK estimate and the adjusted one for overall non-vaccine strains could give an indication of between-strain competition.
The a priori partition of strains into vaccine and non-vaccine strains is important in choosing the appropriate reference set of colonisation states ( ). Some non-vaccine pneumococcal serotypes, however, may be affected by the vaccine through cross-immunity (Dagan et al., 2002). In this case, we showed the comparison set can be restricted to a smaller set of non-vaccine strains or even to the uncolonised state. The same argument can be applied in trials in which the control vaccine is another pneumococcal vaccine with a possibly different composition of vaccine-strains compared to the experimental vaccine.
A potential caveat in all approaches to estimating vaccine efficacy for multiple strains is the sensitivity of detection is less than perfect. In this study, the robustness of the suggested aggregation was shown with regard to complete insensitivity to detect double colonisation. However, the situation could be different with differential sensitivity across strains and under asymmetric competition between strains.
The proposed method of cross-sectional measurement of hazard ratios has a close connection to risk-set sampling in nested case-control studies. In particular, hazard ratios can be estimated by selecting controls from those at risk of the event of interest at the incident times of the cases (Wacholder et al., 1998). In the current setting with many states, the at-risk set corresponds to states from which there are arrows to the target states (Lubin, 1986). Specifically, although the setting in this paper is prospective, it can be cast as a case-control study. The “cases” are prevalent, i.e., subjects colonised with the target strains, while the “controls” are those in states from which a transition to the target state is possible or, more generally, states upon which the exposure (vaccine) does not have a direct effect. In contrast to nested case-control studies, cases themselves must not be included in the risk set.
Open questions for future research include incorporating asymmetric competition, heterogeneous response to vaccination, and allowing hazard rates of colonisation to change during the study. In the meantime, we have demonstrated under broadly general conditions, cross-sectional data can be used to estimate the protective effect of vaccination on colonisation.
Supplementary Material
Supplementary Data
NIHMS409126-supplement-Supplementary_Data.pdf (93.2KB, pdf)
Acknowledgments
We thank Mikko Kosola, Sangita Kulathinal and Juha Mehtälä for useful discussions while preparing the manuscript. This study is part of the research of the PneumoCarr Consortium, funded by the Bill and Melinda Gates Foundation through the Grand Challenges in Global Health Initiative. MEH was supported by NIAID R37-AI032042.
Footnotes
Supplementary Materials
Web Appendices A, B and C referenced in Sections 3, 4, 5 and 6 are available under the Paper information link at the Biometrics website
Contributor Information
Kari Auranen, Department of Vaccination and Immune Protection, National Institute for Health and Welfare, Finland.
Hanna Rinta-Kokko, Department of Vaccination and Immune Protection, National Institute for Health and Welfare, Finland.
M. Elizabeth Halloran, Center for Statistics and Quantitative Infectious Diseases, Vaccine and Infectious Disease Division, Fred Hutchinson Cancer Research Center and Department of Biostatistics, University of Washington.
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Summary
1. Introduction
2. Models of colonisation and vaccine effects
3. Overall vaccine efficacy for multiple strains
4. Cross-sectional estimation of vaccine efficacy against colonisation
5. The aggregated model and sufficient conditions
6. Simulations, sensitivity analyses, and data analyses
7. Discussion
Supplementary Material
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A Problem Solver's Paradise : Titu's Lemma
Community » Blogs » A Problem Solver's Paradise » Titu's Lemma
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A Problem Solver's Paradise ===========================
Titu's Lemma
by Potla, Mar 13, 2009, 11:19 AM
Titu's Lemma
It is just a variant of Cauchy Schwartz inequality. It states that for and we will have:
Generalisation
We have for and :
Hint for proof
Just use Cauchy
Applications
Using Titu's lemma one can solve a variety of problems.A list of such problems:
Prove that for positive reals we have:
(NESSBITT)
2.For positive unequal reals we have:
3.Prove that
for all reals
4. are positive numbers such that Prove that:
5.For positive reals we have:
6. Prove that
7. Prove that:
8.Let for . Prove that:
P/S
Any questions might be submitted to me through private messages.
Solutions to these problems(which are correct) can be submitted here.
[edited]!
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I'll a post a quick proof for Nessbit with the Andreescu lemma.
Making squares and using Titu's lemma we have;
Which is equivalent to;
Because
by Papalakis, Mar 13, 2009, 1:08 PM
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Equality cannot hold because it implies tha t
.
This post has been edited 1 time. Last edited by Agr_94_Math, Mar 19, 2009, 7:32 AM
by Agr_94_Math, Mar 14, 2009, 11:12 AM
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.
.
by Agr_94_Math, Mar 14, 2009, 11:28 AM
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Both of you are correct with your problems, I would like to see more solutions!
by Potla, Mar 18, 2009, 8:30 AM
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I mentioned about the CS proof.
Here is another one with PMI (Induction):
It is obviously true for . For ,
........(1)
(check it yourself)
Assume, for , its true. It suffices to show that it is true for .
Or, it suffices to show that :
(from our induction hypothesis)
Which immediately follows from (1)!
.
by Potla, Mar 18, 2009, 8:44 AM
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Agr_Math wrote:
Why will equality not hold?
by Potla, Mar 18, 2009, 10:01 AM
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problem #3.(Aplications of Titu's Lemma )
Applying Titu's Lemma ,and using that
$ a^2 + b^2 \geq \left(\frac {a}{\sqrt {2}} + \frac {b}{\sqrt {2}}}\right)^2$
we obtain the desired inequality
by enndb0x, Jun 9, 2009, 5:46 PM
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by youxia2, Feb 27, 2011, 3:15 PM
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I am posting the solution of the IMO problem well that is too well known but still...............
The inequality is
. when and positive reals.
Now we put .
And the inequality is transformed to
.
Now we apply the Titu's lemma,
because .Hence we are done .
This post has been edited 4 times. Last edited by mathdebam, Jan 28, 2014, 5:20 AM
by mathdebam, Jan 28, 2014, 5:16 AM
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by harekrishna, Feb 8, 2010, 6:58 PM
ANNOUNCEMENT
My Regional Tests are tomorrow (29 th Nov.), and after that I have my school exams, so it seems as if this blog will not be updated for around a week. Sorry for the inconvenience.
Sincerely,
Potla
(creator and owner of this blog )
by Potla, Nov 27, 2009, 7:23 PM
Nice. Keep working
by mathson, Oct 6, 2009, 11:29 AM
Not at all.
by Potla, Sep 4, 2009, 10:35 AM
Wow! You are very smart.
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good going
by turikachi, May 16, 2009, 3:40 AM
Because of my exams..... I must do some work now.
by Potla, Apr 9, 2009, 9:51 AM
well, why no activity here?
by Ramchandran, Apr 6, 2009, 1:46 PM
Thanks for all of yours' kind( )coordination, but I cannot update this bulky( : ) blog till my exams are over, so I want the contrib to continue their job (I am making peine a contrib too).......
by Potla, Mar 3, 2009, 12:56 PM
Really nice solutions.......
However Iliked ochas and murgis problem more because of its simplicity.
however mathias DK sol is just awe strucking
by UKO, Feb 14, 2009, 4:47 AM
@ mathias DK really your proof is beautifull
I didnot wanted to be rude .never mind.
by murgi, Feb 14, 2009, 4:12 AM
@murgi: I just wanted to provide a more beatiful proof.
by Mathias_DK, Feb 13, 2009, 11:08 PM
I couldnot understand why such acomplicated proof is necessary?
when there exits such a simple proof.
by murgi, Feb 13, 2009, 5:11 PM
From now on, this is the place of problem solvers.
Each problem for the day will bring 10 points to the first one to give a solution. Better solutions with ideas other than the ones before will also bring 10 points to the user. Meanwhile, incorrect problems will lose 5 points, so be careful while answering.
by Potla, Feb 13, 2009, 10:28 AM
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Thermal Expansion Coeffic...
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How do different materials compare in thermal expansion coefficients?
Mustafa Youssouf
Teacher Especially English Language. at Teacher (General Studies) (2016–present)
·
11mo
Thermal expansion coefficients measure how much a material expands or contracts with temperature changes. Different materials exhibit varying coefficients, which can significantly affect their performance in structural applications. Here’s a comparison of the thermal expansion coefficients of common materials:
1. Metals
Aluminum: ~22-24 x 10⁻⁶ /°C
Steel: ~11-13 x 10⁻⁶ /°C
Copper: ~16-17 x 10⁻⁶ /°C
Brass: ~18-20 x 10⁻⁶ /°C
2. Plastics and Polymers
Polyethylene (PE): ~100-200 x 10⁻⁶ /°C
Polyvinyl Chloride (PVC): ~50-75 x 10⁻⁶ /°C
Polystyrene: ~70-
Thermal expansion coefficients measure how much a material expands or contracts with temperature changes. Different materials exhibit varying coefficients, which can significantly affect their performance in structural applications. Here’s a comparison of the thermal expansion coefficients of common materials:
1. Metals
Aluminum: ~22-24 x 10⁻⁶ /°C
Steel: ~11-13 x 10⁻⁶ /°C
Copper: ~16-17 x 10⁻⁶ /°C
Brass: ~18-20 x 10⁻⁶ /°C
2. Plastics and Polymers
Polyethylene (PE): ~100-200 x 10⁻⁶ /°C
Polyvinyl Chloride (PVC): ~50-75 x 10⁻⁶ /°C
Polystyrene: ~70-90 x 10⁻⁶ /°C
Epoxy: ~30-60 x 10⁻⁶ /°C
3. Ceramics
Porcelain: ~3-5 x 10⁻⁶ /°C
Alumina: ~7-8 x 10⁻⁶ /°C
Silicon Carbide: ~4.5-5 x 10⁻⁶ /°C
4. Composites
Carbon Fiber Reinforced Polymer (CFRP): ~0-5 x 10⁻⁶ /°C (depends on the matrix)
Glass Fiber Reinforced Polymer (GFRP): ~20-30 x 10⁻⁶ /°C (depends on the matrix)
5. Wood
Softwood: ~3-5 x 10⁻⁶ /°C (along the grain)
Hardwood: ~4-6 x 10⁻⁶ /°C (along the grain)
6. Concrete
Concrete: ~10-12 x 10⁻⁶ /°C (varies with composition)
Summary Table
| Material | Thermal Expansion Coefficient (x 10⁻⁶ /°C) |
| Aluminum | 22-24 |
| Steel | 11-13 |
| Copper | 16-17 |
| Brass | 18-20 |
| Polyethylene | 100-200 |
| PVC | 50-75 |
| Polystyrene | 70-90 |
| Epoxy | 30-60 |
| Porcelain | 3-5 |
| Alumina | 7-8 |
| Silicon Carbide | 4.5-5 |
| CFRP | 0-5 |
| GFRP | 20-30 |
| Softwood | 3-5 (along the grain) |
| Hardwood | 4-6 (along the grain) |
| Concrete | 10-12 |
Conclusion
Understanding the thermal expansion coefficients of materials is essential for engineering and design, especially in applications where temperature fluctuations can lead to stress and deformation. Selecting materials with compatible thermal expansion properties can help prevent structural issues and improve overall performance.
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Related questions
Do any materials have a zero thermal expansion coefficient?
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How do I know the thermal linear expansion coefficient (TLE) of temperature for different steel grades?
What material has the highest thermal expansion coefficient?
How does the choice of materials affect the thermal expansion coefficient of CPUs?
Girija Devi
Former Retired Asst Exe Engineer Civil at Kerala State Electricity Board (1994–2022)
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Author has 6K answers and 5.1M answer views
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Mar 11
Related
What material has the highest thermal expansion coefficient?
Comparison - thermal expansion coefficients
Aluminum has the highest thermal expansion coefficient of all metals. This means that aluminum changes its shape the most when the temperature changes.
A thermal expansion coefficient is related to how much a material changes in size per degree change in temperature; essentially, it measures the fractional change in size of an object with a change in temperature, indicating how much a material expands or contracts when heated or cooled.
This coefficient is related to other coefficients like the coefficient of linear expansion, coefficient of superficial
Comparison - thermal expansion coefficients
Aluminum has the highest thermal expansion coefficient of all metals. This means that aluminum changes its shape the most when the temperature changes.
A thermal expansion coefficient is related to how much a material changes in size per degree change in temperature; essentially, it measures the fractional change in size of an object with a change in temperature, indicating how much a material expands or contracts when heated or cooled.
This coefficient is related to other coefficients like the coefficient of linear expansion, coefficient of superficial expansion (area expansion), and coefficient of volumetric expansion.
Key points about the relationship:
Linear expansion coefficient (α): Represents the fractional change in length per unit temperature change.
Represents change in length per degree temperature change.
The linear thermal expansion coefficient of aluminum is 23–25 x 10–6 per degree Celsius (°C)
Superficial expansion coefficient (β): Approximately twice the linear expansion coefficient, representing the fractional change in area per unit temperature change.
Represents change in area per degree temperature change.
The area thermal expansion coefficient for aluminum is approximately 48 x 10^-6 per degree Celsius; this means that for every degree Celsius increase in temperature, the area of an aluminum object will increase by 48 parts per million of its original area.
Volumetric expansion coefficient (γ): Approximately three times the linear expansion coefficient, representing the fractional change in volume per unit temperature change.
Represents change in volume per degree temperature change.
The volumetric thermal expansion coefficient of aluminum represents the fractional change in volume of aluminum per degree Celsius (or Fahrenheit) change in temperature; essentially, how much the volume of aluminum increases for every degree it is heated up.
Approximately 69 x 10^-6 per degree Celsius.
The volumetric thermal expansion coefficient is roughly three times the linear thermal expansion coefficient of Aluminum.
Felix Chen
Materials scientist, reliability engineer, amateur violinist
·
Author has 487 answers and 2.4M answer views
·
5y
Related
What material has the highest thermal expansion coefficient?
You can check this out visually:
Since stronger interatomic bonding would be expected to better resist the temperature induced separation of atoms, it’s no surprise that materials with higher Young’s modulus, which reflects bond strength, exhibit lower CTEs (coefficient of thermal expansion). That’s the trend in the above diagram as well. Thus, it can be seen that of conventional solid materials polymers exhibit the highest CTEs, generally. In fact, in a work task a few years back we measured the CTE of a polymer as several hundred microstrain per degrees C. That’s near the top range of the ab
Footnotes
Materials Selection in Mechanical Design
You can check this out visually:
Since stronger interatomic bonding would be expected to better resist the temperature induced separation of atoms, it’s no surprise that materials with higher Young’s modulus, which reflects bond strength, exhibit lower CTEs (coefficient of thermal expansion). That’s the trend in the above diagram as well. Thus, it can be seen that of conventional solid materials polymers exhibit the highest CTEs, generally. In fact, in a work task a few years back we measured the CTE of a polymer as several hundred microstrain per degrees C. That’s near the top range of the above chart.
In a few of my previous answers I talk more about CTE and material type:
Felix Chen's answer to Why are the coefficients of linear expansion of metals, alloys, polymers and ceramics significantly different?
Felix Chen's answer to Why is a low thermal expansion related to a low modulus?
Footnotes
Materials Selection in Mechanical Design
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Satya Parkash Sud
M.Sc. in Physics & Nuclear Physics, University of Delhi (Graduated 1962)
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What will happen if a bimetallic strip is made of 2 materials with the same coefficient of thermal expansion?
Question: What will happen if a bimetallic strip is made of two materials haveing the same coefficient of thermal expansion.
Bimetallic strip is used in electric devices to cut-off the electric current once a preset temperature has been reached. When the strip is heated beyond the pre-set temperature, it does not remain straight, as one strip bends more than the other, due to unequal thermal expansion coefficients and the contact is broken, stopping flow of current. Due to stopping of current, the temperature starts to fall, straightening the strip, again making the contact and making the curre
Question: What will happen if a bimetallic strip is made of two materials haveing the same coefficient of thermal expansion.
Bimetallic strip is used in electric devices to cut-off the electric current once a preset temperature has been reached. When the strip is heated beyond the pre-set temperature, it does not remain straight, as one strip bends more than the other, due to unequal thermal expansion coefficients and the contact is broken, stopping flow of current. Due to stopping of current, the temperature starts to fall, straightening the strip, again making the contact and making the current flow. This is how the bimetallic strip works.
Bimetallic strips are used in automatic electric iron, storage water heaters. There could be many more uses for such devices but these two came to mind.
In case the two metals have the same coefficient of thermal expansion, the strip will not do its job of regulating the temperature by making and breaking the electric contact.
Akash Jaiswal
Studying science
·
8y
Related
What does thermal expansion signify? How do we relate the thermal expansion equation for different materials?
The expansion occurring due to applied heat on a body is know as “Thermal Expansion”. When we apply heat to a body the kinetic energy of each of its molecules increases. This increase in kinetic energy results into more number of mutual collisions of molecules. Now when the collisions will occur possessing higher kinetic energy than earlier will make larger space around it. This practically leads to expansion.
Now, relating it to equation part:
Every substance have their own unique arrangement and packaging depending upon the force of attraction acting between molecules. So, the one with lesser
The expansion occurring due to applied heat on a body is know as “Thermal Expansion”. When we apply heat to a body the kinetic energy of each of its molecules increases. This increase in kinetic energy results into more number of mutual collisions of molecules. Now when the collisions will occur possessing higher kinetic energy than earlier will make larger space around it. This practically leads to expansion.
Now, relating it to equation part:
Every substance have their own unique arrangement and packaging depending upon the force of attraction acting between molecules. So, the one with lesser attraction will make more space in surrounding with lesser kinetic energy than the one with higher force of attraction. So, experimentally the coefficient of thermal expansion has been determined for different metals.
for example: coefficient of thermal expansion of copper is [math]16.610^-6 m /mK.[/math] That means if you apply 1 kelvin of heat on a copper rod of length 1 m, it will expand [math]16.610^-6 [/math]m more.
Expansion ∞ Length Lo (i) Note: ∞ is for proportionality sign.
Suppose an iron rod of 10 m and other of 1 m, obviously the larger one will expand more. (other functions remain same)
Expansion ∞ Change in Temperature - Δ (ii)
The more the temperature the higher kinetic energy the molecules posses. Therefore more expansion.
Therefore, on combining Expansion = α L0 ΔT Here α is the thermal expansion of the material.
Now adding L0 on both sides and taking L0 common on the right side (to get the new length of the rod)
L= L0(1+αΔT)
This is for linear expansion, similarly you can relate for superficial and cubical expansion.
Since, you have just asked relation, I think that is enough.
Thanks.
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Related questions
Can a material have more than one coefficient of thermal expansion?
By how much does the coefficient of thermal expansion differ between steel and concrete?
What is an example of a material with a negative thermal expansion coefficient?
How might one go about reducing the coefficient of thermal expansion of a material?
How the coefficient of thermal expansion effects on materials?
Ron Brown
Decades of teaching physics to undergrads
·
Author has 13.6K answers and 84.2M answer views
·
Updated Mar 6
Related
What affects the coefficient of thermal expansion?
This is a great question, because thermal expansion is simultaneously ubiquitous and a tiny effect with big consequences. But it is also a subtle concept.
The glib answer is that it is the properties of the materials themselves that affect the thermal expansion coefficient for any material. But that doesn’t really answer the question, does it? What is it about the properties of the materials that controls their expansion? And to get to that, we need to think small - even though the most noticeable and significant thermal expansion effects are typically on very larger objects - like bridges, bui
This is a great question, because thermal expansion is simultaneously ubiquitous and a tiny effect with big consequences. But it is also a subtle concept.
The glib answer is that it is the properties of the materials themselves that affect the thermal expansion coefficient for any material. But that doesn’t really answer the question, does it? What is it about the properties of the materials that controls their expansion? And to get to that, we need to think small - even though the most noticeable and significant thermal expansion effects are typically on very larger objects - like bridges, buildings, and railroad tracks.
Like most things, it’s way more complicated than can be quickly explained, so I’m going to limit the discussion to just thinking about solids - in fact, crystalline solids. Although the principle is not limited to that, it is easier to think about.
And it’s helpful to have a “model”. So imagine a solid as a collection of atoms in some arrangement connected by tiny invisible springs - and it’s easier to visualize in two dimensions. (I know, that fundamentally doesn’t make sense, since springs are made up of atoms and are not invisible. But try.)
The thing is, the atoms that make up a solid are not fixed in position. Even in a single crystal of some solid, the atoms are vibrating around an equilibrium position relative to one another held together by their atomic interactions (which we are visualizing as the invisible springs). As more thermal energy is absorbed by a solid, the more the atoms vibrate. But that, by itself, does not explain thermal expansion.
Solids form because it is energetically favorable for them to do so. That is, there are atomic or molecular interactions that allow the atoms or molecules to coalesce into solids, but that also prevent them from collapsing in on themselves. That can be represented by a curve representing the potential energy of interaction between neighboring atoms or molecules.
The minimum in such an interaction curve represents the equilibrium separation between near-neighbors if the atoms were not vibrating.
So two things are important about this. The stiffness of the solid depends on how sharp (not a technical term) the curvature about the minimum is. The sharper the curve, the stiffer the material - diamond is more rigid than lead, for example. And that has to do with the bonding mechanisms between those atoms, the nature of the solid.
But the thermal expansion coefficient depends on the asymmetry of the curve near the minimum. The reason is that as thermal energy is absorbed raising the temperature of the solid, the atoms vibrate through a larger amplitude. But if the curve were symmetric, that would not change the average separation of the atoms, so the solid would not expand.
If the curve is asymmetric as shown above, as energy is absorbed and the atoms vibrate through a larger amplitude, their average separation increases slightly. The effect of that over an entire chunk of the solid is for the outside dimensions to increase. It is interesting that stiffer solids typically have smaller thermal expansion coefficients.
The principle isn’t limited to just crystalline solids, but to all metals and alloys, as well as to liquids. Most liquids - water, gasoline, etc., - have much larger volume expansion coefficients than most solids, sometimes by a couple of orders of magnitude.
We think of, say, steel as being very rigid. But the expansion has to be accounted for in large structures. The expansion joints between the separate sections of something like the Golden Gate Bridge, for example, have to be such that they could accommodate the length of the bridge to change by as much as a meter from the coldest to the hottest conceivable temperatures the bridge might be subjected to in order to prevent buckling!
Jesse Berezovsky
Professor of Physics, Case Western Reserve University
·
Upvoted by
Justin Dragna
, PhD in chemistry from UT Austin and
Mark Barton
, PhD physicist with University of Glasgow · Author has 143 answers and 2.3M answer views
·
11y
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Why do different materials have different specific heat capacities?
In many cases, the difference in molar heat capacity between materials is an intrinsically quantum effect. As Kit Kilgour points out, it is not surprising that different materials have different specific heat capacities, since specific heat capacity is "per unit mass" and different materials have different numbers of particles per unit mass. What is more interesting, I think, is why the molar heat capacity (or "heat capacity per particle") differs for different solids.
In classical thermodynamics, one expects the energy of a system to be [math]k_B T/2[/math] for each "degree of freedom". ([math]k_B[/math] is just a
In many cases, the difference in molar heat capacity between materials is an intrinsically quantum effect. As Kit Kilgour points out, it is not surprising that different materials have different specific heat capacities, since specific heat capacity is "per unit mass" and different materials have different numbers of particles per unit mass. What is more interesting, I think, is why the molar heat capacity (or "heat capacity per particle") differs for different solids.
In classical thermodynamics, one expects the energy of a system to be [math]k_B T/2[/math] for each "degree of freedom". ([math]k_B[/math] is just a unit conversion factor between the standard units of temperature and energy). In a solid, the primary contribution to the heat capacity comes from vibration of the atoms that make up the crystal lattice. For a system of N particles that each vibrate about some fixed position, each particle has six degrees of freedom (they can move up/down, forward/back, or left/right, and for each direction, the potential and kinetic energy both count as a separate degree of freedom). So classically, one would expect the vibrations to have total energy [math]U = 3 N k_B T[/math]. The heat capacity per particle (at constant volume) is just the derivative of U/N with respect to T, so [math]C = 3 k_B[/math]. Plugging in numbers, we get a molar heat capacity of [math]C = 3 k_B N_A \approx 25[/math] J/(mol K). If you look up molar heat capacities of some materials at room temperature, you find many that are right around this number. E.g. gold (25.4), sodium (28.2), zinc (25.4), bismuth (25.5), and many more (all numbers in J/(mol K)). But then there are exceptions: beryllium (16.4), boron (11.1), silicon (19.8). Carbon is the most extreme counter-example, with molar heat capacity of 8.5 J/(mol K) as graphite, and 6.1 /(mol K) as diamond. So why do these materials not have the classically expected heat capacity? This is where quantum physics comes in!
According to quantum physics, things can't vibrate with just any amount of energy. Specifically, for an object vibrating at frequency f, the energy comes in steps [math] \Delta E = hf[/math], where h is a constant -- Planck's constant. So the higher the frequency, the larger the separation between energy levels. At low frequencies, when the energy separation is much less than the thermal energy [math]k_B T[/math], the quantization doesn't really matter, and you just get energy [math]k_B T/2[/math] per degree of freedom. If you look at higher and higher frequencies, eventually the separation between energy levels becomes greater than [math]k_BT[/math]. When this happens, it becomes increasingly unlikely for the particles to be able to vibrate at that frequency, even with the minimum amount of energy. That is, the particles essentially can't vibrate at the highest frequencies. Materials have a particular maximum frequency at which the particles can vibrate, which depends on the density and speed of sound in that material. In many materials, the energy level spacing at the maximum frequency is less than thermal energy at room temperature. In that case, the classical approximation works well. But in materials where the particles are particularly light and/or tightly bound, the energy levels at the highest frequencies are spaced far enough apart that the vibration at these frequencies are strongly suppressed at room temperature, leading to lower molar heat capacity.
These types of quantum effects show up in similar ways in more complicated materials as well, where there are additional degrees of freedom, such as rotation. These degrees of freedom are also quantized, so their contribution to the heat capacity behaves classically at temperatures above the quantization energy, but are suppressed at lower temperatures.
I stated above that vibration of the atoms in a crystal is the main contribution to the heat capacity. But in metals, there are also a lot of free electrons around (often as many or more as there are atoms). Classically, each of those electrons has 3 degrees of translational freedom, so you would expect an additional [math]\frac{3}{2} N k_B[/math] contribution to the heat capacity for N free electrons. In reality, the electrons in a metal do contribute to the heat capacity, but much much less than this. So much less that at room temperature, the heat capacity is really dominated by the vibration of the atoms, and you can ignore the electrons. This is another quantum effect. The thing going on here is that because the electrons are spatially confined within the material, the energy of the possible electron states is quantized, and because electrons are fermions, only one electron can be in each state. What happens is that there are enough electrons in there that the lower energy quantum states become completely filled, so if you increase the temperature only those electrons with the very highest energy have any higher energy states to be promoted into. So only a small fraction of the free electrons can accept extra energy, leading to a much lower contribution to the heat capacity than one would expect from classical physics.
Jason Bryant
Engineer and expert user of the interwebs
·
9y
Related
Can a material have more than one coefficient of thermal expansion?
It’s probably true to say that there are no materials that have exactly one coefficient of thermal expansion (CTE). At different temperatures, or over different temperature ranges, the CTE will be different. For some materials the change in CTE can be dramatic, but for others it only changes slightly.
The CTE for 304 stainless, for example, is 17.3 um/m-C from 0–100C. From 0–315C it is slightly higher at 17.8 um/m-C. And, if you heat it from 0–650C it is 18.7 um/m-C. That range of values is not that big and many sources will only give you a single value.
Some plastics will have very different CT
It’s probably true to say that there are no materials that have exactly one coefficient of thermal expansion (CTE). At different temperatures, or over different temperature ranges, the CTE will be different. For some materials the change in CTE can be dramatic, but for others it only changes slightly.
The CTE for 304 stainless, for example, is 17.3 um/m-C from 0–100C. From 0–315C it is slightly higher at 17.8 um/m-C. And, if you heat it from 0–650C it is 18.7 um/m-C. That range of values is not that big and many sources will only give you a single value.
Some plastics will have very different CTEs at different temperatures. It can be quite complicated! I grabbed this graph from the Sandia National Laboratories website:
Charles Chicklis
Former Retired Physics Teacher
·
3y
Related
What are the different types of thermal expansion coefficients?
Alpha is the coefficient of linear expansion.
Beta is the coefficient of areal expansion.
Gamma is the coefficient of volume expansion.
Beta=2×alpha
Gamma=3×alpha
Mahesh Ghadge
Masters in Mechanical Engineering
·
8y
Related
Do any materials have a zero thermal expansion coefficient?
ZERODUR is a glass ceramics manufactured by SCHOTT AG; a manufacturing company of glass ceramics based in Germany.The thermal expansion coefficient of Zerodur is 0 ± 0.007×e-6 K^-1 in the range of 0 to 50 degree Celsius, which is the lowest till date for any material and because of having such a low thermal expansion coefficient which is as good as zero, it has many applications in aerospace and astronomy. To know more refer:
PhD in Physics & Thermoelectrics, University of Florida (Graduated 2017)
·
Author has 599 answers and 338.3K answer views
·
7y
Related
What is the thermal expansion coefficient of air?
Assuming the volume of air to be an ideal gas of temperature T, its volumetric coefficient of thermal expansion $\alpha_V$ is given by the ideal gas law as,
Assuming the volume of air to be an ideal gas of temperature T, its volumetric coefficient of thermal expansion $\alpha_V$ is given by the ideal gas law as,
Gustavo Marques Hobold
Master's student
·
Upvoted by
Gary Hiel
, College Chem Prof and former Industrial Organic Chemist · Author has 261 answers and 1.9M answer views
·
11y
Related
Why do specific heat capacity of objects differ?
Because objects with different molecular structures have different ways of storing energy. Since the specific heat is a macroscopic "counting" property which counts how much energy certain materials can store, they are different for different materials.
For example: the gas argon is monoatomic because it is composed by several molecules of a single atom (Ar). A single atom has only "three ways" of storing energy, which correspond to moving faster in the three directions we know (translational kinetic energy). A more complex gas, oxygen, is composed of many molecules of two atoms of oxygen (O2)
Because objects with different molecular structures have different ways of storing energy. Since the specific heat is a macroscopic "counting" property which counts how much energy certain materials can store, they are different for different materials.
For example: the gas argon is monoatomic because it is composed by several molecules of a single atom (Ar). A single atom has only "three ways" of storing energy, which correspond to moving faster in the three directions we know (translational kinetic energy). A more complex gas, oxygen, is composed of many molecules of two atoms of oxygen (O2) and this molecule has more ways of storing energy, mainly by rotating (rotational kinetic energy) and moving around (translational kinetic energy). Therefore, the specific energy of the gas oxygen should be higher than argon's because it has more ways of storing energy. Bigger molecules, as you can imagine, have many other ways of storing energy. Water, for example, can rotate in several ways, move around, vibrate and interact with other molecules to store energy, so its specific heat is even higher.
Jousef Murad
Community Manager at SimScale (2017–present)
·
Author has 88 answers and 634.5K answer views
·
8y
Related
What is the difference between instantaneous and the secant coefficient of thermal expansion? Which should be used when?
If we look at a plot of thermal strain vs. temperature, the average CTE is a measure of the thermal strain generated due a change in temperature from a specific reference temperature to the current temperature. The instantaneous CTE represents the thermal strain generated due to an infinitesimal change in temperature around the current temperature.
You can see that average CTEs generated using 1 reference temperature cannot be used with a different reference temperature. For example, if I know the thermal strain generated due to heating a part up from 70 degrees to 400 degrees, I cannot use tha
If we look at a plot of thermal strain vs. temperature, the average CTE is a measure of the thermal strain generated due a change in temperature from a specific reference temperature to the current temperature. The instantaneous CTE represents the thermal strain generated due to an infinitesimal change in temperature around the current temperature.
You can see that average CTEs generated using 1 reference temperature cannot be used with a different reference temperature. For example, if I know the thermal strain generated due to heating a part up from 70 degrees to 400 degrees, I cannot use that same CTE to calculate the thermal strain generated in a cool down from 730 degrees to 400 degrees, even though the final temperature and the delta temperature are the same.
You MUST make sure that any average temperature dependant CTE’s entered are corrected for the operating reference temperature. Many FEA codes can do this conversion automatically, for example, the MPAMOD command in ANSYS will convert average CTE from one reference temperature to another.
If you have instantaneous data instead of average data, there is no conversion needed. Instantaneous data is valid for any reference temperature.
So, be sure to understand what your CTE data represents before using it in a finite element model!
Source: Using Coefficients of Thermal Expansion in a Finite Element Model
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What are Mathematical Combinations and How To Calculate Them
What are Mathematical Combinations and How To Calculate Them
Written by
Indeed Editorial Team
Updated July 24, 2025
Combinations are mathematical figures that statisticians, data analysts, software engineers and other technical professionals often use to represent an unordered set of items in a series of arrangements. You can apply combinations and their associated formulas to many areas, including information technology, health care, finance and accounting. Learning more about combinations and how to determine them can help you succeed in a data-centered role.Keep reading to discover what combinations are and learn formulas, tips, and examples to help you calculate how many combinations you have.
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What is a combination?
A combination refers to the number of arrangements you can create when taking a sample of values or items from a bigger set. The combinations you can form show how many subsets you can make from the entire set of items. In a mathematical combination, the order of items is unimportant. This means that you can use the values from the set to group combinations in any order, although some combinations result in an ordered sequence, resulting in a permutation. Additionally, combinations can be repeating, like this:(0, 0, 1, 1, 2, 3)They can also be non-repeating, like this:(0, 1, 2, 3, 4)Here’s a real-life example of a combination in use:A software engineer encounters a situation where they can choose three potential applications out of a list of 20 options. Using a formula, they determine how many potential combinations of applications they could select from within the 20 options. This helps them learn more about their available choices and guides their decision-making process.Related: Everything You Need To Know About Predictive Analytics
Formula for finding how many combinations you have
You can use the following formula to calculate the number of ways in which you can get an arrangement of non-repeating items (r) from a larger set of distinct items (n) when the order is unimportant:C(n, r) = (n!) / [(r!) x (n - r)!]The formula for calculating combinations also requires the computation of factorials, which are the products of all positive integers equal to and less than the number you’re computing. The factorial within the combination formula shows as a (!), which represents the factorial functions you can apply as part of your calculation.Related: How To Calculate Probability in Excel (With an Example)
How to calculate combinations
Use the formula for calculating combinations:C(n, r) = (n!) / [(r!) x (n - r)!]Then follow these four steps to calculate how many combinations you can obtain from a sample set:
1. Determine your r and n values
Find your r and n values by choosing a smaller set of items from a larger set.For example, assume you have eight books on a shelf, and you want to choose four books to read. In the formula, the (r) represents the sample of the four books you choose and the (n) represents the larger set of eight books. When you determine these values, substitute them for the (r) and (n) variables in the formula:C(n, r) = (8!) / [(4!) x (8 - 4)!]Related: 20 Jobs for Mathematics Degree Holders
2. Subtract your r value from your n value
When you identify your sample set and larger set for the r and n variables in the formula, subtract these two values. Using the example of n = 30 books and r = 10 books in the previous step, find the difference:C(n, r) = (8!) / [(4!) x (8 - 4)!] =C(n, r) = (8!) / [(4!) x (4)!]Related: How To Calculate Percent
3. Expand on the factorials
When you've simplified the expressions within the formula, you can start calculating each factorial in the problem. In the example with the larger set of eight books, the sample set of four books and the difference you find when subtracting these values, expand each factorial in the formula:8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 and 4! x 4! = 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1
Related: 18 Highest Paying Math Jobs That You Can Pursue
4. Cancel like terms and divide
Before dividing your factorials in the formula, you can cancel out the like terms between them since these values connect through division. Between 8! and (4! x 4!), the common terms include 4, 3, 2, 1 on both the top and bottom of the fraction portion of the equation. Canceling these terms out results in:C(n, r) = (8 x 7 x 6 x 5) / (4 x 3 x 2 x 1) =C(n, r) = (1,680) / (24) =C(n, r) = 70 The result is 70, which represents how many combinations you can get when choosing four books from a set of eight when n ≥ r ≥ 0. This means that all values you work with can be greater than or equal to zero, and your (n) value can be greater than or equal to your (r) value.Related: How To Calculate Ratios (With Example)
Combinations vs. permutations
Combinations can have both repeating and non-repeating arrangements, but the order that you find these arrangements doesn’t matter. With permutations, though, the order becomes important, as permutations must occur in an ordered set. One way to remember the difference between these two concepts is that a permutation is an ordered combination.As with combinations, permutations can also have repeating or non-repeating values within a set. For example, assume you want to find the combination to a lock. The combination to the lock represents a permutation in which the order of the numbers is important for the combination to work. You may have a repeating permutation for the lock combination (such as the combination 4, 4, 4), or you may have a non-repeating permutation (as in 3, 4, 5).Related: 44 Probability Interview Questions
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Permutation calculation example
To calculate combinations when the order matters, as with permutations, you can use two formulas:
Repeating permutation= n^r: In this formula, (n) represents different types of an item and (r) represents the smaller set of types you choose from (n).
Non-repeating permutation = (n!) / (n - r)!: In this formula, (n!) is the factorial function of the entire set from which you choose (r) number of items.
Using these formulas, you can find the permutations in the following example problem:Assume you want to find the combination to a lock, and you have six numbers from which to choose. Six represents your (n) value in both formulas. If the lock requires three numbers for the combination, three becomes your (r) value or the smaller subset you choose from the entire set. To find how many combinations you can obtain with repeating values, use the formula n^r:n^r = (6) ^ (3) =6 x 6 x 6 = 216 This means there are 216 possible combinations you can find that have repeating values. However, if the lock combination has non-repeating values, you use the formula (n!) / (n - r)! to calculate the possible combinations for each of the three items you choose from the six values in the set:(n!) / (n - r)! =(6!) / (6 - 3)! =(6 x 5 x 4 x 3 x 2 x 1) / (3)! =(6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1) =6 x 5 x 4 / 1 =120 / 1 = 120 This result indicates there are 120 possible combinations for a lock with values arranged into permutations.Similar to calculating combinations you can simply reduce the factorials that are in common between each factor you divide in the formula. These formulas can help you better understand the possible combinations or permutations within a larger set.
The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers
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2013 AMC 10B Problems/Problem 16
Contents
[hide]
1 Problem
2 Solution 1 ( mass points)
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6
8 See also
Problem
In triangle , medians and intersect at , , , and . What is the area of ?
Solution 1 ( mass points)
Let us use mass points: Assign mass . Thus, because is the midpoint of , also has a mass of . Similarly, has a mass of . and each have a mass of because they are between and and and respectively. Note that the mass of is twice the mass of , so must be twice as long as . PD has length , so has length and has length . Similarly, is twice and , so and . Now note that triangle is a right triangle with the right angle . Since the diagonals of quadrilaterals , and , are perpendicular, the area of is
Solution 2
Note that triangle is a right triangle, and that the four angles (angles and ) that have point are all right angles. Using the fact that the centroid () divides each median in a ratio, and . Quadrilateral is now just four right triangles. The area is
Solution 3
From the solution above, we can find that the lengths of the diagonals are and . Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is
Solution 4
From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
Solution 5
We know that , and using median properties. So Now we try to find . Since , then the side lengths of are twice as long as since and are midpoints. Therefore, . It suffices to compute . Notice that is a Pythagorean Triple, so . This implies , and then . Finally, .
~CoolJupiter
Solution 6
As from Solution 4, we find the area of to be . Because , the altitude perpendicular to . Also, because , is similar to with side length ratio , so and the altitude perpendicular to . The altitude of trapezoid is then and the bases are and . So, we use the formula for the area of a trapezoid to find the area of
See also
2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15Followed by
Problem 17
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 10 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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Beta Distribution
Basic Concepts
Definition 1: For the binomial distribution the number of successes x is a random variable and the number of trials n and the probability of success p on any single trial are parameters (i.e. constants). Instead, we would now like to view the probability of success on any single trial as the random variable, and the number of trials n and the total number of successes in n trials as constants.
Let α = # of successes in n trials and β = # of failures in n trials (and so α + β = n). The probability density function (pdf) for x = the probability of success on any single trial is given by
This is a special case of the pdf of the beta distribution
where Γ is the gamma function.
Note that in the general case, α + β does not have to be a positive integer, although α and β do have to be positive numbers and x must be between 0 and 1.
Chart
A chart of the beta distribution for β = 8 and α = 2, 4 and 6 is displayed in Figure 1.
Figure 1 – Beta Distribution
Actually, the beta distribution can take a great variety of shapes, as shown in Figure 2.
Figure 2 – Other shapes of beta distribution
Properties
Key statistical properties of the beta distribution are:
Figure 3 – Statistical properties of the beta distribution
The mode is (a) any value in (0,1) for α, β = 1, (b) 0 or 1 for α, β < 1, (c) 0 if α ≤ 1 and β > 1 and (d) 1 if β ≤ 1 and α > 1.
Excel Worksheet Functions
Excel Functions: Excel provides the following functions:
BETA.DIST(x, α, β, cum) = the pdf of the beta function f(x) when cum = FALSE and the corresponding cumulative distribution function F(x) when cum = TRUE.
BETA.INV(p, α, β) = x such that BETA.DIST(x, α, β, TRUE) = p. Thus BETA.INV is the inverse of the cdf of the beta distribution.
These functions are not available in versions of Excel prior to Excel 2010. Instead, these versions of Excel use BETADIST(x,α,β), which is equivalent to BETA.DIST(x,α,β,TRUE), and BETAINV, which is equivalent to BETA.INV.
Observations
Note the following equality when s and n are positive integers with n > s:
BETA.DIST(p, s, n-s, TRUE) = 1 – BINOM.DIST(s-1, n-1, p, TRUE)
The cdf of the beta distribution is also called the regularized incomplete beta function, denoted Ix(α,β). Thus, Ix(α,β) = BETA.DIST(x, α, β, TRUE).
Example
Example 1: A lottery organization claims that at least one out of every ten people wins. Of the last 500 lottery tickets sold 37 were winners. Based on this sample, what is the probability that the lottery organization’s claim is true: namely that players have at least a 10% probability of buying a winning ticket? What is the 95% confidence interval?
To answer the first question we use the cumulative beta distribution function as follows:
BETA.DIST(.1, 37, 463, TRUE) = 98.1%
This represents that organization’s claim is false (i.e. less than 10% probability of success). The probability that the organization’s claim is true is only 100% – 98.1% = 1.9%.
The lower bound of the 95% confidence interval is
BETA.INV(.025, 37, 463) = 5.3%
The upper bound of the 95% confidence interval is
BETA.INV(.975, 37, 463) = 9.8%
Since 10% is not in the 95% confidence (5.3%, 9.8%), we conclude (with 95% confidence) that the lottery’s claim is not accurate.
Real Statistics Worksheet Functions
Real Statistics Function: The Real Statistics Resource Pack provides the following function:
BETA(α, β) = the beta function = Γ(α)Γ(β)/Γ(α+β)
Thus, the pdf of the beta distribution is
Four-parameter Distribution
Observation: The two-parameter version of the beta distribution, as described above, is only defined for values of x between 0 and 1. There is also a four-parameter version of the distribution for which x is defined for all x between a and b where a < b.
The cdf for the four-parameter beta distribution at x is F(z) where F is the cdf for the two-parameter distribution and z = (x–a)/(b–a). The pdf at x is
Excel Functions: Excel provides the following functions to support the four-parameter version of the beta distribution.
BETA.DIST(x, α, β, cum, a, b) = the pdf of the beta function f(x) when cum = FALSE and the corresponding cumulative distribution function F(x) when cum = TRUE.
BETA.INV(p, α, β, a, b) = x such that BETA.DIST(x, α, β, TRUE_, a, b_) = p; i.e. the inverse of the cdf of the beta distribution.
These functions are not available in versions of Excel before Excel 2010. Instead, these versions of Excel use BETADIST(x, α, β, a, b), which is equivalent to BETA.DIST(x, α, β, TRUE_, a, b_), and BETAINV, which is equivalent to BETA.INV.
Examples Workbook
Click here to download the Excel workbook with the examples described on this webpage.
References
Taboga, M. (2017) Beta distribution. Lectures on probability theory and mathematical statistics, Third edition. Kindle Direct Publishing
Wikipedia (2020) Beta distribution
51 thoughts on “Beta Distribution”
Steven Cabrera
January 9, 2024 at 1:44 am
Is the Real Statistics BETA function compatible with the first formula shown on this page (the formula right under “A/B testing: binary outcomes”)?:
Thank you.
Reply
Charles
January 9, 2024 at 9:34 am
Hello Steven,
The Real Statistics BETA function is equivalent to the B function on that page.
Charles
Reply
LM
October 12, 2022 at 3:23 pm
Thanks for sharing. Could we not have tested the lottery organisation’s claim by just building a binomial model with p = 0.1, calculating P(X ≤ 37), and rejecting the null hypothesis at a chosen significance level, say 5%?
Reply
Charles
October 12, 2022 at 9:39 pm
LM,
First of all, the scenario is not for p = .1, but for any value of p >\= .1.
Charles
Reply
Ronald
November 26, 2020 at 10:51 am
Hi Charles,
I have the following set of data (18) and trying to find the best fit assuming Beta distribution:
2 2.4 2.6 2.9 3.1 3.9 3.9 3.9 4.1 4.1 4.2 4.6 4.6 4.7 4.9 4.9 5 5
Please can you assist. Thanks a lot.
Ronald
Reply
Charles
November 26, 2020 at 5:21 pm
Ronald,
See
Two methods are described for the beta distribution. Usually, the MLE approach is better.
Charles
Reply
Arnel Mark John M. Cristobal
January 23, 2020 at 3:38 pm
Hi charles im not too much good in statistics, however i am very obsessed with non normal distribution. Since i am developing an excel file which i want to be very dynamic and sensitive with the skewed plot. I have some questions;
1. Could be the beta distribution able to show or represent a negative skewed graph? (i learnt gamma distribution only show positive skewed)
2. If i have a set of data (assumed it is negative skewed) ranging from cell A1:A1000, how can i compute for a”alpha” and B”beta” to be filled in the “=beta.dist(x,a,B,cum)”? (i learnt that “a = mean^2 / variance” and “b = variance / mean” which fits a gives me a good graph in the funtion gamma.dist in excel, however using it in beta.dist gives me an error result)
Reply
Arnel Mark John M. Cristobal
January 23, 2020 at 3:55 pm
In connection with question number 2, i forgot to tell that the range cell A1:A1001 is a set of ungrouped raw data (assumed to be skewed).
Cell B1 “=average(a1:a1001)
Cell B2 “=stdev.s(a1:a1001)
Then, on cell C1:C21 i grouped them into 20 classes
Cell C1 “=B1-(10B2)
Cell C2 “=C1+$B$1
Drag down C2 upto C21
Now on D1:D21 i want to get their “=beta.dist(C1,a,B,cum)
a=?
B=?
Reply
Charles
January 23, 2020 at 4:32 pm
If the approach suggested in my previous response is not appropriate, then perhaps you can get the answer using Solver.
Charles
Reply
Charles
January 23, 2020 at 4:19 pm
1. Using a for alpha and b for beta, the skewness of the beta distribution is 2(b-a)Sqrt(a+b+1)/(a+b+2)Sqrt(ab)). Thus, you can choose appropriate values of a and b to obtain the type of skewness that you desire (if it exists).
2. The Real Statistics website describes two methods for fitting a beta distribution to some data. See the following webpages:
If an error is returned it might mean that the data can’t be fit by a beta distribution
Charles
Reply
Arnel Mark John M. Cristobal
January 24, 2020 at 4:32 am
Thank you so much charles
Another question, could you site any distribution that may fit with the negatives skewed histogram
Reply
Charles
January 25, 2020 at 7:33 pm
See
Charles
Reply
Skitskraj
June 19, 2018 at 12:56 pm
Hi, this is probably trivial to people who know statistics well, but can you please explain/give reference to the following statement in the text:
The probability density function (pdf) for x = the probability of success on any single trial is given by
Reply
Charles
June 28, 2018 at 5:51 pm
See the following webpage for an explanation:
The following webpage may also be helpful:
Charles
Reply
Matija
June 4, 2018 at 6:43 pm
Hi,
when calculating the lower and upper bounds -> if the sample size is small, one should use the following formulas:
alpha – confidence interval
k – number of successes
n – number of trials
lower: BETAINV((1-alpha)/2, k, n-k+1)
upper: BETAINV((1+alpha)/2, k+1, n-k)
however, i’m not sure, if any changes should be applied to the BETADIST function.
Reply
Charles
June 4, 2018 at 7:35 pm
Matijia,
What is the reference for the statement “if the sample size is small, one should use the following formulas…”?
Charles
Reply
VANKUDOTH VEERANNA
December 10, 2021 at 7:27 am
i want to find out activity weightage and Quantity , through calendardate , here I will create report like below
start date finish date activity calender
weightage
01/09/2021
are given , 02/01/20 20/04/20 94.4
to create formule in Beta distribution
Reply
Charles
December 10, 2021 at 9:07 am
Sorry, but I don’t understand your question.
Charles
Reply
Andrew
July 13, 2018 at 5:24 pm
Matijia,
Thank you for providing these equations. Using them, I was able to match the behavior of the Matlab binofit() function.
Andrew
Reply
Raj
January 3, 2018 at 10:30 pm
Thank you Charles ..
However May i know what is the application of this charts
regards
Raj
Reply
Charles
January 3, 2018 at 10:39 pm
Raj,
To provide a sense of what the distribution looks like and what is the effect of changing a parameter value.
Charles
Reply
Raj
January 3, 2018 at 1:48 pm
How did we get the graphs in the Beta distribution. can you please explain
Reply
Charles
January 3, 2018 at 2:43 pm
Raj,
Enter 2 in cell B1, 4 in cell C1 and 6 in cell D1.
Next enter 0 in cell A2 and the formula =A2+.02 in cell A3. Highlight the range A3:A51 and press Ctrl-D.
Now insert the formula =BETA.DIST(B2,B$1,8,FALSE) in cell B2. Highlight range B2:D51 and press Crtl-R and then Ctrl-D.
This completes all the data. You now create the chart by highlighting range and selecting Insert > Charts > Line Chart.
You can then add titles and make other modifications as described on the webpage
Excel Charting
Charles
Reply
Lam
December 2, 2017 at 2:49 am
I used Exact Excel method to calculate the two sided CI (95% Confidence Level) as below. It is equivalent to “Clopper Pearson” method.
The upper bound is BETAINV(0.975, 38, 463) =10.1%. The lower bound is BETAINV(.025, 37, 464) = 5.3%.
In this case, 10% is in the CI (5.3%, 10.1%), we can conclude that the lottery’s claim is accurate.
In sum, the exact method uses α=38 instead of 37 used in the upper bound. In contrast, using the exact method draws a different conclusion although the two set of confidence limits are very close. There is statistical significance but no practical significance.
Reply
Fred
October 11, 2017 at 2:50 pm
Charles,
First, thanks for devoting time to set up that website, it is a goldmine!
In your lottery example (nber 1) above, could we also use the Binomial Distribution to compute the probability to observe 37 or less winning tickets if the underlying probability is 10%?
So, =BINOM.DIST(37,500,0.1,1)?
Many thanks in advance Charles,
Fred
Reply
Charles
October 14, 2017 at 5:56 pm
Fred, BINOM.DIST(37,500,.1,1) = .02743, which means that the probability of getting 37 or fewer winning tickets out of 500 is 2.743% when the probability of getting a winning ticket (on any one ticket) is 10%. This doesn’t answer question 1.
Charles
Reply
Fred
October 15, 2017 at 2:26 pm
Thank you! I was looking at it from the point of view of hypothesis testing, We would reject H0: p=0.1 because p-value=0.02743 <0.05?
You should really self-publish an e-book version of the website on Amazon.
Thanks again!
Fred
Reply
R Singh
February 28, 2017 at 7:04 pm
Hello there,
For some reason, the equations on all webpages are not displayed. May be some background maintenance is underway; just wanted to bring to your attention.
Thanks!
Reply
Charles
March 1, 2017 at 7:55 am
Thanks for bringing this to my attention.
I don’t know why the equations were not displayed. I see that they are displayed now (at least on my computer).
Charles
Reply
Karl Buennagel
March 21, 2016 at 6:19 pm
Nice job Charles. That is a sweet explanation. I saved the webpage. Thanks! If this is a book, I need a copy!
Reply
Charles
March 21, 2016 at 8:54 pm
Karl,
There will be a series of books shortly.
Charles
Reply
Adam
March 9, 2016 at 12:14 am
I am a little confused how Bayesian updating is carried out for the following set-up:
I have developed a logistic regression model which provides % probability of a 1 or 0. The 1 or 0 is then compared to what actually happened and a Beta Distribution is updated for each new success or failure which provides % probability of successes (1s).
I now which to use my Logistic Regression model which new data to provide a new % estimate of a 1. This in turn now needs to be updated with the previously calculated probability from the Beta distribution calculated on the previous time step.
How should I use Bayes to update the current Logistic regression probity with the actual % wins/losses? I am confused as which one is the prior and psoterior?
Thanks for the help.
Adam
Reply
Charles
March 9, 2016 at 8:41 am
Sorry Adam, but I really don’t understand the scenario that you are describing.
Charles
Reply
Anthony
July 20, 2016 at 7:59 pm
I believe Adam was referring to how the Beta distribution is often used iteratively in Bayesian stats. It can be explained as the probability of probabilities. E.g. what are the chances of having a final sample mean of 0.3 if the current sample mean is 0.27 with 2 out of 9 iterations remaining.
The usual example is baseball, see but your lottery example works.
In summary, if we approach the goal at a faster rate than expected then we are more likely to meet or exceed that goal. And if we fall behind the less likely we are to are to meet it.
Reply
Charles
July 22, 2016 at 6:03 pm
Anthony,
Thanks for your explanation.
Charles
Reply
mitrananda
December 28, 2015 at 9:20 pm
how the alpha and beta values are to be taken for a problem.
please explain me.
thank you
Reply
Charles
December 29, 2015 at 10:56 pm
For the problem described on the referenced webpage, α = # of successes in n trials and β = # of failures in n trials (and so α + β = n).
Charles
Reply
John Vanneck
November 17, 2015 at 11:39 am
Hi,
When there are 0 successes or n successes (out of n trials), the formula returns #NUM!
Do you have a work around for this? As in the situation where we had 100 trials and 0 successes we could safely assume that the probability of success was close to 0 – but this formula would not give an interval.
Reply
John Vanneck
November 17, 2015 at 11:40 am
oops – I forgot to mention that I am talking about estimating confidence intervals using the beta.inv(p,alpha,beta) formula. Thank you
Reply
Charles
November 28, 2015 at 3:15 pm
John,
Note that BETA.DIST(p, s, n-s, True) = 1 – BINOM.DIST(s-1, n-1, p, True).
Thus, BETA.DIST(p, 0, 100, True) = 1 – BINOM.DIST(-1, 99, p, True).
This means that the problem you are addressing is equivalent to asking (the binomial distribution question) what is the probability of getting -1 successes in 99 trials (for any value of p = probability of success on any single trial)? This is clearly 0 with a confidence interval consisting only of the point zero. Thus the confidence interval you are looking for is [1,1], i.e. the point 1.
Charles
Reply
Rick
June 15, 2015 at 4:50 pm
“Thus the probability that the organization’s claim is true is 100% – 98.1% = 1.9%”
This sure sounds like the inverse probability error.
Pr(.074 | π = .1) = .019
does not mean
Pr(π = .1 | .074) = .019
Would it not be better to say that obtaining p < 37/500 = .074 would occur only 1.9% of the time if the organization’s claim (H0: π = .1) were true?
Rick
Reply
Charles
June 25, 2015 at 4:36 pm
Rick,
While you are right to be cautious, I believe the logic used is correct.
Pr(π < .1 | # of successes in 500 trials = 37 and # of failures in 500 trials) = BETADIST(.1, 37, 463) = 98.1% Thus Pr(π >\= .1 | # of successes in 500 trials = 37 and # of failures in 500 trials) = 1 – 98.1% = 1.9%
Charles
Reply
Charles
June 25, 2015 at 7:15 pm
Rick,
While you are right to be cautious, I believe the logic used is correct.
Pr(π < .1 | # of successes in 500 trials = 37 and # of failures in 500 trials) = BETADIST(.1, 37, 463) = 98.1% Thus Pr(π >\= .1 | # of successes in 500 trials = 37 and # of failures in 500 trials) = 1 – 98.1% = 1.9%
Charles
Reply
Matthias
September 3, 2017 at 11:29 am
I noticed that in this example, assuming 50 wins and 450 losses, so exactly 10%, yields:
BETADIST(.1, 50, 450) = 51.6%
and increasing the sample size to e.g. 5000 out of 45000 or higher, only leads to values closer and closer to 50%.
With the stated logic there would always be at least a 50% chance the claimed chance of 10% wins is a fraud.
Could it be that the correct calculation for 50 out 500 is the following:
2(51.6% – 50%) = 3.2% chance of a lie
And with your example 37 out of 463:
2(98.1% – 50%) = 96.2% chance of a lie
This way the whole range from 0% to 100% truth or lie is possible and entering the promised 10% wins approaches 0% chance of a lie with increasing sample size.
Not sure about this, though.
Thank you for the helpful article!
Reply
Charles
September 13, 2017 at 12:06 pm
Matthias,
I don’t know what you mean by “exactly 10%” in your first sentence. I also don’t know what you mean by “a lie”. Sorry, but I also don’t understand the calculations. Let me review a few things.
BETADIST(.1, 50, 450) is the cumulative probability distribution function and is equal to 51.59%. This means that given that we have observed 50 winners in 500 lottery tickets, the probability that any single ticket has up to a 10% chance of winning is 51.59%. Note that I have said “up to a 10% chance of winning” and not “exactly a 10% chance of winning”. All this is difficult to interpret since we have two levels of probability.
The chance that any single ticket has more than a 10% of winning is therefore 48.41%. Thus, it is more likely that any single ticket is a winner less than 10% of the time than more than 10% of the time, but the difference is close (and so my be due to chance).
Note that if you increase the sample size to 5000, but keep the 50 fixed, then BETADIST(.1,50,4950) = 1, which is what you would expect since it seems pretty unlikely that 10% of the tickets are winners given that only 50 out of 5,000 were winners. In fact, note that BETADIST(.01,50,4950) = 51.86%.
Charles
Reply
Justin
March 28, 2015 at 11:38 am
Good and simple explanation of beta distribution compared to 99% sites I looked.Could you tell why we place (n-1)! in the numerator when the number of trials will always be an whole number.It might be certain to have (\alpha-1)! and (\beta-1)! in the denominator since and might be real numbers.Also why isn’t the power of x and (x-1) in the numerator \alpha and \beta instead of (\alpha-1) and (\beta-1) given in the expression?
Reply
Charles
April 2, 2015 at 10:45 am
Justin,
The only time I need to use the beta distribution on the website is when the alpha and beta values are integers, although the beta distribution is used for many other purposes, including cases where the alpha and beta parameters are not integers. The distribution uses the gamma function. You see a number of instances of some integer minus 1. The reason for this is that Gamma(n) = (n-1)! when n is a positive integer.
Charles
Reply
Richard
October 27, 2015 at 4:05 pm
I like your derivation (whenever I have seen this done previously, it has been through repeated application of integration by parts on the cdf, and is much more complicated), but I have a similar problem to Justin. To go from the pdf
f(x) = n!/k!/(n-k)!x^k(1-x)^(n-k)
I would have to set a-1=k and b-1=n-k, which gives
f(x) = n!/(a-1)!/(b-1)!x^(a-1)(1-x)^(b-1)
and is therefore not the same as the first of your equations above (by a factor of n). Do you see where I’m going wrong?
Reply
Charles
November 3, 2015 at 4:29 pm
Richard,
First note that Gamma(m) = (m-1)! for any positive integer m. Thus for any positive integers alpha and beta, (alpha-1)! = Gamma(alpha) and (beta-1)! = Gamma(beta). Now let n = alpha + beta. Then (n-1)! = Gamma(n) = Gamma(alpha+beta).
Putting the pieces together yields (n-1)!/((alpha-1)!(beta-1)!) = Gamma(alpha+beta)/(Gamma(alpha)Gamma(beta)).
This is not the same thing as deriving the cdf from the pdf.
Charles
Reply
Sadam Hussain Channa
December 26, 2017 at 6:34 pm
I used Beta Distribution Equation in Excel but result not same of Beta.Dist…Why
Charles
December 27, 2017 at 9:49 am
Sadam,
BETADIST calculates the cumulative distribution function (cdf) F(x), while the formula is for the probability density function (pdf) f(x).
BETA.DIST(x,alpha,beta,FALSE) will calculate the pdf.
Charles
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http://www.geometryexpressions.com/downloads/Conics%20and%20Loci.pdf
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Conics and Loci Studies in Precalculus Using Geometry Expressions 2008 Saltire Software Incorporated Page 2 of 54 Conics and Loci Unit Overview Many traditional studies of conic sections center on the implicit formulas and how they can be used to quickly sketch graphs. These skills are quickly becoming outdated as graphing technologies evolve. Geometric definitions of conic sections are often treated as a separate topic, sometimes even in a separate course. The intent of this unit is to look at conic sections through three lenses: Definitions of conic sections as a locus of points Parametric equations of conic sections, in terms of trigonometric functions Implicit equations of conic sections The main goal of this unit is to make connections between these three representations of the conic sections. At the end of the unit, students will have geometric definitions, parametric equations, and implicit equations for circles, ellipses and hyperbolas, as well as a geometric definition and implicit equation for parabolas. Lesson 1: Introducing Loci • The concept of a locus of points is introduced. Circles, angle bisectors, and perpendicular bisectors are among the loci that are studied. While interesting in its own right, most of this lesson gets students acquainted with the features of Geometry Expressions that they will need throughout the unit. Lesson 2: Loci: Circles • The definition of a circle as a locus of points is presented. Connections are made to the unit circle, the Pythagorean Theorem, and to the distance formula. Lesson 3: The Ellipse • An ellipse is constructed using the geometric definition. Students use Geometry Expressions’ symbolic capabilities to generate the parametric and implicit equations. Lesson 4 Envelope Curves • The concept of a locus of lines and the resulting envelope curve is used to re-define the ellipse. This definition is extended to introduce hyperbolas. Lesson 5 Inside-Out Ellipses • Lesson 4 is used as a jumping-off point for this lesson on hyperbolas. The geometric definition, parametric equation, and implicit equation are developed through comparison with ellipses. 2008 Saltire Software Incorporated Page 3 of 54 Lesson 6 Eccentricity and Parabolas • Parabolas are introduced as the locus of points equidistant from a line and a point. The concept of eccentricity is presented to expand this definition to include hyperbolas and ellipses. Topics outside the scope of this unit Topics outside the scope of this unit include: • A discussion on why circles, ellipses, hyperbolas, and parabolas are called “conic sections.” • Using implicit equation to sketch graphs of conics. • Using the discriminant to determine the type of conic section. • Rotation of axis to eliminate the xy term • An exhaustive discussion on the special properties of parabolas. Conics and Loci Lesson 1: Introducing Loci Level: PreCalculus Time required: 60 minutes 2008 Saltire Software Incorporated Page 4 of 54 Learning Objectives This lesson is the introduction to conic sections. Conic sections (parabolas, hyperbolas, ellipses, circles) can all be described as a locus of points. In this lesson, students will become re-acquainted with some familiar loci, and with how Geometry Expressions can be used to explore loci. Math Objectives • Learn the concept of a “locus.” Technology Objectives • Create loci with Geometry Expressions Math Prerequisites • Parallel lines, perpendicular bisectors and angle bisectors • Parametric Functions Technology Prerequisites • None Materials • Computers with Geometry Expressions Conics and Loci Lesson 1: Introducing Loci Level: PreCalculus Time required: 60 minutes 2008 Saltire Software Incorporated Page 5 of 54 Overview for the Teacher 1. Part one introduces the students to the software, and to the idea of a locus. Watch to see if students are changing the value of a – it should remain constant. Otherwise, they might be led to believe that the locus is a family of concentric circles. The Variables Tool Panel can be used to lock variable a. Click on the entry for a, and then click on the lock icon. Note that θ measures the rotation of the segment counter-clockwise from the horizontal. However, if the student draws the segment from A to B, the A is placed at the origin. If the segment is drawn from B to A, then B is placed at the origin. The result is a circle with center A and radius a, as seen in Diagram 1. 2. The difficult part of question 2 is making a case for creating point A. To create a locus of points, Geometry Expressions requires a Parametric Variable, which seems artificial in this case. Geometry Expressions will not look for the other parallel line. It uses the position of P as a cue to determine which points to draw. The result is shown in Diagram 2. Students can get the other half of the locus by duplicating their work on the other side of the line, or by using the Reflection tool in the Construct Tool Panel. The desired result is two lines parallel to the original line, each a distance of d from the original line. 3. Students are encouraged to deduce how to use Geometry Expressions more autonomously as the lessons progress. P A ? d a d x 0 , y 0 Diagram 3 Diagram 1 B A θ x0,y0 a A P a d x0,y0 Y=b0+X·m0 Diagram 2 Conics and Loci Lesson 1: Introducing Loci Level: PreCalculus Time required: 60 minutes 2008 Saltire Software Incorporated Page 6 of 54 If students put the point outside of the parallel lines, it will move one of the lines so that they coincide. It is especially important to lock variable d for this part, since that will keep the lines a constant distance apart – they won’t wiggle around. The locus of points equidistant from two parallel lines is a parallel line exactly half-way between the two lines. For intersecting lines, students may try to create an additional constraint for their parametric variable. Most likely, they will get a message box telling them that they have too many constraints. Remind them that they can use distance d as the parametric variable for their locus. The locus of points equidistant from intersecting lines bisects the angle formed by the lines. 4. Students are expected to create this locus from start to finish. The result is the perpendicular bisector of the line segment. 5. In summary, a locus is a collection of points meeting a given set of criteria. The locus can be described geometrically, for example though distances from fixed points or shapes. The next lesson in this unit will examine the circle locus more closely. It will create a link between the geometric construction and algebraic equations of conics. Both parametric and implicit forms will be examined. B C A a a Diagram 5 P d Y=b0+X·m0 d Y=b1+X·m1 Diagram 4 2008 Saltire Software Incorporated Page 7 of 54 Loci A Locus is a set of all the points that meet a particular description. “What are all the points a certain distance from a fixed point?” “If I tie my dog to a stake, what are the boundaries that he can’t cross?” 1. Find the locus of points equidistant from a fixed point (like all the places a dog can go if it is tied to a stake). Open a new drawing with Geometry Expressions Draw point A. Constrain its location to (x0, y0). Click on the point, and then click on Draw point B. Constrain the distance between the points to a. Hold the shift key and select each point. Then click on Draw a line segment connecting the two points. Constrain the direction of the line segment to θ. Choose θ in the Variables Tool Panel. Use the slider bar to see the locations that point B can occupy. Construct the locus of the points, using θ for the parametric variable. Select point B Click on Construct locus Choose θ for the Parametric Variable. How would you describe this locus? 2. What is the locus of points equidistant from a line (perhaps the path a dog creates along the edge of a fence)? Open a new drawing with Geometry Expressions. Draw a line. Constrain the line’s implicit equation. You can leave the equation at its default setting. Draw point P so that it is not on the line. Constrain its distance from the line to be d. Name: ______ Date: _____ Sketch the locus of points equidistant from a fixed point. 2008 Saltire Software Incorporated Page 8 of 54 You can lock variables by selecting them in the Variables Tool Panel, and then clicking the lock icon. Lock variable d. Drag point P to see its locus – all of the points that are d units from the line. Geometry Expressions will draw the locus, but it wants a point of reference that changes. We’ll ask it to draw each point that is d units from the line, but at all the different distances from some fixed point. So, we need to draw a fixed point. Draw point A anywhere. Constrain its coordinates. Constrain the distance from A to P to t. This constraint is called a parametric variable. To see the graph of the locus, Select point P. Construct its locus . Use t for its Parametric Variable. Are there any points that are distance d away from the line, but that weren’t drawn? Computers aren’t always able to do a complete job, based on your description. Unlock variable d, and try dragging P again and see what happens. Are there any other locations possible for P? Sketch them in your diagram. 3. What is the locus of points equidistant from two lines? There are two cases . Case number 1: two parallel lines: Draw a line. Constrain one of the line’s equations. Draw a second line, and constrain the two lines to be parallel Draw point P between the two lines. Constrain the distance from P to each line to be d. Think about how you created a parametric variable for your locus in part 2. Create a parametric variable, and create the locus. What is the locus of points equidistant from two lines, if the lines are parallel? Sketch the locus of points equidistant from a line. Sketch the locus of points equidistant from two parallel lines. 2008 Saltire Software Incorporated Page 9 of 54 Case number 2: Two intersecting lines: Draw intersecting lines, and constrain both of their equations. Draw point P, constrained to be distance d from each line. For parallel lines, the distance from the point to the lines was always the same. For intersecting lines, the distance changes. That means you can use d as your parametric variable when you create the locus. Describe the locus in your words. What is the special name for this locus? 4. What is the locus of points equidistant from the endpoints of a line segment? Use the line segment tool to draw a line segment. Explore the locus of points equidistant from the endpoints of the line segment. Describe the locus in your words. What is the special name for this locus? 5. Summarize what you have learned in this lesson. What is the definition of a locus? What are some different ways that a locus can be described? Locus of points equidistant from two intersecting lines. Locus of points equidistant from the endpoints of a line segment. Loci and Conics Lesson 2: The Circle Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 10 of 54 Learning Objectives Students are now acquainted with the idea of “locus,” and how Geometry Expressions can be used to explore loci. Now, they will look more closely at the circle, defined as the locus of points on a plane equidistant from a fixed point. In particular, they will find the parametric and implicit equations of circles. The approach is to make sense of the parametric equation in terms of the unit circle, and the implicit equation in terms of the distance formula. Review of these two concepts may be beneficial before the start of the lesson. Math Objectives • Learn or re-enforce the general parametric and implicit equations of a circle. • Connect translations and dilations to the general equations. • Connect center and radius to the general equations. Technology Objectives • Use Geometry Expressions to find equations of curves. • Use Geometry Expressions to translate and dilate figures. Math Prerequisites • Distance formula • Algebra, including factoring parts of an expression and completing the square. • Unit Circle • Pythagorean identity • Translations and dilations • Parametric functions vs. implicit equations Technology Prerequisites • Knowledge of Geometry Expressions from previous lessons. Materials • Computers with Geometry Expressions. Loci and Conics Lesson 2: The Circle Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 11 of 54 Overview for the Teacher 1. Question one gets the students back into using Geometry Expressions to find the locus of a point. Diagram 1 exhibits typical results. A common error is to use r (the default) as the parametric variable instead of changing it to θ. Students are asked why they didn’t just use the circle tool. Expected results are that the definition of a circle as a locus of points was reinforced. 2. In Diagram 2 you can see the general parametric equation of a circle, where d is the radius and (h, k) is the center. If students are getting numerical constants, they are calculating the real equation instead of the symbolic equation. Help them to select the symbolic tab and try again. After changing the constraints so that the center is (0,0) and the radius is 1, the parametric equation will be ( ) ( ) cos sin X Y θ θ = = If students are not getting this, they may be changing the value of the d in the Variable Tool Panel, rather than changing the constraint itself to 1. 3. Diagram 3 shows the results of finding the implicit equation for the circle. The simplification process involves some grouping and simple factoring. The end result is the distance formula: ( ) ( ) 2 2 2 r X h Y k = − + − , which is frequently written ( ) ( ) 2 2 2 r x h y k = − + − Diagram 1 B A r θ (h,k) A B z1⇒ X=h+r·cos(θ) Y=k+r·sin(θ) r θ (h,k) Diagram 2 Diagram 3 A B z0⇒X 2+Y 2-2·X·h+h 2-2·Y·k+k 2-r 2=0 r θ (h,k) Loci and Conics Lesson 2: The Circle Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 12 of 54 Changing the center to (0,0) and the radius to 1 results in 2 2 1 0 X Y −+ + = or 2 2 1 X Y + = Exceptional students will begin making connections between the equation of a circle, the distance formula, and the Pythagorean Theorem at this point. Most will likely have been exposed to these connections in a previous course. You may wish to highlight these relationships at this time. 4. The result is the Pythagorean Identity: 2 2 cos sin 1 θ θ + = 5. Translating and dilating the unit circle will restore us to the general form. Note that the center will be (u0, v0) instead of (h, k). If students find this confusing, they can change the variables used in the vector to (h, k). Encourage students to change the vector by dragging D, especially if they cannot see the whole picture on the screen. If your students are not comfortable with completing the square, then the generalization of the implicit equation may be a demonstration. 6. If students get incorrect results for their dilations, check to see that they have chosen the center of the translated circle as their center of dilation. If they choose some other point, their dilated circle will not be concentric with the translated circle, and their equations will be wrong. See diagram 6 for typical correct results. 7. In summary: The general parametric form of the equation of a circle is cos sin X h r Y h r θ θ = + = + (x0, y0) is the center of the circle and d is the radius. The general implicit formula of a circle is ( ) ( ) 2 2 2 r X h Y k = − + − Again, (h, k) is the center of the circle and d is the radius. B A C A' D B' z0⇒u0,v 0 θ 1 u0 v 0 (0,0) Diagram 5 B A C A' D B'' B' z0⇒u0,v 0 θ 1 a u0 v 0 (0,0) Diagram 6 Loci and Conics Lesson 2: The Circle Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 13 of 54 It is important that students have these three concepts, as they will be repeated for the rest of the conic sections: 1. Description of the curve as a locus of points. 2. Parametric equation of the curve. 3. Implicit equation of the curve. Subsequent lessons will search for further attributes of the curve being studied. For example, the study of ellipses will include the focus, major axis and minor axis. 2008 Saltire Software Incorporated Page 14 of 54 Loci: Circles In the last lesson, you were introduced to the idea of a “locus.” A locus is a collection of points that meet a particular description. For example, the locus of points equidistant from a fixed point is a circle. 1. Reconstruct the locus of points equidistant from a fixed point. Open Geometry Expressions. Create point A, and constrain its location to (h, k). Create point B, and constrain its distance from point A to be r. Create line segment AB – be sure to draw starting at point A and ending at point B – and constrain its direction to be θ. Create the locus of points d units from point A. Use θ as your parameter. You could have just used the circle tool if all you wanted to do was to draw a circle. What did you learn about circles by doing it this way? 2. Often, a locus can be described with equations describing its x and y coordinates. This type of algebraic description is called a Parametric Equation. X and Y are described as functions of a third variable, called a parameter. Click on the circle you created in step one. Click on Calculate Symbolic Parametric equation. . Geometry Expressions has just given you the General Parametric Equation of a circle. Write it in the box to the right. Recall the Unit Circle. How are the coordinates of points on the unit circle defined in terms of cosine and sine? Change the constraint on the center to (0,0) and change the radius to 1. What does this do to the Parametric equation? Name: ____ Date: ____ Sketch the locus of points equidistant from a fixed point. Write the general parametric equation of a circle 2008 Saltire Software Incorporated Page 15 of 54 3. You can use CTRL-Z to undo your changes to the constraints on your drawing. Do so repeatedly, until the radius is r and the center is (h, k). Then use Geometry Expressions to create the implicit equation. Select the circle. Click on Calculate Implicit equation . The result looks confusing at first, but you can clean it up: Add d2 to both sides. Group the X and x0 terms. Group the Y and y0 terms. What formula is beginning to emerge? Factor the X and x0 terms. Factor the Y and y0 terms. The result is the General equation of a circle. Change the constraints as you did in part 2. The result is the implicit equation for the unit circle. Write it here: 4. Using the equations for the unit circle, substitute the two parametric equations into the implicit equation. What relationship do you get? 5. We are going to subject our unit circle to some transformations. First, translate the unit circle. Create a vector. Constrain the components of the vector to 0 0 u v . Select the circle and its center (use shift and click on each). Click Construct Translation Click the vector. Simplify the Implicit Equation here, to get the General Equation of a circle. Translation of a circle 2008 Saltire Software Incorporated Page 16 of 54 What are the coordinates of the center of the translated circle? Click on the center. Click on Calculate Symbolic Coordinates. Find the parametric equation of the translated circle. Where do the coordinates of the center appear in the equations? Find the implicit equation of the translated circle. Use the method of completing the square to simplify the implicit equation. 6. Now, dilate the circle that you translated. Click on the circle. Click on Construct Dilation. Click on the center of the translated circle. Type r for your scale factor. What is the radius of the dilated circle? Find the parametric equation of the translated/dilated circle. Where does a, the length of the radius appear in the equations? Find the implicit equation of the translated/dilated circle. Simplify with completing the square, as you did in part 5. Sketch the translation and dilation of the unit circle. 2008 Saltire Software Incorporated Page 17 of 54 7. Summary The general parametric form of the equation of a circle is: Where is the center of the circle and _ is the radius of the circle. The general implicit form of the equation of a circle is: Where __ is the center of the circle and is the radius of the circle. Loci and Conics Lesson 3: The Ellipse Level: Precalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 18 of 54 Learning Objectives In this lesson, students will generalize their knowledge of the circle to the ellipse. The parametric and implicit equations of an ellipse will be generated, as will two important properties of the ellipse: that the sum of the distances from any point to the foci is equal to the major diameter, and the Pythagorean Property of Ellipses. Math Objectives • Understand the geometric definition of an ellipse. • Generate the parametric and implicit equations for an ellipse. • Locate the foci, given the equation of an ellipse. • Discover relationships between the parameters of an ellipse. Technology Objectives • Use Geometry Expressions to create a more complex locus of points. • Find evidence for equivalence using Geometry Expressions. Math Prerequisites • Pythagorean Theorem • Translations • Parametric functions and implicit equations • Sine and Cosine Technology Prerequisites • Knowledge of Geometry Expressions from previous lessons. Materials • Computers, with Geometry Expressions. Loci and Conics Lesson 3: The Ellipse Level: Precalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 19 of 54 Overview for the Teacher 1. Diagram 1 represents a typical result for question 1. If students are getting semicircles with center at point F1, they are creating a locus in terms of t instead of d. Only half an ellipse is shown because the entire ellipse cannot be generated as a function of d. Points below the foci will have the same distances as points above. An ellipse is like a circle in that it is a curve based on distances from fixed points. It is different in that it “has different radii.” 2. An appropriate result would be cos sin X a Y b θ θ = = . Transposing the sine and cosine functions will have no effect on the final curve. Note that the transposed version is also the “sample” parametric function given by the software. Using the transposed version yields the same results, graphically. It just changes the starting place and direction that the ellipse is graphed. Encourage students to change values for a and b to get an ellipse that is not just a circle. A sample is shown in Diagram 2 3. Some assumptions are made here about the symmetrical nature of an ellipse. You may wish to explore these assumptions with the class at this time. Desired solutions: a. The distance from F1 to P is 2a – m or 2c + m b. The distance from F2 to P is m c. The sum is therefore 2a or 2c + 2m. d. The width of the ellipse is 2a e. t = 2a. Remind students that they need to type 2a . 4. The constraint from F2 to P is 2a – d. Diagram 3 shows expected results. X=a·cos(T) Y=b·sin(T) Diagram 2 F1 P F2 d -d+t Diagram 1 Diagram 3 2 4 -2 2 -2 O P F2 F1 d X=a·cos(T) Y=b·sin(T) 2·a-d Loci and Conics Lesson 3: The Ellipse Level: Precalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 20 of 54 5. Desired solutions: a. The sum of the distances is 2a. b. The distance from F2 to P is equal to a, since the three points form an isosceles triangle. c. Using the Pythagorean theorem, a2 = b2 + c2, so 2 2 c a b = − d. Point P will now appear to be on the ellipse., as shown in Diagram 4. 6. In both instances, the implicit equation will be 2 2 2 2 2 2 0 Y a X b a b ⋅ + ⋅ − ⋅ = 7. The general parametric function that is generated is ( ) ( ) 0 0 cos sin X u a T Y v b T = + = + The implicit formula is 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 2 2 0 Y a X b a b Xb u b u Ya v a v + − − + − + = 8. Steps are as follows: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 2 2 0 0 2 2 2 2 2 2 0 0 2 2 2 2 2 2 2 2 0 0 2 2 2 2 0 2 2 2 2 1 Y a X b a b Xb u b u Ya v a v X b Xb u b u Y a Ya v a v a b b X Xu u a Y Yv v a b b X u a Y v a b b X u a Y v a b a b a b a b X u Y v a b + − − + − + = − + + − + = − + + − + = − + − = − − + = − − + = 9. Results as follows: a. If a = b, then the result is a circle. b. If a < b, then the ellipse is taller than it is wide. The foci lie on a vertical line rather than a horizontal line. The Pythagorean Property would then be 2 2 2 b a c = + 2 4 -2 2 -2 P F2 F1 O d X=a·cos(T) Y=b·sin(T) a 2-b 2 2·a-d a 2-b 2 Diagram 4 2 4 6 8 -2 -4 2 4 -2 A B z0⇒ X=u0+a·cos(T) Y=v 0+b·sin(T) z1⇒Y2·a2+X 2·b2-a2·b2-2·X·b2·u0+b2·u0 2-2·Y·a2·v 0+a2·v 0 2=0 X=a·cos(T) Y=b·sin(T) u0 v 0 Diagram 5 Loci and Conics Lesson 3: The Ellipse Level: Precalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 21 of 54 10. Summary: The general parametric form of the equation of an ellipse is ( ) ( ) 0 0 cos sin X u a T Y v b T = + = + where (u0, v0) is the center of the ellipse and a and b are the radii of the ellipse. The general implicit form ot the equation of an ellipse is ( ) ( ) 2 2 0 0 2 2 1 X u Y v a b − − + = where (u0, v0) is the center of the ellipse. If a > b ,then 2a is the major diameter and 2b is the minor diameter. If b < a, then 2b is the major diameter and 2a is the minor diameter. The sum of the distances from any point on the ellipse to the two foci is 2a The distance from the center of the ellipse to either focus follows the equation 2 2 2 c a b = + 2008 Saltire Software Incorporated Page 22 of 54 The Ellipse In the last lesson, you found equations for a circle: the locus of points equidistant from a fixed point. An ellipse is defined as all the points such that the sum of the distance from two fixed points is a constant. . Each of the two points is called a focus (the plural of “focus” is “foci”). 1. Create an ellipse using the definition above. Open a new Geometry Expressions drawing. Create two points, and name them F1 and F2. Constrain the coordinates of these points. Create a third point, and name it P. Constrain the distance from P to F1 to be d. Constrain the distance from P to F2 to be t – d (see Diagram 1 to understand why!) Lock the value of t in the Variable Panel, but keep d unlocked. Find the locus of P with parameter d. If you drag P around, you can see that the locus forms part of an ellipse. To get more of the ellipse: Double click on the curve. Change the Start Value and End Value for d The most you can get is half of the ellipse, because the software assumes you know which side P is on – you drew it there! To get the rest of the ellipse: Draw a line segment from F1 to F2. Select the locus curve. Click on construct reflection and then click on the segment. Some of the ellipse may still be missing. How is an ellipse like a circle? How is it different? Name: ______ Date: ________ d t – d P F1 F2 D + (t –d) = a constant Diagram 1 Sketch of the ellipse 2008 Saltire Software Incorporated Page 23 of 54 Before continuing, make sure Geometry Express is set to Radians. In the Edit Menu Select preferences. Click on the Math icon at the left. Under Math, change Angle Mode to radians. 2. Recall that the general parametric equation for a circle with the center at the origin is ( ) ( ) cos sin X r T Y r T = = . Predict the general parametric equation for an ellipse. Test your prediction with Geometry Expressions Open a new Geometry Expressions drawing. Click on the Function tool in the Draw tool panel. Type in your prediction to see if you are right. Make additional guesses if you need to. 3. The curve you created in part 2 looks like an ellipse, but is it really an ellipse? If it is, we will be able to find its foci, and the constant sum of distances from the foci. a. In Diagram 2, how far is it from focus F1 to point P? b. How far is it from focus F2 to point P? c. What is the sum of distances from P to the two foci? d. What is the horizontal width of the ellipse? e. Write an expression for your answer to part c, in terms of distance a. 4. Open a new Geometry Expressions drawing, and create this parametric function: cos( ) sin( ) X a T Y b T = = Use the Variable Tool Panel to change the values of a and b so that a is greater than b. Lock variable a and b. a m m F1 F2 P Diagram 2 c c 2008 Saltire Software Incorporated Page 24 of 54 Add three points to your Geometry Expressions drawing. First, turn on the axes. Draw F1 and F2 on the x-axis. Draw P so that it is not on either axis, nor is it on the curve. Constrain the distance from F1 to P to be d. What should the constraint from F2 to P be? Review the results from part 3 to help you decide. 5. Refer to Diagram 3 to find the positions of F1 and F2. The triangle shown is an isosceles triangle, with P at the vertex. a. If P is on the ellipse, what is the sum of the distances from F1 to P and from F2 to P (your solution to 3c)? b. Given that the triangle is isosceles, what is the distance from F1 to point P? c. Use the Pythagorean Theorem to write an expression for the distance from the origin to point F2. d. In your Geometry Expressions drawing, constrain the distance from F1 to the origin to your answer to 5c. Do the same for the distance from F2 to the origin. (You will need to draw a point at the origin first). NOTE: If you want to type a square root in to Geometry Expressions, use sqrt, and if you want to type in an exponent, use ^. For example, 2 2 a b + can be typed: sqrt(a^2 + b^2) e. Does P appear to fall on the ellipse? Drag it around. Does it stay on the ellipse? b a F1 F2 P Diagram 3 2008 Saltire Software Incorporated Page 25 of 54 6. If the curve in your Geometry Expressions drawing is truly an ellipse, then its implicit equation will match the locus of point P. Select the curve, and click on the Calculate Implicit Equation icon. Now, hide the curve. Select the curve. Right click on the curve. Choose hide. Create the locus of point P with respect to d, and calculate its implicit equation. Restore the original curve by clicking on Show All in the View menu. Are the two implicit equations the same? How do they differ, if at all? 7. How does a translation affect the equation of an ellipse? Open a new drawing and use Draw Function to create a new ellipse. Choose Parametric for the type and enter ( ) ( ) cos sin X a T Y b T = = . Create a vector, and constrain it to its default values, 0 0 u v . Translate the ellipse. Select the ellipse Click on Construct Translation Click on the vector. Calculate the parametric equation of the new ellipse, and record it in the box. Calculate the implicit equation of the new ellipse. General Parametric Equation of an Ellipse 2008 Saltire Software Incorporated Page 26 of 54 8. The General form for the implicit equation of an ellipse is ( ) ( ) 2 2 0 0 2 2 1 x u y v a b − − + = . Verify that your implicit equation is equivalent to the general form. 9. In part 6, you found a relationship known as “The Pythagorean Property for Ellipses” 2 2 2 a b c = + a is half the horizontal axis of the ellipse b is half the vertical axis of the ellipse c is the distance from the center of the ellipse to each focus a. What happens if a = b? b. Is it possible for a < b? How would you need to modify the Pythagorean Property for the ellipse in Diagram 4? In any ellipse, the larger of 2a and 2b is called the major axis. The smaller of 2a and 2b is called the minor axis. If a > b, then the foci lie on a horizontal line. If a < b, then the foci lie on a vertical line. If a = b, then the ellipse is actually a circle. a Diagram 4 b 2008 Saltire Software Incorporated Page 27 of 54 10. Summary: The general parametric form of the equation of an ellipse is: where __ is the center of the ellipse, is the horizontal radius of the ellipse, and _and _ are the radii of the ellipse. The general implicit form of the equation of an ellipse is: where is the center of the ellipse. If _, then ___ is the major diameter and ___ is the minor diameter. If _, then ___ is the major diameter and ___ is the minor diameter. The sum of the distances from any point on the ellipse to the two foci is _. The distance from the center of the ellipse to either focus follows the equation _. ( , ) Loci and Conics Lesson 4: Conics and Envelope Curves Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 28 of 54 Learning Objectives Students begin by looking at an envelope curve that generates an ellipse. The curve is modified to become a hyperbola, thereby introducing that concept. Math Objectives • Extend the idea of a locus of points to a locus of lines. • Find a definition of the ellipse as an envelope curve. • Find definitions for hyperbola, first as a locus of lines, and then as an envelope curve. Technology Objectives • Use Geometry Expressions to create an envelope curve • Use Geometry Expressions as an aide to creating geometric proof. Math Prerequisites • Two-column triangle congruence proofs Technology Prerequisites • Skills with Geometry Expressions, as developed in previous lessons. Materials • Computer with Geometry Expressions Loci and Conics Lesson 4: Conics and Envelope Curves Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 29 of 54 Overview for the Teacher 1. The purpose of the first part of the lesson is to become familiar with the envelope curve created by a locus of lines, and the related Gemetry Expressions capabilities. It is important that students follow directions closely for part 1. Some will attempt to draw in all of the lines, copying the diagram in the student master. This is not really helpful for the rest of the lesson. An error that is more difficult to diagonose may occur if students have their axis turned on, use the origin for point A, and place points B and C on the axes. When they try to set points M and N proportional along the curve, they will run into an error. That is because the software knows M and N are on the axis, but does not know they are bounded to the segments. The easy fix is probably to turn off the axis and start over. 2. Diagram 2 shows the expected result. If the envelope curve does not show up, it is most likely because the student has chosen the wrong parametric variable for the locus. The parameter should be θ. Another possibility is that the domain for θ is incorrect. By default, it is assigned values between 0 and 6.28 (2π). Generating the locus of point E with parameter θ will result in the same curve. 3. You may choose whether to do this proof as a class, guide them through it, or let individuals or groups complete the proof, depending on what is right for your class. Statement Reason ED BC ⊥ Given ∠EDC ≅ ∠EDB Right angles are congruent BD CD ≅ Definition of bisector ED ED ≅ Reflexive property of congruence ∆EBD ≅ ∆ECD Side-Angle-Side EB EC ≅ CPCTC AE + EB = AE + EC Additive property of equality A C M N B t t Diagram 1 5 -5 5 -5 E B A C θ r (a,0) Diagram 2 Loci and Conics Lesson 4: Conics and Envelope Curves Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 30 of 54 AE +EB = r, the radius of the circle Given AE + EC = r, the radius of the circle Substitution AE + AC is a constant Definition of a radius The locus is an ellipse Definition of an ellipse 4. Most of the assumptions that Geometry Expression makes when it is asked to “use assumptions” have to do with whether one value is greater than another. In this case, the drawing indicates a < r, and that is the basic assumption that is made. If this is not assumed, Geometry Expressions uses absolute values in place of the assumption, and the symbolics are not as fully simplified. Watch students to make sure they type the expression in correctly, and with the right subscripts. When testing the lesson with current software, the expression resolved to r. It is possible that, depending on how the drawing is constructed, the expression will not be completely simplified. In that case, help students to finish the simplification process themselves. Diagram 3 shows typical student results. Diagram 3 5 10 15 20 -5 5 10 -5 E A C B z1⇒ a--a 2·cos(θ) 2·(r-a·cos(θ)) + r 2·cos(θ) 2·(r-a·cos(θ)) 2 + -a 2·sin(θ) 2·r-2·a·cos(θ) + r 2·sin(θ) 2·r-2·a·cos(θ) 2 z2: z0+z1⇒ r | 2·a·r·cos(θ)a·cos(θ) z0⇒ -a 2+r 2 2·(r-a·cos(θ)) | a 2\a·cos(θ) θ r (a,0) Loci and Conics Lesson 4: Conics and Envelope Curves Level: Precalculus Time required: 90 minutes 2008 Saltire Software Incorporated Page 31 of 54 5. The reason that students need to delete the calculated distances first is that Geometry Expressions is assuming that r > a, and we are about to change that. It is best if point E appears on the branch of the hyperbola that is near point C. Students can toggle the value of θ to make this happen. Diagram 4 includes the expression for the sum of the two lengths. The successful student will try calculating the difference of the two lengths. Some will reverse the order, resulting in –r instead of r. This corresponds to E being on the other branch of the hyperbola. 6. The result for part 6 is a circle with center (0,0) and radius r/2. While this is not a dramatic or important result, it does satisfy the need mathematicians have for “completeness.” The next lesson in this series will develop parametric and implicit functions for hyperbolas. The implicit function is quite similar to the ellipse, but the parametric formulas are a little more surprising. 5 10 15 20 25 -5 5 10 -5 E A C B z0⇒ a--a 2·cos(θ) 2·(r-a·cos(θ)) + r 2·cos(θ) 2·(r-a·cos(θ)) 2 + -a 2·sin(θ) 2·r-2·a·cos(θ) + r 2·sin(θ) 2·r-2·a·cos(θ) 2 z1: z2+z0⇒ 2·a 2-2·a·r·cos(θ) 2·(-r+a·cos(θ)) | 2·a·r·cos(θ)r 2 z2⇒ a 2-r 2 2·(-r+a·cos(θ)) | a 2>r 2 | r r which agrees with point C being outside the circle. Is the sum of the distances still a constant, or is the expression more complex? Try changing the plus sign to something else. Double click on the expression z2 to edit it. When the result is a constant, you can complete the definition of a hyperbola: A hyperbola is the locus of points such that the __ of the distances from the foci is a constant. 6. What happens if point C is the center of the circle? Constrain the coordinates of point C to be (0,0). Calculate AE as you did in parts 4 and 5 (CE and AE are the same segment now). How would you describe this locus of points? Loci and Conics Lesson 5: Inside-out Ellipses Level: PreCalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 36 of 54 Lesson 5: Inside Out Ellipses Learning Objectives What, exactly, is a hyperbola? What are its characteristics and defining equations? How does it relate to the unit circle, and to ellipses? These are the primary objectives of this lesson. Math Objectives • Find out what a hyperbola is, and how it is related to an ellipses. • Learn about characteristics of a hyperbola, particularly its foci and its asymptotes • Discover General and parametric equations for hyperbolas. Technology Objectives • Use Geometry Expressions to draw hyperbolas, using the geometric definition, the parametric equation, and the implicit equation. • Use Geometry Expressions to connect these three disparate representations of a hyperbola. Math Prerequisites • Definitions of secant and tangent, both in terms of right triangles and in terms of sine and cosine. • Pythagorean identity. • Solving simple algebra equations. • Knowledge of circles and ellipses, as provided in earlier lessons of this unit. Materials • Computer with Geometry Expressions. Loci and Conics Lesson 5: Inside-out Ellipses Level: PreCalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 37 of 54 Overview for the Teacher The student title for this lesson is “Inside-out Ellipses.” This presages the next lesson, which is on eccentricity. The main goal for this lesson is to get a feel for hyperbolas and how they relate to ellipses and circles. Ultimately, the students will have a geometric definition of the hyperbola as a locus of points, and algebraic definitions in parametric form and in implicit form. 1. In part 1, students use the parametric and implicit equations of the unit circle to reproduce the Pythagorean Identity, ( ) ( ) 2 2 sin cos 1 θ θ + = . Diagram 1 shows what students should be seeing. 2. Here is the sequence for question 2. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 cos sin 1 cos 1 sin cos sin 1 cos cos cos 1 sec tan T T T T T T T T T T T + = = − = − = − Thus, x will be sec (T) and y will be tan (T) Diagram 2 shows the hyperbola and its parametric function. 3. The asymptotes are going to be used to find the foci of this hyperbola. They are also interesting in their own right. Some students will still have the hyperbola on the same drawing as the unit circle, and others will not. Makes sure they have a both the hyperbola and the unit circle when they proceed to part 4. 1 2 -1 1 -1 A B z0⇒ X=cos(t) Y=sin(t) z1⇒-1+X 2+Y 2=0 1 Diagram 1 2 -2 2 X= 1 cos(T) Y=tan(T) Diagram 2 Diagram 3 2 -2 2 -2 X= a cos(T) Y=b·tan(T) X+Y·-1=0 X+Y=0 Loci and Conics Lesson 5: Inside-out Ellipses Level: PreCalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 38 of 54 4. A small leap of faith is needed in this step, since no proof is offered that the foci can be constructed in this manner. Rather, the foci are used to create a hyperbola that results in a match with this one. Diagram 4 shows student results before all the lines are hidden. 5. The answer to “What is AC – CD” is 2, so the distances from the foci to the point are d and d + 2. The point falls on the existing hyperbola. The equation for the locus is 2 2 1 0 X Y − + = . Some students may need help recognizing this as 2 2 1 X Y − = , our starting point. 6. Changing a stretches the hyperbola horizontally, but since this also moves the foci towards or away from the origin, the hyperbola changes vertically too. Changing b only stretches the hyperbola vertically since the foci are not affected. The ellipse is tangent to the hyperbola at its vertices. The line b y x a = is one of the asymptotes for the hyperbola. The other asymptote is b y x a = − You may wish to point out that b a and b a − have opposite signs, but are not reciprocals. Therefore, the asymptotes are not perpendicular See Diagram 5 for results. 7. The parametric function for a hyperbola translated by 0 0 u v is shown in Diagram 6. 2 -2 2 -2 B F G E A D H I X·-1+Y=0 1 X+Y=0 X= 1 cos(T) Y=tan(T) Diagram 4 2 -2 2 -2 X=a·cos(T) Y=b·sin(T) X= a cos(T) Y=b·tan(T) -X·b a +Y=0 X·b a +Y=0 Diagram 5 2 4 6 2 -2 B A z0⇒ X=u0+ a cos(T) Y=v 0+b·tan(T) X= a cos(T) Y=b·tan(T) u0 v 0 Diagram 6 Loci and Conics Lesson 5: Inside-out Ellipses Level: PreCalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 39 of 54 2 2 Y= 1+X 2 Diagram 7 8. Inductive reasoning is used at this point because it reinforces the roles of the different parameters. You may wish to derive the general implicit equation in class as well. Start with the trig identity derived earlier: 2 2 sec tan 1 T T − = . Solve each of the parametric equations for sec T and tan T, respectively. Substitute the results into the identity. Make a few algebraic adjustments, and you are finished. The general equation for a hyperbola is ( ) ( ) 2 2 0 0 2 2 1 X u Y v a b − − − = a stretches the hyperbola horizontally b stretches the hyperbola vertically u0 is the horizontal translation v0 is the vertical translation 9. Students should graph the implicit function 2 1 y x = + . They will see the top half of a hyperbola that opens up, as seen in Diagram 7. The other half of the hyperbola is produces by 2 1 y x = − + . A drawing of the parametric form is shown in Diagram 8. Both parts of the hyperbola are drawn because no plus-or-minus is required in the parametric form. 2 2 -2 X=a·tan(T) Y= b cos(T) Diagram 8 Loci and Conics Lesson 5: Inside-out Ellipses Level: PreCalculus Time required: 120 minutes 2008 Saltire Software Incorporated Page 40 of 54 10. Summary: A hyperbola is the locus of points for which the difference of the distances from the foci is a constant. The asymptotes of the hyperbola have slopes b a and b a − . If a hyperbola opens to the left and to the right: Its parametric equation is: ( ) ( ) 0 0 sec tan X a T u Y b T v = + = + Its implicit equation is: ( ) ( ) 2 2 0 0 2 2 1 X u Y v a b − − − = If a hyperbola opens up and down: Its parametric equation is: ( ) ( ) 0 0 tan sec X a T u Y b T v = + = + Its implicit equation is: ( ) ( ) 2 2 0 0 2 2 1 Y u X v a b − − − = 2008 Saltire Software Incorporated Page 41 of 54 Inside-out Ellipses In the last lesson, we turned an ellipse inside-out and ended up with a hyperbola. In this lesson, we are going to find general equations for hyperbolas. Make sure Geometry Expressions is in Radians before you proceed. 1. Recall the parametric and implicit formulas for the Unit Circle. Open Geometry Expressions. Create a circle with center at the origin, and with radius 1. Calculate the symbolic parametric equation for the circle. Calculate the symbolic implicit equation for the circle. Now, substitute the parametric equations into the implicit equation. What identity did you create? 2. What do you suppose the graph of 2 2 1 x y − = would look like? First, we need to find an identity in the form 2 – 2 = 1. Start with the identity you recalled in part one. Subtract sin2 (T) from both sides. Divide each term by cos2 (T) Compare your result to the pattern: 2 2 1 x y − = What is the trig expression that is standing in for x ? What is the trig expression that is standing in for y? Note: for the next step, if you need to enter sec(x), use 1/cos(x) csc(x), use 1/sin(x) cot(x), use 1/tan(x) Draw a function in Geometry Expressions Select Parametric type Type your trig expression in for x and y. Your result should appear to be a hyperbola. But is it? Name: ______ Date: ________ Equations for the unit circle Deriving the unit hyperbola 2008 Saltire Software Incorporated Page 42 of 54 The asymptotes of the “hyperbola” that you just created are the lines y = x and y = -x. Add them to your drawing Create a line. Select the line, and click on Constrain Implicit Equation . Type in y = x Repeat for y = -x 3. Remember that a hyperbola is the locus of points such that the difference of the distances from the foci is a constant.. To verify whether our “hyperbola” is actually a hyperbola, first we need to find its foci. If you don’t have the unit circle on the same drawing as your “hyperbola,” construct it now. Construct the points of intersection for the circle and each asymptote. Select the circle and Select the line Click on Construct Intersection Construct perpendiculars at each intersection point. Select the line and the point. Click on Construct Perpendicular . The points where the perpendiculars intersect the x-axis are the foci of the hyperbola. Construct points at those intersections Select the perpendicular and the x-axis. Click on Construct Intersection. Clean up your drawing by hiding all of the straight lines. Click on the line Right click Select Hide from the pop-up menu. Sketch of the hyperbola and its asymptotes 2008 Saltire Software Incorporated Page 43 of 54 4. We’ve found the foci, so now all we need to do is find the constant difference. Use diagram 1 to help you find it A and D are foci. Consider point C on the hyperbola. How far is it from C to D (length CD)? How far is it from A to C (length AC)? What is AC – CD? On your Geometry Expressions drawing, draw a point that is not on the hyperbola or on the circle. Constrain the distance from one of the foci to your point to be d. What did you get for AC – CD in Diagram 1? Add that value to d and use it to constrain the distance from the other focus to your point. What happens to your point? Hide the hyperbola. Create the locus for your point, using d as a parameter. Adjust the start and end values if you think its necessary (you can change them later by double-clicking on the locus curve). Find the symbolic implicit equation of the hyperbola, and record it here: 5. So far we have found the parametric and implicit equations of the unit hyperbola. But what are the general equations? Start with a new Geometry Expressions drawing. Create a unit hyperbola using parametric equations. Now, modify the parametric equations like this: ( ) ( ) 1 cos tan X a T Y b T = = a a a a C D A B (-1,0) (1,0) Diagram 1 2008 Saltire Software Incorporated Page 44 of 54 Change the value of a with the slider bar on the Variable Tool Panel. What happens? Change the value of b with the slider bar on the Variable Tool Panel. What happens? Create the ellipse with this parametric equation: ( ) ( ) cos sin X a T Y b T = = How are the hyperbola and the ellipse related? Create a line and constrain its implicit equation to be y = b/a x. This line is one of the asymptotes of the hyperbola. What do you think is the equation of the other asymptote? Test your conjecture on your Geometry Expressions drawing until you get it right. 6. How does a translation affect the equation of the hyperbola? Delete or hide the ellipse and the asymptotes from your drawing. To translate the hyperbola: Create a vector and constrain it to the default settings. Select the hyperbola. Click on the Construct Translation tool. Click on the vector to translate the hyperbola. Select the new hyperbola Click on Calculate Symbolic Parametric equation and record the results. 2008 Saltire Software Incorporated Page 45 of 54 7. Use inductive reasoning to surmise the general implicit equation of a hyperbola, and complete the table Unit Circle Unit Hyperbola ( ) ( ) cos sin X T Y T = = 2 2 1 x y + = ( ) ( ) sec tan X T Y T = = 2 2 1 x y − = General Ellipse General Hyperbola ( ) ( ) 0 0 cos sin X a T u Y b T v = + = + ( ) ( ) 2 2 0 0 2 2 1 X u Y v a b − − + = ( ) ( ) 0 0 sec tan X a T u Y b T v = + = + In the general equation of a hyperbola: a results in _________ b results in _________ u0 results in _________ v0 results in _________ 2008 Saltire Software Incorporated Page 46 of 54 8. Subtraction is not commutative, so what is the graph of 2 2 1 y x − = ? Solve the equation for y. Open a new Geometry Express file and graph the result. Remember to type sqrt for square root, and to put the radicand in parenthesis. Is the entire graph shown, or is some missing? How is this hyperbola different from the others? We got this graph by exchanging x and y. Create a hyperbola with parametric equations, but exchange the x and y. What is the result? 9. Summary A hyperbola is he locus of points _______ __________ If a hyperbola opens to the left and to the right: Its parametric equation is: Its implicit equation is: If a hyperbola opens up and down: Its parametric equation is: Its implicit equation is: The asymptotes of the hyperbola have slopes __ and __. Solve for y 2 2 1 y x − = Conics and Loci Lesson 6: Eccentricity Level: Precalculus Time required: 60 minutes 2008 Saltire Software Incorporated Page 47 of 54 Learning Objectives So far, the unit has neglected parabolas. In this lesson, we begin with the definition of a parabola as the locus of points equidistant from a point (the focus) and a line (the directrix). Next, the definition is changed to include eccentricity, the ratio of the two distances, resulting in new definitions for ellipses and hyperbolas. Math Objectives • Define the parabola as the locus of points equidistant from a point (the focus) and a line (the directrix). • Introduce the concept of eccentricity. • Define ellipses and hyperbolas in terms of focus, directrix, and eccentricity. • Find the limit of the locus as e goes to 0, and compare with the limit of the equation as e goes to 0. Technology Objectives • Use Geometry Expressions to create a locus of points and derive its equation. Math Prerequisites • Knowledge of conic sections presented earlier in this unit. • Ratios • Equations of parabolas • Factoring by grouping Technology Prerequisites • Geometry Expressions, as learned in this unit. Materials • Computers with Geometry Expressions. Conics and Loci Lesson 6: Eccentricity Level: Precalculus Time required: 60 minutes 2008 Saltire Software Incorporated Page 48 of 54 Overview for the Teacher 1. The geometric definition of a parabola is the locus of points equidistant from a line (the directrix) and a point (the focus). Diagram 1 shows sample student work. 2. Some students will try to generate the symbolic implicit equation for the curve before nailing down some of the constraints. Make sure they follow the steps to lead them to a simpler result. Diagram 2 shows the parabola with the x-axis for the directrix, and the focus on the y-axis. 3. In part 3, the drawing will be essentially the same, but the equation will change to ( ) 2 2 2 2 2 1 0 X Yc c Y e − + + − = The process of substituting 1 for e will help students see the role that e plays in the equation as well as in the diagram. Caution: Some students may think that that e represents Euler’s constant. It does not. Diagram 3 shows a curve with eccentricity less than 1: an ellipse. The X2 and Y2 coefficients will have the same sign in this case. P' F P C0+X·A0+Y·B0=0 d d x0,y0 Diagram 1 Diagram 3 F P' P z0⇒X 2-2·Y·c+c 2+Y 2· 1-e 2 =0 (0,c) d·e d Y=0 Diagram 2 F P' P z0⇒-X 2+2·Y·c-c 2=0 (0,c) d d Y=0 Conics and Loci Lesson 6: Eccentricity Level: Precalculus Time required: 60 minutes 2008 Saltire Software Incorporated Page 49 of 54 Diagram 4 shows a curve with eccentricity greater than 1, a hyperbola. Some students may think they have returned to a parabola – help them see that the Y2 term in the equation contradicts this. For a hyperbola, the X2 and Y2 terms have opposite signs. This corresponds with the general equations they have already discovered. 4. The focus-directrix model breaks down for e = 0; but the equation generated resembles a circle, but it is a circle with radius 0. 5. Conclusion: If e > 1, thenX2 and Y2 coefficients have opposite signs and the curve is a hyperbola. If e = 1, then Y2 term is eliminated and the curve is a parabola. If 0 < e < 1, then X2 and Y2 coefficients have the same sign and the curve is an ellipse. As e approaches 0, the X2 and Y2 coefficients become the same the curve becomes closer and closer to a circle. F P' P z0⇒X 2-2·Y·c+c 2+Y 2· 1-e 2 =0 (0,c) d·e d Y=0 Diagram 4 2008 Saltire Software Incorporated Page 50 of 54 Eccentricity The final locus that we will look at in this unit is the locus of points equidistant from a fixed line and a fixed point. From there, we’ll see what happens if the ratio of those distances is something other than one. 1. What is the locus of points equidistant from a fixed line and a fixed point? Open a new Geometry Expressions file. Create a line, and constrain its implicit equation to the default. Select the line Click on Constrain Implicit Function This line is called “the directrix.” Create a point that isn’t on the line, and name it F. Constrain its coordinates to the default. This point is called “the focus.” Create another point that isn’t on the line. Label it P Constrain the distance from P to F to be d. Constrain the distance from P to the directrix to also be d. Create the locus of point P with d as the parameter. Set the start value to 0 and the end value to 20. To see the whole curve Construct a perpendicular from F to the directrix. Select the locus. Reflect it across the perpendicular. What type of curve do you think this is? 2. Calculate the Symbolic implicit equation for the locus To make this simpler, we are first going to rotate and translate the locus curve. Remember that rotation and translation preserve the size and shape of a figure. Constrain the implicit equation of the directrix to y = 0. Constrain the coordinates of F to (0, c). Name: ______ Date: _____ 2008 Saltire Software Incorporated Page 51 of 54 Solve the equation for y Do you recognize the form of the equation? What type of curve is this? Does your answer agree with your guess in part 1? 3. What if the two distances you used to create the locus were different? Change the distance from P to F to be de. The value e, which is the ratio of the two distances, is called “the eccentricity.” Set the value of e to be 1, and you have the same curve as in part 2. Write down the implicit equation for the locus. Substitute e = 1. What is the result? If e = 1, then the curve is ________ Sketch your curve when e = 1 F 2008 Saltire Software Incorporated Page 52 of 54 Use the Variable Tool panel to change the eccentricity to a value that is less than 1. Remember that the general implicit form for an ellipse is ( ) ( ) 2 2 0 0 2 2 1 x x y y a b − − + = and the general implicit form for a hyperbola is ( ) ( ) 2 2 0 0 2 2 1 x x y y a b − − − = Is the Y2 coefficient positive or negative? Based on this, what type of curve do you think this is? Why? If e < 1, then the curve is _______ Now, set the value of e to be greater than 1. Is the Y2 coefficient positive or negative? What type of curve do you think this is? If e > 0, then the curve is ______ Sketch your curve when e < 1 F Sketch your curve when e > 1 F 2008 Saltire Software Incorporated Page 53 of 54 4. What curve do you expect if the eccentricity is equal to 0? Use the slider bar on the Variable Tool Panel to move e towards 0. If your curve disappears, make values for d and c closer together. As e gets closer to 0, what happens to PF? As e gets closer to 0, what happens to the shape of the curve? Substitute 0 in for e in the equation. ( ) 2 2 2 2 2 1 0 X Yc c Y e − + + − = Put the equation into ( ) ( ) 2 2 2 0 0 x u y v r − + − = form. What is the center of the circle? What is the radius of the circle? As the eccentricity gets closer to circle, the radius also gets closer to zero. Therefore the eccentricity model cannot be used to define a circle. It only defined ellipses, hyperbolas, and parabolas. 2008 Saltire Software Incorporated Page 54 of 54 5. Conclusion The fixed line is called the directrix. The fixed point is called the focus. The curves we studied are the locus of points for which the distance from the directrix and the distance from the focus had the same ratio. That ratio is called the eccentricity, e. The curves are all of the form ( ) 2 2 2 2 2 1 0 X Yc c Y e − + + − = If e > 1, thenX2 and Y2 coefficients have opposite signs and the curve is a ___. If e = 1, then the Y2 term is eliminated and the curve is a ___. If 0 < e < 1, then X2 and Y2 coefficients have the same sign and the curve is an ___. As e approaches 0, the X2 and Y2 coefficients become the same the curve becomes closer and closer to ______
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Home›Fiber Optic Tutorials›Stokes Polarization Parameters
Stokes Polarization Parameters
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Limitations of Polarization Ellipse
As introduced in the article of "polarization states" and Jones Matrix (click to learn more), polarization ellipse is very useful since we can study the various states of polarized light from a single equation.
Linearly polarized light and circularly polarized light are shown to be just special cases of polarization ellipse.
But, there two serious limitations of polarization ellipse (also Jones matrix calculus):
Lights are of very high frequencies, a single period of vibration is in a time interval of the order 10-15 sec. We cannot actually observe the polarization ellipse.
The polarization ellipse can only describe completely polarized light. It can not be used to describe partially polarized nor unpolarized light. But in practice, most lights are partially polarized.
Polarization ellipse is an idealization of the true behavior of light; it is only correct at any given instant of time which we cannot observe directly.
Thus, in practice we have to describe light with observed or measured quantities. We can only use time averaged values of the optical field that we can observe.
in 1852, Sir George Gabriel Stokes (1819 - 1903,mathematician, physicist, politician and theologian of Ireland), found that any state of polarized light could be completely described by four measurable quantities:
The total intensity of the light (polarized + unpolarized)S 0
The intensity of linear horizontal or vertical polarization S 1
The intensity of linear +45° or -45° polarization S 2
The intensity of right or left circular polarization S 3
The intensity of the polarized part of the lightwave is given by:
The total intensity S 0 is
Thus:
The Derivation of Stokes Polarization Parameters
We consider two plane waves that are orthogonal to each other at a point in space (x, y, z). Without losing generality, we can set z = 0.
These two waves which may not necessarily be monochromatic can be written in equations:
where E 0x(t), E 0y(t) are the instantaneous amplitudes, ω is the instantaneous angular frequency,δ x(t),δ y(t) are the instantaneous phase factors.
Comparing to the super fast vibrations of the waves (cosinusoids), the amplitudes E 0x(t), E 0y(t) and phase factors δ x(t),δ y(t) fluctuate very slowly. Here, we are considering monochromatic waves, and their amplitudes and phase factors are constant, not changing with time.So we can use them as E0x, E0y,δ x,δ y.
By removing the term ωt in (1) and (2), we then get the familiar polarization ellipse:
where δ =δ y-δ x
In order to represent (3) in terms of the observables of the optical field, we must take an average over the time of observation.
Since the vibration is so fast, the observation time can be seen as infinite. But, since Ex(t) and Ey(t) are both periodical, we can actually average (3) only over a single oscillation period.
The time average is represented by the angular brackets<...>, so (3) can written as:
where
We then multiple (4) by 4E 0x 2 E 0y 2, and then get:
From (1) and (2), we can find the average values of Equation (6) using Equation (5)
Substituting (7), (8), (9) into (6) and we get:
Since we want to express the final result in terms of intensity, we can add and subtract the quantity E 0x 4 + E 0y 4 to the left-hand side of (10); doing this yields perfect squares.
Then we can get:
We write the quantities inside the parentheses as:
Note that S0, S1, S2, S3are time-averaged quantities performed over a time intervalτ D that is the characteristic time constant of the detection process.
We then rewrite (11) as:
The four equations (12), (13), (14) and (15) are the Stokes polarization parameters for a plane wave.
Note:
Stokes parameters are expressed in terms of intensities (which we can measure)
Stokes parameters are real quantities (instead of complex numbers as in Jones matrices)
If we have partially polarized light, equation (12) is still true for very short time intervals since the amplitudes and phases fluctuate slowly.
Based on Schwartz's inequality, we can say that for any state of polarized light:
In (17), the equality is true for completely polarized light, and inequality is true for partially polarized or unpolarized light.
Stokes parameters and polarization ellipse
From this article about polarization states and polarization ellipse, we can see the orientation angle ψ of the polarization ellipse is given by:
When we take a close look at (12), (13), (14) and (15), we find out that if we divide (14) by (13),ψ can be written as expressions of Stokes parameters as:
Also from the same article, we see that another important parameter is the ellipticity angle χ of the polarization ellipse, which is given by:
After some calculation, the final result of χ is expressed with amplitudes and phase factor as:
We found out that we can divide(15) by (12) and rewrite χ as an expression of Stokes parameters:
Degree of Polarization (DOP) in terms of Stokes Parameters
We can use Stokes parameters to describe the degree of polarization for any state of polarization (completely polarized, partially polarized, and unpolarized).
The Degree of Polarization P is defined as (based on light intensity):
where Ipol is the intensity of the sum of the polarization components, and Itot is the total intensity of the beam.
So the meaning of P is:
P = 1 ---> completely polarized light
P = 0 ---> unpolarized light
0 < P < 1 ---> partially polarized light
Complex Amplitudes representation of Stokes Parameters
We usually use complex amplitudes to represent the real optical amplitudes. So equation (1) and (2) can be written as:
where complex amplitudes Ex, Ey are defined as:
We now can get the Stokes parameters for a plane wave from the formulas below:
where E x and E y are the complex conjugates of E x and E y respectively.
From now on, we will use the above definitions expressed in complex amplitudes for the Stokes parameters.
Stokes Parameters Examples for Completely Polarized Light
We now discuss Stokes parameters for some special cases.
1. Linear Horizontally Polarized Light (LHP)
For this case, there is no vertical field component, so E 0y = 0. From (12), (13), (14) and (15) we get:
2. Linear Vertically Polarized Light (LVP)
For this case, there is no horizontal field component, so E 0x = 0. From (12) to (15) we get:
3. Linear +45° Polarized Light (L +45)
The condition for obtaining linear +45° polarized light are:
E 0x =E 0y = E 0
δ = δ y - δ x = 0
This means this polarization is a superposition of in-phase, equal-amplitude horizontal and vertical fields. With (12) to (15), we get:
4. Linear -45° Polarized Light (L -45)
The condition for obtaining linear -45° polarized light are:
E 0x =E 0y = E 0
δ = δ y - δ x = 180°
We then get:
5. Right Circularly Polarized Light (RCP)
The condition for obtaining right circularly polarized light are:
E 0x =E 0y = E 0
δ = δ y - δ x = 90°
We then get:
6. Left Circularly Polarized Light (LCP)
The condition for obtaining left circularly polarized light are:
E 0x =E 0y = E 0
δ = δ y - δ x = - 90°
We then get:
7. Elliptically Polarized Light
The Stokes parameters for general elliptically polarized light are as the definition:
Stokes Parameters for Unpolarized Light
In order to understand Stokes parameters for unpolarized and partially polarized light, please read this tutorial on how to measure Stokes parameters.
From the article about how to measure Stokes parameters, we saw that the intensity I(φ,θ) of a light beam emerging from the retarder/polarizer pair was:
Unpolarized light has no preference over any particularly linear polarization direction as shown in the previous picture.
The intensityI(φ,θ)is unaffected by either the presence of a retarder (φ) or rotation of the linear polarizer (θ), and there is no polarized part in the intensity. Thus we have:
And the Stokes vector for unpolarized light is:
Read this article for more info about Stokes vector.
Stokes Parameters for Partially Polarized Light
Partially polarized light is a mixture of completely polarized light and unpolarized light.
According to (23), the Degree of Polarization P is defined as:
The Stokes vector for partially polarized light is:
For completely polarized light, P = 1, and the above equation reduces to the Stokes vector for elliptically polarized light.
Similarly, for unpolarized light, P = 0, and the above equation reduces to the Stokes vector for unpolarized light.
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College Physics 2e
13.3 The Ideal Gas Law
College Physics 2e13.3 The Ideal Gas Law
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Highlights
Table of contents
Preface
1 Introduction: The Nature of Science and Physics
Introduction to Science and the Realm of Physics, Physical Quantities, and Units
1.1 Physics: An Introduction
1.2 Physical Quantities and Units
1.3 Accuracy, Precision, and Significant Figures
1.4 Approximation
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
2 Kinematics
Introduction to One-Dimensional Kinematics
2.1 Displacement
2.2 Vectors, Scalars, and Coordinate Systems
2.3 Time, Velocity, and Speed
2.4 Acceleration
2.5 Motion Equations for Constant Acceleration in One Dimension
2.6 Problem-Solving Basics for One-Dimensional Kinematics
2.7 Falling Objects
2.8 Graphical Analysis of One-Dimensional Motion
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
3 Two-Dimensional Kinematics
Introduction to Two-Dimensional Kinematics
3.1 Kinematics in Two Dimensions: An Introduction
3.2 Vector Addition and Subtraction: Graphical Methods
3.3 Vector Addition and Subtraction: Analytical Methods
3.4 Projectile Motion
3.5 Addition of Velocities
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
4 Dynamics: Force and Newton's Laws of Motion
Introduction to Dynamics: Newton’s Laws of Motion
4.1 Development of Force Concept
4.2 Newton’s First Law of Motion: Inertia
4.3 Newton’s Second Law of Motion: Concept of a System
4.4 Newton’s Third Law of Motion: Symmetry in Forces
4.5 Normal, Tension, and Other Examples of Forces
4.6 Problem-Solving Strategies
4.7 Further Applications of Newton’s Laws of Motion
4.8 Extended Topic: The Four Basic Forces—An Introduction
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
Introduction: Further Applications of Newton’s Laws
5.1 Friction
5.2 Drag Forces
5.3 Elasticity: Stress and Strain
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
6 Uniform Circular Motion and Gravitation
Introduction to Uniform Circular Motion and Gravitation
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
6.3 Centripetal Force
6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
6.5 Newton’s Universal Law of Gravitation
6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
7 Work, Energy, and Energy Resources
Introduction to Work, Energy, and Energy Resources
7.1 Work: The Scientific Definition
7.2 Kinetic Energy and the Work-Energy Theorem
7.3 Gravitational Potential Energy
7.4 Conservative Forces and Potential Energy
7.5 Nonconservative Forces
7.6 Conservation of Energy
7.7 Power
7.8 Work, Energy, and Power in Humans
7.9 World Energy Use
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
8 Linear Momentum and Collisions
Introduction to Linear Momentum and Collisions
8.1 Linear Momentum and Force
8.2 Impulse
8.3 Conservation of Momentum
8.4 Elastic Collisions in One Dimension
8.5 Inelastic Collisions in One Dimension
8.6 Collisions of Point Masses in Two Dimensions
8.7 Introduction to Rocket Propulsion
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
9 Statics and Torque
Introduction to Statics and Torque
9.1 The First Condition for Equilibrium
9.2 The Second Condition for Equilibrium
9.3 Stability
9.4 Applications of Statics, Including Problem-Solving Strategies
9.5 Simple Machines
9.6 Forces and Torques in Muscles and Joints
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
10 Rotational Motion and Angular Momentum
Introduction to Rotational Motion and Angular Momentum
10.1 Angular Acceleration
10.2 Kinematics of Rotational Motion
10.3 Dynamics of Rotational Motion: Rotational Inertia
10.4 Rotational Kinetic Energy: Work and Energy Revisited
10.5 Angular Momentum and Its Conservation
10.6 Collisions of Extended Bodies in Two Dimensions
10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
11 Fluid Statics
Introduction to Fluid Statics
11.1 What Is a Fluid?
11.2 Density
11.3 Pressure
11.4 Variation of Pressure with Depth in a Fluid
11.5 Pascal’s Principle
11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
11.7 Archimedes’ Principle
11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
11.9 Pressures in the Body
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
12 Fluid Dynamics and Its Biological and Medical Applications
Introduction to Fluid Dynamics and Its Biological and Medical Applications
12.1 Flow Rate and Its Relation to Velocity
12.2 Bernoulli’s Equation
12.3 The Most General Applications of Bernoulli’s Equation
12.4 Viscosity and Laminar Flow; Poiseuille’s Law
12.5 The Onset of Turbulence
12.6 Motion of an Object in a Viscous Fluid
12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
13 Temperature, Kinetic Theory, and the Gas Laws
Introduction to Temperature, Kinetic Theory, and the Gas Laws
13.1 Temperature
13.2 Thermal Expansion of Solids and Liquids
13.3 The Ideal Gas Law
13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
13.5 Phase Changes
13.6 Humidity, Evaporation, and Boiling
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
14 Heat and Heat Transfer Methods
Introduction to Heat and Heat Transfer Methods
14.1 Heat
14.2 Temperature Change and Heat Capacity
14.3 Phase Change and Latent Heat
14.4 Heat Transfer Methods
14.5 Conduction
14.6 Convection
14.7 Radiation
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
15 Thermodynamics
Introduction to Thermodynamics
15.1 The First Law of Thermodynamics
15.2 The First Law of Thermodynamics and Some Simple Processes
15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
16 Oscillatory Motion and Waves
Introduction to Oscillatory Motion and Waves
16.1 Hooke’s Law: Stress and Strain Revisited
16.2 Period and Frequency in Oscillations
16.3 Simple Harmonic Motion: A Special Periodic Motion
16.4 The Simple Pendulum
16.5 Energy and the Simple Harmonic Oscillator
16.6 Uniform Circular Motion and Simple Harmonic Motion
16.7 Damped Harmonic Motion
16.8 Forced Oscillations and Resonance
16.9 Waves
16.10 Superposition and Interference
16.11 Energy in Waves: Intensity
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
17 Physics of Hearing
Introduction to the Physics of Hearing
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
17.3 Sound Intensity and Sound Level
17.4 Doppler Effect and Sonic Booms
17.5 Sound Interference and Resonance: Standing Waves in Air Columns
17.6 Hearing
17.7 Ultrasound
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
18 Electric Charge and Electric Field
Introduction to Electric Charge and Electric Field
18.1 Static Electricity and Charge: Conservation of Charge
18.2 Conductors and Insulators
18.3 Coulomb’s Law
18.4 Electric Field: Concept of a Field Revisited
18.5 Electric Field Lines: Multiple Charges
18.6 Electric Forces in Biology
18.7 Conductors and Electric Fields in Static Equilibrium
18.8 Applications of Electrostatics
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
19 Electric Potential and Electric Field
Introduction to Electric Potential and Electric Energy
19.1 Electric Potential Energy: Potential Difference
19.2 Electric Potential in a Uniform Electric Field
19.3 Electrical Potential Due to a Point Charge
19.4 Equipotential Lines
19.5 Capacitors and Dielectrics
19.6 Capacitors in Series and Parallel
19.7 Energy Stored in Capacitors
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
20 Electric Current, Resistance, and Ohm's Law
Introduction to Electric Current, Resistance, and Ohm's Law
20.1 Current
20.2 Ohm’s Law: Resistance and Simple Circuits
20.3 Resistance and Resistivity
20.4 Electric Power and Energy
20.5 Alternating Current versus Direct Current
20.6 Electric Hazards and the Human Body
20.7 Nerve Conduction–Electrocardiograms
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
21 Circuits and DC Instruments
Introduction to Circuits and DC Instruments
21.1 Resistors in Series and Parallel
21.2 Electromotive Force: Terminal Voltage
21.3 Kirchhoff’s Rules
21.4 DC Voltmeters and Ammeters
21.5 Null Measurements
21.6 DC Circuits Containing Resistors and Capacitors
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
22 Magnetism
Introduction to Magnetism
22.1 Magnets
22.2 Ferromagnets and Electromagnets
22.3 Magnetic Fields and Magnetic Field Lines
22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
22.6 The Hall Effect
22.7 Magnetic Force on a Current-Carrying Conductor
22.8 Torque on a Current Loop: Motors and Meters
22.9 Magnetic Fields Produced by Currents: Ampere’s Law
22.10 Magnetic Force between Two Parallel Conductors
22.11 More Applications of Magnetism
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
23.1 Induced Emf and Magnetic Flux
23.2 Faraday’s Law of Induction: Lenz’s Law
23.3 Motional Emf
23.4 Eddy Currents and Magnetic Damping
23.5 Electric Generators
23.6 Back Emf
23.7 Transformers
23.8 Electrical Safety: Systems and Devices
23.9 Inductance
23.10 RL Circuits
23.11 Reactance, Inductive and Capacitive
23.12 RLC Series AC Circuits
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
24 Electromagnetic Waves
Introduction to Electromagnetic Waves
24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
24.2 Production of Electromagnetic Waves
24.3 The Electromagnetic Spectrum
24.4 Energy in Electromagnetic Waves
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
25 Geometric Optics
Introduction to Geometric Optics
25.1 The Ray Aspect of Light
25.2 The Law of Reflection
25.3 The Law of Refraction
25.4 Total Internal Reflection
25.5 Dispersion: The Rainbow and Prisms
25.6 Image Formation by Lenses
25.7 Image Formation by Mirrors
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
26 Vision and Optical Instruments
Introduction to Vision and Optical Instruments
26.1 Physics of the Eye
26.2 Vision Correction
26.3 Color and Color Vision
26.4 Microscopes
26.5 Telescopes
26.6 Aberrations
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
27 Wave Optics
Introduction to Wave Optics
27.1 The Wave Aspect of Light: Interference
27.2 Huygens's Principle: Diffraction
27.3 Young’s Double Slit Experiment
27.4 Multiple Slit Diffraction
27.5 Single Slit Diffraction
27.6 Limits of Resolution: The Rayleigh Criterion
27.7 Thin Film Interference
27.8 Polarization
27.9 Extended Topic Microscopy Enhanced by the Wave Characteristics of Light
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
28 Special Relativity
Introduction to Special Relativity
28.1 Einstein’s Postulates
28.2 Simultaneity And Time Dilation
28.3 Length Contraction
28.4 Relativistic Addition of Velocities
28.5 Relativistic Momentum
28.6 Relativistic Energy
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
29 Quantum Physics
Introduction to Quantum Physics
29.1 Quantization of Energy
29.2 The Photoelectric Effect
29.3 Photon Energies and the Electromagnetic Spectrum
29.4 Photon Momentum
29.5 The Particle-Wave Duality
29.6 The Wave Nature of Matter
29.7 Probability: The Heisenberg Uncertainty Principle
29.8 The Particle-Wave Duality Reviewed
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
30 Atomic Physics
Introduction to Atomic Physics
30.1 Discovery of the Atom
30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
30.3 Bohr’s Theory of the Hydrogen Atom
30.4 X Rays: Atomic Origins and Applications
30.5 Applications of Atomic Excitations and De-Excitations
30.6 The Wave Nature of Matter Causes Quantization
30.7 Patterns in Spectra Reveal More Quantization
30.8 Quantum Numbers and Rules
30.9 The Pauli Exclusion Principle
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
31 Radioactivity and Nuclear Physics
Introduction to Radioactivity and Nuclear Physics
31.1 Nuclear Radioactivity
31.2 Radiation Detection and Detectors
31.3 Substructure of the Nucleus
31.4 Nuclear Decay and Conservation Laws
31.5 Half-Life and Activity
31.6 Binding Energy
31.7 Tunneling
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
32 Medical Applications of Nuclear Physics
Introduction to Applications of Nuclear Physics
32.1 Diagnostics and Medical Imaging
32.2 Biological Effects of Ionizing Radiation
32.3 Therapeutic Uses of Ionizing Radiation
32.4 Food Irradiation
32.5 Fusion
32.6 Fission
32.7 Nuclear Weapons
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
33 Particle Physics
Introduction to Particle Physics
33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
33.2 The Four Basic Forces
33.3 Accelerators Create Matter from Energy
33.4 Particles, Patterns, and Conservation Laws
33.5 Quarks: Is That All There Is?
33.6 GUTs: The Unification of Forces
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
34 Frontiers of Physics
Introduction to Frontiers of Physics
34.1 Cosmology and Particle Physics
34.2 General Relativity and Quantum Gravity
34.3 Superstrings
34.4 Dark Matter and Closure
34.5 Complexity and Chaos
34.6 High-temperature Superconductors
34.7 Some Questions We Know to Ask
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
A | Atomic Masses
B | Selected Radioactive Isotopes
C | Useful Information
D | Glossary of Key Symbols and Notation
Answer Key
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Chapter 25
Chapter 26
Chapter 27
Chapter 28
Chapter 29
Chapter 30
Chapter 31
Chapter 32
Chapter 33
Chapter 34
Index
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
State the ideal gas law in terms of molecules and in terms of moles.
Use the ideal gas law to calculate pressure change, temperature change, volume change, or the number of molecules or moles in a given volume.
Use Avogadro’s number to convert between number of molecules and number of moles.
Figure 13.17 The air inside this hot air balloon flying over Putrajaya, Malaysia, is hotter than the ambient air. As a result, the balloon experiences a buoyant force pushing it upward that is larger than its weight. (credit: Kevin Poh, Flickr)
In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, N2N2N2, and oxygen, O2O2O2, are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas, but note that this discussion also applies to monatomic gases, such as helium.)
Gases are easily compressed. We can see evidence of this in Table 13.2, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same βββ. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.
The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure 13.18. Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.
Figure 13.18 Atoms and molecules in a gas are typically widely separated, as shown. Because the forces between them are quite weak at these distances, the properties of a gas depend more on the number of atoms per unit volume and on temperature than on the type of atom.
To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See Figure 13.19.)
Figure 13.19 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion and the pressure increases with more air. (c) Once the tire is inflated, its pressure increases with temperature.
In many common circumstances, including, for example, room temperature air, the gas particles have negligible volume and do not interact with each other, aside from perfectly elastic collisions. In such cases, the gas is called an ideal gas, and the relationship between the pressure, volume, and temperature is given by the equation called the ideal gas law. An equation such as the ideal gas law, which relates behavior of a physical system in terms of its thermodynamic properties, is called an equation of state.
Ideal Gas Law
The ideal gas law states that
PV\=NkT,PV\=NkT,PV\=NkT,
13.18
where PPP is the absolute pressure of a gas, VVV is the volume it occupies, NNN is the number of atoms and molecules in the gas, and TTT is its absolute temperature. The constant kkk is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value
k\=1.38×10−23J/K.k\=1.38×10−23J/K.k\=1.38×10−23 J/K.
13.19
The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product PVPVPV is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of VVV. The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that NNN is the total number of atoms and molecules, independent of the type of gas.)
Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure PPP is essentially equal to atmospheric pressure, and the volume VVV increases in direct proportion to the number of atoms and molecules NNN put into the tire. Once the volume of the tire is constant, the equation PV\=NkTPV\=NkTPV\=NkT predicts that the pressure should increase in proportion to the number N of atoms and molecules.
Example 13.6
Calculating Pressure Changes Due to Temperature Changes: Tire Pressure
Suppose your bicycle tire is fully inflated, with an absolute pressure of 7.00×105Pa7.00×105Pa7.00×105 Pa (a gauge pressure of just under 90.0lb/in290.0lb/in290.0lb/in2) at a temperature of 18.0ºC18.0ºC18.0ºC. What is the pressure after its temperature has risen to 35.0ºC35.0ºC35.0ºC? Assume that there are no appreciable leaks or changes in volume.
Strategy
The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.
We know the initial pressure P0\=7.00×105PaP0\=7.00×105PaP0\=7.00×105 Pa, the initial temperature T0\=18.0ºCT0\=18.0ºCT0\=18.0ºC, and the final temperature Tf\=35.0ºCTf\=35.0ºCTf\=35.0ºC. We must find the final pressure PfPfPf. How can we use the equation PV\=NkTPV\=NkTPV\=NkT? At first, it may seem that not enough information is given, because the volume VVV and number of atoms NNN are not specified. What we can do is use the equation twice: P0V0\=NkT0P0V0\=NkT0P0V0\=NkT0 and PfVf\=NkTfPfVf\=NkTfPfVf\=NkTf. If we divide PfVfPfVfPfVf by P0V0P0V0P0V0 we can come up with an equation that allows us to solve for PfPfPf.
PfVfP0V0\=NfkTfN0kT0PfVfP0V0\=NfkTfN0kT0PfVfP0V0\=NfkTfN0kT0
13.20
Since the volume is constant, VfVfVf and V0V0V0 are the same and they cancel out. The same is true for NfNfNf and N0N0N0, and kkk, which is a constant. Therefore,
PfP0\=TfT0.PfP0\=TfT0.PfP0\=TfT0.
13.21
We can then rearrange this to solve for PfPfPf:
Pf\=P0TfT0,Pf\=P0TfT0,Pf\=P0TfT0,
13.22
where the temperature must be in units of kelvins, because T0T0T0 and TfTfTf are absolute temperatures.
Solution
1. Convert temperatures from Celsius to Kelvin.
T0\=(18.0+273)K\=291 KTf\=(35.0+273)K\=308 KT0\=18.0+273K\=291 KTf\=35.0+273K\=308 KT0\=18.0+273 K\=291 KTf\=35.0+273 K\=308 K
13.23
2. Substitute the known values into the equation.
Pf\=P0TfT0\=7.00×105Pa(308 K291 K)\=7.41×105PaPf\=P0TfT0\=7.00×105Pa308 K291 K\=7.41×105PaPf\=P0TfT0\=7.00×105 Pa308 K291 K\=7.41×105Pa
13.24
Discussion
The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.
Making Connections: Take-Home Experiment—Refrigerating a Balloon
Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why?
Example 13.7
Calculating the Number of Molecules in a Cubic Meter of Gas
How many molecules are in a typical object, such as air in a tire? We can use the ideal gas law to give us an idea of how large NNN typically is.
Calculate the number of molecules in a cubic meter of air at standard temperature and pressure (STP), which is defined to be 0ºC0ºC0ºC and atmospheric pressure.
Strategy
Because pressure, volume, and temperature are all specified, we can use the ideal gas law PV\=NkTPV\=NkTPV\=NkT, to find NNN.
Solution
1. Identify the knowns.
TPVk\=\=\=\=0ºC\=273 K1.01×105Pa1.00m31.38×10−23J/KT\=0ºC\=273 KP\=1.01×105PaV\=1.00m3k\=1.38×10−23J/KT\=0ºC\=273 KP\=1.01×105 PaV\=1.00 m3k\=1.38×10−23 J/K
13.25
2. Identify the unknown: number of molecules, NNN.
3. Rearrange the ideal gas law to solve for NNN.
PV\=NkTN\=PVkTPV\=NkTN\=PVkTPV\=NkTN\=PVkT
13.26
4. Substitute the known values into the equation and solve for NNN.
N\=PVkT\=(1.01×105Pa)(1.00 m3)(1.38×10−23J/K)(273 K)\=2.68×1025moleculesN\=PVkT\=1.01×105Pa1.00 m31.38×10−23J/K273 K\=2.68×1025moleculesN\=PVkT\=1.01×105 Pa1.00 m31.38×10−23 J/K273 K\=2.68×1025molecules
13.27
Discussion
This number is undeniably large, considering that a gas is mostly empty space. NNN is huge, even in small volumes. For example, 1cm31cm31 cm3 of a gas at STP has 2.68×10192.68×10192.68×1019 molecules in it. Once again, note that NNN is the same for all types or mixtures of gases.
Moles and Avogadro’s Number
It is sometimes convenient to work with a unit other than molecules when measuring the amount of substance. A mole (abbreviated mol) is defined to be the amount of a substance that contains as many atoms or molecules as there are atoms in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s number(NA)(NA)(NA), in recognition of Italian scientist Amedeo Avogadro (1776–1856). He developed the concept of the mole, based on the hypothesis that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules. That is, the number is independent of the type of gas. This hypothesis has been confirmed, and the value of Avogadro’s number is
NA\=6.02×1023mol−1.NA\=6.02×1023mol−1.NA\=6.02×1023mol−1.
13.28
Avogadro’s Number
One mole always contains 6.02×10236.02×10236.02×1023 particles (atoms or molecules), independent of the element or substance. A mole of any substance has a mass in grams equal to its molecular (molar) mass, which can be calculated by multiplying the number of moles of the substance by its atomic mass. The atomic masses of elements are given in the periodic table of elements and in Appendix A
NA\=6.02×1023mol−1NA\=6.02×1023mol−1NA\=6.02×1023mol−1
13.29
Figure 13.20 How big is a mole? On a macroscopic level, one mole of table tennis balls would cover the Earth to a depth of about 40 km.
Check Your Understanding
The active ingredient in a Tylenol pill is 325 mg of acetaminophen (C8H9NO2)(C8H9NO2)(C8H9NO2). Find the molar mass of acetaminophen, and from this, the number of moles and the number of molecules of acetaminophen in a single pill.
Solution
We first need to calculate the molar mass (the mass of one mole) of acetaminophen. To do this, we need to multiply the number of atoms of each element by the element’s atomic mass.
(8 moles of carbon)(12 grams/mole)+(9 moles hydrogen)(1 gram/mole)+(1 mole nitrogen)(14 grams/mole)+(2 moles oxygen)(16 grams/mole)\=151 g(8 moles of carbon)(12 grams/mole)+(9 moles hydrogen)(1 gram/mole)+(1 mole nitrogen)(14 grams/mole)+(2 moles oxygen)(16 grams/mole)\=151 g(8 moles of carbon)(12 grams/mole)+(9 moles hydrogen)(1 gram/mole)+(1 mole nitrogen)(14 grams/mole)+(2 moles oxygen)(16 grams/mole)\= 151 g
13.30
Then we need to calculate the number of moles in 325 mg.
(325 mg151 grams/mole)(1 gram1000 mg)\=2.15×10−3moles325 mg151 grams/mole1 gram1000 mg\=2.15×10−3moles325 mg151 grams/mole1 gram1000 mg\=2.15×10−3moles
13.31
Then use Avogadro’s number to calculate the number of molecules.
N\=(2.15×10−3moles)(6.02×1023molecules/mole)\=1.30×1021moleculesN\=2.15×10−3moles6.02×1023molecules/mole\=1.30×1021moleculesN\=2.15×10−3moles6.02×1023molecules/mole\=1.30×1021molecules
13.32
Example 13.8
Calculating Moles per Cubic Meter and Liters per Mole
Calculate: (a) the number of moles in 1.00m31.00m31.00 m3 of gas at STP, and (b) the number of liters of gas per mole at STP.
Strategy and Solution
(a) We are asked to find the number of moles per cubic meter, and we know from Example 13.7 that the number of molecules per cubic meter at STP is 2.68×10252.68×10252.68×1025. The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let nnn stand for the number of moles,
nmol/m3\=Nmolecules/m36.02×1023molecules/mol\=2.68×1025molecules/m36.02×1023molecules/mol\=44.5mol/m3.nmol/m3\=Nmolecules/m36.02×1023molecules/mol\=2.68×1025molecules/m36.02×1023molecules/mol\=44.5mol/m3.nmol/m3\=Nmolecules/m36.02×1023molecules/mol\=2.68×1025molecules/m36.02×1023molecules/mol\=44.5mol/m3.
13.33
(b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain
(103L/m3)44.5mol/m3\=22.5L/mol.103L/m344.5mol/m3\=22.5L/mol.103L/m344.5mol/m3\=22.5L/mol.
13.34
Discussion
This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas.
The (average) molar weight of dry air (approximately 80% N2N2N2 and 20% O2O2O2) at STP is M\=28.8g/mol.M\=28.8g/mol.M\=28.8g/mol. Thus the mass of one cubic meter of air is 1.28 kg. The density of dry room temperature air is about 10% lower. If a living room has dimensions 5m×5 m×3 m,5m×5 m×3 m,5 m×5 m×3 m, the mass of air inside the room is around 90 kg, which is the typical mass of a human.
Check Your Understanding
The density of air at standard conditions (P\=1atm(P\=1atm(P\=1atm and T\=20ºC)T\=20ºC)T\=20ºC) is 1.20kg/m31.20kg/m31.20 kg/m3. At what pressure is the density 0.60kg/m30.60kg/m30.60kg/m3 if the temperature and number of molecules are kept constant?
Solution
The best way to approach this question is to think about what is happening. If the density drops to half its original value and no molecules are lost, then the volume must double. If we look at the equation PV\=NkTPV\=NkTPV\=NkT, we see that when the temperature is constant, the pressure is inversely proportional to volume. Therefore, if the volume doubles, the pressure must drop to half its original value, and Pf\=0.50atm.Pf\=0.50atm.Pf\=0.50 atm.
The Ideal Gas Law Restated Using Moles
A very common expression of the ideal gas law uses the number of moles, nnn, rather than the number of atoms and molecules, NNN. We start from the ideal gas law,
PV\=NkT,PV\=NkT,PV\=NkT,
13.35
and multiply and divide the equation by Avogadro’s number NANANA. This gives
PV\=NNANAkT.PV\=NNANAkT.PV\=NNANAkT.
13.36
Note that n\=N/NAn\=N/NAn\=N/NA is the number of moles. We define the universal gas constant R\=NAkR\=NAkR\=NAk, and obtain the ideal gas law in terms of moles.
Ideal Gas Law (in terms of moles)
The ideal gas law (in terms of moles) is
PV\=nRT.PV\=nRT.PV\=nRT.
13.37
The numerical value of RRR in SI units is
R\=NAk\=(6.02×1023mol−1)(1.38×10−23J/K)\=8.31J/(mol⋅K).R\=NAk\=6.02×1023mol−11.38×10−23J/K\=8.31J/(mol⋅K).R\=NAk\=6.02×1023mol−11.38×10−23J/K\=8.31J/(mol⋅K).
13.38
In other units,
RR\=\=1.99cal/(mol⋅K)0.0821L⋅atm/(mol⋅K).R\=1.99cal/(mol⋅K)R\=0.0821L⋅atm/(mol⋅K).R\=1.99 cal/(mol⋅K)R\=0.0821L⋅atm/(mol⋅K).
13.39
You can use whichever form of RRR is most convenient for a particular problem.
Example 13.9
Calculating Number of Moles: Gas in a Bike Tire
How many moles of gas are in a bike tire with a volume of 2.00×10–3m3(2.00 L),2.00×10–3m3(2.00 L),2.00×10–3m3(2.00 L), a pressure of 7.00×105Pa7.00×105Pa7.00×105Pa (a gauge pressure of just under 90.0lb/in290.0lb/in290.0lb/in2), and at a temperature of 18.0ºC18.0ºC18.0ºC?
Strategy
Identify the knowns and unknowns, and choose an equation to solve for the unknown. In this case, we solve the ideal gas law, PV\=nRTPV\=nRTPV\=nRT, for the number of moles nnn.
Solution
1. Identify the knowns.
PVTR\=\=\=\=7.00×105Pa2.00×10−3m318.0ºC\=291 K8.31J/mol⋅KP\=7.00×105PaV\=2.00×10−3m3T\=18.0ºC\=291 KR\=8.31J/mol⋅KP\=7.00×105PaV\=2.00×10−3m3T\=18.0ºC\=291 KR\=8.31J/mol⋅K
13.40
2. Rearrange the equation to solve for nnn and substitute known values.
n\=\=PVRT\=(7.00×105Pa)(2.00×10−3m3)(8.31J/mol⋅K)(291K)0.579moln\=PVRT\=7.00×105Pa2.00×10−3m38.31J/mol⋅K291K\=0.579moln\=PVRT\=7.00×105Pa2.00×10−3m38.31J/mol⋅K291 K\= 0.579mol
13.41
Discussion
The most convenient choice for RRR in this case is 8.31J/mol⋅K,8.31J/mol⋅K,8.31 J/mol⋅K, because our known quantities are in SI units. The pressure and temperature are obtained from the initial conditions in Example 13.6, but we would get the same answer if we used the final values.
The ideal gas law can be considered to be another manifestation of the law of conservation of energy (see Conservation of Energy). Work done on a gas results in an increase in its energy, increasing pressure and/or temperature. This increased energy can also be viewed as increased internal kinetic energy, given the gas’s atoms and molecules.
The Ideal Gas Law and Energy
Let us now examine the role of energy in the behavior of gases. When you inflate a bike tire by hand, you do work by repeatedly exerting a force through a distance. This energy goes into increasing the pressure of air inside the tire and increasing the temperature of the pump and the air.
The ideal gas law is closely related to energy: the dimensions on both sides are those of energy, with units of joules when using SI units. The right-hand side of the ideal gas law in PV\=NkTPV\=NkTPV\=NkT is NkTNkTNkT. This term is proportional to the amount of translational kinetic energy of NNN atoms or molecules at an absolute temperature TTT, as we shall see formally in Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature. The left-hand side of the ideal gas law is PVPVPV, which also has the units of joules. Pressure is force per unit area, so pressure multiplied by volume is force times displacement, or energy. The important point is that there is energy in a gas related to both its pressure and its volume. The energy can be changed when the gas is doing work as it expands—something we explore in Heat and Heat Transfer Methods—similar to what occurs in gasoline or steam engines and turbines.
Problem-Solving Strategy: The Ideal Gas Law
Step 1 Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal.
Step 2 Make a list of what quantities are given, or can be inferred from the problem as stated (identify the known quantities). Convert known values into proper SI units (K for temperature, Pa for pressure, m3m3m3 for volume, molecules for NNN, and moles for nnn).
Step 3 Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful.
Step 4 Determine whether the number of molecules or the number of moles is known, in order to decide which form of the ideal gas law to use. The first form is PV\=NkTPV\=NkTPV\=NkT and involves NNN, the number of atoms or molecules. The second form is PV\=nRTPV\=nRTPV\=nRT and involves nnn, the number of moles.
Step 5 Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed.
Step 6 Substitute the known quantities, along with their units, into the appropriate equation, and obtain numerical solutions complete with units. Be certain to use absolute temperature and absolute pressure.
Step 7 Check the answer to see if it is reasonable: Does it make sense?
Check Your Understanding
Liquids and solids have densities about 1000 times greater than gases. Explain how this implies that the distances between atoms and molecules in gases are about 10 times greater than the size of their atoms and molecules.
Solution
Atoms and molecules are close together in solids and liquids. In gases they are separated by empty space. Thus gases have lower densities than liquids and solids. Density is mass per unit volume, and volume is related to the size of a body (such as a sphere) cubed. So if the distance between atoms and molecules increases by a factor of 10, then the volume occupied increases by a factor of 1000, and the density decreases by a factor of 1000.
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2022 AIME I Problems/Problem 6
Contents
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1 Problem
2 Solution 1
3 Solution 2 (Rigorous)
4 Solution 3
5 Solution 4
6 See Also
Problem
Find the number of ordered pairs of integers such that the sequenceis strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Solution 1
Since and cannot be an arithmetic progression, or can never be . Since and cannot be an arithmetic progression, and can never be . Since , there are ways to choose and with these two restrictions in mind.
However, there are still specific invalid cases counted in these pairs . Since cannot form an arithmetic progression, . cannot be an arithmetic progression, so ; however, since this pair was not counted in our , we do not need to subtract it off. cannot form an arithmetic progression, so . cannot form an arithmetic progression, so . cannot form an arithmetic progression, ; however, since this pair was not counted in our (since we disallowed or to be ), we do not to subtract it off.
Also, the sequences , , , , and will never be arithmetic, since that would require and to be non-integers.
So, we need to subtract off progressions from the we counted, to get our final answer of .
~ ihatemath123
Solution 2 (Rigorous)
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of , which is impossible.)
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.
Finally, we count the sequences that are something like (one of 3,4,5,), , (one of 30, 40, 50). If this is to be the case, then let be the starting value in the sequence. The sequence will be ; We see that if we subtract the largest term by the smallest term we have , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are . Of these, only the last is invalid because it gives , larger than our bounds . Therefore, we subtract from this case.
Our final answer is
~KingRavi
Solution 3
Denote .
Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .
Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .
Hence, is a subset of , such that there exists an arithmetic sequence that has 4 terms and includes both and .
Hence, this problem asks us to compute
First, we compute .
We have .
Second, we compute .
: .
We have . Thus, the number of solutions is 22.
: .
We have . Thus, the number of solutions is 9.
Thus, .
Third, we compute .
In , we have . However, because , we have . Thus, .
This implies . Note that belongs in .
Thus, .
Fourth, we compute .
: In the arithmetic sequence, the two numbers beyond and are on the same side of and .
Hence, . Therefore, the number solutions in this case is 3.
: In the arithmetic sequence, the two numbers beyond and are on the opposite sides of and .
: The arithmetic sequence is .
Hence, .
: The arithmetic sequence is .
Hence, .
: The arithmetic sequence is .
Hence, . However, the sequence is not strictly increasing.
Putting two cases together,
Therefore,
~Steven Chen (www.professorchenedu.com)
Solution 4
divide cases into .(Notice that can't be equal to , that's why I divide them into two parts. There are three cases that arithmetic sequence forms: .(NOTICE that IS NOT A VALID SEQUENCE!) So when , there are possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)
When , there are ways.
In all, there are possible sequences.
~bluesoul
See Also
2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5Followed by
Problem 7
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
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This document contains calculations of gas volumes at standard temperature and pressure (STP) using the molar volume of gases (24 dm3). Several examples are shown: 1) 10 dm3 of a gas is calc…
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0 9/1 2/2 0 2 3, 1 2:0 1 K a i I N O U E-T I E W - C h e m s h e e t s-G C S E-1 1 1 2-G a s-v o l u m e s-2.p d f | D o c H u b h t t p s://d o c h u b.c o m/6 2 1 4-h v m d x/A L z m Z B 7 w M 1 z Y g y N w X 8 J 5 6 0/k a i-i n o u e-t i e w-c h e m s h e e t s-g c s e-1 1 1 2-g a s-v o l u m e s-2-p d f 1/1
K a i
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UPGRADE
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10dm3 = 0.4167mols 24 0.4167 / 2 = 0.2084 mols 0.2084 24 = 5 dm3 0.2 dm3 = 0.008333 0.008333 0.0250 0.0250 24 = 0.6 dm3 0.16 dm3 = 0.006667 0.006667 2 = 0.001333 0.001333 24 = 0.032 0 dm3 4 = 0.1667 24 0.1667 6.5 = 4.168 4.168 24 = 100 dm3 mols C3H8 = 10/ 44 = 0.2273 0.2273 5 = 1.137 mols volume O2 = 1.137 24 = 27.3 dm3 0.12 = 0.005 24 mols KF = 0.01 mass KF = 0.01 58 = 0.058 g
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Finding the Image of a Function
Travis Collier
641 subscribers
189 likes
Description
27030 views
Posted: 14 Feb 2021
Finding the Image of a Function, Image Set
Algebra 2
4 comments
Transcript:
my series on functions today i'm going to be finding the image of a function so when we talk about a function generally we're talking about a function called f and f is going to map x to y f stands for function and x is going to be the domain while y is the codomain codomain now sometimes when we talk about functions we'll talk about images of functions so we can talk about images of a subset of the domain so if a is a subset of x we say that f of a is the image of a so what we're doing is we're taking a subset of the domain which is called a a is going to be a subset we say that f of a is going to be the image of a so f of a is going to be the set of all images of elements from a so let's say a1 a2 a3 all the way to a n are elements of a now because they're all elements of a they're also all elements of x since a is just a subset of x so if all of these elements belong to the domain then we can put them in the function to get the images so then that means the images are going to be f of a1 f of a2 f of a3 all the way to f of a n these are called the images the images so when we talk about f of a f of a what we're doing is we're taking every single element that's in a and we're getting the set of all the images so f of a we're going to be taking every single element of a and we're going to be putting it through the function and that is what's known as the image of a function okay so let's do an example let's say we have a function that maps the whole numbers to the whole numbers and let's let that function be defined by f of x is equal to the absolute value of x the absolute value of x and now what we're going to do is we're going to define a set x we're going to let x be the set containing the numbers negative 1. 0 1 and 2. so this is a four element set called x and what we want to do is if we know that this is going to be the set x we want to find f of x f of capital x so what we want to do is we want to find the image of x well the image of x is going to be f of x which is going to be the set f of negative one f of zero f of one and f of two and then we close up the set so what this is is we're taking all the elements from set x and we're feeding them all through the function f now if you recall f of x is defined to be the absolute value of whatever you put into it that's the absolute value function so what we're going to do is we're going to take the absolute value of all four of these numbers so we're going to take the absolute value of negative 1 and we're going to take the absolute value of 0 we're going to take the absolute value of positive 1 and we're going to take the absolute value of 2. so working this out the absolute value of negative one would just be positive one that's the distance from negative one to zero that's absolute value the absolute value of zero is of course going to be zero the absolute value of positive one is one since the number one is one away from zero and the absolute value of positive two is going to be positive two since positive 2 is 2 away from 0. so we get the set containing 0 1 and 2. so we can better rewrite this as 0 1 2 and that's because the 1 gets repeated right here so we're just going to limit that to the one only showing up once in this set so that's the image of x 0 1 2. so when you want to find the image what you want to do is you want to take all the numbers from the set that you're given and you want to feed them into the function until you find the image of whatever set you're trying to find the image of so i hope this helps
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Electric Force Between Two Hanging Charges
Dot Physics
51700 subscribers
138 likes
Description
10408 views
Posted: 14 Apr 2020
Physics Explained
Chapter 1: The Electric Field
In this video: Two charges are hanging by strings. What is the value of the charge?
Related videos:
Introduction to Vectors
Universal Gravity
Superposition of gravitational forces
Link to calculation code
16 comments
Transcript:
okay so here's the situation I have two charges and they have the same charge with a little tiny like pith balls let's say something something very light and I charge them up with the same amount of charge and they're supported by these strings but they're held the strings are held near each other so that the two balls repel each other and the question is what's the charge on the two charges so I don't know the charge I picked a mass of 1 gram point zero zero one kilograms the length of the string is 9 centimeters in this angles 10 degrees this is actually a application of Coulomb's law but it's also a forces problem so this charge is in equilibrium and that means I can draw a force diagram for that it looks like this I have the gravitational force pulling down I have the tension pulling up at an angle like that and then I have a repulsive Coulomb force from this other charge right here I'll call it FC and those forces have to add up to zero so let's just put our axes here now if that angle is 10 degrees then this angle is 10 degrees and that's going to be important ok so if it's in equilibrium I can set up my F net x equals zero and f netic y is equal to zero let's just do F net y equals y so in the Y direction let's do that one first because it doesn't have that Coulomb force in the Y direction I have a component of the tension pulling up and the gravitational force blowing down so this is my tension in the Y direction that's a Y and this is my attention in the X direction so that's the adjacent side of this triangle and that's where I told you this to be important so this is going to be equal to T cosine theta minus mg equals zero so I can go ahead and solve for the tension I can add mg to both sides and divide by cosine theta I get T equals mg divided by the cosine of theta okay that's the magnitude the tension now I can look at the what the X forces in the X direction I have a component of the tension T X and the Coulomb force so it's going to be T X is going to be T sine theta negative plus the Coulomb force equals zero so the Coulomb force is going to be equal to T sine theta but I know T so I can say FC is going to be M mg sine theta over cosine theta and that's equal to mg tangent theta okay so now I'm moving over around all around over here but I'm gonna box it out so B okay so now let's go that way now I know the Coulomb force I know I need to know this distance right here so this distance is from here to there it's going to be L times the sine of theta because this is a right triangle that's L and that's the opposite side so the distance R is going to be 2 L sine theta now I can use that in my Coulomb equation FC equals I'm just using the magnitude now I'm not doing the vector version K q1 times q2 but they're the same value so this is gonna be Q squared over R squared system e4l squared sine squared theta and I can set this equal to this M G tangent theta now I want to solve for Q so I'm going to multiply everything by this and divide by that and take the square root so Q equals the square root of M G tangent theta 4 l squared sine squared theta I know that looks crazy all that over K and remember K you can't see remember that's where you should remember that you can't see K K is 9 times 10 to the ninth Newton's meter squared per Coulomb squared okay so I'm not going to put this number in because I don't want to but I'm going to use these values and I will put them in to this equation down here and I'll post a link to a Python code that does the calculation because you should use Python as your calculator because it's awesome ok that's it
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Rounding 3-Digit Numbers to the Nearest Ten, plus Estimation
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Rounding 3-Digit Numbers to the Nearest Ten, plus Estimation A free lesson for 3rd grade math ============================================================================================== Rounding to the nearest ten This number line has the whole tens marked (720, 730, and so on). Let's think about all of the numbers between 740 and 750. Which of them are nearer to 740 than to 750? Which ones are nearer to 750? Rounding a number to the nearest ten means finding which whole ten the number is closest to. We use the symbol≈ when rounding. Read it as “is approximately” or “is about”. Read the examples below. They are also illustrated on the number line with arrows. 731 ≈ 730 (731 is approximately 730)767 ≈ 770 (767 is about 770)724 ≈ 720 (724 is approximately 720) Numbers that end in 1, 2, 3, or 4 are rounded down to the previous whole ten. Numbers that end in 5, 6, 7, 8, and 9 are rounded up to the next whole ten. Notice that numbers ending in 5 are rounded up even though actually they are as far from the previous whole ten as they are from the next whole ten. For example, 855 is equally far away from 850 as it is from 860, but 855 ≈ 860 when rounding to the nearest ten. 1.Round the numbers to the nearest whole ten. Use the number line to help. a. 243 ≈ _b. 287 ≈ _c. 251 ≈ _d. 298 ≈ _ e. 266 ≈ _f. 214 ≈ _g. 255 ≈ _h. 295 ≈ _ i. 307 ≈ _j. 302 ≈ _k. 276 ≈ _l. 242 ≈ _ Whole hundreds are whole tens, too Remember that 100, 200, 300 and all of the other whole hundreds are also whole tens. Why? Remember 100 is 10 tens. 200 would be 20 tens. So, when rounding 603 to the nearest whole ten, we get 603 ≈ 600. And 799 ≈ 800. 2. Round the numbers to the nearest ten. a. 402 ≈ _ 897 ≈ _b. 396 ≈ _ 393 ≈ _c. 804 ≈ _ 805 ≈ _d. 97 ≈ _ 997 ≈ _ Estimation When we use rounded numbers to calculate with, that is called estimation. Mrs. Spencer, the principal, bought books and notebooks for $547, pencils for $38, and erasers for $31. We can estimate her total bill by rounding these numbers to the nearest ten, and adding: $550 + $40 + $30 = $620. Her total bill is about $620. 3. Estimate the total bill. a. a computer, $296, and desk, $188 computer about $__ desk about $_ total bill about $ ___________b. a tennis racket, $123, and balls, $38 racket about $__ balls about $__ total bill about $ _____ c. a movie, $29, MP3 player, $99, and pants, $32 total bill about $ _____d. a printer, $65, paper $13, and printer ink $117 total bill about $ _____ 4. Write the previous and the next whole ten, then round the number. a. __, 472, _ 472 ≈ ______b. __, 829, _ 829 ≈ ______c. __, 514, _ 514 ≈ ______ d. __, 317, _ 317 ≈ ______e. __, 608, _ 608 ≈ ______f. __, 455, _ 455 ≈ ______ 5. Anna wants to buy a bicycle for $129. Round this number to the nearest ten. She earns $30 each week. How much does she earn in two weeks? In three weeks? In how many weeks does she have enough to buy the bicycle? 6. Estimate, using rounded numbers, the total distance all the way around this triangular path in the forest. Also find the real distance by adding the original numbers in columns. Estimate: _______ ft 7. Round each number to the nearest whole ten. Place each answer in the cross-number puzzle.a. c. b. c.d.e. e. Across: a.633 b.796 c.447 d.54 e.306 Down: a.655 b.819 c.397 d.512 e.911 _This lesson is taken from Maria Miller's book Math Mammoth Add & Subtract 3, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller.. Math Lessons menu Place Value Grade 1 Using a 100-bead abacus in elementary math Teaching tens and ones Practicing with two-digit numbers Counting in groups of ten Skip-counting practice (0-100) Comparing 2-digit numbers Cents and dimes Grade 2 Three-digit numbers Comparing 3-digit numbers Grade 3 Place value with thousands Comparing 4-digit numbers Rounding & estimating Rounding to the nearest 100 Grade 4 Place value - big numbers Add & subtract lessons Grade 1 Missing addend concept (0-10) Addition facts when the sum is 6 Addition & subtraction connection Grade 2 Fact families & basic addition/subtraction facts Sums that go over over the next ten Add/subtract whole tens (0-100) Add a 2-digit number and a single-digit number mentally Add 2-digit numbers mentally Regrouping in addition Regrouping twice in addition Regrouping or borrowing in subtraction Grade 3 Mental subtraction strategies Rounding & estimating Multiplication Grade 3 Multiplication concept as repeated addition Multiplication on number line Commutative Multiply by zero Word problems Order of operations Structured drill for multiplication tables Drilling tables of 2, 3, 5, or 10 Drilling tables of 4, 11, 9 Grade 4 Multiplying by whole tens & hundreds Distributive property Partial products - the easy way Partial products - video lesson Multiplication algorithm Multiplication Algorithm — Two-Digit Multiplier Scales problems - video lesson Estimation when multiplying Division Grade 3 Division as making groups Division/multiplication connection Division is repeated subtraction Zero in division Division that is not exact (remainder) Divisibility Grade 4 How to teach long division Long division as repeated subtraction Why long division works Zero in dividend Remainder & long division Two-digit divisor Review of division topics Divisibility Divisibility within 0-1000 Divisibility rules Prime factorization 1 Prime factorization 2 Sieve of Eratosthenes Fraction Lessons Understanding fractions Finding fractional parts with division Mixed numbers Fractions to mixed numbers and vv. 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Size of monochromatic components in local edge colorings - ScienceDirect
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Volume 308, Issue 12, 28 June 2008, Pages 2620-2622
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Size of monochromatic components in local edge colorings
Author links open overlay panelAndrás Gyárfás a, Gábor N.Sárközy a b 1
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An edge coloring of a graph is a local r coloring if the edges incident to any vertex are colored with at most r distinct colors. We determine the size of the largest monochromatic component that must occur in any local r coloring of a complete graph or a complete bipartite graph.
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Local colorings
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An easy exercise—in fact a note of Paul Erdős—is that in every 2-coloring of the edges of K n there is a monochromatic connected subgraph on n vertices. For three colors the analogue problem was solved in , . The problem was rediscovered in . The generalization of this for r colors is proved by the first author : if the edges of K n are colored with r colors then there is a monochromatic connected component with at least n/(r-1) vertices. This result also follows from a more general result of Füredi . The result is sharp if r-1 is a prime power and r-1 divides n. For sharp results when r-1 does not divide n and r is small, see . Generalization of the problem for hypergraphs is treated in . Recently some results are obtained for the case when connectivity is replaced by k-connectivity , .
We show how the answer changes if r-coloring is replaced by local r-coloring, where the number of colors can be larger than r, but the requirement is that edges incident to any vertex are colored with at most r colors. Ramsey numbers in local r-colorings have been introduced in , .
Let f(n,r) denote the largest m such that in every local r-coloring of the edges of K n there is a monochromatic connected subgraph with m vertices. This function has been also defined implicitly in , where mixed Ramsey numbers introduced. In particular, RM(T n,G) was defined as the minimum m such that in any edge coloring of K m there is either a monochromatic tree on n vertices or a totally multicolored copy of G. The special case when G is a forest on four edges were treated in . Since the requirement of forbidding a multicolored K 1,4 is equivalent to local 3-colorings, RM(T n,K 1,4)⩽(7 n/3) follows from the case r=3 of our main result, settling a question raised in .
Theorem 1
f(n,r)⩾rn/(r 2-r+1)with equality if a finite plane of order r-1 exists and r 2-r+1 divides n.
To show equality in the claimed case, consider the points of a finite plane of order r-1 as the vertices of a complete graph and color each pair of vertices by the line going through them. Then replace each vertex i by a k-element set A i so that the A i's are pairwise disjoint. The coloring is extended naturally with the proviso that the edges within A i's are colored with some color among the colors that were incident to vertex i. The result is a locally r-colored K n where n=k(r 2-r+1) and the largest monochromatic connected subgraph has kr=nr/(r 2-r+1) vertices.
We give two proofs for Theorem 1. One is based on the following result, perhaps interesting in its own. A double star is a tree obtained from two vertex disjoint stars by connecting their centers.
Theorem 2
Assume that the edges of a complete bipartite graph G=[A,B]are colored so that the edges incident to any vertex of A are colored with at most p colors and the edges incident to any vertex of B are colored with at most q colors. Then there exists a monochromatic connected subgraph H with at least|A|/q+|B|/p vertices. In fact, H can be selected as a double star.
Corollary 1
If the edges of a complete bipartite graph G are locally r-colored, there exists a monochromatic connected subgraph (in fact a double star) with at least|V(G)|/r vertices.
The special case of Corollary 1, when local r-colorings are replaced by usual r-colorings was proved in (without the remark about the double star). A considerably simpler proof (that gives the stronger result about the double star) was given by Liu et al. . We use their method to prove Theorem 2.
Proof of Theorem 2
Let d i(v) denote the degree of v in color i. For any edge ab of color i, a∈A,b∈B, set c(a,b)=d i(a)+d i(b). Let I(v) denote the set of colors on the edges incident to v∈V(G). Then, by using the Cauchy–Swartz inequality and the local coloring conditions, we get∑ab∈E(G)c(a,b)=∑a∈A∑i∈I(a)d i 2(a)+∑b∈B∑i∈I(b)d i 2(b)⩾|A|p∑a∈A∑i∈I(a)d i(a)|A|p 2+|B|q∑b∈B∑i∈I(b)d i(b)|B|q 2=|A||B||B|p+|A|q,therefore for some a∈A,b∈B, c(a,b)⩾|A|/q+|B|/p. Since the edges incident to a or b in the color of ab span a monochromatic connected subgraph with c(a,b) vertices, Theorem 2 follows.□
Proof of Theorem 1
If any monochromatic, say red component C satisfies |C|⩾rn/(r 2-r+1), we have nothing to prove. Otherwise apply Theorem 2 for the complete bipartite graph [A,B]=[V(C),V(G)⧹V(C)]. The edges incident to any v∈A are colored with at most p=r-1 colors and the edges incident to any v∈B are colored with at most q=r colors. Thus, by Theorem 2, there is a monochromatic component of size at least|A|/q+|B|/p=|A|r+n-|A|r-1=n r-1-|A|1 r-1-1 r⩾⩾n 1 r-1-r r 2-r+1 1 r(r-1)=rn r 2-r+1.□
Our second proof for Theorem 1 applies a result of Füredi . Assume that the edges of K n are locally r-colored. Consider the hypergraph H whose vertices are the vertices of K n and whose edges are the vertex sets of the connected monochromatic components. In the dual of H, H, every edge has at most r vertices and each pair of edges has a nonempty intersection. Füredi proved that in such hypergraphs the fractional transversal number, τ(H)⩽r-1+1/r. Using well-known elementary facts, |E(H)|D(H)⩽ν(H)=τ(H)⩽r-1+1 r,where D is the maximum degree of H. Thus we have r|E(H)|/(r 2-r+1)⩽D(H). Noting that |E(H)|=n and D(H) equals to the maximum size of an edge in H, i.e. the maximum size of a connected component in the local r-coloring, the inequality of Theorem 1 follows.
Recommended articles
References
B. Andrásfai Remarks on a paper of Gerencsér and Gyárfás Ann. Univ. Sci. Eötvös Budapest, 13 (1970), pp. 103-107 Google Scholar
A. Bialostocki, P. Dierker, W. Voxman, Either a graph or its complement is connected: a continuing saga, manuscript, 2001. Google Scholar
A. Bialostocki, W. Voxman, Monochromatic trees and a rainbow forest on four edges, preprint. Google Scholar
J. Bierbrauer, A. Brandis On generalized Ramsey numbers for trees Combinatorica, 5 (1985), pp. 95-107 View in ScopusGoogle Scholar
J. Bierbrauer, A. Gyárfás On (n,k)-colorings of complete graphs Congressus Num., 58 (1987), pp. 123-139 Google Scholar
B. Bollobás, A. Gyárfás, Highly connected monochromatic subgraphs, Discrete Math., to appear. Google Scholar
Z. Füredi Maximum degree and fractional matchings in uniform hypergraphs Combinatorica, 1 (1981), pp. 155-162 View in ScopusGoogle Scholar
Z. Füredi, A. Gyárfás Covering t-element sets by partitions Eur. J. Combinatorics, 12 (1991), pp. 483-489 View PDFView articleView in ScopusGoogle Scholar
L. Gerencsér, A. Gyárfás On Ramsey type problems Ann. Univ. Sci. Eötvös, Budapest, 10 (1967), pp. 167-170 Google Scholar
A. Gyárfás Partition coverings and blocking sets in hypergraphs (in Hungarian) Commun. Comput. Automat. Inst. Hungar. Acad. Sci., 71 (1977), p. 62 Google Scholar
A. Gyárfás, J. Lehel, R.H. Schelp, Zs. Tuza Ramsey numbers for local colorings Graphs Combin., 3 (1987), pp. 267-277 View in ScopusGoogle Scholar
A. Gyárfás, J. Lehel, J. Nesetril, V. Rödl, R.H. Schelp, Zs. Tuza Local k-colorings of graphs and hypergraphs J. Combin. Theory B, 43 (1987), pp. 127-139 View PDFView articleView in ScopusGoogle Scholar
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Cited by (5)
Monochromatic Partitions In Local Edge Colorings
2020, Acta Mathematica Hungarica
### Large Monochromatic Triple Stars in Edge Colourings
2015, Journal of Graph Theory
### Note on 2-edge-colorings of complete graphs with small monochromatic k-connected subgraphs
2014, Applied Mathematics
### Large monochromatic components in edge colorings of graphs: A survey
2011, Progress in Mathematics
### Size of monochromatic double stars in edge colorings
2008, Graphs and Combinatorics
1
Research supported in part by the National Science Foundation under Grant no. DMS-0456401.
Copyright © 2007 Elsevier B.V. All rights reserved.
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Convert inequalities into interval notations step by step
Frequently Asked Questions (FAQ)
What are the types of interval notation?
There are two types of interval notation: closed interval notation and open interval notation.
What is a closed interval notation?
A closed interval notation is a way of representing a set of numbers that includes all the numbers in the interval between two given numbers. In this notation, the numbers at the endpoints of the interval are included in the set. The notation for a closed interval is typically of the form [a,b], where a and b are the endpoints of the interval.
What is an open interval notation?
An open interval notation is a way of representing a set of numbers that includes all the numbers in the interval between two given numbers, but does not include the numbers at the endpoints of the interval. The notation for an open interval is typically of the form (a,b), where a and b are the endpoints of the interval.
What does ∩ mean in interval notation?
In interval notation, the symbol ∩ is used to represent the intersection of two intervals. The intersection of two intervals is the set of all values that are common to both intervals.
How do you find interval notation?
To find interval notation for a set of numbers, identify the minimum and maximum values of the set, and then use the appropriate symbols to represent the set. To express a set of numbers that includes both the minimum and maximum values, use square brackets [ ] for the endpoints of the set. To express a set of numbers that does not include the minimum or maximum values, use parentheses ( ) for the endpoints of the set.
What is interval notation?
Interval notation is a way to represent a set of real numbers on the number line. It consists of two numbers separated by a comma, and the numbers are enclosed in either parentheses or square brackets.
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How does momentum thrust mechanically act on combustion chambers and nozzles in a jet propulsion?
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I'm asking this question since I can't intuitively understand how the acceleration of mass flow and therefore the increase of its momentum can actually mechanically act on the combustion chamber walls and nozzle walls to induce momentum thrust.
For example, when it comes to pressure thrust induced by pressure recovery at various diffuser sections within a propulsion system, it is easy to understand that the parallel component of the force (parallel to the axis of total thrust vector) acting on the diffuser walls of the inlet, compressor diffuser, compressor stator blades, etc will induce an asymmetric load and therefore result in forward thrust. It is also intuitively understandable how the propeller/fan/compressor of gas turbine engines will induce thrust, since it is mechanically altering the mass flow, giving it more momentum and as a result of conservation of momentum, an opposing momentum, i.e. momentum thrust will occur on the individual propeller/fan/compressor blade.
Though, how does the reaction force actually act on combustion chamber/afterburner or nozzle walls? For example, from what I understand the combustion chamber and the afterburner of the gas turbine engine is a constant pressure device so it's not an increase in pressure load acting on the engine compartment that is resulting in thrust when it comes to combustion chambers of gas turbines. The way I understand combustion is that it adds heat energy to the mass flow, and subsequently some of that energy will do work and get converted into kinetic energy resulting in increased mass flow velocity. I can understand that somehow, the expanded gas would be interacting with the chamber wall mechanically resulting in increase of momentum and reactionary force, but it's just not intuitive enough and I'd like to have some better explanation to it.
As for the nozzle, according to NASA's beginner's guide to aeronautics nozzle is not doing any thermodynamic work and there's no pressure change throughout the nozzle. Though this seems a bit counterintuitive to me since my understanding is that a nozzle decreases the pressure of mass flow and increases its velocity. For such reason it is also not understandable to me how the nozzle exit speed greatly influences total thrust when there is no thermodynamic work involved (i.e. conversion of thermal energy into kinetic work, increasing momentum); it must mean that I'm not understanding some basic fundamentals correctly.
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edited Jul 31, 2023 at 0:13
Bergi
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Welcome to Aviation.SE. Interesting question, and something that I didn't realize I don't have an explanation for, until I see it asked like that. Nicely done!Ralph J –Ralph J♦ 2023-07-29 23:15:04 +00:00 Commented Jul 29, 2023 at 23:15
@RalphJ Thank you. I've been searching for the answer of this question but came to no avail through quite a few literatures. It is somewhat counterintuitive to me so hopefully I can get some answers!MK.s –MK.s 2023-07-30 02:21:44 +00:00 Commented Jul 30, 2023 at 2:21
Related: On which point(s) in a jet engine does the reaction force act?mins –mins 2023-07-30 11:05:49 +00:00 Commented Jul 30, 2023 at 11:05
@sophit Yes, thank you, this is exactly the answer I was looking for. So would it be correct if I think of shear force (action) induced thrust (reaction) interacting with combustion chamber rigid body in the same way how a boundary layer induces drag through shear, but reveresed?MK.s –MK.s 2023-07-30 15:21:19 +00:00 Commented Jul 30, 2023 at 15:21
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It all comes down to where you draw your control volume.
At one extreme you can draw your control volume as a simple rectangle box 'around' the engine, but with a good bit of space between the boundary of the volume and the engine. The 'top' and 'bottom' boundaries of the control volume are streamlines -- no mass flows through them. The 'front' of the control volume is the inflow -- it is far enough ahead of the engine that flow entering the control volume has freestream properties. The 'back' of the control volume is the outflow. It has two portions -- the flow that went around the engine (the boundary is far enough away that this again has freestream properties) -- and the flow that went through the engine.
With this control volume, all of the control volume boundaries have equal pressure on both sides, so there is no pressure contribution to thrust. However, the flow that goes through the engine has a change in momentum, which leaves us with T=m˙(V j−V inf)T=m˙(V j−V inf).
Where V j V j is the jet velocity, T T is thrust, m˙m˙ is the mass flow through the engine.
At another extreme, you could shrink the control volume down such that it coincides with all of the surfaces of the engine. In this case, the only mass flow that crosses the boundary is the fuel flow coming in from the injectors. If they point aft, they will contribute some thrust, but they could also be arranged to contribute nothing to thrust. To obtain the thrust, you need to perform a detailed integration of all shear and pressure forces on the boundary on every surface. These surfaces point in complex directions, so to do this integral, you have to be really careful.
Both of these approaches (and any in-between) will give the same answer. Every action has an equal and opposite reaction. Things that happen entirely 'inside' the control volume can be ignored -- they cancel themselves out such that all we need to worry about is what happens to the control volume boundary.
In practice, it is much simpler to draw the control volume far away. It makes all the math & bookkeeping much easier. It gets us the overall answer -- but without a ton of details (what is the force on an individual compressor blade).
What matters is consistency. You can't draw your control volume one place -- and then try to reconcile that part of the boundary with something that happens on the interior of the control volume.
This also works in the opposite manner. Instead of a control volume, you can think about a control mass -- a small chunk of fluid moving through an engine. We can think of it as a tiny cube. It starts off with freestream properties. The only forces that act on this control mass are pressures and shear on the six faces. The control mass will accelerate due to these forces. The energy of the control mass can change due to heat transfer in or out of each face. If the control mass is up against a boundary, then the pressure on that face will equal the pressure on the boundary.
The acceleration of the flow is intimately tied to the pressures in the flow and on the surfaces that interact with the flow.
In a simple model of a rocket, the combustion chamber is at constant pressure, but that is certainly not true in the nozzle. The nozzle expands the flow and accelerates it. Fundamentally the nozzle converts the thermodynamic energy from combustion into kinetic energy of the exhaust gas. As it does this, the gasses expand from high pressure in the combustion chamber to ambient pressure at the nozzle exit. A jet engine adds a turbine between the combustor and nozzle in order to extract work to drive the compressor (and anything else). But the turbine is also a flow expander (like the nozzle).
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edited Jul 30, 2023 at 10:31
Peter Kämpf
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answered Jul 30, 2023 at 4:27
Rob McDonaldRob McDonald
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Thank you for a thorough answer regarding the methodology. In answering my question I think setting a control volume that has boundaries that matches exactly with engine surfaces, as you've suggested, would be most helpful. Also, you greatly helped by suggesting that I should look into shear; I've considered that shear in this case would probably only contribute to the rearward thrust for some reason, although I'm probably wrong with that assumption.MK.s –MK.s 2023-07-30 06:27:35 +00:00 Commented Jul 30, 2023 at 6:27
I'm glad I've reminded you about shear -- but it isn't a significant player in propulsion. In fact, it will almost always amount to a loss (source of inefficiency). Most propulsion analysis can be done in terms of quasi 1D flow -- where there is only one component of velocity (x), but the cross sectional area can change (A). When doing Quasi 1D analysis by hand, we usually work from station to station -- engine inlet, compressor face/exit, burner inlet/exit, turbine inlet/exit, engine nozzle, etc. By hand, we don't pay close attention to the details of the area change between stations.Rob McDonald –Rob McDonald 2023-07-30 18:43:34 +00:00 Commented Jul 30, 2023 at 18:43
However, Quasi 1D can also be applied in much more detail, breaking each engine component into many steps and then subjected to numerical methods. In either case, F=P dA. Force is pressure acting on the change in area. The force on the air is equal and opposite the force on the walls of the tube.Rob McDonald –Rob McDonald 2023-07-30 18:51:06 +00:00 Commented Jul 30, 2023 at 18:51
Oh, so does that mean you could disregard of mass flow itself and just calculate using given pressure and rate of area change subject to pressure at any given part of the engince? Also, what you've described about shear is how I've also initially understood, that shear caused at boundary layer at the surface of each engine components would lead to surface drag. Though after reading your post and comment from Sophit above, I've thought that maybe shear is how momentum thrust is derived, especially in isobaric process where there is no area change. Am I understanding this wrongly?MK.s –MK.s 2023-07-30 19:43:27 +00:00 Commented Jul 30, 2023 at 19:43
If you knew the pressure everywhere in an engine, you could calculate thrust by integrating pressures (and you would need to know the fuel flow rate to the injectors). This is the idea of the 'shrunk down' control volume approach. The shears are there, but they are a viscous effect that always act against motion - they are not essential to making thrust. With the shrunken control volume, there is no fluid in the control volume -- so there is no air to accelerate. We ignore it. With the big control volume, there are no surfaces for pressure to act on, so we ignore them.Rob McDonald –Rob McDonald 2023-07-31 00:05:17 +00:00 Commented Jul 31, 2023 at 0:05
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how does the reaction force actually act on combustion chamber/afterburner or nozzle walls?
Let's first clarify what the reaction force is: It is the force which accelerates a mass of burning fuel-air mixture while it passes through the combustion chamber.
What causes this acceleration? In Newtonian physics, a force is needed to accelerate a mass. Here, acceleration happens due to the isobaric heating which expands the burning fuel-air mixture. In an isochoric process (volume stays constant), no acceleration occurs but pressure rises. Here, the burner can is open at both ends and surrounded by colder air, so the volume of the burning fuel-air mixture is free to expand in all directions. However, there is high-pressure air flowing in from the diffusor which blocks the expansion in forward direction and to the sides. However, downstream expansion is still possible, so that is where the gas expands to, accelerating in the process.
In one sentence, the force which accelerates the gas in the combustion chamber is the pressure force of the air flowing in from the compressor and diffusor.
To be precise: The flow in the combustion chamber can best be approximated as a subsonic Rayleigh flow, so a small pressure loss can be observed along the flow direction as the burning fuel-air mixture heats up and its stagnation pressure drops along the way. Also, friction will cause another pressure loss.
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answered Jul 30, 2023 at 10:50
Peter KämpfPeter Kämpf
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Thank you so much. I've realized talking with another person that I was too focused in the word "constant pressure" (or as you point out that combustion in a constant diameter combustion chamber is isobaric) and ignored the fact that pressure drop at turbine inlet nozzle would naturally induce the mass flow to a rearward direction. As for the other part of the question, I think Sophit and Rob gave me the answer, that shear forces could act as an action force that results in forward acceleration of engine rigid body from reaction force. These two clarifications answer the question I've asked.MK.s –MK.s 2023-07-30 15:27:16 +00:00 Commented Jul 30, 2023 at 15:27
@MK.s I wouldn't put so much importance on shear. Pressure is the dominant source of the propulsive force. By far.Peter Kämpf –Peter Kämpf 2023-07-31 09:52:27 +00:00 Commented Jul 31, 2023 at 9:52
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From a fluid dynamics point of view, forces can be transmitted either through pressure or viscosity (aka friction aka shear).
The pressure contribution is by far the biggest one and can be simply understood by looking at the following picture (taken from this answer):
This picture is about the combustion chamber of a rocket, but the basic idea is the same: the horizontal force is given by the pressure acting on all the vertical areas of the chamber. As pointed out in another answer, the combustion in a turbojet/turbofan happens at more or less constant pressure so that the biggest contribution come from the vertical end of the combustion chamber and of the nozzle.
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edited Jul 31, 2023 at 10:28
answered Jul 31, 2023 at 10:20
sophitsophit
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Thanks for the answer, though my question would then be how an afterburner works. As you'd know the afterburner fuel injector/flamholder is separated from the turbine exit diffuser cone and afterburners also "doesn't apply additional pressure forward of the fuel injector". Shouldn't that mean there is no "plane that is normal to the nozzle axis" where the thermodynamic work of the afterburner can act on? Or is it that, although there is no pressure increase within the afterburner chamber, additional force is still being transmitted to the diffuser cone?MK.s –MK.s 2023-07-31 18:25:34 +00:00 Commented Jul 31, 2023 at 18:25
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https://physics.stackexchange.com/questions/283017/is-there-an-easy-way-to-remember-which-formulas-use-kelvin-vs-which-ones-use-cel
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Is there an easy way to remember which formulas use Kelvin vs which ones use Celsius?
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$\begingroup$
I know that the ones with change in temperature don't matter since the increments are the same, but I'm not sure if I can memorize which ones require $K$ and which ones require $^{\circ}C$. Is it simpler than I think?
To clarify, I'm talking about formulas like Power for conductivity/radiation, dealing with ideal gases ($PV=nRT$), things like that.
thermodynamics
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edited Oct 3, 2016 at 14:21
Qmechanic♦
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$\begingroup$ I have never seen an equation in which using Celsius made any sense. Can you give an example? $\endgroup$
DanielSank
– DanielSank
2016-09-29 02:24:57 +00:00
Commented Sep 29, 2016 at 2:24
$\begingroup$ There's a speed of sound in air formula that uses Celsius. That's the only one I can think of off the top of my head. $\endgroup$
M. Enns
– M. Enns
2016-09-29 02:26:53 +00:00
Commented Sep 29, 2016 at 2:26
$\begingroup$ Well, I could have sworn there were some constants or formulas that used C but I'm probably wrong. So is it pretty much guaranteed that a formula, especially when multiplying, will use K? $\endgroup$
hhh
– hhh
2016-09-29 02:39:38 +00:00
Commented Sep 29, 2016 at 2:39
$\begingroup$ @DanielSank: Can you give an example?. Easy: $q=mc\Delta T$, $\Delta T$ in Celsius is fine. $\endgroup$
Gert
– Gert
2016-09-29 05:36:41 +00:00
Commented Sep 29, 2016 at 5:36
$\begingroup$ @Gert of course, as noted by OP equations with temperature difference work whether we use Kelvin or Celsius. I meant to ask for any other cases. $\endgroup$
DanielSank
– DanielSank
2016-09-29 06:38:13 +00:00
Commented Sep 29, 2016 at 6:38
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2 Answers 2
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$\begingroup$
You can always use Kelvin units for temperature. As far as I know there are no formulas which rely on Celcius units. You can use Celcius units for things like Newton's law of cooling which deal with the difference in temperatures. However, that is only because the difference between two temperatures in Celcius is the same as the difference in Kelvin. In short: always use Kelvin.
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answered Sep 29, 2016 at 3:43
Suzu HiroseSuzu Hirose
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$\begingroup$ In fact, there are very few equations which rely on any system of units. $\endgroup$
DanielSank
– DanielSank
2016-09-29 06:38:34 +00:00
Commented Sep 29, 2016 at 6:38
$\begingroup$ @DanielSank The problem is that Celsius isn't really a unit: if we used 'units' for mass where the zero was a mass of 1kg, we'd run into similarly awful problems that we do with Celsius. $\endgroup$
user107153
– user107153
2016-09-29 07:14:54 +00:00
Commented Sep 29, 2016 at 7:14
$\begingroup$ @tfb Agreed. Do note, however, the the very notion of writing equations "in a system of units" is almost always total nonsense. $\endgroup$
DanielSank
– DanielSank
2016-09-29 07:17:20 +00:00
Commented Sep 29, 2016 at 7:17
$\begingroup$ @tfb You need to get down a lab a bit more... en.wikipedia.org/wiki/Callendar%E2%80%93Van_Dusen_equation $\endgroup$
user56903
– user56903
2016-09-29 09:49:57 +00:00
Commented Sep 29, 2016 at 9:49
$\begingroup$ @DirkBruere I could write a similar equation for a quantity which was any surjective function of temperature: not all of those quantities are good units. $\endgroup$
user107153
– user107153
2016-09-29 12:43:53 +00:00
Commented Sep 29, 2016 at 12:43
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$\begingroup$
If you are multiplying or dividing you need absolute (Kelvin)
If you are adding or subtracting you just need the same on both sides.
For example: if the room temperature is 20C and it rises 2C it is now 22C.
If you have a gas cylinder at 20C and want to heat it until the pressure doubles you can't heat it to just 40C. You must instead convert to (273K+20K) 2 = 586K = 313C
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edited Sep 29, 2016 at 15:09
answered Sep 29, 2016 at 2:57
Martin BeckettMartin Beckett
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$\begingroup$ Wrong. If you are adding temperatures (when would you do that?) you cannot use Celcius. You can only use Celcius when you are considering differences, that is, subtracting. $\endgroup$
Suzu Hirose
– Suzu Hirose
2016-09-29 03:42:41 +00:00
Commented Sep 29, 2016 at 3:42
1
$\begingroup$ @SuzuHirose if the temperature in the room is 20C and it rises by 2C it is now 22C, there is no need to convert this to 293K+2K=295K. $\endgroup$
Martin Beckett
– Martin Beckett
2016-09-29 15:05:22 +00:00
Commented Sep 29, 2016 at 15:05
2
$\begingroup$ You're not adding 22+20. $\endgroup$
Suzu Hirose
– Suzu Hirose
2016-09-29 15:07:34 +00:00
Commented Sep 29, 2016 at 15:07
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https://byjus.com/chemistry/group-18-elements-characteristics/
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Members of group 18 in the modern periodic table are known as the noble gases.
They are:
Helium (He)
Neon (Ne)
Argon (Ar)
Krypton (Kr)
Xenon (Xe)
Radon (Rn)
The members of the group have eight electrons in their outermost orbit (except helium which has two electrons). Thus, they have a stable configuration. Group 18 elements are gases and chemically unreactive, which means they don’t form many compounds. Thus, the elements are known as inert gases. Like the other group elements, noble gas elements also exhibit trends in their physical and chemical properties. The general characteristics of noble gases are discussed below.
Neon Electronic Configuration
Characteristics of Group 18 Elements:
Electronic configuration of noble gases:
Members of group 18 have eight valence electrons, i.e., they have eight electrons in their outermost orbit (except helium). Thus, they exhibit a stable octet configuration. But helium exhibits a duplet configuration. The general configuration of the noble gas family is given as ns2np6 (except helium which has 1s2).
2. #### Atomic radii of noble gases:
The members of group 18 have very small atomic radii. Atomic radii of noble gases increase down the group with an increase in atomic number due to the addition of new shells.
3. #### Ionization enthalpy of noble gases:
Noble gases have eight valence electrons, that is, they have eight electrons in their outermost orbit (except helium). Thus, they exhibit stable octet or duplet configuration. Hence, the elements of group 18 exhibit very high ionization enthalpies. Ionization enthalpy of noble gases decreases down the group due to the increase in their atomic size.
4. #### Electron gain enthalpy of noble gases:
Members of the group have eight valence electrons (except helium). Thus, they exhibit stable octet or duplet configuration. Hence, elements of group 18 exhibit very large positive values of electron gain enthalpy.
To learn more about general characteristics of noble gases download BYJU’S – The Learning App.
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William H. Greene, Econometric Analysis, Seventh Edition, Pearson International, 2011
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electrostatics - Why do we use a cylinder as a Gaussian surface for infinitely long charged wire? - Physics Stack Exchange
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Why do we use a cylinder as a Gaussian surface for infinitely long charged wire?
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Why do we use a cylinder as a Gaussian surface for infinitely long charged wire and not some other shape like cube?
electrostatics
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edited Jun 25, 2019 at 11:56
Qmechanic♦
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Pranav KPranav K
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8 Have you tried doing it with a cube?knzhou –knzhou 2019-06-25 11:59:15 +00:00 Commented Jun 25, 2019 at 11:59
As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!Josh Hoffmann –Josh Hoffmann 2019-06-25 12:03:35 +00:00 Commented Jun 25, 2019 at 12:03
@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.kushal –kushal 2019-06-25 14:57:10 +00:00 Commented Jun 25, 2019 at 14:57
@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.Joshua –Joshua 2019-06-25 19:40:56 +00:00 Commented Jun 25, 2019 at 19:40
@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?kushal –kushal 2019-06-26 14:07:43 +00:00 Commented Jun 26, 2019 at 14:07
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Gauss' theorem would apply to a cube or other shape. You can write
∫E⋅d S=Q ϵ 0∫E⋅d S=Q ϵ 0
where the surface has any shape you like. However, the next step is to do the integral. How can we evaluate E⋅d S E⋅d S if we don't even know the angle between E E and d S d S? Nor do we know how the size of E E relates to its distance from the middle of the cube or whatever. This is where the cylinder comes into play. If we choose a cylinder centred on the line charge then we can argue from the rotational and translational symmmetry that E E must be radially outwards at the curved surface, and also that the size of E E will be the same everywhere around the curved surface. It is only because we can make such a claim that we can proceed to do the integral. This neat way to perform the integral wouldn't work for a some other shape of surface.
Of course, if you were treating a different problem then you would choose your surface to suit that problem.
By the way, it is also interesting to note that Gauss's law on its own does not tell you the whole solution here, because it cannot rule out that the field may also have a further contribution having zero divergence so not contributing to the flux integral. To be specific, it does not rule out there may be a non-zero component of E E in the direction in loops around the line charge. To rule that out you must invoke some further information, which could be either Coulomb's law, or the third Maxwell equation which describes Faraday's law of induction. In the present case there is no changing magnetic flux and therefore the line integral of E E around any loop is zero, which rules out the possibility of a further component to E E. If your professor has omitted to mention this further piece of reasoning then you may like to inquire about it!
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edited Jun 25, 2019 at 15:26
answered Jun 25, 2019 at 10:11
Andrew SteaneAndrew Steane
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Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.kushal –kushal 2019-06-25 14:56:12 +00:00 Commented Jun 25, 2019 at 14:56
Note that you also need reflection symmetry (not just rotational and translational) to argue that E E is radial; otherwise, it could have a component parallel to the wire.Michael Seifert –Michael Seifert 2019-06-25 17:39:21 +00:00 Commented Jun 25, 2019 at 17:39
@MichaelSeifert yes: quite right, thanks; I forgot to mention it Andrew Steane –Andrew Steane 2019-06-25 20:51:23 +00:00 Commented Jun 25, 2019 at 20:51
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As mentioned by others, any Gaussian surface can be used. There are some rules of thumb that are helpful. If you have some element of your system that extends uniformly and infinitely in some direction, then you want your Gaussian surface to be uniform in that direction as well. This suggests you want a cylinder of some cross section. The math is easier if the tangents of the surface are at a constant angle relative to the vectors pointing radially outward from the wire. This suggests you want the tangent to be perpendicular to the radii, so a circular cross section.
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answered Jun 25, 2019 at 16:29
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As a matter of fact you can choose any arbitrary shape to be your gaussian surface, as long as the charges are inside it. But just because we can, should we? Gauss Law can be thought of geometric approach to coulombs law. We use the geometric symmetries of the charge distribution to make our problem simpler.
Remember, Gauss' Law has a dot product of E and dS. If you want, you can take a cube centered around the line charge, but then you have to integrate over all the angles of E and dS making it more cumbersome. You can reduce this effort, by cunningly choosing such a surface where the electric field is always parallel to dS or perpendicular to the surface. That way, you don't have to include any angle consideration as before.
So in conclusion, you can choose any gaussian surface. But carefully choosing one makes our life easy. That's all.
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edited Jun 27, 2019 at 7:24
answered Jun 25, 2019 at 10:35
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Key Concepts of Graphing Linear Functions to Know for Algebra 1
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Sum of Cubes, 3 Ways
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In this long-awaited Josh Academy video, we take a look at 3 ingenious ways to derive the sum of the first n cube numbers: 1³ + 2³ + 3³ +... + n³. First, we look at a friendly, understandable solution using techniques from discrete calculus. After we discover a strange relationship between the sum of cubes and sum of natural numbers, we make sense of it with a visual argument. Finally, we tackle the problem from an abstract, graduate-level perspective by employing the use of the all-important generating functions.
NOTE: At 16:57, Josh meant to write n = 0 to ∞, not i. At 22:05, Josh meant to write over (1-x)^5, not ^4.
DISCRETE CONVOLUTION:
SUM OF FIRST N SQUARES, 3 WAYS:
SUM OF FIRST N NATURALS, 3 WAYS:
Playlist: The Essence of Generating Functions:
Please comment or email joshacademytutoring@gmail.com for any questions/concerns, tutoring and video requests, and if any mistakes/inaccuracies were spotted. Josh Academy holds itself to the highest standards, and relies on your feedback!
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Transcript:
well folks here I have the sum of the first n cubes and this has been a video that's been long requested on the channel and there's been a long break in between um there's been a lot of things that I've been working on and doing in the past few months that I couldn't really devote all the time to the channel but thank you so much for over 30,000 total views on the channel and over 730 subscribers we're almost there at 1 th and we'll see if we can get there before the end of the year so let's Dive Right In the sum of the first n cubes as expressed in Sigma notation looks like I go from 1 to n of I cubed and this is basically the same thing as saying 1 cubed + 2 cubed + 3 Cub + 4 cubed all the way up to n cubed and we're going to find a nice closed form solution to this and the first method I want to introduce is that of discrete calculus all right so method one is called discrete calculus now discrete calculus is calculus Without Limits essentially what this means is functions which are continuous F ofx so remember how F Maps real numbers to real numbers these now become sequences all right a subn which map integer values so integer inputs to real number values they can be integers they can be real numbers so for example you could have a subn is equal to N squared so a Sub 0 is 0 a of 1 is 1 squar a of two is 2 squ so on and so forth so I want to introduce an operation on sequences because now that we keep in mind that sequences are now our functions the the operator is called the forward difference which is Delta right here and it's defined to be the next term in the sequence minus the current term this is called the forward difference all right operator and this's a connection between this and continuous calculus and I'll show you what this connection is it's basically analogous to the derivative of a function as being the limit as Delta X goes to zero of what's called a difference quotient for a reason it's Delta y over Delta X and so you can see that Delta Y and Delta X are both in fact forward differences it's the difference between the another term the next term in the sequence minus the current term and you're basically making that difference smaller and smaller um with the limit in continuous sense now I want to introduce something called the falling power this is called n to the falling K with a little underline and it's defined to be n n -1 N - 2 n- 3 all the way down to nus k + 1 depending on whatever K is and I bring this up because it has a very interesting uh property that works with the forward difference essentially when you take the difference of a falling power Delta of n the K you get KNN to the falling K minus one and this should remind you of the derivative power rule DDX X the Nal NX to the nus1 for the continuous form of a um function and so this is called the power rule for differencing so power rule and I bring this up as well and I bring up one more point after this which are all going to be used together to find the sum of the first n cubes it's the fundamental theorem of discret calculus and it looks like this you can see the sum as n or your argument goes from A to B of the Delta of a sequence U is equal to that sequence originally U evaluated at b + 1 in that situation minus U at a and it's important to not get confused here the upper bound of the summation is in fact always B and then you're evaluating at b + 1 in the result and this is very similar to the fundamental theorem of continuous calculus the integral from A to B of the derivative of x or F ofx is equal to F of B minus F of a so you can see the an analogy over here all right so now we're going to use this all together to find the sum of the first n cubes and how we're going to do this well you can see that the ultimate result is going to be leveraging the fundamental theorem of discret calculus we're going to use a sum as I goes from 1 to n of the Delta of some sequence we're looking for the sequence whose Delta is going to be I cubed and we're going to do this by breaking down I cubed into a sum of following powers so let me show you what I mean sum as I goes from 1 to n of I cubed well I cubed can be written by the definition of our following powers in the following way I I -1 i - 2 all right and we need I cubed and we have to make up for this by subtracting and adding back terms that offset this entire term this boils down to be I cubed and then you have the um minus i - 2 I I so it's be - 3 i^ 2 and then add 2 I back so we have to essentially add back 3 i^ 2 and subtract 2 I to make this entire top statement true now what we want to do is Express I to the following we can we have to express I squar as a Su of falling Powers so what is that i^ 2 is I i - 1 and then we have to add back an i because you have a minus I over here and so what we're going to do here is remember that by the definition of our folling power I to the first Power continuously or in the normal Power sense is equal to I to the falling one and that's a fact that you don't have to worry about kind of showing or determining you can just verify that with the definition so this is ultimately equal to this by definition term is i to the following 3 like this and then we can add to that 3 I i - 1 over here is i to the following 2 so 3 I to the following two and then we have to add 3 I but we subtract 2 I so we get ultimately in the end positive 1 I over here which is going to be just I to the following one so this means that essentially the sum as I goes from 1 to n of I cubed is the sum as I goes from 1 to n of I to the following 3 3 + 3 I to the following 2 plus I to the following 1 and by the fal of discreet calculus this is the Delta of some function now which function is that well the function whose inverse Delta is going to be the um kind of inverse Delta of this is going to be our result U and so you can see back to the power rule if the Delta of n to the falling K is k n to the falling K minus one then the inverse Delta of n to the falling k is equal to or also known as the anti- difference it should be n the k + 1 falling all over k + 1 and so we're going to use this anti- difference over here and so if I reproduce this below um let me try like this we can see that this is equal to the sum as I goes from 1 to n of the Delta of the following situation we have 1 14 I to the falling 4 plus 3 1/3 so those cancel out the coefficients cancel out over here so you get just I to the falling 3 plus 12 I to the falling 2 and the Delta is a linear operator which means you can essentially distribute it over sums and constants so this holds true and now by the fundamental the of discreet calculus you just simply get 1/4 what's I to the following for it's I i - 1 i - 2 i - 3 all right plus I to the following 3 is I i - 1 i - 2 and you add that to 12 I i - 1 and when you evaluate all of this by the final the of discreet calculus the lower bound is going to be IAL 1 and the upper bound is going to be n plus1 important to note but look at what happens at I equal 1 every term here has an i -1 in it so they all go to zero so we can ignore that um one term and just simply look at what happens at n plus one so what do we get we get 1/4 and we can actually factor out some terms here we can Factor an i i - one out of every term so what we're going to get here is um I i - one and then we have in square brackets over here I - 2 i - 3 over 4 like this and we add to that I - 2 and we add back just2 over here all right and again we're evaluating this just at the top n + one and what we can do here as well is factor out a 1/4 and I'll show you why because it makes things a little easier I by I -1 over 4 times this done out is going to be i^ 2 - 2 i - 3 i i 2 - 5 I + 6 we to multiply by four over here to make up for that 1/4 Factor so it's going to be + 4 i - 8 and then 1/ 12 multiply by four because we factored out 1/4 is just going to be plus two and so what happens here you might see is that the six here and the two add together and cancel with the eight and then you have essentially i^ 2 and - 5 I + 4 I is going to be minus I and i^ 2 - I is going to be equal to I i - 1 which gives us the result over here to equal i^ 2 i - 1^ 2 over 4 and evaluated at n + 1 we get the ultimate result that the sum as I goes from 1 to n of I cubed or the sum of the first n cubes is equal to n^ 2 n + 1^ 2 all over 4 and that ladies and gentlemen is the answer and for those of you who have maybe seen my other video on the of the first n natural numbers I'll card that as well right now you might notice that this is actually equal to n + 1 n / 2^ 2 which is equal to the sum of the first n natural numbers squared and so this is a very strange relationship how would this work it's this is this just a coincidence well my second method of proof of this sum of cubes formula will show you that this is actually motivated by a very beautiful visual argument so our visual proof I'm going to go all the way back up here is going to involve literally imagining this as a sum of cubes so number two is a visual method so we have here the unit cube one side here this is one cubed I'm going to try and uh color code it a little bit so we have one cubed over here as Volume Plus these are each side length one but they add up to two so here we have a 2 cubed and you add that to 3 cubed like this and you add it all the way up to the N Cube okay we have n cubed now what I'm going to do here is going to involve a little bit of visual interpretation it's very it's not very easy to show on the screen but I want you to imagine breaking these up so that each of these segments has a height of one and so what we're going to do is we're going to split this in an interesting way I'm going to reproduce this a little bit up top so that we can make it um maybe a little easier to compare but I want you to look at this current diagram and look at how we can split the top part as such so the unit cube stays the same but then the second square or cube well we take it in half we split it in half down the middle so we have one 2X two with height one uh kind of square and we can split the other top half into two like this and I'll show you why we do this soon then the third um fre cubed we can split it down the sections in the middle to get three um 3x3 squares essentially and we can do this for every other Cube number we do up until n cubed where we can split it into different ways as well either like the three or like the two depending on whether it's even or odd and what I'm going to do here is going to be very interesting now and this is a very creative argument that I found elsewhere I did not come up with this myself even though I wish I would so if we move this Cube or these two little unit cubes from the two ^2 over here and we get it and add it like this notice how we get a 3x3 square and it's already lined up diagonally with this with this 3x3 Square so if we throw this 3x3 Square next to it and then the other 3x3 Square in the other missing Gap we have a 6x6 square and notice how we can move these little cutdown segments together in such a way that we ultimately will get a large Square which has a height of one unit and then what do we have to do in order to find the volume we know that the height is one so that's going to be just the area of whatever this is and so what is this going to be well over here notice how this Square's side length is going to be one over here plus 2 + 3 plus all the way up to n so this is going to be 1 + 2 + 3 all the way up to n and then the other side is going to be the exact same thing 1 + 2 + 3 all the way up to the N so 1 + 2 + 3 all the way up to n and notice how no matter how we packed it whether we have even numbers split with the top part even split further to create a square over here and then we have the Cub so the odd terms also break nicely and create squares each time then the area of this entire shape is going to be the square of the sum of the first n natural numbers so sum of i as I goes from 1 to n quantity squared is therefore equal to the sum of cubes and that is a very interesting and nice video visual argument for this fact so that was just a very um again I'm repeating myself but I cannot stress enough how clever and how beautiful this visual argument is and I want you to appreciate it um if you see this all right now let's go to the final method and this going to take a little bit of time this is going to involve what's called generating functions so method three we have generating functions GFS I'll just say so what is what is a generating function a generating function is a way of encoding a sequence gaban that we've seen already in discret calculus as a coefficient of an infinite polinomial so every term of the sequence is associated with its own power of X as as a dummy variable so essentially this is going to equal g x 0 which is 1 plus G1 X 1 plus G2 x^2 all the way up to infinity infinity so why do I bring this up well there's an operation on gen generating functions called the discrete convolution where if a ofx is a generating function encoding a sequence as subn so I'm going to say a ofx encodes the sum of a subn x the n and B of X encodes the sum of an encoded sequence B subn like this then the convolution of the two generating functions which is basically the same thing as multiplying them in this context gives you another generating function C of X like this which we've shown in another video which is carded below as well for your convenience when it comes out where it's going to have coefficients rather than a subn or b subn or their product an interesting result where it's the sum of these sort of two oppositely oriented column wise products of the coefficients A subi and B sub I so you get essentially this formula that the coefficient of the X to the nth term in the convolution is going to be this term over here the sum as I goes from 0 to n of a sub I and B subn minus I and so we read this notation as the coefficient in square brackets of the X to the nth term in the generating function a star B of X now you can see that if we want the sum of the first n CU or of the first n cubes sum as I goes from0 to n of I cubed and notice how I when it's I is zero this is zero so this is the exact same thing as saying the sum as I goes from 1 to n there's no difference so we can call this all S subn as a new sequence then what we want is we can find a generating function perhaps whose coefficients are going to be SN using this formula and so what this involves is in the simplest way a sequence a sub I which is going to be n cubed or in this situation I cubed so we got to find the generating function that encodes the cube numbers and then B subn minus I should just be all ones then and the function that encodes all ones is going to be given by the sum as I goes from 0 to Infinity of just one x to the n and we know this to be the geometric series 1 1 - x so all we got to do is find the generating function a of X that encodes the cube numbers multiply it by 1 over 1 - x and then expand that to see if we can find a coefficient that gives us a nice form of the sum so let's do that so we know from the video on the sum of the first and natural numbers or actually um from the um generating function episode two that I have which I can card right now as well for your convenience that they we have a following result the short form X over 1 - x^2 is equal to the sum as n goes from 0 to Infinity of NX to the n and we got this by differentiating the geometric series to get NX to the nus one and then multiplying by X so essentially x times the derivative with respect to X of the entire term and applying that to the other side gives us this result and so likewise um or sorry this should be um x to the N because we multiply by X the offset the N minus one and now what we have to do is repeat this process a couple more times so we do xddx times xddx of this entire expression to get ourselves n cubed x to the N so let's do that first we want to take the derivative with respect to X of this expression over here and what we get here is going to be the following we get by the quotient rule frime G so 1 - x 2us G Prime is going to be 2 1 - x -1 so it becomes 2 1 - x X over G ^2 so 1 - x 4th power and so you can see that we have um and sorry this should not be X squ it should be 1 - x^ s so then what we get here is we can cross out one of these each like this and so we get is 1 - x + 2x is going to be just 1 + x over 1 - x cub and then to multiply this by X to offset the X the nus1 term that's going to ultimately happen when we differentiate this we can find that x 1 + x or this in the numerator becomes x + x^2 like this gives us the generating function of the squares so this is going to equal to sum as n goes from 0 to Infinity of n^ 2 x the n and so of course we just do this one more time so we hit this entire expression with an xddx x the derivative with respect to X of x + x^2 over 1 - x cubed is going to equal x the derivative is going to be well frime is going to be 2x + 1 and G is going to be 1 - x cubed minus G Prime F so you have minus you have three hopping down 1 - x^2 quantity multiply by Chain rule minus one just adds a one here and we have x + x^2 as the original top function and we divide the whole thing by 1 - x Cub 2 which becomes 1 - x 6 and once again we see here that we can cross out some numbers for example 1 - xity squ crosses out here we get a one power on top here and to get rid of two of them over here we get four so what we get is going to be equal to um x 2x + 1 1 X which we will deal with in a minute and you have plus 3x + 3x^2 like this and you divide this entire thing by 1 - x 4th like this and so then we can expand and distribute the X so we have x well this becomes 2x + 1 - 2x^2 - x so this becomes um 2x + 1 - 2x^ 2 - x + 3x + 3x^2 all over 1 - x to 4th power and so we can see right here that this - 2x^2 cancels with the three and just gets a one then you have 2X + 3x is going to be 5x - x is 4X so this all becomes 4X and then multiplying the X across the 1 4X and X2 gives us the following generating function x + 4x^2 + x cubed all over 1 - x 4 this is our G ofx generating function which will then encode the sum or not the sum just the cube numbers and cubed x the n and now to get the sum of the first n cubes we have to perform the convolution we just multiply this entire thing by 1 over 1 - x and so when we after we hit it with a 1 over 1 - x we get X x + 4x^2 + x Cub all over 1 - x 4 should equal the following function the sum as n goes from 0 to Infinity of remember we shortcut the sum of the first n cubes as s subn x to the N so then what we're essentially looking for is the coefficient of the X to the nth term in this geometric or not geometric the generating function which we want to call P of X so we're looking for the coefficient of the X to the nth term in P of X like this and so what is that well I'm going to bring in another result that we found from a video on discret convolutions which again will be carded um for your convenience when it comes out but it's the following results and this is just something that we have to derive ourselves it's going to be the following let me just reproduce that Bel over here one/ 1 - x to the K power is going to equal the sum as n goes from 0 to Infinity of n Plus K -1 choose K -1 X the n and so we can see over here that P of X has some resemblance here we can break this down I'm going to reproduce this once once again into the following form this is going to equal x and this should be um if I hit it with one 1 over 1 - x it should be X the 5th power excuse me like this then we have X the 5th power here make sure you keep track of those Powers so you have x 1 1 - x 5th + 4x^2 1 1 - x 5 + x cubed 1 1 - x 5 like this and so we can express this as that following sum in this situation we have K is equal to 5 so then n + K - 1 is going to be simply n + 4 and K - 1 becomes 4 so we're going to replace this here x the sum as n goes from 0 to Infinity of n + 4 choose 4 x to the n and then we can reproduce this again I'm just going to um reproduce this three times because we have those expressions and so we're going to have this added to 4x^2 time the sum and we add that to X cubed time the sum and notice how in this situation we get that X is a constant with respect to the AR that's being summed up n so we can move this x into the summation and get X the n + 1 in this first term then we can move the 4x2 or just the x s term in to get X the n + 2 inside the infinite Series so n plus 2 and we can finally do that with the X Cub term and instead of x to the N we now have X the n + 3 and remember that this is all going to be the result P of X and since we're looking for the co efficient of the X to the nth term in P of X we need to find out when all of these terms equal n together for the first term it's when n is going to equal n minus1 or when n is set to n minus one in the summation in the second term it's going to be when n is set to n minus 2 and in the third term it's when n is set to n minus 3 because when you plug these in you get x to the n x to the n x to the n and so that means the coefficient of the X to the nth term should just be the sum of all those coefficients so what do we get here when we get n minus one what becomes N - 1 + 4 choose 4 we add that to 4 nus 2 + 4 choose 4 or I keep adding it sometimes I forget that this a combination because I'm used to seeing two numbers divide with a bar but it's not there's no bar here it's not a fraction so you have N - 2 + 4 choose four and we add that to N - 3 + 4 choose 4 and so this ultimately boils down to n + 3 choose 4 or+ 4 n + 2 choose 4 plus n+1 choose 4 and you can verify yourself using the definition of a combination or um otherwise that this will in fact give you the same result as what we saw for the other two problems where it's equal to n^ 2 N - 1 2 or n + 1 2 right over four all right and you can see this right away because when you have a choose four you're multiplying four times and so you have a kind of fourth order term and you can see that in the result on on the bottom we have fourth order n n squ and then n squ becomes n 4th so you can see how it's already going in the right direction but this was the third way and this is a very interesting way as well I really enjoy generating functions and you can check out the series that we have on geometric series and GE and generating functions as a whole in the link below and card it right now for the playlist so this is the sum of the first end cubes I hope you enjoy these three ways let me know if you want to see three more ways if you want to see the sum of the first and natural numbers three more ways or squares check out those videos as well and thanks so much once again for all the support I'll see you in the next video
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Rising Entropy
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Short and sweet proof of the f(xy) = f(x) + f(y) logarithmic property
If you want a continuously differentiable function f(x) from the reals to the reals that has the property that for all real x and y, f(xy) = f(x) + f(y), then this function must take the form f(x) = k log(x) for some real k.
A proof of this just popped into my head in the shower. (As always with shower-proofs, it was slightly wrong, but I worked it out and got it right after coming out).
I haven’t seen it anywhere before, and it’s a lot simpler than previous proofs that I’ve encountered.
Here goes:
f(xy) = f(x) + f(y)
differentiate w.r.t. x…
f'(xy) y = f'(x)
differentiate w.r.t. y…
f”(xy) xy + f'(xy) = 0
rearrange, and rename xy to z…
f”(z) = -f'(z)/z
solve for f'(z) with standard 1st order DE techniques…
df’/f’ = – dz/z
log(f’) = -log(z) + constant
f’ = constant/z
integrate to get f…
f(z) = k log(z) for some constant k
And that’s the whole proof!
As for why this is interesting to me… the equation f(xy) = f(x) + f(y) is very easy to arrive at in constructing functions with desirable features. In words, it means that you want the function’s outputs to be additive when the inputs are multiplicative.
One example of this, which I’ve written about before, is formally quantifying our intuitive notion of surprise. We formalize surprise by asking the question: How surprised should you be if you observe an event that you thought had a probability P? In other words, we treat surprise as a function that takes in a probability and returns a scalar value.
We can lay down a few intuitive desideratum for our formalization of surprise, and one such desideratum is that for independent events E and F, our surprise at them both happening should just be the sum of the surprise at each one individually. In other words, we want surprise to be additive for independent events E and F.
But if E and F are independent, then the joint probability P(E, F) is just the product of the individual probabilities: P(E, F) = P(E) P(F). In other words, we want our outputs to be additive, when our inputs are multiplicative!
This automatically gives us that the form of our surprise function must be k log(z). To spell it out explicitly…
Desideratum: Surprise(P(E, F)) = Surprise(P(E)) + Surprise(P(F))
But P(E,F) = P(E) P(F), so…
Surprise(P(E) P(F)) = Surprise(P(E)) + Surprise(P(F))
Renaming P(E) to x and P(F) to y…
Surprise(xy) = Surprise(x) + Surprise(y)
Thus, by the above proof…
Surprise(x) = k log(x) for some constant k
That’s a pretty strong constraint for some fairly weak inputs!
That’s basically why I find this interesting: it’s a strong constraint that comes out of an intuitively weak condition.
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5 thoughts on “Short and sweet proof of the f(xy) = f(x) + f(y) logarithmic property”
Take a longer shower. You wanted to prove “continuous f satisfying property… is of the form…” and promptly differentiated.
Ah, you’re right. Changed to C1, thanks for the correction.
I just thought of a simpler one. Define g(x) = f(e^(x)). Then it’s easy to see that g is additive, hence if f continuous, g must be linear (linear functions are the only continuous additive ones – this follows easily by looking at h(r)/r for r rational and h continuous+additive). So invert x = log(u) to get the result.
I just thought of a simpler one. Define g(x) = f(e^(x)). Then it’s easy to see that g is additive, hence if f continuous, g must be linear (linear functions are the only continuous additive ones – this follows easily by looking at h(r)/r for r rational and h continuous+additive). So invert x = log(u) to get the result.
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lagrangian formalism - Why is it possible to neglect higher order terms in the variation of the action? - Physics Stack Exchange
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Why is it possible to neglect higher order terms in the variation of the action?
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In order to get the Euler-Lagrange equations, we should find the variation of the action δ S δ S and to neglect higher-order terms:
δ S=∫L(q+δ q,q′+δ q′,t)d t−∫L(q,q′,t)d t+O[(δ q)2]
I have two questions:
Why is it legal to neglect the higher-order terms?
If we get the Euler-Lagrange equations from a first-order approximation, doesn't it means that the equations themselves are only an approximation?
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edited Oct 26, 2022 at 14:41
Tobias Fünke
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asked Oct 26, 2022 at 13:19
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5 It is no approximation. This is calculus of variations. Similarly, if you want to find a stationary point of a (real) function f, what do you do/which equation do you have to solve?Tobias Fünke –Tobias Fünke 2022-10-26 13:20:23 +00:00 Commented Oct 26, 2022 at 13:20
8 You're not "neglecting" higher order terms. Like the extremum of a function is fully determined by the first derivative, so the extremum of the integral is fully determined by the first variation.ZeroTheHero –ZeroTheHero 2022-10-26 13:21:01 +00:00 Commented Oct 26, 2022 at 13:21
1 The stationary action principle can be stated in terms of the functional derivative viz. δ S δ q=0 or in terms of the functional differential viz. δ S=0; but as the above links show, δ S is defined as a definite integral whose integrand is proportional to δ S δ q, which is not quite what you've done.J.G. –J.G. 2022-10-26 13:37:36 +00:00 Commented Oct 26, 2022 at 13:37
Just to rephrase what others have said in a slightly different way: knowing the first order variation for an arbitrary starting path q is much more information than knowing the first order variation for one specific path q. The former is like knowing the entire first derivative of a function, the latter is only knowing the first derivative at a single point. Knowing the first derivative at all points is enough information to find stationary points.Andrew –Andrew 2022-10-26 18:30:31 +00:00 Commented Oct 26, 2022 at 18:30
@ZeroTheHero. Interesting. You may want to make a formal answer using your idea. It would be nice to show why variational calculus is only interested in the first order term.Sedumjoy –Sedumjoy 2023-09-15 19:30:47 +00:00 Commented Sep 15, 2023 at 19:30
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The following is not mathematical rigorous, but a sketch of the central ideas.
To start, let us consider the case of a (differentiable) real scalar function f:R⟶R. We say that x is a stationary point if and only if f′(x)=0. The Taylor expansion of f around a point x is given by:
f(x+h)=f(x)+f′(x)h+O(h 2).
Now if x is a stationary point, we have f′(x)=0, which regarding equation (1) yields
f(x+h)−f(x)=O(h 2).
What this means is that small changes of x induce only changes to second (and higher) order in f if x is a stationary point. Conversely, if equation (2) holds, then x is a stationary point, which can be seen by dividing (2) with h and taking the limit h→0, yielding f′(x)=0.
This also means that you can find the stationary points of f by (Taylor) expanding it in terms of h and set the terms proportional to h to zero. But we don't neglect any terms or approximate anything.
The very same line of thought applies to functionals: Indeed, consider a functional S:F⟶R, with a suitable chosen space of functions F. We define the n-th functional derivative as δ n S[f][η]:=d n S[f+ϵ η]d ϵ n|ϵ=0, where ϵ∈R and η denotes a suitable function. We can further define a Taylor series as follows:
S[f+h η]=S[f]+∑n=1 h n n!δ n S[f][η]
We say that S is stationary at f if the first functional derivative vanishes for all η, or, equivalently, S does not change to first order for a small change of f, i.e. for all η we have S[f+h η]−S[f]=O(h 2).
Again, note that (5) is equivalent to say that the functional derivative of S at f vanishes.
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edited Oct 27, 2022 at 6:31
answered Oct 26, 2022 at 14:27
Tobias FünkeTobias Fünke
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Is (5) equivalent to saying that the functional derivative of S at f vanishes because when you take the limit h→0, the RHS of (5) will vanish?physics_fan_123 –physics_fan_123 2022-10-26 18:59:35 +00:00 Commented Oct 26, 2022 at 18:59
@physics_fan_123 Well, if you divide by h and take the limit h→0, then the LHS is simply the functional derivative, cf. equation (3) for n=1. On the RHS there are only terms at least proportional to h, so they'll vanish in the limit.Tobias Fünke –Tobias Fünke 2022-10-26 19:07:09 +00:00 Commented Oct 26, 2022 at 19:07
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We want to fund the function y(x) that will minimize I=∫b a d x F(x,y(x),y′(x)), where the integrand depends x, y(x) and also on the derivative y′(x). Obviously, more general integrands, which could depend on higher derivatives of y are possible but they are not needed for our immediate purpose, so we'll stick to the next simplest case, at least as far as derivatives are concerned.
We introduce Y(x,ε)=y(x)+ε γ(x), with γ(a)=γ(b)=0, so that the integral now becomes a parametrized integral I(ε)=∫b a d x F(x,Y(x,ε),Y′(x,ε)). Moreover, the function y(x) is assumed to be the function that minimizes (1). As the function γ(x) is basically arbitrary, the function Y(y,x,ϵ) is basically also arbitrary. Moreover, it is clear that Y(x,0)=y(x).
Since the integral is now a function of the parameter ϵ, we can just take d I(ϵ)/d ϵ and find where this derivative is 0. Moreover, if we set ϵ=0 at that point, we know that Y will collapse to y, which is the function we're searching for.
Using the chain rule for partial derivatives, we have d I(ε)d ε=∫b a d x(∂F(x,Y(x,ε),Y′(x,ε))∂Y∂Y(x,ε)∂ε+∂F(x,Y(x,ε),Y′(x,ε))∂Y′∂Y′(x,ε)∂ε), and (d I(ε)d ε)ε=0=∫b a d x(∂F(x,y(x),y′(x))∂y γ(x)+∂F(x,y(x),y′(x))∂y′d γ(x)d x), where we use the fact that Y(x,0)=y(x) at ϵ=0.
When y′′(x) is continuous, we can integrate the second term by part and obtain, using γ(a)=γ(b)=0,% ∫b a d x∂F(x,y(x),y′(x))∂y′d γ(x)d x=(∂F(x,y(x),y′(x))∂y′γ(x))|b a−∫b a d x(d d x∂F(x,y(x),y′(x))∂y′)γ(x).
The first part on the right of the equality disappears by virtue of the boundary conditions, and the second part can be combined with the first term in Eqn. (2) to obtain (d I(ε)d ε)ε=0=∫b a d x(∂F(x,y(x),y′(x))∂y−d d x∂F(x,y(x),y′(x))∂y′)γ(x). Again, since γ(x) is arbitrary, we have (d I(ε)d ε)ε=0=0⟹∂F(x,y(x),y′(x))∂y−d d x∂F(x,y(x),y′(x))∂y′=0. The symbol (d I(ε)d ε)ε=0≡δ I So really one never needs to expand anything to "beyond first order" since the problem is really one of considering the first derivative of a parametrized integral since we're only interested in the extrema of the integral.
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answered Sep 24, 2023 at 16:06
ZeroTheHeroZeroTheHero
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Thank you for following up on my suggestion. This is a really nice job of it.Sedumjoy –Sedumjoy 2023-09-24 17:40:08 +00:00 Commented Sep 24, 2023 at 17:40
1 @Sedumjoy no worries. The English could be improved but I don't think there's any significant error in the math.ZeroTheHero –ZeroTheHero 2023-09-24 19:57:06 +00:00 Commented Sep 24, 2023 at 19:57
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Evidence that the human cell cycle is a series of uncoupled, memoryless phases - PMC
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Mol Syst Biol
. 2019 Mar 19;15(3):e8604. doi: 10.15252/msb.20188604
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Evidence that the human cell cycle is a series of uncoupled, memoryless phases
Hui Xiao Chao
Hui Xiao Chao
1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
2 Curriculum for Bioinformatics and Computational Biology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
Find articles by Hui Xiao Chao
1,2, Randy I Fakhreddin
Randy I Fakhreddin
1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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1, Hristo K Shimerov
Hristo K Shimerov
1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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1, Katarzyna M Kedziora
Katarzyna M Kedziora
1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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1, Rashmi J Kumar
Rashmi J Kumar
1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
3 Curriculum in Genetics and Molecular Biology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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1,3, Joanna Perez
Joanna Perez
4 Department of Biochemistry and Biophysics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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4, Juanita C Limas
Juanita C Limas
5 Department of Pharmacology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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5, Gavin D Grant
Gavin D Grant
4 Department of Biochemistry and Biophysics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
6 Lineberger Comprehensive Cancer Center, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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4,6, Jeanette Gowen Cook
Jeanette Gowen Cook
4 Department of Biochemistry and Biophysics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
6 Lineberger Comprehensive Cancer Center, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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4,6, Gaorav P Gupta
Gaorav P Gupta
6 Lineberger Comprehensive Cancer Center, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
7 Department of Radiation Oncology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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6,7, Jeremy E Purvis
Jeremy E Purvis
1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
2 Curriculum for Bioinformatics and Computational Biology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
3 Curriculum in Genetics and Molecular Biology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
6 Lineberger Comprehensive Cancer Center, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
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1 Department of Genetics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
2 Curriculum for Bioinformatics and Computational Biology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
3 Curriculum in Genetics and Molecular Biology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
4 Department of Biochemistry and Biophysics, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
5 Department of Pharmacology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
6 Lineberger Comprehensive Cancer Center, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
7 Department of Radiation Oncology, University of North Carolina at Chapel Hill, Chapel Hill, NC, USA
Corresponding author. Tel: +1 919 962 4923; E‐mail: jeremy_purvis@med.unc.edu
✉
Corresponding author.
Received 2018 Aug 9; Revised 2019 Feb 7; Accepted 2019 Feb 8; Collection date 2019 Mar.
© 2019 The Authors. Published under the terms of the CC BY 4.0 license
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PMCID: PMC6423720 PMID: 30886052
Abstract
The cell cycle is canonically described as a series of four consecutive phases: G1, S, G2, and M. In single cells, the duration of each phase varies, but the quantitative laws that govern phase durations are not well understood. Using time‐lapse microscopy, we found that each phase duration follows an Erlang distribution and is statistically independent from other phases. We challenged this observation by perturbing phase durations through oncogene activation, inhibition of DNA synthesis, reduced temperature, and DNA damage. Despite large changes in durations in cell populations, phase durations remained uncoupled in individual cells. These results suggested that the independence of phase durations may arise from a large number of molecular factors that each exerts a minor influence on the rate of cell cycle progression. We tested this model by experimentally forcing phase coupling through inhibition of cyclin‐dependent kinase 2 (CDK2) or overexpression of cyclin D. Our work provides an explanation for the historical observation that phase durations are both inherited and independent and suggests how cell cycle progression may be altered in disease states.
Keywords: cell cycle, cell‐to‐cell variability, computational systems biology, Erlang model, single‐cell dynamics
Subject Categories: Cell Cycle, Quantitative Biology & Dynamical Systems
Introduction
The discovery that DNA synthesis occurs during a well‐defined period of time between cell divisions (Howard & Pelc, 1951) led to the development of the canonical four‐stage cell cycle model comprising G1, S, G2, and M phases. It has long been known that the durations of these phases can vary considerably across cell types (Dawson et al, 1965). For example, stem cells and immune cells have relatively brief G1 durations compared to somatic cells (Becker et al, 2006; Kareta et al, 2015; Kinjyo et al, 2015). Phase durations can also change under certain environmental stresses such as starvation, which lengthens G1 (Cooper, 2003), or DNA damage, which mainly prolongs G1 and G2 (Arora et al, 2017; Chao et al, 2017). Furthermore, examination of individual cells has revealed that phase durations vary even among clonal cells under similar environmental conditions (Dawson et al, 1965). These apparently stochastic differences in cell cycle durations were originally attributed exclusively to the G1 phase (Zetterberg & Larsson, 1985). However, more recent studies in multiple cell types have revealed that S and G2 also contribute significant variability to total cell cycle duration (Dowling et al, 2014; Weber et al, 2014; Zhang et al, 2017). Collectively, these studies have revealed that differences in cell cycle durations are an inherent property of individual cells and raise the fundamental question of how these durations are determined.
Over the past 50 years, multiple models have been put forth to explain the differences in cell cycle phase durations among individual cells. By measuring the time between consecutive cell divisions in unsynchronized cells, Smith and Martin proposed a probabilistic model in which the cell cycle is composed of a random part (“A‐state”) that includes most of G1, and a determinate part (“B‐phase”) composed of the combined S‐G2‐M phases and the remaining duration of G1 (Smith & Martin, 1973). The widely accepted implication of this model is that variability in total cell cycle duration stems mostly from G1 and that the durations of the A‐state and B‐phase are uncorrelated (since one is fixed and the other is random). However, a more recent body of work using time‐lapse fluorescence microscopy suggests that cell cycle phase durations may in fact be correlated. Using the FUCCI fluorescent reporter system (Sakaue‐Sawano et al, 2008) to estimate the onset of S phase in proliferating mouse lymphocytes, the duration of the combined S‐G2‐M phase was reported to be proportional to the total cell cycle duration (Dowling et al, 2014; Sandler et al, 2015). This so‐called “stretched” cell cycle model suggests that S‐G2‐M contributes a substantial amount of variation to total cell cycle duration and claims that a persistent molecular factor may affect progression through multiple phases. As a counterexample to the stretched model, Araujo et al (2016) showed that the duration of M phase is not correlated with total cell cycle length and is instead “temporally insulated” from upstream events. Unifying these disparate observations and interpretations will require a physical model that can explain the quantitative relationships between phase durations in proliferating cells.
The possibility that certain phases are coupled is supported by the observation that many biochemical processes are known to exert control over more than one phase. For example, expression of the E2F family of transcription factors, which target genes involved in the G1/S and G2/M transitions and replication, influences the durations of G1, S, and G2 (Helin, 1998; Ishida et al, 2001; Reis & Edgar, 2004; Dong et al, 2014, 2018). Furthermore, certain stress signals, such as those evoked by DNA damage, can be transmitted from one phase to the next or even inherited from a mother cell's G2 to the daughter cell's G1 (Arora et al, 2017; Yang et al, 2017). The existence of molecular factors that control phase durations is also consistent with the observation that sister cells show strong correlations in their phase durations (Froese, 1964; Sandler et al, 2015). Recent quantification of G1 and S‐G2‐M in mouse lymphoblasts showed that G1 itself is heritable and highly correlated between sisters (Sandler et al, 2015). Reconciling the heritable nature of phase durations with the question of phase coupling is necessary for building a comprehensive picture of cell cycle progression in individual cells.
Here, we report precise measurements of G1, S, G2, and M phase durations in three human cell types. We find that each phase operates according to a distinct timescale, and we detect no evidence of coupling among phases. Instead, phase progression can be accurately modeled as a sequence of memoryless steps in which the duration of each phase is independent of previous phase durations. This lack of correlation holds even when phase durations are altered by external stresses, although, under certain conditions of extreme perturbation or defective checkpoints, phase coupling can be introduced. To explain these observations, we propose a mathematical model in which a large number of heritable factors can each weakly couple the durations of individual phases, but in ensemble, the phases are effectively uncoupled. This quantitative description of cell cycle progression provides a new conceptual framework for studying diseases in which cell cycle progress is dysregulated.
Results
Cell cycle phase durations are uncoupled under unstressed conditions
We examined cell cycle progression in three human cell types: a non‐transformed cell line (hTERT RPE‐1, abbreviated RPE), a transformed osteosarcoma cell line (U2OS), and an embryonic stem cell line (H9). RPE cells are non‐transformed human epithelial cells immortalized with telomerase reverse transcriptase with intact cell cycle regulators (Bodnar et al, 1998); U2OS cells are transformed cancer cell line with near triploidy and an unstable G1 checkpoint (Diller et al, 1990; Stott et al, 1998; Forbes et al, 2017). H9 cells are derived from human blastocysts (Thomson et al, 1998) and exhibit rapid proliferation characterized by a shortened G1 duration (Becker et al, 2006). We used the proliferating cell nuclear antigen (PCNA)‐mCherry fluorescent reporter to quantify, for each cell, the duration of G1, S, G2, and M, and, implicitly, the entire cell cycle duration (Chao et al, 2017; Fig1A, Appendix Fig S1). It has been firmly established in previous studies that, during S phase, PCNA is loaded at DNA replication forks and forms foci in well‐described punctate patterns (Madsen & Celis, 1985; Kennedy et al, 2000; Leonhardt, 2000; Wilson et al, 2016; Chao et al, 2017). PCNA localization is precisely correlated with DNA replication and thus is a bona fide marker of S phase (Madsen & Celis, 1985; Leonhardt, 2000; Burgess et al, 2012; Wilson et al, 2016; Chao et al, 2017; Zerjatke et al, 2017). The transition from diffuse to punctate (G1/S) and from punctate back to diffuse (S/G2) was readily detectable between consecutive frames of time‐lapse imaging by both manual and automated procedures (Barr et al, 2017; Appendix Fig S2A). The G2/M transition was easily identified by nuclear envelope breakdown (Kennedy et al, 2000; Araujo et al, 2016), and the M/G1 transition was recorded as the first frame after telophase (Spencer et al, 2013; Chao et al, 2017).
Figure 1. Variation and lack of correlation among cell cycle phase durations in single human cells.
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Diagram of the cell cycle composed of G1, S, G2, and M phases (not to scale). Phase durations were quantified by time‐lapse fluorescence microscopy using a PCNA‐mCherry reporter to identify four discrete events during the lifetime of an individual cell (see main text and Materials and Methods). Images were acquired every 10 min.
Mean phase durations in RPE, U2OS, and H9 cell lines. Error bars represent standard deviations.
Coefficient of variation (CV) of phase durations.
Percentage of the total variation in cell cycle duration contributed by individual phases.
Correlations between individual cell cycle phase durations.
Data information: Sample sizes were adequate to detect correlations (see Materials and Methods). n=125 (RPE), 130 (U2OS), 113 (H9). R 2, square of Pearson correlation coefficient.Source data are available online for this figure.
As expected, we found that G1 showed the most variability among cell types, ranging from 2.1 h in H9 to 7.9 h in RPE. In contrast, the durations of S (7.6–10.1 h), G2 (3.4–4.0 h), and M (~0.5 h) were relatively consistent (Fig1B). In the RPE and U2OS populations, we did not observe a significant number of p27‐ or p21‐positive cells (Appendix Fig S3), suggesting that quiescent cells arising from contact inhibition, serum starvation, or endogenous DNA damage did not contribute significantly to the measured distribution of G1 durations (Oki et al, 2015; Barr et al, 2017). When looking among individual cells, however, both G1 and G2 durations showed substantial variability within each cell type (Fig1C and D). S phase showed the most narrow distribution of durations (Cameron & Greulich, 1963) with a consistent coefficient of variance across cell types, even in the near‐triploid U2OS. Thus, G1 is the most variable duration across cell types, whereas G1 and G2 are both highly variable among individual cells within a cell type.
We then asked whether any of the phase durations were correlated in individual cells. Correlation between phase durations would indicate the existence of “cellular memory” of the progression rate that persists for more than one cell cycle phase, as would be expected from previous studies (Dowling et al, 2014; Araujo et al, 2016). We compared the durations of G1, S, and G2 phases only since the duration of M phase (~30 min) was significantly shorter than the other phases; similar to the image sampling rate (10 min); and contributed little variance to the total cell cycle duration (Fig1D; Araujo et al, 2016). Surprisingly, we detected neither a significant (P<0.01) nor strong (R 2>0.1) correlation between any pair of phase durations under basal conditions (Fig1E, Appendix Fig S4). This lack of correlation was not due to measurement error because we were able to readily detect correlations between phase durations in sister cells for every cell type (Appendix Fig S5A and B), as reported previously (Minor & Smith, 1974; Dowling et al, 2014; Sandler et al, 2015). Furthermore, the lack of correlation was not due to the sampling frequency chosen, as the correlation coefficients did not depend on the sampling frequency in the range that was used (Appendix Fig S6). To confirm these results, we validated the lack of correlation among cell cycle phases in an independent fluorescent reporter system (Grant et al, 2018; Appendix Fig S2B and C, and Table S1). A statistical power analysis revealed that our sample size would be adequate to detect significant correlations, if present (Materials and Methods). Finally, we note that many phases showed pronounced variability, indicating that the lack of correlation was not due to a lack of variability under basal conditions. Thus, contrary to previous claims that cell cycle phases are correlated, we find no evidence for phase coupling within a cell cycle for three distinct human cell types.
Recent single‐cell studies have provided strong evidence that cellular memories of growth and stress signals can be passed on from mother to daughter cells and alter the daughter's cell cycle progress (Arora et al, 2017; Barr et al, 2017; Yang et al, 2017). To examine whether these memories can, in addition, lead to inter‐generational coupling of phase durations, we examined the correlation between the mother cells’ G2 and their daughter cells’ G1. Although we were able to reproduce the prolongation in daughter cell G1 after DNA damage stress in the mother cell, we did not observe coupling between mother G2 and daughter G1 durations at the single‐cell level (Appendix Fig S5C). Thus, while memory of stress can prolong a daughter cell's G1 phase, these molecular factors appear to be G1‐specific and do not affect the duration of the mother cell's G2 in a correlated manner. Taken together, these results suggest that factors determining the duration of a given cell cycle phase do not significantly affect the duration of the previous, or next, cell cycle phase.
Each cell cycle phase follows an Erlang distribution with a characteristic rate and number of steps
The observed independence of phase durations suggests that each phase may be subject to a unique rate‐governing process. We therefore examined the probability distributions of the phase durations in order to define the underlying stochastic processes driving them. All phases followed a similarly shaped distribution characterized by a minimum duration time and skewed right tail (Fig2A). This distribution immediately ruled out a one‐step stochastic process, which would be expected to produce an exponential distribution of phase durations (Smith & Martin, 1973). Instead, each distribution of phase durations resembled an Erlang distribution, which represents the sum of k Poisson processes with rate λ (Fig2B). The Erlang distribution was originally developed to describe the waiting time before a series of telephone calls is handled by an operator (Erlang, 1909). In its application to the cell cycle, each phase can be thought of as a series of steps that proceeds at some fundamental rate (Chao et al, 2017; Yates et al, 2017). Conceptually, the steps simply refer to some sequence of events in a cell cycle phase that need to be completed in order to proceed to the next phase. These events could be, for example, the sequential degradation of proteins (Coleman et al, 2015) or the stepwise accumulation of a molecular factor (Ghusinga et al, 2016; Garmendia‐Torres et al, 2018) that must reach a threshold in order to complete the phase. The total amount of time needed to complete all steps in the phase has an Erlang distribution (Soltani et al, 2016). This model does not claim that each cell cycle phase is, in actuality, merely a series of exactly k steps. Rather, the Erlang model provides a concise, phenomenological description of cell cycle progression that has a simple and relevant biological interpretation: Each cell cycle phase is a multistep biochemical process that must be completed in order to advance to the next phase (Murray & Kirschner, 1989). Similar mathematical models have been proposed to describe the “microstates” of stem cell differentiation, a sequential biological process that undergoes a discrete number of observable state transitions (Stumpf et al, 2017). In contrast to the differentiation process, however, our model fitting suggested that a single rate parameter for all cell cycle phases was unable to fit the data well (Appendix Fig S7A), suggesting that each cell cycle phase is controlled by distinct rate‐governing mechanisms.
Figure 2. Erlang model of cell cycle progression.
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Distributions of cell cycle phase durations for RPE, U2OS, and H9 cells using single‐cell measurements of phase duration reported in Fig1. Black curves represent fits to Erlang distribution.
Erlang model of cell cycle progression. Each phase consists of a distinct number of steps, k. Each step is a Poisson process with rate parameter, λ. After fitting each phase to the Erlang distribution, we were able to accurately simulate all phase durations except for M phase (2‐sided Kolmogorov–Smirnov test for difference between measured and simulated distributions, Appendix Fig S7B and C).
Fitted shape parameter, k, representing the number of steps for each phase. Error bars represent std from 1,000 bootstraps.
Normalized shape parameter, k, for G1, S, and G2 phases in RPE, U2OS, and H9 cells. Bar height represents the fraction of total cell cycle steps spent in each phase. Error bars represent std from 1,000 bootstraps.
Fitted rate parameter, λ, representing the progression rate of each step within a cell cycle phase. Error bars represent std from 1,000 bootstraps.
Rate parameter λ for each phase, shown by cell type.
By fitting the experimentally measured distributions of phase durations for each cell type, we obtained two parameters for each phase: a shape parameter, k, which represents the number of steps within a phase; and a rate, λ, which represents how quickly on average the step is completed (Fig2A, black curves). Using the estimated parameters, we were able to accurately simulate the cell cycle phase durations under basal conditions for all phases except for M phase, for which the time resolution of measurement was low (10 min) compared to the average duration (~30 min; Appendix Fig S7B and C). The fitted parameters were robust to the sampling frequency used in our experiments (Appendix Fig S7D). When we compared the shapes and rates across cell types, several interesting observations emerged. First, the number of steps was high (k=43–128) for S phase but low for both G1 and G2 (k<20; Fig2C). In addition, although the absolute number of steps differed across cell types, the proportions of steps for each phase were highly conserved, especially for RPE and U2OS (Fig2D). In addition, the rate parameters generally followed the trends of the step parameters across cell types, with high λ corresponding to high k (Fig2C and E). This trend suggests that, regardless of the cell cycle phase, each cell type had a different set of kinetic parameters for cell cycle progression. RPE cell cycle kinetics were better fitted with higher rates through more numerous steps, followed by U2OS, then by H9 with slower rates and fewer steps. The one exception to this pattern was G1 in H9 (Fig2D and F), which is consistent with the unusually short G1 duration in embryonic stem cells (White & Dalton, 2005; Becker et al, 2006; Matson et al, 2017). Although this analysis makes no claims about the actual molecular mechanisms that control phase durations, it further supports the hypothesis that each cell cycle phase obeys a unique rate‐governing process and is therefore consistent with the observation that phase durations are uncorrelated.
Cell cycle phase durations remain uncoupled even when phase durations are altered
To further test whether phase durations are independently controlled, we introduced a series of perturbations in the non‐transformed RPE cell line. Our goal was to alter the durations of specific phases and ask whether subsequent phases remained uncoupled. We first specifically perturbed G1 length by inducing oncogene activation (Fig3A). Overexpressing the oncogene Myc strongly and specifically shortened G1 by 55% (Fig3B, Appendix Fig S8A) without strongly affecting other phases. These large changes in G1 duration did not introduce phase coupling among individual cells (Fig3C, Appendix Fig S8B–D). We next targeted S phase by introducing replication stress with aphidicolin, an inhibitor of DNA polymerase (Fig3D). Aphidicolin specifically prolonged S phase while leaving G2 duration unchanged (Fig3E, Appendix Fig S8E), and there was no evidence of coupling between S phase and G2 (Fig3F, Appendix Fig S8F). Recent studies have shown that replication stress can prolong the G1 duration in the following cell cycle generation (Arora et al, 2017; Barr et al, 2017; Mankouri et al, 2013; Appendix Fig S8G). However, we found that prolonged S phase in the treated cells and prolonged G1 duration in the daughter cells were still uncoupled (Appendix Fig S8H). We next asked whether phases could become coupled by perturbing multiple phases. We prolonged all phases by incubating cells at 34°C (Fig3G). Each phase lengthened by a similar proportion (Fig3H, Appendix Fig S8I). Surprisingly, even though all phases lengthened proportionally in response to lower temperature, the phase durations remained uncoupled at the single‐cell level (Fig3I, Appendix Fig S8J). Similarly, shortening all phases by incubating cells at 40°C did not induce phase coupling (Appendix Fig S8K–O) with the exception of very weak correlation between G1 and G2 (R 2=0.078, P=0.002).
Figure 3. Lack of coupling among cell cycle phases under perturbation.
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Schematic of shortening G1 by myc overexpression. RPE cells infected with retrovirus harboring a tamoxifen‐inducible myc overexpression construct.
Shift in phase durations of RPE cells overexpressing Myc.
Pairwise correlation between cell cycle phase durations of RPE cells overexpressing Myc.
Schematic of prolonging S phase by replication stress using aphidicolin. Asynchronously proliferating RPE cells were treated with 50 ng/ml aphidicolin for 8 h, washed with PBS, and then replenished with fresh media. Only cells whose S phase overlapped with the 8‐h treatment window for at least 1.8 h were analyzed.
Shift in phase durations of RPE cells treated with 50 ng/ml aphidicolin.
Pairwise correlation between cell cycle phase durations under aphidicolin treatment.
Schematic of prolonging all phases by incubating cells at 34°C.
Shift in phase durations of RPE cells incubated at 34°C.
Pairwise correlation between phase durations for cells incubated at 34°C.
Schematic of prolonging G1 by DNA damage using NCS. Asynchronously proliferating RPE mother cells were treated with 25 ng/ml NCS, and their daughter cells were analyzed for a full cell cycle.
Shift in phase durations of RPE cells treated with NCS.
Pairwise correlation between phase durations for cells treated with NCS.
Data information: In panels (B, E, H, and K), boxplots representing the distributions of phase durations in untreated cells are underlaid for comparison. Horizontal lines: median; box ranges: 25 th to 75 th percentiles; error bars: 1.5 interquartile away from the box range. P<1×10−5; P<1×10−10; P<1×10−20, 2‐sided Kolmogorov–Smirnov test. Number of cells: Myc, n=116; aphidicolin, n=115; 34°C, n=122; NCS, n=119.Source data are available online for this figure.
We next introduced the DNA damaging agent neocarzinostatin (NCS) to mother cells and measured the phase durations for daughter cells (Fig3J, Appendix Fig S9A). Recent work in human cells has shown that DNA damage signaling in the mother cell's G2 can persist through mitosis to lengthen the duration of G1, suggesting that coupling of maternal G2 and daughter G1 could potentially arise under genotoxic stress (Arora et al, 2017; Barr et al, 2017; Yang et al, 2017). As expected, at the highest NCS dosages that permitted cells to finish a cell cycle without permanent arrest, we confirmed that DNA damage significantly lengthened G1 in daughter cells (Fig3K, Appendix Fig S8P). However, we found no strong correlations between phase durations, with the possible exception of the daughters’ G1 and G2, which showed a weakly significant correlation (R 2=0.063, P=0.006; Fig3L, Appendix Fig S8Q). We then asked whether perturbing the duration of G2 in the mother could lead to a correlated G1 duration in the daughter cells. We found that DNA damage induced at different phases led to different responses in the daughter cells (Appendix Fig S9B) and that DNA damage in the mother cell's S phase prolonged both the mother's G2 phase and their daughters’ G1 phases in a dose‐dependent manner (Appendix Fig S9C). However, these prolonged phase durations were still uncorrelated (Appendix Fig S9D), implying that lengthening of phase durations caused by external factors is modified by intrinsic cell properties, for example, differences in checkpoint efficiency. Therefore, whichever factors determined G2 duration did not necessarily determine G1 duration. Thus, although increasing levels of DNA damage increased both G2 and the subsequent G1, a mother with a prolonged G2 did not necessarily have daughters with long G1 durations, indicating that there was no phase coupling between G2 and the subsequent G1 at the single‐cell level. We further observed no effect of NCS treatment in the mother cells on the S and G2 phases in the daughter cells (Appendix Fig S9B). In addition, there was no coupling between S and G2 durations in cells damaged during S phase (Appendix Fig S9E and F). In summary, DNA damage incurred in mother cells lengthens phase durations in the daughter cells but does not couple the durations of cell cycle phases either within or across cell cycle generations.
Thus far, our results suggest that the rate of progression for each cell cycle phase is controlled in an independent manner that leads to uncoupling between phase durations. We find no evidence of proportionality between phases as would be expected by a stretched cell cycle model (Dowling et al, 2014), although we can reproduce the presence of a strong linear correlation between individual phase durations and the total cell cycle duration (Appendix Fig S10A and B, Materials and Methods). Such a correlation is expected, however, since any two independent random variables will be correlated to their sum. When comparing phase duration to total cell cycle duration, the R 2 value merely represents the proportion of variance in total cell cycle duration explained by a given phase (Fig1D, Appendix Fig S10C). It is neither an indication of coupling nor of proportional stretching between phases; such a claim requires direct comparison between phases. Interestingly, we found that the combined S‐G2‐M duration accounted for a relatively small part of total variability in RPE cells, whereas S‐G2‐M accounted for the majority of total variability in H9 cell cycle duration (Appendix Fig S10D). Thus, the RPE cell type is more consistent with the Smith–Martin model in which G1 accounts for most of the variability in cell cycle duration (Smith & Martin, 1973). In contrast, rapidly proliferating H9 cells are most similar to the lymphocytes that form the basis for the stretched model (Dowling et al, 2014) in which variability in cell cycle duration stems primarily from S and G2 due to the relatively short duration of G1 in those cells.
A model for heritable factors governing independence of phase duration
We next sought to reconcile our model of independent phase progression with previous observations concerning the heritability of cell cycle phase durations. It has long been known that sibling cells show strong correlations in total cell cycle duration as well as the durations of individual phases (Minor & Smith, 1974; Dowling et al, 2014; Kinjyo et al, 2015; Sandler et al, 2015). These observations strongly suggest the existence of heritable factors that influence the rate of cell cycle progression. However, this observation raises an obvious paradox: If cells retain factors that control the durations of cell cycle phases, how can consecutive phases be uncoupled and memoryless? To reconcile these two observations, we considered three models in which heritable factors might control phase durations. In the first model, which we refer to as the “one‐for‐all” model, a single heritable factor influences the duration of all phases (Fig4A). Under this model, all phases should be strongly correlated because each phase is under a common control. However, the observed lack of coupling between phases (Figs1E and 3) is inconsistent with this model.
Figure 4. A model for heritable factors governing the rate of cell cycle phase progression.
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Alternative models for inheritance of molecular factors governing the durations of cell cycle phases.
Simulation of the “strength of coupling” as a function of the number of unique phase‐coupling molecule types under the many‐for‐all model. Each simulation generated 200 cells for which an R 2 value was calculated. R 2 values were averaged across 200 simulations. The shaded area represents the standard deviation of R 2 across the simulations.
Simulation of coupling strength as a function of the number of unique phase‐coupling and phase‐specific factors. Phase‐coupling factors have shared control over a pair of cell cycle phases, whereas phase‐specific factors affect only one cell cycle phase. Strength of coupling is represented by mean R 2 value as in panel (B).
Same as in (B), but simulating the effect of perturbing a single phase‐coupling factor by significantly increasing its abundance or activity. Perturbation was simulated by increasing the abundance of a phase‐coupling factor by 10‐fold.
Same as in (C), but simulating the effect of increasing a phase‐coupling factor by 10‐fold (see Materials and Methods).
Schematic of prolonging all phases by adding CDK2 inhibitor. RPE cells were treated with 2 μM CVT‐313 and the durations of each phase were quantified for a full cell cycle.
Shifts in phase durations for RPE cells treated with 2 μM CVT‐313. A boxplot representing the distributions of durations in untreated cells is underlaid for comparison. Horizontal lines: median; box ranges: 25 th to 75 th percentiles; error bars: 1.5 interquartile away from the box range. P<1×10−5, 2‐sided Kolmogorov–Smirnov test. (n=117 cells).
Pairwise correlation between cell cycle phase durations upon treatment with CVT‐313. P indicates P‐value from Student's t‐test for Pearson correlation coefficient.
Schematic of shortening phases by overexpression of cyclin D. Cyclin D was overexpressed in RPE cells and the durations of each phase were quantified for a full cell cycle.
Shifts in phase durations for RPE cells overexpressing cyclin D. A boxplot representing the distributions of durations in untreated cells is underlaid for comparison. Horizontal lines: median; box ranges: 25 th to 75 th percentiles; error bars: 1.5 interquartile away from the box range. P<1×10−5, 2‐sided Kolmogorov–Smirnov test. (n=113 cells).
Pairwise correlation between cell cycle phase durations upon overexpression of cyclin D. P indicates P‐value from Student's t‐test for Pearson correlation coefficient.
Source data are available online for this figure.
A second model, called “one‐for‐each”, entails that each cell cycle phase has its own rate‐determining factor and that these heritable factors propagate independently to daughter cells (Fig4A). Under the one‐for‐each model, each cell cycle phase proceeds independent of previous phases, which is consistent with our results. However, this model contradicts several well‐established findings regarding molecular factors that control multiple phase durations. For example, cells that have elevated E2F activity, which controls both the entry into S phase and DNA replication, are expected to progress through both G1 and S more rapidly (Dong et al, 2014, 2018). In contrast, cells with high Cdt1 expression, which functions to license origins for replication, finish G1 early but have a prolonged S phase (Arentson et al, 2002; Pozo & Cook, 2016). To accommodate this existing knowledge, therefore, we considered a model that contains numerous types of heritable factors that can each control multiple phase durations in potentially different directions. We called this the “many‐for‐all” model (Fig4A). Under this model, each phase is under shared control by multiple types of molecular factors. Because each factor individually has a coupling effect, the net effect of a group of such factors could potentially lead to coupling of cell cycle phase durations.
To explore under what conditions such heritable factors would lead to phase coupling under the many‐for‐all model, we computationally modeled the coupling between two phases under shared control as a function of the number of unique factor types (Materials and Methods). Simulation results revealed that the coupling between phases weakened as the number of unique coupling factor types increased (Fig4B). Intuitively, this uncoupling effect arises as the net effect of numerous heritable factors dilutes the effect of individual coupling factors, preventing any single coupling factor from dominating control over phase durations. In addition, introducing more phase‐specific factors, which only affect a single phase, would further uncouple the phases by diluting the coupling factors’ effects (Fig4C). Because we observed no correlation between cell cycle phase durations under basal or perturbed conditions, our experimental results are consistent with the regime of numerous factor types under many‐for‐all model of cell cycle phase progression.
We gained further insight into the inheritance of phase‐coupling factors by analyzing sister cell pairs. Because sister cells share similar amounts of heritable factors due to shared cytoplasmic and genetic content (Rosenfeld et al, 2005; Rohn et al, 2014), all of the models above would be expected to produce correlations between sister cells’ phase durations. However, in order to achieve the observed phase uncoupling in individual cells, the distribution of each type of heritable factor to daughter cells must be independent of the others (Materials and Methods). If factors segregate independently, then even in sister cells—for which phase durations are highly correlated—the noise for each cell cycle phase length is expected to be uncoupled between sisters. For example, the differences between G1 durations in sisters would not be expected to correlate with differences between S durations. To support this hypothesis, we show that even though cell cycle phase durations are highly correlated between sisters (Appendix Fig S5A and B), there is no correlation between the differences in sibling cells’ phase durations for any pair of phases (Appendix Fig S11A). Thus, phase durations appear to be controlled by a large number of heritable factors that segregate independently during cell division.
Perturbation of a single factor leads to coupling between cell cycle phase durations
According to the many‐for‐all model of heritable factors, no single factor dominates the coupling effect among phase durations. However, we hypothesized that cell cycle phases could be forced to show coupling by increasing the level, or activity, of a single molecular factor that controls more than one phase, so that the effect of this factor becomes dominant. To explore this possibility, we computationally simulated the effect of increasing the abundance of a single molecular factor (Materials and Methods). Simulations showed that increasing one factor's net coupling effect could indeed introduce coupling between phase durations (Fig4D and E). We then performed two experiments to test this prediction. First, we introduced a negative regulator of cell cycle progression, the cyclin‐dependent kinase 2 (CDK2) inhibitor CVT‐313 (Brooks et al, 1997), and measured the resulting phase durations (Fig4F). Treating cells with CVT‐313 resembles—but is not identical to—increasing the abundance of a negative cell cycle regulator such as p21 protein, which acts during multiple phases and is a potent inhibitor of CDK2 (Akiyama et al, 1992; Hu et al, 2001; Wadler, 2001; Woo & Poon, 2003). As expected, treatment with the CDK2 inhibitor prolonged all cell cycle phases, having the strongest effect on G1 (Fig4G, Appendix Fig S11B and C). Moreover, consistent with the model's prediction, introducing high levels of CDK2 inhibitor also introduced a strong correlation between each pair of phase durations (Fig4H, Appendix Fig S11D). As a second test of the model's prediction, we overexpressed cyclin D, which is known to positively control the G1/S transition and G2 progression (Fig4I; Guo et al, 2002; Gabrielli et al, 1999; Yang et al, 2006). Increased cyclin D levels led to a strong and significant shortening of both G1 and G2, as well as coupling between G1 and G2 durations (Fig4J and K, Appendix Fig S11E–H). It is both notable and consistent with the model that increases in the CDK2 inhibitor or cyclin D—both single molecular factors that affect multiple phases—led to coupling of phase durations. In contrast, changes in temperature, which alters the activities of a large number of molecular factors, did not lead to coupling.
Uncoupling between cell cycle phases is disrupted in a cancer cell line
The above results in a non‐transformed cell type suggest that cell cycle phases are not intrinsically isolated, but that their phase durations only appear uncoupled due to a large number of factors sharing control over multiple phases. It is therefore possible that this balance of molecular control is disrupted in other cell types that possess disproportionate abundances of cell cycle control factors. To examine phase coupling in cells with dysregulated cell cycle control, we measured whether coupling occurred after perturbing specific phases in U2OS cells, which are known to have a compromised G1 checkpoint (Diller et al, 1990; Stott et al, 1998). We used NCS to induce DNA damage in mother cells and quantified the daughter cells’ phase durations (Fig5A). As in RPE cells, only G1 was significantly prolonged by the DNA damage induced in the mother cells (Fig5B, Appendix Fig S12A). However, unlike in RPE cells, DNA damage induced in U2OS mother cells introduced coupling between daughters’ G1 and S durations and resulted in a positive correlation (Fig5C, Appendix Fig S12B), even in the absence of an increase in S phase duration at the population level. S and G2‐M durations remained uncoupled in cells damaged during S phase (Appendix Fig S12C–E). We next perturbed S phase with aphidicolin to induce replication stress in U2OS cells (Fig5D). As in RPE cells, we observed significant prolongation of S phase duration only (Fig5E, Appendix Fig S12F). In contrast to RPE cells, however, S phase lengthening was coupled to longer durations of both G1 and G2 (Fig5F, Appendix Fig S12G). Interestingly, the duration of G1—which elapsed before the perturbed S phase and was unaffected—predetermined a cell's sensitivity to aphidicolin; that is, longer G1 durations predicted a longer perturbed S phase. This result suggests that one or more phase‐controlling factors contribute to both G1 phase progression and S phase progression but only become rate‐limiting for S phase progression under replication stress. For example, a higher level of endogenous DNA damage in G1 may “leak” through the compromised G1 checkpoint, and these existing damage signals could exacerbate the effect of replication stress in lengthening S phase. In addition to intra‐generation coupling, we observed coupling between the G1 duration of the daughter cell and all phase durations of mother cells (Appendix Fig S12H). Another prediction of the many‐for‐all model is that a perturbation promiscuously affecting many factors would not introduce coupling despite a dysregulated cell cycle. Consistent with this prediction, low‐temperature perturbation prolonged cell cycle phases without introducing phase coupling (Appendix Fig S12I–M). Taken together, these results show that in cells with defective G1/S checkpoints, coupling between cell cycle phases—most strongly between G1 and S—can be revealed under cellular stress.
Figure 5. Stress‐induced coupling of cell cycle phases in a cancer cell line.
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Schematic of prolonging G1 by DNA damage using NCS. Asynchronously proliferating U2OS mother cells were treated with 100 ng/ml NCS, and their daughter cells were analyzed for a full cell cycle.
Shift in phase durations of U2OS cells treated with NCS.
Pairwise correlation between phase durations for U2OS cells treated with NCS. P indicates P‐value from Student's t‐test for Pearson correlation coefficient.
Schematic of prolonging S phase by replication stress using aphidicolin. Asynchronously proliferating U2OS cells were treated with 50 ng/ml aphidicolin for 8 h, washed with PBS, and then replenished with fresh media. Only cells whose S phase overlapped with the 8‐h treatment window were analyzed.
Shift in phase durations of U2OS cells treated with 50 ng/ml aphidicolin.
Pairwise correlation between cell cycle phase durations under aphidicolin treatment. P indicates P‐value from Student's t‐test for Pearson correlation coefficient.
Data information: In panels (B and E), boxplots representing the distributions of phase durations in untreated cells are underlaid for comparison. Horizontal lines: median; box ranges: 25 th to 75 th percentiles; error bars: 1.5 interquartile away from the box range. P<1×10−5; P<1×10−20, 2‐sided Kolmogorov–Smirnov test. Number of cells: NCS, n=114; aphidicolin, n=153.Source data are available online for this figure.
Discussion
Our understanding of cell cycle progression is built largely upon accumulated knowledge of the molecular mechanisms that act during each phase. Various computational models have been developed to integrate these mechanisms into a quantitative framework. These models have provided invaluable insights into the cell cycle's temporal organization (Orlando et al, 2008; Tyson & Novak, 2008), adaptation to stress (Heldt et al, 2018), role in cellular decision making (Spencer et al, 2013; Dong et al, 2014; Cappell et al, 2016), and irreversible nature (Novak et al, 2007). However, these models do not account for the observed distribution of cell cycle phase durations in single cells nor do they explain how phase durations can be both heritable and independent. Such a systems‐level understanding of the governing principles regulating cell cycle progression could provide a new conceptual framework to guide further experimental work and therapeutic efforts (Ryl et al, 2017). In this study, we developed a phenomenological model for cell cycle progression informed by precise measurements of G1, S, G2, and M durations in three human cell types. We found that cells with functional cell cycle checkpoints progress through each phase at a rate that is independent of previous phase durations. This independence between phases, which can be modeled as a sequence of memoryless processes, can be explained by the presence of many molecular factors that each contributes a small effect over one or more phase durations.
The lack of correlation between phase durations was unexpected and would seem to disagree with previous experimental results and theoretical models. We offer several explanations for this discrepancy. First, many previous data supporting cell cycle phase coupling relied on correlations between the total cell cycle duration and a part of the total duration (Dowling et al, 2014). As noted above, such a relationship does not necessarily imply coupling (Appendix Fig S10). In contrast, we directly measured the degree of coupling between individual phases and found no evidence of significant coupling. Second, previous studies examined different cell types (e.g., mouse lymphocytes) that could have different profiles of factors controlling cell cycle progression such as elevated endogenous DNA damage levels (Adelman et al, 1988) or different activities of major cell cycle regulators such as p53 or the retinoblastoma protein, Rb (Rangarajan & Weinberg, 2003). According to the many‐for‐all model, observation of coupling between cell cycle phases would imply that phase‐controlling factors are relatively more abundant in certain cell types. Third, we employ a more accurate method of measuring cell cycle phase durations based on appearance and disappearance of PCNA foci during S phase and validate these results with an orthogonal measurement of phase duration (Grant et al, 2018; Appendix Fig S2B and C; Chao et al, 2017; Grant et al, 2018). Previous studies employed the FUCCI reporter system to distinguish G1 and S‐G2‐M cells, but this system is known to give unclear cell cycle phase boundaries (Wilson et al, 2016; Grant et al, 2018). Fourth, it is possible that previous studies analyzed a mixed population of cells (e.g., cells at different stages of differentiation or maturation) that could result in correlation between phase durations across the different cell types (Roccio et al, 2013). To avoid this problem, we analyzed three clonal cell lines under steady‐state growth conditions. Finally, our results indicate that certain stressful growth conditions may introduce coupling between cell cycle phases, such as high levels of environmental stress (Fig5C and F) or upregulation of potent cell cycle inhibitors such as p21, which inhibits CDK2 activity (Fig4H; Harper et al, 1995).
Our model explains why phase coupling may not always be observed despite the fact that certain coupling factors are known to exist. For example, a “sizer” model for cell cycle control, in which cell size constrains cell cycle progression through the G1/S and G2/M transitions, would predict a correlation between phase durations in single cells since cell size is a continuously changing quantity, and thus, it is “inherited” from one phase to the next (Donnan & John, 1983; Ferrell et al, 2011; R Jones et al, 2017; Garmendia‐Torres et al, 2018). Similarly, we might also expect coupling to arise due to the ordered CDK substrate sensitivity that is observed across phases (Nurse, 1980; Swaffer et al, 2016). However, according to the many‐for‐all model, in which each phase‐controlling factor has shared control over multiple phases, the net effect of many such factors leads to uncoupling between cell cycle phase durations due to a large number effect. Under this model, perturbations that influence more phase‐length factor types are expected to show less coupling, whereas perturbations that influence one or a few factors are expected to introduce stronger coupling. Consistent with this prediction, we find the degree of coupling is the weakest under the most phase‐specific perturbation with replication stress (Fig3F) and under the most promiscuous perturbation of reduced temperature (Fig3I), in which presumably all biochemical processes are slowed. Increased temperature generally accelerates biochemical processes, but it may have an additional detrimental effect (e.g., protein denaturation) on a smaller set of factors that are more sensitive to high temperature (Spiess et al, 1999). Myc overexpression leads to global changes in transcription for a broad spectrum of cellular functions (Dang, 1999; Li et al, 2003), and DNA damage induces the DNA damage response (DDR) network, which is also an ensemble of components (Arcas et al, 2014). Both perturbations involve large numbers of phase‐controlling factors that can preserve the diversity of factors and lead to mild phase coupling. In contrast, perturbing a single phase‐controlling factor by increasing CDK2 inhibitor or cyclin D levels introduced the strongest coupling among cell cycle phases. This observation echoes the previous observation that induced lengthening of one gap phase in Drosophila leads to accelerated progress through the subsequent gap phase via E2F1 regulation (Reis & Edgar, 2004), although further work is required to determine whether E2F1‐altered phases are actually coupled in single cells. Recent work in yeast suggests that certain cell cycle phase durations can show coupling (Garmendia‐Torres et al, 2018). This observation may be explained by the fewer number of cell cycle regulators in yeast (Malumbres & Barbacid, 2009; Lim & Kaldis, 2013) or a more dominant role for cell size control (Garmendia‐Torres et al, 2018), or both. Thus, our model harmonizes with other descriptions of cell cycle progression by providing a framework for predicting when phase couplings may occur. More generally, the many‐for‐all model is consistent with the behavior of other signaling cascades in which multiple factors each exert a partial contribution to the overall cell‐to‐cell heterogeneity (Wagner et al, 2007; Chang et al, 2008; Cohen et al, 2008; Spencer et al, 2009).
Phase coupling may be an indicator of dysregulated cell cycle control in human cells. We found that in the U2OS cell line, which harbors known defects in the G1 checkpoint (Diller et al, 1990; Stott et al, 1998; Kleiblova et al, 2013), stresses such as DNA damage and replication stress introduced phase coupling. Under these conditions, stress signals such as ATM and ATR operate as phase‐controlling factors that impede cell cycle progression. Functional checkpoints normally detect stress signals and wait for the signals to resolve before allowing cell cycle progression to resume, making each phase effectively “insulated” from the previous phase and producing a memoryless process. Without a fully functional checkpoint, however, the memory of stress signal levels could be transmitted from one phase to the next and lead to phase coupling (Lukas et al, 2011; Burrell et al, 2013). Coupling may also arise in cancerous cells through oncogene activation or tumor suppressor loss. Oncogene activation leads to overexpression and thus dominance of a few phase‐controlling factors, whereas tumor suppressor loss decreases the pool of phase‐controlling factor types (Roumeliotis et al, 2017), both of which could lead to imbalance in the competing pool of factors and susceptibility to phase coupling under stress. Cancer cells are characterized by genome instability and defective DDR pathways and are often over‐reliant on the remaining intact part of the DDR network such as ATM and Chk1 (O'Connor, 2015; Jackson & Helleday, 2016; Brown et al, 2017). When further DNA damage or replication stress is incurred, these few critical components are further induced, which may lead to dominating control over other factors and phase coupling. Further work is required to determine whether phase coupling is a common feature among cancer cell types.
Finally, our work is consistent with the observation that the memory of cell cycle duration is lost when a cell divides, as evidenced by the lack of correlation between mother and daughter phase durations (Froese, 1964; Absher & Cristofalo, 1984; Sandler et al, 2015; Barr et al, 2017). Work by Sandler et al suggests that this apparent stochasticity is driven by underlying deterministic factors that operate on a different timescale than the cell cycle. They propose a “kicked” model in which an out‐of‐phase, external deterministic factor leads to a lack of correlation between consecutive cell cycles. Consistent with these observations, our results suggest that, in cells with intact cell cycle regulation, memory of cell cycle phase durations is not only lost over generations but also within a single cell's lifetime between consecutive cell cycle phases. In keeping with this trend, Barr et al (2017) found strong correlations between p21 level and G2 duration in mother cells; between p21 level and G1 duration in daughter cells; and between p21 levels in mother and daughter cells. Given these relationships, one would expect that mother G2 and daughter G1 durations would be coupled. Surprisingly, however, no correlation was observed between mother's G2 and daughters’ G1 durations (Barr et al, 2017). This paradoxical finding is not only consistent with our experimental results, but can also be explained by the many‐for‐all model: Although p21 has a strong effect on multiple cell cycle phase durations and can be inherited across phases and generations, numerous other factors can dilute p21's effect and result in no coupling across consequent phases. Taken together, the emerging theme for governance of cell cycle progression is that durations may be strongly coupled between temporally concurrent, but not consecutive, cell cycle phases.
Materials and Methods
Cell culture
hTERT retinal pigment epithelial cells (RPE) were obtained from the ATCC (ATCC® CRL‐4000™) and cultured in DMEM supplemented with 10% fetal calf serum (FBS) and penicillin/streptomycin. U2OS cells were obtained from the laboratory of Dr. Yue Xiong and cultured in DMEM supplemented with 10% fetal bovine serum (FBS) and penicillin/streptomycin (Gibco). WA09 (H9) hES cell line was purchased from WiCell (Wisconsin) and maintained in mTeSR1 (85850, StemCell Technologies) on growth factor reduced Matrigel (354230, BD). Cells were authenticated by STR profiling (ATCC, Manassas, VA) and confirmed to be free of mycoplasma. Cells were passaged using trypsin (25300054, Gibco) for RPE and U2OS or ReLeSR™ (05872, StemCell Technologies) for H9 as needed. When required, the medium was supplemented with selective antibiotics (2 μg/ml puromycin for RPE and U2OS; 0.5 μg/ml puromycin for H9; A1113803, Gibco).
Chemical and genetic perturbation of the cell cycle phases
For NCS treatment, medium was replaced with fresh medium supplemented with neocarzinostatin (N9162, Sigma‐Aldrich) during experiments. For myc overexpression, RPE cells were infected with fresh retrovirus containing MSCV‐Myc‐ER‐IRES‐GFP and 1 μl polybrene. Cells were subsequently passaged 48 of post‐infection and seeded onto a glass‐bottom plate for imaging. 16 h prior to imaging, tamoxifen was added at a final concentration of 50 nM. For aphidicolin treatment, medium was replaced with fresh medium supplemented with aphidicolin (A0781, Sigma‐Aldrich) for 8 h during experiments, washed off once with PBS, and then replenished with imaging media described below. For CDK2 inhibition, cells were treated with 2 μM CVT‐313 (221445, Santa Cruz) prior to starting the imaging.
Cell line construction
The construction of the pLenti‐PGK‐Puro‐TK‐NLS‐mCherry‐PCNA plasmid was described in our previous publication (Chao et al, 2017). The plasmid was stably expressed into RPE, U2OS, and H9 cells by first transfecting the plasmid into 293T cells to generate replication‐defective viral particles using standard protocols (TR‐1003 EMD Millipore), which were used to stably transduce the RPE, U2OS, and H9 cell lines. The cells were maintained in selective media and hand‐picked to generate a clonal population.
The MSCV‐Myc‐ER‐IRES‐GFP was made by cloning the Myc‐ER from pBabe‐puro‐Myc‐Er into MSCV‐IRES GFP. pBabe‐puro‐myc‐ER was a gift from Wafik El‐Deiry (Addgene plasmid # 19128; Ricci et al, 2004). MSCV‐IRES‐GFP was a gift from Tannishtha Reya (Addgene plasmid # 20672). The cloned plasmid was then sequenced and verified.
Stable RPE and U2OS cell lines expressing PCNA‐mTurq2 and PIP‐FUCCI (PIP‐mVenus+Gem1‐110‐mCherry) were created by antibiotic selection of transduced cells (Grant et al, 2018). Briefly, the PIP‐mVenus sensor was built by adding a fluorescent tag and nuclear localization signal (NLS) to a PIP motif (17 aa) of Cdt1 protein.
The RPE cell line with Dox‐inducible cyclin D1 was generated using the pInducer20 plasmid (Meerbrey et al, 2011). First, PCNA‐mTurq was stably expressed in RPE1‐hTERT cells by viral transduction and cells were sorted for medium expression. The pInducer20 plasmid harboring the cyclin D1 cDNA was transfected into 293T cells to generate replication‐defective virus which was used to transduce the target cells followed by manual clonal selection and screening for appropriate cyclin D1 expression.
The DHB‐mCherry reporter was a gift from S. Spencer (Spencer et al, 2013). The plasmid was stably expressed into RPE cells by first transfecting the plasmid into 293T cells to generate replication‐defective viral particles using standard protocols (TR‐1003 EMD Millipore), which were used to stably infect the RPE cell lines. The cells were maintained in selective media and hand‐picked to generate a clonal population.
Time‐lapse microscopy
Prior to microscopy, RPE and U2OS cells were plated in poly‐d‐lysine‐coated glass‐bottom plates (Cellvis) with FluoroBrite™ DMEM (Invitrogen) supplemented with 10% FBS, 4 mM l‐glutamine, and penicillin/streptomycin. H9 cells were plated in Matrigel‐coated glass‐bottom plates with phenol red‐free DMEM/F‐12 (Invitrogen) supplemented with 1× mTeSR1 supplement (85852, StemCell Technologies). Fluorescence images were acquired using a Nikon Ti Eclipse inverted microscope with a Nikon Plan Apochromat Lambda 40× objective with a numerical aperture of 0.95 using an Andor Zyla 4.2 sCMOS detector. In addition, we employed the Nikon Perfect Focus System (PFS) in order to maintain focus of live cells throughout the entire acquisition period. The microscope was surrounded by a custom enclosure (Okolabs) in order to maintain constant temperature (37°C) and atmosphere (5% CO 2). The filter set used for mCherry was as follows: 560/40 nm; 585 nm; 630/75 nm (excitation; beam splitter; emission filter; Chroma). Images were acquired every 10 min for RPE and H9 cells and every 10 or 20 min for U2OS cells in the mCherry channel. We acquired 2‐by‐2 stitched large image for RPE cell. NIS‐Elements AR software was used for image acquisition and analysis.
Image analysis
Images were sampled every 10 min. Image analysis on the cell cycle phase was performed by manually tracking each cell and recording the frame at which PCNA foci appeared (G1/S) or disappeared (S/G2) and nuclear envelope breakdown (G2/M) using ImageJ to quantify the durations of each cell cycle phase. This provided reliable measurement of phase durations with a measurement error of one time frame (±10 min). In addition, due to the nature of time‐lapse imaging, there was an uncertainty regarding when the phase transition occurred within the 10‐min time frame.
Image and data analysis was performed in Fiji (Schindelin et al, 2012; version 1.51n, ImageJ NIH) and MATLAB (R2017b, MathWorks). Images from time‐lapse experiments (16 bit) were processed with rolling ball background subtraction algorithm prior to analysis. PCNA channel was selected for segmentation of nuclear regions of interest (ROIs) and tracking of individual cells by in‐house developed ImageJ scripts with a user‐assisted approach. Briefly, user‐defined tracks were used for local automated segmentation of ROIs based on intensity thresholding followed by morphological operations to define an oval shape and a watershed algorithm to separate adjacent nuclei. Defined ROIs were used to analyze all fluorescent channels.
PCNA pattern (PCNA variance) was defined within nuclear ROIs from which nucleoli (dark regions) were eliminated using Remove Outliers algorithm (ImageJ). Images were smoothed with a Gaussian filter (sigma=1×) and then processed with a variance filter (sigma=2×) to enhance PCNA pattern. Intensity and standard deviation of variance images were measured within 70% central region of defined ROIs to avoid edge artifacts. Beginning and end of S phase were defined as a transition from low to high and high to low variance, respectively, and detected automatically from a one‐dimensional signal.
Detection of S phase based on PIP‐FUCCI reporter: PIP‐mVenus signal was used exclusively to detect S phase boundaries as the sensor shows high levels in G1 and G2 phases and is rapidly and efficiently degraded in S phase. Beginning of S phase was defined as a point of 50% loss of G1 level of the sensor, while beginning of G2 was defined as an increase (2% of maximum signal) over S phase level.
In silico mapping of cell cycle progression in individual cells
We quantified the cell cycle phase durations of our cell lines by imaging asynchronously dividing cells. During the entire life of each individual cell, we took five time point measurements: the time of cell birth (t birth), the onset of S phase (t s_onset), the end of S phase (t s_end), the time of nuclear envelope breakdown (NEB, t m_start), and the time of telophase (t telophase), which were manually identified from the PCNA‐mCherry reporter. These five time points allowed for quantifying the durations of four cell cycle phases: G1, S, G2, and M phases.
Statistical analysis and sample size
Sample size was calculated based on Type I error rate of 0.2, Type II error rate of 0.01, and R 2=0.1 to prevent false negative correlation, which resulted in 112 cells per condition (Hulley et al, 2013). Nonparametric bootstrap was performed with 10,000 iterations to calculate the distribution of correlation coefficients for each condition and the percentage of iterations with no significant correlation (R 2<0.1).
Immunofluorescence
Cells were fixed with 4% paraformaldehyde for 10 min, permeabilized with 0.5% Triton X‐100 for 10 min, and stained overnight at 4°C with anti‐p27 KIP 1 antibody [Y236] (Abcam ab32034) or anti‐p21 antibody (CST #2947). Primary antibodies were visualized using a secondary antibody conjugated to Alexa Fluor‐594 and imaged with appropriate filters. EdU incorporation and staining were performed using the Click‐iT™ EdU kit (Invitrogen C10337).
Western blot
For myc overexpression, cell pellets were harvested from RPE cells infected with retroviral Myc‐ER on day 3 post‐infection in culture. Cells were pelleted at 1,000 RPM for 5 min, and lysed with 2× Laemmli buffer and water, boiled at 95°C for 5 min. Prior to loading for Western blot, protein levels were quantified using Qubit, and Qubit high‐sensitivity protein quantification assay (Thermo Fisher). Loading samples were prepared by taking 50 μg of protein per sample. 5% BME and water were added and water for total sample volume of 30 μl. Protein was separated on 4–20% Mini PROTEANTGX gels from Bio‐Rad, optimized for proteins up to 200 kDa. Blot was transferred onto a PVDF membrane and incubated with c‐Myc primary antibody (Santa Cruz Antibody Cat # sc‐40) overnight at 4°C. After washes, blot was incubated with secondary antibody (rabbit anti‐mouse HRP conjugate; Jackson Labs Cat # 315‐035‐003) for 2 h. Substrate ECL (Bio‐Rad 1705060) was added for 5 min. Blot was imaged using ChemiDoc imaging systems after a 10‐s exposure time.
For cyclin D1 detection, cells were collected by trypsinization, washed with 1× phosphate buffer solution (PBS) and then centrifuged at 1,700×g for 3 min. For total protein lysates, cells were lysed on ice for 20 min in CSK buffer (300 mM sucrose, 100 mM NaCl, 3 mM MgCl 2, 10 mM PIPES pH 7.0) with 0.5% Triton X‐100 and protease and phosphatase inhibitors (0.1 mM AEBSF, 1 mg/ml pepstatin A, 1 mg/ml leupeptin, 1 mg/ml aprotinin, 10 mg/ml phosvitin, 1 mM β‐glycerol phosphate, 1 mM Na‐orthovanadate). Cells were centrifuged at 13,000×g at 4°C for 5 min, and then, the supernatants were transferred to a new tube for a Bradford assay (Bio‐Rad, Hercules, CA) using a BSA standard curve. Immunoblotting samples were diluted with SDS loading buffer (final: 1% SDS, 2.5% 2‐mercaptoethanol, 0.1% bromophenol blue, 50 mM Tris pH 6.8, 10% glycerol) and boiled. Samples were separated on SDS–PAGE gels, and then, the proteins transferred onto polyvinylidene difluoride membranes (PVDF; Thermo Fisher, Waltham, MA). Membranes were blocked at room temperature for 1 h in 5% milk in Tris‐buffered saline‐0.1% Tween‐20 (TBST) and then incubated in primary antibody overnight at 4°C in 2.5% milk in 1× TBST with 0.01% sodium azide (anti‐cyclin D1 sc‐753; Santa Cruz Biotechnologies 1:2,000). Blots were washed with 1× TBST, incubated in HRP‐conjugated secondary antibody (Jackson ImmunoResearch) in 2.5% milk in 1× TBST for 1 h, washed with 1× TBST, incubated with ECL Prime (Amersham, United Kingdom), and scanned with a ChemiDoc (Bio‐Rad). Equal protein loading was verified by Ponceau S staining (Sigma‐Aldrich).
Cell cycle progression model simulations and parameter fitting
Fitting with the simple Markovian model with a single rate parameter
All simulations and parameter fitting were performed using MATLAB. The durations of cell cycle phase—G1, S, G2, and M—under basal conditions were together fitted to four Erlang distributions with the same rate (λ) parameter. The shape (k) parameters were restricted to positive integer and were allowed to vary for each cell cycle phase. The fitting was performed by maximizing the likelihood of observing the experimental data using the fminsearch function in MATLAB.
Under the Erlang distribution, the probability of observing a cell of a particular cell cycle phase, for example, G1's, duration x,f(x;k, λ) is
where ∆T is the measurement interval.
Then, the probability of observing a cell of four cell cycle phase durations x,f(x;k, λ) is
The shape and rate parameters were determined by solving for the maximal likelihood of observing the experimental data:
where x i is the i th cell in the experimental data, and n is the total number of observed cells.
Fitting with the Erlang model with independent rate parameters
The Erlang model provides a simple and biologically relevant framework for modeling the cell cycle phase progression. The associated Erlang distribution is a special case of the gamma distribution with integer k, but the Erlang model can be simulated with the Gillespie stochastic algorithm. Under the Erlang model, the durations of each cell cycle phase —G1, S, G2, or M —under basal conditions were independently fitted to an Erlang distribution (Fig2A). For each cell cycle phase, we fit the experimental distribution of cell cycle phase durations to obtain the shape (k) parameter and the rate (λ). For each cell cycle phase, the shape and rate parameters were independently determined by solving for the maximal likelihood of observing the experimental data of each phase:
where x i is the i th cell in the experimental data, and n is the total number of observed cells.
Simulation of cell cycle phase transition
After the fitting with the Erlang model, we obtained two parameters for each cell cycle phase and each cell line. Using the estimated parameters, we simulated the progression of cell cycle phase using the Gillespie stochastic algorithm. Alternatively, because the Erlang distribution is a special case of the gamma distribution with integer scale parameter, we can generate the phase durations from a gamma distribution in MATLAB:
For the normal distribution model, parameters for each cell cycle phase were independently chosen according to the mean (μ) and variance (σ 2) of the experimental cell cycle phase durations’ distributions. The cell cycle phase durations were then simulated from a normal distribution.
Erlang distribution as an approximation of the hypoexponential distribution
Our Erlang model describes the cell cycle phase progression as a series of sub‐phase transitions with the same rate λ. The relevant biological interpretation of the Erlang model is that each cell cycle phase can be viewed as a multistep biochemical process that needs to be completed sequentially in order to advance to the next cell cycle phase. Biologically, the rate of each sub‐phase transition could be different from one another. A model that can account for this flexibility is the hypoexponential distribution, or the generalized Erlang distribution, which allows the rate parameter of each transition to be different. However, the Welch–Satterthwaite equation provides a good approximation of the generic sum of multiple Erlang distributions as one Erlang distribution (Satterthwaite, 1946; Welch, 1947):
where the k i and θ i are the shape and scale parameters for the i th individual Erlang distribution, and the sum of i Erlang distributions can be approximated by a gamma distribution with only two parameters: gamma(k sum, θ sum). The approximated Erlang distribution will be chosen to be the closest distribution to gamma(k sum, θ sum), but with an integer k.
The many‐for‐all model of heritable factors governing cell cycle progression rate
Many‐for‐all model with only phase‐coupling factors
The many‐for‐all model for heritable factors assumes that there are physical factors, called “phase‐length factor”, inside the cells that control the rate of cell cycle phase progression. In addition, the levels of these factors can fluctuate throughout the cell cycle but are evenly distributed among sibling cells during mitosis so that sibling cells share similar amounts of the heritable factor. Each type of phase‐length factor has shared control among two or more cell cycle phases, exerting an effect (a) on multiple cell cycle phase durations by influencing the rates of cell cycle phase progression. The magnitude of the factor effect is proportional to the amount of factor (copy number of molecules). Take G1 and S phase for example, the rate of G1 progression is dependent on the sum of effects among every factor:
where λ 0,G1 is the average progression rate of G1, γ is the fraction of progression rate subjected to the control of phase‐length factors, a G1,i is the effect coefficient of factor type i,n i is the copy number of factor type i, and m is the total number of different factor types. The effect coefficients were assumed to follow a normal distribution with mean zero:
Similarly for S phase:
σ was chosen to be 0.01 to generate cell cycle phase distributions that resembled experimental data. The copy numbers of each factor type (n i) for each cell were assumed to follow a Poisson distribution (Shahrezaei & Swain, 2008; Pendar et al, 2013), with mean abundance following a lognormal distribution of μ=1,000 and σ=0.6 (Ghaemmaghami et al, 2003; Furusawa et al, 2005; Eriksson & Fenyö, 2007).
Modeling the factor copy number with a normal distribution with variance equals the mean did not affect the results:
Hence, the G1 duration is
Similarly for S phase:
The Pearson correlation coefficients were then calculated by generating 200 cells with G1 and S phase durations using the simulation framework described above (Fig4B).
Many‐for‐all model with both phase‐coupling factors and phase‐specific factors
In addition to the phase‐coupling factors, which has shared control among two or more cell cycle phases, we took into account the presence of phase‐specific factors, which affect only one specific cell cycle phase. Take G1 and S phase for example, for G1‐specific factors, a S,i=0. For S‐specific factors, a G1,i=0. The rate of G1 progression is dependent on the sum of effects among every factor, including both the phase‐coupling factors and the phase‐specific factors.
where λ 0,G1 is the average progression rate of G1, γ is the fraction of progression rate subjected to the control of phase‐length factors, a G1,i is the effect coefficient of factor type i,n i is the copy number of factor type i, and m is the total number of different factor types. The effect coefficients were assumed to follow a normal distribution with mean zero:
for phase‐coupling factors, and equals zero for S phase‐specific factors.
for phase‐coupling factors, and equals zero for G1 phase‐specific factors.
σ was chosen to be 0.01 to generate cell cycle phase distributions that resembled experimental data. The copy numbers of each factor type (n i) for each cell were assumed to follow a Poisson distribution (Shahrezaei & Swain, 2008; Pendar et al, 2013), with mean abundance following a lognormal distribution of μ=1,000 and σ=0.6 (Ghaemmaghami et al, 2003; Furusawa et al, 2005; Eriksson & Fenyö, 2007).
Modeling the factor copy number with a normal distribution with variance equals to mean did not affect the results.
Hence, the G1 duration is
Similarly for S phase:
The Pearson correlation coefficients were then calculated by generating 200 cells with G1 and S phase durations using the simulation framework described above (Fig4C).
Perturbation of a single phase‐coupling factor
The effect of perturbing a single factor was modeled by choosing the type of phase‐coupling factor that had the largest product of effect coefficients on two phases; that is, find i that maximizes (a G1,i×a S,i). After i was determined, the abundance of that factor was increased by 10‐fold, that is, n’i=10 n i.
The cell cycle phase durations were simulated similarly as above, except for the increased value of n i calculated above (Fig4D and E).
Requirement of independent assortment of heritable factor into daughter cells
For sibling cells, the factor abundance is assumed to be strongly correlated; that is, the correlation coefficient between the copy number for each factor type i (ρ n 1 i,n 2 i) is large. n 1 i and n 2 i represent the copy numbers of factor type i for the two sibling cells.
Thus, the difference in cell cycle phase duration between the two sibling cells can be expressed as a function of:
The segregation of each factor during cell division is not independently distributed, but correlated; that is, if ∆n’i s are correlated, then we can rewrite
where γ i is the proportionality terms between ∆n’i s plus the noise term. Under this condition, ∆G1 and ∆S would be correlated. Our results show no correlation in the differences in cell cycle phase durations between sibling cells, suggesting that the propagation of factors into daughter cells is not interdependent.
Nonparametric bootstrap simulation
Nonparametric bootstrap with consideration of movie image frequency
The distribution of Pearson correlation coefficient was simulated by nonparametric bootstrap. Specifically, N cells were selected without replacement, where N is the experimental sample size of that experimental condition. For each cell, a noise term accounting for (i) measurement accuracy (±1 frame) and (ii) measurement uncertainty was added to the beginning time point and to the ending time point of the phase. Measurement accuracy accounts for the ability to identify cell cycle phase based on PCNA morphology within 1 frame of accuracy (±1 frame). Measurement uncertainty accounts for the nature that imaging was not performed continuously, but was performed every 10 min (20 min for some cases), and it is impossible exactly when the phase transition actually happened within this 10‐min window.
Explicitly, the noise terms for the begin frame and end frame both are
where Uniform (−0.5, 0.5) accounts for the measurement uncertainty, T is the time interval between each image, and δ=3 is the frame error accounting for measurement accuracy of ±1 frame.
Therefore, each cell's phase duration was added a noise term:
For each phase and each cell, the noise was generated independently, and the phase durations with noise were used to calculate the Pearson correlation coefficient. The bootstrap was performed on these cells 100 times, and this process was iterated 100 times to generate 10,000 R values for each condition.
Correlation between independent random variables and the sum
It can be shown that two independent random variables are both correlated to their sum.
Let
The Pearson correlation coefficient between the part B and the sum A can be written as
where cov(A,B) is the covariance between A and B, σ is the variance. Then,
Since B and C are independent random variables, cov(C, B)=0.
Therefore, ρ A,B is positive and scales with the proportion of A's variance contributed by B.
Author contributions
HXC constructed the PCNA‐mCherry reporter cell lines. KMK and GDG constructed the PIP‐FUCCI reporter cell lines. HXC, JGC, GPG, and JEP designed the experiments. HXC, RIF, HKS, KMK, GDG, and RJK performed live‐cell imaging and experiments. HXC, RIF, HKS, KMK, GDG, and RJK conducted image analysis and cell tracking. JP and JCL constructed the cyclin D construct cell line and performed validation experiments. HXC performed computational modeling and analysis. HXC wrote the manuscript with contributions from all authors.
Conflict of interest
The authors declare that they have no conflict of interest.
Supporting information
Appendix
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Code EV1
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Source Data for Appendix
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Review Process File
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Source Data for Figure 1
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Source Data for Figure 3
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Source Data for Figure 4
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Source Data for Figure 5
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Acknowledgements
We thank Samuel Wolff for guidance on experiments and microscopy; Amy House for technical assistance and training; Rebecca Ward for critical feedback on the manuscript; Pankaj Mehta and Robert Corty for helpful conversations and feedbacks; Po‐Hao Huang for brainstorming ideas and titles; and members of the Purvis Lab for helpful discussions and technical suggestions. This work was supported by the National Institutes of Health research grants DP2‐HD091800 (J.E.P.), GM083024 (J.G.C.), GM102413 (J.G.C.), T32CA009156 (G.D.G.), and training fellowship F30‐CA213876 (H.X.C.); HHMI (Howard Hughes Medical Institute) Gilliam Fellowship GT10886 (J.C.L.); PREP grant R25GM089569 (J.P.), a Medical Research Grant from the W.M. Keck Foundation (J.E.P. and J.G.C.); the Loken Stem Cell Fund; and the North Carolina University Cancer Research Fund.
Mol Syst Biol. (2019) 15: e8604
Data availability
The scripts that generated the simulations are provided as Code EV1. Raw images have been deposited in BioStudies ( under accession S‐BSST230.
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Code EV1
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Source Data for Figure 1
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Data Availability Statement
The scripts that generated the simulations are provided as Code EV1. Raw images have been deposited in BioStudies ( under accession S‐BSST230.
Articles from Molecular Systems Biology are provided here courtesy of Nature Publishing Group
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Math 121-002: symmetries of graphs, function operations a.k.a. new functions from old, 9-28-22
James Cook Math
21200 subscribers
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Posted: 28 Sep 2022
Transcript:
let me begin I will I will start dearly father we thank you for this day again I thank you for the students I pray that you died just guide our steps today help us to learn a little bit more about functions and graphing and all that and it's great you'd encourage the students as they're working on this class to just do their best each day Lord For Your Glory renew my prayer Lord Jesus amen all right so um anybody still missing like a test or something do I have I not returned anybody's testing here you get your test already okay but do you want your paper back ah but you want the one I see because I think Olivia right here's the because you want to know where you know uh-huh all right well there's that I hope everyone's top padded now your day is not complete until you're top headed right yeah yeah so I'm glad I could provide you that service the excitement of top heading all right so talk a little bit more about properties of graphs so this won't take long um let me just tell you kind of where we're going um so today I'm going to talk about a little bit more about properties of graphs I'm going to describe some particular examples that illustrate some words all right and that's once we're done with that I'll talk about operations on functions how to add subtract multiply divide and I think compose functions so we'll do that and that'll probably just be today then like next class I will start talking about how to graph and analyze parabolas all right so I think that's more or less what I'm going to do and then so anyway long story short we're going to get pretty far ahead of the schedule that's in the course planner all right so that's my my intention anyway is to get a fair amount of head before the next test so we'll have a little bit of time for the material to kind of sink in before test two just so you know all right so example one let's just do one kind of like we did last time how about this what if I have y equals to 3 plus 1 over x minus 2 squared what would that look like let's graph this so this year this is just another example from last class to get us started so I look at this I think okay so I can take y equals 1 over x squared right I can take that graph I can do what I can take that I can shift right to right that gives me y equals 1 over x minus 2 squared right and then I can shift up three right and that will be the graph so that's that's how I can assemble this formula I can shift right by two shift up by three like that and so if I understand you know the basic graph here which is the volcano graph right then I can I can build my graph two transformations of graphs okay so do that here so the volcano graph looks like this right it's got a vertical asymptote of x equals to zero right and a horizontal asymptote of y equals to zero that's the starting thing then I shift right to so now my vertical asymptote shifts over to right the horizontal asymptote stays put and so my graph then looks like these right and then finally I shift up three when I shift up three that keeps the vertical asymptotes just right where it is at two right but my horizontal asymptote moves up three so this is three this is two and finishing product here the graph looks something like this all right so sorry I chose a poor scale here let me make extend my axes to make it look a little bit better so there you go this is the graph y equals 3 plus 1 over x minus 2 squared all right so um I'm going to use this graph in my discussion going forward because this graph is something which doesn't have actually a lot of the properties I'd like to talk about Okay so um it's a useful counter example I suppose so let's talk about um even what's an even function what's an odd function all right and then let's also talk about a discontinuous function so I want to talk about these terms and what that means graphically all right so first of all I'm going to give just a quick definition of even and odd what does it mean for a function to be even this means that f of minus X is equal to f of x for each X in its domain right an odd function means that F of minus X is equal to minus f of x for each X in its domain right now discontinuous is more subtle to Define carefully so I'm just going to give you a heuristic picture of what we mean by discontinuous discontinuous means there's a jump in the graph right or another way you could say discontinuous would be as you're graphing the function you have to lift your pen off the paper in order to graph it that's what a lot of times people will say so let me try to give you examples of even functions and odd functions yeah so like example two we could look at say y equals um x squared right so here that's the parabola right so if I that's my f of x f of x equals to x squared okay is this even is this Oddity I think it's even right so if we look at F of minus x what happens we get minus X quantity squared which is equal to x squared which is equal to f of x so because squaring the minus makes it go away we get F of minus X is equal to f of x for every X right so this is an even function what's another way you can talk about even function you can say it has what is symmetric it's symmetric with respect to what I heard the x-axis but I would say that it's not symmetric with respect to the x-axis because it's decidedly on the upper half well um let me think about it for a second so what do you guys think Eve is it even with respect to the x-axis I mean is it symmetric with respect to the x-axis or is it symmetric with respect to the y-axis I think we say it's symmetric with respect to the y-axis because when we talk about symmetry it's talking symmetry is judged in reference to a reflection right so the point here is it's it's symmetric with respect to the y-axis and the reason for that is that if we reflect the graph over the y-axis right if you can imagine just mirroring the graph over the y-axis like this it comes back to itself right so if you can imagine flipping the graph around the y-axis it's just the same graph again so it's symmetric with respect to the y-axis um example three and this example three doesn't actually fit into the function discussion okay so truth truth and advertising here if we look at x equals to Y squared now that's this is technically something else right this is giving us y y squared equals to X which gives me Y is equal to plus or minus the square root of x right Y is equal to plus or minus square root of x so here if I was to graph I have to graph not one function but actually 2 in this example all right so the upper half is the square root function the lower half right is the what's the lower half it's y equals to minus the square root right but if I did this right so technically this is not the graph of a function because it fails the what the vertical line test right but if we look at this right here this the graph has graph which is symmetric with respect to what with respect to this one is symmetric with respect to the x-axis so this would be an example of something symmetric with respect to the x-axis see because you can you can reflect the graph over the x-axis and it comes back to itself this is what we would say is symmetric with respect to the x-axis so because that means it's not a function if you're symmetric with respect to the x-axis that means that you have two parts two parts of your function that have different y values in the same x value so that's why we're not going to talk about that for the most part the reason I'm talking about is because you guys one of you said a symmetric with respect to the x-axis so I felt obligated to talk about that but I don't know if this is actually in your homework anyway symmetric with respect to the y-axis it's characteristic of an even function example four let's see here suppose we have y equals f of x equals to oh I don't know X cubed um minus X so I'm going to get ahead of the story a little bit here but the way to graph this guy is to go okay so that's x times x squared minus 1 right which is by the way x times X plus one times x minus 1 factoring right so as it happens the graph of this guy has x-intercepts at zero and at one and at minus one and and so the graph of this this character looks something like this so that's y equals to x cubed minus X do you guys like time travel okay good because this is like next Monday doing this more systematically for more examples that's the kind of thing we're doing Monday how to graph polynomials okay so anyway you're not really supposed to understand all this quite yet but we'll do more of it the issue here is I wanted to give you an example where we could see is this function even is it odd what do you think how can we check if the function's odd well we can check F of minus x what would that be we would have minus X quantity cubed minus a minus X what's that give me so if we if we Cube minus one we get minus again right and this gives me plus X right so I could factor out that minus X like that right and so what I have is I've got minus f of x again so in fact this function is an odd function all right this is an odd function and so what do people how do people describe the geometry of an odd function well they say they say that such a graph is symmetric with respect to the origin that's what people say symmetric with respect to the origin okay so this is it's an odd graph it's symmetric with respect to origin so symmetric with respect to the origin means that if the point like a comma B is on the graph then the point minus a comma minus B is also on the graph see that like on this graph at this point is a comma B then there's also a corresponding Point down here minus a comma minus B so if there's always this pairing of points and their negatives then it's symmetric with respect to the origin if you think the terminology symmetric with respect to the origin is odd well I tend to agree with you but anyway it's what we say in contrast the even function you see the pairing here if you have the point a comma B on the even function what's the corresponding the mirror point not minus a comma it's minus a comma the B stays the same all right so like there's this kind of pairing and an even graph right there's that kind of pairing in an odd graph let's go back to example one can you thanks yeah so example one is this even is this odd well yeah people would a lot of people would say it's discontinuous at the uh people would say there's a discontinuity at the vertical asymptote a lot of people would say that that's true um but is it even or is it odd like if you pick if I pick my point here right like a comma B if that's a you know typical typical point right there do you have the other point over here do you have a comma you know minus a comma B is that in the graph no right what about this metric with respect to the origin is this point down here oops you know so clearly this Point's not over here right clearly that point down here is also not in it right I mean you can see that this graph is neither even nor odd right most graphs are like that right even graphs and odd graphs are special cases yeah if you look at your standard example sheet I gave you last time there's other ways you could be failed to fail to be even or odd right example four if we look at you know y equals f of x equals the square root of x what's that thing look like it looks like this right that's the graph see this this problem this graph you can't even pose the question really properly is f of x equal to F of minus X right because if x is in the domain minus X is not so this example this function doesn't even have a domain which is symmetric about the origin right in order to even ask the question of even or odd you notice that you need the property for the function to be that both X and minus X are in the domain right so like this I can't even I can't even entertain the question is it even or odd so people would say that that's neither even nor Odd as well okay but for a different reason um here's a question suppose you start with this right I hope you agree with me it's neither even nor odd and now you want to extend to a case-wise defined function which will be even all right so here's the question is how would you extend you know f of x to G of x and it's going to have two cases all right it's going to have square root of x if x is greater than or equal to zero and it's going to have something over here if x is less than or equal to zero how would you fill in that blank to make the function even the function y equals g of X even so let me draw the graph of y equals g of X we want it to be even right so we want this graph to be mirrored on the other side right so you want it like this yeah you want to take the given f of x and mirror it over like that what would you what would you fill in the blank there minus the square root of x see that won't work that's a good guess but why won't that work if you plug in like -1 into that formula what happens you get the square root minus the square root of -1 which is well that's minus I right so the right idea but instead of putting the minus outside the square root we need to put it inside the square root and that's okay like for example G of minus 4 in your definition would be the square root of minus a minus four which would be the square root of 4 which would be 2. right in other words this point here 4 2. gets mirrored over to this point over here minus 4 2. it's got that reflection across the y-axis property so this is a game that you can play Given half of a function how would you mirror it in order to make the new function even or odd right now your minus squared of X would have been perfectly good if instead of asking the question so I I want to extend f of x to G of X and I I said it out loud but I didn't write it on the board we want even we want it to be even all right what if instead I had said I want the new function to be odd then what you said would have been right so if I wanted to extend extend F to odd function let's call it h where H of X has two cases yeah it's got the first case fill in the blank it's got the second case that's already given square root of x for X greater than equal to zero so if you were going to play that game then to fill in that blank we'd use your formula we could put minus the square root of x right there and then if we look at the graph of this h of X it would look like oops my bad it would look like the given the square root function right and then oh man no that wouldn't work either junk that's no good we need to combine your answer and my answer for this one so we need to put minus the square root of minus X here to make it an odd to make it an odd extension see because if I do that the graph y equals minus the minus the square root of x that would be like this this over here is y equals minus the square root of minus X it's the square graph reflected across the y-axis and then reflected across the x-axis again this H would be symmetric with respect to the origin yep where would you find when would you negative um I'm sure there are applications where you end up taking the square root of minus a minus thing like any time you've got like the square root of an absolute value that'll end up happening I don't know if I know any examples which would be um interesting for here myself oh you talk to yourself it's all right I also have to check if you guys can hear my thoughts as well though see if I see if you can hear the voices in my head as well I guess that means I said them I mean given current technology anyway right okay so one more example here example five oh six thank you to take them oh I did oops let me fix that so this was five wasn't it thank you so example six this one it's a weird example but this is called the um so you could say y equals I'll use this notation this is the so-called ceiling of x let me just illustrate how it works by showing you some numbers so like what is the ceiling of 2.2 what's the ceiling of minus 3.1 so what you do is it's just round up you always just round up to the next integer so like the ceiling of 2.2 is 3. the ceiling of minus 3.1 is minus three hours go up up in the number line the ceiling of 0.8 is what it's it's one all right so like each number each no each range of real numbers I always just take it and you go to the next integer all right yeah yeah right right here so what's up okay so the way this looks then is so like here to round if we had like minus 0.5 that rounds up to zero okay so graphically what you're talking about is like this now there's an open circle there and then open circle at zero so like the um the ceiling of zero is zero wait a minute did I say it's zero well I guess I'm saying you know I am actually not totally sure what the definition is in terms of like standard um you know I said this the ceiling just a second as I have it graphed yeah let's say the ceiling of zero so let's say any the ceiling function of any integer is just the integer so like ceiling of zero zero ceiling of one is one all right and then this all right like this the open circle means it's not defined the value of the graph is not to be fine there so like here this is one right so like the ceiling of one we said is equal to one okay so this is this is a pretty this is a pretty weird example right you've really never seen anything quite like this in terms of formulas right this is pretty far divorced from y equals x squared Y equals X cubed y equals the square root of x the volcano function the reciprocal graph all the standard examples right um because this this example is really from like number Theory all right so it's it's something else um I'd say it's home really is more in like discrete mathematics or something like that yeah but this this is a nice example for the discussion of discontinuous because it's got all kinds of discontinuities right it's got a discontinuity at where every single whole number or every single integer it's got a discontinuity see it jumps right so this has got infinitely many discontinuities one for each you know each integer so this is my example of a discontinuous graph this one's also yep this one's also interesting so if I call this f of x what's the domain what's the domain of f of x so the domain is the domain is everything because we can always round up to the next integer or itself right that's defined for any real number so the domain is everything minus infinity to Infinity what's the range so the range is kind of funny right the range is what it's got zero it's got one it's got two it's got three it's got minus one it's got minus two it's got minus three and it's got nothing else in between those numbers right so the range is actually this set you know zero plus or minus one plus or minus two and so forth and so on forever and ever yeah in other words the range of f of x is the set of integers we have a notation for this typically it's this funny looking Z so it's the set of integers all right Okay so that's my what I've got to say here about these terminologies for graphs I'm going to yep well we we never we don't write square bracket around plus or minus infinity because plus our minus infinity are not real numbers so that is an idea that's to express the concept that we're going on and on Without End but we don't want to say that infinity or minus infinity themselves are real numbers now there is something called the very long line where people actually throw in well maybe that's trying to remember what's called but there is a concept of the real numbers with infinity and minus infinity stuck onto them and that is a number system that people study but we don't do that in here you still you're you have a question about it or you're just oh cool I will I will leave it unharmed Okay so I think um what I'm about to talk about guys I think is much easier graphic is hard if you don't know it already you like graphing these sorts of qualitative descriptions of shapes you know there's really something to learn there and once you learn it you have it but before you do it's really kind of hard to have a it's hard to have a halfway understanding of what even means or a halfway understanding of what odd means or a halfway understanding of what discontinuous means you know I mean like you kind of either get it or you don't right but I think what we're about to cover everyone's going to understand it's pretty mechanical all right so this is called operations on functions so sometimes I'll also call this AKA new functions from old and so what I mean by that is you're given function f and G all right all right you're given two functions function f and function G you can create new functions or you can say construct if you like reconstruct new functions f plus G F minus G um a constant times f and F times G and F over G and even F composed with G all right as follows let me write under those symbols the names of each one of these okay some of the names are like totally unsurprising this is the sum of the function right sum difference this is scalar multiple or you could just say it's a well I mean that's the best term I have scale scale or multiple here the C is just a number Okay C is a number F times G this is the product of two functions F over G is the quotient of two functions and the last one which is by far the most challenging thing here in this list this is called the composite of f and g and I'd add where G is the inside function and F is the outside function right so there's an inside and there's an outside in which is first and which is last matters right like for example a real world example of function compositions you put on your socks and you put on your shoes right one of these you do before the other right which is the inside function socks right you put on socks first and then you put on shoes somebody here Barefoot with sandals get out is there anybody no it's getting kind of cold okay so I've just this is just all name calling how do you actually Define them all of these things are defined quote unquote Point wise what does that mean f plus G of X is defined to be f of x plus G of x F minus G of X guess what it's f of x minus G of x a constant times f of X guess what it is the constant times f of x so here I'm defining the rules for the new functions I'm giving you the formula for the new function in terms of the function the old function F times G of x is equal to f of x times G of X like that F over G of X is equal to f of x over G of x I would argue that everything I just wrote in the blue relatively simple and not on you know understandable I think what I'm about to write is more you know this is watch out and so the main issue here is to not confuse the product with the composite composite's different composite F composed with G of x is defined to be F of G of x F of G of X this is composition it might help for me to show you the composition the other way like what would G composed with f of x look like it's a little bit different right that would be equal to G of f of x right in contrast f f g of x these are just numbers f of x times G of X is just a number right so do numbers commute numbers do commute right so I can rewrite this as G of x times f of x and so you see this is just equal to GF of x so we can multiply a function f times function G and that's going to same is going to be the same as multiplying G times F in contrast the composition of two functions could be really different in different orders yep absolutely I'll get out of the way you got your telephoto lens yeah I'm sorry about that I should I should make a habit of not writing that far over or is it the glare I don't know sometimes the glare are you can I I can erase now no problem please do ask about that sort of thing I don't mind okay so let's make up some formulas and let's calculate these things yeah let's see here now I forgot what example I was on I'm on seven thank you all right so suppose you got f of x equals to let's say x squared plus one and you've got G of X is equal to oh I don't know square root of x all right let's calculate the sum the difference the product the quotient and the composite for this okay what does it look like so what's f plus G of x n is x squared plus one plus the square root of x you just add the formulas I bet you don't really mean you need me to teach you this right you probably just already knew it right I think students just instinctually would do this even if you didn't tell them same for subtraction yeah and product see so that would be the product of f times G it's x squared plus 1 times the square root of x now notice that just to illustrate again these are just these are just numbers right for a given value of x so you could commute these right you could commute these and rewrite that as square root of x times x squared plus 1. and so then it's manifest it should be manifestfully clear that that's G times f of x yeah so that's G of x times f of x right which is by definition g f of x so my point to you is function multiplication commutes like it doesn't matter which order you do it in I guess the same is not true for division right g f over G is not the same as G over f is it f f over G of X would be x squared plus one over the square root of x right if you wanted to we could look at G over F right well G over f of x look like square root of x divided by x squared plus one there are strings attached right I haven't really talked about attaching the strings let's say but when I talk about the quotient of two functions when is it defined what do you need for the function for the quotient function to be defined like this this whatever we put in the denominator we can't be what zero right so this course is relatively simple in terms of things you can't do what do you have to watch out for you have to watch out for division by zero right and the square root of a negative number those are the big ones watch out for those we'll add to that list one more thing at the end but that's pretty much it for now over here notice that we're we're we're we're we're we're fine because x squared plus 1 is never zero right x squared plus one is prime that's never there's no zero to that that polynomial the smallest this is is one so the domain of G over F actually is a lot bigger of course we're still limited by the square root of x right square root of x forces X to be what non-negative right we can't boot like a negative number into square root square root not without adjusting the input to the square root okay so what else what's left F composed of the G right let's try that one so what's F composed with g of X how's that work so I'll write it out this is f of G of X so it's F of the square root of x and so how does that work so let's um so what we do is we take yeah so we take this the square root and we plug it into the the pattern for ACT F right so you do square root of f square root of x quantity squared plus one so the square root of x squared is just I mean you can simplify this formula right this is just X plus one for what it's worth so this is the composite of f with g what is the composite of G with f so we're looking at G of f of x which is x squared plus 1. so that means we have to plug in x squared plus 1 into the formula for G so that would be the square root of x squared plus 1. are those equal no those are not equal right so G composed with f is not equal to F this is not equal to um f composed of G so there's a big big difference between product of functions and compositive functions all right example eight you're given F of 2 equals to 7 F of rather G of 2 equals to 6 and F rather G of 7 . is equal to 13. all right suppose you're just given that that's enough information to calculate the sum the difference the product the quotient and even a composite of these two functions let me show you how I don't even have a formula right there's no formula I just know this information so if I know that much information I can calculate f plus G of 2. right because it's F of 2 plus G of 2. which is what seven yeah seven plus six which is 13. now let me make this number something other than 13 because it's distracted I didn't mean for that coincidence let me let me make that a 15 there that should help us not I don't want I don't want you to identify those two numbers they don't mean any it was a coincidence okay um what's F my f minus G of 2. 7 minus six right it's one what's F times G of 2 what's F over G of 2 right it's what it's F of 2 times G of 2 which is what six times seven right this is f of 2 over G of 2. which is seven over six right now let me ask you a slightly trickier question which composite can we calculate here can I which one can I calculate can I calculate F composed of a g of 2. can I calculate G composed with f of 2. so F composed of G of 2 means F of G of 2. and this means G of f of 2. so what was G of 2 well it's 6 right so here we have F of 6. and here we have G of what seven so which one of these can I calculate with what I've currently given the latter one right the second one here because we're I only I know G of 7 is 15 right so since I know G of 7 is 15 I can tell that g composed with f of 2 is actually equal to G of 7 which is by the way 15. how about F of 6 can we calculate F composed of G of 2 with what's given in this problem as it stands the answer is no we don't have enough information right so I would have to add information so this is your class I'll let you add the information what do you want to make F of 6 equal to F of 6 is 12. okay so with that addition the answer is then 12. right so G of 2 is 6. so it's just this yeah there's it's not if you're if you don't see it yet you're not far from it it's not much what's the other way you could ask this question I'm always looking for new ways to confuse the students right so like a way to make this more confusing would be to give the information about the graphs give the information about the functions in terms of graphs right here's the graph of f here's the graph of G then you can answer questions about the sum difference product quotient composite of those functions if you have enough information on the graph all right so with the remaining two minutes let me give you a notational strategy so some of you you you kind of hung with me for these calculations here but maybe it would be better if you had a notational scheme all right so here's here's one thing you could do going back to example seven if I wanted to calculate f um composed of a g of X I could think about this as just F of G all right where g equals to G of X right just to just drop the X for a second and if you think about F of G what you can do is just do the formula for f with G replacing X so what's that it's G squared plus one right and then you and then you go oh well that's G of X is the square root of x squared plus 1. so this is a notational trick the same thing for the other ones you composed with f of X we could look at as just like G of f where f is equal to f of x you know so what's G of f well here's G so I do the square root of f right and then that is the square root of x squared plus 1. so this is I think a helpful just simple substitution idea that could help you find Composites more reliably because maybe what I what I've written down here in Black is more mathematically correct in some sense but I don't think it's as helpful to the student right like everybody understands the idea just take X and replace it with f or just take X and replace it with G like I wrote in the red or the blue here so anyway thanks guys I'll see you what Friday hey maybe if we get a far enough ahead we can not meet
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How Do You Write Inequalities in Set Builder Notation?
Convert these two solutions into set-builder notation: 1st Solution: "t is the set of all numbers, such that t is greater than 80" 2nd Solution: "y is the set of all numbers, such that y is greater than 24 and less than or equal to 53"
Summary
Set-builder notation has three parts
In math, anything contained within the curly braces is known as a set
The vertical line '|' represents the words "such that"
The inequality symbol tells us we are dealing with an inequality
The '<' symbol means LESS THAN
The '>' symbol means GREATER THAN
The '≤' symbol means LESS THAN OR EQUAL TO
Because the 2nd solution is a compound inequality, we can write the set-builder notation more than one way
Notes
There are three necessary parts to remember for set-builder notation
There are three necessary parts to remember for set-builder notation
Enclosing the set in curly braces is one part to remember
There are three necessary parts to remember for set-builder notation
Enclosing the set in curly braces is one part to remember
Curly braces look like '{ }'
There are three necessary parts to remember for set-builder notation
Enclosing the set in curly braces is one part to remember
Curly braces look like '{ }'
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
'x' is our variable here
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
'x' is our variable here
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
'x' is our variable here
The vertical line '|' means "such that"
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
'x' is our variable here
The vertical line '|' means "such that", and is what actually "defines" the variable 'x' in this case
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
'x' is our variable here
The vertical line '|' means "such that", and is what actually "defines" the variable 'x' in this case
There are three necessary parts to remember for set-builder notation
Identifying the variable that represents the set of numbers is a necessary part of set-builder notation
'x' is our variable here
The vertical line '|' means "such that", and is what actually "defines" the variable 'x' in this case
The specifics of what makes the set unique come after the vertical line
There are three necessary parts to remember for set-builder notation
The inequality that defines the numbers contained in the set 'x' is a necessary part to remember
'x' is our variable representing the set's numbers
There are three necessary parts to remember for set-builder notation
The inequality that defines the numbers contained in the set 'x' is a necessary part to remember
'x' is our variable representing the set's numbers
'y' is also a variable here
The inequality defining the set is 'x<y'
The '<' symbol means LESS THAN
There are three necessary parts to remember for set-builder notation
The inequality that defines the numbers contained in the set 'x' is a necessary part to remember
'x' is our variable representing the set's numbers
'y' is also a variable here
The inequality defining the set is 'x<y'
The '<' symbol means LESS THAN
Putting it all together, we get '{x | x<y}'
There are three necessary parts to remember for set-builder notation
The inequality that defines the numbers contained in the set 'x' is a necessary part to remember
'x' is our variable representing the set's numbers
'y' is also a variable here
The inequality defining the set is 'x<y'
The '<' symbol means LESS THAN
Putting it all together, we get '{x | x<y}'
The variable represents all the numbers included in the set
The vertical line '|' says we're going to define the variable
The inequality is what actually defines what numbers will be included in the set
The variable represents all the numbers included in the set
The vertical line '|' says we're going to define the variable
The inequality is what actually defines what numbers will be included in the set
The variable, vertical line, and inequality are all contained in the curly braces, since we're defining a set
Let's look at two example solutions
Our variable representing the set of numbers is 't' in this case
The phrase "t is GREATER THAN 80" can be rewritten as 't>80'
Remember that curly braces are used when dealing with sets
Our variable representing the set of numbers is 't' in this case
The vertical line '|' says we're going to define what numbers 't' represents in the set
The inequality is what actually defines what numbers will be included in the set
Our inequality is 't>80', meaning that this set will include all numbers greater than 80
The notation is: '{t | t>80}'
The phrase "y is GREATER THAN 24 and LESS THAN OR EQUAL TO 53" can be rewritten as 'y>24 AND y≤53'
Our variable representing the set of numbers is 'y' in this case
Remember that curly braces are used when dealing with sets
Our variable representing the set of numbers is 'y' in this case
The vertical line '|' says we're going to define what numbers 'y' represents in the set
We have a compound inequality, so we have two inequalities to deal with in this example
Our variable representing the set of numbers is 'y' in this case
The '>' symbol means GREATER THAN
So 'y>24' goes after the '{y |' part in the set-builder notation
We use 'AND' to link the two inequalities together and form a compound inequality
So 'AND' goes after the '{y | y>24' part in the set-builder notation
Our variable representing the set of numbers is 'y' in this case
The '≤' symbol means LESS THAN OR EQUAL TO
Don't forget to put a curly brace on the end to close the set!
Remember that curly braces are used when dealing with sets
The notation is: '{y | y>24 AND y≤53}'
We can simplify the notation slightly by writing the notation in a different way
We can simplify the notation slightly by writing the notation in a different way
Remember that curly braces are used when dealing with sets
The order of terms in an inequality can be flipped around as long as you remember to flip the inequality symbol, too!
The order of terms in an inequality can be flipped around as long as you remember to flip the inequality symbol, too!
The order of terms in an inequality can be flipped around as long as you remember to flip the inequality symbol, too!
Here, we're looking at the first half of our compound inequality, 'y>24'
So 'y>24' can also be written as '24<y'
The order of terms in an inequality can be flipped around as long as you remember to flip the inequality symbol, too!
Here, we're looking at the first half of our compound inequality, 'y>24'
So 'y>24' can also be written as '24<y'
Our second inequality is 'y≤53'
We can combine the two inequalities by dropping the 'AND' and dropping a 'y'
The '≤' symbol means LESS THAN OR EQUAL TO
So this will give reduce our compound inequality to: '24<y≤53'
Don't forget to put a curly brace on the end to close the set!
The notation is: '{y | 24<y≤53}'
We went from '{y | y>24 AND y≤53}' to '{y | 24<y≤53}'
We went from '{y | y>24 AND y≤53}' to '{y | 24<y≤53}'
Remember that curly braces are used when dealing with sets
We use variables to represent numbers in the set
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Math Tutor Explains: What are the Factors of 289? | Thinkster Math
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Factors
What Are The Factors Of 289
Math Tutor Explains: What are the Factors of 289?
As your math tutor, I’m here to help you understand factors! The factors of 289 are any whole numbers that can be multiplied together to equal exactly 289. In other words, finding the factors of 289 is like breaking it down into all the smaller numbers that, when multiplied, give you 289. Let’s explore this step by step!
Factors of 289: As Taught by a Math Tutor
Methods
What are the Factors of 289?
As your math tutor, I’m here to guide you through the different types of factors of 289. Understanding factors is key to mastering multiplication, division, and prime numbers. Here’s a breakdown:
• Factors of 289: 1, 17, 289
• Sum of Factors of 289: 307
• Negative Factors of 289: -1, -17, -289
• Prime Factors of 289: 17
• Prime Factorization of 289: 17^2
There are two main ways a math tutor would explain how to find the factors of 289: using factor pairs and prime factorization. Let’s explore both!
The Factor Pairs of 289
As your math tutor, I’m here to help you break down factor pairs of 289 step by step!
Factor pairs of 289 are any two numbers that, when multiplied together, equal 289. The question to ask is “what two numbers multiplied together equal 289?” Every factor can be paired with another factor, and multiplying the two will result in 289.
To find the factor pairs of 289, follow these steps:
Step 1:
Find the smallest prime number that is larger than 1, and is a factor of 289. For reference, the first prime numbers to check are 2, 3, 5, 7, 11, and 13. In this case, the smallest factor that’s a prime number larger than 1 is 17.
Step 2:
Divide 289 by the smallest prime factor, in this case, 17:
289 ÷ 17 = 17
17 and 17 will make a new factor pair.
Step 3:
Repeat Steps 1 and 2, using 17 as the new focus. Find the smallest prime factor that isn’t 1, and divide 17 by that number. In this case, 17 is the new smallest prime factor:
17 ÷ 17 = 1
Remember that this new factor pair is only for the factors of 17, not 289. So, to finish the factor pair for 289, you’d multiply 17 and 17 before pairing with 1:
17 x 17 = 289
Step 4:
Repeat this process until there are no longer any prime factors larger than one to divide by. At the end, you should have the full list of factor pairs.
Here are all the factor pairs for 289:
(1, 289), (17, 17)
So, to list all the factors of 289: 1, 17, 289
The negative factors of 289 would be: -1, -17, -289
Now you’ve got it! A math tutor would always encourage you to practice with different numbers to reinforce your understanding of factor pairs. Try another one!
Prime Factorization of 289
To find the prime factorization of 289, we break it down step by step until only prime factors remain. Then, we express 289 as a product of these prime factors multiplied together. Let’s go through the process and simplify it like a math tutor would!
The process of finding the prime factorization of 289 only has a few differences from the above method of finding the factors of 289. Instead of ensuring we find the right factor pairs, we continue to factor each step until we are left with only the list of smallest prime factors greater than 1.
Here are the steps for finding the prime factorization of 289:
Step 1:
Find the smallest prime number that is larger than 1, and is a factor of 289. For reference, the first prime numbers to check are 2, 3, 5, 7, 11, and 13. In this case, the smallest factor that’s a prime number larger than 1 is 17.
Step 2:
Divide 289 by the smallest prime factor, in this case, 17
289 ÷ 17 = 17
17 becomes the first number in our prime factorization.
Step 3:
Repeat Steps 1 and 2, using 17 as the new focus. Find the smallest prime factor that isn’t 1, and divide 17 by that number. The smallest prime factor you pick for 17 will then be the next prime factor. If you keep repeating this process, there will be a point where there will be no more prime factors left, which leaves you with the prime factors for prime factorization.
So, the unique prime factors of 289 are: 17
Math Tutor Suggests: Find the Factors of Other Numbers
Practice your factoring skills by exploring how to factor other numbers, like the ones below:
Factors of 31 - The factors of 31 are 1, 31
Factors of 80 - The factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
Factors of 90 - The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Factors of 64 - The factors of 64 are 1, 2, 4, 8, 16, 32, 64
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Article 2. Units of Measure
Article 2. Units of Measure
© Darrel R Barnette, Digital to Definitive LLC 2024
Synopsis: Whenever physical measurements are made or equations are used to determine performance metrics, it is imperative to use appropriate reference units of measure such that the results are dimensionally correct and the results make physical sense.The units of measure commonly used in the US are based on the US Customary system, which shares developmental roots with the British Imperial system, with units in both systems being of ancient Roman decent in many instances. It is therefore not surprising given the long history of development of the longbow in England and the more recent development of the compound bow in the US that use of these British-Influenced measurement systems persist in the sport of Archery. In this article we will review the US Customary system of units as it relates to the sport of Archery and provide sample calculations to facilitate proper use of the units when performing measurements and subsequent calculations. Particular focus is given to the correct interpretation and use of the pound unit which can describe either force or mass and therefore can be a point of confusion for many archers.
Before we dive into the science of arrow flight, we would do well to start off by establishing a firm foundation of how to quantify the physical values and performance metrics that we wish to measure or calculate. It would make no sense to describe an arrow shaft as 30 long for instance without appending some standard unit of measure to the number in order to quantify it. A “unit of measure” is an accepted standard that facilitates comparison between an unknown quantity and a known standard reference quantity. For example, if we measure the length of an arrow shaft in inches, we would know what that length is because the “inch” unit is a defined standard reference length. Of course there are multiple standard units of measure for length such as feet, yards, centimeters, etc. that we could have chosen to describe the length of the arrow shaft. In general utilizing known reference standards allows us to
Easily quantify the physical characteristics of an object,
Add universally understandable meaning to the measurement, and
Facilitate conversion from one unit of measure to another.
The problem is that US Customary units, unlike units of the Metric System, are a hodgepodge of units that are not systematic multiples of each other, and the names of the units are not interrelated. Table 1 highlights some comparisons between the two systems. Although length units are illustrated, mass and volume units have similar disparities. Notice in particular that unit prefixes in the Metric System (Kilo=1000, Centi=100, Milli=1/1000, Micro=1/100000) signify the multipliers of the base unit (Meter). There’s no such consistency in the US Customary units.
Many people feel that the Metric System is the better system because of its consistency and its worldwide adoption as the standard system of measurement (Ref. 2). However, given the centuries of history of the bow and arrow in England (Ref. 3) and that fact that US Customary units were derived from the British Imperial System of units, it’s not hard to understand why the sport of Archery has traditionally used these units. Hence going forward, we will focus on using US Customary units for our examples and calculations (except for one small foray into the British Gravitational System of units, which we will discuss next).
So far, we have established the need for units, and discussed the standard units that we will be using. Next we must address an apparent inconsistency issue with US Customary units and show by example how we can easily navigate around the issue. By far, the unit in the US Customary system that trips people up the most is the “pound” unit. That’s because the “pound” unit can describe either a unit of force or a unit of mass. Let’s dive into this a bit because in archery, almost every performance metric we want to quantify involves either force or mass, or both.
Even though the “pound” unit name is the same for both force and for mass, the quantity being measured is not the same and they cannot be directly set equal to each other. An object will have the same mass whether it’s located on the Moon or on the Earth. But its weight (a force) will be different at those two locations because the gravitational attraction of the object on the Earth is different (stronger) than the gravitational attraction acting on the same object on the Moon. To alleviate (or add to) the confusion, scientists have come up with more descriptive names to clutter up the unit. If a “pound” describes a unit of force, the name “pound force” (abbreviated as lb f) is used. If a “pound” is to describe a unit of mass, the name “pound mass” (abbreviated as lb m or confusingly, just lb in most common literature) is used. A pound force is not the same thing as a pound mass. The confusion is a long standing one even among engineers. For background, Ref. 1 provides a short historical perspective for the interested reader of how our systems of units have developed over the centuries.
To elaborate, when we weigh an arrow on a digital scale, the scale can only measure weight (not mass), but it is calibrated to automatically give you the answer in units of mass. When you weigh a “450 grain” arrow, the display will show 450 grains (a unit of mass) because the scale reads the weight (the force of gravity pulling down on the arrow’s mass), and an internal program inside the scale converts that reading into the equivalent mass for the amount of gravitational force that’s measured. If you took that same arrow, and the same digital scale to the Moon without recalibrating it and weigh the arrow there, the scale would not correctly display the arrow mass in grains even though we know that mass is the same regardless of whether the arrow is on the Moon or on Earth. Perhaps an even clearer example would be if you just used your finger to push down on the scale. Obviously the scale is not measuring the mass of your finger, even though its display is showing units of mass (such as grains). The scale is measuring force and translating it to mass units.Similarly if the scale has the option of displaying in terms of pounds, those units will most likely be in pounds mass and not pounds force. Adding to the confusion is that the pound mass unit is often displayed on the scale as merely “lb” and not “lb m“, because by definition (Ref. 1) the symbol “lb” without a subscript means pounds mass.
Let’s examine a common calculation to show how a person could get in a bind using pounds mass as a unit of measure.
PROBLEM STATEMENT: Calculate the kinetic energy of an arrow whose mass is 450 grains just after it flies through the chronograph screens positioned close to the bow. The chronograph reads the velocity as 300 ft/sec.
Express your answer in units of ft lb f (ironically, commonly called foot pounds when spoken).
SOLUTION ATTEMPT:
Weigh the arrow on a scale: 450 grains
Convert the 450 grains to units of lb m :
Plug the mass and velocity values into the equation for kinetic energy:
While this answer is technically correct, it does not match the units of ft·lb f that we commonly associate with kinetic energy.The calculation is incomplete as it stands, and without more information, we are stuck with an answer we have calculated that we can’t do much with.
I’ll confess that during my first three years as a college student, I did not understand how to finish calculating the kinetic energy of a body in motion. Whatever my profs were teaching involving some obscure gravitational constant term g c that was incorporated into the equations to make the numbers come out right didn’t register with me. So I just calculated everything in Metric units like every other engineering student was taught to do. At that time in the US (1980s), there was a big government push to try to convert over to the Metric System, so the Profs just tended to skip over US Customary units with a bit of hand waiving anyway. Unfortunately by doing so, a lot of misunderstanding of (and a lack of respect for) the US Customary system was inadvertently ingrained into the minds of engineering students of my generation. My guess is that this situation is even worse today on most US college campuses. Contrary to my education, I found great value throughout my career in being able to use both the Metric system and the US Customary system interchangeably. In particular, among the respected old-timers that I worked with over the years, almost none of them used the Metric system, and they tended to call out young engineers who used only the Metric system because there was a lack of what they would call “intuition” involved. The old timers instinctively knew what an inch looked like, how much a pound of mass weighed, or how fast 60 miles/hr was. The young engineers would correctly report their results in millimeters, Newtons, and kilometers/hr because it was easy to do so, but they did not have a feel for what those numbers represented. In other words, students knew the Metric system by rote memorization, but had no intuition about what the units represented or how to convert to units that the old-timers historically used and had a feel for. In essence, they were correctly speaking a language, but they were not speaking the correct language for the situation at hand. I believe this is the essence of the costly mistake described in footnote 2 above.
Speaking of old timers and their experience, one day at the start of my 4 th year of college, the assistant prof, a Brigadier General Richard Drury (Ret’d) strolled up to the chalk board and drew the following diagram as shown in Fig. 1. It was a revelation that I still recall some 40 years later:
On the far left of the chalk board he drew a very small box. He pointed to it and said, “This is a grain.” Then he drew a bigger box next to it and said, “This is a pound mass. There are 7000 grains in pound mass.” And finally he drew a larger box next to the second one, and said, “This is a slug. There are 32.174 pounds mass in a slug. The units of a slug are pounds force seconds squared per foot.” I don’t recall ever hearing of a slug before that day except for the snail-without-a-shell variety. It turns out that a slug is a derived unit of mass in the British Gravitation System (Ref. 5), and it is defined as that quantity of mass that is accelerated to 1 ft/s 2 when a net force of one pound force(lb f) is exerted on it. That quantity of mass turns out to be 32.174 Ib m. The slug is a derived unit of mass that comes from Newton’s second law of motion. We will talk a lot more about Newton’s second law in future articles so that we don’t digress too much here. For now let’s return to the goal of calculating the kinetic energy of our arrow and see how using the slug as a unit of mass can help us do that. Let’s start with the same problem statement as before:
PROBLEM STATEMENT: Calculate the kinetic energy of an arrow whose mass is 450 grains at the point it flies through the chronograph screens positioned close to the bow. The chronograph reads the velocity as 300 ft/sec.
Express your answer in units of ft lb f.
SOLUTION:
Weigh the arrow on a scale: 450 grains
Convert the 450 grains to lb m and then to slugs (shown in blue):
Plug the mass and velocity values into the equation for kinetic energy:
When we use slugs rather than lb m as the mass unit, the calculation will automatically be of the final form we desire without a separate conversion constant since the slug unit inherently expresses the relationship between pounds mass and pounds force. Think of the lb m unit as a gateway between grains and slugs and always convert any mass measurement from grains to slugs. Pretty soon, you will get used to automatically dividing the mass in grains by 7000 and then again by 32.174 to end up with units of lb f s 2/ft. I find it easier to remember the two step conversion values rather than simply trying to divide the number of grains by the product of 32.1747000=225218, because mentally it helps me remember the meaning behind each conversion step, just as outlined by Mr. Drury in Fig. 1.
Let’s look at another example:
PROBLEM STATEMENT: How much does 1 lb m nominally weigh on Earth?
SOLUTION: We know from Newton’s 2 nd Law that the weight of an object is equal to its mass multiplied by the Earth’s local gravitational acceleration (Ref. 7) that is acting upon that mass. If we just want to use a nominal constant value for the gravitational acceleration, we can represent that by g 0. In equation form this can be expressed as:
where brackets have been placed around the mass symbol above to delineate where the conversion from 1b m to slugs occurs in the formula below. Therefore
Note that this DOES NOT say that 1lb f = 1lb m. What it does say is that 1lb f =1lb m X 32.174 ft/s 2 on Earth. Why does the value 32.174 appear twice in this equation? Recall that the slug unit is defined as that amount of mass that is accelerated to 1 ft/s 2 when a net force of one pound force(lb f) is exerted on it. That amount of mass turns out to be 32.174 lb m and this value appearing inside the brackets does not change with location. However, the value 32.174 that is outside of the brackets is the nominal acceleration due to gravity on Earth, and this term does vary with location. If we wanted to calculate the weight of an object on the Moon having a mass of 1 lb m for instance, it is this value outside of the brackets that would need to change. In fact let’s do just that and see what we get. The standard acceleration due to gravity at the surface of the Moon (Ref. 8) is g 0=5.309 ft/s 2. Therefore we can determine that the weight of 1 lb m on the surface of the moon is:
So the weight of 1 lb m on Earth is 1 lb f, but the weight of 1 lb m on the Moon is .165 lb f, which is approximately six times less.
As a final topic, let’s put together a simple table of units that are commonly used in Archery. Table 2 lists units for Length, Mass, Force, and Time that we will be using in future articles. We will refer back to this table as necessary to avoid redefining the units each time we want to use them.
In summary, by historical precedence the engineering units commonly used in archery in the US at the present time are the US Customary units, with the addition of the slug as defined as a unit of mass in the British Gravitational System. The focus of this particular article was to help clarify the difference between the pound unit when used to describe a force, versus the pound unit when used to describe a mass. Calculations were then used to illustrate a method of mass-units conversion from grains to slugs by utilizing the pounds mass unit only as a stepping stone in the process. Future articles will rely upon this conversion process to quantify arrow behavior at launch, in flight, and during penetration.
References
Units tend to grow more precise in definition with advancing technology. See Ref. 1, for example.
A particularly costly mistake (>$500M) of unit conversion error between US Customary unit and the Metric System made by US engineers of my generation :
He always had students refer to him as just “Mr. Drury”. I had no idea of his true background until writing this article.
A very old reference unit of mass, which was quantified by the weight of 1 grain of barley. See Ref. 4.
Weight varies slightly with location on Earth. See Ref. 6.
At least one reference states that the slug is the base unit in the US Customary system. For example see Ref. 9.
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© Darrel R Barnette, Digital to Definitive LLC 2024. Unauthorized use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used provided that full and clear credit is given to Darrel R Barnette, Digital to Definitive LLC with appropriate and specific direction to the original content.
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DUSTIN APPLE Oct 23, 2024 I’m very interested in the explanation of how KE is only a unit of potential & Slugs are a unit of ability. While speed is sexy the difference of 1/2 a slug is tremendous. Until we understand Slugs we can not begin to peel back the layers of resistance created by broadhead profiles, angle of slip, & ferrule designs.
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Article 2. Units of Measure
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https://www.physics.usyd.edu.au/~helenj/Thermal/PDF/thermal7.pdf
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The first law of thermodynamics continued Lecture 7 Pre-reading: §19.5 Where we are The pressure p, volume V, and temperature T are related by an equation of state. For an ideal gas, pV = nRT = NkT For an ideal gas, the temperature T is is a direct measure of the average kinetic energy of its molecules: and vrms = p (v2)av = r 3kT m = r 3RT M KEtr = 3 2nRT = 3 2NkT Where we are We define the internal energy of a system: For an ideal gas, i.e. the internal energy depends only on its temperature U KE PE = + ∑ ∑ Random chaotic motion interaction between atoms & molecules 2 f U N k T = Where we are By considering adding heat to a fixed volume of an ideal gas, we showed and so, from the definition of heat capacity we have that for any ideal gas. Change in internal energy: Q = f 2 Nk∆T = f 2 nR∆T Q = nC∆T CV = f 2 R ∆U = nCV ∆T Heat capacity of an ideal gas Now consider adding heat to an ideal gas at constant pressure. By definition, and So from we get or It takes greater heat input to raise the temperature of a gas a given amount at constant pressure than constant volume YF §19.4 Q = nCp∆T W = p∆V = nR∆T ∆U = Q −W nCV ∆T = nCp∆T −nR∆T Cp = CV + R Ratio of heat capacities Look at the ratio of these heat capacities: we have and so For a monatomic gas, so and YF §19.4 R f CV 2 = R f R C C V p 2 2 + = + = 1 > = V p C C γ CV = 3 2R Cp = 3 2R + R = 5 2R γ = Cp CV = 5 2R 3 2R = 5 3 = 1.67 Problem An ideal gas is enclosed in a cylinder which has a movable piston. The gas is heated, resulting in an increase in temperature of the gas, and work is done by the gas on the piston so that the pressure remains constant. a) Is the work done by the gas positive, negative or zero? Explain b) From a microscopic view, how is the internal energy of the gas molecules affected? c) Is the heat less than, greater than or equal to the work? Explain. The first law of thermodynamics The total change in internal energy of a system is the sum of the heat added to it and the work done on it. (conservation of energy). ΔU depends only on the initial and final states, i.e. does not depend on the path, and does not depend on the kind of process that occurs. Two special cases of the first law are worth mentioning. YF §19.4 ∆U = Q −W Isolated processes An isolated process is one where the system does no work, and there is no heat flow: W = Q = 0 and hence U2 = U1 = ΔU = 0 so the internal energy of an isolated system is constant. YF §19.5 Cyclic processes A cyclic process is one where the system returns to its original state. Since the final state is the same as the initial state, the total energy change must be zero; so U2 = U1 and Q = W If work is done by the system, then an equal quantity of heat must have flowed into the system as Q; but in general Q and W do not have to be zero. YF §19.4 Thermodynamic processes Now we want to think in detail about different thermodynamic processes. For an ideal gas, we can represent these processes using pV-diagrams. In each case, we want to relate the change in internal energy, the heat added, and the work done. Thermodynamic processes We can identify four specific kinds of thermodynamic processes that often occur: • Adiabatic – no heat transfer into or out of the system • Isochoric – occurs at constant volume • Isobaric – occurs at constant pressure • Isothermal – occurs at constant temperature YF §19.5 Adiabatic processes An adiabatic process is one there is no heat flow: Q = 0 A process can be adiabatic if the system is thermally insulated, or if it is so rapid that there is not enough time for heat to flow. For an adiabatic process, U2 – U1 = ΔU = –W YF §19.5 Isochoric processes An isochoric process is a constant volume process. With no volume change, there is no work: W = 0 All the energy added as heat remains in the system as an increase in internal energy. For an isochoric process, U2 – U1 = ΔU = Q YF §19.5 Isochoric processes Isochoric process for an ideal gas: ΔV = 0 W = 0; ΔU = Q = n CV ΔT YF §19.5 Isothermals pV = constant 0 20 40 60 80 100 120 140 160 180 200 0.00 0.05 0.10 0.15 0.20 0.25 volume V (m3) pressure p (kPa) 100 K 400 K 800 K 1 2 W = 0 Isochoric process 1 to 2: Q > 0 T 1 < T 2 ΔT > 0 ΔU > 0 Isobaric processes An isobaric process is a constant pressure process. In general, none of ΔU, Q or W is zero, but: W = p(V2 – V1) YF §19.5 Isothermals pV = constant 0 20 40 60 80 100 120 140 160 180 200 0.00 0.05 0.10 0.15 0.20 0.25 volume V (m3) pressure p (kPa) 100 K 400 K 800 K 1 2 W Isobaric process Isobaric processes Isobaric process for an ideal gas: Δp = 0 W = p ΔV, Q = n CP ΔT, ΔU = Q – W = n CV ΔT YF §19.5 T2>T1 so ΔU > 0, W > 0, Q > 0, W < Q Isothermal processes An isothermal process is a constant temperature process. For a process to be isothermal, any heat flow must occur slowly enough that thermal equilibrium is maintained. In general, none of ΔU, Q or W is zero. YF §19.5 Isothermals pV = constant 0 20 40 60 80 100 120 140 160 180 200 0.00 0.05 0.10 0.15 0.20 0.25 volume V (m3) pressure p (kPa) 100 K 400 K 800 K 1 2 W Isothermal process Isothermal processes Isothermal process for an ideal gas: ΔT = 0 T1 = T2, p1V1 = p2V2 ; W is the area under the isothermal curve. YF §19.5 T1= T2 , p1V1 = p2 V2 Adiabatic process Adiabatic process for an ideal gas: Q = 0 so ΔU = –W As gas expands from Va to Vb, it does positive work so U drops and T drops. So if a is on an isotherm with temperature T + dT, point b is on an isotherm with lower temperature T è adiabat is steeper. YF §19.5 Adiabatic process Recall that for an ideal gas, ΔU = nCVΔT, and W = p ΔV. Then since ΔU = –W, we have nCVΔT = –p ΔV Using the ideal gas equation, we can show that where YF §19.5 TV γ−1 = constant pV γ = constant W = CV R (p1V1 −p2V2) = 1 γ −1(p1V1 −p2V2) γ = CP CV Isothermals pV = constant 0 20 40 60 80 100 120 140 160 180 200 0.00 0.05 0.10 0.15 0.20 0.25 volume V (m3) pressure p (kPa) 100 K 400 K 800 K 1 2 W Adiabatic process Adiabatic process for an ideal gas: ΔT = 0 T1 = T2, p1V1 = p2V2 ; W is the area under the isothermal curve. Adiabatic processes YF §19.5 1 to 2: Q = 0 T1 > T2, W > 0, ΔU < 0 W A cylinder with a movable piston contains 0.25 mole of monatomic idea gas at 2.40×105 Pa and 355 K. The ideal gas first expands isobarically to twice its original volume. It is then compressed adiabatically back to its original volume, and finally it is cooled isochorically to its original pressure. a) Compute the temperature after the adiabatic compression. b) Compute the total work done by gas on the piston during the whole process. Take Problem 1 1.
.
47 .
12 − − = K mol J CV 67 .
1 = γ Next lecture The second law of thermodynamics Read: YF §20.1
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381
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https://math.stackexchange.com/questions/1462247/point-normal-equation-of-plane
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Point normal equation of plane
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Im doing an homeassignment in linear algebra! This is the question I'm having problems with!
The plane M contains two lines;
l1 : (x, y, z) = t(2, 1, 0), t R,
l2 : (x, y, z) = t(0, 1, 2), t R.
Determine a point-normal equation of the plane M and compute the acute angle between M and the line;
l3 : (x, y, z) = t(2, 1, 1), t R. (ON-system assumed.)
I have taken the crossproduct of l1 and l2 and got that computed to be: (-2, -4, 2) , which in my understanding is n (the normal of the plane?) But them I am stuck.I can't understand if the plane contains fully the two lines, or only points from the lines?? From what i've learned two lines are too much information for a plane in 3-dim? Three points or one line and one point is enough.?? Also why is the lines only described in parametric (x, y, x) = t(0,1,2). What does the parameter exactly mean, is it a direction vector? usually is lines not described with: {x-3 = t, y+6 = 3t, z+5 = 2t}...I am probably not understanding the concept completely, I have been trying to do this question for hours now and I really need some help to get me going. I also know that for shortest distance I want the angle of line and plane to be pi/2, since then cos(theta) will be 0. I can probably also use this somehow?? Grateful if I could get some concrete tips and tricks how to do this and what it all really means...
From where do I get the point p0 and p, are they from the lines in the plane or from the line i will compute the acute angle of l-M??
Have I calculated n (normal) to plane correctly??
Why are the lines only described in parameter t??
linear-algebra
vectors
plane-curves
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edited Oct 3, 2015 at 15:27
John Alexiou
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asked Oct 3, 2015 at 11:10
Elin HägglundElin Hägglund
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$\begingroup$ Maybe I could calculate the acute angel from cos(phi) = |m||n|/squareroot(m^2)squareroot(n^2) ?? But how do I get the equation of the plane in point normal form? Is'nt it better to have it in the form Ax+By+Cz = D ?? $\endgroup$
Elin Hägglund
– Elin Hägglund
2015-10-03 11:24:34 +00:00
Commented Oct 3, 2015 at 11:24
$\begingroup$ Please use block-quotes (> block-quoted text) to distinguish what is asked in the assignment what are your thoughts. See example above, but I didn't know if l3 is part of the answer or part of the question (see math.stackexchange.com/editing-help#simple-blockquotes) $\endgroup$
John Alexiou
– John Alexiou
2015-10-03 11:45:59 +00:00
Commented Oct 3, 2015 at 11:45
$\begingroup$ Oh sorry. L3 is part of the question and not the answer. I'll try to fix my writing style :) so it's more understandable. Thanks for tips. $\endgroup$
Elin Hägglund
– Elin Hägglund
2015-10-03 12:55:21 +00:00
Commented Oct 3, 2015 at 12:55
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2 Answers 2
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You're right that it's generally impossible to find a plane that contains two given lines. In fact, it's possible only if the two given lines intersect. Fortunately, the two lines $l_1$ and $l_2$ in your problem do intersect; in fact, they intersect at the origin $(0,0,0)$.
To see this, look at the equations. The line $l_1$ has equation $(x,y,z) = t(2,-1,0)$. This says that every point $(x,y,z)$ on the line is some multiple ($t$) of the vector $(2,-1,0)$. In particular, when $t=0$, you get $(x,y,z) = (0,0,0)$. So, the line passes through the origin and has direction $(2,-1,0)$.
If we let $\mathbf{p}_0 = (0,0,0)$, then $l_1$ has equation $$ (x,y,z) = \mathbf{p}_0 +t(2,-1,0) = (0+2t, 0-t, 0) $$ which might be the kind of line equation you were expecting.
Similarly, the line $l_2$ passes through the origin and has direction $(0,1,2)$.
The plane containing $l_1$ and $l_2$ has normal vector $\mathbf{n} = (-2, -4,2)$, which you correctly computed as a cross product, and it also passes through the origin. So its equation is $(x,y,z).\mathbf{n} = 0$, i.e. $-2x-4y+2z=0$. The equation $x +2y -z =0$ is simpler, and gives you the same plane.
To get the angle between line $l_3$ and the plane, we can get the angle between the line and the plane normal, and then subtract from 90 degrees.
The line $l_3$ is in the direction of the vector $\mathbf{v} = (2,1,1)$, and the plane normal is $\mathbf{n} = (-2, -4,2)$. The angle $\theta$ between these two vectors is given by $$ \cos\theta = \frac {\mathbf{v} \cdot \mathbf{n}} {|\mathbf{v}| \cdot |\mathbf{n}|} = \frac {(2)(-2) + (1)(-4)+(1)(2)} {\sqrt{2^2 + 1^2 + 1^2}\sqrt{(-2)^2 + (-4)^2 + 2^2}} = \frac{-6}{12} $$ Then the angle between the plane and $l_3$ is $90- \theta$.
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edited Oct 4, 2015 at 2:19
answered Oct 3, 2015 at 11:29
bubbabubba
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$\begingroup$ Thank you so much! This was really helpful. Now I know that I was on the right path with planes being described by 3 points and that I computed the correct crossproduct for n of the plane. I was not completely sure that it was the right one, but I assumed so, since with my information that was what I could do. I'm going to read this thoroughly and see if I can finish this. If I still have something unclear and undone, I'll post some more questions til I understand it fully. Thank you again for taking time to help me :) $\endgroup$
Elin Hägglund
– Elin Hägglund
2015-10-03 11:45:33 +00:00
Commented Oct 3, 2015 at 11:45
$\begingroup$ I computed the point normal form as: (-2, -4, 2)(-2t, 0, 2t) = (4t, 0, 4t). Is this right. I want to compute the angle but I am a little unsure, do I use cos(theta) = (2,1,1)(-2,-4.2)/ (point normal eq. of plane?) $\endgroup$
Elin Hägglund
– Elin Hägglund
2015-10-03 12:48:34 +00:00
Commented Oct 3, 2015 at 12:48
$\begingroup$ The equation of the plane is $-2x -4y +2z = 0$. Not entirely sure what you mean by "point-normal" form. $\endgroup$
bubba
– bubba
2015-10-03 13:51:49 +00:00
Commented Oct 3, 2015 at 13:51
$\begingroup$ Regarding the angle, see the addtions to my answer. $\endgroup$
bubba
– bubba
2015-10-04 02:12:18 +00:00
Commented Oct 4, 2015 at 2:12
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Take the cross product of l1 and l2 directions to find the plane normal vector (non-unit vector)
$$ \vec{n} = \vec{e}_1 \times \vec{e}_2 = (2,-1,0) \times (0,1,2) = (-2,-4,2) $$
The point-normal equation for a plane is $$\vec{n} \cdot (\vec{r}-\vec{r}_0) =0$$ which works of any scalar multiple of $\vec{n}$ (no need to normalize to a unit vector) and $\vec{r}_0$ as any point on the plane. Might as well take $t=0$ for l1 to give $\vec{r}_0 = (0,0,0)$. So the plane equation is
$$ -2x-4y+2z=0 $$
The acute angle $\theta$ of the plane to the line l3 is found by noting that it is equal to 90° minus the angle $\varphi$ between the line and the plane normal. This is found by the inner product rule
$$ \vec{n} \cdot \vec{e} = | \vec{n}| |\vec{e}| \cos\varphi $$
with $\vec{n}=(-2,-4,2)$ and $\vec{e}=(2,1,1)$
$$\varphi = \cos^{-1} \left( \frac{\vec{n}\cdot\vec{e}}{|\vec{n}| |\vec{e}|} \right) = \left( \frac{(-2,-4,2)\cdot(2,1,1)}{ (2\sqrt{6})(\sqrt{6})} \right) = \cos^{-1} \left( \frac{-6}{12} \right) = 120°$$
Since the angle is > 90° we have to flip the direction of the line to get $\vec{e} = (-2,-1,-1)$ and
$$ \varphi = \cos^{-1} \left( \frac{6}{12} \right) = 30° $$
So the angle of the line l3 to the plane is $$\boxed{\theta = 90° - \varphi = 60°}$$
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answered Oct 3, 2015 at 23:48
John AlexiouJohn Alexiou
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$\begingroup$ How do I know i need to flip the direction of the line? Can I just do that without changing the information somehow. I would never have thought of that. And also cos^-1(6/12) = 60 degrees right, so then theta = 90-60 = 30 degrees?? :) Thanx for this, very helpful! $\endgroup$
Elin Hägglund
– Elin Hägglund
2015-10-06 12:38:33 +00:00
Commented Oct 6, 2015 at 12:38
$\begingroup$ If you don't flip the answer will be -30° which doesn't have any physical meaning. You can flip the direction of the line, or take the absolute value of the angle (the result would be the same). Yes, on the other points. $\endgroup$
John Alexiou
– John Alexiou
2015-10-07 00:38:00 +00:00
Commented Oct 7, 2015 at 0:38
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Two numbers are selected at random (without replacement) from first 7 natural numbers. If X denotes the smaller of the two numbers obtained, find the probability distribution of X. Also, find mean of the distribution.
