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8.2.1
8.2.2
8.2.1
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8.2: The Hydrogen Atom
Last updated
: Mar 26, 2025
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8.1: Prelude to Atomic Structure
8.3: Orbital Magnetic Dipole Moment of the Electron
Page ID
: 4534
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Learning Objectives
By the end of this section, you will be able to:
Describe the hydrogen atom in terms of
wave function
,
probability density
,
total energy
, and orbital angular momentum
Identify the physical significance of each of the quantum numbers (n, l, m) of the hydrogen atom
Distinguish between the Bohr and Schrödinger models of the atom
Use quantum numbers to calculate important information about the hydrogen atom
The hydrogen atom is the simplest atom in nature and, therefore, a good starting point to study atoms and atomic structure. The hydrogen atom consists of a single negatively charged electron that moves about a positively charged proton (Figure 8.2.18.2.1). In Bohr’s model, the electron is pulled around the proton in a perfectly circular orbit by an attractive Coulomb force. The proton is approximately 1800 times more massive than the electron, so the proton moves very little in response to the force on the proton by the electron. (This is analogous to the Earth-Sun system, where the Sun moves very little in response to the force exerted on it by Earth.) An explanation of this effect using Newton’s laws is given in Photons and Matter Waves.
With the assumption of a fixed proton, we focus on the motion of the electron.
In the electric field of the proton, the potential energy of the electron is
U(r)=−ke2r,
U(r)=−ke2r,
where k=1/4πϵ0k=1/4πϵ0 and rr is the distance between the electron and the proton. As we saw earlier, the force on an object is equal to the negative of the gradient (or slope) of the potential energy function. For the special case of a hydrogen atom, the force between the electron and proton is an attractive Coulomb force.
Notice that the potential energy function U(r)U(r) does not vary in time. As a result, Schrödinger’s equation of the hydrogen atom reduces to two simpler equations: one that depends only on space (x, y, z) and another that depends only on time (t). (The separation of a
wave function
into space- and time-dependent parts for time-independent potential energy functions is discussed in Quantum Mechanics.) We are most interested in the space-dependent equation:
−ℏ2me(∂2ψ∂x2+∂2ψ∂y2+∂2ψ∂z2)−ke2rψ=Eψ,
−ℏ2me(∂2ψ∂x2+∂2ψ∂y2+∂2ψ∂z2)−ke2rψ=Eψ,
where ψ=psi(x,y,z)ψ=psi(x,y,z) is the three-dimensional
wave function
of the electron, meme is the mass of the electron, and EE is the
total energy
of the electron. Recall that the total
wave function
Ψ(x,y,z,t)Ψ(x,y,z,t), is the product of the space-dependent
wave function
ψ=ψ(x,y,z)ψ=ψ(x,y,z) and the time-dependent
wave function
φ=φ(t)φ=φ(t).
In addition to being time-independent, U(r)U(r) is also spherically symmetrical. This suggests that we may solve Schrödinger’s equation more easily if we express it in terms of the spherical coordinates (r,θ,ϕr,θ,ϕ) instead of rectangular coordinates (x,y,zx,y,z). A spherical coordinate system is shown in Figure 8.2.28.2.2. In spherical coordinates, the variable rr is the radial coordinate, θθ is the polar angle (relative to the vertical z-axis), and ϕϕ is the azimuthal angle (relative to the x-axis). The relationship between spherical and rectangular coordinates is x=rsinθcosϕx=rsinθcosϕ, y=rsinθsinϕy=rsinθsinϕ, z=rcosθz=rcosθ.