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To solve the problem, we need to find the probability distribution of the random variable X, which represents the smaller of the two randomly selected numbers from the first 7 natural numbers (1 to 7). We will also calculate the mean of this distribution. 1. Identify the Sample Space: The first 7 natural numbers are: S={1,2,3,4,5,6,7} We need to select 2 numbers from this set without replacement. 2. Determine the Total Number of Ways to Choose 2 Numbers: The total number of ways to choose 2 numbers from 7 is given by the combination formula: Total choices=(72)=7×62×1=21 3. Possible Values of X: Since X is defined as the smaller of the two selected numbers, the possible values of X are: X∈{1,2,3,4,5,6} (Note that X cannot be 7 because there would be no larger number to pair with it.) 4. Calculate the Probability Distribution: We will calculate P(X=x) for each possible value of X: - For X=1: The second number can be any of {2, 3, 4, 5, 6, 7} (6 choices). P(X=1)=621 - For X=2: The second number can be any of {3, 4, 5, 6, 7} (5 choices). P(X=2)=521 - For X=3: The second number can be any of {4, 5, 6, 7} (4 choices). P(X=3)=421 - For X=4: The second number can be any of {5, 6, 7} (3 choices). P(X=4)=321 - For X=5: The second number can be any of {6, 7} (2 choices). P(X=5)=221 - For X=6: The second number can only be 7 (1 choice). P(X=6)=121 5. Summarize the Probability Distribution: The probability distribution of X is: P(X=1)=621P(X=2)=521P(X=3)=421P(X=4)=321P(X=5)=221P(X=6)=121 6. Calculate the Mean of the Distribution: The mean E(X) is calculated as: E(X)=∑(xi⋅P(X=xi)) E(X)=1⋅621+2⋅521+3⋅421+4⋅321+5⋅221+6⋅121 =6+10+12+12+10+621=5621=83 Final Answers: - The probability distribution of X is: P(X=1)=621,P(X=2)=521,P(X=3)=421,P(X=4)=321,P(X=5)=221,P(X=6)=121 - The mean of the distribution is: E(X)=83
To solve the problem, we need to find the probability distribution of the random variable X, which represents the smaller of the two randomly selected numbers from the first 7 natural numbers (1 to 7). We will also calculate the mean of this distribution.
Identify the Sample Space:
The first 7 natural numbers are:
S={1,2,3,4,5,6,7}
We need to select 2 numbers from this set without replacement.
Determine the Total Number of Ways to Choose 2 Numbers:
The total number of ways to choose 2 numbers from 7 is given by the combination formula:
Total choices=(72)=7×62×1=21
Possible Values of X:
Since X is defined as the smaller of the two selected numbers, the possible values of X are:
X∈{1,2,3,4,5,6}
(Note that X cannot be 7 because there would be no larger number to pair with it.)
Calculate the Probability Distribution:
We will calculate P(X=x) for each possible value of X:
For X=1:
The second number can be any of {2, 3, 4, 5, 6, 7} (6 choices).
P(X=1)=621
For X=2:
The second number can be any of {3, 4, 5, 6, 7} (5 choices).
P(X=2)=521
For X=3:
The second number can be any of {4, 5, 6, 7} (4 choices).
P(X=3)=421
For X=4:
The second number can be any of {5, 6, 7} (3 choices).
P(X=4)=321
For X=5:
The second number can be any of {6, 7} (2 choices).
P(X=5)=221
For X=6:
The second number can only be 7 (1 choice).
P(X=6)=121
Summarize the Probability Distribution:
The probability distribution of X is:
P(X=1)=621P(X=2)=521P(X=3)=421P(X=4)=321P(X=5)=221P(X=6)=121
Calculate the Mean of the Distribution:
The mean E(X) is calculated as:
E(X)=∑(xi⋅P(X=xi))
E(X)=1⋅621+2⋅521+3⋅421+4⋅321+5⋅221+6⋅121
=6+10+12+12+10+621=5621=83
Final Answers:
- The probability distribution of X is:
P(X=1)=621,P(X=2)=521,P(X=3)=421,P(X=4)=321,P(X=5)=221,P(X=6)=121
- The mean of the distribution is:
E(X)=83
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Step by step solution :
Step 1 :
Equation at the end of step 1 :
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 2a2-a-6
The first term is, 2a2 its coefficient is 2 .
The middle term is, -a its coefficient is -1 .
The last term, "the constant", is -6
Step-1 : Multiply the coefficient of the first term by the constant 2 • -6 = -12
Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is -1 .
| | | | | | | |
--- --- ---
| | -12 | + | 1 | = | -11 | |
| | -6 | + | 2 | = | -4 | |
| | -4 | + | 3 | = | -1 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and 3
2a2 - 4a + 3a - 6
Step-4 : Add up the first 2 terms, pulling out like factors :
2a • (a-2)
Add up the last 2 terms, pulling out common factors :
3 • (a-2)
Step-5 : Add up the four terms of step 4 :
(2a+3) • (a-2)
Which is the desired factorization
Equation at the end of step 2 :
Step 3 :
Theory - Roots of a product :
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
3.2 Solve : a-2 = 0
Add 2 to both sides of the equation :
a = 2
Solving a Single Variable Equation :
3.3 Solve : 2a+3 = 0
Subtract 3 from both sides of the equation :
2a = -3
Divide both sides of the equation by 2:
a = -3/2 = -1.500
Supplement : Solving Quadratic Equation Directly
Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Parabola, Finding the Vertex :
4.1 Find the Vertex of y = 2a2-a-6
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 2 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Aa2+Ba+C,the a -coordinate of the vertex is given by -B/(2A) . In our case the a coordinate is 0.2500
Plugging into the parabola formula 0.2500 for a we can calculate the y -coordinate :
y = 2.0 0.25 0.25 - 1.0 0.25 - 6.0
or y = -6.125
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 2a2-a-6
Axis of Symmetry (dashed) {a}={ 0.25}
Vertex at {a,y} = { 0.25,-6.12}
a -Intercepts (Roots) :
Root 1 at {a,y} = {-1.50, 0.00}
Root 2 at {a,y} = { 2.00, 0.00}
Solve Quadratic Equation by Completing The Square
4.2 Solving 2a2-a-6 = 0 by Completing The Square .
Divide both sides of the equation by 2 to have 1 as the coefficient of the first term :
a2-(1/2)a-3 = 0
Add 3 to both side of the equation :
a2-(1/2)a = 3
Now the clever bit: Take the coefficient of a , which is 1/2 , divide by two, giving 1/4 , and finally square it giving 1/16
Add 1/16 to both sides of the equation :
On the right hand side we have :
3 + 1/16 or, (3/1)+(1/16)
The common denominator of the two fractions is 16 Adding (48/16)+(1/16) gives 49/16
So adding to both sides we finally get :
a2-(1/2)a+(1/16) = 49/16
Adding 1/16 has completed the left hand side into a perfect square :
a2-(1/2)a+(1/16) =
(a-(1/4)) • (a-(1/4)) =
(a-(1/4))2
Things which are equal to the same thing are also equal to one another. Since
a2-(1/2)a+(1/16) = 49/16 and
a2-(1/2)a+(1/16) = (a-(1/4))2
then, according to the law of transitivity,
(a-(1/4))2 = 49/16
We'll refer to this Equation as Eq. #4.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(a-(1/4))2 is
(a-(1/4))2/2 =
(a-(1/4))1 =
a-(1/4)
Now, applying the Square Root Principle to Eq. #4.2.1 we get:
a-(1/4) = √ 49/16
Add 1/4 to both sides to obtain:
a = 1/4 + √ 49/16
Since a square root has two values, one positive and the other negative
a2 - (1/2)a - 3 = 0
has two solutions:
a = 1/4 + √ 49/16
or
a = 1/4 - √ 49/16
Note that √ 49/16 can be written as
√ 49 / √ 16 which is 7 / 4
Solve Quadratic Equation using the Quadratic Formula
4.3 Solving 2a2-a-6 = 0 by the Quadratic Formula .
According to the Quadratic Formula, a , the solution for Aa2+Ba+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
a = ————————
2A
In our case, A = 2
B = -1
C = -6
Accordingly, B2 - 4AC =
1 - (-48) =
49
Applying the quadratic formula :
1 ± √ 49
a = —————
4
Can √ 49 be simplified ?
Yes! The prime factorization of 49 is
7•7
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 49 = √ 7•7 =
± 7 • √ 1 =
± 7
So now we are looking at:
a = ( 1 ± 7) / 4
Two real solutions:
a =(1+√49)/4=(1+7)/4= 2.000
or:
a =(1-√49)/4=(1-7)/4= -1.500
Two solutions were found :
How did we do?
Why learn this
Terms and topics
Related links
Latest Related Drills Solved
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Back to the Table of Contents
A Review of Basic Geometry - Lesson 6
Classifying Polygons by Symmetry
Lesson Overview
Reflection-Symmetric Figures
Hierarchy of Triangles, Symmetric Triangles (Isosceles, Equilateral)
Incenter, Circumcenter, Orthocenter, and Centroid
Hierarchy of Quadrilaterals
Kites and Their Properties
Trapezoids and Their Properties
Parallelograms and Their Properties
Rotational Symmetric Figures
Regular Polygons
Schedules with Regular Polygons
Noether's Theorem, Homework
Reflection-Symmetric Figures
A plane figure is reflection-symmetric if and only if there is a line which reflects the figure onto itself. This line is a symmetry line for the figure. |
The capital letters A, B, C, D, E, H, I, K, M, O, T, U, V, W, X, and Y are often written as reflection symmetric figures. Some are symmetric about a horizontal line (BCDEHIKOX) whereas others are symmetric about a vertical line (AHIMOTUVWXY). As you can see since some are in both lists (HIOX), there may be more than one line of symmetry. A challenge would be to find words such as DIXIE or COOKBOOK composed entirely of letters with a horizontal line of symmetry or MOM, WAXY, YOUTH (written vertically!) composed entirely of letters with a vertical line of symmetry. After collecting enough of these words you might make them into a crossword puzzle (for extra credit)!
Our textbook states and proves what they call the Flip-Flop Theorem: (reflection is symmetric).
| |
| If F and G are points/figures, and rl(F)=G, then rl(G)=F. |
From this it can be proved that every segment has two lines of symmetry: itself and its perpendicular bisector. This is the same as the letter I discussed above. Angles only have one line of symmetry: the angle bisector which causes one ray to reflect onto the other ray. A circle has infinitely many lines of symmetry (no matter which way you draw the diameter, the semicircles are reflections of each other). The section concludes with the following important result.
If a figure is symmetric, then any pair of corresponding parts under the symmetry are congruent. |
Rorschach inkblots and logos commonly are reflective-symmetric. These symmetries will be useful when applied to various polygons. Symmetry is also important in algebra. The function y=x2 defines a parabola in which the sign of x doesn't matter. This makes it an even function (the exponent of 2 is another clue).
Symmetric Triangles (Isosceles and Equilateral)
Triangles, as mentioned in Numbers lesson 11 and Geometry lesson 2, can be classified either by the number of sides with the same length (0 is scalene, 2 or more is isosceles, all 3 is equilateral) or by the largest angle (acute, right, obtuse). A hierarchy chart combining both situations is given at the left. Due to the overlap, hierarchy charts for either situation are typically given instead. Note: a right/acute/obtuse triangle might be either scalene or isosceles.
Also, our definition of isosceles includes and does not exclude the equilateral triangle. Just as there are special names associated with the sides of a right triangle (hypotenuse and legs), there are special names associated with the angles and sides of an isosceles triangle. The angle determined by the two equal sides is called the vertex angle. The side opposite the vertex angle is called the base. The two angles opposite the equal sides are the base angles (and are equal). These can also be described as the angles at the endpoints of the base.
Three important theorems are as follows. Certain terms will be defined further below.
The line containing the bisector of the vertex angle of an isosceles triangle is a symmetry line for the triangle. |
In an isosceles triangle, the bisector of the vertex angle, the perpendicular bisector of the base, and the median to the base determine the same line. |
If a triangle has two congruent sides, then the angles opposite them are congruent. |
This last theorem is generally known as the Isosceles Triangle Base Angle Tneorem and commonly stated as: in an isosceles triangle, base angles are equal. The converse (if two angles of a triangle are congruent, then the sides opposite them are congruent) is also true but awaits chapter 7. Sometimes this goes by its historic Latin name: pons asinorum, or bridge of asses/fools, due to the proof diagram that was used.
Every equilateral triangle has three lines of symmetry. These are the bisectors of the angles/sides. |
If a triangle is equilateral, then it is equiangular. |
A corollary (a theorem which logically follows immediately from another theorem) is that the angles of an equilateral triangle are all 60°. Although the line of symmetry of an isosceles triangle is an angle bisector, median, perpendicular bisector, and an altitude, in most triangles, these lines are different.
Incenter, Circumcenter, Orthocenter, and Centroid
We will defined various important auxiliary lines which may be constructed on a polygon. We will also discuss some specific applications of these lines to triangles.
A ray is an angle bisector if and only if it forms two angles of equal measure with the sides of the angle. |
The three angle bisectors of a triangle are coincident at the incenter. |
The incenter is equidistant (distance r) from all three sides of a triangle. Thus if a circle were drawn with the incenter as its center with a radius r, it would be inscribed in the triangle.
A segment, ray, or line is a perpendicular bisector (of a segment) if and only if it contains the midpoint of the segment and is perpendicular to the segment. |
The three perpendicular bisectors of a triangle are coincident at the circumcenter. |
The circumcenter is equidistant (distance r) from all three vertices of a triangle. Thus if a circle were drawn with the circumcenter as its center with a radius r, it would circumscribed the triangle.
A segment is an altitude if and only if it is perpendicular to the line containing the side opposite a vertex and contains that vertex. |
Altitude can also refer to the length of the segment described above. Trapezoids also have altitudes. In addition, the height of the 3-dimensional objects: prisms, cylinders, pyramids, and cones is termed altitude.
The three altitudes of a triangle are coincident at the orthocenter. |
The orthocenter need not be in the interior of a triangle. It will be located inside only if the triangle is acute. If the triangle is right, the orthocenter will be on the hypotenuse. If the triangle is obtuse, the orthocenter will be outside the triangle.
A segment is a median of a triangle if and only if it connects one vertex to the midpoint of the opposite side. |
The three medians of a triangle are coincident at the centroid. |
If the triangle were made of a uniformly dense material, the centroid would be the center of mass or center of gravity of the triangle. A thin solid object in this shape would balance at this point. Thus if a triangle is hung by a vertex, a line toward the local point of graviational attraction (local nadir or straight down) would describe a median and go through the centroid. Medians also have another important property.
Medians always divide each other into a 1:2 ratio, with the larger portion (2/3) toward the vertex and the smaller portion (1/3) toward the opposite side. |
The circumcenter, orthocenter, and centroid are always collinear. This line is called the Euler Line. |
In an isosceles triangle, all four of these points are collinear. In an equilateral triangle, all four are coincident.
An interesting construction is the Nine-point circle, the circle which goes through the midpoints of each side, the base of each altitude, as well as the midpoint of the segments between the orthocenter and each vertex.
Types of Quadrilaterals
Quadrilaterals can be classify by the lengths of their sides and how many pairs of sides are parallel. Familiarize yourself with the hierarchy chart at left. Remember that any property held by all figures of one type is held by all types connected below it.
Kites and Their Properties
A quadrilateral is a kite if and only if it has two distinct pair of consecutive sides congruent. |
This name should be familiar from the shape of the scientific instrument allegedly used by Benjamin Franklin, which are now used primarily as toys. An arrowhead, or the Startrek chevron is typically in the shape of a nonconvex kite. Another common name for this shape is dart. The vertices shared by the congruent sides are ends. The line containing the ends of a kite is a symmetry line for a kite. The symmetry line for a kite bisects the angles at the ends of the kite. The symmetry diagonal of a kite is a perpendicular bisector of the other diagonal.
Trapezoids and Their Properties
A quadrilateral is a trapezoid if and only if it has at least one pair of sides parallel. |
Some books define trapezoids as having exactly one pair of parallel sides, so beware. The parallel sides in a trapezoid are called bases. In a trapezoid, consecutive angles between pairs of parallel sides are supplementary.
A trapezoid is an isosceles trapezoid if and only if it has base angles which are congruent. |
If follows directly that the sides opposite the congruent angles in an isosceles trapezoid are congruent. In an isosceles trapezoid, the perpendicular bisector of one base is also the other base's perpendicular bisector. This bisector is thus also a line of symmetry.
One of my favorite questions uses an isosceles trapozoid. If we give the height of an isosceles trapozoid, as well as the length of its two bases, it is possible to find its perimeter. Example: suppose we know a certain isosceles trapozoid has bases of 10 and 16 with height 4. We know that right triangles are formed outside the rectangular region defined by the height and shorter base. The right triangles have a base of 3 and height of 4 thus hypotenuse of 5. Hence the perimeter is 36. The triangles formed need not be integer and we will continue with area in a later lesson.
Parallelograms and Their Properties
A quadrilateral is a parallelogram if and only if both pair of sides are parallel. |
Thus a parallelogram not only is a trapezoid, but also is an isosceles trapezoid.
A quadrilateral is a rectangle if and only if all angles are congruent. |
Thus a rectangle is an equiangular quadrilateral. Each angle is 90° or right. Since alternate interior angles are equal we can easily deduce that the sides are also parallel. Thus a rectangle is a parallelogram. Both perpendicular bisectors of the sides of a rectangle are symmetry lines.
A quadrilateral is a rhombus if and only if all sides are congruent. |
Thus a rhombus is an equilateral quadrilateral. A rhombus is both a kite and a parallelogram—both in two different ways. The diagonals of a rhombus are perpendicular bisectors of each other. In Kindergarten you likely learned of the shape called diamond. However, this shape is not very specific in that it could be either a (non-square) rhombus or a kite (the more traditional style carbonic crystals are cut). Thus we will not use this shape designation.
A quadrilateral is a square if and only if all sides and all angles are congruent. |
Please note that a square is a rectangle and a square is a rhombus. It is also a kite, parallelogram, and a trapezoid. Unfortunately, many of you learned otherwise as early as preschool and it takes five times as long to unteach something as to teach it! Also note that most quadrilaterals do not belong to any of these special catagories! There are other classifications. For example, quadrilaterals that can be inscribed in a circle are cyclic.
Rotational Symmetry
A plane figure F is rotation-symmetric if and only if there is a rotation (strictly) between 0° and 360° such that R(F)=F. The center of R is the center of symmetry for F. |
Notice that 0° and 360° are excluded (nothing happens).
A figure is said to have n-fold rotational symmetry if n rotations each of magnitude 360°/n produce an identical figure. The last rotation returns it to its original position. A figure can have rotational and reflective properties separately or together. A figure with reflective properties can have rotational symmetry if and only if the lines of symmetry intersect.
Going back to our letter examples, H, I, O, X had two intersecting lines of symmetric and thus both reflective and rotational symmetry. The letters N, S, and Z have rotational symmetry but not reflective symmetry. Certain letters (M & W, b & q, d & p, n & u, h & y? or 4??) rotate into the other of the pair!
Regular Polygons
A regular polygon is a convex polygon whose angles are all congruent and whose sides are all congruent. The regular polygons with three and four sides have the special names equilateral triangle and square, respectively. Otherwise, they are simply called regular pentagon, regular hexagon, etc., regular n-gon.
A polygon can be either equilateral or equiangular without, necessarily, being both (regular). The rectangle is an example of an equiangular quadrilateral, and the rhombus is an example of an equilateral quadrilateral. Neither has to be a square. However, for 3-gons, as stated in the theorem above, an equilateral triangle must also be equianglar.
In any regular polygon, a point termed the center is equidistant from all vertices. |
The distance from the midpoint of a side of a regular polygon to the center is the apothem. |
The apothem may also refer to the segment with length as describe above. The apothem is often used in some formulas. For example, the area of a regular polygon is A=asn/2, where a is the apothem, s is the length of each side, and n is the number of sides. Since p=sn or the perimeter is the number of sides times the length of each side, it can also be written A=ap/2.
Every regular n-gon has n lines of symmetry and n-fold rotational symmetry. The lines of symmetry are all either angle bisectors or perpendicular bisectors of each side (or both if n is odd). |
The measure of the internal angles of a regular n-gon can be found as follows. Triangulate the polygon by drawing the n-3 diagonals from one vertex to all other vertices. This divides the n-gon into n-2 triangles. The angles of any triangle sum to 180°. Thus the internal angles of any n-gon sum to (n-2)•180°. The internal angles of a regular n-gon will be (n-2)•180°/n. We gave the formula for the number of diagonals of an n-gon in lesson 2 as n•(n-3)/2. You might consider further how many different lengths these diagonals might be, especially in a regular polygon.
Schedules with Regular Polygons
Regular polygons can be used to form a round robin tournament schedule, a tournament where each team plays each other team once. There are other methods, but none quite so straight-forward. To use this algorithm, use a regular polygon with n or n-1 sides, whichever is odd. Place the teams at the vertices. If one is left over, place that team in the middle. Draw parallel diagonals connecting pairs of teams (a pairing). (These may also be chords on a circle.) The one left over will either get a bye (not play this round), if odd; or pairs with the team in the center, if even. Record these pairings as round one. Rotate the diagonals/chords to the next vertex. Record these pairings as round two. When you have gone halfway around, you will have all possible pairings, or a complete round robin tournament schedule.
Noether's Theorem
In 1905 Emily Noether published a theorem linking conservation laws in physics to symmetries in the mathematical formulation. The fact that energy and momentum are conserved are related to symmetries in time and space. Similarly, the conservation of angular momentum is related to rotational symmetry. Other symmetries are discrete instead of continuous, such as the three known as CPT: charge conjugation, parity, and time reversal. It was long believed that particles and antiparticles obeyed the same laws of physics (C), that physics in a mirror universe was the same (P), and that physics did not depend on a direction in time (T). We now know that P is maximally violated˜there are only left-handed neutrinos˜and C is also violated by the weak force. The combined operation of CP is violated about twice in every one thousand times. The standard model of physics assumes the combined operation of CPT is conserved. However, CP violation implies T violation which has not been found—yet.
BACK | HOMEWORK | ACTIVITY | CONTINUE |
e-mail: calkins@andrews.edu
voice/mail: 269 471-6629; BCM&S Smith Hall 106; Andrews University;
fax/classroom: 269 471-6646; Smith Hall 100; Berrien Springs, MI, 49104-0140
home: 269 473-2572; 610 N. Main St.; Berrien Springs, MI 49103-1013
URL:
Copyright ©2000-05, Keith G. Calkins. Revised on or after January 12, 2006.
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trigonometry - Using the identity $\tan\theta = \tan(\theta-180^\circ)$ to find values of $\theta$ such that $\tan\theta=\tan 20^\circ$ - Mathematics Stack Exchange
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Using the identity tan θ=tan(θ−180∘)tanθ=tan(θ−180∘) to find values of θ θ such that tan θ=tan 20∘tanθ=tan20∘
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This is question 1c of a list of related items.
State the value(s) of θ θ in the range 0∘0∘ to 360∘360∘ so that the following is true:
tan θ=tan 20∘tanθ=tan20∘
Here is the answer (from the list of answers):
θ=20∘;θ=180∘+20∘=200∘θ=20∘;θ=180∘+20∘=200∘
I am using the trig identity for tan, the one where
tan θ=tan(θ−180∘)tanθ=tan(θ−180∘)
If θ=20∘θ=20∘ for the question, then tan(θ−180∘)tan(θ−180∘) is tan(20∘−180∘)tan(20∘−180∘), which is tan(−160∘)tan(−160∘), which is taking me to a completely different direction than the solution.
I would appreciate it if someone could explain the steps of using this trig identity to determine which other angles have the same tan ratio as 20∘20∘.
trigonometry
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edited Aug 26, 2020 at 18:56
Blue
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asked Aug 26, 2020 at 18:47
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Hi Bushra, welcome to MSE! As a general tip, it is better to typeset the question in full than to link to images. In answer to your question, you do indeed want to use that the tan tan graph repeats every 180 deg, but plugging θ=20 θ=20 directly into that formula is going the wrong way. What if you set θ−180=20 θ−180=20?DanLewis3264 –DanLewis3264 2020-08-26 18:53:37 +00:00 Commented Aug 26, 2020 at 18:53
In which two quadrants, do both sin sin and cos cos have the same sign? Why do you think you are away from the answer? −θ−θ can be written as 2 π−θ 2 π−θ. So what will −160 0−160 0 be?Math Lover –Math Lover 2020-08-26 19:00:01 +00:00 Commented Aug 26, 2020 at 19:00
Hi @bounceback. Thank you for your help. Whilst I understood your comment and applied it to this question, I don't seem to understand how the solution was derived for question 1(f) using the same logic. It would be an incredible help if you explained the solution for this one. Thank you.Bee –Bee 2020-08-26 21:15:02 +00:00 Commented Aug 26, 2020 at 21:15
Use that tan tan is an odd function: tan(−x)=−tan(x)tan(−x)=−tan(x). Then argue as before DanLewis3264 –DanLewis3264 2020-08-26 23:15:51 +00:00 Commented Aug 26, 2020 at 23:15
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Hint:period of tangent function is 180∘,tan 20∘=tan(180+20)∘,tan(20−180)∘=tan(−160)∘,200∘180∘,tan20∘=tan(180+20)∘,tan(20−180)∘=tan(−160)∘,200∘ is anticlockwise rotation, −160∘−160∘ is clockwise rotation.
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edited Aug 27, 2020 at 18:24
amWhy
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answered Aug 26, 2020 at 18:59
Lion HeartLion Heart
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Hi. Thank you for your comment. So is the period of the sine and cosine function 180 also?Bee –Bee 2020-08-26 19:15:50 +00:00 Commented Aug 26, 2020 at 19:15
1 Period of sine and cosine is 360, tangent and cotangent is180 Lion Heart –Lion Heart 2020-08-26 19:17:39 +00:00 Commented Aug 26, 2020 at 19:17
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Since tan θ=tan(n π+θ)tanθ=tan(n π+θ), where n n is any integer and θ=π 9 θ=π 9 in our case. So, we have 0≤n π+π 9≤2 π⟹0≤(9 n+1)π≤18 π 0≤n π+π 9≤2 π⟹0≤(9 n+1)π≤18 π. Now you can clearly see that only n=0 n=0 and n=1 n=1 satisfy this bound, hence the only answers are θ=π 9,10 π 9 θ=π 9,10 π 9.
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answered Aug 26, 2020 at 19:05
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θ=20∘;θ=20∘±n⋅180∘;θ=20∘;θ=20∘±n⋅180∘;
Other angles are
=20∘,200∘,380∘,460∘=20∘,200∘,380∘,460∘
We can add or subtract n⋅π n⋅π ( n n positive or negative integer) to 20∘20∘ getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding 180∘180∘.
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answered Aug 26, 2020 at 19:12
NarasimhamNarasimham
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You suggest we use that
tan θ=tan(θ−180∘).tanθ=tan(θ−180∘).
Equivalently, if we set ϕ=θ−180∘ϕ=θ−180∘, the above equation reads
tan(ϕ+180∘)=tan ϕ.tan(ϕ+180∘)=tanϕ.
To summarize: the tangent function is 180∘180∘-periodic: it repeats every 180∘180∘. So the possible answers are …,−340∘,−160∘,20∘,200∘,380∘,……,−340∘,−160∘,20∘,200∘,380∘,…
Narasimham's answer shows how to express this in a single equation with n n varying, ABCD does the same in the language of radians (180∘=π 180∘=π radians).
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answered Aug 26, 2020 at 19:18
DanLewis3264DanLewis3264
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Contents
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1 Medical uses
1.1 Duration of action
2 Side effects
2.1 Cardiovascular
2.2 Bronchospasm
2.3 Seizures
3 Pharmacokinetics
3.1 Intramuscular function parameters
4 History
5 References
6 External links
Atracurium besilate
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From Wikipedia, the free encyclopedia
Chemical compound
Pharmaceutical compound
Atracurium besilate
| Clinical data |
| Trade names | Tracrium, Acurium |
| Other names | Atracurium besylate |
| AHFS/Drugs.com | Monograph |
| Pregnancycategory | C |
| Routes ofadministration | IV |
| ATC code | M03AC04 (WHO) |
| Legal status |
| Legal status | In general: ℞ (Prescription only) |
| Pharmacokinetic data |
| Bioavailability | 100% (IV) |
| Protein binding | 82% |
| Metabolism | Hofmann elimination (retro-Michael addition) and ester hydrolysis by nonspecific esterases |
| Elimination half-life | 17–21 minutes |
| Identifiers |
| IUPAC name 2,2'-{1,5-Pentanediylbis[oxy(3-oxo-3,1-propanediyl)]}bis[1-(3,4-dimethoxybenzyl)-6,7-dimethoxy-2-methyl-1,2,3,4-tetrahydroisoquinolinium] dibenzenesulfonate |
| CAS Number | 64228-81-5Y |
| PubChem CID | 47320 |
| DrugBank | DB00732Y |
| ChemSpider | 43068Y |
| UNII | 40AX66P76P |
| KEGG | D00758 |
| ChEBI | CHEBI:2915Y |
| ChEMBL | ChEMBL1200527Y |
| CompTox Dashboard (EPA) | DTXSID6022630 |
| ECHA InfoCard | 100.058.840 |
| Chemical and physical data |
| Formula | C65H82N2O18S2 |
| Molar mass | 1243.49 g·mol−1 |
| 3D model (JSmol) | Interactive image |
| Melting point | 85 to 90 °C (185 to 194 °F) |
| SMILES C[N+]1(CCc2cc(c(cc2C1Cc3ccc(c(c3)OC)OC)OC)OC)CCC(=O)OCCCCCOC(=O)CC[N+]4(CCc5cc(c(cc5C4Cc6ccc(c(c6)OC)OC)OC)OC)C.c1ccc(cc1)S(=O)(=O)[O-].c1ccc(cc1)S(=O)(=O)[O-] |
| InChI InChI=1S/C53H72N2O12.2C6H6O3S/c1-54(22-18-38-32-48(62-7)50(64-9)34-40(38)42(54)28-36-14-16-44(58-3)46(30-36)60-5)24-20-52(56)66-26-12-11-13-27-67-53(57)21-25-55(2)23-19-39-33-49(63-8)51(65-10)35-41(39)43(55)29-37-15-17-45(59-4)47(31-37)61-6;27-10(8,9)6-4-2-1-3-5-6/h14-17,30-35,42-43H,11-13,18-29H2,1-10H3;21-5H,(H,7,8,9)/q+2;;/p-2Y Key:XXZSQOVSEBAPGS-UHFFFAOYSA-LY |
| NY (what is this?) (verify) |
Atracurium besilate, also known as atracurium besylate, is a medication used in addition to other medications to provide skeletal muscle relaxation during surgery or mechanical ventilation. It can also be used to help with endotracheal intubation but suxamethonium (succinylcholine) is generally preferred if this needs to be done quickly. It is given by injection into a vein. Effects are greatest at about 4 minutes and last for up to an hour.
Common side effects include flushing of the skin and low blood pressure. Serious side effects may include allergic reactions; however, it has not been associated with malignant hyperthermia. Prolonged paralysis may occur in people with conditions like myasthenia gravis. It is unclear if use in pregnancy is safe for the baby. Atracurium is in the neuromuscular-blocker family of medications and is of the non-depolarizing type. It works by blocking the action of acetylcholine on skeletal muscles.
Atracurium was approved for medical use in the United States in 1983. It is on the World Health Organization's List of Essential Medicines. Atracurium is available as a generic medication.
Medical uses
[edit]
Atracurium is a medication used in addition to other medications in to provide skeletal muscle relaxation during surgery or mechanical ventilation. It can be used to help with endotracheal intubation but takes up to 2.5 minutes to result in appropriate intubating conditions.
Duration of action
[edit]
Neuromuscular-blocking agents can be classified in accordance to their duration of pharmacological action, defined as follows:
Classification of neuromuscular-blocking agents by duration of pharmacological action (minutes)
| Parameter | Ultra-short Duration | Short Duration | Intermediate Duration | Long Duration |
| Clinical Duration (Time from injection to T25% recovery) | 6-8 | 12-20 | 30-45 | >60 |
| Recovery Time (Time from injection to T95% recovery) | <15 | 25-30 | 50-70 | 90-180 |
| Recovery Index (T25%-T75% recovery slope) | 2-3 | 6 | 10-15 | >30 |
Side effects
[edit]
Cardiovascular
[edit]
| |
| This section needs additional citations for verification. Please help improve this article by adding citations to reliable sources in this section. Unsourced material may be challenged and removed. (December 2022) (Learn how and when to remove this message) |
The tetrahydroisoquinolinium class of neuromuscular blocking agents, in general, is associated with histamine release upon rapid administration of a bolus intravenous injection. There are some exceptions to this rule; cisatracurium (Nimbex), for example, is one such agent that does not elicit histamine release even up to 5×ED95 doses. The liberation of histamine is a dose-dependent phenomenon such that, with increasing doses administered at the same rate, there is a greater propensity for eliciting histamine release and its ensuing sequelae. Most commonly, the histamine release following administration of these agents is associated with observable cutaneous flushing (facial face and arms, commonly), hypotension and a consequent reflex tachycardia. These sequelae are very transient effects: the total duration of the cardiovascular effects is no more than one to two minutes, while the facial flush may take around 3–4 minutes to dissipate. Because these effects are so transient, there is no reason to administer adjunctive therapy to ameliorate either the cutaneous or the cardiovascular effects.
Bronchospasm
[edit]
Bronchospasm has been reported on occasion with the use of atracurium. However, this particular undesirable effect does not appear to be observed nearly as often as that seen with rapacuronium, which led to the latter's withdrawal of approval for clinical use worldwide.
The issue of bronchospasm acquired prominence in the neuromuscular-blocking agents arena after the withdrawal from clinical use of rapacuronium (Raplon - a steroidal neuromuscular-blocking agent marketed by Organon) in 2001 after several serious events of bronchospasm, including five unexplained fatalities, following its administration. Bronchospasm was not an unknown phenomenon prior to rapacuronium: occasional reports of bronchospasm have been noted also with the prototypical agents, tubocurarine and succinylcholine, as well as alcuronium, pancuronium, vecuronium, and gallamine.
Seizures
[edit]
Seizures rarely occur.
Because atracurium undergoes Hofmann elimination as a primary route of chemodegradation, one of the major metabolites from this process is laudanosine, a tertiary amino alkaloid reported to be a modest CNS stimulant with epileptogenic activity and cardiovascular effects such a hypotension and bradycardia. As part of the then fierce marketing battle between the competing pharmaceutical companies (Burroughs Wellcome Co. and Organon, Inc.) with their respective products, erroneous information was quickly and subtly disseminated very shortly after the clinical introduction of atracurium that the clinical use of atracurium was likely to result in a terrible tragedy because of the significant clinical hazard by way of frank seizures induced by the laudanosine by-product - the posited hypothesis being that the laudanosine produced from the chemodegradation of parent atracurium would cross the blood–brain barrier in sufficiently high enough concentrations that lead to epileptogenic foci. Fortunately, both for the public and for atracurium, rapid initial investigations irrefutably failed to find any overt or EEG evidence for a connection between atracurium administration and epileptogenic activity. Indeed, because laudanosine is cleared primarily via renal excretion, a cat study modelling anephric patients went so far as to corroborate that EEG changes, when observed, were evident only at plasma concentrations 8 to 10 times greater than those observed in humans during infusions of atracurium. Thus, the cat study predicted that, following atracurium administration in an anephric patient, laudanosine accumulation and related CNS or cardiovascular toxicity were unlikely - a prediction that correlated very well with a study in patients with kidney failure and undergoing cadaveric renal transplantation. Furthermore, almost a decade later, work by Cardone et al.. confirmed that, in fact, it is the steroidal neuromuscular-blocking agents pancuronium and vecuronium that, when introduced directly into the CNS, were likely to cause acute excitement and seizures, owing to accumulation of cytosolic calcium caused by activation of acetylcholine receptor ion channels. Unlike the two steroidal agents, neither atracurium nor laudanosine caused such accumulation of intracellular calcium. Just over two decades later with availability of atracurium, there is little doubt that laudanosine accumulation and related toxicity will likely never be seen with the doses of atracurium that are generally used.
Laudanosine is also a metabolite of cisatracurium that, because of its identical structure to atracurium, undergoes chemodegradation via Hofmann elimination in vivo. Plasma concentrations of laudanosine generated are lower when cisatracurium is used.
Pharmacokinetics
[edit]
Atracurium is susceptible to degradation by Hofmann elimination and ester hydrolysis as components of the in vivo metabolic processes. The initial in vitro studies appeared to indicate a major role for ester hydrolysis but, with accumulation of clinical data over time, the preponderance of evidence indicated that Hofmann elimination at physiological pH is the major degradation pathway vindicating the premise for the design of atracurium to undergo an organ-independent metabolism.
Hofmann elimination is a temperature- and pH-dependent process, and therefore atracurium's rate of degradation in vivo is highly influenced by body pH and temperature: An increase in body pH favors the elimination process, whereas a decrease in temperature slows down the process. Otherwise, the breakdown process is unaffected by the level of plasma esterase activity, obesity, age, or by the status of renal or hepatic function. On the other hand, excretion of the metabolite, laudanosine, and, to a small extent, atracurium itself is dependent on hepatic and renal functions that tend to be less efficient in the elderly population. The pharmaceutical presentation is a mixture of all ten possible stereoisomers. Although there are four stereocentres, which could give 16 structures, there is a plane of symmetry running through the centre of the diester bridge, and so 6 meso structures (structures that can be superimposed by having the opposite configuration then 180° rotation) are formed. This reduces the number from sixteen to ten. There are three cis-cis isomers (an enantiomeric pair and a meso structure), four cis-trans isomers (two enantiomeric pairs), and three trans-trans isomers (an enantiomeric pair and a meso structure). The proportions of cis−cis, cis−trans, and trans−trans isomers are in the ratio of 10.5 :6.2 :1. [cis-cis isomers ≈ 58% cis-trans isomers ≈ 36% trans-trans isomers ≈ 6%]. One of the three cis-cis structures is marketed as a single-isomer preparation, cisatracurium (trade name Nimbex); it has the configuration 1R, 2R, 1′R, 2′R at the four stereocentres. The beta-blocking drug nebivolol has ten similar structures with 4 stereocentres and a plane of symmetry, but only two are presented in the pharmaceutical preparation.
Intramuscular function parameters
[edit]
ED95: the dose of any given intramuscular-blocking agent required to produce 95% suppression of muscle twitch (e.g., the abductor pollicis) response with balanced anesthesia
Clinical duration: difference in time between time of injection and time to 25% recovery from neuromuscular block
Train-of-four (TOF) response: stimulated muscle twitch response in trains of four when stimuli are applied in a burst of four as opposed to a single stimulus, equal depression in depolarizing and fading response with non-depolarizing blocker.
25%-75% recovery index: an indicator of the rate of skeletal muscle recovery - essentially, the difference in time between the time to recovery to 25% and time to recovery to 75% of baseline value
T4:T1 ≥ 0.7: a 70% ratio of the fourth twitch to the first twitch in a TOF - provides a measure of the recovery of neuromuscular function
T4:T1 ≥ 0.9: a 90% ratio of the fourth twitch to the first twitch in a TOF - provides a measure of the full recovery of neuromuscular function
History
[edit]
Atracurium besilate was first made in 1974 by George H. Dewar, a pharmacist and a medicinal chemistry doctoral candidate in John B. Stenlake's medicinal chemistry research group in the Department of Pharmacy at Strathclyde University, Scotland. Dewar first named this compound "33A74" before its eventual emergence in the clinic as atracurium. Atracurium was the culmination of a rational approach to drug design to produce the first non-depolarizing non-steroidal skeletal muscle relaxant that undergoes chemodegradation in vivo. The term chemodegradation was coined by Roger D. Waigh, Ph.D., also a pharmacist and a postdoctoral researcher in Stenlake's research group. Atracurium was licensed by Strathclyde University to the Wellcome Foundation UK, which developed the drug (then known as BW 33A) and its introduction to first human trials in 1979, and then eventually to its first introduction (as a mixture of all ten stereoisomers) into clinical anesthetic practice in the UK, in 1983, under the tradename of Tracrium.
The premise to the design of atracurium and several of its congeners stemmed from the knowledge that a bis-quaternary structure is essential for neuromuscular-blocking activity: ideally, therefore, a chemical entity devoid of this bis-quaternary structure via susceptibility to inactive breakdown products by enzymic-independent processes would prove to be invaluable in the clinical use of a drug with a predictable onset and duration of action. Hofmann elimination provided precisely this basis: It is a chemical process in which a suitably activated quaternary ammonium compound can be degraded by the mildly alkaline conditions present at physiological pH and temperature. In effect, Hofmann elimination is a retro-Michael addition chemical process. It is important to note here that the physiological process of Hofmann elimination differs from the non-physiological Hofmann degradation process: the latter is a chemical reaction in which a quaternary ammonium hydroxide solid salt is heated to 100 °C, or an aqueous solution of the salt is boiled. Regardless of which Hofmann process is referenced, the end-products in both situations will be the same: an alkene and a tertiary amine.
The approach to utilizing Hofmann elimination as a means to promoting biodegradation had its roots in much earlier observations that the quaternary alkaloid petaline (obtained from the Lebanese plant Leontice leontopetalum) readily underwent facile Hofmann elimination to a tertiary amine called leonticine upon passage through a basic (as opposed to an acidic) ion-exchange resin. Stenlake's research group advanced this concept by systematically synthesizing numerous quaternary ammonium β-aminoesters and β-aminoketones and evaluated them for skeletal muscle relaxant activity: one of these compounds, initially labelled as 33A74, eventually led to further clinical development, and came to be known as atracurium.
Atracurium played a part in triggering the 2019 Samoa measles outbreak.
References
[edit]
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^ a b Kent AP, Parker CJ, Hunter JM (1989). "Pharmacokinetics of atracurium and laudanosine in the elderly". British Journal of Anaesthesiology. 63 (6): 661–666. doi:10.1093/bja/63.6.661. PMID 2611066.
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^ Parker CJ, Jones JE, Hunter JM (1988). "Disposition of infusions of atracurium and its metabolite, laudanosine, in patients in renal and respiratory failure in an ITU". British Journal of Anaesthesiology. 61 (5): 531–540. doi:10.1093/bja/61.5.531. PMID 3207525.
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^ Stenlake JB, Waigh RD, Dewar GH, Dhar NC, Hughes R, Chapple DJ, et al. (1984). "Biodegradable neuromuscular blocking agents. Part 6. Stereochemical studies on atracurium and related polyalkylene di-esters". European Journal of Medicinal Chemistry. 19 (5): 441–450.
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External links
[edit]
Neuromuscular+blocking+agents at the U.S. National Library of Medicine Medical Subject Headings (MeSH)
| v t e Skeletal muscle relaxants (M03) |
| Peripherally acting(primarily antinicotinic,NMJ block) | | | | | | | | --- --- --- | | Non-depolarizing | | | | --- | | Curare alkaloids | Alcuronium Dimethyltubocurarine Tubocurarine | | 4° ammonium agents | ultra-short duration: Gantacurium short duration: Rapacuronium Mivacurium Chandonium intermediate duration: Atracurium Cisatracurium Fazadinium Rocuronium Vecuronium long duration: Doxacurium Dimethyltubocurarine Pancuronium Pipecuronium Laudexium Gallamine unsorted: Hexafluronium (Hexafluorenium) | | | Depolarizing | Choline derivatives: Suxamethonium (Succinylcholine) Polyalkylene derivatives: Hexamethonium | | ACh release inhibitors | Botulinum toxin | |
| Centrally acting | | | | --- | | Carbamic acid esters | Carisoprodol Cyclarbamate Difebarbamate Febarbamate Meprobamate Phenprobamate Styramate Tybamate | | Benzodiazepines | Bromazepam Diazepam Clonazepam Flunitrazepam Lorazepam Nitrazepam Temazepam Tetrazepam | | Nonbenzodiazepines | | | Thienodiazepines | | | Quinazolines | | | Anticholinergics (Antimuscarinics) | Cyclobenzaprine Orphenadrine | | Other | Arbaclofen placarbil Baclofen Chlormezanone Chlorphenesin Chlorzoxazone Eperisone Fenyramidol Flopropione Gabapentin GHB Inaperisone Lanperisone Mephenesin Mephenoxalone Metaxalone Methocarbamol Phenibut Pregabalin Pridinol Promoxolane Quinine Silperisone Thiocolchicoside Tizanidine Tolperisone Zoxazolamine | |
| Directly acting | |
| v t e Nicotinic acetylcholine receptor modulators |
| nAChRsTooltip Nicotinic acetylcholine receptors | | | | --- | | Agonists(and PAMsTooltip positive allosteric modulators) | 5-HIAA 6-Chloronicotine A-84,543 A-366,833 A-582,941 A-867,744 ABT-202 ABT-418 ABT-560 ABT-894 Acetylcholine Altinicline Anabasine Anatabine Anatoxin-a AR-R17779 Bephenium hydroxynaphthoate Butinoline Butyrylcholine Carbachol Choline Choline m-bromophenyl ether Cotinine Cytisine Decamethonium Desformylflustrabromine Dianicline Dimethylphenylpiperazinium Epibatidine Epiboxidine Ethanol (alcohol) Ethoxysebacylcholine EVP-4473 EVP-6124 Galantamine GTS-21 Ispronicline Ivermectin JNJ-39393406 Levamisole Lobeline MEM-63,908 (RG-3487) Morantel Nicotine (tobacco) NS-1738 PHA-543,613 PHA-709,829 PNU-120,596 PNU-282,987 Pozanicline Pyrantel Rivanicline RJR-2429 Sazetidine A SB-206553 Sebacylcholine SIB-1508Y SIB-1553A SSR-180,711 Suberyldicholine Suxamethonium (succinylcholine) Suxethonium (succinyldicholine) TC-1698 TC-1734 TC-1827 TC-2216 TC-5214 TC-5619 TC-6683 Tebanicline Tribendimidine Tropisetron UB-165 Varenicline WAY-317,538 XY-4083 | | Antagonists(and NAMsTooltip negative allosteric modulators) | 18-MAC 18-MC α-Neurotoxins (e.g., α-bungarotoxin, α-cobratoxin, α-conotoxin, many others) ABT-126 Alcuronium Allopregnanolone Amantadine Anatruxonium AQW051 Atracurium Barbiturates (e.g., pentobarbital, sodium thiopental) BNC-210 Bungarotoxins (e.g., α-bungarotoxin, κ-bungarotoxin) Bupropion BW284C51 BW-A444 Candocuronium iodide (chandonium iodide) Chlorisondamine Cisatracurium Coclaurine Coronaridine Curare Cyclopropane Dacuronium bromide Decamethonium Dehydronorketamine Desflurane Dextromethorphan Dextropropoxyphene Dextrorphan Diadonium DHβE Dihydrochandonium Dimethyltubocurarine (metocurine) Dioscorine Dipyrandium Dizocilpine (MK-801) Doxacurium Encenicline Enflurane Erythravine Esketamine Fazadinium Gallamine Gantacurium chloride Halothane Hexafluronium Hexamethonium (benzohexonium) Hydroxybupropion Hydroxynorketamine Ibogaine Isoflurane Ketamine Kynurenic acid Laudanosine Laudexium (laudolissin) Levacetylmethadol Levomethadone Malouetine ME-18-MC Mecamylamine Memantine Methadone Methorphan (racemethorphan) Methyllycaconitine Metocurine Mivacurium Morphanol (racemorphan) Neramexane Nitrous oxide Norketamine Pancuronium bromide Pempidine Pentamine Pentolinium Phencyclidine Pipecuronium bromide Progesterone Promegestone Radafaxine Rapacuronium bromide Reboxetine Rocuronium bromide Sevoflurane Stercuronium iodide Surugatoxin Thiocolchicoside Threohydrobupropion Toxiferine Tramadol Trimetaphan camsilate (trimethaphan camsylate) Tropeinium Tubocurarine Vanoxerine Vecuronium bromide Xenon | |
| Precursors(and prodrugs) | Acetyl-coA Adafenoxate Choline (lecithin) Citicoline Cyprodenate Dimethylethanolamine Glycerophosphocholine Meclofenoxate (centrophenoxine) Phosphatidylcholine Phosphatidylethanolamine Phosphorylcholine Pirisudanol |
| See also Receptor/signaling modulators Muscarinic acetylcholine receptor modulators Acetylcholine metabolism/transport modulators |
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1: The Dawn of the Quantum Theory
CH-331 Text
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Thu, 05 Dec 2019 18:30:00 GMT
1.8: The Bohr Theory of the Hydrogen Atom
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Contents
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Campus Bookshelves
Bethune-Cookman University
B-CU:CH-331 Physical Chemistry I
CH-331 Text
CH-331 Text
1: The Dawn of the Quantum Theory
1.8: The Bohr Theory of the Hydrogen Atom
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CH-331 Text
Front Matter
1: The Dawn of the Quantum Theory
1.1: Blackbody Radiation Cannot Be Explained Classically
1.2: Quantum Hypothesis used for Blackbody Radiation Law
1.3: Photoelectric Effect Explained with Quantum Hypothesis
1.4: The Hydrogen Atomic Spectrum
1.5: The Rydberg Formula and the Hydrogen Atomic Spectrum
1.6: Matter Has Wavelike Properties
1.7: de Broglie Waves can be Experimentally Observed
1.8: The Bohr Theory of the Hydrogen Atom
1.9: The Heisenberg Uncertainty Principle
1.E: The Dawn of the Quantum Theory (Exercises)
2: The Classical Wave Equation
3: The Schrödinger Equation & a Particle in a Box
4: Postulates and Principles of Quantum Mechanics
5: The Harmonic Oscillator and the Rigid Rotor
6: The Hydrogen Atom
7: Approximation Methods
8: Multielectron Atoms
9: The Chemical Bond: Diatomic Molecules
MathChapters
Back Matter
1.8: The Bohr Theory of the Hydrogen Atom
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: Dec 5, 2019
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1.7: de Broglie Waves can be Experimentally Observed
1.9: The Heisenberg Uncertainty Principle
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( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
Learning Objectives
Rutherford's Failed Planetary Atom
Conservative Forces can be explained with Potentials
The Bohr Model
Exercise 1.8.1
Exercise 1.8.2
: Rydberg states
The Wave Argument for Quantization
Derivation of the Rydberg Equation from Bohr Model
Exercise 1.8.3
Contributors and Attributions
Learning Objectives
Introduce the fundamentals behind the Bohr Atom and demonstrate it can predict the Rydberg's equation for the atomic spectrum of hydrogen
Rutherford's Failed Planetary Atom
Ernest Rutherford had proposed a model of atoms based on the αα-particle scattering experiments of Hans Geiger and Ernest Marsden. In these experiments helium nuclei (αα-particles) were shot at thin gold metal foils. Most of the particles were not scattered; they passed unchanged through the thin metal foil. Some of the few that were scattered were scattered in the backward direction; i.e. they recoiled. This backward scattering requires that the foil contain heavy particles. When an αα-particle hits one of these heavy particles it simply recoils backward, just like a ball thrown at a brick wall. Since most of the αα-particles don’t get scattered, the heavy particles (the nuclei of the atoms) must occupy only a very small region of the total space of the atom. Most of the space must be empty or occupied by very low-mass particles. These low-mass particles are the electrons that surround the nucleus.
There are some basic problems with the Rutherford model. The Coulomb force that exists between oppositely charge particles means that a positive nucleus and negative electrons should attract each other, and the atom should collapse. To prevent the collapse, the electron was postulated to be orbiting the positive nucleus. The Coulomb force (discussed below) is used to change the direction of the velocity, just as a string pulls a ball in a circular orbit around your head or the gravitational force holds the moon in orbit around the Earth. The origin for this hypothesis that suggests this perspective is plausible is the similarity of gravity and Coulombic interactions. The expression for the force of gravity between two masses (Newton's Law of gravity) is
Fgravity∝m1m2r2
Fgravity∝m1m2r2(1.8.1)
with m1m1 and m2m2 representing the mass of object 1 and 2, respectively and rr representing the distance between the objects centers
The expression for the Coulomb force between two charged species is
FCoulomb∝Q1Q2r2
FCoulomb∝Q1Q2r2(1.8.2)
with Q1Q1 and Q2Q2 representing the charge of object 1 and 2, respectively and rr representing the distance between the objects centers.
However, this analogy has a problem too. An electron going around in a circle is constantly being accelerated because its velocity vector is changing. A charged particle that is being accelerated emits radiation. This property is essentially how a radio transmitter works. A power supply drives electrons up and down a wire and thus transmits energy (electromagnetic radiation) that your radio receiver picks up. The radio then plays the music for you that is encoded in the waveform of the radiated energy.
Figure 1.8.1
: The classical death spiral of an electron around a nucleus. (CC BY-NC; Ümit Kaya via LibreTexts)
If the orbiting electron is generating radiation, it is losing energy. If an orbiting particle loses energy, the radius of the orbit decreases. To conserve angular momentum, the frequency of the orbiting electron increases. The frequency increases continuously as the electron collapses toward the nucleus. Since the frequency of the rotating electron and the frequency of the radiation that is emitted are the same, both change continuously to produce a continuous spectrum and not the observed discrete lines. Furthermore, if one calculates how long it takes for this collapse to occur, one finds that it takes about 10‑1110‑11 seconds. This means that nothing in the world based on the structure of atoms could exist for longer than about 10−1110−11 seconds. Clearly something is terribly wrong with this classical picture, which means that something was missing at that time from the known laws of physics.
Conservative Forces can be explained with Potentials
A conservative force is dependent only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point. When an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken. The potential can be constructed as simple derivatives for 1-D forces:
F=−dVdx
F=−dVdx
or as gradients in 3-D forces
F=−∇V
F=−∇V
where ∇∇ is the vector of partial derivatives
∇=(∂∂x,∂∂y,∂∂z)
∇=(∂∂x,∂∂y,∂∂z)
The most familiar conservative forces are gravity and Coulombic forces.
The Coulomb force law (Equation 1.8.21.8.2) comes from the corresponding Coulomb potential (sometimes call electrostatic potential)
V(r)=kQ1Q2r
V(r)=kQ1Q2r(1.8.3)
and it can be easily verified that the Coulombic force from this interaction (F(r)F(r)) is
F(r)=−dVdr
F(r)=−dVdr(1.8.4)
As rr is varied, the energy will change, so that we have an example of a potential energy curve V(r)V(r) (Figure 1.8.2;left1.8.2;left). If Q1Q1 and Q2Q2 are the same sign, then the curve which is a purely repulsive potential, i.e., the energy increases monotonically as the charges are brought together and decreases monotonically as they are separated. From this, it is easy to see that like charges (charges of the same sign) repel each other.
Figure 1.8.2
: Potential energy curve for the Coulomb interactions between two charges of same sign (left) and opposite signs (right). (CC BY-NC; Ümit Kaya via LibreTexts)
If the charges are of opposite sign, then the curve appears roughly Figure 1.8.2;right1.8.2;right and this is a purely attractive potential. Thus, the energy decreases as the charges are brought together, implying that opposite charges attract
The Bohr Model
It is observed that line spectra discussed in the previous sections show that hydrogen atoms absorb and emit light at only discrete wavelengths. This observation is connected to the discrete nature of the allowed energies of a quantum mechanical system. Quantum mechanics postulates that, in contrast to classical mechanics, the energy of a system can only take on certain discrete values. This leaves us with the question: How do we determine what these allowed discrete energy values are? After all, it seems that Planck's formula for the allowed energies came out of nowhere.
The model we will describe here, due to Niels Bohr in 1913, is an early attempt to predict the allowed energies for single-electron atoms such as HH, He+He+, Li2+Li2+, Be3+Be3+, etc. Although Bohr's reasoning relies on classical concepts and hence, is not a correct explanation, the reasoning is interesting, and so we examine this model for its historical significance.
Figure 1.8.3
: Bohr atom with an electron revolving around a fixed nucleus. (CC BY-NC; Ümit Kaya via LibreTexts)
Consider a nucleus with charge +Ze+Ze and one electron orbiting the nucleus. In this analysis, we will use another representation of the constant kk in Coulomb's law (Equation 1.8.31.8.3), which is more commonly represented in the form:
k=14πϵ0
k=14πϵ0(1.8.5)
where ϵ0ϵ0 is known as the permittivity of free spacewith the numerical value ϵ0=8.8541878×10−12 C2J−1m−1ϵ0=8.8541878×10−12 C2J−1m−1.
The total energy of the electron (the nucleus is assumed to be fixed in space at the origin) is the sum of kinetic and potential energies:
Etotal=p22mekinetic energy−Ze24πϵ0rpotential energy
Etotal=p22mekinetic energy−Ze24πϵ0rpotential energy
The force on the electron is
→F=−Ze24πϵ0r3r
F⃗ =−Ze24πϵ0r3r
and its magnitude is
F=|→F|=Ze24πϵ0r3|r|=Ze24πϵ0r2
F=|F⃗ |=Ze24πϵ0r3|r|=Ze24πϵ0r2
since →F=me→aF⃗ =mea⃗ , the magnitude, it follows that |→F|=me|→a||F⃗ |=me|a⃗ |. If we assume that the orbit is circular, then the acceleration is purely centripetal, so
|a|=v2r
|a|=v2r
where vv is the velocity of the electron. Equating force |F||F| to me|a|me|a|, we obtain
Ze24πϵ0r2=mev2r
Ze24πϵ0r2=mev2r
or
Ze24πϵ0=mev2r
Ze24πϵ0=mev2r
or
Ze2mer4πϵ0=(mevr)2
Ze2mer4πϵ0=(mevr)2(1.8.6)
The reason for writing the equation this way is that the quantity mevrmevr is the classical orbital angular momentum of the electron. Bohr was familiar with Maxwell's theory of classical electromagnetism and knew that in a classical theory, the orbiting electron should radiate energy away and eventually collapse into the nucleus (Figure 1.8.1
). He circumvented this problem by following Planck's idea underlying blackbody radiation and positing that the orbital angular momentum mevrmevr of the electron could only take on specific values
mevr=nℏ
mevr=nℏ(1.8.7)
with n=1,2,3,...n=1,2,3,....
Note that the electron must be in motion, so n=0n=0 is not allowed.
Substituting Equation 1.8.71.8.7 into the Equation 1.8.61.8.6, we find
Ze2mer4πϵ0=n2(ℏ)2
Ze2mer4πϵ0=n2(ℏ)2(1.8.8)
Equation 1.8.8 implies that orbits could only have certain allowed radii
rn=4πϵ0ℏ2Ze2men2=a0Zn2
with n=1,2,3,.... The collection of constants has been defined to be a0
a0=4πϵ0ℏ2e2me
a quantity that is known as the Bohr radius.
We can also calculate the allowed momenta since mevr=nℏ, and p=mev. Thus,
pnrn=nℏpn=nℏrn=ℏZa0n=Ze2me4πϵ0ℏn
From pn and rn, we can calculate the allowed energies from
En=p2n2me−Ze24πϵ0rn
Substituting in the expressions for pn and rn and simplifying gives
En=−Z2e4me32π2ϵ20ℏ21n2=−e4me8ϵ20h2Z2n2
We can redefine a new energy scale by defining the Rydbergas
1 Ry=e4me8ϵ20h2=2.18×10−18 J.
and this simplifies the allowed energies predicted by the Bohr model (Equation 1.8.11) as
En=−(2.18×10−18)Z2n2 J=−Z2n2 Ry
Hence, the energy of the electron in an atom also is quantized. Equation 1.8.12 gives the energies of the electronic states of the hydrogen atom. It is very useful in analyzing spectra to represent these energies graphically in an energy-level diagram. An energy-level diagram has energy plotted on the vertical axis with a horizontal line drawn to locate each energy level (Figure 1.8.4
).
Figure 1.8.4
: Energy levels predicted by the Bohr model of hydrogen (Z=1). (CC BY-NC; Ümit Kaya via LibreTexts)
These turn out to be the correct energy levels, apart from small corrections that cannot be accounted for in this pseudo-classical treatment. Despite the fact that the energies are essentially correct, the Bohr model masks the true quantum nature of the electron, which only emerges from a fully quantum mechanical analysis.
Exercise 1.8.1
Calculate a value for the Bohr radius using Equation 1.8.8 to check that this equation is consistent with the value 52.9 pm. What would the radius be for n=1 in the Li2+ ion?
Answer
: Starting from Equation 1.8.8 and solving for r:
Ze2mer4πϵ0=n2ℏ2r=4n2ℏ2πϵ0Ze2me
with
- e is the fundamental charge: e=1.60217662×10−19C2
- me is the mass of an electron: me=9.10938356×10−31kg
- ϵo is the permittivity of free space: ϵo=8.854×10−12C2N−1m−2
- ℏ is the reduced planks constant: ℏ=1.0546×10−34m2kg/s
For the ground-state of the hydrogen atom: Z=1 and n=1.
r=4ℏ2πϵ0e2me=4(1.0546×10−34m2kg/s)2×π×8.854×10−12C2N−1m−2(1.60217662×10−19C)2(9.10938356×10−31kg)=5.29×10−11m=52.9pm
For the ground-state of the lithium +2 ion: Z=3 and n=1
r=4ℏ2πϵ03e2me=4(1.0546×10−34m2kg/s)2×π×8.854×10−12C2N−1m−23(1.60217662×10−19C)2(9.10938356×10−31kg)=1.76×10−11m=17.6pm
As expected, the Li2+ has a smaller radius than the H atoms because of the increased nuclear charge.
Exercise 1.8.2 : Rydberg states
How do the radii of the hydrogen orbits vary with n? Prepare a graph showing r as a function of n. States of hydrogen atoms with n=200 have been prepared (called Rydberg states). What is the diameter of the atoms in these states?
Answer
: This is a straightforward application of Equation of 1.8.10. The hydrogen atom has only certain allowable radii and these radii can be predicted from the equation that relates them with each n. Note that the electron must be in motion so n=0 is not allowed.
Scatter plot with text Processing Speed on x-axis and Requests per Second on y-axis. Red data points form a curve that rises steeply, indicating an exponential relationship.
This plot shows the relationship of radius as a function of n. Note that at n=1 the radius is not zero. (CC BY-NC; Ümit Kaya via LibreTexts)
4πϵ0=1.113×10−10C2J−1m−1 and ℏ=1.054×10−34Js, also knowing
e=1.602×10−19C with me=9.109×10−31kg
and Z is the nuclear charge, we use this equation directly. A simplification can be made by taking advantage of the fact that
a0=4πϵ0ℏ2e2me
resulting in
rn=a0Zn2
where a0=5.292×10−11m which is the Bohr Radius.
Suppose we want to find the radius where n=200.n2=40000 so plugging in directly we have
rn=(5.292×10−11)(1)(40000)=2.117×10−6m
for the radius of a hydrogen atom with an electron excited to the n=200 state. The diameter is then 4.234×10−6m.
The Wave Argument for Quantization
The above discussion is based off of a classical picture of an orbiting electron with the quantization from the angular momentum (Equation 1.8.7) requirement lifted from Planck's quantization arguments. Hence, only allows certain trajectories are stable (with differing radii). However, as discussed previously, the electron will have a wavelike property also with a de Broglie wavelength λ
λ=hp
Hence, a larger momentum p implies a shorter wavelength. That means as n increases (Equation 1.8.12), the wavelength must also increase; this is a common feature in quantum mechanics and will be often observed. In the Bohr atom, the circular symmetry and the wave property of the electron requires that the electron waves have an integer number of wavelengths (Figure 1.8.1A). If not, then the waves will overlap imperfectly and cancel out (i.e., the electron will cease to exist) as demonstrated in Figure 1.8.1B.
Figure 1.8.5
: Waves on a string have a wavelength related to the length of the string, allowing them to interfere constructively. (A) If we imagine the string bent into a closed circle, we get a rough idea of how electrons in circular orbits can interfere constructively. (B) If the wavelength does not fit into the circumference, the electron interferes destructively; it cannot exist in such an orbit. (CC BY-NC; Ümit Kaya via LibreTexts)
A more detailed discussion of the effect of electron waves in atoms will be discuss in the following chapters.
Derivation of the Rydberg Equation from Bohr Model
Given a prediction of the allowed energies of a system, how could we go about verifying them? The general experimental technique known as spectroscopy permits us to probe the various differences between the allowed energies. Thus, if the prediction of the actual energies, themselves, is correct, we should also be able to predict these differences. Let us assume that we are able to place the electron in Bohr's hydrogen atom into an energy state En for n>1, i.e. one of its so-called excited states. The electron will rapidly return to its lowest energy state, known as the ground state and, in doing so, emit light. The energy carried away by the light is determined by the condition that the total energy is conserved (Figure 1.8.6
).
Figure 1.8.6
: A simple illustration of Bohr's model of the atom, with an electron making quantum leaps. (CC BY-SA 3.0 unported; Kurzon via Wikipedia)
Thus, if ni is the integer that characterizes the initial (excited) state of the electron, and nf is the final state (here we imagine that nf=1, but is applicable in cases that nf<ni, i.e., emission)
Enf=Eni−hν
or
ν=Eni−Enfh=Z2e4me8ϵ20h3(1n2f−1n2i)
We can now identify the Rydberg constant RH with the ratio of constants on the right hand side of Equation 1.8.13
RH=mee48ϵ20h3
Evaluating RH from the fundamental constants in this formula gives a value within 0.5% of that obtained experimentally from the hydrogen atom spectrum.
Thus, by observing the emitted light, we can determine the energy difference between the initial and final energy levels, which results in the emission spectra discussed in Sections 1.4 and 1.5. Different values of nf determine which emission spectrum is observed, and the examples shown in the figure are named after the individuals who first observed them. The figure below shows some of the transitions possible for different nf and ni values discussed previously.
Figure 1.8.7
: Electron Transitions Responsible for the Various Series of Lines Observed in the Emission Spectrum of Hydrogen. The Lyman series of lines is due to transitions from higher-energy orbits to the lowest-energy orbit (n = 1); these transitions release a great deal of energy, corresponding to radiation in the ultraviolet portion of the electromagnetic spectrum. The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation. Orbits are not drawn to scale. (CC BY-NC-SA 3.0; anonymous)
If the atom absorbs light it ends up in an excited state as a result of the absorption. The absorption is only possible for light of certain frequencies, and again, conservation of energy determines what these frequencies are. If light is absorbed, then the final energy Enf will be related to the initial energy Eni with nf>ni by
Enf=Eni+hν
or
ν=Enf−Enih=Z2e4me8ϵ20h3(1n2i−1n2f)
Exercise 1.8.3
Calculate the energy of a photon that is produced when an electron in a hydrogen atom goes from an orbit with n=4 to an orbit with n=1.
What happens to the energy of the photon as the initial value of n approaches infinity?
Answer
: a:
Enf=Eni−hνEphoton=hν=Enf−Eni=Z2e4me8ϵ2oh2(1n2f−1n2i)=e4me8ϵ2oh2(112−142)=2.18×10−18(1−116)=2.04×10−18J
b:
As ni→∞
Ephoton=e4me8ϵ2oh2(1n2f−1n2i)1n2i→0Ephoton→e4me8ϵ2oh2(1n2f)
Bohr’s proposal explained the hydrogen atom spectrum, the origin of the Rydberg formula, and the value of the Rydberg constant. Specifically it demonstrated that the integers in the Rydberg formula are a manifestation of quantization. The energy, the angular momentum, and the radius of the orbiting electron all are quantized. This quantization also parallels the concept of stable orbits in the Bohr model. Only certain values of E, M, and r are possible, and therefore the electron cannot collapse onto the nucleus by continuously radiating energy because it can only have certain energies, and it cannot be in certain regions of space. The electron can only jump from one orbit (quantum state) to another. The quantization means that the orbits are stable, and the electron cannot spiral into the nucleus in spite of the attractive Coulomb force.
Although Bohr’s ideas successfully explained the hydrogen spectrum, they failed when applied to the spectra of other atoms. In addition a profound question remained. Why is angular momentum quantized in units of ℏ? As we shall see, de Broglie had an answer to this question, and this answer led Schrödinger to a general postulate that produces the quantization of angular momentum as a consequence. This quantization is not quite as simple as proposed by Bohr, and we will see that it is not possible to determine the distance of the electron from the nucleus as precisely as Bohr thought. In fact, since the position of the electron in the hydrogen atom is not at all as well defined as a classical orbit (such as the moon orbiting the earth) it is called an orbital. An electron orbital represents or describes the position of the electron around the nucleus in terms of a mathematical function called a wavefunction that yields the probability of positions of the electron.
Contributors and Attributions
Mark Tuckerman (New York University)
David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski ("Quantum States of Atoms and Molecules")
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Python | sympy.ntheory.primetest.is_square() method
Last Updated : 02 Aug, 2019
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With the help of sympy.ntheory.primetest.is_square() method, we can find that weather an integer is perfect square or not that is passed as a parameter in the sympy.ilcm() method and returns a boolean value.
Syntax :sympy.ntheory.primetest.is_square(val)Return :Return boolean value.
Example #1 : In this example we can see that by using sympy.ntheory.primetest.is_square() method, we are able to find that weather an integer if perfect square or not that is passed as parameters. Python3 1=1
```python3
import sympy
from sympy.ntheory.primetest import is_square
Using sympy.ntheory.primetest.is_square() method
gfg = is_square(16)
print(gfg)
Output :True
Example #2 :Python3 1=1python3
import sympy
from sympy.ntheory.primetest import is_square
Using sympy.ntheory.primetest.is_square() method
gfg = is_square(24)
print(gfg)
```
Output :False
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jitender_1998
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Published Time: 2024-06-27T12:12:00+00:00
Process Tolerance: Balancing Quality & Cost | A Comprehensive Guide - SixSigma.us
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Articles
Process Tolerance: Balancing Quality & Cost | A Comprehensive Guide
June 27th, 2024
Ensuring quality while optimizing costs is so important. This delicate balance surfaces through comprehensive knowledge and application of process tolerance – a core concept in quality control and enhancement techniques like Six Sigma.
With over two decades as a Master Black Belt, I’ve seen process tolerancing’s profound impacts firsthand working extensively with major brands.
Insights gained illuminate tolerance examination’s intricate details and role reducing defects, improving consistency and driving cost-efficiency.
Process tolerance defines an acceptable production variation range, establishing standards determining a process’s capability.
It establishes minimum and maximum target value deviations, often shown as +/- or combining average and deviation standards.
Striking this quality-cost balance intricately through effective tolerancing demands deep understanding of approaches, strengths and limitations.
Key Highlights
Understand the fundamentals of process tolerance and its critical role in balancing quality and cost optimization
Gain insights into the three primary approaches to tolerancing: worst-case, statistical, and process tolerancing
Learn how to determine appropriate tolerance levels based on factors like functional limits, usage conditions, process capability studies, and operational windows
Explore the relationship between process tolerance and Six Sigma quality, including defects per million calculations and tolerance specifications
Discover practical applications and case studies highlighting the benefits of process tolerancing in addressing challenges like multiple identical components, off-center processes, and industry-specific scenarios
Master the art of accurate output prediction and achieving quality at lower costs through the effective implementation of process tolerancing principles
Understanding Process Tolerance: Balancing Quality and Cost
At the heart of any manufacturing or production process lies a delicate balance between achieving consistent, high-quality output and minimizing costs.
This balance is largely governed by the concept of process tolerance – a critical factor that sets the standard by which the capability of a process is determined.
What is Process Tolerance?
Process tolerance is a value that defines the acceptable range of variation in the production of goods.
It establishes the minimum and maximum deviations from a target value or specification, often represented by a +/- value or a combination of an average and standard deviation requirement.
For instance, a paint color may require 5 oz (+/- 0.2) of red dye, meaning the end products can have a weight between 4.8 and 5.2 oz and still be considered within tolerance.
Tolerance limits and specification ranges play a crucial role in defining the boundaries of acceptability.
Units falling outside these predetermined limits are considered nonconformances or defects, potentially compromising product quality and functionality.
Importance of Process Tolerance
Implementing appropriate process tolerances is essential for ensuring product consistency, safety, and reliability.
In industries where human safety is paramount, such as aerospace or medical device manufacturing, tolerances are typically set to stricter standards to minimize the risk of defects.
However, excessively tight tolerances can also lead to increased production costs and higher rates of waste, as even minor deviations may result in rejected units.
Finding the right balance between quality requirements and cost optimization is key, and this is where a comprehensive understanding of process tolerance becomes invaluable.
Approaches to Tolerancing
There are three primary approaches to tolerancing: worst-case, statistical, and process tolerancing.
Each method has its strengths, limitations, and applications, and understanding their nuances is crucial for effective tolerance analysis and process control.
Worst-Case Tolerancing
Worst-case tolerancing is a conservative approach that aims to account for the most extreme scenarios.
It assumes that all inputs will simultaneously be at their respective tolerance limits, representing a highly unlikely, if not impossible, situation in most real-world processes.
While this method ensures that the output will remain within its tolerance range as long as the inputs are within their specified limits, it often leads to an overestimation of variation.
This can result in unnecessarily tight tolerances for inputs, driving up costs and potentially increasing waste.
Statistical Tolerancing
Statistical tolerancing, on the other hand, takes a more realistic approach by assuming that processes are centered and that inputs vary independently according to a statistical distribution.
This method relies on techniques like propagation of error and variation transmission analysis to predict the average and standard deviation of the output based on the statistical tolerances of the inputs.
While statistical tolerancing can provide narrower tolerance ranges for outputs, potentially reducing costs, it can also underestimate variation if the assumption of centered processes is violated.
Off-center or unstable processes may result in defective products, even when inputs meet their statistical tolerance requirements.
Process Tolerancing: A Unified Approach
Process tolerancing represents a unified approach that combines the strengths of worst-case and statistical tolerancing while addressing their limitations.
It allows for the specification of both worst-case tolerances (for averages) and statistical tolerances (for variation around the average) within the same analysis.
This flexibility enables accurate representation of input behaviors, whether they are best described by worst-case scenarios, centered processes, or off-center conditions.
By providing a middle ground between the two extremes, process tolerancing offers a powerful tool for predicting output behavior accurately while balancing quality and cost considerations.
Determining Process Tolerance
Selecting the appropriate process tolerance for each input variable is a critical step in ensuring accurate tolerance analysis and effective process control.
The choice depends on various factors, including functional limits, usage conditions, process capability studies, and operational windows.
Worst-Case Tolerance Considerations
Worst-case tolerances are often the most appropriate choice for inputs representing usage or environmental conditions, such as temperature or humidity, where individual units may be continuously exposed to extreme conditions.
These tolerances are also suitable for inputs that are outside the direct control of the designer or manufacturer.
Statistical Tolerance Considerations
Statistical tolerances are well-suited for inputs with automatic controls that keep them centered on the target value, such as die temperature in a heat-sealing process.
Stable processes and in statistical control, as determined by capability studies or control charts, can also be represented by statistical tolerances, provided the assumption of independence between inputs is met.
Process Tolerance Considerations
Process tolerances are particularly useful for manually controlled processes, where an operating window can be maintained around the target value using control charts or other monitoring techniques.
They can also account for deterioration or drift over time, combining the effects of manufacturing variation and long-term changes in a single tolerance specification.
In cases where multiple identical components are used, process tolerances can capture the correlated behavior arising from lot-to-lot variation, providing a more accurate representation than independent statistical tolerances.
Process Tolerance and Six Sigma Quality
The pursuit of process excellence and defect reduction is a central tenet of the Six Sigma methodology, and process tolerance plays a pivotal role in achieving these goals.
Six Sigma Requirements
Six Sigma quality is defined as a measurable characteristic, such as flow rate or seal strength, having no more than 3.4 defects per million units produced.
This stringent requirement translates into specific process tolerance specifications, where the average must be within a certain range, and the standard deviation must not exceed a predetermined value.
Implementing Six Sigma Quality
To meet Six Sigma standards, organizations must implement rigorous process monitoring and control measures.
Control charts are invaluable tools for tracking process performance and ensuring that averages and variations remain within the specified tolerance limits.
Continuous improvement efforts, guided by process tolerancing principles, can help identify opportunities for tightening tolerances, reducing variation, and driving defect rates towards the Six Sigma goal of 3.4 defects per million or fewer.
Conclusion
Process tolerancing emerges crucial for quality and cost optimization pursuits. Its unified approach combining worst-case and statistical methodology strengths offers unparalleled flexibility and accuracy in envisioning output behavior.
The power to tailor specifications to each variable’s unique characteristics – be it functional boundaries, usage situations, process capabilities or operational parameters – equips organizations achieving the ideal balance of high quality, low costs.
Through effectively applying process tolerancing principles, companies unlock perks of reliable planning, defect declines and ultimately, building quality products while keeping costs down.
By embracing this potent methodology, businesses can stay ahead of rivals and pave the road to enduring success in challenging global market environments, where demands constantly evolve.
Process tolerancing demonstrates the payoffs awaiting those committed to performance and efficiency progress through an analytical mindset.
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391
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https://www.chemteam.info/AcidBase/Calc-percent-dissoc-of-weak-acid.html
|
Calculate the percent dissociation of a weak acid, given the pH and Ka
Return to the Acid Base menu
Return to a listing of many types of acid base problems and their solutions
Problem #1: A weak acid has a pKa of 4.994 and the solution pH is 4.523. What percentage of the acid is dissociated?
A comment before discussing the solution: note that the pKa is given, rather than the Ka. The first thing we will need to do is convert the pKa to the Ka. Then, the two values we need to obtain to solve the problem given just above are [H+], which is pretty easy and [HA], which is only a tiny bit more involved.
Solution:
1) Convert pKa to Ka:
Ka = 10¯pKa = 10¯4.994 = 1.0139 x 10¯5
Often the identity of the weak acid is not specified. That is because, with few exceptions, all weak acids behave in the same way and so the same techniques can be used no matter what acid is used in the problem. In cases where no acid is identified, you can use a generic weak acid, signified by the formula HA. Here is the dissociation equation for HA:
HA ⇌ H+ + A¯
Here is the equilibrium expression for that dissociation:
Ka = ([H+] [A¯]) / [HA]
2) The pH will give [H+] (and the [A¯]):
[H+] = 10¯pH = 10¯4.523 = 3.00 x 10¯5 M
Because of the 1:1 molar ratio in the above equation, we know that [A¯] = [H+] = 3.00 x 10¯5 M.
3) This means that the only value left is [HA], so we will use the equilibrium expression to calculate [HA].
1.0139 x 10¯5 = [(3.00 x 10¯5) (3.00 x 10¯5)] / x
x = 8.88 x 10¯5 M
4) Percent dissociation for an acid is [H+] / [HA] and then times 100.
3.00 x 10¯5 / 8.88 x 10¯5 = 33.8%
Problem #2: A solution of acetic acid (Ka = 1.77 x 10¯5) has a pH of 2.876. What is the percent dissociation?
Solution:
1) Calculate the [H+] from the pH:
[H+] = 10¯pH = 10¯2.876 = 1.33 x 10¯3 M
2) From the 1:1 stoichiometry of the chemical equation, we know that the acetate ion concentration, [Ac¯] equals the [H+]. Therefore,
[Ac¯] = 1.33 x 10¯3 M
3) We need to determine [HAc], the acetic acid concentration. We use the Ka expression to determine this value:
1.77 x 10¯5 = [(1.33 x 10¯3) (1.33 x 10¯3)] / x
x = 0.09993 M = 0.100 M
4) Percent dissociation:
(1.33 x 10¯3 / 0.100) times 100 = 1.33%
Comment: the first example is somewhat artifical, in that the percent dissocation is quite high. The second example is more in line with what teachers usually ask. The usual percent dissociation answer is between 1 and 5 per cent. However, the 33.8% answer, while not commonly found in introductory chemistry classes, is possible.
Problem #3: A generic weak acid (formula = HA) has a pKa of 4.401. If the solution pH is 3.495, what percentage of the acid is undissociated?
Solution:
1) Convert pKa to Ka:
Ka = 10¯pKa = 10¯4.401 = 3.97 x 10¯5
2) The pH gives [H+] (and the [A¯]):
[H+] = 10¯pH = 10¯3.495 = 3.20 x 10¯4 M
3) Determine the concentration of the weak acid:
Ka = ([H+] [A¯]) / [HA]
3.97 x 10¯5 = [(3.20 x 10¯4) (3.20 x 10¯4)] / x
x = 0.00258 M
4) Determine percent dissociation:
3.20 x 10¯4 / 0.00258 = 12.4%
5) Determine percent undissociated:
100 - 12.4 = 87.6%
Comment: the calculation technique discussed above determines the percent dissociation. Notice that the above problem asks for the percent undissociated.
Be aware! The problem above goes one step beyond what is normally taught. This might show up as a test question.
Problem #4: A weak acid has a pKa of 4.289. If the solution pH is 3.202, what percentage of the acid is dissociated?
Solution:
1) Convert pKa to Ka:
Ka = 10¯pKa = 10¯4.289 = 5.14 x 10¯5
2) The pH gives [H+] (and the [A¯]):
[H+] = 10¯pH = 10¯3.202 = 6.28 x 10¯4 M
3) Determine the concentration of the weak acid:
Ka = ([H+] [A¯]) / [HA]
5.14 x 10¯5 = [(6.28 x 10¯4) (6.28 x 10¯4)] / x
x = 0.00767284 M (kept a few guard digits on this one
4) Determine percent dissociation:
6.28 x 10¯4 / 0.00767284 = 8.2%
Return to the Acid Base menu
Return to a listing of many types of acid base problems and their solutions
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392
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https://craftknife.blogspot.com/2020/09/homeschool-math-perfect-squares-hiding.html
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Craft Knife: Homeschool Math: Perfect Squares Hiding inside Area Models
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Thursday, September 3, 2020
Homeschool Math: Perfect Squares Hiding inside Area Models
Syd was simplifying radicals the other day, and not having a fun time of it. She was struggling to link the concept of factoring the radicand to simplifying, and I was trying, as usual, to think of hands-on manipulatives that might clarify the process.
I did NOT find a way to model simplifying radicals using manipulatives, alas, but while I was playing around with the decanomial square I DID find a hands-on enrichment that kids who are first learning the concept of perfect squares might enjoy.
I like this little activity because it connects the mathematical definition of the perfect square with the Montessori-style sensorial skill of eyeballing it, or even measuring it by feel. Although you're technically not allowed to eyeball stuff as mathematical proof, pattern recognition via the senses is very important. That's how kids learn to read, for one thing, and it's how IQ tests are built, for another.
Use this activity with a kid who's first learning, or reviewing, the concept of the perfect square. You can do it with paper area models that a kid can draw and color on, or you can do it, as I've done here, with the decanomial square model, which is extra fun because it has pieces you can manipulate. Kids could try to find the largest perfect square(s) that would fit inside the area model, or just find any perfect squares that would--whatever they find fun and you find helpful! Here are some models that show examples:
These first two are when I was still thinking I might figure out a way to model simplifying radicals. I LOVE combining manipulatives with a dry-erase board to help kids connect the model to the algorithm it represents.
For all these examples, I've pulled an area model from our decanomial square, and we're arranging the perfect squares on top of it, leaving, of course, a remainder since the area models aren't themselves perfect squares.
You can write algebraic equations with these, showing how to use the Order of Operations and/or solve for x. For example:
5 + 5 x 5 = 30
or
8^2 + 2^2 + y = 80
You just can't, you know, use them to model how to simplify radicals...
The search continues!
P.S. Here are the resources that I used to help both kids master radicals.
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UP Physics Grade 12
Course: UP Physics Grade 12>Unit 9
Lesson 3: Faraday's law
Faraday's Law Introduction
Faraday's law - existence of induced emf-i
Science>
UP Physics Grade 12>
Electromagnetic induction>
Faraday's law
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Faraday's Law Introduction
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About About this video Transcript
How a current can be induced in a loop of wire by a change in magnetic flux.
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seungheon choo 10 years ago Posted 10 years ago. Direct link to seungheon choo's post “At 3:13, does it always g...” more At 3:13 , does it always go clockwise?
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Christopher Reese 10 years ago Posted 10 years ago. Direct link to Christopher Reese's post “No, there is a technique ...” more No, there is a technique discussed in the previous magnetism videos that you can use to find the direction of the current called the right-hand rule (RHR). The next video on this playlist called "Len'z Law" explains how to determine the direction of the current using the RHR.
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saraiyadeepam 4 years ago Posted 4 years ago. Direct link to saraiyadeepam's post “if a magnet and closed lo...” more if a magnet and closed loop are in motion but they relative velocity=0 will there be an induced emf generated in the coil?
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Mike Rosen 2 years ago Posted 2 years ago. Direct link to Mike Rosen's post “Wouldn't the average flux...” more Wouldn't the average flux in a larger "area" be the same as in the smaller area? Wouldn't this be so because you are increasing the the amount of space for the charge to travel?
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Gilleon Khonglah Ropmay 9 years ago Posted 9 years ago. Direct link to Gilleon Khonglah Ropmay's post “Would Decreasing the flux...” more Would Decreasing the flux also induce a current?
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Mez Cooper 5 years ago Posted 5 years ago. Direct link to Mez Cooper's post “I believe so as there is ...” more I believe so as there is flux when the magnetic field is losing strength. The induced current is probably weaker as a result.
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masani5039 7 years ago Posted 7 years ago. Direct link to masani5039's post “How does moving a magnet ...” more How does moving a magnet help change flux?
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Andrew M 7 years ago Posted 7 years ago. Direct link to Andrew M's post “The whole concept of flux...” more The whole concept of flux was invented basically to describe the effect of moving a magnet. Flux is a measure of the magnetic field's penetration of a particular area of space around a magnet, so of course if you move the magnet, it's field's penetration of that area of space will change.
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JC 9 years ago Posted 9 years ago. Direct link to JC's post “So the right hand rule ap...” more So the right hand rule applies when the current is the cause of the magnetic field ,whereas essentially the direction is opposite when the magnetic field causes the current..... Am I correct? this is really racking my brain.....someone please reply...
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Carla Cristina Almeida 9 years ago Posted 9 years ago. Direct link to Carla Cristina Almeida's post “but if I point my thumb o...” more but if I point my thumb outwards (in the direction of the magnetic vector), my other fingers will curl anticlockwise. what am I missing?
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mahimakashyap222 9 years ago Posted 9 years ago. Direct link to mahimakashyap222's post “Why does this happen that...” more Why does this happen that change is magnetic flux induce current?
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ashesreveal 9 years ago Posted 9 years ago. Direct link to ashesreveal's post “How is the current actual...” more How is the current actually induced in the loop?
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Andrew M 9 years ago Posted 9 years ago. Direct link to Andrew M's post “By the forces applied to ...” more By the forces applied to charges by the electromagnetic field
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TheTheoryOfElla 2 years ago Posted 2 years ago. Direct link to TheTheoryOfElla's post “At 2:59, why does it go i...” more At 2:59 , why does it go in a clockwise direction?
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Charles LaCour 2 years ago Posted 2 years ago. Direct link to Charles LaCour's post “When you solve maxwells f...” more When you solve maxwells field equations for this situation you will find that the force acts in a direction that is perpendicular to the direction of the charge's motion and the direction of the magnetic field. Because of this it would either go in a clockwise or counterclockwise and based on the way we have chosen to labeled positive and negative as well as the direction of the field it turns out to be clockwise.
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Video transcript
[Instructor] In other videos, we talk about how a current flowing through a wire can induce a magnetic field. Now what we're going to talk about in this video is how we can go the other way. How a change in a magnetic field can affect or induce a current in a loop of wire. We're gonna be going from a change, change in, and actually we're gonna focus on magnetic flux. Magnetic. Magnetic flux through a loop. Through a loop. Now we're going to see how this actually induces a current in that loop. So, induces. Induces a current. Induced current. This right over here, all right, this is my attempted drawing a magnetic field and these are the magnetic field lines. They don't look like lines because they're all popping out of the screen. They're moving towards you, towards the viewer so you could view these as the tips of the arrow. Now there's different ways of showing magnetic fields. You could show magnetic field lines like this, you could also use vectors. And when you're using field lines, it's the density of the field lines tell you how strong the magnetic field is. If you got to the right here, they are less dense and so the magnetic field is less strong on the right hand side over here than they are on the left hand side, or at least that's what I'm trying to depict. Now if we did it with vectors we would have bigger arrows pointing out over here than we would over here. But now let's do a loop of, let's do a loop of wire. Let me draw a loop of wire here. Let's say I have a loop of wire that is, this is my loop of wire. My loop of wire. If I just throw that loop of wire and if it's just stationary, it's in this magnetic field and the magnetic field isn't changing, I do have some flux going through the, I guess you can say the surface defined by this wire. And if you're unfamiliar with the term magnetic flux, I encourage you to watch the video on magnetic flux. But if I just have this wire stationary in the magnetic field, nothing is going to happen. But I will be able to induce a current if I change the magnetic flux going through this surface in some way. So for example, right now the magnetic field is pointing out of the screen. If I were to make it even stronger in the direction pointing out of the screen. So, I guess one way to think about it is if I were to, if the change in the magnetic field, the change in the flux were to get even stronger in the outward direction. So I don't know, a good way to, just gonna get even stronger in the outward direction. I'll draw a big arrow there. I guess you could say that these things became even denser in this outward direction. It is actually going to induce a current. And the current that it will induce is going to go, is going to go in that direction and let me draw it a little bit clearer. The current is going to go in that direction. It's going to go in a clockwise direction around this. And that is because that change in the magnetic flux that induces a voltage, an electromotive force that causes this current to flow. That causes the current to flow and there's other ways to change the magnetic flux. If I were to lessen the magnetic field in the outward direction or another way, if the change in the magnetic field were inwards, then the current would go the other direction. But the key here is when I change the flux through this surface defined by this wire, it's going to induce a current. The current isn't going to be there if the thing is, if the magnetic field is stationary and I'm not changing this loop in any way. But as soon as I change the flux in some way, I am going to induce a current. I could also instead of changing the field, I could actually move my coil. I could move it that way. And if I were to move it that way, the flux going through this surface or I guess coming out of this surface will increase. Because if I move this to the left, the magnetic field is denser, I guess it's stronger so there'll be more flux through this area here. So, if you move it that way, you also would have a current like this. Now if you moved it the other way, if it was on the left hand side and you moved it that way, it would also induce a current but now since the flux is lessening in the outward direction, the current would go in the other way. Now there's other ways to change the flux. You could actually change the area of this actual loop if somehow it was made out of some maybe stretchy, stretchy wire somehow. If you increased its, if you would increase. Let me draw it this way. If you were somehow able to stretch it. Stretch it so it contains, so the actual area increases. If you were able to stretch it out so that the actual area increases which would cause the flux in the out of the screen direction to increase even more, that also would induce, that also would induce the current. And so this whole idea of a change in magnetic flux inducing a current, this is the essence and we'll go deeper into it in future videos, this is the essence of Faraday's Law. Faraday's, Faraday's Law. And we'll quantify this more in future videos but it's just the notion that if I have a loop of wire and I have a changing magnetic flux through the loop of wire, that is going to induce a current in that wire.
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ANTENNA THEORY ANTENNA THEORY ANALYSIS AND DESIGN FOURTH EDITION Constantine A. Balanis Cover Image: Courtesy NASA/JPL-Caltech Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved Published by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at
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Library of Congress Cataloging-in-Publication Data: Balanis, Constantine A., 1938– Modern antenna handbook / Constantine A. Balanis.—4th ed.
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ISBN 978-1-118-642060-1 (cloth) 1. Antennas (Electronics) I. Title.
TK7871.6.B354 2016 621.382′4—dc22 2016050162 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 To the memory of my parents, uncle and aunt Στη μν´ ημη των γoν´ εων, τoυ θε´ ιoυ και τη𝜍θε´ ια𝜍μoυ Contents Preface xiii About the Companion Website xix 1 Antennas 1 1.1 Introduction 1 1.2 Types of Antennas 3 1.3 Radiation Mechanism 7 1.4 Current Distribution on a Thin Wire Antenna 15 1.5 Historical Advancement 18 1.6 Multimedia 21 References 22 2 Fundamental Parameters and Figures-of-Merit of Antennas 25 2.1 Introduction 25 2.2 Radiation Pattern 25 2.3 Radiation Power Density 35 2.4 Radiation Intensity 37 2.5 Beamwidth 40 2.6 Directivity 41 2.7 Numerical Techniques 55 2.8 Antenna Efficiency 60 2.9 Gain, Realized Gain 61 2.10 Beam Efficiency 65 2.11 Bandwidth 65 2.12 Polarization 66 2.13 Input Impedance 75 2.14 Antenna Radiation Efficiency 79 2.15 Antenna Vector Effective Length and Equivalent Areas 81 2.16 Maximum Directivity and Maximum Effective Area 86 2.17 Friis Transmission Equation and Radar Range Equation 88 2.18 Antenna Temperature 96 2.19 Multimedia 100 References 103 Problems 105 vii viii CONTENTS 3 Radiation Integrals and Auxiliary Potential Functions 127 3.1 Introduction 127 3.2 The Vector Potential A for an Electric Current Source J 128 3.3 The Vector Potential F for A Magnetic Current Source M 130 3.4 Electric and Magnetic Fields for Electric (J) and Magnetic (M) Current Sources 131 3.5 Solution of the Inhomogeneous Vector Potential Wave Equation 132 3.6 Far-Field Radiation 136 3.7 Duality Theorem 137 3.8 Reciprocity and Reaction Theorems 138 References 143 Problems 143 4 Linear Wire Antennas 145 4.1 Introduction 145 4.2 Infinitesimal Dipole 145 4.3 Small Dipole 155 4.4 Region Separation 158 4.5 Finite Length Dipole 164 4.6 Half-Wavelength Dipole 176 4.7 Linear Elements Near or On Infinite Perfect Electric Conductors (PEC), Perfect Magnetic Conductors (PMC) and Electromagnetic Band-Gap (EBG) Surfaces 179 4.8 Ground Effects 203 4.9 Computer Codes 216 4.10 Multimedia 216 References 218 Problems 220 5 Loop Antennas 235 5.1 Introduction 235 5.2 Small Circular Loop 236 5.3 Circular Loop of Constant Current 250 5.4 Circular Loop with Nonuniform Current 259 5.5 Ground and Earth Curvature Effects for Circular Loops 268 5.6 Polygonal Loop Antennas 269 5.7 Ferrite Loop 270 5.8 Mobile Communication Systems Applications 272 5.9 Multimedia 272 References 275 Problems 277 6 Arrays: Linear, Planar, and Circular 285 6.1 Introduction 285 6.2 Two-Element Array 286 6.3 N-Element Linear Array: Uniform Amplitude and Spacing 293 6.4 N-Element Linear Array: Directivity 312 6.5 Design Procedure 318 6.6 N-Element Linear Array: Three-Dimensional Characteristics 319 6.7 Rectangular-to-Polar Graphical Solution 322 CONTENTS ix 6.8 N-Element Linear Array: Uniform Spacing, Nonuniform Amplitude 323 6.9 Superdirectivity 345 6.10 Planar Array 348 6.11 Design Considerations 360 6.12 Circular Array 363 6.13 Multimedia 367 References 367 Problems 368 7 Antenna Synthesis and Continuous Sources 385 7.1 Introduction 385 7.2 Continuous Sources 386 7.3 Schelkunoff Polynomial Method 387 7.4 Fourier Transform Method 392 7.5 Woodward-Lawson Method 398 7.6 Taylor Line-Source (Tschebyscheff-Error) 404 7.7 Taylor Line-Source (One-Parameter) 408 7.8 Triangular, Cosine, and Cosine-Squared Amplitude Distributions 415 7.9 Line-Source Phase Distributions 416 7.10 Continuous Aperture Sources 417 7.11 Multimedia 420 References 420 Problems 421 8 Integral Equations, Moment Method, and Self and Mutual Impedances 431 8.1 Introduction 431 8.2 Integral Equation Method 432 8.3 Finite Diameter Wires 439 8.4 Moment Method Solution 448 8.5 Self-Impedance 455 8.6 Mutual Impedance Between Linear Elements 463 8.7 Mutual Coupling in Arrays 474 8.8 Multimedia 480 References 480 Problems 482 9 Broadband Dipoles and Matching Techniques 485 9.1 Introduction 485 9.2 Biconical Antenna 487 9.3 Triangular Sheet, Flexible and Conformal Bow-Tie, and Wire Simulation 492 9.4 Vivaldi Antenna 496 9.5 Cylindrical Dipole 500 9.6 Folded Dipole 505 9.7 Discone and Conical Skirt Monopole 512 9.8 Matching Techniques 513 9.9 Multimedia 523 References 524 Problems 525 x CONTENTS 10 Traveling Wave and Broadband Antennas 533 10.1 Introduction 533 10.2 Traveling Wave Antennas 533 10.3 Broadband Antennas 549 10.4 Multimedia 580 References 580 Problems 582 11 Frequency Independent Antennas, Antenna Miniaturization, and Fractal Antennas 591 11.1 Introduction 591 11.2 Theory 592 11.3 Equiangular Spiral Antennas 593 11.4 Log-Periodic Antennas 598 11.5 Fundamental Limits of Electrically Small Antennas 614 11.6 Antenna Miniaturization 619 11.7 Fractal Antennas 627 11.8 Multimedia 633 References 633 Problems 635 12 Aperture Antennas 639 12.1 Introduction 639 12.2 Field Equivalence Principle: Huygens’ Principle 639 12.3 Radiation Equations 645 12.4 Directivity 648 12.5 Rectangular Apertures 648 12.6 Circular Apertures 667 12.7 Design Considerations 675 12.8 Babinet’s Principle 680 12.9 Fourier Transforms in Aperture Antenna Theory 684 12.10 Ground Plane Edge Effects: The Geometrical Theory of Diffraction 702 12.11 Multimedia 707 References 707 Problems 709 13 Horn Antennas 719 13.1 Introduction 719 13.2 E-Plane Sectoral Horn 719 13.3 H-Plane Sectoral Horn 733 13.4 Pyramidal Horn 743 13.5 Conical Horn 756 13.6 Corrugated Horn 761 13.7 Aperture-Matched Horns 766 13.8 Multimode Horns 769 13.9 Dielectric-Loaded Horns 771 13.10 Phase Center 773 13.11 Multimedia 774 References 775 Problems 778 CONTENTS xi 14 Microstrip and Mobile Communications Antennas 783 14.1 Introduction 783 14.2 Rectangular Patch 788 14.3 Circular Patch 815 14.4 Quality Factor, Bandwidth, and Efficiency 823 14.5 Input Impedance 826 14.6 Coupling 827 14.7 Circular Polarization 830 14.8 Arrays and Feed Networks 832 14.9 Antennas for Mobile Communications 837 14.10 Dielectric Resonator Antennas 847 14.11 Multimedia 858 References 862 Problems 867 15 Reflector Antennas 875 15.1 Introduction 875 15.2 Plane Reflector 875 15.3 Corner Reflector 876 15.4 Parabolic Reflector 884 15.5 Spherical Reflector 920 15.6 Multimedia 923 References 923 Problems 925 16 Smart Antennas 931 16.1 Introduction 931 16.2 Smart-Antenna Analogy 931 16.3 Cellular Radio Systems Evolution 933 16.4 Signal Propagation 939 16.5 Smart Antennas’ Benefits 942 16.6 Smart Antennas’ Drawbacks 943 16.7 Antenna 943 16.8 Antenna Beamforming 946 16.9 Mobile Ad hoc Networks (MANETs) 960 16.10 Smart-Antenna System Design, Simulation, and Results 964 16.11 Beamforming, Diversity Combining, Rayleigh-Fading, and Trellis-Coded Modulation 972 16.12 Other Geometries 975 16.13 Multimedia 976 References 976 Problems 980 17 Antenna Measurements 981 17.1 Introduction 981 17.2 Antenna Ranges 982 17.3 Radiation Patterns 1000 17.4 Gain Measurements 1003 17.5 Directivity Measurements 1010 xii CONTENTS 17.6 Radiation Efficiency 1012 17.7 Impedance Measurements 1012 17.8 Current Measurements 1014 17.9 Polarization Measurements 1014 17.10 Scale Model Measurements 1019 References 1024 Appendix I: f(x) = sin(x) x 1027 Appendix II: fN(x) = | | | | sin(Nx) N sin(x) | | | | N = 1, 3, 5, 10, 20 1029 Appendix III: Cosine and Sine Integrals 1031 Appendix IV: Fresnel Integrals 1033 Appendix V: Bessel Functions 1035 Appendix VI: Identities 1041 Appendix VII: Vector Analysis 1045 Appendix VIII: Method of Stationary Phase 1055 Appendix IX: Television, Radio, Telephone, and Radar Frequency Spectrums 1061 Index 1065 Preface The fourth edition of Antenna Theory is designed to meet the needs of electrical engineering and physics students at the senior undergraduate and beginning graduate levels, and those of practicing engineers. The text presumes that the students have knowledge of basic undergraduate electromag-netic theory, including Maxwell’s equations and the wave equation, introductory physics, and dif-ferential and integral calculus. Mathematical techniques required for understanding some advanced topics in the later chapters are incorporated in the individual chapters or are included as appendices.
The book, since its first edition in 1982 and subsequent two editions in 1997 and 2005, has been a pacesetter and trail blazer in updating the contents to keep abreast with advancements in antenna technology. This has been accomplished by: r Introducing new topics r Originating innovative features and multimedia to animate, visualize, illustrate and display radiation characteristics r Providing design equations, procedures and associate software This edition is no exception, as many new topics and features have been added. In particular: r New sections have been introduced on: 1. Flexible and conformal bowtie 2. Vivaldi antenna 3. Antenna miniaturization 4. Antennas for mobile communications 5. Dielectric resonator antennas 6. Scale modeling r Additional MATLAB and JAVA programs have been developed.
r Color and gray scale figures and illustrations have been developed to clearly display and visu-alize antenna radiation characteristics.
r A companion website has been structured by the publisher which houses the MATLAB pro-grams, JAVA-based applets and animations, Power Point notes, and JAVA-based interactive questionnaires. A solutions manual is available only for the instructors that adopt the book as a classroom text.
r Over 100 additional end-of-chapter problems have been included.
While incorporating the above new topics and features in the current edition, the book maintained all of the attractive features of the first three additions, especially the: r Three-dimensional graphs to display the radiation characteristics of antennas. This feature was hailed, at the time of its introduction, as innovative and first of its kind addition in a textbook on antennas.
xiii xiv PREFACE r Advanced topics, such as a chapter on Smart Antennas and a section on Fractal Antennas.
r Multimedia: 1. Power Point notes 2. MATLAB programs 3. FORTRAN programs 4. JAVA-based animations 5. JAVA-based applets 6. JAVA-based end-of-the-chapter questionnaires The book’s main objective is to introduce, in a unified manner, the fundamental principles of antenna theory and to apply them to the analysis, design, and measurements of antennas. Because there are so many methods of analysis and design and a plethora of antenna structures, applications are made to some of the most basic and practical configurations, such as linear dipoles; loops; arrays; broadband, and frequency-independent antennas; aperture antennas; horn antennas; microstrip antennas; and reflector antennas.
A tutorial chapter on Smart Antennas is included to introduce the student in a technology that will advance antenna theory and design, and revolutionize wireless communications. It is based on antenna theory, digital signal processing, networks and communications. MATLAB simulation software has also been included, as well as a plethora of references for additional reading.
Introductory material on analytical methods, such as the Moment Method and Fourier transform (spectral) technique, is also included. These techniques, together with the fundamental principles of antenna theory, can be used to analyze and design almost any antenna configuration. A chapter on antenna measurements introduces state-of-the-art methods used in the measurements of the most basic antenna characteristics (pattern, gain, directivity, radiation efficiency, impedance, current, and polarization) and updates progress made in antenna instrumentation, antenna range design, and scale modeling. Techniques and systems used in near- to far-field measurements and transformations are also discussed.
A sufficient number of topics have been covered, some for the first time in an undergraduate text, so that the book will serve not only as a text but also as a reference for the practicing and design engineer and even the amateur radio buff. These include design procedures, and associated computer programs, for Yagi–Uda and log-periodic arrays, horns, and microstrip patches; synthesis techniques using the Schelkunoff, Fourier transform, Woodward–Lawson, Tschebyscheff, and Taylor meth-ods; radiation characteristics of corrugated, aperture-matched, and multimode horns; analysis and design of rectangular and circular microstrip patches; and matching techniques such as the binomial and Tschebyscheff. Also new sections have been introduced on flexible & conformal bowtie and Vivaldi antennas in Chapter 9, antenna miniaturization in Chapter 11 and expanded scale modeling in Chapter 17.
Chapter 14 has been expanded to include antennas for Mobile Communications. In particular, this new section includes basic concepts and design equations for the Planar Inverted-F Antenna (PIFA), Slot Antenna, Inverted-F Antenna (IFA), Multiband U-type Slot Antenna, and Dielectric Resonator Antennas (DRAs). These are popular internal antennas for mobile devices (smart phones, laptops, pads, tablets, etc.). A MATLAB computer program, referred to as DRA Analysis Design, has been developed to analyze the resonant frequencies of Rectangular, Cylindrical, Hemicylindri-cal, and Hemispherical DRAs using TE and TM modal cavity techniques by modeling the walls as PMCs. Hybrid modes are used to analyze and determine the resonant frequencies and quality fac-tor (Q) of the Cylindrical DRA. The MATLAB program DRA Analysis Design has the capability, using a nonlinear solver, to design (i.e., find the Q, range of values for the dielectric constant, and finally the dimensions of the Cylindrical DRA) once the hybrid mode (TE01𝛿, TM01𝛿or HE11𝛿), frac-tional bandwidth (BW, in %), VSWR and resonant frequency (fr, in GHz) are specified. A detailed procedure to follow the design is outlined in Section 14.10.4.
PREFACE xv The text contains sufficient mathematical detail to enable the average undergraduate electrical engineering and physics students to follow, without difficulty, the flow of analysis and design. A certain amount of analytical detail, rigor, and thoroughness allows many of the topics to be traced to their origin. My experiences as a student, engineer, and teacher have shown that a text for this course must not be a book of unrelated formulas, and it must not resemble a “cookbook.” This book begins with the most elementary material, develops underlying concepts needed for sequential topics, and progresses to more advanced methods and system configurations. Each chapter is subdivided into sections or subsections whose individual headings clearly identify the antenna characteristic(s) discussed, examined, or illustrated.
A distinguished feature of this book is its three-dimensional graphical illustrations from the first edition, which have been expanded and supplemented in the second, third and fourth editions. In the past, antenna texts have displayed the three-dimensional energy radiated by an antenna by a number of separate two-dimensional patterns. With the advent and revolutionary advances in digi-tal computations and graphical displays, an additional dimension has been introduced for the first time in an undergraduate antenna text by displaying the radiated energy of a given radiator by a single three-dimensional graphical illustration. Such an image, formed by the graphical capabilities of the computer and available at most computational facilities, gives a clear view of the energy radi-ated in all space surrounding the antenna. In this fourth edition, almost all of the three-dimensional amplitude radiation patterns, along with many two-dimensional graphs, are depicted in color and gray-scale. This is a new and pacesetting feature adopted, on a large scale, in this edition. It is hoped that this will lead to a better understanding of the underlying principles of radiation and provide a clearer visualization of the pattern formation in all space.
In addition, there is an abundance of general graphical illustrations, design data, references, and an expanded list of end-of-the chapter problems. Many of the principles are illustrated with examples, graphical illustrations, and physical arguments. Although students are often convinced that they understand the principles, difficulties arise when they attempt to use them. An example, especially a graphical illustration, can often better illuminate those principles. As they say, “a picture is worth a thousand words.” Numerical techniques and computer solutions are illustrated and encouraged. A number of MATLAB computer programs are included in the publisher’s website for the book. Each program is interactive and prompts the user to enter the data in a sequential manner. Some of these programs are translations of the FORTRAN ones that were included in the first and second editions. However, many new ones have been developed. Every chapter, other than Chapters 3 and 17, has at least one MATLAB computer program; some have as many as four. The outputs of the MATLAB programs include graphical illustrations and tabulated results. For completeness, the FORTRAN computer programs are also included, although nowdays there is not as much interest in them. The computer programs can be used for analysis and design. Some of them are more of the design type while some of the others are of the analysis type. Associated with each program there is a READ ME file, which summarizes the respective program.
The purpose of the Power Point Lecture Notes is to provide the instructors a copy of the text figures and some of the most important equations of each chapter. They can be used by the instructors in their lectures but may be supplemented with additional narratives. The students can use them to listen to the instructors’ lectures, without having to take detailed notes, but can supplement them in the margins with annotations from the lectures. Each instructor will use the notes in a different way.
The Interactive Questionnaires are intended as reviews of the material in each chapter. The student can use them to review for tests, exams, and so on. For each question, there are three possible answers, but only one is correct. If the reader chooses one of them and it the correct answer, it will so indicate.
However, if the chosen answer is the wrong one, the program will automatically indicate the correct answer. An explanation button is provided, which gives a short narrative on the correct answer or indicates where in the book the correct answer can be found.
xvi PREFACE The Animations can be used to illustrate some of the radiation characteristics, such as amplitude patterns, of some antenna types, like line sources, dipoles, loops, arrays, and horns. The Applets cover more chapters and can be used to examine some of the radiation characteristics (such as amplitude patterns, impedance, bandwidth, etc.) of some of the antennas. This can be accomplished very rapidly without having to resort to the MATLAB programs, which are more detailed.
For course use, the text is intended primarily for a two-semester (or two- or three-quarter) sequence in antenna theory. The first course should be given at the senior undergraduate level, and should cover most of the material in Chapters 1 through 7, and some sections of Chapters 14, 16 and 17. The material in Chapters 8 through 16 should be covered in detail in a beginning graduate-level course. Selected chapters and sections from the book can be covered in a single semester, without loss of continuity. However, it is essential that most of the material in Chapters 2 through 6 be cov-ered in the first course and before proceeding to any more advanced topics. To cover all the material of the text in the proposed time frame would be, in some cases, an ambitious and challenging task.
Sufficient topics have been included, however, to make the text complete and to give the teacher the flexibility to emphasize, deemphasize, or omit sections or chapters. Some of the chapters and sections can be omitted without loss of continuity.
In the entire book, an ej𝜔t time variation is assumed, and it is suppressed. The International Sys-tem of Units, which is an expanded form of the rationalized MKS system, is used in the text. In some cases, the units of length are in meters (or centimeters) and in feet (or inches). Numbers in parentheses () refer to equations, whereas those in brackets [] refer to references. For emphasis, the most important equations, once they are derived, are boxed. In some of the basic chapters, the most important equations are summarized in tables.
I will like to acknowledge the invaluable suggestions from all those that contributed to the first three additions of the book, too numerous to mention here. Their names and contributions are stated in the respective editions. It is my pleasure to acknowledge the suggestions of the reviewers for the fourth edition: Dr. Stuart A. Long of the University of Houston, Dr. Leo Kempel of Michigan State University, and Dr. Cynthia M. Furse of the University of Utah. There have been other con-tributors to this edition, and their contributions are valued and acknowledged. Many graduate and undergraduate students at Arizona State University have written and verified most of the MATLAB computer programs; some of these programs were translated from FORTRAN, which appeared in the first three editions and updated for the fourth edition. However some new MATLAB and JAVA programs have been created, which are included for the first time in the fourth edition. I am indebted to Dr. Alix Rivera-Albino who developed with special care all of the color and gray scale figures and illustrations for the fourth edition and contributed to the manuscript and figures for the Vivaldi and mobile antennas. The author also acknowledges Dr. Razib S. Shishir of Intel, formerly of Ari-zona State University, for the JAVA-based software for the third edition, including the Interactive Questionnaires, Applets and Animations. These have been supplemented with additional ones for the fourth edition. Many thanks to Dr. Stuart A. Long, from the University of Houston, for review-ing the section on DRAs and Dr. Christos Christodoulou, from the University of New Mexico, for reviewing the manuscript on antennas for mobile devices, Dr. Peter J. Bevelacqua of Google for mate-rial related to planar antennas for mobile units, Dr. Arnold Mckinley of University College London (formerly with the Australian National University) for information and computer program related to nonuniform loop antennas, Dr. Steven R. Best of Mitre Corporation for figures on the folded spherical helix, Dr. Edward J. Rothwell, from Michigan State University, for antenna miniaturiza-tion information, Dr. Seong-Ook Park of the Korea Advanced Institute of Science and Technology (KAIST), for the photo and permission of the U-slot antenna, and Dr. Yahia Antar and Dr. Jawad Y. Siddiqui, both from the Royal Military College of Canada, for information related to cylindri-cal DRAs. I would also like to thank Craig R. Birtcher, and my graduate students Dr. Ahmet C.
Durgun (now with Intel), Dr. Nafati Aboserwal (now at the University of Oklahoma), Sivaseethara-man Pandi, Mikal Askarian Amiri, Wengang Chen, Saud Saeed and Anuj Modi, all of Arizona State University, for proofreading of the manuscript and many other contributions to the fourth edition.
PREFACE xvii Special thanks to the companies that contributed photos, illustrations and copyright permissions for the third edition. However, other companies, Samsung, Microsoft and HTC have provided updated photos of their respective smart phones for the fourth edition.
During my 50+ year professional career, I have made many friends and professional colleagues.
The list is too long to be included here, as I fear that I may omit someone. Thank you for your friend-ship, collegiality and comradery. I will like to recognize George C. Barber, Dennis DeCarlo and the entire membership (members, government agencies and companies) of the Advanced Helicopter Electromagnetics (AHE) Program for the 25 years of interest and support. It has been an unprece-dented professional partnership and collaboration. To all my teachers and mentors, thank you. You have been my role models and inspiration.
This journey got started in the middle to the late 1970s, at the early stages of my academic career.
Many may speculate why I have chosen to remain as the sole author and steward for so many years, dating back to first edition in 1982 and then through the subsequent three editions of this book and two editions of the Advanced Engineering Electromagnetics book. I wanted, as long as I was able to accomplish the tasks, to have the books manifest my own fingerprint and reflect my personal philoso-phy, methodology and pedagogy. Also I wanted the manuscript to display continuity and consistency, and to control my own destiny, in terms of material to be included and excluded, revisions, deadlines and timelines. Finally, I wanted to be responsible for the contents of the book. In the words of Frank Sinatra, ‘I did it my way.’ Each edition presented its own challenges, but each time I cherished and looked forward to the mission and venture.
I am also grateful to the staff of John Wiley & Sons, Inc., especially Brett Kurzman, Editor, Alex Castro, Editorial Assistant, and Danielle LaCourciere, Production Editor for this edition. Special thanks to Shikha Sharma, from Aptara, Inc., for supervising the typesetting of the book. Finally I must pay tribute and homage to my family (Helen, Renie, Stephanie, Bill, Pete and Ellie) for their unconditional support, patience, sacrifice, and understanding for the many years of neglect during the completion of all four editions of this book and two editions of the Advanced Engineering Electromagnetics. Each edition has been a pleasant experience although a daunting task.
Constantine A. Balanis Arizona State University Tempe, AZ About the Companion Website There is a student companion website that contains: r PowerPoint Viewgraphs r MATLAB Programs r JAVA Applets r Animations r End-of-Chapter Interactive Questionnaires To access the material on the companion site simply find your unique website redemption code printed on the inside front endpaper of this book. Peel off the sticker and then visit www.wiley.com/go/antennatheory4e to follow the instructions for how to register your pin.
If you have purchased this title as an e-book, Wiley Customer Care will provide your access code for the companion website. Visit to request via the “Live Chat” or “Ask A Question” tabs, within 90 days of purchase, and please have your receipt for verification.
This book is also accompanied by a password protected companion website for instructors only.
This website contains: r Power Point Viewgraphs r MATLAB Programs r JAVA Applets r Animations r End-of-Chapter Interactive Questionnaires r Solutions Manual To access the material on the instructor’s website simply visit www.wiley.com/go/ instructors antennatheory4e and follow the instructions for how to register.
xix CHAPTER1 Antennas 1.1 INTRODUCTION An antenna is defined by Webster’s Dictionary as “a usually metallic device (as a rod or wire) for radiating or receiving radio waves.” The IEEE Standard Definitions of Terms for Antennas (IEEE Std 145–1983)∗defines the antenna or aerial as “a means for radiating or receiving radio waves.” In other words the antenna is the transitional structure between free-space and a guiding device, as shown in Figure 1.1. The guiding device or transmission line may take the form of a coaxial line or a hollow pipe (waveguide), and it is used to transport electromagnetic energy from the transmitting source to the antenna, or from the antenna to the receiver. In the former case, we have a transmitting antenna and in the latter a receiving antenna.
A transmission-line Thevenin equivalent of the antenna system of Figure 1.1 in the transmitting mode is shown in Figure 1.2 where the source is represented by an ideal generator, the transmission line is represented by a line with characteristic impedance Zc, and the antenna is represented by a load ZA [ZA = (RL + Rr) + jXA] connected to the transmission line. The Thevenin and Norton circuit equivalents of the antenna are also shown in Figure 2.27. The load resistance RL is used to represent the conduction and dielectric losses associated with the antenna structure while Rr, referred to as the radiation resistance, is used to represent radiation by the antenna. The reactance XA is used to represent the imaginary part of the impedance associated with radiation by the antenna. This is discussed more in detail in Sections 2.13 and 2.14. Under ideal conditions, energy generated by the source should be totally transferred to the radiation resistance Rr, which is used to represent radiation by the antenna. However, in a practical system there are conduction-dielectric losses due to the lossy nature of the transmission line and the antenna, as well as those due to reflections (mismatch) losses at the interface between the line and the antenna. Taking into account the internal impedance of the source and neglecting line and reflection (mismatch) losses, maximum power is delivered to the antenna under conjugate matching. This is discussed in Section 2.13.
The reflected waves from the interface create, along with the traveling waves from the source toward the antenna, constructive and destructive interference patterns, referred to as standing waves, inside the transmission line which represent pockets of energy concentrations and storage, typical of resonant devices. A typical standing wave pattern is shown dashed in Figure 1.2, while another is exhibited in Figure 1.15. If the antenna system is not properly designed, the transmission line ∗IEEE Transactions on Antennas and Propagation, vols. AP-17, No. 3, May 1969; AP-22, No. 1, January 1974; and AP-31, No. 6, Part II, November 1983.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1 2 ANTENNAS E-field Radiated free-space wave Antenna Transmission line Source Figure 1.1 Antenna as a transition device.
could act to a large degree as an energy storage element instead of as a wave guiding and energy transporting device. If the maximum field intensities of the standing wave are sufficiently large, they can cause arching inside the transmission lines.
The losses due to the line, antenna, and the standing waves are undesirable. The losses due to the line can be minimized by selecting low-loss lines while those of the antenna can be decreased by XA Zg Vg Rr RL Standing wave ZA = (RL + Rr) + jXA Source Antennna Transmission line Figure 1.2 Transmission-line Thevenin equivalent of antenna in transmitting mode.
TYPES OF ANTENNAS 3 reducing the loss resistance represented by RL in Figure 1.2. The standing waves can be reduced, and the energy storage capacity of the line minimized, by matching the impedance of the antenna (load) to the characteristic impedance of the line. This is the same as matching loads to transmission lines, where the load here is the antenna, and is discussed more in detail in Section 9.7. An equivalent similar to that of Figure 1.2 is used to represent the antenna system in the receiving mode where the source is replaced by a receiver. All other parts of the transmission-line equivalent remain the same.
The radiation resistance Rr is used to represent in the receiving mode the transfer of energy from the free-space wave to the antenna. This is discussed in Section 2.13 and represented by the Thevenin and Norton circuit equivalents of Figure 2.27.
In addition to receiving or transmitting energy, an antenna in an advanced wireless system is usually required to optimize or accentuate the radiation energy in some directions and suppress it in others. Thus the antenna must also serve as a directional device in addition to a probing device. It must then take various forms to meet the particular need at hand, and it may be a piece of conducting wire, an aperture, a patch, an assembly of elements (array), a reflector, a lens, and so forth.
For wireless communication systems, the antenna is one of the most critical components. A good design of the antenna can relax system requirements and improve overall system performance. A typical example is the TV for which the overall broadcast reception can be improved by utilizing a high-performance antenna. The antenna serves to a communication system the same purpose that eyes and eyeglasses serve to a human.
The field of antennas is vigorous and dynamic, and over the last 60 years antenna technology has been an indispensable partner of the communications revolution. Many major advances that occurred during this period are in common use today; however, many more issues and challenges are facing us today, especially since the demands for system performances are even greater. Many of the major advances in antenna technology that have been completed in the 1970s through the early 1990s, those that were under way in the early 1990s, and signals of future discoveries and breakthroughs were captured in a special issue of the Proceedings of the IEEE (Vol. 80, No. 1, January 1992) devoted to Antennas. The introductory paper of this special issue provides a carefully structured, elegant discussion of the fundamental principles of radiating elements and has been written as an introduction for the nonspecialist and a review for the expert.
1.2 TYPES OF ANTENNAS We will now introduce and briefly discuss some forms of the various antenna types in order to get a glance as to what will be encountered in the remainder of the book.
1.2.1 Wire Antennas Wire antennas are familiar to the layman because they are seen virtually everywhere—on automo-biles, buildings, ships, aircraft, spacecraft, and so on. There are various shapes of wire antennas such as a straight wire (dipole), loop, and helix which are shown in Figure 1.3. Loop antennas need not only be circular. They may take the form of a rectangle, square, ellipse, or any other configuration.
The circular loop is the most common because of its simplicity in construction. Dipoles are discussed in more detail in Chapter 4, loops in Chapter 5, and helices in Chapter 10.
1.2.2 Aperture Antennas Aperture antennas may be more familiar to the layman today than in the past because of the increasing demand for more sophisticated forms of antennas and the utilization of higher frequencies. Some forms of aperture antennas are shown in Figure 1.4. Antennas of this type are very useful for aircraft and spacecraft applications, because they can be very conveniently flush-mounted on the skin of 4 ANTENNAS Figure 1.3 Wire antenna configurations.
(a) Pyramidal horn (b) Conical horn (c) Rectangular waveguide Figure 1.4 Aperture antenna configurations.
TYPES OF ANTENNAS 5 the aircraft or spacecraft. In addition, they can be covered with a dielectric material to protect them from hazardous conditions of the environment. Waveguide apertures are discussed in more detail in Chapter 12 while horns are examined in Chapter 13.
1.2.3 Microstrip Antennas Microstrip antennas became very popular in the 1970s primarily for spaceborne applications. Today they are used for government and commercial applications. These antennas consist of a metallic patch on a grounded substrate. The metallic patch can take many different configurations, as shown in Figure 14.2. However, the rectangular and circular patches, shown in Figure 1.5, are the most popular because of ease of analysis and fabrication, and their attractive radiation characteristics, especially low cross-polarization radiation. The microstrip antennas are low profile, comformable to planar and nonplanar surfaces, simple and inexpensive to fabricate using modern printed-circuit technology, mechanically robust when mounted on rigid surfaces, compatible with MMIC designs, and very versatile in terms of resonant frequency, polarization, pattern, and impedance. These antennas can be mounted on the surface of high-performance aircraft, spacecraft, satellites, missiles, cars, and even mobile devices. They are discussed in more detail in Chapter 14.
1.2.4 Array Antennas Many applications require radiation characteristics that may not be achievable by a single element.
It may, however, be possible that an aggregate of radiating elements in an electrical and geometrical h h Ground plane (a) Rectangular Patch Substrate L t W Ground plane (b) Circular r t Substrate r Patch a Figure 1.5 Rectangular and circular microstrip (patch) antennas.
6 ANTENNAS Reflectors Directors Feed element (a) Yagi-Uda array (c) Microstrip patch array (d) Slotted-waveguide array Patch r Substrate Ground plane (b) Aperture array Figure 1.6 Typical wire, aperture, and microstrip array configurations.
arrangement (an array) will result in the desired radiation characteristics. The arrangement of the array may be such that the radiation from the elements adds up to give a radiation maximum in a particular direction or directions, minimum in others, or otherwise as desired. Typical examples of arrays are shown in Figure 1.6. Usually the term array is reserved for an arrangement in which the individual radiators are separate as shown in Figures 1.6(a–c). However the same term is also used to describe an assembly of radiators mounted on a continuous structure, shown in Figure 1.6(d).
1.2.5 Reflector Antennas The success in the exploration of outer space has resulted in the advancement of antenna theory.
Because of the need to communicate over great distances, sophisticated forms of antennas had to be used in order to transmit and receive signals that had to travel millions of miles. A very com-mon antenna form for such an application is a parabolic reflector shown in Figures 1.7(a) and (b).
Antennas of this type have been built with diameters of 305 m or even larger. Such large dimensions are needed to achieve the high gain required to transmit or receive signals after millions of miles of travel. Another form of a reflector, although not as common as the parabolic, is the corner reflector, shown in Figure 1.7(c). These antennas are examined in detail in Chapter 15.
1.2.6 Lens Antennas Lenses are primarily used to collimate incident divergent energy to prevent it from spreading in undesired directions. By properly shaping the geometrical configuration and choosing the appropri-ate material of the lenses, they can transform various forms of divergent energy into plane waves.
They can be used in most of the same applications as are the parabolic reflectors, especially at RADIATION MECHANISM 7 Figure 1.7 Typical reflector configurations.
higher frequencies. Their dimensions and weight become exceedingly large at lower frequencies.
Lens antennas are classified according to the material from which they are constructed, or according to their geometrical shape. Some forms are shown in Figure 1.8 .
In summary, an ideal antenna is one that will radiate all the power delivered to it from the trans-mitter in a desired direction or directions. In practice, however, such ideal performances cannot be achieved but may be closely approached. Various types of antennas are available and each type can take different forms in order to achieve the desired radiation characteristics for the particular appli-cation. Throughout the book, the radiation characteristics of most of these antennas are discussed in detail.
1.3 RADIATION MECHANISM One of the first questions that may be asked concerning antennas would be “how is radiation accom-plished?” In other words, how are the electromagnetic fields generated by the source, contained and guided within the transmission line and antenna, and finally “detached” from the antenna to form a free-space wave? The best explanation may be given by an illustration. However, let us first examine some basic sources of radiation.
1.3.1 Single Wire Conducting wires are material whose prominent characteristic is the motion of electric charges and the creation of current. Let us assume that an electric volume charge density, represented by qv 8 ANTENNAS Figure 1.8 Typical lens antenna configurations. (source: L. V. Blake, Antennas, Wiley, New York, 1966).
(coulombs/m3), is distributed uniformly in a circular wire of cross-sectional area A and volume V, as shown in Figure 1.9. The total charge Q within volume V is moving in the z direction with a uniform velocity vz (meters/sec). It can be shown that the current density Jz (amperes/m2) over the cross section of the wire is given by Jz = qvvz (1-1a) If the wire is made of an ideal electric conductor, the current density Js (amperes/m) resides on the surface of the wire and it is given by Js = qsvz (1-1b) z x y V A E Jc +vz Δz l Figure 1.9 Charge uniformly distributed in a circular cross section cylinder wire.
RADIATION MECHANISM 9 where qs (coulombs/m2) is the surface charge density. If the wire is very thin (ideally zero radius), then the current in the wire can be represented by Iz = qlvz (1-1c) where ql (coulombs/m) is the charge per unit length.
Instead of examining all three current densities, we will primarily concentrate on the very thin wire. The conclusions apply to all three. If the current is time varying, then the derivative of the current of (1-1c) can be written as dIz dt = ql dvz dt = qlaz (1-2) where dvz∕dt = az (meters/sec2) is the acceleration. If the wire is of length l, then (1-2) can be written as ldIz dt = lql dvz dt = lqlaz (1-3) Equation (1-3) is the basic relation between current and charge, and it also serves as the fundamental relation of electromagnetic radiation , . It simply states that to create radiation, there must be a time-varying current or an acceleration (or deceleration) of charge. We usually refer to currents in time-harmonic applications while charge is most often mentioned in transients. To create charge acceleration (or deceleration) the wire must be curved, bent, discontinuous, or terminated , .
Periodic charge acceleration (or deceleration) or time-varying current is also created when charge is oscillating in a time-harmonic motion, as shown in Figure 1.17 for a λ∕2 dipole. Therefore: 1. If a charge is not moving, current is not created and there is no radiation.
2. If charge is moving with a uniform velocity: a. There is no radiation if the wire is straight, and infinite in extent.
b. There is radiation if the wire is curved, bent, discontinuous, terminated, or truncated, as shown in Figure 1.10.
3. If charge is oscillating in a time-motion, it radiates even if the wire is straight.
A qualitative understanding of the radiation mechanism may be obtained by considering a pulse source attached to an open-ended conducting wire, which may be connected to the ground through a discrete load at its open end, as shown in Figure 1.10(d). When the wire is initially energized, the charges (free electrons) in the wire are set in motion by the electrical lines of force created by the source. When charges are accelerated in the source-end of the wire and decelerated (negative accel-eration with respect to original motion) during reflection from its end, it is suggested that radiated fields are produced at each end and along the remaining part of the wire, , . Stronger radiation with a more broad frequency spectrum occurs if the pulses are of shorter or more compact duration while continuous time-harmonic oscillating charge produces, ideally, radiation of single frequency determined by the frequency of oscillation. The acceleration of the charges is accomplished by the external source in which forces set the charges in motion and produce the associated field radiated.
The deceleration of the charges at the end of the wire is accomplished by the internal (self) forces associated with the induced field due to the buildup of charge concentration at the ends of the wire.
The internal forces receive energy from the charge buildup as its velocity is reduced to zero at the ends of the wire. Therefore, charge acceleration due to an exciting electric field and deceleration due 10 ANTENNAS (e) Truncated (d) Terminated Ground ZL (c) Discontinuous (b) Bent (a) Curved Figure 1.10 Wire configurations for radiation.
to impedance discontinuities or smooth curves of the wire are mechanisms responsible for electro-magnetic radiation. While both current density (Jc) and charge density (qv) appear as source terms in Maxwell’s equation, charge is viewed as a more fundamental quantity, especially for transient fields. Even though this interpretation of radiation is primarily used for transients, it can be used to explain steady state radiation .
1.3.2 Two-Wires Let us consider a voltage source connected to a two-conductor transmission line which is connected to an antenna. This is shown in Figure 1.11(a). Applying a voltage across the two-conductor trans-mission line creates an electric field between the conductors. The electric field has associated with it electric lines of force which are tangent to the electric field at each point and their strength is proportional to the electric field intensity. The electric lines of force have a tendency to act on the free electrons (easily detachable from the atoms) associated with each conductor and force them to be displaced. The movement of the charges creates a current that in turn creates a magnetic field intensity. Associated with the magnetic field intensity are magnetic lines of force which are tangent to the magnetic field.
RADIATION MECHANISM 11 Figure 1.11 Source, transmission line, antenna, and detachment of electric field lines.
We have accepted that electric field lines start on positive charges and end on negative charges.
They also can start on a positive charge and end at infinity, start at infinity and end on a negative charge, or form closed loops neither starting or ending on any charge. Magnetic field lines always form closed loops encircling current-carrying conductors because physically there are no magnetic charges. In some mathematical formulations, it is often convenient to introduce equivalent magnetic charges and magnetic currents to draw a parallel between solutions involving electric and mag-netic sources.
The electric field lines drawn between the two conductors help to exhibit the distribution of charge.
If we assume that the voltage source is sinusoidal, we expect the electric field between the conduc-tors to also be sinusoidal with a period equal to that of the applied source. The relative magnitude of the electric field intensity is indicated by the density (bunching) of the lines of force with the arrows showing the relative direction (positive or negative). The creation of time-varying electric and mag-netic fields between the conductors forms electromagnetic waves which travel along the transmission line, as shown in Figure 1.11(a). The electromagnetic waves enter the antenna and have associated 12 ANTENNAS Figure 1.12 Electric field lines of free-space wave for a λ∕2 antenna at t = 0, T/8, T/4, and 3T/8. (source: J. D. Kraus, Electromagnetics, 4th ed., McGraw-Hill, New York, 1992. Reprinted with permission of J. D. Kraus and John D. Cowan, Jr.).
with them electric charges and corresponding currents. If we remove part of the antenna structure, as shown in Figure 1.11(b), free-space waves can be formed by “connecting” the open ends of the electric lines (shown dashed). The free-space waves are also periodic but a constant phase point P0 moves outwardly with the speed of light and travels a distance of λ∕2 (to P1) in the time of one-half of a period. It has been shown that close to the antenna the constant phase point P0 moves faster than the speed of light but approaches the speed of light at points far away from the antenna (analo-gous to phase velocity inside a rectangular waveguide). Figure 1.12 displays the creation and travel of free-space waves by a prolate spheroid with λ∕2 interfocal distance where λ is the wavelength.
The free-space waves of a center-fed λ∕2 dipole, except in the immediate vicinity of the antenna, are essentially the same as those of the prolate spheroid.
The question still unanswered is how the guided waves are detached from the antenna to create the free-space waves that are indicated as closed loops in Figures 1.11 and 1.12. Before we attempt to explain that, let us draw a parallel between the guided and free-space waves, and water waves created by the dropping of a pebble in a calm body of water or initiated in some other manner.
Once the disturbance in the water has been initiated, water waves are created which begin to travel outwardly. If the disturbance has been removed, the waves do not stop or extinguish themselves but continue their course of travel. If the disturbance persists, new waves are continuously created which lag in their travel behind the others. The same is true with the electromagnetic waves created by an electric disturbance. If the initial electric disturbance by the source is of a short duration, the created electromagnetic waves travel inside the transmission line, then into the antenna, and finally are radiated as free-space waves, even if the electric source has ceased to exist (as was with the water waves and their generating disturbance). If the electric disturbance is of a continuous nature, electromagnetic waves exist continuously and follow in their travel behind the others. This is shown in Figure 1.13 for a biconical antenna. When the electromagnetic waves are within the transmission line and antenna, their existence is associated with the presence of the charges inside the conductors.
However, when the waves are radiated, they form closed loops and there are no charges to sustain RADIATION MECHANISM 13 Figure 1.13 Electric field lines of free-space wave for biconical antenna.
their existence. This leads us to conclude that electric charges are required to excite the fields but are not needed to sustain them and may exist in their absence. This is in direct analogy with water waves.
1.3.3 Dipole Now let us attempt to explain the mechanism by which the electric lines of force are detached from the antenna to form the free-space waves. This will again be illustrated by an example of a small dipole antenna where the time of travel is negligible. This is only necessary to give a better physical interpretation of the detachment of the lines of force. Although a somewhat simplified mechanism, it does allow one to visualize the creation of the free-space waves. Figure 1.14(a) displays the lines of force created between the arms of a small center-fed dipole in the first quarter of the period during which time the charge has reached its maximum value (assuming a sinusoidal time variation) and the lines have traveled outwardly a radial distance λ∕4. For this example, let us assume that the number of lines formed are three. During the next quarter of the period, the original three lines travel an additional λ∕4 (a total of λ∕2 from the initial point) and the charge density on the conductors begins to diminish. This can be thought of as being accomplished by introducing opposite charges which at the end of the first half of the period have neutralized the charges on the conductors. The lines of force created by the opposite charges are three and travel a distance λ∕4 during the second quarter of the first half, and they are shown dashed in Figure 1.14(b). The end result is that there are three lines of force pointed upward in the first λ∕4 distance and the same number of lines directed downward in the second λ∕4. Since there is no net charge on the antenna, then the lines of force must have been forced to detach themselves from the conductors and to unite together to form closed loops.
This is shown in Figure 1.14(c). In the remaining second half of the period, the same procedure is followed but in the opposite direction. After that, the process is repeated and continues indefinitely and electric field patterns, similar to those of Figure 1.12, are formed.
1.3.4 Computer Animation-Visualization of Radiation Problems A difficulty that students usually confront is that the subject of electromagnetics is rather abstract, and it is hard to visualize electromagnetic wave propagation and interaction. With today’s advanced numerical and computational methods, and animation and visualization software and hardware, this dilemma can, to a large extent, be minimized. To address this problem, we have developed and 14 ANTENNAS Figure 1.14 Formation and detachment of electric field lines for short dipole.
included in this chapter computer programs to animate and visualize three radiation mechanisms.
Descriptions of the computer programs are found in the website created by the publisher for this book. Each problem is solved using the Finite-Difference Time-Domain (FD-TD) method –, a method which solves Maxwell’s equations as a function of time in discrete time steps at discrete points in space. A picture of the fields can then be taken at each time step to create a video which can be viewed as a function of time. Other animation and visualization software, referred to as applets, are included in the book website.
The three radiation problems that are animated and can be visualized using the computer program of this chapter and included in the book website are: a. Infinite length line source (two-dimensional) excited by a single Gaussian pulse and radiating in an unbounded medium.
b. Infinite length line source (two-dimensional) excited by a single Gaussian pulse and radiating inside a perfectly electric conducting (PEC) square cylinder.
c. E-plane sectoral horn (two-dimensional form of Figure 13.2) excited by a continuous cosinu-soidal voltage source and radiating in an unbounded medium.
CURRENT DISTRIBUTION ON A THIN WIRE ANTENNA 15 In order to animate and then visualize each of the three radiation problems, the user needs MATLAB and the MATLAB M-file, found in the publisher’s website for the book, to produce the corresponding FD-TD solution of each radiation problem. For each radiation problem, the M-File executed in MATLAB produces a video by taking a picture of the computational domain every third time step. The video is viewed as a function of time as the wave travels in the computational space.
A. Infinite Line Source in an Unbounded Medium (tm open) The first FD-TD solution is that of an infinite length line source excited by a single time-derivative Gaussian pulse, with a duration of approximately 0.4 nanoseconds, in a two-dimensional TMz-computational domain. The unbounded medium is simulated using a six-layer Berenger Perfectly Matched Layer (PML) Absorbing Boundary Condition (ABC) , to truncate the computa-tional space at a finite distance without, in principle, creating any reflections. Thus, the pulse travels radially outward creating a traveling type of a wavefront. The outward moving wavefronts are easily identified using the coloring scheme for the intensity (or gray scale for black and white monitors) when viewing the video. The video is created by the MATLAB M-File which produces the FD-TD solution by taking a picture of the computational domain every third time step. Each time step is 5 picoseconds while each FD-TD cell is 3 mm on a side. The video is 37 frames long covering 185 picoseconds of elapsed time. The entire computational space is 15.3 cm by 15.3 cm and is mod-eled by 2500 square FD-TD cells (50 × 50), including 6 cells to implement the PML ABC.
B. Infinite Line Source in a PEC Square Cylinder (tm box) This problem is simulated similarly as that of the line source in an unbounded medium, including the characteristics of the pulse. The major difference is that the computational domain of this problem is truncated by PEC walls; therefore there is no need for PML ABC. For this problem the pulse travels in an outward direction and is reflected when it reaches the walls of the cylinder. The reflected pulse, along with the radially outward traveling pulse, interfere constructively and destructively with each other and create a standing type of a wavefront. The peaks and valleys of the modified wavefront can be easily identified when viewing the video, using the colored or gray scale intensity schemes.
Sufficient time is allowed in the video to permit the pulse to travel from the source to the walls of the cylinder, return back to the source, and then return back to the walls of the cylinder. Each time step is 5 picoseconds and each FD-TD cell is 3 mm on a side. The video is 70 frames long covering 350 picoseconds of elapsed time. The square cylinder, and thus the computational space, has a cross section of 15.3 cm by 15.3 cm and is modeled using an area 50 by 50 FD-TD cells.
C. E-Plane Sectoral Horn in an Unbounded Medium (te horn) The E-plane sectoral horn is excited by a cosinusoidal voltage (CW) of 9.84 GHz in a TEz com-putational domain, instead of the Gaussian pulse excitation of the previous two problems. The unbounded medium is implemented using an eight-layer Berenger PML ABC. The computational space is 25.4 cm by 25.4 cm and is modeled using 100 by 100 FD-TD cells (each square cell being 2.54 mm on a side). The video is 70 frames long covering 296 picoseconds of elapsed time and is created by taking a picture every third frame. Each time step is 4.23 picoseconds in duration. The horn has a total flare angle of 52◦and its flared section is 2.62 cm long, is fed by a parallel plate 1 cm wide and 4.06 cm long, and has an aperture of 3.56 cm.
1.4 CURRENT DISTRIBUTION ON A THIN WIRE ANTENNA In the preceding section we discussed the movement of the free electrons on the conductors rep-resenting the transmission line and the antenna. In order to illustrate the creation of the current distribution on a linear dipole, and its subsequent radiation, let us first begin with the geometry of a 16 ANTENNAS Figure 1.15 Current distribution on a lossless two-wire transmission line, flared transmission line, and lin-ear dipole.
lossless two-wire transmission line, as shown in Figure 1.15(a). The movement of the charges cre-ates a traveling wave current, of magnitude I0∕2, along each of the wires. When the current arrives at the end of each of the wires, it undergoes a complete reflection (equal magnitude and 180◦phase reversal). The reflected traveling wave, when combined with the incident traveling wave, forms in each wire a pure standing wave pattern of sinusoidal form as shown in Figure 1.15(a). The current in each wire undergoes a 180◦phase reversal between adjoining half-cycles. This is indicated in Figure 1.15(a) by the reversal of the arrow direction. Radiation from each wire individually occurs because of the time-varying nature of the current and the termination of the wire.
For the two-wire balanced (symmetrical) transmission line, the current in a half-cycle of one wire is of the same magnitude but 180◦out-of-phase from that in the corresponding half-cycle of the other wire. If in addition the spacing between the two wires is very small (s ≪λ), the fields radiated by the current of each wire are essentially cancelled by those of the other. The net result is an almost ideal (and desired) nonradiating transmission line.
CURRENT DISTRIBUTION ON A THIN WIRE ANTENNA 17 As the section of the transmission line between 0 ≤z ≤l∕2 begins to flare, as shown in Fig-ure 1.15(b), it can be assumed that the current distribution is essentially unaltered in form in each of the wires. However, because the two wires of the flared section are not necessarily close to each other, the fields radiated by one do not necessarily cancel those of the other. Therefore, ideally, there is a net radiation by the transmission-line system.
Ultimately the flared section of the transmission line can take the form shown in Figure 1.15(c).
This is the geometry of the widely used dipole antenna. Because of the standing wave current pattern, it is also classified as a standing wave antenna (as contrasted to the traveling wave antennas which will be discussed in detail in Chapter 10). If l < λ, the phase of the current standing wave pattern in each arm is the same throughout its length. In addition, spatially it is oriented in the same direction as that of the other arm as shown in Figure 1.15(c). Thus the fields radiated by the two arms of the dipole (vertical parts of a flared transmission line) will primarily reinforce each other toward most directions of observation (the phase due to the relative position of each small part of each arm must also be included for a complete description of the radiation pattern formation).
If the diameter of each wire is very small (d ≪λ), the ideal standing wave pattern of the current along the arms of the dipole is sinusoidal with a null at the end. However, its overall form depends on the length of each arm. For center-fed dipoles with l ≪λ, l = λ∕2, λ∕2 < l < λ and λ < l < 3λ∕2, the current patterns are illustrated in Figures 1.16(a–d). The current pattern of a very small dipole (usually λ∕50 < l ≤λ∕10) can be approximated by a triangular distribution since sin(kl∕2) ≃kl∕2 when kl/2 is very small. This is illustrated in Figure 1.16(a).
Figure 1.16 Current distribution on linear dipoles.
18 ANTENNAS Figure 1.17 Current distribution on a λ∕2 wire antenna for different times.
Because of its cyclical spatial variations, the current standing wave pattern of a dipole longer than λ(l > λ) undergoes 180◦phase reversals between adjoining half-cycles. Therefore the current in all parts of the dipole does not have the same phase. This is demonstrated graphically in Figure 1.16(d) for λ < l < 3λ∕2. In turn, the fields radiated by some parts of the dipole will not reinforce those of the others. As a result, significant interference and cancelling effects will be noted in the formation of the total radiation pattern. See Figure 4.11 for the pattern of a λ∕2 dipole and Figure 4.7 for that of a 1.25λ dipole.
For a time-harmonic varying system of radian frequency 𝜔= 2𝜋f , the current standing wave patterns of Figure 1.16 represent the maximum current excitation for any time. The current variations, as a function of time, on a λ∕2 center-fed dipole, are shown in Figure 1.17 for 0 ≤t ≤T∕2 where T is the period. These variations can be obtained by multiplying the current standing wave pattern of Figure 1.16(b) by cos(𝜔t).
1.5 HISTORICAL ADVANCEMENT The history of antennas dates back to James Clerk Maxwell who unified the theories of electric-ity and magnetism, and eloquently represented their relations through a set of profound equations best known as Maxwell’s Equations. His work was first published in 1873 . He also showed that light was electromagnetic and that both light and electromagnetic waves travel by wave distur-bances of the same speed. In 1886, Professor Heinrich Rudolph Hertz demonstrated the first wireless electromagnetic system. He was able to produce in his laboratory at a wavelength of 4 m a spark in the gap of a transmitting λ∕2 dipole which was then detected as a spark in the gap of a nearby loop. It was not until 1901 that Guglielmo Marconi was able to send signals over large distances.
He performed, in 1901, the first transatlantic transmission from Poldhu in Cornwall, England, to St.
John’s Newfoundland. His transmitting antenna consisted of 50 vertical wires in the form of a fan connected to ground through a spark transmitter. The wires were supported horizontally by a guyed wire between two 60-m wooden poles. The receiving antenna at St. John’s was a 200-m wire pulled and supported by a kite. This was the dawn of the antenna era.
From Marconi’s inception through the 1940s, antenna technology was primarily centered on wire related radiating elements and frequencies up to about UHF. It was not until World War II that modern antenna technology was launched and new elements (such as waveguide apertures, horns, reflectors) HISTORICAL ADVANCEMENT 19 were primarily introduced. Much of this work is captured in the book by Silver . A contributing factor to this new era was the invention of microwave sources (such as the klystron and magnetron) with frequencies of 1 GHz and above.
While World War II launched a new era in antennas, advances made in computer architecture and technology during the 1960s through the 1990s have had a major impact on the advance of modern antenna technology, and they are expected to have an even greater influence on antenna engineer-ing into the twenty-first century. Beginning primarily in the early 1960s, numerical methods were introduced that allowed previously intractable complex antenna system configurations to be ana-lyzed and designed very accurately. In addition, asymptotic methods for both low frequencies (e.g., Moment Method (MM), Finite-Difference, Finite-Element) and high frequencies (e.g., Geometrical and Physical Theories of Diffraction) were introduced, contributing significantly to the maturity of the antenna field. While in the past antenna design may have been considered a secondary issue in overall system design, today it plays a critical role. In fact, many system successes rely on the design and performance of the antenna. Also, while in the first half of this century antenna technology may have been considered almost a “cut and try” operation, today it is truly an engineering science. Anal-ysis and design methods are such that antenna system performance can be predicted with remarkable accuracy. In fact, many antenna designs proceed directly from the initial design stage to the prototype without intermediate testing. The level of confidence has increased tremendously.
The widespread interest in antennas is reflected by the large number of books written on the subject . These have been classified under four categories: Fundamental, Handbooks, Measure-ments, and Specialized. This is an outstanding collection of books, and it reflects the popularity of the antenna subject, especially since the 1950s. Because of space limitations, only a partial list is included here , , , –, including the first, second and third editions of this book in 1982, 1997, 2005. Some of other books are now out of print.
1.5.1 Antenna Elements Prior to World War II most antenna elements were of the wire type (long wires, dipoles, helices, rhombuses, fans, etc.), and they were used either as single elements or in arrays. During and after World War II, many other radiators, some of which may have been known for some and others of which were relatively new, were put into service. This created a need for better understanding and optimization of their radiation characteristics. Many of these antennas were of the aperture type (such as open-ended waveguides, slots, horns, reflectors, lenses), and they have been used for com-munication, radar, remote sensing, and deep space applications both on airborne and earth-based platforms. Many of these operate in the microwave region and are discussed in Chapters 12, 13, 15 and in .
Prior to the 1950s, antennas with broadband pattern and impedance characteristics had band-widths not much greater than about 2:1. In the 1950s, a breakthrough in antenna evolution was cre-ated which extended the maximum bandwidth to as great as 40:1 or more. Because the geometries of these antennas are specified by angles instead of linear dimensions, they have ideally an infinite bandwidth. Therefore, they are referred to as frequency independent. These antennas are primarily used in the 10–10,000 MHz region in a variety of applications including TV, point-to-point com-munications, feeds for reflectors and lenses, and many others. This class of antennas is discussed in more detail in Chapter 11 and in .
It was not until almost 20 years later that a fundamental new radiating element, which has received a lot of attention and many applications since its inception, was introduced. This occurred in the early 1970s when the microstrip or patch antennas was reported. This element is simple, lightweight, inexpensive, low profile, and conformal to the surface. These antennas are discussed in more detail in Chapter 14 and in .
Major advances in millimeter wave antennas have been made in recent years, including integrated antennas where active and passive circuits are combined with the radiating elements in one compact unit (monolithic form). These antennas are discussed in .
20 ANTENNAS Specific radiation pattern requirements usually cannot be achieved by single antenna elements, because single elements usually have relatively wide radiation patterns and low values of directivity.
To design antennas with very large directivities, it is usually necessary to increase the electrical size of the antenna. This can be accomplished by enlarging the electrical dimensions of the chosen single element. However, mechanical problems are usually associated with very large elements. An alternative way to achieve large directivities, without increasing the size of the individual elements, is to use multiple single elements to form an array. An array is a sampled version of a very large single element. In an array, the mechanical problems of large single elements are traded for the electrical problems associated with the feed networks of arrays. However, with today’s solid-state technology, very efficient and low-cost feed networks can be designed.
Arrays are the most versatile of antenna systems. They find wide applications not only in many spaceborne systems, but in many earthbound missions as well. In most cases, the elements of an array are identical; this is not necessary, but it is often more convenient, simpler, and more practical.
With arrays, it is practical not only to synthesize almost any desired amplitude radiation pattern, but the main lobe can be scanned by controlling the relative phase excitation between the elements. This is most convenient for applications where the antenna system is not readily accessible, especially for spaceborne missions. The beamwidth of the main lobe along with the side lobe level can be controlled by the relative amplitude excitation (distribution) between the elements of the array. In fact, there is a trade-off between the beamwidth and the side lobe level based on the amplitude distribution. Analysis, design, and synthesis of arrays are discussed in Chapters 6 and 7. However, advances in array technology are reported in –.
A new antenna array design referred to as smart antenna, based on basic technology of the 1970s and 1980s, is sparking interest especially for wireless applications. This antenna design, which com-bines antenna technology with that of digital signal processing (DSP), is discussed in some detail in Chapter 16.
1.5.2 Methods of Analysis There is plethora of antenna elements, many of which exhibit intricate configurations. To analyze each as a boundary-value problem and obtain solutions in closed form, the antenna structure must be described by an orthogonal curvilinear coordinate system. This places severe restrictions on the type and number of antenna systems that can be analyzed using such a procedure. Therefore, other exact or approximate methods are often pursued. Two methods that in the last four decades have been preeminent in the analysis of many previously intractable antenna problems are the Integral Equation (IE) method and the Geometrical Theory of Diffraction (GTD).
The Integral Equation method casts the solution to the antenna problem in the form of an integral (hence its name) where the unknown, usually the induced current density, is part of the integrand.
Numerical techniques, such as the Moment Method (MM), are then used to solve for the unknown.
Once the current density is found, the radiation integrals of Chapter 3 are used to find the fields radiated and other systems parameters. This method is most convenient for wire-type antennas and more efficient for structures that are small electrically. One of the first objectives of this method is to formulate the IE for the problem at hand. In general, there are two type of IE’s. One is the Electric Field Integral Equation (EFIE), and it is based on the boundary condition of the total tangential electric field. The other is the Magnetic Field Integral Equation (MFIE), and it is based on the boundary condition that expresses the total electric current density induced on the surface in terms of the incident magnetic field. The MFIE is only valid for closed surfaces. For some problems, it is more convenient to formulate an EFIE, while for others it is more appropriate to use an MFIE. Advances, applications, and numerical issues of these methods are addressed in Chapter 8 and in and .
When the dimensions of the radiating system are many wavelengths, low-frequency methods are not as computationally efficient. However, high-frequency asymptotic techniques can be used to analyze many problems that are otherwise mathematically intractable. One such method that has MULTIMEDIA 21 received considerable attention and application over the years is the GTD/UTD, which is an exten-sion of geometrical optics (GO), and it overcomes some of the limitations of GO by introducing a diffraction mechanism. The Geometrical/Uniform Theory of Diffraction is briefly discussed in Sec-tion 12.10. However, a detailed treatment is found in Chapter 13 of while recent advances and applications are found in and .
For structures that are not convenient to analyze by either of the two methods, a combination of the two is often used. Such a technique is referred to as a hybrid method, and it is described in detail in . Another method, which has received a lot of attention in scattering, is the Finite-Difference Time-Domain (FDTD). This method has also been applied to antenna radiation problems –.
A method that has gained a lot of momentum in its application to antenna problems is the Finite Element Method –.
1.5.3 Some Future Challenges Antenna engineering has enjoyed a very successful period during the 1940s–1990s. Responsible for its success have been the introduction and technological advances of some new elements of radiation, such as aperture antennas, reflectors, frequency independent antennas, and microstrip antennas.
Excitement has been created by the advancement, utilization, and proliferation of Computational ElectoMagentics (CEM) software that provides students, engineers, and scientists with versatile and indispensable tools for modeling, visualizing, animating, and interpreting EM phenomena and characteristics. In addition, with such tools, electrically large structures that are complex and may otherwise be intractable can be designed and analyzed to gain insight into the performance of systems in order to advance and improve their efficiency. Today, antenna engineering is a science based on fundamental principles.
Although a certain level of maturity has been attained, there are many challenging opportunities and problems to be solved. Phased array architecture integrating monolithic MIC technology is still a most challenging problem. Integration of new materials, such as metamaterials , artificial mag-netic conductors and soft/hard surfaces , into antenna technology offers many opportunities to control, discipline, harness and manipulate the EM waves to design devices with desired and funtional characteristics, and improved performance. Computational electromagnetics using supercomputing and parallel computing capabilities will model complex electromagnetic wave interactions, in both the frequency and time domains. Innovative antenna designs, such as those using smart antennas , and multifunction, multiband, ultra wide hand, reconfigurable antennas and antenna systems , to perform complex and demanding system functions remain a challenge. New basic elements are always welcomed and offer refreshing opportunities. New applications include, but are not limited to nanotechnology, wireless communications, direct broadcast satellite systems, global positioning satellites (GPS), high-accuracy airborne navigation, security systems, global weather, earth resource systems, and others. Because of the many new applications, the lower portion of the EM spectrum has been saturated and the designs have been pushed to higher frequencies, including the millimeter wave and terahertz frequency bands.
1.6 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources related to this chapter are included: a. Java-based interactive questionnaire with answers.
b. Three Matlab-based animation-visualization programs designated r tm open 22 ANTENNAS r tm box r te horn which are described in detail in Section 1.3.4 and the corresponding READ ME file in the book website.
c. Power Point (PPT) viewgraphs.
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CHAPTER2 Fundamental Parameters and Figures-of-Merit of Antennas 2.1 INTRODUCTION To describe the performance of an antenna, definitions of various parameters are necessary. Some of the parameters are interrelated and not all of them need be specified for complete description of the antenna performance. Parameter definitions will be given in this chapter. Many of those in quotation marks are from the IEEE Standard Definitions of Terms for Antennas [IEEE Std 145-1993.
Reaffirmed 2004(R2004)]. This is a revision of the IEEE Std 145-1983.∗ 2.2 RADIATION PATTERN An antenna radiation pattern or antenna pattern is defined as “a mathematical function or a graph-ical representation of the radiation properties of the antenna as a function of space coordinates. In most cases, the radiation pattern is determined in the far-field region and is represented as a function of the directional coordinates. Radiation properties include power flux density, radiation intensity, field strength, directivity, phase or polarization.” The radiation property of most concern is the two-or three-dimensional spatial distribution of radiated energy as a function of the observer’s position along a path or surface of constant radius. A convenient set of coordinates is shown in Figure 2.1.
A trace of the received electric (magnetic) field at a constant radius is called the amplitude field pattern. On the other hand, a graph of the spatial variation of the power density along a constant radius is called an amplitude power pattern.
Often the field and power patterns are normalized with respect to their maximum value, yielding normalized field and power patterns. Also, the power pattern is usually plotted on a logarithmic scale or more commonly in decibels (dB). This scale is usually desirable because a logarithmic scale can accentuate in more details those parts of the pattern that have very low values, which later we will refer to as minor lobes. For an antenna, the a. field pattern (in linear scale) typically represents a plot of the magnitude of the electric or magnetic field as a function of the angular space.
∗IEEE Transactions on Antennas and Propagation, Vols. AP-17, No. 3, May 1969; Vol. AP-22, No. 1, January 1974; and Vol. AP-31, No. 6, Part II, November 1983.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 25 26 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Figure 2.1 Coordinate system for antenna analysis.
b. power pattern (in linear scale) typically represents a plot of the square of the magnitude of the electric or magnetic field as a function of the angular space.
c. power pattern (in dB) represents the magnitude of the electric or magnetic field, in decibels, as a function of the angular space.
To demonstrate this, the two-dimensional normalized field pattern (plotted in linear scale), power pattern (plotted in linear scale), and power pattern (plotted on a logarithmic dB scale) of a 10-element linear antenna array of isotropic sources, with a spacing of d = 0.25λ between the elements, are shown in Figure 2.2. In this and subsequent patterns, the plus (+) and minus (−) signs in the lobes indicate the relative polarization (positive or negative) of the amplitude between the various lobes, which changes (alternates) as the nulls are crossed. To find the points where the pattern achieves its half-power (−3 dB points), relative to the maximum value of the pattern, you set the value of the a. field pattern at 0.707 value of its maximum, as shown in Figure 2.2(a) b. power pattern (in a linear scale) at its 0.5 value of its maximum, as shown in Figure 2.2(b) c. power pattern (in dB) at −3 dB value of its maximum, as shown in Figure 2.2(c).
All three patterns yield the same angular separation between the two half-power points, 38.64◦, on their respective patterns, referred to as HPBW and illustrated in Figure 2.2. This is discussed in detail in Section 2.5.
In practice, the three-dimensional pattern is measured and recorded in a series of two-dimensional patterns. However, for most practical applications, a few plots of the pattern as a function of 𝜃for some particular values of 𝜙, plus a few plots as a function of 𝜙for some particular values of 𝜃, give most of the useful and needed information.
2.2.1 Radiation Pattern Lobes Various parts of a radiation pattern are referred to as lobes, which may be subclassified into major or main, minor, side, and back lobes.
RADIATION PATTERN 27 + + Field Pattern (linear scale) 0 9 1 7 0 7 .
0 7 0 7 .
0 HPBW 0.7 0.8 0.9 |Er| 0.5 0.6 --0.3 0.4 + + + + --0 0.1 0.2 + + -0 Power Pattern + + Power Pattern (linear scale) 0.9 1 HPBW 0.7 0.8 0.5 0.5 0.5 0.6 0 2 0.3 0.4 --0 0.1 0.2 --3 dB -3 dB HPBW + + Power Pattern (dB) -5 0 ---15 -10 5 + + -25 -20 15 + + ---35 -30 25 -35 (a) Field pattern (in linear scale) (b) Power pattern (in linear scale) (c) Power pattern (in dB) Figure 2.2 Two-dimensional normalized field pattern (linear scale), power pattern (linear scale), and power pattern (in dB) of a 10-element linear array with a spacing of d = 0.25λ.
A radiation lobe is a “portion of the radiation pattern bounded by regions of relatively weak radiation intensity.” Figure 2.3(a) demonstrates a symmetrical three-dimensional polar pattern with a number of radiation lobes. Some are of greater radiation intensity than others, but all are classified as lobes. Figure 2.3(b) illustrates a linear two-dimensional pattern [one plane of Figure 2.3(a)] where the same pattern characteristics are indicated.
28 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS z Normalized Field Pattern (linear scale) 1 First null beamwidth (FNBW) Major lobe 0.7 0.8 0.9 Half-power beamwidth (HPBW) + + 0.4 0.5 0.6 Side lobe 0.1 0.2 0.3 Minor lobes -+ -+ x y Minor lobes Back lobe --+ + -Minor lobes Normalized Field Pattern (linear scale) 0.9 1 Half-power beamwidth (HPBW) Major lobe Radiation intensity 0.6 0.7 0.8 + + p ( ) First null beamwidth (FNBW) j 0.3 0.4 0.5 HPBW Minor lobes Side lobe 0.1 0.2 FNBW + − − − − Back lobe + + + π π/2 0 π/2 π − − + (a) (b) Figure 2.3 (a) Radiation lobes and beamwidths of an antenna amplitude pattern in polar form. (b) Linear plot of power pattern and its associated lobes and beamwidths.
MATLAB-based computer programs, designated as Polar and Spherical, have been developed and are included in the publisher’s website for this book. These programs can be used to plot the two-dimensional patterns, both polar and semipolar (in linear and dB scales), in polar form and spherical three-dimensional patterns (in linear and dB scales). A description of these programs is found in the publisher’s website for this book. Other programs that have been developed for plotting rectangular and polar plots are those of –.
A major lobe (also called main beam) is defined as “the radiation lobe containing the direction of maximum radiation.” In Figure 2.3 the major lobe is pointing in the 𝜃= 0 direction. In some RADIATION PATTERN 29 Normalized Field Pattern (linear scale) z r E 0.9 1 0.8 Eϕ E 0.6 0.7 θ Eθ 0.4 0.5 0.2 0.3 0.1 y x ϕ Figure 2.4 Normalized three-dimensional amplitude field pattern (in linear scale) of a 10-element linear array antenna with a uniform spacing of d = 0.25λ and progressive phase shift 𝛽= −0.6𝜋between the elements.
antennas, such as split-beam antennas, there may exist more than one major lobe. A minor lobe is any lobe except a major lobe. In Figures 2.3(a) and (b) all the lobes with the exception of the major can be classified as minor lobes. A side lobe is “a radiation lobe in any direction other than the intended lobe.” (Usually a side lobe is adjacent to the main lobe and occupies the hemisphere in the direction of the main beam.) A back lobe is “a radiation lobe whose axis makes an angle of approximately 180◦with respect to the beam of an antenna.” Usually it refers to a minor lobe that occupies the hemisphere in a direction opposite to that of the major (main) lobe.
Minor lobes usually represent radiation in undesired directions, and they should be minimized.
Side lobes are normally the largest of the minor lobes. The level of minor lobes is usually expressed as a ratio of the power density in the lobe in question to that of the major lobe. This ratio is often termed the side lobe ratio or side lobe level. Side lobe levels of −20 dB or smaller are usually not desirable in most applications. Attainment of a side lobe level smaller than −30 dB usually requires very careful design and construction. In most radar systems, low side lobe ratios are very important to minimize false target indications through the side lobes.
A normalized three-dimensional far-field amplitude pattern, plotted on a linear scale, of a 10-element linear antenna array of isotropic sources with a spacing of d = 0.25λ and progressive phase shift 𝛽= −0.6𝜋, between the elements is shown in Figure 2.4. It is evident that this pattern has one major lobe, five minor lobes and one back lobe. The level of the side lobe is about −9 dB relative to the maximum. A detailed presentation of arrays is found in Chapter 6. For an amplitude pattern of an antenna, there would be, in general, three electric-field components (Er, E𝜃, E𝜙) at each observation point on the surface of a sphere of constant radius r = rc, as shown in Figure 2.1. In the far field, the radial Er component for all antennas is zero or vanishingly small compared to either one, or 30 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Figure 2.5 Principal E- and H-plane patterns for a pyramidal horn antenna.
both, of the other two components (see Section 3.6 of Chapter 3). Some antennas, depending on their geometry and also observation distance, may have only one, two, or all three components. In general, the magnitude of the total electric field would be |E| = √ |Er|2 + |E𝜃|2 + |E𝜙|2. The radial distance in Figure 2.4, and similar ones, represents the magnitude of |E|.
2.2.2 Isotropic, Directional, and Omnidirectional Patterns An isotropic radiator is defined as “a hypothetical lossless antenna having equal radiation in all direc-tions.” Although it is ideal and not physically realizable, it is often taken as a reference for express-ing the directive properties of actual antennas. A directional antenna is one “having the property of radiating or receiving electromagnetic waves more effectively in some directions than in others.
This term is usually applied to an antenna whose maximum directivity is significantly greater than that of a half-wave dipole.” Examples of antennas with directional radiation patterns are shown in Figures 2.5 and 2.6. It is seen that the pattern in Figure 2.6 is nondirectional in the azimuth plane [f(𝜙), 𝜃= 𝜋∕2] and directional in the elevation plane [g(𝜃), 𝜙= constant]. This type of a pattern is designated as omnidirectional, and it is defined as one “having an essentially nondirectional pattern in a given plane (in this case in azimuth) and a directional pattern in any orthogonal plane (in this case in elevation).” An omnidirectional pattern is then a special type of a directional pattern.
2.2.3 Principal Patterns For a linearly polarized antenna, performance is often described in terms of its principal E- and H-plane patterns. The E-plane is defined as “the plane containing the electric-field vector and the RADIATION PATTERN 31 Radiation pattern Antenna z x r H E θ ϕ y H E Figure 2.6 Omnidirectional antenna pattern.
direction of maximum radiation,” and the H-plane as “the plane containing the magnetic-field vec-tor and the direction of maximum radiation.” Although it is very difficult to illustrate the principal patterns without considering a specific example, it is the usual practice to orient most antennas so that at least one of the principal plane patterns coincide with one of the geometrical principal planes.
An illustration is shown in Figure 2.5. For this example, the x-z plane (elevation plane; 𝜙= 0) is the principal E-plane and the x-y plane (azimuthal plane; 𝜃= 𝜋∕2) is the principal H-plane. Other coordinate orientations can be selected.
The omnidirectional pattern of Figure 2.6 has an infinite number of principal E-planes (elevation planes; 𝜙= 𝜙c) and one principal H-plane (azimuthal plane; 𝜃= 90◦).
2.2.4 Field Regions The space surrounding an antenna is usually subdivided into three regions: (a) reactive near-field, (b) radiating near-field (Fresnel) and (c) far-field (Fraunhofer) regions as shown in Figure 2.7. These regions are so designated to identify the field structure in each. Although no abrupt changes in the field configurations are noted as the boundaries are crossed, there are distinct differences among them. The boundaries separating these regions are not unique, although various criteria have been established and are commonly used to identify the regions.
Reactive near-field region is defined as “that portion of the near-field region immediately surrounding the antenna wherein the reactive field predominates.” For most antennas, the outer boundary of this region is commonly taken to exist at a distance R < 0.62 √ D3∕λ from the antenna surface, where λ is the wavelength and D is the largest dimension of the antenna. “For a very short dipole, or equivalent radiator, the outer boundary is commonly taken to exist at a distance λ∕2𝜋 from the antenna surface.” Radiating near-field (Fresnel) region is defined as “that region of the field of an antenna between the reactive near-field region and the far-field region wherein radiation fields predominate and wherein the angular field distribution is dependent upon the distance from the antenna. If the antenna has a maximum dimension that is not large compared to the wavelength, this region may not exist. For an antenna focused at infinity, the radiating near-field region is sometimes referred to 32 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Far-field (Fraunhofer) region Radiating 3 1 2 2 0.62 2 D R D R = = near-field (Fresnel) region Reactive near-field region R1 D 1 R2 λ λ Figure 2.7 Field regions of an antenna.
as the Fresnel region on the basis of analogy to optical terminology. If the antenna has a maximum overall dimension which is very small compared to the wavelength, this field region may not exist.” The inner boundary is taken to be the distance R ≥0.62 √ D3∕λ and the outer boundary the distance R < 2D2∕λ where D is the largest∗dimension of the antenna. This criterion is based on a maximum phase error of 𝜋∕8. In this region the field pattern is, in general, a function of the radial distance and the radial field component may be appreciable.
Far-field (Fraunhofer) region is defined as “that region of the field of an antenna where the angu-lar field distribution is essentially independent of the distance from the antenna. If the antenna has a maximum† overall dimension D, the far-field region is commonly taken to exist at distances greater than 2D2∕λ from the antenna, λ being the wavelength. The far-field patterns of certain antennas, such as multibeam reflector antennas, are sensitive to variations in phase over their apertures. For these antennas 2D2∕λ may be inadequate. In physical media, if the antenna has a maximum overall dimen-sion, D, which is large compared to 𝜋∕|𝛾|, the far-field region can be taken to begin approximately at a distance equal to |𝛾|D2∕𝜋from the antenna, 𝛾being the propagation constant in the medium. For an antenna focused at infinity, the far-field region is sometimes referred to as the Fraunhofer region on the basis of analogy to optical terminology.” In this region, the field components are essentially trans-verse and the angular distribution is independent of the radial distance where the measurements are made. The inner boundary is taken to be the radial distance R = 2D2∕λ and the outer one at infinity.
The amplitude pattern of an antenna, as the observation distance is varied from the reactive near field to the far field, changes in shape because of variations of the fields, both magnitude and phase. A typical progression of the shape of an antenna, with the largest dimension D, is shown in Figure 2.8.
∗To be valid, D must also be large compared to the wavelength (D > λ).
†To be valid, D must also be large compared to the wavelength (D > λ).
RADIATION PATTERN 33 Reactive Near-field Radiating Near-field Far-Field Fraunhofer D Field Distribution Fresnel Figure 2.8 Typical changes of antenna amplitude pattern shape from reactive near field toward the far field.
(source: Y. Rahmat-Samii, L. I. Williams, and R. G. Yoccarino, The UCLA Bi-polar Planar-Near-Field Antenna Measurement and Diagnostics Range,” IEEE Antennas & Propagation Magazine, Vol. 37, No. 6, December 1995 c ⃝1995 IEEE).
It is apparent that in the reactive near-field region the pattern is more spread out and nearly uniform, with slight variations. As the observation is moved to the radiating near-field region (Fresnel), the pattern begins to smooth and form lobes. In the far-field region (Fraunhofer), the pattern is well formed, usually consisting of few minor lobes and one, or more, major lobes.
To illustrate the pattern variation as a function of radial distance beyond the minimum 2D2∕λ far-field distance, in Figure 2.9 we have included three patterns of a parabolic reflector calculated at distances of R = 2D2∕λ, 4D2∕λ, and infinity . It is observed that the patterns are almost iden-tical, except for some differences in the pattern structure around the first null and at a level below 25 dB. Because infinite distances are not realizable in practice, the most commonly used criterion for minimum distance of far-field observations is 2D2∕λ.
2.2.5 Radian and Steradian The measure of a plane angle is a radian. One radian is defined as the plane angle with its vertex at the center of a circle of radius r that is subtended by an arc whose length is r. A graphical illustration is shown in Figure 2.10(a). Since the circumference of a circle of radius r is C = 2𝜋r, there are 2𝜋 rad (2𝜋r∕r) in a full circle.
The measure of a solid angle is a steradian. One steradian is defined as the solid angle with its vertex at the center of a sphere of radius r that is subtended by a spherical surface area equal to that of a square with each side of length r. A graphical illustration is shown in Figure 2.10(b). Since the area of a sphere of radius r is A = 4𝜋r2, there are 4𝜋sr (4𝜋r2∕r2) in a closed sphere.
The infinitesimal area dA on the surface of a sphere of radius r, shown in Figure 2.1, is given by dA = r2 sin 𝜃d𝜃d𝜙 (m2) (2-1) Therefore, the element of solid angle dΩ of a sphere can be written as dΩ = dA r2 = sin 𝜃d𝜃d𝜙 (sr) (2-2) 34 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Figure 2.9 Calculated radiation patterns of a paraboloid antenna for different distances from the antenna.
(source: J. S. Hollis, T. J. Lyon, and L. Clayton, Jr. (eds.), Microwave Antenna Measurements, Scientific-Atlanta, Inc., July 1970).
r r 1 rad (a) Radian (b) Steradian r r Equivalent in area Area = r2 Area = r2 1 sr r Figure 2.10 Geometrical arrangements for defining a radian and a steradian.
RADIATION POWER DENSITY 35 Example 2.1 For a sphere of radius r, find the solid angle ΩA (in square radians or steradians) of a spheri-cal cap on the surface of the sphere over the north-pole region defined by spherical angles of 0 ≤𝜃≤30◦, 0 ≤𝜙≤360◦. Refer to Figures 2.1 and 2.10. Do this a. exactly.
b. using ΩA ≈ΔΘ1 ⋅ΔΘ2, where ΔΘ1 and ΔΘ2 are two perpendicular angular separations of the spherical cap passing through the north pole.
Compare the two.
Solution: a. Using (2-2), we can write that ΩA = ∫ 360◦ 0 ∫ 30◦ 0 dΩ = ∫ 2𝜋 0 ∫ 𝜋∕6 0 sin 𝜃d𝜃d𝜙= ∫ 2𝜋 0 d𝜙∫ 𝜋∕6 0 sin 𝜃d𝜃 = 2𝜋[−cos 𝜃]|𝜋∕6 0 = 2𝜋[−0.867 + 1] = 2𝜋(0.133) = 0.83566 b. ΩA ≈ΔΘ1 ⋅ΔΘ2 ΔΘ1=ΔΘ2 ⏞ ⏞ ⏞ = (ΔΘ1)2 = 𝜋 3 (𝜋 3 ) = 𝜋2 9 = 1.09662 It is apparent that the approximate beam solid angle is about 31.23% in error.
2.3 RADIATION POWER DENSITY Electromagnetic waves are used to transport information through a wireless medium or a guiding structure, from one point to the other. It is then natural to assume that power and energy are associated with electromagnetic fields. The quantity used to describe the power associated with an electromag-netic wave is the instantaneous Poynting vector defined as 𝓦= 𝓔× 𝓗 (2-3) 𝓦= instantaneous Poynting vector (W/m2) 𝓔= instantaneous electric-field intensity (V/m) 𝓗= instantaneous magnetic-field intensity (A/m) Note that script letters are used to denote instantaneous fields and quantities, while roman letters are used to represent their complex counterparts.
Since the Poynting vector is a power density, the total power crossing a closed surface can be obtained by integrating the normal component of the Poynting vector over the entire surface. In equation form 𝒫= ∯ S 𝓦⋅ds = ∯ S 𝓦⋅̂ n da (2-4) 36 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 𝒫= instantaneous total power (W) ̂ n = unit vector normal to the surface da = infinitesimal area of the closed surface (m2) For applications of time-varying fields, it is often more desirable to find the average power density which is obtained by integrating the instantaneous Poynting vector over one period and dividing by the period. For time-harmonic variations of the form ej𝜔t, we define the complex fields E and H which are related to their instantaneous counterparts 𝓔and 𝓗by 𝓔(x, y, z; t) = Re[E(x, y, z)ej𝜔t] (2-5) 𝓗(x, y, z; t) = Re[H(x, y, z)ej𝜔t] (2-6) Using the definitions of (2-5) and (2-6) and the identity Re[Eej𝜔t] = 1 2[Eej𝜔t + E∗e−j𝜔t], (2-3) can be written as 𝓦= 𝓔× 𝓗= 1 2Re[E × H∗] + 1 2Re[E × Hej2𝜔t] (2-7) The first term of (2-7) is not a function of time, and the time variations of the second are twice the given frequency. The time average Poynting vector (average power density) can be written as Wav(x, y, z) = [𝓦(x, y, z; t)]av = 1 2Re[E × H∗] (W/m2) (2-8) The 1 2 factor appears in (2-7) and (2-8) because the E and H fields represent peak values, and it should be omitted for RMS values.
A close observation of (2-8) may raise a question. If the real part of (E × H∗)∕2 represents the average (real) power density, what does the imaginary part of the same quantity represent? At this point it will be very natural to assume that the imaginary part must represent the reactive (stored) power density associated with the electromagnetic fields. In later chapters, it will be shown that the power density associated with the electromagnetic fields of an antenna in its far-field region is predominately real and will be referred to as radiation density.
Based upon the definition of (2-8), the average power radiated by an antenna (radiated power) can be written as Prad = Pav = ∯ S Wrad ⋅ds = ∯ S Wav ⋅̂ n da = 1 2 ∯ S Re(E × H∗) ⋅ds (2-9) The power pattern of the antenna, whose definition was discussed in Section 2.2, is just a measure, as a function of direction, of the average power density radiated by the antenna. The observations are usually made on a large sphere of constant radius extending into the far field. In practice, absolute power patterns are usually not desired. However, the performance of the antenna is measured in terms of the gain (to be discussed in a subsequent section) and in terms of relative power patterns.
Three-dimensional patterns cannot be measured, but they can be constructed with a number of two-dimensional cuts.
RADIATION INTENSITY 37 Example 2.2 The radial component of the radiated power density of an antenna is given by Wrad = ̂ arWr = ̂ arA0 sin 𝜃 r2 (W/m2) where A0 is the peak value of the power density, 𝜃is the usual spherical coordinate, and ̂ ar is the radial unit vector. Determine the total radiated power.
Solution: For a closed surface, a sphere of radius r is chosen. To find the total radiated power, the radial component of the power density is integrated over its surface. Thus Prad = ∯ S Wrad ⋅̂ n da = ∫ 2𝜋 0 ∫ 𝜋 0 ( ̂ arA0 sin 𝜃 r2 ) ⋅(̂ arr2 sin 𝜃d𝜃d𝜙) = 𝜋2A0 (W) A three-dimensional normalized plot of the average power density at a distance of r = 1 m is shown in Figure 2.6.
An isotropic radiator is an ideal source that radiates equally in all directions. Although it does not exist in practice, it provides a convenient isotropic reference with which to compare other antennas.
Because of its symmetric radiation, its Poynting vector will not be a function of the spherical coordi-nate angles 𝜃and 𝜙. In addition, it will have only a radial component. Thus the total power radiated by it is given by Prad = ∯ S W0 ⋅ds = ∫ 2𝜋 0 ∫ 𝜋 0 [̂ arW0(r)] ⋅[̂ arr2 sin 𝜃d𝜃d𝜙] = 4𝜋r2W0 (2-10) and the power density by W0 = ̂ arW0 = ̂ ar ( Prad 4𝜋r2 ) (W/m2) (2-11) which is uniformly distributed over the surface of a sphere of radius r.
2.4 RADIATION INTENSITY Radiation intensity in a given direction is defined as “the power radiated from an antenna per unit solid angle.” The radiation intensity is a far-field parameter, and it can be obtained by simply mul-tiplying the radiation density by the square of the distance. In mathematical form it is expressed as U = r2Wrad (2-12) where U = radiation intensity (W/unit solid angle) Wrad = radiation density (W/m2) 38 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS The radiation intensity is also related to the far-zone electric field of an antenna, referring to Figure 2.4, by U(𝜃, 𝜙) = r2 2𝜂|E(r, 𝜃, 𝜙)|2 ≃r2 2𝜂 [ |E𝜃(r, 𝜃, 𝜙)|2 + |E𝜙(r, 𝜃, 𝜙)|2] ≃1 2𝜂 [ |E◦ 𝜃(𝜃, 𝜙)|2 + |E◦ 𝜙(𝜃, 𝜙)|2] (2-12a) where E(r, 𝜃, 𝜙) = far-zone electric-field intensity of the antenna = E◦(𝜃, 𝜙)e−jkr r E𝜃, E𝜙= far-zone electric-field components of the antenna 𝜂= intrinsic impedance of the medium The radial electric-field component (Er) is assumed, if present, to be small in the far zone. Thus the power pattern is also a measure of the radiation intensity.
The total power is obtained by integrating the radiation intensity, as given by (2-12), over the entire solid angle of 4𝜋. Thus Prad = ∯ Ω U dΩ = ∫ 2𝜋 0 ∫ 𝜋 0 U sin 𝜃d𝜃d𝜙 (2-13) where dΩ = element of solid angle = sin 𝜃d𝜃d𝜙.
Example 2.3 For the problem of Example 2.2, find the total radiated power using (2-13).
Solution: Using (2-12) U = r2Wrad = A0 sin 𝜃 and by (2-13) Prad = ∫ 2𝜋 0 ∫ 𝜋 0 U sin 𝜃d𝜃d𝜙= A0 ∫ 2𝜋 0 ∫ 𝜋 0 sin2 𝜃d𝜃d𝜙= 𝜋2A0 which is the same as that obtained in Example 2.2. A three-dimensional plot of the relative radi-ation intensity is also represented by Figure 2.6.
For an isotropic source U will be independent of the angles 𝜃and 𝜙, as was the case for Wrad.
Thus (2-13) can be written as Prad = ∯ Ω U0 dΩ = U0 ∯ Ω dΩ = 4𝜋U0 (2-14) or the radiation intensity of an isotropic source as U0 = Prad 4𝜋 (2-15) RADIATION INTENSITY 39 U(θ ,ϕ ) Normalized Field Pattern (linear scale) 1 0.8 0.9 0.6 0.7 0.3 0.4 0.5 0.1 0.2 Normalized Field Pattern (linear scale) 1 z z y x 0.8 0.9 HPBW = 28 65° 0.6 0.7 HPBW .
FNBW = 60° 0 3 0.4 0.5 FNBW 0.5 0.5 0.1 0.2 (a) Three-dimensional (b) Two-dimensional Figure 2.11 Three- and two-dimensional power patterns (in linear scale) of U(𝜃) = cos2(𝜃) cos2(3𝜃).
40 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2.5 BEAMWIDTH Associated with the pattern of an antenna is a parameter designated as beamwidth. The beamwidth of a pattern is defined as the angular separation between two identical points on opposite side of the pattern maximum. In an antenna pattern, there are a number of beamwidths. One of the most widely used beamwidths is the Half-Power Beamwidth (HPBW), which is defined by IEEE as: “In a plane containing the direction of the maximum of a beam, the angle between the two directions in which the radiation intensity is one-half value of the beam.” This is demonstrated in Figure 2.2.
Another important beamwidth is the angular separation between the first nulls of the pattern, and it is referred to as the First-Null Beamwidth (FNBW). Both the HPBW and FNBW are demonstrated for the pattern in Figure 2.11 for the pattern of Example 2.4. Other beamwidths are those where the pattern is −10 dB from the maximum, or any other value. However, in practice, the term beamwidth, with no other identification, usually refers to HPBW.
The beamwidth of an antenna is a very important figure of merit and often is used as a trade-off between it and the side lobe level; that is, as the beamwidth decreases, the side lobe increases and vice versa. In addition, the beamwidth of the antenna is also used to describe the resolution capabilities of the antenna to distinguish between two adjacent radiating sources or radar targets. The most common resolution criterion states that the resolution capability of an antenna to distinguish between two sources is equal to half the first-null beamwidth (FNBW/2), which is usually used to approximate the half-power beamwidth (HPBW) , . That is, two sources separated by angular distances equal or greater than FNBW/2 ≈HPBW of an antenna with a uniform distribution can be resolved. If the separation is smaller, then the antenna will tend to smooth the angular separation distance.
Example 2.4 The normalized radiation intensity of an antenna is represented by U(𝜃) = cos2(𝜃) cos2(3𝜃), (0 ≤𝜃≤90◦, 0◦≤𝜙≤360◦) The three- and two-dimensional plots of this, plotted in a linear scale, are shown in Figure 2.11.
Find the a. half-power beamwidth HPBW (in radians and degrees) b. first-null beamwidth FNBW (in radians and degrees) Solution: a. Since the U(𝜃) represents the power pattern, to find the half-power beamwidth you set the function equal to half of its maximum, or U(𝜃)|𝜃=𝜃h = cos2(𝜃) cos2(3𝜃)|𝜃=𝜃h = 0.5 ⇒cos 𝜃h cos 3𝜃h = 0.707 𝜃h = cos−1 ( 0.707 cos 3𝜃h ) Since this is an equation with transcendental functions, it can be solved iteratively. After a few iterations, it is found that 𝜃h ≈0.25 radians = 14.325◦ DIRECTIVITY 41 Since the function U(𝜃) is symmetrical about the maximum at 𝜃= 0, then the HPBW is HPBW = 2𝜃h ≈0.50 radians = 28.65◦ b. To find the first-null beamwidth (FNBW), you set the U(𝜃) equal to zero, or U(𝜃)|𝜃=𝜃n = cos2(𝜃) cos2(3𝜃)|𝜃=𝜃n = 0 This leads to two solutions for 𝜃n.
cos 𝜃n = 0 ⇒𝜃n = cos−1(0) = 𝜋 2 radians = 90◦ cos 3𝜃n = 0 ⇒𝜃n = 1 3 cos−1(0) = 𝜋 6 radians = 30◦ The one with the smallest value leads to the FNBW. Again, because of the symmetry of the pattern, the FNBW is FNBW = 2𝜃n = 𝜋 3 radians = 60◦ 2.6 DIRECTIVITY In the 1983 version of the IEEE Standard Definitions of Terms for Antennas, there has been a substan-tive change in the definition of directivity, compared to the definition of the 1973 version. Basically the term directivity in the new 1983 version has been used to replace the term directive gain of the old 1973 version. In the new 1983 version the term directive gain has been deprecated. According to the authors of the new 1983 standards, “this change brings this standard in line with common usage among antenna engineers and with other international standards, notably those of the International Electrotechnical Commission (IEC).” Therefore directivity of an antenna defined as “the ratio of the radiation intensity in a given direction from the antenna to the radiation intensity averaged over all directions. The average radiation intensity is equal to the total power radiated by the antenna divided by 4𝜋. If the direction is not specified, the direction of maximum radiation intensity is implied.” Stated more simply, the directivity of a nonisotropic source is equal to the ratio of its radiation inten-sity in a given direction over that of an isotropic source. In mathematical form, using (2-15), it can be written as D = U U0 = 4𝜋U Prad (2-16) If the direction is not specified, it implies the direction of maximum radiation intensity (maximum directivity) expressed as Dmax = D0 = U|max U0 = Umax U0 = 4𝜋Umax Prad (2-16a) 42 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS D = directivity (dimensionless) D0 = maximum directivity (dimensionless) U = radiation intensity (W/unit solid angle) Umax = maximum radiation intensity (W/unit solid angle) U0 = radiation intensity of isotropic source (W/unit solid angle) Prad = total radiated power (W) For an isotropic source, it is very obvious from (2-16) or (2-16a) that the directivity is unity since U, Umax, and U0 are all equal to each other.
For antennas with orthogonal polarization components, we define the partial directivity of an antenna for a given polarization in a given direction as “that part of the radiation intensity corre-sponding to a given polarization divided by the total radiation intensity averaged over all directions.” With this definition for the partial directivity, then in a given direction “the total directivity is the sum of the partial directivities for any two orthogonal polarizations.” For a spherical coordinate system, the total maximum directivity D0 for the orthogonal 𝜃and 𝜙components of an antenna can be writ-ten as D0 = D𝜃+ D𝜙 (2-17) while the partial directivities D𝜃and D𝜙are expressed as D𝜃= 4𝜋U𝜃 (Prad)𝜃+ (Prad)𝜙 (2-17a) D𝜙= 4𝜋U𝜙 (Prad)𝜃+ (Prad)𝜙 (2-17b) where U𝜃= radiation intensity in a given direction contained in 𝜃field component U𝜙= radiation intensity in a given direction contained in 𝜙field component (Prad)𝜃= radiated power in all directions contained in 𝜃field component (Prad)𝜙= radiated power in all directions contained in 𝜙field component Example 2.5 As an illustration, find the maximum directivity of the antenna whose radiation intensity is that of Example 2.2. Write an expression for the directivity as a function of the directional angles 𝜃 and 𝜙.
Solution: The radiation intensity is given by U = r2Wrad = A0 sin 𝜃 The maximum radiation is directed along 𝜃= 𝜋∕2. Thus Umax = A0 DIRECTIVITY 43 In Example 2.2 it was found that Prad = 𝜋2A0 Using (2-16a), we find that the maximum directivity is equal to D0 = 4𝜋Umax Prad = 4 𝜋= 1.27 Since the radiation intensity is only a function of 𝜃, the directivity as a function of the directional angles is represented by D = D0 sin 𝜃= 1.27 sin 𝜃 Before proceeding with a more general discussion of directivity, it may be proper at this time to consider another example, compute its directivity, compare it with that of the previous example, and comment on what it actually represents. This may give the reader a better understanding and appreciation of the directivity.
Example 2.6 The radial component of the radiated power density of an infinitesimal linear dipole of length l ≪λ is given by Wav = ̂ arWr = ̂ arA0 sin2 𝜃 r2 (W/m2) where A0 is the peak value of the power density, 𝜃is the usual spherical coordinate, and ̂ ar is the radial unit vector. Determine the maximum directivity of the antenna and express the directivity as a function of the directional angles 𝜃and 𝜙.
Solution: The radiation intensity is given by U = r2Wr = A0 sin2 𝜃 The maximum radiation is directed along 𝜃= 𝜋∕2. Thus Umax = A0 The total radiated power is given by Prad = ∯ Ω U dΩ = A0 ∫ 2𝜋 0 ∫ 𝜋 0 sin2 𝜃sin 𝜃d𝜃d𝜙= A0 (8𝜋 3 ) Using (2-16a), we find that the maximum directivity is equal to D0 = 4𝜋Umax Prad = 4𝜋A0 8𝜋 3 (A0) = 3 2 44 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS which is greater than 1.27 found in Example 2.5. Thus the directivity is represented by D = D0 sin2 𝜃= 1.5 sin2 𝜃 At this time it will be proper to comment on the results of Examples 2.5 and 2.6. To better understand the discussion, we have plotted in Figure 2.12 the relative radiation intensities of Example 2.5 (U = A0 sin 𝜃) and Example 2.6 (U = A0 sin2 𝜃) where A0 was set equal to unity. We see that both patterns are omnidirectional but that of Example 2.6 has more directional characteristics (is narrower) in the elevation plane. Since the directivity is a “figure of merit” describing how well the radiator directs energy in a certain direction, it should be convincing from Figure 2.12 that the directivity of Example 2.6 should be higher than that of Example 2.5.
To demonstrate the significance of directivity, let us consider another example; in particular let us examine the directivity of a half-wavelength dipole (l = λ∕2), which is derived in Section 4.6 of Chapter 4 and can be approximated by D = D0 sin3 𝜃= 1.67 sin3 𝜃 (2-18) since it can be shown that [see Figure 4.12(b)] sin3 𝜃≃ ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 (2-18a) where 𝜃is measured from the axis along the length of the dipole. The values represented by (2-18) and those of an isotropic source (D = 1) are plotted two- and three-dimensionally in Fig-ure 2.13(a,b). For the three-dimensional graphical representation of Figure 2.13(b), at each obser-vation point only the largest value of the two directivities is plotted. It is apparent that when r 1.0 z y x U = sin2 U = sin 1.0 θ θ θ Figure 2.12 Three-dimensional radiation intensity patterns. (source: P. Lorrain and D. R. Corson, Electro-magnetic Fields and Waves, 2nd ed., W. H. Freeman and Co. Copyright c ⃝1970).
DIRECTIVITY 45 Directivity (dimensionless) 2 z θ 1 4 1.6 1.8 D=1 1.0 1 1.2 .
D=1.67sin3(θ ) 0 4 0.6 0.8 y x 1.67 1.67 0 0.2 .
Directivity (dimensionless) 1.8 1.2 1.4 1.6 1 0.4 0.6 0.8 0 0.2 y x 1.67 D (isotropic)=1 D (dipole)=1 67sin3( ) 57.44° 57.44° D (dipole)=1.67sin3( ) 122.56° 122.56° 57.44° 122.56° (a) Two-dimensional (b) Three-dimensional Figure 2.13 Two- and three-dimensional directivity patterns of a λ∕2 dipole. (source: C. A. Balanis, “Antenna Theory: A Review.” Proc. IEEE, Vol. 80, No. 1. January 1992. c ⃝1992 IEEE).
sin−1(1∕1.67)1∕3 = 57.44◦< 𝜃< 122.56◦, the dipole radiator has greater directivity (greater inten-sity concentration) in those directions than that of an isotropic source. Outside this range of angles, the isotropic radiator has higher directivity (more intense radiation). The maximum directivity of the dipole (relative to the isotropic radiator) occurs when 𝜃= 𝜋∕2, and it is 1.67 (or 2.23 dB) more intense than that of the isotropic radiator (with the same radiated power).
46 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS The three-dimensional pattern of Figure 2.13(b), and similar ones, are included throughout the book to represent the three-dimensional radiation characteristics of antennas. These patterns are plotted to visualize the three-dimensional radiation pattern of the antenna. These three-dimensional programs, along with the others, can be used effectively toward the design and synthesis of antennas, especially arrays, as demonstrated in and . A MATLAB-based program, designated as Spher-ical, is also included in the publisher’s website to produce these and similar three-dimensional plots.
The directivity of an isotropic source is unity since its power is radiated equally well in all direc-tions. For all other sources, the maximum directivity will always be greater than unity, and it is a relative “figure of merit” which gives an indication of the directional properties of the antenna as compared with those of an isotropic source. In equation form, this is indicated in (2-16a). The directivity can be smaller than unity; in fact it can be equal to zero. For Examples 2.5 and 2.6, the directivity is equal to zero in the 𝜃= 0 direction. The values of directivity will be equal to or greater than zero and equal to or less than the maximum directivity (0 ≤D ≤D0).
A more general expression for the directivity can be developed to include sources with radiation patterns that may be functions of both spherical coordinate angles 𝜃and 𝜙. In the previous exam-ples we considered intensities that were represented by only one coordinate angle 𝜃, in order not to obscure the fundamental concepts by the mathematical details. So it may now be proper, since the basic definitions have been illustrated by simple examples, to formulate the more general expres-sions.
Let the radiation intensity of an antenna be of the form U = B0F(𝜃, 𝜙) ≃1 2𝜂 [ |E0 𝜃(𝜃, 𝜙)|2 + |E0 𝜙(𝜃, 𝜙)|2] (2-19) where B0 is a constant, and E0 𝜃and E0 𝜙are the antenna’s far-zone electric-field components. The maximum value of (2-19) is given by Umax = B0F(𝜃, 𝜙)|max = B0Fmax(𝜃, 𝜙) (2-19a) The total radiated power is found using Prad = ∯ Ω U(𝜃, 𝜙) dΩ = B0 ∫ 2𝜋 0 ∫ 𝜋 0 F(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-20) We now write the general expression for the directivity and maximum directivity using (2-16) and (2-16a), respectively, as D(𝜃, 𝜙) = 4𝜋 F(𝜃, 𝜙) ∫ 2𝜋 0 ∫ 𝜋 0 F(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-21) D0 = 4𝜋 F(𝜃, 𝜙)|max ∫ 2𝜋 0 ∫ 𝜋 0 F(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-22) DIRECTIVITY 47 Equation (2-22) can also be written as D0 = 4𝜋 [ ∫ 2𝜋 0 ∫ 𝜋 0 F(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 ] / F(𝜃, 𝜙)|max = 4𝜋 ΩA (2-23) where ΩA is the beam solid angle, and it is given by ΩA = 1 F(𝜃, 𝜙)|max ∫ 2𝜋 0 ∫ 𝜋 0 F(𝜃, 𝜙) sin 𝜃d𝜃d𝜙= ∫ 2𝜋 0 ∫ 𝜋 0 Fn(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-24) Fn(𝜃, 𝜙) = F(𝜃, 𝜙) F(𝜃, 𝜙)|max (2-25) Dividing by F(𝜃, 𝜙)|max merely normalizes the radiation intensity F(𝜃, 𝜙), and it makes its max-imum value unity.
The beam solid angle ΩA is defined as the solid angle through which all the power of the antenna would flow if its radiation intensity is constant (and equal to the maximum value of U) for all angles within ΩA.
2.6.1 Directional Patterns Instead of using the exact expression of (2-23) to compute the directivity, it is often convenient to derive simpler expressions, even if they are approximate, to compute the directivity. These can also be used for design purposes. For antennas with one narrow major lobe and very negligible minor lobes, the beam solid angle is approximately equal to the product of the half-power beamwidths in two perpendicular planes shown in Figure 2.14(a). For a rotationally symmetric pattern, the half-power beamwidths in any two perpendicular planes are the same, as illustrated in Figure 2.14(b).
z ΩA 1r θ 2r θ 1r θ 2r θ = y (a) Nonsymmetrical pattern x z ΩA y (b) Symmetrical pattern x Figure 2.14 Beam solid angles for nonsymmetrical and symmetrical radiation patterns.
48 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS With this approximation, (2-23) can be approximated by D0 = 4𝜋 ΩA ≃ 4𝜋 Θ1rΘ2r (2-26) The beam solid angle ΩA has been approximated by ΩA ≃Θ1rΘ2r (2-26a) where Θ1r = half-power beamwidth in one plane (rad) Θ2r = half-power beamwidth in a plane at a right angle to the other (rad) If the beamwidths are known in degrees, (2-26) can be written as D0 ≃4𝜋(180∕𝜋)2 Θ1dΘ2d = 41,253 Θ1dΘ2d (2-27) where Θ1d = half-power beamwidth in one plane (degrees) Θ2d = half-power beamwidth in a plane at a right angle to the other (degrees) For planar arrays, a better approximation to (2-27) is D0 ≃ 32,400 ΩA(degrees)2 = 32,400 Θ1dΘ2d (2-27a) The validity of (2-26) and (2-27) is based on a pattern that has only one major lobe and any minor lobes, if present, should be of very low intensity. For a pattern with two identical major lobes, the value of the maximum directivity using (2-26) or (2-27) will be twice its actual value. For patterns with significant minor lobes, the values of maximum directivity obtained using (2-26) or (2-27), which neglect any minor lobes, will usually be too high.
Example 2.7 The radiation intensity of the major lobe of many antennas can be adequately represented by U = B0 cos4 𝜃 where B0 is the maximum radiation intensity. The radiation intensity exists only in the upper hemisphere (0 ≤𝜃≤𝜋∕2, 0 ≤𝜙≤2𝜋), and it is shown in Figure 2.15.
Find the a. beam solid angle; exact and approximate.
b. maximum directivity; exact using (2-23) and approximate using (2-26).
DIRECTIVITY 49 Solution: The half-power point of the pattern occurs at 𝜃= 32.765◦. Thus the beamwidth in the 𝜃direction is 65.53◦or Θ1r = 1.1437 rads z Normalized Field Pattern (linear scale) 1 0.7 0.8 0.9 U = cos4 θ θ θ 1r 0.5 0.6 θ 2r y 0.2 0.3 0.4 x 0.1 ϕ Figure 2.15 Radiation intensity pattern of the form U = cos4 𝜃in the upper hemisphere.
Since the pattern is independent of the 𝜙coordinate, the beamwidth in the other plane is also equal to Θ2r = 1.1437 rads a. Beam solid angle ΩA: Exact: Using (2-24), (2-25) ΩA = ∫ 360◦ 0 ∫ 90◦ 0 cos4 𝜃dΩ = ∫ 2𝜋 0 ∫ 𝜋∕2 0 cos4 𝜃sin 𝜃d𝜃d𝜙 = ∫ 2𝜋 0 d𝜙∫ 𝜋∕2 0 cos4 𝜃sin 𝜃d𝜃 = 2𝜋∫ 𝜋∕2 0 cos4 𝜃sin 𝜃d𝜃= 2𝜋 5 steradians Approximate: Using (2-26a) ΩA ≈Θ1rΘ2r = 1.1437(1.1437) = (1.1437)2 = 1.308 steradians 50 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS b. Directivity D0: Exact: D0 = 4𝜋 ΩA = 4𝜋(5) 2𝜋 = 10 (dimensionless) = 10 dB The same exact answer is obtained using (2-16a).
Approximate: D0 ≈4𝜋 ΩA = 4𝜋 1.308 = 9.61 (dimensionless) = 9.83 dB The exact maximum directivity is 10 and its approximate value, using (2-26), is 9.61. Even better approximations can be obtained if the patterns have much narrower beamwidths, which will be demonstrated later in this section.
Many times it is desirable to express the directivity in decibels (dB) instead of dimensionless quanti-ties. The expressions for converting the dimensionless quantities of directivity and maximum direc-tivity to decibels (dB) are D(dB) = 10 log10[D(dimensionless)] (2-28a) D0(dB) = 10 log10[D0(dimensionless)] (2-28b) It has also been proposed that the maximum directivity of an antenna can also be obtained approximately by using the formula 1 D0 = 1 2 ( 1 D1 + 1 D2 ) (2-29) where D1 ≃ 1 [ 1 2 ln 2 ∫ Θ1r∕2 0 sin 𝜃d𝜃 ] ≃16 ln 2 Θ2 1r (2-29a) D2 ≃ 1 [ 1 2 ln 2 ∫ Θ2r∕2 0 sin 𝜃d𝜃 ] ≃16 ln 2 Θ2 2r (2-29b) Θ1r and Θ2r are the half-power beamwidths (in radians) of the E- and H-planes, respectively. The formula of (2-29) will be referred to as the arithmetic mean of the maximum directivity. Using (2-29a) and (2-29b) we can write (2-29) as 1 D0 ≃ 1 2 ln 2 ( Θ2 1r 16 + Θ2 2r 16 ) = Θ2 1r + Θ2 2r 32 ln 2 (2-30) or D0 ≃ 32 ln 2 Θ2 1r + Θ2 2r = 22.181 Θ2 1r + Θ2 2r (2-30a) D0 ≃22.181(180∕𝜋)2 Θ2 1d + Θ2 2d = 72,815 Θ2 1d + Θ2 2d (2-30b) DIRECTIVITY 51 TABLE 2.1 Comparison of Exact and Approximate Values of Maximum Directivity for U = 𝐜𝐨𝐬n 𝜽 Power Patterns Exact Kraus Tai and Pereira Equation Equation Kraus Equation Tai and Pereira n (2-22) (2-26) % Error (2-30a) % Error 1 4 2.86 −28.50 2.53 −36.75 2 6 5.09 −15.27 4.49 −25.17 3 8 7.35 −8.12 6.48 −19.00 4 10 9.61 −3.90 8.48 −15.20 5 12 11.87 −1.08 10.47 −12.75 6 14 14.13 +0.93 12.46 −11.00 7 16 16.39 +2.48 14.47 −9.56 8 18 18.66 +3.68 16.47 −8.50 9 20 20.93 +4.64 18.47 −7.65 10 22 23.19 +5.41 20.47 −6.96 11.28 24.56 26.08 +6.24 23.02 −6.24 15 32 34.52 +7.88 30.46 −4.81 20 42 45.89 +9.26 40.46 −3.67 where Θ1d and Θ2d are the half-power beamwidths in degrees. Equation (2-30a) is to be contrasted with (2-26) while (2-30b) should be compared with (2-27).
In order to make an evaluation and comparison of the accuracies of (2-26) and (2-30a), examples whose radiation intensities (power patterns) can be represented by U(𝜃, 𝜙) = { B0 cosn(𝜃) 0 ≤𝜃≤𝜋∕2, 0 ≤𝜙≤2𝜋 0 elsewhere (2-31) where n = 1 −10, 11.28, 15, and 20 are considered. The maximum directivities were computed using (2-26) and (2-30a) and compared with the exact values as obtained using (2-23). The results are shown in Table 2.1. From the comparisons it is evident that the error due to Tai & Pereira’s formula is always negative (i.e., it predicts lower values of maximum directivity than the exact ones) and monotonically decreases as n increases (the pattern becomes more narrow). However, the error due to Kraus’ formula is negative for small values of n and positive for large values of n. For small values of n the error due to Kraus’ formula is negative and positive for large values of n; the error is zero when n = 5.497 ≃5.5 (half-power beamwidth of 56.35◦). In addition, for symmetrically rota-tional patterns the absolute error due to the two approximate formulas is identical when n = 11.28, which corresponds to a half-power beamwidth of 39.77◦. From these observations we conclude that, Kraus’ formula is more accurate for small values of n (broader patterns) while Tai & Pereira’s is more accurate for large values of n (narrower patterns). Based on absolute error and symmetrically rotational patterns, Kraus’ formula leads to smaller error for n < 11.28 (half-power beamwidth greater than 39.77◦) while Tai & Pereira’s leads to smaller error for n > 11.28 (half-power beamwidth smaller than 39.77◦). The results are shown plotted in Figure 2.16 for 0 < n ≤450.
2.6.2 Omnidirectional Patterns Some antennas (such as dipoles, loops, broadside arrays) exhibit omnidirectional patterns, as illus-trated by the three-dimensional patterns in Figure 2.17 (a,b). As single-lobe directional patterns can be approximated by (2-31), omnidirectional patterns can often be approximated by U = | sinn(𝜃)| 0 ≤𝜃≤𝜋, 0 ≤𝜙≤2𝜋 (2-32) 52 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 0 50 100 150 200 250 300 350 400 450 0 100 200 300 400 500 600 700 800 900 1000 20.00 23.01 24.77 26.02 26.99 27.78 28.45 29.03 29.54 30.00 Directivity Do (dimensionless) Directivity Do (dB) n 180.0 19.0 13.5 11.0 9.5 8.5 7.8 7.2 6.7 6.4 HPBW (degrees) Do (exact) Do (Kraus)(Eq. 2-26) Do (Tai & Pereira)(Eq. 2-30a) U = cosnθ Figure 2.16 Comparison of exact and approximate values of directivity for directional U = cosn 𝜃power pat-terns.
where n represents both integer and noninteger values. The directivity of antennas with patterns represented by (2-32) can be determined in closed form using the definition of (2-16a). However, as was done for the single-lobe patterns of Figure 2.14, approximate directivity formulas have been derived , for antennas with omnidirectional patterns similar to the ones shown in Figure 2.17 whose main lobe is approximated by (2-32). The approximate directivity formula for an omnidirec-tional pattern as a function of the pattern half-power beamwidth (in degrees), which is reported by McDonald in , was derived based on the array factor of a broadside collinear array [see Sec-tion 6.4.1 and (6-38a)] and is given by D0 ≃ 101 HPBW (degrees) −0.0027 [HPBW (degrees)]2 (2-33a) However, that reported by Pozar in is derived based on the exact values obtained using (2-32) and then representing the data in closed-form using curve-fitting, and it is given by D0 ≃−172.4 + 191 √ 0.818 + 1∕HPBW (degrees) (2-33b) The approximate formula of (2-33a) should, in general, be more accurate for omnidirectional pat-terns with minor lobes, as shown in Figure 2.17(a), while (2-33b) should be more accurate for omni-directional patterns with minor lobes of very low intensity (ideally no minor lobes), as shown in Figure 2.17(b).
The approximate formulas of (2-33a) and (2-33b) can be used to design omnidirectional antennas with specified radiation pattern characteristics. To facilitate this procedure, the directivity of antennas with omnidirectional patterns approximated by (2-32) is plotted in Figure 2.18 versus n and the DIRECTIVITY 53 z x y Normalized Field Pattern (linear scale) 1 0.8 0.9 0.6 0.7 0.3 0.4 0.5 0.1 0.2 z x y Normalized Field Pattern (linear scale) 1 0.8 0.9 0.6 0.7 0.3 0.4 0.5 0.1 0.2 (a) With minor lobes (b) Without minor lobes Figure 2.17 Omnidirectional patterns with and without minor lobes.
0 50 100 150 200 250 300 350 400 450 0 2 4 6 8 10 12 14 16 18 20 3.01 6.02 7.78 9.03 10.00 10.79 11.46 12.04 12.55 30.01 Directivity Do (dimensionless) Directivity Do (dB) n 180.0 19.0 13.5 11.0 9.5 8.5 7.8 7.2 6.7 6.4 HPBW (degrees) Do (exact) Do (McDonald)(Eq.2-33a) Do (Pozar)(Eq. 2-33b) U = sinnθ Figure 2.18 Comparison of exact and approximate values of directivity for omnidirectional U = sinn 𝜃 power patterns.
54 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS half-power beamwidth (in degrees). Three curves are plotted in Figure 2.18; one using (2-16a) and referred as exact, one using (2-33a) and denoted as McDonald, and the third using (2-33b) and denoted as Pozar. Thus, the curves of Figure 2.18 can be used for design purposes, as follows: a. Specify the desired directivity and determine the value of n and half-power beamwidth of the omnidirectional antenna pattern, or b. Specify the desired value of n or half-power beamwidth and determine the directivity of the omnidirectional antenna pattern.
To demonstrate the procedure, an example is taken.
Example 2.8 Design an antenna with omnidirectional amplitude pattern with a half-power beamwidth of 90◦.
Express its radiation intensity by U = sinn 𝜃. Determine the value of n and attempt to identify elements that exhibit such a pattern. Determine the directivity of the antenna using (2-16a), (2-33a), and (2-33b).
Solution: Since the half-power beamwidth is 90◦, the angle at which the half-power point occurs is 𝜃= 45◦. Thus U(𝜃= 45◦) = 0.5 = sinn(45◦) = (0.707)n or n = 2 Therefore, the radiation intensity of the omnidirectional antenna is represented by U = sin2 𝜃.
An infinitesimal dipole (see Chapter 4) or a small circular loop (see Chapter 5) are two antennas which possess such a pattern.
Using the definition of (2-16a), the exact directivity is Umax = 1 Prad = ∫ 2𝜋 0 ∫ 𝜋 0 sin2 𝜃sin 𝜃d𝜃d𝜙= 8𝜋 3 D0 = 4𝜋 8𝜋∕3 = 3 2 = 1.761 dB Since the half-power beamwidth is equal to 90◦, then the directivity based on (2-33a) is equal to D0 = 101 90 −0.0027(90)2 = 1.4825 = 1.71 dB while that based on (2-33b) is equal to D0 = −172.4 + 191 √ 0.818 + 1∕90 = 1.516 = 1.807 dB The value of n and the three values of the directivity can also be obtained using Figure 2.18, although they may not be as accurate as those given above because they have to be taken off the graph. However, the curves can be used for other problems.
NUMERICAL TECHNIQUES 55 2.7 NUMERICAL TECHNIQUES For most practical antennas, their radiation patterns are so complex that closed-form mathemati-cal expressions are not available. Even in those cases where expressions are available, their form is so complex that integration to find the radiated power, required to compute the maximum directiv-ity, cannot be performed. Instead of using the approximate expressions of Kraus, Tai and Pereira, McDonald, or Pozar alternate and more accurate techniques may be desirable. With the high-speed computer systems now available, the answer may be to apply numerical methods.
Let us assume that the radiation intensity of a given antenna is separable, and it is given by U = B0f (𝜃)g(𝜙) (2-34) where B0 is a constant. The directivity for such a system is given by D0 = 4𝜋Umax Prad (2-35) where Prad = B0 ∫ 2𝜋 0 { ∫ 𝜋 0 f (𝜃)g(𝜙) sin 𝜃d𝜃 } d𝜙 (2-36) which can also be written as Prad = B0 ∫ 2𝜋 0 g(𝜙) { ∫ 𝜋 0 f(𝜃) sin 𝜃d𝜃 } d𝜙 (2-37) If the integrations in (2-37) cannot be performed analytically, then from integral calculus we can write a series approximation ∫ 𝜋 0 f (𝜃) sin 𝜃d𝜃= N ∑ i=1 [f(𝜃i) sin 𝜃i]Δ𝜃i (2-38) For N uniform divisions over the 𝜋interval, Δ𝜃i = 𝜋 N (2-38a) Referring to Figure 2.19, 𝜃i can take many different forms. Two schemes are shown in Figure 2.19 such that 𝜃i = i ( 𝜋 N ) , i = 1, 2, 3, … , N (2-38b) or 𝜃i = 𝜋 2N + (i −1) 𝜋 N , i = 1, 2, 3, … , N (2-38c) In the former case, 𝜃i is taken at the trailing edge of each division; in the latter case, 𝜃i is selected at the middle of each division. The scheme that is more desirable will depend upon the problem under investigation. Many other schemes are available.
56 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS In a similar manner, we can write for the 𝜙variations that ∫ 2𝜋 0 g(𝜙) d𝜙= M ∑ j=1 g(𝜙j)Δ𝜙j (2-39) where for M uniform divisions Δ𝜙j = 2𝜋 M (2-39a) Again referring to Figure 2.19 𝜙j = j (2𝜋 M ) , j = 1, 2, 3, … , M (2-39b) or 𝜙j = 2𝜋 2M + (j −1)2𝜋 M , j = 1, 2, 3, … , M (2-39c) Combining (2-38), (2-38a), (2-39), and (2-39a) we can write (2-37) as Prad = B0 ( 𝜋 N ) (2𝜋 M ) M ∑ j=1 { g(𝜙j) [ N ∑ i=1 f (𝜃i) sin 𝜃i ]} (2-40) The double summation of (2-40) is performed by adding for each value of j(j = 1, 2, 3, … , M) all values of i(i = 1, 2, 3, … , N). In a computer program flowchart, this can be performed by a loop within a loop. Physically, (2-40) can be interpreted by referring to Figure 2.19. It simply states that Figure 2.19 Digitization scheme of pattern in spherical coordinates.
NUMERICAL TECHNIQUES 57 for each value of g(𝜙) at the azimuthal angle 𝜙= 𝜙j, the values of f(𝜃) sin 𝜃are added for all values of 𝜃= 𝜃i(i = 1, 2, 3, … , N). The values of 𝜃i and 𝜙j can be determined by using either of the forms as given by (2-38b) or (2-38c) and (2-39b) or (2-39c).
Since the 𝜃and 𝜙variations are separable, (2-40) can also be written as Prad = B0 ( 𝜋 N ) (2𝜋 M ) [ M ∑ j=1 g(𝜙j) ] [ N ∑ i=1 f (𝜃i) sin 𝜃i ] (2-41) in which case each summation can be performed separately.
If the 𝜃and 𝜙variations are not separable, and the radiation intensity is given by U = B0F(𝜃, 𝜙) (2-42) the digital form of the radiated power can be written as Prad = B0 ( 𝜋 N ) (2𝜋 M ) M ∑ j=1 [ N ∑ i=1 F(𝜃i, 𝜙j) sin 𝜃i ] (2-43) 𝜃i and 𝜙j take different forms, two of which were introduced and are shown pictorially in Figure 2.19.
The evaluation and physical interpretation of (2-43) is similar to that of (2-40).
To examine the accuracy of the technique, two examples will be considered.
Example 2.9(a) The radiation intensity of an antenna is given by U(𝜃, 𝜙) = ⎧ ⎪ ⎨ ⎪ ⎩ B0 sin 𝜃sin2 𝜙, 0 ≤𝜃≤𝜋, 0 ≤𝜙≤𝜋 0 elsewhere The three-dimensional pattern of U(𝜃, 𝜙) is shown in Figure 2.20.
Determine the maximum directivity numerically by using (2-41) with 𝜃i and 𝜙j of (2-38b) and (2-39b), respectively. Compare it with the exact value.
Solution: Let us divide the 𝜃and 𝜙intervals each into 18 equals segments (N = M = 18).
Since 0 ≤𝜙≤𝜋, then Δ𝜙j = 𝜋∕M and (2-41) reduces to Prad = B0 ( 𝜋 18 )2 [ 18 ∑ j=1 sin2 𝜙j ] [ 18 ∑ i=1 sin2 𝜃i ] with 𝜃i = i ( 𝜋 18 ) = i(10◦), i = 1, 2, 3, … , 18 𝜙j = j ( 𝜋 18 ) = j(10◦), j = 1, 2, 3, … , 18 58 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2 0 sin( )sin ( ) 0 B U θ θ ϕ ϕ π ⎧ ⎧ ⎪ ⎨ = ≤ ≤ 0 π ≤ ≤ ⎨ ⎩ Normalized Field Pattern (linearscale) z 0 Elsewhere ⎪ ⎩ 0.8 0.9 1 i ( ) / 0.6 0.7 sin(θ ); ϕ = /2 sin2( ); = π/2 0.3 0.4 0.5 y sin (ϕ ); θ /2 0 0.1 0.2 x = π π Figure 2.20 Three-dimensional pattern of the radiation of Examples 2.9(a,b).
Thus Prad = B0 ( 𝜋 18 )2 [sin2(10◦) + sin2(20◦) + ⋯+ sin2(180◦)]2 Prad = B0 ( 𝜋 18 )2 (9)2 = B0 ( 𝜋2 4 ) and D0 = 4𝜋Umax Prad = 4𝜋 𝜋2∕4 = 16 𝜋= 5.0929 The exact value is given by Prad = B0 ∫ 𝜋 0 sin2 𝜙d𝜙∫ 𝜋 0 sin2 𝜃d𝜃= 𝜋 2 (𝜋 2 ) B0 = 𝜋2 4 B0 and D0 = 4𝜋Umax Prad = 4𝜋 𝜋2∕4 = 16 𝜋= 5.0929 which is the same as the value obtained numerically!
NUMERICAL TECHNIQUES 59 Example 2.9(b) Given the same radiation intensity as that in Example 2.9(a), determine the directivity using (2-41) with 𝜃i and 𝜙j of (2-38c) and (2-39c).
Solution: Again using 18 divisions in each interval, we can write (2-41) as Prad = B0 ( 𝜋 18 )2 [ 18 ∑ j=1 sin2 𝜙j ] [ 18 ∑ i=1 sin2 𝜃i ] with 𝜃i = 𝜋 36 + (i −1) 𝜋 18 = 5◦+ (i −1)10◦, i = 1, 2, 3, … , 18 𝜙j = 𝜋 36 + (j −1) 𝜋 18 = 5◦+ (j −1)10◦, j = 1, 2, 3, … , 18 Because of the symmetry of the divisions about the 𝜃= 𝜋∕2 and 𝜙= 𝜋∕2 angles, we can write Prad = B0 ( 𝜋 18 )2 [ 2 9 ∑ j=1 sin2 𝜙j ] [ 2 9 ∑ i=1 sin2 𝜃i ] Prad = B0 ( 𝜋 18 )2 4[sin2(5◦) + sin2(15◦) + ⋯+ sin2(85◦)]2 Prad = B0 ( 𝜋 18 )2 4(4.5)2 = B0 ( 𝜋 18 )2 (81) = B0 ( 𝜋2 4 ) which is identical to that of the previous example. Thus D0 = 4𝜋Umax Prad = 4𝜋 𝜋2∕4 = 16 𝜋= 5.0929 which again is equal to the exact value!
It is interesting to note that decreasing the number of divisions (M and/or N) to 9, 6, 4, and even 2 leads to the same answer, which also happens to be the exact value! To demonstrate as to why the number of divisions does not affect the answer for this pattern, let us refer to Figure 2.21 where we have plotted the sin2 𝜙function and divided the 0◦≤𝜙≤180◦interval into six divisions. The exact value of the directivity uses the area under the solid curve. Doing the problem numerically, we find the area under the rectangles, which is shown shaded. Because of the symmetrical nature of the function, it can be shown that the shaded area in section #1 (included in the numerical evaluation) is equal to the blank area in section #1′ (left out by the numerical method). The same is true for the areas in sections #2 and #2′, and #3 and #3′. Thus, there is a one-to-one compensation. Similar justification is applicable for the other number of divisions.
It should be emphasized that all functions, even though they may contain some symmetry, do not give the same answers independent of the number of divisions. As a matter of fact, in most cases the answer only approaches the exact value as the number of divisions is increased to a large number.
60 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 1.0 0.8 0.6 0.4 0.2 0 30° 60° #1 #2 #3 #1´ #2´ #3´ 90° (degrees) sin2 120° 150° 180° ϕ ϕ Figure 2.21 Digitized form of sin2 𝜙function.
A MATLAB and FORTRAN computer program called Directivity has been developed to com-pute the maximum directivity of any antenna whose radiation intensity is U = F(𝜃, 𝜙) based on the formulation of (2-43). The intensity function F does not have to be a function of both 𝜃and 𝜙. The numerical evaluations are made at the trailing edge, as defined by (2-38b) and (2-39b). The program is included in the publisher’s website for this book. It contains a subroutine for which the intensity factor U = F(𝜃, 𝜙) for the required application must be specified by the user. As an illustration, the antenna intensity U = sin 𝜃sin2 𝜙has been inserted in the subroutine. In addition, the upper and lower limits of 𝜃and 𝜙must be specified for each application of the same pattern.
2.8 ANTENNA EFFICIENCY Associated with an antenna are a number of efficiencies and can be defined using Figure 2.22. The total antenna efficiency e0 is used to take into account losses at the input terminals and within the structure of the antenna. Such losses may be due, referring to Figure 2.22(b), to 1. reflections because of the mismatch between the transmission line and the antenna 2. I2R losses (conduction and dielectric) In general, the overall efficiency can be written as e0 = ereced (2-44) where e0 = total efficiency (dimensionless) er = reflection (mismatch) efficiency = (1 −|Γ|2) (dimensionless) GAIN, REALIZED GAIN 61 ec = conduction efficiency (dimensionless) ed = dielectric efficiency (dimensionless) Γ = voltage reflection coefficient at the input terminals of the antenna [Γ = (Zin −Z0)∕(Zin + Z0) where Zin = antenna input impedance, Z0 = characteristic impedance of the transmission line] VSWR = voltage standing wave ratio = 1 + |Γ| 1 −|Γ| Figure 2.22 Reference terminals and losses of an antenna.
Usually ec and ed are very difficult to compute, but they can be determined experimentally. Even by measurements they cannot be separated, and it is usually more convenient to write (2-44) as e0 = erecd = ecd(1 −|Γ|2) (2-45) where ecd = eced = antenna radiation efficiency, which is used to relate the gain and directivity.
2.9 GAIN, REALIZED GAIN Another useful figure-of-merit describing the performance of an antenna is the gain. Although the gain of the antenna is closely related to the directivity, it is a measure that takes into account the efficiency of the antenna as well as its directional capabilities. Remember that directivity is a measure that describes only the directional properties of the antenna, and it is therefore controlled only by the pattern.
Gain of an antenna (in a given direction) is defined as “the ratio of the intensity, in a given direc-tion, to the radiation intensity that would be obtained if the power accepted by the antenna were radiated isotropically. The radiation intensity corresponding to the isotropically radiated power is equal to the power accepted (input) by the antenna divided by 4𝜋.” In equation form this can be 62 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS expressed as Gain = 4𝜋 radiation intensity total input (accepted) power = 4𝜋U(𝜃, 𝜙) Pin (dimensionless) (2-46) In most cases we deal with relative gain, which is defined as “the ratio of the power gain in a given direction to the power gain of a reference antenna in its referenced direction.” The power input must be the same for both antennas. The reference antenna is usually a dipole, horn, or any other antenna whose gain can be calculated or it is known. In most cases, however, the reference antenna is a lossless isotropic source. Thus G = 4𝜋U(𝜃, 𝜙) Pin(lossless isotropic source) (dimensionless) (2-46a) When the direction is not stated, the power gain is usually taken in the direction of maximum radi-ation.
Referring to Figure 2.22(a), we can write that the total radiated power (Prad) is related to the total input power (Pin) by Prad = ecdPin (2-47) where ecd is the antenna radiation efficiency (dimensionless) which is defined in (2-44), (2-45) and Section 2.14 by (2-90). According to the IEEE Standards, “gain does not include losses arising from impedance mismatches (reflection losses) and polarization mismatches (losses).” In this edition of the book we define two gains; one, referred to as gain (G), and the other, referred to as realized gain (Gre), that also takes into account the reflection/mismatch losses represented in both (2-44) and (2-45).
Using (2-47) reduces (2-46a) to G(𝜃, 𝜙) = ecd [ 4𝜋U(𝜃, 𝜙) Prad ] (2-48) which is related to the directivity of (2-16) and (2-21) by G(𝜃, 𝜙) = ecdD(𝜃, 𝜙) (2-49) In a similar manner, the maximum value of the gain is related to the maximum directivity of (2-16a) and (2-23) by G0 = G(𝜃, 𝜙)|max = ecdD(𝜃, 𝜙)|max = ecdD0 (2-49a) While (2-47) does take into account the losses of the antenna element itself, it does not take into account the losses when the antenna element is connected to a transmission line, as shown in Figure 2.22. These connection losses are usually referred to as reflections (mismatch) losses, and they are taken into account by introducing a reflection (mismatch) efficiency er, which is related to the reflection coefficient as represented in (2-45) or er = (1 −|Γ|2). Thus, we can introduce a realized gain Gre that takes into account the reflection/mismatch losses (due to the connection of the antenna GAIN, REALIZED GAIN 63 element to the transmission line), and it can be written as Gre(𝜃, 𝜙) = erG(𝜃, 𝜙) = (1 −|Γ|2)G(𝜃, 𝜙) = erecdD(𝜃, 𝜙) = eoD(𝜃, 𝜙) (2-49b) where eo is the overall efficiency as defined in (2-44), (2-45). Similarly, the maximum realized gain Gre0 of (2-49a) is related to the maximum directivity D0 by Gre0 = Gre(𝜃, 𝜙)|max = erG(𝜃, 𝜙)|max = (1 −|Γ|2)G(𝜃, 𝜙)|max = erecdD(𝜃, 𝜙)|max = eoD(𝜃, 𝜙)|max = eoD0 (2-49c) If the antenna is matched to the transmission line, that is, the antenna input impedance Zin is equal to the characteristic impedance Zc of the line (|Γ| = 0), then the two gains are equal (Gre = G).
As was done with the directivity, we can define the partial gain of an antenna for a given polariza-tion in a given direction as “that part of the radiation intensity corresponding to a given polarization divided by the total radiation intensity that would be obtained if the power accepted by the antenna were radiated isotropically.” With this definition for the partial gain, then, in a given direction, “the total gain is the sum of the partial gains for any two orthogonal polarizations.” For a spherical coor-dinate system, the total maximum gain G0 for the orthogonal 𝜃and 𝜙components of an antenna can be written, in a similar form as was the maximum directivity in (2-17)–(2-17b), as G0 = G𝜃+ G𝜙 (2-50) while the partial gains G𝜃and G𝜙are expressed as G𝜃= 4𝜋U𝜃 Pin (2-50a) G𝜙= 4𝜋U𝜙 Pin (2-50b) where U𝜃= radiation intensity in a given direction contained in E𝜃field component U𝜙= radiation intensity in a given direction contained in E𝜙field component Pin = total input (accepted) power For many practical antennas an approximate formula for the gain, corresponding to (2-27) or (2-27a) for the directivity, is G0 ≃30,000 Θ1dΘ2d (2-51) In practice, whenever the term “gain” is used, it usually refers to the maximum gain as defined by (2-49a) or (2-49c).
Usually the gain is given in terms of decibels, instead of the dimensionless quantity of (2-49a).
The conversion formula is G0(dB) = 10 log10[ecdD0 (dimensionless)] (2-52) 64 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Example 2.10 A lossless resonant half-wavelength dipole antenna, with input impedance of 73 ohms, is con-nected to a transmission line whose characteristic impedance is 50 ohms. Assuming that the pattern of the antenna is given approximately by U = B0 sin3 𝜃 find the maximum realized gain of this antenna.
Solution: Let us first compute the maximum directivity of the antenna. For this U|max = Umax = B0 Prad = ∫ 2𝜋 0 ∫ 𝜋 0 U(𝜃, 𝜙) sin 𝜃d𝜃d𝜙= 2𝜋B0 ∫ 𝜋 0 sin4 𝜃d𝜃= B0 ( 3𝜋2 4 ) D0 = 4𝜋Umax Prad = 16 3𝜋= 1.697 Since the antenna was stated to be lossless, then the radiation efficiency ecd = 1.
Thus, the total maximum gain is equal to G0 = ecdD0 = 1(1.697) = 1.697 G0(dB) = 10 log10(1.697) = 2.297 which is identical to the directivity because the antenna is lossless.
There is another loss factor which is not taken into account in the gain. That is the loss due to reflection or mismatch losses between the antenna (load) and the transmission line. This loss is accounted for by the reflection efficiency of (2-44) or (2-45), and it is equal to er = (1 −|Γ|2) = ( 1 − | | | | 73 −50 73 + 50 | | | | 2) = 0.965 er(dB) = 10 log10(0.965) = −0.155 Therefore the overall efficiency is e0 = erecd = 0.965 e0(dB) = −0.155 Thus, the overall losses are equal to 0.155 dB. The maximum realized gain is equal to Gre0 = e0D0 = 0.965(1.697) = 1.6376 Gre0(dB) = 10 log10(1.6376) = 2.142 The gain in dB can also be obtained by converting the directivity and radiation efficiency in dB and then adding them. Thus, ecd(dB) = 10 log10(1.0) = 0 D0(dB) = 10 log10(1.697) = 2.297 G0(dB) = ecd(dB) + D0(dB) = 2.297 which is the same as obtained previously. The same procedure can be used for the realized gain.
BANDWIDTH 65 2.10 BEAM EFFICIENCY Another parameter that is frequently used to judge the quality of transmitting and receiving antennas is the beam efficiency. For an antenna with its major lobe directed along the z-axis (𝜃= 0), as shown in Figure 2.1(a), the beam efficiency (BE) is defined by BE = power transmitted (received) within cone angle 𝜃1 power transmitted (received) by the antenna (dimensionless) (2-53) where 𝜃1 is the half-angle of the cone within which the percentage of the total power is to be found.
Equation (2-53) can be written as BE = ∫ 2𝜋 0 ∫ 𝜃1 0 U(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 ∫ 2𝜋 0 ∫ 𝜋 0 U(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-54) If 𝜃1 is chosen as the angle where the first null or minimum occurs (see Figure 2.1), then the beam efficiency will indicate the amount of power in the major lobe compared to the total power. A very high beam efficiency (between the nulls or minima), usually in the high 90s, is necessary for anten-nas used in radiometry, astronomy, radar, and other applications where received signals through the minor lobes must be minimized. The beam efficiencies of some typical rectangular and circular aperture antennas will be discussed in Chapter 12.
2.11 BANDWIDTH The bandwidth of an antenna is defined as “the range of frequencies within which the performance of the antenna, with respect to some characteristic, conforms to a specified standard.” The bandwidth can be considered to be the range of frequencies, on either side of a center frequency (usually the resonance frequency for a dipole), where the antenna characteristics (such as input impedance, pat-tern, beamwidth, polarization, side lobe level, gain, beam direction, radiation efficiency) are within an acceptable value of those at the center frequency.
r For broadband antennas, the bandwidth is usually expressed as the ratio of the upper-to-lower frequencies of acceptable operation. For example, a 10:1 bandwidth indicates that the upper frequency is 10 times greater than the lower.
r For narrowband antennas, the bandwidth is expressed as a percentage of the frequency dif-ference (upper minus lower) over the center frequency of the bandwidth. For example, a 5% bandwidth indicates that the frequency range of acceptable operation is 5% of the bandwidth center frequency.
Because the characteristics (input impedance, pattern, gain, polarization, etc.) of an antenna do not necessarily vary in the same manner or are even critically affected by the frequency, there is no unique characterization of the bandwidth. The specifications are set in each case to meet the needs of the particular application. Usually there is a distinction made between pattern and input impedance variations. Accordingly pattern bandwidth and impedance bandwidth are used to emphasize this dis-tinction. Associated with pattern bandwidth are gain, side lobe level, beamwidth, polarization, and beam direction while input impedance and radiation efficiency are related to impedance bandwidth.
For example, the pattern of a linear dipole with overall length less than a half-wavelength (l < λ∕2) 66 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS is basically insensitive to frequency. The limiting factor for this antenna is its impedance, and its bandwidth can be formulated in terms of the Q. The Q of antennas or arrays with dimensions large compared to the wavelength, excluding superdirective designs, is near unity. Therefore the band-width is usually formulated in terms of beamwidth, side lobe level, and pattern characteristics. For intermediate length antennas, the bandwidth may be limited by either pattern or impedance vari-ations, depending upon the particular application. For these antennas, a 2:1 bandwidth indicates a good design. For others, large bandwidths are needed. Antennas with very large bandwidths (like 40:1 or greater) have been designed in recent years. These are known as frequency independent antennas, and they are discussed in Chapter 11.
The above discussion presumes that the coupling networks (transformers, baluns, etc.) and/or the dimensions of the antenna are not altered in any manner as the frequency is changed. It is possible to increase the acceptable frequency range of a narrowband antenna if proper adjustments can be made on the critical dimensions of the antenna and/or on the coupling networks as the frequency is changed. Although not an easy or possible task in general, there are applications where this can be accomplished. The most common examples are the old whip antenna of a car radio and the “rabbit ears” of a television. Both usually have adjustable lengths which can be used to tune the antenna for better reception. Antennas of this type, whose adjustments are made to modify their radiation characteristics, are often referred to as reconfigurable antennas.
2.12 POLARIZATION Polarization of an antenna in a given direction is defined as “the polarization of the wave transmitted (radiated) by the antenna. Note: When the direction is not stated, the polarization is taken to be the polarization in the direction of maximum gain.” In practice, polarization of the radiated energy varies with the direction from the center of the antenna, so that different parts of the pattern may have different polarizations.
Polarization of a radiated wave is defined as “that property of an electromagnetic wave describing the time-varying direction and relative magnitude of the electric-field vector; specifically, the figure traced as a function of time by the extremity of the vector at a fixed location in space, and the sense in which it is traced, as observed along the direction of propagation.” Polarization then is the curve traced by the end point of the arrow (vector) representing the instantaneous electric field. The field must be observed along the direction of propagation. A typical trace as a function of time is shown in Figures 2.23(a) and (b). Animation of the these two traces can be performed using the MATLAB computer program Polarization Diagram Ellipse Animation found in the publisher’s website for this book.
The polarization of a wave can be defined in terms of a wave radiated (transmitted) or received by an antenna in a given direction. The polarization of a wave radiated by an antenna in a specified direction at a point in the far field is defined as “the polarization of the (locally) plane wave which is used to represent the radiated wave at that point. At any point in the far field of an antenna the radiated wave can be represented by a plane wave whose electric-field strength is the same as that of the wave and whose direction of propagation is in the radial direction from the antenna. As the radial distance approaches infinity, the radius of curvature of the radiated wave’s phase front also approaches infinity and thus in any specified direction the wave appears locally as a plane wave.” This is a far-field characteristic of waves radiated by all practical antennas, and it is illustrated analytically in Section 3.6 of Chapter 3. The polarization of a wave received by an antenna is defined as the “polarization of a plane wave, incident from a given direction and having a given power flux density, which results in maximum available power at the antenna terminals.” Polarization may be classified as linear, circular, or elliptical. If the vector that describes the electric field at a point in space as a function of time is always directed along a line, the field is said to be linearly polarized. In general, however, the figure that the electric field traces is an ellipse, and the field is said to be elliptically polarized. Linear and circular polarizations are special cases of elliptical, POLARIZATION 67 (a) Rotation of wave (b) Polarization ellipse z Eyo Exo Minor axis Major axis OB OA τ ωt ωt 0 0 2π 4π 6π 6π 4π 2π x x x y y Figure 2.23 Rotation of a plane electromagnetic wave and its polarization ellipse at z = 0 as a function of time.
and they can be obtained when the ellipse becomes a straight line or a circle, respectively. The figure of the electric field is traced in a clockwise (CW) or counterclockwise (CCW) sense. Clockwise rotation of the electric-field vector is also designated as right-hand polarization and counterclockwise as left-hand polarization.
In general, the polarization characteristics of an antenna can be represented by its polarization pattern whose one definition is “the spatial distribution of the polarizations of a field vector excited (radiated) by an antenna taken over its radiation sphere. When describing the polarizations over the radiation sphere, or portion of it, reference lines shall be specified over the sphere, in order to measure the tilt angles (see tilt angle) of the polarization ellipses and the direction of polarization for linear polarizations. An obvious choice, though by no means the only one, is a family of lines tangent at each point on the sphere to either the 𝜃or 𝜙coordinate line associated with a spherical coordinate system of the radiation sphere. At each point on the radiation sphere the polarization is usually resolved into a pair of orthogonal polarizations, the co-polarization and cross polarization. To accomplish this, the co-polarization must be specified at each point on the radiation sphere.” “Co-polarization represents the polarization the antenna is intended to radiate (receive) while cross-polarization represents the polarization orthogonal to a specified polarization, which is usually the co-polarization.” 68 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS “For certain linearly polarized antennas, it is common practice to define the co-polarization in the following manner: First specify the orientation of the co-polar electric-field vector at a pole of the radiation sphere. Then, for all other directions of interest (points on the radiation sphere), require that the angle that the co-polar electric-field vector makes with each great circle line through the pole remain constant over that circle, the angle being that at the pole.” “In practice, the axis of the antenna’s main beam should be directed along the polar axis of the radiation sphere. The antenna is then appropriately oriented about this axis to align the direction of its polarization with that of the defined co-polarization at the pole.” “This manner of defining co-polarization can be extended to the case of elliptical polarization by defining the constant angles using the major axes of the polarization ellipses rather than the co-polar electric-field vector. The sense of polarization (rotation) must also be specified.” The polarization of the wave radiated by the antenna can also be represented on the Poincar´ e sphere –. Each point on the Poincar´ e sphere represents a unique polarization. The north pole represents left circular polarization, the south pole represents right circular, and points along the equator represent linear polarization of different tilt angles. All other points on the Poincar´ e sphere represent elliptical polarization. For details, see Figure 17.24 of Chapter 17.
The polarization of an antenna is measured using techniques described in Chapter 17.
2.12.1 Linear, Circular, and Elliptical Polarizations The instantaneous field of a plane wave, traveling in the negative z direction, can be written as 𝓔(z; t) = ̂ axℰx(z; t) + ̂ ayℰy(z; t) (2-55) According to (2-5), the instantaneous components are related to their complex counterparts by ℰx(z; t) = Re[Ex −ej(𝜔t+kz)] = Re[Exoej(𝜔t+kz+𝜙x)] = Exo cos(𝜔t + kz + 𝜙x) (2-56) ℰy(z; t) = Re[Ey −ej(𝜔t+kz)] = Re[Eyoej(𝜔t+kz+𝜙y)] = Eyo cos(𝜔t + kz + 𝜙y) (2-57) where Exo and Eyo are, respectively, the maximum magnitudes of the x and y components.
A. Linear Polarization For the wave to have linear polarization, the time-phase difference between the two components must be Δ𝜙= 𝜙y −𝜙x = n𝜋, n = 0, 1, 2, 3, … (2-58) B. Circular Polarization Circular polarization can be achieved only when the magnitudes of the two components are the same and the time-phase difference between them is odd multiples of 𝜋∕2. That is, |ℰx| = |ℰy| ➱Exo = Eyo (2-59) Δ𝜙= 𝜙y −𝜙x = ⎧ ⎪ ⎨ ⎪ ⎩ + ( 1 2 + 2n)𝜋, n = 0, 1, 2, … for CW (2-60) −( 1 2 + 2n)𝜋, n = 0, 1, 2, … for CCW (2-61) SUMMARY 69 If the direction of wave propagation is reversed (i.e., +z direction), the phases in (2-60) and (2-61) for CW and CCW rotation must be interchanged.
C. Elliptical Polarization Elliptical polarization can be attained only when the time-phase difference between the two com-ponents is odd multiples of 𝜋∕2 and their magnitudes are not the same or when the time-phase difference between the two components is not equal to multiples of 𝜋∕2 (irrespective of their mag-nitudes). That is, |ℰx| ≠|ℰy| ➱Exo ≠Eyo when Δ𝜙= 𝜙y −𝜙x = n = 0, 1, 2, … ⎧ ⎪ ⎨ ⎪ ⎩ + ( 1 2 + 2n)𝜋 for CW (2-62a) −( 1 2 + 2n)𝜋 for CCW (2-62b) or Δ𝜙= 𝜙y −𝜙x ≠±n 2𝜋= n = 0, 1, 2, 3, … ⎧ ⎪ ⎨ ⎪ ⎩ > 0 for CW (2-63) < 0 for CCW (2-64) For elliptical polarization, the curve traced at a given position as a function of time is, in general, a tilted ellipse, as shown in Figure 2.23(b). The ratio of the major axis to the minor axis is referred to as the axial ratio (AR), and it is equal to AR = major axis minor axis = OA OB, 1 ≤AR ≤∞ (2-65) where OA = [ 1 2{E2 xo + E2 yo + [E4 xo + E4 yo + 2E2 xoE2 yo cos(2Δ𝜙)]1∕2} ]1∕2 (2-66) OB = [ 1 2{E2 xo + E2 yo −[E4 xo + E4 yo + 2E2 xoE2 yo cos(2Δ𝜙)]1∕2} ]1∕2 (2-67) The tilt of the ellipse, relative to the y axis, is represented by the angle 𝜏given by 𝜏= 𝜋 2 −1 2 tan−1 [ 2ExoEyo E2 xo −E2 yo cos(Δ𝜙) ] (2-68) When the ellipse is aligned with the principal axes [𝜏= n𝜋∕2, n = 0, 1, 2, …], the major (minor) axis is equal to Exo(Eyo) or Eyo(Exo) and the axial ratio is equal to Exo∕Eyo or Eyo∕Exo.
SUMMARY We will summarize the preceding discussion on polarization by stating the general characteristics, and the necessary and sufficient conditions that the wave must have in order to possess linear, circular or elliptical polarization.
70 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Linear Polarization A time-harmonic wave is linearly polarized, at a given point in space, if the electric-field (or magnetic-field) vector at that point is always oriented along the same straight line at every instant of time. This is accomplished if the field vector (electric or magnetic) possesses: a. Only one component, or b. Two orthogonal linear components that are in time phase or 180◦(or multiples of 180◦) out-of-phase.
Circular Polarization A time-harmonic wave is circularly polarized, at a given point in space, if the electric (or magnetic) field vector at that point traces a circle as a function of time.
The necessary and sufficient conditions to accomplish this are if the field vector (electric or mag-netic) possesses all of the following: a. The field must have two orthogonal linear components, and b. The two components must have the same magnitude, and c. The two components must have a time-phase difference of odd multiples of 90◦.
The sense of rotation is always determined by rotating the phase-leading component toward the phase-lagging component and observing the field rotation as the wave is viewed as it travels away from the observer. If the rotation is clockwise, the wave is right-hand (or clockwise) circularly polar-ized; if the rotation is counterclockwise, the wave is left-hand (or counterclockwise) circularly polar-ized. The rotation of the phase-leading component toward the phase-lagging component should be performed along the angular separation between the two components that is less than 180◦. Phases equal to or greater than 0◦and less than 180◦should be considered leading whereas those equal to or greater than 180◦and less than 360◦should be considered lagging.
Elliptical Polarization A time-harmonic wave is elliptically polarized if the tip of the field vector (electric or magnetic) traces an elliptical locus in space. At various instants of time the field vector changes continuously with time at such a manner as to describe an elliptical locus. It is right-hand (clockwise) elliptically polarized if the field vector rotates clockwise, and it is left-hand (counter-clockwise) elliptically polarized if the field vector of the ellipse rotates counterclockwise . The sense of rotation is determined using the same rules as for the circular polarization. In addition to the sense of rotation, elliptically polarized waves are also specified by their axial ratio whose magnitude is the ratio of the major to the minor axis.
A wave is elliptically polarized if it is not linearly or circularly polarized. Although linear and circular polarizations are special cases of elliptical, usually in practice elliptical polarization refers to other than linear or circular. The necessary and sufficient conditions to accomplish this are if the field vector (electric or magnetic) possesses all of the following: a. The field must have two orthogonal linear components, and b. The two components can be of the same or different magnitude.
c. (1) If the two components are not of the same magnitude, the time-phase difference between the two components must not be 0◦or multiples of 180◦(because it will then be linear). (2) If the two components are of the same magnitude, the time-phase difference between the two components must not be odd multiples of 90◦(because it will then be circular).
SUMMARY 71 If the wave is elliptically polarized with two components not of the same magnitude but with odd multiples of 90◦time-phase difference, the polarization ellipse will not be tilted but it will be aligned with the principal axes of the field components. The major axis of the ellipse will align with the axis of the field component which is larger of the two, while the minor axis of the ellipse will align with the axis of the field component which is smaller of the two.
2.12.2 Polarization Loss Factor and Efficiency In general, the polarization of the receiving antenna will not be the same as the polarization of the incoming (incident) wave. This is commonly stated as “polarization mismatch.” The amount of power extracted by the antenna from the incoming signal will not be maximum because of the polarization loss. Assuming that the electric field of the incoming wave can be written as Ei = ̂ ρwEi (2-69) where ̂ ρw is the unit vector of the wave, and the polarization of the electric field of the receiving antenna can be expressed as Ea = ̂ ρaEa (2-70) where ̂ ρa is its unit vector (polarization vector), the polarization loss can be taken into account by introducing a polarization loss factor (PLF). It is defined, based on the polarization of the antenna in its transmitting mode, as PLF = | ̂ ρw ⋅̂ ρa|2 = | cos 𝜓p|2 (dimensionless) (2-71) where 𝜓p is the angle between the two unit vectors. The relative alignment of the polarization of the incoming wave and of the antenna is shown in Figure 2.24. If the antenna is polarization matched, its PLF will be unity and the antenna will extract maximum power from the incoming wave.
Another figure of merit that is used to describe the polarization characteristics of a wave and that of an antenna is the polarization efficiency (polarization mismatch or loss factor) which is defined as “the ratio of the power received by an antenna from a given plane wave of arbitrary polarization to the power that would be received by the same antenna from a plane wave of the same power flux density and direction of propagation, whose state of polarization has been adjusted for a maximum received power.” This is similar to the PLF and it is expressed as pe = |𝓵e ⋅Einc|2 |𝓵e|2|Einc|2 (2-71a) a ^ w p ψ ψ ^ Figure 2.24 Polarization unit vectors of incident wave ( ̂ ρw) and antenna ( ̂ ρa), and polarization loss fac-tor (PLF).
72 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS where 𝓵e = vector effective length of the antenna Einc = incident electric field The vector effective length 𝓵e of the antenna has not yet been defined, and it is introduced in Section 2.15. It is a vector that describes the polarization characteristics of the antenna. Both the PLF and pe lead to the same answers.
The conjugate (∗) is not used in (2-71) or (2-71a) so that a right-hand circularly polarized incident wave (when viewed in its direction of propagation) is matched to right-hand circularly polarized receiving antenna (when its polarization is determined in the transmitting mode). Similarly, a left-hand circularly polarized wave will be matched to a left-hand circularly polarized antenna.
To illustrate the principle of polarization mismatch, two examples are considered.
Example 2.11 The electric field of a linearly polarized electromagnetic wave given by Ei = ̂ axE0(x, y)e−jkz is incident upon a linearly polarized antenna whose electric-field polarization is expressed as Ea ≃(̂ ax + ̂ ay)E(r, 𝜃, 𝜙) Find the polarization loss factor (PLF).
Solution: For the incident wave ̂ ρw = ̂ ax and for the antenna ̂ ρa = 1 √ 2 (̂ ax + ̂ ay) The PLF is then equal to PLF = |̂ ρw ⋅̂ ρa|2 = |̂ ax ⋅ 1 √ 2 (̂ ax + ̂ ay)|2 = 1 2 which in dB is equal to PLF (dB) = 10 log10 PLF (dimensionless) = 10 log10(0.5) = −3 Even though in Example 2.11 both the incoming wave and the antenna are linearly polarized, there is a 3-dB loss in extracted power because the polarization of the incoming wave is not aligned with the polarization of the antenna. If the polarization of the incoming wave is orthogonal to the polarization of the antenna, then there will be no power extracted by the antenna from the incoming wave and the PLF will be zero or −∞dB. In Figures 2.25(a,b) we illustrate the polarization loss factors (PLF) of two types of antennas: wires and apertures.
We now want to consider an example where the polarization of the antenna and the incoming wave are described in terms of complex polarization vectors.
SUMMARY 73 ψ ψ p ψp PLF = | (aligned) w • a|2 = 1 PLF = | (orthogonal) w • a|2 = 0 PLF = | (rotated) (a) PLF for transmitting and receiving aperture antennas w • p a|2 = cos2 ψ PLF = | (aligned) w • a|2 = 1 PLF = | (orthogonal) w • a|2 = 0 PLF = | (rotated) (b) PLF for transmitting and receiving linear wire antennas w • p a|2 = cos2 ψp ψp ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ Figure 2.25 Polarization loss factors (PLF) for aperture and linear wire antennas.
Example 2.12 A right-hand (clockwise) circularly polarized wave radiated by an antenna, placed at some dis-tance away from the origin of a spherical coordinate system, is traveling in the inward radial direction at an angle (𝜃, 𝜙) and it is impinging upon a right-hand circularly polarized receiving antenna placed at the origin (see Figures 2.1 and 17.23 for the geometry of the coordinate sys-tem). The polarization of the receiving antenna is defined in the transmitting mode, as desired by the definition of the IEEE. Assuming the polarization of the incident wave is represented by Ew = (̂ a𝜃+ ĵ a𝜙)E(r, 𝜃, 𝜙) Determine the polarization loss factor (PLF).
Solution: The polarization of the incident right-hand circularly polarized wave traveling along the −r radial direction is described by the unit vector ̂ ρw = ( ̂ a𝜃+ ĵ a𝜙 √ 2 ) while that of the receiving antenna, in the transmitting mode, is represented by the unit vector ̂ ρa = ( ̂ a𝜃−ĵ a𝜙 √ 2 ) 74 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Therefore the polarization loss factor is PLF = | ̂ ρw ⋅̂ ρa|2 = 1 4|1 + 1|2 = 1 = 0 dB Since the polarization of the incoming wave matches (including the sense of rotation) the polar-ization of the receiving antenna, there should not be any losses. Obviously the answer matches the expectation.
Based upon the definitions of the wave transmitted and received by an antenna, the polarization of an antenna in the receiving mode is related to that in the transmitting mode as follows: 1. “In the same plane of polarization, the polarization ellipses have the same axial ratio, the same sense of polarization (rotation) and the same spatial orientation.
2. “Since their senses of polarization and spatial orientation are specified by viewing their polar-ization ellipses in the respective directions in which they are propagating, one should note that: a. Although their senses of polarization are the same, they would appear to be opposite if both waves were viewed in the same direction.
b. Their tilt angles are such that they are the negative of one another with respect to a common reference.” Since the polarization of an antenna will almost always be defined in its transmitting mode, according to the IEEE Std 145-1993, “the receiving polarization may be used to specify the polar-ization characteristic of a nonreciprocal antenna which may transmit and receive arbitrarily different polarizations.” The polarization loss must always be taken into account in the link calculations design of a com-munication system because in some cases it may be a very critical factor. Link calculations of com-munication systems for outer space explorations are very stringent because of limitations in space-craft weight. In such cases, power is a limiting consideration. The design must properly take into account all loss factors to ensure a successful operation of the system.
An antenna that is elliptically polarized is that composed of two crossed dipoles, as shown in Figure 2.26. The two crossed dipoles provide the two orthogonal field components that are not Figure 2.26 Geometry of elliptically polarized cross-dipole antenna.
INPUT IMPEDANCE 75 necessarily of the same field intensity toward all observation angles. If the two dipoles are identical, the field intensity of each along zenith (perpendicular to the plane of the two dipoles) would be of the same intensity. Also, if the two dipoles were fed with a 90◦degree time-phase difference (phase quadrature), the polarization along zenith would be circular and elliptical toward other directions.
One way to obtain the 90◦time-phase difference Δ𝜙between the two orthogonal field components, radiated respectively by the two dipoles, is by feeding one of the two dipoles with a transmission line which is λ∕4 longer or shorter than that of the other [Δ𝜙= kΔ𝓁= (2𝜋∕λ)(λ∕4) = 𝜋∕2]. One of the lengths (longer or shorter) will provide right-hand (CW) rotation while the other will provide left-hand (CCW) rotation.
A MATLAB computer program Polarization Propag is included at the end of the chapter, and it computes the Poinar´ e sphere angles and the polarization of the wave radiated by an antenna and traveling in an infinite homogeneous medium.
2.13 INPUT IMPEDANCE Input impedance is defined as “the impedance presented by an antenna at its terminals or the ratio of the voltage to current at a pair of terminals or the ratio of the appropriate components of the electric to magnetic fields at a point.” In this section we are primarily interested in the input impedance at a pair of terminals which are the input terminals of the antenna. In Figure 2.27(a) these terminals I¢ g Generator (Zg) RL XA Vg Rg Rr Gg Bg BA Gr GL Xg Figure 2.27 Transmitting antenna and its equivalent circuits.
76 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS are designated as a −b. The ratio of the voltage to current at these terminals, with no load attached, defines the impedance of the antenna as ZA = RA + jXA (2-72) where ZA = antenna impedance at terminals a–b (ohms) RA = antenna resistance at terminals a–b (ohms) XA = antenna reactance at terminals a–b (ohms) In general the resistive part of (2-72) consists of two components; that is RA = Rr + RL (2-73) where Rr = radiation resistance of the antenna RL = loss resistance of the antenna The radiation resistance will be considered in more detail in later chapters, and it will be illustrated with examples.
If we assume that the antenna is attached to a generator with internal impedance Zg = Rg + jXg (2-74) where Rg = resistance of generator impedance (ohms) Xg = reactance of generator impedance (ohms) and the antenna is used in the transmitting mode, we can represent the antenna and generator by an equivalent circuit∗shown in Figure 2.27(b). To find the amount of power delivered to Rr for radiation and the amount dissipated in RL as heat (I2RL∕2), we first find the current developed within the loop which is given by Ig = Vg Zt = Vg ZA + Zg = Vg (Rr + RL + Rg) + j(XA + Xg) (A) (2-75) and its magnitude by |Ig| = |Vg| [(Rr + RL + Rg)2 + (XA + Xg)2]1∕2 (2-75a) where Vg is the peak generator voltage. The power delivered to the antenna for radiation is given by Pr = 1 2|Ig|2Rr = |Vg|2 2 [ Rr (Rr + RL + Rg)2 + (XA + Xg)2 ] (W) (2-76) ∗This circuit can be used to represent small and simple antennas. It cannot be used for antennas with lossy dielectric or antennas over lossy ground because their loss resistance cannot be represented in series with the radiation resistance.
INPUT IMPEDANCE 77 and that dissipated as heat by PL = 1 2|Ig|2RL = |Vg|2 2 [ RL (Rr + RL + Rg)2 + (XA + Xg)2 ] (W) (2-77) The remaining power is dissipated as heat on the internal resistance Rg of the generator, and it is given by Pg = |Vg|2 2 [ Rg (Rr + RL + Rg)2 + (XA + Xg)2 ] (W) (2-78) The maximum power delivered to the antenna occurs when we have conjugate matching; that is when Rr + RL = Rg (2-79) XA = −Xg (2-80) For this case Pr = |Vg|2 2 [ Rr 4(Rr + RL)2 ] = |Vg|2 8 [ Rr (Rr + RL)2 ] (2-81) PL = |Vg|2 8 [ RL (Rr + RL)2 ] (2-82) Pg = |Vg|2 8 [ Rg (Rr + RL)2 ] = |Vg|2 8 [ 1 Rr + RL ] = |Vg|2 8Rg (2-83) From (2-81)–(2-83), it is clear that Pg = Pr + PL = |Vg|2 8 [ Rg (Rr + RL)2 ] = |Vg|2 8 [ Rr + RL (Rr + RL)2 ] (2-84) The power supplied by the generator during conjugate matching is Ps = 1 2VgI∗ g = 1 2Vg [ V∗ g 2(Rr + RL) ] = |Vg|2 4 [ 1 Rr + RL ] (W) (2-85) Of the power that is provided by the generator, half is dissipated as heat in the internal resis-tance (Rg) of the generator and the other half is delivered to the antenna. This only happens when we have conjugate matching. Of the power that is delivered to the antenna, part is radi-ated through the mechanism provided by the radiation resistance and the other is dissipated as heat which influences part of the overall efficiency of the antenna. If the antenna is lossless and matched to the transmission line (eo = 1), then half of the total power supplied by the gener-ator is radiated by the antenna during conjugate matching, and the other half is dissipated as heat in the generator. Thus, to radiate half of the available power through Rr you must dis-sipate the other half as heat in the generator through Rg. These twopowers are, respectively, 78 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS IT¢ Load (ZT) RL XA VT Rr RT XT GT Gr GL BA BT Figure 2.28 Antenna and its equivalent circuits in the receiving mode.
analogous to the power transferred to the load and the power scattered by the antenna in the receiving mode. In Figure 2.27 it is assumed that the generator is directly connected to the antenna. If there is a transmission line between the two, which is usually the case, then Zg represents the equivalent impedance of the generator transferred to the input terminals of the antenna using the impedance transfer equation. If, in addition, the transmission line is lossy, then the available power to be radi-ated by the antenna will be reduced by the losses of the transmission line. Figure 2.27(c) illustrates the Norton equivalent of the antenna and its source in the transmitting mode.
The use of the antenna in the receiving mode is shown in Figure 2.28(a). The incident wave impinges upon the antenna, and it induces a voltage VT which is analogous to Vg of the transmitting mode. The Thevenin equivalent circuit of the antenna and its load is shown in Figure 2.28(b) and the Norton equivalent in Figure 2.28(c). The discussion for the antenna and its load in the receiving mode parallels that for the transmitting mode, and it will not be repeated here in detail. Some of the results will be summarized in order to discuss some subtle points. Following a procedure similar to that for the antenna in the transmitting mode, it can be shown using Figure 2.28 that in the receiving mode under conjugate matching (Rr + RL = RT and XA = −XT) the powers delivered to RT, Rr, and ANTENNA RADIATION EFFICIENCY 79 RL are given, respectively, by PT = |VT|2 8 [ RT (Rr + RL)2 ] = |VT|2 8 ( 1 Rr + RL ) = |VT|2 8RT (2-86) Pr = |VT|2 2 [ Rr 4(Rr + RL)2 ] = |VT|2 8 [ Rr (Rr + RL)2 ] (2-87) PL = |VT|2 8 [ RL (Rr + RL)2 ] (2-88) while the induced (collected or captured) is Pc = 1 2VTI∗ T = 1 2VT [ V∗ T 2(Rr + RL) ] = |VT|2 4 ( 1 Rr + RL ) (2-89) These are analogous, respectively, to (2-81)–(2-83) and (2-85). The power Pr of (2-87) delivered to Rr is referred to as scattered (or reradiated) power. It is clear through (2-86)–(2-89) that under conjugate matching of the total power collected or captured [Pc of (2-89)] half is delivered to the load RT [PT of (2-86)] and the other half is scattered or reradiated through Rr [Pr of (2-87)] and dissipated as heat through RL [PL of (2-88)]. If the losses are zero (RL = 0), then half of the captured power is delivered to the load and the other half is scattered. This indicates that in order to deliver half of the power to the load you must scatter the other half. This becomes important when discussing effective equivalent areas and aperture efficiencies, especially for high directivity aperture antennas such as waveguides, horns, and reflectors with aperture efficiencies as high as 80 to 90%. Aperture efficiency (𝜀ap) is defined by (2-100) and is the ratio of the maximum effective area to the physical area. The effective area is used to determine the power delivered to the load, which under conjugate matching is only one-half of that intercepted; the other half is scattered and dissipated as heat. For a lossless antenna (RL = 0) under conjugate matching, the maximum value of the effective area is equal to the physical area (𝜀ap = 1) and the scattering area is also equal to the physical area. Thus half of the power is delivered to the load and the other half is scattered. Using (2-86) to (2-89) we conclude that even though the aperture efficiencies are higher than 50% (they can be as large as 100%) all of the power that is captured by the antenna is not delivered to the load but it includes that which is scattered plus dissipated as heat by the antenna. The most that can be delivered to the load is only half of that captured and that is only under conjugate matching and lossless transmission line.
The input impedance of an antenna is generally a function of frequency. Thus the antenna will be matched to the interconnecting transmission line and other associated equipment only within a bandwidth. In addition, the input impedance of the antenna depends on many factors including its geometry, its method of excitation, and its proximity to surrounding objects. Because of their complex geometries, only a limited number of practical antennas have been investigated analytically.
For many others, the input impedance has been determined experimentally.
2.14 ANTENNA RADIATION EFFICIENCY The antenna efficiency that takes into account the reflection, conduction, and dielectric losses was discussed in Section 2.8. The conduction and dielectric losses of an antenna are very difficult to compute and in most cases they are measured. Even with measurements, they are difficult to separate and they are usually lumped together to form the ecd efficiency. The resistance RL is used to represent the conduction-dielectric losses.
80 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS The conduction-dielectric efficiency ecd is defined as the ratio of the power delivered to the radi-ation resistance Rr to the power delivered to Rr and RL. Using (2-76) and (2-77), the radiation efficiency can be written as ecd = [ Rr RL + Rr ] (dimensionless) (2-90) For a metal rod of length l and uniform cross-sectional area A, the dc resistance is given by Rdc = 1 𝜎 l A (ohms) (2-90a) If the skin depth 𝛿[𝛿= √ 2∕(𝜔𝜇0𝜎)] of the metal is very small compared to the smallest diagonal of the cross section of the rod, the current is confined to a thin layer near the conductor surface.
Therefore the high-frequency resistance can be written, based on a uniform current distribution, as Rhf = l PRs = l P √𝜔𝜇0 2𝜎 (ohms) (2-90b) where P is the perimeter of the cross section of the rod (P = C = 2𝜋b for a circular wire of radius b), Rs is the conductor surface resistance, 𝜔is the angular frequency, 𝜇0 is the permeability of free-space, and 𝜎is the conductivity of the metal.
Example 2.13 A resonant half-wavelength dipole is made out of copper (𝜎= 5.7 × 107S/m) wire. Determine the conduction-dielectric (radiation) efficiency of the dipole antenna at f = 100 MHz if the radius of the wire b is 3 × 10−4λ, and the radiation resistance of the λ∕2 dipole is 73 ohms.
Solution: At f = 108 Hz λ = v f = 3 × 108 108 = 3 m l = λ 2 = 3 2m C = 2𝜋b = 2𝜋(3 × 10−4)λ = 6𝜋× 10−4λ For a λ∕2 dipole with a sinusoidal current distribution RL = 1 2Rhf where Rhf is given by (2-90b).
See Problem 2.52. Therefore, RL = 1 2Rhf = 0.25 6𝜋× 10−4 √ 𝜋(108)(4𝜋× 10−7) 5.7 × 107 = 0.349 ohms Thus, ecd(dimensionless) = 73 73 + 0.349 = 0.9952 = 99.52% ecd(dB) = 10 log10(0.9905) = −0.02 ANTENNA VECTOR EFFECTIVE LENGTH AND EQUIVALENT AREAS 81 2.15 ANTENNA VECTOR EFFECTIVE LENGTH AND EQUIVALENT AREAS An antenna in the receiving mode, whether it is in the form of a wire, horn, aperture, array, dielectric rod, etc., is used to capture (collect) electromagnetic waves and to extract power from them, as shown in Figures 2.29(a) and (b). For each antenna, an equivalent length and a number of equivalent areas can then be defined.
These equivalent quantities are used to describe the receiving characteristics of an antenna, whether it be a linear or an aperture type, when a wave is incident upon the antenna.
2.15.1 Vector Effective Length The effective length of an antenna, whether it be a linear or an aperture antenna, is a quantity that is used to determine the voltage induced on the open-circuit terminals of the antenna when a wave impinges upon it. The vector effective length 𝓵e for an antenna is usually a complex vector quantity ZT Direction of propagation (a) Dipole antenna in receiving mode (b) Aperture antenna in receiving mode E-field of plane wave E-field of plane wave A l/2 l/2 ψ ψ Direction of propagation Receiver Figure 2.29 Uniform plane wave incident upon dipole and aperture antennas.
82 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS represented by 𝓵e(𝜃, 𝜙) = ̂ a𝜃l𝜃(𝜃, 𝜙) + ̂ a𝜙l𝜙(𝜃, 𝜙) (2-91) It should be noted that it is also referred to as the effective height. It is a far-field quantity and it is related to the far-zone field Ea radiated by the antenna, with current Iin in its terminals, by – Ea = ̂ a𝜃E𝜃+ ̂ a𝜙E𝜙= −j𝜂kIin 4𝜋r𝓵ee−jkr (2-92) The effective length represents the antenna in its transmitting and receiving modes, and it is par-ticularly useful in relating the open-circuit voltage Voc of receiving antennas. This relation can be expressed as Voc = Ei ⋅𝓵e (2-93) where Voc = open-circuit voltage at antenna terminals Ei = incident electric field 𝓵e = vector effective length In (2-93) Voc can be thought of as the voltage induced in a linear antenna of length 𝓁e when 𝓵e and Ei are linearly polarized , . From the relation of (2-93) the effective length of a linearly polarized antenna receiving a plane wave in a given direction is defined as “the ratio of the magnitude of the open-circuit voltage developed at the terminals of the antenna to the magnitude of the electric-field strength in the direction of the antenna polarization. Alternatively, the effective length is the length of a thin straight conductor oriented perpendicular to the given direction and parallel to the antenna polarization, having a uniform current equal to that at the antenna terminals and producing the same far-field strength as the antenna in that direction.” In addition, as shown in Section 2.12.2, the antenna vector effective length is used to determine the polarization efficiency of the antenna. To illustrate the usefulness of the vector effective length, let us consider an example.
Example 2.14 The far-zone field radiated by a small dipole of length l < λ∕10 and with a triangular current distribution, as shown in Figure 4.4, is derived in Section 4.3 of Chapter 4 and it is given by (4-36a), or Ea = ̂ a𝜃j𝜂kIinle−jkr 8𝜋r sin 𝜃 Determine the vector effective length of the antenna.
Solution: According to (2-92), the vector effective length is 𝓵e = −̂ a𝜃 l 2 sin 𝜃 This indicates, as it should, that the effective length is a function of the direction angle 𝜃, and its maximum occurs when 𝜃= 90◦. This tells us that the maximum open-circuit voltage at the dipole ANTENNA VECTOR EFFECTIVE LENGTH AND EQUIVALENT AREAS 83 terminals occurs when the incident direction of the wave of Figure 2.29(a) impinging upon the small dipole antenna is normal to the axis (length) of the dipole (𝜃= 90◦). This is expected since the dipole has a radiation pattern whose maximum is in the 𝜃= 90◦. In addition, the effective length of the dipole to produce the same output open-circuit voltage is only half (50%) of its physical length if it were replaced by a thin conductor having a uniform current distribution (it can be shown that the maximum effective length of an element with an ideal uniform current distribution is equal to its physical length).
2.15.2 Antenna Equivalent Areas With each antenna, we can associate a number of equivalent areas. These are used to describe the power capturing characteristics of the antenna when a wave impinges on it. One of these equivalent areas is the effective area (aperture), which in a given direction is defined as “the ratio of the available power at the terminals of a receiving antenna to the power flux density of a plane wave incident on the antenna from that direction, the wave being polarization-matched to the antenna. If the direction is not specified, the direction of maximum radiation intensity is implied.” In equation form it is written as Ae = PT Wi = |IT|2RT∕2 Wi (2-94) where Ae = effective area (effective aperture) (m2) PT = power delivered to the load (W) Wi = power density of incident wave (W/m2) The effective aperture is the area which when multiplied by the incident power density gives the power delivered to the load. Using the equivalent of Figure 2.28, we can write (2-94) as Ae = |VT|2 2Wi [ RT (Rr + RL + RT)2 + (XA + XT)2 ] (2-95) Under conditions of maximum power transfer (conjugate matching), Rr + RL = RT and XA = −XT, the effective area of (2-95) reduces to the maximum effective aperture given by Aem = |VT|2 8Wi [ RT (RL + Rr)2 ] = |VT|2 8Wi [ 1 Rr + RL ] (2-96) When (2-96) is multiplied by the incident power density, it leads to the maximum power delivered to the load of (2-86).
All of the power that is intercepted, collected, or captured by an antenna is not delivered to the load, as we have seen using the equivalent circuit of Figure 2.28. In fact, under conjugate matching only half of the captured power is delivered to the load; the other half is scattered and dissipated as heat. Therefore to account for the scattered and dissipated power we need to define, in addition to the effective area, the scattering, loss and capture equivalent areas. In equation form these can be defined similarly to (2-94)–(2-96) for the effective area.
84 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS The scattering area is defined as the equivalent area when multiplied by the incident power density is equal to the scattered or reradiated power. Under conjugate matching this is written, similar to (2-96), as As = |VT|2 8Wi [ Rr (RL + Rr)2 ] (2-97) which when multiplied by the incident power density gives the scattering power of (2-87).
The loss area is defined as the equivalent area, which when multiplied by the incident power density leads to the power dissipated as heat through RL. Under conjugate matching this is written, similar to (2-96), as AL = |VT|2 8Wi [ RL (RL + Rr)2 ] (2-98) which when multiplied by the incident power density gives the dissipated power of (2-88).
Finally the capture area is defined as the equivalent area, which when multiplied by the incident power density leads to the total power captured, collected, or intercepted by the antenna. Under conjugate matching this is written, similar to (2-96), as Ac = |VT|2 8Wi [RT + Rr + RL (RL + Rr)2 ] (2-99) When (2-99) is multiplied by the incident power density, it leads to the captured power of (2-89). In general, the total capture area is equal to the sum of the other three, or Capture Area = Effective Area + Scattering Area + Loss Area This is apparent under conjugate matching using (2-96)–(2-99). However, it holds even under non-conjugate matching conditions.
Now that the equivalent areas have been defined, let us introduce the aperture efficiency 𝜀ap of an antenna, which is defined as the ratio of the maximum effective area Aem of the antenna to its physical area Ap, or 𝜀ap = Aem Ap = maximum effective area physical area (2-100) For aperture type antennas, such as waveguides, horns, and reflectors, the maximum effective area cannot exceed the physical area but it can equal it (Aem ≤Ap or 0 ≤𝜀ap ≤1). Therefore the maxi-mum value of the aperture efficiency cannot exceed unity (100%). For a lossless antenna (RL = 0) the maximum value of the scattering area is also equal to the physical area. Therefore even though the aperture efficiency is greater than 50%, for a lossless antenna under conjugate matching only half of the captured power is delivered to the load and the other half is scattered.
We can also introduce a partial effective area of an antenna for a given polarization in a given direction, which is defined as “the ratio of the available power at the terminals of a receiving antenna to the power flux density of a plane wave incident on the antenna from that direction and with a specified polarization differing from the receiving polarization of the antenna.” The effective area of an antenna is not necessarily the same as the physical aperture. It will be shown in later chapters that aperture antennas with uniform amplitude and phase field distributions have maximum effective areas equal to the physical areas; they are smaller for nonuniform field ANTENNA VECTOR EFFECTIVE LENGTH AND EQUIVALENT AREAS 85 distributions. In addition, the maximum effective area of wire antennas is greater than the physical area (if taken as the area of a cross section of the wire when split lengthwise along its diameter).
Thus the wire antenna can capture much more power than is intercepted by its physical size! This should not come as a surprise. If the wire antenna would only capture the power incident on its physical size, it would be almost useless. So electrically, the wire antenna looks much bigger than its physical stature.
To illustrate the concept of effective area, especially as applied to a wire antenna, let us consider an example. In later chapters, we will consider examples of aperture antennas.
Example 2.15 A uniform plane wave is incident upon a very short lossless dipole (l ≪λ), as shown in Fig-ure 2.29(a). Find the maximum effective area assuming that the radiation resistance of the dipole is Rr = 80(𝜋l∕λ)2, and the incident field is linearly polarized along the axis of the dipole.
Solution: For RL = 0, the maximum effective area of (2-96) reduces to Aem = |VT|2 8Wi [ 1 Rr ] Since the dipole is very short, the induced current can be assumed to be constant and of uniform phase. The induced voltage is VT = El where VT = induced voltage on the dipole E = electric field of incident wave l = length of dipole For a uniform plane wave, the incident power density can be written as Wi = E2 2𝜂 where 𝜂is the intrinsic impedance of the medium (≃120𝜋ohms for a free-space medium). Thus Aem = (El)2 8(E2∕2𝜂)(80𝜋2l2∕λ2) = 3λ2 8𝜋= 0.119λ2 The above value is only valid for a lossless antenna (the losses of a short dipole are usually signif-icant). If the loss resistance is equal to the radiation resistance (RL = Rr) and the sum of the two is equal to the load (receiver) resistance (RT = Rr + RL = 2Rr), then the effective area is only one-half of the maximum effective area given above.
Let us now examine the significance of the effective area. From Example 2.15, the maximum effective area of a short dipole with l ≪λ was equal to Aem = 0.119λ2. Typical antennas that fall under this category are dipoles whose lengths are l ≤λ∕50. For demonstration, let us assume that l = λ∕50. Because Aem = 0.119λ2 = lwe = (λ∕50)we, the maximum effective electrical width of this dipole is we = 5.95λ. Typical physical diameters (widths) of wires used for dipoles may be 86 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS R Transmitter Receiver Atm, Dt Arm, Dr #1 #2 Direction of propagation of wave Figure 2.30 Two antennas separated by a distance R.
about wp = λ∕300. Thus the maximum effective width we is about 1,785 times larger than its phys-ical width.
2.16 MAXIMUM DIRECTIVITY AND MAXIMUM EFFECTIVE AREA To derive the relationship between directivity and maximum effective area, the geometrical arrange-ment of Figure 2.30 is chosen. Antenna 1 is used as a transmitter and 2 as a receiver. The effective areas and directivities of each are designated as At, Ar and Dt, Dr. If antenna 1 were isotropic, its radiated power density at a distance R would be W0 = Pt 4𝜋R2 (2-101) where Pt is the total radiated power. Because of the directive properties of the antenna, its actual density is Wt = W0Dt = PtDt 4𝜋R2 (2-102) The power collected (received) by the antenna and transferred to the load would be Pr = WtAr = PtDtAr 4𝜋R2 (2-103) or DtAr = Pr Pt (4𝜋R2) (2-103a) If antenna 2 is used as a transmitter, 1 as a receiver, and the intervening medium is linear, passive, and isotropic, we can write that DrAt = Pr Pt (4𝜋R2) (2-104) MAXIMUM DIRECTIVITY AND MAXIMUM EFFECTIVE AREA 87 Equating (2-103a) and (2-104) reduces to Dt At = Dr Ar (2-105) Increasing the directivity of an antenna increases its effective area in direct proportion. Thus, (2-105) can be written as D0t Atm = D0r Arm (2-106) where Atm and Arm (D0t and D0r) are the maximum effective areas (directivities) of antennas 1 and 2, respectively.
If antenna 1 is isotropic, then D0t = 1 and its maximum effective area can be expressed as Atm = Arm D0r (2-107) Equation (2-107) states that the maximum effective area of an isotropic source is equal to the ratio of the maximum effective area to the maximum directivity of any other source. For example, let the other antenna be a very short (l ≪λ) dipole whose effective area (0.119λ2 from Example 2.15) and maximum directivity (1.5) are known.
The maximum effective area of the isotropic source is then equal to Atm = Arm D0r = 0.119λ2 1.5 = λ2 4𝜋 (2-108) Using (2-108), we can write (2-107) as Arm = D0rAtm = D0r ( λ2 4𝜋 ) (2-109) In general then, the maximum effective aperture (Aem) of any antenna is related to its maximum directivity (D0) by Aem = λ2 4𝜋D0 (2-110) Thus, when (2-110) is multiplied by the power density of the incident wave it leads to the maximum power that can be delivered to the load. This assumes that there are no conduction-dielectric losses (radiation efficiency ecd is unity), the antenna is matched to the load (reflection efficiency er is unity), and the polarization of the impinging wave matches that of the antenna (polarization loss factor PLF and polarization efficiency pe are unity). If there are losses associated with an antenna, its maximum effective aperture of (2-110) must be modified to account for conduction-dielectric losses (radiation efficiency). Thus, Aem = ecd ( λ2 4𝜋 ) D0 (2-111) The maximum value of (2-111) assumes that the antenna is matched to the load and the incoming wave is polarization-matched to the antenna. If reflection and polarization losses are also included, 88 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS then the maximum effective area of (2-111) is represented by Aem = e0 ( λ2 4𝜋 ) D0| ̂ ρw ⋅̂ ρa|2 = ecd(1 −|Γ|2) ( λ2 4𝜋 ) D0| ̂ ρw ⋅̂ ρa|2 (2-112) 2.17 FRIIS TRANSMISSION EQUATION AND RADAR RANGE EQUATION The analysis and design of radar and communications systems often require the use of the Friis Transmission Equation and the Radar Range Equation. Because of the importance of the two equations, a few pages will be devoted for their derivation.
2.17.1 Friis Transmission Equation The Friis Transmission Equation relates the power received to the power transmitted between two antennas separated by a distance R > 2D2∕λ, where D is the largest dimension of either antenna.
Referring to Figure 2.31, let us assume that the transmitting antenna is initially isotropic. If the input power at the terminals of the transmitting antenna is Pt, then its isotropic power density W0 at distance R from the antenna is W0 = et Pt 4𝜋R2 (2-113) where et is the radiation efficiency of the transmitting antenna. For a nonisotropic transmitting antenna, the power density of (2-113) in the direction 𝜃t, 𝜙t can be written as Wt = PtGt(𝜃t, 𝜙t) 4𝜋R2 = et PtDt(𝜃t, 𝜙t) 4𝜋R2 (2-114) where Gt(𝜃t, 𝜙t) is the gain and Dt(𝜃t, 𝜙t) is the directivity of the transmitting antenna in the direction 𝜃t, 𝜙t. Since the effective area Ar of the receiving antenna is related to its efficiency er and directivity Dr by Ar = erDr(𝜃r, 𝜙r) ( λ2 4𝜋 ) (2-115) (θr, ϕr) (θt, ϕt) Transmitting antenna (Pt, Gt, Dt, ecdt, Γt, t) ^ Receiving antenna (P r, Gr, Dr, ecdr, Γr, r) ^ R Figure 2.31 Geometrical orientation of transmitting and receiving antennas for Friis transmission equation.
FRIIS TRANSMISSION EQUATION AND RADAR RANGE EQUATION 89 the amount of power Pr collected by the receiving antenna can be written, using (2-114) and (2-115), as Pr = erDr(𝜃r, 𝜙r) λ2 4𝜋Wt = eter λ2Dt(𝜃t, 𝜙t)Dr(𝜃r, 𝜙r)Pt (4𝜋R)2 | ̂ ρt ⋅̂ ρr|2 (2-116) or the ratio of the received to the input power as Pr Pt = eter λ2Dt(𝜃t, 𝜙t)Dr(𝜃r, 𝜙r) (4𝜋R)2 (2-117) The power received based on (2-117) assumes that the transmitting and receiving antennas are matched to their respective lines or loads (reflection efficiencies are unity) and the polarization of the receiving antenna is polarization-matched to the impinging wave (polarization loss factor and polarization efficiency are unity). If these two factors are also included, then the ratio of the received to the input power of (2-117) is represented by Pr Pt = ecdtecdr(1 −|Γt|2)(1 −|Γr|2) ( λ 4𝜋R ) 2 Dt(𝜃t, 𝜙t)Dr(𝜃r, 𝜙r)| ̂ ρt ⋅̂ ρr|2 (2-118) Example 2.16 Two lossless X-band (8.2–12.4 GHz) horn antennas are separated by a distance of 100λ. The reflection coefficients at the terminals of the transmitting and receiving antennas are 0.1 and 0.2, respectively. The maximum directivities of the transmitting and receiving antennas (over isotropic) are 16 dB and 20 dB, respectively. Assuming that the input power in the lossless transmission line connected to the transmitting antenna is 2 W, and the antennas are aligned for maximum radiation between them and are polarization-matched, find the power delivered to the load of the receiver.
Solution: For this problem ecdt = ecdr = 1 because the antennas are lossless.
| ̂ ρt ⋅̂ ρr|2 = 1 because the antennas are polarization-matched Dt = D0t} because the antennas are aligned for Dr = D0r maximum radiation between them D0t = 16 dB ➱39.81 (dimensionless) D0r = 20 dB ➱100 (dimensionless) Using (2-118), we can write Pr = [1 −(0.1)2][1 −(0.2)2][λ∕4𝜋(100λ)]2(39.81)(100)(2) = 4.777 mW 90 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS For reflection and polarization-matched antennas aligned for maximum directional radiation and reception, (2-118) reduces to Pr Pt = ( λ 4𝜋R )2 G0tG0r (2-119) Equations (2-117), (2-118), or (2-119) are known as the Friis Transmission Equation, and it relates the power Pr (delivered to the receiver load) to the input power of the transmitting antenna Pt. The term (λ∕4𝜋R)2 is called the free-space loss factor, and it takes into account the losses due to the spherical spreading of the energy by the antenna.
2.17.2 Radar Range Equation Now let us assume that the transmitted power is incident upon a target, as shown in Figure 2.32.
We now introduce a quantity known as the radar cross section or echo area (𝜎) of a target which is defined as the area intercepting that amount of power which, when scattered isotropically, produces at the receiver a density which is equal to that scattered by the actual target . In equation form lim R→∞ [ 𝜎Wi 4𝜋R2 ] = Ws (2-120) or 𝜎= lim R→∞ [ 4𝜋R2 Ws Wi ] = lim R→∞ [ 4𝜋R2 |Es|2 |Ei|2 ] = lim R→∞ [ 4𝜋R2 |Hs|2 |Hi|2 ] (2-120a) where 𝜎= radar cross section or echo area (m2) R = observation distance from target (m) Wi = incident power density (W/m2) Ws = scattered power density (W/m2) Ei (Es) = incident (scattered) electric field (V/m) Hi (Hs) = incident (scattered) magnetic field (A/m) Any of the definitions in (2-120a) can be used to derive the radar cross section of any antenna or target. For some polarization one of the definitions based either on the power density, electric field, or magnetic field may simplify the derivation, although all should give the same answers .
Using the definition of radar cross section, we can consider that the transmitted power incident upon the target is initially captured and then it is reradiated isotropically, insofar as the receiver is concerned. The amount of captured power Pc is obtained by multiplying the incident power density of (2-114) by the radar cross section 𝜎, or Pc = 𝜎Wt = 𝜎PtGt(𝜃t, 𝜙t) 4𝜋R2 1 = et𝜎PtDt(𝜃t, 𝜙t) 4𝜋R2 1 (2-121) FRIIS TRANSMISSION EQUATION AND RADAR RANGE EQUATION 91 Incident wave Transmitting antenna (Pt, Gt, Dt, ecdt, Γt, t) Scattered wave ( w) Targetσ Receiving antenna (P r, Gr, Dr, ecdr, Γr, r) R1 R2 ^ ^ ^ Figure 2.32 Geometrical arrangement of transmitter, target, and receiver for radar range equation.
The power captured by the target is reradiated isotropically, and the scattered power density can be written as Ws = Pc 4𝜋R2 2 = ecdt𝜎PtDt(𝜃t, 𝜙t) (4𝜋R1R2)2 (2-122) The amount of power delivered to the receiver load is given by Pr = ArWs = ecdtecdr𝜎PtDt(𝜃t, 𝜙t)Dr(𝜃r, 𝜙r) 4𝜋 ( λ 4𝜋R1R2 )2 (2-123) where Ar is the effective area of the receiving antenna as defined by (2-115).
Equation (2-123) can be written as the ratio of the received power to the input power, or Pr Pt = ecdtecdr𝜎Dt(𝜃t, 𝜙t)Dr(𝜃r, 𝜙r) 4𝜋 ( λ 4𝜋R1R2 )2 (2-124) Expression (2-124) is used to relate the received power to the input power, and it takes into account only conduction-dielectric losses (radiation efficiency) of the transmitting and receiving antennas. It does not include reflection losses (reflection efficiency) and polarization losses (polarization loss fac-tor or polarization efficiency). If these two losses are also included, then (2-124) must be expressed as Pr Pt = ecdtecdr(1 −|Γt|2)(1 −|Γr|2)𝜎Dt(𝜃t, 𝜙t)Dr(𝜃r, 𝜙r) 4𝜋 × ( λ 4𝜋R1R2 )2 | ̂ ρw ⋅̂ ρr|2 (2-125) where ̂ ρw = polarization unit vector of the scattered waves ̂ ρr = polarization unit vector of the receiving antenna 92 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS For polarization-matched antennas aligned for maximum directional radiation and reception, (2-125) reduces to Pr Pt = 𝜎G0tG0r 4𝜋 [ λ 4𝜋R1R2 ]2 (2-126) Equation (2-124), or (2-125) or (2-126) is known as the Radar Range Equation. It relates the power Pr (delivered to the receiver load) to the input power Pt transmitted by an antenna, after it has been scattered by a target with a radar cross section (echo area) of 𝜎.
2.17.3 Antenna Radar Cross Section The radar cross section, usually referred to as RCS, is a far-field parameter, which is used to char-acterize the scattering properties of a radar target. For a target, there is monostatic or backscatter-ing RCS when the transmitter and receiver of Figure 2.32 are at the same location, and a bistatic RCS when the transmitter and receiver are not at the same location. In designing low-observable or low-profile (stealth) targets, it is the parameter that you attempt to minimize. For complex tar-gets (such as aircraft, spacecraft, missiles, ships, tanks, automobiles) it is a complex parameter to derive. In general, the RCS of a target is a function of the polarization of the incident wave, the angle of incidence, the angle of observation, the geometry of the target, the electrical properties of the target, and the frequency of operation. The units of RCS of three-dimensional targets are meters squared (m2) or for normalized values decibels per squared meter (dBsm) or RCS per squared wavelength in decibels (RCS∕λ2 in dB). Representative values of some typical targets are shown in Table 2.2 . Although the frequency was not stated , these numbers could be representative at X-band.
The RCS of a target can be controlled using primarily two basic methods: shaping and the use of materials. Shaping is used to attempt to direct the scattered energy toward directions other than the desired. However, for many targets shaping has to be compromised in order to meet other requirements, such as aerodynamic specifications for flying targets. Materials is used to trap the incident energy within the target and to dissipate part of the energy as heat or to direct it toward directions other than the desired. Usually both methods, shaping and materials, are used together in order to optimize the performance of a radar target. One of the “golden rules” to observe in order to TABLE 2.2 RCS of Some Typical Targets Typical RCSs Object RCS (m2) RCS (dBsm) Pickup truck 200 23 Automobile 100 20 Jumbo jet airliner 100 20 Large bomber or commercial jet 40 16 Cabin cruiser boat 10 10 Large fighter aircraft 6 7.78 Small fighter aircraft or four-passenger jet 2 3 Adult male 1 0 Conventional winged missile 0.5 −3 Bird 0.01 −20 Insect 0.00001 −50 Advanced tactical fighter 0.000001 −60 FRIIS TRANSMISSION EQUATION AND RADAR RANGE EQUATION 93 achieve low RCS is to “round corners, avoid flat and concave surfaces, and use material treatment in flare spots.” There are many methods of analysis to predict the RCS of a target , –. Some of them are full-wave methods, others are designated as asymptotic methods, either low-frequency or high-frequency, and some are considered as numerical methods. The methods of analysis are often contingent upon the shape, size, and material composition of the target. Some targets, because of their geometrical complexity, are often simplified and are decomposed into a number of basic shapes (such as strips, plates, cylinders, cones, wedges) which when put together represent a very good replica of the actual target. This has been used extensively and proved to be a very good approach. The topic is very extensive to be treated here in any detail, and the reader is referred to the literature , –. There is a plethora of references but because of space limitations, only a limited number is included here to get the reader started on the subject.
Antennas individually are radar targets which many exhibit large radar cross section. In many applications, antennas are mounted on the surface of other complex targets (such as aircraft, space-craft, satellites, missiles, automobiles), and become part of the overall radar target. In such configu-rations, many antennas, especially aperture types (such as waveguides, horns) become large contrib-utors to the total RCS, monostatic or bistatic, of the target. Therefore, in designing low-observable targets, the antenna type, location and contributions become an important consideration of the over-all design.
The scattering and transmitting (radiation) characteristics of an antenna are related –.
There are various methods which can be used to analyze the fields scattered by an antenna. The summary here parallels that in , –. In general, the electric field scattered by an antenna with a load impedance ZL can be expressed by Es(ZL) = Es(0) −Is It ZL ZL + ZA Et (2-127) where Es(ZL) = electric field scattered by antenna with a load ZL Es(0) = electric field scattered by short-circuited antenna (ZL = 0) Is = short-circuited current induced by the incident field on the antenna with ZL = 0 It = antenna current in transmitting mode ZA = RA + jXA = antenna input impedance Et = electric field radiated by the antenna in transmitting mode Green expressed the field scattered by an antenna terminated with a load ZL in a more convenient form which allows it to be separated into the structural and antenna mode scattering terms , –. This is accomplished by assuming that the antenna is loaded with a conjugate-matched impedance (ZL = Z∗ A).
r The structural scattering term is introduced by the currents induced on the surface of the antenna by the incident field when the antenna is conjugate-matched, and it is independent of the load impedance.
r The antenna mode scattering term is only a function of the radiation characteristics of the antenna, and its scattering pattern is the square of the antenna radiation pattern.
The antenna mode depends on the power absorbed by the load of a lossless antenna and the power that is radiated by the antenna due to a load mismatch. This term vanishes when the antenna is conjugate-matched.
94 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS In general, the field scattered by an antenna loaded with an impedance ZL is related to the field radiated by the antenna in the transmitting mode in three different ways.
r First, the field scattered by an antenna when it is loaded with an impedance ZL is equal to the field scattered by the antenna when it is short-circuited (ZL = 0) minus a term related to the antenna reflection coefficient and the field transmitted by the antenna.
r Second, the field scattered by an antenna when it is terminated with an impedance ZL is equal to the field scattered by the antenna when it is conjugate-matched with an impedance Z∗ A minus the field transmitted (radiated) times the conjugate reflection coefficient.
r Third, the field scattered by the antenna when it is terminated with an impedance ZL is equal to the field scattered by the antenna when it is matched with an impedance ZA minus the field transmitted (radiated) times the reflection coefficient weighted by the ratio of two currents (Im∕It, Im = scattering current when antenna is matched with an impedance ZL, It = antenna current in the transmitting mode).
It can be shown that the total radar cross section of an antenna terminated with a load ZL can be written as 𝜎= | √ 𝜎s −(1 + ΓA) √ 𝜎aej𝜙r|2 (2-128) where 𝜎= total RCS with antenna terminated with ZL 𝜎s = RCS due to structural term 𝜎a = RCS due to antenna mode term 𝜙r = relative phase between the structural and antenna mode terms If the antenna is short-circuited (ΓA = −1), then according to (2-128) 𝜎short = 𝜎s (2-129) If the antenna is open-circuited (ΓA = +1), then according to (2-128) 𝜎open = | √ 𝜎s −2 √ 𝜎aej𝜙r|2 = 𝜎residual (2-130) Lastly, if the antenna is matched ZL = ZA(ΓA = 0), then according to (2-128) 𝜎match = | √ 𝜎s − √ 𝜎aej𝜙r|2 (2-131) Therefore, under matched conditions, according to (2-131), the range of values (minimum to maxi-mum) of the radar cross section is | √ 𝜎s − √ 𝜎a| ≤𝜎≤| √ 𝜎s + √ 𝜎a| (2-132) The minimum value occurs when the two RCSs are in phase while the maximum occurs when they are out of phase.
FRIIS TRANSMISSION EQUATION AND RADAR RANGE EQUATION 95 Example 2.17 The structural RCS of a resonant wire dipole is in phase and its magnitude is slightly greater than four times that of the antenna mode. Relate the short-circuited, open-circuited, and matched RCSs to that of the antenna mode.
Solution: Using (2-129) 𝜎short = 4𝜎antenna Using (2-130) 𝜎open = 2𝜎antenna(0) = 0 or very small The matched value is obtained using (2-131), or 𝜎match = 𝜎antenna To produce a zero RCS, (2-128) must vanish. This is accomplished if Re(ΓA) = −1 + cos 𝜙r √ 𝜎s∕𝜎a (2-133a) Im(ΓA) = −sin 𝜙r √ 𝜎s∕𝜎a (2-133b) Assuming positive values of resistances, the real value of ΓA cannot be greater than unity. There-fore there are some cases where the RCS cannot be reduced to zero by choosing ZL. Because ZA can be complex, there is no limit on the imaginary part of ΓA.
In general, the structural and antenna mode scattering terms are very difficult to predict and usually require that the antenna is solved as a boundary-value problem. However, these two terms have been obtained experimentally utilizing the Smith chart –.
For a monostatic system the receiving and transmitting antennas are collocated. In addition, if the antennas are identical (G0r = G0t = G0) and are polarization-matched (̂ ρr = ̂ ρt = 1), the total radar cross section of the antenna for backscattering can be written as 𝜎= λ2 0 4𝜋G2 0|A −Γ∗|2 (2-134) where A is a complex parameter independent of the load.
If the antenna is a thin dipole, then A ≃1 and (2-134) reduces to 𝜎≃ λ2 0 4𝜋G2 0|1 −Γ∗|2 = λ2 0 4𝜋G2 0 | | | | | 1 − ZL −Z∗ A ZL + ZA | | | | | = λ2 0 4𝜋G2 0 | | | | 2RA ZL + ZA | | | | 2 (2-135) If in addition we assume that the dipole is resonant and its length is l = λ0∕2 and is short-circuited (ZL = 0), then the normalized radar cross section of (2-135) is equal to 𝜎 λ2 0 ≃ G2 0 𝜋= (1.643)2 𝜋 = 0.8593 ≃0.86 (2-136) 96 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS –90 –60 –30 0 30 Incidence angle (degrees) E-plane RCS (dBsm) 60 90 –80 –70 –60 –50 –40 –30 –20 –10 Matched-load Short-circuited Open-circuited Half-wavelength dipole 3.7465 cm long ´ 0.2362 cm diameter Frequency = 4.02 GHz Figure 2.33 E-plane monostatic RCS (𝜎𝜃𝜃) versus incidence angle for a half-wavelength dipole.
which agrees with experimental corresponding maximum monostatic value of Figure 2.33 and those reported in the literature , .
Shown in Figure 2.33 is the measured E-plane monostatic RCS of a half-wavelength dipole when it is matched to a load, short-circuited (straight wire) and open-circuited (gap at the feed). The aspect angle is measured from the normal to the wire. As expected, the RCS is a function of the observation (aspect) angle. Also it is apparent that there are appreciable differences between the three responses.
For the short-circuited case, the maximum value is approximately −24 dBsm which closely agrees with the computed value of −22.5 dBsm using (2-136). Similar responses for the monostatic RCS of a pyramidal horn are shown in Figure 2.34(a) for the E-plane and in Figure 2.34(b) for the H-plane. The antenna is a commercial X-band (8.2-12.4 GHz) 20-dB standard gain horn with aperture dimension of 9.2 cm by 12.4 cm. The length of the horn is 25.6 cm. As for the dipole, there are differences between the three responses for each plane. It is seen that the short-circuited response exhibits the largest return.
Antenna RCS from scale model measurements and microstrip patches , have been reported.
2.18 ANTENNA TEMPERATURE Every object with a physical temperature above absolute zero (0 K = −273◦C) radiates energy .
The amount of energy radiated is usually represented by an equivalent temperature TB, better known as brightness temperature, and it is defined as TB(𝜃, 𝜙) = 𝜀(𝜃, 𝜙)Tm = (1 −|Γ|2)Tm (2-137) ANTENNA TEMPERATURE 97 –45 –30 –15 0 15 Incidence angle (degrees) RCS (dBsm) 30 45 –35 –25 –15 –5 5 15 Matched-load Short-circuited Open-circuited Standard gain pyramidal horn E-plane Frequency = 10 GHz –45 –30 –15 0 15 Incidence angle (degrees) (b) H-plane ( ) (a) E-plane ( ) RCS (dBsm) 30 45 –40 –30 –20 –10 0 10 Matched-load Short-circuited Open-circuited Standard gain pyramidal horn H-plane Frequency = 10 GHz σθθ σϕϕ Figure 2.34 E- and H-plane monostatic RCS versus incidence angle for a pyramidal horn antenna.
98 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS where TB = brightness temperature (equivalent temperature; K) 𝜀= emissivity (dimensionless) Tm = molecular (physical) temperature (K) Γ(𝜃, 𝜙) = reflection coefficient of the surface for the polarization of the wave Since the values of emissivity are 0 ≤𝜀≤1, the maximum value the brightness temperature can achieve is equal to the molecular temperature. Usually the emissivity is a function of the frequency of operation, polarization of the emitted energy, and molecular structure of the object. Some of the better natural emitters of energy at microwave frequencies are (a) the ground with equivalent temperature of about 300 K and (b) the sky with equivalent temperature of about 5 K when looking toward zenith and about 100–150 K toward the horizon.
The brightness temperature emitted by the different sources is intercepted by antennas, and it appears at their terminals as an antenna temperature. The temperature appearing at the terminals of an antenna is that given by (2-137), after it is weighted by the gain pattern of the antenna. In equation form, this can be written as TA = ∫ 2𝜋 0 ∫ 𝜋 0 TB(𝜃, 𝜙)G(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 ∫ 2𝜋 0 ∫ 𝜋 0 G(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-138) where TA = antenna temperature (effective noise temperature of the antenna radiation resistance; K) G(𝜃, 𝜙) = gain (power) pattern of the antenna Assuming no losses or other contributions between the antenna and the receiver, the noise power transferred to the receiver is given by Pr = kTAΔf (2-139) where Pr = antenna noise power (W) k = Boltzmann’s constant (1.38 × 10−23 J/K) TA = antenna temperature (K) Δf = bandwidth (Hz) If the antenna and transmission line are maintained at certain physical temperatures, and the transmission line between the antenna and receiver is lossy, the antenna temperature TA as seen by the receiver through (2-139) must be modified to include the other contributions and the line losses. If the antenna itself is maintained at a certain physical temperature Tp and a transmission line of length l, constant physical temperature T0 throughout its length, and uniform attenuation of 𝛼(Np/unit length) is used to connect an antenna to a receiver, as shown in Figure 2.35, the effective antenna temperature at the receiver terminals is given by Ta = TAe−2𝛼l + TAPe−2𝛼l + T0(1 −e−2𝛼l) (2-140) ANTENNA TEMPERATURE 99 T0 Tr Ta TB TP Ts = Ta + Tr TA + TAP Transmission line Receiver Antenna Emitting source l Figure 2.35 Antenna, transmission line, and receiver arrangement for system noise power calculation.
where TAP = ( 1 eA −1 ) Tp (2-140a) Ta = antenna temperature at the receiver terminals (K) TA = antenna noise temperature at the antenna terminals (2-138) (K) TAP = antenna temperature at the antenna terminals due to physical temperature (2-140a) (K) Tp = antenna physical temperature (K) 𝛼= attenuation coefficient of transmission line (Np/m) eA = thermal efficiency of antenna (dimensionless) l = length of transmission line (m) T0 = physical temperature of the transmission line (K) The antenna noise power of (2-139) must also be modified and written as Pr = kTaΔf (2-141) where Ta is the antenna temperature at the receiver input as given by (2-140).
If the receiver itself has a certain noise temperature Tr (due to thermal noise in the receiver components), the system noise power at the receiver terminals is given by Ps = k(Ta + Tr)Δf = kTsΔf (2-142) where Ps = system noise power (at receiver terminals) Ta = antenna noise temperature (at receiver terminals) Tr = receiver noise temperature (at receiver terminals) Ts = Ta + Tr = effective system noise temperature (at receiver terminals) A graphical relation of all the parameters is shown in Figure 2.35. The effective system noise temperature Ts of radio astronomy antennas and receivers varies from very few degrees (typically ≃ 10 K) to thousands of Kelvins depending upon the type of antenna, receiver, and frequency of opera-tion. Antenna temperature changes at the antenna terminals, due to variations in the target emissions, may be as small as a fraction of one degree. To detect such changes, the receiver must be very sen-sitive and be able to differentiate changes of a fraction of a degree.
100 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS Example 2.18 The effective antenna temperature of a target at the input terminals of the antenna is 150 K.
Assuming that the antenna is maintained at a thermal temperature of 300 K and has a thermal efficiency of 99% and it is connected to a receiver through an X-band (8.2–12.4 GHz) rectangular waveguide of 10 m (loss of waveguide = 0.13 dB/m) and at a temperature of 300 K, find the effective antenna temperature at the receiver terminals.
Solution: We first convert the attenuation coefficient from dB to Np by 𝛼(dB/m) = 20(log10 e)𝛼(Np/m) = 20(0.434)𝛼(Np/m) = 8.68𝛼(Np/m). Thus 𝛼(Np/m) = 𝛼(dB/m)∕8.68 = 0.13∕8.68 = 0.0149. The effective antenna temperature at the receiver terminals can be written, using (2-140a) and (2-140), as TAP = 300 ( 1 0.99 −1 ) = 3.03 Ta = 150e−0.149(2) + 3.03e−0.149(2) + 300[1 −e−0.149(2)] = 111.345 + 2.249 + 77.31 = 190.904 K The results of the above example illustrate that the antenna temperature at the input terminals of the antenna and at the terminals of the receiver can differ by quite a few degrees. For a smaller transmission line or a transmission line with much smaller losses, the difference can be reduced appreciably and can be as small as a fraction of a degree.
A summary of the pertinent parameters and associated formulas and equation numbers for this chapter are listed in Table 2.3.
2.19 MULTIMEDIA In the website created by the publisher for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Java-based applet for computing and displaying graphically the directivity of an antenna.
c. Matlab and Fortran computer program, designated Directivity, for computing the directivity of an antenna. A description of this program is in the READ ME file in the publisher’s website for this book.
d. Matlab plotting computer programs: r 2-D Polar (designated as Polar). This program can be used to plot the two-dimensional patterns, in both polar and semipolar form (in linear or dB scale), of an antenna.
r 3-D Spherical. This program (designated as Spherical) can be used to plot the three-dimensional pattern (in linear or dB scale) of an antenna in spherical form.
r Polarization Diagram Ellipse Animation: Animates the 3-D polarization diagram of a rotating electric field vector [Figure 2.23(a)]. It also animates the 2-D polarization ellipse [Figure 2.23(b)] for linear, circular and elliptical polarized waves, and sense of rotation. It also computes the axial ratio (AR).
r Polarization Propag: Computes the Poincar´ e sphere angles, and thus the polarization wave travelling in an infinite homogeneous medium.
A description of these programs is in the corresponding READ ME files in the publisher’s website for this book.
e. Power Point (PPT) viewgraphs, in multicolor.
MULTIMEDIA 101 TABLE 2.3 Summary of Important Parameters and Associated Formulas and Equation Numbers Equation Parameter Formula Number Infinitesimal area of sphere dA = r2 sin 𝜃d𝜃d𝜙 (2-1) Elemental solid angle of sphere dΩ = sin 𝜃d𝜃d𝜙 (2-2) Average power density Wav = 1 2Re[E × H∗] (2-8) Radiated power/average radiated power Prad = Pav = ∯ S Wav ⋅ds = 1 2 ∯ S Re[E × H∗] ⋅ds (2-9) Radiation density of isotropic radiator W0 = Prad 4𝜋r2 (2-11) Radiation intensity (far field) U = r2Wrad = B0F(𝜃, 𝜙) ≃r2 2𝜂 × [|E𝜃(r, 𝜃, 𝜙)|2 + |E𝜙(r, 𝜃, 𝜙)|2] (2-12), (2-12a) Directivity D(𝜃, 𝜙) D = U U0 = 4𝜋U Prad = 4𝜋 ΩA (2-16), (2-23) Beam solid angle ΩA ΩA = ∫ 2𝜋 0 ∫ 𝜋 0 Fn(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 Fn(𝜃, 𝜙) = F(𝜃, 𝜙) |F(𝜃, 𝜙)|max (2-24) (2-25) Maximum directivity D0 Dmax = D0 = Umax U0 = 4𝜋Umax Prad (2-16a) Partial directivities D𝜃, D𝜙 D0 = D𝜃+ D𝜙 D𝜃= 4𝜋U𝜃 Prad = 4𝜋U𝜃 (Prad)𝜃+ (Prad)𝜙 D𝜙= 4𝜋U𝜙 Prad = 4𝜋U𝜙 (Prad)𝜃+ (Prad)𝜙 (2-17) (2-17a) (2-17b) Approximate maximum directivity (one main lobe pattern) D0 ≃ 4𝜋 Θ1rΘ2r = 41,253 Θ1dΘ2d (Kraus) D0 ≃ 32 ln 2 Θ2 1r + Θ2 2r = 22.181 Θ2 1r + Θ2 2r = 72,815 Θ2 1d + Θ2 2d (Tai-Pereira) (2-26), (2-27) (2-30), (2-30a), (2-30b) Approximate maximum directivity (omnidirectional pattern) D0 ≃ 101 HPBW(degrees) −0.0027[HPBW(degrees)]2 (McDonald) D0 ≃−172.4 + 191 √ 0.818 + 1 HPBW(degrees) (Pozar) (2-33a) (2-33b) (continued overleaf) 102 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS TABLE 2.3 (continued) Equation Parameter Formula Number Gain G(𝜃, 𝜙) G = 4𝜋U(𝜃, 𝜙) Pin = ecd [4𝜋U(𝜃, 𝜙) Prad ] = ecdD(𝜃, 𝜙) Prad = ecdPin (2-46), (2-47) (2-49) Antenna radiation efficiency ecd ecd = Rr Rr + RL (2-90) Loss resistance RL (straight wire/uniform current) RL = Rhf = l P √𝜔𝜇0 2𝜎 (2-90b) Loss resistance RL (straight wire/ λ∕2 dipole) RL = l 2P √𝜔𝜇0 2𝜎 Maximum gain G0 G0 = ecdDmax = ecdD0 (2-49a) Partial gains G𝜃, G𝜙 G0 = G𝜃+ G𝜙 G𝜃= 4𝜋U𝜃 Pin , G𝜙= 4𝜋U𝜙 Pin (2-50) (2-50a), (2-50b) Realized gain Gre Gre = erG(𝜃, 𝜙) = erecdD(𝜃, 𝜙) = (1 −|Γ|2)ecdD(𝜃, 𝜙) = e0D(𝜃, 𝜙) (2-49a) (2-49b) Total antenna efficiency e0 e0 = ereced = erecd = (1 −|Γ|2)ecd (2-52) Reflection efficiency er er = (1 −|Γ|2) (2-45) Beam efficiency BE BE = ∫ 2𝜋 0 ∫ 𝜃1 0 U(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 ∫ 2𝜋 0 ∫ 𝜋 0 U(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-54) Polarization loss factor (PLF) PLF = |̂ ρw ⋅̂ ρa|2 (2-71) Vector effective length 𝓵e(𝜃, 𝜙) 𝓵e(𝜃, 𝜙) = ̂ a𝜃l𝜃(𝜃, 𝜙) + ̂ a𝜙l𝜙(𝜃, 𝜙) (2-91) Polarization efficiency pe pe = |𝓵e ⋅Einc|2 |𝓵e|2|Einc|2 (2-71a) Antenna impedance ZA ZA = RA + jXA = (Rr + RL) + jXA (2-72), (2-73) Maximum effective area Aem Aem = |VT|2 8Wi [ 1 Rr + RL ] = ecd ( λ2 4𝜋 ) D0| ̂ ρw ⋅̂ ρa|2 = ( λ2 4𝜋 ) G0|̂ ρw ⋅̂ ρa|2 (2-96), (2-111), (2-112) Aperture efficiency 𝜀ap 𝜀ap = Aem Ap = maximum effective area physical area (2-100) Friis transmission equation Pr Pt = ( λ 4𝜋R )2 G0tG0r| ̂ ρt ⋅̂ ρr|2 (2-118), (2-119) Radar range equation Pr Pt = 𝜎G0tG0r 4𝜋 [ λ 4𝜋R1R2 ]2 |̂ ρw ⋅̂ ρr|2 (2-125), (2-126) REFERENCES 103 TABLE 2.3 (continued) Equation Parameter Formula Number Radar cross section (RCS) (m2) 𝜎= lim R→∞ [ 4𝜋R2 Ws Wi ] = lim R→∞ [ 4𝜋R2 |Es|2 |Ei|2 ] = lim R→∞ [ 4𝜋R2 |Hs|2 |Hi|2 ] (2-120a) Brightness temperature TB(𝜃, 𝜙) (K) TB(𝜃, 𝜙) = 𝜀(𝜃, 𝜙)Tm = (1 −|Γ|2)Tm (2-144) Antenna temperature TA (K) TA = ∫ 2𝜋 0 ∫ 𝜋 0 TB(𝜃, 𝜙)G(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 ∫ 2𝜋 0𝜋 ∫ 𝜋 0 G(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (2-145) REFERENCES 1. A. Z. Elsherbeni and C. D. Taylor Jr., “Antenna Pattern Plotter,” Copyright c ⃝1995, Electrical Engineering Department, The University of Mississippi, University, MS.
2. W. R. Scott Jr., “A General Program for Plotting Three-dimensional Antenna Patterns,” IEEE Antennas and Propagation Society Newsletter, pp. 6–11, December 1989.
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4. J. S. Hollis, T. J. Lyon, and L. Clayton Jr. (eds.), Microwave Antenna Measurements, Scientific-Atlanta, Inc., July 1970.
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8. J. Romeu and R. Pujol, “Array,” Software Book in Electromagnetics, Vol. II, Chapter 12, CAEME Center for Multimedia Education, University of Utah, pp. 467–481, 1995.
9. R. S. Elliott, “Beamwidth and Directivity of Large Scanning Arrays,” Last of Two Parts, The Microwave Journal, pp. 74–82, January 1964.
10. C.-T. Tai and C. S. Pereira, “An Approximate Formula for Calculating the Directivity of an Antenna,” IEEE Trans. Antennas Propagat., Vol. AP-24, No. 2, pp. 235–236, March 1976.
11. N. A. McDonald, “Approximate Relationship Between Directivity and Beamwidth for Broadside Collinear Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-26, No. 2, pp. 340–341, March 1978.
12. D. M. Pozar, “Directivity of Omnidirectional Antennas,” IEEE Antennas Propagat. Mag., Vol. 35, No. 5, pp. 50–51, October 1993.
13. C. A. Balanis, Advanced Engineering Electromagnetics, Second Edition, John Wiley and Sons, New York, 2012.
14. H. Poincar´ e, Theorie Mathematique de la Limiere, Georges Carre, Paris, France, 1892.
15. G. A. Deschamps, “Part II—Geometrical Representation of the Polarization of a Plane Electromagnetic Wave,” Proc. IRE, Vol. 39, pp. 540–544, May 1951.
16. E. F. Bolinder, “Geometrical Analysis of Partially Polarized Electromagnetic Waves,” IEEE Trans. Anten-nas Propagat., Vol. AP-15, No. 1, pp. 37–40, January 1967.
17. G. A. Deschamps and P. E. Mast, “Poincar´ e Sphere Representation of Partially Polarized Fields,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 4, pp. 474–478, July 1973.
104 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 18. G. Sinclair, “The Transmission and Reflection of Elliptically Polarized Waves,” Proc. IRE, Vol. 38, pp. 148–151, February 1950.
19. C. A. Balanis, “Antenna Theory: A Review,” Proc. IEEE, Vol. 80, No. 1, pp. 7–23, January 1992.
20. R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill Book Co., New York, 1985.
21. M. I. Skolnik, Radar Systems, Chapter 2, McGraw-Hill Book Co., New York, 1962.
22. J. A. Adam, “How to Design an “Invisible’ Aircraft,” IEEE Spectrum, pp. 26–31, April 1988.
23. G. T. Ruck, D. E. Barrick, W. D. Stuart, and C. K. Krichbaum, Radar Cross Section Handbook, Vols. 1, 2, Plenum Press, New York, 1970.
24. M. I. Skolnik (Ed.), Radar Handbook, Chapter 27, Section 6, McGraw-Hill Book Co., New York, pp. 27-19–27-40, 1970.
25. J. W. Crispin, Jr. and K. M. Siegel, Methods of Radar Cross Section Analysis, Academic Press, Inc., New York, 1968.
26. J. J. Bowman, T. B. A. Senior, and P. L. Uslenghi (Eds.), Electromagnetic and Acoustic Scattering by Simple Shapes, Amsterdam, The Netherland: North-Holland, 1969.
27. E. F. Knott, M. T. Turley, and J. F. Shaeffer, Radar Cross Section, Artech House, Inc., Norwood, MA, 1985.
28. A. K. Bhattacharya and D. L. Sengupta, Radar Cross Section Analysis and Control, Artech House, Inc., Norwood, MA, 1991.
29. A. F. Maffett, Topics for a Statistical Description of Radar Cross Section, John Wiley and Sons, New York, 1989.
30. Special issue, Proc. IEEE, Vol. 53, No. 8, August 1965.
31. Special issue, Proc. IEEE, Vol. 77, No. 5, May 1989.
32. Special issue, IEEE Trans. Antennas Propagat., Vol. 37, No. 5, May 1989.
33. W. R. Stone (ed.), Radar Cross Sections of Complex Objects, IEEE Press, New York, 1989.
34. A. F. Stevenson, “Relations Between the Transmitting and Receiving Properties of Antennas,” Q. Appl.
Math., pp. 369–384, January 1948.
35. R. F. Harrington, “Theory of Loaded Scatterers,” Proc. IEE (British), Vol. 111, pp. 617– 623, April 1964.
36. R. E. Collin, “The Receiving Antenna,” in Antenna Theory, Part I, (R. E. Collin and F. J. Zucker, Eds.), McGraw-Hill Book Co., 1969.
37. R. B. Green, “The Effect of Antenna Installations on the Echo Area of an Object,” Report No. 1109-3, ElectroScience Laboratory, Ohio State University, Columbus, OH, September 1961.
38. R. B. Green “Scattering from Conjugate-Matched Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-14, No. 1, pp. 17–21, January 1966.
39. R. J. Garbacz, “The Determination of Antenna Parameters by Scattering Cross-Section Measurements, III. Antenna Scattering Cross Section,” Report No. 1223-10, Antenna Laboratory, Ohio State University, November 1962.
40. R. C. Hansen, “Relationships Between Antennas as Scatterers and as Radiators,” Proc. IEEE, Vol. 77, No.
5, pp. 659–662, May 1989.
41. S. H. Dike and D. D. King, “Absorption Gain and Backscattering Cross Section of the Cylindrical Antenna,” Proc. IRE, Vol. 40, 1952.
42. J. Sevick, “Experimental and Theoretical Results on the Backscattering Cross Section of Coupled Anten-nas,” Tech. Report No. 150, Cruft Laboratory, Harvard University, May 1952.
43. D. L. Moffatt, “Determination of Antenna Scattering Properties From Model Measurements,” Report No.
1223-12, Antenna Laboratory, Ohio State University, January 1964.
44. J. T. Aberle, Analysis of Probe-Fed Circular Microstrip Antennas, PhD Dissertation, University of Mass., Amherst, MA, 1989.
45. J. T. Aberle, D. M. Pozar, and C. R. Birtcher, “Evaluation of Input Impedance and Radar Cross Section of Probe-Fed Microstrip Patch Elements Using an Accurate Feed Model,” IEEE Trans. Antennas Propagat., Vol. 39, No. 12, pp. 1691–1696, December 1991.
PROBLEMS 105 PROBLEMS 2.1.
An antenna has a beam solid angle that is equivalent to a trapezoidal patch (patch with 4 sides, 2 of which are parallel to each other) on the surface of a sphere of radius r. The angular space of the patch on the surface of the sphere extends between 𝜋∕6 ≤𝜃≤𝜋∕3(30◦≤𝜃≤ 60◦) in latitude and 𝜋∕4 ≤𝜙≤𝜋∕3(45◦≤𝜙≤60◦) in longitude. Find the following: (a) Equivalent beam solid angle [which is equal to number of square radians/steradians or (degrees)2] of the patch [in square radians/steradians and in (degrees)2].
r Exact.
r Approximate using ΩA = ΔΘ ⋅ΔΦ = (𝜃2 −𝜃1) ⋅(𝜙2 −𝜙1). Compare with the exact.
(b) Corresponding antenna maximum directivities of part a (dimensionless and in dB).
2.2.
Derive (2-7) given the definitions of (2-5) and (2-6) 2.3.
A hypothetical isotropic antenna is radiating in free-space. At a distance of 100 m from the antenna, the total electric field (E𝜃) is measured to be 5 V/m. Find the (a) power density (Wrad) (b) power radiated (Prad) 2.4.
Find the half-power beamwidth (HPBW) and first-null beamwidth (FNBW), in radians and degrees, for the following normalized radiation intensities: (a) U(𝜃) = cos 𝜃 (b) U(𝜃) = cos2 𝜃 (c) U(𝜃) = cos(2𝜃) (d) U(𝜃) = cos2(2𝜃) (e) U(𝜃) = cos(3𝜃) (f) U(𝜃) = cos2(3𝜃) ⎫ ⎪ ⎬ ⎪ ⎭ (0 ≤𝜃≤90◦, 0 ≤𝜙≤360◦) 2.5.
Find the half-power beamwidth (HPBW) and first-null beamwidth (FNBW), in radians and degrees, for the following normalized radiation intensities: (a) U(𝜃) = cos 𝜃cos(2𝜃) (b) U(𝜃) = cos2 𝜃cos2(2𝜃) (c) U(𝜃) = cos(𝜃) cos(3𝜃) (d) U(𝜃) = cos2(𝜃) cos2(3𝜃) (e) U(𝜃) = cos(2𝜃) cos(3𝜃) (f) U(𝜃) = cos2(2𝜃) cos2(3𝜃) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (0 ≤𝜃≤90◦, 0 ≤𝜙≤360◦) 2.6.
The maximum radiation intensity of a 90% efficiency antenna is 200 mW/unit solid angle.
Find the directivity and gain (dimensionless and in dB) when the (a) input power is 125.66 mW (b) radiated power is 125.66 mW 2.7.
The power radiated by a lossless antenna is 10 watts. The directional characteristics of the antenna are represented by the radiation intensity of (a) U = Bo cos2 𝜃 (b) U = Bo cos3 𝜃 } (watts/unit solid angle) (0 ≤𝜃≤𝜋∕2, 0 ≤𝜙≤2𝜋) For each, find the (a) maximum power density (in watts/square meter) at a distance of 1,000 m (assume far-field distance). Specify the angle where this occurs.
(b) exact and approximate beam solid angle ΩA.
(c) directivity, exact and approximate, of the antenna (dimensionless and in dB).
(d) gain, exact and approximate, of the antenna (dimensionless and in dB).
106 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2.8.
The approximate far zone normalized electric field radiated by a resonant linear dipole antenna used in wireless mobile units, positioned symmetrically at the origin along the z-axis, is given by Ea ≃̂ a𝜃Ea sin1.5 𝜃e−jkr r , 0◦≤𝜃≤180◦ 0◦≤𝜃≤360◦ where Ea is a constant and r is the spherical radial distance measured from the origin of the coordinate system. Determine the: (a) Exact maximum directivity (dimensionless and in dB).
(b) Half-power beamwidth (in degrees) (c) Approximate maximum directivity (dimensionless and in dB). Indicate which approxi-mate formula you are using and why?
(d) Approximate maximum directivity (dimensionless and in dB) using another approximate formula. Indicate which other formula you are using and why?
(e) Maximum directivity (dimensionless and in dB) using the computer program Directivity.
2.9.
You are an antenna engineer and you are asked to design a high directivity/gain antenna for a space-borne communication system operating at 10 GHz. The specifications of the antenna are such that its pattern consists basically of one major lobe and, for simplicity, no minor lobes (if there are any minor lobes they are of such very low intensity and you can assume they are negligible/zero). Also it is desired that the pattern is symmetrical in the azimuthal plane. In order to meet the desired objectives, the main lobe of the pattern should have a half-power beamwidth of 10 degrees. In order to expedite the design, it is assumed that the major lobe of the normalized radiation intensity of the antenna is approximated by U(𝜃, 𝜙) = cosn(𝜃) and it exists only in the upper hemisphere (0 ≤𝜃≤𝜋∕2, 0 ≤𝜙≤2𝜋). Determine the: (a) Value of n (not necessarily an integer) to meet the specifications of the major lobe. Keep 5 significant figures in your calculations.
(b) Exact maximum directivity of the antenna (dimensionless and in dB).
(c) Approximate maximum directivity of the antenna based on Kraus’ formula (dimension-less and in dB).
(d) Approximate maximum directivity of the antenna based on Tai & Pereira’s formula (dimensionless and in dB).
2.10.
In target-search ground-mapping radars it is desirable to have echo power received from a target, of constant cross section, to be independent of its range. For one such application, the desirable radiation intensity of the antenna is given by U(𝜃, 𝜙) = { 1 0◦≤𝜃< 20◦ 0.342 csc(𝜃) 20◦≤𝜃< 60◦ 0 60◦≤𝜃≤180◦ } 0◦≤𝜙≤360◦ Find the directivity (in dB) using the exact formula.
PROBLEMS 107 2.11.
A beam antenna has half-power beamwidths of 30◦and 35◦in perpendicular planes inter-secting at the maximum of the mainbeam. Find its approximate maximum effective aperture (in λ2) using: (a) Kraus’ (b) Tai and Pereira’s formulas.
The minor lobes are very small and can be neglected.
2.12.
The normalized radiation intensity of a given antenna is given by (a) U = sin 𝜃sin 𝜙 (b) U = sin 𝜃sin2 𝜙 (c) U = sin 𝜃sin3 𝜙 (d) U = sin2 𝜃sin 𝜙 (e) U = sin2 𝜃sin2 𝜙 (f) U = sin2 𝜃sin3 𝜙 The intensity exists only in the 0 ≤𝜃≤𝜋, 0 ≤𝜙≤𝜋region, and it is zero elsewhere.
Find the (a) exact directivity (dimensionless and in dB).
(b) azimuthal and elevation plane half-power beamwidths (in degrees).
2.13.
The normalized radiation intensity radiated by an antenna is given by U(𝜃, 𝜙) = ⎧ ⎪ ⎨ ⎪ ⎩ sin 𝜃cos2 𝜙 0◦≤𝜃≤180◦ 90◦≤𝜃≤270◦ 0 Elsewhere The maximum of the radiation intensity occurs towards 𝜃= 90◦and 𝜙= 180◦. Find the: (a) Exact maximum directivity (dimensionless and in dB).
(b) Half-power beamwidth (in degrees) in the principal azimuth (horizontal) plane.
(c) Half-power beamwidth (in degrees) in the principal elevation (vertical) plane.
(d) Maximum directivity (dimensionless and in dB) using an appropriate approximate method that you know. Indicate which one you are using.
2.14.
Find the directivity (dimensionless and in dB) for the antenna of Problem 2.12 using (a) Kraus’ approximate formula (2-26) (b) Tai and Pereira’s approximate formula (2-30a) 2.15.
For Problem 2.10, determine the approximate directivity (in dB) using (a) Kraus’ formula (b) Tai and Pereira’s formula.
2.16.
The normalized radiation intensity of an antenna is rotationally symmetric in 𝜙, and it is represented by U = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 0◦≤𝜃< 30◦ 0.5 30◦≤𝜃< 60◦ 0.1 60◦≤𝜃< 90◦ 0 90◦≤𝜃≤180◦ (a) What is the directivity (above isotropic) of the antenna (in dB)?
(b) What is the directivity (above an infinitesimal dipole) of the antenna (in dB)?
108 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2.17.
The radiation intensity of an antenna is given by U(𝜃, 𝜙) = cos4 𝜃sin2 𝜙 for 0 ≤𝜃≤𝜋∕2 and 0 ≤𝜙≤2𝜋(i.e., in the upper half-space). It is zero in the lower half-space.
Find the (a) exact directivity (dimensionless and in dB) (b) elevation plane half-power beamwidth (in degrees) 2.18.
The normalized radiation intensity of an antenna is symmetric, and it can be approxi-mated by U(𝜃) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 0◦≤𝜃< 30◦ cos(𝜃) 0.866 30◦≤𝜃< 90◦ 0 90◦≤𝜃≤180◦ and it is independent of 𝜙. Find the (a) exact directivity by integrating the function (b) approximate directivity using Kraus’ formula 2.19.
The maximum gain of a horn antenna is +20 dB, while the gain of its first sidelobe is −15 dB.
What is the difference in gain between the maximum and first sidelobe: (a) in dB (b) as a ratio of the field intensities.
2.20.
The normalized radiation intensity of an antenna is approximated by U = sin 𝜃 where 0 ≤𝜃≤𝜋, and 0 ≤𝜙≤2𝜋. Determine the directivity using the (a) exact formula (b) formulas of (2-33a) by McDonald and (2-33b) by Pozar (c) computer program Directivity of this chapter.
2.21.
Repeat Problem 2.20 for a λ∕2 dipole whose normalized intensity is approximated by U ≃sin3 𝜃 Compare the value with that of (4-91) or 1.643 (2.156 dB).
2.22.
The radiation intensity of a circular loop of radius a and of constant current is given by U = J2 1(ka sin 𝜃), 0 ≤𝜃≤𝜋 and 0 ≤𝜙≤2𝜋 where J1(x) is the Bessel function of order 1. For a loop with radii of a = λ∕10 and λ∕20, determine the directivity using the: (a) formulas (2-33a) by McDonald and (2-33b) by Pozar.
(b) computer program Directivity of this chapter.
Compare the answers with that of a very small loop represented by 1.5 or 1.76 dB.
PROBLEMS 109 2.23.
Find the directivity (dimensionless and in dB) for the antenna of Problem 2.12 using numer-ical techniques with 10◦uniform divisions and with the field evaluated at the (a) midpoint (b) trailing edge of each division.
2.24.
Compute the directivity values of Problem 2.12 using the computer program Directivity of this chapter.
2.25.
The far-zone electric-field intensity (array factor) of an end-fire two-element array antenna, placed along the z-axis and radiating into free-space, is given by E = cos [𝜋 4 (cos 𝜃−1) ] e−jkr r , 0 ≤𝜃≤𝜋 Find the directivity using (a) Kraus’ approximate formula (b) the computer program Directivity of this chapter.
2.26.
Repeat Problem 2.25 when E = cos [𝜋 4 (cos 𝜃+ 1) ] e−jkr r , 0 ≤𝜃≤𝜋 2.27.
The radiation intensity is represented by U = { U0 sin(𝜋sin 𝜃), 0 ≤𝜃≤𝜋∕2 and 0 ≤𝜙≤2𝜋 0 elsewhere Find the directivity (a) exactly (b) using the computer program Directivity of this chapter.
2.28.
The approximate far-zone electric field radiated by a very thin wire circular loop of radius a, positioned symmetrically about the z-axis and with its area parallel to the xy-plane, is given by E𝜙≃C0 sin1.5 𝜃ejkr r where C0 is a constant. Determine the: (a) Exact directivity (dimensionless and in dB).
(b) Approximate directivity (dimensionless and in dB) using an approximate but appropriate formula (state the formula you are using).
2.29.
The radiation intensity of an aperture antenna, mounted on an infinite ground plane with z perpendicular to the aperture, is rotationally symmetric (not a function of 𝜙), and it is given by U = [sin(𝜋sin 𝜃) 𝜋sin 𝜃 ]2 Find the approximate directivity (dimensionless and in dB) using (a) numerical integration. Use the Directivity computer program of this chapter.
(b) Kraus’ formula (c) Tai and Pereira’s formula.
110 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2.30.
The normalized far-zone field pattern of an antenna is given by E = ⎧ ⎪ ⎨ ⎪ ⎩ (sin 𝜃cos2 𝜙)1∕2 0 ≤𝜃≤𝜋and 0 ≤𝜙≤𝜋∕2, 3𝜋∕2 ≤𝜙≤2𝜋 0 elsewhere Find the directivity using (a) the exact expression (b) Kraus’ approximate formula (c) Tai and Pereira’s approximate formula (d) the computer program Directivity of this chapter 2.31.
The normalized field pattern of the main beam of a conical horn antenna, mounted on an infinite ground plane with z perpendicular to the aperture, is given by J1(ka sin 𝜃) sin 𝜃 where a is its radius at the aperture. Assuming that a = λ, find the (a) half-power beamwidth (b) directivity using Kraus’ approximate formula 2.32.
A base station cellular communication systems lossless antenna has a maximum gain of 16 dB (above isotropic) at 1,900 MHz. Assuming the input power to the antenna is 8 watts, what is the maximum radiated power density (in watts/cm2) at a distance of 100 meters?
This will determine the safe level for human exposure to electromagnetic radiation.
2.33.
A uniform plane wave, of a form similar to (2-55), is traveling in the positive z-direction.
Find the polarization (linear, circular, or elliptical), sense of rotation (CW or CCW), axial ratio (AR), and tilt angle 𝜏(in degrees) when (a) Ex = Ey, Δ𝜙= 𝜙y −𝜙x = 0 (b) Ex ≠Ey, Δ𝜙= 𝜙y −𝜙x = 0 (c) Ex = Ey, Δ𝜙= 𝜙y −𝜙x = 𝜋∕2 (d) Ex = Ey, Δ𝜙= 𝜙y −𝜙x = −𝜋∕2 (e) Ex = Ey, Δ𝜙= 𝜙y −𝜙x = 𝜋∕4 (f) Ex = Ey, Δ𝜙= 𝜙y −𝜙x = −𝜋∕4 (g) Ex = 0.5Ey, Δ𝜙= 𝜙y −𝜙x = 𝜋∕2 (h) Ex = 0.5Ey, Δ𝜙= 𝜙y −𝜙x = −𝜋∕2 In all cases, justify the answer.
2.34.
Derive (2-66), (2-67), and (2-68).
2.35.
Write a general expression for the polarization loss factor (PLF) of two linearly polarized antennas if (a) both lie in the same plane (b) both do not lie in the same plane 2.36.
A linearly polarized wave traveling in the positive z-direction is incident upon a circularly polarized antenna. Find the polarization loss factor PLF (dimensionless and in dB) when the antenna is (based upon its transmission mode operation) (a) right-handed (CW) (b) left-handed (CCW) 2.37.
A 300 MHz uniform plane wave, traveling along the x-axis in the negative x-direction, whose electric field is given by Ew = Eo(ĵ ay + 3̂ az)e+jkx PROBLEMS 111 where Eo is a real constant, impinges upon a dipole antenna that is placed at the origin and whose electric field radiated toward the x-axis in the positive x-direction is given by Ea = Ea(̂ ay + 2̂ az)e−jkx where Ea is a real constant. Determine the following: (a) Polarization of the incident wave (including axial ratio and sense of rotation, if any).
You must justify (state why?).
(b) Polarization of the antenna (including axial ratio and sense of rotation, if any). You must justify (state why?).
(c) Polarization loss factor (dimensionless and in dB).
Incident Wave Antenna x z y 2.38.
The electric field of a uniform plane wave traveling along the negative z-direction is given by Ei w = (̂ ax + ĵ ay)E0e+jkz and is incident upon a receiving antenna placed at the origin and whose radiated electric field, toward the incident wave, is given by Ea = (̂ ax + 2̂ ay)E1 e−jkr r where E0 and E1 are constants.
Determine the following: (a) Polarization of the incident wave, and why?
(b) Sense of rotation of the incident wave.
(c) Polarization of the antenna, and why?
(d) Sense of rotation of the antenna polarization.
(e) Losses (dimensionless and in dB) due to polarization mismatch between the incident wave and the antenna.
2.39.
A spherical wave travelling in free-space along the negative y-axis, and whose electric field is given by Ew = (4̂ az + j2̂ ax)Ew e+jky y where Ew is a constant, impinges upon a λ∕2 dipole antennas positioned at the origin of the coordinate system whose normalized electric field antenna radiation characteristics along the y-direction are given by Ea = ̂ azEa e−jky y where Ea is a constant.
112 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS z x y Ew /2 dipole λ Assuming for this problem the receiving l = λ∕2 dipole is lossless and it is perfectly-matched to the connecting transmission line, determine the: (a) Polarization of the impinging wave (linear, circular or elliptical). Why?
(b) Sense of rotation (CW or CCW), if any, of the impinging wave. Why?
(c) Axial Ratio (AR) of the impinging wave. Why?
(d) Polarization Loss Factor (PLF) between the impinging wave and the receiving λ∕2 dipole antenna (dimensionless and in dB).
2.40.
A ground-based helical antenna is placed at the origin of a coordinate system and it is used as a receiving antenna. The normalized far-zone electric-field pattern of the helical antenna in the transmitting mode is represented in the direction 𝜃o, 𝜙o by Ea = E0(ĵ a𝜃+ 2̂ a𝜙) fo(𝜃o, 𝜙o)e−jkr r The far-zone electric field transmitted by an antenna on a flying aircraft towards 𝜃o, 𝜙o, which is received by the ground-based helical antenna, is represented by Ew = E1(2̂ a𝜃+ ĵ a𝜙) f1(𝜃o, 𝜙o)e+jkr r Determine the following: (a) Polarization (linear, circular, or elliptical) of the helical antenna in the transmitting mode. State also the sense of rotation, if any.
(b) Polarization (linear, circular, or elliptical) of the incoming wave that impinges upon the helical antenna. State also the sense of rotation, if any.
(c) Polarization loss (dimensionless and in dB) due to match/mismatch of the polarizations of the antenna and incoming wave.
2.41.
A circularly polarized wave, traveling in the positive z-direction, is incident upon a circularly polarized antenna. Find the polarization loss factor PLF (dimensionless and in dB) for right-hand (CW) and left-hand (CCW) wave and antenna.
2.42.
The electric field radiated by a rectangular aperture, mounted on an infinite ground plane with z perpendicular to the aperture, is given by E = [̂ a𝜃cos 𝜙−̂ a𝜙sin 𝜙cos 𝜃] f (r, 𝜃, 𝜙) where f(r, 𝜃, 𝜙) is a scalar function which describes the field variation of the antenna.
Assuming that the receiving antenna is linearly polarized along the x-axis, find the polar-ization loss factor (PLF).
PROBLEMS 113 2.43.
A circularly polarized wave, traveling in the positive z-direction, is received by an ellipti-cally polarized antenna whose reception characteristics near the main lobe are given approx-imately by Ea ≃[2̂ ax + ĵ ay] f(r, 𝜃, 𝜙) Find the polarization loss factor PLF (dimensionless and in dB) when the incident wave is (a) right-hand (CW) (b) left-hand (CCW) circularly polarized. Repeat the problem when Ea ≃[2̂ ax −ĵ ay] f(r, 𝜃, 𝜙) In each case, what is the polarization of the antenna? How does it match with that of the wave?
2.44.
A linearly polarized wave traveling in the negative z-direction has a tilt angle (𝜏) of 45◦. It is incident upon an antenna whose polarization characteristics are given by ̂ ρa = 4̂ ax + ĵ ay √ 17 Find the polarization loss factor PLF (dimensionless and in dB).
2.45.
An elliptically polarized wave traveling in the negative z-direction is received by a circularly polarized antenna whose main lobe is along the 𝜃= 0 direction. The unit vector describing the polarization of the incident wave is given by ̂ ρw = 2̂ ax + ĵ ay √ 5 Find the polarization loss factor PLF (dimensionless and in dB) when the wave that would be transmitted by the antenna is (a) right-hand CP (b) left-hand CP 2.46.
A CW circularly polarized uniform plane wave is traveling in the positive z-direction. Find the polarization loss factor PLF (dimensionless and in dB) assuming the receiving antenna (in its transmitting mode) is (a) CW circularly polarized (b) CCW circularly polarized 2.47.
The polarization of the field radiated by a helical antenna which is placed at the origin of a spherical coordinate system, which is used as a receiving antenna, is given by Ea = (2̂ a𝜃+ j4̂ a𝜙)Ea e−jkr r where Ea is a constant. The polarization of an incoming wave, which is received by the helical antenna, is given by Ew = (j4̂ a𝜃+ 2̂ a𝜙)Ew e+jkr r 114 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS where Ew is a constant. Determine the: (a) Polarization of the helical antenna (linear, circular, elliptical).
(b) The sense of rotation of the polarization of the helical antenna (CW or CCW).
(c) Axial Ratio of the polarization of the helical antenna (AR).
(d) Polarization of the incoming wave (linear, circular, elliptical).
(e) The sense of rotation of the incoming wave (CW or CCW).
(f) Axial Ratio of the incoming wave (AR).
(g) Polarization Loss factor (PLF) between the polarization of the helical antenna and that of the incoming wave (dimensionless and in dB).
2.48.
A linearly polarized uniform plane wave traveling in the positive z-direction, with a power density of 10 milliwatts per square meter, is incident upon a CW circularly polarized antenna whose gain is 10 dB at 10 GHz. Find the (a) maximum effective area of the antenna (in square meters) (b) power (in watts) that will be delivered to a load attached directly to the terminals of the antenna.
2.49.
A wave, whose electric field is given by Ew ≃̂ azEw e+jky y where Ew is a constant, is traveling along the -y axis and impinges upon the maximum inten-sity direction of the λ∕2 dipole, whose maximum electric field is given by Ea | | | |max = ̂ a𝜃Ea sin1.5 𝜃e−jkr r | | | | max 𝜃=90◦,r=y ̂ a𝜃=−̂ az = −̂ azEa e−jky y where Ea is a constant. Assuming the incoming wave has, at a frequency of 10 GHz, a power density of 100 mwatts/cm2, determine at 10 GHz the: (a) Polarization Loss Factor (PLF) (dimensionless and in dB) (b) Maximum power (in watts) that can be delivered to a load connected to the λ∕2 receiving antenna whose input impedance is Zin = 73 + j42.5 Its total loss resistance is 5 ohms and the antenna is connected to a transmission line with characteristic impedance of 50 ohms.
2.50.
A linearly polarized plane wave traveling along the negative z-axis is incident upon an ellip-tically polarized antenna (either CW or CCW). The axial ratio of the antenna polarization ellipse is 2:1 and its major axis coincides with the principal x-axis. Find the polarization loss factor (PLF) assuming the incident wave is linearly polarized in the (a) x-direction (b) y-direction 2.51.
A wave traveling normally outward from the page (toward the reader) is the resultant of two elliptically polarized waves, one with components of E given by: ℰ′ y = 3 cos 𝜔t; ℰ′ x = 7 cos ( 𝜔t + 𝜋 2 ) PROBLEMS 115 and the other with components given by: ℰ′′ y = 2 cos 𝜔t; ℰ′′ x = 3 cos ( 𝜔t −𝜋 2 ) (a) What is the axial ratio of the resultant wave?
(b) Does the resultant vector E rotate clockwise or counterclockwise?
2.52.
A linearly polarized antenna lying in the x-y plane is used to determine the polarization axial ratio of incoming plane waves traveling in the negative z-direction. The polarization of the antenna is described by the unit vector ̂ ρa = ̂ ax cos 𝜓+ ̂ ay sin 𝜓 0 50 100 150 200 250 300 350 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (deg) (a) PLF versus PLF ψ (b) PLF versus ψ (c) PLF versus ψ ψ ψ 0 50 100 150 200 250 300 350 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (deg) PLF ψ 0 50 100 150 200 250 300 350 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (deg) PLF where 𝜓is an angle describing the orientation in the x-y plane of the receiving antenna.
Above are the polarization loss factor (PLF) versus receiving antenna orientation curves obtained for three different incident plane waves. For each curve determine the axial ratio of the incident plane wave.
2.53.
A λ∕2 dipole, with a total loss resistance of 1 ohm, is connected to a generator whose internal impedance is 50 + j25 ohms. Assuming that the peak voltage of the generator is 2 V and the impedance of the dipole, excluding the loss resistance, is 73 + j42.5 ohms, find the power (a) supplied by the source (real) (c) dissipated by the antenna (b) radiated by the antenna 116 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2.54.
The antenna and generator of Problem 2.53 are connected via a 50-ohm λ∕2-long lossless transmission line. Find the power (a) supplied by the source (real) (c) dissipated by the antenna (b) radiated by the antenna 2.55.
An antenna with a radiation resistance of 48 ohms, a loss resistance of 2 ohms, and a reac-tance of 50 ohms is connected to a generator with open-circuit voltage of 10 V and internal impedance of 50 ohms via a λ∕4-long transmission line with characteristic impedance of 100 ohms.
(a) Draw the equivalent circuit (b) Determine the power supplied by the generator (c) Determine the power radiated by the antenna 2.56.
A transmitter, with an internal impedance Z0 (real), is connected to an antenna through a lossless transmission line of length l and characteristic impedance Z0. Find a simple expres-sion for the ratio between the antenna gain and its realized gain.
Antenna Transmission line x = 0 x Zo Vs Zin Zo Transmitter l V(x) = A [e–jkx + Γ(0)e+jkx] I(x) = [e–jkx – Γ(0)e+jkx] A Z0 Vs = strength of voltage source Zin = Rin + jXin = input impedance of the antenna Z0 = R0 = characteristic impedance of the line Paccepted = power accepted by the antenna {Paccepted = Re[V(0)I∗(0)]} Pavailable = power delivered to a matched load [i.e., Zin = Z∗ 0 = Z0] 2.57.
The input reactance of an infinitesimal linear dipole of length λ∕60 and radius a = λ∕200 is given by Xin ≃−120[ln(l∕2a) −1] tan(kl∕2) Assuming the wire of the dipole is copper with a conductivity of 5.7 × 107S/m, determine at f = 1 GHz the (a) loss resistance (b) radiation resistance (c) radiation efficiency (d) VSWR when the antenna is connected to a 50-ohm line 2.58.
A dipole antenna consists of a circular wire of length l. Assuming the current distribution on the wire is cosinusoidal, i.e., Iz(z) = I0 cos (𝜋 l z′) −l∕2 ≤z′ ≤l∕2 where I0 is a constant, derive an expression for the loss resistance RL, which is one-half of (2-90b).
PROBLEMS 117 2.59.
The E-field pattern of an antenna, independent of 𝜙, varies as follows: E = { 1 0◦≤𝜃≤45◦ 0 45◦< 𝜃≤90◦ 1 2 90◦< 𝜃≤180◦ (a) What is the directivity of this antenna?
(b) What is the radiation resistance of the antenna at 200 m from it if the electric field is equal to 10 V/m (rms) for 𝜃= 0◦at that distance and the terminal current is 5 A (rms)?
2.60.
The approximate far-zone normalized electric field radiated by a ground-based end-fire heli-cal antenna used in communication systems for space-borne applications, positioned sym-metrically at the origin along the z-axis, is given by Ea ≃ ⎧ ⎪ ⎨ ⎪ ⎩ ̂ a𝜃Ea cos3 𝜃e−jkr r , 0 ≤𝜃≤90◦ 0 ≤𝜙≤360◦ 0 Elshewhere where Ea is a constant and r is the spherical radial distance measured from the origin of the coordinate system.
Determine the: (a) Half power beamwidth (HPBW) (in degrees) (b) Exact maximum directivity (dimensionless and in dB).
(c) Approximate maximum directivity (dimensionless and in dB) using the most accurate approximate formula for this problem.
Indicate which most accurate approximate formula you are using and why?
(d) Maximum directivity (dimensionless and in dB) using the computer program Directivity.
(e) Maximum effective area (Aem)(in λ2), based on the exact directivity.
2.61.
The far-zone field radiated by a rectangular aperture mounted on a ground plane, with dimensions a and b and uniform aperture distribution, is given by (see Table 12.1) E ≈̂ a𝜃E𝜃+ ̂ a𝜙E𝜙 E𝜃= C sin 𝜙sin X X sin Y Y E𝜙= C cos 𝜃cos 𝜙sin X X sin Y Y ⎫ ⎪ ⎬ ⎪ ⎭ X = ka 2 sin 𝜃cos 𝜙; 0 ≤𝜃≤90◦ Y = kb 2 sin 𝜃sin 𝜙; 0 ≤𝜙≤360◦ where C is a constant. For an aperture with a = 3λ, b = 2λ, determine the (a) maximum partial directivities D𝜃, D𝜙(dimensionless and in dB) and (b) total maximum directivity Do (dimensionless and in dB). Compare with that computed using the equation in Table 12.1.
Use the computer program Directivity of this chapter.
118 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS 2.62.
Repeat Problem 2.61 when the aperture distribution is that of the dominant TE10 mode of a rectangular waveguide, or from Table 12.1 E ≈̂ a𝜃E𝜃+ ̂ a𝜙E𝜙 E𝜃= −𝜋 2 C sin 𝜙 cos X (X)2 − (𝜋 2 )2 sin Y Y E𝜙= −𝜋 2 C cos 𝜃cos 𝜙 cos X (X)2 − (𝜋 2 )2 sin Y Y ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ X = ka 2 sin 𝜃cos 𝜙 Y = kb 2 sin 𝜃sin 𝜙 2.63.
Repeat Problem 2.62 when the aperture dimensions are those of an X-band rectangular waveguide with a = 2.286 cm (0.9 in.), b = 1.016 cm (0.4 in.) and frequency of operation is 10 GHz.
2.64.
Repeat Problem 2.61 for a circular aperture with a uniform distribution and whose far-zone fields are, from Table 12.2 E ≈̂ a𝜃E𝜃+ ̂ a𝜙E𝜙 E𝜃= jC1 sin 𝜙J1(Z) Z E𝜙= jC1 cos 𝜃cos 𝜙J1(Z) Z ⎫ ⎪ ⎬ ⎪ ⎭ Z = ka sin 𝜃; 0 ≤𝜃≤90◦ 0 ≤𝜙≤360◦ where C1 is a constant and J1(Z) is the Bessel function of the first kind. Assume a = 1.5λ.
2.65.
Repeat Problem 2.64 when the aperture distribution is that of the dominant TE11 mode of a circular waveguide, or from Table 12.2 E ≈̂ a𝜃E𝜃+ ̂ a𝜙E𝜙 E𝜃= C2 sin 𝜙J1(Z) Z E𝜙= C2 cos 𝜃cos 𝜙 J′ z(Z) (1) − ( Z 𝜒′ 11 )2 ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ Z = ka sin 𝜃; 0 ≤𝜃≤90◦ J′ z(Z) = Jo(Z) 0 ≤𝜙≤360◦ −J1(Z)∕Z; where C2 is a constant, J′ 1(Z) is the derivative of J1(Z), 𝜒′ 11 = 1.841 is the first zero of J′ 1(Z), and Jo(Z) is the Bessel function of the first find of order zero.
2.66.
Repeat 2.65 when the radius of the aperture is a = 1.143 cm (0.45 in.) and the frequency of operation is 10 GHz.
2.67.
The normalized far-field total electric field radiated by an antenna, consisting of an infinites-imal vertical dipole (oriented along the z-axis) plus a small circular loop (parallel to the xy-plane), placed at the origin of a spherical coordinate system is given by: Ea = (̂ a𝜃+ j2̂ a𝜙) sin 𝜃Eo e−jkr r ; (0◦≤𝜃≤180◦, 0◦≤𝜙≤360◦) PROBLEMS 119 Determine the: (a) Polarization of the wave (linear, circular or elliptical). Justify it. If circular, state the sense of rotation. If elliptical, state the sense of rotation and the axial ratio (AR).
(b) partial maximum directivities (D0)𝜃and (D0)𝜙(dimensionless and in dB).
(c) Total maximum directivity Do (dimensionless and in dB).
Hint: For this problem, there are two different and distinct partial directivities.
z a dipole loop y x r θ ϕ 2.68.
A 1-m long dipole antenna is driven by a 150 MHz source having a source resistance of 50 ohms and a voltage of 100 V. If the ohmic resistance of the antennas is given by RL = 0.625 ohms, find the: (a) Current going into the antenna (Iant); (b) Power dissipated by the antenna (c) Power radiated by the antenna; (d) Radiation efficiency of the antenna 2.69.
The field radiated by an infinitesimal dipole of very small length (l ≤λ∕50), and of uniform current distribution Io, is given by (4-26a) or E = ̂ a𝜃E𝜃≈̂ a𝜃j𝜂kIol 4𝜋re−jkr sin 𝜃 Determine the (a) vector effective length (b) maximum value of the vector effective length. Specify the angle.
(c) ratio of the maximum effective length to the physical length l.
2.70.
The field radiated by a half-wavelength dipole (l = λ∕2), with a sinusoidal current distribu-tion, is given by (4-84) or E = ̂ a𝜃E𝜃≈̂ a𝜃j𝜂Io 2𝜋re−jkr ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ where Io is the maximum current. Determine the (a) vector effective length (b) maximum value of the vector effective length. Specify the angle.
(c) ratio of the maximum effective length to the physical length l.
2.71.
A uniform plane wave, of 10−3watts/cm2 power density, is incident upon an infinitesimal dipole of length l = λ∕50 and uniform current distribution, as shown in Figure 2.29(a). For 120 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS a frequency of 10 GHz, determine the maximum open-circuited voltage at the terminals of the antenna. See Problem 2.69.
2.72.
Repeat Problem 2.71 for a small dipole with triangular current distribution and length l = λ∕10. See Example 2.14.
2.73.
Repeat Problem 2.71 for a half-wavelength dipole (𝓁= λ∕2) with sinusoidal current distri-bution. See Problem 2.70.
2.74.
Show that the effective length of a linear antenna can be written as le = √ Ae|Zt|2 𝜂RT which for a lossless antenna and maximum power transfer reduces to le = 2 √ AemRr 𝜂 Ae and Aem represent, respectively, the effective and maximum effective apertures of the antenna while 𝜂is the intrinsic impedance of the medium.
2.75.
An antenna has a maximum effective aperture of 2.147 m2 at its operating frequency of 100 MHz. It has no conduction or dielectric losses. The input impedance of the antenna itself is 75 ohms, and it is connected to a 50-ohm transmission line. Find the directivity of the antenna system (“system” meaning includes any effects of connection to the transmission line). Assume no polarization losses.
2.76.
A small circular parabolic reflector, often referred to as dish, is now being advertised as a TV antenna for direct broadcast. Assuming the diameter of the antenna is 1 meter, the frequency of operation is 3 GHz, and its aperture efficiency is 68%, determine the following: (a) Physical area of the reflector (in m2).
(b) Maximum effective area of the antenna (in m2).
(c) Maximum directivity (dimensionless and in dB).
(d) Maximum power (in watts) that can be delivered to the TV if the power density of the wave incident upon the antenna is 10 𝜇watts∕m2. Assume no losses between the incident wave and the receiver (TV).
2.77.
A uniform plane wave, with a power density of 10 mwatts/cm2, is impinging upon a half wavelength dipole at an angle normal/perpendicular to the axis of the dipole. Determine the: (a) Maximum effective area (in λ2) of the lossless dipole element. Assume the dipole has a directivity of 2.148 dB, it is polarized matched to the incident wave and it is mismatched, with reflection coefficieitn of 0.2. to the transmission line it is connected.
(b) Physical area (in λ2). Assume the physical area of the dipole is equal to its lengthwise cross sectional area; the dipole diameter is λ/300.
(c) Aperture efficiency (in %).
(d) Maximum power the dipole will interecept and deliver to a load. Assume a frequency of operation of 1 GHz.
2.78.
An incoming wave, with a uniform power density equal to 10−3 W/m2 is incident normally upon a lossless horn antenna whose directivity is 20 dB. At a frequency of 10 GHz, deter-mine the very maximum possible power that can be expected to be delivered to a receiver PROBLEMS 121 or a load connected to the horn antenna. There are no losses between the antenna and the receiver or load.
2.79.
You are a communication/antenna engineer and you are asked to determine whether the sig-nal received by a space borne communication system operating at 10 GHz will be of suffi-cient strength to be detected by the receiver. The power density incident upon the receiving antenna of the space borne system is 10 × 10−3 Watts/cm2. The polarization of the inci-dent wave is right-hand circularly polarized while that of the receiving antenna is linearly polarized. The maximum directivity of the receiving space borne antenna is 12 dB, its input impednace is 100 ohms, and its radiation efficiency is 75%. The characteristic impednace of the transmission line from the receiving space borne antenna to the space borne receiver has a characteistic impedance of 50 ohms. Determine the: (a) maximum effective area (in cm2) of the antenna assuming no losses.
(b) maximum effective area (in cm2) of the antenna taking into account all the stated losses.
Identify each of the losses in %.
(c) Maximum power (in Watts) delivered to the receiver taking into account all of the stated losses.
2.80.
A linearly polarized aperture antenna, with a uniform field distribution over its area, is used as a receiving antenna. The antenna physical area over its aperture is 10 cm2, and it is oper-ating at 10 GHz. The antenna is illuminated with a circularly polarized plane wave whose incident power density is 10 mwatts/cm2. Assuming the antenna element itself is lossless, determine its (a) gain (dimensionless and in dB).
(b) maximum power (in watts) that can be delivered to a load connected to the antenna.
Assume no other losses between the antenna and the load.
2.81.
A uniform plane wave traveling along the negative z-axis, and whose normalized electric field is given by Ew = (ĵ ax + 2̂ ay)e+jkz it impinges upon an antenna whose polarization along the z-axis is represented by the nor-malized electric field of Ea = ĵ aye−jkz The power density of the wave which impinges upon the antenna is 10 mwatts/λ2. The antenna has a maximum derectivity of 2.15 dB along the z-direction.
Determine the: (a) Polarization of the impinging/incident wave, including it Axial Ratio and sense of rota-tion, if any.
(b) Polarization of the antenna, including its Axial Ratio and sense of rotatio, if any.
(c) Very maximum efective area (in λ2) of the antenna assuming it has no losses of any kind.