The factor rsinθrsinθ is the magnitude of a vector formed by the projection of the polar vector onto the xy-plane. Also, the coordinates of x and y are obtained by projecting this vector onto the x- and y-axes, respectively. The inverse transformation gives
r=√x2+y2+z2θ=cos−1(zr),ϕ=cos−1(x√x2+y2)
Schrödinger’s wave equation for the hydrogen atom in spherical coordinates is discussed in more advanced courses in modern physics, so we do not consider it in detail here. However, due to the spherical symmetry of U(r), this equation reduces to three simpler equations: one for each of the three coordinates (r, θ, and ϕ). Solutions to the time-independent
wave function
are written as a product of three functions:
ψ(r,θ,ϕ)=R(r)Θ(θ)Φ(ϕ),
where R is the radial function dependent on the radial coordinate r only; Θ is the polar function dependent on the polar coordinate θ only; and Φ is the phi function of ϕ only. Valid solutions to Schrödinger’s equation ψ(r,θ,ϕ) are labeled by the quantum numbers n, l, and m.
n:
principal quantum number
l: angular momentum
quantum number
m: angular momentum projection
quantum number
(The reasons for these names will be explained in the next section.) The radial function R depends only on n and l; the polar function Θ depends only on l and m; and the phi function Φ depends only on m. The dependence of each function on quantum numbers is indicated with subscripts:
ψnlm(r,θ,ϕ)=Rnl(r)Θlm(θ)Φm(ϕ).
Not all sets of quantum numbers (n, l, m) are possible. For example, the orbital angular
quantum number
l can never be greater or equal to the
principal quantum number
n(l<n). Specifically, we have
n=1,2,3,...
l=0,1,2,...,(n−1)
m=−l,(−l+1),...,0,...,(+l−1),+l
Notice that for the ground state, n=1, l=0, and m=0. In other words, there is only one quantum state with the
wave function
for n=1, and it is ψ100. However, for n=2, we have
l=0,m=0
and
l=1,m=−1,0,1.
Therefore, the allowed states for the n=2 state are ψ200, ψ21−1, ψ210, and ψ211. Example wave functions for the hydrogen atom are given in Table 8.2.1. Note that some of these expressions contain the letter i, which represents √−1. When probabilities are calculated, these complex numbers do not appear in the final answer.
8.2.1: Wave Functions of the Hydrogen Atom
| n=1,l=0,ml=0 | ψ100=1√π1a3/20e−r/a0 |
| n=2,l=0,ml=0 | ψ200=14√2π1a3/20(2−ra0)e−r/2a0 |
| n=2,l=1,ml=−1 | ψ21−1=18√π1a3/20ra0e−r/2a0sinθe−iϕ |
| n=2,l=1,ml=0 | ψ210=14√2π1a3/20ra0e−r/2a0cosθ |
| n=2,l=1,ml=1 | ψ211=18√π1a3/20ra0e−r/2a0sinθeiϕ |
Physical Significance of the Quantum Numbers
Each of the three quantum numbers of the hydrogen atom (n, l, m) is associated with a different physical quantity.
Principal Quantum Number
The
principal quantum number
n is associated with the
total energy
of the electron, En. According to Schrödinger’s equation:
En=−(mek2e42ℏ2)(1n2)=−E0(1n2),
where E0=−13.6eV. Notice that this expression is identical to that of Bohr’s model. As in the Bohr model, the electron in a particular state of energy does not radiate.
Example 8.2.1: How Many Possible States?
For the hydrogen atom, how many possible quantum states correspond to the principal number n=3? What are the energies of these states?
Strategy
For a hydrogen atom of a given energy, the number of allowed states depends on its orbital angular momentum. We can count these states for each value of the
principal quantum number
, n=1,2,3. However, the
total energy
depends on the
principal quantum number
only, which means that we can use Equation 8.2.1 and the number of states counted.
Solution
If n=3, the allowed values of l are 0, 1, and 2. If l=0, m=0 (1 state). If l=1, m=−1,0,1 (3 states); and if l=2, m=−2,−1,0,1,2 (5 states). In total, there are 1 + 3 + 5 = 9 allowed states. Because the
total energy
depends only on the
principal quantum number
, n=3, the energy of each of these states is
En3=−E0(1n2)=−13.6eV9=−1.51eV.
Significance
An electron in a hydrogen atom can occupy many different angular momentum states with the very same energy. As the orbital angular momentum increases, the number of the allowed states with the same energy increases.
Angular Momentum Orbital Quantum Number
The angular momentum orbital
quantum number l is associated with the orbital angular momentum of the electron in a hydrogen atom. Quantum theory tells us that when the hydrogen atom is in the state ψnlm, the magnitude of its orbital angular momentum is
L=√l(l+1)ℏ,
where l=0,1,2,...,(n−1).
This result is slightly different from that found with Bohr’s theory, which quantizes angular momentum according to the rule L=n, where n=1,2,3,...
Spectroscopic Notation
Quantum states with different values of orbital angular momentum are distinguished using spectroscopic notation (Table 8.2.2). The designations s, p, d, and f result from early historical attempts to classify atomic spectral lines. (The letters stand for sharp, principal, diffuse, and fundamental, respectively.) After f, the letters continue alphabetically.
The ground state of hydrogen is designated as the 1s state, where “1” indicates the energy level (n=1) and “s” indicates the orbital angular momentum state (l=0). When n=2, l can be either 0 or 1. The n=2, l=0 state is designated “2s.” The n=2, l=1 state is designated “2p.” When n=3, l can be 0, 1, or 2, and the states are 3s, 3p, and 3d, respectively. Notation for other quantum states is given in Table 8.2.3.
Table 8.2.2: Spectroscopic Notation and Orbital Angular Momentum
| Orbital Quantum Number l | Angular Momentum | State | Spectroscopic Name |
| 0 | 0 | s | Sharp |
| 1 | √2h | p | Principal |
| 2 | √6h | d | Diffuse |
| 3 | √12h | f | Fundamental |
| 4 | √20h | g | |
| 5 | √30h | h | |
Angular Momentum Projection Quantum Number
The angular momentum projection
quantum number m is associated with the azimuthal angle ϕ (see Figure 8.2.2) and is related to the z-component of orbital angular momentum of an electron in a hydrogen atom. This component is given by
Lz=mℏ,
where m=−l,−l+1,...,0,...,+l−1,l.
The z-component of angular momentum is related to the magnitude of angular momentum by
Lz=Lcosθ,
where θ is the angle between the angular momentum vector and the z-axis. Note that the direction of the z-axis is determined by experiment - that is, along any direction, the experimenter decides to measure the angular momentum. For example, the z-direction might correspond to the direction of an external magnetic field. The relationship between Lz and L is given in Figure 8.2.3.
Table 8.2.3: Spectroscopic Description of Quantum States
| | l=0 | l=1 | l=2 | l=3 | l=4 | l=5 |
| n=1 | 1s | | | | | |
| n=2 | 2s | 2p | | | | |
| n=3 | 3s | 3p | 3d | | | |
| n=4 | 4s | 4p | 4d | 4f | | |
| n=5 | 5s | 5p | 5d | 5f | 5g | |
| n=6 | 6s | 6p | 6d | 6f | 6g | 6h |
The quantization of Lz is equivalent to the quantization of θ. Substituting √l(l+1)ℏ for L and m for Lz into this equation, we find
mℏ=√l(l+1)ℏcosθ.
Thus, the angle θ is quantized with the particular values
θ=cos−1(m√l(l+1)).
Notice that both the polar angle (θ) and the projection of the angular momentum vector onto an arbitrary z-axis (Lz) are quantized.
The quantization of the polar angle for the l=3 state is shown in Figure 8.2.4. The orbital angular momentum vector lies somewhere on the surface of a cone with an opening angle θ relative to the z-axis (unless m=0, in which case θ=90o and the vector points are perpendicular to the z-axis).
A detailed study of angular momentum reveals that we cannot know all three components simultaneously. In the previous section, the z-component of orbital angular momentum has definite values that depend on the
quantum number
m. This implies that we cannot know both x- and y-components of angular momentum, Lx and Ly, with certainty. As a result, the precise direction of the orbital angular momentum vector is unknown.
Example 8.2.2: What Are the Allowed Directions?
Calculate the angles that the angular momentum vector →L can make with the z-axis for l=1, as shown in Figure 8.2.5.
Strategy
The vectors →L and →Lz (in the z-direction) form a right triangle, where →L is the hypotenuse and →Lz is the adjacent side. The ratio of Lz to |→L| is the cosine of the angle of interest. The magnitudes L=|→L| and Lz are given by
L=√l(l+1)ℏ
and
Lz=mℏ.