(d) Maximum effective area (in λ2) assuming the antenna is losseless, but it possesses a reflection coefficeint of 0.5 to the transmission line to which it is connected. Include any other losses that should be accounted for.
122 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS (e) Maximum power the antenna will intercept and deliver to a load/receiver connected to its trasmission line, taking into account all the losses associated with this problem. Assume a frequency of operation of 1 GHz.
z y x Impinging wave Antenna 2.82.
The far-zone power density radiated by a helical antenna can be approximated by Wrad = Wave ≈̂ arCo 1 r2 cos4 𝜃 The radiated power density is symmetrical with respect to 𝜙, and it exists only in the upper hemisphere (0 ≤𝜃≤𝜋∕2, 0 ≤𝜙≤2𝜋); Co is a constant.
Determine the following: (a) Power radiated by the antenna (in watts).
(b) Maximum directivity of the antenna (dimensionless and in dB) (c) Direction (in degrees) along which the maximum directivity occurs.
(d) Maximum effective area (in m2) at 1 GHz.
(e) Maximum power (in watts) received by the antenna, assuming no losses, at 1 GHz when the antenna is used as a receiver and the incident power density is 10 mwatts/m2.
2.83.
The antenna used at a base station is a λ∕2 dipole which has a maximum directivity of 2.286 dB. The power radiated by the λ∕2 dipole is 10 watts. Assume an operating frequency of 1,900 MHz: (a) Determine the maximum power density (in watts/cm2) radiated by the λ∕2 dipole at a distance of 1,000 meters.
(b) Assuming the receving antenna used in a mobile unit, such as an automobile, is a λ∕4 monopole, with a maximum directivity of 5.286 dB, determine the maximum effective area (in cm2) of the monopole antenna.
(c) If the receiving mobile unit, the automobile, is at a distance of 1,000 meters from the base station, what is the maximum power (in watts) that can be received by the antenna (λ/4 monopole), which is mounted on the top of the automobile, and delivered to a matched load/receiver. Assume the antennas are polarization matched, and that there are no losses of any kind in both antennas, including no matching/reflection losses.
2.84.
For an X-band (8.2–12.4 GHz) rectangular horn, with aperture dimensions of 5.5 cm and 7.4 cm, find its maximum effective aperture (in cm2) when its gain (over isotropic) is (a) 14.8 dB at 8.2 GHz (b) 16.5 dB at 10.3 GHz (c) 18.0 dB at 12.4 GHz 2.85.
A base station is installed near your neighborhood. One of the concerns of the residents liv-ing nearby is the exposure to electromagnetic radiation. The input power inside the transmis-sion line feeding the base station antenna is 100 Watts while the omnidirectional radiation amplitude pattern of the base station antenna can be approximated by U(𝜃, 𝜙) = Bo sin(𝜃) 0 ≤𝜃≤180◦, 0 ≤𝜙≤360◦ PROBLEMS 123 where Bo is a constant. The characteristic impedance of the transmission line feeding the base station antenna is 75 ohms while the input impedance of the base station antenna is 100 ohms. The radiation (conduction/dielectric) efficiency of the base station antenna is 50%.
Determine the: (a) Reflection/mismatch efficiency of the antenna (in %) (b) Total efficiency (in %) of the antenna (c) Value of Bo. Must do the integration in closed form and show the details.
(d) Maximum exact directivity (dimensionless and in dB) (e) Maximum power density (in Watts/cm2) at a distance of 1,000 meters. This may repre-sent the distance from the base station to your house.
2.86.
For Problem 2.61 compute the (a) maximum effective area (in λ2) using the computer program Directivity of this chapter.
Compare with that computed using the equation in Table 12.1.
(b) aperture efficiencies of part (a). Are they smaller or larger than unity and why?
2.87.
Repeat Problem 2.86 for Problem 2.62.
2.88.
Repeat Problem 2.86 for Problem 2.63.
2.89.
Repeat Problem 2.86 for Problem 2.64. Compare with those in Table 12.2.
2.90.
Repeat Problem 2.86 for Problem 2.65. Compare with those in Table 12.2.
2.91.
Repeat Problem 2.86 for Problem 2.66. Compare with those in Table 12.2.
2.92.
A 30-dB, right-circularly polarized antenna in a radio link radiates (in the negative z-direction) 5 W of power at 2 GHz. The receiving antenna has an impedance mismatch at its terminals, which leads to a VSWR of 2. The receiving antenna is about 95% effi-cient and has a field pattern near the beam maximum (in the positive z-direction) given by Er = (2̂ ax + ĵ ay)Fr(𝜃, 𝜙). The distance between the two antennas is 4,000 km, and the receiving antenna is required to deliver 10−14 W to the receiver. Determine the maximum effective aperture of the receiving antenna.
2.93.
The radiation intensity of an antenna can be approximated by U(𝜃, 𝜙) = { cos4(𝜃) 0◦≤𝜃< 90◦ 0 90◦≤𝜃≤180◦ with 0◦≤𝜙≤360◦ Determine the maximum effective aperture (in m2) of the antenna if its frequency of oper-ation is f = 10 GHz.
2.94.
A communication satellite is in stationary (synchronous) orbit about the earth (assume altitude of 22,300 statute miles). Its transmitter generates 8.0 W. Assume the transmit-ting antenna is isotropic. Its signal is received by the 210-ft diameter tracking paraboloidal antenna on the earth at the NASA tracking station at Goldstone, California. Also assume no resistive losses in either antenna, perfect polarization match, and perfect impedance match at both antennas. At a frequency of 2 GHz, determine the: (a) power density (in watts/m2) incident on the receiving antenna.
(b) power received by the ground-based antenna whose gain is 60 dB.
2.95.
A lossless (ecd = 1) antenna is operating at 100 MHz and its maximum effective aperture is 0.7162 m2 at this frequency. The input impedance of this antenna is 75 ohms, and it is 124 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS attached to a 50-ohm transmission line. Find the directivity (dimensionless) of this antenna if it is polarization-matched.
2.96.
A resonant, lossless (ecd = 1.0) half-wavelength dipole antenna, having a directivity of 2.156 dB, has an input impedance of 73 ohms and is connected to a lossless, 50 ohms transmission line. A wave, having the same polarization as the antenna, is incident upon the antenna with a power density of 5 W/m2 at a frequency of 10 MHz. Find the received power available at the end of the transmission line.
2.97.
Two X-band (8.2–12.4 GHz) rectangular horns, with aperture dimensions of 5.5 cm and 7.4 cm and each with a gain of 16.3 dB (over isotropic) at 10 GHz, are used as transmitting and receiving antennas. Assuming that the input power is 200 mW, the VSWR of each is 1.1, the conduction-dielectric efficiency is 100%, and the antennas are polarization-matched, find the maximum received power when the horns are separated in air by (a) 5 m (b) 50 m (c) 500 m 2.98.
Transmitting and receiving antennas operating at 1 GHz with gains (over isotropic) of 20 and 15 dB, respectively, are separated by a distance of 1 km. Find the maximum power delivered to the load when the input power is 150 W. Assume that the (a) antennas are polarization-matched (b) transmitting antenna is circularly polarized (either right- or left-hand) and the receiving antenna is linearly polarized.
2.99.
Two lossless, polarization-matched antennas are aligned for maximum radiation between them, and are separated by a distance of 50λ. The antennas are matched to their transmission lines and have directivities of 20 dB. Assuming that the power at the input terminals of the transmitting antenna is 10 W, find the power at the terminals of the receiving antenna.
2.100.
Repeat Problem 2.99 for two antennas with 30 dB directivities and separated by 100λ. The power at the input terminals is 20 W.
2.101.
Transmitting and receiving antennas operating at 1 GHz with gains of 20 and 15 dB, respec-tively, are separated by a distance of 1 km. Find the power delivered to the load when the input power is 150 W. Assume the PLF = 1.
2.102.
A series of microwave repeater links operating at 10 GHz are used to relay television sig-nals into a valley that is surrounded by steep mountain ranges. Each repeater consists of a receiver, transmitter, antennas, and associated equipment. The transmitting and receiving antennas are identical horns, each having gain over isotropic of 15 dB. The repeaters are separated in distance by 10 km. For acceptable signal-to-noise ratio, the power received at each repeater must be greater than 10 nW. Loss due to polarization mismatch is not expected to exceed 3 dB. Assume matched loads and free-space propagation conditions. Determine the minimum transmitter power that should be used.
2.103.
A one-way communication system, operating at 100 MHz, uses two identical λ∕2 vertical, resonant, and lossless dipole antennas as transmitting and receiving elements separated by 10 km. In order for the signal to be detected by the receiver, the power level at the receiver terminals must be at least 1 μW. Each antenna is connected to the transmitter and receiver by a lossless 50-Ω transmission line. Assuming the antennas are polarization-matched and are aligned so that the maximum intensity of one is directed toward the maximum radiation intensity of the other, determine the minimum power that must be generated by the trans-mitter so that the signal will be detected by the receiver. Account for the proper losses from the transmitter to the receiver.
PROBLEMS 125 2.104.
In a long-range microwave communication system operating at 9 GHz, the transmitting and receiving antennas are identical, and they are separated by 10,000 m. To meet the signal-to-noise ratio of the receiver, the received power must be at least 10 μW. Assuming the two antennas are aligned for maximum reception to each other, including being polarization-matched, what should the gains (in dB) of the transmitting and receiving antennas be when the input power to the transmitting antenna is 10 W?
2.105.
A mobile wireless communication system operating at 2 GHz utilizes two antennas, one at the base station and the other at the mobile unit, which are separated by 16 kilometers.
The transmitting antenna, at the base station, is circularly-polarized while the receiving antenna, at the mobile station, is linearly polarized. The maximum gain of the transmitting antenna is 20 dB while the gain of the receiving antennas is unknown. The input power to the transmitting antenna is 100 watts and the power received at the receiver, which is connected to the receiving antenna, is 5 nanowatts. Assuming that the two antennas are aligned so that the maximum of one is directed toward the maximum of the other, and also assuming no reflection/mismatch losses at the transmitter or the receiver, what is the maximum gain of the receiving antenna (dimensions and in dB)?
2.106.
A rectangular X-band horn, with aperture dimensions of 5.5 cm and 7.4 cm and a gain of 16.3 dB (over isotropic) at 10 GHz, is used to transmit and receive energy scattered from a perfectly conducting sphere of radius a = 5λ. Find the maximum scattered power delivered to the load when the distance between the horn and the sphere is (a) 200λ (b) 500λ Assume that the input power is 200 mW, and the radar cross section is equal to the geo-metrical cross section.
2.107.
A radar antenna, used for both transmitting and receiving, has a gain of 150 (dimensionless) at its operating frequency of 5 GHz. It transmits 100 kW, and is aligned for maximum direc-tional radiation and reception to a target 1 km away having a radar cross section of 3 m2.
The received signal matches the polarization of the transmitted signal. Find the received power.
2.108.
In an experiment to determine the radar cross section of a Tomahawk cruise missile, a 100 W, 10 GHz signal was transmitted toward the target, and the received power was measured to be –160 dB. The same antenna, whose gain was 80 (dimensionless), was used for both transmitting and receiving. The polarizations of both signals were identical (PLF = 1), and the distance between the antenna and missile was 104 m. What is the radar cross section of the cruise missile?
2.109.
Repeat Problem 2.108 for a radar system with 100 W, 3 GHz transmitted signal, –160 dB received signal, an antenna with a gain of 80 (dimensionless), and separation between the antenna and target of 104 m.
2.110.
The maximum radar cross section of a resonant linear λ∕2 dipole is approximately 0.86λ2.
For a monostatic system (i.e., transmitter and receiver at the same location), find the received power (in W) if the transmitted power is 100 W, the distance of the dipole from the trans-mitting and receiving antennas is 100 m, the gain of the transmitting and receiving antennas is 15 dB each, and the frequency of operation is 3 GHz. Assume a polarization loss factor of −1 dB.
2.111.
The effective antenna temperature of an antenna looking toward zenith is approximately 5 K. Assuming that the temperature of the transmission line (waveguide) is 72◦F, find the 126 FUNDAMENTAL PARAMETERS AND FIGURES-OF-MERIT OF ANTENNAS effective temperature at the receiver terminals when the attenuation of the transmission line is 4 dB/100 ft and its length is (a) 2 ft (b) 100 ft Compare it to a receiver noise temperature of about 54 K.
2.112.
Derive (2-140). Begin with an expression that assumes that the physical temperature and the attenuation of the transmission line are not constant.
CHAPTER3 Radiation Integrals and Auxiliary Potential Functions 3.1 INTRODUCTION In the analysis of radiation problems, the usual procedure is to specify the sources and then require the fields radiated by the sources. This is in contrast to the synthesis problem where the radiated fields are specified, and we are required to determine the sources.
It is a very common practice in the analysis procedure to introduce auxiliary functions, known as vector potentials, which will aid in the solution of the problems. The most common vector potential functions are the A (magnetic vector potential) and F (electric vector potential). Another pair is the Hertz potentials 𝚷e and 𝚷h. Although the electric and magnetic field intensities (E and H) represent physically measurable quantities, among most engineers the potentials are strictly mathematical tools. The introduction of the potentials often simplifies the solution even though it may require determination of additional functions. While it is possible to determine the E and H fields directly from the source-current densities J and M, as shown in Figure 3.1, it is usually much simpler to find the auxiliary potential functions first and then determine the E and H. This two-step procedure is also shown in Figure 3.1.
The one-step procedure, through path 1, relates the E and H fields to J and M by integral relations.
The two-step procedure, through path 2, relates the A and F (or 𝚷e and 𝚷h) potentials to J and M by integral relations. The E and H are then determined simply by differentiating A and F (or 𝚷e and 𝚷h). Although the two-step procedure requires both integration and differentiation, where path 1 requires only integration, the integrands in the two-step procedure are much simpler.
The most difficult operation in the two-step procedure is the integration to determine A and F (or 𝚷e and 𝚷h). Once the vector potentials are known, then E and H can always be determined because any well-behaved function, no matter how complex, can always be differentiated.
The integration required to determine the potential functions is restricted over the bounds of the sources J and M. This will result in the A and F (or 𝚷e and 𝚷h) to be functions of the observation point coordinates; the differentiation to determine E and H must be done in terms of the observation point coordinates. The integration in the one-step procedure also requires that its limits be determined by the bounds of the sources.
The vector Hertz potential 𝚷e is analogous to A and 𝚷h is analogous to F. The functional relation between them is a proportionality constant which is a function of the frequency and the constitu-tive parameters of the medium. In the solution of a problem, only one set, A and F or 𝚷e and 𝚷h, Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 127 128 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS Sources J, M Vector potentials A, F or Πe, Πh Radiated fields E, H Integration path 1 Differentiation path 2 Integration path 2 Figure 3.1 Block diagram for computing fields radiated by electric and magnetic sources.
is required. The author prefers the use of A and F, which will be used throughout the book. The derivation of the functional relations between A and 𝚷e, and F and 𝚷h are assigned at the end of the chapter as problems. (Problems 3.1 and 3.2).
3.2 THE VECTOR POTENTIAL A FOR AN ELECTRIC CURRENT SOURCE J The vector potential A is useful in solving for the EM field generated by a given harmonic electric current J. The magnetic flux B is always solenoidal; that is, 𝛁⋅B = 0. Therefore, it can be repre-sented as the curl of another vector because it obeys the vector identity 𝛁⋅𝛁× A = 0 (3-1) where A is an arbitrary vector. Thus we define BA = 𝜇HA = 𝛁× A (3-2) or HA = 1 𝜇𝛁× A (3-2a) where subscript A indicates the field due to the A potential. Substituting (3-2a) into Maxwell’s curl equation 𝛁× EA = −j𝜔𝜇HA (3-3) reduces it to 𝛁× EA = −j𝜔𝜇HA = −j𝜔𝛁× A (3-4) which can also be written as 𝛁× [EA + j𝜔A] = 0 (3-5) THE VECTOR POTENTIAL A FOR AN ELECTRIC CURRENT SOURCE J 129 From the vector identity 𝛁× (−𝛁𝜙e) = 0 (3-6) and (3-5), it follows that EA + j𝜔A = −𝛁𝜙e (3-7) or EA = −𝛁𝜙e −j𝜔A (3-7a) The scalar function 𝜙e represents an arbitrary electric scalar potential which is a function of position.
Taking the curl of both sides of (3-2) and using the vector identity 𝛁× 𝛁× A = 𝛁(𝛁⋅A) −𝛁2A (3-8) reduces it to 𝛁× (𝜇HA) = 𝛁(𝛁⋅A) −𝛁2A (3-8a) For a homogeneous medium, (3-8a) reduces to 𝜇𝛁× HA = 𝛁(𝛁⋅A) −𝛁2A (3-9) Equating Maxwell’s equation 𝛁× HA = J + j𝜔𝜀EA (3-10) to (3-9) leads to 𝜇J + j𝜔𝜇𝜀EA = 𝛁(𝛁⋅A) −𝛁2A (3-11) Substituting (3-7a) into (3-11) reduces it to 𝛁2A + k2A = −𝜇J + 𝛁(𝛁⋅A) + 𝛁(j𝜔𝜇𝜀𝜙e) = −𝜇J + 𝛁(𝛁⋅A + j𝜔𝜇𝜀𝜙e) (3-12) where k2 = 𝜔2𝜇𝜀.
In (3-2), the curl of A was defined. Now we are at liberty to define the divergence of A, which is independent of its curl. In order to simplify (3-12), let 𝛁⋅A = −j𝜔𝜀𝜇𝜙e ➱𝜙e = − 1 j𝜔𝜇𝜀𝛁⋅A (3-13) which is known as the Lorentz condition. Substituting (3-13) into (3-12) leads to 𝛁2A + k2A = −𝜇J (3-14) 130 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS In addition, (3-7a) reduces to EA = −𝛁𝜙e −j𝜔A = −j𝜔A −j 1 𝜔𝜇𝜀𝛁(𝛁⋅A) (3-15) Once A is known, HA can be found from (3-2a) and EA from (3-15). EA can just as easily be found from Maxwell’s equation (3-10) with J = 0. It will be shown later how to find A in terms of the current density J. It will be a solution to the inhomogeneous Helmholtz equation of (3-14).
3.3 THE VECTOR POTENTIAL F FOR A MAGNETIC CURRENT SOURCE M Although magnetic currents appear to be physically unrealizable, equivalent magnetic currents arise when we use the volume or the surface equivalence theorems. The fields generated by a harmonic magnetic current in a homogeneous region, with J = 0 but M ≠0, must satisfy 𝛁⋅D = 0. Therefore, EF can be expressed as the curl of the vector potential F by EF = −1 𝜀𝛁× F (3-16) Substituting (3-16) into Maxwell’s curl equation 𝛁× HF = j𝜔𝜀EF (3-17) reduces it to 𝛁× (HF + j𝜔F) = 0 (3-18) From the vector identity of (3-6), it follows that HF = −𝛁𝜙m −j𝜔F (3-19) where 𝜙m represents an arbitrary magnetic scalar potential which is a function of position. Taking the curl of (3-16) 𝛁× EF = −1 𝜀𝛁× 𝛁× F = −1 𝜀[𝛁𝛁⋅F −𝛁2F] (3-20) and equating it to Maxwell’s equation 𝛁× EF = −M −j𝜔𝜇HF (3-21) leads to 𝛁2F + j𝜔𝜇𝜀HF = 𝛁𝛁⋅F −𝜀M (3-22) Substituting (3-19) into (3-22) reduces it to 𝛁2F + k2F = −𝜀M + 𝛁(𝛁⋅F) + 𝛁(j𝜔𝜇𝜀𝜙m) (3-23) ELECTRIC AND MAGNETIC FIELDS FOR ELECTRIC (J) AND MAGNETIC (M) CURRENT SOURCES 131 By letting 𝛁⋅F = −j𝜔𝜇𝜀𝜙m ➱𝜙m = − 1 j𝜔𝜇𝜀𝛁⋅F (3-24) reduces (3-23) to 𝛁2F + k2F = −𝜀M (3-25) and (3-19) to HF = −j𝜔F − j 𝜔𝜇𝜀𝛁(𝛁⋅F) (3-26) Once F is known, EF can be found from (3-16) and HF from (3-26) or (3-21) with M = 0. It will be shown later how to find F once M is known. It will be a solution to the inhomogeneous Helmholtz equation of (3-25).
3.4 ELECTRIC AND MAGNETIC FIELDS FOR ELECTRIC (J) AND MAGNETIC (M) CURRENT SOURCES In the previous two sections we have developed equations that can be used to find the electric and magnetic fields generated by an electric current source J and a magnetic current source M. The procedure requires that the auxiliary potential functions A and F generated, respectively, by J and M are found first. In turn, the corresponding electric and magnetic fields are then determined (EA, HA due to A and EF, HF due to F). The total fields are then obtained by the superposition of the individual fields due to A and F (J and M).
In summary form, the procedure that can be used to find the fields is as follows: Summary 1. Specify J and M (electric and magnetic current density sources).
2. a. Find A (due to J) using A = 𝜇 4𝜋∫∫∫ V Je−jkR R dv′ (3-27) which is a solution of the inhomogeneous vector wave equation of (3-14).
b. Find F (due to M) using F = 𝜀 4𝜋∫∫∫ V Me−jkR R dv′ (3-28) 132 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS which is a solution of the inhomogeneous vector wave equation of (3-25). In (3-27) and (3-28), k2 = 𝜔2𝜇𝜀and R is the distance from any point in the source to the observation point. In a latter section, we will demonstrate that (3-27) is a solution to (3-14) as (3-28) is to (3-25).
3. a. Find HA using (3-2a) and EA using (3-15). EA can also be found using Maxwell’s equation of (3-10) with J = 0.
b. Find EF using (3-16) and HF using (3-26). HF can also be found using Maxwell’s equation of (3-21) with M = 0.
4. The total fields are then determined by E = EA + EF = −j𝜔A −j 1 𝜔𝜇𝜀𝛁(𝛁⋅A) −1 𝜀𝛁× F (3-29) or E = EA + EF = 1 j𝜔𝜀𝛁× HA −1 𝜀𝛁× F (3-29a) and H = HA + HF = 1 𝜇𝛁× A −j𝜔F −j 1 𝜔𝜇𝜀𝛁(𝛁⋅F) (3-30) or H = HA + HF = 1 𝜇𝛁× A − 1 j𝜔𝜇𝛁× EF (3-30a) Whether (3-15) or (3-10) is used to find EA and (3-26) or (3-21) to find HF depends largely upon the problem. In many instances one may be more complex than the other or vice versa. In computing fields in the far-zone, it will be easier to use (3-15) for EA and (3-26) for HF because, as it will be shown, the second term in each expression becomes negligible in that region.
3.5 SOLUTION OF THE INHOMOGENEOUS VECTOR POTENTIAL WAVE EQUATION In the previous section we indicated that the solution of the inhomogeneous vector wave equation of (3-14) is (3-27).
To derive it, let us assume that a source with current density Jz, which in the limit is an infinitesimal source, is placed at the origin of a x, y, z coordinate system, as shown in Figure 3.2(a). Since the current density is directed along the z-axis (Jz), only an Az component will exist. Thus we can write (3-14) as 𝛁2Az + k2Az = −𝜇Jz (3-31) SOLUTION OF THE INHOMOGENEOUS VECTOR POTENTIAL WAVE EQUATION 133 Figure 3.2 Coordinate systems for computing fields radiated by sources.
At points removed from the source (Jz = 0), the wave equation reduces to 𝛁2Az + k2Az = 0 (3-32) Since in the limit the source is a point, it requires that Az is not a function of direction (𝜃and 𝜙); in a spherical coordinate system, Az = Az(r) where r is the radial distance. Thus (3-32) can be written as 𝛁2Az(r) + k2Az(r) = 1 r2 𝜕 𝜕r [ r2 𝜕Az(r) 𝜕r ] + k2Az(r) = 0 (3-33) which when expanded reduces to d2Az(r) dr2 + 2 r dAz(r) dr + k2Az(r) = 0 (3-34) The partial derivative has been replaced by the ordinary derivative since Az is only a function of the radial coordinate.
134 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS The differential equation of (3-34) has two independent solutions Az1 = C1 e−jkr r (3-35) Az2 = C2 e+jkr r (3-36) Equation (3-35) represents an outwardly (in the radial direction) traveling wave and (3-36) describes an inwardly traveling wave (assuming an ej𝜔t time variation). For this problem, the source is placed at the origin with the radiated fields traveling in the outward radial direction. Therefore, we choose the solution of (3-35), or Az = Az1 = C1 e−jkr r (3-37) In the static case (𝜔= 0, k = 0), (3-37) simplifies to Az = C1 r (3-38) which is a solution to the wave equation of (3-32), (3-33), or (3-34) when k = 0. Thus at points removed from the source, the time-varying and the static solutions of (3-37) and (3-38) differ only by the e−jkr factor; or the time-varying solution of (3-37) can be obtained by multiplying the static solution of (3-38) by e−jkr.
In the presence of the source (Jz ≠0) and k = 0, the wave equation of (3-31) reduces to 𝛁2Az = −𝜇Jz (3-39) This equation is recognized to be Poisson’s equation whose solution is widely documented. The most familiar equation with Poisson’s form is that relating the scalar electric potential 𝜙to the electric charge density 𝜌. This is given by 𝛁2𝜙= −𝜌 𝜺 (3-40) whose solution is 𝜙= 1 4𝜋𝜀∫∫∫ V 𝜌 r dv′ (3-41) where r is the distance from any point on the charge density to the observation point. Since (3-39) is similar in form to (3-40), its solution is similar to (3-41), or Az = 𝜇 4𝜋∫∫∫ V Jz r dv′ (3-42) SOLUTION OF THE INHOMOGENEOUS VECTOR POTENTIAL WAVE EQUATION 135 Equation (3-42) represents the solution to (3-31) when k = 0 (static case). Using the comparative analogy between (3-37) and (3-38), the time-varying solution of (3-31) can be obtained by multiply-ing the static solution of (3-42) by e−jkr. Thus Az = 𝜇 4𝜋∫∫∫ V Jz e−jkr r dv′ (3-43) which is a solution to (3-31).
If the current densities were in the x- and y-directions (Jx and Jy), the wave equation for each would reduce to 𝛁2Ax + k2Ax = −𝜇Jx (3-44) 𝛁2Ay + k2Ay = −𝜇Jy (3-45) with corresponding solutions similar in form to (3-43), or Ax = 𝜇 4𝜋∫∫∫ V Jx e−jkr r dv′ (3-46) Ay = 𝜇 4𝜋∫∫∫ V Jy e−jkr r dv′ (3-47) The solutions of (3-43), (3-46), and (3-47) allow us to write the solution to the vector wave equa-tion of (3-14) as A = 𝜇 4𝜋∫∫∫ V Je−jkr r dv′ (3-48) If the source is removed from the origin and placed at a position represented by the primed coor-dinates (x′, y′, z′), as shown in Figure 3.2(b), (3-48) can be written as A(x, y, z) = 𝜇 4𝜋∫∫∫ V J(x′, y′, z′)e−jkR R dv′ (3-49) where the primed coordinates represent the source, the unprimed the observation point, and R the distance from any point on the source to the observation point. In a similar fashion we can show that the solution of (3-25) is given by F(x, y, z) = 𝜺 4𝜋∫∫∫ V M(x′, y′, z′)e−jkR R dv′ (3-50) 136 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS If J and M represent linear densities (m−1), (3-49) and (3-50) reduce to surface integrals, or A = 𝜇 4𝜋∫∫ S Js(x′, y′, z′)e−jkR R ds′ (3-51) F = 𝜀 4𝜋∫∫ S Ms(x′, y′, z′)e−jkR R ds′ (3-52) For electric and magnetic currents Ie and Im, (3-51) and (3-52) reduce to line integrals of the form A = 𝜇 4𝜋∫C Ie(x′, y′, z′)e−jkR R dl′ (3-53) F = 𝜺 4𝜋∫C Im(x′, y′, z′)e−jkR R dl′ (3-54) 3.6 FAR-FIELD RADIATION The fields radiated by antennas of finite dimensions are spherical waves. For these radiators, a general solution to the vector wave equation of (3-14) in spherical components, each as a function of r, 𝜃, 𝜙, takes the general form of A = ̂ arAr(r, 𝜃, 𝜙) + ̂ a𝜃A𝜃(r, 𝜃, 𝜙) + ̂ a𝜙A𝜙(r, 𝜃, 𝜙) (3-55) The amplitude variations of r in each component of (3-55) are of the form 1∕rn, n = 1, 2, … , .
Neglecting higher order terms of 1∕rn(1∕rn = 0, n = 2, 3, …) reduces (3-55) to A ≃[̂ arA′ r(𝜃, 𝜙) + ̂ a𝜃A′ 𝜃𝜃, 𝜙) + ̂ a𝜙A′ 𝜙(𝜃, 𝜙)]e−jkr r , r →∞ (3-56) The r variations are separable from those of 𝜃and 𝜙. This will be demonstrated in the chapters that follow by many examples.
Substituting (3-56) into (3-15) reduces it to E = 1 r {−j𝜔e−jkr[̂ ar(0) + ̂ a𝜃A′ 𝜃(𝜃, 𝜙) + ̂ a𝜙A′ 𝜙(𝜃, 𝜙)]} + 1 r2 {⋯} + ⋯ (3-57) The radial E-field component has no 1/r terms, because its contributions from the first and second terms of (3-15) cancel each other.
Similarly, by using (3-56), we can write (3-2a) as H = 1 r { j𝜔 𝜂e−jkr[̂ ar(0) + ̂ a𝜃A′ 𝜙(𝜃, 𝜙) −̂ a𝜙A′ 𝜃(𝜃, 𝜙)] } + 1 r2 {⋯} + ⋯ (3-57a) where 𝜂= √ 𝜇∕𝜺is the intrinsic impedance of the medium.
DUALITY THEOREM 137 Neglecting higher order terms of 1∕rn, the radiated E- and H-fields have only 𝜃and 𝜙components.
They can be expressed as Far-Field Region Er ≃0 E𝜃≃−j𝜔A𝜃 E𝜙≃−j𝜔A𝜙 ⎫ ⎪ ⎬ ⎪ ⎭ ➱ EA ≃−j𝜔A (for the 𝜃and 𝜙components only since Er ≃0) (3-58a) Hr ≃0 H𝜃≃+j𝜔 𝜂A𝜙= − E𝜙 𝜂 H𝜙≃−j𝜔 𝜂A𝜃= +E𝜃 𝜂 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ➱ HA ≃̂ ar 𝜂× EA = −j𝜔 𝜂̂ ar × A (for the 𝜃and 𝜙components only since Hr ≃0) (3-58b) Radial field components exist only for higher order terms of 1∕rn.
In a similar manner, the far-zone fields due to a magnetic source M (potential F) can be written as Far-Field Region Hr ≃0 H𝜃≃−j𝜔F𝜃 H𝜙≃−j𝜔F𝜙 ⎫ ⎪ ⎬ ⎪ ⎭ ➱ HF ≃−j𝜔F (for the 𝜃and 𝜙components only since Hr ≃0) (3-59a) Er ≃0 E𝜃≃−j𝜔𝜂F𝜙= 𝜂H𝜙 E𝜙≃+j𝜔𝜂F𝜃= −𝜂H𝜃 ⎫ ⎪ ⎬ ⎪ ⎭ ➱ EF = −𝜂̂ ar × HF = j𝜔𝜂̂ ar × F (for the 𝜃and 𝜙components only since Er ≃0) (3-59b) Simply stated, the corresponding far-zone E- and H-field components are orthogonal to each other and form TEM (to r) mode fields. This is a very useful relation, and it will be adopted in the chapters that follow for the solution of the far-zone radiated fields. The far-zone (far-field) region for a radiator is defined in Figures 2.7 and 2.8. Its smallest radial distance is 2D2∕λ where D is the largest dimension of the radiator.
3.7 DUALITY THEOREM When two equations that describe the behavior of two different variables are of the same mathe-matical form, their solutions will also be identical. The variables in the two equations that occupy identical positions are known as dual quantities and a solution of one can be formed by a systematic interchange of symbols to the other. This concept is known as the duality theorem.
Comparing Equations (3-2a), (3-3), (3-10), (3-14), and (3-15) to (3-16), (3-17), (3-21), (3-25), and (3-26), respectively, it is evident that they are to each other dual equations and their variables dual quantities. Thus knowing the solutions to one set (i.e., J ≠0, M = 0), the solution to the other 138 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS TABLE 3.1 Dual Equations for Electric (J) and Magnetic (M) Current Sources Electric Sources Magnetic Sources (J ≠0, M = 0) (J = 0, M ≠0) 𝛁× EA = −j𝜔𝜇HA 𝛁× HF = j𝜔𝜺EF 𝛁× HA = J + j𝜔𝜺EA −𝛁× EF = M + j𝜔𝜇HF 𝛁2A + k2A = −𝜇J 𝛁2F + k2F = −𝜺M A = 𝜇 4𝜋∫∫∫ V Je−jkR R dv′ F = 𝜀 4𝜋∫∫∫ V Me−jkR R dv′ HA = 1 𝜇𝛁× A EF = −1 𝜀𝛁× F EA = −j𝜔A HF = −j𝜔F −j 1 𝜔𝜇𝜀𝛁(𝛁⋅A) −j 1 𝜔𝜇𝜀𝛁(𝛁⋅F) set (J = 0, M ≠0) can be formed by a proper interchange of quantities. The dual equations and their dual quantities are listed, respectively in Tables 3.1 and 3.2 for electric and magnetic sources.
Duality only serves as a guide to form mathematical solutions. It can be used in an abstract manner to explain the motion of magnetic charges giving rise to magnetic currents, when compared to their dual quantities of moving electric charges creating electric currents. It must, however, be emphasized that this is purely mathematical in nature since it is known, as of today, that there are no magnetic charges or currents in nature.
3.8 RECIPROCITY AND REACTION THEOREMS We are all well familiar with the reciprocity theorem, as applied to circuits, which states that “in any network composed of linear, bilateral, lumped elements, if one places a constant current (voltage) generator between two nodes (in any branch) and places a voltage (current) meter between any other two nodes (in any other branch), makes observation of the meter reading, then interchanges the locations of the source and the meter, the meter reading will be unchanged” . We want now to discuss the reciprocity theorem as it applies to electromagnetic theory. This is done best by the use of Maxwell’s equations.
TABLE 3.2 Dual Quantities for Electric (J) and Magnetic (M) Current Sources Electric Sources Magnetic Sources (J ≠0, M = 0) (J = 0, M ≠0) EA HF HA −EF J M A F 𝜺 𝜇 𝜇 𝜺 k k 𝜂 1∕𝜂 1∕𝜂 𝜂 RECIPROCITY AND REACTION THEOREMS 139 Let us assume that within a linear and isotropic medium, but not necessarily homogeneous, there exist two sets of sources J1, M1, and J2, M2 which are allowed to radiate simultaneously or individ-ually inside the same medium at the same frequency and produce fields E1, H1 and E2, H2, respec-tively. It can be shown , that the sources and fields satisfy −𝛁⋅(E1 × H2 −E2 × H1) = E1 ⋅J2 + H2 ⋅M1 −E2 ⋅J1 −H1 ⋅M2 (3-60) which is called the Lorentz Reciprocity Theorem in differential form.
Taking a volume integral of both sides of (3-60) and using the divergence theorem on the left side, we can write it as −∯ S (E1 × H2 −E2 × H1) ⋅ds′ = ∫∫∫ V (E1 ⋅J2 + H2 ⋅M1 −E2 ⋅J1 −H1 ⋅M2) dv′ (3-61) which is designated as the Lorentz Reciprocity Theorem in integral form.
For a source-free (J1 = J2 = M1 = M2 = 0) region, (3-60) and (3-61) reduce, respectively, to 𝛁⋅(E1 × H2 −E2 × H1) = 0 (3-62) and ∯ S (E1 × H2 −E2 × H1) ⋅ds′ = 0 (3-63) Equations (3-62) and (3-63) are special cases of the Lorentz Reciprocity Theorem and must be sat-isfied in source-free regions.
As an example of where (3-62) and (3-63) may be applied and what they would represent, consider a section of a waveguide where two different modes exist with fields E1, H1 and E2, H2. For the expressions of the fields for the two modes to be valid, they must satisfy (3-62) and/or (3-63).
Another useful form of (3-61) is to consider that the fields (E1, H1, E2, H2) and the sources (J1, M1, J2, M2) are within a medium that is enclosed by a sphere of infinite radius. Assume that the sources are positioned within a finite region and that the fields are observed in the far field (ide-ally at infinity). Then the left side of (3-61) is equal to zero, or ∯ S (E1 × H2 −E2 × H1) ⋅ds′ = 0 (3-64) which reduces (3-61) to ∫∫∫ V (E1 ⋅J2 + H2 ⋅M1 −E2 ⋅J1 −H1 ⋅M2) dv′ = 0 (3-65) 140 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS Equation (3-65) can also be written as ∫∫∫ V (E1 ⋅J2 −H1 ⋅M2) dv′ = ∫∫∫ V (E2 ⋅J1 −H2 ⋅M1) dv′ (3-66) The reciprocity theorem, as expressed by (3-66), is the most useful form.
A close observation of (3-61) reveals that it does not, in general, represent relations of power because no conjugates appear. The same is true for the special cases represented by (3-63) and (3-66). Each of the integrals in (3-66) can be interpreted as a coupling between a set of fields and a set of sources, which produce another set of fields. This coupling has been defined as Reaction and each of the integrals in (3-66) are denoted by ⟨1, 2⟩= ∫∫∫ V (E1 ⋅J2 −H1 ⋅M2) dv (3-67) ⟨2, 1⟩= ∫∫∫ V (E2 ⋅J1 −H2 ⋅M1) dv (3-68) The relation ⟨1, 2⟩of (3-67) relates the reaction (coupling) of fields (E1, H1), which are pro-duced by sources J1, M1 to sources (J2, M2), which produce fields E2, H2; ⟨2, 1⟩relates the reaction (coupling) of fields (E2, H2) to sources (J1, M1). For reciprocity to hold, it requires that the reaction (coupling) of one set of sources with the corresponding fields of another set of sources must be equal to the reaction (coupling) of the second set of sources with the corresponding fields of the first set of sources, and vice versa. In equation form, it is written as ⟨1, 2⟩= ⟨2, 1⟩ (3-69) 3.8.1 Reciprocity for Two Antennas There are many applications of the reciprocity theorem. To demonstrate its potential, an antenna example will be considered. Two antennas, whose input impedances are Z1 and Z2, are separated by a linear and isotropic (but not necessarily homogeneous) medium, as shown in Figure 3.3. One antenna (#1) is used as a transmitter and the other (#2) as a receiver. The equivalent network of each antenna is given in Figure 3.4. The internal impedance of the generator Zg is assumed to be the conjugate of the impedance of antenna #1 (Zg = Z∗ 1 = R1 −jX1) while the load impedance ZL is equal to the conjugate of the impedance of antenna #2 (ZL = Z∗ 2 = R2 −jX2). These assumptions are made only for convenience.
Figure 3.3 Transmitting and receiving antenna systems.
RECIPROCITY AND REACTION THEOREMS 141 Figure 3.4 Two-antenna system with conjugate loads.
The power delivered by the generator to antenna #1 is given by (2-83) or P1 = 1 2Re[V1I∗ 1] = 1 2Re [( VgZ1 Z1 + Zg ) V∗ g (Z1 + Zg)∗ ] = |Vg|2 8R1 (3-70) If the transfer admittance of the combined network consisting of the generator impedance, antennas, and load impedance is Y21, the current through the load is VgY21 and the power delivered to the load is P2 = 1 2Re[Z2(VgY21)(VgY21)∗] = 1 2R2|Vg|2|Y21|2 (3-71) The ratio of (3-71) to (3-70) is P2 P1 = 4R1R2|Y21|2 (3-72) In a similar manner, we can show that when antenna #2 is transmitting and #1 is receiving, the power ratio of P1∕P2 is given by P1 P2 = 4R2R1|Y12|2 (3-73) Under conditions of reciprocity (Y12 = Y21), the power delivered in either direction is the same.
3.8.2 Reciprocity for Antenna Radiation Patterns The radiation pattern is a very important antenna characteristic. Although it is usually most conve-nient and practical to measure the pattern in the receiving mode, it is identical, because of reciprocity, to that of the transmitting mode.
Reciprocity for antenna patterns is general provided the materials used for the antennas and feeds, and the media of wave propagation are linear. Nonlinear devices, such as diodes, can make the antenna system nonreciprocal. The antennas can be of any shape or size, and they do not have to be matched to their corresponding feed lines or loads provided there is a distinct single propagating mode at each port. The only other restriction for reciprocity to hold is for the antennas in the transmit and receive modes to be polarization matched, including the sense of rotation. This is necessary so that the antennas can transmit and receive the same field components, and thus total power. If the antenna that is used as a probe to measure the fields radiated by the antenna under test is not of the same polarization, then in some situations the transmit and receive patterns can still be the same. For example, if the transmit antenna is circularly polarized and the probe antenna is linearly polarized, 142 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS Figure 3.5 Antenna arrangement for pattern measurements and reciprocity theorem.
then if the linearly polarized probe antenna is used twice and it is oriented one time to measure the 𝜃-component and the other the 𝜙-component, then the sum of the two components can represent the pattern of the circularly polarized antenna in either the transmit or receive modes. During this procedure, the power level and sensitivities must be held constant.
To detail the procedure and foundation of pattern measurements and reciprocity, let us refer to Figures 3.5(a) and (b). The antenna under test is #1 while the probe antenna (#2) is oriented to transmit or receive maximum radiation. The voltages and currents V1, I1 at terminals 1–1 of antenna #1 and V2, I2 at terminals 2–2 of antenna #2 are related by V1 = Z11I1 + Z12I2 V2 = Z21I1 + Z22I2 (3-74) where Z11 = self-impedance of antenna #1 Z22 = self-impedance of antenna #2 Z12, Z21 = mutual impedances between antennas #1 and #2 If a current I1 is applied at the terminals 1–1 and voltage V2 (designated as V2oc) is measured at the open (I2 = 0) terminals of antenna #2, then an equal voltage V1oc will be measured at the open (I1 = 0) terminals of antenna #1 provided the current I2 of antenna #2 is equal to I1. In equation form, we can write Z21 = V2oc I1 | | | |I2=0 (3-75a) Z12 = V1oc I2 | | | |I1=0 (3-75b) If the medium between the two antennas is linear, passive, isotropic, and the waves monochro-matic, then because of reciprocity Z21 = V2oc I1 | | | |I2=0 = V1oc I2 | | | |I1=0 = Z12 (3-76) PROBLEMS 143 If in addition I1 = I2, then V2oc = V1oc (3-77) The above are valid for any position and any configuration of operation between the two antennas.
Reciprocity will now be reviewed for two modes of operation. In one mode, antenna #1 is held stationary while #2 is allowed to move on the surface of a constant radius sphere, as shown in Fig-ure 3.5(a). In the other mode, antenna #2 is maintained stationary while #1 pivots about a point, as shown in Figure 3.5(b).
In the mode of Figure 3.5(a), antenna #1 can be used either as a transmitter or receiver. In the transmitting mode, while antenna #2 is moving on the constant radius sphere surface, the open ter-minal voltage V2oc is measured. In the receiving mode, the open terminal voltage V1oc is recorded.
The three-dimensional plots of V2oc and V1oc, as a function of 𝜃and 𝜙, have been defined in Sec-tion 2.2 as field patterns. Since the three-dimensional graph of V2oc is identical to that of V1oc (due to reciprocity), the transmitting (V2oc) and receiving (V1oc) field patterns are also equal. The same conclusion can be arrived at if antenna #2 is allowed to remain stationary while #1 rotates, as shown in Figure 3.5(b).
The conditions of reciprocity hold whether antenna #1 is used as a transmitter and #2 as a receiver or antenna #2 as a transmitter and #1 as a receiver. In practice, the most convenient mode of operation is that of Figure 3.5(b) with the test antenna used as a receiver. Antenna #2 is usually placed in the far-field of the test antenna (#1), and vice versa, in order that its radiated fields are plane waves in the vicinity of #1.
The receiving mode of operation of Figure 3.5(b) for the test antenna is most widely used to mea-sure antenna patterns because the transmitting equipment is, in most cases, bulky and heavy while the receiver is small and lightweight. In some cases, the receiver is nothing more than a simple diode detector. The transmitting equipment usually consists of sources and amplifiers. To make precise measurements, especially at microwave frequencies, it is necessary to have frequency and power stabilities. Therefore, the equipment must be placed on stable and vibration-free platforms. This can best be accomplished by allowing the transmitting equipment to be held stationary and the receiving equipment to rotate.
An excellent manuscript on test procedures for antenna measurements of amplitude, phase, impedance, polarization, gain, directivity, efficiency, and others has been published by IEEE .
A condensed summary of it is found in , and a review is presented in Chapter 17 of this text.
REFERENCES 1. R. F. Harrington, Time-Harmonic Electromagnetic Fields, McGraw-Hill, New York, 1961.
2. C. A. Balanis, Advanced Engineering Electromagnetics, Second edition, John Wiley & Sons, New York, 2012.
3. P. E. Mayes, personal communication.
4. V. H. Rumsey, “The Reaction Concept in Electromagnetic Theory,” Physical Review, Series 2, Vol. 94, No.
6, June 15, 1954, pp. 1483–1491.
5. IEEE Standard Test Procedures for Antennas, IEEE Std 149–1979, IEEE, Inc., New York, 1979.
6. W. H. Kummer and E. S. Gillespie, “Antenna Measurements–1978,” Proc. IEEE, Vol. 66, No. 4, April 1978, pp. 483–507.
PROBLEMS 3.1.
If He = j𝜔𝜺𝛁× 𝚷e, where 𝚷e is the electric Hertzian potential, show that (a) 𝛁2𝚷e + k2𝚷e = j 1 𝜔𝜺J (b) Ee = k2𝚷e + 𝛁(𝛁⋅𝚷e) (c) 𝚷e = −j 1 𝜔𝜇𝜀A 144 RADIATION INTEGRALS AND AUXILIARY POTENTIAL FUNCTIONS 3.2.
If Eh = −j𝜔𝜇𝛁× 𝚷h, where 𝚷h is the magnetic Hertzian potential, show that (a) 𝛁2𝚷h + k2𝚷h = j 1 𝜔𝜇M (b) Hh = k2𝚷h + 𝛁(𝛁⋅𝚷h) (c) 𝚷h = −j 1 𝜔𝜇𝜀F 3.3.
Verify that (3-35) and (3-36) are solutions to (3-34).
3.4.
Show that (3-42) is a solution to (3-39) and (3-43) is a solution to (3-31).
3.5.
Verify (3-57) and (3-57a).
3.6.
Derive (3-60) and (3-61).
CHAPTER4 Linear Wire Antennas 4.1 INTRODUCTION Wire antennas, linear or curved, are some of the oldest, simplest, cheapest, and in many cases the most versatile for many applications. It should not then come as a surprise to the reader that we begin our analysis of antennas by considering some of the oldest, simplest, and most basic configurations.
Initially we will try to minimize the complexity of the antenna structure and geometry to keep the mathematical details to a minimum.
4.2 INFINITESIMAL DIPOLE An infinitesimal linear wire (l ≪λ) is positioned symmetrically at the origin of the coordinate system and oriented along the z axis, as shown in Figure 4.1(a). Although infinitesimal dipoles are not very practical, they are used to represent capacitor-plate (also referred to as top-hat-loaded) antennas. In addition, they are utilized as building blocks of more complex geometries. The end plates are used to provide capacitive loading in order to maintain the current on the dipole nearly uniform. Since the end plates are assumed to be small, their radiation is usually negligible. The wire, in addition to being very small (l ≪λ), is very thin (a ≪λ). The spatial variation of the current is assumed to be constant and given by I(z′) = ̂ azI0 (4-1) where I0 = constant.
4.2.1 Radiated Fields To find the fields radiated by the current element, the two-step procedure of Figure 3.1 is used. It will be required to determine first A and F and then find the E and H. The functional relation between A and the source J is given by (3-49), (3-51), or (3-53). Similar relations are available for F and M, as given by (3-50), (3-52), and (3-54).
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 145 146 LINEAR WIRE ANTENNAS end-plate Figure 4.1 Geometrical arrangement of an infinitesimal dipole and its associated electric-field components on a spherical surface.
Since the source only carries an electric current Ie, Im and the potential function F are zero. To find A we write A(x, y, z) = 𝜇 4𝜋∫C Ie(x′, y′, z′)e−jkR R dl′ (4-2) where (x, y, z) represent the observation point coordinates, (x′, y′, z′) represent the coordinates of the source, R is the distance from any point on the source to the observation point, and path C is along the length of the source. For the problem of Figure 4.1 Ie(x′, y′, z′) = ̂ azI0 (4-3a) x′ = y′ = z′ = 0 (infinitesimal dipole) (4-3b) INFINITESIMAL DIPOLE 147 R = √ (x −x′)2 + (y −y′)2 + (z −z′)2 = √ x2 + y2 + z2 = r = constant (4-3c) dl′ = dz′ (4-3d) so we can write (4-2) as A(x, y, z) = ̂ az 𝜇I0 4𝜋re−jkr ∫ +l∕2 −l∕2 dz′ = ̂ az 𝜇I0l 4𝜋r e−jkr (4-4) The next step of the procedure is to find HA using (3-2a) and then EA using (3-15) or (3-10) with J = 0. To do this, it is often much simpler to transform (4-4) from rectangular to spherical components and then use (3-2a) and (3-15) or (3-10) in spherical coordinates to find H and E.
The transformation between rectangular and spherical components is given, in matrix form, by (VII-12a) (see Appendix VII) [ Ar A𝜃 A𝜙 ] = [ sin 𝜃cos 𝜙 sin 𝜃sin 𝜙 cos 𝜃 cos 𝜃cos 𝜙 cos 𝜃sin 𝜙 −sin 𝜃 −sin 𝜙 cos 𝜙 0 ] [ Ax Ay Az ] (4-5) For this problem, Ax = Ay = 0, so (4-5) using (4-4) reduces to Ar = Az cos 𝜃= 𝜇I0le−jkr 4𝜋r cos 𝜃 (4-6a) A𝜃= −Az sin 𝜃= −𝜇I0le−jkr 4𝜋r sin 𝜃 (4-6b) A𝜙= 0 (4-6c) Using the symmetry of the problem (no 𝜙variations), (3-2a) can be expanded in spherical coor-dinates and written in simplified form as H = ̂ a𝜙 1 𝜇r [ 𝜕 𝜕r(rA𝜃) −𝜕Ar 𝜕𝜃 ] (4-7) Substituting (4-6a)–(4-6c) into (4-7) reduces it to Hr = H𝜃= 0 H𝜙= jkI0l sin 𝜃 4𝜋r [ 1 + 1 jkr ] e−jkr (4-8a) (4-8b) The electric field E can now be found using (3-15) or (3-10) with J = 0. That is, E = EA = −j𝜔A −j 1 𝜔𝜇𝜀𝛁(𝛁⋅A) = 1 j𝜔𝜀𝛁× H (4-9) 148 LINEAR WIRE ANTENNAS Substituting (4-6a)–(4-6c) or (4-8a)–(4-8b) into (4-9) reduces it to Er = 𝜂I0l cos 𝜃 2𝜋r2 [ 1 + 1 jkr ] e−jkr E𝜃= j𝜂kI0l sin 𝜃 4𝜋r [ 1 + 1 jkr − 1 (kr)2 ] e−jkr E𝜙= 0 (4-10a) (4-10b) (4-10c) The E- and H-field components are valid everywhere, except on the source itself, and they are sketched in Figure 4.1(b) on the surface of a sphere of radius r. It is a straightforward exercise to verify Equations (4-10a)–(4-10c), and this is left as an exercise to the reader (Prob. 4.14).
4.2.2 Power Density and Radiation Resistance The input impedance of an antenna, which consists of real and imaginary parts, was discussed in Section 2.13. For a lossless antenna, the real part of the input impedance was designated as radiation resistance. It is through the mechanism of the radiation resistance that power is transferred from the guided wave to the free-space wave. To find the input resistance for a lossless antenna, the Poynting vector is formed in terms of the E- and H-fields radiated by the antenna. By integrating the Poynting vector over a closed surface (usually a sphere of constant radius), the total power radiated by the source is found. The real part of it is related to the input resistance.
For the infinitesimal dipole, the complex Poynting vector can be written using (4-8a)–(4-8b) and (4-10a)–(4-10c) as W = 1 2(E × H∗) = 1 2(̂ arEr + ̂ a𝜃E𝜃) × (̂ a𝜙H∗ 𝜙) = 1 2(̂ arE𝜃H𝜙 ∗−̂ a𝜃ErH𝜙 ∗) (4-11) whose radial Wr and transverse W𝜃components are given, respectively, by Wr = 𝜂 8 | | | | I0l λ | | | | 2 sin2 𝜃 r2 [ 1 −j 1 (kr)3 ] (4-12a) W𝜃= j𝜂k|I0l|2 cos 𝜃sin 𝜃 16𝜋2r3 [ 1 + 1 (kr)2 ] (4-12b) The complex power moving in the radial direction is obtained by integrating (4-11)– (4-12b) over a closed sphere of radius r. Thus it can be written as P = ∯ S W ⋅ds = ∫ 2𝜋 0 ∫ 𝜋 0 (̂ arWr + ̂ a𝜃W𝜃) ⋅̂ arr2 sin 𝜃d𝜃d𝜙 (4-13) which reduces to P = ∫ 2𝜋 0 ∫ 𝜋 0 Wrr2 sin 𝜃d𝜃d𝜙= 𝜂𝜋 3 | | | | I0l λ | | | | 2 [ 1 −j 1 (kr)3 ] (4-14) INFINITESIMAL DIPOLE 149 The transverse component W𝜃of the power density does not contribute to the integral. Thus (4-14) does not represent the total complex power radiated by the antenna. Since W𝜃, as given by (4-12b), is purely imaginary, it will not contribute to any real radiated power. However, it does contribute to the imaginary (reactive) power which along with the second term of (4-14) can be used to determine the total reactive power of the antenna. The reactive power density, which is most dominant for small values of kr, has both radial and transverse components. It merely changes between outward and inward directions to form a standing wave at a rate of twice per cycle. It also moves in the transverse direction as suggested by (4-12b).
Equation (4-13), which gives the real and imaginary power that is moving outwardly, can also be written as P = 1 2 ∫∫ S E × H∗⋅ds = 𝜂 (𝜋 3 ) | | | | I0l λ | | | | 2 [ 1 −j 1 (kr)3 ] = Prad + j2𝜔( ̃ Wm −̃ We) (4-15) where P = power (in radial direction) Prad = time-average power radiated ̃ Wm = time-average magnetic energy density (in radial direction) ̃ We = time-average electric energy density (in radial direction) 2𝜔( ̃ Wm −̃ We) = time-average imaginary (reactive) power (in radial direction) From (4-14) Prad = 𝜂 (𝜋 3 ) | | | | I0l λ | | | | 2 (4-16) and 2𝜔( ̃ Wm −̃ We) = −𝜂 (𝜋 3 ) | | | | I0l λ | | | | 2 1 (kr)3 (4-17) It is clear from (4-17) that the radial electric energy must be larger than the radial magnetic energy.
For large values of kr (kr ≫1 or r ≫λ), the reactive power diminishes and vanishes when kr = ∞.
Since the antenna radiates its real power through the radiation resistance, for the infinitesimal dipole it is found by equating (4-16) to Prad = 𝜂 (𝜋 3 ) | | | | I0l λ | | | | 2 = 1 2|I0|2Rr (4-18) where Rr is the radiation resistance. Equation (4-18) reduces to Rr = 𝜂 (2𝜋 3 ) ( l λ )2 = 80𝜋2 ( l λ )2 (4-19) for a free-space medium (𝜂≃120𝜋). It should be pointed out that the radiation resistance of (4-19) represents the total radiation resistance since (4-12b) does not contribute to it.
150 LINEAR WIRE ANTENNAS For a wire antenna to be classified as an infinitesimal dipole, its overall length must be very small (usually l ≤λ∕50).
Example 4.1 Find the radiation resistance of an infinitesimal dipole whose overall length is l = λ∕50.
Solution: Using (4-19) Rr = 80𝜋2 ( l λ )2 = 80𝜋2 ( 1 50 )2 = 0.316 ohms Since the radiation resistance of an infinitesimal dipole is about 0.3 ohms, it will present a very large mismatch when connected to practical transmission lines, many of which have characteristic impedances of 50 or 75 ohms. The reflection efficiency (er) and hence the overall efficiency (e0) will be very small.
The reactance of an infinitesimal dipole is capacitive. This can be illustrated by considering the dipole as a flared open-circuited transmission line, as discussed in Section 1.4. Since the input impedance of an open-circuited transmission line a distance l/2 from its open end is given by Zin = −jZc cot (𝛽l∕2), where Zc is its characteristic impedance, it will always be negative (capacitive) for l ≪λ.
4.2.3 Radian Distance and Radian Sphere The E- and H-fields for the infinitesimal dipole, as represented by (4-8a)–(4-8b) and (4-10a)–(4-10c), are valid everywhere (except on the source itself). An inspection of these equations reveals the following: a. At a distance r = λ∕2𝜋(or kr = 1), which is referred to as the radian distance, the magnitude of the first and second terms within the brackets of (4-8b) and (4-10a) is the same. Also at the radian distance the magnitude of all three terms within the brackets of (4-10b) is identical; the only term that contributes to the total field is the second, because the first and third terms cancel each other. This is illustrated in Figure 4.2.
b. At distances less than the radian distance r < λ∕2𝜋(kr < 1), the magnitude of the second term within the brackets of (4-8b) and (4-10a) is greater than the first term and begins to dominate as r ≪λ∕2𝜋. For (4-10b) and r < λ∕2𝜋, the magnitude of the third term within the brackets is greater than the magnitude of the first and second terms while the magnitude of the second term is greater than that of the first one; each of these terms begins to dominate as r ≪λ∕2𝜋.
This is illustrated in Figure 4.2. The region r < λ∕2𝜋(kr < 1) is referred to as the near-field region, and the energy in that region is basically imaginary (stored).
c. At distances greater than the radian distance r > λ∕2𝜋(kr > 1), the first term within the brackets of (4-8b) and (4-10a) is greater than the magnitude of the second term and begins to dominate as r ≫λ∕2𝜋(kr ≫1). For (4-10b) and r > λ∕2𝜋, the first term within the brackets is greater than the magnitude of the second and third terms while the magni-tude of the second term is greater than that of the third; each of these terms begins to dominate as r ≫λ∕2𝜋. This is illustrated in Figure 4.2. The region r > λ∕2𝜋(kr > 1) is referred to as the intermediate-field region while that for r ≫λ∕2𝜋(kr ≫1) is referred to as the far-field region, and the energy in that region is basically real (radi-ated).
INFINITESIMAL DIPOLE 151 10–2 2 5 2 5 10–1 radian distance (r = 1/2 ) Energy basically real (radiated) Energy basically imaginary (stored) 1 10 102 10–2 10–1 1 10 2 5 2 5 102 ⎜1.0/(kr)⎜2 = ⎜ /(2 r)⎜2 ⎜1.0/(kr)⎜ = ⎜ /(2 r)⎜ 1.0 2 5 2 5 2 5 2 5 Magnitude r ( ) λ λ λ π π π Figure 4.2 Magnitude variation, as a function of the radial distance, of the field terms radiated by an infinites-imal dipole.
d. The sphere with radius equal to the radian distance (r = λ∕2𝜋) is referred as the radian sphere, and it defines the region within which the reactive power density is greater than the radiated power density –. For an antenna, the radian sphere represents the volume occupied mainly by the stored energy of the antenna’s electric and magnetic fields. Outside the radian sphere the radiated power density is greater than the reactive power density and begins to dominate as r ≫λ∕2𝜋. Therefore the radian sphere can be used as a reference, and it defines the transition between stored energy pulsating primarily in the ±𝜃direction [represented by (4-12b)] and energy radiating in the radial (r) direction [represented by the first term of (4-12a); the second term represents stored energy pulsating inwardly and outwardly in the radial (r) direction]. Similar behavior, where the power density near the antenna is primarily reactive and far away is primarily real, is exhibited by all antennas, although not exactly at the radian distance.
4.2.4 Near-Field (kr ≪1) Region An inspection of (4-8a)–(4-8b) and (4-10a)–(4-10c) reveals that for kr ≪λ or r ≪λ∕2𝜋they can be reduced in much simpler form and can be approximated by Er ≃−j𝜂I0le−jkr 2𝜋kr3 cos 𝜃 E𝜃≃−j𝜂I0le−jkr 4𝜋kr3 sin 𝜃 E𝜙= Hr = H𝜃= 0 H𝜙≃I0le−jkr 4𝜋r2 sin 𝜃 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ kr ≪1 (4-20a) (4-20b) (4-20c) (4-20d) 152 LINEAR WIRE ANTENNAS The E-field components, Er and E𝜃, are in time-phase but they are in time-phase quadrature with the H-field component H𝜙; therefore there is no time-average power flow associated with them. This is demonstrated by forming the time-average power density as Wav = 1 2Re[E × H∗] = 1 2Re[̂ arE𝜃H∗ 𝜙−̂ a𝜃ErH∗ 𝜙] (4-21) which by using (4-20a)–(4-20d) reduces to Wav = 1 2Re [ −̂ arj𝜂 k | | | | I0l 4𝜋 | | | | 2 sin2 𝜃 r5 + ̂ a𝜃j𝜂 k |I0l|2 8𝜋2 sin 𝜃cos 𝜃 r5 ] = 0 (4-22) The condition of kr ≪1 can be satisfied at moderate distances away from the antenna provided that the frequency of operation is very low. Equations (4-20a) and (4-20b) are similar to those of a static electric dipole and (4-20d) to that of a static current element. Thus we usually refer to (4-20a)– (4-20d) as the quasistationary fields.
4.2.5 Intermediate-Field (kr > 1) Region As the values of kr begin to increase and become greater than unity, the terms that were dominant for kr ≪1 become smaller and eventually vanish. For moderate values of kr the E-field components lose their in-phase condition and approach time-phase quadrature. Since their magnitude is not the same, in general, they form a rotating vector whose extremity traces an ellipse. This is analogous to the polarization problem except that the vector rotates in a plane parallel to the direction of prop-agation and is usually referred to as the cross field. At these intermediate values of kr, the E𝜃and H𝜙components approach time-phase, which is an indication of the formation of time-average power flow in the outward (radial) direction (radiation phenomenon).
As the values of kr become moderate (kr > 1), the field expressions can be approximated again but in a different form. In contrast to the region where kr ≪1, the first term within the brackets in (4-8b) and (4-10a) becomes more dominant and the second term can be neglected. The same is true for (4-10b) where the second and third terms become less dominant than the first. Thus we can write for kr > 1 Er ≃𝜂I0le−jkr 2𝜋r2 cos 𝜃 E𝜃≃j𝜂kI0le−jkr 4𝜋r sin 𝜃 E𝜙= Hr = H𝜃= 0 H𝜙≃jkI0le−jkr 4𝜋r sin 𝜃 ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ kr > 1 (4-23a) (4-23b) (4-23c) (4-23d) The total electric field is given by E = ̂ arEr + ̂ a𝜃E𝜃 (4-24) whose magnitude can be written as |E| = √ |Er|2 + |E𝜃|2 (4-25) INFINITESIMAL DIPOLE 153 4.2.6 Far-Field (kr ≫1) Region Since (4-23a)–(4-23d) are valid only for values of kr > 1 (r > λ), then Er will be smaller than E𝜃 because Er is inversely proportional to r2 where E𝜃is inversely proportional to r. In a region where kr ≫1, (4-23a)–(4-23d) can be simplified and approximated by E𝜃≃j𝜂kI0le−jkr 4𝜋r sin 𝜃 Er ≃E𝜙= Hr = H𝜃= 0 H𝜙≃jkI0le−jkr 4𝜋r sin 𝜃 ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ kr ≫1 (4-26a) (4-26b) (4-26c) The ratio of E𝜃to H𝜙is equal to Zw = E𝜃 H𝜙 ≃𝜂 (4-27) where Zw = wave impedance 𝜂= intrinsic impedance (377 ≃120𝜋ohms for free-space) The E- and H-field components are perpendicular to each other, transverse to the radial direction of propagation, and the r variations are separable from those of 𝜃and 𝜙. The shape of the pattern is not a function of the radial distance r, and the fields form a Transverse ElectroMagnetic (TEM) wave whose wave impedance is equal to the intrinsic impedance of the medium. As it will become even more evident in later chapters, this relationship is applicable in the far-field region of all antennas of finite dimensions. Equations (4-26a)–(4-26c) can also be derived using the procedure outlined and relationships developed in Section 3.6. This is left as an exercise to the reader (Prob. 4.15).
Example 4.2 For an infinitesimal dipole determine and interpret the vector effective length [see Section 2.15, Figure 2.29(a)]. At what incidence angle does the open-circuit maximum voltage occurs at the output terminals of the dipole if the electric-field intensity of the incident wave is 10 mV/m? The length of the dipole is 10 cm.
Solution: Using (4-26a) and the effective length as defined by (2-92), we can write that E𝜃= j𝜂kI0le−jkr 4𝜋r sin 𝜃= −̂ a𝜃j𝜂kI0e−jkr 4𝜋r ⋅(−̂ a𝜃l sin 𝜃) = −̂ a𝜃j𝜂kI0e−jkr 4𝜋r ⋅𝓵e Therefore, the effective length is 𝓵e = −̂ a𝜃l sin 𝜃 154 LINEAR WIRE ANTENNAS whose maximum value occurs when 𝜃= 90◦, and it is equal to l. Therefore, to achieve maximum output the wave must be incident upon the dipole at a normal incidence angle (𝜃= 90◦).
The open-circuit maximum voltage is equal to Voc|max = |Ei ⋅𝓵e|max = |̂ a𝜃10 × 10−3 ⋅(−̂ a𝜃l sin 𝜃)|max = 10 × 10−3l = 10−3 volts 4.2.7 Directivity The real power Prad radiated by the dipole was found in Section 4.2.2, as given by (4-16). The same expression can be obtained by first forming the average power density, using (4-26a)–(4-26c).
That is, Wav = 1 2Re(E × H∗) = ̂ ar 1 2𝜂|E𝜃|2 = ̂ ar 𝜂 2 | | | | kI0l 4𝜋 | | | | 2 sin2 𝜃 r2 (4-28) Integrating (4-28) over a closed sphere of radius r reduces it to (4-16). This is left as an exercise to the reader (Prob. 4.15).
Associated with the average power density of (4-28) is a radiation intensity U which is given by U = r2 Wav = 𝜂 2 (kI0l 4𝜋 )2 sin2 𝜃= r2 2𝜂|E𝜃(r, 𝜃, 𝜙)|2 (4-29) and it conforms with (2-12a). The normalized pattern of (4-29) is shown in Figure 4.3. The maximum value occurs at 𝜃= 𝜋∕2 and it is equal to Umax = 𝜂 2 (kI0l 4𝜋 )2 (4-30) Using (4-16) and (4-30), the directivity reduces to D0 = 4𝜋Umax Prad = 3 2 (4-31) and the maximum effective aperture to Aem = ( λ2 4𝜋 ) D0 = 3λ2 8𝜋 (4-32) The radiation resistance of the dipole can be obtained by the definition of (4-18). Since the radiated power obtained by integrating (4-28) over a closed sphere is the same as that of (4-16), the radiation resistance using it will also be the same as obtained previously and given by (4-19).
Integrating the complex Poynting vector over a closed sphere, as was done in (4-13), results in the power (real and imaginary) directed in the radial direction. Any transverse components of power density, as given by (4-12b), will not be captured by the integration even though they are part of the SMALL DIPOLE 155 a 1 y 1 Dipole antenna Radiation pattern U = sin2 z x θ θ ϕ Figure 4.3 Three-dimensional radiation pattern of infinitesimal dipole.
overall power. Because of this limitation, this method cannot be used to derive the input reactance of the antenna.
The procedure that can be used to derive the far-zone electric and magnetic fields radiated by an antenna, along with some of the most important parameters/figures of merit that are used to describe the performance of an antenna, are summarized in Table 4.1.
4.3 SMALL DIPOLE The creation of the current distribution on a thin wire was discussed in Section 1.4, and it was illustrated with some examples in Figure 1.16. The radiation properties of an infinitesimal dipole, which is usually taken to have a length l ≤λ∕50, were discussed in the previous section. Its current distribution was assumed to be constant. Although a constant current distribution is not realizable (other than top-hat-loaded elements), it is a mathematical quantity that is used to represent actual current distributions of antennas that have been incremented into many small lengths.
A better approximation of the current distribution of wire antennas, whose lengths are usu-ally λ∕50 < l ≤λ∕10, is the triangular variation of Figure 1.16(a). The sinusoidal variations of Figures 1.16(b)–(c) are more accurate representations of the current distribution of any length wire antenna.
The most convenient geometrical arrangement for the analysis of a dipole is usually to have it positioned symmetrically about the origin with its length directed along the z-axis, as shown in Figure 4.4(a). This is not necessary, but it is usually the most convenient. The current distribution of a small dipole (λ∕50 < l ≤λ∕10) is shown in Figure 4.4(b), and it is given by Ie(x′, y′, z′) = ⎧ ⎪ ⎨ ⎪ ⎩ ̂ azI0 ( 1 −2 l z′) , 0 ≤z′ ≤l∕2 ̂ azI0 ( 1 + 2 l z′) , −l∕2 ≤z′ ≤0 (4-33) where I0 = constant.
156 LINEAR WIRE ANTENNAS TABLE 4.1 Summary of Procedure to Determine the Far-Field Radiation Characteristics of an Antenna 1.
Specify electric and/or magnetic current densities J, M [physical or equivalent (see Chapter 3, Figure 3.1)] 2.
Determine vector potential components A𝜃, A𝜙and/or F𝜃, F𝜙using (3-46)–(3-54) in far field 3.
Find far-zone E and H radiated fields (E𝜃, E𝜙; H𝜃, H𝜙) using (3-58a)–(3-58b) 4.
Form either a.
Wrad(r, 𝜃, 𝜙) = Wav(r, 𝜃, 𝜙) = 1 2Re[E × H∗] ≃1 2Re [(̂ a𝜃E𝜃+ ̂ a𝜙E𝜙) × (̂ a𝜃H∗ 𝜃+ ̂ a𝜙H∗ 𝜙)] Wrad(r, 𝜃, 𝜙) = ̂ ar 1 2 [ |E𝜃|2 + |E𝜙|2 𝜂 ] = ̂ ar 1 r2 |f(𝜃, 𝜙)|2 or b. U(𝜃, 𝜙) = r2 Wrad(r, 𝜃, 𝜙) = |f(𝜃, 𝜙)|2 5.
Determine either a. Prad = ∫ 2𝜋 0 ∫ 𝜋 0 Wrad(r, 𝜃, 𝜙)r2 sin 𝜃d𝜃d𝜙 or b. Prad = ∫ 2𝜋 0 ∫ 𝜋 0 U(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 6.
Find directivity using D(𝜃, 𝜙) = U(𝜃, 𝜙) U0 = 4𝜋U(𝜃, 𝜙) Prad D0 = Dmax = D(𝜃, 𝜙)|max = U(𝜃, 𝜙)|max U0 = 4𝜋U(𝜃, 𝜙)|max Prad 7.
Form normalized power amplitude pattern: Pn(𝜃, 𝜙) = U(𝜃, 𝜙) Umax 8.
Determine radiation and input resistance: Rr = 2Prad |I0|2 ; Rin = Rr sin2 (kl 2 ) 9.
Determine maximum effective area Aem = λ2 4𝜋D0 SMALL DIPOLE 157 Figure 4.4 Geometrical arrangement of dipole and current distribution.
Following the procedure established in the previous section, the vector potential of (4-2) can be written using (4-33) as A(x, y, z) = 𝜇 4𝜋 [ ̂ az ∫ 0 −l∕2 I0 ( 1 + 2 l z′) e−jkR R dz′ + ̂ az ∫ l∕2 0 I0 ( 1 −2 l z′) e−jkR R dz′ ] (4-34) Because the overall length of the dipole is very small (usually l ≤λ∕10), the values of R for different values of z′ along the length of the wire (−l∕2 ≤z′ ≤l∕2) are not much different from r. Thus R can be approximated by R ≃r throughout the integration path. The maximum phase error in (4-34) by allowing R = r for λ∕50 < l ≤λ∕10, will be kl∕2 = 𝜋∕10 rad = 18◦for l = λ∕10. Smaller values will occur for the other lengths. As it will be shown in the next section, this amount of phase error is usually considered negligible and has very little effect on the overall radiation characteristics.
158 LINEAR WIRE ANTENNAS Performing the integration, (4-34) reduces to A = ̂ azAz = ̂ az 1 2 [𝜇I0le−jkr 4𝜋r ] (4-35) which is one-half of that obtained in the previous section for the infinitesimal dipole and given by (4-4).
The potential function given by (4-35) becomes a more accurate approximation as kr →∞. This is also the region of most practical interest, and it has been designated as the far-field region. Since the potential function for the triangular distribution is one-half of the corresponding one for the constant (uniform) current distribution, the corresponding fields of the former are one-half of the latter. Thus we can write the E- and H-fields radiated by a small dipole as E𝜃≃j𝜂kI0le−jkr 8𝜋r sin 𝜃 Er ≃E𝜙= Hr = H𝜃= 0 H𝜙≃jkI0le−jkr 8𝜋r sin 𝜃 ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ kr ≫1 (4-36a) (4-36b) (4-36c) with the wave impedance equal, as before, to (4-27).
Since the directivity of an antenna is controlled by the relative shape of the field or power pattern, the directivity, and maximum effective area of this antenna are the same as the ones with the constant current distribution given by (4-31) and (4-32), respectively.
The radiation resistance of the antenna is strongly dependent upon the current distribution. Using the procedure established for the infinitesimal dipole, it can be shown that for the small dipole its radiated power is one-fourth ( 1 4) of (4-18). Thus the radiation resistance reduces to Rr = 2Prad |I0|2 = 20𝜋2 ( l λ )2 (4-37) which is also one-fourth ( 1 4) of that obtained for the infinitesimal dipole as given by (4-19). Their relative patterns (shapes) are the same and are shown in Figure 4.3.
4.4 REGION SEPARATION Before we attempt to solve for the fields radiated by a finite dipole of any length, it would be very desirable to discuss the separation of the space surrounding an antenna into three regions; namely, the reactive near-field, radiating near-field (Fresnel) and the far-field (Fraunhofer) which were intro-duced briefly in Section 2.2.4. This is necessary because for a dipole antenna of any length and any current distribution, it will become increasingly difficult to solve for the fields everywhere. Approx-imations can be made, especially for the far-field (Fraunhofer) region, which is usually the one of most practical interest, to simplify the formulation to yield closed form solutions. The same approx-imations used to simplify the formulation of the fields radiated by a finite dipole are also used to formulate the fields radiated by most practical antennas. So it will be very important to introduce them properly and understand their implications upon the solution.
REGION SEPARATION 159 The difficulties in obtaining closed form solutions that are valid everywhere for any practical antenna stem from the inability to perform the integration of A(x, y, z) = 𝜇 4𝜋∫C Ie(x′, y′, z′)e−jkR R dl′ (4-38) where R = √ (x −x′)2 + (y −y′)2 + (z −z′)2 (4-38a) For a finite dipole with sinusoidal current distribution, the integral of (4-38) can be reduced to a closed form that is valid everywhere! This will be shown in Chapter 8. The length R is defined as the distance from any point on the source to the observation point. The integral of (4-38) was used to solve for the fields of infinitesimal and small dipoles in Sections 4.1 and 4.2. However in the first case (infinitesimal dipole) R = r and in the second case (small dipole) R was approximated by r(R ≃r) because the length of the dipole was restricted to be l ≤λ∕10. The major simplification of (4-38) will be in the approximation of R.
A very thin dipole of finite length l is symmetrically positioned about the origin with its length directed along the z-axis, as shown in Figure 4.5(a). Because the wire is assumed to be very thin (x′ = y′ = 0), we can write (4-38) as R = √ (x −x′)2 + (y −y′)2 + (z −z′)2 = √ x2 + y2 + (z −z′)2 (4-39) which when expanded can be written as R = √ (x2 + y2 + z2) + (−2zz′ + z′2) = √ r2 + (−2rz′ cos 𝜃+ z′2) (4-40) where r2 = x2 + y2 + z2 (4-40a) z = r cos 𝜃 (4-40b) Using the binomial expansion, we can write (4-40) in a series as R = r −z′ cos 𝜃+ 1 r ( z′2 2 sin2 𝜃 ) + 1 r2 ( z′3 2 cos 𝜃sin2 𝜃 ) + ⋯ (4-41) whose higher order terms become less significant provided r ≫z′.
160 LINEAR WIRE ANTENNAS Figure 4.5 Finite dipole geometry and far-field approximations.
4.4.1 Far-Field (Fraunhofer) Region The most convenient simplification of (4-41), other than R ≃r, will be to approximate it by its first two terms, or R ≃r −z′ cos 𝜃 (4-42) The most significant neglected term of (4-41) is the third whose maximum value is 1 r ( z′2 2 sin2 𝜃 ) max = z′2 2r when 𝜃= 𝜋∕2 (4-43) When (4-43) attains its maximum value, the fourth term of (4-41) vanishes because 𝜃= 𝜋∕2. It can be shown that the higher order terms not shown in (4-41) also vanish. Therefore approximating (4-41) by (4-42) introduces a maximum error given by (4-43).
REGION SEPARATION 161 It has been shown by many investigators through numerous examples that for most practical antennas, with overall lengths greater than a wavelength (l > λ), a maximum total phase error of 𝜋∕8 rad (22.5◦) is not very detrimental in the analytical formulations. Using that as a criterion we can write, using (4-43), that the maximum phase error should always be k(z′)2 2r ≤𝜋 8 (4-44) which for −l∕2 ≤z′ ≤l∕2 reduces to r ≥2 ( l2 λ ) (4-45) Equation (4-45) simply states that to maintain the maximum phase error of an antenna equal to or less than 𝜋∕8 rad (22.5◦), the observation distance r must equal or be greater than 2l2∕λ where l is the largest∗dimension of the antenna structure. The usual simplification for the far-field region is to approximate the R in the exponential (e−jkR) of (4-38) by (4-42) and the R in the denominator of (4-38) by R ≃r. These simplifications are designated as the far-field approximations and are usually denoted in the literature as Far-field Approximations R ≃r −z′ cos 𝜃 for phase terms R ≃r for amplitude terms (4-46) provided r satisfies (4-45).
It may be advisable to illustrate the approximation (4-46) geometrically. For R ≃r −z′ cos 𝜃, where 𝜃is the angle measured from the z-axis, the radial vectors R and r must be parallel to each other, as shown in Figure 4.5(b). For any other antenna whose maximum dimension is D, the approx-imation of (4-46) is valid provided the observations are made at a distance r ≥2D2 λ (4-47) For an aperture antenna the maximum dimension is taken to be its diagonal.
For most practical antennas, whose overall length is large compared to the wavelength, these are adequate approximations which have been shown by many investigators through numerous examples to give valid results in pattern predictions. Some discrepancies are evident in regions of low intensity (usually below −25 dB). This is illustrated in Figure 2.9 where the patterns of a paraboloidal antenna for R = ∞and R = 2D2∕λ differ at levels below −25 dB. Allowing R to have a value of R = 4D2∕λ gives better results.
It would seem that the approximation of R in (4-46) for the amplitude is more severe than that for the phase. However a close observation reveals this is not the case. Since the observations are made at a distance where r is very large, any small error in the approximation of the denominator (amplitude) will not make much difference in the answer. However, because of the periodic nature of the phase (repeats every 2𝜋rad), it can be a major fraction of a period. The best way to illustrate it will be to consider an example.
∗Provided the overall length (l) of the antenna is large compared to the wavelength [see IEEE Standard Definitions of Terms for Antennas, IEEE Std (145-1983)].
162 LINEAR WIRE ANTENNAS Example 4.3 For an antenna with an overall length l = 5λ, the observations are made at r = 60λ. Find the errors in phase and amplitude using (4-46).
Solution: For 𝜃= 90◦, z′ = 2.5λ, and r = 60λ, (4-40) reduces to R1 = λ √ (60)2 + (2.5)2 = 60.052λ and (4-46) to R2 = r = 60λ Therefore the phase difference is Δ𝜙= kΔR = 2𝜋 λ (R1 −R2) = 2𝜋(0.052) = 0.327 rad = 18.74◦ which in an appreciable fraction (≃1 20) of a full period (360◦).
The difference of the inverse values of R is 1 R2 −1 R1 = 1 λ ( 1 60 − 1 60.052 ) = 1.44 × 10−5 λ which should always be a very small value in amplitude.
4.4.2 Radiating Near-Field (Fresnel) Region If the observation point is chosen to be smaller than r = 2l2∕λ, the maximum phase error by the approximation of (4-46) is greater than 𝜋∕8 rad (22.5◦) which may be undesirable in many applica-tions. If it is necessary to choose observation distances smaller than (4-45), another term (the third) in the series solution of (4-41) must be retained to maintain a maximum phase error of 𝜋∕8 rad (22.5◦). Doing this, the infinite series of (4-41) can be approximated by R ≃r −z′ cos 𝜃+ 1 r ( z′2 2 sin2 𝜃 ) (4-48) The most significant term that we are neglecting from the infinite series of (4-41) is the fourth. To find the maximum phase error introduced by the omission of the next most significant term, the angle 𝜃at which this occurs must be found. To do this, the neglected term is differentiated with respect to 𝜃and the result is set equal to zero. Thus 𝜕 𝜕𝜃 [ 1 r2 ( z′3 2 cos 𝜃sin2 𝜃 )] = z′3 2r2 sin 𝜃[−sin2 𝜃+ 2 cos2 𝜃] = 0 (4-49) The angle 𝜃= 0 is not chosen as a solution because for that value the fourth term is equal to zero.
In other words, 𝜃= 0 gives the minimum error. The maximum error occurs when the second term of (4-49) vanishes; that is when [−sin2 𝜃+ 2 cos2 𝜃]𝜃=𝜃1 = 0 (4-50) REGION SEPARATION 163 or 𝜃1 = tan−1(± √ 2) (4-50a) If the maximum phase error is allowed to be equal or less than 𝜋∕8 rad, the distance r at which this occurs can be found from kz′3 2r2 cos 𝜃sin2 𝜃 | | | | z′=l∕2 𝜃=tan−1 √ 2 = 𝜋 λ l3 8r2 ( 1 √ 3 ) (2 3 ) = 𝜋 12 √ 3 ( l3 λr2 ) ≤𝜋 8 (4-51) which reduces to r2 ≥ 2 3 √ 3 ( l3 λ ) = 0.385 ( l3 λ ) (4-52) or r ≥0.62 √ l3∕λ (4-52a) A value of r greater than that of (4-52a) will lead to an error less than 𝜋∕8 rad (22.5◦). Thus the region where the first three terms of (4-41) are significant, and the omission of the fourth introduces a maximum phase error of 𝜋∕8 rad (22.5◦), is defined by 2l2∕λ > r ≥0.62 √ l3∕λ (4-53) where l is the length of the antenna. This region is designated as radiating near-field because the radiating power density is greater than the reactive power density and the field pattern (its shape) is a function of the radial distance r. This region is also called the Fresnel region because the field expressions in this region reduce to Fresnel integrals.
The discussion has centered around the finite length antenna of length l with the observation considered to be a point source. If the antenna is not a line source, l in (4-53) must represent the largest dimension of the antenna (which for an aperture is the diagonal). Also if the transmitting antenna has maximum length lt and the receiving antenna has maximum length lr, then the sum of lt and lr must be used in place of l in (4-53).
The boundaries for separating the far-field (Fraunhofer), the radiating near-field (Fresnel), and the reactive near-field regions are not very rigid. Other criteria have also been established but the ones introduced here are the most “popular.” Also the fields, as the boundaries from one region to the other are crossed, do not change abruptly but undergo a very gradual transition.
4.4.3 Reactive Near-Field Region If the distance of observation is smaller than the inner boundary of the Fresnel region, this region is usually designated as reactive near-field with inner and outer boundaries defined by 0.62 √ l3∕λ > r > 0 (4-54) where l is the length of the antenna. In this region the reactive power density predominates, as was demonstrated in Section 4.1 for the infinitesimal dipole.
164 LINEAR WIRE ANTENNAS In summary, the space surrounding an antenna is divided into three regions whose boundaries are determined by reactive near-field [0.62 √ D3∕λ > r > 0] radiating near-field (Fresnel) [2D2∕λ > r ≥0.62 √ D3∕λ] far-field (Fraunhofer) [∞≥r ≥2D2∕λ] (4-55a) (4-55b) (4-55c) where D is the largest dimension of the antenna (D = l for a wire antenna).
4.5 FINITE LENGTH DIPOLE The techniques that were developed previously can also be used to analyze the radiation character-istics of a linear dipole of any length. To reduce the mathematical complexities, it will be assumed in this chapter that the dipole has a negligible diameter (ideally zero). This is a good approximation provided the diameter is considerably smaller than the operating wavelength. Finite radii dipoles will be analyzed in Chapters 8 and 9.
4.5.1 Current Distribution For a very thin dipole (ideally zero diameter), the current distribution can be written, to a good approximation, as Ie(x′ = 0, y′ = 0, z′) = ⎧ ⎪ ⎨ ⎪ ⎩ ̂ azI0 sin [ k ( l 2 −z′)] , 0 ≤z′ ≤l∕2 ̂ azI0 sin [ k ( l 2 + z′)] , −l∕2 ≤z′ ≤0 (4-56) This distribution assumes that the antenna is center-fed and the current vanishes at the end points (z′ = ±l∕2). Experimentally it has been verified that the current in a center-fed wire antenna has sinusoidal form with nulls at the end points. For l = λ∕2 and λ∕2 < l < λ the current distribution of (4-56) is shown plotted in Figures 1.16(b) and 1.12(c), respectively. The geometry of the antenna is that shown in Figure 4.5.
4.5.2 Radiated Fields: Element Factor, Space Factor, and Pattern Multiplication For the current distribution of (4-56) it will be shown in Chapter 8 that closed form expressions for the E- and H-fields can be obtained which are valid in all regions (any observation point except on the source itself). In general, however, this is not the case. Usually we are limited to the far-field region, because of the mathematical complications provided in the integration of the vector potential A of (4-2). Since closed form solutions, which are valid everywhere, cannot be obtained for many antennas, the observations will be restricted to the far-field region. This will be done first in order to illustrate the procedure. In some cases, even in that region it may become impossible to obtain closed form solutions.
The finite dipole antenna of Figure 4.5 is subdivided into a number of infinitesimal dipoles of length Δz′. As the number of subdivisions is increased, each infinitesimal dipole approaches a length dz′. For an infinitesimal dipole of length dz′ positioned along the z-axis at z′, the electric and magnetic FINITE LENGTH DIPOLE 165 field components in the far field are given, using (4-26a)–(4-26c), as dE𝜃≃j𝜂kIe(x′, y′, z′)e−jkR 4𝜋R sin 𝜃dz′ (4-57a) dEr ≃dE𝜙= dHr = dH𝜃= 0 (4-57b) dH𝜙≃jkIe(x′, y′, z′)e−jkR 4𝜋R sin 𝜃dz′ (4-57c) where R is given by (4-39) or (4-40).
Using the far-field approximations given by (4-46), (4-57a) can be written as dE𝜃≃j𝜂kIe(x′, y′, z′)e−jkr 4𝜋r sin 𝜃e+jkz′ cos 𝜃dz′ (4-58) Summing the contributions from all the infinitesimal elements, the summation reduces, in the limit, to an integration. Thus E𝜃= ∫ +l∕2 −l∕2 dE𝜃= j𝜂ke−jkr 4𝜋r sin 𝜃 [ ∫ +l∕2 −l∕2 Ie(x′, y′, z′)ejkz′ cos 𝜃dz′ ] (4-58a) The factor outside the brackets is designated as the element factor and that within the brackets as the space factor. For this antenna, the element factor is equal to the field of a unit length infinitesimal dipole located at a reference point (the origin). In general, the element factor depends on the type of current and its direction of flow while the space factor is a function of the current distribution along the source.
The total field of the antenna is equal to the product of the element and space factors. This is referred to as pattern multiplication for continuously distributed sources (see also Chapter 7), and it can be written as total field = (element factor) × (space factor) (4-59) The pattern multiplication for continuous sources is analogous to the pattern multiplication of (6-5) for discrete-element antennas (arrays).
For the current distribution of (4-56), (4-58a) can be written as E𝜃≃j𝜂kI0e−jkr 4𝜋r sin 𝜃 { ∫ 0 −l∕2 sin [ k ( l 2 + z′)] e+jkz′ cos 𝜃dz′ + ∫ +l∕2 0 sin [ k ( l 2 −z′)] e+jkz′ cos 𝜃dz′ } (4-60) Each one of the integrals in (4-60) can be integrated using ∫e𝛼x sin(𝛽x + 𝛾) dx = e𝛼x 𝛼2 + 𝛽2 [𝛼sin(𝛽x + 𝛾) −𝛽cos(𝛽x + 𝛾)] (4-61) 166 LINEAR WIRE ANTENNAS where 𝛼= ±jk cos 𝜃 (4-61a) 𝛽= ±k (4-61b) 𝛾= kl∕2 (4-61c) After some mathematical manipulations, (4-60) takes the form of E𝜃≃j𝜂I0e−jkr 2𝜋r ⎡ ⎢ ⎢ ⎢ ⎣ cos (kl 2 cos 𝜃 ) −cos (kl 2 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ (4-62a) In a similar manner, or by using the established relationship between the E𝜃and H𝜙in the far field as given by (3-58b) or (4-27), the total H𝜙component can be written as H𝜙≃E𝜃 𝜂≃jI0e−jkr 2𝜋r ⎡ ⎢ ⎢ ⎢ ⎣ cos (kl 2 cos 𝜃 ) −cos (kl 2 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ (4-62b) 4.5.3 Power Density, Radiation Intensity, and Radiation Resistance For the dipole, the average Poynting vector can be written as Wav = 1 2Re[E × H∗] = 1 2Re[̂ a𝜃E𝜃× ̂ a𝜙H∗ 𝜙] = 1 2Re [ ̂ a𝜃E𝜃× ̂ a𝜙 E∗ 𝜃 𝜂 ] Wav = ̂ arWav = ̂ ar 1 2𝜂|E𝜃|2 = ̂ ar𝜂|I0|2 8𝜋2r2 ⎡ ⎢ ⎢ ⎢ ⎣ cos (kl 2 cos 𝜃 ) −cos (kl 2 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 (4-63) and the radiation intensity as U = r2 Wav = 𝜂|I0|2 8𝜋2 ⎡ ⎢ ⎢ ⎢ ⎣ cos (kl 2 cos 𝜃 ) −cos (kl 2 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 (4-64) The normalized (to 0 dB) elevation power patterns, as given by (4-64) for l = λ∕4, λ∕2, 3λ∕4, and λ are shown plotted in Figure 4.6. The current distribution of each is given by (4-56). The power patterns for an infinitesimal dipole l ≪λ (U ∼sin2 𝜃) is also included for comparison. As the length of the antenna increases, the beam becomes narrower. Because of that, the directivity should also FINITE LENGTH DIPOLE 167 Figure 4.6 Elevation plane amplitude patterns for a thin dipole with sinusoidal current distribution (l = λ∕50, λ∕4, λ∕2, 3λ∕4, λ).
increase with length. It is found that the 3-dB beamwidth of each is equal to l ≪λ l = λ∕4 l = λ∕2 l = 3λ∕4 l = λ 3-dB beamwidth = 90◦ 3-dB beamwidth = 87◦ 3-dB beamwidth = 78◦ 3-dB beamwidth = 64◦ 3-dB beamwidth = 47.8◦ (4-65) As the length of the dipole increases beyond one wavelength (l > λ), the number of lobes begin to increase. The normalized power pattern for a dipole with l = 1.25λ is shown in Figure 4.7. In 168 LINEAR WIRE ANTENNAS z Normalized Field Pattern (dB) 0 --–15 –10 –5 + + –25 –20 x y -+ + ---–35 (a) Three-dimensional (b) Two-dimensional –30 θ θ Normalized Field Pattern (dB) –5 0 --–15 –10 + + –25 –20 30° 60° 90° 120° 150° 180° 150° 120° 90° 60° 30° 0° --–30 –35 Figure 4.7 Three- and two-dimensional amplitude patterns for a thin dipole of l = 1.25λ and sinusoidal current distribution.
Figure 4.7(a) the three-dimensional pattern in color is illustrated, while in Figure 4.7(b) the two-dimensional (elevation pattern) in color is depicted. For the three-dimensional illustration, a 90◦ angular section of the pattern has been omitted to illustrate the elevation plane directional pattern variations. The current distribution for the dipoles with l = λ∕4, λ∕2, λ, 3λ∕2, and 2λ, as given by (4-56), is shown in Figure 4.8.
FINITE LENGTH DIPOLE 169 Figure 4.8 Current distributions along the length of a linear wire antenna.
To find the total power radiated, the average Poynting vector of (4-63) is integrated over a sphere of radius r. Thus Prad = ∯ S Wav ⋅ds = ∫ 2𝜋 0 ∫ 𝜋 0 ̂ arWav ⋅̂ arr2 sin 𝜃d𝜃d𝜙 = ∫ 2𝜋 0 ∫ 𝜋 0 Wavr2 sin 𝜃d𝜃d𝜙 (4-66) Using (4-63), we can write (4-66) as Prad = ∫ 2𝜋 0 ∫ 𝜋 0 Wavr2 sin 𝜃d𝜃d𝜙 = 𝜂|I0|2 4𝜋∫ 𝜋 0 [ cos (kl 2 cos 𝜃 ) −cos (kl 2 )]2 sin 𝜃 d𝜃 (4-67) 170 LINEAR WIRE ANTENNAS After some extensive mathematical manipulations, it can be shown that (4-67) reduces to Prad = 𝜂|I0|2 4𝜋{C + ln(kl) −Ci(kl) + 1 2 sin(kl)[Si(2kl) −2Si(kl)] + 1 2 cos(kl)[C + ln(kl∕2) + Ci(2kl) −2Ci(kl)]} (4-68) where C = 0.5772 (Euler’s constant) and Ci(x) and Si(x) are the cosine and sine integrals (see Appendix III) given by Ci(x) = −∫ ∞ x cos y y dy = ∫ x ∞ cos y y dy (4-68a) Si(x) = ∫ x 0 sin y y dy (4-68b) The derivation of (4-68) from (4-67) is assigned as a problem at the end of the chapter (Prob. 4.22).
Ci(x) is related to Cin(x) by Cin(x) = ln(𝛾x) −Ci(x) = ln(𝛾) + ln(x) −Ci(x) = 0.5772 + ln(x) −Ci(x) (4-69) where Cin(x) = ∫ x 0 (1 −cos y y ) dy (4-69a) Ci(x), Si(x) and Cin(x) are tabulated in Appendix III.
The radiation resistance can be obtained using (4-18) and (4-68) and can be written as Rr = 2Prad |I0|2 = 𝜂 2𝜋{C + ln(kl) −Ci(kl) + 1 2 sin(kl) × [Si(2kl) −2Si(kl)] + 1 2 cos(kl) × [C + ln(kl∕2) + Ci(2kl) −2Ci(kl)]} (4-70) Shown in Figure 4.9(a) is a plot of Rr as a function of l (in wavelengths) when the antenna is radiating into free-space (𝜂≃120𝜋).
The imaginary part of the impedance cannot be derived using the same method as the real part because, as was explained in Section 4.2.2, the integration over a closed sphere in (4-13) does not capture the imaginary power contributed by the transverse component W𝜃of the power density.
Therefore, the EMF method is used in Chapter 8 as an alternative approach. Using the EMF method, FINITE LENGTH DIPOLE 171 0.0 0.5 1.0 1.5 (a) Resistance 2.0 2.5 3.0 0 100 200 300 400 500 600 700 800 900 1000 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Rr, Rin (ohms) D0 (dimensionless) Dipole length l (wavelengths) D0 Rr Rin 3.0 2.5 a = 10–5 λ a = 10–2 λ a = 10–3 λ a = 10–4 λ 2.0 1.5 1.0 Dipole length (wavelengths) (b) Reactance (to current maximum) Reactance Xm (ohms) 0.5 0 –600 –400 –200 200 400 600 800 0 Figure 4.9 Radiation resistance and reactance, input resistance and directivity of a thin dipole with sinusoidal current distribution.
the imaginary part of the impedance, relative to the current maximum, is given by (8-57b) or Xm = 𝜂 4𝜋 { 2Si(kl) + cos(kl)[2Si(kl) −Si(2kl)] −sin(kl) [ 2Ci(kl) −Ci(2kl) −Ci ( 2ka2 l )]} (4-70a) An approximate form of (4-57b) for small dipoles is given by (8-59).
Ideally, the radius of the wire does not affect the input resistance, as is indicated by (4-70). How-ever, in practice, it does, although the wire radius is not as significant as it is for the input reactance.
172 LINEAR WIRE ANTENNAS To examine the effect the radius has on the values of the reactance, its values, as given by (4-70a), are plotted in Figure 4.9(b) for a = 10−5λ, 10−4λ, 10−3λ, and 10−2λ. The overall length of the wire is taken to be 0 < l ≤3λ. The same ones are displayed in Figure 8.17, and they are derived in Chapter 8 based on the EMF method. It is apparent that the reactance can be reduced to zero provided that the overall length is slightly less than nλ/2, n = 1, 3, ..., or slightly greater than nλ∕2, n = 2, 4, …. This is often done, in practice, for the l ≃λ/2 because the input resistance is close to 50 ohms, an almost ideal match for the widely used 50-ohm lines. How much smaller than λ/2 should it be reduced depends on the radius of the wire; the thicker the radius, the more there needs to be cut off. Typ-ical dipole lengths for the first resonance range around λ ≃(0.46–0.48)λ. For very small radii, the reactance for l = λ∕2 equals 42.5 ohms.
4.5.4 Directivity As was illustrated in Figure 4.6, the radiation pattern of a dipole becomes more directional as its length increases. When the overall length is greater than one wavelength, the number of lobes increases and the antenna loses its directional properties. The parameter that is used as a “figure of merit” for the directional properties of the antenna is the directivity which was defined in Section 2.6.
The directivity was defined mathematically by (2-22), or D0 = 4𝜋 F(𝜃, 𝜙)|max ∫ 2𝜋 0 ∫ 𝜋 0 F(𝜃, 𝜙) sin 𝜃d𝜃d𝜙 (4-71) where F(𝜃, 𝜙) is related to the radiation intensity U by (2-19), or U = B0F(𝜃, 𝜙) (4-72) From (4-64), the dipole antenna of length l has F(𝜃, 𝜙) = F(𝜃) = ⎡ ⎢ ⎢ ⎢ ⎣ cos (kl 2 cos 𝜃 ) −cos (kl 2 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 (4-73) and B0 = 𝜂|I0|2 8𝜋2 (4-73a) Because the pattern is not a function of 𝜙, (4-71) reduces to D0 = 2F(𝜃)|max ∫ 𝜋 0 F(𝜃) sin 𝜃d𝜃 (4-74) Equation (4-74) can be written, using (4-67), (4-68), and (4-73), as D0 = 2F(𝜃)|max Q (4-75) FINITE LENGTH DIPOLE 173 where Q = {C + ln(kl) −Ci(kl) + 1 2 sin(kl)[Si(2kl) −2Si(kl)] + 1 2 cos(kl)[C + ln(kl∕2) + Ci(2kl) −2Ci(kl)]} (4-75a) The maximum value of F(𝜃) varies and depends upon the length of the dipole.
Values of the directivity, as given by (4-75) and (4-75a), have been obtained for 0 < l ≤3λ and are shown plotted in Figure 4.9. The corresponding values of the maximum effective aperture are related to the directivity by Aem = λ2 4𝜋D0 (4-76) 4.5.5 Input Resistance In Section 2.13 the input impedance was defined as “the ratio of the voltage to current at a pair of terminals or the ratio of the appropriate components of the electric to magnetic fields at a point.” The real part of the input impedance was defined as the input resistance which for a lossless antenna reduces to the radiation resistance, a result of the radiation of real power.
In Section 4.2.2, the radiation resistance of an infinitesimal dipole was derived using the definition of (4-18). The radiation resistance of a dipole of length l with sinusoidal current distribution, of the form given by (4-56), is expressed by (4-70). By this definition, the radiation resistance is referred to the maximum current which for some lengths (l = λ∕4, 3λ∕4, λ, etc.) does not occur at the input terminals of the antenna (see Figure 4.8). To refer the radiation resistance to the input terminals of the antenna, the antenna itself is first assumed to be lossless (RL = 0). Then the power at the input terminals is equated to the power at the current maximum.
Referring to Figure 4.10, we can write |Iin|2 2 Rin = |I0|2 2 Rr (4-77) or Rin = [ I0 Iin ]2 Rr (4-77a) where Rin = radiation resistance at input (feed) terminals Rr = radiation resistance at current maximum Eq. (4-70) I0 = current maximum Iin = current at input terminals For a dipole of length l, the current at the input terminals (Iin) is related to the current maximum (I0) referring to Figure 4.10, by Iin = I0 sin (kl 2 ) (4-78) 174 LINEAR WIRE ANTENNAS Figure 4.10 Current distribution of a linear wire antenna when current maximum does not occur at the input terminals.
Thus the input radiation resistance of (4-77a) can be written as Rin = Rr sin2 (kl 2 ) (4-79) Values of Rin for 0 < l ≤3λ are shown in Figure 4.9(a).
To compute the radiation resistance (in ohms), directivity (dimensionless and in dB), and input resistance (in ohms) for a dipole of length l, a MATLAB and FORTRAN computer program has been developed. The program is based on the definitions of each as given by (4-70), (4-71), and (4-79). The radiated power Prad is computed by numerically integrating (over a closed sphere) the radiation intensity of (4-72)–(4-73a). The program, both in MATLAB and FORTRAN, is included in the publisher’s website for this book. The length of the dipole (in wavelengths) must be inserted as an input.
When the overall length of the antenna is a multiple of λ (i.e., l = nλ, n = 1, 2, 3, …), it is apparent from (4-56) and from Figure 4.8 that Iin = 0. That is, Iin = I0 sin [ k ( l 2 ± z′)]| | | | z′=0 l=nλ,n=0,1,2,… = 0 (4-80) which indicates that the radiation resistance at the input terminals, as given by (4-77a) or (4-79) is infinite. In practice this is not the case because the current distribution does not follow an exact sinu-soidal distribution, especially at the feed point. It has, however, very high values (see Figure 4.11).
Two of the primary factors which contribute to the nonsinusoidal current distribution on an actual wire antenna are the nonzero radius of the wire and finite gap spacing at the terminals.
The radiation resistance and input resistance, as predicted, respectively, by (4-70) and (4-79), are based on the ideal current distribution of (4-56) and do not account for the finite radius of the FINITE LENGTH DIPOLE 175 wire or the gap spacing at the feed. Although the radius of the wire does not strongly influence the resistances, the gap spacing at the feed does play a significant role especially when the current at and near the feed point is small.
4.5.6 Finite Feed Gap To analytically account for a nonzero current at the feed point for antennas with a finite gap at the terminals, Schelkunoff and Friis have changed the current of (4-56) by including a quadrature term in the distribution. The additional term is inserted to take into account the effects of radiation on the antenna current distribution. In other words, once the antenna is excited by the “ideal” current distribution of (4-56), electric and magnetic fields are generated which in turn disturb the “ideal” current distribution. This reaction is included by modifying (4-56) to Ie(x′, y′, z′) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ̂ az { I0 sin [ k ( l 2 −z′)] + jpI0 [ cos(kz′) −cos (k 2l )]} , 0 ≤z′ ≤l∕2 ̂ az { I0 sin [ k ( l 2 + z′)] + jpI0 [ cos(kz′) −cos (k 2l )]} , −l∕2 ≤z′ ≤0 (4-81) where p is a coefficient that is dependent upon the overall length of the antenna and the gap spacing at the terminals. The values of p become smaller as the radius of the wire and the gap decrease.
When l = λ∕2, Ie(x′, y′, z′) = ̂ azI0(1 + jp) cos(kz′) 0 ≤|z′| ≤λ∕4 (4-82) and for l = λ Ie(x′, y′, z′) = { ̂ azI0{sin(kz′) + jp[1 + cos(kz′)]} 0 ≤z′ ≤λ∕2 ̂ azI0{−sin(kz′) + jp[1 + cos(kz′)]} −λ∕2 ≤z′ ≤0 (4-83) Thus for l = λ∕2 the shape of the current is not changed while for l = λ it is modified by the second term which is more dominant for small values of z′. The current distribution based on (4-83) is displayed in Figure 4.11.
The variations of the current distribution and impedances, especially of wire-type antennas, as a function of the radius of the wire and feed gap spacing can be easily taken into account by using advanced computational methods and numerical techniques, especially Integral Equations and Moment Method –, which are introduced in Chapter 8.
To illustrate the point, the current distribution of an l = λ∕2 and l = λ dipole has been computed using an integral equation formulation with a moment method numerical solution, and it is shown in Figure 8.13(b) where it is compared with the ideal distribution of (4-56) and other available data.
For the moment method solution, a gap at the feed has been inserted. As expected and illustrated in Figure 8.13(b), the current distribution for the l = λ∕2 dipole based on (4-56) is not that differ-ent from that based on the moment method. This is also illustrated by (4-82). Therefore the input resistance based on these two methods will not be that different. However, for the l = λ dipole, the current distribution based on (4-56) is quite different, especially at and near the feed point, compared to that based on the moment method, as shown in Figure 8.13(b). This is expected since the current distribution based on the ideal current distribution is zero at the feed point; for practical antennas it is very small. Therefore the gap at the feed plays an important role on the current distribution at and near the feed point. In turn, the values of the input resistance based on the two methods will be quite different, since there is a significant difference in the current between the two methods. This is discussed further in Chapter 8.
176 LINEAR WIRE ANTENNAS Zo = 400 p(Zo = 200) = 0.249 p(Zo = 400) = 0.124 p(l = ) = 199.1 4Zo l/2 l/2 I0|sin(kz')| – for z' ≤ 0 + for z' ≥ 0 I0p[1 + cos(kz')] |Ie| = I0|±sin(kz') + jp [1 + cos(kz')]| Zo = 200 Zo = 400 Zo = 200 I0 | Ie | λ Figure 4.11 Typical current distribution for a l = λ dipole with finite gap based on (4-83).
4.6 HALF-WAVELENGTH DIPOLE One of the most commonly used antennas is the half-wavelength (l = λ∕2) dipole. Because its radi-ation resistance is 73 ohms, which is very near the 50-ohm or 75-ohm characteristic impedances of some transmission lines, its matching to the line is simplified especially at resonance. Because of its wide acceptance in practice, we will examine in a little more detail its radiation characteristics.
The electric and magnetic field components of a half-wavelength dipole can be obtained from (4-62a) and (4-62b) by letting l = λ∕2. Doing this, they reduce to E𝜃≃j𝜂I0e−jkr 2𝜋r ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ (4-84) HALF-WAVELENGTH DIPOLE 177 H𝜙≃jI0e−jkr 2𝜋r ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ (4-85) In turn, the time-average power density and radiation intensity can be written, respectively, as Wav = 𝜂|I0|2 8𝜋2r2 ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 ≃𝜂|I0|2 8𝜋2r2 sin3 𝜃 (4-86) and U = r2 Wav = 𝜂|I0|2 8𝜋2 ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 ≃𝜂|I0|2 8𝜋2 sin3 𝜃 (4-87) whose two-dimensional pattern is shown plotted in Figure 4.6 while the three-dimensional pattern is depicted in Figure 4.12a. For the three-dimensional pattern of Figure 4.12a, a 90◦angular sector has been removed to illustrate the figure-eight elevation plane pattern variations.
The radiation intensity of the λ/2 dipole can be approximated by a sine function with integer exponent of three, as represented in (4-87); that is, U ∼sin3𝜃. Actually, noninteger exponent values, slightly less than three, match the exact pattern even better. This is indicated in Figure 4.12(b) where a sine function, with exponent values of 2.6 and 2.8, is used to plot the normalized pattern of (4-87) and to compare it to that of sin3𝜃. It is apparent that a noninteger exponent of nearly 2.6 for the sine function is a better match to the exact pattern.
The total power radiated can be obtained as a special case of (4-67), or Prad = 𝜂|I0|2 4𝜋∫ 𝜋 0 cos2 (𝜋 2 cos 𝜃 ) sin 𝜃 d𝜃 (4-88) which when integrated reduces, as a special case of (4-68), to Prad = 𝜂|I0|2 8𝜋∫ 2𝜋 0 (1 −cos y y ) dy = 𝜂|I0|2 8𝜋Cin(2𝜋) (4-89) By the definition of Cin(x), as given by (4-69), Cin(2𝜋) is equal to Cin(2𝜋) = 0.5772 + ln(2𝜋) −Ci(2𝜋) = 0.5772 + 1.838 −(−0.02) ≃2.435 (4-90) where Ci(2𝜋) is obtained from the tables in Appendix III.
178 LINEAR WIRE ANTENNAS Normalized Field Pattern (linear scale) 1 z θ 0.7 0.8 0.9 0.4 0.5 0.6 0.1 0.2 0.3 y 1.0 1.0 (a) (b) 0 x ϕ /2 dipole 30° 0° θ θ Relative Power (dB) –10 –20 –30 Exact U~sin3 U~sin2.8 U~sin2.6 30° 60° 60° 90° 90° 120° 120° 150° 150° 180° λ Figure 4.12 Three- and two-dimensional patterns of a λ∕2 dipole (a) three-dimensional pattern of a λ∕2 dipole. (b) comparison of two-dimensional patterns for a λ∕2 dipole.
LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 179 Using (4-87), (4-89), and (4-90), the maximum directivity of the half-wavelength dipole reduces to D0 = 4𝜋Umax Prad = 4𝜋 U|𝜃=𝜋∕2 Prad = 4 Cin(2𝜋) = 4 2.435 ≃1.643 (4-91) The corresponding maximum effective area is equal to Aem = λ2 4𝜋D0 = λ2 4𝜋(1.643) ≃0.13λ2 (4-92) and the radiation resistance, for a free-space medium (𝜂≃120𝜋), is given by Rr = 2Prad |I0|2 = 𝜂 4𝜋Cin(2𝜋) = 30(2.435) ≃73 (4-93) The radiation resistance of (4-93) is also the radiation resistance at the input terminals (input resistance) since the current maximum for a dipole of l = λ∕2 occurs at the input terminals (see Figure 4.8). As it will be shown in Chapter 8, the imaginary part (reactance) associated with the input impedance of a dipole is a function of its length (for l = λ∕2, it is equal to j42.5). Thus the total input impedance for l = λ∕2 is equal to Zin = 73 + j42.5 (4-93a) To reduce the imaginary part of the input impedance to zero, the antenna is matched or reduced in length until the reactance vanishes. The latter is most commonly used in practice for half-wavelength dipoles.
Depending on the radius of the wire, the length of the dipole for first resonance is about l = 0.47λ to 0.48λ; the thinner the wire, the closer the length is to 0.48λ. Thus, for thicker wires, a larger segment of the wire has to be removed from λ∕2 to achieve resonance. The variations of the reactance as a function of the dipole length l, for different wire radii, are displayed in Figures 4.9(b) and 8.17. A summary of the dipole directivity, gain and realized gain are listed in Table 4.2.
4.7 LINEAR ELEMENTS NEAR OR ON INFINITE PERFECT ELECTRIC CONDUCTORS (PEC), PERFECT MAGNETIC CONDUCTORS (PMC) AND ELECTROMAGNETIC BAND-GAP (EBG) SURFACES Thus far we have considered the radiation characteristics of antennas radiating into an unbounded medium. The presence of an obstacle, especially when it is near the radiating element, can signif-icantly alter the overall radiation properties of the antenna system. In practice the most common obstacle that is always present, even in the absence of anything else, is the ground. Any energy from the radiating element directed toward the ground undergoes a reflection. The amount of reflected energy and its direction are controlled by the geometry and constitutive parameters of the ground.
In general, the ground is a lossy medium (𝜎≠0) whose effective conductivity increases with fre-quency. Therefore it should be expected to act as a very good conductor above a certain frequency, depending primarily upon its composition and moisture content. To simplify the analysis, it will first 180 LINEAR WIRE ANTENNAS TABLE 4.2 Summary of Dipole Directivity, Gain and Realized Gain (Resonant XA = 0; f = 100 MHz; 𝝈= 5.7 × 107 S/m; Zc = 50; b = 3 × 10−4𝛌) l = λ∕50 l = λ∕10 l = λ∕2 l = λ Rhf 0.0279 0.2792 0.698 1.3692 RL 0.0279 0.1396 0.349 0.6981 Rr 0.3158 1.9739 73 199 Rin 0.3158 1.9739 73 ∞ ecd 0.9188 0.9339 0.9952 0.9965 (−0.368 dB) (−0.296 dB) (−0.021 dB) (−0.015 dB) D0 1.5 1.5 1.6409 2.411 (1.761 dB) (1.761 dB) (2.151 dB) (3.822 dB) G0 1.3782 1.4009 1.6331 2.4026 (1.393 dB) (1.464 dB) (2.13 dB) (3.807 dB) Γ −0.9863 −0.9189 0.18929 1 er 0.0271 0.1556 0.9642 0 (−15.67 dB) (−8.08 dB) (−0.158 dB) (−∞dB) Gre0 0.0374 0.2181 1.5746 0 (−14.27 dB) (−6.613 dB) (1.972 dB) (−∞dB) be assumed that the ground is a perfect electric conductor, flat, and infinite in extent. The effects of finite conductivity and earth curvature will be incorporated later. The same procedure can also be used to investigate the characteristics of any radiating element near any other infinite, flat, perfect electric conductor. Although infinite structures are not realistic, the developed procedures can be used to simulate very large (electrically) obstacles. The effects that finite dimensions have on the radiation properties of a radiating element can be conveniently accounted for by the use of the Geo-metrical Theory of Diffraction (Chapter 12, Section 12.10) and/or the Moment Method (Chapter 8, Section 8.4).
Magnetic conductors are nonphysical, meaning they do not exist in nature. However, in recent years, techniques have been developed to synthesize and fabricate materials which exhibit interest-ing, attractive, and exciting electromagnetic properties. These properties have captured the attention and imagination of leading engineers and scientists in academia, industry, and government. When such materials are further integrated with electromagnetic devices and interact with electromagnetic waves, they exhibit some unique and intriguing characteristics and phenomena. For example, they can be used to control, advance, and optimize the performance of antennas, microwave components and circuits, transmission lines, scatterers, and optical devices. A brief introduction to engineered synthesized magnetic surfaces, especially as used as ground planes, follows in Section 4.7.1.
4.7.1 Ground Planes: Electric and Magnetic Surfaces that exhibit ideal electric conducting properties, and accordingly satisfy the electromag-netic boundary conditions such that the tangential components of the electric field vanish over their surface, are usually referred to as Perfect Electric Conductors (PEC). Such surfaces exist in nature and metals, with electric conductivities on the order of 107–108, are good approximations for most electrical characteristics, especially in their utilization as ground planes for antenna applications.
The conductivity can be increased even further by applying superconductivity technology .
In comparison, materials that exhibit ideal magnetic conductivities such that the tangential com-ponents of the magnetic field vanish over their surface, although used previously as equivalents in electromagnetic boundary value problems, do not exist in nature and are nonphysical . Yet, in recent years, technologies have been developed to synthesize and fabricate materials which exhibit LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 181 interesting, attractive and exciting electromagnetic properties that have captured the attention and imagination of leading engineers and scientists from academia, industry and government. When such materials are integrated with electromagnetic devices and interact with electromagnetic waves, they exibit some unique and intriguing characteristics and phenomena which can be used, for example, to control, advance, and optimize the performance of antennas, microwave components and circuits, transmission lines, scatterers, and optical devices. Examples as to how the amplitude patterns of a monopole and an aperture are influenced and controlled by such artificially synthesized surfaces are illustrated in .
In general, materials that do not exist in nature, but can be artificially synthesized, are referred to as metamaterials (beyond materials; meta in Greek for beyond/after) . Such synthesized surfaces behave as nearly magnetic conductors only over a limited frequency range, and this limited range is often referred to as band-gap, although technologies are being pursued to advance and extend their frequency range –. There have been many other designations for such materials as well: r AMC (artificial magnetic conductors) r AIS (artificial impedance surfaces) r EES (engineered electromagnetic surfaces) r PBG (photonic band-gap) r EBG (electromagnetic band-gap) r HIS (high impedance surfaces) There are too many different types of synthesized magnetic conductors to list them all here; a number of them are mentioned in , –. An extensive discussion and full list can be found in .
A surface can be synthesized to exhibit nearly magnetic properties by modifying its geometry and/or adding other layers, so that the surface waves and/or the phase of the reflection coefficient of the modified surface can be controlled. Although the magnitude of the reflection coefficient will also be affected, it is the phase that primarily has the largest impact, especially when the surface is used as a ground plane. While an ideal PMC surface introduces, through its image, a zero-phase shift in the reflected field, in contrast to a PEC, which presents a 180◦phase shift, the reflection phase of an EBG surface can, in general, vary from −180◦to +180◦, which makes the EBG more versatile and unique .
One of the first and most widely utilized PMC surfaces is that shown in Figure 4.13. This surface consists of an array of periodic patches of different shapes, in this case, squares, placed above a very thin substrate (which can be air) and connected to the ground plane by posts through vias, if an actual substrate is utilized. The height of the substrate is usually less than a tenth of a wavelength (h < λ/10).
The vias are necessary to suppress surface waves within the substrate. This surface structure is also referred to as EBG and PBG. It is a practical form of engineered textured surfaces or metamaterials.
Because of the directional characteristics of EBG/PBG structures, integration of antenna elements with such structures can have some unique characteristics. A semi-empirical model of the mushroom EBG surface in Figure 4.13 was developed in , ; also reported in . While EBG surfaces exhibit similar characteristics to PMCs when radiating elements are mounted on them, they have the additional ability to suppress surface waves of low-profile antenna designs, such as microstrip arrays. The surface waves introduced in microstrip arrays primarily travel within the substrate and are instrumental in developing coupling between the array elements. This can limit the beam scan-ning capabilities of the microstrip arrays; ultimately, surface waves and coupling may even lead to scan blindness (as discussed in Chapter 14, Section 14.8). When a plane wave is normally incident upon a surface, such as that of Figure 4.13 with a surface impedance Zs, the +90◦to −90◦phase vari-ation is also evident when the magnitude of the surface impedance exceeds the free-space intrinsic impedance 𝜂0 [21, 22]. An EBG surface that does not include the vias does not suppress the surface waves, even though its reflection phase changes between +180◦and −180◦.
182 LINEAR WIRE ANTENNAS Figure 4.13 Geometry of PMC textured surface of square patches . (a) Perspective view. (b) Top view.
(c) Side view.
The performance of the mushroom PMC surface is verified based on the specified and obtained geometrical dimensions. The plane wave normal incidence reflection phase variations of S11 of the mushroom textured surface of square patches of Figure 4.13, between +90◦and −90◦, are shown in Figure 4.14 where they are compared with the results based on the design equations of Section 8.8.4 of . Very good agreement is apparent between the two. The simulated data indicate a bandwidth of 3.9 GHz (fl= 10.35 GHz and fh = 14.25 GHz), compared to the specified one of 4 GHz (fl = 10 GHz and fh = 14 GHz), a center frequency of 12.15 GHz (compared to 12 GHz), and a fractional bandwidth of 0.321 (compared to 0.333). Overall, the performance indicates the design to be very favorable.
4.7.2 Image Theory To analyze the performance of an antenna near an infinite plane conductor, virtual sources (images) will be introduced to account for the reflections. As the name implies, these are not real sources but imaginary ones, which when combined with the real sources, form an equivalent system. For analysis purposes only, the equivalent system gives the same radiated field on and above the conductor as the actual system itself. Below the conductor, the equivalent system does not give the correct field.
However, in this region the field is zero and there is no need for the equivalent.
To begin the discussion, let us assume that a vertical electric dipole is placed a distance h above an infinite, flat, perfect electric conductor as shown in Figure 4.15(a). The arrow indicates the polarity of the source. Energy from the actual source is radiated in all directions in a manner determined by its unbounded medium directional properties. For an observation point P1, there is a direct wave.
In addition, a wave from the actual source radiated toward point R1 of the interface undergoes a reflection. The direction is determined by the law of reflection (𝜃i 1 = 𝜃r 1) which assures that the energy LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 183 Figure 4.14 Phase of reflection coefficient S11 of PMC textured surface with square patches simulated using HFSS and design equations .
in homogeneous media travels in straight lines along the shortest paths. This wave will pass through the observation point P1. By extending its actual path below the interface, it will seem to originate from a virtual source positioned a distance h below the boundary. For another observation point P2 the point of reflection is R2, but the virtual source is the same as before. The same is concluded for all other observation points above the interface.
The amount of reflection is generally determined by the respective constitutive parameters of the media below and above the interface. For a perfect electric conductor below the interface, the incident wave is completely reflected and the field below the boundary is zero. According to the boundary conditions, the tangential components of the electric field must vanish at all points along the interface.
Thus for an incident electric field with vertical polarization shown by the arrows, the polarization of the reflected waves must be as indicated in the figure to satisfy the boundary conditions. To excite the polarization of the reflected waves, the virtual source must also be vertical and with a polarity in the same direction as that of the actual source (thus a reflection coefficient of +1).
Another orientation of the source will be to have the radiating element in a horizontal position, as shown in Figure 4.27. Following a procedure similar to that of the vertical dipole, the virtual source (image) is also placed a distance h below the interface but with a 180◦polarity difference relative to the actual source (thus a reflection coefficient of −1).
In addition to electric sources, nonphysical equivalent “magnetic” sources and magnetic conduc-tors have been introduced to aid in the analyses of electromagnetic boundary-value problems. Fig-ure 4.16(a) displays the sources and their images for an electric plane conductor. The single arrow indicates an electric element and the double a magnetic one. The direction of the arrow identifies the polarity. Since many problems can be solved using duality, Figure 4.16(b) illustrates the sources and their images when the obstacle is an infinite, flat, perfect “magnetic” conductor.
4.7.3 Vertical Electric Dipole The analysis procedure for vertical and horizontal electric and magnetic elements near infinite elec-tric and magnetic plane conductors, using image theory,was illustrated graphically in the previous 184 LINEAR WIRE ANTENNAS Actual source Direct Direct Reflected Virtual source (image) (a) Vertical electric dipole Reflected = ∞ σ θ i 2 θ r 2 θ r 1 θ i 1 P1 P2 h h R2 R1 Direct (b) Field components at point of reflection Reflected = ∞ σ h h n ^ Er2 Er1 E 2 θ E 1 θ 0, 0 μ Figure 4.15 Vertical electric dipole above an infinite, flat, perfect electric conductor.
section. Based on the graphical model of Figure 4.15, the mathematical expressions for the fields of a vertical linear element near a perfect electric conductor will now be developed. For simplicity, only far-field observations will be considered.
Referring to the geometry of Figure 4.15(a), the far-zone direct component of the electric field of the infinitesimal dipole of length l, constant current I0, and observation point P is given according to (4-26a) by Ed 𝜃= j𝜂kI0le−jkr1 4𝜋r1 sin 𝜃1 (4-94) LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 185 Figure 4.16 Electric and magnetic sources and their images near electric (PEC) and magnetic (PMC) con-ductors.
The reflected component can be accounted for by the introduction of the virtual source (image), as shown in Figure 4.14(a), and it can be written as Er 𝜃= jRv𝜂kI0le−jkr2 4𝜋r2 sin 𝜃2 (4-95) or Er 𝜃= j𝜂kI0le−jkr2 4𝜋r2 sin 𝜃2 (4-95a) since the reflection coefficient Rv is equal to unity.
The total field above the interface (z ≥0) is equal to the sum of the direct and reflected components as given by (4-94) and (4-95a). Since a field cannot exist inside a perfect electric conductor, it is equal to zero below the interface. To simplify the expression for the total electric field, it is referred to the origin of the coordinate system (z = 0).
186 LINEAR WIRE ANTENNAS In general, we can write that r1 = [r2 + h2 −2rh cos 𝜃]1∕2 (4-96a) r2 = [r2 + h2 −2rh cos(𝜋−𝜃)]1∕2 (4-96b) For far-field observations (r ≫h), (4-96a) and (4-96b) reduce using the binomial expansion to r1 ≃r −h cos 𝜃 (4-97a) r2 ≃r + h cos 𝜃 (4-97b) As shown in Figure 4.17(b), geometrically (4-97a) and (4-97b) represent parallel lines. Since the amplitude variations are not as critical r1 ≃r2 ≃r for amplitude variations (4-98) Using (4-97a)–(4-98), the sum of (4-94) and (4-95a) can be written as E𝜃≃j𝜂kI0le−jkr 4𝜋r sin 𝜃[2 cos(kh cos 𝜃)] z ≥0 E𝜃= 0 z < 0 ⎫ ⎪ ⎬ ⎪ ⎭ (4-99) It is evident that the total electric field is equal to the product of the field of a single source positioned symmetrically about the origin and a factor [within the brackets in (4-99)] which is a function of the antenna height (h) and the observation angle (𝜃). This is referred to as pattern multiplication and the factor is known as the array factor [see also (6-5)]. This will be developed and discussed in more detail and for more complex configurations in Chapter 6.
The shape and amplitude of the field is not only controlled by the field of the single element but also by the positioning of the element relative to the ground. To examine the field variations as a function of the height h, the normalized (to 0 dB) power patterns for h = 0, λ∕8, λ∕4, 3λ∕8, λ∕2, and λ have been plotted in Figure 4.18. Because of symmetry, only half of each pattern is shown. For h > λ∕4 more minor lobes, in addition to the major ones, are formed. As h attains values greater than λ, an even greater number of minor lobes is introduced. These are shown in Figure 4.19 for h = 2λ and 5λ. The introduction of the additional lobes in Figure 4.19 is usually called scalloping.
In general, the total number of lobes is equal to the integer that is closest to number of lobes ≃2h λ + 1 (4-100) Since the total field of the antenna system is different from that of a single element, the directivity and radiation resistance are also different. To derive expressions for them, we first find the total radiated power over the upper hemisphere of radius r using Prad = ∯ S Wav ⋅ds = 1 2𝜂∫ 2𝜋 0 ∫ 𝜋∕2 0 |E𝜃|2r2 sin 𝜃d𝜃d𝜙 = 𝜋 𝜂∫ 𝜋∕2 0 |E𝜃|2r2 sin 𝜃d𝜃 (4-101) LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 187 z P y x r1 1 θ 2 θ θ ϕ ψ r2 (a) Vertical electric dipole above ground plane r h h σ = ∞ z y x r1 i r2 r (b) Far-field observations h h σ = ∞ θ θ ϕ ψ Figure 4.17 Vertical electric dipole above infinite perfect electric conductor.
which simplifies, with the aid of (4-99), to Prad = 𝜋𝜂 | | | | I0l λ | | | | 2 [ 1 3 −cos(2kh) (2kh)2 + sin(2kh) (2kh)3 ] (4-102) As kh →∞the radiated power, as given by (4-102), is equal to that of an isolated element. How-ever, for kh →0, it can be shown by expanding the sine and cosine functions into series that the power is twice that of an isolated element. Using (4-99), the radiation intensity can be written as U = r2 Wav = r2 ( 1 2𝜂|E𝜃|2 ) = 𝜂 2 | | | | I0l λ | | | | 2 sin2 𝜃cos2(kh cos 𝜃) (4-103) 188 LINEAR WIRE ANTENNAS 0° θ θ 30° 60° 90° 90° 60° 30° h = 0 h = /8 h = /4 h = 3 /8 h = /2 h = 30 20 10 30 10 20 Relative power (dB down) λ λ λ λ λ Figure 4.18 Elevation plane amplitude patterns of a vertical infinitesimal electric dipole for different heights above an infinite perfect electric conductor.
The maximum value of (4-103) occurs at 𝜃= 𝜋∕2 and is given, excluding kh →∞, by Umax = U|𝜃=𝜋∕2 = 𝜂 2 | | | | I0l λ | | | | 2 (4-103a) 0° θ θ 30° 60° 90° 90° 60° 30° h = 2 h = 5 30 20 10 30 10 20 Relative power (dB down) + + + + + + + + + − − − − − − − λ λ Figure 4.19 Elevation plane amplitude patterns of a vertical infinitesimal electric dipole for heights of 2λ and 5λ above an infinite perfect electric conductor.
LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 189 0° θ θ 30° 60° 90° 90° 60° 30° h = 0.4585 30 20 10 20 10 30 Relative power (dB down) + + − − λ Figure 4.20 Elevation plane amplitude pattern of a vertical infinitesimal electric dipole at a height of 0.4585λ above an infinite perfect electric conductor.
which is four times greater than that of an isolated element. With (4-102) and (4-103a), the directivity can be written as D0 = 4𝜋Umax Prad = 2 [ 1 3 −cos(2kh) (2kh)2 + sin(2kh) (2kh)3 ] (4-104) whose value for kh = 0 is 3. The maximum value occurs when kh = 2.881 (h = 0.4585λ), and it is equal to 6.566 which is greater than four times that of an isolated element (1.5). The pattern for h = 0.4585λ is shown plotted in Figure 4.20 while the directivity, as given by (4-104), is displayed in Figure 4.21 for 0 ≤h ≤5λ.
Using (4-102), the radiation resistance can be written as Rr = 2Prad |I0|2 = 2𝜋𝜂 ( l λ )2 [ 1 3 −cos(2kh) (2kh)2 + sin(2kh) (2kh)3 ] (4-105) whose value for kh →∞is the same and for kh = 0 is twice that of the isolated element as given by (4-19). When kh = 0, the value of Rr as given by (4-105) is only one-half the value of an l′ = 2l isolated element according to (4-19). The radiation resistance, as given by (4-105), is plotted in Figure 4.19 for 0 ≤h ≤5λ when l = λ∕50 and the element is radiating into free-space (𝜂≃120𝜋). It can be compared to the value of Rr = 0.316 ohms for the isolated element of Example 4.1.
In practice, a wide use has been made of a quarter-wavelength monopole (l = λ∕4) mounted above a ground plane, and fed by a coaxial line, as shown in Figure 4.22(a). For analysis pur-poses, a λ∕4 image is introduced and it forms the λ∕2 equivalent of Figure 4.22(b). It should be emphasized that the λ∕2 equivalent of Figure 4.22(b) gives the correct field values for the actual system of Figure 4.22(a) only above the interface (z ≥0, 0 ≤𝜃≤𝜋∕2). Thus, the far-zone electric and magnetic fields for the λ∕4 monopole above the ground plane are given, respectively, by (4-84) and (4-85).
190 LINEAR WIRE ANTENNAS Figure 4.21 Directivity and radiation resistance of a vertical infinitesimal electric dipole as a function of its height above an infinite perfect electric conductor.
Figure 4.22 Quarter-wavelength monopole on an infinite perfect electric conductor.
LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 191 σ = 0.01 S/m σ = 0.10 S/m σ σ = 1.00 S/m f = 200 MHz Rm Xm a = 10–5 r = 10.0 σ = 10.0 S/m PEC ( = ∞) No ground 73 42.5 0 20 40 60 80 100 120 0.2 0.4 Height h (wavelengths) Input impedance Zim (ohms) 0.6 0.8 1.0 0 /2 h λ λ Figure 4.23 Input impedance of a vertical λ∕2 dipole above a flat lossy electric conducting surface.
From the discussions of the resistance of an infinitesimal dipole above a ground plane for kh = 0, it follows that the input impedance of a λ∕4 monopole above a ground plane is equal to one-half that of an isolated λ∕2 dipole. Thus, referred to the current maximum, the input impedance Zim is given by Zim (monopole) = 1 2Zim (dipole) = 1 2[73 + j42.5] = 36.5 + j21.25 (4-106) where 73 + j42.5 is the input impedance (and also the impedance referred to the current maximum) of a λ∕2 dipole as given by (4-93a).
The same procedure can be followed for any other length. The input impedance Zim = Rim + jXim (referred to the current maximum) of a vertical λ∕2 dipole placed near a flat lossy electric conductor, as a function of height above the ground plane, is plotted in Figure 4.23, for 0 ≤h ≤λ. Conductivity values considered were 10−2, 10−1, 1, 10 S/m, and infinity (PEC). It is apparent that the conductivity does not strongly influence the impedance values. The conductivity values used are representative of dry to wet earth. It is observed that the values of the resistance and reactance approach, as the height increases, the corresponding ones of the isolated element (73 ohms for the resistance and 42.5 ohms for the reactance).
4.7.4 Approximate Formulas for Rapid Calculations and Design Although the input resistance of a dipole of any length can be computed using (4-70) and (4-79), while that of the corresponding monopole using (4-106), very good answers can be obtained using simpler but approximate expressions. Defining G as G = kl∕2 for dipole (4-107a) G = kl for monopole (4-107b) 192 LINEAR WIRE ANTENNAS where l is the total length of each respective element, it has been shown that the input resistance of the dipole and monopole can be computed approximately using 0 < G < 𝜋∕4 (maximum input resistance of dipole is less than 12.337 ohms) Rin (dipole) = 20G2 0 < l < λ∕4 (4-108a) Rin (monopole) = 10G2 0 < l < λ∕8 (4-108b) 𝜋∕4 ≤G < 𝜋∕2 (maximum input resistance of dipole is less than 76.383 ohms) Rin (dipole) = 24.7G2.5 λ∕4 ≤l < λ∕2 (4-109a) Rin (monopole) = 12.35G2.5 λ∕8 ≤l < λ∕4 (4-109b) 𝜋∕2 ≤G < 2 (maximum input resistance of dipole is less than 200.53 ohms) Rin (dipole) = 11.14G4.17 λ∕2 ≤l < 0.6366λ (4-110a) Rin (monopole) = 5.57G4.17 λ∕4 ≤l < 0.3183λ (4-110b) Besides being much simpler in form, these formulas are much more convenient in design (syn-thesis) problems where the input resistance is given and it is desired to determine the length of the element. These formulas can be verified by plotting the actual resistance versus length on a log–log scale and observe the slope of the line . For example, the slope of the line for values of G up to about 𝜋∕4 ≃0.75 is 2.
Example 4.4 Determine the length of a dipole whose input resistance is 50 ohms. Verify the answer.
Solution: Using (4-109a) 50 = 24.7G2.5 or G = 1.3259 = kl∕2 Therefore l = 0.422λ Using (4-70) and (4-79) Rin for 0.422λ is 45.816 ohms, which closely agrees with the desired value of 50 ohms. To obtain 50 ohms using (4-70) and (4-79), l = 0.4363λ.
4.7.5 Mobile Communication Devices and Antennas for Mobile Communication Systems The cellular era officially began in March 1982 when the FCC (Federal Communication Commis-sion) gave communication carriers the official go-ahead to develop cellular technology. On March LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 193 6, 1983, Motorola officially unveiled the DynaTAC 8000X cellular telephone, which then weighted around 1.75 pounds (0.79 kg) and was nearly 13 inches (33 cm) long, and the race began. Since then, there has been an explosion in the advancement, miniaturization, and utilization of wireless commu-nication devices, especially cell phones, smartphones, tablets and pads. While in 1998 there were worldwide about 200 million of cellular handset units sold, the figure grew to nearly 750 million units in 2006 and to nearly 1,000 million in 2013; over 1 billion in 2015. Many companies played key roles in this evolution; most prominent among them were Motorola, Qualcomm, Nokia, Eric-sson, Apple, Samsung, LG, Huawei, and Lenovo. During this period, these devices provided vast services in the exchange of information, via emails, text messaging, news, stock quotes, weather, traveling maps, GPS, TV, search engines, just to name a few. Antenna technology, the “eyes and ears” of wireless communication systems, led this evolution, starting from the design and utiliza-tion of external radiating elements (primarily monopole and dipole type), such as in the DynaTAC, to embedded elements (primarily planar elements), such as microstrip, IFA, and PIFA employed in almost all of today’s smartphones, tablets, pads, and other mobile units. In this chapter we introduce the basic radiation characteristics of dipoles and monopoles while those of planar elements, such as microstrips and PIFAs, are discussed in Chapter 14.
The dipole and monopole are two of the most widely used antennas for wireless mobile com-munication systems –. An array of dipole elements is extensively used as an antenna at the base station of a land mobile system while the monopole, because of its broadband characteristics and simple construction, is perhaps the most common antenna element for portable equipment, such as cellular telephones, cordless telephones, automobiles, trains, etc. The radiation efficiency and gain characteristics of both of these elements are strongly influenced by their electrical length which is related to the frequency of operation. In a handheld unit, such as a cellular telephone, the posi-tion of the monopole element on the unit influences the pattern while it does not strongly affect the input impedance and resonant frequency. In addition to its use in mobile communication systems, the quarter-wavelength monopole is very popular in many other applications. An alternative to the monopole for the handheld unit is the loop, which is discussed in Chapter 5. Other elements include the inverted F, planar inverted F antenna (PIFA), microstrip (patch), spiral, and others –.
The variation of the input impedance, real and imaginary parts, of a vertical monopole antenna mounted on an experimental unit, simulating a cellular telephone, are shown in Fig-ure 4.24(a,b) –. It is apparent that the first resonance, around 1,000 MHz, is of the series type with slowly varying values of impedance versus frequency, and of desirable magnitude, for practical implementation. For frequencies below the first resonance, the impedance is capacitive (imaginary part is negative), as is typical of linear elements of small lengths (see Figure 4.9); above the first resonance, the impedance is inductive (positive imaginary part). The second resonance, around 1,500 MHz, is of the parallel type (antiresonance) with large and rapid changes in the val-ues of the impedance. These values and variation of impedance are usually undesirable for practical implementation. The order of the types of resonance (series vs. parallel) can be interchanged by choosing another element, such as a loop, as illustrated in Chapter 5, Section 5.8, Figure 5.20 .
The radiation amplitude patterns are those of a typical dipole with intensity in the lower hemisphere.
Examples of monopole type antennas used in cellular and cordless telephones, walkie-talkies, and CB radios are shown in Figure 4.25. The monopoles used in these units are either stationary, as it was in the first cell phone (Motorola DynaTAC) introduced in 1982–1984, or retractable/telescopic.
The length of the retractable/telescopic monopole, such as the one used in the Motorola StarTAC and in others, is varied during operation to improve the radiation characteristics, such as the amplitude pattern and especially the input impedance. During nonusage, the element is usually retracted within the body of the device to prevent it from damage. Units that do not utilize a visible monopole type of antenna, especially in modern smart phones and similar devices like pads and tablets, some of which are shown in Figure 4.25, use embedded/internal type of antenna element. One such embed-ded/internal element that is often used is a Planar Inverted F Antenna (PIFA) , which will be discussed in Chapter 14; there are others. Many of the stationary monopoles are often covered with 194 LINEAR WIRE ANTENNAS 900 800 700 600 500 400 300 200 100 0 500 1000 1500 2000 2500 Frequency (MHz) (a) real part Resistance (Ohms) 3000 3500 4000 4500 5000 8.33 cm 1 cm 6 cm 3 cm 10 cm x y z Centered Offset (1 cm from edge) 400 200 0 −200 −400 −600 −800 −1000 500 1000 1500 2000 2500 Frequency (MHz) (b) imaginary Reactance (Ohms) 3000 3500 4000 4500 5000 8.33 cm 1 cm 6 cm 3 cm 10 cm x y z Centered Offset (1 cm from edge) Figure 4.24 Input impedance, real and imaginary parts, of a vertical monopole mounted on an experimental cellular telephone device.
a dielectric cover. Within the cover, there is typically a straight wire. However, another design that is often used is a helix antenna (see Chapter 10, Section 10.3.1) with a very small circumference and overall length so that the helix operates in the normal mode, whose relative pattern is exhibited in Figure 10.14(a) and which resembles that of a straight-wire monopole. The helix is used, in lieu of a straight wire, because it can be designed to have larger input resistance, which is more attractive for matching to typical feed lines, such as a coaxial line (see Problem 10.20).
LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 195 Figure 4.25 Examples of external and embedded/internal antennas used in commercial cellular and CB radios.
(source: Reproduced with permissions from Motorola, Inc. c ⃝Motorola, Inc.; Nokia; Samsung c ⃝Samsung; Microsoft; HTC; Midland Radio Corporation c ⃝Midland Radio Corporation).
An antenna configuration that is widely used as a base-station antenna for mobile communication and is seen almost everywhere is shown in Figure 4.26. It is a triangular array configuration consisting of twelve dipoles, with four dipoles on each side of the triangle. Each four-element array, on each side of the triangle, is used to cover an angular sector of 120◦, forming what is usually referred to as a sectoral array [see Section 16.3.1(B) and Figure 16.6(a)].
4.7.6 Horizontal Electric Dipole Another dipole configuration is when the linear element is placed horizontally relative to the infinite electric ground plane, as shown in Figure 4.27. The analysis procedure of this is identical to the 196 LINEAR WIRE ANTENNAS Figure 4.26 Triangular array of dipoles used as a sectoral base-station antenna for mobile communication.
Actual source Direct Direct Reflected Virtual source (image) Reflected = ∞ i 2 r 2 r 1 i 1 P1 P2 h h R2 R1 θ θ θ θ σ Figure 4.27 Horizontal electric dipole, and its associated image, above an infinite, flat, perfect electric con-ductor.
LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 197 one of the vertical dipole. Introducing an image and assuming far-field observations, as shown in Figure 4.28(a,b), the direct component can be written as Ed 𝜓= j𝜂kI0le−jkr1 4𝜋r1 sin 𝜓 (4-111) and the reflected one by Er 𝜓= jRh𝜂kI0le−jkr2 4𝜋r2 sin 𝜓 (4-111a) or Er 𝜓= −j𝜂kI0le−jkr2 4𝜋r2 sin 𝜓 (4-111b) since the reflection coefficient is equal to Rh = −1.
To find the angle 𝜓, which is measured from the y-axis toward the observation point, we first form cos 𝜓= ̂ ay ⋅̂ ar = ̂ ay ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = sin 𝜃sin 𝜙 (4-112) from which we find sin 𝜓= √ 1 −cos2 𝜓= √ 1 −sin2 𝜃sin2 𝜙 (4-113) Since for far-field observations r1 ≃r −h cos 𝜃 r2 ≃r + h cos 𝜃 } for phase variations (4-114a) r1 ≃r2 ≃r for amplitude variations (4-114b) the total field, which is valid only above the ground plane (z ≥h; 0 ≤𝜃≤𝜋∕2, 0 ≤𝜙≤2𝜋), can be written as E𝜓= Ed 𝜓+ Er 𝜓= j𝜂kI0le−jkr 4𝜋r √ 1 −sin2 𝜃sin2 𝜙[2j sin(kh cos 𝜃)] (4-115) Equation (4-115) again consists of the product of the field of a single isolated element placed sym-metrically at the origin and a factor (within the brackets) known as the array factor. This again is the pattern multiplication rule of (6-5), which is discussed in more detail in Chapter 6.
The general electric field expression E𝜓of (4-111), when the horizontal dipole is placed at the origin of the coordinate system of Figure 4.28(a), reduces in the principal planes (xy, yz, xz) to the 198 LINEAR WIRE ANTENNAS θ z y x r1 1 2 r2 (a) Horizontal electric dipole above ground plane r h h z y x r1 i r2 r (b) Far-field observations h h θ θ ψ ψ ϕ θ θ ϕ Figure 4.28 Horizontal electric dipole above an infinite perfect electric conductor.
spherical components E𝜃and E𝜙, as follows: xy plane (𝜃= 90◦) : E𝜓= ⎧ ⎪ ⎨ ⎪ ⎩ E𝜃= 0 E𝜙= −j 𝜂kIole−jkr 4𝜋r √ 1 −sin2 𝜃sin2 𝜙 | | | | |𝜃=90◦ = −j𝜔𝜇Iole−jkr 4𝜋r cos 𝜙 (4-116a) yz plane (𝜙= 90◦) : E𝜓= ⎧ ⎪ ⎨ ⎪ ⎩ E𝜃= −j 𝜂kIole−jkr 4𝜋r √ 1 −sin2 𝜃sin2 𝜙 | | | | |𝜙=90◦ = −j𝜔𝜇Iole−jkr 4𝜋r cos 𝜃 E𝜙= 0 (4-116b) LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 199 xz plane (𝜙= 0◦) : E𝜓= ⎧ ⎪ ⎨ ⎪ ⎩ E𝜃= 0 E𝜙= −j𝜔𝜇Iole−jkr 4𝜋r = constant (isotropic) (4-116c) These components match those of Example 4.5 that follows, which are derived based on the vector potential approach. Also, based on (4-116a)–(4-116c), it is easier to decide on the shape of the amplitude pattern and ascertain the polarization of the wave in the three principal planes.
Example 4.5 Using the vector potential A and the procedure outlined in Section 3.6 of Chapter 3, derive the far-zone spherical electric and magnetic field components of a horizontal infinitesimal dipole placed at the origin of the coordinate system of Figure 4.1.
Solution: Using (4-4), but for a horizontal infinitesimal dipole of uniform current directed along the y-axis, the corresponding vector potential can be written as A = ̂ ay 𝜇I0le−jkr 4𝜋r with the corresponding spherical components, using the rectangular to spherical components transformation of (4-5), expressed as A𝜃= Ay cos 𝜃sin 𝜙= 𝜇I0le−jkr 4𝜋r cos 𝜃sin 𝜙 A𝜙= Ay cos 𝜙= 𝜇I0le−jkr 4𝜋r cos 𝜙 Using (3-58a) and (3-58b), we can write the corresponding far-zone electric and magnetic field components as E𝜃≅−j𝜔A𝜃= −j𝜔𝜇I0le−jkr 4𝜋r cos 𝜃sin 𝜙 E𝜙≅−j𝜔A𝜙= −j𝜔𝜇I0le−jkr 4𝜋r cos 𝜙 H𝜃≅− E𝜙 𝜂= j𝜔𝜇I0le−jkr 4𝜋𝜂r cos 𝜙 H𝜙≅+E𝜃 𝜂= +j𝜔𝜇I0le−jkr 4𝜋𝜂r cos 𝜃sin 𝜙 Although the electric-field components, and thus the magnetic field components, take a different analytical form than (4-111), the patterns are the same.
To examine the variations of the total field as a function of the element height above the ground plane, the two-dimensional elevation plane patterns (normalized to 0 dB) for 𝜙= 90◦(y-z plane) when h = 0, λ∕8, λ∕4, 3λ∕8, λ∕2, and λ are plotted in Figure 4.29. Since this antenna system is not symmetric with respect to the z axis, the azimuthal plane (x-y plane) pattern is not isotropic.
200 LINEAR WIRE ANTENNAS 0° 30° 60° 90° 90° 60° 30° 20 10 10 Relative power (dB down) h = 0 (free-space) h = /8 h = /4 h = 3 /8 h = /2 h = 30 20 30 θ θ λ λ λ λ λ Figure 4.29 Elevation plane (𝜙= 90◦) amplitude patterns of a horizontal infinitesimal electric dipole for different heights above an infinite perfect electric conductor.
To obtain a better visualization of the radiation intensity in all directions above the interface, the three-dimensional pattern for h = λ is shown plotted in Figure 4.30. The radial distance on the x-y plane represents the elevation angle 𝜃from 0◦to 90◦, and the z-axis represents the normalized ampli-tude of the radiation field intensity from 0 to 1. The azimuthal angle 𝜙(0 ≤𝜙≤2𝜋) is measured from the x- toward the y-axis on the x-y plane.
As the height increases beyond one wavelength (h > λ), a larger number of lobes is again formed.
This is illustrated in Figure 4.31 for h = 2λ and 5λ. The scalloping effect is evident here, as in Figure 4-19 for the vertical dipole. The total number of lobes is equal to the integer that most closely R l ti A lit d Normalized Field Relative Amplitude 1.0 Normalized Field Pattern (linear scale) 0.9 1 0.6 0.7 0.8 0 3 0.4 0.5 0.1 0.2 0.3 90° 90° x-z plane ( = 0°) y-z plane ( = 90°) 0 x-z plane ( 0 ) ϕ ϕ θ θ Figure 4.30 Three-dimensional amplitude pattern of an infinitesimal horizontal dipole a distance h = λ above an infinite perfect electric conductor.
LINEAR ELEMENTS NEAR OR ON INFINITE PEC, PMC AND EBG SURFACES 201 0° θ θ 30° 60° 90° 90° 60° 30° h = 2 h = 5 30 20 10 10 + − − − − − − − + + + + + Relative power (dB down) 30 20 λ λ Figure 4.31 Elevation plane (𝜙= 90◦) amplitude patterns of a horizontal infinitesimal electric dipole for heights 2λ and 5λ above an infinite perfect electric conductor.
is equal to number of lobes ≃2 (h λ ) (4-117) with unity being the smallest number.
Following a procedure similar to the one performed for the vertical dipole, the radiated power can be written as Prad = 𝜂𝜋 2 | | | | I0l λ | | | | 2 [ 2 3 −sin(2kh) 2kh −cos(2kh) (2kh)2 + sin(2kh) (2kh)3 ] (4-118) and the radiation resistance as Rr = 𝜂𝜋 ( l λ )2 [ 2 3 −sin(2kh) 2kh −cos(2kh) (2kh)2 + sin(2kh) (2kh)3 ] (4-119) By expanding the sine and cosine functions into series, it can be shown that (4-119) reduces for small values of kh to Rr kh→0 = 𝜂𝜋 ( l λ )2 [ 2 3 −2 3 + 8 15 (2𝜋h λ )2] = 𝜂32𝜋3 15 ( l λ )2 (h λ )2 (4-120) For kh →∞, (4-119) reduces to that of an isolated element. The radiation resistance, as given by (4-119), is plotted in Figure 4.32 for 0 ≤h ≤5λ when l = λ∕50 and the antenna is radiating into free-space (𝜂≃120𝜋).
The radiation intensity is given by U ≃r2 2𝜂|E𝜓|2 = 𝜂 2 | | | | I0l λ | | | | 2 (1 −sin2 𝜃sin2 𝜙) sin2(kh cos 𝜃) (4-121) 202 LINEAR WIRE ANTENNAS 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0.0 0.2 0.3 0.4 0.5 4 5 6 7 8 Radiation resistance (ohms) Directivity (dimensionless) Height h (wavelengths) Radiation resistance Directivity Figure 4.32 Radiation resistance and maximum directivity of a horizontal infinitesimal electric dipole as a function of its height above an infinite perfect electric conductor.
The maximum value of (4-121) depends on the value of kh (whether kh ≤𝜋∕2, h ≤λ∕4 or kh > 𝜋∕2, h > λ∕4). It can be shown that the maximum of (4-121) is: Umax = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝜂 2 | | | | I0l λ | | | | 2 sin2(kh) 𝜂 2 | | | | I0l λ | | | | 2 kh ≤𝜋∕2 (h ≤λ∕4) (𝜃= 0◦) kh > 𝜋∕2 (h > λ∕4) [𝜙= 0◦and sin(kh cos 𝜃max) = 1 or 𝜃max = cos−1(𝜋∕2kh)] (4-122a) (4-122b) Using (4-118) and (4-122a), (4-122b), the directivity can be written as D0 = 4𝜋Umax Prad = ⎧ ⎪ ⎨ ⎪ ⎩ 4 sin2(kh) R(kh) kh ≤𝜋∕2 (h ≤λ∕4) 4 R(kh) kh > 𝜋∕2 (h > λ∕4) (4-123a) (4-123b) where R(kh) = [ 2 3 −sin(2kh) 2kh −cos(2kh) (2kh)2 + sin(2kh) (2kh)3 ] (4-123c) For small values of kh (kh →0), (4-123a) reduces to D0 kh→0 = 4 sin2(kh) [2 3 −2 3 + 8 15(kh)2 ] = 7.5 (sin kh kh )2 (4-124) For h = 0 the element is shorted and it does not radiate. The directivity, as given by (4-123a)–(4-123b) is plotted for 0 ≤h ≤5λ in Figure 4.32. It exhibits a maximum value of 7.5 for small values of h. Maximum values of slightly greater than 6 occur when h ≃(0.615 + n∕2)λ, n = 1, 2, 3, … .
GROUND EFFECTS 203 σ = 0.01 S/m σ = 0.10 S/m σ = 1.00 S/m f = 200 MHz Rm Xm a = 10–5 r = 10.0 σ = 10.0 S/m No ground 73 42.5 0 20 40 60 80 100 120 0.2 0.4 Height h (wavelengths) Input impedance Zim (ohms) 0.6 0.8 1.0 0 /2 h σ PEC ( = ∞) λ λ Figure 4.33 Input impedance of a horizontal λ∕2 above a flat lossy electric conducting surface.
The input impedance Zim = Rim + jXim (referred to the current maximum) of a horizontal λ∕2 dipole above a flat lossy electric conductor is shown plotted in Figure 4.33 for 0 ≤h ≤λ. Conduc-tivities of 10−2, 10−1, 1, 10 S/m, and infinity (PEC) were considered. It is apparent that the con-ductivity does have a more pronounced effect on the impedance values, compared to those of the vertical dipole shown in Figure 4.23. The conductivity values used are representative of those of the dry to wet earth. The values of the resistance and reactance approach, as the height increases, the corresponding values of the isolated element (73 ohms for the resistance and 42.5 ohms for the reactance).
4.8 GROUND EFFECTS In the previous two sections the variations of the radiation characteristics (pattern, radiation resis-tance, directivity) of infinitesimal vertical and horizontal linear elements were examined when they were placed above plane perfect electric conductors. Although ideal electric conductors (𝜎= ∞) are not realizable, their effects can be used as guidelines for good conductors (𝜎≫𝜔𝜀, where 𝜀is the permittivity of the medium).
One obstacle that is not an ideal conductor, and it is always present in any antenna system, is the ground (earth). In addition, the earth is not a plane surface. To simplify the analysis, however, the earth will initially be assumed to be flat. For pattern analysis, this is a very good engineering approximation provided the radius of the earth is large compared to the wavelength and the obser-vation angles are greater than about 57.3∕(ka)1∕3 degrees from grazing (a is the earth radius) .
Usually these angles are greater than about 3◦.
In general, the characteristics of an antenna at low (LF) and medium (MF) frequencies are pro-foundly influenced by the lossy earth. This is particularly evident in the input resistance. When the antenna is located at a height that is small compared to the skin depth of the conducting earth, the input resistance may even be greater than its free-space values . This leads to antennas with very low efficiencies. Improvements in the efficiency can be obtained by placing radial wires or metallic disks on the ground.
204 LINEAR WIRE ANTENNAS The analytical procedures that are introduced to examine the ground effects are based on the geo-metrical optics models of the previous sections. The image (virtual) source is again placed a distance h below the interface to account for the reflection. However, for each polarization nonunity reflection coefficients are introduced which, in general, will be a function of the angles of incidence and the constitutive parameters of the two media. Although plane wave reflection coefficients are used, even though spherical waves are radiated by the source, the error is small for conducting media . The spherical nature of the wavefront begins to dominate the reflection phenomenon at grazing angles (i.e., as the point of reflection approaches the horizon) . If the height (h) of the antenna above the interface is much less than the skin depth 𝛿[𝛿= √ 2∕(𝜔𝜇𝜎)] of the ground, the image depth h below the interface should be increased by a complex distance 𝛿(1 −j).
The geometrical optics formulations are valid provided the sources are located inside the lossless medium. When the sources are placed within the ground, the formulations should include possible surface-wave contributions. Exact boundary-value solutions, based on Sommerfeld integral formu-lations, are available . However they are too complex to be included in an introductory chapter.
4.8.1 Vertical Electric Dipole The field radiated by an electric infinitesimal dipole when placed above the ground can be obtained by referring to the geometry of Figures 4.17(a) and (b). Assuming the earth is flat and the observations are made in the far field, the direct component of the field is given by (4-94) and the reflected component by (4-95a) where the reflection coefficient Rv is given by Rv = 𝜂0 cos 𝜃i −𝜂1 cos 𝜃t 𝜂0 cos 𝜃i + 𝜂1 cos 𝜃t = −R|| (4-125) where R|| is the reflection coefficient for parallel polarization and 𝜂0 = √𝜇0 𝜀0 = intrinsic impedance of free-space (air) 𝜂1 = √ j𝜔𝜇1 𝜎1 + j𝜔𝜀1 = intrinsic impedance of the ground 𝜃i = angle of incidence (relative to the normal) 𝜃t = angle of refraction (relative to the normal) The angles 𝜃i and 𝜃t are related by Snell’s law of refraction 𝛾0 sin 𝜃i = 𝛾1 sin 𝜃t (4-126) where 𝛾0 = jk0 = propagation constant for free-space (air) k0 = phase constant for free-space (air) 𝛾1 = (𝛼1 + jk1) = propagation constant for the ground 𝛼1 = attenuation constant for the ground k1 = phase constant for the ground Using the far-field approximations of (4-97a)–(4-98), the total electric field above the ground (z ≥0) can be written as E𝜃= j𝜂kI0le−jkr 4𝜋r sin 𝜃[ejkh cos 𝜃+ Rve−jkh cos 𝜃] z ≥0 (4-127) where Rv is given by (4-125).
GROUND EFFECTS 205 The permittivity and conductivity of the earth are strong functions of the ground’s geological con-stituents, especially its moisture. Typical values for the relative permittivity 𝜀r (dielectric constant) are in the range of 5–100 and for the conductivity 𝜎in the range of 10−4 −10 S/m.
Plots of the magnitude and phase of the reflection coefficient Rv, as a function of the incidence angle 𝜃i at a frequency of f = 1 GHz, are shown in Figure 4.34(a, b), respectively. As is appar-ent in Figure 4.34(a), for this polarization the reflection coefficient vanishes at the angles called Brewster angles . At these angles the phase undergoes an 180◦phase jump, as illustrated in Figure 4.34(b).
Normalized (to 0 dB) patterns of an infinitesimal dipole placed above the ground with height h = λ∕4, above a flat interface, are shown plotted in Figure 4.35. In the presence of the ground, the radiation toward the vertical direction (80◦> 𝜃> 0◦) is more intense than that for the perfect electric conductor, but it vanishes for grazing angles (𝜃= 90◦). The null field toward the horizon (𝜃= 90◦) is formed because the reflection coefficient Rv approaches −1 as 𝜃i →90◦. Thus the ground effects on the pattern of a vertically polarized antenna are significantly different from those of a perfect conductor.
Significant changes also occur in the impedance. Because the formulation for the impedance is much more complex , it will not be presented here. Graphical illustrations for the impedance change of a vertical dipole placed a height h above a homogeneous lossy half-space, as compared to those in free-space, can be found in .
4.8.2 Horizontal Electric Dipole The analytical formulation of the horizontal dipole above the ground can also be obtained in a similar manner as for the vertical electric dipole. Referring to Figure 4.28(a) and (b), the direct component is given by (4-111) and the reflected by (4-111a) where the reflection coefficient Rh is given by Rh = { R⟂for 𝜙= 0◦, 180◦plane R|| for 𝜙= 90◦, 270◦plane (4-128) where R|| is the reflection coefficient for parallel polarization, as given by (4-125), and R⟂is the reflection coefficient for perpendicular polarization given by R⟂= 𝜂1 cos 𝜃i −𝜂0 cos 𝜃t 𝜂1 cos 𝜃i + 𝜂0 cos 𝜃t (4-128a) The angles 𝜃i and 𝜃t are again related by Snell’s law of refraction as given by (4-126).
Using the far-field approximations of (4-114a) and (4-114b), the total field above the ground (z ≥h) can be written as E𝜓= j𝜂kI0e−jkr 4𝜋r √ 1 −sin2 𝜃sin2 𝜙[ejkh cos 𝜃+ Rhe−jkh cos 𝜃], z ≥h (4-129) where Rh is given by (4-128).
To give an insight why R⊥is used in the 𝜙= 0◦, 180◦plane and why R|| is used in the 𝜙= 90◦, 270◦plane, the reader is referred to electric field of (4-115), which is decomposed into the far-zone spherical electric field components E𝜃and E𝜙represented by (4-116a), (4-116b) and (4-116c) in the principal planes. From these expresssions, the polarization of the electric field radi-ated by the horizontal dipole, and that reflected by the PEC ground plane, is more apparent; thus the use of the appropriate reflection coeffients R⊥and R|| in the respective principal planes.
Plots of the magnitude and phase of the reflection coefficient Rh, as a function of the incidence angle 𝜃i on the xz(𝜙= 0◦) plane, are shown in Figure 4.36(a, b), respectively. It is apparent from Figure 4.36(a) that for this polarization the reflection coefficient does not vanish, but rather mono-tonically increases as the angle increases . Similarly, the corresponding phase remains nearly constant and at 180◦. The corresponding normalized amplitude patterns in this plane (𝜙= 0◦) of a horizontal electric dipole placed at a height h = λ∕4 above an interface are shown in Figure 4.37(a), 206 LINEAR WIRE ANTENNAS 0 10 20 30 40 50 60 70 80 90 0 0.2 0.4 0.6 0.8 1 1.2 Magnitude of Rv PEC Earth ( r = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) r σ r r σ i (degrees) θ (a) Magnitude ( = 90° plane) ϕ 0 10 20 30 40 50 60 70 80 90 –200 –150 –100 –50 0 50 100 150 200 Phase of Rv PEC Earth ( = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) r r σ r r σ i (degrees) θ (b) Phase ( = 90° plane) ϕ Figure 4.34 Magnitude and phase variations of reflection coefficient Rv, as a function of incidence angle 𝜃i, for f = 1 GHz.
GROUND EFFECTS 207 –30 –20 –10 0 30 150 60 120 90 90 120 60 150 30 180 0 PEC Earth ( = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) r r σ r r σ Figure 4.35 Elevation plane (𝜙= 0◦, 90◦planes) amplitude patterns of an infinitesimal vertical dipole above an interface. (h = λ∕4, f = 1 GHz).
whose shape is basically the same for the cases examined. This result is attributed to monotonic and small variations of the amplitude and basically constant phase, for all the cases examined, of the corresponding reflection coefficient Rh as a function of the angle of incidence, as displayed in Figure 4.36(a, b).
For the yz(𝜙= 90◦) plane, the magnitude and phase variations of the reflection coefficient Rh, as a function of the incidence angle 𝜃i, are the same as those for the vertical polarization coefficient Rv, as are displayed in Figure 4.34(a, b). For this plane (𝜙= 90◦), the corresponding normalized amplitude pattern of a horizontal infinitesimal electric dipole placed at a height h = λ/4 above the interface are shown in Figure 4.37(b), whose shape is basically the same for the cases examined.
This can be attributed to the small amplitude and phase variation of the reflection coefficient Rh/Rv in this plane for incidence angle up to the Brewster angle, as displayed in Figure 4.34(a, b), and it is due to the small field intensity of the horizontal dipole near and greater than the Brewster angle and as the observation angle approached 90◦; the field intensity vanishes at 𝜃= 90◦.
4.8.3 PEC, PMC and EBG Surfaces In general, PEC, PMC, and EBG surfaces individually possess attractive characteristics, but these surfaces also exhibit shortcomings when electromagnetic radiating elements are mounted on them, especially when the designs are judged based on aerodynamic, stealth, and conformal criteria. For example, when an electric element is mounted vertically on a PEC surface, radiation and system efficiency get reinforced; however, its low-profile geometry is undesirable for aerodynamic, stealth, and conformal designs. Yet, when the same electric radiating element is placed horizontally on a PEC surface, its radiation efficiency suffers because, at h = 0, the image possesses a 180◦phase shift and its radiation cancels the radiation of the actual electric element. While the height stays small electrically, the radiation efficiency is low. For the radiation to attain maximum efficiency at a direction normal to the surface, the horizontal element must be placed at a height λ∕4 above 208 LINEAR WIRE ANTENNAS 0 10 20 30 40 50 60 70 80 90 0 0.2 0.4 0.6 0.8 1 1.2 Magnitude of Rh PEC Earth ( = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) 0 10 20 30 40 50 60 70 80 90 –200 –150 –100 –50 0 50 100 150 200 Phase of Rh PEC Earth ( = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) r r σ r r σ r r r r σ σ i (degrees) θ i (degrees) θ (a) Magnitude ( = 0° plane) ϕ (b) Phase ( = 0° plane) ϕ Figure 4.36 Magnitude and phase variations of the reflection coefficient Rh, as a function of incidence angle 𝜃i, for f = 1 GHz.
GROUND EFFECTS 209 –30 –20 –10 0 30 150 60 120 90 90 120 60 150 30 180 0 –30 –20 –10 0 30 150 60 120 90 90 120 60 150 30 180 0 PEC Earth ( = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) r r σ r r σ PEC Earth ( = 5) Lossy Earth ( = 5, = 0.01 S/m) Ocean ( = 81) Lossy Ocean ( = 81, = 4 S/m) r r σ r r σ (a) = 0° ϕ (b) = 90° ϕ Figure 4.37 Elevation plane (𝜙= 0◦and 𝜙= 90◦) amplitude patterns of an infinitesimal dipole above an interface (h = λ∕4, f = 1 GHz).
210 LINEAR WIRE ANTENNAS Figure 4.38 Geometry and S11 of horizontal dipole above PEC, PMC and EBG surfaces. (a) Geometry ( Reprinted with permission from John Wiley & Sons, Inc.) (b) Reflection coefficient (source: c ⃝) 2003 IEEE).
the surface. Such an arrangement is especially not desirable on space-borne platforms because of aerodynamic considerations. Furthermore, for stealth type of targets, configurations that are visible to radar can create a large radar cross section (RCS) signature, which is why low-profile designs are desirable for aerodynamic, stealth, and conformal applications. When the same electric radiating element is placed horizontally on a PMC surface, its image then has a low profile and 0◦phase, which reinforces the radiation of the actual electric element. The characteristics of vertical and horizontal electric elements placed vertically and horizontally on PEC and PMC surfaces are based on image theory, and they are visually contrasted in Figure 4.16.
Whether a PEC, PMC, or EBG surface outperforms the others as a ground plane depends on the application. This is best illustrated by a basic example. In Figure 4.38, a 0.4λ12 dipole (λ12 is the free-space wavelength at f= 12 GHz) is placed horizontally above PEC, PMC, and EBG surfaces.
The EBG surface has a height of 0.04λ12. The dipole is placed at a height h of 0.06λ12(h = 0.06λ12) above a λ12× λ12 PEC, PMC square surface, which in turn means that the dipole is placed at a height of 0.02λ12 above the EBG surface. The S11 of this system (based on a 50-ohm line impedance) was simulated, using the FDTD method, over a frequency range of 10–18 GHz , and the results are shown in Figure 4.38(b). From these results, it is clear that the EBG surface (which has a reflection phase variation from +180◦to −180◦; see an example in Figure 4.14) exhibits a best return loss of −27 dB while the PMC (which has a reflection phase of 0◦) has a best return loss of −7.2 dB and the PEC (which has a reflection phase of 180◦) has a best return loss of only −3.5 dB. For the PMC sur-face, the return loss is influenced by the mutual coupling, due to the close proximity between the main element and its in-phase image, whereas for the PEC the return loss is influenced by the 180◦phase reversal, which severely impacts the radiation efficiency. In this example, the EBG surface, because of its +180◦to −180◦phase variation over the frequency band-gap of the EBG design, outperforms the PEC and PMC and serves as a good ground plane. The phase characteristics of the PEC and PMC surfaces are constant (out-of-phase and in-phase, respectively) over the entire frequency range.
GROUND EFFECTS 211 4.8.4 Earth Curvature Antenna pattern measurements on aircraft can be made using either scale models or full scale in-flight. Scale model measurements usually are made indoors using electromagnetic anechoic cham-bers, as described in Chapter 17. The indoor facilities provide a controlled environment, and all-weather capability, security, and minimize electromagnetic interference. However, scale model mea-surements may not always simulate real-life outdoor conditions, such as the reflecting surface of seawater. Therefore full-scale model measurements may be necessary. For in-flight measurements, reflecting surfaces, such as earth and seawater, introduce reflections, which usually interfere with the direct signal. These unwanted signals are usually referred to as multipath. Therefore the total mea-sured signal in an outdoor system configuration is the combination of the direct signal and that due to multipath, and usually it cannot be easily separated in its parts using measuring techniques. Since the desired signal is that due to the direct path, it is necessary to subtract from the total response the contributions due to multipath. This can be accomplished by developing analytical models to predict the contributions due to multipath, which can then be subtracted from the total signal in order to be left with the desired direct path signal. In this section we will briefly describe techniques that have been used to accomplish this , .
The analytical formulations of Sections 4.8.1 and 4.8.2 for the patterns of vertical and horizontal dipoles assume that the earth is flat. This is a good approximation provided the curvature of the earth is large compared to the wavelength and the angle of observation is greater than about 3◦ from grazing [or more accurately greater than about 57.3∕(ka)1∕3 degrees, where a is the radius of the earth] from grazing . The curvature of the earth has a tendency to spread out (weaken, diffuse, diverge) the reflected energy more than a corresponding flat surface. The spreading of the reflected energy from a curved surface as compared to that from a flat surface is taken into account by introducing a divergence factor D , , , defined as D = divergence factor = reflected field from curved surface reflected field from flat surface (4-130) The formula for D can be derived using purely geometrical considerations. It is accomplished by comparing the ray energy density in a small cone reflected from a sphere near the principal point of reflection with the energy density the rays (within the same cone) would have if they were reflected from a plane surface. Based on the geometrical optics energy conservation law for a bundle of rays within a cone, the reflected rays within the cone will subtend a circle on a perpendicular plane for reflections from a flat surface, as shown in Figure 4.39(a). However, according to the geometry of Figure 4.39(b), it will subtend an ellipse for a spherical reflecting surface. Therefore the divergence factor of (4-130) can also be defined as D = Er s Er f = [ area contained in circle area contained in ellipse ]1∕2 (4-131) where Er s = reflected field from spherical surface Er f = reflected field from flat surface Using the geometry of Figure 4.40, the divergence factor can be written as and D = √ 𝜌r 1𝜌r 2 (𝜌r 1 + s)(𝜌r 2 + s) s′ s′ + s (4-132) 212 LINEAR WIRE ANTENNAS Figure 4.39 Reflection from flat and spherical surfaces.
where 𝜌r 1 and 𝜌r 2 are the principal radii of curvature of the reflected wavefront at the point of reflection and are given, according to the geometry of Figure 4.40, by 1 𝜌r 1 = 1 s′ + 1 𝜌sin 𝜓+ √ 1 (𝜌sin 𝜓)2 −4 a2 (4-132a) GROUND EFFECTS 213 h'1 h'2 h1 O' O h2 E Source Observation s' s d d2 d2 d1 ψ ψ α β γ Figure 4.40 Geometry for reflections from a spherical surface.
1 𝜌r 2 = 1 s′ + 1 𝜌sin 𝜓− √ 1 (𝜌sin 𝜓)2 −4 a2 (4-132b) 𝜌= a 1 + sin2 𝜓 (4-132c) A simplified form of the divergence factor is that of D ≅ [ 1 + 2s′s a(s′ + s) sin 𝜓 ]−1∕2 ⋅ [ 1 + 2s′s a(s′ + s) ]−1∕2 (4-133) Both (4-132) and (4-133) take into account the earth curvature in two orthogonal planes.
Assuming that the divergence of rays in the azimuthal plane (plane vertical to the page) is negli-gible (two-dimensional case), the divergence factor can be written as D ≃ [ 1 + 2 ss′ ad tan 𝜓 ]−1∕2 (4-134) where 𝜓is the grazing angle. Thus the divergence factor of (4-134) takes into account energy spread-ing primarily in the elevation plane. According to Figure 4.40 h′ 1 = height of the source above the earth (with respect to the tangent at the point of reflection) h′ 2 = height of the observation point above the earth (with respect to the tangent at the point of reflection) d = range (along the surface of the earth) between the source and the observation point 214 LINEAR WIRE ANTENNAS a = radius of the earth (3,959 mi). Usually a 4 3 radius (≃5,280 mi) is used.
𝜓= reflection angle (with respect to the tangent at the point of reflection).
d1 = distance (along the surface of the earth) from the source to the reflection point d2 = distance (along the surface of the earth) from the observation point to the reflection point The divergence factor D can be included in the formulation of the fields radiated by a vertical or a horizontal dipole, in the presence of the earth, by modifying (4-127) and (4-129) and writing them, respectively, as E𝜃= j𝜂kI0le−jkr 4𝜋r sin 𝜃[ejkh cos 𝜃+ DRve−jkh cos 𝜃] (4-135a) E𝜓= j𝜂kI0le−jkr 4𝜋r √ 1 −sin2 𝜃sin2 𝜙[ejkh cos 𝜃+ DRhe−jkh cos 𝜃] (4-135b) While the previous formulations are valid for smooth surfaces, they can still be used with rough surfaces, provided the surface geometry satisfies the Rayleigh criterion and hm < λ 8 sin 𝜓 (4-136) where hm is the maximum height of the surface roughness. Since the dividing line between a smooth and a rough surface is not that well defined, (4-136) should only be used as a guideline.
The coherent contributions due to scattering by a surface with Gaussian rough surface statistics can be approximately accounted for by modifying the vertical and horizontal polarization smooth surface reflection coefficients of (4-125) and (4-128) and express them as Rs v,h = R0 v,he−2(k0h0 cos 𝜃i)2 (4-137) where Rs v,h = reflection coefficient of a rough surface for either vertical or horizontal polarization R0 v,h = reflection coefficient of a smooth surface for either vertical (4-125) or horizontal (4-128) polarization h2 0 = mean-square roughness height A slightly rough surface is defined as one whose rms height is much smaller than the wavelength, while a very rough surface is defined as one whose rms height is much greater than the wavelength.
Plots of the divergence factor as a function of the grazing angle 𝜓(or as a function of the observa-tion point h′ 2) for different source heights are shown in Figure 4.41. It is observed that the divergence factor is somewhat different and smaller than unity for small grazing angles, and it approaches unity as the grazing angle becomes larger. The variations of D displayed in Figure 4.41 are typical but not unique. For different positions of the source and observation point, the variations will be somewhat different. More detailed information on the variation of the divergence factor and its effect on the overall field pattern is available .
The most difficult task usually involves the determination of the reflection point from a knowledge of the heights of the source and observation points, and the range d between them. Procedures to do this have been developed , –.
Using the analytical model developed here, computations were performed to see how well the predictions compared with measurements. For the computations it was assumed that the reflecting GROUND EFFECTS 215 Figure 4.41 Divergence factor for a 4/3 radius earth (ae = 5,280 mi = 8,497.3 km) as a function of grazing angle 𝜓.
surface is seawater possessing a dielectric constant of 81 and a conductivity of 4.64 S/m , .
To account for atmospheric refraction, a 4/3 earth was assumed , , so the atmo-sphere above the earth can be considered homogeneous with propagation occurring along straight lines.
For computations using the earth as the reflecting surface, all three divergence factors of (4-132)–(4-134) gave the same results. However, for nonspherical reflecting surfaces and for those Range (nautical miles) Height gain (dB) 14 4 6 8 10 12 –18 –15 –12 –9 –6 –3 0 3 6 Measured Calculated Frequency = 167.5 MHz Aircraft altitude = 5000 ft Antenna height = 62 ft Polarization = vertical Figure 4.42 Measured and calculated height gain over the ocean (𝜀r = 81, 𝜎= 4.64 S/m) for vertical polar-ization.
216 LINEAR WIRE ANTENNAS with smaller radii of curvature, the divergence factor of (4-132) is slightly superior followed by (4-133) and then by (4-134). In Figure 4.42 we display and compare the predicted and measured height gain versus range d (4 < d < 14 nautical miles) for a vertical-vertical polarization system configuration at a frequency of 167.5 MHz. The height gain is defined as the ratio of the total field in the presence of the earth divided by the total field in the absence of the earth. A good agreement is noted between the two. The peaks and nulls are formed by constructive and destructive interferences between the direct and reflected components. If the reflecting surface were perfectly conducting, the maximum height gain would be 2 (6 dB). Because the modeled reflecting surface of Figure 4.42 was seawater with a dielectric constant of 81 and a conductivity of 4.64 S/m, the maximum height gain is less than 6 dB. The measurements were taken by aircraft and facilities of the Naval Air Warfare Center, Patuxent River, MD. Additional measurements were made but are not included here; they can be found in and .
A summary of the pertinent parameters, and associated formulas and equation numbers for this chapter are listed in Table 4.3.
4.9 COMPUTER CODES There are many computer codes that have been developed to analyze wire-type linear antennas, such as the dipole, and they are too numerous to mention here. One simple program to characterize the radiation characteristics of a dipole, designated as Dipole (both in FORTRAN and MATLAB), is included in the publisher’s website for this book. Another much more advanced program, desig-nated as the Numerical Electromagnetics Code (NEC), is a user-oriented software developed by Lawrence Livermore National Laboratory . It is a Method of Moments (MoM) code for ana-lyzing the interaction of electromagnetic waves with arbitrary structures consisting of conducting wires and surfaces. In the 1970s and 1980s the NEC was the most widely distributed and used electromagnetics code. Included with the distribution are graphics programs for generating plots of the structure, antenna patterns, and impedance. There are also other commercial software that are based on the NEC. A compact version of the NEC is the MININEC (Mini-Numerical Elec-tromagnetics Code) –. The MININEC is more convenient for the analysis of wire-type antennas.
4.10 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter.
a. Java-based interactive questionnaire, with answers.
b. Java-based applet for computing and displaying the radiation characteristics of a dipole.
c. Java-based visualization/animation for displaying the radiation characteristics of a dipole of different lengths.
d. Matlab and Fortran computer program, designated Dipole, for computing the radiation char-acteristics of a dipole. The description of the program is found in the corresponding READ ME file in the publisher’s website for this book.
e. Matlab computer program, designated Ground−Reflections, to compute the amplitude and phase variations of the reflection coefficients and the corresponding amplitude pattern of ver-tical and horizontal dipoles placed at a height h, above a planar interface.
f. Power Point (PPT) viewgraphs, in multicolor.
MULTIMEDIA 217 TABLE 4.3 Summary of Important Parameters and Associated Formulas and Equation Numbers for a Dipole in the Far Field Parameter Formula Equation Number Infinitesimal Dipole (l ≤λ∕50) Normalized power pattern U = (E𝜃n)2 = C0 sin2 𝜃 (4-29) Radiation resistance Rr Rr = 𝜂 (2𝜋 3 ) ( l λ )2 = 80𝜋2 ( l λ )2 (4-19) Input resistance Rin Rin = Rr = 𝜂 (2𝜋 3 ) ( l λ )2 = 80𝜋2 ( l λ )2 (4-19) Wave impedance Zw Zw = E𝜃 H𝜙 ≃𝜂= 377 ohms Directivity D0 D0 = 3 2 = 1.761 dB (4-31) Maximum effective area Aem Aem = 3λ2 8𝜋 (4-32) Vector effective length 𝓁e 𝓵e = −̂ a𝜃l sin 𝜃 (2-92) |𝓵e|max = λ Example 4.2 Half-power beamwidth HPBW = 90◦ (4-65) Loss resistance RL RL = l P √𝜔𝜇0 2𝜎= l 2𝜋b √𝜔𝜇0 2𝜎 (2-90b) Small Dipole (λ∕50 < l ≤λ∕10) Normalized power pattern U = (E𝜃n)2 = C1 sin2 𝜃 (4-36a) Radiation resistance Rr Rr = 20𝜋2 ( l λ )2 (4-37) Input resistance Rin Rin = Rr = 20𝜋2 ( l λ )2 (4-37) Wave impedance Zw Zw = E𝜃 H𝜙 ≃𝜂= 377 ohms (4-36a), (4-36c) Directivity D0 D0 = 3 2 = 1.761 dB Maximum effective area Aem Aem = 3λ2 8𝜋 Vector effective length 𝓁e 𝓵e = −̂ a𝜃 l 2 sin 𝜃 (2-92) |𝓵e|max = l 2 (4-36a) Half-power beamwidth HPBW = 90◦ (4-65) Half Wavelength Dipole (l = λ∕2) Normalized power pattern U = (E𝜃n)2 = C2 ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 ≃C2 sin3 𝜃 (4-87) Radiation resistance Rr Rr = 𝜂 4𝜋Cin(2𝜋) ≃73 ohms (4-93) (continued overleaf) 218 LINEAR WIRE ANTENNAS TABLE 4.3 (continued) Parameter Formula Equation Number Input resistance Rin Rin = Rr = 𝜂 4𝜋Cin(2𝜋) ≃73 ohms (4-79), (4-93) Input impedance Zin Zin = 73 + j42.5 (4-93a) Wave impedance Zw Zw = E𝜃 H𝜙 ≃𝜂= 377 ohms Directivity D0 D0 = 4 Cin(2𝜋) ≃1.643 = 2.156 dB (4-91) Vector effective length 𝓁e 𝓵e = −̂ a𝜃 λ 𝜋 cos (𝜋 2 cos 𝜃 ) sin 𝜃 (2-91) |𝓵e|max = λ 𝜋= 0.3183λ (4-84) Half-power beamwidth HPBW = 78◦ (4-65) Loss resistance RL RL = l 2P √𝜔𝜇0 2𝜎= l 4𝜋b √𝜔𝜇0 2𝜎 Example (2-13) Quarter-Wavelength Monopole (l = λ∕4) Normalized power pattern U = (E𝜃n)2 = C2 ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 ≃C2 sin3 𝜃 (4-87) Radiation resistance Rr Rr = 𝜂 8𝜋Cin(2𝜋) ≃36.5 ohms (4-106) Input resistance Rin Rin = Rr = 𝜂 8𝜋Cin(2𝜋) ≃36.5 ohms (4-106) Input impedance Zin Zin = 36.5 + j21.25 (4-106) Wave impedance Zw Zw = E𝜃 H𝜙 ≃𝜂= 377 ohms Directivity D0 D0 = 3.286 = 5.167 dB Vector effective length 𝓁e 𝓵e = −̂ a𝜃 λ 𝜋cos (𝜋 2 cos 𝜃 ) (2-91) |𝓵e|max = λ 𝜋= 0.3183λ (4-84) REFERENCES 1. W. A. Wheeler, “The Spherical Coil as an Inductor, Shield, or Antenna,” Proc. IRE, Vol. 46, pp. 1595–1602, September 1958 (correction, Vol. 48, p. 328, March 1960).
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22. D. Sievenpiper, “Artificial Impedance Surfaces,” Chapter 15, in Modern Antenna Handbook, C. A. Balanis (editor), John Wiley & Sons, pp. 737–777, 2008.
23. F. Yang and Y. Rahmat-Samii, “Reflection Phase Characterization of the EBG Ground Plane for Low Profile Wire Antenna Applications,” IEEE Trans. Antenna Propagat., Vol. 51, No. 10, pp. 2691–2703, October 2003.
24. R. F. Schwartz, “Input Impedance of a Dipole or Monopole,” Microwave J., Vol. 15, No. 12, p. 22, December 1972.
25. K. Fujimoto and J. R. James, Mobile Antenna Systems Handbook, Artech House, Norwood, MA, 1994.
26. M. A. Jensen and Y. Rahmat-Samii, “Performance Analysis of Antennas for Hand-Held Transceivers Using FDTD,” IEEE Trans. Antennas Propagat., Vol. 42, No. 8, pp. 1106–1113, August 1994.
27. M. A. Jensen and Y. Rahmat-Samii, “EM Interaction of Handset Antennas and a Human in Personal Com-munications,” Proc. IEEE, Vol. 83, No. 1, pp. 7–17, January 1995.
28. K. D. Katsibas, “Analysis and Design of Mobile Antennas for Handheld Units,” Master’s Thesis, Arizona State University, Tempe, AZ, August 1996.
29. K. D. Katsibas, C. A. Balanis, P. A. Tirkas, and C. R. Birtcher, “Folded Loop Antenna for Mobile Handheld Units,” IEEE Trans. Antennas Propagat., Vol. 46, No. 2, pp. 260–266, February 1998.
30. R. E. Collin and F. J. Zucker (Eds.), Antenna Theory Part 2, Chapters 23 and 24 (by J. R. Wait), McGraw-Hill, New York, 1969.
31. P. R. Bannister, “Image Theory Results for the Mutual Impedance of Crossing Earth Return Circuits,” IEEE Trans. Electromagn. Compat., Vol. 15, No. 4, pp. 158–160, 1973.
220 LINEAR WIRE ANTENNAS 32. D. E. Kerr, Propagation of Short Radio Waves, MIT Radiation Laboratory Series, McGraw-Hill, New York, 1951, Vol. 13, pp. 98–109, 112–122, 396–444.
33. L. E. Vogler and J. L. Noble, “Curves of Input Impedance Change due to Ground for Dipole Antennas,” U.S. National Bureau of Standards, Monograph 72, January 31, 1964.
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35. C. A. Balanis, R. Hartenstein, and D. DeCarlo, “Multipath Interference for In-Flight Antenna Measure-ments,” IEEE Trans. Antennas Propagat., Vol. AP-32, No. 1, pp. 100–104, January 1984.
36. J. R. Wait and A. M. Conda, “Pattern of an Antenna on a Curved Lossy Surface,” IRE Trans. Antennas Propagat., Vol. AP-6, No. 4, pp. 348–359, October 1958.
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42. G. J. Burke and A. J. Poggio, “Numerical Electromagnetics Code (NEC)-Method of Moments,” Technical Document 116, Naval Ocean Systems Center, San Diego, CA, January 1981.
43. A. J. Julian, J. M. Logan, and J. W. Rockway, “MININEC: A Mini-Numerical Electromagnetics Code,” Technical Document 516, Naval Ocean Systems Center, San Diego, CA, September 6,1982.
44. J. Rockway, J. Logan, D. Tam, and S. Li, The MININEC SYSTEM: Microcomputer Analysis of Wire Anten-nas, Artech House, Inc., Norwood, MA, 1988.
PROBLEMS 4.1.
A horizontal infinitesimal electric dipole of constant current I0 is placed symmetrically about the origin and directed along the x-axis. Derive the (a) far-zone fields radiated by the dipole (b) directivity of the antenna 4.2.
Repeat Problem 4.1 for a horizontal infinitesimal electric dipole directed along the y-axis.
4.3.
Repeat Problem 4.1 using the procedure of Example 4.5.
4.4.
For Example 4.5, (a) formulate an expression for the directivity.
(b) determine the radiated power.
(c) determine the maximum directivity by integrating the radiated power. Compare with that of Problem 4.2 or any other infinitesimal dipole.
(d) determine the maximum directivity using the computer program Dipole; compare with that of part (c).
4.5.
For Problem 4.1 determine the polarization of the radiated far-zone electric fields (E𝜃, E𝜙) and normalized amplitude pattern in the following planes: (a) 𝜙= 0◦ (b) 𝜙= 90◦ (c) 𝜃= 90◦ PROBLEMS 221 4.6.
Repeat Problem 4.5 for the horizontal infinitesimal electric dipole of Problem 4.2, which is directed along the y-axis.
4.7.
For Problem 4.3, determine the polarization of the radiated far-zone fields (E𝜃, E𝜙) in the following planes: (a) 𝜙= 0◦ (b) 𝜙= 90◦ (c) 𝜃= 90◦ Compare with those of Problem 4.5.
4.8.
For Example 4.5, determine the polarization of the radiated far-zone fields (E𝜃, E𝜙) in the following planes: (a) 𝜙= 0◦ (b) 𝜙= 90◦ (c) 𝜃= 90◦ Compare with those of Problem 4.6.
4.9.
An infinitesimal magnetic dipole of constant current Im and length l is symmetrically placed about the origin along the z-axis. Find the (a) spherical E- and H-field components radiated by the dipole in all space (b) directivity of the antenna 4.10.
For the infinitesimal magnetic dipole of Problem 4.9, find the far-zone fields when the element is placed along the (a) x-axis, (b) y-axis 4.11.
An infinitesimal electric dipole is centered at the origin and lies on the x-y plane along a line which is at an angle of 45◦with respect to the x-axis. Find the far-zone electric and magnetic fields radiated. The answer should be a function of spherical coordinates.
4.12.
Repeat Problem 4.11 for an infinitesimal magnetic dipole.
4.13.
An infinitesimal electric dipole of length l and constant current Io is placed symmetrically about the origin and it is tilted at an angle of 45◦on the yz-plane. Using the vector potential approach, determine for the infinitesimal dipole the: (a) Far-zone electric and magnetic fields (Er, E𝜃, E𝜙, Hr, H𝜃, H𝜙) in terms of the spherical coordinates r, 𝜃, 𝜙. For example, E𝜃(r, 𝜃, 𝜙). The same for the other components.
(b) Directivity (dimensionless and in dB).
(c) Polarization of the radiated fields (linear circular or elliptical).
45° z x y 4.14.
Derive (4-10a)–(4-10c) using (4-8a)–(4-9).
4.15.
Derive the radiated power of (4-16) by forming the average power density, using (4-26a)– (4-26c), and integrating it over a sphere of radius r.
4.16.
Derive the far-zone fields of an infinitesimal electric dipole, of length l and constant current I0, using (4-4) and the procedure outlined in Section 3.6. Compare the results with (4-26a)– (4-26c).
4.17.
Derive the fifth term of (4-41).
4.18.
For an antenna with a maximum linear dimension of D, find the inner and outer boundaries of the Fresnel region so that the maximum phase error does not exceed (a) 𝜋∕16 rad (b) 𝜋∕4 rad (c) 18◦ (d) 15◦ 222 LINEAR WIRE ANTENNAS 4.19.
The boundaries of the far-field (Fraunhofer) and Fresnel regions were selected based on a maximum phase error of 22.5◦, which occur, respectively, at directions of 90◦and 54.74◦ from the axis along the largest dimension of the antenna. For an antenna of maximum length of 5λ, what do these maximum phase errors reduce to at an angle of 30◦from the axis along the length of the antenna? Assume that the phase error in each case is totally contributed by the respective first higher order term that is being neglected in the infinite series expansion of the distance from the source to the observation point.
4.20.
The current distribution on a terminated and matched long linear (traveling wave) antenna of length l, positioned along the z-axis and fed at its one end, is given by I = ̂ azI0e−jkz′, 0 ≤z′ ≤l where I0 is a constant. Derive expressions for the (a) far-zone spherical electric and magnetic field components (b) radiation power density 4.21.
A line source of infinite length and constant current I0 is positioned along the z-axis. Find the (a) vector potential A (b) cylindrical E- and H-field components radiated Hint: ∫ +∞ −∞ e−j𝛽 √ b2+t2 √ b2 + t2 dt = −j𝜋H0 (2)(𝛽b) where H0 (2)(𝛼x) is the Hankel function of the second kind of order zero.
4.22.
Show that (4-67) reduces to (4-68) and (4-88) to (4-89).
4.23.
A thin linear dipole of length l is placed symmetrically about the z-axis. Find the far-zone spherical electric and magnetic components radiated by the dipole whose current distribution can be approximated by (a) Iz(z′) = ⎧ ⎪ ⎨ ⎪ ⎩ I0 ( 1 + 2 l z′) , −l∕2 ≤z′ ≤0 I0 ( 1 −2 l z′) , 0 ≤z′ ≤l∕2 (b) Iz(z′) = I0 cos (𝜋 l z′) , −l∕2 ≤z′ ≤l∕2 (c) Iz(z′) = I0 cos2 (𝜋 l z′) , −l∕2 ≤z′ ≤l∕2 4.24.
A center-fed electric dipole of length l is attached to a balanced lossless transmission line whose characteristic impedance is 50 ohms. Assuming the dipole is resonant at the given length, find the input VSWR when (a) l = λ∕4 (b) l = λ∕2 (c) l = 3λ∕4 (d) l = λ 4.25.
Use the equations in the book or the computer program of this chapter. Find the radiation efficiency of resonant linear electric dipoles of length (a) l = λ∕50 (b) l = λ∕4 (c) l = λ∕2 (d) l = λ PROBLEMS 223 Assume that each dipole is made out of copper [𝜎= 5.7 × 107 S/m], has a radius of 10−4λ, and is operating at f = 10 MHz. Use the computer program Dipole of this chapter to find the radiation resistances.
4.26.
Write the far-zone electric and magnetic fields radiated by a magnetic dipole of l = λ∕2 aligned with the z-axis. Assume a sinusoidal magnetic current with maximum value Im0.
4.27.
A resonant center-fed dipole is connected to a 50-ohm line. It is desired to maintain the input VSWR = 2.
(a) What should the largest input resistance of the dipole be to maintain the VSWR = 2?
(b) What should the length (in wavelengths) of the dipole be to meet the specification?
(c) What is the radiation resistance of the dipole?
4.28.
The radiation field of a particular antenna is given by: E = ̂ a𝜃j𝜔𝜇k sin 𝜃I0A1e−jkr 4𝜋r + ̂ a𝜙𝜔𝜇sin 𝜃I0A2e−jkr 2𝜋r The values A1 and A2 depend on the antenna geometry. Obtain an expression for the radiation resistance. What is the polarization of the antenna?
4.29.
The approximate far zone electric field radiated by a very thin wire linear dipole of length l, positioned symmetrically along the z-axis, is given by E𝜃≃Co sin1.5 𝜃e−jkr r where Co is a constant. Determine the: (a) Exact directivity (dimensionless and in dB).
(b) Approximate directivity (dimensionless and in dB) using an approximate but appropriate formula (state the formula you are using).
(c) Length of the dipole (in wavelengths).
(d) Input impedance of the dipole. Assume the wire radius a is very small (a ≪λ).
4.30.
For a λ∕2 dipole placed symmetrical along the z-axis, determine the (a) vector effective height (b) maximum value (magnitude) of the vector effective height (c) ratio (in percent) of the maximum value (magnitude) of the vector effective height to its total length (d) maximum open-circuit output voltage when a uniform plane wave with an electric field of Ei|𝜃=90◦= −̂ a𝜃10−3 volts/wavelength impinges at broadside incidence on the dipole.
4.31.
A base-station cellular communication system utilizes arrays of λ∕2 dipoles as transmitting and receiving antennas. Assuming that each element is lossless and that the input power to each of the λ∕2 dipoles is 1 watt, determine at 1,900 MHz and a distance of 5 km the maximum (a) radiation intensity. Specify also the units.
(b) radiation density (in watts∕m2) for each λ∕2 dipole. This determines the safe level for human exposure to EM radiation.
224 LINEAR WIRE ANTENNAS 4.32.
A λ∕2 dipole situated with its center at the origin radiates a time-averaged power of 600 W at a frequency of 300 MHz. A second λ∕2 dipole is placed with its center at a point P(r, 𝜃, 𝜙), where r = 200 m, 𝜃= 90◦, 𝜙= 40◦. It is oriented so that its axis is parallel to that of the transmitting antenna. What is the available power at the terminals of the second (receiving) dipole?
4.33.
A half-wave dipole is radiating into free-space. The coordinate system is defined so that the origin is at the center of the dipole and the z-axis is aligned with the dipole. Input power to the dipole is 100 W. Assuming an overall efficiency of 50%, find the power density (in W/m2) at r = 500 m, 𝜃= 60◦, 𝜙= 0◦.
4.34.
A small dipole of length l = λ∕20 and of wire radius a = λ∕400 is fed symmetrically, and it is used as a communications antenna at the lower end of the VHF band (f = 30 MHz). The antenna is made of perfect electric conductor (PEC). The input reactance of the dipole is given by Xin = −j120[ln(l∕2a) −1] tan (𝜋l λ ) Determine the following: (a) Input impedance of the antenna. State whether it is inductive or capacitive.
(b) Radiation efficiency (in percent).
(c) Capacitor (in farads) or inductor (in henries) that must be connected in series with the dipole at the feed in order to resonate the element. Specify which element is used and its value.
4.35.
A half-wavelength (l = λ∕2) dipole is connected to a transmission line with a characteristic impedance of 75 ohms. Determine the following: (a) Reflection coefficient. Magnitude and phase (in degrees).
(b) VSWR.
It is now desired to resonate the dipole using, in series, an inductor or capacitor. At a fre-quency of 100 MHz, determine: (c) What kind of an element, inductor or capacitor, is needed to resonate the dipole?
(d) What is the inductance or capacitance?
(e) The new VSWR of the resonant dipole.
4.36.
A λ∕2 dipole is connected to a 50-ohm lossless transmission line. It is desired to resonate the element at 300 MHz by placing an inductor or capacitor in parallel/shunt at its feed points.
(a) What is the reflection coefficient and VSWR of the dipole before the insertion of the parallel/shunt element?
(b) What kind of an element is needed, inductor or capacitor, and what is its value in order to resonate the dipole?
(c) What is the new reflection coefficient and VSWR inside the transmission line after the insertion of the parallel/shunt element?
4.37.
A λ∕2 dipole is used as a radiating element while it is connected to a 50-ohm lossless transmis-sion line. It is desired to resonate the element at 1.9 GHz by placing in series capacitor(s) or inductor(s) (whichever are appropriate) at its input terminals. Determine the following: PROBLEMS 225 (a) VSWR inside the transmission line before the dipole is resonated [before the capaci-tor(s) or inductor(s) are placed in series].
(b) Total single capacitance CT (in farads) or inductance LT (in henries) that must be placed in series with the element at its input terminals in order to resonate it. (See diagram a).
(c) Individual two capacitances Co (in farads) or inductances Lo (in henries) that must be placed in series with the element at its input terminals in order to resonate it. We need to use two capacitors or two inductors to keep the system balanced by placing in series one with each arm of the dipole (see diagram b).
(d) VSWR after the element is resonated with capacitor(s) or inductor(s).
CT/LT 50 Ohms (a) Co/Lo Co/Lo 50 Ohms (b) 4.38.
The input impedance of a λ∕2 dipole, assuming the input (feed) terminals are at the center of the dipole, is equal to 73 + j42.5. Assuming the dipole is lossless, find the (a) input impedance (real and imaginary parts) assuming the input (feed) terminals have been shifted to a point on the dipole which is λ∕8 from either end point of the dipole length (b) capacitive or inductive reactance that must be placed across the new input terminals of part (a) so that the dipole is self-resonant (c) VSWR at the new input terminals when the self-resonant dipole of part (b) is connected to a “twin-lead” 300-ohm line 4.39.
A linear half-wavelength dipole is operating at a frequency of 1 GHz; determine the capacitance or inductance that must be placed across (in parallel) the input terminals of the dipole so that the antenna becomes resonant (make the total input impedance real).
What is then the VSWR of the resonant half-wavelength dipole when it is connected to a 50-ohm line?
4.40.
A folded dipole (whose length is l = λ∕2 and spacing s between the two parallel length is much smaller than λ, s ≪λ), acts as an impedance transformer with a turns ratio of 2:1; i.e., its impedance is 4 times greater than that of a regular λ∕2 dipole. Such a folded dipole was a basic feed element of a Yagi-Uda antenna, which was popular TV antenna prior to the cable, especially for 2-3 TV channels. Assuming the wire radius of the folded dipole is very small compared to the wavelength (a ≪λ): (a) Write an expression for the input impedance of the folded dipole.
(b) If we want to resonate the folded dipole, what kind of an element (inductor or capacitor) shall we place in paral-lel across the input terminals of the folded dipole; i.e., which of the two is appropriate to accomplish the task?
(c) At a frequency of 100 MHz, what is the value of the inductor or capacitor, whichever is appropriate to res-onate the folded dipole of Part b?
Transmission Line Input 2a s l 226 LINEAR WIRE ANTENNAS (d) After the folded dipole has been resonated, what is its: r Input impedance?
r Input reflection coefficient (assume a twin-lead transmission line connected to it with a characteristic impedance of 300 ohms)?
r VSWR?
4.41.
The field radiated by an infinitesimal electric dipole, placed along the z-axis a distance s along the x-axis, is incident upon a waveguide aperture antenna of dimensions a and b, mounted on an infinite ground plane, as shown in the figure. The normalized electric field radiated by the aperture in the E-plane (x-z plane; 𝜙= 0◦) is given by E = −̂ a𝜃j𝜔𝜇bI0e−jkr 4𝜋r sin (kb 2 cos 𝜃 ) kb 2 cos 𝜃 z y x b r dipole = ∞ a θ ϕ σ s Assuming the dipole and aperture antennas are in the far field of each other, determine the polarization loss (in dB) between the two antennas.
4.42.
We are given the following information about antenna A: (a) When A is transmitting, its radiated far-field expression for the E field is given by: Ea(z) = E0 e−jkz 4𝜋z ( ̂ ax + ĵ ay √ 2 ) V∕m (b) When A is receiving an incident plane wave given by: E1(z) = ̂ ayejkz V∕m its open-circuit voltage is V1 = 4ej20◦V.
If we use the same antenna to receive a second incident plane given by: E2(z) = 10(2̂ ax + ̂ ayEj30◦)ejkz V∕m find its received open-circuit voltage V2.
PROBLEMS 227 4.43.
A 3-cm long dipole carries a phasor current I0 = 10ej60A. Assuming that λ = 5 cm, determine the E- and H-fields at 10 cm away from the dipole and at 𝜃= 45◦.
4.44.
The radiation resistance of a thin, lossless linear electric dipole of length l = 0.6λ is 120 ohms.
What is the input resistance?
4.45.
A lossless, resonant, center-fed 3λ∕4 linear dipole, radiating in free-space is attached to a balanced, lossless transmission line whose characteristic impedance is 300 ohms. Assuming a = 0.03λ, calculate the: (a) radiation resistance (referred to the current maximum) (b) input impedance (referred to the input terminals) (c) VSWR on the transmission line For parts (a) and (b) use the computer program Dipole at the end of the chapter.
4.46.
Repeat Problem 4.45 for a center-fed 5λ∕8 dipole.
4.47.
A dipole antenna, with a triangular current distribution, is used for communication with sub-marines at a frequency of 150 kHz. The overall length of the dipole is 200 m, and its radius is 1 m. Assume a loss resistance of 2 ohms in series with the radiation resistance of the antenna.
(a) Evaluate the input impedance of the antenna including the loss resistance. The input reac-tance can be approximated by Xin = −120[ln(l∕2a) −1] tan(𝜋l∕λ) (b) Evaluate the radiation efficiency of the antenna.
(c) Evaluate the radiation power factor of the antenna.
(d) Design a conjugate-matching network to provide a perfect match between the antenna and a 50-ohm transmission line. Give the value of the series reactance X and the turns ratio n of the ideal transformer.
(e) Assuming a conjugate match, evaluate the instantaneous 2:1 VSWR bandwidth of the antenna.
4.48.
A uniform plane wave traveling along the negative z-axis given by y x Incident Wave crossed-dipole antenna z Ew = (2âx − jây)e+jkzEo impinges upon an crossed-dipole antenna consisting of two identical dipoles, one directed along the x-axis and the other directed along the y-axis, both fed with the same amplitude.
The y-directed dipole is fed with a 90◦phase lead compared to the x-directed dipole.
(a) Write an expression for the polarization unit vector of the incident wave.
(b) Write an expression for the polarization unit vector of the receiving antenna along the + z-axis.
(c) For the incident wave, state the following: 1. Polarization (linear, circular, elliptical) and axial ratio.
2. Rotation of the polarization vector (CW, CCW).
228 LINEAR WIRE ANTENNAS (d) For the receiving antenna, state the following: 1. Polarization (linear, circular, elliptical) and axial ratio.
2. Rotation of the polarization vector (CW, CCW).
(e) Determine the polarization loss factor (dimensionless and in dB) between the incident wave and the receiving antenna.
4.49.
A half-wavelength (l = λ∕2) dipole, positioned symmetrically about the origin along the z-axis, is used as a receiving antenna. A 300 MHz uniform plane wave, traveling along the x-axis in the negative x direction, impinges upon the λ∕2 dipole. The incident plane wave has a power density of 2μ watts/m2, and its electric field is given by Ei w = (3̂ az + ĵ ay)E0e+jkx where E0 is a constant. Determine the following: (a) Polarization of the incident wave (including its axial ratio and sense of rotation, if appli-cable).
(b) Polarization of the antenna toward the x-axis (including its axial ratio and sense of direc-tion, if applicable).
(c) Polarization losses (in dB) between the antenna and the incoming wave (assume far-zone fields for the antenna).
(d) Maximum power (in watts) that can be delivered to a matched load connected to the λ∕2 dipole (assume no other losses).
4.50.
Derive (4-102) using (4-99).
4.51.
Determine the smallest height that an infinitesimal vertical electric dipole of l = λ∕50 must be placed above an electric ground plane so that its pattern has only one null (aside from the null toward the vertical), and it occurs at 30◦from the vertical. For that height, find the directivity and radiation resistance.
4.52.
A λ∕50 linear dipole is placed vertically at a height h = 2λ above an infinite electric ground plane. Determine the angles (in degrees) where all the nulls of its pattern occur.
4.53.
A linear infinitesimal dipole of length l and constant current is placed vertically a distance h above an infinite electric ground plane. Find the first five smallest heights (in ascending order) so that a null is formed (for each height) in the far-field pattern at an angle of 60◦from the vertical.
4.54.
A vertical infinitesimal linear dipole is placed a distance h = 3λ∕2 above an infinite perfectly conducting flat ground plane. Determine the (a) angle(s) (in degrees from the vertical) where the array factor of the system will achieve its maximum value (b) angle (in degrees from the vertical) where the maximum of the total field will occur (c) relative (compared to its maximum) field strength (in dB) of the total field at the angles where the array factor of the system achieves its maximum value (as obtained in part a).
4.55.
An infinitesimal dipole of length l is placed a distance s from an air-conductor interface and at an angle of 𝜃= 60◦from the vertical axis, as shown in the fig-ure. Determine the location and direction of the image source which can be used to account for reflections.
Be very clear in indicating the location and direction of the image. Your answer can be in the form of a very clear sketch.
60° s 0, 0 μ ε ∞ σ = ∞ + ∞ – PROBLEMS 229 4.56.
An infinitesimal magnetic dipole of length l, directed along the z-axis, is placed at a height h above a Perfect Electric Conductor (PEC).
(a) Write (you do not have to derive it, as long as it is correct) an expression for the normal-ized Array Factor of the equivalent system.
(b) For a height h = 0.5λ, find all the nulls in terms of the angle theta (0◦≤𝜃≤180◦) (in degrees).
(c) If it is desired to place a null at 𝜃= 60◦, find the two smallest heights, (other than h = 0) h (in λ) that will accomplish this.
4.57.
It is desired to design an antenna system, which utilizes a vertical infinitesimal dipole of length l placed a height h above a flat, perfect electric conductor of infinite extent. The design specifications require that the pattern of the array factor of the source and its image has only one maximum, and that maximum is pointed at an angle of 60◦from the vertical. Determine (in wavelengths) the height of the source to achieve this desired design specification.
4.58.
A very short (l ≤λ∕50) vertical electric dipole is mounted on a pole a height h above the ground, which is assumed to be flat, perfectly conducting, and of infinite extent. The dipole is used as a transmitting antenna in a VHF (f = 50 MHz) ground-to-air communication system.
In order for the communication system transmitting antenna signal not to interfere with a nearby radio station, it is necessary to place a null in the vertical dipole system pattern at an angle of 80◦from the vertical. What should the shortest height (in meters) of the dipole be to achieve the desired specifications?
4.59.
A half-wavelength dipole is placed vertically on an infinite electric ground plane. Assuming that the dipole is fed at its base, find the (a) radiation impedance (referred to the current maximum) (b) input impedance (referred to the input terminals) (c) VSWR when the antenna is connected to a lossless 50-ohm transmission line.
4.60.
A lossless half-wavelength (l = λ∕2) dipole operating at 1 GHz, with an ideal sinusoidal current distribution, is placed horizontally a height h above a flat, smooth and infinite in extent perfect electric conductor (PEC).
(a) Write an expression for the normalized array factor. Assume angle 𝜃is measured from the vertical to the ground plane.
(b) For a height h = 1.5λ, determine all the physical angles 𝜃(0◦≤𝜃≤90◦) where the array factor achieves its maximum value.
(c) For the same height h = 1.5λ, find all the physical angles 𝜃(0◦≤𝜃≤90◦) where the array factor has null(s).
4.61.
A resonant vertical λ∕8 monopole, mounted on an infinite flat Perfect Electric Conductor (PEC), is connected to a lossless transmission line. It is desired to maintain the maximum reflection coefficient inside the transmission line to 0.2. Determine the: (a) Total far-zone electric field radiated by the λ∕8 monopole on and above the PEC.
(b) Input resistance of the monopole.
(c) The desired characteristic impedance of the transmission line to maintain the maximum reflection coefficient to 0.2.
4.62.
A vertical λ∕2 dipole is the radiating element in a circular array used for over-the-horizon communication system operating at 1 GHz. The circular array (center of the dipoles) is placed at a height h above the ground that is assumed to be flat, perfect electric conducting, and infinite in extent.
230 LINEAR WIRE ANTENNAS (a) In order for the array not to be interfered with by another communication system that is operating in the same frequency, it is desired to place only one null in the elevation pattern of the array factor of a single vertical λ∕2 dipole at an angle of 𝜃= 30◦from zenith (axis of the dipole). Determine the smallest nonzero height h (in meters) above the ground at which the center of the dipole must be placed to accomplish this.
(b) If the height (at its center) of the vertical dipole is 0.3 m above ground, determine all the angles 𝜃from zenith (in degrees) where all the 1. null(s) of the array factor of a single dipole in the elevation plane will be directed toward.
2. main maximum (maxima) of the array factor of a single dipole in the elevation plane will be directed toward.
4.63.
A vertical λ∕2 dipole antenna is used as a ground-to-air, over-the-horizon communication antenna at the VHF band (f = 200 MHz). The antenna is elevated at a height h (measured from its center/feed point) above ground (assume the ground is flat, smooth, and perfect elec-tric conductor extending to infinity). In order to avoid interference with other simultaneously operating communication systems, it is desired to place a null in the far-field amplitude pat-tern of the antenna system at an angle of 60◦from the vertical.
Determine the three smallest physical/nontrivial heights (in meters at 200 MHz) above the ground at which the antenna can be placed to meet the desired pattern specifications.
4.64.
A ground-based, resonant, lossless linear vertical half-wavelength dipole (of length l = λ∕2), elevated at a height h about PEC (perfect electric conducting) ground plane, is used as the antenna for a ground-based communication system. It is expected that some interfer-ers/threats to the ground-based communication system will appear at a height 1,000 meters and at a horizontal distance of 1,000 meters from the ground-based antenna, as shown in the figure below.
To eliminate the presence of the interferers/threats to the operation of the ground-based communication system, we want to choose the height h of the ground-based system above the PEC ground so that we place an ideal null in the angular direction 𝜃of the interferers as measured from the reference of the coordinate system.
Determine, at a frequency of 300 MHz: (a) The normalized AF (array factor) of the equivalent antenna system that is valid in all space on and above the PEC ground plane.
Interferers/ Threats 1,000 m 1,000 m PEC h ~ λ/2 dipole θ ψ PROBLEMS 231 (b) The two smallest heights of h (in meters), from the smallest to largest, that we can place the ground-based antenna and achieve the goal; i.e., to place a null in the 𝜃direction and eliminate the presence of the interferers/threats to the operation of the ground-based system. Assume far-field observations.
4.65.
A base-station cellular communication systems lossless antenna, which is placed in a resi-dential area of a city, has a maximum gain of 16 dB (above isotropic) toward the residential area at 1,900 MHz. Assuming the input power to the antenna is 8 watts, what is the (a) maximum radiated power density (in watts∕cm2) at a distance of 100 m (line of sight) from the base station to the residential area? This will determine the safe level for human exposure to electromagnetic radiation.
(b) power (in watts) received at that point of the residential area by a cellular telephone whose antenna is a lossless λ∕4 vertical monopole and whose maximum value of the ampli-tude pattern is directed toward the maximum incident power density. Assume the λ∕4 monopole is mounted on an infinite ground plane.
4.66.
A vertical λ∕4 monopole is used as the antenna on a cellular telephone operating at 1.9 GHz.
Even though the monopole is mounted on a box-type cellular telephone, for simplicity pur-poses, assume here that it is mounted on a perfectly electric conducting (PEC) ground plane.
Assuming an incident maximum power density of 10−6 watts/m2, state or determine, for the monopole’s omnidirectional pattern, the (a) maximum directivity (dimensionless and in dB). You must state the rationale or method you are using to find the directivity.
(b) maximum power that can be delivered to the cellular telephone receiver. Assume no losses.
4.67.
A vertical, infinitesimal in length (l = λ∕50), monopole is placed on top of a police car, and it is used as an antenna for an emergency radio receiver system operating at 10 MHz. Consider that the top of the police car to be an infinite and planar PEC.The sensitivity (minimum power) of the system receiver, to be able to detect an incoming signal, is 10 𝜇watts. Assume the incoming signal is circularly polarized and it incident from a horizontal direction (grazing angle; 𝜃= 90◦), what is the: (a) Maximum directivity of the monopole in the presence of the PEC (dimensionless and in dB)?
(b) Minimum required power density (in watts/cm2) of the incoming signal to be detected by the radio receiver?
Incoming Wave ~ = 90° x z θ PEC λ/150 y 232 LINEAR WIRE ANTENNAS 4.68.
A homeowner uses a CB antenna mounted on the top of his house. Let us assume that the operating frequency is 900 MHz and the radiated power is 1,000 watts. In order not to be exposed to a long-term microwave radiation, there have been some standards, although con-troversial, developed that set the maximum safe power density that humans can be exposed to and not be subject to any harmful effects. Let us assume that the maximum safe power density of long-term human RF exposure is 10−3 watts∕cm2 or 10 watts∕m2. Assuming no losses, determine the shortest distance (in meters) from the CB antenna you must be in order not to exceed the safe level of power density exposure. Assume that the CB antenna is radiating into free-space and it is (a) an isotropic radiator.
(b) a λ∕4 monopole mounted on an infinite PEC and radiating towards its maximum.
4.69.
Derive (4-118) using (4-115).
4.70.
An infinitesimal horizontal electric dipole of length l = λ∕50 is placed parallel to the y-axis a height h above an infinite electric ground plane.
(a) Find the smallest height h (excluding h = 0) that the antenna must be elevated so that a null in the 𝜙= 90◦plane will be formed at an angle of 𝜃= 45◦from the vertical axis.
(b) For the height of part (a), determine the (1) radiation resistance and (2) directivity (for 𝜃= 0◦) of the antenna system.
4.71.
A horizontal λ∕50 infinitesimal dipole of constant current and length l is placed parallel to the y-axis a distance h = 0.707λ above an infinite electric ground plane. Find all the nulls formed by the antenna system in the 𝜙= 90◦plane.
4.72.
An infinitesimal electric dipole of length l = λ∕50 is placed horizontally at a height of h = 2λ above a flat, smooth, perfect electric conducting plane which extends to infinity. It is desired to measure its far-field radiation characteristics (e.g. amplitude pattern, phase pattern, polarization pattern, etc.). The system is operating at 300 MHz. What should the minimum radius (in meters) of the circle be where the measurements should be carried out? The radius should be measured from the origin of the coordinate system, which is taken at the interface between the actual source and image.
4.73.
An infinitesimal magnetic dipole is placed vertically a distance h above an infinite, perfectly conducting electric ground plane. Derive the far-zone fields radiated by the element above the ground plane.
4.74.
Repeat Problem 4.73 for an electric dipole above an infinite, perfectly conducting magnetic ground plane.
4.75.
A λ∕50 infinitesimal linear electric dipole operating at 500 MHz is placed horizontally a height h above a simulated flat, smooth and infinite in extent perfect magnetic conductor (PMC). Determine/write the: (a) Array factor for the system that can be used to determine the far-zone field on and above the PMC. Justify as to why you chose that array factor.
(b) Smallest height h (in λ and in cm) that the λ∕50 electric dipole must be placed so that the total far-zone amplitude pattern has a null at an angle of 𝜃= 60◦from the normal/vertical to the PMC interface.
4.76.
Repeat Problem 4.73 for a magnetic dipole above an infinite, perfectly conducting magnetic ground plane.
4.77.
An infinitesimal vertical electric dipole is placed at height h above an infinite PMC (perfect magnetic conductor) ground plane.
PROBLEMS 233 (a) Find the smallest height h (excluding h = 0) to which the antenna must be elevated so that a null is formed at an angle 𝜃= 60◦from the vertical axis (b) For the value of h found in part (a), determine 1. the directive gain of the antenna in the 𝜃= 45◦direction 2. the radiation resistance of the antenna normalized to the intrinsic impedance of the medium above the ground plane Assume that the length of the antenna is l = λ∕100.
4.78.
A vertical λ∕2 dipole, operating at 1 GHz, is placed a distance of 5 m (with respect to the tangent at the point of reflections) above the earth. Find the total field at a point 20 km from the source (d = 20 × 103 m), at a height of 1,000 m (with respect to the tangent) above the ground. Use a 4∕3 radius earth and assume that the electrical parameters of the earth are 𝜀r = 5, 𝜎= 10−2 S/m.
4.79.
A ground-based, resonant, lossless linear vertical half-wavelength dipole (of length l = λ∕2), is used to communicate with a space-borne lossless, resonant, linear wave-wavelength dipole (of length l = λ∕2). Both dipoles are oriented along the z axis. While one dipole is assumed to be at ground level, the other is elevated at a height of 1,000 meter; the two dipoles are separated horizontally by 1,000 meters, as shown below.
Assuming the input power (in the 50-ohm transmission line) feeding the dipole at the ground level is 100 mwatts, determine, at a frequency of 3 GHz, the power (in watts) received in the 50-ohm transmission line which is connected to the space borne dipole.
Assume both dipoles are radiating in an unbounded free space and each is in the far-field of the other.
z z 1,000 m λ/2 dipole Zc = 50 ohms Zc = 50 ohms λ/2 dipole 1,000 m ~ ~ 4.80.
Two astronauts equipped with handheld radios land on different parts of a large asteroid. The radios are identical and transmit 5 W average power at 300 MHz. Assume the asteroid is a smooth sphere with physical radius of 1,000 km, has no atmosphere, and consists of a lossless dielectric material with relative permittivity 𝜀r = 9. Assume that the radios’ antennas can be modeled as vertical infinitesimal electric dipoles. Determine the signal power (in microwatts) received by each radio from the other, if the astronauts are separated by a range (distance along the asteroid’s surface) of 2 km, and hold their radios vertically at heights of 1.5 m above the asteroid’s surface.
Additional Information Required to Answer this Question: Prior to landing on the asteroid the astronauts calibrated their radios. Separating themselves in outer space by 10 km, the astronauts found the received signal power at each radio from the other was 10 microwatts, when both antennas were oriented in the same direction.
234 LINEAR WIRE ANTENNAS 4.81.
A satellite S transmits an electromagnetic wave, at 10 GHz, via its transmitting antenna. The characteristics of the satellite-based transmitter are: (a) The power radiated from the satellite antenna is 10 W.
(b) The distance between the satellite antenna and a point A on the earth’s surface is 3.7 × 107 m, and (c) The satellite transmitting antenna directivity in the direction SA is 50 dB. Ignoring ground effects, 1. Determine the magnitude of the E-field at A.
2. If the receiver at point A is a λ∕2 dipole, what would be the voltage reading at the terminals of the antenna?
4.82.
Derive (4-132) based on geometrical optics as presented in section 13.2 of .
CHAPTER5 Loop Antennas 5.1 INTRODUCTION Another simple, inexpensive, and very versatile antenna type is the loop antenna. Loop antennas take many different forms such as a rectangle, square, triangle, ellipse, circle, and many other configura-tions. Because of the simplicity in analysis and construction, the circular loop is the most popular and has received the widest attention. It will be shown that a small loop (circular or square) is equivalent to an infinitesimal magnetic dipole whose axis is perpendicular to the plane of the loop. That is, the fields radiated by an electrically small circular or square loop are of the same mathematical form as those radiated by an infinitesimal magnetic dipole.
Loop antennas are usually classified into two categories, electrically small and electrically large.
Electrically small antennas are those whose overall length (circumference) is usually less than about one-tenth of a wavelength (C < λ∕10). However, electrically large loops are those whose circum-ference is about a free-space wavelength (C ∼λ). Most of the applications of loop antennas are in the HF (3–30 MHz), VHF (30–300 MHz), and UHF (300–3,000 MHz) bands. When used as field probes, they find applications even in the microwave frequency range.
Loop antennas with electrically small circumferences or perimeters have small radiation resis-tances that are usually smaller than their loss resistances. Thus they are very poor radiators, and they are seldom employed for transmission in radio communication. When they are used in any such application, it is usually in the receiving mode, such as in portable radios and pagers, where antenna efficiency is not as important as the signal-to-noise ratio. They are also used as probes for field measurements and as directional antennas for radiowave navigation. The field pattern of electrically small antennas of any shape (circular, elliptical, rectangular, square, etc.) is similar to that of an infinitesimal dipole with a null perpendicular to the plane of the loop and with its maximum along the plane of the loop. As the overall length of the loop increases and its circumference approaches one free-space wavelength, the maximum of the pattern shifts from the plane of the loop to the axis of the loop which is perpendicular to its plane.
The radiation resistance of the loop can be increased, and made comparable to the characteris-tic impedance of practical transmission lines, by increasing (electrically) its perimeter and/or the number of turns. Another way to increase the radiation resistance of the loop is to insert, within its circumference or perimeter, a ferrite core of very high permeability which will raise the magnetic field intensity and hence the radiation resistance. This forms the so-called ferrite loop.
Electrically large loops are used primarily in directional arrays, such as in helical antennas (see Section 10.3.1), Yagi-Uda arrays (see Section 10.3.3), quad arrays (see Section 10.3.4), and so on.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 235 236 LOOP ANTENNAS (a) Single element (b) Array of eight elements Figure 5.1 Commercial loop antenna as a single vertical element and in the form of an eight-element linear array. (Courtesy: TCI, A Dielectric Company).
For these and other similar applications, the maximum radiation is directed toward the axis of the loop forming an end-fire antenna. To achieve such directional pattern characteristics, the circumference (perimeter) of the loop should be about one free-space wavelength. The proper phasing between turns enhances the overall directional properties.
Loop antennas can be used as single elements, as shown in Figure 5.1(a), whose plane of its area is perpendicular to the ground. The relative orientation of the loop can be in other directions, including its plane being parallel relative to the ground. Thus, its mounting orientation will determine its radiation characteristics relative to the ground. Loops are also used in arrays of various forms.
The particular array configuration will determine its overall pattern and radiation characteristics.
One form of arraying is shown in Figure 5.1(b), where eight loops of Figure 5.1(a) are placed to form a linear array of eight vertical elements.
5.2 SMALL CIRCULAR LOOP The most convenient geometrical arrangement for the field analysis of a loop antenna is to position the antenna symmetrically on the x-y plane, at z = 0, as shown in Figure 5.2(a). The wire is assumed to be very thin and the current spatial distribution is given by I𝜙= I0 (5-1) where I0 is a constant. Although this type of current distribution is accurate only for a loop antenna with a very small circumference, a more complex distribution makes the mathematical formulation quite cumbersome.
5.2.1 Radiated Fields To find the fields radiated by the loop, the same procedure is followed as for the linear dipole. The potential function A given by (3-53) as A(x, y, z) = 𝜇 4𝜋∫C Ie(x′, y′, z′)e−jkR R dl′ (5-2) SMALL CIRCULAR LOOP 237 Figure 5.2 Geometrical arrangement for loop antenna analysis.
is first evaluated. Referring to Figure 5.2(a), R is the distance from any point on the loop to the observation point and dl′ is an infinitesimal section of the loop antenna. In general, the current spatial distribution Ie(x′, y′, z′) can be written as Ie(x′, y′, z′) = ̂ axIx(x′, y′, z′) + ̂ ayIy(x′, y′, z′) + ̂ azIz(x′, y′, z′) (5-3) whose form is more convenient for linear geometries. For the circular-loop antenna of Figure 5.2(a), whose current is directed along a circular path, it would be more convenient to write the rectan-gular current components of (5-3) in terms of the cylindrical components using the transformation (see Appendix VII) [ Ix Iy Iz ] = [ cos 𝜙′ −sin 𝜙′ 0 sin 𝜙′ cos 𝜙′ 0 0 0 1 ] [ I𝜌 I𝜙 Iz ] (5-4) 238 LOOP ANTENNAS which when expanded can be written as Ix = I𝜌cos 𝜙′ −I𝜙sin 𝜙′ Iy = I𝜌sin 𝜙′ + I𝜙cos 𝜙′ Iz = Iz ⎫ ⎪ ⎬ ⎪ ⎭ (5-5) Since the radiated fields are usually determined in spherical components, the rectangular unit vectors of (5-3) are transformed to spherical unit vectors using the transformation matrix given by (4-5). That is, ̂ ax = ̂ ar sin 𝜃cos 𝜙+ ̂ a𝜃cos 𝜃cos 𝜙−̂ a𝜙sin 𝜙 ̂ ay = ̂ ar sin 𝜃sin 𝜙+ ̂ a𝜃cos 𝜃sin 𝜙+ ̂ a𝜙cos 𝜙 ̂ az = ̂ ar cos 𝜃 −̂ a𝜃sin 𝜃 ⎫ ⎪ ⎬ ⎪ ⎭ (5-6) Substituting (5-5) and (5-6) in (5-3) reduces it to Ie = ̂ ar[I𝜌sin 𝜃cos(𝜙−𝜙′) + I𝜙sin 𝜃sin(𝜙−𝜙′) + Iz cos 𝜃] + ̂ a𝜃[I𝜌cos 𝜃cos(𝜙−𝜙′) + I𝜙cos 𝜃sin(𝜙−𝜙′) −Iz sin 𝜃] + ̂ a𝜙[−I𝜌sin(𝜙−𝜙′) + I𝜙cos(𝜙−𝜙′)] (5-7) allowing for I𝜌, I𝜙and Iz current components.
It should be emphasized that the source coordinates are designated as primed (𝜌′, 𝜙′, z′) and the observation coordinates as unprimed (r, 𝜃, 𝜙). For a very thin wire radius circular loop, the current is flowing in the 𝜙direction (I𝜙) so that (5-7) reduces to Ie = ̂ arI𝜙sin 𝜃sin(𝜙−𝜙′) + ̂ a𝜃I𝜙cos 𝜃sin(𝜙−𝜙′) + ̂ a𝜙I𝜙cos(𝜙−𝜙′) (5-8) The distance R, from any point on the loop to the observation point, can be written as R = √ (x −x′)2 + (y −y′)2 + (z −z′)2 (5-9) Since x = r sin 𝜃cos 𝜙 y = r sin 𝜃sin 𝜙 z = r cos 𝜃 x2 + y2 + z2 = r2 (5-10) x′ = a cos 𝜙′ y′ = a sin 𝜙′ z′ = 0 x′2 + y′2 + z′2 = a2 SMALL CIRCULAR LOOP 239 (5-9) reduces to R = √ r2 + a2 −2ar sin 𝜃cos(𝜙−𝜙′) (5-11) By referring to Figure 5.2(a), the differential element length is given by dl′ = a d𝜙′ (5-12) Using (5-8), (5-11), and (5-12), the 𝜙-component of (5-2) can be written as A𝜙= a𝜇 4𝜋∫ 2𝜋 0 I𝜙cos(𝜙−𝜙′) e−jk √ r2+a2−2ar sin 𝜃cos(𝜙−𝜙′) √ r2 + a2 −2ar sin 𝜃cos(𝜙−𝜙′) d𝜙′ (5-13) Since the spatial current I𝜙as given by (5-1) is constant, the field radiated by the loop will not be a function of the observation angle 𝜙. Thus any observation angle 𝜙can be chosen; for simplicity 𝜙= 0. Therefore (5-13) reduces to A𝜙= a𝜇I0 4𝜋∫ 2𝜋 0 cos 𝜙′ e−jk √ r2+a2−2ar sin 𝜃cos 𝜙′ √ r2 + a2 −2ar sin 𝜃cos 𝜙′ d𝜙′ (5-14) The integration of (5-14), for very thin circular loop of any radius, can be carried out and is rep-resented by a complex infinite series whose real part contains complete elliptic integrals of the first and second kind while the imaginary part consists of elementary functions . This treatment is only valid provided the observation distance is greater than the radius of the loop (r > a). Another very detailed and systematic treatment is that of , which is valid for any observation distance (r < a, r > a) except when the observation point is on the loop itself (r = a, 𝜃= 𝜋∕2). The develop-ment in , has been applied to circular loops whose current distribution is uniform, cosinusoidal, and Fourier cosine series. Asymptotic expansions have been presented in , to find simplified and approximate forms for far-field observations.
Both treatments, –, are too complex to be presented here. The reader is referred to the literature. In this chapter a method will be presented that approximates the integration of (5-14). For small loops, the function f = e−jk √ r2+a2−2ar sin 𝜃cos 𝜙′ √ r2 + a2 −2ar sin 𝜃cos 𝜙′ (5-15) which is part of the integrand of (5-14), can be expanded in a Maclaurin series in a using f = f(0) + f ′(0)a + 1 2!f ′′(0)a2 + ⋯+ 1 (n −1)!f (n−1)(0)an−1 + ⋯ (5-15a) 240 LOOP ANTENNAS where f ′(0) = 𝜕f∕𝜕a|a=0, f ′′(0) = 𝜕2f∕𝜕a2|a=0, and so forth. Taking into account only the first two terms of (5-15a), or f(0) = e−jkr r (5-15b) f ′(0) = (jk r + 1 r2 ) e−jkr sin 𝜃cos 𝜙′ (5-15c) f ≃ [ 1 r + a (jk r + 1 r2 ) sin 𝜃cos 𝜙′ ] e−jkr (5-15d) reduces (5-14) to A𝜙≃a𝜇I0 4𝜋∫ 2𝜋 0 cos 𝜙′ [ 1 r + a (jk r + 1 r2 ) sin 𝜃cos 𝜙′ ] e−jkr d𝜙′ A𝜙≃a2𝜇I0 4 e−jkr (jk r + 1 r2 ) sin 𝜃 (5-16) In a similar manner, the r- and 𝜃-components of (5-2) can be written as Ar ≃a𝜇I0 4𝜋sin 𝜃∫ 2𝜋 0 sin 𝜙′ [ 1 r + a (jk r + 1 r2 ) sin 𝜃cos 𝜙′ ] e−jkr d𝜙′ (5-16a) A𝜃≃−a𝜇I0 4𝜋cos 𝜃∫ 2𝜋 0 sin 𝜙′ [ 1 r + a (jk r + 1 r2 ) sin 𝜃cos 𝜙′ ] e−jkr d𝜙′ (5-16b) which when integrated reduce to zero. Thus A ≃̂ a𝜙A𝜙= ̂ a𝜙 a2𝜇I0 4 e−jkr [jk r + 1 r2 ] sin 𝜃 = ̂ a𝜙jk𝜇a2I0 sin 𝜃 4r [ 1 + 1 jkr ] e−jkr (5-17) Substituting (5-17) into (3-2a) reduces the magnetic field components to Hr = jka2I0 cos 𝜃 2r2 [ 1 + 1 jkr ] e−jkr H𝜃= −(ka)2I0 sin 𝜃 4r [ 1 + 1 jkr − 1 (kr)2 ] e−jkr H𝜙= 0 (5-18a) (5-18b) (5-18c) SMALL CIRCULAR LOOP 241 Using (3-15) or (3-10) with J = 0, the corresponding electric-field components can be written as Er = E𝜃= 0 E𝜙= 𝜂(ka)2I0 sin 𝜃 4r [ 1 + 1 jkr ] e−jkr (5-19a) (5-19b) 5.2.2 Small Loop and Infinitesimal Magnetic Dipole A comparison of (5-18a)–(5-19b) with those of the infinitesimal magnetic dipole indicates that they have similar forms. In fact, the electric and magnetic field components of an infinitesimal magnetic dipole of length l and constant “magnetic” spatial current Im are given by Er = E𝜃= H𝜙= 0 (5-20a) E𝜙= −jkIml sin 𝜃 4𝜋r [ 1 + 1 jkr ] e−jkr (5-20b) Hr = Iml cos 𝜃 2𝜋𝜂r2 [ 1 + 1 jkr ] e−jkr (5-20c) H𝜃= jkIml sin 𝜃 4𝜋𝜂r [ 1 + 1 jkr − 1 (kr)2 ] e−jkr (5-20d) These can be obtained, using duality, from the fields of an infinitesimal electric dipole, (4-8a)– (4-10c). When (5-20a)–(5-20d) are compared with (5-18a)–(5-19b), they indicate that a magnetic dipole of magnetic moment Iml is equivalent to a small electric loop of radius a and constant electric current I0 provided that Iml = jS𝜔𝜇I0 (5-21) where S = 𝜋a2 (area of the loop). Thus, for analysis purposes, the small electric loop can be replaced by a small linear magnetic dipole of constant current. The geometrical equivalence is illustrated in Figure 5.2(a) where the magnetic dipole is directed along the z-axis which is also perpendicular to the plane of the loop.
5.2.3 Power Density and Radiation Resistance The fields radiated by a small loop, as given by (5-18a)–(5-19b), are valid everywhere except at the origin. As was discussed in Section 4.1 for the infinitesimal dipole, the power in the region very close to the antenna (near field, kr ≪1) is predominantly reactive and in the far field (kr ≫1) is predominantly real. To illustrate this for the loop, the complex power density W = 1 2(E × H∗) = 1 2[(̂ a𝜙E𝜙) × (̂ arH∗ r + ̂ a𝜃H∗ 𝜃)] = 1 2(−̂ arE𝜙H∗ 𝜃+ ̂ a𝜃E𝜙H∗ r ) (5-22) 242 LOOP ANTENNAS is first formed. When (5-22) is integrated over a closed sphere, only its radial component given by Wr = 𝜂(ka)4 32 |I0|2 sin2 𝜃 r2 [ 1 + j 1 (kr)3 ] (5-22a) contributes to the complex power Pr. Thus Pr = ∯ S W⋅ds = 𝜂(ka)4 32 |I0|2 ∫ 2𝜋 0 ∫ 𝜋 0 [ 1 + j 1 (kr)3 ] sin3 𝜃d𝜃d𝜙 (5-23) which reduces to Pr = 𝜂 ( 𝜋 12 ) (ka)4|I0|2 [ 1 + j 1 (kr)3 ] (5-23a) and whose real part is equal to Prad = 𝜂 ( 𝜋 12 ) (ka)4|I0|2 (5-23b) For small values of kr(kr ≪1), the second term within the brackets of (5-23a) is dominant which makes the power mainly reactive. In the far field (kr ≫1), the second term within the brackets diminishes, which makes the power real. A comparison between (5-23a) with (4-14) indicates a difference in sign between the terms within the brackets. Whereas for the infinitesimal dipole the radial power density in the near field is capacitive, for the small loop it is inductive. This is illustrated in Figure 4.21 for the dipole and in Figures 5.13 and 5.20 for the loop.
The radiation resistance of the loop is found by equating (5-23b) to |I0|2Rr∕2. Doing this, the radiation resistance can be written as Rr = 𝜂 (𝜋 6 ) (k2a2)2 = 𝜂2𝜋 3 (kS λ )2 = 20𝜋2 (C λ )4 ≃31,171 ( S2 λ4 ) (5-24) where S = 𝜋a2 is the area and C = 2𝜋a is the circumference of the loop. The last form of (5-24) holds for loops of other configurations, such as rectangular, elliptical, etc. (See Problem 5.30).
The radiation resistance as given by (5-24) is only for a single-turn loop. If the loop antenna has N turns wound so that the magnetic field passes through all the loops, the radiation resistance is equal to that of single turn multiplied by N2. That is, Rr = 𝜂 (2𝜋 3 ) (kS λ )2 N2 = 20𝜋2 (C λ )4 N2 ≃31,171 N2 ( S2 λ4 ) (5-24a) Even though the radiation resistance of a single-turn loop may be small, the overall value can be increased by including many turns. This is a very desirable and practical mechanism that is not available for the infinitesimal dipole.
SMALL CIRCULAR LOOP 243 Example 5.1 Find the radiation resistance of a single-turn and an eight-turn small circular loop. The radius of the loop is λ∕25 and the medium is free-space.
Solution: S = 𝜋a2 = 𝜋 ( λ 25 )2 = 𝜋λ2 625 Rr (single turn) = 120𝜋 (2𝜋 3 ) ( 2𝜋2 625 )2 = 0.788 ohms Rr (8 turns) = 0.788(8)2 = 50.43 ohms The radiation and loss resistances of an antenna determine the radiation efficiency, as defined by (2-90). The loss resistance of a single-turn small loop is, in general, much larger than its radiation resistance; thus the corresponding radiation efficiencies are very low and depend on the loss resis-tance. To increase the radiation efficiency, multiturn loops are often employed. However, because the current distribution in a multiturn loop is quite complex, great confidence has not yet been placed in analytical methods for determining the radiation efficiency. Therefore greater reliance has been placed on experimental procedures. Two experimental techniques that can be used to measure the radiation efficiency of a small multiturn loop are those that are usually referred to as the Wheeler method and the Q method .
Usually it is assumed that the loss resistance of a small loop is the same as that of a straight wire whose length is equal to the circumference of the loop, and it is computed using (2-90b). Although this assumption is adequate for single-turn loops, it is not valid for multiturn loops. In a multiturn loop, the current is not uniformly distributed around the wire but depends on the skin and proximity effects . In fact, for close spacings between turns, the contribution to the loss resistance due to the proximity effect can be larger than that due to the skin effect.
The total ohmic resistance for an N-turn circular-loop antenna with loop radius a, wire radius b, and loop separation 2c, shown in Figure 5.3(a) is given by Rohmic = Na b Rs (Rp R0 + 1 ) (5-25) where Rs = √𝜔𝜇0 2𝜎= surface impedance of conductor Rp = ohmic resistance per unit length due to proximity effect R0 = NRs 2𝜋b = ohmic skin effect resistance per unit length (ohms/m) The ratio of Rp∕R0 has been computed as a function of the spacing c/b for loops with 2 ≤N ≤8 and it is shown plotted in Figure 5.3(b). It is evident that for close spacing the ohmic resistance is twice as large as that in the absence of the proximity effect (Rp∕R0 = 0).
244 LOOP ANTENNAS Figure 5.3 N-turn circular loop and ohmic resistance due to proximity effect. (source: G. S. Smith, “Radia-tion Efficiency of Electrically Small Multiturn Loop Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-20, No. 5, September, pp. 656–657. 1972 c ⃝1972 IEEE).
Example 5.2 Find the radiation efficiency of a single-turn and an eight-turn small circular loop at f = 100 MHz. The radius of the loop is λ∕25, the radius of the wire is 10−4λ, and the turns are spaced 4 × 10−4λ apart. Assume the wire is copper with a conductivity of 5.7 × 107(S/m) and the antenna is radiating into free-space.
Solution: From Example 5.1 Rr (single turn) = 0.788 ohms Rr (8 turns) = 50.43 ohms The loss resistance for a single turn is given, according to (2-90b), by RL = Rhf = a b √𝜔𝜇0 2𝜎= 1 25(10−4) √ 𝜋(108)(4𝜋× 10−7) 5.7 × 107 = 1.053 ohms SMALL CIRCULAR LOOP 245 and the radiation efficiency, according to (2-90), by ecd = 0.788 0.788 + 1.053 = 0.428 = 42.8% From Figure 5.3(b) Rp R0 = 0.38 and from (5-25) RL = Rohmic = 8 25(10−4) √ 𝜋(108)(4𝜋× 10−7) 5.7 × 107 (1.38) = 11.62 Thus ecd = 50.43 50.43 + 11.62 = 0.813 = 81.3% 5.2.4 Near-Field (kr ≪1) Region The expressions for the fields, as given by (5-18a)–(5-19b), can be simplified if the observations are made in the near field (kr ≪1). As for the infinitesimal dipole, the predominant term in each expression for the field in the near-zone region is the last one within the parentheses of (5-18a)– (5-19b). Thus for kr ≪1 Hr ≃a2I0e−jkr 2r3 cos 𝜃 H𝜃≃a2I0e−jkr 4r3 sin 𝜃 H𝜙= Er = E𝜃= 0 E𝜙≃−ja2kI0e−jkr 4r2 sin 𝜃 ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ kr ≪1 (5-26a) (5-26b) (5-26c) (5-26d) The two H-field components are in time-phase. However, they are in time quadrature with those of the electric field. This indicates that the average power (real power) is zero, as is for the infinitesimal electric dipole. The condition of kr ≪1 can be satisfied at moderate distances away from the antenna provided the frequency of operation is very low. The fields of (5-26a)–(5-26d) are usually referred to as quasi-stationary.
5.2.5 Far-Field (kr ≫1) Region The other space of interest where the fields can be approximated is the far-field (kr ≫1) region. In contrast to the near field, the dominant term in (5-18a)–(5-19b) for kr ≫1 is the first one within the parentheses. Since for kr > 1 the Hr component will be inversely proportional to r2 whereas H𝜃 246 LOOP ANTENNAS will be inversely proportional to r. For large values of kr(kr ≫1), the Hr component will be small compared to H𝜃. Thus it can be assumed that it is approximately equal to zero. Therefore for kr ≫1, H𝜃≃−k2a2I0e−jkr 4r sin 𝜃= −𝜋SI0e−jkr λ2r sin 𝜃 E𝜙≃𝜂k2a2I0e−jkr 4r sin 𝜃= 𝜂𝜋SI0e−jkr λ2r sin 𝜃 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ kr ≫1 Hr ≃H𝜙= Er = E𝜃= 0 (5-27a) (5-27b) (5-27c) where S = 𝜋a2 is the geometrical area of the loop.
Forming the ratio of −E𝜙∕H𝜃, the wave impedance can be written as Zw = − E𝜙 H𝜃 ≃𝜂 (5-28) where Zw = wave impedance 𝜂= intrinsic impedance As for the infinitesimal dipole, the E- and H-field components of the loop in the far-field (kr ≫1) region are perpendicular to each other and transverse to the direction of propagation. They form a Transverse Electro Magnetic (TEM) field whose wave impedance is equal to the intrinsic impedance of the medium. Equations (5-27a)–(5-27c) can also be derived using the procedure outlined and relationships developed in Section 3.6. This is left as an exercise to the reader (Problem 5.9).
5.2.6 Radiation Intensity and Directivity The real power Prad radiated by the loop was found in Section 5.2.3 and is given by (5-23b). The same expression can be obtained by forming the average power density, using (5-27a)–(5-27c), and integrating it over a closed sphere of radius r. This is left as an exercise to the reader (Problem 5.8).
Associated with the radiated power Prad is an average power density Wav. It has only a radial com-ponent Wr which is related to the radiation intensity U by U = r2Wr = 𝜂 2 ( k2a2 4 )2 |I0|2 sin2 𝜃= r2 2𝜂|E𝜙(r, 𝜃, 𝜙)|2 (5-29) and it conforms to (2-12a). The normalized pattern of the loop, as given by (5-29), is identical to that of the infinitesimal dipole shown in Figure 4.3. The maximum value occurs at 𝜃= 𝜋∕2, and it is given by Umax = U|𝜃=𝜋∕2 = 𝜂 2 ( k2a2 4 )2 |I0|2 (5-30) Using (5-30) and (5-23b), the directivity of the loop can be written as D0 = 4𝜋Umax Prad = 3 2 (5-31) SMALL CIRCULAR LOOP 247 and its maximum effective area as Aem = ( λ2 4𝜋 ) D0 = 3λ2 8𝜋 (5-32) It is observed that the directivity, and as a result the maximum effective area, of a small loop is the same as that of an infinitesimal electric dipole. This should be expected since their patterns are identical.
The far-field expressions for a small loop, as given by (5-27a)–(5-27c), will be obtained by another procedure in the next section. In that section a loop of any radius but of constant current will be analyzed. The small loop far-field expressions will then be obtained as a special case of that problem.
Example 5.3 The radius of a small loop of constant current is λ∕25. Find the physical area of the loop and compare it with its maximum effective aperture.
Solution: S (physical) = 𝜋a2 = 𝜋 ( λ 25 )2 = 𝜋λ2 625 = 5.03 × 10−3λ2 Aem = 3λ2 8𝜋= 0.119λ2 Aem S = 0.119λ2 5.03 × 10−3λ2 = 23.66 Electrically the loop is about 24 times larger than its physical size, which should not be surprising.
To be effective, a small loop must be larger electrically than its physical size.
5.2.7 Equivalent Circuit A small loop is primarily inductive, and it can be represented by a lumped element equivalent circuit similar to those of Figure 2.28.
A. Transmitting Mode The equivalent circuit for its input impedance when the loop is used as a transmitting antenna is that shown in Figure 5.4. This is similar to the equivalent circuit of Figure 2.28(b). Therefore its input impedance Zin is represented by Zin = Rin + jXin = (Rr + RL) + j(XA + Xi) (5-33) where Rr = radiation resistance as given by (5-24) RL = loss resistance of loop conductor XA = external inductive reactance of loop antenna = 𝜔LA Xi = internal high-frequency reactance of loop conductor = 𝜔Li 248 LOOP ANTENNAS Zg a b Rr XA Xi Cr RL Zin ' Zin + – Vg Figure 5.4 Equivalent circuit of loop antenna in transmitting mode.
In Figure 5.4 the capacitor Cr is used in parallel to (5-33) to resonate the antenna; it can also be used to represent distributed stray capacitances. In order to determine the capacitance of Cr at resonance, it is easier to represent (5-33) by its equivalent admittance Yin of Yin = Gin + jBin = 1 Zin = 1 Rin + jXin (5-34) where Gin = Rin R2 in + X2 in (5-34a) Bin = − Xin R2 in + X2 in (5-34b) At resonance, the susceptance Br of the capacitor Cr must be chosen to eliminate the imaginary part Bin of (5-34) given by (5-34b). This is accomplished by choosing Cr according to Cr = Br 2𝜋f = −Bin 2𝜋f = 1 2𝜋f Xin R2 in + X2 in (5-35) Under resonance the input impedance Z′ in is then equal to Z′ in = R′ in = 1 Gin = R2 in + X2 in Rin = Rin + X2 in Rin (5-36) The loss resistance RL of the loop conductor can be computed using techniques illustrated in Example 5.2. The inductive reactance XA of the loop is computed using the inductance LA of: Circular loop of radius a and wire radius b: LA = 𝜇0a [ ln (8a b ) −2 ] (5-37a) Square loop with sides a and wire radius b: LA = 2𝜇0 a 𝜋 [ ln (a b ) −0.774 ] (5-37b) SMALL CIRCULAR LOOP 249 Zin ' ZL 2 1 + – Voc + – VL 2b 1 2 a x y z ψ Hi Ei θ φ' (a) Plane wave incident on a receiving loop (G.S. Smith, “Loop Antennas,” Copyright © 1984, McGraw-Hill, Inc. Permission by McGraw-Hill, Inc.) (b) Thevenin equivalent i i Figure 5.5 Loop antenna and its equivalent in receiving mode.
The internal reactance of the loop conductor Xi can be found using the internal inductance Li of the loop which for a single turn can be approximated by Li = l 𝜔P √𝜔𝜇0 2𝜎= a 𝜔b √𝜔𝜇0 2𝜎 Circular loop (5-38a) Li = l 𝜔P √𝜔𝜇0 2𝜎= 2a 𝜔𝜋b √𝜔𝜇0 2𝜎 Square loop (5-38b) where l is the length and P is the perimeter (circumference) of the cross section of the wire of the loop.
B. Receiving Mode The loop antenna is often used as a receiving antenna or as a probe to measure magnetic flux density.
Therefore when a plane wave impinges upon it, as shown in Figure 5.5(a), an open-circuit voltage develops across its terminals. This open-circuit voltage is related according to (2-93) to its vector effective length and incident electric field. This open-circuit voltage is proportional to the incident magnetic flux density Bi z, which is normal to the plane of the loop. Assuming the incident field is uniform over the plane of the loop, the open-circuit voltage for a single-turn loop can be written as Voc = j𝜔𝜋a2Bi z (5-39) Defining in Figure 5.5(a) the plane of incidence as the plane formed by the z axis and radial vector, then the open-circuit voltage of (5-39) can be related to the magnitude of the incident magnetic and electric fields by Voc = j𝜔𝜋a2𝜇0Hi cos 𝜓i sin 𝜃i = jk0𝜋a2Ei cos 𝜓i sin 𝜃i (5-39a) where 𝜓i is the angle between the direction of the magnetic field of the incident plane wave and the plane of incidence, as shown in Figure 5.5(a).
250 LOOP ANTENNAS Since the open-circuit voltage is also related to the vector effective length by (2-93), then the effective length for a single-turn loop can be written as 𝓁e = ̂ a𝜙le = ̂ a𝜙jk0𝜋a2 cos 𝜓i sin 𝜃i = ̂ a𝜙jk0S cos 𝜓i sin 𝜃i (5-40) where S is the area of the loop. The factor cos 𝜓i sin 𝜃i is introduced because the open-circuit voltage is proportional to the magnetic flux density component Bi z which is normal to the plane of the loop.
When a load impedance ZL is connected to the output terminals of the loop as shown in Figure 5.5(b), the voltage VL across the load impedance ZL is related to the input impedance Z′ in of Figure 5.5(b) and the open-circuit voltage of (5-39a) by VL = Voc ZL Z′ in + ZL (5-41) 5.3 CIRCULAR LOOP OF CONSTANT CURRENT Let us now reconsider the loop antenna of Figure 5.2(a) but with a radius that may not necessarily be small. The current in the loop will again be assumed to be constant, as given by (5-1). For this current distribution, the vector potential is given by (5-14). The integration in (5-14) is quite complex, as is indicated right after (5-14). However, if the observation are restricted in the far-field (r ≫a) region, the small radius approximation is not needed to simplify the integration of (5-14).
Although the uniform current distribution along the perimeter of the loop is only valid provided the circumference is less than about 0.2λ (radius less than about 0.032λ), the procedure developed here for a constant current can be followed to find the far-zone fields of any size loop with not necessarily uniform current.
5.3.1 Radiated Fields To find the fields in the far-field region, the distance R can be approximated by R = √ r2 + a2 −2ar sin 𝜃cos 𝜙′ ≃ √ r2 −2ar sin 𝜃cos 𝜙′ for r ≫a (5-42) which can be reduced, using the binomial expansion, to R ≃r √ 1 −2a r sin 𝜃cos 𝜙′ = r −a sin 𝜃cos 𝜙′ = r −a cos 𝜓0 for phase terms R ≃r for amplitude terms ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (5-43) since cos 𝜓0 = ̂ a′ 𝜌⋅̂ ar|𝜙=0 = (̂ ax cos 𝜙′ + ̂ ay sin 𝜙′) ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃)|𝜙= 0 = sin 𝜃cos 𝜙′ (5-43a) CIRCULAR LOOP OF CONSTANT CURRENT 251 Figure 5.6 Geometry for far-field analysis of a loop antenna.
The geometrical relation between R and r, for any observation angle 𝜙in the far-field region, is shown in Figure 5.2(b). For observations at 𝜙= 0, it simplifies to that given by (5-43) and shown in Figure 5.6. Thus (5-14) can be simplified to A𝜙≃a𝜇I0e−jkr 4𝜋r ∫ 2𝜋 0 cos 𝜙′e+jka sin 𝜃cos 𝜙′ d𝜙′ (5-44) and it can be separated into two terms as A𝜙≃a𝜇I0e−jkr 4𝜋r [ ∫ 𝜋 0 cos 𝜙′e+jka sin 𝜃cos 𝜙′ d𝜙′ + ∫ 2𝜋 𝜋 cos 𝜙′e+jka sin 𝜃cos 𝜙′ d𝜙′ ] (5-45) The second term within the brackets can be rewritten by making a change of variable of the form 𝜙′ = 𝜙′′ + 𝜋 (5-46) Thus (5-45) can also be written as A𝜙≃a𝜇I0e−jkr 4𝜋r [ ∫ 𝜋 0 cos 𝜙′e+jka sin 𝜃cos 𝜙′ d𝜙′ −∫ 𝜋 0 cos 𝜙′′e−jka sin 𝜃cos 𝜙′′ d𝜙′′ ] (5-47) Each of the integrals in (5-47) can be integrated by the formula (see Appendix V) 𝜋jnJn(z) = ∫ 𝜋 0 cos(n𝜙)e+jz cos 𝜙d𝜙 (5-48) 252 LOOP ANTENNAS where Jn(z) is the Bessel function of the first kind of order n. Using (5-48) reduces (5-47) to A𝜙≃a𝜇I0e−jkr 4𝜋r [𝜋jJ1(ka sin 𝜃) −𝜋jJ1(−ka sin 𝜃)] (5-49) The Bessel function of the first kind and order n is defined (see Appendix V) by the infinite series Jn(z) = ∞ ∑ m=0 (−1)m(z∕2)n+2m m!(m + n)!
(5-50) By a simple substitution into (5-50), it can be shown that Jn(−z) = (−1)nJn(z) (5-51) which for n = 1 is equal to J1(−z) = −J1(z) (5-52) Using (5-52) we can write (5-49) as A𝜙≃ja𝜇I0e−jkr 2r J1(ka sin 𝜃) (5-53) The next step is to find the E- and H-fields associated with the vector potential of (5-53). Since (5-53) is only valid for far-field observations, the procedure outlined in Section 3.6 can be used. The vector potential A, as given by (5-53), is of the form suggested by (3-56). That is, the r variations are separable from those of 𝜃and 𝜙. Therefore, according to (3-58a)–(3-58b) and (5-53) Er ≃E𝜃= 0 E𝜙≃ak𝜂I0e−jkr 2r J1(ka sin 𝜃) Hr ≃H𝜙= 0 H𝜃≃− E𝜙 𝜂= −akI0e−jkr 2r J1(ka sin 𝜃) (5-54a) (5-54b) (5-54c) (5-54d) 5.3.2 Power Density, Radiation Intensity, Radiation Resistance, and Directivity The next objective for this problem will be to find the power density, radiation intensity, radiation resistance, and directivity. To do this, the time-average power density is formed. That is, Wav = 1 2Re[E × H∗] = 1 2Re[̂ a𝜙E𝜙× ̂ a𝜃H∗ 𝜃] = ̂ ar 1 2𝜂|E𝜙|2 (5-55) which can be written using (5-54b) as Wav = ̂ arWr = ̂ ar (a𝜔𝜇)2|I0|2 8𝜂r2 J1 2(ka sin 𝜃) (5-56) CIRCULAR LOOP OF CONSTANT CURRENT 253 Figure 5.7 Elevation plane amplitude patterns for a circular loop of constant current (a = 0.1λ, 0.2λ, and 0.5λ).
with the radiation intensity given by U = r2Wr = (a𝜔𝜇)2|I0|2 8𝜂 J1 2(ka sin 𝜃) (5-57) The 2-D radiation patterns for a = λ∕10, λ∕5, and λ∕2, based on a uniform current distribution, are shown in Figure 5.7. These patterns indicate that the field radiated by the loop along its axis (𝜃= 0◦) is zero. Also the shape of these patterns is similar to that of a linear dipole with l ≤λ (a figure-eight shape). As the radius a increases beyond 0.5λ, the field intensity along the plane of the loop (𝜃= 90◦) diminishes and eventually it forms a null when a ≃0.61λ. This is left as an exercise to the reader for verification (Prob. 5.21). Beyond a = 0.61λ, the radiation along the plane of the loop begins to intensify and the pattern attains a multilobe form.
Three-dimensional patterns for loop circumferences of C = 0.1λ and 5λ, assuming a uniform current distribution, are shown in Figure 5.8. It is apparent that for the 0.1λ circumference the pattern 254 LOOP ANTENNAS Normalized amplitude pattern (dB) z –10 –5 0 –20 –15 x (a) C = 0.1 λ (b) C = 5 λ y –35 –30 –25 z Normalized amplitude pattern (dB) –5 0 –15 –10 –25 –20 x y –35 –30 Figure 5.8 Three-dimensional amplitude patterns of a circular loop with constant current distribution.
CIRCULAR LOOP OF CONSTANT CURRENT 255 is basically that of figure eight (sin 𝜃), while for the 5λ loop it exhibits multiple lobes. The multiple lobes in a large loop begin to form when the circumference exceeds about 3.83λ (radius exceeds about 0.61λ); see Problem 5.21.
The patterns represented by (5-57) (some of them are illustrated in Figure 5.7) assume that the current distribution, no matter what the loop size, is constant. This is not a valid assumption if the loop circumference C(C = 2𝜋a) exceeds about 0.2λ (i.e., a > 0.032λ) . For radii much greater than about 0.032λ, the current variation along the circumference of the loop begins to attain a distribution that is best represented by a Fourier series . Although a most common assumption is that the current distribution is nearly cosinusoidal, it is not physical and satisfactory particularly near the driving point of the antenna.
A uniform and nonuniform in-phase current distribution can be attained on a loop antenna even if the radius is large. To accomplish this, the loop is subdivided into sections, with each section/arc of the loop fed with a different feed line; all feed lines are typically fed from a common feed source. Such an arrangement, although more complex, can approximate either uniform or nonuniform in-phase current distribution.
It has been shown that when the circumference of the loop is about one wavelength (C ≃λ), its maximum radiation based on a nonuniform current distribution is along its axis (𝜃= 0◦, 180◦) which is perpendicular to the plane of the loop. This will also be discussed in Section 5.4 that fol-lows. Feature of the loop antenna has been utilized to design Yagi-Uda arrays whose basic elements (feed, directors, and reflectors) are circular loops –. Because of its many applications, the one-wavelength circumference circular-loop antenna is considered as fundamental as a half-wavelength dipole.
The radiated power can be written using (5-56) as Prad = ∫∫ S Wav ⋅ds = 𝜋(a𝜔𝜇)2|I0|2 4𝜂 ∫ 𝜋 0 J1 2(ka sin 𝜃) sin 𝜃d𝜃 (5-58) The integral in (5-58) can be rewritten as ∫ 𝜋 0 J1 2(ka sin 𝜃) sin 𝜃d𝜃= 1 ka ∫ 2ka 0 J2(x) dx (5-59) The evaluation of the integral of (5-59) has been the subject of papers –. In these refer-ences, along with some additional corrections, the integral of (5-59) Q(1) 11 (ka) = 1 2 ∫ 𝜋 0 J2 1(ka sin 𝜃) sin 𝜃d𝜃= 1 2ka ∫ 2ka 0 J2(x) dx (5-59a) can be represented by a series of Bessel functions Q(1) 11 (ka) = 1 ka ∞ ∑ m=0 J2m+3(2ka) (5-59b) where Jm(x) is the Bessel function of the first kind, mth order. This is a highly convergent series (typically no more than 2ka terms are necessary), and its numerical evaluation is very efficient.
Approximations to (5-59) can be made depending upon the values of the upper limit (large or small radii of the loop).
256 LOOP ANTENNAS A. Large Loop Approximation (a ≥λ∕2) To evaluate (5-59), the first approximation will be to assume that the radius of the loop is large (a ≥λ∕2). For that case, the integral in (5-59) can be approximated by ∫ 𝜋 0 J1 2(ka sin 𝜃) sin 𝜃d𝜃= 1 ka ∫ 2ka 0 J2(x) dx ≃1 ka (5-60) and (5-58) by Prad ≃𝜋(a𝜔𝜇)2|I0|2 4𝜂(ka) (5-61) The maximum radiation intensity occurs when ka sin 𝜃= 1.84 so that U|max = (a𝜔𝜇)2|I0|2 8𝜂 J1 2(ka sin 𝜃)|ka sin 𝜃=1.84 = (a𝜔𝜇)2|I0|2 8𝜂 (0.582)2 (5-62) Thus Rr = 2Prad |I0|2 = 2𝜋(a𝜔𝜇)2 4𝜂(ka) = 𝜂 (𝜋 2 ) ka = 60𝜋2(ka) = 60𝜋2 (C λ ) (5-63a) D0 = 4𝜋Umax Prad = 4𝜋ka(0.582)2 2𝜋 = 2ka(0.582)2 = 0.677 (C λ ) (5-63b) Aem = λ2 4𝜋D0 = λ2 4𝜋 [ 0.677 (C λ )] = 5.39 × 10−2λC (5-63c) where C(circumference) = 2𝜋a and 𝜂≃120𝜋.
B. Intermediate Loop Approximation (λ∕6𝜋≤a < λ∕2) If the radius of the loop is λ∕(6𝜋) = 0.053λ ≤a < λ∕2, the integral of (5-59) for Q(1) 11 (ka) is approx-imated by (5-59a) and (5-59b), and the radiation resistance and directivity can be expressed, respec-tively, as Rr = 2Prad |I0|2 = 𝜂𝜋(ka)2Q(1) 11 (ka) (5-64a) D0 = 4𝜋Umax Prad = Fm(ka) Q(1) 11 (ka) (5-64b) where Fm(ka) = J2 1(ka sin 𝜃)|max = ⎧ ⎪ ⎨ ⎪ ⎩ J2 1(1.840) = (0.582)2 = 0.339 ka > 1.840 (a > 0.293λ) J2 1(ka) ka < 1.840 (a < 0.293λ) (5-64c) (5-64d) C. Small Loop Approximation (a < λ∕6𝜋) If the radius of the loop is small (a < λ∕6𝜋), the expressions for the fields as given by (5-54a)– (5-54d) can be simplified. To do this, the Bessel function J1(ka sin 𝜃) is expanded, by the definition CIRCULAR LOOP OF CONSTANT CURRENT 257 of (5-50), in an infinite series of the form (see Appendix V) J1(ka sin 𝜃) = 1 2(ka sin 𝜃) −1 16(ka sin 𝜃)3 + ⋯ (5-65) For small values of ka(ka < 1 3), (5-65) can be approximated by its first term, or J1(ka sin 𝜃) ≃ka sin 𝜃 2 (5-65a) Thus (5-54a)–(5-54d) can be written as Er ≃E𝜃= 0 E𝜙≃a2𝜔𝜇kI0e−jkr 4r sin 𝜃= 𝜂a2k2I0e−jkr 4r sin 𝜃 Hr ≃H𝜙= 0 H𝜃≃−a2𝜔𝜇kI0e−jkr 4𝜂r sin 𝜃= −a2k2I0e−jkr 4r sin 𝜃 ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ a < λ∕6𝜋 (5-66a) (5-66b) (5-66c) (5-66d) which are identical to those of (5-27a)–(5-27c). Thus the expressions for the radiation resistance, radiation intensity, directivity, and maximum effective aperture are those given by (5-24), (5-29), (5-31), and (5-32).
To demonstrate the variation of the radiation resistance as a function of the radius a of the loop, it is plotted in Figure 5.9 for λ∕100 ≤a ≤λ∕30 using (5-24), based on the approximation of (5-65a).
Figure 5.9 Radiation resistance for a constant current circular-loop antenna based on the approximation of (5-65a).
258 LOOP ANTENNAS Figure 5.10 Radiation resistance and directivity for circular loop of constant current. (source: E. A. Wolff, Antenna Analysis, Wiley, New York, 1966).
It is evident that the values are extremely low (less than 1 ohm), and they are usually smaller than the loss resistances of the wires. These radiation resistances also lead to large mismatch losses when connected to practical transmission lines of 50 or 75 ohms. To increase the radiation resistance, it would require multiple turns as suggested by (5-24a). This, however, also increases the loss resistance which contributes to the inefficiency of the antenna. A plot of the radiation resistance for 0 < ka = C∕λ < 20, based on the evaluation of (5-59) by numerical techniques, is shown in Figure 5.10. The dashed line represents (5-63a) based on the large loop approximation of (5-60) and the dotted (⋅⋅⋅⋅⋅) represents (5-24) based on the small loop approximation of (5-65a).
In addition to the real part of the input impedance, there is also an imaginary component which would increase the mismatch losses, even if the real part is equal to the characteristic impedance of the lossless transmission line. However, the imaginary component can always, in principle at least, be eliminated by connecting a reactive element (capacitive or inductive) across the terminals of the loop, as shown in Figure 5.4, to make the antenna a resonant circuit.
To facilitate the computations for the directivity and radiation resistance of a circular loop with a constant current distribution, a MATLAB and FORTRAN computer program has been developed.
The program utilizes (5-62) and (5-58) to compute the directivity [(5-58) is integrated numeri-cally]. The program requires as an input the radius of the loop (in wavelengths). A Bessel function CIRCULAR LOOP WITH NONUNIFORM CURRENT 259 subroutine is contained within the FORTRAN program. A listing of the program is included in the CD attached with the book.
5.4 CIRCULAR LOOP WITH NONUNIFORM CURRENT The analysis in the previous sections was based on a uniform current, which would be a valid approx-imation when the radius of the loop is small electrically (usually about a < 0.032λ). As the dimen-sions of the loop increase, the current variations along the circumference of the loop must be taken into account.
A common assumption is a cosinusoidal variation for the current distribution , . This, however, is not a physical and satisfactory representation, particularly near the driving point . A better distribution would be to represent the current by a Fourier series, based on a delta gap voltage V across an infinitesimal gap at 𝜙′ = 0 on the loop, and represented by –: I(𝜙′) = ∞ ∑ n = −∞ Inejn𝜙′ = ∞ ∑ n=0 In cos(n𝜙′) = V𝛿(𝜙′) j𝜋𝜂0 [ 1 c0 + 2 ∞ ∑ n =1 cos ( n𝜙′) cn ] (5-67) where 𝛿(𝜙′) = delta function, 𝜂0 = 377 ohms, and 𝜙′ is measured from the feed point of the loop along the circumference, as shown in the inset of Figure 5.11.
A complete analysis of the fields radiated by a loop with nonuniform current distribution of (5-67) is somewhat complex and quite lengthy , . Instead of attempting to include the ana-lytical formulations, which are advanced but well documented in the cited references, a summary will be presented along with number of graphical illustrations of numerical and experimental data.
These curves can be used in facilitating designs.
Based on the current distribution of (5-67), it is shown in that the far-zone electric fields radiated by the loop are represented by E𝜃≈−V cot 𝜃 2𝜋 e−jkr r ∞ ∑ n=1 n(j)n cn sin(n𝜙)Jn(ka sin 𝜃) (5-68a) E𝜙≈−Vka 2𝜋 e−jkr r ∞ ∑ n=0 (j)n cn cos(n𝜙)J′ n(ka sin 𝜃) (5-68b) where cn is evaluated using (5-70a) of page 262.
To illustrate that the current distribution of a wire loop antenna is not uniform unless its radius is very small, the magnitude and phase of it have been plotted in Figure 5.11 as a function of 𝜙′ (in degrees). The loop circumference C is ka = C∕λ = 0.05, 0.1, 0.2, 0.3, and 0.4 and the wire radius was chosen so that Ω = 2 ln(2𝜋a∕b) = 10. It is apparent that for ka = 0.2 the current is nearly uni-form. For ka = 0.3 the variations are slightly greater and become even larger as ka increases. On the basis of these results, loops much larger than ka = 0.2 (radius much greater than 0.032λ) cannot be considered small.
The maximum of the far-field radiation pattern shifts from 𝜃= 90◦[x-y plane; Figure 5.8(a) for small loop, C = 0.1λ, with uniform current] to 𝜃= 0◦, 180◦(for large loops with nonuniform cur-rent). Evidence that this shift occurs was computed using 3-D and 2-D patterns (elevation plane for 𝜙= 0◦, 45◦, and 90◦), based on (5-68a) and (5-68b), for a circular loop of C = λ. These results are displayed in Figure 5.12. As is apparent from the 3-D and 2-D patterns of Figure 5.12, the maximum is along 𝜃= 0◦and 180◦. The three 2-D elevation plane patterns of Figure 5.12(b) are not identical, as they should not be, because the current distribution is not uniform along its circumference; they 260 LOOP ANTENNAS 0 50 100 (a) Magnitude (b) Phase 150 200 250 300 350 0 2 4 6 8 10 12 14 16 18 Current amplitude (mA) ka = 0.05 ka = 0.1 ka = 0.2 ka = 0.3 ka = 0.4 Ω = 10 (degrees) ϕ′ 0 50 100 150 200 250 300 350 –91 –90 –89 –88 –87 –86 –85 Current phase (degrees) ka = 0.05 ka = 0.1 ka = 0.2 ka = 0.3 ka = 0.4 Ω = 10 (degrees) ϕ′ Figure 5.11 Current magnitude and phase distributions on small circular loop antennas.
are identical if the current distribution is uniform for loops of small radii (radius less than about 0.032λ).
As was indicated above, the maximum of the pattern for a loop antenna shifts from the plane of the loop (𝜃= 90◦) to its axis (𝜃= 0◦, 180◦) as the circumference of the loop approaches one wavelength, as the current changes from uniform to nonuniform. Based on the nonuniform current distribution of (5-67), the directivity of the loop along 𝜃= 0◦has been computed, and it is plotted in Figure 5.13 versus the circumference of the loop in wavelengths. The maximum directivity is about 4.63 dB, CIRCULAR LOOP WITH NONUNIFORM CURRENT 261 Normalized radiation pattern (dB) z 0 –2 –4 –6 –8 –10 –12 –14 –16 –18 –20 x y (a) 3-D (b) 2-D (elevation) –30 –20 –10 0 30 150 60 120 90 90 120 60 150 30 180 0 = 0° ϕ θ θ = 45° ϕ = 90° ϕ Figure 5.12 Far-field normalized three- and two-dimensional amplitude patterns for a loop with C = λ and Ω = 10.
262 LOOP ANTENNAS 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 –3 –2 –1 0 1 2 3 4 5 Directivity (dB) Ω = 8 Ω = 10 Ω = 12 Ω = 20 = ka (circumference in wavelengths) C/ λ D (dB) 8 1.48 4.626 10 1.45 4.592 12 1.43 4.523 20 1.39 4.354 = ka C/ λ Ω 8 1 3.344 10 1 3.412 12 1 3.442 20 1 3.476 = ka C/ λ D (dB) Ω Figure 5.13 Directivity of circular-loop antenna for 𝜃= 0, 𝜋versus electrical size (C∕λ).
and it occurs when the circumference is about 1.48λ. For a one-wavelength circumference, which is usually the optimum design for a helical antenna, the directivity is about 3.476 dB for Ω = 20. It is also apparent that the directivity is basically independent of the radius of the wire, as long as the circumference is equal or less than about 1.3 wavelengths; there are differences in directivity as a function of the wire radius for greater circumferences.
While the maximum directivity toward 𝜃= 0◦, 180◦is displayed in Figure 5.13, the overall maximum directivity may not occur along 𝜃= 0◦, 180◦for all radii of the loop. For example, for small radii loops the maximum directivity occurs along 𝜃= 90◦, and it is equal to 1.5 (1.76 dB).
To find the overall maximum directivity, a search procedure was employed over all observation angles (0◦≤𝜃≤180◦, 0◦≤𝜙≤360◦) for Ω = 8 −12, and the results are displayed in Figure 5.14.
As can be seen, the maximum overall directivity of Ω = 8 −12 occurs for Ω = 8, C = 1.6λ, it is equal to 4.88 dB, which is about 0.25 dB greater than the corresponding one along 𝜃= 0◦, 180◦of Figure 5.13.
The input impedance Zin at 𝜙′ = 0 is given by Zin = j𝜋𝜂0 1 c0 + 2 ∞ ∑ n=1 1 Cn = 1 1 Z0 + ∞ ∑ n=1 1 Zn (5-69) where Z0 = j𝜋𝜂0c0 and Zn = j𝜋𝜂0 (Cn∕2). In both (5.67) and (5-69), the maximum number of terms needed for the infinite summation to converge is the maximum value of either 5 or 3ka; that is, the max number is max[5, 3ka] .
When the loop is a perfect electric conductor and the wire radius b is much smaller than the radius a of the loop (b ≪a), the coefficients are represented by , : cn = c−n = ka (Nn+1 + Nn−1 2 ) −n2 kaNn (5-70a) Nn = N−n = 1 𝜋 [ K0 (nb a ) I0 (nb a ) + Cn ] −1 2 ∫ 2ka 0 [Ω2n(x) + jJ2n(x)] for n > 1 (5-70b) CIRCULAR LOOP WITH NONUNIFORM CURRENT 263 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1 1.5 2 2.5 3 3.5 4 4.5 5 Maximum directivity (dB) Ω = 9 Ω = 10 Ω = 11 Ω = 12 = ka (circumference in wavelengths) C/ λ Ω = 8 Figure 5.14 Circular loop maximum directivity as a function of circumference.
Cn = ln(4n) + 𝛾−2 p−1 ∑ p=0 1∕(2p + 1) and 𝛾= 0.5772 (Euler’s constant). The zero order term for Nn reduces to N0 = 1 𝜋ln (8a b ) −1 2 [ ∫ 2ka 0 [Ω0(x) + jJ0(x)]dx ] (5-71) where Ωn(x) : Lommel–Weber function Jn(x) : Bessel function of the first kind I0(x) = 1 + O(x2) ≈1: modified Bessel function of the first kind K0(x) = −(ln(x∕2) + 𝛾)I0(x) + O(x2) ≈−ln(x∕2) −𝛾: modified Bessel function of the second kind The preceding equations are commonly used to find the input impedance of a circular loop antenna. Recently, an RCL representation was proposed to solve this problem using an equivalent circuit model , . One of the objectives of this new method is to demonstrate that each natural mode of the loop can be represented as a series of resonant circuits; thus, the overall performance of the circular loop is obtained by combining the modal impedances in parallel. In using this analogy, it is easier to understand intuitively the impedance of a circular loop antenna.
Computed impedances from (5-67), based on the Fourier series representation of the current, are shown plotted in Figure 5.15. The input resistance and reactance are plotted as a function of the circumference C (in wavelengths) for 0 ≤ka = C∕λ ≤2.5. The diameter of the wire was chosen so that Ω = 2 ln(2𝜋a∕b) = 8, 9, 10, 11, and 12. It is apparent that the first antiresonance occurs when the circumference of the loop is about λ∕2, and it is extremely sharp. It is also noted that as the loop wire increases in thickness, there is a rapid disappearance of the resonances. As a matter of fact, for Ω < 9 there is only one antiresonance point. These curves (for C > λ) are similar, both qualitatively and quantitatively, to those of a linear dipole. The major difference is that the loop is more capacitive (by about 130 ohms) than a dipole. This shift in reactance allows the dipole to have several resonances and antiresonances while moderately thick loops (Ω < 9) have only one antiresonance. Also small loops are primarily inductive while small dipoles are primarily capacitive. The resistance curves for the loop and the dipole are very similar.
264 LOOP ANTENNAS 0 0.5 1 1.5 (a) Resistance (b) Reactance 2 2.5 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Resistance (ohms) Ω = 9 Ω = 10 Ω = 11 Ω = 12 Ω = 8 = ka (circumference in wavelengths) C/ λ 0 0.5 1 1.5 2 2.5 –700 –600 –500 –400 –300 –200 –100 0 100 200 300 Reactance (ohms) Ω = 9 Ω = 10 Ω = 11 Ω = 12 Ω = 8 = ka (circumference in wavelengths) C/ λ Figure 5.15 Input impedance of circular-loop antennas.
To verify the analytical formulations and the numerical computations, loop antennas were built and measurements of impedance were made . The measurements were conducted using a half-loop over an image plane, and it was driven by a two-wire line. An excellent agreement between theory and experiment was indicated everywhere except near resonances where computed conduc-tance curves were slightly higher than those measured. This is expected since ohmic losses were not taken into account in the analytical formulation. It was also noted that the measured susceptance curve was slightly displaced vertically by a constant value. This can be attributed to the “end effect” of the experimental feeding line and the “slice generator” used in the analytical modeling of the feed.
For a dipole, the correction to the analytical model is usually a negative capacitance in shunt with CIRCULAR LOOP WITH NONUNIFORM CURRENT 265 the antenna . A similar correction for the loop would result in a better agreement between the computed and measured susceptances. Computations for a half-loop above a ground plane were also performed by J. E. Jones using the Moment Method.
The radiation resistance and maximum directivity of a loop antenna with a cosinusoidal current distribution I𝜙(𝜙) = I0 cos 𝜙was derived in , and evaluated by integrating in far-zone fields.
Doing this, the values are plotted, respectively, in Figures 5.16(a,b) where they are compared with those based on a uniform and nonuniform current distribution.
0 1 2 3 4 5 6 7 8 9 10 0 1000 2000 3000 4000 5000 6000 7000 8000 C/λ = ka (circumference in wavelengths) C/ λ = ka (circumference in wavelengths) (a) Radiation resistance Rr (b) Maximum directivity D0 Radiation resistance (Ohms) Uniform Current Cosinusoidal Current Nonuniform Current (Ω = 10) 0 1 2 3 4 5 6 7 8 9 10 0 2 4 6 8 10 12 14 Maximum directivity (dB) Uniform Current Cosinusoidal Current Nonuniform Current ((Ω = 10) Figure 5.16 Radiation resistance (Rr) and maximum directivity (D0) of a circular loop with uniform, cosinu-soidal and nonuniform current distributions.
266 LOOP ANTENNAS A general Matlab computer program Circular Loop Nonuniform has been developed to com-pute the following current distributions for uniform, cosinusoidal, and Fourier series: r Current distribution, based on (5-67) r Input impedance, based on (5-69) r Far-field amplitude radiation pattern, based on (5-68a), (5-68b) r Directivity pattern (in dB) r Maximum directivity (dimensionless and in dB) The uniform (n = 0) and cosinusoidal (n = 1) current distributions are treated as special cases of the Fourier series distribution. The Matlab program was advanced by the Matlab programs written by A. F. McKinley and associates, based on , , and made available to this author. The programs by A. F. McKinley are more general while the one in this book is included to primarily aid the reader in computing the parameters listed above.
5.4.1 Arrays In addition to being used as single elements and in arrays, as shown in Figure 5.1(a,b), there are some other classic arrays of loop configurations. Two of the most popular arrays of loop antennas are the helical antenna and the Yagi-Uda array. The loop is also widely used to form a solenoid which in conjunction with a ferrite cylindrical rod within its circumference is used as a receiving antenna and as a tuning element, especially in transistor radios. This is discussed in Section 5.7.
The helical antenna, which is discussed in more detail in Section 10.3.1, is a wire antenna, which is wound in the form of a helix, as shown in Figure 10.13. It is shown that it can be modeled approx-imately by a series of loops and vertical dipoles, as shown in Figure 10.15. The helical antenna possesses in general elliptical polarization, but it can be designed to achieve nearly circular polar-ization. There are two primary modes of operation for a helix, the normal mode and the axial mode.
The helix operates in its normal mode when its overall length is small compared to the wavelength, and it has a pattern with a null along its axis and the maximum along the plane of the loop. This pattern (figure-eight type in the elevation plane) is similar to that of a dipole or a small loop. A heli-cal antenna operating in the normal mode is sometimes used as a monopole antenna for mobile cell and cordless telephones, and it is usually covered with a plastic cover. This helix monopole is used because its input impedance is larger than that of a regular monopole and more attractive for matching to typical transmission lines used as feed lines, such as a coaxial line (see Problem 10.18).
The helix operates in the axial mode when the circumference of the loop is between 3∕4λ < C < 4∕3λ with an optimum design when the circumference is nearly one wavelength. When the circumference of the loop approaches one wavelength, the maximum of the pattern is along its axis.
In addition, the phasing among the turns is such that overall the helix forms an end-fire antenna with attractive impedance and polarization characteristics (see Example 10.1). In general, the helix is a popular communication antenna in the VHF and UHF bands.
The Yagi-Uda antenna is primarily an array of linear dipoles with one element serving as the feed while the others act as parasitic. However this arrangement has been extended to include arrays of loop antennas, as shown in Figure 10.30. As for the helical antenna, in order for this array to perform as an end-fire array, the circumference of each of the elements is near one wavelength. More details can be found in Section 10.3.4 and especially in –. A special case is the quad antenna which is very popular amongst ham radio operators. It consists of two square loops, one serving as the excitation while the other is acting as a reflector; there are no directors. The overall perimeter of each loop is one wavelength.
CIRCULAR LOOP WITH NONUNIFORM CURRENT 267 5.4.2 Design Procedure The design of small loops is based on the equations for the radiation resistance (5-24), (5-24a), direc-tivity (5-31), maximum effective aperture (5-32), resonance capacitance (5-35), resonance input impedance (5-36) and inductance (5-37a), (5-37b). In order to resonate the element, the capaci-tor Cr of Figure 5.4 is chosen based on (5-35) so as to cancel out the imaginary part of the input impedance Zin.
For large loops with a nonuniform current distribution, the design is accomplished using the curves of Figure 5.13 for the axial directivity and those of Figure 5.15 for the impedance. To resonate the loop, usually a capacitor in parallel or an inductor in series is added, depending on the radius of the loop and that of the wire.
Example 5.4 Design a resonant loop antenna to operate at 100 MHz so that the pattern maximum is along the axis of the loop. Determine the radius of the loop and that of the wire (in meters), the axial directivity (in dB), and the parallel lumped element (capacitor in parallel or inductor in series) that must be used in order to resonate the antenna.
Solution: In order for the pattern maximum to be along the axis of the loop, the circumference of the loop must be large compared to the wavelength. Therefore the current distribution will be nonuniform. To accomplish this, Figure 5.15 should be used. There is not only one unique design which meets the specifications, but there are many designs that can accomplish the goal.
One design is to select a circumference where the loop is self resonant, and there is no need for a resonant capacitor. For example, referring to Figure 5.15(b) and choosing an Ω = 12, the cir-cumference of the loop is nearly 1.089λ. Since the free-space wavelength at 100 MHz is 3 meters, then the circumference is circumference ≃1.089(3) = 3.267 meters while the radius of the loop is radius = a = 3.267 2𝜋 = 0.52 meters The radius of the wire is obtained using Ω = 12 = 2 ln (2𝜋a b ) or a b = 64.2077 Therefore the radius of the wire is b = a 64.2077 = 0.52 64.2077 = 0.8099 cm = 8.099 × 10−3meters Using Figure 5.13, the axial directivity for this design is approximately 3.7 dB. Using Figure 5.15(a), the input impedance is approximately Zin = Z′ in ≃149 ohms 268 LOOP ANTENNAS Since the antenna chosen is self resonant, there is no need for a lumped element to resonate the radiator.
Another design will be to use another circumference where the loop is not self resonant. This will necessitate the use of a capacitor Cr to resonate the antenna. This is left as an end of the chapter exercise.
5.5 GROUND AND EARTH CURVATURE EFFECTS FOR CIRCULAR LOOPS The presence of a lossy medium can drastically alter the performance of a circular loop. The parame-ters mostly affected are the pattern, directivity, input impedance, and antenna efficiency. The amount of energy dissipated as heat by the lossy medium directly affects the antenna efficiency. As for the linear elements, geometrical optics techniques can be used to analyze the radiation characteristics of loops in the presence of conducting surfaces. The reflections are taken into account by introducing appropriate image (virtual) sources. Divergence factors are introduced to take into account the effects of the ground curvature. Because the techniques are identical to the formulations of Section 4.8, they will not be repeated here. The reader is directed to that section for the details. It should be pointed out, however, that a horizontal loop has horizontal polarization in contrast to the vertical polarization of a vertical electric dipole. Exact boundary-value solutions, based on Sommerfeld integral formulations, are available . However they are too complex to be included in an introductory chapter.
By placing the loop above a reflector, the pattern is made unidirectional and the directivity is increased. To simplify the problem, initially the variations of the axial directivity (𝜃= 0◦) of a cir-cular loop with a circumference of one wavelength (ka = 1) when placed horizontally a height h above an infinite in extent perfect electric conductor are examined as a function of the height above the ground plane. These were obtained using image theory and the array factor of two loops, and they are shown for 10 < Ω < 20 in Figure 5.17 , . Since only one curve is shown for 10 < Ω < 20, it is evident that the directivity variations as a function of the height are not strongly dependent on the radius of the wire of the loop. It is also apparent that for 0.05λ < h < 0.2λ and 0.65λ < h < 0.75λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 2 4 6 8 10 12 h/ Directivity (dB) λ θ Reflector = 0 h s s Theory, infinite reflector Ω = 10 – 20 Figure 5.17 Directivity of circular-loop antenna, C = ka = 1, for 𝜃= 0 versus distance from reflector h∕λ.
Theoretical curve is for infinite planar reflector. (source: G. S. Smith, “Loop Antennas,” Chapter 5 of Antenna Engineering Handbook, 1984, c ⃝1984 McGraw-Hill, Inc. Permission by McGraw-Hill, Inc).
POLYGONAL LOOP ANTENNAS 269 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 –150 –100 –50 0 50 100 150 200 h/ Resistance, R; Reactance, X (ohms) 15 20 Ω = 10 12 Theory, infinite reflector R X λ Figure 5.18 Input impedance of circular-loop antenna C = ka = 1 versus distance from reflector h∕λ. Theo-retical curves are for infinite planar reflector; measured points are for square reflector. (source: G. S. Smith, “Loop Antennas,” Chapter 5 of Antenna Engineering Handbook, 1984, c ⃝1984, McGraw-Hill, Inc. Permission by McGraw-Hill, Inc).
the directivity is about 9 dB. For the same size loop, the corresponding variations of the impedance as a function of the height are shown in Figure 5.18 , . While the directivity variations are not strongly influenced by the radius of the wire, the variations of the impedance do show a dependence on the radius of the wire of the loop for 10 < Ω < 20.
A qualitative criterion that can be used to judge the antenna performance is the ratio of the radia-tion resistance in free-space to that in the presence of the homogeneous lossy medium . This is a straightforward but very tedious approach. A much simpler method is to find directly the self-impedance changes (real and imaginary) that result from the presence of the conducting medium.
Since a small horizontal circular loop is equivalent to a small vertical magnetic dipole (see Sec-tion 5.2.2), computations were carried out for a vertical magnetic dipole placed a height h above a homogeneous lossy half-space. The changes in the self-impedance, normalized with respect to the free-space radiation resistance R0 given by (5-24), are found in . Significant changes, compared to those of a perfect conductor, are introduced by the presence of the ground.
The effects that a stratified lossy half-space have on the characteristics of a horizontal small cir-cular loop have also been investigated and documented . It was found that when a resonant loop is close to the interface, the changes in the input admittance as a function of the antenna height and the electrical properties of the lossy medium were very pronounced. This suggests that a reso-nant loop can be used effectively to sense and to determine the electrical properties of an unknown geological structure.
5.6 POLYGONAL LOOP ANTENNAS The most attractive polygonal loop antennas are the square, rectangular, triangular, and rhombic.
These antennas can be used for practical applications such as for aircraft, missiles, and communi-cations systems. However, because of their more complex structure, theoretical analyses seem to be 270 LOOP ANTENNAS unsuccessful . Thus the application of these antennas has received much less attention. However design curves, computed using the Moment Method, do exist and can be used to design polyg-onal loop antennas for practical applications. Usually the circular loop has been used in the UHF range because of its higher directivity while triangular and square loops have been applied in the HF and UHF bands because of advantages in their mechanical construction. Broadband impedance characteristics can be obtained from the different polygonal loops.
The input impedance (Z = R + jX) variations, for the following four configurations are found in : r Top-driven triangular r Base-driven triangular r Rectangular r Rhombic If the appropriate shape and feed point are chosen, a polygonal loop will have broadband impedance characteristics and be matched to the 50-ohm lines. From the four configurations listed above, the two most attractive configurations are the triangular loop (isosceles triangle) and the rectangular loop with a height/length = 0.5 ratio.
5.7 FERRITE LOOP Because the loss resistance is comparable to the radiation resistance, electrically small loops are very poor radiators and are seldom used in the transmitting mode. However, they are often used for receiving signals, such as in radios and pagers, where the signal-to-noise ratio is much more important than the efficiency.
5.7.1 Radiation Resistance The radiation resistance, and in turn the antenna efficiency, can be raised by increasing the circumfer-ence of the loop. Another way to increase the radiation resistance, without increasing the electrical dimensions of the antenna, would be to insert within its circumference a ferrite core that has a ten-dency to increase the magnetic flux, the magnetic field, the open-circuit voltage, and in turn the radiation resistance of the loop , . This is the so-called ferrite loop and the ferrite material can be a rod of very few inches in length. The radiation resistance of the ferrite loop is given by Rf Rr = (𝜇ce 𝜇0 )2 = 𝜇2 cer (5-72) where Rf = radiation resistance of ferrite loop Rr = radiation resistance of air core loop 𝜇ce = effective permeability of ferrite core 𝜇0 = permeability of free-space 𝜇cer = relative effective permeability of ferrite core Using (5-24), the radiation resistance of (5-72) for a single-turn small ferrite loop can be written as Rf = 20𝜋2 (C λ )4 (𝜇ce 𝜇0 )2 = 20𝜋2 (C λ )4 𝜇2 cer (5-73) FERRITE LOOP 271 and for an N-turn loop, using (5-24a), as Rf = 20𝜋2 (C λ )4 (𝜇ce 𝜇0 )2 N2 = 20𝜋2 (C λ )4 𝜇2 cerN2 (5-74) The relative effective permeability of the ferrite core 𝜇cer is related to the relative intrinsic per-meability of the unbounded ferrite material 𝜇fr(𝜇fr = 𝜇f ∕𝜇0) by 𝜇cer = 𝜇ce 𝜇0 = 𝜇fr 1 + D(𝜇fr −1) (5-75) where D is the demagnetization factor which has been found experimentally for different core geometries, as shown in Figure 5.19. For most ferrite material, the relative intrinsic permeability 𝜇fr is very large (𝜇fr ≫1) so that the relative effective permeability of the ferrite core 𝜇cer is approximately inversely proportional to the demagnetization factor, or 𝜇cer ∼1∕D = D−1. In general, the demag-netization factor is a function of the geometry of the ferrite core. For example, the demagnetization factor for a sphere is D = 1 3 while that for an ellipsoid of length 2l and radius a, such that l ≫a, is D = (a l )2 [ ln (2l a ) −1 ] , l ≫a (5-75a) 5.7.2 Ferrite-Loaded Receiving Loop Because of their smallness, ferrite loop antennas of few turns wound around a small ferrite rod are used as antennas, especially in the older generation pocket transistor radios. The antenna is usu-ally connected in parallel with the RF amplifier tuning capacitance and, in addition to acting as an antenna, it furnishes the necessary inductance to form a tuned circuit. Because the inductance is Figure 5.19 Demagnetization factor as a function of core length/diameter ratio.
(source: E. A. Wolff, Antenna Analysis, Wiley, New York, 1966).
272 LOOP ANTENNAS obtained with only few turns, the loss resistance is kept small. Thus the Q is usually very high, and it results in high selectivity and greater induced voltage.
The equivalent circuit for a ferrite-loaded loop antenna is similar to that of Figure 5.4 except that a loss resistance RM, in addition to RL, is needed to account for the power losses in the ferrite core.
Expressions for the loss resistance RM and inductance LA for the ferrite-loaded loop of N turns can be found in and depend on some empirical factors which are determined from an average of experimental results. The inductance Li is the same as that of the unloaded loop.
5.8 MOBILE COMMUNICATION SYSTEMS APPLICATIONS As was indicated in Section 4.7.4 of Chapter 4, the monopole was one of the most widely used elements for handheld units of mobile communication systems. An alternative to the monopole is the loop, –, which has been often used in pagers but has found very few applications in handheld transceivers. This is probably due to loop’s high resistance and inductive reactance which are more difficult to match to standard feed lines. The fact that loop antennas are more immune to noise makes them more attractive for an interfering and fading environment, like that of mobile communication systems. In addition, loop antennas become more viable candidates for wireless communication systems which utilize devices operating at higher frequency bands, particularly in designs where balanced amplifiers must interface with the antenna. Relative to top side of the handheld unit, such as the telephone, the loop can be placed either horizontally or vertically –. Either configuration presents attractive radiation characteristics for land-based mobile systems.
The radiation characteristics, normalized pattern and input impedance, of a monopole and vertical loop mounted on an experimental mobile handheld device were examined in –. The loop was in the form of a folded configuration mounted vertically on the handheld conducting device with its one end either grounded or ungrounded to the device. The predicted and measured input impedance of the folded loop, when its terminating end was grounded to the box, are displayed in Figure 5.20(a,b). It is evident that the first resonance, around 900 MHz, of the folded loop is of the parallel type (antiresonance) with a very high, and rapidly changing versus frequency, resistance, and reactance. These values and variations of impedance are usually undesirable for practical imple-mentation. For frequencies below the first resonance, the impedance is inductive (imaginary part is positive), as is typical of small loop antennas (see Figure 5.15); above the first resonance, the impedance is capacitive (negative imaginary part). The second resonance, around 2,100 MHz, is of the series type with slowly varying values of impedance, and of desirable magnitude, for practical implementation. The resonance forms (parallel vs. series) can be interchanged if the terminating end of the folded loop is ungrounded with the element then operating as an L monopole – and exhibiting the same resonance behavior as that of a monopole mounted on the device (see Chapter 4, Section 4.7.5, Figure 4.24). Even though the radiating element is a loop whose plane is vertical to the box, the amplitude pattern, in both cases (loop and L), is similar and nearly omnidirectional as that of the monopole of Figure 4.24 because the PEC box is also part of the radiating system.
A summary of the pertinent parameters and associated formulas and equation numbers for this chapter are listed in Table 5.1.
5.9 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Java-based applet for computing and displaying the radiation characteristics of a loop.
c. Java-based animation of loop amplitude pattern.
MULTIMEDIA 273 2000 1600 1200 800 400 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 −400 −800 −1200 −1600 Frequency (MHz) Frequency (MHz) (b) Imaginary part 1 cm 1 cm 2 cm 4 cm 3 cm 10 cm 10 cm 2 cm 1 cm 1 cm 3 cm 4 cm 6 cm x y z MEASUREMENTS FDTD MEASUREMENTS FDTD Reactance (Ohms) Reactance (Ohms) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 3000 2800 2600 2400 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 0 6 cm x y z (a) Real part Figure 5.20 Input impedance, real and imaginary parts of a wire folded loop mounted vertically on a con-ducting mobile hand-held unit (source: K. D. Katsibas, et al., “Folded Loop Antenna for Mobile Hand-Held Units,” IEEE Transactions Antennas Propagat., Vol. 46, No. 2, February 1998, pp. 260–266. c ⃝1998 IEEE).
d. Matlab computer program, designated Circular Loop Uniform, for computing the radiation characteristics of a loop. A description of the program is found in the READ ME file of the corresponding program in the publisher’s website for this book.
e. The Matlab computer program Circular Loop Nonuniform can be used to compute the radi-ation characteristics (current distribution, input impedance, amplitude pattern, and directivity pattern) of a circular loop with uniform cosinusoidal and Fourier series current distributions.
f. Power Point (PPT) viewgraphs, in multicolor.
274 LOOP ANTENNAS TABLE 5.1 Summary of Important Parameters, and Associated Formulas and Equation Numbers for Loop in Far Field Equation Parameter Formula Number Small Circular Loop (a < λ∕6𝜋, C < λ∕3) (Uniform Current) Normalized power pattern U = |E𝜙n|2 = C0 sin2 𝜃 (5-27b) Wave impedance Zw Zw = − E𝜙 H𝜃 ≃𝜂= 377 Ohms (5-28) Directivity D0 D0 = 3 2 = 1.761 dB (5-31) Maximum effective area Aem Aem = 3λ2 8𝜋 (5-32) Radiation resistance Rr (one turn) Rr = 20𝜋2 (C λ )4 (5-24) Radiation resistance Rr (N turns) Rr = 20𝜋2 (C λ )4 N2 (5-24a) Input resistance Rin Rin = Rr = 20𝜋2 (C λ )4 (5-24) Loss resistance RL (one turn) RL = l P √𝜔𝜇0 2𝜎= C 2𝜋b √𝜔𝜇0 2𝜎 (2-90b) Loss resistance RL (N turns) RL = Na b Rs (Rp R0 + 1 ) (5-25) Circular loop external inductance LA LA = 𝜇0a [ ln (8a b ) −2 ] (5-37a) Square loop external inductance LA LA = 2𝜇0 a 𝜋 [ ln (a b ) −0.774 ] (5-37b) Circular loop internal inductance Li Li = a 𝜔b √𝜔𝜇0 2𝜎 (5-38a) Square loop internal inductance Li LA = 2a 𝜔𝜋b √𝜔𝜇0 2𝜎 (5-38b) Vector effective length 𝓵e 𝓵e = ̂ a𝜙jk0𝜋a2 cos 𝜓i sin 𝜃i (5-40) Half-power beamwidth HPBW = 90◦ (4-65) Large Circular Loop (a ≥λ∕2, C ≥3.14λ) (Uniform Current) Normalized power pattern U = |E𝜙n|2 = C1J2 1(ka sin 𝜃) (5-57) Wave impedance Zw Zw = − E𝜙 H𝜃 ≃𝜂= 377 Ohms (5-28) Directivity D0 (a > λ∕2) D0 = 0.677 (C λ ) (5-63b) Maximum effective area Aem (a > λ∕2) Aem = λ2 4𝜋 [ 0.677 (C λ )] (5-63c) Radiation resistance (a > λ∕2), (one turn) Rr = 60𝜋2 (C λ ) (5-63a) Input resistance (a > λ∕2), (one turn) Rin = Rr = 60𝜋2 (C λ ) (5-63a) Loss resistance RL (one turn) RL = l P √𝜔𝜇0 2𝜎= C 2𝜋b √𝜔𝜇0 2𝜎 (2-90b) REFERENCES 275 TABLE 5.1 (continued) Equation Parameter Formula Number Loss resistance RL (N turns) RL = Na b Rs (Rp R0 + 1 ) (5-25) External inductance LA LA = 𝜇0a [ ln (8a b ) −2 ] (5-37a) Internal inductance Li Li = a 𝜔b √𝜔𝜇0 2𝜎 (5-38a) Vector effective length 𝓵e 𝓵e = ̂ a𝜙jk0𝜋a2 cos 𝜓i sin 𝜃i (5-40) Ferrite Circular Loop (a < λ∕6𝜋, C < λ∕3) (uniform current) Radiation resistance Rf (one turn) Rf = 20𝜋2 (C λ )4 𝜇2 cer (5-73) 𝜇cer = 𝜇fr 1 + D(𝜇fr −1) (5-75) Radiation resistance Rf (N turns) Rf = 20𝜋2 (C λ )4 𝜇2 cerN2 (5-74) Ellipsoid: D = (a l )2 [ ln (2l a ) −1 ] Demagnetizing factor D l ≫a (5-75a) Sphere: D = 1 3 REFERENCES 1. P. L. Overfelt, “Near Fields of the Constant Current Thin Circular Loop Antenna of Arbitrary Radius,” IEEE Trans. Antennas Propagat., Vol. 44, No. 2, February 1996, pp. 166–171.
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9. J. E. Storer, “Impedance of Thin-Wire Loop Antennas,” AIEE Trans., (Part I. Communication and Elec-tronics), Vol. 75, Nov. 1956, pp. 606–619.
10. S. Adachi and Y. Mushiake, “Studies of Large Circular Loop Antenna,” Sci. Rep. Research Institute of Tohoku University (RITU), B, Vol. 9, No. 2, 1957, pp. 79–103.
11. S. Ito, N. Inagaki, and T. Sekiguchi, “An Investigation of the Array of Circular-Loop Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-19, No. 4, July 1971, pp. 469–476.
12. A. Shoamanesh and L. Shafai, “Properties of Coaxial Yagi Loop Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-26, No. 4, July 1978, pp. 547–550.
276 LOOP ANTENNAS 13. A. Shoamanesh and L. Shafai, “Design Data for Coaxial Yagi Array of Circular Loops,” IEEE Trans.
Antennas Propagat., Vol. AP-27, September 1979, pp. 711–713.
14. D. DeMaw (ed.), The Radio Amateur’s Handbook, American Radio Relay League, 56th ed., 1979, pp. 20–18.
15. G. N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge University Press, London, 1922.
16. S. V. Savov, “An Efficient Solution of a Class of Integrals Arising in Antenna Theory,” IEEE Antennas Propagat. Mag., Vol. 44, No. 5, October 2002, pp. 98–101.
17. J. D. Mahony, “Circular Microstrip-Patch Directivity Revisited: An Easily Computable Exact Expression,” IEEE Antennas Propagat. Mag., Vol. 45, No. 1, February 2003, pp. 120–122.
18. J. D. Mahony, “A Comment on Q-Type Integrals and Their Use in Expressions for Radiated Power,” IEEE Antennas Propagat. Mag., Vol. 45, No. 3, June 2003, pp. 127–138.
19. S. V. Savov, “A Comment on the Radiation Resistance,” IEEE Antennas Propagat. Mag., Vol. 45, No. 3, June 2003, p. 129.
20. I. Gradshteyn and I. Ryzhik, Tables of Integrals, Series and Products, Academic Press, New York, 1965.
21. J. E. Lindsay, Jr., “A Circular Loop Antenna with Non-Uniform Current Distribution,” IRE Trans. Antennas Propagat., Vol. AP-8, No. 4, July 1960, pp. 438–441.
22. E. A. Wolff, Antenna Analysis, Wiley, New York, 1966.
23. H. C. Pocklington, “Electrical Oscillations in Wire,” Cambridge Philosophical Society Proceedings, London, England, Vol. 9, 1897, p. 324.
24. E. Hall´ en, “Theoretical Investigations into the Transmitting and Receiving Qualities of Antennae,” Nova Acta Regiae Soc. Sci. Upsaliensis, Ser. IV, No. 4, pp. 1–44, 1938.
25. J. E. Storer, “Impedance of Thin-Wire Loop Antennas,” AIEE Trans., (Part I. Communication and Elec-tronics), Vol. 75, pp. 606–619, Nov. 1956.
26. T. T. Wu, “Theory of Thin Circular Antenna,” J. Math Phys., Vol. 3, pp. 1301–1304, Nov.–Dec. 1962.
27. L. D. Licking and D. E. Merewether, “An Analysis of Thin-Wire Circular Loop Antennas of Arbitrary Size,” Report No. SC-RR-70-433, Sandia National Lab., Albuquerque, NM, Aug. 1970.
28. A. F. McKinley, T. P. White, I. S. Maksymov, and K. R. Catchpole, “The Analytical Basis for the Reso-nances and Anti-Resonances of Loop Antennas and Meta-Material Ring Resonators,” Journal of Applied Physics, Vol. 112, No. 9, pp. 094911-094911-9, Nov. 2012.
29. A. F. McKinley, T. P. White, K. R. Catchpole, “Theory of the Circular Loop Antenna in the Terahertz, Infrared, and Optical Regions,” Journal of Applied Physics, Vol. 114, No. 4, pp. 044317-044317-10, 2013.
30. R. King, “Theory of Antennas Driven from Two-Wire Line,” J. Appl. Phys., Vol. 20, 1949, p. 832.
31. D. G. Fink (ed.), Electronics Engineers’Handbook, Section 18, “Antennas” (by W. F. Croswell), McGraw-Hill, New York, pp. 18–22.
32. K. Iizuka, R. W. P. King, and C. W. Harrison, Jr., “Self- and Mutual Admittances of Two Identical Circular Loop Antennas in a Conducting Medium and in Air,” IEEE Trans. Antennas Propagat., Vol. AP-14, No. 4, July 1966, pp. 440–450.
33. R. E. Collin and F. J. Zucher (eds.), Antenna Theory Part 2, Chapter 23 (by J. R. Wait), McGraw-Hill, New York, 1969.
34. J. R. Wait, “Possible Influence of the Ionosphere on the Impedance of a Ground-Based Antenna,” J. Res.
Natl. Bur. Std. (U.S.), Vol. 66D, September–October 1962, pp. 563–569.
35. L. E. Vogler and J. L. Noble, “Curves of Input Impedance Change Due to Ground for Dipole Antennas,” U.S. National Bureau of Standards, Monograph 72, January 31, 1964.
36. D. C. Chang, “Characteristics of a Horizontal Circular Loop Antenna over a Multilayered, Dissipative Half-Space,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 6, November 1973, pp. 871–874.
37. R. W. P. King, “Theory of the Center-Driven Square Loop Antenna,” IRE Trans. Antennas Propagat., Vol. AP-4, No. 4, July 1956, p. 393.
38. T. Tsukiji and S. Tou, “On Polygonal Loop Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-28, No. 4, July 1980, pp. 571–575.
39. M. A. Islam, “A Theoretical Treatment of Low-Frequency Loop Antennas with Permeable Cores,” IEEE Trans. Antennas Propagat., Vol. AP-11, No. 2, March 1963, pp. 162–169.
PROBLEMS 277 40. V. H. Rumsey and W. L. Weeks, “Electrically Small Ferrite Loaded Loop Antennas,” IRE Convention Rec., Vol. 4, Part 1, 1956, pp. 165–170.
41. K. Fujimoto and J. R. James, Mobile Antenna Systems Handbook, Artech House, Norwood, MA, 1994.
42. M. A. Jensen and Y. Rahmat-Samii, “Performance Analysis of Antennas for Hand-Held Transceivers Using FDTD,” IEEE Trans. Antennas Propagat., Vol. 42, No. 8, August 1994, pp. 1106–1113.
43. M. A. Jensen and Y. Rahmat-Samii, “EM Interaction of Handset Antennas and a Human in Personal Com-munications,” Proc. IEEE, Vol. 83, No. 1, January 1995, pp. 7–17.
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45. C. A. Balanis, K. D. Katsibas, P. A. Tirkas, and C. R. Birtcher, “Loop Antenna for Mobile and Per-sonal Communication Systems,” IEEE International Vehicular Technology Conference (IEEE VTC ’97), Phoenix, AZ, May 5–7, 1997.
46. K. D. Katsibas, C. A. Balanis, P. A. Tirkas, and C. R. Birtcher, “Folded Loop Antenna for Mobile Handheld Units,” IEEE Trans. Antennas Propagat., Vol. 46, No. 2, February 1998, pp. 260–266.
PROBLEMS 5.1.
Derive (a) (5-18a)–(5-18c) using (5-17) and (3-2a) (b) (5-19a)–(5-19b) using (5-18a)–(5-18c) 5.2.
Write the fields of an infinitesimal linear magnetic dipole of constant current Im, length l, and positioned along the z-axis. Use the fields of an infinitesimal electric dipole, (4-8a)–(4-10c), and apply the principle of duality. Compare with (5-20a)–(5-20d).
5.3.
A circular loop, of loop radius λ∕30 and wire radius λ∕1000, is used as a transmitting/ receiving antenna in a back-pack radio communication system at 10 MHz. The wire of the loop is made of copper with a conductivity of 5.7 × 107 S∕m. Assuming the antenna is radi-ating in free space, determine the (a) radiation resistance of the loop; (b) loss resistance of the loop (assume that its value is the same as if the wire were straight); (c) input resistance; (d) input impedance; (e) radiation efficiency.
5.4.
A small circular loop with a uniform current distribution, and with its classical omnidirec-tional pattern, is used as a receiving antenna. Determine the maximum directivity (dimen-sionless and in dB) using: (a) Exact method.
(b) An approximate method appropriate for this pattern. Specify the method used.
(c) Another approximate method appropriate for this pattern. Specify the method used.
Hint: For the approximate methods, the word omnidirectional is a clue.
5.5.
A N-turn resonant circular loop with a uniform current distribution and with a circumfer-ence of λ∕4, is fed by a lossless balanced twin-lead transmission line with a characteristic impedance of 300 ohms. Neglecting proximity effects, determine the (a) closest integer number of turns so that the input impedance is nearly 300 ohms; (b) input impedance of the antenna; (c) reflection coefficient; (d) VSWR inside the transmission line.
5.6.
A small circular loop with circumference C < λ∕20 is used as a receiving antenna. A uniform plane wave traveling along the x-axis and toward the positive (+) x direction (as shown in the 278 LOOP ANTENNAS figure), whose electric field is given by Ei w = (̂ ay + 2̂ az)e−jkx is incident upon the antenna. Determine the (a) polarization of the incident wave. Justify your answer.
(b) axial ratio of the polarization ellipse of the incident wave.
(c) polarization of the loop antenna toward the x-axis.
(d) polarization loss factor (dimensionless and in dB).
(e) maximum power at 1 GHz that can be deliv-ered to a load connected to the antenna, if the power density of the above inci-dent wave is 5 mwatts∕cm2. Assume no other losses.
z y x r ϕ Ei w θ a Hint: ̂ a𝜙= −̂ ax sin 𝜙+ ̂ ay cos 𝜙 5.7.
A combination of a horizontal small loop of uniform current and a vertical infinitesimal dipole, as shown in the figure below, are used as one antenna. The currents in the two elements are adjusted so that the magnitudes of the corresponding far-zone electric field components radiated by each element are equal. For the entire antenna system, loop plus dipole: (a) Write an expression for the normalized electric filed radi-ated by the combination of the two elements.
(b) State the polarization of the entire antenna system (linear, circular, elliptical).
(c) Determine the polarization loss factor (dimensionless and in dB) if a linearly polarized wave, coming from any direc-tion, is incident upon this antenna which is used as a receiv-ing antenna.
x y z a 5.8.
Find the radiation efficiency of a single-turn and a four-turn circular loop each of radius λ∕(10𝜋) and operating at 10 MHz. The radius of the wire is 10−3λ and the turns are spaced 3 × 10−3λ apart. Assume the wire is copper with a conductivity of 5.7 × 107 S/m, and the antenna is radiating into free-space.
5.9.
Find the power radiated by a small loop by forming the average power density, using (5-27a)–(5-27c), and integrating over a sphere of radius r. Compare the answer with (5-23b).
5.10.
For a small loop of constant current, derive its far-zone fields using (5-17) and the procedure outlined and relationships developed in Section 3.6. Compare the answers with (5-27a)– (5-27c).
5.11.
A single-turn resonant circular loop with a λ∕8𝜋radius is made of copper wire with a wire radius of 10−4λ∕2𝜋and conductivity of 5.7 × 107 S/m. For a frequency of 100 MHz, deter-mine, assuming uniform current, the (a) radiation efficiency (assume the wire is straight); (b) maximum gain of the antenna (dimensionless and in dB).
5.12.
A horizontal, one-turn, loop antenna with a circumference of C = πλ is radiating in free space, and it used as a ground-based receiving antenna for an over-the-horizon communica-tion system. Assuming the current distribution is uniform, determine the (a) Maximum directivity of the antenna (dimensionless and in dB).
(b) Loss resistance of the wire of the loop. Assume the wire is straight, has a radius of 10−4λ, a conductivity of 107 S/m, and the loop is operating at 100 MHz.
PROBLEMS 279 (c) Radiation efficiency of the loop (in %).
(d) Maximum gain (dimensionless and in dB) of the loop.
5.13.
Design a lossless resonant circular loop operating at 10 MHz so that its single-turn radiation resistance is 0.73 ohms. The resonant loop is to be connected to a matched load through a balanced “twin-lead” 300-ohm transmission line.
(a) Determine the radius of the loop (in meters and wavelengths).
(b) To minimize the matching reflections between the resonant loop and the 300-ohm trans-mission line, determine the closest number of integer turns the loop must have.
(c) For the loop of part b, determine the maximum power that can be expected to be delivered to a receiver matched load if the incident wave is polarization matched to the lossless resonant loop. The power density of the incident wave is 10−6 watts/m2.
5.14.
A resonant six-turn loop of closely spaced turns is operating at 50 MHz. The radius of the loop is λ∕30, and the loop is connected to a 50-ohm transmission line. The radius of the wire is λ∕300, its conductivity is 𝜎= 5.7 × 107 S/m, and the spacing between the turns is λ∕100.
Determine the (a) directivity of the antenna (in dB) (b) radiation efficiency taking into account the proximity effects of the turns (c) reflection efficiency (d) gain of the antenna (in dB) 5.15.
A horizontal, lossless, one-turn circular loop of circumference C = λ, with a nonuniform current distribution, is radiating in free space. The Tar-field pattern of the antenna can be approximated by E𝜙≃C0 cos2 𝜃e−jkr r } 0◦≤𝜃≤90◦ 0◦≤𝜙≤360◦ where C0 is a constant and 𝜃is measured from the normal to the plane/area of the loop.
Determine the (a) Maximum exact directivity (dimensionless and in dB) of the antenna.
(b) Approximate input impedance of the loop.
(c) Input reflection coefficient when the antenna is connected to a balanced “twin-lead” trans-mission line with a characteristic impedance of 300 ohms.
(d) Maximum gain of the loop (dimensionless and in dB).
(e) Maximum absolute gain of the loop (dimensionless and in dB).
5.16.
Find the radiation efficiency (in percent) of an eight-turn circular-loop antenna operating at 30 MHz. The radius of each turn is a = 15 cm, the radius of the wire is b = 1 mm, and the spacing between turns is 2c = 3.6 mm. Assume the wire is copper (𝜎= 5.7 × 107 S/m), and the antenna is radiating into free-space. Account for the proximity effect.
5.17.
A very small circular loop of radius a(a < λ∕6𝜋) and constant current I0 is symmetrically placed about the origin at x = 0 and with the plane of its area parallel to the y-z plane. Find the (a) spherical E- and H-field components radiated by the loop in the far zone (b) directivity of the antenna 5.18.
Repeat Problem 5.17 when the plane of the loop is parallel to the x-z plane at y = 0.
5.19.
Using the computer program of this chapter, compute the radiation resistance and the direc-tivity of a circular loop of constant current with a radius of (a) a = λ∕50 (b) a = λ∕10 (c) a = λ∕4 (d) a = λ∕2 280 LOOP ANTENNAS 5.20.
A constant current circular loop of radius a = 5λ∕4 is placed on the x-y plane. Find the two smallest angles (excluding 𝜃= 0◦) where a null is formed in the far-field pattern.
5.21.
Design a circular loop of constant current such that its field intensity vanishes only at 𝜃= 0◦(𝜃= 180◦) and 90◦. Find its (a) radius (b) radiation resistance (c) directivity 5.22.
Design a constant current circular loop so that its first minimum, aside from 𝜃= 0◦, in its far-field pattern is at 30◦from a normal to the plane of the loop. Find the (a) smallest radius of the antenna (in wavelengths) (b) relative (to the maximum) radiation intensity (in dB) in the plane of the loop 5.23.
Design a constant current circular loop so that its pattern has a null in the plane of the loop, and two nulls above and two nulls below the plane of the loop. Find the (a) radius of the loop (b) angles where the nulls occur 5.24.
A constant current circular loop is placed on the x-y plane. Find the far-field position, relative to that of the loop, that a linearly polarized probe antenna must have so that the polarization loss factor (PLF) is maximized.
5.25.
A very small (a ≪λ) circular loop of constant current is placed a distance h above an infinite electric ground plane. Assuming z is perpendicular to the ground plane, find the total far-zone field radiated by the loop when its plane is parallel to the (a) x-z plane (b) y-z plane 5.26.
A very small loop antenna (a ≪λ∕30) of constant current is placed a height h above a flat, perfectly conducting ground plane of infinite extent. The area plane of the loop is parallel to the interface (x-y plane). For far-field observations (a) find the total electric field radiated by the loop in the presence of the ground plane (b) all the angles (in degrees) from the vertical to the interface where the total field will vanish when the height is λ (c) the smallest nonzero height (in λ) such that the total far-zone field exhibits a null at an angle of 60◦from the vertical z y x h a I 0, 0 = ∞ μ σ 5.27.
The antenna of the VOR (VHF Omni Range) airport guidance system consists of a small radius a circular loop (a ≪λ, so that its current is uniform). The circular loop is placed on a plane horizontal and at a height h above an ideal planar and infinite in extent synthesized PMC (perfect magnetic conductor) ground plane. To make sure the antenna remains operational at all times, the antenna is placed at a height h = 0.75λ, above the PMC ground plane.
Assuming far-field observations, determine (a) The normalized array factor of the equivalent antenna system that is valid in all space on and above the PMC ground plane.
(b) All the angles 𝜃(in degrees) that th AF (array factor) of the equivalent system will achieve its maximum radiation and allow safe operation of the VOR navigation system.
h VOR Loop Antenna PMC θ PROBLEMS 281 5.28.
A very small circular loop, of radius a and constant current I0, is placed a height h above an infinite and flat Perfect Magnetic Conductor (PMC). The area of the loop is parallel to the PMC, which is on the xy-plane; the z-axis is perpendicular to the PMC interface. Determine the (a) Total far-zone electric field radiated by the loop in the presence of the PMC.
(b) Smallest height h (in wavelengths) so that the total field pattern possesses simultaneously nulls only at 𝜃= 0◦and 30◦.
5.29.
A small circular loop, with its area parallel to the x-z plane, is placed a height h above an infinite flat per-fectly electric conducting ground plane. Determine (a) the array factor for the equivalent problem which allows you to find the total field on and above the ground plane (b) angle(s) 𝜃(in degrees) where the array factor will vanish when the loop is placed at a height λ∕2 above the ground plane z y x h a = ∞ σ 5.30.
A small circular electric loop, of uniform current I0, is placed horizontally/parallel at zero height (h = 0) above a Perfect Magnetic Conductor (PMC), and it is used as a receiving antenna at a frequency of 100 MHz. The circumference of the loop is C = λ∕20. Assuming the wire radius is very small (b ≪λ), determine the: (a) Maximum directivity (dimensionless and in dB). Justify your answer.
(b) Maximum effective area (in cm2).
(c) Maximum power (in watts) that can be delivered to a matched load, connected to the loop, when a circularly polarized wave, with an power density of 10−4 watts/cm2, is incident (in the direction of maximum directivity) upon the loop. Assume no other losses.
5.31.
A small circular loop with its area parallel to the x-z plane is placed at a height h above an infinite perfectly conducting ground plane, as shown in the figure for Problem 5.29. Deter-mine the (a) array factor for the equivalent problem which will allow you to find the total field on and above the ground plane.
(b) two smallest heights h (in λ) greater than h = 0 (i.e., h > 0) that will form a maximum on the magnitude of the array factor toward 𝜃= 0◦.
5.32.
The emergency radio police system of Problem 1 (f = 10 MHz) now uses a very small circular loop of constant current distribution. The circular loop is placed horizontally, as shown below, a height h above the top of the police car. Consider that the top of the police car to be an infinite and planar artificial PMC surface. The sensitivity (minimum power) of the system receiver, to be able to detect an incoming signal, is 10 𝜇watts. Assuming the incoming signal is circularly polarized and it incident from a horizontal direction (grazing angle; 𝜃= 90◦), what is the (a) Smallest obvious height h (in λ) of the loop above the PMC to maximize the directivity?
(b) What is this maximum directivity, using the smallest height h from part a, (dimensionless and in dB)?
(c) Minimum power density (in watts/cm2) of the incoming signal to be detected by the radio receiver, using the smallest height h from part a?
282 LOOP ANTENNAS z a = 90° Incoming Wave (PMC) y x h θ 5.33.
For the loop of Problem 5.25(a), find the smallest height h so that a null is formed in the y-z plane at an angle of 45◦above the ground plane.
5.34.
A circular loop with a radius of a = λ∕20𝜋 is placed vertically a height h above a PEC ground plane, as shown in the figure (the yz-plane of its area is perpendicular to the ground plane). The height h is measured from the center of the loop. For this configu-ration, determine the (a) Normalized array factor. Indicate how you obtained it. You do not need to derive it as long as you explain the rationale.
(b) Smallest height (in wavelengths) that the loop must be placed above the ground plane to introduce the first null in the array factor at an angle of 𝜃= 60 degrees from the vertical direction.
y x PEC z h a θ 5.35.
The transmitting and receiving antennas of a wireless communication system consist, respec-tively, of a small horizontal circular loop (with radius a ≪λ, so that its current is uniform) and an ideal infinitesimal electric dipole (l ≪λ∕50). The two antennas are at the same level and separated by a distance d so that one is in the far-field of the other. Assuming both radiate in an unbounded infinite free-space medium: Determine the PLF (polarization loss factor, dimensionless and in dB) of the two-antenna communication system for two different dipole orientations: i.e., when the linear dipole is oriented along the (a) z direction (b) y direction In both cases, the plane (area) of the loop lies on the horizontal plane (parallel to the xy-plane); i.e., the loop does NOT change orientation; stays the same for both dipole orientations.
z z-oriented dipole y-oriented dipole y x d Horizontal Loop θ ϕ PROBLEMS 283 5.36.
A small single-turn circular loop of radius a = 0.05λ is operating at 300 MHz. Assuming the radius of the wire is 10−4λ, determine the (a) loss resistance (b) radiation resistance (c) loop inductance Show that the loop inductive reactance is much greater than the loss resistance and radiation resistance indicating that a small loop acts primarily as an inductor.
5.37.
Determine the radiation resistance of a single-turn small loop, assuming the geometrical shape of the loop is (a) rectangular with dimensions a and b (a, b ≪λ) (b) elliptical with major axis a and minor axis b (a, b ≪λ) 5.38.
A one-turn small circular loop is used as a radiating element for a VHF (f = 100 MHz) com-munications system. The circumference of the loop is C = λ∕20 while the radius of the wire is λ∕400. Determine, using a wire conductivity of 𝜎= 5.7 × 107 S/m, the (a) input resistance of the wire for a single turn.
(b) input reactance of the loop. Is it inductive or capacitive? Be specific.
(c) inductance (in henries) or capacitance (in farads) that can be placed in series with the loop at the feed to resonate the antenna at f = 100 MHz; choose the element that will accomplish the desired objective.
5.39.
Show that for the rectangular loop the radiation resistance is represented by Rr = 31,171 ( a2b2 λ4 ) while for the elliptical loop is represented by Rr = 31,171 ( 𝜋2a2b2 16λ4 ) 5.40.
Assuming the direction of the magnetic field of the incident plane wave coincides with the plane of incidence, derive the effective length of a small circular loop of radius a based on the definition of (2-92). Show that its effective length is (S = 𝜋a2) 𝓵e = ̂ a𝜙jkS sin(𝜃) 5.41.
A circular loop of nonconstant current distribution, with circumference of 1.4λ, is attached to a 300-ohm line. Assuming the radius of the wire is 1.555 × 10−2λ, find the (a) input impedance of the loop (b) VSWR of the system (c) inductance or capacitance that must be placed across the feed points so that the loop becomes resonant at f = 100 MHz.
5.42.
A very popular antenna for amateur radio operators is a square loop antenna (referred to as quad antenna) whose circumference is one wavelength. Assuming the radiation characteris-tics of the square loop are well represented by those of a circular loop: (a) What is the input impedance (real and imaginary parts) of the antenna?
(b) What element (inductor or capacitor), and of what value, must be placed in series with the loop at the feed point to resonate the radiating element at a frequency of 1 GHz?
(c) What is the input VSWR, having the inductor or capacitor in place, if the loop is con-nected to a 78-ohm coaxial cable?
284 LOOP ANTENNAS 5.43.
A circular loop of nonuniform current, circumference C = λ, and wire radius b = 2.47875 × 10−3λ, is used for end-fire (over-the-head; toward zenith; 𝜃= 0◦) communication. The loop is connected to a 75-ohm transmission line. Determine the (a) Approximate input impedance (real and imaginary parts). To get total credit, state how or where you got the answer. You do not necessarily have to compute it. Equations (5-37a)–(5-38) are valid only for uniform current distribution.
(b) Is the input impedance capacitive or inductive?
(c) What kind of a lumped element, capacitor or inductor, must be placed in parallel to resonate the loop?
(d) At a frequency of 500 MHz, what is the capacitance or inductance of the parallel lumped element?
(e) What is the new input impedance of the resonated loop (in the presence of the parallel capacitor or inductor)?
(f) What is the input VSWR of the resonated loop (in the presence of the parallel capacitor or inductor)?
5.44.
Design circular loops of wire radius b, which resonate at the first resonance. Find (a) four values of a/b where the first resonance occurs (a is the radius of the loop) (b) the circumference of the loops and the corresponding radii of the wires for the antennas of part (a).
5.45.
Using (5-54b) the asymptotic form of (5-65a) for small argument, show that the radiation resistance Rr for a small radius (a ≪λ) loop of uniform current is given by Rr = 20𝜋2(ka)4 = 20𝜋2 (C λ )4 5.46.
Consider a circular loop of wire of radius a on the x-y plane and centered about the origin.
Assume the current on the loop is given by I𝜙(𝜙′) = I0 cos(𝜙′) (a) Show that the far-zone electric field of the loop is given by E𝜃= j𝜂ka 2 I0 e−jkr r J1(ka sin 𝜃) ka sin 𝜃 cos 𝜃sin 𝜙 E𝜙= j𝜂ka 2 I0 e−jkr r J1 ′(ka sin 𝜃) cos 𝜙 J1 ′(x) = dJ1(x) dx (b) Evaluate the radiation intensity U(𝜃, 𝜙) in the direction 𝜃= 0 and 𝜙= 𝜋 2 as a function of ka.
CHAPTER6 Arrays: Linear, Planar, and Circular 6.1 INTRODUCTION In the previous chapter, the radiation characteristics of single-element antennas were discussed and analyzed. Usually the radiation pattern of a single element is relatively wide, and each element pro-vides low values of directivity (gain). In many applications it is necessary to design antennas with very directive characteristics (very high gains) to meet the demands of long distance communication.
This can only be accomplished by increasing the electrical size of the antenna.
Enlarging the dimensions of single elements often leads to more directive characteristics. Another way to enlarge the dimensions of the antenna, without necessarily increasing the size of the individual elements, is to form an assembly of radiating elements in an electrical and geometrical configuration.
This new antenna, formed by multielements, is referred to as an array. In most cases, the elements of an array are identical. This is not necessary, but it is often convenient, simpler, and more practical.
The individual elements of an array may be of any form (wires, apertures, etc.).
The total field of the array is determined by the vector addition of the fields radiated by the individual elements. This assumes that the current in each element is the same as that of the isolated element (neglecting coupling). This is usually not the case and depends on the separation between the elements. To provide very directive patterns, it is necessary that the fields from the elements of the array interfere constructively (add) in the desired directions and interfere destructively (cancel each other) in the remaining space. Ideally this can be accomplished, but practically it is only approached.
In an array of identical elements, there are at least five controls that can be used to shape the overall pattern of the antenna. These are: 1. the geometrical configuration of the overall array (linear, circular, rectangular, spherical, etc.) 2. the relative displacement between the elements 3. the excitation amplitude of the individual elements 4. the excitation phase of the individual elements 5. the relative pattern of the individual elements The influence that each one of the above has on the overall radiation characteristics will be the subject of this chapter. In many cases the techniques will be illustrated with examples.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 285 286 ARRAYS: LINEAR, PLANAR, AND CIRCULAR There are a plethora of antenna arrays used for personal, commercial, and military applications utilizing different elements including dipoles, loops, apertures, microstrips, horns, reflectors, and so on. Arrays of dipoles are shown in Figures 4.26, 10.19, and 11.15. The one in Figure 4.26 is an array that is widely used as a base-station antenna for mobile communication. It is a triangular array consisting of twelve dipoles, with four dipoles on each side of the triangle. Each four-element array, on each side of the triangle, is basically used to cover an angular sector of 120◦forming what is usually referred to as a sectoral array. The one in Figure 10.19 is a classic array of dipoles, referred to as the Yagi-Uda array, and it is primarily used for TV and amateur radio applications. The array of Figure 11.12 is also an array of dipoles, which is referred to as the log-periodic antenna, which is primarily used for TV reception and has wider bandwidth than the Yagi-Uda array but slightly smaller directivity. An array of loops is shown in Figure 5.1 and one utilizing microstrips as elements is displayed in Figure 14.35. An advanced array design of slots, used in the AWACS, is shown in Figure 6.29.
The simplest and one of the most practical arrays is formed by placing the elements along a line.
To simplify the presentation and give a better physical interpretation of the techniques, a two-element array will first be considered. The analysis of an N-element array will then follow. Two-dimensional analysis will be the subject at first. In latter sections, three-dimensional techniques will be introduced.
6.2 TWO-ELEMENT ARRAY Let us assume that the antenna under investigation is an array of two infinitesimal horizontal dipoles positioned along the z-axis, as shown in Figure 6.1(a). The total field radiated by the two elements, assuming no coupling between the elements, is equal to the sum of the two and in the y-z plane it is given by Et = E1 + E2 = ̂ a𝜃j𝜂kI0l 4𝜋 { e−j[kr1−(𝛽∕2)] r1 cos 𝜃1 + e−j[kr2+(𝛽∕2)] r2 cos 𝜃2 } (6-1) where 𝛽is the difference in phase excitation between the elements. The magnitude excitation of the radiators is identical. Assuming far-field observations and referring to Figure 6.1(b), 𝜃1 ≃𝜃2 ≃𝜃 (6-2a) r1 ≃r −d 2 cos 𝜃 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ for phase variations (6-2b) r2 ≃r + d 2 cos 𝜃 r1 ≃r2 ≃r for amplitude variations (6-2c) Equation 6-1 reduces to Et = ̂ a𝜃j𝜂kI0le−jkr 4𝜋r cos 𝜃[e+j(kd cos 𝜃+𝛽)∕2 + e−j(kd cos 𝜃+𝛽)∕2] Et = ̂ a𝜃j𝜂kI0le−jkr 4𝜋r cos 𝜃 { 2 cos [1 2(kd cos 𝜃+ 𝛽) ]} (6-3) It is apparent from (6-3) that the total field of the array is equal to the field of a single element positioned at the origin multiplied by a factor which is widely referred to as the array factor. Thus TWO-ELEMENT ARRAY 287 Figure 6.1 Geometry of a two-element array positioned along the z-axis.
for the two-element array of constant amplitude, the array factor is given by AF = 2 cos[ 1 2(kd cos 𝜃+ 𝛽)] (6-4) which in normalized form can be written as (AF)n = cos[ 1 2(kd cos 𝜃+ 𝛽)] (6-4a) The array factor is a function of the geometry of the array and the excitation phase. By varying the separation d and/or the phase 𝛽between the elements, the characteristics of the array factor and of the total field of the array can be controlled.
It has been illustrated that the far-zone field of a uniform two-element array of identical elements is equal to the product of the field of a single element, at a selected reference point (usually the origin), and the array factor of that array. That is, E(total) = [E(single element at reference point)] × [array factor] (6-5) 288 ARRAYS: LINEAR, PLANAR, AND CIRCULAR This is referred to as pattern multiplication for arrays of identical elements, and it is analogous to the pattern multiplication of (4-59) for continuous sources. Although it has been illustrated only for an array of two elements, each of identical magnitude, it is also valid for arrays with any number of identical elements which do not necessarily have identical magnitudes, phases, and/or spacings between them. This will be demonstrated in this chapter by a number of different arrays.
Each array has its own array factor. The array factor, in general, is a function of the number of elements, their geometrical arrangement, their relative magnitudes, their relative phases, and their spacings. The array factor will be of simpler form if the elements have identical amplitudes, phases, and spacings. Since the array factor does not depend on the directional characteristics of the radiating elements themselves, it can be formulated by replacing the actual elements with isotropic (point) sources. Once the array factor has been derived using the point-source array, the total field of the actual array is obtained by the use of (6-5). Each point-source is assumed to have the amplitude, phase, and location of the corresponding element it is replacing.
In order to synthesize the total pattern of an array, the designer is not only required to select the proper radiating elements but the geometry (positioning) and excitation of the individual elements.
To illustrate the principles, let us consider some examples.
Example 6.1 Given the array of Figures 6.1(a) and (b), find the nulls of the total field when d = λ∕4 and a. 𝛽= 0 b. 𝛽= +𝜋 2 c. 𝛽= −𝜋 2 Solution: a. 𝛽= 0 The normalized field is given by Etn = cos 𝜃cos (𝜋 4 cos 𝜃 ) The nulls are obtained by setting the total field equal to zero, or Etn = cos 𝜃cos (𝜋 4 cos 𝜃 ) |𝜃=𝜃n = 0 Thus cos 𝜃n = 0 ➱𝜃n = 90◦ and cos (𝜋 4 cos 𝜃n ) = 0 ➱𝜋 4 cos 𝜃n = 𝜋 2 , −𝜋 2 ➱𝜃n = does not exist The only null occurs at 𝜃= 90◦and is due to the pattern of the individual elements. The array factor does not contribute any additional nulls because there is not enough separation between the elements to introduce a phase difference of 180◦between the elements, for any observation angle.
TWO-ELEMENT ARRAY 289 b. 𝛽= +𝜋 2 The normalized field is given by Etn = cos 𝜃cos [𝜋 4 (cos 𝜃+ 1) ] The nulls are found from Etn = cos 𝜃cos [𝜋 4 (cos 𝜃+ 1) ] |𝜃=𝜃n = 0 Thus cos 𝜃n = 0 ➱𝜃n = 90◦ and cos [𝜋 4 (cos 𝜃+ 1) ] |𝜃=𝜃n = 0 ➱𝜋 4 (cos 𝜃n + 1) = 𝜋 2 ➱𝜃n = 0◦ and ➱𝜋 4 (cos 𝜃n + 1) = −𝜋 2 ➱𝜃n = does not exist The nulls of the array occur at 𝜃= 90◦and 0◦. The null at 0◦is introduced by the arrangement of the elements (array factor). This can also be shown by physical reasoning, as shown in Fig-ure 6.2(a). The element in the negative z-axis has an initial phase lag of 90◦relative to the other element. As the wave from that element travels toward the positive z-axis (𝜃= 0◦direction), it undergoes an additional 90◦phase retardation when it arrives at the other element on the positive z-axis. Thus there is a total of 180◦phase difference between the waves of the two elements when travel is toward the positive z-axis (𝜃= 0◦). The waves of the two elements are in phase when they travel in the negative z-axis (𝜃= 180◦), as shown in Figure 6.2(b).
z #1 #2 e–j /4 e–j3 /4 /8 /8 e j /4 = 0° = 180° z #1 #2 e–j /4 Δ = 90° /8 /8 e j /4 θ = 0° θ θ = 180° θ (b) = 180° direction θ (a) = 0° direction θ ϕ Δ = 90° ϕ π π π π π λ λ λ λ Figure 6.2 Phase accumulation for two-element array for null formation toward 𝜃= 0◦and 180◦.
290 ARRAYS: LINEAR, PLANAR, AND CIRCULAR c. 𝛽= −𝜋 2 The normalized field is given by Etn = cos 𝜃cos [𝜋 4 (cos 𝜃−1) ] and the nulls by Etn = cos 𝜃cos [𝜋 4 (cos 𝜃−1) ] |𝜃=𝜃n = 0 Thus cos 𝜃n = 0 ➱𝜃n = 90◦ and cos [𝜋 4 (cos 𝜃n −1) ] = 0 ➱𝜋 4 (cos 𝜃n −1) = 𝜋 2 ➱𝜃n = does not exist and ➱𝜋 4 (cos 𝜃n −1) = −𝜋 2 ➱𝜃n = 180◦ The nulls occur at 90◦and 180◦. The element at the positive z-axis has a phase lag of 90◦relative to the other, and the phase difference is 180◦when travel is restricted toward the negative z-axis.
There is no phase difference when the waves travel toward the positive z-axis. A diagram similar to that of Figure 6.2 can be used to illustrate this case.
To better illustrate the pattern multiplication rule, the normalized patterns of the single ele-ment, the array factor, and the total array for each of the above array examples are shown in Fig-ures 6.3, 6.4(a), and 6.4(b). In each figure, the total pattern of the array is obtained by multiplying the pattern of the single element by that of the array factor. In each case, the pattern is normalized to its own maximum. Since the array factor for the example of Figure 6.3 is nearly isotropic (within 3 dB), the element pattern and the total pattern are almost identical in shape. The largest magnitude difference between the two is about 3 dB, and for each case it occurs toward the direction along which the phases of the two elements are in phase quadrature (90◦out of phase). For Figure 6.3 this occurs along 𝜃= 0◦while for Figures 6.4(a,b) this occurs along 𝜃= 90◦. Because the array factor for Figure 6.4(a) is of cardioid form, its corresponding element and total patterns are considerably different. In the total pattern, the null at 𝜃= 90◦is due to the element pattern while that toward 𝜃= 0◦is due to the array factor. Similar results are displayed in Figure 6.4(b).
Example 6.2 Consider an array of two identical infinitesimal dipoles oriented as shown in Figures 6.1(a) and (b). For a separation d and phase excitation difference 𝛽between the elements, find the angles of observation where the nulls of the array occur. The magnitude excitation of the elements is the same.
TWO-ELEMENT ARRAY 291 Figure 6.3 Element, array factor, and total field patterns of a two-element array of infinitesimal horizontal dipoles with identical phase excitation (𝛽= 0◦, d = λ∕4).
Solution: The normalized total field of the array is given by (6-3) as Etn = cos 𝜃cos[ 1 2(kd cos 𝜃+ 𝛽)] To find the nulls, the field is set equal to zero, or Etn = cos 𝜃cos[ 1 2(kd cos 𝜃+ 𝛽)]|𝜃=𝜃n = 0 Thus cos 𝜃n = 0 ➱𝜃n = 90◦ 292 ARRAYS: LINEAR, PLANAR, AND CIRCULAR (a) Figure 6.4 Pattern multiplication of element, array factor, and total array patterns of a two-element array of infinitesimal horizontal dipoles with (a) 𝛽= +90◦, d = λ∕4. (continued) and cos [1 2(kd cos 𝜃n + 𝛽) ] = 0 ➱1 2(kd cos 𝜃n + 𝛽) = ± (2n + 1 2 ) 𝜋 ➱𝜃n = cos−1 ( λ 2𝜋d [−𝛽± (2n + 1)𝜋] ) , n = 0, 1, 2, … The null at 𝜃= 90◦is attributed to the pattern of the individual elements of the array while the remaining ones are due to the formation of the array. For no phase difference between the elements (𝛽= 0), the separation d must be equal or greater than half a wavelength (d ≥λ∕2) in order for at least one null, due to the array, to occur.
N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 293 (b) Figure 6.4 (Continued) (b) 𝛽= −90◦, d = λ∕4.
6.3 N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING Now that the arraying of elements has been introduced and it was illustrated by the two-element array, let us generalize the method to include N elements. Referring to the geometry of Figure 6.5(a), let us assume that all the elements have identical amplitudes but each succeeding element has a 𝛽 progressive phase lead current excitation relative to the preceding one (𝛽represents the phase by which the current in each element leads the current of the preceding element). An array of identical elements all of identical magnitude and each with a progressive phase is referred to as a uniform array. The array factor can be obtained by considering the elements to be point sources. If the actual elements are not isotropic sources, the total field can be formed by multiplying the array factor of the isotropic sources by the field of a single element. This is the pattern multiplication rule of (6-5), and it applies only for arrays of identical elements. The array factor is given by AF = 1 + e+j(kd cos 𝜃+𝛽) + e+j2(kd cos 𝜃+𝛽) + ⋯+ ej(N−1)(kd cos 𝜃+𝛽) AF = N ∑ n=1 ej(n−1)(kd cos 𝜃+𝛽) (6-6) 294 ARRAYS: LINEAR, PLANAR, AND CIRCULAR d cos 1 2 3 4 N z y d d d d cos d cos rN r4 r3 r2 r1 (a) Geometry 2 2 3 #N #4 #3 #2 #1 1 (N – 1) 1 3 1 1 1 0° (AF) (b) Phasor diagram Ψ Ψ Ψ Ψ Ψ Ψ Ψ θ θ θ θ θ θ θ θ Figure 6.5 Far-field geometry and phasor diagram of N-element array of isotropic sources positioned along the z-axis.
which can be written as AF = N ∑ n=1 ej(n−1)𝜓 where 𝜓= kd cos 𝜃+ 𝛽 (6-7) (6-7a) Since the total array factor for the uniform array is a summation of exponentials, it can be rep-resented by the vector sum of N phasors each of unit amplitude and progressive phase 𝜓relative to the previous one. Graphically this is illustrated by the phasor diagram in Figure 6.5(b). It is appar-ent from the phasor diagram that the amplitude and phase of the AF can be controlled in uniform arrays by properly selecting the relative phase 𝜓between the elements; in nonuniform arrays, the amplitude as well as the phase can be used to control the formation and distribution of the total array factor.
The array factor of (6-7) can also be expressed in an alternate, compact and closed form whose functions and their distributions are more recognizable. This is accomplished as follows.
Multiplying both sides of (6-7) by ej𝜓, it can be written as (AF)ej𝜓= ej𝜓+ ej2𝜓+ ej3𝜓+ ⋯+ ej(N−1)𝜓+ ejN𝜓 (6-8) Subtracting (6-7) from (6-8) reduces to AF(ej𝜓−1) = (−1 + ejN𝜓) (6-9) N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 295 which can also be written as AF = [ ejN𝜓−1 ej𝜓−1 ] = ej[(N−1)∕2]𝜓 [ ej(N∕2)𝜓−e−j(N∕2)𝜓 ej(1∕2)𝜓−e−j(1∕2)𝜓 ] = ej[(N−1)∕2]𝜓 ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 𝜓 ) sin (1 2𝜓 ) ⎤ ⎥ ⎥ ⎥ ⎦ (6-10) If the reference point is the physical center of the array, the array factor of (6-10) reduces to AF = ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 𝜓 ) sin (1 2𝜓 ) ⎤ ⎥ ⎥ ⎥ ⎦ (6-10a) For small values of 𝜓, the above expression can be approximated by AF ≃ ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 𝜓 ) 𝜓 2 ⎤ ⎥ ⎥ ⎥ ⎦ (6-10b) The maximum value of (6-10a) or (6-10b) is equal to N. To normalize the array factors so that the maximum value of each is equal to unity, (6-10a) and (6-10b) are written in normalized form as (see Appendix II) (AF)n = 1 N ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 𝜓 ) sin (1 2𝜓 ) ⎤ ⎥ ⎥ ⎥ ⎦ (6-10c) and (see Appendix I) (AF)n ≃ ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 𝜓 ) N 2 𝜓 ⎤ ⎥ ⎥ ⎥ ⎦ (6-10d) To find the nulls of the array, (6-10c) or (6-10d) is set equal to zero. That is, sin (N 2 𝜓 ) = 0 ➱N 2 𝜓|𝜃=𝜃n = ±n𝜋➱𝜃n = cos−1 [ λ 2𝜋d ( −𝛽± 2n N 𝜋 )] n = 1, 2, 3, … n ≠N, 2N, 3N, … with (6-10c) (6-11) For n = N, 2N, 3N, …, (6-10c) attains its maximum values because it reduces to a sin(0)/0 form. The values of n determine the order of the nulls (first, second, etc.). For a zero to exist, the argument of 296 ARRAYS: LINEAR, PLANAR, AND CIRCULAR the arccosine cannot exceed unity. Thus the number of nulls that can exist will be a function of the element separation d and the phase excitation difference 𝛽.
The maximum values of (6-10c) occur when 𝜓 2 = 1 2(kd cos 𝜃+ 𝛽)|𝜃=𝜃m = ±m𝜋➱𝜃m = cos−1 [ λ 2𝜋d(−𝛽± 2m𝜋) ] m = 0, 1, 2, … (6-12) The array factor of (6-10d) has only one maximum and occurs when m = 0 in (6-12). That is, 𝜃m = cos−1 ( λ𝛽 2𝜋d ) (6-13) which is the observation angle that makes 𝜓= 0.
The 3-dB point for the array factor of (6-10d) occurs when (see Appendix I) N 2 𝜓= N 2 (kd cos 𝜃+ 𝛽)|𝜃=𝜃h = ±1.391 ➱𝜃h = cos−1 [ λ 2𝜋d ( −𝛽± 2.782 N )] (6-14) which can also be written as 𝜃h = 𝜋 2 −sin−1 [ λ 2𝜋d ( −𝛽± 2.782 N )] (6-14a) For large values of d(d ≫λ), it reduces to 𝜃h ≃ [𝜋 2 − λ 2𝜋d ( −𝛽± 2.782 N )] (6-14b) The half-power beamwidth Θh can be found once the angles of the first maximum (𝜃m) and the half-power point (𝜃h) are determined. For a symmetrical pattern Θh = 2|𝜃m −𝜃h| (6-14c) For the array factor of (6-10d), there are secondary maxima (maxima of minor lobes) which occur approximately when the numerator of (6-10d) attains its maximum value. That is, sin (N 2 𝜓 ) = sin [N 2 (kd cos 𝜃+ 𝛽) ] |𝜃=𝜃s ≃±1 ➱N 2 (kd cos 𝜃+ 𝛽)|𝜃=𝜃s ≃± (2s + 1 2 ) 𝜋➱𝜃s ≃cos−1 { λ 2𝜋d [ −𝛽± (2s + 1 N ) 𝜋 ]} , s = 1, 2, 3, … (6-15) which can also be written as 𝜃s ≃𝜋 2 −sin−1 { λ 2𝜋d [ −𝛽± (2s + 1 N ) 𝜋 ]} , s = 1, 2, 3, … (6-15a) N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 297 For large values of d(d ≫λ), it reduces to 𝜃s ≃𝜋 2 − λ 2𝜋d [ −𝛽± (2s + 1 N ) 𝜋 ] , s = 1, 2, 3, … (6-15b) The maximum of the first minor lobe of (6-10c) occurs approximately when (see Appendix I) N 2 𝜓= N 2 (kd cos 𝜃+ 𝛽)|𝜃=𝜃s ≃± (3𝜋 2 ) (6-16) or when 𝜃s = cos−1 { λ 2𝜋d [ −𝛽± 3𝜋 N ]} (6-16a) At that point, the magnitude of (6-10d) reduces to (AF)n ≃ ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 𝜓 ) N 2 𝜓 ⎤ ⎥ ⎥ ⎥ ⎦𝜃=𝜃s s=1 = 2 3𝜋= 0.212 (6-17) which in dB is equal to (AF)n = 20 log10 ( 2 3𝜋 ) = −13.46 dB (6-17a) Thus the maximum of the first minor lobe of the array factor of (6-10d) is 13.46 dB down from the maximum at the major lobe. More accurate expressions for the angle, beamwidth, and magnitude of first minor lobe of the array factor of (6-10d) can be obtained. These will be discussed in Chapter 12.
6.3.1 Broadside Array In many applications it is desirable to have the maximum radiation of an array directed normal to the axis of the array [broadside; 𝜃0 = 90◦of Figure 6.5(a)]. To optimize the design, the maxima of the single element and of the array factor should both be directed toward 𝜃0 = 90◦. The requirements of the single elements can be accomplished by the judicious choice of the radiators, and those of the array factor by the proper separation and excitation of the individual radiators. In this section, the requirements that allow the array factor to “radiate” efficiently broadside will be developed.
Referring to (6-10c) or (6-10d), the first maximum of the array factor occurs when 𝜓= kd cos 𝜃+ 𝛽= 0 (6-18) Since it is desired to have the first maximum directed toward 𝜃0 = 90◦, then 𝜓= kd cos 𝜃+ 𝛽|𝜃=90◦= 𝛽= 0 (6-18a) Thus to have the maximum of the array factor of a uniform linear array directed broadside to the axis of the array, it is necessary that all the elements have the same phase excitation (in addition to the same amplitude excitation). The separation between the elements can be of any value. To ensure that there are no principal maxima in other directions, which are referred to as grating lobes, 298 ARRAYS: LINEAR, PLANAR, AND CIRCULAR the separation between the elements should not be equal to multiples of a wavelength (d ≠nλ, n = 1, 2, 3 …) when 𝛽= 0. If d = nλ, n = 1, 2, 3, … and 𝛽= 0, then 𝜓= kd cos 𝜃+ 𝛽| |d=nλ 𝛽=0 n=1,2,3,… = 2𝜋n cos 𝜃| |𝜃=0◦,180◦= ±2n𝜋 (6-19) This value of 𝜓when substituted in (6-10c) makes the array factor attain its maximum value. Thus for a uniform array with 𝛽= 0 and d = nλ, in addition to having the maxima of the array factor directed broadside (𝜃0 = 90◦) to the axis of the array, there are additional maxima directed along the axis (𝜃0 = 0◦, 180◦) of the array (end-fire radiation).
One of the objectives in many designs is to avoid multiple maxima, in addition to the main max-imum, which are referred to as grating lobes. Often it may be required to select the largest spac-ing between the elements but with no grating lobes. To avoid any grating lobe, the largest spacing between the elements should be less than one wavelength (dmax < λ).
To illustrate the method, the three-dimensional array factor of a 10-element (N = 10) uniform array with 𝛽= 0 and d = λ∕4 is shown plotted in Figure 6.6(a). A 90◦angular sector has been x y z Normalized Field Pattern (linear scale) (a) Broadside ( = 0, d = λ/4) β (b) Broadside/end-fire ( = 0, d = λ) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x y z Normalized Field Pattern (linear scale) 0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 β Figure 6.6 Three-dimensional amplitude patterns for broadside, and broadside/end-fire arrays (N = 10).
N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 299 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 d = /4 d = θ θ λ λ Figure 6.7 Array factor patterns of a 10-element uniform amplitude broadside array (N = 10, 𝛽= 0).
removed for better view of the pattern distribution in the elevation plane. The only maximum occurs at broadside (𝜃0 = 90◦). To form a comparison, the three-dimensional pattern of the same array but with d = λ is also plotted in Figure 6.6(b). For this pattern, in addition to the maximum at 𝜃0 = 90◦, there are additional maxima directed toward 𝜃0 = 0◦, 180◦. The corresponding two-dimensional patterns of Figures 6.6(a,b) are shown in Figure 6.7.
If the spacing between the elements is chosen between λ < d < 2λ, then the maximum of Fig-ure 6.6 toward 𝜃0 = 0◦shifts toward the angular region 0◦< 𝜃0 < 90◦while the maximum toward 𝜃0 = 180◦shifts toward 90◦< 𝜃0 < 180◦. When d = 2λ, there are maxima toward 0◦, 60◦, 90◦, 120◦ and 180◦.
In Tables 6.1 and 6.2 the expressions for the nulls, maxima, half-power points, minor lobe max-ima, and beamwidths for broadside arrays have been listed. They are derived from (6-10c)–(6-16a).
6.3.2 Ordinary End-Fire Array Instead of having the maximum radiation broadside to the axis of the array, it may be desirable to direct it along the axis of the array (end-fire). As a matter of fact, it may be necessary that it radiates toward only one direction (either 𝜃0 = 0◦or 180◦of Figure 6.5).
To direct the first maximum toward 𝜃0 = 0◦, 𝜓= kd cos 𝜃+ 𝛽|𝜃=0◦= kd + 𝛽= 0 ➱𝛽= −kd (6-20a) If the first maximum is desired toward 𝜃0 = 180◦, then 𝜓= kd cos 𝜃+ 𝛽|𝜃=180◦= −kd + 𝛽= 0 ➱𝛽= kd (6-20b) 300 ARRAYS: LINEAR, PLANAR, AND CIRCULAR TABLE 6.1 Nulls, Maxima, Half-Power Points, and Minor Lobe Maxima for Uniform Amplitude Broadside Arrays NULLS 𝜃n = cos−1 ( ± n N λ d ) n = 1, 2, 3, … n ≠N, 2N, 3N, … MAXIMA 𝜃m = cos−1 ( ±mλ d ) m = 0, 1, 2, … HALF-POWER POINTS 𝜃h ≃cos−1 ( ±1.391λ 𝜋Nd ) 𝜋d∕λ ≪1 MINOR LOBE MAXIMA 𝜃s ≃cos−1 [ ± λ 2d (2s + 1 N )] s = 1, 2, 3, … 𝜋d∕λ ≪1 Thus end-fire radiation is accomplished when 𝛽= −kd (for 𝜃0 = 0◦) or 𝛽= kd (for 𝜃0 = 180◦).
If the element separation is d = λ∕2, end-fire radiation exists simultaneously in both direc-tions (𝜃0 = 0◦and 𝜃0 = 180◦). If the element spacing is a multiple of a wavelength (d = nλ, n = 1, 2, 3, …), then in addition to having end-fire radiation in both directions, there also exist max-ima in the broadside directions. Thus for d = nλ, n = 1, 2, 3, … there exist four maxima; two in the broadside directions and two along the axis of the array. To have only one end-fire maximum and to avoid any grating lobes, the maximum spacing between the elements should be less than dmax < λ∕2.
The three-dimensional radiation patterns of a 10-element (N = 10) array with d = λ∕4, 𝛽= +kd are plotted in Figure 6.8. When 𝛽= −kd, the maximum is directed along 𝜃0 = 0◦and the three-dimensional pattern is shown in Figure 6.8(a). However, when 𝛽= +kd, the maximum is oriented toward 𝜃0 = 180◦, and the three-dimensional pattern is shown in Figure 6.8(b). The two-dimensional patterns of Figures 6.8(a,b) are shown in Figure 6.9. To form a comparison, the array factor of the same array (N = 10) but with d = λ and 𝛽= −kd has been calculated. Its pattern is identical to that of a broadside array with N = 10, d = λ, and it is shown plotted in Figure 6.7. It is seen that there are four maxima; two broadside and two along the axis of the array.
The expressions for the nulls, maxima, half-power points, minor lobe maxima, and beamwidths, as applied to ordinary end-fire arrays, are listed in Tables 6.3 and 6.4.
TABLE 6.2 Beamwidths for Uniform Amplitude Broadside Arrays FIRST-NULL BEAMWIDTH (FNBW) Θn = 2 [𝜋 2 −cos−1 ( λ Nd )] HALF-POWER BEAMWIDTH (HPBW) Θh ≃2 [𝜋 2 −cos−1 (1.391λ 𝜋Nd )] 𝜋d∕λ ≪1 FIRST SIDE LOBE BEAMWIDTH (FSLBW) Θs ≃2 [𝜋 2 −cos−1 ( 3λ 2dN )] 𝜋d∕λ ≪1 N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 301 x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (a) 0 = 0° θ (b) 0 = 180° θ Figure 6.8 Three-dimensional amplitude patterns for end-fire arrays toward 𝜃0 = 0◦and 180◦(N = 10, d = λ∕4).
302 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 = +kd β = –kd β θ θ Figure 6.9 Array factor patterns of a 10-element uniform amplitude end-fire array (N = 10, d = λ∕4).
6.3.3 Phased (Scanning) Array In the previous two sections it was shown how to direct the major radiation from an array, by con-trolling the phase excitation between the elements, in directions normal (broadside) and along the axis (end fire) of the array. It is then logical to assume that the maximum radiation can be ori-ented in any direction to form a scanning array. The procedure is similar to that of the previous two sections.
Let us assume that the maximum radiation of the array is required to be oriented at an angle 𝜃0(0◦≤𝜃0 ≤180◦). To accomplish this, the phase excitation 𝛽between the elements must be TABLE 6.3 Nulls, Maxima, Half-Power Points, and Minor Lobe Maxima for Uniform Amplitude Ordinary End-Fire Arrays NULLS 𝜃n = cos−1 ( 1 −nλ Nd ) n = 1, 2, 3, … n ≠N, 2N, 3N, … MAXIMA 𝜃m = cos−1 ( 1 −mλ d ) m = 0, 1, 2, … HALF-POWER POINTS 𝜃h ≃cos−1 ( 1 −1.391λ 𝜋dN ) 𝜋d∕λ ≪1 MINOR LOBE MAXIMA 𝜃s ≃cos−1 [ 1 −(2s + 1)λ 2Nd ] s = 1, 2, 3, … 𝜋d∕λ ≪1 N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 303 TABLE 6.4 Beamwidths for Uniform Amplitude Ordinary End-Fire Arrays FIRST-NULL BEAMWIDTH (FNBW) Θn = 2 cos−1 ( 1 −λ Nd ) HALF-POWER BEAMWIDTH (HPBW) Θh ≃2 cos−1 ( 1 −1.391λ 𝜋dN ) 𝜋d∕λ ≪1 FIRST SIDE LOBE BEAMWIDTH (FSLBW) Θs ≃2 cos−1 ( 1 −3λ 2Nd ) 𝜋d∕λ ≪1 adjusted so that 𝜓= kd cos 𝜃+ 𝛽|𝜃=𝜃0 = kd cos 𝜃0 + 𝛽= 0 ➱𝛽= −kd cos 𝜃0 (6-21) Thus by controlling the progressive phase difference between the elements, the maximum radiation can be squinted in any desired direction to form a scanning array. This is the basic principle of electronic scanning phased array operation. Since in phased array technology the scanning must be continuous, the system should be capable of continuously varying the progressive phase between the elements. In practice, this is accomplished electronically by the use of ferrite or diode phase shifters. For ferrite phase shifters, the phase shift is controlled by the magnetic field within the ferrite, which in turn is controlled by the amount of current flowing through the wires wrapped around the phase shifter.
For diode phase shifter using balanced, hybrid-coupled varactors, the actual phase shift is con-trolled either by varying the analog bias dc voltage (typically 0–30 volts) or by a digital command through a digital-to-analog (D/A) converter –.
Shown in Figure 6.10 is an incremental switched-line PIN-diode phase shifter –. This design is simple, straightforward, lightweight, and high speed. The lines of lengths l1 and l2 are switched on and off by controlling the bias of the PIN diodes, using two single-pole double-throw switches, as illustrated in Figure 6.10. The differential phase shift, provided by switching on and off the two paths, is given by Δ𝜙= k(l2 −l1) (6-21a) l1 l2 In Out Figure 6.10 Incremental switched-line phase shifter using PIN diodes. (source: D.M. Pozar, Microwave Engineering, John Wiley & Sons, Inc. 2004).
304 ARRAYS: LINEAR, PLANAR, AND CIRCULAR By properly choosing l1 and l2, and the operating frequency, the differential phase shift (in degrees) provided by each incremental line phase shifter can be as small as desired, and it determines the resolution of the phase shifter. The design of an entire phase shifter typically utilizes several such incremental phase shifters to cover the entire range (0 −180◦) of phase. However, the switched-line phase shifter, as well as many other ones, are usually designed for binary phase shifts of Δ𝜙= 180◦, 90◦, 45◦, 22.5◦, etc. . There are other designs of PIN-diode phase shifters, includ-ing those that utilize open-circuited stubs and reactive elements . The basic designs of a phase shifter utilizing PIN diodes are typically classified into three categories: switched line, loaded line, and reflection type . The loaded-line phase shifter can be used for phase shifts generally 45◦or smaller. Phase shifters that utilize PIN diodes are not ideal switches since the PIN diodes usually possess finite series resistance and reactance that can contribute significant insertion loss if several of them are used. These phase shifters can also be used as time-delay devices.
To demonstrate the principle of scanning, the three-dimensional radiation pattern of a 10-element array, with a separation of λ∕4 between the elements and with the maximum squinted in the 𝜃0 = 60◦direction, is plotted in Figure 6.11(a). The corresponding two-dimensional pattern is shown in Figure 6.11(b).
The half-power beamwidth of the scanning array is obtained using (6-14) with 𝛽= −kd cos 𝜃0.
Using the minus sign in the argument of the inverse cosine function in (6-14) to represent one angle of the half-power beamwidth and the plus sign to represent the other angle, then the total beamwidth is the difference between these two angles and can be written as Θh = cos−1 [ λ 2𝜋d ( kd cos 𝜃0 −2.782 N )] −cos−1 [ λ 2𝜋d ( kd cos 𝜃0 + 2.782 N )] = cos−1 ( cos 𝜃0 −2.782 Nkd ) −cos−1 ( cos 𝜃0 + 2.782 Nkd ) (6-22) Since N = (L + d)∕d, (6-22) reduces to Θh = cos−1 [ cos 𝜃0 −0.443 λ (L + d) ] −cos−1 [ cos 𝜃0 + 0.443 λ (L + d) ] (6-22a) where L is the length of the array. Equation (6-22a) can also be used to compute the half-power beamwidth of a broadside array. However, it is not valid for an end-fire array. A plot of the half-power beamwidth (in degrees) as a function of the array length is shown in Figure 6.12. These curves are valid for broadside, ordinary end-fire, and scanning uniform arrays (constant magnitude but with progressive phase shift). In a later section it will be shown that the curves of Figure 6.12 can be used, in conjunction with a beam broadening factor , to compute the directivity of nonuniform amplitude arrays.
6.3.4 Hansen-Woodyard End-Fire Array The conditions for an ordinary end-fire array were discussed in Section 6.3.2. It was concluded that the maximum radiation can be directed along the axis of the uniform array by allowing the progressive phase shift 𝛽between elements to be equal to (6-20a) for 𝜃0 = 0◦and (6-20b) for 𝜃0 = 180◦.
To enhance the directivity of an end-fire array without destroying any of the other characteristics, Hansen and Woodyard in 1938 proposed that the required phase shift between closely spaced N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 305 x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 = 60° = 60° θ θ θ θ (a) Three-dimensional (b) Two-dimensional Figure 6.11 Three- and two-dimensional array factor patterns of a 10-element uniform amplitude scanning array (N = 10, 𝛽= −kd cos 𝜃0, 𝜃0 = 60◦, d = λ∕4).
306 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.12 Half-power beamwidth for broadside, ordinary end-fire, and scanning uniform linear arrays.
(source: R. S. Elliott, “Beamwidth and Directivity of Large Scanning Arrays,” First of Two Parts, The Microwave Journal, December 1963).
elements of a very long array† should be 𝛽= − ( kd + 2.92 N ) ≃− ( kd + 𝜋 N ) ⇒for maximum in 𝜃0 = 0◦ (6-23a) 𝛽= + ( kd + 2.92 N ) ≃+ ( kd + 𝜋 N ) ⇒for maximum in 𝜃0 = 180◦ (6-23b) These requirements are known today as the Hansen-Woodyard conditions for end-fire radiation.
They lead to a larger directivity than the conditions given by (6-20a) and (6-20b). It should be pointed out, however, that these conditions do not necessarily yield the maximum possible directivity. In fact, the maximum may not even occur at 𝜃0 = 0◦or 180◦, its value found using (6-10c) or (6-10d) may not be unity, and the side lobe level may not be −13.46 dB. Both of them, maxima and side lobe levels, depend on the number of array elements, as will be illustrated.
To realize the increase in directivity as a result of the Hansen-Woodyard conditions, it is necessary that, in addition to the conditions of (6-23a) and (6-23b), |𝜓| assumes values of For maximum radiation along 𝜃0 = 0◦ |𝜓| = |kd cos 𝜃+ 𝛽|𝜃=0◦= 𝜋 N and |𝜓| = |kd cos 𝜃+ 𝛽|𝜃=180◦≃𝜋 (6-24a) †In principle, the Hansen-Woodyard condition was derived for an infinitely long antenna with continuous distribution. It thus gives good results for very long, finite length discrete arrays with closely spaced elements.
N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 307 For maximum radiation along 𝜃0 = 180◦ |𝜓| = |kd cos 𝜃+ 𝛽|𝜃=180◦= 𝜋 N and |𝜓| = |kd cos 𝜃+ 𝛽|𝜃=0◦≃𝜋 (6-24b) The condition of |𝜓| = 𝜋∕N in (6-24a) or (6-24b) is realized by the use of (6-23a) or (6-23b), respec-tively. Care must be exercised in meeting the requirement of |𝜓| ≃𝜋for each array. For an array of N elements, the condition of |𝜓| ≃𝜋is satisfied by using (6-23a) for 𝜃= 0◦, (6-23b) for 𝜃= 180◦, and choosing for each a spacing of d = (N −1 N ) λ 4 (6-25) If the number of elements is large, (6-25) can be approximated by d ≃λ 4 (6-25a) Thus for a large uniform array, the Hansen-Woodyard condition can only yield an improved direc-tivity provided the spacing between the elements is approximately λ∕4.
This is also illustrated in Figure 6.13 where the 3-D field patterns of the ordinary and the Hansen-Woodyard end-fire designs, for N = 10 and d = λ∕4, are placed next to each other. It is apparent that the major lobe of the ordinary end-fire is wider (HPBW = 74◦) than that of the Hansen-Woodyard (HPBW = 37◦); thus, higher directivity for the Hansen-Woodyard. However, the side lobe of the ordinary end-fire is lower (about −13.5 dB) compared to that of the Hansen-Woodyard, which is about −8.9 dB. The lower side lobe by the ordinary end-fire is not sufficient to offset the benefit from the narrower beamwidth of the Hansen-Woodyard that leads to the higher directivity. A comparison between the ordinary and Hansen-Woodyard end-fire array patterns is also illustrated in Figure 10.16 for the design of a helical antenna.
To make the comparisons more meaningful, the directivities for each of the patterns of Fig-ures 6.13 have been calculated, using numerical integration, and it is found that they are equal to 11 and 19, respectively. Thus the Hansen-Woodyard conditions realize a 73% increase in directivity for this case.
As will be shown in Section 6.4 and listed in Table 6.8, the directivity of a Hansen-Woodyard end-fire array is always approximately 1.805 times (or 2.56 dB) greater than the directivity of an ordinary end-fire array. The increase in directivity of the pattern in Figure 6.13 for the Hansen-Woodyard design is at the expense of an increase of about 4 dB in side lobe level. Therefore in the design of an array, there is a trade-off between directivity (or half-power beamwidth) and side lobe level.
To show that (6-23a) and (6-23b) do not lead to improved directivities over those of (6-20a) and (6-20b) if (6-24a) and (6-24b) are not satisfied, the pattern for the same array (N = 10) with d = λ∕4(𝛽= −3𝜋∕5) and d = λ∕2(𝛽= −11𝜋∕10) are plotted in Figure 6.14. Even though the d = λ∕2 pattern exhibits a very narrow lobe in the 𝜃0 = 0◦direction, its back lobes are larger than its main lobe. The d = λ∕2 pattern fails to realize a larger directivity because the necessary |𝜓|𝜃=180◦≃𝜋 condition of (6-24a) is not satisfied. That is, |𝜓| = |(kd cos 𝜃+ 𝛽)| 𝜃=180◦ 𝛽=−(kd+𝜋∕N) = | −(2kd + 𝜋∕N)| d=λ∕2 N=10 = 2.1𝜋 (6-26) which is not equal to 𝜋as required by (6-24a). Similar results occur for spacings other than those specified by (6-25) or (6-25a).
To better understand and appreciate the Hansen-Woodyard conditions, a succinct derivation of (6-23a) will be outlined. The procedure is identical to that reported by Hansen and Woodyard in their classic paper .
308 ARRAYS: LINEAR, PLANAR, AND CIRCULAR x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (a) Ordinary (b) Hansen-Woodyard Figure 6.13 Three-dimensional patterns for ordinary and Hansen-Woodyard end-fire designs (N = 10, d = λ∕4).
The array factor of an N-element array is given by (6-10c) as (AF)n = 1 N ⎧ ⎪ ⎨ ⎪ ⎩ sin [N 2 (kd cos 𝜃+ 𝛽) ] sin [1 2(kd cos 𝜃+ 𝛽) ] ⎫ ⎪ ⎬ ⎪ ⎭ (6-27) N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 309 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 d = λ/4 d = λ/2 θ θ Figure 6.14 Array factor patterns of a 10-element uniform amplitude Hansen-Woodyard end-fire array [N = 10, 𝛽= −(kd + 𝜋∕N)].
and approximated, for small values of 𝜓(𝜓= kd cos 𝜃+ 𝛽), by (6-10d) or (AF)n ≃ sin [N 2 (kd cos 𝜃+ 𝛽) ] [N 2 (kd cos 𝜃+ 𝛽) ] (6-27a) If the progressive phase shift between the elements is equal to 𝛽= −pd (6-28) where p is a constant, (6-27a) can be written as (AF)n = {sin[q(k cos 𝜃−p)] q(k cos 𝜃−p) } = [sin(Z) Z ] (6-29) where q = Nd 2 (6-29a) Z = q(k cos 𝜃−p) (6-29b) The radiation intensity can be written as U(𝜃) = [(AF)n]2 = [sin(Z) Z ]2 (6-30) 310 ARRAYS: LINEAR, PLANAR, AND CIRCULAR whose value at 𝜃= 0◦is equal to U(𝜃)|𝜃=0◦= {sin[q(k cos 𝜃−p)] q(k cos 𝜃−p) }2| | | | |𝜃=0◦ = {sin[q(k −p)] q(k −p) }2 (6-30a) Dividing (6-30) by (6-30a), so that the value of the array factor is equal to unity at 𝜃= 0◦, leads to U(𝜃)n = { q(k −p) sin[q(k −p)] sin[q(k cos 𝜃−p)] [q(k cos 𝜃−p)] }2 = [ 𝜐 sin(𝜐) sin(Z) Z ]2 (6-31) where 𝜐= q(k −p) (6-31a) Z = q(k cos 𝜃−p) (6-31b) The directivity of the array factor can be evaluated using D0 = 4𝜋Umax Prad = Umax U0 (6-32) where U0 is the average radiation intensity and it is given by U0 = Prad 4𝜋= 1 4𝜋∫ 2𝜋 0 ∫ 𝜋 0 U(𝜃) sin 𝜃d𝜃d𝜙 = 1 2 [ 𝜐 sin(𝜐) ]2 ∫ 𝜋 0 [sin(Z) Z ]2 sin 𝜃d𝜃 (6-33) By using (6-31a) and (6-31b), (6-33) can be written as U0 = 1 2 [ q(k −p) sin[q(k −p)] ]2 ∫ 𝜋 0 [sin[q(k cos 𝜃−p)] q(k cos 𝜃−p) ]2 sin 𝜃d𝜃 (6-33a) To maximize the directivity, as given by (6-32), (6-33a) must be minimized. Performing the inte-gration, (6-33a) reduces to U0 = 1 2kq [ 𝜐 sin(𝜐) ]2 [ 𝜋 2 + [cos(2𝜐) −1] 2𝜐 + Si(2𝜐) ] = 1 2kqg(𝜐) (6-34) where 𝜐= q(k −p) (6-34a) Si(z) = ∫ z 0 sin t t dt (6-34b) g(𝜐) = [ 𝜐 sin(𝜐) ]2 [ 𝜋 2 + [cos(2𝜐) −1] 2𝜐 + Si(2𝜐) ] (6-34c) N-ELEMENT LINEAR ARRAY: UNIFORM AMPLITUDE AND SPACING 311 0 –0.5 –1.0 –1.5 –2.0 0.25 0.50 0.75 0.871 –1.46 1.00 1.25 1.50 1.75 υ g (υ) Figure 6.15 Variation of g(𝜐) (see Eq. 6-34c) as a function of 𝜐.
The function g(𝜐) is plotted in Figure 6.15 and its minimum value occurs when 𝜐= q(k −p) = Nd 2 (k −p) = −1.46 (6-35) Thus 𝛽= −pd = − ( kd + 2.92 N ) (6-36) which is the condition for end-fire radiation with improved directivity (Hansen-Woodyard condition) along 𝜃0 = 0◦, as given by (6-23a). Similar procedures can be followed to establish (6-23b).
Ordinarily, (6-36) is approximated by 𝛽= − ( kd + 2.92 N ) ≃− ( kd + 𝜋 N ) (6-36a) with not too much relaxation in the condition since the curve of Figure 6.15 is broad around the minimum point 𝜐= −1.46. Its value at 𝜐= −1.57 is almost the same as the minimum at 𝜐= −1.46.
The expressions for the nulls, maxima, half-power points, minor lobe maxima, and beamwidths are listed in Tables 6.5 and 6.6.
For the broadside, end-fire and scanning linear designs, there is a maximum spacing dmax that should not be exceeded to maintain in the amplitude pattern either one or two maxima. A second maximum of the array factor, if it exists, will begin to appear at 𝜃= 0◦as the separation between the elements increases and the 𝜓of (6-21) approaches 2𝜋. This is indicated in Figure 6.7 for the broad-side array with d = λ when a second maximum appears in both of the end-fire directions (𝜃o = 0◦ and 180◦). The maximum spacing separation dmax is intended to keep a second lobe from appear-ing, whose amplitude would equal the amplitude of the main lobe (referred to as grating lobe). This 312 ARRAYS: LINEAR, PLANAR, AND CIRCULAR TABLE 6.5 Nulls, Maxima, Half-Power Points, and Minor Lobe Maxima for Uniform Amplitude Hansen-Woodyard End-Fire Arrays NULLS 𝜃n = cos−1 [ 1 + (1 −2n) λ 2dN ] n = 1, 2, 3, … n ≠N, 2N, 3N, … MAXIMA 𝜃m = cos−1 { 1 + [1 −(2m + 1)] λ 2Nd } m = 1, 2, 3, … 𝜋d∕λ ≪1 HALF-POWER POINTS 𝜃h = cos−1 ( 1 −0.1398 λ Nd ) 𝜋d∕λ ≪1 N large MINOR LOBE MAXIMA 𝜃s = cos−1 ( 1 −sλ Nd ) s = 1, 2, 3, … 𝜋d∕λ ≪1 TABLE 6.6 Beamwidths for Uniform Amplitude Hansen-Woodyard End-Fire Arrays FIRST-NULL BEAMWIDTH (FNBW) Θn = 2 cos−1 ( 1 − λ 2dN ) HALF-POWER BEAMWIDTH (HPBW) Θh = 2 cos−1 ( 1 −0.1398 λ Nd ) 𝜋d∕λ ≪1 N large FIRST SIDE LOBE BEAMWIDTH (FSLBW) Θs = 2 cos−1 ( 1 −λ Nd ) 𝜋d∕λ ≪1 separation can be obtained by equating 𝜓of (6-21), with 𝜃= 0◦, to less than 2𝜋, or 𝜓= kd(cos 𝜃+ | cos 𝜃o|)| 𝜃=0 d=dmax < 2𝜋⇒dmax < λ (1 + | cos 𝜃o|) (6-37) where 𝜃o is the scan angle of the main lobe, which allows scanning toward the broadside (𝜃o = 90◦), in both end-fire directions (𝜃o = 0◦, 180◦) and all other angles (0◦≤𝜃o ≤180◦). Table 6.7 lists the maximum element spacing dmax for the various linear and planar arrays, uniform and nonuniform, in order to maintain either one or two amplitude maxima. The dmax for broadside, end-fire and scanning arrays are the same as those obtained from (6-37).
6.4 N-ELEMENT LINEAR ARRAY: DIRECTIVITY The criteria that must be met to achieve broadside and end-fire radiation by a uniform linear array of N elements were discussed in the previous section. It would be instructive to investigate the direc-tivity of each of the arrays, since it represents a figure of merit on the operation of the system.
N-ELEMENT LINEAR ARRAY: DIRECTIVITY 313 TABLE 6.7 Maximum Element Spacing d𝐦𝐚𝐱to Maintain Either One or Two Amplitude Maxima of a Linear Array Array Distribution Type Direction of Maximum Element Spacing Linear Uniform Broadside 𝜃0 = 90◦only dmax < λ 𝜃0 = 0◦, 90◦, 180◦ simultaneously d = λ Linear Uniform Ordinary end-fire 𝜃0 = 0◦only dmax < λ∕2 𝜃0 = 180◦only dmax < λ∕2 𝜃0 = 0◦, 90◦, 180◦ simultaneously d = λ Linear Uniform Hansen-Woodyard end-fire 𝜃0 = 0◦only d ≃λ∕4 𝜃0 = 180◦only d ≃λ∕4 Linear Uniform Scanning 𝜃0 = 𝜃max dmax < λ 0 < 𝜃0 < 180◦ Linear Nonuniform Binomial 𝜃0 = 90◦only dmax < λ 𝜃0 = 0◦, 90◦, 180◦ simultaneously d = λ Linear Nonuniform Dolph-Tschebyscheff 𝜃0 = 90◦only dmax ≤λ 𝜋cos−1 ( −1 zo ) 𝜃0 = 0◦, 90◦, 180◦ simultaneously d = λ Planar Uniform Planar 𝜃0 = 0◦only dmax < λ 𝜃0 = 0◦, 90◦and 180◦; 𝜙0 = 0◦, 90◦, 180◦, 270◦ simultaneously d = λ 6.4.1 Broadside Array As a result of the criteria for broadside radiation given by (6-18a), the array factor for this form of the array reduces to (AF)n = 1 N ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 kd cos 𝜃 ) sin (1 2kd cos 𝜃 ) ⎤ ⎥ ⎥ ⎥ ⎦ (6-38) which for a small spacing between the elements (d ≪λ) can be approximated by (AF)n ≃ ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 kd cos 𝜃 ) (N 2 kd cos 𝜃 ) ⎤ ⎥ ⎥ ⎥ ⎦ (6-38a) 314 ARRAYS: LINEAR, PLANAR, AND CIRCULAR The radiation intensity can be written as U(𝜃) = [(AF)n]2 = ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 kd cos 𝜃 ) N 2 kd cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 = [sin(Z) Z ]2 (6-39) Z = N 2 kd cos 𝜃 (6-39a) The directivity can be obtained using (6-32) where Umax of (6-39) is equal to unity (Umax = 1) and it occurs at 𝜃= 90◦. The average value U0 of the intensity reduces to U0 = 1 4𝜋Prad = 1 2 ∫ 𝜋 0 [sin(Z) Z ]2 sin 𝜃d𝜃 = 1 2 ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎣ sin (N 2 kd cos 𝜃 ) N 2 kd cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ 2 sin 𝜃d𝜃 (6-40) By making a change of variable, that is, Z = N 2 kd cos 𝜃 (6-40a) dZ = −N 2 kd sin 𝜃d𝜃 (6-40b) (6-40) can be written as U0 = −1 Nkd ∫ −Nkd∕2 +Nkd∕2 [sin Z Z ]2 dZ = 1 Nkd ∫ +Nkd∕2 −Nkd∕2 [sin Z Z ]2 dZ (6-41) For a large array (Nkd∕2 →large), (6-41) can be approximated by extending the limits to infinity.
That is, U0 = 1 Nkd ∫ +Nkd∕2 −Nkd∕2 [sin Z Z ]2 dZ ≃ 1 Nkd ∫ +∞ −∞ [sin Z Z ]2 dZ (6-41a) Since ∫ +∞ −∞ [sin(Z) Z ]2 dZ = 𝜋 (6-41b) (6-41a) reduces to U0 ≃ 𝜋 Nkd (6-41c) The directivity of (6-32) can now be written as D0 = Umax U0 ≃Nkd 𝜋 = 2N (d λ ) (6-42) N-ELEMENT LINEAR ARRAY: DIRECTIVITY 315 Using L = (N −1)d (6-43) where L is the overall length of the array, (6-42) can be expressed as D0 ≃2N (d λ ) ≃2 ( 1 + L d ) (d λ ) (6-44) which for a large array (L ≫d) reduces to D0 ≃2N (d λ ) = 2 ( 1 + L d ) (d λ ) L ≫d ≃ 2 (L λ ) (6-44a) Example 6.3 Given a linear, broadside, uniform array of 10 isotropic elements (N = 10) with a separation of λ∕4(d = λ∕4) between the elements, find the directivity of the array.
Solution: Using (6-44a) D0 ≃2N (d λ ) = 5 (dimensionless) = 10 log10(5) = 6.99 dB 6.4.2 Ordinary End-Fire Array For an end-fire array, with the maximum radiation in the 𝜃0 = 0◦direction, the array factor is given by (AF)n = ⎡ ⎢ ⎢ ⎢ ⎣ sin [N 2 kd(cos 𝜃−1) ] N sin [1 2kd(cos 𝜃−1) ] ⎤ ⎥ ⎥ ⎥ ⎦ (6-45) which, for a small spacing between the elements (d ≪λ), can be approximated by (AF)n ≃ ⎡ ⎢ ⎢ ⎢ ⎣ sin [N 2 kd(cos 𝜃−1) ] [N 2 kd(cos 𝜃−1) ] ⎤ ⎥ ⎥ ⎥ ⎦ (6-45a) The corresponding radiation intensity can be written as U(𝜃) = [(AF)n]2 = ⎡ ⎢ ⎢ ⎢ ⎣ sin [N 2 kd(cos 𝜃−1) ] N 2 kd(cos 𝜃−1) ⎤ ⎥ ⎥ ⎥ ⎦ 2 = [sin(Z) Z ]2 (6-46) Z = N 2 kd(cos 𝜃−1) (6-46a) 316 ARRAYS: LINEAR, PLANAR, AND CIRCULAR whose maximum value is unity (Umax = 1) and it occurs at 𝜃= 0◦. The average value of the radiation intensity is given by U0 = 1 4𝜋∫ 2𝜋 0 ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎣ sin [N 2 kd(cos 𝜃−1) ] N 2 kd(cos 𝜃−1) ⎤ ⎥ ⎥ ⎥ ⎦ 2 sin 𝜃d𝜃d𝜙 = 1 2 ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎣ sin [N 2 kd(cos 𝜃−1) ] N 2 kd(cos 𝜃−1) ⎤ ⎥ ⎥ ⎥ ⎦ 2 sin 𝜃d𝜃 (6-47) By letting Z = N 2 kd(cos 𝜃−1) (6-47a) dZ = −N 2 kd sin 𝜃d𝜃 (6-47b) (6-47) can be written as U0 = −1 Nkd ∫ −Nkd 0 [sin(Z) Z ]2 dZ = 1 Nkd ∫ Nkd 0 [sin(Z) Z ]2 dZ (6-48) For a large array (Nkd →large), (6-48) can be approximated by extending the limits to infinity.
That is, U0 = 1 Nkd ∫ Nkd 0 [sin(Z) Z ]2 dZ ≃ 1 Nkd ∫ ∞ 0 [sin(Z) Z ]2 dZ (6-48a) Using (6-41b) reduces (6-48a) to U0 ≃ 𝜋 2Nkd (6-48b) and the directivity to D0 = Umax U0 ≃2Nkd 𝜋 = 4N (d λ ) (6-49) Another form of (6-49), using (6-43), is D0 ≃4N (d λ ) = 4 ( 1 + L d ) (d λ ) (6-49a) which for a large array (L ≫d) reduces to D0 ≃4N (d λ ) = 4 ( 1 + L d ) (d λ ) L ≫d ≃ 4 (L λ ) (6-49b) It should be noted that the directivity of the end-fire array, as given by (6-49)–(6-49b), is twice that for the broadside array as given by (6-42)–(6-44a).
N-ELEMENT LINEAR ARRAY: DIRECTIVITY 317 Example 6.4 Given a linear, end-fire, uniform array of 10 elements (N = 10) with a separation of λ∕4(d = λ∕4) between the elements, find the directivity of the array factor. This array is identical to the broadside array of Example 6.3.
Solution: Using (6-49) D0 ≃4N (d λ ) = 10 (dimensionless) = 10 log10(10) = 10 dB This value for the directivity (D0 = 10) is approximate, based on the validity of (6-48a). However, it compares very favorably with the value of D0 = 10.05 obtained by numerically integrating (6-45) using the Directivity computer program of Chapter 2.
6.4.3 Hansen-Woodyard End-Fire Array For an end-fire array with improved directivity (Hansen-Woodyard designs) and maximum radiation in the 𝜃0 = 0◦direction, the radiation intensity (for small spacing between the elements, d ≪λ) is given by (6-31)–(6-31b). The maximum radiation intensity is unity (Umax = 1), and the average radiation intensity is given by (6-34) where q and 𝜐are defined, respectively, by (6-29a) and (6-34a).
Using (6-29a), (6-34a), (6-35), and (6-37), the radiation intensity of (6-34) reduces to U0 = 1 Nkd (𝜋 2 )2 [𝜋 2 + 2 𝜋−1.8515 ] = 0.871 Nkd (6-50) which can also be written as U0 = 0.871 Nkd = 1.742 2Nkd = 0.554 ( 𝜋 2Nkd ) (6-50a) The average value of the radiation intensity, as given by (6-50a), is 0.554 times that for the ordi-nary end-fire array of (6-48b). Thus the directivity can be expressed, using (6-50a), as D0 = Umax U0 = 1 0.554 [2Nkd 𝜋 ] = 1.805 [ 4N (d λ )] (6-51) which is 1.805 times that of the ordinary end-fire array as given by (6-49). Using (6-43), (6-51) can also be written as D0 = 1.805 [ 4N (d λ )] = 1.805 [ 4 ( 1 + L d ) d λ ] (6-51a) which for a large array (L ≫d) reduces to D0 = 1.805 [ 4N (d λ )] = 1.805 [ 4 ( 1 + L d ) (d λ )] ≃1.805 [ 4 (L λ )] (6-51b) 318 ARRAYS: LINEAR, PLANAR, AND CIRCULAR TABLE 6.8 Directivities for Broadside and End-Fire Arrays Array Directivity BROADSIDE D0 = 2N (d λ ) = 2 ( 1 + L d ) d λ ≃2 (L λ ) N𝜋d∕λ →∞, L ≫d END-FIRE (ORDINARY) D0 = 4N (d λ ) = 4 ( 1 + L d ) d λ ≃4 (L λ ) Only one maximum (𝜃0 = 0◦or 180◦) 2N𝜋d∕λ →∞, L ≫d D0 = 2N (d λ ) = 2 ( 1 + L d ) d λ ≃2 (L λ ) Two maxima (𝜃0 = 0◦and 180◦) END-FIRE (HANSEN-WOODYARD) D0 = 1.805 [ 4N (d λ )] = 1.805 [ 4 ( 1 + L d ) d λ ] = 1.805 [ 4 (L λ )] 2N𝜋d∕λ →∞, L ≫d Example 6.5 Given a linear, end-fire (with improved directivity) Hansen-Woodyard, uniform array of 10 ele-ments (N = 10) with a separation of λ∕4(d = λ∕4) between the elements, find the directivity of the array factor. This array is identical to that of Examples 6.3 (broadside) and 6.4 (ordinary end-fire), and it is used for comparison.
Solution: Using (6-51b) D0 = 1.805 [ 4N (d λ )] = 18.05 (dimensionless) = 10 log10(18.05) = 12.56 dB The value of this directivity (D0 = 18.05) is 1.805 times greater than that of Example 6.4 (ordinary end-fire) and 3.61 times greater than that found in Example 6.3 (broadside).
Table 6.8 lists the directivities for broadside, ordinary end fire, and Hansen-Woodyard arrays.
6.5 DESIGN PROCEDURE In the design of any antenna system, the most important design parameters are usually the number of elements, spacing between the elements, excitation (amplitude and phase), half-power beamwidth, directivity, and side lobe level. In a design procedure some of these parameters are specified and the others are then determined.
The parameters that are specified and those that are determined vary among designs. For a uniform array, other than for the Hansen-Woodyard end-fire, the side lobe is always approximately −13.5 dB.
For the Hansen-Woodyard end-fire array the side lobe level is somewhat compromised above the −13.5 dB in order to gain about 1.805 (or 2.56 dB) in directivity. The order in which the other parameters are specified and determined varies among designs. For each of the uniform linear arrays that have been discussed, equations and some graphs have been presented which can be used to determine the half-power beamwidth and directivity, once the number of elements and spacing (or the total length of the array) are specified. In fact, some of the equations have been boxed or listed in tables. This may be considered more of an analysis procedure. The other approach is to specify the half-power beamwidth or directivity and to determine most of the other parameters. This can be viewed more as a design approach, and can be accomplished to a large extent with equations or N-ELEMENT LINEAR ARRAY: THREE-DIMENSIONAL CHARACTERISTICS 319 graphs that have been presented. More exact values can be obtained, if necessary, using iterative or numerical methods.
Example 6.6 Design a uniform linear scanning array whose maximum of the array factor is 30◦from the axis of the array (𝜃0 = 30◦). The desired half-power beamwidth is 2◦while the spacing between the elements is λ∕4. Determine the excitation of the elements (amplitude and phase), length of the array (in wavelengths), number of elements, and directivity (in dB).
Solution: Since the desired design is a uniform linear scanning array, the amplitude excitation is uniform. However, the progressive phase between the elements is, using (6-21) 𝛽= −kd cos 𝜃0 = −2𝜋 λ (λ 4 ) cos(30◦) = −1.36 radians = −77.94◦ The length of the array is obtained using an iterative procedure of (6-22) or its graphical solu-tion of Figure 6.12. Using the graph of Figure 6.12 for a scan angle of 30◦and 2◦half-power beamwidth, the approximate length plus one spacing (L + d) of the array is 50λ. For the 50λ length plus one spacing dimension from Figure 6.12 and 30◦scan angle, (6-22) leads to a half-power beamwidth of 2.03◦, which is very close to the desired value of 2◦. Therefore, the length of the array for a spacing of λ∕4 is 49.75λ.
Since the length of the array is 49.75λ and the spacing between the elements is λ∕4, the total number of elements is N = L d + 1 = (L + d d ) = 50 1∕4 = 200 The directivity of the array is obtained using the radiation intensity and the computer program Directivity of Chapter 2, and it is equal to 100.72 or 20.03 dB.
6.6 N-ELEMENT LINEAR ARRAY: THREE-DIMENSIONAL CHARACTERISTICS Up to now, the two-dimensional array factor of an N-element linear array has been investigated.
Although in practice only two-dimensional patterns can be measured, a collection of them can be used to reconstruct the three-dimensional characteristics of an array. It would then be instructive to examine the three-dimensional patterns of an array of elements. Emphasis will be placed on the array factor.
6.6.1 N-Elements Along Z-Axis A linear array of N isotropic elements are positioned along the z-axis and are separated by a distance d, as shown in Figure 6.5(a). The amplitude excitation of each element is an and there exists a progressive phase excitation 𝛽between the elements. For far-field observations, the array factor can be written according to (6-6) as AF = N ∑ n=1 anej(n−1)(kd cos 𝛾+𝛽) = N ∑ n=1 anej(n−1)𝜓 (6-52) 𝜓= kd cos 𝛾+ 𝛽 (6-52a) 320 ARRAYS: LINEAR, PLANAR, AND CIRCULAR where the an’s are the amplitude excitation coefficients and 𝛾is the angle between the axis of the array (z-axis) and the radial vector from the origin to the observation point.
In general, the angle 𝛾can be obtained from the dot product of a unit vector along the axis of the array with a unit vector directed toward the observation point. For the geometry of Figure 6.5(a) cos 𝛾= ̂ az ⋅̂ ar = ̂ az ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = cos 𝜃➱𝛾= 𝜃 (6-53) Thus (6-52) along with (6-53) is identical to (6-6), because the system of Figure 6.5(a) possesses a symmetry around the z-axis (no 𝜙variations). This is not the case when the elements are placed along any of the other axes, as will be shown next.
6.6.2 N-Elements Along X- or Y-Axis To demonstrate the simplicity that a judicious coordinate system and geometry can provide in the solution of a problem, let us consider an array of N isotropic elements along the x-axis, as shown in Figure 6.16. The far-zone array factor for this array is identical in form to that of Figure 6.5(a) except for the phase factor 𝜓. For this geometry cos 𝛾= ̂ ax ⋅̂ ar = ̂ ax ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = sin 𝜃cos 𝜙 (6-54) cos 𝛾= sin 𝜃cos 𝜙➱𝛾= cos−1(sin 𝜃cos 𝜙) (6-54a) The array factor of this array is also given by (6-52) but with 𝛾defined by (6-54a). For this system, the array factor is a function of both angles (𝜃and 𝜙).
Figure 6.16 Linear array of N isotropic elements positioned along the x-axis.
N-ELEMENT LINEAR ARRAY: THREE-DIMENSIONAL CHARACTERISTICS 321 In a similar manner, the array factor for N isotropic elements placed along the y-axis is that of (6-52) but with 𝛾defined by cos 𝛾= ̂ ay ⋅̂ ar = sin 𝜃sin 𝜙➱𝛾= cos−1(sin 𝜃sin 𝜙) (6-55) Physically placing the elements along the z-, x-, or y-axis does not change the characteristics of the array. Numerically they yield identical patterns even though their mathematical forms are different.
Example 6.7 Two half-wavelength dipole (l = λ∕2) are positioned along the x-axis and are separated by a distance d, as shown in Figure 6.17. The lengths of the dipoles are parallel to the z-axis. Find the total field of the array. Assume uniform amplitude excitation and a progressive phase difference of 𝛽.
z x y x z y λ/2 (a) = 0° β (b) = ± 180° β λ/2 Figure 6.17 Three-dimensional patterns for two λ∕2 dipoles spaced λ∕2. (source: P. Lorrain and D. R. Corson, Electromagnetic Fields and Waves, 2nd ed., W. H. Freeman and Co., Copyright c ⃝1970).
322 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Solution: The field pattern of a single element placed at the origin is given by (4-84) as E𝜃= j𝜂I0e−jkr 2𝜋r ⎡ ⎢ ⎢ ⎢ ⎣ cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ Using (6-52), (6-54a), and (6-10c), the array factor can be written as (AF)n = sin(kd sin 𝜃cos 𝜙+ 𝛽) 2 sin [1 2(kd sin 𝜃cos 𝜙+ 𝛽) ] The total field of the array is then given, using the pattern multiplication rule of (6-5), by E𝜃t = E𝜃⋅(AF)n = j𝜂I0e−jkr 2𝜋r cos (𝜋 2 cos 𝜃 ) sin 𝜃 ⎡ ⎢ ⎢ ⎢ ⎣ sin(kd sin 𝜃cos 𝜙+ 𝛽) 2 sin [1 2(kd sin 𝜃cos 𝜙+ 𝛽) ] ⎤ ⎥ ⎥ ⎥ ⎦ To illustrate the techniques, the three-dimensional patterns of the two-element array of Exam-ple 6.7 have been sketched in Figures 6.17(a) and (b). For both, the element separation is λ∕2(d = λ∕2). For the pattern of Figure 6.17(a), the phase excitation between the elements is identical (𝛽= 0).
In addition to the nulls in the 𝜃= 0◦direction, provided by the individual elements of the array, there are additional nulls along the x-axis (𝜃= 𝜋∕2, 𝜙= 0 and 𝜙= 𝜋) provided by the formation of the array. The 180◦phase difference required to form the nulls along the x-axis is a result of the separa-tion of the elements [kd = (2𝜋∕λ)(λ∕2) = 𝜋].
To form a comparison, the three-dimensional pattern of the same array but with a 180◦phase excitation (𝛽= 180◦) between the elements is sketched in Figure 6.17(b). The overall pattern of this array is quite different from that shown in Figure 6.17(a). In addition to the nulls along the z-axis (𝜃= 0◦) provided by the individual elements, there are nulls along the y-axis formed by the 180◦ excitation phase difference.
6.7 RECTANGULAR-TO-POLAR GRAPHICAL SOLUTION In antenna theory, many solutions are of the form f(𝜁) = f(C cos 𝛾+ 𝛿) (6-56) where C and 𝛿are constants and 𝛾is a variable. For example, the approximate array factor of an N-element, uniform amplitude linear array [Equation (6-10d)] is that of a sin(𝜁)∕𝜁form with 𝜁= C cos 𝛾+ 𝛿= N 2 𝜓= N 2 (kd cos 𝜃+ 𝛽) (6-57) where C = N 2 kd (6-57a) 𝛿= N 2 𝛽 (6-57b) N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 323 Usually the f(𝜁) function can be sketched as a function of 𝜁in rectilinear coordinates. Since 𝜁in (6-57) has no physical analog, in many instances it is desired that a graphical representation of |f(𝜁)| be obtained as a function of the physically observable angle 𝜃. This can be constructed graphically from the rectilinear graph, and it forms a polar plot.
The procedure that must be followed in the construction of the polar graph is as follows: 1. Plot, using rectilinear coordinates, the function |f (𝜁)|.
2. a. Draw a circle with radius C and with its center on the abscissa at 𝜁= 𝛿.
b. Draw vertical lines to the abscissa so that they will intersect the circle.
c. From the center of the circle, draw radial lines through the points on the circle intersected by the vertical lines.
d. Along the radial lines, mark off corresponding magnitudes from the linear plot.
e. Connect all points to form a continuous graph.
To better illustrate the procedure, the polar graph of the function f(𝜁) = sin (N 2 𝜓 ) N sin (𝜓 2 ), 𝜁= 5𝜋 2 cos 𝜃−5𝜋 4 (6-58) has been constructed in Figure 6.18. The function f(𝜁) of (6-58) represents the array factor of a 10-element (N = 10) uniform linear array with a spacing of λ∕4(d = λ∕4) and progressive phase shift of −𝜋∕4(𝛽= −𝜋∕4) between the elements. The constructed graph can be compared with its exact form shown in Figure 6.11.
From the construction of Figure 6.18, it is evident that the angle at which the maximum is directed is controlled by the radius of the circle C and the variable 𝛿. For the array factor of Figure 6.18, the radius C is a function of the number of elements (N) and the spacing between the elements (d).
In turn, 𝛿is a function of the number of elements (N) and the progressive phase shift between the elements (𝛽). Making 𝛿= 0 directs the maximum toward 𝜃= 90◦(broadside array). The part of the linear graph that is used to construct the polar plot is determined by the radius of the circle and the relative position of its center along the abscissa. The usable part of the linear graph is referred to as the visible region and the remaining part as the invisible region. Only the visible region of the linear graph is related to the physically observable angle 𝜃(hence its name).
6.8 N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE The theory to analyze linear arrays with uniform spacing, uniform amplitude, and a progressive phase between the elements was introduced in the previous sections of this chapter. A number of numerical and graphical solutions were used to illustrate some of the principles. In this section, broadside arrays with uniform spacing but nonuniform amplitude distribution will be considered.
Most of the discussion will be directed toward binomial and Dolph-Tschebyscheff broadside arrays (also spelled Tchebyscheff or Chebyshev).
Of the three distributions (uniform, binomial, and Tschebyscheff), a uniform amplitude array yields the smallest half-power beamwidth. It is followed, in order, by the Dolph-Tschebyscheff and binomial arrays. In contrast, binomial arrays usually possess the smallest side lobes followed, in order, by the Dolph-Tschebyscheff and uniform arrays. As a matter of fact, binomial arrays with 324 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.18 Rectangular-to-polar plot graphical solution.
element spacing equal or less than λ∕2 have no side lobes. It is apparent that the designer must compromise between side lobe level and beamwidth.
A criterion that can be used to judge the relative beamwidth and side lobe level of one design to another is the amplitude distribution (tapering) along the source. It has been shown analytically that for a given side lobe level the Dolph-Tschebyscheff array produces the smallest beamwidth between the first nulls. Conversely, for a given beamwidth between the first nulls, the Dolph-Tschebyscheff design leads to the smallest possible side lobe level.
Uniform arrays usually possess the largest directivity. However, superdirective (or super gain as most people refer to them) antennas possess directivities higher than those of a uniform array . Although a certain amount of superdirectivity is practically possible, superdirective arrays usu-ally require very large currents with opposite phases between adjacent elements. Thus the net total current and efficiency of each array are very small compared to the corresponding values of an indi-vidual element.
N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 325 Figure 6.19 Nonuniform amplitude arrays of even and odd number of elements.
Before introducing design methods for specific nonuniform amplitude distributions, let us first derive the array factor.
6.8.1 Array Factor An array of an even number of isotropic elements 2M (where M is an integer) is positioned sym-metrically along the z-axis, as shown in Figure 6.19(a). The separation between the elements is d, and M elements are placed on each side of the origin. Assuming that the amplitude excitation is symmetrical about the origin, the array factor for a nonuniform amplitude broadside array can be written as (AF)2M = a1e+j(1∕2)kd cos 𝜃+ a2e+j(3∕2)kd cos 𝜃+ ⋯ + aMe+j[(2M−1)∕2]kd cos 𝜃 + a1e−j(1∕2)kd cos 𝜃+ a2e−j(3∕2)kd cos 𝜃+ ⋯ + aMe−j[(2M−1)∕2]kd cos 𝜃 (AF)2M = 2 M ∑ n=1 an cos [(2n −1) 2 kd cos 𝜃 ] (6-59) 326 ARRAYS: LINEAR, PLANAR, AND CIRCULAR which in normalized form reduces to (AF)2M = M ∑ n=1 an cos [(2n −1) 2 kd cos 𝜃 ] (6-59a) where an’s are the excitation coefficients of the array elements.
If the total number of isotropic elements of the array is odd 2M + 1 (where M is an integer), as shown in Figure 6.19(b), the array factor can be written as (AF)2M+1 = 2a1 + a2e+jkd cos 𝜃+ a3ej2kd cos 𝜃+ ⋯+ aM+1ejMkd cos 𝜃 + a2e−jkd cos 𝜃+ a3e−j2kd cos 𝜃+ ⋯+ aM+1e−jMkd cos 𝜃 (AF)2M+1 = 2 M+1 ∑ n=1 an cos[(n −1)kd cos 𝜃] (6-60) which in normalized form reduces to (AF)2M+1 = M+1 ∑ n=1 an cos[(n −1)kd cos 𝜃] (6-60a) The amplitude excitation of the center element is 2a1.
Equations (6-59a) and (6-60a) can be written in normalized form as (AF)2M(even) = M ∑ n=1 an cos[(2n −1)u] (AF)2M+1(odd) = M+1 ∑ n=1 an cos[2(n −1)u] where u = 𝜋d λ cos 𝜃 (6-61a) (6-61b) (6-61c) The next step will be to determine the values of the excitation coefficients (an’s).
6.8.2 Binomial Array The array factor for the binomial array is represented by (6-61a)–(6-61c) where the an’s are the excitation coefficients which will now be derived.
A. Excitation Coefficients To determine the excitation coefficients of a binomial array, J. S. Stone suggested that the function (1 + x)m−1 be written in a series, using the binomial expansion, as (1 + x)m−1 = 1 + (m −1)x + (m −1)(m −2) 2!
x2 + (m −1)(m −2)(m −3) 3!
x3 + ⋯ (6-62) N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 327 The positive coefficients of the series expansion for different values of m are m = 1 1 m = 2 1 1 m = 3 1 2 1 m = 4 1 3 3 1 m = 5 1 4 6 4 1 m = 6 1 5 10 10 5 1 m = 7 1 6 15 20 15 6 1 m = 8 1 7 21 35 35 21 7 1 m = 9 1 8 28 56 70 56 28 8 1 m = 10 1 9 36 84 126 126 84 36 9 1 (6-63) The above represents Pascal’s triangle. If the values of m are used to represent the number of elements of the array, then the coefficients of the expansion represent the relative amplitudes of the elements.
Since the coefficients are determined from a binomial series expansion, the array is known as a binomial array.
Referring to (6-61a), (6-61b), and (6-63), the amplitude coefficients for the following arrays are: 1. Two elements (2M = 2) a1 = 1 2. Three elements (2M + 1 = 3) 2a1 = 2 ➱a1 = 1 a2 = 1 3. Four elements (2M = 4) a1 = 3 a2 = 1 4. Five elements (2M + 1 = 5) 2a1 = 6 ➱a1 = 3 a2 = 4 a3 = 1 The coefficients for other arrays can be determined in a similar manner.
B. Design Procedure One of the objectives of any method is its use in a design. For the binomial method, as for any other nonuniform array method, one of the requirements is the amplitude excitation coefficients for a given number of elements. This can be accomplished using either (6-62) or the Pascal triangle of (6-63) or extensions of it. Other figures of merit are the directivity, half-power beamwidth and side lobe level. It already has been stated that binomial arrays do not exhibit any minor lobes provided the spacing between the elements is equal or less than one-half of a wavelength. Unfortunately, 328 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.20 Array factor power patterns for a 10-element broadside binomial array with N = 10 and d = λ∕4, λ∕2, 3λ∕4, and λ.
closed-form expressions for the directivity and half-power beamwidth for binomial arrays of any spacing between the elements are not available. However, because the design using a λ∕2 spacing leads to a pattern with no minor lobes, approximate closed-form expressions for the half-power beamwidth and maximum directivity for the d = λ∕2 spacing only have been derived in terms of the numbers of elements or the length of the array, and they are given, respectively, by HPBW(d = λ∕2) ≃ 1.06 √ N −1 = 1.06 √ 2L∕λ = 0.75 √ L∕λ (6-64) D0 = 2 ∫ 𝜋 0 [ cos (𝜋 2 cos 𝜃 )]2(N−1) sin 𝜃d𝜃 (6-65) N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 329 D0 = (2N −2)(2N −4) ⋯2 (2N −3)(2N −5) ⋯1 (6-65a) D0 ≃1.77 √ N = 1.77 √ 1 + 2L∕λ (6-65b) These expressions can be used effectively to design binomial arrays with a desired half-power beamwidth or directivity. The value of the directivity as obtained using (6-65) to (6-65b) can be compared with the value using the array factor and the computer program Directivity of Chapter 2.
To illustrate the method, the patterns of a 10-element binomial array (2M = 10) with spacings between the elements of λ∕4, λ∕2, 3λ∕4, and λ, respectively, have been plotted in Figure 6.20.
The patterns are plotted using (6-61a) and (6-61c) with the coefficients of a1 = 126, a2 = 84, a3 = 36, a4 = 9, and a5 = 1. It is observed that there are no minor lobes for the arrays with spacings of λ∕4 and λ∕2 between the elements. While binomial arrays have very low level minor lobes, they exhibit larger beamwidths (compared to uniform and Dolph-Tschebyscheff designs). A major prac-tical disadvantage of binomial arrays is the wide variations between the amplitudes of the different elements of an array, especially for an array with a large number of elements. This leads to very low efficiencies for the feed network, and it makes the method not very desirable in practice. For exam-ple, the relative amplitude coefficient of the end elements of a 10-element array is 1 while that of the center element is 126. Practically, it would be difficult to obtain and maintain such large amplitude variations among the elements. They would also lead to very inefficient antenna systems. Because the magnitude distribution is monotonically decreasing from the center toward the edges and the magnitude of the extreme elements is negligible compared to those toward the center, a very low side lobe level is expected.
Table 6.7 lists the maximum element spacing dmax for the various linear and planar arrays, includ-ing binomial arrays, in order to maintain either one or two amplitude maxima.
Example 6.8 For a 10-element binomial array with a spacing of λ∕2 between the elements, whose amplitude pattern is displayed in Figure 6.20, determine the half-power beamwidth (in degrees) and the maximum directivity (in dB). Compare the answers with other available data.
Solution: Using (6-64), the half-power beamwidth is equal to HPBW ≃ 1.06 √ 10 −1 = 1.06 3 = 0.353 radians = 20.23◦ The value obtained using the array factor, whose pattern is shown in Figure 6.20, is 20.5◦which compares well with the approximate value.
Using (6-65a), the value of the directivity is equal for N = 10 D0 = 5.392 = 7.32 dB while the value obtained using (6-65b) is D0 = 1.77 √ 10 = 5.597 = 7.48 dB The value obtained using the array factor and the computer program Directivity is D0 = 5.392 (dimensionless) = 7.32 dB. These values compare favorably with each other.
330 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 6.8.3 Dolph-Tschebyscheff Array: Broadside Another array, with many practical applications, is the Dolph-Tschebyscheff array. The method was originally introduced by Dolph and investigated afterward by others –. It is pri-marily a compromise between uniform and binomial arrays. Its excitation coefficients are related to Tschebyscheff polynomials. A Dolph-Tschebyscheff array with no side lobes (or side lobes of −∞dB) reduces to the binomial design. The excitation coefficients for this case, as obtained by both methods, would be identical.
A. Array Factor Referring to (6-61a) and (6-61b), the array factor of an array of even or odd number of elements with symmetric amplitude excitation is nothing more than a summation of M or M + 1 cosine terms.
The largest harmonic of the cosine terms is one less than the total number of elements of the array.
Each cosine term, whose argument is an integer times a fundamental frequency, can be rewritten as a series of cosine functions with the fundamental frequency as the argument. That is, m = 0 cos(mu) = 1 m = 1 cos(mu) = cos u m = 2 cos(mu) = cos(2u) = 2 cos2 u −1 m = 3 cos(mu) = cos(3u) = 4 cos3 u −3 cos u m = 4 cos(mu) = cos(4u) = 8 cos4 u −8 cos2 u + 1 m = 5 cos(mu) = cos(5u) = 16 cos5 u −20 cos3 u + 5 cos u m = 6 cos(mu) = cos(6u) = 32 cos6 u −48 cos4 u + 18 cos2 u −1 m = 7 cos(mu) = cos(7u) = 64 cos7 u −112 cos5 u + 56 cos3 u −7 cos u m = 8 cos(mu) = cos(8u) = 128 cos8 u −256 cos6 u + 160 cos4 u −32 cos2 u + 1 m = 9 cos(mu) = cos(9u) = 256 cos9 u −576 cos7 u + 432 cos5 u −120 cos3 u + 9 cos u (6-66) The above are obtained by the use of Euler’s formula [eju]m = (cos u + j sin u)m = ejmu = cos(mu) + j sin(mu) (6-67) and the trigonometric identity sin2 u = 1 −cos2 u.
If we let z = cos u (6-68) (6-66) can be written as m = 0 cos(mu) = 1 = T0(z) m = 1 cos(mu) = z = T1(z) m = 2 cos(mu) = 2z2 −1 = T2(z) m = 3 cos(mu) = 4z3 −3z = T3(z) m = 4 cos(mu) = 8z4 −8z2 + 1 = T4(z) m = 5 cos(mu) = 16z5 −20z3 + 5z = T5(z) m = 6 cos(mu) = 32z6 −48z4 + 18z2 −1 = T6(z) m = 7 cos(mu) = 64z7 −112z5 + 56z3 −7z = T7(z) m = 8 cos(mu) = 128z8 −256z6 + 160z4 −32z2 + 1 = T8(z) m = 9 cos(mu) = 256z9 −576z7 + 432z5 −120z3 + 9z = T9(z) (6-69) N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 331 and each is related to a Tschebyscheff (Chebyshev) polynomial Tm(z). These relations between the cosine functions and the Tschebyscheff polynomials are valid only in the −1 ≤z ≤+1 range.
Because | cos(mu)| ≤1, each Tschebyscheff polynomial is |Tm(z)| ≤1 for −1 ≤z ≤+1. For |z| > 1, the Tschebyscheff polynomials are related to the hyperbolic cosine functions.
The recursion formula for Tschebyscheff polynomials is Tm(z) = 2zTm−1(z) −Tm−2(z) (6-70) It can be used to find one Tschebyscheff polynomial if the polynomials of the previous two orders are known. Each polynomial can also be computed using Tm(z) = cos[m cos−1(z)] −1 ≤z ≤+1 Tm(z) = cosh[m cosh−1(z)]† z < −1, z > +1 (6-71a) (6-71b) In Figure 6.21 the first six Tschebyscheff polynomials have been plotted. The following properties of the polynomials are of interest: 1. All polynomials, of any order, pass through the point (1, 1).
2. Within the range −1 ≤z ≤1, the polynomials have values within −1 to +1.
3. All roots occur within −1 ≤z ≤1, and all maxima and minima have values of +1 and −1, respectively.
Since the array factor of an even or odd number of elements is a summation of cosine terms whose form is the same as the Tschebyscheff polynomials, the unknown coefficients of the array factor can be determined by equating the series representing the cosine terms of the array factor to the appropriate Tschebyscheff polynomial. The order of the polynomial should be one less than the total number of elements of the array.
The design procedure will be outlined first, and it will be illustrated with an example. In outlining the procedure, it will be assumed that the number of elements, spacing between the elements, and ratio of major-to-minor lobe intensity (R0) are known. The requirements will be to determine the excitation coefficients and the array factor of a Dolph-Tschebyscheff array.
B. Array Design Statement. Design a broadside Dolph-Tschebyscheff array of 2M or 2M + 1 elements with spacing d between the elements. The side lobes are R0 dB below the maximum of the major lobe. Find the excitation coefficients and form the array factor.
Procedure a. Select the appropriate array factor as given by (6-61a) or (6-61b).
b. Expand the array factor. Replace each cos(mu) function (m = 0, 1, 2, 3, …) by its appropriate series expansion found in (6-66).
c. Determine the point z = z0 such that Tm(z0) = R0 (voltage ratio). The order m of the Tschebyscheff polynomial is always one less than the total number of elements. The design procedure requires that the Tschebyscheff polynomial in the −1 ≤z ≤z1, where z1 is the null †x = cosh−1(y) = ln[y ± (y2 −1)1∕2] 332 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.21 Tschebyscheff polynomials of orders zero through five.
nearest to z = +1, be used to represent the minor lobes of the array. The major lobe of the pattern is formed from the remaining part of the polynomial up to point z0(z1 < z ≤z0).
d. Substitute cos(u) = z z0 (6-72) in the array factor of step b. The cos(u) is replaced by z∕z0, and not by z, so that (6-72) would be valid for |z| ≤|z0|. At |z| = |z0|, (6-72) attains its maximum value of unity.
e. Equate the array factor from step b, after substitution of (6-72), to a Tm(z) from (6-69). The Tm(z) chosen should be of order m where m is an integer equal to one less than the total number of elements of the designed array. This will allow the determination of the excitation coeffi-cients an’s.
f. Write the array factor of (6-61a) or (6-61b) using the coefficients found in step e.
N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 333 Example 6.9 Design a broadside Dolph-Tschebyscheff array of 10 elements with spacing d between the ele-ments and with a major-to-minor lobe ratio of 26 dB. Find the excitation coefficients and form the array factor.
Solution: 1. The array factor is given by (6-61a) and (6-61c). That is, (AF)2M = M=5 ∑ n=1 an cos[(2n −1)u] u = 𝜋d λ cos 𝜃 2. When expanded, the array factor can be written as (AF)10 = a1 cos(u) + a2 cos(3u) + a3 cos(5u) + a4 cos(7u) + a5 cos(9u) Replace cos(u), cos(3u), cos(5u), cos(7u), and cos(9u) by their series expansions found in (6-66).
3. R0 (dB) = 26 = 20 log10(R0) or R0 (voltage ratio) = 20. Determine z0 by equating R0 to T9(z0). Thus R0 = 20 = T9(z0) = cosh[9 cosh−1(z0)] or z0 = cosh[ 1 9 cosh−1(20)] = 1.0851 Another equation which can, in general, be used to find z0 and does not require hyper-bolic functions is z0 = 1 2 [( R0 + √ R2 0 −1 )1∕P + ( R0 − √ R2 0 −1 )1∕P] (6-73) where P is an integer equal to one less than the number of array elements (in this case P = 9). R0 = H0∕H1 and z0 are identified in Figure 6.22.
4. Substitute cos(u) = z z0 = z 1.0851 in the array factor found in step 2.
334 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.22 Tschebyscheff polynomial of order nine (a) amplitude (b) magnitude.
5. Equate the array factor of step 2, after the substitution from step 4, to T9(z). The polynomial T9(z) is shown plotted in Figure 6.22. Thus (AF)10 = z[(a1 −3a2 + 5a3 −7a4 + 9a5)∕z0] + z3[(4a2 −20a3 + 56a4 −120a5)∕z0 3] + z5[(16a3 −112a4 + 432a5)∕z0 5] + z7[(64a4 −576a5)∕z0 7] + z9[(256a5)∕z09] = 9z −120z3 + 432z5 −576z7 + 256z9 N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 335 Matching similar terms allows the determination of the an’s. That is, 256a5∕z9 0 = 256 ➱a5 = 2.0860 (64a4 −576a5)∕z0 7 = −576 ➱a4 = 2.8308 (16a3 −112a4 + 432a5)∕z0 5 = 432 ➱a3 = 4.1184 (4a2 −20a3 + 56a4 −120a5)∕z0 3 = −120 ➱a2 = 5.2073 (a1 −3a2 + 5a3 −7a4 + 9a5)∕z0 = 9 ➱a1 = 5.8377 In normalized form, the an coefficients can be written as a5 = 1 a5 = 0.357 a4 = 1.357 a4 = 0.485 a3 = 1.974 or a3 = 0.706 a2 = 2.496 a2 = 0.890 a1 = 2.798 a1 = 1 The first (left) set is normalized with respect to the amplitude of the elements at the edge while the other (right) is normalized with respect to the amplitude of the center element.
6. Using the first (left) set of normalized coefficients, the array factor can be written as (AF)10 = 2.798 cos(u) + 2.496 cos(3u) + 1.974 cos(5u) + 1.357 cos(7u) + cos(9u) where u = [(𝜋d∕λ) cos 𝜃].
The array factor patterns of Example 6.9 for d = λ∕4 and λ∕2 are shown plotted in Figure 6.23.
Since the spacing is less than λ(d < λ), maxima exist only at broadside (𝜃0 = 90◦). However when the spacing is equal to λ(d = λ), two more maxima appear (one toward 𝜃0 = 0◦and the other toward 𝜃0 = 180◦). For d = λ the array has four maxima, and it acts as an end-fire as well as a broadside array.
Table 6.7 lists the maximum element spacing dmax for the various linear and planar arrays, including Dolph-Tschebyscheff arrays, in order to maintain either one or two amplitude maxima.
To better illustrate how the pattern of a Dolph-Tschebyscheff array is formed from the Tschebyscheff polynomial, let us again consider the 10-element array whose corresponding Tschebyscheff polynomial is of order 9 and is shown plotted in Figure 6.22. The abscissa of Fig-ure 6.22, in terms of the spacing between the elements (d) and the angle 𝜃, is given by (6-72) or z = z0 cos u = z0 cos (𝜋d λ cos 𝜃 ) = 1.0851 cos (𝜋d λ cos 𝜃 ) (6-74) For d = λ∕4, λ∕2, 3λ∕4, and λ the values of z for angles from 𝜃= 0◦to 90◦to 180◦are shown tabulated in Table 6.9. Referring to Table 6.9 and Figure 6.22, it is interesting to discuss the pattern formation for the different spacings.
336 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 d = λ/4 d = λ/2 θ θ Figure 6.23 Array factor power pattern of a N = 10 element broadside Dolph-Tschebyscheff array.
1. d = λ∕4, N = 10, R0 = 20 At 𝜃= 0◦the value of z is equal to 0.7673 (point A). As 𝜃attains larger values, z increases until it reaches its maximum value of 1.0851 for 𝜃= 90◦. Beyond 90◦, z begins to decrease and reaches its original value of 0.7673 for 𝜃= 180◦. Thus for d = λ∕4, only the Tschebyscheff polynomial between the values 0.7673 ≤z ≤1.0851(A ≤z ≤z0) is used to form the pattern of the array factor.
TABLE 6.9 Values of the Abscissa z as a Function of 𝜽for a 10-Element Dolph-Tschebyscheff Array with R0 = 20 d = 𝛌∕4 d = 𝛌∕2 d = 3𝛌∕4 d = 𝛌 𝜽 z (Eq. 6-74) z (Eq. 6-74) z (Eq. 6-74) z (Eq. 6-74) 0◦ 0.7673 0.0 −0.7673 −1.0851 10◦ 0.7764 0.0259 −0.7394 −1.0839 20◦ 0.8028 0.1026 −0.6509 −1.0657 30◦ 0.8436 0.2267 −0.4912 −0.9904 40◦ 0.8945 0.3899 −0.2518 −0.8049 50◦ 0.9497 0.5774 0.0610 −0.4706 60◦ 1.0025 0.7673 0.4153 0.0 70◦ 1.0462 0.9323 0.7514 0.5167 80◦ 1.0750 1.0450 0.9956 0.9276 90◦ 1.0851 1.0851 1.0851 1.0851 100◦ 1.0750 1.0450 0.9956 0.9276 110◦ 1.0462 0.9323 0.7514 0.5167 120◦ 1.0025 0.7673 0.4153 0.0 130◦ 0.9497 0.5774 0.0610 −0.4706 140◦ 0.8945 0.3899 −0.2518 −0.8049 150◦ 0.8436 0.2267 −0.4912 −0.9904 160◦ 0.8028 0.1026 −0.6509 −1.0657 170◦ 0.7764 0.0259 −0.7394 −1.0839 180◦ 0.7673 0.0 −0.7673 −1.0851 N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 337 2. d = λ∕2, N = 10, R0 = 20 At 𝜃= 0◦the value of z is equal to 0 (point B). As 𝜃becomes larger, z increases until it reaches its maximum value of 1.0851 for 𝜃= 90◦. Beyond 90◦, z decreases and comes back to the original point for 𝜃= 180◦. For d = λ∕2, a larger part of the Tschebyscheff polynomial is used (0 ≤z ≤1.0851; B ≤z ≤z0).
3. d = 3λ∕4, N = 10, R0 = 20 For this spacing, the value of z for 𝜃= 0◦is −0.7673 (point C), and it increases as 𝜃becomes larger. It attains its maximum value of 1.0851 at 𝜃= 90◦. Beyond 90◦, it traces back to its original value (−0.7673 ≤z ≤z0; C ≤z ≤z0).
4. d = λ, N = 10, R0 = 20 As the spacing increases, a larger portion of the Tschebyscheff polynomial is used to form the pattern of the array factor. When d = λ, the value of z for 𝜃= 0◦is equal to −1.0851 (point D) which in magnitude is equal to the maximum value of z. As 𝜃attains values larger than 0◦, z increases until it reaches its maximum value of 1.0851 for 𝜃= 90◦. At that point the polynomial (and thus the array factor) again reaches its maximum value. Beyond 𝜃= 90◦, z and in turn the polynomial and array factor retrace their values (−1.0851 ≤z ≤+1.0851; D ≤z ≤z0). For d = λ there are four maxima, and a broadside and an end-fire array have been formed simultaneously.
It is often desired in some Dolph-Tschebyscheff designs to take advantage of the largest possible spacing between the elements while maintaining the same level of all minor lobes, including the one toward 𝜃= 0◦and 180◦. In general, as well as in Example 6.8 (Figure 6.20), the only minor lobe that can exceed the level of the others, when the spacing exceeds a certain maximum spacing between the elements, is the one toward end fire (𝜃= 0◦or 180◦or z = −1 in Figure 6.21 or Fig-ure 6.22). The maximum spacing which can be used while meeting the requirements is obtained using (6-72) or z = z0 cos(u) = z0 cos (𝜋d λ cos 𝜃 ) (6-75) The requirement not to introduce a minor lobe with a level exceeding the others is accomplished by utilizing the Tschebyscheff polynomial up to, but not going beyond z = −1. Therefore, for 𝜃= 0◦ or 180◦ −1 ≥z0 cos (𝜋dmax λ ) (6-76) or dmax ≤λ 𝜋cos−1 ( −1 z0 ) (6-76a) Equation (6-76a) provides the maximum spacing dmax. With this separation, for a given broadside Tschebyscheff array with fixed number of N elements and specified side lobe level, the array factor maintains the same side lobe level for all of its minor lobes. There is another element separation dopt, referred to as optimum separation, which, for a broadside Tschebyscheff array with fixed number of elements and side lobe level, leads to smallest possible HPBW, and it is dopt = λ [ 1 −cos−1(1∕𝛾) 𝜋 ] (6-76b) 338 ARRAYS: LINEAR, PLANAR, AND CIRCULAR given by 𝛾= cosh [( 1 N −1 ) ln ( R + √ R2 −1 )] (6-76c) where R represents the side lobe level (as a voltage ratio).
Design curves for dmax = dmin (from −10 dB to −100 dB), for a broadside Dolph-Tschebyscheff array with elements N = 10, 20, 30, and 40, are displayed in Figure 6.24(b). As expected, the max-imum/optimum element separation, for each array with fixed number of elements, it gets smaller as the side lobe level decreases. Also the design curves for dmax = dmin as a function of the number of elements, N = 3 −20 for side lobe levels −30, −40, −50, and −60 dB, are displayed in Figure 6.24(b). As expected, for each of the side lobe levels, the maximum/optimum element separation increases as the number of elements increase.
The excitation coefficients of a Dolph-Tschebyscheff array can be derived using various doc-umented techniques – and others. One method, whose results are suitable for computer calculations, is that by Barbiere . The coefficients using this method can be obtained using an = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ M ∑ q=n (−1)M−q(z0)2q−1 (q + M −2)!(2M −1) (q −n)!(q + n −1)!(M −q)!
for even 2M elements n = 1, 2, … M (6-77a) M+1 ∑ q=n (−1)M−q+1(z0)2(q−1) (q + M −2)!(2M) 𝜀n(q −n)!(q + n −2)!(M −q + 1)!
for odd 2M + 1 elements n = 1, 2, … M + 1 (6-77b) where 𝜀n = { 2 n = 1 1 n ≠1 C. Beamwidth and Directivity For large Dolph-Tschebyscheff arrays scanned not too close to end-fire and with side lobes in the range from −20 to −60 dB, the half-power beamwidth and directivity can be found by introducing a beam broadening factor given approximately by f = 1 + 0.636 { 2 R0 cosh [√ (cosh−1 R0)2 −𝜋2 ]}2 (6-78) where R0 is the major-to-side lobe voltage ratio. The beam broadening factor is plotted in Fig-ure 6.25(a) as a function of side lobe level (in dB).
The half-power beamwidth of a Dolph-Tschebyscheff array can be determined by 1. calculating the beamwidth of a uniform array (of the same number of elements and spacing) using (6-22a) or reading it off Figure 6.12 2. multiplying the beamwidth of part (1) by the appropriate beam broadening factor f computed using (6-78) or reading it off Figure 6.25(a) The same procedure can be used to determine the beamwidth of arrays with a cosine-on-a-pedestal distribution .
N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 339 (a) Maximum/optimum element separation as a function of side lobe level for different number of elements (b) Maximum/optimum element separation as a function of number of elements for different side lobe levels N=10 0.95 Side Lobe Level (dB) –10 –20 –30 –40 –50 –60 –70 –80 –90 –100 dmax/dopt (λ) dmax/dopt (λ) 1 0.95 0.85 0.75 0.65 0.9 0.8 0.7 0.85 0.75 0.65 0.55 0.9 0.8 0.7 0.6 0.5 Number of elements (N) 2 4 6 8 10 12 14 16 18 20 N=20 N=30 N=40 –30 dB –40 dB –50 dB –60 dB Figure 6.24 Maximum/optimum element seperation for broadside Dolph-Tschebyscheff array as a function of side lobe level and number of elements.
340 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.25 Beam broadening factor and directivity of Tschebyscheff arrays. (source: R. S. Elliott, “Beamwidth and Directivity of Large Scanning Arrays,” First of Two Parts, The Microwave Journal, Decem-ber 1963).
The beam broadening factor f can also be used to determine the directivity of large Dolph-Tschebyscheff arrays, scanned near broadside, with side lobes in the −20 to −60 dB range .
That is, D0 = 2R2 0 1 + (R2 0 −1)f λ (L + d) (6-79) which is shown plotted in Figure 6.25(b) as a function of L + d (in wavelengths).
N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 341 From the data in Figure 6.25(b) it can be concluded that: 1. The directivity of a Dolph-Tschebyscheff array, with a given side lobe level, increases as the array size or number of elements increases.
2. For a given array length, or a given number of elements in the array, the directivity does not necessarily increase as the side lobe level decreases. As a matter of fact, a −15 dB side lobe array has smaller directivity than a −20 dB side lobe array (see Figure 6.27). This may not be the case for all other side lobe levels.
The beamwidth and the directivity of an array depend linearly, but not necessarily at the same rate, on the overall length or total number of elements of the array. Therefore, the beamwidth and directivity must be related to each other. For a uniform broadside array this relation is D0 = 101.5 Θd (6-80) where Θd is the 3-dB beamwidth (in degrees). The above relation can be used as a good approxima-tion between beamwidth and directivity for most linear broadside arrays with practical distributions (including the Dolph-Tschebyscheff array). Equation (6-80) states that for a linear broadside array the product of the 3-dB beamwidth and the directivity is approximately equal to 100. This is analo-gous to the product of the gain and bandwidth for electronic amplifiers.
D. Design The design of a Dolph-Tschebyscheff array is very similar to those of other methods. Usually a certain number of parameters is specified, and the remaining are obtained following a certain pro-cedure. In this section we will outline an alternate method that can be used, in addition to the one outlined and followed in Example 6.9, to design a Dolph-Tschebyscheff array. This method leads to the excitation coefficients more directly.
Specify a. The side lobe level (in dB).
b. The number of elements.
Design Procedure a. Transform the side lobe level from decibels to a voltage ratio using R0(Voltage Ratio) = [R0(VR)] = 10R0(dB)∕20 (6-81) b. Calculate P, which also represents the order of the Tschebyscheff polynomial, using P = number of elements −1 c. Determine z0 using (6-73) or z0 = cosh [ 1 P cosh−1[R0(VR)] ] (6-82) d. Calculate the excitation coefficients using (6-77a) or (6-77b).
e. Determine the beam broadening factor using (6-78).
342 ARRAYS: LINEAR, PLANAR, AND CIRCULAR f. Calculate using (6-22a) the half-power beamwidth of a uniform array with the same number of elements and spacing between them.
g. Find the half-power beamwidth of the Tschebyscheff array by multiplying the half-power beamwidth of the uniform array by the beam broadening factor.
h. The maximum spacing between the elements should not exceed that of (6-76a).
i. Determine the directivity using (6-79).
j. The number of minor lobes for the three-dimensional pattern on either side of the main maxi-mum (0◦≤𝜃≤90◦), using the maximum permissible spacing, is equal to N −1.
k. Calculate the array factor using (6-61a) or (6-61b).
This procedure leads to the same results as any other.
Example 6.10 Calculate the half-power beamwidth and the directivity for the Dolph-Tschebyscheff array of Example 6.9 for a spacing of λ∕2 between the elements.
Solution: For Example 6.9, R0 = 26 dB ➱R0 = 20 (voltage ratio) Using (6-78) or Figure 6.24(a), the beam broadening factor f is equal to f = 1.079 According to (6-22a) or Figure 6.12, the beamwidth of a uniform broadside array with L + d = 5λ is equal to Θh = 10.17◦ Thus the beamwidth of a Dolph-Tschebyscheff array is equal to Θh = 10.17◦f = 10.17◦(1.079) = 10.97◦ The directivity can be obtained using (6-79), and it is equal to D0 = 2(20)2 1 + [(20)2 −1]1.079 5 = 9.18 (dimensionless) = 9.63 dB which closely agrees with the results of Figure 6.24(b).
In designing nonuniform arrays, the amplitude distribution between the elements is used to con-trol the side lobe level. Shown in Figure 6.26 are the excitation amplitude distributions of Dolph-Tschebyscheff arrays each with N = 10 elements, uniform element spacing of d = λ∕4, and different side lobe levels. It is observed that as the side lobe level increases the distribution from the center ele-ment(s) toward those at the edges is smoother and monotonically decreases for all levels except that of −20 dB. For this particular design (N = 10, d = 0.25λ), the smallest side lobe level, which still maintains a monotonic amplitude distribution from the center toward the edges, is about −21.05 dB, which is also displayed in Figure 6.26. Smaller side lobe levels than −20 dB will lead to even more abrupt amplitude distribution at the edges.
N-ELEMENT LINEAR ARRAY: UNIFORM SPACING, NONUNIFORM AMPLITUDE 343 Side lobe level = 80 dB Side lobe level = 60 dB Side lobe level = 40 dB Side lobe level = 30 dB Side lobe level = 21.05 dB Side lobe level = 20 dB 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −1 NORMALIZED EXCITATION COEFFICIENTS −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 ARRAY LENGTH ( ) λ Figure 6.26 Amplitude distribution, for different side lobe levels, of a Dolph-Tschebyscheff array with N = 10, d = λ∕4.
In designing nonuniform arrays, there is a compromise between side lobe level and half-power beamwidth/directivity. While the side lobe level decreases, the half-power beamwidth (HPBW) decreases and the directivity usually increases. This is demonstrated in Figure 6.27 for a 10-element Dolph-Tschebyscheff linear array with a uniform spacing of λ∕4 between the elements. Similar trends can be expected for other designs.
7 6 DIRECTIVITY (dB) HPBW (degrees) 5 20 30 40 50 10 4 8 101 102 SIDE LOBE LEVEL (dB) 103 HPBW DIRECTIVITY Figure 6.27 Directivity and half-power beamwidth versus side lobe level for a Dolph-Tschebyscheff array of N = 10, d = λ∕4.
344 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 6.8.4 Tschebysheff Design: Scanning In Section 6.8.3, the formulation for the broadside (𝜃0 = 90◦) array factor design was developed.
The Tschebyscheff design can be extended to allow for scanning the main beam in other directions (0◦≤𝜃0 ≤180◦) while maintaining all the minor lobes at the same level. The approach is to utilize (6-21), which allows the phase excitation for scanning the main beam toward any angle 𝜃0 in the angular range of 0◦≤𝜃0 ≤180◦. By doing this, we can rewrite (6-21) as 𝜓= kd(cos 𝜃−cos 𝜃0) (6-83) while (6-75) can be expressed as z = z0 cos[kd(cos 𝜃−cos 𝜃0)] = z0 cos [𝜋d λ (cos 𝜃−cos 𝜃0) ] (6-84) The maximum spacing dmax, while allowing all minor lobes at the same level, can be derived in a manner similar to (6-75)–(6-76a) but utilizing (6-84) in this case. As before, the requirement not to introduce a minor lobe with a level exceeding the others is accomplished by utilizing the Tschebyscheff polynomial up to, but not beyond z = −1 in Figure 6.21. Doing this, (6-76) can be written for 𝜃= 0◦or 180◦, while allowing 0◦≤𝜃0 ≤180◦, as −1 ≥z0 cos [𝜋dmax λ (1 −cos 𝜃0) ] (6-85) When solved for dmax, equation (6-85) reduces to dmax ≤ λ 𝜋(1 −cos 𝜃0) cos−1 ( −1 z0 ) (6-85a) To illustrate the design procedure, an example is used.
Example 6.11 Design a 10-element (N = 10) scanning Tschebyscheff array that meets the following specifica-tions: r The maximum of the main beam directed toward 𝜃= 60◦(𝜃0 = 60◦).
r All minor lobes are maintained at the same level of −26 dB over the entire angular range of 0◦≤𝜃≤180◦.
r Use the maximum allowable spacing dmax between the elements.
r Plot the pattern and compare it with that of the corresponding design but with the maximum of the main beam directed toward 𝜃= 90◦(broadside design; 𝜃0 = 90◦).
Solution: r Using (6-82), z0 = 1.0851, which is the same as for the broadside design of Example 6.9.
r Using (6-85a), dmax = 0.5821λ while for the broadside design of Example 6.9 it is dmax = 0.8731λ.
r The amplitude excitation coefficients are the same as those of Example 6.9.
SUPERDIRECTIVITY 345 r The normalized amplitude pattern for the scanning design displayed in Figure 6.28, where it is compared with the pattern of the corresponding broadside design. As expected, the pattern for the scanned design is asymmetrical about the 𝜃= 90◦direction while that of the broadside design is symmetrical. Also, it is evident that both designs maintain the −26 dB side lobe level over the entire 0◦≤𝜃≤180◦angular range.
Array Factor Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Array Factor Pattern (dB) –35 180° 150° 150° 120° 90° 60° 30° 0° 30° 60° 90° 120° –30 –25 –20 –15 –10 –5 0 N = 10, R0 = - 26 dB Tschebyscheff: Scanning (d = dmax= 0.5821 ) Tschebyscheff: Broadside (d = dmax = 0.8731 ) θ θ λ λ Figure 6.28 Normalized amplitude patterns of 10-element Tschebyscheff design for scanning and broad-side arrays.
An interactive MATLAB and FORTRAN computer program entitled Arrays has been developed, and it performs the analysis for uniform and nonuniform linear arrays, and uniform planar arrays.
The MATLAB version of the program also analyzes uniform circular arrays. The description of the program is provided in the corresponding READ ME file in the publisher’s website for this book.
6.9 SUPERDIRECTIVITY Antennas whose directivities are much larger than the directivity of a reference antenna of the same size are known as superdirective antennas. Thus a superdirective array is one whose directivity is larger than that of a reference array (usually a uniform array of the same length). In an array, superdi-rectivity is accomplished by inserting more elements within a fixed length (decreasing the spacing).
Doing this leads eventually to very large magnitudes and rapid changes of phase in the excitation coefficients of the elements of the array. Thus adjacent elements have very large and oppositely directed currents. This necessitates a very precise adjustment of their values. Associated with this are increases in reactive power (relative to the radiated power) and the Q of the array.
346 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 6.9.1 Efficiency and Directivity Because of the very large currents in the elements of superdirective arrays, the ohmic losses increases and the antenna efficiency decreases very sharply. Although practically the ohmic losses can be reduced by the use of superconductive materials, there is no easy solution for the precise adjustment of the amplitudes and phases of the array elements. High radiation efficiency superdirective arrays can be designed utilizing array functions that are insensitive to changes in element values .
In practice, superdirective arrays are usually referred to as super gain. However, super gain is a misnomer because such antennas have actual overall gains (because of very low efficiencies) less than uniform arrays of the same length. Although significant superdirectivity is very difficult and usually very impractical, a moderate amount can be accomplished. Superdirective antennas are very intriguing, and they have received much attention in the literature.
The length of the array is usually the limiting factor to the directivity of an array. Schelkunoff pointed out that theoretically very high directivities can be obtained from linear end-fire arrays.
Bowkamp and de Bruijn , however, concluded that theoretically there is no limit in the directivity of a linear antenna. More specifically, Riblet showed that Dolph-Tschebyscheff arrays with element spacing less than λ∕2 can yield any desired directivity. A numerical example of a Dolph-Tschebyscheff array of nine elements, λ∕32 spacing between the elements (total length of λ∕4), and a 1/19.5 (−25.8 dB) side lobe level was carried out by Yaru . It was found that to produce a directivity of 8.5 times greater than that of a single element, the currents on the individual elements must be on the order of 14 × 106 amperes and their values adjusted to an accuracy of better than one part in 1011. The maximum radiation intensity produced by such an array is equivalent to that of a single element with a current of only 19.5 × 10−3 amperes. If the elements of such an array are 1-cm diameter, of copper, λ∕2 dipoles operating at 10 MHz, the efficiency of the array is less than 10−14%.
6.9.2 Designs with Constraints To make the designs more practical, applications that warrant some superdirectivity should incorpo-rate constraints. One constraint is based on the sensitivity factor, and it was utilized for the design of superdirective arrays . The sensitivity factor (designated as K) is an important parameter which is related to the electrical and mechanical tolerances of an antenna, and it can be used to describe its performance (especially its practical implementation). For an N-element array, such as that shown in Figure 6.5(a), it can be written as K = N ∑ n=1 |an|2 | | | | | | N ∑ n=1 ane−jkrn | | | | | | 2 (6-86) where an is the current excitation of the nth element, and r′ n is the distance from the nth element to the far-field observation point (in the direction of maximum radiation).
In practice, the excitation coefficients and the positioning of the elements, which result in a desired pattern, cannot be achieved as specified. A certain amount of error, both electrical and mechanical, will always be present. Therefore the desired pattern will not be realized exactly, as required. How-ever, if the design is accomplished based on specified constraints, the realized pattern will approxi-mate the desired one within a specified deviation.
SUPERDIRECTIVITY 347 To derive design constraints, the realized current excitation coefficients cn’s are related to the desired ones an’s by cn = an + 𝛼nan = an(1 + 𝛼n) (6-86a) where 𝛼nan represents the error in the nth excitation coefficient. The mean square value of 𝛼n is denoted by 𝜀2 = ⟨|𝛼n|⟩2 (6-86b) To take into account the error associated with the positioning of the elements, we introduce 𝛿2 = (k𝜎)2 3 (6-86c) where 𝜎is the root-mean-square value of the element position error. Combining (6-86b) and (6-86c) reduces to Δ2 = 𝛿2 + 𝜀2 (6-86d) where Δ is a measure of the combined electrical and mechanical errors.
For uncorrelated errors KΔ2 = average radiation intensity of realized pattern maximum radiation intensity of desired pattern If the realized pattern is to be very close to the desired one, then KΔ2 ≪1 ➱Δ ≪ 1 √ K (6-86e) Equation (6-86e) can be rewritten, by introducing a safety factor S, as Δ = 1 √ SK (6-86f) S is chosen large enough so that (6-86e) is satisfied. When Δ is multiplied by 100, 100Δ represents the percent tolerance for combined electrical and mechanical errors.
The choice of the value of S depends largely on the required accuracy between the desired and realized patterns. For example, if the focus is primarily on the realization of the main beam, a value of S = 10 will probably be satisfactory. For side lobes of 20 dB down, S should be about 1,000. In general, an approximate value of S should be chosen according to S ≃10 × 10b∕10 (6-86g) where b represents the pattern level (in dB down) whose shape is to be accurately realized.
The above method can be used to design, with the safety factor K constrained to a certain value, arrays with maximum directivity. Usually one first plots, for each selected excitation distribution and positioning of the elements, the directivity D of the array under investigation versus the correspond-ing sensitivity factor K (using 6-86) of the same array. The design usually begins with the excitation and positioning of a uniform array (i.e., uniform amplitudes, a progressive phase, and equally spaced 348 ARRAYS: LINEAR, PLANAR, AND CIRCULAR elements). The directivity associated with it is designated as D0 while the corresponding sensitivity factor, computed using (6-86), is equal to K0 = 1∕N.
As the design deviates from that of the uniform array and becomes superdirective, the values of the directivity increase monotonically with increases in K. Eventually a maximum directivity is attained (designated as Dmax), and it corresponds to a K = Kmax; beyond that point (K > Kmax), the directivity decreases monotonically. The antenna designer should then select the design for which D0 < D < Dmax and K0 = 1∕N < K < Kmax.
The value of D is chosen subject to the constraint that K is a certain number whose corresponding tolerance error Δ of (6-86f), for the desired safety factor S, can be achieved practically. Tolerance errors of less than about 0.3 percent are usually not achievable in practice. In general, the designer must trade-off between directivity and sensitivity factor; larger D’s (provided D ≤Dmax) result in larger K’s (K ≤Kmax), and vice versa.
A number of constrained designs can be found in . For example, an array of cylindrical monopoles above an infinite and perfectly conducting ground plane was designed for optimum direc-tivity at f = 30 MHz, with a constraint on the sensitivity factor. The spacing d between the elements was maintained uniform.
For a four-element array, it was found that for d = 0.3λ the maximum directivity was 14.5 dB and occurred at a sensitivity factor of K = 1. However for d = 0.1λ the maximum directivity was up to 15.8 dB, with the corresponding sensitivity factor up to about 103. At K0 = 1∕N = 1∕4, the directivities for d = 0.3λ and 0.1λ were about 11.3 and 8 dB, respectively. When the sensitivity factor was maintained constant and equal to K = 1, the directivity for d = 0.3λ was 14.5 dB and only 11.6 dB for d = 0.1λ. It should be noted that the directivity of a single monopole above an infinite ground plane is twice that of the corresponding dipole in free-space and equal to about 3.25 (or about 5.1 dB).
6.10 PLANAR ARRAY In addition to placing elements along a line (to form a linear array), individual radiators can be posi-tioned along a rectangular grid to form a rectangular or planar array. Planar arrays provide additional variables which can be used to control and shape the pattern of the array. Planar arrays are more ver-satile and can provide more symmetrical patterns with lower side lobes. In addition, they can be used to scan the main beam of the antenna toward any point in space. Applications include tracking radar, search radar, remote sensing, communications, and many others.
A planar array of slots, used in the Airborne Warning and Control System (AWACS), is shown in Figure 6.29. It utilizes rectangular waveguide sticks placed vertically, with slots on the narrow wall of the waveguides. The system has 360◦view of the area, and at operating altitudes can detect targets hundreds of kilometers away. It is usually mounted at a height above the fuselage of an aircraft.
6.10.1 Array Factor To derive the array factor for a planar array, let us refer to Figure 6.30. If M elements are initially placed along the x-axis, as shown in Figure 6.30(a), the array factor of it can be written according to (6-52) and (6-54) as AF = M ∑ m=1 Im1ej(m−1)(kdx sin 𝜃cos 𝜙+𝛽x) (6-87) where Im1 is the excitation coefficient of each element. The spacing and progressive phase shift between the elements along the x-axis are represented, respectively, by dx and 𝛽x. If N such arrays are placed next to each other in the y-direction, a distance dy apart and with a progressive phase 𝛽y, PLANAR ARRAY 349 Figure 6.29 AWACS antenna array of waveguide slots. (photograph courtesy: Northrop Grumman Corporation).
a rectangular array will be formed as shown in Figure 6.30(b). The array factor for the entire planar array can be written as AF = N ∑ n=1 I1n [ M ∑ m=1 Im1ej(m−1)(kdx sin 𝜃cos 𝜙+𝛽x) ] ej(n−1)(kdy sin 𝜃sin 𝜙+𝛽y) (6-87a) or AF = SxmSyn (6-88) where Sxm = M ∑ m=1 Im1ej(m−1)(kdx sin 𝜃cos 𝜙+𝛽x) (6-88a) Syn = N ∑ n=1 I1nej(n−1)(kdy sin 𝜃sin 𝜙+𝛽y) (6-88b) Equation (6-88) indicates that the pattern of a rectangular array is the product of the array factors of the arrays in the x- and y-directions.
350 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.30 Linear and planar array geometries.
If the amplitude excitation coefficients of the elements of the array in the y-direction are propor-tional to those along the x, the amplitude of the (m, n)th element can be written as Imn = Im1I1n (6-89) If in addition the amplitude excitation of the entire array is uniform (Imn = I0), (6-87a) can be expressed as AF = I0 M ∑ m=1 ej(m−1)(kdx sin 𝜃cos 𝜙+𝛽x) N ∑ n=1 ej(n−1)(kdy sin 𝜃sin 𝜙+𝛽y) (6-90) PLANAR ARRAY 351 According to (6-6), (6-10), and (6-10c), the normalized form of (6-90) can also be written as AFn(𝜃, 𝜙) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 M sin (M 2 𝜓x ) sin (𝜓x 2 ) ⎫ ⎪ ⎬ ⎪ ⎭ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 N sin (N 2 𝜓y ) sin (𝜓y 2 ) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ where 𝜓x = kdx sin 𝜃cos 𝜙+ 𝛽x 𝜓y = kdy sin 𝜃sin 𝜙+ 𝛽y (6-91) (6-91a) (6-91b) When the spacing between the elements is equal or greater than λ∕2, multiple maxima of equal magnitude can be formed. The principal maximum is referred to as the major lobe and the remaining as the grating lobes. A grating lobe is defined as “a lobe, other than the main lobe, produced by an array antenna when the inter element spacing is sufficiently large to permit the in-phase addition of radiated fields in more than one direction.” To form or avoid grating lobes in a rectangular array, the same principles must be satisfied as for a linear array. To avoid grating lobes in the x-z and y-z planes, the spacing between the elements in the x- and y-directions, respectively, must be less than λ∕2 (dx < λ∕2 and dy < λ∕2).
Table 6.7 lists the maximum element spacing dmax (for either dx, dy or both) for the various uni-form and nonuniform arrays, including planar arrays, in order to maintain either one or two ampli-tude maxima.
For a rectangular array, the major lobe and grating lobes of Sxm and Syn in (6-88a) and (6-88b) are located at kdx sin 𝜃cos 𝜙+ 𝛽x = ±2m𝜋 m = 0, 1, 2, … (6-92a) kdy sin 𝜃sin 𝜙+ 𝛽y = ±2n𝜋 n = 0, 1, 2, … (6-92b) The phases 𝛽x and 𝛽y are independent of each other, and they can be adjusted so that the main beam of Sxm is not the same as that of Syn. However, in most practical applications it is required that the conical main beams of Sxm and Syn intersect and their maxima be directed toward the same direction.
If it is desired to have only one main beam that is directed along 𝜃= 𝜃0 and 𝜙= 𝜙0, the progressive phase shift between the elements in the x- and y-directions must be equal to 𝛽x = −kdx sin 𝜃0 cos 𝜙0 𝛽y = −kdy sin 𝜃0 sin 𝜙0 (6-93a) (6-93b) When solved simultaneously, (6-93a) and (6-93b) can also be expressed as tan 𝜙0 = 𝛽ydx 𝛽xdy (6-94a) sin2 𝜃0 = ( 𝛽x kdx )2 + ( 𝛽y kdy )2 (6-94b) 352 ARRAYS: LINEAR, PLANAR, AND CIRCULAR The principal maximum (m = n = 0) and the grating lobes can be located by kdx(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0) = ±2m𝜋, m = 0, 1, 2, … (6-95a) kdy(sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0) = ±2n𝜋, n = 0, 1, 2, … (6-95b) or sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0 = ±mλ dx , m = 0, 1, 2, … (6-96a) sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0 = ±nλ dy , n = 0, 1, 2, … (6-96b) which, when solved simultaneously, reduce to 𝜙= tan−1 [ sin 𝜃0 sin 𝜙0 ± nλ∕dy sin 𝜃0 cos 𝜙0 ± mλ∕dx ] (6-97a) and 𝜃= sin−1 [sin 𝜃0 cos 𝜙0 ± mλ∕dx cos 𝜙 ] = sin−1 [sin 𝜃0 sin 𝜙0 ± nλ∕dy sin 𝜙 ] (6-97b) In order for a true grating lobe to occur, both forms of (6-97b) must be satisfied simultaneously (i.e., lead to the same 𝜃value).
To demonstrate the principles of planar array theory, the three-dimensional pattern of a 5 × 5 element array of uniform amplitude, 𝛽x = 𝛽y = 0, and dx = dy = λ∕4, is shown in Figure 6.31. The maximum is oriented along 𝜃0 = 0◦and only the pattern above the x-y plane is shown. An identical pattern is formed in the lower hemisphere which can be diminished by the use of a ground plane.
To examine the pattern variation as a function of the element spacing, the three-dimensional pat-tern of the same 5 × 5 element array of isotropic sources with dx = dy = λ∕2 and 𝛽x = 𝛽y = 0 is dis-played in Figure 6.32. As contrasted with Figure 6.31, the pattern of Figure 6.32 exhibits complete minor lobes in all planes. Figure 6.33 displays the corresponding two-dimensional elevation patterns with cuts at 𝜙= 0◦(x-z plane), 𝜙= 90◦(y-z plane), and 𝜙= 45◦. The two principal patterns (𝜙= 0◦ and 𝜙= 90◦) are identical. The patterns of Figures 6.31 and 6.32 display a fourfold symmetry.
As discussed previously, arrays possess wide versatility in their radiation characteristics. The most common characteristic of an array is its scanning mechanism. To illustrate that, the three-dimensional pattern of the same 5 × 5 element array, with its maximum oriented along the 𝜃0 = 30◦, 𝜙0 = 45◦, is plotted in Figure 6.34. In Figure 6.34(a) is plotted in ‘cylindrical’ format while in Figure 6.34(b) is plotted in ‘spherical’ format. The element spacing is dx = dy = λ∕2. The maximum is found in the first quadrant of the upper hemisphere. The two-dimensional patterns are shown in Figure 6.35, and they exhibit only a twofold symmetry. The principal-plane pattern (𝜙= 0◦or 𝜙= 90◦) is normalized relative to the maximum which occurs at 𝜃0 = 30◦, 𝜙0 = 45◦. Its maximum along the principal planes (𝜙= 0◦or 𝜙= 90◦) occurs when 𝜃= 21◦and it is 17.37 dB down from the maximum at 𝜃0 = 30◦, 𝜙0 = 45◦.
To illustrate the formation of the grating lobes, when the spacing between the elements is large, the three-dimensional pattern of the 5 × 5 element array with dx = dy = λ and 𝛽x = 𝛽y = 0 are dis-played in Figure 6.36. Its corresponding two-dimensional elevation patterns at 𝜙= 0◦(𝜙= 90◦) and PLANAR ARRAY 353 Relative Amplitude x-z plane ( = 0°) y-z plane ( = 90°) Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ θ 90° 90° ϕ ϕ dx = dy = λ/4 x = y = 0 M = N = 5 β β Figure 6.31 Three-dimensional antenna pattern of a planar array of isotropic elements with a spacing of dx = dy = λ∕4, and equal amplitude and phase excitations.
𝜙= 45◦are exhibited in Figure 6.37. Besides the maxima along 𝜃= 0◦and 𝜃= 180◦, additional maxima with equal intensity, referred to as grating lobes, appear along the principal planes (x-z and y-z planes) when 𝜃= 90◦. Further increase of the spacing to dx = dy = 2λ would result in additional grating lobes.
Relative Amplitude Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ θ 90° 90° x-z plane ( = 0°) y-z plane ( = 90°) ϕ ϕ dx = dy = λ/2 x = y = 0 M = N = 5 β β Figure 6.32 Three-dimensional antenna pattern of a planar array of isotropic elements with a spacing of dx = dy = λ∕2, and equal amplitude and phase excitations.
354 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 = (x-z plane) = 90° (y-z plane) = 45° dx = dy = λ/2 x = y = 0 M = N = 5 Antenna Pattern (dB) –40 –35 –30 –25 –20 –15 –10 –5 0 θ θ ϕ ϕ ϕ 0° β β Figure 6.33 Two-dimensional antenna patterns of a planar array of isotropic elements with a spacing of dx = dy = λ∕2, and equal amplitude and phase excitations.
The array factor of the planar array has been derived assuming that each element is an isotropic source. If the antenna is an array of identical elements, the total field can be obtained by applying the pattern multiplication rule of (6-5) in a manner similar as for the linear array.
When only the central element of a large planar array is excited and the others are passively ter-minated, it has been observed experimentally that additional nulls in the pattern of the element are developed which are not accounted for by theory which does not include coupling. The nulls were observed to become deeper and narrower as the number of elements surrounding the excited element increased and approached a large array. These effects became more noticeable for arrays of open waveguides. It has been demonstrated that dips at angles interior to grating lobes are formed by coupling through surface wave propagation. The coupling decays very slowly with dis-tance, so that even distant elements from the driven elements experience substantial parasitic exci-tation. The angles where these large variations occur can be placed outside scan angles of interest by choosing smaller element spacing than would be used in the absence of such coupling. Because of the complexity of the problem, it will not be pursued here any further but the interested reader is referred to the published literature.
6.10.2 Beamwidth The task of finding the beamwidth of nonuniform amplitude planar arrays is quite formidable.
Instead, a very simple procedure will be outlined which can be used to compute these parameters for large arrays whose maximum is not scanned too far off broadside. The method utilizes results of a uniform linear array and the beam broadening factor of the amplitude distribution.
PLANAR ARRAY 355 Relative Amplitude Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x-z plane ( = 0°) y-z plane ( = 90°) θ θ 90° 90° ϕ ϕ x y z 0 = 45° 0 = 30° Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ ϕ (a) in ‘cylindrical’ format (b) in ‘spherical’ format dx = dy = λ/2 x = y = – / 2 2 0 = 30°, 0 = 45° M = N = 5 β β θ ϕ π dx = dy = λ/2 x = y = – / 2 2 0 = 30°, 0 = 45° M = N = 5 β β θ ϕ π Figure 6.34 Three-dimensional antenna patterns of a planar array of isotropic elements with a spacing of dx = dy = λ∕2, equal amplitude, and progressive phase excitation.
356 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 = 0° (x-z plane) = 90° (y-z plane) = 45° dx = dy = λ/2 x = y = – / 2 2 0 = 30°, 0 = 45° M = N = 5 Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ ϕ ϕ ϕ β β θ ϕ π Figure 6.35 Two-dimensional antenna patterns of a planar array of isotropic elements with a spacing of dx = dy = λ∕2, equal amplitude, and progressive phase excitation.
Relative Amplitude Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x-z plane ( = 0°) y-z plane ( = 90°) θ θ 90° 90° ϕ ϕ dx = dy = x = y = 0 M = N = 5 β β λ Figure 6.36 Three-dimensional antenna pattern of a planar array of isotropic elements with a spacing of dx = dy = λ, and equal amplitude and phase excitations.
PLANAR ARRAY 357 Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 = 0° (x-z plane) = 90° (y-z plane) = 45° dx = dy = x = y = 0 M = N = 5 Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ ϕ ϕ ϕ β β λ Figure 6.37 Two-dimensional antenna patterns of a planar array of isotropic elements with a spacing of dx = dy = λ, and equal amplitude and phase excitations.
The maximum of the conical main beam of the array is assumed to be directed toward 𝜃0, 𝜙0 as shown in Figure 6.38. To define a beamwidth, two planes are chosen. One is the elevation plane defined by the angle 𝜙= 𝜙0 and the other is a plane that is perpendicular to it. The corresponding half-power beamwidth of each is designated, respectively, by Θh and Ψh. For example, if the array maximum is pointing along 𝜃0 = 𝜋∕2 and 𝜙0 = 𝜋∕2, Θh represents the beamwidth in the y-z plane and Ψh, the beamwidth in the x-y plane.
For a large array, with its maximum near broadside, the elevation plane half-power beamwidth Θh is given approximately by Θh = √ 1 cos2 𝜃0[Θ−2 x0 cos2 𝜙0 + Θ−2 y0 sin2 𝜙0] (6-98) where Θx0 represents the half-power beamwidth of a broadside linear array of M elements. Similarly, Θy0 represents the half-power beamwidth of a broadside array of N elements.
The values of Θx0 and Θy0 can be obtained by using previous results. For a uniform distribu-tion, for example, the values of Θx0 and Θy0 can be obtained by using, respectively, the lengths (Lx + dx)∕λ and (Ly + dy)∕λ and reading the values from the broadside curve of Figure 6.12. For a Tschebyscheff distribution, the values of Θx0 and Θy0 are obtained by multiplying each uniform dis-tribution value by the beam broadening factor of (6-78) or Figure 6.25(a). The same concept can be 358 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Figure 6.38 Half-power beamwidths for a conical main beam oriented toward 𝜃= 𝜃0, 𝜙= 𝜙0. (source: R. S. Elliott, “Beamwidth and Directivity of Large Scanning Arrays,” Last of Two Parts, The Microwave Jour-nal, January 1964.).
used to obtain the beamwidth of other distributions as long as their corresponding beam broadening factors are available.
For a square array (M = N, Θx0 = Θy0), (6-95) reduces to Θh = Θx0 sec 𝜃0 = Θy0 sec 𝜃0 (6-98a) Equation (6-98a) indicates that for 𝜃0 > 0 the beamwidth increases proportionally to sec 𝜃0 = 1∕cos 𝜃0. The broadening of the beamwidth by sec 𝜃0, as 𝜃0 increases, is consistent with the reduction by cos 𝜃0 of the projected area of the array in the pointing direction.
The half-power beamwidth Ψh, in the plane that is perpendicular to the 𝜙= 𝜙0 elevation, is given by Ψh = √ 1 Θ−2 x0 sin2 𝜙0 + Θ−2 y0 cos2 𝜙0 (6-99) and it does not depend on 𝜃0. For a square array, (6-99) reduces to Ψh = Θx0 = Θy0 (6-99a) The values of Θx0 and Θy0 are the same as in (6-98) and (6-98a).
PLANAR ARRAY 359 For a planar array, it is useful to define a beam solid angle ΩA by ΩA = ΘhΨh (6-100) as it was done in (2-23), (2-24), and (2-26a). Using (6-98) and (6-99), (6-100) can be expressed as ΩA = Θx0Θy0 sec 𝜃0 [ sin2 𝜙0 + Θ2 y0 Θ2 x0 cos2 𝜙0 ]1∕2 [ sin2 𝜙0 + Θ2 x0 Θ2 y0 cos2 𝜙0 ]1∕2 (6-101) 6.10.3 Directivity The directivity of the array factor AF(𝜃, 𝜙) whose major beam is pointing in the 𝜃= 𝜃0 and 𝜙= 𝜙0 direction, can be obtained by employing the definition of (2-22) and writing it as D0 = 4𝜋[AF(𝜃0, 𝜙0)][AF(𝜃0, 𝜙0)]∗|max ∫ 2𝜋 0 ∫ 𝜋 0 [AF(𝜃, 𝜙)][AF(𝜃, 𝜙)]∗sin 𝜃d𝜃d𝜙 (6-102) A novel method has been introduced for integrating the terms of the directivity expression for isotropic and conical patterns.
As in the case of the beamwidth, the task of evaluating (6-102) for nonuniform amplitude distri-bution is formidable. Instead, a very simple procedure will be outlined to compute the directivity of a planar array using data from linear arrays.
It should be pointed out that the directivity of an array with bidirectional characteristics (two-sided pattern in free space) would be half the directivity of the same array with unidirectional (one-sided pattern) elements (e.g., dipoles over ground plane).
For large planar arrays, which are nearly broadside, the directivity reduces to D0 = 𝜋cos 𝜃0DxDy (6-103) where Dx and Dy are the directivities of broadside linear arrays each, respectively, of length and number of elements Lx, M and Ly, N. The factor cos 𝜃0 accounts for the decrease of the directivity because of the decrease of the projected area of the array. Each of the values, Dx and Dy, can be obtained by using (6-79) with the appropriate beam broadening factor f. For Tschebyscheff arrays, Dx and Dy can be obtained using (6-78) or Figure 6-25(a) and (6-79). Alternatively, they can be obtained using the graphical data of Figure 6.25(b).
For most practical amplitude distributions, the directivity of (6-103) is related to the beam solid angle of the same array by D0 ≃ 𝜋2 ΩA(rads2) = 32,400 ΩA(degrees2) (6-104) where ΩA is expressed in square radians or square degrees. Equation (6-104) should be compared with (2-26) or (2-27) given by Kraus.
360 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Example 6.12 Compute the half-power beamwidths, beam solid angle, and directivity of a planar square array of 100 isotropic elements (10 × 10). Assume a Tschebyscheff distribution, λ∕2 spacing between the elements, −26 dB side lobe level, and the maximum oriented along 𝜃0 = 30◦, 𝜙0 = 45◦.
Solution: Since in the x- and y-directions Lx + dx = Ly + dy = 5λ and each is equal to L + d of Example 6.10, then Θx0 = Θy0 = 10.97◦ According to (6-98a) Θh = Θx0 sec 𝜃0 = 10.97◦sec (30◦) = 12.67◦ and (6-99a) Ψh = Θx0 = 10.97◦ and (6-100) ΩA = ΘhΨh = 12.67(10.97) = 138.96 (degrees2) The directivity can be obtained using (6-103). Since the array is square, Dx = Dy, each one is equal to the directivity of Example 6.10. Thus D0 = 𝜋cos(30◦)(9.18)(9.18) = 229.28 (dimensionless) = 23.60 dB Using (6-104) D0 ≃ 32,400 ΩA(degrees2) = 32,400 138.96 = 233.16 (dimensionless) = 23.67 dB Obviously we have an excellent agreement.
An interactive MATLAB and FORTRAN computer program entitled Arrays has been developed, and it performs the analysis for uniform and nonuniform linear arrays, and uniform planar arrays.
The MATLAB version of the program also analyzes uniform circular arrays. The description of the program is provided in the corresponding READ ME file.
6.11 DESIGN CONSIDERATIONS Antenna arrays can be designed to control their radiation characteristics by properly selecting the phase and/or amplitude distribution between the elements. It has already been shown that a control of the phase can significantly alter the radiation pattern of an array. In fact, the principle of scan-ning arrays, where the maximum of the array pattern can be pointed in different directions, is based primarily on control of the phase excitation of the elements. In addition, it has been shown that a DESIGN CONSIDERATIONS 361 proper amplitude excitation taper between the elements can be used to control the beamwidth and side lobe level. Typically the level of the minor lobes can be controlled by tapering the distribution across the array; the smoother the taper from the center of the array toward the edges, the lower the side lobe level and the larger the half-power beamwidth, and conversely. Therefore a very smooth taper, such as that represented by a binomial distribution or others, would result in very low side lobe but larger half-power beamwidth. In contrast, an abrupt distribution, such as that of uniform illumi-nation, exhibits the smaller half-power beamwidth but the highest side lobe level (about −13.5 dB).
Therefore, if it is desired to achieve simultaneously both a very low side lobe level, as well as a small half-power beamwidth, a compromise design has to be selected. The Dolph-Tschebyscheff design of Section 6.8.3 is one such distribution. There are other designs that can be used effectively to achieve a good compromise between side lobe level and beamwidth. Two such examples are the Taylor Line-Source (Tschebyscheff-Error) and the Taylor Line-Source (One-Parameter). These are discussed in detail in Sections 7.6 and 7.7 of Chapter 7, respectively. Both of these are very similar to the Dolph-Tschebyscheff, with primarily the following exceptions.
For the Taylor Tschebyscheff-Error design, the number of minor lobes with the same level can be controlled as part of the design; the level of the remaining one is monotonically decreasing.
This is in contrast to the Dolph-Tschebyscheff where all the minor lobes are of the same level.
Therefore, given the same side lobe level, the half-power beamwidth of the Taylor Tschebyscheff-Error is slightly greater than that of the Dolph-Tschebyscheff. For the Taylor One-Parameter design, the level of the first minor lobe (closest to the major lobe) is controlled as part of the design; the level of the remaining ones are monotonically decreasing. Therefore, given the same side lobe level, the half-power beamwidth of the Taylor One-Parameter is slightly greater than that of the Taylor Tschebyscheff-Error, which in turn is slightly greater than that of the Dolph-Tschebyscheff design.
More details of these two methods, and other ones, can be found in Chapter 7. However there are some other characteristics that can be used to design arrays.
Uniform arrays are usually preferred in design of direct-radiating active-planar arrays with a large number of elements . One design consideration in satellite antennas is the beamwidth which can be used to determine the “footprint” area of the coverage. It is important to relate the beamwidth to the size of the antenna. In addition, it is also important to maximize the directivity of the antenna within the angular sector defined by the beamwidth, especially at the edge-of-the-coverage (EOC) . For engineering design purposes, closed-form expressions would be desirable.
To relate the half-power beamwidth, or any other beamwidth, to the length of the array in closed form, it is easier to represent the uniform array with a large number of elements as an aperture. The normalized array factor for a rectangular array is that of (6-91). For broadside radiation (𝜃0 = 0◦) and small spacings between the elements (dx ≪λ and dy ≪λ), (6-91) can be used to approximate the pattern of a uniform illuminated aperture. In one principal plane (i.e., x-z plane; 𝜙= 0◦) of Figure 6.30, (6-101) reduces for small element spacing and large number of elements to (AF)n(𝜃, 𝜙= 0) = 1 M sin (Mkdx 2 sin 𝜃 ) sin (kdx 2 sin 𝜃 ) ≃ sin (Mkdx 2 sin 𝜃 ) Mkdx 2 sin 𝜃 ≃ sin (kLx 2 sin 𝜃 ) kLx 2 sin 𝜃 (6-105) where Lx is the length of the array in the x direction. The array factor of (6-105) can be used to repre-sent the field in a principal plane of a uniform aperture (see Sections 12.5.1, 12.5.2 and Table 12.1).
Since the maximum effective area of a uniform array is equal to its physical area Aem = Ap [see (12-37)], the maximum directivity is equal to D0 = 4𝜋 λ2 Aem = 4𝜋 λ2 Ap = 4𝜋 λ2 LxLy (6-106) 362 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Therefore the normalized power pattern in the xz-plane, multiplied by the maximum directivity, can be written as the product of (6-105) and (6-106), and it can be expressed as P(𝜃, 𝜙= 0) = (4𝜋LxLy λ2 ) ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kLx 2 sin 𝜃 ) kLx 2 sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 (6-107) The maximum of (6-107) occurs when 𝜃= 0◦. However, for any other angle 𝜃= 𝜃c, the maximum of the pattern occurs when sin (kLx 2 sin 𝜃c ) = 1 (6-108) or Lx = 𝜋 k sin 𝜃c = λ 2 sin 𝜃c (6-108a) Therefore to maximize the directivity at the edge 𝜃= 𝜃c of a given angular sector 0◦≤𝜃≤𝜃c, the optimum aperture dimension must be chosen according to (6-108a). Doing otherwise leads to a decrease in directivity at the edge-of-the-coverage.
For a square aperture (Ly = Lx) the maximum value of the normalized power pattern of (6-107) occurs when 𝜃= 0◦, and it is equal to P(𝜃= 0◦)|max = 4𝜋 (Lx λ )2 (6-109) while that at the edge of the covering, using the optimum dimension, is P(𝜃= 𝜃c) = 4𝜋 (Lx λ )2 ( 2 𝜋 )2 (6-110) Therefore the value of the directivity at the edge of the desired coverage (𝜃= 𝜃c), relative to its maximum value at 𝜃= 0◦, is P(𝜃= 𝜃c) P(𝜃= 0◦) = ( 2 𝜋 )2 = 0.4053 (dimensionless) = −3.92 dB (6-111) Thus the variation of the directivity within the desired coverage (0◦≤𝜃≤𝜃c) is less than 4 dB.
If, for example, the length of the array for a maximum half-power beamwidth coverage is changed from the optimum or chosen to be optimized at another angle, then the directivity at the edge of the half-power beamwidth is reduced from the optimum.
Similar expressions have been derived for circular apertures with uniform, parabolic and parabolic with −10 dB cosine-on-a-pedestal distributions , and they can be found in Chapter 12, Sec-tion 12.7.
CIRCULAR ARRAY 363 Figure 6.39 Geometry of an N-element circular array.
6.12 CIRCULAR ARRAY The circular array, in which the elements are placed in a circular ring, is an array configuration of very practical interest. Over the years, applications span radio direction finding, air and space navigation, underground propagation, radar, sonar, and many other systems. More recently, circular arrays have been proposed for wireless communication, and in particular for smart antennas .
For more details see Section 16.12.
6.12.1 Array Factor Referring to Figure 6.39, let us assume that N isotropic elements are equally spaced on the x-y plane along a circular ring of the radius a. The normalized field of the array can be written as En(r, 𝜃, 𝜙) = N ∑ n=1 an e−jkRn Rn (6-112) where Rn is the distance from the nth element to the observation point. In general Rn = (r2 + a2 −2ar cos 𝜓)1∕2 (6-112a) which for r ≫a reduces to Rn ≃r −a cos 𝜓n = r −a(̂ a𝜌⋅̂ ar) = r −a sin 𝜃cos(𝜙−𝜙n) (6-112b) 364 ARRAYS: LINEAR, PLANAR, AND CIRCULAR where ̂ a𝜌⋅̂ ar = (̂ ax cos 𝜙n + ̂ ay sin 𝜙n) ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = sin 𝜃cos(𝜙−𝜙n) (6-112c) Thus (6-112) reduces, assuming that for amplitude variations Rn ≃r, to En(r, 𝜃, 𝜙) = e−jkr r N ∑ n=1 ane+jka sin 𝜃cos(𝜙−𝜙n) (6-113) where an = excitation coefficients (amplitude and phase) of nth element 𝜙n = 2𝜋 ( n N ) = angular position of nth element on x-y plane In general, the excitation coefficient of the nth element can be written as an = Inej𝛼n (6-114) where In = amplitude excitation of the nth element 𝛼n = phase excitation (relative to the array center) of the nth element With (6-114), (6-113) can be expressed as En(r, 𝜃, 𝜙) = e−jkr r [AF(𝜃, 𝜙)] (6-115) where AF(𝜃, 𝜙) = N ∑ n=1 Inej[ka sin 𝜃cos(𝜙−𝜙n)+𝛼n] (6-115a) Equation (6-115a) represents the array factor of a circular array of N equally spaced elements.
To direct the peak of the main beam in the (𝜃0, 𝜙0) direction, the phase excitation of the nth element can be chosen to be 𝛼n = −ka sin 𝜃0 cos(𝜙0 −𝜙n) (6-116) Thus the array factor of (6-115a) can be written as AF(𝜃, 𝜙) = N ∑ n=1 Inejka[sin 𝜃cos(𝜙−𝜙n)−sin 𝜃0 cos(𝜙0−𝜙n)] = N ∑ n=1 Inejka[cos 𝜓−cos 𝜓0) (6-117) CIRCULAR ARRAY 365 To reduce (6-117) to a simpler form, we define 𝜌0 as 𝜌0 = a[(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0)2 + (sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0)2]1∕2 (6-118) Thus the exponential in (6-117) takes the form of ka(cos 𝜓−cos 𝜓0) = k𝜌0[sin 𝜃cos(𝜙−𝜙n) −sin 𝜃0 cos(𝜙0 −𝜙n)] [(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0)2 + (sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0)2]1∕2 (6-119) which when expanded reduces to ka(cos 𝜓−cos 𝜓0) = k𝜌0 {cos 𝜙n(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0) + sin 𝜙n(sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0) [(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0)2 + (sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0)2]1∕2 } (6-119a) Defining cos 𝜉= sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0 [(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0)2 + (sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0)2]1∕2 (6-120) then sin 𝜉= [1 −cos2 𝜉]1∕2 = sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0 [(sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0)2 + (sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0)2]1∕2 (6-121) Thus (6-119a) and (6-117) can be rewritten, respectively, as ka(cos 𝜓−cos 𝜓0) = k𝜌0(cos 𝜙n cos 𝜉+ sin 𝜙n sin 𝜉) = k𝜌0 cos(𝜙n −𝜉) (6-122) AF(𝜃, 𝜙) = N ∑ n=1 Inejka(cos 𝜓−cos 𝜓0) = N ∑ n=1 Inejk𝜌0 cos(𝜙n−𝜉) where 𝜉= tan−1 [ sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0 sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0 ] (6-123) (6-123a) and 𝜌0 is defined by (6-118).
Equations (6-123), (6-118), and (6-123a) can be used to calculate the array factor once N, In, a, 𝜃0, and 𝜙0 are specified. This is usually very time consuming, even for moderately large values of N.
The three-dimensional pattern of the array factor for a 10-element uniform circular array of ka = 10 is shown in Figure 6.40. The corresponding two-dimensional principal-plane patterns are displayed in Figure 6.41. As the radius of the array becomes very large, the directivity of a uniform circular array approaches the value of N, where N is equal to the number of elements. An excellent discussion on circular arrays can be found in .
366 ARRAYS: LINEAR, PLANAR, AND CIRCULAR Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Relative Amplitude 90° 90° θ θ y-z plane ( = 90°) ϕ x-z plane ( = 0°) ϕ Figure 6.40 Three-dimensional amplitude pattern of the array factor for a uniform circular array of N = 10 elements (C∕λ = ka = 10).
Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 x-z plane ( = 0°) y-z plane ( = 90°) Antenna Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ ϕ ϕ Figure 6.41 Principal-plane amplitude patterns of the array factor for a uniform circular array of N = 10 elements (C∕λ = ka = 10).
REFERENCES 367 For a uniform amplitude excitation of each element (In = I0), (6-123) can be written as AF(𝜃, 𝜙) = NI0 +∞ ∑ m=−∞ JmN(k𝜌0)ejmN(𝜋∕2−𝜉) (6-124) where Jp(x) is the Bessel function of the first kind (see Appendix V). The part of the array factor associated with the zero order Bessel function J0(k𝜌0) is called the principal term and the remaining terms are noted as the residuals. For a circular array with a large number of elements, the term J0(k𝜌0) alone can be used to approximate the two-dimensional principal-plane patterns. The remaining terms in (6-124) contribute negligibly because Bessel functions of larger orders are very small.
The MATLAB computer program Arrays, which is used to compute radiation characteristics of planar and circular arrays, does compute the radiation patterns of a circular array based on (6-123) and (6-124). The one based on (6-124) computes two patterns; one based on the principal term and the other based on the principal term plus two residual terms.
6.13 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Java-based applet for computing and displaying the radiation characteristics, directivity, and pattern of uniform and nonuniform linear arrays.
c. Java-based pattern animation of uniform and nonuniform linear arrays.
d. Matlab and Fortran computer program, designated as Arrays, for computing the radiation characteristics linear, planar, and circular arrays.
e. Power Point (PPT) viewgraphs, in multicolor.
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5. W. W. Hansen and J. R. Woodyard, “A New Principle in Directional Antenna Design,” Proc. IRE, Vol. 26, No. 3, March 1938, pp. 333–345.
6. J. S. Stone, United States Patents No. 1,643,323 and No. 1,715,433.
7. C. L. Dolph, “A Current Distribution for Broadside Arrays Which Optimizes the Relationship Between Beamwidth and Side-Lobe Level,” Proc. IRE and Waves and Electrons, June 1946.
8. N. Yaru, “A Note on Super-Gain Arrays,” Proc. IRE, Vol. 39, September 1951, pp. 1081–1085.
9. L. J. Ricardi, “Radiation Properties of the Binomial Array,” Microwave J., Vol. 15, No. 12, December 1972, pp. 20–21.
10. H. J. Riblet, Discussion on “A Current Distribution for Broadside Arrays Which Optimizes the Relationship Between Beamwidth and Side-Lobe Level,” Proc. IRE, May 1947, pp. 489–492.
11. D. Barbiere, “A Method for Calculating the Current Distribution of Tschebyscheff Arrays,” Proc. IRE, January 1952, pp. 78–82.
368 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 12. R. J. Stegen, “Excitation Coefficients and Beamwidths of Tschebyscheff Arrays,” Proc. IRE, November 1953, pp. 1671–1674.
13. C. J. Drane, Jr., “Useful Approximations for the Directivity and Beamwidth of Large Scanning Dolph-Chebyshev Arrays,” Proc. IEEE, November 1968, pp. 1779–1787.
14. A. Safaai-Jazi, “A New Formulation for the Design of Chebyshev Arrays,” IEEE Trans. Antennas Propa-gat., Vol. 42, No. 3, March 1994, pp. 439–443.
15. M. M. Dawoud and A. P. Anderson, “Design of Superdirective Arrays with High Radiation Efficiency.” IEEE Trans. Antennas Propagat., Vol. AP-26. No. 6, January 1978, pp. 819–823.
16. S. A. Schelkunoff, “A Mathematical Theory of Linear Arrays,” Bell System Tech. Journal, Vol. 22, January 1943, pp. 80–87.
17. C. J. Bowkamp and N. G. de Bruijn, “The Problem of Optimum Antenna Current Distribution,” Phillips Res. Rept., Vol. 1, January 1946, pp. 135–158.
18. E. H. Newman, J. H. Richmond, and C. H. Walter, “Superdirective Receiving Arrays,” IEEE Trans. Anten-nas Propagat., Vol. AP-26, No. 5, September 1978, pp. 629–635.
19. J. L. Allen, “On Surface-Wave Coupling Between Elements of Large Arrays,” IEEE Trans. Antennas Prop-agat., Vol. AP-13, No. 4, July 1965, pp. 638–639.
20. R. H. T. Bates, “Mode Theory Approach to Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-13, No. 2, March 1965, pp. 321–322.
21. R. S. Elliott, “Beamwidth and Directivity of Large Scanning Arrays,” Last of Two Parts, Microwave Jour-nal, January 1964, pp. 74–82.
22. B. J. Forman, “A Novel Directivity Expression for Planar Antenna Arrays,” Radio Science, Vol. 5, No. 7, July 1970, pp. 1077–1083.
23. K. Praba, “Optimal Aperture for Maximum Edge-of-Coverage (EOC) Directivity,” IEEE Antennas Prop-agat. Magazine, Vol. 36, No. 3, June 1994, pp. 72–74.
24. P. Ioannides and C. A. Balanis, “Uniform Circular Arrays for Smart Antennas,” IEEE Trans. Antennas and Propagat. Society International Symposium, Monterey, CA, June 20–25, 2004, Vol. 3, pp. 2796–2799.
25. M. T. Ma, Theory and Application of Antenna Arrays, Wiley, 1974, Chapter 3, pp. 191–202.
PROBLEMS 6.1.
Three isotropic sources, with spacing d between them, are placed along the z-axis. The exci-tation coefficient of each outside element is unity while that of the center element is 2. For a spacing of d = λ∕4 between the elements, find the (a) array factor (b) angles (in degrees) where the nulls of the pattern occur (0◦≤𝜃≤180◦) (c) angles (in degrees) where the maxima of the pattern occur (0◦≤𝜃≤180◦) (d) directivity using the computer program Directivity of Chapter 2.
6.2.
Two very short dipoles (“infinitesimal”) of equal length are equidistant from the origin with their centers lying on the y-axis, and oriented parallel to the z-axis. They are excited with currents of equal amplitude. The current in dipole 1 (at y = −d∕2) leads the current in dipole 2 (at y = +d∕2) by 90◦in phase. The spacing between dipoles is one quarter wavelength. To simplify the notation, let E0 equal the maximum magnitude of the far field at distance r due to either source alone.
(a) Derive expressions for the following six principal-plane patterns: 1. |E𝜃(𝜃)| for 𝜙= 0◦ 2. |E𝜃(𝜃)| for 𝜙= 90◦ 3. |E𝜃(𝜙)| for 𝜃= 90◦ 4. |E𝜙(𝜃)| for 𝜙= 0◦ 5. |E𝜙(𝜃)| for 𝜙= 90◦ 6. |E𝜙(𝜙)| for 𝜃= 90◦ (b) Sketch the six field patterns.
PROBLEMS 369 6.3.
A three-element array of isotropic sources has the phase and magnitude relationships shown.
The spacing between the elements is d = λ∕2.
(a) Find the array factor.
(b) Find all the nulls.
2 #1 #3 +1 –j z y –1 d d 6.4.
Repeat Problem 6.3 when the excitation coefficients for elements #1, #2 and #3 are, respec-tively, +1, +j and −j.
6.5.
Four isotropic sources are placed along the z-axis as shown. Assuming that the amplitudes of elements #1 and #2 are +1 and the amplitudes of elements #3 and #4 are −1 (or 180 degrees out of phase with #1 and #2), find (a) the array factor in simplified form (b) all the nulls when d = λ∕2 #1 #4 z y #2 #3 d d/2 d/2 d 6.6.
A uniform linear broadside array of 4 elements are placed along the z-axis each a distance d apart.
(a) Write the normalized array factor in simplified form.
(b) For a separation of d = 3λ∕8 between the elements, determine the: 1. Approximate half-power beamwidth (in degrees).
2. Approximate directivity (dimensionless and in dB).
6.7.
Three isotropic elements of equal excitation phase are placed along the y-axis, as shown in the figure. If the relative amplitude of #1 is +2 and of #2 and #3 is +1, find a simplified expression for the three-dimensional unnormalized array factor.
3 #2 #1 d d z x y 370 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 6.8.
Design a uniform broadside linear array of N elements placed along the z-axis with a uniform spacing d = λ∕10 between the elements. Determine the closest integer number of elements so that in the elevation plane the (a) Half-power beamwidth of the array factor is approximately 60◦.
(b) First-null beamwidth of the array factor is 60◦.
6.9.
A uniform array of 3 elements is designed so that its maximum is directed toward broadside.
The spacing between the elements is λ∕2. For the array factor of the antenna, determine (a) all the angles (in degrees) where the nulls will occur.
(b) all the angles (in degrees) where all the maxima will occur.
(c) the half-power beamwidth (in degrees).
(d) directivity (dimensionless and in dB).
(e) the relative value (in dB) of the magnitude of the array factor toward end-fire (𝜃0 = 0◦) compared to that toward broadside (𝜃0 = 90◦).
6.10.
Design a two-element uniform array of isotropic sources, positioned along the z-axis a dis-tance λ∕4 apart, so that its only maximum occurs along 𝜃0 = 0◦. Assuming ordinary end-fire conditions, find the (a) relative phase excitation of each element (b) array factor of the array (c) directivity using the computer program Directivity of Chapter 2. Compare it with Kraus’ approximate formula.
6.11.
Repeat the design of Problem 6.10 so that its only maximum occurs along 𝜃= 180◦.
6.12.
Design a four-element ordinary end-fire array with the elements placed along the z-axis a distance d apart. For a spacing of d = λ∕2 between the elements find the (a) progressive phase excitation between the elements to accomplish this (b) angles (in degrees) where the nulls of the array factor occur (c) angles (in degrees) where the maximum of the array factor occur (d) beamwidth (in degrees) between the first nulls of the array factor (e) directivity (in dB) of the array factor. Verify using the computer program Directivity of Chapter 2.
6.13.
Design an ordinary end-fire uniform linear array with only one maximum so that its directivity is 20 dB (above isotropic). The spacing between the elements is λ∕4, and its length is much greater than the spacing. Determine the (a) number of elements (b) overall length of the array (in wavelengths) (c) approximate half-power beamwidth (in degrees) (d) amplitude level (compared to the maximum of the major lobe) of the first minor lobe (in dB) (e) progressive phase shift between the elements (in degrees).
6.14.
Design a uniform ordinary end-fire linear array of 8 elements placed along the z-axis so that the maximum amplitude of the array factor is oriented in different directions. Determine the range (in λ) of the spacing d between the elements when the main maximum of the array factor is directed toward (a) 𝜃0 = 0◦only; (b) 𝜃0 = 180◦only; (c) 𝜃0 = 0◦and 180◦only; (d) 𝜃0 = 0◦, 90◦, and 180◦only.
PROBLEMS 371 6.15.
It is desired to design a uniform ordinary end-fire array of 6 elements with a maximum toward 𝜃0 = 0◦and 𝜃0 = 180◦, simultaneously. Determine the (a) smallest separation between the elements (in λ).
(b) excitation progressive phase shift (in degrees) that should be used (c) approximate directivity of the array (dimensionless and in dB) (d) relative value (in dB) of the magnitude of the array factor toward broadside (𝜃0 = 90◦) compared to that toward the maximum (𝜃0 = 0◦or 180◦).
6.16.
Redesign the end-fire uniform array of Problem 6.13 in order to increase its directivity while maintaining the same, as in Problem 6.13, uniformity, number of elements, spacing between them, and end-fire radiation.
(a) What different from the design of Problem 6.13 are you going to do to achieve this? Be very specific, and give values.
(b) By how many decibels (maximum) can you increase the directivity, compared to the design of Problem 6.13?
(c) Are you expecting the half-power beamwidth to increase or decrease? Why increase or decrease and by how much?
(d) What antenna figure of merit will be degraded by this design? Be very specific in naming it, and why is it degraded?
6.17.
Ten isotropic elements are placed along the z-axis. Design a Hansen-Woodyard end-fire array with the maximum directed toward 𝜃0 = 180◦. Find the: (a) desired spacing (b) progressive phase shift 𝛽(in radians) (c) location of all the nulls (in degrees) (d) first-null beamwidth (in degrees) (e) directivity; verify using the computer program Directivity of Chapter 2.
6.18.
Design a uniform ordinary end-fire array of 6 elements placed along the z-axis and with the maximum of the array factor directed only along 𝜃0 = 0◦(end-fire only in one direction).
Determine the (a) maximum spacing (in λ) that can be used between the elements.
(b) maximum directivity (in dB) of the array factor using the maximum allowable spacing.
If the array was designed to be a Hansen-Woodyard end-fire array of the same number of elements, what would the following parameters be for the new array?
(c) directivity (in dB).
(d) Spacing (in λ) between the elements.
6.19.
Design a uniform linear array of elements placed long the z-axis with a uniform spacing of 0.2λ between the elements. It is desired that the array factor has end-fire radiation along the 𝜃= 0◦direction only and it achieves its maximum possible directivity as we presently know.
Determine the: (a) Linear array design that will achieve the desired specifications. State the name of the design.
(b) Number of elements required for the design.
(c) Half-power beamwidth (in degrees) of the array factor.
(d) Directivity (dimensionless and in dB) of the array factor.
(e) Approximate side lobe level (in dB) of the array factor for this design.
6.20.
It is desired to design a linear uniform end-fire array that will maximize its directivity along the 𝜃= 0◦direction only. The array elements are all placed along the z-axis with a uniform spacing d between them. The desired maximum directivity is 9.5545 dB. Determine the: 372 ARRAYS: LINEAR, PLANAR, AND CIRCULAR (a) Array design; state its name.
(b) Number of elements.
(c) Exact spacing between the elements (in λ).
(d) Exact progressive phase difference between the elements (in degrees).
(e) Relative exact phase excitation of each of the elements (in degrees).
Take element #1 as a reference (0 degrees).
6.21.
An array of 10 isotropic elements are placed along the z-axis a distance d apart. Assum-ing uniform distribution, find the progressive phase (in degrees), half-power beamwidth (in degrees), first-null beamwidth (in degrees), first side lobe level maximum beamwidth (in degrees), relative side lobe level maximum (in dB), and directivity (in dB) (using equations and the computer program Directivity of Chapter 2, and compare) for (a) broadside (b) ordinary end-fire (c) Hansen-Woodyard end-fire arrays when the spacing between the elements is d = λ∕4.
6.22.
Find the beamwidth and directivity of a 10-element uniform scanning array of isotropic sources placed along the z-axis. The spacing between the elements is λ∕4 and the maximum is directed at 45◦from its axis.
6.23.
It is desired to design a linear adaptive array of 6 isotropic elements placed along the z-axis a distance λ∕2 apart. The linear adaptive array is to be able to receive a signal-of-interest (SOI), (i.e., maximum) from a direction of 𝜃= 30◦form the axis (z-axis) of the array. No other specifications are placed upon the design. Choose the simplest type of a linear array design that will accomplish this.
(a) State the array design; i.e., give its name.
(b) What should be the normalized amplitude excitation of each element?
(c) What should be the progressive phase difference (in degrees) between the elements.
6.24.
It is desired to design a linear adaptive array of 6 isotropic elements placed along the z-axis a distance λ∕2 apart. The linear adaptive array is to be able to simultaneously (all the SOI and SNOIs at the same time): r Receive a signal-of-interest (SOI), (i.e., maximum) from a direction of 𝜃m = 90◦(perpen-dicular to the z axiz) r SNOIs (interferers; nulls) from 𝜃n = ⎧ ⎪ ⎨ ⎪ ⎩ 70.529◦and 109.471◦ 48.189◦and 131.811◦ 0◦and 180◦ ⎫ ⎪ ⎬ ⎪ ⎭ (0◦≤𝜃≤180◦) Choose the simplest type of a linear array design that will accomplish placing the SOI and all the SNOIs all at the same time.
(a) State the array design; i.e., give its name.
(b) What should be the normalized amplitude excitation of each element?
(c) Verify (compute) all the angles of 𝜃(in degrees) of the SOI and the SNOIs.
(d) What is the progressive phase difference (in degrees) between the elements.
6.25.
Show that in order for a uniform array of N elements not to have any minor lobes, the spacing and the progressive phase shift between the elements must be (a) d = λ∕N, 𝛽= 0 for a broadside array.
(b) d = λ∕(2N), 𝛽= ±kd for an ordinary end-fire array.
PROBLEMS 373 6.26.
A uniform array of 20 isotropic elements is placed along the z-axis a distance λ∕4 apart with a progressive phase shift of 𝛽rad. Calculate 𝛽(give the answer in radians) for the following array designs: (a) broadside (b) end-fire with maximum at 𝜃0 = 0◦ (c) end-fire with maximum at 𝜃0 = 180◦ (d) phased array with maximum aimed at 𝜃0 = 30◦ (e) Hansen-Woodyard with maximum at 𝜃0 = 0◦ (f) Hansen-Woodyard with maximum at 𝜃0 = 180◦ 6.27.
Design a 19-element uniform linear scanning array with a spacing of λ∕4 between the ele-ments.
(a) What is the progressive phase excitation between the elements so that the maximum of the array factor is 30◦from the line where the elements are placed?
(b) What is the half-power beamwidth (in degrees) of the array factor of part a?
(c) What is the value (in dB) of the maximum of the first minor lobe?
Verify using the computer program Arrays of this chapter.
6.28.
For a uniform broadside linear array of 10 isotropic elements, determine the approximate directivity (in dB) when the spacing between the elements is (a) λ∕4 (b) λ∕2 (c) 3λ∕4 (d) λ Compare the values with those obtained using the computer program Arrays.
6.29.
The maximum distance d between the elements in a linear scanning array to suppress grating lobes is dmax = λ 1 + | cos(𝜃0)| where 𝜃0 is the direction of the pattern maximum. What is the maximum distance between the elements, without introducing grating lobes, when the array is designed to scan to maximum angles of (a) 𝜃0 = 30◦ (b) 𝜃0 = 45◦ (c) 𝜃0 = 60◦ 6.30.
An array of 4 isotropic sources is formed by placing one at the origin, and one along the x-, y-, and z-axes a distance d from the origin. Find the array factor for all space. The excitation coefficient of each element is identical.
6.31.
The normalized array factor of a linear array of discrete elements placed along the z-axis can be approximated by (AF)n ≈cos 𝜃 } 0 ≤𝜃≤90◦ 0 ≤𝜙≤360◦ Assume now that the same physical linear array, with the same number of elements, is placed along the y-axis, and it is radiating in the 0 ≤𝜃≤180◦, 0 ≤𝜙≤180◦angular space. For the array with its elements along the y-axis (a) Write the new approximate array factor.
(b) Find the half-power beamwidth (in degrees) in the two principal planes.
374 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 1. xy–plane 2. yz–plane (c) Exact directivity (dimensionless and in dB) based on the approximate expression for the array factor.
6.32.
Repeat Problem 6.31 for an array factor of (AF)n ≈cos2 𝜃 } 0 ≤𝜃≤90◦ 0 ≤𝜙≤360◦ 6.33.
Design a linear array of isotropic elements placed along the z-axis such that the nulls of the array factor occur at 𝜃= 0◦and 𝜃= 45◦. Assume that the elements are spaced a distance of λ∕4 apart and that 𝛽= 0◦.
(a) Sketch and label the visible region on the unit circle (b) Find the required number of elements (c) Determine their excitation coefficients 6.34.
Design a linear array of isotropic elements placed along the z-axis such that the zeros of the array factor occur at 𝜃= 10◦, 70◦, and 110◦. Assume that the elements are spaced a distance of λ∕4 apart and that 𝛽= 45◦.
(a) Sketch and label the visible region on the unit circle (b) Find the required number of elements (c) Determine their excitation coefficients 6.35.
Repeat Problem 6.34 so that the nulls occur at 𝜃= 0◦, 50◦and 100◦. Assume a spacing of λ∕5 and 𝛽= 0◦between the elements.
6.36.
Design a three-element binomial array of isotropic elements positioned along the z-axis a distance d apart. Find the (a) normalized excitation coefficients (b) array factor (c) nulls of the array factor for d = λ (d) maxima of the array factor for d = λ 6.37.
Show that a three-element binomial array with a spacing of d ≤λ∕2 between the elements does not have a side lobe.
6.38.
Four isotropic sources are placed symmetrically along the z-axis a distance d apart. Design a binomial array. Find the (a) normalized excitation coefficients (b) array factor (c) angles (in degrees) where the array factor nulls occur when d = 3λ∕4 6.39.
Five isotropic sources are placed symmetrically along the z-axis, each separated from its neighbor by an electrical distance kd = 5𝜋∕4. For a binomial array, find the (a) excitation coefficients (b) array factor (c) normalized power pattern (d) angles (in degrees) where the nulls (if any) occur Verify parts of the problem using the computer program Arrays.
6.40.
Design a four-element binomial array of λ∕2 dipoles, placed symmetrically along the x-axis a distance d apart. The length of each dipole is parallel to the z-axis.
(a) Find the normalized excitation coefficients.
(b) Write the array factor for all space.
(c) Write expressions for the E-fields for all space.
PROBLEMS 375 6.41.
Repeat the design of Problem 6.40 when the λ∕2 dipoles are placed along the y-axis.
6.42.
Design a broadside binomial array of six elements placed along the z-axis separated by a distance d = λ∕2.
(a) Find the amplitude excitation coefficients (an’s).
(b) What is the progressive phase excitation between the elements?
(c) Write the array factor.
(d) Now assume that the elements are λ∕4 dipoles oriented in the z-direction. Write the expression for the electric field vector in the far field.
Verify parts of the problem using the computer program Arrays.
6.43.
Repeat Problem 6.42 for an array of seven elements.
6.44.
Five isotropic elements, with spacing d between them, are placed along the z-axis. For a binomial amplitude distribution, (a) write the array factor in its most simplified form (b) compute the directivity (in dB) and compare using the computer program Arrays of this chapter (d = λ∕2) (c) find the nulls of the array when d = λ(0◦≤𝜃≤180◦) 6.45.
A typical base station that you see as you travel around the city consists of an equilat-eral/triangular array of dipoles. Assume that each side of the equilateral triangle consists of three dipoles. Let us assume that each of the dipoles, at a frequency of 1.9 GHz, is λ∕2 in length. The dipoles are placed along the y-axis, are separated by a distance of λ∕2, and are pointing along the z-axis. The center element is placed at the origin and the other two are placed one on each side of the center element. Assuming that the elements are fed in phase and are designed for a broadside binomial amplitude distribution: (a) Determine the total normalized amplitude excitation coefficient of each element.
(b) Write an expression for the normalized array factor.
(c) Determine the maximum directivity (dimensionless and in dB) of the array factor when d = λ∕2.
(d) State the directivity (dimensionless and in dB) of each individual element.
(e) Making an educated guess, what would you expect the very maximum directivity (dimen-sionless and in dB) of the entire 3-element array, which takes into account the element pattern and the array factor, could not exceed?
6.46.
A nonuniform linear array has 3 elements placed symmetrically along the z-axis and spaced d = λ∕4 apart, and all are fed with the same phase. However, the total amplitude excitation coefficients of the elements are as follows: r 2 for the center element r Unity for each of the edge elements 376 ARRAYS: LINEAR, PLANAR, AND CIRCULAR For the array factor of the array, determine the: (a) Angle 𝜃(in degrees) where the maximum of the main lobe occurs.
(b) Angles 𝜃(in degrees) where the 2 half-power points of the main lobe occur.
(c) Half-power beamwidth (in degrees) of the main lobe.
(d) Approximate maximum directivity (dimensionless and in dB).
6.47.
Design a five-element binomial array with elements placed along the z-axis.
(a) Derive the excitation coefficients.
(b) Write a simplified expression for the array factor.
(c) For a spacing of d = 3λ∕4, determine all the angles 𝜃(in degrees) where the array factor possesses nulls.
(d) For a spacing of d = 3λ∕4, determine all the angles 𝜃(in degrees) where the array factor possesses main maxima.
6.48.
Design a five-element binomial array with the elements placed along the z-axis. It is desired that the amplitude pattern of the array factor has nulls only at 𝜃= 0◦and 180◦, one major lobe with the maximum at 𝜃0 = 90◦, and no minor lobes. To meet the requirements of this array, determine the: (a) Spacing between the elements (in λ).
(b) Total amplitude excitation coefficient of each element.
(c) Directivity (dimensionless and in dB).
(d) Half-power beamwidth (in degrees).
(e) Verify the design using the computer program Arrays.
6.49.
It is desired to design a binomial array with a uniform spacing between the elements of λ∕2 placed along the z-axis, and with an elevation half-power beamwidth for its array factor of 15.18 degrees. To accomplish this, determine the: (a) Number of elements.
(b) Directivity (dimensionless and in dB).
(c) Sidelobe level of the array factor (in dB).
6.50.
It is to design a linear nommiform broadside array whose directivity is 6 dB and whose pattern has nulls only at 𝜃= 0◦and 180◦. Assume the elements are placed along the z-axis.
(a) From the nonuniform designs you have been exposed to, select one that will accomplish this. State its name or distribution.
(b) Determine the closest integer number of elements.
(c) State the total amplitude excitation of each element.
(d) What is the spacing between the elements (in λ).
(e) Determine the half-power beamwidth (in degrees).
6.51.
It is desired to design two separate broadside square planar arrays (with 9x9 elements; a total of 81 elements) with the elements placed along the xy-plane; the spacing between the elements in both directions is λ∕2. One of the designs is a uniform design and the other one is a binomial design. Assume that both arrays only radiate in the upper hemisphere (above the xy-plane).
(a) For the Uniform planar array design: Determine, for the entire planar array, the: r Half-power beamwidth in the elevation plane (in degrees).
r Half-power beamwidth in a plane perpendicular to the elevation plane (in degrees).
r Maximum directivity (dimensionless and in dB).
PROBLEMS 377 (b) For the Binomial planar array design: Determine, for the entire planar array, the: r Half-power beamwidth in the elevation plane (in degrees).
r Half-power beamwidth in a plane perpendicular to the elevation plane (in degrees).
r Maximum directivity (dimensionless and in dB).
(c) Is the half-power beamwidth of the binomial array design, compared to that of the uniform design: r Smaller or larger? Why? Justify the answer. Is this what you expected?
(d) Is the maximum directivity of the binomial array design, compared to that of the uniform design: r Smaller or larger? Why? Justify the answer. Is this what you expected?
6.52.
Design a nonuniform binomial broadside linear array of N elements, with a uniform spacing d between the elements, which is desired to have no minor lobes.
(a) Determine the largest spacing d (in λ).
(b) Find the closest integer number of elements so that the half-power beamwidth is 18◦.
(c) Compare the directivities (in dB) of the uniform and binomial broadside arrays with the same number of elements and spacing between them. Which is smaller and by how many dB?
6.53.
It is desired to design a broadside uniform linear array with the elements placed along the x-axis. The design requires a total length (edge-to-edge) of the array is 4λ and the spacing between the elements must be λ∕2. Determine the: (a) Number of elements of the array.
(b) Half-power beamwidth (in degrees) for the uniform broadside array.
(c) Maximum directivity (dimensionless and in dB) for the uniform broadside array.
(d) Side lobe level (in dB) of the uniform broadside array.
(e) Total excitation coefficients if the linear array is a binomial broadside design (normalize the coefficients so that those of the edge elements are unity).
(f) Half-power beamwidth (in degrees) if the linear array is a binomial broadside design.
(g) Maximum directivity (dimensionless and in dB) if the array is a binomial broadside design.
(h) Sidelobe level (in dB) if the linear array is a binomial broadside design.
6.54.
Design a nonuniform broadside binomial array of three elements that its pattern has one major lobe and no minor lobes in 0◦≤𝜃≤180◦. It is also required that the pattern exhibits nulls toward 𝜃= 0◦and 𝜃= 180◦. Determine the: (a) Normalized total amplitude coefficient of each of the three elements (the ones at the edges to be of unity amplitude).
(b) Half-power beamwidth (in degrees).
(c) Directivity (dimensionless and in dB).
6.55.
The normalized array factor of a nonuniform linear broadside array of N elements, with a uniform spacing of d between them, is given by AF(𝜃) = 2 cos2 (kd 2 cos 𝜃 ) = 2 cos2(u), u = 𝜋d λ cos 𝜃 (a) Determine the number of elements of the array with this specific array factor.
(b) What is the normalized (relative to the ones at the edge) total excitation coefficient of each of the elements?
378 ARRAYS: LINEAR, PLANAR, AND CIRCULAR (c) What is the specific name of this nonuniform classic broadside array design?
(d) For d = 3λ∕4, determine all the angles (0◦≤0 ≤180◦) where the array factor exhibits nulls.
6.56.
The array factor of a nonuniform linear array, with the elements placed along the z-axis and with a uniform spacing d among the elements, is given by AF = 1 + cos ( 2𝜋d λ cos 𝜃 ) Determine the: (a) Total number of elements of the array.
(b) Total excitation coefficient of each element. Identify the total value of the coefficient with the appropriate element.
(c) All the nulls (in degrees; 0◦≤𝜃≤180◦) of the array factor when the spacing between the elements is d = 3λ∕4.
(d) All the major maxima (in degrees; 0◦≤𝜃≤180◦) of the array factor when the spacing between the elements is d = 3λ∕4.
6.57.
A 4-element array of isotropic elements are placed along the z-axis, symmetrically about the origin. The separation between all of the adjoining elements is d. The normalized amplitude excitation of the two inner most elements is 3 while that of the two outer most elements (at each of the two outer edges of the array) is unity. All of the elements are excited by the same phase (𝛽= 0).
(a) Write the array factor in simplified form. You may want to refer to (6-66).
(b) Assuming a spacing of d = λ∕2 between the elements, determine analytically: r All the nulls (0◦≤𝜃≤180◦) in degrees.
r All the main maxima (0◦≤𝜃≤180◦) in degrees.
r Half-power beamwidth (in degrees).
r Directivity (dimensionless and in dB).
6.58.
Design a four-element nonuniform linear broadside array with a uniform spacing between the elements. It is desired that the array factor pattern has no minor lobes and the maximum permissible spacing between the elements is selected. The selected spacing should be greater than λ∕4.
(a) Select the appropriate nonuniform array design. State its name.
(b) Determine the maximum permissible spacing (in λ) between the elements.
(c) Determine the normalized amplitude excitation coefficients.
(d) Write the array factor assuming the elements are placed along the x-axis.
(e) Write the array factor assuming the elements are placed along the y-axis.
(f) Half-power beamwidth (in degrees) of the array factor.
(g) Directivity (dimensionless and in dB) of the array factor.
6.59.
The normalized Array Factor of a broadside array is given by (AF)n = 3 cos (𝜋d λ sin 𝜃cos 𝜙 ) + cos ( 3𝜋d λ sin 𝜃cos 𝜙 ) which can also be written, using trigonometric identities, as (AF)n = 4 [ cos (𝜋d λ sin 𝜃cos 𝜙 )]3 = 4 cos3 (𝜋d λ sin 𝜃cos 𝜙 ) PROBLEMS 379 where d is the spacing between the elements, and 𝜃and 𝜙are the standard spherical coordinate angles.
(a) What type of an array we have (linear, square, rectangular, circular, other)?
(b) Along what axis (x, y or z) are the elements positioned?
(c) What is the amplitude distribution of the elements along the respective axis (uniform, binomial, Tschebysheff, cosine, cosine squared, other)?
(d) How many array elements are there along the respective axis?
(e) Assuming d = λ∕2, determine the: r Approximate half-power beam width (in degrees) of the array, assuming isotropic elements.
r Exact directivity (dimensionless and in dB) of the array, assuming isotropic elements (dimensionless and in dB).
r Approximate directivity (dimensionless and in dB) of the array, using another/alternate formula/expression which uses the half-power beamwidth information. Stale which one you are using and why. Be specific.
r Silelobe level (in dB).
6.60.
Show that the: (a) Maximum spacing dmax of (6-76a) reduces to the optimum spacing dopt of (6-76b).
(b) Optimum spacing dopt of (6-76b) reduces to the maximum spacing dmax of (6-76a).
6.61.
Repeat the design of Problem 6.36 for a Dolph-Tschebyscheff array with a side lobe level of −20 dB.
6.62.
Design a three-element, −40 dB side lobe level Dolph-Tschebyscheff array of isotropic ele-ments placed symmetrically along the z-axis. Find the (a) amplitude excitation coefficients (b) array factor (c) angles where the nulls occur for d = 3λ∕4(0◦≤𝜃≤180◦) (d) directivity for d = 3λ∕4 (e) half-power beamwidth for d = 3λ∕4 6.63.
Design a four-element, −40 dB side lobe level Dolph-Tschebyscheff array of isotropic ele-ments placed symmetrically about the z-axis. Find the (a) amplitude excitation coefficients (b) array factor (c) angles where the nulls occur for d = 3λ∕4.
Are all of the minor lobes of the same level? Why not? What needs to be changed to make them of the same level?
Verify parts of the problem using the computer program Arrays.
6.64.
Repeat the design of Problem 6.63 for a five-element, −20 dB Dolph-Tschebyscheff array.
6.65.
Repeat the design of Problem 6.63 for a six-element, −20 dB Dolph-Tschebyscheff array.
6.66.
Repeat the design of Problem 6.40 for a Dolph-Tschebyscheff distribution of −40 dB side lobe level and λ∕4 spacing between the elements. In addition, find the (a) directivity of the entire array (b) half-power beamwidths of the entire array in the x-y and y-z planes 6.67.
Repeat the design of Problem 6.41 for a Dolph-Tschebyscheff distribution of −40 dB side lobe level and λ∕4 spacing between the elements. In addition, find the (a) directivity of the entire array (b) half-power beamwidths of the entire array in the x-y and y-z planes 380 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 6.68.
Design a five-element, −40 dB side lobe level Dolph-Tschebyscheff array of isotropic ele-ments. The elements are placed along the x-axis with a spacing of λ∕4 between them. Deter-mine the: (a) normalized amplitude coefficients (b) array factor (c) directivity (d) half-power beamwidth 6.69.
The total length of a discrete-element array is 4λ. For a −30 dB side lobe level Dolph-Tschebyscheff design and a spacing of λ∕2 between the elements along the z-axis, find the: (a) number of elements (b) excitation coefficients (c) directivity (d) half-power beamwidth 6.70.
Design a broadside three-element, −26 dB side lobe level Dolph-Tschebyscheff array of iso-topic sources placed along the z-axis. For this design, find the: (a) normalized excitation coefficients (b) array factor (c) nulls of the array factor when d = λ∕2 (in degrees) (d) maxima of the array factor when d = λ∕2 (in degrees) (e) HPBW beamwidth (in degrees) of the array factor when d = λ∕2 (f) directivity (in dB) of the array factor when d = λ∕2 (g) verify the design using the computer program Arrays.
6.71.
Design a broadside uniform array, with its elements placed along the z axis, in order the directivity of the array factor is 33 dB (above isotropic). Assuming the spacing between the elements is λ∕16, and it is very small compared to the overall length of the array, deter-mine the: (a) closest number of integer elements to achieve this.
(b) overall length of the array (in wavelengths).
(c) half-power beamwidth (in degrees).
(d) amplitude level (in dB) of the maximum of the first minor lobe compared to the maximum of the major lobe.
6.72.
The design of Problem 6.71 needs to be changed to a nonuniform Dolph-Tschebyscheff in order to lower the side lobe amplitude level to −30 dB, while maintaining the same number of elements and spacing. For the new nonuniform design, what is the: (a) half-power beamwidth (in degrees).
(b) directivity (in dB).
6.73.
Design a Dolph-Tschebyscheff linear array of 6 elements with uniform spacing between them. The array factor must meet the following specifications: 1. −40 dB side lobe level.
2. Minor lobes from 0◦≤𝜃≤90◦all of the same level.
3. Largest allowable spacing between the elements (in wavelengths) and still meet above specifications.
Determine: (a) excitation coefficients, normalized so that the ones of the edge elements is unity.
(b) maximum allowable spacing (in wavelengths) between the elements and still meet spec-ifications.
(c) plot (in 1◦increments) the normalized (max = 0 dB) array factor (in dB). Check to see that the array factor meets the specifications. If not, find out what is wrong with it.
Verify parts of the problem using the computer program Arrays.
PROBLEMS 381 6.74.
Design the array factor of a three-element Dolph-Tschebyscheff broadside array with a side lobe level of −40 dB. Determine the (a) normalized amplitude coefficients.
(b) maximum allowable spacing (in λ) to maintain the same sidelobe level for all minor lobes.
(c) approximate half-power beamwidth (in degrees) using the spacing from part b.
(d) approximate directivity (in dB) using the spacing from part b.
6.75.
Design a Dolph-Tschebyscheff broadside array of 5 elements with a −30 dB sidelobe level.
(a) Determine the normalized amplitude excitation coefficients. Make the ones at the edges of the array unity.
(b) Determine the maximum spacing between the elements (in λ) so that all sidelobes are maintained at the same level of −30 dB.
(c) For the spacing of Part (b), determine the half-power beamwidth (in degrees). Compare it to that of a uniform array of the same number of elements and spacing.
(d) For the spacing of Part b, determine the directivity (dimensionless and in dB).
6.76.
It is desired to design a Dolph-Tschebyscheff nonuniform linear broadside array. The desired array should have 20 elements with a uniform spacing between them. The required sidelobe level −40 dB down from the maximum. Determine the: (a) Maximum uniform spacing that can be used between the elements and still maintain a constant sidelobe level of −40 dB for all minor lobes.
(b) Half-power beamwidth (in degrees) of a uniform broadside linear array of the same num-ber of elements and spacing as the Dolph-Tschebyscheff array. Assume d = λ∕2.
(c) Half-power beamwidth (in degrees) of the Dolph-Tschebyscheff array with d = λ∕2.
(d) Directivity of the Dolph-Tschebyscheff array of d = λ∕2 (dimensionless and in dB).
(e) Directivity of the uniform broadside array of d = λ∕2 (dimensionless and in dB).
6.77.
A nonuniform Dolph-Tschebyscheff linear broadside array of 4 elements are placed along the z-axis each a distance d apart. For a separation of d = 3λ∕8 between the elements and a −27.959 dB sidelobe level, determine analytically (not graphically) the: (a) Maximum element separation dmax (in λ) that can be used to maintain the constant −27.959 dB sidelobe level?
(b) Approximate half-power beamwidth (in degrees) using d = 3λ∕8.
(c) Approximate directivity (dimensionless and in dB) using d = 3λ∕8.
6.78.
Design a broadside linear Dolph-Tschebyscheff array with the elements placed along the z-axis so that its array factor pattern, using the largest possible spacing between the elements while still maintaining the same sidelobe level, has 9 minor lobes on each side of the three-dimensional pattern. The desired sidelobe level is −60 dB. To accomplish this, determine the: (a) Order of the Tschebyscheff polynomial.
(b) Number of elements.
(c) Maximum spacing between the elements (in λ).
6.79.
It is desired to design a broadside Tschebyscheff linear array of N = 10 elements, placed along the z-axis, with a uniform spacing of d = λ∕2 between the elements and with a uniform sidelobe level of −26 dB from the main maximum. Determine the: (a) Progressive phase (in degrees) excitation between the elements.
(b) Number of complete minor lobes in the elevation plane between 0◦≤𝜃≤90◦.
(c) Maximum spacing between the elements to maintain the same sidelobe level over all the minor lobes.
382 ARRAYS: LINEAR, PLANAR, AND CIRCULAR 6.80.
It is desired to design a −25 dB broadside Dolph-Tschebyscheff array of 6 elements with a spacing of d = λ∕4 between the elements. Determine the: (a) HPBW (in degrees).
(b) Maximum directivity (dimensionless and in dB).
6.81.
A Dolph-Tschebyscheff broadside array of 6 elements, 50-dB sidelobe level, and with a spac-ing of λ∕2 between the elements is designed to operate at 9 GHz. For the array factor of the antenna, determine the: (a) Half-power beamwidth (in degrees). How much narrower or wider (in degrees) is this half-power beamwidth compared to that of a uniform array of the same number of ele-ments and spacing? Justify your answer. Do you expect it? Why?
(b) Directivity of array factor (in dB). How much smaller or larger (in dB) is this directivity (in dB) compared to that of a uniform array of the same number of elements and spacing?
Justify you answer. Do you expect it? Why?
6.82.
Dolph-Tschebyscheff designs are practical because you can perform the design to meet the required specifications (like the selection of the number of elements, spacing between the elements, and to maintain all the minor lobes at a required uniform/constant level). Such a design leads to amplitude coefficients, which for modest sidelobe levels usually do not vary more than 4:1, and can be achieved with efficient feed designs. This is usually not the case for binomial designs where the amplitude excitation coefficients usually vary more than 100:1, especially when a large number of elements (equal or greater than 10) are required. To demonstrate this, design two separate broadside linear arrays, with each design of 9 elements placed along the z-axis. The spacing between the elements is λ/2 and a sidelobe level of −30 dB for the Dolph-Tschebyscheff design.
(a) For the binomial broadside linear array design: Determine the: r Sidelobe level (in dB).
r Half-power beamwidth (in degrees).
r Maximum directivity (dimensionless and in dB).
(b) For the Dolph-Tschebyscheff broadside linear array design: Determine the: r Half-power beamwidth (in degrees).
r Maximum directivity (dimensionless and in dB).
(c) Is the half-power beamwidth of the binomial broadside linear array design, compared to that of the Dolph-Tschebysceff broadside linear array design: r Smaller or larger? Why? Justify the answer. Is this what you expected?
(d) Is the maximum directivity of the binomial broadside linear array design, compared to that of the Dolph-Tschebyscheff broadside linea design: r Smaller or larger? Why? Justify the answer. Is this what you expected?
6.83.
In the design of antenna arrays, with a spacing of d ≤λ∕2, there is a choice between uniform, binomial, cosine-squared, and Dolph-Tschebyscheff (of −25 dB sidelobe level) distributions.
If it is desired to: (a) Select the design distributions with the smallest half-power beamwidths, place the anten-nas in order of smaller-to-larger half-power beamwidths.
(b) Select the design distributions with the lowest sidelobe level, place the antennas in order of lower-to-higher sidelobe level.
6.84.
It is desired to design a planar square array with uniform illumination so that its approximate half-power beamwidth is 1◦, when the main beam maximum, is pointed in some direction 𝜃0. Determine the total dimension (in λ) on each side of the square array when its maximum is directed toward (z-axis is perpendicular to the plane of the array): PROBLEMS 383 (a) Broadside (𝜃0 = 0◦); (b) 𝜃0 = 60◦from broadside.
Treat the planar array in each plane as a source with a continuous distribution (like an aperture), and assume it is large in terms of a wavelength.
6.85.
In high-performance radar arrays low-sidelobes are very desirable. In a particular application it is desired to design a broadside linear array which maintains all the sidelobes at the same level of −30 dB. The number of elements must be 3 and the spacing between them must be λ∕4.
(a) State the design that will meet the specifications.
(b) What are the amplitude excitations of the elements?
(c) What is the half-power beamwidth (in degrees) of the main lobe?
(d) What is the directivity (in dB) of the array?
6.86.
Design a nonuniform amplitude broadside linear array of 5 elements. The total length of the array is 2λ. To meet the sidelobe and half-power beamwidth specifications, the amplitude excitations of the elements must be that of a cosine-on-a-pedestal distribution represented by Amplitude distribution = 1 + cos(𝜋xn∕L) where xn is the position of the nth element (in terms of L) measured from the center of the array. Determine the amplitude excitation coefficients an’s of the five elements. Assume uni-form spacing between the elements and the end elements are located at the edges of the array length.
6.87.
It is desired to design a uniform square scanning array whose elevation half-power beamwidth is 2◦. Determine the minimum dimensions of the array when the scan maximum angle is (a) 𝜃0 = 30◦ (b) 𝜃0 = 45◦ (c) 𝜃0 = 60◦ 6.88.
Determine the azimuthal and elevation angles of the grating lobes for a 10 × 10 element uniform planar array when the spacing between the elements is λ. The maximum of the main beam is directed toward 𝜃0 = 60◦, 𝜙0 = 90◦and the array is located on the x-y plane.
6.89.
Four isotropic elements are placed on the xy-plane, and all four are excited with the same amplitude; uniform array. Two of the elements are placed along the x-axis, symmetrically about the z-axis, with a total separation of dx between them, and a phase excitation of +𝛽x between them. The other two elements are placed along the y-axis, symmetrically about the z-axis, with a total separation of dy between them, and a phase excitation of 𝛽y between them.
For far-field observations, (a) Write (you do not have to derive it) the normalized array factor, in simplified form (sine and/or cosines only), for only the two elements along the x-axis.
(b) Write (you do not have to derive it) the normalized array factor, in simplified form (sine and/or cosines only), for only the two elements along the y-axis.
(c) Write (you do not have to derive it) the normalized array factor, in simplified form (sine and/or cosines only), of the four elements along the x- and y- axes. You can use superpo-sition.
(d) Without having to derive anything but based on the fundamental phasing principles that we covered in class and discussed in the book, what should the phases 𝛽x and 𝛽y be (in degrees) so that there is (assuming dx = dy = λ∕4): r A maximum along the +z direction; i.e., perpendicular to the xy-plane.
r A minimum along the +z direction; i.e., perpendicular to the xy-plane.
384 ARRAYS: LINEAR, PLANAR, AND CIRCULAR y x dy dx 6.90.
Design a 10 × 8 (10 in the x-direction and 8 in the y) element uniform planar array so that the main maximum is oriented along 𝜃0 = 10◦, 𝜙0 = 90◦. For a spacing of dx = dy = λ∕8 between the elements, find the (a) progressive phase shift between the elements in the x and y directions (b) directivity of the array (c) half-power beamwidths (in two perpendicular planes) of the array.
Verify the design using the computer program Arrays of this chapter.
6.91.
It is desired to design a nonuniform binomial planar array of 4x4 elements placed along the x-y plane with a uniform spacing of λ∕2 between the elements along both the x- and y-directions. The z-axis is perpendicular to the array. During the operation, the array factor is scanned with its maximum directed along the 𝜃o = 30◦, 𝜙o = 45◦direction. Determine the: (a) Half-power beamwidths (in degrees) of the array amplitude pattern in two mutually per-pendicular planes that both pass through the maximum of the array factor.
(b) Exact directivity (dimensionless and in dB) of the array factor.
(c) Approximate directivity (dimensionless and in dB) of the array factor. To get credit, select the most appropriate formula to compute the approximate directivity. State which one you are using.
6.92.
The main beam maximum of a 10 × 10 planar array of isotropic elements (100 elements) is directed toward 𝜃0 = 10◦and 𝜙0 = 45◦. Find the directivity, beamwidths (in two perpendic-ular planes), and beam solid angle for a Tschebyscheff distribution design with side lobes of −26 dB. The array is placed on the x-y plane and the elements are equally spaced with d = λ∕4. It should be noted that an array with bidirectional (two-sided pattern) elements would have a directivity which would be half of that of the same array but with unidirec-tional (one-sided pattern) elements. Verify the design using the computer program Arrays of this chapter.
6.93.
Repeat Problem 6.90 for a Tschebyscheff distribution array of −30 dB side lobes.
6.94.
In the design of uniform linear arrays, the maximum usually occurs at 𝜃= 𝜃0 at the design frequency f = f0, which has been used to determine the progressive phase between the ele-ments. As the frequency shifts from the designed center frequency f0 to fh, the maximum amplitude of the array factor at f = fh is 0.707 the normalized maximum amplitude of unity at f = f0. The frequency fh is referred to as the half-power frequency, and it is used to deter-mine the frequency bandwidth over which the pattern maximum varies over an amplitude of 3 dB. Using the array factor of a linear uniform array, derive an expression for the 3-dB frequency bandwidth in terms of the length L of the array and the scan angle 𝜃0.
CHAPTER7 Antenna Synthesis and Continuous Sources 7.1 INTRODUCTION Thus far in the book we have concentrated primarily on the analysis and design of antennas. In the analysis problem an antenna model is chosen, and it is analyzed for its radiation characteristics (pattern, directivity, impedance, beamwidth, efficiency, polarization, and bandwidth). This is usually accomplished by initially specifying the current distribution of the antenna, and then analyzing it using standard procedures. If the antenna current is not known, it can usually be determined from integral equation formulations. Numerical techniques, such as the Moment Method of Chapter 8, can be used to numerically solve the integral equations.
In practice, it is often necessary to design an antenna system that will yield desired radiation characteristics. For example, a very common request is to design an antenna whose far-field pattern possesses nulls in certain directions. Other common requests are for the pattern to exhibit a desired distribution, narrow beamwidth and low sidelobes, decaying minor lobes, and so forth. The task, in general, is to find not only the antenna configuration but also its geometrical dimensions and excita-tion distribution. The designed system should yield, either exactly or approximately, an acceptable radiation pattern, and it should satisfy other system constraints. This method of design is usually referred to as synthesis. Although synthesis, in its broadest definition, usually refers to antenna pat-tern synthesis, it is often used interchangeably with design. Since design methods have been outlined and illustrated previously, as in Chapter 6, in this chapter we want to introduce and illustrate antenna pattern synthesis methods.
Antenna pattern synthesis usually requires that first an approximate analytical model is chosen to represent, either exactly or approximately, the desired pattern. The second step is to match the analytical model to a physical antenna model. Generally speaking, antenna pattern synthesis can be classified into three categories. One group requires that the antenna patterns possess nulls in desired directions. The method introduced by Schelkunoff can be used to accomplish this; it will be discussed in Section 7.3. Another category requires that the patterns exhibit a desired distribution in the entire visible region. This is referred to as beam shaping, and it can be accomplished using the Fourier transform and the Woodward-Lawson , methods. They will be discussed and illustrated in Sections 7.4 and 7.5, respectively. A third group includes techniques that produce pat-terns with narrow beams and low sidelobes. Some methods that accomplish this have already been discussed; namely the binomial method (Section 6.8.2) and the Dolph-Tschebyscheff method (also spelled Tchebyscheff or Chebyshev) of Section 6.8.3. Other techniques that belong to this family Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 385 386 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES are the Taylor line-source (Tschebyscheff-error) and the Taylor line-source (one parameter) .
They will be outlined and illustrated in Sections 7.6 and 7.7, respectively.
The synthesis methods will be utilized to design line-sources and linear arrays whose space factors [as defined by (4-58a)] and array factors [as defined by (6-52)] will yield desired far-field radiation patterns. The total pattern is formed by multiplying the space factor (or array factor) by the element factor (or element pattern) as dictated by (4-59) [or (6-5)]. For very narrow beam patterns, the total pattern is nearly the same as the space factor or array factor. This is demonstrated by the dipole antenna (l ≪λ) of Figure 4.3 whose element factor, as given by (4-58a), is sin 𝜃; for values of 𝜃near 90◦(𝜃≃90◦), sin 𝜃≃1.
The synthesis techniques will be followed with a brief discussion of some very popular line-source distributions (triangular, cosine, cosine-squared) and continuous aperture distributions (rectangular and circular).
7.2 CONTINUOUS SOURCES Very long (in terms of a wavelength) arrays of discrete elements usually are more difficult to imple-ment, more costly, and have narrower bandwidths. For such applications, antennas with continuous distributions would be convenient to use. A very long wire and a large reflector represent, respec-tively, antennas with continuous line and aperture distributions. Continuous distribution antennas usually have larger sidelobes, are more difficult to scan, and in general, they are not as versatile as arrays of discrete elements. The characteristics of continuously distributed sources can be approxi-mated by discrete-element arrays, and vice-versa, and their development follows and parallels that of discrete-element arrays.
7.2.1 Line-Source Continuous line-source distributions are functions of only one coordinate, and they can be used to approximate linear arrays of discrete elements and vice-versa.
The array factor of a discrete-element array, placed along the z-axis, is given by (6-52) and (6-52a). As the number of elements increases in a fixed-length array, the source approaches a con-tinuous distribution. In the limit, the array factor summation reduces to an integral. For a continuous distribution, the factor that corresponds to the array factor is known as the space factor. For a line-source distribution of length l placed symmetrically along the z-axis as shown in Figure 7.1(a), the space factor (SF) is given by SF(𝜃) = ∫ +l∕2 −l∕2 In(z′)ej[kz′ cos 𝜃+𝜙n(z′)] dz′ (7-1) where In(z′) and 𝜙n(z′) represent, respectively, the amplitude and phase distributions along the source. For a uniform phase distribution 𝜙n(z′) = 0.
Equation (7-1) is a finite one-dimensional Fourier transform relating the far-field pattern of the source to its excitation distribution. Two-dimensional Fourier transforms are used to represent the space factors for two-dimensional source distributions. These relations are results of the angular spectrum concept for plane waves, introduced first by Booker and Clemmow , and it relates the angular spectrum of a wave to the excitation distribution of the source.
For a continuous source distribution, the total field is given by the product of the element and space factors as defined in (4-59). This is analogous to the pattern multiplication of (6-5) for arrays.
The type of current and its direction of flow on a source determine the element factor. For a finite length linear dipole, for example, the total field of (4-58a) is obtained by summing the contributions SCHELKUNOFF POLYNOMIAL METHOD 387 Figure 7.1 Continuous and discrete linear sources.
of small infinitesimal elements which are used to represent the entire dipole. In the limit, as the infinitesimal lengths become very small, the summation reduces to an integration. In (4-58a), the factor outside the brackets is the element factor and the one within the brackets is the space factor and corresponds to (7-1).
7.2.2 Discretization of Continuous Sources The radiation characteristics of continuous sources can be approximated by discrete-element arrays, and vice-versa. This is illustrated in Figure 7.1(b) whereby discrete elements, with a spacing d between them, are placed along the length of the continuous source. Smaller spacings between the elements yield better approximations, and they can even capture the fine details of the continuous distribution radiation characteristics. For example, the continuous line-source distribution In(z′) of (7-1) can be approximated by a discrete-element array whose element excitation coefficients, at the specified element positions within −l∕2 ≤z′ ≤l∕2, are determined by the sampling of In(z′)ej𝜙n(z′).
The radiation pattern of the digitized discrete-element array will approximate the pattern of the con-tinuous source.
The technique can be used for the discretization of any continuous distribution. The accuracy increases as the element spacing decreases; in the limit, the two patterns will be identical. For large element spacing, the patterns of the two antennas will not match well. To avoid this, another method known as root-matching can be used . Instead of sampling the continuous current distribution to determine the element excitation coefficients, the root-matching method requires that the nulls of the continuous distribution pattern also appear in the initial pattern of the discrete-element array. If the synthesized pattern using this method still does not yield (within an acceptable accuracy) the desired pattern, a perturbation technique can then be applied to the distribution of the discrete-element array to improve its accuracy.
7.3 SCHELKUNOFF POLYNOMIAL METHOD A method that is conducive to the synthesis of arrays whose patterns possess nulls in desired direc-tions is that introduced by Schelkunoff . To complete the design, this method requires information 388 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES on the number of nulls and their locations. The number of elements and their excitation coefficients are then derived. The analytical formulation of the technique follows.
Referring to Figure 6.5(a), the array factor for an N-element, equally spaced, nonuniform ampli-tude, and progressive phase excitation is given by (6-52) as AF = N ∑ n=1 anej(n−1)(kd cos 𝜃+𝛽) = N ∑ n=1 anej(n−1)𝜓 (7-2) where an accounts for the nonuniform amplitude excitation of each element. The spacing between the elements is d and 𝛽is the progressive phase shift.
Letting z = x + jy = ej𝜓= ej(kd cos 𝜃+𝛽) (7-3) we can rewrite (7-2) as AF = N ∑ n=1 anzn−1 = a1 + a2z + a3z2 + ⋯+ aNzN−1 (7-4) which is a polynomial of degree (N −1). From the mathematics of complex variables and algebra, any polynomial of degree (N −1) has (N −1) roots and can be expressed as a product of (N −1) linear terms. Thus we can write (7-4) as AF = an(z −z1)(z −z2)(z −z3) ⋯(z −zN−1) (7-5) where z1, z2, z3, … , zN−1 are the roots, which may be complex, of the polynomial. The magnitude of (7-5) can be expressed as |AF| = |an||z −z1||z −z2||z −z3| ⋯|z −zN−1| (7-6) Some very interesting observations can be drawn from (7-6) which can be used judiciously for the analysis and synthesis of arrays. Before tackling that phase of the problem, let us first return and examine the properties of (7-3).
The complex variable z of (7-3) can be written in another form as z = |z|ej𝜓= |z| ∠𝜓= 1 ∠𝜓 (7-7) 𝜓= kd cos 𝜃+ 𝛽= 2𝜋 λ d cos 𝜃+ 𝛽 (7-7a) It is clear that for any value of d, 𝜃, or 𝛽the magnitude of z lies always on a unit circle; however its phase depends upon d, 𝜃, and 𝛽. For 𝛽= 0, we have plotted in Figures 7-2(a)–(d) the value of z, magnitude and phase, as 𝜃takes values of 0 to 𝜋rad. It is observed that for d = λ∕8 the values of z, for all the physically observable angles of 𝜃, only exist over the part of the circle shown in Figure 7.2(a).
Any values of z outside that arc are not realizable by any physical observation angle 𝜃for the spacing d = λ∕8. We refer to the realizable part of the circle as the visible region and the remaining as the invisible region. In Figure 7.2(a) we also observe the path of the z values as 𝜃changes from 0◦to 180◦.
SCHELKUNOFF POLYNOMIAL METHOD 389 x jy ψ λ λ λ λ Figure 7.2 Visible Region (VR) and Invisible Region (IR) boundaries for complex variable z when 𝛽= 0.
In Figures 7.2(b)–(d) we have plotted the values of z when the spacing between the elements is λ∕4, λ∕2, and 3λ∕4. It is obvious that the visible region can be extended by increasing the spacing between the elements. It requires a spacing of at least λ∕2 to encompass, at least once, the entire circle. Any spacing greater than λ∕2 leads to multiple values for z. In Figure 7.2(d) we have double values for z for half of the circle when d = 3λ∕4.
To demonstrate the versatility of the arrays, in Figures 7.3(a)–(d) we have plotted the values of z for the same spacings as in Figure 7.2(a)–(d) but with a 𝛽= 𝜋∕4. A comparison between the corresponding figures indicates that the overall visible region for each spacing has not changed but its relative position on the unit circle has rotated counterclockwise by an amount equal to 𝛽.
We can conclude then that the overall extent of the visible region can be controlled by the spacing between the elements and its relative position on the unit circle by the progressive phase excitation of the elements. These two can be used effectively in the design of the array factors.
Now let us return to (7-6). The magnitude of the array factor, its form as shown in (7-6), has a geometrical interpretation. For a given value of z in the visible region of the unit circle, corresponding 390 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES Figure 7.3 Visible Region (VR) and Invisible Region (IR) boundaries for complex variable z when 𝛽= 𝜋∕4.
to a value of 𝜃as determined by (7-3), |AF| is proportional to the product of the distances between z and z1, z2, z3, … , zN−1, the roots of AF. In addition, apart from a constant, the phase of AF is equal to the sum of the phases between z and each of the zeros (roots). This is best demonstrated geometrically in Figure 7.4(a). If all the roots z1, z2, z3, … , zN−1 are located in the visible region of the unit circle, then each one corresponds to a null in the pattern of |AF| because as 𝜃changes z changes and eventually passes through each of the zn’s. When it does, the length between z and that zn is zero and (7-6) vanishes. When all the zeros (roots) are not in the visible region of the unit circle, but some lie outside it and/or any other point not on the unit circle, then only those zeros on the visible region will contribute to the nulls of the pattern. This is shown geometrically in Figure 7.4(b).
If no zeros exist in the visible region of the unit circle, then that particular array factor has no nulls for any value of 𝜃. However, if a given zero lies on the unit circle but not in its visible region, that zero can be included in the pattern by changing the phase excitation 𝛽so that the visible region is rotated until it encompasses that root. Doing this, and not changing d, may exclude some other zero(s).
To demonstrate all the principles, we will consider an example along with some computations.
SCHELKUNOFF POLYNOMIAL METHOD 391 Figure 7.4 Array factor roots within and outside unit circle, and visible and invisible regions.
Example 7.1 Design a linear array with a spacing between the elements of d = λ∕4 such that it has zeros at 𝜃= 0◦, 90◦, and 180◦. Determine the number of elements, their excitation, and plot the derived pattern. Use Schelkunoff’s method.
Solution: For a spacing of λ∕4 between the elements and a phase shift 𝛽= 0◦, the visible region is shown in Figure 7.2(b). If the desired zeros of the array factor must occur at 𝜃= 0◦, 90◦, and 180◦, then these correspond to z = j, 1, −j on the unit circle. Thus a normalized form of the array factor is given by AF = (z −z1)(z −z2)(z −z3) = (z −j)(z −1)(z + j) AF = z3 −z2 + z −1 Referring to (7-4), the above array factor and the desired radiation characteristics can be realized when there are four elements and their excitation coefficients are equal to a1 = −1 a2 = +1 a3 = −1 a4 = +1 To illustrate the method, we plotted in Figure 7.5 the pattern of that array; it clearly meets the desired specifications. Because of the symmetry of the array, the pattern of the left hemisphere is identical to that of the right.
392 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES Figure 7.5 Amplitude radiation pattern of a four-element array of isotropic sources with a spacing of λ∕4 between them, zero degrees progressive phase shift, and zeros at 𝜃= 0◦, 90◦, and 180◦.
7.4 FOURIER TRANSFORM METHOD This method can be used to determine, given a complete description of the desired pattern, the exci-tation distribution of a continuous or a discrete source antenna system. The derived excitation will yield, either exactly or approximately, the desired antenna pattern. The pattern synthesis using this method is referred to as beam shaping.
7.4.1 Line-Source For a continuous line-source distribution of length l, as shown in Figure 7.1, the normalized space factor of (7-1) can be written as SF(𝜃) = ∫ l∕2 −l∕2 I(z′)ej(k cos 𝜃−kz)z′ dz′ = ∫ l∕2 −l∕2 I(z′)ej𝜉z′ dz′ (7-8) 𝜉= k cos 𝜃−kz ➱𝜃= cos−1 (𝜉+ kz k ) (7-8a) FOURIER TRANSFORM METHOD 393 where kz is the excitation phase constant of the source. For a normalized uniform current distribution of the form I(z′) = I0∕l, (7-8) reduces to SF(𝜃) = I0 sin [ kl 2 ( cos 𝜃−kz k )] kl 2 ( cos 𝜃−kz k ) (7-9) The observation angle 𝜃of (7-9) will have real values (visible region) provided that −(k + kz) ≤𝜉≤ (k −kz) as obtained from (7-8a).
Since the current distribution of (7-8) extends only over −l∕2 ≤z′ ≤l∕2 (and it is zero outside it), the limits can be extended to infinity and (7-8) can be written as SF(𝜃) = SF(𝜉) = ∫ +∞ −∞ I(z′)ej𝜉z′ dz′ (7-10a) The form of (7-10a) is a Fourier transform, and it relates the excitation distribution I(z′) of a contin-uous source to its far-field space factor SF(𝜃). The transform pair of (7-10a) is given by I(z′) = 1 2𝜋∫ +∞ −∞ SF(𝜉)e−jz′𝜉d𝜉= 1 2𝜋∫ +∞ −∞ SF(𝜃)e−jz′𝜉d𝜉 (7-10b) Whether (7-10a) represents the direct transform and (7-10b) the inverse transform, or vice-versa, does not matter here. The most important thing is that the excitation distribution and the far-field space factor are related by Fourier transforms.
Equation (7-10b) indicates that if SF(𝜃) represents the desired pattern, the excitation distribution I(z′) that will yield the exact desired pattern must in general exist for all values of z′(−∞≤z′ ≤∞).
Since physically only sources of finite dimensions are realizable, the excitation distribution of (7-10b) is usually truncated at z′ = ±l∕2 (beyond z′ = ±l∕2 it is set to zero). Thus the approximate source distribution is given by Ia(z′) ≃ ⎧ ⎪ ⎨ ⎪ ⎩ I(z′) = 1 2𝜋∫ +∞ −∞ SF(𝜉)e−jz′𝜉d𝜉 −l∕2 ≤z′ ≤l∕2 0 elsewhere (7-11) and it yields an approximate pattern SF(𝜃)a. The approximate pattern is used to represent, within certain error, the desired pattern SF(𝜃)d. Thus SF(𝜃)d ≃SF(𝜃)a = ∫ l∕2 −l∕2 Ia(z′)ej𝜉z′ dz′ (7-12) It can be shown that, over all values of 𝜉, the synthesized approximate pattern SF(𝜃)a yields the least-mean-square error or deviation from the desired pattern SF(𝜃)d. However that criterion is not satisfied when the values of 𝜉are restricted only in the visible region , .
To illustrate the principles of this design method, an example is taken.
394 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES Example 7.2 Determine the current distribution and the approximate radiation pattern of a line-source placed along the z-axis whose desired radiation pattern is symmetrical about 𝜃= 𝜋∕2, and it is given by SF(𝜃) = { 1 𝜋∕4 ≤𝜃≤3𝜋∕4 0 elsewhere This is referred to as a sectoral pattern, and it is widely used in radar search and communication applications.
Solution: Since the pattern is symmetrical, kz = 0. The values of 𝜉, as determined by (7-8a), are given by k∕ √ 2 ≥𝜉≥−k∕ √ 2. In turn, the current distribution is given by (7-10b) or I(z′) = 1 2𝜋∫ +∞ −∞ SF(𝜉)e−jz′𝜉d𝜉 = 1 2𝜋∫ k∕ √ 2 −k∕ √ 2 e−jz′𝜉d𝜉= k 𝜋 √ 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ sin ( kz′ √ 2 ) kz′ √ 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ and it exists over all values of z′(−∞≤z′ ≤∞). Over the extent of the line-source, the current distribution is approximated by Ia(z′) ≃I(z′), −l∕2 ≤z′ ≤l∕2 If the derived current distribution I(z′) is used in conjunction with (7-10a) and it is assumed to exist over all values of z′, the exact and desired sectoral pattern will result. If however it is truncated at z′ = ±l∕2 (and assumed to be zero outside), then the desired pattern is approximated by (7-12) or SF(𝜃)d ≃SF(𝜃)a = ∫ l∕2 −l∕2 Ia(z′)ej𝜉z′ dz′ = 1 𝜋 { Si [ l λ𝜋 ( cos 𝜃+ 1 √ 2 )] −Si [ l λ𝜋 ( cos 𝜃− 1 √ 2 )]} where Si(x) is the sine integral of (4-68b).
The approximate current distribution (normalized so that its maximum is unity) is plotted in Figure 7.6(a) for l = 5λ and l = 10λ. The corresponding approximate normalized patterns are shown in Figure 7.6(b) where they are compared with the desired pattern. A very good reconstruc-tion is indicated. The longer line-source (l = 10λ) provides a better realization. The sidelobes are about 0.102 (−19.83 dB) for l = 5λ and 0.081 (−21.83 dB) for l = 10λ (relative to the pattern at 𝜃= 90◦).
FOURIER TRANSFORM METHOD 395 1.0 0.5 –5 –4 –3 –2 –1 0 1 2 3 4 5 Normalized current I (z′) Source position z′(in λ) Line-source (l = 5λ) Desired pattern Line-source (l = 10λ) Line-source (l = 5λ) Line-source (l = 10λ) Line-array (l = 5λ, 10λ) (b) Space factor 50 90 0.5 1.0 130 170 10 Observation angle θ (degrees) Space factor |SF (θ)| (a) Current distribution Figure 7.6 Normalized current distribution, desired pattern, and synthesized patterns using the Fourier transform method.
7.4.2 Linear Array The array factor of an N-element linear array of equally spaced elements and nonuniform excitation is given by (7-2). If the reference point is taken at the physical center of the array, the array factor can also be written as Odd Number of Elements (N = 2M + 1) AF(𝜃) = AF(𝜓) = M ∑ m=−M amejm𝜓 (7-13a) 396 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES Even Number of Elements (N = 2M) AF(𝜃) = AF(𝜓) = −1 ∑ m=−M amej[(2m+1)∕2]𝜓+ M ∑ m=1 amej[(2m−1)∕2]𝜓 (7-13b) where 𝜓= kd cos 𝜃+ 𝛽 (7-13c) For an odd number of elements (N = 2M + 1), the elements are placed at z′ m = md, m = 0, ±1, ±2, … , ±M (7-13d) and for an even number (N = 2M) at z′ m = ⎧ ⎪ ⎨ ⎪ ⎩ 2m −1 2 d, 1 ≤m ≤M 2m + 1 2 d, −M ≤m ≤−1 (7-13e) An odd number of elements must be utilized to synthesize a desired pattern whose average value, over all angles, is not equal to zero. The m = 0 term of (7-13a) is analogous to the d.c. term in a Fourier series expansion of functions whose average value is not zero.
In general, the array factor of an antenna is a periodic function of 𝜓, and it must repeat for every 2𝜋radians. In order for the array factor to satisfy the periodicity requirements for real values of 𝜃(visible region), then 2kd = 2𝜋or d = λ∕2. The periodicity and visible region requirement of d = λ∕2 can be relaxed; in fact, it can be made d < λ∕2. However, the array factor AF(𝜓) must be made pseudoperiodic by using fill-in functions, as is customarily done in Fourier series analysis. Such a construction leads to nonunique solutions, because each new fill-in function will result in a different solution. In addition, spacings smaller than λ∕2 lead to superdirective arrays that are undesirable and impractical. If d > λ∕2, the derived patterns exhibit undesired grating lobes; in addition, they must be restricted to satisfy the periodicity requirements.
If AF(𝜓) represents the desired array factor, the excitation coefficients of the array can be obtained by the Fourier formula of Odd Number of Elements (N = 2M + 1) am = 1 T ∫ T∕2 −T∕2 AF(𝜓)e−jm𝜓d𝜓= 1 2𝜋∫ 𝜋 −𝜋 AF(𝜓)e−jm𝜓d𝜓 −M ≤m ≤M (7-14a) FOURIER TRANSFORM METHOD 397 Even Number of Elements (N = 2M) am = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 T ∫ T∕2 −T∕2 AF(𝜓)e−j[(2m+1)∕2]𝜓d𝜓 = 1 2𝜋∫ 𝜋 −𝜋 AF(𝜓)e−j[(2m+1)∕2]𝜓d𝜓 −M ≤m ≤−1 1 T ∫ T∕2 −T∕2 AF(𝜓)e−j[(2m−1)∕2]𝜓d𝜓 = 1 2𝜋∫ 𝜋 −𝜋 AF(𝜓)e−j[(2m−1)∕2]𝜓d𝜓 1 ≤m ≤M (7-14b) (7-14c) Simplifications in the forms of (7-13a)–(7-13b) and (7-14a)–(7-14c) can be obtained when the exci-tations are symmetrical about the physical center of the array.
Example 7.3 Determine the excitation coefficients and the resultant pattern for a broadside discrete-element array whose array factor closely approximates the desired symmetrical sectoral pattern of Exam-ple 7.2. Use 11 elements with a spacing of d = λ∕2 between them. Repeat the design for 21 elements.
Solution: Since the array is broadside, the progressive phase shift between the elements as required by (6-18a) is zero (𝛽= 0). Since the pattern is nonzero only for 𝜋∕4 ≤𝜃≤3𝜋∕4, the corresponding values of 𝜓are obtained from (7-13c) or 𝜋∕ √ 2 ≥𝜓≥−𝜋∕ √ 2. The excitation coefficients are obtained from (7-14a) or am = 1 2𝜋∫ 𝜋∕ √ 2 −𝜋∕ √ 2 e−jm𝜓d𝜓= 1 √ 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ sin ( m𝜋 √ 2 ) m𝜋 √ 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ and they are symmetrical about the physical center of the array [am(−z′ m) = am(z′ m)]. The corre-sponding array factor is given by (7-13a).
The normalized excitation coefficients are a0 = 1.0000 a±4 = 0.0578 a±8 = −0.0496 a±1 = 0.3582 a±5 = −0.0895 a±9 = 0.0455 a±2 = −0.2170 a±6 = 0.0518 a±10 = −0.0100 a±3 = 0.0558 a±7 = 0.0101 They are displayed graphically by a dot (∙) in Figure 7.6(a) where they are compared with the continuous current distribution of Example 7.2. It is apparent that at the element positions, the line-source and linear array excitation values are identical. This is expected since the two antennas are of the same length (for N = 11, d = λ∕2 ➱l = 5λ and for N = 21, d = λ∕2 ➱l = 10λ).
398 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES 1.0 0.5 Observation angle θ (degrees) 10 Desired pattern Linear array (N = 11, d = λ/2) Linear array (N = 21, d = λ/2) 50 90 130 170 Array factor | AF (θ)| Figure 7.7 Desired array factor and synthesized normalized patterns for linear array of 11 and 21 elements using the Fourier transform method.
The corresponding normalized array factors are displayed in Figure 7.7. As it should be expected, the larger array (N = 21, d = λ∕2) provides a better reconstruction of the desired pat-tern. The sidelobe levels, relative to the value of the pattern at 𝜃= 90◦, are 0.061 (−24.29 dB) for N = 11 and 0.108 (−19.33 dB) for N = 21.
Discrete-element linear arrays only approximate continuous line-sources. Therefore, their pat-terns shown in Figure 7.7 do not approximate as well the desired pattern as the corresponding patterns of the line-source distributions shown in Figure 7.6(b).
Whenever the desired pattern contains discontinuities or its values in a given region change very rapidly, the reconstruction pattern will exhibit oscillatory overshoots which are referred to as Gibbs’ phenomena. Since the desired sectoral patterns of Examples 7.2 and 7.3 are discontinuous at 𝜃= 𝜋∕4 and 3𝜋∕4, the reconstructed patterns displayed in Figures 7.6(b) and 7.7 exhibit these oscillatory overshoots.
7.5 WOODWARD-LAWSON METHOD A very popular antenna pattern synthesis method used for beam shaping was introduced by Wood-ward and Lawson , , . The synthesis is accomplished by sampling the desired pattern at various discrete locations. Associated with each pattern sample is a harmonic current of uniform amplitude distribution and uniform progressive phase, whose corresponding field is referred to as a composing function. For a line-source, each composing function is of an bm sin(𝜓m)∕𝜓m form WOODWARD-LAWSON METHOD 399 whereas for a linear array it takes an bm sin(N𝜙m)∕N sin(𝜙m) form. The excitation coefficient bm of each harmonic current is such that its field strength is equal to the amplitude of the desired pattern at its corresponding sampled point. The total excitation of the source is comprised of a finite summation of space harmonics. The corresponding synthesized pattern is represented by a finite summation of composing functions with each term representing the field of a current harmonic with uniform amplitude distribution and uniform progressive phase.
The formation of the overall pattern using the Woodward-Lawson method is accomplished as follows. The first composing function produces a pattern whose main beam placement is determined by the value of its uniform progressive phase while its innermost sidelobe level is about −13.5 dB; the level of the remaining sidelobes monotonically decreases. The second composing function has also a similar pattern except that its uniform progressive phase is adjusted so that its main lobe maximum coincides with the innermost null of the first composing function. This results in the filling-in of the innermost null of the pattern of the first composing function; the amount of filling-in is controlled by the amplitude excitation of the second composing function. Similarly, the uniform progressive phase of the third composing function is adjusted so that the maximum of its main lobe occurs at the second innermost null of the first composing function; it also results in filling-in of the second innermost null of the first composing function. This process continues with the remaining finite number of composing functions.
The Woodward-Lawson method is simple, elegant, and provides insight into the process of pattern synthesis. However, because the pattern of each composing function perturbs the entire pattern to be synthesized, it lacks local control over the sidelobe level in the unshaped region of the entire pattern. In 1988 and 1989 a spirited and welcomed dialogue developed concerning the Woodward-Lawson method –. The dialogue centered whether the Woodward-Lawson method should be taught and even appear in textbooks, and whether it should be replaced by an alternate method which overcomes some of the shortcomings of the Woodward-Lawson method. The alternate method of is more of a numerical and iterative extension of Schelkunoff’s polynomial method which may be of greater practical value because it provides superior beamshape and pattern control. One of the distinctions of the two methods is that the Woodward-Lawson method deals with the synthesis of field patterns while that of deals with the synthesis of power patterns.
The analytical formulation of this method is similar to the Shannon sampling theorem used in communications which states that “if a function g(t) is band-limited, with its highest frequency being fh, the function g(t) can be reconstructed using samples taken at a frequency fs. To faithfully reproduce the original function g(t), the sampling frequency fs should be at least twice the high-est frequency fh (fs ≥2fh) or the function should be sampled at points separated by no more than Δt ≤1∕fs ≥1∕(2fh) = Th∕2 where Th is the period of the highest frequency fh.” In a similar manner, the radiation pattern of an antenna can be synthesized by sampling functions whose samples are separated by λ∕l rad, where l is the length of the source , .
7.5.1 Line-Source Let the current distribution of a continuous source be represented, within −l∕2 ≤z′ ≤l∕2, by a finite summation of normalized sources each of constant amplitude and linear phase of the form im(z′) = bm l e−jkz′ cos 𝜃m −l∕2 ≤z′ ≤l∕2 (7-15) As it will be shown later, 𝜃m represents the angles where the desired pattern is sampled. The total current I(z′) is given by a finite summation of 2M (even samples) or 2M + 1 (odd samples) current 400 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES sources each of the form of (7-15). Thus I(z′) = 1 l M ∑ m=−M bme−jkz′ cos 𝜃m (7-16) where m = ±1, ±2, … , ±M (for 2M even number of samples) (7-16a) m = 0, ±1, ±2, … , ±M (for 2M + 1 odd number of samples) (7-16b) For simplicity use odd number of samples.
Associated with each current source of (7-15) is a corresponding field pattern of the form given by (7-9) or sm(𝜃) = bm ⎧ ⎪ ⎨ ⎪ ⎩ sin [kl 2 (cos 𝜃−cos 𝜃m) ] kl 2 (cos 𝜃−cos 𝜃m) ⎫ ⎪ ⎬ ⎪ ⎭ (7-17) whose maximum occurs when 𝜃= 𝜃m. The total pattern is obtained by summing 2M (even samples) or 2M + 1 (odd samples) terms each of the form given by (7-17). Thus SF(𝜃) = M ∑ m=−M bm ⎧ ⎪ ⎨ ⎪ ⎩ sin [kl 2 (cos 𝜃−cos 𝜃m) ] kl 2 (cos 𝜃−cos 𝜃m) ⎫ ⎪ ⎬ ⎪ ⎭ (7-18) The maximum of each individual term in (7-18) occurs when 𝜃= 𝜃m, and it is equal to SF(𝜃= 𝜃m). In addition, when one term in (7-18) attains its maximum value at its sample at 𝜃= 𝜃m, all other terms of (7-18) which are associated with the other samples are zero at 𝜃= 𝜃m. In other words, all sampling terms (composing functions) of (7-18) are zero at all sampling points other than at their own. Thus at each sampling point the total field is equal to that of the sample. This is one of the most appealing properties of this method. If the desired space factor is sampled at 𝜃= 𝜃m, the excitation coefficients bm can be made equal to its value at the sample points 𝜃m.
Thus bm = SF(𝜃= 𝜃m)d (7-19) The reconstructed pattern is then given by (7-18), and it approximates closely the desired pattern.
In order for the synthesized pattern to satisfy the periodicity requirements of 2𝜋for real val-ues of 𝜃(visible region) and to faithfully reconstruct the desired pattern, each sample should be separated by kz′Δ||z′|=l = 2𝜋➱Δ = λ l (7-19a) WOODWARD-LAWSON METHOD 401 The location of each sample is given by cos 𝜃m = mΔ = m (λ l ) , m = 0, ±1, ±2, ..
for odd samples cos 𝜃m = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (2m −1) 2 Δ = (2m −1) 2 (λ l ) , m = +1, +2, ..
for even samples (2m + 1) 2 Δ = (2m + 1) 2 (λ l ) , m = −1, −2, ..
for even samples (7-19b) (7-19c) Therefore, M should be the closest integer to M = l∕λ.
As long as the location of each sample is determined by (7-19b or 7-19c), the pattern value at the sample points is determined solely by that of one sample and it is not correlated to the field of the other samples.
Example 7.4 Repeat the design of Example 7.2 for l = 5λ using odd samples and the Woodward-Lawson line-source synthesis method.
Solution: Since l = 5λ, M = 5 and the sampling separation is 0.2. The total number of sam-pling points is 11. The angles where the sampling is performed are given, according to (7-19b), by 𝜃m = cos−1 ( mλ l ) = cos−1(0.2m), m = 0, ±1, … , ±5 The angles and the excitation coefficients at the sample points are listed below.
m 𝜃m bm = SF(𝜃m)d m 𝜃m bm = SF(𝜃m)d 0 90◦ 1 1 78.46◦ 1 −1 101.54◦ 1 2 66.42◦ 1 −2 113.58◦ 1 3 53.13◦ 1 −3 126.87◦ 1 4 36.87◦ 0 −4 143.13◦ 0 5 0◦ 0 −5 180◦ 0 The computed pattern is shown in Figure 7.8(a) where it is compared with the desired pattern. A good reconstruction is indicated. The sidelobe level, relative to the value of the pattern at 𝜃= 90◦, is 0.160 (−15.92 dB).
To demonstrate the synthesis of the pattern using the sampling concept, we have plotted in Figure 7.8(b) all seven nonzero composing functions sm(𝜃) used for the reconstruction of the l = 5λ line-source pattern of Figure 7.8(a). Each nonzero sm(𝜃) composing function was computed using (7-17) for m = 0, ±1, ±2, ±3. It is evident that at each sampling point all the composing 402 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES functions are zero, except the one that represents that sample. Thus the value of the desired pattern at each sampling point is determined solely by the maximum value of a single composing function. The angles where the composing functions attain their maximum values are listed in the previous table.
0.5 90 135 180 45 0 (a) Normalized amplitude patterns Normalized magnitude 1.0 (b) Composing functions for line-source (l = 5 ) 1.0 Desired pattern Line-source (l = 5 ) Composing functions sm ( ), m = 0, ±1, ±2, ±3 ⎮SF ( )⎮ Desired pattern Line-source (l = 5λ) Linear array (N = 10, d = /2) ⎮SF ( )⎮ ⎮AF ( )⎮ Observation angle (degrees) θ θ θ θ θ λ λ λ Figure 7.8 Desired and synthesized patterns, and composing functions for Woodward-Lawson designs.
WOODWARD-LAWSON METHOD 403 7.5.2 Linear Array The Woodward-Lawson method can also be implemented to synthesize discrete linear arrays. The technique is similar to the Woodward-Lawson method for line-sources except that the pattern of each sample, as given by (7-17), is replaced by the array factor of a uniform array as given by (6-10c).
The pattern of each sample can be written as fm(𝜃) = bm sin [N 2 kd(cos 𝜃−cos 𝜃m) ] N sin [1 2kd(cos 𝜃−cos 𝜃m) ] (7-20) l = Nd assumes the array is equal to the length of the line-source (for this design only, the length l of the line includes a distance d/2 beyond each end element). The total array factor can be written as a superposition of 2M + 1 sampling terms (as was done for the line-source) each of the form of (7-20). Thus AF(𝜃) = M ∑ m=−M bm sin [N 2 kd(cos 𝜃−cos 𝜃m) ] N sin [1 2kd(cos 𝜃−cos 𝜃m) ] (7-21) As for the line-sources, the excitation coefficients of the array elements at the sample points are equal to the value of the desired array factor at the sample points. That is, bm = AF(𝜃= 𝜃m)d (7-22) The sample points are taken at cos 𝜃m = mΔ = m (λ l ) , m = 0, ±1, ±2, ..
for odd samples cos 𝜃m = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (2m −1) 2 Δ = (2m −1) 2 ( λ Nd ) , m = +1, +2, ..
for even samples (2m + 1) 2 Δ = (2m + 1) 2 ( λ Nd ) , m = −1, −2, ..
for even samples (7-23a) (7-23b) The normalized excitation coefficient of each array element, required to give the desired pattern, is given by an(z′) = 1 N M ∑ m=−M bme−jkzn′ cos 𝜃m (7-24) 404 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES where zn ′ indicates the position of the nth element (element in question) symmetrically placed about the geometrical center of the array.
Example 7.5 Repeat the design of Example 7.4 for a linear array of 10 elements using the Woodward-Lawson method with odd samples and an element spacing of d = λ∕2.
Solution: According to (7-19), (7-19b), (7-22) and (7-23a), the excitation coefficients of the array at the sampling points are the same as those of the line-source. Using the values of bm as listed in Example 7.4, the computed array factor pattern using (7-21) is shown in Figure 7.8(a).
A good synthesis of the desired pattern is displayed. The sidelobe level, relative to the pattern value at 𝜃= 90◦, is 0.221 (−13.1 dB). The agreement between the line-source and the linear array Woodward-Lawson designs are also good.
The normalized pattern of the symmetrical discrete array can also be generated using the array factor of (6-61a) or (6-61b), where the normalized excitation coefficients an’s of the array ele-ments are obtained using (7-24). For this example, the excitation coefficients of the 10-element array, along with their symmetrical position, are listed below. To achieve the normalized ampli-tude pattern of unity at 𝜃= 90◦in Figure 7.8(a), the array factor of (6-61a) must be multiplied by 1∕Σan = 1∕0.4482 = 2.2312.
Element Number Element Position Excitation Coefficient n z′n an ±1 ±0.25λ 0.6080 ±2 ±0.75λ −0.1295 ±3 ±1.25λ 0.0000 ±4 ±1.75λ 0.0660 ±5 ±2.25λ −0.0963 In general, the Fourier transform synthesis method yields reconstructed patterns whose mean-square error (or deviation) from the desired pattern is a minimum. However, the Woodward-Lawson synthesis method reconstructs patterns whose values at the sampled points are identical to the ones of the desired pattern; it does not have any control of the pattern between the sample points, and it does not yield a pattern with least-mean-square deviation.
Ruze points out that the least-mean-square error design is not necessarily the best. The particular application will dictate the preference between the two. However, the Fourier transform method is best suited for reconstruction of desired patterns which are analytically simple and which allow the integrations to be performed in closed form. Today, with the advancements in high-speed computers, this is not a major restriction since the integration can be performed (with high efficiency) numerically. In contrast, the Woodward-Lawson method is more flexible, and it can be used to synthesize any desired pattern. In fact, it can even be used to reconstruct patterns which, because of their complicated nature, cannot be expressed analytically. Measured patterns, either of analog or digital form, can also be synthesized using the Woodward-Lawson method.
7.6 TAYLOR LINE-SOURCE (TSCHEBYSCHEFF-ERROR) In Chapter 6 we discussed the classic Dolph-Tschebyscheff array design which yields, for a given sidelobe level, the smallest possible first-null beamwidth (or the smallest possible sidelobe level for a given first-null beamwidth). Another classic design that is closely related to it, but is more applicable TAYLOR LINE-SOURCE (TSCHEBYSCHEFF-ERROR) 405 for continuous distributions, is that by Taylor (this method is different from that by Taylor which will be discussed in the next section).
The Taylor design yields a pattern that is an optimum compromise between beamwidth and sidelobe level. In an ideal design, the minor lobes are maintained at an equal and specific level.
Since the minor lobes are of equal ripple and extend to infinity, this implies an infinite power. More realistically, however, the technique as introduced by Taylor leads to a pattern whose first few minor lobes (closest to the main lobe) are maintained at an equal and specified level; the remaining lobes decay monotonically. Practically, even the level of the closest minor lobes exhibits a slight monotonic decay. This decay is a function of the space u over which these minor lobes are required to be maintained at an equal level. As this space increases, the rate of decay of the closest minor lobes decreases. For a very large space of u (over which the closest minor lobes are required to have an equal ripple), the rate of decay is negligible. It should be pointed out, however, that the other method by Taylor (of Section 7.7) yields minor lobes, all of which decay monotonically.
The details of the analytical formulation are somewhat complex (for the average reader) and lengthy, and they will not be included here. The interested reader is referred to the literature , .
Instead, a succinct outline of the salient points of the method and of the design procedure will be included. The design is for far-field patterns, and it is based on the formulation of (7-1).
Ideally the normalized space factor that yields a pattern with equal-ripple minor lobes is given by SF(𝜃) = cosh[ √ (𝜋A)2 −u2] cosh(𝜋A) (7-25) u = 𝜋l λ cos 𝜃 (7-25a) whose maximum value occurs when u = 0. The constant A is related to the maximum desired sidelobe level R0 by cosh(𝜋A) = R0 (voltage ratio) (7-26) The space factor of (7-25) can be derived from the Dolph-Tschebyscheff array formulation of Section 6.8.3, if the number of elements of the array are allowed to become infinite.
Since (7-25) is ideal and cannot be realized physically, Taylor suggested that it be approxi-mated (within a certain error) by a space factor comprised of a product of factors whose roots are the zeros of the pattern. Because of its approximation to the ideal Tschebyscheff design, it is also referred to as Tschebyscheff-error. The Taylor space factor is given by SF(u, A, n) = sin(u) u n−1 ∏ n=1 [ 1 − ( u un )2] n−1 ∏ n=1 [ 1 − ( u n𝜋 )2] (7-27) u = 𝜋v = 𝜋l λ cos 𝜃 (7-27a) un = 𝜋vn = 𝜋l λ cos 𝜃n (7-27b) 406 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES where 𝜃n represents the locations of the nulls. The parameter n is a constant chosen by the designer so that the minor lobes for |v| = |u∕𝜋| ≤n are maintained at a nearly constant voltage level of 1∕R0 while for |v| = |u∕𝜋| > n the envelope, through the maxima of the remaining minor lobes, decays at a rate of 1∕v = 𝜋∕u. In addition, the nulls of the pattern for |v| ≥n occur at integer values of v.
In general, there are n −1 inner nulls for |v| < n and an infinite number of outer nulls for |v| ≥n.
To provide a smooth transition between the inner and the outer nulls (at the expense of slight beam broadening), Taylor introduced a parameter 𝜎. It is usually referred to as the scaling factor, and it spaces the inner nulls so that they blend smoothly with the outer ones. In addition, it is the factor by which the beamwidth of the Taylor design is greater than that of the Dolph-Tschebyscheff, and it is given by 𝜎= n √ A2 + ( n −1 2 )2 (7-28) The location of the nulls are obtained using un = 𝜋vn = 𝜋l λ cos 𝜃n = ⎧ ⎪ ⎨ ⎪ ⎩ ±𝜋𝜎 √ A2 + ( n −1 2 )2 1 ≤n < n ±n𝜋 n ≤n ≤∞ (7-29) The normalized line-source distribution, which yields the desired pattern, is given by I(z′) = λ l ⎡ ⎢ ⎢ ⎣ 1 + 2 n−1 ∑ p=1 SF(p, A, n) cos ( 2𝜋pz′ l )⎤ ⎥ ⎥ ⎦ (7-30) The coefficients SF(p, A, n) represent samples of the Taylor pattern, and they can be obtained from (7-27) with u = 𝜋p. They can also be found using SF(p, A, n) = ⎧ ⎪ ⎨ ⎪ ⎩ [(n −1)!]2 (n −1 + p)!(n −1 −p)!
n−1 ∏ m=1 [ 1 − (𝜋p um )2] |p| < n 0 |p| ≥n (7-30a) with SF(−p, A, n) = SF(p, A, n).
The half-power beamwidth is given approximately by Θ0 ≃2 sin−1 ⎧ ⎪ ⎨ ⎪ ⎩ λ𝜎 𝜋l ⎡ ⎢ ⎢ ⎣ (cosh−1R0)2 − ( cosh−1 R0 √ 2 )2⎤ ⎥ ⎥ ⎦ 1∕2⎫ ⎪ ⎬ ⎪ ⎭ (7-31) 7.6.1 Design Procedure To initiate a Taylor design, you must 1. specify the normalized maximum tolerable sidelobe level 1∕R0 of the pattern.
TAYLOR LINE-SOURCE (TSCHEBYSCHEFF-ERROR) 407 2. choose a positive integer value for n such that for |v| = |(l∕λ) cos 𝜃| ≤n the normalized level of the minor lobes is nearly constant at 1∕R0. For |v| > n, the minor lobes decrease monotonically.
In addition, for |v| < n there exist (n −1) nulls. The position of all the nulls is found using (7-29). Small values of n yield source distributions which are maximum at the center and monotonically decrease toward the edges. In contrast, large values of n result in sources which are peaked simultaneously at the center and at the edges, and they yield sharper main beams.
Therefore, very small and very large values of n should be avoided. Typically, the value of n should be at least 3 and at least 6 for designs with sidelobes of −25 and −40 dB, respectively.
To complete the design, you do the following: 1. Determine A using (7-26), 𝜎using (7-28), and the nulls using (7-29).
2. Compute the space factor using (7-27), the source distribution using (7-30) and (7-30a), and the half-power beamwidth using (7-31).
Example 7.6 Design a −20 dB Taylor, Tschebyscheff-error, distribution line-source with n = 5. Plot the pattern and the current distribution for l = 7λ(−7 ≤v = u∕𝜋≤7).
Solution: For a −20 dB sidelobe level R0 (voltage ratio) = 10 Using (7-26) A = 1 𝜋cosh−1(10) = 0.95277 and by (7-28) 𝜎= 5 √ (0.95277)2 + (5 −0.5)2 = 1.0871 The nulls are given by (7-29) or vn = un∕𝜋= ±1.17, ±1.932, ±2.91, ±3.943, ±5.00, ±6.00, ±7.00, … The corresponding null angles for l = 7λ are 𝜃n = 80.38◦(99.62◦), 73.98◦(106.02◦), 65.45◦(114.55◦), 55.71◦(124.29◦), 44.41◦(135.59◦), and 31.00◦(149.00◦) The half-power beamwidth for l = 7λ is found using (7-31), or Θ0 ≃7.95◦ The source distribution, as computed using (7-30) and (7-30a), is displayed in Figure 7.9(a). The corresponding radiation pattern for −7 ≤v = u∕𝜋≤7 (0◦≤𝜃≤180◦for l = 7λ) is shown in Figure 7.9(b).
All the computed parameters compare well with results reported in and .
408 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES Normalized current I(z′) Normalized amplitude (dB down) Position on Source (z′) (b) Space factor 1.0 0.8 0.6 0.4 –l/2 (–3.5λ) –l/4 (–1.75λ) v = (l/λ) cos θ 0 l/4 (1.75λ) l/2 (3.5λ) 0.2 Observation angle θ (l = 7λ) (a) Current distribution 10 20 30 –7 180° 149° 124.8° 106.6° 90° 73.4° 55.2° 31° 0° –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 Figure 7.9 Normalized current distribution and far-field space factor pattern for a −20 dB sidelobe and n = 5 Taylor (Tschebyscheff-error) line-source of l = 7λ.
7.7 TAYLOR LINE-SOURCE (ONE-PARAMETER) The Dolph-Tschebyscheff array design of Section 6.8.3 yields minor lobes of equal intensity while the Taylor (Tschebyscheff-error) produces a pattern whose inner minor lobes are maintained at a constant level and the remaining ones decrease monotonically. For some applications, such as radar and low-noise systems, it is desirable to sacrifice some beamwidth and low inner minor lobes to have TAYLOR LINE-SOURCE (ONE-PARAMETER) 409 all the minor lobes decay as the angle increases on either side of the main beam. In radar applications this is preferable because interfering or spurious signals would be reduced further when they try to enter through the decaying minor lobes. Thus any significant contributions from interfering signals would be through the pattern in the vicinity of the major lobe. Since in practice it is easier to maintain pattern symmetry around the main lobe, it is also possible to recognize that such signals are false targets. In low-noise applications, it is also desirable to have minor lobes that decay away from the main beam in order to diminish the radiation accepted through them from the relatively “hot” ground.
A continuous line-source distribution that yields decaying minor lobes and, in addition, controls the amplitude of the sidelobe is that introduced by Taylor in an unpublished classic memorandum.
It is referred to as the Taylor (one-parameter) design and its source distribution is given by In(z′) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ J0 ⎡ ⎢ ⎢ ⎣ j𝜋B √ 1 − ( 2z′ l )2⎤ ⎥ ⎥ ⎦ −l∕2 ≤z′ ≤+ l∕2 0 elsewhere (7-32) where J0 is the Bessel function of the first kind of order zero, l is the total length of the continuous source [see Figure 7.1(a)], and B is a constant to be determined from the specified sidelobe level.
The space factor associated with (7-32) can be obtained by using (7-1). After some intricate math-ematical manipulations, utilizing Gegenbauer’s finite integral and Gegenbauer polynomials , the space factor for a Taylor amplitude distribution line-source with uniform phase [𝜙n(z′) = 𝜙0 = 0] can be written as SF(𝜃) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ lsinh[ √ (𝜋B)2 −u2] √ (𝜋B)2 −u2 , u2 < (𝜋B)2 lsin[ √ u2 −(𝜋B)2] √ u2 −(𝜋B)2 , u2 > (𝜋B)2 (7-33) where u = 𝜋l λ cos 𝜃 (7-33a) B = constant determined from sidelobe level l = line-source dimension The derivation of (7-33) is assigned as an exercise to the reader (Problem 7.28). When (𝜋B)2 > u2, (7-33) represents the region near the main lobe. The minor lobes are represented by (𝜋B)2 < u2 in (7-33). Either form of (7-33) can be obtained from the other by knowing that (see Appendix VI) sin(jx) = j sinh(x) sinh(jx) = j sin(x) (7-34) 410 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES When u = 0 (𝜃= 𝜋∕2 and maximum radiation), the normalized pattern height is equal to (SF)max = sinh(𝜋B) 𝜋B = H0 (7-35) For u2 ≫(𝜋B)2, the normalized form of (7-33) reduces to SF(𝜃) = sin[ √ u2 −(𝜋B)2] √ u2 −(𝜋B)2 ≃sin(u) u u ≫𝜋B (7-36) and it is identical to the pattern of a uniform distribution. The maximum height H1 of the sidelobe of (7-36) is H1 = 0.217233 (or 13.2 dB down from the maximum), and it occurs when (see Appendix I) [u2 −(𝜋B)2]1∕2 ≃u = 4.494 (7-37) Using (7-35), the maximum voltage height of the sidelobe (relative to the maximum H0 of the major lobe) is equal to H1 H0 = 1 R0 = 0.217233 sinh(𝜋B)∕(𝜋B) (7-38) or R0 = 1 0.217233 sinh(𝜋B) 𝜋B = 4.603sinh(𝜋B) 𝜋B (7-38a) Equation (7-38a) can be used to find the constant B when the intensity ratio R0 of the major-to-the-sidelobe is specified. Values of B for typical sidelobe levels are Sidelobe Level (dB) −10 −15 −20 −25 −30 −35 −40 B j0.4597 0.3558 0.7386 1.0229 1.2761 1.5136 1.7415 The disadvantage of designing an array with decaying minor lobes as compared to a design with equal minor lobe level (Dolph-Tschebyscheff), is that it yields about 12 to 15% greater half-power beamwidth. However such a loss in beamwidth is a small penalty to pay when the extreme minor lobes decrease as 1/u.
To illustrate the principles, let us consider an example.
Example 7.7 Given a continuous line-source, whose total length is 4λ, design a Taylor, one-parameter, distri-bution array whose sidelobe is 30 dB down from the maximum of the major lobe.
a. Find the constant B.
b. Plot the pattern (in dB) of the continuous line-source distribution.
c. For a spacing of λ∕4 between the elements, find the number of discrete isotropic elements needed to approximate the continuous source. Assume that the two extreme elements are placed at the edges of the continuous line source.
TAYLOR LINE-SOURCE (ONE-PARAMETER) 411 d. Find the normalized coefficients of the discrete array of part (c).
e. Write the array factor of the discrete array of parts (c) and (d).
f. Plot the array factor (in dB) of the discrete array of part (e).
g. For a corresponding Dolph-Tschebyscheff array, find the normalized coefficients of the discrete elements.
h. Compare the patterns of the Taylor continuous line-source distribution and discretized array, and the corresponding Dolph-Tschebyscheff discrete-element array.
Solution: For a −30 dB maximum sidelobe, the voltage ratio of the major-to-the-sidelobe level is equal to 30 = 20 log10 (R0) ➱R0 = 31.62 a. The constant B is obtained using (7-38a) or R0 = 31.62 = 4.603sinh(𝜋B) 𝜋B ➱B = 1.2761 b. The normalized space factor pattern is obtained using (7-33), and it is shown plotted in Figure 7.10.
c. For d = λ∕4 and with elements at the extremes, the number of elements is 17.
Continuous (l = 4 ) Discretized (l = 4 , d = /4, N = 17) Amplitude Pattern (dB) –80 –70 –60 –50 –40 –20 –10 0 Amplitude Pattern (dB) –80 –70 –60 –50 –40 –30 –20 –10 0 θ θ λ λ λ 60° 90° 120° 150° 180° 150° 120° 90° 30° 60° 30° 0° Figure 7.10 Far-field amplitude patterns of continuous and discretized Taylor (one-parameter) dis-tributions.
412 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES d. The coefficients are obtained using (7-32). Since we have an odd number of elements, their positioning and excitation coefficients are those shown in Figure 6.19(b). Thus the total excitation coefficient of the center element is 2a1 = In(z′)|z′=0 = J0(j4.009) = 11.400 ➱a1 = 5.70 The coefficients of the elements on either side of the center element are identical (because of symmetry), and they are obtained from a2 = I(z′)|z′=±λ∕4 = J0(j3.977) = 11.106 The coefficients of the other elements are obtained in a similar manner, and they are given by a3 = 10.192 a4 = 8.889 a5 = 7.195 a6 = 5.426 a7 = 3.694 a8 = 2.202 a9 = 1.000 e. The array factor is given by (6-61b) and (6-61c), or (AF)17 = 9 ∑ n=1 an cos[2(n −1)u] u = 𝜋d λ cos 𝜃= 𝜋 4 cos 𝜃 where the coefficients (an’s) are those found in part (d).
f. The normalized pattern (in dB) of the discretized distribution (discrete-element array) is shown in Figure 7.10.
g. The normalized coefficients of a 17-element Dolph-Tschebyscheff array, with −30 dB side-lobes, are obtained using the method outlined in the Design Section of Section 6.8.3 and are given by Unnormalized Normalized a1 = 2.858 a1n = 1.680 a2 = 5.597 a2n = 3.290 a3 = 5.249 a3n = 3.086 a4 = 4.706 a4n = 2.767 a5 = 4.022 a5n = 2.364 a6 = 3.258 a6n = 1.915 a7 = 2.481 a7n = 1.459 a8 = 1.750 a8n = 1.029 a9 = 1.701 a9n = 1.000 TAYLOR LINE-SOURCE (ONE-PARAMETER) 413 As with the discretized Taylor distribution array, the coefficients are symmetrical, and the form of the array factor is that given in part (e).
h. The normalized pattern (in dB) is plotted in Figure 7.11 where it is compared with that of the discretized Taylor distribution. From the patterns in Figures 7.10 and 7.11, it can be concluded that 1. the main lobe of the continuous line-source Taylor design is well approximated by the discretized distribution with a λ∕4 spacing between the elements. Even the minor lobes are well represented, and a better approximation can be obtained with more elements and smaller spacing between them.
2. the Taylor distribution array pattern has a wider main lobe than the corresponding Dolph-Tschebyscheff, but it displays decreasing minor lobes away from the main beam.
Dolph-Tschebyscheff Taylor (one-parameter); Discretized Amplitude Pattern (dB) –80 –70 –60 –50 –40 –30 –20 –10 30° 60° 90° 120° 150° 180° 150° 120° 90° 60° 30° 0° 0 Amplitude Pattern (dB) –80 –70 –60 –50 –40 –30 –20 –10 0 θ θ Figure 7.11 Far-field amplitude patterns of Taylor (discretized) and Dolph-Tschebyscheff distributions (l = 4λ, d = λ∕4, N = 17).
A larger spacing between the elements does not approximate the continuous distribution as accu-rately. The design of Taylor and Dolph-Tschebyscheff arrays for l = 4λ and d = λ∕2(N = 9) is assigned as a problem at the end of the chapter (Problem 7.38).
To qualitatively assess the performance between uniform, binomial, Dolph-Tschebyscheff, and Taylor (one-parameter) array designs, the amplitude distribution of each has been plotted in Fig-ure 7.12(a). It is assumed that l = 4λ, d = λ∕4, N = 17, and the maximum sidelobe is 30 dB down.
The coefficients are normalized with respect to the amplitude of the corresponding element at the center of that array.
414 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES Figure 7.12 Amplitude distribution of nonuniform amplitude linear arrays.
The binomial design possesses the smoothest amplitude distribution (between 1 and 0) from the center to the edges (the amplitude toward the edges is vanishingly small). Because of this characteristic, the binomial array displays the smallest sidelobes followed, in order, by the Taylor, Tschebyscheff, and the uniform arrays. In contrast, the uniform array possesses the smallest half-power beamwidth followed, in order, by the Tschebyscheff, Taylor, and binomial arrays. As a rule of thumb, the array with the smoothest amplitude distribution (from the center to the edges) has the smallest sidelobes and the larger half-power beamwidths. The best design is a trade-off between sidelobe level and beamwidth.
TRIANGULAR, COSINE, AND COSINE-SQUARED AMPLITUDE DISTRIBUTIONS 415 TABLE 7.1 Radiation Characteristics for Line-Sources and Linear Arrays with Uniform, Triangular, Cosine, and Cosine-Squared Distributions Distribution Uniform Triangular Cosine Cosine-Squared Distribution In (analytical) I0 I1 ( 1 −2 l |z′| ) I2 cos (𝜋 l z′) I3 cos2 (𝜋 l z′) Distribution (graphical) Space factor (SF) u = (𝜋l λ ) cos 𝜃 I0lsin(u) u I1 l 2 ⎡ ⎢ ⎢ ⎢ ⎣ sin (u 2 ) u 2 ⎤ ⎥ ⎥ ⎥ ⎦ 2 I2l𝜋 2 cos(u) (𝜋∕2)2 −u2 I3 l 2 sin(u) u [ 𝜋2 𝜋2 −u2 ] Space factor |SF| Half-power beamwidth (degrees) l ≫λ 50.6 (l∕λ) 73.4 (l∕λ) 68.8 (l∕λ) 83.2 (l∕λ) First-null beamwidth (degrees) l ≫λ 114.6 (l∕λ) 229.2 (l∕λ) 171.9 (l∕λ) 229.2 (l∕λ) First sidelobe max. (to main max.) (dB) −13.2 −26.4 −23.2 −31.5 Directivity factor (l large) 2 ( l λ ) 0.75 [ 2 ( l λ )] 0.810 [ 2 ( l λ )] 0.667 [ 2 ( l λ )] A MATLAB computer program entitled Synthesis has been developed, and it performs synthe-sis using the Schelkunoff, Fourier transform, Woodward-Lawson, Taylor (Tschebyscheff-error) and Taylor (one-parameter) methods. The program is included in the publisher’s website for this the book. The description of the program is provided in the corresponding READ ME file.
7.8 TRIANGULAR, COSINE, AND COSINE-SQUARED AMPLITUDE DISTRIBUTIONS Some other very common and simple line-source amplitude distributions are those of the triangular, cosine, cosine-squared, cosine on-a-pedestal, cosine-squared on-a-pedestal, Gaussian, inverse taper, and edge. Instead of including many details, the pattern, half-power beamwidth, first-null beamwidth, magnitude of sidelobes, and directivity for uniform, triangular, cosine, and cosine-squared amplitude distributions (with constant phase) are summarized in Table 7.1 , .
The normalized coefficients for a uniform, triangular, cosine, and cosine-squared arrays of l = 4λ, d = λ∕4, N = 17 are shown plotted in Figure 7.12(b). The array with the smallest sidelobes and the larger half-power beamwidth is the cosine-squared, because it possesses the smoothest distribu-tion. It is followed, in order, by the triangular, cosine, and uniform distributions. This is verified by examining the characteristics in Table 7.1.
Cosine on-a-pedestal distribution is obtained by the superposition of the uniform and the cosine distributions. Thus it can be represented by In(z′) = ⎧ ⎪ ⎨ ⎪ ⎩ I0 + I2 cos (𝜋 l z′) , −l∕2 ≤z′ ≤l∕2 0 elsewhere (7-39) 416 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES where I0 and I2 are constants. The space factor pattern of such a distribution is obtained by the addition of the patterns of the uniform and the cosine distributions found in Table 7.1. That is, SF(𝜃) = I0lsin(u) u + I2 𝜋l 2 cos u (𝜋∕2)2 −u2 (7-40) A similar procedure is used to represent and analyze a cosine-squared on-a-pedestal distribution.
7.9 LINE-SOURCE PHASE DISTRIBUTIONS The amplitude distributions of the previous section were assumed to have uniform phase variations throughout the physical extent of the source. Practical radiators (such as reflectors, lenses, horns, etc.) have nonuniform phase fronts caused by one or more of the following: 1. displacement of the reflector feed from the focus 2. distortion of the reflector or lens surface 3. feeds whose wave fronts are not ideally cylindrical or spherical (as they are usually presumed to be) 4. physical geometry of the radiator These are usually referred to phase errors, and they are more evident in radiators with tilted beams.
To simplify the analytical formulations, most of the phase fronts are represented with linear, quadratic, or cubic distributions. Each of the phase distributions can be associated with each of the amplitude distributions. In (7-1), the phase distribution of the source is represented by 𝜙n(z′). For linear, quadratic, and cubic phase variations, 𝜙n(z′) takes the form of linear: 𝜙1(z′) = 𝛽1 2 l z′, −l∕2 ≤z′ ≤l∕2 quadratic: 𝜙2(z′) = 𝛽2 (2 l )2 z′2, −l∕2 ≤z′ ≤l∕2 cubic: 𝜙3(z′) = 𝛽3 (2 l )3 z′3, −l∕2 ≤z′ ≤l∕2 (7-41a) (7-41b) (7-41c) and it is shown plotted in Figure 7.13. The quadratic distribution is used to represent the phase varia-tions at the aperture of a horn and of defocused (along the symmetry axis) reflector and lens antennas.
The space factor patterns corresponding to the phase distributions of (7-41a)–(7-41c) can be obtained by using (7-1). Because the analytical formulations become lengthy and complex, espe-cially for the quadratic and cubic distributions, they will not be included here. Instead, a general guideline of their effects will be summarized , .
Linear phase distributions have a tendency to tilt the main beam of an antenna by an angle 𝜃0 and to form an asymmetrical pattern. The pattern of this distribution can be obtained by replacing the u (for uniform phase) in Table 7.1 by (u −𝜃0). In general, the half-power beamwidth of the tilted pattern is increased by 1/cos 𝜃0 while the directivity is decreased by cos 𝜃0. This becomes more apparent by realizing that the projected length of the line-source toward the maximum is reduced by cos 𝜃0. Thus the effective length of the source is reduced.
Quadratic phase errors lead primarily to a reduction of directivity, and an increase in sidelobe level on either side of the main lobe. The symmetry of the original pattern is maintained. In addition, for moderate phase variations, ideal nulls in the patterns disappear. Thus the minor lobes blend into each other and into the main beam, and they represent shoulders of the main beam instead of appearing as CONTINUOUS APERTURE SOURCES 417 Figure 7.13 Linear, quadratic, and cubic phase variations.
separate lobes. Analytical formulations for quadratic phase distributions are introduced in Chapter 13 on horn antennas.
Cubic phase distributions introduce not only a tilt in the beam but also decrease the directivity. The newly formed patterns are asymmetrical. The minor lobes on one side are increased in magnitude and those on the other side are reduced in intensity.
7.10 CONTINUOUS APERTURE SOURCES Space factors for aperture (two-dimensional) sources can be introduced in a similar manner as in Section 7.2.1 for line-sources.
418 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES 7.10.1 Rectangular Aperture Referring to the geometry of Figure 6.28(b), the space factor for a two-dimensional rectangular distribution along the x-y plane is given by SF = ∫ ly∕2 −ly∕2 ∫ lx∕2 −lx∕2 An(x′, y′)ej[kx′ sin 𝜃cos 𝜙+ky′ sin 𝜃sin 𝜙+𝜙n(x′,y′)] dx′ dy′ (7-42) where lx and ly are, respectively, the linear dimensions of the rectangular aperture along the x and y axes. An(x′, y′) and 𝜙n(x′, y′) represent, respectively, the amplitude and phase distributions on the aperture.
For many practical antennas (such as waveguides, horns, etc.) the aperture distribution (amplitude and phase) is separable. That is, An(x′, y′) = Ix(x′)Iy(y′) (7-42a) 𝜙n(x′, y′) = 𝜙x(x′) + 𝜙y(y′) (7-42b) so that (7-42) can be written as SF = SxSy (7-43) where Sx = ∫ lx∕2 −lx∕2 Ix(x′)ej[kx′ sin 𝜃cos 𝜙+𝜙x(x′)] dx′ (7-43a) Sy = ∫ ly∕2 −ly∕2 Iy(y′)ej[ky′ sin 𝜃sin 𝜙+𝜙y(y′)] dy′ (7-43b) which is analogous to the array factor of (6-85)–(6-85b) for discrete-element arrays.
The evaluation of (7-42) can be accomplished either analytically or graphically. If the distribution is separable, as in (7-42a) and (7-42b), the evaluation can be performed using the results of a line-source distribution.
The total field of the aperture antenna is equal to the product of the element and space factors.
As for the line-sources, the element factor for apertures depends on the type of equivalent current density and its orientation.
7.10.2 Circular Aperture The space factor for a circular aperture can be obtained in a similar manner as for the rectangular distribution. Referring to the geometry of Figure 6.37, the space factor for a circular aperture with radius a can be written as SF(𝜃, 𝜙) = ∫ 2𝜋 0 ∫ a 0 An(𝜌′, 𝜙′)ej[k𝜌′ sin 𝜃cos(𝜙−𝜙′)+𝜁n(𝜌′,𝜙′)]𝜌′ d𝜌′ d𝜙′ (7-44) CONTINUOUS APERTURE SOURCES 419 TABLE 7.2 Radiation Characteristics for Circular Apertures and Circular Planar Arrays with Circular Symmetry and Tapered Distribution Distribution Uniform Radial Taper Radial Taper Squared Distribution (analytical) I0 [ 1 − (𝜌′ a )2]0 I1 [ 1 − (𝜌′ a )2]1 I2 [ 1 − (𝜌′ a )2]2 Distribution (graphical) Space factor (SF) u = ( 2𝜋a λ ) sin 𝜃 I02𝜋a2 J1(u) u I14𝜋a2 J2(u) u I216𝜋a2 J3(u) u Half-power beamwidth (degrees) a ≫λ 29.2 (a∕λ) 36.4 (a∕λ) 42.1 (a∕λ) First-null beamwidth (degrees) a ≫λ 69.9 (a∕λ) 93.4 (a∕λ) 116.3 (a∕λ) First sidelobe max. (to main max.) −17.6 dB −24.6 dB −30.6 dB Directivity factor (2𝜋a λ )2 0.75 (2𝜋a λ )2 0.56 (2𝜋a λ )2 where 𝜌′ is the radial distance (0 ≤𝜌′ ≤a), 𝜙′ is the azimuthal angle over the aperture (0 ≤𝜙′ ≤2𝜋 for 0 ≤𝜌′ ≤a), and An(𝜌′, 𝜙′) and 𝜁n(𝜌′, 𝜙′) represent, respectively, the amplitude and phase dis-tributions over the aperture. Equation (7-44) is analogous to the array factor of (6-112a) for dis-crete elements.
If the aperture distribution has uniform phase [𝜁n(𝜌′, 𝜙′) = 𝜁0 = 0] and azimuthal amplitude sym-metry [An(𝜌′, 𝜙′) = An(𝜌′)], (7-44) reduces, by using (5-48), to SF(𝜃) = 2𝜋∫ a 0 An(𝜌′)J0(k𝜌′ sin 𝜃)𝜌′ d𝜌′ (7-45) where J0(x) is the Bessel function of the first kind and of order zero.
Many practical antennas, such as a parabolic reflector, have distributions that taper toward the edges of the apertures. These distributions can be approximated reasonably well by functions of the form An(𝜌′) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [ 1 − (𝜌′ a )2]n 0 ≤𝜌′ ≤a, n = 0, 1, 2, 3, … 0 elsewhere (7-46) For n = 0, (7-46) reduces to a uniform distribution.
The radiation characteristics of circular apertures or planar circular arrays with distributions (7-46) with n = 0, 1, 2 are shown tabulated in Table 7.2 . It is apparent, as before, that dis-tributions with lower taper toward the edges (larger values of n) have smaller sidelobes but larger beamwidths. In design, a compromise between sidelobe level and beamwidth is necessary.
420 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES 7.11 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Java-based applet for computing and displaying the synthesis characteristics of Schelkunoff, Woodward-Lawson and Tschebyscheff-error designs.
c. Matlab computer program, designated Synthesis, for computing and displaying the radiation characteristics of r Schelkunoff r Fourier transform (line-source and linear array) r Woodward-Lawson (line-source and linear array) r Taylor (Tschebyscheff-error and One-parameter) synthesis designs.
d. Power Point (PPT) viewgraphs, in multicolor.
REFERENCES 1. S. A. Schelkunoff, “A Mathematical Theory of Linear Arrays,” Bell System Technical Journal, Vol. 22, pp. 80–107, 1943.
2. H. G. Booker and P. C. Clemmow, “The Concept of an Angular Spectrum of Plane Waves, and Its Relation to That of Polar Diagram and Aperture Distribution,” Proc. IEE (London), Paper No. 922, Radio Section, Vol. 97, pt. III, pp. 11–17, January 1950.
3. P. M. Woodward, “A Method for Calculating the Field over a Plane Aperture Required to Produce a Given Polar Diagram,” J. IEE, Vol. 93, pt. IIIA, pp. 1554–1558, 1946.
4. P. M. Woodward and J. D. Lawson, “The Theoretical Precision with Which an Arbitrary Radiation-Pattern May be Obtained from a Source of a Finite Size,” J. IEE, Vol. 95, pt. III, No. 37, pp. 363–370, September 1948.
5. T. T. Taylor, “Design of Line-Source Antennas for Narrow Beamwidth and Low Sidelobes,” IRE Trans.
Antennas Propagat., Vol. AP-3, No. 1, pp. 16–28, January 1955.
6. T. T. Taylor, “One Parameter Family of Line-Sources Producing Modified Sin(𝜋u)∕𝜋u Patterns,” Hughes Aircraft Co. Tech. Mem. 324, Culver City, Calif., Contract AF 19(604)-262-F-14, September 4, 1953.
7. R. S. Elliott, “On Discretizing Continuous Aperture Distributions,” IEEE Trans. Antennas Propagat., Vol. AP-25, No. 5, pp. 617–621, September 1977.
8. R. C. Hansen (ed.), Microwave Scanning Antennas, Vol. I, Academic Press, New York, 1964, p. 56.
9. J. Ruze, “Physical Limitations on Antennas,” MIT Research Lab., Electronics Tech. Rept. 248, October 30, 1952.
10. M. I. Skolnik, Introduction to Radar Systems, McGraw-Hill, New York, 1962, pp. 320–330.
11. R. S. Elliott, “Criticisms of the Woodward-Lawson Method,” IEEE Antennas Propagation Society Newslet-ter, Vol. 30, p. 43, June 1988.
12. H. Steyskal, “The Woodward-Lawson Method: A Second Opinion,” IEEE Antennas Propagation Society Newsletter, Vol. 30, p. 48, October 1988.
13. R. S. Elliott, “More on the Woodward-Lawson Method,” IEEE Antennas Propagation Society Newsletter, Vol. 30, pp. 28–29, December 1988.
14. H. Steyskal, “The Woodward-Lawson Method-To Bury or Not to Bury,” IEEE Antennas Propagation Soci-ety Newsletter, Vol. 31, pp. 35–36, February 1989.
15. H. J. Orchard, R. S. Elliott, and G. J. Stern. “Optimizing the Synthesis of Shaped Beam Antenna Patterns,” IEE Proceedings, Part H, pp. 63–68, 1985.
PROBLEMS 421 16. R. S. Elliott, “Design of Line-Source Antennas for Narrow Beamwidth and Asymmetric Low Sidelobes,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 1, pp. 100–107, January 1975.
17. G. N. Watson, A Treatise on the Theory of Bessel Functions, 2nd. Ed., Cambridge University Press, London, pp. 50 and 379, 1966.
18. S. Silver (ed.), Microwave Antenna Theory and Design, MIT Radiation Laboratory Series, Vol. 12, McGraw-Hill, New York, 1965, Chapter 6, pp. 169–199.
19. R. C. Johnson and H. Jasik (eds.), Antenna Engineering Handbook, 2nd. Ed., McGraw-Hill, New York, 1984, pp. 2–14 to 2-41.
PROBLEMS 7.1.
A three-element array is placed along the z-axis. Assuming the spacing between the elements is d = λ∕4 and the relative amplitude excitation is equal to a1 = 1, a2 = 2, a3 = 1, (a) find the angles where the array factor vanishes when 𝛽= 0, 𝜋∕2, 𝜋, and 3𝜋∕2 (b) plot the relative pattern for each array factor Use Schelkunoff’s method.
7.2.
Design a linear array of isotropic elements placed along the z-axis such that the zeros of the array factor occur at 𝜃= 0◦, 60◦, and 120◦. Assume that the elements are spaced λ∕4 apart and that the progressive phase shift between them is 0◦.
(a) Find the required number of elements.
(b) Determine their excitation coefficients.
(c) Write the array factor.
(d) Plot the array factor pattern to verify the validity of the design.
Verify using the computer program Synthesis.
7.3.
To minimize interference between the operational system, whose antenna is a linear array with elements placed along the z-axis, and other undesired sources of radiation, it is required that nulls be placed at elevation angles of 𝜃= 0◦, 60◦, 120◦, and 180◦. The elements will be separated with a uniform spacing of λ∕4. Choose a synthesis method that will allow you to design such an array that will meet the requirements of the amplitude pattern of the array factor. To meet the requirements: (a) Specify the synthesis method you will use.
(b) Determine the number of elements.
(c) Find the excitation coefficients.
7.4.
It is desired to synthesize a discrete array of vertical infinitesimal dipoles placed along the z-axis with a spacing of d = λ∕2 between the adjacent elements. It is desired for the array factor to have nulls along 𝜃= 60◦, 90◦, and 120◦. Assume there is no initial progressive phase excitation between the elements. To achieve this, determine: (a) number of elements.
(b) excitation coefficients.
(c) angles (in degrees) of all the nulls of the entire array (including of the actual elements).
7.5.
It is desired to synthesize a linear array of elements with spacing d = 3λ∕8. It is important that the array factor (AF) exhibits nulls along 𝜃= 0, 90, and 180 degrees. Assume there is no initial progressive phase excitation between the elements (i.e., 𝛽= 0). To achieve this design, determine: (a) The number of elements 422 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES (b) The excitation coefficients (amplitude and phase) If the design allows the progressive phase shift (𝛽) to change, while maintaining the spac-ing constant (d = 3λ∕8), (c) What would it be the range of possible values for the progressive phase shift to cause the null at 𝜃= 90 degrees disappear (to place its corresponding root outside the visible region)?
7.6.
The z-plane array factor of an array of isotropic elements placed along the z-axis is given by AF = z(z4 −1) Determine the (a) number of elements of the array. If there are any elements with zero excitation coefficients (null elements), so indicate (b) position of each element (including that of null elements) along the z axis (c) magnitude and phase (in degrees) excitation of each element (d) angles where the pattern vanishes when the total array length (including null elements) is 2λ Verify using the computer program Synthesis.
7.7.
Repeat Problem 7.6 when AF = z(z3 −1) 7.8.
The z-plane array factor of an array of isotropic elements placed along the z-axis is given by (assume 𝛽= 0) AF(z) = (z + 1)3 Determine the (a) Number of elements of the discrete array to have such an array factor.
(b) Normalized excitation coefficients of each of the elements of the array (the ones at the edges to be unity).
(c) Classical name of the array design with these excitation coefficients.
(d) Angles in theta (𝜃in degrees) of all the nulls of the array factor when the spacing d between the elements is d = λ∕2.
(e) Half-power beamwidth (in degrees) of the array factor when d = λ∕2.
(f) Maximum directivity (dimensionless and in dB) of the array factor when d = λ∕2.
7.9.
The Array Factor in the complex z-plane (z = x + jy) of a linear array, with its elements placed along the z-axis, is given by: AF(z) = (z2 + 1)(z −1) Assuming a phase excitation of 𝛽= 45◦and a spacing of d = λ∕8 between the elements, determine analytically, the: (a) Actual number of elements that contribute in the visible region (0 ≤𝜃≤180◦).
(b) Excitation coefficients of these actual number of elements.
(c) ALL the angles 𝜃(in degrees) in the visible region (0 ≤𝜃≤180◦) where the nulls will occur.
PROBLEMS 423 7.10.
The z-plane array factor of an array of isotropic elements placed along the z-axis is given by (assume 𝛽= 0) AF(z) = (z + 1)4 Determine the (a) Number of elements of the discrete array to have such an array factor.
(b) Normalized excitation coefficients of each of the elements of the array (the ones at the edges to be unity).
(c) Classical name of the array design with these excitation coefficients.
(d) Angles in theta (𝜃in degrees) of all the nulls of the array factor when the spacing d between the elements is 1. d = λ0∕4 2. d = λ0∕2 (e) Half-power beamwidth (in degrees) of the array factor when d = λ0∕2.
(f) Maximum directivity (dimensionless and in dB) of the array factor when d = λ0∕2.
7.11.
The desired array factor in complex form (z-plane) of an array, with the elements along the z-axis, is given by AF(z) = (z4 − √ 2z3 + 2z2 − √ 2z + 1) = (z2 + 1)(z2 − √ 2z + 1) = (z2 + 1) [( z2 − √ 2z + 1 2 ) + 1 2 ] = (z2 + 1) ⎡ ⎢ ⎢ ⎣ ( z − 1 √ 2 )2 + 1 2 )⎤ ⎥ ⎥ ⎦ where z = x + jy in the complex z-plane.
(a) Determine the number of elements that you will need to realize this array factor.
(b) Determine all the roots of this array factor on the unity circle of the complex plane.
(c) For a spacing of d = λ∕4 and zero initial phase (𝛽= 0), determine all the angles 𝜃(in degrees) where this pattern possesses nulls.
7.12.
The z-plane (z = x + jy) array factor of a linear array of elements placed along the z-axis, with a uniform spacing d between them and with 𝛽= 0, is given by AF = z(z2 + 1) Determine, analytically, the (a) number of elements of the array; (b) excitation coefficients; (c) all the roots of the array factor in the visible region only (0 ≤𝜃≤180◦) when d = λ∕4; (d) all the nulls of the array factor (in degrees) in the visible region only (0 ≤𝜃≤180◦) when d = λ∕4.
Verify using the computer program Synthesis.
7.13.
The array factor, in the complex z-plane (z = x + jy), of a linear array with elements placed along the z-axis is given by AF = z4 −1 424 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES For a uniform spacing of d = λ∕4 between the elements and assuming a phase excitation of 𝛽= 0, determine, analytically: (a) The number of elements of the array, including those that may have null excitation.
(b) The excitation coefficient of each element, including any if they are zero.
(c) All the roots of the array factor.
(d) All the roots of the array factor in the visible region.
(e) All the nulls of the array factor (in degrees) only in the visible region (0◦≤𝜃≤180◦).
7.14.
The roots of the array factor of a linear array in the z-plane (z = x + jy) array, with the ele-ments placed along the z-axis and with a uniform spacing of d = λ∕2 between them, are given by z1 = +1, z2 = −1, z3 = +j, z4 = −j Assuming a phase excitation of 𝛽= 0, determine, analytically, the: (a) Array factor.
(b) Number of elements of the array.
(c) Excitation coefficients of each element. Indicate any that may be zero/null.
(d) All the nulls of the array factor (in degrees) only in the visible (0 ≤𝜃≤180◦).
7.15.
It is desired to design a linear array, with the elements along the z-axis, with a spacing of d = λ∕2 and a progressive phase excitation of 𝛽= 0◦between the elements. It is also neces-sary that the array exhibits nulls along 𝜃= 60◦, 90◦and 120◦.
(a) Select an array design that will meet these specifications (give its name).
(b) Determine the array factor (in expanded polynomial form) expressed in the complex z-plane.
(c) How many elements are necessary to meet the desired specifications?
(d) State the amplitude excitation coefficients for each of the array elements.
7.16.
Repeat Example 7.2 when SF(𝜃) = { 1 40◦≤𝜃≤140◦ 0 elsewhere Verify using the computer program Synthesis.
7.17.
Repeat the Fourier transform design of Example 7.2 for a line-source along the z-axis whose sectoral pattern is given by SF(𝜃) = { 1 60◦≤𝜃≤120◦ 0 elsewhere Use l = 5λ and 10λ. Compare the reconstructed patterns with the desired one.
Verify using the computer program Synthesis.
7.18.
Repeat the Fourier transform design of Problem 7.17 for a linear array with a spacing of d = λ∕2 between the elements and (a) N = 11 elements (b) N = 21 elements PROBLEMS 425 7.19.
Repeat the design of Problem 7.17 using the Woodward-Lawson method for line sources.
7.20.
Repeat the design of Problem 7.18 using the Woodward-Lawson method for linear arrays for N = 10, 20.
7.21.
Design, using the Woodward-Lawson method, a line-source of l = 5λ whose space factor pattern is given by SF(𝜃) = sin3(𝜃) 0◦≤𝜃≤180◦ Determine the current distribution and compare the reconstructed pattern with the desired pat-tern. Verify using the computer program Synthesis.
7.22.
The desired Space Factor SF of a line-source of l = 3λ is the sectoral pattern of SF = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 30𝜃−2, 60◦≤𝜃≤90◦ −1 30𝜃+ 4, 90◦≤𝜃≤120◦ (𝜃is in degrees) 0 elsewhere SF 1 60° 90° 120° θ It is required to synthesize the desired pattern using the Woodward–Lawson method with odd samples. To accomplish this, determine analytically when l = 3λ, the (a) number of samples; (b) all the angles (in degrees) where the SF is sampled; (c) all the excitation coefficients bm at the sampling points.
Verify using the computer program Synthesis.
7.23.
The space factor of a linear array of discrete elements placed along the z-axis, with a phase of 𝛽= 0 between the elements, is given by the polynomial SF = (z2 + 1)(z + 1) Determine the: (a) Number of elements necessary.
(b) Corresponding excitation coefficients of the elements.
(c) All the nulls in 𝜃(in degrees) (0◦≤𝜃≤180◦) when the spacing between the elements is d = λ∕4.
(d) All the nulls in 𝜃(in degrees) (0◦≤𝜃≤180◦) when the spacing between the elements is d = λ∕2.
7.24.
Repeat Problem 7.22 using the Woodward-Lawson method with even samples.
426 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES 7.25.
Using the Woodward-Lawson method, design a line source with its length along the z-axis, whose Space Factor is represented by the normalized triangular distribution shown below.
Assuming the length of the line source is l = 4λ and the observation angular region is 0◦≤ 𝜃≤180◦: (a) Write expressions, in terms of the obser-vation angle 𝜃(in degrees), that represent the Space Factor that is displayed graph-ically below.
(b) Assuming odd samples, determine the angles 𝜃(in degrees) where the Space Factor should be sampled.
(c) Determine the normalized amplitude of the Space Factor at the sampled angles.
1.0 SF( ) 0° 90° 180° θ θ 7.26.
Repeat the design of Problem 7.21 for a linear array of N = 10 elements with a spacing of d = λ∕2 between them.
7.27.
It is desired to synthesize the space factor of a line source, of length l = 3λ, whose normalized ideal space factor is SF = sin2(𝜃) Using the Woodward-Lawson method and assuming even samples, determine analytically the: (a) Total number of samples.
(b) All the angles (in degrees) where the SF should be sampled.
(c) All the excitation coefficients bm of the corresponding composing functions at the sam-pling points. If there are any excitation coefficients which are null (zero), state them.
7.28.
In target-search, grounding-mapping radars, and in airport beacons it is desirable to have the echo power received from a target, of constant cross section, to be independent of its range R.
Generally, the far-zone field radiated by an antenna is given by |E(R, 𝜃, 𝜙)| = C0 |F(𝜃, 𝜙)| R where C0 is a constant. According to the geometry of the figure R = h∕sin(𝜃) = h csc (𝜃) For a constant value of 𝜙, the radiated field expression reduces to |E(R, 𝜃, 𝜙= 𝜙0)| = C0 |F(𝜃, 𝜙= 𝜙0)| R = C1 |f(𝜃)| R A constant value of field strength can be maintained provided the radar is flying at a con-stant altitude h and the far-field antenna pattern is equal to f (𝜃) = C2 csc (𝜃) This is referred to as a cosecant pattern, and it is used to compensate for the range variations.
For very narrow beam antennas, the total pattern is approximately equal to the space or array PROBLEMS 427 factor. Design a line-source, using the Woodward-Lawson method, whose space factor is given by SF(𝜃) = { 0.342 csc (𝜃), 20◦≤𝜃≤60◦ 0 elsewhere Plot the synthesized pattern for l = 20λ, and compare it with the desired pattern. Verify using the computer program Synthesis.
7.29.
Repeat the design of Problem 7.28 for a linear array of N = 41 elements with a spacing of d = λ∕2 between them.
7.30.
For some radar search applications, it is more desirable to have an antenna which has a square beam for 0 ≤𝜃≤𝜃0, a cosecant pattern for 𝜃0 ≤𝜃≤𝜃m, and it is zero elsewhere. Design a line-source, using the Woodward-Lawson method, with a space factor of SF(𝜃) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 15◦≤𝜃< 20◦ 0.342 csc (𝜃) 20◦≤𝜃≤60◦ 0 elsewhere Plot the reconstructed pattern for l = 20λ, and compare it with the desired pattern.
7.31.
To maximize the aperture efficiency, and thus the directivity, of a paraboloidal (parabola of revolution) reflector, the desired normalized space factor power pattern g(𝜃) of the feed antenna is given by SF(𝜃) = g(𝜃) = sec4 (90◦−𝜃 2 ) 60◦≤𝜃≤120◦ It is desired to synthesize the desired antenna feed pattern using the Woodward-Lawson method with a line source of total length 3λ. Assuming odd samples, determine analytically the: (a) Number of samples.
(b) All the angles (in degrees) where the SF should be sampled.
(c) All the normalized excitation coefficients bm of the corresponding composing functions at the sampling points. If there are any excitation coefficients which are null (zero), so indicate.
60° 1.0 sec4 90° 120° θ SF( ) = g( ) θ θ 2 90° – θ 7.32.
Repeat the design of Problem 7.30, using the Woodward-Lawson method, for a linear array of 41 elements with a spacing of d = λ∕2 between them.
428 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES 7.33.
Repeat Problem 7.22 for a linear array of l = 3λ with a spacing of d = 0.5λ between the elements. Replacing the space factor (SF) of Problem 7.22 with an array factor (AF) and using the Woodward-Lawson method with odd samples, determine analytically the (a) number of elements; (b) number of samples; (c) angles (in degrees) where the AF is sampled; (d) all the excitation coefficients bm at the sampling points.
(e) all the normalized excitation coefficients an of the elements.
Verify using the computer program Synthesis.
7.34.
Repeat Problem 7.33 using the Woodward-Lawson method with even samples.
7.35.
Design a Taylor (Tschebyscheff-error) line-source with a (a) −25 dB sidelobe level and n = 5 (b) −20 dB sidelobe level and n = 10 For each, find the half-power beamwidth and plot the normalized current distribution and the reconstructed pattern when l = 10λ. Verify using the computer program Synthesis.
7.36.
Derive (7-33) using (7-1), (7-32), and Gegenbauer’s finite integral and polynomials.
7.37.
Design a Taylor (Tschebyscheff Error) line source, with its length placed along the z-axis, so that the sidelobe level of the pattern for the 2 inner nulls is maintained ideally at −40 dB.
Determine the: (a) Constant A so that the sidelobes of the line source are maintained at the −40 dB level.
(b) Scaling factor so that the inner most nulls blend with the outer most.
(c) Angles 𝜃(in degrees) (0◦≤𝜃≤180◦) where ALL the inner and outer most nulls occur.
Assume the length l of the line source is l = 3λ.
7.38.
Repeat the design of Example 7.7 for an array with l = 4λ, d = λ∕2, N = 9.
7.39.
Using a spacing of d = λ∕4 between the elements, determine for a Taylor Line-Source (Tschebyscheff-error) of −30 dB, for n = 4 and an equivalent length equal to that of 20 dis-crete elements: (a) The half-power beamwidth (in degrees).
(b) State whether the half-power beamwidth in Part a is larger or smaller than that of an equivalent design for a Dolph-Tschebyscheff array with the same sidelobe level. Explain as to why one is larger or smaller than the other.
(c) The number of complete innermost minor lobes that would have approximately equal sidelobe level.
7.40.
It is desired to synthesize a Taylor (Tschebyscheff-Error) line-source continuous distribution source with a −26 dB side lobe level. The total length of the array is 5λ, and it is desired to maintain −26 dB side lobe level for n = 3.
Determine the half-power beamwidth (in degrees) of the: (a) Taylor (Tschebyscheff-Error) design.
(b) Corresponding Dolph-Tschebyscheff array design with the same side lobe level.
7.41.
Design a broadside five-element, −40 dB sidelobe level Taylor (one-parameter) distribution array of isotropic sources. The elements are placed along the x-axis with a spacing of λ∕4 between them. Determine the (a) normalized excitation coefficients (amplitude and phase) of each element (b) array factor Verify using the computer program Synthesis.
PROBLEMS 429 7.42.
Given a continuous line-source, whose total length is 4λ, design a symmetrical Taylor One-Parameter distribution array whose sidelobe is 35 dB down from the maximum of the major lobe.
(a) Find the constant B.
(b) For a spacing of λ between the elements, find the number of discrete elements needed to approximate the continuous source. Assume that the two extreme elements are placed at the edges of the continuous line source.
(c) Find the total normalized coefficients (edge elements to be unity) of the discrete array of part (c). Identify the position of the corresponding elements.
(d) Write the array factor of the discrete array of parts (c) and (d).
Verify using the computer program Synthesis.
7.43.
Derive the space factors for uniform, triangular, cosine, and cosine-squared line-source con-tinuous distributions. Compare with the results in Table 7.1.
7.44.
Compute the half-power beamwidth, first-null beamwidth, first sidelobe level (in dB), and directivity of a linear array of closely spaced elements with overall length of 4λ when its amplitude distribution is (a) uniform (b) triangular (c) cosine (d) cosine-squared 7.45.
Design a continuous line-source of length l = 5λ whose half-power beamwidth of the space factor is 16.64◦. Determine the: (a) Amplitude tapering over the length of the line source that will lead to the desired half-power beamwidth; i.e., state the desired amplitude distribution.
(b) Corresponding side lobe level (in dB) of the space factor.
(c) Corresponding directivity (dimensionless and in dB) of the space factor.
7.46.
A line source of total length l, with a continuous triangular current distribution, is center-fed to achieve the desired excitation. Assuming the total length is l = 3λ, calculate (you do not have to derive them) the: (a) Half-power beamwidth (in degrees).
(b) Side lobe level (in dB) (c) Directivity (dimensionless and in dB) (d) Maximum effective area (in λ2). Assume no losses.
(e) Aperture efficiency (in %) when the diameter of the wire representing the line source is λ∕300. Assume that the physical area of the line source is the cross section of the wire along its length.
(f) Maximum power delivered to a load connected to the line source when the incident wave is circularly polarized with an incident power density of 10−3 Watts/cm2. Assume the operating frequency is 1 GHz.
7.47.
A nonuniform linear array with a triangular symmetrical distribution consists of seven dis-crete elements placed λ∕2 apart. The total length (edge-to-edge with elements at the edges) of the array is l = 3λ. Determine the following: (a) Normalized amplitude excitation coefficients (maximum is unity).
(b) Half-power beamwidth (in degrees).
(c) Maximum directivity (dimensionless and in dB) for the triangular distribution.
(d) Maximum directivity (dimensionless and in dB) if the distribution were uniform.
How does it compare with that of Part c? Is it smaller or larger, and by how many dB?
430 ANTENNA SYNTHESIS AND CONTINUOUS SOURCES (e) Maximum effective length (assuming that the distribution of the discrete array is the same as that of a continuous line-source) compared to physical length l: 1. If the distribution is triangular.
2. If the distribution is uniform.
7.48.
Synthesize a 7-element nonuniform array with symmetrical amplitude excitation and with uniform spacing between the elements. The desired amplitude distribution across the elements (relative to the center of the array) is cos2 [i.e., cos2(𝜋x′∕L)] where L is the total length of the array (the end elements are at the edges of the array length) and x′ is the position of each element relative to the center of the array. The overall length L of the array is 3λ. The end elements are at the edges of the array length. Determine the: (a) Spacing between the elements (in λ).
(b) Normalized total amplitude coefficients of each of the elements (normalize them so that the amplitude of the center element is unity). Identify each of the element position and its corresponding normalized amplitude coefficient.
(c) Approximate half-power beamwidth (in degrees).
(d) Approximate maximum directivity (in dB).
7.49.
Derive the space factors for the uniform radial taper, and radial taper-squared circular aperture continuous distributions. Compare with the results in Table 7.2.
7.50.
Design a nonuniform line source, placed along the z-axis, of total length of 4λ. It is desired that the major lobe is at broadside (𝜃= 90◦) and the side lobe of the first minor lobe of the space factor is approximately −23 dB from the maximum of the major lobe.
Choose the easiest design which will satisfy the requirement of −23 dB side lobe level of the first minor lobe.
(a) Determine the nonuniform design/distribution. State its name.
(b) If the continuous line source is approximated by a discrete array of 9 elements (end-to-end), determine the spacing (in λ) between the elements.
(c) For part b, determine the normalized amplitude excitation coefficients of the elements (normalized to the value of the element at the center; make the total amplitude excitation of the center element equal to unity).
(d) Determine the half-power beamwidth (in degrees).
(e) Compute the directivity (dimensionless and in dB).
7.51.
Compute the half-power beamwidth, first-null beamwidth, first sidelobe level (in dB), and gain factor of a circular planar array of closely spaced elements, with radius of 2λ when its amplitude distribution is (a) uniform (b) radial taper (c) radial taper-squared.
CHAPTER8 Integral Equations, Moment Method, and Self and Mutual Impedances 8.1 INTRODUCTION In Chapter 2 it was shown, by the Th´ evenin and Norton equivalent circuits of Figures 2.27 and 2.28, that an antenna can be represented by an equivalent impedance ZA[ZA = (Rr + RL) + jXA]. The equivalent impedance is attached across two terminals (terminals a −b in Figures 2.27 and 2.28) which are used to connect the antenna to a generator, receiver, or transmission line. In general, this impedance is called the driving-point impedance. However, when the antenna is radiating in an unbounded medium, in the absence of any other interfering elements or objects, the driving-point impedance is the same as the self-impedance of the antenna. In practice, however, there is always the ground whose presence must be taken into account in determining the antenna driving-point impedance. The self- and driving-point impedances each have, in general, a real and an imaginary part. The real part is designated as the resistance and the imaginary part is called the reactance.
The impedance of an antenna depends on many factors including its frequency of operation, its geometry, its method of excitation, and its proximity to the surrounding objects. Because of their complex geometries, only a limited number of practical antennas have been investigated analytically.
For many others, the input impedance has been determined experimentally.
The impedance of an antenna at a point is defined as the ratio of the electric to the magnetic fields at that point; alternatively, at a pair of terminals, it is defined as the ratio of the voltage to the current across those terminals. There are many methods that can be used to calculate the impedance of an antenna . Generally, these can be classified into three categories: (1) the boundary-value method, (2) the transmission-line method, and (3) the Poynting vector method. Extensive and brief discussions and comparisons of these methods have been reported , .
The boundary-value approach is the most basic, and it treats the antenna as a boundary-value problem. The solution to this is obtained by enforcing the boundary conditions (usually that the tan-gential electric-field components vanish at the conducting surface). In turn, the current distribution and finally the impedance (ratio of applied emf to current) are determined, with no assumptions as to their distribution, as solutions to the problem. The principal disadvantage of this method is that it has limited applications. It can only be applied and solved exactly on simplified geometrical shapes where the scalar wave equation is separable.
The transmission-line method, which has been used extensively by Schelkunoff , treats the antenna as a transmission line, and it is most convenient for the biconical antenna. Since it utilizes Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 431 432 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES tangential electric-field boundary conditions for its solution, this technique may also be classified as a boundary-value method.
The basic approach to the Poynting vector method is to integrate the Poynting vector (power density) over a closed surface. The closed surface chosen is usually either a sphere of a very large radius r (r ≥2D2∕λ where D is the largest dimension of the antenna) or a surface that coincides with the surface of the antenna. The large sphere closed surface method has been introduced in Chapters 4 and 5, but it lends itself to calculations only of the real part of the antenna impedance (radiation resistance). The method that utilizes the antenna surface has been designated as the induced emf method, and it has been utilized - for the calculation of antenna impedances.
The impedance of an antenna can also be found using an integral equation with a numerical tech-nique solution, which is widely referred to as the Integral Equation Method of Moments –.
This method, which in the late 1960s was extended to include electromagnetic problems, is ana-lytically simple, it is versatile, but it requires large amounts of computation. The limitation of this technique is usually the speed and storage capacity of the computer.
In this chapter the integral equation method, with a Moment Method numerical solution, will be introduced and used first to find the self- and driving-point impedances, and mutual impedance of wire type of antennas. This method casts the solution for the induced current in the form of an integral (hence its name) where the unknown induced current density is part of the integrand. Numerical techniques, such as the Moment Method –, can then be used to solve the current density. In particular two classical integral equations for linear elements, Pocklington’s and Hall´ en’s Integral Equations, will be introduced. This approach is very general, and it can be used with today’s modern computational methods and equipment to compute the characteristics of complex configurations of antenna elements, including skewed arrangements. For special cases, closed-form expressions for the self, driving-point, and mutual impedances will be presented using the induced emf method.
This method is limited to classical geometries, such as straight wires and arrays of collinear and parallel straight wires.
8.2 INTEGRAL EQUATION METHOD The objective of the Integral Equation (IE) method for radiation or scattering is to cast the solution for the unknown current density, which is induced on the surface of the radiator/scatterer, in the form of an integral equation where the unknown induced current density is part of the integrand. The integral equation is then solved for the unknown induced current density using numerical techniques such as the Moment Method (MM). To demonstrate this technique, we will initially consider some specific problems. For introduction, we will start with an electrostatics problem and follow it with time-harmonic problems.
8.2.1 Electrostatic Charge Distribution In electrostatics, the problem of finding the potential that is due to a given charge distribution is often considered. In physical situations, however, it is seldom possible to specify a charge distribution.
Whereas we may connect a conducting body to a voltage source, and thus specify the potential throughout the body, the distribution of charge is obvious only for a few rotationally symmetric geometries. In this section we will consider an integral equation approach to solve for the electric charge distribution once the electric potential is specified. Some of the material here and in other sections is drawn from , .
From statics we know that a linear electric charge distribution 𝜌(r′) creates a scalar electric poten-tial, V(r), according to V(r) = 1 4𝜋𝜀0 ∫source (charge) 𝜌(r′) R dl′ (8-1) INTEGRAL EQUATION METHOD 433 x y z x y z V = 1 V (a) Straight wire (b) Segmented wire Δ y1 y2 ym yN – 1 yN rm 2a l r' Rm Figure 8.1 Straight wire of constant potential and its segmentation.
where r′(x′, y′, z′) denotes the source coordinates, r(x, y, z) denotes the observation coordinates, dl′ is the path of integration, and R is the distance from any one point on the source to the observation point, which is generally represented by R(r, r′) = |r −r′| = √ (x −x′)2 + (y −y′)2 + (z −z′)2 (8-1a) We see that (8-1) may be used to calculate the potentials that are due to any known line charge density. However, the charge distribution on most configurations of practical interest, i.e., complex geometries, is not usually known, even when the potential on the source is given. It is the nontrivial problem of determining the charge distribution, for a specified potential, that is to be solved here using an integral equation-numerical solution approach.
A. Finite Straight Wire Consider a straight wire of length l and radius a, placed along the y axis, as shown in Figure 8-1(a).
The wire is given a normalized constant electric potential of 1 V.
Note that (8-1) is valid everywhere, including on the wire itself (Vwire = 1 V). Thus, choosing the observation along the wire axis (x = z = 0) and representing the charge density on the surface of the wire, (8-1) can be expressed as 1 = 1 4𝜋𝜀0 ∫ l 0 𝜌(y′) R(y, y′)dy′, 0 ≤y ≤l (8-2) 434 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES where R(y, y′) = R(r, r′)|x=z=0 = √ (y −y′)2 + [(x′)2 + (z′)2] = √ (y −y′)2 + a2 (8-2a) The observation point is chosen along the wire axis and the charge density is represented along the surface of the wire to avoid R(y, y′) = 0, which would introduce a singularity in the integrand of (8-2).
It is necessary to solve (8-2) for the unknown 𝜌(y′) (an inversion problem). Equation (8-2) is an integral equation that can be used to find the charge density 𝜌(y′) based on the 1 V potential. The solution may be reached numerically by reducing (8-2) to a series of linear algebraic equations that may be solved by conventional matrix equation techniques. To facilitate this, let us approximate the unknown charge distribution 𝜌(y′) by an expansion of N known terms with constant, but unknown, coefficients, that is, 𝜌(y′) = N ∑ n=1 angn(y′) (8-3) Thus, (8-2) may be written, using (8-3), as 4𝜋𝜀0 = ∫ l 0 1 R(y, y′) [ N ∑ n=1 angn(y′) ] dy′ (8-4) Because (8-4) is a nonsingular integral, its integration and summation can be interchanged, and it can be written as 4𝜋𝜀0 = N ∑ n=1 an ∫ l 0 gn(y′) √ (y −y′)2 + a2 dy′ (8-4a) The wire is now divided into N uniform segments, each of length Δ = l∕N, as illustrated in Figure 8.1(b). The gn(y′) functions in the expansion (8-3) are chosen for their ability to accurately model the unknown quantity, while minimizing computation. They are often referred to as basis (or expansion) functions, and they will be discussed further in Section 8.4.1. To avoid complexity in this solution, subdomain piecewise constant (or “pulse”) functions will be used. These functions, shown in Figure 8.8, are defined to be of a constant value over one segment and zero elsewhere, or gn(y′) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 0 y′ < (n −1)Δ 1 (n −1)Δ ≤y′ ≤nΔ 0 nΔ < y′ (8-5) Many other basis functions are possible, some of which will be introduced later in Section 8.4.1.
INTEGRAL EQUATION METHOD 435 Replacing y in (8-4) by a fixed point such as ym, results in an integrand that is solely a function of y′, so the integral may be evaluated. Obviously, (8-4) leads to one equation with N unknowns an written as 4𝜋𝜀0 = a1 ∫ Δ 0 g1(y′) R(ym, y′) dy′ + a2 ∫ 2Δ Δ g2(y′) R(ym, y′) dy′ + ⋯ + an ∫ nΔ (n−1)Δ gn(y′) R(ym, y′) dy′ + ⋯+ aN ∫ l (N−1)Δ gN(y′) R(ym, y′) dy′ (8-6) In order to obtain a solution for these N amplitude constants, N linearly independent equations are necessary. These equations may be produced by choosing N observation points ym each at the center of each Δ length element as shown in Figure 8.1(b). This results in one equation of the form of (8-6) corresponding to each observation point. For N such points, we can reduce (8-6) to 4𝜋𝜀0 = a1 ∫ Δ 0 g1(y′) R(y1, y′) dy′ + ⋯+ aN ∫ l (N−1)Δ gN(y′) R(y1, y′) dy′ ⋮ (8-6a) 4𝜋𝜀0 = a1 ∫ Δ 0 g1(y′) R(yN, y′) dy′ + ⋯+ aN ∫ l (N−1)Δ gN(y′) R(yN, y′) dy′ We may write (8-6a) more concisely using matrix notation as [Vm] = [Zmn][In] (8-7) where each Zmn term is equal to Zmn = ∫ l 0 gn(y′) √ (ym −y′)2 + a2 dy′ = ∫ nΔ (n−1)Δ 1 √ (ym −y′)2 + a2 dy′ (8-7a) and [In] = [an] (8-7b) [Vm] = [4𝜋𝜀0].
(8-7c) The Vm column matrix has all terms equal to 4𝜋𝜀0, and the In = an values are the unknown charge distribution coefficients. Solving (8-7) for [In] gives [In] = [an] = [Zmn]−1[Vm] (8-8) Either (8-7) or (8-8) may readily be solved on a digital computer by using any of a number of matrix inversion or equation solving routines. Whereas the integral involved here may be evalu-ated in closed form by making appropriate approximations, this is not usually possible with more complicated problems. Efficient numerical integral computer subroutines are commonly available in easy-to-use forms.
436 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES One closed-form evaluation of (8-7a) is to reduce the integral and represent it by Zmn = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 2 ln ( Δ 2 + √ a2 + (Δ 2 )2 a ) m = n (8-9a) ln { d+ mn + [(d+ mn)2 + a2]1∕2 d− mn + [(d− mn)2 + a2]1∕2 } m ≠n but |m −n| ≤2 (8-9b) ln (d+ mn d− mn ) |m −n| > 2 (8-9c) where d+ mn = lm + Δ 2 (8-9d) d− mn = lm −Δ 2 (8-9e) lm is the distance between the mth matching point and the center of the nth source point.
In summary, the solution of (8-2) for the charge distribution on a wire has been accomplished by approximating the unknown with some basis functions, dividing the wire into segments, and then sequentially enforcing (8-2) at the center of each segment to form a set of linear equations.
Even for the relatively simple straight wire geometry we have discussed, the exact form of the charge distribution is not intuitively apparent. To illustrate the principles of the numerical solution, an example is now presented.
Example 8.1 A 1-m long straight wire of radius a = 0.001 m is maintained at a constant potential of 1 V.
Determine the linear charge distribution on the wire by dividing the length into 5 and 20 uniform segments. Assume subdomain pulse basis functions.
Solution: 1. N = 5. When the 1-m long wire is divided into five uniform segments each of length Δ = 0.2 m, (8-7) reduces to ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 10.60 1.10 0.51 0.34 0.25 1.10 10.60 1.10 0.51 0.34 0.51 1.10 10.60 1.10 0.51 0.34 0.51 1.10 10.60 1.10 0.25 0.34 0.51 1.10 10.60 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ a1 a2 a3 a4 a5 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ 1.11 × 10−10 1.11 × 10−10 ⋮ 1.11 × 10−10 ⎤ ⎥ ⎥ ⎥ ⎦ Inverting this matrix leads to the amplitude coefficients and subsequent charge distribu-tion of a1 = 8.81pC∕m a2 = 8.09pC∕m a3 = 7.97pC∕m a4 = 8.09pC∕m a5 = 8.81pC∕m The charge distribution is shown in Figure 8.2(a).
INTEGRAL EQUATION METHOD 437 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Length (m) (a) N = 5 (b) N = 20 Charge density (pC/m) 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Length (m) Charge density (pC/m) Figure 8.2 Charge distribution on a 1-m straight wire at 1 V.
2. N = 20. Increasing the number of segments to 20 results in a much smoother distribution, as shown plotted in Figure 8.2(b). As more segments are used, a better approximation of the actual charge distribution is attained, which has smaller discontinuities over the length of the wire.
B. Bent Wire In order to illustrate the solution of a more complex structure, let us analyze a body composed of two noncollinear straight wires; that is, a bent wire. If a straight wire is bent, the charge distribution will be altered, although the solution to find it will differ only slightly from the straight wire case.
We will assume a bend of angle 𝛼, which remains on the yz-plane, as shown in Figure 8.3.
For the first segment l1 of the wire, the distance R can be represented by (8-2a). However, for the second segment l2 we can express the distance as R = √ (y −y′)2 + (z −z′)2 (8-10) 438 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES x y z V = 1 V 2a l1 l2 α Figure 8.3 Geometry for bent wire.
Also because of the bend, the integral in (8-7a) must be separated into two parts of Zmn = ∫ l1 0 𝜌n(l′ 1) R dl′ 1 + ∫ l2 0 𝜌n(l′ 2) R dl′ 2 (8-11) where l1 and l2 are measured along the corresponding straight sections from their left ends.
Example 8.2 Repeat Example 8.1 assuming that the wire has been bent 90◦at its midpoint. Subdivide the wire into 20 uniform segments.
Solution: The charge distribution for this case, calculated using (8-10) and (8-11), is plotted in Figure 8.4 for N = 20 segments. Note that the charge is relatively more concentrated near the ends of this structure than was the case for a straight wire of Figure 8.2(b). Further, the overall density, and thus capacitance, on the structure has decreased.
7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Length (m) Charge density (pC/m) Figure 8.4 Charge distribution on a 1-m bent wire (𝛼= 90◦, N = 20).
FINITE DIAMETER WIRES 439 Arbitrary wire configurations, including numerous ends and even curved sections, may be ana-lyzed by the methods already outlined here. As with the simple bent wire, the only alterations gen-erally necessary are those required to describe the geometry analytically.
8.2.2 Integral Equation Equation (8-2), for the 1 V potential on a wire of length l, is an integral equation, which can be used to solve for the charge distribution. Numerically this is accomplished using a method, which is usually referred to as Moment Method or Method of Moments –. To solve (8-2) numerically the unknown charge density 𝜌(y′) is represented by N terms, as given by (8-3). In (8-3) gn(y′) are a set of N known functions, usually referred to as basis or expansion functions, while an represents a set of N constant, but unknown, coefficients. The basis or expansion functions are chosen to best represent the unknown charge distribution.
Equation (8-2) is valid at every point on the wire. By enforcing (8-2) at N discrete but different points on the wire, the integral equation of (8-2) is reduced to a set of N linearly independent alge-braic equations, as given by (8-6a). This set is generalized by (8-7)–(8-7c), which is solved for the unknown coefficients an by (8-8) using matrix inversion techniques. Since the system of N linear equations each with N unknowns, as given by (8-6a)–(8-8), was derived by applying the bound-ary condition (constant 1 V potential) at N discrete points on the wire, the technique is referred to as point-matching (or collocation) method , . Thus, by finding the elements of the [V] and [Z], and then the inverse [Z]−1 matrices, we can then determine the coefficients an of the [I] matrix using (8-8). This in turn allows us to approximate the charge distribution 𝜌(y′) using (8-3). This was demonstrated by Examples 8.1 and 8.2 for the straight and bent wires, respectively.
In general, there are many forms of integral equations. For time-harmonic electromagnetics, two of the most popular integral equations are the electric-field integral equation (EFIE) and the mag-netic field integral equation (MFIE) . The EFIE enforces the boundary condition on the tangen-tial electric field while the MFIE enforces the boundary condition on the tangential components of the magnetic field. The EFIE is valid for both closed or open surfaces while the MFIE is valid for closed surfaces. These integral equations can be used for both radiation and scattering problems.
Two- and three-dimensional EFIE and MFIE equations for TE and TM polarizations are derived and demonstrated in . For radiation problems, especially wire antennas, two popular EFIEs are the Pocklington Integral Equation and the Hall´ en Integral Equation. Both of these will be discussed and demonstrated in the section that follows.
8.3 FINITE DIAMETER WIRES In this section we want to derive and apply two classic three-dimensional integral equations, referred to as Pocklington’s integrodifferential equation and Hall´ en’s integral equation –, that can be used most conveniently to find the current distribution on conducting wires. Hall´ en’s equation is usually restricted to the use of a delta-gap voltage source model at the feed of a wire antenna.
Pocklington’s equation, however, is more general and it is adaptable to many types of feed sources (through alteration of its excitation function or excitation matrix), including a magnetic frill . In addition, Hall´ en’s equation requires the inversion of an N + 1 order matrix (where N is the number of divisions of the wire) while Pocklington’s equation requires the inversion of an N order matrix.
For very thin wires, the current distribution is usually assumed to be of sinusoidal form as given by (4-56). For finite diameter wires (usually diameters d of d > 0.05λ), the sinusoidal current distri-bution is representative but not accurate. To find a more accurate current distribution on a cylindrical wire, an integral equation is usually derived and solved. Previously, solutions to the integral equation were obtained using iterative methods ; presently, it is most convenient to use moment method techniques –.
440 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES y y x z z 2a l/2 l/2 i y x z 2a l/2 l/2 Ei θ θ p Incident Scattered θs Iz(z') (a) Geometry (b) Equivalent current Figure 8.5 Uniform plane wave obliquely incident on a conducting wire.
If we know the voltage at the feed terminals of a wire antenna and find the current distribution, the input impedance and radiation pattern can then be obtained. Similarly, if a wave impinges upon the surface of a wire scatterer, it induces a current density that in turn is used to find the scattered field.
Whereas the linear wire is simple, most of the information presented here can be readily extended to more complicated structures.
8.3.1 Pocklington’s Integral Equation To derive Pocklington’s integral equation, refer to Figure 8.5. Although this derivation is general, it can be used either when the wire is a scatterer or an antenna. Let us assume that an incident wave impinges on the surface of a conducting wire, as shown in Figure 8.5(a), and it is referred to as the incident electric field Ei(r). When the wire is an antenna, the incident field is produced by the feed source at the gap, as shown in Figure 8.7. Part of the incident field impinges on the wire and induces on its surface a linear current density Js (amperes per meter). The induced current density Js reradiates and produces an electric field that is referred to as the scattered electric field Es(r).
Therefore, at any point in space the total electric field Et(r) is the sum of the incident and scattered fields, or Et(r) = Ei(r) + Es(r) (8-12) where Et(r) = total electric field Ei(r) = incident electric field Es(r) = scattered electric field When the observation point is moved to the surface of the wire (r = rs) and the wire is per-fectly conducting, the total tangential electric field vanishes. In cylindrical coordinates, the electric field radiated by the dipole has a radial component (E𝜌) and a tangential component (Ez). These are FINITE DIAMETER WIRES 441 represented by (8-52a) and (8-52b). Therefore on the surface of the wire the tangential component of (8-12) reduces to Et z(r = rs) = Ei z(r = rs) + Es z(r = rs) = 0 (8-13) or Es z(r = rs) = −Ei z(r = rs) (8-13a) In general, the scattered electric field generated by the induced current density Js is given by (3-15), or Es(r) = −j𝜔A −j 1 𝜔𝜇𝜀∇(∇⋅A) = −j 1 𝜔𝜇𝜀[k2A + ∇(∇⋅A)] (8-14) However, for observations at the wire surface only the z component of (8-14) is needed, and we can write it as Es z(r) = −j 1 𝜔𝜇𝜀 ( k2Az + 𝜕2Az 𝜕z2 ) (8-15) According to (3-51) and neglecting edge effects Az = 𝜇 4𝜋∫∫S Jz e−jkR R ds′ = 𝜇 4𝜋∫ +l∕2 −l∕2 ∫ 2𝜋 0 Jz e−jkR R a d𝜙′ dz′ (8-16) If the wire is very thin, the current density Jz is not a function of the azimuthal angle 𝜙, and we can write it as 2𝜋aJz = Iz(z′) ➱Jz = 1 2𝜋aIz(z′) (8-17) where Iz(z′) is assumed to be an equivalent filament line-source current located a radial distance 𝜌= a from the z axis, as shown in Figure 8.6(a). Thus (8-16) reduces to Az = 𝜇 4𝜋∫ +l∕2 −l∕2 [ 1 2𝜋a ∫ 2𝜋 0 Iz(z′)e−jkR R a d𝜙′ ] dz′ (8-18) R = √ (x −x′)2 + (y −y′)2 + (z −z′)2 = √ (𝜌2 + a2 −2𝜌a cos(𝜙−𝜙′) + (z −z′)2 (8-18a) where 𝜌is the radial distance to the observation point and a is the radius.
442 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 2a l/2 l/2 (a) On the surface Gap 2a l/2 l/2 (b) Along the center Gap z'n Iz(z') z'n zm zm Iz(z') Figure 8.6 Dipole segmentation and its equivalent current.
Because of the symmetry of the scatterer, the observations are not a function of 𝜙. For simplicity, let us then choose 𝜙= 0. For observations on the surface 𝜌= a of the scatterer (8-18) and (8-18a) reduce to Az(𝜌= a) = 𝜇∫ +l∕2 −l∕2 Iz(z′) ( 1 2𝜋∫ 2𝜋 0 e−jkR 4𝜋R d𝜙′ ) dz′ = 𝜇∫ +l∕2 −l∕2 Iz(z′)G(z, z′) dz′ (8-19) G(z, z′) = 1 2𝜋∫ 2𝜋 0 e−jkR 4𝜋R d𝜙′ (8-19a) R(𝜌= a) = √ 4a2 sin2 (𝜙′ 2 ) + (z −z′)2 (8-19b) Thus for observations on the surface 𝜌= a of the scatterer, the z component of the scattered electric field can be expressed as Es z(𝜌= a) = −j 1 𝜔𝜀 ( k2 + d2 dz2 ) ∫ +l∕2 −l∕2 Iz(z′)G(z, z′) dz′ (8-20) which by using (8-13a) reduces to −j 1 𝜔𝜀 ( d2 dz2 + k2 ) ∫ +l∕2 −l∕2 Iz(z′)G(z, z′) dz′ = −Ei z(𝜌= a) (8-21) FINITE DIAMETER WIRES 443 or ( d2 dz2 + k2 ) ∫ +l∕2 −l∕2 Iz(z′)G(z, z′) dz′ = −j𝜔𝜀Ei z(𝜌= a) (8-21a) Interchanging integration with differentiation, we can rewrite (8-21a) as ∫ +l∕2 −l∕2 Iz(z′) [( 𝜕2 𝜕z2 + k2 ) G(z, z′) ] dz′ = −j𝜔𝜀Ei z(𝜌= a) (8-22) where G(z, z′) is given by (8-19a).
Equation (8-22) is referred to as Pocklington’s integral equation , and it can be used to deter-mine the equivalent filamentary line-source current of the wire, and thus current density on the wire, by knowing the incident field on the surface of the wire.
If we assume that the wire is very thin (a ≪λ), (8-19a) reduces to G(z, z′) = G(R) = e−jkR 4𝜋R (8-23) Substituting (8-23) into (8-22) reduces it to ∫ +l∕2 −l∕2 Iz(z′) [( 𝜕 𝜕z2 + k2 ) e−jkR 4𝜋R ] dz′ = ∫ +l∕2 −l∕2 Iz(z′) 1 4𝜋 { 𝜕 𝜕z [ 𝜕 𝜕z ( e−jkR R )] + k2 e−jkR R } dz′ = −j𝜔𝜀Ei z(𝜌= a) (8-24a) We apply the chain rule to (8-24a) and that R = √ a2 + (z −z′)2 in order to rewrite (8-24a) as ∫ +l∕2 −l∕2 Iz(z′) 1 4𝜋 { 𝜕 𝜕z [ 𝜕 𝜕R ( e−jkR R ) 𝜕 𝜕z (√ a2 + (z −z′)2 )] + k2 e−jkR R } dz′ = −j𝜔𝜀Ei z(𝜌= a) (8-24b) After taking partial derivatives within the brackets of the integrand, we can rewrite the first part of the integrand as 𝜕 𝜕z [ 𝜕 𝜕R ( e−jkR R ) 𝜕 𝜕z (√ a2 + (z −z′)2 )] = 𝜕 𝜕z [ −e−jkR(z −z′) (1 + jkR R3 )] = e−jkR R5 {−R2(1 + jkR) −[jkR(z −z′)2 −(1 + jkR)(3 + jkR)(z −z′)2]} (8-24c) We use (8-24c) and (z −z′)2 = R2 −a2 to express (8-24a) in a more compact and convenient form as ∫ +l∕2 −l∕2 Iz(z′) e−jkR 4𝜋R5 [(1 + jkR)(2R2 −3a2) + (kaR)2] dz′ = −j𝜔𝜀Ei z(𝜌= a) (8-25) 444 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES where for observations along the center of the wire (𝜌= 0) R = √ a2 + (z −z′)2 (8-25a) In (8-22) or (8-25), Iz(z′) represents the equivalent filamentary line-source current located on the surface of the wire, as shown in Figure 8.5(b), and it is obtained by knowing the incident electric field on the surface of the wire. By point-matching techniques, this is solved by matching the boundary conditions at discrete points on the surface of the wire. Often it is easier to choose the matching points to be at the interior of the wire, especially along the axis as shown in Figure 8.6(a), where Iz(z′) is located on the surface of the wire. By reciprocity, the configuration of Figure 8.6(a) is analogous to that of Figure 8.6(b) where the equivalent filamentary line-source current is assumed to be located along the center axis of the wire and the matching points are selected on the surface of the wire.
Either of the two configurations can be used to determine the equivalent filamentary line-source current Iz(z′); the choice is left to the individual.
8.3.2 Hall ´ en’s Integral Equation Referring again to Figure 8.5(a), let us assume that the length of the cylinder is much larger than its radius (l ≫a) and its radius is much smaller than the wavelength (a ≪λ) so that the effects of the end faces of the cylinder can be neglected. Therefore the boundary conditions for a wire with infinite conductivity are those of vanishing total tangential E fields on the surface of the cylinder and vanishing current at the ends of the cylinder [Iz(z′ = ±l∕2) = 0].
Since only an electric current density flows on the cylinder and it is directed along the z axis (J = ̂ azJz), then according to (3-14) and (3-51) A = ̂ azAz(z′), which for small radii is assumed to be only a function of z′. Thus (3-15) reduces to Et z = −j𝜔Az −j 1 𝜔𝜇𝜀 𝜕2Az 𝜕z2 = −j 1 𝜔𝜇𝜀 [ d2Az dz2 + 𝜔2𝜇𝜀Az ] (8-26) Since the total tangential electric field Et z vanishes on the surface of the PEC cylinder, (8-26) reduces to d2Az dz2 + k2Az = 0 (8-26a) The differential equation of (8-26a) is of the homogeneous form, and its solution for Az is Az = −j √ 𝜇𝜀 [ B1 cos(kz) + C1 sin(kz) ] (8-27) Because the current density on the cylindrical wire is symmetrical, meaning Jz(+z) = Jz(−z), the potential Az is also symmetrical, Az(+z) = Az(−z). Hence (8-27) reduces to Az = −j√𝜇𝜀[B1 cos(kz) + C1 sin(k |z|)] (8-27a) FINITE DIAMETER WIRES 445 where B1 and C1 are constants. To determine C1, do the following: r Differentiate Az of (8-27a) with respect to z and let z →0 ⇒lim z→0 (𝜕Az 𝜕z ) = −j√𝜇𝜀kC1 r Use ∇⋅A = −j𝜔𝜇𝜀V ⇒ 𝜕Az 𝜕z = −j𝜔𝜇𝜀V r Set 𝜕Az(+z) 𝜕z = −j𝜔𝜇𝜀V(+z) and 𝜕Az(−z) 𝜕z = −j𝜔𝜇𝜀V(−z) r Since V(+z) = V(−z) ⇒ Vi = 2 lim z→0 [V(+z)] = j 2 𝜔𝜇𝜀lim z→0 [𝜕Az(+z) 𝜕z ] r Thus, Vi = j 2 𝜔𝜇𝜀 ( −j √ 𝜇𝜀kC1 ) = 2C1 ⇒ C1 = Vi 2 (8-27b) For a current-carrying wire, its potential is also given by (3-53). Equating (8-27a) to (3-53) leads to ∫ +l∕2 −l∕2 Iz(z′)e−jkR 4𝜋R dz′ = −j √𝜀 𝜇[B1 cos(kz) + C1 sin(k|z|)] (8-28) The constant B1 is determined from the boundary condition that requires the current to vanish at the end points of the wire.
Equation (8-28) is referred to as Hall´ en’s integral equation for a perfectly conducting wire. It was derived by solving the differential equation (3-15) or (8-26a) with the enforcement of the appropriate boundary conditions.
8.3.3 Source Modeling Let us assume that the wire of Figure 8.5 is symmetrically fed by a voltage source, as shown in Figure 8.7(a), and the element is acting as a dipole antenna. To use, for example, Pocklington’s integrodifferential equation (8-22) or (8-25) we need to know how to express Ei z(𝜌= a). Tradi-tionally there have been two methods used to model the excitation to represent Ei z(𝜌= a, 0 ≤𝜙≤ 2𝜋, −l∕2 ≤z ≤+l∕2) at all points on the surface of the dipole: One is referred to as the delta-gap excitation and the other as the equivalent magnetic ring current (better known as magnetic-frill generator) .
A. Delta Gap The delta-gap source modeling is the simplest and most widely used of the two, but it is also the least accurate, especially for impedances. Usually it is most accurate for smaller width gaps. Using the delta gap, it is assumed that the excitation voltage at the feed terminals is of a constant Vi value and zero elsewhere. Therefore the incident electric field Ei z(𝜌= a, 0 ≤𝜙≤2𝜋, −l∕2 ≤z ≤+l∕2) is also a constant (Vs∕Δ where Δ is the gap width) over the feed gap and zero elsewhere; hence the name delta gap. For the delta-gap model, the feed gap Δ of Figure 8.7(a) is replaced by a narrow band of strips of equivalent magnetic current density of Mg = −̂ n × Ei = −̂ a𝜌× ̂ az Vs Δ = ̂ a𝜙 Vs Δ −Δ 2 ≤z′ ≤Δ 2 (8-29) The magnetic current density Mg is sketched in Figure 8.7(a).
446 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 2a l/2 l/2 2a (b) Segmentation and gap modeling (a) Dipole l/2 l/2 dz' ϕ θ ρ R r Mg x x y b y z z z' Δ Magnetic frill Mf Figure 8.7 Cylindrical dipole, its segmentation, and gap modeling.
B. Magnetic-Frill Generator The magnetic-frill generator was introduced to calculate the near- as well as the far-zone fields from coaxial apertures . To use this model, the feed gap is replaced with a circumferentially directed magnetic current density that exists over an annular aperture with inner radius a, which is usually chosen to be the radius of the wire, and an outer radius b, as shown in Figure 8.7(b). Since the dipole is usually fed by transmission lines, the outer radius b of the equivalent annular aperture of the magnetic-frill generator is found using the expression for the characteristic impedance of the transmission line.
Over the annular aperture of the magnetic-frill generator, the electric field is represented by the TEM mode field distribution of a coaxial transmission line given by Ef = ̂ a𝜌 Vs 2𝜌′ ln(b∕a) a ≤𝜌′ ≤b (8-30) where Vs is the voltage supplied by the source. The 1/2 factor is used because it is assumed that the source impedance is matched to the input impedance of the antenna. The 1/2 should be replaced by unity if the voltage Vi present at the input connection to the antenna is used, instead of the voltage Vs supplied by the source.
Therefore the corresponding equivalent magnetic current density Mf for the magnetic-frill gen-erator used to represent the aperture is equal to Mf = −2̂ n × Ef = −2̂ az × ̂ a𝜌E𝜌= −̂ a𝜙 Vs 𝜌′ ln(b∕a) a ≤𝜌′ ≤b (8-31) FINITE DIAMETER WIRES 447 The fields generated by the magnetic-frill generator of (8-31) on the surface of the wire are found by using Ei z ( 𝜌= a, 0 ≤𝜙≤2𝜋, −l 2 ≤z ≤l 2 ) ≃−Vs ( k(b2 −a2)e−jkR0 8 ln(b∕a)R2 0 { 2 [ 1 kR0 + j ( 1 −b2 −a2 2R2 0 )] + a2 R0 [( 1 kR0 + j ( 1 −(b2 + a2) 2R2 0 )) ( −jk −2 R0 ) + ( −1 kR2 0 + jb2 + a2 R3 0 )]}) (8-32) where R0 = √ z2 + a2 (8-32a) The fields generated on the surface of the wire computed using (8-32) can be approximated by those found along the axis (𝜌= 0). Doing this leads to a simpler expression of the form Ei z ( 𝜌= 0, −l 2 ≤z ≤l 2 ) = − Vs 2 ln(b∕a) [ e−jkR1 R1 −e−jkR2 R2 ] (8-33) where R1 = √ z2 + a2 (8-33a) R2 = √ z2 + b2 (8-33b) To compare the results using the two-source modelings (delta-gap and magnetic-frill generator), an example is performed.
Example 8.3 For a center-fed linear dipole of l = 0.47λ and a = 0.005λ, determine the induced voltage along the length of the dipole based on the incident electric field of the magnetic frill of (8-33). Subdi-vide the wire into 21 segments (N = 21). Compare the induced voltage distribution based on the magnetic frill to that of the delta gap. Assume a 50-ohm characteristic impedance with free-space between the conductors for the annular feed.
Solution: Since the characteristic impedance of the annular aperture is 50 ohms, then Zc = √𝜇0 𝜀0 ln(b∕a) 2𝜋 = 50 ➱b a = 2.3 Subdividing the total length (l = 0.47λ) of the dipole to 21 segments makes Δ = 0.47λ 21 = 0.0224λ 448 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES Using (8-33) to compute Ei z, the corresponding induced voltages, obtained by multiplying the value of −Ei z at each segment by the length of the segment are listed in Table 8.1, where they are compared with those of the delta gap. In Table 8.1, n = 1 represents the outermost segment and n = 11 represents the center segment. Because of the symmetry, only values for the center segment and half of the other segments are shown. Although the two distributions are not iden-tical, the magnetic-frill distribution voltages decay quite rapidly away from the center segment and they very quickly reach almost vanishing values.
TABLE 8.1 Unnormalized and Normalized Dipole Induced Voltage† Differences for Delta-Gap and Magnetic-Frill Generator (l = 0.47𝛌, a = 0.005𝛌, N = 21) Delta-Gap Voltage Magnetic-Frill Generator Voltage Segment Number n Unnormalized Normalized Unnormalized Normalized 1 0 0 1.11 × 10−4 −26.03◦ 7.30 × 10−5 −26.03◦ 2 0 0 1.42 × 10−4 −20.87◦ 9.34 × 10−5 −20.87◦ 3 0 0 1.89 × 10−4 −16.13◦ 1.24 × 10−4 −16.13◦ 4 0 0 2.62 × 10−4 −11.90◦ 1.72 × 10−4 −11.90◦ 5 0 0 3.88 × 10−4 −8.23◦ 2.55 × 10−4 −8.23◦ 6 0 0 6.23 × 10−4 −5.22◦ 4.10 × 10−4 −5.22◦ 7 0 0 1.14 × 10−3 −2.91◦ 7.50 × 10−4 −2.91◦ 8 0 0 2.52 × 10−3 −1.33◦ 1.66 × 10−3 −1.33◦ 9 0 0 7.89 × 10−3 −0.43◦ 5.19 × 10−3 −0.43◦ 10 0 0 5.25 × 10−2 −0.06◦ 3.46 × 10−2 −0.06◦ 11 1 1 1.52 0◦ 1.0 0◦ †Voltage differences as defined here represent the product of the incident electric field at the center of each segment and the corresponding segment length.
8.4 MOMENT METHOD SOLUTION Equations (8-22), (8-25), and (8-28) each has the form of F(g) = h (8-34) where F is a known linear operator, h is a known excitation function, and g is the response function.
For (8-22) F is an integrodifferential operator while for (8-25) and (8-28) it is an integral operator.
The objective here is to determine g once F and h are specified.
While the inverse problem is often intractable in closed form, the linearity of the operator F makes a numerical solution possible. One technique, known as the Moment Method – requires that the unknown response function be expanded as a linear combination of N terms and written as g(z′) ≃a1g1(z′) + a2g2(z′) + ⋯+ aNgN(z′) = N ∑ n=1 angn(z′) (8-35) Each an is an unknown constant and each gn(z′) is a known function usually referred to as a basis or expansion function. The domain of the gn(z′) functions is the same as that of g(z′). Substituting MOMENT METHOD SOLUTION 449 (8-35) into (8-34) and using the linearity of the F operator reduces (8-34) to N ∑ n=1 anF(gn) = h (8-36) The basis functions gn are chosen so that each F(gn) in (8-36) can be evaluated conveniently, preferably in closed form or at the very least numerically. The only task remaining then is to find the an unknown constants.
Expansion of (8-36) leads to one equation with N unknowns. It alone is not sufficient to determine the N unknown an (n = 1, 2, … , N) constants. To resolve the N constants, it is necessary to have N linearly independent equations. This can be accomplished by evaluating (8-36) (e.g., applying boundary conditions) at N different points. This is referred to as point-matching (or collocation).
Doing this, (8-36) takes the form of N ∑ n=1 InF(gn) = hm, m = 1, 2, … , N (8-37) In matrix form, (8-37) can be expressed as [Zmn][In] = [Vm] (8-38) where Zmn = F(gn) (8-38a) In = an (8-38b) Vm = hm (8-38c) The unknown coefficients an can be found by solving (8-38) using matrix inversion techniques, or [In] = [Zmn]−1[Vm] (8-39) 8.4.1 Basis (Expansion) Functions One very important step in any numerical solution is the choice of basis functions. In general, one chooses as basis functions the set that has the ability to accurately represent and resemble the antici-pated unknown function, while minimizing the computational effort required to employ it –.
Do not choose basis functions with smoother properties than the unknown being represented.
Theoretically, there are many possible basis sets. However, only a limited number are used in practice. These sets may be divided into two general classes. The first class consists of subdomain functions, which are nonzero only over a part of the domain of the function g(x′); its domain is the surface of the structure. The second class contains entire-domain functions that exist over the entire domain of the unknown function. The entire-domain basis function expansion is analogous to the well-known Fourier series expansion method.
A. Subdomain Functions Of the two types of basis functions, subdomain functions are the most common. Unlike entire-domain bases, they may be used without prior knowledge of the nature of the function that they must represent.
450 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES x0 x1 x2 x3 xN x 1 g2 (x') x0 x1 x2 x3 xN x a3g3 (x') a2g2 (x') a1g1 (x') a3g3 (x') a2g2 (x') a1g1 (x') Σangn (x') x0 x1 x2 x3 xN x n (a) Single (b) Multiple (c) Function representation Figure 8.8 Piecewise constant subdomain functions.
The subdomain approach involves subdivision of the structure into N nonoverlapping segments, as illustrated on the axis in Figure 8.8(a). For clarity, the segments are shown here to be collinear and of equal length, although neither condition is necessary. The basis functions are defined in conjunction with the limits of one or more of the segments.
Perhaps the most common of these basis functions is the conceptually simple piecewise constant, or “pulse” function, shown in Figure 8.8(a). It is defined by Piecewise Constant gn(x′) = { 1 x′ n−1 ≤x′ ≤x′ n 0 elsewhere (8-40) Once the associated coefficients are determined, this function will produce a staircase representation of the unknown function, similar to that in Figures 8.8(b) and (c).
Another common basis set is the piecewise linear, or “triangle,” functions seen in Figure 8.9(a).
These are defined by Piecewise Linear gn(x′) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ x′ −x′ n−1 x′ n −x′ n−1 x′ n−1 ≤x′ ≤x′ n x′ n+1 −x′ x′ n+1 −x′ n x′ n ≤x′ ≤x′ n+1 0 elsewhere (8-41) MOMENT METHOD SOLUTION 451 x0 x1 x2 xN xN + 1 x 1 g2 (x') x0 x aNgN (x') a2g2 (x') a1g1 (x') Σangn (x') x0 x n x1 x2 xN xN + 1 x1 x2 xN xN + 1 (a) Single (b) Multiple (c) Function representation Figure 8.9 Piecewise linear subdomain functions.
and are seen to cover two segments, and overlap adjacent functions [Figure 8.9(b)]. The resulting representation [Figure 8.9(c)] is smoother than that for “pulses,” but at the cost of increased compu-tational complexity.
Increasing the sophistication of subdomain basis functions beyond the level of the “triangle” may not be warranted by the possible improvement in accuracy. However, there are cases where more specialized functions are useful for other reasons. For example, some integral operators may be evaluated without numerical integration when their integrands are multiplied by a sin(kx′) or cos(kx′) function, where x′ is the variable of integration. In such examples, considerable advantages in computation time and resistance to errors can be gained by using basis functions like the piecewise sinusoid of Figure 8.10 or truncated cosine of Figure 8.11. These functions are defined, respectively, by Piecewise Sinusoid gn(x′) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ sin[k(x′ −x′ n−1)] sin[k(x′ n −x′ n−1)] x′ n−1 ≤x′ ≤x′ n sin[k(x′ n+1 −x′)] sin[k(x′ n+1 −x′ n)] x′ n ≤x′ ≤x′ n+1 0 elsewhere (8-42) Truncated Cosine gn(x′) = ⎧ ⎪ ⎨ ⎪ ⎩ cos [ k ( x′ − x′ n −x′ n−1 2 )] x′ n−1 ≤x′ ≤x′ n 0 elsewhere (8-43) 452 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES x0 x1 x2 xN xN + 1 x 1.0 g2 (x') x0 x aNgN (x') a2g2 (x') a1g1 (x') Σangn (x') x0 x n x1 x2 xN xN + 1 x1 x2 xN xN + 1 (a) Single (b) Multiple (c) Function representation Figure 8.10 Piecewise sinusoids subdomain functions.
x0 x1 x2 x3 xN x 1.0 g2 (x') x0 x a3g3 (x') a2g2 (x') a1g1 (x') Σangn (x') x0 x n x1 x2 x3 xN x1 x2 x3 xN (a) Single (b) Multiple (c) Function representation Figure 8.11 Truncated cosines subdomain functions.
MOMENT METHOD SOLUTION 453 B. Entire-Domain Functions Entire-domain basis functions, as their name implies, are defined and are nonzero over the entire length of the structure being considered. Thus no segmentation is involved in their use.
A common entire-domain basis set is that of sinusoidal functions, where Entire Domain gn(x′) = cos [(2n −1)𝜋x′ l ] −l 2 ≤x′ ≤l 2 (8-44) Note that this basis set would be particularly useful for modeling the current distribution on a wire dipole, which is known to have primarily sinusoidal distribution. The main advantage of entire-domain basis functions lies in problems where the unknown function is assumed a priori to fol-low a known pattern. Such entire-domain functions may render an acceptable representation of the unknown while using far fewer terms in the expansion of (8-35) than would be necessary for subdo-main bases. Representation of a function by entire-domain cosine and/or sine functions is similar to the Fourier series expansion of arbitrary functions.
Because we are constrained to use a finite number of functions (or modes, as they are sometimes called), entire-domain basis functions usually have difficulty in modeling arbitrary or complicated unknown functions.
Entire-domain basis functions, sets like (8-44), can be generated using Tschebyscheff, Maclaurin, Legendre, and Hermite polynomials, or other convenient functions.
8.4.2 Weighting (Testing) Functions To improve the point-matching solution of (8-37), (8-38), or (8-39) an inner product ⟨w, g⟩can be defined which is a scalar operation satisfying the laws of ⟨w, g⟩= ⟨g, w⟩ (8-45a) ⟨bf + cg, w⟩= b⟨f, w⟩+ c⟨g, w⟩ (8-45b) ⟨g∗, g⟩> 0 if g ≠0 (8-45c) ⟨g∗, g⟩= 0 if g = 0 (8-45d) where b and c are scalars and the asterisk (∗) indicates complex conjugation. A typical, but not unique, inner product is ⟨w, g⟩= ∫∫s w∗⋅g ds (8-46) where the w’s are the weighting (testing) functions and S is the surface of the structure being ana-lyzed. Note that the functions w and g can be vectors. This technique is better known as the Moment Method or Method of Moments (MM) , .
The collocation (point-matching) method is a numerical technique whose solutions satisfy the electromagnetic boundary conditions (e.g., vanishing tangential electric fields on the surface of an electric conductor) only at discrete points. Between these points the boundary conditions may not be satisfied, and we define the deviation as a residual (e.g., residual = ΔE|tan = E(scattered)|tan + E(incident)|tan ≠0 on the surface of an electric conductor). For a half-wavelength dipole, a typical residual is shown in Figure 8.12(a) for pulse basis functions and point-matching and Figure 8.12(b) exhibits the residual for piecewise sinusoids-Galerkin method . As expected, the pulse basis 454 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 0.0 0.1 0.2 0.3 0.4 0.5 10–4 10–3 10–2 10–1 100 101 102 103 104 End Center z/l z/l (a) Pulse point-matching Tangential E-field (V/m) 0.0 0.1 0.2 0.3 0.4 0.5 10–4 10–3 10–2 10–1 100 101 102 103 104 End Center (b) Piecewise sinusoids-Galerkin Tangential E-field (V/m) Figure 8.12 Tangential electric field on the conducting surface of a λ∕2 dipole. (source: E. K. Miller and F.
J. Deadrick, “Some computational aspects of thin-wire modeling” in Numerical and Asymptotic Techniques in Electromagnetics, 1975, Springer-Verlag).
SELF-IMPEDANCE 455 point-matching exhibits the most ill-behaved residual and the piecewise sinusoids-Galerkin method indicates an improved residual. To minimize the residual in such a way that its overall average over the entire structure approaches zero, the method of weighted residuals is utilized in conjunction with the inner product of (8-46). This technique, referred to as the Moment Method (MM), does not lead to a vanishing residual at every point on the surface of a conductor, but it forces the boundary conditions to be satisfied in an average sense over the entire surface.
The choice of weighting functions is important in that the elements of {wn} must be linearly independent, –, , . Further, it will generally be advantageous to choose weighting functions that minimize the computations required to evaluate the inner product.
The condition of linear independence between elements and the advantage of computational sim-plicity are also important characteristics of basis functions. Because of this, similar types of functions are often used for both weighting and expansion. A particular choice of functions may be to let the weighting and basis function be the same, that is, wn = gn. This technique is known as Galerkin’s method .
It should be noted that there are N2 terms to be evaluated in (8-46), where N represents the number of weighting (or testing) functions. Each term usually requires two or more integrations; at least one to evaluate each F(gn) and one to perform the inner products of (8-46). When these integrations are to be done numerically, as is often the case, vast amounts of computation time may be necessary. There is, however, a unique set of weighting functions that reduce the number of required integrations. This is the set of Dirac delta weighting functions [wm] = [𝛿(p −pm)] = [𝛿(p −p1), 𝛿(p −p2), …] (8-47) where p specifies a position with respect to some reference (origin) and pm represents a point at which the boundary condition is enforced.
Using (8-47) reduces (8-46) to one integration; the one represented by F(gn). This simplification may lead to solutions that would be impractical if other weighting functions were used. Physically, the use of Dirac delta weighting functions is seen as the relaxation of boundary conditions so that they are enforced only at discrete points on the surface of the structure, hence the name point-matching.
An important consideration when using point-matching is the positioning of the N matching points (pm). While equally-spaced points often yield good results, much depends on the basis func-tions used. When using subsectional basis functions in conjunction with point-matching, one match point should be placed on each segment (to maintain linear independence). Placing the points at the center of the segments usually produces the best results. It is important that a match point does not coincide with the “peak” of a triangle or a similar discontinuous function, where the basis function is not differentiably continuous. This may cause errors in some situations.
8.5 SELF-IMPEDANCE The input impedance of an antenna is a very important parameter, and it is used to determine the efficiency of the antenna. In Section 4.5 the real part of the impedance (referred either to the current at the feed terminals or to the current maximum) was found. At that time, because of mathematical complexities, no attempt was made to find the imaginary part (reactance) of the impedance. In this section the self-impedance of a linear element will be examined using both the Integral Equation-Moment Method and the induced emf method. The real and imaginary parts of the impedance will be found using both methods.
8.5.1 Integral Equation-Moment Method To use this method to find the self-impedance of a dipole, the first thing to do is to solve the inte-gral equation for the current distribution. This is accomplished using either Pocklington’s Integral 456 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES equation of (8-22) or (8-25) or Hall´ en’s integral equation of (8-28). For Pocklington’s integral equa-tion you can use either the delta-gap voltage excitation of (8-29) or the magnetic-frill model of (8-32) or (8-33). Hall´ en’s integral equation is based on the delta-gap model of (8-29).
Once the current distribution is found, using either or both of the integral equations, then the self (input) impedance is determined using the ratio of the voltage to current, or Zin = Vin Iin (8-48) A computer program MOM (Method of Moments) has been developed based on Pocklington’s and Hall´ en’s integral equations, and it is found in the publisher’s website for the book. Pocklington’s uses both the delta-gap and magnetic-frill models while Hall´ en’s uses only the delta-gap feed model. Both, however, use piecewise constant subdomain functions and point-matching. The program computes the current distribution, normalized amplitude radiation pattern, and input impedance. The user must specify the length of the wire, its radius (both in wavelengths), and the type of feed modeling (delta-gap or magnetic-frill) and the number of segments.
To demonstrate the procedure and compare the results using the two-source modelings (delta-gap and magnetic-frill generator) for Pocklington’s integral equation, an example is performed.
Example 8.4 Assume a center-fed linear dipole of l = 0.47λ and a = 0.005λ. This is the same element of Example 8.3.
1. Determine the normalized current distribution over the length of the dipole using N = 21 segments to subdivide the length. Plot the current distribution.
2. Determine the input impedance using segments of N = 7, 11, 21, 29, 41, 51, 61, 71, and 79.
Use Pocklington’s integrodifferential equation (8-25) with piecewise constant subdomain basis functions and point-matching to solve the problem, model the gap with one segment, and use both the delta-gap and magnetic-frill generator to model the excitation. Use (8-33) for the magnetic-frill generator. Because the current at the ends of the wire vanishes, the piecewise constant sub-domain basis functions are not the most judicious choice. However, because of their simplicity, they are chosen here to illustrate the principles even though the results are not the most accurate.
Assume that the characteristic impedance of the annular aperture is 50 ohms and the excitation voltage Vi is 1 V.
Solution: 1. The voltage distribution was found in Example 8.3, and it is listed in Table 8.1. The cor-responding normalized currents obtained using (8-25) with piecewise constant pulse func-tions and point-matching technique for both the delta-gap and magnetic-frill generator are shown plotted in Figure 8.13(a). It is apparent that the two distributions are almost identical in shape, and they resemble that of the ideal sinusoidal current distribution which is more valid for very thin wires and very small gaps. The distributions obtained using Pockling-ton’s integral equation do not vanish at the ends because of the use of piecewise constant subdomain basis functions.
SELF-IMPEDANCE 457 0.25 0.50 0.75 1.0 1 2 3 4 5 6 7 8 9 10 11 (a) l = 0.47 10 9 8 7 6 5 4 3 2 1 Hallen (delta-gap) ' Pocklington (magnetic-frill) Pocklington (delta-gap) Sinusoidal l/2 l/2 In l = 0.47 N = 21 a = 0.005 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.0 0.2 0.4 0.6 0.8 1.0 Distance from dipole center (wavelengths) Normalized current magnitude Sinusoidal a = 10–5 a = 10–3 (b) l = /2 and l = dipole 0.5 dipole λ λ λ λ λ λ λ λ λ Figure 8.13 Current distribution on a dipole antenna.
2. The input impedances using both the delta-gap and the magnetic-frill generator are shown listed in Table 8.2. It is evident that the values begin to stabilize and compare favorably to each other once 61 or more segments are used.
TABLE 8.2 Dipole Input Impedance for Delta-Gap and Magnetic-Frill Generator Using Pocklington’s Integral Equation (l = 0.47𝛌, a = 0.005𝛌) N Delta Gap Magnetic Frill 7 122.8 + j113.9 26.8 + j24.9 11 94.2 + j49.0 32.0 + j16.7 21 77.7 −j0.8 47.1 −j0.2 29 75.4 −j6.6 57.4 −j4.5 41 75.9 −j2.4 68.0 −j1.0 51 77.2 + j2.4 73.1 + j4.0 61 78.6 + j6.1 76.2 + j8.5 71 79.9 + j7.9 77.9 + j11.2 79 80.4 + j8.8 78.8 + j12.9 458 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES Figure 8.14 Dipole geometry for near-field analysis.
To further illustrate the point on the variation of the current distribution on a dipole, it has been computed by Moment Method and plotted in Figure 8.13(b) for l = λ∕2 and l = λ for wire radii of a = 10−5λ and 10−3λ where it is compared with that based on the sinusoidal distribution. It is apparent that the radius of the wire does not influence to a large extent the distribution of the l = λ∕2 dipole. However it has a profound effect on the current distribution of the l = λ dipole at and near the feed point. Therefore the input impedance of the l = λ dipole is quite different for the three cases of Figure 8.13(b), since the zero current at the center of the sinusoidal distribution predicts an infinite impedance. In practice, the impedance is not infinite but is very large.
8.5.2 Induced EMF Method The induced emf method is a classical method to compute the self and mutual impedances –, . The method is basically limited to straight, parallel, and in echelon elements, and it is more difficult to account accurately for the radius of the wires as well as the gaps at the feeds. How-ever it leads to closed-form solutions which provide very good design data. From the analysis of the infinitesimal dipole in Section 4.2, it was shown that the imaginary part of the power density, which contributes to the imaginary power, is dominant in the near-zone of the element and becomes negligible in the far-field. Thus, near-fields of an antenna are required to find its input reactance.
SELF-IMPEDANCE 459 A. Near-Field of Dipole In Chapter 4 the far-zone electric and magnetic fields radiated by a finite length dipole with a sinu-soidal current distribution were found. The observations were restricted in the far-field in order to reduce the mathematical complexities. The expressions of these fields were used to derive the radia-tion resistance and the input resistance of the dipole. However, when the input reactance and/or the mutual impedance between elements are desired, the near-fields of the element must be known. It is the intent here to highlight the derivation.
The fields are derived based on the geometry of Figure 8.14. The procedure is identical to that used in Section 4.2.1 for the infinitesimal dipole. The major difference is that the integrations are much more difficult. To minimize long derivations involving complex integrations, only the procedure will be outlined and the final results will be given. The derivation is left as an end of the chapter problems.
The details can also be found in the first edition of this book.
To derive the fields, the first thing is to specify the sinusoidal current distribution for a finite dipole which is that of (4-56). Once that is done, then the vector potential A of (4-2) is determined. Then the magnetic field is determined using (3-2a), or H = 1 𝜇𝛁× A = −̂ a𝜙 1 𝜇 𝜕Az 𝜕𝜌 (8-49) It is recommended that cylindrical coordinates are used. By following this procedure and after some lengthy analytical details, it can be shown by referring to Figure 8.14(b) that the magnetic field radiated by the dipole is H = ̂ a𝜙H𝜙= −̂ a𝜙 I0 4𝜋j 1 y [ e−jkR1 + e−jkR2 −2 cos (kl 2 ) e−jkr] (8-50) where r = √ x2 + y2 + z2 = √ 𝜌2 + z2 (8-50a) R1 = √ x2 + y2 + ( z −l 2 )2 = √ 𝜌2 + ( z −l 2 )2 (8-50b) R2 = √ x2 + y2 + ( z + l 2 )2 = √ 𝜌2 + ( z + l 2 )2 (8-50c) The corresponding electric field is found using Maxwell’s equation of E = 1 j𝜔𝜀𝛁× H (8-51) Once this is done, it can be shown that the electric field radiated by the dipole is E = ̂ a𝜌E𝜌+ ̂ azEz = −̂ a𝜌 1 j𝜔𝜀 𝜕H𝜙 𝜕z + ̂ az 1 j𝜔𝜀 1 𝜌 𝜕 𝜕𝜌(𝜌H𝜙) (8-52) where E𝜌= Ey = j 𝜂I0 4𝜋y [( z −l 2 ) e−jkR1 R1 + ( z + l 2 ) e−jkR2 R2 −2z cos (kl 2 ) e−jkr r ] (8-52a) Ez = −j𝜂I0 4𝜋 [ e−jkR1 R1 + e−jkR2 R2 −2 cos (kl 2 ) e−jkr r ] (8-52b) 460 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES a z' Jzadϕ dz' Feed Figure 8.15 Uniform linear current density over cylindrical surface of wire.
It should be noted that the last term in (8-50), (8-52a), and (8-52b) vanishes when the over-all length of the element is an integral number of odd half-wavelengths (l = nλ∕2, n = 1, 3, 5, …) because cos(kl∕2) = cos(n𝜋∕2) = 0 for n = 1, 3, 5, … .
The fields of (8-50), (8-52a), and (8-52b) were derived assuming a zero radius wire. In practice all wire antennas have a finite radius which in most cases is very small electrically (typically less than λ∕200). Therefore the fields of (8-50), (8-52a), and (8-52b) are good approximations for finite, but small, radius dipoles.
B. Self-Impedance The technique, which is used in this chapter to derive closed-form expressions for the self- and driving-point impedances of finite linear dipoles, is known as the induced emf method. The general approach of this method is to form the Poynting vector using (8-50), (8-52a), and (8-52b), and to integrate it over the surface that coincides with the surface of the antenna (linear dipole) itself. How-ever, the same results can be obtained using a slightly different approach, as will be demonstrated here. The expressions derived using this method are more valid for small radii dipoles. Expressions, which are more accurate for larger radii dipoles, were derived in the previous section based on the Integral Equation-Moment Method.
To find the input impedance of a linear dipole of finite length and radius, shown in Figure 8.15, the tangential electric-field component on the surface of the wire is needed. This was derived previously and is represented by (8-52b). Based on the current distribution and tangential electric field along the surface of the wire, the induced potential developed at the terminals of the dipole based on the maximum current is given by Vm = ∫ +l∕2 −l∕2 dVm = −1 Im ∫ +l∕2 −l∕2 Iz(𝜌= a, z = z′)Ez(𝜌= a, z = z′) dz′ (8-53) where Im is the maximum current. The input impedance (referred to at the current maximum Im) is defined as Zm = Vm Im (8-54) SELF-IMPEDANCE 461 and can be expressed using (8-53) as Zm = −1 I2 m ∫ l∕2 −l∕2 Iz(𝜌= a, z = z′)Ez(𝜌= a, z = z′) dz′ (8-54a) Equation (8-54a) can also be obtained by forming the complex power density, integrating it over the surface of the antenna, and then relating the complex power to the terminal and induced voltages . The integration can be performed either over the gap at the terminals or over the surface of the conducting wire.
For a wire dipole, the total current Iz is uniformly distributed around the surface of the wire, and it forms a linear current sheet Jz. The current is concentrated primarily over a very small thickness of the conductor, as shown in Figure 8.15, and it is given, based on (4-56), by Iz = 2𝜋aJz = Im sin [ k ( l 2 −|z′| )] (8-55) Therefore (8-54a) can be written as Zm = −1 Im ∫ l∕2 −l∕2 sin [ k ( l 2 −|z′| )] Ez(𝜌= a, z = z′) dz′ (8-56) For simplicity, it is assumed that the E-field produced on the surface of the wire by a current sheet is the same as if the current were concentrated along a filament placed along the axis of the wire.
Then the E-field used in (8-56) is the one obtained along a line parallel to the wire at a distance 𝜌= a from the filament.
Letting Im = Io and substituting (8-52b) into (8-56) it can be shown, after some lengthy but straightforward manipulations, that the real and imaginary parts of the input impedance (referred to at the current maximum) can be expressed as Rr = Rm = 𝜂 2𝜋 { C + ln(kl) −Ci(kl) + 1 2 sin(kl)[Si(2kl) −2Si(kl)] + 1 2 cos(kl)[C + ln(kl∕2) + Ci(2kl) −2Ci(kl)] } (8-57a) Xm = 𝜂 4𝜋 { 2Si(kl) + cos(kl)[2Si(kl) −Si(2kl)] −sin(kl) [ 2Ci(kl) −Ci(2kl) −Ci ( 2ka2 l )]} (8-57b) where C = 0.5772 (Euler’s constant), and Si(x) and Ci(x) are the sine and cosine integrals of Appendix III. Equation (8-57a) is identical to (4-70). In deriving (8-57a) it was assumed that the radius of the wire is negligible (in this case set to zero), and it has little effect on the overall answer.
This is a valid assumption provided l ≫a, and it has been confirmed by other methods.
The input resistance and input reactance (referred to at the current at the input terminals) can be obtained by a transfer relation given by (4-79), or Rin = ( I0 Iin )2 Rr = Rr sin2(kl∕2) (8-58a) Xin = ( I0 Iin )2 Xm = Xm sin2(kl∕2) (8-58b) 462 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 0.0 0.5 1.0 1.5 (a) Resistance (b) Reactance 2.0 2.5 3.0 0 200 400 600 800 1000 l / Resistance (ohms) Rin Rm = Rr 0.0 0.5 1.0 1.5 2.0 2.5 3.0 –1500 –1000 –500 0 500 1000 1500 l / Reactance (ohms) Xin Xm a = 10–5 λ λ λ Figure 8.16 Self-resistance and self-reactance of dipole antenna with wire radius of 10−5 λ.
For a small dipole the input reactance is given by Xin = Xm = −120[ln(l∕2a) −1] tan(kl∕2) (8-59) while its input resistance and radiation resistance are given by (4-37). Plots of the self-impedance, both resistance and reactance, based on (8-57a), (8-57b) and (8-58a), (8-58b) for 0 ≤l ≤3λ are shown in Figures 8.16(a,b). The radius of the wire is 10−5λ. It is evident that when the length of the wire is multiples of a wavelength the resistances and reactances become infinite; in practice they are large.
Ideally the radius of the wire does not affect the input resistance, as is indicated by (8-57a).
However in practice it does have an effect, although it is not as significant as it is for the input reactance. To examine the effect the radius has on the values of the reactance, its values as given by (8-57b) have been plotted in Figure 8.17 for a = 10−5λ, 10−4λ, 10−3λ, and 10−2λ. The overall length of the wire is taken to be 0 < l ≤3λ. It is apparent that the reactance can be reduced to zero provided MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS 463 Figure 8.17 Reactance (referred to the current maximum) of linear dipole with sinusoidal current distribution for different wire radii.
the overall length is slightly less than nλ∕2, n = 1, 3, … , or slightly greater than nλ∕2, n = 2, 4, … .
This is commonly done in practice for l ≃λ∕2 because the input resistance is close to 50 ohms, an almost ideal match for the widely used 50-ohm lines. For small radii, the reactance for l = λ∕2 is equal to 42.5 ohms.
From (8-57b) it is also evident that when l = nλ∕2, n = 1, 2, 3, … , the terms within the last bracket do not contribute because sin(kl) = sin(n𝜋) = 0. Thus for dipoles whose overall length is an integral number of half-wavelengths, the radius has no effect on the antenna reactance. This is illustrated in Figure 8.17 by the intersection points of the curves.
Example 8.5 Using the induced emf method, compute the input reactance for a linear dipole whose lengths are nλ∕2, where n = 1 −6.
Solution: The input reactance for a linear dipole based on the induced emf method is given by (8-57b) whose values are equal to 42.5 for λ∕2, 125.4 for λ, 45.5 for 3λ∕2, 133.1 for 2λ, 46.2 for 5λ∕2, and 135.8 for 3λ.
8.6 MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS In the previous section, the input impedance of a linear dipole was derived when the element was radiating into an unbounded medium. The presence of an obstacle, which could be another ele-ment, would alter the current distribution, the field radiated, and in turn the input impedance of the antenna. Thus the antenna performance depends not only on its own current but also on the current of neighboring elements. For resonant elements with no current excitation of their own, there could be a substantial current induced by radiation from another source. These are known as parasitic ele-ments, as in the case of a Yagi-Uda antenna (see Section 10.3.3), and play an important role in the 464 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES [Z] Z11 – Z12 Z22 – Z21 Z12, Z21 Figure 8.18 Two-port network and its T-equivalent.
overall performance of the entire antenna system. The antenna designer, therefore, must take into account the interaction and mutual effects between elements. The input impedance of the antenna in the presence of the other elements or obstacles, which will be referred to as driving-point impedance, depends upon the self-impedance (input impedance in the absence of any obstacle or other element) and the mutual impedance between the driven element and the other obstacles or elements.
To simplify the analysis, it is assumed that the antenna system consists of two elements. The system can be represented by a two-port (four-terminal) network, as shown in Figure 8.18, and by the voltage-current relations V1 = Z11I1 + Z12I2 V2 = Z21I1 + Z22I2 ⎫ ⎪ ⎬ ⎪ ⎭ (8-60) where Z11 = V1 I1 | | | |I2=0 (8-60a) is the input impedance at port #1 with port #2 open-circuited, Z12 = V1 I2 | | | |I1=0 (8-60b) is the mutual impedance at port #1 due to a current at port #2 (with port #1 open-circuited), Z21 = V2 I1 | | | |I2=0 (8-60c) MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS 465 is the mutual impedance at port #2 due to a current in port #1 (with port #2 open-circuited), Z22 = V2 I2 | | | |I1=0 (8-60d) is the input impedance at port #2 with port #1 open-circuited. For a reciprocal network, Z12 = Z21.
The impedances Z11 and Z22 are the input impedances of antennas 1 and 2, respectively, when each is radiating in an unbounded medium. The presence of another element modifies the input impedance and the extent and nature of the effects depends upon (1) the antenna type, (2) the relative placement of the elements, and (3) the type of feed used to excite the elements.
Equation (8-60) can also be expressed as Z1d = V1 I1 = Z11 + Z12 (I2 I1 ) (8-61a) Z2d = V2 I2 = Z22 + Z21 (I1 I2 ) (8-61b) Z1d and Z2d represent the driving-point impedances of antennas 1 and 2, respectively. Each driving-point impedance depends upon the current ratio I1∕I2, the mutual impedance, and the self-input impedance (when radiating into an unbounded medium). When attempting to match any antenna, it is the driving-point impedance that must be matched. It is, therefore, apparent that the mutual impedance plays an important role in the performance of an antenna and should be investigated.
However, the analysis associated with it is usually quite complex and only simplified models can be examined with the induced emf method. Integral Equation-Moment Method techniques can be used for more complex geometries, including skewed arrangements of elements.
Referring to Figure 8.19, the induced open-circuit voltage in antenna 2, referred to its current at the input terminals, due to radiation from antenna 1 is given by V21 = −1 I2i ∫ l2∕2 −l2∕2 Ez21(z′)I2(z′) dz′ (8-62) where Ez21(z′) = E-field component radiated by antenna 1, which is parallel to antenna 2 I2(z′) = current distribution along antenna 2 Therefore the mutual impedance of (8-60c), (referred to at the input current I1i of antenna 1), is expressed as Z21i = V21 I1i = − 1 I1iI2i ∫ l2∕2 −l2∕2 Ez21(z′)I2(z′) dz′ (8-63) 8.6.1 Integral Equation-Moment Method To use this method to find the mutual impedance based on (8-63), an integral equation must be formed to find Ez21, which is the field radiated by antenna 1 at any point on antenna 2. This integral equation must be a function of the unknown current on antenna 1, and it can be derived using a procedure similar to that used to form Pocklington’s Integral Equation of (8-22) or (8-25), or Hall´ en’s Integral Equation of (8-28). The unknown current of antenna 1 can be represented by a series of finite number of terms with N unknown coefficients and a set of known (chosen) basis functions.
The current I2(z) must also be expanded into a finite series of N terms with N unknown coefficients 466 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES l1/2 l1/2 h d y y' z' z l2/2 l2/2 R1 r R2 x' θ x Figure 8.19 Dipole positioning for mutual coupling.
and a set of N chosen basis functions. Once each of them is formulated, then they can be used interactively to reduce (8-63) into an N × N set of linearly independent equations to find the mutual impedance.
To accomplish this requires a lengthy formulation and computer programming. The process usu-ally requires numerical integrations or special functions for the evaluation of the impedance matri-ces of Ez21 and the integral of (8-63). There are national computer codes, such as the Numer-ical Electromagnetics Code (NEC) and the simplified version Mini Numerical Electromagnetics Code (MININEC), for the evaluation of the radiation characteristics, including impedances, of wire antennas –. Both of these are based on an Integral Equation-Moment Method formulation.
Information concerning these two codes follows. There are other codes; however, these two seem to be popular, especially for wire type antennas.
Another procedure that has been suggested to include mutual effects in arrays of linear elements is to use a convergent iterative algorithm , . This method can be used in conjunction with a calculator , and it has been used to calculate impedances, patterns, and directivities of arrays of dipoles .
A. Numerical Electromagnetics Code (NEC) The Numerical Electromagnetics Code (NEC) is a user-oriented program developed at Lawrence Livermore National Laboratory. It is a moment method code for analyzing the interaction of electro-magnetic waves with arbitrary structures consisting of conducting wires and surfaces. It combines an integral equation for smooth surfaces with one for wires to provide convenient and accurate mod-eling for a wide range of applications. The code can model nonradiating networks and transmission lines, perfect and imperfect conductors, lumped element loading, and perfect and imperfect conduct-ing ground planes. It uses the electric-field integral equation (EFIE) for thin wires and the magnetic field integral equation (MFIE) for surfaces. The excitation can be either an applied voltage source or an incident plane wave. The program computes induced currents and charges, near- and far-zone electric and magnetic fields, radar cross section, impedances or admittances, gain and directivity, power budget, and antenna-to-antenna coupling.
MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS 467 B. Mini-Numerical Electromagnetics Code (MININEC) The Mini-Numerical Electromagnetics Code (MININEC) , is a user-oriented compact ver-sion of the NEC developed at the Naval Ocean Systems Center (NOSC) [now Space and Naval Warfare Systems Command (SPAWAR)]. It is also a moment method code, but coded in BASIC, and has retained the most frequently used options of the NEC. It is intended to be used in mini, micro, and personal computers, as well as work stations, and it is most convenient to analyze wire antennas. It computes currents, and near- and far-field patterns. It also optimizes the feed excitation voltages that yield desired radiation patterns.
8.6.2 Induced EMF Method The induced emf method is also based on (8-63) except that I2(z′) is assumed to be the ideal current distribution of (4-56) or (8-55) while Ez21(z′) is the electric field of (8-52b). Using (8-55) and (8-52b), we can express (8-63) as Z21i = V21 I1i = j𝜂I1mI2m 4𝜋I1iI2i ∫ l2∕2 −l2∕2 sin [ k (l2 2 −|z′| )] [ e−jkR1 R1 + e−jkR2 R2 −2 cos ( kl1 2 ) e−jkr r ] dz′ (8-64) where r, R1, and R2 are given, respectively, by (8-50a), (8-50b) and (8-50c) but with y = d and l = l1.
I1m, I2m and I1i, I2i represent, respectively, the maximum and input currents for antennas 1 and 2. By referring each of the maximum currents to those at the input using (4-78) and assuming free-space, we can write (8-64) as Z21i = j 30 sin (kl1 2 ) sin (kl2 2 ) ∫ l2∕2 −l2∕2 sin [ k (l2 2 −|z′| )] [ e−jkR1 R1 + e−jkR2 R2 −2 cos ( kl1 2 ) e−jkr r ] dz′ (8-65) The mutual impedance referred to the current maxima is related to that at the input of (8-65) by Z21m = Z21i sin (kl1 2 ) sin (kl2 2 ) (8-66) which for identical elements (l1 = l2 = l) reduces to Z21m = Z21i sin2 (kl 2 ) (8-67) whose real and imaginary parts are related by R21m = R21i sin2 (kl 2 ) (8-67a) X21m = X21i sin2 (kl 2 ) (8-67b) 468 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES Figure 8.20 Dipole configuration of two identical elements for mutual impedance computations.
For a two-element array of linear dipoles, there are three classic configurations for which closed-form solutions for (8-65), in terms of sine and cosine integrals, are obtained . These are shown in Figure 8.20, and they are referred to as the side-by-side [Figure 8.20(a)], collinear [Figure 8.20(b)], and parallel-in-echelon [Figure 8.20(c)]. For two identical elements (each with odd multiples of λ∕2, l = nλ∕2, n = 1, 3, 5, …) (8-67) reduces for each arrangement to the expressions that follow. Expressions for linear elements of any length are much more complex and can be found in .
A MATLAB and FORTRAN computer program referred to as Impedances, based on (8-68a)– (8-70i), is included in the publisher’s website for the book.
Side-by-Side Configuration [Figure 8.20(a)] R21m = 𝜂 4𝜋[2Ci(u0) −Ci(u1) −Ci(u2)] (8-68a) X21m = −𝜂 4𝜋[2Si(u0) −Si(u1) −Si(u2)] (8-68b) u0 = kd (8-68c) u1 = k( √ d2 + l2 + l) (8-68d) u2 = k( √ d2 + l2 −l) (8-68e) Collinear Configuration [Figure 8-20(b)] R21m = −𝜂 8𝜋cos(v0)[−2Ci(2v0) + Ci(v2) + Ci(v1) −ln(v3)] + 𝜂 8𝜋sin(v0)[2Si(2v0) −Si(v2) −Si(v1)] (8-69a) X21m = −𝜂 8𝜋cos(v0)[2Si(2v0) −Si(v2) −Si(v1)] + 𝜂 8𝜋sin(v0)[2Ci(2v0) −Ci(v2) −Ci(v1) −ln(v3)] (8-69b) MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS 469 v0 = kh (8-69c) v1 = 2k(h + l) (8-69d) v2 = 2k(h −l) (8-69e) v3 = (h2 −l2)∕h2 (8-69f) Parallel-in-Echelon Configuration [Figure 8.20(c)] R21m = −𝜂 8𝜋cos(w0)[−2Ci(w1) −2Ci(w′ 1) + Ci(w2) + Ci(w′ 2) + Ci(w3) + Ci(w′ 3)] + 𝜂 8𝜋sin(w0)[2Si(w1) −2Si(w′ 1) −Si(w2) + Si(w′ 2) −Si(w3) + Si(w′ 3)] (8-70a) X21m = −𝜂 8𝜋cos(w0)[2Si(w1) + 2Si(w′ 1) −Si(w2) −Si(w′ 2) −Si(w3) −Si(w′ 3)] + 𝜂 8𝜋sin(w0)[2Ci(w1) −2Ci(w′ 1) −Ci(w2) + Ci(w′ 2) −Ci(w3) + Ci(w′ 3)] (8-70b) w0 = kh (8-70c) w1 = k( √ d2 + h2 + h) (8-70d) w′ 1 = k( √ d2 + h2 −h) (8-70e) w2 = k[ √ d2 + (h −l)2 + (h −l)] (8-70f) w′ 2 = k[ √ d2 + (h −l)2 −(h −l)] (8-70g) w3 = k[ √ d2 + (h + l)2 + (h + l)] (8-70h) w′ 3 = k[ √ d2 + (h + l)2 −(h + l)] (8-70i) The mutual impedance, referred to the current maximum, based on the induced emf method of a side-by-side and a collinear arrangement of two half-wavelength dipoles is shown plotted in Figure 8.21. It is apparent that the side-by-side arrangement exhibits larger mutual effects since the antennas are placed in the direction of maximum radiation. The data is compared with those based on the Moment Method/NEC using a wire with a radius of 10−5λ. A very good agreement is indi-cated between the two sets because a wire with a radius of 10−5λ for the MM/NEC is considered very thin. Variations as a function of the radius of the wire for both the side-by-side and collinear arrange-ments using the MM/NEC are shown, respectively, in Figures 8.22(a,b). Similar sets of data were computed for the parallel-in-echelon arrangement of Figure 8.20(c), and they are shown, respec-tively, in Figures 8.23(a) and 8.23(b) for d = λ∕2, 0 ≤h ≤λ and h = λ∕2, 0 < d < λ for wire radii of 10−5λ. Again a very good agreement between the induced emf and Moment Method/NEC data is indicated.
470 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 0.0 0.5 1.0 1.5 2.0 2.5 3.0 –60 –40 –20 0 20 40 60 80 100 d/ (a) Side-by-side Mutual impedance Z21m (ohms) R21m (EMF) R21m (MoM) X21m (EMF) X21m (MoM) R21m X21m R21m X21m a = 10–5 0.0 0.5 1.0 1.5 (b) Collinear 2.0 2.5 3.0 –10 0 10 20 30 s/ Mutual impedance Z21m (ohms) R21m (EMF) R21m (MoM) X21m (EMF) X21m (MoM) a = 10–5λ λ λ λ Figure 8.21 Mutual impedance of two side-by-side and collinear λ∕2 dipoles using the moment method and induced emf method.
Example 8.6 Two identical linear half-wavelength dipoles are placed in a side-by-side arrangement, as shown in Figure 8.20(a). Assuming that the separation between the elements is d = 0.35λ, find the driving-point impedance of each.
Solution: Using (8-61a) Z1d = V1 I1 = Z11 + Z12 (I2 I1 ) Since the dipoles are identical, I1 = I2. Thus Z1d = Z11 + Z12 MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS 471 From Figure 8.21(a) Z12 ≃25 −j38 Since Z11 = 73 + j42.5 Z1d reduces to Z1d ≃98 + j4.5 which is also equal to Z2d of (8-61b).
0.00 0.50 1.00 1.50 2.00 2.50 3.00 –40.0 –20.0 0.0 20.0 40.0 60.0 80.0 Separation d (wavelengths) (a) Side-by-side Mutual impedance Z21m (ohms) a = 10–5 a = 10–4 a = 10–3 R21m X21m 0.00 0.50 1.00 1.50 2.00 2.50 3.00 –40.0 –20.0 0.0 20.0 40.0 60.0 80.0 Separation s (wavelengths) (b) Collinear Mutual impedance Z21m (ohms) a = 10–5 a = 10–4 a = 10–3 R21m X21m λ λ λ λ λ λ Figure 8.22 Variations of mutual impedance as a function of wire radius for side-by-side and collinear λ∕2 dipole arrangements.
472 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 0.0 0.2 0.4 0.6 0.8 1.0 –40 –30 –20 –10 0 10 20 30 h/ (a) d = 0.5λ Mutual impedance Z21m (ohms) R21m (EMF) R21m (MoM) R21m X21m (EMF) X21m (MoM) X21m R21m X21m a = 10–5 0.0 0.2 0.4 0.6 0.8 1.0 –40 –30 –20 –10 0 10 20 30 40 50 60 d/ (b) h = 0.5 Mutual impedance Z21m (ohms) R21m (EMF) R21m (MoM) X21m (EMF) X21m (MoM) a = 10–5 λ λ λ λ λ Figure 8.23 Mutual impedance for two parallel-in-echelon λ∕2 dipoles with wire radii of 10−5 λ.
As discussed in Chapter 2, Section 2.13, maximum power transfer between the generator and the transmitting antenna occurs when their impedances are conjugate-matched. The same is necessary for the receiver and receiving antenna. This ensures maximum power transfer between the transmitter and receiver, when there is no interaction between the antennas. In practice, the input impedance of one antenna depends on the load connected to the other antenna. Under those conditions, the matched loads and maximum coupling can be computed using the Linville method , which is used in rf amplifier design. This technique has been incorporated into the NEC . Using this method, the maximum coupling Cmax is computed using Cmax = 1 L[1 −(1 −L2)1∕2] (8-71) MUTUAL IMPEDANCE BETWEEN LINEAR ELEMENTS 473 where L = |Y12Y21| 2Re(Y11)Re(Y22) −Re(Y12Y21) (8-71a) To ensure maximum coupling, the admittance of the matched load on the receiving antenna should be YL = [1 −𝜌 1 + 𝜌+ 1 ] Re(Y22) −Y22 (8-72) where 𝜌= Cmax(Y12Y21)∗ |Y12Y21| (8-72a) and Cmax computed using (8-71). The corresponding input admittance of the transmitting antenna is Yin = Y11 −Y21Y12 YL + Y22 (8-73) Based on (8-71)–(8-73), maximum coupling for two half-wavelength dipoles in side-by-side and collinear arrangements as a function of the element separation (d for side-by-side and s for collinear) was computed using the NEC, and it is shown in Figure 8.24. As expected, the side-by-side arrangement exhibits much stronger coupling, since the elements are placed along the direction of their respective maximum radiation. Similar curves can be computed for the parallel-in-echelon arrangement.
0.00 0.50 1.00 1.50 2.00 2.50 3.00 –60.0 –50.0 –40.0 –30.0 –20.0 –10.0 0.0 Separation d, s (wavelengths) Maximum coupling Cmax (dB) Collinear Side-by-side Figure 8.24 Maximum coupling for the two λ∕2 dipoles in side-by-side and collinear arrangements as a function of separation.
474 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 8.7 MUTUAL COUPLING IN ARRAYS When two antennas are near each other, whether one and/or both are transmitting or receiving, some of the energy that is primarily intended for one ends up at the other. The amount depends primarily on the a. radiation characteristics of each b. relative separation between them c. relative orientation of each There are many different mechanisms that can cause this interchange of energy. For example, even if both antennas are transmitting, some of the energy radiated from each will be received by the other because of the nonideal directional characteristics of practical antennas. Part of the incident energy on one or both antennas may be rescattered in different directions allowing them to behave as secondary transmitters. This interchange of energy is known as “mutual coupling,” and in many cases it complicates the analysis and design of an antenna. Furthermore, for most practical config-urations, mutual coupling is difficult to predict analytically but must be taken into account because of its significant contribution. Because the mutual effects of any antenna configuration cannot be generalized, in this section we want first to briefly introduce them in a qualitative manner and then examine their general influence on the behavior of the radiation characteristics of the antenna.
8.7.1 Coupling in the Transmitting Mode To simplify the discussion, let us assume that two antennas m and n of an array are positioned relative to each other as shown in Figure 8.25(a). The procedure can be extended to a number of elements. If a source is attached to antenna n, the generated energy traveling toward the antenna labeled as (0) will be radiated into space (1) and toward the mth antenna (2). The energy incident on the mth antenna sets up currents which have a tendency to rescatter some of the energy (3) and allow the remaining to travel toward the generator of m (4). Some of the rescattered energy (3) may be redirected back toward antenna n (5). This process can continue indefinitely. The same process would take place if antenna m is excited and antenna n is the parasitic element. If both antennas, m and n, are excited simultaneously, the radiated and rescattered fields by and from each must be added vectorially to arrive at the total field at any observation point. Thus, “the total contribution to the far-field pattern of a particular element in the array depends not only upon the excitation furnished by its own gen-erator (the direct excitation) but upon the total parasitic excitation as well, which depends upon the couplings from and the excitation of the other generators .” The wave directed from the n to the m antenna and finally toward its generator (4) adds vectorially to the incident and reflected waves of the m antenna itself, thus enhancing the existing standing wave pattern within m. For coherent excitations, the coupled wave (4) due to source n differs from the reflected one in m only in phase and amplitude. The manner in which these two waves interact depends on the coupling between them and the excitation of each. It is evident then that the vector sum of these two waves will influence the input impedance looking in at the terminals of antenna m and will be a function of the position and excitation of antenna n. This coupling effect is commonly modeled as a change in the apparent driving impedance of the elements and it is usually referred to as mutual impedance variation.
To demonstrate the usefulness of the driving impedance variation, let us assume that a set of elements in an array are excited. For a given element in the array, the generator impedance that is optimum in the sense of maximizing the radiated power for that element is that which would be a conjugate impedance match at the element terminals. This is accomplished by setting a reflected wave which is equal in amplitude and phase to the backwards traveling waves induced due to the coupling. Even though this is not the generator impedance which is a match to a given element when all other elements are not excited, it does achieve maximum power transfer.
MUTUAL COUPLING IN ARRAYS 475 4 5 2 3 3 1 1 z Antenna m 0 z Antenna n (a) Transmitting 1 3 0 0 2 z Antenna m 4 z Antenna n (b) Receiving Incident wave front Figure 8.25 Transmitting mode coupling paths between antennas m and n. (Reprinted with permission of MIT Lincoln Laboratory, Lexington, MA.) To minimize confusion, let us adopt the following terminology : 1. Antenna impedance: The impedance looking into a single isolated element.
2. Passive driving impedance: The impedance looking into a single element of an array with all other elements of the array passively terminated in their normal generator impedance unless otherwise specified.
3. Active driving impedance: The impedance looking into a single element of an array with all other elements of the array excited.
Since the passive driving impedance is of minor practical importance and differs only slightly from the antenna impedance, the term driving impedance will be used to indicate active driving impedance, unless otherwise specified.
Since the driving impedance for a given element is a function of the placement and excitation of the other elements, then optimum generator impedance that maximizes array radiation efficiency 476 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES (gain) varies with array excitation. These changes, with element excitation variations, constitute one of the principal aggravations in electronic scanning arrays.
8.7.2 Coupling in the Receiving Mode To illustrate the coupling mechanism in the receiving mode, let us again assume that the antenna system consists of two passively loaded elements of a large number, as shown in Figure 8.25(b).
Assume that a plane wave (0) is incident, and it strikes antenna m first where it causes current flow.
Part of the incident wave will be rescattered into space as (2), the other will be directed toward antenna n as (3) where it will add vectorially with the incident wave (0), and part will travel into its feed as (1). It is then evident that the amount of energy received by each element of an antenna array is the vector sum of the direct waves and those that are coupled to it parasitically from the other elements.
The amount of energy that is absorbed and reradiated by each element depends on its match to its terminating impedance. Thus, the amount of energy received by any element depends upon its terminating impedance as well as that of the other elements. In order to maximize the amount of energy extracted from an incident wave, we like to minimize the total backscattered (2) energy into space by properly choosing the terminating impedance. This actually can be accomplished by mismatching the receiver itself relative to the antenna so that the reflected wave back to the antenna (4) is cancelled by the rescattered wave, had the receiver been matched to the actual impedance of each antenna.
As a result of the previous discussion, it is evident that mutual coupling plays an important role in the performance of an antenna. However, the analysis and understanding of it may not be that simple.
8.7.3 Mutual Coupling on Array Performance The effects of the mutual coupling on the performance of an array depends upon the a. antenna type and its design parameters b. relative positioning of the elements in the array c. feed of the array elements d. scan volume of the array These design parameters influence the performance of the antenna array by varying its element impedance, reflection coefficients, and overall antenna pattern. In a finite-element array, the mul-tipath routes the energy follows because of mutual coupling will alter the pattern in the absence of these interactions. However, for a very large regular array (array with elements placed at regular intervals on a grid and of sufficient numbers so that edge effects can be ignored), the relative shape of the pattern will be the same with and without coupling interactions. It will only require a scaling up or down in amplitude while preserving the shape. This, however, is not true for irregular placed elements or for small regular arrays where edge effects become dominant.
8.7.4 Coupling in an Infinite Regular Array The analysis and understanding of coupling can be considerably simplified by considering an infinite regular array. Although such an array is not physically realizable, it does provide an insight and, in many cases, answers of practical importance. For the infinite regular array we assume that a. all the elements are placed at regular intervals b. all the elements are identical MUTUAL COUPLING IN ARRAYS 477 c. all the elements have uniform (equal) amplitude excitation d. there can be a linear relative phasing between the elements in the two orthogonal directions The geometry of such an array is shown in Figure 6.28 with its array factor given by (6-87) or (6-88). This simplified model will be used to analyze the coupling and describes fairly accurately the behavior of most elements in arrays of modest to large size placed on flat or slowly curved surfaces with smoothly varying amplitude and phase taper.
To assess the behavior of the element driving impedance as a function of scan angle, we can write the terminal voltage of any one element in terms of the currents flowing in the others, assuming a single-mode operation, as Vmn = ∑ p ∑ q Zmn,pqIpq (8-74) where Zmn,pq defines the terminal voltage at antenna mn due to a unity current in element pq when the current in all the other elements is zero. Thus the Zmn,pq terms represent the mutual impedances when the indices mn and pq are not identical. The driving impedance of the mnth element is defined as ZDmn = Vmn Imn = ∑ p ∑ q Zmn,pq Ipq Imn (8-75) Since we assumed that the amplitude excitation of the elements of the array was uniform and the phase linear, we can write that Ipq = I00ej(p𝛽x+q𝛽y) (8-76a) Imn = I00ej(m𝛽x+n𝛽y) (8-76b) Thus, (8-75) reduces to ZDmn = ∑ p ∑ q Zmn,pqej(p−m)𝛽xej(q−n)𝛽y (8-77) It is evident that the driving-point impedance of mn element is equal to the vector sum of the ele-ment self impedance (mn = pq) and the phased mutual impedances between it and the other ele-ments (mn ≠pq). The element self-impedance (mn = pq) is obtained when all other elements are open-circuited so that the current at their feed points is zero [Ipq(pq ≠mn) = 0]. For most practical antennas, physically this is almost equivalent to removing the pq elements and finding the impedance of a single isolated element.
A consequence of the mutual coupling problem is the change of the array input reflection coef-ficient, and thus the input impedance, with scan angle. To demonstrate the variations of the input reflection coefficient, and thus of the input impedance, of an infinite array as a function of scan angle, the input reflection coefficient of an infinite array of circular microstrip patches matched at broadside is shown in Figure 8.26 for the E-plane and H-plane . The variations are due mainly to coupling between the elements. The variations are more pronounced for the E-plane than for the H-plane. For microstrip patches, coupling is attributed to space waves (with 1/r radial variations), higher order waves (with 1∕𝜌2 radial variations), surface waves (with 1∕𝜌2 radial variations), and leaky waves [with exp(−λ𝜌)∕𝜌1∕2 radial variations]. As is shown in Chapter 14 and Figures 14.41, 14.42, the variations of the reflection coefficient can be reduced by suppressing the surface waves supported by the substrate using cavities to back the patches . The variations of the reflection coefficient as a function of scan angle can lead, due to large values of the reflection coefficient (ideally unity), to 478 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES 0 10 20 30 40 50 60 70 80 90 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Input reflection coefficient ⏐Γin⏐ 0 (degrees) E-plane H-plane θ Radius of patch = a = 0.156 0 Height of substrate = h = 0.08 0 Dielectric constant of substrate = r = 2.5 Element separation = dx = dy = 0.5 0 λ λ λ Figure 8.26 Typical magnitude of input reflection coefficient versus scan angle in E- and H-planes for infinite array of microstrip patches (courtesy J. T. Aberle and F. Zavosh).
what is usually referred as array scan blindness –. This is evident for the E-plane near 72◦– 73◦and is due to excitation in that plane of a leaky-wave mode, which is not as strongly excited as the scan angle increases beyond those values. Scan blindness is reached at a scan angle of 90◦. Also there can be degradation of side lobe level and main beam shape due to the large variations of the reflection coefficient.
Scan blindness is attributed to slow waves which are supported by the structure of the antenna array. These structures may take the form of dielectric layers (such as radomes, superstrates, and substrates) over the face of the array, or metallic grids or fence structures. The scan blindness has been referred to as a “forced surface wave” , , or a “leaky wave” , resonant response of the slow wave structure by the phased array. For the microstrip arrays, the substrate layer supports a slow surface wave which contributes to scan blindness .