Solution
We are given l=1, so m can be +1, 0, or +1. Thus, L has the value given by
L=√l(l+1)ℏ=√2ℏ.
The quantity Lz can have three values, given by Lz=mlℏ.
Lz={ℏ,if ml=+10,if ml=0ℏ,if ml=−1
As you can see in Figure 8.2.5, cosθ=Lz/L, so for m=+1, we have
cosθ1=LzL=ℏ√2ℏ=1√2=0.707
Thus,
θ1=cos−10.707=45.0°.
Similarly, for m=0, we find cosθ2=0; this gives
θ2=cos−10=90.0°.
Then for ml=−1:
cosθ3=LZL=−ℏ√2ℏ=−1√2=−0.707,
so that
θ3=cos−1(−0.707)=135.0°.
Significance
The angles are consistent with the figure. Only the angle relative to the z-axis is quantized. L can point in any direction as long as it makes the proper angle with the z-axis. Thus, the angular momentum vectors lie on cones, as illustrated. To see how the
correspondence principle
holds here, consider that the smallest angle (θ1 in the example) is for the maximum value of ml, namely ml=l. For that smallest angle,
cosθ=LzL=l√l(l+1),
which approaches 1 as l becomes very large. If cosθ=1, then θ=0º. Furthermore, for large l, there are many values of ml, so that all angles become possible as l gets very large.
Exercise 8.2.1
Can the magnitude Lz ever be equal to L?
Answer
: No. The
quantum number
m=−l,−l+l,...,0,...,l−1,l. Thus, the magnitude of Lz is always less than L because <√l(l+1)
Using the Wave Function to Make Predictions
As we saw earlier, we can use quantum mechanics to make predictions about physical events by the use of probability statements. It is therefore proper to state, “An electron is located within this volume with this probability at this time,” but not, “An electron is located at the position (x, y, z) at this time.” To determine the probability of finding an electron in a hydrogen atom in a particular region of space, it is necessary to integrate the
probability density
(|ψ_{nlm}|^2)_ over that region:
Probability=∫volume|ψnlm|2dV,
where dV is an infinitesimal volume element. If this integral is computed for all space, the result is 1, because the probability of the particle to be located somewhere is 100% (the normalization condition). In a more advanced course on modern physics, you will find that |ψnlm|2=ψ∗nlmψnlm, where ψ∗nlm is the complex conjugate. This eliminates the occurrences i=√−1 in the above calculation.
Consider an electron in a state of zero angular momentum (l=0). In this case, the electron’s
wave function
depends only on the radial coordinate r. (Refer to the states ψ100 and ψ200 in Table 8.2.1.) The infinitesimal volume element corresponds to a spherical shell of radius r and infinitesimal thickness dr, written as
dV=4πr2dr.
The probability of finding the electron in the region r to r+dr (“at approximately r”) is
P(r)dr=|ψn00|24πr2dr.
Here P(r) is called the radial probability density function (a probability per unit length). For an electron in the ground state of hydrogen, the probability of finding an electron in the region r to r+dr is
|ψn00|24πr2dr=(4/a3))r2exp(−2r/a0)dr,
where a0=0.5 angstroms. The
radial probability density function
P(r) is plotted in Figure 8.2.6. The area under the curve between any two radial positions, say r1 and r2, gives the probability of finding the electron in that radial range. To find the most probable radial position, we set the first derivative of this function to zero (dP/dr=0) and solve for r. The most probable radial position is not equal to the average or
expectation value
of the radial position because |ψn00|2 is not symmetrical about its peak value.
If the electron has orbital angular momentum (l≠0), then the wave functions representing the electron depend on the angles θ and ϕ; that is, ψnlm=ψnlm(r,θ,ϕ). Atomic orbitals for three states with n=2 and l=1 are shown in Figure 8.2.7. An atomic orbital is a region in space that encloses a certain percentage (usually 90%) of the electron probability. (Sometimes atomic orbitals are referred to as “clouds” of probability.) Notice that these distributions are pronounced in certain directions. This directionality is important to chemists when they analyze how atoms are bound together to form molecules.