8.7.5 Active Element Pattern in an Array In Figure 8.26, it was demonstrated how the reflection coefficient, and thus the input impedance, in an array of microstrip patches is impacted by the mutual impedance, due primarily to the surface waves, as the array is scanned. In fact, if the array is not designed properly, the reflection coefficient can reach unity, leading to scan blindness.
Similarly, the pattern of a single element of an array of element is influenced by the presence of the other elements. In Chapter 6, we used the pattern multiplication rule to find the total field of an array of identical elements by simply multiplying the element pattern by the array factor, as represented by (6-5). The pattern multiplication rule does not take into account the mutual coupling between the elements, and it is only an approximation. It has been demonstrated in that the pattern of an element in an array, in the presence of the other elements, referred to as the active element pattern is influenced by the presence of the other elements, and it is not necessarily the same as the pattern of the element in the absence of the others, referred to as the isolated element pattern.
In , the impact of the pattern of an element in an infinite array, by the presence of the other elements, is derived rigorously using scattering parameters. However, in the same article, the author derives the same relationship using a more simplistic and less rigorous but uses more intuitive argu-ments based on basic concepts, even those derived in Chapter 2. It will be the less rigorous but more MUTUAL COUPLING IN ARRAYS 479 intuitive derivation that we will outline here. The reader is referred to to for the more rigorous derivation.
Let us consider an infinite uniform array of aperture elements. The maximum effective aper-ture, and in turn the maximum gain, from a single isolated rectangular aperture element with dimensions a and b, assuming no losses, can be written according to (2-110), (12-37), and (12-40) as Aem = 𝜀apAp = 𝜀ap(ab) = λ2 4𝜋G0 (8-78) G0 = (4𝜋 λ2 Aem ) = 4𝜋 λ2 (𝜀apAp) = 4𝜋 λ2 (𝜀apab) (8-78a) If we consider a large uniform array of N elements, the maximum gain G0(𝜃0, 𝜙0) available from a single isolated element scanned at an angle 𝜃0, 𝜙0 is, according to (6-100), G0(𝜃0, 𝜙0) = (4𝜋 λ2 Aem ) = 4𝜋 λ2 (𝜀apAp cos 𝜃0) = 4𝜋 λ2 𝜀ap (ab) cos 𝜃0 (8-79) while the overall maximum gain Ga 0(𝜃0, 𝜙0) of the entire array of N elements can be expressed, assuming no coupling and using superposition, as Ga 0(𝜃0, 𝜙0) = (4𝜋 λ2 Aem ) = 4𝜋 λ2 (𝜀apAp cos 𝜃0) = 4𝜋 λ2 𝜀ap (ab) cos 𝜃0 (8-79a) The realized maximum gain Gre0(𝜃0, 𝜙0) of the array, accounting for reflection/mismatch loss, is according to (2-49b) or (2-49c) Gre0(𝜃0, 𝜙0) = 4𝜋 λ2 N [ 1 −|Γi(𝜃0, 𝜙0)|2] ab cos 𝜃0 (8-80) Note that (8-80) assumes identical reflections at each element, and that |Γi(𝜃0, 𝜙0)| is the magnitude of the active reflection coefficient of the ith element of a fully excited phased array.
Now, if the maximum gain Ge 0(𝜃0, 𝜙0) of the active element is defined as the gain Gi 0(𝜃0, 𝜙0) of a singly excited isolated element of the array, with all the other elements match-terminated, we can rewrite it as Ge 0(𝜃0, 𝜙0) = Gi 0(𝜃0, 𝜙0) [ 1 −|Γi(𝜃0, 𝜙0)|2] (8-81) After the superposition, the gain of the fully excited array in the direction of the main beam can now be written, taking into account coupling, as Ga re0(𝜃0, 𝜙0) = NGe 0(𝜃0, 𝜙0) (8-82) Based on the above, the realized active element gain pattern Ge re(𝜃, 𝜙) of an array element in the presence of the other array elements, taking into account coupling, can be related to the gain pattern Gi(𝜃, 𝜙) of an isolated array element pattern by Ge re(𝜃, 𝜙) = Gi(𝜃, 𝜙) [1 −|Γi(𝜃, 𝜙)|2] (8-83) Figure 8.27 displays the E-plane gain patterns of an isolated Gi(𝜃) and an active Ge re(𝜃), based on relationship of (8-83), printed dipole element in an infinite microstrip array . The active element 480 INTEGRAL EQUATIONS, MOMENT METHOD, AND SELF AND MUTUAL IMPEDANCES Substrate Substr Element sp 0 0.0 0.5 1.0 1.5 2.0 Element gain (dB) 2.5 3.0 3.5 10 20 30 40 Theta (degrees) 50 60 70 80 90 a height = 0.1 9 rate r = 2.55 cing: dx = dy= 0.5 [Gr e e( )] Active Isolated 9 0 d 0 θ [Gi ( )] θ λ λ Figure 8.27 E-plane element gain patterns of an Infinite Planar Array of microstrip dipoles. (source c ⃝1944 IEEE).
pattern displays a null around 46◦, due to scan blindness (reflection coefficient nearly unity) influ-enced primarily by the surface waves of the printed array, as was illustrated in Figure 8.26. The gain pattern of the isolated element does not display a null, and the isolated element pattern is quite dif-ferent than the active element pattern. This comparison indicates that mutual coupling in the array, due to surface waves within the substrate of the microstrip array, impacts the performance of the array, and thus, it should be taken into account.
8.8 MULTIMEDIA In the publisher’s website for the book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab and Fortran Method of Moment computer program, designated MOM, for comput-ing and displaying the radiation characteristics of dipoles, using Hall´ en’s and Pocklington’s integral equations.
c. Matlab and Fortran computer program, designated Impedance, for computing the self and mutual impedance of linear elements.
d. Power Point (PPT) viewgraphs, in multicolor.
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PROBLEMS 8.1.
Derive Pocklington’s integral equation (8-25) using (8-22)–(8-24c).
8.2.
Show that the incident tangential electric field (Ei z) generated on the surface of a wire of radius a by a magnetic field generator of (8-31) is given by (8-32).
PROBLEMS 483 8.3.
Reduce (8-32) to (8-33) valid only along the z axis (𝜌= 0).
8.4.
For the center-fed dipole of Example 8.3 write the [Z] matrix for N = 21 using for the gap the delta-gap generator and the magnetic-frill generator. Use the computer program MOM (Pocklington) of this chapter.
8.5.
For an infinitesimal center-fed dipole of l = λ∕50 and radius a = 0.005λ, derive the input impedance using Pocklington’s integral equation with piecewise constant subdomain basis functions and point-matching. Use N = 21 and model the gap as a delta-gap generator and as a magnetic-frill generator. Use the MOM (Pocklington) computer program of this chapter.
8.6.
Using the MOM (Hall´ en) computer program at the end of the chapter, compute the input impedance of a λ∕4 and 3λ∕4 dipole with an l/d ratio of l/d = 50 and 25. Use 20 subsec-tions. Compare the results with the impedances of a dipole with l/d = 109. Plot the current distribution and the far-field pattern of each dipole.
8.7.
Derive (8-50)-(8-52b) using (8-49), (3-2a), and (4-56).
8.8.
For a linear dipole with sinusoidal current distribution, radiating in free-space, find, using tab-ulated data or subroutines for the sine and cosine integrals, the radiation Zim and the input Zin impedances when a = λ∕20. Verify using the computer program Impedance of this chapter.
(a) l = λ∕4 (b) l = λ∕2 (c) l = 3λ∕4 (d) l = λ 8.9.
A λ∕2 dipole of finite radius is not self-resonant. However, if the dipole is somewhat less than λ∕2, it becomes self-resonant. For a dipole with radius of a = λ∕200 radiating in free-space, find the (a) nearest length by which the λ∕2 dipole becomes self-resonant (b) radiation resistance (referred to the current maximum) of the new resonant dipole (c) input resistance (d) VSWR when the dipole is connected to a 50-ohm line 8.10.
Find the length, at the first resonance, of linear dipoles with wire radii of (a) 10−5λ (b) 10−4λ (c) 10−3λ (d) 10−2λ Compute the radiation resistance of each.
8.11.
A quarter-wavelength monopole of radius a = 10−2λ is placed upon an infinite ground plane.
Determine the (a) impedance of the monopole (b) length by which it must be shortened to become self-resonant (first resonance) (c) impedance of the monopole when its length is that given in part b.
(d) VSWR when the monopole of part b is connected to a 50-ohm line.
8.12.
For two half-wavelength dipoles radiating in free-space, compute (using equations, not curves) the mutual impedance Z21m referred to the current maximum for (a) side-by-side arrangement with d = λ∕4 (b) collinear configuration with s = λ∕4 Verify using the computer program Impedance of this chapter.
8.13.
Two identical linear λ∕2 dipoles are placed in a collinear arrangement a distance s = 0.35λ apart. Find the driving-point impedance of each. Verify using the computer program Impedance of this chapter.
8.14.
Two identical linear λ∕2 dipoles are placed in a collinear arrangement. Find the spacings between them so that the driving-point impedance of each has the smallest reactive part.
CHAPTER9 Broadband Dipoles and Matching Techniques 9.1 INTRODUCTION In Chapter 4 the radiation properties (pattern, directivity, input impedance, mutual impedance, etc.) of very thin-wire antennas were investigated by assuming that the current distribution, which in most cases is nearly sinusoidal, is known. In practice, infinitely thin (electrically) wires are not realizable but can be approximated. In addition, their radiation characteristics (such as pattern, impedance, gain, etc.) are very sensitive to frequency. The degree to which they change as a function of frequency depends on the antenna bandwidth. For applications that require coverage over a broad range of frequencies, such as television reception of all channels, wide-band antennas are needed. There are numerous antenna configurations, especially of arrays, that can be used to produce wide bandwidths.
Some simple and inexpensive dipole configurations, including the conical and cylindrical dipoles, can be used to accomplish this to some degree.
For a finite diameter wire (usually d > 0.05λ) the current distribution may not be sinusoidal and its effect on the radiation pattern of the antenna is usually negligible. However, it has been shown that the current distribution has a pronounced effect on the input impedance of the wire antenna, especially when its length is such that a near null in current occurs at its input terminals. The effects are much less severe when a near current maximum occurs at the input terminals.
Historically there have been three methods that have been used to take into account the finite conductor thickness. The first method treats the problem as boundary-value problem , the second as a tapered transmission line or electromagnetic horn , and the third finds the current distribution on the wire from an integral equation . The boundary-value approach is well suited for idealis-tic symmetrical geometries (e.g., ellipsoids, prolate spheroids) which cannot be used effectively to approximate more practical geometries such as the cylinder. The method expresses the fields in terms of an infinite series of free oscillations or natural modes whose coefficients are chosen to satisfy the conditions of the driving source. For the assumed idealized configurations, the method does lead to very reliable data, but it is very difficult to know how to approximate more practical geometries (such as a cylinder) by the more idealized configurations (such as the prolate spheroid). For these reasons the boundary-value method is not very practical and will not be pursued any further in this text.
In the second method Schelkunoff represents the antenna as a two-wire uniformly tapered trans-mission line, each wire of conical geometry, to form a biconical antenna. Its solution is obtained by applying transmission-line theory (incident and reflected waves), so well known to the average engineer. The analysis begins by first finding the radiated fields which in turn are used, in conjunction with transmission-line theory, to find the input impedance.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 485 486 BROADBAND DIPOLES AND MATCHING TECHNIQUES (a) classic (narrow BW) (b) biconical (intermediate BW) (c) tapered (intermediate BW) (d) hemispherical (wide BW) r r r r Figure 9.1 Dipole configurations and associated qualitative bandwidths (BW).
For the third technique, the main objectives are to find the current distribution on the antenna and in turn the input impedance. These were accomplished by Hall´ en by deriving an integral equation for the current distribution whose approximate solution, of different orders, was obtained by iter-ation and application of boundary conditions. Once a solution for the current is formed, the input impedance is determined by knowing the applied voltage at the feed terminals.
The details of the second method will follow in summary form. The integral equation technique of Hall´ en, along with that of Pocklington, form the basis of Moment Method techniques which were discussed in Chapter 8.
One of the main objectives in the design of an antenna is to broadband its characteris-tics—increase its bandwidth. Usually, this is a daunting task, especially if the specifications are quite ambitious. Typically, the response of each antenna, versus frequency, can be classified qualitatively into three categories: narrowband, intermediate band, and wide band. In Chapter 11, Section 11.5, it is pointed out that the bandwidth of an antenna (which can be enclosed within a sphere of radius r) can be improved only if the antenna utilizes efficiently, with its geometrical configuration, the available volume within the sphere. In Figure 9.1, we exhibit four different dipole configurations, starting with the classic dipole in Figure 9.1(a) and concluding with the hemispherical dipole of Figure 9.1(d). If we were to examine the frequency characteristics of a dipole on the basis of this fundamental principle, we can qualitatively categorize the frequency response of the different dipole configurations of Figure 9.1 into three groups; narrowband, intermediate band, and wide band. The same can be concluded for the geometries of the four monopole geometries of Figure 9.2. It should BICONICAL ANTENNA 487 (c) tapered (intermediate BW) (a) classical (narrow BW) (b) conical (intermediate BW) (d) hemispherical (wide BW) r r r r Figure 9.2 Monopole configurations and associated qualitative bandwidths (BW).
be pointed out that the third configuration (tapered) in each figure will provide the best reflection (matching) efficiency when fed from traditional transmission lines. Although in each of the previous two figures, Figures 9.1 and 9.2, the last two configurations (d and e in each) exhibit the most broad-band characteristics, usually these geometries are not as convenient and economical for practical implementation. However, any derivatives of these geometries, especially two-dimensional simula-tions, are configurations that may be used to broadband the frequency characteristics.
Since the biconical antenna of Figure 9.1(b) and derivatives of it are classic configurations with practical applications, they will be examined in some detail in the section that follows.
9.2 BICONICAL ANTENNA One simple configuration that can be used to achieve broadband characteristics is the biconical antenna formed by placing two cones of infinite extent together, as shown in Figure 9.3(a). This can be thought to represent a uniformly tapered transmission line. The application of a voltage Vi at the input terminals will produce outgoing spherical waves, as shown in Figure 9.3(b), which in turn produce at any point (r, 𝜃= 𝜃c, 𝜙) a current I along the surface of the cone and voltage V between the cones (Figure 9.4). These can then be used to find the characteristic impedance of the transmis-sion line, which is also equal to the input impedance of an infinite geometry. Modifications to this expression, to take into account the finite lengths of the cones, will be made using transmission-line analogy.
9.2.1 Radiated Fields The analysis begins by first finding the radiated E- and H-fields between the cones, assuming dom-inant TEM mode excitation (E and H are transverse to the direction of propagation). Once these are 488 BROADBAND DIPOLES AND MATCHING TECHNIQUES z y x r Transmission feed line (a) Biconical geometry (b) Spherical waves Vi /2 α α/2 θ ϕ Figure 9.3 Biconical antenna geometry and radiated spherical waves.
Vi V(r) V(r) I(r) I(r) Hϕ Eθ r Figure 9.4 Electric and magnetic fields, and associated voltages and currents, for a biconical antenna.
BICONICAL ANTENNA 489 determined for any point (r, 𝜃, 𝜙), the voltage V and current I at any point on the surface of the cone (r, 𝜃= 𝜃c, 𝜙) will be formed. From Faraday’s law we can write that ∇× E = −j𝜔𝜇H (9-1) which when expanded in spherical coordinates and assuming that the E-field has only an E𝜃com-ponent independent of 𝜙, reduces to ∇× E = ̂ a𝜙 1 r 𝜕 𝜕r(rE𝜃) = −j𝜔𝜇(̂ arHr + ̂ a𝜃H𝜃+ ̂ a𝜙H𝜙) (9-2) Since H only has an H𝜙component, necessary to form the TEM mode with E𝜃, (9-2) can be written as 1 r 𝜕 𝜕r(rE𝜃) = −j𝜔𝜇H𝜙 (9-2a) From Ampere’s law we have that ∇× H = +j𝜔𝜀E (9-3) which when expanded in spherical coordinates, and assuming only E𝜃and H𝜙components indepen-dent of 𝜙, reduces to ̂ ar 1 r2 sin 𝜃 [ 𝜕 𝜕𝜃(r sin 𝜃H𝜙) ] −̂ a𝜃 1 r sin 𝜃 [ 𝜕 𝜕r (r sin 𝜃H𝜙) ] = +j𝜔𝜀(̂ a𝜃E𝜃) (9-4) which can also be written as 𝜕 𝜕𝜃(r sin 𝜃H𝜙) = 0 (9-4a) 1 r sin 𝜃 𝜕 𝜕r(r sin 𝜃H𝜙) = −j𝜔𝜀E𝜃 (9-4b) Rewriting (9-4b) as 1 r 𝜕 𝜕r(rH𝜙) = −j𝜔𝜀E𝜃 (9-5) and substituting it into (9-2a) we form a differential equation for H𝜙as − 1 j𝜔𝜀r 𝜕 𝜕r [ 𝜕 𝜕r(rH𝜙) ] = −j𝜔𝜇H𝜙 (9-6) or 𝜕2 𝜕r2 (rH𝜙) = −𝜔2𝜇𝜀(rH𝜙) = −k2(rH𝜙) (9-6a) A solution for (9-6a) must be obtained to satisfy (9-4a). To meet the condition of (9-4a), the 𝜃vari-ations of H𝜙must be of the form H𝜙= f(r) sin 𝜃 (9-7) 490 BROADBAND DIPOLES AND MATCHING TECHNIQUES A solution of (9-6a), which also meets the requirements of (9-7) and represents an outward traveling wave, is H𝜙= H0 sin 𝜃 e−jkr r (9-8) where f(r) = H0 e−jkr r (9-8a) An inward traveling wave is also a solution but does not apply to the infinitely long structure.
Since the field is of TEM mode, the electric field is related to the magnetic field by the intrinsic impedance, and we can write it as E𝜃= 𝜂H𝜙= 𝜂H0 sin 𝜃 e−jkr r (9-9) In Figure 9.4(a) we have sketched the electric and magnetic field lines in the space between the two conical structures. The voltage produced between two corresponding points on the cones, a distance r from the origin, is found by V(r) = ∫ 𝜋−𝛼∕2 𝛼∕2 E ⋅dl = ∫ 𝜋−𝛼∕2 𝛼∕2 (̂ a𝜃E𝜃) ⋅(̂ a𝜃rd𝜃) = ∫ 𝜋−𝛼∕2 𝛼∕2 E𝜃r d𝜃 (9-10) or by using (9-9) V(r) = 𝜂H0e−jkr ∫ 𝜋−𝛼∕2 𝛼∕2 d𝜃 sin 𝜃= 𝜂H0e−jkr ln [cot(𝛼∕4) tan(𝛼∕4) ] V(r) = 2𝜂H0e−jkr ln [ cot (𝛼 4 )] (9-10a) The current on the surface of the cones, a distance r from the origin, is found by using (9-8) as I(r) = ∫ 2𝜋 0 H𝜙r sin 𝜃d𝜙= H0e−jkr ∫ 2𝜋 0 d𝜙= 2𝜋H0e−jkr (9-11) In Figure 9.4(b) we have sketched the voltage and current at a distance r from the origin.
9.2.2 Input Impedance A. Infinite Cones Using the voltage of (9-10a) and the current of (9-11), we can write the characteristic impedance as Zc = V(r) I(r) = 𝜂 𝜋ln [ cot (𝛼 4 )] (9-12) Since the characteristic impedance is not a function of the radial distance r, it also represents the input impedance at the antenna feed terminals of the infinite structure. For a free-space medium, BICONICAL ANTENNA 491 (9-12) reduces to Zc = Zin = 120 ln [ cot (𝛼 4 )] (9-12a) which is a pure resistance. For small cone angles Zin = 𝜂 𝜋ln [ cot (𝛼 4 )] = 𝜂 𝜋ln [ 1 tan(𝛼∕4) ] ≃𝜂 𝜋ln ( 4 𝛼 ) (9-12b) Variations of Zin as a function of the half-cone angle 𝛼∕2 are shown plotted in Figure 9.5(a) for 0◦< 𝛼∕2 ≤90◦and in Figure 9.5(b) in an expanded scale for 0◦< 𝛼∕2 ≤2◦. Although the half-cone angle is not very critical in the design, it is usually chosen so that the characteristic impedance of the biconical configuration is nearly the same as that of the transmission line to which it will be attached. Small angle biconical antennas are not very practical but wide-angle configurations (30◦< 𝛼∕2 < 60◦) are frequently used as broadband antennas.
The radiation resistance of (9-12) can also be obtained by first finding the total radiated power Prad = ∯ S Wav ⋅ds = ∫ 2𝜋 0 ∫ 𝜋−𝛼∕2 𝛼∕2 |E|2 2𝜂r2 sin 𝜃d𝜃d𝜙= 𝜋𝜂|H0|2 ∫ 𝜋−𝛼∕2 0 d𝜃 sin 𝜃 Prad = 2𝜋𝜂|H0|2 ln [ cot (𝛼 4 )] (9-13) and by using (9-11) evaluated at r = 0 we form Rr = 2Prad [I(0)]2 = 𝜂 𝜋ln [ cot (𝛼 4 )] (9-14) which is identical to (9-12).
B. Finite Cones The input impedance of (9-12) or (9-14) is for an infinitely long structure. To take into account the finite dimensions in determining the input impedance, Schelkunoff has devised an ingenious method where he assumes that for a finite length cone (r = l∕2) some of the energy along the surface of the cone is reflected while the remaining is radiated. Near the equator most of the energy is radi-ated. This can be viewed as a load impedance connected across the ends of the cones. The electrical equivalent is a transmission line of characteristic impedance Zc terminated in a load impedance ZL.
Values of the input impedance (resistance and reactance) have been computed, for small angle cones, in where it is shown the antenna becomes more broadband (its resistance and reactance variations are less severe) as the cone angle increases.
The biconical antenna represents one of the canonical problems in antenna theory, and its model is well suited for examining general characteristics of dipole-type antennas.
C. Unipole Whenever one of the cones is mounted on an infinite plane conductor (i.e., the lower cone is replaced by a ground plane), it forms a unipole and its input impedance is one-half of the two-cone structure, as is the case of a λ∕4 monopole compared to that of a dipole, and it is represented by (4-106).
Input impedances for unipoles of various cone angles as a function of the antenna length l have been 492 BROADBAND DIPOLES AND MATCHING TECHNIQUES 2 2 2 2 2 in in in in Figure 9.5 Input impedance of an infinitely long biconical antenna radiating in free-space.
measured . Radiation patterns of biconical dipoles fed by coaxial lines have been computed by Papas and King .
9.3 TRIANGULAR SHEET, FLEXIBLE AND CONFORMAL BOW-TIE, AND WIRE SIMULATION Because of their broadband characteristics, biconical antennas have been employed for many years in the VHF and UHF frequency ranges. However, the solid or shell biconical structure is so massive for most frequencies of operation that it is impractical to use. Because of its attractive radiation characteristics, compared to those of other single antennas, realistic variations to its mechanical structure have been sought while retaining as many of the desired electrical features as possible.
TRIANGULAR SHEET, FLEXIBLE AND CONFORMAL BOW-TIE, AND WIRE SIMULATION 493 Figure 9.6 Triangular sheet, bow-tie, and wire simulation of biconical antenna.
Geometrical approximations of the solid or shell conical unipole or biconical antenna are the trian-gular sheet and bow-tie antennas shown in Figures 9.6(a, b), respectively, each fabricated from sheet metal. The triangular sheet has been investigated experimentally by Brown and Woodward . Each of these antennas can also be simulated by a wire along the periphery of its surface, which reduces significantly the weight and wind resistance of the structure. This was done in using the Method of Moments. In order to simulate better the attractive surface of revolution of a biconical antenna of low-mass structures, multielement intersecting wire bow-ties were employed, as shown in Figure 9.6(c). It has been shown that eight or more intersecting wire-constructed bow-ties can approximate reasonably well the radiation characteristics of solid conical body-of-revolution antenna.
The bow-tie of Figure 9.6(b) was investigated more in depth in , using a flexible thin sub-strate . The substrate is a thin plastic (heat-stabilized PEN), allowing the antenna to be flexible.
The substrate is covered with a very thin silicon nitride layer, which is the gate dielectric . The conducting material used for the feed network, balun, and the antenna element is aluminum. Fig-ure 9.7 illustrates a simplified model for the flexible substrate that best approximates the electrical properties of the actual substrate, also referred to as plastic in .
A bow-tie design, shown in Figure 9.8(a) along with its balun, was selected as the design because of its basic geometry, broadband characteristics, and variety of applications when compared to a linear wire and a printed dipole. Furthermore, bow-tie antennas are expected to be more directive than conventional dipole antennas because of the larger radiating area . They are also used in size reduction applications of patch antennas to achieve lower operating frequencies without increasing the overall patch area.
The coplanar bow-tie – requires a balanced feed network, so that the antenna can be fed by a microstrip line or a coplanar waveguide with the use of a balun. The design procedure of a printed Figure 9.7 Simplified geometry PEN (polyethylene naphthalate) plastic. (source: c ⃝2011 IEEE).
494 BROADBAND DIPOLES AND MATCHING TECHNIQUES Figure 9.8 Flexible bow-tie antenna on a thin plastic (PEN) substrate at 7.66 GHz. (source: c ⃝2011 IEEE).
bow-tie antenna is similar to the design of rectangular microstrip patches. There is a set of design equations, which are obtained by modifying the semi-empirical design equations for rectangular patches of Chapter 14. The resonant frequency of a bow-tie patch, for the dominant TM10 mode, can be obtained using the equations that follow , , .
(fr)TM 10 = 1 2√𝜀reff L ( 1.152 Rt ) (9-15a) Rt = L 2 (W + 2ΔL) + (Wc + 2ΔL) (W + 2ΔL) + (S + 2ΔL) (9-15b) ΔL h = 0.412 (𝜀reff + 0.3) ( Wi h + 0.264 ) (𝜀reff −0.258) ( Wi h + 0.813 ) (9-15c) 𝜀reff = 𝜀r + 1 2 + 𝜀r −1 2 [ 1 + 12 h Wi ]−1∕2 (9-15d) Wi = (W + Wc 2 ) (9-15e) In this set of design equations, the thickness, relative and effective permittivity of the substrate, are denoted by h, 𝜀r, and 𝜀reff , respectively. The other geometrical parameters are defined in Figure 9.8(a). Although these equations were primarily derived for microstrip patch type bow-ties, they were used to obtain an initial design of a coplanar bow-tie antenna. Afterward, the antenna design TRIANGULAR SHEET, FLEXIBLE AND CONFORMAL BOW-TIE, AND WIRE SIMULATION 495 (a) Photo of bow-tie in flexible mode (b) S11 of flexible bow-tie Figure 9.9 Photo and S11 of flexible bow-tie antenna. (source: c ⃝2011 IEEE).
was fine-tuned and finalized by numerical simulations . There was a slight difference between the initial and the final values because of the approximate form of the design equations and the existence of the microstrip to coplanar feed network balun.
The balun introduces an 180◦phase difference between the coupled microstrip lines near the cen-ter frequency. The length of the phase shifter is a very important parameter for the balun design. The lengths of the two branches of the microstrip line should be adjusted such that their difference is equal to one-quarter of the guided wavelength at the center frequency . Another critical param-eter is the gap between the coplanar strip lines. This gap can be adjusted to optimize the balun’s performance .
Based on these equations and procedure, a bow-tie was designed, using the flexible substrate of Figure 9.7, to resonate at 7.66 GHz; the overall dimensions are indicated in Figure 9.8(b). A plot of the current density at 7.66 GHz is displayed in Figure 9.8(c), which is most intense along its edges.
A photo of the bow tie, in its flexible mode, is shown in Figure 9.9(a). Its return loss is shown in Figure 9.9(b). It is worth to note that the bandwidth of the bow-tie, with respect to −15 dB (VSWR < 1.5:1) return loss, was decreased from 15% to 8.75% after the inclusion of the balun. This is due to the rapid change of the phase shift, introduced by the balun, with respect to frequency. This observation verifies that the balun is the critical device in the design of the bandwidth, and that the balun determines the bandwidth of the overall design.
In addition to return loss, amplitude radiation patterns of the antennas were simulated, both 3-D and 2-D, and they are displayed in Figure 9.10. The 2-D patterns, each normalized to its own maximum, were simulated in three planes: principal E-plane (y-z plane), secondary E-plane (x-z plane), and H-plane (x-y plane), and they were compared with measurements. The secondary E-plane (x-z plane) is defined as the one along which the E-field is parallel to it but does not pass through the overall field maximum. It can be seen that the measured radiation patterns are in excellent agreement with the simulated ones in all of the three planes. Although the pattern in the secondary E-plane is very close to the pattern of an ideal dipole, the patterns in the principal E- and H-planes are noticeably distorted. The back lobes of the patterns are approximately 10 dB lower compared to the forward lobes. This difference is due to the presence of the ground plane, which “pushes” the radiation pattern against itself and toward the bow-tie. This structure can also be considered as a 2-element Yagi-Uda antenna , which is composed of the bow-tie dipole and the ground plane. Yagi-Uda antennas are discussed in detail in Section 10.3.3. The ground plane acts as the reflector of the Yagi antenna, which has a decreased backward radiation. Therefore, the direction of the peak gain is away from the ground plane.
496 BROADBAND DIPOLES AND MATCHING TECHNIQUES (a) 3-D pattern (b) y-z (principal E-plane) (c) x-z (secondary E-plane) (d) x-y (H-plane) Figure 9.10 Normalized 3-D and 2-D amplitude patterns of bow-tie at 7.66 GHz. (source: c ⃝2011 IEEE).
In addition to the solid flexible bow-tie, an outline bow-tie was designed, simulated, and mea-sured, where the center part of the metallic structure was removed , . The outline bow-tie resonated at a lower frequency (7.4 GHz, instead of 7.66 GHz for the solid) because electrically was larger, due to the removal of the center part of the metallic structure where the current density is less intense and not critical to the performance of the radiating bow-tie. For more details, the reader is directed to , .
9.4 VIVALDI ANTENNA The Vivaldi antenna is a broadband end-fire traveling wave type introduced by Gibson in 1979 .
Its basic geometry is shown in Figure 9.11, and it usually is implemented on a substrate with the Vivaldi design etched on the upper cladding of the substrate. The basic structure consists of a λs∕4 uniform slot that is connected to an exponentially tapered slot; the subscript s is used to identify the slot. The slot is excited/fed by a microstrip transmission line from the undersurface of the substrate, as shown in Figure 9.11. An alternate design is to use a resonant area, typically either square or circular, instead of the λs∕4 uniform slot, which is also usually excited by a microstrip line. The resonant area is connected to an exponentially tapered slot, with or without a transmission line.
The Vivaldi antenna is sometimes referred to as tapered slot antenna (TSA), flared or notch antenna, end-fire slot, and other related names. Because of its planar structure, it exhibits attractive VIVALDI ANTENNA 497 Vivaldi microstrip transmission line px Ae ± L max W TL W min W 4 s m 4 (a) 3-D view (b) Top view r substrate microstrip transmission line h max W (c) Right front view z y x Vivaldi H-Plane E-Plane z λ λ Figure 9.11 Vivaldi antenna geometry.
geometrical characteristics, especially to be integrated with MMICs (Monolithic Microwave Inte-grated Circuits). The Vivaldi design is usually low cost, and it possesses excellent radiation charac-teristics, such as high gain, broadband performance, constant beamwidth, and low side lobes. The directivity of Vivaldi antennas increases as the length L of the antenna increases, achieving gains up to 17 dB . When first presented by Gibson , it had achieved, as a single element, an impedance bandwidth from below 2 GHz to above 40 GHz with a gain of 10 dB and side lobes of −20 dB. Subsequent wide bandwidth applications have used the Vivaldi antenna geometry in arrays , , including active electronically scanned arrays (AESAs).
One may question as to where the name came from, and why the name of Antonio Vivaldi, a composer from the Baroque period, is associated with an antenna. The question is best answered in , where according to an article on the Pharos JRA focal plane array, it “received its name from a resemblance to the shape of a cello or violin, instruments used by Antonio Vivaldi, the designer’s favorite composer.” Gibson himself was a musician, composer, and teacher of music.
Vivaldi antennas are categorized as broadband or frequency independent antennas, with band-width up to 6:1 ; even 10:1 or greater for VSWR < 2 (S11 < −10 dB). For arrays, the main lobe of the pattern is nearly proportional to cos(𝜃), and it is basically maintained for scan angles up to about 50◦–60◦. The bandwidth is limited by the opening width Wmin and the aperture width Wmax of the antenna.
This antenna is usually incorporated with MMICs; thus, the bandwidth is also limited by the transition between the microstrip line (which connects to the MMICs) and the slot line of the antenna, a subject of interest in . The proper design of this transition results in a broad-band performance that matches the antenna performance . The simpler transition is a λm∕4 open microstrip and uniform λs∕4 slot line, where λm and λs are the wavelengths at the cen-ter frequency; the subscript m is used to identify the microstrip line. A coplanar waveguide feed can also be used, and it presents a wider bandwidth. Also, baluns can be used with Vivaldi antennas to make them more compatible when connected to strip lines and microstrips. A com-monly used balun is that adapted from Knorr’s microstrip-to-slot transition , . Over the entire bandwidth, the beamwidth is nearly constant . Furthermore, Vivaldi antennas exhibit 498 BROADBAND DIPOLES AND MATCHING TECHNIQUES a symmetric end-fire beam ; that is, the beamwidth is approximately the same in both the E-plane (parallel to the substrate) and the H-plane (perpendicular to the substrate). As the length of the antenna increases, the beamwidth narrows .
As can be observed in Figure 9.11, the antenna is an exponentially tapered slot cut in a thin film of metal that is supported by a substrate. The exponential taper can be defined by y(x) = ±Aepx (9-16) where y is the half separation of the slot and x is the position across the length of the antenna, A is half of the opening width Wmin, and p is the taper rate. Larger values of the rate of change p improve the low-frequency resistance but simultaneously create larger variations in resistance and reactance over the entire band. Therefore, for wide bandwidth applications, a compromise is usually required between the taper rate p and the square resonant/cavity area . The taper rate has a significant impact on the bandwidth and beamwidth of the antenna. In general, as the taper rate increases, the beamwidth in the E-plane increases, the beamwidth in the H-plane decreases, and the bandwidth increases.
Parametric studies have shown that the optimal performance is achieved when the length L is greater than one wavelength at the lowest frequency. The opening width Wmin is based on the highest frequency and the aperture width Wmax influences the lower frequency . Furthermore, the value of the aperture width Wmax typically should, based on parametric examinations, be in the range Wmax1 and Wmax2, where Wmax1 ≈λ0 (9-17) and Wmax2 ≈λmin 2 (9-18) such that Wmax1 < Wmax < Wmax2, where λmin is the wavelength at the minimum frequency and λ0 is the wavelength at the center frequency .
The feed of Vivaldi antennas is usually a microstrip line that connects it to the MMICs, as illus-trated in Figure 9.11. The microstrip line is printed on a different layer inside the substrate, and as mentioned before, the transition between the microstrip line and the slot line of the antenna should be designed properly so as not to introduce a significant limitation in the bandwidth.
Example 9.1 Design a Vivaldi antenna with a center frequency of 10 GHz using a substrate with permittivity of 2.33 and thickness of 0.508 mm. Assume that the minimum frequency is 4 GHz.
Solution: With (9-17) and (9-18) used as a guideline, the range of values for the aperture width is Wmax1 < Wmax < Wmax2 Wmax1 = λ0 = 3 × 108 (10 × 109) √ 2.33 = 19.65 ≈20 mm Wmax2 = λmin 2 = 3 × 108 2(4 × 109) √ 2.33 = 24.57 ≈25 mm VIVALDI ANTENNA 499 Through simulations, an aperture width of Wmax = 25 mm led to the best performance and was used in the design. The length L was selected to be 27 mm, Wmin = 2A = 0.1 mm, and p = 0.204.
Based on these dimensions, the S11 of the design is illustrated in Figure 9.12. Using S11 = −10 dB as a criterion, the low and high frequencies are 8.6 GHz and 23.9 GHz, respectively, leading to a bandwidth of 2.77: 1. However the antenna operates below 8.6 GHz but is not as well matched.
–45 –40 –35 –30 –25 –20 –15 –10 –5 S11 (dB) 0 2.5 0 5 7.5 10 12.5 Frequency (GHz) BW (–10dB) = 2.77 : 1 15 17.5 20 22.5 f = 23.9 GHz f = 8.6 GHz 25 Figure 9.12 S11 for the Vivaldi antenna.
The 3-D radiation pattern and the 2-D patterns in the E- and H-planes, are shown in Figures 9.13 and 9.14, respectively.
Figure 9.13 3-D radiation pattern for the Vivaldi antenna at 10 GHz.
500 BROADBAND DIPOLES AND MATCHING TECHNIQUES (a) E-plane (b) H-plane θ θ ϕ -15 dB -10 dB -5 dB 0 dB 30° 210° 60° 240° 90° 270° 120° 300° 150° 330° 180° 0° -15 dB -10 dB -5 dB 0 dB 30° 150° 60° 120° 90° 90° 120° 60° 150° 30° 180° 0° Figure 9.14 2-D patterns for the Vivaldi antenna at 10 GHz.
The current density distribution at 10 GHz, associated with the design of Example 9.1 is displayed in Figure 9.15.
Figure 9.15 Current density for the Vivaldi antenna at 10 GHz.
9.5 CYLINDRICAL DIPOLE Another simple and inexpensive antenna whose radiation characteristics are frequency dependent is a cylindrical dipole (i.e., a wire of finite diameter and length) of the form shown in Figure 9.16.
CYLINDRICAL DIPOLE 501 l/2 l/2 θ θ d = 2a x z′ dz′ ρ r y R A B Figure 9.16 Center-fed cylindrical antenna configuration.
Thick dipoles are considered broadband while thin dipoles are more narrowband. This geometry can be considered to be a special form of the biconical antenna when 𝛼= 0◦. A thorough analysis of the current, impedance, pattern, and other radiation characteristics can be performed using the Moment Method. With that technique the antenna is analyzed in terms of integral formulations of the Hall´ en and Pocklington type which can be evaluated quite efficiently by the Moment Method.
The analytical formulation of the Moment Method has been presented in Chapter 8. In this section we want to present, in summary form, some of its performance characteristics.
9.5.1 Bandwidth As has been pointed out previously, a very thin linear dipole has very narrowband input impedance characteristics. Any small perturbations in the operating frequency will result in large changes in its operational behavior. One method by which its acceptable operational bandwidth can be enlarged will be to decrease the l/d ratio. For a given antenna, this can be accomplished by holding the length the same and increasing the diameter of the wire. For example, an antenna with a l/d ≃5,000 has an acceptable bandwidth of about 3%, which is a small fraction of the center frequency. An antenna of the same length but with a l/d ≃260 has a bandwidth of about 30%.
9.5.2 Input Impedance The input impedance (resistance and reactance) of a very thin dipole of length l and diameter d can be computed using (8-60a)–(8-61b). As the radius of the wire increases, these equations become inaccurate. However, using integral equation analyses along with the Moment Method of Chapter 8, input impedances can be computed for wires with different l/d ratios. In general, it has been observed that for a given length wire its impedance variations become less sensitive as a function of frequency as the l/d ratio decreases. Thus more broadband characteristics can be obtained by increasing the 502 BROADBAND DIPOLES AND MATCHING TECHNIQUES 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 0 200 400 600 800 1000 Input resistance (ohms) Dipole length (wavelengths) (a) Input resistance l/d = 25 (Moment method) l/d = 50 (Moment method) l/d = 104 (Sinusoidal current) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 –600 –400 –200 0 200 400 Input reactance (ohms) Dipole length (wavelengths) (b) Input reactance l/d = 25 (Moment method) l/d = 50 (Moment method) l/d = 104 (Sinusoidal current) Figure 9.17 (a) Input resistance and (b) reactance of wire dipoles.
diameter of a given wire. To demonstrate this, in Figures 9.17(a) and (b) we have plotted, as a function of electrical length (as a function of frequency for a constant physical length), the input resistance and reactance of dipoles with l∕d = 104(Ω = 19.81), 50(Ω = 9.21), and 25(Ω = 7.824) where Ω = 2 ln(2l∕d) and d is the diameter. For l∕d = 104 the values were computed using (8-60a) and (8-61a) and then transferred to the input terminals by (8-60b) and (8-61b), respectively. The others were computed using the Moment Method techniques of Chapter 8. It is noted that the variations of each are less pronounced as the l/d ratio decreases, thus providing greater bandwidth.
Measured input resistances and reactances for a wide range of constant l/d ratios have been reported . These curves are for a cylindrical antenna driven by a coaxial cable mounted on a large ground plane on the earth’s surface. Thus they represent half of the input impedance of a center-fed cylindrical dipole radiating in free-space. The variations of the antenna’s electri-cal length were obtained by varying the frequency while the length-to-diameter (l/d) ratio was held constant.
CYLINDRICAL DIPOLE 503 TABLE 9.1 Cylindrical Dipole Resonances First Resonance Second Resonance Third Resonance Fourth Resonance LENGTH 0.48λF 0.96λF 1.44λF 1.92λF RESISTANCE (ohms) 67 R2 n 67 95 R2 n 95 F = l∕2a 1 + l∕2a; Rn = 150 log10(l∕2a) 9.5.3 Resonance and Ground Plane Simulation The imaginary part of the input impedance of a linear dipole can be eliminated by making the total length, l, of the wire slightly less than an integral number of half-wavelengths (i.e., l slight less than nλ∕2, n = 1, 3, …) or slightly greater than an integral number of wavelengths (i.e., l slightly greater than nλ, n = 1, 2, …). The amount of reduction or increase in length, is a function of the radius of the wire, and it can be determined for thin wires iteratively using (8-60b) and (8-61b). At the resonance length, the resistance can then be determined using (8-60a) and (8-61a). Empirical equations for approximating the length, impedance, and the order of resonance of the cylindrical dipoles are found in Table 9.1 . Rn is called the natural resistance and represents the geometric mean resistance at an odd resonance and at the next higher even resonance. For a cylindrical stub above a ground plane, as shown in Figure 9.11, the corresponding values are listed in Table 9.2 .
To reduce the wind resistance, to simplify the design, and to minimize the costs, the ground plane of Figure 9.18(a) is often simulated, especially at low frequencies, by crossed wires as shown in Figure 9.18(b). Usually only two crossed wires (four radials) are employed. A larger number of radials results in a better simulation of the ground plane. Ground planes are also simulated by wire mesh. The spacing between the wires is usually selected to be equal or smaller than λ∕10. The flat or shaped reflecting surfaces for UHF educational TV are usually realized approximately by using wire mesh.
9.5.4 Radiation Patterns The theory for the patterns of infinitesimally thin wires was developed in Chapter 4. Although accu-rate patterns for finite diameter wires can be computed using current distributions obtained by the Moment Method of Chapter 8, the patterns calculated using ideal sinusoidal current distributions, valid for infinitely small diameters, provide a good first-order approximation even for relatively thick cylinders. To illustrate this, in Figure 9.19 we have plotted the relative patterns for l = 3λ∕2 with l∕d = 104(Ω = 19.81), 50(Ω = 9.21), 25(Ω = 6.44), and 8.7(Ω = 5.71), where Ω = 2 ln(2l∕d) and d is the diameter. For l∕d = 104 the current distribution was assumed to be purely sinusoidal, as given by (4-56); for the others, the Moment Method techniques of Chapter 8 were used. The patterns were computed using the Moment Method formulations outlined in Section 8.4. It is noted that the TABLE 9.2 Cylindrical Stub Resonances First Resonance Second Resonance Third Resonance Fourth Resonance LENGTH 0.24λF′ 0.48λF′ 0.72λF′ 0.96λF′ RESISTANCE (ohms) 34 (Rn ′)2 34 48 (Rn ′)2 48 F′ = l∕a 1 + l∕a; Rn ′ = 75 log10(l∕a) 504 BROADBAND DIPOLES AND MATCHING TECHNIQUES Figure 9.18 Cylindrical monopole above circular solid and wire-simulated ground planes.
pattern is essentially unaffected by the thickness of the wire in regions of intense radiation. However, as the radius of the wire increases, the minor lobes diminish in intensity and the nulls are filled by low-level radiation. The same characteristics have been observed for other length dipoles such as l = λ∕2, λ and 2λ. The input impedance for the l = λ∕2 and l = 3λ∕2 dipoles, with l∕d = 104, 50, and 25, is equal to l = 𝛌∕2 l = 3𝛌∕2 Zin(l∕d = 104) = 73 + j42.5 Zin(l∕d = 104) = 105.49 + j45.54 Zin(l∕d = 50) = 85.8 + j54.9 Zin(l∕d = 50) = 103.3 + j9.2 (9-19) Zin(l∕d = 25) = 88.4 + j27.5 Zin(l∕d = 25) = 106.8 + j4.9 9.5.5 Equivalent Radii Up to now, the formulations for the current distribution and the input impedance assume that the cross section of the wire is constant and of radius a. An electrical equivalent radius can be obtained for some uniform wires of noncircular cross section. This is demonstrated in Table 9.3 where the actual cross sections and their equivalent radii are illustrated.
The equivalent radius concept can be used to approximate the antenna or scattering characteristics of electrically small wires of arbitrary cross sections. It is accomplished by replacing the noncircu-lar cross-section wire with a circular wire whose radius is the “equivalent” radius of the noncircular cross section. In electrostatics, the equivalent radius represents the radius of a circular wire whose FOLDED DIPOLE 505 Figure 9.19 Amplitude radiation patterns of a 3λ∕2 dipole of various thicknesses.
capacitance is equal to that of the noncircular geometry. This definition can be used at all frequen-cies provided the wire remains electrically small. The circle with equivalent radius lies between the circles which circumscribe and inscribe the geometry and which together bound the noncircular cross section.
9.6 FOLDED DIPOLE To achieve good directional pattern characteristics and at the same time provide good matching to practical coaxial lines with 50- or 75-ohm characteristic impedances, the length of a single wire element is usually chosen to be λ∕4 ≤l < λ. The most widely used dipole is that whose overall length is l ≃λ∕2, and which has an input impedance of Zin ≃73 + j42.5 and directivity of D0 ≃1.643. In practice, there are other very common transmission lines whose characteristic impedance is much 506 BROADBAND DIPOLES AND MATCHING TECHNIQUES TABLE 9.3 Conductor Geometrical Shapes and their Equivalent Circular Cylinder Radii Geometrical Shape Electrical Equivalent Radius ae = 0.25a ae ≃0.2(a + b) ae = 0.59a ae = 1 2(a + b) ln ae ≃ 1 (S1 + S2)2 × [S2 1 ln a1 + S2 2 ln a2 + 2S1S2 ln s] S1, S2 = peripheries of conductors C1, C2 a1, a2 = equivalent radii of conductors C1, C2 higher than 50 or 75 ohms. For example, a “twin-lead” transmission line (usually two parallel wires separated by about 5 16 in. and embedded in a low-loss plastic material used for support and spacing) is widely used for TV applications and has a characteristic impedance of about 300 ohms.
In order to provide good matching characteristics, variations of the single dipole element must be used. One simple geometry that can achieve this is a folded wire which forms a very thin (s ≪λ) rect-angular loop as shown in Figure 9.20(a). This antenna, when the spacing between the two larger sides is very small (usually s < 0.05λ), is known as a folded dipole and it serves as a step-up impedance transformer (approximately by a factor of 4 when l = λ∕2) of the single-element impedance. Thus FOLDED DIPOLE 507 Figure 9.20 Folded dipole and its equivalent transmission-line and antenna mode models. (source: G. A.
Thiele, E. P. Ekelman, Jr., and L. W. Henderson, “On the Accuracy of the Transmission Line Model for Folded Dipole,” IEEE Trans. Antennas Propagat., Vol. AP-28, No. 5, pp. 700–703, September 1980. c ⃝(1980) IEEE).
when l = λ∕2 and the antenna is resonant, impedances on the order of about 300 ohms can be achieved, and it would be ideal for connections to “twin-lead” transmission lines.
A folded dipole operates basically as a balanced system, and it can be analyzed by assuming that its current is decomposed into two distinct modes: a transmission-line mode [Figure 9.20(b)] and an antenna mode [Figure 9.20(c)]. This type of an analytic model can be used to predict accurately the input impedance provided the longer parallel wires are close together electrically (s ≪λ).
To derive an equation for the input impedance, let us refer to the modeling of Figure 9.20. For the transmission-line mode of Figure 9.20(b), the input impedance at the terminals a −b or e −f, looking toward the shorted ends, is obtained from the impedance transfer equation Zt = Z0 [ZL + jZ0 tan(kl′) Z0 + jZL tan(kl′) ] l′=l∕2 ZL=0 = jZ0 tan ( k l 2 ) (9-20) where Z0 is the characteristic impedance of a two-wire transmission line Z0 = 𝜂 𝜋cosh−1 (s∕2 a ) = 𝜂 𝜋ln [ s∕2 + √ (s∕2)2 −a2 a ] (9-21) which can be approximated for s∕2 ≫a by Z0 = 𝜂 𝜋ln [ s∕2 + √ (s∕2)2 −a2 a ] ≃𝜂 𝜋ln ( s a ) = 0.733𝜂log10 ( s a ) (9-21a) Since the voltage between the points a and b is V/2, and it is applied to a transmission line of length l/2, the transmission-line current is given by It = V∕2 Zt (9-22) 508 BROADBAND DIPOLES AND MATCHING TECHNIQUES For the antenna mode of Figure 9.20(c), the generator points c −d and g −h are each at the same potential and can be connected, without loss of generality, to form a dipole. Each leg of the dipole is formed by a pair of closely spaced wires (s ≪λ) extending from the feed (c −d or g −h) to the shorted end. Thus the current for the antenna mode is given by Ia = V∕2 Zd (9-23) where Zd is the input impedance of a linear dipole of length l and diameter d computed using (8-57a)–(8-58b). For the configuration of Figure 9.20(c), the radius that is used to compute Zd for the dipole can be either the half-spacing between the wires (s/2) or an equivalent radius ae. The equivalent radius ae is related to the actual wire radius a by (from Table 9.3) ln(ae) = 1 2 ln(a) + 1 2 ln(s) = ln(a) + 1 2 ln ( s a ) = ln √ as (9-24) or ae = √ as (9-24a) It should be expected that the equivalent radius yields the most accurate results.
The total current on the feed leg (left side) of the folded dipole of Figure 9.20(a) is given by Iin = It + Ia 2 = V 2Zt + V 4Zd = V(2Zd + Zt) 4ZtZd (9-25) and the input impedance at the feed by Zin = V Iin = 2Zt(4Zd) 2Zt + 4Zd = 4ZtZd 2Zd + Zt (9-26) Based on (9-26), the folded dipole behaves as the equivalent of Figure 9.21(a) in which the antenna mode impedance is stepped up by a ratio of four. The transformed impedance is then placed in shunt with twice the impedance of the nonradiating (transmission-line) mode to result in the input impedance.
When l = λ∕2, it can be shown that (9-26) reduces to Zin = 4Zd (9-27) or that the impedance of the folded dipole is four times greater than that of an isolated dipole of the same length as one of its sides. This is left as an exercise for the reader (Prob. 9.9).
The impedance relation of (9-27) for the l = λ∕2 can also be derived by referring to Figure 9.22.
Since for a folded dipole the two vertical arms are closely spaced (s ≪λ), the current distribution in each is identical as shown in Figure 9.22(a). The equivalent of the folded dipole of Figure 9.22(a) is the ordinary dipole of Figure 9.22(b). Comparing the folded dipole to the ordinary dipole, it is appar-ent that the currents of the two closely spaced and identical arms of the folded dipole are equal to the one current of the ordinary dipole, or 2If = Id (9-28) FOLDED DIPOLE 509 Figure 9.21 Equivalent circuits for two-element and N-element (with equal radii elements) folded dipoles.
where If is the current of the folded dipole and Id is the current of the ordinary dipole. Also the input power of the two dipoles are identical, or Pf ≡1 2I2 f Zf = Pd ≡1 2I2 dZd (9-29) Substituting (9-28) into (9-29) leads to Zf = 4Zd (9-30) where Zf is the impedance of the folded dipole while Zd is the impedance of the ordinary dipole.
Equation (9-30) is identical to (9-27).
To better understand the impedance transformation of closely spaced conductors (of equal diam-eter) and forming a multielement folded dipole, let us refer to its equivalent circuit in Figure 9.21(b).
+ + /2 s → 0 If (a) Folded dipole + /2 Id (b) Regular dipole λ λ Figure 9.22 Folded dipole and equivalent regular dipole.
510 BROADBAND DIPOLES AND MATCHING TECHNIQUES For N elements, the equivalent voltage at the center of each conductor is V/N and the current in each is In, n = 1, 2, 3, … , N. Thus the voltage across the first conductor can be represented by V N = N ∑ n=1 InZ1n (9-31) where Z1n represents the self or mutual impedance between the first and nth element. Because the elements are closely spaced In ≃I1 and Z1n ≃Z11 (9-32) for all values of n = 1, 2, … , N. Using (9-32), we can write (9-31) as V N = N ∑ n=1 InZ1n ≃I1 N ∑ n=1 Z1n ≃NI1Z11 (9-33) or Zin = V I1 ≃N2Z11 = N2Zr (9-33a) since the self-impedance Z11 of the first element is the same as its impedance Zr in the absence of the other elements. Additional impedance step-up of a single dipole can be obtained by introducing more elements. For a three-element folded dipole with elements of identical diameters and of l ≃λ∕2, the input impedance would be about nine times greater than that of an isolated element or about 650 ohms. Greater step-up transformations can be obtained by adding more elements; in practice, they are seldom needed. Many other geometrical configurations of a folded dipole can be obtained which would contribute different values of input impedances. Small variations in impedance can be obtained by using elements of slightly different diameters and/or lengths.
To test the validity of the transmission-line model for the folded dipole, a number of computations were made and compared with data obtained by the Moment Method, which is considered to be more accurate. In Figures 9.23(a) and (b) the input resistance and reactance for a two-element folded dipole is plotted as a function of l∕λ when the diameter of each wire is d = 2a = 0.001λ and the spacing between the elements is s = 0.00613λ. The characteristic impedance of such a transmission line is 300 ohms. The equivalent radius was used in the calculations of Zd. An excellent agreement is indicated between the results of the transmission-line model and the Moment Method. Computations and comparisons for other spacings (s = 0.0213λ, Z0 = 450 ohms and s = 0.0742λ, Z0 = 600 ohms) but with elements of the same diameter (d = 0.001λ) have been made . It has been shown that as the spacing between the wires increased, the results of the transmission-line model began to dis-agree with those of the Moment Method. For a given spacing, the accuracy for the characteristic impedance, and in turn for the input impedance, can be improved by increasing the diameter of the wires. The characteristic impedance of a transmission line, as given by (9-21) or (9-21a), depends not on the spacing but on the spacing-to-diameter (s/d) ratio, which is more accurate for smaller s/d. Computations were also made whereby the equivalent radius was not used. The comparisons of these results indicated larger disagreements, thus concluding the necessity of the equivalent radius, especially for the larger wire-to-wire spacings.
A two-element folded dipole is widely used, along with “twin-lead” line, as feed element of TV antennas such as Yagi-Uda antennas, which are discussed in detail in Section 10.3.3. Although the impedance of an isolated folded dipole may be around 300 ohms, its value will be somewhat FOLDED DIPOLE 511 Figure 9.23 Input resistance and reactance of folded dipole. (source: G. A. Thiele, E. P. Ekelman, Jr., and L.
W. Henderson, “On the Accuracy of the Transmission Line Model for Folded Dipole,” IEEE Trans. Antennas Propagat., Vol. AP-28, No. 5, pp. 700–703, September 1980. c ⃝(1980) IEEE).
different when it is used as an element in an array or with a reflector. The folded dipole has better bandwidth characteristics than a single dipole of the same size. Its geometrical arrangement tends to behave as a short parallel stub line which attempts to cancel the off resonance reactance of a single dipole. The folded dipole can be thought to have a bandwidth which is the same as that of a single dipole but with an equivalent radius (a < ae < s∕2).
512 BROADBAND DIPOLES AND MATCHING TECHNIQUES Symmetrical and asymmetrical planar folded dipoles can also be designed and constructed using strips which can be fabricated using printed-circuit technology . The input impedance can be varied over a wide range of values by adjusting the width of the strips. In addition, the impedance can be adjusted to match the characteristic impedance of printed-circuit transmission lines with four-to-one impedance ratios.
A MATLAB computer program, entitled Folded, has been developed to perform the design of a folded dipole. The description of the program is found in the corresponding READ ME file included in the publisher’s website for the book.
9.7 DISCONE AND CONICAL SKIRT MONOPOLE There are innumerable variations to the basic geometrical configurations of cones and dipoles, some of which have already been discussed, to obtain broadband characteristics. Two other common radi-ators that meet this characteristic are the conical skirt monopole and the discone antenna shown in Figures 9.24(a) and (b), respectively.
For each antenna, the overall pattern is essentially the same as that of a linear dipole of length l < λ (i.e., a solid of revolution formed by the rotation of a figure-eight) whereas in the horizontal (azimuthal) plane it is nearly omnidirectional. The polarization of each is vertical. Each antenna because of its simple mechanical design, ease of installation, and attractive broadband characteris-tics has wide applications in the VHF (30–300 MHz) and UHF (300 MHz–3 GHz) spectrum for broadcast, television, and communication applications.
The discone antenna is formed by a disk and a cone. The disk is attached to the center conductor of the coaxial feed line, and it is perpendicular to its axis. The cone is connected at its apex to the outer shield of the coaxial line. The geometrical dimensions and the frequency of operation of two designs are shown in Table 9.4.
In general, the impedance and pattern variations of a discone as a function of frequency are much less severe than those of a dipole of fixed length l. The performance of this antenna as a function of frequency is similar to a high-pass filter. Below an effective cutoff frequency it becomes inefficient, and it produces severe standing waves in the feed line. At cutoff, the slant height of the cone is approximately λ∕4.
Figure 9.24 Conical skirt monopole, discone, and wire-simulated cone surface.
MATCHING TECHNIQUES 513 TABLE 9.4 Frequency and Dimensions of Two Designs Frequency (MHz) A (cm) B (cm) C (cm) 90 45.72 60.96 50.80 200 22.86 31.75 35.56 Measured elevation (vertical) plane radiation patterns from 250 to 650 MHz, at 50-MHz intervals, have been published for a discone with a cutoff frequency of 200 MHz. No major changes in the “figure-eight” shape of the patterns were evident other than at the high-frequency range where the pattern began to turn downward somewhat.
The conical skirt monopole is similar to the discone except that the disk is replaced by a monopole of length usually λ∕4. Its general behavior also resembles that of the discone. Another way to view the conical skirt monopole is with a λ∕4 monopole mounted above a finite ground plane. The plane has been tilted downward to allow more radiation toward and below the horizontal plane.
To reduce the weight and wind resistance of the cone, its solid surface can be simulated by radial wires, as shown in Figure 9.24(c). This is a usual practice in the simulation of finite size ground planes for monopole antennas. The lengths of the wires used to simulate the ground plane are on the order of about λ∕4 or greater.
9.8 MATCHING TECHNIQUES The operation of an antenna system over a frequency range is not completely dependent upon the frequency response of the antenna element itself but rather on the frequency characteristics of the transmission line–antenna element combination. In practice, the characteristic impedance of the transmission line is usually real whereas that of the antenna element is complex. Also the variation of each as a function of frequency is not the same. Thus efficient coupling-matching networks must be designed which attempt to couple-match the characteristics of the two devices over the desired frequency range.
There are many coupling-matching networks that can be used to connect the transmission line to the antenna element and which can be designed to provide acceptable frequency characteristics.
Only a limited number will be introduced here.
9.8.1 Stub-Matching Ideal matching at a given frequency can be accomplished by placing a short- or open-circuited shunt stub a distance s from the transmission-line–antenna element connection, as shown in Figure 9.25(a).
Assuming a real characteristic impedance, the length s is controlled so as to make the real part of the antenna element impedance equal to the characteristic impedance. The length l of the shunt line is varied until the susceptance of the stub is equal in magnitude but opposite in phase to the line input susceptance at the point of the transmission line–shunt element connection. The matching procedure is illustrated best graphically with the use of a Smith chart. Analytical methods, on which the Smith chart graphical solution is based, can also be used. The short-circuited stub is more practical because an equivalent short can be created by a pin connection in a coaxial cable or a slider in a waveguide.
This preserves the overall length of the stub line for matchings which may require longer length stubs.
A single stub with a variable length l cannot always match all antenna (load) impedances. A double-stub arrangement positioned a fixed distance s from the load, with the length of each stub variable and separated by a constant length d, will match a greater range of antenna impedances.
However, a triple-stub configuration will always match all loads.
514 BROADBAND DIPOLES AND MATCHING TECHNIQUES Figure 9.25 Matching and microstrip techniques.
Excellent treatments of the analytical and graphical methods for the single-, double-, triple-stub, and other matching techniques are presented in and . The higher-order stub arrangements provide more broad and less sensitive matchings (to frequency variations) but are more complex to implement. Usually a compromise is chosen, such as the double-stub.
9.8.2 Quarter-Wavelength Transformer A. Single Section Another technique that can be used to match the antenna to the transmission line is to use a λ∕4 transformer. If the impedance of the antenna is real, the transformer is attached directly to the load.
However if the antenna impedance is complex, the transformer is placed a distance s0 away from the MATCHING TECHNIQUES 515 antenna, as shown in Figure 9.25(b). The distance s0 is chosen so that the input impedance toward the load at s0 is real and designated as Rin. To provide a match, the transformer characteristic impedance Z1 should be Z1 = √ RinZ0, where Z0 is the characteristic impedance (real) of the input transmission line. The transformer is usually another transmission line with the desired characteristic impedance.
Because the characteristic impedances of most off-the-shelf transmission lines are limited in range and values, the quarter-wavelength transformer technique is most suitable when used with microstrip transmission lines. In microstrips, the characteristic impedance can be changed by simply varying the width of the center conductor.
B. Multiple Sections Matchings that are less sensitive to frequency variations and that provide broader bandwidths, require multiple λ∕4 sections. In fact the number and characteristic impedance of each section can be designed so that the reflection coefficient follows, within the desired frequency bandwidth, pre-scribed variations which are symmetrical about the center frequency. The antenna (load) impedance will again be assumed to be real; if not, the antenna element must be connected to the transformer at a point s0 along the transmission line where the input impedance is real.
Referring to Figure 9.25(c), the total input reflection coefficient Γin for an N-section quarter-wavelength transformer with RL > Z0 can be written approximately as Γin(f ) ≃𝜌0 + 𝜌1e−j2𝜃+ 𝜌2e−j4𝜃+ ⋯+ 𝜌Ne−j2N𝜃 = N ∑ n=0 𝜌ne−j2n𝜃 (9-34) where 𝜌n = Zn+1 −Zn Zn+1 + Zn (9-34a) 𝜃= kΔl = 2𝜋 λ (λ0 4 ) = 𝜋 2 ( f f0 ) (9-34b) In (9-34), 𝜌n represents the reflection coefficient at the junction of two infinite lines with character-istic impedances Zn and Zn+1, f0 represents the designed center frequency, and f the operating fre-quency. Equation (9-34) is valid provided the 𝜌n’s at each junction are small (RL ≃Z0). If RL < Z0, the 𝜌n’s should be replaced by −𝜌n’s. For a real load impedance, the 𝜌n’s and Zn’s will also be real.
For a symmetrical transformer (𝜌0 = 𝜌N, 𝜌1 = 𝜌N−1, etc.), (9-34) reduces to Γin(f ) ≃2e−jN𝜃[𝜌0 cos N𝜃+ 𝜌1 cos(N −2)𝜃+ 𝜌2 cos(N −4)𝜃+ ⋯] (9-35) The last term in (9-35) should be 𝜌[(N−1)∕2] cos 𝜃for N = odd integer (9-35a) 1 2𝜌(N∕2) for N = even integer (9-35b) C. Binomial Design One technique, used to design an N-section λ∕4 transformer, requires that the input reflection coeffi-cient of (9-34) have maximally flat passband characteristics. For this method, the junction reflection 516 BROADBAND DIPOLES AND MATCHING TECHNIQUES coefficients (𝜌n’s) are derived using the binomial expansion. Doing this, we can equate (9-34) to Γin(f) = N ∑ n=0 𝜌ne−j2n𝜃= e−jN𝜃RL −Z0 RL + Z0 cosN(𝜃) = 2−N RL −Z0 RL + Z0 N ∑ n=0 CN n e−j2n𝜃 (9-36) where CN n = N!
(N −n)!n!, n = 0, 1, 2, … , N (9-36a) From (9-34) 𝜌n = 2−N RL −Z0 RL + Z0 Cn N (9-37) For this type of design, the fractional bandwidth Δf∕f0 is given by Δf f0 = 2(f0 −fm) f0 = 2 ( 1 −fm f0 ) = 2 ( 1 −2 𝜋𝜃m ) (9-38) Since 𝜃m = 2𝜋 λm (λ0 4 ) = 𝜋 2 (fm f0 ) (9-39) (9-38) reduces using (9-36) to Δf f0 = 2 −4 𝜋cos−1 [ 𝜌m (RL −Z0)∕(RL + Z0) ]1∕N (9-40) where 𝜌m is the maximum value of reflection coefficient which can be tolerated within the bandwidth.
The usual design procedure is to specify the 1. load impedance (RL) 2. input characteristic impedance (Z0) 3. number of sections (N) 4. maximum tolerable reflection coefficient (𝜌m) [or fractional bandwidth (Δf∕f0)] and to find the 1. characteristic impedance of each section 2. fractional bandwidth [or maximum tolerable reflection coefficient (𝜌m)] To illustrate the principle, let us consider an example.
MATCHING TECHNIQUES 517 Example 9.2 A linear dipole with an input impedance of 70 + j37 is connected to a 50-ohm line. Design a two-section λ∕4 binomial transformer by specifying the characteristic impedance of each section to match the antenna to the line at f = f0. If the input impedance (at the point the transformer is connected) is assumed to remain constant as a function of frequency, determine the maximum reflection coefficient and VSWR within a fractional bandwidth of 0.375.
Solution: Since the antenna impedance is not real, the antenna must be connected to the trans-former through a transmission line of length s0. Assuming a 50-ohm characteristic impedance for that section of the transmission line, the input impedance at s0 = 0.062λ is real and equal to 100 ohms. Using (9-36a) and (9-37) 𝜌n = 2−N RL −Z0 RL + Z0 Cn N = 2−N RL −Z0 RL + Z0 N!
(N −n)!n!
which for N = 2, RL = 100, Z0 = 50 n = 0 : 𝜌0 = Z1 −Z0 Z1 + Z0 = 1 12 ➱Z1 = 1.182Z0 = 59.09 n = 1 : 𝜌1 = Z2 −Z1 Z2 + Z1 = 1 6 ➱Z2 = 1.399Z1 = 82.73 For a fractional bandwidth of 0.375 (𝜃m = 1.276 rad = 73.12◦) we can write, using (9-40) Δf f0 = 0.375 = 2 −4 𝜋cos−1 [ 𝜌m (RL −Z0)∕(RL + Z0) ]1∕2 which for RL = 100 and Z0 = 50 gives 𝜌m = 0.028 The maximum voltage standing wave ratio is VSWRm = 1 + 𝜌m 1 −𝜌m = 1.058 The magnitude of the reflection coefficient is given by (9-36) as |Γin| = 𝜌in = | | | | RL −Z0 RL + Z0 | | | | cos2 𝜃= 1 3 cos2 [ 2𝜋 λ (λ0 4 )] = 1 3 cos2 [ 𝜋 2 ( f f0 )] which is shown plotted in Figure 9.26, and it is compared with the response of a single-section λ∕4 transformer and that of a two-section Tschebyscheff design whose maximum 𝜌m = 0.0147.
518 BROADBAND DIPOLES AND MATCHING TECHNIQUES 0.3 0.25 0.2 0.15 0.1 0.05 00 0.2 0.4 0.6 0.8 1 Relative frequency (f / f0) Input reflection coefficient Γin 1.2 1.4 1.6 1.8 2 Tschebyscheff (N = 2) Binomial (N = 2) Single section ρm = 0.028 (Binomial) ρm = 0.0147 (Tschebyscheff) RL = 100, Z0 = 50 f / f0 = 0.375 Figure 9.26 Responses of single-section, and two-section binomial and Tschebyscheff quarter-wavelength transformers.
Microstrip designs are ideally suited for antenna arrays, as shown in Figure 9.25(d). In gen-eral the characteristic impedance of a microstrip line, whose top and end views are shown in Fig-ures 9.25(e) and (f), respectively, is given by Zc = 87 √ 𝜀r + 1.41 ln ( 5.98h 0.8w + t ) for h < 0.8w (9-41) where 𝜀r = dielectric constant of dielectric substrate (board material) h = height of substrate w = width of microstrip center conductor t = thickness of microstrip center conductor Thus for constant values of 𝜀r, h, and t, the characteristic impedance can be changed by simply varying the width (w) of the center conductor.
D. Tschebyscheff Design The reflection coefficient can be made to vary within the bandwidth in an oscillatory manner and have equal-ripple characteristics. This can be accomplished by making Γin behave according to a Tschebyscheff polynomial. For the Tschebyscheff design, the equation that corresponds to (9-36) is Γin(f ) = e−jN𝜃RL −Z0 RL + Z0 TN(sec 𝜃m cos 𝜃) TN(sec 𝜃m) (9-42) where TN(x) is the Tschebyscheff polynomial of order N.
MATCHING TECHNIQUES 519 The maximum allowable reflection coefficient occurs at the edges of the passband where 𝜃= 𝜃m and TN(sec 𝜃m cos 𝜃)|𝜃=𝜃m = 1. Thus 𝜌m = | | | | RL −Z0 RL + Z0 1 TN(sec 𝜃m) | | | | (9-43) or |TN(sec 𝜃m)| = | | | | 1 𝜌m RL −Z0 RL + Z0 | | | | (9-43a) Using (9-43), we can write (9-42) as Γin(f ) = e−jN𝜃𝜌mTN(sec 𝜃m cos 𝜃) (9-44) and its magnitude as |Γin(f)| = |𝜌mTN(sec 𝜃m cos 𝜃)| (9-44a) For this type of design, the fractional bandwidth Δf∕fo is also given by (9-38).
To be physical, 𝜌m must be smaller than the reflection coefficient when there is no matching.
Therefore, from (9-43) 𝜌m = |Γm| = | | | | RL −Z0 RL + Z0 1 TN(sec𝜃m) | | | | < | | | | RL −Z0 RL + Z0 | | | | ➱|TN(sec 𝜃m)| > 1 (9-45) The Tschebyscheff polynomial can be represented by either (6-71a) or (6-71b). Since |TN(sec 𝜃m)| > 1, then using (6-71b) we can express it as TN(sec 𝜃m) = cosh[N cosh−1(sec 𝜃m)] (9-46) or using (9-43a) as |TN(sec 𝜃m)| = | cosh[N cosh−1(sec 𝜃m)]| = | | | | 1 𝜌m RL −Z0 RL + Z0 | | | | (9-46a) Thus sec 𝜃m = cosh [ 1 N cosh−1 (| | | | 1 𝜌m RL −Z0 RL + Z0 | | | | )] (9-47) 520 BROADBAND DIPOLES AND MATCHING TECHNIQUES Figure 9.27 T-match and Gamma match.
or 𝜃m = sec−1 { cosh [ 1 N cosh−1 (| | | | 1 𝜌m RL −Z0 RL + Z0 | | | | )]} (9-47a) Using (9-44), we can write the reflection coefficient of (9-35) as Γin(𝜃) = 2e−jN𝜃[𝜌0 cos(N𝜃) + 𝜌1 cos(N −2)𝜃+ ⋯] = e−jN𝜃𝜌mTN(sec 𝜃m cos 𝜃) (9-48) For a given N, replace TN(sec 𝜃m cos 𝜃) by its polynomial series expansion of (6-69) and then match terms. The usual procedure for the Tschebyscheff design is the same as that of the binomial as outlined previously.
The first few Tschebyscheff polynomials are given by (6-69). For z = sec 𝜃m cos 𝜃, the first three polynomials reduce to T1(sec 𝜃m cos 𝜃) = sec 𝜃m cos 𝜃 T2(sec 𝜃m cos 𝜃) = 2(sec 𝜃m cos 𝜃)2 −1 = sec2 𝜃m cos 2𝜃+ (sec2 𝜃m −1) T3(sec 𝜃m cos 𝜃) = 4(sec 𝜃m cos 𝜃)3 −3(sec 𝜃m cos 𝜃) = sec3 𝜃m cos 3𝜃+ 3(sec3 𝜃m −sec 𝜃m) cos 𝜃 (9-49) The design of Example 9.2 using a Tschebyscheff transformer is assigned as an exercise to the reader (Prob. 9.24). However its response is shown plotted in Figure 9.26 for comparison whose maximum 𝜌m = 0.0147 compared to 𝜌m = 0.028 for the binomial design.
In general, the multiple sections (either binomial or Tschebyscheff) provide greater bandwidths than a single section. As the number of sections increases, the bandwidth also increases. The advan-tage of the binomial design is that the reflection coefficient values within the bandwidth mono-tonically decrease from both ends toward the center. Thus the values are always smaller than an acceptable and designed value that occurs at the “skirts” of the bandwidth. For the Tschebyscheff MATCHING TECHNIQUES 521 design, the reflection coefficient values within the designed bandwidth are equal or smaller than an acceptable and designed value. The number of times the reflection coefficient reaches the maximum ripple value within the bandwidth is determined by the number of sections. In fact, for an even num-ber of sections the reflection coefficient at the designed center frequency is equal to its maximum allowable value, while for an odd number of sections it is zero. For a maximum tolerable reflection coefficient, the N-section Tschebyscheff transformer provides a larger bandwidth than a correspond-ing N-section binomial design, or for a given bandwidth the maximum tolerable reflection coefficient is smaller for a Tschebyscheff design.
A MATLAB computer program, entitled Quarterwave, has been developed to perform the design of binomial and Tschebyscheff quarter-wavelength impedance transformer designs. The description of the program is found in the corresponding READ ME file included in the publisher’s website for the book.
Other matching techniques, especially for dipole type antennas, include the T-match and Gamma match. These are discussed in detail in the previous three editions of this book.
9.8.3 Baluns and Transformers A twin-lead transmission line (two parallel-conductor line) is a symmetrical line whereas a coaxial cable is inherently unbalanced. Because the inner and outer (inside and outside parts of it) conduc-tors of the coax are not coupled to the antenna in the same way, they provide the unbalance. The result is a net current flow to ground on the outside part of the outer conductor. This is shown in Figure 9.28(a) where an electrical equivalent is also indicated. The amount of current flow I3 on the outside surface of the outer conductor is determined by the impedance Zg from the outer shield to ground. If Zg can be made very large, I3 can be reduced significantly. Devices that can be used to balance inherently unbalanced systems, by canceling or choking the outside current, are known as baluns (balance to unbalance).
One type of a balun is that shown in Figure 9.28(b), referred to usually as a bazooka balun.
Mechanically it requires that a λ∕4 in length metal sleeve, and shorted at its one end, encapsulates the coaxial line. Electrically the input impedance at the open end of this λ∕4 shorted transmission line, which is equivalent to Zg, will be very large (ideally infinity). Thus the current I3 will be choked, if not completely eliminated, and the system will be nearly balanced.
Another type of a balun is that shown in Figure 9.28(c). It requires that one end of a λ∕4 section of a transmission line be connected to the outside shield of the main coaxial line while the other is connected to the side of the dipole which is attached to the center conductor. This balun is used to cancel the flow of I3. The operation of it can be explained as follows: In Figure 9.28(a) the voltages between each side of the dipole and the ground are equal in magnitude but 180◦out of phase, thus producing a current flow on the outside of the coaxial line. If the two currents I1 and I2 are equal in magnitude, I3 would be zero. Since arm #2 of the dipole is connected directly to the shield of the coax while arm #1 is weakly coupled to it, it produces a much larger current I2. Thus there is relatively little cancellation in the two currents.
The two currents, I1 and I2, can be made equal in magnitude if the center conductor of the coax is connected directly to the outer shield. If this connection is made directly at the antenna terminals, the transmission line and the antenna would be short-circuited, thus eliminating any radiation. However, the indirect parallel-conductor connection of Figure 9.28(c) provides the desired current cancellation without eliminating the radiation. The current flow on the outer shield of the main line is canceled at the bottom end of the λ∕4 section (where the two join together) by the equal in magnitude, but opposite in phase, current in the λ∕4 section of the auxiliary line. Ideally then there is no current flow in the outer surface of the outer shield of the remaining part of the main coaxial line. It should be stated that the parallel auxiliary line need not be made λ∕4 in length to achieve the balance. It is made λ∕4 to prevent the upsetting of the normal operation of the antenna.
522 BROADBAND DIPOLES AND MATCHING TECHNIQUES Figure 9.28 Balun configurations.
A compact construction of the balun in Figure 9.28(c) is that in Figure 9.28(d). The outside metal sleeve is split and a portion of it is removed on opposite sides. The remaining opposite parts of the outer sleeve represent electrically the two shorted λ∕4 parallel transmission lines of Figure 9.28(c).
Another balun is that of Figure 9.8(a,b) used to feed the flexible bow–tie antenna. All of the baluns shown in Figure 9.28 are narrowband devices.
Devices can be constructed which provide not only balancing but also step-up impedance trans-formations. One such device is the λ∕4 coaxial balun, with a 4:1 impedance transformation, of Fig-ure 9.29(a). The U-shaped section of the coaxial line must be λ∕2 long .
MULTIMEDIA 523 Figure 9.29 Balun and ferrite core transformers.
Because all the baluns-impedance transformers that were discussed so far are narrowband devices, the bandwidth can be increased by employing ferrite cores in their construction . Two such designs, one a 4:1 or 1:4 transformer and the other a 1:1 balun, are shown in Figures 9.29(b,c). The ferrite core has a tendency to maintain high impedance levels over a wide frequency range .
A good design and construction can provide bandwidths of 8 or even 10 to 1. Coil coaxial baluns, constructed by coiling the coaxial line itself to form a balun , can provide bandwidths of 2 or 3 to 1.
9.9 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab computer program, designated Quarterwave, for computing and displaying the char-acteristics of binomial and Tschebyscheff quarter-wavelength impedance matching designs.
c. Matlab computer programs, designated Folded for computing the characteristics of a folded dipole.
d. Power Point (PPT) viewgraphs, in multicolor.
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PROBLEMS 9.1.
A 300-ohm “twin-lead” transmission line is attached to a biconical antenna.
(a) Determine the cone angle that will match the line to an infinite length biconical antenna.
(b) For the cone angle of part (a), determine the two smallest cone lengths that will resonate the antenna.
(c) For the cone angle and cone lengths from part (b), what is the input VSWR?
9.2.
Determine the first two resonant lengths, and the corresponding diameters and input resis-tances, for dipoles with l∕d = 25, 50, and 104 using (a) the data in Figures 9.17(a) and 9.17(b) (b) Table 9.1 9.3.
Design a resonant cylindrical stub monopole of length l, diameter d, and l/d of 50. Find the length (in λ), diameter (in λ), and the input resistance (in ohms) at the first four resonances.
9.4.
A linear dipole of l∕d = 25, 50, and 104 is attached to a 50-ohm line. Determine the VSWR of each l/d when (a) l = λ∕2 (b) l = λ (c) l = 3λ∕2 526 BROADBAND DIPOLES AND MATCHING TECHNIQUES 9.5.
Find the equivalent circular radius ae for a (a) very thin flat plate of width λ∕10 (b) square wire with sides of λ∕10 (c) rectangular wire with sides of λ∕10 and λ∕100 (d) elliptical wire with major and minor axes of λ∕10 and λ∕20 (e) twin-lead transmission line with wire radii of 1.466×10−2 cm and separation of 0.8 cm 9.6.
Compute the characteristic impedance of a two-wire transmission line with wire diameter of d = 10−3λ and center-to-center spacings of (a) 6.13×10−3λ (b) 2.13×10−2λ (c) 7.42×10−2λ 9.7.
From the expressions listed in Table 9.3, show that the “equivalent” radius ae of the circular two-wire side-by-side parallel arrangemenet can be written as ln(ae) ≈ 1 (a′ + a2)[(a′)2 ln a′ + (a)2 ln a + 2a′a ln s] where a and a′ are, respectively, the radii of the two wires and s is the center-to-center sepra-tion between the two parallel wires.
9.8.
To increase its bandwidth, a λ∕4 monopole antenna operating at 1 GHz is made of two side-by-side copper wires (𝜎= 5.7×107 S/m) of circular cross section. The wires at each end of the arm are connected (shorted) electrically together. The radius of each wire is λ∕200 and the separation between them is λ∕50.
(a) What is the effective radius (in meters) of the two wires? Compare with the physical radius of each wire (in meters).
(b) What is the high-frequency loss resistance of each wire? What is the total loss resistance of the two together in a side-by-side shorted at the ends arrangement? What is the loss resistance based on the effective radius?
(c) What is the radiation efficiency of one wire by itself? Compare with that of the two together in a side-by-side arrangement. What is the radiation efficiency based on the loss resistance of the effective radius?
9.9.
Show that the input impedance of a two-element folded dipole of l = λ∕2 is four times greater than that of an isolated element of the same length.
9.10.
Design a two-element folded dipole with wire diameter of 10−3λ and center-to-center spac-ing of 6.13×10−3λ.
(a) Determine its shortest length for resonance.
(b) Compute the VSWR at the first resonance when it is attached to a 300-ohm line.
9.11.
A two-element folded dipole of identical wires has an l∕d = 500 and a center-to-center spac-ing of 6.13×10−3λ between the wires. Compute the (a) approximate length of a single wire at its first resonance (b) diameter of the wire at the first resonance (c) characteristic impedance of the folded dipole transmission line (d) input impedance of the transmission line mode model (e) input impedance of the folded dipole using as the radius of the antenna mode (1) the radius of the wire a, (2) the equivalent radius ae of the wires, (3) half of the center-to-center spacing (s/2). Compare the results.
PROBLEMS 527 9.12.
The input impedance of a 0.47λ folded dipole operating at 10 MHz is Zin = 306 + j75.3 To resonate the element, it is proposed to place a lumped element in shunt (parallel) at the input terminals where the impedance is measured.
(a) What kind of an element (capacitor or inductor) should be used to accomplish the requirement?
(b) What is the value of the element (in Farads or Henries)?
(c) What is the new value of the input impedance?
(d) What is the VSWR when the resonant antenna is connected to a 300-ohm line?
Verify using the computer program Folded.
l s Input terminals 9.13.
A half-wavelength, two-element symmetrical folded dipole is connected to a 300-ohm “twin-lead” transmission line. In order for the input impedance of the dipole to be real, an energy storage lumped element is placed across its input terminals. Determine, assuming f = 100 MHz, the (a) capacitance or inductance of the element that must be placed across the terminals.
(b) VSWR at the terminals of the transmission line taking into account the dipole and the energy storage element.
Verify using the computer program Folded.
9.14.
A half-wavelength, two-element symmetrical folded dipole whose each wire has a radius of 10−3λ is connected to a 300-ohm “twin-lead” transmission line. The center-to-center spac-ing of the two wires is 4×10−3λ. In order for the input impedance of the dipole to be real, determine, assuming f = 100 MHz, the (a) total capacitance CT that must be placed in series at the input terminals.
(b) capacitance C (two of them) that must be placed in series at each of the input terminals of the dipole in order to keep the antenna symmetrical.
(c) VSWR at the terminals of the transmission line connected to the dipole through the two capacitances.
Verify using the computer program Folded.
9.15.
An l = 0.47λ folded dipole, whose wire radius is 5×10−3λ, is fed by a “twin lead” transmis-sion line with a 300-ohm characteristic impedance. The center-to-center spacing of the two side-by-side wires of the dipole is s = 0.025λ. The dipole is operating at f = 10 MHz. The input impedance of the “regular” dipole of l = 0.47λ is Zd = 79 + j13.
(a) Determine the (i) Input impedance of the folded dipole.
(ii) Amplification factor of the input impedance of the folded dipole, compared to the corresponding value of the regular dipole.
(iii) Input reflection coefficient.
(iv) Input VSWR.
(b) To resonate the folded dipole and keep the system balanced, two capacitors (each C) are connected each symmetrically in series at the input terminals of the folded dipole.
(i) What should C be to resonate the dipole?
(ii) What is the new reflection coefficient at the input terminals of the “twin-lead” line?
(iii) What is the new VSWR?
528 BROADBAND DIPOLES AND MATCHING TECHNIQUES 9.16.
To reduce the input impedance of a regular quarter-wavelength (λ∕4) monopole, it was decided to design a folded quarter-wavelength (λ∕4) monopole, as shown in the figure.
Assuming that the radius of the monopole wire is very thin (a ≪λ) and the spacing between the wires is also very small (s ≪λ): (a) Determine the input impedance of the regular monopole.
(b) Determine the input impedance of the folded monopole.
(c) What kind of a lumped element, inductor or capacitor, must be placed in series in order to resonate (eliminate the imaginary part of the impedance) the folded dipolc?
(d) What is the value of the scries inductor or capacitor at a frequency of 100 MHz?
(e) After the folded monopole is resonated, what is the input impedance of the folded monopole assuming it is connected to a coaxial line with a characteristic impedance of 75 ohms?
(a) regular monopole s (b) folded monopole /4 /4 λ λ 9.17.
A folded dipole of length λ0∕2 is composed of two wires, each of radii λ0∕200 and λ0∕300, respectively, and separated by a distance of λ0∕100. The folded dipole is connected to a 300-ohm transmission line. Determine the following: (a) Input impedance of the dipole.
(b) What capacitance or inductance must be placed in series to the transmission line at the feed point so that the dipole is resonant at fo = 600 MHz.
(c) Input reflection coefficient, assuming the resonant dipole is connected to a 300-ohm transmission line.
(d) Input VSWR of the resonant dipole when connected to the 300-ohm transmission line.
9.18.
A folded dipole antenna operating at 100 MHz with identical wires in both arms, and with overall length of each arm being 0.48λ, is connected to a 300-ohm twin-lead line. The radius of each wire is a = 5×10−4λ while the center-to-center separation is s = 5×10−3λ. Deter-mine the (a) Approximate length (in λ) of the regular dipole, at the first resonance, in the absence of the other wire.
(b) Input impedance of the single wire resonant regular dipole.
(c) Input impedance of the folded dipole at the length of the first resonance of the single ele-ment.
(d) Capacitance (in F/m) or inductance (in H/m), whichever is appropriate, that must be placed in series with the element at the feed to resonate the folded dipole. To keep the element balanced, place two elements, one in each arm.
(e) VSWR of the resonant folded dipole.
PROBLEMS 529 9.19.
A folded dipole made of 3 “legs” each with a total length λ∕2, as shown in the figure, is connected to a 300-ohm transmission line. Assuming the wires are very thin and the spacing between the legs is small compared to the length, determine the: (a) Input impedance of the folded dipole, real and imaginary parts (in the absence of any resonant element at the input).
(b) Lumped element, capacitor or inductor (whichever is appropriate), that must be placed in parallel at the input (as shown in the figure dashed across the input) to res-onate the folded dipole at a frequency of 300 MHz.
(c) New input impedance after the element is resonated; take into account the capacitor/inductor.
(d) VSWR at the input of the resonant folded dipole.
λ/2 Zin 9.20.
A folded monopole made up of 3 “legs” each with a total length λ∕4, as shown in the figure, is connected to a 75-ohm transmission line. Assuming the wires are very thin and the spacing between the legs is small compared to the length, determine the: (a) Input impedance of the folded monopole, real and imaginary parts (in the absence of any resonant element at the input); figure (a) below.
(b) Total iumped capacitance/inductance, (whichever is appropriate), that must be placed in parallel at the input to resonate the folded 3-wire monopole at 300 MHz.
(c) Individual lumped capacitance (C) or inductance (L) (shown in the figure dashed across the input) of Each of two identical capacitance/inductance elements placed in parallel to resonate the folded monopole and keep it symmetrical at 300 MHz; figure (b) below.
State the capacitance/inductance of Each of the 2 identical elements.
(d) New input impedance after the element is resonated; take into account the presence of the 2 identical capacitors/inductors.
(e) VSWR at the input of the resonant folded monopole, taking into account the presence of the two lumped elements. Input transmission line has 75-ohms characteristic impedance.
(a) (b) s /4 /4 λ λ 9.21.
A λ∕2 folded dipole is used as the feed element for a Yagi-Uda array whose input impedance, when the array is fed by a regular dipole, is 30 + j3. The folded dipole used is not symmet-rical (i.e., the two wires are not of the same radius). One of the wires has a radius of 10−4λ and the other (the one that is connected to the feed) has a radius of 0.5×10−4λ. The center-to-center spacing between the two dipoles is 2×10−4λ. Determine the (a) turns ratio of the equivalent-circuit impedance transformer (b) effective impedance of the regular dipole when it is transferred from the secondary to the primary of the equivalent-circuit impedance transformer (c) input impedance of the folded dipole.
9.22.
A three-element folded dipole, whose length of each element is λ∕2, is used as the feed element of a Yagi-Uda array.
(a) Determine the input impedance of the three-element folded dipole; real and imaginary parts.
530 BROADBAND DIPOLES AND MATCHING TECHNIQUES (b) What is the magnitude of the input reflection coefficient and VSWR before the insert of the capacitors/inductors to resonate the antenna.
(c) In order to resonate the antenna (cancel the imaginary part of the input impedance), lumped element (capacitor or inductor, whichever is appropriate) is placed in series (one in each arm to keep the system balanced), as shown in the box below. Indicate which kind of an element, capacitor or inductor, will be appropriate.
Justify your selection.
(d) For a frequency of 15 MHz, determine the value of capacitor C or inductor L of Part c.
(e) What is the new input impedance after the antenna is resonated? Account for the presence of the series capacitors or inductors when you are finding the new input impedance.
(f) What is the VSWR if the resonated antenna is con-nected to a 300-ohm “twin-lead” transmission line?
C / L Zc = 300 Ohms C / L /2 λ 9.23.
It is desired to design two-section Tschebyscheff impedance transformer to match an antenna, whose impedance is ZL = 185.56 + j50, to a lossless input line whose charac-teristic impedance is Z0 = 50 ohms. In order to accomplish this, the load impedance has to be real. Therefore, referring to page 516, Figure 9.25 (b) of the book, the 2-section Tschebyscheff impedance transformer has to be connected at a distance s0 from the load impedance [Figure 9.25(b) shows only one-section but in this problem we want two-sections]. Given that the required distance is s0 = 0.011475λ, to convert the complex load impedance ZL = 186.56 + j50 to a real input impedance of 200 ohms at s0 = 0.011475λ, determine the: (a) Magnitude of the reflection coefficient and VSWR at a distance s0 = 0.011475λ, looking toward the load before the insertion of the quarter-wavelength impedance transformer.
Assume a characteristic impedance of 50 ohms.
(b) Fractional bandwidth of the quarter-wavelength impedance transformer for a maximum tolerable reflection coefficient of 𝜌m = 0.3333.
(c) Intrinsic reflection coefficients Γ0, Γ1 between the sections of the quarter-wavelength impedance transformer.
(d) Characteristic impedances Z1, Z2 of the two-section Tschcbyschcff quarter-wavelength impedance transformer (in ohms).
9.24.
Repeat the design of Example 9.1 using a Tschebyscheff transformer.
9.25.
Repeat the design of Example 9.1 for a three-section (a) binomial transformer (b) Tschebyscheff transformer Verify using the computer program Quarterwave.
9.26.
The radiation resistance of a center-fed resonant l = 0.723λ linear dipole is Rr = 175.35 ohms. It is desired to connect the dipole to a transmission line so that thr system meets certain specifications.
(a) Determine the magnitude of the maximum reflection coefficient so that the VSWR does not exceed 3.
(b) To achieve the specifications of Part a, determine the characteristic impedance of the input line. Assume that RL > Z0 where RL is the load resistance.
PROBLEMS 531 (c) To reduce the reflection coefficient so and its maximum tolerable value is one-half of that in Part (a), a three-section quarter-wavelength binomial impedance transformer is going to be placed between the load impedance and the input line. Determine the fractional bandwidth (in %) over which the system can now be operated so that it does not exceed the new value of the maximum tolerable reflection coefficient [i.e.; not to exceed one-half of that of Part (a)].
(d) Repeat Part (c) for a 3-section quarter-wavelength Tschebyscheff design.
9.27.
A self-resonant (first resonance) half-wavelength dipole of radius a = 10−3λ is connected to a 300-ohm “twin-lead” line through a three-section binomial impedance transformer. Deter-mine the impedances of the three-section binomial transformer required to match the reso-nant dipole to the “twin-lead” line. Verify using the computer program Quarterwave.
9.28.
Design a three-section binomial transformer matching a 200-ohm load to a 100-ohm input transmission line. The maximum tolerable reflection coefficient is 0.3. Determine the: (a) Intrinsic reflection coefficients at each intersection of one line to the other line.
(b) Characteristic impedances of each line between the input line and the load.
(c) VSWR at the input in the absence of the impedance transformer.
(d) Fractional bandwidth, in the presence of the impedance transformer, over which the system can operate without exceeding the maximum tolerable reflection coefficient.
9.29.
Consider a center-fed thin-wire dipole with wire radius a = 0.005λ.
(a) Determine the shortest resonant length l (in wavelengths) and corresponding input resis-tance Rin of the antenna using assumed sinusoidal current distribution.
(b) Design a two-section Tschebyscheff quarter-wavelength transformer to match the antenna to a 75-ohm transmission line. Design the transformer to achieve an equal-ripple response to Rin over a fractional bandwidth of 0.25.
(c) Compare the performance of the matching network in Part (b) to an ideal transformer by plotting the input reflection coefficient magnitude versus normalized frequency for 0 ≤f ∕f0 ≤2 for both cases.
Verify using the computer program Quarterwave.
CHAPTER10 Traveling Wave and Broadband Antennas 10.1 INTRODUCTION In the previous chapters we have presented the details of classical methods that are used to analyze the radiation characteristics of some of the simplest and most common forms of antennas (i.e., infinitely thin linear and circular wires, broadband dipoles, and arrays). In practice there is a myriad of antenna configurations, and it would be almost impossible to consider all of them in this book. In addition, many of these antennas have bizarre types of geometries and it would be almost impractical, if not even impossible, to investigate each in detail. However, the general performance behavior of some of them will be presented in this chapter with a minimum of analytical formulations. Today, comprehensive analytical formulations are available for most of them, but they would require so much space that it would be impractical to include them in this book.
10.2 TRAVELING WAVE ANTENNAS In Chapter 4, center-fed linear wire antennas were discussed whose amplitude current distribu-tion was 1. constant for infinitesimal dipoles (l ≤λ∕50) 2. linear (triangular) for short dipoles (λ∕50 < l ≤λ∕10) 3. sinusoidal for long dipoles (l > λ∕10) In all cases the phase distribution was assumed to be constant. The sinusoidal current distribution of long open-ended linear antennas is a standing wave constructed by two waves of equal amplitude and 180◦phase difference at the open end traveling in opposite directions along its length. The voltage distribution has also a standing wave pattern except that it has maxima (loops) at the end of the line instead of nulls (nodes) as the current. In each pattern, the maxima and minima repeat every integral number of half wavelengths. There is also a λ∕4 spacing between a null and a maximum in each of the wave patterns. The current and voltage distributions on open-ended wire antennas are similar to the standing wave patterns on open-ended transmission lines. Linear antennas that exhibit current and voltage standing wave patterns formed by reflections from the open end of the wire are referred to as standing wave or resonant antennas.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 533 534 TRAVELING WAVE AND BROADBAND ANTENNAS θ Reflected (backward) Ground Incident (forward) Wire h d Ib If RL Zc 0, z (a) Long wire above ground and radiation pattern m θ μ 0 0, μ 0 0, μ 0 0, μ 0 θm (b) Equivalent: Source and image h h Source Image (c) Wave tilt Ground Wire Direction of energy flow h θ Ei τ Figure 10.1 Beverage (long-wire) antenna above ground.
Antennas can be designed which have traveling wave (uniform) patterns in current and voltage.
This can be achieved by properly terminating the antenna wire so that the reflections are minimized if not completely eliminated. An example of such an antenna is a long wire that runs horizontal to the earth, as shown in Figure 10.1. The input terminals consist of the ground and one end of the wire.
This configuration is known as Beverage or wave antenna. There are many other configurations of traveling wave antennas. In general, all antennas whose current and voltage distributions can be represented by one or more traveling waves, usually in the same direction, are referred to as traveling wave or nonresonant antennas. A progressive phase pattern is usually associated with the current and voltage distributions.
Standing wave antennas, such as the dipole, can be analyzed as traveling wave antennas with waves propagating in opposite directions (forward and backward) and represented by traveling wave currents If and Ib in Figure 10.1(a). Besides the long-wire antenna there are many examples of traveling wave antennas such as dielectric rod (polyrod), helix, and various surface wave antennas.
Aperture antennas, such as reflectors and horns, can also be treated as traveling wave antennas.
In addition, arrays of closely spaced radiators (usually less than λ∕2 apart) can also be analyzed as traveling wave antennas by approximating their current or field distribution by a continuous traveling wave. Yagi-Uda, log-periodic, and slots and holes in a waveguide are some examples of discrete-element traveling wave antennas. In general, a traveling wave antenna is usually one that is associated with radiation from a continuous source. An excellent book on traveling wave antennas is one by C.
H. Walter .
A traveling wave may be classified as a slow wave if its phase velocity vp(vp = 𝜔∕k, 𝜔= wave angular frequency, k = wave phase constant) is equal or smaller than the velocity of light c in free-space (vp∕c ≤1). A fast wave is one whose phase velocity is greater than the speed of light (vp∕c > 1).
In general, there are two types of traveling wave antennas. One is the surface wave antenna defined as “an antenna which radiates power flow from discontinuities in the structure that interrupt a bound TRAVELING WAVE ANTENNAS 535 (a) Upper wall—longitudinal (b) Sidewall—inclined Matched load Matched load Input y x z a z x b b ψ a y Figure 10.2 Leaky-wave waveguide slots; upper (broad) and side (narrow) walls.
wave on the antenna surface.”∗A surface wave antenna is, in general, a slow wave structure whose phase velocity of the traveling wave is equal to or less than the speed of light in free-space (vp∕c ≤1).
For slow wave structures radiation takes place only at nonuniformities, curvatures, and disconti-nuities (see Figure 1.10). Discontinuities can be either discrete or distributed. One type of discrete discontinuity on a surface wave antenna is a transmission line terminated in an unmatched load, as shown in Figure 10.1(a). A distributed surface wave antenna can be analyzed in terms of the varia-tion of the amplitude and phase of the current along its structure. In general, power flows parallel to the structure, except when losses are present, and for plane structures the fields decay exponentially away from the antenna. Most of the surface wave antennas are end-fire or near-end-fire radiators.
Practical configurations include line, planar surface, curved, and modulated structures.
Another traveling wave antenna is a leaky-wave antenna defined as “an antenna that couples power in small increments per unit length, either continuously or discretely, from a traveling wave struc-ture to free-space”† Leaky-wave antennas continuously lose energy due to radiation, as shown in Figure 10.2 by a slotted rectangular waveguide. The fields decay along the structure in the direction of wave travel and increase in others. Most of them are fast wave structures.
10.2.1 Long Wire An example of a slow wave traveling antenna is a long wire, as shown in Figure 10.1. An antenna is usually classified as a long wire antenna if it is a straight conductor with a length ∗“IEEE Standard Definitions of Terms for Antennas” (IEEE Std 145-1983), IEEE Trans. Antennas and Propagat., Vol. AP-31, No. 6, Part II, Nov. 1983.
†Ibid.
536 TRAVELING WAVE AND BROADBAND ANTENNAS Figure 10.3 Long-wire antenna.
from one to many wavelengths. A long wire antenna has the distinction of being the first traveling wave antenna.
The long wire of Figure 10.1(a), in the presence of the ground, can be analyzed approximately using the equivalent of Figure 10.1(b) where an image is introduced to take into account the presence of the ground. The magnitude and phase of the image are determined using the reflection coefficient for horizontal polarization as given by (4-128). The height h of the antenna above the ground must be chosen so that the reflected wave (or wave from the image), which includes the phase due to reflection is in phase with the direct wave at the angles of desired maximum radiation. However, for typical electrical constitutive parameters of the earth, and especially for observation angles near grazing, the reflection coefficient for horizontal polarization is approximately −1. Therefore the total field radiated by the wire in the presence of the ground can be found by multiplying the field radiated by the wire in free space by the array factor of a two-element array, as was done in Section 4.8.2 and represented by (4-129).
The objective now is to find the field radiated by the long wire in free space. This is accom-plished by referring to Figure 10.3. As the wave travels along the wire from the source toward the load, it continuously leaks energy. This can be represented by an attenuation coefficient.
Therefore the current distribution of the forward traveling wave along the structure can be represented by If = ̂ azIz(z′)e−𝛾(z′)z′ = ̂ azI0e−[𝛼(z′)+jkz(z′)]z′ (10-1) where 𝛾(z′) is the propagation coefficient [𝛾(z′) = 𝛼(z′) + jkz(z′) where 𝛼(z′) is the attenuation con-stant (nepers/meter) while kz(z′) is the phase constant (radians/meter) associated with the traveling wave]. In addition to the losses due to leakage, there are wire and ground losses. The attenuation fac-tor 𝛼(z′) can also be used to take into account the ohmic losses of the wire as well as ground losses.
However, these, especially the ohmic losses, are usually very small and for simplicity are neglected.
In addition, when the radiating medium is air, the loss of energy in a long wire (l ≫λ) due to leakage is also usually very small, and it can also be neglected. Therefore the current distribution of (10-1) can be approximated by I = ̂ azI(z′)e−jkzz′ = ̂ azI0e−jkzz′ (10-1a) TRAVELING WAVE ANTENNAS 537 where I(z′) = I0 is assumed to be constant. Using techniques outlined and used in Chapter 4, it can be easily shown that in the far field Er ≃E𝜙= Hr = H𝜃= 0 E𝜃≃j𝜂klI0e−jkr 4𝜋r e−j(kl∕2)(K−cos 𝜃) sin 𝜃sin[(kl∕2)(cos 𝜃−K)] (kl∕2)(cos 𝜃−K) H𝜙≃E𝜃 𝜂 (10-2a) (10-2b) (10-2c) where K is used to represent the ratio of the phase constant of the wave along the transmission line (kz) to that of free-space (k), or K = kz k = λ λg (10-3) λg = wavelength of the wave along the transmission line Assuming a perfect electric conductor for the ground, the total field for Figure 10.1(a) is obtained by multiplying each of (10-2a)–(10-2c) by the array factor sin(kh sin 𝜃).
For kz = k(K = 1) the time-average power density can be written as Wav = Wrad = ̂ ar𝜂|I0|2 8𝜋2r2 sin2 𝜃 (cos 𝜃−1)2 sin2 [kl 2 (cos 𝜃−1) ] (10-4) which reduces to Wav = Wrad = ̂ ar𝜂|I0|2 8𝜋2r2 cot2 (𝜃 2 ) sin2 [kl 2 (cos 𝜃−1) ] (10-5) From (10-5) it is evident that the power distribution of a wire antenna of length l is a multilobe pattern whose number of lobes depends upon its length. Assuming that l is very large such that the variations in the sine function of (10-5) are more rapid than those of the cotangent, the peaks of the lobes occur approximately when sin2 [kl 2 (cos 𝜃−1) ] 𝜃=𝜃m = 1 (10-6) or kl 2 (cos 𝜃m −1) = ± (2m + 1 2 ) 𝜋, m = 0, 1, 2, 3, … (10-6a) The angles where the peaks occur are given by 𝜃m = cos−1 [ 1 ± λ 2l(2m + 1) ] , m = 0, 1, 2, 3, … (10-7) The angle where the maximum of the major lobe occurs is given by m = 0 (or 2m + 1 = 1). As l becomes very large (l ≫λ), the angle of the maximum of the major lobe approaches zero degrees and the structure becomes a near-end-fire array.
538 TRAVELING WAVE AND BROADBAND ANTENNAS In finding the values of the maxima, the variations of the cotangent term in (10-5) were assumed to be negligible (as compared to those of the sine term). If the effects of the cotangent term were to be included, then the values of the 2m + 1 term in (10-7) should be 2m + 1 = 0.742, 2.93, 4.96, 6.97, 8.99, 11, 13, … (10-8) (instead of 1, 3, 5, 7, 9, …) for the first, second, third, and so forth maxima. The approximate values approach those of the exact for the higher order lobes.
In a similar manner, the nulls of the pattern can be found and occur when sin2 [kl 2 (cos 𝜃−1) ] 𝜃=𝜃n = 0 (10-9) or kl 2 (cos 𝜃n −1) = ±n𝜋, n = 1, 2, 3, 4, … (10-9a) The angles where the nulls occur are given by 𝜃n = cos−1 ( 1 ± nλ l ) , n = 1, 2, 3, 4, … (10-10) for the first, second, third, and so forth nulls.
The total radiated power can be found by integrating (10-5) over a closed sphere of radius r and reduces to Prad = ∯ S Wrad ⋅ds = 𝜂 4𝜋|I0|2 [ 1.415 + ln (kl 𝜋 ) −Ci(2kl) + sin(2kl) 2kl ] (10-11) where Ci(x) is the cosine integral of (4-68a). The radiation resistance is then found to be Rr = 2Prad |I0|2 = 𝜂 2𝜋 [ 1.415 + ln (kl 𝜋 ) −Ci(2kl) + sin(2kl) 2kl ] (10-12) Using (10-5) and (10-11) the directivity can be written as D0 = 4𝜋Umax Prad = 2 cot2 [1 2 cos−1 ( 1 −0.371λ l )] 1.415 + ln (2l λ ) −Ci(2kl) + sin(2kl) 2kl (10-13) A. Amplitude Patterns, Maxima, and Nulls To verify some of the derivations and illustrate some of the principles, a number of computations were made. Shown in Figure 10.4(a) is the three-dimensional pattern of a traveling wire antenna with length l = 5λ while in Figure 10.4(b) is the three-dimensional pattern for a standing wave wire antenna with length l = 5λ. The corresponding two-dimensional patterns are shown in Figure 10.5.
The pattern of Figure 10.4(a) is formed by the forward traveling wave current If = I1e−jkz of Fig-ure 10.1(a) while that of Figure 10.4(b) is formed by the forward If plus backward Ib traveling TRAVELING WAVE ANTENNAS 539 (a) Traveling wave x z Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y θ θ x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (b) Standing wave Figure 10.4 Three-dimensional free-space amplitude pattern for traveling and standing wave wire antennas of l = 5λ.
wave currents of Figure 10.1(a). The two currents If and Ib together form a standing wave; that is, Is = If + Ib = I1e−jkz −I2e+jkz = −2jI0 sin(kz) when I2 = I1 = I0. As expected, for the traveling wave antenna of Figure 10.4(a) there is maximum radiation in the forward direction while for the standing wave antenna of Figure 10.4(b) there is maximum radiation in the forward and backward directions. The lobe near the axis of the wire in the directions of travel is the largest. The magnitudes of the other lobes from the main decrease progressively, with an envelope proportional to cot2(𝜃∕2), toward the other direction. The traveling wave antenna is used when it is desired to radiate or receive predominantly from one direction. As the length of the wire increases, the maximum of the main lobe shifts closer toward the axis and the number of lobes increase. This is illustrated in Figure 10.6 for a traveling wave wire antenna with l = 5λ and 10λ. The angles of the maxima of the first four lobes, computed using (10-8), are plotted in Figure 10.7(a) for 0.5λ ≤l ≤10λ. The corresponding angles of the first four nulls, computed using (10-10), are shown in Figure 10.7(b) for 0.5λ ≤l ≤10λ.
These curves can be used effectively to design long wires when the direction of the maximum or null is desired.
540 TRAVELING WAVE AND BROADBAND ANTENNAS Standing Wave Traveling Wave Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 30° 150° 60° 120° 90° 90° 120° 60° 150° 30° 180° 0° Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Figure 10.5 Two-dimensional free-space amplitude pattern for traveling and standing wave wire antennas of l = 5λ.
l = 5 l = 10 Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 30° 150° 60° 120° 90° 90° 120° 60° 150° 30° 180° 0° Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 λ λ Figure 10.6 Two-dimensional free-space amplitude pattern for traveling wave wire antenna of l = 5λ and 10λ.
TRAVELING WAVE ANTENNAS 541 1 2 3 4 5 6 7 8 9 10 0 20 40 60 80 100 120 140 160 180 Length l (wavelengths) (a) Maxima Angle m (degrees) θ 2m + 1 = 4.96 2m + 1 = 6.97 2m + 1 = 0.742 2m + 1 = 2.93 1 2 3 4 5 6 7 8 9 10 0 20 40 60 80 100 120 140 160 180 Length l (wavelengths) (b) Nulls Angle n (degrees) θ n = 1 n = 2 n = 3 n = 4 Figure 10.7 Angles versus length of wire antenna where maxima and nulls occur.
B. Input Impedance For traveling wave wire antennas the radiation in the opposite direction from the maximum is sup-pressed by reducing, if not completely eliminating, the current reflected from the end of the wire.
This is accomplished by increasing the diameter of the wire or more successfully by properly ter-minating it to the ground, as shown in Figure 10.1. Ideally a complete elimination of the reflections (perfect match) can only be accomplished if the antenna is elevated only at small heights (compared to the wavelength) above the ground, and it is terminated by a resistive load. The value of the load resistor, to achieve the impedance match, is equal to the characteristic impedance of the wire near the ground (which is found using image theory). For a wire with diameter d and height h above the ground, an approximate value of the termination resistance is obtained from RL = 138 log10 ( 4h d ) (10-14) 542 TRAVELING WAVE AND BROADBAND ANTENNAS To achieve a reflection-free termination, the load resistor can be adjusted about this value (usu-ally about 200–300 ohms) until there is no standing wave on the antenna wire. Therefore the input impedance is the same as the load impedance or the characteristic impedance of the line, as given by (10-14).
If the antenna is not properly terminated, waves reflected from the load traveling in the opposite direction from the incident waves create a standing wave pattern. Therefore the input impedance of the line is not equal to the load impedance. To find the input impedance, the transmission line impedance transfer equation of (9-18) can be used. Doing this we can write that the impedance at the input terminals of Figure 10.1(a) is Zin(l) = Zc [RL + jZc tan(𝛽l) Zc + jRL tan(𝛽l) ] (10-15) C. Polarization A long-wire antenna is linearly polarized, and it is always parallel to the plane formed by the wire and radial vector from the center of the wire to the observation point. The direction of the linear polarization is not the same in all parts of the pattern, but it is perpendicular to the radial vector (and parallel to the plane formed by it and the wire). Thus the wire antenna of Fig-ure 10.1, when its height above the ground is small compared to the wavelength and its main beam is near the ground, is not an effective element for horizontal polarization. Instead it is usu-ally used to transmit or receive waves that have an appreciable vector component in the vertical plane. This is what is known as a Beverage antenna which is used more as a receiving rather than a transmitting element because of its poor radiation efficiency due to power absorbed in the load resistor.
When a TEM wave travels parallel to an air-conductor interface, it creates a forward wave tilt which is determined by applying the boundary conditions on the tangential fields along the interface.
The amount of tilt is a function of the constitutive parameters of the ground. If the conductor is a perfect electric conductor (PEC), then the wave tilt is zero because the tangential electric field van-ishes along the PEC. The wave tilt increases with frequency and with ground resistivity. Therefore, for a Beverage wire antenna, shown in Figure 10.1(c) in the receiving mode, reception is influenced by the tilt angle of the incident vertically polarized wavefront, which is formed by the losses of the local ground. The electric-field vector of the incident wavefront produces an electric force that is parallel to the wire, which in turn induces a current in the wire. The current flows in the wire toward the receiver, and it is reinforced up to a certain point along the wire by the advancing wavefront. The wave along the wire is transverse magnetic.
D. Resonant Wires Resonant wire antennas are formed when the load impedance of Figure 10.1(a) is not matched to the characteristic impedance of the line. This causes reflections which with the incident wave form a standing wave. Resonant antennas, including the dipole, were examined in detail in Chapter 4, and the electric and magnetic field components of a center-fed wire of total length l are given, respec-tively, by (4-62a) and (4-62b). Other radiation characteristics (including directivity, radiation resis-tance, maximum effective area, etc.) are found in Chapter 4.
Resonant antennas can also be formed using long wires. It can be shown that for resonant long wires with lengths odd multiple of half wavelength (l = nλ∕2, n = 1, 3, 5, …), the radiation resistance is given approximately (within 0.5 ohms) by , Rr = 73 + 69 log10(n) (10-16) TRAVELING WAVE ANTENNAS 543 This expression gives a very good answer even for n = 1. For the same elements, the angle of max-imum radiation is given by 𝜃max = cos−1 (n −1 n ) (10-17) This formula is more accurate for small values of n, although it gives good results even for large values of n. It can also be shown that the maximum directivity is related to the radiation resistance by D0 = 120 Rr sin2 𝜃max (10-18) The values based on (10-18) are within 0.5 dB from those based on (4-75). It is apparent that all three expressions, (10-16)–(10-18), lead to very good results for the half-wavelength dipole (n = 1).
Long-wire antennas (both resonant and nonresonant) are very simple, economical, and effec-tive directional antennas with many uses for transmitting and receiving waves in the MF (300 KHz–3 MHz) and HF (3–30 MHz) ranges. Their properties can be enhanced when used in arrays.
A MATLAB computer program, entitled Beverage, has been developed to analyze the radiation characteristics of a long-wire antenna. The description of the program is found in the corresponding READ ME file included in the CD attached to the book.
10.2.2 V Antenna For some applications a single long-wire antenna is not very practical because (1) its directivity may be low, (2) its side lobes may be high, and (3) its main beam is inclined at an angle, which is controlled by its length. These and other drawbacks of single long-wire antennas can be overcome by utilizing an array of wires.
One very practical array of long wires is the V antenna formed by using two wires each with one of its ends connected to a feed line as shown in Figure 10.8(a). In most applications, the plane formed by the legs of the V is parallel to the ground leading to a horizontal V array whose principal polarization is parallel to the ground and the plane of the V. Because of increased sidelobes, the directivity of ordinary linear dipoles begins to diminish for lengths greater than about 1.25λ, as shown in Figure 4.9. However by adjusting the included angle of a V-dipole, its directivity can be made greater and its side lobes smaller than those of a corresponding linear dipole. Designs for maximum directivity usually require smaller included angles for longer V’s.
Most V antennas are symmetrical (𝜃1 = 𝜃2 = 𝜃0 and l1 = l2 = l). Also V antennas can be designed to have unidirectional or bidirectional radiation patterns, as shown in Figures 10.8(b) and (c), respec-tively. To achieve the unidirectional characteristics, the wires of the V antenna must be nonresonant which can be accomplished by minimizing if not completely eliminating reflections from the ends of the wire. The reflected waves can be reduced by making the inclined wires of the V relatively thick. In theory, the reflections can even be eliminated by properly terminating the open ends of the V leading to a purely traveling wave antenna. One way of terminating the V antenna will be to attach a load, usually a resistor equal in value to the open end characteristic impedance of the V-wire transmission line, as shown in Figure 10.9(a). The terminating resistance can also be divided in half and each half connected to the ground leading to the termination of Figure 10.9(b). If the length of each leg of the V is very long (typically l > 5λ), there will be sufficient leakage of the field along each leg that when the wave reaches the end of the V it will be sufficiently reduced that there will not necessarily be a need for a termination. Of course, termination with a load is not possible without a ground plane.
544 TRAVELING WAVE AND BROADBAND ANTENNAS Feed line (a) V antenna l1 l2 1 θ 2 θ Feed line 1 (b) Unidirectional 1m θ 1m θ 2m θ 2m θ 2 0 θ 2 3 4 Feed line (c) Bidirectional 1 2 5 6 7 8 3 4 Figure 10.8 Unidirectional and bidirectional V antennas.
The patterns of the individual wires of the V antenna are conical in form and are inclined at an angle from their corresponding axes. The angle of inclination is determined by the length of each wire. For the patterns of each leg of a symmetrical V antenna to add in the direction of the line bisecting the angle of the V and to form one major lobe, the total included angle 2𝜃0 of the V should be equal to 2𝜃m, which is twice the angle that the cone of maximum radiation of each wire makes with its axis. When this is done, beams 2 and 3 of Figure 10.8(b) are aligned and add constructively. Similarly for Figure 10.8(c), beams 2 and 3 are aligned and add constructively in the forward direction, while beams 5 and 8 are aligned and add constructively in the backward direction. If the total included angle of the V is greater than 2𝜃m(2𝜃0 > 2𝜃m) the main lobe is split into two distinct beams. However, if TRAVELING WAVE ANTENNAS 545 Figure 10.9 Terminated V antennas.
2𝜃0 < 2𝜃m, then the maximum of the single major lobe is still along the plane that bisects the V but it is tilted upward from the plane of the V. This may be a desired designed characteristic when the antenna is required to transmit waves upward toward the ionosphere for optimum reflection or to receive signals reflected downward by the ionosphere. For optimum operation, typically the included angle is chosen to be approximately 𝜃0 ≃0.8𝜃m. When this is done, the reinforcement of the fields from the two legs of the V lead to a total directivity for the V of approximately twice the directivity of one leg of the V.
For a symmetrical V antenna with legs each of length l, there is an optimum included angle which leads to the largest directivity. Design data for optimum included angles of V dipoles were computed using Moment Method techniques and are shown in Figure 10.10(a). The correspond-ing directivities are shown in Figure 10.10(b). In each figure the dots (⋅) represent values computed using the Moment Method while the solid curves represent second- or third-order polynomials fitted through the computed data. The polynomials for optimum included angles and maximum directivi-ties are given by 2𝜃0 = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ −149.3 ( l λ )3 + 603.4 ( l λ )2 −809.5 ( l λ ) + 443.6 for 0.5 ≤l∕λ ≤1.5 13.39 ( l λ )2 −78.27 ( l λ ) + 169.77 for 1.5 ≤l∕λ ≤3 D0 = 2.94 ( l λ ) + 1.15 for 0.5 ≤l∕λ ≤3 (10-19a) (10-19b) (10-20) The dashed curves represent data obtained from empirical formulas . The corresponding input impedances of the V’s are slightly smaller than those of straight dipoles.
546 TRAVELING WAVE AND BROADBAND ANTENNAS Figure 10.10 Optimum included angle for maximum directivity as a function of arm length for V dipoles.
(source: G. A. Thiele and E. P. Ekelman, Jr., “Design Formulas for Vee Dipoles,” IEEE Trans. Antennas Prop-agat., Vol. AP-28, pp. 588–590, July 1980. c ⃝(1980) IEEE).
Another form of a V antenna is shown in the insert of Figure 10.11(a). The V is formed by a monopole wire, bent at an angle over a ground plane, and by its image shown dashed. The included angle of the V as well as the length can be used to tune the antenna. For included angles greater than 120◦(2𝜃0 > 120◦), the antenna exhibits primarily vertical polarization with radiation patterns almost identical to those of straight dipoles. As the included angle becomes smaller than about 120◦, a horizontally polarized field component is excited which tends to fill the pattern toward the horizontal direction, making it a very attractive communication antenna for aircraft. The computed impedance of the ground plane and free-space V configurations obtained by the Moment Method is shown plotted in Figure 10.11(a).
TRAVELING WAVE ANTENNAS 547 Figure 10.11 Computed impedance (R + jX) of V and bent wire antennas above ground. (source: D. G. Fink (ed.), Electronics Engineer’s Handbook, Chapter 18 (by W. F. Croswell), McGraw-Hill, New York, 1975).
Another practical form of a dipole antenna, particularly useful for airplane or ground-plane appli-cations, is the 90◦bent wire configuration of Figure 10.11(b). The computed impedance of the antenna, obtained also by the Moment Method , is shown plotted in Figure 10.11(b). This antenna can be tuned by adjusting its perpendicular and parallel lengths h1 and h2. The radiation pattern in the plane of the antenna is nearly omnidirectional for h1 ≤0.1λ. For h1 > 0.1λ the pattern approaches that of vertical λ∕2 dipole.
548 TRAVELING WAVE AND BROADBAND ANTENNAS V I I (a) Rhombus formed by 2 V's 2 0 θ 7 8 5 6 + – l (b) Rhombus formed by inverted vertical V over ground 2 1 4 3 2 3 4 1 Figure 10.12 Rhombic antenna configurations.
10.2.3 Rhombic Antenna A. Geometry and Radiation Characteristics Two V antennas can be connected at their open ends to form a diamond or rhombic antenna, as shown in Figure 10.12(a). The antenna is usually terminated at one end in a resistor, usually about 600–800 ohms, in order to reduce if not eliminate reflections. However, if each leg is long enough (typically greater than 5λ) sufficient leakage occurs along each leg that the wave that reaches the far end of the rhombus is sufficiently reduced that it may not be necessary to terminate the rhom-bus. To achieve the single main lobe, beams 2, 3, 6, and 7 are aligned and add constructively. The other end is used to feed the antenna. Another configuration of a rhombus is that of Figure 10.12(b) which is formed by an inverted V and its image (shown dashed). The inverted V is connected to the ground through a resistor. As with the V antennas, the pattern of rhombic antennas can be con-trolled by varying the element lengths, angles between elements, and the plane of the rhombus.
Rhombic antennas are usually preferred over V’s for nonresonant and unidirectional pattern appli-cations because they are less difficult to terminate. Additional directivity and reduction in side lobes can be obtained by stacking, vertically or horizontally, a number of rhombic and/or V antennas to form arrays.
The field radiated by a rhombus can be found by adding the fields radiated by its four legs.
For a symmetrical rhombus with equal legs, this can be accomplished using array theory and pat-tern multiplication. When this is done, a number of design equations can be derived –. For this design, the plane formed by the rhombus is placed parallel and a height h above a perfect electric conductor.
BROADBAND ANTENNAS 549 B. Design Equations Let us assume that it is desired to design a rhombus such that the maximum of the main lobe of the pattern, in a plane which bisects the V of the rhombus, is directed at an angle 𝜓0 above the ground plane. The design can be optimized if the height h is selected according to hm λ0 = m 4 cos(90◦−𝜓0), m = 1, 3, 5, … (10-21) with m = 1 representing the minimum height.
The minimum optimum length of each leg of a symmetrical rhombus must be selected accord-ing to l λ0 = 0.371 1 −sin(90◦−𝜓0) cos 𝜃0 (10-22) The best choice for the included angle of the rhombus is selected to satisfy 𝜃0 = cos−1[sin(90◦−𝜓0)] (10-23) 10.3 BROADBAND ANTENNAS In Chapter 9 broadband dipole antennas were discussed. There are numerous other antenna designs that exhibit greater broadband characteristics than those of the dipoles. Some of these antenna can also provide circular polarization, a desired extra feature for many applications. In this section we want to discuss briefly some of the most popular broadband antennas.
10.3.1 Helical Antenna Another basic, simple, and practical configuration of an electromagnetic radiator is that of a con-ducting wire wound in the form of a screw thread forming a helix, as shown in Figure 10.13. In most cases the helix is used with a ground plane. The ground plane can take different forms. One is for the ground to be flat, as shown in Figure 10.13. Typically the diameter of the ground plane should be at least 3λ∕4. However, the ground plane can also be cupped in the form of a cylindrical cavity (see Figure 10.17) or in the form of a frustrum cavity . In addition, the helix is usually connected to the center conductor of a coaxial transmission line at the feed point with the outer conductor of the line attached to the ground plane.
The geometrical configuration of a helix consists usually of N turns, diameter D and spacing S between each turn. The total length of the antenna is L = NS while the total length of the wire is Ln = NL0 = N √ S2 + C2 where L0 = √ S2 + C2 is the length of the wire between each turn and C = 𝜋D is the circumference of the helix. Another important parameter is the pitch angle 𝛼which is the angle formed by a line tangent to the helix wire and a plane perpendicular to the helix axis. The pitch angle is defined by 𝛼= tan−1 ( S 𝜋D ) = tan−1 ( S C ) (10-24) When 𝛼= 0◦, then the winding is flattened and the helix reduces to a loop antenna of N turns.
On the other hand, when 𝛼= 90◦then the helix reduces to a linear wire. When 0◦< 𝛼< 90◦, then 550 TRAVELING WAVE AND BROADBAND ANTENNAS Figure 10.13 Helical antenna with ground plane.
a true helix is formed with a circumference greater than zero but less than the circumference when the helix is reduced to a loop (𝛼= 0◦).
The radiation characteristics of the antenna can be varied by controlling the size of its geometrical properties compared to the wavelength. The input impedance is critically dependent upon the pitch angle and the size of the conducting wire, especially near the feed point, and it can be adjusted by controlling their values. The general polarization of the antenna is elliptical. However circular and linear polarizations can be achieved over different frequency ranges.
The helical antenna can operate in many modes; however the two principal ones are the normal (broadside) and the axial (end-fire) modes. The three-dimensional amplitude patterns representative of a helix operating, respectively, in the normal (broadside) and axial (end-fire) modes are shown in Figure 10.14. The one representing the normal mode, Figure 10.14(a), has its maximum in a plane normal to the axis and is nearly null along the axis. The pattern is similar in shape to that of a small dipole or circular loop. The pattern representative of the axial mode, Figure 10.14(b), has its maximum along the axis of the helix, and it is similar to that of an end-fire array. More details are in the sections that follow. The axial (end-fire) mode is usually the most practical because it can achieve circular polarization over a wider bandwidth (usually 2:1) and it is more efficient.
Because an elliptically polarized antenna can be represented as the sum of two orthogonal linear components in time-phase quadrature, a helix can always receive a signal transmitted from a rotating linearly polarized antenna. Therefore helices are usually positioned on the ground for space telemetry applications of satellites, space probes, and ballistic missiles to transmit or receive signals that have undergone Faraday rotation by traveling through the ionosphere.
A. Normal Mode In the normal mode of operation the field radiated by the antenna is maximum in a plane normal to the helix axis and minimum along its axis, as shown sketched in Figure 10.14(a), which is a figure-eight rotated about its axis similar to that of a linear dipole of l < λ0 or a small loop (a ≪λ0).
BROADBAND ANTENNAS 551 (a) Normal (broadside) mode (b) End-fire (axial) mode x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 10.14 Three-dimensional normalized amplitude linear power patterns for normal and end-fire modes helical designs.
To achieve the normal mode of operation, the dimensions of the helix are usually small compared to the wavelength (i.e., NL0 ≪λ0).
The geometry of the helix reduces to a loop of diameter D when the pitch angle approaches zero and to a linear wire of length S when it approaches 90◦. Since the limiting geometries of the helix are a loop and a dipole, the far field radiated by a small helix in the normal mode can be described in terms of E𝜃and E𝜙components of the dipole and loop, respectively. In the normal mode, the helix of Figure 10.15(a) can be simulated approximately by N small loops and N short dipoles connected together in series as shown in Figure 10.14(b). The fields are obtained by superposition of the fields from these elemental radiators. The planes of the loops are parallel to each other and perpendicular to the axes of the vertical dipoles. The axes of the loops and dipoles coincide with the axis of the helix.
Since in the normal mode the helix dimensions are small, the current throughout its length can be assumed to be constant and its relative far-field pattern to be independent of the number of loops and short dipoles. Thus its operation can be described accurately by the sum of the fields radiated 552 TRAVELING WAVE AND BROADBAND ANTENNAS Figure 10.15 Normal (broadside) mode for helical antenna and its equivalent.
by a small loop of diameter D and a short dipole of length S, with its axis perpendicular to the plane of the loop, and each with the same constant current distribution.
The far-zone electric field radiated by a short dipole of length S and constant current I0 is E𝜃, and it is given by (4-26a) as E𝜃= j𝜂kI0Se−jkr 4𝜋r sin 𝜃 (10-25) where l is being replaced by S. In addition the electric field radiated by a loop is E𝜙, and it is given by (5-27b) as E𝜙= 𝜂k2(D∕2)2I0e−jkr 4r sin 𝜃 (10-26) where D/2 is substituted for a. A comparison of (10-25) and (10-26) indicates that the two compo-nents are in time-phase quadrature, a necessary but not sufficient condition for circular or elliptical polarization.
The ratio of the magnitudes of the E𝜃and E𝜙components is defined here as the axial ratio (AR), and it is given by AR = |E𝜃| |E𝜙| = 4S 𝜋kD2 = 2λS (𝜋D)2 (10-27) By varying the D and/or S the axial ratio attains values of 0 ≤AR ≤∞. The value of AR = 0 is a special case and occurs when E𝜃= 0 leading to a linearly polarized wave of horizontal polarization (the helix is a loop). When AR = ∞, E𝜙= 0 and the radiated wave is linearly polarized with vertical polarization (the helix is a vertical dipole). Another special case is the one when AR is unity (AR = 1) BROADBAND ANTENNAS 553 and occurs when 2λ0S (𝜋D)2 = 1 (10-28) or C = 𝜋D = √ 2Sλ0 (10-28a) for which tan 𝛼= S 𝜋D = 𝜋D 2λ0 (10-29) When the dimensional parameters of the helix satisfy the above relation, the radiated field is circu-larly polarized in all directions other than 𝜃= 0◦where the fields vanish.
When the dimensions of the helix do not satisfy any of the above special cases, the field radiated by the antenna is not circularly polarized. The progression of polarization change can be described geometrically by beginning with the pitch angle of zero degrees (𝛼= 0◦), which reduces the helix to a loop with linear horizontal polarization. As 𝛼increases, the polarization becomes elliptical with the major axis being horizontally polarized. When 𝛼, is such that C∕λ0 = √ 2S∕λ0, AR = 1 and we have circular polarization. For greater values of 𝛼, the polarization again becomes elliptical but with the major axis vertically polarized. Finally when 𝛼= 90◦the helix reduces to a linearly polarized vertical dipole.
To achieve the normal mode of operation, it has been assumed that the current throughout the length of the helix is of constant magnitude and phase. This is satisfied to a large extent provided the total length of the helix wire NL0 is very small compared to the wavelength (Ln ≪λ0) and its end is terminated properly to reduce multiple reflections. Because of the critical dependence of its radiation characteristics on its geometrical dimensions, which must be very small compared to the wavelength, this mode of operation is very narrow in bandwidth and its radiation efficiency is very small. Practically this mode of operation is limited, and it is seldom utilized.
B. End-Fire Mode A more practical mode of operation, which can be generated with great ease, is the end-fire or axial mode. In this mode of operation, there is only one major lobe and its maximum radiation intensity is along the axis of the helix, as shown in Figure 10.14(b). The minor lobes are at oblique angles to the axis.
To excite this mode, the diameter D and spacing S must be large fractions of the wavelength. To achieve circular polarization, primarily in the major lobe, the circumference of the helix must be in the 3 4 < C∕λ0 < 4 3 range (with C∕λ0 = 1 near optimum), and the spacing about S ≃λ0∕4. The pitch angle is usually 12◦≤𝛼≤14◦. Most often the antenna is used in conjunction with a ground plane, whose diameter is at least λ0∕2, and it is fed by a coaxial line. However, other types of feeds (such as waveguides and dielectric rods) are possible, especially at microwave frequencies. The dimensions of the helix for this mode of operation are not as critical, thus resulting in a greater bandwidth.
C. Design Procedure The terminal impedance of a helix radiating in the end-fire mode is nearly resistive with values between 100 and 200 ohms. Smaller values, even near 50 ohms, can be obtained by properly design-ing the feed. Empirical expressions, based on a large number of measurements, have been derived , 554 TRAVELING WAVE AND BROADBAND ANTENNAS and they are used to determine a number of parameters. The input impedance (purely resistive) is obtained by R ≃140 ( C λ0 ) (10-30) which is accurate to about ±20%, the half-power beamwidth by HPBW (degrees) ≃ 52λ3∕2 0 C √ NS (10-31) the beamwidth between nulls by FNBW (degrees) ≃ 115λ3∕2 0 C √ NS (10-32) the directivity by D0 (dimensionless) ≃15N C2S λ3 0 (10-33) the axial ratio (for the condition of increased directivity) by AR = 2N + 1 2N (10-34) and the normalized far-field pattern by E = sin ( 𝜋 2N ) cos 𝜃sin[(N∕2)𝜓] sin[𝜓∕2] where 𝜓= k0 ( S cos 𝜃−L0 p ) p = L0∕λ0 S∕λ0 + 1 For ordinary end-fire radiation p = L0∕λ0 S∕λ0 + (2N + 1 2N ) For Hansen-Woodyard end-fire radiation (10-35) (10-35a) (10-35b) (10-35c) All these relations are approximately valid provided 12◦< 𝛼< 14◦, 3 4 < C∕λ0 < 4 3, and N > 3.
BROADBAND ANTENNAS 555 The far-field pattern of the helix, as given by (10-35), has been developed by assuming that the helix consists of an array of N identical turns (each of nonuniform current and identical to that of the others), a uniform spacing S between them, and the elements are placed along the z-axis. The cos 𝜃 term in (10-35) represents the field pattern of a single turn, and the last term in (10-35) is the array factor of a uniform array of N elements. The total field is obtained by multiplying the field from one turn with the array factor (pattern multiplication).
The value of p in (10-35a) is the ratio of the velocity with which the wave travels along the helix wire to that in free space, and it is selected according to (10-35b) for ordinary end-fire radiation or (10-35c) for Hansen-Woodyard end-fire radiation. These are derived as follows.
For ordinary end-fire the relative phase 𝜓among the various turns of the helix (elements of the array) is given by (6-7a), or 𝜓= k0S cos 𝜃+ 𝛽 (10-36) where d = S is the spacing between the turns of the helix. For an end-fire design, the radiation from each one of the turns along 𝜃= 0◦must be in phase. Since the wave along the helix wire between turns travels a distance L0 with a wave velocity v = pv0 (p < 1 where v0 is the wave velocity in free space) and the desired maximum radiation is along 𝜃= 0◦, then (10-36) for ordinary end-fire radiation is equal to 𝜓= (k0S cos 𝜃−kL0)𝜃=0◦= k0 ( S −L0 p ) = −2𝜋m, m = 0, 1, 2, … (10-37) Solving (10-37) for p leads to p = L0∕λ0 S∕λ0 + m (10-38) For m = 0 and p = 1, L0 = S. This corresponds to a straight wire (𝛼= 90◦), and not a helix. Therefore the next value is m = 1, and it corresponds to the first transmission mode for a helix. Substituting m = 1 in (10-38) leads to p = L0∕λ0 S∕λ0 + 1 (10-38a) which is that of (10-35b).
In a similar manner, it can be shown that for Hansen-Woodyard end-fire radiation (10-37) is equal to 𝜓= (k0S cos 𝜃−kL0)𝜃=0◦= k0 ( S −L0 p ) = − ( 2𝜋m + 𝜋 N ) , m = 0, 1, 2, … (10-39) which when solved for p leads to p = L0∕λ0 S∕λ0 + (2mN + 1 2N ) (10-40) 556 TRAVELING WAVE AND BROADBAND ANTENNAS For m = 1, (10-40) reduces to p = L0∕λ0 S∕λ0 + (2N + 1 2N ) (10-40a) which is identical to (10-35c).
Example 10.1 Design a 10-turn helix to operate in the axial mode. For an optimum design, 1. Determine the: a. Circumference (in λo), pitch angle (in degrees), and separation between turns (in λo) b. Relative (to free space) wave velocity along the wire of the helix for: (i) Ordinary end-fire design (ii) Hansen-Woodyard end-fire design c. Half-power beamwidth of the main lobe (in degrees) d. Directivity (in dB) using: (i) A formula (ii) The computer program Directivity of Chapter 2 e. Axial ratio (dimensionless and in dB) 2. Plot the normalized three-dimensional linear power pattern for the ordinary and Hansen-Woodyard designs.
Solution: 1. a. For an optimum design C ≃λo, 𝛼≃13◦⇒S = C tan 𝛼= λo tan(13◦) = 0.231λo b. The length of a single turn is Lo = √ S2 + C2 = λo √ (0.231)2 + (1)2 = 1.0263λo Therefore the relative wave velocity is: (i) Ordinary end-fire: p = 𝜈h 𝜈o = Lo∕λo So∕λo + 1 = 1.0263 0.231 + 1 = 0.8337 (ii) Hansen-Woodyard end-fire: p = 𝜈h 𝜈o = Lo∕λo So∕λo + (2N + 1 2N ) = 1.0263 0.231 + 21∕20 = 0.8012 BROADBAND ANTENNAS 557 c. The half-power beamwidth according to (10-31) is HPBW ≃52λ3∕2 o C √ NS = 52 1 √ 10(0.231) = 34.2135◦ d. The directivity is: (i) Using (10-33): Do ≃15N C2S λ3 o = 15(10)(1)2(0.231) = 34.65 (dimensionless) = 15.397 dB x z y (a) Ordinary end-fire (b) Hansen-Woodyard end-fire Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x z y Amplitude Pattern (linear scale) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 10.16 Three-dimensional normalized amplitude linear power patterns for helical ordinary (p = 0.8337) and Hansen-Woodyard (p = 0.8012) end-fire designs.
558 TRAVELING WAVE AND BROADBAND ANTENNAS (ii) Using the computer program Directivity and (10-35): a. ordinary end-fire (p = 0.8337): Do = 12.678 (dimensionless) = 11.03 dB b. H-W end-fire (p = 0.8012): Do = 26.36 (dimensionless) = 14.21 dB e. The axial ratio according to (10-34) is: AR = 2N + 1 2N = 20 + 1 20 = 1.05 (dimensionless) = 0.21 dB 2. The three-dimensional linear power patterns for the two end-fire designs, ordinary and Hansen-Woodyard, are shown in Figure 10.16.
D. Feed Design The nominal impedance of a helical antenna operating in the axial mode, computed using (10-30), is 100–200 ohms. However, many practical transmission lines (such as a coax) have characteristic impedance of about 50 ohms. In order to provide a better match, the input impedance of the helix must be reduced to near that value. There may be a number of ways by which this can be accomplished.
One way to effectively control the input impedance of the helix is to properly design the first 1/4 turn of the helix which is next to the feed , . To bring the input impedance of the helix from nearly 150 ohms down to 50 ohms, the wire of the first 1/4 turn should be flat in the form of a strip and the transition into a helix should be very gradual. This is accomplished by making the wire from the feed, at the beginning of the formation of the helix, in the form of a strip of width w by flattening it and nearly touching the ground plane which is covered with a dielectric slab of height h = w 377 √𝜀rZ0 −2 (10-41) where w = width of strip conductor of the helix starting at the feed 𝜀r = dielectric constant of the dielectric slab covering the ground plane Z0 = characteristic impedance of the input transmission line Typically the strip configuration of the helix transitions from the strip to the regular circular wire and the designed pitch angle of the helix very gradually within the first 1/4–1/2 turn.
This modification decreases the characteristic impedance of the conductor-ground plane effective transmission line, and it provides a lower impedance over a substantial but reduced bandwidth. For example, a 50-ohm helix has a VSWR of less than 2:1 over a 40% bandwidth compared to a 70% bandwidth for a 140-ohm helix. In addition, the 50-ohm helix has a VSWR of less than 1.2:1 over a 12% bandwidth as contrasted to a 20% bandwidth for one of 140 ohms.
A simple and effective way of increasing the thickness of the conductor near the feed point will be to bond a thin metal strip to the helix conductor . For example, a metal strip 70-mm wide was used to provide a 50-ohm impedance in a helix whose conducting wire was 13-mm in diameter and it was operating at 230.77 MHz.
A commercially available helix with a cupped ground plane is shown in Figure 10.17. It is right-hand circularly-polarized (RHCP) operating between 100–160 MHz with a gain of about 6 dB at 100 MHz and 12.8 dB at 160 MHz. The right-hand winding of the wire is clearly shown in the BROADBAND ANTENNAS 559 Figure 10.17 Commercial helix with a cupped ground plane. (Courtesy: Seavey Engineering Associates, Inc, Pembroke, MA).
photo. The axial ratio is about 8 dB at 100 MHz and 2 dB at 160 MHz. The maximum VSWR in the stated operating frequency, relative to a 50-ohm line, does not exceed 3:1.
A MATLAB computer program, entitled Helix, has been developed to analyze and design a helical antenna. The description of the program is found in the corresponding READ ME file included in the CD attached to the book.
10.3.2 Electric-Magnetic Dipole It has been shown in the previous section that the circular polarization of a helical antenna operating in the normal mode was achieved by assuming that the geometry of the helix is represented by a number of horizontal small loops and vertical infinitesimal dipoles. It would then seem reasonable that an antenna with only one loop and a single vertical dipole would, in theory, represent a radiator with an elliptical polarization. Ideally circular polarization, in all space, can be achieved if the current in each element can be controlled, by dividing the available power between the dipole and the loop, so that the magnitude of the field intensity radiated by each is equal.
Experimental models of such an antenna were designed and built one operating around 350 MHz and the other near 1.2 GHz. A sketch of one of them is shown in Figure 10.18. The mea-sured VSWR in the 1.15–1.32 GHz frequency range was less than 2:1.
This type of an antenna is very useful in UHF communication networks where considerable amount of fading may exist. In such cases the fading of the horizontal and vertical components are affected differently and will not vary in the same manner. Hopefully, even in severe cases, there will always be one component all the time which is being affected less than the other, thus provid-ing continuous communication. The same results would apply in VHF and/or UHF broadcasting. In addition, a transmitting antenna of this type would also provide the versatility to receive with hori-zontally or vertically polarized elements, providing a convenience in the architectural design of the receiving station.
10.3.3 Yagi-Uda Array of Linear Elements Another very practical radiator in the HF (3–30 MHz), VHF (30–300 MHz), and UHF (300–3,000 MHz) ranges is the Yagi-Uda antenna. This antenna consists of a number of linear dipole 560 TRAVELING WAVE AND BROADBAND ANTENNAS Figure 10.18 Electric-magnetic dipole configuration. (source: A. G. Kandoian, “Three New Antenna Types and Their Applications,” Proc. IRE, Vol. 34, pp. 70W–75W, February 1946. c ⃝(1946) IEEE).
elements, as shown in Figure 10.19, one of which is energized directly by a feed transmission line while the others act as parasitic radiators whose currents are induced by mutual coupling. A com-mon feed element for a Yagi-Uda antenna is a folded dipole. This radiator is exclusively designed to operate as an end-fire array, and it is accomplished by having the parasitic elements in the forward beam act as directors while those in the rear act as reflectors. Yagi designated the row of directors as a “wave canal.” The Yagi-Uda array has been widely used as a home TV antenna; so it should be familiar to most of the readers, if not to the general public.
N – 1 N – 2 1 2a 2 Directors Driven (energized) element Reflector N x y z yn ln sn Figure 10.19 Yagi-Uda antenna configuration.
BROADBAND ANTENNAS 561 The original design and operating principles of this radiator were first described in Japanese in articles published in the Journal of I.E.E. of Japan by S. Uda of the Tohoku Imperial University in Japan . In a later, but more widely circulated and read article , one of Professor Uda’s colleagues, H. Yagi, described the operation of the same radiator in English. This paper has been considered a classic, and it was reprinted in 1984 in its original form in the Proceedings of the IEEE as part of IEEE’s centennial celebration. Despite the fact that Yagi in his English written paper acknowledged the work of Professor Uda on beam radiators at a wavelength of 4.4 m, it became customary throughout the world to refer to this radiator as a Yagi antenna, a generic term in the antenna dictionary. However, in order for the name to reflect more appropriately the contributions of both inventors, it should be called a Yagi-Uda antenna, a name that will be adopted in this book.
Although the work of Uda and Yagi was done in the early 1920s and published in the middle 1920s, full acclaim in the United States was not received until 1928 when Yagi visited the United States and presented papers at meetings of the Institute of Radio Engineers (IRE) in New York, Washington, and Hartford. In addition, his work was published in the Proceedings of IRE, June 1928, where J. H.
Dellinger, Chief of Radio Division, Bureau of Standards, Washington, D.C., and himself a pioneer of radio waves, characterized it as “exceptionally fundamental” and wrote “I have never listened to a paper that I felt so sure was destined to be a classic.” So true!!
In 1984, IEEE celebrated its centennial year (1884–1984). Actually, IEEE was formed in 1963 when the IRE and AIEE united to form IEEE. During 1984, the Proceedings of the IEEE republished some classic papers, in their original form, in the different areas of electrical engineering that had appeared previously either in the Proceeding of the IRE or IEEE. In antennas, the only paper that was republished was that by Yagi . Not only that, in 1997, the Proceedings of the IEEE republished for the second time the original paper by Yagi , . That in itself tells us something of the impact this particular classic antenna design had on the electrical engineering profession.
The Yagi-Uda antenna has received exhaustive analytical and experimental investigations in the open literature and elsewhere. It would be impractical to list all the contributors, many of whom we may not be aware. However, we will attempt to summarize the salient point of the analysis, describe the general operation of the radiator, and present some design data.
To achieve the end-fire beam formation, the parasitic elements in the direction of the beam are somewhat smaller in length than the feed element. Typically the driven element is resonant with its length slightly less than λ∕2 (usually 0.45–0.49λ) whereas the lengths of the directors should be about 0.4 to 0.45λ. However, the directors are not necessarily of the same length and/or diameter.
The separation between the directors is typically 0.3 to 0.4λ, and it is not necessarily uniform for optimum designs. It has been shown experimentally that for a Yagi-Uda array of 6λ total length the overall gain was independent of director spacing up to about 0.3λ. A significant drop (5–7 dB) in gain was noted for director spacings greater than 0.3λ. For that antenna, the gain was also independent of the radii of the directors up to about 0.024λ. The length of the reflector is somewhat greater than that of the feed. In addition, the separation between the driven element and the reflector is somewhat smaller than the spacing between the driven element and the nearest director, and it is found to be near optimum at 0.25λ.
Since the length of each director is smaller than its corresponding resonant length, the impedance of each is capacitive and its current leads the induced emf. Similarly the impedances of the reflec-tors is inductive and the phases of the currents lag those of the induced emfs. The total phase of the currents in the directors and reflectors is not determined solely by their lengths but also by their spacing to the adjacent elements. Thus, properly spaced elements with lengths slightly less than their corresponding resonant lengths (less than λ∕2) act as directors because they form an array with cur-rents approximately equal in magnitude and with equal progressive phase shifts which will reinforce the field of the energized element toward the directors. Similarly, a properly spaced element with a length of λ∕2 or slightly greater will act as a reflector. Thus a Yagi-Uda array may be regarded as a structure supporting a traveling wave whose performance is determined by the current distribution in each element and the phase velocity of the traveling wave. It should be noted that the previous 562 TRAVELING WAVE AND BROADBAND ANTENNAS discussion on the lengths of the directors, reflectors, and driven elements is based on the first reso-nance. Higher resonances are available near lengths of λ, 3λ∕2, and so forth, but are seldom used.
In practice, the major role of the reflector is played by the first element next to the one ener-gized, and very little in the performance of a Yagi-Uda antenna is gained if more than one (at the most two) elements are used as reflectors. However, considerable improvements can be achieved if more directors are added to the array. Practically there is a limit beyond which very little is gained by the addition of more directors because of the progressive reduction in magnitude of the induced currents on the more extreme elements. Usually most antennas have about 6 to 12 directors. How-ever, many arrays have been designed and built with 30 to 40 elements. Array lengths on the order of 6λ have been mentioned as typical. A gain (relative to isotropic) of about 5 to 9 per wave-length is typical for such arrays, which would make the overall gain on the order of about 30 to 54 (14.8–17.3 dB) typical.
The radiation characteristics that are usually of interest in a Yagi-Uda antenna are the forward and backward gains, input impedance, bandwidth, front-to-back ratio, and magnitude of minor lobes. The lengths and diameters of the directors and reflectors, as well as their respective spacings, determine the optimum characteristics. For a number of years optimum designs were accomplished experimen-tally. However, with the advent of high-speed computers many different numerical techniques, based on analytical formulations, have been utilized to derive the geometrical dimensions of the array for optimum operational performance. Usually Yagi-Uda arrays have low input impedance and rela-tively narrow bandwidth (on the order of about 2%). Improvements in both can be achieved at the expense of others (such as gain, magnitude of minor lobes, etc.). Usually a compromise is made, and it depends on the particular design. One way to increase the input impedance without affecting the performance of other parameters is to use an impedance step-up element as a feed (such as a two-element folded dipole with a step-up ratio of about 4). Front-to-back ratios of about 30 (≃15 dB) can be achieved at wider than optimum element spacings, but they usually are compromised somewhat to improve other desirable characteristics.
The Yagi-Uda array can be summarized by saying that its performance can be considered in three parts: 1. the reflector-feeder arrangement 2. the feeder 3. the rows of directors It has been concluded, numerically and experimentally, that the reflector spacing and size have (1) negligible effects on the forward gain and (2) large effects on the backward gain (front-to-back ratio) and input impedance, and they can be used to control or optimize antenna parameters without affecting the gain significantly. The feeder length and radius has a small effect on the forward gain but a large effect on the backward gain and input impedance. Its geometry is usually chosen to con-trol the input impedance that most commonly is made real (resonant element). The size and spacing of the directors have a large effect on the forward gain, backward gain, and input impedance, and they are considered to be the most critical elements of the array.
Yagi-Uda arrays are quite common in practice because they are lightweight, simple to build, low-cost, and provide moderately desirable characteristics (including a unidirectional beam) for many applications. The design for a small number of elements (typically five or six) is simple but the design becomes quite critical if a large number of elements are used to achieve a high directivity. To increase the directivity of a Yagi-Uda array or to reduce the beamwidth in the E-plane, several rows of Yagi-Uda arrays can be used to form a curtain antenna. To neutralize the effects of the feed transmission line, an odd number of rows is usually used.
A. Theory: Integral Equation-Moment Method There have been many experimental , investigations and analytical – formulations of the Yagi-Uda array. A method based on rigorous integral equations for the electric field BROADBAND ANTENNAS 563 radiated by the elements in the array will be presented and it will be used to describe the complex current distributions on all the elements, the phase velocity, and the corresponding radiation patterns.
The method is similar to that of , which is based on Pocklington’s integral equation of (8-24) while the one presented here follows that of but is based on Pocklington’s integral equation of (8-22) and formulated by Tirkas . Mutual interactions are also included and, in principle, there are no restrictions on the number of elements. However, for computational purposes, point-matching numerical methods, based on the techniques of Section 8.4, are used to evaluate and satisfy the integral equation at discrete points on the axis of each element rather than everywhere on the surface of every element. The number of discrete points where boundary conditions are matched must be sufficient in number to allow the computed data to compare well with experimental results.
The theory is based on Pocklington’s integral equation of (8-22) for the total field generated by an electric current source radiating in an unbounded free-space, or ∫ +l∕2 −l∕2 I(z′) ( 𝜕2 𝜕z2 + k2 ) e−jkR R dz′ = j4𝜋𝜔𝜀0Et z (10-42) where R = √ (x −x′)2 + (y −y′)2 + (z −z′)2 (10-42a) Since 𝜕2 𝜕z2 ( e−jkR R ) = 𝜕2 𝜕z′2 ( e−jkR R ) (10-43) (10-42) reduces to ∫ +l∕2 −l∕2 I(z′) 𝜕2 𝜕z′2 ( e−jkR R ) dz′ + k2 ∫ +l∕2 −l∕2 I(z′)e−jkR R dz′ = j4𝜋𝜔𝜀0Et z (10-44) We will now concentrate in the integration of the first term of (10-44). Integrating the first term of (10-44) by parts where u = I(z′) (10-45) du = dI(z′) dz′ dz′ (10-45a) dv = 𝜕2 𝜕z′2 ( e−jkR R ) dz′ = 𝜕 𝜕z′ [ 𝜕 𝜕z′ ( e−jkR R )] dz′ (10-46) v = 𝜕 𝜕z′ ( e−jkR R ) (10-46a) reduces it to ∫ +l∕2 −l∕2 I(z′) 𝜕2 𝜕z′2 ( e−jkR R ) dz′ = I(z′) [ 𝜕 𝜕z′ ( e−jkR R )] | | | | +l∕2 −l∕2 −∫ +l∕2 −l∕2 𝜕 𝜕z′ ( e−jkR R ) dI(z′) dz′ dz′ (10-47) 564 TRAVELING WAVE AND BROADBAND ANTENNAS Since we require that the current at the ends of each wire vanish [i.e., Iz(z′ = +l∕2) = Iz(z′ = −l∕2) = 0], (10-47) reduces to ∫ +l∕2 −l∕2 I(z′) 𝜕2 𝜕z′2 ( e−jkR R ) dz′ = −∫ +l∕2 −l∕2 𝜕 𝜕z′ ( e−jkR R ) dz′ dI(z′) dz′ (10-48) Integrating (10-48) by parts where u = dI(z′) dz′ (10-49) du = d2I(z′) dz′2 dz′ (10-49a) dv = 𝜕 𝜕z′ ( e−jkR R ) dz′ (10-50) v = e−jkR R (10-50a) reduces (10-48) to ∫ +l∕2 −l∕2 I(z′) 𝜕2 𝜕z′2 ( e−jkR R ) dz′ = −dI(z′) dz′ e−jkR R | | | | +l∕2 −l∕2 + ∫ +l∕2 −l∕2 d2I(z′) dz′2 e−jkR R dz′ (10-51) When (10-51) is substituted for the first term of (10-44) reduces it to −dI(z′) dz′ e−jkR R | | | | +l∕2 −l∕2 + ∫ +l∕2 −l∕2 [ k2I(z′) + d2I(z′) dz′2 ] e−jkR R dz′ = j4𝜋𝜔𝜀0Et z (10-52) For small diameter wires the current on each element can be approximated by a finite series of odd-ordered even modes. Thus, the current on the nth element can be written as a Fourier series expansion of the form In(z′) = M ∑ m=1 Inm cos [ (2m −1)𝜋z′ ln ] (10-53) where Inm represents the complex current coefficient of mode m on element n and ln represents the corresponding length of the n element. Taking the first and second derivatives of (10-53) and substituting them, along with (10-53), into (10-52) reduces it to M ∑ m=1 Inm { (2m −1)𝜋 ln sin [ (2m −1) 𝜋z′ n ln ] e−jkR R | | | | +ln∕2 −ln∕2 + [ k2 −(2m −1)2𝜋2 l2 n ] × ∫ +ln∕2 −ln∕2 cos [ (2m −1) 𝜋z′ n ln ] e−jkR R dz′ n } = j4𝜋𝜔𝜀0Et z (10-54) BROADBAND ANTENNAS 565 Since the cosine is an even function, (10-54) can be reduced by integrating over only 0 ≤z′ ≤l∕2 to M ∑ m=1 Inm { (−1)m+1 (2m −1)𝜋 ln G2 ( x, x′, y, y′∕z, ln 2 ) + [ k2 −(2m −1)2𝜋2 l2 n ] × ∫ ln∕2 0 G2(x, x′, y, y′∕z, z′ n) cos [(2m −1)𝜋z′ n ln ] dz′ n } = j4𝜋𝜔𝜀0Et z (10-55) where G2(x, x′, y, y′∕z, z′ n) = e−jkR− R− + e−jkR+ R+ (10-55a) R± = √ (x −x′)2 + (y −y′)2 + a2 + (z ± z′)2 (10-55b) n = 1, 2, 3, … , N N = total number of elements where R± is the distance from the center of each wire radius to the center of any other wire, as shown in Figure 10.20(a).
The integral equation of (10-55) is valid for each element, and it assumes that the number M of current modes is the same for each element. To apply the Moment Method solution to the integral equation of (10-55), each wire element is subdivided in M segments. On each element, other than the driven element, the matching is done at the center of the wire, and it requires that Et z of (10-55) vanishes at each matching point of each segment [i.e., Et z(z = zi) = 0], as shown in Figure 10.20(b).
On the driven element the matching is done on the surface of the wire, and it requires that Et z of (10-55) vanishes at M −1 points, even though there are m modes, and it excludes the segment at the feed as shown in Figure 10.20(c). This generates M −1 equations. The Mth equation on the feed element is generated by the constraint that the normalized current for all M modes at the feed point (z′ = 0) of the driven element is equal to unity , , or M ∑ m=1 Inm(z′ = 0) | | | |n=N = 1 (10-56) Based on the above procedure, a system of linear equations is generated by taking into account the interaction of a. each mode in each wire segment with each segment on the same wire.
b. each mode in each wire segment with each segment on the other wires.
This system of linear equations is then solved to find the complex amplitude coefficients of the current distribution in each wire as represented by (10-53). This is demonstrated in for a three-element array (one director, one reflector, and the driven element) with two modes in each wire.
566 TRAVELING WAVE AND BROADBAND ANTENNAS y z 2a R si x zN z4 z3 z2 z1 zN – 1 li ln si + 1 (x, y, z) (x', y', z') y z ......
......
z4 z3 z2 z1 zN – 2 zN – 1 y z 1 amp.
(a) Separation distance (b) Parasitic elements (c) Driven element Figure 10.20 Geometry of Yagi-Uda array for Moment Method formulation (source: G. A. Thiele, “Yagi-Uda Type Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-17, No. 1, pp. 24–31, January 1969. c ⃝ (1969) IEEE).
B. Far-Field Pattern Once the current distribution is found, the far-zone field generated by each element can be found using the techniques outlined in Chapter 3. The total field of the entire Yagi-Uda array is obtained by summing the contributions from each.
Using the procedure outlined in Chapter 3, Section 3.6, the far-zone electric field generated by the M modes of the nth element oriented parallel to the z axis is given by E𝜃n ≃−j𝜔A𝜃n (10-57) A𝜃n ≃−𝜇e−jkr 4𝜋r sin 𝜃∫ +ln∕2 −ln∕2 Inejk(xn sin 𝜃cos 𝜙+yn sin 𝜃sin 𝜙+z′ n cos 𝜃) dz′ n ≃−𝜇e−jkr 4𝜋r sin 𝜃 [ ejk(xn sin 𝜃cos 𝜙+yn sin 𝜃sin 𝜙) ∫ +ln∕2 −ln∕2 Inejkz′ n cos 𝜃dz′ n ] (10-57a) BROADBAND ANTENNAS 567 where xn, yn represent the position of the nth element. The total field is then obtained by summing the contributions from each of the N elements, and it can be written as E𝜃= N ∑ n=1 E𝜃n = −j𝜔A𝜃 (10-58) A𝜃= N ∑ n=1 A𝜃n = −𝜇e−jkr 4𝜋r sin 𝜃 N ∑ n=1 { ejk(xn sin 𝜃cos 𝜙+yn sin 𝜃sin 𝜙) × [ ∫ +ln∕2 −ln∕2 Inejkz′ n cos 𝜃dz′ n ]} (10-58a) For each wire, the current is represented by (10-53). Therefore the last integral in (10-58a) can be written as ∫ +ln∕2 −ln∕2 Inejkz′ n cos 𝜃dz′ n = M ∑ m=1 Inm ∫ +ln∕2 −ln∕2 cos [(2m −1)𝜋z′ n ln ] ejkz′ n cos 𝜃dz′ n (10-59) Since the cosine is an even function, (10-59) can also be expressed as ∫ +ln∕2 −ln∕2 Inejkz′ n cos 𝜃dz′ n = M ∑ m=1 Inm ∫ +ln∕2 0 2 cos [(2m −1)𝜋z′ n ln ] × [ ejkz′ n cos 𝜃+ e−jkz′ n cos 𝜃 2 ] dz′ n = M ∑ m=1 Inm ∫ +ln∕2 0 2 cos [(2m −1)𝜋z′ n ln ] × cos(kz′ n cos 𝜃)dz′ n (10-60) Using the trigonometric identity 2 cos(𝛼) cos(𝛽) = cos(𝛼+ 𝛽) + cos(𝛼−𝛽) (10-61) (10-60) can be rewritten as ∫ +ln∕2 −ln∕2 Inejkz′ n cos 𝜃dz′ n = M ∑ m=1 Inm { ∫ +ln∕2 0 cos [(2m −1)𝜋 ln + k cos 𝜃 ] z′ n dz′ n + ∫ +ln∕2 0 cos [(2m −1)𝜋 ln −k cos 𝜃 ] z′ n dz′ n (10-62) 568 TRAVELING WAVE AND BROADBAND ANTENNAS Since ∫ 𝛼∕2 0 cos[(b ± c)z] dz = 𝛼 2 sin [ (b ± c)𝛼 2 ] (b ± c)𝛼 2 (10-63) (10-62) can be reduced to ∫ +ln∕2 −ln∕2 Inejkz′ n cos 𝜃dz′ n = M ∑ m=1 Inm [sin(Z+) Z+ + sin(Z−) Z− ] ln 2 (10-64) Z+ = [(2m −1)𝜋 ln + k cos 𝜃 ] ln 2 (10-64a) Z−= [(2m −1)𝜋 ln −k cos 𝜃 ] ln 2 (10-64b) Thus, the total field represented by (10-58) and (10-58a) can be written as E𝜃= N ∑ n=1 E𝜃n = −j𝜔A (10-65) A𝜃= N ∑ n=1 A𝜃n = −𝜇e−jkr 4𝜋r sin 𝜃 N ∑ n=1 { ejk(xn sin 𝜃cos 𝜙+yn sin 𝜃sin 𝜙) × M ∑ m=1 Inm [sin(Z+) Z+ + sin(Z−) Z− ]} ln 2 (10-65a) There have been other analyses , based on the integral equation formulation that allows the conversion to algebraic equations. In order not to belabor further the analytical formulations, which in call cases are complicated because of the antenna structure, it is appropriate at this time to present some results and indicate design procedures.
C. Computer Program and Results Based on the preceding formulation, a MATLAB and FORTRAN computer program entitled Yagi Uda has been developed that computes the E- and H-plane patterns, their correspond-ing half-power beamwidths, and the directivity of the Yagi-Uda array. The program is described in the corresponding READ ME file, and both are included in the publisher’s website for the book.
The input parameters include the total number of elements (N), the number of current modes in each element (M), the length of each element, and the spacing between the elements. The program assumes one reflector, one driven element, and N −2 directors. For the development of the formula-tion and computer program, the numbering system (n = 1, 2, … , N) for the elements begins with the first director (n = 1), second director (n = 2), etc. The reflector is represented by the next to the last element (n = N −1), while the driven element is designated as the last element (n = N), as shown in Figure 10.19.
One Yagi-Uda array design is considered here, which is the same as one of the two included in ; the other ones are assigned as end of the chapter problems. The patterns, beamwidths, and directivities were computed based on the computer program developed here.
BROADBAND ANTENNAS 569 Example 10.2 Design a Yagi-Uda array of 15 elements (13 directors, one reflector, and the exciter). Compute and plot the E- and H-plane patterns, normalized current at the center of each element, and directivity and front-to-back ratio as a function of reflector spacing and director spacing. Use the computer program Yagi Uda of this chapter. The dimensions of the array are as follows: N = total number of elements = 15 number of directors = 13 number of reflectors = 1 number of exciters = 1 total length of reflector = 0.5λ total length of feeder = 0.47λ total length of each director = 0.406λ spacing between reflector and feeder = 0.25λ spacing between adjacent directors = 0.34λ a = radius of wires = 0.003λ Solution: Using the computer program of this chapter, the computed E- and H-plane patterns of this design are shown in Figure 10.21. The corresponding beamwidths are: E-plane (Θe = 26.98◦), H-plane (Θh = 27.96◦) while the directivity is 14.64 dB. A plot of the current at the center of each element versus position of the element is shown in Figure 10.22; the current of the feed element at its center is unity, as required by (10-56). One important figure-of-merit in E-plane H-plane Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 60° 120° 30° 150° 0° 180° 30° 150° 60° 120° 90° 90° Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Figure 10.21 E- and H-plane amplitude patterns of 15-element Yagi-Uda array.
570 TRAVELING WAVE AND BROADBAND ANTENNAS 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 0.0 0.2 0.4 0.6 0.8 1.0 Element current amplitude Element number Figure 10.22 Normalized current at the center of each element of a 15-element Yagi-Uda array.
Yagi-Uda array is the front-to-back ratio of the pattern [20 log10 E(𝜃= 90◦, 𝜙= 90◦)∕E(𝜃= 90◦, 𝜙= 270◦)] as a function of the spacing of the reflector with respect to the feeder. This, along with the directivity, is shown in Figure 10.23 for spacing from 0.1λ–0.5λ. For this design, the maximum front-to-back ratio occurs for a reflector spacing of about 0.23λ while the directivity is monotonically decreasing very gradually, from about 15.2 dB at a spacing of 0.1λ down to about 10.4 dB at a spacing of 0.5λ.
0.1 0.2 0.3 0.4 0.5 0 5 10 15 20 25 30 35 40 Directivity (D0) Front-to-Back ratio Decibels (dB) Reflector spacing ( ) λ Figure 10.23 Directivity and front-to-back ratio, as a function of reflector spacing, of a 15-element Yagi-Uda array.
Another important parametric investigation is the variation of the front-to-back ratio and direc-tivity as a function of director spacing. This is shown in Figure 10.24 for spacings from 0.1λ to 0.5λ. It is apparent that the directivity exhibits a slight increase from about 12 dB at a spacing of BROADBAND ANTENNAS 571 about 0.1λ to about 15.3 dB at a spacing of about 0.45λ. A steep drop in directivity occurs for spacings greater than about 0.45λ. This agrees with the conclusion arrived at in and that large reductions in directivity occur in Yagi-Uda array designs for spacings greater than about 0.4λ. For this design, the variations of the front-to-back ratio are much more sensitive as a func-tion of director spacing, as shown in Figure 10.24; excursions on the order of 20–25 dB are evident for changes in spacing of about 0.05λ. Such variations in front-to-back ratio as shown in Figure 10.24, as a function of director spacing, are not evident in Yagi-Uda array designs with a smaller number of elements. However, they are even more pronounced for designs with a larger number of elements. Both of these are demonstrated in design problems assigned at the end of the chapter.
0.1 0.2 0.3 0.4 0.5 0 5 10 15 20 25 30 35 40 Directivity (D0) Front-to-back ratio Decibels (dB) Directors spacing (λ) Figure 10.24 Directivity and front-to-back ratio, as a function of director spacing, for 15-element Yagi-Uda array.
D. Optimization The radiation characteristics of the array can be adjusted by controlling the geometrical parameters of the array. This was demonstrated in Figures 10.23 and 10.24 for the 15-element array using uniform lengths and making uniform variations in spacings. However, these and other array characteristics can be optimized by using nonuniform director lengths and spacings between the directors. For example, the spacing between the directors can be varied while holding the reflector–exciter spacing and the lengths of all elements constant. Such a procedure was used by Cheng and Chen to optimize the directivity of a six-element (four-director, reflector, exciter) array using a perturbational technique. The results of the initial and the optimized (perturbed) array are shown in Table 10.1. For the same array, they allowed all the spacings to vary while maintaining constant all other parameters.
The results are shown in Table 10.2.
Another optimization procedure is to maintain the spacings between all the elements constant and vary the lengths so as to optimize the directivity. The results of a six-element array are shown in Table 10.3. The ultimate optimization is to vary both the spacings and lengths. This was accom-plished by Chen and Cheng whereby they first optimized the array by varying the spacing, while maintaining the lengths constant. This was followed, on the same array, with perturbations in the lengths while maintaining the optimized spacings constant. The results of this procedure are shown in Table 10.4 with the corresponding H-plane (𝜃= 𝜋∕2, 𝜙) far-field patterns shown in Figure 10.25.
572 TRAVELING WAVE AND BROADBAND ANTENNAS TABLE 10.1 Directivity Optimization for Six-Element Yagi-Uda Array (Perturbation of Director Spacings), l1 = 0.51𝛌, l2 = 0.50𝛌, l3 = l4 = l5 = l6 = 0.43𝛌, a = 0.003369𝛌 s21∕𝛌 s32∕𝛌 s43∕𝛌 s54∕𝛌 s65∕𝛌 Directivity (dB) Initial array 0.250 0.310 0.310 0.310 0.310 11.21 Optimized array 0.250 0.336 0.398 0.310 0.407 12.87 (source: D. K. Cheng and C. A. Chen, “Optimum Spacings for Yagi-Uda Arrays,” IEEE Trans.
Antennas Propag., Vol. AP-21, pp. 615–623, September 1973. c ⃝(1973) IEEE).
TABLE 10.2 Directivity Optimization for Six-Element Yagi-Uda Array (Perturbation of All Element Spacings), l1 = 0.51𝛌, l2 = 0.50𝛌, l3 = l4 = l5 = l6 = 0.43𝛌, a = 0.003369𝛌 s21∕𝛌 s32∕𝛌 s43∕𝛌 s54∕𝛌 s65∕𝛌 Directivity (dB) Initial array 0.280 0.310 0.310 0.310 0.310 10.92 Optimized array 0.250 0.352 0.355 0.354 0.373 12.89 (source: D. K. Cheng and C. A. Chen, “Optimum Spacings for Yagi-Uda Arrays,” IEEE Trans.
Antennas Propag., Vol. AP-21, pp. 615–623, September 1973. c ⃝(1973) IEEE).
In all, improvements in directivity and front-to-back ratio are noted. The ideal optimization will be to allow the lengths and spacings to vary simultaneously. Such an optimization was not performed in or , although it could have been done iteratively by repeating the procedure.
Another parameter that was investigated for the directivity-optimized Yagi-Uda antenna was the frequency bandwidth . The results of such a procedure are shown in Figure 10.26. The antenna was a six-element array optimized at a center frequency f0. The array was designed, using space perturbations on all the elements, to yield an optimum directivity at f0. The geometrical parameters are listed in Table 10.2. The 3-dB bandwidth seems to be almost the same for the initial and the opti-mized arrays. The rapid decrease in the directivity of the initial and optimized arrays at frequencies higher than f0 and nearly constant values below f0 may be attributed to the structure of the antenna which can support a “traveling wave” at f < f0 but not at f > f0. It has thus been suggested that an increase in the bandwidth can be achieved if the geometrical dimensions of the antenna are chosen slightly smaller than the optimum.
E. Input Impedance and Matching Techniques The input impedance of a Yagi-Uda array, measured at the center of the driven element, is usu-ally small and it is strongly influenced by the spacing between the reflector and feed element. For a 13-element array using a resonant driven element, the measured input impedances are listed in Table 10.5 . Some of these values are low for matching to a 50-, 78-, or 300-ohm transmis-sion lines.
TABLE 10.3 Directivity Optimization for Six-Element Yagi-Uda Array (Perturbation of All Element Lengths), s21 = 0.250𝛌, s32 = s43 = s54 = s65 = 0.310𝛌, a = 0.003369𝛌 l1∕𝛌 l2∕𝛌 l3∕𝛌 l4∕𝛌 l5∕𝛌 l6∕𝛌 Directivity (dB) Initial array 0.510 0.490 0.430 0.430 0.430 0.430 10.93 Length-perturbed array 0.472 0.456 0.438 0.444 0.432 0.404 12.16 (source: C. A. Chen and D. K. Cheng, “Optimum Element Lengths for Yagi-Uda Arrays,” IEEE Trans. Antennas Propag., Vol. AP-23, pp. 8–15, January 1975. c ⃝(1975) IEEE).
TABLE 10.4 Directivity Optimization for Six-Element Yagi-Uda Array (Perturbation of Director Spacings and All Element Lengths), a = 0.003369𝛌 Directivity l1∕𝛌 l2∕𝛌 l3∕𝛌 l4∕𝛌 l5∕𝛌 l6∕𝛌 s21∕𝛌 s32∕𝛌 s43∕𝛌 s54∕𝛌 s65∕𝛌 (dB) Initial array 0.510 0.490 0.430 0.430 0.430 0.430 0.250 0.310 0.310 0.310 0.310 10.93 Array after spacing perturbation 0.510 0.490 0.430 0.430 0.430 0.430 0.250 0.289 0.406 0.323 0.422 12.83 Optimum array after spacing and length perturbation 0.472 0.452 0.436 0.430 0.434 0.430 0.250 0.289 0.406 0.323 0.422 13.41 (source: C. A. Chen and D. K. Cheng, “Optimum Element Lengths for Yagi-Uda Arrays,” IEEE Trans. Antennas Propag., Vol. AP-23, pp. 8–15, January 1975. c ⃝(1975) IEEE).
573 574 TRAVELING WAVE AND BROADBAND ANTENNAS Figure 10.25 Normalized amplitude antenna patterns of initial, perturbed, and optimum six-element Yagi-Uda arrays (Table 10.4). (source: C. A. Chen and D. K. Cheng, “Optimum Element Lengths for Yagi-Uda Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-23, pp. 8–15, January 1975. c ⃝ (1975) IEEE).
Figure 10.26 Bandwidth of initial and optimum six-element Yagi-Uda array with perturbation of all element spacings (Table 10.2). (source: N. K. Takla and L.-C. Shen, “Bandwidth of a Yagi Array with Optimum Direc-tivity,” IEEE Trans. Antennas Propagat., Vol. AP-25, pp. 913–914, November 1977. c ⃝(1977) IEEE).
BROADBAND ANTENNAS 575 TABLE 10.5 Input Impedance of a 15-Element Yagi-Uda Array (Reflector Length = 0.5λ; Director Spacing = 0.34λ; Director Length = 0.406λ) Reflector Spacing (s21∕𝛌) Input Impedance (ohms) 0.25 62 0.18 50 0.15 32 0.13 22 0.10 12 There are many techniques that can be used to match a Yagi-Uda array to a transmission line and eventually to the receiver, which in many cases is a television set which has a large impedance (on the order of 300 ohms). Two common matching techniques are the use of the folded dipole, of Section 9.5, as a driven element and simultaneously as an impedance transformer, and the Gamma-match of Section 9.7.4. Which one of the two is used depends primarily on the transmission line from the antenna to the receiver.
The coaxial cable is now widely used as the primary transmission line for television, especially with the wide spread and use of cable TV; in fact, most television sets are already prewired with coaxial cable connections. Therefore, if the coax with a characteristic impedance of about 78 ohms is the transmission line used from the Yagi-Uda antenna to the receiver and since the input impedance of the antenna is typically 30–70 ohms (as illustrated in Table 10.5), the Gamma-match is the most prudent matching technique to use. This has been widely used in commercial designs where a clamp is usually employed to vary the position of the short to achieve a best match.
If, however, a “twin-lead” line with a characteristic impedance of about 300 ohms is used as the transmission line from the antenna to the receiver, as was used widely some years ago, then it would be most prudent to use a folded dipole as the driven element which acts as a step-up impedance transformer of about 4:1 (4:1) when the length of the element is exactly λ∕2. This technique is also widely used in commercial designs.
Another way to explain the end-fire beam formation and whether the parameters of the Yagi-Uda array are properly adjusted for optimum directivity is by drawing a vector diagram of the progres-sive phase delay from element to element. If the current amplitudes throughout the array are equal, the total phase delay for maximum directivity should be about 180◦, as is required by the Hansen-Woodyard criteria for improved end-fire radiation. Since the currents in a Yagi-Uda array are not equal in all the elements, the phase velocity of the traveling wave along the antenna structure is not the same from element-to-element but it is always slower than the velocity of light and faster than the corresponding velocity for a Hansen-Woodyard design. For a Yagi-Uda array, the decrease in the phase velocity is a function of the increase in total array length.
In general then, the phase velocity, and in turn the phase shift, of a traveling wave in a Yagi-Uda array structure is controlled by the geometrical dimensions of the array and its elements, and it is not uniform from element to element.
F. Design Procedure A government document has been published which provides extensive data of experimental investigations carried out by the National Bureau of Standards to determine how parasitic element diameter, element length, spacings between elements, supporting booms of different cross-sectional areas, various reflectors, and overall length affect the measured gain. Numerous graphical data is included to facilitate the design of different length antennas to yield maximum gain. In addition, design criteria are presented for stacking Yagi-Uda arrays either one above the other or side by side.
576 TRAVELING WAVE AND BROADBAND ANTENNAS TABLE 10.6 Optimized Uncompensated Lengths of Parasitic Elements for Yagi-Uda Antennas of Six Different Lengths d∕𝛌= 0.0085 Length of Yagi-Uda (in wavelengths) s12 = 0.2𝛌 0.4 0.8 1.20 2.2 3.2 4.2 LENGTH OF REFLECTOR (l1∕λ) 0.482 0.482 0.482 0.482 0.482 0.475 LENGTH OF DIRECTORS, λ l3 0.442 0.428 0.428 0.432 0.428 0.424 l4 0.424 0.420 0.415 0.420 0.424 l5 0.428 0.420 0.407 0.407 0.420 l6 0.428 0.398 0.398 0.407 l7 0.390 0.394 0.403 l8 0.390 0.390 0.398 l9 0.390 0.386 0.394 l10 0.390 0.386 0.390 l11 0.398 0.386 0.390 l12 0.407 0.386 0.390 l13 0.386 0.390 l14 0.386 0.390 l15 0.386 0.390 l16 0.386 l17 0.386 SPACING BETWEEN DIRECTORS (sik∕λ) 0.20 0.20 0.25 0.20 0.20 0.308 DIRECTIVITY RELATIVE TO HALF-WAVE DIPOLE (dB) 7.1 9.2 10.2 12.25 13.4 14.2 DESIGN CURVE (SEE FIGURE 10.27) (A) (B) (B) (C) (B) (D) (source: Peter P. Viezbicke, Yagi Antenna Design, NBS Technical Note 688, December 1976).
A step-by-step design procedure has been established in determining the geometrical parameters of a Yagi-Uda array for a desired directivity (over that of a λ∕2 dipole mounted at the same height above ground). The included graphs can only be used to design arrays with overall lengths (from reflector element to last director) of 0.4, 0.8, 1.2, 2.2, 3.2, and 4.2λ with corresponding directivities of 7.1, 9.2, 10.2, 12.25, 13.4, and 14.2 dB, respectively, and with a diameter-to-wavelength ratio of 0.001 ≤d∕λ ≤0.04. Although the graphs do not cover all possible designs, they do accommodate most practical requests. The driven element used to derive the data was a λ∕2 folded dipole, and the measurements were carried out at f = 400 MHz. To make the reader aware of the procedure, it will be outlined by the use of an example. The procedure is identical for all other designs at frequencies where included data can accommodate the specifications.
The basis of the design is the data included in 1. Table 10.6 which represents optimized antenna parameters for six different lengths and for a d∕λ = 0.0085 2. Figure 10.27 which represents uncompensated director and reflector lengths for 0.001 ≤ d∕λ ≤0.04 3. Figure 10.28 which provides compensation length increase for all the parasitic elements (direc-tors and reflectors) as a function of boom-to-wavelength ratio 0.001 ≤D∕λ ≤0.04 BROADBAND ANTENNAS 577 0.001 0.002 0.003 0.01 0.02 0.03 0.04 0.36 0.40 0.45 0.50 Element diameter d (wavelengths) l1′ = 0.485 l1″ = 0.482 Element length li (wavelengths) A, B, C D Reflector A C B D Directors l4'' = 0.424 0.00424 0.0085 l4' = 0.438 l3'' = l5'' = 0.428 l3' = l5' = 0.442 Δl Δl λ λ λ λ λ λ Figure 10.27 Design curves to determine element lengths of Yagi-Uda arrays. (source: P. P. Viezbicke, “Yagi Antenna Design,” NBS Technical Note 688, U.S. Department of Commerce/National Bureau of Standards, December 1976).
The specified information is usually the center frequency, antenna directivity, d∕λ and D∕λ ratios, and it is required to find the optimum parasitic element lengths (directors and reflectors). The spacing between the directors is uniform but not the same for all designs. However, there is only one reflector and its spacing is s = 0.2λ for all designs.
Figure 10.28 Increase in optimum length of parasitic elements as a function of metal boom diameter. (source: P. P. Viezbicke, “Yagi Antenna Design,” NBS Technical Note 688, U.S. Department of Commerce/National Bureau of Standards, December 1976).
578 TRAVELING WAVE AND BROADBAND ANTENNAS Example 10.3 Design a Yagi-Uda array with a directivity (relative to a λ∕2 dipole at the same height above ground) of 9.2 dB at f0 = 50.1 MHz. The desired diameter of the parasitic elements is 2.54 cm and of the metal supporting boom 5.1 cm. Find the element spacings, lengths, and total array length.
Solution: a. At f0 = 50.1 MHz the wavelength is λ = 5.988 m = 598.8 cm. Thus d∕λ = 2.54∕598.8 = 4.24 × 10−3 and D∕λ = 5.1∕598.8 = 8.52 × 10−3.
b. From Table 10.6, the desired array would have a total of five elements (three directors, one reflector, one feeder). For a d∕λ = 0.0085 ratio the optimum uncompensated lengths would be those shown in the second column of Table 10.6 (l3 = l5 = 0.428λ, l4 = 0.424λ, and l1 = 0.482λ). The overall antenna length would be L = (0.6 + 0.2)λ = 0.8λ, the spacing between directors 0.2λ, and the reflector spacing 0.2λ. It is now desired to find the optimum lengths of the parasitic elements for a d∕λ = 0.00424.
c. Plot the optimized lengths from Table 10.6 (l′′ 3 = l′′ 5 = 0.428λ, l′′ 4 = 0.424λ, and l′′ 1 = 0.482λ) on Figure 10.27 and mark them by a dot (⋅).
d. In Figure 10.27 draw a vertical line through d∕λ = 0.00424 intersecting curves (B) at direc-tor uncompensated lengths l′ 3 = l′ 5 = 0.442λ and reflector length l′ 1 = 0.485λ. Mark these points by an x.
e. With a divider, measure the distance (Δl) along director curve (B) between points l′′ 3 = l′′ 5 = 0.428λ and l′′ 4 = 0.424λ. Transpose this distance from the point l′ 3 = l′ 5 = 0.442λ on curve (B), established in step (d) and marked by an x, downward along the curveand determine the uncompensated length l′ 4 = 0.438λ. Thus the boom uncompensated lengths of the array at f0 = 50.1 MHz are l′ 3 = l′ 5 = 0.442λ l′ 4 = 0.438λ l′ 1 = 0.485λ f. Correct the element lengths to compensate for the boom diameter. From Figure 10.28, a boom diameter-to-wavelength ratio of 0.00852 requires a fractional length increase in each element of about 0.005λ. Thus the final lengths of the elements should be l3 = l5 = (0.442 + 0.005)λ = 0.447λ l4 = (0.438 + 0.005)λ = 0.443λ l1 = (0.485 + 0.005)λ = 0.490λ The design data were derived from measurements carried out on a nonconducting Plexiglas boom mounted 3λ above the ground. The driven element was a λ∕2 folded dipole matched to a 50-ohm line by a double-stub tuner. All parasitic elements were constructed from aluminum tubing. Using Plexiglas booms, the data were repeatable and represented the same values as air-dielectric booms.
However that was not the case for wooden booms because of differences in the moisture, which had a direct affect on the gain. Data on metal booms was also repeatable provided the element lengths were increased to compensate for the metal boom structure.
BROADBAND ANTENNAS 579 Figure 10.29 Commercial Yagi-Uda dipole TV array. (Courtesy: Winegard Company, Burlington, IA).
A commercial Yagi-Uda antenna is shown in Figure 10.29. It is a TV antenna designed primarily for channels 2–13. Its gain (over a dipole) ranges from 4.4 dB for channel 2 to 7.3 dB for channel 13, and it is designed for 300 ohms impedance.
10.3.4 Yagi-Uda Array of Loops A side from the dipole, the loop antenna is one of the most basic antenna elements. The pattern of a very small loop is similar to that of a very small dipole and in the far-field region it has a null along its axis. As the circumference of the loop increases, the radiation along its axis increases and reaches near maximum at about one wavelength . Thus loops can be used as the basic elements, instead of the linear dipoles, to form a Yagi-Uda array as shown in Figure 10.30. By properly choosing the dimensions of the loops and their spacing, they can form a unidirectional beam along the axis of the loops and the array.
It has been shown that the radiation characteristics of a two-element loop array, one driven element and a parasitic reflector, resulted in the elimination of corona problems at high altitudes . In addition, the radiation characteristics of loop arrays mounted above ground are less affected by the electrical properties of the soil, as compared with those of dipoles . A two-element loop array also resulted in a 1.8 dB higher gain than a corresponding array of two dipoles . A two-element Figure 10.30 Yagi-Uda array of circular loops.
580 TRAVELING WAVE AND BROADBAND ANTENNAS array of square loops (a feeder and a reflector) in a boxlike construction is called a “cubical quad” or simply a “quad” antenna, and it is very popular in amateur radio applications . The sides of each square loop are λ∕4 (perimeter of λ), and the loops are usually supported by a fiberglass or bamboo cross-arm assembly.
The general performance of a loop Yagi-Uda array is controlled by the same geometrical param-eters (reflector, feeder, and director sizes, and spacing between elements), and it is influenced in the same manner as an array of dipoles –.
In a numerical parametric study of coaxial Yagi-Uda arrays of circular loops of 2 to 10 directors, it has been found that the optimum parameters for maximum forward gain were 1. circumference of feeder 2𝜋b2 ≃1.1λ, where b2 is its radius. This radius was chosen so that the input impedance for an isolated element is purely resistive.
2. circumference of the reflector 2𝜋b1 ≃1.05λ, where b1 is its radius. The size of the reflector does not strongly influence the forward gain but has a major effect on the backward gain and input impedance.
3. feeder–reflector spacing of about 0.1λ. Because it has negligible effect on the forward gain, it can be used to control the backward gain and/or the input impedance.
4. circumference of directors 2𝜋b ≃0.7λ, where b is the radius of any director and it was chosen to be the same for all. When the circumference approached a value of one wavelength, the array exhibited its cutoff properties.
5. spacing of directors of about 0.25λ, and it was uniform for all.
The radius a of all the elements was retained constant and was chosen to satisfy Ω = 2 ln(2𝜋b2∕a) = 11 where b2 is the radius of the feeder.
While most of the Yagi-Uda designs have been implemented using dipoles, and some using loops, as the primary elements, there have been Yagi-Uda designs using slots and microstrip patch elements . The microstrip design was primarily configured for low-angle satellite reception for mobile communications .
10.4 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab computer programs, designated r Beverage r Helix for computing and displaying the radiation characteristics of beverage and helical antennas.
c. Matlab and Fortran computer program, designated Yagi Uda, for computing and displaying the radiation characteristics of a Yagi-Uda array design.
d. Power Point (PPT) viewgraphs, in multicolor.
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2. J. D. Kraus, Electromagnetics, McGraw-Hill Book Co., New York, 1992.
3. J. G. Brainerd et al., Ultra-High-Frequency Techniques, Van Nostrand, New York, 1942.
REFERENCES 581 4. L. V. Blake, Antennas, John Wiley and Sons, New York, 1966.
5. G. A. Thiele and E. P. Ekelman, Jr., “Design Formulas for Vee Dipoles,” IEEE Trans. Antennas Propagat., Vol. AP-28, No. 4, pp. 588–590, July 1980.
6. W. L. Weeks, Antenna Engineering, McGraw-Hill, New York, 1968, pp. 140–142.
7. D. G. Fink (ed.), Electronics Engineers’ Handbook, Chapter 18 (by W. F. Croswell), McGraw-Hill, New York, 1975.
8. J. D. Kraus, Antennas, McGraw-Hill, New York, 1988.
9. A. A. de Carvallo, “On the Design of Some Rhombic Antenna Arrays,” IRE Trans. Antennas Propagat., Vol. AP-7, No. 1, pp. 39–46, January 1959.
10. E. Bruce, A. C. Beck, and L. R. Lowry, “Horizontal Rhombic Antennas,” Proc. IRE, Vol. 23, pp. 24–26, January 1935.
11. R. S. Elliott, Antenna Theory and Design, Prentice-Hall, Englewood Cliffs, New Jersey, 1981.
12. J. D. Kraus, “A 50-Ohm Input Impedance for Helical Beam Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-25, No. 6, p. 913, November 1977.
13. A. G. Kandoian, “Three New Antenna Types and Their Applications,” Proc. IRE, Vol. 34, pp. 70W–75W, February 1946.
14. S. Uda, “Wireless Beam of Short Electric Waves,” J. IEE (Japan), pp. 273–282, March 1926, and pp. 1209–1219, November 1927.
15. H. Yagi, “Beam Transmission of Ultra Short Waves,” Proc. IRE, Vol. 26, pp. 715–741, June 1928. Also Proc. IEEE, Vol. 72, No. 5, pp. 634–645, May 1984; Proc. IEEE, Vol. 85, No. 11, pp. 1864–1874, Novem-ber 1997.
16. D. M. Pozar, “Beam Transmission of Ultra Short Waves: An Introduction to the Classic Paper by H. Yagi,” Proc. IEEE, Vol. 85, No. 11, pp. 1857–1863, November 1997.
17. R. M. Fishender and E. R. Wiblin, “Design of Yagi Aerials,” Proc. IEE (London), pt. 3, Vol. 96, pp. 5–12, January 1949.
18. C. C. Lee and L.-C. Shen, “Coupled Yagi Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-25, No. 6, pp. 889–891, November 1977.
19. H. W. Ehrenspeck and H. Poehler, “A New Method for Obtaining Maximum Gain from Yagi Antennas,” IRE Trans. Antennas Propagat., Vol. AP-7, pp. 379–386, October 1959.
20. H. E. Green, “Design Data for Short and Medium Length Yagi-Uda Arrays,” Elec. Engrg. Trans. Inst.
Engrgs. (Australia), pp. 1–8, March 1966.
21. W. Wilkinshaw, “Theoretical Treatment of Short Yagi Aerials,” Proc. IEE (London), pt. 3, Vol. 93, p. 598, 1946.
22. R. J. Mailloux, “The Long Yagi-Uda Array,” IEEE Trans. Antennas Propagat., Vol. AP-14, pp. 128–137, March 1966.
23. D. Kajfez, “Nonlinear Optimization Reduces the Sidelobes of Yagi Antennas,” IEEE Trans. Antennas Prop-agat., Vol. AP-21, No. 5, pp. 714–715, September 1973.
24. D. Kajfez, “Nonlinear Optimization Extends the Bandwidth of Yagi Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-23, pp. 287–289, March 1975.
25. G. A. Thiele, “Analysis of Yagi-Uda Type Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-17, No.
1, pp. 24–31, January 1969.
26. P. A. Tirkas, Private communication.
27. G. A. Thiele, “Calculation of the Current Distribution on a Thin Linear Antenna,” IEEE Trans. Antennas Propagat., Vol. AP-14, No. 5, pp. 648–649, September 1966.
28. D. K. Cheng and C. A. Chen, “Optimum Spacings for Yagi-Uda Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 5, pp. 615–623, September 1973.
29. C. A. Chen and D. K. Cheng, “Optimum Element Lengths for Yagi-Uda Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 1, pp. 8–15, January 1975.
30. N. K. Takla and L.-C. Shen, “Bandwidth of a Yagi Array with Optimum Directivity,” IEEE Trans. Antennas Propagat., Vol. AP-25, No. 6, pp. 913–914, November 1977.
582 TRAVELING WAVE AND BROADBAND ANTENNAS 31. P. P. Viezbicke, “Yagi Antenna Design,” NBS Technical Note 688, U.S. Department of Commerce/National Bureau of Standards, December 1968.
32. S. Adachi and Y. Mushiake, “Studies of Large Circular Loop Antennas,” Sci. Rep. Research Institute of Tohoku University (RITU), B, Vol. 9, No. 2, pp. 79–103, 1957.
33. J. E. Lindsay, Jr., “A Parasitic End-Fire Array of Circular Loop Elements,” IEEE Trans. Antennas Propa-gat., Vol. AP-15, No. 5, pp. 697–698, September 1967, 34. E. Ledinegg, W. Paponsek, and H. L. Brueckmann, “Low-Frequency Loop Antenna Arrays: Ground Reac-tion and Mutual Interaction,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 1, pp. 1–8, January 1973.
35. D. DeMaw (ed.), The Radio Amateur’s Handbook, American Radio Relay League, Newington, CT, 56th ed., 1979.
36. A. Shoamanesh and L. Shafai, “Properties of Coaxial Yagi Loop Arrays,” IEEE Trans. Antennas Propagat., Vol. AP-26, No. 4, pp. 547–550, July 1978.
37. L. C. Shen and G. W. Raffoul, “Optimum Design of Yagi Array of Loops,” IEEE Trans. Antennas Propagat., Vol. AP-22, No. 11, pp. 829–830, November 1974.
38. A. Shoamanesh and L. Shafai, “Design Data for Coaxial Yagi Array of Circular Loops,” IEEE Trans.
Antennas Propagat., Vol. AP-27, No. 5, pp. 711–713, September 1979.
39. R. J. Coe and G. Held, “A Parasitic Slot Array,” IEEE Trans. Antennas Propagat., Vol. AP-12, No. 1, pp. 10–16, January 1964.
40. J. Huang and A. C. Densmore, “Microstrip Yagi Antenna for Mobile Satellite Vehicle Application,” IEEE Trans. Antennas Propagat., Vol. AP-39, No. 6, pp. 1024–1030, July 1991.
PROBLEMS 10.1.
Given the current distribution of (10-1a), show that the (a) far-zone electric field intensity is given by (10-2a) and (10-2b) (b) average power density is given by (10-4) and (10-5) (c) radiated power is given by (10-11) 10.2.
Determine the phase velocity (compared to free-space) of the wave on a Beverage antenna (terminated long wire) of length l = 50λ so that the maximum occurs at angles of (a) 10◦ (b) 20◦ from the axis of the wire.
10.3.
The current distribution on a terminated and matched long linear (traveling wave) antenna of length l, positioned along the x-axis and fed at its one end, is given by I = ̂ axI0e−jkx′, 0 ≤x′ ≤1 Find the far field electric and magnetic field components in standard spherical coordinates.
10.4.
A long linear (traveling wave) antenna of length l, positioned along the z-axis and fed at the z = 0 end, is terminated in a load at the z = l end. There is a nonzero reflection at the load such that the current distribution on the wire is given by I(z) = I0 e−jkz + Rejkz 1 + R , 0 ≤z ≤l Determine as a function of R and l the (a) far-zone spherical electric-field components (b) radiation intensity in the 𝜃= 𝜋∕2 direction PROBLEMS 583 10.5.
Design a Beverage antenna so that the first maximum occurs at 10◦from its axis. Assuming the phase velocity of the wave on the line is the same as that of free-space, find the (a) lengths (exact and approximate) to accomplish that (b) angles (exact and approximate) where the next six maxima occur (c) angles (exact and approximate) where the nulls, between the maxima found in parts (a) and (b), occur (d) radiation resistance using the exact and approximate lengths (e) directivity using the exact and approximate lengths Verify using the computer program Beverage.
10.6.
It is desired to place the first maximum of a long wire traveling wave antenna at an angle of 25◦from the axis of the wire. For the wire antenna, find the (a) exact required length (b) radiation resistance (c) directivity (in dB) The wire is radiating into free space.
10.7.
Compute the directivity of a long wire with lengths of l = 2λ and 3λ.
Verify using the computer program Beverage.
10.8.
A long wire of diameter d is placed (in the air) at a height h above the ground.
(a) Find its characteristic impedance assuming h ≫d.
(b) Compare this value with (10-14).
10.9.
A long resonant wire is placed vertically so that the axis of the wire is pointed along zenith (𝜃= 0◦). Design the wire so that its radiation resistance is 138.843 ohms. Determine the approximate: (a) Length of the wire (in odd number of λ∕2) (b) Angle (in degrees), measured from zenith, where the first maximum of the amplitude pattern will occur.
(c) Maximum directivity (in dB).
10.10.
Beverage (long-wire) antennas are used for over-the-horizon communication where the maximum of the main beam is pointed few degrees above the horizon. Assuming the wire antenna of length l and radius λ∕200 is placed horizontally parallel to the z-axis a height h = λ∕20 above a flat, perfect electric conducting plane of infinite extent (x-axis is perpen-dicular to the ground plane).
(a) Derive the array factor for the equivalent two-element array.
(b) What is the normalized total electric field of the wire in the presence of the conduct-ing plane?
(c) What value of load resistance should be placed at the terminating end to eliminate any reflections and not create a standing wave?
10.11.
It is desired to design a very long resonant (standing wave) wire for over-the-horizon com-munication system, with a length equal to odd multiple of a half of a wavelength, so that its first maximum is 10◦from the axis of the wire. To meet the design requirements, determine the resonant (standing wave) antenna’s (a) approximate length (in λ).
(b) radiation resistance.
(c) directivity (in dB).
To make the resonant (standing wave) antenna a traveling wave antenna, it will be ter-minated with a load resistance. If the wire antenna’s diameter is λ∕400 and it is placed horizontally a height λ∕20 above an infinite perfectly conducting flat ground plane, (d) what should the load resistance be to accomplish this?
584 TRAVELING WAVE AND BROADBAND ANTENNAS 10.12.
Compute the optimum directivities of a V antenna with leg lengths of l = 2λ and l = 3λ.
Compare these values with those of Problem 10.7.
10.13.
Design a symmetrical V antenna so that its optimum directivity is 8 dB. Find the lengths of each leg (in λ) and the total included angle of the V (in degrees).
10.14.
Repeat the design of Problem 10.13 for an optimum directivity of 5 dB.
10.15.
It is desired to design a V-dipole with a maximized directivity. The length of each arm is 0.5λ (overall length of the entire V-dipole is λ). To meet the requirements of the design, what is the (a) total included angle of the V-dipole (in degrees)?
(b) directivity (in dB)?
(c) gain (in dB) if the overall antenna efficiency is 35%?
10.16.
It is desired to design a V-dipole with optimum dimensions so at to maximize its directivity.
The desired maximum directivity is 5.257 dB. Determine the: (a) Length (in λ) of each arm of the dipole.
(b) Included angle (in degrees) of the V-dipole.
(c) Approximate directivity (In dB) of a regular dipole (with an included angle of 180◦) of the same overall length as the V-dipole.
(d) Which of the dipole designs (V or 180◦) exhibits the larger directivity and by how many dB?
10.17.
Ten identical elements of V antennas are placed along the z-axis to form a uniform broadside array. Each element is designed to have a maximum directivity of 9 dB. Assuming each element is placed so that its maximum is also broadside (𝜃= 90◦) and the elements are spaced λ∕4 apart, find the (a) arm length of each V (in λ) (b) included angle (in degrees) of each V (c) approximate total directivity of the array (in dB).
10.18.
Design a resonant 90◦bent, λ∕4 long, 0.25 × 10−3λ radius wire antenna placed above a ground plane. Find the (a) height where the bent must be made (b) input resistance of the antenna (c) VSWR when the antenna is connected to a 50-ohm line.
10.19.
Design a five-turn helical antenna which at 400 MHz operates in the normal mode. The spacing between turns is λ0∕50. It is desired that the antenna possesses circular polarization.
Determine the (a) circumference of the helix (in λ0 and in meters) (b) length of a single turn (in λ0 and in meters) (c) overall length of the entire helix (in λ0 and in meters) (d) pitch angle (in degrees).
10.20.
A helical antenna of 4 turns is operated in the normal mode at a frequency of 880 MHz and is used as an antenna for a wireless cellular telephone. The length L of the helical antenna is 5.7 cm and the diameter of each turn is 0.5 cm. Determine the: (a) Spacing S (in λ0) between the turns.
(b) Length L0 (in λ0) of each turn.
(c) Overall length Ln (in λ0) of entire helix.
PROBLEMS 585 (d) Axial ratio of the helix (dimensionless).
(e) On the basis of the answer in Part d, the primary dominant component (E𝜃or E𝜙) of the far-zone field radiated by the helix.
(f) Primary polarization of the helix (vertical or horizontal) and why? Does the antenna primarily radiate as a linear vertical wire antenna or as a horizontal loop? Why?
Explain.
(g) Radiation resistance of the helical antenna assuming that it can be determined using Rr ≈640 ( L λ0 )2 (h) Radiation resistance of a single straight wire monopole of length L (the same L as that of the helix) mounted above an infinite ground plane.
(i) On the basis of the values of Parts g and h, what can you say about which antenna is preferable to be used as an antenna for a cellular telephone and why? Explain.
10.21.
Helical antennas are often used for space applications where the polarization of the other antenna is not necessarily constant at all times. To assure uninterrupted communication, the helical antenna is designed to produce nearly circular polarization. For a 10-turn helix and axial mode operation, determine the following: (a) radius (in λ0), pitch angle (in degrees), and separation between turns (in λ) of the helix for optimum operation.
(b) half-power beamwidth (in degrees).
(c) directivity (in dB).
(d) axial ratio (dimensionless).
(e) minimum loss (in dB) of the received signal if the other antenna of the communication system is linearly polarized: 1. in the vertical direction; 2. at a 45◦tilt relative to the vertical.
10.22.
Design a nine-turn helical antenna operating in the axial mode so that the input impedance is about 110 ohms. The required directivity is 10 dB (above isotropic). For the helix, determine the approximate: (a) circumference (in λ0).
(b) spacing between the turns (in λ0).
(c) half-power beamwidth (in degrees).
Assuming a symmetrical azimuthal pattern: (d) determine the directivity (in dB) using Kraus’formula. Compare with the desired.
10.23.
Design a helical antenna with optimum dimensions, maximum 17.16 dB (above isotropic) directivity for optimized end-fire radiation, and circular polarization. To achieve the desired specifications, determine the: (a) Optimized pitch angle (in degrees), circumference (in λo) and spacing between turns (in λo) (b) Nearest integer number of turns and corresponding axial ratio (dimemionless).
(c) Normalized phase velocity (only one of them; the appropriate one) of the wave as it travels on the wire of the helix.
(d) Assuming the helix can be approximated by the equivalent model of Figure 10.15(b), does the design meet the necessary Hansen-Woodyard (H-W) spacing betweenturns 586 TRAVELING WAVE AND BROADBAND ANTENNAS condition for maximum end-fire radiation? Justify it by indicating whether the spacing between the turns meets or does not meet the desired conditions for optimized end-fire radiation. State the necessary spacing (in λo) for H-W conditions.
10.24.
Design a helical antenna which operates in the axial (end-fire) mode with an axial ratio (AR) of 1.1. For optimum design, determine the: (a) Number of turns (N), circumference (in λ) and 𝛼(in degrees).
(b) Half-power beamwidth (in degrees).
(c) Directivity (dimensionless and in dB).
(d) Directivity (dimensionless and in dB) using another simple, closed form expression which is a function of the HPBW. Write the expression for this other formula and give its name. Assume the amplitude pattern of the helical antenna operating in the axial (end-fire) mode is rotationally symmetric around the z-axis; not a function of 𝜙when referring to Figure 10.14(b).
(e) Directivity (dimensionless and in dB) using another simple, closed form expression which is a function of the HPBW; aside from those used in Parts d and e. Write the expression for this other formula and give its name. Assume the amplitude pattern of the helical antenna operating in the axial (end-fire) mode is rotationally symmetric around the z-axis; not a function of 𝜙when referring to Figure 10.14(b).
(f) Of the directivities found in parts d and e, which one agrees closer with the one in part c, and why?
10.25.
It is desired to design an optimum end-fire helical antenna radiating in the axial mode at 100 MHz whose polarization axial ratio is 1.1. Determine the: (a) directivity (dimensionless and in dB).
(b) half-power beamwidth (in degrees).
(c) input impedance.
(d) VSWR when connected to a 50-ohm line.
(e) wave velocity of the wave traveling along the helix for an ordinary end-fire radiation (in m/sec).
10.26.
Design an optimum configuration of a helical antenna to operate in the axial/end-fire mode at a frequency of 500 MHz. The helix should have 10 turns and operates in the end-fire mode with the largest potential directivity.
(a) State the name of the end-fire design that will accomplish the required specifications.
(b) Give the dimensions of the circumference (in λo), pitch angle (in degrees) and spacing between turns (in λo).
(c) Approximate side lobe level (in dB) of the array factor part only of the overall pattern.
(d) Phase velocity of wave along the helical wire (in m/sec).
(e) Half-power beamwidth (in degrees).
(f) Directivity (in dB) using two different equations/methods (i.e., two different values of directivity). State the equations/methods you are using to do the calculation.
10.27.
Design a five-turn helical antenna which at 300 MHz operates in the axial mode and pos-sesses circular polarization in the major lobe. Determine the (a) near optimum circumference (in λ0 and in meters) (b) spacing (in λ0 and in meters) for near optimum pitch angle design (c) input impedance PROBLEMS 587 (d) half-power beamwidth (in degrees), first-null beamwidth (in degrees), directivity (dimensionless and in dB), and axial ratio (e) VSWR when the antenna is connected to 50- and 75-ohm coaxial lines 10.28.
For civilian L1 GPS applications (f = 1,575.42 MHz) it is desired to design and use an optimum configuration helical antenna with right-hand circular polarization and with an Axial Ratio of 1.1. The antenna should operate in the end-fire/axial mode. To accomplish this, using optimum geometrical parameters, determine the: (a) Pitch angle (in degrees) and the number of turns.
(b) Circumference (in λ) and spacing between turns (in λ).
(c) Half-power beamwidth and first-null beamwidth (both in degrees).
(d) Directivity (dimensionless and in dB).
(e) Phase velocity (in meters/sec) the wave travels around the helix, for both ordinary end-fire and Hansen-Woodyard designs.
10.29.
For Problem 10.27, plot the normalized polar amplitude pattern (in dB) assuming phas-ing for (a) ordinary end-fire (b) Hansen-Woodyard end-fire (c) p = 1 10.30.
For Problem 10.27, compute the directivity (in dB) using the computer program Directivity of Chapter 2 assuming phasing for (a) ordinary end-fire (b) Hansen-Woodyard end-fire (c) p = 1 Compare with the value obtained using (10-33).
10.31.
Repeat the design of Problem 10.27 at a frequency of 500 MHz.
10.32.
Design an end-fire right-hand circularly polarized helix having a half-power beamwidth of 45◦, pitch angle of 13◦, and a circumference of 60 cm at a frequency of 500 MHz. Deter-mine the (a) turns needed (b) directivity (in dB) (c) axial ratio (d) lower and upper frequencies of the bandwidth over which the required parameters remain relatively constant (e) input impedance at the center frequency and the edges of the band from part (d).
10.33.
An end-fire, 20-turn helical antenna, designed for increased directivity, is used as a receiv-ing antenna for a satellite communication system operating at 1 GHz. The desired inter-cepted power to be delivered to the receiver is 1 watt based on a maximum incident power density of 1 mwatt/cm2 of a linearly-polarized incident wave. Assuming no other losses, determine the helix’s: (a) Axial ratio (dimensionless) (b) Approximate directivity (dimensionless and in dB) (c) Approximate gain (dimensionless and in dB) (d) circumference (in λo), pitch angle (in degrees), spacing between turns (in λo), and length of each turn (in λo).
(e) normalized velocity, compared to that of free space.
10.34.
Design a helical antenna with a directivity of 15 dB that is operating in the axial mode and whose polarization is nearly circular. The spacing between the turns is λ0∕10. Determine the (a) Number of turns.
(b) Axial ratio, both as an dimensionless quantity and in dB.
588 TRAVELING WAVE AND BROADBAND ANTENNAS (c) Directivity (in dB) based on Kraus’formula and Tai & Pereira’s formula. How do they compare with the desired value?
(d) Progressive phase shifts (in degrees) between the turns to achieve the axial mode radi-ation.
10.35.
Design a 10-turn helical antenna so that at the center frequency of 10 GHz, the circumference of each turn is 0.95λ0. Assuming a pitch angle of 14◦, determine the (a) mode in which the antenna operates (b) half-power beamwidth (in degrees) (c) directivity (in dB). Compare this answer with what you get using Kraus’formula.
10.36.
A lossless 10-turn helical antenna with a circumference of one wavelength is connected to a 78-ohm coaxial line, and it is used as a transmitting antenna in a 500-MHz spacecraft communication system. The spacing between turns is λ0∕10. The power in the coaxial line from the transmitter is 5 watts. Assuming the antenna is lossless: (a) What is radiated power?
(b) If the antenna were isotropic, what would the power density (watts/m2) be at a distance of 10 kilometers?
(c) What is the power density (in watts/m2) at the same distance when the transmitting antenna is the 10-turn helix and the observations are made along the maximum of the major lobe?
(d) If at 10 kilometers along the maximum of the major lobe an identical 10-turn helix was placed as a receiving antenna, which was polarization-matched to the incoming wave, what is the maximum power (in watts) that can be received?
10.37.
A 20-turn helical antenna operating in the axial mode is used as a transmitting antenna in a 500-MHz long distance communication system. The receiving antenna is a linearly polar-ized element. Because the transmitting and receiving elements are not matched in polar-ization, approximately how many dB of losses are introduced because of polarization mis-match?
10.38.
A Yagi-Uda array of linear elements is used as a TV antenna receiving primarily channel 8 whose center frequency is 183 MHz. With a regular resonant λ∕2 dipole as the feed element in the array, the input impedance is approximately 68 ohms. The antenna is to be connected to the TV using a “twin-lead” line with a characteristic impedance of nearly 300 ohms. At the center frequency of 183 MHz (a) what should the smallest input impedance of the array be if it is desired to maintain a VSWR equal or less than 1.1?
(b) what is the best way to modify the present feed to meet the desired VSWR speci-fications? Be very specific in explaining why your recommendation will accomplish this.
10.39.
The input impedance of a Yagi-Uda array design for TV reception using a folded dipole as the feed element is 300 + j25. The antenna will be connected to the TV receiver with a lossless “twin-lead” line with a characteristic impedance of 300 ohms. To eliminate the imaginary part of the input impedance, a shorted transmission line with a characteristic impedance of 300 ohms will be connected in parallel to the antenna at the feed points of the folded feed dipole element.
(a) Determine the reactance of the shorted transmission line, which will be connected in parallel to the folded dipole at its feed, that will be required to accomplish this.
(b) State whether the reactance of the shorted transmission line in part a is inductive or capacitive.
PROBLEMS 589 (c) What is the shortest length (in λ) of the shorted transmission line?
(d) What is the VSWR in the “twin-lead” line from the antenna to the TV receiver?
10.40.
Evaluate approximately the effect of the spacing between the director and driven element in the three-element Yagi-Uda array shown in the accompanying figure. Assume that the far-zone (radiated) field of the antenna is given by E𝜃= sin 𝜃[1 −e−j𝜋∕8e−jks12 sin 𝜃sin 𝜙−ej𝜋∕8e+jks23 sin 𝜃sin 𝜙] where s12 is the spacing between the reflector and the driven element, and s23 is the spacing between the director and the driven element. For this problem, set s12 = 0.2λ and let s23 = 0.15λ, 0.20λ, 0.25λ.
(a) Generate polar plots of the radiation power patterns in both E- and H-planes. Normalize the power pattern to its value for 𝜃= 𝜋∕2, 𝜙= 𝜋∕2. Generate two plots, one for E-plane and one for H-plane.
(b) Compute the front-to-back ratio (FBR) in the E-plane given by FBR|E−plane = Pn ( 𝜃= 𝜋 2 , 𝜙= 𝜋 2 ) Pn ( 𝜃= 𝜋 2 , 𝜙= −𝜋 2 ) Driven element Reflector Director s12 x y s23 Leave your answers for both parts in terms of numbers, not dB.
10.41.
Analyze a 27-element Yagi-Uda array, using the computer program Yagi Uda of this chap-ter, having the following specifications: N = total number of elements = 27 Number of directors = 25 Number of reflectors = 1 Total length of reflector = 0.5λ Total length of feeder = 0.47λ Total length of each director = 0.406λ Spacing between reflector and feeder = 0.125λ Spacing between adjacent directors = 0.34λ a = radius of wires = 0.003λ Use 8 modes for each element. Compute the (a) far-field E- and H-plane amplitude patterns (in dB).
(b) directivity of the array (in dB).
(c) E-plane half-power beamwidth (in degrees).
(d) H-plane half-power beamwidth (in degrees).
(e) E-plane front-to-back ratio (in dB).
(f) H-plane front to back ratio (in dB).
590 TRAVELING WAVE AND BROADBAND ANTENNAS 10.42.
Repeat the analysis of Problem 10.41 for a three-element array having the following speci-fications: N = total number of elements = 3 Number of directors = 1 Number of reflectors = 1 Total length of reflector = 0.5λ Total length of feeder = 0.475λ Total length of director = 0.45λ Spacing between reflector and feeder = 0.2λ Spacing between feeder and director = 0.16λ a = radius of wires = 0.005λ Use 8 modes for each element.
10.43.
Design a Yagi-Uda array of linear dipoles to cover all the VHF TV channels (starting with 54 MHz for channel 2 and ending with 216 MHz for channel 13. See Appendix IX). Per-form the design at f0 = 216 MHz. Since the gain is not affected appreciably at f < f0, as Figure 10.26 indicates, this design should accommodate all frequencies below 216 MHz.
The gain of the antenna should be 14.4 dB (above isotropic). The elements and the support-ing boom should be made of aluminum tubing with outside diameters of 3 8in.(≃0.95 cm) and 3 4in.(≃1.90 cm), respectively. Find the number of elements, their lengths and spacings, and the total length of the array (in λ, meters, and feet).
10.44.
Repeat the design of Problem 10.43 for each of the following: (a) VHF-TV channels 2–6 (54–88 MHz. See Appendix IX) (b) VHF-TV channels 7–13 (174–216 MHz. See Appendix IX) 10.45.
Design a Yagi-Uda antenna to cover the entire FM band of 88–108 MHz (100 channels spaced at 200 KHz apart. See Appendix IX). The desired gain is 12.35 dB (above isotropic).
Perform the design at f0 = 108 MHz. The elements and the supporting boom should be made of aluminum tubing with outside diameters of 3 8in.(≃0.95 cm) and 3 4in.(≃1.90 cm), respectively. Find the number of elements, their lengths and spacings, and the total length of the array (in λ, meters, and feet).
10.46.
Design a Yagi-Uda antenna to cover the UHF TV channels (512–806 MHz. See Appendix IX). The desired gain is 12.35 dB (above isotropic). Perform the design at f0 = 806 MHz.
The elements and the supporting boom should be made of wire with outside diameters of 3 32 in.(≃0.2375 cm) and 3 16 in.(≃0.475 cm), respectively. Find the number of elements, their lengths and spacings, and the total length of the array (in λ, meters, and feet).
CHAPTER11 Frequency Independent Antennas, Antenna Miniaturization, and Fractal Antennas 11.1 INTRODUCTION The numerous applications of electromagnetics to the advances of technology have necessitated the exploration and utilization of most of the electromagnetic spectrum. In addition, the advent of broadband systems have demanded the design of broadband radiators. The use of simple, small, lightweight, and economical antennas, designed to operate over the entire frequency band of a given system, would be most desirable. Although in practice all the desired features and benefits cannot usually be derived from a single radiator, most can effectively be accommodated. Previous to the 1950s, antennas with broadband pattern and impedance characteristics had bandwidths not greater than about 2:1. In the 1950s, a breakthrough in antenna evolution was made which extended the bandwidth to as great as 40:1 or more. The antennas introduced by the breakthrough were referred to as frequency independent, and they had geometries that were specified by angles. These antennas are primarily used in the 10–10,000 MHz region in a variety of practical applications such as TV, point-to-point communication, feeds for reflectors and lenses, and so forth.
In antenna scale modeling, characteristics such as impedance, pattern, polarization, and so forth, are invariant to a change of the physical size if a similar change is also made in the operating fre-quency or wavelength. For example, if all the physical dimensions are reduced by a factor of two, the performance of the antenna will remain unchanged if the operating frequency is increased by a factor of two. In other words, the performance is invariant if the electrical dimensions remain unchanged. This is the principle on which antenna scale model measurements are made. For a com-plete and thorough discussion of scaling, the reader is referred to Section 17.10 entitled “Scale Model Measurements.” The scaling characteristics of antenna model measurements also indicate that if the shape of the antenna were completely specified by angles, its performance would have to be independent of frequency . The infinite biconical dipole of Figure 9.1 is one such structure. To make infinite structures more practical, the designs usually require that the current on the structure decrease with distance away from the input terminals. After a certain point the current is negligible, and the struc-ture beyond that point to infinity can be truncated and removed. Practically then the truncated antenna has a lower cutoff frequency above which it radiation characteristics are the same as those of the infinite structure. The lower cutoff frequency is that for which the current at the point of truncation becomes negligible. The upper cutoff is limited to frequencies for which the dimensions of the feed Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 591 592 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS transmission line cease to look like a “point” (usually about λ2∕8 where λ2 is the wavelength at the highest desirable frequency). Practical bandwidths are on the order of about 40:1. Even higher ratios (i.e., 1,000:1) can be achieved in antenna design but they are not necessary, since they would far exceed the bandwidths of receivers and transmitters.
Even though the shape of the biconical antenna can be completely specified by angles, the current on its structure does not diminish with distance away from the input terminals, and its pattern does not have a limiting form with frequency. This can be seen by examining the current distribution as given by (9-11). It is evident that there are phase but no amplitude variations with the radial distance r. Thus the biconical structure cannot be truncated to form a frequency independent antenna. In practice, however, antenna shapes exist which satisfy the general shape equation, as proposed by Rumsey , to have frequency independent characteristics in pattern, impedance, polarization, and so forth, and with current distribution which diminishes rapidly.
Rumsey’s general equation will first be developed, and it will be used as the unifying concept to link the major forms of frequency independent antennas. Classical shapes of such antennas include the equiangular geometries of planar and conical spiral structures –, and the logarithmically periodic structures , .
Fundamental limitations in electrically small antennas will be discussed in Section 11.5. These will be derived using spherical mode theory, with the antenna enclosed in a virtual sphere. Mini-mum Q curves, which place limits on the achievable bandwidth, will be included. Fractal antennas, discussed in Section 11.5, is one class whose design is based on this fundamental principle.
11.2 THEORY The analytical treatment of frequency independent antennas presented here parallels that introduced by Rumsey and simplified by Elliott for three-dimensional configurations.
We begin by assuming that an antenna, whose geometry is best described by the spherical coor-dinates (r, 𝜃, 𝜙), has both terminals infinitely close to the origin and each is symmetrically disposed along the 𝜃= 0, 𝜋-axes. It is assumed that the antenna is perfectly conducting, it is surrounded by an infinite homogeneous and isotropic medium, and its surface or an edge on its surface is described by a curve r = F(𝜃, 𝜙) (11-1) where r represents the distance along the surface or edge. If the antenna is to be scaled to a frequency that is K times lower than the original frequency, the antenna’s physical surface must be made K times greater to maintain the same electrical dimensions. Thus the new surface is described by S9 r′ = KF(𝜃, 𝜙) (11-2) The new and old surfaces are identical; that is, not only are they similar but they are also congruent (if both surfaces are infinite). Congruence can be established only by rotation in 𝜙. Translation is not allowed because the terminals of both surfaces are at the origin. Rotation in 𝜃is prohibited because both terminals are symmetrically disposed along the 𝜃= 0, 𝜋-axes.
For the second antenna to achieve congruence with the first, it must be rotated by an angle C so that KF(𝜃, 𝜙) = F(𝜃, 𝜙+ C) (11-3) The angle of rotation C depends on K but neither depends on 𝜃or 𝜙. Physical congruence implies that the original antenna electrically would behave the same at both frequencies. However the radiation EQUIANGULAR SPIRAL ANTENNAS 593 pattern will be rotated azimuthally through an angle C. For unrestricted values of K(0 ≤K ≤∞), the pattern will rotate by C in 𝜙with frequency, because C depends on K but its shape will be unaltered.
Thus the impedance and pattern will be frequency independent.
To obtain the functional representation of F(𝜃, 𝜙), both sides of (11-3) are differentiated with respect to C to yield d dC[KF(𝜃, 𝜙)] = dK dC F(𝜃, 𝜙) = 𝜕 𝜕C[F(𝜃, 𝜙+ C)] = 𝜕 𝜕(𝜙+ C)[F(𝜃, 𝜙+ C)] (11-4) and with respect to 𝜙to give 𝜕 𝜕𝜙[KF(𝜃, 𝜙)] = K 𝜕F(𝜃, 𝜙) 𝜕𝜙 = 𝜕 𝜕𝜙[F(𝜃, 𝜙+ C)] = 𝜕 𝜕(𝜙+ C)[F(𝜃, 𝜙+ C)] (11-5) Equating (11-5) to (11-4) yields dK dC F(𝜃, 𝜙) = K 𝜕F(𝜃, 𝜙) 𝜕𝜙 (11-6) Using (11-1) we can write (11-6) as 1 K dK dC = 1 r 𝜕r 𝜕𝜙 (11-7) Since the left side of (11-7) is independent of 𝜃and 𝜙, a general solution for the surface r = F(𝜃, 𝜙) of the antenna is r = F(𝜃, 𝜙) = ea𝜙f (𝜃) where a = 1 K dK dC (11-8) (11-8a) and f(𝜃) is a completely arbitrary function.
Thus for any antenna to have frequency independent characteristics, its surface must be described by (11-8). This can be accomplished by specifying the function f(𝜃) or its derivatives. Subsequently, interesting, practical, and extremely useful antenna configurations will be introduced whose surfaces are described by (11-8).
11.3 EQUIANGULAR SPIRAL ANTENNAS The equiangular spiral is one geometrical configuration whose surface can be described by angles.
It thus fulfills all the requirements for shapes that can be used to design frequency independent antennas. Since a curve along its surface extends to infinity, it is necessary to designate the length of the arm to specify a finite size antenna. The lowest frequency of operation occurs when the total arm 594 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS length is comparable to the wavelength . For all frequencies above this, the pattern and impedance characteristics are frequency independent.
11.3.1 Planar Spiral The shape of an equiangular plane spiral curve can be derived by letting f(𝜃) in (11-8) be f(𝜃) = A𝛿 (𝜋 2 −𝜃 ) (11-9) where A is a constant and 𝛿is the Dirac delta function. Using (11-9) reduces (11-8) to r|𝜃=𝜋∕2 = 𝜌= ⎧ ⎪ ⎨ ⎪ ⎩ Aea𝜙= 𝜌0ea(𝜙−𝜙0) 𝜃= 𝜋∕2 0 elsewhere (11-10) where A = 𝜌0e−a𝜙0 (11-10a) In wavelengths, (11-10) can be written as 𝜌λ = 𝜌 λ = A λ ea𝜙= Aea[𝜙−ln(λ)∕a] = Aea(𝜙−𝜙1) (11-11) where 𝜙1 = 1 a ln(λ) (11-11a) Another form of (11-10) is 𝜙= 1 a ln ( 𝜌 A ) = tan 𝜓ln ( 𝜌 A ) = tan 𝜓(ln 𝜌−ln A) (11-12) where 1/a is the rate of expansion of the spiral and 𝜓is the angle between the radial distance 𝜌and the tangent to the spiral, as shown in Figure 11.1(a).
It is evident from (11-11) that changing the wavelength is equivalent to varying 𝜙0 which results in nothing more than a pure rotation of the infinite structure pattern. Within limitations imposed by the arm length, similar characteristics have been observed for finite structures. The same result can be concluded by examining (11-12). Increasing the logarithm of the frequency (ln f ) by C0 is equivalent to rotating the structure by C0 tan 𝜓. As a result, the pattern is merely rotated but otherwise unaltered.
Thus we have frequency independent antennas.
The total length L of the spiral can be calculated by L = ∫ 𝜌1 𝜌0 [ 𝜌2 (d𝜙 d𝜌 )2 + 1 ]1∕2 d𝜌 (11-13) EQUIANGULAR SPIRAL ANTENNAS 595 Figure 11.1 Spiral wire antennas.
which reduces, using (11-10), to L = (𝜌1 −𝜌0) √ 1 + 1 a2 (11-14) where 𝜌0 and 𝜌1 represent the inner and outer radii of the spiral.
Various geometrical arrangements of the spiral have been used to form different antenna systems.
If 𝜙0 in (11-10) is 0 and 𝜋, the spiral wire antenna takes the form of Figure 11.1(b). The arrangements of Figures 11.1(c) and 11.1(d) are each obtained when 𝜙0 = 0, 𝜋∕2, 𝜋, and 3𝜋∕2. Numerous other combinations are possible.
An equiangular metallic solid surface, designated as P, can be created by defining the curves of its edges, using (11-10), as 𝜌2 = 𝜌′ 2ea𝜙 (11-15a) 𝜌3 = 𝜌′ 3ea𝜙= 𝜌′ 2ea(𝜙−𝛿) (11-15b) where 𝜌′ 3 = 𝜌′ 2e−a𝛿 (11-15c) 596 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS such that K = 𝜌3 𝜌2 = e−a𝛿< 1 (11-16) The two curves, which specify the edges of the conducting surface, are of identical relative shape with one magnified relative to the other or rotated by an angle 𝛿with respect to the other. The magnification or rotation allows the arm of conductor P to have a finite width, as shown in Figure 11.2(a).
The metallic arm of a second conductor, designated as Q, can be defined by 𝜌4 = 𝜌′ 4ea𝜙= 𝜌′ 2ea(𝜙−𝜋) (11-17) where 𝜌′ 4 = 𝜌′ 2e−a𝜋 (11-17a) 𝜌5 = 𝜌′ 5ea𝜙= 𝜌′ 4ea(𝜙−𝛿) = 𝜌′ 2ea(𝜙−𝜋−𝛿) (11-18) where 𝜌′ 5 = 𝜌′ 4e−a𝛿= 𝜌′ 2e−a(𝜋+𝛿) (11-18a) The system composed of the two conducting arms, P and Q, constitutes a balanced system, and it is shown in Figure 11.2(a). The finite size of the structure is specified by the fixed spiraling length L0 along the centerline of the arm. The entire structure can be completely specified by the rotation angle 𝛿, the arm length L0, the rate of spiral 1/a, and the terminal size 𝜌′ 2. However, it has been found that most characteristics can be described adequately by only three; that is, L0, 𝜌′ 2, and K = e−a𝛿 as given by (11-16). In addition each arm is usually tapered at its end, shown by dashed lines in Figure 11.2(a), to provide a better matching termination.
The previous analytical formulations can be used to describe two different antennas. One antenna would consist of two metallic arms suspended in free-space, as shown in Figure 11.2(a), and the other of a spiraling slot on a large conducting plane, as shown in Figure 11.2(b). The second is also usu-ally tapered to provide better matching termination. The slot antenna is the most practical, because it can be conveniently fed by a balanced coaxial arrangement to maintain its overall balancing.
The antenna in Figure 11.2(a) with 𝛿= 𝜋∕2 is self-complementary, as defined by Babinet’s princi-ple , and its input impedance for an infinite structure should be Zs = Zc = 188.5 ≃60𝜋ohms (for discussion of Babinet’s Principle see Section 12.8). Experimentally, measured mean input Q P Feed (b) Spiral slot L0 Q P y Feed (a) Spiral plate x ρ 4 ρ 5 ρ 2 ρ 3 Figure 11.2 Spiral plate and slot antennas.
EQUIANGULAR SPIRAL ANTENNAS 597 impedances were found to be only about 164 ohms. The difference between theory and exper-iment is attributed to the finite arm length, finite thickness of the plate, and nonideal feeding conditions.
Spiral slot antennas, with good radiation characteristics, can be built with one-half to three turns.
The most optimum design seems to be that with 1.25 to 1.5 turns with an overall length equal to or greater than one wavelength. The rate of expansion should not exceed about 10 per turn. The patterns are bidirectional, single lobed, broadside (maximum normal to the plane), and must vanish along the directions occupied by the infinite structure. The wave is circularly polarized near the axis of the main lobe over the usable part of the bandwidth. For a fixed cut, the beamwidth will vary with frequency since the pattern rotates. Typical variations are on the order of 10◦. In general, however, slot antennas with more broad arms and/or more tightly wound spirals exhibit smoother and more uniform patterns with smaller variations in beamwidth with frequency. For symmetrical structures, the pattern is also symmetrical with no tilt to the lobe structure.
To maintain the symmetrical characteristics, the antenna must be fed by an electrically and geo-metrically balanced line. One method that achieves geometrical balancing requires that the coax is embedded into one of the arms of the spiral. To maintain symmetry, a dummy cable is usually placed into the other arm. No appreciable currents flow on the feed cables because of the rapid attenuation of the fields along the spiral. If the feed line is electrically unbalanced, a balun must be used. This limits the bandwidth of the system.
The polarization of the radiated wave is controlled by the length of the arms. For very low fre-quencies, such that the total arm length is small compared to the wavelength, the radiated field is linearly polarized. As the frequency increases, the wave becomes elliptically polarized and eventu-ally achieves circular polarization. Since the pattern is essentially unaltered through this frequency range, the polarization change with frequency can be used as a convenient criterion to select the lower cutoff frequency of the usable bandwidth. In many practical cases, this is chosen to be the point where the axial ratio is equal or less than 2 to 1, and it occurs typically when the overall arm length is about one wavelength. A typical variation in axial ratio of the on-axis field as a function of frequency for a one-turn slot antenna is shown in Figure 11.3. The off-axis radiated field has nearly circular polarization over a smaller part of the bandwidth. In addition to the limitation imposed on the bandwidth by the overall length of the arms, another critical factor that can extend or reduce the bandwidth is the construction precision of the feed.
Figure 11.3 On-axis polarization as a function of frequency for one-turn spiral slot. (source: J. D. Dyson, “The Equiangular Spiral Antenna,” IRE Trans. Antennas Propagat., Vol. AP-7, pp. 181–187, April 1959.
c ⃝(1959) IEEE).
598 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS The input impedance of a balanced equiangular slot antenna converges rapidly as the frequency is increased, and it remains reasonably constant for frequencies for which the arm length is greater than about one wavelength. Measured values for a 700–2,500 MHz antenna were about 75–100 ohms with VSWR’s of less than 2 to 1 for 50-ohm lines.
For slot antennas radiating in free-space, without dielectric material or cavity backing, typical measured efficiencies are about 98% for arm lengths equal to or greater than one wavelength. Rapid decreases are observed for shorter arms.
11.3.2 Conical Spiral The shape of a nonplanar spiral can be described by defining the derivative of f (𝜃) to be df d𝜃= f ′(𝜃) = A𝛿(𝛽−𝜃) (11-19) in which 𝛽is allowed to take any value in the range 0 ≤𝛽≤𝜋. For a given value of 𝛽, (11-19) in conjunction with (11-8) describes a spiral wrapped on a conical surface. The edges of one conical spiral surface are defined by r2 = r′ 2e(a sin 𝜃0)𝜙= r′ 2eb𝜙 (11-20a) r3 = r′ 3ea sin 𝜃0𝜙= r′ 2ea sin 𝜃0(𝜙−𝛿) (11-20b) where r′ 3 = r′ 2e−(a sin 𝜃0)𝛿 (11-20c) and 𝜃0 is half of the total included cone angle. Larger values of 𝜃0 in 0 ≤𝜃≤𝜋∕2 represent less tightly wound spirals. These equations correspond to (11-15a)–(11-15c) for the planar surface. The second arm of a balanced system can be defined by shifting each of (11-20a)–(11-20c) by 180◦, as was done for the planar surface by (11-17)–(11-18a). A conical spiral metal strip antenna of elliptical polarization is shown in Figure 11.4.
The conducting conical spiral surface can be constructed conveniently by forming, using printed-circuit techniques, the conical arms on the dielectric cone which is also used as a support. The feed cable can be bonded to the metal arms which are wrapped around the cone. Symmetry can be preserved by observing the same precautions, like the use of a dummy cable, as was done for the planar surface.
A distinct difference between the planar and conical spirals is that the latter provides unidirec-tional radiation (single lobe) toward the apex of the cone with the maximum along the axis. Circular polarization and relatively constant impedances are preserved over large bandwidths. Smoother pat-terns have been observed for unidirectional designs. Conical spirals can be used in conjunction with a ground plane, with a reduction in bandwidth when they are flush mounted on the plane.
11.4 LOG-PERIODIC ANTENNAS Another type of an antenna configuration, which closely parallels the frequency independent concept, is the log-periodic structure introduced by DuHamel and Isbell . Because the entire shape of it cannot be solely specified by angles, it is not truly frequency independent.
LOG-PERIODIC ANTENNAS 599 Figure 11.4 Conical spiral metal strip antenna. (source: Antennas, Antenna Masts and Mounting Adaptors, American Electronic Laboratories, Inc., Lansdale, Pa., Catalog 7.5M-7-79. Courtesy of American Electronic Laboratories, Inc., Montgomeryville, PA 18936 USA).
11.4.1 Planar and Wire Surfaces A planar log-periodic structure is shown in Figure 11.5(a). It consists of a metal strip whose edges are specified by the angle 𝛼∕2. However, in order to specify the length from the origin to any point on the structure, a distance characteristic must be included.
In spherical coordinates (r, 𝜃, 𝜙) the shape of the structure can be written as 𝜃= periodic function of [b ln(r)] (11-21) An example of it would be 𝜃= 𝜃0 sin [ b ln ( r r0 )] (11-22) It is evident from (11-22) that the values of 𝜃are repeated whenever the logarithm of the radial frequency ln(𝜔) = ln(2𝜋f) differs by 2𝜋∕b. The performance of the system is then periodic as a function of the logarithm of the frequency; thus the name logarithmic-periodic or log-periodic.
A typical log-periodic antenna configuration is shown in Figure 11.5(b). It consists of two copla-nar arms of the Figure 11.5(a) geometry. The pattern is unidirectional toward the apex of the cone 600 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Figure 11.5 Typical metal strip log-periodic configuration and antenna structure.
formed by the two arms, and it is linearly polarized. Although the patterns of this and other log-periodic structures are not completely frequency independent, the amplitude variations of certain designs are very slight. Thus, practically, they are frequency independent.
Log-periodic wire antennas were introduced by DuHamel . While investigating the current distribution on log-periodic surface structures of the form shown in Figure 11.6(a), he discovered that the fields on the conductors attenuated very sharply with distance. This suggested that perhaps Figure 11.6 Planar and wire logarithmically periodic antennas.
LOG-PERIODIC ANTENNAS 601 Figure 11.7 Planar and wire trapezoidal toothed log-periodic antennas.
there was a strong current concentration at or near the edges of the conductors. Thus, removing part of the inner surface to form a wire antenna as shown in Figure 11.6(b), should not seriously degrade the performance of the antenna. To verify this, a wire antenna, with geometrical shape identical to the pattern formed by the edges of the conducting surface, was built and it was investigated experi-mentally. As predicted, it was found that the performance of this antenna was almost identical to that of Figure 11.6(a); thus the discovery of a much simpler, lighter in weight, cheaper, and less wind resistant antenna. The fact that, for a planar bow-tie antenna, the removal of the inner surface of Figure 11.6(a), and also of Figure 11.7(a), does not impact significantly the radiation characteristics was also verified through simulations and measurements in . The current distribution on the sur-face of a solid planar bow-tie is shown in Figure 9.8. As is seen in Figure 9.8, the most intense current density is along the perimeter edges; thus, the removal of the inner surface, to form an “outline” bow tie, did not affect significantly the performance of the bow-tie . Nonplanar geometries in the form of a V, formed by bending one arm relative to the other, are also widely used.
Figure 11.8 Linearly polarized flush-mounted cavity-backed log-periodic slot antenna and typical gain char-acteristics. (source: Antennas, Antenna Masts and Mounting Adaptors, American Electronic Laboratories, Inc., Lansdale, Pa., Catalog 7.5M-7-79. Courtesy of American Electronic Laboratories, Inc., Montgomeryville, PA 18936 USA).
602 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS If the wires or the edges of the plates are linear (instead of curved), the geometries of Figure 11.6 reduce, respectively, to the trapezoidal tooth log-periodic structures of Figure 11.7. These simplifi-cations result in more convenient fabrication geometries with basically no loss in operational per-formance. There are numerous other bizarre but practical configurations of log-periodic structures, including log-periodic arrays.
If the geometries of Figure 11.6 use uniform periodic teeth, we define the geometric ratio of the log-periodic structure by 𝜏= Rn Rn+1 (11-23) and the width of the antenna slot by 𝜒= rn Rn+1 (11-24) The geometric ratio 𝜏of (11-23) defines the period of operation. For example, if two frequencies f1 and f2 are one period apart, they are related to the geometric ratio 𝜏by 𝜏= f1 f2 , f2 > f1 (11-25) Extensive studies on the performance of the antenna of Figure 11.6(b) as a function of 𝛼, 𝛽, 𝜏, and 𝜒, have been performed . In general, these structures performed almost as well as the planar and conical structures. The only major difference is that the log-periodic configurations are linearly polarized instead of circular.
A commercial lightweight, cavity-backed, linearly polarized, flush-mounted log-periodic slot antenna and its associated gain characteristics are shown in Figures 11.8(a) and (b). Typical elec-trical characteristics are: VSWR—2:1; E-plane beamwidth—70◦; H-plane beamwidth—70◦. The maximum diameter of the cavity is about 2.4 in. (6.1 cm), the depth is 1.75 in. (4.445 cm), and the weight is near 5 oz (0.14 kg).
11.4.2 Dipole Array To the layman, the most recognized log-periodic antenna structure is the configuration introduced by Isbell which is shown in Figure 11.9(a). It consists of a sequence of side-by-side parallel linear dipoles forming a coplanar array. Although this antenna has slightly smaller directivities than the Yagi–Uda array (7–12 dB), they are achievable and maintained over much wider bandwidths. There are, however, major differences between them.
While the geometrical dimensions of the Yagi–Uda array elements do not follow any set pattern, the lengths (ln’s), spacings (Rn’s), diameters (dn’s), and even gap spacings at dipole centers (sn’s) of the log-periodic array increase logarithmically as defined by the inverse of the geometric ratio 𝜏.
That is, 1 𝜏= l2 l1 = ln+1 ln = R2 R1 = Rn+1 Rn = d2 d1 = dn+1 dn = s2 s1 = sn+1 sn (11-26) LOG-PERIODIC ANTENNAS 603 (a) Dipole array dn+1 sn+1 ln+1 Rn+1 dn ln sn Rn x z s y 2α (b) Straight connection (c) Crisscross connection (d) Coaxial connection Figure 11.9 Log-periodic dipole array and associated connections.
Another parameter that is usually associated with a log-periodic dipole array is the spacing factor 𝜎 defined by 𝜎= Rn+1 −Rn 2ln+1 (11-26a) 604 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Straight lines through the dipole ends meet to form an angle 2𝛼which is a characteristic of frequency independent structures.
Because it is usually very difficult to obtain wires or tubing of many different diameters and to maintain tolerances of very small gap spacings, constant dimensions in these can be used. These relatively minor factors will not sufficiently degrade the overall performance.
While only one element of the Yagi–Uda array is directly energized by the feed line, while the others operate in a parasitic mode, all the elements of the log-periodic array are connected. There are two basic methods, as shown in Figures 11.9(b) and 11.9(c), which could be used to connect and feed the elements of a log-periodic dipole array. In both cases the antenna is fed at the small end of the structure.
The currents in the elements of Figure 11.9(b) have the same phase relationship as the terminal phases. If in addition the elements are closely spaced, the phase progression of the currents is to the right. This produces an end-fire beam in the direction of the longer elements and interference effects to the pattern result.
It was recognized that by mechanically crisscrossing or transposing the feed between adjacent elements, as shown in Figure 11.9(c), a 180◦phase is added to the terminal of each element. Since the phase between the adjacent closely spaced short elements is almost in opposition, very little energy is radiated by them and their interference effects are negligible. However, at the same time, the longer and larger spaced elements radiate. The mechanical phase reversal between these elements produces a phase progression so that the energy is beamed end fire in the direction of the shorter elements.
The most active elements for this feed arrangement are those that are near resonant with a combined radiation pattern toward the vertex of the array.
The feed arrangement of Figure 11.9(c) is convenient provided the input feed line is a balanced line like the two-conductor transmission line. Using a coaxial cable as a feed line, a practical method to achieve the 180◦phase reversal between adjacent elements is shown in Figure 11.9(d). This feed arrangement provides a built-in broadband balun resulting in a balanced overall system. The elements and the feeder line of this array are usually made of piping. The coaxial cable is brought to the feed through the hollow part of one of the feeder-line pipes. While the outside conductor of the coax is connected to that conductor at the feed, its inner conductor is extended and it is connected to the other pipe of the feeder line.
If the geometrical pattern of the log-periodic array, as defined by (11-26), is to be maintained to achieve a truly log-periodic configuration, an infinite structure would result. However, to be useful as a practical broadband radiator, the structure is truncated at both ends. This limits the frequency of operation to a given bandwidth.
The cutoff frequencies of the truncated structure can be determined by the electrical lengths of the longest and shortest elements of the structure. The lower cutoff frequency occurs approximately when the longest element is λ∕2; however, the high cutoff frequency occurs when the shortest ele-ment is nearly λ∕2 only when the active region is very narrow. Usually it extends beyond that element.
The active region of the log-periodic dipole array is near the elements whose lengths are nearly or slightly smaller than λ∕2. The role of active elements is passed from the longer to the shorter ele-ments as the frequency increases. Also the energy from the shorter active elements traveling toward the longer inactive elements decreases very rapidly so that a negligible amount is reflected from the truncated end. The movement of the active region of the antenna, and its associated phase center, is an undesirable characteristic in the design of feeds for reflector antennas (see Chapter 15). For this reason, log-periodic arrays are not widely used as feeds for reflectors.
The decrease of energy toward the longer inactive elements is demonstrated in Figure 11.10(a).
The curves represent typical computed and measured transmission-line voltages (amplitude and phase) on a log-periodic dipole array as a function of distance from its apex. These are feeder-line voltages at the base of the elements of an array with 𝜏= 0.95, 𝜎= 0.0564, N = 13, and ln∕dn = 177. The frequency of operation is such that element No. 10 is λ∕2. The amplitude voltage is nearly constant from the first (the feed) to the eighth element while the corresponding phase is LOG-PERIODIC ANTENNAS 605 uniformly progressive. Very rapid decreases in amplitude and nonlinear phase variations are noted beyond the eighth element.
The region of constant voltage along the structure is referred to as the transmission region, because it resembles that of a matched transmission line. Along the structure, there is about 150◦phase change for every λ∕4 free-space length of transmission line. This indicates that the phase velocity of the wave traveling along the structure is 𝜐p = 0.6𝜐0, where 𝜐0 is the free-space velocity. The smaller velocity results from the shunt capacitive loading of the line by the smaller elements. The loading is almost constant per unit length because there are larger spacings between the longer elements.
The corresponding current distribution is shown in Figure 11.10(b). It is noted that the rapid decrease in voltage is associated with strong current excitation of elements 7–10 followed by a rapid decline. The region of high current excitation is designated as the active region, and it encompasses 4 to 5 elements for this design. The voltage and current excitations of the longer elements (beyond the ninth) are relatively small, reassuring that the truncated larger end of the structure is not affecting the performance. The smaller elements, because of their length, are not excited effectively. As the frequency changes, the relative voltage and current patterns remain essentially the same, but they move toward the direction of the active region.
There is a linear increase in current phase, especially in the active region, from the shorter to the longer elements. This phase shift progression is opposite in direction to that of an unloaded line.
It suggests that on the log-periodic antenna structure there is a wave that travels toward the feed forming a unidirectional end-fire pattern toward the vertex.
The radiated wave of a single log-periodic dipole array is linearly polarized, and it has horizon-tal polarization when the plane of the antenna is parallel to the ground. Bidirectional patterns and circular polarization can be obtained by phasing multiple log-periodic dipole arrays. For these, the overall effective phase center can be maintained at the feed.
If the input impedance of a log-periodic antenna is plotted as a function of frequency, it will be repetitive. However, if it is plotted as a function of the logarithm of the frequency, it will be periodic (not necessarily sinusoidal) with each cycle being exactly identical to the preceding one.
Hence the name log-periodic, because the variations are periodic with respect to the logarithm of the frequency. A typical variation of the impedance as a function of frequency is shown in Figure 11.11.
Other parameters that undergo similar variations are the pattern, directivity, beamwidth, and side lobe level.
The periodicity of the structure does not ensure broadband operation. However, if the variations of the impedance, pattern, directivity, and so forth within one cycle are made sufficiently small and acceptable for the corresponding bandwidth of the cycle, broadband characteristics are ensured within acceptable limits of variation. The total bandwidth is determined by the number of repetitive cycles for the given truncated structure.
The relative frequency span Δ of each cycle is determined by the geometric ratio as defined by (11-25) and (11-26).∗Taking the logarithm of both sides in (11-25) reduces to Δ = ln(f2) −ln(f1) = ln (1 𝜏 ) (11-27) The variations that occur within a given cycle (f1 ≤f ≤f2 = f1∕𝜏) will repeat identically at other cycles of the bandwidth defined by f1∕𝜏n−1 ≤f ≤f1∕𝜏n, n = 1, 2, 3, ….
Typical designs of log-periodic dipole arrays have apex half angles of 10◦≤𝛼≤45◦and 0.95 ≥ 𝜏≥0.7. There is a relation between the values of 𝛼and 𝜏. As 𝛼increases, the corresponding 𝜏values decrease, and vice versa. Larger values of 𝛼or smaller values of 𝜏result in more compact designs which require smaller number of elements separated by larger distances. In contrast, smaller values ∗In some cases, the impedance (but not the pattern) may vary with a period which is one-half of (11-27). That is, Δ = 1 2 ln(1∕𝜏).
Figure 11.10 Measured and computed voltage and current distributions on a log-periodic dipole array of 13 elements with frequency such that l10 = λ∕2. (source: R. L. Carrel, “Analysis and Design of the Log-Periodic Dipole Antenna,” Ph.D. Dissertation, Elec. Eng.
Dept., University of Illinois, 1961, University Microfilms, Inc., Ann Arbor, Michigan).
606 LOG-PERIODIC ANTENNAS 607 Figure 11.11 Typical input impedance variation of a log-periodic antenna as a function of the logarithm of the frequency.
of 𝛼or larger values of 𝜏require a larger number of elements that are closer together. For this type of a design, there are more elements in the active region which are nearly λ∕2. Therefore the variations of the impedance and other characteristics as a function of frequency are smaller, because of the smoother transition between the elements, and the gains are larger.
Experimental models of log-periodic dipole arrays have been built and measurements were made . The input impedances (purely resistive) and corresponding directivities (above isotropic) for three different designs are listed in Table 11.1. Larger directivities can be achieved by array-ing multiple log-periodic dipole arrays. There are other configurations of log-periodic dipole array designs, including those with V instead of linear elements . This array provides moderate band-widths with good directivities at the higher frequencies, and it is widely used as a single TV antenna covering the entire frequency spectrum from the lowest VHF channel (54 MHz) to the highest UHF (806 MHz). Typical gain, VSWR, and E- and H-plane half-power beamwidths of commercial log-periodic dipole arrays are shown in Figures 11.12(a), (b), (c), respectively. The overall length of each of these antennas is about 105 in. (266.70 cm) while the largest element in each has an overall length of about 122 in. (309.88 cm). The weight of each antenna is about 31 lb (≃14 kg).
TABLE 11.1 Input Resistances (Rin in ohms) and Directivities (dB above isotropic) for Log-Periodic Dipole Arrays 𝝉= 0.81 𝝉= 0.89 𝝉= 0.95 𝜶 Rin(ohms) D0(dB) Rin(ohms) D0(dB) Rin(ohms) D0(dB) 10 98 — 82 9.8 77.5 10.7 12.5 — — 77 — — — 15 — 7.2 — — — — 17.5 — — 76 7.7 62 8.8 20 — — 74 — — — 25 — — 63 7.2 — 8.0 30 80 — 64 — 54 — 35 — — 56 6.5 — — 45 65 5.2 59 6.2 — — (source: D. E. Isbell, “Log Periodic Dipole Arrays,” IRE Trans. Antennas Propagat., Vol. AP-8, pp. 260–267, May 1960. c ⃝(1960) IEEE.) 608 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Figure 11.12 Typical gain, VSWR, and half-power beamwidth of commercial log-periodic dipole arrays.
(source: Antennas, Antenna Masts and Mounting Adaptors, American Electronic Laboratories, Inc., Lansdale, Pa., Catalog 7.5M-7-79. Courtesy of American Electronic Laboratories, Inc., Montgomeryville, PA 18936 USA).
11.4.3 Design of Dipole Array The ultimate goal of any antenna configuration is the design that meets certain specifications. Proba-bly the most introductory, complete, and practical design procedure for a log-periodic dipole array is that by Carrel . To aid in the design, he has a set of curves and nomographs. The general config-uration of a log-periodic array is described in terms of the design parameters 𝜏, 𝛼, and 𝜎related by 𝛼= tan−1 [1 −𝜏 4𝜎 ] (11-28) LOG-PERIODIC ANTENNAS 609 1.0 0.96 0.92 0.88 0.84 0.80 0.76 0.04 0.08 0.06 0.10 0.14 0.18 0.20 0.12 0.16 0.22 Relative spacing Scale factor Optimum 11 dB 10.5 dB 10 dB 9.5 dB 9 dB 8.5 dB 8 dB 7.5 dB 7 dB 6.5 dB σ τ σ Figure 11.13 Computed contours of constant directivity versus 𝜎and 𝜏for log-periodic dipole arrays.
(source: R. L. Carrel, “Analysis and Design of the Log-Periodic Dipole Antenna,” Ph.D. Dissertation, Elec.
Eng. Dept., University of Illinois, 1961, University Microfilms, Inc., Ann Arbor Michigan). Note: The initial curves led to designs whose directivities are 1–2 dB too high. They have been reduced by an average of 1 dB (see P. C. Butson and G. T. Thompson, “A Note on the Calculation of the Gain of Log-Periodic Dipole Antennas,” IEEE Trans. Antennas Propagat., AP-24, pp. 105–106, January 1976).
Once two of them are specified, the other can be found. Directivity (in dB) contour curves as a function of 𝜏for various values of 𝜎are shown in Figure 11.13.
The original directivity contour curves in are in error because the expression for the E-plane field pattern in is in error. To correct the error, the leading sin(𝜃) function in front of the summation sign of equation 47 in should be in the denominator and not in the numerator [i.e., replace sin 𝜃by 1∕sin(𝜃)] . The influence of this error in the contours of Figure 11.13 is variable and leads to 1–2 dB higher directivities. However it has been suggested that, as an average, the directivity of each original contour curve be reduced by about 1 dB. This has been implemented already, and the curves in Figure 11.13 are more accurate as they now appear.
A. Design Equations In this section a number of equations will be introduced that can be used to design a log-periodic dipole array.
While the bandwidth of the system determines the lengths of the shortest and longest elements of the structure, the width of the active region depends on the specific design. Carrel has introduced a semiempirical equation to calculate the bandwidth of the active region Bar related to 𝛼and 𝜏by Bar = 1.1 + 7.7(1 −𝜏)2 cot 𝛼 (11-29) In practice a slightly larger bandwidth (Bs) is usually designed than that which is required (B). The two are related by Bs = BBar = B[1.1 + 7.7(1 −𝜏)2 cot 𝛼] (11-30) 610 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS where Bs = designed bandwidth B = desired bandwidth Bar = active region bandwidth The total length of the structure L, from the shortest (lmin) to the longest (lmax) element, is given by L = λmax 4 ( 1 −1 Bs ) cot 𝛼 where λmax = 2lmax = 𝜐 fmin (11-31) (11-31a) From the geometry of the system, the number of elements are determined by N = 1 + ln(Bs) ln(1∕𝜏) (11-32) The center-to-center spacing s of the feeder-line conductors can be determined by specifying the required input impedance (assumed to be real), and the diameter of the dipole elements and the feeder-line conductors. To accomplish this, we first define an average characteristic impedance of the elements given by Za = 120 [ ln ( ln dn ) −2.25 ] (11-33) where ln∕dn is the length-to-diameter ratio of the nth element of the array. For an ideal log-periodic design, this ratio should be the same for all the elements of the array. Practically, however, the ele-ments are usually divided into one, two, three, or more groups with all the elements in each group having the same diameter but not the same length. The number of groups is determined by the total number of elements of the array. Usually three groups (for the small, middle, and large elements) should be sufficient.
The effective loading of the dipole elements on the input line is characterized by the graphs shown in Figure 11.14 where 𝜎′ = 𝜎∕ √ 𝜏= relative mean spacing Za = average characteristic impedance of the elements Rin = input impedance (real) Z0 = characteristic impedance of the feeder line The center-to-center spacing s between the two rods of the feeder line, each of identical diameter d, is determined by s = d cosh ( Z0 120 ) (11-34) LOG-PERIODIC ANTENNAS 611 Figure 11.14 Relative characteristic impedance of a feeder line as a function of relative characteristic impedance of dipole element. (source: R. L. Carrel, “Analysis and Design of the Log-Periodic Dipole Antenna,” Ph.D. Dissertation, Elec. Eng. Dept., University of Illinois, 1961, University Microfilms, Inc., Ann Arbor, Michigan).
B. Design Procedure A design procedure is outlined here, based on the equations introduced above and in the previous page, and assumes that the directivity (in dB), input impedance Rin (real), diameter of elements of feeder line (d), and the lower and upper frequencies (B = fmax∕fmin) of the bandwidth are specified.
It then proceeds as follows: 1. Given D0 (dB), determine 𝜎and 𝜏from Figure 11.13.
2. Determine 𝛼using (11-28).
3. Determine Bar using (11-29) and Bs using (11-30).
4. Find L using (11-31) and N using (11-32).
5. Determine Za using (11-33) and 𝜎′ = 𝜎∕ √ 𝜏.
6. Determine Z0∕Rin using Figure 11.14.
7. Find s using (11-34).
Example 11.1 Design a log-periodic dipole antenna, of the form shown in Figure 11.9(d), to cover all the VHF TV channels (starting with 54 MHz for channel 2 and ending with 216 MHz for chan-nel 13. See Appendix IX.) The desired directivity is 8 dB and the input impedance is 50 ohms (ideal for a match to 50-ohm coaxial cable). The elements should be made of aluminum tub-ing with 3 4 in. (1.9 cm) outside diameter for the largest element and the feeder line and 3 16 in.
(0.48 cm) for the smallest element. These diameters yield identical l/d ratios for the smallest and largest elements.
Solution: 1. From Figure 11.13, for D0 = 8 dB the optimum 𝜎is 𝜎= 0.157 and the corresponding 𝜏is 𝜏= 0.865.
2. Using (11-28) 𝛼= tan−1 [ 1 −0.865 4(0.157) ] = 12.13◦≃12◦ 612 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS 3. Using (11-29) Bar = 1.1 + 7.7(1 −0.865)2 cot(12.13◦) = 1.753 and from (11-30) Bs = BBar = 216 54 (1.753) = 4(1.753) = 7.01 4. Using (11-31a) λmax = 𝜐 fmin = 3 × 108 54 × 106 = 5.556 m (18.227 ft) From (11-31) L = 5.556 4 ( 1 − 1 7.01 ) cot(12.13◦) = 5.541 m (18.178 ft) and from (11-32) N = 1 + ln(7.01) ln(1∕0.865) = 14.43 (14 or 15 elements) 5. 𝜎′ = 𝜎 √ 𝜏 = 0.157 √ 0.865 = 0.169 At the lowest frequency lmax = λmax 2 = 18.227 2 = 9.1135 ft lmax dmax = 9.1135(12) 0.75 = 145.816 Using (11-33) Za = 120[ln(145.816) −2.25] = 327.88 ohms Thus Za Rin = 327.88 50 = 6.558 6. From Figure 11.14 Z0 ≃1.2Rin = 1.2(50) = 60 ohms 7. Using (11-34), assuming the feeder line conductor is made of the same size tubing as the largest element of the array, the center-to-center spacing of the feeder conductors is s = 3 4 cosh ( 60 120 ) = 0.846 ≃0.85 in.
which allows for a 0.1 in. separation between their conducting surfaces.
LOG-PERIODIC ANTENNAS 613 Figure 11.15 Commercial log-periodic dipole antenna of 21 elements. (Courtesy: Antenna Research Asso-ciates, Inc., Beltsville, MD).
For such a high-gain antenna, this is obviously a good practical design. If a lower gain is specified and designed for, a smaller length will result.
A commercial log-periodic dipole antenna of 21 elements is shown in Figure 11.15. The antenna is designed to operate in the 100–1,100 MHz with a gain of about 6 dBi, a VSWR (to a 50-ohm line) of 2:1 maximum, a front-to-back ratio of about 20 dB, and with typical half-power beamwidths of about: E-plane (75◦) and H-plane (120◦).
C. Design and Analysis Computer Program A computer program Log-Periodic Dipole Array (log perd) has been developed in FORTRAN and translated in MATLAB, based primarily on the design equations of (11-28)–(11-34), and Figures 11.13 and 11.14, to design a log-periodic dipole array whose geometry is shown in Figure 11.9(a). Although most of the program is based on the same design equations as outlined in the design subsection, this program takes into account more design specifications than those included in the previous design procedure, and it is more elaborate. Once the design is completed, the computer program can be used to analyze the design of the antenna. The description is found in the computer disc available with this book. The program has been developed based on input specifications, which are listed in the program. It can be used as a design tool to determine the geometry of the array (including the number of elements and their corresponding lengths, diameters, and positions) along with the radiation characteristics of the array (including input impedance, VSWR, directivity, front-to-back ratio, E- and H-plane patterns, etc.) based on desired specifications. The input data includes the desired directivity, lower and upper frequency of the operating band, length-to-diameter ratio of the elements, characteristic impedance of the input transmission line, desired input impedance, termination (load) impedance, etc. These and others are listed in the program included in the CD.
The program assumes that the current distribution on each antenna element is sinusoidal. This approximation would be very accurate if the elements were very far from each other. However, in the active region the elements are usually separated by a distance of about 0.1λ when 𝛼= 15◦ and 𝜏= 0.9. Referring to Figure 8.21, one can see that two λ∕2 dipoles separated by 0.1λ have a mutual impedance (almost real) of about 70 ohms. If this mutual impedance is high compared to the resistance of the transmission line (not the characteristic impedance), then the primary method of coupling energy to each antenna will be through the transmission line. If the mutual impedance is 614 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS high compared to the self-impedance of each element, then the effect on the radiation pattern should be small. In practice, this is usually the case, and the approximation is relatively good. However, an integral equation formulation with a Moment Method numerical solution would be more accurate.
The program uses (8-60a) for the self-resistance and (8-60b) for the self-reactance. It uses (8-68) for the mutual impedance, which for the side-by-side configuration reduces to the sine and cosine integrals in , similar in form to (8-71a)–(8-71e) for the l = λ∕2 dipole.
The geometry of the designed log-periodic dipole array is that of Figure 11.9, except that the program also allows for an input transmission line (connected to the first/shortest element), a ter-mination transmission line (extending beyond the last/longest element), and a termination (load) impedance. The length of the input transmission line changes the phase of computed data (such as voltage, current, reflection coefficient, etc.) while its characteristic impedance is used to calculate the VSWR, which in turn affects the input impedance measured at the source. The voltages and currents are found based on the admittance method of Kyle . The termination transmission line and the termination (load) impedance allow for the insertion of a matching section whose primary purpose is to absorb any energy which manages to continue past the active region. Without the termination (load) impedance, this energy would be reflected along the transmission line back into the active region where it would affect the radiation characteristics of the array design and performance.
In designing the array, the user has the choice to select 𝜎and 𝜏(but not the directivity) or to select the directivity (but not 𝜎and 𝜏). In the latter case, the program finds 𝜎and 𝜏by assuming an optimum design as defined by the dashed line of Figure 11.13. For the geometry of the array, the program assumes that the elements are placed along the z-axis (with the shortest at z = 0 and the longest along the positive z-axis). Each linear element of the array is directed along the y-axis (i.e., the array lies on the yz-plane). The angle 𝜃is measured from the z axis toward the xy-plane while angle 𝜙is measured from the x-axis (which is normal to the plane of the array) toward the y-axis along the xy-plane. The E-plane of the array is the yz-plane (𝜙= 90◦, 270◦; 0◦≤𝜃≤180◦) while the H-plane is the xz-plane (𝜙= 0◦, 180◦; 0◦≤𝜃≤180◦).
11.5 FUNDAMENTAL LIMITS OF ELECTRICALLY SMALL ANTENNAS In all areas of electrical engineering, especially in electronic devices and computers, the attention has been shifted toward miniaturization. Electromagnetics, and antennas in particular, are of no excep-tion. A large emphasis in the last few years has been placed toward electrically small antennas, including printed board designs. However, there are fundamental limits as to how small the antenna elements can be made. The basic limitations are imposed by the free-space wavelength to which the antenna element must couple to, which has not been or is expected to be miniaturized .
An excellent paper on the fundamental limits in antennas has been published , and revisited in , and most of the material in this section is drawn from . It reviews the limits of electri-cally small, superdirective, super-resolution, and high-gain antennas. The limits on electrically small antennas are derived by assuming that the entire antenna structure (with a largest linear dimension of 2a), and its transmission line and oscillator are all enclosed within a sphere of radius a as shown in Figure 11.16(a). Because of the arbitrary current or source distribution of the antenna inside the sphere, its radiated field outside the sphere is expressed as a complete set of orthogonal spherical vector waves or modes. For vertically polarized omnidirectional antennas, only TMm0 circularly symmetric (no azimuthal variations) modes are required. Each mode is used to represent a spher-ical wave which propagates in the outward radial direction. This approach was introduced first by Chu , and it was followed by Harrington . Earlier papers on the fundamental limitations and performance of small antennas were published by Wheeler , . He derived the limits of a small dipole and a small loop (used as a magnetic dipole) from the limitations of a capacitor and an inductor, respectively. The capacitor and inductor were chosen to occupy, respectively, volumes equal to those of the dipole and the loop.
FUNDAMENTAL LIMITS OF ELECTRICALLY SMALL ANTENNAS 615 Using the mathematical formulation introduced by Chu , the source or current distribution of the antenna system inside the sphere is not uniquely determined by the field distribution outside the sphere. Since it is possible to determine an infinite number of different source or current distributions inside the sphere, for a given field configuration outside the sphere, Chu confined his interest to the most favorable source distribution and its corresponding antenna structure that could exist within the sphere. This approach was taken to minimize the details and to simplify the task of identifying the antenna structure. It was also assumed that the desired current or source distribution minimizes the amount of energy stored inside the sphere so that the input impedance at a given frequency is resistive.
Because the spherical wave modes outside the sphere are orthogonal, the total energy (electric or magnetic) outside the sphere and the complex power transmitted across the closed spherical sur-face are equal, respectively, to the sum of the energies and complex powers associated with each corresponding spherical mode. Therefore there is no coupling, in energy or power, between any two modes outside the sphere. As a result, the space outside the sphere can be replaced by a number of independent equivalent circuits as shown in Figure 11.16(b). The number of equivalent circuits is equal to the number of spherical wave modes outside the sphere, plus one. The terminals of each equivalent circuit are connected to a box which represents the inside of the sphere, and from inside the box a pair of terminals are drawn to represent the input terminals. Using this procedure, the antenna space problem has been reduced to one of equivalent circuits.
The radiated power of the antenna is calculated from the propagating modes while all modes contribute to the reactive power. When the sphere (which encloses the antenna element) becomes Antenna structure Biconical antenna within a sphere (after C. J. Chu ) (a) Equivalent network of a vertically-polarized omnidirectional antenna (after C. J. Chu ) (b) Equivalent circuit for N spherical modes (c) ZN IN a I1 Z1 C1 C2 CN RN LN L2 L1 Z2 Z3 I2 I3 Input Input Coupling network representing antenna structure Figure 11.16 Antenna within a sphere of radius a, and its equivalent circuit modeling. (source: C. J. Chu, “Physical Limitations of Omnidirectional Antennas,” J. Appl. Phys., Vol. 19, pp. 1163–1175, December 1948).
616 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS very small, there exist no propagating modes. Therefore the Q of the system becomes very large since all modes are evanescent (below cutoff) and contribute very little power. However, unlike closed waveguides, each evanescent mode here has a real part (even though it is very small).
For a lossless antenna (radiation efficiency ecd = 100%), the equivalent circuit of each spherical mode is a single network section with a series C and a shunt L. The total circuit is a ladder network of L −C sections (one for each mode) with a final shunt resistive load, as shown in Figure 11.16(c).
The resistive load is used to represent the normalized antenna radiation resistance.
From this circuit structure, the input impedance is found. The Q of each mode is formed by the ratio of its stored to its radiated energy. When several modes are supported, the Q is formed from the contributions of all the modes.
It has been shown that the higher order modes within a sphere of radius a become evanescent when ka < 1. Therefore the Q of the system, for the lowest order TM10 mode, reduces to , Q = 1 + 2(ka)2 (ka)3[1 + (ka)2] ka≪1 ≃ 1 (ka)3 (11-35a) or for the lowest TM10 mode, according to , to Q = 1 (ka)3 + 1 ka = [1 + (ka)2] (ka)3 ka≪1 ≃ 1 (ka)3 (11-35b) Both (11-35a) and (11-35b) lead to the same results when ka ≪1.
When two modes are excited, one TE and the other TM, the values of Q are halved. Equations (11-35a) and (11-35b), which relate the lowest achievable Q to the largest linear dimension of an electrically small antenna, are independent of the geometrical configuration of the antenna within the sphere of radius a. The shape of the radiating element within the bounds of the sphere only determines whether TE, TM, or TE and TM modes are excited. Therefore (11-35a) and (11-35b) represent the fundamental limit on the electrical size of an antenna. In practice, this limit is only approached but is never exceeded or even equaled.
The losses of an antenna can be taken into account by including a loss resistance in series with the radiation resistance, as shown by the equivalent circuits of Figures 2.27(b) and 2.28(b). This influences the Q of the system and the antenna radiation efficiency as given by (2-90).
Computed values of Q versus kr for idealized antennas enclosed within a sphere of radius a, and with radiation efficiencies of ecd = 100, 50, 10, and 5, are shown plotted in Figure 11.17. These curves represent the minimum values of Q that can be obtained from an antenna whose structure can be enclosed within a sphere of radius a and whose radiated field, outside the sphere, can be represented by a single spherical wave mode.
For antennas with equivalent circuits of fixed values, the fractional bandwidth is related to the Q of the system by fractional bandwidth = FBW = Δf f0 = 1 Q (11-36) where f0 = center frequency Δf = bandwidth The relationship of (11-36) is valid for Q ≫1 since the equivalent resonant circuit with fixed values is a good approximation for an antenna. For values of Q < 2, (11-36) is not accurate.
FUNDAMENTAL LIMITS OF ELECTRICALLY SMALL ANTENNAS 617 Dipole Spherical Helix (4-arm folded) 100 60 40 20 10 6 4 2 1 0.1 0.3 0.5 0.7 Sphere size, ka Quality factor, Q 0.9 1.1 1.3 1.5 ecd = 100% 50% 10% 5% Dipole Goubau Spherical Helix (4-arm folded) Figure 11.17 Fundamental limits of Q versus antenna size (enclosed within a sphere of radius a) for single-mode antennas of various radiation efficiencies. (source: R. C. Hansen, “Fundamental Limitations in Anten-nas,” Proc. IEEE, Vol. 69, No. 2, February 1981. c ⃝(1981) IEEE).
To compare the results of the minimum Q curves of Figure 11.17 with values of practical antenna structures, data points for a small linear dipole and a Goubau antenna are included in the same figure. For a small linear dipole of length l and wire radius b, its impedance is given by (4-37) and (8-62) Zin ≃20𝜋2 ( l λ )2 −j120 [ ln ( l 2b ) −1 ] tan ( 𝜋l λ ) (11-37) and its corresponding Q by Q ≃ [ ln ( l 2b ) −1 ] ( 𝜋l λ )2 tan ( 𝜋l λ ) (11-38) The computed Q values of the small dipole were for kl∕2 = ka ≃0.62 and 1.04 with l∕2b = l∕d = 50, and of the Goubau antenna were for ka ≃1.04.
It is apparent that the Q’s of the dipole are much higher than the corresponding values of the minimum Q curves even for the 100% efficient antennas. However the Goubau antenna, of the same 618 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS (a) 3-D view (b) Top view Figure 11.18 Three-dimensional and top views of a four-arm folded helix: (source: c ⃝2004 IEEE).
radius sphere, demonstrates a lower value of Q and approaches the values of the 100% minimum Q curve. This indicates that the fractional bandwidth of the Goubau antenna, which is inversely proportional to its Q as defined by (11-36), is higher than that of a dipole enclosed within the same radius sphere. In turn, the bandwidth of an idealized antenna, enclosed within the same sphere, is even larger.
From the above, it is concluded that the bandwidth of an antenna (which can be closed within a sphere of radius a) can be improved only if the antenna utilizes efficiently, with its geometrical configuration, the available volume within the sphere. The dipole, being a one-dimensional structure, is a poor utilizer of the available volume within the sphere. However a Goubau antenna, being a clover leaf dipole with coupling loops over a ground plane (or a double cover leaf dipole without a ground plane), is a more effective design for utilizing the available three-dimensional space within the sphere. A design, such as that of a spiral, that utilizes the space even more efficiently than the Goubau antenna would possess a lower Q and a higher fractional bandwidth. Ultimately, the values would approach the minimum Q curves. In practice, these curves are only approached but are never exceeded or even equaled.
A geometry of a radiating element that utilizes efficiently the available space within a sphere of radius a, and improves the bandwidth, is that of a folded spherical helix. Three-dimensional and top views of a four-arm folded helix is shown in Figure 11.18 .
The resonant properties of a four-arm folded spherical helical antenna, of the form shown in Figure 11.18, are listed on Table 11.2. Two points of its corresponding Q, from , are indicated in Figure 11.17 where they have approached the fundamental limit curve. It has been concluded in that the radiation properties of a folded spherical helical antenna are functions of: r Number of turns r Number of helical arms TABLE 11.2 Resonant performance properties of a four-arm folded spherical helix antenna No. of Turns Arm Length (cm) fR (MHz) RA (Ohms) Efficiency (%) Q 1∕2 17 515.8 87.6 99.6 5.6 1 30.9 300.3 43.1 98.6 32 11∕2 45.07 210 23.62 97.6 88 (source: c ⃝2004 IEEE).
ANTENNA MINIATURIZATION 619 Increasing the number of arms, while maintaining the same sphere radius, decreases the correspond-ing Q and increases the bandwidth. By properly selecting these parameters, the spherical helices can be designed to : r Be electrically small (ka < 0.5) r Be self-resonant r Have high radiation efficiency r Have Qs within 1.5 times the fundamental limit r Have radiation resistance, at resonance, near 50 ohm While the Qs are within 1.5 the fundamental limit, the absolute value of the Qs can be high because of the small values of ka at which these antennas are self-resonant. Any increase in the bandwidth or decrease in the Q beyond their corresponding limits, while maintaining the same volume, would be at the expense of reducing the antenna efficiency, as indicated in Figure 11.17.
Other examples of dipole and monopole antennas that illustrate the principle of efficient utiliza-tion of the space within a sphere in order to increase the bandwidth (lower the Q) are shown in Figures 9.1 and 9.2. The qualitative bandwidths are also indicated in these figures.
11.6 ANTENNA MINIATURIZATION Antenna miniaturization is a process by, the antenna designer can exercise to reduce the dimensions of the antenna while maintaining, within reasonable limits, the original antenna radiation characteristics. The effort to shrink the physical dimensions of an antenna, without sig-nificant performance degradation, has continued for over half a century. However, this is not an easy task, and most often some compromises have had to be accepted between reduction in dimensions and antenna performance. Over the years, many creative techniques have been developed to minia-turize such basic antenna elements as dipoles, monopoles, loops, and slots microstrip/patches, as well as many other radiating elements. An excellent review paper of these attempts is . This sec-tion follows and reports on some of the miniaturization experimentations. In the examples reported here and in , miniaturization involves modifying the geometry of the structure, by adding com-ponents, or by altering the material characteristics. It should be mentioned that at this time, although the performance characteristics of the miniaturized design, most likely is not as good as that of the original antenna, the result is still better than that of an unmodified antenna reduced by the same size; otherwise, miniaturization is really not achieved, even though the dimensions are reduced but the radiation characteristics are considerably degraded. This is why small antennas, such as the infinitesimal linear dipole of Figure 4.1 (l ≤λ∕50), have radiation characteristics (e.g., impedance, efficiency, gain, bandwidth) that are not very attractive. Therefore, small antennas are differentiated from miniaturized antennas, and as is defined in , for an antenna to be miniaturized, it must be: “An antenna from a well-established category that has been reduced in size while preserving the fidelity of at least one performance characteristic.” Miniaturization, as already mentioned, is accomplished by altering the geometry of the structure, and possibly adding components or integrating materials into the structure. The performance of the revised design does not need to be as good as that of the original version, but it must be better than a version of the original antenna that had been reduced in size by the same amount as the miniaturized antenna. Therefore, in the design techniques reported in and some here, no active elements are considered that will enhance the performance of the designs.
A recent technique that has received considerable attention for miniaturizing antenna design is the integration of metamaterials into the antenna structures. However, the many papers that have reported on such designs base their results only on simulations. Numerous references pertaining to 620 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS such work are provided in , and the designs reported there are those that have been fabricated and their radiation characteristics have been verified with experimental data. Nevertheless, because metamaterial techniques are beyond the scope of this book, they will not be covered here. The reader is encouraged to refer to for more details and references.
Some of the antenna characteristics that can be used to judge the merits of miniaturization, not all inclusive, are as follows: r Antenna input impedance r Return loss r Impedance bandwidth r Fractional bandwidth r Impedance matching/reflection r Radiation efficiency r Directivity r Gain r Realized gain r Resonant Q Numerous techniques have been devised to miniaturize existing types of antennas have varying lev-els of complexity and ingenuity. The examples of miniaturization included here, particularly those related to dipole and monopole antennas, are also reported in ; a discussion of patch antennas as reported in is provided later in this book, in Section 14.9.
11.6.1 Monopole Antenna Monopole antennas, as well as dipoles, are basic elements whose analysis and radiation character-istics are discussed in Chapter 4.
A monopole antenna is resonant (real input impedance) for a length (height) slightly smaller than l = λo∕4. For this reason, miniaturization is can succeed for designs in which the length is signifi-cantly smaller than λo∕4. A few such basic designs will be reported here.
A. Impedance Loading One of the simplest designs used to miniaturize monopoles and dipoles is to add lumped impedance elements along the antenna’s length. This is referred to as impedance loading. One such approach, reported in 1963 by Harrison , was to add lumped inductor coils at various points along a monopole, as shown in Figure 11.19. It is reported in , that by inserting a coil of a given Q, the antenna was made considerable shorter. However, the incorporation and integration of coil(s) in Figure 11.19 Cylindrical monopole loaded with lumped inductor .
ANTENNA MINIATURIZATION 621 the design impacted the antenna’s performance, decreasing the radiation resistance, reflection coef-ficient, radiation efficiency, and realized gain. It was in fact discovered that, as the radiating element is made smaller, due to the insertion of a coil, the radiation resistance decreases, the bandwidth narrows and the radiation efficiency decreases. To illustrate the trade-offs of miniaturization due to impedance loading, we give an example below.
Example 11.2 Consider an electrically small resonant monopole of length (height) of l = λo∕(10𝜋) = 3.183 × 10−2λo, radius a = 2.229 × 10−4λo [2ln(2l/a) = 12.5] that is connected to a 50-ohm line. A coil, with a Q = 300, is placed in its middle, as in Figure 11.19. In accord with , the coil’s: r radiation resistance is Rr = 20.15 ohms r radiation efficiency ecd (accounting for the coil’s losses) is ecd = 19.85%. (No other losses on the element itself are included.) Assuming that the maximum directivity Do of the shortened monopole, loaded with the coil, is the same as that of a regular short monopole (Do = 3 dimensionless = 4.77 dB, l ≪λo), the objective is to determine the following components: a. Reflection coefficient Γ b. Return loss S11 (in dB) c. Maximum realized gain Gre (in dB) d. Maximum realized gain Gre (in dB) when the coil is lossless (Q = ∞), ecd = 100%, and whose radiation resistance is Rr = 2.37 ohms e. Compare the realized gain of parts c and d with those of a regular λo∕4 resonant, lossless monopole Solution: a. Γ = 20.15−50 20.15+50 = −0.4255 b. S11 = 20 log10 |Γ| = −20 log10 |−0.4255| = −7.422 dB c. The maximum directivity of a regular infinitesimal (l ≪λo) monopole is Do = 2(1.5) = 3.
Therefore the realized gain of the short monopole, with the coil in its middle, is Gre0 = 10 log10 [ ecd ( 1 −|Γ|2) Do ] = 10 log10 [ 0.1985 ( 1 −|−0.4255|2) 3 ] = −3.119 dB d. For a very short (l ≪λo) monopole with a lossless coil (Q = ∞, ecd = 100%), Rr = 2.37 Γ = 2.37 −50 2.37 + 50 = −0.9095 Gre0 = 10 log10 [ ecd ( 1 −|Γ|2) Do ] = 10 log10 [ 1 ( 1 −|−0.9095|2) 3 ] = −2.853 dB e. The realized gain of a regular lossless (ecd = 1) resonant λo∕4 monopole, based on the radiation resistance of Rr = 73∕2 = 36.5, is Γ = Rr −50 Rr + 50 = 36.5 −50 36.5 + 50 = −0.1561 Gre0 = 10 log10 [ecd (1 −|Γ|2) Do ] = 10 log10 [1 (1 −|−0.1561|2) 3] = 4.664 dB 622 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Therefore, while a coil-loaded monopole of length (height) l = λo∕(10𝜋) is physically shortened, compared to a regular λo∕4 monopole, by a factor of λo∕4 λo∕(10𝜋) = 10𝜋 4 = 7.85 its realized gain has been reduced by: a. l = λo∕(10𝜋) monopole with coil of Q = 300: |4.664 −(−3.119)| = 7.783 dB b. l = λo∕(10𝜋) monopole with coil of Q = ∞: |4.664 −(−2.853)| = 7.517 dB So, there is a trade-off of small size, by a factor of 7.85, at the expense of reduction in real-ized gain by 7.783 dB and 7.517 dB, respectively. Whether such a trade-off is acceptable will depend on the application and antenna engineer. However, the example illustrates the trade-offs of miniaturization; length reduction versus reduction in realized gain.
B. Materials Loading Several authors have attempted at miniaturizating dipole and monopole antennas by covering them with dielectric sleeves and spheres . James and Henderson provide a detailed account of such material loading and the impact on the antenna’s pattern, bandwidth, input impedance, and radiation efficiency. To illustrate the trade-offs of this type of miniaturization, we select an example from , of a monopole loaded with a cylindrical dielectric cover, as shown in Figure 11.20.
Figure 11.20 Cylindrical monopole loaded covered with a cylindrical dielectric cover .
Example 11.3 The resonant monopole, shown in Figure 11.20, has length (height) l = 23 mm and radius a = 0.75 mm, and it is surrounded by a dielectric cylinder of the same length (height) and radius b = 23.5 mm. The dielectric cylinder is filled with pure water (𝜀r = 73, 𝜇r = 1, tan 𝛿e = 0.02), and it was measured to be resonant at 381 MHz (λo = 30 × 1010∕381 × 106 = 787.4 mm; λo∕4 = 196.85 mm, λ∕4 = 23 mm; a reduction in length/height by a factor of √ 73 = 8.5).
The radiation resistance and radiation efficiency were measured to be, respectively, Rr = 0.4 ohm ANTENNA MINIATURIZATION 623 and er = 12.6% . Assuming that the monopole is connected to a 50-ohm line and the direc-tivity of the dielectric loaded monopole is the same as a regular monopole, determine for the dielectric loaded monopole the following components: a. Reflection coefficient Γ b. Return loss S11 (in dB) c. Maximum realized gain Gre (in dB) d. Reduction of maximum realized gain Gre (in dB) of resonant dielectric loaded monopole compared to a resonant lossless regular infinitesimal monopole e. Reduction of length (height) factor of dielectric loaded monopole f. Increased of volume of dielectric loaded monopole Solution: a. Since the radiation resistance of the dielectric covered resonant monopole is 0.4 ohm, its reflection coefficient is Γ = (0.4 −50)∕(0.4 + 50) = −0.9841.
b. S11 = 20 log10 |Γ| = −20 log10 |−0.9841| = −0.139dB.
c. Since the length (height) of the monopole is l = 23 = λ∕4 mm, the maximum directivity is Do = 2(1.643) = 3.286. Therefore, the realized gain of the dielectric monopole is Gre0 = 10 log10 [𝜀r (1 −|Γ|2) Do ] = 10 log10 [0.126 (1 −|−0.9841|2) 3.286] = −18.85 dB d. The realized gain of a regular lossless resonant λo∕4 monopole is Γ = 36.5 −50 36.5 + 50 = −0.1561 Gre0 = 10 log10 [ ecd ( 1 −|Γ|2) Do ] = 10 log10 [ 1 ( 1 −|−0.1561|2) 3.286 ] = 5.06 dB Therefore, the reduction of the realized gain of the dielectric loaded monopole, compared to an infinitesimal monopole is 5.06 −(−18.85) = 24.01 dB e. Since the dielectric constant of the dielectric cover is 73, the wavelength λ in the dielectric, compared to the free-space wavelength λo, has be reduced by a factor of λo λ = √ 73 = 8.544 factor Thus, the dielectric-loaded monopole has been shortened by a factor of 8.544.
f. The volume of the dielectric monopole, compared to the volume of the free-space monopole, is V Vo = 𝜋b2(l) 𝜋a2(lo) = (23.5 0.74 )2 ( 1 √ 73 ) = 118 Therefore, while the height of the dielectric loaded monopole has been reduced by a fac-tor of 8.544, its volume has been increased by a factor of 118, and its realized gain has been reduced by 23.9 dB. This does not seem to be a good trade-off of miniaturization, considering the large reduction in realized gain.
624 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS A second example in concerns a monopole of length (height) of l = 160 mm and radius a = 0.6 mm, resonant at 253 MHz (lo = λo∕4 = 1,185.77∕4 = 296.44 mm; l = 160 mm; reduction in length (height) = 296.44/160 = 1.85) surrounded by a cylinder of water of radius b = 11.5 mm. For this example, the reduction of length (height) is only 1.85, but there is a considerable increase in volume by a factor of 199: V Vo = (11.5 0.6 )2 × ( l lo ) = (11.5 0.6 )2 × ( 160 296.44 ) = 198.3 ≈199 which is much greater than the 118 factor of Example 11.3. However, because the measured radiation efficiency was ecd = 96.6% and the radiation resistance was measured to be 8.5 ohms (|Γ| = 0.7094, S11 = −2.9822 dB), the realized gain is Gre0 = 10 log10 [ ecd ( 1 −|Γ|2) Do ] = 10 log10 [ 0.966 ( 1 −|−0.7094|2) 3.286 ] = 1.978 dB In this case, the radiation efficiency is much higher, which leads a positive realized gain. Therefore, for this example, miniaturization leads to more positive trade-offs.
C. Folding Another technique that can be used to miniaturize a dipole is to fold its arm and create a compact structure. There are two ways to do this: r The arm’s linear physical length could be shortened and its current path length maintained; this way, its resonance is maintained.
r The arm’s linear length could be increased and its current path length also increased; this way, its resonance is decreased.
However, in both approaches, the radiation characteristics (radiation resistance, bandwidth, efficiency) might be compromised. As a result, many of the attempts to miniaturize dipoles and monopoles were by bending the wire into different compact shapes, including zig-zag, which results in a reduction in length but an increase in area . In , there are discussed a number of folding shapes that were tried and the trade-offs between length reduction and radiation efficiency that ensued.
In , there is report of a meandering shape of a dipole that included measurements of radia-tion resistance, bandwidth, and efficiency. This was accomplished by folding the monopole N times into U shapes with 90◦corners to form planar structures of width 2W and length l, as shown in Figure 11.21.
Figure 11.21 Monopole with meandering pattern. (source: c ⃝1999 IEEE).
ANTENNA MINIATURIZATION 625 One such example reported in involved a resonant linear monopole of length l ≈λo∕4 ≈ 135 mm, radius a = 0.4 mm, resonant frequency of 545 MHz (λo = 550.46 mm), radiation resis-tance of 36.5 ohms (|Γ| = 0.156, based on a line characteristic impedance of 50 ohms), radiation efficiency of ecd = 99.1%, and a fractional bandwidth 9.5% [based on S11 = −10 dB (VSWR = 2)]. This monopole was physically miniaturized by folding it into meandering patterns, of the form shown in Figure 11.21, but with each of the various values of N occupying a surface of width W = 2.7 mm and length 𝓁= 45 mm.
To judge the merits of the folding, a reduction factor 𝛽was introduced and defined as 𝛽= L∕l (where L is the length of a conventional monopole and l is the length of the length of the folded monopole). It was determined, through experimental data, that with N = 2 the antenna resonated at 922 MHz (λ = 325.38 mm; λ∕4 = 81.34 mm). This folding corresponded to a physical length reduction of about 𝛽= L∕l = 81.34∕45 = 1.8 but an increase in width by a factor of 2W/2a = 5.4/0.8 = 6.75. For the reduced folding geometry, the measured radiation resistance was Rr = 13 ohms (|Γ| = 0.5873, S11 = −4.62 dB, based a 50-ohm line), the radiation efficiency was 96.7%, and the fractional bandwidth was 3%. Based on these figures of merit, it can be stated that the radi-ation efficiency had likely not been drastically reduced, but both the radiation resistance and the bandwidth, were significantly compromised. Ultimately, and the realized gain was reduced to Gre0 = 10 log10 [er (1 −|Γ|2) Do ] = 10 log10 [0.967 (1 −|−0.5873|2) 2(1.643)] = 3.18 dB Compared to the realized gain of Gre0 = 5.02 dB of a resonant λ∕4 monopole (Rr = 36.5, ecd = 99.1%, |Γ| = 0.1561, 50-omh line), the realized gain of the N = 2 folded monopole was reduced by about 2 dB. Since the radiation efficiency was reduced slightly (from 99.1% to 96.7%) and the realized gain was reduced by about 2 dB, this could be considered a miniaturized design, based on the definition of miniaturization stated in this section.
It is shown in that as N increases, as indicated in Figure 11.21, the meandering pattern becomes more complex and the reduction factor is minimized; however, the radiation efficiency and bandwidth increase. Another example of folding geometry from , for the case where N = 14, is as follows: r Length reduction factor is only 𝛽= 1∕0.75 = 1.33.
r Radiation efficiency is 98%.
r Fractional bandwidth is 8%.
r Radiation resistance is 23.5 ohms (|Γ| = 0.3605, S11 = −8.86 dB; based on 50-ohm line).
Its corresponding realized gain is Gre0 = 10 log10 [er (1 −|Γ|2) Do ] = 10 log10 [0.98 (1 −|−0.3605|2) 2(1.643)] = 4.47 dB which is reduced by only 0.6 dB compared to an ideal resonant lossless λ∕4 monopole.
Two additional folding geometries are shown in Figure 11.22 . As is evident in these two meandering patternsshown in the figure, the currents in the horizontal lengths oppose each other, basically canceling each other, whereas the vertical currents reinforce each other. Thus the radia-tion due to the horizontal sections basically cancels and that of the vertical section reinforces. This design usually leads to a reduction of the effective or net-inductance and in the overall electrical length; thus, resonance occurs at a higher frequency. Also, it should be stated that the vertical sec-tions are primarily responsible for the radiation resistance and bandwidth, which are usually small, as in small vertical straight-wire monopoles discussed in Section 4.7. However, in Figure 11.22(b), the currents in the helical meandering pattern reinforce each other in all parts, leading to increases of the effective or net-inductance and the overall electrical length; thus, resonance occurs at a lower frequency, as discussed in the design of helical antennas in Section 10.3.1. For example, it is stated in , that for designs of Figure 11.22 with same overall height (10 cm), same total wire length 626 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Opposing Feed Point (a) Rectangular meandering (b) Helical meandering Feed Point Reinforcing Reinforcing Figure 11.22 Monopole with meandering pattern. ( Reprinted with permission from John Wiley & Sons, Inc.).
(30 cm), same conductor diameter (1 mm), and same overall cylindrical diameter (3.3 cm), the geom-etry of Figure 11.22(a) resonates at 361 MHz while that of the helix Figure 11.22(b) (operating in the normal/broadside mode) resonates at 312.6 MHz; a clear indication that the geometry of Figure 11.22(b) is electrically longer; resonates at a lower frequency. It should be further noted that the spac-ing between the turns plays a key role in the reinforcing or opposing current vectors. Because of its compactness, despite some of its shortcomings, the design of Figure 11.22(a) is an attractive design that has had wide use in wireless mobile units and in external cell phone antennas. The frequency ranges of both designs in Figure 11.22 are considered to be electrically small and their radiation patterns donut shaped (null along the axis), compared to the patterns of Figure 4.15 (of a small monopole) and of Figure 10.14(a) (of the broadside/normal mode helical antenna). For more details on the electrical performances of these and other designs, including some other coupling and decou-pling configurations that result in resonances at higher frequencies, the reader is referred to .
11.6.2 Patch Antennas There are numerous techniques that are used to miniaturize planar type antennas, such as microstrips.
One such technique is to use a shorting pin/strip, as shown in Figure 14.44, leading to a microstrip of length nearly λ∕4 (instead of λ∕2); is a physical reduction by a factor of 2. Similar methods are utilized in the design of PIFA and IFA antennas (as discussed later, respectively, in Sections 14.9.1 and 14.9.3). The reader is also referred to for more on patch/microstrip antenna miniaturization techniques.
11.6.3 Antenna Miniaturization Using Metamaterials Artificial impedance surfaces, also known as engineered electromagnetic surfaces, have been devel-oped over the last few decades to alter the impedance boundary conditions of the surface and thus control the radiation characteristics, such as radiation efficiency and pattern, of antenna elements placed at or near them, or the scattering of impinging electromagnetic waves. When electromagnetic FRACTAL ANTENNAS 627 waves interact with surfaces that exhibit geometrical periodicity, they result in some interesting and exciting characteristics, which typically have numerous applications that have captured the attention and imagination of engineers and scientists. Using a “broad brush” designation, these metamateri-als have included applications of dielectric and metallic structures of waveguides, resonators, filters, antennas, and other devices.
Another antenna miniaturization technique that appears to have much potential has involved attempts to incorporate metamaterials into the structures. Even though many papers have reported on such attempts, the reported designs are based only on simulations. The reader is referred to , for the numerous references on such designs, but only reported are designs that have been fabricated and their radiation characteristics have verified with experimental data. Otherwise, because metamaterial techniques are beyond the scope of this book, they will not be reported here.
11.7 FRACTAL ANTENNAS One of the main objectives in wireless communication systems is the design of wideband, or even multiband, low profile, small antennas. Applications of such antennas include, but are not limited to, personal communication systems, small satellite communication terminals, unmanned aerial vehi-cles, and many more. In order to meet the specification that the antenna be small, some severe limitations are placed on the design, which must meet the fundamental limits of electrically small antennas discussed in the previous section. It was pointed out in the previous section as well as in Section 9.1 that the bandwidth of an antenna enclosed in a sphere of radius a can be improved only if the antenna utilizes efficiently, with its geometrical configuration, the available volume within the sphere. For planar geometries, the bandwidth of the antenna can be improved as the geometry of the antenna best utilizes the available planar area of a circle of radius a that encloses the antenna.
It was illustrated in Figure 11.17 that a Goubau antenna exhibits a lower Q, and thus a larger band-width, than a small linear cylindrical dipole that can be enclosed in a sphere of the same radius. An even lower Q is achieved using a 4-arm folded spherical helix of Figure 11.18, also indicated in Figure 11.17. Both fo these were achieved because a Goubau antenna utilizes the available volume of the sphere more efficiently. Another antenna that accomplishes the same objective, compared to a cylindrical dipole, is a biconical antenna, as illustrated in Figure 11.16. Ideally a biconical antenna of Figure 11.16 that would exhibit the smallest Q (largest bandwidth) would be one whose included angle of each cone is 180◦, or a hemispherical dipole. Other dipole and monopole configurations, as well as their qualitative bandwidths, are shown in Figures 9.1 and 9.2.
Another antenna that can meet the requirements of utilizing the available space within a sphere of radius a more efficiently is a fractal antenna. Fractal antennas are based on the concept of a fractal, which is a recursively generated geometry that has fractional dimensions, as pioneered and advanced by Benoit B. Mandelbrot . Mandelbrot coined the term fractal and investigated the relationship between fractals and nature using discoveries made by Gaston Julia, Pierre Fatou, and Felix Hausdorff –. He was able to show that many fractals exist in nature and can be used to accurately model certain phenomena. In addition, he was able to introduce new fractals to model more complex structures, including trees and mountains, that possess an inherent self-similarity and self-affinity in their geometrical shape.
Fractal concepts have been applied to many branches of science and engineering, including fractal electrodynamics for radiation, propagation, and scattering –. These fractal concepts have been extended to antenna theory and design, and there have been many studies and implementations of different fractal antenna elements and arrays –, and many others. The theory of the basics and advances of fractal antennas is well documented in some of the stated references and others, and it will not be repeated here because of space limitations. Instead, a qualitative discussion will be given here. The interested reader is directed to the references for more details.
628 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Generator Indentation Width Initiator (a) Minkowski island (SOURCE: © 2002 IEEE) (SOURCE: © 2000 IEEE) Generator (b) Koch loop Figure 11.23 Iterative generation process of fractals.
Fractals can be classified in two categories: deterministic and random. Deterministic, such as the von Koch snowflake and the Sierpinski gaskets, are those that are generated of several scaled-down and rotated copies of themselves . Such fractals can be generated using computer graphics requiring particular mapping that is repeated over and over using a recursive algorithm. Random fractals also contain elements of randomness that allow simulation of natural phenomena. Procedures and algorithms for generating fractals, both deterministic and random, can be found in .
Fractal geometries can best be described and generated using an iterative process that leads to self-similar and self-affinity structures as outlined in , . The process can best be illustrated graphically as shown for the two different geometries in Figure 11.23(a,b). Figure 11.23(a) exhibits what is referred to as the Minkowski island fractal , while Figure 11.23(b) illustrates the Koch fractal loop . The geometry generating process of a fractal begins with a basic geometry referred to as the initiator, which in Figure 11.23(a) is a Euclidean square while that of Figure 11.23(b) is a Euclidean triangle. In Figure 11.23(a), each of the four straight sides of the square is replaced with a generator that is shown at the bottom of the figure. The first three generated iterations are displayed.
In Figure 11.23(b), the middle third of each side of the triangle is replaced with its own generator.
The first four generated iterations are displayed. The process and generator of Figure 11.23(b) can also be used to generate the Koch dipole that will be discussed later in this section. The trend of the fractal antenna geometry can be deduced by observing several iterations of the process. The final fractal geometry is a curve with an infinitely intricate underlying structure such that, no matter how closely the structure is viewed, the fundamental building blocks cannot be differentiated because they are scaled versions of the initiator. Fractal geometries are used to represent structures in nature, such as trees, plants, mountain ranges, clouds, waves, and so on. Mathematically generated fractal geometries of plants are shown in Figure 11.24. The theory and design of fractal antenna arrays is described in .
Another classic fractal is the Sierpinski gasket , . This fractal can be generated with the Pascal triangle of (6-63) using the following procedure , . Consider an equilateral triangular grid of nodes, as shown in Figure 11.25(a). Starting from the top, each row is labeled n = 1, 2, 3, … , FRACTAL ANTENNAS 629 Figure 11.24 Fractals used to represent plants in nature.
and each row contains n nodes. A number is assigned to each node for identification purposes. If all the nodes whose number is divisible by a prime number p(p = 2, 3, 5, …) are deleted, the result is a self-similar fractal referred to as the Sierpinski gasket of Mod-p. For example, if the nodes in Figure 11.25(a) whose numbers are divisible by 2 are deleted, the result is a Sierpinski gasket of Mod-2. If the nodes whose number is divisible by 3 are deleted, a Sierpinski gasket of Mod-3 is (a) Pascal's triangle and Sierpinski gasket Mod-2 (b) Mod-3 (c) Mod-5 Figure 11.25 Pascal’s triangle and Sierpinski gaskets (Mod-2, 3, 5). (source: c ⃝2001 IEEE).
630 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS Equal Radius Circle Loop Koch Loop Figure 11.26 Circular and Koch loops of equal radii. (source: c ⃝2002 IEEE).
obtained, as shown in Figure 11.25(b). Sierpinski gaskets can be used as elements in monopoles and dipoles having geometries whose peripheries are similar to the cross section of conical monopoles and biconical dipoles. The Sierpinski gaskets exhibit favorable radiation characteristics in terms of resonance, impedance, directivity, pattern, and so on, just like the other fractals.
Fractals antennas exhibit space-filling properties that can be used to miniaturize classic antenna elements, such as dipoles and loops, and overcome some of the limitations of small antennas. The line that is used to represent the fractal can meander in such a way as to effectively fill the available space, leading to curves that are electrically long but compacted into a small physical space. This is part of the fundamental limit of small antennas, discussed in Section 11.5 and represented by (11-35a) and (11-35b) (see Figure 11.17), which leads to smaller Qs/larger bandwidths. It also results in antenna elements that, although are compacted in small space, can be resonate and exhibit input resistances that are much greater than the classic geometries of dipoles, loops, etc.
To demonstrate this, let us consider two classic geometries: a circular loop and a linear dipole.
Both of these are discussed in Chapters 5 and 4, respectively, and both, when they are small elec-trically, exhibit small input resistances and must be large in order to resonate. For an ideal circular loop, as discussed in detail in Chapter 5, the input impedance is very small (usually around 1 ohm) if the loop is electrically small. This can be overcome by using a Koch loop of Figure 11.23(b). In fact, in Figure 11.26 we display a Koch loop of four iterations, shown also in Figure 11.23(b), circum-scribed by a circular loop of equal radius. In Figure 11.55, we compare the input resistance below resonance of these two geometries for a circumference ranging from 0.15λ to 0.27λ (for the circular loop) and 0.39λ to 0.7λ for the Koch loop. While the input impedance of a small circular loop is very small (around 1.33 ohms at about 0.265λ in circumference for the circular loop), the input resistance of the Koch fractal loop of equal radius (radius ≈0.04218λ) is 35 ohms. Both antennas are below resonance and would require matching for the reactive part.
For an ideal linear dipole, the first resonance occurs when the overall length is λ∕2 (as shown in Figure 8.17), which for some frequencies can be physically large. The length can be miniaturized using fractal dipoles, such as the Koch dipole and other similar geometries. To generate the Koch dipole, we apply the iterative generating procedure using the generator of Figure 11.23(b), as shown on the top part of Figure 11.28. This procedure can be extended to generate quasi-fractal tree dipoles and 3D quasi-fractal tree dipoles, as also illustrated in the second and third parts of Figure 11.28, respectively.
In Figure 11.29, we exhibit the resonant frequency for the first five iterations of each fractal dipole of Figure 11.28. It is apparent that the higher iterative geometries exhibit lower resonant frequen-cies, as if the overall length of the dipole was large electrically. The resonant frequencies plotted in FRACTAL ANTENNAS 631 0.16 0.18 0.2 0.22 0.24 0.26 0 5 10 15 20 25 30 35 40 45 50 Perimeter of circular loop (in λ) Input Resistance (ohms) Circular Loop Antenna Koch Fractal Loop Antenna Figure 11.27 Input resistance of circular loop and Koch fractal loop vs. perimeter. (source: c ⃝2000 IEEE).
Figure 11.29, which correspond to decrease in the length of the dipole, were computed when the reac-tance was zero and the input impedance was approximately 50 ohms at equal frequencies. It should also be pointed out that the major decrease of the resonant frequency occurs for the 3D fractal dipole, corresponding to about 40% reduction, after five iterations over the equivalent tree dipole . It should also be stated that most of the miniaturization benefits of the fractal dipole occur within the first five iterations with very little changes in the characteristics of the dipole occurring with minimal increases in the complexity afterwards.
As was indicated in Section 11.6, the theoretical limit on the lowest Q of an antenna of any shape is represented by (11-35a) and (11-35b) and plotted in Figure 11.17. Because of the space-filling properties of fractal antennas, their Q can be lower than that of a classic cylindrical dipole, as was Fractal Tree Dipole 3D Fractal Tree Dipole Koch Dipole 3 4 Iteration 0 1 2 5 Figure 11.28 Fractal dipole geometries for Koch, tree, and 3D-tree fractal dipoles for five iterations.
(source: c ⃝2002 IEEE).
632 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS 0 1 2 3 4 5 Fractal Iteration 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 Resonant Frequency (MHz) Koch Dipole Fractal Tree Dipole 3D Fractal Tree Dipole Figure 11.29 Resonant frequency for first five iterations of Koch, tree, and 3D-tree fractal dipoles.
(source: c ⃝2002 IEEE).
the case for Goubau and spherical helix antennas illustrated in Figure 11.17. To demonstrate that, we have plotted in Figure 11.30 the Q of a Koch dipole of the first five iterations, including the zero-order linear dipole. The curve representing the fundamental limit is also included. It is apparent that each of the Koch dipoles exhibits lower Qs/higher bandwidths, compared to the classic linear dipole, as the order of iteration increases. The Qs can be lowered even more using the fractal tree and especially the 3D fractal dipole that exhibits the most space-filling properties . The Qs shown in Figure 11.30 were obtained using Method of Moments .
0.6 0.8 1 1.2 1.4 1.6 0 10 20 30 40 50 60 kl Q K0 Q K1 Q K2 Q K3 Q K4 Q K5 Q Fundamental Limit Fundamental Limit Figure 11.30 Quality factor Q for Koch dipole of up to five iterations as a function of kh (h is maximum length/height of dipole). (source: c ⃝2002 IEEE).
REFERENCES 633 The concepts discussed above for loops and dipoles can be extended to other antenna elements, including microstrip/patch antennas . The reader is directed to the references.
11.8 MULTIMEDIA In the publisher’s website for this book the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab and Fortran computer program, designated log perd, for computing and displaying the radiation characteristics of a log-periodic linear dipole array design.
c. Power Point (PPT) viewgraphs, in multicolor.
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11. R. L. Carrel, “Analysis and Design of the Log-Periodic Dipole Antenna,” Ph.D. Dissertation, Elec. Eng.
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13. P. C. Budson and G. T. Thompson, “A Note on the Calculation of the Gain of Log-Periodic Dipole Anten-nas,” IEEE Trans. Antennas Propagat., AP-24, pp. 105–106, January 1976.
14. H. E. King, “Mutual Impedance of Unequal Length Antennas in Echelon,” IRE Trans. Antennas Propagat., Vol. AP-5, pp. 306–313, July 1957.
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634 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS 18. L. J. Chu, “Physical Limitations of Omnidirectional Antennas,” J. Appl. Phys., Vol. 19, pp. 1163–1175, December 1948.
19. R. F. Harrington, “Effect of Antenna Size on Gain, Bandwidth, and Efficiency,” J. Res. Nat. Bur. Stand.-D, Radio Propagat., Vol. 64D, pp. 1–12, January–February 1960.
20. H. A. Wheeler, “Fundamental Limitations of Small Antennas,” Proc. IRE, pp. 1479–1488, December 1947.
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22. G. Goubau, “Multi-element Monopole Antennas,” Proc. Workshop on Electrically Small Antennas ECOM, Ft. Monmouth, NJ, pp. 63–67, May 1976.
23. S. R. Best, “The Radiation Properties of Electrically Small Folded Spherical Helix Antennas,” IEEE Trans.
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24. E. J. Rothwell and R. O. Ouedraogo, “Antenna Miniaturization: Definitions, Concepts and a Review with Emphasis on Metamaterials,” J. Electromagnetic Waves Appl., Vol. 28, No. 17, pp. 2089–2123, 2014.
25. C. W. Harrison, “Monopole with Inductive Loading,” IEEE Trans. Antennas Propagat., Vol. 11, No. 4, pp. 394–400, July 1963.
26. J. L. Birchfield and W. R. Free, “Dielectrically Loaded Short Antennas,” IEEE Trans. Antennas Propagat., Vol. 22, No. 3, pp. 471–472, May 1974.
27. M. S. Smith, “Properties of Dielectrically Loaded Antennas,” Proc. IEE, Vol. 124, No. 10, pp. 837–839, 1977.
28. J. R. James and A. Henderson, “Electrically Short Monopole Antennas with Dielectric or Ferrite Coatings,” Proc. IEE, Vol. 125, No. 9, pp. 793–803, 1978.
29. H. Nakano, H. Togami, A. Yoshizawa and J. Yamauchi, “Shortening Ratios of Modified Dipole Antennas,” IEEE Trans. Antennas Propagat., Vol. 32, No. 4, pp. 385–386, April 1984.
30. J. M. Gonzalez-Arbesu, S. Blanch, and J. Romeu, “Shortening Ratios of Modified Dipole Antenna,” IEEE Antennas Wireless Propagat. Lett., Vol. 2, pp. 147–150, 2003.
31. R. Rashed and C.-T. Tai, “A New Class of Resonant Antennas,” IEEE Trans. Antennas Propagat., Vol. 39, No. 9, pp. 1428–1430, Sept. 1991.
32. S. S. Best, “Small and Fractal Antennas,” Chapter 10, pp. 475–528 in Modern Antenna Handbook (C. A. Balanis, editor), John Wiley & Sons, Inc., 2008.
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34. B. B. Mandelbrot, The Fractal Geometry of Nature, Freeman, 1983.
35. X. Yang, J. Chiochetti, D. Papadopoulos, and L. Susman, “Fractal Antenna Elements and Arrays,” Appl.
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36. K. Falconer, Fractal Geometry: Mathematical Foundations and Applications, John Wiley & Sons, New York, 1990.
37. H. Lauwerier, Fractals: Endlessly Repeated Geometrical Figures, Princeton University Press, Princeton, NJ, 1991.
38. J.-F. Gouyet, Physics and Fractal Structures, Springer, New York, 1996.
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PROBLEMS 635 44. D. H. Werner, “An Overview of Fractal Electrodynamics Research,” Proceedings of the 11th Annual Review of Progress in Applied Computational Electromagnetics (ACES), Vol. II, Naval Postgraduate School, Monterey, CA, pp. 964–969, March 1955.
45. D. L. Jaggard, “Fractal Electrodynamics: From Super Antennas to Superlattices,” in J. L. Vehel, E. Lutton, and C. Tricot (eds.), Fractals in Engineering, Springer-Verlag, New York, pp. 204–221, 1997.
46. Y. Kim and D. L. Jaggard, “The Fractal Random Array,” Proc. IEEE, Vol. 74, No. 9, pp. 1278–1280, 1986.
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48. N. Cohen, “Fractal Antennas Part 2: A Discussion of Relevant, but Disparate Qualities,” Communications Quarterly, Summer, pp. 53–66, Summer 1996.
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50. C. Puente, J. Romeu, R. Pous, and A. Cardama, “On the Behavior of the Sierpinski Multiband Antenna,” IEEE Trans. Antennas Propagat., Vol. 46, No. 4, pp. 517–524, April 1998.
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PROBLEMS 11.1.
Design a symmetrical two-wire plane spiral (𝜙0 = 0, 𝜋) at f = 10 MHz with total feed ter-minal separation of 10−3λ. The total length of each spiral should be one wavelength and each wire should be of one turn.
(a) Determine the rate of spiral of each wire.
(b) Find the radius (in λ and in meters) of each spiral at its terminal point.
(c) Plot the geometric shape of one wire. Use meters for its length.
11.2.
Verify (11-28).
11.3.
Design log-periodic dipole arrays, of the form shown in Figure 11.9(d), each with directiv-ities of 9 dB, input impedance of 75 ohms, and each with the following additional specifi-cations: Cover the (see Appendix IX) (a) VHF TV channels 2–13 (54–216 MHz). Use aluminum tubing with outside diame-ters of 3 4 in. (1.905 cm) and 3 16 in. (0.476 cm) for the largest and smallest elements, respectively.
636 FREQUENCY INDEPENDENT ANTENNAS, ANTENNA MINIATURIZATION, AND FRACTAL ANTENNAS (b) VHF TV channels 2–6 (54–88 MHz). Use diameters of 1.905 and 1.1169 cm for the largest and smallest elements, respectively.
(c) VHF TV channels 7–13 (174–216 MHz). Use diameters of 0.6 and 0.476 cm for the largest and smallest elements, respectively.
(d) UHF TV channels (512–806 MHz). The largest and smallest elements should have diameters of 0.2 and 0.128 cm, respectively.
(e) FM band of 88–108 MHz (100 channels at 200 KHz apart). The largest and smallest elements should have diameters of 1.169 and 0.9525 cm, respectively.
In each design, the feeder line should have the same diameter as the largest element.
11.4.
For each design in Problem 11.3, determine the (a) span of each period over which the radiation characteristics will vary slightly (b) number of periods (cycles) within the desired bandwidth 11.5.
Using the log perd computer program and Appendix IX, design an array which covers the VHF television band. Design the antenna for 7 dBi gain optimized in terms of 𝜎−𝜏.
The antenna should be matched to a 75-ohm coaxial input cable. For this problem, set the input line length to 0 meters, the source resistance to 0 ohms, the termination line length to 0 meters, the termination impedance to 100 Kohms, the length-to-diameter ratio to 40, and the boom diameter to 10 cm. To make the actual input impedance 75 ohms, one must iteratively find the optimal desired input impedance.
(a) Plot the gain, magnitude of the input impedance, and VSWR versus frequency from 30 MHz to 400 MHz.
(b) Based on the ripples in the plot of gain versus frequency, what is 𝜏? Compare this value to the value calculated by the computer program.
(c) Why does the gain decrease rapidly for frequencies less than the lower design frequency yet decrease very slowly for frequencies higher than the upper design frequency?
11.6.
For the antenna of Problem 11.5, replace the 100-Kohm load with a 75-ohm resistor.
(a) Plot the gain, magnitude of the input impedance, and VSWR versus frequency from 30 MHz to 400 MHz.
(b) What does the termination resistor do which makes this antenna an improvement over the antenna of Problem 11.5?
11.7.
For the antenna of Problem 11.5, replace the 100-Kohm termination (load) with a 75-ohm resistor and make the source resistance 10 ohms. This resistance represents the internal resistance of the power supply as well as losses in the input line.
(a) Plot the gain versus frequency from 30 MHz to 400 MHz.
(b) What is the antenna efficiency of this antenna?
(c) Based on your result from parts (a) and (b), what should the gain versus frequency plot look like for Problem 11.6?
11.8.
Design a log-periodic dipole array which operates from 470 MHz to 806 MHz (UHF band) with 8 dBi gain. This antenna should be matched to a 50-ohm cable of length 2 meters with no source resistance. The termination should be left open. Select the length-to-diameter ratio to be 25. At 600 MHz, do the following. Use the computer program log perd of this chapter.
(a) Plot the E- and H-plane patterns.
(b) Calculate the E- and H-plane half-power beamwidths.
(c) Find the front-to-back ratio.
(d) Why does the E-plane pattern have deep nulls while the H-plane pattern does not?
PROBLEMS 637 11.9.
The overall length of a small linear dipole antenna (like a biconical antenna, or cylindrical dipole, or any other) is λ∕𝜋. Assuming the antenna is 100% efficient, what is: (a) The smallest possible value of Q for an antenna of such a length? Practically it will be larger than that value.
(b) The largest fractional bandwidth (Δf∕f0 where f0 is the center frequency)?
11.10.
For a small (l = λ∕10) dipole with radius a = λ∕500 operating at a center frequency of 500 MHz: (a) Approximate maximum possible fractional bandwidth.
(b) Lower and upper frequencies (in MHz) of the operating bandwidth.
(c) The radius of the sphere (in λ) which encloses the dipole and which separates the region where the energy within the sphere is basically imaginary and outside the sphere is basically read.
11.11.
It is desired to design a 100% efficient biconical dipole antenna whose overall length is λ∕20.
The design guidelines specify a need to optimize the frequency response (bandwidth). To accomplish this, the quality factor Q of the antenna should be minimized. In order to get some indications as to the fundamental limits of the design: (a) What is the lowest possible limit of the Q for this size antenna?
(b) In order to approach this lower fundamental limit, should the included angle of the biconical antenna be made larger or smaller, and why?
11.12.
It is desired to design a very small, 100% efficient spiral antenna whose largest fractional bandwidth is 10%. Determine the following: (a) The smallest possible quality factor (Q) that the spiral can ever have.
(b) The largest possible diameter (in λ) the spiral can ever have.
11.13.
Small antennas are very highly reactive radiating elements and exhibit very high quality factors and very small bandwidths. Assuming the antenna is 50% efficient and its fractional bandwidth is 1% (a) determine the largest overall linear dimension (in λ) of the antenna; (b) would, in practice, the measured.
1. Q be smaller or larger than the one designed based on the 1% fractional bandwidth, and why?
2. fractional bandwidth be smaller or larger than the designed one of 1%, and why?
11.14.
Consider a small biconical antenna with 100% efficiency and a fractional bandwidth of 2%.
(a) Determine the largest possible overall dimension (in λ) of the antenna.
(b) Does the Q become smaller or larger with increasing cone angle? Why?
(c) Does the fractional bandwidth become smaller or larger with increasing cone angle? Why?
CHAPTER12 Aperture Antennas 12.1 INTRODUCTION Aperture antennas are most common at microwave frequencies. There are many different geometrical configurations of an aperture antenna with some of the most popular shown in Figure 1.4. They may take the form of a waveguide or a horn whose aperture may be square, rectangular, circular, elliptical, or any other configuration. Aperture antennas are very practical for space applications, because they can be flush mounted on the surface of the spacecraft or aircraft. Their opening can be covered with a dielectric material to protect them from environmental conditions. This type of mounting does not disturb the aerodynamic profile of the craft, which in high-speed applications is critical.
In this chapter, the mathematical tools will be developed to analyze the radiation characteristics of aperture antennas. The concepts will be demonstrated by examples and illustrations. Because they are the most practical, emphasis will be given to the rectangular and circular configurations. Due to mathematical complexities, the observations will be restricted to the far-field region. The edge effects, due to the finite size of the ground plane to which the aperture is mounted, can be taken into account by using diffraction methods such as the Geometrical Theory of Diffraction, better known as GTD. This is discussed briefly and only qualitatively in Section 12.10.
The radiation characteristics of wire antennas can be determined once the current distribution on the wire is known. For many configurations, however, the current distribution is not known exactly and only physical intuition or experimental measurements can provide a reasonable approximation to it. This is even more evident in aperture antennas (slits, slots, waveguides, horns, reflectors, lenses). It is therefore expedient to have alternate methods to compute the radiation characteristics of antennas.
Emphasis will be placed on techniques that for their solution rely primarily not on the current distri-bution but on reasonable approximations of the fields on or in the vicinity of the antenna structure.
One such technique is the Field Equivalence Principle.
12.2 FIELD EQUIVALENCE PRINCIPLE: HUYGENS’ PRINCIPLE The field equivalence is a principle by which actual sources, such as an antenna and transmitter, are replaced by equivalent sources. The fictitious sources are said to be equivalent within a region Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 639 640 APERTURE ANTENNAS because they produce the same fields within that region. The formulations of scattering and diffrac-tion problems by the equivalence principle are more suggestive to approximations.
The field equivalence was introduced in 1936 by S. A. Schelkunoff , , and it is a more rigor-ous formulation of Huygens’ principle which states that “each point on a primary wavefront can be considered to be a new source of a secondary spherical wave and that a secondary wavefront can be constructed as the envelope of these secondary spherical waves .” The equivalence principle is based on the uniqueness theorem which states that “a field in a lossy region is uniquely specified by the sources within the region plus the tangential components of the electric field over the boundary, or the tangential components of the magnetic field over the boundary, or the former over part of the boundary and the latter over the rest of the boundary , .” The field in a lossless medium is considered to be the limit, as the losses go to zero, of the corresponding field in a lossy medium.
Thus if the tangential electric and magnetic fields are completely known over a closed surface, the fields in the source-free region can be determined.
By the equivalence principle, the fields outside an imaginary closed surface are obtained by plac-ing over the closed surface suitable electric- and magnetic-current densities which satisfy the bound-ary conditions. The current densities are selected so that the fields inside the closed surface are zero and outside they are equal to the radiation produced by the actual sources. Thus the technique can be used to obtain the fields radiated outside a closed surface by sources enclosed within it. The for-mulation is exact but requires integration over the closed surface. The degree of accuracy depends on the knowledge of the tangential components of the fields over the closed surface.
In most applications, the closed surface is selected so that most of it coincides with the conducting parts of the physical structure. This is preferred because the vanishing of the tangential electric field components over the conducting parts of the surface reduces the physical limits of integration.
The equivalence principle is developed by considering an actual radiating source, which electri-cally is represented by current densities J1 and M1, as shown in Figure 12.1(a). The source radiates fields E1 and H1 everywhere. However, it is desired to develop a method that will yield the fields outside a closed surface. To accomplish this, a closed surface S is chosen, shown dashed in Fig-ure 12.1(a), which encloses the current densities J1 and M1. The volume within S is denoted by V1 and outside S by V2. The primary task will be to replace the original problem, shown in Fig-ure 12.1(a), by an equivalent one which yields the same fields E1 and H1 outside S (within V2). The formulation of the problem can be aided eminently if the closed surface is judiciously chosen so that fields over most, if not the entire surface, are known a priori.
An equivalent problem of Figure 12.1(a) is shown in Figure 12.1(b). The original sources J1 and M1 are removed, and we assume that there exist fields E and H inside S and fields E1 and H1 outside of S. For these fields to exist within and outside S, they must satisfy the boundary conditions on the tangential electric and magnetic field components. Thus on the imaginary surface S there must exist S V2 V1 V1 E1, H1 (a) Actual problem (b) Equivalent problem E1, H1 E, H J1 M1 V2 n Js = n [H1 – H] Ms = –n [E1 – E] E1, H1 S μ1, 1 μ1, 1 μ1, 1 μ1, 1 ^ ^ ^ Figure 12.1 Actual and equivalent models.
FIELD EQUIVALENCE PRINCIPLE: HUYGENS’ PRINCIPLE 641 the equivalent sources Js = ̂ n × [H1 −H] (12-1) Ms = −̂ n × [E1 −E] (12-2) and they radiate into an unbounded space (same medium everywhere). The current densities of (12-1) and (12-2) are said to be equivalent only within V2, because they produce the original fields (E1, H1) only outside S. Fields E, H, different from the originals (E1, H1), result within V1. Since the currents of (12-1) and (12-2) radiate in an unbounded space, the fields can be determined using (3-27)–(3-30a) and the geometry of Figure 12.2(a). In Figure 12.2(a), R is the distance from any point on the surface S, where Js and Ms exist, to the observation point.
Figure 12.2 Coordinate system for aperture antenna analysis.
642 APERTURE ANTENNAS V1 (a) Love's equivalent 0, 0 V2 n Js = n H1 Ms = –n E1 E1, H1 E1, H1 S V1 (b) Electric conductor equivalent 0, 0 V2 Ms = –n E1 E1, H1 S V1 (c) Magnetic conductor equivalent 0, 0 V2 Js = n H1 S Electric conductor Magnetic conductor ^ ^ ^ ^ ^ Figure 12.3 Equivalence principle models.
So far, the tangential components of both E and H have been used in setting up the equivalent problem. From electromagnetic uniqueness concepts, it is known that the tangential components of only E or H are needed to determine the fields. It will be demonstrated that equivalent problems can be found which require only the magnetic-current densities (tangential E) or only electric current densities (tangential H). This requires modifications to the equivalent problem of Figure 12.1(b).
Since the fields E, H within S can be anything (this is not the region of interest), it can be assumed that they are zero. In that case the equivalent problem of Figure 12.1(b) reduces to that of Fig-ure 12.3(a) with the equivalent current densities being equal to Js = ̂ n × (H1 −H)|H=0 = ̂ n × H1 (12-3) Ms = −̂ n × (E1 −E)|E=0 = −̂ n × E1 (12-4) This form of the field equivalence principle is known as Love’s Equivalence Principle , . Since the current densities of (12-3) and (12-4) radiate in an unbounded medium (same 𝜇, 𝜀everywhere), they can be used in conjunction with (3-27)–(3-30a) to find the fields everywhere.
Love’s Equivalence Principle of Figure 12.3(a) produces a null field within the imaginary surface S. Since the value of the E = H = 0 within S cannot be disturbed if the properties of the medium within it are changed, let us assume that it is replaced by a perfect electric conductor (𝜎= ∞). The introduction of the perfect conductor will have an effect on the equivalent source Js, and it will pro-hibit the use of (3-27)–(3-30a) because the current densities no longer radiate into an unbounded medium. Imagine that the geometrical configuration of the electric conductor is identical to the pro-file of the imaginary surface S, over which Js and Ms exist. As the electric conductor takes its place, SUMMARY 643 Ms = –n E1 Ms σ = ∞ Ms = –n E1 Ms = –2n E1 μ, μ, μ, μ, μ, (a) (b) (c) ^ ^ ^ Figure 12.4 Equivalent models for magnetic source radiation near a perfect electric conductor.
as shown in Figure 12.3(b), according to the uniqueness theorem , the equivalent of Figure 12.3(a) reduces to that of Figure 12.3(b). Only a magnetic current density Ms (tangential components of the electric field) are necessary over the entire S, and it radiates in the presence of the electric conduc-tor producing outside S the original fields E1, H1. Within S the fields are zero but, as before, this is not a region of interest. The difficulty in trying to use the equivalent problem of Figure 12.3(b) is that (3-27)–(3-30a) cannot be used, because the current densities do not radiate into an unbounded medium. The problem of a magnetic current density radiating in the presence of an electric conduct-ing surface must be solved. So it seems that the equivalent problem is just as difficult as the original problem itself.
Before some special simple geometries are considered and some suggestions are made for approximating complex geometries, let us introduce another equivalent problem. Referring to Fig-ure 12.3(a), let us assume that instead of placing a perfect electric conductor within S we introduce a perfect magnetic conductor (PMC). Again, according to the uniqueness theorem , the equivalent problem of Figure 12.3(a) reduces to that shown in Figure 12.3(c) (requires only a Js over the entire surface S, i.e., tangential components of the magnetic field). As was with the equivalent problem of Figure 12.3(b), (3-27)–(3-30a) cannot be used with Figure 12.3(c) and the problem is just as difficult as that of Figure 12.3(b) or the original of Figure 12.1(a).
To begin to see the utility of the field equivalence principle, especially that of Figure 12.3(b), let us assume that the surface of the electric conductor is flat and extends to infinity as shown in Figure 12.4(a). For this geometry, the problem is to determine how a magnetic source radiates in the presence of a flat electric conductor. From image theory, this problem reduces to that of Fig-ure 12.4(b) where an imaginary magnetic source is introduced on the side of the conductor and takes its place (remove conductor). Since the imaginary source is in the same direction as the equivalent source, the equivalent problem of Figure 12.4(b) reduces to that of Figure 12.4(c). The magnetic current density is doubled, it radiates in an unbounded medium, and (3-27)–(3-30a) can be used.
The equivalent problem of Figure 12.4(c) yields the correct E, H fields to the right side of the inter-face. If the surface of the obstacle is not flat and infinite, but its curvature is large compared to the wavelength, a good approximation is the equivalent problem of Figure 12.3(c).
SUMMARY In the analysis of electromagnetic problems, many times it is easier to form equivalent problems that yield the same solution within a region of interest. This is the case for scattering, diffraction, and 644 APERTURE ANTENNAS aperture antenna problems. In this chapter, the main interest is in aperture antennas. The concepts will be demonstrated with examples.
The steps that must be used to form an equivalent and solve an aperture problem are as follows: 1. Select an imaginary surface that encloses the actual sources (the aperture). The surface must be judiciously chosen so that the tangential components of the electric and/or the magnetic field are known, exactly or approximately, over its entire span. In many cases this surface is a flat plane extending to infinity.
2. Over the imaginary surface form equivalent current densities Js, Ms which take one of the following forms: a. Js and Ms over S assuming that the E- and H-fields within S are not zero.
b. or Js and Ms over S assuming that the E- and H-fields within S are zero (Love’s theorem) c. or Ms over S (Js = 0) assuming that within S the medium is a perfect electric conductor d. or Js over S (Ms = 0) assuming that within S the medium is a perfect magnetic conductor.
3. Solve the equivalent problem. For forms (a) and (b), (3-27)–(3-30a) can be used. For form (c), the problem of a magnetic current source next to a perfect electric conductor must be solved [(3-27)–(3-30a) cannot be used directly, because the current density does not radiate into an unbounded medium]. If the electric conductor is an infinite flat plane the problem can be solved exactly by image theory. For form (d), the problem of an electric current source next to a perfect magnetic conductor must be solved. Again (3-27)–(3-30a) cannot be used directly. If the magnetic conductor is an infinite flat plane, the problem can be solved exactly by image theory.
To demonstrate the usefulness and application of the field equivalence theorem to aperture antenna theory, an example is considered.
Example 12.1 A waveguide aperture is mounted on an infinite ground plane, as shown in Figure 12.5(a). Assum-ing that the tangential components of the electric field over the aperture are known, and are given by Ea, find an equivalent problem that will yield the same fields E, H radiated by the aperture to the right side of the interface.
Solution: First an imaginary closed surface is chosen. For this problem it is appropriate to select a flat plane extending from minus infinity to plus infinity, as shown in Figure 12.5(b). Over the infinite plane, the equivalent current densities Js and Ms are formed. Since the tangential components of E do not exist outside the aperture, because of vanishing boundary conditions, the magnetic current density Ms is only nonzero over the aperture. The electric current density Js is nonzero everywhere and is yet unknown. Now let us assume that an imaginary flat elec-tric conductor approaches the surface S, and it shorts out the current density Js everywhere. Ms exists only over the space occupied originally by the aperture, and it radiates in the presence of the conductor [see Figure 12.5(c)]. By image theory, the conductor can be removed and replaced by an imaginary (equivalent) source Ms as shown in Figure 12.5(d), which is analogous to Fig-ure 12.4(b). Finally, the equivalent problem of Figure 12.5(d) reduces to that of Figure 12.5(e), which is analogous to that of Figure 12.4(c). The original problem has been reduced to a very simple equivalent and (3-27)–(3-30a) can be utilized for its solution.
RADIATION EQUATIONS 645 σ = ∞ σ = ∞ σ = ∞ Waveguide Ea S (a) (b) (c) (d) (e) S S S S σ = ∞ σ = ∞ Js Ms = –n Ea Js Ms = –n Ea Ms = –n Ea Js = 0 Ms = –n Ea Js = 0 Ms = –2n Ea Js = 0 Ms = 0 Js Ms = 0 Js = 0 Ms = 0 Js = 0 Ms = 0 Js = 0 Ms = 0 Ms = 0 Js = 0 Ms = 0 Js = 0 Ms = 0 Js = 0 (image) n ^ n ^ n ^ n ^ n ^ ^ ^ ^ ^ ^ μ, μ, μ, μ, μ, μ, μ, μ, Figure 12.5 Equivalent models for waveguide aperture mounted on an infinite flat electric ground plane.
In this chapter the theory will be developed to compute the fields radiated by an aperture, like that shown in Figure 12.5(a), making use of its equivalent of Figure 12.5(e). For other problems, their equivalent forms will not necessarily be the same as that shown in Figure 12.5(e).
12.3 RADIATION EQUATIONS In Chapter 3 and in the previous section it was stated that the fields radiated by sources Js and Ms in an unbounded medium can be computed by using (3-27)–(3-30a) where the integration must be per-formed over the entire surface occupied by Js and Ms. These equations yield valid solutions for all observation points , . For most problems, the main difficulty is the inability to perform the inte-grations in (3-27) and (3-28). However for far-field observations, the complexity of the formulation can be reduced.
As was shown in Section 4.4.1, for far-field observations R can most commonly be approxi-mated by R ≃r −r′ cos 𝜓 for phase variations (12-5a) R ≃r for amplitude variations (12-5b) where 𝜓is the angle between the vectors r and r′, as shown in Figure 12.2(b). The primed coordinates (x′, y′, z′, or r′, 𝜃′, 𝜙′) indicate the space occupied by the sources Js and Ms, over which integration must be performed. The unprimed coordinates (x, y, z or r, 𝜃, 𝜙) represent the observation point.
Geometrically the approximation of (12-5a) assumes that the vectors R and r are parallel, as shown in Figure 12.2(b).
646 APERTURE ANTENNAS Using (12-5a) and (12-5b), (3-27) and (3-28) can be written as A = 𝜇 4𝜋∫∫ S Js e−jkR R ds′ ≃𝜇e−jkr 4𝜋r N (12-6) N = ∫∫ S Jsejkr′ cos 𝜓ds′ (12-6a) F = 𝜀 4𝜋∫∫ S Ms e−jkR R ds′ ≃𝜀e−jkr 4𝜋r L (12-7) L = ∫∫ S Msejkr′ cos 𝜓ds′ (12-7a) In Section 3.6 it was shown that in the far-field only the 𝜃and 𝜙components of the E- and H-fields are dominant. Although the radial components are not necessarily zero, they are negligible compared to the 𝜃and 𝜙components. Using (3-58a)–(3-59b), the EA of (3-29) and HF of (3-30) can be written as (EA)𝜃≃−j𝜔A𝜃 (12-8a) (EA)𝜙≃−j𝜔A𝜙 (12-8b) (HF)𝜃≃−j𝜔F𝜃 (12-8c) (HF)𝜙≃−j𝜔F𝜙 (12-8d) and the EF of (3-29) and HA of (3-30), with the aid of (12-8a)–(12-8d), as (EF)𝜃≃+𝜂(HF)𝜙= −j𝜔𝜂F𝜙 (12-9a) (EF)𝜙≃−𝜂(HF)𝜃= +j𝜔𝜂F𝜃 (12-9b) (HA)𝜃≃− (EA)𝜙 𝜂 = +j𝜔 A𝜙 𝜂 (12-9c) (HA)𝜙≃+(EA)𝜃 𝜂 = −j𝜔A𝜃 𝜂 (12-9d) Combining (12-8a)–(12-8d) with (12-9a)–(12-9d), and making use of (12-6)–(12-7a) the total E- and H-fields can be written as Er ≃0 E𝜃≃−jke−jkr 4𝜋r (L𝜙+ 𝜂N𝜃) E𝜙≃+jke−jkr 4𝜋r (L𝜃−𝜂N𝜙) Hr ≃0 H𝜃≃jke−jkr 4𝜋r ( N𝜃−L𝜃 𝜂 ) H𝜙≃−jke−jkr 4𝜋r ( N𝜃+ L𝜙 𝜂 ) (12-10a) (12-10b) (12-10c) (12-10d) (12-10e) (12-10f) SUMMARY 647 The N𝜃, N𝜙, L𝜃, and L𝜙can be obtained from (12-6a) and (12-7a). That is, N = ∫∫ S Jse+jkr′ cos 𝜓ds′ = ∫∫ S (̂ axJx + ̂ ayJy + ̂ azJz)e+jkr′ cos 𝜓ds′ (12-11a) L = ∫∫ S Mse+jkr′ cos 𝜓ds′ = ∫∫ S (̂ axMx + ̂ ayMy + ̂ azMz)e+jkr′ cos 𝜓ds′ (12-11b) Using the rectangular-to-spherical component transformation, obtained by taking the inverse (in this case also the transpose) of (4-5), (12-11a) and (12-11b) reduce for the 𝜃and 𝜙components to N𝜃= ∫∫ S [Jx cos 𝜃cos 𝜙+ Jy cos 𝜃sin 𝜙−Jz sin 𝜃]e+jkr′ cos 𝜓ds′ N𝜙= ∫∫ S [−Jx sin 𝜙+ Jy cos 𝜙]e+jkr′ cos 𝜓ds′ L𝜃= ∫∫ S [Mx cos 𝜃cos 𝜙+ My cos 𝜃sin 𝜙−Mz sin 𝜃]e+jkr′ cos 𝜓ds′ L𝜙= ∫∫ S [−Mx sin 𝜙+ My cos 𝜙]e+jkr′ cos 𝜓ds′ (12-12a) (12-12b) (12-12c) (12-12d) SUMMARY To summarize the results, the procedure that must be followed to solve a problem using the radiation integrals will be outlined. Figures 12.2(a) and 12.2(b) are used to indicate the geometry.
1. Select a closed surface over which the total electric and magnetic fields Ea and Ha are known.
2. Form the equivalent current densities Js and Ms over S using (12-3) and (12-4) with H1 = Ha and E1 = Ea.
3. Determine the A and F potentials using (12-6)–(12-7a) where the integration is over the closed surface S.
4. Determine the radiated E- and H-fields using (3-29) and (3-30).
The above steps are valid for all regions (near-field and far-field) outside the surface S. If, however, the observation point is in the far-field, steps 3 and 4 can be replaced by 3′ and 4′. That is, 3′. Determine N𝜃, N𝜙, L𝜃and L𝜙using (12-12a)–(12-12d).
4′. Determine the radiated E- and H-fields using (12-10a)–(12-10f).
Some of the steps outlined above can be reduced by a judicious choice of the equivalent model. In the remaining sections of this chapter, the techniques will be applied and demonstrated with examples of rectangular and circular apertures.
648 APERTURE ANTENNAS 12.4 DIRECTIVITY The directivity of an aperture can be found in a manner similar to that of other antennas. The primary task is to formulate the radiation intensity U(𝜃, 𝜙), using the far-zone electric and magnetic field components, as given by (2-12a) or U(𝜃, 𝜙) = 1 2Re[(̂ a𝜃E𝜃+ ̂ a𝜙E𝜙) × (̂ a𝜃H𝜃+ ̂ a𝜙H𝜙)∗] = 1 2𝜂(|E0 𝜃|2 + |E0 𝜙|2) (12-13) which in normalized form reduces to Un(𝜃, 𝜙) = (|E0 𝜃(𝜃, 𝜙)|2 + |E0 𝜙(𝜃, 𝜙)|2) = B0F(𝜃, 𝜙) (12-13a) The directive properties can then be found using (2-19)–(2-22).
Because the radiation intensity U(𝜃, 𝜙) for each aperture antenna will be of a different form, a general equation for the directivity cannot be formed. However, a general MATLAB and FORTRAN computer program, designated as Directivity, has been written to compute the directivity of any antenna, including an aperture, once the radiation intensity is specified. The program is based on the formulations of (12-13a), (2-19)–(2-20), and (2-22), and it is included in Chapter 2. In the main program, it requires the lower and upper limits on 𝜃and 𝜙. The radiation intensity for the antenna in question must be specified in the subroutine U(𝜃, 𝜙, F) of the program.
Expressions for the directivity of some simple aperture antennas, rectangular and circular, will be derived in later sections of this chapter.
12.5 RECTANGULAR APERTURES In practice, the rectangular aperture is probably the most common microwave antenna. Because of its configuration, the rectangular coordinate system is the most convenient system to express the fields at the aperture and to perform the integration. Shown in Figure 12.6 are the three most common and convenient coordinate positions used for the solution of an aperture antenna. In Figure 12.6(a) the aperture lies on the y-z plane, in Figure 12.6(b) on the x-z plane, and in Figure 12.6(c) on the x-y plane. For a given field distribution, the analytical forms for the fields for each of the arrangements are not the same. However the computed values will be the same, since the physical problem is identical in all cases.
For each of the geometries shown in Figure 12.6, the only difference in the analysis is in the formulation of 1. the components of the equivalent current densities (Jx, Jy, Jz, Mx, My, Mz) 2. the difference in paths from the source to the observation point (r′ cos 𝜓) 3. the differential area ds′ In general, the nonzero components of Js and Ms are Jy, Jz, My, Mz [Figure 12.6(a)] (12-14a) Jx, Jz, Mx, Mz [Figure 12.6(b)] (12-14b) Jx, Jy, Mx, My [Figure 12.6(c)] (12-14c) RECTANGULAR APERTURES 649 Figure 12.6 Rectangular aperture positions for antenna system analysis.
The differential paths take the form of r′ cos 𝜓= r′ ⋅̂ ar = (̂ ayy′ + ̂ azz′) ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = y′ sin 𝜃sin 𝜙+ z′ cos 𝜃 [Figure 12.6(a)] (12-15a) r′ cos 𝜓= r′ ⋅̂ ar = (̂ axx′ + ̂ azz′) ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = x′ sin 𝜃cos 𝜙+ z′ cos 𝜃 [Figure 12.6(b)] (12-15b) r′ cos 𝜓= r′ ⋅̂ ar = (̂ axx′ + ̂ ayy′) ⋅(̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃) = x′ sin 𝜃cos 𝜙+ y′ sin 𝜃sin 𝜙 [Figure 12.6(c)] (12-15c) 650 APERTURE ANTENNAS Figure 12.7 Rectangular aperture on an infinite electric ground plane.
and the differential areas are represented by ds′ = dy′ dz′ [Figure 12.6(a)] ds′ = dx′ dz′ [Figure 12.6(b)] ds′ = dx′ dy′ [Figure 12.6(c)] (12-16a) (12-16b) (12-16c) 12.5.1 Uniform Distribution on an Infinite Ground Plane The first aperture examined is a rectangular aperture mounted on an infinite ground plane, as shown in Figure 12.7. To reduce the mathematical complexities, initially the field over the opening is assumed to be constant and given by Ea = ̂ ayE0 −a∕2 ≤x′ ≤a∕2, −b∕2 ≤y′ ≤b∕2 (12-17) where E0 is a constant. The task is to find the fields radiated by it, the pattern beamwidths, the side lobe levels of the pattern, and the directivity. To accomplish these, the equivalent will be formed first.
A. Equivalent To form the equivalent, a closed surface is chosen which extends from −∞to +∞on the x-y plane.
Since the physical problem of Figure 12.7 is identical to that of Figure 12.5(a), its equivalents are RECTANGULAR APERTURES 651 those of Figures 12.5(a)–(e). Using the equivalent of Figure 12.5(e) Ms = { −2̂ n × Ea = −2̂ az × ̂ ayE0 = +̂ ax2E0 −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 0 elsewhere (12-18) Js = 0 everywhere B. Radiation Fields: Element and Space Factors The far-zone fields radiated by the aperture of Figure 12.7 can be found by using (12-10a)–(12-10f), (12-12a)–(12-12d), (12-14c), (12-15c), (12-16c), and (12-18). Thus, N𝜃= N𝜙= 0 (12-19) L𝜃= ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 [Mx cos 𝜃cos 𝜙]ejk(x′ sin 𝜃cos 𝜙+y′ sin 𝜃sin 𝜙) dx′ dy′ L𝜃= cos 𝜃cos 𝜙 [ ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Mxejk(x′ sin 𝜃cos 𝜙+y′ sin 𝜃sin 𝜙) dx′ dy′ ] (12-19a) In (12-19a), the integral within the brackets represents the space factor for a two-dimensional distribution. It is analogous to the space factor of (4-58a) for a line source (one-dimensional distri-bution). For the L𝜃component of the vector potential F, the element factor is equal to the product of the factor outside the brackets in (12-19a) and the factor outside the brackets in (12-10c). The total field is equal to the product of the element and space factors, as defined by (4-59), and expressed in (12-10b) and (12-10c).
Using the integral ∫ +c∕2 −c∕2 ej𝛼z dz = c ⎡ ⎢ ⎢ ⎢ ⎣ sin (𝛼 2 c ) 𝛼 2 c ⎤ ⎥ ⎥ ⎥ ⎦ (12-20) (12-19a) reduces to L𝜃= 2abE0 [ cos 𝜃cos 𝜙 (sin X X ) (sin Y Y )] (12-21) where X = ka 2 sin 𝜃cos 𝜙 (12-21a) Y = kb 2 sin 𝜃sin 𝜙 (12-21b) Similarly it can be shown that L𝜙= −2abE0 [ sin 𝜙 (sin X X ) (sin Y Y )] (12-22) 652 APERTURE ANTENNAS Substituting (12-19), (12-21), and (12-22) into (12-10a)–(12-10f), the fields radiated by the aperture can be written as Er = 0 (12-23a) E𝜃= jabkE0e−jkr 2𝜋r [ sin 𝜙 (sin X X ) (sin Y Y )] (12-23b) E𝜙= jabkE0e−jkr 2𝜋r [ cos 𝜃cos 𝜙 (sin X X ) (sin Y Y )] (12-23c) Hr = 0 (12-23d) H𝜃= − E𝜙 𝜂 (12-23e) H𝜙= +E𝜃 𝜂 (12-23f) Equations (12-23a)–(12-23f) represent the three-dimensional distributions of the far-zone fields radiated by the aperture. Experimentally only two-dimensional plots can be measured. To reconstruct experimentally a three-dimensional plot, a series of two-dimensional plots must be made. In many applications, however, only a pair of two-dimensional plots are usually sufficient. These are the principal E- and H-plane patterns whose definition was stated in Section 2.2.3 and illustrated in Figure 2.3.
For the problem in Figure 12.7, the E-plane pattern is on the y-z plane (𝜙= 𝜋∕2) and the H-plane is on the x-z plane (𝜙= 0). Thus E−Plane (𝝓= 𝝅∕2) Er = E𝜙= 0 (12-24a) E𝜃= jabkE0e−jkr 2𝜋r ⎡ ⎢ ⎢ ⎢ ⎣ sin (kb 2 sin 𝜃 ) kb 2 sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ (12-24b) H−Plane (𝝓= 0) Er = E𝜃= 0 (12-25a) E𝜙= jabkE0e−jkr 2𝜋r ⎧ ⎪ ⎨ ⎪ ⎩ cos 𝜃 ⎡ ⎢ ⎢ ⎢ ⎣ sin (ka 2 sin 𝜃 ) ka 2 sin 𝜃 ⎤ ⎥ ⎥ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭ (12-25b) To demonstrate the techniques, three-dimensional patterns have been plotted in Figures 12.8 and 12.9. The dimensions of the aperture are indicated in each figure. Multiple lobes appear, because the dimensions of the aperture are greater than one wavelength. The number of lobes increases as the dimensions increase. For the aperture whose dimensions are a = 3λ and b = 2λ (Figure 12.8), there are a total of five lobes in the principal H-plane and three lobes in the principal E-plane. The pattern in the H-plane is only a function of the dimension a whereas that in the E-plane is only influenced RECTANGULAR APERTURES 653 Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ θ ϕ Relative Amplitude 90° 90° H-plane (x-z plane, = 0°) ϕ E-plane (y-z plane, = 90°) Figure 12.8 Three-dimensional field pattern of a constant field rectangular aperture mounted on an infinite ground plane (a = 3λ, b = 2λ).
Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ Relative Amplitude θ 90° 90° H-plane (x-z plane, ϕ = 0°) E-plane (y-z plane, ϕ = 90°) Figure 12.9 Three-dimensional field pattern of a constant field square aperture mounted on an infinite ground plane (a = b = 3λ).
654 APERTURE ANTENNAS Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 E-plane H-plane θ θ Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Figure 12.10 E- and H-plane amplitude patterns for uniform distribution aperture mounted on an infinite ground plane (a = 3λ, b = 2λ).
by b. In the E-plane, the side lobe formed on each side of the major lobe is a result of λ < b ≤2λ. In the H-plane, the first minor lobe on each side of the major lobe is formed when λ < a ≤2λ and the second side lobe when 2λ < a ≤3λ. Additional lobes are formed when one or both of the aperture dimensions increase. This is illustrated in Figure 12.9 for an aperture with a = b = 3λ.
The two-dimensional principal plane patterns for the aperture with a = 3λ, b = 2λ are shown in Figure 12.10. For this, and for all other size apertures mounted on an infinite ground plane, the H-plane patterns along the ground plane vanish. This is dictated by the boundary conditions. The E-plane patterns, in general, do not have to vanish along the ground plane, unless the dimension of the aperture in that plane (in this case b) is a multiple of a wavelength.
The patterns computed above assumed that the aperture was mounted on an infinite ground plane.
In practice, infinite ground planes are not realizable, but they can be approximated by large structures.
Edge effects, on the patterns of apertures mounted on finite size ground planes, can be accounted for by diffraction techniques. They will be introduced and illustrated in Section 12.9. Computed results, which include diffractions, agree extremely well with measurements –.
C. Beamwidths For the E-plane pattern given by (12-24b), the maximum radiation is directed along the z-axis (𝜃= 0). The nulls (zeros) occur when kb 2 sin 𝜃|𝜃=𝜃n = n𝜋, n = 1, 2, 3, … (12-26) RECTANGULAR APERTURES 655 or at the angles of 𝜃n = sin−1 (2n𝜋 kb ) = sin−1 (nλ b ) rad (12-26a) = 57.3 sin−1 (nλ b ) degrees, n = 1, 2, 3, … If b ≫nλ, (12-26a) reduces approximately to 𝜃n ≃nλ b rad = 57.3 (nλ b ) degrees, n = 1, 2, 3, … (12-26b) The total beamwidth between nulls is given by Θn = 2𝜃n = 2 sin−1 (nλ b ) rad = 114.6 sin−1 (nλ b ) degrees, n = 1, 2, 3, … (12-27) or approximately (for large apertures, b ≫nλ) by Θn ≃2nλ b rad = 114.6 (nλ b ) degrees, n = 1, 2, 3, … (12-27a) The first-null beamwidth (FNBW) is obtained by letting n = 1.
The half-power point occurs when (see Appendix I) kb 2 sin 𝜃|𝜃=𝜃h = 1.391 (12-28) or at an angle of 𝜃h = sin−1 (2.782 kb ) = sin−1 (0.443λ b ) rad = 57.3 sin−1 (0.443λ b ) degrees (12-28a) If b ≫0.443λ, (12-28a) reduces approximately to 𝜃h ≃ ( 0.443λ b ) rad = 25.38 (λ b ) degrees (12-28b) Thus the total half-power beamwidth (HPBW) is given by Θh = 2𝜃h = 2 sin−1 (0.443λ b ) rad = 114.6 sin−1 (0.443λ b ) degrees (12-29) 656 APERTURE ANTENNAS or approximately (when b ≫0.443λ) by Θh ≃ ( 0.886λ b ) rad = 50.8 (λ b ) degrees (12-29a) The maximum of the first side lobe occurs when (see Appendix I) kb 2 sin 𝜃|𝜃=𝜃s = 4.494 (12-30) or at an angle of 𝜃s = sin−1 (8.988 kb ) = sin−1 (1.43λ b ) rad = 57.3 sin−1 (1.43λ b ) degrees (12-30a) If b ≫1.43λ, (12-30a) reduces to 𝜃s ≃1.43 (λ b ) rad = 81.9 (λ b ) degrees (12-30b) The total beamwidth between first side lobes (FSLBW) is given by Θs = 2𝜃s = 2 sin−1 (1.43λ b ) rad = 114.6 sin−1 (1.43λ b ) degrees (12-30c) or approximately (when b ≫1.43λ) by Θs ≃2.86 (λ b ) rad = 163.8 (λ b ) degrees (12-30d) D. Side Lobe Level The maximum of (12-24b) at the first side lobe is given by (see Appendix I) |E𝜃(𝜃= 𝜃s)| = | | | | sin(4.494) 4.494 | | | | = 0.217 = −13.26 dB (12-31) which is 13.26 dB down from the maximum of the main lobe.
An approximate value of the maximum of the first side lobe can be obtained by assuming that the maximum of (12-24b) occurs when its numerator is maximum. That is, when kb 2 sin 𝜃|𝜃=𝜃s ≃3𝜋 2 (12-32) Thus, |E𝜃(𝜃= 𝜃s)| ≃ 1 3𝜋∕2 = 0.212 = −13.47 dB (12-33) These values are very close to the exact ones given by (12-31).
RECTANGULAR APERTURES 657 A similar procedure can be followed to find the nulls, 3-dB points, beamwidth between nulls and 3-dB points, angle where the maximum of first side lobe occurs, and its magnitude at that point for the H-plane pattern of (12-25b). A comparison between the E- and H-plane patterns of (12-24b) and (12-25b) shows that they are similar in form except for the additional cos 𝜃term that appears in (12-25b). An examination of the terms in (12-25b) reveals that the cos 𝜃term is a much slower varying function than the sin(ka sin 𝜃∕2)∕(ka sin 𝜃∕2) term, especially when a is large.
As a first approximation, (12-26)–(12-33), with b replaced by a, can also be used for the H-plane.
More accurate expressions can be obtained by also including the cos 𝜃term. In regions well removed from the major lobe, the inclusion of the cos 𝜃term becomes more essential for accurate results.
E. Directivity The directivity for the aperture can be found using (12-23a)–(12-23c), (12-13)–(12-13a), and (2-19)–(2-22). The analytical details using this procedure, especially the integration to compute the radiated power (Prad), are more cumbersome.
Because the aperture is mounted on an infinite ground plane, an alternate and much simpler method can be used to compute the radiated power. The average power density is first formed using the fields at the aperture, and it is then integrated over the physical bounds of the opening. The inte-gration is confined to the physical bounds of the opening. Using Figure 12.7 and assuming that the magnetic field at the aperture is given by Ha = −̂ ax E0 𝜂 (12-34) where 𝜂is the intrinsic impedance, the radiated power reduces to Prad = ∯ S Wav ⋅ds = |E0|2 2𝜂 ∫∫ Sa ds = ab|E0|2 2𝜂 (12-35) The maximum radiation intensity (Umax), using the fields of (12-23a)–(12-23b), occurs toward 𝜃= 0◦and it is equal to Umax = (ab λ )2 |E0|2 2𝜂 (12-36) Thus the directivity is equal to D0 = 4𝜋Umax Prad = 4𝜋 λ2 ab = 4𝜋 λ2 Ap = 4𝜋 λ2 Aem (12-37) where Ap = physical area of the aperture Aem = maximum effective area of the aperture Using the definition of (2-110), it is shown that the physical and maximum effective areas of a constant distribution aperture are equal.
The beamwidths, side lobe levels, and directivity of this and other apertures are summarized in Table 12.1.
TABLE 12.1 Equivalents, Fields, Beamwidths, Side Lobe Levels, and Directivities of Rectangular Apertures Uniform Distribution Aperture Uniform Distribution Aperture TE10-Mode Distribution Aperture on Ground Plane in Free-Space on Ground Plane Aperture distribution of tangen-tial components (analytical) Ea = ̂ ayE0 } −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 Ea = ̂ ayE0 Ha = −̂ ax E0 𝜂 ⎫ ⎪ ⎬ ⎪ ⎭ −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 Ea = ̂ ayE0 cos (𝜋 a x′) } −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 Aperture distribution of tangen-tial components (graphical) Equivalent Ms = { −2̂ n × Ea 0 } −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 elsewhere Js = 0 everywhere Ms = −̂ n × Ea Js = ̂ n × Ha ⎫ ⎪ ⎬ ⎪ ⎭ −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 Ms ≃Js ≃0 elsewhere Ms = { −2̂ n × Ea 0 } −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 elsewhere Js = 0 everywhere Far-zone fields Er = Hr = 0 Er = Hr = 0 Er = Hr = 0 X = ka 2 sin 𝜃cos 𝜙 E𝜃= C sin 𝜙sin X X sin Y Y E𝜃= C 2 sin 𝜙(1 + cos 𝜃)sin X X sin Y Y E𝜃= −𝜋 2 C sin 𝜙 cos X (X)2 − (𝜋 2 )2 sin Y Y Y = kb 2 sin 𝜃sin 𝜙 E𝜙= C cos 𝜃cos 𝜙sin X X sin Y Y E𝜙= C 2 cos 𝜙(1 + cos 𝜃)sin X X sin Y Y E𝜙= −𝜋 2 C cos 𝜃cos 𝜙 cos X (X)2 − (𝜋 2 )2 sin Y Y C = jabkE0e−jkr 2𝜋r H𝜃= −E𝜙∕𝜂 H𝜙= E𝜃∕𝜂 H𝜃= −E𝜙∕𝜂 H𝜙= E𝜃∕𝜂 H𝜃= −E𝜙∕𝜂 H𝜙= E𝜃∕𝜂 658 Half-power beamwidth (degrees) E-plane b ≫λ 50.8 b∕λ 50.8 b∕λ 50.8 b∕λ H-plane a ≫λ 50.8 a∕λ 50.8 a∕λ 68.8 a∕λ First null beamwidth (degrees) E-plane b ≫λ 114.6 b∕λ 114.6 b∕λ 114.6 b∕λ H-plane a ≫λ 114.6 a∕λ 114.6 a∕λ 171.9 a∕λ First side lobe max. (to main max.) (dB) E-plane −13.26 −13.26 −13.26 H-plane −13.26 a ≫λ −13.26 a ≫λ −23 a ≫λ Directivity D0 (dimensionless) 4𝜋 λ2 (area) = 4𝜋 (ab λ2 ) 4𝜋 λ2 (area) = 4𝜋 (ab λ2 ) 8 𝜋2 [ 4𝜋 (ab λ2 )] = 0.81 [ 4𝜋 (ab λ2 )] 659 660 APERTURE ANTENNAS Example 12.2 A rectangular aperture with a constant field distribution, with a = 3λ and b = 2λ, is mounted on an infinite ground plane. Compute the a. FNBW in the E-plane b. HPBW in the E-plane c. FSLBW in the E-plane d. FSLMM in the E-plane e. directivity using (12-37) f. directivity using the computer program Directivity at the end of Chapter 2, the fields of (12-23a)–(12-23f), and the formulation of Section 12.4 Solution: a. Using (12-27) Θ1 = 114.6 sin−1( 1 2) = 114.6(0.524) = 60◦ b. Using (12-29) Θh = 114.6 sin−1 (0.443 2 ) = 114.6(0.223) = 25.6◦ c. Using (12-30c) Θs = 2𝜃s = 114.6 sin−1 (1.43 2 ) = 114.6(0.796) = 91.3◦ d. Using (12-31) |E𝜃|𝜃=𝜃s = 0.217 ≃−13.26 dB e. Using (12-37) D0 = 4𝜋(3)(2) = 75.4 = 18.77 dB f. Using the computer program at the end of Chapter 2 D0 ≃80.4 = 19.05 dB The difference in directivity values using (12-37) and the computer program is not attributed to the accuracy of the numerical method. The main contributor is the aperture tangential magnetic field of (12-34), which was assumed to be related to the aperture tangential electric field by the intrinsic impedance. Although this is a good assumption for large size apertures, it is not exact.
Therefore the directivity value computed using the computer program should be considered to be the more accurate.
RECTANGULAR APERTURES 661 12.5.2 Uniform Distribution in Space The second aperture examined is that of Figure 12.7 when it is not mounted on an infinite ground plane. The field distribution is given by Ea = ̂ ayE0 Ha = −̂ ax E0 𝜂 ⎫ ⎪ ⎬ ⎪ ⎭ −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 (12-38) where E0 is a constant. The geometry of the opening for this problem is identical to the previous one. However the equivalents and radiated fields are different, because this time the aperture is not mounted on an infinite ground plane.
A. Equivalent To form the equivalent, a closed surface is chosen which again extends from −∞to +∞on the x-y plane. Over the entire surface Js and Ms are formed. The difficulty encountered in this problem is that both Js and Ms are not zero outside the opening, and expressions for them are not known there.
The replacement of the semi-infinite medium to the left of the boundary (negative z) by an imaginary electric or magnetic conductor only eliminates one or the other current densities (Js or Ms) but not both. Thus, even though an exact equivalent for this problem exists in principle, it cannot be used practically because the fields outside the opening are not known a priori. We are therefore forced to adopt an approximate equivalent.
The usual and most accurate relaxation is to assume that both Ea and Ha (and in turn Ms and Js) exist over the opening but are zero outside it. It has been shown, by comparison with measurements and other available data, that this approximate equivalent yields the best results.
B. Radiated Fields Using a procedure similar to that of the previous section, the radiation characteristics of this aperture can be derived. A summary of them is shown in Table 12.1.
The field components of this aperture are identical in form to those of the aperture when it is mounted on an infinite ground plane if the (1 + cos 𝜃) term in each component is replaced by 2.
Thus for small values of 𝜃(in the main lobe and especially near its maximum), the patterns of the two apertures are almost identical. This procedure can be used, in general, to relate the fields of an aperture when it is and it is not mounted on an infinite ground plane. However, the coordinate system chosen must have the z-axis perpendicular to the aperture.
A three-dimensional pattern for an aperture with a = 3λ, b = 2λ was computed, and it is shown in Figure 12.11. The dimensions of this aperture are the same as those of Figure 12.8. However the angular limits over which the radiated fields now exist have been extended to 0◦≤𝜃≤180◦.
Although the general structures of the two patterns are similar, they are not identical. Because of the enlarged space over which fields now exist, additional minor lobes are formed.
C. Beamwidths and Side Lobe Levels To find the beamwidths and the angle at which the maximum of the side lobe occurs, it is usually assumed that the (1 + cos 𝜃) term is a much slower varying function than the sin(ka sin 𝜃∕2)∕(ka sin 𝜃∕2) or the sin(kb sin 𝜃∕2)∕(kb sin 𝜃∕2) terms. This is an approximation, and it is more valid for large apertures (large a and/or b) and for angles near the main maximum. More accurate results can be obtained by considering the (1 + cos 𝜃) term. Thus (12-26)–(12-33) can be used, to a good approximation, to compute the beamwidths and side lobe level. A summary is included in Table 12.1.
662 APERTURE ANTENNAS Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ Relative Amplitude 180° 180° H-plane (x-z plane, ϕ θ ϕ = 0°) E-plane (y-z plane, = 90°) Figure 12.11 Three-dimensional field pattern of a constant field rectangular aperture (a = 3λ, b = 2λ).
D. Directivity Although the physical geometry of the opening of this problem is identical to that of Section 12.5.1, their directivities are not identical. This is evident by examining their far-zone field expressions or by realizing that the fields outside the aperture along the x-y plane are not exactly the same.
To derive an exact expression for the directivity of this aperture would be a very difficult task.
Since the patterns of the apertures are nearly the same, especially at the main lobe, their directivities are almost the same. To verify this, an example is taken.
Example 12.3 Repeat the problem of Example 12.2 for an aperture that is not mounted on an infinite ground plane.
Solution: Since the E-plane patterns of the two apertures are identical, the FNBW, HPBW, FSLBW, and FSLMM are the same. The directivities as computed by (12-37), are also the same.
Since the fields radiated by the two apertures are not identical, their directivities computed using the far-zone fields will not be exactly the same. Therefore for this problem D0 ≃81.16(dimensionless) = 19.09 dB As with Example 12.2, the directivities computed using (12-37) and the computer program do not agree exactly. For this problem, however, neither one is exact. For (12-37), it has been assumed that the aperture tangential magnetic field is related to the aperture tangential electric field by the intrinsic impedance 𝜂. This relationship is good but not exact. For the computer program, the formulation is based on the equivalent of this section where the fields outside the aperture were assumed to be negligible. Again this is a good assumption for some problems, but it is not exact.
RECTANGULAR APERTURES 663 A summary of the radiation characteristics of this aperture is included in Table 12.1 where it is compared with that of other apertures.
12.5.3 TE10-Mode Distribution on an Infinite Ground Plane In practice, a commonly used aperture antenna is that of a rectangular waveguide mounted on an infinite ground plane. At the opening, the field is usually approximated by the dominant TE10-mode. Thus Ea = ̂ ayE0 cos (𝜋 a x′) { −a∕2 ≤x′ ≤+a∕2 −b∕2 ≤y′ ≤+b∕2 (12-39) A. Equivalent, Radiated Fields, Beamwidths, and Side Lobe Levels Because the physical geometry of this antenna is identical to that of Figure 12.7, their equivalents and the procedure to analyze each one are identical. They differ only in the field distribution over the aperture.
The details of the analytical formulation are not included. However, a summary of its radia-tion characteristics is included in Table 12.1. The E-plane pattern of this aperture is identical in form (with the exception of a normalization factor) to the E-plane of the aperture of Section 12.5.1.
This is expected, since the TE10-mode field distribution along the E-plane (y-z plane) is also a con-stant. That is not the case for the H-plane or at all other points removed from the principal planes.
To demonstrate that, a three-dimensional pattern for the TE10-mode aperture with a = 3λ, b = 2λ was computed and it is shown in Figure 12.12. This pattern should be compared with that of Figure 12.8.
The expressions for the beamwidths and side lobe level in the E-plane are identical to those given by (12-26)–(12-33). However those for the H-plane are more complex, and a simple procedure is Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Relative Amplitude ° 0 9 90° H-plane (x-z plane, = 0°) E-plane (y-z plane, = 90°) θ ϕ ϕ θ Figure 12.12 Three-dimensional field pattern of a TE10-mode rectangular waveguide mounted on an infinite ground plane (a = 3λ, b = 2λ).
664 APERTURE ANTENNAS Figure 12.13 E-plane beamwidths and first side lobe relative maximum magnitude for TE10-mode rectangular waveguide mounted on an infinite ground plane.
not available. Computations for the HPBW, FNBW, FSLBW, FSLMM in the E- and H-planes were made, and they are shown graphically in Figures 12.13 and 12.14.
When the same aperture is not mounted on a ground plane, the far-zone fields do not have to be re-derived but rather can be written by inspection. This is accomplished by introducing appropriately, in each of the field components (E𝜃and E𝜙) of the fourth column of Table 12.1, a (1 + cos 𝜃)/2 factor, as is done for the fields of the two apertures in the second and third columns. This factor is appropriate when the z-axis is perpendicular to the plane of the aperture. Other similar factors will have to be used when either the x-axis or y-axis is perpendicular to the plane of the aperture.
B. Directivity and Aperture Efficiency The directivity of this aperture is found in the same manner as that of the uniform distribution aper-ture of Section 12.5.1. Using the aperture electric field of (12-39), and assuming that the aperture magnetic field is related to the electric field by the intrinsic impedance 𝜂, the radiated power can be written as Prad = ∯ S Wav ⋅ds = ab|E0|2 4𝜂 (12-39a) The maximum radiation intensity occurs at 𝜃= 0◦, and it is given by Umax = 8 𝜋2 (ab λ )2 |E0|2 4𝜂 (12-39b) RECTANGULAR APERTURES 665 Figure 12.14 H-plane beamwidths and first side lobe relative maximum magnitude for TE10-mode rectangular waveguide mounted on an infinite ground plane.
Thus the directivity is equal to D0 = 8 𝜋2 [ ab (4𝜋 λ2 )] = 0.81 [ ab (4𝜋 λ2 )] = 0.81Ap (4𝜋 λ2 ) = Aem (4𝜋 λ2 ) (12-39c) In general, the maximum effective area Aem is related to the physical area Ap by Aem = 𝜀apAp, 0 ≤𝜀ap ≤1 (12-40) where 𝜀ap is the aperture efficiency. For this problem 𝜀ap = 8∕𝜋2 ≃0.81. The aperture efficiency is a figure of merit which indicates how efficiently the physical area of the antenna is utilized. Typi-cally, aperture antennas have aperture efficiencies from about 30% to 90%, horns from 35% to 80% (optimum gain horns have 𝜀ap ≃50%), and circular reflectors from 50% to 80%.
For reflectors, the aperture efficiency is a function of many factors. The most prominent are the spillover, amplitude taper, phase distribution, polarization uniformity, blockage, and surface random errors. These are discussed in detail in Section 15.4.1 of Chapter 15.
666 APERTURE ANTENNAS Figure 12.15 Beam efficiency versus half-cone angle 𝜃1, for a square aperture with different field distributions.
The aperture is not mounted on an infinite ground plane. (source: D. G. Fink (ed.), Electronics Engineers’ Handbook, Section 18 (by W. F. Croswell), McGraw-Hill, New York, 1975).
12.5.4 Beam Efficiency The beam efficiency for an antenna was introduced in Section 2.10 and was defined by (2-53). When the aperture is mounted on the x-y plane, the beam efficiency can be calculated using (2-54). The beam efficiency can be used to judge the ability of the antenna to discriminate between signals received through its main lobe and those through the minor lobes. Beam efficiencies for rectangular apertures with different aperture field distributions are plotted, versus the half-cone angle 𝜃1, in Figure 12.15 . The uniform field distribution aperture has the least ability to discriminate between main lobe and minor lobe signals. The aperture radiates in an unbounded medium, and it is not mounted on an infinite ground plane. The lower abscissa scale is in terms of 𝜃1 (in degrees), and it should be used only when a = b = 20λ. The upper abscissa scale is in terms of u[u = (ka∕2) sin 𝜃1 = (kb∕2) sin 𝜃1], and it should be used for any square aperture.
Example 12.4 Determine the beam efficiency, within a cone of half-angle 𝜃1 = 10◦, for a square aperture with uniform field distribution and with a. a = b = 20λ b. a = b = 3λ CIRCULAR APERTURES 667 Solution: The solution is carried out using the curves of Figure 12.15.
a. When a = b = 20λ, the lower abscissa scale can be used. For 𝜃1 = 10◦, the efficiency for the uniform aperture is about 94%.
b. For a = b = 3λ and 𝜃1 = 10◦ u = ka 2 sin 𝜃1 = 3𝜋sin(10◦) = 1.64 Using the upper abscissa scale, the efficiency for the uniform aperture at u = 1.64 is about 58%.
An antenna array of slotted rectangular waveguides used for the AWACS airborne system is shown in Figure 6.27. It utilizes waveguide sticks, with slits on their narrow wall.
A MATLAB computer program, designated as Aperture, has been developed to compute and display different radiation characteristics of rectangular and circular apertures. The description of the program is found in the corresponding READ ME file included in the publisher’s website for this book.
12.6 CIRCULAR APERTURES A widely used microwave antenna is the circular aperture. One of the attractive features of this configuration is its simplicity in construction. In addition, closed form expressions for the fields of all the modes that can exist over the aperture can be obtained.
The procedure followed to determine the fields radiated by a circular aperture is identical to that of the rectangular, as summarized in Section 12.3. The primary differences lie in the formulation of the equivalent current densities (Jx, Jy, Jz, Mx, My, Mz), the differential paths from the source to the observation point (r′ cos 𝜓), and the differential area (ds′). Before an example is considered, these differences will be reformulated for the circular aperture.
Because of the circular profile of the aperture, it is often convenient and desirable to adopt cylindri-cal coordinates for the solution of the fields. In most cases, therefore, the electric and magnetic field components over the circular opening will be known in cylindrical form; that is, E𝜌, E𝜙, Ez, H𝜌, H𝜙, and Hz. Thus the components of the equivalent current densities Ms and Js would also be conve-niently expressed in cylindrical form (M𝜌, M𝜙, Mz, J𝜌, J𝜙, Jz). In addition, the required integration over the aperture to find N𝜃, N𝜙, L𝜃, and L𝜙of (12-12a)–(12-12d) should also be done in cylindrical coordinates. It is then desirable to reformulate r′ cos 𝜓and ds′, as given by (12-15a)–(12-16c).
The most convenient position for placing the aperture is that shown in Figure 12.16 (aperture on x-y plane). The transformation between the rectangular and cylindrical components of Js is given by (see Appendix VII) [ Jx Jy Jz ] = [ cos 𝜙′ −sin 𝜙′ 0 sin 𝜙′ cos 𝜙′ 0 0 0 1 ] [ J𝜌 J𝜙 Jz ] (12-41a) A similar transformation exists for the components of Ms. The rectangular and cylindrical coordi-nates are related by (see Appendix VII) x′ = 𝜌′ cos 𝜙′ y′ = 𝜌′ sin 𝜙′ (12-41b) z′ = z′ 668 APERTURE ANTENNAS z y x a r′ = p′ p′d ′ ϕ r R dp′ θ θ′ ψ ϕ′ ϕ Figure 12.16 Circular aperture mounted on an infinite ground plane.
Using (12-41a), (12-12a)–(12-12d) can be written as N𝜃= ∫∫ S [J𝜌cos 𝜃cos(𝜙−𝜙′) + J𝜙cos 𝜃sin(𝜙−𝜙′) −Jz sin 𝜃] × e+jkr′ cos 𝜓ds′ N𝜙= ∫∫ S [−J𝜌sin(𝜙−𝜙′) + J𝜙cos(𝜙−𝜙′)]e+jkr′ cos 𝜓ds′ L𝜃= ∫∫ S [M𝜌cos 𝜃cos(𝜙−𝜙′) + M𝜙cos 𝜃sin(𝜙−𝜙′) −Mz sin 𝜃] × e+jkr′ cos 𝜓ds′ L𝜙= ∫∫ S [−M𝜌sin(𝜙−𝜙′) + M𝜙cos(𝜙−𝜙′)]e+jkr′ cos 𝜓ds′ (12-42a) (12-42b) (12-42c) (12-42d) where r′ cos 𝜓and ds′ can be written, using (12-15c) and (12-41b), as r′ cos 𝜓= x′ sin 𝜃cos 𝜙+ y′ sin 𝜃sin 𝜙= 𝜌′ sin 𝜃cos(𝜙−𝜙′) ds′ = dx′ dy′ = 𝜌′ d𝜌′ d𝜙′ (12-43a) (12-43b) CIRCULAR APERTURES 669 In summary, for a circular aperture antenna the fields radiated can be obtained by either of the fol-lowing: 1. If the fields over the aperture are known in rectangular components, use the same procedure as for the rectangular aperture with (12-43a) and (12-43b) substituted in (12-12a)–(12-12d).
2. If the fields over the aperture are known in cylindrical components, use the same procedure as for the rectangular aperture with (12-42a)–(12-42d), along with (12-43a) and (12-43b), taking the place of (12-12a)–(12-12d).
12.6.1 Uniform Distribution on an Infinite Ground Plane To demonstrate the methods, the field radiated by a circular aperture mounted on an infinite ground plane will be formulated. To simplify the mathematical details, the field over the aperture is assumed to be constant and given by Ea = ̂ ayE0 𝜌′ ≤a (12-44) where E0 is a constant.
A. Equivalent and Radiation Fields The equivalent problem of this is identical to that of Figure 12.7. That is, Ms = { −2̂ n × Ea = ̂ ax2E0 𝜌′ ≤a 0 elsewhere ⎫ ⎪ ⎬ ⎪ ⎭ Js = 0 everywhere (12-45) Thus, N𝜃= N𝜙= 0 (12-46) L𝜃= 2E0 cos 𝜃cos 𝜙∫ a 0 𝜌′ [ ∫ 2𝜋 0 e+jk𝜌′ sin 𝜃cos(𝜙−𝜙′) d𝜙′ ] d𝜌′ (12-47) Because ∫ 2𝜋 0 e+jk𝜌′ sin 𝜃cos(𝜙−𝜙′) d𝜙′ = 2𝜋J0(k𝜌′ sin 𝜃) (12-48) (12-47) can be written as L𝜃= 4𝜋E0 cos 𝜃cos 𝜙∫ a 0 J0(k𝜌′ sin 𝜃)𝜌′ d𝜌′ (12-49) where J0(t) is the Bessel function of the first kind of order zero. Making the substitution t = k𝜌′ sin 𝜃 dt = k sin 𝜃d𝜌′ (12-49a) 670 APERTURE ANTENNAS reduces (12-49) to L𝜃= 4𝜋E0 cos 𝜃cos 𝜙 (k sin 𝜃)2 ∫ ka sin 𝜃 0 tJ0(t) dt (12-49b) Since ∫ 𝛽 0 zJ0(z) dz = zJ1(z) | | | | 𝛽 0 = 𝛽J1(𝛽) (12-50) where J1(𝛽) is the Bessel function of order one, (12-49b) takes the form of L𝜃= 4𝜋a2E0 { cos 𝜃cos 𝜙 [J1(ka sin 𝜃) ka sin 𝜃 ]} (12-51) Similarly L𝜙= −4𝜋a2E0 sin 𝜙 [J1(ka sin 𝜃) ka sin 𝜃 ] (12-52) Using (12-46), (12-51), and (12-52), the electric field components of (12-10a)–(12-10c) can be written as Er = 0 (12-53a) E𝜃= jka2E0e−jkr r { sin 𝜙 [J1(ka sin 𝜃) ka sin 𝜃 ]} (12-53b) E𝜙= jka2E0e−jkr r { cos 𝜃cos 𝜙 [J1(ka sin 𝜃) ka sin 𝜃 ]} (12-53c) In the principal E- and H-planes, the electric field components simplify to E-Plane (𝝓= 𝝅∕2) Er = E𝜙= 0 (12-54a) E𝜃= jka2E0e−jkr r [J1(ka sin 𝜃) ka sin 𝜃 ] (12-54b) H-Plane (𝝓= 0) Er = E𝜃= 0 (12-55a) E𝜙= jka2E0e−jkr r { cos 𝜃 [J1(ka sin 𝜃) ka sin 𝜃 ]} (12-55b) CIRCULAR APERTURES 671 θ θ ϕ ϕ Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Relative Amplitude 90° 90° H-plane (x-z plane, = 0°) E-plane (y-z plane, = 90°) Figure 12.17 Three-dimensional field pattern of a constant field circular aperture mounted on an infinite ground plane (a = 1.5λ).
A three-dimensional pattern has been computed for the constant field circular aperture of a = 1.5λ, and it is shown in Figure 12.17. The pattern of Figure 12.17 seems to be symmetrical. However closer observation, especially through the two-dimensional E- and H-plane patterns, will reveal that not to be the case. It does, however, possess characteristics that are almost symmetrical.
B. Beamwidth, Side Lobe Level, and Directivity Exact expressions for the beamwidths and side lobe levels cannot be obtained easily. However approximate expressions are available, and they are shown tabulated in Table 12.2. More exact data can be obtained by numerical methods.
Since the field distribution over the aperture is constant, the directivity is given by D0 = 4𝜋 λ2 Aem = 4𝜋 λ2 Ap = 4𝜋 λ2 (𝜋a2) = (2𝜋a λ )2 = (C λ )2 (12-56) since the maximum effective area Aem is equal to the physical area Ap of the aperture [as shown for the rectangular aperture in (12-37)].
A summary of the radiation parameters of this aperture is included in Table 12.2.
12.6.2 TE11-Mode Distribution on an Infinite Ground Plane A very practical antenna is a circular waveguide of radius a mounted on an infinite ground plane, as shown in Figure 12.16. However, the field distribution over the aperture is usually that of the TABLE 12.2 Equivalents, Fields, Beamwidths, Side Lobe Levels, and Directivities of Circular Apertures Uniform Distribution Aperture TE11-Mode Distribution Aperture on Ground Plane on Ground Plane Aperture distribution of tangen-tial components (analytical) Ea = ̂ ayE0 𝜌′ ≤a Ea = ̂ a𝜌E𝜌+ ̂ a𝜙E𝜙 E𝜌= E0J1(𝜒′ 11𝜌′∕a) sin 𝜙′∕𝜌′ E𝜙= E0J1 ′(𝜒′ 11𝜌′∕a) cos 𝜙′ ⎫ ⎪ ⎬ ⎪ ⎭ 𝜌′ ≤a 𝜒′ 11 = 1.841 ′ = 𝜕 𝜕𝜌′ Aperture distribution of tangential components (graphical) Equivalent Ms = ⎧ ⎪ ⎨ ⎪ ⎩ −2̂ n × Ea 𝜌′ ≤a 0 elsewhere Js = 0 everywhere Ms = ⎧ ⎪ ⎨ ⎪ ⎩ −2̂ n × Ea 𝜌′ ≤a 0 elsewhere Js = 0 everywhere Far-zone fields Er = Hr = 0 Er = Hr = 0 Z = ka sin 𝜃 E𝜃= jC1 sin 𝜙J1(Z) Z E𝜃= C2 sin 𝜙J1(Z) Z C1 = jka2E0e−jkr r E𝜙= jC1 cos 𝜃cos 𝜙J1(Z) Z E𝜙= C2 cos 𝜃cos 𝜙 J1 ′(Z) 1 −(Z∕𝜒′ 11)2 C2 = j kaE0J1(𝜒′ 11)e−jkr r H𝜃= −E𝜙∕𝜂 H𝜃= −E𝜙∕𝜂 𝜒′ 11 = 1.841 H𝜙= E𝜃∕𝜂 H𝜙= E𝜃∕𝜂 J1 ′(Z) = J0(Z) −J1(Z)∕Z 672 Half-power beamwidth (degrees) E-plane a ≫λ 29.2 a∕λ 29.2 a∕λ H-plane a ≫λ 29.2 a∕λ 37.0 a∕λ First null beamwidth (degrees) E-plane a ≫λ 69.9 a∕λ 69.9 a∕λ H-plane a ≫λ 69.9 a∕λ 98.0 a∕λ First side lobe max. (to main max.) (dB) E-plane −17.6 −17.6 H-plane −17.6 −26.2 Directivity D0 (dimensionless) 4𝜋 λ2 (area) = 4𝜋 λ2 (𝜋a2) = (2𝜋a λ )2 0.836 (2𝜋a λ )2 = 10.5𝜋 (a λ )2 673 674 APERTURE ANTENNAS Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Relative Amplitude ° 0 9 ° 0 9 θ θ H-plane (x-z plane, = 0°) E-plane (y-z plane, = 90°) ϕ ϕ Figure 12.18 Three-dimensional field pattern of a TE11-mode circular waveguide mounted on an infinite ground plane (a = 1.5λ).
dominant TE11-mode for a circular waveguide given by E𝜌= E0 𝜌′ J1 ( 𝜒′ 11 a 𝜌′ ) sin 𝜙′ E𝜙= E0 𝜕 𝜕𝜌′ [ J1 ( 𝜒′ 11 a 𝜌′ )] cos 𝜙′ ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ Ez = 0 𝜒′ 11 = 1.841 (12-57) The analysis of this problem is assigned at the end of this chapter as an exercise to the reader (Prob-lem 12.35). However, a three-dimensional pattern for a = 1.5λ was calculated, and it is shown in Figure 12.18. This pattern should be compared with that of Figure 12.17 for the constant aperture field distribution.
The beamwidths and the side lobe levels in the E- and H-planes are different, and exact closed-form expressions cannot be obtained. However, they can be calculated using iterative methods, and the data are shown in Figures 12.19 and 12.20 for the E- and H-planes, respectively.
A summary of all the radiation characteristics is included in Table 12.2. When the same aper-tures of Table 12.2 are not mounted on a ground plane, the far-zone fields do not have to be re-derived but rather can be written by inspection. This is accomplished by introducing appropriately, in each of the field components (E𝜃and E𝜙) of the second and third columns of Table 12.2, a (1 + cos 𝜃)/2 factor, as was done for the fields of the two apertures in the second and third columns of Table 12.1.
DESIGN CONSIDERATIONS 675 Figure 12.19 E-plane beamwidths and first side lobe relative maximum magnitude for TE11-mode circular aperture mounted on an infinite ground plane.
12.6.3 Beam Efficiency Beam efficiency, as defined by (2-53) and calculated by (2-54), for circular apertures not mounted on infinite ground planes is shown in Figure 12.21 . The lower abscissa scale (in degrees) is in terms of the half-cone angle 𝜃1 (in degrees), and it should be used only when the radius of the aperture is 20λ(a = 20λ). The upper abscissa scale is in terms of u(u = ka sin 𝜃1), and it should be used for any radius circular aperture.
The procedure for finding the beam efficiency of a circular aperture is similar to that of a rect-angular aperture as discussed in Section 12.5.4, illustrated in Figure 12.15, and demonstrated by Example 12.4.
A MATLAB computer program, designated as Aperture, has been developed to compute and display different radiation characteristics of rectangular and circular apertures. The description of the program is found in the corresponding READ ME file included in the publisher’s website for this book.
12.7 DESIGN CONSIDERATIONS As is the case for arrays, aperture antennas can be designed to control their radiation characteris-tics. Typically the level of the minor lobes can be controlled by tapering the distribution across the 676 APERTURE ANTENNAS Figure 12.20 H-plane beamwidths and first side lobe relative maximum magnitude for TE11-mode circular waveguide mounted on an infinite ground plane.
aperture; the smoother the taper from the center of the aperture toward the edge, the lower the side lobe level and the larger the half-power beamwidth, and conversely. Therefore a very smooth taper, such as that represented by a binomial distribution or others, would result in very low side lobes but larger half-power beamwidths. In contrast, an abrupt distribution, such as that of uniform illumina-tion, exhibits the smaller half-power beamwidth but the highest side lobe level (about - 13.5 dB).
Therefore if it is desired to achieve simultaneously both a very low sidelobe level, as well as a small half-power beamwidth, a compromise has to be made. Typically an intermediate taper, such as that of a Tschebyscheff distribution or any other similar one, will have to be selected. This has been dis-cussed in detail both in Chapter 6 for arrays and in Chapter 7 for continuous sources. These can be used to design continuous distributions for apertures.
Aperture antennas, both rectangular and circular, can also be designed for satellite applications where the beamwidth can be used to determine the “footprint” area of the coverage. In such designs, it is important to relate the beamwidth to the size of the aperture. In addition, it is also important to maximize the directivity of the antennas within a desired angular sector defined by the beamwidth, especially at the edge of coverage (EOC) . This can be accomplished, using approximate closed-form expressions, as outlined in . This procedure was used in Section 6.11 of Chapter 6 for arrays, and it is applicable for apertures, both rectangular and circular.
DESIGN CONSIDERATIONS 677 Figure 12.21 Beam efficiency versus half-cone angle 𝜃1, for a circular aperture with different field distribu-tions. The aperture is not mounted on an infinite ground plane. (source: D. G. Fink (ed.), Electronics Engineers’ Handbook, Section 18 (by W. F. Croswell), McGraw-Hill, New York, 1975).
12.7.1 Rectangular Aperture For a rectangular aperture, of dimensions a and b, and with a uniform distribution, the procedure to determine the optimum aperture dimensions a,b to maximize the directivity at an edge angle 𝜃c of a given angular sector (0 ≤𝜃≤𝜃c) is identical to that outlined in Section 6.11. Thus to determine the optimum dimension b of the aperture so that the directivity is maximum at an edge-of-coverage angle 𝜃ce of an angular sector 0 ≤𝜃≤𝜃ce in the E-plane is given by (6-105a), or E-Plane: b = λ 2 sin 𝜃ce (12-58a) Similarly for the H-plane, the optimum dimension a is determined by H-Plane: a = λ 2 sin 𝜃ch (12-58b) where 𝜃ch is the angle, in the H-plane, at the edge-of-coverage (EOC) angular sector where the directivity needs to be maximized.
Since the aperture antenna is uniformly illuminated, the directivity of (6-103) based on the opti-mum dimensions of (12-58a) and (12-58b) is D0 = 4𝜋 λ2 Aem = 4𝜋 λ2 Ap = 4𝜋 λ2 ( λ 2 sin 𝜃ce ) ( λ 2 sin 𝜃ch ) (12-59) 678 APERTURE ANTENNAS 12.7.2 Circular Aperture A procedure similar to that for the rectangular aperture can be used for the circular aperture. In fact, it can be used for circular apertures with uniform distributions as well as tapered (parabolic or parabolic with a pedestal) .
For a circular aperture with uniform distribution, the normalized power pattern multiplied by the maximum directivity can be written as P(𝜃) = (2𝜋a)2 {2J1(ka sin 𝜃) ka sin 𝜃 }2 (12-60) The maximum value of (12-60) occurs when 𝜃= 0. However, for any other angle 𝜃= 𝜃c, the maxi-mum of the pattern occurs when ka sin 𝜃c = 1.841 (12-61) or a = 1.841λ 2𝜋sin 𝜃c = λ 3.413 sin 𝜃c (12-61a) Therefore to maximize the directivity at the edge 𝜃= 𝜃c of a given angular sector 0 ≤𝜃≤𝜃c, the optimum radius of the uniformly illuminated circular aperture must be chosen according to (12-61a).
The maximum value of (12-60), which occurs at 𝜃= 0, is equal to P(𝜃= 0)|max = (2𝜋a)2 (12-62) while at the edge of the angular sector (𝜃= 𝜃c) is equal to P(𝜃= 𝜃c) = (2𝜋a)2 {2(0.5818) 1.841 }2 = (2𝜋a)2(0.3995) (12-63) Therefore the value of the directivity at the edge of the desired coverage (𝜃= 𝜃c), relative to its maximum value at 𝜃= 0, is P(𝜃= 𝜃c) P(𝜃= 0) = 0.3995 = −3.985 dB (12-64) Since the aperture is uniformly illuminated, the directivity based on the optimum radius of (12-61a) is D0 = 4𝜋 λ2 Ap = 4𝜋 λ2 𝜋 ( 1.841λ 2𝜋sin 𝜃c )2 = 3.3893 sin2 𝜃c = 1.079𝜋 sin2 𝜃c (12-65) A similar procedure can be followed for circular apertures with radial taper (parabolic) and radial taper squared of Table 7.2, as well as radial taper (parabolic) with pedestal. The characteristics of these, along with those of the uniform, are listed in Table 12.3.
DESIGN CONSIDERATIONS 679 TABLE 12.3 Edge-of-Coverage (EOC) Designs for Square and Circular Apertures Size Square: Side EOC Directivity Aperture Distribution Circular: Radius Directivity (relative to peak) Square Uniform λ 2 sin(𝜃c) 𝜋 sin2(𝜃c) −3.920 dB Circular Uniform λ 3.413 sin(𝜃c) 1.086𝜋 sin2(𝜃c) −3.985 dB Circular Parabolic taper λ 2.732 sin(𝜃c) 1.263𝜋 sin2(𝜃c) −4.069 dB Circular Parabolic taper with −10 dB pedestal λ 3.064 sin(𝜃c) 1.227𝜋 sin2(𝜃c) −4.034 dB (source: K. Praba, “Optimal Aperture for Maximum Edge-of-Coverage (EOC) Directivity,” IEEE Antennas & Propagation Magazine, Vol. 36, No. 3, pp. 72–74, June 1994. c ⃝(1994) IEEE) Example 12.5 It is desired to design an aperture antenna, with uniform illumination, so that the directivity is maximized at an angle 30◦from the normal to the aperture. Determine the optimum dimension and its associated directivity when the aperture is a. Square b. Circular Solution: For a square aperture 𝜃ce = 𝜃ch. Therefore the optimum dimension, according to (12-58a) or (12-58b), is a = b = λ 2 sin(30◦) = λ while the directivity, according to (12-59), is D0 = 𝜋 sin2 𝜃c = 𝜋 sin2(30◦) = 12.5664 = 10.992 dB The directivity at 𝜃= 30◦is −3.920 dB from the maximum at 𝜃= 0◦, or 7.072 dB.
For a circular aperture the optimum radius, according to (12-61a), is a = λ 3.413 sin(30◦) = λ 3.413(0.5) = 0.586λ while the directivity, according to (12-65), is D0 = 1.079𝜋 sin2 𝜃c = 1.079𝜋 sin2(30◦) = 13.559 = 11.32 dB The directivity at 𝜃= 30◦is −3.985 dB from the maximum at 𝜃= 0◦, or 7.365 dB.
680 APERTURE ANTENNAS 12.8 BABINET’S PRINCIPLE Now that wire and aperture antennas have been analyzed, one may inquire as to whether there is any relationship between them. This can be answered better by first introducing Babinet’s principle which in optics states that when the field behind a screen with an opening is added to the field of a complementary structure, the sum is equal to the field when there is no screen. Babinet’s principle in optics does not consider polarization, which is so vital in antenna theory; it deals primarily with absorbing screens. An extension of Babinet’s principle, which includes polarization and the more practical conducting screens, was introduced by Booker , . Referring to Figure 12.22(a), let us assume that an electric source J radiates into an unbounded medium of intrinsic impedance 𝜂= (𝜇∕𝜀)1∕2 and produces at point P the fields E0, H0. The same fields can be obtained by combining the fields when the electric source radiates in a medium with intrinsic impedance 𝜂= (𝜇∕𝜀)1∕2 in the presence of PEC PEC PMC Figure 12.22 Electric source in an unbounded medium and Babinet’s principle equivalents.
BABINET’S PRINCIPLE 681 1. an infinite, planar, very thin, perfect electric conductor with an opening Sa, which produces at P the fields Ee, He [Figure 12.22(b)] 2. a flat, very thin, perfect magnetic conductor Sa, which produces at P the fields Em, Hm [Fig-ure 12.22(c)].
That is, E0 = Ee + Em H0 = He + Hm (12-66a) The field produced by the source in Figure 12.22(a) can also be obtained by combining the fields of 1. an electric source J radiating in a medium with intrinsic impedance 𝜂= (𝜇∕𝜀)1∕2 in the pres-ence of an infinite, planar, very thin, perfect electric conductor Sa, which produces at P the fields Ee, He [Figure 12.22(b)] 2. a magnetic source M radiating in a medium with intrinsic impedance 𝜂d = (𝜀∕𝜇)1∕2 in the presence of a flat, very thin, perfect electric conductor Sa, which produces at P the fields Ed, Hd [Figure 12.22(d)] That is, E0 = Ee + Hd H0 = He −Ed (12-66b) The dual of Figure 12.22(d) is more easily realized in practice than that of Figure 12.22(c).
To obtain Figure 12.22(d) from Figure 12.22(c), J is replaced by M, Em by Hd, Hm by −Ed, 𝜀by 𝜇, and 𝜇by 𝜀. This is a form of duality often used in electromagnetics (see Section 3.7, Table 3.2). The electric screen with the opening in Figure 12.22(b) and the electric conductor of Figure 12.22(d) are also dual. They are usually referred to as complementary structures, because when combined they form a single solid screen with no overlaps. A proof of Babinet’s principle and its extension can be found in the literature .
Using Booker’s extension it can be shown , by referring to Figure 12.23, that if a screen and its complement are immersed in a medium with an intrinsic impedance 𝜂and have terminal Zs Zc a b Transmission feed line Transmission feed line (b) Complementary dipole (a) Screen with opening Perfect electric conductor Perfect electric conductor Current line Figure 12.23 Opening on a screen and its complementary dipole.
682 APERTURE ANTENNAS impedances of Zs and Zc, respectively, the impedances are related by ZsZc = 𝜂2 4 (12-67) To obtain the impedance Zc of the complement (dipole) in a practical arrangement, a gap must be introduced to represent the feed points. In addition, the far-zone fields radiated by the opening on the screen (E𝜃s, E𝜙s, H𝜃s, H𝜙s) are related to the far-zone fields of the complement (E𝜃c, E𝜙c, H𝜃c, H𝜙c) by E𝜃s = H𝜃c, E𝜙s = H𝜙c, H𝜃s = −E𝜃c 𝜂02 , H𝜙s = − E𝜙c 𝜂02 (12-68) Infinite, flat, very thin conductors are not realizable in practice but can be closely approximated. If a slot is cut into a plane conductor that is large compared to the wavelength and the dimensions of the slot, the behavior predicted by Babinet’s principle can be realized to a high degree. The impedance properties of the slot may not be affected as much by the finite dimensions of the plane as would be its pattern. The slot of Figure 12.23(a) will also radiate on both sides of the screen. Unidirectional radiation can be obtained by placing a backing (box or cavity) behind the slot, forming a so-called cavity-backed slot whose radiation properties (impedance and pattern) are determined by the dimen-sions of the cavity.
To demonstrate the application of Babinet’s principle, an example is considered.
Example 12.6 A very thin half-wavelength slot is cut on an infinite, planar, very thin, perfectly conducting electric screen as shown in Figure 12.24(a). Find its input impedance. Assume it is radiating into free-space.
w λ/2 Perfect electric conductor Transmission feed line (a) λ/2 thin slot (w 0) w λ/2 Electric conductor Transmission feed line (b) λ/2 flat thin dipole (w 0) Figure 12.24 Half-wavelength thin slot on an electric screen and its complement.
BABINET’S PRINCIPLE 683 Solution: From Babinet’s principle and its extension we know that a very thin half-wavelength dipole, shown in Figure 12.24(b), is the complementary structure to the slot. From Chapter 4, the terminal (input) impedance of the dipole is Zc = 73 + j42.5. Thus the terminal (input) impedance of the slot, using (12-67), is given by Zs = 𝜂02 4Zc ≃ (376.7)2 4(73 + j42.5) ≃35,475.72 73 + j42.5 Zs ≃362.95 −j211.31 The slot of Figure 12.24(a) can be made to resonate by choosing the dimensions of its complement (dipole) so that it is also resonant. The pattern of the slot is identical in shape to that of the dipole except that the E- and H-fields are interchanged. When a vertical slot is mounted on a vertical screen, as shown in Figure 12.25(a), its electric field is horizontally polarized while that of the dipole is w E E E E H H H H x y z λ/2 (a) λ/2 slot on a screen (b) λ/2 flat dipole w E E E E E E H H H H H H x y z λ/2 Figure 12.25 Radiation fields of a λ∕2 slot on a screen and of a λ∕2 flat dipole. (source: J. D. Kraus, Antennas, McGraw-Hill, New York, 1988, Chapter 13).
684 APERTURE ANTENNAS vertically polarized [Fig. 12.25(b)]. Changing the angular orientation of the slot or screen will change the polarization.
The slot antenna, as a cavity-backed design, has been utilized in a variety of law enforcement applications. Its main advantage is that it can be fabricated and concealed within metallic objects, and with a small transmitter it can provide covert communications. There are various methods of feeding a slot antenna . For proper operation, the cavity depth must be equal to odd multiples of λg∕4, where λg is the guide wavelength.
12.9 FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY Previously the spatial domain analysis of aperture antennas was introduced, and it was applied to rect-angular and circular apertures radiating in an infinite, homogeneous, lossless medium. The analysis of aperture antennas mounted on infinite ground planes, covered with lossless and/or lossy dielectric media, becomes too complex when it is attempted in the spatial domain. Considerable simplification can result with the utility of the frequency (spectral) domain.
12.9.1 Fourier Transforms-Spectral Domain From Fourier series analysis, any periodic function f(x) with a period T can be represented by a Fourier series of cosine and sine terms. If the function f(x) is aperiodic and exists only in the interval of 0 < x < T, a Fourier series can be formed by constructing, in a number of ways, a periodic func-tion. The Fourier series for the constructed periodic function represents the actual aperiodic function f(x) only in the interval 0 < x < T. Outside this space, the aperiodic function f (x) is zero and the series representation is not needed. A Fourier series for f(x) converges to the values of f(x) at each point of continuity and to the midpoint of its values at each discontinuity.
In addition, f (x) can also be represented as a superposition of discrete complex exponentials of the form f (x) = +∞ ∑ n=−∞ cne−j(2n𝜋∕T)x (12-69) cn = 1 T ∫ T 0 f (x)e+j(2n𝜋∕T)x dx (12-69a) or of continuous complex exponentials of the form f(x) = 1 2𝜋∫ +∞ −∞ ℱ(𝜔)e−jx𝜔d𝜔 −∞< 𝜔< +∞ (12-70a) whose inverse is given by ℱ(𝜔) = ∫ +∞ −∞ f(x)e+j𝜔x dx −∞< x < +∞ (12-70b) The integral operation in (12-70a) is referred to as the direct transformation and that of (12-70b) as the inverse transformation and both form a transform pair.
Another useful identity is Parseval’s theorem, which for the transform pair, can be written as ∫ +∞ −∞ f(x)g∗(x) dx = 1 2𝜋∫ +∞ −∞ ℱ(𝜔)𝒢∗(𝜔) d𝜔 (12-71) where ∗indicates complex conjugate.
FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 685 From the definitions of (12-70a), (12-70b) and (12-71), the Fourier transforms can be expanded to two dimensions and can be written as f (x, y) = 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ ℱ(𝜔1, 𝜔2)e−j(𝜔1x+𝜔2y) d𝜔1 d𝜔2 (12-72a) ℱ(𝜔1, 𝜔2) = ∫ +∞ −∞∫ +∞ −∞ f(x, y)e+j(𝜔1x+𝜔2y) dx dy (12-72b) ∫ +∞ −∞∫ +∞ −∞ f(x, y)g∗(x, y) dx dy = 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ ℱ(𝜔1, 𝜔2)𝒢∗(𝜔1, 𝜔2) d𝜔1 d𝜔2 (12-72c) The process can be continued to n dimensions.
The definitions, theorems, and principles introduced will be utilized in the sections that follow to analyze the radiation characteristics of aperture antennas mounted on infinite ground planes.
12.9.2 Radiated Fields To apply Fourier transforms (spectral techniques) to the analysis of aperture antennas, let us consider a rectangular aperture of dimensions a and b mounted on an infinite ground plane, as shown in Figure 12.26. In the source-free region (z > 0), the field E(x, y, z) of a monochromatic wave radiated by the aperture can be written as a superposition of plane waves (all of the same frequency, different amplitudes, and traveling in different directions) of the form f(kx, ky)e−jk⋅r , . The function f(kx, ky) is the vector amplitude of the wave, and kx and ky are the spectral frequencies which extend over the entire frequency spectrum (−∞≤kx, ky ≤∞). Thus the field E(x, y, z) can be written as E(x, y, z) = 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ f(kx, ky)e−jk⋅r dkx dky (12-73) r y x b z y h x z = (a) Uncovered aperture (b) Dielectric-covered aperture a , θ ϕ μ σ ∞ Figure 12.26 Rectangular apertures mounted on infinite ground planes.
686 APERTURE ANTENNAS according to the definition of (12-72a). The object of a plane wave expansion is to determine the unknown amplitudes f(kx, ky) and the direction of propagation of the plane waves. Since r = ̂ axx + ̂ ayy + ̂ azz (12-74) and the propagation factor k (often referred to as the vector wavenumber) can be defined as k = ̂ axkx + ̂ ayky + ̂ azkz (12-75) (12-73) can be written as E(x, y, z) = 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ [f(kx, ky)e−jkzz]e−j(kxx+kyy) dkx dky.
(12-76) The part of the integrand within the brackets can be regarded as the transform of E(x, y, z). This allows us to write the transform pair as E(x, y, z) = 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ ℰ ℰ ℰ(kx, ky, z)e−j(kxx+kyy) dkx dky (12-77a) ℰ ℰ ℰ(kx, ky, z) = ∫ +∞ −∞∫ +∞ −∞ E(x, y, z)e+j(kxx+kyy) dx dy (12-77b) where ℰ ℰ ℰ(kx, ky, z) = f(kx, ky)e−jkzz (12-77c) In principle then, according to (12-77a) and (12-77b) the fields radiated by an aperture E(x, y, z) can be found provided its transform ℰ ℰ ℰ(kx, ky, z) is known. To this point the transform field ℰ ℰ ℰ(kx, ky, z) can only be found provided the actual field E(x, y, z) is known a priori. In other words, the answer must be known beforehand! However, as it will be seen from what follows, if the transform field at z = 0 ℰ ℰ ℰ(kx, ky, z = 0) = f(kx, ky) (12-78) is formed, it will be sufficient to determine E(x, y, z). To form the transform ℰ ℰ ℰ(kx, ky, z = 0) = f(kx, ky), it will be necessary and sufficient to know only the tangential components of the E-field at z = 0. For the problem of Figure 12.26(a), the tangential components of the E-field at z = 0 exist only over the bounds of the aperture (they vanish outside it because of the presence of the infinite ground plane).
In general f(kx, ky) = ̂ axfx(kx, ky) + ̂ ayfy(kx, ky) + ̂ azfz(kx, ky) (12-79) which can also be written as f(kx, ky) = ft(kx, ky) + ̂ azfz(kx, ky) (12-79a) ft(kx, ky) = ̂ axfx(kx, ky) + ̂ ayfy(kx, ky) (12-79b) FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 687 For aperture antennas positioned along the x-y plane, the only components of f(kx, ky) that need to be found are fx and fy. As will be shown in what follows, fz can be found once fx and fy are known.
This is a further simplification of the problem. The functions fx and fy are found, using (12-77a) and (12-77b), provided the tangential components of the E-field over the aperture (Exa and Eya) are specified. The solution of (12-77c) is valid provided the z variations of E(kx, ky, z) are separable. In addition, in the source-free region the field E(x, y, z) of (12-77a) must satisfy the homogeneous vector wave equation. These allow us to relate the propagation constant kz to kx, ky and k = (𝜔√𝜇𝜀), by k2 z = k2 −(k2 x + k2 y) (12-80) or kz = { + [k2 −(k2 x + k2 y)]1∕2 when k2 ≥k2 x + k2 y (12-80a) −j[(k2 x + k2 y) −k2]1∕2 when k2 < k2 x + k2 y (12-80b) This is left as an exercise to the reader. The form of kz as given by (12-80a) contributes to the prop-agating waves (radiation field) of (12-76) and (12-77a) whereas that of (12-80b) contributes to the evanescent waves. Since the field in the far zone of the antenna is of the radiation type, its contribu-tion comes from the part of the kx, ky spectrum which satisfies (12-80a). The values of kx and ky in (12-80)–(12-80b) are analogous to the eigenvalues for the fields inside a rectangular waveguide .
In addition, kz is analogous to the propagation constant for waveguides which is used to define cutoff.
To find the relation between fz and fx, fy, we proceed as follows. In the source-free region (z > 0) the field E(x, y, z), in addition to satisfying the vector wave equation, must also be solenoidal so that ∇⋅E(x, y, z) = ∇⋅ { 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ f(kx, ky)e−jk⋅r dkx dky } = 0 (12-81) Interchanging differentiation with integration and using the vector identity ∇⋅(𝛼A) = 𝛼∇⋅A + A ⋅∇𝛼 (12-82) reduces (12-81) to 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ [f ⋅∇(e−jk⋅r)] dkx dky = 0 (12-83) since ∇⋅f(kx, ky) = 0. Equation (12-83) is satisfied provided that f ⋅∇e−jk⋅r = −jf ⋅ke−jk⋅r = 0 (12-84) or f ⋅k = (ft + ̂ azfz) ⋅k = 0 (12-84a) or fz = −ft ⋅k kz = − (fxkx + fyky) kz (12-84b) From (12-84b) it is evident that fz can be formed once fx and fy are known.
688 APERTURE ANTENNAS All three components of f (fx, fy and fz) can be found, using (12-77b) and (12-78), provided the two components of E (Ex, Ey) at z = 0, which is the plane of the aperture and ground plane of Fig-ure 12.26(a), are known. Because Ex and Ey along the z = 0 plane are zero outside the bounds of the aperture (|x| > a∕2, |y| > b∕2), (12-77b) and (12-78) reduce for fx and fy to fx(kx, ky) = ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Exa(x′, y′, z′ = 0)e+j(kxx′+kyy′) dx′ dy′ (12-85a) fy(kx, ky) = ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Eya(x′, y′, z′ = 0)e+j(kxx′+kyy′) dx′ dy′ (12-85b) where primes indicate source points. Exa(x′, y′, z′ = 0) and Eya(x′, y′, z′ = 0), which represent the tangential components of the electric field over the aperture, are the only fields that need to be known.
Once fx and fy are found by using (12-85a) and (12-85b), fz and ℰ ℰ ℰ(kx, ky, z) can be formed using (12-84b) and (12-77c), respectively. Thus, the solution for E(x, y, z) for the aperture in Figure 12.26(a) is given by E(x, y, z) = 1 4𝜋2 ⎧ ⎪ ⎨ ⎪ ⎩ ∫∫ k2 x+k2 y≤k2 kz=[k2−(k2 x+k2 y)]1∕2 ℰ(kx, ky, z)e−j(kxx+kyy) dkx dky + ∫∫ k2 x+k2 y>k2 kz=−j[(k2 x+k2 y)−k2]1∕2 ℰ(kx, ky, z)e−j(kxx+kyy) dkx dky ⎫ ⎪ ⎬ ⎪ ⎭ (12-86) ℰ ℰ ℰ(kx, ky, z) = [ ̂ axfx + ̂ ayfy −̂ az (fxkx + fyky kz )] e−jkzz (12-86a) where fx and fy are given by (12-85a) and (12-85b).
In summary, the field radiated by the aperture of Figure 12.26(a) can be found by the follow-ing procedure: 1. Specify the tangential components of the E-field (Exa and Eya) over the bounds of the aperture.
2. Find fx and fy using (12-85a) and (12-85b), respectively.
3. Find fz using (12-84b).
4. Find ℰ ℰ ℰ(kx, ky, z) using (12-86a).
5. Formulate E(x, y, z) using (12-86).
This completes the solution for E(x, y, z). However, as is evident from (12-86), the integration is quite difficult even for the simplest of problems. However, if the observations are restricted in the far-field region, many simplifications in performing the integrations can result. This was apparent in Chapters 4, 5 and in others. In many practical problems, the far zone is usually the region of greatest importance. Since it is also known that for all antennas the fields in the far zone are primarily of the radiated type (propagating waves), then only the first integral in (12-86) contributes in that region.
FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 689 In the next section, our attention is directed toward the evaluation of (12-86a) or (12-73) in the far-zone region (large values of kr). This is accomplished by evaluating (12-73) asymptotically for large values of kr by the method of Stationary Phase , .
To complete the formulation of the radiated fields in all regions, let us outline the procedure to find H(x, y, z). From Maxwell’s equations H(x, y, z) = −1 j𝜔𝜇∇× E(x, y, z) = −1 j𝜔𝜇∇× [ 1 4𝜋2 ∫ +∞ −∞∫ +∞ −∞ f(kx, ky)e−jk⋅r dkx dky ] (12-87) Interchanging integration with differentiation and using the vector identity ∇× (𝛼A) = 𝛼∇× A + (∇𝛼) × A (12-88) reduces (12-87) to H(x, y, z) = − 1 4𝜋2k𝜂∫ +∞ −∞∫ +∞ −∞ (f × k)e−jk⋅r dkx dky (12-89) since ∇× f(kx, ky) = 0 and ∇(e−jk⋅r) = −jke−jk⋅r from (12-84).
12.9.3 Asymptotic Evaluation of Radiated Field The main objective in this section is the evaluation of (12-73) or (12-86a) for observations made in the far-field. For most practical antennas, the field distribution on the aperture is such that an exact evaluation of (12-73) in closed form is not possible. However, if the observations are restricted to the far-field region (large kr), the integral evaluation becomes less complex. This was apparent in Chapters 4, 5, and others. The integral of (12-73) will be evaluated asymptotically for large values of kr using the method of Stationary Phase (Appendix VIII) , .
The stationary phase method assumes that the main contribution to the integral of (12-73) comes from values of kx and ky where k ⋅r does not change for first order changes in kx and ky. That is to say k ⋅r remains stationary at those points. For the other values of kx and ky, k ⋅r changes very rapidly and the function e−jk⋅r oscillates very rapidly between the values of +1 and −1. Assum-ing that f(kx, ky) is a slowly varying function of kx and ky, the integrand of (12-73) oscillates very rapidly outside the stationary points so that the contribution to the integral from that region is negli-gible. As the observation point approaches infinity, the contributions to the integral from the region outside the stationary points is zero. For practical applications, the observation point cannot be at infinity. However, it will be assumed to be far enough such that the major contributions come from the stationary points.
The first task in the asymptotic evaluation of (12-73) is to find the stationary points of k ⋅r. For that, k ⋅r is written as k ⋅r = (̂ axkx + ̂ ayky + ̂ azkz)⋅‚ arr (12-90) Using the inverse transformation of (4-5), (12-90) can be written as k ⋅r = r(kx sin 𝜃cos 𝜙+ ky sin 𝜃sin 𝜙+ kz cos 𝜃) (12-91) 690 APERTURE ANTENNAS which reduces, using (12-80a) to k ⋅r = r[kx sin 𝜃cos 𝜙+ ky sin 𝜃sin 𝜙+ √ k2 −k2 x −k2 y cos 𝜃] (12-92) The stationary points can be found by 𝜕(k ⋅r) 𝜕kx = 0 (12-93a) 𝜕(k ⋅r) 𝜕ky = 0 (12-93b) Using (12-92) and (12-80), (12-93a) and (12-93b) reduce to 𝜕(k ⋅r) 𝜕kx = r ( sin 𝜃cos 𝜙−kx kz cos 𝜃 ) = 0 (12-94a) 𝜕(k ⋅r) 𝜕ky = r ( sin 𝜃sin 𝜙− ky kz cos 𝜃 ) = 0 (12-94b) whose solutions are given, respectively, by kx = kz sin 𝜃cos 𝜙 cos 𝜃 (12-95a) ky = kz sin 𝜃sin 𝜙 cos 𝜃 (12-95b) Using (12-95a) and (12-95b), (12-80) can be written as k2 = k2 z + k2 x + k2 y = k2 z ( 1 + sin2 𝜃 cos2 𝜃 ) (12-96) which reduces for kz to kz = k cos 𝜃 (12-97) With the aid of (12-97), the stationary point of (12-95a) and (12-95b) simplify to kx = k sin 𝜃cos 𝜙= k1 (12-98a) ky = k sin 𝜃sin 𝜙= k2 (12-98b) The function k ⋅r can be expanded into a Taylor series, about the stationary point k1, k2, and it can be approximated by the zero, first, and second order terms. That is, k ⋅r ≃k ⋅r | | | |k1,k2 + 𝜕(k ⋅r) 𝜕kx | | | |k1,k2 (kx −k1) + 𝜕(k ⋅r) 𝜕ky | | | |k1,k2 (ky −k2) + 1 2 𝜕2(k ⋅r) 𝜕k2 x | | | |k1,k2 (kx −k1)2 + 1 2 𝜕2(k ⋅r) 𝜕k2 y | | | |k1,k2 (ky −k2)2 + 𝜕2(k ⋅r) 𝜕kx𝜕ky | | | |k1,k2 (kx −k1)(ky −k2) (12-99) FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 691 Since the second and third terms vanish at the stationary point kx = k1 and ky = k2, (12-99) can be expressed as k ⋅r = k ⋅r| | | |k1,k2 −A𝜉2 −B𝜂2 −C𝜉𝜂 (12-100) where A = −1 2 𝜕2(k ⋅r) 𝜕k2 x | | | |k1,k2 (12-100a) B = −1 2 𝜕2(k ⋅r) 𝜕k2 y | | | |k1,k2 (12-100b) C = −𝜕2(k ⋅r) 𝜕kx𝜕ky | | | |k1,k2 (12-100c) 𝜉= (kx −k1) (12-100d) 𝜂= (ky −k2) (12-100e) Using (12-97)-(12-98b), (12-90) reduces to k ⋅r| | | |k1,k2 = kr (12-101) Similarly, with the aid of (12-92), A, B, and C can be written, after a few manipulations, as A = −1 2 𝜕2(k ⋅r) 𝜕k2 x | | | |k1,k2 = r 2k ( 1 + sin2 𝜃cos2 𝜙 cos2 𝜃 ) (12-102a) B = −1 2 𝜕2(k ⋅r) 𝜕k2 y | | | |k1,k2 = r 2k ( 1 + sin2 𝜃sin2 𝜙 cos2 𝜃 ) (12-102b) C = −𝜕2(k ⋅r) 𝜕kx𝜕ky | | | |k1,k2 = r k sin2 𝜃 cos2 𝜃cos 𝜙sin 𝜙 (12-102c) Thus (12-73) can be approximated around the stationary point kx = k1 and ky = k2, which contributes mostly to the integral, by E(x, y, z) ≃ 1 4𝜋2 ∫∫ S1,2 f(kx = k1, ky = k2)e−j(kr−A𝜉2−B𝜂2−C𝜉𝜂) d𝜉d𝜂 (12-103) or E(x, y, z) ≃ 1 4𝜋2 f(k1, k2)e−jkr ∫∫ S1,2 e+j(A𝜉2+B𝜂2+C𝜉𝜂) d𝜉d𝜂 (12-103a) where S1,2 is the surface near the stationary point.
692 APERTURE ANTENNAS The integral of (12-103a) can be evaluated with the method of Stationary Phase. That is, (see Appendix VIII) ∫∫ S1,2 ej(A𝜉2+B𝜂2+C𝜉𝜂) d𝜉d𝜂= j 2𝜋𝛿 √ |4AB −C2| (12-104) 𝛿= ⎧ ⎪ ⎨ ⎪ ⎩ +1 if 4AB > C2 and A > 0 −1 if 4AB > C2 and A < 0 −j if 4AB < C2 (12-104a) With the aid of (12-102a)-(12-102c), the factor 4AB −C2 is 4AB −C2 = ( r k cos 𝜃 )2 (12-105) Since 4AB > C2 and A > 0, (12-103) reduces to ∫∫ S1,2 ej(A𝜉2+B𝜂2+C𝜉𝜂) d𝜉d𝜂= j2𝜋k r cos 𝜃 (12-106) and (12-103a) to E(r, 𝜃, 𝜙) ≃jke−jkr 2𝜋r [cos 𝜃f(k1 = k sin 𝜃cos 𝜙, k2 = k sin 𝜃sin 𝜙)] (12-107) In the far-field region, only the 𝜃and 𝜙components of the electric and magnetic fields are dom-inant. Therefore, the E𝜃and E𝜙components of (12-107) can be written in terms of fx and fy. With the aid of (12-84b), f can be expressed as f = ̂ axfx + ̂ ayfy + ̂ azfz = [ ̂ axfx + ̂ ayfy −̂ az (fxkx + fyky) kz ] (12-108) At the stationary point (kx = k1 = k sin 𝜃cos 𝜙, ky = k2 = k sin 𝜃sin 𝜙, kz = k cos 𝜃), (12-108) reduces to f(k1, k2) = [ ̂ axfx + ̂ ayfy −̂ az sin 𝜃 cos 𝜃(fx cos 𝜙+ fy sin 𝜙) ] (12-109) Using the inverse transformation of (4-5), the 𝜃and 𝜙components of f can be written as f𝜃= fx cos 𝜙+ fy sin 𝜙 cos 𝜃 (12-110a) f𝜙= −fx sin 𝜙+ fy cos 𝜙 (12-110b) FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 693 The E-field of (12-107) reduces, for the 𝜃and 𝜙components, to E(r, 𝜃, 𝜙) ≃jke−jkr 2𝜋r [̂ a𝜃(fx cos 𝜙+ fy sin 𝜙) + ̂ a𝜙cos 𝜃(−fx sin 𝜙+ fy cos 𝜙)] (12-111) and the H-field to H(r, 𝜃, 𝜙) = √𝜀 𝜇[̂ ar × E(r, 𝜃, 𝜙)] (12-112) where from (12-85a) and (12-85b) fx(kx = k1, ky = k2) = ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Exa(x′, y′, z′ = 0)ejk(x′ sin 𝜃cos 𝜙+y′ sin 𝜃sin 𝜙) dx′ dy′ (12-113a) fy(kx = k1, ky = k2) = ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Eya(x′, y′, z′ = 0)ejk(x′ sin 𝜃cos 𝜙+y′ sin 𝜃sin 𝜙) dx′ dy′ (12-113b) To illustrate the frequency domain (spectral) techniques, the problem of a uniform illuminated aperture, which was previously analyzed in Section 12.5.1 using spatial methods, will be solved again using transform methods.
Example 12.7 A rectangular aperture of dimensions a and b is mounted on an infinite ground plane, as shown in Figure 12.26(a). Find the field radiated by it assuming that over the opening the electric field is given by Ea = ̂ ayE0, −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 where E0 is a constant.
Solution: From (12-113a) and (12-113b) fx = 0 fy = E0 ∫ +b∕2 −b∕2 ejky′ sin 𝜃sin 𝜙dy′ ∫ +a∕2 −a∕2 ejkx′ sin 𝜃cos 𝜙dx′ 694 APERTURE ANTENNAS which, when integrated, reduces to fy = abE0 (sin X X ) (sin Y Y ) X = ka 2 sin 𝜃cos 𝜙 Y = kb 2 sin 𝜃sin 𝜙 The 𝜃and 𝜙components of (12-111) can be written as E𝜃= jabkE0e−jkr 2𝜋r { sin 𝜙 [sin X X ] [sin Y Y ]} E𝜙= jabkE0e−jkr 2𝜋r { cos 𝜃cos 𝜙 [sin X X ] [sin Y Y ]} which are identical to those of (12-23b) and (12-23c), respectively.
12.9.4 Dielectric-Covered Apertures The transform (spectral) technique can easily be extended to determine the field radiated by dielectric-covered apertures , . For the sake of brevity, the details will not be included here.
However, it can be shown that for a single lossless dielectric sheet cover of thickness h, dielectric constant 𝜀r, unity relative permeability, and free-space phase constant k0, the far-zone radiated field E𝜃, E𝜙of the covered aperture of Figure 12.26(b) are related to E0 𝜃, E0 𝜙of the uncovered aperture of Figure 12.26(a) by E𝜃(r, 𝜃, 𝜙) = f(𝜃)E0 𝜃(r, 𝜃, 𝜙) (12-114a) E𝜙(r, 𝜃, 𝜙) = g(𝜃)E0 𝜙(r, 𝜃, 𝜙) (12-114b) where E𝜃, E𝜙= field components of dielectric-covered aperture [Fig. 12.26(b)] E0 𝜃, E0 𝜙= field components of uncovered aperture [Fig. 12.26(a)] f (𝜃) = ejk0h cos 𝜃 cos 𝜓+ jZh sin 𝜓 (12-114c) g(𝜃) = ejk0h cos 𝜃 cos 𝜓+ jZe sin 𝜓 (12-114d) 𝜓= k0h √ 𝜀r −sin2 𝜃 (12-114e) FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 695 Ze = cos 𝜃 √ 𝜀r −sin2 𝜃 (12-114f) Zh = √ 𝜀r −sin2 𝜃 𝜀r cos 𝜃 (12-114g) The above relations do not include surface wave contributions which can be taken into account but are beyond the scope of this section .
To investigate the effect of the dielectric sheet, far-zone principal E- and H-plane patterns were computed for a rectangular waveguide shown in Figure 12.26(b). The waveguide was covered with a single dielectric sheet, was operating in the dominant TE10 mode, and was mounted on an infinite ground plane. The E- and H-plane patterns are shown in Figure 12.27(a) and 12.27(b), respectively.
In the E-plane patterns, it is evident that the surface impedance of the modified ground plane forces the normal electric field component to vanish along the surface (𝜃= 𝜋∕2). This is similar to the effects experienced by the patterns of the vertical dipole above ground shown in Figure 4.31. Since the H-plane patterns have vanishing characteristics when the aperture is radiating in free-space, the presence of the dielectric sheet has a very small overall effect. This is similar to the effects experienced by the patterns of a horizontal dipole above ground shown in Figure 4.32. However, both the E- and H-plane patterns become more broad near the surface, and more narrow elsewhere, as the thickness increases.
12.9.5 Aperture Admittance Another parameter of interest, especially when the antenna is used as a diagnostic tool, is its termi-nating impedance or admittance. In this section, using Fourier transform (spectral) techniques, the admittance of an aperture antenna mounted on an infinite ground plane and radiating into free-space will be formulated. Computations will be presented for a parallel-plate waveguide. The techniques can best be presented by considering a specific antenna configuration and field distribution. Similar steps can be used for any other geometry and field distribution.
The geometrical arrangement of the aperture antenna under consideration is shown in Fig-ure 12.26(a). It consists of a rectangular waveguide mounted on an infinite ground plane. It is assumed that the field distribution, above cutoff, is that given by the TE10 mode, or Ea = ̂ ayE0 cos (𝜋 a x′) −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 (12-115) where E0 is a constant. The aperture admittance is defined as Ya = 2P∗ |V|2 (12-116) where P∗= conjugate of complex power transmitted by the aperture V = aperture reference voltage.
696 APERTURE ANTENNAS 0° 30° 60° 90° (a) E-plane 90° 60° 30° h = 0.250 h = 0.500 0 0 0 0 Free-space h = 0.125 E-plane r = 2.1 b = 0.42 h = 0.250 h = 0.500 0 0 0 0 Free-space h = 0.125 H-plane r = 2.1 a = 0.945 10 20 30 Relative power (dB down) 0° 30° 60° 90° (b) H-plane 90° 60° 30° 10 20 30 Relative power (dB down) θ θ θ θ λ λ λ λ λ λ λ λ Figure 12.27 Amplitude radiation patterns of a dielectric-covered waveguide mounted on an infinite ground plane and with a TE10-mode aperture field distribution.
The complex power transmitted by the aperture can be written as P = 1 2 ∫∫ Sa [E(x′, y′, z′ = 0) × H∗(x′, y′, z′ = 0)] ⋅̂ az dx′ dy′ (12-117) where Sa is the aperture of the antenna. E(x′, y′, z′ = 0) and H(x′, y′, z′ = 0) represent the total elec-tric and magnetic fields at the aperture including those of the modes which operate below cutoff and contribute to the imaginary power. For the field distribution given by (12-115), (12-117) reduces to P = −1 2 ∫∫ Sa [Ey(x′, y′, z′ = 0)H∗ x (x′, y′, z′ = 0)] dx′ dy′ (12-117a) FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 697 The amplitude coefficients of all modes that can exist within the waveguide, propagating and nonpropagating, can be evaluated provided the total tangential E- and/or H-field at any point within the waveguide is known. Assuming that (12-115) represents the total tangential E-field, it allows the determination of all mode coefficients. Even though this can be accomplished, the formulation of (12-117a) in the spatial domain becomes rather complex .
An alternate and simpler method in the formulation of the aperture admittance is to use Fourier transforms. By Parseval’s theorem of (12-72c), (12-117a) can be written as P = −1 2 ∫ +∞ −∞∫ +∞ −∞ Ey(x′, y′, z′ = 0)H∗ x(x′, y′, z′ = 0) dx′ dy′ = −1 8𝜋2 ∫ +∞ −∞∫ +∞ −∞ ℰy(kx, ky)ℋ∗ x (kx, ky) dkx dky (12-118) where the limits of the first integral have been extended to infinity since Ey(x′, y′, z′ = 0) vanishes outside the physical bounds of the aperture. ℰy(kx, ky) and ℋx(kx, ky) are the Fourier transforms of the aperture E- and H-fields, respectively.
The transform ℰ(kx, ky, z = 0) is obtained from (12-78) while ℋ(kx, ky, z = 0) can be written, by referring to (12-89), as ℋ(kx, ky, z = 0) = −1 k𝜂(f × k) (12-119) For the problem at hand, the transforms ℰy and ℋx are given by ℰy(kx, ky) = fy(kx, ky) (12-120) ℋx(kx, ky) = −1 k𝜂 ( kz + k2 y kz ) fy = −1 k𝜂 ( k2 −k2 x kz ) fy (12-121) Using (12-77b) and (12-115), (12-120) reduces to fy(kx, ky) = E0 ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 cos (𝜋 a x′) ej(kxx′+kyy′) dx′ dy′ fy(kx, ky) = (𝜋ab 2 ) E0 [ cos X (𝜋∕2)2 −(X)2 ] [sin Y Y ] (12-122) where X = kxa 2 (12-122a) Y = kyb 2 (12-122b) 698 APERTURE ANTENNAS Substituting (12-120)–(12-122b) into (12-118) leads to P = (𝜋abE0)2 32𝜋2k𝜂∫ +∞ −∞∫ +∞ −∞ { (k2 −k2 x) k∗ z [ cos X (𝜋∕2)2 −(X)2 ]2 [sin Y Y ]2} dkx dky (12-123) If the reference aperture voltage is given by V = ab √ 2 E0 (12-124) the aperture admittance can be written as Ya = 2P∗ |V|2 = 1 8k𝜂∫ +∞ −∞ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 ∫ +∞ −∞ (k2 −k2 x) kz × ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (𝜋 2 )2 − (kxa 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dkx ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ dky (12-125) where kz is given by (12-80a) and (12-80b). As stated before, the values of kz as given by (12-80a) contribute to the radiated (real) power and those of (12-80b) contribute to the reactive (imaginary) power. Referring to Figure 12.28, values of kx and ky within the circle contribute to the aperture conductance, and the space is referred to as the visible region. Values of kx and ky outside the cir-cle contribute to the aperture susceptance and constitute the invisible region. Thus (12-125) can be separated into its real and imaginary parts, and it can be written as Ya = Ga + jBa (12-126) Visible region Invisible region Invisible region Invisible region Invisible region k ky kx Figure 12.28 Visible and invisible regions in k-space.
FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 699 Ga = 1 4k𝜂∫ k 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∫ √ k2−k2 y 0 (k2 −k2 x) [k2 −(k2 x + k2 y)]1∕2 × ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (𝜋 2 )2 − (kxa 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dkx ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ dky (12-126a) Ba = − 1 4k𝜂 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∫ k 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∫ ∞ √ k2−k2 y (k2 x −k2) [(k2 x + k2 y) −k2]1∕2 × ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (𝜋 2 )2 − (kxa 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dkx ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ dky + ∫ ∞ k ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∫ ∞ 0 (k2 x −k2) [(k2 x + k2 y) −k2]1∕2 × ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (𝜋 2 )2 − (kxa 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dkx ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ dky ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (12-126b) The first term in (12-126b) takes into account the contributions from the strip outside the circle for which ky < k, and the second term includes the remaining space outside the circle.
The numerical evaluation of (12-126a) and (12-126b) is complex and will not be attempted here.
Computations for the admittance of rectangular apertures radiating into lossless and lossy half spaces have been carried out and appear in the literature –. Various ingenious techniques have been used to evaluate these integrals.
Because of the complicated nature of (12-126a) and (12-126b) to obtain numerical data, a simpler configuration will be considered as an example.
700 APERTURE ANTENNAS Example 12.8 A parallel plate waveguide (slot) is mounted on an infinite ground plane, as shown in Fig-ure 12.29. Assuming the total electric field at the aperture is given by Ea = ̂ ayE0 −b∕2 ≤y′ ≤b∕2 where E0 is a constant, find the aperture admittance assuming the aperture voltage is given by V = bE0.
Solution: This problem bears a very close similarity to that of Figure 12.26(a), and most of the results of this example can be obtained almost directly from the previous formulation. Since the problem is two-dimensional, (12-120)–(12-122) reduce to ℰy(ky) = fy(ky) = bE0 sin (kyb 2 ) kyb 2 ℋx(ky) = k 𝜂 fy kz = kbE0 𝜂kz sin (kyb 2 ) kyb 2 y x z b E-field Figure 12.29 Parallel-plate waveguide geometry and aperture field distribution.
and (12-123) to P = (bE0)2k 4𝜋𝜂 ∫ +∞ −∞ 1 k∗ z ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky FOURIER TRANSFORMS IN APERTURE ANTENNA THEORY 701 Since the aperture voltage is given by V = bE0, the aperture slot admittance (per unit length along the x-direction) of (12-125) can be written as Ya = k 2𝜋𝜂∫ +∞ −∞ 1 kz ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky and the aperture slot conductance and susceptance of (12-126a) and (12-126b) as Ga = k 2𝜋𝜂∫ k −k 1 √ k2 −k2 y ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky = k 𝜋𝜂∫ k 0 1 √ k2 −k2 y ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky Ba = k 2𝜋𝜂 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∫ −k −∞ 1 √ k2 y −k2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky + ∫ ∞ k 1 √ k2 y −k2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ Ba = k 𝜋𝜂∫ ∞ k 1 √ k2 y −k2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (kyb 2 ) kyb 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 dky If w = b 2ky the expressions for the slot conductance and susceptance reduce to Ga = 2 𝜂λ ∫ kb∕2 0 1 √ (kb∕2)2 −w2 (sin w w )2 dw Ba = 2 𝜂λ ∫ ∞ kb∕2 1 √ w2 −(kb∕2)2 (sin w w )2 dw The admittance will always be capacitive since Ba is positive.
702 APERTURE ANTENNAS The expressions for the slot conductance and susceptance (per unit length along the x-direction) reduce for small values of kb to Ga ≃𝜋 𝜂λ [ 1 −(kb)2 24 ] ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ b λ < 1 10 Ba ≃𝜋 𝜂λ[1 −0.636 ln (kb)] and for large values of kb to Ga ≃1 𝜂b ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ b λ > 1 Ba ≃λ 𝜂 ( 1 𝜋b )2 [ 1 −1 2 √ λ b cos (2b λ + 1 4 ) 𝜋 ] Normalized values of λGa and λBa as a function of b∕λ for an aperture radiating into free space are shown plotted in Figure 12.30.
0 0.2 0.4 0.6 0.8 1.0 0 2 4 6 8 10 Ga b/ Ba Ga, Ba (10–3) λ λ λ λ λ Figure 12.30 Normalized values of conductance and susceptance of narrow slot.
12.10 GROUND PLANE EDGE EFFECTS: THE GEOMETRICAL THEORY OF DIFFRACTION Infinite size (physically and/or electrically) ground planes are not realizable in practice, but they can be approximated closely by very large structures. The radiation characteristics of antennas (cur-rent distribution, pattern, impedance, etc.) mounted on finite size ground planes can be modified considerably, especially in regions of very low intensity, by the effects of the edges. The ground plane edge diffractions for an aperture antenna are illustrated graphically in Figure 12.31. For these GROUND PLANE EDGE EFFECTS: THE GEOMETRICAL THEORY OF DIFFRACTION 703 Figure 12.31 Diffraction mechanisms for an aperture mounted on a finite size ground plane (diffractions at upper-lower and left-right edges of the ground plane).
problems, rigorous solutions do not exist unless the object’s surface can be described by curvilinear coordinates. Presently there are two methods that can be used conveniently to account for the edge effects. One technique is the Moment Method (MM) discussed in Chapter 8 and the other is the Geometrical Theory of Diffraction (GTD) .
The Moment Method describes the solution in the form of an integral, and it can be used to handle arbitrary shapes. It mostly requires the use of a digital computer for numerical computations and, because of capacity limitations of computers, it is most computationally efficient for objects that are small electrically. Therefore, it is usually referred to as a low-frequency asymptotic method.
When the dimensions of the radiating object are large compared to the wavelength, high-frequency asymptotic techniques can be used to analyze many otherwise not mathematically tractable prob-lems. One such technique, which has received considerable attention in the past few years, is the Geometrical Theory of Diffraction (GTD) which was originally developed by Keller . The GTD is an extension of the classical Geometrical Optics (GO; direct, reflected, and refracted rays), and it overcomes some of the limitations of GO by introducing a diffraction mechanism .
The diffracted field, which is determined by a generalization of Fermat’s principle , , is initiated at points on the surface of the object where there is a discontinuity in the incident GO field (incident and reflected shadow boundaries). The phase of the field on a diffracted ray is assumed to be equal to the product of the optical length of the ray (from some reference point) and the phase constant of the medium. Appropriate phase jumps must be added as a ray passes through caustics.∗ The amplitude is assumed to vary in accordance with the principle of conservation of energy in a narrow tube of rays. The initial value of the field on a diffracted ray is determined from the incident field with the aid of an appropriate diffraction coefficient (which, in general, is a dyadic for electro-magnetic fields). The diffraction coefficient is usually determined from the asymptotic solutions of the simplest boundary-value problems which have the same local geometry at the points of diffrac-tion as the object(s) of investigation. Geometries of this type are referred to as canonical problems.
∗A caustic is a point or a line through which all the rays of a wave pass. Examples of it are the focal point of a paraboloid (parabola of revolution) and the focal line of a parabolic cylinder. The field at the caustic is infinite because, in principle, an infinite number of rays pass through it.
704 APERTURE ANTENNAS Figure 12.32 Aperture geometry in principal E-plane (𝜙= 𝜋∕2).
Figure 12.33 Principal E-plane amplitude patterns of an aperture antenna mounted on a finite size ground plane.
GROUND PLANE EDGE EFFECTS: THE GEOMETRICAL THEORY OF DIFFRACTION 705 Figure 12.34 Measured and computed principal elevation plane amplitude patterns of a λ∕4 monopole above infinite and finite square ground planes.
One of the simplest geometries is a conducting wedge , . Another is that of a conducting, smooth, and convex surface –.
The primary objective in using the GTD to solve complicated geometries is to resolve each such problem into smaller components –, . The partitioning is made so that each smaller com-ponent represents a canonical geometry of a known solution. These techniques have also been applied for the modeling and analysis of antennas on airplanes , and they have combined both wedge and smooth conducting surface diffractions , . The ultimate solution is a superposition of the contributions from each canonical problem.
Some of the advantages of GTD are 1. It is simple to use.
2. It can be used to solve complicated problems that do not have exact solutions.
3. It provides physical insight into the radiation and scattering mechanisms from the various parts of the structure.
706 APERTURE ANTENNAS Figure 12.35 Measured and computed principal elevation plane amplitude patterns of a λ∕4 monopole (blade) above a circular ground plane.
4. It yields accurate results which compare extremely well with experiments and other methods.
5. It can be combined with other techniques such as the Moment Method .
The derivation of the diffraction coefficients for a conducting wedge and their application are lengthy, and will not be repeated here. An extensive and detailed treatment of over 100 pages, for both antennas and scattering, can be found in . However, to demonstrate the versatility and poten-tial of the GTD, three examples are considered. The first is the E-plane pattern of a rectangular aperture of dimensions a,b mounted on a finite size ground plane, as shown in Figure 12.31. The GTD formulation along the E-plane includes the direct radiation and the fields diffracted by the two edges of the ground plane, as shown in Figure 12.32. The computed E-plane pattern along with the measured one are shown in Figure 12.33; an excellent agreement is indicated.
REFERENCES 707 The two other examples considered here are the elevation pattern of a λ∕4 monopole mounted on square and circular ground planes. The diffraction mechanism on the principal planes for these is the same as that of the aperture, which is shown in Figure 12.32. The corresponding principal elevation plane pattern of the monopole on the square ground plane is displayed in Figure 12.34 while that on the circular one is exhibited in Figure 12.35. For each case an excellent agreement is indicated with the measurements. It should be noted that the minor lobes near the symmetry axis (𝜃= 0◦and 𝜃= 180◦) for the circular ground plane of Figure 12.35 are more intense than the corresponding ones for the square ground plane of Figure 12.34. These effects are due to the ring-source radiation by the rim of the circular ground plane toward the symmetry axis , .
12.11 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab computer program, designated as Aperture, for computing and displaying the radia-tion characteristics of rectangular and circular apertures.
c. Power Point (PPT) viewgraphs, in multicolor.
REFERENCES 1. S. A. Schelkunoff, “Some Equivalence Theorems of Electromagnetics and Their Application to Radiation Problems,” Bell Syst. Tech. J. Vol. 15, pp. 92–112, 1936.
2. C. A. Balanis, Advanced Engineering Electromagnetics, 2nd edition, John Wiley and Sons, New York, 2012.
3. C. Huygens, Traite De La Lumiere, Leyden, 1690. Translated into English by S. P. Thompson, London, 1912, Reprinted by The University of Chicago Press.
4. J. D. Kraus and K. R. Carver, Electromagnetics (Second edition), McGraw-Hill, New York, 1973, pp. 464–467.
5. R. F. Harrington, Time-Harmonic Electromagnetic Fields, McGraw-Hill, New York, 1961, pp. 100–103, 143–263, 365–367.
6. A. E. H. Love, “The Integration of the Equations of Propagation of Electric Waves,” Philos. Trans. R. Soc.
London, Ser. A, Vol. 197, pp. 1–45, 1901.
7. R. Mittra (ed.), Computer Techniques for Electromagnetics, Pergamon Press, New York, 1973, pp. 9–13.
8. C. A. Balanis and L. Peters, Jr., “Equatorial Plane Pattern of an Axial-TEM Slot on a Finite Size Ground Plane,” IEEE Trans. Antennas Propagat., Vol. AP-17, No. 3, pp. 351–353, May 1969.
9. C. A. Balanis, “Pattern Distortion Due to Edge Diffractions,” IEEE Trans. Antennas Propagat., Vol. AP-18, No. 4, pp. 561–563, July 1970.
10. C. R. Cockrell and P. H. Pathak, “Diffraction Theory Techniques Applied to Aperture Antennas on Finite Circular and Square Ground Planes,” IEEE Trans. Antennas Propagat., Vol. AP-22, No. 3, pp. 443–448, May 1974.
11. D. G. Fink (ed.), Electronics Engineers’ Handbook, Section 18 (Antennas by W. F. Croswell), McGraw-Hill, New York, 1975.
12. K. Praba, “Optimal Aperture for Maximum Edge-of-Coverage (EOC) Directivity,” IEEE Antennas Prop-agation Magazine, Vol. 36, No. 3, pp. 72–74, June 1994.
13. H. G. Booker, “Slot Aerials and Their Relation to Complementary Wire Aerials,” J. Inst. Elect. Eng., part III A, pp. 620–626, 1946.
708 APERTURE ANTENNAS 14. E. C. Jordan and K. G. Balmain, Electromagnetic Waves and Radiating Systems, Prentice-Hall, Inc., Engle-wood Cliffs, NJ, 1968.
15. J. D. Kraus, Antennas, Chapter 13, McGraw-Hill, New York, 1988.
16. H. G. Booker and P. C. Clemmow, “The Concept of an Angular Spectrum of Plane Waves, and its Relation to that of Polar Diagram and Aperture Distribution,” Proc. IEE (London), Vol. 97, part III, pp. 11–17, January 1950.
17. G. Borgiotti, “Fourier Transforms Method of Aperture Antennas,” Alta Freq., Vol. 32, pp. 196–204, November 1963.
18. L. B. Felsen and N. Marcuvitz, Radiation and Scattering of Waves, Prentice-Hall, Englewood Cliffs, NJ, 1973.
19. R. E. Collin and F. J. Zucker, Antenna Theory: Part 1, Chapter 3, McGraw-Hill Book Co., New York, 1969.
20. C. M. Knop and G. I. Cohn, “Radiation from an Aperture in a Coated Plane,” Radio Sci. J. Res., Vol. 68D, No. 4, pp. 363–378, April 1964.
21. F. L. Whetten, “Dielectric Coated Meandering Leaky-Wave Long Slot Antennas,” Ph.D. Dissertation, Ari-zona State University, May 1993.
22. M. H. Cohen, T. H. Crowley, and C. A. Lewis, “The Aperture Admittance of a Rectangular Waveguide Radi-ating Into Half-Space,” Antenna Lab., Ohio State University, Rept. 339-22, Contract W 33-038 ac21114, November 14, 1951.
23. R. T. Compton, “The Admittance of Aperture Antennas Radiating Into Lossy Media,” Antenna Lab., Ohio State University Research Foundation, Rept. 1691-5, March 15, 1964.
24. A. T. Villeneuve, “Admittance of a Waveguide Radiating into a Plasma Environment,” IEEE Trans. Anten-nas Propagat., Vol. AP-13, No. 1, pp. 115–121, January 1965.
25. J. Galejs, “Admittance of a Waveguide Radiating Into a Stratified Plasma,” IEEE Trans. Antennas Propa-gat., Vol. AP-13, No. 1, pp. 64–70, January 1965.
26. M. C. Bailey and C. T. Swift, “Input Admittance of a Circular Waveguide Aperture Covered by a Dielectric Slab,” IEEE Trans. Antennas Propagat., Vol. AP-16, No. 4, pp. 386–391, July 1968.
27. W. F. Croswell, W. C. Taylor, C. T. Swift, and C. R. Cockrell, “The Input Admittance of a Rectangular Waveguide-Fed Aperture Under an Inhomogeneous Plasma: Theory and Experiment,” IEEE Trans. Anten-nas Propagat., Vol. AP-16, No. 4, pp. 475–487, July 1968.
28. R. F. Harrington, Field Computation by Moment Methods, Macmillan Co., New York, 1968.
29. J. B. Keller, “Geometrical Theory of Diffraction,” Journal Optical Society of America, Vol. 52, No. 2, pp. 116–130, February 1962.
30. R. G. Kouyoumjian, “The Geometrical Theory of Diffraction and Its Applications,” in Numeri-cal and Asymptotic Techniques in Electromagnetics (R. Mittra, ed.), Springer-Verlag, New York, 1975.
31. R. G. Kouyoumjian and P. H. Pathak, “A Uniform Geometrical Theory of Diffraction for an Edge in a Perfectly Conducting Surface,” Proc. IEEE, Vol. 62, No. 11, pp. 1448–1461, November 1974.
32. D. L. Hutchins, “Asymptotic Series Describing the Diffraction of a Plane Wave by a Two-Dimensional Wedge of Arbitrary Angle,” Ph.D. Dissertation, The Ohio State University, Dept. of Electrical Engineering, 1967.
33. P. H. Pathak and R. G. Kouyoumjian, “An Analysis of the Radiation from Apertures on Curved Sur-faces by the Geometrical Theory of Diffraction, Proc. IEEE, Vol. 62, No. 11, pp. 1438–1447, November 1974.
34. G. L. James, Geometrical Theory of Diffraction for Electromagnetic Waves, Peter Peregrinus, Ltd., Steve-nage, Herts., England, 1976.
35. C. A. Balanis and L. Peters, Jr., “Analysis of Aperture Radiation from an Axially Slotted Circular Con-ducting Cylinder Using GTD,” IEEE Trans. Antennas Propagat., Vol. AP-17, No. 1, pp. 93–97, January 1969.
36. C. A. Balanis and Y.-B. Cheng, “Antenna Radiation and Modeling for Microwave Landing System,” IEEE Trans. Antennas Propagat., Vol. AP-24, No. 4, pp. 490–497, July 1976.
PROBLEMS 709 37. W. D. Burnside, C. L. Yu, and R. J. Marhefka, “A Technique to Combine the Geometrical Theory of Diffrac-tion and the Moment Method,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 4, pp. 551–558, July 1975.
PROBLEMS 12.1.
A uniform plane wave traveling in the +z direction, whose magnetic field is expressed as Hi = ̂ ayH0e−jkz z ≤0 impinges upon an aperture on an infinite, flat, perfect elec-tric conductor whose cross section is indicated in the fig-ure.
(a) State the equivalent that must be used to determine the field radiated by the aperture to the right of the conductor (z > 0).
(b) Assuming the aperture dimension in the y direction is b, determine the far-zone fields for z > 0.
μ x z y a/2 a/2 0, 0 12.2.
Repeat Problem 12.1 when the incident magnetic field is polarized in the x direction.
12.3.
Repeat Problem 12.1 when the incident electric field is polarized in the y direction.
12.4.
Repeat Problem 12.1 when the incident electric field is polarized in the x direction.
12.5.
A perpendicularly polarized plane wave is obliquely incident upon an aperture, with dimension a and b, on a perfectly electric conducting ground plane of infinite extent, as shown in the figure. Assuming the field over the aperture is given by the incident field (ignore diffractions from the edges of the aperture), find the far-zone spherical components of the fields for x > 0.
y Hi Ei x r z a/2 a/2 ϕ0 ϕ 12.6.
Repeat Problem 12.5 for a parallelly polarized plane wave (when the incident magnetic field is polarized in the z direction, i.e., the incident magnetic field is perpendicular to the x-y plane while the incident electric field is parallel to the x-y plane).
12.7.
A narrow rectangular slot of size L by W is mounted on an infinite ground plane that covers the x-y plane. The tangential field over the aperture is given by Ea = ̂ ayE0e−jk0x′√ 2∕2 Using the equivalence principle and image theory, we can replace the aperture and infinite ground plane with an equivalent magnetic current radiating in free-space. Determine the (a) appropriate equivalent (b) far-zone spherical electric field components for z > 0 (c) direction (𝜃, 𝜙) in which the radiation intensity is maximum W z x y L 710 APERTURE ANTENNAS 12.8.
A rectangular aperture, of dimensions a and b, is mounted on an infinite ground plane, as shown in Figure 12.6(a). Assuming the tangential field over the aperture is given by Ea = ̂ azE0 −a∕2 ≤y′ ≤a∕2, −b∕2 ≤z′ ≤b∕2 find the far-zone spherical electric and magnetic field components radiated by the aperture.
12.9.
Repeat Problem 12.8 when the same aperture is analyzed using the coordinate system of Figure 12.6(b). The tangential aperture field distribution is given by Ea = ̂ axE0 −b∕2 ≤x′ ≤b∕2, −a∕2 ≤z′ ≤a∕2 12.10.
Repeat Problem 12.8 when the aperture field is given by Ea = ̂ azE0 cos (𝜋 a y′) −a∕2 ≤y′ ≤a∕2, −b∕2 ≤z′ ≤b∕2 12.11.
Repeat Problem 12.9 when the aperture field distribution is given by Ea = ̂ axE0 cos (𝜋 a z′) −b∕2 ≤x′ ≤b∕2, −a∕2 ≤z′ ≤a∕2 12.12.
Find the fields radiated by the apertures of Problems (a) 12.8 (b) 12.9 (c) 12.10 (d) 12.11 when each of the apertures with their associated field distributions are not mounted on a ground plane. Assume the tangential H-field at the aperture is related to the E-field by the intrinsic impedance.
12.13.
Find the fields radiated by the rectangular aperture of Section 12.5.3 when it is not mounted on an infinite ground plane.
12.14.
For the rectangular aperture of Section 12.5.3 (with a = 4λ, b = 3λ), compute the (a) E-plane beamwidth (in degrees) between the maxima of the second minor lobe (b) E-plane amplitude (in dB) of the maximum of the second minor lobe (relative to the maximum of the major lobe) (c) approximate directivity of the antenna using Kraus’ formula. Compare it with the value obtained using the expression in Table 12.1.
12.15.
A rectangular X-band (8.2–12.4 GHz) waveguide (with inside dimensions of 0.9 in by 0.4 in) operating in the dominant TE10 mode at 10 GHz is mounted on an infinite ground plane and used as a receiving antenna. This antenna is connected to a matched lossless transmission line and a matched load is attached to the transmission line. Determine the: (a) Directivity (dimensionless and in dB) using: 1. the most accurate formula that is available to you in class.
2. Kraus’ formula.
(b) Maximum power (in watts) that can be delivered to the load when a uniform plane wave with a power density of 10 mW/cm2 is incident upon the antenna at normal incidence.
Neglect any losses.
12.16.
A lossless aperture antenna has a gain of 11 dB and overall physical area of 2λ2.
(a) What is the aperture efficiency of this antenna (in %)?
(b) Assuming the antenna is matched to a lossless transmission line that in turn is connected to a load that is also matched to the transmission line, what is the maximum power that PROBLEMS 711 can be delivered to the load if the incident power density at the antenna aperture is 10 × 10−3 watts/cm2? The frequency of operation is 10 GHz.
12.17.
For the rectangular aperture of Section 12.5.1 with a = b = 3λ, compute the directivity using (12-37) and the computer program Aperture.
12.18.
For the rectangular aperture of Section 12.5.2 with a = b = 3λ, compute the directivity using (12-37) and the computer program Aperture.
12.19.
Compute the directivity of the aperture of Section 12.5.3, using the computer program Aperture, when (a) a = 3λ, b = 2λ (b) a = b = 3λ 12.20.
Repeat Problem 12.19 when the aperture is not mounted on an infinite ground plane.
12.21.
For the rectangular aperture of Section 12.5.3 with a = 3λ, b = 2λ, compute the (a) E-plane half-power beamwidth (b) H-plane half-power beamwidth (c) E-plane first-null beamwidth (d) H-plane first-null beamwidth (e) E-plane first side lobe maximum (relative to main maximum) (f) H-plane first side lobe maximum (relative to main maximum) using the formulas of Table 12.1. Compare the results with the data from Figures 12.13 and 12.14. Verify using the computer program Aperture.
12.22.
A square waveguide aperture, of dimensions a = b and lying on the x-y plane, is radiating into free-space. Assuming a cos(𝜋x′∕a) by cos(𝜋y′∕b) distribution over the aperture, find the dimensions of the aperture (in wavelengths) so that the beam efficiency within a 37◦ total included angle cone is 90%.
12.23.
Verify (12-39a), (12-39b), (12-39c), and (12-40).
12.24.
A rectangular aperture mounted on an infi-nite ground plane has aperture electric field distributions and corresponding effi-ciencies of Field Aperture Distribution Efficiency (a) Triangular 75% (b) Cosine square 66.67% What are the corresponding directives (in dB) if the dimensions of the aperture are a = λ∕2 and b = λ∕4?
a b Triangular cos2 12.25.
The physical area of an aperture antenna operating at 10 GHz is 200 cm2 while its directivity is 23 dB. Assuming the antenna has an overall radiation efficiency of 90% and it is perfectly matched to the input transmission line, find the aperture efficiency of the antenna.
12.26.
Two X-band (8.2–12.4 GHz) rectangular waveguides, each operating in the dominant TE10-mode, are used, respectively, as transmitting and receiving antennas in a long distance com-munication system. The dimensions of each waveguide are a = 2.286 cm (0.9 in.) and b = 1.016 cm (0.4 in.) and the center frequency of operation is 10 GHz. Assuming the waveguides are separated by 10 kilometers and they are positioned for maximum radiation and reception toward each other, and the radiated power is 1 watt, find the: (a) Incident power density at the receiving antenna (b) Maximum power that can be delivered to a matched load 712 APERTURE ANTENNAS Assume the antennas are lossless, are polarization matched, and each is mounted on an infinite ground plane.
12.27.
The normalized far-zone electric field radiated in the E-plane (x-z plane; 𝜙= 0◦) by a waveg-uide aperture antenna of dimensions a and b, mounted on an infinite ground plane as shown in the figure, is given by E = −̂ a𝜃j𝜔𝜇bI0e−jkr 4𝜋r sin (kb 2 cos 𝜃 ) kb 2 cos 𝜃 Determine in the E-plane the: (a) Vector effective length of the antenna.
(b) Maximum value of the effective length.
State the value of 𝜃(in degrees) which maximizes the effective length.
∞ σ z x y a b = 12.28.
A uniform plane wave is incident upon an X-band rectangular waveguide, with dimensions of 2.286 cm and 1.016 cm, mounted on an infinite ground plane. Assuming the waveguide is operating in the dominant TE10 mode, determine the maximum power that can be delivered to a matched load. The frequency is 10 GHz and the power density of the incident plane wave is 10−4 watts/m2.
12.29.
Compute the aperture efficiency of a rectangular aperture, mounted on an infinite ground plane as shown in Figure 12.7, with an E-field aperture distribution directed toward y but with variations (a) triangular in the x and uniform in the y (b) cosine-squared in the x and uniform in the y (c) cosine in the x and cosine in the y (d) cosine-squared in both the x and y directions.
How do they compare with those of a cosine distribution?
12.30.
An X-band (8.2–12.4 GHz) WR 90 rectangular waveguide, with inner dimensions of 0.9 in. (2.286 cm) and 0.4 in. (1.016 cm), is mounted on an infinite ground plane. Assuming the waveguide is operating in the dominant TE10-mode, find its directivity at f = 10 GHz using the (a) computer program Aperture (b) formula in Table 12.1 Compare the answers.
12.31.
An X-band (8.2–12.4 GHz) rectangular waveguide, with internal dimensions of a = 2.286 cm and b = 1.016 cm, and mounted on an infinite PEC, is used as a ground-based receiving antenna for a satellite borne communication system. The power radiated by the satellite antenna is 10 watts, the satellite is at a distance of 100 kilometers from the ground-based antenna, and its antenna is assumed to have isotropic radiation characteristics. Assum-ing an operating frequency of 10 GHz and dominant TE10 mode field configuration, deter-mine the maximum power (in watts) that can be intercepted by the receiving antenna when the incident wave is perpendicularly (normal to the aperture) incident on the receiving antenna and its polarization is: (a) Linearly polarized. Assume no other losses.
(b) Circularly polarized. Assume no other losses.
The maximum radiation of the ground-based receiving antenna is directed toward the satel-lite borne communication system.
PROBLEMS 713 12.32.
Repeat Problem 12.30 at f = 20 GHz for a K-band (18–26.5 GHz) WR 42 rectangular waveguide with inner dimensions of 0.42 in. (1.067 cm) and 0.17 in. (0.432 cm).
12.33.
A rectangular waveguide, of dimensions a and b [as shown in Figure 12.6(c)], is mounted on an infinite in extent planar Perfect Magnetic Conductor (PMC) ground plane. The aperture lies on the x-y plane, with the z-axis perpendicular to the aperture. To simplify the formu-lation, the tangential electric and magnetic field components over the aperture are given by: Ea = ̂ ayEo Ha = −̂ ax Eo 𝜂 ⎫ ⎪ ⎬ ⎪ ⎭ −a∕2 ≤x′ ≤+a∕2 −b∕2 ≤y′ ≤+b∕2 where 𝜂is the intrinsic impedance of free space.
(a) Based on the Field Equivalence Principle (FEP), also known as Huygen’s Principle, determine the exact vector surface equivalent current densities Js and Ms that will pro-vide a unique and exact solution to the actual physical radiation problem.
You do NOT have to derive the equivalent surface current densities Js and Ms.
Also you must give a proper explanation how you arrived at them using the FEP and any other theorem(s).
(b) Determine the far-field electric fields radiated by the aperture.
The electric fields must be expressed in simplified form (using sin X∕X, sin Y∕Y and sine/cosine functions, as is done in the text book).
12.34.
Four rectangular X-band waveguides of dimensions a = 0.9 in. (2.286 cm) and b = 0.4 in.
(1.016 cm) and each operating on the dominant TE10-mode, are mounted on an infinite ground plane so that their apertures and the ground plane coincide with the x-y plane. The apertures form a linear array, are placed along the x-axis with a center-to-center separation of d = 0.85λ apart, and they are fed so that they form a broadside Dolph–Tschebyscheff array of −30 dB minor lobes. Assuming a center frequency of 10 GHz, determine the overall directivity of the array in decibels.
12.35.
Sixty-four (64) X-band rectangular waveguides are mounted so that the aperture of each is mounted on an infinite ground plane that coincides with the x-y plane, and all together form an 8 × 8 = 64 planar array. Each waveguide has dimensions of a = 0.9 in. (2.286 cm), b = 0.4 in. (1.086 cm) and the center-to-center spacing between the waveguides is dx = dy = 0.85λ. Assuming a TE10-mode operation for each waveguide, a center frequency of 10 GHz, and the waveguides are fed to form a uniform broadside planar array, find the directivity of the total array.
12.36.
Find the far-zone fields radiated when the circular aperture of Section 12.6.1 is not mounted on an infinite ground plane.
12.37.
Derive the far-zone fields when the circular aperture of Section 12.6.2 (a) is; (b) is not mounted on an infinite ground plane.
12.38.
A circular waveguide (not mounted on a ground plane), operating in the dominant TE11 mode, is used as an antenna radiating in free-space. Write in simplified form the normalized far-zone electric field components radiated by the waveguide antenna. You do not have to derive them.
714 APERTURE ANTENNAS 12.39.
A circular waveguide of radius a = 1.125 cm, NOT mounted on an infinite PEC ground plane and operating in the dominant TE11 mode at a frequency of 10 GHz, is used as a receiving antenna. On the basis of the approximate equivalent, determine the following: (a) Far-zone electric and magnetic radiated fields (you do not have to derive them). Specify the angular limits (lower and higher in degrees) on the observation angles 𝜃and 𝜙.
(b) Maximum power (in watts) that can be delivered to a receiver (load) assuming the receiver (load) is matched to the transmission line that connects the antenna and the receiver (load). Assume that the transmission line has a characteristic impedance of 300 ohms while the antenna has an input impedance of 350 +j 400 ohms. Assume no other losses. The maximum power density of the wave impinging upon the antenna is 100 watts∕m2.
12.40.
A ground-based and airborne communication system operates at X-Band (f = 10 GHz). The receiving ground-based system utilizes a circular waveguide, with a radius of a = 2 cm, and it mounted on an infinite PEC ground plane. The transmitting airborne system utilizes a rectangular waveguide with dimensions of a = 2.286 cm and b = 1.016 cm, and it is also mounted on an infinite PEC ground plane. Assuming that the two systems are aligned so that the maximum radiation from airborne system is directed toward the maximum reception of the ground-based system, the power radiated by the airborne system is 100 watts, the distance between the two systems is 100 kilometers, and assuming no conduction/dielectric or reflection/mismatch losses in either of the two systems, determine the: (a) Maximum directivity (dimensionless and in dB) of the transmitting antenna.
(b) Maximum power density (in watts/cm2) radiated by the airborne system and impinging at the ground-based system, which is at 100 kilometers.
(c) Maximum effective area (in cm2) of the ground-based system; do not include any polar-ization mismatch at this point.
(d) Maximum power (in watts) delivered to the ground-based system if the: r Incident wave and the ground based antenna are polarization matched (no other losses).
r Incident wave is circularly polarized while the ground based antenna is linearly polarized (no other losses).
12.41.
A lossless circular aperture antenna has a gain of 15 dB and overall physical area of 25 cm2.
The frequency of operation is 10 GHz.
(a) What is the aperture efficiency of the antenna (in %)?
(b) What is the power PL delivered to a matched load given that the power density of the inci-dent wave at the antenna aperture is 30 mW/cm2? Assume ideal conditions (no losses).
12.42.
A lossless circular aperture antenna operating on the dominant TE11-mode and mounted on an infinite ground plane has an overall gain of 9 dB. Determine the following: (a) Physical area (in λ2) of the antenna.
(b) Maximum effective/equivalent area (in λ2) of the antenna.
(c) Aperture efficiency (in percent).
(d) How much more or less efficient (in percent) is this antenna with a TE11-mode distri-bution compared with the same antenna but with a uniform field distribution over its aperture. State which one is more or less efficient.
12.43.
For the circular aperture of Section 12.6.1, compute its directivity, using the computer program Aperture of this chapter, when its radius is PROBLEMS 715 (a) a = 0.5λ (b) a = 1.5λ (c) a = 3.0λ Compare the results with data from Table 12.2.
12.44.
Repeat Problem 12.43 when the circular aperture of Section 12.6.1 is not mounted on an infinite ground plane. Compare the results with those of Problem 12.43.
12.45.
For the circular aperture of Problem 12.37, compute the directivity, using the computer program Aperture of this chapter, when its radius is (a) a = 0.5λ (b) a = 1.5λ (c) a = 3.0λ Compare the results with data from Table 12.2.
12.46.
For the circular aperture of Section 12.6.2 with a = 1.5λ, compute the (a) E-plane half-power beamwidth (b) H-plane half-power beamwidth (c) E-plane first-null beamwidth (d) H-plane first-null beamwidth (e) E-plane first side lobe maximum (relative to main maximum) (f) H-plane first side lobe maximum (relative to main maximum) using the formulas of Table 12.2. Compare the results with the data from Figures 12.19 and 12.20. Verify using the computer program Aperture.
12.47.
A circular waveguide, of radius a [as shown in Figure 12.16], is not mounted on a PEC ground plane; it is radiating in free-space with no ground plane. The aperture lies on the x-y plane, with the z-axis perpendicular to the aperture. To simplify the formulation, the tangential electric and magnetic field components over the aperture are given by: Ea = ̂ ayEo Ha = −̂ ax Eo 𝜂 ⎫ ⎪ ⎬ ⎪ ⎭ 0 ≤𝜌′ ≤a∕2 where 𝜂is the intrinsic impedance of free space.
(a) Detrmine/write the approximate far-zone electric and magnetic field components radi-ated by the aperture, based on the approximate Field Equivalence Principle (FEP), also known as Huygen’s Principle. You do not have to derive the expressions for the electric and magnetic fields, but you must indicate how you got them.
The expressions for the fields must be written in simplified form, just as the examples in the book.
(b) For an aperture radius of a = 3λ, determine the: r E-plane half-power beamwidth (in degrees).
r H-plane half-power beamwidth (in degrees).
r Directivity (dimensions and in dB).
r Directivity (dimensionless and in dB) using an appropriate alternate/another expres-sion that uses the half-power beamwidth information.
State which alternate expression you are using and why?
12.48.
A circular aperture of radius a is mounted on an infinite electric ground plane. Assuming the opening is on the x-y plane and its field distribution is given by (a) Ea = ̂ ayE0 [ 1 − (𝜌′ a )2] , 𝜌′ ≤a 716 APERTURE ANTENNAS (b) Ea = ̂ ayE0 [ 1 − (𝜌′ a )2]2 , 𝜌′ ≤a find the far-zone electric and magnetic field components radiated by the antenna.
12.49.
Repeat Problem 12.48 when the electric field is given by Ea = ̂ ayE0[1 −(𝜌′∕a)], 𝜌′ ≤a Find only the radiation vectors L and N. Work as far as you can. If you find you cannot com-plete the solution in closed form, state clearly why you cannot. Simplify as much as possible.
12.50.
A coaxial line of inner and outer radii a and b, respectively, is mounted on an infinite electric ground plane. Assuming that the electric field over the aperture of the coax is Ea = −̂ a𝜌 V 𝜀ln(b∕a) 1 𝜌′ , a ≤𝜌′ ≤b z y x b a where V is the applied voltage and 𝜀is the permittivity of the coax medium, find the far-zone spherical electric and magnetic field components radiated by the antenna.
12.51.
It is desired to design a circular aperture antenna with a field distribution over its opening of E = C[1 −(𝜌′∕a)2] where C is a constant, a its radius, and 𝜌′ any point on the aperture, such that its beam efficiency within a 60◦total included angle cone is 90%. Find its radius in wavelengths.
12.52.
For the antenna of Problem 12.51, find its efficiency within a 40◦total included angle cone when its radius is 2λ.
12.53.
Design square apertures with uniform illumination so that the directivity at 60◦from the normal is maximized relative to that at 𝜃= 0◦. Determine the: (a) Dimensions of the aperture (in λ) (b) Maximum directivity (in dB) (c) Directivity (in dB) at 60◦from the maximum 12.54.
Design a circular aperture with uniform illumination so that the directivity at 60◦from the normal is maximized relative to that at 𝜃= 0◦. Determine the (a) Radius of the aperture (in λ) (b) Maximum directivity (in dB) (c) Directivity (in dB) at 60◦from the maximum 12.55.
Repeat Problem 12.50 for a circular aperture with a parabolic distribution.
PROBLEMS 717 12.56.
Repeat Problem 12.50 for a circular aperture with a parabolic taper on 10 dB pedestal.
12.57.
Derive the edge-of-coverage (EOC) design characteristics for a circular aperture with a parabolic taper.
12.58.
Design a rectangular aperture of uniform illumination so that its directivity in the E- and H-planes is maximized (relative to its maximum value), respectively, at angles of 30◦and 45◦from the normal to the aperture.
(a) Determine the optimum dimensions (in λ) of the aperture.
(b) What is the maximum directivity (in dB) of the aperture and at what angle(s) (in degrees) will this occur?
(c) What is the directivity (in dB) of the aperture along the E-plane at 30◦from the normal to the aperture?
(d) What is the directivity (in dB) of the aperture along the H-plane at 45◦from the normal to the aperture?
12.59.
Design a circular aperture with uniform distribution so that its directivity at an angle 𝜃= 35◦from the normal to the aperture is maximized relative to its maximum value at 𝜃= 0◦.
Specifically, (a) find the optimum radius (in λ) of the aperture.
(b) what is the maximum directivity (in dB) at 𝜃= 0◦?
(c) what is the directivity (in dB) at 𝜃= 35◦?
12.60.
A vertical dipole is radiating into a free-space medium and produces fields E0 and H0. Illus-trate alternate methods for obtaining the same fields using Babinet’s principle and extensions of it.
12.61.
(a) (1) Sketch the six principal-plane patterns, and (2) define the direction of E and H along the three principal axes and at 45◦to the axes, for a thin slot one-half wavelength long, cut in a conducting sheet which has infinite conductivity and extending to infinity, and open on both sides. Inside dimensions of the slot are approximately 0.5λ by 0.1λ.
Assume that the width (0.1λ) of the slot is small compared to a wavelength. Assume a coordinate system such that the conducting plane lies on the x-y plane with the larger dimension of the slot parallel to the y-axis.
(b) Sketch the six approximate principal-plane patterns E𝜃(𝜙= 0◦), E𝜙(𝜙= 0◦), E𝜃(𝜙= 90◦), E𝜙(𝜙= 90◦), E𝜃(𝜃= 90◦), E𝜙(𝜃= 90◦).
12.62.
A very thin circular annular slot with circumference of one wavelength is cut on a very thin, infinite, flat, perfectly electric conducting plate. The slot is radiating into free-space. What is the impedance (real and imaginary parts) of the slot?
a 12.63.
Repeat Example 12.7 for a rectangular aperture with an electric field distribution of Ea = ̂ ayE0 cos (𝜋 a x′) , −a∕2 ≤x′ ≤a∕2 −b∕2 ≤y′ ≤b∕2 718 APERTURE ANTENNAS 12.64.
Two identical very thin (b = λ∕20) parallel-plate waveguides (slots), each mounted on an infinite ground plane, as shown in Figure 12.29 of the book for one of them, are separated by a distance of λg∕2 where λg is the parallel-plate waveguide (transmission line) that is connecting the two slots. Assuming each slot is of width W = 10 cm, the parallel-plate waveguide (transmission line) is filled with air, the slots are radiating in free-space and are operating at 10 GHz: (a) What is the admittance Ya of one slot in the absence of the other (both real and imagi-nary parts)?
(b) What is the total input impedance Zin of both slots together when looking in at the input of one of them in the presence of the other (both real and imaginary parts)?
Ya Ya Zc Zin Parallel Plate Transmission Line λg 2 CHAPTER13 Horn Antennas 13.1 INTRODUCTION One of the simplest and probably the most widely used microwave antenna is the horn. Its existence and early use dates back to the late 1800s. Although neglected somewhat in the early 1900s, its revival began in the late 1930s from the interest in microwaves and waveguide transmission lines during the period of World War II. Since that time a number of articles have been written describing its radiation mechanism, optimization design methods, and applications. Many of the articles published since 1939 which deal with the fundamental theory, operating principles, and designs of a horn as a radiator can be found in a book of reprinted papers and chapters in handbooks , .
The horn is widely used as a feed element for large radio astronomy, satellite tracking, and communication dishes found installed throughout the world. In addition to its utility as a feed for reflectors and lenses, it is a common element of phased arrays and serves as a universal standard for calibration and gain measurements of other high-gain antennas. Its widespread applicability stems from its simplicity in construction, ease of excitation, versatility, large gain, and preferred overall performance.
An electromagnetic horn can take many different forms, four of which are shown in Figure 13.1.
The horn is nothing more than a hollow pipe of different cross sections, which has been tapered (flared) to a larger opening. The type, direction, and amount of taper (flare) can have a profound effect on the overall performance of the element as a radiator. In this chapter, the fundamental theory of horn antennas will be examined. In addition, data will be presented that can be used to understand better the operation of a horn and its design as an efficient radiator.
13.2 E-PLANE SECTORAL HORN The E-plane sectoral horn is one whose opening is flared in the direction of the E-field, and it is shown in Figure 13.2(a). A more detailed geometry is shown in Figure 13.2(b).
13.2.1 Aperture Fields The horn can be treated as an aperture antenna. To find its radiation characteristics, the equivalent principle techniques developed in Chapter 12 can be utilized. To develop an exact equivalent of it, Portions of this chapter on aperture-matched horns, multimode horns, and dielectric-loaded horns were first published by the author in , Copyright 1988, reprinted by permission of Van Nostrand Reinhold Co.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 719 720 HORN ANTENNAS Figure 13.1 Typical electromagnetic horn antenna configurations.
it is necessary that the tangential electric and magnetic field components over a closed surface are known. The closed surface that is usually selected is an infinite plane that coincides with the aperture of the horn. When the horn is not mounted on an infinite ground plane, the fields outside the aperture are not known and an exact equivalent cannot be formed. However, the usual approximation is to assume that the fields outside the aperture are zero, as was done for the aperture of Section 12.5.2.
The fields at the aperture of the horn can be found by treating the horn as a radial waveg-uide –. The fields within the horn can be expressed in terms of cylindrical TE and TM wave functions which include Hankel functions. This method finds the fields not only at the aperture of the horn but also within the horn. The process is straightforward but laborious, and it will not be included here. However, it is assigned as an exercise at the end of the chapter (Problem 13.1).
It can be shown that if the (1) fields of the feed waveguide are those of its dominant TE10 mode and (2) horn length is large compared to the aperture dimensions, the lowest order mode fields at the aperture of the horn are given by E′ z = E′ x = H′ y = 0 (13-1a) E′ y(x′, y′) ≃E1 cos (𝜋 a x′) e−j[ky′2∕(2𝜌1)] (13-1b) H′ z(x′, y′) ≃jE1 ( 𝜋 ka𝜂 ) sin (𝜋 a x′) e−j[ky′2∕(2𝜌1)] (13-1c) H′ x(x′, y′) ≃−E1 𝜂cos (𝜋 a x′) e−j[ky′2∕(2𝜌1)] (13-1d) 𝜌1 = 𝜌e cos 𝜓e (13-1e) where E1 is a constant. The primes are used to indicate the fields at the aperture of the horn. The expressions are similar to the fields of a TE10-mode for a rectangular waveguide with aperture E-PLANE SECTORAL HORN 721 a a y y' x' z z' x b b1 (a) E-plane horn y z' x' z x b b1/2 (b) E-plane view y' e e 1 1 ψ ρ ρ ψ ρ ρ ψ δ Figure 13.2 E-plane horn and coordinate system.
dimensions of a and b1(b1 > a). The only difference is the complex exponential term which is used here to represent the quadratic phase variations of the fields over the aperture of the horn.
The necessity of the quadratic phase term in (13-1b)–(13-1d) can be illustrated geometrically.
Referring to Figure 13.2(b), let us assume that at the imaginary apex of the horn (shown dashed) there exists a line source radiating cylindrical waves. As the waves travel in the outward radial direction, the constant phase fronts are cylindrical. At any point y′ at the aperture of the horn, the phase of the field will not be the same as that at the origin (y′ = 0). The phase is different because the wave has traveled different distances from the apex to the aperture. The difference in path of travel, designated as 𝛿(y′), can be obtained by referring to Figure 13.2(b). For any point y′ [𝜌1 + 𝛿(y′)]2 = 𝜌2 1 + (y′)2 (13-2) or 𝛿(y′) = −𝜌1 + [𝜌2 1 + (y′)2]1∕2 = −𝜌1 + 𝜌1 [ 1 + ( y′ 𝜌1 )2]1∕2 (13-2a) which is referred to as the spherical phase term.
722 HORN ANTENNAS Using the binomial expansion and retaining only the first two terms of it, (13-2a) reduces to 𝛿(y′) ≃−𝜌1 + 𝜌1 [ 1 + 1 2 ( y′ 𝜌1 )2] = 1 2 (y′2 𝜌1 ) (13-2b) when (13-2b) is multiplied by the phase factor k, the result is identical to the quadratic phase term in (13-1b)–(13-1d).
The quadratic phase variation for the fields of the dominant mode at the aperture of a horn antenna has been a standard for many years, and it has been chosen because it yields in most practical cases very good results. Because of its simplicity, it leads to closed form expressions, in terms of sine and cosine Fresnel integrals, for the radiation characteristics (far-zone fields, directivity, etc.) of the horn.
It has been shown recently that using the more accurate expression of (13-2a) for the phase, error variations and numerical integration yield basically the same directivities as using the approximate expression of (13-2b) for large aperture horns (b1 of Figures 13.2 or a1 of Figure 13.10 greater than 50λ) or small peak aperture phase error (S = 𝜌e −𝜌1 of Figure 13.2 or T = 𝜌h −𝜌2 of Figure 13.10 less than 0.2λ). However, for intermediate aperture sizes (5λ ≤b1 or a1 ≤8λ) or intermediate peak aperture phase errors (0.2λ ≤S or T ≤0.6λ) the more accurate expression of (13-2a) for the phase variation yields directivities which are somewhat higher (by as much as a few tenths of a decibel) than those obtained using (13-2b). Also it has been shown using a full-wave Moment Method analysis of the horn that as the horn dimensions become large the amplitude distribution at the aperture of the horn contains higher-order modes than the TE10 mode and the phase distribution at the aperture approaches the parabolic phase front.
Example 13.1 Design an E-plane sectoral horn so that the maximum phase deviation at the aperture of the horn is 56.72◦. The dimensions of the horn are a = 0.5λ, b = 0.25λ, b1 = 2.75λ.
Solution: Using (13-2b) Δ𝜙|max = k𝛿(y′)|y′=b1∕2 = k(b1∕2)2 2𝜌1 = 56.72 ( 𝜋 180 ) or 𝜌1 = (2.75 2 )2 180 56.72λ = 6λ The total flare angle of the horn should be equal to 2𝜓e = 2 tan−1 (b1∕2 𝜌1 ) = 2 tan−1 (2.75∕2 6 ) = 25.81◦ 13.2.2 Radiated Fields To find the fields radiated by the horn, only the tangential components of the E- and/or H-fields over a closed surface must be known. The closed surface is chosen to coincide with an infinite plane passing E-PLANE SECTORAL HORN 723 through the mouth of the horn. To solve for the fields, the approximate equivalent of Section 12.5.2 is used. That is, Jy = −E1 𝜂cos (𝜋 a x′) e−jk𝛿(y′) Mx = E1 cos (𝜋 a x′) e−jk𝛿(y′) ⎫ ⎪ ⎬ ⎪ ⎭ −a∕2 ≤x′ ≤a∕2 −b1∕2 ≤y′ ≤b1∕2 (13-3) and Js = Ms = 0 elsewhere (13-3a) Using (12-12a) N𝜃= −E1 𝜂cos 𝜃sin 𝜙I1I2 (13-4) where I1 = ∫ +a∕2 −a∕2 cos (𝜋 a x′) ejkx′ sin 𝜃cos 𝜙dx′ = − (𝜋a 2 ) ⎡ ⎢ ⎢ ⎢ ⎣ cos (ka 2 sin 𝜃cos 𝜙 ) (ka 2 sin 𝜃cos 𝜙 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎦ (13-4a) I2 = ∫ +b1∕2 −b1∕2 e−jk[𝛿(y′)−y′ sin 𝜃sin 𝜙] dy′ (13-4b) The integral of (13-4b) can also be evaluated in terms of cosine and sine Fresnel integrals. To do this, I2 can be written, by completing the square, as I2 = ∫ +b1∕2 −b1∕2 e−j[ky′2∕(2𝜌1)−kyy′] dy′ = ej(k2 y𝜌1∕2k) ∫ +b1∕2 −b1∕2 e−j[(ky′−ky𝜌1)2∕2k𝜌1] dy′ (13-5) where ky = k sin 𝜃sin 𝜙 (13-5a) Making a change of variable √ 𝜋 2 t = √ 1 2k𝜌1 (ky′ −ky𝜌1) (13-6a) t = √ 1 𝜋k𝜌1 (ky′ −ky𝜌1) (13-6b) dt = √ k 𝜋𝜌1 dy′ (13-6c) 724 HORN ANTENNAS reduces (13-5) to I2 = √𝜋𝜌1 k ej(k2 y𝜌1∕2k) ∫ t2 t1 e−j(𝜋∕2)t2 dt = √𝜋𝜌1 k ej(k2 y𝜌1∕2k) ∫ t2 t1 [ cos (𝜋 2 t2) −j sin (𝜋 2 t2)] dt (13-7) and takes the form of I2 = √𝜋𝜌1 k ej(k2 y𝜌1∕2k){[C(t2) −C(t1)] −j[S(t2) −S(t1)]} (13-8) where t1 = √ 1 𝜋k𝜌1 ( −kb1 2 −ky𝜌1 ) (13-8a) t2 = √ 1 𝜋k𝜌1 (kb1 2 −ky𝜌1 ) (13-8b) C(x) = ∫ x 0 cos (𝜋 2 t2) dt (13-8c) S(x) = ∫ x 0 sin (𝜋 2 t2) dt (13-8d) C(x) and S(x) are known as the cosine and sine Fresnel integrals and are well tabulated (see Appendix IV). Computer subroutines are also available for efficient numerical evaluation of each , .
Using (13-4a) and (13-8), (13-4) can be written as N𝜃= E1 𝜋a 2 √𝜋𝜌1 k ej(k2 y𝜌1∕2k) × ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ cos 𝜃sin 𝜙 𝜂 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (kxa 2 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ F(t1, t2) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (13-9) where kx = k sin 𝜃cos 𝜙 (13-9a) ky = k sin 𝜃sin 𝜙 (13-9b) F(t1, t2) = [C(t2) −C(t1)] −j[S(t2) −S(t1)] (13-9c) E-PLANE SECTORAL HORN 725 In a similar manner, N𝜙, L𝜃, L𝜙of (12-12b)–(12-12d) reduce to N𝜙= E1 𝜋a 2 √𝜋𝜌1 k ej(k2 y𝜌1∕2k) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ cos 𝜙 𝜂 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (kxa 2 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ F(t1, t2) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (13-10a) L𝜃= E1 𝜋a 2 √𝜋𝜌1 k ej(k2 y𝜌1∕2k) × ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −cos 𝜃cos 𝜙 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (kxa 2 ) 2 − (𝜋 2 ) 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ F(t1, t2) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (13-10b) L𝜙= E1 𝜋a 2 √𝜋𝜌1 k ej(k2 y𝜌1∕2k) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ sin 𝜙 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (kxa 2 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ F(t1, t2) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (13-10c) The electric field components radiated by the horn can be obtained by using (12-10a)–(12-10c), and (13-9)–(13-10c). Thus, Er = 0 (13-11a) E𝜃= −j a √ 𝜋k𝜌1E1e−jkr 8r × ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ej(k2 y𝜌1∕2k) sin 𝜙(1 + cos 𝜃) ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (kxa 2 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ F(t1, t2) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (13-11b) E𝜙= −j a √ 𝜋k𝜌1E1e−jkr 8r × ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ej(k2 y𝜌1∕2k) cos 𝜙(cos 𝜃+ 1) ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos (kxa 2 ) (kxa 2 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ F(t1, t2) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (13-11c) where t1, t2, kx, ky, and F(t1, t2) are given, respectively, by (13-8a), (13-8b), (13-9a), (13-9b), and (13-9c). The corresponding H-field components are obtained using (12-10d)–(12-10f).
726 HORN ANTENNAS In the principal E- and H-planes, the electric field reduces to E-Plane (𝝓= 𝜋∕2) Er = E𝜙= 0 (13-12a) E𝜃= −j a √ 𝜋k𝜌1E1e−jkr 8r { −ej(k𝜌1 sin2 𝜃∕2) ( 2 𝜋 )2 (1 + cos 𝜃)F(t′ 1, t′ 2) } (13-12b) t′ 1 = √ k 𝜋𝜌1 ( −b1 2 −𝜌1 sin 𝜃 ) (13-12c) t′ 2 = √ k 𝜋𝜌1 ( +b1 2 −𝜌1 sin 𝜃 ) (13-12d) H-Plane (𝝓= 0) Er = E𝜃= 0 (13-13a) E𝜙= −j a √ 𝜋k𝜌1E1e−jkr 8r ⎧ ⎪ ⎨ ⎪ ⎩ (1 + cos 𝜃) ⎡ ⎢ ⎢ ⎢ ⎣ cos (ka 2 sin 𝜃 ) (ka 2 sin 𝜃 )2 − (𝜋 2 )2 ⎤ ⎥ ⎥ ⎥ ⎦ F(t′′ 1 , t′′ 2 ) ⎫ ⎪ ⎬ ⎪ ⎭ (13-13b) t′′ 1 = −b1 2 √ k 𝜋𝜌1 (13-13c) t′′ 2 = +b1 2 √ k 𝜋𝜌1 (13-13d) To better understand the performance of an E-plane sectoral horn and gain some insight into its performance as an efficient radiator, a three-dimensional normalized field pattern has been plotted in Figure 13.3 utilizing (13-11a)–(13-11c). As expected, the E-plane pattern is much narrower than the H-plane because of the flaring and larger dimensions of the horn in that direction. Figure 13.3 pro-vides an excellent visual view of the overall radiation performance of the horn. To display additional details, the corresponding normalized E- and H-plane patterns (in dB) are illustrated in Figure 13.4.
These patterns also illustrate the narrowness of the E-plane and provide information on the relative levels of the pattern in those two planes.
To examine the behavior of the pattern as a function of flaring, the E-plane patterns for a horn antenna with 𝜌1 = 15λ and with flare angles of 20◦≤2𝜓e ≤35◦are plotted in Figure 13.5. A total of four patterns is illustrated. Since each pattern is symmetrical, only half of each pattern is displayed.
For small included angles, the pattern becomes narrower as the flare increases. Eventually the pattern begins to widen, becomes flatter around the main lobe, and the phase tapering at the aperture is such that even the main maximum does not occur on axis. This is illustrated in Figure 13.5 by the pattern with 2𝜓e = 35◦. As the flaring is extended beyond that point, the flatness (with certain allowable rip-ple) increases and eventually the main maximum returns again on axis. It is also observed that as the flaring increases, the pattern exhibits much sharper cutoff characteristics. In practice, to compensate for the phase taper at the opening, a lens is usually placed at the aperture making the pattern of the horn always narrower as its flare increases.
E-PLANE SECTORAL HORN 727 Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ Relative Amplitude θ 180° 180° H-plane (x-z plane, ϕ = 0°) E-plane (y-z plane, ϕ = 90°) Figure 13.3 Three-dimensional field pattern of E-plane sectoral horn (𝜌1 = 6λ, b1 = 2.75λ, a = 0.5λ).
Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 E-plane H-plane θ θ Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Figure 13.4 E- and H-plane patterns of an E-plane sectoral horn.
728 HORN ANTENNAS 2ψ = 20° 2ψ = 25° 2ψ = 30° 2ψ = 35° Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ Figure 13.5 E-plane patterns of E-plane sectoral horn for constant length and different included angles.
Similar pattern variations occur as the length of the horn is varied while the flare angle is held constant. As the length increases, the pattern begins to broaden and eventually becomes flatter (with a ripple). Beyond a certain length, the main maximum does not even occur on axis, and the pattern continues to broaden and to become flatter (within an allowable ripple) until the maximum returns on axis. The process continues indefinitely.
13.2.3 Directivity The directivity is one of the parameters that is often used as a figure of merit to describe the perfor-mance of an antenna. To find the directivity, the maximum radiation is formed. That is, Umax = U(𝜃, 𝜙)|max = r2 2𝜂|E|2 max (13-14) For most horn antennas |E|max is directed nearly along the z-axis (𝜃= 0◦). Thus, |E|max = √ |E𝜃|2 max + |E𝜙|2 max = 2a √ 𝜋k𝜌1 𝜋2r |E1||F(t)| (13-15) E-PLANE SECTORAL HORN 729 Using (13-11b), (13-11c), and (13-9c) |E𝜃|max = 2a √ 𝜋k𝜌1 𝜋2r |E1 sin 𝜙F(t)| (13-15a) |E𝜙|max = 2a √ 𝜋k𝜌1 𝜋2r |E1 cos 𝜙F(t)| (13-15b) F(t) = [C(t) −jS(t)] (13-15c) t = b1 2 √ k 𝜋𝜌1 = b1 √ 2λ𝜌1 (13-15d) since kx = ky = 0 (13-15e) t1 = −t = −b1 2 √ k 𝜋𝜌1 = − b1 √ 2λ𝜌1 (13-15f) t2 = +t = +b1 2 √ k 𝜋𝜌1 = b1 √ 2λ𝜌1 (13-15g) C(−t) = −C(t) (13-15h) S(−t) = −S(t) (13-15i) Thus Umax = r2 2𝜂|E|2 max = 2a2k𝜌1 𝜂𝜋3 |E1|2|F(t)|2 = 4a2𝜌1|E1|2 𝜂λ𝜋2 |F(t)|2 (13-16) where |F(t)|2 = [ C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )] (13-16a) The total power radiated can be found by simply integrating the average power density over the aperture of the horn. Using (13-1a)–(13-1d) Prad = 1 2 ∫∫ S0 Re(E′ × H′∗) ⋅ds = 1 2𝜂∫ +b1∕2 −b1∕2 ∫ +a∕2 −a∕2 |E1|2 cos2 (𝜋 a x′) dx′ dy′ (13-17) which reduces to Prad = |E1|2 b1a 4𝜂 (13-17a) 730 HORN ANTENNAS Using (13-16) and (13-17a), the directivity for the E-plane horn can be written as DE = 4𝜋Umax Prad = 64a𝜌1 𝜋λb1 |F(t)|2 = 64a𝜌1 𝜋λb1 [ C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )] (13-18) The overall performance of an antenna system can often be judged by its beamwidth and/or its directivity. The half-power beamwidth (HPBW), as a function of flare angle, for different horn lengths is shown in Figure 13.6. In addition, the directivity (normalized with respect to the constant aperture dimension a) is displayed in Figure 13.7. For a given length, the horn exhibits a monotonic decrease in half-power beamwidth and an increase in directivity up to a certain flare. Beyond that point a monotonic increase in beamwidth and decrease in directivity is indicated followed by rises and falls. The increase in beamwidth and decrease in directivity beyond a certain flare indicate the broadening of the main beam.
Figure 13.6 Half-power beamwidth of E-plane sectoral horn as a function of included angle and for different lengths.
E-PLANE SECTORAL HORN 731 Figure 13.7 Normalized directivity of E-plane sectoral horn as a function of aperture size and for different lengths.
If the values of b1 (in λ), which correspond to the maximum directivities in Figure 13.7, are plotted versus their corresponding values of 𝜌1 (in λ), it can be shown that each optimum directivity occurs when b1 ≃ √ 2λ𝜌1 (13-18a) with a corresponding value of s equal to s|b1 = √ 2λ𝜌1 = sop = b2 1 8λ𝜌1 | | | | |b1= √ 2λ𝜌1 = 1 4 (13-18b) The classic expression of (13-18) for the directivity of an E-plane horn has been the standard for many years. However, it has been shown that this expression may not always yield very accurate val-ues for the on-axis directivity. A more accurate expression for the maximum on-axis directivity based on an exact open-ended parallel-plate waveguide analysis has been derived, and it yields a modifica-tion to the on-axis value of (13-18), which provides sufficient accuracy for most designs , .
Using (13-18a), the modified formula for the on-axis value of (13-18) can be written as , DE(max) = 16ab1 λ2(1 + λg∕λ) [ C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )] e 𝜋a λ ( 1−λ λg ) (13-18c) where λg is the guide wavelength in the feed waveguide for the dominant TE10 mode. Predicted values based on (13-18) and (13-18c) have been compared with measurements and it was found that (13-18c) yielded results which were closer to the measured values .
732 HORN ANTENNAS Figure 13.8 GE as a function of B. (source: Adopted from data by E. H. Braun, “Some Data for the Design of Electromagnetic Horns,” IRE Trans. Antennas Propagat., Vol. AP-4, No. 1, January 1956. c ⃝1956 IEEE).
The directivity of an E-plane sectoral horn can also be computed by using the following proce-dure .
1. Calculate B by B = b1 λ √ 50 𝜌e∕λ (13-19a) 2. Using this value of B, find the corresponding value of GE from Figure 13.8. If, however, the value of B is smaller than 2, compute GE using GE = 32 𝜋B (13-19b) 3. Calculate DE by using the value of GE from Figure 13.8 or from (13-19b). Thus DE = a λ GE √ 50 𝜌e∕λ (13-19c) Example 13.2 An E-plane sectoral horn has dimensions of a = 0.5λ, b = 0.25λ, b1 = 2.75λ, and 𝜌1 = 6λ. Com-pute the directivity using (13-18) and (13-19c). Compare the answers.
Solution: For this horn b1 √ 2λ𝜌1 = 2.75 √ 2(6) = 0.794 H-PLANE SECTORAL HORN 733 Therefore (from Appendix IV) [C(0.794)]2 = (0.72)2 = 0.518 [S(0.794)]2 = (0.24)2 = 0.0576 Using (13-18) DE = 64(0.5)6 2.75𝜋(0.518 + 0.0576) = 12.79 = 11.07 dB To compute the directivity using (13-19c), the following parameters are evaluated: 𝜌e = λ √ (6)2 + (2.75 2 )2 = 6.1555λ √ 50 𝜌e∕λ = √ 50 6.1555 = 2.85 B = 2.75(2.85) = 7.84 For B = 7.84, GE = 73.5 from Figure 13.8. Thus, using (13-19c) DE = 0.5(73.5) 2.85 = 12.89 = 11.10 dB Obviously an excellent agreement between the results of (13-18) and (13-19c).
13.3 H-PLANE SECTORAL HORN Flaring the dimensions of a rectangular waveguide in the direction of the H-field, while keeping the other constant, forms an H-plane sectoral horn shown in Figure 13.1(b). A more detailed geometry is shown in Figure 13.9.
The analysis procedure for this horn is similar to that for the E-plane horn, which was outlined in the previous section. Instead of including all the details of the formulation, a summary of each radiation characteristic will be given.
13.3.1 Aperture Fields The fields at the aperture of the horn can be found by treating the horn as a radial waveguide forming an imaginary apex shown dashed in Figure 13.9. Using this method, it can be shown that at the aperture of the horn E′ x = H′ y = 0 (13-20a) E′ y(x′) = E2 cos ( 𝜋 a1 x′ ) e−jk𝛿(x′) (13-20b) H′ x(x′) = −E2 𝜂cos ( 𝜋 a1 x′ ) e−jk𝛿(x′) (13-20c) 𝛿(x′) = 1 2 ( x′2 𝜌2 ) (13-20d) 𝜌2 = 𝜌h cos 𝜓h (13-20e) 734 HORN ANTENNAS Figure 13.9 H-plane sectoral horn and coordinate system.
13.3.2 Radiated Fields The fields radiated by the horn can be found by first formulating the equivalent current densities Js and Ms. Using (13-20a)–(13-20c), it can be shown that over the aperture of the horn Jx = Jz = My = Mz = 0 (13-21a) Jy = −E2 𝜂cos ( 𝜋 a1 x′ ) e−jk𝛿(x′) (13-21b) Mx = E2 cos ( 𝜋 a1 x′ ) e−jk𝛿(x′) (13-21c) and they are assumed to be zero elsewhere. Thus (12-11a) can be expressed as N𝜃= ∫∫ S Jy cos 𝜃sin 𝜙e+jkr′ cos 𝜓ds′ = −E2 𝜂cos 𝜃sin 𝜙I1I2 (13-22) H-PLANE SECTORAL HORN 735 where I1 = ∫ +b∕2 −b∕2 e+jky′ sin 𝜃sin 𝜙dy′ = b ⎡ ⎢ ⎢ ⎢ ⎣ sin (kb 2 sin 𝜃sin 𝜙 ) kb 2 sin 𝜃sin 𝜙 ⎤ ⎥ ⎥ ⎥ ⎦ (13-22a) I2 = ∫ +a1∕2 −a1∕2 cos ( 𝜋 a1 x′ ) e−jk[𝛿(x′)−x′ sin 𝜃cos 𝜙] dx′ (13-22b) By rewriting cos[(𝜋∕a1)x′] as cos ( 𝜋 a1 x′ ) = [ ej(𝜋∕a1)x′ + e−j(𝜋∕a1)x′ 2 ] (13-23) (13-22b) can be expressed as I2 = I′ 2 + I′′ 2 (13-24) where I′ 2 = 1 2 √𝜋𝜌2 k ej(k′2 x 𝜌2∕2k){[C(t′ 2) −C(t′ 1)] −j[S(t′ 2) −S(t′ 1)]} (13-25) t′ 1 = √ 1 𝜋k𝜌2 ( −ka1 2 −k′ x𝜌2 ) (13-25a) t′ 2 = √ 1 𝜋k𝜌2 ( +ka1 2 −k′ x𝜌2 ) (13-25b) k′ x = k sin 𝜃cos 𝜙+ 𝜋 a1 (13-25c) I′′ 2 = 1 2 √𝜋𝜌2 k ej(k′′2 x 𝜌2∕2k){[C(t′′ 2 ) −C(t′′ 1 )] −j[S(t′′ 2 ) −S(t′′ 1 )]} (13-26) t′′ 1 = √ 1 𝜋k𝜌2 ( −ka1 2 −k′′ x 𝜌2 ) (13-26a) t′′ 2 = √ 1 𝜋k𝜌2 ( +ka1 2 −k′′ x 𝜌2 ) (13-26b) k′′ x = k sin 𝜃cos 𝜙−𝜋 a1 (13-26c) 736 HORN ANTENNAS C(x) and S(x) are the cosine and sine Fresnel integrals of (13-8c) and (13-8d), and they are well tabulated (see Appendix IV).
With the aid of (13-22a), (13-24), (13-25), and (13-26), (13-22) reduces to N𝜃= −E2 b 2 √𝜋𝜌2 k {cos 𝜃sin 𝜙 𝜂 sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-27) F(t1, t2) = [C(t2) −C(t1)] −j[S(t2) −S(t1)] (13-27a) f1 = k′2 x 𝜌2 2k (13-27b) f2 = k′′2 x 𝜌2 2k (13-27c) Y = kb 2 sin 𝜃sin 𝜙 (13-27d) In a similar manner, N𝜙, L𝜃, and L𝜙of (12-12b)–(12-12d) can be written as N𝜙= −E2 b 2 √𝜋𝜌2 k {cos 𝜙 𝜂 sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-28a) L𝜃= E2 b 2 √𝜋𝜌2 k { cos 𝜃cos 𝜙sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-28b) L𝜙= −E2 b 2 √𝜋𝜌2 k { sin 𝜙sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-28c) The far-zone electric field components of (12-10a)–(12-10c) can then be expressed as Er = 0 (13-29a) E𝜃= jE2 b 8 √ k𝜌2 𝜋 e−jkr r × { sin 𝜙(1 + cos 𝜃)sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-29b) E𝜙= jE2 b 8 √ k𝜌2 𝜋 e−jkr r × { cos 𝜙(cos 𝜃+ 1)sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-29c) H-PLANE SECTORAL HORN 737 The electric field in the principal E- and H-planes reduces to E-Plane (𝝓= 𝝅∕2) Er = E𝜙= 0 (13-30a) E𝜃= jE2 b 8 √ k𝜌2 𝜋 e−jkr r × { (1 + cos 𝜃)sin Y Y [ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )] } (13-30b) Y = kb 2 sin 𝜃 (13-30c) k′ x = 𝜋 a1 (13-30d) k′′ x = −𝜋 a1 (13-30e) H-Plane (𝝓= 0) Er = E𝜃= 0 (13-31a) E𝜙= jE2 b 8 √ k𝜌2 𝜋 e−jkr r × {(cos 𝜃+ 1)[ejf 1F(t′ 1, t′ 2) + ejf 2F(t′′ 1 , t′′ 2 )]} (13-31b) k′ x = k sin 𝜃+ 𝜋 a1 (13-31c) k′′ x = k sin 𝜃−𝜋 a1 (13-31d) with f1, f2, F(t′ 1, t′ 2), F(t′′ 1 , t′′ 2 ), t′ 1, t′ 2, t′′ 1 , and t′′ 2 as defined previously.
Computations similar to those for the E-plane sectoral horn were also performed for the H-plane sectoral horn. A three-dimensional field pattern of an H-plane sectoral horn is shown in Figure 13.10.
Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 θ θ Relative Amplitude 180° 180° H-plane (x-z plane, ϕ = 0°) E-plane (y-z plane, ϕ = 90°) Figure 13.10 Three-dimensional field pattern of an H-plane sectoral horn (𝜌2 = 6λ, a1 = 5.5λ, b = 0.25λ).
738 HORN ANTENNAS Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 H-plane E-plane Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ Figure 13.11 E- and H-plane patterns of H-plane sectoral horn.
Its corresponding E- and H-plane patterns are displayed in Figure 13.11. This horn exhibits narrow pattern characteristics in the flared H-plane.
Normalized H-plane patterns for a given length horn (𝜌2 = 12λ) and different flare angles are shown in Figure 13.12. A total of four patterns is illustrated. Since each pattern is symmetrical, only half of each pattern is displayed. As the included angle is increased, the pattern begins to become narrower up to a given flare. Beyond that point the pattern begins to broaden, attributed primarily to the phase taper (phase error) across the aperture of the horn. To correct this, a lens is usually placed at the horn aperture, which would yield narrower patterns as the flare angle is increased. Similar pattern variations are evident when the flare angle of the horn is maintained fixed while its length is varied.
13.3.3 Directivity To find the directivity of the H-plane sectoral horn, a procedure similar to that for the E-plane is used. As for the E-plane sectoral horn, the maximum radiation is directed nearly along the z-axis (𝜃= 0◦). Thus |E𝜃|max = |E2| b 4r √ 2𝜌2 λ | sin 𝜙{[C(t′ 2) + C(t′′ 2 ) −C(t′ 1) −C(t′′ 1 )] −j[S(t′ 2) + S(t′′ 2 ) −S(t′ 1) −S(t′′ 1 )]}| (13-32) t′ 1 = √ 1 𝜋k𝜌2 ( −ka1 2 −𝜋 a1 𝜌2 ) (13-32a) t′ 2 = √ 1 𝜋k𝜌2 ( +ka1 2 −𝜋 a1 𝜌2 ) (13-32b) H-PLANE SECTORAL HORN 739 2 = 15° = 20° = 25° = 30° Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ ψh 2ψh 2ψh 2ψh Figure 13.12 H-plane patterns of H-plane sectoral horn for constant length and different included angles.
t′′ 1 = √ 1 𝜋k𝜌2 ( −ka1 2 + 𝜋 a1 𝜌2 ) = −t′ 2 = v (13-32c) t′′ 2 = √ 1 𝜋k𝜌2 ( +ka1 2 + 𝜋 a1 𝜌2 ) = −t′ 1 = u (13-32d) Since C(−x) = −C(x) (13-33a) S(−x) = −S(x) (13-33b) |E𝜃|max = |E2|b r √𝜌2 2λ | | sin 𝜙{[C(u) −C(v)] −j[S(u) −S(v)]}| | (13-34) u = t′′ 2 = −t′ 1 = √ 1 𝜋k𝜌2 ( +ka1 2 + 𝜋 a1 𝜌2 ) = 1 √ 2 (√ λ𝜌2 a1 + a1 √ λ𝜌2 ) (13-34a) v = t′′ 1 = −t′ 2 = √ 1 𝜋k𝜌2 ( −ka1 2 + 𝜋 a1 𝜌2 ) = 1 √ 2 (√ λ𝜌2 a1 − a1 √ λ𝜌2 ) (13-34b) Similarly |E𝜙|max = |E2|b r √𝜌2 2λ | | cos 𝜙{[C(u) −C(v)] −j[S(u) −S(v)]}| | (13-35) 740 HORN ANTENNAS Thus |E|max = √ |E𝜃|2 max + |E𝜙|2 max = |E2|b r √𝜌2 2λ{[C(u) −C(v)]2 + [S(u) −S(v)]2}1∕2 (13-36) Umax = |E2|2 b2𝜌2 4𝜂λ {[C(u) −C(v)]2 + [S(u) −S(v)]2} (13-37) The total power radiated can be obtained by simply integrating the average power density over the mouth of the horn, and it is given by Prad = |E2|2 ba1 4𝜂 (13-38) Using (13-37) and (13-38), the directivity for the H-plane sectoral horn can be written as DH = 4𝜋Umax Prad = 4𝜋b𝜌2 a1λ × {[C(u) −C(v)]2 + [S(u) −S(v)]2} (13-39) where u = 1 √ 2 (√ λ𝜌2 a1 + a1 √ λ𝜌2 ) v = 1 √ 2 (√ λ𝜌2 a1 − a1 √ λ𝜌2 ) (13-39a) (13-39b) The half-power beamwidth (HPBW) as a function of flare angle is plotted in Figure 13.13. The normalized directivity (relative to the constant aperture dimension b) for different horn lengths, as a function of aperture dimension a1, is displayed in Figure 13.14. As for the E-plane sectoral horn, the HPBW exhibits a monotonic decrease and the directivity a monotonic increase up to a given flare; beyond that, the trends are reversed.
If the values of a1 (in λ), which correspond to the maximum directivities in Figure 13.14, are plotted versus their corresponding values of 𝜌2 (in λ), it can be shown that each optimum directivity occurs when a1 ≃ √ 3λ𝜌2 (13-39c) with a corresponding value of t equal to t|a1= √ 3λ𝜌2 = top = a2 1 8λ𝜌2 | | | | |a1= √ 3λ𝜌2 = 3 8 (13-39d) H-PLANE SECTORAL HORN 741 Figure 13.13 Half-power beamwidth of H-plane sectoral horn as a function of included angle and for differ-ent lengths.
The directivity of an H-plane sectoral horn can also be computed by using the following proce-dure .
1. Calculate A by A = a1 λ √ 50 𝜌h∕λ (13-40a) 2. Using this value of A, find the corresponding value of GH from Figure 13.15. If the value of A is smaller than 2, then compute GH using GH = 32 𝜋A (13-40b) 3. Calculate DH by using the value of GH from Figure 13.15 or from (13-40b). Thus DH = b λ GH √ 50 𝜌h∕λ (13-40c) This is the actual directivity of the horn.
742 HORN ANTENNAS Figure 13.14 Normalized directivity of H-plane sectoral horn as a function of aperture size and for differ-ent lengths.
Figure 13.15 GH as a function of A. (source: Adopted from data by E. H. Braun, “Some Data for the Design of Electromagnetic Horns,” IRE Trans. Antennas Propagat., Vol. AP-4, No. 1, January 1956. c ⃝1956 IEEE).
PYRAMIDAL HORN 743 Example 13.3 An H-plane sectoral horn has dimensions of a = 0.5λ, b = 0.25λ, a1 = 5.5λ, and 𝜌2 = 6λ. Com-pute the directivity using (13-39) and (13-40c). Compare the answers.
Solution: For this horn u = 1 √ 2 (√ 6 5.5 + 5.5 √ 6 ) = 1.9 v = 1 √ 2 (√ 6 5.5 −5.5 √ 6 ) = −1.273 Therefore (from Appendix IV) C(1.9) = 0.394 C(−1.273) = −C(1.273) = −0.659 S(1.9) = 0.373 S(−1.273) = −S(1.273) = −0.669 Using (13-39) DH = 4𝜋(0.25)6 5.5 [(0.394 + 0.659)2 + (0.373 + 0.669)2] DH = 7.52 = 8.763 dB To compute the directivity using (13-40c), the following parameters are computed: 𝜌h = λ √ (6)2 + (5.5∕2)2 = 6.6λ √ 50 𝜌h∕λ = √ 50 6.6 = 2.7524 A = 5.5(2.7524) = 15.14 For A = 15.14, GH = 91.8 from Figure 13.15. Thus, using (13-40c) DH = 0.25(91.8) 2.7524 = 8.338 = 9.21 dB Although there is a good agreement between the results of (13-39) and (13-40c), they do not compare as well as those of Example 13.2.
13.4 PYRAMIDAL HORN The most widely used horn is the one which is flared in both directions, as shown in Figure 13.16. It is widely referred to as a pyramidal horn, and its radiation characteristics are essentially a combination of the E- and H-plane sectoral horns.
744 HORN ANTENNAS Figure 13.16 Pyramidal horn and coordinate system.
13.4.1 Aperture Fields, Equivalent, and Radiated Fields To simplify the analysis and to maintain a modeling that leads to computations that have been shown to correlate well with experimental data, the tangential components of the E- and H-fields over the aperture of the horn are approximated by E′ y(x′, y′) = E0 cos ( 𝜋 a1 x′ ) e−j[k(x′2∕𝜌2+y′2∕𝜌1)∕2] (13-41a) H′ x(x′, y′) = −E0 𝜂cos ( 𝜋 a1 x′ ) e−j[k(x′2∕𝜌2+y′2∕𝜌1)∕2] (13-41b) PYRAMIDAL HORN 745 and the equivalent current densities by Jy(x′, y′) = −E0 𝜂cos ( 𝜋 a1 x′ ) e−j[k(x′2∕𝜌2+y′2∕𝜌1)∕2] (13-42a) Mx(x′, y′) = E0 cos ( 𝜋 a1 x′ ) e−j[k(x′2∕𝜌2+y′2∕𝜌1∕2] (13-42b) The above expressions contain a cosinusoidal amplitude distribution in the x′ direction and quadratic phase variations in both the x′ and y′ directions, similar to those of the sectoral E- and H-plane horns.
The N𝜃, N𝜙, L𝜃and L𝜙can now be formulated as before, and it can be shown that they are given by N𝜃= −E0 𝜂cos 𝜃sin 𝜙I1I2 (13-43a) N𝜙= −E0 𝜂cos 𝜙I1I2 (13-43b) L𝜃= E0 cos 𝜃cos 𝜙I1I2 (13-43c) L𝜙= −E0 sin 𝜙I1I2 (13-43d) where I1 = ∫ +a1∕2 −a1∕2 cos ( 𝜋 a1 x′ ) e−jk[x′2∕(2𝜌2)−x′sin𝜃cos𝜙] dx′ (13-43e) I2 = ∫ +b1∕2 −b1∕2 e−jk[y′2∕(2𝜌1)−y′sin𝜃sin𝜙] dy′ (13-43f) Using (13-22b), (13-24), (13-25), and (13-26), (13-43e) can be expressed as I1 = 1 2 √𝜋𝜌2 k (ej(k′2 x 𝜌2∕2k){[C(t′ 2) −C(t′ 1)] −j[S(t′ 2) −S(t′ 1)]} + ej(k′′2 x 𝜌2∕2k){[C(t′′ 2 ) −C(t′′ 1 )] −j[S(t′′ 2 ) −S(t′′ 1 )]}) (13-44) where t′ 1, t′ 2, k′ x, t′′ 1 , t′′ 2 , and k′′ x are given by (13-25a)–(13-25c) and (13-26a)–(13-26c). Similarly, using (13-5)–(13-8d), I2 of (13-43f) can be written as I2 = √𝜋𝜌1 k ej(k2 y𝜌1∕2k){[C(t2) −C(t1)] −j[S(t2) −S(t1)]} (13-45) where ky, t1, and t2 are given by (13-5a), (13-8a), and (13-8b).
746 HORN ANTENNAS Combining (13-43a)–(13-43d), the far-zone E- and H-field components of (12-10a)–(12-10c) reduce to Er = 0 E𝜃= −jkejkr 4𝜋r [L𝜙+ 𝜂N𝜃] = jkE0e−jkr 4𝜋r [sin 𝜙(1 + cos 𝜃)I1I2] E𝜙= +jke−jkr 4𝜋r [L𝜃−𝜂N𝜙] = jkE0e−jkr 4𝜋r [cos 𝜙(cos 𝜃+ 1)I1I2] (13-46a) (13-46b) (13-46c) where I1 and I2 are given by (13-44) and (13-45), respectively.
The fields radiated by a pyramidal horn, as given by (13-46a)–(13-46c), are valid for all angles of observation. An examination of these equations reveals that the principal E-plane pattern (𝜙= 𝜋∕2) of a pyramidal horn, aside from a normalization factor, is identical to the E-plane pattern of an E-plane sectoral horn. Similarly the H-plane (𝜙= 0) is identical to that of an H-plane sectoral horn.
Therefore the pattern of a pyramidal horn is very narrow in both principal planes and, in fact, in all planes. This is illustrated in Figure 13.17. The corresponding E-plane pattern is shown in Figure 13.4 and the H-plane pattern in Figure 13.11.
To demonstrate that the maximum radiation for a pyramidal horn is not necessarily directed along its axis, the three-dimensional field pattern for a horn with 𝜌1 = 𝜌2 = 6λ, a1 = 12λ, b1 = 6λ, a = 0.50λ and b = 0.25λ is displayed in Figure 13.18. The corresponding two-dimensional E- and H-plane patterns are shown in Figure 13.19. The maximum does not occur on Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Relative Amplitude 180° 180° H-plane (x-z plane, = 0°) E-plane (y-z plane, = 90°) θ θ ϕ ϕ Figure 13.17 Three-dimensional field pattern of a pyramidal horn (𝜌1 = 𝜌2 = 6λ, a1 = 5.5λ, b1 = 2.75λ, a = 0.5λ, b = 0.25λ).
PYRAMIDAL HORN 747 Normalized Antenna Pattern (linear scale) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Relative Amplitude 180° 180° θ θ H-plane (x-z plane, = 0°) E-plane (y-z plane, = 90°) ϕ ϕ Figure 13.18 Three-dimensional field pattern of a pyramidal horn with maximum not on axis (𝜌1 = 𝜌2 = 6λ, a1 = 12λ, b1 = 6λ, a = 0.5λ, b = 0.25λ).
axis because the phase error taper at the aperture is such that the rays emanating from the different parts of the aperture toward the axis are not in phase and do not add constructively.
To physically construct a pyramidal horn, the dimension pe of Figure 13.16(b) given by pe = (b1 −b) [(𝜌e b1 )2 −1 4 ]1∕2 (13-47a) should be equal to the dimension ph of Figure 13.16(c) given by ph = (a1 −a) [(𝜌h a1 )2 −1 4 ]1∕2 (13-47b) The dimensions chosen for Figures 13.17 and 13.18 do satisfy these requirements. For the horn of Figure 13.17, 𝜌e = 6.1555λ, 𝜌h = 6.6λ, and pe = ph = 5.4544λ, whereas for that of Figure 13.18, 𝜌e = 6.7082λ, 𝜌h = 8.4853λ, and pe = ph = 5.75λ. The fields of (13-46a)–(13-46c) provide accu-rate patterns for angular regions near the main lobe and its closest minor lobes. To accurately predict the field intensity of the pyramidal and other horns, especially in the minor lobes, diffraction tech-niques can be utilized –. These methods take into account diffractions that occur near the aperture edges of the horn. The diffraction contributions become more dominant in regions where the radiation of (13-46a)–(13-46c) is of very low intensity.
In addition to the previous methods, the horn antenna has been examined using full-wave analyses, such as the Method of Moments (MoM) and the Finite-Difference Time-Domain (FDTD) .
These methods yield more accurate results in all regions, and they are able to include many of the 748 HORN ANTENNAS Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 E-plane H-plane Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 θ θ Figure 13.19 E- and H-plane amplitude patterns of a pyramidal horn with maximum not on axis.
other features of the horn, such as its wall thickness, etc. Predicted patterns based on these methods compare extremely well with measurements, even in regions of very low intensity (such as the back lobes). An example of such a comparison is made in Figure 13.20(a,b) for the E- and H-plane pat-terns of a 20-dB standard-gain horn whose Method of Moment predicted values are compared with measured patterns and with predicted values based on (13-46a) and (13-46c), which in Figure 13.20 are labeled as approximate. It is apparent that the MoM predicted patterns compare extremely well with the measured data.
All of the patterns presented previously represent the main polarization of the field radiated by the antenna (referred to as copolarized or copol). If the horn is symmetrical and it is excited in the dominant mode, ideally there should be no field component radiated by the antenna which is orthogonal to the main polarization (referred to as cross polarization or cross-pol), especially in the principal planes. However, in practice, either because of nonsymmetries, defects in construction and/or excitation of higher-order modes, all antennas exhibit cross-polarized components. These cross-pol components are usually of very low intensity compared to those of the primary polarization.
For good designs, these should be 30 dB or more below the copolarized fields and are difficult to measure accurately or be symmetrical, as they should be in some cases.
13.4.2 Directivity As for the E- and H-plane sectoral horns, the directivity of the pyramidal configuration is vital to the antenna designer. The maximum radiation of the pyramidal horn is directed nearly along the z-axis (𝜃= 0◦). It is a very simple exercise to show that |E𝜃|max, |E𝜙|max, and in turn Umax can be written, using (13-46b) and (13-46c), as |E𝜃|max = |E0 sin 𝜙| √𝜌1𝜌2 r {[C(u) −C(v)]2 + [S(u) −S(v)]2}1∕2 × { C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )}1∕2 (13-48a) PYRAMIDAL HORN 749 180 150 120 90 60 30 0 30 60 90 120 150 180 –35 –25 –15 –5 5 15 25 θ Moment method Approximate Measured Gain (dB) (degrees) (a) E-plane a1 = 4.87" b1 = 3.62" pe = ph = 10.06" a = 0.9", b = 0.4" a × b a1 × b1 a1 = 4.87" b1 = 3.62" pe = ph = 10.06" a = 0.9", b = 0.4" a1 × b1 θ 180 150 120 90 60 30 0 30 60 90 120 150 180 –55 –35 –45 –25 –15 –5 5 15 25 θ Moment method Approximate Measured Gain (dB) (degrees) (b) H-plane a × b θ pe = ph pe = ph Figure 13.20 Comparison of E- and H-plane patterns for 20-dB standard-gain horn at 10 GHz.
|E𝜙|max = |E0 cos 𝜙| √𝜌1𝜌2 r {[C(u) −C(v)]2 + [S(u) −S(v)]2}1∕2 × { C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )}1∕2 (13-48b) Umax = r2 2𝜂|E|2 max = |E0|2 𝜌1𝜌2 2𝜂{[C(u) −C(v)]2 + [S(u) −S(v)]2} × { C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )} (13-48c) where u and 𝜈are defined by (13-39a) and (13-39b).
750 HORN ANTENNAS Since Prad = |E0|2 a1b1 4𝜂 (13-49) the directivity of the pyramidal horn can be written as Dp = 4𝜋Umax Prad = 8𝜋𝜌1𝜌2 a1b1 {[C(u) −C(v)]2 + [S(u) −S(v)]2} × { C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )} (13-50) which reduces to Dp = 𝜋λ2 32abDEDH (13-50a) where DE and DH are the directivities of the E- and H-plane sectoral horns as given by (13-18) and (13-39), respectively. This is a well-known relationship and has been used extensively in the design of pyramidal horns.
The directivity (in dB) of a pyramidal horn, over isotropic, can also be approximated by Dp(dB) = 10 [ 1.008 + log10 (a1b1 λ2 )] −(Le + Lh) (13-51) where Le and Lh represent, respectively, the losses (in dB) due to phase errors in the E- and H-planes of the horn which are found plotted in Figure 13.21.
The directivity of a pyramidal horn can also be calculated by doing the following .
1. Calculate A = a1 λ √ 50 𝜌h∕λ (13-52a) B = b1 λ √ 50 𝜌e∕λ (13-52b) 2. Using A and B, find GH and GE, respectively, from Figures 13.15 and 13.8. If the values of either A or B or both are smaller than 2, then calculate GE and/or GH by GE = 32 𝜋B (13-52c) GH = 32 𝜋A (13-52d) PYRAMIDAL HORN 751 Figure 13.21 Loss figures for E- and H-planes due to phase errors. (source: W. C. Jakes, in H. Jasik (ed.), Antenna Engineering Handbook, McGraw-Hill, New York, 1961).
3. Calculate Dp by using the values of GE and GH from Figures 13.8 and 13.15 or from (13-52c) and (13-52d). Thus Dp = GEGH 32 𝜋 √ 50 𝜌e∕λ √ 50 𝜌h∕λ = GEGH 10.1859 √ 50 𝜌e∕λ √ 50 𝜌h∕λ = λ2𝜋 32abDEDH (13-52e) where DE and DH are, respectively, the directivities of (13-19c) and (13-40c). This is the actual directivity of the horn. The above procedure has led to results accurate to within 0.01 dB for a horn with 𝜌e = 𝜌h = 50λ.
A commercial X-band (8.2–12.4 GHz) horn is that shown in Figure 13.22. It is a lightweight precision horn antenna, which is usually cast of aluminum, and it can be used as a 1. standard for calibrating other antennas 2. feed for reflectors and lenses 3. pickup (probe) horn for sampling power 4. receiving and/or transmitting antenna.
752 HORN ANTENNAS Figure 13.22 Typical standard gain X-band (8.2–12.4 GHz) pyramidal horn and its gain characteristics.
(Courtesy of The NARDA Microwave Corporation).
It possesses an exponential taper, and its dimensions and typical gain characteristics are indicated in the figure. The half-power beamwidth in both the E- and H-planes is about 28◦while the side lobes in the E- and H-planes are, respectively, about 13 and 20 dB down.
Gains of the horn antenna which were measured, predicted, and provided by the manufacturer, whose amplitude patterns are shown in Figure 13.20, are displayed in Figure 13.23. A very good agreement amongst all three sets is indicated.
8 9 10 11 12 20 25 15 Measured Calculated Provided by mfg Standard gain horn Gain (dBi) Frequency (GHz) Figure 13.23 Gains of the pyramidal horn which were measured, computed, and provided by the manufac-turer. The amplitude patterns of the horn are shown in Figure 13.22.
PYRAMIDAL HORN 753 Example 13.4 A pyramidal horn has dimensions of 𝜌1 = 𝜌2 = 6λ, a1 = 5.5λ, b1 = 2.75λ, a = 0.5λ, and b = 0.25λ.
a. Check to see if such a horn can be constructed physically.
b. Compute the directivity using (13-50a), (13-51), and (13-52e).
Solution: From Examples 13.2 and 13.3.
𝜌e = 6.1555λ 𝜌h = 6.6λ Thus pe = (2.75 −0.25)λ √(6.1555 2.75 )2 −1 4 = 5.454λ ph = (5.5 −0.5)λ √(6.6 5.5 )2 −1 4 = 5.454λ Therefore the horn can be constructed physically.
The directivity can be computed by utilizing the results of Examples 13.2 and 13.3. Using (13-50a) with the values of DE and DH computed using, respectively, (13-18) and (13-39) gives Dp = 𝜋λ2 32abDEDH = 𝜋 32(0.5)(0.25)(12.79)(7.52) = 75.54 = 18.78 dB Utilizing the values of DE and DH computed using, respectively, (13-19c) and (13-40c), the directivity of (13-52e) is equal to Dp = 𝜋λ2 32abDEDH = 𝜋 32(0.5)0.25(12.89)(8.338) = 84.41 = 19.26 dB For this horn s = b2 1 8λ𝜌1 = (2.75)2 8(6) = 0.1575 t = a2 1 8λ𝜌2 = (5.5)2 8(6) = 0.63 For these values of s and t LE = 0.20 dB LH = 2.75 dB from Figure 13.21. Using (13-51) Dp = 10{1.008 + log10[5.5(2.75)]} −(0.20 + 2.75) = 18.93 dB The agreement is best between the directivities of (13-50a) and (13-51).
754 HORN ANTENNAS A MATLAB and FORTRAN computer program entitled Analysis has been developed to analyze the radiation characteristics of a pyramidal horn and the directivities of the corresponding E- and H-plane sectoral horns. The program and the READ ME file are found in the publisher’s website for this book.
13.4.3 Design Procedure The pyramidal horn is widely used as a standard to make gain measurements of other antennas (see Section 17.4, Chapter 17), and as such it is often referred to as a standard gain horn. To design a pyra-midal horn, one usually knows the desired gain G0 and the dimensions a, b of the rectangular feed waveguide. The objective of the design is to determine the remaining dimensions (a1, b1, 𝜌e, 𝜌h, Pe, and Ph) that will lead to an optimum gain. The procedure that follows can be used to accomplish this , .
The design equations are derived by first selecting values of b1 and a1 that lead, respectively, to optimum directivities for the E- and H-plane sectoral horns using (13-18a) and (13-39c). Since the overall efficiency (including both the antenna and aperture efficiencies) of a horn antenna is about 50% , , the gain of the antenna can be related to its physical area. Thus it can be written using (12-37c), (12-38), and (13-18a), (13-39c) as G0 = 1 2 4𝜋 λ2 (a1b1) = 2𝜋 λ2 √ 3λ𝜌2 √ 2λ𝜌1 ≃2𝜋 λ2 √ 3λ𝜌h √ 2λ𝜌e (13-53) since for long horns 𝜌2 ≃𝜌h and 𝜌1 ≃𝜌e. For a pyramidal horn to be physically realizable, Pe and Ph of (13-47a) and (13-47b) must be equal. Using this equality, it can be shown that (13-53) reduces to (√ 2𝜒−b λ )2 (2𝜒−1) = ( G0 2𝜋 √ 3 2𝜋 1 √𝜒 −a λ )2 ( G2 0 6𝜋3 1 𝜒−1 ) (13-54) where 𝜌e λ = 𝜒 𝜌h λ = G2 0 8𝜋3 ( 1 𝜒 ) (13-54a) (13-54b) Equation (13-54) is the horn-design equation.
1. As a first step of the design, find the value of 𝜒which satisfies (13-54) for a desired gain G0 (dimensionless). Use an iterative technique and begin with a trial value of 𝜒(trial) = 𝜒1 = G0 2𝜋 √ 2𝜋 (13-55) 2. Once the correct 𝜒has been found, determine 𝜌e and 𝜌h using (13-54a) and (13-54b), respec-tively.
PYRAMIDAL HORN 755 3. Find the corresponding values of a1 and b1 using (13-18a) and (13-39c) or a1 = √ 3λ𝜌2 ≃ √ 3λ𝜌h = G0 2𝜋 √ 3 2𝜋𝜒λ (13-56a) b1 = √ 2λ𝜌1 ≃ √ 2λ𝜌e = √ 2𝜒λ (13-56b) 4. The values of pe and ph can be found using (13-47a) and (13-47b).
A MATLAB and FORTRAN computer program entitled Design has been developed to accomplish this, and it is included in the publisher’s website for this book.
Example 13.5 Design an optimum gain X-band (8.2–12.4 GHz) pyramidal horn so that its gain (above isotropic) at f = 11 GHz is 22.6 dB. The horn is fed by a WR 90 rectangular waveguide with inner dimen-sions of a = 0.9 in. (2.286 cm) and b = 0.4 in. (1.016 cm).
Solution: Convert the gain G0 from dB to a dimensionless quantity. Thus G0(dB) = 22.6 = 10 log10 G0 ⇒G0 = 102.26 = 181.97 Since f = 11 GHz, λ = 2.7273 cm and a = 0.8382λ b = 0.3725λ 1. The initial value of 𝜒is taken, using (13-55), as 𝜒1 = 181.97 2𝜋 √ 2𝜋 = 11.5539 which does not satisfy (13-54) for the desired design specifications. After a few iterations, a more accurate value is 𝜒= 11.1157.
2. Using (13-54a) and (13-54b) 𝜌e = 11.1157λ = 30.316 cm = 11.935 in.
𝜌h = 12.0094λ = 32.753 cm = 12.895 in.
3. The corresponding values of a1 and b1 are a1 = 6.002λ = 16.370 cm = 6.445 in.
b1 = 4.715λ = 12.859 cm = 5.063 in.
4. The values of pe and ph are equal to pe = ph = 10.005λ = 27.286 cm = 10.743 in.
The same values are obtained using the computer program at the end of this chapter.
756 HORN ANTENNAS The derived design parameters agree closely with those of a commercial gain horn available in the market. As a check, the gain of the designed horn was computed using (13-50a) and (13-51), assuming an antenna efficiency et of 100%, and (13-53). The values were G0 ≃D0 = 22.4 dB for (13-52a) G0 ≃D0 = 22.1 dB for (13-53) G0 = 22.5 dB for (13-55) All three computed values agree closely with the designed value of 22.6 dB.
The previous formulations for all three horn configurations (E-plane, H-plane, and pyramidal) are based on the use of the quadratic phase term of (13-2b) instead of the spherical phase term of (13-2a). This was necessary so that the integrations can be performed and expressed in terms of cosine and sine Fresnel integrals. In order to examine the differences using the spherical phase term instead of the quadratic, especially as it relates to the directivity, a numerical integration was used in . It was found that the directivity of a pyramidal horn of Figure 13.18 using the spherical phase term, instead of the quadratic r is basically the same for 1. large aperture horns (a1, b1 > 50λ); 2. small peak aperture phase errors (S = 𝜌e −𝜌1 < 0.2λ or T = 𝜌h −𝜌2 < 0.2λ of Figure 13.16).
r is slightly higher (more improved) by as much as 0.6 dB for 1. intermediate apertures (5λ ≤a1 ≤8λ or 5λ ≤b1 ≤8λ); 2. intermediate peak aperture phase errors (0.2λ ≤S = 𝜌e −𝜌1 ≤0.6λ or 0.2λ ≤T = 𝜌h −𝜌2 ≤0.6λ of Figure 13.16).
r can be lower, especially for E-plane horns, for 1. large peak aperture phase errors (S = 𝜌e −𝜌1 > 0.6λ or T = 𝜌h −𝜌2 > 0.6λ of Figure 13.16).
13.5 CONICAL HORN Another very practical microwave antenna is the conical horn shown in Figure 13.24 with a photo of one in Figure 13.25. While the pyramidal, E-, and H-plane sectoral horns are usually fed by a rectangular waveguide, the feed of a conical horn is often a circular waveguide.
The first rigorous treatment of the fields radiated by a conical horn is that of Schorr and Beck .
The modes within the horn are found by introducing a spherical coordinate system and are in terms of spherical Bessel functions and Legendre polynomials. The analysis is too involved and will not be attempted here. However data, in the form of curves have been presented which give a qualitative description of the performance of a conical horn.
Despite its popularity and wide range of applications, the directivity and amplitude patterns of a conical horn have not received the same attention as those of the pyramidal horn. For the classic curves by Gray and Schelkunoff, which were reported in and included in as a reference in many books and papers, there has not been a clear documentation whether the reported directivity curves were computed based on quadratic or spherical phase distributions. This and other issues related CONICAL HORN 757 Figure 13.24 Geometry of conical horn.
to the conical horn, like loss figures and amplitude patterns, are addressed in . Figure 13.26 displays a set of directivity curves, versus diameter of horn aperture, for different axial lengths L.
The curves were computed based on quadratic (QPD) and spherical (SPD) phase distributions, and they are compared with those based on modal solution (MS) as well as those reported in . It is clear that the directivities obtained based on the SPD are in closer agreement with those of the MS as well as those of Gray and Schelkunoff, which were reported in .
Based on the numerical data of Figure 13.26 and using linear curve fitting in the least square sense, a set of approximate equations were obtained for the optimum maximum directivity (Dc)opt and axial length L based on the optimum horn line of Figure 13.26 . These equations can be used for designing optimum directivity conical horn, and they are represented by (Dc)opt ≈15.9749 (L λ ) + 1.7209 (13-57a) (Dc)opt ≈5.1572 (dm λ )2 −0.6451 (dm λ ) + 1.3645 (13-57b) L ≈0.3232 (dm λ )2 −0.0475 (dm λ ) + 0.0052 (13-57c) Figure 13.25 Photo of an X-band commercial conical horn (L = 7.147λ, 2𝜓c = 35◦).
758 HORN ANTENNAS Directivity Dc (dB) Spherical Phase Distribution (SPD) Quadratic Phase Distribution (QPD) Modal Solution (MS) 6 8 10 12 14 16 18 20 22 24 1 2 3 4 5 6 7 8 9 10 Diameter of horn aperture dm (λ) optimum horn line L=λ L=2 λ L=4 λ L=6 λ L=8 λ L=10 λ Figure 13.26 Directivity of a conical horn as a function of aperture diameter for different axial horn lengths.
(source: c ⃝2013 IEEE).
These expressions are used to determine the dimensions of the conical horn once the optimum direc-tivity of the conical horn is specified.
Referring to Figure 13.26, it is apparent that the behavior of a conical horn is similar to that of a pyramidal or a sectoral horn. As the flare angle increases, the directivity for a given length horn increases until it reaches a maximum beyond which it begins to decrease. The decrease is a result of the dominance of the quadratic phase error at the aperture. In the same figure, an optimum directivity line is indicated.
The results of Figure 13.26 behave as those of Figures 13.7 and 13.14. When the horn aperture (dm) is held constant and its length (L) is allowed to vary, the maximum directivity is obtained when the flare angle is zero (𝜓c = 0 or L = ∞). This is equivalent to a circular waveguide of diameter dm. As for the pyramidal and sectoral horns, a lens is usually placed at the aperture of the con-ical horn to compensate for its quadratic phase error. The result is a narrower pattern as its flare increases.
The directivity (in dB) of a conical horn, with an aperture efficiency of 𝜀ap and aperture circum-ference C, can be computed using Dc(dB) = 10 log10 [ 𝜀ap 4𝜋 λ2 (𝜋a2) ] = 10 log10 (C λ )2 −L(s) (13-58) CONICAL HORN 759 where a is the radius of the horn at the aperture and L(s) = −10 log10(𝜀ap) (13-58a) The first term in (13-58) represents the directivity of a uniform circular aperture whereas the second term, represented by (13-58a), is a correction figure to account for the loss in directivity due to the aperture efficiency. Usually the term in (13-58a) is referred to as loss figure which can be computed (in dB) using the advanced expressions of L(s) ≈ { 0.5030 + 5.1123s −7.1138s2 + 23.1401s3, L ≤3λ (13-58b) 0.7853 −0.3976s + 13.112s2 + 3.901s3, L > 3λ (13-58c) where s is the maximum phase deviation (in wavelengths), and it is given by s = d2 m 8λl (13-58d) The directivity of a conical horn is optimum when its diameter is equal to dm ≃ √ 3lλ (13-59) which corresponds to a maximum aperture phase deviation of s = 3∕8 (wavelengths) and a loss figure of about 2.9 dB (or an aperture efficiency of about 51%).
A set of figures based on (13-58b) and (13-58c), and their impact on the maximum directivity of a conical horn, are displayed in Figure 13.27(a,b), respectively, where they are compared with exact data. It is apparent that the advanced expressions of (13-58b) and (13-58c) provide a good approximation to both the loss figure and directivity of the conical horn over an extended range of aperture phase deviation represented by s.
Far-field amplitude patterns, E-plane and H-plane, over an extended dynamic range of 0–60 dB for the X-band conical horn of Figure 13.25, are displayed in Figure 13.28 where they are compared with predictions based on GTD/UTD, simulations and measurements . Excellent agreement is indicated everywhere, including in the back lobe region.
A summary of all the pertinent formulas and equation numbers that can be used to compute the directivity of E-plane, H-plane, pyramidal, and conical horns is listed in Table 13.1.
0 35 30 25 20 15 10 5 0 0 –5 5 10 15 20 25 300 0.5 Exact (13-58b) (13-58c) Exact (13-58b) (13-58c) Maximum aperture phase deviation s (wavelengths) Maximum aperture phase deviation s (wavelengths) Directivity Dc (dB) Loss figure L (dB) (a) Loss figure (b) Maximum directivity 1 1.5 0.5 1 1.5 L = 1.5 L = 1.5 L = 50 L = 50 λ λ λ λ Figure 13.27 Conical horn loss factor and maximum directivity as a function of maximum phase deviations (L = 50λ, L = 1.5λ). (source: c ⃝2013 IEEE).
760 HORN ANTENNAS Figure 13.28 Normalized far-field amplitude patterns of X-band conical horn, of Figure 13.25, at f = 10.5 GHz (L = 7.147λ, 2𝜓c = 35◦). (source: c ⃝2013 IEEE).
CORRUGATED HORN 761 TABLE 13.1 Directivity Formulas and Equation Numbers for Horns Horn Type Directivity Equation Number E-plane DE = 64a𝜌1 𝜋λb1 [ C2 ( b1 √ 2λ𝜌1 ) + S2 ( b1 √ 2λ𝜌1 )] (13-18) E-plane (alternate) DE = a λ GE √ 50 𝜌e∕λ , B = b1 λ √ 50 𝜌e∕λ (13-19c), (13-19a) GE = 32 𝜋B if B < 2 GE from Figure 13.8 if B ≥2 (13-19b) Figure 13.8 H-plane DH = 4𝜋b𝜌2 a1λ {[C(u) −C(𝜈)]2 + [S(u) −S(𝜈)]2} (13-39) u = 1 √ 2 (√ λ𝜌2 a1 + a1 √ λ𝜌2 ) , 𝜈= 1 √ 2 (√ λ𝜌2 a1 − a1 √ λ𝜌2 ) (13-39a), (13-39b) H-plane (alternate) DH = b λ GH √ 50 𝜌h∕λ , A = a1 λ √ 50 𝜌h∕λ (13-40c), (13-40a) GH = 32 𝜋A if A < 2 GH from Figure 13.15 if A ≥2 (13-40b) Figure 13.15 Pyramidal DP = 𝜋λ2 32abDEDH (13-50a), (13-18), (13-39) Pyramidal (alternate) DP(dB) = 10 [ 1.008 + log10 (a1b1 λ2 )] −(Le + Lh) (13-51) Le, Lh Figure 13.21 Pyramidal (alternate) DP = GEGH 32 𝜋 √ 50 𝜌e∕λ √ 50 𝜌h∕λ = λ2𝜋 32abDEDH (13-52e), (13-50a) (13-19c), (13-40c) Conical Dc(dB) = 10 log10 [ 𝜀ap 4𝜋 λ2 (𝜋a2) ] = 10 log10 (C λ )2 −L(s) (13-58) L(s) = −10 log10(𝜀ap) ≃LF(s) ≈ { 0.5030 + 5.1123s −7.1138s2 + 23.1401s3, L ≤3λ 0.7853 −0.3976s + 13.112s2 + 3.901s3, L > 3λ (13-58b), (13-58c) s = d2 m 8λl, dm ≃ √ 3lλ (13-58d), (13-59) 13.6 CORRUGATED HORN The large emphasis placed on horn antenna research in the 1960s was inspired by the need to reduce spillover efficiency and cross-polarization losses and increase aperture efficiencies of large reflectors used in radio astronomy and satellite communications. In the 1970s, high-efficiency and rotation-ally symmetric antennas were needed in microwave radiometry. Using conventional feeds, aperture 762 HORN ANTENNAS (a) Corrugated horn a1 x' z' y' b1 a b w t d c (b) E-plane view b1/2 b1/2 b ψ Figure 13.29 Pyramidal horn with corrugations in the E-plane.
efficiencies of 50–60% were obtained. However, efficiencies of the order of 75–80% can be obtained with improved feed systems utilizing corrugated horns.
The aperture techniques introduced in Chapter 12 can be used to compute the pattern of a horn antenna and would yield accurate results only around the main lobe and the first few minor lobes.
The antenna pattern structure in the back lobe region is strongly influenced by diffractions from the edges, especially from those that are perpendicular to the E-field at the horn aperture. The diffractions lead to undesirable radiation not only in the back lobes but also in the main lobe and in the minor lobes. However, they dominate only in low-intensity regions.
In 1964, Kay realized that grooves on the walls of a horn antenna would present the same boundary conditions to all polarizations and would taper the field distribution at the aperture in all the planes. The creation of the same boundary conditions on all four walls would eliminate the spu-rious diffractions at the edges of the aperture. For a square aperture, this would lead to an almost rotationally symmetric pattern with equal E- and H-plane beamwidths. A corrugated (grooved) pyra-midal horn, with corrugations in the E-plane walls, is shown in Figure 13.29(a) with a side view in Figure 13.29(b). Since diffractions at the edges of the aperture in the H-plane are minimal, corruga-tions are usually not placed on the walls of that plane. Corrugations can also be placed in a conical horn forming a conical corrugated horn, also referred to in as a scalar horn. However, instead of the corrugations being formed as shown in Figure 13.30(a), practically it is much easier to machine them to have the profile shown in Figure 13.30(b).
A photograph of a corrugated conical horn, often referred to as a scalar horn, is shown in Figure 13.31. This type of a horn is widely used as a feed of reflector antennas, especially of the Cassegrain (dual reflector) configuration of Figures 15.9 and 15.30.
CORRUGATED HORN 763 a1 a1 2a t w d c (a) Corrugations perpendicular to surface a1 a1 2a t w d c ψ (b) Corrugations perpendicular to axis ψ Figure 13.30 Side view profiles of conical corrugated horns.
To form an effective corrugated surface, it usually requires 10 or more slots (corrugations) per wavelength . To simplify the analysis of an infinite corrugated surface, the following assumptions are usually required: 1. The teeth of the corrugations are vanishingly thin.
2. Reflections from the base of the slot are only those of a TEM mode.
Figure 13.31 Corrugated conical (scalar) horn. (courtesy: March Microwave Systems, B.V., The Netherlands).
764 HORN ANTENNAS The second assumption is satisfied provided the width of the corrugation (w) is small compared to the free-space wavelength (λ0) and the slot depth (d) (usually w < λ0∕10). For a corrugated surface satisfying the above assumptions, its approximate surface reactance is given by , X = w w + t √𝜇0 𝜀0 tan(k0 d) when w w + t ≃1 (13-60) (13-60a) which can be satisfied provided t ≤w∕10.
The surface reactance of a corrugated surface, used on the walls of a horn, must be capacitive in order for the surface to force to zero the tangential magnetic field parallel to the edge at the wall.
Such a surface will not support surface waves, will prevent illumination of the E-plane edges, and will diminish diffractions. This can be accomplished, according to (13-60), if λ0∕4 < d < λ0∕2 or more generally when (2n + 1)λ0∕4 < d < (n + 1)λ0∕2. Even though the cutoff depth is also a function of the slot width w, its influence is negligible if w < λ0∕10 and λ0∕4 < d < λ0∕2.
The effect of the corrugations on the walls of a horn is to modify the electric field distribution in the E-plane from uniform (at the waveguide-horn junction) to cosine (at the aperture). Through measurements, it has been shown that the transition from uniform to cosine distribution takes place almost at the onset of the corrugations. For a horn of about 45 corrugations, the cosine distribution has been established by the fifth corrugation (from the onset) and the spherical phase front by the fifteenth .
Referring to Figure 13.29(a), the field distribution at the aperture can be written as E′ y(x′, y′) = E0 cos ( 𝜋 a1 x′ ) cos ( 𝜋 b1 y′ ) e−j[k(x′2∕𝜌2+y′2∕𝜌1)∕2] (13-61a) H′ x(x′, y′) = −E0 𝜂cos ( 𝜋 a1 x′ ) cos ( 𝜋 b1 y′ ) e−j[k(x′2∕𝜌2+y′2∕𝜌1)∕2] (13-61b) corresponding to (13-41a) and (13-41b) of the uncorrugated pyramidal horn. Using the above dis-tributions, the fields radiated by the horn can be computed in a manner analogous to that of the pyramidal horn of Section 13.4. Patterns have been computed and compare very well with measure-ments .
In Figure 13.32(a) the measured E-plane patterns of an uncorrugated square pyramidal horn (referred to as the control horn) and a corrugated square pyramidal horn are shown. The aperture size on each side was 3.5 in. (2.96λ at 10 GHz) and the total flare angle in each plane was 50◦. It is evident that the levels of the minor lobes and back lobes are much lower for the corrugated horn than those of the control horn. However the corrugated horn also exhibits a wider main beam for small angles; thus a larger 3-dB beamwidth (HPBW) but a lower 10-dB beamwidth. This is attributed to the absence of the diffracted fields from the edges of the corrugated horn which, for nearly on-axis observations, add to the direct wave contribution because of their in-phase relationship. The fact that the on-axis far-fields of the direct and diffracted fields are nearly in phase is also evident from the pronounced on-axis maximum of the control horn. The E- and H-plane patterns of the corrugated horn are almost identical to those of Figure 13.32(a) over the frequency range from 8 to 14 GHz.
These suggest that the main beam in the E-plane can be obtained from known H-plane patterns of horn antennas.
CORRUGATED HORN 765 0 30 60 90 120 150 180 –50 –40 –30 –20 –10 0 Observation angle (degrees) Amplitude (dB) Frequency = 10 GHz 2 e = 2 h = 50° Aperture- matched horn (computed) Corrugated horn (measured) (a) E-plane (2.96 × 2.96 ) (b) E-plane (8.2 × 8.2 ) (c) E-plane back lobe level (2.96 × 2.96 ) (d) E-plane half-power beamwidth (2.96 × 2.96 ) Conventional horn (measured) 180 120 60 0 60 120 180 –60 –40 –20 0 Observation angle (degrees) Amplitude (dB) Large conventional horn (measured) Large corrugated horn (measured) Frequency = 10 GHz 2 e = 34° 2 h = 31° 7 8 9 10 11 12 13 14 15 16 –50 –40 –30 –20 Back lobe level (dB) Conventional horn (measured) Aperture-matched horn (computed) Corrugated horn (measured) Frequency (GHz) 7 8 9 10 11 12 13 14 15 10 20 30 Half-power beamwidth (degrees) Frequency (GHz) Corrugated horn E-plane (measured) Aperture-matched horn E-plane (computed) All horns H-plane (measured) Conventional horn E-plane (measured) ψ ψ ψ ψ λ λ λ λ λ λ λ λ Figure 13.32 Radiation characteristics of conventional (control), corrugated, and aperture-matched pyramidal horns. (source: (a), (c), (d). W. D. Burnside and C. W. Chuang, “An Aperture-Matched Horn Design,” IEEE Trans. Antennas Propagat., Vol. AP-30, No. 4, pp. 790–796, July 1982. c ⃝1982 IEEE (b) R. E. Laurie and L. Peters, Jr., “Modifications of Horn Antennas for Low Side Lobe Levels,” IEEE Trans. Antennas Propagat., Vol. AP-14, No. 5, pp. 605–610, September 1966. c ⃝1966 IEEE).
766 HORN ANTENNAS In Figure 13.32(b) the measured E-plane patterns of larger control and corrugated square pyra-midal horns, having an aperture of 9.7 in. on each side (8.2λ at 10 GHz) and included angles of 34◦and 31◦in the E- and H-planes, respectively, are shown. For this geometry, the pattern of the corrugated horn is narrower and its minor and back lobes are much lower than those of the corre-sponding control horn. The saddle formed on the main lobe of the control horn is attributed to the out-of-phase relations between the direct and diffracted rays. The diffracted rays are nearly absent from the corrugated horn and the minimum on-axis field is eliminated. The control horn is a thick-edged horn which has the same interior dimensions as the corrugated horn. The H-plane pattern of the corrugated horn is almost identical to the H-plane pattern of the corresponding control horn.
In Figures 13.32(c) and 13.32(d) the back lobe level and the 3-dB beamwidth for the smaller size control and corrugated horns, whose E-plane patterns are shown in Figure 13.32(a), are plotted as a function of frequency. All the observations made previously for that horn are well evident in these figures.
The presence of the corrugations, especially near the waveguide-horn junction, can affect the impedance and VSWR of the antenna. The usual practice is to begin the corrugations at a small distance away from the junction. This leads to low VSWR’s over a broad band. Previously it was indicated that the width w of the corrugations must be small (usually w < λ0∕10) to approximate a corrugated surface. This would cause corona and other breakdown phenomena. However the large corrugated horn, whose E-plane pattern is shown in Figure 13.32(b), has been used in a system whose peak power was 20 kW at 10 GHz with no evidence of any breakdown phenomena.
The design concepts of the pyramidal corrugated horn can be extended to include circumfer-entially corrugated conical horns, as shown in Figures 13.30 and 13.31. Several designs of coni-cal corrugated horns were investigated in terms of pattern symmetry, low cross polarization, low side lobe levels, circular polarization, axial ratio, and phase center –. For small flare angles (𝜓c less than about 20◦to 25◦) the slots can be machined perpendicular to the axis of the horn, as shown in Figure 13.30(b), and the grooves can be considered sections of parallel-plate TEM-mode waveguides of depth d. For large flare angles, however, the slots should be constructed perpendicular to the surface of the horn, as shown in Figure 13.30(a) and implemented in the design of Figure 13.31.
The groove arrangement of Figure 13.30(b) is usually preferred because it is easier to fabricate.
13.7 APERTURE-MATCHED HORNS A horn which provides significantly better performance than an ordinary horn (in terms of pattern, impedance, and frequency characteristics) is that shown in Figure 13.33(a), which is referred to as an aperture-matched horn . The main modification to the ordinary (conventional) horn, which we refer to here as the control horn, consists of the attachment of curved-surface sections to the outside of the aperture edges, which reduces the diffractions that occur at the sharp edges of the aperture and provides smooth matching sections between the horn modes and the free-space radiation.
In contrast to the corrugated horn, which is complex and costly and reduces the diffractions at the edges of the aperture by minimizing the incident field, the aperture-matched horn reduces the diffractions by modifying the structure (without sacrificing size, weight, bandwidth, and cost) so that the diffraction coefficient is minimized. The basic concepts were originally investigated using elliptic cylinder sections, as shown in Figure 13.33(b); however, other convex curved surfaces, which smoothly blend to the ordinary horn geometry at the attachment point, will lead to similar improve-ments. This modification in geometry can be used in a wide variety of horns, and includes E-plane, H-plane, pyramidal, and conical horns. Bandwidths of 2:1 can be attained easily with aperture-matched horns having elliptical, circular, or other curved surfaces. The radii of curvature of the curved surfaces used in experimental models ranged over 1.69λ ≤a ≤8.47λ with a = b and b = 2a. Good results can be obtained by using circular cylindrical surfaces with 2.5λ ≤a ≤5λ.
APERTURE-MATCHED HORNS 767 Curved surface section Curved surface section Horn Modified throat Waveguide R Attached curved surface Horn Curved surface is mounted flush with horn walls Attached curved surface a b a b (a) Basic geometry (b) Elliptical cylinder curved surface (d) Modified throat Edge diffracted Geometrical optics Edge diffracted (c) Diffraction mechanism Attached curved surface Horn Figure 13.33 Geometry and diffraction mechanism of an aperture-matched horn. (source: W. D. Burnside and C. W. Chuang, “An Aperture-Matched Horn Design,” IEEE Trans. Antennas Propagat., Vol. AP-30, No. 4, pp. 790–796, July 1982. c ⃝1982 IEEE).
The basic radiation mechanism of such a horn is shown in Figure 13.33(c). The introduction of the curved sections at the edges does not eliminate diffractions; instead it substitutes edge diffractions by curved-surface diffractions which have a tendency to provide an essentially undisturbed energy flow across the junction, around the curved surface, and into free-space. Compared with conventional horns, this radiation mechanism leads to smoother patterns with greatly reduced black lobes and negligible reflections back into the horn. The size, weight, and construction costs of the aperture-matched horn are usually somewhat larger and can be held to a minimum if half (one-half sections of an ellipse) or quadrant (one-fourth sections of an ellipse) sections are used instead of the complete closed surfaces.
768 HORN ANTENNAS To illustrate the improvements provided by the aperture-matched horns, the E-plane pattern, back lobe level, and half-power beamwidth of a pyramidal 2.96λ × 2.96λ horn were computed and compared with the measured data of corresponding control and corrugated horns. The data are shown in Figures 13.32(a,c,d). It is evident by examining the patterns of Figure 13.32(a) that the aperture-matched horn provides a smoother pattern and lower back lobe level than conventional horns (referred to here as control horn); however, it does not provide, for the wide minor lobes, the same reduction as the corrugated horn. To achieve nearly the same E-plane pattern for all three horns, the overall horn size must be increased. If the modifications for the aperture-matched and corrugated horns were only made in the E-plane edges, the H-plane patterns for all three horns would be virtu-ally the same except that the back lobe level of the aperture-matched and corrugated horns would be greatly reduced.
The back lobe level of the same three horns (control, corrugated, and aperture matched) are shown in Figure 13.32(c). The corrugated horn has lower back lobe intensity at the lower end of the frequency band, while the aperture-matched horn exhibits superior performance at the high end.
However, both the corrugated and aperture-matched horns exhibit superior back lobe level character-istics to the control (conventional) horn throughout almost the entire frequency band. The half-power beamwidth characteristics of the same three horns are displayed in Figure 13.32(d). Because the con-trol (conventional) horn has uniform distribution across the complete aperture plane, compared with the tapered distributions for the corrugated and aperture-matched horns, it possesses the smallest beamwidth almost over the entire frequency band.
In a conventional horn the VSWR and antenna impedance are primarily influenced by the throat and aperture reflections. Using the aperture-matched horn geometry of Figure 13.33(a), the aperture reflections toward the inside of the horn are greatly reduced. Therefore the only remaining dominant factors are the throat reflections. To reduce the throat reflections it has been suggested that a smooth curved surface be used to connect the waveguide and horn walls, as shown in Figure 13.33(d). Such a transition has been applied in the design and construction of a commercial X-band (8.2–12.4 GHz) pyramidal horn (see Figure 13.22), whose tapering is of an exponential nature. The VSWRs mea-sured in the 8–12 GHz frequency band using the conventional exponential X-band horn (shown in Figure 13.22), with and without curved sections at its aperture, are shown in Figure 13.34.
The matched sections used to create the aperture-matched horn were small cylinder sections. The VSWR’s for the conventional horn are very small (less than 1.1) throughout the frequency band because the throat reflections are negligible compared with the aperture reflections. It is evident, however, that the VSWR’s of the corresponding aperture-matched horn are much superior to those of the conventional horn because both the throat and aperture reflections are very minimal.
The basic design of the aperture-matched horn can be extended to include corrugations on its inside surface . This type of design enjoys the advantages presented by both the aperture-matched 8 9 10 11 12 1.00 1.05 1.10 1.15 Aperture-matched NARDA horn Original NARDA horn VSWR Frequency (GHz) Figure 13.34 Measured VSWR for exponentially tapered pyramidal horns (conventional and aperture matched). (source: W. D. Burnside and C. W. Chuang, “An Aperture-Matched Horn Design,” IEEE Trans.
Antennas Propagat., Vol. AP-30, No. 4, pp. 790–796, July 1982. c ⃝1982 IEEE).
MULTIMODE HORNS 769 and corrugated horns with cross-polarized components of less than −45 dB over a significant part of the bandwidth. Because of its excellent cross-polarization characteristics, this horn is recommended for use as a reference and for frequency reuse applications in both satellite and terrestrial applications.
13.8 MULTIMODE HORNS Over the years there has been a need in many applications for horn antennas which provide sym-metric patterns in all planes, phase center coincidence for the electric and magnetic planes, and side lobe suppression. All of these are attractive features for designs of optimum reflector systems and monopulse radar systems. Side lobe reduction is a desired attribute for horn radiators utilized in antenna range, anechoic chamber, and standard gain applications, while pattern plane symmetry is a valuable feature for polarization diversity.
Pyramidal horns have traditionally been used over the years, with good success, in many of these applications. Such radiators, however, possess nonsymmetric beamwidths and undesirable side lobe levels, especially in the E-plane. Conical horns, operating in the dominant TE11 mode, have a tapered aperture distribution in the E-plane. Thus, they exhibit more symmetric electric- and magnetic-plane beamwidths and reduced side lobes than do the pyramidal horns. One of the main drawbacks of a conical horn is its relative incompatibility with rectangular waveguides.
To remove some of the deficiencies of pyramidal and conical horns and further improve some of their attractive characteristics, corrugations were introduced on the interior walls of the waveguides, which lead to the corrugated horns that were discussed in a previous section of this chapter. In some other cases designs were suggested to improve the beamwidth equalization in all planes and reduce side lobe levels by utilizing horn structures with multiple-mode excitations. These have been designated as multimode horns, and some of the designs will be discussed briefly here. For more information the reader should refer to the cited references.
One design of a multimode horn is the “diagonal” horn , shown in Figure 13.35, all of whose cross sections are square and whose internal fields consist of a superposition of TE10 and TE01 modes in a square waveguide. For small flare angles, the field structure within the horn is such that the E-field a a x y (a) x y (b) a a 1.5 1.5 1.0 1.0 0.7 0.7 0.5 0.5 0.2 0.2 k k Figure 13.35 Electric field configuration inside square diagonal horn. (a) Two coexisting equal orthogonal modes. (b) Result of combining the two modes shown in (a). (After Love reprinted with permission of Microwave Journal, Vol. V, No. 3, March 1962).
770 HORN ANTENNAS vector is parallel to one of the diagonals. Although it is not a multimode horn in the true sense of the word, because it does not make use of higher-order TE and TM modes, it does possess the desirable attributes of the usual multimode horns, such as equal beamwidths and suppressed beamwidths and side lobes in the E- and H-planes which are nearly equal to those in the principal planes. These attractive features are accomplished, however, at the expense of pairs of cross-polarized lobes in the intercardinal planes which make such a horn unattractive for applications where a high degree of polarization purity is required.
Diagonal horns have been designed, built, and tested such that the 3-, 10-, and 30-dB beamwidths are nearly equal not only in the principal E- and H-planes, but also in the 45◦and 135◦ planes. Although the theoretical limit of the side lobe level in the principal planes is 31.5 dB down, side lobes of at least 30 dB down have been observed in those planes. Despite a theoretically predicted level of −19.2 dB in the ±45◦planes, side lobes with level of −23 to −27 dB have been observed.
The principal deficiency in the side lobe structure appears in the ±45◦-plane cross-polarized lobes whose intensity is only 16 dB down; despite this, the overall horn efficiency remains high. Com-pared with diagonal horns, conventional pyramidal square horns have H-plane beamwidths which are about 35% wider than those in the E-plane, and side lobe levels in the E-plane which are only 12 to 13 dB down (although those in the H-plane are usually acceptable).
For applications which require optimum performance with narrow beamwidths, lenses are usu-ally recommended for use in conjunction with diagonal horns. Diagonal horns can also be converted to radiate circular polarization by inserting a differential phase shifter inside the feed guide whose cross section is circular and adjusted so that it produces phase quadrature between the two orthogo-nal modes.
Another multimode horn which exhibits suppressed side lobes, equal beamwidths, and reduces cross polarization is the dual-mode conical horn . Basically this horn is designed so that diffrac-tions at the aperture edges of the horn, especially those in the E-plane, are minimized by reducing the fields incident on the aperture edges and consequently the associated diffractions. This is accom-plished by utilizing a conical horn which at its throat region is excited in both the dominant TE11 and higher-order TM11 mode. A discontinuity is introduced at a position within the horn where two modes exist. The horn length is adjusted so that the superposition of the relative amplitudes of the two modes at the edges of the aperture is very small compared with the maximum aperture field mag-nitude. In addition, the dimensions of the horn are controlled so that the total phase at the aperture is such that, in conjunction with the desired amplitude distribution, it leads to side lobe suppression, beamwidth equalization, and phase center coincidence.
Qualitatively the pattern formation of a dual-mode conical horn operating in the TE11 and TM11 modes is accomplished by utilizing a pair of modes which have radiation functions with the same argument. However, one of the modes, in this case the TM11 mode, contains an additional envelope factor which varies very rapidly in the main beam region and remains relatively constant at large angles. Thus, it is possible to control the two modes in such a way that their fields cancel in all direc-tions except within the main beam. The TM11 mode exhibits a null in its far-field pattern. Therefore a dual-mode conical horn possesses less axial gain than a conventional dominant-mode conical horn of the same aperture size. Because of that, dual-mode horns render better characteristics and are more attractive for applications where pattern plane symmetry and side lobe reduction are more important than maximum aperture efficiency. A most important application of a dual-mode horn is as a feed of Cassegrain reflector systems.
Dual-mode conical horns have been designed, built, and tested with relatively good success in their performance. Generally, however, diagonal horns would be good competitors for the dual-mode horns if it were not for the undesirable characteristics (especially the cross-polarized components) that they exhibit in the very important 45◦and 135◦planes. Improved performance can be obtained from dual-mode horns if additional higher-order modes (such as the TE12, TE13 and TM12) are excited and if their relative amplitudes and phases can be properly controlled.
Computed maximum aperture efficiencies of paraboloidal reflectors, using such horns as feeds, have DIELECTRIC-LOADED HORNS 771 reached 90% contrasted with efficiencies of about 76% for reflector systems using conventional dominant-mode horn feeds. In practice the actual maximum efficiency achieved is determined by the number of modes that can be excited and the degree to which their relative amplitudes and phases can be controlled.
The techniques of the dual-mode and multimode conical horns can be extended to the design of horns with rectangular cross sections. In fact a multimode pyramidal horn has been designed, built, and tested to be used as a feed in a low-noise Cassegrain monopulse system . This rectan-gular pyramidal horn utilizes additional higher-order modes to provide monopulse capability, side lobe suppression in both the E- and H-planes, and beamwidth equalization. Specifically the various pattern modes for the monopulse system are formed in a single horn as follows: a. Sum: Utilizes TE10 + TE30 instead of only TE10. When the relative amplitude and phase exci-tations of the higher-order TE30 mode are properly adjusted, they provide side lobe cancellation at the second minor lobe of the TE10-mode pattern b. E-Plane Difference: Utilizes TE11 + TM11 modes c. H-Plane Difference: Utilizes TE20 mode In its input, the horn of contained a four-guide monopulse bridge circuitry, a multimode matching section, a difference mode phasing section, and a sum mode excitation and control section.
To illustrate the general concept, in Figure 13.36(a–c) are plots of three-dimensional patterns of the sum, E-plane difference, and H-plane difference modes which utilize, respectively, the TE10 + TE30, TE11 + TM11, and TE20 modes. The relative excitation between the modes has been controlled so that each pattern utilizing multiple modes in its formation displays its most attractive features for its function.
13.9 DIELECTRIC-LOADED HORNS Over the years much effort has been devoted to enhance the antenna and aperture efficiencies of aperture antennas, especially for those that serve as feeds for reflectors (such as the horn). One technique that was proposed and then investigated was to use dielectric guiding structures, referred to as Dielguides , between the primary feed and the reflector (or subreflector). The technique is simple and inexpensive to implement and provides broadband, highly efficient, and low-noise antenna feeds. The method negates the compromise between taper and spillover efficiencies, and it is based on the principle of internal reflections, which has been utilized frequently in optics. Its role bears a very close resemblance to that of a lens, and it is an extension of the classical parabolic-shaped lens to other geometrical shapes.
Another method that has been used to control the radiation pattern of electromagnetic horns is to insert totally within them various shapes of dielectric material (wedges, slabs, etc.) – to control in a predictable manner not only the phase distribution over the aperture, as is usually done by using the classical parabolic lenses, but also to change the power (amplitude) distribution over the aperture. The control of the amplitude and phase distributions over the aperture are essential in the design of very low side lobe antenna patterns.
Symmetrical loading of the H-plane walls has also been utilized, by proper parameter selection, to create a dominant longitudinal section electric (LSE) mode and to enhance the aperture efficiency and pattern-shaping capabilities of symmetrically loaded horns . The method is simple and inex-pensive, and it can also be utilized to realize high efficiency from small horns which can be used in limited scan arrays. Aperture efficiencies on the order of 92 to 96% have been attained, in contrast to values of 81% for unloaded horns.
A similar technique has been suggested to symmetrically load the E-plane walls of rectangu-lar horns – and eventually to line all four of its walls with dielectric slabs. Other similar 772 HORN ANTENNAS Figure 13.36 Three-dimensional sum and difference (E- and H-planes) field patterns of a monopulse pyra-midal horn. (source: C. A. Balanis, “Horn Antennas,” in Antenna Handbook (Y. T. Lo and S. W. Lee, eds.), c ⃝1988, Van Nostrand Reinhold Co., Inc.).
PHASE CENTER 773 Figure 13.36 (Continued) techniques have been suggested, and a summary of these and other classical papers dealing with dielectric-loaded horns can be found in .
13.10 PHASE CENTER Each far-zone field component radiated by an antenna can be written, in general, as Eu = ̂ uE(𝜃, 𝜙)ej𝜓(𝜃,𝜙) e−jkr r (13-62) where ̂ u is a unit vector. The terms E(𝜃, 𝜙) and 𝜓(𝜃, 𝜙) represent, respectively, the (𝜃, 𝜙) variations of the amplitude and phase.
In navigation, tracking, homing, landing, and other aircraft and aerospace systems it is usually desirable to assign to the antenna a reference point such that for a given frequency, 𝜓(𝜃, 𝜙) of (13-63) is independent of 𝜃and 𝜙(i.e., 𝜓(𝜃, 𝜙) = constant). The reference point which makes 𝜓(𝜃, 𝜙) independent of 𝜃and 𝜙is known as the phase center of the antenna –. When refer-enced to the phase center, the fields radiated by the antenna are spherical waves with ideal spherical wave fronts or equiphase surfaces. Therefore a phase center is a reference point from which radiation is said to emanate, and radiated fields measured on the surface of a sphere whose center coincides with the phase center have the same phase.
For practical antennas such as arrays, reflectors, and others, a single unique phase center valid for all values of 𝜃and 𝜙does not exist; for most, however, their phase center moves along a surface, and its position depends on the observation point. However, in many antenna systems a reference point can be found such that 𝜓(𝜃, 𝜙) = constant, or nearly so, over most of the angular space, especially 774 HORN ANTENNAS over the main lobe. When the phase center position variation is sufficiently small, that point is usually referred to as the apparent phase center.
The need for the phase center can best be explained by examining the radiation characteristics of a paraboloidal reflector (parabola of revolution). Plane waves incident on a paraboloidal reflector focus at a single point which is known as the focal point. Conversely, spherical waves emanating from the focal point are reflected by the paraboloidal surface and form plane waves. Thus in the receiving mode all the energy is collected at a single point. In the transmitting mode, ideal plane waves are formed if the radiated waves have spherical wavefronts and emanate from a single point.
In practice, no antenna is a point source with ideal spherical equiphases. Many of them, however, contain a point from which their radiation, over most of the angular space, seems to have spherical wavefronts. When such an antenna is used as a feed for a reflector, its phase center must be placed at the focal point. Deviations of the feed from the focal point of the reflector lead to phase errors which result in significant gain reductions of the antenna, as illustrated in Section 15.4.1(G) of Chapter 15.
The analytical formulations for locating the phase center of an antenna are usually very laborious and exist only for a limited number of configurations –. Experimental techniques , are available to locate the phase center of an antenna. The one reported in is also discussed in some detail in , and it will not be repeated here. The interested reader is referred to and .
The horn is a microwave antenna that is widely used as a feed for reflectors . To perform as an efficient feed for reflectors, it is imperative that its phase center is known and it is located at the focal point of the reflector. Instead of presenting analytical formulations for the phase center of a horn, graphical data will be included to illustrate typical phase centers.
Usually the phase center of a horn is not located at its mouth (throat) or at its aperture but mostly between its imaginary apex point and its aperture. The exact location depends on the dimensions of the horn, especially on its flare angle. For large flare angles the phase center is closer to the apex. As the flare angle of the horn becomes smaller, the phase center moves toward the aperture of the horn.
Computed phase centers for an E-plane and an H-plane sectoral horn are displayed in Figure 13.37(a,b). It is apparent that for small flare angles the E- and H-plane phase centers are identical. Although each specific design has its own phase center, the data of Figure 13.37(a,b) are typical. If the E- and H-plane phase centers of a pyramidal horn are not identical, its phase center can be taken to be the average of the two.
Phase center nomographs for conical corrugated and uncorrugated (TE11-mode) horns are avail-able , and they can be found in and . The procedure to use these in order to locate a phase center is documented in and , and it is not repeated here. The interested reader is referred to where examples are also illustrated.
13.11 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab and Fortran computer programs, designated r Analysis r Design for computing and displaying the analysis and design characteristics of a pyramidal horn.
c. Matlab-based animation-visualization program, designated te horn, that can be used to ani-mate and visualize the radiation of a two-dimensional horn. A detailed description of this program is provided in Section 1.3.4 and in the corresponding READ ME file of Chapter 1.
d. Power Point (PPT) viewgraphs, in multicolor.
REFERENCES 775 0 5 10 15 20 –1 0 1 2 (a) E-plane H-plane E-plane Flare angle e (degrees) Phase center Z0 (wavelengths) e = 5λ , a = 0.7λ 0 5 10 15 20 –1 0 1 2 (b) H-plane E-plane H-plane Flare angle h (degrees) Phase center Z0 (wavelengths) h = 5λ , a = 0.7λ b a b a h e Z0 e Z0 h ψ ρ ψ ψ ρ ψ ρ ρ Figure 13.37 Phase center location, as a function of flare angle, for E- and H-plane sectoral horns. (Adapted from Hu ).
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22. A. P. King, “The Radiation Characteristics of Conical Horn Antennas,” Proc. IRE, Vol. 38, pp. 249–251, March 1950.
23. N. A. Aboserwal, C. A. Balanis and C. R. Birtcher, “Conical Horn: Gain and Amplitude Patterns,” IEEE Trans. Antenna Propagat., Vol. 61, No. 7, pp. 3427–3433, July 2013.
24. A. F. Kay, “The Scalar Feed,” AFCRL Rep. 64–347, AD601609, March 1964.
25. R. E. Lawrie and L. Peters, Jr., “Modifications of Horn Antennas for Low Side Lobe Levels,” IEEE Trans.
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26. R. S. Elliott, “On the Theory of Corrugated Plane Surfaces,” IRE Trans. Antennas Propagat., Vol. AP-2, No. 2, pp. 71–81, April 1954.
27. C. A. Mentzer and L. Peters, Jr., “Properties of Cutoff Corrugated Surfaces for Corrugated Horn Design,” IEEE Trans. Antennas Propagat., Vol. AP-22, No. 2, pp. 191–196, March 1974.
28. C. A. Mentzer and L. Peters, Jr., “Pattern Analysis of Corrugated Horn Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-24, No. 3, pp. 304–309, May 1976.
29. M. J. Al-Hakkak and Y. T. Lo, “Circular Waveguides and Horns with Anisotropic and Corrugated Bound-aries,” Antenna Laboratory Report No. 73-3, Department of Electrical Engineering, University of Illinois, Urbana, January 1973.
30. B. MacA. Thomas, G. L. James and K. J. Greene, “Design of Wide-Band Corrugated Conical Horns for Cassegrain Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-34, No. 6, pp. 750–757, June 1986.
REFERENCES 777 31. B. MacA. Thomas, “Design of Corrugated Conical Horns,” IEEE Trans. Antennas Propagat., Vol. AP-26, No. 2, pp. 367–372, March 1978.
32. B. MacA. Thomas and K. J. Greene, “A Curved-Aperture Corrugated Horn Having Very Low Cross-Polar Performance,” IEEE Trans. Antennas Propagat., Vol. AP-30, No. 6, pp. 1068–1072, November 1982.
33. G. L. James, “TE11-to-HE11 Mode Converters for Small-Angle Corrugated Horns,” IEEE Trans. Antennas Propagat., Vol. AP-30, No. 6, pp. 1057–1062, November 1982.
34. K. Tomiyasu, “Conversion of TE11 mode by a Large-Diameter Conical Junction,” IEEE Trans. Microwave Theory Tech., Vol. MTT-17, No. 5, pp. 277–279, May 1969.
35. B. MacA. Thomas, “Mode Conversion Using Circumferentially Corrugated Cylindrical Waveguide,” Elec-tron. Lett., Vol. 8, pp. 394–396, 1972.
36. J. K. M. Jansen and M. E. J. Jeuken, “Surface Waves in Corrugated Conical Horn,” Electronic Letters, Vol. 8, pp. 342–344, 1972.
37. Y. Tacheichi, T. Hashimoto, and F. Takeda, “The Ring-Loaded Corrugated Waveguide,” IEEE Trans.
Microwave Theory Tech., Vol. MTT-19, No. 12, pp. 947–950, December 1971.
38. F. Takeda and T. Hashimoto, “Broadbanding of Corrugated Conical Horns by Means of the Ring-Loaded Corrugated Waveguide Structure,” IEEE Trans. Antennas Propagat., Vol. AP-24, No. 6, pp. 786–792, November 1976.
39. W. D. Burnside and C. W. Chuang, “An Aperture-Matched Horn Design,” IEEE Trans. Antennas Propagat., Vol. AP-30, No. 4, pp. 790–796, July 1982.
40. A. W. Love, “The Diagonal Horn Antenna,” Microwave Journal, Vol. V, pp. 117–122, March 1962.
41. P. D. Potter, “A New Horn Antenna with Suppressed Side Lobes and Equal Beamwidths,” Microwave Journal, pp. 71–78, June 1963.
42. P. D. Potter and A. C. Ludwig, “Beamshaping by Use of Higher-Order Modes in Conical Horns,” Northeast Electron. Res. Eng. Mtg., pp. 92–93, November 1963.
43. P. A. Jensen, “A Low-Noise Multimode Cassegrain Monopulse with Polarization Diversity,” Northeast Electron. Res. Eng. Mtg., pp. 94–95, November 1963.
44. H. E. Bartlett and R. E. Moseley, “Dielguides—Highly Efficient Low-Noise Antenna Feeds,” Microwave Journal, Vol. 9, pp. 53–58, December 1966.
45. L. L. Oh, S. Y. Peng, and C. D. Lunden, “Effects of Dielectrics on the Radiation Patterns of an Electro-magnetic Horn,” IEEE Trans. Antennas Propagat., Vol. AP-18, No. 4, pp. 553–556, July 1970.
46. G. N. Tsandoulas and W. D. Fitzgerald, “Aperture Efficiency Enhancement in Dielectrically Loaded Horns,” IEEE Trans. Antennas Propagat., Vol. AP-20, No. 1, pp. 69–74, January 1972.
47. R. Baldwin and P. A. McInnes, “Radiation Patterns of Dielectric Loaded Rectangular Horns,” IEEE Trans.
Antennas Propagat., Vol. AP-21, No. 3, pp. 375–376, May 1973.
48. C. M. Knop, Y. B. Cheng, and E. L. Osterlag, “On the Fields in a Conical Horn Having an Arbitrary Wall Impedance,” IEEE Trans. Antennas Propagat., Vol. AP-34, No. 9, pp. 1092–1098, September 1986.
49. J. J. H. Wang, V. K. Tripp, and R. P. Zimmer, “Magnetically Coated Horn for Low Sidelobes and Low Cross-Polarization,” IEE Proc., Vol. 136, pp. 132–138, April 1989.
50. J. J. H. Wang, V. K. Tripp, and J. E. Tehan, “The Magnetically Coated Conducting Surface as a Dual Conductor and Its Application to Antennas and Microwaves,” IEEE Trans. Antennas Propagat., Vol.
AP-38, No. 7, pp. 1069–1077, July 1990.
51. K. Liu and C. A. Balanis, “Analysis of Horn Antennas with Impedance Walls,” 1990 IEEE Antennas and Propagation Symposium Digest, Vol. I, pp. 1184–1187, May 7–11, 1990, Dallas, TX.
52. K. Liu and C. A. Balanis, “Low-Loss Material Coating for Horn Antenna Beam Shaping,” 1991 IEEE Antennas and Propagation Symposium Digest, Vol. 3, pp. 1664–1667, June 24–28, 1991, London, Ontario, Canada.
53. P. A. Tirkas, “Finite-Difference Time-Domain for Aperture Antenna Radiation,” PhD Dissertation, Dept.
of Electrical Engineering, Arizona State University, December 1993.
54. Y. Y. Hu, “A Method of Determining Phase Centers and Its Applications to Electromagnetic Horns,” Journal of the Franklin Institute, Vol. 271, pp. 31–39, January 1961.
778 HORN ANTENNAS 55. E. R. Nagelberg, “Fresnel Region Phase Centers of Circular Aperture Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-13, No. 3, pp. 479–480, May 1965.
56. I. Ohtera and H. Ujiie, “Nomographs for Phase Centers of Conical Corrugated and TE11 Mode Horns,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 6, pp. 858–859, November 1975.
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PROBLEMS 13.1.
Derive (13-1a)–(13-1e) by treating the E-plane horn as a radial waveguide.
13.2.
Design an E-plane horn such that the maximum phase difference between two points at the aperture, one at the center and the other at the edge, is 120◦. Assuming that the maximum length along its wall (𝜌e), measured from the aperture to its apex, is 10λ, find the (a) maximum total flare angle of the horn (b) largest dimension of the horn at the aperture (c) directivity of the horn (dimensionless and in dB) (d) gain of the antenna (in dB) when the reflection coefficient within the waveguide feeding the horn is 0.2. Assume only mismatch losses. The waveguide feeding the horn has dimensions of 0.5λ and 0.25λ 13.3.
For an E-plane horn with 𝜌1 = 6λ, b1 = 3.47λ, and a = 0.5λ, (a) compute (in dB) its pattern at 𝜃= 0◦, 10◦, and 20◦.
(b) compute its directivity using (13-18) and (13-19c). Compare the answers.
13.4.
Repeat Problem 13.3 for 𝜌1 = 6λ, b1 = 6λ, and a = 0.5λ.
13.5.
For an E-plane sectoral horn, plot b1 (in λ) versus 𝜌1 (in λ) using (13-18a). Verify, using the data of Figure 13.7, that the maximum directivities occur when (13-18a) is satisfied.
13.6.
For an E-plane sectoral horn with 𝜌1 = 20λ, a = 0.5λ (a) find its optimum aperture dimensions for maximum normalized directivity (b) compute the total flare angle of the horn (c) compute its directivity, using (13-18), and compare it with the graphical answer (d) find its half-power beamwidth (in degrees) (e) compute the directivity using (13-19c) 13.7.
An E-plane horn is fed by an X-band WR 90 rectangular waveguide with inner dimensions of 0.9 in. (2.286 cm) and b = 0.4 in. (1.016 cm). Design the horn so that its maximum directivity at f = 11 GHz is 30 (14.77 dB).
13.8.
Design an optimum directivity E-plane sectoral horn whose axial length is 𝜌1 = 10λ. The horn is operating at X-band with a desired center frequency equal to f = 10 GHz. The dimen-sions of the feed waveguide are a = 0.9 in. (2.286 cm) and b = 0.4 in. (1.016 cm). Assuming a 100% efficient horn (e0 = 1), find the (a) horn aperture dimensions b1 and 𝜌e (in λ), and flare half-angle 𝜓e (in degrees) PROBLEMS 779 (b) directivity DE (in dB) using (13-19c) (c) aperture efficiency (d) largest phase difference (in degrees) between center of horn at the aperture and any point on the horn aperture along the principal E-plane.
13.9.
Derive (13-20a)–(13-20e) by treating the H-plane horn as a radial waveguide.
13.10.
For an H-plane sectoral horn with 𝜌2 = 6λ, a1 = 6λ, and b = 0.25λ compute the (a) directivity (in dB) using (13-39), (13-40c) and compare the answers (b) normalized field strength (in dB) at 𝜃= 30◦, 45◦, and 90◦.
13.11.
For an H-plane sectoral horn, plot a1 (in λ) versus 𝜌2 (in λ) using (13-39c). Verify, using the data of Figure 13.14, that the maximum directivities occur when (13-39c) is satisfied.
13.12.
Design an H-plane horn so that its maximum directivity at f = 10 GHz is 13.25 dB.
The horn is fed with a standard X-band waveguide with dimensions a = 2.286 cm and b = 1.016 cm. Determine: (a) the horn aperture dimension a1 (in cm).
(b) the axial length 𝜌2 of the horn (in cm).
(c) the flare angle of the horn (in degrees).
13.13.
An H-plane sectoral horn is fed by an X-band WR 90 rectangular waveguide with dimen-sions of a = 0.9 in. (2.286 cm) and b = 0.4 in. (1.016 cm). Design the horn so that its max-imum directivity at f = 11 GHz is 16.3 (12.12 dB).
13.14.
Repeat the design of Problem 13.8 for an H-plane sectoral horn where axial length is also 𝜌2 = 10λ. The feed waveguide dimensions and center frequency of operation are the same as in Problem 13.8. Assuming an 100% efficient horn (e0 = 1), find the (a) horn aperture dimensions a1 and 𝜌h (in λ), and the flare half-angle 𝜓h (in degrees) (b) directivity DH (in dB) using (13-40c) (c) aperture efficiency (d) largest phase difference (in degrees) between the center of the horn at the aperture and any point on the horn aperture along the principal H-plane.
13.15.
Show that (13-47a) and (13-47b) must be satisfied in order for a pyramidal horn to be phys-ically realizable.
13.16.
A standard-gain X-band (8.2–12.4 GHz) pyramidal horn has dimensions of 𝜌1 ≃13.5 in.
(34.29 cm), 𝜌2 ≃14.2 in. (36.07 cm), a1 = 7.65 in. (19.43 cm), b1 = 5.65 in. (14.35 cm), a = 0.9 in. (2.286 cm), and b = 0.4 in. (1.016 cm).
(a) Check to see if such a horn can be constructed physically.
(b) Compute the directivity (in dB) at f = 8.2, 10.3, 12.4 GHz using for each (13-50a), (13-51), and (13-52e). Compare the answers. Verify with the computer program Analysis of this chapter.
13.17.
A standard-gain X-band (8.2–12.4 GHz) pyramidal horn has dimensions of 𝜌1 ≃5.3 in.
(13.46 cm), 𝜌2 ≃6.2 in. (15.75 cm), a1 = 3.09 in. (7.85 cm), b1 = 2.34 in. (5.94 cm), a = 0.9 in. (2.286 cm), and b = 0.4 in. (1.016 cm).
(a) Check to see if such a horn can be constructed physically.
(b) Compute the directivity (in dB) at f = 8.2, 10.3, 12.4 GHz using for each (13-50a), (13-51), and (13-52e). Compare the computed answers with the gains of Figure 13.22.
Verify with the computer program Analysis of this chapter.
780 HORN ANTENNAS 13.18.
A lossless linearly polarized pyramidal horn antenna is used as a receiver in a microwave communications system operating at 10 GHz. Over the aperture of the horn, the incident wave of the communications system is uniform and circularly polarized with a total power density of 10 mW/λ2. The pyramidal horn has been designed for optimum gain. The dimen-sions of the horn at the aperture are 4λo by 2.5λo. Determine the (a) Approximate aperture efficiency of the optimum gain horn (in %).
(b) Maximum directivity of the horn (in dB).
(c) Maximum power (in mW) that can be delivered to the receiver that is assumed to be matched to the transmission line that connects the antenna to the receiver. Assume no other losses.
13.19.
Repeat the design of the optimum X-band pyramidal horn of Example 13.4 so that the gain at f = 11 GHz is 17.05 dB.
13.20.
It is desired to design an optimum directivity pyramidal horn antenna. The length of the horn from its interior apex is 𝜌1 = 𝜌2 = 9λ. The horn is fed by an X-band waveguide whose interior dimensions are 0.5λ by 0.22λ.
(a) To accomplish this, what should the aperture dimensions (in λ) of the horn be?
(b) What is the directivity (in dB) of the horn?
13.21.
Design a pyramidal horn antenna with optimum gain at a frequency of 10 GHz. The overall length of the antenna from the imaginary vertex of the horn to the center of the aperture is 10λ and is nearly the same in both planes. Determine the (a) Aperture dimensions of the horn (in cm).
(b) Gain of the antenna (in dB) (c) Aperture efficiency of the antenna (in %). Assume the reflection, conduction, and dielec-tric losses of the antenna are negligible.
(d) Power delivered to a matched load when the incident power density is 10 μwatts/m2.
13.22.
Design an optimum gain C-band (3.95–5.85 GHz) pyramidal horn so that its gain at f = 4.90 GHz is 20.0 dBi. The horn is fed by a WR 187 rectangular waveguide with inner dimen-sions of a = 1.872 in. (4.755 cm) and b = 0.872 in. (2.215 cm). Refer to Figure 13.16 for the horn geometry. Determine in cm, the remaining dimensions of the horn: 𝜌e, 𝜌h, a1, b1, pe, and ph. Verify using the computer program Design of this chapter.
13.23.
It is desired to design a pyramidal horn for optimum directivity with total flare angles of 43◦in the E-plane and 50◦in the H-plane. Assuming the dimensions of the feed rectangular waveguide are a = 0.5λ and b = 0.25λ, determinet the (a) Horn dimensions at its aperture (a1 and b1; both in λ).
(b) Other horn dimensions 𝜌1, 𝜌2, 𝜌e and 𝜌h (all in λ).
(c) Verify that this horn is physically realizable.
13.24.
For a conical horn, plot dm (in λ) versus l (in λ) using (13-59). Verify, using the data of Figure 13.26, that the maximum directivities occur when (13-59) is satisfied.
13.25.
A conical horn has dimensions of L = 19.5 in. (49.53 cm), dm = 15 in. (38.10 cm), and d = 2.875 in. (7.3025 cm).
(a) Find the frequency (in GHz) which will result in maximum directivity for this horn.
What is that directivity (in dB)?
(b) Find the directivity (in dB) at 2.5 and 5 GHz.
(c) Compute the cutoff frequency (in GHz) of the TE11-mode which can exist inside the circular waveguide that is used to feed the horn.
PROBLEMS 781 13.26.
It is desired to design an optimum directivity conical horn antenna of circular cross section whose overall slanted length l is 10λ. Determine the (a) geometrical dimensions of the conical horn [radius (in λ), diameter (in λ), total flare angle (in degrees)].
(b) aperture efficiency of the horn (in %).
(c) directivity of the horn (dimensionless and in dB).
13.27.
Design an optimum directivity conical horn, using (13-58)–(13-59), so that its direc-tivity (above isotropic) at f = 11 GHz is 22.6 dB. Check your design with the data in Figure 13.26. Compare the design dimensions with those of the pyramidal horn of Example 13.5.
13.28.
Design an optimum directivity conical horn so that its maximum directivity is 20 dB (above isotropic). Determine the (a) Diameter (in λ) of the horn at its aperture.
(b) Length (in λ) of the horn from its virtual apex to the edge of the aperture.
(c) Length l (in λ) of the horn from its virtual apex to the edge of the aperture.
(d) Total flare angle (in degrees) of the horn.
(e) Aperture efficiency (in %) of the horn.
Refer to Figure 13.24 for the geometry of the problem.
13.29.
Design an optimum directivity conical horn so that its directivity at 10 GHz (above a standard-gain horn of 15 dB directivity) is 5 dB. Determine the horn diameter (in cm) and its flare angle (in degrees).
13.30.
Design a conical horn, for an optimum directivity, when the total included angle is 50◦.
Calculate (a) The diameter of the horn at its aperture (in λ).
(b) The aperture efficiency (in %).
(c) What should the expected aperture efficiency (in %) be for maximum directivity? How does the answer in Part b compares with the expected value?
(d) The directivity of the horn (dimensionless and in dB).
13.31.
It is desired to design optimum gain, lossless (ecd = 1) pyramidal (as shown in Fig. 13.16a) and conical (as shown in Fig. 13.24) horns. Assuming the area of each horn at its respective aperture is 20 𝜋λ2, determine the gain of the optimum gain horn (dimensionless and in dB) for the (a) Pyramidal horn of Figure 13.16a.
(b) (b) Conical horn of Figure 13.24.
13.32.
As part of a 10-GHz microwave communication system, you purchase a horn antenna that is said to have a directivity of 75 (dimensionless). The conduction and dielectric losses of the antenna are negligible, and the horn is polarization matched to the incoming sig-nal. A standing wave meter indicates a voltage reflection coefficient of 0.1 at the antenna-waveguide junction.
(a) Calculate the maximum effective aperture of the horn.
(b) If an impinging wave with a uniform power density of 1 μwatts/m2 is incident upon the horn, what is the maximum power delivered to a load which is connected and matched to the lossless waveguide?
782 HORN ANTENNAS 13.33.
For an X-band pyramidal corrugated horn operating at 10.3 GHz, find the (a) smallest lower and upper limits of the corrugation depths (in cm) (b) width w of each corrugation (in cm) (c) width t of each corrugation tooth (in cm) 13.34.
Find the E- and H-plane phase centers (in λ) of (a) an E-plane (𝜌e = 5λ, a = 0.7λ) (b) an H-plane (𝜌h = 5λ, a = 0.7λ) sectoral horn with a total included angle of 30◦.
CHAPTER14 Microstrip and Mobile Communications Antennas 14.1 INTRODUCTION In high-performance aircraft, spacecraft, satellite, and missile applications, where size, weight, cost, performance, ease of installation, and aerodynamic profile are constraints, low-profile antennas may be required. Presently there are many other government and commercial applications, such as mobile radio and wireless communications, that have similar specifications. To meet these requirements, microstrip antennas – can be used. These antennas are low profile, conformable to planar and nonplanar surfaces, simple and inexpensive to manufacture using modern printed-circuit tech-nology, mechanically robust when mounted on rigid surfaces, compatible with MMIC designs, and when the particular patch shape and mode are selected, they are very versatile in terms of resonant frequency, polarization, pattern, and impedance. In addition, by adding loads between the patch and the ground plane, such as pins and varactor diodes, adaptive elements with variable resonant fre-quency, impedance, polarization, and pattern can be designed , –.
Major operational disadvantages of microstrip antennas are their low efficiency, low power, high Q (sometimes in excess of 100), poor polarization purity, poor scan performance, spurious feed radi-ation and very narrow frequency bandwidth, which is typically only a fraction of a percent or at most a few percent. In some applications, such as in government security systems, narrow bandwidths are desirable. However, there are methods, such as increasing the height of the substrate, that can be used to extend the efficiency (to as large as 90 percent if surface waves are not included) and bandwidth (up to about 35 percent) . However, as the height increases, surface waves are introduced which usu-ally are not desirable because they extract power from the total available for direct radiation (space waves). The surface waves travel within the substrate and they are scattered at bends and surface discontinuities, such as the truncation of the dielectric and ground plane –, and degrade the antenna pattern and polarization characteristics. Surface waves can be eliminated, while maintaining large bandwidths, by using cavities , . Stacking, as well as other methods, of microstrip ele-ments can also be used to increase the bandwidth , –. In addition, microstrip antennas also exhibit large electromagnetic signatures at certain frequencies outside the operating band, are rather large physically at VHF and possibly UHF frequencies, and in large arrays there is a trade-off between bandwidth and scan volume –.
Many commercial substrates are available for use in the design and fabrication of microstrip type antennas. Some common substrates are listed in Table 14.1, along with the most pertinent parameters (name of company, substrate name, thickness, frequency range, dielectric constant, and loss tangent).
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 783 784 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS TABLE 14.1 Typical Substrates and Their Parameters Thickness Frequency Company Substrate (mm) (GHz) 𝜺r tan𝜹 Rogers Corporation Duroid® 5880 0.127 0 – 40 2.20 0.0009 RO 3003 1.575 0 – 40 3.00 0.0010 RO 3010 3.175 0 – 10 10.2 0.0022 RO 4350 0.168 0 – 10 3.48 0.0037 0.508 1.524 — FR4 0.05 – 100 0.001 4.70 — DuPont HK 04J 0.025 0.001 3.50 0.005 Isola IS 410 0.05 – 3.2 0.1 5.40 0.035 Arlon DiClad 870 0.091 0 – 10 2.33 0.0013 Polyflon Polyguide 0.102 0 – 10 2.32 0.0005 Neltec NH 9320 3.175 0 – 10 3.20 0.0024 Taconic RF-60A 0.102 0 – 10 6.15 0.0038 This is only a small sample; some companies, especially the Rogers Corporation, have a variety of substrate types to choose from. The Rogers substrates are usually referred to as PTFE (polyte-trafluorethylene), which are woven glass laminates, and are very popular for microstrip designs. FR4 is another very popular substrate.
14.1.1 Basic Characteristics Microstrip antennas received considerable attention starting in the 1970s, although the idea of a microstrip antenna can be traced to 1953 and a patent in 1955 . Microstrip antennas, as shown in Figure 14.1(a), consist of a very thin (t ≪λ0, where λ0 is the free-space wavelength) metallic strip (patch) placed a small fraction of a wavelength (h ≪λ0, usually 0.003λ0 ≤h ≤0.05λ0) above x y t h Ground plane Radiating slot #1 Radiating slot #2 Ground plane (a) Microstrip antenna (b) Side view (c) Coordinate system for each radiating slot Substrate L W r L x y z W θ θ ϕ ϕ (r, , ) z h Patch h r Figure 14.1 Microstrip antenna and coordinate system.
INTRODUCTION 785 (a) Square (b) Rectangular (c) Dipole (d) Circular (e) Elliptical (f) Triangular (g) Disc sector (h) Circular ring (i) Ring sector Figure 14.2 Representative shapes of microstrip patch elements.
a ground plane. The microstrip patch is designed so its pattern maximum is normal to the patch (broadside radiator). This is accomplished by properly choosing the mode (field configuration) of excitation beneath the patch. End-fire radiation can also be accomplished by judicious mode selec-tion. For a rectangular patch, the length L of the element is usually λ0∕3 < L < λ0∕2. The strip (patch) and the ground plane are separated by a dielectric sheet (referred to as the substrate), as shown in Figure 14.1(a).
There are numerous substrates that can be used for the design of microstrip antennas, and their dielectric constants are usually in the range of 2.2 ≤𝜀r ≤12. The ones that are most desirable for good antenna performance are thick substrates whose dielectric constant is in the lower end of the range because they provide better efficiency, larger bandwidth, loosely bound fields for radiation into space, but at the expense of larger element size . Thin substrates with higher dielectric constants are desirable for microwave circuitry because they require tightly bound fields to minimize undesired radiation and coupling, and lead to smaller element sizes; however, because of their greater losses, they are less efficient and have relatively smaller bandwidths . Since microstrip antennas are often integrated with other microwave circuitry, a compromise has to be reached between good antenna performance and circuit design.
Often microstrip antennas are also referred to as patch antennas. The radiating elements and the feed lines are usually photoetched on the dielectric substrate. The radiating patch may be square, rectangular, thin strip (dipole), circular, elliptical, triangular, or any other configuration. These and others are illustrated in Figure 14.2. Square, rectangular, dipole (strip), and circular are the most common because of ease of analysis and fabrication, and their attractive radiation characteristics, especially low cross-polarization radiation. Microstrip dipoles are attractive because they inherently possess a large bandwidth and occupy less space, which makes them attractive for arrays , , , . Linear and circular polarizations can be achieved with either single elements or arrays of microstrip antennas. Arrays of microstrip elements, with single or multiple feeds, may also be used to introduce scanning capabilities and achieve greater directivities. These will be discussed in later sections.
14.1.2 Feeding Methods There are many configurations that can be used to feed microstrip antennas. The four most popular are the microstrip line, coaxial probe, aperture coupling, and proximity coupling , , , , , –. These are displayed in Figure 14.3. One set of equivalent circuits for each one of these is shown in Figure 14.4. The microstrip feed line is also a conducting strip, usually of much smaller width compared to the patch. The microstrip-line feed is easy to fabricate, simple to match by controlling the inset position and rather simple to model. However as the substrate thickness 786 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Ground plane (a) Microstrip line feed Substrate L W r Patch h (b) Probe feed Circular microstrip patch Dielectric substrate r Ground plane Coaxial connector r1 r2 (c) Aperture-coupled feed Patch Slot Microstrip line r1 r2 (d) Proximity-coupled feed Patch Microstrip line Figure 14.3 Typical feeds for microstrip antennas.
increases, surface waves and spurious feed radiation increase, which for practical designs limit the bandwidth (typically 2–5%).
Coaxial-line feeds, where the inner conductor of the coax is attached to the radiation patch while the outer conductor is connected to the ground plane, are also widely used. The coaxial probe feed is also easy to fabricate and match, and it has low spurious radiation. However, it also has narrow bandwidth and it is more difficult to model, especially for thick substrates (h > 0.02λ0).
Both the microstrip feed line and the probe possess inherent asymmetries which generate higher order modes which produce cross-polarized radiation. To overcome some of these problems, INTRODUCTION 787 (a) Microstrip line (b) Probe (c) Aperture-coupled (d) Proximity-coupled Figure 14.4 Equivalent circuits for typical feeds of Figure 14.3.
noncontacting aperture-coupling feeds, as shown in Figures 14.3(c,d), have been introduced. The aperture coupling of Figure 14.3(c) is the most difficult of all four to fabricate and it also has nar-row bandwidth. However, it is somewhat easier to model and has moderate spurious radiation. The aperture coupling consists of two substrates separated by a ground plane. On the bottom side of the lower substrate there is a microstrip feed line whose energy is coupled to the patch through a slot on the ground plane separating the two substrates. This arrangement allows independent optimization of the feed mechanism and the radiating element. Typically a high dielectric material is used for the bottom substrate, and thick low dielectric constant material for the top substrate. The ground plane between the substrates also isolates the feed from the radiating element and minimizes interference of spurious radiation for pattern formation and polarization purity. For this design, the substrate elec-trical parameters, feed line width, and slot size and position can be used to optimize the design .
Typically matching is performed by controlling the width of the feed line and the length of the slot.
The coupling through the slot can be modeled using the theory of Bethe , which is also used to account for coupling through a small aperture in a conducting plane. This theory has been success-fully used to analyze waveguide couplers using coupling through holes . In this theory the slot is represented by an equivalent normal electric dipole to account for the normal component (to the slot) of the electric field, and an equivalent horizontal magnetic dipole to account for the tangential component (to the slot) magnetic field. If the slot is centered below the patch, where ideally for the dominant mode the electric field is zero while the magnetic field is maximum, the magnetic coupling will dominate. Doing this also leads to good polarization purity and no cross-polarized radiation in the principal planes . Of the four feeds described here, the proximity coupling has the largest bandwidth (as high as 13 percent), is somewhat easy to model and has low spurious radiation. How-ever its fabrication is somewhat more difficult. The length of the feeding stub and the width-to-line ratio of the patch can be used to control the match .
14.1.3 Methods of Analysis There are many methods of analysis for microstrip antennas. The most popular models are the transmission-line , , cavity , , , , and full wave (which include primar-ily integral equations/Moment Method) , , –. The transmission-line model is the easiest of all, it gives good physical insight, but is less accurate and it is more difficult to model coupling . Compared to the transmission-line model, the cavity model is more accurate but at the same time more complex. However, it also gives good physical insight and is rather difficult 788 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS to model coupling, although it has been used successfully , , . In general when applied properly, the full-wave models are very accurate, very versatile, and can treat single elements, finite and infinite arrays, stacked elements, arbitrary shaped elements, and coupling. However they are the most complex models and usually give less physical insight. In this chapter we will cover the transmission-line and cavity models only. However results and design curves from full-wave mod-els will also be included. Since they are the most popular and practical, in this chapter the only two patch configurations that will be considered are the rectangular and circular. Representative radiation characteristics of some other configurations will be included.
14.2 RECTANGULAR PATCH The rectangular patch is by far the most widely used configuration. It is very easy to analyze using both the transmission-line and cavity models, which are most accurate for thin substrates . We begin with the transmission-line model because it is easier to illustrate.
14.2.1 Transmission-Line Model It was indicated earlier that the transmission-line model is the easiest of all but it yields the least accurate results and it lacks the versatility. However, it does shed some physical insight. As it will be demonstrated in Section 14.2.2 using the cavity model, a rectangular microstrip antenna can be represented as an array of two radiating narrow apertures (slots), each of width W and height h, separated by a distance L. Basically the transmission-line model represents the microstrip antenna by two slots, separated by a low-impedance Zc transmission line of length L.
A. Fringing Effects Because the dimensions of the patch are finite along the length and width, the fields at the edges of the patch undergo fringing. This is illustrated along the length in Figures 14.1(a,b) for the two radiating slots of the microstrip antenna. The same applies along the width. The amount of fringing is a function of the dimensions of the patch and the height of the substrate. For the principal E-plane (xy-plane) fringing is a function of the ratio of the length of the patch L to the height h of the substrate (L/h) and the dielectric constant 𝜀r of the substrate. Since for microstrip antennas L∕h ≫1, fringing is reduced; however, it must be taken into account because it influences the resonant frequency of the antenna. The same applies for the width.
For a microstrip line shown in Figure 14.5(a), typical electric field lines are shown in Fig-ure 14.5(b). This is a nonhomogeneous line of two dielectrics; typically the substrate and air. As can be seen, most of the electric field lines reside in the substrate and parts of some lines exist in air. As W∕h ≫1 and 𝜀r ≫1, the electric field lines concentrate mostly in the substrate. Fringing in this case makes the microstrip line look wider electrically compared to its physical dimensions.
Since some of the waves travel in the substrate and some in air, an effective dielectric constant 𝜀reff is introduced to account for fringing and the wave propagation in the line.
To introduce the effective dielectric constant, let us assume that the center conductor of the microstrip line with its original dimensions and height above the ground plane is embedded into one dielectric, as shown in Figure 14.5(c). The effective dielectric constant is defined as the dielectric constant of the uniform dielectric material so that the line of Figure 14.5(c) has identical electrical characteristics, particularly propagation constant, as the actual line of Figure 14.5(a). For a line with air above the substrate, the effective dielectric constant has values in the range of 1 < 𝜀reff < 𝜀r. For most applications where the dielectric constant of the substrate is much greater than unity (𝜀r ≫1), the value of 𝜀reff will be closer to the value of the actual dielectric constant 𝜀r of the substrate. The effective dielectric constant is also a function of frequency. As the frequency of operation increases, most of the electric field lines concentrate in the substrate. Therefore the microstrip line behaves RECTANGULAR PATCH 789 (a) Microstrip line (b) Electric field lines h W t r (c) Effective dielectric constant t reff W h Figure 14.5 Microstrip line and its electric field lines, and effective dielectric constant geometry.
more like a homogeneous line of one dielectric (only the substrate), and the effective dielectric con-stant approaches the value of the dielectric constant of the substrate. Typical variations, as a function of frequency, of the effective dielectric constant for a microstrip line with three different substrates are shown in Figure 14.6.
For low frequencies the effective dielectric constant is essentially constant. At intermediate fre-quencies its values begin to monotonically increase and eventually approach the values of the dielec-tric constant of the substrate. The initial values (at low frequencies) of the effective dielectric constant are referred to as the static values, and they are given by W∕h > 1 𝜀reff = 𝜀r + 1 2 + 𝜀r −1 2 [ 1 + 12 h W ]−1∕2 (14-1) 9 10 11 12 13 0 2 4 6 8 10 12 Log frequency Effective dielectric constant ( reff) W = 0.125″ = 0.3175 cm h = 0.050″ = 0.1270 cm r = 10.2 r = 6.80 r = 2.33 r = 10.2 r = 6.80 r = 2.33 Figure 14.6 Effective dielectric constant versus frequency for typical substrates.
790 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS W L (a) Top view (b) Side view Patch r Δ L Δ L h Figure 14.7 Physical and effective lengths of rectangular microstrip patch.
B. Effective Length, Resonant Frequency, and Effective Width Because of the fringing effects, electrically the patch of the microstrip antenna looks greater than its physical dimensions. For the principal E-plane (xy-plane), this is demonstrated in Figure 14.7 where the dimensions of the patch along its length have been extended on each end by a distance ΔL, which is a function of the effective dielectric constant 𝜀reff and the width-to-height ratio (W/h).
A very popular and practical approximate relation for the normalized extension of the length is ΔL h = 0.412 (𝜀reff + 0.3) (W h + 0.264 ) (𝜀reff −0.258) (W h + 0.8 ) (14-2) Since the length of the patch has been extended by ΔL on each side, the effective length of the patch is now (L = λ∕2 for dominant TM010 mode with no fringing) Leff = L + 2ΔL (14-3) For the dominant TM010 mode, the resonant frequency of the microstrip antenna is a function of its length. Usually it is given by (fr)010 = 1 2L√𝜀r √𝜇0𝜀0 = 𝜐0 2L√𝜀r (14-4) where 𝜐0 is the speed of light in free space. Since (14-4) does not account for fringing, it must be modified to include edge effects and should be computed using (frc)010 = 1 2Leff √𝜀reff √𝜇0𝜀0 = 1 2(L + 2ΔL)√𝜀reff √𝜇0𝜀0 = q 1 2L√𝜀r √𝜇0𝜀0 = q 𝜐0 2L√𝜀r (14-5) RECTANGULAR PATCH 791 where q = (frc)010 (fr)010 (14-5a) The q factor is referred to as the fringe factor (length reduction factor). As the substrate height increases, fringing also increases and leads to larger separations between the radiating edges and lower resonant frequencies. The designed resonant frequency, based on fringing, is lower as the patch looks longer, as indicated in Figure 14.7. The resonant frequency decrease due to fringing is usually 2–6%.
C. Design Based on the simplified formulation that has been described, a design procedure is outlined which leads to practical designs of rectangular microstrip antennas. The procedure assumes that the spec-ified information includes the dielectric constant of the substrate (𝜀r), the resonant frequency (fr), and the height of the substrate h. The procedure is as follows: Specify: 𝜀r, fr (in Hz), and h Determine: W, L Design procedure: 1. For an efficient radiator, a practical width that leads to good radiation efficiencies is W = 1 2fr √𝜇0𝜀0 √ 2 𝜀r + 1 = 𝜐0 2fr √ 2 𝜀r + 1 (14-6) where 𝜐0 is the free-space velocity of light.
2. Determine the effective dielectric constant of the microstrip antenna using (14-1).
3. Once W is found using (14-6), determine the extension of the length ΔL using (14-2).
4. The actual length of the patch can now be determined by solving (14-5) for L, or L = 1 2fr √𝜀reff √𝜇0𝜀0 −2ΔL (14-7) Typical lengths of microstrip patches vary between L ≈(0.47 −0.49) λo √𝜀r = (0.47 −0.49)λd (14-7a) where λd is the wavelength in the dielectric. The smaller the dielectric constant of the substrate, the larger is the fringing; thus, the length of the microstrip patch is smaller. In contrast, the larger the dielectric constant, the more tightly the fields are held within the substrate; thus, the fringing is smaller and the length is longer and closer to half-wavelength in the dielectric.
792 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Example 14.1 Design a rectangular microstrip antenna using a substrate (RT/duroid 5880) with dielectric con-stant of 2.2, h = 0.1588 cm (0.0625 inches) so as to resonate at 10 GHz.
Solution: Using (14-6), the width W of the patch is W = 30 2(10) √ 2 2.2 + 1 = 1.186 cm (0.467 in) The effective dielectric constant of the patch is found using (14-1), or 𝜀reff = 2.2 + 1 2 + 2.2 −1 2 ( 1 + 120.1588 1.186 )−1∕2 = 1.972 The extended incremental length of the patch ΔL is, using (14-2) ΔL = 0.1588(0.412) (1.972 + 0.3) ( 1.186 0.1588 + 0.264 ) (1.972 −0.258) ( 1.186 0.1588 + 0.8 ) = 0.081 cm (0.032 in) The actual length L of the patch is found using (14-3), or L = λ 2 −2ΔL = 30 2(10) √ 1.972 −2(0.081) = 0.906 cm (0.357 in) Finally the effective length is Le = L + 2ΔL = λ 2 = 1.068 cm (0.421 in) An experimental rectangular patch based on this design was built and tested. It is probe fed from underneath by a coaxial line and is shown in Figure 14.8(a). Its principal E- and H-plane patterns are displayed in Figure 14.21(c,d) for fo = 9.8 GHz.
(a) Rectangular (b) Circular Figure 14.8 Experimental models of rectangular and circular patches based, respectively, on the designs of Examples 14.1 and 14.4.
RECTANGULAR PATCH 793 D. Conductance Each radiating slot is represented by a parallel equivalent admittance Y (with conductance G and susceptance B). This is shown in Figure 14.9. The slots are labeled as #1 and #2. The equivalent admittance of slot #1, based on an infinitely wide, uniform slot, is derived in Example 12.8 of Chap-ter 12, and it is given by Y1 = G1 + jB1 (14-8) where for a slot of finite width W G1 = W 120λ0 [ 1 −1 24(k0h)2] h λ0 < 1 10 (14-8a) B1 = W 120λ0 [1 −0.636 ln(k0h)] h λ0 < 1 10 (14-8b) Since slot #2 is identical to slot #1, its equivalent admittance is Y2 = Y1, G2 = G1, B2 = B1 (14-9) W L (a) Rectangular patch (b) Transmission model equivalent B2 G2 B1 YC G1 Figure 14.9 Rectangular microstrip patch and its equivalent circuit transmission-line model.
The conductance of a single slot can also be obtained by using the field expression derived by the cavity model. In general, the conductance is defined as G1 = 2Prad |V0|2 (14-10) Using the electric field of (14-41), the radiated power is written as Prad = |V0|2 2𝜋𝜂0 ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃d𝜃 (14-11) Therefore the conductance of (14-10) can be expressed as G1 = I1 120𝜋2 (14-12) 794 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS where I1 = ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃d𝜃 = −2 + cos(X) + XSi(X) + sin(X) X (14-12a) X = k0W (14-12b) Asymptotic values of (14-12) and (14-12a) are G1 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 90 ( W λ0 )2 W ≪λ0 1 120 ( W λ0 ) W ≫λ0 (14-13) The values of (14-13) for W ≫λ0 are identical to those given by (14-8a) for h ≪λ0. A plot of G as a function of W∕λ0 is shown in Figure 14.10.
E. Resonant Input Resistance The total admittance at slot #1 (input admittance) is obtained by transferring the admittance of slot #2 from the output terminals to input terminals using the admittance transformation equation of transmission lines , , . Ideally the two slots should be separated by λ∕2 where λ is the wavelength in the dielectric (substrate). However, because of fringing the length of the patch is electrically longer than the actual length. Therefore the actual separation of the two slots is slightly less than λ∕2. If the reduction of the length is properly chosen using (14-2) (typically 0.47λ < L < 0.49λ), the transformed admittance of slot #2 becomes ̃ Y2 = ̃ G2 + j̃ B2 = G1 −jB1 (14-14) or ̃ G2 = G1 (14-14a) ̃ B2 = −B1 (14-14b) Therefore the total resonant input admittance is real and is given by Yin = Y1 + ̃ Y2 = 2G1 (14-15) Since the total input admittance is real, the resonant input impedance is also real, or Zin = 1 Yin = Rin = 1 2G1 (14-16) RECTANGULAR PATCH 795 10–3 10–2 10–1 1 10+1 10+2 10–8 10–7 10–6 10–5 10–4 10–3 10–2 10–1 1 Slot width W/λo Conductance G1 (S) Figure 14.10 Slot conductance as a function of slot width.
The resonant input resistance, as given by (14-16), does not take into account mutual effects between the slots. This can be accomplished by modifying (14-16) to Rin = 1 2(G1 ± G12) (14-17) where the plus (+) sign is used for modes with odd (antisymmetric) resonant voltage distribution beneath the patch and between the slots while the minus (−) sign is used for modes with even (sym-metric) resonant voltage distribution. The mutual conductance is defined, in terms of the far-zone fields, as G12 = 1 |V0|2 Re ∫∫ S E1 × H∗ 2 ⋅ds (14-18) where E1 is the electric field radiated by slot #1, H2 is the magnetic field radiated by slot #2, V0 is the voltage across the slot, and the integration is performed over a sphere of large radius. It can be shown that G12 can be calculated using , G12 = 1 120𝜋2 ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 J0(k0L sin 𝜃) sin3 𝜃d𝜃 (14-18a) where J0 is the Bessel function of the first kind of order zero. For typical microstrip antennas, the mutual conductance obtained using (14-18a) is small compared to the self conductance G1 of (14-8a) or (14-12).
796 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS An alternate approximate expression for the input impedance, Rin, for a resonant patch is Rin = 90 (𝜀r)2 𝜀r −1 ( L W ) (14-18b) This expression is valid for thin substrates (h ≪λ0). It was derived on approximations of rigorously developed formulations, including Sommerfeld integrals, for thin grounded substrates. Equation (14-17), and (14-18b) give reasonable results for the input resistance Rin, although they are not identical.
As shown by (14-8a) and (14-17), the input resistance is not strongly dependent upon the substrate height h. In fact for very small values of h, such that k0h ≪1, the input resistance is not dependent on h. Modal-expansion analysis also reveals that the input resistance is not strongly influenced by the substrate height h. It is apparent from (14-8a) and (14-17) that the resonant input resistance can be decreased by increasing the width W of the patch. This is acceptable as long as the ratio of W/L does not exceed 2 because the aperture efficiency of a single patch begins to drop, as W/L increases beyond 2.
F. Matching Techniques The resonant input resistance, as calculated by (14-17), is referenced at slot #1. However, it has been shown that the resonant input resistance can be changed by using an inset feed, recessed a distance y0 from slot #1, as shown in Figure 14.11(a). This technique can be used effectively to match the W L (a) Recessed microstrip-line feed y0 W0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.9 0.8 1.0 0.0 0.25 0.50 0.75 1.0 yo/L (b) Normalized input resistance Rin(y = yo)/Rin(y = 0) Figure 14.11 Recessed microstrip-line feed and variation of normalized input resistance.
RECTANGULAR PATCH 797 patch antenna using a microstrip-line feed whose characteristic impedance is given by Zc = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 60 √𝜀reff ln [ 8h W0 + W0 4h ] , 120𝜋 √𝜀reff [W0 h + 1.393 + 0.667 ln (W0 h + 1.444 )], W0 h ≤1 W0 h > 1 (14-19a) (14-19b) where W0 is the width of the microstrip line, as shown in Figure 14.11. Using modal-expansion analysis, the input resistance for the inset feed is given approximately by , Rin(y = y0) = 1 2(G1 ± G12) [ cos2 (𝜋 Ly0 ) + G2 1 + B2 1 Y2 c sin2 (𝜋 Ly0 ) −B1 Yc sin (2𝜋 L y0 )] (14-20) where Yc = 1∕Zc. Since for most typical microstrips G1∕Yc ≪1 and B1∕Yc ≪1, (14-20) reduces to Rin(y = y0) = 1 2(G1 ± G12) cos2 (𝜋 Ly0 ) = Rin(y = 0) cos2 (𝜋 Ly0 ) (14-20a) A plot of the normalized value of (14-20a) is shown in Figure 14.11(b).
The values obtained using (14-20) agree fairly well with experimental data. However, the inset feed introduces a physical notch, which in turn introduces a junction capacitance. The physical notch and its corresponding junction capacitance influence slightly the resonance frequency, which typi-cally may vary by about 1%. It is apparent from (14-20a) and Figure 14.11(b) that the maximum value occurs at the edge of the slot (y0 = 0) where the voltage is maximum and the current is min-imum; typical values are in the 150–300 ohms. The minimum value (zero) occurs at the center of the patch (y0 = L∕2) where the voltage is zero and the current is maximum. As the inset feed point moves from the edge toward the center of the patch the resonant input impedance decreases mono-tonically and reaches zero at the center. When the value of the inset feed point approaches the center of the patch (y0 = L∕2), the cos2(𝜋y0∕L) function varies very rapidly; therefore the input resistance also changes rapidly with the position of the feed point. To maintain very accurate values, a close tolerance must be preserved.
Other matching techniques, aside from the recessed microstrip of Figure 14.11, are the coupled recessed microstrip and the λ∕4 impedance transformer of Figure 14.12(a,b). The Rin in Figure 14.12(b) is the input resistance at the leading edge of the resonant patch; it must be real.
Example 14.2 A microstrip antenna with overall dimensions of L = 0.906 cm (0.357 inches) and W = 1.186 cm (0.467 inches), substrate with height h = 0.1588 cm (0.0625 inches) and dielectric constant of 𝜀r = 2.2, is operating at 10 GHz. Find: a. The input impedance.
b. The position of the inset feed point where the input impedance is 50 ohms.
798 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Solution: λ0 = 30 10 = 3 cm Using (14-12) and (14-12a) G1 = 0.00157 siemens which compares with G1 = 0.00328 using (14-8a). Using (14-18a) G12 = 6.1683 × 10−4 Using (14-17) with the (+) sign because of the odd field distribution between the radiating slots for the dominant TM010 mode Rin = 228.3508 ohms.
Since the input impedance at the leading radiating edge of the patch is 228.3508 ohms while the desired impedance is 50 ohms, the inset feed point distance y0 is obtained using (14-20a). Thus 50 = 228.3508 cos2 (𝜋 Ly0 ) or y0 = 0.3126 cm (0.123 inches) Microstrip Transmission Line r Substrate W y0 L (a) Coupled (b) λ/4 impedance transformer Microstrip Transmission Line r Substrate L W λ/4 Zc Z1 1 in c R Z Z = Figure 14.12 Alternate feeding techniques of microstrip antenna for impedance matching.
14.2.2 Cavity Model Microstrip antennas resemble dielectric-loaded cavities, and they exhibit higher order resonances.
The normalized fields within the dielectric substrate (between the patch and the ground plane) can be found more accurately by treating that region as a cavity bounded by electric conductors (above and below it) and by magnetic walls (to simulate an open circuit) along the perimeter of the patch. This RECTANGULAR PATCH 799 + + + + + + + + – – – – – – – – – W Jb Jt r + + + + + + + + – – – – – – – – h + Figure 14.13 Charge distribution and current density creation on microstrip patch.
is an approximate model, which in principle leads to a reactive input impedance (of zero or infinite value of resonance), and it does not radiate any power. However, assuming that the actual fields are approximate to those generated by such a model, the computed pattern, input admittance, and resonant frequencies compare well with measurements , , . This is an accepted approach, and it is similar to the perturbation methods which have been very successful in the analysis of waveguides, cavities, and radiators .
To shed some insight into the cavity model, let us attempt to present a physical interpretation into the formation of the fields within the cavity and radiation through its side walls. When the microstrip patch is energized, a charge distribution is established on the upper and lower surfaces of the patch, as well as on the surface of the ground plane, as shown in Figure 14.13. The charge distribution is controlled by two mechanisms; an attractive and a repulsive mechanism . The attractive mechanism is between the corresponding opposite charges on the bottom side of the patch and the ground plane, which tends to maintain the charge concentration on the bottom of the patch.
The repulsive mechanism is between like charges on the bottom surface of the patch, which tends to push some charges from the bottom of the patch, around its edges, to its top surface. The movement of these charges creates corresponding current densities Jb and Jt, at the bottom and top surfaces of the patch, respectively, as shown in Figure 14.13. Since for most practical microstrips the height-to-width ratio is very small, the attractive mechanism dominates and most of the charge concentration and current flow remain underneath the patch. A small amount of current flows around the edges of the patch to its top surface. However, this current flow decreases as the height-to-width ratio decreases. In the limit, the current flow to the top would be zero, which ideally would not create any tangential magnetic field components to the edges of the patch. This would allow the four side walls to be modeled as perfect magnetic conducting surfaces which ideally would not disturb the magnetic field and, in turn, the electric field distributions beneath the patch. Since in practice there is a finite height-to-width ratio, although small, the tangential magnetic fields at the edges would not be exactly zero. However, since they will be small, a good approximation to the cavity model is to treat the side walls as perfectly magnetic conducting. This model produces good normalized electric and magnetic field distributions (modes) beneath the patch.
The fringing field is another mechanism that can be used to explain why the microstrip antenna radiates. Consider the side view of a patch antenna shown in Figure 14.13. Because the current at the end of the patch is zero (open circuit), the current is maximum at the center of the half-wave patch and ideally zero at the beginning of the patch. This low current value at the feed explains in part why the impedance is high when fed at the leading edge, as shown in Figure 14.11(a). Because the patch antenna can be ideally viewed as an open-circuited transmission line, the voltage reflection coefficient of the voltage is +1; therefore the voltage and current are out of phase. Hence, at the end of the patch, the voltage is maximum. At the center of the patch (a quarter-wavelength away from the edges), the voltage must be minimum. Basically, taking into account the finite dimensions of the cavity, the fields underneath the patch resemble those of Figure 14.12(b), which roughly displays the fringing of the fields around the edges.
Based on the field structure in Figure 14.14, it is the fringing fields that are responsible for the radiation. Note that the fringing fields at the leading and trailing edges (left and right of Figure 14.14) of the patch antenna are both in the same direction (in this case y direction). Hence, the fringing 800 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS L (a) No Fringing (b) Fringing L Total No fringing Fringing h r r Figure 14.14 Side view of microstrip antenna, without and with fringing.
E-fields on the leading and trailing edges of the microstrip antenna add up in phase and produce the radiation of the microstrip antenna. The current also adds up in phase on the patch; however, an equal current but with opposite direction is on the ground plane (as illustrated in Figure 14.13), which cancels the radiation. This also explains why the microstrip antenna radiates but the microstrip transmission line does not. The microstrip antenna’s radiation arises from the fringing fields, which are due to the favorable voltage distribution; hence, the radiation is attributed due to the voltage and not the current. The patch antenna is, therefore, a “voltage radiator,” as opposed to wire antennas, which are “current radiators.” Furthermore, note that the smaller the dielectric constant 𝜀r of the substrate is, the fringing field is greater; that is, the field “bows” away further from the patch. Therefore, using a smaller permittivity for the microstrip yields more efficient radiation. In contrast, when no radiation is desired, as in transmission lines, a high value of 𝜀r is desired. The high 𝜀r causes the fields to become more tightly bound (less fringing), resulting in a desired smaller radiation efficiency (ideally zero).
If the microstrip antenna were treated only as a cavity, it would not be sufficient to find the abso-lute amplitudes of the electric and magnetic fields. In fact by treating the walls of the cavity, as well as the material within it as lossless, the cavity would not radiate and its input impedance would be purely reactive. Also the function representing the impedance would only have real poles. To account for radiation, a loss mechanism has to be introduced. In Figures 2.27 and 2.28 of Chap-ter 2, this was taken into account by the radiation resistance Rr and loss resistance RL. These two resistances allow the input impedance to be complex and for its function to have complex poles; the imaginary poles representing, through Rr and RL, the radiation and conduction-dielectric losses. To make the microstrip lossy using the cavity model, which would then represent an antenna, the loss is taken into account by introducing an effective loss tangent 𝛿eff. The effective loss tangent is chosen appropriately to represent the loss mechanism of the cavity, which now behaves as an antenna and is taken as the reciprocal of the antenna quality factor Q (𝛿eff = 1∕Q).
Because the thickness of the microstrip is usually very small, the waves generated within the dielectric substrate (between the patch and the ground plane) undergo considerable reflections when they arrive at the edge of the patch. Therefore only a small fraction of the incident energy is radiated; thus the antenna is considered to be very inefficient. The fields beneath the patch form standing waves that can be represented by cosinusoidal wave functions. Since the height of the substrate is very small (h ≪λ where λ is the wavelength within the dielectric), the field variations along the height will be considered constant. In addition, because of the very small substrate height, the fringing of the fields along the edges of the patch are also very small whereby the electric field is nearly normal to the surface of the patch. Therefore only TMx field configurations will be considered within the cavity.
While the top and bottom walls of the cavity are perfectly electric conducting, the four side walls will be modeled as perfectly conducting magnetic walls (tangential magnetic fields vanish along those four walls).
A. Field Configurations (modes)—TMx The field configurations within the cavity can be found using the vector potential approach described in detail in Chapter 8 of . Referring to Figure 14.15, the volume beneath the patch can be treated as a rectangular cavity loaded with a dielectric material with dielectric constant 𝜀r. The dielectric RECTANGULAR PATCH 801 L h x z y W r Figure 14.15 Rectangular microstrip patch geometry.
material of the substrate is assumed to be truncated and not extended beyond the edges of the patch.
The vector potential Ax must satisfy the homogeneous wave equation of ∇2Ax + k2Ax = 0 (14-21) whose solution is written in general, using the separation of variables, as Ax = [A1 cos(kxx) + B1 sin(kxx)][A2 cos(kyy) + B2 sin(kyy)] ⋅[A3 cos(kzz) + B3 sin(kzz)] (14-22) where kx, ky and kz are the wavenumbers along the x, y, and z directions, respectively. These will be determined subject to the boundary conditions. The electric and magnetic fields within the cavity are related to the vector potential Ax by Ex = −j 1 𝜔𝜇𝜀 ( 𝜕2 𝜕x2 + k2 ) Ax Hx = 0 Ey = −j 1 𝜔𝜇𝜀 𝜕2Ax 𝜕x𝜕y Hy = 1 𝜇 𝜕Ax 𝜕z Ez = −j 1 𝜔𝜇𝜀 𝜕2Ax 𝜕x𝜕z Hz = −1 𝜇 𝜕Ax 𝜕y (14-23) subject to the boundary conditions of Ey(x′ = 0, 0 ≤y′ ≤L, 0 ≤z′ ≤W) = Ey(x′ = h, 0 ≤y′ ≤L, 0 ≤z′ ≤W) = 0 Hy(0 ≤x′ ≤h, 0 ≤y′ ≤L, z′ = 0) (14-24) = Hy(0 ≤x′ ≤h, 0 ≤y′ ≤L, z′ = W) = 0 Hz(0 ≤x′ ≤h, y′ = 0, 0 ≤z′ ≤W) = Hz(0 ≤x′ ≤h, y′ = L, 0 ≤z′ ≤W) = 0 The primed coordinates x′, y′, z′ are used to represent the fields within the cavity.
802 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Applying the boundary conditions Ey(x′ = 0, 0 ≤y′ ≤L, 0 ≤z′ ≤W) = 0 and Ey(x′ = h, 0 ≤ y′ ≤L, 0 ≤z′ ≤W) = 0, it can be shown that B1 = 0 and kx = m𝜋 h , m = 0, 1, 2, … (14-25) Similarly, applying the boundary conditions Hy(0 ≤x′ ≤h, 0 ≤y′ ≤L, z′ = 0) = 0 and Hy(0 ≤x′ ≤ h, 0 ≤y′ ≤L, z′ = W) = 0, it can be shown that B3 = 0 and kz = p𝜋 W , p = 0, 1, 2, … (14-26) Finally, applying the boundary conditions Hz(0 ≤x′ ≤h, y′ = 0, 0 ≤z′ ≤W) = 0 and Hz(0 ≤x′ ≤ h, y′ = L, 0 ≤z′ ≤W) = 0, it can be shown that B2 = 0 and ky = n𝜋 L , n = 0, 1, 2, . … (14-27) Thus the final form for the vector potential Ax within the cavity is Ax = Amnp cos(kxx′) cos(kyy′) cos(kzz′) (14-28) where Amnp represents the amplitude coefficients of each mnp mode. The wavenumbers kx, ky, kz are equal to kx = (m𝜋 h ) , m = 0, 1, 2, … ky = (n𝜋 L ) , n = 0, 1, 2, … kz = (p𝜋 W ) , p = 0, 1, 2, … ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ m = n = p ≠0 (14-29) where m, n, p represent, respectively, the number of half-cycle field variations along the x, y, z direc-tions.
Since the wavenumbers kx, ky, and kz are subject to the constraint equation k2 x + k2 y + k2 z = (m𝜋 h )2 + (n𝜋 L )2 + (p𝜋 W )2 = k2 r = 𝜔2 r𝜇𝜀 (14-30) the resonant frequencies for the cavity are given by (fr)mnp = 1 2𝜋√𝜇𝜀 √(m𝜋 h )2 + (n𝜋 L )2 + (p𝜋 W )2 (14-31) RECTANGULAR PATCH 803 Substituting (14-28) into (14-23), the electric and magnetic fields within the cavity are written as Ex = −j (k2 −k2 x) 𝜔𝜇𝜀 Amnp cos(kxx′) cos(kyy′) cos(kzz′) Ey = −j kxky 𝜔𝜇𝜀Amnp sin(kxx′) sin(kyy′) cos(kzz′) Ez = −j kxkz 𝜔𝜇𝜀Amnp sin(kxx′) cos(kyy′) sin(kzz′) (14-32) Hx = 0 Hy = −kz 𝜇Amnp cos(kxx′) cos(kyy′) sin(kzz′) Hz = ky 𝜇Amnp cos(kxx′) sin(kyy′) cos(kzz′) To determine the dominant mode with the lowest resonance, we need to examine the resonant frequencies. The mode with the lowest order resonant frequency is referred to as the dominant mode.
Placing the resonant frequencies in ascending order determines the order of the modes of operation.
For all microstrip antennas h ≪L and h ≪W. If L > W > h, the mode with the lowest frequency (dominant mode) is the TMx 010 whose resonant frequency is given by (fr)010 = 1 2L√𝜇𝜀 = 𝜐0 2L√𝜀r (14-33) where 𝜐0 is the speed of light in free-space. If in addition L > W > L∕2 > h, the next higher order (second) mode is the TMx 001 whose resonant frequency is given by (fr)001 = 1 2W√𝜇𝜀 = 𝜐0 2W√𝜀r (14-34) If, however, L > L∕2 > W > h, the second order mode is the TMx 020, instead of the TMx 001, whose resonant frequency is given by (fr)020 = 1 L√𝜇𝜀 = 𝜐0 L√𝜀r (14-35) If W > L > h, the dominant mode is the TMx 001 whose resonant frequency is given by (14-34) while if W > W∕2 > L > h the second order mode is the TMx 002. Based upon (14-32), the distribution of the tangential electric field along the side walls of the cavity for the TMx 010, TMx 001, TMx 020 and TMx 002 is as shown, respectively, in Figure 14.16.
In all of the preceding discussion, it was assumed that there is no fringing of the fields along the edges of the cavity. This is not totally valid, but it is a good assumption. However, fringing effects and their influence were discussed previously, and they should be taken into account in determining the resonant frequency. This was done in (14-5) for the dominant TMx 010 mode.
B. Equivalent Current Densities It has been shown using the cavity model that the microstrip antenna can be modeled reasonably well by a dielectric-loaded cavity with two perfectly conducting electric walls (top and bottom), and four perfectly conducting magnetic walls (sidewalls). It is assumed that the material of the substrate 804 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS L h W L h W L h W L h W TMx 010 (a) TMx 010 TMx 001 (b) TM x 001 TMx 020 (c) TM x 020 TMx 002 (d) TM x 002 Figure 14.16 Field configurations (modes) for rectangular microstrip patch.
is truncated and does not extend beyond the edges of the patch. The four sidewalls represent four narrow apertures (slots) through which radiation takes place. Using the Field Equivalence Principle (Huygens’ Principle) of Section 12.2 of Chapter 12, the microstrip patch is represented by an equiv-alent electric current density Jt at the top surface of the patch to account for the presence of the patch (there is also a current density Jb at the bottom of the patch which is not needed for this model). The four side slots are represented by the equivalent electric current density Js and equivalent magnetic current density Ms, as shown in Figure 14.17(a), each represented by Js = ̂ n × Ha (14-36) and Ms = −̂ n × Ea (14-37) where Ea and Ha represent, respectively, the electric and magnetic fields at the slots.
Because it was shown for microstrip antennas with very small height-to-width ratio that the cur-rent density Jt at the top of the patch is much smaller than the current density Jb at the bottom of the patch, it will be assumed it is negligible here and it will be set to zero. Also it was argued that the tangential magnetic fields along the edges of the patch are very small, ideally zero. Therefore the corresponding equivalent electric current density Js will be very small (ideally zero), and it will be set to zero here. Thus the only nonzero current density is the equivalent magnetic current density Ms of (14-37) along the side periphery of the cavity radiating in the presence of the ground plane, as shown in Figure 14.17(b). The presence of the ground plane can be taken into account by image theory which will double the equivalent magnetic current density of (14-37). Therefore the final equivalent is a magnetic current density of twice (14-37) or Ms = −2̂ n × Ea (14-38) around the side periphery of the patch radiating into free-space, as shown in Figure 14.17(c).
RECTANGULAR PATCH 805 W Jt Js, Ms (a) Js, Ms with ground plane L W Jt ≅ 0 Js = 0, Ms (b) Js = 0, Ms with ground plane L W Ms = –2n Ea (c) Ms with no ground plane ^ Js, Ms Js = 0, Ms L Ms = –2n Ea ^ Figure 14.17 Equivalent current densities on four sides of rectangular microstrip patch.
It was shown, using the transmission-line model, that the microstrip antenna can be represented by two radiating slots along the length of the patch (each of width W and height h). Similarly it will be shown here also that while there are a total of four slots representing the microstrip antenna, only two (the radiating slots) account for most of the radiation; the fields radiated by the other two, which are separated by the width W of the patch, cancel along the principal planes. Therefore the same two slots, separated by the length of the patch, are referred to here also as radiating slots. The slots are separated by a very low-impedance parallel-plate transmission line of length L, which acts as a trans-former. The length of the transmission line is approximately λ∕2, where λ is the guide wavelength in the substrate, in order for the fields at the aperture of the two slots to have opposite polarization.
This is illustrated in Figures 14.1(a) and 14.16(a). The two slots form a two-element array with a spacing of λ∕2 between the elements. It will be shown here that in a direction perpendicular to the ground plane the components of the field add in phase and give a maximum radiation normal to the patch; thus it is a broadside antenna.
Assuming that the dominant mode within the cavity is the TMx 010 mode, the electric and magnetic field components reduce from (14-32) to Ex = E0 cos (𝜋 Ly′) Hz = H0 sin (𝜋 Ly′) (14-39) Ey = Ez = Hx = Hy = 0 where E0 = −j𝜔A010 and H0 = (𝜋∕𝜇L)A010. The electric field structure within the substrate and between the radiating element and the ground plane is sketched in Figures 14.1(a,b) and 14.16(a).
806 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS L W M2 n2 y x z ^ n1 ^ θ #2 #1 E2 ϕ E1 M1 Figure 14.18 Rectangular microstrip patch radiating slots and equivalent magnetic current densities.
It undergoes a phase reversal along the length but it is uniform along its width. The phase reversal along the length is necessary for the antenna to have broadside radiation characteristics.
Using the equivalence principle of Section 12.2, each slot radiates the same fields as a magnetic dipole with current density Ms equal to (14-38). By referring to Figure 14.18 the equivalent magnetic current densities along the two slots, each of width W and height h, are both of the same magnitude and of the same phase. Therefore these two slots form a two-element array with the sources (current densities) of the same magnitude and phase, and separated by L. Thus these two sources will add in a direction normal to the patch and ground plane forming a broadside pattern. This is illustrated in Figures 14.19(a) where the normalized radiation pattern of each slot in the principal E-plane is sketched individually along with the total pattern of the two. In the H-plane, the normalized pattern of each slot and of the two together is the same, as shown in Figure 14.19(b).
The equivalent current densities for the other two slots, each of length L and height h, are shown in Figure 14.20. Since the current densities on each wall are of the same magnitude but of opposite direction, the fields radiated by these two slots cancel each other in the principal H-plane. Also since corresponding slots on opposite walls are 180◦out of phase, the corresponding radiations cancel each other in the principal E-plane. This will be shown analytically. The radiation from these two side walls in nonprincipal planes is small compared to the other two side walls. Therefore these two slots are usually referred to as nonradiating slots.
C. Fields Radiated—TMx 010 Mode To find the fields radiated by each slot, we follow a procedure similar to that used to analyze the aperture in Section 12.5.1. The total field is the sum of the two-element array with each element Total #2 #1 (a) E-plane ϕ x x y z #1, #2, Total (b) H-plane θ Figure 14.19 Typical E- and H-plane patterns of each microstrip patch slot, and of the two together.
RECTANGULAR PATCH 807 L x z y W h Ms Ms Ms Ms Figure 14.20 Current densities on nonradiating slots of rectangular microstrip patch.
representing one of the slots. Since the slots are identical, this is accomplished by using an array factor for the two slots.
Radiating Slots Following a procedure similar to that used to analyze the aperture in Sec-tion 12.5.1, the far-zone electric fields radiated by each slot, using the equivalent current densities of (14-38), are written as Er ≃E𝜃≃0 (14-40a) E𝜙= +jk0hWE0e−jk0r 2𝜋r { sin 𝜃sin(X) X sin(Z) Z } (14-40b) where X = k0h 2 sin 𝜃cos 𝜙 (14-40c) Z = k0W 2 cos 𝜃 (14-40d) For very small heights (k0h ≪1), (14-40b) reduces to E𝜙≃+jV0e−jk0r 𝜋r ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ sin 𝜃 sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (14-41) where V0 = hE0.
According to the theory of Chapter 6, the array factor for the two elements, of the same magnitude and phase, separated by a distance Le along the y direction is (AF)y = 2 cos (k0Le 2 sin 𝜃sin 𝜙 ) (14-42) 808 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS where Le is the effective length of (14-3). Thus, the total electric field for the two slots (also for the microstrip antenna) is Et 𝜙= +jk0hWE0e−jk0r 𝜋r { sin 𝜃sin(X) X sin(Z) Z } × cos (k0Le 2 sin 𝜃sin 𝜙 ) (14-43) where X = k0h 2 sin 𝜃cos 𝜙 (14-43a) Z = k0W 2 cos 𝜃 (14-43b) For small values of h (k0h ≪1), (14-43) reduces to Et 𝜙≃+j2V0e−jk0r 𝜋r ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ sin 𝜃 sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ cos (k0Le 2 sin 𝜃sin 𝜙 ) (14-44) where V0 = hE0 is the voltage across the slot.
E-Plane (𝜽= 90◦, 0◦≤𝝓≤90◦and 270◦≤𝝓≤360◦) For the microstrip antenna, the x-y plane (𝜃= 90◦, 0◦≤𝜙≤90◦and 270◦≤𝜙≤360◦) is the principal E-plane. For this plane, the expressions for the radiated fields of (14-43)–(14-43b) reduce to Et 𝜙= +jk0WV0e−jk0r 𝜋r ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ sin (k0h 2 cos 𝜙 ) k0h 2 cos 𝜙 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ cos (k0Le 2 sin 𝜙 ) (14-45) H-Plane (𝝓= 0◦, 0◦≤𝜽≤180◦) The principal H-plane of the microstrip antenna is the x-z plane (𝜙= 0◦, 0◦≤𝜃≤180◦), and the expressions for the radiated fields of (14-43)–(14-43b) reduce to Et 𝜙≃+jk0WV0e−jk0r 𝜋r ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ sin 𝜃 sin (k0h 2 sin 𝜃 ) k0h 2 sin 𝜃 sin (k0W 2 cos 𝜃 ) k0W 2 cos 𝜃 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (14-46) RECTANGULAR PATCH 809 (a) 3-D (b) S11 (c) E-plane (θ = 90°) (d) H-plane (ϕ = 0°) –45 –40 –35 –30 –25 –20 –15 –10 –5 0 9 9.25 9.5 9.75 10 10.25 10.5 10.75 11 S11 (dB) Frequency (GHz) Simulations Measurements ϕ -30 dB -20 dB -10 dB 0 dB 330° 150° 300° 120° 270° 90° 240° 60° 210° 30° 180° 0° Simulations Measurements Cavity Model θ θ -30 dB -20 dB -10 dB 0 dB 60° 120° 30° 150° 0° 180° 30° 150° 60° 120° 90° 90° Simulations Measurements Cavity Model Figure 14.21 Normalized 3D and 2D patterns and S11 of rectangular microstrip patch (L = 0.906 cm, W = 1.186 cm, h = 0.1588 cm, y0 = 0.203 cm, 𝜀r = 2.2, f0 = 9.8 GHz).
To illustrate the modeling of the microstrip using the cavity model, the 3-D, S11 and 2-D princi-pal E- and H-plane patterns have been computed at f0 = 9.8 GHz for the rectangular microstrip of Example 14.1 and Figure 14.8(a). These are displayed in Figure 14.21 where they are compared with measurements. A good agreement is indicated. The ground plane was 10 cm × 10 cm. Edge effects can be taken into account using diffraction theory , . The noted asymmetry in the measured and simulated patterns is due to the feed which is not symmetrically positioned along the E-plane.
The simulation accounts for the position of the feed, while the cavity model does not account for it. The pattern for 0◦≤𝜙≤180◦[left half in Figure 14.21(c)] corresponds to observation angles which lie on the same side of the patch as does the feed probe.
The presence of the dielectric-covered ground plane modifies the reflection coefficient, which influences the magnitude and phase of the image. This is similar to the ground effects discussed in Section 4.8 of Chapter 4. To account for the dielectric, the reflection coefficient for vertical polar-ization of +1 must be replaced by the reflection coefficient of (4-125) while the reflection coef-ficient for horizontal polarization of −1 must be replaced by the reflection coefficient of (4-128).
810 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Basically the introduction of the reflection coefficients of (4-125) and (4-128) to account for the dielectric cover of the ground plane is to modify the boundary conditions of the perfect conductor to one with an impedance surface. The result is for (4-125) to modify the shape of the pattern in the E-plane of the microstrip antenna, primarily for observation angles near grazing (near the ground plane), as was done in Figure 4.35 for the lossy earth. Similar changes are expected for the microstrip antenna. The changes in the pattern near grazing come from the fact that, for the perfect conductor, the reflection coefficient for vertical polarization is +1 for all observation angles. However for the dielectric-covered ground plane (impedance surface), the reflection coefficient of (4-125) is nearly +1 for observation angles far away from grazing but begins to change very rapidly near grazing and becomes −1 at grazing ; thus the formation of an ideal null at grazing.
Similarly the reflection coefficient of (4-128) should basically control the pattern primarily in the H-plane. However, because the reflection coefficient for horizontal polarization for a perfect conduc-tor is −1 for all observation angles while that of (4-128) for the dielectric-covered ground plane is nearly −1 for all observation angles, the shape of the pattern in the H-plane is basically unaltered by the presence of the dielectric cover . This is illustrated in Figure 4.37 for the earth. The pattern also exhibits a null along the ground plane. Similar changes are expected for the microstrip antenna.
Nonradiating Slots The fields radiated by the so-called nonradiating slots, each of effective length Le and height h, are found using the same procedure as for the two radiating slots. Using the fields of (14-39), the equivalent magnetic current density of one of the nonradiating slots facing the +z axis is Ms = −2̂ n × Ea = ̂ ay2E0 cos ( 𝜋 Le y′ ) (14-47) and it is sketched in Figure 14.20. A similar one is facing the −z axis. Using the same procedure as for the radiating slots, the normalized far-zone electric field components radiated by each slot are given by E𝜃= −k0hLeE0e−jk0r 2𝜋r { Y cos 𝜙sin X X cos Y (Y)2 −(𝜋∕2)2 } ej(X+Y) (14-48a) E𝜙= k0hLeE0e−jk0r 2𝜋r { Y cos 𝜃sin 𝜙sin X X cos Y (Y)2 −(𝜋∕2)2 } ej(X+Y) (14-48b) where X = k0h 2 sin 𝜃cos 𝜙 (14-48c) Y = k0Le 2 sin 𝜃sin 𝜙 (14-48d) Since the two nonradiating slots form an array of two elements, of the same magnitude but of opposite phase, separated along the z axis by a distance W, the array factor is (AF)z = 2j sin (k0W 2 cos 𝜃 ) (14-49) Therefore the total far-zone electric field is given by the product of each of (14-48a) and (14-48b) with the array factor of (14-49).
RECTANGULAR PATCH 811 In the H-plane (𝜙= 0◦, 0◦≤𝜃≤180◦), (14-48a) and (14-48b) are zero because the fields radi-ated by each quarter cycle of each slot are cancelled by the fields radiated by the other quarter.
Similarly in the E-plane (𝜃= 90◦, 0◦≤𝜙≤90◦and 270◦≤𝜙≤360◦) the total fields are also zero because (14-49) vanishes. This implies that the fields radiated by each slot are cancelled by the fields radiated by the other. The nonradiation in the principal planes by these two slots was discussed ear-lier and demonstrated by the current densities in Figure 14.20. However, these two slots do radiate away from the principal planes, but their field intensity in these other planes is small compared to that radiated by the two radiating slots such that it is usually neglected. Therefore they are referred to as nonradiating slots.
14.2.3 Directivity As for every other antenna, the directivity is one of the most important figures-of-merit whose defi-nition is given by (2-16a) or D0 = Umax U0 = 4𝜋Umax Prad (14-50) Single Slot (k0h ≪1) Using the electric field of (14-41), the maximum radiation intensity and radiated power can be written, respectively, as Umax = |V0|2 2𝜂0𝜋2 ( 𝜋W λ0 )2 (14-51) Prad = |V0|2 2𝜂0𝜋∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃d𝜃 (14-52) Therefore, the directivity of a single slot can be expressed as D0 = ( 2𝜋W λ0 )2 1 I1 (14-53) where I1 = ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃d𝜃 = [ −2 + cos(X) + XSi(X) + sin(X) X ] (14-53a) X = k0W (14-53b) 812 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS 0.25 0.35 0.45 0.55 0.65 0.75 4.00 5.00 6.00 7.00 8.00 9.00 Directivity (dB) Width W of patch ( 0) h = 0.01 0 (1 slot) h = 0.05 0 (1 slot) h = 0.01 0 (2 slots) h = 0.05 0 (2 slots) ( r = 2.25, L = d/2 = 0/(2√2.25) = 0/3) λ λ λ λ λ λ λ λ Figure 14.22 Computed directivity of one and two slots as a function of the slot width.
Asymptotically the values of (14-53) vary as D0 = ⎧ ⎪ ⎨ ⎪ ⎩ 3.3(dimensionless) = 5.2 dB W ≪λ0 4 ( W λ0 ) W ≫λ0 (14-54) The directivity of a single slot can be computed using (14-53) and (14-53a). In addition, it can also be computed using (14-41) and the computer program Directivity of Chapter 2. Since both are based on the same formulas, they should give the same results. Plots of the directivity of a single slot for h = 0.01λ0 and 0.05λ0 as a function of the width of the slot are shown in Figure 14.22. It is evident that the directivity of a single slot is not influenced strongly by the height of the substrate, as long as it is maintained electrically small.
Two Slots (k0h ≪1) For two slots, using (14-44), the directivity can be written as D2 = ( 2𝜋W λ0 )2 𝜋 I2 = 2 15Grad ( W λ0 )2 (14-55) where Grad is the radiation conductance and I2 = ∫ 𝜋 0 ∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃cos2 (k0Le 2 sin 𝜃sin 𝜙 ) d𝜃d𝜙 (14-55a) RECTANGULAR PATCH 813 The total broadside directivity D2 for the two radiating slots, separated by the dominant TMx 010 mode field (antisymmetric voltage distribution), can also be written as , D2 = D0DAF = D0 2 1 + g12 (14-56) DAF = 2 1 + g12 g12≪1 ≃ 2 (14-56a) where D0 = directivity of single slot [as given by (14-53) and (14-53a)] DAF = directivity of array factor AF [ AF = cos (k0Le 2 sin 𝜃sin 𝜙 )] g12 = normalized mutual conductance = G12∕G1 This can also be justified using the array theory of Chapter 6. The normalized mutual conductance g12 can be obtained using (14-12), (14-12a), and (14-18a). Computed values based on (14-18a) show that usually g12 ≪1; thus (14-56a) is usually a good approximation to (14-56).
Asymptotically the directivity of two slots (microstrip antenna) can be expressed as D2 = ⎧ ⎪ ⎨ ⎪ ⎩ 6.6(dimensionless) = 8.2 dB W ≪λ0 8 ( W λ0 ) W ≫λ0 (14-57) The directivity of the microstrip antenna can now be computed using (14-55) and (14-55a). In addi-tion, it can also be computed using (14-44) and the computer program Directivity of Chapter 2.
Since they are based on the same formulas, they should give the same results. Plots of directivity of a microstrip antenna, modeled by two slots, for h = 0.01λ0 and 0.05λ0 are shown plotted as a func-tion of the width of the patch (W∕λ0) in Figure 14.22. It is evident that the directivity is not a strong function of the height, as long as the height is maintained electrically small. About 2 dB difference is indicated between the directivity of one and two slots. A typical plot of the directivity of a patch for a fixed resonant frequency as a function of the substrate height (h∕λ0), for two different dielectrics, is shown in Figure 14.23.
The directivity of the slots also can be approximated by Kraus’s, (2-26), and Tai & Pereira’s, (2-30a), formulas in terms of the E- and H-plane beamwidths, which can be approximated by ΘE ≃2 sin−1 √ 7.03λ2 0 4(3L2 e + h2)𝜋2 (14-58) ΘH ≃2 sin−1 √ 1 2 + k0W (14-59) The values of the directivities obtained using (14-58) and (14-59) along with either (2-26) or (2-30a) will not be very accurate since the beamwidths, especially in the E-plane, are very large. However, they can serve as guidelines.
814 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS 0.00 .01 .02 .03 .04 .05 0.00 2.00 4.00 6.00 8.00 10.00 Directivity (dB) Substrate height h/ 0 r = 2.55 r = 10.2 λ Figure 14.23 Directivity variations as a function of substrate height for a square microstrip patch antenna.
(Courtesy of D. M. Pozar).
Example 14.3 For the rectangular microstrip antenna of Examples 14.1 and 14.2, with overall dimensions of L = 0.906 cm and W = 1.186 cm, substrate height h = 0.1588 cm, and dielectric constant of 𝜀r = 2.2, center frequency of 10 GHz, find the directivity based on (14-56) and (14-56a). Compare with the values obtained using (14-55) and (14-55a).
Solution: From the solution of Example 14.2 G1 = 0.00157 Siemens G12 = 6.1683 × 10−4 Siemens g12 = G12∕G1 = 0.3921 Using (14-56a) DAF = 2 1 + g12 = 2 1 + 0.3921 = 1.4367 = 1.5736 dB Using (14-53) and (14-53a) I1 = 1.863 D0 = ( 2𝜋W λ0 )2 1 I1 = 3.312 = 5.201 dB According to (14-56) D2 = D0DAF = 3.312(1.4367) = 4.7584 = 6.7746 dB CIRCULAR PATCH 815 Using (14-55a) I2 = 3.59801 Finally, using (14-55) D2 = ( 2𝜋W λ0 )2 𝜋 I2 = 5.3873 (dimensionless) = 7.314 dB There is about 0.5 dB difference between the directivities computed using (14-55) and (14-56); the one based on (14-55) is more accurate.
A MATLAB and FORTRAN computer program, designated as Microstrip, has been developed to design and compute the radiation characteristics of rectangular and circular microstrip patch anten-nas. The description of the program is found in the corresponding READ ME file included in the publisher’s website for this book.
14.3 CIRCULAR PATCH Other than the rectangular patch, the next most popular configuration is the circular patch or disk, as shown in Figure 14.24. It also has received a lot of attention not only as a single element , , , , , , but also in arrays and . The modes supported by the circular patch antenna can be found by treating the patch, ground plane, and the material between the two as a circular cavity. As with the rectangular patch, the modes that are supported primarily by a circular microstrip antenna whose substrate height is small (h ≪λ) are TMz where z is taken perpendicular to the patch. As far as the dimensions of the patch, there are two degrees of freedom to control (length and width) for the rectangular microstrip antenna. Therefore the order of the modes can be changed by changing the relative dimensions of the width and length of the patch (width-to-length ratio).
However, for the circular patch there is only one degree of freedom to control (radius of the patch).
Doing this does not change the order of the modes; however, it does change the absolute value of the resonant frequency of each .
ϕ r h ϕ ρ a ' ' z y x Figure 14.24 Geometry of circular microstrip patch antenna.
816 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Other than using full-wave analysis , , , the circular patch antenna can only be ana-lyzed conveniently using the cavity model , , . This can be accomplished using a proce-dure similar to that for the rectangular patch but now using cylindrical coordinates . The cavity is composed of two perfect electric conductors at the top and bottom to represent the patch and the ground plane, and by a cylindrical perfect magnetic conductor around the circular periphery of the cavity. The dielectric material of the substrate is assumed to be truncated beyond the extent of the patch.
14.3.1 Electric and Magnetic Fields—TMz mnp To find the fields within the cavity, we use the vector potential approach. For TMz we need to first find the magnetic vector potential Az, which must satisfy, in cylindrical coordinates, the homogeneous wave equation of ∇2Az(𝜌, 𝜙, z) + k2Az(𝜌, 𝜙, z) = 0.
(14-60) It can be shown that for TMz modes, whose electric and magnetic fields are related to the vector potential Az by E𝜌= −j 1 𝜔𝜇𝜀 𝜕2Az 𝜕𝜌𝜕z H𝜌= 1 𝜇 1 𝜌 𝜕Az 𝜕𝜙 E𝜙= −j 1 𝜔𝜇𝜀 1 𝜌 𝜕2Az 𝜕𝜙𝜕z H𝜙= −1 𝜇 𝜕Az 𝜕𝜌 Ez = −j 1 𝜔𝜇𝜀 ( 𝜕2 𝜕z2 + k2 ) Az Hz = 0 (14-61) subject to the boundary conditions of E𝜌(0 ≤𝜌′ ≤a, 0 ≤𝜙′ ≤2𝜋, z′ = 0) = 0 E𝜌(0 ≤𝜌′ ≤a, 0 ≤𝜙′ ≤2𝜋, z′ = h) = 0 H𝜙(𝜌′ = a, 0 ≤𝜙′ ≤2𝜋, 0 ≤z′ ≤h) = 0 (14-62) the magnetic vector potential Az reduces to Az = BmnpJm(k𝜌𝜌′)[A2 cos(m𝜙′) + B2 sin(m𝜙′)] cos(kzz′) (14-63) with the constraint equation of (k𝜌)2 + (kz)2 = k2 r = 𝜔2 r𝜇𝜀 (14-63a) CIRCULAR PATCH 817 The primed cylindrical coordinates 𝜌′, 𝜙′, z′ are used to represent the fields within the cavity while Jm(x) is the Bessel function of the first kind of order m, and k𝜌= 𝜒′ mn∕a (14-63b) kz = p𝜋 h (14-63c) m = 0, 1, 2, … (14-63d) n = 1, 2, 3, … (14-63e) p = 0, 1, 2, … (14-63f) In (14-63b) 𝜒′ mn represents the zeroes of the derivative of the Bessel function Jm(x), and they determine the order of the resonant frequencies. The first four values of 𝜒′ mn, in ascending order, are 𝜒′ 11 = 1.8412 𝜒′ 21 = 3.0542 𝜒′ 01 = 3.8318 (14-64) 𝜒′ 31 = 4.2012 14.3.2 Resonant Frequencies The resonant frequencies of the cavity, and thus of the microstrip antenna, are found using (14-63a)–(14-63f). Since for most typical microstrip antennas the substrate height h is very small (typically h < 0.05λ0), the fields along z are essentially constant and are presented in (14-63f) by p = 0 and in (14-63c) by kz = 0. Therefore the resonant frequencies for the TMz mn0 modes can be written using (14-63a) as (fr)mn0 = 1 2𝜋√𝜇𝜀 (𝜒′ mn a ) (14-65) Based on the values of (14-64), the first four modes, in ascending order, are TMz 110, TMz 210, TMz 010, and TMz 310. The dominant mode is the TMz 110 whose resonant frequency is (fr)110 = 1.8412 2𝜋a√𝜇𝜀 = 1.8412𝜐0 2𝜋a√𝜀r (14-66) where 𝜐0 is the speed of light in free-space.
The resonant frequency of (14-66) does not take into account fringing. As was shown for the rectangular patch, and illustrated in Figure 14.7, fringing makes the patch look electrically larger and it was taken into account by introducing a length correction factor given by (14-2). Similarly for the circular patch a correction is introduced by using an effective radius ae, to replace the actual radius a, given by ae = a { 1 + 2h 𝜋a𝜀r [ ln (𝜋a 2h ) + 1.7726 ]}1∕2 (14-67) 818 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Therefore the resonant frequency of (14-66) for the dominant TMz 110 should be modified by using (14-67) and expressed as (frc)110 = 1.8412𝜐0 2𝜋ae √𝜀r = 8.791 × 109 ae √𝜀r ⇒ae = 8.791 × 109 frc √𝜀r (14-68) 14.3.3 Design Based on the cavity model formulation, a design procedure is outlined which leads to practical designs of circular microstrip antennas for the dominant TMz 110 mode. The procedure assumes that the specified information includes the dielectric constant of the substrate (𝜀r), the resonant frequency (fr) and the height of the substrate h. The procedure is as follows: Specify 𝜀r, fr(in Hz), and h (in cm) Determine The actual radius a of the patch.
Design Procedure A first-order approximation to the solution of (14-67) for a is to find ae. This is accomplished by substituting (14-68) for ae on the left side of (14-67) and for a only within the brackets {..} on the right side of (14-67). Doing this and solving for a leads to a = F { 1 + 2h 𝜋𝜀rF [ ln (𝜋F 2h ) + 1.7726 ]}1∕2 (14-69) where F = 8.791 × 109 fr √𝜀r (14-69a) Remember that h in (14-69) must be in cm.
Example 14.4 Design a circular microstrip antenna using a substrate (RT/duroid 5880) with a dielectric constant of 2.2, h = 0.1588 cm (0.0625 in.) so as to resonate at 10 GHz.
Solution: Using (14-69a) F = 8.791 × 109 10 × 109√ 2.2 = 0.593 Therefore using (14-69) a = F { 1 + 2h 𝜋𝜀rF [ ln (𝜋F 2h ) + 1.7726 ]}1∕2 = 0.525 cm (0.207 in.) CIRCULAR PATCH 819 An experimental circular patch based on this design was built and tested. It is probe fed from underneath by a coaxial line and is shown in Figure 14.8(b). Its 3-D, S11 and principal E- and H-plane patterns are displayed in Figure 14.25, where they are compared with simulations, mea-surements and cavity model. A very good agreement is indicated.
–45 –40 –35 –30 –25 –20 –15 –10 –5 0 9 9.25 9.5 9.75 10 10.25 10.5 10.75 11 S11 (dB) Frequency (GHz) Simulations Measurements θ θ -30 dB -20 dB -10 dB 0 dB 30° 150° 60° 120° 90° 90° 120° 60° 150° 30° 180° 0° Simulations Measurements Cavity Model θ θ -30 dB -20 dB -10 dB 0 dB 30° 150° 60° 120° 90° 90° 120° 60° 150° 30° 180° 0° Simulations Measurements Cavity Model (a) 3-D (b) S11 (c) E-plane (ϕ = 0°) (d) H-plane (ϕ = 90°) Figure 14.25 Normalized 3D and 2D patterns and S11 of circular microstrip patch (a = 0.525 cm, ae = 0.598 cm, h = 0.1588 cm, 𝜌f = 0.2 cm, 𝜀r = 2.2, fo = 10 GHz).
14.3.4 Equivalent Current Densities and Fields Radiated As was done for the rectangular patch using the cavity model, the fields radiated by the circular patch can be found by using the Equivalence Principle whereby the circumferential wall of the cavity is replaced by an equivalent magnetic current density of (14-38) as shown in Figure 14.26. Based on (14-61)–(14-63) and assuming a TMz 110 mode field distribution beneath the patch, the normalized electric and magnetic fields within the cavity for the cosine azimuthal variations can be written as E𝜌= E𝜙= Hz = 0 (14-70a) Ez = E0J1(k𝜌′) cos 𝜙′ (14-70b) 820 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS h z y x ϕ ρ a ' ' r Mϕ Figure 14.26 Cavity model and equivalent magnetic current density for circular microstrip patch antenna.
H𝜌= j E0 𝜔𝜇0 1 𝜌J1(k𝜌′) sin 𝜙′ (14-70c) H𝜙= j E0 𝜔𝜇0 J′ 1(k𝜌′) cos 𝜙′ (14-70d) where ′ = 𝜕∕𝜕𝜌and 𝜙′ is the azimuthal angle along the perimeter of the patch.
Based on (14-70b) evaluated at the electrical equivalent edge of the disk (𝜌′ = ae), the magnetic current density of (14-38) can be written as Ms = −2̂ n × Ea|𝜌′=ae = ̂ a𝜙2E0J1(kae) cos 𝜙′ (14-71) Since the height of the substrate is very small and the current density of (14-71) is uniform along the z direction, we can approximate (14-71) by a filamentary magnetic current of Im = hMs = ̂ a𝜙2hE0J1(kae) cos 𝜙′ = ̂ a𝜙2V0 cos 𝜙′ (14-71a) where V0 = hE0J1(kae) at 𝜙′ = 0.
Using (14-71a) the microstrip antenna can be treated as a circular loop. Referring to Chapter 5 for the loop and using the radiation equations of Sections 12.3 and 12.6, we can write that , Er = 0 (14-72a) E𝜃= −jk0aeV0e−jk0r 2r {cos 𝜙J′ 02} (14-72b) E𝜙= jk0aeV0e−jk0r 2r {cos 𝜃sin 𝜙J02} (14-72c) J′ 02 = J0(k0ae sin 𝜃) −J2(k0ae sin 𝜃) (14-72d) J02 = J0(k0ae sin 𝜃) + J2(k0ae sin 𝜃) (14-72e) CIRCULAR PATCH 821 where ae is the effective radius as given by (14-67). The fields in the principal planes reduce to: E-plane (𝝓= 0◦, 180◦, 0◦≤𝜽≤90◦) E𝜃= jk0aeV0e−jk0r 2r [J′ 02] (14-73a) E𝜙= 0 (14-73b) H-plane (𝝓= 90◦, 270◦, 0◦≤𝜽≤90◦) E𝜃= 0 (14-74a) E𝜙= jk0aeV0e−jk0r 2r [cos 𝜃J02] (14-74b) Patterns have been computed for the circular patch of Example 14.4, Figure 14.8(b) based on (14-73a)–(14-74b), and they are shown in Figure 14.25 where they are compared with mea-surements and simulated patterns. The noted asymmetry in the measured and simulated patterns is due to the feed which is not symmetrically positioned along the E-plane. The simulations accounts for the position of the feed, while the cavity model does not account for it. The pattern for the left half of Figure 14.25(c) corresponds to observation angles which lie on the same side of the patch as does the feed probe. The ground plane was 10 cm × 10 cm.
14.3.5 Conductance and Directivity The conductance due to the radiated power and directivity of the circular microstrip patch antenna can be computed using their respective definitions of (14-10) and (14-50). For each we need the radiated power, which based on the fields of (14-72b) and (14-72c) of the cavity model can be expressed as Prad = |V0|2 (k0ae)2 960 ∫ 𝜋∕2 0 [J′2 02 + cos2 𝜃J2 02] sin 𝜃d𝜃 (14-75) Therefore the conductance across the gap between the patch and the ground plane at 𝜙′ = 0◦ based on (14-10) and (14-75) can be written as Grad = (k0ae)2 480 ∫ 𝜋∕2 0 [J′2 02 + cos2 𝜃J2 02] sin 𝜃d𝜃 (14-76) A plot of the conductance of (14-76) for the TMz 110 mode is shown in Figure 14.27. While the conductance of (14-76) accounts for the losses due to radiation, it does not take into account losses due to conduction (ohmic) and dielectric losses, which each can be expressed as Gc = 𝜀mo𝜋(𝜋𝜇0fr)−3∕2 4h2√ 𝜎 [(kae)2 −m2] (14-77) Gd = 𝜀mo tan 𝛿 4𝜇0hfr [(kae)2 −m2] (14-78) 822 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 10 20 30 Grad (10–3) ae/λo Figure 14.27 Radiation conductance versus effective radius for circular microstrip patch operating in domi-nant TMz 110 mode.
where 𝜀mo = 2 for m = 0, 𝜀mo = 1 for m ≠0, and fr represents the resonant frequency of the mn0 mode. Thus, the total conductance can be written as Gt = Grad + Gc + Gd (14-79) Based on (14-50), (14-72b), (14-72c), (14-75) and (14-76), the directivity for the slot at 𝜃= 0◦ can be expressed as D0 = (k0ae)2 120Grad (14-80) A plot of the directivity of the dominant TMz 110 mode as a function of the radius of the disk is shown plotted in Figure 14.28. For very small values of the radius the directivity approaches 3 (4.8 dB), which is equivalent of that of a slot above a ground plane and it agrees with the value of (14-54) for W ≪λ0.
14.3.6 Resonant Input Resistance As was the case for the rectangular patch antenna, the input impedance of a circular patch at res-onance is real. The input power is independent of the feed-point position along the circumference.
Taken the reference of the feed at 𝜙′ = 0◦, the input resistance at any radial distance 𝜌′ = 𝜌0 from the center of the patch, for the dominant TM11 mode (the one that does not have a zero in the amplitude pattern normal to the patch), can be written as Rin(𝜌′ = 𝜌0) = 1 Gt J2 1(k𝜌0) J2 1(kae) (14-81) where Gt is the total conductance due to radiation, conduction (ohmic) and dielectric losses, as given by (14-79). As was the case with the rectangular patch, the resonant input resistance of a circular QUALITY FACTOR, BANDWIDTH, AND EFFICIENCY 823 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 4 6 8 10 12 Do (dB) ae/λo Figure 14.28 Directivity versus effective radius for circular microstrip patch antenna operating in dominant TMz 110 mode.
patch with an inset feed, which is usually a probe, can be written as Rin(𝜌′ = 𝜌0) = Rin(𝜌′ = ae) J2 1(k𝜌0) J2 1(kae) (14-82) Rin(𝜌′ = ae) = 1 Gt (14-82a) where Gt is given by (14-79). This is analogous to (14-20a) for the rectangular patch.
A MATLAB and FORTRAN computer program, designated as Microstrip, has been developed to design and compute the radiation characteristics of rectangular and circular microstrip patch anten-nas. The description of the program is found in the corresponding READ ME file included in the publisher’s website for this book.
14.4 QUALITY FACTOR, BANDWIDTH, AND EFFICIENCY The quality factor, bandwidth, and efficiency are antenna figures-of-merit, which are interrelated, and there is no complete freedom to independently optimize each one. Therefore there is always a trade-off between them in arriving at an optimum antenna performance. Often, however, there is a desire to optimize one of them while reducing the performance of the other.
The quality factor is a figure-of-merit that is representative of the antenna losses. Typically there are radiation, conduction (ohmic), dielectric and surface wave losses. Therefore the total quality factor Qt is influenced by all of these losses and is, in general, written as 1 Qt = 1 Qrad + 1 Qc + 1 Qd + 1 Qsw (14-83) where Qt = total quality factor Qrad = quality factor due to radiation (space wave) losses 824 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Qc = quality factor due to conduction (ohmic) losses Qd = quality factor due to dielectric losses Qsw = quality factor due to surface waves For very thin substrates, the losses due to surface waves are very small and can be neglected. How-ever, for thicker substrates they need to be taken into account . These losses can also be eliminated by using cavities and .
For very thin substrates (h ≪λ0) of arbitrary shapes (including rectangular and circular), there are approximate formulas to represent the quality factors of the various losses , . These can be expressed as Qc = h √ 𝜋f𝜇𝜎 (14-84) Qd = 1 tan 𝛿 (14-85) Qrad = 2𝜔𝜀r hGt∕lK (14-86) where tan 𝛿is the loss tangent of the substrate material, 𝜎is the conductivity of the conductors associated with the patch and ground plane, Gt∕l is the total conductance per unit length of the radiating aperture and K = ∫∫ area |E|2 dA ∮perimeter |E|2 dl (14-86a) For a rectangular aperture operating in the dominant TMx 010 mode K = L 4 (14-87a) Gt∕l = Grad W (14-87b) The Qrad as represented by (14-86) is inversely proportional to the height of the substrate, and for very thin substrates is usually the dominant factor.
The fractional bandwidth of the antenna is inversely proportional to the Qt of the antenna, and it is defined by (11-36) or Δf f0 = 1 Qt (14-88) However, (14-88) may not be as useful because it does not take into account impedance matching at the input terminals of the antenna. A more meaningful definition of the fractional bandwidth is over a band of frequencies where the VSWR at the input terminals is equal to or less than a desired maximum value, assuming that the VSWR is unity at the design frequency. A modified form QUALITY FACTOR, BANDWIDTH, AND EFFICIENCY 825 of (14-88) that takes into account the impedance matching is Δf f0 = VSWR −1 Qt √ VSWR (14-88a) In general the bandwidth is proportional to the volume, which for a rectangular microstrip antenna at a constant resonant frequency can be expressed as BW ∼volume = area ⋅height = length ⋅width ⋅height ∼ 1 √𝜀r 1 √𝜀r √𝜀r = 1 √𝜀r (14-89) An approximate expression for the bandwidth (for VSWR ≤2, |Γ| ≤1∕3), provided that the surface wave power is much smaller than the radiated (space-wave) power , is BW = 16 3 √ 2 [𝜀r −1 (𝜀r)2 ] h λo (W L ) = 3.771 [𝜀r −1 (𝜀r)2 ] h λo (W L ) (14-89a) The expression is valid for h ≪λo as it is derived based on approximations of rigorously devel-oped formulas, including Sommerfeld integrals for thin grounded substrates.
Therefore, according to (14-89), the bandwidth is inversely proportional to the square root of the dielectric constant of the substrate. A typical variation of the bandwidth for a microstrip antenna as a function of the normalized height of the substrate, for two different substrates, is shown in Figure 14.29. It is evident that the bandwidth increases as the substrate height increases.
The radiation efficiency of an antenna is expressed by (2-90), and it is defined as the power radiated over the input power. It can also be expressed in terms of the quality factors, which for 0.00 0.02 0.04 0.06 0.08 0.10 0.00 0.20 0.40 0.60 0.80 1.00 15 10 5 0 Efficiency Percent bandwidth Substrate height h/λo r = 2.2 r = 10 r = 10 r = 2.2 BW BW ecdsw ecdsw Figure 14.29 Efficiency and bandwidth versus substrate height at constant resonant frequency for rectangular microstrip patch for two different substrates. (source: D. M. Pozar, “Microstrip Antennas,” Proc. IEEE, Vol.
80, No. 1, January 1992. c ⃝1992 IEEE).
826 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS a microstrip antenna can be written as ecdsw = 1∕Qrad 1∕Qt = Qt Qrad (14-90) where Qt is given by (14-83). Typical variations of the efficiency as a function of the substrate height for a microstrip antenna, with two different substrates, are shown in Figure 14.29.
14.5 INPUT IMPEDANCE In the previous sections of this chapter, we derived approximate expressions for the resonant input resistance for both rectangular and circular microstrip antennas. Also, approximate expressions were stated which describe the variation of the resonant input resistance as a function of the inset-feed position, which can be used effectively to match the antenna element to the input transmission line.
In general, the input impedance is complex and it includes both a resonant and a nonresonant part which is usually reactive. Both the real and imaginary parts of the impedance vary as a function of frequency, and a typical variation is shown in Figure 14.30. Ideally both the resistance and reactance exhibit symmetry about the resonant frequency, and the reactance at resonance is equal to the average of sum of its maximum value (which is positive) and its minimum value (which is negative).
Typically the feed reactance is very small, compared to the resonant resistance, for very thin sub-strates. However, for thick elements the reactance may be significant and needs to be taken into account in impedance matching and in determining the resonant frequency of a loaded element .
The variations of the feed reactance as a function of position can be intuitively explained by con-sidering the cavity model for a rectangular patch with its four side perfect magnetic conducting walls , . As far as the impedance is concerned, the magnetic walls can be taken into account by introducing multiple images with current flow in the same direction as the actual feed. When the 1,200 1,225 1,250 –20 0 20 40 Resistance, reactance (ohms) Frequency (MHz) Resistance, R Reactance, X Peak resonant resistance Xf = feed reactance = Xmin + Xmax 2 Xmax Xmin Xf Figure 14.30 Typical variation of resistance and reactance of rectangular microstrip antenna versus frequency (Electromagnetics, Vol. 3, Nos. 3 and 4, p. 33, W. F. Richards, J. R. Zinecker, and R. D. Clark, Taylor & Francis, Washington, D.C. Reproduced by permission. All rights reserved).
COUPLING 827 feed point is far away from one of the edges, the magnetic field associated with the images and that of the actual feed do not overlap strongly. Therefore the inductance associated with the magnetic energy density stored within a small testing volume near the feed will be primarily due to the current of the actual feed. However, when the feed is at one of the edges, the feed and one of the images, which accounts for the magnetic wall at that edge, coincide. Thus, the associated magnetic field stored energy of the equivalent circuit doubles while the respective stored magnetic energy density quadruples. However, because the volume in the testing region of the patch is only half from that when the feed was far removed from the edge, the net stored magnetic density is only double of that of the feed alone. Thus, the associated inductance and reactance, when the feed is at the edge, is twice that when the feed is far removed from the edge. When the feed is at a corner, there will be three images in the testing volume of the patch, in addition to the actual feed, to take into account the edges that form the corner. Using the same argument as above, the associated inductance and reactance for a feed at a corner is four times that when the feed is removed from an edge or a corner.
Thus, the largest reactance (about a factor of four larger) is when the feed is at or near a corner while the smallest is when the feed is far removed from an edge or a corner.
Although such an argument predicts the relative variations (trends) of the reactance as a function of position, they do predict very accurately the absolute values especially when the feed is at or very near an edge. In fact it overestimates the values for feeds right on the edge; the actual values predicted by the cavity model with perfect magnetic conducting walls are smaller . A formula that has been suggested to approximate the feed reactance, which does not take into account any images, is xf ≃−𝜂kh 2𝜋 [ ln (kd 4 ) + 0.577 ] (14-91) where d is the diameter of the feed probe. More accurate predictions of the input impedance, based on full-wave models, have been made for circular patches where an attachment current mode is introduced to match the current distribution of the probe to that of the patch .
14.6 COUPLING The coupling between two or more microstrip antenna elements can be taken into account easily using full-wave analyses. However, it is more difficult to do using the transmission-line and cavity models, although successful attempts have been made using the transmission-line model and the cavity model , . It can be shown that coupling between two patches, as is coupling between two aperture or two wire antennas, is a function of the position of one element relative to the other.
This has been demonstrated in Figure 4.23 for a vertical half-wavelength dipole above a ground plane and in Figure 4.33 for a horizontal half-wavelength dipole above a ground plane. From these two, the ground effects are more pronounced for the horizontal dipole. Also, mutual effects have been discussed in Chapter 8 for the three different arrangements of dipoles, as shown in Figure 8.20 whose side-by-side arrangement exhibits the largest variations of mutual impedance.
For two rectangular microstrip patches the coupling for two side-by-side elements is a func-tion of the relative alignment. When the elements are positioned collinearly along the E-plane, this arrangement is referred to as the E-plane, as shown in Figure 14.31(a); when the elements are posi-tioned collinearly along the H-plane, this arrangement is referred to as the H-plane, as shown in Figure 14.31(b). For an edge-to-edge separation of s, the E-plane exhibits the smallest coupling isolation for very small spacing (typically s < 0.10λ0) while the H-plane exhibits the smallest cou-pling for large spacing (typically s > 0.10λ0). The spacing at which one plane coupling overtakes the other one depends on the electrical properties and geometrical dimensions of the microstrip antenna.
Typical variations are shown in Figure 14.32.
828 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS E E E E s s (b) H-plane (a) E-plane Figure 14.31 E- and H-plane arrangements of microstrip patch antennas.
In general, mutual coupling is primarily attributed to the fields that exist along the air-dielectric interface. The fields can be decomposed to space waves (with 1∕𝜌radial variations), higher order waves (with 1∕𝜌2 radial variations), surface waves (with 1∕𝜌1∕2 radial variations), and leaky waves [with exp(−λ𝜌)∕𝜌1∕2 radial variations] , . Because of the spherical radial variation, space (1∕𝜌) and higher order waves (1∕𝜌2) are most dominant for very small spacing while surface waves, because of their 1∕𝜌1∕2 radial variations are dominant for large separations. Surface waves exist and propagate within the dielectric, and their excitation is a function of the thickness of the substrate .
In a given direction, the lowest order (dominant) surface wave mode is TM(odd) with zero cutoff frequency followed by a TE(even), and alternatively by TM(odd) and TE(even) modes. For a rectan-gular microstrip patch, the fields are TM in a direction of propagation along the E-plane and TE in a direction of propagation along the H-plane. Since for the E-plane arrangement of Figure 14.31(a) the elements are placed collinearly along the E-plane where the fields in the space between the elements are primarily TM, there is a stronger surface wave excitation (based on a single dominant surface wave mode) between the elements, and the coupling is larger. However for the H-plane arrangement of Figure 14.31(b), the fields in the space between the elements are primarily TE and there is not a strong dominant mode surface wave excitation; therefore there is less coupling between the ele-ments. This does change as the thickness of the substrate increases which allows higher order TE surface wave excitation.
0 0.25 0.50 0.75 1.00 1.25 –40 –30 –20 –10 0 s/λo |S12|2 (dB) Measured—Carver Calculated—Pozar E-plane E-plane H-plane H-plane L L s s W W E E E E Figure 14.32 Measured and calculated mutual coupling between two coax-fed microstrip antennas, for both E-plane and H-plane coupling, (W = 10.57 cm, L = 6.55 cm, h = 0.1588 cm, 𝜀r = 2.55, fr = 1,410 MHz).
(source: D. M. Pozar, “Input Impedance and Mutual Coupling of Rectangular Microstrip Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-30, No. 6, November 1982. c ⃝1982 IEEE).
COUPLING 829 0.0 0.5 1.0 1.5 2.0 2.5 3.0 –0.002 –0.001 0.0 0.001 0.002 0.003 s/λo G12 (siemens) E-plane H-plane Figure 14.33 E- and H-plane mutual conductance versus patch separation for rectangular microstrip patch antennas (W = 1.186 cm, L = 0.906 cm, 𝜀r = 2.2, λ0 = 3 cm).
The mutual conductance between two rectangular microstrip patches has also been found using the basic definition of conductance given by (14-18), the far fields based on the cavity model, and the array theory of Chapter 6. For the E-plane arrangement of Figure 14.31(a) and for the odd mode field distribution beneath the patch, which is representative of the dominant mode, the mutual conductance is G12 = 1 𝜋 √𝜀 𝜇∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃 { 2J0 ( Y λ0 2𝜋sin 𝜃 ) +J0 ( Y + L λ0 2𝜋sin 𝜃 ) + J0 ( Y −L λ0 2𝜋sin 𝜃 )} d𝜃 (14-92) where Y is the center-to-center separation between the slots and J0 is the Bessel function of the first kind of order zero. The first term in (14-92) represents the mutual conductance of two slots separated by a distance X along the E-plane while the second and third terms represent, respectively, the conductances of two slots separated along the E-plane by distances Y + L and Y −L. Typical normalized results are shown by the solid curve in Figure 14.33.
For the H-plane arrangement of Figure 14.31(b) and for the odd mode field distribution beneath the patch, which is representative of the dominant mode, the mutual conductance is G12 = 2 𝜋 √𝜀 𝜇∫ 𝜋 0 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ sin (k0W 2 cos 𝜃 ) cos 𝜃 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2 sin3 𝜃cos ( Z λ0 2𝜋cos 𝜃 ) ⋅ { 1 + J0 ( L λ0 2𝜋sin 𝜃 )} d𝜃 (14-93) where Z is the center-to-center separation between the slots and J0 is the Bessel function of the first kind of order zero. The first term in (14-93) represents twice the mutual conductance of two slots 830 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS separated along the H-plane by a distance Z while the second term represents twice the conductance between two slots separated along the E-plane by a distance L and along the H-plane by a distance Z.
Typical normalized results are shown by the dashed curve in Figure 14.31. By comparing the results of Figure 14.31 it is clear that the mutual conductance for the H-plane arrangement, as expected, decreases with distance faster than that of the E-plane. Also it is observed that the mutual conduc-tance for the E-plane arrangement is higher for wider elements while it is lower for wider elements for the H-plane arrangement.
14.7 CIRCULAR POLARIZATION The patch elements that we discussed so far, both the rectangular and the circular, radiate primarily linearly polarized waves if conventional feeds are used with no modifications. However, circular and elliptical polarizations can be obtained using various feed arrangements or slight modifications made to the elements. We will discuss here some of these arrangements.
Circular polarization can be obtained if two orthogonal modes are excited with a 90◦time-phase difference between them. This can be accomplished by adjusting the physical dimensions of the patch and using either single, or two, or more feeds. There have been some suggestions made and reported in the literature using single patches. For a square patch element, the easiest way to excite ideally circular polarization is to feed the element at two adjacent edges, as shown in Figures 14.34(a,b), to excite the two orthogonal modes; the TMx 010 with the feed at one edge and the TMx 001 with the feed at the other edge. The quadrature phase difference is obtained by feeding the element with a 90◦power divider or 90◦hybrid. Examples of arrays of linear elements that generate circular polarization are discussed in .
For a circular patch, circular polarization for the TMz 110 mode is achieved by using two feeds with proper angular separation. An example is shown in Figure 14.34(c) using two coax feeds separated by 90◦which generate fields that are orthogonal to each other under the patch, as well as outside the patch. Also with this two-probe arrangement, each probe is always positioned at a point where the field generated by the other probe exhibits a null; therefore there is very little mutual coupling between the two probes. To achieve circular polarization, it is also required that the two feeds are fed in such a manner that there is 90◦time-phase difference between the fields of the two; this is achieved through the use of a 90◦hybrid, as shown in Figure 14.34(c). The shorting pin is placed at the center of the patch to ground the patch to the ground plane which is not necessary for circular polarization but is used to suppress modes with no 𝜙variations and also may improve the quality of circular polarization.
For higher order modes, the spacing between the two feeds to achieve circular polarization is different. This is illustrated in Figure 14.34(d) and tabulated in Table 14.2, for the TMz 110 [same as in Figure 14.32(c)], TMz 210, TMz 310, and TMz 410 modes . However to preserve symmetry and minimize cross polarization, especially for relatively thick substrates, two additional feed probes located diametrically opposite of the original poles are usually recommended. The additional probes are used to suppress the neighboring (adjacent) modes which usually have the next highest mag-nitudes . For the even modes (TMz 210 and TMz 410), the four feed probes should have phases of 0◦, 90◦, 0◦and 90◦while the odd modes (TMz 110 and TMz 310) should have phases of 0◦, 90◦, 180◦ and 270◦, as shown in Figure 14.34(d) .
To overcome the complexities inherent in dual-feed arrangements, circular polarization can also be achieved with a single feed. One way to accomplish this is to feed the patch at a single point to excite two orthogonal degenerate modes (of some resonant frequency) of ideally equal amplitudes.
By introducing then a proper asymmetry in the cavity, the degeneracy can be removed with one mode increasing with frequency while the orthogonal mode will be decreasing with frequency by the same amount. Since the two modes will have slightly different frequencies, by proper design the field of one mode can lead by 45◦while that of the other can lag by 45◦resulting in a 90◦phase difference CIRCULAR POLARIZATION 831 necessary for circular polarization . To achieve this, several arrangements have been suggested.
This can be illustrated with a square patch fed as shown in Figure 14.35 .
The resonant frequencies f1 and f2 of the bandwidth of (14-88a) associated with the two lengths L and W of a rectangular microstrip are f1 = f0 √ 1 + 1∕Qt (14-94a) f2 = f0 √ 1 + 1∕Qt (14-94b) where f0 is the center frequency. Feeding the element along the diagonal starting at the lower left corner toward the upper right corner, shown dashed in Figure 14.35(b), yields ideally left-hand circu-lar polarization at broadside. Right-hand circular polarization can be achieved by feeding along the opposite diagonal, which starts at the lower right corner and proceeds toward the upper left corner, shown dashed in Figure 14.35(c). Instead of moving the feed point each time to change the modes in order to change the type of circular polarization, varactor diodes can be used to adjust the capac-itance and bias, which effectively shifts by electrical means the apparent physical location of the feed point.
Example 14.5 The fractional bandwidth at a center frequency of 10 GHz of a rectangular patch antenna whose substrate is RT/duroid 5880 (𝜀r = 2.2) with height h = 0.1588 cm is about 5% for a VSWR of 2:1. Within that bandwidth, find resonant frequencies associated with the two lengths of the rectangular patch antenna, and the relative ratio of the two lengths.
Solution: The total quality factor Qt of the patch antenna is found using (14-88a) or Qt = 1 0.05 √ 2 = 14.14 Using (14-94a) and (14-94b) f1 = 10 × 109 √ 1 + 1∕14.14 = 9.664 GHz f2 = 10 × 109√ 1 + 1∕14.14 = 10.348 GHz The relative ratio of the two lengths from Figure 14.35(a) L W = 1 + 1 Qt = 1 + 1 14.14 = 1.07 which makes the patch nearly square.
There are some other practical ways of achieving nearly circular polarization. For a square patch, this can be accomplished by cutting very thin slots as shown in Figures 14.36(a,b) with dimensions c = L 2.72 = W 2.72 (14-95a) d = c 10 = L 27.2 = W 27.2 (14-95b) 832 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS /4 Power divider Square patch (a) Square patch driven at adjacent sides through a power divider 90° hybrid Square patch (b) Square patch driven at adjacent sides through a 90° hybrid 90° hybrid Coaxial connector Coaxial connector Center of disk grounded to ground plane = 90° x y (c) Circular patch fed with coax (d) Circular patch feed arrangements for TM110 and higher order modes a TM110 0° 180° 270° 90° 90° TM410 0° 0° 90° 90° 67.5° TM210 0° 0° 90° 90° 45° TM310 0° 180° 90° 30° 270° z λ α Figure 14.34 Rectangular and circular patch arrangements for circular polarization. (source: J. Huang, “Cir-cularly Polarized Conical Patterns from Circular Microstrip Antennas,” IEEE Trans. Antennas Propagat., Vol.
AP-32, No. 9, Sept. 1984. c ⃝1984 IEEE).
An alternative way is to trim the ends of two opposite corners of a square patch and feed at points 1 or 3, as shown in Figure 14.37(a). Circular polarization can also be achieved with a circular patch by making it slightly elliptical or by adding tabs, as shown in Figure 14.37(b).
TABLE 14.2 Feed Probe Angular Spacing of Different Modes for Circular Polarization (after ) TM110 TM210 TM310 TM410 TM510 TM610 45◦ 30◦ 22.5◦ 18◦, 54◦ 15◦, 45◦ 𝛼 90◦ or or or or or 135◦ 90◦ 67.5◦ 90◦ 75◦ 14.8 ARRAYS AND FEED NETWORKS Microstrip antennas are used not only as single elements but are very popular in arrays , , , , , , , –, and –. As discussed in Chapter 6, arrays are very versatile and are used, among other things, to synthesize a required pattern that cannot be achieved with a single element. In addition, they are used to scan the beam of an antenna system, increase ARRAYS AND FEED NETWORKS 833 (a) Nearly square patch Nearly square patch L ≅ W 1 + 1 Qt W L y z L W Feed point (y', z') (b) Left-hand circular (LHC) y z L W Feed point (y', z') (c) Right-hand circular (RHC) Figure 14.35 Single-feed arrangements for circular polarization of rectangular microstrip patches.
the directivity, and perform various other functions which would be difficult with any one single element. The elements can be fed by a single line, as shown in Figure 14.38(a), or by multiple lines in a feed network arrangement, as shown in Figure 14.38(b). The first is referred to as a series-feed network while the second is referred to as a corporate-feed network.
The corporate-feed network is used to provide power splits of 2n (i.e., n = 2, 4, 8, 16, 32, etc.).
This is accomplished by using either tapered lines, as shown in Figure 14.39(a), to match 100-ohm patch elements to a 50-ohm input or using quarter-wavelength impedance transformers, as shown in Figure 14.39(b) . The design of single- and multiple-section quarter-wavelength impedance transformers is discussed in Section 9.8.
W = L W = L (a) Right-hand c d Square patch W = L W = L (b) Left-hand c d Square patch Figure 14.36 Circular polarization for square patch with thin slots on patch (c = W∕2.72 = L∕2.72, d = c∕10 = W∕27.2 = L∕27.2).
834 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS W L (a) Trimmed square (L = W) (b) Elliptical with tabs b a 45° 1 3 Figure 14.37 Circular polarization by trimming opposite corners of a square patch and by making circular patch slightly elliptical and adding tabs.
Series-fed arrays can be conveniently fabricated using photolithography for both the radiating elements and the feed network. However, this technique is limited to arrays with a fixed beam or those which are scanned by varying the frequency, but it can be applied to linear and planar arrays with single or dual polarization. Also any changes in one of the elements or feed lines affects the performance of the others. Therefore in a design it is important to be able to take into account these and other effects, such as mutual coupling, and internal reflections.
Corporate-fed arrays are general and versatile. With this method the designer has more control of the feed of each element (amplitude and phase) and it is ideal for scanning phased arrays, multi-beam arrays, or shaped-beam arrays. As discussed in Chapter 6, the phase of each element can be controlled using phase shifters while the amplitude can be adjusted using either amplifiers or attenu-ators. An electronically-steered phased array (ATDRSS) of 10 × 10 rectangular microstrip elements, operating in the 2–2.3 GHz frequency range and used for space-to-space communications, is shown in Figure 14.40.
Those who have been designing and testing microstrip arrays indicate that radiation from the feed line, using either a series or corporate-feed network, is a serious problem that limits the cross-polarization and side lobe level of the arrays . Both cross-polarization and side lobe levels can be improved by isolating the feed network from the radiating face of the array. This can be accomplished using either probe feeds or aperture coupling.
Arrays can be analyzed using the theory of Chapter 6. However, such an approach does not take into account mutual coupling effects, which for microstrip patches can be significant. Therefore for more accurate results, full-wave solutions must be performed. In microstrip arrays , as in any other array , mutual coupling between elements can introduce scan blindness which limits, for a certain maximum reflection coefficient, the angular volume over which the arrays can be scanned. For microstrip antennas, this scan limitation is strongly influenced by surface waves within the substrate.
This scan angular volume can be extended by eliminating surface waves. One way to do this is to use (b) Corporate feed (a) Series feed Figure 14.38 Feed arrangements for microstrip patch arrays.
ARRAYS AND FEED NETWORKS 835 4 100 Ω 100 Ω 100 Ω 100 Ω 50 Ω 50 Ω 70 Ω 70 Ω 50 Ω input 50 Ω input 100 Ω 100 Ω 70 Ω 70 Ω 50 Ω 50 Ω 100 Ω 100 Ω 100 Ω 50 Ω 50 Ω 70 Ω 70 Ω 50 Ω 50 Ω 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω (a) Tapered lines (b) /4 transformers 4 λ λ λ Figure 14.39 Tapered and λ∕4 impedance transformer lines to match 100-ohm patches to a 50-ohm line.
(source: R. E. Munson, “Conformal Microstrip Antennas and Microstrip Phased Arrays,” IEEE Trans. Anten-nas Propagat., Vol. AP-22, No. 1, January 1974. c ⃝1974 IEEE).
cavities in conjunction with microstrip elements , . Figure 14.41 shows an array of circular patches backed by either circular or rectangular cavities. It has been shown that the presence of cavities, either circular or rectangular, can have a pronounced enhancement in the E-plane scan volume, especially for thicker substrates . The H-plane scan volume is not strongly enhanced.
However the shape of the cavity, circular or rectangular, does not strongly influence the results.
Typical results for broadside-matched reflection coefficient infinite array of circular patches, with a Figure 14.40 Antenna array of 10 × 10 rectangular microstrip patches, 2–2.3 GHz, for space-to-space com-munications. (Courtesy: Ball Aerospace & Technologies Corp.).
836 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS c a b 2c 2c dy dx (a) Top view (b) Side view 2c 2b 2a 2r0 0, h μ Figure 14.41 Array of circular patches backed by circular cavities. (Courtesy J. T. Aberle and F. Zavosh).
substrate 0.08λ0 thick and backed by circular and rectangular cavities, are shown in Figure 14.42 for the E-plane and H-plane. The broadside-matched reflection coefficient Γ(𝜃, 𝜙) is defined as Γ(𝜃, 𝜙) = Zin(𝜃, 𝜙) −Zin(0, 0) Zin(𝜃, 𝜙) + Z∗ in(0, 0) (14-96) 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 0.00 0.20 0.40 0.60 0.80 1.00 Conv. patch (E-plane) Circ. cavity (E-plane) Rect. cavity (E-plane) Conv. patch (H-plane) Circ. cavity (H-plane) Rect. cavity (H-plane) 2:1 VSWR 2:1 VSWR a = 0.156 o Scan angle o (degrees) θ Reflection coefficient |Γin| b = 0.195 o c = 0.25 o dx = dy = 0.5 o r = 2.5 h = 0.08 o r0 = 0.004 o E-plane (conventional) H-plane (cavities) E-plane (cavities) H-plane (conventional) λ λ λ λ λ λ Figure 14.42 E- and H-plane broadside-matched input reflection coefficient versus scan angle for infinite array of circular microstrip patches with and without cavities. (Courtesy J. T. Aberle and F. Zavosh).
ANTENNAS FOR MOBILE COMMUNICATIONS 837 Figure 14.43 Dipole phased array on a PBG surface (source: c ⃝2004 IEEE). (a) 3 × 3 array. (b) Dipole between 2 × 2 unit cell.
where Zin(𝜃, 𝜙) is the input impedance when the main beam is scanned toward an angle (𝜃, 𝜙). The results are compared with those of a conventional cavity (noncavity backed). It is apparent that there is a significant scan enhancement for the E-plane, especially for a VSWR of about 2:1. H-plane enhancement occurs for reflection coefficients greater than about 0.60. For the conventional array, the E-plane response exhibits a large reflection coefficient, which approaches unity, near a scan angle of 𝜃0 = 72.5◦. This is evidence of scan blindness which ideally occurs when the reflection coefficient is unity, and it is attributed to the coupling between the array elements due to leaky waves . Scan blindness occurs for both the E- and H-planes at grazing incidence (𝜃0 = 90◦).
Another way to minimize the surface waves, without the use of cavities, is to mount the patches on EBG textured surfaces with vias, as shown in Figure 4.13(a) and Figure 14.43(a) for a 3 × 3 array. A 2 × 2 unit cell of EBG surface, with a dipole in its middle, is displayed in Figure 14.43(b) . Such surfaces have the ability to control and minimize surface waves in substrates. A microstrip dipole element, placed within a textured high-impedance surface, was designed, simulated, fabricated, and measured . It consists of a dipole patch of length 9.766 mm placed in the middle of 4 × 4 EBG unit cells; each cell had dimensions of w = l = 1.22 mm and a separation gap between them of g = 1.66 mm. The substrate had a dielectric constant of 𝜀r = 2.2 and a height of h = 4.771 mm. The scanned magnitude of the simulated reflection coefficient at 13 GHz of such a design is shown in Figure 14-44. The band-gap frequency range was 9.7–15.1 GHz. The E-plane curves are presented by the E curves, the H-plane by the H curves and the diagonal (45◦) plane by the D curves. It is clear that using conventional substrates, the reflection coefficient varies as a function of the scan angle, and in fact creates scan bindness around 50◦. However, when the same elements were placed on a textured high-impendance surface, the reflection coefficient was reduced, especially in the E-plane, and the scan blindness was eliminated.
14.9 ANTENNAS FOR MOBILE COMMUNICATIONS For conventional rectangular microstrips, of the form shown in Figure 14.1(a), their length L has to be slightly less than λ∕2 for them to operate properly. For many applications, especially at lower frequencies and mobile units, this length may be too long. However, the length of the microstrip, whether it is fed by a microstrip line or a probe, can be reduced to basically one-half of its size of the 838 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Figure 14.44 Reflection coefficient of phase array of dipoles on a regular and 4 × 4 until cell PBG substrate.
(source: c ⃝2004 IEEE).
conventional design, by using a shorting sheet or pin at its end, as shown in Figure 14.45(a,b), respec-tively, and they may be referred to as λ∕4 microstrips. Their radiation characteristics are similar to those of conventional λ∕2 microstrip designs.
14.9.1 Planar Inverted-F Antenna (PIFA) Planar Inverted-F Antenna (PIFA) is a very popular design in mobile communications –. Its name is due to the resemblance of a letter F with its face down in its side view. The intrinsic PIFA is basically a derivative of the λ/4 rectangular microstrip of Figure 14.45.
Among its advantages are the following: r Low backward radiation reduces the specific absorption rate (SAR) compared to other antenna types when used in mobile applications.
r Both vertical and horizontal polarizations can be received.
substrate r microstrip transmission line Patch shorting sheet h ground plane L ≈λ/4 x z y W w sp substrate r Patch shorting sheet h ground plane feed patch x z y L ≈λ/4 y0 W wsp (a) Microstrip feed (b) Probe feed Figure 14.45 Rectangular λ∕4 microstrips.
ANTENNAS FOR MOBILE COMMUNICATIONS 839 r It can be easily integrated in mobile devices.
r It is light weight.
r It is conformal.
r It is easy to design.
r It is low cost.
r It is reliable.
To understand its performance, it is important to understand the λ∕2 patch antenna. The current distribution of a λ∕2 antenna has a peak on the center of the patch, and it goes to zero at the edges.
The voltage is out of phase from the current, starting from a positive at one edge to the negative at the other edge or vice versa. Therefore, at the center of the λ∕2 patch, the voltage is zero and the current reaches its maximum. The zero voltage at the center of the λ∕2 patch means a short circuit from the patch to the ground plane. If a shorting pin is physically placed at the center of a λ∕2 microstrip patch and half of the antenna is removed, the current and voltage will be basically the same as the ones for a λ∕2 antenna; thus, the antenna will radiate . By reducing the size to λ∕4, the disadvantages, compared to the λ∕2 patch, include a reduction in its bandwidth and gain. The PIFA design is shown in Figure 14.46.
The resonant frequency of the PIFA antenna is determined by the length of the patch L, the width of the patch W, the width of the shorting sheet ws, and the height of the substrate h. Basically, the height h is small, and it can be neglected. If the height is neglected, the main restriction in the design is a quarter-wavelength distance between the pin and the opposite edge of the patch. Thus, if the shorting sheet extends along the entire width of the patch, the distance L should be equal to quarter wavelength. Alternatively, if the shorting sheet is only a pin, then L + W ≈λ∕4. In summary, the design equation is L + W −ws = λ 4 + h ➱ L = −W + ws + λ 4 + h (14-97) r Substrate Ground Plane PIFA Shorting Lh y0 y0 Le ws L L PIFA E-Plane H-Plane Feed Feed Substrate Ground Plane Shorting sheet/pin Substrate Shorting sheet/pin W sheet/pin Feed h L W y0 x z y ws (a) 3D view (b) Top view (c) Side view r r Figure 14.46 Planar Inverted-F Antenna (PIFA) design.
840 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS where λ is the wavelength in the dielectric. For the special cases, (14-97) reduces to L = λ 4 + h for ws = W (14-97a) L ≈−W + λ 4 + h for ws ≈0 (14-97b) Equation (14-97) can be solved for the resonant frequency and expressed as f = v0 4(L + W −ws −h)√𝜀r (10-98) where v0 is the free-space speed of light and 𝜀r is the relative permittivity (dielectric constant) of the substrate.
The input impedance of the PIFA is controlled by the distance from the shorting sheet (pin) to the feed point. The magnitude of the input impedance decreases as this distance y0 decreases, and vice versa. To match the antenna to a 50-ohm transmission line, the feed should be closer to the shorting sheet (pin) than to the open end of the PIFA (i.e., y0 < L∕2).
The radiation pattern is a hybrid between a patch antenna and an Inverted-F Antenna (IFA). Sev-eral modifications, such as U-slots, can be incorporated to the PIFA to obtain multiple resonances required in mobile communications , .
Example 14.6 Design a PIFA antenna using an air substrate (𝜀r = 1) and height h = 0.448 cm to resonate at 1.8 GHz. Assume W = 0.773L, ws = 0.428W = 0.331L.
r Determine L, W, ws (in cm).
r Simulate the design for S11, 3-D, and principal plane 2-D amplitude patterns.
Solution: Using (14-97), write L + 0.773L −0.331L = λ 4 + h ⇒L = 1 1.442 (λ 4 + h ) L = 1 1.442 ( 30 × 109 4 √ 1(1.8 × 109) + 0.448 ) = 4.6147 1.442 = 3.2 cm = 0.192λ W = 0.773L = 2.474 cm = 0.1484λ ws = 0.331L = 1.059 cm = 0.0636λ A PIFA design, based on this example, was simulated. The S11, 3-D and 2-D E-plane and H-plane patterns are shown in Figures 14.47 and 14.48, respectively.
ANTENNAS FOR MOBILE COMMUNICATIONS 841 –60 –50 –40 –30 –20 –10 0 1.86 1.84 1.82 1.8 1.78 1.76 1.74 S11 (dB) Frequency (GHz) Figure 14.47 S11 for a PIFA design: f = 1.8 GHz, h = 0.048 cm, L = 0.192λ = 3.2 cm, W = 0.148λ = 2.474 cm, ws = 0.0636λ = 1.06 cm, y0 = 0.57 cm.
0° 30° 30° 90° 90° 150° 150° 180° 120° 120° 60° 60° E-Plane H-Plane –30 dB–20 dB–10 dB 0 dB E-Plane (a) 3-D (b) E-plane (solid) and H-plane (dashed) H-Plane Figure 14.48 Normalized 3-D and 2-D (E-plane; xy and H-plane;yz) radiation patterns for a PIFA design of Example 14.6: f = 1.8 GHz, h = 0.448 cm, L = 0.192λ = 3.2 cm, W = 0.1484λ = 2.474 cm, ws = 0.0636λ = 1.059 cm, y0 = 0.57 cm.
14.9.2 Slot Antenna A slot antenna is a radiating element which typically is formed by cutting an opening on a ground plane, as is indicated in Figures 12.6 and 12.7, and repeated here in Figure 14.49.
Usually, this opening is referred to as a slot instead of an aperture, and the length L in Figure 14.49 should be much longer than the width W. For proper radiation, the length L of the slot should be around half a wavelength (L ≈λ∕2) while the width W typically should be W ≤(0.05 −0.1)λ.
The slot is usually excited by a voltage source placed symmetrically in the middle of the slot, at λ∕4 from each of its edges. For such symmetrical excitation, the voltage reaches its maximum at the center (λ∕4 from the edges), and it is minimum at the edges, as shown in Figure 14.50(a). In contrast, the current is negative at one edge, reaches zero at the center (λ∕4 from the edges), and it 842 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Substrate V L W Ground Plane r Figure 14.49 Slot antenna with voltage excitation.
is positive at the other edge, as shown in Figure 14.50(b) or vice-versa in polarity. The center of the slot antenna, where the current is zero, can be seen as an open circuit.
A slot antenna is the complement of a dipole type of a radiator, where the arms of the dipole are PEC strips, as shown in Figure 12.24 for a L = λ∕2 length slot and dipole, and where the slot excitation at the center is replaced by that of the dipole. The impedance of the dipole in free space, Zd, is related to that of the slot in free space, Zs, by Babinet’s principle of Section 12.8, and it is equal to (12-67) or ZdZs = 𝜂2 o 4 (14-99) where 𝜂o is the free-space intrinsic impedance. Knowing either Zd or Zs, you can determine the other by (14-99).
The slot antenna of Figure 14.49 is a bidirectional radiator; that is, it radiates both in the forward and backward directions of the ground plane. Typically, to convert it to a unidirectional radiator in the + A Amps 0 Amps –A Amps V Volts 0 Volts L Voltage across aperture (a) Voltage distribution (b) Current distribution Current on top surface of slot W Figure 14.50 Voltage and current distribution of a L = λ∕2 slot antenna.
ANTENNAS FOR MOBILE COMMUNICATIONS 843 forward direction, a cavity is inserted in its back side, reducing it to what it is usually referred to as a Cavity-Backed Slot (CBS) antenna –. If the slot antenna is cut to half, to physically place an open circuit, it reduces to an IFA, which will be discussed next, whose performance is ideally similar to the original slot antenna.
Example 14.7 A very thin slot (W = λ∕10) with length L = λ∕2 is cut on a PEC ground plane. Determine ana-lytically its approximate input impedance Zs. Verify your result by simulations using a CEM software. Assume that the slot is radiating in free space.
Solution: The slot (L = λ/2, W = λ/10) is, according to Figure 12.24, the complementary structure of a thin strip half-wavelength (l = λ∕2) dipole, whose ideal thin-wire input impedance is, according to (4-93a), Zd = 73 + j42.5 Using Babinet’s principle of Section 12.8 and (12-67) or (14-99), the input impedance of the slot Zs can be represented approximately by Zs ≈𝜂o 4Zd = 367.7 4(73 + j42.5) = 362.95 −j211.31 Using a commercial CEM software, the simulated input impedance of a slot Zs with L = λ∕2, W = λ∕10 on a 5λ × 5λ PEC ground plane is Zs = 339.38 −j78.85 While the input resistance of 339.38 ohms is close to the analytical value of 362.95 ohms, the simulated reactance (capacitive) of −78.85 ohms is considerably different from the analytical of −211.31 ohms. However, as the width W of the slot decreases, the simulated capacitive reactance becomes more negative and approaches the analytical one.
14.9.3 Inverted-F Antenna (IFA) A planar Inverted-F Antenna (IFA) consists of a thin arm or wire shorted at one of the ends to the ground plane , , , as shown in Figure 14.51. The length of the arm should be nearly λ∕4. The position of the feed with respect to the shorting pin controls the input impedance. As the feed becomes closer to the shorting pin, the input impedance is reduced. Thus, to obtain a 50-ohm input impedance (used in most practical designs), the feed should be closer to the shorting pin than to the open end.
Although the IFA looks like the PIFA (with a wire or thin arm instead of a planar patch), its performance is different. The performance of the IFA is similar to that of the slot antenna . The length of the IFA should be half of a slot antenna, as mentioned in the previous section. If the slot antenna is halved, and physically create an open circuit, it reduces to an IFA whose performance is ideally similar to that of the original slot antenna. The shorting pin introduces to the input impedance an inductance Ls while the open end introduces a capacitance Cs . To obtain the impedance at resonance, the two reactive components should cancel out, leaving only the radiation resistance Rs. The width W of the shorting pin is very small compared to the wavelength (W ≪λ), usually W ≤(0.05 −0.1)λ.
844 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Substrate Arm Arm Substrate Substrate Ground Plane W V Shorting Pin (a) 3-D view (b) Top view (c) Side view Ground Plane L ≈/4 L ≈/4 z h V W x y Ground Plane Shorting pin r λ λ r r Figure 14.51 Inverted-F Antenna (IFA) design.
Example 14.8 Design a planar IFA antenna using Rogers RT/duroid 5870 substrate (dielectric constant of 2.33) and thickness of 0.1524 cm to resonate at 1.8 GHz.
r Determine the length L (in cm) r Simulate the design for S11, and 3-D and 2-D amplitude patterns.
Solution: The length of the IFA antenna should be equal to L = λ 4 = 30 × 109 4 √ 2.33(1.8 × 109) = 2.73 cm An IFA based on this design was simulated using W = 0.06λ = 0.655 cm. The S11, 3-D and 2-D E-plane and H-plane patterns are shown in Figures 14.52 and 14.53, respectively. It is apparent –40 –35 –30 –25 –20 –15 –10 –5 0 1.86 1.84 1.82 1.8 1.78 1.76 1.74 S11 (dB) Frequency (GHz) Figure 14.52 S11 for the IFA design of Example 14.8: f = 1.8 GHz, W = 0.006λ = 0.655 cm, L = 0.25λ = 2.73 cm. Rogers RT/duroid 5870 substrate, h = 0.1524 cm.
ANTENNAS FOR MOBILE COMMUNICATIONS 845 that the pattern of the IFA is nearly ominidirectional, as it should be, since the slot is the com-plement of the dipole and their patterns (slot and dipole) are, according to Babinet’s principle, ideally identical as indicated in Figure 12.25 with the E and H fields reversed.
–15 dB –10 dB –5 dB 0 dB 30° 150° 60° 120° 90° 90° 120° 60° 150° 30° 180° 0° E-Plane H-Plane H-Plane E-Plane x y z (a) 3-D (b) E-plane (solid) and H-plane (dashed) Figure 14.53 Normalized 3-D and 2-D (E-plane;yz and H-plane;xz) radiation patterns for the IFA design of Example 14.8: f = 1.8 GHz, W = 0.06λ = 0.655 cm, L = 0.25λ = 2.73 cm. Rogers RT/duroid 5870 sub-strate, h = 0.1524 cm.
While the IFA configuration of Figure 14.51 is considered a planar design, which is most widely used for mobile devices, there is another IFA design, which is wire-type and nonplanar , as shown in Figure 14.54.
Ground Plane z w L h1 h t y x IFA Feed Substrate (a) 3D view r L t IFA Feed Substrate (b) Top view r w L h1 h t Feed Ground Plane Substrate (c) Side view r Figure 14.54 Wire-type Inverted-F Antenna (IFA).
846 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS Ground Plane Substrate Feed Figure 14.55 Three-band U-shot microstrip/patch antenna.
14.9.4 Multiband Antennas for Mobile Units Today’s mobile units, such as the smartphone, provide a plethora of services, from phone calls, emails, messaging, news, sports, music, to maps, just to name a few. Also these units have become smaller and slimmer, and less room is available to accommodate efficiently all these services. Since it is impossible to dedicate an individual antenna design for each of the services, antenna design engineers face major challenges to come up with designs whereby each individual antenna structure can accommodate multiple services. This requires that each physical antenna structure resonate at multiple frequencies, leading to multiband designs.
One planar antenna structure that can resonate at multiple frequencies is a rectangular microstrip with multiple slots, in the form of a U, placed on its patch. Each U slot is designed to resonate at a given frequency, and this is accomplished by making the overall length of each slot nearly λ∕2 . A three-band U-shape microstrip slot design is shown in Figure 14.55. The first resonance is attributed to the rectangular patch and the other two to the two U slots. Observe that the slots are separated from each other and the edges of the patch so that there is no major coupling between them.
A U-shape two-band planar antenna is shown in Figure 14.56. It is designed to operate at two bands and to accommodate two cellular services; the outer/longer slot to service the GSM band GSM: 874–958 MHz BW = 84 MHz DCS: 1,711–1,943 MHz BW = 232 MHz Figure 14.56 Two-band (GSM and DCS) U-shot antenna (courtesy of Prof. Seong-Ook Park).
DIELECTRIC RESONATOR ANTENNAS 847 (874–958 MHz), with a bandwidth of 84 MHz (about 9%), and the inner/smaller one to accommodate the DCS band (1,711–1,943 MHz), with a bandwidth of 232 MHz (nearly 13%).
14.10 DIELECTRIC RESONATOR ANTENNAS The dielectric resonator antenna followed the introduction and development of dielectric resonators, which are nonmetalized dielectric objects (spheres, hemispheres, cylinders, disks, parallelepipeds, etc.) that can function as energy storage devices due to a high dielectric constant (usually ceramic) and a high quality factor Q. Electromagnetic energy is injected into the block of dielectric material creating electromagnetic waves that bounce back and forth between the walls of the cavity creating standing waves. For the dielectric resonator to function as a resonant cavity, the dielectric constant of the material must be large (usually 50 or greater). Under these conditions, the dielectric–air interface acts almost as an open circuit and causes internal reflections that confine the energy in the dielectric material, thus creating a resonant structure. This effect can be expressed as an approximation, using the plane wave reflection coefficient Γ at a dielectric–air interface of Γ = 𝜂0 −𝜂1 𝜂0 + 𝜂1 𝜇1=𝜇0 ⏞ ⏞ ⏞ = √ 𝜇0∕𝜀0 − √ 𝜇0∕𝜀1 √ 𝜇0∕𝜀0 + √ 𝜇0∕𝜀1 = √ 𝜀1∕𝜀0 −1 √ 𝜀1∕𝜀0 + 1 = √𝜀r −1 √𝜀r + 1 𝜀r→large ⏞ ⏞ ⏞ = +1 (14-100) Thus, the coefficient Γ approaches +1 as the dielectric constant becomes very large. Given these conditions, the dielectric–air interface can further be approximated by an ideal nonphysical perfect magnetic conductor (PMC) surface, which requires that the tangential components of the magnetic field to vanish (in contrast to the perfect electric conductor, PEC, which requires that the tangential electric field components to vanish). This is a well-known and widely used technique in solving boundary-value electromagnetic problems. It is, however, a first-order approximation, although it usually leads to reasonable results. The magnetic wall model has been used to analyze both dielectric waveguides and dielectric resonant cavities, and was used in Sections 14.2.2 and 14.3 to analyze rectangular and circular microstrip antennas.
Dielectric resonators were introduced in 1939 by Richtmyer , but for almost 25 years his the-oretical work failed to generate any continuous and prolonged interest. The introduction in the 1960s of material, such as rutile, of high dielectric constant (around 100) renewed the interest in dielectric resonators –, and the work of Van Bladel introduced a detailed theory on the modes of the dielectric resonator . However, the poor temperature stability of rutile resulted in large resonant frequency changes and prevented the development of practical microwave components. In the 1970s, low-loss and temperature-stable ceramics, such as barium tetratitanate and (Zr–Sn)TiO4, were intro-duced and were used for the design of high performance microwave components such as filters and oscillators. Because dielectric resonators are small, lightweight, temperature stable, of high Q, and low cost, they are ideal for the design and fabrication of monolithic microwave integrated circuits (MMICs) and general semiconductor devices. Thus, dielectric resonators have replaced traditional waveguide resonators, especially in MIC applications, and have implementations in the millimeter wave region.
Dielectric Resonator Antennas (DRAs) also consist of blocks of ceramic materials. DRAs are mounted on a PEC ground plane, and are commonly used at microwave frequencies and above. They were reported initially and introduced as practical radiating elements in the early 1980s by Long et al. in – for the cylindrical, rectangular, and hemispherical geometries, respectively.
Since then, there have been numerous advances and publications on DRAs both as single elements and as elements in arrays, which have been documented in –, and many others. For the blocks of dielectric material to act as radiators, their dielectric constants must be moderate, in the range of 5–30, to allow some of the energy from inside the cavity to escape through the walls of the cavity and create radiation.