A slightly different representation of the
wave function
is given in Figure 8.2.8. In this case, light and dark regions indicate locations of relatively high and low probability, respectively. In contrast to the Bohr model of the hydrogen atom, the electron does not move around the proton nucleus in a well-defined path. Indeed, the uncertainty principle makes it impossible to know how the electron gets from one place to another.
8.1: Prelude to Atomic Structure
8.3: Orbital Magnetic Dipole Moment of the Electron
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Image: The parallelogram law, or commutative law, of vector addition
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1 Switching Circuits & Logic Design Jie-Hong Roland Jiang 江介宏 Department of Electrical Engineering National Taiwan University Fall 2013 2 §7 Multi-Level Gate Circuits 3 Outline Multi-level gate circuits NAND and NOR gates Design of two-level circuits using NAND and NOR gates Design of multi-level NAND- and NOR-gate circuits Circuit conversion using alternative gate symbols Design of two-level, multiple-output circuits Multiple-output NAND and NOR circuits 4 Multi-Level Gate Circuits The number of levels of gates The maximum number of gates cascaded in series between a circuit input and output Inverters which are connected directly to input variables do not count (assume variables and their complements are available as circuit inputs) SOP (POS) correspond to AND-OR (OR-AND) two-level gate circuits AND-OR 2-level circuit composed of a level of AND gates followed by an OR gate at the output OR-AND 2-level circuit composed of a level of OR gates followed by an AND gate at the output OR-AND-OR 3-level circuit composed of a level of OR gates followed by a level of AND gates followed by an OR gate at the output Circuit of AND and OR gates No particular order of the gates 5 Multi-Level Gate Circuits AND-OR-AND-OR 4-level realization of Z 4 levels, 6 gates, 13 gate inputs A B C D E F G H Z Z=(AB+C)(D+E+FG)+H 2 2 3 2 2 2 level 4 level 3 level 1 level 2 6 Multi-Level Gate Circuits OR-AND-OR 3-level realization of Z Partially multiplying out Z = (AB+C)[(D+E)+FG]+H 3 levels, 6 gates, 19 gate inputs Z=(AB(D+E)+C(D+E)+ABFG+CFG+H 5 3 2 level 3 level 1 level 2 A G C D E F G H Z B C B A F 3 4 2 same gate 7 Multi-Level Gate Circuits Drawing the tree diagram of an expression helps determine the realization costs #level circuit delay #gates, #gate inputs circuit area Different expressions of a Boolean function provide different tradeoffs between delay and area 8 Multi-Level Gate Circuits Find a circuit of AND and OR gates realizing f(a,b,c,d) = m(1,5,6,10,13,14) 1 0 1 0 01 1 0 1 0 11 0 1 01 0 0 11 10 00 10 00 1 0 0 0 ab cd f = a'c'd+bc'd+bcd'+acd' a' f c' d b c' d b c d' a c d' 2 levels, 5 gates, 16 gate inputs 9 Multi-Level Gate Circuits Example (cont’d) 1 0 1 0 01 1 0 1 0 11 0 1 01 0 0 11 10 00 10 00 1 0 0 0 ab cd f = a'c'd+bc'd+bcd'+acd' = c'd(a'+b)+cd'(a+b) a' f c' d b b c d' a 3 levels, 5 gates, 12 gate inputs 10 Multi-Level Gate Circuits Example (cont’d) 1 0 1 0 01 1 0 1 0 11 0 1 01 0 0 11 10 00 10 00 1 0 0 0 ab cd f' = c'd'+ab'c'+cd+a'b'c f = (c+d)(a'+b+c)(c'+d')(a+b+c') c f d b a' c d' a b c' c' 2 levels, 5 gates, 14 gate inputs 11 Multi-Level Gate Circuits Example (cont’d) 1 0 1 0 01 1 0 1 0 11 0 1 01 0 0 11 10 00 10 00 1 0 0 0 ab cd f = (c+d)(a'+b+c)(c'+d')(a+b+c') = (c+a'd+bd)(c'+ad'+bd') f d' d' d b d a' b a c' c 3 levels, 7 gates, 16 gate inputs 12 Multi-Level Gate Circuits To be sure of obtaining a minimum solution, we have to find both the circuit with the AND-gate output and the one with the OR-gate output If the expression for f' has n levels with an AND-gate (OR-gate) output, its complement is an n-level expression for f with an OR-gate (AND-gate) output 13 NAND and NOR Gates NAND AND-NOT gate 3-input NAND: F = (ABC)' = A'+B'+C' n-input NAND: F = (X1X2…Xn)' = X1'+X2'+...+Xn' NOR OR-NOT gate 3-input NOR: F = (A+B+C)' = A'B'C' n-input NOR: F = (X1+X2+…+Xn)' = X1'X2'...Xn' A B C F A B C F X1 F X2 Xn … A B C F A B C F X1 F X2 Xn … 14 NAND and NOR Gates A set of logic operations is functionally complete if any Boolean function can be expressed in terms of the set of operations {AND, OR, NOT} is functionally complete Any Boolean function can be expressed in SOP form, which uses only the AND, OR, NOT operations Any set of logic operations that can realize AND, OR, NOT is also functionally complete {AND, NOT} is functionally complete since OR can be realized using AND and NOT as shown below X Y' (X'Y')'=X+Y Y X' X'Y' 15 NAND and NOR Gates {NAND} is functionally complete NOT: (XX)' = X' AND: ((AB)')' = AB OR: (A'B')' = A+B X X' A AB B (AB)' A B A' B' (A'B')'=A+B 16 NAND and NOR Gates How to show whether or not a set of logic operations is functionally complete?
1.
Write out a minimum SOP expression for the function realized by each gate 2.
If no complement appears in any of these expressions, then NOT cannot be realized 3.
Otherwise, NOT can be realized by an appropriate choice of inputs to the corresponding gate (assume 0 and 1 are available as gate inputs) 4.
Try to realize AND or OR (now with NOT available) 17 NAND and NOR Gates Exercises Show that the sets {NOR} and {OR, NOT} are functionally complete Is the majority gate major functionally complete?
major(A,B,C) = 1 iff at least two of A, B, C are 1 Is the minority gate minor functionally complete?
minor(A,B,C) = 1 iff at most one of A, B, C is 1 Is {} functionally complete?
AB is true iff A is false (0), or both A and B are true (1) Does the assumption “0 and 1 are available as gate inputs” make a difference?
18 Two-Level Circuit Design Using NAND and NOR Gates A 2-level circuit composed of AND, OR gates can be converted to a circuit composed of NAND, NOR gates By using F = (F')' and DeMorgan’s laws Example 1 F = A+BC'+B'CD (AND-OR) = [(A+BC'+B'CD)']' = [A'(BC')'(B'CD)']' (NAND-NAND) = [A'(B'+C)(B+C'+D')]' (OR-NAND) = A+(B'+C)'+(B+C'+D')' (NOR-OR) 19 Two-Level Circuit Design Using NAND and NOR Gates F=A+BC'+B'CD F=[A'(BC')'(B'CD)']' F=[A'(B'+C)(B+C'+D')]' D F B C' C B' A F B C' D C B' A' F B' C D' C' B A' F=A+(B'+C)'+(B+C'+D')' F B' C D' C' B A AND-OR NAND-NAND NOR-OR OR-NAND 20 Two-Level Circuit Design Using NAND and NOR Gates Example 2 F = (A+B+C)(A+B'+C')(A+C'+D) (OR-AND) = {[(A+B+C)(A+B'+C')(A+C'+D)]'}' = [(A+B+C)'+(A+B'+C')'+(A+C'+D)']' (NOR-NOR) = [(A'B'C')+(A'BC)+(A'CD')]' (AND-NOR) = (A'B'C')'(A'BC)'(A'CD')' (NAND-AND) 21 Two-Level Circuit Design Using NAND and NOR Gates F=(A+B+C)(A+B'+C')(A+C'+D) F=[(A+B+C)'+ (A+B'+C')'+(A+C'+D)']' OR-AND NOR-NOR NAND-AND AND-NOR F=(A'B'C'+A'BC+A'CD')' F=(A'B'C')' (A'BC)'(A'CD')' F A C B A C' B' A D C' A' F C B B' C' A' A' C D F B' C' D C' A A A B C F C' C B A' B' A' A' C D' 22 Two-Level Circuit Design Using NAND and NOR Gates Among the 16 two-level forms: The following 8 are generic (can realize all switching functions): AND-OR, AND-NOR, OR-AND, OR-NAND, NAND-AND, NAND-NAND, NOR-OR, NOR-NOR The following 8 are degenerate (cannot realize all functions): AND-AND, AND-NAND, OR-OR, OR-NOR, NAND-OR, NAND-NOR, NOR-AND, NOR-NAND E.g., NAND-NOR form can realize only a product of literals and not a sum of products F = [(ab)'+(cd)'+e]' = abcde' a b d c e 23 Two-Level Circuit Design Using NAND and NOR Gates NAND-NAND and NOR-NOR are the most widely used forms in integrated circuits Procedure for minimum NAND-NAND (NOR-NOR) implementation 1.