848 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS In general, DRAs have a number of desirable features: r Are efficient radiators, especially when compared to microstrips antennas; they do not generate surface waves and they are open on all sides, other than the side mounted on the PEC ground plane, whereas microstrip antennas radiate basically through their leading and trailing sides, as modeled in Figures 14.9 and 14.18.
r Are proportional in size to the wavelength in the dielectric, which, as is the case with microstrip antennas, decreases as the dielectric constant increases [λ = λo∕√𝜀r, λo = free space wave-length, 𝜀r = dielectric constant].
r Are wideband (typically BW ≈10%), especially as compared to microstrip antennas, with some geometries achieving bandwidths of 50% or even higher. In cases where the Q is very high, their bandwidth can be reduced to single digits.
r Have high power capability, as the dielectric constant is high.
r Are versatile, as different shapes can be used to excite various modes with different radiation characteristics.
r Can be excited by different feeds, such as probes (e.g., coaxial line), microstrip lines, slots, and coplanar waveguides, as is also the case for microstrip antennas; some of them are illustrated in Figure 14.3.
r Can be of low profile, as for some geometries their height h can be as small as 0.05λ.
r Can be used in a wide range of applications, including mobile wireless communication, satellite systems, GPS, and indoor communications.
r Can range in frequency of operation from about 1 to 60 GHz, and higher. The low frequency is limited by the size and weight of the DRAs.
r Exhibit lower losses at the millimeter wavelength region because at these frequencies metallic radiators become more inefficient due to skin effect while DRAs are made purely of dielectrics with, ideally, no losses.
r Are low cost and lightweight.
14.10.1 Basic DRA Geometries Four basic DRA geometries are the rectangular, cylindrical, hemicylindrical, and hemispherical ones shown, respectively, in Figures 14.57 to 14.60. In each illustration, a probe feed line is also indicated whose position yo, 𝜌o, ro is relative to the center of the cavity, and can be adjusted to improve the matching of the DRA to the feed line; for a coaxial line, this is typically 50 ohms. Although there are other configurations, these four are considered basic and are most commonly used. They also can be analyzed for their resonant frequencies and radiated fields by using the cavity model.
(a) Perspective z y x b Rectangular DRA Feed probe c Ground Plane y0 a θ >>1 ϕ θ (b) Side view c b z Ground Plane y l Feed line Feed probe d >>1 y0 r r Figure 14.57 Rectangular DRA with a probe feeding line.
DIELECTRIC RESONATOR ANTENNAS 849 (b) Side view z y ρ0 l h d Ground Plane 2a Feed line Feed probe θ >>1 (a) Perspective Ground Plane Feed probe Cylindrical DRA a z h x y θ ϕ >>1 0 ρ r r Figure 14.58 Cylindrical DRA with a probe feeding line.
(a) Perspective Ground Plane h h y x x h a z 0 ρ ϕ r >>1 (b) Side view Feed line Ground Plane y x l a ρ0 d r >>1 Figure 14.59 Hemicylindrical DRA with a probe feeding line.
(a) Perspective θ ϕ a x y z Ground Plane Hemispherical DRA Feed probe >>1 r r0 (b) Side view θ Ground Plane Feed probe Feed line >>1 r z y r0 l a d Figure 14.60 Hemispherical DRA with a probe feeding line.
14.10.2 Methods of Analysis and Design These, and similar geometries, can be analyzed and designed using various methods such as: r Cavity modal model where the open sides are treated as PMC while the ground plane is treated as a PEC.
r Full-wave solutions, such as: 1. Integral equations (IE)/MoM 2. Finite-difference time-domain (FDTD) 850 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS 3. Finite element method (FEM) 4. Transmission line method (TLM) 5. Hybrid methods 6. Perturbation methods Although the cavity modal model (magnetic wall modeling of the open sides) does not always lead to the most accurate data, it will be utilized initially here, as a first-order approximation, because it is simple, instructive, and provides physical insight. It was used, as a first-order approximation, for the rectangular and circular microstrip antennas in Sections 14.2.2 and 14.3. Refinements upon the cavity modal model can be implemented to improve the designs. It has been demonstrated that this model is accurate for high dielectric constants, especially for resonant frequencies and radiation patterns , though it may not be necessarily accurate for the input impedance. Design equations based on other hybrid mode models will also be included.
14.10.3 Cavity Model Resonant Frequencies (TE and TM Modes) The procedure for predicting the radiation characteristics using the cavity model was outlined in Section 14.2.2 for the TMx modes for the rectangular microstrip antenna, and in Section 14.3 for the TMz modes for the circular microstrip antenna. A similar procedure can be used for TE modes, but it is beyond the scope of this book. A detailed procedure for all modes, TEM, TM, TE, and hybrid modes, is provided in Chapter 6 of . The TE and TM mode resonant frequencies, for the rectangular, cylindrical, hemicylindrical, and hemispherical are summarized here.
A. Rectangular DRA of Figure 14.57 (mnp modes: m for x variations; n for y variations; p for z variations).
(fr)TEz mnp = 1 2𝜋√𝜇d𝜀d √(m𝜋 a )2 + (n𝜋 b )2 + [ (2p −1) ( 𝜋 2c )]2 }m = 1, 2, 3 … n = 1, 2, 3, … p = 1, 2, 3, … (14-101a) (fr)TMz mnp = 1 2𝜋√𝜇d𝜀d √(m𝜋 a )2 + (n𝜋 b )2 + [ (2p −1) ( 𝜋 2c )]2 } m = 0, 1, 2 … n = 0, 1, 2, … p = 1, 2, 3, … ⎫ ⎪ ⎬ ⎪ ⎭ m = n ≠0 (14-101b) r For a > b > 2c and c > a > b, the TEz mnp or TMz mnp mode with the smallest resonant frequency (fr)mnp is the TMz 101 (m = 1, n = 0, p = 1) with (fr)TMz 101 of (fr)TMz 101 = 1 2√𝜇d𝜀d √(1 a )2 + ( 1 2c )2 = vo 2√𝜇r𝜀r √(1 a )2 + ( 1 2c )2 (14-102a) where vo is the speed of light.
r However for c > b > a, the lowest order TEz mnp or TMz mnp mode with the smallest resonant (fr)mnp is the TMz 011(m = 0, n = 1, p = 1) with (fr)TMz 011 of (fr)TMz 011 = 1 2√𝜇d𝜀d √(1 b )2 + ( 1 2c )2 = vo 2√𝜇r𝜀r √(1 b )2 + ( 1 2c )2 (14-102b) DIELECTRIC RESONATOR ANTENNAS 851 B. Cylindrical DRA of Figure 14.58 (mnp modes: m for azimuthal 𝜙variations, n for radial 𝜌vari-ations, p for height z variations) (fr)TEz mnp = 1 2𝜋√𝜇d𝜀d √(𝜒mn a )2 + [ (2p + 1) ( 𝜋 2h )]2 } m = 0, 1, 2 … n = 1, 2, 3, … p = 0, 1, 2, 3, … (14-103a) (fr)TMz mnp = 1 2𝜋√𝜇d𝜀d √(𝜒′ mn a )2 + [ (2p + 1) ( 𝜋 2h )]2⎫ ⎪ ⎬ ⎪ ⎭ m = 0, 1, 2 … n = 1, 2, 3, … p = 0, 1, 2, 3, … (14-103b) where 𝜒mn and 𝜒′ mn are the n = 1, 2, …, zeros of the Bessel function Jm and its derivative J′ m, respec-tively. These are listed, respectively in Tables 9.2 and 9.1 of Chapter 9 of . The five smallest values of each are 𝜒01 = 2.4049, 𝜒11 = 3.8318, 𝜒21 = 5.1357, 𝜒02 = 5.5201, 𝜒31 = 6.3802 𝜒′ 11 = 1.8412, 𝜒′ 21 = 3.0542, 𝜒′ 01 = 3.8318, 𝜒′ 31 = 4.2012, 𝜒′ 31 = 5.3175 r Based on these values of 𝜒mn and 𝜒′ mn, the TEz mnp and TMz mnp modes with the smallest resonant frequency (fr)mnp are the (fr)TEz 010 = 1 2𝜋√𝜇d𝜀d √(2.4049 a )2 + ( 𝜋 2h )2 } (14-104a) (fr)TMz 110 = 1 2𝜋√𝜇d𝜀d √(1.8412 a )2 + ( 𝜋 2h )2 } (14-104b) Of these two resonant frequencies, the smaller of the two is the TMz 110 (m = 1, n = 1, p = 0) mode whose resonant frequency is that of (14-104b).
C. Hemicylindrical DRA of Figure 14.59 (mnp modes: m for azimuthal 𝜙variations, n for radial 𝜌 variations, p for height z variations) (fr)TEz mnp = 1 2𝜋√𝜇d𝜀d √√(𝜒mn a )2 + (p𝜋 h )2⎫ ⎪ ⎬ ⎪ ⎭ m = 0, 1, 2 … n = 1, 2, 3, … p = 0, 1, 2, 3, … (14-105a) (fr)TMz mnp = 1 2𝜋√𝜇d𝜀d √ √ √ √ √(𝜒′ mn a )2 + (p𝜋 h )2⎫ ⎪ ⎬ ⎪ ⎭ m = 1, 2, 3, … n = 1, 2, 3, … p = 1, 2, 3, … (14-105b) r Based on these values of 𝜒mn and 𝜒′ mn, the TEz mnp and TMz mnp modes with the smallest resonant frequency (fr)mnp are the (fr)TEz 010 = 1 2𝜋√𝜇d𝜀d (𝜒01 a ) = 2.4049 2𝜋a√𝜇d𝜀d (14-106a) (fr)TMz 111 = 1 2𝜋√𝜇d𝜀d √ √ √ √ √ ( 𝜒′ 11 a )2 + (𝜋 h )2 = 1 2𝜋√𝜇d𝜀d √(1.841 a )2 + (𝜋 h )2 (14-106b) 852 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS The TEz mnp or TMz mnp mode with the smallest resonant frequency is the: r For h∕a < 2.031: TEz 010(m = 0, n = 1, p = 0) with a resonant frequency of (14-106a).
r For h∕a > 2.031: TMz 111(m = 1, n = 1, p = 1) with a resonant frequency of (14-106b).
D. Hemispherical DRA of Figure 14.60 (mnp modes: m for azimuthal 𝜙variations, n for elevation 𝜃variations; p for radial r variations) (fr)TEr mnp(even, odd) = 𝜁np 2𝜋a√𝜇d𝜀d (14-107a) (fr)TMr mnp (even, odd) = 𝜁′ np 2𝜋a√𝜇d𝜀d (14-107b) (fr)TEr 011(even) = (fr)TEr 111(even, odd) = 𝜁11 2𝜋a√𝜇d𝜀d = 4.493 2𝜋a√𝜇d𝜀d (14-107c) (fr)TMr 011 (even) = (fr)TMr 111 (even, odd) = 𝜁′ 11 2𝜋a√𝜇d𝜀d = 2.744 2𝜋a√𝜇d𝜀d (14-107d) where 𝜁np and 𝜁′ np are, for a given n, the p = 1, 2, ...zeroes of the spherical Bessel function ̂ Jn and its derivative ̂ J′ n , respectively. These are listed, respectively, in Tables 10.1 and 10.2 of Chapter 10 of .
The five smallest values of each are: 𝜁11 = 4.493, 𝜁21 = 5.763, 𝜁31 = 6.988, 𝜁12 = 7.725, 𝜁41 = 8.183 𝜁′ 11 = 2.744, 𝜁′ 21 = 3.870, 𝜁′ 31 = 4.973, 𝜁′ 41 = 6.062, 𝜁′ 12 = 6.117 r Based upon these values of 𝜁mn and 𝜁′ mn, the TEr mnp or TMr mnp mode with the smallest resonant frequency (fr)mnp, having a three-fold degeneracy, is the TMr 011 (even: m = 0, n = 1, p = 1) = TMr 111 (even: m = 1, n = 1, p = 1) = TMr 111 (odd: m = 1, n = 1, p = 1) whose resonant fre-quency (fr)TMr 011 (even) = (fr)TMr 111 (even) = (fr)TMr 111 (odd) is that of (14-107d).
14.10.4 Hybrid Modes: Resonant Frequencies and Quality Factors As indicated previously, the formulations based on the cavity PMC/PEC models of Section 14.10.3 are first-order approximations, especially for the resonant frequency and far-zone radiated fields.
More accurate expressions for the resonant frequencies (fr) and radiation quality factors (Q) of the cylindrical DRA of Figure 14.58 have been derived for TEmn𝛿, TMmn𝛿and hybrid HEmn𝛿modes, which are the most common modes for this DRA geometry. The notation here is equivalent to that of Section 9.5.2 of , adopted based on the notation of , where the first integer subscript m (m = 0, 1, 2, ...) represents azimuthal 𝜙field variations, the second integer subscript n (n = 1, 2, 3, ...) represents radial 𝜌field variations, while the third subscript 𝛿(𝛿= noninteger) represents z field variations. The noninteger value of 𝛿indicates that the field variations along the z direction are not full periods or the length of the DRA is not integer multiple of λd/2, where λd is the wavelength in the DRA dielectric. In , the dielectric resonator has been modeled using two PEC flat plates, each placed a distance h from each end of the cylindrical dielectric resonator, as shown in Figure 9.15 of . The value of 𝛿depends on the dielectric constant 𝜀r of the DRA and the proximity h1, h2 of each of the two PEC plates from the ends of the DRA , as illustrated in Figure 9.15 of .
It has been shown in , that the field variations within the cylindrical DRA for the TEmn𝛿and TMmn𝛿are azimuthally symmetric (not 𝜙dependent) while those of the hybrid HEmn𝛿 are 𝜙dependent, as shown, respectively, in Figures 14.61 (a,b,c).
DIELECTRIC RESONATOR ANTENNAS 853 Z (a) TE01 Z H-field E-field Z (b) TM01 (c) HE11 0 r 0 r 0 r δ δ δ Figure 14.61 Field configurations within a cylindrical DRA for the TE01𝛿, TM01𝛿, and hybrid HE11𝛿modes [source: A. A. Kishk and Y. M. M. Antar, “Dielectric Resonator Antennas,” Chapter 17 in Antenna Engineering Handbook, J. L. Volakis (ed.), c ⃝McGraw-Hill Book Co., 2007].
Based on these more accurate TEmn𝛿, TMmn𝛿, and hybrid HEmn𝛿modes, closed-form expres-sions for the resonant frequencies and Q have been derived, for the cylindrical DRA of Figure 14.58, through extensive numerical and experimental investigations and curve fitting routines. One set of these expressions, based on , , is listed here; another set can be found in .
TE01𝛿Mode : fr = vo 2𝜋a ( 2.327 √ 𝜀r + 1 ) [ 1 + 0.2123 (a h ) −0.00898 (a h )2] ; 0.125 ≤a h ≤5 (14-108a) Q = 0.078192 (𝜀r )1.27 ⎡ ⎢ ⎢ ⎢ ⎣ 1 + 17.31 (h a ) −21.57 (h a )2 +10.86 (h a )3 −1.98 (h a )4 ⎤ ⎥ ⎥ ⎥ ⎦ ; 0.5 ≤a h ≤5 (14-108b) TM01𝛿Mode : fr = vo 2𝜋a ( 1 √ 𝜀r + 2 ) √[ (03.83)2 + (𝜋 2 )2 (a h )2] ; 0.125 ≤a h ≤5 (14-109a) Q = 0.00872 ( 𝜀r )0.888 e0.03975𝜀r [ 1 −(0.3 −0.2z) (38 −𝜀r 28 )] × (9.498z + 2, 058.33z4.3226e−3.501z) 0.125 ≤a h ≤5 (14-109b) HE11𝛿Mode : fr = vo 2𝜋a ( 6.324 √ 𝜀r + 2 ) (0.27 + 0.18z + 0.005z2) ; 0.125 ≤a h ≤5 (14-110a) Q = 0.01z (𝜀r )1.3 [ 1 + 100e−2.05 ( 0.5z−0.0125z2)] ; 0.5 ≤a h ≤5 (14-110b) 854 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS TABLE 14.3 Updated Predicted and Measured Resonant Frequencies and Qs 𝜺r = 38, a = 0.6415 cm, h = 0.2810 cm Measured Theoretical % Measured Theoretical % Mode fr (GHz) fr(GHz) Difference Q-Factor Q-Factor Difference TE01𝛿 3.97 3.988 0.45% 46.2 41.92 9.3% TM01𝛿 6.13 6.175 0.73% 72.1 46.34 35.7% HE11𝛿 5.18 5.262 1.58% 30.2 31.24 3.4% 𝜺r = 79, a = 0.5145 cm, h = 0.2255 cm Measured Theoretical % Measured Theoretical % Mode fr (GHz) fr(GHz) Difference Q-Factor Q-Factor Difference TE01𝛿 3.48 3.47 2.9% 114.7 106.21 7.4% TM01𝛿 5.41 5.41 0.0% 336.7 349.61 3.8% HE11𝛿 4.56 4.61 1.1% 76.4 80.94 5.9% It should be pointed out that the Qs based on the equations above are small, near the lower limits of a∕h; larger values are attained at the middle and upper range of a∕h ratios. Table 14.2 compares predicted, based on (14-108a) and (14-110b), and measured, based on , resonant frequencies and Qs of two isolated cylindrical DRAs, of dielectric constants 𝜀r = 38 and 𝜀r = 79. It should be noted that the predicted values in Table 14.3 are slightly different than those from Table 4.2 of . The original data for the measured Qs are based on radar cross sections measurements from . The resonant frequencies based on (14-108a), (14-109a), and (14-110a) are less than 3% from measured ones while the differences for the Qs, based on (14-108b), (14-109b), and (14-110b), are somewhat larger than the corresponding ones based on measurements but, except for one case, always smaller than 10%.
A Matlab computer program, DRA Analysis Design, has been developed and it can be used to: r Analyze the four DRAs of Figures 14.57–14.60 r Design the cylindrical DRA of Figure 14.58 .........................................................................................................
r In the analysis part of the program, once the radius a, height h and dielectric constant 𝜀r are specified, the following are computed: 1. Dominant mode resonant frequencies, of any of the four DRA geometries of Figures 14.57 to 14.60, based on the modal solution expressions of (14-101a)–(14-107d). The computed resonant frequencies are: r Lowest-order 5 TEz, 5 TMz , and 5 TEz/TMz modes of each of the cubic, cylindrical, and hemicylindrical DRAs of Figures 14.57 to 14.59.
r Lowest-order 3 degenerate TEr modes of the hemispherical DRA of Figure 14.60.
2. The resonant frequencies and Qs of the cylindrical resonator of Figure 14.58 for all three modes (TE01𝛿, TM01𝛿, or HE11𝛿), based on (14-108a)–(14-110b). Example 14.9 is an analysis exercise.
r The design part of the program is for a cylindrical DRA of Figure 14.58 which, once the fol-lowing are specified: 1. Mode (TE01𝛿, TM01𝛿, or HE11𝛿) 2. Fractional bandwidth (BW, in %) 3. VSWR 4. Resonant frequency (fr, in GHz) DIELECTRIC RESONATOR ANTENNAS 855 the program performs the following: 1. Computes the Q of the cavity [based on (14-88a)].
2. Prompts the user to select a desired dielectric constant from the following: r From a range of values for the TE01𝛿mode.
r Greater than some minimum value for the TM01𝛿mode.
r From a range of values for the HE11𝛿mode.
These ranges and values for the dielectric constant are determined, once the Q is computed using (14-88a), by solving using a nonlinear procedure as follows: r TE01𝛿mode: (14-108b) for 0.5 ≤a∕h ≤5.
r TM01𝛿mode: (14-109b) for 0.125 ≤a∕h ≤5.
r HE11𝛿mode: (14-110b) for 0.5 ≤a∕h ≤5.
Otherwise, the stated design specifications cannot be met.
3. At this point, the Q, resonant frequency fr and dielectric constant 𝜀r have been decided step by step. To complete the design, based on the stated specifications, the computer program (using a nonlinear procedure) determines the dimensions of the cylindrical DRA (radius a and height h, both in cm) by solving simultaneously: r TE01𝛿mode: (14-108a) and (14-108b).
r TM01𝛿mode: (14-109a) and (14-109b).
r HE11𝛿mode: (14-110a) and (14-110b).
Example 14.10 provides a design exercise based on this procedure.
14.10.5 Radiated Fields The far-zone radiated fields, for each of the configurations in Figures 14.57 to 14.60, can be found by formulating the equivalent surface magnetic current density Ms on the open surface of each of the four geometries and then by using the radiation equations. This process was implemented in Chapter 12. In Chapter 12 and this chapter the procedure is used to determine the far-zone fields radiated of the following geometries: r Rectangular apertures in Section 12.3 r Circular apertures in Section 12.6 r Rectangular microstrip antennas in Section 14.2.2(C) r Circular microstrips in 14.3.4 The same method was applied successfully in for cylindrical DRAs where the predicted pat-terns compared well with measurements.
As has been shown in , two-dimensional normalized amplitude patterns, in the respective principal planes, of the electric fields radiated by the cylindrical DRA of Figure 14.58 for the HE11𝛿 and M01𝛿modes resemble, respectively, those in Figure 14.62(a,b). The patterns of Figure 14.62(a) for the HE11𝛿were found to be typical of those of a horizontal magnetic dipole, and those of Figure 14.62(b) for the TM01𝛿mode to be typical of those of an infinitesimal vertical electric monopole , as represented in Figure 4.15 of .
In addition, it has been shown in that typical ideal two-dimensional normalized amplitude patterns, along the principle planes, of the fields radiated by the hemicylindrical DRA of Figure 14.59, for the TE01𝛿mode, look like those in Figure 14.63. The patterns of Figure 14.63 for the TE01𝛿mode were found to be typical of those radiated by a short horizontal magnetic dipole.
The radiation characteristics (resonant frequency, radiation pattern, and input impedance) of a cylindrical DRA were examined analytically, numerically, and experimentally in . In , the dominant mode, as computed using (14-103a)–(14-104b), was determined to be the TMz 110 followed 856 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS –40 dB E0 (xz-plane) (a) HE11 mode (E , xz-plane; E , yz-plane) δ θ ϕ (b) TM01 mode (E , xz-plane; E , yz-plane) δ θ θ E0 (yz-plane) –30 –20 –10 0 –40 dB –30 –20 –10 0 90° 90° 90° 90° 60° 60° 60° 60° 30° 30° 30° θ θ θ θ 30° 0° E0 (xz-plane) E0 (yz-plane) 0° Figure 14.62 Normalized amplitude patterns for the cylindrical DRA of Figure 14.58 for the HE11𝛿and TM01𝛿 modes [source: A. A. Kishk and Y. M. M. Antar, “Dielectric Resonator Antennas,” Chapter 17 in Antenna Engineering Handbook, J. L. Volakis (ed.), c ⃝McGraw-Hill Book Co., 2007].
–40 dB –30 –20 –10 0 0° 180° 30° 150° 60° ϕ E (xy-plane) ϕ E (yz-plane) ϕ θ 120° or 90° TE01 mode (E , xy-plane; E , yz-plane) δ ϕ ϕ Figure 14.63 Normalized amplitude patterns for the hemicylindrical DRA of Figure 14.58 for the TE01𝛿mode [source: A. A. Kishk and Y. M. M. Antar, “Dielectric Resonator Antennas,” Chapter 17 in Antenna Engineering Handbook, J. L. Volakis (ed.), c ⃝McGraw-Hill Book Co., 2007].
by either the TMz 111 for small a∕h ratios or the TEz 010 for large h∕a ratios. For the cylindrical DRA of Figure 14.58, the far-field patterns for large h∕a ratios match those of the HE11𝛿hybrid mode of Figure 14.62(a), with the maximum radiation along the z axis (𝜃= 0◦). However, for small h∕a ratios, a null begins to form along 𝜃= 0◦.
Example 14.9: Analysis Given a cylindrical DRA of Figure 14.58 with a = 0.5 cm, h = 0.3 cm (a∕h = 1.67), l = 0.38 cm, d = 0.05 cm, and probe position 𝜙′ = 90◦, 𝜌o = 0.36 cm.
a. Using commercial software, simulate and plot versus frequency the: r S11 r Input impedance (real and imaginary parts) b. Compute the resonant frequencies of the dominant TEz and TMz modes based on (14-104a) and (14-104b) for: r 𝜀r = 8.9 r 𝜀r = 89 DIELECTRIC RESONATOR ANTENNAS 857 c. Compute the resonant frequencies of the TE01𝛿and TM01𝛿modes, and hybrid HE11𝛿mode for: r 𝜀r = 8.9 r 𝜀r = 89 d. Compute the Qs of the TE01𝛿and TM01𝛿modes, and hybrid HE11𝛿mode for: r 𝜀r = 8.9 r 𝜀r = 89 e. Compute the 3-D and 2-D principal E-plane and H-plane patterns for 𝜀r = 8.9 based on the resonant frequency of the lowest S11 of part a.
Solution: a. The DRA, with the subject dimensions, has been simulated and the S11 and input impedance vs. frequency are plotted in Figure 14.64(a,b). The resonant frequency, based on the lowest S11, is 10. 69 GHz.
b. Based on (14-104a) and (14-104b), the resonant frequencies are: r 𝜀r = 8.9 : fr(TE010) = 11.38 GHz, fr(TM110) = 10.24 GHz r 𝜀r = 89 : fr(TE010) = 3.60 GHz, fr(TM110) = 3.24 GHz –20 –18 –16 –14 –12 –10 –8 –6 –4 –2 0 8 S11 (dB) 9.5 9 8.5 Frequency (GHz) BW (–10 dB) =7.02% 11 10.5 10 12.5 12 11.5 13 –20 –10 0 10 20 30 40 50 60 70 80 8 S11 (dB) 9.5 9 8.5 Frequency (GHz) 11.5 11 10.5 10 Resistance Reactance 13 12.5 12 90° 150° 120° 60° 30° 0° 180° –30 dB –20 dB 30° 150° –10 dB 0 dB 60° 90° 120° Eθ (a) S11 (b) Input impedance (c) 3D (d) 2D (E-plane, : ; H-plane, : ) ϕ = 90° ϕ = 0° Eθ Eϕ Eϕ Figure 14.64 S11, input impedance, and 3-D and 2-D normalized amplitude patterns (f= 10.69 GHz) of Example 4.9.
858 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS c. Based on (14-108a), (14-109a), and (14-110a), the resonant frequencies are: r 𝜀r = 8.9: fr(TE01𝛿) = 9.385 GHz, fr(TM01𝛿) = 13.419 GHz, fr(HE11𝛿) = 8.564 GHz r 𝜀r = 89: fr(TE01𝛿) = 3.113 GHz, fr(TM01𝛿) = 4.644 GHz, fr(HE11𝛿) = 2.964 GHz d. Based on (14-108b), (14-109b), and (14-110b), the Qs are: r 𝜀r = 8.9: Q =(TE01𝛿) = 7.17, Q(TM01𝛿) = 6.325, Q(HE11𝛿) = 5.8864 r 𝜀r = 89: Q(TE01𝛿) = 133.512, Q(TM01𝛿) = 1,071.95, Q(HE11𝛿) = 117.45 e. The corresponding 3-D and 2-D amplitude patterns are displayed, respectively, in Figure 14.64(c,d), and they match those of . Note that the feed in the present model of Figure 14.57 is at 𝜙′ = 90◦while that of Figure 1 of is at 𝜙′ = 0◦; therefore the E-plane and H-plane are interchanged, as they should be.
Example 14.10: Design Design a cylindrical DRA, with geometry of Figure 14.58, to operate on the TE01𝛿mode and to exhibit a fractional bandwidth of 2.887%, VSWR = 3 and resonant frequency fr = 10 GHz. Use the Design part of the Matlab computer program DRA Analysis Design to: r Determine the Q.
r Select a dielectric constant 𝜀r (greater than some minimum value).
r Determine the dimensions a and h (both in cm).
Solution: The Matlab computer program DRA Analysis Design computes the following: r Q = 40 based on (14-88a) r Dielectric constant range: 34 < 𝜀r < 48.65. Select 𝜀r = 38.
r Based on the answers for the Q and dielectric constant 𝜀r, the radius and height are a = 0.2653 cm, h = 0.1020 cm When the dimensions a = 0.2653 cm, h = 0.1020 cm, and 𝜀r = 38 are used in the Analysis part of the Matlab program, they lead, for the TE01𝛿mode, to a resonant frequency fr = 10 GHz and a Q = 40, which verifies the design procedure.
A summary of the pertinent parameters, and associated formulas and equation numbers for this chapter are listed in Table 14.4.
14.11 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab and Fortran computer program, designated Microstrip, for computing and displaying the radiation characteristics of rectangular and circular microstrip antennas.
c. Matlab computer program, designated DRA Analysis Design, which: r Analyzes a DRA cavity, as in Figures 14.57 to 14.60, once the dimensions and dielectric constant are specified.
MULTIMEDIA 859 r Designs a cylindrical DRA, as in Figure 14.58, once the mode (TE01𝛿, TM01𝛿, or HE11𝛿), fractional bandwidth (BW), VSWR, and resonant frequency are specified.
d. Power Point (PPT) viewgraphs, in multicolor.
TABLE 14.4 Summary of Important Parameters and Associated Formulas and Equation Numbers Equation Parameter Formula Number Transmission-Line Model-Rectangular Patch Effective dielectric constant 𝜀reff (W∕h ≫1) 𝜀reff = 𝜀r + 1 2 + 𝜀r −1 2 [ 1 + 12 h W ]−1∕2 (14-1) Effective length Leff Leff = L + 2ΔL (14-3) Normalized extension length ΔL∕h ΔL h = 0.412 (𝜀reff + 0.3) [W h + 0.264 ] (𝜀reff −0.258) [W h + 0.8 ] (14-2) Resonant frequency; dominant mode (L > W) (no fringing) (fr)010 = 1 2L√𝜀r √𝜇o𝜀o (14-4) Resonant frequency; dominant mode (L > W) (with fringing) (frc)010 = 1 2Leff √𝜀reff √𝜇o𝜀o (14-5) Slot conductance G1 G1 = W 120λo [ 1 −1 24(koh)2] , h λo < 1 10 (14-8a) Slot susceptance B1 B1 = W 120λo [1 −0.636 ln(koh)], h λo < 1 10 (14-8b) Input slot resistance Rin (at resonance; no coupling) (14-16) Rin = 1 2G1 Input slot resistance Rin (at resonance; with coupling) (14-17) Rin = 1 2(G1 ± G12) + for modes with odd symmetry −for modes with even symmetry Input slot resistance Rin (at resonance; with coupling) Rin = 90 (𝜀r)2 𝜀r −1 ( L W ) (14-18b) Input resistance Rin(y = yo) (no coupling) Rin(y = yo) = Rin(y = 0) cos2 (𝜋 L yo ) = 1 2G1 cos2 (𝜋 L yo ) (14-20a) Input resistance Rin(y = yo) (with coupling) Rin(y = yo) = Rin(y = 0) cos2 (𝜋 L yo ) = 1 2(G1 ± G12) cos2 (𝜋 L yo ) (14-20a) (continued overleaf) 860 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS TABLE 14.4 (continued) Equation Parameter Formula Number Cavity Model-Rectangular Patch Resonant frequency (frc)010; dominant mode (L > W) (no fringing) (frc)010 = 1 2L√𝜀r √𝜇o𝜀o (14-33) Resonant frequency (fr)010; dominant mode (L > W) (with fringing) (fr)010 = 1 2Leff √𝜀reff √𝜇o𝜀o (14-5) Resonant frequency (fr)001; dominant mode (L > W > L∕2 > h) (no fringing) (fr)001 = 1 2W√𝜀r √𝜇o𝜀o (14-34) Resonant frequency (fr)020; dominant mode (L > L∕2 > h); (no fringing) (fr)020 = 1 L√𝜀r √𝜇o𝜀o (14-35) Total electric field Et 𝜙 Et 𝜙= E𝜙(single slot) × AF (14-40a)–(14-41), (14-43) Array factor (AF)y (AF)y = 2 cos (koLe 2 sin 𝜃sin 𝜙 ) (14-42) Directivity Do (single slot) Do = ⎧ ⎪ ⎨ ⎪ ⎩ 3.3 (dimensionless) = 5.2 dB; W ≪λo 4 ( W λo ) ; W ≫λo (14-54) Directivity Do (two slots) Do = ⎧ ⎪ ⎨ ⎪ ⎩ 6.6 (dimensionless) = 8.2 dB; W ≪λo 8 ( W λo ) ; W ≫λo (14-57) Cavity Model-Circular Patch Resonant frequency (fr)110; dominant mode TM110 mode; (no fringing) (fr)110 = 1.8412 2𝜋a√𝜀r √𝜇o𝜀o (14-66) Resonant frequency (frc)110; dominant mode TM110 mode; (with fringing) (frc)110 = 1.8412 2𝜋ae √𝜀r √𝜇o𝜀o (14-68) Effective radius ae ae = a { 1 + 2h 𝜋a𝜀r [ ln (𝜋a 2h ) + 1.7726 ]}1∕2 (14-68) (continued) MULTIMEDIA 861 TABLE 14.4 (continued) Equation Parameter Formula Number Physical radius a a = F { 1 + 2h 𝜋𝜀rF [ ln (𝜋F 2h ) + 1.7726 ]}1∕2 F = 8.791 × 109 fr √𝜀r ; (h in cm) (14-69) (14-69a) Directivity Do Do = (koae)2 120Grad (14-80) Radiation conductance Grad Grad = (koae)2 480 ∫ 𝜋∕2 0 [(J′ 02)2 + cos2 𝜃(J02)2] sin 𝜃d𝜃 (14-76) J′ 02 = Jo(koae sin 𝜃) −J2(koae sin 𝜃) (14-72d) J02 = Jo(koae sin 𝜃) + J2(koae sin 𝜃) (14-72e) Input resistance Rin(𝜌′ = 𝜌o) Rin(𝜌′ = 𝜌o) = Rin(𝜌′ = ae) J2 1(k𝜌o) J2 1(kae) (14-82) Rin(𝜌′ = ae) = 1 Gt (14-82a) Gt = Grad + Gc + Gd (14-79) Gc = 𝜀mo𝜋(𝜋𝜇ofr)−3∕2 4h2√ 𝜎 [(kae)2 −m2] (14-77) Gd = 𝜀mo tan 𝛿 4𝜇ohf r [(kae)2 −m2] (14-78) where for mn0 mode (m = n = 1 for dominant mode) 𝜀mo = 2 for m = 0 𝜀mo = 1 for m ≠0 Total quality factor Qt 1 Qt = 1 Qrad + 1 Qc + 1 Qd + 1 Qsw For h ≪λo Qc = h √ 𝜋f𝜇𝜎; Qd = 1 tan 𝛿 Qrad = 2𝜔𝜀r hGt∕lK; K = ∫∫ area |E|2 dA ∮perimeter |E|2 dl (14-83) (14-84), (14-85) (14-86), (14-86a) Fractional bandwidth Δf fo Δf fo = VSWR −1 Qt √ VSWR (14-88a) 862 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS REFERENCES 1. G. A. Deschamps, “Microstrip Microwave Antennas,” Presented at the Third USAF Symposium on Anten-nas, 1953.
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PROBLEMS 14.1.
A microstrip line is used as a feed line to a microstrip patch. The substrate of the line is alu-mina (𝜀r ≃10) while the dimensions of the line are w∕h = 1.2 and t∕h = 0. Determine the effective dielectric constant and characteristic impedance of the line. Compare the computed characteristic impedance to that of a 50-ohm line.
14.2.
A microstrip transmission line of beryllium oxide (𝜀r ≃6.8) has a width-to-height ratio of w∕h = 1.5. Assuming that the thickness-to-height ratio is t∕h = 0, determine: (a) effective dielectric constant (b) characteristic impedance of the line 14.3.
A microstrip line, which is open at one end and extends to infinity toward the other end, has a center conductor width = 0.4λo, substrate height of 0.05λo, and it is operating at 10 GHz.
The dielectric constant of the substrate is 2.25. This type of microstrip line is used to con-struct rectangular patch antennas. Determine the following: (a) The input admittance (real and imaginary parts) of the microstrip line at the leading open edge. Is it capacitive or inductive?
(b) What kind of a lumped element (capacitor or inductor) can be placed at the leading open edge between the center conductor of the line and its ground plane to resonate the admittance? What is the value of the lumped element?
(c) The new input impedance, taking into account the presence of the lumped element.
14.4.
Design a rectangular microstrip antenna so that it will resonate at 2 GHz. The idealistic lossless substrate (RT/Duroid 6010.2) has a dielectric constant of 10.2 and a height of 0.05 in. (0.127 cm).
(a) Determine the physical dimensions (width and length) of the patch (in cm).
(b) Approximate range of lengths (in cm) between the two radiating slots of the rectangu-lar patch, if we want the input impedance (taking into account both radiating slots) to be real.
(c) What is the real input impedance of Part b? Neglect coupling.
(d) Location (in cm from the leading radiating slot) of a coaxial feed so that the total input impedance is 150 ohms.
868 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS 14.5.
Design a rectangular microstrip antenna to resonate at 9 GHz using a substrate with a dielec-tric constant of 2.56. Determine the following: (a) Directivity of a single radiating slot (dimensionless and in dB). Use the cavity model.
(b) Approximate directivity of the entire patch (dimensions and in dB). Use the cavity model and neglect coupling between the two slots.
14.6.
It is desired to design rectangular microstrip patch antenna to resonate at a frequency of f = 10 GHz. Using a substrate with a dielectric constant of 4 (𝜀r = 4) and a height of h = 0.25 cm, determine the: (a) Width of the microstrip patch (in cm).
(b) Effective dielectric constant (𝜀reff ) of the substrate.
(c) Effective length Le of the patch (in cm).
(d) Physical length L of the patch (in cm).
14.7.
A rectangular microstrip antenna was designed, without taking into account fringing effects from any of the four edges of the patch, to operate at a center frequency of 4.6 GHz. The width of the patch was chosen to be W = 1.6046 cm and the substrate had a height of 0.45 cm and a dielectric constant of 6.8. However, when the patch was tested, it was found to resonate at a frequency of 4.046 GHz!
(a) Find the physical length L of the patch (in cm).
(b) Why did the patch resonate at 4.046 GHz, instead of the designed frequency of 4.6 GHz?
Verify the new resonant frequency. Must justify your answer mathematically. Show that the measured resonant frequency is correct.
14.8.
Cellular and mobile telephony, using earth-based repeaters, has received wide acceptance and has become an essential means of communication for business, even for the household.
Cellular telephony by satellites is the wave of the future and communication systems are being designed for that purpose. The present allocated frequency band for satellites is at L-band (≃1.6 GHz). Various antennas are being examined for that purpose; one candidate is the microstrip patch antenna. Design a rectangular microstrip patch antenna, based on the dominant mode, that can be mounted on the roof of a car to be used for satellite cellular telephone. The designed center frequency is 1.6 GHz, the dielectric constant of the substrate is 10.2 (i.e., RT/duroid), and the thickness of the substrate is 0.127 cm. Determine the (a) dimensions of the rectangular patch (in cm) (b) resonant input impedance, assuming no coupling between the two radiating slots (c) mutual conductance between the two radiating slots of the patch (d) resonant input impedance, taking into account coupling (e) position of the feed to match the patch antenna to a 75-ohm line 14.9.
Repeat the design of Problem 14.8 using a substrate with a dielectric constant of 2.2 (i.e., RT/duroid 5880) and with a height of 0.1575 cm. Are the new dimensions of the patch realistic for the roof of a personal car?
14.10.
Design a rectangular microstrip patch with dimensions W and L, over a single substrate, whose center frequency is 10 GHz. The dielectric constant of the substrate is 10.2 and the height of the substrate is 0.127 cm (0.050 in.). Determine the physical dimensions W and L (in cm) of the patch, taking into account field fringing.
14.11.
Using the transmission-line model of Figure 14.9(b), derive (14-14)–(14-15).
14.12.
To take into account coupling between the two radiating slots of a rectangular microstrip patch, the resonant input resistance is represented by (14-17). Justify, explain, and/or show PROBLEMS 869 why the plus (+) sign is used for modes with odd (antisymmetric) resonant voltage distribu-tions beneath the patch while the minus (−) sign is used for modes with even (symmetric) resonant voltage distributions.
14.13.
Show that for typical rectangular microstrip patches G1∕Yc ≪1 and B1∕Yc ≪1 so that (14-20) reduces to (14-20a).
14.14.
A rectangular microstrip patch antenna is operating at 10 GHz with 𝜀r = 10.2 and dimen-sions of length L = 0.4097 cm, width W = 0.634 cm, and substrate height h = 0.127 cm. It is desired to feed the patch using a probe feed. Neglecting mutual coupling, calculate: (a) What is the input impedance of the patch at one of the radiating edges based on the transmission-line model?
(b) At what distance y0 (in cm) from one of the radiating edges should the coax feed be placed so that the input impedance is 50 ohms?
14.15.
A rectangular microstrip patch antenna, whose input impedance is 152.44 ohms at its leading radiating edge, is fed by a microstrip line as shown in Figure 14.11. Assuming the width of the feeding line is W0 = 0.2984 cm, the height of the substrate is 0.1575 cm and the dielectric constant of the substrate is 2.2, at what distance y0 should the microstrip patch antenna be fed so as to have a perfect match between the line and the radiating element?
The overall microstrip patch element length is 0.9068 cm.
14.16.
The rectangular microstrip patch of Example 14.2 is fed by a microstrip transmission line of Figure 14.5. In order to reduce reflections at the inset feed point between the line and the patch element, design the microstrip line so that its characteristic impedance matches that of the radiating element.
14.17.
Repeat the design of Example 14.2 so that the input impedance of the radiating patch at the feed point is: (a) 75 ohms (b) 100 ohms Then, assuming the feed line is a microstrip line, determine the dimensions of the line so that its characteristic impedance matches that of the radiating patch.
14.18.
A rectangular microstrip patch antenna has dimensions of L = 0.906 cm, W = 1.186 cm, and h = 0.1575 cm. The dielectric constant of the substrate is 𝜀r = 2.2. Using the geometry of Figure 14.15 and assuming no fringing, determine the resonant frequency of the first 4 TMz 0np modes, in order of ascending resonant frequency.
14.19.
Derive the TMz mnp field configurations (modes) for the rectangular microstrip patch based on the geometry of Fig-ure P14.19. Determine the: (a) eigenvalues (b) resonant frequency (fr)mnp for the mnp mode.
(c) dominant mode if L > W > h (d) resonant frequency of the dominant mode.
y P14.19 x z L h W r 14.20.
A microstrip antenna with a rectangular patch, with dimensions L = 5 cm and W = 2 cm, and a substrate with h = 0.1568 cm and a dielectric constant of 𝜀r = 2.2, is used in a wireless communications system. Accounting for fringing: 870 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS (a) Identify the second-order TMx mnp mode. Selection of the wrong mode will make all the other answers wrong, and no credit will be given for answers of the other parts of this problem for the wrong mode.
(b) Compute the resonant frequency of the 2nd-order TMx mnp mode.
(c) Calculate the resonant input resistance (assume no coupling) of the 2nd-order TMx mnp mode.
(d) Compute the resonant input resistance, of the 2nd-order TMx mnp mode, when the mutual conductance of this mode is 0.24921 × 10−3 Siemens.
14.21.
Repeat Problem 14.19 for the TMy mnp modes based x z y L h W r P14.21 on the geometry of Figure P14.21.
14.22.
Derive the array factor of (14-42).
14.23.
Assuming the coordinate system for the rectangu-lar microstrip patch is that of Problem 14.19 (Fig-ure P14.19), derive based on the cavity model the (a) far-zone electric field radiated by one of the radiating slots of the patch (b) array factor for the two radiating slots of the patch (c) far-zone total electric field radiated by both of the radiating slots 14.24.
Repeat Problem 14.23 for the rectangular patch geometry of Problem 14.21 (Figure P14.21).
14.25.
Determine the directivity (in dB) of the rectangular microstrip patch of Example 14.3 using (a) Kraus’ approximate formula (b) Tai & Pereira’s approximate formula 14.26.
Derive the directivity (in dB) of the rectangular microstrip patch of Problem 14.8.
14.27.
Derive the directivity (in dB) of the rectangular microstrip patch of Problem 14.9.
14.28.
For a circular microstrip patch antenna operating in the dominant TMz 110 mode, derive the far-zone electric fields radiated by the patch based on the cavity model.
14.29.
Using the cavity model, derive the TMz mnp resonant frequencies for a microstrip patch whose shape is that of a half of a circular patch (semicircle).
14.30.
Repeat Problem 14.29 for a 90◦circular disc (angular sector of 90◦) microstrip patch.
14.31.
Repeat Problem 14.29 for the circular sector microstrip patch antenna whose geometry is shown in Figure P14.31.
y z z y x x a h f f 0 Top view Side view Feed s, s a μ = ∞ σ ϕ ϕ ρ P14.31 PROBLEMS 871 14.32.
Repeat Problem 14.29 for the annular microstrip patch antenna whose geometry is shown in Figure P14.30.
z y x h Side view s, s f b b a a μ ρ x Top view Feed y z f b a = ∞ σ ϕ ρ ρ P14.32 14.33.
Repeat Problem 14.29 for the annular sector microstrip patch antenna whose geometry is shown in Figure P14.31.
Side view z y x h b a s, s μ Top view y z x b a Feed 0 f f = ∞ σ ϕ ϕ ρ P14.33 14.34.
Repeat the design of Problem 14.8 for a circular microstrip patch antenna operating in the dominant TMz 110 mode. Use 𝜎= 107 S/m and tan 𝛿= 0.0018.
14.35.
Repeat the design of Problem 14.9 for a circular microstrip patch antenna operating in the dominant TMz 110 mode. Use 𝜎= 107 S/m and tan 𝛿= 0.0018.
14.36.
For ground-based cellular telephony, the desired pattern coverage is omnidirectional and similar to that of a monopole (with a null toward zenith, 𝜃= 0◦). This can be accomplished using a circular microstrip patch antenna operating in a higher order mode, such as the TMz 210. Assuming the desired resonant frequency is 900 MHz, design a circular microstrip patch antenna operating in the TMz 210 mode. Assuming a substrate with a dielectric constant of 10.2 and a height of 0.127 cm: (a) Derive an expression for the resonant frequency of the TMz 210 mode; (b) Determine the radius of the circular patch (in cm). Neglect fringing.
14.37.
Microstrip (patch) antennas are usually designed so that the maximum of the amplitude radiation pattern is perpendicular to the patch. However for ground-based cellular telephony, the pattern of the antenna should usually match that of a vertical monopole with a null towards zenith (𝜃= 0◦). This can be accomplished if a circular patch is selected and it is excited at a higher-order mode, such as the TMz 210 mode.
Assuming a TMz 210 mode (NOT dominant TMz 210 mode), the desired operating frequency, without taking into account fringing, is 1.9 GHz, the substrate has a dielectric constant of 10.2 and its height is 0.127 cm. Determine the: (a) Physical radius of the circular patch (in cm). Neglect fringing.
872 MICROSTRIP AND MOBILE COMMUNICATIONS ANTENNAS (b) Effective radius of the circular patch (in cm). Account for fringing.
(c) New resonant frequency (in GHz) taking into account fringing.
P.S. The design procedure of Section 14.3.3 is for the TMz 110 and should not be used.
Other equations should be used. However Equation (14-67) is still valid for the TMz 210.
14.38.
For ground-based cellular telephony, the desired pattern coverage is omnidirectional and similar to that of a monopole (with a null toward zenith). This can be accomplished using circular microstrip patch antennas operating in higher order modes, such as the TMz 210, TMz 310, TMz 410, etc. Assuming that the desired resonant frequency is 900 MHz, design a circular microstrip patch antenna operating in the TMz 210 mode. Assuming a sub-strate with a dielectric constant of 10.2 and a height of 0.127 cm: (a) Derive an expression for the resonant frequency.
(b) Determine the radius of the circular patch. Neglect fringing.
(c) Derive expressions for the far-zone radiated fields.
(d) Plot the normalized E- and H-plane amplitude patterns (in dB).
(e) Plot the normalized azimuthal (x-y plane) amplitude pattern (in dB).
(f) Determine the directivity (in dB) using the Directivity computer program of Chapter 2.
14.39.
Repeat Problem 14.38 for the TMz 310 mode.
14.40.
Repeat Problem 14.38 for the TMz 410 mode.
14.41.
The diameter of a typical probe feed for a microstrip patch antenna is d = 0.1 cm. At f = 10 GHz, determine the feed reactance assuming a substrate with a dielectric constant of 2.2 and height of 0.1575 cm.
14.42.
Determine the impedance of a single-section quarter-wavelength impedance transformer to match a 100-ohm patch element to a 50-ohm microstrip line. Determine the dimen-sions of the line assuming a substrate with a dielectric constant of 2.2 and a height of 0.1575 cm.
14.43.
Repeat the design of Problem 14.42 using a two-section binomial transformer. Determine the dimensions of each section of the transformer.
14.44.
Repeat the design of Problem 14.42 using a two-section Tschebyscheff transformer. Deter-mine the dimensions of each section of the transformer.
14.45.
A very thin (w⪡λ) half-guide wavelength slot, as shown in the Figure P14.45, is fed by a dipole of length slightly less than λg∕2 (the ends of the dipole do not touch the ground plane forming the slot). Determine the impedance of the slot.
Ground Ground Dipole g/2 w<< λ λ P14.45 14.46.
For a cubic dielectric resonator of Figure 14.57, with dimensions of a = 1 cm, b = 1 cm, and c = 0.3 cm, deter-mine the resonant frequencies for the first five TEz and/or TMz modes when: (a) 𝜀r = 8.9 (b) 𝜀r = 89. Use Matlab computer program DRA Analysis Design.
14.47.
Repeat Problem 14.46 for a hemicylindrical dielectric resonator of Figure 14.59 with dimen-sions of a = 0.3 cm and h = 1 cm.
14.48.
Repeat Problem 14.46 for the three dominant degenerate modes of a hemispherical dielectric resonator of Figure 14.60 with radius of 0.3 cm.
PROBLEMS 873 14.49.
Design a cylindrical resonator of Figure 14.58, to operate in the TM01𝛿mode and to exhibit a fractional bandwidth of 2.887%, VSWR = 3 and resonant frequency fr = 10 GHz. Deter-mine the: (a) Q of the cavity (b) Minimum dielectric constant to achieve the design (c) Choose a dielectric constant of 38 (d) Determine the dimensions a and h (both in cm) based on a dielectric constant of 38.
Use Matlab computer program DRA Analysis Design.
14.50.
Design a cylindrical resonator of Figure 14.58, to operate in the HE11𝛿mode and to exhibit a fractional bandwidth of 2.887%, VSWR = 3 and resonant frequency fr = 10 GHz. Determine the: (a) Q of the cavity (b) Range of dielectric constants to achieve the design (c) Choose a dielectric constant of 38 (d) Determine the dimensions a and h (both in cm) based on a dielectric constant of 38.
Use Matlab computer program DRA Analysis Design.
CHAPTER15 Reflector Antennas 15.1 INTRODUCTION Reflector antennas, in one form or another, have been in use since the discovery of electromagnetic wave propagation in 1888 by Hertz. However the fine art of analyzing and designing reflectors of many various geometrical shapes did not forge ahead until the days of World War II when numerous radar applications evolved. Subsequent demands of reflectors for use in radio astronomy, microwave communication, and satellite tracking resulted in spectacular progress in the development of sophis-ticated analytical and experimental techniques in shaping the reflector surfaces and optimizing illu-mination over their apertures so as to maximize the gain. The use of reflector antennas for deep-space communication, such as in the space program and especially their deployment on the surface of the moon, resulted in establishing the reflector antenna almost as a household word during the 1960s.
Although reflector antennas take many geometrical configurations, some of the most popular shapes are the plane, corner, and curved reflectors (especially the paraboloid), as shown in Figure 15.1, each of which will be discussed in this chapter. Many articles on various phases of the analysis and design of curved reflectors have been published and some of the most referenced can be found in a book of reprinted papers .
15.2 PLANE REFLECTOR The simplest type of reflector is a plane reflector introduced to direct energy in a desired direction.
The arrangement is that shown in Figure 15.1(a) which has been extensively analyzed in Section 4.7 when the radiating source is a vertical or horizontal linear element. It has been clearly demonstrated that the polarization of the radiating source and its position relative to the reflecting surface can be used to control the radiating properties (pattern, impedance, directivity) of the overall system.
Image theory has been used to analyze the radiating characteristics of such a system. Although the infinite dimensions of the plane reflector are idealized, the results can be used as approximations for electrically large surfaces. The perturbations introduced by keeping the dimensions finite can be accounted for by using special methods such as the Geometrical Theory of Diffraction – which was introduced in Section 12.10.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 875 876 REFLECTOR ANTENNAS Figure 15.1 Geometrical configuration for some reflector systems.
15.3 CORNER REFLECTOR To better collimate the energy in the forward direction, the geometrical shape of the plane reflector itself must be changed so as to prohibit radiation in the back and side directions. One arrangement which accomplishes that consists of two plane reflectors joined so as to form a corner, as shown in Figures 15.1(b) and in 15.2(a). This is known as the corner reflector. Because of its simplicity in construction, it has many unique applications. For example, if the reflector is used as a passive target for radar or communication applications, it will return the signal exactly in the same direction as it received it when its included angle is 90◦. This is illustrated geometrically in Figure 15.2(b).
Because of this unique feature, military ships and vehicles are designed with minimum sharp corners to reduce their detection by enemy radar. Corner reflectors are also widely used as receiving elements for home television.
In most practical applications, the included angle formed by the plates is usually 90◦; however other angles are sometimes used. To maintain a given system efficiency, the spacing between the vertex and the feed element must increase as the included angle of the reflector decreases, and vice versa. For reflectors with infinite sides, the gain increases as the included angle between the planes decreases. This, however, may not be true for finite size plates. For simplicity, in this chapter it will be assumed that the plates themselves are infinite in extent (l = ∞). However, since in practice the dimensions must be finite, guidelines on the size of the aperture (Da), length (l), and height (h) will be given.
The feed element for a corner reflector is almost always a dipole or an array of collinear dipoles placed parallel to the vertex a distance s away, as shown in a perspective view in Figure 15.2(c).
Greater bandwidth is obtained when the feed elements are cylindrical or biconical dipoles instead of thin wires. In many applications, especially when the wavelength is large compared to tolerable CORNER REFLECTOR 877 Figure 15.2 Side and perspective views of solid and wire-grid corner reflectors.
physical dimensions, the surfaces of the corner reflector are frequently made of grid wires rather than solid sheet metal, as shown in Figure 15.2(d). One of the reasons for doing that is to reduce wind resistance and overall system weight. The spacing (g) between wires is made a small fraction of a wavelength (usually g ≤λ∕10). For wires that are parallel to the length of the dipole, as is the case for the arrangement of Figure 15.2(d), the reflectivity of the grid-wire surface is as good as that of a solid surface.
In practice, the aperture of the corner reflector (Da) is usually made between one and two wave-lengths (λ < Da < 2λ). The length of the sides of a 90◦corner reflector is most commonly taken to be about twice the distance from the vertex to the feed (l ≃2s). For reflectors with smaller included angles, the sides are made larger. The feed-to-vertex distance (s) is usually taken to be between λ∕3 and 2λ∕3(λ∕3 < s < 2λ∕3). For each reflector, there is an optimum feed-to-vertex spacing. If the spacing becomes too small, the radiation resistance decreases and becomes comparable to the loss resistance of the system which leads to an inefficient antenna. For very large spacing, the system pro-duces undesirable multiple lobes, and it loses its directional characteristics. It has been experimen-tally observed that increasing the size of the sides does not greatly affect the beamwidth and direc-tivity, but it increases the bandwidth and radiation resistance. The main lobe is somewhat broader for reflectors with finite sides compared to that of infinite dimensions. The height (h) of the reflector is usually taken to be about 1.2 to 1.5 times greater than the total length of the feed element, in order to reduce radiation toward the back region from the ends.
878 REFLECTOR ANTENNAS Figure 15.3 Corner reflectors and their images (with perpendicularly polarized feeds) for angles of 90◦, 60◦, 45◦, and 30◦.
The analysis for the field radiated by a source in the presence of a corner reflector is facilitated when the included angle (𝛼) of the reflector is 𝛼= 𝜋∕n, where n is an integer (𝛼= 𝜋, 𝜋∕2, 𝜋∕3, 𝜋∕4, etc.). For those cases (𝛼= 180◦, 90◦, 60◦, 45◦, etc.) it is possible to find a system of images, which when properly placed in the absence of the reflector plates, form an array that yields the same field within the space formed by the reflector plates as the actual system. The number of images, polarity, and position of each is controlled by the included angle of the corner reflector and the polarization of the feed element. In Figure 15.3 we display the geometrical and electrical arrangement of the images for corner reflectors with included angles of 90◦, 60◦, 45◦, and 30◦and a feed with perpendicular polarization. The procedure for finding the number, location, and polarity of the images is demon-strated graphically in Figure 15.4 for a corner reflector with a 90◦included angle. It is assumed that the feed element is a linear dipole placed parallel to the vertex. A similar procedure can be followed for all other reflectors with an included angle of 𝛼= 180◦∕n, where n is an integer.
15.3.1 90◦Corner Reflector The first corner reflector to be analyzed is the one with an included angle of 90◦. Because its radiation characteristics are the most attractive, it has become the most popular.
Referring to the reflector of Figure 15.2(c) with its images in Figure 15.4(b), the total field of the system can be derived by summing the contributions from the feed and its images. Thus E(r, 𝜃, 𝜙) = E1(r1, 𝜃, 𝜙) + E2(r2, 𝜃, 𝜙) + E3(r3, 𝜃, 𝜙) + E4(r4, 𝜃, 𝜙) (15-1) CORNER REFLECTOR 879 Figure 15.4 Geometrical placement and electrical polarity of images for a 90◦corner reflector with a parallel polarized feed.
In the far-zone, the normalized scalar field can be written as E(r, 𝜃, 𝜙) = f(𝜃, 𝜙)e−jkr1 r1 −f (𝜃, 𝜙)e−jkr2 r2 + f(𝜃, 𝜙)e−jkr3 r3 −f(𝜃, 𝜙)e−jkr4 r4 E(r, 𝜃, 𝜙) = [e+jks cos 𝜓1 −e+jks cos 𝜓2 + e+jks cos 𝜓3 −e+jks cos 𝜓4]f(𝜃, 𝜙)e−jkr r (15-2) where cos 𝜓1 = ̂ ax⋅̂ ar = sin 𝜃cos 𝜙 (15-2a) cos 𝜓2 = ̂ ay⋅̂ ar = sin 𝜃sin 𝜙 (15-2b) cos 𝜓3 = −̂ ax⋅̂ ar = −sin 𝜃cos 𝜙 (15-2c) cos 𝜓4 = −̂ ay⋅̂ ar = −sin 𝜃sin 𝜙 (15-2d) since ̂ ar = ̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃. Equation (15-2) can also be written, using (15-2a)–(15-2d), as E(r, 𝜃, 𝜙) = 2[cos(ks sin 𝜃cos 𝜙) −cos(ks sin 𝜃sin 𝜙)] f(𝜃, 𝜙)e−jkr r (15-3) 880 REFLECTOR ANTENNAS where for 𝛼= 𝜋∕2 = 90◦ 0 ≤𝜃≤𝜋, 0 ≤𝜙≤𝛼∕2 2𝜋−𝛼∕2 ≤𝜙≤2𝜋 (15-3a) Letting the field of a single isolated (radiating in free-space) element to be E0 = f(𝜃, 𝜙)e−jkr r (15-4) (15-3) can be rewritten as E E0 = AF(𝜃, 𝜙) = 2[cos(ks sin 𝜃cos 𝜙) −cos(ks sin 𝜃sin 𝜙)] (15-5) Equation (15-5) represents not only the ratio of the total field to that of an isolated element at the origin but also the array factor of the entire reflector system. In the azimuthal plane (𝜃= 𝜋∕2), (15-5) reduces to E E0 = AF(𝜃= 𝜋∕2, 𝜙) = 2[cos(ks cos 𝜙) −cos(ks sin 𝜙)] (15-6) To gain some insight into the performance of a corner reflector, in Figure 15.5 we display the normalized patterns for an 𝛼= 90◦corner reflector for spacings of s = 0.1λ, 0.7λ, 0.8λ, 0.9λ, and 1.0λ. It is evident that for the small spacings the pattern consists of a single major lobe whereas multiple lobes appear for the larger spacings (s > 0.7λ). For s = λ the pattern exhibits two lobes separated by a null along the 𝜙= 0◦axis.
Another parameter of performance for the corner reflector is the field strength along the symmetry axis (𝜃= 90◦, 𝜙= 0◦) as a function of feed-to-vertex distance s . The normalized (relative to the field of a single isolated element) absolute field strength |E∕E0| as a function of s∕λ(0 ≤s ≤10λ) for 𝛼= 90◦is shown plotted in Figure 15.6. It is apparent that the first field strength peak is achieved when s = 0.5λ, and it is equal to 4. The field is also periodic with a period of Δs∕λ = 1.
15.3.2 Other Corner Reflectors A similar procedure can be used to derive the array factors and total fields for all other corner reflec-tors with included angles of 𝛼= 180◦∕n. Referring to Figure 15.3, it can be shown that the array factors for 𝛼= 60◦, 45◦, and 30◦can be written as 𝛼= 60◦ AF(𝜃, 𝜙) = 4 sin (X 2 ) [ cos (X 2 ) −cos (√ 3Y 2 )] (15-7) 𝛼= 45◦ AF(𝜃, 𝜙) = 2 [ cos(X) + cos(Y) −2 cos ( X √ 2 ) cos ( Y √ 2 )] (15-8) CORNER REFLECTOR 881 Figure 15.5 Normalized radiation amplitude patterns for 𝛼= 90◦corner reflector.
𝛼= 30◦ AF(𝜃, 𝜙) = 2 [ cos(X) −2 cos (√ 3 2 X ) cos (Y 2 ) −cos(Y) + 2 cos (X 2 ) cos (√ 3 2 Y )] (15-9) where X = ks sin 𝜃cos 𝜙 Y = ks sin 𝜃sin 𝜙 (15-9a) (15-9b) These are assigned, at the end of the chapter, as exercises to the reader (Problem 15.2).
For a corner reflector with an included angle of 𝛼= 180◦∕n, n = 1, 2, 3, … , the number of images is equal to N = (360∕𝛼) −1 = 2n −1.
882 REFLECTOR ANTENNAS Figure 15.6 Relative field strength along the axis (𝜃= 90◦, 𝜙= 0◦) of an 𝛼= 90◦corner reflector as a function of feed-to-vertex spacing.
It has also been shown by using long filament wires as feeds, that the azimuthal plane (𝜃= 𝜋∕2) array factor for corner reflectors with 𝛼= 180◦∕n, where n is an integer, can also be written as n = even (n = 2, 4, 6,…) AF(𝜙) = 4n(−1)n∕2[Jn(ks) cos(n𝜙) + J3n(ks) cos(3n𝜙) + J5n(ks) cos(5n𝜙) + ⋯] (15-10a) n = odd (n = 1, 3, 5,…) AF(𝜙) = 4nj(−1)(n−1)∕2[Jn(ks) cos(n𝜙) −J3n(ks) cos(3n𝜙) + J5n(ks) cos(5n𝜙) + ⋯] (15-10b) where Jm(x) is the Bessel function of the first kind of order m (see Appendix V).
When n is not an integer, the field must be found by retaining a sufficient number of terms of the infinite series. It has also been shown that for all values of n = m (integral or fractional) that the field can be written as AF(𝜙) = 4m[ejm𝜋∕2Jm(ks) cos(m𝜙) + ej3m𝜋∕2J3m(ks) cos(3m𝜙) + ⋯] (15-11) The array factor for a corner reflector, as given by (15-10a)–(15-11), has a form that is similar to the array factor for a uniform circular array, as given by (6-121). This should be expected since the feed sources and their images in Figure 15.3 form a circular array. The number of images increase as the included angle of the corner reflector decreases.
CORNER REFLECTOR 883 Patterns have been computed for corner reflectors with included angles of 60◦, 45◦, and 30◦. It has been found that these corner reflectors have also single-lobed patterns for the smaller values of s, and they become narrower as the included angle decreases. Multiple lobes begin to appear when s ≃0.95λ for 𝛼= 60◦ s ≃1.2λ for 𝛼= 45◦ s ≃2.5λ for 𝛼= 30◦ The field strength along the axis of symmetry (𝜃= 90◦, 𝜙= 0◦) as a function of the feed-to-vertex distance s, has been computed for reflectors with included angles of 𝛼= 60◦, 45◦, and 30◦.
The results for 𝛼= 45◦are shown in Figure 15.7 for 0 ≤s ≤10λ.
For reflectors with 𝛼= 90◦and 60◦, the normalized field strength is periodic with periods of λ and 2λ, respectively. However, for the 45◦and 30◦reflectors the normalized field is not periodic but rather “almost periodic” or “pseudoperiodic” . For the 45◦and 30◦reflectors the arguments of the trigonometric functions representing the arrays factors, and given by (15-8)–(15-9b), are related by irrational numbers and therefore the arrays factors do not repeat. However, when plotted they look very similar. Therefore when examined only graphically, the observer erroneously may conclude that the patterns are periodic (because they look so much the same). However, when the array factors are examined analytically it is concluded that the functions are not periodic but rather nearly peri-odic. The field variations are “nearly similar” in form in the range Δs ≃16.69λ for the 𝛼= 45◦and Δs ≃30λ for the 𝛼= 30◦. Therefore the array factors of (15-8) and (15-9) belong to the class of nearly periodic functions .
It has also been found that the maximum field strength increases as the included angle of the reflector decreases. This is expected since a smaller angle reflector exhibits better directional char-acteristics because of the narrowness of its angle. The maximum values of |E∕E0| for 𝛼= 60◦, 45◦, Figure 15.7 Relative field strength along the axis (𝜃= 90◦, 𝜙= 0◦) for an 𝛼= 45◦corner reflector as a func-tion of feed-to-vertex spacing.
884 REFLECTOR ANTENNAS and 30◦are approximately 5.2, 8, and 9, respectively. The first field strength peak, but not necessarily the ultimate maximum, is achieved when s ≃0.65λ for 𝛼= 60◦ s ≃0.85λ for 𝛼= 45◦ s ≃1.20λ for 𝛼= 30◦ 15.4 PARABOLIC REFLECTOR The overall radiation characteristics (antenna pattern, antenna efficiency, polarization discrimination, etc.) of a reflector can be improved if the structural configuration of its surface is upgraded. It has been shown by geometrical optics that if a beam of parallel rays is incident upon a reflector whose geometrical shape is a parabola, the radiation will converge (focus) at a spot which is known as the focal point. In the same manner, if a point source is placed at the focal point, the rays reflected by a parabolic reflector will emerge as a parallel beam. This is one form of the principle of reciprocity, and it is demonstrated geometrically in Figure 15.1(c). The symmetrical point on the parabolic surface is known as the vertex. Rays that emerge in a parallel formation are usually said to be collimated.
In practice, collimation is often used to describe the highly directional characteristics of an antenna even though the emanating rays are not exactly parallel. Since the transmitter (receiver) is placed at the focal point of the parabola, the configuration is usually known as front fed.
The disadvantage of the front-fed arrangement is that the transmission line from the feed must usually be long enough to reach the transmitting or the receiving equipment, which is usually placed behind or below the reflector. This may necessitate the use of long transmission lines whose losses may not be tolerable in many applications, especially in low-noise receiving systems. In some appli-cations, the transmitting or receiving equipment is placed at the focal point to avoid the need for long transmission lines. However, in some of these applications, especially for transmission that may require large amplifiers and for low-noise receiving systems where cooling and weatherproofing may be necessary, the equipment may be too heavy and bulky and will provide undesirable blockage.
Another arrangement that avoids placing the feed (transmitter and/or receiver) at the focal point is that shown in Figure 15.1(d), and it is known as the Cassegrain feed. Through geometrical optics, Cassegrain, a famous astronomer (hence its name), showed that incident parallel rays can be focused to a point by utilizing two reflectors. To accomplish this, the main (primary) reflector must be a parabola, the secondary reflector (subreflector) a hyperbola, and the feed placed along the axis of the parabola usually at or near the vertex. Cassegrain used this scheme to construct optical telescopes, and then its design was copied for use in radio frequency systems. For this arrangement, the rays that emanate from the feed illuminate the subreflector and are reflected by it in the direction of the primary reflector, as if they originated at the focal point of the parabola (primary reflector). The rays are then reflected by the primary reflector and are converted to parallel rays, provided the primary reflector is a parabola and the subreflector is a hyperbola. Diffractions occur at the edges of the subreflector and primary reflector, and they must be taken into account to accurately predict the overall system pattern, especially in regions of low intensity –. Even in regions of high intensity, diffractions must be included if an accurate formation of the fine ripple structure of the pattern is desired. With the Cassegrain-feed arrangement, the transmitting and/or receiving equipment can be placed behind the primary reflector. This scheme makes the system relatively more accessible for servicing and adjustments.
A parabolic reflector can take two different forms. One configuration is that of the parabolic right cylinder, shown in Figure 15.8(a), whose energy is collimated at a line that is parallel to the axis of the cylinder through the focal point of the reflector. The most widely used feed for this type of a PARABOLIC REFLECTOR 885 Reflector (parabolic cylinder) Feed (dipole) Feed (horn) Reflector (paraboloid) (a) Parabolic right cylinder (b) Paraboloid Figure 15.8 Parabolic right cylinder and paraboloid.
reflector is a linear dipole, a linear array, or a slotted waveguide. The other reflector configuration is that of Figure 15.8(b) which is formed by rotating the parabola around its axis, and it is referred to as a paraboloid (parabola of revolution). A pyramidal or a conical horn has been widely utilized as a feed for this arrangement.
There are many other types of reflectors whose analysis is widely documented in the literature –. The spherical reflector, for example, has been utilized for radioastronomy and small earth-station applications, because its beam can be efficiently scanned by moving its feed. An example of that is the 1,000-ft (305-m) diameter spherical reflector at Arecibo, Puerto Rico whose primary surface is built into the ground and scanning of the beam is accomplished by movement of the feed.
For spherical reflectors a substantial blockage may be provided by the feed leading to unacceptable minor lobe levels, in addition to the inherent reduction in gain and less favorable cross-polarization discrimination.
To eliminate some of the deficiencies of the symmetric configurations, offset-parabolic reflector designs have been developed for single- and dual-reflector systems . Because of the asymmetry of the system, the analysis is more complex. However the advent and advances of computer technol-ogy have made the modeling and optimization of the offset-reflector designs available and conve-nient. Offset-reflector designs reduce aperture blocking and VSWR. In addition, they lead to the use of larger f/d ratios while maintaining acceptable structural rigidity, which provide an opportunity for improved feed pattern shaping and better suppression of cross-polarized radiation emanating from the feed. However, offset-reflector configurations generate cross-polarized antenna radiation when illuminated by a linearly polarized primary feed. Circularly polarized feeds eliminate depolarization, but they lead to squinting of the main beam from boresight. In addition, the structural asymmetry of the system is usually considered a major drawback.
Paraboloidal reflectors are the most widely used large aperture ground-based antennas . At the time of its construction, the world’s largest fully steerable reflector was the 100-m diameter radio telescope of the Max Planck Institute for Radioastronomy at Effelsberg, West Germany, while the largest in the United States was the 64-m diameter reflector at Goldstone, California built primarily for deep-space applications. When fed efficiently from the focal point, paraboloidal reflec-tors produce a high-gain pencil beam with low side lobes and good cross-polarization discrimination characteristics. This type of an antenna is widely used for low-noise applications, such as in radioas-tronomy, and it is considered a good compromise between performance and cost. To build a large reflector requires not only a large financial budget but also a difficult structural undertaking, because it must withstand severe weather conditions.
Cassegrain designs, employing dual-reflector surfaces, are used in applications where pattern con-trol is essential, such as in satellite ground-based systems, and have efficiencies of 65–80%. They 886 REFLECTOR ANTENNAS Figure 15.9 Shaped 10-m earth-station dual-reflector antenna. (Courtesy Andrew Corp).
supersede the performance of the single-reflector front-fed arrangement by about 10%. Using geo-metrical optics, the classical Cassegrain configuration, consisting of a paraboloid and a hyperboloid, is designed to achieve a uniform phase front in the aperture of the paraboloid. By employing good feed designs, this arrangement can achieve lower spillover and more uniform illumination of the main reflector. In addition, slight shaping of one or both of the dual-reflector’s surfaces can lead to an aper-ture with almost uniform amplitude and phase with a substantial enhancement in gain . These are referred to as shaped reflectors. Shaping techniques have been employed in dual-reflectors used in earth-station applications. An example is the 10-m earth-station dual-reflector antenna, shown in Figure 15.9, whose main reflector and subreflector are shaped.
For many years horns or waveguides, operating in a single mode, were used as feeds for reflector antennas. However because of radioastronomy and earth-station applications, considerable efforts have been placed in designing more efficient feeds to illuminate either the main reflector or the subreflector. It has been found that corrugated horns that support hybrid mode fields (combination of TE and TM modes) can be used as desirable feeds. Such feed elements match efficiently the fields of the feeds with the desired focal distribution produced by the reflector, and they can reduce cross-polarization. Dielectric cylinders and cones are other antenna structures that support hybrid PARABOLIC REFLECTOR 887 modes . Their structural configuration can also be used to support the subreflector and to provide attractive performance figures.
There are primarily two techniques that can be used to analyze the performance of a reflector system . One technique is the aperture distribution method and the other the current distribution method. Both techniques will be introduced to show the similarities and differences.
15.4.1 Front-Fed Parabolic Reflector Parabolic cylinders have widely been used as high-gain apertures fed by line sources. The analy-sis of a parabolic cylinder (single curved) reflector is similar, but considerably simpler than that of a paraboloidal (double curved) reflector. The principal characteristics of aperture amplitude, phase, and polarization for a parabolic cylinder, as contrasted to those of a paraboloid, are as follows: 1. The amplitude taper, due to variations in distance from the feed to the surface of the reflector, is proportional to 1∕𝜌in a cylinder compared to 1∕r2 in a paraboloid.
2. The focal region, where incident plane waves converge, is a line source for a cylinder and a point source for a paraboloid.
3. When the fields of the feed are linearly polarized parallel to the axis of the cylinder, no cross-polarized components are produced by the parabolic cylinder. That is not the case for a paraboloid.
Generally, parabolic cylinders, as compared to paraboloids, (1) are mechanically simpler to build, (2) provide larger aperture blockage, and (3) do not possess the attractive characteristics of a paraboloid. In this chapter, only paraboloidal reflectors will be examined.
A. Surface Geometry The surface of a paraboloidal reflector is formed by rotating a parabola about its axis. Its surface must be a paraboloid of revolution so that rays emanating from the focus of the reflector are transformed into plane waves. The design is based on optical techniques, and it does not take into account any deformations (diffractions) from the rim of the reflector. Referring to Figure 15.10 and choosing a plane perpendicular to the axis of the reflector through the focus, it follows that OP + PQ = constant = 2f (15-12) Since OP = r′ (15-13) PQ = r′ cos 𝜃′ (15-12) can be written as r′(1 + cos 𝜃′) = 2f (15-14) or r′ = 2f 1 + cos 𝜃′ = f sec2 ( 𝜃′ 2 ) 𝜃≤𝜃0 (15-14a) 888 REFLECTOR ANTENNAS Figure 15.10 Two-dimensional configuration of a paraboloidal reflector.
Since a paraboloid is a parabola of revolution (about its axis), (15-14a) is also the equation of a paraboloid in terms of the spherical coordinates r′, 𝜃′, 𝜙′. Because of its rotational symmetry, there are no variations with respect to 𝜙′.
Equation (15-14a) can also be written in terms of the rectangular coordinates x′, y′, z′. That is, r′ + r′ cos 𝜃′ = √ (x′)2 + (y′)2 + (z′)2 + z′ = 2f (15-15) or (x′)2 + (y′)2 = 4f (f −z′) with (x′)2 + (y′)2 ≤(d∕2)2 (15-15a) In the analysis of parabolic reflectors, it is desirable to find a unit vector that is normal to the local tangent at the surface reflection point. To do this, (15-14a) is first expressed as f −r′ cos2 ( 𝜃′ 2 ) = S = 0 (15-16) and then a gradient is taken to form a normal to the surface. That is, N = ∇ [ f −r′ cos2 ( 𝜃′ 2 )] = ̂ a′ r 𝜕S 𝜕r′ + ̂ a′ 𝜃 1 r′ 𝜕S 𝜕𝜃′ = −̂ a′ r cos2 ( 𝜃′ 2 ) + ̂ a′ 𝜃cos ( 𝜃′ 2 ) sin ( 𝜃′ 2 ) (15-17) PARABOLIC REFLECTOR 889 A unit vector, normal to S, is formed from (15-17) as ̂ n = N |N| = −̂ a′ r cos ( 𝜃′ 2 ) + ̂ a′ 𝜃sin ( 𝜃′ 2 ) (15-18) To find the angle between the unit vector ̂ n which is normal to the surface at the reflection point, and a vector directed from the focus to the reflection point, we form cos 𝛼= −̂ a′ r⋅̂ n = −̂ a′ r ⋅ [ −̂ a′ r cos ( 𝜃′ 2 ) + ̂ a′ 𝜃sin ( 𝜃′ 2 )] = cos ( 𝜃′ 2 ) (15-19) In a similar manner we can find the angle between the unit vector ̂ n and the z-axis. That is, cos 𝛽= −̂ az ⋅̂ n = −̂ az ⋅ [ −̂ a′ r cos ( 𝜃′ 2 ) + ̂ a′ 𝜃sin ( 𝜃′ 2 )] (15-20) Using the transformation of (4-5), (15-20) can be written as cos 𝛽= −(̂ a′ r cos 𝜃′ −̂ a′ 𝜃sin 𝜃′) ⋅ [ −̂ a′ r cos ( 𝜃′ 2 ) + ̂ a′ 𝜃sin ( 𝜃′ 2 )] = cos ( 𝜃′ 2 ) (15-21) which is identical to 𝛼of (15-19). This is nothing more than a verification of Snell’s law of reflection at each differential area of the surface, which has been assumed to be flat locally.
Another expression that is usually very prominent in the analysis of reflectors is that relating the subtended angle 𝜃0 to the f/d ratio. From the geometry of Figure 15.10 𝜃0 = tan−1 (d∕2 z0 ) (15-22) where z0 is the distance along the axis of the reflector from the focal point to the edge of the rim.
From (15-15a) z0 = f −x0 2 + y0 2 4f = f −(d∕2)2 4f = f −d2 16f (15-23) Substituting (15-23) into (15-22) reduces it to 𝜃0 = tan−1 | | | | | | | | | d 2 f −d2 16f | | | | | | | | | = tan−1 | | | | | | | | | | 1 2 ( f d ) ( f d )2 −1 16 | | | | | | | | | | (15-24) 890 REFLECTOR ANTENNAS It can also be shown that another form of (15-24) is f = (d 4 ) cot (𝜃0 2 ) (15-25) B. Induced Current Density To determine the radiation characteristics (pattern, gain, efficiency, polarization, etc.) of a parabolic reflector, the current density induced on its surface must be known.
The current density Js can be determined by using Js = ̂ n × H = ̂ n × (Hi + Hr) (15-26) where Hi and Hr represent, respectively, the incident and reflected magnetic field components eval-uated at the surface of the conductor, and ̂ n is a unit vector normal to the surface. If the reflecting surface can be approximated by an infinite plane surface (this condition is met locally for a parabola), then by the method of images ̂ n × Hi = ̂ n × Hr (15-27) and (15-26) reduces to Js = ̂ n × (Hi + Hr) = 2̂ n × Hi = 2̂ n × Hr (15-28) The current density approximation of (15-28) is known as the physical-optics approximation, and it is valid when the transverse dimensions of the reflector, radius of curvature of the reflecting object, and the radius of curvature of the incident wave are large compared to a wavelength.
If the reflecting surface is in the far-field of the source generating the incident waves, then (15-28) can also be written as Js = 2̂ n × Hi ≃2 𝜂[̂ n × (̂ si × Ei)] (15-29) or Js = 2̂ n × Hr ≃2 𝜂[̂ n × (̂ sr × Er)] (15-29a) where 𝜂is the intrinsic impedance of the medium, ̂ si and ̂ sr are radial unit vectors along the ray paths of the incident and reflected waves (as shown in Figure 15.11), and Ei and Er are the incident and reflected electric fields.
C. Aperture Distribution Method It was pointed out earlier that the two most commonly used techniques in analyzing the radiation characteristics of reflectors are the aperture distribution and the current distribution methods.
For the aperture distribution method, the field reflected by the surface of the paraboloid is first found over a plane which is normal to the axis of the reflector. Geometrical optics techniques (ray tracing) are usually employed to accomplish this. In most cases, the plane is taken through the focal point, and it is designated as the aperture plane, as shown in Figure 15.12. Equivalent sources are PARABOLIC REFLECTOR 891 β α Γ S2 S1 st ds′ sr n ^ ^ ^ Figure 15.11 Reflecting surface with boundary Γ.
then formed over that plane. Usually it is assumed that the equivalent sources are zero outside the projected area of the reflector on the aperture plane. These equivalent sources are then used to com-pute the radiated fields utilizing the aperture techniques of Chapter 12.
For the current distribution method, the physical optics approximation of the induced current density Js given by (15-28) (Js ≃2̂ n × Hi where Hi is the incident magnetic field and ̂ n is a unit vector normal to the reflector surface) is formulated over the illuminated side of the reflector (S1) of Figure 15.11. This current density is then integrated over the surface of the reflector to yield the far-zone radiation fields.
For the reflector of Figure 15.11, approximations that are common to both methods are: 1. The current density is zero on the shadow side (S2) of the reflector.
2. The discontinuity of the current density over the rim (Γ) of the reflector is neglected.
3. Direct radiation from the feed and aperture blockage by the feed are neglected.
Figure 15.12 Three-dimensional geometry of a paraboloidal reflector system.
892 REFLECTOR ANTENNAS These approximations lead to accurate results, using either technique, for the radiated fields on the main beam and nearby minor lobes. To predict the pattern more accurately in all regions, espe-cially the far minor lobes, geometrical diffraction techniques – can be applied. Because of the level of the material, it will not be included here. The interested reader can refer to the literature.
The advantage of the aperture distribution method is that the integration over the aperture plane can be performed with equal ease for any feed pattern or feed position . The integration over the surface of the reflector as required for the current distribution method, becomes quite complex and time consuming when the feed pattern is asymmetrical and/or the feed is placed off-axis.
Let us assume that a y-polarized source with a gain function of Gf (𝜃′, 𝜙′) is placed at the focal point of a paraboloidal reflector. The radiation intensity of this source is given by U(𝜃′, 𝜙′) = Pt 4𝜋Gf (𝜃′, 𝜙′) (15-30) where Pt is the total radiated power. Referring to Figure 15.12, at a point r′ in the far-zone of the source U(𝜃′, 𝜙′) = 1 2Re[E◦(𝜃′, 𝜙′) × H◦∗(𝜃′, 𝜙′)] = 1 2𝜂|E◦(𝜃′, 𝜙′)|2 (15-31) or |E◦(𝜃′, 𝜙′)| = [2𝜂U(𝜃′, 𝜙′)]1∕2 = [ 𝜂Pt 2𝜋Gf (𝜃′, 𝜙′) ]1∕2 (15-31a) The incident field, with a direction perpendicular to the radial distance, can then be written as Ei(r′, 𝜃′, 𝜙′) = ̂ ei [√ 𝜇 𝜀 Pt 2𝜋Gf (𝜃′, 𝜙′) ]1∕2 e−jkr′ r′ = ̂ eiC1 √ Gf (𝜃′, 𝜙′)e−jkr′ r′ (15-32) C1 = (𝜇 𝜀 )1∕4 ( Pt 2𝜋 )1∕2 (15-32a) where ̂ ei is a unit vector perpendicular to ̂ a′ r and parallel to the plane formed by ̂ a′ r and ̂ ay, as shown in Figure 15.13.
It can be shown that on the surface of the reflector Js = 2 √𝜀 𝜇[̂ n × (̂ si × Ei)] = 2 √𝜀 𝜇C1 √ Gf (𝜃′, 𝜙′)e−jkr′ r′ u (15-33) where u = ̂ n × (̂ a′ r × ̂ ei) = (̂ n ⋅̂ ei)̂ a′ r −(̂ n ⋅̂ a′ r)̂ ei (15-33a) PARABOLIC REFLECTOR 893 Figure 15.13 Unit vector alignment for a paraboloidal reflector system.
which reduces to u = [ −̂ ax sin 𝜃′ sin ( 𝜃′ 2 ) sin 𝜙′ cos 𝜙′ + ̂ ay cos ( 𝜃′ 2 ) (sin2 𝜙′ cos 𝜃′ + cos2 𝜙′) −̂ az cos 𝜃′ sin 𝜙′ sin ( 𝜃′ 2 )] /√ 1 −sin2 𝜃′ sin2 𝜙′ (15-34) To find the aperture field Eap at the plane through the focal point, due to the reflector currents of (15-33), the reflected field Er at r′ (the reflection point) is first found. This is of the form Er = ̂ erC1 √ Gf (𝜃′, 𝜙′)e−jkr′ r′ (15-35) where ̂ er is a unit vector depicting the polarization of the reflected field. From (15-29a) Js = 2 √𝜀 𝜇[̂ n × (̂ sr × Er)] (15-36) Because ̂ sr = −̂ az, (15-36) can be written, using (15-35), as Js = 2 √𝜀 𝜇C1 √ Gf (𝜃′, 𝜙′)e−jkr′ r′ u (15-37) where u = ̂ n × (−̂ az × ̂ er) = −̂ az(̂ n ⋅̂ er) −̂ er cos ( 𝜃′ 2 ) (15-37a) 894 REFLECTOR ANTENNAS Since u in (15-37) and (15-37a) is the same as that of (15-33)–(15-34), it can be shown through some extensive mathematical manipulations that ̂ er = ̂ ax sin 𝜙′ cos 𝜙′(1 −cos 𝜃′) −̂ ay(sin2 𝜙′ cos 𝜃′ + cos2 𝜙′) √ 1 −sin2 𝜃′ sin2 𝜙′ (15-38) Thus the field Er at the point of reflection r′ is given by (15-35) where ̂ er is given by (15-38). At the plane passing through the focal point, the field is given by Eap = ̂ erC1 √ Gf (𝜃′, 𝜙′)e−jkr′(1+cos 𝜃′) r′ = ̂ axExa + ̂ ayEya (15-39) where Exa and Eya represent the x- and y-components of the reflected field over the aperture. Since the field from the reflector to the aperture plane is a plane wave, no correction in amplitude is needed to account for amplitude spreading.
Using the reflected electric field components (Exa and Eya) as given by (15-39), an equivalent is formed at the aperture plane. That is, J′ s = ̂ n × Ha = −̂ az × ( ̂ ax Eay 𝜂 −̂ ay Eax 𝜂 ) = −̂ ax Eax 𝜂 −̂ ay Eay 𝜂 (15-40a) M′ s = −̂ n × Ea = +̂ az × (̂ axEax + ̂ ayEay) = −̂ axEay + ̂ ayEax (15-40b) The radiated field can be computed using the (15-40a), (15-40b), and the formulations of Section 12.3. The integration is restricted only over the projected cross-sectional area S0 of the reflec-tor at the aperture plane shown dashed in Figure 15.12. That is, E𝜃s = jke−jkr 4𝜋r (1 −cos 𝜃) ∫∫ S0 (−Eax cos 𝜙−Eay sin 𝜙) × ejk(x′ sin 𝜃cos 𝜙+y′ sin 𝜃sin 𝜙) dx′ dy′ E𝜙s = jke−jkr 4𝜋r (1 −cos 𝜃) ∫∫ S0 (−Eax sin 𝜙+ Eay cos 𝜙) × ejk(x′ sin 𝜃cos 𝜙+y′ sin 𝜃sin 𝜙) dx′ dy′ (15-41a) (15-41b) The aperture distribution method has been used to compute, using efficient numerical integration techniques, the radiation patterns of paraboloidal and spherical reflectors. The fields given by (15-41a) and (15-41b) represent only the secondary pattern due to scattering from the reflector.
The total pattern of the system is represented by the sum of secondary pattern and the primary pattern of the feed element. For most feeds (such as horns), the primary pattern in the boresight (forward) direction of the reflector is of very low intensity and usually can be neglected.
To demonstrate the utility of the techniques, the principal E- and H-plane secondary patterns of a 35 GHz reflector, with an f∕d ≃0.82 [f = 8.062 in. (20.48 cm), d = 9.84 in. (24.99 cm)] and fed by a conical dual-mode horn, were computed and they are displayed in Figure 15.14. Since the feed horn has identical E- and H-plane patterns and the reflector is fed symmetrically, the reflector E- and H-plane patterns are also identical and do not possess any cross-polarized components.
To simultaneously display the field intensity associated with each point in the aperture plane of the reflector, a computer generated plot was developed . The field point locations, showing quantized contours of constant amplitude in the aperture plane, are illustrated in Figure 15.15. The PARABOLIC REFLECTOR 895 Figure 15.14 Principal E- or H-plane pattern of a symmetrical front-fed paraboloidal reflector. (Courtesy M. C. Bailey, NASA Langley Research Center).
Figure 15.15 Field point locations of constant amplitude contours in the aperture plane of a symmetrical front-fed paraboloidal reflector. (Courtesy M. C. Bailey, NASA Langley Research Center).
896 REFLECTOR ANTENNAS Figure 15.16 Principal (y-direction) and cross-polarization (x-direction) components of a paraboloidal reflec-tor. (source: S. Silver (ed.), Microwave Antenna Theory and Design (MIT Radiation Lab. Series, Vol. 12), McGraw-Hill, New York, 1949).
reflector system has an f∕d ≃0.82 with the same physical dimensions [ f = 8.062 in. (20.48 cm), d = 9.84 in. (24.99 cm)] and the same feed as the principal pattern of Figure 15.14. One symbol is used to represent the amplitude level of each 3-dB region. The field intensity within the bounds of the reflector aperture plane is within the 0–15 dB range.
D. Cross-Polarization The field reflected by the paraboloid, as represented by (15-35) and (15-38) of the aperture distribu-tion method, contains x- and y-polarized components when the incident field is y-polarized. The y-component is designated as the principal polarization and the x-component as the cross-polarization. This is illustrated in Figure 15.16. It is also evident that symmetrical (with respect to the principal planes) cross-polarized components are 180◦out of phase with one another. However for very narrow beam reflectors or for angles near the boresight axis (𝜃′ ≃0◦), the cross-polarized x-component diminishes and it vanishes on axis (𝜃′ = 0◦). A similar procedure can be used to show that for an incident x-polarized field, the reflecting surface decomposes the wave to a y-polarized field, in addition to its x-polarized component.
An interesting observation about the polarization phenomenon of a parabolic reflector can be made if we first assume that the feed element is an infinitesimal electric dipole (l ≪λ) with its length along the y-axis. For that feed, the field reflected by the reflector is given by (15-35) where from (4-114) C1 √ Gf (𝜃′, 𝜙′) = j𝜂kI0l 4𝜋sin 𝜓= j𝜂kI0l 4𝜋 √ 1 −cos2 𝜓 = j𝜂kI0l 4𝜋 √ 1 −sin2 𝜃′ sin2 𝜙′ (15-42) The angle 𝜓is measured from the y-axis toward the observation point.
When (15-42) is inserted in (15-35), we can write with the aid of (15-38) that Er = [̂ ax sin 𝜙′ cos 𝜙′(1 −cos 𝜃′) −̂ ay(sin2 𝜙′ cos 𝜃′ + cos2 𝜙′)] × j𝜂kI0l 4𝜋 e−jkr′ r′ (15-43) Now let us assume that an infinitesimal magnetic dipole, with its length along the x-axis (or a small loop with its area parallel to the y-z plane) and with a magnetic moment of −̂ axM, is placed at PARABOLIC REFLECTOR 897 Figure 15.17 Electric and magnetic dipole fields combined to form a Huygens’ source with ideal feed polar-ization for reflector. (source: A. W. Love, “Some Highlights in Reflector Antenna Development,” Radio Sci-ence, Vol. 11, Nos. 8, 9, August–September 1976).
the focal point and used as a feed. It can be shown – that the field reflected by the reflector has x- and y-components. However the x-component has a reverse sign to the x-component of the electric dipole feed. By adjusting the ratio of the electric to the magnetic dipole moments to be equal to √ 𝜇∕𝜀, the two cross-polarized reflected components (x-components) can be made equal in magnitude and for their sum to vanish (because of reverse signs). Thus a cross electric and magnetic dipole combination located at the focal point of a paraboloid can be used to induce currents on the surface of the reflector which are parallel everywhere. This is illustrated graphically in Figure 15.17.
The direction of the induced current flow determines the far-field polarization of the antenna. Thus for the crossed electric and magnetic dipole combination feed, the far-field radiation is free of cross-polarization. This type of feed is “ideal” in that it does not require that the surface of the reflector be solid but can be formed by closely spaced parallel conductors. Because of its ideal characteristics, it is usually referred to as a Huygens’source.
E. Current Distribution Method The current distribution method was introduced as a technique that can be used to better approxi-mate, as compared to the geometrical optics (ray-tracing) method, the field scattered from a surface.
Usually the main difficulty in applying this method is the approximation of the current density over the surface of the scatterer.
To analyze the reflector using this technique, we refer to the radiation integrals and auxiliary potential functions formulations of Chapter 3. While the two-step procedure of Figure 3.1 often simplifies the solution of most problems, the one-step formulation of Figure 3.1 is most convenient for the reflectors.
Using the potential function methods outlined in Chapter 3, and referring to the coordinate system of Figure 12.2(a), it can be shown that the E- and H-fields radiated by the sources J and M can be written as E = EA + EF = −j 1 4𝜋𝜔𝜀∫V [(J ⋅∇)∇+ k2J + j𝜔𝜀M × ∇]e−jkR R dv′ (15-44a) H = HA + HF = −j 1 4𝜋𝜔𝜇∫V [(M ⋅∇)∇+ k2M −j𝜔𝜇J × ∇]e−jkR R dv′ (15-44b) which for far-field observations reduce, according to the coordinate system of Figure 12.2(b), to E ≃−j 𝜔𝜇 4𝜋re−jkr ∫V [ J −(J ⋅̂ ar)̂ ar + √𝜀 𝜇M × ̂ ar ] e+jkr′⋅̂ ar dv′ (15-45a) H ≃−j 𝜔𝜀 4𝜋re−jkr ∫V [ M −(M ⋅̂ ar)̂ ar − √ 𝜇 𝜀J × ̂ ar ] e+jkr′⋅̂ ar dv′ (15-45b) 898 REFLECTOR ANTENNAS Figure 15.18 Geometrical arrangement of reflecting surface.
If the current distributions are induced by electric and magnetic fields incident on a perfect elec-tric conducting (𝜎= ∞) surface shown in Figure 15.18, the fields created by these currents are referred to as scattered fields. If the conducting surface is closed, the far-zone fields are obtained from (15-45a) and (15-45b) by letting M = 0 and reducing the volume integral to a surface integral with the surface current density J replaced by the linear current density Js. Thus Es = −j 𝜔𝜇 4𝜋r e−jkr ∯ S [Js −(Js⋅̂ ar)̂ ar]e+jkr′⋅̂ ar ds′ Hs = +j 𝜔√𝜇𝜀 4𝜋r e−jkr ∯ S [Js × ̂ ar]e+jkr′⋅̂ ar ds′ (15-46a) (15-46b) The electric and magnetic fields scattered by the closed surface of the reflector of Figure 15.11, and given by (15-46a) and (15-46b), are valid provided the source-density functions (current and charge) satisfy the equation of continuity. This would be satisfied if the scattering object is a smooth closed surface. For the geometry of Figure 15.11, the current distribution is discontinuous across the boundary Γ (being zero over the shadow area S2) which divides the illuminated (S1) and shadow (S2) areas. It can be shown that the equation of continuity can be satisfied if an appropriate line-source distribution of charge is introduced along the boundary Γ. Therefore the total scattered field would be the sum of the (1) surface currents over the illuminated area, (2) surface charges over the illuminated area, and (3) line-charge distribution over the boundary Γ.
The contributions from the surface charge density are taken into account by the current distri-bution through the equation of continuity. However it can be shown that in the far-zone the contribution due to line-charge distribution cancels out the longitudinal component introduced by the surface current and charge distributions. Since in the far-zone the field components are pre-dominantly transverse, the contribution due to the line-charge distribution need not be included and (15-46a)–(15-46b) can be applied to an open surface.
In this section, (15-46a) and (15-46b) will be used to calculate the field scattered from the surface of a parabolic reflector. Generally the field radiated by the currents on the shadow region of the reflector is very small compared to the total field, and the currents and field can be set equal to zero.
The field scattered by the illuminated (concave) side of the parabolic reflector can be formulated, using the current distribution method, by (15-46a) and (15-46b) when the integration is restricted over the illuminated area.
PARABOLIC REFLECTOR 899 The total field of the system can be obtained by a superposition of the radiation from the primary source in directions greater than 𝜃0(𝜃> 𝜃0) and that scattered by the surface as obtained by using either the aperture distribution or the current distribution method.
Generally edge effects are neglected. However the inclusion of diffracted fields – from the rim of the reflector not only introduce fields in the shadow region of the reflector, but also mod-ify those present in the transition and lit regions. Any discontinuities introduced by geometrical optics methods along the transition region (between lit and shadow regions) are removed by the diffracted components.
The far-zone electric field of a parabolic reflector, neglecting the direct radiation, is given by (15-46a). When expanded, (15-46a) reduces, by referring to the geometry of Figure 15.18, to the two components of E𝜃= −j 𝜔𝜇 4𝜋re−jkr ∫∫ S1 ̂ a𝜃⋅Jse+jkr′⋅̂ ar ds′ E𝜙= −j 𝜔𝜇 4𝜋re−jkr ∫∫ S1 ̂ a𝜙⋅Jse+jkr′⋅̂ ar ds′ (15-47a) (15-47b) According to the geometry of Figure 15.19 ds′ = dW dN = (r′ sin 𝜃′ d𝜙′) [ r′ sec ( 𝜃′ 2 ) d𝜃′ ] = (r′)2 sin 𝜃′ sec ( 𝜃′ 2 ) d𝜃′ d𝜙′ (15-48) since dW = r′ sin 𝜃′ d𝜙′ (15-48a) dH = −̂ a′ r ⋅dN = −̂ a′ r⋅̂ n dN = −̂ a′ r ⋅ [ −̂ a′ r cos ( 𝜃′ 2 ) + ̂ a′ 𝜃sin ( 𝜃′ 2 )] dN = cos ( 𝜃′ 2 ) dN (15-48b) dN = sec ( 𝜃′ 2 ) dH = sec ( 𝜃′ 2 ) r′ d𝜃′ = r′ sec ( 𝜃′ 2 ) d𝜃′ (15-48c) Therefore, it can be shown that (15-47a) and (15-47b) can be expressed, with the aid of (15-37), (15-37a), and (15-48), as [ E𝜃 E𝜙 ] = −j 𝜔𝜇 2𝜋r √𝜀 𝜇C1e−jkr [ ̂ a𝜃⋅I ̂ a𝜙⋅I ] = −j𝜔𝜇e−jkr 2𝜋r [√𝜀 𝜇 Pt 2𝜋 ]1∕2 [ ̂ a𝜃⋅I ̂ a𝜙⋅I ] (15-49) where I = It + Iz (15-49a) 900 REFLECTOR ANTENNAS Figure 15.19 Projected cross section and side view of reflector.
It = −∫ 2𝜋 0 ∫ 𝜃0 0 ̂ er cos ( 𝜃′ 2 ) √ Gf (𝜃′, 𝜙′) r′ e−jkr′[1−sin 𝜃′ sin 𝜃cos(𝜙′−𝜙)−cos 𝜃′ cos 𝜃] × (r′)2 sin 𝜃′ sec ( 𝜃′ 2 ) d𝜃′ d𝜙′ (15-49b) Iz = −̂ az ∫ 2𝜋 0 ∫ 𝜃0 0 (̂ n ⋅̂ er) √ Gf (𝜃′, 𝜙′) r′ e−jkr′[1−sin 𝜃′ sin 𝜃cos(𝜙′−𝜙)−cos 𝜃′ cos 𝜃] × (r′)2 sin 𝜃′ sec ( 𝜃′ 2 ) d𝜃′ d𝜙′ (15-49c) By comparing (15-49) with (15-35), the radiated field components formulated by the aperture distribution and current distribution methods lead to similar results provided the Iz contribution of (15-49c) is neglected. As the ratio of the aperture diameter to wavelength (d∕λ) increases, the cur-rent distribution method results reduce to those of the aperture distribution and the angular pattern becomes more narrow.
PARABOLIC REFLECTOR 901 For variations near the 𝜃= 𝜋region, the Iz contribution becomes negligible because ̂ a𝜃⋅[−̂ az(̂ n ⋅̂ er)] = [̂ ax sin 𝜃cos 𝜙+ ̂ ay cos 𝜃sin 𝜙−̂ az sin 𝜃] ⋅[−̂ az(̂ n ⋅̂ er)] = (̂ n ⋅̂ er) sin 𝜃 (15-50a) ̂ a𝜙⋅[−̂ az(̂ n ⋅̂ er)] = [−̂ ax sin 𝜙+ ̂ ay cos 𝜙] ⋅[−̂ az(̂ n ⋅̂ er)] = 0 (15-50b) F. Directivity and Aperture Efficiency In the design of antennas, the directivity is a very important figure of merit. The purpose of this section will be to examine the dependence of the directivity and aperture efficiency on the primary-feed pattern Gf (𝜃′, 𝜙′) and f/d ratio (or the included angle 2𝜃0) of the reflector. To simplify the analysis, it will be assumed that the feed pattern Gf (𝜃′, 𝜙′) is circularly symmetric (not a function of 𝜙′) and that Gf (𝜃′) = 0 for 𝜃′ > 90◦.
The secondary pattern (formed by the surface of the reflector) is given by (15-49). Approximating the I of (15-49a) by It, the total E-field in the 𝜃= 𝜋direction is given by either E𝜃or E𝜙of (15-49).
Assuming the feed is circularly symmetric, linearly polarized in the y-direction, and by neglecting cross-polarized contributions, it can be shown with the aid of (15-14a) that (15-49) reduces to E(r, 𝜃= 𝜋) = −j2𝜔𝜇f r [√𝜀 𝜇 Pt 2𝜋 ]1∕2 e−jk(r+2f) ∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′ (15-51) The power intensity (power/unit solid angle) in the forward direction U(𝜃= 𝜋) is given by U(𝜃= 𝜋) = 1 2r2 √𝜀 𝜇|E(r, 𝜃= 𝜋)|2 (15-52) which by using (15-51) reduces to U(𝜃= 𝜋) = 16𝜋2 λ2 f 2 Pt 4𝜋 | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 (15-52a) The antenna directivity in the forward direction can be written, using (15-52a), as D0 = 4𝜋U(𝜃= 𝜋) Pt = U(𝜃= 𝜋) Pt∕4𝜋 = 16𝜋2 λ2 f 2 | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 (15-53) The focal length is related to the angular spectrum and aperture diameter d by (15-25). Thus (15-53) reduces to D0 = (𝜋d λ )2 { cot2 (𝜃0 2 ) | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2} (15-54) The factor (𝜋d∕λ)2 is the directivity of a uniformly illuminated constant phase aperture; the remain-ing part is the aperture efficiency defined as 𝜀ap = cot2 (𝜃0 2 ) | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 (15-55) 902 REFLECTOR ANTENNAS It is apparent by examining (15-55) that the aperture efficiency is a function of the subtended angle (𝜃0) and the feed pattern Gf (𝜃′) of the reflector. Thus for a given feed pattern, all paraboloids with the same f/d ratio have identical aperture efficiency.
To illustrate the variation of the aperture efficiency as a function of the feed pattern and the angular extent of the reflector, Silver considered a class of feeds whose patterns are defined by Gf (𝜃′) = ⎧ ⎪ ⎨ ⎪ ⎩ G0 (n) cosn(𝜃′) 0 ≤𝜃′ ≤𝜋∕2 0 𝜋∕2 < 𝜃′ ≤𝜋 (15-56) where G0 (n) is a constant for a given value of n. Although idealistic, these patterns were chosen because (1) closed form solutions can be obtained, and (2) they often are used to represent a major part of the main lobe of many practical antennas. The intensity in the back region (𝜋∕2 < 𝜃′ ≤𝜋) was assumed to be zero in order to avoid interference between the direct radiation from the feed and scattered radiation from the reflector.
The constant G0 (n) can be determined from the relation ∯ S Gf (𝜃′) dΩ = ∯ S Gf (𝜃′) sin 𝜃′ d𝜃′ d𝜙′ = 4𝜋 (15-57) which for (15-56) becomes G0 (n) ∫ 𝜋∕2 0 cosn 𝜃′ sin 𝜃′ d𝜃′ = 2 ➱G0 (n) = 2(n + 1) (15-58) Substituting (15-56) and (15-58) into (15-55) leads, for the even values of n = 2 through n = 8, to 𝜀ap(n = 2) = 24 { sin2 (𝜃0 2 ) + ln [ cos (𝜃0 2 )]}2 cot2 (𝜃0 2 ) (15-59a) 𝜀ap(n = 4) = 40 { sin4 (𝜃0 2 ) + ln [ cos (𝜃0 2 )]}2 cot2 (𝜃0 2 ) (15-59b) 𝜀ap(n = 6) = 14 { 2 ln [ cos (𝜃0 2 )] + [1 −cos(𝜃0)]3 3 + 1 2 sin2(𝜃0) }2 cot2 (𝜃0 2 ) (15-59c) 𝜀ap(n = 8) = 18 {1 −cos4(𝜃0) 4 −2 ln [ cos (𝜃0 2 )] −[1 −cos(𝜃0)]3 3 −1 2 sin2(𝜃0) }2 cot2 (𝜃0 2 ) (15-59d) The variations of (15-59a)–(15-59d), as a function of the angular aperture of the reflector 𝜃0 or the f/d ratio, are shown plotted in Figure 15.20. It is apparent, from the graphical illustration, that for a given feed pattern (n = constant) 1. There is only one reflector with a given angular aperture or f/d ratio which leads to a maximum aperture efficiency.
2. Each maximum aperture efficiency is in the neighborhood of 82–83%.
PARABOLIC REFLECTOR 903 Figure 15.20 Aperture, and taper and spillover efficiencies as a function of the reflector half-angle 𝜃0 (or f/d ratio) for different feed patterns.
3. Each maximum aperture efficiency, for any one of the given patterns, is almost the same as that of any of the others.
4. As the feed pattern becomes more directive (n increases), the angular aperture of the reflector that leads to the maximum efficiency is smaller.
The aperture efficiency is generally the product of the 1. fraction of the total power that is radiated by the feed, intercepted, and collimated by the reflecting surface (generally known as spillover efficiency 𝜀s) 2. uniformity of the amplitude distribution of the feed pattern over the surface of the reflector (generally known as taper efficiency 𝜀t) 904 REFLECTOR ANTENNAS 3. phase uniformity of the field over the aperture plane (generally known as phase efficiency 𝜀p) 4. polarization uniformity of the field over the aperture plane (generally known as polarization efficiency 𝜀x) 5. blockage efficiency 𝜀b 6. random error efficiency 𝜀r over the reflector surface Thus in general 𝜀ap = 𝜀s𝜀t𝜀p𝜀x𝜀b𝜀r (15-60) For feeds with symmetrical patterns 𝜀s = ∫ 𝜃0 0 Gf (𝜃′) sin 𝜃′ d𝜃′ ∫ 𝜋 0 Gf (𝜃′) sin 𝜃′ d𝜃′ (15-61) 𝜀t = 2 cot2 (𝜃0 2 ) | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 ∫ 𝜃0 0 Gf (𝜃′) sin 𝜃′ d𝜃′ (15-62) which by using (15-25) can also be written as 𝜀t = 32 ( f d )2 | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 ∫ 𝜃0 0 Gf (𝜃′) sin 𝜃′ d𝜃′ (15-62a) Thus 1. 100(1 −𝜀s) = percent power loss due to energy from feed spilling past the main reflector.
2. 100(1 −𝜀t) = percent power loss due to nonuniform amplitude distribution over the reflec-tor surface.
3. 100(1 −𝜀p) = percent power loss if the field over the aperture plane is not in phase everywhere.
4. 100(1 −𝜀x) = percent power loss if there are cross-polarized fields over the antenna aper-ture plane.
5. 100(1 −𝜀b) = percent power loss due to blockage provided by the feed or supporting struts (also by subreflector for a dual reflector).
6. 100(1 −𝜀r) = percent power loss due to random errors over the reflector surface.
An additional factor that reduces the antenna gain is the attenuation in the antenna feed and associated transmission line.
PARABOLIC REFLECTOR 905 For feeds with 1. symmetrical patterns 2. aligned phase centers 3. no cross-polarized field components 4. no blockage 5. no random surface error the two main factors that contribute to the aperture efficiency are the spillover and nonuniform ampli-tude distribution losses. Because these losses depend primarily on the feed pattern, a compromise between spillover and taper efficiency must emerge. Very high spillover efficiency can be achieved by a narrow beam pattern with low minor lobes at the expense of a very low taper efficiency.
Example 15.1 Show that a parabolic reflector using a point source as a feed, as shown in Figure 15.21, creates an amplitude taper proportional to ( f r′ ) = cos4 ( 𝜃′ 2 ) and that the feed pattern has to be equal to Gf (𝜃′) = G0 sec4 ( 𝜃′ 2 ) = sec4 ( 𝜃′ 2 ) to produce uniform illumination.
Solution: The field radiated by a point source varies as e−jkr r Therefore, a field radiated by a point source at the focal point of a parabolic reflector, which acts as a feed, creates an amplitude taper across the reflector due to the space factor (r′∕f), by referring to Figure 15.21, which compares the field at the vertex to that at any other point 𝜃′ on the reflector surface.
According to (15-14) or (15-14a) r′ f = 2 1 + cos(𝜃′) = sec2 ( 𝜃′ 2 ) ⇒ ( f r′ ) = 1 sec2(𝜃′∕2) = cos ( 𝜃′ 2 ) The power amplitude taper created across the reflector is therefore ( f r′ )2 = cos4 ( 𝜃′ 2 ) To compensate for this and obtain uniform illumination, the feed power must be Gf (𝜃′) = 1 cos4(𝜃′∕2) = sec4 ( 𝜃′ 2 ) , 𝜃′ ≤𝜃0 906 REFLECTOR ANTENNAS 0 f f r′ Δr 0 θ θ ′ Figure 15.21 Geometry of a parabola (solid) and that of a circle (dashed) for aperture illumination.
Through Example 15.1, uniform illumination and ideal taper, spillover and aperture efficiencies can be obtained when the feed power pattern is Gf (𝜃′) = ⎧ ⎪ ⎨ ⎪ ⎩ G0 sec4 ( 𝜃′ 2 ) 0 ≤𝜃′ ≤𝜃0 0 𝜃′ > 𝜃0 (15-63) whose normalized distribution is shown plotted in Figure 15.22. To accomplish this, the necessary value of G0 is derived in Example 15.2. Although such a pattern is “ideal” and impractical to achieve, much effort has been devoted to develop feed designs which attempt to approximate it .
Figure 15.22 Normalized gain pattern of feed for uniform amplitude illumination of paraboloidal reflector with a total subtended angle of 80◦.
PARABOLIC REFLECTOR 907 Example 15.2 Show that a feed pattern of Gf (𝜃′) = ⎧ ⎪ ⎨ ⎪ ⎩ G0 sec4 ( 𝜃′ 2 ) 0 ≤𝜃′ ≤𝜃0 0 𝜃′ > 𝜃0 in conjunction with a parabolic reflector, leads to an ideal aperture efficiency of 𝜀ap = 1. Deter-mine the value of G0 that will accomplish this.
Solution: The aperture efficiency of (15-55) reduces, using the given feed pattern, to 𝜀ap = cot2 (𝜃0 2 ) | | | | |∫ 𝜃0 0 √ Gf (𝜃′) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 = cot2 (𝜃0 2 ) G0 | | | | |∫ 𝜃0 0 sec2 ( 𝜃′ 2 ) tan ( 𝜃′ 2 ) d𝜃′| | | | | 2 𝜀ap = cot2 (𝜃0 2 ) G0|I|2 where I = ∫ 𝜃0 0 sec2 ( 𝜃′ 2 ) tan ( 𝜃′ 2 ) d𝜃′ = ∫ 𝜃0 0 1 cos2(𝜃′∕2) sin(𝜃′∕2) cos(𝜃′∕2) d𝜃′ = ∫ 𝜃0 0 sin(𝜃′∕2) cos3(𝜃′∕2) d𝜃′ I = [ 1 cos2(𝜃0∕2) −1 ] = [ sec2 (𝜃0 2 ) −1 ] = tan2 (𝜃0 2 ) Thus 𝜀ap = G0 cot2 (𝜃0 2 ) tan4 (𝜃0 2 ) = G0 tan2 (𝜃0 2 ) To insure also an ideal spillover efficiency (𝜀s = 1), all the power that is radiated by the feed must be contained within the angular region subtended by the edges of the reflector. Using (15-57), and assuming symmetry in the pattern with respect to 𝜙, we can write that ∫ 2𝜋 0 ∫ 𝜋 0 Gf (𝜃′) sin 𝜃′ d𝜃′ d𝜙′ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ dΩ = 2𝜋∫ 𝜋 0 Gf (𝜃′) sin 𝜃′ d𝜃′ = 4𝜋 ∫ 𝜋 0 Gf (𝜃′) sin 𝜃′ d𝜃′ = 2 Using the given feed pattern Gf (𝜃′) = G0 sec4 ( 𝜃′ 2 ) 0 ≤𝜃′ ≤𝜃0 908 REFLECTOR ANTENNAS we can write that ∫ 𝜃0 0 G0 sec4 ( 𝜃′ 2 ) sin 𝜃′ d𝜃′ = G0 ∫ 𝜃0 0 sec4 ( 𝜃′ 2 ) sin 𝜃′ d𝜃′ = 2 G0 = 2 ∫𝜃0 0 sec4(𝜃′∕2) sin 𝜃′ d𝜃′ = 2 2 tan2(𝜃0∕2) = cot2 (𝜃0 2 ) Thus the total aperture efficiency, taking into account uniform illumination and total spillover control, is 𝜀ap = G0 cot2 (𝜃0 2 ) tan4 (𝜃0 2 ) = G0 tan2 (𝜃0 2 ) = cot2 (𝜃0 2 ) tan2 (𝜃0 2 ) = 1 To develop guidelines for designing practical feeds which result in high aperture efficiencies, it is instructive to examine the relative field strength at the edges of the reflector’s bounds (𝜃′ = 𝜃0) for patterns that lead to optimum efficiencies. For the patterns of (15-56), when used with reflectors that result in optimum efficiencies as demonstrated graphically in Figure 15.20, the relative field strength at the edges of their angular bounds (𝜃′ = 𝜃0) is shown plotted in Figure 15.23. Thus for n = 2 the field strength of the pattern at 𝜃′ = 𝜃0 is 8 dB down from the maximum. As the pattern becomes more narrow (n increases), the relative field strength at the edges for maximum efficiency is further reduced as illustrated in Figure 15.20. Since for n = 2 through n = 10 the field strength is between 8 to 10.5 dB down, for most practical feeds the figure used is 9–10 dB.
Figure 15.23 Relative field strength of feed pattern along reflector edge bounds as a function of primary-feed pattern number (cosn 𝜃). (source: S. Silver (ed.), Microwave Antenna Theory and Design (MIT Radiation Lab.
Series, Vol. 12), McGraw-Hill, New York, 1949).
PARABOLIC REFLECTOR 909 TABLE 15.1 Aperture Efficiency and Field Strength at the Edge of Reflector, Relative to That at the Vertex, due to the Feed Pattern (Feed) and Path Length (Path) between the Edge and the Vertex Feed (dB) Path (dB) Total (dB) n 𝜺ap 𝜽0 (deg) f∕d 10 log[G(𝜽0)] 10 log(r′∕f)2 Feed + Path 2 0.829 66 0.385 −7.8137 −3.056 −10.87 4 0.8196 53.6 0.496 −8.8215 −1.959 −10.942 6 0.8171 46.2 0.5861 −9.4937 −1.439 −10.93 8 0.8161 41.2 0.6651 −9.7776 1.137 −10.914 Another parameter to examine for the patterns of (15-56), when used with reflectors that lead to optimum efficiency, is the amplitude taper or illumination of the main aperture of the reflector which is defined as the ratio of the field strength at the edge of the reflector surface to that at the vertex.
The aperture illumination is a function of the feed pattern and the f/d ratio of the reflector. To obtain that, the ratio of the angular variation of the pattern toward the two points [Gf (𝜃′ = 0)∕Gf (𝜃′ = 𝜃0)] is multiplied by the space attenuation factor (r0∕f)2, where f is the focal distance of the reflector and r0 is the distance from the focal point to the edge of the reflector. For each of the patterns, the reflector edge illumination for maximum efficiency is nearly 11 dB down from that at the vertex.
The details for 2 ≤n ≤8 are shown in Table 15.1.
The results obtained with the idealized patterns of (15-56) should only be taken as typical, because it was assumed that 1. the field intensity for 𝜃′ > 90◦was zero 2. the feed was placed at the phase center of the system 3. the patterns were symmetrical 4. there were no cross-polarized field components 5. there was no blockage 6. there were no random errors at the surface of the reflector Each factor can have a significant effect on the efficiency, and each has received much attention which is well documented in the open literature .
In practice, maximum reflector efficiencies are in the 65–80% range. To demonstrate that, paraboloidal reflector efficiencies for square corrugated horns feeds were computed, and they are shown plotted in Figure 15.24. The corresponding amplitude taper and spillover efficiencies for the aperture efficiencies of Figures 15.20(a) and 15.24 are displayed, respectively, in Figures 15.20(b) and 15.25. For the data of Figures 15.24 and 15.25, each horn had aperture dimensions of 8λ × 8λ, their patterns were assumed to be symmetrical (by averaging the E- and H-planes), and they were computed using the techniques of Section 13.6. From the plotted data, it is apparent that the maxi-mum aperture efficiency for each feed pattern is in the range of 74–79%, and that the product of the taper and spillover efficiencies is approximately equal to the total aperture efficiency.
We would be remiss if we left the discussion of this section without reporting the gain of some of the largest reflectors that exist around the world . The gains are shown in Figure 15.26 and include the 1,000-ft (305-m) diameter spherical reflector at Arecibo, Puerto Rico, the 100-m radio telescope at Effelsberg, West Germany, the 64-m reflector at Goldstone, California, the 22-m reflector at Krim, USSR, and the 12-m telescope at Kitt Peak, Arizona. The dashed portions of the curves indicate extrapolated values. For the Arecibo reflector, two curves are shown. The 215-m diameter curve is for a reduced aperture of the large reflector (305-m) for which a line feed at 1,415 MHz was designed .
910 REFLECTOR ANTENNAS Figure 15.24 Parabolic reflector aperture efficiency as a function of angular aperture for 8λ × 8λ square cor-rugated horn feed with total flare angles of 2𝜓0 = 70◦, 85◦, and 100◦.
G. Phase Errors Any departure of the phase, over the aperture of the antenna, from uniform can lead to a significant diminution of its directivity . For a paraboloidal reflector system, phase errors result from 1. displacement (defocusing) of the feed phase center from the focal point 2. deviation of the reflector surface from a parabolic shape or random errors at the surface of the reflector 3. departure of the feed wavefronts from spherical shape Figure 15.25 Parabolic reflector taper and spillover efficiencies as a function of reflector aperture for different corrugated horn feeds.
PARABOLIC REFLECTOR 911 Figure 15.26 Gains of some worldwide large reflector antennas. (source: A. W. Love, “Some Highlights in Reflector Antenna Development,” Radio Science, Vol. 11, Nos. 8, 9, August–September 1976).
The defocusing effect can be reduced by first locating the phase center of the feed antenna and then placing it at the focal point of the reflector. In Chapter 13 (Section 13.10) it was shown that the phase center for horn antennas, which are widely utilized as feeds for reflectors, is located between the aperture of the horn and the apex formed by the intersection of the inclined walls of the horn .
Very simple expressions have been derived to predict the loss in directivity for rectangular and circular apertures when the peak values of the aperture phase deviation is known. When the phase errors are assumed to be relatively small, it is not necessary to know the exact amplitude or phase distribution function over the aperture.
Assuming the maximum radiation occurs along the axis of the reflector, and that the maximum phase deviation over the aperture of the reflector can be represented by |Δ𝜙(𝜌)| = |𝜙(𝜌) −𝜙(𝜌)| ≤m, 𝜌≤a (15-64) where 𝜙(𝜌) is the aperture phase function and 𝜙(𝜌) is its average value, then the ratio of the directivity with (D) and without (D0) phase errors can be written as D D0 = directivity with phase error directivity without phase error ≥ ( 1 −m2 2 )2 (15-65) and the maximum fractional reduction in directivity as ΔD D0 = D0 −D D0 ≤m2 ( 1 −m2 4 ) (15-66) Relatively simple expressions have also been derived to compute the maximum possible change in half-power beamwidth.
912 REFLECTOR ANTENNAS Example 15.3 A 10-m diameter reflector, with an f∕d ratio of 0.5, is operating at f = 3 GHz. The reflector is fed with an antenna whose primary pattern is symmetrical and which can be approximated by Gf (𝜃′) = 6 cos2 𝜃′. Find its (a) aperture efficiency (b) overall directivity (c) spillover and taper efficiencies (d) directivity when the maximum aperture phase deviation is 𝜋∕8 rad Solution: Using (15-24), half of the subtended angle of the reflector is equal to 𝜃0 = tan−1 ⎡ ⎢ ⎢ ⎢ ⎣ 0.5(0.5) (0.5)2 −1 16 ⎤ ⎥ ⎥ ⎥ ⎦ = 53.13◦ (a) The aperture efficiency is obtained using (15-59a). Thus 𝜀ap = 24{sin2(26.57◦) + ln[cos(26.57◦)]}2 cot2(26.57◦) = 0.75 = 75% which agrees with the data of Figure 15.20.
(b) The overall directivity is obtained by (15-54), or D = 0.75[𝜋(100)]2 = 74,022.03 = 48.69 dB (c) The spillover efficiency is computed using (15-61) where the upper limit of the integral in the denominator has been replaced by 𝜋∕2. Thus 𝜀s = ∫ 53.13◦ 0 cos2 𝜃′ sin 𝜃′ d𝜃′ ∫ 90◦ 0 cos2 𝜃′ sin 𝜃′ d𝜃′ = 2 cos3 𝜃′|53.13◦ 0 2 cos3 𝜃′|90◦ 0 = 0.784 = 78.4% In a similar manner, the taper efficiency is computed using (15-62). Since the numerator in (15-62) is identical in form to the aperture efficiency of (15-55), the taper efficiency can be found by multiplying (15-59a) by 2 and dividing by the denominator of (15-62). Thus 𝜀t = 2(0.75) 1.568 = 0.9566 = 95.66% The product of 𝜀s and 𝜀t is equal to 𝜀s𝜀t = 0.784(0.9566) = 0.75 and it is identical to the total aperture efficiency computed above.
PARABOLIC REFLECTOR 913 (d) The directivity for a maximum phase error of m = 𝜋∕8 = 0.3927 rad can be computed using (15-65). Thus D D0 ≥ ( 1 −m2 2 )2 = [ 1 −(0.3927)2 2 ]2 = 0.8517 = −0.69 dB or D ≥0.8517D0 = 0.8517(74,022.03) = 63,046.94 = 48.0 dB.
Surface roughness effects on the directivity of the antenna were first examined by Ruze where he indicated that for any reflector antenna there is a wavelength (λmax) at which the directivity reaches a maximum. This wavelength depends on the RMS deviation (𝜎) of the reflector surface from an ideal paraboloid. For a random roughness of Gaussian distribution, with correlation interval large compared to the wavelength, they are related by λmax = 4𝜋𝜎 (15-67) Thus the directivity of the antenna, given by (15-54), is modified to include surface roughness and can be written as D = (𝜋d λ )2 𝜀ape−(4𝜋𝜎∕λ)2 (15-68) Using (15-67), the maximum directivity of (15-68) can be written as Dmax = 102q𝜀ap ( e−1 16 ) (15-69) where q is the index of smoothness defined by d 𝜎= 10+q (15-70) In decibels, (15-69) reduces to Dmax(dB) = 20q −16.38 + 10 log10(𝜀ap) (15-71) For an aperture efficiency of unity (𝜀ap = 1), the directivity of (15-68) is plotted in Figure 15.27, as a function of (d∕λ), for values of q = 3.5, 4.0, and 4.5. It is apparent that for each value of q and a given reflector diameter d, there is a maximum wavelength where the directivity reaches a maximum value. This maximum wavelength is given by (15-67).
H. Feed Design The widespread use of paraboloidal reflectors has stimulated interest in the development of feeds to improve the aperture efficiency and to provide greater discrimination against noise radiation from the ground. This can be accomplished by developing design techniques that permit the synthesis of feed patterns with any desired distribution over the bounds of the reflector, rapid cutoff at its edges, and very low minor lobes in all the other space. In recent years, the two main problems that concerned feed designers were aperture efficiency and low cross-polarization.
914 REFLECTOR ANTENNAS Figure 15.27 Reflector surface roughness effects on antenna directivity. (source: A. W. Love, “Some High-lights in Reflector Antenna Development,” Radio Science, Vol. 11, Nos. 8, 9, August–September 1976).
In the receiving mode, an ideal feed and a matched load would be one that would absorb all the energy intercepted by the aperture when uniform and linearly polarized plane waves are normally incident upon it. The feed field structure must be made to match the focal region field structure formed by the reflecting, scattering, and diffracting characteristics of the reflector. By reciprocity, an ideal feed in the transmitting mode would be one that radiates only within the solid angle of the aperture and establishes within it an identical outward traveling wave. For this ideal feed system, the transmitting and receiving mode field structures within the focal region are identical with only the direction of propagation reversed.
An optical analysis indicates that the focal region fields, formed by the reflection of linearly polar-ized plane waves incident normally on an axially symmetric reflector, are represented by the well-known Airy rings described mathematically by the amplitude distribution intensity of [2J1(u)∕u]2.
This description is incomplete, because it is a scalar solution, and it does not take into account polar-ization effects. In addition, it is valid only for reflectors with large f/d ratios, which are commonly used in optical systems, and it would be significantly in error for reflectors with f/d ratios of 0.25 to 0.5, which are commonly used in microwave applications.
A vector solution has been developed which indicates that the fields near the focal region can be expressed by hybrid TE and TM modes propagating along the axis of the reflector. This repre-sentation provides a clear physical picture for the understanding of the focal region field formation.
The boundary conditions of the hybrid modes indicate that these field structures can be represented by a spectrum of hybrid waves that are linear combinations of TE1n and TM1n modes of circular waveguides.
A single hollow pipe cannot simultaneously satisfy both the TE and TM modes because of the different radial periodicities. However, it has been shown that λ∕4 deep annular slots on the inner surface of a circular pipe force the boundary conditions on E and H to be the same and provide a single anisotropic reactance surface which satisfies the boundary conditions on both TE and TM modes. This provided the genesis of hybrid mode waveguide radiators and corrugated horns . Corrugated horns, whose aperture size and flare angle are such that at least 180◦phase error over their aperture is assured, are known as “scalar” horns . Design data for uncorrugated horns that can be used to maximize the aperture efficiency or to produce maximum power transmission to the feed have been calculated and are shown in graphical form in Figure 15.28.
PARABOLIC REFLECTOR 915 Figure 15.28 Optimum pyramidal horn dimensions versus f/d ratio for various horn lengths.
A FORTRAN software designated Reflector for computer-aided analysis and design of reflector antennas has been developed . A converted MATLAB version of Reflector is included in the publisher’s website for this book. The program computes the radiation of a parabolic reflector. In addition it provides spatial and spectral methods to compute radiation due to an aperture distribution.
The software package can also be used to investigate the directivity, sidelobe level, polarization, and near-to-far-zone fields.
15.4.2 Cassegrain Reflectors To improve the performance of large ground-based microwave reflector antennas for satellite track-ing and communication, it has been proposed that a two-reflector system be utilized. The arrange-ment suggested was the Cassegrain dual-reflector system of Figure 15.1(d), which was often utilized in the design of optical telescopes and it was named after its inventor. To achieve the desired collimation characteristics, the larger (main) reflector must be a paraboloid and the smaller (sec-ondary) a hyperboloid. The use of a second reflector, which is usually referred to as the subreflector or subdish, gives an additional degree of freedom for achieving good performance in a number of different applications. For an accurate description of its performance, diffraction techniques must be used to take into account diffractions from the edges of the subreflector, especially when its diameter is small .
In general, the Cassegrain arrangement provides a variety of benefits, such as the 1. ability to place the feed in a convenient location 2. reduction of spillover and minor lobe radiation 3. ability to obtain an equivalent focal length much greater than the physical length 4. capability for scanning and/or broadening of the beam by moving one of the reflecting surfaces 916 REFLECTOR ANTENNAS To achieve good radiation characteristics, the subreflector or subdish must be several, at least a few, wavelengths in diameter. However, its presence introduces shadowing which is the principal limitation of its use as a microwave antenna. The shadowing can significantly degrade the gain of the system, unless the main reflector is several wavelengths in diameter. Therefore the Cassegrain is usually attractive for applications that require gains of 40 dB or greater. There are, however, a variety of techniques that can be used to minimize aperture blocking by the subreflector. Some of them are (1) minimum blocking with simple Cassegrain, and (2) twisting Cassegrains for least blocking.
The first comprehensive published analysis of the Cassegrain arrangement as a microwave antenna is that by Hannan . He uses geometrical optics to derive the geometrical shape of the reflecting surfaces, and he introduces the equivalence concepts of the virtual feed and the equivalent parabola. Although his analysis does not predict fine details, it does give reasonably good results.
Refinements to his analysis have been introduced –.
To improve the aperture efficiency, suitable modifications to the geometrical shape of the reflect-ing surfaces have been suggested –. The reshaping of the reflecting surfaces is used to gener-ate desirable amplitude and phase distributions over one or both of the reflectors. The resultant system is usually referred to as shaped dual reflector. The reflector antenna of Figure 15.9 is such a system.
Shaped reflector surfaces, generated using analytical models, are illustrated in . It also has been suggested that a flange is placed around the subreflector to improve the aperture efficiency.
Since many reflectors have dimensions and radii of curvature large compared to the operating wavelength, they were traditionally designed based on geometrical optics (GO) . Both the single-and double-reflector (Cassegrain) systems were designed to convert a spherical wave at the source (focal point) into a plane wave. Therefore the reflecting surfaces of both reflector systems were primarily selected to convert the phase of the wavefront from spherical to planar. However, because of the variable radius of curvature at each point of reflection, the magnitude of the reflected field is also changed due to spatial attenuation or amplitude spreading factor or divergence factor (4-131) of Section 4.8.3, which are functions of the radius of curvature of the surface at the point of reflection. This ultimately leads to amplitude taper of the wavefront at the aperture plane. This is usually undesirable, and it can sometimes be compensated to some extent by the design of the pattern of the feed element or of the reflecting surface.
For a shaped-dual reflector system, there are two surfaces or degrees of freedom that can be utilized to compensate for any variations in the phase and amplitude of the field at the aperture plane. To determine how each surface may be reshaped to control the phase and/or amplitude of the field at the aperture plane, let us use geometrical optics and assume that the field radiated by the feed (pattern) is represented, both in amplitude and phase, by a bundle of rays which has a well-defined periphery. This bundle of rays is initially intercepted by the subreflector and then by the main reflector. Ultimately the output, after two reflections, is also a bundle of rays with prescribed phase and amplitude distributions, and a prescribed periphery, as shown in Figure 15.29(a) . It has been shown in that for a two-reflector system with high magnification (i.e., large ratio of main reflector diameter to subreflector diameter) that over the aperture plane the a. amplitude distribution is controlled largely by the subreflector curvature.
b. phase distribution is controlled largely by the curvature of the main reflector.
Therefore in a Cassegrain two-reflector system the reshaping of the paraboloid main reflector can be used to optimize the phase distribution while the reshaping of the hyperboloid subreflector can be used to control the amplitude distribution. This was used effectively in to design a shaped two-reflector system whose field reflected by the subreflector had a nonspherical phase but a csc4(𝜃∕2) amplitude pattern. However, the output from the main reflector had a perfect plane wave phase front and a very good approximate uniform amplitude distribution, as shown in Figure 15.29(b).
Because a comprehensive treatment of this arrangement can be very lengthy, only a brief introduc-tion of the system will be presented here. The interested reader is referred to the referenced literature.
PARABOLIC REFLECTOR 917 Figure 15.29 Geometrical optics for the reshaping and synthesis of the reflectors of a Cassegrain system.
(source: R. Mittra and V. Galindo-Israel “Shaped Dual Reflector Synthesis,” IEEE Antennas and Propagation Society Newsletter, Vol. 22, No. 4, pp. 5–9, August 1980. c ⃝(1980) IEEE).
A. Classical Cassegrain Form The operation of the Cassegrain arrangement can be introduced by referring to Figure 15.1(d) and assuming the system is operating in the receiving or transmitting mode. To illustrate the principle, a receiving mode is adopted.
Let us assume that energy, in the form of parallel rays, is incident upon the reflector system.
Energy intercepted by the main reflector, which has a large concave surface, is reflected toward the subreflector. Energy collected by the convex surface of the subdish is reflected by it, and it is directed toward the vertex of the main dish. If the incident rays are parallel, the main reflector is a paraboloid, and the subreflector is a hyperboloid, then the collected bundle of rays is focused at a single point.
The receiver is then placed at this focusing point.
A similar procedure can be used to describe the system in the transmitting mode. The feed is placed at the focusing point, and it is usually sufficiently small so that the subdish is located in its far-field region. In addition, the subreflector is large enough that it intercepts most of the radiation from the feed. Using the geometrical arrangement of the paraboloid and the hyperboloid, the rays reflected by the main dish will be parallel. The amplitude taper of the emerging rays is determined by the feed pattern and the tapering effect of the geometry.
The geometry of the classical Cassegrain system, employing a concave paraboloid as the main dish and a convex hyperboloid as the subreflector, is simple and it can be described completely 918 REFLECTOR ANTENNAS Figure 15.30 Virtual-feed and equivalent parabola concepts. (source: P. W. Hannan, “Microwave Anten-nas Derived from the Cassegrain Telescope,” IRE Trans. Antennas Propagat. Vol. AP-9, No. 2, March 1961.
c ⃝(1961) IEEE).
by only four independent parameters (two for each reflector). The analytical details can be found in .
To aid in the understanding and in predicting the essential performance of a Cassegrain, the con-cept of virtual feed is useful. By this principle, the real feed and the subreflector are replaced by an equivalent system which consists of a virtual feed located at the focal point of the main reflector, as shown by the dashed lines of Figure 15.30(a). For analysis purposes then, the new system is a single-reflector arrangement with the original main dish, a different feed, and no subreflector.
PARABOLIC REFLECTOR 919 The configuration of the virtual feed can be determined by finding the optical image of the real feed. This technique is only accurate when examining the effective aperture of the feed and when the dimensions of the real and virtual feeds are larger than a wavelength. In fact, for the classical Cassegrain arrangement of Figure 15.30(a), the virtual feed has a smaller effective aperture, and a corresponding broader beamwidth, than the real feed. The increase in beamwidth is a result of the convex curvature of the subreflector, and it can be determined by equating the ratio of the virtual to real-feed beamwidths to the ratio of the angles 𝜃v∕𝜃r.
The ability to obtain a different effective aperture for the virtual feed as compared to that of the real feed is useful in many applications such as in a monopulse antenna . To maintain efficient and wideband performance and effective utilization of the main aperture, this system requires a large feed aperture, a corresponding long focal length, and a large antenna structure. The antenna dimensions can be maintained relatively small by employing the Cassegrain configuration which utilizes a large feed and a short focal length for the main reflector.
Although the concept of virtual feed can furnish useful qualitative information for a Cassegrain system, it is not convenient for an accurate quantitative analysis. Some of the limitations of the virtual-feed concept can be overcome by the concept of the equivalent parabola .
By the technique of the equivalent parabola, the main dish and the subreflector are replaced by an equivalent focusing surface at a certain distance from the real focal point. This surface is shown dashed in Figure 15.30(b), and it is defined as “the locus of intersection of incoming rays par-allel to the antenna axis with the extension of the corresponding rays converging toward the real focal point.” Based on simple geometrical optics ray tracing, the equivalent focusing surface for a Cassegrain configuration is a paraboloid whose focal length equals the distance from its vertex to the real focal point. This equivalent system also reduces to a single-reflector arrangement, which has the same feed but a different main reflector, and it is accurate when the subreflector is only a few wavelengths in diameter. More accurate results can be obtained by including diffraction patterns. It also has the capability to focus toward the real focal point an incoming plane wave, incident from the opposite direction, in exactly the same manner as the actual main dish and the subreflector.
B. Cassegrain and Gregorian Forms In addition to the classical Cassegrain forms, there are other configurations that employ a variety of main reflector and subreflector surfaces and include concave, convex, and flat shapes . In one form, the main dish is held invariant while its feed beamwidth progressively increases and the axial dimensions of the antenna progressively decrease. In another form, the feed beamwidth is held invariant while the main reflector becomes progressively flatter and the axial dimensions progres-sively increase.
One of the other reflector arrangements is the classical Gregorian design of Figure 15.31 where the main reflector is a parabola while the subreflector is a concave ellipse. The focal point is Elliptical concave reflector Equivalent surface Parabolic reflector Feed Figure 15.31 Gregorian reflector arrangement and its equivalent surface.
920 REFLECTOR ANTENNAS between the main reflector and subreflector. Its equivalent parabola is shown dashed in Figure 15.31.
When the overall size and the feed beamwidth of the classical Gregorian, of Figure 15.31, are identi-cal to those of the classical Cassegrain, of Figure 15.30, the Gregorian requires a shorter focal length for the main dish .
15.5 SPHERICAL REFLECTOR The discussion and results presented in the previous sections illustrate that a paraboloidal reflector is an ideal collimating device. However, despite all of its advantages and many applications it is severely handicapped in angular scanning. Although scanning can be accomplished by (1) mechan-ical rotation of the entire structure, and (2) displacement of the feed alone, it is thwarted by the large mechanical moment of inertia in the first case and by the large coma and astigmatism in the second.
By contrast, the spherical reflector can make an ideal wide-angle scanner because of its perfectly symmetrical geometrical configuration. However, it is plagued by poor inherent collimating prop-erties. If, for example, a point source is placed at the focus of the sphere, it does not produce plane waves. The departure of the reflected wavefront from a plane wave is known as spherical aberration, and it depends on the diameter and focal length of the sphere. By reciprocity, plane waves incident on a spherical reflector surface parallel to its axis do not converge at the focal point. However a spher-ical reflector has the capability of focusing plane waves incident at various angles by translating and orientating the feed and by illuminating different parts of the structural geometry. The 1,000-ft diameter reflector at Arecibo, Puerto Rico is a spherical reflector whose surface is built into the earth and the scanning is accomplished by movement of the feed.
The focusing characteristics of a typical spherical reflector is illustrated in Figure 15.32 for three rays. The point F in the figure is the paraxial focus, and it is equal to one-half the radius of the Figure 15.32 Spherical reflector geometry and rays that form a caustic.
SPHERICAL REFLECTOR 921 sphere. The caustic∗surface is an epicycloid and is generated by the reflection of parallel rays.
A degenerate line FV of this caustic is parallel to the incident rays and extends from the paraxial focus to the vertex of the reflector. If one draws a ray diagram of plane waves incident within a 120◦cone, it will be shown that all energy must pass through the line FV. Thus, the line FV can be used for the placement of the feed for the interception of plane waves incident parallel to the axial line. It can thus be said that a spherical reflector possesses a line focus instead of a point.
However, amplitude and phase corrections must be made in order to realize maximum antenna efficiency.
Ashmead and Pippard proposed to reduce spherical aberration and to minimize path error by placing a point-source feed not at the paraxial focus F (half radius of the sphere), as taught in optics, but displaced slightly toward the reflector. For an aperture of diameter d, the correct location for placing a point source is a distance f0 from the vertex such that the maximum path error value is Δmax ≃ d4 2000f0 3 (15-72) and the maximum phase error does not differ from a paraboloid by more than one-eighth of a wave-length. This, however, leads to large f/d and to poor area utilization. A similar result was obtained by Li . He stated that the total phase error (sum of maximum absolute values of positive and negative phase errors) over an aperture of radius a is least when the phase error at the edge of the aperture is zero. Thus the optimum focal length is fop = 1 4(R + √ R2 −a2) (15-73) where R = radius of the spherical reflector a = radius of the utilized aperture Thus when R = 2a, the optimum focal length is 0.4665R and the corresponding total phase error, using the formula found in , is 0.02643(R∕λ) rad. Even though the optimum focal length leads to minimum total phase error over a prescribed aperture, it does not yield the best radiation pattern when the illumination is not uniform. For a tapered amplitude distribution, the focal length that yields the best radiation pattern will be somewhat longer, and in practice, it is usually determined by experiment. Thus for a given maximum aperture size there exists a maximum value of total allowable phase error, and it is given by ( a R )4 max = 14.7(Δ∕λ)total (R∕λ) (15-74) where (Δ∕λ) is the total phase error in wavelengths.
∗A caustic is a point, a line, or a surface through which all the rays in a bundle pass and where the intensity is infinite. The caustic also represents the geometrical loci of all the centers of curvature of the wave surfaces. Examples of it include the focal line for cylindrical parabolic reflector and the focal point of a paraboloidal reflector.
922 REFLECTOR ANTENNAS Example 15.4 A spherical reflector has a 10-ft diameter. If at 11.2 GHz the maximum allowable phase error is λ∕16, find the maximum permissible aperture.
Solution: At f = 11.2 GHz λ = 0.08788 ft ( a R )4 max = 14.7 ( 1∕16 56.8957 ) = 0.01615 a4 ≃10.09 a = 1.78 ft To overcome the shortcoming of a point feed and minimize spherical aberration, Spencer, Sletten, and Walsh were the first to propose the use of a line-source feed. Instead of a continuous line source, a set of discrete feed elements can be used to reduce spherical aberration when they are properly placed along the axis in the vicinity of the paraxial focus. The number of elements, their position, and the portion of the reflector surface which they illuminate is dictated by the allowable wavefront distortion, size, and curvature of the reflector. This is shown in Figure 15.33 . A single feed located near the paraxial focus will illuminate the central portion of the reflector. If the reflector is large, additional feed elements along the axis toward the vertex will be needed to minimize the phase errors in the aperture. The ultimate feed design will be a compromise between a single element and a line-source distribution.
An extensive effort has been placed on the analysis and experiment of spherical reflectors, and most of it can be found well documented in a book of reprinted papers . In addition, a number of two-dimensional patterns and aperture plane constant amplitude contours, for symmetrical and offset feeds, have been computed .
R R 2 Paraxial focus Figure 15.33 Reflector illumination by feed sections placed between paraxial focus and vertex. (source: A. C. Schell, “The Diffraction Theory of Large-Aperture Spherical Reflector Antennas,” IRE Trans. Antennas Propagat., Vol. AP-11, No. 4, July 1963. c ⃝(1963) IEEE).
REFERENCES 923 15.6 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab computer program, designated Reflector, for computing and displaying the radiation characteristics of a paraboloidal reflector.
c. Power Point (PPT) viewgraphs, in multicolor.
REFERENCES 1. A. W. Love (ed.), Reflector Antennas, IEEE Press, New York, 1978.
2. Y. Obha, “On the Radiation of a Corner Reflector Finite in Width,” IEEE Trans. Antennas Propagat., Vol. AP-11, No. 2, pp. 127–132, March 1963.
3. C. A. Balanis and L. Peters, Jr., “Equatorial Plane Pattern of an Axial-TEM Slot on a Finite Size Ground Plane,” IEEE Trans. Antennas Propagat., Vol. AP-17, No. 3, pp. 351–353, May 1969.
4. C. A. Balanis, “Pattern Distortion Due to Edge Diffractions,” IEEE Trans. Antennas Propagat., Vol. AP-18, No. 4, pp. 551–563, July 1970.
5. C. A. Balanis, “Analysis of an Array of Line Sources Above a Finite Ground Plane,” IEEE Trans. Antennas Propagat., Vol. AP-19, No. 2, pp. 181–185, March 1971.
6. D. Proctor, “Graphs Simplify Corner Reflector Antenna Design,” Microwaves, Vol. 14, No. 7, pp. 48–52, July 1975.
7. E. B. Moullin, Radio Aerials, Oxford University Press, 1949, Chapters 1 and 3.
8. R. E. Paley and N. Wiener, Fourier Transforms in the Complex Domain, American Mathematical Society, Providence, R.I., p. 116, 1934.
9. P. A. J. Ratnasiri, R. G. Kouyoumjian, and P. H. Pathak, “The Wide Angle Side Lobes of Reflector Antennas,” ElectroScience Laboratory, The Ohio State University, Technical Report 2183-1, March 23, 1970.
10. G. L. James and V. Kerdemelidis, “Reflector Antenna Radiation Pattern Analysis by Equivalent Edge Cur-rents,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 1, pp. 19–24, January 1973.
11. C. A. Mentzer and L. Peters, Jr., “A GTD Analysis of the Far-out Side Lobes of Cassegrain Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 5, pp. 702–709, September 1975.
12. L. M. LaLonde and D. E. Harris, “A High Performance Line Source Feed for the AIO Spherical Reflector,” IEEE Trans. Antennas Propagat., Vol. AP-18, No. 1, pp. 41–48, January 1970.
13. A. W. Rudge, “Offset-Parabolic-Reflector Antennas: A Review,” Proc. IEEE, Vol. 66, No. 12, pp. 1592–1618, December 1978.
14. P. J. B. Clarricoats and G. T. Poulton, “High-Efficiency Microwave Reflector Antennas—A Review,” Proc.
IEEE, Vol. 65, No. 10, pp. 1470–1502, October 1977.
15. O. Hachenberg, B. H. Grahl, and R. Wielebinski, “The 100-Meter Radio Telescope at Effelsberg,” Proc.
IEEE, Vol. 69, No. 9, pp. 1288–1295, 1973.
16. P. D. Potter, W. D. Merrick, and A. C. Ludwig, “Big Antenna Systems for Deep-Space Communications,” Astronaut. Aeronaut., pp. 84–95, October 1966.
17. S. Silver (ed.), Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949 (MIT Radiation Lab. Series, Vol. 12).
18. J. F. Kauffman, W. F. Croswell, and L. J. Jowers, “Analysis of the Radiation Patterns of Reflector Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-24, No. 1, pp. 53–65, January 1976.
19. R. E. Collin and F. J. Zucker (eds.), Antenna Theory Part II, McGraw-Hill, New York, pp. 36–48, 1969.
924 REFLECTOR ANTENNAS 20. P. K. Agrawal, J. F. Kauffman, and W. F. Croswell, “Calculated Scan Characteristics of a Large Spher-ical Reflector Antenna,” IEEE Trans. Antennas Propagat., Vol. AP-27, No. 3, pp. 430–431, May 1979.
21. E. M. T. Jones, “Paraboloid Reflector and Hyperboloid Lens Antennas,” IRE Trans. Antennas Propagat., Vol. AP-2, No. 3, pp. 119–127, July 1954.
22. I. Koffman, “Feed Polarization for Parallel Currents in Reflectors Generated by Conic Sections,” IEEE Trans. Antennas Propagat., Vol. AP-14, No. 1, pp. 37–40, January 1966.
23. A. W. Love, “Some Highlights in Reflector Antenna Development,” Radio Sci., Vol. 11, Nos. 8, 9, pp. 671–684, August–September 1976.
24. D. K. Cheng, “Effect of Arbitrary Phase Errors on the Gain and Beamwidth Characteristics of Radiation Pattern,” IRE Trans. Antennas Propagat., Vol. AP-3, No. 3, pp. 145–147, July 1955.
25. Y. Y. Hu, “A Method of Determining Phase Centers and Its Application to Electromagnetic Horns,” J. Franklin Inst., pp. 31–39, January 1961.
26. J. Ruze, “The Effect of Aperture Errors on the Antenna Radiation Pattern,” Nuevo Cimento Suppl., Vol. 9, No. 3, pp. 364–380, 1952.
27. H. C. Minnett and B. MacA. Thomas, “Fields in the Image Space of Symmetrical Focusing Reflectors,” Proc. IEE, Vol. 115, pp. 1419–1430, October 1968.
28. G. F. Koch, “Coaxial Feeds for High Aperture Efficiency and Low Spillover of Paraboloidal Reflector Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 2, pp. 164–169, March 1973.
29. R. E. Lawrie and L. Peters, Jr., “Modifications of Horn Antennas for Low Side Lobe Levels,” IEEE Trans.
Antennas Propagat., Vol. AP-14, No. 5, pp. 605–610, September 1966.
30. A. J. Simmons and A. F. Kay, “The Scalar Feed-A High Performance Feed for Large Paraboloid Reflectors,” Design and Construction of Large Steerable Aerials, IEE Conf. Publ. 21, pp. 213–217, 1966.
31. W. M. Truman and C. A. Balanis, “Optimum Design of Horn Feeds for Reflector Antennas,” IEEE Trans.
Antennas Propagat., Vol. AP-22, No. 4, pp. 585–586, July 1974.
32. B. Houshmand, B. Fu, and Y. Rahmat-Samii, “Reflector Antenna Analysis Software,” Vol. II, Chapter 11, CAEME Center for Multimedia Education, University of Utah, pp. 447–465, 1995.
33. P. W. Hannan, “Microwave Antennas Derived from the Cassegrain Telescope,” IRE Trans. Antennas Prop-agat., Vol. AP-9, No. 2, pp. 140–153, March 1961.
34. W. V. T. Rusch, “Scattering from a Hyperboloidal Reflector in a Cassegrain Feed System,” IEEE Trans.
Antennas Propagat., Vol. AP-11, No. 4, pp. 414–421, July 1963.
35. P. D. Potter, “Application of Spherical Wave Theory to Cassegrainian-Fed Paraboloids,” IEEE Trans. Anten-nas Propagat., Vol. AP-15, No. 6, pp. 727–736, November 1967.
36. W. C. Wong, “On the Equivalent Parabola Technique to Predict the Performance Characteristics of a Cassegrain System with an Offset Feed,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 3, pp. 335–339, May 1973.
37. V. Galindo, “Design of Dual-Reflector Antennas with Arbitrary Phase and Amplitude Distributions,” IEEE Trans. Antennas Propagat., Vol. AP-12, No. 4, pp. 403–408, July 1964.
38. W. F. Williams, “High Efficiency Antenna Reflector,” Microwave Journal, Vol. 8, pp. 79–82, July 1965.
39. G. W. Collins, “Shaping of Subreflectors in Cassegrainian Antennas for Maximum Aperture Efficiency,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 3, pp. 309–313, May 1973.
40. C. A. Balanis, Advanced Engineering Electromagnetics, John Wiley & Sons, Inc., New York, pp. 744–764, 1989.
41. R. Mittra and V. Galindo-Israel, “Shaped Dual Reflector Synthesis,” IEEE Antennas Propagation Society Newsletter, Vol. 22, No. 4, pp. 5–9, August 1980.
42. K. A. Green, “Modified Cassegrain Antenna for Arbitrary Aperture Illumination,” IEEE Trans. Antennas Propagat., Vol. AP-11, No. 5, pp. 589–590, September 1963.
43. J. Ashmead and A. B. Pippard, “The Use of Spherical Reflectors as Microwave Scanning Aerials,” J. Inst.
Elect. Eng., Vol. 93, Part III-A, pp. 627–632, 1946.
44. T. Li, “A Study of Spherical Reflectors as Wide-Angle Scanning Antennas,” IRE Trans. Antennas Propa-gat., Vol. AP-7, No. 3, pp. 223–226, July 1959.
PROBLEMS 925 45. R. C. Spencer, C. J. Sletten, and J. E. Walsh, “Correction of Spherical Aberration by a Phased Line Source,” Proceedings National Electronics Conference, Vol. 5, pp. 320–333, 1949.
46. A. C. Schell, “The Diffraction Theory of Large-Aperture Spherical Reflector Antennas,” IEEE Trans.
Antennas Propagat., Vol. AP-11, No. 4, pp. 428–432, July 1963.
PROBLEMS 15.1.
An infinite line source, of constant electric current I0, is placed a distance s above a flat and infinite electric ground plane. Derive the array factor.
15.2.
For corner reflectors with included angles of 𝛼= 60◦, 45◦, and 30◦: (a) Derive the array factors of (15-7)–(15-9b).
(b) Plot the field strength along the axis (𝜃= 90◦, 𝜙= 0◦) as a function of the feed-to-vertex spacing, 0 ≤s∕λ ≤10.
15.3.
Consider a corner reflector with an included angle of 𝛼= 36◦.
(a) Derive the array factor.
(b) Plot the relative field strength along the axis (𝜃= 90◦, 𝜙= 0◦) as a function of the feed-to-vertex spacing s, for 0 ≤s∕λ ≤10.
(c) Determine the spacing that yields the first maximum possible field strength along the axis. For this spacing, what is the ratio of the field strength of the corner reflector along the axis to the field strength of the feed element alone?
(d) For the spacing in part c, plot the normalized power pattern in the azimuthal plane (𝜃= 90◦).
15.4.
A 60◦corner reflector, in conjunction with a λ∕2 dipole feed, is used in a radar tracking system. One of the requirements for such a system is that the antenna, in one of its modes of operation, has a null along the forward symmetry axis. In order to accomplish this, what should be the feed spacing from the vertex (in wavelengths)? Give all the possible values of the feed-to-vertex spacing.
15.5.
For a parabolic reflector, derive (15-25) which relates the f/d ratio to its subtended angle 𝜃0.
15.6.
Show that for a parabolic reflector (a) 0 ≤f ∕d ≤0.25 relates to 180◦≥𝜃0 ≥90◦ (b) 0.25 ≤f∕d ≤∞relates to 90◦≥𝜃0 ≥0◦ 15.7.
The diameter of a paraboloidal reflector antenna (dish), used for public television stations, is 10 meters. Find the far-zone distance if the antenna is used at 2 and 4 GHz.
15.8.
Show that the directivity of a uniformly illuminated circular aperture of diameter d is equal to (𝜋d∕λ)2.
15.9.
Verify (15-33) and (15-33a).
15.10.
The field radiated by a paraboloidal reflector with an f/d ratio of 0.5 is given by E = (̂ ax + ̂ ay sin 𝜙cos 𝜙)f(r, 𝜃, 𝜙) where the x-component is the co-pol and the y-component is the cross-pol.
(a) At what observation angle(s) (in degrees) (0◦–180◦) is the cross-pol minimum? What is the minimum value?
(b) At what observation angle(s) (in degrees) (0◦–180◦) is the cross-pol maximum? What is the maximum value?
926 REFLECTOR ANTENNAS (c) What is the polarization loss factor when the receiving antenna is linearly polarized in the x-direction.
(d) What is the polarization loss factor when the receiving antenna is linearly polarized in the y-direction.
(e) What should the polarization of the receiving antenna be in order to eliminate the losses due to polarization? Write an expression for the polarization of the receiving antenna to achieve this.
15.11.
Verify (15-49) and (15-54).
15.12.
A small parabolic reflector (dish) of revolution, referred to as a paraboloid, is now being advertised as a TV antenna for direct broadcast. Assuming the diameter of the reflector is 1 meter, determine at 3 GHz the directivity (in dB) of the antenna if the feed is such that (a) the illumination over the aperture is uniform (ideal) (b) the taper efficiency is 80% while the spillover efficiency is 85%. Assume no other losses.
What is the total aperture efficiency of the antenna (in dB)?
15.13.
The 140-ft (42.672-m) paraboloidal reflector at the National Radio Astronomy Observatory, Green Bank, W. Va, has an f/d ratio of 0.4284. Determine the (a) subtended angle of the reflector (b) aperture efficiency assuming the feed pattern is symmetrical and its gain pattern is given by 2 cos2(𝜃′∕2), where 𝜃′ is measured from the axis of the reflector (c) directivity of the entire system when the antenna is operating at 10 GHz, and it is illu-minated by the feed pattern of part (b) (d) directivity of the entire system at 10 GHz when the reflector is illuminated by the feed pattern of part (b) and the maximum aperture phase deviation is 𝜋∕16 rad 15.14.
A paraboloidal reflector has an f/d ratio of 0.38. Determine (a) which cosn 𝜃′ symmetrical feed pattern will maximize its aperture efficiency (b) the directivity of the reflector when the focal length is 10λ (c) the value of the feed pattern in dB (relative to the main maximum) along the edges of the reflector.
15.15.
The diameter of an educational TV station reflector is 10 meters. It is desired to design the reflector at 1 GHz with a f/d ratio of 0.5. The pattern of the feed is given by Gf = { 3.4286 cos4(𝜃′∕2) 0 ≤𝜃′ ≤𝜋∕2 0 Elsewhere Assume a symmetrical pattern in the 𝜙′ direction. Determine the following: (a) Total subtended angle of the reflector (in degrees) (b) Aperture efficiency (in %) (c) Directivity of the reflector (dimensionless and in dB).
15.16.
The feed pattern of a paraboloidal reflector, with a diameter of 10 meters and a f/d ratio of 0.433, is rotationally symmetric and it is given by: Gf (𝜃′) = { 2.667 cos2 ( 𝜃′∕2 ) 0◦≤𝜃′ ≤90◦ 0 Elsewhere PROBLEMS 927 Calculate the: (a) Total subtended angle of the reflector (in degrees).
(b) Aperture efficiency (in %).
(c) Maximum power (in Watts) that can be delivered, in the receiving mode, to a load con-nected to the reflector when the power density of an incident wave is 10−6 Watts/cm2.
15.17.
A 10-m diameter earth-based paraboloidal reflector antenna, which is used for satellite tele-vision, has a f∕d ratio of 0.433. Assuming the amplitude gain pattern of the feed is rotation-ally symmetric in 𝜙′ and is given by Gf (𝜃′, 𝜙′) = { G0 cos2(𝜃′∕2) 0◦≤𝜃′ ≤90◦ 0 Elsewhere where G0 is a constant, determine the (a) total subtended angle of the reflector (in degrees) (b) value of G0.
(c) aperture efficiency (in %).
15.18.
A 10-m diameter earth-based paraboloidal reflector antenna, which is used for satellite tele-vision, has a total subtended angle of 120◦. Assuming the amplitude gain pattern of the feed is rotationally symmetric in 𝜙and is given by Gf (𝜃′) = { G0 cos2(𝜃′∕2) 0◦≤𝜃′ ≤90◦ 0 Elsewhere where G0 is a constant, determine the (a) value of G0 (b) taper efficiency (in %).
Compare the aperture efficiency of the above pattern with that of Gf (𝜃′) = { G0 cos2(𝜃′) 0◦≤𝜃′ ≤90◦ 0 Elsewhere Which of the two patterns’ aperture efficiency is lower, and why. Explain the why by com-paring qualitatively the corresponding taper and spillover efficiencies, and thus the aper-ture efficiency, of the two feed patterns; do this without calculating the taper and spillover efficiencies.
15.19.
The symmetrical feed pattern for a paraboloidal reflector is given by Gf = ⎧ ⎪ ⎨ ⎪ ⎩ G0 cos4 ( 𝜃′ 2 ) 0 ≤𝜃′ ≤𝜋∕2 0 Elsewhere where G0 is a constant.
(a) Evaluate the constant G0.
(b) Derive an expression for the aperture efficiency.
(c) Find the subtended angle of the reflector that will maximize the aperture efficiency.
What is the maximum aperture efficiency?
928 REFLECTOR ANTENNAS 15.20.
A paraboloidal reflector is operating at a frequency of 5 GHz. It is 8 meters in diameter, with an f/d ratio of 0.25. It is fed with an antenna whose primary pattern is symmetrical and which can be approximated by Gf = { 10 cos4 𝜃′ 0 ≤𝜃′ ≤𝜋∕2 0 Elsewhere Find its (a) aperture efficiency (b) overall directivity (c) spillover efficiency (d) taper efficiency 15.21.
A paraboloidal reflector operating at 10 GHz, with a diameter of 10 meters and a total subtended angle of 80◦, is fed with an antenna whose gain pattern is given by: Gf (𝜃′) = { G0 cos8(𝜃′) 0◦≤𝜃′ ≤90◦ 0 Elsewhere (a) Calculate the: 1. Aperture efficiency (in %). No graphical solution.
2. Value of G0.
(b) Determine the (you can use graphical solution): 1. Taper efficiency (in %).
2. Spillover efficiency (in %).
(c) Calculate the directivity (dimensionless and in dB).
15.22.
A parabolic reflector has a diameter of 10 meters and has an included angle of 𝜃0 = 30◦.
The directivity at the operating frequency of 25 GHz is 5,420,000. The phase efficiency, polarization efficiency, blockage efficiency, and random error efficiency are all 100%. The feed has a phi-symmetric pattern given by Gf = { G0 cos10 𝜃′ 0◦≤𝜃′ ≤90◦ 0 Elsewhere Find the taper, spillover, and overall efficiencies.
15.23.
A 10-meter diameter paraboloidal reflector is used as a TV satellite antenna. The focus-to-diameter ratio of the reflector is 0.536 and the pattern of the feed in the forward region can be approximated by cos2(𝜃′). Over the area of the reflector, the incident power density from the satellite can be approximated by a uniform plane wave with a power density of 10 μwatts/m2. At a center frequency of 9 GHz: (a) What is the maximum directivity of the reflector (in dB)?
(b) Assuming no losses of any kind, what is the maximum power that can be delivered to a TV receiver connected to the reflector through a lossless transmission line?
15.24.
A 3-meter diameter parabolic reflector is used as a receiving antenna for satellite television reception at 5 GHz. The reflector is connected to the television receiver through a 78-ohm coaxial cable. The aperture efficiency is approximately 75%. Assuming the maximum inci-dent power density from the satellite is 10 microwatts/square meter and the incident wave is polarization-matched to the reflector antenna, what is the: (a) Directivity of the antenna (in dB) PROBLEMS 929 (b) Maximum power (in watts) that can be delivered to the receiving TV? Assume no losses.
(c) Power (in watts) delivered to the receiving TV if the reflection coefficient at the trans-mission line/receiving TV terminal junction is 0.2. Assume no other losses.
15.25.
A reflector antenna with a total subtended angle of 120◦is illuminated at 10 GHz with a specially designed feed so that its aperture efficiency is nearly unity. The focal distance of the reflector is 5 meters. Assuming the radiation pattern is nearly symmetric, determine the: (a) Half-power beamwidth (in degrees).
(b) Sidelobe level (in dB).
(c) Directivity (in dB).
(d) Directivity (in dB) based on Kraus’ formula and Tai & Pereira’s formula.
(e) Loss in directivity (in dB) if the surface has rms random roughness of 0.64 mm.
15.26.
A front-fed paraboloidal reflector with an f/d radio of 0.357, whose diameter is 10 meters and which operates at 10 GHz, is fed by an antenna whose power pattern is rotationally symmetric and it is represented by cos2(𝜃∕2). All the power is radiated in the forward region (0◦≤𝜃′ ≤90◦) of the feed. Determine the (a) spillover efficiency (b) taper efficiency (c) overall aperture efficiency (d) directivity of the reflector (in dB) (e) directivity of the reflector (in dB) if the RMS, reflector surface deviation from an ideal paraboloid is λ∕100.
15.27.
A three-dimensional paraboloidal reflector, as shown in Figure 15.12 with radius a at its aperture, has a aperture field distribution A(𝜌′) which is represented by A(𝜌′) = A0 [ 1 − (𝜌′ a )2] a ρ′ where A0 = constant 𝜌′ = radial distance from the center of aperture (0 ≤𝜌′ ≤a) a = radius of the aperture Determine, for an aperture radius of a = 50λ, the: (a) Aperture efficiency (in %).
(b) Directivity (dimensionless and in dB) of the reflector.
(c) Half-power beamwidth (in degrees).
(d) Approximate directivity (dimensionless and in dB) using another appropriate formula at your availability.
15.28.
A uniform plane wave at a frequency of 10 GHz and with a power density of 10 𝜇watts/cm2 is incident upon a three-dimensional paraboloidal reflector, with a f ∕d = 0.43, which is used as a receiving antenna. The antenna used as a feed, which is placed at the focal point of the reflector, has a rotationally symmetric pattern (not a function of 𝜙) and it is given by Gf (𝜃′) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ cot2 (𝜃0 2 ) cos4 ( 𝜃′ 2 ) 0 ≤𝜃′ ≤𝜃0 0 𝜃′ ≥𝜃0 930 REFLECTOR ANTENNAS Assuming the reflector has a diameter of 10 meters, determine the: (a) Total subtended angle (in degrees) of the reflector.
(b) Spillover efficiency (in %); do not have to derive.
(c) Taper efficiency (in %) do not have to derive.
(d) Aperture efficiency (in %); do not have to derive.
(e) Maximum directivity (in dB) of the reflector.
(f) Maximum power (in watts) delivered to a load connected to the feed. Assume no other losses.
Refer to Figure 15.10 for the geometry of the problem.
15.29.
Design pyramidal horn antennas that will maximize the aperture efficiency or produce max-imum power transmission to the feed, for paraboloidal reflectors with f/d ratios of (a) 0.50 (b) 0.75 (c) 1.00 15.30.
A three-dimensional paraboloidal reflector, like the one shown in Figure 15.12, is fed in such a way that it creates, after the field radiated by the feed if reflected by the reflector surface, over its aperture (over the area of the reflector which includes the rim with radius a, as shown in Figure 15.12), a normalized tapered electric field distribution given by: Normalized Aperture Electric Field Distribution = [ 1 − (𝜌′ a )2] , 0 ≤𝜌′ ≤a where 𝜌′ = distance from the center of the aperture towards the reflector circumference/ rim/edge, as shown in Figure 15.12. What is the: (a) Aperture efficiency 𝜀ap (in %) of the aperture of radius a with the above aperture field distribution.
(b) Directivity of the reflector (dimensionless and in dB) when its radius is 20λ.
CHAPTER16 Smart Antennas 16.1 INTRODUCTION Over the last decade, wireless technology has grown at a formidable rate, thereby creating new and improved services at lower costs. This has resulted in an increase in airtime usage and in the number of subscribers. The most practical solution to this problem is to use spatial processing. As Andrew Viterbi, founder of Qualcomm Inc., clearly stated: “Spatial processing remains as the most promising, if not the last frontier, in the evolution of multiple access systems” .
Spatial processing is the central idea of adaptive antennas or smart-antenna systems. Although it might seem that adaptive antennas have been recently discovered, they date back to World War II with the conventional Bartlett beamformer . It is only of today’s advancement in powerful low-cost digital signal processors, general-purpose processors (and ASICs—Application-Specific Integrated Circuits), as well as innovative software-based signal-processing techniques (algorithms), that smart-antenna systems have received enormous interest worldwide. In fact, many overviews and tutorials – have emerged, and a great deal of research is being done on the adaptive and direction-of-arrival (DOA) algorithms for smart-antenna systems , . As the number of users and the demand for wireless services increases at an exponential rate, the need for wider coverage area and higher transmission quality rises. Smart-antenna systems provide a solution to this problem.
This chapter presents an introduction and general overview of smart-antenna systems. First, it gives the reader an insight on smart-antenna systems using the human auditory system as an analogy.
Then, it presents the purpose for smart antennas by introducing the cellular radio system and its evolution. A brief overview of signal propagation is given to emphasize the need for smart-antenna systems. These topics are followed by antenna array theory, time of arrival and adaptive digital processing algorithms, mutual coupling, mobile ad hoc networks, antenna design, and simulation and impact of antenna designs on network capacity/throughput and communication channel bit-error-rate (BER). Material of this chapter is extracted from –. More extensive discussion and details on each topic can be found in and other sources.
16.2 SMART-ANTENNA ANALOGY The functionality of many engineering systems is readily understood when it is related to our human body system . Therefore, to give an insight into how a smart-antenna system works, let us imagine Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 931 932 SMART ANTENNAS Desired speaker Desired speaker Unwanted speaker Unwanted speaker DSP w1 w2 (a) Human analogy (b) Electrical equivalent Figure 16.1 Smart-antenna analogy. (a) Human analogy ; (b) Electrical equivalent.
two persons carrying on a conversation inside a dark room [refer to Figure 16.1(a)]. The listener among the two persons is capable of determining the location of the speaker as he moves about the room because the voice of the speaker arrives at each acoustic sensor, the ear, at a different time. The human signal processor, the brain, computes the direction of the speaker from the time differences or delays of the voice received by the two ears. Afterward, the brain adds the strength of the signals from each ear so as to focus on the sound of the computed direction. Furthermore, if additional speakers join in the conversation, the brain can tune out unwanted interferers and concentrate on one conversation at a time. Conversely, the listener can respond back to the same direction of the desired speaker by orienting the transmitter (mouth) toward the speaker.
Electrical smart-antenna systems work the same way using two antennas instead of the two ears and a digital signal processor instead of a brain [refer to Figure 16.1(b)]. Therefore, after the digital signal processor measures the time delays from each antenna element, it computes the direction of arrival (DOA) of the signal-of-interest (SOI), and then it adjusts the excitations (gains and phases of the signals) to produce a radiation pattern that focuses on the SOI while, ideally, tuning out any signal-not-of-interest (SNOI).
CELLULAR RADIO SYSTEMS EVOLUTION 933 16.3 CELLULAR RADIO SYSTEMS EVOLUTION Maintaining capacity has always been a challenge as the number of services and subscribers increased. To achieve the capacity demand required by the growing number of subscribers, cellular radio systems had to evolve throughout the years. To justify the need for smart-antenna systems in the current cellular system structure, a brief history on the evolution of the cellular radio systems is presented. For more in-depth details refer to –.
16.3.1 Omnidirectional Systems Since the early days, system designers knew that capacity was going to be a problem, especially when the number of channels or frequencies allotted by the Federal Communications Commission (FCC) was limited. Therefore, to achieve the capacity required for thousands of subscribers, a suitable cellular structure had to be designed; an example of the resulting structure is depicted in Figure 16.2.
Each shaded hexagonal area in Figure 16.2 represents a small geographical area named cell with maximum radius R . At the center of each cell resides a base station equipped with an omni-directional antenna with a given band of frequencies. Base stations in adjacent cells are assigned frequency bands that contain different frequencies compared to the neighboring cells. By limiting the coverage area to within the boundaries of a cell, the same band of frequencies may be used to cover different cells that are separated from one another by distances large enough to keep inter-ference levels below the threshold of the others. The design process of selecting and allocating the same bands of frequencies to different cells of cellular base stations within a system is referred to as frequency reuse . This is shown in Figure 16.2 by the repeating shaded pattern or clusters ; cells having the same shaded pattern use the same frequency spectrum. In the first cellular radio systems deployed, each base station was equipped with an omnidirectional antenna with a typical amplitude pattern as that shown in Figure 16.3. Because only a small percentage of the total energy reached the desired user, the remaining energy was radiated in undesired directions. As the number of users increased, so did the interference, thereby reducing capacity. An immediate solution to this problem was to subdivide a cell into smaller cells; this technique is referred to as cell splitting .
R cell D Figure 16.2 Typical cellular structure with 7 cells reuse pattern.
934 SMART ANTENNAS Radiation pattern Antenna z x r H E y H E θ ϕ Figure 16.3 Omnidirectional antenna pattern.
A. Cell Splitting Cell splitting , as shown in Figure 16.4, subdivides a congested cell into smaller cells called microcells, each with its own base station and a corresponding reduction in antenna height and trans-mitter power. Cell splitting improves capacity by decreasing the cell radius R and keeping the D∕R ratio unchanged; D is the distance between the centers of the clusters. The disadvantages of cell splitting are costs incurred from the installation of new base stations, the increase in the number of handoffs (the process of transferring communication from one base station to another when the mobile unit travels from one cell to another), and a higher processing load per subscriber.
B. Sectorized Systems As the demand for wireless service grew even higher, the number of frequencies assigned to a cell eventually became insufficient to support the required number of subscribers. Thus, a cellular design technique was needed to provide more frequencies per coverage area. This technique is referred to as cell sectoring , where a single omnidirectional antenna is replaced at the base station with several directional antennas. Typically, a cell is sectorized into three sectors of 120◦each , as shown in Figure 16.5.
Cell Microcell Figure 16.4 Cell splitting.
CELLULAR RADIO SYSTEMS EVOLUTION 935 Figure 16.5 Sectorized base-station antenna.
In sectoring, capacity is improved while keeping the cell radius unchanged and reducing the D∕R ratio. In other words, by reducing the number of cells in a cluster and thus increasing the frequency reuse, capacity improvement is achieved. However, in order to accomplish this, it is necessary to reduce the relative interference without decreasing the transmitting power. The cochannel interfer-ence in such cellular systems is reduced since only two neighboring cells interfere instead of six for the omnidirectional case , (see Figure 16.6). The penalty for improved signal-to-interference (S∕I) ratio and capacity is an increase in the number of antennas at the base station, and a decrease in trunking efficiency , due to channel sectoring at the base station –. Trunking (a) Sectoring (a) Omnidirectional D 120° Figure 16.6 Cochannel interference comparison between (a) sectoring, and (b) omnidirectional.
936 SMART ANTENNAS efficiency is a measure of the number of users that can be offered service with a particular configu-ration of fixed number of frequencies.
16.3.2 Smart-Antenna Systems Despite its benefits, cell sectoring did not provide the solution needed for the capacity problem.
Therefore, the system designers began to look into a system that could dynamically sectorize a cell.
Hence, they began to examine smart antennas.
Many refer to smart-antenna systems as smart antennas, but in reality antennas are not smart; it is the digital signal processing, along with the antennas, which makes the system smart. Although it might seem that smart-antenna systems is a new technology, the fundamental theory of smart (adap-tive) antennas is not new , . In fact, it has been applied in defense-related systems since World War II. Until recent years, with the emergence of powerful low-cost digital signal processors (DSPs), general-purpose processors (and Application-Specific Integrated Circuits—ASICs), as well as inno-vative signal-processing algorithms, smart-antenna systems have become practical for commercial use .
Smart-antenna systems are basically an extension of cell sectoring in which the sector coverage is composed of multiple beams . This is achieved by the use of antenna arrays, and the number of beams in the sector (e.g., 120◦) is a function of the array geometry. Because smart antennas can focus their radiation pattern toward the desired users while rejecting unwanted interferences, they can provide greater coverage area for each base station. Moreover, because smart antennas have a higher rejection interference, and therefore lower bit error rate (BER), they can provide a substantial capacity improvement. These systems can generally be classified as either Switched-Beam or Adaptive Array , , .
A. Switched-Beam Systems A switched-beam system is a system that can choose from one of many predefined patterns in order to enhance the received signal (see Figure 16.7), and it is obviously an extension of cell sectoring as each sector is subdivided into smaller sectors. As the mobile unit moves throughout the cell, the switched-beam system detects the signal strength, chooses the appropriate predefined beam pattern, 120° User Figure 16.7 Switched-beam system.
CELLULAR RADIO SYSTEMS EVOLUTION 937 (a) (b) Figure 16.8 Comparison of (a) switched-beam scheme, and (b) adaptive array scheme.
and continually switches the beams as necessary. The overall goal of the switched-beam system is to increase the gain according to the location of the user. However, since the beams are fixed, the intended user may not be in the center of any given main beam. If there is an interferer near the center of the active beam, it may be enhanced more than the desired user .
B. Adaptive Array Systems Adaptive array systems , provide more degrees of freedom since they have the ability to adapt in real time the radiation pattern to the RF signal environment. In other words, they can direct the main beam toward the pilot signal or SOI while suppressing the antenna pattern in the direction of the interferers or SNOIs. To put it simply, adaptive array systems can customize an appropriate radi-ation pattern for each individual user. This is far superior to the performance of a switched-beam system, as shown in Figure 16.8. This figure shows that not only the switched-beam system may not able to place the desired signal at the maximum of the main lobe but also it exhibits the inability to fully reject the interferers. Because of the ability to control the overall radiation pattern in a greater coverage area for each cell site, as illustrated in Figure 16.9, adaptive array systems greatly increase capacity. Figure 16.9 shows a comparison, in terms of relative coverage area, of conventional sec-torized switched-beam and adaptive arrays. In the presence of a low-level interference, both types Switched Beam Adaptive Array (a) Low Interference Environment Conventional Sectoring Switched Beam Adaptive Array (b) High Interference Environment Figure 16.9 Relative coverage area comparison among sectorized systems, switched-beam systems, and adap-tive array systems in (a) low interference environment, and (b) high interference environment .
938 SMART ANTENNAS A/D A/D • • • • • • • • • • • • Adaptive Algorithm ∑ w0 wM-1 DSP Output Signal DOA RCVR/Downconvert to Baseband RCVR/Downconvert to Baseband Figure 16.10 Functional block diagram of an adaptive array system.
of smart antennas provide significant gains over the conventional sectored systems. However, when a high-level interference is present, the interference rejection capability of the adaptive systems pro-vides significantly more coverage than either the conventional or switched-beam system .
Adaptive array systems can locate and track signals (users and interferers) and dynamically adjust the antenna pattern to enhance reception while minimizing interference using signal-processing algo-rithms. A functional block diagram of such a system is shown in Figure 16.10. This figure shows that after the system downconverts the received signals to baseband and digitizes them, it locates the SOI using the direction-of-arrival (DOA) algorithm, and it continuously tracks the SOI and SNOIs by dynamically changing the weights (amplitudes and phases of the signals). Basically, the DOA computes the direction of arrival of all signals by computing the time delays between the antenna elements, and afterward the adaptive algorithm, using a cost function, computes the appropriate weights that result in an optimum radiation pattern. The details of how the time delays and the weights are computed are discussed later in this chapter. Because adaptive arrays are generally more digital-processing intensive and require a complete RF portion of the transceiver behind each antenna element, they tend to be more costly than switched-beam systems.
C. Spatial Division Multiple Access (SDMA) The ultimate goal in the development of cellular radio systems is SDMA , . SDMA is among the most-sophisticated utilization of smart-antenna technology; its advanced spatial-processing capa-bility enables it to locate many users, creating different beam for each user, as shown in Figure 16.11.
This means that more than one user can be allocated to the same physical communication channel in the same cell simultaneously, with only an angle separation. This is accomplished by having N par-allel beamformers at the base station operating independently, where each beamformer has its own adaptive beamforming algorithm to control its own set of weights and its own direction-of-arrival algorithm (DOA) to determine the time delay of each user’s signal (see Figure 16.12) , .
Each beamformer creates a maximum toward its desired user while nulling or attenuating the other users. This technology dramatically improves the interference suppression capability while greatly increasing frequency reuse, resulting in increased capacity and reduced infrastructure cost. Basically, capacity is increased not only through intercell frequency reuse but also through intracell frequency reuse .
SIGNAL PROPAGATION 939 User #2 User #1 Figure 16.11 SDMA multibeam system.
16.4 SIGNAL PROPAGATION Up until now, the problem of capacity has been associated solely with cochannel interference and with the depletion of channels due to the high number of users. However, multipath fad-ing and delay spread also play a role in reducing system capacity , . Fortunately, because of the ability of smart-antenna systems to adapt to the signal environment, they are able to Beamformer 1 • • • • • • • • • • • • • • • • • • Adaptive Algorithm + w0 w1 DSP y1 DOA wM-1 Beamformer 2 y2 Beamformer N yN • • • Figure 16.12 SDMA system block diagram , .
940 SMART ANTENNAS Direct signal Multipath signal Multipath signal Figure 16.13 Multipath environment.
considerably reduce delay spread and multipath fading, thereby increasing capacity. This section gives a brief overview on signal propagation; for an in-depth study of the subject, the reader is referred to , , .
The signal generated by the user mobile device is omnidirectional in nature; therefore, it causes the signal to be reflected by structures, such as buildings. Ultimately, this results in the arrival of multiple delayed versions (multipath) of the main (direct) signals at the base station, as depicted in Figure 16.13. This condition is referred to as multipath , . In general, these multiple delayed signals do not match in phase because of the difference in path length at the base station, as shown by the example in Figure 16.14 . Because smart-antenna systems can tailor themselves to the signal environment, they can exploit or reject the reflected signals depending whether the signals are delayed copies of the SOI or the SNOIs. This is an advantage because smart antennas are not only capable of extracting information from the direct path of the SOI but they can also extract information from the reflected version of the SOI while rejecting all interferers or SNOIs. Therefore, because of this ability to manage multipath signals, smart-antenna systems improve link quality. As the signals are delayed, the phases of the multipath signal components can combine destructively over a narrow Figure 16.14 Two out-of-phase multipath signals .
SIGNAL PROPAGATION 941 Figure 16.15 Fade effect on a user signal .
bandwidth, leading to fading of the received signal level. This results in a reduction of the signal strength, and a representative example is illustrated in Figure 16.15 .
One type of fading is Rayleigh fading or fast fading , . Fading is constantly changing, and it is a three-dimensional (3-D) phenomenon that creates fade zones . These fade zones are usually small, and they tend to periodically attenuate the received signal (i.e., degrade it in quality) as the users pass through them. Although fading, in general, is a difficult problem, smart-antenna systems perform better than earlier systems, unless fading is severe. Figure 16.16 shows lighter shaded area as a representation of fade zones in a multipath environment.
Occasionally, the multipath signals are 180◦out of phase, as shown in Figure 16.17. This multi-path problem is called phase cancellation . When this happens, a call cannot be maintained for a long period of time, and it is dropped. In digital signals, the effect of multipath causes a condi-tion called delay spread. In other words, the symbols representing the bits collide with one another causing intersymbol interference (ISI), as shown in Figure 16.18. When this occurs, the bit error rate (BER) rises, and a noticeable degradation in quality is observed .
Finally, another signal propagation problem is cochannel interference . This occurs when a user’s signal interferes with a cell having the same set of frequencies. Omnidirectional cells suffer more from cochannel interference than do sectorized cells, and smart-antenna systems, because of their ability to tune out cochannel interference, perform at their best under such environment ; this comparison is shown in Figure 16.19.
Figure 16.16 Fade zones .
942 SMART ANTENNAS Direct signal Reflected signal Figure 16.17 Phase cancellation.
Figure 16.18 Delay spread and ISI.
16.5 SMART ANTENNAS’ BENEFITS The primary reason for the growing interest in smart-antenna systems is the capacity increase. In densely populated areas, mobile systems are usually interference-limited, meaning that the inter-ference from other users is the main source of noise in the system. This means that the signal-to-interference ratio (SIR) is much smaller than the signal-to-noise ratio (SNR). In general, smart anten-nas will, by simultaneously increasing the useful received signal level and lowering the interference level, increase the SIR.
Another benefit that smart-antenna systems provide is range increase. Because smart antennas are more directional than omnidirectional and sectorized antennas, a range increase potential is available Omnidirectional base station Sectorized base station 120° Smart antenna D Figure 16.19 Cochannel interference comparison of the different systems .
ANTENNA 943 . In other words, smart antennas are able to focus their energy toward the intended users, instead of directing it in other unnecessary directions (wasting) like omnidirectional antennas do. This means that base stations can be placed further apart, leading to a more cost-efficient development. Therefore, in rural and sparsely populated areas, where radio coverage rather than capacity is more important, smart-antenna systems are also well suited .
Another added advantage of smart-antenna systems is security. In a society that becomes more dependent on conducting business and transmitting personal information, security is an impor-tant issue. Smart antennas make it more difficult to tap a connection because the intruder must be positioned in the same direction as the user as seen from the base station to successfully tap a connection .
Finally, because of the spatial detection nature of smart-antenna systems, they can be used to locate humans in emergencies or for any other location-specific service .
16.6 SMART ANTENNAS’ DRAWBACKS While smart antennas provide many benefits, they do suffer from certain drawbacks. For example, their transceivers are much more complex than traditional base station transceivers. The antenna needs separate transceiver chains for each array antenna element and accurate real-time calibra-tion for each of them . Moreover, the antenna beamforming is computationally intensive, which means that smart-antenna base stations must be equipped with very powerful digital signal proces-sors. This tends to increase the system costs in the short term, but since the benefits outweigh the costs, it will be less expensive in the long run. For a smart antenna to have pattern-adaptive capabil-ities and reasonable gain, an array of antenna elements is necessary.
16.7 ANTENNA One essential component of a smart-antenna system is its sensors or antenna elements. Just as in humans the ears are the transducers that convert acoustic waves into electrochemical impulses, antenna elements convert electromagnetic waves into electrical impulses. These antenna elements play an important role in shaping and scanning the radiation pattern and constraining the adaptive algorithm used by the digital signal processor. There are a plethora of antenna elements that can be selected to form an adaptive array. This includes classic radiators such as dipoles, monopoles, loops, apertures, horns, reflectors, microstrips, and so on. Thus, a good antenna designer should consider the type of antenna element that is best suited for the application.
One element that meets the requirements and capabilities of a mobile device is that of an array of printed elements. There are a number of printed element geometries, patches as they are usu-ally referred to, some shown in Figure 14.2. The two most popular are the rectangular, discussed in detail in Section 14.2, and the circular, discussed in detail in Section 14.3. Such an array possesses the attributes to provide the necessary bandwidth, scanning capabilities, beamwidth, and sidelobe level. Furthermore, it is a low-cost technology suitable for commercial products, in addition to being lightweight and conformal to surfaces. The analysis and design of microstrip/patch antennas is dis-cussed in detail in Chapter 14. There are a number of analysis methods as well as software packages; one is that of .
16.7.1 Array Design The main beam of a larger array can resolve, because of its narrower beamwidth, the signals-of-interest (SOIs) more accurately and allows the smart-antenna system to reject more signals-not-of-interest (SNOIs). Although this may seem attractive for a smart-antenna system, it has two main 944 SMART ANTENNAS z y ...
...
d 1 2 3 M/2 M/2 3 2 1 d x r ϕ θ Figure 16.20 Linear array with elements along the y-axis.
disadvantages. One disadvantage is that it tends to increase the cost and the complexity of the hard-ware implementation, and the other is that it increases the convergence time for the adaptive algo-rithms, thereby reducing valuable bandwidth. Therefore, this issue is resolved on the basis of the analysis of the network throughput that will be discussed later in this chapter.
The array configuration that is well suited for mobile communication is usually a planar array.
The linear array configuration is not as attractive because of its inability to scan in 3-D space. On the other hand, a planar array can scan the main beam in any direction of 𝜃(elevation) and 𝜙(azimuth), as discussed in Chapter 6. Initially, a linear array will be analyzed to demonstrate some of the basic principles of array theory; eventually, most of the effort will be on planar arrays.
16.7.2 Linear Array The array factor of a linear array of M (even) identical elements with uniform spacing positioned symmetrically along the y-axis, as shown in Figure 16.20, can be written on the basis of the theory of Chapter 6, as (AF)M = w1e+j(1∕2)𝜓1 + w2e+j(3∕2)𝜓2 + ⋯+ wM∕2e+j[(M−1)∕2]𝜓M∕2 + w1e−j(1∕2)𝜓1 + w2e−j(3∕2)𝜓2 + ⋯+ wM∕2e−j[(M−1)∕2]𝜓M∕2 (16-1) Simplifying and normalizing (16-1), the array factor for an even number of elements with uniform spacing along the y-axis reduces to (AF)M = M∕2 ∑ n=1 wn cos[(2n −1)𝜓n] (16-2) ANTENNA 945 Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 Uniform Dolph-Tschebyscheff Figure 16.21 Array factors of an eight-element linear array (interelement spacing of 0.5λ).
where 𝜓n = 𝜋d λ sin 𝜃sin 𝜙+ 𝛽n (16-2a) In (16-2) and (16-2a), wn and 𝛽n represent, respectively, the amplitude and phase excitations of the individual elements. While in Chapter 6 the amplitude coefficients were represented by an, in signal-processing beamforming it is most common to represent them by wn, and it will be adopted in this chapter. The amplitude coefficients wn control primarily the shape of the pattern and the major-to-minor lobe level while the phase excitations control primarily the scanning capabilities of the array. Tapered amplitude distributions exhibit wider beamwidths but lower sidelobes. Therefore, an antenna designer can choose different amplitude distributions to conform to the application spec-ifications. This is shown in Figure 16.21, where the array factor of a uniform linear array is com-pared with the array factor of a Dolph–Tchebyscheff linear array. As discussed in Chapter 6, Dolph–Tchebyscheff arrays maintain all their minor lobes at the same level while compromising slightly on a wider half-power beamwidth.
16.7.3 Planar Array As mentioned earlier, linear arrays lack the ability to scan in 3-D space, and since it is necessary for portable devices to scan the main beam in any direction of 𝜃(elevation) and 𝜙(azimuth), planar arrays are more attractive for these mobile devices. Let us assume that we have M × N identical elements, M and N being even, with uniform spacing positioned symmetrical in the xy-plane as shown in Figure 16.22.
The array factor for this planar array with its maximum along 𝜃0, 𝜙0, for an even number of elements in each direction can be written as [AF(𝜃, 𝜙)]M×N = 4 M∕2 ∑ m=1 N∕2 ∑ n=1 wmn cos[(2m −1)u] cos[(2n −1)v] (16-3) 946 SMART ANTENNAS • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • y ( , ) z x dy dx • θ θ ϕ ϕ Figure 16.22 Planar array with uniformly spaced elements.
where u = 𝜋dx λ (sin 𝜃cos 𝜙−sin 𝜃0 cos 𝜙0) (16-3a) v = 𝜋dy λ (sin 𝜃sin 𝜙−sin 𝜃0 sin 𝜙0) (16-3b) In (16-3), wmn is the amplitude excitation of the individual element. For separable distributions wmn = wmwn. However, for nonseparable distributions, wmn ≠wmwn. This means that for an M × N planar array, only M + N excitation values need to be computed from a separable distribution, while M × N values are needed from a nonseparable distribution .
16.8 ANTENNA BEAMFORMING Intelligence, based on the definition of the Webster’s dictionary, is the ability to apply knowledge and to manipulate one’s environment. Consequently, the amount of intelligence a system has depends on the information collected, how it gains knowledge from the processed information, and its ability to apply this knowledge. In smart-antenna systems, this knowledge is gained and applied via algorithms processed by a digital signal processor (DSP) as shown in Figure 16.10. The objectives of a DSP are to estimate 1. the direction of arrival (DOA) of all impinging signals, and 2. the appropriate weights to ideally steer the maximum radiation of the antenna pattern toward the SOI and to place nulls toward the SNOI.
Hence, the work on smart antennas promotes research in adaptive signal-processing algorithms such as DOA and adaptive beamforming. The DOA estimation involves a correlation analysis of the array signals followed by eigenanalysis and signal/noise subspace formation while in adaptive beamform-ing the goal is to adapt the beam by adjusting the magnitude and phase of each antenna element such that a desirable pattern is formed .
This section presents a brief overview of direction-of-arrival algorithms followed by a review of adaptive beamforming algorithms. The adaptive beamforming begins with a simple example to give the reader an insight on the basics of adaptive beamforming.
ANTENNA BEAMFORMING 947 z r1 s(t) r2 (m,n) x dy ndy y mdx dx { dmn a az ^ a ax ^ a a ^ a ar ^ a ay Δr ^ θ ψ ϕ ρ Figure 16.23 M × N planar array with graphical representation of the time delay.
16.8.1 Overview of Direction-Of-Arrival (DOA) Algorithms After the antenna array receives the incoming signals from all directions, the DOA algorithm deter-mines the directions of all incoming signals based on the time delays. To compute the time delays, consider an M × N planar array with interelement spacing dx along the x-axis and dy along the y-axis as shown in Figure 16.23. When an incoming wave, carrying a baseband signal s(t) impinges at an angle (𝜃, 𝜙) on the antenna array, it produces time delays relative to the other antenna elements.
These time delays depend on the antenna geometry, number of elements, and interelement spacing.
For the rectangular grid array of Figure 16.23, the time delay of the signal s(t) at the (m, n)th element, relative to the reference element (0, 0) at the origin, is 𝜏mn = Δr v0 (16-4) where Δr and v0 represent respectively the differential distance and the speed of the light in free-space. The differential distance, Δr, is computed using Δr = dmn cos(𝜓) (16-5) where dmn is the distance from the origin and the (m, n)th element, and 𝜓is the angle between the radial unit vector from the origin to the (m, n)th element and the radial unit vector in the direction of the incoming signal s(t). Subsequently, dmn and cos(𝜓) are determined using dmn = √ m2 d2 x + n2 d2 y (16-6) cos(𝜓) = ̂ ar ⋅̂ a𝜌 |̂ ar| |̂ a𝜌| (16-7) 948 SMART ANTENNAS where ̂ ar and ̂ a𝜌are, respectively, the unit vectors along the direction of the incoming signal s(t) and along the distance dmn to the (m, n)th element. Thus, the unit vectors (i.e., ̂ ar and ̂ a𝜌) are expressed as ̂ ar = ̂ ax sin 𝜃cos 𝜙+ ̂ ay sin 𝜃sin 𝜙+ ̂ az cos 𝜃 (16-8a) ̂ a𝜌= ̂ axmdx + ̂ ayndy √ m2 d2 x + n2 d2 y (16-8b) where ̂ ax, ̂ ay, and ̂ az are, respectively, the unit vectors along the x-, y-, and z-axis. Finally, substitut-ing (16-5)–(16-8b) into (16-4), the time delay of the (m, n)th element, with respect to the element at the origin [i.e., (0, 0)], is written as 𝜏mn = mdx sin 𝜃cos 𝜙+ ndy sin 𝜃sin 𝜙 v0 (16-9) DOA estimation techniques can be categorized on the basis of the data analysis and implementa-tion into four different areas: conventional methods, subspace-based methods, maximum likelihood methods, and integrated methods, which combine property-restoral techniques with subspace-based approach .
Conventional methods for DOA estimation are based on the concepts of beamforming and null steering and do not exploit the statistics of the received signal. In this technique, the DOA of all the signals is determined from the peaks of the output power spectrum obtained from steering the beam in all possible directions. Examples of conventional methods are the delay-and-sum method (classical beamformer method or Fourier method) and Capon’s minimum variance method. One major disadvantage of the delay-and-sum method is its poor resolution; that is, the width of the main beam and the height of the sidelobes limit its ability to separate closely spaced signals . On the other hand, Capon’s minimum variance technique tries to overcome the poor resolution problem associated with the delay-and-sum method, and in fact, it gives a significant improvement. Although it provides better resolution, Capon’s method fails when the SNOIs are correlated with the SOI.
Unlike conventional methods, subspace methods exploit the structure of the received data, result-ing in a dramatic improvement in resolution. Two main algorithms that fall into this category are the MUltiple SIgnal Classification (MUSIC) algorithm and the Estimation of Signal Parameters via Rotational Invariance Technique (ESPRIT). In 1979, Schmidt proposed the conventional MUSIC algorithm that exploited the eigenstructure of the input covariance matrix . This algorithm pro-vides information about the number of incident signals, DOA of each signal, strengths and cross correlations between incident signals, and noise powers. Like many algorithms, the conventional MUSIC possesses drawbacks. One of the drawbacks is that it requires very precise and accurate array calibration. Another drawback is that, if the impinging signals are highly correlated, it fails because the covariance matrix of the received signals becomes singular. And lastly, it is computa-tionally intensive.
To improve the conventional MUSIC algorithm further, several attempts were made to increase its resolution performance and decrease its computational complexity. In 1983, Barabell developed the Root-MUSIC algorithm based on polynomial rooting and provided higher resolution; its drawback was that it was applicable only to uniformly spaced linear arrays . In 1989, Schmidt proposed the Cyclic MUSIC, a selective direction finding algorithm, which exploited the spectral coherence properties of the received signal and made it possible to resolve signals spaced more closely than the resolution threshold of the array. Moreover, the Cyclic MUSIC also avoids the requirements that the total number of signals impinging on the array must be less than the number of sensor elements . Then, in 1994, Xu presented the Fast Subspace Decomposition (FSD) technique to decrease ANTENNA BEAMFORMING 949 the computational complexity of the MUSIC algorithm . In a signal environment with multipath, where the signals received are highly correlated, the performance of MUSIC degrades severely. To overcome such a detriment, a technique called spatial smoothing was applied to the covariance matrix , .
The ESPRIT algorithm is another subspace-based DOA estimation technique originally proposed by Roy . Because ESPRIT has several advantages over MUSIC, such as that it 1. is less computationally intensive, 2. requires much less storage, 3. does not involve an exhaustive search through all possible steering vectors to estimate the DOA, and 4. does not require the calibration of the array, it has become the algorithm of choice. It is also used in the computer program, designated as DOA, which is found in the publisher’s website for this book, to determine the directions of arrival for linear and planar array designs of isotropic sources. Since its conception, the ESPRIT has evolved into the 2-D Unitary ESPRIT and the Equirotational Stack ESPRIT (ES-ESPRIT) , a more accurate version of the ESPRIT. The corresponding READ ME file explains the details of the program.
Maximum Likelihood (ML) techniques were some of the first techniques to be investigated for DOA estimation, but they are less popular than suboptimal subspace techniques because ML methods are computationally intensive. However, ML techniques outperform the subspace-based techniques in low SNR and in correlated signal environment .
The final category of DOA algorithms is the integrated technique that combines the property-restoral method with the subspace-based approach. A property-restoral technique is the Iterative Least Squares Projection Based Constant Modulus Algorithm (ILSP-CMA), a data-efficient and cost-efficient approach that is used to detect the envelope of the received signals and overcome many of the problems associated with the Multistage CMA algorithms .
In 1995, Xu and Lin proposed a new scheme that integrated ILSP-CMA and the subspace DOA approach. With an M-element antenna array, this scheme can estimate up to 2M2/3 DOAs of direct path and multipath signals while a conventional DOA can resolve no more than M DOAs. In 1996, Muhamed and Rappaport showed improvement in performance using an integrated DOA over the conventional ESPRIT when they combined the subspace-based techniques, such as ESPRIT and MUSIC, with the ILSP-CMA.
To show that the DOA can be determined on the basis of the time delays, the DOA of a two-element array will be derived.
Example 16.1 Derive the DOA of a two-element array. Show that the angle of arrival/incidence can be deter-mined on the basis of time delays and geometry of the system.
Solution: On the basis of the geometry of Figure 16.24, the time difference of the signal arriv-ing at the two elements can be written as Δt = (t1 −t2) = Δd v0 = d cos(𝜃) v0 where v0 is the speed of light in free-space. This equation can be rewritten as cos(𝜃) = v0 d Δt = v0 d (t1 −t2) 950 SMART ANTENNAS or 𝜃= cos−1 ( d v0 Δt ) = cos−1 [ d v0 (t1 −t2) ] This clearly demonstrates that the angle of incidence 𝜃(direction of arrival) can be determined knowing the time delay between the two elements (Δt = t1 −t2), and the geometry of the antenna array (in this case a linear array of two elements with a spacing d between the elements).
Incoming signal #1 d .cos( ) #2 d θ θ Figure 16.24 Incoming signal on a two-element array.
16.8.2 Adaptive Beamforming As depicted in Figure 16.10, the information supplied by the DOA algorithm is processed by means of an adaptive algorithm to ideally steer the maximum radiation of the antenna pattern toward the SOI and place nulls in the pattern toward the SNOIs. This is only necessary for DOA-based adaptive beamforming algorithms. However, for reference (or training) based adaptive beamforming algo-rithms, like the Least Mean Square (LMS) , that is used in this chapter, the adaptive beam-forming algorithm does not need the DOA information but instead uses the reference signal, or training sequence, to adjust the magnitudes and phases of each weight to match the time delays created by the impinging signals into the array. In essence, this requires solving a linear system of normal equations. The main reason why it is generally undesirable to solve the normal equations directly is because the signal environment is constantly changing. Before reviewing the most com-mon adaptive algorithm used in smart antennas, an example is given, based on , to illustrate the basic concept of how the weights are computed to satisfy certain requirements of the pattern, especially the formation of nulls.
Example 16.2 Determine the complex weights of a two-element linear array, half-wavelength apart, to receive a desired signal of certain magnitude (unity) at 𝜃0 = 0◦while tuning out an interferer (SNOI) at 𝜃1 = 30◦, as shown in Figure 16.25. The elements of the array in Figure 16.25 are assumed to be, for simplicity, isotropic and the impinging signals are sinusoids.
ANTENNA BEAMFORMING 951 Desired p(t) = Pejw0t Interference n(t) = Ne jw0t Figure 16.25 Two-element array receiving a desired signal at 𝜃0 = 0◦and an SNOI at 𝜃0 = 30◦.
Solution: The output y(t) of the array due to the desired signal p(t) is analyzed first, followed by the output due to the interferer n(t). Thus, the output y(t) of the array due to the desired signal p(t) is y(t) = Pej𝜔0t( ̇ w1 + ̇ w2) For the output y(t) to be equal (unity) only to the desired signal, p(t), it is necessary that ̇ w1 + ̇ w2 = 1 On the other hand, the output y(t) due to the interfering signal n(t) is given as y(t) = Nej(𝜔0t−𝜋∕4) ̇ w1 + Nej(𝜔0t+𝜋∕4) ̇ w2 where −𝜋/4 and +𝜋/4 terms in the above equation are due to the phase lag and lead, respec-tively, in reference to the array midpoint of the impinging interfering signal (see Figure 16.26 for details). Because ej(𝜔0t−𝜋∕4) = ej𝜔0t √ 2 (1 −j) ej(𝜔0t+𝜋∕4) = ej𝜔0t √ 2 (1 + j), the output y(t) can be rewritten as y(t) = Nej𝜔0t [√ 2 2 (1 −j) ̇ w1 + √ 2 2 (1 + j) ̇ w2 ] Therefore, for the output response y(t) to be zero (i.e., reject totally the interference), it is neces-sary that √ 2 2 (1 −j) ̇ w1 + √ 2 2 (1 + j) ̇ w2 = 0 952 SMART ANTENNAS 30° 30° 30° k(λ/4)sin(30°) = /4 π k(λ/4)sin(30°) = /4 n(t) λ/4 λ/4 #1 #2 π Figure 16.26 Phase lag and lead computations due to signal n(t).
Solving simultaneously the linear system of two complex equations, the second one of this exam-ple and the previous one, for ̇ w1 and ̇ w2, yields ̇ w1 = w′ 1 + jw′′ 1 = 1 2 −j1 2 ̇ w2 = w′ 2 + jw′′ 2 = 1 2 + j1 2 Thus, the above values of ̇ w1 and ̇ w2 are the optimum weights that guarantee the maximum signal-to-interference ratio (SIR) for a desired signal at 𝜃0 = 0◦and an interferer at 𝜃1 = 30◦.
Figure 16.27 shows, by the solid line, the array factor obtained on the basis of the weights derived above.
100 10–1 10–2 10–3 Amplitude 10–4 0 10 20 30 40 50 60 70 80 90 Observation angle (degrees) Without coupling With coupling With coupling (normalized) Figure 16.27 Comparison of array factors in the absence and in the presence of mutual coupling.
ANTENNA BEAMFORMING 953 If the signal environment is stationary, the weights are easily computed by solving the normal equations as shown in this example. However, in practice, the signal environment is dynamic or time varying, and therefore, the weights need to be computed with adaptive methods. In Section 16.8.4, one of the optimal beamforming techniques and adaptive algorithms (Least Mean Square—LMS) used in smart-antenna systems , , is reviewed; it is the one used in this chapter. There are others, and the interested reader is referred to the literature , –.
16.8.3 Mutual Coupling When the radiating elements in the array are in the vicinity of each other, the radiation character-istics, such as the impedance and radiation pattern, of an excited antenna element is influenced by the presence of the others. This effect is known as mutual coupling, and it can have a deleterious impact on the performance of a smart-antenna array. Mutual coupling usually causes the maximum and nulls of the radiation pattern to shift and to fill the nulls; consequently, the DOA algorithm and the beamforming algorithm produce inaccurate results unless this effect is taken into account. Fur-thermore, this detrimental effect intensifies as the interelement spacing is reduced , . This will not be discussed in this chapter. The interested reader is referred to the literature , .
However, a simple example follows to illustrate the effects of coupling on adaptive beamforming.
Example 16.3 To illustrate the effects of mutual coupling in beamforming, Example 16.2 is repeated here. How-ever, this time, mutual coupling is considered. Let us reconsider the two-element linear array, with half-wavelength spacing, receiving a desired signal at 𝜃0 = 0◦while tuning out an interferer (SNOI) at 𝜃1 = 30◦in the presence of mutual coupling. The elements of the array in Figure 16.28 are assumed to be, for simplicity, isotropic and the impinging signals are sinusoids.
Desired p(t) = Pe jw0t Interference n(t) = Ne jw0t Figure 16.28 Two-element array receiving an SOI at 𝜃0 = 0◦and an SNOI at 𝜃1 = 30◦in the presence of mutual coupling.
Solution: The output y(t) of the array due to the desired signal p(t) is analyzed first, followed by the output due to the interferer n(t). Thus, the output y(t) of the array due to p(t) is y(t) = Pej𝜔0t[(c11 + c12) ̇ ̃ w1 + (c22 + c21) ̇ ̃ w2] 954 SMART ANTENNAS where c11, c12, c21, and c22 represent, respectively, the mutual coefficients that describe how an element is affected because of the presence of another. Therefore, for the output y(t) to be equal only to the desired signal, p(t), it is necessary that (c11 + c12) ̇ ̃ w1 + (c22 + c21) ̇ ̃ w2 = 1 On the other hand, the output y(t) due to the interfering signal n(t) is given as y(t) = Nej(𝜔0t−𝜋∕4)(c11 ̇ ̃ w1 + c21 ̇ ̃ w2) + Nej(𝜔0t+𝜋∕4)(c12 ̇ ̃ w1 + c22 ̇ ̃ w2) where −𝜋/4 and +𝜋/4 terms are due to the phase lag and lead, respectively, in reference to the array midpoint of the impinging interfering signal (see Figure 16.26 for details). Because ej(𝜔0t−𝜋∕4) = ej𝜔0t √ 2 (1 −j) ej(𝜔0t+𝜋∕4) = ej𝜔0t √ 2 (1 + j), the output y(t) can be rewritten as y(t) = Nej𝜔0t { [ √ 2 2 (1 −j)c11 + √ 2 2 (1 + j)c12 ] ̇ ̃ w1 + [√ 2 2 (1 −j)c21 + √ 2 2 (1 + j)c22 ] ̇ ̃ w2 } Therefore, for the output response to be zero (i.e., reject totally the interference), it is necessary that [√ 2 2 (1 −j)c11 + √ 2 2 (1 + j)c12 ] ̇ ̃ w1 + [√ 2 2 (1 −j)c21 + √ 2 2 (1 + j)c22 ] ̇ ̃ w2 = 0 Solving simultaneously the linear system of complex equations, the second one of this example and the previous one, for ̇ w1 and ̇ w2, yields ̇ ̃ w1 = 1 2 c22 −c21 c22c11 −c12c21 −j1 2 c22 + c21 c22c11 −c12c21 ̇ ̃ w2 = 1 2 c11 −c12 c22c11 −c12c21 + j1 2 c11 + c12 c22c11 −c12c21 Note that the above computed weights in the presence of mutual coupling are related to those weights in the absence of mutual coupling by ̇ ̃ w1 = ̇ w1 ( c22 c22c11 −c12c21 −j c21 c22c11 −c12c21 ) ̇ ̃ w2 = ̇ w2 ( c11 c22c11 −c12c21 + j c12 c22c11 −c12c21 ) ANTENNA BEAMFORMING 955 where ̇ w1 and ̇ w2 are the computed weights in the absence of mutual coupling as derived in Example 16.2. Let us assume that the values for c11, c12, c21, and c22 are given, on the basis of the formulation in , by c11 = c22 = 2.37 + j0.340 c12 = c21 = −0.130 −j0.0517 Then, the computed altered weights in the presence of mutual coupling, using the mutual coupling coefficients above and the weights of Example 16.2, are ̇ ̃ w1 = ̃ w′ 1 + j ̃ w′′ 1 = 0.189 −j0.223 ̇ ̃ w2 = ̃ w′ 2 + j ̃ w′′ 2 = 0.250 + j0.167 On the basis of these new weights, ̇ ̃ w1 and ̇ ̃ w2, the computed patterns with coupling are dis-played and compared with that without coupling in Figure 16.27. The one with coupling is also normalized so that its value at 𝜃= 0◦is unity. It is apparent that mutual coupling plays a signif-icant role in the formation of the pattern. For example, in the presence of coupling, the pattern minimum is at 𝜃1 = 32.5◦at a level of nearly −35 dB (from the maximum), while in the absence of coupling the null is at 𝜃1 = 30◦at a level nearly −∞dB.
16.8.4 Optimal Beamforming Techniques In optimal beamforming techniques, a weight vector that minimizes a cost function is determined.
Typically, this cost function, related with a performance measure, is inversely associated with the quality of the signal at the array output, so that when the cost function is minimized, the quality of the signal is maximized at the array output . The most commonly used optimally beamforming techniques or performance measures are the Minimum Mean Square Error (MMSE), Maximum Signal-to-Noise Ratio (MSNR), and Minimum (noise) Variance (MV).
A. Minimum Mean Square Error (MMSE) Criterion One of the most widely used performance measures in computing the optimum weights is to mini-mize the MSE cost function. The solution of this function leads to a special class of optimum filters called Wiener filters , . In order to derive the weights based on the MMSE criterion, the error, 𝜀k, between the desired signal, dk, and the output signal of the array, yk, is written as 𝜀k = dk −wHxk (16-10) Therefore, the MSE based cost function can be written as JMSE(E[𝜀2 k]) = E[(dk −wHxk)2] = E[d2 k −2dkwHxk + wHxkxH k w] = d2 k −2wHE[dkxk] + wHE[xkxH k ]w (16-11) 956 SMART ANTENNAS where E[⋅] represents the expected value of [⋅]. Because E[dkxk] and E[xkxT k ] in (16-11) are the cross correlation, rxd, and the covariance, Rxx, respectively, (16-11) can be rewritten as JMSE(E[𝜀2 k]) = d2 k −2wHrxd + wHRxxw (16-12) In order to minimize the cost function (i.e., to minimize the MSE) in (16-12) with respect to the weights, one must compute the gradient, which is achieved by taking the derivative with respect to the weights and setting it equal to zero; that is, min w {JMSE(E[𝜀2 k])} ⇒𝜕 𝜕w{JMSE(E[𝜀2 k])} = 0 (16-13) Taking the derivative in (16-13) and solving in terms of the weights, w, yields wopt = R−1 xx rxd (16-14) Equation (16-14) is the so-called Wiener solution, which is the optimal antenna array weight vector, wopt, in the MMSE sense.
B. Least Mean Square (LMS) Algorithm One of the simplest algorithms that is commonly used to adapt the weights is the Least Mean Square (LMS) algorithm . The LMS algorithm is a low complexity algorithm that requires no direct matrix inversion and no memory. Moreover, it is an approximation of the steepest descent method using an estimator of the gradient instead of the actual value of the gradient, since computation of the actual value of the gradient is impossible because it would require knowledge of the incoming signals a priori. Therefore, at each iteration in the adaptive process, the estimate of the gradient is of the form ⌢ ∇[J(w)]k = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 𝜕J(w) 𝜕w0 ⋮ 𝜕J(w) 𝜕wL ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ (16-15) where the J(w) is the cost function of (16-12) to be minimized. Hence, according to the method of steepest descent, the iterative equation that updates the weights at each iteration is wk+1 = wk −𝜇 ⌢ ∇[J(w)]k (16-16) where 𝜇is the step size related to the rate of convergence. This simplifies the calculations to be per-formed considerably and allows algorithms, like the LMS algorithm, to be used in real-time appli-cations.
To summarize, the LMS algorithm minimizes the MSE cost function, and it solves the Wiener–Hopf equation, represented by (16-14), iteratively without the need for matrix inversion.
Thus, the LMS algorithm computes the weights iteratively as wk+1 = wk + 2𝜇xk(dk −xT k wk) (16-17) In order to assure convergence of the weights, wk, the step size 𝜇is bounded by the condition 0 < 𝜇< 1 λmax (16-18) ANTENNA BEAMFORMING 957 Figure 16.29 Implementation of the LMS algorithm based on the Wiener-Hopf equation .
where λmax is the maximum eigenvalue of the covariance matrix, Rxx. The main disadvantage of the LMS algorithm is that it tends to converge slowly, especially in noisy environments. A block diagram of the implementation of this (LMS) algorithm is shown in Figure 16.29 where d(k) is the reference/desired signal.
An interactive MATLAB computer program entitled Smart is included in the publisher’s website for this book, and it can be used to perform beamforming based on the LMS algorithm. One part of the program, designated Linear, is used for linear arrays of N isotropic elements while the other part, designated Planar, is used for planar arrays of M × N isotropic elements. The description of each is found in the READ ME file of the corresponding program.
To demonstrate the beamforming capabilities of the LMS algorithm, let us consider two examples.
Example 16.4 For an eight-element (M = 8) linear array of isotropic elements with a spacing of d = 0.5λ between them, as shown in Figure 16.20, it is desired to form a pattern where the pattern maxi-mum (SOI) is at 𝜃0 = 20◦. There are no requirements on any desired nulls (SNOIs) at any specific angles. Determine the relative amplitude (w’s) and phase (𝛽’s) excitation coefficients of the ele-ments using the following: 1. LMS beamforming algorithm 2. Classical method described in Chapter 6. Compare the results (amplitude, phase, and pat-tern) of the two methods.
Solution: Using the LMS algorithm, the obtained normalized amplitude and phase excitation coefficients are listed in Table 16.1 while the corresponding pattern, after 55 iterations, is exhib-ited in Figure 16.30. The beamformed pattern does meet the desired requirement of having a maximum at 𝜃0 = 20◦. Using the classical method of a scanning array detailed in Section 6.3.3, the obtained amplitude and phase excitation coefficients are also listed in Table 16.1 while the corresponding pattern is also exhibited in Figure 16.30.
By comparing the results of the two methods, it is apparent that the two methods lead to basically identical results in amplitude and phase excitation and corresponding patterns (they are basically indistinguishable). This indicates that the LMS algorithm, using only an SOI, converges to the element excitations, and corresponding pattern, of a uniform linear array.
958 SMART ANTENNAS TABLE 16.1 Amplitude (w’s) and Phase (𝜷’s) Excitation Coefficients of an Eight-Element (M = 8) Array Using the Classical Method and LMS Algorithm (d = 0.5𝛌, SOI = 20◦, 𝝁= 0.01, 55 iterations) Classical LMS (i = 55) Element w 𝜷(degrees) w 𝜷(degrees) 1 1.0000 0 1.0000 0 2 1.0000 −61.56 1.0000 −61.56 3 1.0000 −123.12 1.0000 −123.13 4 1.0000 −184.69 1.0000 −184.69 5 1.0000 −246.25 1.0000 −246.25 6 1.0000 −307.82 1.0000 −307.82 7 1.0000 −369.38 1.0000 −369.38 8 1.0000 −430.95 1.0000 −430.95 Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 SOI: 0 = 20° θ 0° 30° 90° 90° 60° 60° 30° Figure 16.30 Normalized pattern of an eight-element (M = 8) linear array of isotropic elements with a spacing of d = 0.5λ using both the classical method and LMS algorithm (𝜇= 0.01, 55 iterations).
The requirements of the beamformed pattern of the linear array of Example 16.4 were that only a maximum (SOI) was placed at a specific angle, with no requirements on forming nulls (SNOIs) at specific angles. This was accomplished using both the LMS algorithm and the classical method of Section 6.3.3. However, if both a maximum and a null are desired simultaneously at specific angles, the classical method of Section 6.3.3 cannot be used. In contrast, the beamforming method using the LMS algorithm, or any other algorithm, has that versatility.
To demonstrate the capability of the LMS algorithm for pattern beamforming, with both maxima and nulls, let us consider another example.
Example 16.5 For the eight-element linear array of isotropic elements of Example 16.4, with d = 0.5λ spac-ing between them, it is now desired to place a maximum (SOI) at 𝜃0 = 20◦(as was the case ANTENNA BEAMFORMING 959 for Example 16.4) and also to simultaneously place a null (SNOI) at 𝜃1 = 45◦. This cannot be accomplished with the classical method of Section 6.3.3. Therefore, perform this using the LMS algorithm.
1. Determine the normalized amplitude (w’s) and phase (𝛽’s) excitation coefficients of the ele-ments.
2. Plot the beamformed pattern.
Solution: Using the LMS algorithm, the obtained normalized excitation amplitude and phase coefficients, after 81 iterations, are listed in Table 16.2. The corresponding beamformed pattern is shown in Figure 16.31.
It is evident that the obtained pattern does meet the desired specifications. It is also observed that the amplitude excitation coefficients are symmetrical about the center of the array (4–5th element). A similar symmetry has also been observed in beamforming using circular arrays.
TABLE 16.2 Amplitude (w’s) and Phase (𝜷’s) Excitation Coefficients of an Eight-Element (M = 8) Array Using the LMS Algorithm (d = 0.5𝛌, SOI = 20◦, SNOI = 45◦, 𝝁= 0.01, 81 Iterations) LMS (i = 81) Element w 𝜷(degrees) 1 1.0000 −11.62 2 0.8982 −57.05 3 1.1384 −109.98 4 1.3760 −178.77 5 1.3760 −252.21 6 1.1384 −321.01 7 0.8982 −373.94 8 1.0000 −419.37 Amplitude Pattern (dB) –35 –30 –25 –20 –15 –10 –5 0 SOI: 0 = 20° SNOI: 0° 30° 60° 90° 90° 60° 30° 1 = 45° θ θ Figure 16.31 Normalized pattern of an eight-element (M = 8) linear array of isotropic elements with a spacing of d = 0.5λ using both the classical method and LMS algorithm (𝜇= 0.01, 81 iterations).
960 SMART ANTENNAS 16.9 MOBILE AD HOC NETWORKS (MANETs) As the necessity of exchanging and sharing data increases in our daily life, users demand ubiq-uitous, easy connectivity, and fast networks whether they are at work, at home, or on the move.
Moreover, these users are interested in interconnecting all their personal electronic devices (PEDs) in an ad hoc fashion. This type of network is referred to as Mobile Ad hoc NETwork (MANET).
However, MANETs are not only limited to civilian use but also to disaster recovery (such as fire, flood, and earthquake), law enforcement (such as crowd control, search, and rescue), and tactical communications (such as soldiers coordinating moves in a battlefield) . Thus, research in the area of mobile ad hoc networking has received wide interest. The ultimate objective is to design high capacity MANETs employing smart antennas.
This section gives a brief overview of MANETs and discusses MANETs employing smart-antenna systems. In particular, a description of the network layout used for the results and simu-lations of Sections 16.10.5 and 16.10.6 is presented. Moreover, the protocol used by the MANETs employing smart antennas is described.
16.9.1 Overview of Mobile Ad hoc NETworks (MANETs) A MANET consists of a collection of wireless mobile stations (nodes) forming a dynamic network whose topology changes continuously and randomly, and its internodal connectivities are managed without the aid of any centralized administration. In contrast, cellular networks are managed by a centralized administration or base station controller (BSC) where each node is connected to a fixed base station. Moreover, cellular networks provide single hop connectivity between a node and a fixed base station while MANETs provide multihop connectivity between Node A (source) and Node B (destination), as illustrated in Figure 16.32. This figure shows an example of two nodes, Node A and Node B, that desire to exchange data and are at some distance apart. Because they are out of radio range with each other, it is necessary for Node A to use the neighboring or intermediate nodes in forwarding its data packets to Node B. In other words, the data packets from Node A are passed onto the neighboring nodes in a series of single hops until the data packets reach Node B.
This type of interaction among nodes is referred to as peer to peer and follows a set of rules for communication, referred to as protocol. Because the responsibilities of organizing and controlling this network are distributed among the nodes, efficient routing protocols must operate in a distributed A B Figure 16.32 Multihop example of a MANET.
MOBILE AD HOC NETWORKS (MANETs) 961 manner and must be topology independent . Communication among the nodes takes place using one of the multiple access techniques: Time Division Multiple Access (TDMA), Frequency Division Multiple Access (FDMA), Code Division Multiple Access (CDMA), and Space Division Multiple Access (SDMA) .
The main advantage of MANETs is that they do not rely on extensive and expensive installations of fixed base stations throughout the usage area. Moreover, with the availability of multiple routes from the source node to the destination node, MANETs can perform route selection based on various metrics such as robustness and energy cost, thus optimizing network capacity and energy consump-tion of the individual node . In other words, if the route with the smallest distances among the intermediate nodes from the source node to the destination node is considered, it would result in a low-capacity network because the many hops produce a long network delay and a higher possibility of a link failure (less robust network). However, because of the small distances among the intermedi-ate nodes, each node or wireless device requires a small amount of energy to propagate its signal to its neighbor, thereby saving battery power. On the other hand, if the route with the farthest distances among the intermediate nodes was selected, it would result in a higher capacity network but with the most energy consumption. Therefore, this trade-off promotes research on algorithms that optimize network capacity, represented by throughput, and energy consumption based on the route selection.
The main disadvantage of MANETs is that the network capacity drops considerably as the network size (number of nodes) increases. For a more in-depth study on the area of MANETs, the reader is referred to , .
The next section discusses mobile ad hoc networks employing smart-antenna systems to improve network throughput. Furthermore, it discusses the protocol, based on the Medium Access Control (MAC) protocol of the IEEE 802.11 Wireless Local Area Network (WLAN) standard, used to allow the nodes to access the channel in a TDMA environment.
16.9.2 MANETs Employing Smart-Antenna Systems The ability of smart antennas to direct their radiation energy toward the direction of the intended node while suppressing interference can significantly increase the network capacity compared to a network equipped with omnidirectional antennas because they allow the communication channel to be reused. In other words, nodes with omnidirectional antennas keep the neighboring nodes on standby during their transmission while nodes with smart antennas focus only on the desired nodes and allow the neighboring nodes to communicate (refer to Figure 16.33). Therefore, smart antennas together with efficient access protocols can provide high capacity as well as robustness and reliability to mobile ad hoc networks. This section discusses the MANET layout used for the design of a high network throughput smart-antenna system. Also, included in this section is a discussion of the MAC protocol used by the MANET.
A. The Wireless Network A MANET of 55 nodes uniformly distributed in an area of 1,000 × 600 m2, as shown in Figure 16.34, is implemented using OPNET TM Modeler/Radio® tool , a simulation software package devel-oped by OPNET Technologies, Inc. to analyze, design, and implement communication networks, devices, and protocols. The nodes of the wireless network are equipped with four planar subarrays to cover all possible directions as shown in Figure 16.35. Specifically, each planar subarray in a node covers a sector of 90◦(from −45◦to +45◦) relative to broadside (𝜃= 0◦). For example, in the multihop network of Figure 16.35, Node A communicates with Node B with a beam scanned at 𝜃0, and Node B communicates with Node C with a beam scanned at 𝜃1. The data traffic through each of these nodes is modeled to follow a Poisson distribution, and each node changes position at random every time it transfers two consecutive packets (payload), to model the nodes’ mobility. The length of each payload packet is 1024 bits. For more details on the MANET layout refer to , while 962 SMART ANTENNAS Load (# packets) Throughput (# packets) Load (# packets) Throughput (# packets) (a) Low throughput with omnidirectional antennas (b) High throughput with adaptive antennas Figure 16.33 Capacity comparison of a network employing omnidirectional antennas and a network with smart antennas.
the design of the smart-antenna system (i.e., antenna and adaptive signal processing) is presented in Section 16.10.
B. The Protocol The protocol used for the MANET of Figure 16.35 is based on the MAC protocol of IEEE 802.11 Wireless Local Area Network standard and incorporates all necessary features to allow nodes to 1,000 m 600 m Figure 16.34 MANET layout of 55 nodes used for the design of a high throughput smart-antenna system.
MOBILE AD HOC NETWORKS (MANETs) 963 Node A Node B Node C 0 0 θ θ 1 θ 1 θ Figure 16.35 Example of three nodes communicating with each other.
access the channel in a TDMA environment. Moreover, it facilitates the use of smart antennas and increases the spatial reuse of TDMA slots, thereby increasing the capacity of the network. The pro-tocol timing diagram is shown in Figure 16.36. A brief explanation of the protocol follows.
a. Channel Access—When a source node (SRC) has a packet to transmit to a destination node (DEST), which is within radio range in a single hop, it first senses the state of the communica-tion channel. If the communication channel is busy, SRC waits until it becomes idle. Once the communication channel is idle, SRC sends a Request-To-Send (RTS) signal using the antenna array in nearly omnidirectional mode. This is possible by having only one active element of each subarray of Figure 16.35. When DEST receives successfully the RTS signal, it transmits in nearly omnidirectional mode a Clear-To-Send (CTS) signal to SRC.
RTS RTS ACK RXTRN DATA (With Smart Antenna) TXTRN CTS Figure 16.36 MAC Protocol timing diagram (based on IEEE 802.11).
964 SMART ANTENNAS b. Beamforming—When all neighboring nodes receive the CTS signal sent by DEST, they wait or standby until beamforming is complete before accessing the channel. After SRC receives the CTS signal, it transmits the training packet for DEST (RXTRN) in nearly omnidirectional mode. DEST receives the RXTRN, determines the angle of arrival of the RXTRN signal using the DOA algorithm and computes the complex weights to steer the beam toward SRC. Then, DEST sends the training packet for SRC (TXTRN) in smart-antenna mode (i.e., using the com-puted weights for the array directed towards DEST). Finally, after SRC receives the TXTRN signal, it also computes the angle of arrival of the TXTRN signal using the DOA algorithm and the complex weights to steer the beam toward DEST.
c. Data Transfer—Once SRC and DEST have directed their radiation pattern towards one another, the beamforming period is complete, and data transferring (DATA) begins. At this time, the channel is freed for the other neighboring nodes to start transmitting their RTS sig-nals. If the neighboring nodes’ transmission causes interference above a predefined threshold to SRC and DEST, a new set of weights are computed by SRC and DEST to place nulls toward the direction of the interfering nodes (i.e., SNOIs). At the completion of the DATA packet, the acknowledged (ACK) signal is transmitted by DEST, acknowledging the successful reception of the DATA packet.
Note that the training packets, RXTRN and TXTRN, are variable and are part of the payload (refer to Figure 16.36); thus, their lengths affect the length of the DATA packet, which is important because it is the packet that carries information. In other words, if the DATA packet is short, less information is transmitted and thus low network throughput results. Conversely, if the DATA packet is long, more information is transmitted and thus high network throughput is achieved. The influence of the beamforming period or training sequence length on the overall network throughput is investigated in Section 16.10.6.
Another point to clarify is that if omnidirectional antennas were used instead of smart antennas, the communication channel would be blocked throughout DATA transmission (refer to Figure 16.36).
In other words, no node within radio range of SRC and DEST would be allowed to transmit in order to avoid packet collision (i.e., interference). Thus, this infers that smart antennas increase network capacity by allowing the communication channel to be reused.
Simulation results of MANETs using smart antennas, described in this section, are presented in Sections 16.10.5 and 16.10.6.
16.10 SMART-ANTENNA SYSTEM DESIGN, SIMULATION, AND RESULTS This section illustrates the design process of a smart-antenna system. The first step of this process is to choose and design an antenna element (in this chapter, a microstrip patch). Then, these designed antenna elements are combined in a particular configuration to form an array. The array config-uration, the interelement spacing, and reference signal or training sequence are used in the LMS algorithm to compute the appropriate complex weights. Finally, these weights are tested using the Ensemble® software to verify the overall design. The design process just described is docu-mented below with the aid of some preliminary results. Network capacity for various antenna pat-terns is evaluated in order to guide the antenna design for high network capacity.
16.10.1 Design Process The objective in this chapter is to introduce a process to design a smart-antenna system for a portable device. The proposed design process is composed of several steps or objectives. The first step of this process is to design an antenna suitable for the network/communication requirements such as the SMART-ANTENNA SYSTEM DESIGN, SIMULATION, AND RESULTS 965 required beamwidth to maintain a tolerable bit error rate (BER) or throughput. The antenna design constitutes of a single-element design (e.g., dimensions, material, and geometry), and array design (e.g., configuration, interelement spacing, and number of elements). To optimize the antenna design, the Ensemble® , a MoM (Method of Moments) simulation software package, analyzes the design and computes the S-parameters, Z-parameters, and far-field amplitude patterns.
After the antenna design step is complete, the second step is to select an adaptive algorithm that minimizes the MSE. In addition, a sidelobe control technique is implemented to prevent environmen-tal noise or clutter noise from being received by the high sidelobes and reduce the overall system performance. Because of its complexity, the details of this step will be omitted. The interested reader is referred to , .
After the adaptive algorithm has determined the complex weights that scan the beam toward the direction of the SOI, and place the nulls toward the direction of the SNOIs, these complex weights are entered in Ensemble® as the final step of the design process to verify the overall design.
Specifically, the final step consists of comparing the far-field amplitude patterns produced by Ensem-ble® using the computed complex weights with the cavity model far-field patterns (i.e., using the array factor produced by the LMS algorithm’s weights and the single-element pattern).
16.10.2 Single Element—Microstrip Patch Design The first step of the design process is to design a single element. As mentioned earlier, because microstrip patch antennas are inexpensive, lightweight, conformal, easy to manufacture, and ver-satile, they are the most suitable type of elements for portable devices. A rectangular microstrip patch antenna is considered in this chapter. It is designed to operate at a frequency of 20 GHz (i.e., fr = 20 GHz) using silicon as a substrate material with a dielectric constant of 11.7 (i.e., 𝜀r = 11.7), a loss tangent of 0.04 (i.e., tan 𝛿= 0.04), a thickness of 0.300 mm (i.e., h = 0.300 mm), and an input impedance of 50 ohms (i.e., Rin = 50 ohms).
16.10.3 Rectangular Patch The rectangular patch is by far the most widely used configuration. It has very attractive radiation characteristics and low cross-polarization radiation . In this chapter, the rectangular patch is ana-lyzed and designed using the cavity model, and then its design is optimized using the IE/MoM with the aid of Ensemble® that computes S-parameters for microstrip and planar microwave struc-tures.
The first step of the design procedure of a rectangular patch antenna is to compute its physical dimensions. The physical width, W, is calculated using (14-6) while its physical length, L, is com-puted using (14-7) and (14-1) to (14-2). Afterward, the probe location or excitation feed point is determined using (14-20a) in order to match the impedance (i.e., Rin = 50 ohms). The computed geometry values obtained from the cavity model are summarized in Figure 16.37.
After the rectangular patch antenna is designed using the cavity model, its design is verified with Ensemble® . As expected, because the cavity model is not as accurate as the IE/MoM, the location and the physical dimensions of the probe of the rectangular patch need to be taken into account and optimized in order for the element to resonate at 20 GHz with an input resistance of 50 ohms. The optimized design parameters of the rectangular patch antenna using the Ensemble® are also shown in Figure 16.37.
The magnitude of S11 versus frequency (also referred to as return loss) is shown, as a verification of the design, in Figure 16.38. It shows that the rectangular patch antenna is indeed resonating at 20 GHz with a return loss of −21.5 dB, and its −3 dB and −10 dB bandwidths are 0.74 GHz and 0.25 GHz, respectively. Because the radio’s power amplifier tends to reduce its output power, or 966 SMART ANTENNAS Cavity Model Optimized ® Ensemble W 2.976 mm 2.247 mm L 2.129 mm 2.062 mm yo 1.488 mm 1.164 mm xo 1.256 mm 0.794 mm h 0.300 mm 0.300 mm 11.7, Si 11.7, Si fr 20 GHz 20 GHz r y W L h x0 y0 x z θ ϕ Figure 16.37 Rectangular patch antenna design using the cavity model.
worse become unstable if the VSWR is too large, a stricter definition of the antenna bandwidth is recommended. Thus, less than the −10 dB bandwidth is used.
The far-field amplitude patterns along the E-plane (i.e., E𝜃when 𝜙= 0◦) and the H-plane (i.e., E𝜙when 𝜙= 90◦) from Ensemble® are compared with the cavity model patterns, and they are shown in Figure 16.39. The H-plane far-field patterns from the two models match almost perfectly; however, the E-plane far-field pattern, based on Ensemble® , shows a maximum at 𝜃= 5◦and a null at 𝜃= 90◦. The reasons for these discrepancies between the two patterns are that the cavity model (the least accurate of the two methods) 1. does not take into account the probe location, and that it 2. assumes the dielectric material of the substrate is truncated and does not cover the ground plane beyond the edges of the patch (see Chapter 14, Section 14.2.2).
19 19.5 20 20.5 21 –25 –20 –15 –10 –5 0 Frequency (GHz) |S11| (dB) –10 dB BW 0.25GHz –3 dB BW 0.74 GHz Figure 16.38 S11 (return loss) for the 20 GHz rectangular patch antenna. (source: c ⃝2002 IEEE).
SMART-ANTENNA SYSTEM DESIGN, SIMULATION, AND RESULTS 967 ϕ Ensemble® Cavity Model 0 90 60 30 30 60 90 −10 −20 −30 dB −30 −20 −10 0 0 90 60 30 30 60 90 0 (a) E-plane ( = 0°) 0 −10 −20 −30 dB −−30 −20 −10 0 (b) H-plane ( = 90°) ϕ Figure 16.39 Rectangular patch antenna amplitude patterns using the cavity model and Ensemble® .
(source: c ⃝2002 IEEE).
16.10.4 Array Design To electronically scan a radiation pattern in a given direction, an array of elements arranged in a spe-cific configuration is essential. Two configurations are chosen here: the linear array and the planar array. Although linear arrays lack the ability to scan in 3-D space, planar arrays can scan the main beam in any direction of 𝜃(elevation) and 𝜙(azimuth). Consequently, a planar array is best suited for portable devices that require to communicate in any direction. However, in this chapter, a linear array is designed and analyzed initially for simplicity to illustrate some important features. Then, a planar array is designed to be eventually incorporated in the smart-antenna system for a wireless and mobile device.
A. Linear Array Design Following the design of the individual rectangular patch antenna, a linear array of eight microstrip patches with interelement spacing of λ/2 (half wavelength), where λ is 1.5 cm (based on the reso-nance frequency of 20 GHz), is designed (see Figure 16.40). The reasons for choosing interelement spacing of λ/2 are as follows. To combat fading, the interelement spacing of at least λ/2 is necessary 968 SMART ANTENNAS fr r 11.7, Si tan 0.04 y0 1.164 mm x0 0.794 mm h 0.300 mm L 2.062 mm W 2.247 mm dx 7.500 mm Dx 54.562 mm 20 GHz x y L W z Dx dx y W L h x0 0 E-plane; x-z plane H-plane; y-z plane θ ϕ δ Figure 16.40 Design of an eight-element linear array with rectangular patches.
so that the signals received from different antenna elements are (almost) independent in a rich scatter-ing environment (more precisely, in a uniform scattering environment , ). In such cases, the antenna arrays provide performance improvement through spatial diversity. However, to avoid grat-ing lobes (multiple maxima), the interelement spacing should not exceed one wavelength. But most important, to avoid aliasing and causing of nulls to be misplaced, the interelement spacing should be less or equal to λ/2 (the Nyquist rate) . Thus, to satisfy all three conditions, the interelement spacing of λ/2 (half wavelength) is chosen.
The total amplitude radiation patterns of the eight-element linear array based on the cavity model are represented, neglecting coupling, by the product of the element pattern (static pattern) and the array factor (dynamic pattern). In Chapter 6 this is referred to as pattern multiplication, and it is represented by (6-5). The array factor is dynamic in the sense that it can be controlled by the complex weights, wn. The amplitudes of these complex weights control primarily the shape of the pattern and the major-to-minor lobe ratio while the phases control primarily the scanning capability of the array.
B. Planar Array Design The planar arrays designed in this section are composed of only rectangular patches. Planar arrays of 4 × 4 and 8 × 8 elements of rectangular microstrip patches with interelement spacing of λ/2 (half y0 W L h x0 y L W x z Dx Dy dx dy Dx 54.562 mm Dy 54.747 mm dx 7.500 mm dy 7.500 mm W 2.247 mm L 2.062 mm h 0.300 mm x0 0.794 mm y0 1.124 mm tan 0.04 r 11.7, Si f 20 GHz δ Figure 16.41 Design of an 8 × 8 planar array with rectangular patches. (source: c ⃝2002 IEEE).
SMART-ANTENNA SYSTEM DESIGN, SIMULATION, AND RESULTS 969 Figure 16.42 Average network throughput versus network load for different array sizes and excitations.
(source: c ⃝2002 IEEE).
wavelength), where λ is 1.5 cm, are designed. The design and dimensions of the 8 × 8 planar array are given in Figure 16.41. Those of the 4 × 4 design are similar.
16.10.5 4 × 4 Planar Array versus 8 × 8 Planar Array After the design and analysis of the 4 × 4 and 8 × 8 planar arrays with rectangular patch elements, it is necessary to determine which array configuration is most attractive for a wireless device in a Mobile Ad hoc NETwork (MANET) described in Section 16.9. As observed in Chapter 6, the number of array elements affects the beamwidth of a radiation pattern. That is, when more elements are used in an array (larger size array), the narrower is the main beam. Furthermore, the narrower beamwidth will resolve more accurately the SOIs and the SNOIs. However, these results and inferences do not quantify the overall performance of the network. Therefore, an ad hoc network of 55 nodes (see Figure 16.34) equipped with the 4 × 4 and 8 × 8 planar arrays is simulated using OPNET TM Modeler/Radio® tool , and the average network throughput (Gavg) is measured.
Figure 16.42 shows the average network throughput versus network load (i.e., Gavg vs. Lavg) for the different nonadaptive antenna patterns; both uniform and Tschebyscheff (−26 dB). This figure indicates that the throughput for the case of the 8 × 8 array size is greater compared to the 4 × 4 array size, and also the Tschebyscheff arrays provide slightly greater throughput than their respec-tive uniform arrays. These are attributed, respectively, to smaller beamwidths of the 8 × 8 arrays (compared to the 4 × 4 arrays) and lower sidelobes of the Tschebyscheff arrays (compared to the uniform arrays). In both cases, the smaller beamwidths and lower sidelobes lead to lower cochannel interference. Thus, on the basis of the network throughput results, the 8 × 8 array configuration is chosen for the remaining part of this chapter.
16.10.6 Adaptive Beamforming The previous section described nonadaptive beamforming where the array factors of the array were obtained from complex weights that did not depend on the signal environment. However, the array factors of this section are produced from complex weights or excitations that depend on the signal 970 SMART ANTENNAS SNOI SOI −80 −60 −40 −20 −10 −20 −30 −40 −50 −60 −70 −80 0 0 20 40 60 80 48 dB (degrees) Magnitude (dB) LMS Algorithm SOI: ⇒ = 0° SNOI: ⇒ = 45°, = 0° Tschebyscheff (−26 dB sidelobes) 8 × 8 array, d = 0.5 λ θ θ θ ϕ Figure 16.43 E-plane (𝜙= 0◦) radiation pattern comparison for the nonadaptive beamformer and adaptive beamformer.
environment. This technique is referred to as adaptive beamforming where a digital signal proces-sor (DSP) computes the complex weights using an adaptive algorithm that generate an array factor for an optimal signal-to-interference ratio (SIR). Specifically, this results in an array pattern where ideally the maximum of the pattern is placed toward the source or SOI while nulling the interferers or SNOIs. Because of this, adaptive beamformers tend to be more costly owing to the extra hard-ware compared to nonadaptive beamformers. Thus, the question may be, how much improvement do adaptive beamformers provide over nonadaptive ones?
In order to answer this question, the overall network throughput of an adaptive beamformer is compared to the throughput of a nonadaptive beamformer. Each beamformer is equipped with an 8 × 8 planar array and receives an SOI at broadside (i.e., 𝜃0 = 0◦) and an SNOI at 𝜃1 = 45◦, 𝜙1 = 0◦. The radiation pattern used for the nonadaptive beamformer is a Tschebyscheff design of -26 dB and for the adaptive beamformer it is the LMS algorithm [discussed in Section 16.8.4(B)] generated pattern.
The E-plane (𝜙= 0◦) of each beamformer is compared in Figure 16.43. The adaptive beamformer or the LMS algorithm generated pattern shown in Figure 16.43 has the maximum toward the SOI and the null toward the SNOI while the nonadaptive Tschebyscheff design beamformer has only the maximum toward the SOI. Also, the intensity toward the SNOI for the Tschebyscheff design is 48 dB higher than that of the LMS algorithm generated pattern. Thus, by attenuating the strength of the interferer more, it would be expected that the throughput of the adaptive beamformer will be higher than the throughput of the nonadaptive beamformer since the cochannel interference is reduced and the SIR is higher. In fact, as shown in Figure 16.44, where the throughput of the adaptive beamformer is compared to the throughput of the Tschebyscheff (−26 dB sidelobe level) nonadapted antenna pattern, the LMS beamforming algorithm pattern leads to a higher throughput even though the Tschebyscheff pattern exhibits much lower minor lobes.
To investigate the influence of the beamforming period or training sequence length (refer to Fig-ure 16.36) on the network efficiency, the network throughput and the network delay are plotted, respectively, in Figure 16.45 and Figure 16.46 for various beamforming periods or training sequence lengths. It is apparent that the network throughput drops and the network delay increases significantly with increasing beamforming period or training sequence length. These figures show that, for this SMART-ANTENNA SYSTEM DESIGN, SIMULATION, AND RESULTS 971 Figure 16.44 Network throughput comparison of the adaptive and nonadaptive beamformer. (source: c ⃝2002 IEEE).
particular network, for beamforming periods or training sequence lengths larger than 20% of the payload, results in a network throughput and network delay similar to a network of omnidirectional antennas. Therefore, this suggests that smart antennas may be not as effective for networks with long beamforming periods or large training sequence lengths.
Figure 16.45 Beamforming period or training sequence length effects on the network throughput. (source: c ⃝2002 IEEE).
972 SMART ANTENNAS 0 5 10 15 20 25 0 2 4 6 8 10 12 14 Train - 6% Train - 10% Train - 20% Ominidirectional Load (# packets) Delay (# packets) Figure 16.46 Beamforming period or training sequence length effects on the network delay. (source: c ⃝2002 IEEE).
16.11 BEAMFORMING, DIVERSITY COMBINING, RAYLEIGH-FADING, AND TRELLIS-CODED MODULATION In this section, adaptive antenna arrays using the LMS algorithm are used to evaluate the performance of a communication channel using the Bit Error Rate (BER) as criterion. The adaptive array used is an 8 × 8 planar array whose patterns are beamformed so that the SOIs and SNOIs are in directions listed in Table 16.3. The BER of the channel is evaluated initially assuming only an SOI and then an SOI plus an SNOI; then the performance of the two is compared. For all the cases considered in this section, the desired (SOI) and interfering signals (SNOI) are assumed to be synchronized, which is considered as a worst-case assumption. It is also assumed that the desired and interfering signals have equal power. For the LMS algorithm, the length of the training sequence was selected to be 60 symbols, and it is transmitted every data sequence of length 940 symbols (total 1,000 symbols; i.e., 6% overhead). The symbol rate chosen was 100 Hz (symbol duration of T = 1∕100 = 0.01 = 10 msec) , .
The BER of a communication channel with Binary Phase Shift Keying (BPSK) modulation is first evaluated where the received signal is corrupted with an Additive White Gaussian Noise (AWGN).
The results are shown in Figure 16.47 with the set of curves marked uncoded. It is apparent that the adaptive antenna array using the LMS algorithm can suppress one interferer without any performance loss over an AWGN channel.
To improve the performance of the system, trellis-coded modulation (TCM) schemes are used together with adaptive arrays –. In this scheme, the source bits are mapped to channel symbols using a TCM scheme, and the symbols are interleaved using a pseudorandom interleaver TABLE 16.3 Signals and Their Directions SOI SNOI DOA 𝜃0 𝜙0 𝜃1 𝜙1 0◦ 0◦ 45◦ 0◦ BEAMFORMING, DIVERSITY COMBINING, RAYLEIGH-FADING, AND TRELLIS-CODED MODULATION 973 Figure 16.47 BER for uncoded BPSK over AWGN channel and trellis-coded QPSK modulation based on eight-state trellis encoder. (source: c ⃝2002 IEEE).
to uncorrelate the consecutive symbols to prevent bursty errors. The signal received by the adaptive antenna array consists of a faded version of the desired signal and a number of interfering signals plus AWGN. The receiver combines the signals from each antenna element using the LMS algorithm. A trellis-coded Quadrature Phase Shift Keying (QPSK) modulation scheme based on an eight-state trellis encoder was considered . The performance of the TCM QPSK system in terms of BER is also displayed in Figure 16.47 with the set of curves indicated as TCM. It is again observed that the adaptive antenna array using the LMS algorithm can suppress one interferer without loss in performance. By comparing the two sets of data, uncoded versus TCM coded, it is apparent that the QPSK TCM system over AWGN is better than that of the uncoded BPSK system over AWGN channel by about 1.5 dB at a BER of 10−5.
Wireless communication systems are characterized by time-varying multipath channels, which are typically modeled as “fading channels.” Fading, if not mitigated by powerful signal-processing and communication techniques, degrades the performance of a wireless system dramatically. In order to combat fading, the receiver is typically provided with multiple replicas of the transmitted signal . If the replicas fade (almost) independently of each other, then the transmitted information will be recovered with high probability since all the replicas will not typically fade simultaneously. An effective diversity technique is space diversity where multiple transmit and/or receive antennas are used. At the receiver, if the separation between antenna elements is at least λ/2, the signal received from different antenna elements are (almost) independent in a rich scattering environment (more pre-cisely, in a uniform scattering environment) and hence, in such cases, the antenna arrays provide performance improvement through spatial diversity. For the channels considered in this section, a Rayleigh-fading channel is assumed.
In order to simulate the fading channel, a filtered Gaussian model is used with a first-order low-pass filter. The BER results of the LMS algorithm of the uncoded system over a Rayleigh flat fading AWGN channel are displayed in Figure 16.48. The BER results indicate that when the Doppler spread of the channel is fm = 0.1 Hz (fmT = 0.001), the performance of the system degraded about 4 dB if one equal power interferer is present compared to the case of no interferers. If the channel 974 SMART ANTENNAS Figure 16.48 BER over uncoded Rayleigh-fading channel with Doppler spreads of fm = 0.1 Hz (fmT = 0.001) and fm = 0.2 Hz (fmT = 0.002). (source: c ⃝2002 IEEE).
faded more rapidly, it is observed that the LMS algorithm performs poorly. For example, the per-formance of the system over the channel with fm = 0.2 Hz ( fmT = 0.002) Doppler spread degraded about 4 dB at a BER of 10−4 compared to the case when the Doppler spread was 0.1 Hz. An error floor for the BER was observed for SNRs larger than about 18 dB. For a relatively faster fading in the presence of an equal power interferer, the performance of the system degrades dramatically, imply-ing that the performance of the adaptive algorithm depends highly on the fading rate. Furthermore, Figure 16.49 BER for trellis-coded QPSK modulation over Rayleigh-fading channel with Doppler spreads of fm = 0.1 Hz (fmT = 0.001) and fm = 0.2 Hz (fmT = 0.002). (source: c ⃝2002 IEEE).
OTHER GEOMETRIES 975 1 2 ^ n N a dc ar ^ a n n x y z r n n + 1 N −1 Δrn ϕ ϕ ψ ρ θ Figure 16.50 Geometry of uniform circular array of N elements.
if the convergence rate of the LMS algorithm is not sufficiently high to track the variations over a rapidly fading channel, adaptive algorithms with faster convergence should be employed.
The same system but with TCM coding was then analyzed over a Rayleigh-fading channel and the BER results for Doppler spreads of fm = 0.1 Hz (fmT = 0.001) and fm = 0.2 Hz (fmT = 0.002), are shown in Figure 16.49. When the TCM coded data of Figure 16.49 is compared with that of uncoded system of Figure 16.48, it is observed that when the Doppler spread is 0.2 Hz and there is one interferer, there is still a reducible error floor on the BER. However, the error floor is reduced compared to the uncoded BPSK case. It can be concluded that the TCM scheme provides some coding advantage in addition to spatial diversity advantage.
16.12 OTHER GEOMETRIES Until now, the investigation of smart antennas suitable for wireless communication systems has involved primarily rectilinear arrays: uniform linear arrays (RLAs) and uniform rectangular arrays (URAs). Different algorithms have been proposed for the estimation of the direction of arrivals 1 2 ^ mn Nmn 1st ring mth ring am dcm ar ^ a mn mn x y z r m,n m,n + 1 . . .
Nm −1 Δrmn θ ϕ ϕ ψ ρ Figure 16.51 Geometry of uniform planar circular array.
976 SMART ANTENNAS (DOAs) of signals arriving to the array and several adaptive techniques have been examined for the shaping of the radiation pattern under different constraints imposed by the wireless environment.
Another general antenna configuration that can be used for pattern beamforming is that of circular topologies. Two such geometries are shown in Figures 16.50 and 16.51: a uniform circular array (UCA) of Figure 16.50 and a uniform planar circular array (UPCA) of Figure 16.51. In the literature for adaptive antennas, not as much attention has been devoted so far to circular configurations despite their ability to offer some advantages. An obvious advantage results from the azimuthal symmetry of the UCA geometry. Because of the fact that a UCA does not have edge elements, directional patterns synthesized by this geometry can be electronically scanned in the azimuthal plane without a significant change in beam shape.
Because of space limitations, the analysis and beamforming capabilities of these configurations will not be pursued further here. The reader is referred to the literature – and others.
16.13 MULTIMEDIA In the publisher’s website for this book, the following multimedia resources are included for the review, understanding, and visualization of the material of this chapter: a. Java-based interactive questionnaire, with answers.
b. Matlab computer program, designated DOA, for computing the direction of arrival of linear and planar arrays.
c. Matlab computer program, designated Smart, for computing and displaying the radiation characteristics of linear and planar smart-antenna beamforming designs.
d. Power Point (PPT) viewgraphs, in multicolor.
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980 SMART ANTENNAS PROBLEMS 16.1.
For a linear array of 8 elements with a spacing of d = 0.5λ, determine the DOA, using the DOA computer program, assuming an SOI at 𝜃0 = 0◦and no SNOIs.
16.2.
Repeat Problem 16.1 assuming an SOI at 𝜃0 = 0◦and an SNOI at 𝜃1 = 60◦.
16.3.
For a planar array of 8 × 8 elements with a spacing of dx = dy = 0.5λ, determine the DOA, using the DOA computer program, assuming an SOI at 𝜃0 = 20◦, 𝜙0 = 90◦and no SNOIs.
16.4.
Repeat Problem 16.3 assuming an SOI at 𝜃0 = 20◦, 𝜙0 = 90◦and two SNOIs, one at 𝜃1 = 60◦, 𝜙1 = 180◦and the other at 𝜃2 = 45◦, 𝜙2 = 270◦.
16.5.
Repeat Example 16.2 for a desired signal at 𝜃0 = 0◦and an interference signal at 45◦.
Assume no coupling. Plot the pattern.
16.6.
Repeat Example 16.2 for a desired signal at 𝜃0 = 0◦and an interference signal at 60◦.
Assume no coupling. Plot the pattern.
16.7.
Repeat Example 16.3 for a desired signal at 𝜃0 = 0◦and an interference signal at 45◦.
Assume coupling and use the same coupling coefficients. Plot the pattern.
16.8.
Repeat Example 16.3 for a desired signal at 𝜃0 = 0◦and an interference signal at 60◦.
Assume coupling and use the same coupling coefficients. Plot the pattern.
16.9.
Using the LMS algorithm, beamform the pattern of the array factor of a linear array of 10 elements, with a uniform λ/2 spacing between them, so the maximum (SOI) of the array is broadside (𝜃0 = 0◦). Assume no requirements on the nulls (SNOIs).
16.10.
Repeat Problem 16.9 but with the maximum (SOI) of the array toward 𝜃0 = 30◦. Assume no requirements on the nulls (SNOIs).
16.11.
Repeat Problem 16.9 with the maximum (SOI) of the array toward 𝜃0 = 0◦and a null (SNOI) toward 𝜃1 = 30◦.
16.12.
Repeat Problem 16.10 with the maximum (SOI) of the array toward 𝜃0 = 30◦and a null (SNOI) toward 𝜃1 = 60◦.
16.13.
Using the LMS algorithm, beamform the pattern of the array factor of a planar array of 10 × 10 elements, with a uniform λ/2 spacing between them so that the maximum (SOI) of the array is broadside (𝜃0 = 0◦). Assume no requirements on the nulls (SNOIs).
16.14.
Repeat Problem 16.13 but with the maximum (SOI) of the array toward 𝜙0 = 45◦, 𝜃0 = 30◦.
Assume no requirements on the nulls (SNOIs).
16.15.
Repeat Problem 16.13 with the maximum (SOI) of the array toward 𝜃0 = 0◦and a null (SNOI) toward 𝜙1 = 45◦, 𝜃1 = 30◦.
16.16.
Repeat Problem 16.14 with the maximum (SOI) of the array toward 𝜙0 = 45◦, 𝜃0 = 30◦and a null (SNOI) toward 𝜙1 = 45◦, 𝜃1 = 60◦.
CHAPTER17 Antenna Measurements 17.1 INTRODUCTION In the previous sixteen chapters, analytical methods have been outlined which can be used to analyze, synthesize, and numerically compute the radiation characteristics of antennas. Often many antennas, because of their complex structural configurations and excitation methods, cannot be investigated analytically. Although the number of radiators that fall into this category has diminished, because special analytical methods (such as the GTD, Moment Method, Finite-Difference Time-Domain and Finite Element) have been developed during the past few years, there are still a fair number that have not been examined analytically. In addition, experimental results are often needed to validate theoretical data.
As was discussed in Chapter 3, Section 3.8.2, it is usually most convenient to perform antenna measurements with the test antenna in its receiving mode. If the test antenna is reciprocal, the receiving mode characteristics (gain, radiation pattern, etc.) are identical to those transmitted by the antenna. The ideal condition for measuring far-field radiation characteristics then, is the illumination of the test antenna by plane waves: uniform amplitude and phase. Although this ideal condition is not achievable, it can be approximated by separating the test antenna from the illumination source by a large distance on an outdoor range. At large radii, the curvature of the spherical phasefront produced by the source antenna is small over the test antenna aperture. If the separation distance is equal to the inner boundary of the far-field region, 2D2∕λ, then the maximum phase error of the incident field from an ideal plane wave is about 22.5◦, as shown in Figure 17.1. In addition to phasefront curva-ture due to finite separation distances, reflections from the ground and nearby objects are possible sources of degradation of the test antenna illumination.
Experimental investigations suffer from a number of drawbacks such as: 1. For pattern measurements, the distance to the far-field region (r > 2D2∕λ) is too long even for outside ranges. It also becomes difficult to keep unwanted reflections from the ground and the surrounding objects below acceptable levels.
2. In many cases, it may be impractical to move the antenna from the operating environment to the measuring site.
3. For some antennas, such as phased arrays, the time required to measure the necessary charac-teristics may be enormous.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 981 982 ANTENNA MEASUREMENTS Spherical phasefronts Source antenna Amplitude pattern Planar surface 22.5° Test antenna D 2D2 λ λ Figure 17.1 Phase error at the edges of a test antenna in the far-field when illuminated by a spherical wave.
4. Outside measuring systems provide an uncontrolled environment, and they do not possess an all-weather capability.
5. Enclosed measuring systems usually cannot accommodate large antenna systems (such as ships, aircraft, large spacecraft, etc.).
6. Measurement techniques, in general, are expensive.
Some of the above shortcomings can be overcome by using special techniques, such as indoor mea-surements, far-field pattern prediction from near-field measurements –, scale model measure-ments, and automated commercial equipment specifically designed for antenna measurements and utilizing computer assisted techniques.
Because of the accelerated progress made in aerospace/defense related systems (with increas-ingly small design margins), more accurate measurement methods were necessary. To accommo-date these requirements, improved instrumentation and measuring techniques were developed which include tapered anechoic chambers , compact and extrapolation ranges , near-field probing techniques –, improved polarization techniques and swept-frequency measurements , indi-rect measurements of antenna characteristics, and automated test systems.
The parameters that often best describe an antenna system’s performance are the pattern (ampli-tude and phase), gain, directivity, efficiency, impedance, current distribution, and polarization. Each of these topics will be addressed briefly in this chapter. A more extensive and exhaustive treatment of these and other topics can be found in the IEEE Standard Test Procedures for Antennas , in a summarized journal paper , and in a book on microwave antenna measurements . Most of the material in this chapter is drawn from these three sources. The author recommends that the IEEE publication on test procedures for antennas becomes part of the library of every practicing antenna and microwave engineer.
17.2 ANTENNA RANGES The testing and evaluation of antennas are performed in antenna ranges. Antenna facilities are cat-egorized as outdoor and indoor ranges, and limitations are associated with both of them. Outdoor ranges are not protected from environmental conditions whereas indoor facilities are limited by space restrictions. Because some of the antenna characteristics are measured in the receiving mode and require far-field criteria, the ideal field incident upon the test antenna should be a uniform plane wave.
To meet this specification, a large space is usually required and it limits the value of indoor facilities.
ANTENNA RANGES 983 Figure 17.2 Geometrical arrangement for reflection range. (source: L. H. Hemming and R. A. Heaton, “Antenna Gain Calibration on a Ground Reflection Range,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 4, pp. 532–537, July 1973. c ⃝(1973) IEEE).
17.2.1 Reflection Ranges In general, there are two basic types of antenna ranges: the reflection and the free-space ranges. The reflection ranges, if judiciously designed , can create a constructive interference in the region of the test antenna which is referred to as the “quiet zone.” This is accomplished by designing the ranges so that specular reflections from the ground, as shown in Figure 17.2, combine constructively with direct rays.
Usually it is desirable for the illuminating field to have a small and symmetric amplitude taper.
This can be achieved by adjusting the transmitting antenna height while maintaining constant that of the receiving antenna. These ranges are of the outdoor type, where the ground is the reflecting surface, and they are usually employed in the UHF region for measurements of patterns of moderately broad antennas. They are also used for systems operating in the UHF to the 16-GHz frequency region.
17.2.2 Free-Space Ranges Free-space ranges are designed to suppress the contributions from the surrounding environment and include elevated ranges, slant ranges , anechoic chambers, compact ranges , and near-field ranges .
A. Elevated Ranges Elevated ranges are usually designed to operate mostly over smooth terrains. The antennas are mounted on towers or roofs of adjacent buildings. These ranges are used to test physically large antennas. A geometrical configuration is shown in Figure 17.3(a). The contributions from the sur-rounding environment are usually reduced or eliminated by 1. carefully selecting the directivity and side lobe level of the source antenna 2. clearing the line-of-sight between the antennas 3. redirecting or absorbing any energy that is reflected from the range surface and/or from any obstacles that cannot be removed 4. utilizing special signal-processing techniques such as modulation tagging of the desired signal or by using short pulses 984 ANTENNA MEASUREMENTS (a) Elevated (after ) (b) Slant (after P. W. Arnold ) 30 m Test antenna Test antenna Source antenna Image Source antenna Fiberglass tower α 45° 30 m h hr ht D R α Figure 17.3 Geometries of elevated and slant ranges. (sources: IEEE Standard Test Procedures for Anten-nas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley; and P. W. Arnold, “The ‘Slant’ Antenna Range,” IEEE Trans. Antennas Propagat., Vol. AP-14, No. 5, pp. 658–659, September 1966.
c ⃝(1966) IEEE).
In some applications, such as between adjacent mountains or hilltops, the ground terrain may be irregular. For these cases, it is more difficult to locate the specular reflection points (points that reflect energy toward the test antenna). To take into account the irregular surface, scaled drawings of the vertical profile of the range are usually constructed from data obtained from the U.S. Geological Survey. The maps show ground contours , and they give sufficient details which can be used to locate the specular reflection points, determine the level of energy reflected toward the test antenna, and make corrections if it is excessive.
B. Slant Ranges Slant ranges are designed so that the test antenna, along with its positioner, are mounted at a fixed height on a nonconducting tower while the source (transmitting) antenna is placed near the ground, as shown in Figure 17.3(b). The source antenna is positioned so that the pattern maximum, of its free-space radiation, is oriented toward the center of the test antenna. The first null is usually directed toward the ground specular reflection point to suppress reflected signals. Slant ranges, in general, are more compact than elevated ranges in that they require less land.
C. Anechoic Chambers To provide a controlled environment, an all-weather capability, and security, and to minimize electro-magnetic interference, indoor anechoic chambers have been developed as an alternative to outdoor ANTENNA RANGES 985 testing. By this method, the testing is performed inside a chamber having walls that are covered with RF absorbers. The availability of commercial high-quality RF absorbing material, with improved electrical characteristics, has provided the impetus for the development and proliferation of ane-choic chambers. Anechoic chambers are mostly utilized in the microwave region, but materials have been developed which provide a reflection coefficient of −40 dB at normal incidence at fre-quencies as low as 100 MHz. In general, as the operating frequency is lowered, the thickness of RF absorbing material must be increased to maintain a given level of reflectivity performance. An RF absorber that meets the minimum electrical requirements at the lower frequencies usually possesses improved performance at higher frequencies.
Presently there are two basic types of anechoic chamber designs: the rectangular and the tapered chamber. The design of each is based on geometrical optics techniques, and each attempts to reduce or to minimize specular reflections. The geometrical configuration of each, with specular reflection points depicted, is shown in Figures 17.4(a) and 17.4(b).
The rectangular chamber is usually designed to simulate free-space conditions and maximize the volume of the quiet zone. The design takes into account the pattern and location of the source, the frequency of operation, and it assumes that the receiving antenna at the test point is isotropic.
Reflected energy is minimized by the use of high-quality RF absorbers. Despite the use of RF absorb-ing material, significant specular reflections can occur, especially at large angles of incidence.
Tapered anechoic chambers take the form of a pyramidal horn. They begin with a tapered chamber which leads to a rectangular configuration at the test region, as shown in Figure 17.4(b). At the lower end of the frequency band at which the chamber is designed, the source is usually placed near the apex so that the reflections from the side walls, which contribute to the illuminating fields Figure 17.4 Rectangular and tapered anechoic chambers and the corresponding side-wall specular reflections.
(source: W. H. Kummer and E. S. Gillespie, “Antenna Measurements—1978,” Proc. IEEE, Vol. 66, No. 4, pp. 483–507, April 1978. c ⃝(1978) IEEE).
986 ANTENNA MEASUREMENTS in the region of the test antenna, occur near the source antenna. For such paths, the phase difference between the direct radiation and that reflected from the walls near the source can be made very small by properly locating the source antenna near the apex. Thus the direct and reflected rays near the test antenna region add vectorially and provide a relatively smooth amplitude illumination taper. This can be illustrated by ray-tracing techniques.
As the frequency of operation increases, it becomes increasingly difficult to place the source sufficiently close to the apex that the phase difference between the direct and specularly reflected rays can be maintained below an acceptable level. For such applications, reflections from the walls of the chamber are suppressed by using high-gain source antennas whose radiation toward the walls is minimal. In addition, the source is moved away from the apex, and it is placed closer to the end of the tapering section so as to simulate a rectangular chamber.
17.2.3 Compact Ranges Microwave antenna measurements require that the radiator under test be illuminated by a uniform plane wave. This is usually achieved only in the far-field region, which in many cases dictates very large distances. The requirement of an ideal plane wave illumination can be nearly achieved by utilizing a compact range.
A Compact Antenna Test Range (CATR) is a collimating device which generates nearly planar wavefronts in a very short distance (typically 10–20 meters) compared to the 2D2∕λ (minimum) dis-tance required to produce the same size test region using the standard system configuration of testing shown in Figure 17.1. Some attempts have been made to use dielectric lenses as collimators , but generally the name compact antenna test range refers to one or more curved metal reflectors which perform the collimating function. Compact antenna test ranges are essentially very large reflector antennas designed to optimize the planar characteristics of the fields in the near field of the aperture.
Compact range configurations are often designated according to their analogous reflector antenna configurations: parabolic, Cassegrain, Gregorian, and so forth.
One compact range configuration is that shown in Figure 17.5 where a source antenna is used as an offset feed that illuminates a paraboloidal reflector, which converts the impinging spherical waves to plane waves . Geometrical Optics (GO) is used in Figure 17.5 to illustrate general CATR oper-ation. The rays from a feed antenna can, over the main beam, be viewed as emanating from a point at its phase center. When the phase center of the feed is located at the prime focus of a parabolic reflec-tor, all rays that are reflected by the reflector and arrive at a plane transverse to the axis of the parabola have traveled an equal distance. See Chapter 15, Section 15.4 for details. Therefore, the field at the aperture of the reflector has a uniform phase; i.e., that of a plane wave. In addition to Geometrical + Test zone Vertex Parabolic surface Feed CATR reflector Figure 17.5 A Compact Antenna Test Range (CATR) synthesizes planar phase fronts by collimating spherical waves with a section of paraboloidal reflector.
ANTENNA RANGES 987 Optics, analysis and design of CATRs have been performed with a number of other analytical meth-ods. Compact range test zone fields have been predicted by the Method of Moments (MoM), but at high frequencies, the large electrical size of the CATR system makes the use of MoM, Finite-Difference Time-Domain (FD-TD), and Finite Element Method (FEM) impractical. High-frequency techniques, however, are well suited for compact range analysis because the fields of interest are near the specular reflection direction, and the reflector is electrically large. The Geometrical Theory of Diffraction (GTD) is, in principle, an appropriate technique, but it is difficult to implement for serrated-edge reflectors due to the large number of diffracting edges. To date, Physical Optics (PO) is probably the most practical and efficient method of predicting the performance of CATRs , .
The major drawbacks of compact ranges are aperture blockage, direct radiation from the source to the test antenna, diffractions from the edges of the reflector and feed support, depolarization coupling between the two antennas, and wall reflections. The use of an offset feed eliminates aperture blockage and reduces diffractions. Direct radiation and diffractions can be reduced further if a reflector with a long focal length is chosen. With such a reflector, the feed can then be mounted below the test antenna and the depolarization effects associated with curved surfaces are reduced. Undesirable radiation toward the test antenna can also be minimized by the use of high-quality absorbing material. These and other concerns will be discussed briefly.
A. CATR Performance A perfect plane wave would be produced by a CATR if the reflector has an ideal parabolic curvature, is infinite in size and is fed by a point source located at its focus. Of course CATR reflectors are of finite size, and their surfaces have imperfections; thus the test zone fields they produce can only approximate plane waves. Although there are different configurations of CATR, their test zone fields have some common characteristics. The usable portion of the test zone consists of nearly planar wavefronts and is referred to as the “quiet zone.” Outside the quiet zone, the amplitude of the fields decreases rapidly as a function of distance transverse to the range axis. The size of the quiet zone is typically about 50%–60% of the dimensions of the main reflector. Although the electromagnetic field in the quiet zone is often a very good approximation, it is not a “perfect” plane wave. The imperfections of the fields in the quiet zone from an ideal plane wave are usually represented by phase errors, and ripple and taper amplitude components. These discrepancies from an ideal plane wave, that occur over a specified test zone dimension, are the primary figures of merit of CATRs.
For most applications phase deviations of less than 10◦, peak-to-peak amplitude ripples of less than 1 dB, and amplitude tapers of less than 1 dB are considered adequate. More stringent quiet-zone specifications may be required to measure, within acceptable error levels, low-side lobe antennas and low-observable scatterers. The sources of quiet-zone taper and ripple are well known, but their minimization is a source of much debate.
Amplitude taper across the quiet zone can be attributed to two sources: the feed pattern and space-attenuation. That portion of the radiation pattern of the feed antenna which illuminates the CATR reflector is directly mirrored into the quiet zone. For example, if the 3-dB beamwidth of the feed is equal to about 60% of the angle formed by lines from the reflector edges to the focal point, then the feed will contribute 3 dB of quiet-zone amplitude taper. In general, as the directivity of the feed antenna increases, quiet-zone amplitude taper increases. Usually, low-gain feed antennas are designed to add less than a few tenths of a dB of amplitude taper. The 1∕r2 space-attenuation occurs with the spherical spreading of the uncollimated radiation from the feed. Although the total path from the feed to the quiet zone is a constant, the distance from the feed to the reflector varies. These differences in the propagation distances from the feed to various points across the reflector surface cause amplitude taper in the quiet zone due to space-attenuation. This taper is asymmetric in the plane of the feed offset.
Amplitude and phase ripple are primarily caused by diffractions from the edges of the reflec-tor. The diffracted fields are spread in all directions which, along with the specular reflected signal, form constructive and destructive interference patterns in the quiet zone, as shown in Figure 17.6(a).
988 ANTENNA MEASUREMENTS Feed Feed Field point Diffracted ray Edge diffracted rays Edge diffracted rays Diffracted ray Reflected ray Fields reflected from curved surface Fields reflected from curved surface Plane wave from parabola Parabolic reflector Curved surface diffracted rays Curved surface diffracted rays Curved edge modification Curved edge modification (b) Rolled-edge (a) Knife-edge Parabolic reflector Figure 17.6 Amplitude and phase ripple in the quiet-zone fields produced by a compact antenna test range caused by the phasor sum of the reflected and diffracted rays from the reflector. (source: W. D. Burnside, M. C. Gilreath, B. M. Kent and G. L. Clerici, “Curved Edge Modification of Compact Range Reflectors,” IEEE Trans. Antennas Propagat., Vol. AP-35, No. 2, pp. 176–182, February 1987. c ⃝(1987) IEEE).
Considerable research has been done on reflector edge terminations in an effort to minimize quiet-zone ripple. Reflector edge treatments are the physical analogs of windowing functions used in Fourier transforms. Edge treatments reduce the discontinuity of the reflector/free-space boundary, caused by the finite size of the reflector, by providing a gradually tapered transition. Common reflec-tor edge treatments include serrations and rolled edges, as shown in Figure 17.7(a,b). The serrated edge of a reflector tapers the amplitude of the reflected fields near the edge. An alternate interpreta-tion of the effects of serrations is based on edge diffraction. Serrations produce many low-amplitude diffractions as opposed to, for example, the large-amplitude diffractions that would be generated by the four straight edges and corners of a rectangular knife-edged reflector. These small diffractions are quasi-randomized in location and direction; hence, they are likely to have cancellations in the quiet zone. Although most serrated-edge CATRs have triangular serrations, curving the edges of each serration can result in improved performance at high frequencies . A number of blended, rolled-edge treatments have been suggested as alternatives to serrations, and have been implemented ANTENNA RANGES 989 (a) Front view of a serrated-edge CATR reflector.
(b) Side view of a rolled-edge CATR reflector.
Figure 17.7 Two common CATR reflector edge treatments that are used to reduce the diffracted fields in the quiet zone.
to gradually redirect energy away from the quiet zone, as shown in Figure 17.6(b) –. In these designs, the concave parabolic surface of the reflector is blended into a convex surface which wraps around the edges of the reflector and terminates behind it. The predicted quiet-zone fields produced by a knife-edged reflector compared to those produced by a rolled-edged reflector are shown, respec-tively, in Figures 17.8(a,b) and demonstrate the effectiveness of this edge treatment. Another method of reducing quiet-zone ripple is to taper the illumination amplitude near the reflector edges. This can be accomplished with a high-gain feed or the feed can consist of an array of small elements designed so that a null in the feed pattern occurs at the reflector edges –. Finally, the surface currents on the reflector can be terminated gradually at the edges by tapering the conductivity and/or the impedance of the reflector via the application of lossy material.
The frequency of operation of a CATR is determined by the size of the reflector and its sur-face accuracy. The low-frequency limit is usually encountered when the reflector is about 25 to 30 wavelengths in diameter . Quiet-zone ripple becomes large at the low-frequency limit. At high frequencies, reflector surface imperfections contribute to the quiet-zone ripple. A rule of thumb used in the design of CATRs is that the surface must deviate less than about 0.007λ from that of a true paraboloid . Since the effects of reflector surface imperfections are additive, dual reflector sys-tems must maintain twice the surface precision of a single reflector system to operate at the same frequency. Many CATR systems operate typically from 1 GHz to 100 GHz.
B. CATR Designs Four reflector configurations that have been commercially developed will be briefly discussed: the single paraboloid, the dual parabolic-cylinder, the dual shaped-reflector, and the single parabolic-cylinder systems. The first three configurations are relatively common fully collimating compact ranges; the fourth is a hybrid approach which combines aspects of compact range technology with near-field/far-field (NF/FF) techniques.
The single paraboloidal reflector CATR design was illustrated in Figure 17.5. As with all compact range designs, the feed antenna is offset by some angle from the propagation direction of the col-limated energy. This is done to eliminate blockage and to reduce scattering of the collimated fields by the feed. To achieve this offset, the reflector is a sector of a paraboloid that does not include the vertex. This design is referred to as a “virtual vertex” compact range. With only one reflector, the paraboloidal CATR has a minimum number of surfaces and edges that can be sources of quiet-zone ripple. Feed spillover into the quiet zone is also low with this design since the feed antenna is pointed almost directly away from the test zone. On the other hand, it is more difficult and costly to produce a high-precision surface that is curved in two planes (three-dimensional) compared to producing a reflector that is curved in only one plane (two-dimensional). In addition, it has been reported that the single paraboloidal reflector design depolarizes the incident fields to a greater degree than other CATR designs. This is due to the relatively low f/d ratio needed to simultaneously maintain the feed antenna between the test zone and the reflector while keeping the test zone as close as possible to the reflector aperture .
990 ANTENNA MEASUREMENTS 0 5 10 15 –7 –6 –5 –4 –3 –2 –1 0 Radial distance (feet) (a) Knife-edge (b) Rolled-edge ρ Radial distance (feet) Field magnitude (dB) Parabolic reflector edge GTD Moment method 0 5 10 15 –7 –6 –5 –4 –3 –2 –1 0 Field magnitude (dB) Parabolic reflector edge GTD Moment method ρ Figure 17.8 Predicted quiet-zone field amplitude versus transverse distance for knife-edge and rolled-edge reflectors. (source: W. D. Burnside, M. C. Gilreath, B. M. Kent, and G. L. Clerici, “Curved Edge Modification of Compact Range Reflectors,” IEEE Trans. Antennas Propagat., Vol. AP-35, No. 2, pp. 176–182, February 1987. c ⃝(1987) IEEE).
The dual parabolic-cylinder reflector concept is illustrated in Figure 17.9, and it consists of two parabolic cylinders arranged so that one is curved in one plane (vertical or horizontal) while the other is curved in the orthogonal plane. The spherical phase fronts radiated by the feed antenna are collimated first in the horizontal or vertical plane by the first reflector, then are collimated in the orthogonal plane by the second reflector . Because the boresight of the feed antenna is directed at almost 90◦to the plane wave propagation direction, direct illumination of the test zone by the feed can be relatively high. In practice, quiet-zone contamination from feed spillover is virtually eliminated through the use of range gating. Relatively low cross polarization is produced with this design because the doubly folded optics results in a long focal length main reflector.
The dual shaped-reflector CATR, shown schematically in Figure 17.10, is similar in design to a Cassegrain antenna, but the reflector surfaces are altered from the classical parabolic/hyperbolic shapes. An iterative design process is used to determine the shapes of the subreflector and main reflector needed to yield the desired quiet-zone performance. The shape of the subreflector maps the ANTENNA RANGES 991 Test zone Feed First reflector (vertical collimation) Second reflector (horizontal collimation) Figure 17.9 Dual parabolic-cylinder compact range collimates the fields in one plane with first reflector and then collimates the fields in the orthogonal plane with second reflector.
high-gain feed pattern into a nearly optimum illumination of the main reflector. An almost uniform energy density illuminates the central part of the main reflector while the amplitude tapers toward the reflector edges. This design results in a very high illumination efficiency (the power of the collimated quiet-zone fields relative to the system input power) . Two of the consequences of this high illumination efficiency are (1) the reduction of spillover into the chamber reduces range clutter, and (2) the increased RF power delivered to the target increases system sensitivity.
The single parabolic-cylinder reflector system is essentially half of the dual parabolic-cylinder CATR. The reflector has a parabolic curvature in the vertical plane and is flat in the horizontal plane.
This semicompact antenna test range collimates the fields only in the vertical plane, producing a quiet zone which consists of cylindrical waves, as shown in Figure 17.11 –. Such a compact range configuration is utilized in the ElectroMagnetic Anechoic Chamber (EMAC) at Arizona State University , .
This Single-Plane Collimating Range (SPCR) approach results in a number of advantages and compromises compared to conventional CATR systems and near-field/far-field (NF/FF) systems.
For antennas that are small compared to the curvature of the cylindrical phasefront, far-field radi-ation patterns can be measured directly. Because of the folded optics, the radius of the cylindrical phasefront produced by the SPCR is larger than the radius of the spherical phasefront obtainable by separating the source antenna from the test antenna in a direct illumination configuration within the same anechoic chamber. Thus, with the SPCR it is possible to measure, directly, the far-field patterns of larger antennas compared to those directly measurable on an indoor far-field range. When the size of the antenna is significant relative to the curvature of the cylindrical phasefront, a NF/FF transfor-mation is used to obtain the far-field pattern. However, because the fields are collimated in the vertical plane, only a one-dimensional transformation is required. This greatly simplifies the transformation algorithm. Most importantly, there is a one-to-one correlation between a single azimuthal pattern cut measured in the near-field, and the predicted far-field pattern. The data acquisition time is identical to that of conventional CATRs, and the NF/FF calculation time is nearly negligible. Another advantage Test zone Feed Main reflector Subreflector Figure 17.10 Dual shaped-reflector compact range analogous to a Cassegrain system.
992 ANTENNA MEASUREMENTS Figure 17.11 ASU Single-Plane Collimating Range (SPCR) produces a cylindrical wave in the quiet zone (artist rendering by Michael Hagelberg).
of the SPCR is the size of the quiet zone. In the vertical plane, the quiet-zone dimension compared to the SPCR reflector is similar to that of conventional CATRs (about 50% to 60%). However, in the horizontal plane, the quiet zone is nearly 100% of the horizontal dimension of the reflector. For a given size anechoic chamber and reflector, targets having much larger horizontal dimensions (yaw patterns of aircraft, for example) can be measured using the SPCR than is possible using a conven-tional CATR. The SPCR system is relatively inexpensive; the manufacturer estimates that its cost is about 60% of conventional CATR systems.
In addition to the added complexity of NF/FF transformation considerations, this cylindrical wave approach has other disadvantages compared to conventional CATR designs. Because the quiet-zone fields are expanding cylindrically as they propagate along the axis of the range, a large portion of the anechoic chamber is directly illuminated. This should be carefully considered in the design of the side walls of the anechoic chamber to control range clutter. Also, some measurement sensitivity is sacrificed for the same reason.
Compact antenna test ranges enable the measurement of full-sized antennas in very short dis-tances, usually within the controlled environment of an anechoic chamber. A compact antenna test range can be used to accomplish any type of antenna testing (including radiation patterns, gain, efficiency, etc.) that can be performed on an outdoor facility.
17.2.4 Near-Field/Far-Field Methods The dimensions of a conventional test range can be reduced by making measurements in the near field, and then using analytical methods to transform the measured near-field data to compute the far-field radiation characteristics –, . These are referred to as near-field to far-field (NF/FF) methods. Such techniques are usually used to measure patterns, and they are often performed indoors.
Therefore, they provide a controlled environment and an all-weather capability, the measuring ANTENNA RANGES 993 system is time and cost effective, and the computed patterns are as accurate as those measured in a far-field range. However, such methods require more complex and expensive systems, more exten-sive calibration procedures, more sophisticated computer software, and the patterns are not obtained in real time.
The near-field measured data (usually amplitude and phase distributions) are measured by a scan-ning field probe over a preselected surface which may be a plane, a cylinder, or a sphere. The mea-sured data are then transformed to the far-field using analytical Fourier transform methods. The complexity of the analytical transformation increases from the planar to the cylindrical, and from the cylindrical to the spherical surfaces. The choice is primarily determined by the antenna to be measured.
In general, the planar system is better suited for high-gain antennas, especially planar phased arrays, and it requires the least amount of computations and no movement of the antenna. Although the cylindrical system requires more computations than the planar, for many antennas its measuring, positioning, and probe equipment are the least expensive. The spherical system requires the most expensive computation, and antenna and probe positioning equipment, which can become quite sig-nificant for large antenna systems. This system is best suited for measurements of low-gain and omnidirectional antennas.
Generally, implementation of NF/FF transformation techniques begins with measuring the mag-nitude and phase of the tangential electric field components radiated by the test antenna at regular intervals over a well-defined surface in the near field. By the principle of modal expansion, the sam-pled E-field data is used to determine the amplitude and phase of an angular spectrum of plane, cylindrical, or spherical waves. Expressing the total field of the test antenna in terms of a modal expansion allows the calculation of the field at any distance from the antenna. Solving for the fields at an infinite distance results in the far-field pattern.
A consideration of the general case of scanning with ideal probes over an arbitrary surface reveals that the choice of scanning surfaces is limited. Morse and Feshbach show that derivation of the far-zone vector field from the near-field depends on vector wave functions that are orthogonal to that surface. Planar, circular cylindrical, spherical, elliptic cylindrical, parabolic cylindrical, and conical are the six coordinate systems that support orthogonal vector wave solutions. The first three coordinate systems are conducive to convenient data acquisition, but the last three require scanning on an elliptic cylinder, a parabolic cylinder, or a sphere in conical coordinates . Thus, the three NF/FF techniques that have been developed and are widely used are based on planar, cylindrical, and spherical near-field scanning surfaces.
Acquisition of planar near-field data is usually conducted over a rectangular x-y grid, as shown in Figure 17.12(a), with a maximum near-field sample spacing of Δx = Δy = λ∕2 . It is also (c) Spherical scanning (b) Cylindrical scanning (a) Planar scanning b a L Δz Δy Δx Δθ ϕ ϕ Δ Δ Figure 17.12 Three near-field scanning surfaces that permit convenient data acquisition (planar, cylindrical, and spherical).
994 ANTENNA MEASUREMENTS Probe Test antenna Figure 17.13 Probe compensation of near-field measurements due to nonisotropic radiation pattern of the probe.
possible to acquire the near-field measurements on a plane-polar grid or a bipolar grid . The test antenna is held stationary while the probe (typically an open-ended waveguide or small horn) is moved to each grid location on the plane. As the probe location varies, its orientation relative to the test antenna changes, as illustrated in Figure 17.13. This directive property of the probe, as well as its polarization, must be taken into account using the technique of probe compensation , . Probe compensation methods use the well-known Lorentz reciprocity theorem to couple the far-zone fields of the test antenna to those of the measuring probe.
The principal advantage of the planar near-field to far-field transformation, over the cylindrical and spherical techniques, is its mathematical simplicity. Furthermore, the planar transformation is suitable for applying the computationally efficient Fast Fourier Transform (FFT) algorithm .
Assuming that the number of near-field data points is 2n (or artificially padded to that number with points of zero value) where n is a positive integer, the full planar far-field transformation can be computed in a time proportional to (ka)2 log2(ka) where a is the radius of the smallest circle that inscribes the test antenna . Planar NF/FF techniques are well suited for measuring antennas which have low backlobes. These include directional antennas such as horns, reflector antennas, planar arrays, and so forth. The primary disadvantage of probing the near-field on a planar surface to calculate the far-field is that the resulting far-field pattern is over a limited angular span. If the planar scanning surface is of infinite extent, one complete hemisphere of the far-field can be computed.
A complete set of near-field measurements over a cylindrical surface includes the information needed to compute complete azimuthal patterns for all elevation angles, excluding the conical regions at the top and bottom of the cylinder axis. Since the numerical integrations can be performed with the FFT, the cylindrical transformation exhibits numerical efficiencies and proportional computa-tion times similar to those of the planar transformation. The angular modal expansion, however, is expressed in terms of Hankel functions, which can be more difficult to calculate, especially for large orders.
The cylindrical scanning grid is shown in Figure 17.12(b). The maximum angular and vertical sample spacing is Δ𝜙= λ 2(a + λ) (17-1) and Δz = λ∕2 (17-2) where λ is the wavelength and a is the radius of the smallest cylinder that encloses the test antenna.
ANTENNA RANGES 995 x z y Probe Test antenna Linear positioner Azimuthal positioner Probe Test antenna AZ/EL positioner (a) Cylindrical (b) Spherical ϕ ϕ θ Figure 17.14 Schematic representation of typical cylindrical and spherical surface near-field positioning systems.
A typical cylindrical scanning system is illustrated in Figure 17.14(a). The azimuthal location of the antenna is held constant while the fields are probed at discrete locations in the vertical direction at some fixed distance from the antenna. At the completion of each vertical scan, the test antenna is rotated to the next angular position. The orientation of the probe with respect to the test antenna changes as the vertical location of the probe changes, thus a probe correction is generally required as in the planar case. In addition to directional antennas, the radiation patterns of antennas with narrow patterns along the vertical axis (horizontal fan beam antennas and vertical dipoles for example) can be predicted efficiently with the cylindrical NF/FF technique.
The information obtained by scanning the near-field radiation over a spherical surface enclosing a test antenna makes possible the most complete prediction of the far-field radiation pattern. The spherical scanning grid is illustrated in Figure 17.12(c). When sampled at the rate of Δ𝜃= λ 2(a + λ) (17-3) and Δ𝜙= λ 2(a + λ) (17-4) all of the spatial radiation characteristics of the test antenna are included in the transformation. Any far-field pattern cut can be computed from a complete near-field measurement with the spherical scanning scheme. Typically, a spherical scan is accomplished by fixing the location and orientation of the probe and varying the angular orientation of the test antenna with a dual-axis positioner, as shown in Figure 17.14(b). Since the probe is always pointed directly toward the test antenna, probe correction can be neglected for sufficiently large scan radii . However, in general, probe correction is necessary.
The primary drawback of the spherical scanning technique lies in the mathematical transforma-tion. A significant portion of the transformation cannot be accomplished via FFTs. Numerical inte-grations, matrix operations, and simultaneous solution of equations are required. This increases the 996 ANTENNA MEASUREMENTS computational time and difficulty of the transformation considerably over those of the planar and cylindrical transformations.
A. Modal-Expansion Method for Planar Systems The mathematical formulations of the planar NF/FF system are based on the plane wave (modal) expansion using Fourier transform (spectral) techniques. Simply stated, any monochromatic, but otherwise arbitrary, wave can be represented as a superposition of plane waves traveling in differ-ent directions, with different amplitudes, but all of the same frequency. The objective of the plane wave expansion is to determine the unknown amplitudes and directions of propagation of the plane waves. The results comprise what is referred to as a modal expansion of the arbitrary wave. Simi-larly, cylindrical wave and spherical wave expansions are used to determine far-field patterns from fields measured in the near field over cylindrical and spherical surfaces, respectively.
The relationships between the near-zone E-field measurements and the far-zone fields for planar systems follow from the transform (spectral) techniques of Chapter 12, Section 12.9, represented by (12-73)–(12-75), or E(x, y, z) = 1 4𝜋2 ∫ ∞ −∞∫ ∞ −∞ f(kx, ky)e−jk ⋅r dkx dky (17-5) where f(kx, ky) = ̂ ax fx(kx, ky) + ̂ ay fy(kx, ky) + ̂ az fz(kx, ky) (17-5a) k = ̂ axkx + ̂ ayky + ̂ azkz (17-5b) r = ̂ axx + ̂ ayy + ̂ azz (17-5c) where f(kx, ky) represents the plane wave spectrum of the field. The x and y components of the electric field measured over a plane surface (z = 0) from (17-5) are Exa(x, y, z = 0) = 1 4𝜋2 ∫ ∞ −∞∫ ∞ −∞ fx(kx, ky)e−j(kxx+kyy) dkx dky (17-6a) Eya(x, y, z = 0) = 1 4𝜋2 ∫ ∞ −∞∫ ∞ −∞ fy(kx, ky)e−j(kxx+kyy) dkx dky (17-6b) The x and y components of the plane wave spectrum, fx(kx, ky) and fy(kx, ky), are determined in terms of the near-zone electric field from the Fourier transforms of (17-6a) and (17-6b) as given by (12-85a), (12-85b), or fx(kx, ky) = ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Exa(x′, y′, z′ = 0)e+j(kxx′+kyy′) dx′ dy′ (17-7a) fy(kx, ky) = ∫ +b∕2 −b∕2 ∫ +a∕2 −a∕2 Eya(x′, y′, z′ = 0)e+j(kxx′+kyy′) dx′ dy′ (17-7b) The far-field pattern of the antenna, in terms of the plane wave spectrum function f, is then that of (12-107) E(r, 𝜃, 𝜙) ≃jke−jkr 2𝜋r [cos 𝜃f(kx, ky)] (17-8) ANTENNA RANGES 997 or (12-111) E𝜃(r, 𝜃, 𝜙) ≃jke−jkr 2𝜋r (fx cos 𝜙+ fy sin 𝜙) (17-9a) E𝜙(r, 𝜃, 𝜙) ≃jke−jkr 2𝜋r cos 𝜃(−fx sin 𝜙+ fy cos 𝜙) (17-9b) The procedure then, to determine the far-zone field from near-field measurements, is as follows: 1. Measure the electric field components Exa(x′, y′, z′ = 0) and Eya(x′, y′, z′ = 0) in the near field.
2. Find the plane wave spectrum functions fx and fy using (17-7a) and (17-7b).
3. Determine the far-zone electric field using (17-8) or (17-9a) and (17-9b).
Similar procedures are used for cylindrical and spherical measuring systems except that the constant surfaces are, respectively, cylinders and spheres. However, their corresponding analytical expressions have different forms.
It is apparent once again, from another application problem, that if the tangential field components are known along a plane, the plane wave spectrum can be found, which in turn permits the evaluation of the field at any point. The computations become more convenient if the evaluation is restricted to the far-field region.
B. Measurements and Computations The experimental procedure requires that a plane surface, a distance z0 from the test antenna, be selected where measurements are made as shown in Figure 17.12(a). The distance z0 should be at least two or three wavelengths away from the test antenna to be out of its reactive near-field region.
The plane over which measurements are made is usually divided into a rectangular grid of M × N points spaced Δx and Δy apart and defined by the coordinates (mΔx, nΔy, 0) where −M∕2 ≤m ≤ M∕2 −1 and −N∕2 ≤n ≤N∕2 −1. The values of M and N are determined by the linear dimensions of the sampling plane divided by the sampling space. To compute the far-field pattern, it requires that both polarization components of the near field are measured. This can be accomplished by a simple rotation of a linear probe about the longitudinal axis or by the use of a dual-polarized probe. The probe used to make the measurements must not be circularly polarized, and it must not have nulls in the angular region of space over which the test antenna pattern is determined because the probe correction coefficients become infinite.
The measurements are carried out until the signal at the edges of the plane is of very low intensity, usually about 45 dB below the largest signal level within the measuring plane. Defining a and b the width and height, respectively, of the measuring plane, M and N are determined using M = a Δx + 1 (17-10a) N = b Δy + 1 (17-10b) The sampling points on the measuring grid are chosen to be less than λ∕2 in order to satisfy the Nyquist sampling criterion. If the plane z = 0 is located in the far field of the source, the sample 998 ANTENNA MEASUREMENTS spacings can increase to their maximum value of λ∕2. Usually the rectangular lattice points are separated by the grid spacings of Δx = 𝜋 kxo (17-11a) Δy = 𝜋 kyo (17-11b) where kxo and kyo are real numbers and represent the largest magnitudes of kx and ky, respectively, such that f(kx, ky) ≃0 for |kx| > kxo or |ky| > kyo.
At the grid sample points, the tangential electric field components, Ex and Ey, are recorded. The subscripts x and y represent, respectively, the two polarizations of the probe. The procedure for probe compensation is neglected here. A previously performed characterization of the probe is used to compensate for its directional effects in what is essentially an application of its “transfer function.” The electric field components over the entire plane can be reconstructed from the samples taken at the grid points, and each is given by Exa(x, y, z = 0) ≃ N∕2−1 ∑ n=−N∕2 M∕2−1 ∑ m=−M∕2 Ex(mΔx, nΔy, 0) sin(kxox −m𝜋) kxox −m𝜋 sin(kyoy −n𝜋) kyoy −n𝜋 (17-12a) Eya(x, y, z = 0) ≃ N∕2−1 ∑ n=−N∕2 M∕2−1 ∑ m=−M∕2 Ey(mΔx, nΔy, 0) sin(kxox −m𝜋) kxox −m𝜋 sin(kyoy −n𝜋) kyoy −n𝜋 (17-12b) Using (17-12a) and (17-12b), fx and fy of (17-7a) and (17-7b) can be evaluated, using a FFT algorithm, at the set of wavenumbers explicitly defined by the discrete Fourier transform and given by kx = 2𝜋m MΔx, −M 2 ≤m ≤M 2 −1 (17-13a) ky = 2𝜋n NΔy, −N 2 ≤n ≤N 2 −1 (17-13b) The wavenumber spectrum points are equal to the number of points in the near-field distribution, and the maximum wavenumber coordinate of the wavenumber spectrum is inversely proportional to the near-field sampling spacing. While the maximum sampling spacing is λ∕2, there is no minimum spacing restrictions. However, there are no advantages to increasing the near-field sample points by decreasing the sample spacing. The decreased sample spacing will increase the limits of the wavenumber spectrum points, which are in the large evanescent mode region, and do not contribute to increased resolution of the far-field pattern.
Increased resolution in the far-field power pattern can be obtained by adding artificial data sam-pling points (with zero value) at the outer extremities of the near-field distribution. This artificially increases the number of sample points without decreasing the sample spacing. Since the sample spacing remains fixed, the wavenumber limits also stay fixed. The additional wavenumber spectrum points are all within the original wavenumber limits and lead to increased resolution in the computed far-field patterns.
To validate the techniques, numerous comparisons between computed far-field patterns, from near-field measurements, and measured far-field patterns have been made. In Figure 17.15, the com-puted and measured sum and difference far-field azimuth plane patterns for a four-foot diameter ANTENNA RANGES 999 –16 –8 0 8 16 –40 –30 –20 –10 0 Scientific Atlanta (far-field range) Georgia Tech (far-field range) Georgia Tech (near-field range) Relative gain (dB) Azimuth angle (degrees) (a) Sum mode –16 –8 0 8 16 –40 –30 –20 –10 0 Relative gain (dB) Azimuth angle (degrees) (b) Difference mode Figure 17.15 Measured and computed sum and difference mode principal-plane far-field patterns for a four-foot parabolic reflector. (source: E. D. Joy, W. M. Leach, Jr., G. P. Rodrique, and D. T. Paris, “Applica-tions of Probe Compensated Near-Field Measurements,” IEEE Trans. Antennas Propagat., Vol. AP-26, No. 3, pp. 379–389, May 1978. c ⃝(1978) IEEE).
parabolic reflector with a nominal gain of 30 dB are displayed . Two measured far-field patterns were obtained on two different high-quality far-field ranges, one at the Georgia Institute of Tech-nology and the other at Scientific-Atlanta. The third trace represents the computed far-field pattern from near-field measurements made at Georgia Tech. There are some minor discrepancies between the two measured far-field patterns which were probably caused by extraneous range reflections. The best agreement is between the computed far-field pattern and the one measured at Scientific-Atlanta.
Many other comparisons have been made with similar success. The limited results shown here, and the many others published in the literature , – clearly demonstrate the capability of the near-field technique.
The near-field technique provides the antenna designers information not previously available to them. For example, if a given far-field pattern does not meet required specifications, it is possible to use near-field data to pinpoint the cause . Near-field measurements can be applied also to antenna 1000 ANTENNA MEASUREMENTS analysis and diagnostic tasks , and it is most attractive when efficient near-field data collection and transformation methods are employed.
17.3 RADIATION PATTERNS The radiation patterns (amplitude and phase), polarization, and gain of an antenna, which are used to characterize its radiation capabilities, are measured on the surface of a constant radius sphere. Any position on the sphere is identified using the standard spherical coordinate system of Figure 17.16.
Since the radial distance is maintained fixed, only the two angular coordinates (𝜃, 𝜙) are needed for positional identification. A representation of the radiation characteristics of the radiator as a function of 𝜃and 𝜙for a constant radial distance and frequency is defined as the pattern of the antenna.
In general, the pattern of an antenna is three-dimensional. Because it is impractical to measure a three-dimensional pattern, a number of two-dimensional patterns, as defined in Section 2.2, are measured. They are used to construct a three-dimensional pattern. The number of two-dimensional patterns needed to construct faithfully a three-dimensional graph is determined by the functional requirements of the description, and the available time and funds. The minimum number of two-dimensional patterns is two, and they are usually chosen to represent the orthogonal principal E- and H-plane patterns, as defined in Section 2.2.
A two-dimensional pattern is also referred to as a pattern cut, and it is obtained by fixing one of the angles (𝜃or 𝜙) while varying the other. For example, by referring to Figure 17.16, pattern cuts can be obtained by fixing 𝜙j(0 ≤𝜙j ≤2𝜋) and varying 𝜃(0 ≤𝜃≤𝜋). These are referred to as elevation patterns, and they are also displayed in Figure 2.19. Similarly, 𝜃can be maintained fixed (0 ≤𝜃i ≤𝜋) while 𝜙is varied (0 ≤𝜙≤2𝜋). These are designated as azimuthal patterns. Part (0 ≤𝜙≤𝜋∕2) of the 𝜃i = 𝜋∕2 azimuthal pattern is displayed in Figure 2.19.
Direction = 180° θ = 0° θ = 90° θ = 0° ϕ = 90° θ = 270° ϕ = 90° θ = 90° ϕ Eθ Eϕ y z x Antenna position ϕ θ ( , ) θ ϕ Equator Figure 17.16 Spherical coordinate system geometry. (source: IEEE Standard Test Procedures for Antennas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley).
RADIATION PATTERNS 1001 The patterns of an antenna can be measured in the transmitting or receiving mode. The mode is dictated by the application. However, if the radiator is reciprocal, as is the case for most practical antennas, then either the transmitting or receiving mode can be utilized. For such cases, the receiving mode is selected. The analytical formulations upon which an amplitude pattern is based, along with the advantages and disadvantages for making measurements in the transmitting or receiving mode, are found in Section 3.8.2. The analytical basis of a phase pattern is discussed in Section 13.10.
Unless otherwise specified, it will be assumed here that the measurements are performed in the receiving mode.
17.3.1 Instrumentation The instrumentation required to accomplish a measuring task depends largely on the functional requirements of the design. An antenna-range instrumentation must be designed to operate over a wide range of frequencies, and it usually can be classified into five categories : 1. source antenna and transmitting system 2. receiving system 3. positioning system 4. recording system 5. data-processing system A block diagram of a system that possesses these capabilities is shown in Figure 17.17.
Receiver Positioner control Test positioner indicators Source tower indicators Source tower control Source control Pattern recorder Figure 17.17 Instrumentation for typical antenna-range measuring system. (source: IEEE Standard Test Pro-cedures for Antennas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley).
1002 ANTENNA MEASUREMENTS The source antennas are usually log-periodic antennas for frequencies below 1 GHz, families of parabolas with broadband feeds for frequencies above 400 MHz, and even large horn antennas.
The system must be capable of controlling the polarization. Continuous rotation of the polarization can be accomplished by mounting a linearly polarized source antenna on a polarization positioner.
Antennas with circular polarization can also be designed, such as crossed log-periodic arrays, which are often used in measurements.
The transmitting RF source must be selected so that it has frequency control, frequency sta-bility, spectral purity, power level, and modulation. The receiving system could be as simple as a bolometer detector, followed possibly by an amplifier, and a recorder. More elaborate and expensive receiving systems that provide greater sensitivity, precision, and dynamic range can be designed. One such system is a heterodyne receiving system , which uses double conversion and phase locking, which can be used for amplitude measurements. A dual-channel heterodyne system design is also available , and it can be used for phase measurements.
To achieve the desired plane cuts, the mounting structures of the system must have the capability to rotate in various planes. This can be accomplished by utilizing rotational mounts (pedestals), two of which are shown in Figure 17.18. Tower-model elevation-over-azimuth pedestals are also available .
There are primarily two types of recorders; one that provides a linear (rectangular) plot and the other a polar plot. The polar plots are most popular because they provide a better visualization of the radiation distribution in space. Usually the recording equipment is designed to graph the relative pattern. Absolute patterns are obtained by making, in addition, gain measurements which will be discussed in the next section. The recording instrumentation is usually calibrated to record relative field or power patterns. Power pattern calibrations are in decibels with dynamic ranges of 0–60 dB.
For most applications, a 40-dB dynamic range is usually adequate and it provides sufficient resolution to examine the pattern structure of the main lobe and the minor lobes.
In an indoor antenna range, the recording equipment is usually placed in a room that adjoins the anechoic chamber. To provide an interference free environment, the chamber is closed during S O O x z −y A θ Figure 17.18 Azimuth-over-elevation and elevation-over-azimuth rotational mounts. (source: IEEE Stan-dard Test Procedures for Antennas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley).
GAIN MEASUREMENTS 1003 measurements. To monitor the procedures, windows or closed-circuit TVs are utilized. In addition, the recording equipment is connected, through synchronous servo-amplifier systems, to the rotational mounts (pedestals) using the traditional system shown in Figure 17.19(a). The system can record rectangular or polar plots. Position references are recorded simultaneously with measurements, and they are used for angular positional identification. As the rotational mount moves, the pattern is graphed simultaneously by the recorder on a moving chart. One of the axes of the chart is used to record the amplitude of the pattern while the other identifies the relative position of the radiator. A modern configuration to measure antenna and RCS patterns, using a network analyzer and being computer automated, is shown in Figure 17.19(b).
17.3.2 Amplitude Pattern The total amplitude pattern of an antenna is described by the vector sum of the two orthogonally polarized radiated field components. The pattern on a conventional antenna range can be measured using the system of Figure 17.17 or Figure 17.19 with an appropriate detector. The receiver may be a simple bolometer (followed possibly by an amplifier), a double-conversion phase-locking hetero-dyne system [7, Fig. 14], or any other design.
In many applications, the movement of the antenna to the antenna range can significantly alter the operational environment. Therefore, in some cases, antenna pattern measurements must be made in situ to preserve the environmental performance characteristics. A typical system arrangement that can be used to accomplish this is shown in Figure 17.20. The source is mounted on an airborne vehicle, which is maneuvered through space around the test antenna and in its far field, to produce a plane wave and to provide the desired pattern cuts. The tracking device provides to the recording equipment the angular position data of the source relative to a reference direction. The measurements can be conducted either by a point-by-point or by a continuous method. Usually the continuous technique is preferred.
17.3.3 Phase Measurements Phase measurements are based on the analytical formulations of Section 13.10. The phase pattern of the field, in the direction of the unit vector ̂ u, is given by the 𝜓(𝜃, 𝜙) phase function of (13-63). For linear polarization, ̂ u is real, and it may represent ̂ a𝜃or ̂ a𝜙in the direction of 𝜃or 𝜙.
The phase of an antenna is periodic, and it is defined in multiples of 360◦. In addition, the phase is a relative quantity, and a reference must be provided during measurements for comparison.
Two basic system techniques that can be used to measure phase patterns at short and long dis-tances from the antenna are shown respectively, in Figures 17.21(a) and 17.21(b). For the design of Figure 17.21(a), a reference signal is coupled from the transmission line, and it is used to compare, in an appropriate network, the phase of the received signal. For large distances, this method does not permit a direct comparison between the reference and the received signal. In these cases, the arrangement of Figure 17.21(b) can be used in which the signal from the source antenna is received simultaneously by a fixed antenna and the antenna under test. The phase pattern is recorded as the antenna under test is rotated while the fixed antenna serves as a reference. The phase measuring circuit may be the dual-channel heterodyne system [7, Figure 15].
17.4 GAIN MEASUREMENTS The most important figure of merit that describes the performance of a radiator is the gain. There are various techniques and antenna ranges that are used to measure the gain. The choice of either depends largely on the frequency of operation.
Usually free-space ranges are used to measure the gain above 1 GHz. In addition, microwave techniques, which utilize waveguide components, can be used. At lower frequencies, it is more Modulator Pen system Pen system Chart system Turntable system Selective filter amplifier RF generator R.F. source Network anlyzer Positioner controller Computer Plotter Printer CRT Data storage Network Mixers RF amplifier Precision variable attenuator Figure 17.19 Block diagrams of typical instrumentations for measuring rectangular and polar antenna and RCS patterns.
1004 GAIN MEASUREMENTS 1005 Figure 17.20 System arrangement for in situ antenna pattern measurements. (source: IEEE Standard Test Procedures for Antennas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley).
difficult to simulate free-space conditions because of the longer wavelengths. Therefore between 0.1–1 GHz, ground-reflection ranges are utilized. Scale models can also be used in this frequency range. However, since the conductivity and loss factors of the structures cannot be scaled conve-niently, the efficiency of the full scale model must be found by other methods to determine the gain of the antenna. This is accomplished by multiplying the directivity by the efficiency to result in the gain. Below 0.1 GHz, directive antennas are physically large and the ground effects become increas-ingly pronounced. Usually the gain at these frequencies is measured in situ. Antenna gains are not usually measured at frequencies below 1 MHz. Instead, measurements are conducted on the field strength of the ground wave radiated by the antenna.
Figure 17.21 Near-field and far-field phase pattern measuring systems. (source: IEEE Standard Test Proce-dures for Antennas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley).
1006 ANTENNA MEASUREMENTS Usually there are two basic methods that can be used to measure the gain of an electromag-netic radiator: realized-gain and gain-transfer (or gain-comparison) measurements. The realized-gain method is used to calibrate antennas that can then be used as standards for gain measurements, and it requires no a priori knowledge of the gains of the antennas. Gain-transfer methods must be used in conjunction with standard gain antennas to determine the realized gain of the antenna under test.
The two antennas that are most widely used and universally accepted as gain standards are the resonant λ∕2 dipole (with a gain of about 2.1 dB) and the pyramidal horn antenna (with a gain ranging from 12–25 dB). Both antennas possess linear polarizations. The dipole, in free-space, exhibits a high degree of polarization purity. However, because of its broad pattern, its polarization may be suspect in other than reflection-free environments. Pyramidal horns usually possess, in free-space, slightly elliptical polarization (axial ratio of about 40 to infinite dB). However, because of their very directive patterns, they are less affected by the surrounding environment.
17.4.1 Realized-Gain Measurements There are a number of techniques that can be employed to make realized-gain measurements. A very brief review of each will be included here. More details can be found in –. All of these methods are based on the Friis transmission formula [as given by (2-118)] which assumes that the measuring system employs, each time, two antennas (as shown in Figure 2.31). The antennas are separated by a distance R, and it must satisfy the far-field criterion of each antenna. For polarization matched antennas, aligned for maximum directional radiation, (2-118) reduces to (2-119).
A. Two-Antenna Method Equation (2-119) can be written in a logarithmic decibel form as (G0t)dB + (G0r)dB = 20 log10 (4𝜋R λ ) + 10 log10 (Pr Pt ) (17-14) where (G0t)dB = gain of the transmitting antenna (dB) (G0r)dB = gain of the receiving antenna (dB) Pr = received power (W) Pt = transmitted power (W) R = antenna separation (m) λ = operating wavelength (m) If the transmitting and receiving antennas are identical (G0t = G0r), (17-14) reduces to (G0t)dB = (G0r)dB = 1 2 [ 20 log10 (4𝜋R λ ) + 10 log10 (Pr Pt )] (17-15) By measuring R, λ, and the ratio of Pr∕Pt, the gain of the antenna can be found. At a given frequency, this can be accomplished using the system of Figure 17.22(a). The system is simple and the procedure straightforward. For continuous multifrequency measurements, such as for broadband antennas, the swept-frequency instrumentation of Figure 17.22(b) can be utilized.
GAIN MEASUREMENTS 1007 Figure 17.22 Typical two- and three-antenna measuring systems for single- and swept-frequency measure-ments. (source: J. S. Hollis, T. J. Lyon, and L. Clayton, Jr., Microwave Antenna Measurements, Scientific-Atlanta, Inc., Atlanta, Georgia, July 1970).
B. Three-Antenna Method If the two antennas in the measuring system are not identical, three antennas (a, b, c) must be employed and three measurements must be made (using all combinations of the three) to deter-mine the gain of each of the three. Three equations (one for each combination) can be written, and each takes the form of (17-14). Thus (a-b Combination) (Ga)dB + (Gb)dB = 20 log10 (4𝜋R λ ) + 10 log10 (Prb Pta ) (17-16a) (a-c Combination) (Ga)dB + (Gc)dB = 20 log10 (4𝜋R λ ) + 10 log10 (Prc Pta ) (17-16b) 1008 ANTENNA MEASUREMENTS (b-c Combination) (Gb)dB + (Gc)dB = 20 log10 (4𝜋R λ ) + 10 log10 (Prc Ptb ) (17-16c) From these three equations, the gains (Ga)dB, (Gb)dB, and (Gc)dB can be determined provided R, λ, and the ratios of Prb∕Pta, Prc∕Pta, and Prc∕Ptb are measured.
The two- and three-antenna methods are both subject to errors. Care must be utilized so 1. the system is frequency stable 2. the antennas meet the far-field criteria 3. the antennas are aligned for boresight radiation 4. all the components are impedance and polarization matched 5. there is a minimum of proximity effects and multipath interference Impedance and polarization errors can be accounted for by measuring the appropriate complex reflection coefficients and polarizations and then correcting accordingly the measured power ratios.
The details for these corrections can be found in , . There are no rigorous methods to account for proximity effects and multipath interference. These, however, can be minimized by maintaining the antenna separation by at least a distance of 2D2∕λ, as is required by the far-field criteria, and by utilizing RF absorbers to reduce unwanted reflections. The interference pattern that is created by the multiple reflections from the antennas themselves, especially at small separations, is more difficult to remove. It usually manifests itself as a cyclic variation in the measured antenna gain as a function of separation.
C. Extrapolation Method The extrapolation method is an realized-gain method, which can be used with the three-antenna method, and it was developed to rigorously account for possible errors due to proximity, mul-tipath, and nonidentical antennas. If none of the antennas used in the measurements are circularly polarized, the method yields the gains and polarizations of all three antennas. If only one antenna is circularly polarized, this method yields only the gain and polarization of the circularly polarized antenna. The method fails if two or more antennas are circularly polarized.
The method requires both amplitude and phase measurements when the gain and the polarization of the antennas are to be determined. For the determination of gains, amplitude measurements are sufficient. The details of this method can be found in , .
D. Ground-Reflection Range Method A method that can be used to measure the gain of moderately broad-beam antennas, usually for fre-quencies below 1 GHz, has been reported . The method takes into account the specular reflections from the ground (using the system geometry of Figure 17.2), and it can be used with some restric-tions and modifications with the two- or three-antenna methods. As described here, the method is applicable to linear antennas that couple only the electric field. Modifications must be made for loop radiators. Using this method, it is recommended that the linear vertical radiators be placed in a horizontal position when measurements are made. This is desired because the reflection coeffi-cient of the earth, as a function of incidence angle, varies very rapidly for vertically polarized waves.
Smoother variations are exhibited for horizontally polarized fields. Circularly and elliptically polar-ized antennas are excluded, because the earth exhibits different reflective properties for vertical and horizontal fields.
To make measurements using this technique, the system geometry of Figure 17.2 is utilized.
Usually it is desirable that the height of the receiving antenna hr be much smaller than the range GAIN MEASUREMENTS 1009 R0 (hr ≪R0). Also the height of the transmitting antenna is adjusted so that the field of the receiving antenna occurs at the first maximum nearest to the ground. Doing this, each of the gain equations of the two- or three-antenna methods take the form of (Ga)dB + (Gb)dB = 20 log10 (4𝜋RD λ ) + 10 log10 (Pr Pt ) −20 log10 (√ DADB + rRD RR ) (17-17) where DA and DB are the directivities (relative to their respective maximum values) along RD, and they can be determined from amplitude patterns measured prior to the gain measurements. RD, RR, λ, and Pr∕Pt are also measured. The only quantity that needs to be determined is the factor r which is a function of the radiation patterns of the antennas, the frequency of operation, and the electrical and geometrical properties of the antenna range.
The factor r can be found by first repeating the above measurements but with the transmitting antenna height adjusted so that the field at the receiving antenna is minimized. The quantities mea-sured with this geometry are designated by the same letters as before but with a prime (′) to distin-guish them from those of the previous measurement.
By measuring or determining the parameters 1. RR, RD, Pr, DA, and DB at a height of the transmitting antenna such that the receiving antenna is at the first maximum of the pattern 2. R′ R, R′ D, P′ r, D′ A, and D′ B at a height of the transmitting antenna such that the receiving antenna is at a field minimum it can be shown that r can be determined from r = ( RRR′ R RDR′ D ) ⎡ ⎢ ⎢ ⎢ ⎣ √ (Pr∕P′ r)(D′ AD′ B)RD − √ DADBR′ D √(Pr∕P′ r)RR + R′ R ⎤ ⎥ ⎥ ⎥ ⎦ (17-18) Now all parameters included in (17-17) can either be measured or computed from measurements.
The free-space range system of Figure 17.22(a) can be used to perform these measurements.
17.4.2 Gain-Transfer (Gain-Comparison) Measurements The method most commonly used to measure the gain of an antenna is the gain-transfer method.
This technique utilizes a gain standard (with a known gain) to determine realized gains. Initially rel-ative gain measurements are performed, which when compared with the known gain of the standard antenna, yield realized values. The method can be used with free-space and reflection ranges, and for in situ measurements.
The procedure requires two sets of measurements. In one set, using the test antenna as the receiv-ing antenna, the received power (PT) into a matched load is recorded. In the other set, the test antenna is replaced by the standard gain antenna and the received power (PS) into a matched load is recorded.
In both sets, the geometrical arrangement is maintained intact (other than replacing the receiving antennas), and the input power is maintained the same.
1010 ANTENNA MEASUREMENTS Writing two equations of the form of (17-14) or (17-17), for free-space or reflection ranges, it can be shown that they reduce to (GT)dB = (GS)dB + 10 log10 (PT PS ) (17-19) where (GT)dB and (GS)dB are the gains (in dB) of the test and standard gain antennas.
System disturbance during replacement of the receiving antennas can be minimized by mounting the two receiving antennas back-to-back on either side of the axis of an azimuth positioner and connecting both of them to the load through a common switch. One antenna can replace the other by a simple, but very precise, 180◦rotation of the positioner. Connection to the load can be interchanged by proper movement of the switch.
If the test antenna is not too dissimilar from the standard gain antenna, this method is less affected by proximity effects and multipath interference. Impedance and polarization mismatches can be corrected by making proper complex reflection coefficient and polarization measurements .
If the test antenna is circularly or elliptically polarized, gain measurements using the gain-transfer method can be accomplished by at least two different methods. One way would be to design a stan-dard gain antenna that possesses circular or elliptical polarization. This approach would be attractive in mass productions of power-gain measurements of circularly or elliptically polarized antennas.
The other approach would be to measure the gain with two orthogonal linearly polarized standard gain antennas. Since circularly and elliptically polarized waves can be decomposed to linear (vertical and horizontal) components, the total power of the wave can be separated into two orthogonal linearly polarized components. Thus the total gain of the circularly or elliptically polarized test antenna can be written as (GT)dB = 10 log10(GTV + GTH) (17-20) GTV and GTH are, respectively, the partial power gains with respect to vertical-linear and horizontal-linear polarizations.
GTV is obtained, using (17-19), by performing a gain-transfer measurement with the standard gain antenna possessing vertical polarization. The measurements are repeated with the standard gain antenna oriented for horizontal polarization. This allows the determination of GTH. Usually a single linearly polarized standard gain antenna (a linear λ∕2 resonant dipole or a pyramidal horn) can be used, by rotating it by 90◦, to provide both vertical and horizontal polarizations. This approach is very convenient, especially if the antenna possesses good polarization purity in the two orthogonal planes.
The techniques outlined above yield good results provided the transmitting and standard gain antennas exhibit good linear polarization purity. Errors will be introduced if either one of them pos-sesses a polarization with a finite axial ratio. In addition, these techniques are accurate if the tests can be performed in a free-space, a ground-reflection, or an extrapolation range. These requirements place a low-frequency limit of 50 MHz.
Below 50 MHz, the ground has a large effect on the radiation characteristics of the antenna, and it must be taken into account. It usually requires that the measurements are performed on full scale models and in situ. Techniques that can be used to measure the gain of large HF antennas have been devised –.
17.5 DIRECTIVITY MEASUREMENTS If the directivity of the antenna cannot be found using solely analytical techniques, it can be computed using measurements of its radiation pattern. One of the methods is based on the approximate expres-sions of (2-27) by Kraus or (2-30b) by Tai and Pereira, whereas the other relies on the numerical DIRECTIVITY MEASUREMENTS 1011 techniques that were developed in Section 2.7. The computations can be performed very efficiently and economically with modern computational facilities and numerical techniques.
The simplest, but least accurate, method requires that the following procedure is adopted: 1. Measure the two principal E- and H-plane patterns of the test antenna.
2. Determine the half-power beamwidths (in degrees) of the E- and H-plane patterns.
3. Compute the directivity using either (2-27) or (2-30b).
The method is usually employed to obtain rough estimates of directivity. It is more accurate when the pattern exhibits only one major lobe, and its minor lobes are negligible.
The other method requires that the directivity be computed using (2-35) where Prad is evaluated numerically using (2-43). The F(𝜃i, 𝜙j) function represents the radiation intensity or radiation pat-tern, as defined by (2-42), and it will be obtained by measurements. Umax in (2-35) represents the maximum radiation intensity of F(𝜃, 𝜙) in all space, as obtained by the measurements.
The radiation pattern is measured by sampling the field over a sphere of radius r. The pattern is measured in two-dimensional plane cuts with 𝜙j constant (0 ≤𝜙j ≤2𝜋) and 𝜃variable (0 ≤𝜃≤𝜋), as shown in Figure 2.19, or with 𝜃i fixed (0 ≤𝜃i ≤𝜋) and 𝜙variable (0 ≤𝜙≤2𝜋). The first are referred to as elevation or great-circle cuts, whereas the second represent azimuthal or conical cuts.
Either measuring method can be used. Equation (2-43) is written in a form that is most convenient for elevation or great-circle cuts. However, it can be rewritten to accommodate azimuthal or conical cuts.
The spacing between measuring points is largely controlled by the directive properties of the antenna and the desired accuracy. The method is most accurate for broad-beam antennas. However, with the computer facilities and the numerical methods now available, this method is very attractive even for highly directional antennas. To maintain a given accuracy, the number of sampling points must increase as the pattern becomes more directional. The pattern data is recorded digitally on tape, and it can be entered into a computer at a later time. If on-line computer facilities are available, the measurements can be automated to provide essentially real-time computations.
The above discussion assumes that all the radiated power is contained in a single polariza-tion, and the measuring probe possesses that polarization. If the antenna is polarized such that the field is represented by both 𝜃and 𝜙components, the partial directivities D𝜃(𝜃, 𝜙) and D𝜙(𝜃, 𝜙) of (2-17)–(2-17b) must each be found. This is accomplished from pattern measurements with the probe positioned, respectively, to sample the 𝜃and 𝜙components. The total directivity is then given by (2-17)–(2-17b), or D0 = D𝜃+ D𝜙 (17-21) where D𝜃= 4𝜋U𝜃 (Prad)𝜃+ (Prad)𝜙 (17-21a) D𝜙= 4𝜋U𝜙 (Prad)𝜃+ (Prad)𝜙 (17-21b) U𝜃, (Prad)𝜃and U𝜙, (Prad)𝜙represent the radiation intensity and radiated power as contained in the two orthogonal 𝜃and 𝜙field components, respectively.
The same technique can be used to measure the field intensity and to compute the directivity of any antenna that possesses two orthogonal polarizations. Many antennas have only one polarization (𝜃or 𝜙). This is usually accomplished by design and/or proper selection of the coordinate sys-tem. In this case, the desired polarization is defined as the primary polarization. Ideally, the other 1012 ANTENNA MEASUREMENTS polarization should be zero. However, in practice, it is nonvanishing, but it is very small. Usually it is referred to as the cross polarization, and for good designs it is usually below −40 dB.
The directivity of circularly or elliptically polarized antennas can also be measured. Precautions must be taken as to which component represents the primary polarization and which the cross-polarization contribution.
17.6 RADIATION EFFICIENCY The radiation efficiency is defined as the ratio of the total power radiated by the antenna to the total power accepted by the antenna at its input terminals during radiation. System factors, such as impedance and/or polarization mismatches, do not contribute to the radiation efficiency because it is an inherent property of the antenna.
The radiation efficiency can also be defined, using the direction of maximum radiation as refer-ence, as radiation efficiency = gain directivity (17-22) where directivity and gain, as defined in Sections 2.6 and 2.9, imply that they are measured or com-puted in the direction of maximum radiation. Using techniques that were outlined in Sections 17.4 and 17.5 for the measurements of the gain and directivity, the radiation efficiency can then be com-puted using (17-22).
If the antenna is very small and simple, it can be represented as a series network as shown in Figures 2.27(b) or 2.28(b). For antennas that can be represented by such a series network, the radi-ation efficiency can also be defined by (2-90) and it can be computed by another method . For these antennas, the real part of the input impedance is equal to the total antenna resistance which consists of the radiation resistance and the loss resistance.
The radiation resistance accounts for the radiated power. For many simple antennas (dipoles, loops, etc.), it can be found by analytically or numerically integrating the pattern, relating it to the radiated power and to the radiation resistance by a relation similar to (4-18). The loss resistance accounts for the dissipated power, and it is found by measuring the input impedance (input resistance −radiation resistance = loss resistance).
Because the loss resistance of antennas coated with lossy dielectrics or antennas over lossy ground cannot be represented in series with the radiation resistance, this method cannot be used to determine their radiation efficiency. The details of this method can be found in .
17.7 IMPEDANCE MEASUREMENTS Associated with an antenna, there are two types of impedances: a self and a mutual impedance.
When the antenna is radiating into an unbounded medium and there is no coupling between it and other antennas or surrounding obstacles, the self-impedance is also the driving-point impedance of the antenna. If there is coupling between the antenna under test and other sources or obstacles, the driving-point impedance is a function of its self-impedance and the mutual impedances between it and the other sources or obstacles. In practice, the driving-point impedance is usually referred to as the input impedance. The definitions and the analytical formulations that underlie the self, mutual, and input impedances are presented in Chapter 8.
To attain maximum power transfer between a source or a source-transmission line and an antenna (or between an antenna and a receiver or transmission line-receiver), a conjugate match is usually desired. In some applications, this may not be the most ideal match. For example, in some receiving systems, minimum noise is attained if the antenna impedance is lower than the load impedance.
IMPEDANCE MEASUREMENTS 1013 However, in some transmitting systems, maximum power transfer is attained if the antenna impedance is greater than the load impedance. If conjugate matching does not exist, the power lost can be computed using Plost Pavailable = | | | | | Zant −Z∗ cct Zant + Zcct | | | | | 2 (17-23) where Zant = input impedance of the antenna Zcct = input impedance of the circuits which are connected to the antenna at its input terminals When a transmission line is associated with the system, as is usually the case, the matching can be performed at either end of the line. In practice, however, the matching is performed near the antenna terminals, because it usually minimizes line losses and voltage peaks in the line and maximizes the useful bandwidth of the system.
In a mismatched system, the degree of mismatch determines the amount of incident or available power which is reflected at the input antenna terminals into the line. The degree of mismatch is a function of the antenna input impedance and the characteristic impedance of the line. These are related to the input reflection coefficient and the input VSWR at the antenna input terminals by the standard transmission-line relationships of Prefl Pinc = |Γ|2 = |Zant −Zc|2 |Zant + Zc|2 = | | | | VSWR −1 VSWR + 1 | | | | 2 (17-24) where Γ = |Γ|ej𝛾= voltage reflection coefficient at the antenna input terminals VSWR = voltage standing wave ratio at the antenna input terminals Zc = characteristic impedance of the transmission line Equation (17-24) shows a direct relationship between the antenna input impedance (Zant) and the VSWR. In fact, if Zant is known, the VSWR can be computed using (17-24). In practice, however, that is not the case. What is usually measured is the VSWR, and it alone does not provide sufficient information to uniquely determine the complex input impedance. To overcome this, the usual pro-cedure is to measure the VSWR, and to compute the magnitude of the reflection coefficient using (17-24). The phase of the reflection coefficient can be determined by locating a voltage maximum or a voltage minimum (from the antenna input terminals) in the transmission line. Since in practice the minima can be measured more accurately than the maxima, they are usually preferred. In addition, the first minimum is usually chosen unless the distance from it to the input terminals is too small to measure accurately. The phase 𝛾of the reflection coefficient is then computed using 𝛾= 2𝛽xn ± (2n −1)𝜋= 4𝜋 λg xn ± (2n −1)𝜋, n = 1, 2, 3, … (17-25) where n = the voltage minimum from the input terminals (i.e., n = 1 is used to locate the first voltage minimum) xn = distance from the input terminals to the nth voltage minimum λg = wavelength measured inside the input transmission line (it is twice the distance between two voltage minima or two voltage maxima) 1014 ANTENNA MEASUREMENTS Once the reflection coefficient is completely described by its magnitude and phase, it can be used to determine the antenna impedance by Zant = Zc [1 + Γ 1 −Γ ] = Zc [1 + |Γ|ej𝛾 1 −|Γ|ej𝛾 ] (17-26) Other methods, utilizing impedance bridges, slotted lines, and broadband swept-frequency network analyzers, can be utilized to determine the antenna impedance –.
The input impedance is generally a function of frequency, geometry, method of excitation, and proximity to its surrounding objects. Because of its strong dependence on the environment, it should usually be measured in situ unless the antenna possesses very narrow beam characteristics.
Mutual impedances, which take into account interaction effects, are usually best described and measured by the cross-coupling coefficients Smn of the device’s (antenna’s) scattering matrix.
The coefficients of the scattering matrix can then be related to the coefficients of the impedance matrix .
17.8 CURRENT MEASUREMENTS The current distribution along an antenna is another very important antenna parameter. A complete description of its amplitude and phase permit the calculation of the radiation pattern.
There are a number of techniques that can be used to measure the current distribution –.
One of the simplest methods requires that a small sampling probe, usually a small loop, be placed near the radiator. On the sampling probe, a current is induced which is proportional to the current of the test antenna.
The indicating meter can be connected to the loop in many different ways . If the wavelength is very long, the meter can be consolidated into one unit with the measuring loop. At smaller wave-lengths, the meter can be connected to a crystal rectifier. In order not to disturb the field distribution near the radiator, the rectifier is attached to the meter using long leads. To reduce the interaction between the measuring instrumentation and the test antenna and to minimize induced currents on the leads, the wires are wound on a dielectric support rod to form a helical choke. Usually the diam-eter of each turn, and spacing between them, is about λ∕50. The dielectric rod can also be used as a support for the loop. To prevent a dc short circuit on the crystal rectifier, a bypass capacitor is placed along the circumference of the loop.
There are many other methods, some of them more elaborate and accurate, and the interested reader can refer to the literature –.
17.9 POLARIZATION MEASUREMENTS The polarization of a wave was defined in Section 2.12 as the curve traced by the instantaneous electric field, at a given frequency, in a plane perpendicular to the direction of wave travel. The far-field polarization of an antenna is usually measured at distances where the field radiated by the antenna forms, in a small region, a plane wave that propagates in the outward radial direction.
In a similar manner, the polarization of the antenna is defined as the curve traced by the instan-taneous electric field radiated by the antenna in a plane perpendicular to the radial direction, as shown in Figure 17.23(a). The locus is usually an ellipse. In a spherical coordinate system, which is usually adopted in antennas, the polarization ellipse is formed by the orthogonal electric field components of E𝜃and E𝜙. The sense of rotation, also referred to as the sense of polarization, is defined by the sense of rotation of the wave as it is observed along the direction of propagation [see Figure 17.23(b)].
POLARIZATION MEASUREMENTS 1015 = 0° θ = 90° θ = 0° ϕ = 90° θ = 90° ϕ Antenna location (a) Polarization ellipse Direction of propagation Direction of propagation Right hand Clockwise (b) Sense of rotation ϕ θ τ r αr ϕ α θ α ˆ ˆ ˆ Figure 17.23 Polarization ellipse and sense of rotation for antenna coordinate system. (source: IEEE Stan-dard Test Procedures for Antennas, IEEE Std 149–1979, published by IEEE, Inc., 1979, distributed by Wiley).
The general polarization of an antenna is characterized by the axial ratio (AR), the sense of rota-tion (CW or CCW, RH or LH), and the tilt angle 𝜏. The tilt angle is used to identify the spatial orientation of the ellipse, and it is usually measured clockwise from the reference direction. This is demonstrated in Figure 17.23(a) where 𝜏is measured clockwise with respect to ̂ a𝜃, for a wave traveling in the outward radial direction.
Care must be exercised in the characterization of the polarization of a receiving antenna. If the tilt angle of an incident wave that is polarization matched to the receiving antenna is 𝜏m, it is related to the tilt angle 𝜏t of a wave transmitted by the same antenna by 𝜏t = 180◦−𝜏m (17-27) if a single coordinate system and one direction of view are used to characterize the polarization. If the receiving antenna has a polarization that is different from that of the incident wave, the polarization loss factor (PLF) of Section 2.12.2 can be used to account for the polarization mismatch losses.
1016 ANTENNA MEASUREMENTS LHC Poles represent circular polarizations Upper hemisphere left-hand sense 45° Linear Longitude represents tilt angle Lower hemisphere right-hand sense Equator represents linear polarizations Latitude represents axial ratio RHC H Figure 17.24 Polarization representation on the Poincar´ e sphere. (source: W. H. Kummer and E. S. Gillespie, “Antenna Measurements—1978,” Proc. IEEE, Vol. 66, No. 4, pp. 483–507, April 1978. c ⃝(1978) IEEE).
The polarization of a wave and/or an antenna can best be displayed and visualized on the surface of a Poincar´ e sphere . Each polarization occupies a unique point on the sphere, as shown in Figure 17.24. If one of the two points on the Poincar´ e sphere is used to define the polarization of the incident wave and the other the polarization of the receiving antenna, the angular separation can be used to determine the polarization losses. The procedure requires that the complex polarization ratios of each are determined, and they are used to compute the polarization efficiency in a number of different ways. The details of this procedure are well documented, and they can be found in , .
Practically it is very difficult to design radiators that maintain the same polarization state in all parts of their pattern. A complete description requires a number of measurements in all parts of the pattern. The number of measurements is determined by the required degree of polarization description.
There are a number of techniques that can be used to measure the polarization state of a radia-tor , , and they can be classified into three main categories: 1. Those that yield partial polarization information. They do not yield a unique point on the Poincar´ e sphere.
2. Those that yield complete polarization information but require a polarization standard for com-parison. They are referred to as comparison methods.
3. Those that yield complete polarization information and require no a priori polarization knowl-edge or no polarization standard. They are designated as absolute methods.
The method selected depends on such factors as the type of antenna, the required accuracy, and the time and funds available. A complete description requires not only the polarization ellipse (axial ratio and tilt angle), but also its sense of rotation (CW or CCW, RH or LH).
In this text, a method will be discussed which can be used to determine the polarization ellipse (axial ratio and tilt angle) of an antenna but not its sense of rotation. This technique is referred to as the polarization-pattern method. The sense of polarization or rotation can be found by performing auxiliary measurements or by using other methods .
To perform the measurements, the antenna under test can be used either in the transmitting or in the receiving mode. Usually the transmitting mode is adopted. The method requires that a linearly polarized antenna, usually a dipole, be used to probe the polarization in the plane that contains the direction of the desired polarization. The arrangement is shown in Figure 17.25(a). The dipole is POLARIZATION MEASUREMENTS 1017 Figure 17.25 Polarization measuring system and typical patterns.
rotated in the plane of the polarization, which is taken to be normal to the direction of the incident field, and the output voltage of the probe is recorded.
If the test antenna is linearly polarized, the output voltage response will be proportional to sin 𝜓 (which is the far-zone field pattern of an infinitesimal dipole). The pattern forms a figure-eight, as shown in Figure 17.25(b), where 𝜓is the rotation angle of the probe relative to a reference direction.
For an elliptically polarized test antenna, the nulls of the figure-eight are filled and a dumbbell polar-ization curve (usually referred to as a polarization pattern) is generated, as shown in Figure 17.25(b).
The dashed curve represents the polarization ellipse.
The polarization ellipse is tangent to the polarization pattern, and it can be used to determine the axial ratio and the tilt angle of the test antenna. The polarization pattern will be a circle, as shown in Figure 17.25(b), if the test antenna is circularly polarized. Ideally, this process must be repeated at every point of the antenna pattern. Usually it is performed at a number of points that describe sufficiently well the polarization of the antenna at the major and the minor lobes.
In some cases the polarization needs to be known over an entire plane. The axial ratio part of the polarization state can be measured using the arrangement of Figure 17.25(a) where the test probe antenna is used usually as a source while the polarization pattern of the test antenna is being recorded while the test antenna is rotated over the desired plane. This arrangement does not yield the tilt angle or sense of rotation of the polarization state. In order to obtain the desired polarization pattern, the rate of rotation of the linear probe antenna (usually a dipole) is much greater than the rotation rate of the positioner over which the test antenna is mounted and rotated to allow, ideally, the probe antenna to measure the polarization response of the test antenna at that direction before moving to another angle. When this is performed over an entire plane, a typical pattern recorded in decibels is shown in Figure 17.26 , and it is referred as the axial ratio pattern. It is apparent that the axial ratio pattern can be inscribed by inner and outer envelopes. At any given angle, the ratio of the outer and inner 1018 ANTENNA MEASUREMENTS Figure 17.26 Pattern of a circularly polarized test antenna taken with a rotating, linearly polarized, source antenna. (source: E. S. Gillespie, “Measurement of Antenna Radiation Characteristics on Far-Field Ranges,” in Antenna Handbook (Y. T. Lo & S. W. Lee, eds.), 1988, c ⃝Van Nostrand Reinhold Co., Inc).
envelope responses represent the axial ratio. If the pattern is recorded in decibels, the axial ratio is the difference between the outer and inner envelopes (in dB); zero dB difference represents circular polarization (axial ratio of unity). Therefore the polarization pattern of Figure 17.26 indicates that the test antenna it represents is nearly circularly polarized (within 1 dB; axial ratio less than 1.122) at and near 𝜃= 0◦and deviates from that almost monotonically at greater angles (typically by about 7 dB maximum; maximum axial ratio of about 2.24).
The sense of rotation can be determined by performing auxiliary measurements. One method requires that the responses of two circularly polarized antennas, one responsive to CW and the other to CCW rotation, be compared . The rotation sense of the test antenna corresponds to the sense of polarization of the antenna which produced the more intense response.
Another method would be to use a dual-polarized probe antenna, such as a dual-polarized horn, and to record simultaneously the amplitude polarization pattern and the relative phase between the two orthogonal polarizations. This is referred to as the phase-amplitude method, and it can be accom-plished using the instrumentation of Figure 17.27. Double-conversion phase-locked receivers can be used to perform the amplitude and phase comparison measurements.
Figure 17.27 System configuration for measurements of polarization amplitude and phase. (source: W. H.
Kummer and E. S. Gillespie, “Antenna Measurements—1978,” Proc. IEEE, Vol. 66, No. 4, pp. 483–507, April 1978. c ⃝(1978) IEEE).
SCALE MODEL MEASUREMENTS 1019 Another absolute polarization method, which can be used to completely describe the polarization of a test antenna, is referred to as the three-antenna method , . As its name implies, it requires three antennas, two of which must not be circularly polarized. There are a number of transfer meth-ods , , but they require calibration standards for complete description of the polarization state.
17.10 SCALE MODEL MEASUREMENTS In many applications (such as with antennas on ships, aircraft, large spacecraft, etc.), the antenna and its supporting structure are so immense in weight and/or size that they cannot be moved or accommodated by the facilities of a measuring antenna range. In addition, a movement of the struc-ture to an antenna range can eliminate or introduce environmental effects. To satisfy these system requirements, in situ measurements are usually executed.
A technique that can be used to perform antenna measurements associated with large structures is geometrical scale modeling. Geometrical modeling is employed to 1. physically accommodate, within small ranges or enclosures, measurements that can be related to large structures 2. provide experimental control over the measurements 3. minimize costs associated with large structures and corresponding experimental paramet-ric studies Geometrical scale modeling by a factor of n (n smaller or greater than unity) requires the scaling indicated in Table 17.1. The primed parameters represent the scaled model while the unprimed rep-resent the full scale model. For a geometrical scale model, all the linear dimensions of the antenna and its associated structure are divided by n whereas the operating frequency and the conductivity of the antenna material and its structure are multiplied by n. In practice, the scaling factor n is usually chosen greater than unity.
Ideal scale modeling for antenna measurements requires exact replicas, both physically and elec-trically, of the full scale structures. In practice, however, this is closely approximated. The most dif-ficult scaling is that of the conductivity. If the full scale model possesses excellent conductors, even better conductors will be required in the scaled models. At microwave and millimeter wave frequen-cies this can be accomplished by utilizing clean polished surfaces, free of films and other residues.
Geometrical scaling is often used for pattern measurements. However, it can also be employed to measure gain, directivity, radiation efficiency, input and mutual impedances, and so forth. For gain measurements, the inability to properly scale the conductivity can be overcome by measuring the directivity and the antenna efficiency and multiplying the two to determine the gain. Scalings that permit additional parameter changes are available . The changes must satisfy the theorem of similitude.
TABLE 17.1 Geometrical Scale Model Scaled Parameters Unchanged Parameters Length: l′ = l∕n Permittivity: 𝜀′ = 𝜀 Time: t′ = t∕n Permeability: 𝜇′ = 𝜇 Wavelength: λ′ = λ∕n Velocity: 𝜐′ = 𝜐 Capacitance: C′ = C∕n Impedance: Z′ = Z Inductance: L′ = L∕n Antenna gain: G0 ′ = G0 Echo area (RCS): Ae ′ = Ae∕n2 Frequency: f ′ = nf Conductivity: 𝜎′ = n𝜎 1020 ANTENNA MEASUREMENTS To illustrate the scaling for Gain (Amplitude) and Echo Area (RCS), two examples are used.
17.10.1 Gain (Amplitude) Measurements, Simulations and Comparisons For the gain, the absolute amplitude radiation patterns of a λ∕4 monopole placed at the belly (bottom side) of a generic scale model helicopter [see Figure 17.28(a)], whose dimensions are about 1/10 the size of a full-scale helicopter. The absolute amplitude patterns of the λ∕4 monopole on the scale model generic helicopter were measured at 7 GHz along the three principal planes; pitch, roll and yaw. In addition, the same patterns were simulated on a full-scale helicopter (10 times larger) but at a frequency of 700 MHz (1/10 the measured frequency) of the same helicopter geometry. The geometry (b) Pitch plane (d) Yaw plane (d) Roll plane (a) Scale model helicopter (10:1 scale) 10 0 –10 –20 0° 30° 60° 90° 120° 150° 180° 150° 120° 90° 60° 30° 10 0 –10 –20 0° 330° 300° 270° 240° 210° 180° 150° 120° 90° 60° 30° 10 0 –10 –20 0° 30° 60° 90° 120° 150° 180° 150° 120° 90° 60° 30° Full-Scale Prediction (700 MHz) Scale Model Measurement (7 GHz) Figure 17.28 Scale model helicopter, scale model measurements (7 GHz) and full-scale model simulations (700 MHz) of a λ∕4 monopole on the belly of airframe.
SCALE MODEL MEASUREMENTS 1021 of the generic scale helicopter, along with the measured and simulated patterns, is indicated in Figure 17.28. It is evident that there is, as expected, a correct scaling between the measured ampli-tude (gain) patterns on the 1/10 scale model but at a frequency of 7 GHz (increased by a factor of 10 since the size of the scale model was 1/10 of the full-scale) and the simulated patterns at 700 MHz (a factor of 1/10 of the measured) but on a full-scale model (larger by a factor of 10). The maximum gain is about 6 dB, which is basically what is expected from a λ∕4 monopole. In addition, there is an excellent comparison between the respective two sets of patterns (measured and simulated) considering the complexity of the airframe.
17.10.2 Echo Area (RCS) Measurements, Simulations and Comparisons In addition to the scaled and full-scale models’ gain (amplitude) radiation patterns, measured and simulated, the procedure was repeated for scaled and full-scale Echo Area (RCS) monostatic pat-terns, both measured and simulated. The geometry of the radar target to perform the monostatic RCS measurements and simulations is a square (a = b) PEC flat plate, whose configuration and polariza-tion designations are shown in Figure 17.29. The scaling factor used for the RCS was 3:1 for both the measurements and simulations; the frequencies for both the measurements and simulations were 15 GHz (scaled) and 5 GHz (full-scale). The dimensions of the full-scale (large) square plate were 30 cm × 30 cm while for the scaled (small) square plate were 10 cm × 10 cm.
The vertical polarization [Figure 17.29(b)], measured and simulated, monostatic RCS patterns (in dBsm) of the scaled and full-scale RCS patterns along the principal y-z plane are displayed in Figure 17.30(a), while those for the horizontal polarization [Figure 17.29(c)] along the principal y-z plane are displayed in Figure 17.30(b). An excellent agreement is indicated between the mea-sured and simulated patterns, for both the full-scale (large) and scaled (small) plates. In fact, all the curves for the full-scale and for the scaled plates, for both the vertical and horizontal polarizations, are so close to each other that they are indistinguishable from one another. The horizontal (soft) polarization patterns of Figure 14.30(b) follow, as they should, a nearly sin(q)/q distribution based on physical optics and due to the very weak first-order diffractions from the edges of the plate for this polarization . For the vertical polarization patterns of Figure 17.30(a) the agreement between measurements and simulations is also excellent. However for the vertical polarization the RCS patterns do not follow the sin(q)/q distribution, especially at the far minor lobes, because the first-order diffractions for this polarization (vertical; hard) are more intense and impact the overall distribution to be different from sin(q)/q.
The maximum monostatic RCS of a flat plate of any geometry, for either vertical or horizon-tal polarization (both are identical based on physical optics), occurs at normal incidence and it is represented by RCSmax = 4𝜋 (Area of plate λ )2 (17-28) Based on (17-28), the maximum monostatic RCS of the full-scale (large) square plate of 30 cm × 30 cm at 5 GHz is: RCS (full-scale)max = 4𝜋 [30(30) 6(100) ]2 = 28.274 (m2) = 14.51 dBsm (17-28a) while the simulated maximum of Figures 17.30(a) and 17.30(b) is 14.6 dBsm [basically the same as that of (17.28a)]. For the scaled (small) square plate, of 10 cm × 10 cm at 15 GHz, the maximum 1022 ANTENNA MEASUREMENTS (a) Geometry of PEC square plate (b) Vertical polarization (c) Horizontal polarization Ei, Hi Es, Hs z y x b/2 b/2 a/2 a/2 θi θs ϕs ϕi Es, Hs Ei Hi z y x b/2 b/2 θs θi Es, Hs Ei Hi z y x b/2 b/2 θs θi Figure 17.29 Flat PEC plate geometry and polarization designations for RCS (echo area) measurements and simulations .
monostatic RCS based on (17-28) is: RCS (scaled)max = 4𝜋 [10(10) 2(100) ]2 = 3.142 (m2) = 4.97 dBsm (17-28b) while the simulated maximum of Figures 17.30(a) and 17.30(b) is 5.10 dBsm [again basically the same as that of (17.28b)].
SCALE MODEL MEASUREMENTS 1023 20 10 0 –10 –20 –30 Monostatic RCS (dBsm) –40 90° 60° 30° 0° 30° 60° 90° Measurements (full-scale, scaled) Simulations (full-scale, scaled) Simulations (scaled + 9.54 dB) 20 10 0 –10 –20 –30 Monostatic RCS (dBsm) –40 90° 60° 30° 0° 30° 60° 90° Measurements (full-scale, scaled) Simulations (full-scale, scaled) Simulations (scaled + 9.54 dB) (a) Vertical polarization Incidence angle i (degrees) θ Incidence angle i (degrees) θ (b) Horizontal polarization E-Field E-Field Figure 17.30 Vertical and horizontal polarizations measured and simulated monostatic RCS (echo area) pat-terns for full-scale (large; 30 cm × 30 cm) and scaled (small; 10 cm × 10 cm) square flat PEC plates of Figure 17.29.
Thus, according to (17-28a) and (17-28b), the RCS difference between the two PEC flat square plates, full-scale (large) and scale (small), is 28.274∕3.142 = 9 (dimensionless) or 14.5 −4.97 = 9.54 dB. This confirms the scaling of the echo area (RCS) of Table 7.1 which indicates that the scaling factor between the full-scale and scaled models is n2, which for this example is n2 = 32 = 9 (dimensionless), or 10 log10 (9) = 9.54 dB.
1024 ANTENNA MEASUREMENTS It is also apparent from the vertical polarization monostatic RCS patterns in Figure 17.30(a) and the horizontal polarization of 17.30(b) that there is a difference of n2 = 32 = 9 (dimensionless) or 10 log10 (9) = 9.54 dB, between the scaled and full-scale (both measured and simulated) RCS pat-terns; i.e. the full-scale measured and simulated monostatic RCS patterns are 9.54 dB greater than those of the scaled, as they should be according the scaling listed on Table 7.1. In fact, if 9.54 dB is added to the measured and simulated monostatic RCS patterns of the scaled (small) plate mono-static RCS patterns, the adjusted (by +9.54 dB) patterns match those of the full-scale (large) plate, as shown in Figures 17.30(a) and 17.30(b). Again, the agreement is so good that it is difficult to distinguish any differences between any of the patterns for the full-scale and scaled plates. Such comparisons and agreements illustrate and validate the scaling principle for Echo Area (RCS).
It should be pointed out that in both the gain (amplitude) and RCS (echo area), the conductivity was not scaled, but should be by a factor of n [the conductivity of the scale (small) models should be greater by a factor of n compared to that of the full-scale (large) models] to properly account for the scaling according to the listing in Table 17.1. For the measurements, of both the helicopter and radar target, the conductivity was finite but very large (about 5 × 107 S/m) while for the simulations the conductivity of the airframe and plate was assumed infinity. Because the conductivity of the physical targets was very large (on the order of 107 S/m), it did not impact, within measuring accuracy, the amplitude (gain) and RCS (echo area) scaling illustrations and validations.
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18. H. F. Schluper, “Verification Method for the Serration Design of CATR Reflectors,” AMTA Proceedings, pp. 10-9–10-14, Monterey, CA, October 1989.
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20. W. D. Burnside, M. C. Gilreath, B. M. Kent, and G. L. Clerici, “Curved Edge Modification of Compact Range Reflector,” IEEE Trans. Antennas Propagat., Vol. AP-35, No. 2, pp. 176–182, February 1987.
21. M. R. Hurst and P. E. Reed, “Hybrid Compact Radar Range Reflector,” AMTA Proceedings, pp. 8-9–8-13, Monterey, CA, October 1989.
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26. H. F. Schluper, J. Van Damme, and V. J. Vokurka, “Optimized Collimators—Theoretical Performance Limits,” AMTA Proceedings, p. 313, Seattle, WA, October 1987.
27. J. D. Huff, J. H. Cook, Jr., and B. W. Smith, “Recent Developments in Large Compact Range Design,” AMTA Proceedings, pp. 5-39–5-44, Columbus, OH, October 1992.
28. H. F. Schluper and V. J. Vokurka, “Troubleshooting Limitations in Indoor RCS Measurements,” Microwaves & RF, pp. 154–163, May 1987.
29. V. J. Vokurka, “Seeing Double Improves Indoor Range,” Microwaves & RF, pp. 71–76, 94, February 1985.
30. T. Harrison, “A New Approach to Radar Cross-Section Compact Range,” Microwave Journal, pp. 137–145, June 1986.
31. K. W. Lam and V. J. Vokurka, “Hybrid Near-Field/Far-Field Antenna Measurement Techniques,” AMTA Proceedings, pp. 9-29–9-34, Boulder, CO, October 1991.
32. C. R. Birtcher, C. A. Balanis, and V. J. Vokurka, “RCS Measurements, Transformations, and Comparisons Under Cylindrical and Plane Wave Illumination,” IEEE Trans. Antennas Propagat., Vol. AP-42, No. 3, pp. 329–334, March 1994.
33. C. R. Birtcher, C. A. Balanis, and V. J. Vokurka, “Quiet Zone Scan of the Single-Plane Collimating Range,” AMTA Proceedings, pp. 4-37–4-42, Boulder, CO, October 1991.
34. A. D. Yaghjian, “An Overview of Near-Field Antenna Measurements,” IEEE Trans. Antennas Propagat., Vol. AP-34, pp. 30–45, January 1986.
35. P. M. Morse and H. Feshbach, Methods of Theoretical Physics, McGraw-Hill, New York, 1953, Chapter 13.
36. A. V. Oppenheim and R. W. Schaffer, Digital Signal Processing, Prentice-Hall, Englewood Cliffs, NJ, 1975, Chapter 3.
37. Y. Rahmat-Samii, V. Galindo, and R. Mittra, “A Plane-Polar Approach for Far-Field Construction from Near-Field Measurements,” IEEE Trans. Antennas Propagat., Vol. AP-28, No. 3, pp. 216–230, March 1980.
38. L. I. Williams and Y. Rahmat-Samii, “Novel Bi-Polar Planar Near-Field Measurement Scanner at UCLA,” 1991 Int. IEEE/AP-S Symp. Dig., London, Ontario, Canada, June 1991.
39. G. D. Bergland, “A Guided Tour of the Fast Fourier Transform,” IEEE Spectrum, pp. 41–52, July 1969.
40. W. M. Leach, Jr. and D. T. Paris, “Probe Compensated Near-Field Measurements on a Cylinder,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 4, pp. 435–445, July 1973.
41. C. F. Stubenrauch and A. C. Newell, “Some Recent Near-Field Antenna Measurements at NBS,” Microwave J., pp. 37–42, November 1980.
1026 ANTENNA MEASUREMENTS 42. J. Lemanczyk and F. H. Larsen, “Comparison of Near-Field Range Results,” IEEE Trans. Antennas Prop-agat., Vol. AP-36, No. 6, pp. 845–851, June 1988.
43. J. J. Lee, E. M. Ferren, D. P. Woollen, and K. M. Lee, “Near-Field Probe Used as a Diagnostic Tool to Locate Defective Elements in an Array Antenna,” IEEE Trans. Antennas Propagat., Vol. AP-36, No. 6, pp. 884–889, June 1988.
44. Y. Rahmat-Samii and J. Lemanczyk, “Application of Spherical Near-Field Measurements to Microwave Holographic Diagnosis of Antennas,” IEEE Trans. Antennas Propagat., Vol. AP-36, No. 6, pp. 869–878, June 1988.
45. A. C. Newell, R. C. Baird, and P. F. Wacker, “Accurate Measurement of Antenna Gain and Polarization at Reduced Distances by an Extrapolation Technique,” IEEE Trans. Antennas Propagat., Vol. AP-21, No. 4, pp. 418–431, July 1973.
46. L. H. Hemming and R. A. Heaton, “Antenna Gain Calibration on a Ground Reflection Range,” IEEE Trans.
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47. R. G. FitzGerrell, “Gain Measurements of Vertically Polarized Antennas over Imperfect Ground,” IEEE Trans. Antennas Propagat., Vol. AP-15, No. 2, pp. 211–216, March 1967.
48. R. G. FitzGerrell, “The Gain of a Horizontal Half-Wave Dipole over Ground,” IEEE Trans. Antennas Prop-agat., Vol. AP-15, No. 4, pp. 569–571, July 1967.
49. R. G. FitzGerrell, “Limitations on Vertically Polarized Ground-Based Antennas as Gain Standards,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 2, pp. 284–286, March 1975.
50. E. H. Newman, P. Bohley, and C. H. Walter, “Two Methods for the Measurement of Antenna Efficiency,” IEEE Trans. Antennas Propagat., Vol. AP-23, No. 4, pp. 457–461, July 1975.
51. M. Sucher and J. Fox, Handbook of Microwave Measurements, Vol. I, Polytechnic Press of the Polytechnic Institute of Brooklyn, New York, 1963.
52. C. G. Montgomery, Techniques of Microwave Measurements, Vol. II, MIT Radiation Laboratory Series, Vol. 11, McGraw-Hill, New York, 1947, Chapter 8.
53. ANSI/IEEE Std 148–1959 (Reaff 1971).
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55. J. D. Kraus, Antennas, McGraw-Hill, New York, 1988.
56. G. Barzilai, “Experimental Determination of the Distribution of Current and Charge Along Cylindrical Antennas,” Proc. IRE (Waves and Electrons Section), pp. 825–829, July 1949.
57. T. Morita, “Current Distributions on Transmitting and Receiving Antennas,” Proc. IRE, pp. 898–904, August 1950.
58. A. F. Rashid, “Quasi-Near-Zone Field of a Monopole Antenna and the Current Distribution of an Antenna on a Finite Conductive Earth,” IEEE Trans. Antennas Propagat., Vol. AP-18, No. 1, pp. 22–28, January 1970.
59. H. G. Booker, V. H. Rumsey, G. A. Deschamps, M. I. Kales, and J. I. Bonhert, “Techniques for Handling Elliptically Polarized Waves with Special Reference to Antennas,” Proc. IRE, Vol. 39, pp. 533–552, May 1951.
60. E. S. Gillespie, “Measurement of Antenna Radiation Characteristics on Far-Field Ranges,” Chapter 32 in Antenna Handbook (Y. T. Lo and S. W. Lee, eds.), pp. 32-1–32-91, Van Nostrand Reinhold Co., Inc., New York, 1988.