Find a minimum SOP expression of F 2.
Draw the corresponding two-level AND-OR (OR-AND) circuit 3.
Replace all gates with NAND (NOR) gates with interconnections unchanged. Complement the single-literal inputs of the output gate F = l1+l2++P1+P2+ F P1 P2 x1 x2 y2 y1 l2 l1 … … … … F P1' P2' x1 x2 y2 y1 l2' l1' … … … … F = (l1'l2'P1'P2')' 24 Design of Multi-Level NAND- and NOR-Gate Circuits Procedure for designing multi-level NAND-gate circuits 1.
Simplify F 2.
Design a multi-level circuit of AND and OR gates with output gate being OR AND-gate (OR-gate) outputs cannot be used as AND-gate (OR-gate) inputs 3.
Number the levels starting with the output gate as level 1. Replace all gates with NAND gates, leaving interconnections unchanged. Invert any literals which appear as inputs to levels 1, 3, 5, … 25 Design of Multi-Level NAND- and NOR-Gate Circuits Example F1 = a'[b'+c(d+e')+f'g']+hi'j+k F1 b' c g' f' d a' k h i' j e' Level 1 Level 2 Level 3 Level 4 Level 5 F1 b c g' f' d' a' k' h i' j e Level 1 Level 2 Level 3 Level 4 Level 5 26 Circuit Conversion Using Alternative Gate Symbols Alternative gate symbols Useful for circuit analysis and design A' A A' A A B AB A B (AB)' A B A+B A B (A+B)' NOT AND OR NAND NOR 27 Circuit Conversion Using Alternative Gate Symbols NAND gate circuit conversion F C E D B' A Z F C E D B' A Z= (A'+B)C+F'+DE F' C E D B A' Z 1 2 3 4 1 2 3 4 1 2 3 4 A'+B [(A'+B)C]' (DE)' 28 Circuit Conversion Using Alternative Gate Symbols Conversion to NOR gates C D B' A Z F E G C' D B' A Z F E G' Double inversion cancels Complemented input cancels inversion 29 Circuit Conversion Using Alternative Gate Symbols A circuit composed of AND and OR gates can be converted to a circuit composed of NAND and NOR gates, and vice versa By properly adding inverters, removing canceling inverter pairs, and/or negating inputs, we can change a gate type to a desired one 30 Circuit Conversion Using Alternative Gate Symbols Conversion to NAND gates (even if AND and OR gates do not alternate) C D B' A F E C D B' A F E C D' B' A F E' 31 Two-Level, Multiple-Output Circuit Design In circuit design, often we need several Boolean functions rather than one Although every function can be implemented separately, recognizing common gates among these functions can achieve logic sharing and thus reduce area When designing multiple-output circuits, we should try to first minimize the total number of gates required and then minimize gate inputs 32 Two-Level, Multiple-Output Circuit Design Example F1(A,B,C,D) = m(11,12,13,14,15) F2(A,B,C,D) = m(3,7,11,12,13,15) F3(A,B,C,D) = m(3,7,12,13,14,15) 01 1 1 1 1 11 01 1 11 10 00 10 00 AB CD 1 01 1 1 1 11 01 1 1 11 10 00 10 00 AB CD 1 01 1 1 1 1 11 01 1 11 10 00 10 00 AB CD F1 F2 F3 Same product term 33 Two-Level, Multiple-Output Circuit Design Example (cont’d) Separate vs. multiple-output realization C B D F1 A A B C' F2 A C D F3 A' Note that F2 in the multiple-output realization is not a minimum SOP 9 gates, 21 gate inputs 7 gates, 