61. G. Sinclair, “Theory of Models of Electromagnetic Systems,” Proc. IRE, Vol. 36, pp. 1364–1370, Novem-ber 1948.
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APPENDIXI f(x) = sin(x) x Figure I.1 Plot of sin(x)∕x function.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
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Companion Website: www.wiley.com/go/antennatheory4e 1027 APPENDIXII fN(x) = | | | | | sin(Nx) N sin(x) | | | | | N = 1, 3, 5, 10, 20 1.0 0.5 0 1 2 3 4 5 6 N = 1 N = 20 N = 10 N = 5 N = 3 Figure II.1 Curves of | sin(Nx)∕N sin(x)| function.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1029 APPENDIXIII Cosine and Sine Integrals Si(x) = ∫ x 0 sin(𝜏) 𝜏 d𝜏 (III-1) Ci(x) = −∫ ∞ x cos(𝜏) 𝜏 d𝜏= ∫ x ∞ cos(𝜏) 𝜏 d𝜏 (III-2) Cin(x) = ∫ x 0 1 −cos(𝜏) 𝜏 d𝜏 (III-3) Cin(x) = ln(𝛾x) −Ci(x) = ln(𝛾) + ln(x) −Ci(x) Cin(x) = ln(1.781) + ln(x) −Ci(x) = 0.577215665 + ln(x) −Ci(x) (III-4) Also Si(x) = ∞ ∑ k=0 (−1)kx2k+1 (2k + 1)(2k + 1)!
Ci(x) = C + ln(x) + ∞ ∑ k=1 (−1)k x2k 2k(2k)!
Cin(x) = ∞ ∑ k=1 (−1)k+1 x2k 2k(2k)!
Figure III.1 Plots of sine and cosine integrals.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
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Companion Website: www.wiley.com/go/antennatheory4e 1031 APPENDIXIV Fresnel Integrals C0(x) = ∫ x 0 cos(𝜏) √ 2𝜋𝜏 d𝜏 (IV-1) S0(x) = ∫ x 0 sin(𝜏) √ 2𝜋𝜏 d𝜏 (IV-2) C(x) = ∫ x 0 cos (𝜋 2 𝜏2) d𝜏 (IV-3) S(x) = ∫ x 0 sin (𝜋 2 𝜏2) d𝜏 (IV-4) C1(x) = ∫ ∞ x cos(𝜏2) d𝜏 (IV-5) S1(x) = ∫ ∞ x sin(𝜏2) d𝜏 (IV-6) C(x) −jS(x) = ∫ x 0 e−j(𝜋∕2)𝜏2 d𝜏= ∫ (𝜋∕2)x2 0 e−j𝜏 √ 2𝜋𝜏 d𝜏 C(x) −jS(x) = C0 (𝜋 2 x2) −jS0 (𝜋 2 x2) (IV-7) C1(x) −jS1(x) = ∫ ∞ x e−j𝜏2 d𝜏= √ 𝜋 2 ∫ ∞ x2 e−j𝜏 √ 2𝜋𝜏 d𝜏 C1(x) −jS1(x) = √ 𝜋 2 { ∫ ∞ 0 e−j𝜏 √ 2𝜋𝜏 d𝜏−∫ x2 0 e−j𝜏 √ 2𝜋𝜏 d𝜏 } C1(x) −jS1(x) = √ 𝜋 2 {[1 2 −j1 2 ] −[C0(x2) −jS0(x2)] } C1(x) −jS1(x) = √ 𝜋 2 {[1 2 −C0(x2) ] −j [1 2 −S0(x2) ]} (IV-8) Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
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Companion Website: www.wiley.com/go/antennatheory4e 1033 1034 APPENDIX IV 0.8 C(x) S(x) 0.6 0.4 0.2 0 1 2 3 4 5 x 6 7 8 9 10 Figure IV.1 Plots of C(x) and S(x) Fresnel integrals.
0.6 C1(x) S1(x) 0.4 0.2 0 –0.2 –0.4 0 1 2 3 4 5 x 6 7 8 9 10 Figure IV.2 Plots of C1(x) and S1(x) Fresnel integrals.
APPENDIXV Bessel Functions Bessel’s equation can be written as x2 d2y dx2 + xdy dx + (x2 −p2)y = 0 (V-1) Using the method of Frobenius, we can write its solutions as y(x) = A1Jp(x) + B1J−p(x), p ≠0 or integer (V-2) or y(x) = A2Jn(x) + B2Yn(x), p = n = 0 or integer (V-3) where Jp(x) = ∞ ∑ m=0 (−1)m(x∕2)2m+p m!(m + p)!
(V-4) J−p(x) = ∞ ∑ m=0 (−1)m(x∕2)2m−p m!(m −p)!
(V-5) Yp(x) = Jp(x) cos(p𝜋) −J−p(x) sin(p𝜋) (V-6) m! = Γ(m + 1) (V-7) Jp(x) is referred to as the Bessel function of the first kind of order p, Yp(x) as the Bessel function of the second kind of order p, and Γ(x) as the gamma function.
When p = n = integer, using (V-5) and (V-7) it can be shown that J−n(x) = (−1)nJn(x) (V-8) Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
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Companion Website: www.wiley.com/go/antennatheory4e 1035 1036 APPENDIX V and no longer are the two Bessel functions independent of each other. Therefore a second solution is required and it is given by (V-3). It can also be shown that Yn(x) = lim p→n Yp(x) = lim p→n Jp(x) cos(p𝜋) −J−p(x) sin(p𝜋) (V-9) When the argument of the Bessel function is negative and p = n, using (V-4) leads to Jn(−x) = (−1)nJn(x) (V-10) In many applications Bessel functions of small and large arguments are required. Using asymptotic methods, it can be shown that J0(x) ≃1 Y0(x) ≃2 𝜋ln (𝛾x 2 ) 𝛾= 1.781 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ x →0 (V-11) Jp(x) ≃1 p!
( x 2 )p Yp(x) ≃−(p −1)!
𝜋 (2 x )p ⎫ ⎪ ⎬ ⎪ ⎭ x →0 p > 0 (V-12) and Jp(x) ≃ √ 2 𝜋x cos ( x −𝜋 4 −p𝜋 2 ) Yp(x) ≃ √ 2 𝜋x sin ( x −𝜋 4 −p𝜋 2 ) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ x →∞ (V-13) For wave propagation it is often convenient to introduce Hankel functions defined as H(1) p (x) = Jp(x) + jYp(x) (V-14) H(2) p (x) = Jp(x) −jYp(x) (V-15) where H(1) p (x) is the Hankel function of the first kind of order p and H(2) p (x) is the Hankel function of the second kind of order p. For large arguments H(1) p (x) ≃ √ 2 𝜋xej[x−p(𝜋∕2)−𝜋∕4], x →∞ (V-16) H(2) p (x) ≃ √ 2 𝜋xe−j[x−p(𝜋∕2)−𝜋∕4], x →∞ (V-17) APPENDIX V 1037 A derivative can be taken using either d dx[Zp(𝛼x)] = 𝛼Zp−1(𝛼x) −p xZp(𝛼x) (V-18) or d dx[Zp(𝛼x)] = −𝛼Zp+1(𝛼x) + p xZp(𝛼x) (V-19) where Zp can be a Bessel function (Jp, Yp) or a Hankel function [H(1) p or H(2) p ].
A useful identity relating Bessel functions and their derivatives is given by Jp(x)Yp ′(x) −Yp(x)Jp ′(x) = 2 𝜋x (V-20) and it is referred to as the Wronskian. The prime (′) indicates a derivative. Also Jp(x)J′ −p(x) −J−p(x)Jp ′(x) = −2 𝜋x sin(p𝜋) (V-21) Some useful integrals of Bessel functions are ∫xp+1Jp(𝛼x) dx = 1 𝛼xp+1Jp+1(𝛼x) + C (V-22) ∫x1−pJp(𝛼x) dx = −1 𝛼x1−pJp−1(𝛼x) + C (V-23) ∫x3J0(x) dx = x3J1(x) −2x2J2(x) + C (V-24) ∫x6J1(x) dx = x6J2(x) −4x5J3(x) + 8x4J4(x) + C (V-25) ∫J3(x) dx = −J2(x) −2 xJ1(x) + C (V-26) ∫xJ1(x) dx = −xJ0(x) + ∫J0(x) dx + C (V-27) ∫x−1J1(x) dx = −J1(x) + ∫J0(x) dx + C (V-28) ∫J2(x) dx = −2J1(x) + ∫J0(x) dx + C (V-29) ∫xmJn(x) dx = xmJn+1(x) −(m −n −1) ∫xm−1Jn+1(x) dx (V-30) ∫xmJn(x) dx = −xmJn−1(x) + (m + n −1) ∫xm−1Jn−1(x) dx (V-31) J1(x) = 2 𝜋∫ 𝜋∕2 0 sin(x sin 𝜃) sin 𝜃d𝜃 (V-32) 1038 APPENDIX V 1 xJ1(x) = 2 𝜋∫ 𝜋∕2 0 cos(x sin 𝜃) cos2 𝜃d𝜃 (V-33) J2(x) = 2 𝜋∫ 𝜋∕2 0 cos(x sin 𝜃) cos 2𝜃d𝜃 (V-34) Jn(x) = j−n 2𝜋∫ 2𝜋 0 ejx cos 𝜙ejn𝜙d𝜙 (V-35) Jn(x) = j−n 𝜋∫ 𝜋 0 cos(n𝜙)ejx cos 𝜙d𝜙 (V-36) Jn(x) = 1 𝜋∫ 𝜋 0 cos(x sin 𝜙−n𝜙) d𝜙 (V-37) J2n(x) = 2 𝜋∫ 𝜋∕2 0 cos(x sin 𝜙) cos(2n 𝜙) d𝜙 (V-38) J2n(x) = (−1)n 2 𝜋∫ 𝜋∕2 0 cos(x cos 𝜙) cos(2n 𝜙) d𝜙 (V-39) The integrals ∫ x 0 J0(𝜏) d𝜏and ∫ x 0 Y0(𝜏) d𝜏 (V-40) often appear in solutions of problems but cannot be integrated in closed form. Graphs and tables for each, obtained using numerical techniques, are included.
1.0 0.5 –0.5 0 5 10 15 20 25 x 30 35 40 45 50 0 J0(x) J1(x) J2(x) J3(x) Figure V.1 Bessel functions of the first kind [J0(x), J1(x), J2(x), J3(x)].
APPENDIX V 1039 1.0 Y0(x) Y1(x) Y2(x) Y3(x) 0.5 –0.5 –1.0 –1.5 0 5 10 15 20 25 x 30 35 40 45 50 0 Figure V.2 Bessel functions of the second kind [Y0(x), Y1(x), Y2(x), Y3(x)].
Figure V.3 Plot of J1(x)∕x function.
Figure V.4 Plots of ∫ x 0 J0(𝜏)d𝜏and ∫ x 0 Y0(𝜏)d𝜏functions.
APPENDIXVI Identities VI.1 TRIGONOMETRIC 1. Sum or difference: a. sin(x + y) = sin x cos y + cos x sin y b. sin(x −y) = sin x cos y −cos x sin y c. cos(x + y) = cos x cos y −sin x sin y d. cos(x −y) = cos x cos y + sin x sin y e. tan(x + y) = tan x + tan y 1 −tan x tan y f. tan(x −y) = tan x −tan y 1 + tan x tan y g. sin2 x + cos2 x = 1 h. tan2 x −sec2 x = −1 i. cot2 x −csc2x = −1 2. Sum or difference into products: a. sin x + sin y = 2 sin 1 2(x + y) cos 1 2(x −y) b. sin x −sin y = 2 cos 1 2(x + y) sin 1 2(x −y) c. cos x + cos y = 2 cos 1 2(x + y) cos 1 2(x −y) d. cos x −cos y = −2 sin 1 2(x + y) sin 1 2(x −y) 3. Products into sum or difference: a. 2 sin x cos y = sin(x + y) + sin(x −y) b. 2 cos x sin y = sin(x + y) −sin(x −y) c. 2 cos x cos y = cos(x + y) + cos(x −y) d. 2 sin x sin y = −cos(x + y) + cos(x −y) 4. Double and half angles: a. sin 2x = 2 sin x cos x b. cos 2x = cos2 x −sin2 x = 2 cos2 x −1 = 1 −2 sin2 x c. tan 2x = 2 tan x 1 −tan2 x Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1041 1042 APPENDIX VI d. sin 1 2x = ± √ 1 −cos x 2 or 2 sin2 𝜃= 1 −cos 2𝜃 e. cos 1 2x = ± √ 1 + cos x 2 or 2 cos2 𝜃= 1 + cos 2𝜃 f. tan 1 2x = ± √ 1 −cos x 1 + cos x = sin x 1 + cos x = 1 −cos x sin x 5. Series: a. sin x = ejx −e−jx 2j = x −x3 3! + x5 5! −x7 7! + ⋯ b. cos x = ejx + e−jx 2 = 1 −x2 2! + x4 4! −x6 6! + ⋯ c. tan x = ejx −e−jx j(ejx + e−jx) = x + x3 3 + 2x5 15 + 17x7 315 + ⋯ VI.2 HYPERBOLIC 1. Definitions: a. Hyperbolic sine: sinh x = 1 2(ex −e−x) b. Hyperbolic cosine: cosh x = 1 2(ex + e−x) c. Hyperbolic tangent: tanh x = sinh x cosh x d. Hyperbolic cotangent: coth x = 1 tanh x = cosh x sinh x e. Hyperbolic secant: sech x = 1 cosh x f. Hyperbolic cosecant: csch x = 1 sinh x 2. Sum or difference: a. cosh(x + y) = cosh x cosh y + sinh x sinh y b. sinh(x −y) = sinh x cosh y −cosh x sinh y c. cosh(x −y) = cosh x cosh y −sinh x sinh y d. tanh(x + y) = tanh x + tanh y 1 + tanh x tanh y e. tanh(x −y) = tanh x −tanh y 1 −tanh x tanh y f. cosh2 x −sinh2 x = 1 g. tanh2 x + sech2 x = 1 h. coth2 x −csch2 x = 1 i. cosh(x ± jy) = cosh x cos y ± j sinh x sin y j. sinh(x ± jy) = sinh x cos y ± j cosh x sin y 3. Series: a. sinh x = ex −e−x 2 = x + x3 3! + x5 5! + x7 7! + ⋯ b. cosh x = ex + e−x 2 = 1 + x2 2! + x4 4! + x6 6! + ⋯ c. ex = 1 + x + x2 2! + x3 3! + x4 4! + ⋯ APPENDIX VI 1043 VI.3 LOGARITHMIC 1. logb(MN) = logb M + logb N 2. logb(M∕N) = logb M −logb N 3. logb(1∕N) = −logb N 4. logb(Mn) = n logb M 5. logb(M1∕n) = 1 n logb M 6. loga N = logb N ⋅loga b = logb N∕logb a 7. loge N = log10 N ⋅loge 10 = 2.302585 log10 N 8. log10 N = loge N ⋅log10 e = 0.434294 loge N APPENDIXVII Vector Analysis VII.1 VECTOR TRANSFORMATIONS In this appendix we present the vector transformations from rectangular to cylindrical (and vice versa), from cylindrical to spherical (and vice versa), and from rectangular to spherical (and vice versa). The three coordinate systems are shown in Figure VII.1.
VII.1.1 Rectangular to Cylindrical (and Vice Versa) The coordinate transformation from rectangular (x, y, z) to cylindrical (𝜌, 𝜙, z) is given, referring to Figure VII.1(b) x = 𝜌cos 𝜙 y = 𝜌sin 𝜙 z = z } (VII-1) In the rectangular coordinate system, we express a vector A as A = ̂ axAx + ̂ ayAy + ̂ azAz (VII-2) where ̂ ax, ̂ ay, ̂ az are the unit vectors and Ax, Ay, Az are the components of the vector A in the rectan-gular coordinate system. We wish to write A as A = ̂ a𝜌A𝜌+ ̂ a𝜙A𝜙+ ̂ azAz (VII-3) where ̂ a𝜌, ̂ a𝜙, ̂ az are the unit vectors and A𝜌, A𝜙, Az are the vector components in the cylindrical coordinate system. The z-axis is common to both of them.
Referring to Figure VII.2, we can write ̂ ax = ̂ a𝜌cos 𝜙−̂ a𝜙sin 𝜙 ̂ ay = ̂ a𝜌sin 𝜙+ ̂ a𝜙cos 𝜙 ̂ az = ̂ az } (VII-4) Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1045 1046 APPENDIX VII Figure VII.1 Rectangular, cylindrical, and spherical coordinate systems.
Using (VII-4) reduces (VII-2) to A = (̂ a𝜌cos 𝜙−̂ a𝜙sin 𝜙)Ax + (̂ a𝜌sin 𝜙+ ̂ a𝜙cos 𝜙)Ay + ̂ azAz A = ̂ a𝜌(Ax cos 𝜙+ Ay sin 𝜙) + ̂ a𝜙(−Ax sin 𝜙+ Ay cos 𝜙) + ̂ azAz (VII-5) which when compared with (VII-3) leads to A𝜌= Ax cos 𝜙+ Ay sin 𝜙 A𝜙= −Ax sin 𝜙+ Ay cos 𝜙 Az = Az } (VII-6) APPENDIX VII 1047 Figure VII.2 Geometrical representation of transformations between unit vectors of rectangular and cylin-drical coordinate systems.
In matrix form, (VII-6) can be written as ( A𝜌 A𝜙 Az ) = ( cos 𝜙 sin 𝜙 0 −sin 𝜙 cos 𝜙 0 0 0 1 ) ( Ax Ay Az ) (VII-6a) where [A]rc = [ cos 𝜙 sin 𝜙 0 −sin 𝜙 cos 𝜙 0 0 0 1 ] (VII-6b) is the transformation matrix for rectangular-to-cylindrical components.
1048 APPENDIX VII Since [A]rc is an orthonormal matrix (its inverse is equal to its transpose), we can write the trans-formation matrix for cylindrical-to-rectangular components as [A]cr = [A]−1 rc = [A]t rc = [ cos 𝜙 −sin 𝜙 0 sin 𝜙 cos 𝜙 0 0 0 1 ] (VII-7) or ( Ax Ay Az ) = ( cos 𝜙 −sin 𝜙 0 sin 𝜙 cos 𝜙 0 0 0 1 ) ( A𝜌 A𝜙 Az ) (VII-7a) or Ax = A𝜌cos 𝜙−A𝜙sin 𝜙 Ay = A𝜌sin 𝜙+ A𝜙cos 𝜙 Az = Az } (VII-7b) VII.1.2 Cylindrical to Spherical (and Vice Versa) Referring to Figure VII.1(c), we can write that the cylindrical and spherical coordinates are related by 𝜌= r sin 𝜃 z = r cos 𝜃 } (VII-8) In a geometrical approach similar to the one employed in the previous section, we can show that the cylindrical-to-spherical transformation of vector components is given by Ar = A𝜌sin 𝜃+ Az cos 𝜃 A𝜃= A𝜌cos 𝜃−Az sin 𝜃 A𝜙= A𝜙 } (VII-9) or in matrix form by ( Ar A𝜃 A𝜙 ) = ( sin 𝜃 0 cos 𝜃 cos 𝜃 0 −sin 𝜃 0 1 0 ) ( A𝜌 A𝜙 Az ) (VII-9a) Thus the cylindrical-to-spherical transformation matrix can be written as [A]cs = [ sin 𝜃 0 cos 𝜃 cos 𝜃 0 −sin 𝜃 0 1 0 ] (VII-9b) The [A]cs matrix is also orthonormal so that its inverse is given by [A]sc = [A]−1 cs = [A]t cs = [ sin 𝜃 cos 𝜃 0 0 0 1 cos 𝜃 −sin 𝜃 0 ] (VII-10) APPENDIX VII 1049 and the spherical-to-cylindrical transformation is accomplished by ( A𝜌 A𝜙 Az ) = ( sin 𝜃 cos 𝜃 0 0 0 1 cos 𝜃 −sin 𝜃 0 ) ( Ar A𝜃 A𝜙 ) (VII-10a) or A𝜌= Ar sin 𝜃+ A𝜃cos 𝜃 A𝜙= A𝜙 Az = Ar cos 𝜃−A𝜃sin 𝜃 } (VII-10b) This time the component A𝜙and coordinate 𝜙are the same in both systems.
VII.1.3 Rectangular to Spherical (and Vice Versa) Many times it may be required that a transformation be performed directly from rectangular-to-spherical components. By referring to Figure VII.1, we can write that the rectangular and spherical coordinates are related by x = r sin 𝜃cos 𝜙 y = r sin 𝜃sin 𝜙 z = r cos 𝜃 } (VII-11) and the rectangular and spherical components by Ar = Ax sin 𝜃cos 𝜙+ Ay sin 𝜃sin 𝜙+ Az cos 𝜃 A𝜃= Ax cos 𝜃cos 𝜙+ Ay cos 𝜃sin 𝜙−Az sin 𝜃 A𝜙= −Ax sin 𝜙+ Ay cos 𝜙 } (VII-12) which can also be obtained by substituting (VII-6) into (VII-9). In matrix form, (VII-12) can be written as ( Ar A𝜃 A𝜙 ) = ( sin 𝜃cos 𝜙 sin 𝜃sin 𝜙 cos 𝜃 cos 𝜃cos 𝜙 cos 𝜃sin 𝜙 −sin 𝜃 −sin 𝜙 cos 𝜙 0 ) ( Ax Ay Az ) (VII-12a) with the rectangular-to-spherical transformation matrix being [A]rs = ( sin 𝜃cos 𝜙 sin 𝜃sin 𝜙 cos 𝜃 cos 𝜃cos 𝜙 cos 𝜃sin 𝜙 −sin 𝜃 −sin 𝜙 cos 𝜙 0 ) (VII-12b) The transformation matrix of (VII-12b) is also orthonormal so that its inverse can be written as [A]sr = [A]−1 rs = [A]t rs = ( sin 𝜃cos 𝜙 cos 𝜃cos 𝜙 −sin 𝜙 sin 𝜃sin 𝜙 cos 𝜃sin 𝜙 cos 𝜙 cos 𝜃 −sin 𝜃 0 ) (VII-13) 1050 APPENDIX VII and the spherical-to-rectangular components related by ( Ax Ay Az ) = ( sin 𝜃cos 𝜙 cos 𝜃cos 𝜙 −sin 𝜙 sin 𝜃sin 𝜙 cos 𝜃sin 𝜙 cos 𝜙 cos 𝜃 −sin 𝜃 0 ) ( Ar A𝜃 A𝜙 ) (VII-13a) or Ax = Ar sin 𝜃cos 𝜙+ A𝜃cos 𝜃cos 𝜙−A𝜙sin 𝜙 Ay = Ar sin 𝜃sin 𝜙+ A𝜃cos 𝜃sin 𝜙+ A𝜙cos 𝜙 Az = Ar cos 𝜃−A𝜃sin 𝜃 } (VII-13b) VII.2 VECTOR DIFFERENTIAL OPERATORS The differential operators of gradient of a scalar (∇𝜓), divergence of a vector (∇⋅A), curl of a vec-tor (∇× A), Laplacian of a scalar (∇2𝜓), and Laplacian of a vector (∇2A) frequently encountered in electromagnetic field analysis will be listed in the rectangular, cylindrical, and spherical coordi-nate systems.
VII.2.1 Rectangular Coordinates ∇𝜓= ̂ ax 𝜕𝜓 𝜕x + ̂ ay 𝜕𝜓 𝜕y + ̂ az 𝜕𝜓 𝜕z (VII-14) ∇⋅A = 𝜕Ax 𝜕x + 𝜕Ay 𝜕y + 𝜕Az 𝜕z (VII-15) ∇× A = ̂ ax (𝜕Az 𝜕y − 𝜕Ay 𝜕z ) + ̂ ay (𝜕Ax 𝜕z −𝜕Az 𝜕x ) + ̂ az (𝜕Ay 𝜕x −𝜕Ax 𝜕y ) (VII-16) ∇⋅∇𝜓= ∇2𝜓= 𝜕2𝜓 𝜕x2 + 𝜕2𝜓 𝜕y2 + 𝜕2𝜓 𝜕z2 (VII-17) ∇2A = ̂ ax∇2Ax + ̂ ay∇2Ay + ̂ az∇2Az (VII-18) VII.2.2 Cylindrical Coordinates ∇𝜓= ̂ a𝜌 𝜕𝜓 𝜕𝜌+ ̂ a𝜙 1 𝜌 𝜕𝜓 𝜕𝜙+ ̂ az 𝜕𝜓 𝜕z (VII-19) ∇⋅A = 1 𝜌 𝜕 𝜕𝜌(𝜌A𝜌) + 1 𝜌 𝜕A𝜙 𝜕𝜙+ 𝜕Az 𝜕z (VII-20) ∇× A = ̂ a𝜌 ( 1 𝜌 𝜕Az 𝜕𝜙− 𝜕A𝜙 𝜕z ) + ̂ a𝜙 (𝜕A𝜌 𝜕z −𝜕Az 𝜕𝜌 ) + ̂ az ( 1 𝜌 𝜕(𝜌A𝜙) 𝜕𝜌 −1 𝜌 𝜕A𝜌 𝜕𝜙 ) (VII-21) ∇2𝜓= 1 𝜌 𝜕 𝜕𝜌 ( 𝜌𝜕𝜓 𝜕𝜌 ) + 1 𝜌2 𝜕2𝜓 𝜕𝜙2 + 𝜕2𝜓 𝜕z2 (VII-22) ∇2A = ∇(∇⋅A) −∇× ∇× A (VII-23) APPENDIX VII 1051 or in an expanded form ∇2A = ̂ a𝜌 ( 𝜕2A𝜌 𝜕𝜌2 + 1 𝜌 𝜕A𝜌 𝜕𝜌− A𝜌 𝜌2 + 1 𝜌2 𝜕2A𝜌 𝜕𝜙2 −2 𝜌2 𝜕A𝜙 𝜕𝜙+ 𝜕2A𝜌 𝜕z2 ) + ̂ a𝜙 ( 𝜕2A𝜙 𝜕𝜌2 + 1 𝜌 𝜕A𝜙 𝜕𝜌− A𝜙 𝜌2 + 1 𝜌2 𝜕2A𝜙 𝜕𝜙2 + 2 𝜌2 𝜕A𝜌 𝜕𝜙+ 𝜕2A𝜙 𝜕z2 ) + ̂ az ( 𝜕2Az 𝜕𝜌2 + 1 𝜌 𝜕Az 𝜕𝜌+ 1 𝜌2 𝜕2Az 𝜕𝜙2 + 𝜕2Az 𝜕z2 ) (VII-23a) In the cylindrical coordinate system ∇2A ≠̂ a𝜌∇2A𝜌+ ̂ a𝜙∇2A𝜙+ ̂ az∇2Az because the orientation of the unit vectors ̂ a𝜌and ̂ a𝜙varies with the 𝜌and 𝜙coordinates.
VII.2.3 Spherical Coordinates ∇𝜓= ̂ ar 𝜕𝜓 𝜕r + ̂ a𝜃 1 r 𝜕𝜓 𝜕𝜃+ ̂ a𝜙 1 r sin 𝜃 𝜕𝜓 𝜕𝜙 (VII-24) ∇⋅A = 1 r2 𝜕 𝜕r(r2Ar) + 1 r sin 𝜃 𝜕 𝜕𝜃(A𝜃sin 𝜃) + 1 r sin 𝜃 𝜕A𝜙 𝜕𝜙 (VII-25) ∇× A = ̂ ar r sin 𝜃 [ 𝜕 𝜕𝜃(A𝜙sin 𝜃) −𝜕A𝜃 𝜕𝜙 ] + ̂ a𝜃 r [ 1 sin 𝜃 𝜕Ar 𝜕𝜙−𝜕 𝜕r(rA𝜙) ] + ̂ a𝜙 r [ 𝜕 𝜕r(rA𝜃) −𝜕Ar 𝜕𝜃 ] (VII-26) ∇2𝜓= 1 r2 𝜕 𝜕r ( r2 𝜕𝜓 𝜕r ) + 1 r2 sin 𝜃 𝜕 𝜕𝜃 ( sin 𝜃𝜕𝜓 𝜕𝜃 ) + 1 r2 sin2 𝜃 𝜕2𝜓 𝜕𝜙2 (VII-27) ∇2A = ∇(∇⋅A) −∇× ∇× A (VII-28) or in an expanded form ∇2A = ̂ ar (𝜕2Ar 𝜕r2 + 2 r 𝜕Ar 𝜕r −2 r2 Ar + 1 r2 𝜕2Ar 𝜕𝜃2 + cot 𝜃 r2 𝜕Ar 𝜕𝜃+ 1 r2 sin2 𝜃 𝜕2Ar 𝜕𝜙2 −2 r2 𝜕A𝜃 𝜕𝜃−2 cot 𝜃 r2 A𝜃− 2 r2 sin 𝜃 𝜕A𝜙 𝜕𝜙 ) + ̂ a𝜃 (𝜕2A𝜃 𝜕r2 + 2 r 𝜕A𝜃 𝜕r − A𝜃 r2 sin2 𝜃 + 1 r2 𝜕2A𝜃 𝜕𝜃2 + cot 𝜃 r2 𝜕A𝜃 𝜕𝜃 + 1 r2 sin2 𝜃 𝜕2A𝜃 𝜕𝜙2 + 2 r2 𝜕Ar 𝜕𝜃−2 cot 𝜃 r2 sin 𝜃 𝜕A𝜙 𝜕𝜙 ) + ̂ a𝜙 ( 𝜕2A𝜙 𝜕r2 + 2 r 𝜕A𝜙 𝜕r − 1 r2 sin2 𝜃 A𝜙+ 1 r2 𝜕2A𝜙 𝜕𝜃2 + cot 𝜃 r2 𝜕A𝜙 𝜕𝜃 + 1 r2 sin2 𝜃 𝜕2A𝜙 𝜕𝜙2 + 2 r2 sin 𝜃 𝜕Ar 𝜕𝜙+ 2 cot 𝜃 r2 sin 𝜃 𝜕A𝜃 𝜕𝜙 ) (VII-28a) 1052 APPENDIX VII Again note that ∇2A ≠̂ ar∇2Ar + ̂ a𝜃∇2A𝜃+ ̂ a𝜙∇2A𝜙since the orientation of the unit vectors ̂ ar, ̂ a𝜃, and ̂ a𝜙varies with the r, 𝜃, and 𝜙coordinates.
VII.3 VECTOR IDENTITIES VII.3.1 Addition and Multiplication A ⋅A = |A|2 (VII-29) A ⋅A∗= |A|2 (VII-30) A + B = B + A (VII-31) A ⋅B = B ⋅A (VII-32) A × B = −B × A (VII-33) (A + B) ⋅C = A ⋅C + B ⋅C (VII-34) (A + B) × C = A × C + B × C (VII-35) A ⋅B × C = B ⋅C × A = C ⋅A × B (VII-36) A × (B × C) = (A ⋅C)B −(A ⋅B)C (VII-37) (A × B) ⋅(C × D) = A ⋅B × (C × D) = A ⋅(B ⋅DC −B ⋅CD) = (A ⋅C)(B ⋅D) −(A ⋅D)(B ⋅C) (VII-38) (A × B) × (C × D) = (A × B ⋅D)C −(A × B ⋅C)D (VII-39) VII.3.2 Differentiation ∇⋅(∇× A) = 0 (VII-40) ∇× ∇𝜓= 0 (VII-41) ∇(𝜙+ 𝜓) = ∇𝜙+ ∇𝜓 (VII-42) ∇(𝜙𝜓) = 𝜙∇𝜓+ 𝜓∇𝜙 (VII-43) ∇⋅(A + B) = ∇⋅A + ∇⋅B (VII-44) ∇× (A + B) = ∇× A + ∇× B (VII-45) ∇⋅(𝜓A) = A ⋅∇𝜓+ 𝜓∇⋅A (VII-46) ∇× (𝜓A) = ∇𝜓× A + 𝜓∇× A (VII-47) ∇(A ⋅B) = (A ⋅∇)B + (B ⋅∇)A + A × (∇× B) + B × (∇× A) (VII-48) ∇⋅(A × B) = B ⋅∇× A −A ⋅∇× B (VII-49) ∇× (A × B) = A∇⋅B −B∇⋅A + (B ⋅∇)A −(A ⋅∇)B (VII-50) ∇× ∇× A = ∇(∇⋅A) −∇2A (VII-51) APPENDIX VII 1053 VII.3.3 Integration ∮ C A ⋅dl = ∫∫ S (∇× A) ⋅ds Stoke’s theorem (VII-52) ∯ S A ⋅ds = ∫∫∫ V (∇⋅A)dv Divergence theorem (VII-53) ∯ S (̂ n × A)ds = ∫∫∫ V (∇× A)dv (VII-54) ∯ S 𝜓ds = ∫∫∫ V ∇𝜓dv (VII-55) ∮ C 𝜓dl = ∫∫ S ̂ n × ∇𝜓ds (VII-56) APPENDIXVIII Method of Stationary Phase In many problems, the following integral is often encountered and in most cases cannot be integrated exactly: I(k) = ∫ b a ∫ d c F(x, y)ejkf(x,y) dx dy (VIII-1) where k = real f (x, y) = real, independent of k, and nonsingular F(x, y) = may be complex, independent of k, and nonsingular Often, however, the above integral needs to be evaluated only for large values of k, but the task is still formidable. An approximate technique, known as the Method of Stationary Phase, exists that can be used to obtain an approximate asymptotic expression, for large values of k, for the above integral.
The method is justified by the asymptotic approximation of the single integral I′(k) = ∫ b a F(x)ejkf (x) dx (VIII-2) where k = real f (x, y) = real, independent of k, and nonsingular F(x, y) = may be complex, independent of k, and nonsingular which can be extended to include double integrals.
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1055 1056 APPENDIX VIII The asymptotic evaluation of (VIII-1) for large k is based on the following: f(x, y) is a well behaved function and its variation near the stationary points xs, ys determined by 𝜕f 𝜕x | | | | x=xs y=ys ≡f ′ x(xs, ys) = 0 (VIII-3a) 𝜕f 𝜕y | | | | x=xs y=ys ≡f ′ y(xs, ys) = 0 (VIII-3b) is slow varying. Outside these regions, the function f(x, y) varies faster such that the exponential factor exp[jkf(x, y)] of the integrand oscillates very rapidly between the values of +1 and −1, for large values of k. Assuming F(x, y) is everywhere a slow varying function, the contributions to the integral outside the stationary points tend to cancel each other. Thus the only contributors to the integral, in an approximate sense, are the stationary points and their neighborhoods. Thus, we can write (VIII-1) approximately as I(k) ≃∫ +∞ −∞∫ +∞ −∞ F(xs, ys)ejkf (x,y) dx dy = F(xs, ys) ∫ +∞ −∞∫ +∞ −∞ ejkf (x,y) dx dy (VIII-4) where the limits have been extended, for convenience, to infinity since the net contribution outside the stationary points and their near regions is negligible.
In the neighborhood of the stationary points, the function f (x, y) can be approximated by a trun-cated Taylor series f (x, y) ≃f(xs, ys) + 1 2(x −xs)2f ′′ xx(xs, ys) + 1 2(y −ys)2f ′′ yy(xs, ys) + (x −xs)(y −ys)f ′′ xy(xs, ys) (VIII-5) since f ′ x(xs, ys) = f ′ y(xs, ys) = 0 (VIII-6) by (VIII-3a) and (VIII-3b). For convenience, we have adopted the notation 𝜕2f 𝜕x2 | | | | | x=xs y=ys ≡f ′′ xx(xs, ys) (VIII-7a) 𝜕2f 𝜕y2 | | | | | x=xs y=ys ≡f ′′ yy(xs, ys) (VIII-7b) 𝜕2f 𝜕x𝜕y | | | | | x=xs y=ys ≡f ′′ xy(xs, ys) (VIII-7c) For brevity, we write (VIII-5) as f (x, y) ≃f(xs, ys) + A𝜉2 + B𝜂2 + C𝜉𝜂 (VIII-8) APPENDIX VIII 1057 where A = 1 2 f ′′ xx(xs, ys) (VIII-8a) B = 1 2 f ′′ yy(xs, ys) (VIII-8b) C = f ′′ xy(xs, ys) (VIII-8c) 𝜉= (x −xs) (VIII-8d) 𝜂= (y −ys) (VIII-8e) Using (VIII-8)–(VIII-8e) reduces (VIII-4) to I(k) ≃F(xs, ys)ejkf (xs,ys) ∫ +∞ −∞∫ +∞ −∞ e jk(A𝜉2+B𝜂2+C𝜉𝜂) d𝜉d𝜂 (VIII-9) We now write the quadratic factor of the exponential, by a proper rotation of the coordinate axes 𝜉, 𝜂to 𝜇, λ, in a diagonal form as A𝜉2 + B𝜂2 + C𝜉𝜂= A′𝜇2 + B′λ2 (VIII-10) related to A, B, and C by A′ = 1 2[(A + B) + √ (A + B)2 −(4AB −C2)] (VIII-10a) B′ = 1 2[(A + B) − √ (A + B)2 −(4AB −C2)] (VIII-10b) which are found by solving the secular determinant | | | | (A −𝜁) C∕2 C∕2 (B −𝜁) | | | | = 0 (VIII-11) with 𝜁1 = A′ and 𝜁2 = B′. Substituting (VIII-10) into (VIII-9) we can write I(k) ≃F(xs, ys)ejkf(xs,ys) ∫ +∞ −∞∫ +∞ −∞ e jk(A′𝜇2+B′λ2) d𝜇dλ I(k) ≃F(xs, ys)ejkf(xs,ys) ∫ +∞ −∞ e± jk|A′|𝜇2 d𝜇∫ +∞ −∞ e± jk|B′|λ2 dλ (VIII-12) where the signs in the exponents are determined by the signs of A′ and B′, which in turn depend upon A and B, as given in (VIII-10a) and (VIII-10b). The two integrals in (VIII-12) are of the same form and can be evaluated by examining the integral I′′(k) = ∫ +∞ −∞ e± jk|𝛼|t2 dt = 2 ∫ ∞ 0 e± jk|𝛼|t2 dt (VIII-13) where 𝛼can represent either A′ or B′ of (VIII-12). Making a change of variable by letting k|𝛼|t2 = 𝜋 2 𝜏2 (VIII-13a) dt = √ 𝜋 2k|𝛼| d𝜏 (VIII-13b) 1058 APPENDIX VIII we can rewrite (VIII-13) as I′′(k) = 2 √ 𝜋 2k|𝛼| ∫ ∞ 0 e± j 𝜋 2 𝜏2 d𝜏 (VIII-14) The integral is recognized as being the complex Fresnel integral whose value is ∫ ∞ 0 e± j 𝜋 2 𝜏2 d𝜏= 1 2(1 ± j) = 1 √ 2 e± j 𝜋 4 (VIII-15) which can be used to write (VIII-14) as I′′(k) = 2 √ 𝜋 2k|𝛼| ∫ ∞ 0 e± j 𝜋 2 𝜏2 d𝜏= √𝜋 k|𝛼|e± j 𝜋 4 (VIII-16) The result of (VIII-16) can be used to reduce (VIII-12) to I(k) ≃F(xs, ys)e jkf (xs,ys) 𝜋 k √ |A′||B′| e± j 𝜋 4 e± j 𝜋 4 (VIII-17) If A′ and B′ are both positive, then e± j 𝜋 4 e± j 𝜋 4 = e+j 𝜋 2 = +j If A′ and B′ are both negative, then e± j 𝜋 4 e± j 𝜋 4 = e−j 𝜋 2 = −j If A′ and B′ have different signs, then e± j 𝜋 4 e± j 𝜋 4 = 1 Thus, (VIII-17) can be cast into the form I(k) ≃F(xs, ys)e jkf (xs,ys) j𝜋𝛿 k √ |A′||B′| (VIII-18) where 𝛿= ⎧ ⎪ ⎨ ⎪ ⎩ +1 if A′ and B′ are both positive −1 if A′ and B′ are both negative −j if A′ and B′ have different signs (VIII-18a) Examining (VIII-10a) and (VIII-10b), it is clear that a. A′ and B′ are real (because A, B, and C are real) b. A′ + B′ = A + B (VIII-19) c. A′B′ = (4AB −C2)∕4 Using the results of (VIII-19), we reduce (VIII-18) to I(k) ≃F(xs, ys)e jkf (xs,ys) j2𝜋𝛿 k √ |4AB −C2| .
(VIII-20) To determine the signs of A′ and B′, let us refer to (VIII-19).
a. If 4AB > C2, then A and B have the same sign and A′B′ > 0. Thus, A′ and B′ have the same sign.
1. If A > 0 then B > 0 and A′ > 0, B′ > 0 2. If A < 0 then B < 0 and A′ < 0, B′ < 0 APPENDIX VIII 1059 b. If 4AB < C2, then A′B′ < 0, and A′ and B′ have different signs. Summarizing the results we can write that 1. If 4AB > C2 and A > 0, then A′ and B′ are both positive 2. If 4AB > C2 and A < 0, then A′ and B′ are both negative 3. If 4AB < C2, then A′ and B′ have different signs Using the preceding deductions, we can write the sign information of (VIII-18a) as 𝛿= ⎧ ⎪ ⎨ ⎪ ⎩ +1 if 4AB > C2 and A > 0 −1 if 4AB > C2 and A < 0 −j if 4AB < C2 (VIII-21) in the evaluation of the integral in I(k) ≃F(xs, ys)ejkf (xs,ys) ∫ +∞ −∞∫ +∞ −∞ ejk(A𝜉2+B𝜂2+C𝜉𝜂) d𝜉d𝜂 I(k) ≃F(xs, ys)ejkf (xs,ys) j2𝜋𝛿 k √ |4AB −C2| (VIII-22) APPENDIXIX Television, Radio, Telephone, and Radar Frequency Spectrums IX.1 TELEVISION IX.1.1 Very High Frequency (VHF) Channels Channel number 2 3 4 5 6 7 8 9 10 11 12 13 Frequency (MHz) 54 ↑60 ↑66 ↑72 76 ↑82 ↑88 174 ↑180 ↑186 ↑192 ↑198 ↑204 ↑210 ↑216 IX.1.2 Ultra High Frequency (UHF) Channels∗ Channel number 14 15 16 17 18 19 20 ...
82 83 Frequency (MHz) 470 ↑476 ↑482 ↑488 ↑494 ↑500 ↑506 ↑512 … 878 ↑884 ↑890 For both VHF and UHF channels, each channel has a 6-MHz bandwidth. For each channel, the carrier frequency for the video part is equal to the lower frequency of the bandwidth plus 1.25 MHz while the carrier frequency for the audio part is equal to the upper frequency of the bandwidth minus 0.25 MHz.
Examples: Channel 2 (VHF): f0(video) = 54 + 1.25 = 55.25 MHz f0(audio) = 60 −0.25 = 59.75 MHz Channel 14 (UHF): f0(video) = 470 + 1.25 = 471.25 MHz f0(audio) = 476 −0.25 = 475.75 MHz In top ten urban areas in the United States, land mobile is allowed in the first seven UHF TV channels (470–512 MHz).
Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1061 1062 APPENDIX IX IX.2 RADIO IX.2.1 Amplitude Modulation (AM) Radio Number of channels: 107 (each with 10-kHz separation) Frequency range: 535–1605 kHz IX.2.2 Frequency Modulation (FM) Radio Number of channels: 100 (each with 200-kHz separation) Frequency range: 88–108 MHz IX.3 AMATEUR BANDS Band Frequency (MHz) Band Frequency (MHz) 160-m 1.8–2.0 2-m 144.0–148.0 80-m 3.5–4.0 — 220–225 40-m 7.0–7.3 — 420–450 20-m 14.0–14.35 — 1215–1300 15-m 21.0–21.45 — 2300–2450 10-m 28.0–29.7 — 3300–3500 6-m 50.0–54.0 — 5650–5925 IX.4 CELLULAR TELEPHONE IX.4.1 Land Mobile Systems Uplink: MS to BS (mobile station to base station) Downlink: BS to MS (base station to mobile station) System Uplink (MHz) Downlink (MHz) Major Covered Areas CDMA IS-95 824–849 869–894 North America, Korea, China GSM 890–915 935–960 North America, Europe, China, Japan Extended-GSM 880–915 925–960 North America, Europe, China, Japan DCS 1800 1710–1785 1805–1880 Europe US PCS 1900 1850–1910 1930–1990 North America WCDMA 1920–1980 2110–2170 Everywhere CDMA2000 All existing CDMA system frequencies Everywhere Downlink is the channel from the base station to the mobile unit, which is also called forward-link.
Uplink is the channel from the mobile unit to the base station, which is also called reverse-link.
IX.4.2 Cordless Telephone United States of America: 46–49 MHz Digital European Cordless Telecommunications (DECT): 1.880–1.990 GHz APPENDIX IX 1063 IX.5 RADAR IEEE BAND DESIGNATIONS HF (High Frequency): 3–30 MHz VHF (Very High Frequency): 30–300 MHz UHF (Ultra High Frequency): 300–1,000 MHz L-band: 1–2 GHz S-band: 2–4 GHz C-band: 4–8 GHz X-band: 8–12 GHz Ku-band: 12–18 GHz K-band: 18–27 GHz Ka-band: 27–40 GHz Millimeter wave band: 40–300 GHz Index adaptive beamforming, 950–953, 951f, 952f, 969–971, 970f, 971f, 972f amplitude pattern shape, 32–33, 33f amplitude pattern shape, radiation pattern (antenna pattern), 32–33, 33f analysis methods, 20–21 antena, radiation mechanism dipole, 13, 14f E-plane sectoral horn, unbounded medium (te horn), 15 infinite line source, unbounded medium (tm open), 15 radiation problems, computer animated-visualization, 13–15 single wire, 7–10, 8f, 10f two-wires, 10–11, 11f, 12f antenna current distribution, thin wire antenna, 15–18, 16f, 17f, 18f defined, overview, 1 as transition device, 2f transmitting mode, transmission-line Thevenin equivalent, 1, 2f antenna beamforming adaptive beamforming, 950–953, 951f, 952f direction-of-arrival (DOA) algorithms, 947–950, 947f, 950f optimal beamforming techniques, least mean square (LMS) algorithm, 956–959, 957f, 958f, 959f optimal beamforming techniques, minimum mean square error (MMSE) criterion, 955–956 antenna efficiency, 60–61, 61f antenna elements, 19–20 antenna equivalent areas, 83–86 antenna, historical advancement, 18 analysis methods, 20–21 antenna elements, 19–20 future challenges, 21 antenna measurements current measurements, 1014 directivity measurements, 1010–1012 impedance measurements, 1012–1014 overview, 981–982, 982f polarization measurements, 1014–1019, 1015f, 1016f, 1017f, 1018f radiation efficiency, 1012 antenna measurements, antenna ranges, 982 compact ranges, 986–987, 986f compact ranges, CATR designs, 989–992, 991f, 992f compact ranges, CATR performance, 987–989, 988f, 989f, 990f free-space ranges, anechoic chambers, 984–986.
985f free-space ranges, elevated ranges, 983–984, 984f free-space ranges, slant ranges, 984, 984f near-field/far-field methods, measurements and computations, 997–1000, 999f near-field/far-field methods, modal-expansion method for planar systems, 996–997 reflection ranges, 983, 983f antenna measurements, gain measurements, 1003, 1005–1006 realized-gain measurements, extrapolation method, 1008 realized-gain measurements, gain-transfer (gain-comparison), measurements, 1009–1010 realized-gain measurements, ground-reflection range method, 1008–1009 realized-gain measurements, three-antenna method, 1007–1008 realized-gain measurements, two-antenna method, 1006, 1007f antenna measurements, near-field/far-field methods, 992–996, 993f, 994f, 995f Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis.
© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/antennatheory4e 1065 1066 INDEX antenna measurements, radiation patterns, 1000–1001, 1000f amplitude pattern, 1001f, 1003, 1004f, 1005f instrumentation, 1001–1003, 1001f, 1002f phase measurements, 1003, 1005f antenna measurements, scale model measurements, 1019, 1019t echo area (RCS) measurements, simulations and comparisons, 1021–1024, 1022f, 1023f gain (amplitude) measurements, simulations and comparisons, 1020–1021, 1020f antenna miniaturization, 619 folding, 624–626, 624f, 626f metamaterials, 626–627 monopole antenna, impedance loading, 620–622, 620f monopole antenna, materials loading, 622–624, 622f patch antennas, 626 antenna radar cross section (RCS), 92–96, 92t, 96f, 97f antenna radiation efficiency, 80 antenna synthesis, continuous sources continuous aperture sources, 417 continuous aperture sources, circular aperture, 418–419, 419t continuous aperture sources, rectangular aperture, 418 discretization of continuous sources, 387, 387f Fourier transform method, linear array, 395–398, 398f Fourier transform method, line-source, 392–394, 395f line-source, 386–387, 387f line-source phase distributions, 416–417, 417f overview, 385 Schelkunoff polynomial method, 387–391, 389f, 390f, 391f, 392f Taylor line-source (one-parameter), 408–414, 411f, 413f, 414f Taylor line-source (Tschebyscheff-error), 404–405 Taylor line-source (Tschebyscheff-error), design procedure, 406–407, 408f triangular, cosine, cosine-squared amplitude distributions, 415–416, 415t Woodward-Lawson method, 398 Woodward-Lawson method, linear array, 403–404 Woodward-Lawson method, line-source, 399–402, 402f antenna temperature, 96, 98–1, 99f antenna types aperture antennas, 3, 4f, 6f array antennas, 5, 6f lens antennas, 6–7, 8f microstrip antennas, 5, 5f, 6f reflector antennas, 6, 7f wire antennas, 3, 4f, 6f antenna vector effective length, equivalent areas antenna equivalent areas, 83–86 maximum directivity, maximum effective area, 86–88, 86f vector effective length, 81–83, 81f aperture antennas, 3, 4f, 6f Babinet’s principle, 680–684 , 680f, 681f, 682f, 683f directivity, 648 field equivalence principle, Huygen’s principle, 639–645, 640f, 641f, 645f ground plane edge effects, geometrical theory of diffraction, 702–703, 703f, 704f, 705–707, 705f, 706f radiation equations, 645–647 aperture antennas, circular apertures, 667–669, 668f beam efficiency, 675, 677f TE11-mode distribution on infinite ground plane, 671, 672–673t, 674, 674f, 676f uniform distribution on infinite ground plane, 669–671, 671f aperture antennas, design considerations, 676 circular aperture, 678–679, 679t rectangular aperture, 677 aperture antennas, Fourier transforms aperture admittance, 695–702, 698f, 700f asymptotic evaluation of radiated field, 689–694 dielectiric-covered apertures, 694–695, 696f Fourier transforms-spectral domain, 685 radiated fields, 685–689, 685f aperture antennas, rectangular apertures, 648–650, 649f beam efficiency, 666–667, 666f, 677f TE10-mode distribution on infinite ground plane, 663, 663f, 664f, 665f uniform distribution in space, 658–659t, 661–663, 662f uniform distribution on infinite ground plane, 650–652, 650f, 653f, 654–657, 654f, 658–659t, 660 array antennas, 5, 6f arrays, array design, 285. See also mutual coupling in arrays design considerations, 361–362 design procedure, 318–319 linear array design, 967–968, 968f planar array design, 968–969 arrays, circular array array factor, 363–365, 363f, 367366f arrays, feed networks microstrip and mobile communications antennas, 832–837, 835f, 836f, 837f, 838f INDEX 1067 arrays, N-element linear array: directivity, 312 broadside, end-fire arrays, 313–314, 318t Hansen-Woodyard end-fire array, 317–318 ordinary end-fire array, 315–317, 318t arrays, N-element linear array: three-dimensional characteristics N-elements along X- or Y-axis, 320–322, 320f, 321f N-elements along Z-axis, 319–320 arrays, N-element linear array: uniform amplitude/spacing, 293–297, 294f broadside array, 297–299, 298f, 299f Hansen-Woodyard end-fire array, 304–312, 309f, 311f, 312t, 313t ordinary end-fire array, 299–300, 300t, 301f, 303t 302f, 304t phased (scanning) array, 302–304, 303f, 305f, 306f arrays, N-element linear array: uniform spacing, nonuniform amplitude, 323–324 array factor, 325–326, 325f binomial array, design procedure, 327–329, 328f binomial array, excitation coeffecients, 326–327 Dolph-Tschebyscheff array broadside, array design, 331–338, 332f, 334f, 336f, 336t, 339f Dolph-Tschebyscheff array broadside, array factor, 330–331 Dolph-Tschebyscheff array broadside, beamwidth and directivity, 338, 340–341, 340f Dolph-Tschebyscheff array broadside, design, 341–343, 343f Tschebyscheff design, scanning, 344–345, 345f arrays, planar array, 348f array factor, 348–354, 350f, 353f, 354f, 355f, 356f beamwidth, 354, 357–359, 358f directivity, 359–350 arrays, rectangular-to-polar graphical solution, 322–333, 324f arrays, superconductivity, 345 designs with constraints, 346–348 efficiency, directivity, 346 arrays, two-element array, 286–293, 287f, 291f, 292f, 293f Babinets principle, 680–684 baluns, transformers, 521–523, 522f, 523f bandwidth, 65–66 beam efficiency, 65 beamforming, diversity combining, Raleigh-fading, trellis coded modulation, 972–975, 972t, 973f, 974f beamwidth, 40 broadband antennas. See also traveling wave antennas electric-magnetic dipole, 559, 560f helical antenna, 549–550, 550f, 551f Yagi-Uda array of linear elements, 559–561, 560f broadband dipoles, matching techniques. See also cylindrical dipole; folded dipole; matching techniques; triangular sheet, bow-tie, and wire simulation of biconical antenna; Vivaldi antenna biconical antenna, input impedance for unipole, 491–492 biconical antenna, input impedance for finite cones, 491 biconical antenna, input impedance for infinite cones, 490–491, 492f biconical antenna, radiated fields, 487, 488f, 489–490 overview, 485–487, 486f, 487f Cassegrain reflectors, 915–916, 917f Cassegrain and Gregorian forms, 919–920, 919f classical form, 917–919, 918f cellular radio systems evolution omnidirectional systems, 933, 933f omnidirectional systems, cell splitting, 934, 934f omnidirectional systems, sectorized systems, 934–936, 935f smart-antenna systems, adaptive array systems, 937–938, 937f, 938f smart-antenna systems, spatial division multiple access (SDMA), 938–939, 939f smart-antenna systems, switch-beam systems, 936–937, 936f, 937f circular loop of constant current, loop antennas power density, radiation intensity, radiation resistance, directivity, 252–253, 253f, 254f, 255–259, 257f, 258f radiated fields, 250–252, 251f circular loop of nonuniform current, loop antennas, 259–260, 260f, 261f, 262–266, 262f, 263f, 264f, 265f arrays, 266 design procedure, 267–268 circular patch, microstrip and mobile communications antennas, 815, 815f conductance, directivity, 821–822, 822f, 823f design, 818–819, 819f electric and magnetic fields - TMz mnp, 816–817 equivalent current densities, fields related, 819–821, 820f resonant frequencies, 817–818 resonant input resistance, 822–823 coordinate system for antenna analysis, 25–26, 26f corner reflector, 876–878, 876f, 877f, 878f, 880–884, 881f, 882f, 883f cross-polarization, reflector antennas (front-fed parabolic), 896–897, 896f, 897f 1068 INDEX current distribution method, reflector antennas (front-fed parabolic), 897–901, 898f, 900f current distribution, thin wire antenna, 15–18, 16f, 17f, 18f cylindrical dipole. See also folded dipole bandwidth, 501 equivalent radii, 504–505, 506t input impedance, 501–502, 502f radiation patterns, 503–504, 505f resonance, ground plane simulation, 503, 503t, 504f dielectric resonator antennas (DRAs), 847–848 analysis and design methods, 849–850 basic geometries, 848, 848f, 849f cavity model resonant frequencies (TE, TM modes), 850–852 hybrid modes: resonant frequencies, quality factors, 852–853, 853f, 854t radiated fields, 855–859, 856f, 858f 857f dipole array design, 608–609, 609f computer program, 613–614 design equations, 609–612, 611f design procedure, 611–613, 613f dipole, See also linear wire antennas, dipole array, 13, 14f, 602–605, 603f, 606f, 607–608, 607f, 607t, 608f directional patterns, 47–48, 47f, 49f, 50–51, 51t direction-of-arrival (DOA) algorithms, 947–950, 947f, 950f directivity, 41–46, 44f, 45f aperture efficiency, 898f, 900f, 901–909, 903f, 906f, 908f, 909t, 910f, 911f directional patterns, 47–48, 47f, 49f, 50–51, 51t omnidirectional patterns, 51–52, 52f, 53f, 54 discone, conical skirt model, 512–513, 512f, 513t duality theorem, 138–139, 139t electric and magnetic fields, electric (J) and magnetic (M) current sources, 131–132 electrically small antennas, fundamental limits, 614–619, 615f, 617f, 618f, 618t electric-magnetic dipole, 559, 560f electrostatic charge distribution, 432 bent wire, 437–439, 438f finite straight wire, 433–437, 437f 433f E-plane sectoral horn, 720f, 721f aperture fields, 719–722 directivity, 728–733, 730f, 731f, 732f E-plane sectoral horn, unbounded medium (te horn), 15 equiangular spiral antennas, 593–594 conical spiral, 598, 599f planar spiral, 593–598, 596f, 597f far-field radiation, 136–137 feed design, reflector antennas (front-fed parabolic), 913–915, 915f ferrite loop, loop antennas ferrite-loading receiving loop, 271–272 radiation resistance, 270–271, 271f field regions, 31–33, 32f, 33f folded dipole, 505–512, 507f, 509f, 511f 4×4 vs 8×8 planar array, 969 fractal antennas, 627–633, 628f, 629f, 630f, 631f, 632f frequency independent antennas. See also antenna miniaturization; electrically small antennas, fundamental limits; equiangular spiral antennas; fractal antennas; log-periodic antennas overview, 591–592 theory, 592–593 Friis transmission equation, 88–90, 88f future challenges, 21 gain, realized gain, 61–64 ground effects, 203–216 half wavelength dipole, 176–179 Hall´ en’s integral equation, 444–445 helical antenna, 549–550, 550f, 551f design procedure, 553–558, 557f end-fire mode, 553 feed design, 558–559, 559f normal mode, 550–553, 552f horizontal electric dipole, 195–203 horn antennas aperture-matched horns, 766–769, 767f, 768f conical horn, 756–759, 757f, 758f, 759f, 760f, 761t corrugated horn, 761–764, 762f, 763f, 765f, 766 dielectric-loaded horns, 771, 773 multimode horns, 769–771, 769f, 772–773f phase center, 773–774 horn antennas, E-plane sectoral horn, 720f, 721f aperture fields, 719–722 directivity, 728–733, 730f, 731f, 732f horn antennas, H-plane sectoral horn, 734f aperture fields, 733 directivity, 738–741, 741f, 742f, 743 radiated fields, 734–738, 737f, 738f, 739f horn antennas, pyramidal horn, 743, 744f aperture fields, equivalent, radiated fields, 744–748, 746f, 747f, 748f, 749f design procedure, 754–756 directivity, 748–754, 751f, 752f INDEX 1069 H-plane sectoral horn, 734f aperture fields, 733 directivity, 738–741, 741f, 742f, 743 radiated fields, 734–738, 737f, 738f, 739f important parameters, associated formulas, equation numbers, 101–103t induced EMF method, 458 near-field of dipole, 458–460, 458f self-impedance, 460–463, 460f, 462f, 463f induced EMF method, mutual impedance between linear elements, 467–473, 470f 468f, 471f, 473f 472f infinite line source, unbounded medium (tm open), 15 inhomogeneous vector potential wave solution, 132–136, 133f input impedance, 75–79, 75f, 78f integral equation-moment method, 455–457, 457f, 457t Mini-Numerical Electromagnetic Code (MININEC), 467 mutual impedance between linear elements, 465 Numerical Electromagnetic Code (NEC), 466 integral equations, 431. See also moment method solution integral equations, finite diameter wires Hall´ en’s integral equation, 444–445 Pocklington’s integral equation, 440–444, 440f source modeling, delta gap, 445, 446f, 448t source modeling, magnetic-frill generator, 446–448, 446f, 448t integral equations, Integral Equation (IE) method electrostatic charge distribution, 432 electrostatic charge distribution, bent wire, 437–439, 438f electrostatic charge distribution, finite straight wire, 433–437, 437f 433f integral equation, 439 inverted-F antenna (IFA), 843–845, 843f, 845f isotropic, directional, omnidirectional patterns, 30, 31f least mean square (LMS) algorithm, optimal beamforming techniques, 956–959, 957f, 958f, 959f lens antennas, 6–7, 8f linear, circular, elliptical polarization, 68–71 linear elements near/on infinite perfect electric conductors (PEC), perfect magnetic conductors (PMC), and electromagnetic band-gap (EBG) surfaces, 179 ground effects, 203 ground planes: electric, magnetic, 180–182, 182f, 183f horizontal electric dipole, 195, 196f, 197–203, 198f, 200f, 201f , 202f, 203f image theory, 182–183, 184f, 185f mobile communication devices, antennas for mobile communication, 192–195, 195f, 196f rapid calculations and design, approximate formulas, 191–192 vertical electric dipole, 183–187, 184f, 185f, 187f, 188f, 189, 189f, 190f, 191, 191f linear wire antennas. See also linear elements near/on infinite perfect electric conductors (PEC), perfect magnetic conductors (PMC), and electromagnetic band-gap (EBG) surfaces computer codes, 216 dipole in far field, parameters, formulas, equation numbers, 217–218t linear wire antennas, finite length dipole current distribution, 164 directivity, 172–173 input resistance, 173–175, 174f, 176f power density, radiation intensity, radiation resistance, 166–172, 167f, 168f, 169f, 171f radiated fields: element, space, and pattern multiplication, 164–166 linear wire antennas, ground effects, 203 earth curvature, 211–216, 212f, 213f, 215f horizontal electric dipole, 205, 207, 207f, 208f, 209f, 210f PEC, PMC, EBG surfaces, 207, 210 verticle electric dipole, 204–205, 206f linear wire antennas, half-wavelength dipole, 176–177, 178f, 179, 179t linear wire antennas, infinitesimal dipole directivity, 154–155, 155f far-field (kr ≫1)region, 153–154 intermediate-field (kr >1) region, 152 near-field (kr ≪1) region, 151–152 power density, radiation resistance, 148–150 radian distance, radian sphere, 150–151, 151f radiated fields, 145–148, 146f linear wire antennas, region separation far-field (Fraunhofer) region, 160–162 radiating near-field (Fresnel) region, 162–163 reactive near-field region, 163–164 linear wire antennas, small dipole, 155, 156t, 157f, 158 log-periodic antennas, 598 dipole array, 602–605, 603f, 606f, 607–608, 607f, 607t, 608f dipole array design, 608–609, 609f dipole array design, computer program, 613–614 dipole array design, design equations, 609–612, 611f 1070 INDEX log-periodic antennas (Continued) dipole array design, design procedure, 611–613, 613f planar, wire surfaces, 599–602, 600f, 601f long wire amplitude patterns, maxima, nulls, 538–539, 539f, 540f, 541f input impedance, 541–542 resonant wires, 542–543 loop antennas, 235, 236f circular loops ground/earth curvature effects, 268–269, 268f, 269f loop in far field paramaters, formulas, equation numbers, 274–275t mobile communications applications, 272, 273f polygonal loop antennas, 269–270 loop antennas, circular loop of constant current power density, radiation intensity, radiation resistance, directivity, 252–253, 253f, 254f, 255–259, 257f, 258f radiated fields, 250–252, 251f loop antennas, circular loop of nonuniform current, 259–260, 260f, 261f, 262–266, 262f, 263f, 264f, 265f arrays, 266 design procedure, 267–268 loop antennas, ferrite loop ferrite-loading receiving loop, 271–272 radiation resistance, 270–271, 271f loop antennas, small circular loop equivalent circuit, receiving mode, 249–250, 249f equivalent circuit, transmitting mode, 247–249, 248f far-field (kr ≫1) region, 245–246 near-field (kr ≪1) region, 246 power density, radiation resistance, 241–245, 244f radiated fields, 236–241, 237f radiation intensity, directivity, 246–247 small loop, infinitesimal magnetic dipole, 241 matching techniques baluns, transformers, 521–523, 522f, 523f quarter-wavelength transformer, binomial design, 515–518 quarter-wavelength transformer, multiple sections, 515 quarter-wavelength transformer, single section, 514–515 quarter-wavelength transformer, Tschebyscheff design, 518–522, 518f, 520f stub-matching, 513 maximum directivity, maximum effective area, 86–88, 86f microstrip and mobile communications antennas analysis methods, 787–788 arrays, feed networks, 832–837, 835f, 836f, 837f, 838f basic characteristics, 784–785, 784f, 785f circular polarization, 830–832, 832f, 832t, 833f, 834f coupling, 827–830, 828f, 829f dielectric resonator antennas (DRAs), 847–848 feeding methods, 785–787, 786f, 787f input impedance, 826–827, 826f inverted-F antenna (IFA), 843–845, 843f, 845f multiband antennas for mobile units, 846–847, 846f overview, 783–784, 784t parameters, formulas, equation numbers, 859t planar inverted-F antenna (PIFA), 838f, 839–840, 839f, 841f quality factor, bandwidth, efficiency, 823–826, 825f slot antenna, 841–843, 842f microstrip and mobile communications antennas, circular patch, 815, 815f conductance, directivity, 821–822, 822f, 823f design, 818–819, 819f electric and magnetic fields - TMz mnp, 816–817 equivalent current densities, fields radiated, 819–821, 820f resonant frequencies, 817–818 resonant input resistance, 822–823 microstrip and mobile communications antennas, rectangular patch cavity model, 799, 799f cavity model, equivalent current densities, 804–807, 805f, 806f, 807 cavity model, field configurations (modes) - TMx, 800–803, 801f, 804f directivity, double slot (k0h ≪1), 812–815, 814f directivity, single slot k0h ≪1), 811–812, 812f nonradiating slots, 810–811 radiating slots, 807–810, 809f transmission-line model effective length, conductance, 793–794, 793f, 795f transmission-line model effective length, design, 791–792, 792f transmission-line model effective length, matching techniques, 796–799, 796f, 799f transmission-line model effective length, resonant frequency, effective width, 790–791, 790f transmission-line model effective length, resonant input resistance, 794–796 transmission-line model, fringing effects, 788–789, 789f microstrip antennas, 5, 5f, 6f Mini-Numerical Electromagnetic Code (MININEC), 467 INDEX 1071 mobile ad hoc networks (MANETs) MANETs employing smart-antenna systems, protocol, 962–964, 963f MANETs employing smart-antenna systems, wireless network, 961–962, 962f overview, 960–961, 960f moment method solution, 448 basis (expansion) functions, entire-domain functions, 453 basis (expansion) functions, subdomain functions, 449–452, 450f, 451f, 452f weighting (testing) functions, 453, 454f, 455 multiband antennas for mobile units, 846–847, 846f mutual coupling in arrays active element pattern in array, 478–480, 478f, 480f infinite regular array coupling, 476–478 mutual coupling on array performance, 476 receiving mode coupling, 476 transmitting mode coupling, 474–476, 475f mutual impedance between linear elements, 463–465, 464f, 466f. See also self-impedance induced EMF method, 467–473, 470f 468f, 471f, 473f 472f integral equation-moment method, 465 integral equation-moment method, Mini-Numerical Electromagnetic Code (MININEC), 467 integral equation-moment method, Numerical Electromagnetic Code (NEC), 466 mutual impedance between linear elements, integral equation-moment method, 465 90◦corner reflector, 878–880, 879f Numerical Equivalent Code (NEC), 466 omnidirectional patterns, 51–52, 52f, 53f, 54 omnidirectional systems, 933, 933f cell splitting, 934, 934f sectorized systems, 934–936, 935f optimal beamforming techniques least mean square (LMS) algorithm, 956–959, 957f, 958f, 959f minimum mean square error (MMSE) criterion, 955–956 optimal beamforming techniques, minimum mean square error (MMSE) criterion, 955–956 phase errors, reflector antennas (front-fed parabolic), 910–913, 914f planar inverted-F antenna (PIFA), 838f, 839–840, 839f, 841f planar, wire surfaces, 599–602, 600f, 601f plane reflector, 875–876, 876f Pocklington’s integral equation, 440–444, 440f polarization, 66–68, 67f linear, circular, elliptical polarization, 68–71 polarization loss factor (PLF), efficiency, 71–75, 71f, 74f pyramidal horn, 743, 744f aperture fields, equivalent, radiated fields, 744–748, 746f, 747f, 748f, 749f design procedure, 754–756 directivity, 748–754, 751f, 752f quarter-wavelength transformer binomial design, 515–518 multiple sections, 515 single section, 514–515 Tschebyscheff design, 518–522, 518f, 520f radar cross section (RCS), 92–96, 92t, 96f, 97f radar range equation, 90–92, 91f radian, steradian, 33, 34f radiating far-field (Fraunhofer) region, 32, 32f radiating near-field (Fresnel) region, 31–32, 32f radiation integrals, auxilliary potential functions electric and magnetic sources, computing fields block diagram, 128f overview, 127–128 radiation intensity, 37–38, 39f radiation pattern (antenna pattern), 25–26, 26f, 27f amplitude pattern shape, 32–33, 33f coordinate system for antenna analysis, 25–26, 26f field regions, 31–33, 32f, 33f isotropic, directional, omnidirectional patterns, 30, 31f principle patterns, 30–31 radiation pattern lobes, 26–30, 28f radiation pattern lobes, 26–30, 28f radiation patterns, numerical techniques, 55–57, 56f, 58f, 59–60, 60f radiation power density, 35–37 radiation problems, computer animation-visualization, 13–15 reactive near-field region, 31, 32f reciprocity, reaction theorems, 138–140 for antenna radiation patterns, 141–143, 142f for two antennas, 140–141, 140f, 141f rectangular patch, 965–966, 966f, 967f, 968f rectangular patch, microstrip and mobile communications antennas nonradiating slots, 810–811 reflector antennas, 6, 7f 90◦corner reflector, 878–880, 879f corner reflector, 876–878, 876f, 877f, 878f, 880–884, 881f, 882f, 883f plane reflector, 875–876, 876f reflector antennas, front-fed parabolic aperture distribution method, 890–896, 891f, 893f, 895f Cassegrain reflectors, 915–916, 917f 1072 INDEX reflector antennas, front-fed parabolic (Continued) Cassegrain reflectors, Cassegrain and Gregorian forms, 919–920, 919f Cassegrain reflectors, classical form, 917–919, 918f cross-polarization, 896–897, 896f, 897f current distribution method, 897–901, 898f, 900f directivity and aperture efficiency, 898f, 900f, 901–909, 903f, 906f, 908f, 909t, 910f, 911f feed design, 913–915, 915f induced current, 890 phase errors, 910–913, 914f surface geometry, 887–890, 888f reflector antennas, parabolic, 884–887, 885f, 886f reflector antennas, spherical reflector, 920–922, 920f, 922f rhombic antenna geometry, design equations, 549 self-impedance. See also mutual impedances between linear elements induced EMF method, 458 induced EMF method, near-field of dipole, 458–460, 458f induced EMF method, self-impedance, 460–463, 460f, 462f, 463f integral equation-moment method, 455–457, 457f, 457t slot antenna, 841–843, 842f smart antennas antenna, array design, 943 antenna, linear array, 944–945, 944f, 945f antenna, planar array, 945–946, 946f beamforming, diversity combining, Raleigh-fading, trellis code modulation, 972–975, 972t, 973f, 974f other geometries, 975–976, 975f signal propagation, 939–941, 940f, 941f, 942f smart antennas’ benefits, 942–943 smart antennas’ drawbacks, 942–943 smart antenna analogy, 931–932, 932f smart antennas, antenna beamforming, 946 adaptive beamforming, 950–953, 951f, 952f direction-of-arrival (DOA) algorithms, 947–950, 947f, 950f optimal beamforming techniques, least mean square (LMS) algorithm, 956–959, 957f, 958f, 959f optimal beamforming techniques, minimum mean square error (MMSE) criterion, 955–956 smart antennas, cellular radio systems evolution omnidirectional systems, 933, 933f omnidirectional systems, cell splitting, 934, 934f omnidirectional systems, sectorized systems, 934–936, 935f smart-antenna systems, adaptive array systems, 937–938, 937f, 938f smart-antenna systems, spatial division multiple access (SDMA), 938–939, 939f smart-antenna systems, switch-beam systems, 936–937, 936f, 937f smart antennas, mobile ad hoc networks (MANETs) MANETs employing smart-antenna systems, protocol, 962–964, 963f MANETs employing smart-antenna systems, wireless network, 961–962, 962f overview, 960–961, 960f smart antennas, system design/simulation/results 4×4 vs 8×8 planar array, 969 adaptive beamforming, 969–971, 970f, 971f, 972f array design, linear array design, 967–968, 968f array design, planar array design, 968–969 design process, 964–965 rectangular patch, 965–966, 966f, 967f, 968f single element, microstrip design, 965 stub-matching, 513 surface geometry, reflector antennas (front-fed parabolic), 887–890, 888f transmitting mode, transmission-line Thevenin equivalent, 1, 2f traveling wave antennas, 533–535, 534f, 535f long wire amplitude patterns, maxima, nulls, 538–539, 539f, 540f, 541f long wire, input impedance, 541–542 long wire, resonant wires, 542–543 rhombic antenna geometry, design equations, 549 V antenna, 543–547, 544f, 545f, 546f, 547f triangular sheet, bow-tie, and wire simulation of biconical antenna, 492–496, 493f, 494f, 495f, 496f two-wires, 10–11, 11f, 12f V antenna, 543–547, 544f, 545f, 546f, 547f vector effective length, 81–83, 81f vector potential A, electric current source J, 129–130 vector potential F, magnetic current source M, 130–131 vertical electric dipole, 183–195 Vivaldi antenna, 496–499, 497f, 499f, 500f wire antennas, 3, 4f, 6f Yagi-Uda array of linear elements, 559–561, 560f computer program and results, 568–571, 569f, 570f design procedure, 575–578, 576t, 577f, 579f far-field pattern, 566–567 impedance and matching techniques, 572, 575, 575t integral equation-moment method, 562–565, 566f optimization, 571–572, 572t, 573t, 574f Yagi-Uda array of loops, 579–580, 579f WILEY END USER LICENSE AGREEMENT Go to www.wiley.com/go/eula to access Wiley’s ebook EULA.
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If the chord joining t_(1) and t_(2) on the parabola y^(2) = 4ax is a focal chord then | 12 | P...
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If the chord joining t_(1) and t_(2) on the parabola y^(2) = 4ax is a focal chord then
Class: 12
Subject: MATHS
Chapter: PARABOLA
Board:IIT JEE
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Transcript:
without need get instant video solutions to all your maths physics chemistry and biology doubts just click the image of the question crop the question and get instant video solution download doubt and app today hello everyone come to the question if the chord joins t1 and t2 on the parabola y square equals to 4ax is a focal point then four option there are four options which one is the current then we can draw first a parabola this is the parabola and this is the axis and this is the chord this name p q and the focus is a comma zero and parabola equation y square equals to four ax the parameter from a p parameter parametric from a p x equals to a t square then general equation y square goes to 4 x y square equals to 4 a x and put the value of x y square equals to four a into a t square then y square equals to four a square t square for a square t square then y equals to 280 root then parameter parametric form p a t square this is plus minus a t square plus minus 280 plus minus 280 then ps plus sq equals to pq in this term ps ps plus sq equals to pq then slope of ps slope of ps equals to slope of sq then 281 minus 0 by a t1 square minus a equals to 0 minus 2 a t square by a minus a t2 whole square t2 square then two two cancel and they're a common the a t one by t one square minus one equals to minus a t square by a common 1 minus t 2 square a cancel out then t 1 y t square minus 1 equals to minus t2 by 1 sorry this is t2 not t square this is t2 and upper lower multiple minus -1 then t 1 y t 1 square minus 1 equals to t 2 by t 2 square minus 1 then cross multiplication t1 t2 into t1 minus t2 plus 1 t1 minus t2 equals to 0 then t1 minus t2 into t1 t2 plus 1 equals to 0 to university to r t1 t2 plus 1 equals to 0 then t1 t2 equals to minus 1 this is the final answer which on which option is correct first option t1 t2 equals to minus 1 this is the quiet option thank you for class 6-12 itj and neet level trusted by more than 5 crore students download doubt and after today
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Midpoint Formula
The Organic Chemistry Tutor
9850000 subscribers
11818 likes
Description
854403 views
Posted: 24 Jul 2017
This math video explains how to use the midpoint formula to find the midpoint between two points. This video contains examples and practice problems with coordinates that contain fractions.
Triangle Congruence - SSS, SAS, ASA:
CPCTC - 2 Column Proofs With Triangles:
Hypotenuse Leg Theorem - HL Postulate:
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Missing Diagrams - 2 Column Proofs:
Right Angle Theorem - SSS & SAS:
Perpendicular Bisector Theorem:
Altitude, Median, Midpoint & P. Bisector:
Parallel & Perpendicular Lines - Slope:
Indirect Proofs - Example Problems:
Proving Parallel Lines With Proofs:
The Exterior Angle Theorem:
Exterior Angle Inequality Theorem:
Polygons:
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Transcript:
in this video we're going to focus on finding the midpoint between two points so let's say we have point a which is two comma six and point b which is four comma ten what is the midpoint between these two points find the coordinates of the midpoint to find the x coordinate all you need to do is average the two x values here's the formula that you can use it's x one plus x two divided by 2 and the y coordinate is simply the average of the other two y coordinates so if you were to average 2 and 4 what number will you get the midpoint between 2 and four is three so for example if we do x one plus x two let's call this x one and this one x two if we add two and four and then divide by two this will give us 2 plus 4 is 6 and 6 divided by 2 is 3 so that will give us a midpoint of 3. now let's do the same thing with the y values let's call this y1 and y2 the midpoint between 6 and 10 instinctively you can see it's 8 so if you were to add 6 and 10 and then divide by 2 you should get 8. 10 plus 6 is 16 16 divided by 2 is 8. so that's how you can find the midpoint between two points so in this example it's three comma eight so now it's your turn go ahead and find the midpoint between a and b let's say a is three comma five and b is nine comma one so feel free to pause the video and try this out for the sake of practice so let's use the same formula we're gonna call this x1 and this is going to be y1 this is x2 y2 so it's x1 plus x2 that's 3 plus 9 divided by 2. that's going to give us the x coordinate of the midpoint now to find the y coordinate it's y one which is five plus y two which is one divided by two now three plus nine is twelve and five plus one is six twelve divided by two is six six divided by two is three and so this is the midpoint between points a and b as you can see six is right in the middle between three and nine six is the midpoint between three and nine and three is the midpoint between five and one three is exactly in the middle of those two numbers here's another one that you could try it's going to have some negative numbers included in it find a midpoint between these two ordered pairs negative 4 comma 2 and 8 comma negative 6. so let's use the same formula x1 plus x2 divided by 2 comma y1 plus y2 divided by 2. so let's call this x1 y1 this is going to be x2 and that's y2 x1 is negative 4 x2 is positive 8. y1 is two y two is negative six negative four plus eight that's equal to positive four and two plus negative six or simply two minus six that's negative four four divided by two is two negative four divided by two is negative two so [Music] this is the coordinates of the midpoint it's two negative two go ahead and try this one find a midpoint between five three and eight negative seven so x one is five x two is eight y one is three y two is negative seven now five plus eight that's thirteen three plus negative seven that's negative four and we know negative four divided by two is negative two now we can't reduce 13 over two so we're just going to leave it as a fraction so this is the answer for that this example now other times you may have coordinates that contain fractions so let's say if point a is one over three comma two and point b is five comma negative one four so feel free to tackle this problem pause the video and see if you could find the midpoint between these two coordinates or between those two points so let's follow the same procedure but actually let's do it one step at a time let's find the midpoint of the x-coordinate first we know it's going to be x one plus x two divided by two so x one is one over three x two is five now we need to simplify uh this result so how can we simplify this complex fraction what i would recommend doing is multiplying the numerator and the denominator by three two times three is six now on top you need to use the distributive property you got to multiply three by one third which the threes will cancel leaving behind one and then you need to multiply three times five which is fifteen so one plus fifteen is sixteen so you have sixteen divided by six which you can simplify the fraction 16 is 8 times 2 6 is 3 times 2 so we can cancel a 2 and so the x coordinate of the midpoint is 8 over 3 which i'm going to put right here now what we need to do next is to find the midpoint i mean the y coordinate of the midpoint which we know it's going to be y one plus y two divided by two so y one is two y two is negative one fourth so we can just simply write minus one fourth divided by 2. so let's follow the same process to simplify the complex fraction so you want to get rid of this fraction on top and notice that it has a denominator of four therefore you want to multiply the top and the bottom by that denominator so on the bottom we have two times four which is eight on the numerator we gotta multiply two and four which is also eight and then we need to multiply four by negative one fourth so the fours will cancel leaving behind one but there's a negative sign in front of the one so it's negative one eight minus one is seven so the second part is seven over eight so that's how you could find the midpoint if you're given two points that contain fractions now let's work on another example that's similar to the last one but with a little bit more work so let's say point a is one over five comma negative two over three and point b that's going to be let's say 4 over 3 comma 3 over 4. so i want you to find the midpoint between points a and b so we have a lot of fractions but the principle of working this problem is the same as the last one so go ahead and try it take a minute and try this problem so first let's find the x coordinate of the midpoint let's average the x values so the first x value is one over five and the second x value is four over three and then we're going to divide it by two so we're averaging two fractions now what i need to do is get rid of the five and the three so i'm going to multiply the top and the bottom by 15. now what's one-fifth of fifteen one-fifth of fifteen is the same as fifteen divided by five which is three next we need to multiply fifteen by four over three fifteen times four is sixty sixty divided by three is twenty or you could divide first you could say fifteen divided by 3 is 5 5 times 4 is 20. on the denominator we have 2 times 15 which is 30. so 3 plus 20 is 23. so the x coordinate of the midpoint is 23 over 30. now let's do the same for the y coordinates y one is negative two over three y two is three over four let's divide it by two so we need to the top and bottom by 12 to get rid of the denominators of the smaller fractions so the common denominator is 12. so let's multiply 12 by negative two thirds twelve divided by three is four four times two is eight so two thirds of twelve is eight but let's not forget about the negative sign in front of it and then twelve times three fourths twelve divided by four is three three times three is nine two times twelve is twenty four negative eight plus nine is one so it's one over twenty four so this is the midpoint between points a and b you
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Esercizi di Meccanica e Termodinamica corso di Fisica Generale I Ultima revisione: 15 marzo 2024 A cura di: Stefano Lacaprara INFN Sezione di Padova via Marzolo, 8 35131 Padova, Italia Stefano.Lacaprara@pd.infn.it Rilasciato con Licenza Creative Commons CC ⃝ Some Rights Reserved B Y : ⃝ $ \ ⃝ C ⃝ Sommario Questa e una raccolta di esercizi indirizzati al corso di Fisica Generale I, meccanica e termodinamica, corso di laurea in Fisica.
Si tratta di esercizi che propongo e svolgo a lezione durante il corso alla facolt a di Fisica dell’Universita di Padova. Vuole essere di aiuto agli studenti che desiderano provare a fare gli esercizi per conto loro, ma non sostituisce la lezione in aula. In particolare, le soluzioni, che si trovano alla fine dei capitoli, sono il pi u delle volte solo accennate oppure e messo solo il risultato numerico. Questo sia per non andare in competizione con il corso stesso, sia per la noia mortale che e scrivere una soluzione completa di un esercizio per quanto semplice.
Gli esercizi stessi vengono da una varieta di fonti, principalmente vecchi com-piti sia proposti da me sia tramandati, come “memoria del dipartimento”, dagli esercitatori del passato, spesso con modifiche e aggiornamenti.
L’ordine degli esercizi segue pi u o meno lo svolgimento del corso, e richiede, ovviamente, lo studio della teoria, che qui non viene minimamente trattata. Le formule utilizzate per gli esercizi svolti in modo completo sono considerate “date”, e non vengono dimostrate o giustificate, a meno che si tratti di casi particolari non coperti da un normale libro di testo.
La correzione delle bozze e, o meglio dovrebbe essere, una parte impor-tante della stesura di questi esercizi: per quanto abbia fatto attenzione, errori e imprecisioni sono sempre possibili, e anzi vi sar o grato se vorrete segnalarmele.
Ultima nota riguardo alla licenza: questo scritto e rilasciato con la licenza Creative Commons CC ⃝ B Y : ⃝ $ \ ⃝ C ⃝Attribuzione - Non commerciale - Condividi allo stesso modo 3.0. In parole povere, tu sei libero di: riprodurre, distribuire, comuni-care al pubblico, esporre in pubblico, rappresentare (?), eseguire e recitare 1 quest’o-pera, modificare quest’opera, alle seguenti condizioni: Devi attribuire la paternit a dell’opera dall’autore ; Non puoi usare quest’opera per fini commerciali ; Se alteri o trasformi quest’opera, o se la usi per crearne un’altra, puoi distribuire l’opera ri-sultante solo con una licenza identica o equivalente a questa. Trovi tutte il legalese su Creative Commons CC ⃝all indirizzo
1se lo fate, vi voglio venire a vedere! O forse no . . .
S. Lacaprara Es. Meccanica e Termodinamica 2 Indice I Meccanica punto materiale 1 1 Vettori 2 2 Cinematica 5 2.1 Cinematica moto circolare . . . . . . . . . . . . . . . . . . . . . . . . . .
18 3 Dinamica del punto materiale 21 3.1 Dinamica del punto materiale con attrito . . . . . . . . . . . . . . . . . .
29 3.2 Dinamica del punto materiale: energia . . . . . . . . . . . . . . . . . . .
41 3.3 Dinamica del punto materiale: sistemi non inerziali . . . . . . . . . . . .
53 4 Gravitazione 58 5 Dinamica dei sistemi 63 II Meccanica corpo rigido 71 6 Dinamica del corpo rigido 72 III Fluidi 93 7 Fluidi 95 IV Termodinamica 100 8 Calorimetria 101 9 Termodinamica: I principio 104 10 Termodinamica: II principio 110 S. Lacaprara Es. Meccanica e Termodinamica 3 Parte I Meccanica punto materiale 1 Vettori Esercizio 1.1 Dati i vettori ⃗ a,⃗ b di componenti, rispettivamente: ⃗ a = (2, 3, −1) e ⃗ b = (3, 3, 4) in rappresentazione cartesiana (x, y, z), si calcoli: 1. |a|, |b|; 2. ⃗ c = ⃗ a +⃗ b, ⃗ d = ⃗ a −⃗ b 3. |c|, |d|; 4. il prodotto scalare s = ⃗ a ·⃗ b 5. l’angolo θab tra i due vettori; 6. l’angolo θx di ⃗ a con l’asse x; 7. il prodotto vettoriale ⃗ v = ⃗ a ×⃗ b, dimostrando che tale vettore ⃗ v risulta perpendi-colare al piano dove giacciono ⃗ a e ⃗ b.
Esercizio 1.2 Si considerino i due vettori ⃗ v = 4.5ˆ x + 5.1ˆ y −3.0ˆ z e ⃗ u = 2.6ˆ x + 6.1ˆ y + 9.0ˆ z in un sistema di coordinate cartesiane (x, y, z).
Determinare: 1. la rappresentazione di ⃗ v e ⃗ u in coordinate cilindriche (ρ, ϕ, z) 2. la rappresentazione di ⃗ v e ⃗ u in coordinate sferiche (ρ, θ, ϕ) 3. ⃗ v · ⃗ u; 4. ⃗ v × ⃗ u; Esercizio 1.3 Dato un cubo di spigolo l, si consideri un vertice A, e il centri B e C di due facce che non contengono il vertice A.
A B C θ 1. Calcolare l’angolo tra AB e AC Esercizio 1.4 Si considerino due vettori ⃗ u e ⃗ v e la loro somma ⃗ r = ⃗ u + ⃗ v, tali che: |⃗ r| = 30 angolo tra ⃗ r e ⃗ u ´ e θu = 25◦ S. Lacaprara Es. Meccanica e Termodinamica 2 θv = 50◦ 1. fornire una rappresentazione di ⃗ u e ⃗ v Esercizio 1.5 Siano dati in un sistema di riferimento cartesiano, i punti P e Q di coordinate: P = (4, 5, −7), Q = (−3, 6, 12) 1. Calcolare la distanza tra il punto P e la retta passante per Q e parallela a ⃗ v = (4, −1, 3) S. Lacaprara Es. Meccanica e Termodinamica 3 Soluzione dell’esercizio 1.1 1. |a| = 3.7, |b| = 5.8; 2. ⃗ c = (5, 6, 3), ⃗ d = (−1, 0, −5); 3. |c| = 8.4, |d| = 5.1; 4. a · b = 11; 5. θab = cos−1 a·b |a||b| = 59.7◦ 6. cos−1 a·ˆ x |a| = 57.7◦ 7. a × b = (15, −11, −3), (a × b) · a = 0 = (a × b) · b Soluzione dell’esercizio 1.2 1. ⃗ v = (6.8, 48.6◦, −3.0)ρ,ϕ,z, ⃗ u = (6.6, 66.9◦, 9.0)ρ,ϕ,z 2. ⃗ v = (7.4, 113.9◦, 48.6◦)r,θ,ϕ, ⃗ u = (11.2, 36.5◦, 66.9◦)r,θ,ϕ 3. ⃗ v · ⃗ u = 15.8; 4. ⃗ v × ⃗ u = (64.2, −48.3, 14.2)x,y,z = (8.16, 80.0◦, −37.0◦)r,θϕ Soluzione dell’esercizio 1.3 1. AB = (l, l/2, l/2), AC = (l/2, l, l/2) θ = a cos AB·AC |AB|·|AC| = 33.55◦ Soluzione dell’esercizio 1.4 1. ⃗ u = (21.6, 10.1) e ⃗ v = (8.4, −10.1) Si puo’ ragionare cosi’: visto che ⃗ r = ⃗ u + ⃗ v, allora vr + ur = 30, dove v/ur sono le proiezioni di v/u lungo r. vr = |v| cosv idem per u. Inoltre, per la stessa ragione, le componenti di ⃗ u e ⃗ v perpendicolari a ⃗ r hanno somma nulla (cioe’ sono uguali e opposte).
Quindi |v| cosv +|u| cosu = |r| |v| sinv +|u| sinu = 0 Soluzione dell’esercizio 1.5 1. Sia P ′ la proiezione di P sulla retta in questione.
La lunghezza di QP ′ e’: ⃗ PQ ⃗ v/|v| = 28/ √ 26 La lunghezza di PP ′ e’: |PP ′| = p |PQ|2 −|PQ|2 = 19.5 S. Lacaprara Es. Meccanica e Termodinamica 4 2 Cinematica Esercizio 2.1 Una corsa podistica si svolge lungo un percorso lungo Ltot = 8.57 km, con un tratto di andata e ritorno in comune. I piu veloci corrono ad un ritmo di 4’30” per km, mentre i pi u lenti hanno un ritmo di 7’ per km.
Dopo L1 = 4.2 km dalla via, i corridori attraversano un incrocio stradale, e lo riattraversano una seconda volta dopo L2 = 6.3 km dalla via.
1. Velocita massima e minima.
2. Tempo dei pi u veloci e dei piu lenti all’arrivo.
3. Per quanto tempo sar a bloccato l’incrocio?
Esercizio 2.2 Il conduttore di treno che viaggia con velocita v1 vede davanti a s´ e un secondo treno che viaggia con una velocit a minore v2, e frena con una decelerazione a.
1. Determinare la distanza minima d cui deve iniziare a frenare per evitare una collisione; 2. In condizioni limite quando e dove si toccano.
Esercizio 2.3 Un oggetto viene lanciato verso l’alto lungo la verticale e, quando ricade, tocca terra con una velocita v = 79.2m/s.
Sapendo che si conficca nel terreno per h = 23cm, calcolare: 1. l’accelerazione che ferma l’oggetto, supponendo che sia costante; 2. il tempo trascorso fino all’arresto; 3. l’altezza massima raggiunta durante il lancio iniziale.
S. Lacaprara Es. Meccanica e Termodinamica 5 Esercizio 2.4 Un’auto che viaggia a v1 = 80km/h si pu o arrestare in d1 = 57m; se viaggia a v2 = 48km/h si arresta in d2 = 24m. Supponendo che il tempo di reazione del guidatore sia uguale nei due casi, calcolare: 1. l’accelerazione di frenata; 2. il tempo di reazione del guidatore.
Esercizio 2.5 Il moto di un punto materiale e descritto, in un sistema cartesiano, dall’equazione: y = 1.28x −0.31x2 con [y] = [x] = m. Il moto e uniformemente accelerato lungo y e vx(0) = 5.19 m/s = costante.
Calcolare: 1. l’accelerazione ⃗ a; 2. la velocita media tra gli istanti t = 0s e t1 = 1.5s; 3. la velocit a istantanea all’istante t1 = 3s; 4. questo e l’unico moto compatibile con la traiettoria descritta dall’equazione?
Esercizio 2.6 In un sistema di riferimento inerziale O, il moto di un punto materiale e descritto dal vettore posizione: ⃗ r(t) = Aˆ x −Ct3ˆ y + (Dt2 −Bt)ˆ z Determinare: 1. le dimensioni fisiche delle quantita A, B, C, D; 2. la velocit a ⃗ v(t); 3. l’accelerazione ⃗ a(t); 4. descrivere qualitativamente il moto; Lo stesso moto viene descritto da un secondo sistema di riferimento O′, dal vettore posizione: ⃗ r′(t) = Aˆ x −Ct3ˆ y + (Dt2 + Et)ˆ z con A = 1, B = 2, C = 2.5, D = 3.5, E = 5 nelle opportune unita SI.
5. quale e la velocita del sistema O′ rispetto ad O; 6. se il punto materiale risente di forze di inerzia in O′.
Esercizio 2.7 Equazione della balistica Un cannone spara un proiettile con una velocit a iniziale v0, ad un angolo α rispetto al terreno. Si determini, ignorando effetti di attrito con l’aria: S. Lacaprara Es. Meccanica e Termodinamica 6 α ⃗ v0 hmax xmax gittata x y ⃗ g 1. l’equazione della traiettoria; 2. la gittata; 3. l’altezza massima; 4. l’angolo di gittata massima 5. il tempo di volo del proiettile Esercizio 2.8 Tre punti materiali seguono le tre traiettorie in figura, nel campo di gravita della terra. Si determini per quale traiettoria : A B C y x g 1. il tempo di volo e minore; 2. la componente orizzonatale della velocita iniziale e minore; 3. il modulo della velocita iniziale e minore.
Esercizio 2.9 Le migliori prestazioni di sempre nel salto in lungo sono le seguenti: Mike Powell (1991, Tokyo) l = 8.95m, Bob Beamon (1968, Mexico City) l = 8.90m, Carl Lewis (1991, Tokyo) l = 8.87. Supponendo che il modulo della velocita del saltatore al momento dello stacco sia pari a v0 = 9.50m/s, si calcoli, trascurando la resistenza dell’aria: 1. l’altezza massima durante il volo dei saltatori; 2. la migliore prestazione teoricamente possibile.
Esercizio 2.10 Eq. della balistica in salita Si vuole lanciare un sasso il pi u lontano possibile lungo un piano inclinato con un angolo θ rispetto all’orizzontale. Supponendo fissata il modulo della velocita iniziale del sasso si calcoli: 1. l’angolo α rispetto al piano orizzontale per raggiungere la massima distanza lungo il piano inclinato; 2. si verifichi che la soluzione trovata si riduce alla nota equazione della balistica per θ = 0.
S. Lacaprara Es. Meccanica e Termodinamica 7 Esercizio 2.11 1. Dimostrare che la gittata e la stessa per angolo θ = 45◦± δ; 2. calcolare δ per v0 = 30m/s e gittata= 20m.
Esercizio 2.12 Un cacciatore al suolo spara con un fucile contro un’anatra che vola con velocita orizzontale va = 60 km/h ad altezza h = 100m, nell’istante in cui l’anatra gli passa sopra.
Calcolare: 1. la minima velocit a di uscita del proiettile v0 per colpire l’anatra; 2. l’alzo minimo per colpire l’anatra.
Esercizio 2.13 Un aereo viaggia orizzontalmente ad una altezza h = 225m e con una velocita v = 155m/s rispetto al suolo e il pilota deve sganciare un pacco su un bersaglio.
1. Calcolare a quale angolo ϕ rispetto all’orizzonte si deve trovare il bersaglio per ottenere un lancio perfetto; 2. idem, se l’aereo sta scendendo con un angolo α = 15◦? Discutere la soluzione; 3. idem, se l’aereo vola orizzonatale e il bersaglio si muove con velocita’ vt = 14m/s nella stessa direzione e verso; Esercizio 2.14 Una barca si muove vicino alla riva con velocit a ⃗ vB = 10km/h rispetto a terra, parallelamente alla riva stessa. Una bandierina sulla barca forma un angolo αB = 30◦ rispetto alla scia della barca. Una bandierina a terra forma un angolo αT = 45◦sempre rispetto alla scia della barca.
Acqua Terra 30o 45o Determinare: 1. la velocita del vento ⃗ vT rispetto a terra; 2. la velocit a apparente del vento ⃗ vA rispetto alla barca.
Esercizio 2.15 Una barca deve attraversare un fiume di larghezza l = 20m, la cui corrente risulta avere una velocita pari a vx = b(y(l −y)), dove y e la distanza dalla sponda e b = 5 · 10−3m−1s−1. La velocita della barca e vy = 3.6km/h.
Si calcoli: S. Lacaprara Es. Meccanica e Termodinamica 8 1. il tempo necessario alla barca per attraversare il fiume; 2. il punto di arrivo.
Esercizio 2.16 Caduta in un pozzo (G.Cella, Un esercizio al giorno) Un sasso viene lasciato cadere dentro un pozzo, e si sente il suono dell’urto con il fondo dopo t = 2.0 s. Trascurando l’attrito dell’aria, calcolare: 1. la profondita del pozzo; 2. l’errore che viene fatto se si considera la velocit a del suono in aria (vs = 340 m/s) infinita.
3. discutere l’errore del punto precedente Esercizio 2.17 (G.Cella, Un esercizio al giorno) Un’auto parte da ferma con moto uniformemente accelerato, con accelerazione a.
Dopo un tempo t0, un sasso viene lanciato verso l’auto dalla posizione iniziale, con velocita iniziale v0, che si pu o supporre costante nel moto del sasso. Si chiede: 1. quale e la velocit a orizzontale minima v0 perch´ e il sasso colpisca l’auto; 2. si discutano i risultati; 3. quale e l’angolo rispetto al piano cui deve essere lanciato il sasso nelle condizioni del punto 1.?
4. con quale velocit a (relativa) il proiettile colpisce l’auto?
Esercizio 2.18 Moto delle farfalle (G.Cella, Un esercizio al giorno) Si supponga che il moto delle farfalle notturne sia guidato da una sorgente luminosa, e che le farfalle mantengano costante l’angolo tra la direzione della luce e quella del volo.
Se la luce e quella della luna, la farfalla vola in direzione rettilinea, ma se la luce e quella di una fiamma vicina, il volo non finira bene per la farfalla.
La stessa strategia e seguita dai falchi nei confronti della preda 1. descrivere il moto della farfalla nelle condizioni di luce vicina; 2. perch´ e la strategia funziona con luce lontana?
Esercizio 2.19 (M.S.V. es:1.6) Un’auto ha una accelerazione massima a1 = 2 m/s2, e una decelerazione massima (frenata) a2 = −4 m/s2. Calcolare 1. il minimo tempo necessario per percorrere l = 500 m, partendo e arrivando fermi; 2. lo spazio percorso in accelerazione; 3. la velocita massima.
S. Lacaprara Es. Meccanica e Termodinamica 9 Esercizio 2.20 Un bagnino deve soccorrere un bagnante in difficolta’. Il bagnino si trova inizialmente ad una distanza di y1 = 30 m dalla riva del mare e il bagnante si trova a y2 = 50 m dalla riva. La distanza tra i due nella direzione parallela alla riva e’ xtot = 100 m.
Il bagnino e’ in grado di correre sulla spiagga con una velocita’ vs = 10 m/s e nuotare con vm = 1.6 m/s.
1. quale percorso deve scegliere per raggiungere il bagnante nel tempo piu’ breve possibile?
2. quanto tempo ci mette?
S. Lacaprara Es. Meccanica e Termodinamica 10 Soluzione dell’esercizio 2.1 1. vmax = 1km/4′30” = 3.70m/s = 13.3 km/h, vmin = 2.38 m/s = 8.5 km/h ∆v = 1.32 m/s = 4.75 km/h 2. Tprimi = 2316 s = 38′36”, Tultimi = 3600 s = 1 h 3. I primi attraversano l’incrocio la prima volta dopo: T1 = 1135 s Lo riattraversano dopo T2 = 1701 s (sempre dal via).
Gli ultimi impegnano l’incrocio dopo T3 = 1765 s, e lo riattraversano dopo T4 = 2647 s.
Quindi l’incrocio resta occupato da T1 a T2, e poi da T3 a T4 cio e per 1512 s = 25′, e per circa 1 minuto, i corridori si incrociano sul ponte.
Soluzione dell’esercizio 2.2 1. In un sistema di riferimento solidale con il secondo treno: d(t) = −d + (v1 −v2)t −1/2at2 ≥0.
Il contatto avviene quando d(t) = 0 e l’equazione in tal caso non ha soluzioni se ∆= (v1 −v2)2 −2ad ≤0, quindi d ≥(v1−v2)2 2a ovvero a ≥(v1−v2)2 2d .
Da notare che ho usato −1/2at2, quindi l’equazione del moto decelerato, e quindi trovo un limite inferiore per a, cioe a ≥amin. Avrei potuto scrivere l’equazione del moto accelerato +1/2at2 e avrei trovato lo stesso risultato con il segno invertito, a significare che l’accelerazione deve essere negativa (decelerazione) e minore di un valore massimo. a ≤amax < 0 Si pu o risolvere anche in un sistema di riferimento solidale a terra (laboratorio) o solidale con il primo treno. Ovviamente il risultato non cambia.
2. l’equazione del moto −d+ (v1 −v2)t−1/2at2 = 0 in condizioni limite ha soluzione t⋆= 2d v1−v2. Da notare che se il moto fosse stato uniforme, la soluzione sarebbe stata t = d v1−v2. Nel caso accelerato, la velocita relativa iniziale e v1 −v2, mentre quella finale e 0, e varia linearmente (accelerazione costante), da cui il fattore 2.
La posizione del primo treno al momento del contatto e s1(t⋆) = v1t⋆+ 1/2at⋆, quella del secondo e s2(t⋆) = d + v2t⋆e si pu o verificare che coincidono (ovviamen-te).
s = d v1+v2 v1−v2 Casi limite: se v1 = v2 non si scontrano mai. Se v2 = 0 si scontrano in posizione d, cioe dove si trova il secondo treno (fermo).
Che significato fisico ha la soluzione nel caso in cui v2 > v1?
Soluzione dell’esercizio 2.3 1. istante di impatto con terreni: t = 0s v(t = 0) = v0 istante di arresto con terreni: t = t⋆v(t⋆) = 0 v(t⋆) = 0 = v0 −at⋆ x(t⋆) = h = vtt⋆−1/2a(t⋆)2 a = v2 0 2h = 13.6 · 103 m/s; S. Lacaprara Es. Meccanica e Termodinamica 11 2. t⋆= v0 a = 5.8 · 10−3 s; 3. Uso equazione della balistica, il modulo della velocit a iniziale del lancio e uguale al-la velocit a di impatto a terra perch´ e il problema e simmetrico rispetto all’inversione del tempo. hmax = v2 0 2g = 319.7 m.
Soluzione dell’esercizio 2.4 1. Scrivo equazione del moto per posizione e velocit a nei due casi, consideranto come istante finale quelli di arresto.
d1 = v1t0 + v1t1 −1/2a(t1)2 d2 = v2t0 + v2t2 −1/2a(t2)2 0 = v1 −at1 0 = v2 −at2 Risolvendo per a, t0, t1, e t2: a = 5.81 m/s2 t0 = 0.65 s t1 = 3.85 s t2 = 2.96 s Soluzione dell’esercizio 2.5 1. x(t) = vx(0)t, y(t) = 1.28vx(0)t −0.31v2 x(0)t2 = 6.64t −8.35t2 = vy(0) + 1/2ayt2 da cui vy(0) = 6.64 m/s e ay = −16.7 m/s2 ⃗ a = ayˆ y; 2. < ⃗ v 0,3s = ⃗ r(t=1.5s)−⃗ r(0) 1.5s = (5.19, −5.89) m/s |v| = 7.85 m/s, θ = −48.6◦; 3. ⃗ v(3s) = (5.19, −43.5) m/s |v| = 43.8 m/s, θ = −74◦; 4. Un moto qualunque vincolato a quella parabola e descritto dalla stessa equazione.
Soluzione dell’esercizio 2.6 1. A [m], B [m/s], C [m/s3], D [m/s2], E [m/s] 2. ⃗ v(t) = −3Ct2ˆ y + (2Dt −B)ˆ z 3. ⃗ a(t) = −6Ctˆ y + 2Dˆ z 4. Nella direzione ˆ x, il punto e fermo x(t) = A; lungo ˆ zil moto e uniformemente accelerato, con az = 2D, velocit a iniziale v(t = 0)z = −B, e posizione iniziale z(t = 0) = 0.
Lungo ˆ y, il moto ha accelezione non costante a(t)y = −6Ct, velocita e posizione iniziale nulle.
5. ⃗ v = ⃗ v′ + ⃗ vO quindi ⃗ vO = d⃗ r(t) dt −d⃗ r′(t) dt = −(B + E)ˆ z 6. il moto di O′ rispetto a O e rettilineo uniforme, quindi non ci sono forze d’inerzia.
S. Lacaprara Es. Meccanica e Termodinamica 12 Soluzione dell’esercizio 2.7 1. In un sistema cartesiano: x(t) = vx(0)t = v0 cos αt y(t) = vy(0)t −1/2gt2 = v0 sin αt −1/2gt2 Traiettoria: y(x) = tan αx − g 2v2 0 cos2 αx2 2. Gittata: x(α) = 2v2 0 cos α sin α g = v2 0 sin 2α g 3. Altezza massima per un dato α si ha per x = Gittata/2 ymax(α) = v2 0 2g sin2 α = v2 y(0) 2g Altezza massima ymax = v2 0 2g per α = π 2 = 90◦(cannone verticale) 4. la gittata massima si ha per sin 2α = 1 quindi per α = π 4 = 45◦, e vale xmax = v2 0 g .
In questo caso hmax(α = π 4) = v2 0 4g = hmax 2 .
5. t = 2v0 sin α g = 2v0,y g , xmax = v0,xt = v2 0 sin 2α Soluzione dell’esercizio 2.8 1. Dalle equazioni balistica, ymax(α) = v2 y(0) 2g , dato che le altezze massime sono uguali, allora vy(0) sono uguali.
Tempo di volo e tempo per salire e scendere T = 2 · vy(0) g ed e uguale per le tre traiettorie; 2. Dato che tempo di volo e costante, vx(0) e maggiore per la traiettoria con la gittata piu lunga. A < B < C; 3. v0 = q v2 x(0) + v2 y(0) quindi segue lo stesso ordine di vx(0) A < B < C.
Soluzione dell’esercizio 2.9 1. Applico equazioni balistica Saltatore θ ymax Powell 38.0◦ 1.74 m Beamon 37.3◦ 1.69 m Lewis 37.2◦ 1.69 m Da notare che la latitudine e l’altezza di Tokyo rispetto a Citt a del Messico rendendo il valore locale di g leggermente diverso nelle due citta.
Tokyo (35◦, 0 m slm) g = 9.798 m/s2 ; Citt a del Messico (19◦, 2400 m slm) g = 9.776 m/s2. L’altezza di Citta del Messico inoltre, rende l’aria meno densa rispetto a Tokyo, e questo riduce l’attrito con l’aria durante il volo (e la rincorsa) dell’atleta. Questo non e considerato in questo problema, ma e in realt a il fattore piu importante per andare in quota per aumentare le prestazioni atletiche.
2. Se Powell avesse saltato con un angolo di 45◦, avrebbe saltato l = 9.21 m a Tokyo e l = 9.23 m a Mexico City. Ovviamente non e scontato che un saltatore possa cambiare l’angolo di stacco mantenendo costante il modulo della velocita iniziale.
S. Lacaprara Es. Meccanica e Termodinamica 13 Soluzione dell’esercizio 2.10 1.
y(x) = tan αx − g 2v2 0 cos2 αx2 y(x) = tan θx R2 = x2 + y2 dove R e la distanza dall’origine del punto cui cade il sasso. Risolvendo per R R = 2v2 0 g cos θ (tan α −tan θ) cos2 α Ottengo il massimo ponendo dR dα = 0, il che avviene per: (tan α −tan θ) sin 2α = 1 .
2. Se θ = 0, si riduce a sin2 α = 1/2, quindi θ = 45◦ Soluzione dell’esercizio 2.11 1. Banale: gittata= v2 0 g sin 2α, che e simmetica attorno a α = 45◦ 2. δ = 39.7◦ Soluzione dell’esercizio 2.12 1. L’altezza massima raggiunta e hmax = v2 y(0) 2g , quindi vy(0) = √2gh. Per colpire l’anatra, vx(0) = va. v0 = p (v2 x(0) + v2 y(0)) = p v2 a + 2gh = 47.3 m/s 2. Se vx(0) = va tutti gli alzi permettono di colpire l’anatra, posto che l’altezza massima sia sufficiente. tan α = vy(0) vx(0) ≥ √2gh va = 2.65, α ≥69◦ Soluzione dell’esercizio 2.13 1. Equazione traiettoria e: y(x) = h −g 2v2 0 x2 y = 0 se x2 = h tan ϕ, che porge: ϕ = a tan q gh 2v2 0 = 12.1◦ 2. Equazione traiettoria diventa: y(x) = h −tan αx − g 2v2 0 cos2 αx2 imponendo di colpire il bersaglio, si ottiene una equazione di secondo grado, con una soluzione negativa (non fisica) e una positiva: ϕ′ = 21.4◦ 3. Come nel caso iniziale, ma v”0 = v0 −vt. ϕ” = a tan gh 2(v”0)2 = 13.3◦ S. Lacaprara Es. Meccanica e Termodinamica 14 Soluzione dell’esercizio 2.14 1. Uso un sistema di riferimento con ˆ y lungo la direzione del fiume, concorde con la corrente, e ˆ x perpendicolare.
La velocit a rispetto al sistema di riferimento della barca e: ⃗ vB = vbˆ y, quella del vento e ⃗ vT = − √ 2/2vT ˆ x − √ 2/2vT ˆ y, dato che il vendo forma un angolo di 45◦ rispetto alla scia della barca, quindi rispetto −ˆ y.
La velocita del vento rispetto alla barca si ottiene: ⃗ vA = ⃗ vT −⃗ vB, ossia sommando alla velocit a del vento rispetto a terra, la velocita della terra rispetto alla barca, che e l’opposto della velocita della barca rispetto a terra: ⃗ vA = (− √ 2/2vT)ˆ x + (− √ 2/2vT −vB) .
Inoltre, l’angolo tra il vento e la scia della barca, nel sistema di riferimento della barca e 30◦, quindi: tan 30◦= 1/ √ 3 = vA,x vA,y = √ 2/2vT √ 2/2vT + vB Da queste due equazioni ricavo: vT = √ 2 2 + √ 3 3 √ 2 2 vB = √ 3 3 vB = 5.2 m/s nella direzione della bandiera a terra; 2. Conoscendo le due componenti di ⃗ vA, posso ricavarmi vA = 7.4 m/s, nella direzione della bandiera sulla barca.
Soluzione dell’esercizio 2.15 1. Moto perpendicolare alla corrente e indipendente dalla corrente. t1 = l vy = 20s 2. Eq. del moto della barca: vy(t) = vy vx(t) = b(y(t)(l −y(t))) Dalla prima equazione ricavo, integrando: y(t) = vyt, quindi la seconda diventa vx(t) = b(vyt(l −vyt)) che integrata rispetto al tempo, fornisce: x(t) = bvyt2 l 2 −vyt 3 Quindi x(t1) = bl3 6vy = 6.7 m Soluzione dell’esercizio 2.16 1. h = 1/2gt2 = 19.6 m/s 2. ts = h/vs, t = tc + ts = q 2h g + h vs Risolvo per h = 18.5 m. Quindi l’errore e circa 1.1 m, pari a ∼6% S. Lacaprara Es. Meccanica e Termodinamica 15 3. l’errore nel non considerare la velocita del suono come finita, porta ad un sovrasti-ma della profondit a di circa 1 m (bias). Tuttavia un errore del 2% sulla misura del tempo (t = t±0.04 s), porterebbe ad un errore simile. Quindi prima di valutare il bias introdotto dal suono che si propaga, e bene valutare l’incertezza sulla misura del tempo. Se questa viene fatta a mano con un orologio o cronometro e molto facile che l’incertezza su start e stop del cronometro sia ben piu grande di qualche %. In tal caso, una correzione per la velocit a del suono sarebbe del tutto inutile.
Soluzione dell’esercizio 2.17 1. v0 = 2at0, e l’urto avviene a t = 2t0; 2. l’altra soluzione si ha per v0 = 0, l’urto avviene a t = 0 quando l’auto e ancora ferma e si trova nella stessa posizione del sasso, che non viene lanciato (ovvero lanciato con velocit a nulla, cioe fermo); 3. applico eq.
della balistica.
Componente verticale della velocit a: vy 0 = 1/2gt0, quindi tan θ = g 4a 4. data la simmetria della traiettoria parabolica del proiettile, e la stessa velocit a verticale con il quale il proiettile viene lanciato, quindi vurto = gt0 4 . Da notare che questa soluzione si riferisce alla velocita minima del sasso: esistono infinite altre soluzioni con velocit a iniziali maggiori, dove il sasso urta l’auto prima e con velocita relativa superiore. NB. Non provatelo a casa!
Soluzione dell’esercizio 2.18 1. Descrivo moto in coord. polari, con centro sulla luce. Sia α l’angolo tra la direzione di volo e la luce (o la preda): ⃗ v = ˙ Rˆ ur + R ˙ θˆ uθ = −v cos αˆ ur + v sin αˆ uθ da cui: ˙ θ = − ˙ R R tan α che ha come soluzione: R(θ) = R0e −θ tan α, che e una spirale logaritmica.
Se il volo e verso la luce (tan α > 0), allora la distanza diminuisce, e la farfalla “cade” verso la fiamma.
2. Se R0 = ∞, allora il moto diventa rettilineo.
Soluzione dell’esercizio 2.19 1. Sia t1 il tempo di accelerazione, v1 la velocit a all’istante t1, s1 lo spazio percorso durante l’accelerazione, e (t, v, s)2 i corrispondenti valori all’istante finale: v(t) = ( 0 + a1t, t < t1 v1 + a2(t −t1), t > t1 Inoltre v(0) = 0 = v(t2) s(t) = ( 0 + 1/2a1t2, t < t1 s1 + v1(t −t1) + 1/2a2(t −t1)2, t > t1 con s(0) = 0 e s(t2) = l S. Lacaprara Es. Meccanica e Termodinamica 16 Risolvendo il sistema si trova t2: t1 = r 2l a1 1−a1 a2 = 18.25 s t2 = t1 1 −a1 a2 = q 2l(a2−a1) a1a2 = 27.39 s t2 −t1 = 9.14 s 2. l1 = 1/2a1t2 1 = 333 m, l2 = l −l1 = 167 m 3. vmax = v1 = a1t1 = 36.5 m/s In alternativa, molto piu’ semplice: durante l’accelerazione v(s) = √2a1s e idem durante la frenata. Visto che la velocita’ massima e’ pari a vmax = √2a1l1 = √2a2l2 2a1l1 = 2a2l2 l1 + l2 = l da cui si ricava l1 = l 1+a2/a1. Da qui e’ semplice calcolare la vmax = √2a1l1 e il tempo per i due tratti: t1 = vmax/a1, ottenendo gli stessi valori numerici di prima.
Soluzione dell’esercizio 2.20 1. Considere un sistema di riferimento x −y centrato nella posizione iniziale del bagnino. In questo riferimento, la posizione del bagnante e (xtot, y1 + y2). Sia x la coordinata in cui la traiettoria del bagnino arriva alla riva del mare.
Lo spazio percorso sulla spiaggia risulta ls = p x2 + y2 1 e il corrispondente tempo t1 = √ x2+y2 1 vs . Analogamente il tempo percorso nuotando risulta: t2 = √ (xtot−x)2+y2 2 vm .
Il tempo minimo si trova minimizzando la funzione t(x) = t1 + t2 in funzione di x. Il minimo si trova per sin θ1 sin θ2 = vs vm, dove θ1 e’ l’angolo tra la traiettoria sulla spiaggia e la perpedicolare alla riva, sin θ1 = x √ x2+y2 1 e analogamente per θ2.
Quindi ho queste 4 equazioni con 4 incognite, che posso, con pazienza e trigono-metria, risolvere.
y1 = l1 cos θ1 y2 = l2 cos θ2 xtot = l1 sin θ1 + l2 sin θ2 sin θ1/vs = sin θ2/vm S. Lacaprara Es. Meccanica e Termodinamica 17 2.1 Cinematica moto circolare Esercizio 2.21 Una ruota di raggio r1 = 30cm parte da ferma con accelerazione angolare α = 0.4rad/s2.
Trasmette il suo moto mediante una cinghia che non scivola ad una seconda ruota di raggio r2 = 12cm. Calcolare: 1. il tempo ∆t che impiega la ruota piccola a raggiungere una velocit a angolare ω = 300giri/s.
Esercizio 2.22 Un punto materiale si muove lungo una circonferenza di raggio R = 3.6m. Ad un certo istante la sua velocita e v0 = 17m/s e l’accelerazione forma un angolo di θ = 22◦ con la direzione radiale. Determinare: 1. di quanto aumenta il modulo della velocita per unit a di tempo per t = t0; 2. legge oraria del modulo della velocita: discutere i risultati; 3. quanto vale il modulo dell’accelerazione totale per t = t0; 4. legge oraria del modulo dell’accelerazione; Esercizio 2.23 Un bambino e fermo sulla piattaforma di una giostra a r = 2m dall’asse di rotazione.
La giostra parte da ferma con accelerazione angolare costante finch´ e raggiunge la velocita angolare di regime ω = 0.15giri/s dopo n = 2 giri. A questo punto il bambino si mette a camminare in direzione radiale verso il bordo con una velocit a vb = 2km/h.
Calcolare: 1. l’accelerazione ⃗ a e la velocita ⃗ v del bambino: a) durante l’accelerazione della giostra; b) quando la giostra e a regime e il bambino inizia a muoversi.
Esercizio 2.24 Un corpo e appeso tramite una fune inestensibile ad una carrucola di raggio R = 10cm.
Il corpo scende con moto uniformente accelerato tale che: v(t0 = 0s) = 0.04m/s (verso il basso) e ∆h(t1 = 2s) = −0.2m (il corpo e sceso).
1. Calcolare l’accelerazione ⃗ a del bordo della carrucola all’istante t = 5s.
S. Lacaprara Es. Meccanica e Termodinamica 18 Soluzione dell’esercizio 2.21 1. La velocita tangenziale delle due ruote e la stessa.
∆t = r2 r1 ω α = 31.4s Soluzione dell’esercizio 2.22 1. La velocita, di cui e noto il modulo, ha direzione tangente alla circonferenza.
Il testo parla solo di direzione dell’accelerazione, in particolare non e esplicito se il verso e’ diretto verso l’interno o verso l’esterno. Tuttavia sappiamo che si tratta di un moto circolare e che ci deve essere una componente centripeta, quindi il verso dell’accelerazione deve essere diretto verso il centro della circonferenza.
L’accelerazione ha componente centripeta aN e tangenziale aT.
aN = v2 R , tan θ = aT aN .
La variazione di modulo della velocit a e dovuta alla sola aT, visto che aN ruota la velocit a ma non la cambia in modulo. d|v| dt = aT = v2 R tan θ = 32.4m/s2 2. l’equazione del moto si ricava integrando: dv v2 = tan θ R dt, che ha soluzione: v(t) = v0R R−v0 tan θ(t−t0).
Qualitativamente: dato l’angolo costante dell’accelerazione, c’e sempre una com-ponente tangenziale, quindi la velocit a tangenziale aumenta. Di conseguenza deve aumentare l’accelerazione centripeta, che quindi fa aumentare l’accelerazione tan-genziale. Si tratta chiaramente di una situazione non fisica, che porta, in tempi finiti (t = t0 + R v0 tan θ) ad avere una velocita infinita. Un modello pi u fisico avrebbe sicuramente un limite superiore all’accelerazione centripeta (il disco si rompe!).
3. |⃗ a| = p a2 N + a2 T = v2 R √ 1 + tan2 θ = 86.6m/s2 4. a(t) = dv dt = v2 0R √ 1+tan2 θ (R−v0 tan θ(t−t0))2, tenendo conto che c’e anche la componente centripeta: ac = at tan θ Soluzione dell’esercizio 2.23 1. ω0 = 0.942 rad/s, vb = 0.556 m/s ω = αt, 2nπ = 1 2αt2, α = ω2 4nπ = 0.0353 rad/s2 ⃗ r(t) = Rˆ ur ⃗ v(t) = d⃗ r(t) dt = dR dt ˆ ur + R dθ dt ˆ uθ ⃗ a(t) = d⃗ v(t) dt = h d2R dt2 −R dθ dt 2i ˆ ur + h R d2θ dt2 + 2 dR dt dθ dt i ˆ uθ a) dR dt = 0, dθ dt = αt d2θ dt2 = α, quindi: ⃗ v = Rαtˆ uθ = 0.071 · tˆ uθ m/s ⃗ a = −Rω2ˆ ur + Rαˆ uθ = −2.5 · 10−3ˆ ur + 0.071ˆ uθ m/s2 b) dR dt = vb, d2R dt2 = 0, dθ dt = ω d2θ dt2 = 0, quindi: ⃗ v = vbˆ ur + Rωˆ uθ = 0.556ˆ ur + 1.884ˆ uθ m/s ⃗ a = −Rω2ˆ ur + 2vbωˆ uθ = −1.77ˆ ur + 1.05ˆ uθ m/s2 Soluzione dell’esercizio 2.24 S. Lacaprara Es. Meccanica e Termodinamica 19 1. ∆h = v0t + 1 2at2, a = 2 (∆h −v0t) /t2 = 0.06 m/s2 ⃗ a = aˆ uθ + v2/Rˆ ur = aˆ uθ + (v0 + at)2/Rˆ ur, per t = 5s, ⃗ a = 0.06ˆ uθ + 3.4ˆ ur m/s2 S. Lacaprara Es. Meccanica e Termodinamica 20 3 Dinamica del punto materiale Esercizio 3.1 Giro della morte Un carrellino di massa m scende lungo un piano inclinato con angolo α su una rotaia che successivamente forma una ruota completa di raggio R alla fine della discesa (giro della morte). La rotaia fornisce un vincolo unilaterale.
h R θ Determinare: 1. la reazione del vincolo ⃗ N(θ) lungo il giro della morte; 2. la velocit a minima v0 per compiere il giro; 3. l’altezza minima h per compiere il giro.
Esercizio 3.2 Un montacarichi di massa complessiva M = 2000 kg e sollevato da una fune ideale con carico di rottura Tr = 28 · 103 N. Si chiede: 1. la massima accelerazione verticale ⃗ a che e in grado di sopportare.
Esercizio 3.3 Due corpi di massa M1 = 2.3 kg e M2 = 1.2 kg sono appoggiati su un piano orizzontale privo di attrito e sono a contatto tra di loro. Sul corpo M1 viene applicata una forza orizzontale, diretta verso M2 di intensita F = 3.2 N. Si chiede: M1 M2 ⃗ F 1. l’intensit a della forza di contatto F21 tra le due masse; 2. F12, se la forza viene applicata sul corpo M2; 3. l’accelerazione a dei due corpi nei due casi.
4. che succede se c’e attrito con il piano?
Esercizio 3.4 Una cassa di massa M = 110 kg e spinta in salita con velocita costante lungo un piano inclinato privo di attrito, che forma un angolo θ = 34◦con l’orizzontale.
S. Lacaprara Es. Meccanica e Termodinamica 21 ⃗ F M θ Determinare: 1. l’intensit a della forza orizzontale F necessaria; 2. la forza ⃗ N esercitata dal piano inclinato.
Esercizio 3.5 Macchina di Atwood (1784) Due masse M1 e M2, con M2 > M1 sono collegate da una fune ideale ad una puleggia priva di attrito e massa trascurabile. Si chiede: 1. l’accelerazione a1,2 delle due masse; 2. la tensione T della fune che collega le due masse; 3. la reazione vincolare T0 del soffitto che sorregge tutto il sistema.
4. studiare i casi limite M1 M2 Esercizio 3.6 Il sistema in figura e costuito da tre masse M1 = 5.0 kg, M2 = 7.6 kg, M = 9.1 kg, una fune ideale, due carrucole di massa trascurabile, ed il tutto si muove senza attrito.
M M2 M1 Calcolare: 1. l’accelerazione dei tre corpi ai; 2. la tensione della corda T1 e T2 su M1 e M2, rispettivamente; 3. la forza risultante che agisce su M.
Esercizio 3.7 S. Lacaprara Es. Meccanica e Termodinamica 22 Il corpi identici di massa M1 = M2 = 12.3 kg nel sistema in figura sono collegati da una fune ideale e le pulegge hanno massa trascurabile, inoltre il corpo M1 si muove senza attrito sul piano.
Si determini: M1 M2 ⃗ T1 ⃗ T2 1. l’accelerazione dei due corpi a1 e a2; 2. la tensione T1 e T2 delle funi attaccate ai due corpi.
Esercizio 3.8 Si consideri il sistema in figura, dove le funi sono ideali, cos ı come le carrucole. Le masse dei tre corpi sono rispettivamente M1 = 1.0 kg, M2 = 2.0 kg, M3 = 3.0 kg.
Si determini: 1. le accelerazione ai dei tre corpi; 2. le tensioni Ti delle funi; 3. quale dovrebbe essere la massa M3 affinch´ e essa abbia accelerazione nulla; M1 M2 M3 Esercizio 3.9 Si considerino i quattro paranchi disegnati in figura2.
1 2 3 4 Per ciascuno, si determini, considerando le funi e le puleggie ideali: 1. di quanto bisogna tirare l’estremita libera della fune per ottenre un sollevamento ∆h del peso; 2. la forza necessaria per tenere il peso in equilibrio; 3. la tensione del cavo collegato al soffitto.
2L’immagine e stata presa da Wikipedia, l’enciclopedia libera, 11/03/2014 S. Lacaprara Es. Meccanica e Termodinamica 23 Esercizio 3.10 Un uomo di massa Mu = 70 kg si trova su una piattaforma di massa Mp = 20 kg. La piattaforma e collegata tramite una fune inestensibile ad una carrucola priva di massa a sua volta fissata al soffitto. L’uomo afferra l’altra estremit a della fune e cerca di tirarsi verso l’alto.
Si calcoli, in condizioni di equilibrio: 1. la forza con cui l’uomo deve tirare la fune; 2. la reazione vincolare esercitata dalla piattaforma sull’uomo; 3. la forza che esercita il soffitto sulla fune; 4. discutere le condizioni di equilibrio.
5. che succede se la carrucola e’ doppia (o tripla. . . ) Esercizio 3.11 Macchina di Atwood infinita Esercizio proposto da uno studente, non mi sono segnato il nome, sorry!
Si consideri la macchina in figura, che si estende identica all’infinito. Le masse siano tutte uguali m, le carrucole prive di massa e le funi ideali.
Si determini: 1. l’accelerazione della massa piu alta; 2. la forza esercitata dal soffitto.
Esercizio 3.12 Un corpo si muove per caduta e senza attrito su un dosso semicilindrico, partendo dalla sommit a da fermo.
Calcolare 1. il punto dove si stacca dalla superfice del cilindro; 2. e possibile fare lo stesso usando esclusivamente con la cinematica?
3. la distanza dal piede della discesa dove il corpo toca terra.
S. Lacaprara Es. Meccanica e Termodinamica 24 Soluzione dell’esercizio 3.1 1.
⃗ N = ˆ u⊥ m R2 (v2 0 + 3Rg cos θ −2Rg) 2. v0 ≥√5Rg 3. h ≥5 2R Soluzione dell’esercizio 3.2 1. ⃗ F = M⃗ g + ⃗ T = M⃗ a.
Proettando lungo un asse verticale diretto verso l’alto T −Mg = Ma, ovvero a = T M −g ≤Tr M −g = 4.2 m/s2 Soluzione dell’esercizio 3.3 1. Lungo il piano: F −F21 = M1a e R = M2a, da cui F21 = F M1 M1+M2 = 1.1 N 2. idem, con M1 e M2 scambiati: F12 = F M2 M1+M2 = 2.1 N 3. ⃗ a = ⃗ F Mtot = 0.91 m/s2 diretta verso destra; Nel secondo caso, l’accelerazione e la stessa, ma con direzione invertita (verso sinistra).
4. La forza totale esterna che agisce sul sistema e F −µM1 + M2g (attrito in direzione opposta al moto. Se µ > F (M1+M2)g = 0.01 allora i due corpi non si muovono.
Soluzione dell’esercizio 3.4 1. lungo il piano, l’equilibrio delle forze fornisce F cos θ = Mg sin θ F = Mg tan θ = 728 N; 2. ortogonlmente al piano, l’equilibrio fornisce: N = F sin θ + Mg cos θ, N = Mg cos θ = 1300 N.
Soluzione dell’esercizio 3.5 1. Per entrambe le masse vale, scegliendo un asse diretto verso il basso: Mig −T = Miai, inolte a1 = −a2 = a. Quindi a = M1−M2 M1+M2g, che e negativa dato che M2 > M1, quindi la massa piu leggera (M1) risale, come atteso; 2. T = 2M1M2 M1+M2g.
3. Nell’ipotesi di carrucola di massa trascurabile, T0 = 2T = 4M1M2 M1+M2g.
4. Fissando M2, se M1 = 0, allora T0 = 0 se M1 = M2, allora T0 = 2M2g se M1 ≫M2, allora T0 = 4M2g Soluzione dell’esercizio 3.6 S. Lacaprara Es. Meccanica e Termodinamica 25 1. prendendo come riferimento gli assi di movimento dei tre corpi, con verso positivo per M1 che sale verso l’alto: M1a = −M1g + T1 Ma = −T1 + T2 M2a = M2g −T2 da cui si ricava: a = M2−M1 M+M1+M2g = 1.18 m/s2 2. T1 = M1g M+2M2 M+M1+M2 = 65.6 N T2 = M2g M+2M1 M+M1+M2 = 54.9 N 3. FM = T2 −T1 = M(M2−M1) M+M1+M2g = 10.7 N Soluzione dell’esercizio 3.7 1.
M1a1 = T1 M2a2 = M2a2 −T2 Mcac = T2 −2T1 = 0carrucola a1 = 2a2 Da cui si ricava: a2 = M2g 4M1+M2g = 1.96 m/s2 e a1 = 2M1 4M1+M2g = 3.92 m/s2; 2. T1 = 2M1M2 4M1+M2g = 48.3 N; T2 = 2T1 = 96.5 N.
Soluzione dell’esercizio 3.8 1. considero un asse di riferimento verticale diretto verso il basso: M1a1 = M1g −T1 M2a2 = M2g −T1 M3a3 = M3g −T3 a1 + a2 + 2a3 = 0 T3 = 2T1 Da cui ricavo: a1 = −4.0 m/s2 (il corpo 1 sale), a2 = +2.9 m/s2 (corpo 2 scende), e a3 = +0.58 m/s2 (scende); 2. T1 = 13.8 N, T3 = 27.7 N; 3. equilibrio se M3g = 2T1, e T1 = g 1/M1+1/M2, il che fornisce: M3 = 1.33 kg.
Soluzione dell’esercizio 3.9 S. Lacaprara Es. Meccanica e Termodinamica 26 Paranco ∆x F T 1 ∆h Mg 2Mg 2 2∆h Mg 2 3 2Mg 3 3∆h Mg 3 4 3Mg 4 4∆h Mg 4 5 4Mg Soluzione dell’esercizio 3.10 1. 0 = Muau = Mg−R−T e 0 = Mpap = Mp+R−T, quindi T = (Mu+Mp)g 2 = 440 N 2. R = (Mu−Mp)g 2 = 245 N; 3. F = 2T = 880 N; 4. Mu ≥Mp.
Soluzione dell’esercizio 3.11 1. Facciamo alcuno considerazioni: la tensione della fune appesa alla prima carrucola e proporzionale a g; la tensione della prima fune e il doppio di quella della seconda fune; Il sistema appeso al soffitto e del tutto identico al sistema appeso alla secon-da carrucola, con l’unica differenza che la prima carrucola e sottoposta ad accelerazione a. Quindi il sistema appeso alla seconda carrucola sente una gravit a pari a g −a, visto che e non inerziale; la prima considerazione e valida anche per la seconda carrucola, ma conside-rando la “gravita” effettiva.
Quindi T1 g = T2 g−a1 = T1/2 g−a che porge: a = g 2.
Da notare che c’ e anche una seconda soluzione valida, cio´ e T1 = 0, il che equivale al caso limite per cui m = 0. In tal caso il sistema e libero e banale.
2. ma1 = −mg + T1, quindi T1 = 3 2mg, quindi T0 = 3mg.
Si pu o anche calcolare la massa equivalente al sistema infinito di carrucole. meqa1 = meqg −T1, quindi meq = −T1 a1−g = 3m Esercizio molto poco fisico.
Soluzione dell’esercizio 3.12 1. Scompongo il moto in direzione radiale ˆ u⊥e parallela ˆ u∥ad ogni punto della traiettoria. Lungo ˆ u∥il vincolo non agisce, quindi ho solo la gravita: a∥= g cos θˆ u∥.
Integro per ottenere la velocit a in funzione dell’angolo.
a∥= dv dt , ds = vdt, e ds = −Rdθ (si noti il segno −dovuto al fatto che l’angolo e definito rispetto alla direzione orizzontale - notare gli estremi di integrazione successivi - e quindi se il punto materiale scende l’angolo diminuisce).
Mettendo insieme le tre esperessioni di prima si ottiene: a∥ds = −a∥Rdθ = dv dt vdt = vdv S. Lacaprara Es. Meccanica e Termodinamica 27 Z θ θ=π/2 a∥Rdθ = −Rg Z θ θ=π/2 cos θdθ = Rg(1 −sin θ) = v2(θ) 2 Quindi v2(θ) = 2Rg(1 −sin θ) L’accelerazione normale alla traiettoria e quella centripeta necessaria per tenere il corpo attaccato alla superfice del cilindro.
a⊥= v2(θ) R = 2g(1 −sin θ) .
Nel punto in cui il corpo si stacca, tale accelerazione normale coincide con quella fornita dalla gravita, dato che il vincolo non fa pi u nulla.
Quindi 2g(1 −sin θ) = g sin θ, il che fornisce il risultato θ = a sin 2 3 = 42◦ 2. ⃗ F = ⃗ N + m⃗ g = m⃗ a Scompongo lungo ˆ u⊥e ˆ u∥, ottengo come prima a⊥= v2(θ) R = 2g(1 −sin θ).
Quindi lungo la direzione ˆ u⊥: N = mg sin θ −ma⊥= mg(2 −3 sin θ). Il corpo si stacca quando N = 0, quindi θ = a sin 2/3 3. Conosco v(θ) e θ all’istante del distacco.
Successivamente il moto e di tipo parabolico (⃗ g) e le condizioni iniziali sono note: posizione e velocit a.
S. Lacaprara Es. Meccanica e Termodinamica 28 3.1 Dinamica del punto materiale con attrito Esercizio 3.13 Un corpo di massa M = 3.57 kg e appoggiato su un piano orizzontale scabro. Su di esso e applicata una forza FT = 7.68 N, diretta ad un angolo θ = 15◦rispetto al piano.
1. Osservando che il moto e rettilineo uniforme, determinare µ.
Esercizio 3.14 Un corpo si muove in salita lungo un piano inclinato di angolo θ = 30◦. Il piano e scabro ed ha un coefficiente di attrito dinamico µ = 0.2. La velocita iniziale del corpo e v0 = 4.2 m/s, e l’altezza del piano e h = 40 cm.
Si determini: 1. il tempo ∆t impiegato per raggiungere la quota h; 2. il coefficiente µ′ perch´ e la velocit a alla quota h sia nulla.
Esercizio 3.15 I corpi del sistema in figura sono collegati da funi ideali e carrucole prive di massa. Le masse sono M1 = 5 kg, M2 = 7 kg, M3 = 1 kg. Il coefficiente di attrito dinamico tra le masse M2 e M3 vale µD 3 = 0.2.
Determinare: M1 M2 M3 µ3 1. Il coeff. di attrito statico minimo µS 3 perche il sitema sia in equilibrio.
Se la massa M1 si muove verso il basso: 2. le accelerazioni a1,2,3 e le tensioni T1,2; 3. idem se c’ e attrito anche tra M2 e il piano: µD 2 = 0.2; Se la carrucola pi´ u a sinistra e collegata al muro con una molla con k = 300 N/m: 4. quale e il suo allungamento in condizioni statiche; 5. il periodo delle oscillazioni nel caso in cui il coefficiente di attrito statico sia grande; 6. l’allungamento statico se il corpo 2 e 3 scivolano l’uno sull’altro.
Esercizio 3.16 Il sistema in figura e costituito da due masse M1 = 4.5 kg e M2 = 3.2 kg, unite da una fune e da una carrucola ideale.
Tra il piano orizzontale e M1 e presente attrito con un coefficente di attrito dinamico µd = 0.35 e statico µs Calcolare: M2 M1 µ S. Lacaprara Es. Meccanica e Termodinamica 29 1. accelerazione a1 e a2; 2. tensione del filo T; 3. condizione per avere moto; Esercizio 3.17 Ad una carrucola priva di massa sono collegati, tramite una fune ideale, due corpi, di massa m1 = 100 g e m2 = 120 g, inizialmente fermi. Il corpo m1 e libero di scorrere lungo la fune ed e frenato da una forza di attrito Fa costante. Si osserva che dopo un tempo t1 = 0.7 s, il corpo m1 e sceso di h = 60 cm.
Calcolare: 1. Fa; 2. a2; 3. l’allungamento ∆l del tratto di fune tra le due masse all’istante t1.
Esercizio 3.18 curva parabolica Un carrellino di massa M = 12 kg percorre una curva “parabolica” con raggio di curvatura R = 2.8 m, che e inclinata di un angolo θ = 25◦rispetto al piano orizzontale. Si calcoli: 1. la velocita v0 per percorrere la curva senza attrito; 2. la reazione ⃗ N0; Nell’ipotesi che il coefficiente di attrito statico sia µ = 0.13; 3. la reazione N1 se la velocit a e v1 = 4.0 m/s; 4. la forza d’attrito F1 presente; 5. la velocit a massima e minima vmax,min; 6. la reazione Nmax,min nei due casi.
Esercizio 3.19 Un corpo di massa M = 5 kg si trova su un piano scabro inclinato ad angolo θ = 30◦, con coefficiente di attrito dinamico µD = 0.4.
Il corpo, inizialmente fermo, e tirato verso l’alto da una fune con la tensione T minima necessaria per muovere il corpo. Si osserva che dopo aver percorso una distanza ∆x = 2 m la velocit a e v = 3.7 m/s.
Calcolare: 1. il coefficiente di attrito statico µS; 2. calcolare il bilancio energetico durante la salita; 3. la massa m che devo aggiungere a M per avere un moto uniforme, nell’ipotesi che la tensione sia sempre la stessa.
S. Lacaprara Es. Meccanica e Termodinamica 30 Esercizio 3.20 Su un piano scabro e appoggiata una tavola m2 = 3 kg inizialmente in quiete. Il coefficiente di attrito statico e dinamico tra la tavola e il piano sono µS = 0.2, µD = 0.1, rispettivamente. Sulla tavola si trova un secondo corpo m1 = 2 kg, che si muove con velocita v1 = 3 m/s su di essa. Tra il corpo e la tavola vi e attrito, con coefficiente pari a µ1 = 0.6.
m2 m1 µ1 µS,D ⃗ v1 1. Discutere le condizioni per il moto della tavola; 2. Determinare la distanza s1−2 percorsa dal corpo sulla tavola, fino a quando il corpo e fermo rispetto alla tavola; 3. la distanza s2 percorsa dalla tavola rispetto al piano nello stesso periodo; 4. l’energia meccanica dissipata nel processo.
Esercizio 3.21 Del sistema in figura sono note le masse, M1 e M2, i coefficiente d’attrito statico µ1 e µ2 con i rispettivi piani d’appoggio, e che il carrello si muove di moto accelerato verso destra.
M1 M2 µ1 µ2 ⃗ a 1. Si determini i valori dell’accelerazione a affinch e le due masse restino ferme; 2. Si discuta il caso limite M2 ≪M1 oppure M1 ≪M2 3. Si puo risolvere il problema sia nel sistema del laboratorio (inerziale) che in un sistema solidarle con il carello (non inerziale). Verificare che le soluzioni siano le stesse.
Esercizio 3.22 Un corpo si muove di moto circolare uniforme lungo una cir-conferenza di raggio R = 0.80 m con periodo T = 12.56 s.
Da un certo istante e sottoposto ad una forza di attrito viscoso ⃗ F = −k⃗ v, e si arresta dopo un angolo θ2 = 2 rad.
Si chiede: 1. il rapporto k/m; 2. l’accelerazione ⃗ a del corpo all’angolo θ1 = 1 rad.
R θ1 θ2 S. Lacaprara Es. Meccanica e Termodinamica 31 Esercizio 3.23 Il sistema in figura e costituito da una cassa di massa M = 25 kg, appoggiata su un piano inclinato α = 20◦scabro con µ = 0.18, e da una massa m = 4 kg appesa al soffitto della cassa tramite una molla k = 1500 N/m, in equilibrio rispetto alla cassa.
β α Determinare: 1. l’accelerazione a del sistema; 2. l’elongazione x della molla; 3. l’angolo β che la massa m forma con la verticale del soffitto della cassa.
Esercizio 3.24 Una massa M1 = 150 g e collegata tramite una fune ideale ad una seconda massa M2 = 200 g che si trova a riposo su un piano orizzontale liscio. M1 e fissata al pavimento da una molla che inizialmente ha una compressione ∆x0 = 19.6 cm. La massa M2 viene messa in moviemento e ruota con R = 25 cm e ω = 0.5 giri/s.
Determinare: 1. la nuova compressione ∆x1 della molla; Con della sabbia, si rende il piano scabro, e si osserva che la massa M2 si arresta dopo aver percorso una spirale lunga d = 39 cm, e che, dopo l’arresto, la molla e compressa ∆x2 = 18 cm. Si chiede: 2. il coefficente di attrito dinamico di M2.
3. la forza di attrito statico di M2; 4. coefficiente attrito statico minimo µs Esercizio 3.25 Due punti materiali di massa MA = 2 kg e MB = 1 kg, sono appoggiati su un piano inclinato θ = 30◦, scabro, con coefficienti di attrito dinamico rispettivamente µA = 0.1 e µB = 0.2. Inizialemnte sono fermi a contatto, con MB a monte di MA, e vengono lasciati liberi di scivolare. I due corpi sono collegati da una fune ideale lunga l = 1 m.
MA MB µA µB ℓ θ Calcolare: 1. il tempo t necessario perche la fune si estenda completamen-te; 2. l’accelerazione a con la fune tesa 3. la tensione della fune Esercizio 3.26 S. Lacaprara Es. Meccanica e Termodinamica 32 Che forza devo applicare ad una fune avvolta su un palo per tenere sollevata una massa M in presenza di attrito statico tra la fune e il palo?
1. quanti giri di corda devo fare attorno al palo, se µs = 0.5, per tenere sollevata una balena (M = 180 ton), se la massa della corda dalla parte libera e pari a m = 1 kg?
S. Lacaprara Es. Meccanica e Termodinamica 33 Soluzione dell’esercizio 3.13 1.
0 = FT cos θ −µ|N| 0 = −Mg+ FT sin θ + |N| da cui: µ = F cos θ Mg−F sin θ = 0.22.
Soluzione dell’esercizio 3.14 1. lungo il piano inclinato, in salita: ¨ x = −g(sin θ + µ cos θ) ˙ x = v0 −g(sin θ + µ cos θ)t x = v0t −g/2(sin θ + µ cos θ)t2 h sin θ = v0t−g/2(sin θ +µ cos θ)t2, che fornisce due soluzione t1,2 = 0.23 s, 1.04 s, la prima e quella di interesse.
2. Dalla equazione della velocit a, ricavo t, ponendo v = 0, e metto t nell’equazione h sin θ = v0t −g/2(sin θ + µ′ cos θ)t2, ricavando: µ′ = tan θ v2 0 2gh −1 = 0.72.
Soluzione dell’esercizio 3.15 1. Le equazioni della dinamica nel caso generale sono: −M3a = −T2 + Fatt M2a = −T2 −Fatt + T1 M1a = −T1 + M1g dove Fatt e la forza di attrito statico o dinamico, a seconda del caso.
Nel caso statico, il modulo (e neppure il verso) della forza di attrito non e noto, ma si puo usare il fatto che le accelerazioni dei tre corpi sono nulle: condizione necessaria (ma non sufficiente) perch e il sistema sia, appunto, statico.
Risolvendo il sistema, si ottiene: Fatt = M1g/2. Per calcolare il minimo coefficiente di attrito statico, e necessario che la forza d’attrito calcolata sia minore della mssima forza di attrito possibile, cio e che: F max att = µSM3g ≤Fatt = M1g/2, da cui si ricava µS ≥ M1 2M3 = 2.5.
Si ricava anche che in queste condizioni T1 = M1g e T2 = M1g/2.
2. si risolve il sistema di prima usando Fatt = M3gµ3, e si ottiene: a = g(M1−2µ3M3) M1+M2+M3 = 3.47 m/s2, T1 = 31.7 N, T2 = 5.4 N.
S. Lacaprara Es. Meccanica e Termodinamica 34 3. il sistema e simile a quello di prima, ma la seconda equazione diventa: M2a = −T2 −µ3M3g −µ2(M3 + M3)g + T1 da cui si ricava: a = g(M1−2µ3M3−µ2(M2+M3)) M1+M2+M3 = 2.26 m/s2 e poi le tensioni T1 = M1(g −a) = 37.7 N, e T2 = M3(a −µ3g) = 4.43 N 4. Se la condizione e realizzata e si ha una situazione di equilibrio allora la forza della molla deve essere pari a Fel = k∆l = 2T2 = M1g, da cui ∆l = M1g k = 16.3 cm 5. Il sistema e del tutto analogo ad un punto materiale di massa M, appeso ad una molla di costante elastica k, visto che il sistema dei carrellini esercita una forza costante sulla molla, esattamente come nel caso della massa appesa.
Considerando i corpi 2 e 3 come un unico corpo, l’equazione della dinamica per i corpi (2+3) e 1 sono: (M2 + M3)a = T1 −Fel M1a = M1g −T1 Questo mi permette immediatamente di trovare le equazioni del moto del sistema, che saranno di tipo oscillatore armonico (M1 + M2 + M3)¨ x + kx −M1g = 0 In quel caso, la forza costante M1g determina solamente la posizione di equilibrio del moto armonico, mentre il periodo risulta essere T = 2π q M k , dove M = M1 + M2 + M3 e la massa inerziale del punto.
6. come prima: ∆l = 2T2 k = 3.6 cm 7. Discussione sulle condizioni di µs con la molla Si puo studiare il caso in cui la molla non sia inizialmente in tensione (e quindi T2 = 0). In questo caso rimane la forza d’attrito statico tra M2 e M3. Se tale forza e sufficente da tenere ferme le due masse tra di loro, di fatto abbiamo un unico punto materiale di massa: M2 + M3.
In questo caso, sempre nell’istante iniziale quando la molla non ´ e in tensione, a = M1g M1+M2+M3. Visto che su M3 la forza di attrito statico ´ e l’unica forza agente, e al massimo vale µsM3g allora risulta: Fatt = M3a = M3M1g M1+M2+M3 ≤µsM3g, e la condizione su µs risulta µs ≥ M1 M1+M2+M3 = 0.38.
Via via che la molla si allunga esercita una forza elastica di richiamo sempre piu’ grande, proporzionale al suo allungamento. La tensione esercitata sul corpo M3 ´ e pari a 2T2 = Fel, come si ricava considerando la carrucola attaccata alla molla.
Le tensione T2 varia da un valore iniziale nullo (quando la molla e’ a riposo) ad uno massimo quando la molla e al suo massimo allungamento. Nel secondo caso il si-stema e istantaneamente fermo (l’oscillazione sta invertendo la direzione), e quindi e’ del tutto analogo al sistema statico che abbiamo studiato precedentemente.
La condizione per cui i corpi M3 e M2 restino fermi l’uno rispetto all’altro ´ e quella studiata precedentemente µS ≥ M1 2M3 = 2.5 Quindi se la µs > 2.5 i corpi 2 e 3 restano sempre fermi l’uno rispetto all’altro e il sistema e semplice. Se µs < 0.38 i corpi 2 e 3 scorrono l’uno rispetto all’altro S. Lacaprara Es. Meccanica e Termodinamica 35 (e il sistema diventa piu’ complicato).
Se infine µs ha un valore intermedio, i due corpi iniziano fermi e si staccano durante l’oscillazione, e il sistema diventa davvero complicato da risolvere.
Soluzione dell’esercizio 3.16 1. a1 = a2 = g M1−M2µ M1+M2 = 4.3 m/s2 2. T = gM1M2(1+µ) M1+M2 = 24.0 N 3. M1 > µdM2 il sistema continua a muoversi (vero). Se inizialmente fermo, inizia a muoversi se µs < M1 M2 Soluzione dell’esercizio 3.17 1. m1 si muove con accelerazione costante, visto che su di esso agiscono due forze costanti: posso ricavarmi tale accelerazione dall’eq. del moto: h(t1) = 1/2a1t2 1.
Quindi uso l’eq. della dinamica su m1: FR = m1g −Fa = m1a1, e infine ricavo Fa = m1 g −2h t2 1 = 0.735 N 2. Sul corpo a2 sono esercitate due forze: FR = m2g −T = m2g −Fa, quindi a2 = g −Fa/m2 = 3.67 m/s2 3. l’accelerazione del corpo m1 rispetto alla fune e: a′ 1 = −a1 + a2 = 6.12 m/s2, quindi ∆l = 1/2a′ 1t2 1 = 1.5 m In alternativa, si puo calcolare la velocit a finale di m1: v1 = a1t, quella di m2: v2 = a2t, la discesa di m2: h2 = a2t2 2 (la discesa di m1 e h), e fare il bilancio energetico del sistema.
Efin −Eini = L(Fa) −m1gh −m2gh2 + 1 2m1v2 1 + 1 2m2v2 2 −0 = −Fa∆l da cui mi ricavo ∆l.
Soluzione dell’esercizio 3.18 1. in direzione radiale: Mv2 0/R = N0 sin θ, in direzione verticale: 0 = N0 cos θ −Mg, quindi v0 = √Rg tan θ = 3.6 m/s; 2. N0 = Mg cos θ = 130 N; 3. v1 > v0, quindi la forza di attrito e rivolta verso l’interno della curva. Le equazioni della dinamica nelle due direzioni radiale e verticale diventano: N1 sin θ + F1 cos θ = Mv2 1/R N1 cos θ −F1 sin θ −Mg = 0 da cui ricavo: N1 = Mv2 1/R sin θ + Mg cos θ = 135.7 N; S. Lacaprara Es. Meccanica e Termodinamica 36 4. N1 = Mv2 1/R cos θ −Mg sin θ = 12.4 N < µN1 = 17.6 N; 5. stesse equazioni di prima, adesso Fmax = µN1, sempre diretto verso l’interno della curva.
Nmax sin θ + µNmax cos θ = Mv2 1/R Nmax cos θ −µNmax sin θ −Mg = 0 vmax = q gR sin θ+µ cos θ cos θ−µ sin θ = 4.2 m/s Per la vmin la forza di attrito e diretta verso l’esterno. vmin = q gR sin θ−µ cos θ cos θ+µ sin θ = 2.95 m/s 6. Nmax = Mg cos θ−µ sin θ = 138.1 N Nmin = Mg cos θ+µ sin θ = 122.3 N.
Soluzione dell’esercizio 3.19 1. Lungo il piano: T −Mg sin θ −µSMg cos θ = Ma Per muovere il corpo, a ≥0, da cui T = Mg sin θ + µSMg cos θ = 58.5 N.
Durante il moto: Ma = T −Mg sin θ −µDMg cos θ, da cui a = g(µS −µD) cos θ Cinematica: ∆x = v2 2a, quindi µS = µD + v2 2g cos θ∆x = 0.8 Oppure calcolo il lavoro e l’energia cinetica: L = Mg(µS −µD)∆x = 1/2Mv2 con identico risultato.
2. Lavoro positivo e solo quello della tensione: LT = T∆x = 117 N.
La forza d’attrito dissipa LA = µDMg cos θ∆x = 34 J. Parte del lavoro speso serve per alzare la quota del corpo Lg = Mg∆x sin θ = 49 J. Infine, il corpo ha una energia cinetica finale EK = 1/2Mv2 = 34 J.
Il bilancio completo e: LT = LA + Lg + EK.
Attenzione ai segni: la tensione fa lavoro positivo sul sistema, mentre la forza peso e la forza d’attrito fanno un lavoro negativo. Con questi segni, il bilancio energerito e in generale: P L = ∆Ek.
Nel nostro caso ho spostato al secondo membro (cambiando di segno) i lavori negativi fatti sul sistema. In questo modo e pi u evidente quali forze forniscono, tramite il loro lavoro, energia al sistema e quali ne sottraggono.
3. 0 = Ma = T −(M + m)g sin θ −µD(M + m)g cos θ = Mg sin θ + µSMg cos θ − (M + m)g sin θ −µD(M + m)g cos θ, da cui m = M(µS−µD) cos θ sin θ+µD cos θ = 2.05 kg Soluzione dell’esercizio 3.20 1. La tavola si muove se la forza di attrito del corpo con la tavola e maggiore di quella (statica) della tavola con il piano.
µ1mg ≥µS(m1 + m2)g che e verificata nel nostro caso.
S. Lacaprara Es. Meccanica e Termodinamica 37 2. a1 = −µ1g, a2 = +µ1g−µD(m1+m2)g m2 , accelerazione relativa del corpo rispetto alla tavola: a1−2 = a1 −a2 Istante di arresto t⋆= v1 a1−2, s1−2 = v1t⋆+ 1 2a1−2t⋆2 = 0.55 m 3. s2 = a2t⋆2 = 0.30 m 4. WD = µ1m1gs1−2 + µD(m1 + m2)gs2 = 7.2 J Da notare che il primo termine dell’enegia dissipata ha la forma F · ∆s, dove ∆s e lo spazio percorso da m1 rispetto a m2. Possiamo arrivare a questa formulazione in modo pi u chiaro se lavoriamo completamente nel sistema di riferimento del laboratorio. In tale sistema, se voglio calcolare il lavoro della forza d’attrito tra m1 e m2, devo fare la somma del lavoro fatto su m1 (che e negativo) e quello fatto su m2, che invece e’ positivo.
Il primo vale L1 = −µ1m1g(s1), dove s1 e lo spazio percorso da m1 prima di fermarsi nel sistema del laboratiorio: s1 = s(1−2)+s2, quindi L1 = −µ1m1g(s2−1+ s2) Il secondo vale L2 = +µ1m1g(s2) e la somma dei due fornisce L1+2 = −µ1m1gs1−2 = −WD. Per convincersi, vi invito a ragionare su quello che succede se m2 e bloccata nel sistema del laboratorio.
Si pu o anche calcolare WD = ∆EK, ma occorre fare attenzione che all’istante t⋆ la tavola si sta ancora muovendo, quindi l’energia cinetica finale non e nulla.
Soluzione dell’esercizio 3.21 1. Scrivo le equazioni della dinamica per i due corpi. Prendo un riferimento cartesiano inerziale (laboratorio) xy, e arbitrariamente il verso delle forze di attrito statico come se il corpo m1 scendesse, quindi F att 1 diretta verso l’alto, e F att 2 diretta verso sinistra nel disegno (“all’indietro”).
L’accelerazione di m1 ha in generale due componenti: verticale (se cade o risa-le, nulla se ferma) e orizzontale dovuta al moto del carrello (accelerazione a): scompongo quindi le eq. lungo ˆ x e ˆ y per m1 m1aˆ x = m1a =N1 m1aˆ y = 0 = −m1g + T + F att 1 m2a2 = m2a =T −F att 2 dove N1 e la reazione vincolare del piano verticale su m1, cui e dovuta la forza di attrito tra il piano verticale e m1. Si noti che F att sono incognite, sappiamo solo che |F att 1 | ≤µ1N, e |F att 2 | ≤µ2m2g.
Dalle eqauzioni del moto mi ricavo: |F att 1 | = |m1g −T| ≤µ1m1a |F att 2 | = |T −m2a| ≤µ2m2g da cui ricavo, eliminando la tensione T sommando a membro a membro le due equazioni precedenti.
−µ1m1a −µ2m2g ≤m1g −m2a ≤µ1m1a + µ2m2g S. Lacaprara Es. Meccanica e Termodinamica 38 Dalle due disequazioni ricavo l’accelerazione minima e: a ≥(m1−µ2m2)g (m2+µ1m1) e quella massima e: a ≤(m2+µ1m1)g (m2−µ1m1) 2. Si trova che a ≤ g µ1 oppure a ≥−g µ1 . Il secondo caso corrisponde al caso fisico in cui i corpi vengono accelerati verso sinistra (nel disegno), e quindi e fisico solo se il vincolo tra il piano verticale e M1 e bilaterale (per esempio se c’ e una rotaia con un aggancio di qualche tipo). In tal caso la reazione vincolare N1 puo essere diretta anche verso sinistra e causare una forza di attrito statico che pu o binalciare la forza peso.
Se il vincolo e unilaterale (piano verticale senza rotaia), allora non e un caso realizzabile fisicamente.
Il case M2 ≫M1 e pi u semplice, e fornisce semplicemente |a| ≤µ2g.
Se la massa appesa e trascurabile, lo e anche la tensione, quindi sul corpo M2 agisce orizzontalmente solo la forza F 2 att ≤µ2M2g, che deve controbilanciare la forza apparente M2a, da cui si ricava direttamente il limite su a.
Soluzione dell’esercizio 3.22 1. Lungo la direzione tangente: maT = −kv, quindi m dv dt = −kv. La soluzione di questa equazione differenziale e v(t) = v0e−k/mt.
A me interessa non la legge oraria, ma la velocit a in funzione dell’angolo, che posso ricavare da: mdv = −kvdt = −kds = −kRdθ, da cui v(θ) = v0 −kR/m∆θ, con v0 = 2πR/T. La velocita si annulla quando 2πR T = kR∆θ m , da cui ricavo k m = 2π Tθ2 = 0.25 s−1.
Se si ricava la legge oraria, si trova v(t) = v0e−kt/m, e θ(t) = mv0 kR 1 −e−kt/m .
Quindi la velocit a si annulla in un tempo infinito, ma in questo tempo l’ango-lo percorso e finito θ = mv0 kR . Quindi la soluzione del problema e tecnicamente corretta, anche se il problema stesso non e molto fisico.
2. L’accelerazione totale e data dalla compomente tangeziale e da quella centripeta: ⃗ a = −k mvˆ u∥+ v2 R ˆ u⊥= 0.07ˆ u∥+ 0.098ˆ u⊥ Soluzione dell’esercizio 3.23 1. prendo un riferimento parallelo ∥e ortogonale ⊥al piano inclinato.
m : ma∥= mg sin α −kx sin β 0 = −mg cos α + kx cos β M : Ma∥= Mg sin α + kx sin β −µN 0 = −Mg sin α −kx sin β + N Da cui ricavo a∥= g (sin α −µ sin β) = 1.70 m/s2, tan β = g sin α−a∥ g cos α = 0.18, e x = mg cos α k cos β = 2.5 cm.
Soluzione dell’esercizio 3.24 S. Lacaprara Es. Meccanica e Termodinamica 39 1. k = M1g ∆x0 = 7.5 N/m.
T = M2ω2R M1g −T = k∆x1 T = 0.49 N, ∆x1 = M1g−T k = ∆x0 −M2ω2R k = 13 cm 2. µDM2gd = 1 2M2ω2R2 + 1 2k (x2 1 −x2 2) + M1g (∆x2 −∆x1), quindi µD = 0.02 3. Fatt = M1g −k∆x2 = 0.21 N, 4. quindi µS ≥M1g−k∆x2 M2g = k(∆x0−∆x2) M2g = 0.12 Soluzione dell’esercizio 3.25 TODO Soluzione dell’esercizio 3.26 1. Esempio numerico: T = T0e−µ∆θ T0/T µ = 0.2 µ = 0.5 1 giro 3.5 23 2 giri 12.3 535 3 giri 43 1.2·104 4 giri 152 2.8·105 5 giri 535 6.6·106 Per tenere sollevata una (M ∼180 ton) mi basta avvolgere per ≈3.5 giri una fune su un palo con µ = 0.5, applicando una forza di 1 kgp (che potrebbe essere il peso della corda stessa).
S. Lacaprara Es. Meccanica e Termodinamica 40 3.2 Dinamica del punto materiale: energia Esercizio 3.27 Nel sistema in figura un carrello di massa m = 0.25 kg e appoggiato ad una molla (k = 2000 N/m) compressa con ∆x = 5 cm. La molla viene rilasciata, e il carrello percorre un piano orizzontale lungo d = 1.0 m, e quindi uno inclinato di un angolo α = 30◦.
d l α Calcolare il percorso lungo il piano inclinato: 1. se non c’ e attrito; 2. se c’e attrito nel piano orizzontale con µD = 0.25; 3. se c’ e attrito anche nel piano inclinato con µD = 0.25; 4. la compressione massima della molla ∆x′ quando il carrello torna indietro.
Esercizio 3.28 L’asta in figura e inclinata di 45◦rispetto alla verticale (l = 2 m): lungo di essa scorre senza attrito una massa m = 0.8 kg, che e collegata tramite una molla ideale (k = 12 N/m, l0 = 0 m) all’origine degli assi. La massa e ferma quando si trova all’estremit a superiore dell’asta.
Si chiede: m k α ⃗ g ω l l 1. la posizione di equilibrio; 2. il periodo delle piccole oscillazioni; 3. la posizione per cui ha velocita massima; 4. la velocit a all’estremita inferiore dell’asta; 5. che succede se il sistema ruota con ω costante attorno all’asse verticale?
Esercizio 3.29 S. Lacaprara Es. Meccanica e Termodinamica 41 Un’asta lunga L e incernierata ad una estremita, e ruota con ω costante attorno ad un asse. Lungo l’asta, un corpo di massa m e collegato alle due estremita dell’asta stessa da due molle identiche (k, l0).
Determinare: 1. la posizione di equilibrio del corpo; 2. le condizioni per cui l’equilibrio e stabile; 3. la frequenza delle piccole oscillazioni; 4. la reazione vincolare dell’asta sul corpo.
ω x L θ Esercizio 3.30 Il carrello (M = 1.2 kg, h = 12 cm α = 30◦) in figura e spinto verso il basso da un’asta di massa m = 0.5 kg e da una molla (k = 35 N/m,l0 = L), vincolata a muoversi lungo la direzione verticale.
La posizione iniziale del carrello e con l’asta in corrispondenza del bordo piu alto, con carrello fermo.
Determinare, supponendo trascurabili gli attriti: h L M α m 1. la velocit a finale del carrello; 2. la reazione del piano inclinato del carrello quando l’asta e a met a del carrello; 3. la reazione del piano orizzontale su cui e appoggiato il carrello nello stesso istante.
Esercizio 3.31 Un paracadutista di M = 90 kg (compresa l’attrezzatura) si lancia da h = 3000 m.
Ad una quota di h1 = 1000 m apre un paracadute emisferico, A = 72 m2, coefficiente di resistenza Cx = 1.28.
Si assuma: ρaria = 1.2 kg/m3, in caduta libera Cx = 1.1 e sezione trasversale uomo in caduta libera Au ≈0.5 m2.
Determinare: 1. la velocit a quando apre il paracadute; 2. la velocita quando tocca terra; 3. la tensione massima delle funi del paracadute.
Esercizio 3.32 S. Lacaprara Es. Meccanica e Termodinamica 42 Un carrello di massa 135 000 lb e collegato tramite un cavo ideale e una carrucola ad un paracadute emisferico di diametro 34 m. Il carrello e spinto da 4 razzi in grado di fornire una spinta di F = 3.5 · 104 lbf per 10s.
Si chiede: 1. la velocit a limite del sistema; 2. la velocita di discesa del paracadute dopo 10s; 3. la massima tensione del cavo; Fonte: Rocket Sled Parachute Design Verification, NASA Esercizio 3.33 Moto armonico con attrito radente Un corpo di massa m e collegata ad una molla di massa trascurabile e lunghezza a riposo nulla, fissata ad una estremita. Il corpo e libero di muoversi su un piano, dove e presente un coefficente di attrito dinamico µ, e pu o passare da una parte all’altra del punto ove e fissata la molla.
All’istante t = 0s, il corpo e fermo ad una distanza L dal fulcro della molla.
Determinare: 1. l’equazione del moto del corpo; 2. l’energia dissipata dalla forza d’attrito; 3. devo preoccuparmi della forza di attrito statico?
Esercizio 3.34 Una catena lunga L e massa complessiva M, e appoggiata su un piano privo di attrito, in modo tale che una sua parte lunga x0 si trovi su un piano inclinato di un angolo α.
Inizialmente la catena e ferma.
x0 L α Determinare: 1. l’equazione del moto della catena; 2. la velocita della catena nell’istante in cui si trova completamente sul piano incli-nato.
Esercizio 3.35 S. Lacaprara Es. Meccanica e Termodinamica 43 Due masse M1 = 1.2 kg e M2 = 3 kg sono collegate da una fune ideale, la seconda appoggiata ad un piano orizzontale, la prima ad una distanza L = 0.95 m e alla stessa altezza di un chiodo fisso. M1 viene lasciata cadere da ferma, e la fune incontra un secondo chiodo ad un angolo β = 60◦, che si trova ad una distanza d = 67.8 cm dal primo chiodo.
O C A B M1 M2 β φ d L Calcolare: 1. la velocit a di M1 appena prima di incontrare il secondo chiodo (A); 2. la tensione della fune nello stesso istante; 3. la velocita angolare di M1 prima e dopo l’incontro con C; 4. l’angolo ϕ per cui la massa M2 si solleva.
Esercizio 3.36 Due punti materiali si trovano su un piano inclinato con θ = 10◦: hanno massa M1 = 5 kg (a monte) e M2 = 2 kg (a valle) e sono collegato da una molla (k = 10 N/m) a riposo. M1 ha coeff. di attrito statico µs0.3 ed e fermo, mentre M2 non ha attrito e scende con v0 = 0.5 m/s.
M2 M1 µs ℓ θ v0 Calcolare, nell’istante iniziale: 1. accelerazione di M2; 2. forza di attrito si M1; 3. il centro delle oscillazioni di M2 e il periodo; 4. il max/min allungamento della molla; 5. l’allungamento max/min della molla perche M1 resti fermo; 6. la velocit a di M2 e tale da far muovere M1?
Esercizio 3.37 Due masse M1 = 2.0 kg e M2 = 4.0 kg sono collegate da una molla con costante elastica k = 50 N/m disposta verticalmente.
Inizialmente M2 e appoggiata al pavimento, mentre M1 si trova appoggiata sopra di essa, entrambe a riposo. M1 e vincolata a muoversi solo in verticale.
M1 M2 y ⃗ F Determinare: 1. la compressione della molla e reazione vincolare sul piano; Successivamente, la massa M1 viene tirata verso l’alto da una forza costante ⃗ F = 15 N. Calcolare: 2. accelerazione iniziale di M1; 3. la velocit a di M1 quando la molla e a riposo; 4. il massimo allungamento della molla e accelerazione di M1 in quell’istante; S. Lacaprara Es. Meccanica e Termodinamica 44 5. discutere le condizione su M2 perch e resti appoggiata.
Esercizio 3.38 Un filo ideale l0 = 0.66 m e fissato in un pun-to ed e collegato all’alta estreminta ad una molla k = 150 N/m e lunghezza a riposo nulla. Alla molla e fissata una massa m = 25 kg che si muove senza attrito su un piano orizzontale. Inizialmente la mas-sa ha distanza r0 = 1.33 m dal centro e si muove ortogonalmente al raggio.
1. ω0 per avere moto circolare stabile Il moto inizia alla stessa distanza, ma con ω1 = 4 rad/s, sempre con velocita ortogonale al raggio.
2. v2 quando r2 = 2.0 m; 3. ω2 nello stesso istante.
S. Lacaprara Es. Meccanica e Termodinamica 45 Soluzione dell’esercizio 3.27 1. Conservazione energia: 1 2k∆x2 = mgL sin α. L = 2.0 m 2.
1 2k∆x2 −mgµDd = mgL sin α. L = 1.5 m 3.
1 2k∆x2 −mgµDd −mgµDL cos α = mgL sin α−. L = 1.07 m 4. Efin = 2L(Fatt) + Eini, ∆x′ = r ∆x2 −4dmgµD(d+L cos α) k = 1.2 cm.
Devo considerare che al ritorno il mio corpo percorre 1.4 cm in meno che all’andata e quindi che l’energia dissipata al ritorno e minore di quella dell’andata? Discutere.
Soluzione dell’esercizio 3.28 1. Si puo calcolare l’equazione del moto. Prendo come coordinata per descrivere il sistema la posizione della massa lungo l’asta, a partire dal punto medio, diretta verso l’alto, e la chiamo s.
Quindi s = l/2 se la massa si trova all’estremit a superiore dell’asta (asse di rotazione per l’ultimo punto) e s = −l/2 per il punto 4.
Le forze che agiscono sono quella peso, e quella elastica, che sono conservative, e la razione vincolare dell’asta che non compie lavoro.
Se considero le componenti delle due forze lungo s, quindi parallele all’asta, ho per la forza peso: F pesoˆ s = −mg/ √ 2.
Per quella elastica chiamo ϕ l’angolo che la molla forma con la bisettrice del primo quadrante (che corrisponde alla posizione s = 0 della massa). La forza elastica e pari a F el = ks/ sin ϕ, diretta verso il centro del sistema xy di assi coordinati. La sua componente lungo l’asse dell’asta, e quindi lungo s e pari a F el sin ϕ, quindi pari a F elˆ s = −ks. Di fatto e come se ci fosse una molla con la stessa costante elastica k tra la massa m e il punto centrale dell’asta s = 0.
Fatte queste considerazioni, F risˆ s = maˆ s, F peso + F el = m¨ s, da cui posso ricavare l’eq differenziale del moto: ¨ s + k ms + g √ 2 = 0 che ha come soluzione, posto ω2 = k m: s(t) = l −g/ω2 √ 2 cos(ωt) − g √ 2ω2 quindi la poszione di equilibrio risulta: s0 = − g √ 2ω2 = −0.46 m.
Visto che il sistema e conservativo, posso anche cercare i minimi dell’energia potenziale: U(s) = mgs √ 2 + k 2 s2 + l2 2 che ha minimi come prima. Da notare che per l’en potenziale elastica uso tutta la lunghezza della molla lmolla = q s2 + l2 2 , e, dato che nell’energia potenziale della molla compare il quadrato di lmolla: Uel = 1/2kl2 molla, c’e una parte costante e una che dipende da s2, come se ci fosse solo la molla lungo l’asse.
S. Lacaprara Es. Meccanica e Termodinamica 46 2. T = 2π ω = 1.62 s; 3. e la posizione di equilibrio: la velocita massima risulta: v = q 2gl + kl2 2m + mg2 2k = 8.5 m/s 4. Conservazione dell’energia: mgl = 1/2mv2, quindi v = √2gl = 6.26 m/s 5. Mi metto in un sistema di riferimento non inerziale solidale con l’asta. Il problema e esattamente quello di prima a patto che consideri la presenza delle forze non iner-ziali, in questo caso quella cosiddetta centrifuga. Il sistema e ancora conservativo.
Attenzione che la conservazione dell’energia totale nei sistemi non inerziali conti-nua a valere a patto che si consideri il lavoro, o l’energia potenziale, se possibile, delle forze non conservative. Nel nostro caso abbiamo la forza non inerziale che e quella centrifuga, che e conservativa in quanto centrale.
La sua energia potenziale la posso calcolare esplicitamente calcolado il lavoro della forza centrifuga per spostare una massa m da raggio r1 a raggio r2: −∆U = L = Z r2 r1 ⃗ Fcentrifugad⃗ r = Z r2 r1 ω2rdr = ω2r2/2 r2 r1 da cui U(r)centripeta = −ω2r2 2 (da notare il segno −). Di fatto e molto simile a quella potenziale elastica, ma con il segno cambiato, il che e ragionevole, visto che si tratta di una forza che dipende linearmente dalla distanza dal centro (come quella elastica), ma con verso opposto.
l’energia potenziale totale diventa, tenendo conto che quella centripeta dipende dalla distanza al quadrato dall’asse di rotazione, U(s) = mgs √ 2 + ks2 2 −mω2 2 l − √ 2s 2 !2 che ha come minimi: s = 2 √ 2m g + ω2l 2 mω2 2 −k .
Se ω2 = 2k m , non ci sono punti stazionari, se ω e piu elevato, ci sono massimi ma per valori di s non fisici, prima dell’asse di rotazione del sistema.
Soluzione dell’esercizio 3.29 1. La forza elastica delle due molle risulta Fel = k(l0−x)−k(l0−(L−x)) = k(L−2x) Ho usato il fatto che la lunghezza totale delle due molle e costante e pari a L, e che quindi la compressione di una e pari all’estensione dell’altra. Quindi, chiamando x la distanza della massa dal punto in cui l’asta e incernierata, la forza della prima molla e pari a F1 = −k(x −l0), e analogamente quella della seconda e pari a F2 = −k(l0 −(L −x)). Come atteso, per un dato valore di x, le due forze sono opposte (una molla compressa e una estesa). Visto che le due molle sono uguali, le lunghezze a riposo si elidono e la forza totale non dipende da l0.
Mi metto in un sistema di riferimento rotante, per cui devo tenere conto anche delle forze non inerziali, ovvero di quella centrifuga mω2r diretta verso l’esterno.
S. Lacaprara Es. Meccanica e Termodinamica 47 Se mi fossi messo in un sistema inerziale, dovevo tenere conto che la mia massa fa un moto circolare e quindi la sua accelerazione sarebbe stata pari a a = ω2r diretta verso l’interno. Quindi nel sistema non inerziale: Ftot +Fcentripeta = Ftot −mω2r = ma = 0, in quello inerziale Ftot = ma = mω2r, e, ovviamente, l’equazione del moto e la stessa.
L’equazione del moto lungo l’asta risulta: m¨ s = Fel −mg cos θ + mω2x sin2 θ Quindi xeq = mg cos θ−kL mω2 sin2 θ−2k 2. Calcolo l’energia potenziale totale (elastica e gravitazionale): U = 1 2k(x −l0)2 + 1 2k(L −x −l0)2 + mgx cos θ −1 2mω2(x sin θ)2 .
Potevo usare direttamente l’en potenziale per calcolare il punto di equilibrio, cer-cando il minimo. Il vantaggio di usare l’energia potenziale e che posso studiare se il punto di equilibrio e stabile (minimo) o instabile (massimo).
La derivata seconda di d2U/dx2 = 2k −mω2 sin2 θ, e quindi si ha un minimo in xeq se k ≥1 2mω2 sin2 θ.
3. Ho ricavato l’eq del moto lungo l’asta, che e di tipo armonico: ¨ s+ 2k m + ω2 sin2 θ x− kL + Mg cos θ, quindi ω′2 = 2k m + ω2 sin2 θ e T = 2π ω′ Quindi il moto e’ armonico non solo per le piccole oscillazioni, ma per oscillazioni qualsiasi.
4. Deve essere pari alla somma delle componenti delle forze perpendicolari all’a-sta, nel nostro caso quella gravitazionale e quella centrifuga. N = Mg sin θ + Mω2x sin θ cos θ Soluzione dell’esercizio 3.30 1. Conservazione energia mgh + 1 2kh2 = 1 2Mv2 c + 1 2mv2 a = 1 2Mv2 c + 1 2m tan α2v2 c v = q kh2+2mgh M+m tan α2 = 1.11 m/s 2. Equazione della dinamica per asta e carrello: R e reazione del piano inclinato sull’asta, Nasta la forza che tiene l’asta verticale, e Npiano la reazione del piano su cui poggia il carrello: m : m¨ y = mg + kδ −R cos α 0 = −R sin α + Nasta M : M ¨ x = R sin α 0 = Mg −Npiano + R cos α ¨ y = tan α¨ x Risolvendo: R = 7.1 N, Npiano = 17.9 N, Nasta = 3.55 N, ¨ x = 2.96 m/s2 Soluzione dell’esercizio 3.31 S. Lacaprara Es. Meccanica e Termodinamica 48 1. A regime, la velocit a limite risulta: vl = q 2Mg CxρA = 48 m/s, equagliando la forza peso Mg con quella di attrito viscoso ⃗ F = −1 2CxρA|v|⃗ v.
Per capire se in 1000m di caduta la velocita limite viene raggiunta, e necessario risolvere l’equazione del moto. Risolvendo v = v(x), dove x e lo spazio percorso in caduta, si ottiene v2(x) = v2 l 1 −e x λ dove λ = v2 l 2g = 120 m. Quindi la velocit a limite si raggiunge in x ≈3−4λ ∼400 m.
2. L’equazione del moto e la stessa, cambia solo il coefficiente della forza di attrito: vl = 4 m/s e viene raggiunta in pochi m.
3. Impulso F∆t = ∆p = 4000 kgm/s.
Il tempo ∆t e quello di apertura del paracadute: se ∆t ≈5s, Fmedia = 800 N.
Soluzione dell’esercizio 3.32 1. La soluzione e del tutto simile a quella dell’esercizio 3.31 vl = 36 m/s; 2. τ = vl g = 4 s, dopo 2.5τ la velocit a e circa 80% di quella limite; 3. T = 4.36 · 104 N; 4. N = mg + 4F sin θ = 6.4 · 105 N.
Soluzione dell’esercizio 3.33 1. Nel primo semiperiodo: x(t) = L −gµ ω2 cos(ωt) + gµ ω2 , con ω2 = k m.
Nel secondo semiperiodo (dopo che il corpo ha raggiunto l’elongazione massima dalla parte opposta a quella di partenza e inizia a tornare indietro): x(t) = L −3gµ ω2 cos(ωt) −gµ ω2 e cos ı via.
Il moto e quindi una successione di moti armonici con pulsazione costante ω, centro in ± gµ ω2, ampiezza che decresce di 2gµ ω2 dopo ogni semiperiodo.
L’inviluppo di tale moto risulta essere un triangolo.
2. Wd = µmg2A, dove A(n) e l’ampiezza dell’oscillazione nel semiperiodo n-esimo.
3. Via via che l’energia viene dissipata, l’ampiezza dell’oscillazione si riduce, e con essa la massima forza di richiamo che esercita la molla all’estremita dell’oscilla-zione. Quando questa forza di richiamo risulta essere minore della forza di attrito statico, il corpo si ferma. La posizione finale del corpo quindi non e al centro del sistema.
Soluzione dell’esercizio 3.34 S. Lacaprara Es. Meccanica e Termodinamica 49 1. M ¨ x = Mg L sin αx, che ha come soluzione, date le condizioni iniziali: x(t) = x0 2 ekt + e−kt , con k = p g/L sin α.
2. ricavando v(t) = dx(t) dt dalla equazione del moto, si ottiene: v(t⋆) = q g sin α L (L2 −x2 0) Oppure piu semplicemente, dalla conservazione dell’energia si ottiene lo stesso risultato: M L x0g(−1 2x0 sin α) = Mg(−1 2L sin α) + 1 2Mv2 Soluzione dell’esercizio 3.35 1. v = √2gL cos β = 3.05 m/s 2. m1 v2 0 L = −m1g cos β + T, quindi T = m1 v2 0 L + g cos β = 17.6 N 3. ω = v0 L = 4.73 rad/s, ω′ = v0 L−d = 11.2 rad/s 4. Uso eq. dinamica e conservazione dell’energia: T −mg cos ϕ = m v2 1 L −d 1 2mv2 0 −mg(L −d) cos β = 1 2mv2 1 −mg(L −d) cos ϕ Ricavo T(ϕ) dalle equazione di sopra, e richiedo che T ≥m2g, che fornisce: cos ϕ ≥ m2 3m1 −2d cos β 3(L−d), quindi ϕ ≤46◦.
Soluzione dell’esercizio 3.36 1. La molla e a riposo, a2 = g sin θ = 2.0 m/s2 2. La molla e a riposo, Fatt = M1g sin θ = 8.5 N ≤µsM1g cos θ = 14.5 N 3. uso la conservazione dell’energia: 1/2M2v2 0 = 1/2kx2 ∓M2gx sin θ, da cui xM/m = 0.75, −0.67 m, con ⃗ x diretto verso il basso; 4. per esempio scrivo l’energia potenziale totale e trovo il minimo. U(x) = 1/2kx2 − M2gx sin θ, che ha minimo per x = M2g sin θ k = 34 cm. Per il periodo scrivo l’equa-zione differenziale per la posizione ¨ x + k/M2x −g sin θ = 0, e quindi il periodo e T = 2π/ω = 2πp m k = 2.8 s. Da notare che non e vero che la pulsazione e sempre p k/m per tutti i moti armonici, ma dipende dalla geometria del caso specifico, quindi e bene scrivere l’equazione differenziale armonica e ricavarsi la pulsazione da quella.
5. se la molla si allunga: Fatt = M1g sin θ + kxmax ≤µsM1g cos θ, xmax = 60 cm Se si accorcia, cambia il verso della forza elastica, e anche della forza d’attrito statico: xmin = −2.3 m 6. La velocit a minima necessaria per spostare M1 si realizza quando M2 riesce ad allungare/accorciare la molla come al punto precedente, arrivando ferma. Se M2 scende: 1/2M2v2 m = 1/2kx2 max + M2gxmax sin θ, con x = 0.6 da cui: v ≥1.96 m/s.
Se invece sale, (x = 2.3) v ≥4.31 m/s.
S. Lacaprara Es. Meccanica e Termodinamica 50 Soluzione dell’esercizio 3.37 1. In condizioni statiche, la forza della molla sulla massa M1 deve essere pari alla forza peso sulla stessa massa, quindi ∆xs = M1g k = 0.39 m.
Sulla massa M2, la reazione vincolare e pari alla somma vettoriale della forza peso e della forza elastica, che hanno verso concorde. N2 = (M2g+k∆x) = (M1+M2)g = 58.8 N.
2. M1a1 = F −M1g + k∆x = F, quindi a1 = F M1 = 7.5 m/s2 3. Uso conservazione dell’energia, usando come quota di riferimento per l’energia potenziale gravitazionale la quota del corpo M1 nell’istante iniziale: Etot ini =0 + 1/2k∆x2 + 0 Etot fin =1/2M1v2 1 + 0 + M1g∆x Etot fin =Etot ini + LF = Etot ini + F∆x da cui ricavo: v1 = 1.43 m/s 4. uso sempre la conservazione dell’energia, tra la posizione in cui la molla e a riposo e il punto di massimo allungamento, e usando come quota di riferimento quella in cui la molla e a riposo: Etot ini =1/2M1v2 1 + 0 + 0 Etot fin =0 + 1/2k∆x2 max + M1g∆xmax Etot fin =Etot ini + LF = Etot ini + F∆xmax da cui ricavo ∆xmax = 0.21 m. L’altra soluzione e negativa e non fisica.
Visto che il moto e di tipo armonico (abbiamo una molla e forze costanti), l’ac-celerazione nel punto di massima estensione e pari, a meno del segno, a quella del punto di massima compressione della molla, che e l’accelerazione calcolata al punto 2. Quindi a(∆xmax) = −a1 = −7.5 m/s2 5. Quando la molla ha la massima estensione, la forza elastica sulla massa M2 e massima. Quindi basta verificare che in quella condizione M2 rimane appoggiata al piano. La reazione vincolare risulta: N2 = M2g−k∆xmax. La condizione perche M2 non si sollevi e quindi: M2 ≥k∆xmax g = 1.06 kg oppure ∆xmax ≤M2g k = 0.78 m, ed e verificata.
Soluzione dell’esercizio 3.38 1. Forza elastica mi d´ a l’accelerazione centripeta: −k(r0 −l0) = −mω2 0r0, da cui: ω0 = q k(r0−l0) mr0 = 1.73 rad/s 2. Conservazione dell’energia: Eini tot = 1 2m(ω0r1)2 + 1 2k(r0 −l0)2 = 389 J Efin tot = 1 2mv2 2 + 1 2k(r2 −l0)2 = Eini tot , da cui v2 = 4.52 m/s. Attenzione che non conosco la direzione di ⃗ v2, quello che ho trovato e il modulo e non so se sia tangente, radiale o entrambe.
S. Lacaprara Es. Meccanica e Termodinamica 51 3. Conservazione momento angolare rispetto a O. LO = mr2 1ω1 = mr2 2ω2, quindi ω2 = r2 1 r2 2 ω1 = 1.78 rad/s. Volendo ricavo vT 2 = ω2r2 = 3.56 m/s, e quindi capisco che ho anche una componente radiale vR 2 = 2.78 m/s S. Lacaprara Es. Meccanica e Termodinamica 52 3.3 Dinamica del punto materiale: sistemi non inerziali Esercizio 3.39 Una piattaforma di massa Mp = 100 kg si muove su e gi´ u di moto armonico con frequenza ν = 5 Hz, e ampiezza A0 = 5 cm. All’istante t = 0 s si trova nel punto piu basso. Sulla piattaforma e appoggiato un corpo di massa m = 1 kg.
Determinare: 1. la posizione della piattaforma quando il corpo appoggiato sopra si stacca; 2. la velocita del corpo m all’istante del distacco; 3. l’altezza massima che raggiunge il corpo dopo il distacco; 4. il lavoro totale L fatto dalle forze che agiscono sulla piattaforma tra l’istante t = 0 s e quello del distacco di m 5. il lavoro della forza che fa oscillare la piattaforma tra gli stessi istanti.
Esercizio 3.40 Un punto materiale di massa m e posto su una piano orizzontale scabro con µS = 0.3.
Il piano ruota attorno ad un asse ad una distanza R = 20 cm dal punto, con una accelerazione angolare α = 1 rad/s2, partendo da fermo.
Si chiede: 1. l’istante t1 quando il punto materiale inizia a muoversi rispetto al piano, se fosse vincolato a muoversi solo lungo un raggio; 2. idem (t2), se vincolato lungo una circonferenza di raggio R; 3. idem (t3), se fosse non vincolato; 4. descrivere qualitativamente il tipo di moto cui e sottoposto dopo tale istante, nell’ultimo caso.
Esercizio 3.41 Una piattaforma di raggio R = 5 m ruota con ω = 3 giri/minuto attorno al suo asse.
Un punto materiale parte da fermo dal centro con accelerazione a′ costante (rispetto alla piattaforma) e arriva sul bordo in t = 3 s Calcolare, nel sistema di riferimento inerziale del laboratorio: 1. ⃗ vf; 2. ⃗ a1(t1 = 1.5 s) Esercizio 3.42 (M.S.V 3.11) Un disco di raggio R ruota senza attrito su un piano orizzontale con velocit a angolare ω attorno al suo asse. Un corpo di massa m e collegato al centro del disco da una molla di costante elastica k e lunghezza a riposo l0 = 0, vincolato a muoversi lungo un diametro da una guida. Si osserva che quando il corpo dista R/2 dal centro del disco, esso ha velocit a radiale nulla.
S. Lacaprara Es. Meccanica e Termodinamica 53 ω m k Calcolare, nell’ipotesi che il corpo rimanga dentro il raggio del disco: 1. l’equazione del moto; 2. la reazione ⃗ N del vincolo quando il corpo si trova al centro del disco; 3. la velocita ⃗ v di uscita dal disco se l’ipotesi iniziale non e soddisfatta.
Esercizio 3.43 Un uomo di massa M = 70 kg si trova dentro un treno che accelera con accelerazione costante da vi = 10 km/h a vf = 200 km/h in ∆t = 5 min Calcolare: 1. la forza ⃗ F1 che subisce all’istante t1 = 1.0 min 2. la forza ⃗ F ′ 1 che sente nel sistema di riferimento solidale al treno.
All’istante t2 = 3 min il treno percorre con velocita costante vf = 200 km/h una curva con raggio di curvatura r = 200 m e l’uomo lascia cadere un oggetto di massa m = 4.0 g 3. la forza ⃗ F2 nel sistema di riferimento inerziale; 4. la forza ⃗ F ′ 2 nel sistema di riferimento solidale al treno; 5. la forza ⃗ Fo sull’oggetto lasciato cadere nel riferimento inerziale.
S. Lacaprara Es. Meccanica e Termodinamica 54 Soluzione dell’esercizio 3.39 1. Il moto della piattaforma e x(t) = −A0 cos(2πνt).
Le forze su m sono: −mg + N −map = ma dove ap e l’accelerazione della piat-taforma rispetto ad un sistema inerziale (laboratorio) e a e quella del corpo m rispetto alla piattaforma.
Il distacco avviene quando N = 0, quindi quando ap = ¨ x(t) = −g. Si ottiene x(t) = +1 cm.
2. v(t) = ˙ x(t) = 1.54 s; La posso calcolare con l’eq del moto o, meglio, con la legge delle forze vive Uk = 1/2mv2 = LRis = R xdistacco −A0 Frisds = R mads = R −mω2xdx = mω2/2(A2 0 −x2 d) 3. conservazione energia Uk = mg∆h, ∆h = 1.2 m 4. L = R x(t) −A0 ⃗ Ftotd⃗ s = (Mp + m) R x(t) −A0 apds = −(Mp + m)(2πν)2 R x(t) −A0 xdx = 120 J.
Oppure, L = ∆EK = 1 2(Mp + m)v2(t) = 120 J.
Soluzione dell’esercizio 3.40 1. Nel RF’ non inerziale, ho forza centrifuga Fc = mac = mω2R = mRα2t2.
Visto che il moto e circolare accelerato, ho anche una accelerazione tangenziale di RF’ rispetto al laboratorio, quindi una seconda forza non inerziale: FT = maT = mRα.
La forza inerziale totale e quindi ⃗ F = mRα2t2ˆ ur+mRαˆ uT. Quando il |F| ≥µsmg, il corpo comincia a muoversi.
In questo caso conta solo la forza inerziale radiale, quindi mRα2t2 ≥µSmg, t ≥ p µSg α2R = 3.8 s 2. conta solo la componente tangenziale: mRα ≥µSmg, che non e mai vera, quindi il corpo in questo caso sta fermo; 3. devo considerare entrambe le componenti: |F| = Rα p (αt2)2 + 1 ≥µSmg, il che avviene per t ≥ r 1 α q µSg Rα 2 −1 = 3.8 s.
Da notare che per (αt2)2 >> 1, cio´ e non appena t ´ e maggiore di t = 1 s, la com-ponente radiale della forza inerziale domina rispetto a quella tangenziale, quindi non e una sorpresa che t1 = t3.
Soluzione dell’esercizio 3.41 1. Nel RF’ non inerziale solidale con la piattaforma: r(t) = 1/2a′t2, quindi a′ = 1.11 m/s2. ω = 3 · 2π/60 = 0.31 rad/s. ⃗ v′ f = ⃗ a′t = 3.33ˆ ur m/s.
Nel RF del laboratorio: ⃗ vf = ⃗ v′f + ⃗ ω × ⃗ r = 3.33ˆ ur + 1.5ˆ uT m/s 2. Per r1(t1) = 1/2a′t2 1, v′ 1 = a′t1.
⃗ a = ⃗ a0 + ⃗ a′ + d⃗ ω dt × ⃗ r1 + ⃗ ω × (⃗ ω × ⃗ r1) + 2(⃗ ω × ⃗ v′ 1) nel nostro caso: ⃗ a = (a′ −ω2r1)ˆ ur + 2ωv′ 1ˆ uT = 0.99ˆ ur + 1.04ˆ uT m/s2 Soluzione dell’esercizio 3.42 S. Lacaprara Es. Meccanica e Termodinamica 55 1. Nel RF’ non inerziale solidale con il disco: Fr = ma′ r = +mω2r −kr, quindi l’equazione del moto e in generale: ¨ r −r ω2 −k m = 0 La soluzione e limitata, e quindi il corpo rimane dentro il disco, se ω2 −k m ≥0, ossia se la forza risulante e diretta verso l’interno. In tal caso, l’equazione del moto risulta quella di un oscillatore armonico, con pulsazione Ω= q ω2 −k m. Date le condizioni iniziali, e scegliendo t = 0 quando r = R/2 e v = 0, si ottine: r(t) = R/2 cos Ωt 2. Le forze non inerziali sono la forza centrifuga (nulla perch´ e siamo a r = 0) e la forza di Coriol ıs, che vale Fco = 2ωv(r = 0) = ωΩR/2 e questa e anche, in modulo, la reazione del vincolo.
3. Se il moto non e vincolalo, l’equazione della dinamica diventa ¨ r + rΩ′ = 0, con Ω′ = q ω2 −k m.
La soluzione, considerando le condizioni al contorno, e r(t) = r 4 eΩ′t + e−Ω′t .
L’istante in cui il corpo raggiunge r = R, risulta t = ln 2+ √ 3 Ω′ , per cui la velocit a ra-diale di uscita risulta: vr(r = R) = √ 3 2 Ω′R, cui va aggiunta la velocita tangenziale pari a vt(r = R) = ωR.
E’ molto pi u semplice risolvere questo punto, come quello precedente, usando la conservazione dell’energia.
La forza centrifuga e conservativa, dato che e una forza centrale. Quella di Coriolis non compie lavoro perche e sempre ortogonale alla velocita e quindi al moto.
L’energia potenziale centrifuga e U(r) = −mω2r2/2 e si ricava facilmente calco-lando in modo esplicito il lavoro della forza centrifuga L = R r2 r1 ⃗ F · d⃗ s = mω2(r2 2 − r2 1)/2 = −∆U(r).
La conservazione dell’energia fornisce Uk(v = 0) + Uel(R/2) + Ucf(R/2) = Uk(v = v) + Uel(R) + Ucf(R), da cui si ricava v = √ 3 2 RΩ′, cui va sempre aggiuntam vettorialmente, la velocita tangenziale v = ωR.
Soluzione dell’esercizio 3.43 1. Mi metto in un sistema di riferimento inerziale solidale alla stazione. Le forze presenti sono: forza peso, reazione vincolare del pavimento, reazione vincolare orizzontale (attrito statico o schienale della poltrona): ⃗ F1 = −Mgˆ z+M ˆ z+Rˆ x = Maˆ x, a = (vf −vi)/∆t = 0.18 m/s2, R = Ma = 12.6 N.
La forza R orizzontale, esercitata dal treno sul passeggero, e quella che fa accelerare il passeggero con la stessa accelerazione del treno.
2. Nel sistema di riferimento non inerziale solidale al treno, le forze lungo ˆ z sono le stesse del punto precedente, mentre lungo il moto del treno ˆ x, si ha una forza non-inerziale pari a FNI ˆ x = −Maˆ x = −12.6ˆ x N. La risultate di tutte le forze nel SdR del treno e ⃗ F ′ 1 = −Mgˆ z + M ˆ z + Rˆ x + FNI ˆ x = 0, infatti il passeggero e fermo rispetto al treno.
S. Lacaprara Es. Meccanica e Termodinamica 56 3. Oltre alle forze del punto 1, c’e anche una forza centripeta ⃗ Fc = Mv2/rˆ u⊥= M(a · t2 + vi)2/rˆ u⊥= 433ˆ u⊥N, perpendicolare al moto del treno, e diretta verso l’intero della curva, che e sempre fornita dall’interazione del treno con il passeggero (attrito statico con pavimento o reazione della poltrona) 4. Vi e una forza non inerziale radiale: FNI ˆ u⊥= −⃗ Fc. Di nuovo il passeggero non accelera rispetto al treno, quindi la risultante delle forze e nulla.
5. Nel sistema di riferimento inerziale, l’oggetto e sottoposto alla sola forza di gravit a: ⃗ Fo = −mgˆ z Nel sistema di riferimento non inerziale, si hanno anche forze non ineziali: ⃗ F NI o = −m(a + ac) = (+7.2 · 10−4ˆ u∥+ 2.4 · 10−4ˆ u⊥) N Durante la caduta, quando la velocita di caduta e v, l’accelerazione ha anche una componente di Coriolis 2ω × v, dove ω e la velocit a angolare del treno che curva ω = vtreno/r = (a · t + v0)/r, mentre v e la velocit a di caduta dell’oggetto v = g(t −t2).
Tuttavia, la velocia angolare e diretta verticalmente, cosı come la velocit a, quindi l’accelerazione di Coriolis e nulla.
S. Lacaprara Es. Meccanica e Termodinamica 57 4 Gravitazione Esercizio 4.1 La sonda VoyagerII di massa m, e velocit a relavita al sole v = 12 km/s ha usato Giove (massa M ≫m e velocita rispetto al sole V = 13 km/s) come fionda per aumentare la sua velocit a.
Si supponga che VoyagerII sia arrivato con direzione opposta al moto di Giove, e, dopo l’interazione, si allontani in direzione concorde.
1. calcolare la nuova velocita della sonda.
Esempio di effetto “Fionda gravitazionale” per Voyager (I/II) (1977-today). Traiet-toria e velocit a rispetto al sole Missione Cassini-Huygens (1997-2017) Missione BepiColombo (2018-2028) S. Lacaprara Es. Meccanica e Termodinamica 58 BepiColombo on Wikipedia BepiColombo Trajectory animation Esercizio 4.2 Un satellite artificiale e lanciato in modo che ad un certo istante la sua distanza dal centro della terra sia = 7.4 · 106 m e la velocit a v sia perpendicolare ad R.
Calcolare: 1. la massima v per avere orbita chiusa; 2. se v = 0.9vmax, i semiassi della sua orbita 3. il periodo dell’orbita.
Esercizio 4.3 Una astronave si trova su un’orbita circolare a h = 300 km dal suolo terrestre RT = 6.370 · 106m. Accende i motore per un breve istante e aumenta la sua velocita di ∆v = 500 m/s, nella direzione del moto.
Calcolare: 1. vorbitale e periodo T iniziale; 2. la distanza massima dal centro della terra dopo l’impulso (apogeo); 3. idem se l’impulso fosse radiale 4. la velocit a all’apogeo; 5. il ∆V per tornare ad un’orbita circolare all’apogeo 6. il periodo T ′ (ellittica e circolare) 7. il massimo ∆v per continuare ad avere un’orbita chiusa.
Esercizio 4.4 Voglio portare in orbita bassa (LEO) (h = 200 km) un satellite di massa m = 1000 kg, usando un Falcon-9R come vettore (vgas = 3 km/s, Mtot = 480 ton).
L’accensione del razzo dura ∆t = 150 s e l’attrito con l’atmosfera causa un ∆Vatmosferico = 200 m/s. Il decollo avviene da Cape Canaveral (λ = 28.5◦), Rterra = 6370 km. Calcolare: 1. la velocita orbitale che devo raggiungere 2. la variazione di energia potenziale gravitazionale del satellite; 3. la variazione di energia cinetica del satellite.
4. la massa finale del razzo nell’ipotesi se sale un solo stadio per tutto il viaggio; S. Lacaprara Es. Meccanica e Termodinamica 59 5. se sale con due stadi, il primo che porta a MACH 6, e il secondo di massa MII = 80 ton con lo stesso vgas; Esercizio 4.5 Un punto materiale viene lanciato radialmente dalla superficie di un pianeta di massa m2 e raggio r2, con velocit a iniziale v0. Si osserva che si arresta quando r = 2r2.
Un diverso punto materiale e si trova in orbita circolare con raggio r1 = 4.6 r2, attorno ad un secondo pianeta con massa m1.
1. supponendo che la velocit a del secondo punto materiale sia proprio v0, si ricavi il rapporto tra le masse m1/m2 S. Lacaprara Es. Meccanica e Termodinamica 60 Soluzione dell’esercizio 4.1 1. E’ un urto: conservo momento lineare e energia.
mv + MV = mv′ + MV ′ 1 2(mv2 + MV 2) = 1 2(mv′2 + MV ′2) da cui ricavo: v′ = 2MV M+m + v m−M m+M = 2V −v = −38 km/s (facendo l’ipotesi che M ≫m) Soluzione dell’esercizio 4.2 1. Impongo Etot < 0 per avere orbita chiusa, e ottengo v ≤ q 2GM R = q 2R2 T g R = 1.038 · 104 m/s 2. Etot = −GMm 2a da cui a = 1.94 · 107 m.
Da L2 = Gm2M b2 a , b = 1.53 · 107 m 3. da terza legge di Keplero: T = q 2π2a3 GM = 15130 s = 4.2 h.
Soluzione dell’esercizio 4.3 1. v1 = q GM RT +h = 7.75 km/s; T = 5410 s.
2. rapogeo = 8.72 · 106 m, ∆r = 2000 km 3. ∆r′apogeo = 360 km 4. vapogeo = 6.31 km/s 5. vorbitale apogeo = 6.6 km/s 6. da terza legge di Keplero: T = q 2π2a3 GM = 6705 s = 1.86 h. Per orbita circolare 2.3h 7. Impongo Etot ≤0, ∆v ≤v( √ 2 −1) = 3.2 km/s.
Soluzione dell’esercizio 4.4 1. vorb = q GMT RT +h = 7.778 km/s = MACH 22.7 2. ∆Ug = GMm 1 RT − 1 RT +h = 1.902 GJ.
Se l’avessi calcolato con ipotesi di gravita costante: ∆U = mgh = 1.96 GJ Errore 2%.
A h = 200 mg(h) = GMm (RT +h)2 = GM RT 1 −2h RT = g 1 −2h RT = g(1 −6%), quindi quasi uguale a quella sulla terra!
3. ∆Ek = 1 2mv2 orb −1 2mv2 t = 30 GJ, ∆Ek ∼15∆Ug S. Lacaprara Es. Meccanica e Termodinamica 61 4. ∆Vtot = ∆V + ∆Vgravitazionale + ∆Vatm ∆V = vorb −vini = vorb −ωterraRT cos λ = 7780 −407 km/s ∆Vgrav = g∆t = 1470 km/s, quindi ∆Vtot = 9.0 km/s MF = MIe −∆V vgas = 24 ton 5. Se due stadi: ∆VI = (2000 −407) + 700 + 200, ipotizzando che salga fino a h/2 e quindi sia il solo a muoversi in atmosfera. MF = 208 ton, compresa la massa del secondo stadio, quindi 128 + 80 ton.
∆VII = (7780 −2000) + 700, quindi M II F = 9.2 ton Soluzione dell’esercizio 4.5 S. Lacaprara Es. Meccanica e Termodinamica 62 5 Dinamica dei sistemi Esercizio 5.1 Due masse M1 e M2 sono collegate da una molla con costante elastica k e lunghezza a riposo nulla.
Inizialmente M2 e appoggiata al pavimento, mentre M1 si trova appoggiata sopra di essa, entrambe a riposo (disegno in nero).
La massa M1 viene tirata verso l’alto da una forza costante ⃗ F (in rosso) .
Determinare: 1. la minima ⃗ F perche M2 si sollevi dal pavimento; 2. l’equazione del moto di M1,2 dopo tale istante.
m1 m2 y ⃗ F m1 Esercizio 5.2 Due corpi di massa mA = 1 kg e mB = 2 kg si trovano su un piano inclinato α = 30◦, e sono collegati da una molla k = 60 N/m e l0 = 0.2 m. Il corpo mA ha un coefficiente di attrito dinamico µA = 0.6, mentre B scivola senza attrito. Inizialmente i due corpi sono fermi ad una distanza d0 = 0.3 m.
α mA mB k µA Calcolare: 1. il periodo T delle oscillazioni di B rispetto ad A.
2. vB quando la distanza tra le due masse e d1 = 0.1 m, sapendo che A si e nel frattempo spostato di ∆xA = 24 cm e vA = 2.3 m/s; 3. per quale distanza iniziale d′ 0 i due corpi mantengono la stessa distanza lungo la discesa; Esercizio 5.3 Due masse m1 = 2 kg e m2 = 3 kg sono collegate da una molla e da un filo che comprime la molla di ∆l = 0.1 m rispetto alla sua lunghezza di equilibrio e si trovano su un piano orizzontale privo di attrito, inizialmente ferme. All’istante t = 0 s, il filo viene tagliato e si osserva che v1 = 0.5 m/s quando la molla e a riposo.
Calcolare: 1. k della molla; 2. equazione del moto.
Esercizio 5.4 pendolo balistico S. Lacaprara Es. Meccanica e Termodinamica 63 Un proiettile di massa m = 10 g viene sparato su una massa M = 10 kg, appesa al soffitto da una fune ideale di lunghezza l. Il proiettile rimane conficcato dentro la massa dopo l’urto e la massa si solleva fino ad un angolo α Determinare: 1. la relazione tra vm e α; 2. il bilancio energetico durante l’urto.
m M α Esercizio 5.5 Un proiettile m = 10 g viene sparato orizzontalmente con velocita v = 200 m/s contro due blocchi di massa M1 = M2 = 1 kg, appoggiati ad un piano scabro con coefficiente di attrito dinamico µ = 0.6. I due blocchi distano inizialmente d = 10 cm, e dopo l’urto restano incollati.
1. v1 dopo l’urto; 2. v′ 1 quando colpisce M2; 3. v2 dopo secondo urto; 4. x percorso successivamente da M2.
Esercizio 5.6 Una tavola M = 4 kg scivola senza attrito su un piano orizzontale, inizialmente con vT = 0.20 m/s. Sopra la tavola viene lanciato un secondo corpo di massa m = 1.0 kg, con una velocit a, rispetto alla tavola vm,T = 0.80 m/s. Questo secondo corpo subisce una forza di attrito con coefficiente µ = 0.06 mentre si sposta sopra la tavola. Si chiede: ⃗ vT ⃗ vm 1. v′ T della tavola quando m e fermo rispetto ad essa; 2. l’accelerazione della tavola ⃗ aT e quella del corpo ⃗ am durante il moto; 3. il tempo ∆t perch´ e il corpo si fermi rispetto alla tavola; 4. lo spazio ∆s percorso dalla tavola in questo tempo; 5. lo spazio ∆x percorso dal corpo sulla tavola.
Esercizio 5.7 S. Lacaprara Es. Meccanica e Termodinamica 64 Un corpo di massa m = 2 g viene lanciato con v = 1 m/s su un piano orizzontale liscio verso una rampa come in figura, di massa M = 198 g inizialmente ferma.
Calcolare m ⃗ v M 1. la velocit a V della rampa quando m e fermo rispetto ad essa; 2. l’altezza massima h del corpo m; 3. la velocit a v′ M della rampa quando il corpo m e sceso da essa.
4. l’energia dissipata se il corpo m urtasse in modo completamente anaelastico un ostacolo di massa M, restando incastrato.
Esercizio 5.8 Due masse m = 400 g sono unite da un’asta di lunghezza l0 = 30 cm che ruota attorno al suo centro con ω0 = 50 giri/min. La distanza tra le masse viene ridotta a l1 = 20 cm.
Determinare: 1. la nuova velocit a angolare ω1; 2. il lavoro necessario per ridurre la distanza tra le masse; 3. nell’ipotesi che la forza usata per avvicinare le masse sia costante, calcolare tale forza.
Esercizio 5.9 Un uomo di m = 80 kg salta a terra da un carrello di MA = 200 kg, con una velocita orizzontale relativa al carrello pari a v′ = 5 m/s. Il carrello e collegato, tramite una molla ideale, ad un secondo carrello di massa MB = 300 kg, entrambi liberi di muoversi su un piano orizzontare senza attrito, e inzialmente a riposo.
Determinare: 1. la velocita nel laboratorio dell’uomo v, e dei due carrelli VA e VB appena dopo il salto; 2. il lavoro W compiuto dall’uomo per saltare; 3. la k della molla se la sua compressione massima e ∆x = 10 cm; 4. descrivere il moto del sistema dopo il salto.
Esercizio 5.10 Ad una massa m1, libera di muoversi senza attriti su un asse orizzontale, e appeso un pendolo di mas-sa m2 e lunghezza l, libero di oscillare senza attri-ti. Il sistema, inizialmente fermo, viene fatto oscillare muovendo il pendolo dalla posizione di equilibrio.
θ l m2 m1 x S. Lacaprara Es. Meccanica e Termodinamica 65 Calcolare: 1. la posizione di equilibrio; 2. il periodo delle piccole oscillazioni attorno all’equilibrio.
Esercizio 5.11 Su un nastro trasportatore, che si muove con velocit a costante v = 0.96 m/s, viene fatta cadere verticalmente del materiale ad un ritmo di dm/dt = 0.134 kg/s.
Determinare: 1. la forza esterna necessaria per mantenere costante la velocita del nastro; 2. cosa succede se il nastro ha lunghezza finita e alla fine scarica il materiale?
3. la velocit a del nastro v(t) nel caso non vi siano forze esterne, se il nastro ha lunghezza l = 10 m e massa a vuoto m0 = 1 kg.
Esercizio 5.12 (Cannone di galileo) Due palle (massa M e m, con M ≫m) perfettamente elastiche vengono fatte cadere insieme a terra da una altezza h, la piu piccola appoggiata sopra quella pi u grande, dove rimbalzano.
1. si calcoli l’altezza raggiunta dalla pallina di massa m confrontandola con quella che raggiunge se viene fatta rimbalzare per terra da sola.
2. che succede se i palloni sono n, tali che Mn+1 ≫Mn?
Esercizio 5.13 Un cannone e posto su una piattaforma libera di muoversi senza attrito su un piano orizzontale. Il cannone e inclinato di un angolo θ′ = 60◦, la massa della piattaforma e del cannone e M = 1480 kg.
Un proiettile di massa m = 20 kg viene sparato dal cannone, con una velocit a iniziale pari a v′ 0 = 300 m/s.
θ′ ⃗ v′ 0 1. la velocita del carrello rispetto a terra dopo lo sparo V ; 2. la velocit a del proiettile riespetto a terra del proiettile v; 3. l’angolo θ di uscita del proiettile rispetto a terra.
S. Lacaprara Es. Meccanica e Termodinamica 66 Soluzione dell’esercizio 5.1 1. equazioni dinamica per i due corpi: M1¨ y1 = +F −M1g −k(y1 −y2) M2¨ y2 = +N2 −M2g + k(y1 −y2) Nelle condizioni iniziali: y2 = 0, e N2 = 0, ¨ y2 ≥0 per far alzare il corpo M2.
Dalla seconda equazione si ricava che y1 ≥M2g k per far alzare il corpo.
Dalla prima, la soluzione generale, date le condizioni iniziali risulta y1(t) = F−M1g k (1 −cos ωt), con ω2 = k M1. y1 massimo vale ymax 1 = 2F−M1g k ≥M2g k , quindi F ≥M1g + M2g 2 Si puo risolvere anche con la conservazione dell’energia. All’estremo superiore del moto di M1 (dove e fermo) ∆U = 1 2ky2 2 + M1gy1 −0 = Fy1, da cui si ricava lo stesso risutato di prima.
2. le equazioni del moto risultano: ¨ yCM = F M1 + M2 −g µ ¨ ∆y + k∆y − M2F M1 + M2 = 0 con yCM = M1y1+M2y2 M1+M2 , ∆y = y1 −y2, 1 µ = 1 M1 + 1 M2.
Le soluzioni in queste coordinate sono semplici.
yCM(t) = 1 2 F M1 + M2 −g t2 ∆y(t) = A0 cos(ωt + ϕ) + M2F (M1 + M2)k da cui si puo ricavare l’equazione per y1,2(t).
Soluzione dell’esercizio 5.2 1. ω2 = k(mA+mB) mAmB , T = 2π ω = 0.66 s.
2. conservo l’energia: ∆Etot = ∆(Ek + Ug + Uel) = −L.
∆Ek = 1 2 mAv2 A + mBv2 B ∆Uel = k 2 d2 f −d2 i ∆Ug = −(mA∆xA + mB∆xB)g sin α L = −µAmAg cos α∆xA Che fornisce vB = 2.0 m/s S. Lacaprara Es. Meccanica e Termodinamica 67 3. dalle equazioni della dinamica: mAmB ¨ ∆x + k(mA + mB)(∆x) = −mamBgµA cos α + k(mA + mB)l0 , l’accelerazione risulta nulla se la distanza tra le due masse risulta: ∆x = −mamBgµA cos α + k(mA + mB)l0 k(mA + mB) = l0 −mamBgµA cos α k(mA + mB) = 14.3 cm Soluzione dell’esercizio 5.3 1. il sistema e conservativo, si conserva l’energia totale e il momento della quantita di moto. m1v1 = m2v2, 1 2kl2 = 1 2m1v2 1 + 1 2m2v2 2, da cui ricavo: k = m1 m2(m1 + m2) v2 1 l2 = 83.3 N/m; 2. xcm(t) = costante, (x1 −x2) si muove con moto armonico con pulsazione ω = s k 1 m1 + 1 m2 Soluzione dell’esercizio 5.4 1. durante l’urto si conserva la quantit a di moto, dopo si conserva l’energia.
mvm = (M + m)vM, 1 2Mv2 M = Mgl(1 −cos α) quindi vm = p 2gl(1 −cos α) M m 2. Eini = 1 2mv2 m, Efin = 1 2M m2 M2v2 m = 1 2 m2 M v2 m < Eini La differenza si dissipa durante l’urto anaelastico.
Soluzione dell’esercizio 5.5 1. mv = (m + M)v1, v1 = 1.98 m/s; 2. ∆Ek = Lattrito, 1 2(m + M)v2 1 = 1 2(m + M)v′2 1 −µ(M + m)gd, v′ 1 = 1.66 m/s; 3. (m + M)v′ 1 = (m + M + M)v2, quindi v2 = 0.83 m/s; 4.
1 2(m + 2M)v2 2 = (M + M + m)gµx, x = v2 2 2gµ = 5.85 cm.
Soluzione dell’esercizio 5.6 1. la forza d’attrito e interna al sistema, quindi il momento della quantit a di moto si conserva.
MvT + m(vT + va) = (M + m)v′ T, v′ T = 0.36 m/s; 2. m⃗ am = −µgmˆ x, quindi ⃗ am = −µgˆ x = −0.59ˆ x m/s2 M⃗ aT = +µgmˆ x, ⃗ aT = 0.147ˆ x m/s2; 3. Nel riferimento della tavola calcolo il tempo necessario perch´ e il corpo si fermi.
L’accelerazione in questo riferimento ´ e quella relativa: ∆t = vm am −aT = .27 s .
S. Lacaprara Es. Meccanica e Termodinamica 68 4. ∆s = vT∆t + 1 2aT∆t = 5.9 cm.
Soluzione dell’esercizio 5.7 1. momento si conserva: mv = (M + m)V , V = 0.01 m/s; 2. energia si conserva: 1 2mv2 = 1 2(m + M)V 2 + mgh, da cui h = 5 cm; 3. si conservano entrambi: e un urto elastico mv = MV ′ + mv′ e 1 2mv2 = 1 2MV ′2 + 1 2mv′2, da cui V ′ = 2mv M+m = 2 cm/s.
4. W = mgh del punto 2 Soluzione dell’esercizio 5.8 1. Momento angolare iniziale e L0 = 2ml0v0/2 = 2m l0 2 2 ω0 = 1.4·10−2 kgm2/s. Le forze sono centrali, quindi il loro momento rispetto al baricentro e nullo, quindi L0 = L1 = 2m l1 2 2 ω1. Quindi ω1 = l2 0ω0 l2 1 = 11.8 rad/s.
2. W = Efin k −Eini k ; Ek = 1 2Iω2 = 1 2 2m l 2 2 ω2 = 21 2mv2 W = m 4 (l2 1ω2 1 −l2 0ω2 0) = 0.31 J 3. W = 2 R b1/2 b0/2 Fdx = 2(b1 −b0)F quindi F = mω2 0 8 b2 0 b2 1(b1 + b0) = 1.55 N.
NB: la forza totale applicata sulle due masse deve tenere conto anche della forza centripeta necessaria per farle ruotare.
Soluzione dell’esercizio 5.9 1. MB e disaccoppiato da MA a causa della molla, che non trasmette forze impulsive: VB = 0 m/s conservazione q.ta di moto: MAVA+mv = 0, v = v′+VA, da cui: VA = −1.43 m/s, v = 3.57 m/s 2. W = 1/2mv2 + 1/2MAV 2 A −0 = 714 J 3. Si conserva p e E; quando compressione e massima, i due carrelli si muovono alla stessa velocita: MAV ′ A + MBV ′ A = MAVA, 1/2MAV 2 A = 1/2(MA + MB)V ′2 A + 1/2k∆x2, da cui: k = MAMB MA+MB V 2 A ∆x2 = 2.45 · 104 N/m 4. solo forze interne elastiche: VCM = MAVA MA+MB = −0.57 m/s costante. Moto rispetto al centro di massa, osillatore armonico: ω = p µ k = 14.3 rad/s (µ = MAMB MA+MB = 120 kg massa ridotta), T = 2π/ω = 0.44 s.
Soluzione dell’esercizio 5.10 1. Uso come coordinate x di m1 e θ di m2 rispetto alla verticale.
U(x, θ) = −m2gl cos θ, che ha un minimo per θ = 0, ∀x 2. l’energia totale si pu o scrivere come la somma dell’energia dei due punti. Per la massa m1 e semplice, per m2 serve scrivere le sue coordinate in termini di (x, θ) x2 = x + l sin θ, y2 = l cos θ, quindi ⃗ v2 = ( ˙ x + l cos θ ˙ θ, −l sin θ ˙ θ), e quindi l’energia cinetica di m2 risulta: S. Lacaprara Es. Meccanica e Termodinamica 69 Ek,2 = 1 2m2v2 2 = 1 2m2 ( ˙ x + l cos θ ˙ θ)2 + (−l sin θ ˙ θ)2 che, sommata a Ek,1 fornisce: Ek = 1 2(m1 + m2) ˙ x2 + 1 2m2 l2 ˙ θ2 + 2l ˙ θ ˙ x cos θ E’ una espressione complessa, che dipende da due variabile x, θ (e dalle loro deri-vate). Posso espanderla attorno alla posizione di equilibrio che ho ricavato prima (θ = 0), e si riduce a: Ek = 1 2(m1 + m2) ˙ x2 + 1 2m2 l2 ˙ θ2 + 2l ˙ θ ˙ x Sempre complessa, ma almeno dipende solo da ˙ x e ˙ θ (e non da x, θ).
Posso quindi usare la conservazione del momento lineare lungo x, visto che le forze esterne, cioe’ la gravit a e la reazione del vincolo, sono verticali: m1 ˙ x + m2 ˙ x + l ˙ θ cos θ = 0 (conservazione q.ta di moto lungo ˆ x), da cui ricavo che ˙ x = −m2l ˙ θ m1+m2, cio e una relazione tra ˙ x e ˙ θ. Sostiuisco nell’espressione precedente e trovo che, attorno a θ = 0, Ek = 1 2 m1m2 m1+m2l2 ˙ θ2 L’energia potenziale e decisamente pi u semplice: Upot = −m2lg(1 −cos θ) = + m2lg 2 θ2 (sempre sviluppata attorno all’equilibrio).
Mettendo tutto assieme si ricava che: Etot = 1 2 m1m2 m1+m2l2 ˙ θ2 + m2lgθ2 2 + cost.
Per analogia con l’energia totale di un oscillatore armonico: Etot = 1 2m ˙ x2 + 1 2kx2, si ricava: T = 2π q lµ m2g, con µ massa ridotta del sistema.
Per m1 ≫m2, µ →m2 e il periodo e quello di un pendolo normale.
Soluzione dell’esercizio 5.11 1. Fext = dp dt = v dm dt = 0.129 N 2. il materiale che cade dal carrello non cade verticalmente, ma mantiene la stessa velocit a orizzontale che aveva sul carrello, quindi la situazione non cambia e la forza necessaria rimane la stessa.
3. p si conserva, v(t) = v0 m0 m0+ dm dt t fino a quando il materiale inizia a cadere.
Il materiale inizia a cadere quando x(t) = m0v0 dm dt ln m0+ dm dt t m0 = l, cioe quando t∗= m dm dt e l dm dt m0v0 −1 = 22.7 s.
Soluzione dell’esercizio 5.12 1. Immagino che la palla grande arrivi a terra appena prima di quella piccola. Anche se le rilascio da ferme nello stesso istante, piccole perturbazioni nel rilascio faranno s ı che una sia inizialmente piu veloce dell’altra. Se la pi u veloce e quella grande, allora arriva prima al pavimento. Se e quella piccola, allora urtera quella grande durante la discesa e come prima quella grande arriver a per prima al pavimento.
M urta il pavimento, e inverte la sua velocita.
Considero ora un sistema di riferimento solidale a M. In questo sistema, la pallina m si avvicina a M con velocit a relativa 2v, e, dato che M ≫m, dopo l’urto (elastico) si allontana con velocita relativa uguale e opposta.
S. Lacaprara Es. Meccanica e Termodinamica 70 Rispetto al pavimento, la velocit a di uscita di m e quindi 2v + v = 3v, dato che M si sta allontanando dal pavimento con v.
Quindi: h = 1v′2 2g = h0(32) = 9h0.
2. Un ragionamento simile con tre palle fornisce v′ 3 = 7v, con 4 v′ 4 = 15v, il che suggerisce v′ n = (2n+1) (si pu o dimostrare per induzione). Quindi hn = (2n+1)2h.
Quante palle ci vorrebbero per far uscire la pallina in alto dall’attrazione terrestre?
Da notare che nel caso generico con n palle, inizia a diventare difficile che Mn+1 ≫ Mn per tutte le coppie: inoltre risulta semper piu difficile garantire che le palle restino allineate una sopra l’altra durante la caduta. Questo riduce la possibilit a di espandere il cannone multiplo per n alti.
Bibliografia: Mellen, W., “Aligner for elastic collisions of dropped balls”, The Physics Teacher 33, 56 (1995) Cross, R., “Vertical bounce of two vertically aligned balls”, Am. J. Phys.
75, 1009 (2007) Soluzione dell’esercizio 5.13 1. Con ′ si indicano le quantita relative al sistema di riferimento solidale con il carrello, mentre quelle senza apice sono quelle rispetto al sistema di riferiemento a terra.
Se il carrello si muove all’indietro con velocit a V , allora: ⃗ v = ⃗ v′ −⃗ V , Quindi: ⃗ v = (v′ 0 cos θ −V )ˆ x + v′ 0 sin θˆ y.
Lungo l’asse orizzontale non ci sono forze esterne, quindi px si conserva: mv′ 0 cos θ′− MV = 0; combinando le due equazione, si ricava V = Vc = mv′ 0 cos θ′ M+m = 2.0 m/s.
2. ⃗ v = pv2 x + v2 y = 299 m/s 3. tan θ = vy vx = v′ 0 sin θ′ v′ 0 cos θ′−V = 1.76, θ = 60.3◦ Da notare che il rapporto tra le masse M/m ∼75, quindi il rinculo del carrello con il cannone e limitato.
Parte II Meccanica corpo rigido S. Lacaprara Es. Meccanica e Termodinamica 71 6 Dinamica del corpo rigido Esercizio 6.1 Un’asta (Ma = 2 kg, L = 70 cm) e fissata ad una parete verticale ad una estremita, dove e collegata ad una molla torsionale, (k = 0.5 Nm/rad) che e a riposo quando l’asta e verticale.
L’asta viene lasciata cadere da ferma e, quando passa per la ver-ticale, colpisce una massa mp = 1.2 kg, inizialmente ferma, che si muove orizzontalmente con vp = 2.2 m/s dopo l’urto.
Ma L mp Determinare: 1. ω0 dell’asta al momento dell’urto; 2. ω1 dell’asta subito dopo dell’urto; 3. l’energia Wd dissipata nell’urto; 4. l’impulso J trasferito nell’urto al vincolo.
Esercizio 6.2 Un’asta (M = 1 kg, L = 40 cm) libera di muoversi su un piano orizzontale, inizialmente ferma, viene colpita perpendicolarmente da un proiettile (m = 20 g) con velocita v0 = 20 m/s, ad una distanza x0 = 10 cm dal centro. Dopo l’urto il proiettile rimane incastrato.
vp mp M Calcolare: 1. il moto del sistema asta-proiettile dopo l’urto; 2. l’energia dissipata nell’urto.
3. che succede se invece l’asta e vincolata all’ estremita pi u lontana rispetto al punto di impatto del proiettile?
Esercizio 6.3 Un’asta di lunghezza L = 1 m e massa Ma = 1.2 kg, vicolata ad una estremita, viene fatta cadere, da ferma, da un angolo θ = 60◦rispetto alla verticale.
Quando l’asta e verticale, essa urta un piccolo dente all’estre-mita di un disco di raggio R = 0.2 m e massa Md, vincolato a ruotare attorno al suo centro.
L’urto e elastico e la velocita angolare dell’asta dopo l’urto si dimezza.
Ma, L Md, R θ S. Lacaprara Es. Meccanica e Termodinamica 72 1. ωd del disco dopo l’urto; 2. Md del disco.
Esercizio 6.4 Una fune ideale e collegata ad una carrucola di raggio R = 0.5 m e massa M = 2.0 kg, priva di attriti, vincolata a ruotare attorno al suo centro. All’istante t = 0 s, la fune, inizialmente ferma, viene tirata verso il basso con una forza costante F = 9.8 N.
1. l’accelerazione angolare α della carrucola; 2. l’accelerazione ⃗ aA di un punto A posto sul bordo della carrucola, all’istante t = 0.25 s; 3. la reazione del vincolo ⃗ R; 4. α′ e ⃗ R′ se invece di tirare verso il basso la corda gli si appende una massa m con forza peso Fp = F; La massa m si ferma a terra e, con la fune tesa, viene applicato un momento τ = 10 Nm per far salire la massa: 5. ∆Emecc della massa quando h = 1.0 m.
Esercizio 6.5 Una lastra rigida, uniforme, con densita di massa σ = 30 kg/m2, ha la forma in figura, con a = 12 cm.
Essa e vincolata a ruotare attorno all’asse x. Calcolare: a a a a a a a 3a x y E ⃗ g 1. il centro di massa del sistema; 2. il momento d’inerzia rispetto all’asse x.
La lastra viene colpita nel punto E da un punto materiale di massa m = 500 g e v = 50 m/s, perpendicolarmente al piano (x, y), e il proiettile rimane attaccato alla lastra. Calcolare: 3. ω dopo l’urto; 4. l’energia dissipata nell’urto; 5. il periodo delle piccole oscillazioni.
Esercizio 6.6 Un disco di massa M = 2.5 kg e raggio R = 35 cm rotola senza strisciare su un piano orizzontale. Una forza esterna Fext = 125 N viene applicata sul disco in modo tale che la distanza tra la retta di applicazione e il centro del disco sia r = 12 cm, e che la retta formi un angolo θ = 30◦rispetto al piano come in figura.
⃗ F r R θ Cal-colare: S. Lacaprara Es. Meccanica e Termodinamica 73 1. l’accelerazione a del disco; 2. la forza di attrito Fatt tra il disco e il piano; 3. discutere i risultati al variare di r e cos θ.
Esercizio 6.7 Una sfera di raggio R = 14 cm e massa M = 5.6 kg, si muove senza rotolare su un piano orizzontale liscio con v0 = 3.4 m/s. Ad un certo punto il piano diventa scabro, con coefficente di attrito dinamico µ = 0.3: a regime il moto e di puro rotolamento.
Determinare: 1. il tempo necessario per arrivare ad un moto di puro rotolamento; 2. la velocit a di regime; 3. l’energia dissipata; 4. da chi viene dissipata? Discutere.
Esercizio 6.8 Un’asta rigida l = 60 cm e M = 100 g e imperniata ad una estremo O su un piano ver-ticale. Ad una distanza x da O, viene applicata una forza F = 10 N, perpendicolarmente all’asta, per ∆t = 5.0 ms. Calcolare: 1. x perch´ e la reazione del vincolo sia nulla; 2. la velocit a del CM e l’energia cinetica dopo l’impulso; 3. l’angolo massimo θmax; 4. Il periodo delle oscillazioni; 5. vmax e amax.
Esercizio 6.9 Un disco R = 30 cm e M = 30 kg ha un secondo disco di r = 12 cm e m = 12 kg fissato ad una distanza (centro-centro) di h = 15 cm. Il sistema poggia su un piano orizzontale e si muove con un moto di puro rotolamento, con velocita iniziale va = 1.5 m/s, e inizialmente il secondo disco si trova nella posizione pi u bassa.
A B vA h 1. Energia cinetica in A; 2. la velocita del disco nel punto B, dove il secondo disco e nella posizione piu alta; 3. la reazione del piano in A; 4. la reazione del piano in B; Esercizio 6.10 S. Lacaprara Es. Meccanica e Termodinamica 74 Il sistema in figura e in equilibrio: m1 = 12 kg, R1 = 10 cm; m2 = 5 kg, R2 = 5 cm; m3, θ = 30◦.
Il corpo m1 rotola senza strisciare sul piano, e la fune (ideale) e arrotolata attorno ad esso.
La fune non striscia neppure sulla carrucola m2.
Determinare: m1, r1 m2, r2 m3 ⃗ g θ 1. m3; 2. La forza d’attrito su m1 Sia m3 = 4 kg. Calcolare: 3. a3, a1, F (1) att ; 4. reazione vincolare sulla carrucola; 5. l’energia cinetica del sistema quando m3 e sceso di h3 = 1.0 m.
Esercizio 6.11 Un disco M = 1 kg e R = 10 cm, incernierato al centro poggia su un piano orizzontale in quiete. Una sbarretta L = 8 cm e m = 0.25 kg, si muove con velocita v0 = 50 cm/s e colpisce il disco come in figura, rimanendo attaccata finich e il sistema ruota di mezzo giro, quindi si stacca.
M, R m, l v0 Determinare: 1. ω sistema dopo l’urto; 2. impulso trasmesso al vincolo durante l’urto; 3. ∆Ek nell’urto; 4. la forza tra sbarra e disco durante la rotazione; 5. il moto del disco e della sbarra dopo il distacco.
Esercizio 6.12 Un blocco di massa M = 1.5 kg scivola lungo un piano inclinato di δ = 20◦, con attrito dinamico µ = 0.2. Una sbarretta rigida di massa trascurabile collega il blocco ad un cilindro di massa Mc = 2 kg che rotola senza strisciare. Il sistema parte da fermo e percorre una distanza d = 1.5 m Calcolare: 1. la velocita del sistema; 2. la forza di attrito statico tra cilindro e piano; 3. la forza che la sbarretta esercita sui due corpi; 4. per quale angolo δ di inclinazione del piano questa si annulla.
Esercizio 6.13 S. Lacaprara Es. Meccanica e Termodinamica 75 Un carrello di massa complessiva M = 10 kg (comprese le 4 ruote) si muove con velocit a iniziale v0 = 6 m/s su un piano orizzontale. Le ruote hanno ciascuna massa m = 1 kg e raggio r = 50 cm, rotolano senza strisciare. All’istante t = 0s, il carro frena, applicando un momento frenante su ciascuna delle ruote τ = 7.5 Nm costante, le ruote non slittano.
Calcolare: 1. accelerazione angolare α e Fattrito delle ruote; 2. lo spazio di frenate s; 3. lavoro compiuto dai freni Wd; 4. lo spazio di frenata s′ se le ruote si bloccano e µd = 0.2.
Esercizio 6.14 Un disco omogeneo, vincolato al centro, M = 4kg, R = 20cm, ruota con attrito costante τ = 0.5 Nm.
Un secondo disco m = 2kg e r = 10 cm e fissato al primo come in figura. Un perno allineato con i due centri e presente sul bordo del disco, che viene lasciato libero di ruotare.
Il perno urta una massa m3 = 1.5 kg che si trova su un piano scabro µd = 0.2, che si ferma dopo l’urto in ∆l = 50cm.
P M,R m,r m3 Determinare: 1. ω0 prima dell’urto; 2. ω1 dopo l’urto; 3. Impulso del vincolo durante l’urto.
Esercizio 6.15 Una sbarra di lunghezza L = 1 m e massa ma = 6 kg, e incernierata ad una estremit a ad un perno. All’altra estremita si trova un disco di massa Md = 24 kg e raggio R = 0.5 m, incernierato al centro e inizialmente fissato rigidamente alla sbarra. Il sistema si trova fermo ad un angolo θ = 60◦rispetto alla verticale e viene lasciato libero.
Quando passa la verticale, il disco viene lasciato libero di ruotare attorno al suo centro, senza attrito.
L R θ Determinare: 1. ω sistema alla verticale; 2. θ′ max dall’altra parte; 3. Il periodo delle piccole oscillazioni con disco fisso; 4. il periodo delle piccole oscillazioni con disco mobile.
Esercizio 6.16 S. Lacaprara Es. Meccanica e Termodinamica 76 Un’asta di lunghezza L = 1.2 m e massa ma = 5 kg, striscia su un piano scabro con l’estremit a B, mentre l’altra estremita A e fissata al centro di un disco di raggio R = 0.6 cm e massa md = 2 kg, senza attriti. Il disco rotola senza strisciare, e si muove con velocita costante v verso sinistra essendo sottoposto ad un momento τ = 1.8Nm.
Determinare: B A D 1. la forza di attrito su B; Supponendo che l’attrito sia trascurabile: 2. l’accelerazione del sistema; 3. la forza esercitata sull’estremo A.
Esercizio 6.17 Una motocicletta di massa m = 300 kg, con interasse l = 1480 mm, ha il peso ripartito al 55% (Rp) sulla ruota posteriore e 45% (Ra) su quella anteriore e l’altezza del baricentro h = 70 cm.
1. quale e l’accelerazione massima per partire senza impennare, supponendo che le ruote non slittino?
2. supponendo che la coppia massima all’albero sia 103 Nm, il raggio delle ruote 30.5 cm, e il rapporto di trasmissione in prima marcia 2.384 · 45/15, riesce ad impennare?
3. quali sono le reazioni delle ruote sull’asfalto in caso di frenata?
4. quale e la massima coppia frenante che posso applicare all’anteriore perch e il posteriore non si alzi?
Esercizio 6.18 Una asta lunga l e di massa M e incernierata ai due estremi a due assi cartesiani x, y, come in figura. Inizialmente e verticale e ferma: una piccola perturbazione la fa cadere.
θ x y z Calcolare: 1. ω quando e orizzontale; 2. la forza esercitata dai vincoli Nx, Ny; 3. il moto del centro di massa; 4. nell’ipotesi che il vincolo lungo y sia unilaterale, quando si stacca dalla parete.
Esercizio 6.19 S. Lacaprara Es. Meccanica e Termodinamica 77 Un cubo di ferro di spigolo d = 12 cm e massa 13.6 kg, viene spinto orizzontalmente su un piano da una forza costante F = 100 N, applicata nel punto A (spigolo della faccia superiore), e e bloccato da un fermo nello spigolo opposto (O), come in figura.
A O ⃗ F 1. Fmin minima per alzare il cubo; 2. l’accelerazione angolare iniziale α0; 3. la reazione vincolare ⃗ R esercitata dal blocco in O nello stesso istante; 4. il lavoro WF compiuto dalla forza F dall’inizio a quando il blocco inizia a cadere oltre O; 5. la velocita angolare ω in tale istante.
Esercizio 6.20 Calcolare il momento d’inerzia di una sfera rispetto ad un suo diametro (asse z) 1. facendo esplicitamente l’integrale Iz = R V ρ2dm; 2. usando il momento d’inerzia di un disco Idisco = mr2/2; 3. calcolando il momento d’inerzia di un guscio sferico.
Esercizio 6.21 Una sfera m = 100 g si trova su una vasca con i bordi inclinati (θ = 45◦), ad una altezza h = 80 cm. Viene lasciata libera da ferma e rotola senza strisciare verso il fondo della vasca. La superfice della vasca diventa perfettamente liscia a met a del percorso della sfera, con continua il suo moto e risale sulla parete opposta.
h0 Scabro Liscio Si chiede: 1. la forza d’attrito ⃗ F sulla sfera all’inizio della discesa; 2. la velocita v0 con la quale la sfera si muove sul fondo della vasca; 3. l’altezza massima hmax che raggiunge sulla parete opposta; 4. h′ max se il moto resta ovunque di puro rotolamento.
Esercizio 6.22 Un disco omogeneo (R = 40 cm, M = 0.5 kg) e appoggia-to su un piano orizzontale e collegato ad una molla orizzontale (k = 42 Nm/) fissata al suo asse. Il piano e scabro (µd = 0.6, µs sufficiente a garantire un moto di puro rotolamento). La molla e inizialmente a riposo, e il disco si si muove con v0 = 0.8 m/s comprimendo la molla. Calcolare: M,R ⃗ v0 S. Lacaprara Es. Meccanica e Termodinamica 78 1. il massimo accorciamento della molla; 2. la massima Fattrito durante il moto; Quando il disco torna alla posizione iniziale, viene colpito da un proiettile, di massa m = 40 g, che viaggia orizzontalmente in direzione opposta, e che si incastra al centro del disco, arrestandone il moto.
3. la velocita del proiettile; 4. l’energia persa nell’urto; 5. l’accelerazione del CM dopo l’urto; Esercizio 6.23 Due dischi identici (A,B M = 15 kg, R = 20 cm) sono appoggiati l’uno sopra l’altro tramite un sottile anello di raggio R/2. Il coeff. di attrito tra i due dischi e µ = 0.2, e possono ruotare attorno ad un asse comune. Si applica un momento costante τ = 8.8 Nm al disco piu’ basso per NA = 20 giri. Calcolare: 1. il numero di giri fatti dall’altro disco NB; 2. ωA,B dopo NA = 20 giri; 3. W sul sistema dalle forze esterne; 4. la velocita angolare dei dischi quando ruotano insieme; 5. W dissipata; 6. il numero di giri relativi.
Esercizio 6.24 Un disco (R = 20 cm, M = 1 kg) e puo ruotare senza attrito attorno al suo asse. Il disco sostiene una sfera, (r = R/2, M ), che ruota attorno ad un asse verticale, solidale al disco, che si trova ad una distanza d = R/2 dal centro O del disco.
Il disco e in quiete e la sfera ruota con una velocita angolare ω0 = 5 giri/s. Ad un certo istante inizia ad agire sull’asse della sfera un momento di attrito. Determinare: 1. La velocit a angolare del disco quando la sfera e ferma rispetto ad esso; 2. L’energia dissipata dal momento di attrito; 3. L’angolo che la reazione in O forma con la verticale quando la velocit a angolare del disco e ω1; Esercizio 6.25 Un’asta lunga L = 1.5 m, M = 2 kg e appesa ad una estremita ad un asse orizzontale con un manicotto che si muovo liberamente lungo tale asse. Inizialmente l’asta e inclinata di θ0 = 60◦rispetto alla verticale e viene lasciata cadere.
A B y θ0 S. Lacaprara Es. Meccanica e Termodinamica 79 Calcolare: 1. traiettoria del centro di massa della sbarra; 2. l’equazione del moto attorno alla posizione di equilibrio; 3. velocita angolare nel punto pi u basso; 4. reazione vincolare nel punto piu basso.
Esercizio 6.26 Una ruota di raggio r = 25 cm e masas m = 20 kg e appoggiata piano orizzontale ed ad un gradino di altezza h = 10 cm, inzialmente in quiete. Sull’asse della ruota vie-ne applicata una forza costante ed orizzontale F = 300 N.
Si chiede: ⃗ F R h 1. se la forza e sufficiente per far salire il gradino alla ruota; 2. la velocit a della ruota quando e salita sul gradino; 3. la velocit a con la quale colpisce il piano se viene lasciata cadere dalla spigolo del gradino.
Esercizio 6.27 Due dischi di massa m = 0.4 kg e r = 0.3 m sono posti all’e-stremita di una asta M = 0.6 kg, L = 1.0 m, e sono liberi di ruotare attorno ai propri assi A, B. L’asta e libera di ruota-re attorno al suo centro O. Si applica sull’asta un momento τ = 4(θ + π) Nm per 5 giri. Calcolare: A B O 1. la velocita angolare ω0 del sistema; Ad un certo instante, si attiva frizione tra il disco A e il suo asse 2. la velocit a angolare ω1 del sistema a regime; Successivamente, si blocca la rotazione dell’asta, e si fa ruotare B attorno al suo asse con ω1, sempre lasciando bloccato A e si attiva frizione tra B e l’asta 3. la nuova velocita angolare ω2 a regime e l’energia dissipata; 4. il moto del disco B se il suo asse si spezza istantaneamente; 5. le forze sull’asse O dopo il distacco del disco B; S. Lacaprara Es. Meccanica e Termodinamica 80 Soluzione dell’esercizio 6.1 1. Conservazione dell’energia: 1 2Iω2 −Mag L 2 sin θ = kθ2 2 , da cui ω0 = 6.8 rad/s 2. Forze interne, L si conserva. Iω0 = Iω1 + mpvpL; ω1 = ω0 −mvpL MaL2 3 = 1.15 rad/s 3. Wd = 1 2Iω2 0 −1 2Iω2 1 −1 2mpv2 p = 4.44 J; 4. calcolo la variazione del momento della quantit a di moto: ∆p = Maω0L 2 − Maω1L 2 + mpvp = 1.32 Ns. L’impulso J e −∆p Soluzione dell’esercizio 6.2 1. il momento si conserva vCM = mv0 M+m = 0.39 m/s, dove il centro di massa xCM = M0+mx0 M+m = 2 mm dal centro dell’asta.
Anche il momento angolare rispetto al centro di massa si conserva Lini CM = mv0(x0− xCM), Lfin CM = ItotωCM, con Itot = ML2 12 + Mx2 CM + m(x0 −xCM)2 = 1.4 · 10−2 kgm.
Quindi ωCM = 5.14 rad/s.
2. Eini k = 1 2mv2 = 4 J, Efin k = 1 2(M + m)v2 CM + 1 2Itotω2 = 0.26 J. Quindi Wd = ∆Ek = 3.75 J.
3. Si conserva il momento angolare rispetto al vincolo: LA = mv0(L/2 + x0) = (ML2/3 + m(L/2 + x0)2) ω′, da cui ω1 = 1.62 rad/s, e impulso del vincolo durante l’urto pari a J = ω′(L/2 + x0)(M + m) −mv0 = 0.036 in direzione opposta al proiettile.
Soluzione dell’esercizio 6.3 1. Uso conservazione energia per calcolare la velocit a angolare dell’asta prima del-l’urto: 1 2Iaω2 0 = Mag L 2 (1 −cos θ) con Ia = MaL2 3 : ω0 = q 3g L (1 −cos θ) = 3.83 rad/s L’urto e elastico, quindi uso conservazione dell’energia durante l’urto elastico: 1 2Iaω2 0 = 1 2Iaω′ 0 2 + 1 2Idω2 d, con ω′ 0 = ω0/2, ovvero: MaL2 2 ω2 0 = MdRω2 d.
Durante l’urto le forze scambiate tra asta e disco sono uguali e contrarie, quindi anche l’impulso subito dall’asta e quello subito dal disco sono uguali e contrari: Ja = −Jd. Durante l’ulto vi sono anche altre forze impulsive su asta e disco dovute ai rispettivi vincoli, di cui non sappiamo nulla.
Se considero l’asta, il momento delle forze durante l’urto calcolato rispetto al suo vincolo risulta essere LF(t), dove F(t) e la forza, impulsiva, scambiata con il disco durante l’urto.
Se integro nel tempo dell’urto: LJa = Z LF(t)dt = Z τ(t)adt = Z dLa dt dt = Z dLa = ∆La dove si e usata la seconda eq. caridinale della meccanica per l’asta, rispetto al vincolo. L’altra forza impulsiva ignota, quella generata dal vincolo, non fornisce momento.
Un analogo ragionamento si pu o fare per il disco, e si ricava RJd = ∆Ld, dove il momento angolare in questo caso e calcolato rispetto al centro del disco (vincolo).
S. Lacaprara Es. Meccanica e Termodinamica 81 Quindi ∆La L = −∆Ld R con La,d il momento angolare dell’asta rispetto all’estremit a fissa e del disco rispetto al suo centro, rispettivamente.
∆La L = Ia(ω0 −ω1) = MaL 3 (ω0 −ω0 2 ) = MaL2 6 ω0 = ∆Ld R = Idωd = MR2 2 ωd , da cui si ricava MaLω0 = 3MdRωd.
Risolvendo il sistema ottengo ωd = 3L 2Rω0 = 28.8 rad/s e Md = 2 9Ma = 0.27 kg.
Soluzione dell’esercizio 6.4 1. τ = FR = MR2 2 ¨ θ: ¨ θ = 2F MR = 19.6 rad/s2; 2. ⃗ aA = ¨ θRˆ ur + ˙ θ(t) 2Rˆ uT = 9.8ˆ ur + 12ˆ uT usando ˙ θ = ¨ θt 3. Su disco agiscono ⃗ F + M⃗ g + ⃗ R = 0, quindi ⃗ R = −(⃗ F + M⃗ g) = 29.4 N diretta verso l’alto.
4. Il sistema e diverso da prima, perch e adesso abbiamo una seconda massa m con la sua inerzia, di cui dobbiamo tenere conto.
MR2 2 ¨ θ = TR mg −T = m¨ x x = Rθ Da cui ricavo: T = m(g −¨ θR), e ¨ θ = 2mg R(M+2m) = 9.8 rad/s2 e T = MR 2 ¨ θ = 4.9 N.
R′ = −(T + Mg) = 24.5 N sempre diretto verso l’alto.
5. L = R Fdx = R τdθ = τ∆θ = N h R = mgh + 1 2mv2 + 1 2Idω2.
1 2Idω2 = MdR2 2 v2 R2 = 1 2Mv2, quindi v = r τh R −mgh m 2 + M 4 = 3.2 ms.
∆Emecc = Efin mecc −0 = mgh + 1 2mv2 = 14.9 J.
Soluzione dell’esercizio 6.5 1. Penso alla lastra come 4 lastre quadrate identiche di lato a. Il centro di massa e il centro di massa del sistema di 4 punti, di massa a2σ posti nel centro geometrico di ciascuna lastra quadrata.
xcm = 0 (per simmetria) ycm = 5 4a = 15 cm.
2. Momento d’inerzia di una lastra quadrata rispetto ad un lato: I = a4σ 3 . Applico Steiner e ottengo il momento d’inerzia totale: Itot = 1 3 + 7 a4σ = 4.6 · 10−1kgm2.
L’applicazione di Steiner va fatta due volte: la prima volta per ricavare I rispetto al CM sapendo quello rispetto al lato: Ilato = ICM + M(a/2)2. La seconda volta per ricavare I rispetto all’asse di rotazione.
3. Durante l’urto si conserva L rispetto all’asse ˆ x. Lini = 2amv = Lfin = Iω = (Itot + m(2a)2)ω, da cui ω = 80 rad/s.
4. Eini = 1 2mv2, Efin = 1 2 (Itot + m(2a)a) ω2 Wd = Efin −Eini = 385 J.
S. Lacaprara Es. Meccanica e Termodinamica 82 5. (Itot + m(2a)2) ¨ θ = (4a2σ + m)gRcmθ, dove Rcm = ycm4a2σ+2am 4a2σ+m = 17cm T = 2π ω = 0.89s.
Soluzione dell’esercizio 6.6 1.
2. le equazioni della dinamica per il disco sono le seguenti, calcolando il moto del centro di massa (CM) lungo il piano, e i momenti rispetto al CM; la terza equazione si ottiene dall’informazione che il disco rotola senza strisciare: Ma = Fext cos θ −Fatt Icmα = Fextr −FattR α = −a R Risolvendo il sistema: a = Fext cos θ −r R R2 MR2 + ICM = 17.4 m/s2 Fatt = Fext (ICM cos θ + MRr) R2 MR2 + ICM = 64.7 N Soluzione dell’esercizio 6.7 1. Equazioni della dinamica per la sfera: Ma = µMg Icmα = µMgR da cui ricavo: α = 5µg 2R e a = µg, entrambe costanti.
Si avr a moto di puro rotolamento quando v(t) = v0 −at = ω(t)R = αtR, da cui ricavo t = 2v0 7µg = 0.33 s 2. a regime il moto e di puro rotolamento: vregime = v0 −at = 5 7v0 = 2.5 m/s e ωregime = vregime R = 17.8 rad/s 3. ∆Ek = 1 2Mv2 regime + 1 2Iω2 regime −1 2Mv2 0 = 7.9 J 4. Questa energia e dissipata dalla forza di attrito durante la transizione tra il moto di puro scivolamento e quello di puro rotolamente, quando c’e moto relativo tra il piano e il punto di contatto della sfera con il piano stesso. Successivamente il sistema e conservativo.
La velocita del punto di contatto della sfera con il piano e pari a vp = vcm −ωR = v0 −7 2µgt, quindi lo spazio percorso dal punto di contatto rispetto al piano e: ∆xp = v0t −1 2 7 2µg t2 = 1 7 v2 0 µg = 0.48 cm S. Lacaprara Es. Meccanica e Termodinamica 83 L’energia dissipata si pu o anche scrivere: Wd = −∆Ek = 1 7Mv0. Il lavoro della forza d’attrito e L = µMg∆xp = 1 7Mv2 0 QED.
Soluzione dell’esercizio 6.8 1. Se reazione vincolo e nulla: J = R Frisdt = R Fdt = ∆p = Mvcm Quindi vcm = J M Calcolo i momenti rispetto al polo O: R Fxdt = Jx = ∆L = Ml2 3 vcm L/2 = 2 3Mlvcm, da cui ricavo x = 2 3l = 40 cm.
2. vcm = 0.5 m/s 3. mg L 2 (1 −cos θmax) = 1 2IOω2 = 2 3Mv2 cm θmax = 0.34 rad 4. equazione del moto e ¨ θ + 3g 2l θ = 0, quindi T = 1.27 s.
5. vmax per θ = 0 e proprio vcm del punto 2. amax . . .
Soluzione dell’esercizio 6.9 1. EA k = 1 2IAω2 a Ia = MR2 2 + MR2 + mr2 2 + m(R −h)2 = 4.4kgm2 ωA = vA R = 5rad/s, EA k = 55 J.
2. EB k = EA k −∆Ugrav = EA k −2mgh = 20 J = 1/2IB(v2 B/R2).
Quindi: vb = q 2EB KR2 IB = 0.73 m/s.
IB si calcola come nel punto precedente, tenendo conto che adesso il disco piccolo si trova ad una distanza R + h dal punto B.
IB = 1/2MR2 + MR2 + 1/2mr2 + m(R + h)2 = 6.6 kgm2 3. NA = (M + m)g + mω2 Ah = 457 N.
Il secondo addendo e dovuto alla forza centripeta necessaria per far ruotare il disco piccolo con una velocit a angolare ωA attorno al centro del disco grande.
4. stesso conto, ma disco piccolo adesso si trova in alto, quindi la direzione della forza centripeta e opposta.
Per ricavare ωB = q 2EB k IB = 2.4 rad/s NB = (M + m)g −mω2 Bh = 401 N Soluzione dell’esercizio 6.10 1. m3 = 3 kg 2. Fatt = 29.4 N verso l’alto.
3. a1 = a3 2 = 0.45 m/s2 4. dalla eq. della dinamica del corpo m3: T23 = m(g −a3) = 35.7N; per il corpo m2: 1 2m2R2 2 a3 R2 = R2(T12 −T23), da cui T12 = 33.4 N Da cui ricavo N2 = 103.4ˆ y + 32.4ˆ x N 5. Ek = m3gh −m1g h 2 sin θ = 9.8 N.
Soluzione dell’esercizio 6.11 1. Conservo L rispetto al perno: mv(R+L/2) = Iω = MR2 2 + ml2 12 + m(R + L/2)2 ω da cui ω = 1.74 rad/s.
S. Lacaprara Es. Meccanica e Termodinamica 84 2. J = (M + m)vcm −mv0 = (M + m)ω m(R+L/2) M+m −mv0 = −0.064 Ns; 3. ∆Ek = 1 2Iω2 −1 2mv2 0 = −0.016 J; 4. Serve una forza centripeta F = mω2(R + L/2) che e fornita da parte della forza peso e dalla forza di interazione. Quindi: Fint = m (ω2(R + L/2) −g cos θ); 5. Non ci sono forze impulsive durante il distacco: quindi il disco continua a ruotare con ω′, mentre la sbarretta continua a ruotare con ω e il suo centro di massa si muove verso il basso con v = ω′(R+L/2) e il moto e gravitazionale. Si noti che ω′ cambia rispetto ad ω subito dopo l’urto perch e durante la discesa agisce la forza peso sulla sbarretta, accelerando la rotazione. Si puo calcolare con la conservazione dell’energia: ω′ = 6 rad/s (da controllare) Soluzione dell’esercizio 6.12 1. Dalla conservazione dell’energia: 1 2(M + M ′)v2 + 1 2Icω2 −(M + M ′)gd sin δ = −µMg cos δd da cui ricavo v (omega = v/R); 2. Equazione della dinamica per i due corpi, proiettate lungo il moto: Ma = Mg sin δ −µMg cos δ −Fsbarretta M ′a = M ′g sin δ −Fatt stat + Fsbarretta Da cui ricavo (M + M ′)a = (M + M ′)g sin δ −µMg cos δ −Fs = cost e a = v2 2d.
Quindi Fs = (M + M ′)(g sin δ −a) −Mg cos δµd 3. la ricavo da una delle equazione del moto; 4. Calcolando i momenti delle forze sul cilindro rispetto al CM del cilindro: FsR = Icα , quindi Fs = M′R2 2 a R2 = M′a 2 M ′g sin δ −Fs = M ′(g sin δ −a/2) = M ′a, quindi a = 2 3g sin δ Forza nulla quando i due corpi si muovono con la stessa accelerazione Mg(sin δ −µ cos δ) = M ′a = M ′ 2 3g sin δ, da cui ricavo δ Soluzione dell’esercizio 6.13 1. Usando anche condizione di puro rotolamente delle ruote; 4Fattrito =Ma = MαR τfreno −FattritoR =1/2mR2α da cui ricavo α = τfreno 1/2mR2+1/4MR2 = 10 rad/s Fattrito = MαR/4 = 12.5 N 2. Rotolamento ruote: ω = ω0 −αt, quindi il carrello si arresta in t = ω0/α = v0/(Rα) = 1.2 s.
Nel frattempo il carrello percorre: s = v0t −1/2at2 = 3.6 m 3. Posso fare Wd = 1/2Mv2 0 + 4 · 1/2mR2/2ω2 0 = 216 J; Oppure Wd = 4τfrenoθ = 4τfrenos/R S. Lacaprara Es. Meccanica e Termodinamica 85 4. In questo caso la forza di attrito tra ruote e piano e µdMg = Ma, quindi a = µdg s′ = v2 0 2a = 9.2 m; Soluzione dell’esercizio 6.14 1. Conservo energia: mgr −τπ/2 = 1/2Itotω2, Itot = MR2 2 + 3mR2 2 da cui ω = 4.62 rad/s 2. Conservo momento angolare rispetto al centro del disco: Itotω = Itotω′ + m3v′ 3R, v′ 3 = 1.4 m/s; 3. J = mω′r + m3v′ 3 −mωr = 1.34 Ns.
Soluzione dell’esercizio 6.15 1. Itot = mL2 3 + MR2 2 + ML2 Conservo energia: Itotω2 2 = (mL/2 + ML) g(1 −cos θ), da cui ω = 3.02 rad/s; 2.
Itotω2 2 = (mL/2 + ML) g(1 −cos θ′) + Idiscoω2 2 da cui θ′ = 56.5◦ 3. T = 2π q Itot (m+M)dcmg = 2.08 s 4. Ltot = Lasta + Ldisco = Iasta + LM(Lω) + Idiscoωdisco, derivo: dL dt = (Iasta + ML2)α, usando il fatto che ωdisco = cost.
Da cui ricavo: T = 2π q Iasta+ML2 (m+M)dcmg = 1.97 s In altre termini, il disco contribuisce al momento d’inerzia del pendolo composto solo come un punto materiale all’estremita dell’asta, dato che il disco stesso e libero di ruotare.
Soluzione dell’esercizio 6.16 1. equazione in equilibrio FD −FB = (md + ma) = 0 τ −FDR = Iα = 0 Da cui FB = 3 N; 2. equazione senza attrito: FD = (md + ma) τ −FDR = mdR2 2 a R Da cui: a = .375 m/s2 3. Scrivo le equazione della dinamica, con i momenti rispetto al CM dell’asta. Da no-tare che c’e anche la reazione vincolare del piano su cui e appoggiata una estremita dell’asta.
S. Lacaprara Es. Meccanica e Termodinamica 86 Nx =maa Ny + Nb =mag NbL/2 cos θ + NxL/2 sin θ =NyL/2 cos θ Quindi: Nx =maa Ny =ma g 2 + a tan θ 2 = 25.0 N Soluzione dell’esercizio 6.17 1. Chiamo x la distanza del CM dalla ruota anteriore, proiettata sul piano, e quindi L−x e’ quella rispetto alla ruota posteriore. Ricavo x pensando la mia motocicletta come due punti materiali, uno sulla ruota posteriore e uno sull’anteriore, con massa pari a MRp/a rispettivamente, e calcolo la posizione del CM. Alternativamente posso calcolare la II eq. cardinale della meccanica rispetto al punto di contatto della ruota posteriore e rispetto a quella anteriore. Trovo Na + Np = Mg e lNp − Mgx = lMg ∗Rp −Mgx = 0 e , da cui ricavo: x = l(Rp) = l(1 −Ra) = 66.6 cm amax = g L−x h m/s2; 2. La coppia motore si trasmette alla ruota posteriore tramite la trasmissione prima-ria e quella secondaria (catena), quindi il momento applicato alla ruota posteriore e’ pari a τp = 103 · 2.384 · 45/15. Quindi l’accelerazione massima che viene ap-plicata alla moto sara’ pari a F = τp/Rp = Ma, quindi l’accelerazione risulta a = 8 m/s2 La confronto con il risultato del punto precedente e osservo che: a < amax = 11.4 m/s2, quindi no, non ce la fa.
3. NA,P = N 0 A,P ± F h L, dove N 0 A,P e la reazione della strada con veicolo fermo; 4. τmax = FmaxR = Mg x hR Soluzione dell’esercizio 6.18 1. Uso come coordinate (x, y) del centro di massa dell’asta e l’angolo θ con l’asse verticale. Scrivo la seconda equazione cardinale rispetto al C.M.: M ¨ x = Nx M ¨ y = Ny −Mg Ml2 12 ¨ θ = −Nx l 2 cos θ + Ny l 2 sin θ Usando: x = l 2 sin θ e y = l 2 cos θ, e di conseguenza ¨ x = −l 2 sin θ ˙ θ2 + l 2 cos θ¨ θ e ¨ y = −l 2 cos θ ˙ θ2 −l 2 sin θ¨ θ, si ricava: S. Lacaprara Es. Meccanica e Termodinamica 87 ¨ θ = 3g 2l sin θ, ovvero, dato che serve ricavare ω = ˙ θ ωdω = 3g 2l sin θdθ che integrata tra θ = [0, θ], porge: ω2 = 3g l (1 −cos θ), quindi ω(θ = π/2) = q 3g l Si puo risolvere anche conservando l’energia: Mgy + 1 2M ( ˙ x2 + ˙ y2) + 1 2 Ml2 12 ˙ θ2 = Mg l 2 e usando ˙ x, ˙ y come prima, si ottiene direttamente ˙ θ2 = 3g l (1 −cos θ) 2. Nx = −l/2M(sin θ ˙ θ2 −cos θ¨ θ) = 3 2Mg 3 2 cos θ −1 sin θ Nx = 0 per θ = 0 Ny = Mg −Ml 2 (cos θ ˙ θ2 + sin θ¨ θ) = Mg 1 −3 2 cos θ + 3 2 cos2 θ −3 4 sin2 θ Ny = Mg per θ = 0, Ny = Mg 4 per θ = π/2 3. Nx = 0 quando 3 2 cos θ −1 = 0, quindi cos θ = 2 3 Soluzione dell’esercizio 6.19 1. τO = −dF + d/2Mg = Icuboα, quindi Forza minima e: F > Mg/2 = 66.7 N 2. Icubo = 2 3Md2; α = τ/Icubo = 30.8 rad/s2 (orario) 3. lungo ˆ x: +F + Rx = mα 2 d, Rx = −74.9 N lungo ˆ y: −Mg + Ry = mα 2 d, Ry = 158.5 N 4. Inizia a cadere quando e inclinato di pi u di 45◦. Il lavoro e: LF = R ⃗ Fd⃗ s = Fd = 12 N 5. conservazione energia: 1 2IOω2 = LF −Mgd 2 √ 2 −1 , da cui ω = 11.5 rad/s Soluzione dell’esercizio 6.20 1. in coordinate polari, la distanza dall’asse z e ρ = r sin ϕ, e l’elemento di volume e dV = r2 sin ϕdrdθdϕ, e l’elemento di massa e dm = M 4 3 πR3dV . Quindi l’integrale diventa: Iz = M 4 3πR3 Z V ρ2dV = M 4 3πR3 Z R 0 dr Z 2π 0 dθ Z π 0 dϕr2 sin2 ϕr2 sin ϕ = M 4 3πR3·R5 5 ·2π·4 3 = 2 5MR2 2. Se considero un disco di spessore dz ad una quota z, ha un raggio r = √ R2 −z2, e il suo momento d’inerzia rispetto all’asse z e: dIz = M 4 3πR3π(R2 −z2)dz · (R2 −z2) Integrando tra [−R, R], si ottine Iz = Z +R −R dIz = 2 5MR2 S. Lacaprara Es. Meccanica e Termodinamica 88 3. Il momento d’inerzia di un guscio sferico di raggio r e spessore dr si pu o calcolare osservando che deve essere lo stesso per ciascuno dei tre assi cartesiani x, y, z dIz = Z V (x2 + y2)dm = dIx = Z V (z2 + y2)dm = dIy = Z V (z2 + x2)dm , e dIz = dIx + dIy + dIz 3 = 2 3 Z V (x2 + y2 + z2)dm = 2 3r2dm = 2 3r2 M 4 3πR34πr2dr Integrando tra [0, R], si ottiene Iz = Z R 0 2 3r2 M 4 3πR34πr2dr = 2 3 M 4 3πR34πR5 5 = 2 5MR2 Soluzione dell’esercizio 6.21 1. Equazioni cardinali, ˆ x verso il basso, ˆ uθ orario.
Assume che ⃗ Fatt abbia verso opposto a ˆ x (da discutere una volta risolto il sistema). Per l’equazione dei momenti delle forze posso usare indifferentemente due poli: il CM oppure il punto di contatto tra la sfera e il piano: Ma = Mg √ 2 2 −F 2 5MR2α = FR rispetto a CM sfera 2 5MR2 + MR2 α = Mg √ 2 2 R rispetto a punto di contatto sfera/piano α = a R moto puro rotolamento Si ottiene F = √ 2 7 Mg = 0.2 N (verso l’alto), a = 5 √ 2 14 g = 4.95 m/s2 (da confrontare con a = √ 2 2 g = 6.93 m/s2 di un corpo che scivola senza rotolare). Da notare che i risultati sono indipendente dal raggio della sfera.
La forza d’attrito e quindi diretta verso l’alto: ci sono diversi modi per convincer-sene anche intuitivamente.
Per prima cosa, notiamo dal punto successivo che la sfera arriva al piano con una velocit a minore di quella che avrebbe se fosse un punto materiale sottoposto alla sola forza di gravita (v = √2gh), il che vuol dire che la forza risultante che agisce sul suo CM deve essere minore di quella di gravit a, e quindi che la forza d’attrito deve essere diretta in direzione opposta al moto.
Un altro modo e considerare la rotazione della sfera, che e chiaramente oraria. Se considero come polo di rotazione il punto di contatto con il piano inclinato, l’unico momento esterno e quello della forza di gravit a, ed ha la direzione giusta per far ruotare la sfera in senso orario. Se viceversa considero come polo di rotazione il CM, l’unico momento esterno e quello della forza di attrito, che deve essere tale da far ruotare la sfera sempre in verso orario: quindi la forza d’attrito deve essere rivolta verso l’alto.
S. Lacaprara Es. Meccanica e Termodinamica 89 2. Mgh = 1 2Mv2 + 1 2ICMω2 = 7 10Mv2, da cui v = q 10 7 gh = 3.35 m/s.
3. Quando il piano diventa liscio non ci possono pi u essere momenti esterni rispetto al CM (l’unica forza e la gravit a - con braccio nullo - e la reazione vincolare - pure con braccio nullo - ), quindi ω rimane costante.
Attenzione che il punto di contatto non e un buon polo perch e, data l’assenza di forze di attrito, non abbiamo la garanzia che il moto sia di puro rotolamento, anzi, siamo sicuri che non lo sia. Quindi il punto di contatto non e istantaneamente fermo, e quindi se lo volessimo usare come polo, dovremmo tenere conto del fatto che ha velocit a non nulla, e quindi la seconda equazione cardinale avrebbe anche il termine vΩ× pCM ̸= 0.
Quindi 1 2Mv2 = Mghmax, (la componente rotazionale dell’energia cinetica rimane identica): hmax = 57 cm. Minore di quella di partenza perche parte dell’energia iniziale e presente come energia di rotazione.
4. il sistema e simmetrico, quindi h′ max = h = 80 cm.
Soluzione dell’esercizio 6.22 1. La forza di attrito statico non fa lavoro, quindi l’energia si conserva: 1 2Mv2 0 + 1 2 MR2 2 ω2 0 = 1 2k∆2x, ω0 = v0 R (puro rotolamento), da cui: ∆x = v0 q 3M 2k = 0.11 m.
2. Scelgo in modo arbitrario la direzione della forza di attrito statico, del disco con il piano, verso sinistra, ovvero nella direzione del moto iniziale del disco e opposta alla forza elastica (con molla in compressione). Scelgo inoltre il verso positivo della velocit a angolare come antiorario, quindi la condizione di puro rotolamente risulta v = ωR (con il segno positivo). Il momento della forza d’attrito con questa direzione e quindi negativo.
Ma = Fel −Fatt Iα = −RFatt da cui: Fatt = Fel 3 , che e massima quando la molla e massimamente compressa Fatt,max = 1.5 N. Il segno positivo indica che la forza effettiva e veramente diretta verso sinistra, come forse e possibile capire intuitivamente.
3. Fatt non e una forza impulsiva, quindi p si conserva. Mv0 + mv1 = 0, v1 = v0 M m = 10 m/s 4. Il momento angolare rispetto al centro del disco si conserva durante l’urto. Viene persa solo l’energia cinetica di traslazione W = 1 2Mv2 0 + 1 2mv2 1 = 2.2 J.
5. Dopo l’urto il disco striscia per terra, visto che continua a ruotare: infatti il momento angolare del disco non cambia a seguito dell’urto.
L’unica forza e quella d’attrito dinamico Fatt,d = µ(M + m)g, quindi a = Fatt,d M+m = µg = 5.9 m/s Soluzione dell’esercizio 6.23 S. Lacaprara Es. Meccanica e Termodinamica 90 1.
MR2 2 αA =τ −µMgR 2 MR2 2 αA = + µMgR 2 da cui ricavo αA = 19.5 rad/s2 e αB = 9.8 rad/s2.
Visto che i due dischi hanno entrambi un moto uniformemente accelerato, gli angoli percorsi (e quindi i giri) sopo proporzionali alle accelerazione. Quindi NB = αB αA NA = 10 giri 2. tNA = q NA2π αA = 3.6 s; ωA = αAtNA = 70.1 rad/s, ωB = αBtNA = 35.2 rad/s Si potrebbe fare anche con la conservazione dell’energia, usando il fatto che le velocit a angolari sono anche loro proporzionali alle accelerazioni, e che il sistema assorbe energia dal lavoro del momento esterno (vedi punto successivo) e ne dissipa per il lavoro delle forze di attrito Wd = −µMgR/2(NA −NB)2π.
3. W = R τdθ = τ(NA)2π = 1.11 kJ 4. conservo il momento angolare totale rispetto all’asse di rotazione.
Lini = MR2 2 ωA + MR2 2 ωB = τtNA = Lfin = 2MR2 2 ωf ωf = ωA+ωB 2 = 52.5 rad/s 5.
Wd = 21 2 MR2 2 ω2 f − 1 2 MR2 2 ω2 A + 1 2 MR2 2 ω2 B = 173 J 6. dal punto precedente Wd = τatt∆θAB, quindi NAB = ∆θAB 2π = Wd µMgR/2 = 1.87 giri Soluzione dell’esercizio 6.24 1. LO = 2 5M R 2 2 ω0 = Itotω1 dove Itot = MR2 2 + 2 5M R 2 2 + M R 2 2 = 17 20MR2 ω1 = 2 17ω0 = 3.7 rad/s 2. Ediss = 1 2 2 5M R 2 2 ω2 0 −1 2Itotω2 1 = 1.74 J 3. θ = tan−1 Mω2 1 R 2 (M+M)g = 0.07 rad Soluzione dell’esercizio 6.25 1. Forze esterne sono solo verticali, quindi CM oscilla verticalmente tra L/2 e L/2(1− cos θ0); 2. risulta un moto armonico con periodo T = 2π q L 6g = 1 s Partiamo dall’energia totale del sistema, che si conserva, e e la somma di potenziale e cinetica. L’energia potenziale in funzione dell’angolo θ del disegno si pu o scrivere come: U(θ) = mgl 2 (1 −cos θ) S. Lacaprara Es. Meccanica e Termodinamica 91 mentre l’energia cinetica risulta: Ek(θ) = 1 2Mv2 cm + 1 2 ML2 12 ˙ θ2 La v2 cm = ˙ x2 cm + ˙ y2 cm, dove xcm(t) = 0 e costante, mentre ycm = L 2 cos θ, quindi rimane solo la velocit a verticale: ˙ ycm = −L 2 sin θ ˙ θ.
Quindi Ek(θ) = 1 2 ML2 4 sin2 θ ˙ θ2 + 1 2 ML2 12 ˙ θ2 L’energia totale si puo scrivere come Etot(θ, ˙ θ) = mgL 2 (1 −cos θ) + 1 2 ML2 4 sin2 θ ˙ θ2 + 1 2 ML2 12 ˙ θ2 Derivando per il tempo, risulta dEtot dt = ¨ θ cos2 θ + 1 3 −sin θ cos θ ˙ θ2 −2g L cos θ = 0 che e l’equazione del moto nel caso generale (oscillazioni generiche, non piccole), decisamente complessa e impossibile da integrare in modo simbolico.
Invece sviluppando l’energia totale per piccolo oscillazioni attorno a θ = 0 (verti-cale), risulta l’equazione semplificata per piccole oscillazioni: Etot = 1 2 MgL 2 θ2 + 1 2 ML2 12 ˙ θ2 che derivata mi da una eqazione armonica con pulsazione ω2 = MgL/2 ML2/12 = 6g L , quindi T = 2π q L 6g = 1 s 3. Dalla conservazione dell’energia, usando l’espressione del punto precedente, si ottiene ˙ θ = q 12g(1−cos θ0) L = 6.2 rad/s 4. Devo bilanciare la forza peso e anche garantire il moto circolare del CM: N = Mg(7 −6 cos θ0) = 78.5 N diretta verso l’alto.
Soluzione dell’esercizio 6.26 1. Considero i momenti delle forze rispetto al punto di contatto: τtot = F(R − h) −Mg √ 2Rh −h2 = Iα. Per avere sollevamento deve risultare τ > 0, quindi F > Mg √ 2Rh−h2 (R−h) = 261 N, quindi la risposta e si.
2. uso la conservazione dell’energia: Ek = 1/2(mr2/2 + mr2)ω2 = F √ 2Rh −h2 − mgh, dove il primo termine e il lavoro della forza esterna. Risulta ω = 6.56 rad/s e quindi v = ωr = 1.64 m/s 3. di nuovo conservazione dell’energia (senza lavoro forza F): mgh = 1/23/2mr2ω′2, quindi ω′ = 4.57 rad/s e v = ω′r = 1.14 m/s Soluzione dell’esercizio 6.27 S. Lacaprara Es. Meccanica e Termodinamica 92 1. Il lavoro del momento esterno e Wext = Z 10π 0 dθ = 240π2 = 2370 J . Questo si trasforma in energia cinetica del sistema: Ek = 1 2M L2 12ω2 0 + 21 2mL2 4 ω2 0 dato che i dischi non ruotano attorno ai loro assi, ma il loro CM si muove con velocit a vcm = ω0L/2. Quindi ω0 = 137.7 rad/s 2. Il momento angolare totale si conserva, e il disco A ruota attorno al suo CM con la stessa velocita dell’asta.
Lini = ML2 12 + m L2 4 ω0 = Lfin = h ML2 12 + m L2 4 + mr2 2 i ω1, ω1 = 128.5 rad/s.
3. di nuovo conservazione momento angolare: nello stato finale il sistema e un corpo rigido.
Lini = mr2 2 ω1 = Lfin = ML2 12 + 2( mr2 2 + m L2 4 ) ω2, ω2 = 8.1 rad/s 4. Wd = ∆Ek 5. dopo il distacco, sul disco B non agiscono forze ne momenti, quindi vcm = ω2L/2 = 4.0 m/s tangenziale e ω = ω2 6. Devo supportare il sistema asta+disco A Rg = (M + m)g verso l’alto.
Inoltre il CM del sistema non e piu sull’asse di rotazione, e quindi fa un moto circolare e quindi devo fornire una forza centripeta Rc = (M + m)ω2 2rcm dove rcm = mL/2/(M + m) = 20 cm.
Inoltre la forza peso esercita un momento rispetto all’asse perch e il CM e spostato rispetto all’asse. τg = (M + m)grcm che ruota con il sistema.
La soluzione nella versione precedente era sbagliata! scusate Infine, il sistema non e piu simmetrico, quindi il tensore di inerzia potrebbe non essere piu’ diagonale. Se cos ı fosse, allora il momento angolare precederebbe e quindi avrei bisogno di un momento delle forze che precede pure lui.
In questo caso specifico, tuttavia, si puo verificare con un conto esplicito che il tensore d’inerzia rimane diagonale, grazie al fatto che il disco si trova sullo stesso piano dell’asta. Si chiamiamo z l’asse verticale, gli elementi fuori diagonale sono −Σimixi(yi)zi, in particolare propozionale a zi che e la distanza dal piano delle masse mi (nel nostro caso il disco). Essendo z di tutto il disco nulla, allora questi elementi fuori diagonale sono pari a zero, e quindi il tensore e diagonale.
Se invece il disco fosse sopra o sotto, zi ̸= 0 e quindi ci sarebbero elementi non diagonali diversi da zero. In questo caso, visto che ω = ω2 lungo l’asse di rota-zione (z), il momento angolare, che non e parallelo a ω precede attorno all’asse di rotazione, cioe ha una componente sul piano di rotazione. Quindi avrei bisogno di un momento torcente rotante per garantire il moto di precessione. τprec = dL/dt NB. per fornire momenti il supporto dell’asse di rotazione deve essere applicato in due punti, con una distanza non nulla tra di loro (per esempio due cuscinetti uno sopra e uno sotto), in modo da poter generare una coppia di forze uguali e opposte e quindi un momento.
S. Lacaprara Es. Meccanica e Termodinamica 93 Parte III Fluidi S. Lacaprara Es. Meccanica e Termodinamica 94 7 Fluidi Esercizio 7.1 Un rubinetto di sezione S = 1 cm2 si trova sul fondo di una grande cisterna cilindrica aperta superiormente, e profonda H = 4 m. Il getto dal rubinetto esce verso il basso.
Calcolare, trascurando tutti gli attriti: 1. la velocit a di uscita dell’acqua dal rubinetto; 2. la sezione s del getto d’acqua h = 20 cm sotto il rubinetto.
Esercizio 7.2 Un recipiente a forma di prisma, con base quadrata (lato l = 20cm) e superficie laterale a forma di triangolo rettangolo isoscele, appog-giato su una base su un piano orizzontale, e rimpito fino a 3 4 di acqua ed e aperto superiormente. Al centro della faccia quadrata vertica-le e praticato un foro con una valvola che fa uscire Qv = 0.01 l/s: la valvola e inizialmente chiusa.
l 3 4l l l Calcolare, all’apertura della valvola: 1. la velocita di discesa del livello dell’acqua; 2. la velocit a di uscita dell’acqua dal foro; 3. la forza necessaria per tenere fermo il contenitore.
Esercizio 7.3 Un tubicino rigido, di lunghezza l si trova su un piano orizzontale ed e vincolato ad un’estremit a attorno alla quale puo ruotare senza attriti. Dentro il tubicino scorre acqua che esce dall’estremit a libera del tubo con velocita u, perpendicolamente al tubo stesso, da una sezione S, piccola rispetto al tubo.
Il tubo e inizialmente in quiete, e il suo momento d’inerzia rispetto al vincolo e I.
1. calcolare l’equazione del moto del tubo.
Esercizio 7.4 Un serbatoio si trova con il pelo dell’acqua a h = 32 metri rispetto al suolo, e ha un diametro di D = 3m. Esso fornisce acqua ad una casa tramite un tubo di diametro 1” = 2.54 cm, a livello del suolo. La portata del tubo e Q = 2.5 · 10−3m3/s.
Un secondo tubo d = 1/2” arriva al secondo piano h′ = 7.2 m.
Calcolare, trascurando la viscosita dell’acqua: 1. la pressione del tubo principale se il rubinetto principale della casa e chiuso; 2. idem se e completamente aperto.
3. la velocit a di uscita dell’acqua dal secondo tubo; 4. la pressione in questo tubo.
S. Lacaprara Es. Meccanica e Termodinamica 95 Esercizio 7.5 Un recipiente cilindrico, altezza h = 0.6 m e raggio R = 0.3 m, e riempito con un liquido di densit a ρ = 1.1 g/cm3, e ruota attorno al suo asse con ω. A riposo e riempito fino all’altezza h0 = 0.4 m.
Calcolare: 1. ωmax perch´ e il liquido non esca dal recipiente; 2. la pressione sul fondo sull’asse di rotazione pA; 3. la pressione sul fondo sulla parete del recipiente pR.
Esercizio 7.6 Un condotto orizzontale e formato da una parte tronco conica con alla fine attaccato un tubo cilindrico (inbuto). Dentro scorre acqua ρ = 1 · 103kg/m2 grazie ad una sovra-pressione all’ingresso. La sezione del cilindro ha raggio r2 = 1 mm, l’apertura del cono r1 = 2mm. La pressione all’uscita e p3 = 1.025 · 105Pa, mentre la velocit a dell’acqua all’entrata e: v1 = 0.1 m/s. La viscosit a dell’acqua vale η = 1.2 · 10−3kg/(ms), e si puo trascurare nel tratto conico.
1. la portata del condotto Q; 2. le pressioni all’ingresso p1 e alla fine del cono p2.
Esercizio 7.7 Un recipiente contiene una colonna alta h = 50 cm di liquidi con ρ = 0.8 · 103kg/m3.
Alla base c’ e un tubicino lungo l = 1 m con raggio r = 3 mm. Utilizzando vmedia nel tubo, calcolare: 1. ηmax perche il moto sia laminare; 2. la portata del liquido Q se η = ηmax 3. l’energia dissipata per unit a di volume.
Esercizio 7.8 Un tubo di acquedotto e disposto obliquamente con pendenza 1/600, vi scorre acqua in regime laminare sotto gravit a. La portata e Qv = 4.0 · 10−5m3/s.
Calcolare: 1. il diametro del tubo; 2. la velocit a media dell’acqua; 3. l’energia persa da un volume V = 0.15 m3 in l = 10 m.
Esercizio 7.9 Un tubo a U, aperto superiormente, con sezione s = 1 cm2, si trova su un piano verticale, ed e riempito di acqua per una lunghezza totale di l = 1 m. Viene applicata una pressione in uno dei due rami della U si porta l’altezza della colonna d’acqua h = 10 cm S. Lacaprara Es. Meccanica e Termodinamica 96 pi u in alto rispetto all’equilibrio e all’istante t = 0s tale pressione viene riportata a quella atmosferica.
Calcolare: 1. equazione del moto della massa d’acqua; 2. l’equazione del moto se l’acqua e sottoposta ad una forza di attrito F = λv, con λ = 0.6 kg/s.
Esercizio 7.10 Un serbatoio e riempito fino ad altezza H da un liquido di densita ρ. Su una parete, ad una altezza h dalla base, si trova un foro di sezione s0.
Calcolare: 1. la distanza ∆x rispetto al piede del serbatoio dove l’acqua arriva a terra; 2. la sezione s′ del getto d’acqua quando tocca terra; 3. se esiste una seconda altezza h′ per cui l’acqua arrivi alla stessa distanza; 4. l’altezza h” per cui la distanza ∆x” e massima.
Esercizio 7.11 Un recipiente e riempito da mercurio (ρHg = 13.6 · 103 kg/m3) e acqua (ρH2O = 1.0 · 103 kg/m3). Un cubetto di ferro (ρFe = 7.87 · 103 kg/m3) di spigolo l = 6 cm e in equilibrio.
1. Calcolare la posizione del cubetto rispetto all’interfaccia mercurio-acqua.
Esercizio 7.12 Una sferetta di legno, di raggio r = 2 mm e densita ρ = 0.7 g/cm3 viene lasciata libera sul fondo di un recipiente pieno d’acqua di altezza h = 1 m, e raggiunge subito una velocit a costante.
ηacqua = 1.7 · 10−3 kg/(ms), Cx = .5 Determinare: 1. il tipo di forza di attrito che subisce nella salita; 2. l’altezza z che raggiunge dopo essere uscito dalla superficie, trascurando effetti di superficie; 3. l’enegia dissipata per attrito durante la salita.
Esercizio 7.13 Una bolla di sapone, di raggio r1 = 38.2 mm, e collegata ad una cannuccia di diamtero interno d = 1.08 mm e lunghezza l = 11.2 cm. L’estremit a libera della cannuccia e in aria a pressione atmosferica.
Tensione superficiale τsapone = 2.5 · 10−1 N/m, viscosit a aria: η = 1.8 · 10−5 Ns/m2 1. Calcolare il tempo necessario per ridurre la bolla ad un raggio r2 = 21.6 mm.
S. Lacaprara Es. Meccanica e Termodinamica 97 Soluzione dell’esercizio 7.1 1. p0 + ρgH = p0 + 1 2ρv2, v = √2gH = 8.86 m/s 2.
1 2ρv2 −1 2ρv02 = ρgh, quindi v = p 2gh + v2 0. Portata e costante, Q = Sv0 = sv, quindi s = Sv0 √ 2gh+v2 0 = 0.95 cm2 Soluzione dell’esercizio 7.2 1. Vprisma = lh (l+(l−h)) 2 = l2h −lh2 2 dV dt = ˙ h(l2 −lh) = −Qv, quindi: ˙ h = 1 mm/s.
2. vout = q 2g(h −l 2) = 1.0 m/s 3. F = ρQvvout = 1 · 10−5 N Soluzione dell’esercizio 7.3 1. il sistema acqua+tubo e isolato (rispetto al piano orizzontale), quindi Ltot = 0 = Ltubo + Lacqua si conserva.
Considero l’uscita dell’acqua nel tempo dt: 0 = dLz = Idω −(ρSudt)vacqual: nel sistema del laboratorio: vacqua = u −ωl Quindi dω u−lω = ρSul I , da cui integrando ottengo ω(t) = u l 1 −e−ρSul2 I t Da notare che ω tende asintoticamente a u l , quando la velocita dell’acqua che esce dal tubo nel sistema del laboratori sarebbe nulla.
Soluzione dell’esercizio 7.4 1. Sia A il serbatoio, B il tubo da 1” e C quello da 1/2”.
A →B: p0 + ρgh + 1 2ρv2 A = pB + 1 2ρv2 B, con vA = vB 1”2 D2 ≪vB Quindi pB = p0 + ρgh −1 2ρv2 B. vB = Q π(1”/4)2 = 4.9 m/s da cui: pB = 4.03 · 105Pa, 2. se chiuso: p′ B = 4.15 · 105Pa 3. vC = Q ( π 1 2”/2)2 = 19.7 m/s 4. pC = p0 + 1 2ρ(v2 A −v2 C) + ρgh′ = 1.49 · 105Pa Soluzione dell’esercizio 7.5 1. L’energia potenziale per unit a di massa e U = gz −1 2ω2r2, quindi le superfici equipotenziali sono z = 1ω2r2 2g + k. Per calcolare quella di interfaccia con l’aria, si pu o calcolare il volume del liquido quando il sistema e in rotazione, e equagliarlo a quello a riposo V = πR2h0 = R R 0 z(r)r2πdr, da cui si ricava k = h0 −ω2R 4g , quindi l’espressione generale della superfice di interfaccia con l’aria e: z = ω2r2 2g + h0 −ω2R 4g . Quindi zmax = z(r = R) = h0 + ω2R 4g ≤h S. Lacaprara Es. Meccanica e Termodinamica 98 da cui ricavo un ωmax = q 4(h−h0)g R2 = 3.34 rad/s 2. In generale ⃗ ∇p = −ρ⃗ ∇U. Integro lungo un cammino, lungo l’asse, dalla superfice al fondo.
R Γ ⃗ ∇pd⃗ l = ∆p = pA −patm = R Gamma −ρ⃗ ∇Ud⃗ l = −ρ R 0 h −gdz = ρgh, dove h = z(r = 0) = h0 −ω2R 4g pA = patm + ρg h0 −ω2R 4g 3. Posso fare un integrale simile al precedente, ma partendo sulla superficie non sull’asse, ma sul bordo del recipiente, quindi da h = z(r = R) = h0 + ω2R 4g a 0.
p1 = patm + ρg h0 + ω2R 4g Oppure posso partire da p0 e spostarmi radialmente fino al bordo, ottenendo, con un diverso integrale, lo stesso risultato.
Soluzione dell’esercizio 7.6 1. Q = πr2 1v1 = 1.257 · 10−6 m3/s 2. Da Hagen-Poiselle, p2 = p3 + 8ηlv2 r2 2 = 1.0228 · 10−5 Pa.
Da Bernoulli: p1 = p2 + 1 2ρv2 1 r1 r2 2 −1 = 1.0296 · 10−5 Pa Soluzione dell’esercizio 7.7 1. Pressione alla base: ∆p = ρgh, vm = ∆pr2 l8η , da cui posso calcolarmi il numero di Reynolds Re, imporre che sia Re < 1000, e ricavarmi η ≥3.0 · 10−3 kg/ms 2. Qv = vmπr2 = ∆pπr4 l8η = 4.12 · 10−5 m3/s = 41.2 ml/s, vm = 1.46 m/s 3. senza tubo: vout = √2gh = 3.13 m/s quindi Wd/V = 1 2vρ (v2 m −v2 out) = 3.0 · 10−3 J/m3 Soluzione dell’esercizio 7.8 1. Forza diretta verso la discesa e F = SLg h L = Sρgh = S∆p, da cui ∆p = ρgh Uso H-P: r = 4 q 8η πρgθ = 0.11 m 2. vm = Q πr2 = 1.1 · 10−3 m3/s 3. ∆E = Mgh = ρvgLθ = 24 J NB: Re = 126, quindi moto e effettivamente laminare.
Soluzione dell’esercizio 7.9 1. e un moto armonico x(t) = h sin ωt, con ω = q 2g l = 4.43 rad/s 2. e un moto armonico smorzato x(t) = e−3t0.136 sin(3.26t + 0.827) S. Lacaprara Es. Meccanica e Termodinamica 99 Soluzione dell’esercizio 7.10 1. ∆x = 2 p h(H −h) 2.
3. soluzione simmetrica se scambio h′ = H −h 4. ∆x e massimo se h = H/2 Soluzione dell’esercizio 7.11 1. Si dimostra che Bernoulli funziona anche per due liquidi.
La forza sul blocchetto e: F = l2g (ρHgx + ρH2O(l −x)) −l2gρFe da cui si ricava x = 3.2 cm.
Soluzione dell’esercizio 7.12 1. Se valesse la legge di Stokes, vregime = 2r2(ρ0−ρL)g 9η = 1.54 m/s che fornisce un numero di Reynolds Re = 1 · 103 ≫0.6, quindi non posso usarla.
La forza di attrito e quindi F = −1 2Cxρ0π()r2v2 che fornisce una velocit a di regime: vregime = 0.177 m/s 2. h = v2 2g = 1.6 mm 3. la posso calcolare come Wd = Fatth oppure come Wd = Vsferettag(ρO −ρL)h − 1 2msferettav2 Wd = 9.8 · 10−5 J Soluzione dell’esercizio 7.13 1.
Parte IV Termodinamica S. Lacaprara Es. Meccanica e Termodinamica 100 8 Calorimetria Esercizio 8.1 Una massa d’acqua ma = 120 g, Ta = 21◦C, e posta in contatto con una massa di rame mcu = 75 g, Tcu = 89◦C. Il calore specifico dell’acqua e Ca = 1 cal/g◦C, quello del rame Ccu = 0.093 cal/g◦C.
1. Calcolare la temperatura all’equilibrio.
Esercizio 8.2 Una massa d’acqua ma = 140 g si trova a Ta = 21 ◦C (Ca = 1 cal/g◦C) ed e in contatto con del ghiaccio, a Tg = −5◦C (Cg = 0.5 cal/g◦C, λf = 79 cal/g).
La temperatura all’equilibrio e Te = 4◦C 1. calcolare la massa del ghiaccio mg.
Esercizio 8.3 Dentro un contenitore si trova acqua ma = 127.698 g e una massa di rame mr = 57.652 g (Cr = 0.093 cal/g◦C) in equilibrio a Ta = 21.9◦C. Viene introdotta una terza sostanza mx = 58.991 g a Tx = 94.5 ◦C. La temperatura del sistema e misurata da un termometro con una capacit a termica K = 2.2 cal/◦C, che, all’equilibrio, misura Tf = 25.6 ◦C 1. Determinare il calore specifico di x.
Esercizio 8.4 Un recipiente contiene acqua, rame, ghiaccio e un termometro. Il ghiaccio si scioglie.
mr = 57.650 g, Cr = 0.093 ca/g◦C, ma = 108.830 g Ca = 1cal/g◦C, mg = 9.920 g Tg = 0◦C, ktermometro = 2.2 cal/◦C, T0 = 19.4◦C, Tf = 11.8 ◦C 1. Calcolare il calore di fusione specifico del ghiaccio λg.
Esercizio 8.5 (M.S.V. 9.9) Un recipiente rigido e isolato termicamente contiene n = 2 moli di ossigeno O2 ad una pressione p0 = 1 bar, T0 = 300 K. Nel recipiente viene introdotto un blocco di rame (di volume trascurabile) con m = 0.1 kg e T1 = 800 K, Crame = 387 J/kgK e si attende l’equilibrio.
1. calcolare la pressione finale del gas.
Esercizio 8.6 (M.S.V. 9.10) Un cilindro chiuso da un pistone scorrevole (senza attriti) contiene n = 1 moli di acqua allo stato di vapore, T1 = 373 K, p = 0.2 bar. Il pistone viene spinto lentamente fino a che V2 = 0.02 m3, mantenendo costante la temperatura.
S. Lacaprara Es. Meccanica e Termodinamica 101 1. Calcolare la massa di vapore acqueo che condensa.
Esercizio 8.7 (M.S.V. 9.16) Cilindro adiabatico r = 0.1 m e chiuso da pistone di massa M = 5 kg, collegato al fondo da una molla (massa e vol trascurabile) l0 = 0.1 m k = 2 · 103 N/m, e contiene n = 0.6 moli di gas ideale.
All’esterno agisce una pressione p0 = 1 bar e si trova all’equilibrio l = 0.5 m dal fondo.
1. Temperatura del gas Esercizio 8.8 (M.S.V. 9.5) Un proiettile di piombo (cPb = 130 J/kgK) m = 10 g si conficca ad una velocit a di v0 = 200 m/s su un blocco di alluminio (cAl = 896 J/kgK) di massa M = 100 g inizialmente fermo e ci rimane incastrato. Entrambi sono inizialmente a T0 = 300 K.
1. Trascurando il calore scambiato con l’esterno, calcolare la temperatura del sistema.
Esercizio 8.9 Un cilindro di diametro interno d = 4.0 cm contiene aria compressa da pistone di massa m = 13.0 kg, ed e libero di moversi senza attriti. Tutto e immerso in un bagno d’acqua la cui temperatura e controllabile.
Inizialmente T = 20◦C e h1 = 4.0 cm. Poi la temperatura viene portata lentamente a Tf = 100◦C 1. calcolare h2.
S. Lacaprara Es. Meccanica e Termodinamica 102 Soluzione dell’esercizio 8.1 1. ∆QA→Cu = −∆QCu→A, Teq = mACATA+mCuCCuTCu mACA+mCuCCu = 298.38 K Soluzione dell’esercizio 8.2 1. maCa(Ta −TF) = mgCg(T0 −Tg) + mgλg + mgCA(Te −T0), da cui mg = 27.84 g Soluzione dell’esercizio 8.3 1. mxCx(T1 −TF) = (mRCR + maCa + K)(TF −T0), da cui Cx = 0.123 cal/g◦C Soluzione dell’esercizio 8.4 1. mgλg + mgCa(TF −Ti) = (mRCR + maCa + K)(T0 −TF), da cui λF = 77.4 cal/g Soluzione dell’esercizio 8.5 1. TF = m1c1T1+ncV T0 m1C1+ncV = 541.2 K, trasformazione isocora, pF = p0 TF T0 = 1.8 bar Soluzione dell’esercizio 8.6 1. Inzialmente si ha solo vapore acqueo, con volume V1 = nRT1 p1 = 0.15 m3.
Quando la pressione e qualla di vapore saturo (1.013 bar), il volume e V ′ 2 = nRT1 p2 = 0.03 m3 che e maggiore di V2, quindi parte del vapore condensa.
Sia x il numero di moli di acqua che condensa.
V2 = Vliq + Vvapore = xA ρ + (1−x)RT1 pv.s.
, ricavo x = 0.35 moli, quindi massa mliq = xA = 6.3 · 10−3 kg Soluzione dell’esercizio 8.7 1. Tgas = 400 K Soluzione dell’esercizio 8.8 1.
Soluzione dell’esercizio 8.9 1. p = patm + 4mg πd2 e costante.
h2 = h1 T2 T1 = 5.1 cm S. Lacaprara Es. Meccanica e Termodinamica 103 9 Termodinamica: I principio Esercizio 9.1 Un recipiente adiabatico contiene aria (gas ideale) a T = 280 K, ed e chiuso in alto da un pistone con massa M = 5 kg; il volume iniziale del gas e Vgas = 0.05 m3 e la sezione del recipiente e pari a: S = 100 cm2.
1. ngas Viene introdotto un blocco di rame di Mrame = 1 kg, T ′ = 370 K e si attende l’equilibrio: il movimento del pistone e molto lento. Il calore specifico del rame e: Crame = 384 J/kgK.
2. Teq 3. ∆h del pistone.
Esercizio 9.2 Un cilindro con pistone libero di muoversi senza attriti contiene 0.1 moli di aria (considerato gas ideale), T = 20◦C. Il pistone viene compresso lentamente, tanto che l’aria rimane in equilibrio termico con l’ambiente, fino a dimezzare il volume iniziale.
1. Calcolare il lavoro compiuto dall’aria durante la compressione.
Esercizio 9.3 Una quantita di elio (gas ideale) pari a n = 103 moli subisce il ciclo in figura, dove BC e una isoterma. Tut-te le trasformazioni sono reversibili. pA = 1.00 atm, VA = 22.4 m3, pB = 2.00 atm.
A B C V p Calcolare: 1. TA, TB, VC; 2. Il lavoro svolto dal sistema in ogni trasformazione e per l’intero ciclo; 3. Il calore scambiato dal sistema in ogni trasformazione e per l’intero ciclo; 4. il rendimento del ciclo Esercizio 9.4 Un cilindro rigido e adiabatico di sezione S = 1.5·10−3 m2 e diviso in due da un pistone a tenuta, adiabatico, libe-ro di muoversi senza attrito. In uno scomparto c’ e aria, dall’altra parte c’e il vuoto e una molla k = 104 N/m con estremit a fissata al pistone e l’altra al fondo del cilindro, inizialmente a riposo.
l0 k Aria Vuoto S. Lacaprara Es. Meccanica e Termodinamica 104 Il compartimento del gas e lungo l0 = 30 cm. Il gas e inizialmente a p0 = 1 atm e T0 = 27◦C. Si sblocca il pistone e il sistema raggiunge il nuovo stato di equilibrio.
Calcolare: 1. lunghezza finale del cilindro; 2. pressione e temperatura finale del gas.
Esercizio 9.5 Un gasometro con p0 e T0 costanti e collegato ad un contenitore rigido, adiabatico inizialmente vuoto. Si apre il rubinetto e elio entro il contenitore fino a raggiungere la pressione p0.
1. calcolare la temperatura finale del contenitore.
Esercizio 9.6 Un cilindro rigido e adiabatico, di raggio r = 10 cm e chiuso da uno stantuffo libero di scorrere fino a l0 = 1.42 m, dove c’e un blocco. Inizialmente il cilindro contiene 1 mole di elio a T = 0◦C lo stantuffo e esposto (a destra) all’atmosfera (stato A).
A B/C D l0 Si riscalda lentamente la base del cilindro in modo che lo stantuffo si sposti, bloccan-dosi a l0 (stato B) e raggiunga la pressione finale pari a 1.1 patm (stato C).
Infine il pistone si sblocca e si riporta il gas alla pressione iniziale (stato D) con una trasformazione isoterma reversibile.
Calcolare: 1. VA, TB, TC, VD; 2. lavoro fatto nelle tre trasformazioni 3. calore scambiato nelle tre trasformazioni.
Esercizio 9.7 Un compressore, che lavora ciclicamente e reversibilmente, fornisce m = 50 kg di aria compressa ogni ora, prendendola da un ambiente in cui p = 1 atm e T = 280 K.
Calcolare la potenza del compressore nei casi seguenti: 1. compressione adiabatica, sapendeo che pout = 10 atm; 2. compressione dove l’aria compressa ha temperatura T2 = 460 K e la perdita di calore nel compressore e Q2 = 210 J/s. Considero l’aria un gas ideale con CV = 710 J/kgK (calore specifico) e γ = 1.4 Esercizio 9.8 Un recipiente rigido, con pistone bloccato, contiene una bombola di 4l, che occupa 2/3 del volume disponibile, e che e piena di Azoto (N2) a TA = 300K e p0 = 1.3 atm (A).
La bombola si apre e il gas riempie rapidamente il recipiente (B). Raffreddo il recipiente, S. Lacaprara Es. Meccanica e Termodinamica 105 sempre con pistone bloccato, fino a raggiungere una pressione pC (C), tale che sia possibile riportare il gas, muovendo il pistone, tramite una trasformazione adiabatica reversibile, allo stato A.
Calcolare: 1. ∆UA−B; 2. pC; 3. ∆U, Q e L scambiati in B →C e C →A Esercizio 9.9 Una mole di gas ideale biatomico compie un ciclo reversibile, formato da una trasfor-mazione adiabatica AB, una isoterma CD, e due isocore BC e DA.
VA = 20 l, VB = 30 l, TA = 350 K, TC = 325 K.
Calcolare: 1. pA,B,C,D 2. QAB,BC,CD,DA 3. η del ciclo; 4. la temperatura T ′ C che dovrebbe avere lo stato C perche il lavoro totale del ciclio sia nullo.
Esercizio 9.10 Un recipiente cilindrico di alluminio ρ = 2.7 g/cm3, V = 5.67 · 10−3 m3, r = 10 cm contiene n = 0.8 moli di aria T0 = 250 K. Viene ceduto calore da un fornello a gas, con temperatura Tfuoco = 600 K. Il tappo, di spessore s = 1 mm, tiene fino a p = 6 atm.
Quando cede, si stacca in ∆t = 0.01 s.
1. temperatura massima dell’aria; 2. calore ceduto all’aria; 3. v del coperchio.
S. Lacaprara Es. Meccanica e Termodinamica 106 Soluzione dell’esercizio 9.1 1. n = patm + Mg S V RT = 2.3 moli 2. Il calore ceduto dal rame e pari a quello assorbito dall’aria, cambiato di segno. La trasformazione dell’aria e’ isobara: ∆QCu = MrameCrame(T ′ −T”) = −ncP(T” −T) , da cui ricavo: T” = 356.6 K.
In alternativa posso usare il I principio della termodinamica scrivere: ∆Q −L = mc(T ′ −T”) −p(V ” −V ) = ∆U = ncV (T” −T) pV ” = nRT” Da cui ricavo: T” = 356.6 K Di fatto e’ la stessa soluzione, ma la seconda e’ piu’ complessa. In generale posso scrivere ∆Qaria −Laria = ∆U, ovvero: ∆Qaria = −∆Qrame = ncV ∆T + R pdV = ncV ∆T +p∆V , usando il fatto che la trasformazione e’ isobara. Infine posso usare l’eq di stato dei gas ideali per l’aria, e, sempre usando che p = cost, p∆V = nR∆T, quindi −∆Qrame = n(cV + R)∆T = ncP∆T, esattamente come prima.
3. V ” = V T” T , e ∆h = V ”−V S = 1.7 m Soluzione dell’esercizio 9.2 1. la trasformazione e reversibile: L = nRT ln V2 V1 = −169 J.
Soluzione dell’esercizio 9.3 1. TA = pAVA nR = 273 K TB = TA pB pA = 546 K VC = VA TB TA = 44.8 m3 2. LAB = 0 J, LBC = nRTB ln VA VB = 3.15 · 106 J, LCA = pA(VA −VC) = −2.26 · 106 J Lciclo = 8.8 · 105 J 3. ∆QAB = ncv(TB −TA) = ∆QBC = LBC = 3.15 · 106 J ∆QCA = ncp(TA −TA) = −5.67 · 106 J Soluzione dell’esercizio 9.4 1. n = p0V0 RT0 = 0.0183 moli stato finale: p1(l0 + x)S = nRT1 p1S = kx ∆Q −L = 0 −1 2kx2 = ncV (T1 −T0) S. Lacaprara Es. Meccanica e Termodinamica 107 da cui ricavo x1 = 0.014 m, e x2 = −0.26 m. Scarto la seconda perch e avevo fatto l’ipotesi che il gas si espandeva per calcolare il lavoro, quidni risulta l1 = 31.4 cm 2. p = kx S = .95 · 105 Pa, e T = p(l+x)S nR = 298 K. La temperatura finale si puo’ calcolare anche con il I principio −L = −1/2kx2 = ncv∆T da cui ricavo ∆T = −2◦C Soluzione dell’esercizio 9.5 1. Stato iniziale. Gasometro: p0V0 = n0RT0 Stato finale. Gasometro: p0V1 = (n0 −n2)RT0, bombola: poV2 = n2RT2 −L = −p(V1 −V0) = ∆U = n2cV (T2 −T0), da cui T2 = γT0 Soluzione dell’esercizio 9.6 1. VA = 22.4 l (gas STP). la = VA πr2 = 0.71 m. TB = TA VB VA = 546 K, TC = pCVB nR = pCTB pB = 600 K, VD = VCpC/pD = 1.1VC = 49.3 l 2. WAB == p∆V +0 = 2.26·103 J WBC = 0, WCD = ∆QCD = 4.77 J Wtot = 7.0·103J 3. ∆Q = ∆QAB + ∆QBC + ∆QCD = ncp(TB −TA) + ncV (TC −TB) + nRT ln V2 V1 = 5.67 · 103 + 0.68 · 103 + 4.77 · 103 = 11..1 · 103 J Soluzione dell’esercizio 9.7 1. Compressione adiabatica: T1 = T0 p0 p1 1−γ γ = 540.6 K.
L = −mCv(T1 −T0), quindi P = L dt = −dm dt Cv(T1 −T0) = −2590 W 2.
dL dt = d dtQ −∆U = −dQ2 dt −dm dt CV (Taria −T0) = −1995 W In entrambi i casi la potenza del compressore e l’opposto di quella ceduta dal gas.
Soluzione dell’esercizio 9.8 1. ∆UAB = 0 (espansione libera) 2. pC = pA VA VC γ = pA VA VB γ = 0.737 atm 3. C →A adiabatica reversibile. ∆Q = 0, L = −∆U = ncV ∆T B →C, L = 0, ∆Q = ∆U = ncV ∆T Soluzione dell’esercizio 9.9 1.
A: uso eq. gas perfetti. p = 1.45 · 105 Pa, T = 350 K; B: trasfromazione isoterma A →B. p = 0.82 · 105 Pa, T = 296 K; C: uso eq. gas perfetti. p = 0.9 · 105 Pa, T = 325 K; D: trasfromazione isocora D →B. p = 1.35 · 105 Pa, T = 325 K; 2.
AB adiabatica ∆Q = 0; BC ∆Q = ncV ∆TBC = 602.5 J; S. Lacaprara Es. Meccanica e Termodinamica 108 CD ∆Q = L = nRTC ln VD VC = −1095 J; DA ∆Q = ncV ∆TDA = 519.0 J; 3. η = |Qass|−|Qced| |Qass| = 0.024 4. La macchina compie lavoro solo tra A →B e tra C →D. L = ncV (TB −TA) − nRT ′ C ln VC VD = 0, quindi T ′ C = 333 K Soluzione dell’esercizio 9.10 1. Al momento della rottura, lo stato temodinamico del sistema e noto, quindi basta usare l’equazione dei gas perfetti in queste cnodizioni.
Tmax = pmaxV nr = 518 K 2. la trasformazione e isocora. Il calore assorbito e Q = ncV (Tmax −T0) = 4.5 kJ, considerando cV = 5/2R (gas ideale biatomico).
3. durante ∆t agisce una forza F = (pmax −patm)πr2 = 15.7 kN.
Quindi viene generato un impulso J = F∆t = 157 Ns. La massa del tappo di alluminio e m = ρAlsπr2 = 85 g, quindi la sua velocit a risulta v = F∆t m = 1.85 · 103 m/s.
Il raggio del cilindro non e necessario, visto che la sezione appare sia al numeratore (F) che al denominatore (m). La temperatura del fuoco serve solo a verificare che l’esplosione del tappo effettivamente avviene.
S. Lacaprara Es. Meccanica e Termodinamica 109 10 Termodinamica: II principio Esercizio 10.1 Un calorimetro contiene 1 kg di acqua, agitata da palette collegate ad un motore che fornisce potenza P = 20 W. Si aggiunge al calorimetro 10 moli di potassio solido a T0 = 323 K.
Dopo ∆t = 5 min, il solido e l’acqua sono in equilibrio termico.
Il calore scambiato tra il solido e l’acqua e Q1 = 13.4 · 103 J; il calore trasferito dall’acqua all’ambiente e Q2 = 600 J.
Calcolare: 1. ∆U dell’acqua Sapendo che il calore molare del solido e cp = a + bT[K] dove a = 1.3 cal/mole e b = 19.4 · 10−3 cal/moleK, calcolare: 2. la temperatura finale del solido; 3. la variazione di entropia del solido.
Esercizio 10.2 Due masse uguali di acqua m = 100 g sono inizialmente a temperatura T1 = 300K e T2 = 275K. Vengono a contatto e arrivano all’equilibrio termico.
Calcolare, supponendo che siano contenute in recipiente adiabatico: 1. la temperatura di equilibrio; 2. la variazione di entropia; Supponendo che la trasformazione sia reversibile, calcolare: 3. T ′ eq 4. L fatto dal sistema Esercizio 10.3 Un blocco di stagno, di massa m = 1.5 kg, si trova a temperatura ambiente T0 = 290 K, e viene messo a contatto con una sorgente a temperatura di fusione dello stagno TF = 505 K (un saldatore). Il calore specifico dello stagno e CF = 218.6 J/kgK e il calore di fusione sia λF.
1. Calcolare la variazione di entropia dell’universo ∆Su Esercizio 10.4 Un recipiente con pareti adiabatiche e diviso in due parti (A e B) da una membrana anch’essa adiabatica. Lo stesso gas biatomico ideale si trova nei due scomparti: nA = 2 moli, TA = 300 K, VA = 2 · 10−3 m3, nB = 3 moli, TB = 800 K, VB = 4 · 10−3 m3. Si considerino due casi: 1. la membrana viene sostituita con una conduttrice di calore, si attende l’equilibrio termico e quindi viene tolta.
S. Lacaprara Es. Meccanica e Termodinamica 110 2. la membrana viene tolta e il gas si mescola fino a raggiungere l’equilibrio.
Calcolare, nei due casi: 1. Teq 2. ∆Sgas 3. ∆Suniverso Esercizio 10.5 Due moli di gas ideale biatomico si trovano ad una temperatura T0 = 300 K, p0 = 1 · 105 Pa, e subiscono una compressione adiabatica e reversibile fino a riempire un recipiente di volume V ′ 0 = 1 · 10−2 m3.
Successivamente il gas torna alla temperatura iniziale a causa di un isolamento non perfetto del contenitore. Calcolare: 1. pmax 2. Tmax 3. pgas finale 4. L durante la compressione 5. ∆Sgas 6. ∆Suniverso Esercizio 10.6 Una quantita di ghiaccio m = 50 kg si trova ad una temperatura T0 = −5◦C in un congelatore. Il congelatore e in una stanza con temperatura T1 = 30◦C, e le pareti del congelatore perdono dQ dt = 120 cal/s.
Considerando il congelatore una macchina reversibile, calcolare: 1. Potenza P per mantenere costante la temperatura del ghiaccio; 2. ∆Su in 1 ora; 3. la quantita di ghiaccio che fonde se il congelatore resta spento per mezz’ora (CG = 2090 J/kgK, λG = 333.5 kJ/kg); 4. ∆Su in quella mezz’ora Esercizio 10.7 Un gas ideale biatomico si trova in equilibrio con l’ambiente (p0 = 105 Pa, T0 = 300 K) in un cilindro adiabatico di volume V0 = 5 l, dotato di un pistone pure adiabatico che scorre senza attrito. Il cilindro e diviso in due parti da una membrana diatermica con un piccolo foro, che all’inizio e a contatto con il pistone mobile.
Una resistenza fornisce al gas un lavoro W in modo tale da raddoppiarne la pressione (a pistone bloccato).
Successivamente il pistone viene lasciato libero di muoversi fino a che il sistema raggiunge l’equilibrio termico. Calcolare: S. Lacaprara Es. Meccanica e Termodinamica 111 1. ∆Ugas 2. ∆Sgas 3. ∆Uambiente 4. ∆Sambiente Esercizio 10.8 Un gas perfetto monoatomico compie un ciclo formato da 1 →2 adiabatica reversibile; 2 →3 isobara a contatto con sorgente a temperatura T3 3 →4 adiabatica reversibile 4 →1 isobara a contatto con sorgente a temperatura T1 Sono noti: V1 = 3.5 l, p1 = 2.5 atm, p2 = 3.5 atm, V3 = 4.5 l, e la variazione di entropia dell’universo per ogni ciclo ∆S = 2.1 J/K. Calcolare: 1. Q2→3; 2. Q4→1; 3. il rendimento del ciclo η; 4. le temperature nei 4 stati Ti.
Esercizio 10.9 Una macchina termica ciclica reversibile pu o lavorare con tre sorgenti di calore, ad una temperatura T1 = 400 K, T2 = 300 K, e T3 = 200 K. Prima lavora con T1 e T2 e successivamente con T1 e T3.
Il calore totale assorbito dalla sorgente T1 e Qass 1 = 1200 J, mentre il lavoro totale fornito e L = 400 J.
Calcalare: 1. il calore scabiato Q2 e Q3 con le sorgenti T2 e T3; 2. ∆S delle sorgenti; 3. ∆S dell’universo.
Esercizio 10.10 Un contenitore adiabatico e formato da due comparti: il primo, A, che e chiuso da un pistone pure adiabatico che scorre senza attrito, e il secondo B, separato da A da una membrana rigida buona conduttrice di calore.
Il comparto A contiene nA = 1 moli di gas biatomico, ad una tempartura TA = 300 K e pressione pA = 1 atm, mentre B contiene nB = 2 moli di gas monoatomico ad una temperatura TB = 600 K. Il sistema si porta all’equilibrio.
Calcolare: S. Lacaprara Es. Meccanica e Termodinamica 112 1. Tequilibrio 2. ∆VA 3. L 4. ∆SA e ∆SB Esercizio 10.11 Un contenitore e diviso in due parti (A e B) da una membrana diatermica con un rubinetto, inizialmente aperto. Il comparto B e chiuso da un pistone mobile, mentre il comparto A (VA = 10 l) puo essere messo in contatto con l’ambiente (T0, p0).
Inizialmente il gas, ideale biatomico, n = 1 moli e’ in equilibrio con l’ambiente T0 = 300 K.
Il sistema e soggetto ad una compressione adiabatica reversibile grazie al pistone, fino a che il comparto B risulta vuoto. Successivamente il rubinetto viene chiuso e si attende l’equilibrio termico con l’ambiente esterno. Calcolare: 1. Q scambiato con l’ambiente Considerando che il rubinetto perde leggermente, calcolare: 2. Lgas 3. ∆S del gas e dell’ambiente.
Esercizio 10.12 Una macchina termica lavora tra una sorgente a temperatura T0 = 273 K, formata da acqua e ghiaccio in fusione (λF = 80 cal/g) e una a T1 = 300 K. In ogni ciclo il lavoro fornito dalla macchina risulta L = 2000J e ∆Suniverso = 4 J/K.
Calcolare: 1. quanto ghiaccio fonde per ogni ciclo della macchina; 2. il rendimento della macchina; 3. il lavoro che sarebbe fornito dalla macchina se essa fosse reversibile e fondesse, per ogni ciclo, la stessa quantita di ghiaccio di prima.
Esercizio 10.13 Un contenitore adiabatico ha la base a contatto termico con l’esterno e contiene n = 1 moli di gas in condizione STP (p = 1 atm, T = 300 K).
Il gas compie le seguenti trasformazioni: AB espansione isobara con TB = 400 K; BC espansione isoterma reversibile fino a VC = 40 l; CD isocora in contatto termico con TD = 217 K; DA compressione adiabatica reversibile.
S. Lacaprara Es. Meccanica e Termodinamica 113 Calcolare: 1. LAB; 2. se il gas e monoatomico o biatomico; 3. ∆SAB+CD gas ; 4. ∆Sciclo universo; Esercizio 10.14 Due moli di gas monoatomico si trovano in uno stato di equilibrio A con VA = 40 l, TA = 300 K. Partendo da questo stato il gas compie un ciclo termodinamico composta da quattro trasformazioni reversibili: AB compressione adiabatica con lavoro WAB = −5 kJ; BC espansione isobara con calore assorbito QBC = 15 kJ CD espansione isoterma; DA isocora. Determinare: 1. temperatura e pressione in B; 2. temperatura in C e lavoro WBC; 3. rendimento η del ciclo.
Si immagini che CD sia una espansione libera. Calcolare: 4. rendimento η′ del ciclo; 5. ∆Su nel ciclo.
Esercizio 10.15 Un cilindro chiuso, a pareti adiabatiche, ha nel suo interno un pistone adiabatico, che pu´ o scorrere senza attrito nel cilindro. In ciascuna delle due camere ci sono n = 0.50 moli di gas perfetto. Inizialmente i due gas hanno la stessa pressione p0 e la stessa temperatura T0 = 300 K. Il volume utile totale delle due camere e pari a V = 4.0 l . Il calore molare a volume costante del gas e cV = 2R.
Tramite un riscaldatore elettrico si fornisce lentamente calore al gas di sinistra fino a che il suo volume raggiunge il valore V1 = 2.6 l .
Calcolare: 1. Le pressioni p1 e p2, e le temperature T1 e T2 dei due gas alla fine del riscaldamento; 2. il lavoro Wd fatto sul gas di destra, e il calore Qs fornito al gas di sinistra.
3. le variazione di entropia ∆S1 e ∆S2 dei due gas.
Esercizio 10.16 Un frigorifero che lavora alla massima efficienza teoricamente possibile viene utilizzato per trasformare in ghiaccio a Tg = 273 K un massa M = 1 kg di acqua, inizialmente a TA = 300 K, cedendo calore all’ambiente esterno che si trova a temperatura TA.
Sapendo che per trasformare in ghiaccio un grammo di acqua a T = 273 K e necessario estrarre 80 cal (calore latente di fusione del ghiaccio λg = 80 cal/g), determinare: S. Lacaprara Es. Meccanica e Termodinamica 114 1. il calore ceduto dal frigorifero all’ambiente per raffreddare l’acqua da TA a Tg; 2. il lavoro assorbito dal frigorifero per raffreddare l’acqua fino a Tg; 3. il calore ceduto dal frigorifero all’ambiente durante il processo che trasforma l’ac-qua a Tg in ghiaccio; 4. il lavoro complessivamente assorbito dal frigorifero per produrre il ghiaccio a partire dall’acqua a temperatura ambiente; Esercizio 10.17 Due moli di gas perfetto compiono un ciclo reversibile scambiando calore con 3 sorgenti ideali. Nella trasformazione a contatto con T1 = 400 K il gas raddoppia il volume, come anche in quella a contatto con T2 = 300 K.
1. Disegnare la trasformazione 2. Vf/Vi per la trasformazione a contatto con T3 = 100 K; 3. lavoro prodotto in un ciclo della macchina; 4. rendimento della macchina.
Esercizio 10.18 Un contenitore adiabatico VT = 20 l e’ diviso in due da un setto adiabatico libero di scorrere senza attrito. In ogni parte ci sono n = 2 moli di gas ideale monoatomico. Il volume di destra e a contatto con ghiaccio e acqua T0 = 273 K, e il volumen di sinistra si trova a T0.
Riscaldo lentamente il volume di sinistra con una resistenza da P = 10 W, e mi fermo quando m = 10 g del ghiaccio a destra si sono sciolti. λg = 333.5 J/g.
1. Volume finale a destra VA 2. Temperatura del gas a sinistra TB 3. Tempo necessario ∆t per sciogliere il ghiaccio Esercizio 10.19 Una quantita pari a n = 0.2 moli di aria (gas ideale biatomico) effettuano un ciclo Diesel reversibile cos ı composto (il rapporto di compressione r = V1/V3 = 5.0) : 1-2 compressione adiabatica; T1 = 300 k 2-3 espansione isobara; T3 = 3300 k 3-4 espansione adiabatica; 4-1 raffreddamento isocora; 1. disegnare il ciclo nel piano p −V ; 2. ∆Sgas 13 3. T2 S. Lacaprara Es. Meccanica e Termodinamica 115 4. Wciclo 5. rendimento η Esercizio 10.20 Un frigorifero si trova in un ambiente a T1 = 27 ◦C e contiene 10 kg di ghiaccio a T2 = −23 ◦C. Le pareti non sono perfettamente adiabatiche, e il frigo scambia con l’ambiente QB = 500 J/s. Il motore del frigorifero ha una potenza di P = 250 W e si puo considerare reversibile.
Calore specifico ghiaccio c = 2.0 kJ/kg K, calore latente di fusione del ghiaccio: λf = 333.5 kJ/kg.
Calcolare: 1. per che frazione di tempo il motore deve stare acceso per mantenere costante la temperatura del ghiaccio; 2. la variazione di entropia dell’universo in un’ora; 3. quanto ghiaccio fonde se il motore resta spento per un’ora; 4. la variazione di entropia dell’universo in questo caso Esercizio 10.21 Due moli di un gas biatomico ideale compiono il ciclo segiente [ABCDA] AB espansione isobara reversibile VA = 10 l, VB = 30 l, pA = 5 atm; BC espansione isoterma reversibile dove ∆Sgas BC = 8.5 J/K; CD compressione reversibile rappresentata nel piano p −V da una retta, fino allo stato finale: VD = 20 l, TD = TA; DA compressione irreversibile a contatto termico con una sorgente a temperatura TA, lungo la quale il gas subisce un lavoro LDA = 5.0 kJ 1. la rappresentazione del ciclo nel piano p −V 2. il rendimento del ciclo 3. la variazione di entropia del gas lungo DA 4. la variazione dell’entropia dell’universo in un ciclo.
Esercizio 10.22 Un gas ideale monoatomico e contenuto dentro un contenitore rigido adiabatico chiuso da un tappo S = 100 cm2, M = 3 kg libero di muoversi senza attrito: T0 = 300 K, V0 = 10 l, patm = 1 · 105 Pa stato A.
Verso lentamente m = 30 kg di sabbia sopra il tappo st. B Una volta raggiunto l’equilibrio, appoggio il contenitore con sorgente ideale TS = 400 K (fondo diventa diatermico), e si attende l’equilibrio st. C.
Infine, sempre con TS a contatto, tolgo il secchio con la sabbia st. D.
Gas TS M m S. Lacaprara Es. Meccanica e Termodinamica 116 1. il numero di moli di gas 2. la temperatura dello stato B 3. il calore scambiato (con segno) dalla sorgente durante la trasformazione B-C 4. il lavoro fatto dal gas durante la trasformazione CD 5. la variazione di entropia dell’universo dallo stato iniziale a quello finale S. Lacaprara Es. Meccanica e Termodinamica 117 Soluzione dell’esercizio 10.1 1. ∆U = (Q1 −Q2) −(−P∆t) = 18800 J 2. Calore ceduto dal solido: −Q1 = n R Teq T0 (a + bT)dT, ottengo una equazione di secondo grado, che risolvo per Teq = 278 K (la seconda soluzione e negativa e non fisica).
3. ∆S = R Teq T0 n(a+bT) T dT = −44.55 J/K Soluzione dell’esercizio 10.2 1. Teq = T1+T2 2 = 287.5 K 2. ∆Su = ∆Sgas = m1C ln Teq T1 + m2C ln Teq T2 = mC ln T 2 eq T1T2 = 0.795 J/K 3. Se la trasformazione e reversibile, ∆Su = 0. Per fare una trasformazione reversibile posso pensare di avere N serbatoi a temperature intermedie e scambiare calore tramite cicli reversibili con queste N sorgenti fino ad arrivare all’equilibrio.
In alternativa posso pensare a m1 e m2 come due sorgenti, di capacita termica finita, utilizzate da un ciclo reversibile (per esempio un ciclo di Carnot), e far andare il ciclo finch e le temperature delle due sorgenti diventano uguali (condizione di equilibrio).
Con questo setup, noto immediatamente che il ciclo fornisce lavoro verso l’esterno, visto che lavora tra due sorgenti a temperatura diversa, e quindi mi aspetto una situazione di equilibrio diversa da quella del punto precedente, dove non veniva scambiato con l’esterno ne calore n e lavoro.
∆Su = ∆Sciclo + ∆Ssorgenti = 0, quindi ∆Ssorgenti = mC ln T ′2 eq T1T2 = 0, quindi T ′ eq = √T1T2 = 287.23 K.
4. Il lavoro fatto dal sistema e il lavoro fornito dalla macchina di Carnot, quindi L = |Q′ 1| −|Q′ 2| = 217 J.
Alternativamente, posso calcolare l’energia resa inutilizzabile durante la prima trasformazione non revesibile Einu = T2∆Su = 217 J, che va confrontata con quella resa inutilizzabile nella seconda trasformazione (che e evidentemente 0).
La differenza e pari al lavoro che viene fornito all’ambiente dalla trasformazione reversibile.
Soluzione dell’esercizio 10.3 1. All’equilibrio T = TF e lo stagno e completamente sciolto.
∆Sstagno = mCF ln TF T0 + mλF TF ∆Ssorgente = −mCF (TF −T0) TF −mλF TF ∆Suniverso = mc ln TF T0 −mc(TF −T0) TF = 42.2 J/K Soluzione dell’esercizio 10.4 1. In entrambi i casi, l’equilibrio si raggiunge quando il calore ceduto dal gas in B viene assorbito dal gas in A.
Teq = nATA+nBTB nA+nB = 600 K S. Lacaprara Es. Meccanica e Termodinamica 118 2.
1.
Prima fase: ∆Sgas = ∆SA + ∆SB = nAcV ln Teq TA + nBcV ln Teq TB seconda fase: ∆Sgas = ∆SA+∆SB = nAcV ln (VA+VB)γ−1 V γ−1 A +nBcV ln (VA+VB)γ−1 V γ−1 B 2. ∆Sgas = ∆SA+∆SB = nAcV ln Teq(VA+VB)γ−1 TAV γ−1 A +nBcV ln Teq(VA+VB)γ−1 TBV γ−1 B = 39.3 J/K Essendo l’entropia una funzione di stato, e essendo gli stati finali identici nei due casi, la variazione di entropia e la stessa nei due casi.
3. ∆Su = ∆Sgas in entrambi i casi, visto che il contenitore e adiabatico.
Soluzione dell’esercizio 10.5 1. V = nRT0 p0 = 50 l la trasformazione e adiabatica: p′ = p V V ′ γ = 9.5 · 105 Pa 2. T ′ = p′V ′ nR = 571.3 K 3. p” = p′ T0 T ′ = 5 · 105 Pa 4. L = −∆U = −ncv(T ′ −T0) = −11278 J 5. Collego stato iniziale e finale con isoterma reversibile: ∆Sgas = nR ln V ′ V = −26.7 J/K 6. ∆Samb = 0 + ncv(T ′−T) T = 37.6 J/K, quindi ∆Su = 10.9 J/K Soluzione dell’esercizio 10.6 1. La macchina e reversibile, quindi ∆S = Q1 T1 + Q0 T0 = 0, dove Q0 e il calore scam-biato dal ghiaccio e Q1 dall’ambiente. La stessa relazione vale anche per il calore scambiato per unit a di tempo dQ dt .
Quindi: dQ1 dt = −502 W e dQ0 dt = − dQ1 dt T0 T1 = 444 W, quindi Pcongelatore = dQcongelatore dt = −dQambiente dt = −( dQ1 dt + dQ0 dt ) = −58 W; Quindi devo fornire P = 58 W al congelatore dall’ambiente per mantenere 2. ∆Su = ∆SG + ∆S0 = ∆t ∗ 1 TG dQG dt + 1 T0 dQ0 dt = 5 · 10−5 ∗3600 = 0.177 J/K.
Da notare che QG adesso e’ dal punto di vista del ghiaccio (sorgente termica) e non da quello della macchina (congelatore) come prima, quindi il segno e’ opposto.
3. Per sciogliersi, il ghiaccio deve prima raggiungere la temperatura di fusione 273 K.
mGcG(T0 −T273) + λmF = dQ dt ∆t, da cui mF = 1.15 kg 4. ∆SG = cGmG ln T273 T0 + mF λ T273 = 2088 + 1466 = 3555 J/K ∆Samb = dQ dt ∆t T1 = −2983 J/K. Quindi ∆Su = 572 J/K Soluzione dell’esercizio 10.7 1. n = p0V0 RT0 = 0.2.
La prima trasformazione e isocora, la temperatura dello stato finale e: T1 = T0 p0 p1 = 600 k.
La seconda transformazione e pi u complessa: visto che il cilindro e il pistone sono adiabatici si potrebbe pensare che anche la trasformazione sia adiabatica, ma non e cos ı . Infatti la presenza del “piccolo foro” che collega le due parti del contenitore S. Lacaprara Es. Meccanica e Termodinamica 119 non garantisce che la pressione sia uguale dalle due parti. Gia questo ci fa capire che non si tratta di una trasformazione quasi statica, ma di due trasformazioni collegate. Nel volume iniziale il gas esce (portando fuori energia interna), dimi-nuisce la pressione e la temperatura. Nel secondo volume (quello a contatto con il pistone mobile), la pressione e costante, mentre il gas entra, il volume aumenta e la temperatura cambia (inizialmente essendo vuoto, la temperatura non e definita).
La situazione e quindi molto complessa.
Posso invece considerare l’insieme di tutto il gas nei due volumi, e applicare il primo principio della Termodinamica al tutto. ∆U = ∆Q −L, osservando che ∆U = ncv(T2 −T1) [funzione di stato], ∆Q = 0 [recipiente adiabatico], e L = p0(V2 −V0) [lavoro della pressione esterna durante l’espansione del pistone]. Usando inoltre l’equazione di stato dei gas perfetti nello stato finale p2V2 = p0V2 = nRT2, si ottine: T2 = p0V0+ncV T1 ncp = 514.3 K.
considero una trasformazione che collega gli stati iniziali e finali: ∆Ugas = ncv(T2− T0) = 1250 J 2. considero una trasformazione isobara che collega gli stati iniziali (p0, V0, T0) e finali (p0, V2, T2): ∆Sgas = ncp ln T2 T0 = 3.12 J/K 3. ∆Uambiente = Q −L = −Lresistenza −Lpistone = ncV (T1 −T0) + ncV (T2 −T1) = −∆Ugas 4. ∆Samb = 0 visto che il contenitore e adiabatico.
Soluzione dell’esercizio 10.8 1. Q2→3 = ncp(T3 −T2) = cp R (p3V3 −p2V2) = 1454 J, dove V2 = V1 p1 p2 1 γ = 2.86 l 2. Q4→1 = cp R (p4V4 −p1V1) = −1271 J, dove V4 = V3 p3 p4 1 γ = 5.51 l 3. η = |Q23|−|Q41| |Q23| = 0.126; 4. Ricavo n calcolando la variazione di entropia dell’universo in un ciclo: ∆S = −Q41 T1 −Q23 T3 = ncp V4 V1 + V2 V3 −2 (notare che il calore scambiato da una sorgente e l’opposto del calore scambiato dal gas) da cui ricavo n = 0.4836.
T1234 = 220.5 K, 252.3 K, 396.9 K, 346.9 K Soluzione dell’esercizio 10.9 1. chiamo A il primo ciclo (tra T1 e T2) e B quello tra T1 e T3.
Q1 = QA 1 + QB 1 L = LA + LB LA = ηAQA 1 LB = ηAQB 1 S. Lacaprara Es. Meccanica e Termodinamica 120 da cui ricavo QA 1 = 800 J QB 1 = 400 J LA = 200 J LB = 200 J Infine: Q2,3 = LA,B −QA,B 1 = −600/ −200 J 2. ∆S1 = −Q1 T1 = −3 J/K, ∆S2 = −Q2 T2 = +2 J/K, ∆S3 = −Q3 T3 = +1 J/K, quindi ∆Ssorgenti = 0 (macchine reversibili) 3. ∆Su = 0 Soluzione dell’esercizio 10.10 1. Teq = nBcp,BTB+nAcV,ATA nBcp,B+nAcV,A = 438 K 2. ∆V = nAR(Teq−TA) patm = 11.4 l 3. L = patm∆V = 1151 J; 4. ∆SA = nAcp,A ln Teq TA = 11.04 J/K ∆SB = nBcp,B ln Teq TB = −7.82 J/K ∆Stot = 3.22 J/K Soluzione dell’esercizio 10.11 1. Volume iniziale prima della compressione V0 = nRT0 p0 = 24.6 l. La compressione e adiabatica: T1 = T0 V0 VA γ−1 = 430.1 k. Quindi Q = ncv(T1 −T0) = 2705 J.
2. L = p0∆V = 1481 J.
3. ∆Samb,1 = Q T0, ∆Samb,2 = −L T0 (perche’ isoterma). ∆Samb = 4.08 J/k. ∆Sgas = 0 visto che e un ciclo.
Soluzione dell’esercizio 10.12 1. E’ un ciclo, quindi Q1 = Q0 + L, inoltre ∆Su = Q0 T0 −Q1 T1 , da cui: Q0 = 32.36 KJ, Q1 = 34.36 KJ, e mF = Q0 λF = 96.6 g.
2. η = L Q1 = 0.058 3. ηR = 1 −T0 T1 = 0.09, da cui LR = ηR 1−ηRQ0 = 3200 J Soluzione dell’esercizio 10.13 1. LAB = nR(TB −TA) = 831.4 J 2. DA e Adiabatica: TD TA = VA VD γ−1 da cui γ = 1 + ln TD TA ln VA VC = 1.667, quindi il gas e monoatomico S. Lacaprara Es. Meccanica e Termodinamica 121 3. ∆SAB = ncp ln TB TA , ∆SCD = ncv ln TDTC, ∆SAB + ∆SCD = −∆SBC = −1.65 J/K 4. Nelle trasformazioni non reversibili: ∆Samb = QCD TD −QAB TA = 5.32 J/K. Per quelle reversibili ∆Suni = 0.
Quindi ∆Suni = 3.67 J/K.
Soluzione dell’esercizio 10.14 1. pA = nrTA VA = 1.25 · 105 Pa. ∆UAB = ncv(TB −TA) = −WAB, TB = 500 K pB = pA TA TB 1−γ γ = 4.48 · 105 Pa 2. QBC = ncp(TC −TB) TC = 862 K VC = nRTC pC = 31.9 l WBC = pC(VC −VB) = 6.6 KJ 3. QCD = WDC = nR ln VD VC = 3.22 kJ WDA = 0, QDA = ncV (TA −TD) = ncv(TA −TC) = −14 kJ η = 22% 4. Nell’espansione libera Q = W = 0, quindi η = 6.6% 5. Nell’unica trasformazione irreversibile (CD), ∆Sgas = nR ln VD VC = 3.74 J/K, ∆Samb = 0.
Soluzione dell’esercizio 10.15 1. p0 = nRT0 V0 = 6.23 · 105 Pa.
Equilibrio meccanico p1 = p2, e compressione adiabatica p1 = p2 = p0 V0 V2 γ = 1.06 · 106 Pa: quindi T1 = p1 V1 = 666 K T2 = p2 V2nR = 357 K 2. Il gas di destra fa una adiabatica: W = −∆U = −ncv(T2 −T0) = −490 J. Quindi il lavoro fatto sul gas e’ +490 J.
Per il gas a sinistra, applico il primo principio: ∆Q = W + ∆U = W + ncV (T1 − T0) = 3.53 kJ 3. Il gas di destra fa adiabatica reversibile ∆S = 0 Quello di sinistra ∆S = nR ln V1 V0 + ncV ln T1 T0 = 7.7 J/K Soluzione dell’esercizio 10.16 1. Il ciclo e reversibile, quindi ∆Su = 0 = Qc1 TA + Z Tg TA Mc dT T = Qc1 TA + Mc ln Tg TA da cui: Qc1 = −TAMc ln Tg TA = 28.3 kcal = 1.16 kJ 2. Calore ceduto dall’acqua: Qacqua = Mc(Tg −TA), da I principio: W1 = −(Qc1 + Mc(Tg −TA)) = −1.29kcal = −5.4 kJ 3. 0 = Qc2 TA + Qg Tg , da cui: Qc2 = −TA Tg Qg = −TA Tg (−Mλ) = 88kcal = 368 kJ S. Lacaprara Es. Meccanica e Termodinamica 122 4. W2 = −(Qc2 + Qg) = −7.9 kcal = −33.1 kJ, Wtot = W1 + W2 = −9.2 kcal = −38.6 kJ Soluzione dell’esercizio 10.17 1.
Soluzione dell’esercizio 10.18 1. QA = WA, quindi −mλ = nRT0 ln VAVT/2 da cui VA = 1/2e− mλ nRT0 = 4.8 l 2. pB = nRTB VT −VA = pa = nRTA VA , TB = TA VT −VA VA = 865 K 3.
WB = −WA = −nRT0 ln VAVT/2 = mλ QB = ncV (TB −T0) + WB = ncV (TB −T0)+ = mλ Quindi ∆t = QB/P = 1810 s Soluzione dell’esercizio 10.19 1. ∆S13 = ncV ln T3 T1 + nR ln V3 V1 = 9.96 −2.67 J/K 2. (1-2) e adiabatica: T2 = T1 p1 p3 1−γ γ Da eq stato dei gas perfetti p1 p3 = T1V3 T3V1, quindi T2 = T1 T1 rT3 1−γ γ = 943 K.
Oppure: (1-2) e adiabatica, quindi ∆Sgas 13 = ∆Sgas 23 = ncP ln T3T2, T3 = T2e −∆S13 ncP 3. Wciclo = ∆Qciclo. Nelle adiabatiche il calore scambiato e nullo. ∆Q23 = ncP∆T23 e ∆Q41 = ncV ∆T41 Mi manca T4 (34) e’ adiabatica, T4 = T3 V3 V1 γ−1 γ = T3 1 r γ−1 γ = 1733 K Wciclo = 1.37 · 104 −5.95 · 103 = 7.75 · 103 J 4. η = Wciclo ∆Qass = Wciclo ∆Q23 = 0.57 Soluzione dell’esercizio 10.20 1. Potenza trasferita dal frigo all’ambiente: dQA dt = −dQB dt TA TB = 600 W.
Potenza necessaria: Pn = dQA dt + dQB dt = 100 W. Quindi la frazione e f = Pn/P = 40% 2. ∆Sghiaccio = 0, ∆Scompressore = 0, ∆Samb = W T2 ∗3600 = 1200 J/K 3. Per portare ghiaccio a T = 0 ◦C serve: Q0 = Mc(T0 −T1) = 4.6 · 105 J.
Il calore che riceve dall’ambiente e Q′ = 500 ∗3600 = 1.8 · 106 J, quindi il restante Q” = Q′ −Q0 = 1.34 · 106 J scalda l’acqua. M = Q” λf = 4.0 kg 4. ∆S′ amb = Q”/T2 = −6000 J/K. ∆S′ ghiaccio = Mc ln T0 T1 = 1760 J/K, nella fusione ∆S′ fusione = mλf T0 = 4910 J/K. ∆S′ u = 670 J/K Soluzione dell’esercizio 10.21 S. Lacaprara Es. Meccanica e Termodinamica 123 1.
2. TA = pAVA nR = 305 K, TB = 914 K. VC = VBe ∆S nR = 50 l, pC = pBe−∆S nR = 3 atm, pD = 2.5 atm.
LAB = pA(VB−VA) = 10.1 kJ, LBC = T∆S = 7.77 kJ, LCD = (pC+pD) 2 (VD−VC) = −8.4 kJ QAB = ncp(TB −TA) = 35.5 kJ, QBC = LBC.
η = Ltot Qass = 10.4% 3. ∆SDA = nr ln VA VD = −11.5 J/K 4. ∆Suniv = ∆Suniv DA = ∆SDA + QDA TA = −11.5 + 16.4 = 4.9 J/K Soluzione dell’esercizio 10.22 1. n = pV RT = (Mg/S+patm)V RT = 0.41 moli, pA = (Mg/S + patm) = 1.03 · 105 Pa 2.
3.
p V A B C D Adiabatica rev Isobara irr.
Isoterma irr.
La trasformazione AB e una adiabatica (reversibile), quindi TB = T0 p0 pB 1−γ γ = 331 K con γ = 5/3 e pB = (M + m)g/S + patm = 1.32 · 105 Pa 4. Trasformazione isobara (irr): QS = −ncp(TS −TB) = −n5/2R(TS −TB) = 585 J 5. Trasformazione isoterma (irr); calcolo il lavoro contro la pressione del pistone LCD = −(Mg/S + patm)(VD −VC) = −309 J; VC = nRTS pB = 10.3 l VD = nRTS pA = 13.3 l 6. ∆SAB u = 0 J/k; ∆SBC u = ncp ln TS TB −QS TS = 1.61 −1.46 = 0.15 J/K; ∆SCD u = nR ln pC pD + LCD TS = +0.845 −0.773 = 0.072 J/K S. Lacaprara Es. Meccanica e Termodinamica 124 Ringraziamenti S. Lacaprara Es. Meccanica e Termodinamica 125 Ringraziamenti Gli esercizi proposti in queste pagine sono pescati o ispirati da una collezione che mi e stata fornita dal dott. P. Ronchese, che ha fatto esercizi di Fisica Generale I in passato con il prof.F. Bobisut. Senza dubbio anche la sua raccolta pesca da una memoria storica del dipartimenti di fisica, e dal contributo di esercitatori del passato. Altri esercizi derivano da miei appunti durante le lezioni del dott. P. Rossi, che svolgeva gli esercizi di Fisica Generale I per il prof. A. Bettini, quando mi sono laureato io.
Ringrazio inoltre tutti gli studenti che mi hanno segnalato imprecisioni ed errori nei te-sti o nelle soluzioni: Gabriele Labanca, Beatrice Moser, Davide Colucci, Paolo Simonetti, Meneghini Giuseppe, Sabatini Mattia, Matteo Fordiani, Maddalena Bin, Marco Zecchi-nato, Benedetta Spina, Davide Zuliani, Cecilia Antonioli, Edoardo Antonaci, Matteo Caldara, Fagherazzi Marco, Edoardo Brando, Elena Piccoli, Francesco Manzali, Giulia Zanfi, Anna Bison, Porcu Pasquale, Brunello Giacomo, Pierobon Tommaso, Davide Be-nedetti, Leso Aurora, Baldo Anna, Gianmarco Esposto, Lanaro Maria, Elena Del Mastro, Marcello Grenzi, Lai Nicolo’, Mariacristina Fiore, Luca Novelli Un ringraziamento particolare al dott. Stefano Zamuner, che ha svolto il ruolo di tutor degli studenti per il primo anno di svolgimento del corso, per le correzioni e i suggerimenti.
E infine, uno speciale a Paolo Ronchese, dalla cui ordinatissima raccolta ho estratto molti degli esercizi che trovate qui.
S. Lacaprara Es. Meccanica e Termodinamica 126
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Published Time: 2024-11-24T09:20:25.857Z
Measuring players’ consistent xG performances with Coefficient of Variation (CV) | by Marc Lamberts | Medium
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Measuring players’ consistent xG performances with Coefficient of Variation (CV)
Marc Lamberts
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Nov 24, 2024
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In media and online scouting reports, we often look at quality. How good is a player at this and that? We look at the quality of the clubs and the leagues. However, we often overlook other important aspects of data scouting, such as consistency and availability.
We often talk about rating strikers and their prime metrics, one of them being expected goals (xG). We look at totals or per 90 metrics, but how can value per 90 if the numbers don’t look at outliers? If you have three high-scoring games, it will cancel out three non-existent games, and still give an average number. So that’s why I today want to have a look at the consistency of strikers by using xG numbers throughout the season.
Contents
Why do we need this metric?
Data collection and justification
Coefficient of Variation
Methodology
Low Coefficient of Variation
High Coefficient of Variation
Final thoughts
Why do we need this metric?
I’m not exactly creating a new metric, but using a mathematical concept applied to football data. However, there is a use for this in football. Like I said above, we value quality so much, but consistency is key. If you are a consistent scoring player or generating xG consistently — that will mean that players will be valued more for their team.
Using this metric, we can see which players are very consistent in their performances and which players are very inconsistent. We will do this by examining their xG variation throughout the period we are examining.
Corner Possession Index: measuring a team’s possession quality after an attacking corner. ----------------------------------------------------------------------------------------- ### In the past year, I have been looking more and more at data related to set pieces, and corners to be more specific. It… marclamberts.medium.com
Data collection and justification
The data used comes from Opta and was collected on November, 22nd 2024. It has event data from the 2024–2025 Eredivisie season up to matchday 12 and is still all raw data. The data that is collected is individual data for each match and not totals, because we need match-level data to measure consistency.
With the data, we can create new metrics and put the data through our xG-model. In doing so, we can calculate the xG per shot, xG per game and xG totals for each individual player and team. I filter for players who have played more than 5 games so we can track the consistency over a longer period.
Coefficient of Variation
The Coefficient of Variation (CV) is a statistical measure that expresses the standard deviation of a dataset as a proportion of its mean. It is often used to assess the relative variability or consistency of data, particularly when comparing datasets with different units or means.
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For instance, when analysing performance metrics, CV provides insight into how consistently values are distributed around the average. It is particularly useful for comparing datasets where absolute values may differ significantly but relative variability is of interest. The CV is expressed as a ratio or a percentage, making it a versatile tool in understanding proportional variability.
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From this formula, we get a result. This result gives a high or low CV. This can have two meanings:
Low CV:
Indicates that the data points are tightly clustered around the mean.
Suggests greater consistency or predictability.
Example: A CV of 0.1 (or 10%) in a soccer player’s performance metric (e.g., goals per game) indicates stable performance over time.
High CV:
Indicates that the data points are more spread out relative to the mean.
Suggests greater variability or inconsistency.
Methodology
As said above, we first need to calculate the xG values for the whole Eredivisie in the 2024–2025 season. I’ have a model that converts shot values to shot values with xG values based on 400.000 shots taken. We then get a result.
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From having this information which was put through R, we will then focus on the critical part of getting to the CV: calculating the mean and standard deviation. I will use Python to run the calculation and it’s focused on having the PlayerId and the xG value.
import pandas as pd
Step 1: Load Excel file
df = pd.read_excel('EREXG.xlsx')
Step 2: Extract date from the timestamp
df['Date'] = df['Date'].str.split('T').str # Extract everything before 'T'
Step 3: Calculate CV for xG grouped by Player and count the number of games
cv_results = df.groupby('PlayerId').agg(
Mean=('xG', 'mean'),
StdDev=('xG', 'std'),
Games_Played=('Date', 'nunique') # Count unique dates
).reset_index()
Calculate Coefficient of Variation (CV)
cv_results['CV'] = cv_results['StdDev'] / cv_results['Mean']
Step 4: Rank players by CV
cv_results = cv_results.sort_values(by='CV') # Ascending: smaller CV = more consistent
Step 5: Save results to an Excel file
output_file = 'Player_CV_Results_With_Games.xlsx' # Specify the output file name
cv_results.to_excel(output_file, index=False)
print(f"\nResults saved to {output_file}")
As a result, I get an Excel file that shows me the PlayerId, the mean, the standard deviation and the CV. The matches played are also included. Now we have the framework on which we can start analysing.
Low Coefficient of Variation
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In the bargraph above you can see the top 15 players with the lowest coefficient of variation. This means that they have the least variability of their xG performance and therefore are consistent in their games through the first 12 games of the season.
High Coefficient of Variation
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In the bargraph above you can see the top 15 players with the highest coefficient of variation. This means that they have the most variability of their xG performance and therefore aren’t consistent in their games through the first 12 games of the season.
This can have several reasons why these players are featured in the low CV or in the high CV. The most common are position, how the team performs, how strong the opposition is. However it shows consistency or inconsistency, which is the aim of this metric.
Final thoughts
It has been an interesting thought experiment to look at consistency across a series of games for players in the Eredivisie, but there are a few things that I would different the next time:
Look at minutes played rather than matches played as it gives a better idea of on field actions
Filter for positions, this will be more useful for attackers than for defenders
Get a database that looks at a longer period to make a representative judgement of consistency.
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The next time I would also like to answer the question whether consistency is also a matter of playing many games. In other words does it even out over time and you become more consistent? From my small sample size it looks like it as you can see in the scatterplot above.
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spinmatch
Nov 26, 2024
Nice blog you got there about football, also please do follow for more knowledgeable and conversational content
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Percent Increase & Decrease, Population Growth Problems - SAT Math Part 34
The Organic Chemistry Tutor
9800000 subscribers
910 likes
Description
54892 views
Posted: 25 Apr 2019
This SAT Math video tutorial provides a basic introduction into calculating the percent increase and decrease of an event using the percent change formula.
SAT Math Part 35:
SAT Chemistry Part 1:
Percentages Made Easy:
Percentage, Base, & Rate Problems:
Part, Whole, & Percent Proportion:
Percent Increase & Decrease Problems:
Percentage Word Problems:
Percent Error:
Sales Price After Discount:
Sales Tax, Discount, & Original Price:
How To Calculate The Original Price:
Unique Percentage Problem:
Basic Math Quick Review:
Percent to Decimal Conversion:
Percent to Fraction Conversion:
Fractions - Basic Introduction:
Unit Rates, Ratios, & Proportions:
Ratios and Proportions - Word Problems:
Acid Mixture Word Problems:
Translating Words to Algebraic Expressions:
Words to Equations - More Problems:
Consecutive Integers Word Problems:
Age Word Problems - Past, Present, Future:
Upstream and Downstream - Problems:
Distance, Rate, & Time - Word Problems:
Time and Work - Word Problems:
Arithmetic Sequences - Word Problems:
Intersection and Union of Sets:
Venn Diagram - Word Problems:
Probability With Marbles - Word Problems:
Expected Value - Word Problems:
Standard Normal Distribution Problems:
Simple Interest - Word Problems:
Compound Interest Word Problems:
Future Value of an Annuity Math Problem:
Population Growth Word Problems:
Bacteria Growth Word Problems:
Exponential Growth & Decay Problems:
Logistic Growth Word Problems:
Angle of Elevation Word Problems:
Bearings and Direction - Word Problems:
Bearing and Navigation Word Problems:
36 comments
Transcript:
Number 117 number 117 in 2008 the average rent price in a certain city for a one-bedroom apartment was 9.50 by 2018 the average rent price increased to 14.25 what is the percent increase in the rent price over the 10-year period there's a formula in which we can calculate the percent change here it is the percent change is equal to the new price minus the original price divided by the original price times 100 percent so the new price is 14.25 the original price is 950 and we're going to divide it by 950 again so 1425 minus 950. let me type that again that's going to be 475. now let's divide 475 by 950. so that's 0.5 and then let's multiply that by 100 so the rent increased by 50 over this 10-year period real estate tends to go up in value but answer choice c is the correct answer Number 118 118 the price of a stock changed from 70 dollars to forty eight dollars in a one year period what is the percent change of this stock over the one year period so we're going to use the same formula the percent change is going to be equal to the new value which the new value is 48 minus the original value of 70 divided by the original value times 100 so 48 minus 70 that's negative 22 so now let's divide negative 22 by 70 and then let's multiply that by 100 so you should get negative 31.4 and of course this is a rounded answer so answer choice a is correct so the negative sign tells us that the stock decreased in value if we had a positive sign it would indicate a percent increase where the stock would increase in value but a negative sign indicates a percent decrease Number 119 number 119 there are 35 000 cats on a certain island in 2000 each year the population of cats increases by 3 at this rate what will be the approximate population of cats by the year 2024. well let's find out so let's write a formula let's say the population with respect to time is represented by the function p of t now this is going to equal the initial amount times basically 1 plus or minus r raised to the t where t is in years the initial amount of cats p0 is 35 000. r is three percent now notice that it increases by three percent so we're going to use the positive side so this this is going to be one plus r three percent if you divide that by a hundred is equal to point zero three as a decimal so thus we have the formula for the population of the cats it's 35 000 times 1.03 raised to the t now t is 0 corresponding to the year 2000 so by the year 2024 t should be equal to 24. so let's go ahead and plug this in so we're calculating p of 24 which is 35 000 times 1.03 raised to the 24th power and so you should get 71 147.79 so rounding that to nearest whole number that's 71 148 approximately which means answer choice d is the answer Number 120 number 120 in the year 2000 there were 50 000 tigers on a certain continent the population of tigers decreases by 4.5 percent every five years at this rate estimate the number of tigers that will be on the continent by the year 2100 so let's start with this formula again so p of t is equal to p initial times 1 plus or minus r raised to the t now this formula is going to be different because t is basically the number of years however notice that the population decreases by 4.5 every five years so in order for t to be the number of years we need to divide it by n where n is the number of times the population is compounded by a decrease of 4.5 percent so in this example r is going to be negative 4.5 or you could think of rs being positive 4.5 percent but we're going to use the negative sign now the second thing is n is 5 because it's going to change by r every five years now the initial population p zero that's going to be fifty thousand so the formula is going to be fifty thousand times one minus now four point five percent if we divide that by a hundred that's point zero four five and then we're gonna raise it to the t divided by five so we have p of t is equal to fifty thousand one minus point zero four five that's going to be point nine five five and that's raised to the t over five now what is t in this example in a year two thousand t is zero but by the year twenty one hundred t is going to be a hundred so the population decreases by four point five percent every five years so in a 100 year period it's going to decrease by this amount 20 times as you can see 100 divided by 5 is 20. so this is going to be 50 000 times 0.955 raised to the 20th power and so the population by 2100 is going to be 19 908 tigers rounded to the nearest whole number so that's how we can answer this particular type of question answer choice e is the answer Number 121 number 121 the number of bacteria in a sample triples every 20 minutes if there were 200 counts of bacteria initially how many counts of bacteria will there be in two hours well let's find out so the population of bacteria is going to equal the initial amount in this case i'm going to say let's use the letter x raised to the t over n the initial amount is 200 x would be the number of times let's say if it doubles or triples or quadruples because it tr um because it triples x is three t is going to be the time in minutes notice that it triples every 20 minutes so n is going to be 20 or 20 minutes so because n is in minutes t has to be in minutes now we want to find out how much bacteria there's going to be in two hours so we need to convert that to minutes there are 60 minutes in one hour so this is going to be 2 times 60 so that's 120 minutes so that's t so let's calculate the population of bacteria 120 minutes later so it's going to be 200 times 3 raised to the 120 divided by 20. 120 divided by 20 is 6. so in a two hour period the number of bacteria is going to triple six times because it triples every 20 minutes so 200 times 3 raised to the sixth power this is going to be 145 800. so that is the population of the bacteria that's going to be present two hours later so answer choice c is the right answer
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