18 gate inputs C B D F1=AB+ACD A A @ C B D F3=A'CD+AB A A' # A B D C' F2=ABC'+CD C @+# 34 Two-Level, Multiple-Output Circuit Design Example f1(a,b,c,d) = m(2,3,5,7,8,9,10,11,13,15) f2(a,b,c,d) = m(2,3,5,6,7,10,11,14,15) f3(a,b,c,d) = m(6,7,8,9,13,14,15) 1 1 01 1 1 11 1 01 1 1 11 10 00 10 00 1 1 1 ab cd 1 1 1 01 1 1 11 01 1 1 11 10 00 10 00 1 1 ab cd 1 1 01 1 1 1 11 1 01 11 10 00 10 00 1 ab cd f1 f2 f3 35 Two-Level, Multiple-Output Circuit Design Example (cont’d) Separate realization f1 = bd+b'c+ab' f2 = c+a'bd f3 = bc+ab'c'+abd 10 gates, 25 gate inputs Multi-output realization f1 = a'bd+abd+ab'c'+b'c f2 = c+a'bd f3 = bc+ab'c'+abd 8 gates, 22 gate inputs 36 Two-Level, Multiple-Output Circuit Design Example When designing multiple-output circuits, it is sometimes best not to combine a 1 with its adjacent 1’s 1 01 1 1 11 1 1 01 11 10 00 10 00 ab cd 1 1 01 1 1 1 11 01 11 10 00 10 00 ab cd 1 01 1 1 11 1 1 01 11 10 00 10 00 ab cd 1 1 01 1 1 1 11 01 11 10 00 10 00 ab cd f1 f2 Best solution: Solution requires an extra gate: f1 f2 37 Two-Level, Multiple-Output Circuit Design Example The solution with the maximum number of common terms is not necessarily best 1 1 1 01 1 11 01 11 10 00 10 00 1 1 ab cd 1 1 01 1 1 11 1 01 11 10 00 10 00 1 ab cd 1 1 1 01 1 11 01 11 10 00 10 00 1 1 ab cd 1 1 01 1 1 11 1 01 11 10 00 10 00 1 ab cd f1 f2 f1 f2 Solution with maximum # of common terms: (8 gates, 26 inputs) Best solution (no common terms): 7 gates, 18 inputs 38 Two-Level, Multiple-Output Circuit Design Procedure of SOP minimization applies for multiple-output realization with slight modification of the determination of essential prime implicants (EPIs) We find prime implicants that are essential to one of the functions and to the multiple-output realization Some of the prime implicants essential to an individual function may not be essential to the multiple-output realization Recall an EPI is a prime implicant that covers some minterm mi that is not covered by any other prime implicant The minterm mi may appear and be covered by a (different ?) prime implicant in another function E.g., Slide 34: bd is essential to f1, but not for the multiple-output realization Slide 36: c'd is essential to f1 for the multiple-output realization abd is essential to f1, but not for the multiple-output realization Slide 37: a'd', a'bc' are essential to f1 for the multiple-output realization bd' is essential to f2 for the multiple-output realization 39 Multiple-Output NAND and NOR Circuits Procedure for designing single-output, multi-level NAND- and NOR-gate circuits applies to multiple-output circuits If all of the output gates are OR (AND), direct conversion to a NAND-gate (NOR-gate) circuit is possible 40 Multiple-Output NAND and NOR Circuits Multi-level circuit conversion to NOR gates Level 1 Level 2 Level 3 Level 4 F1 d c e' a h f b' F2 g' F1 d c' e' a h' f b' F2 g'
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https://manual.gamemaker.io/monthly/en/GameMaker_Language/GML_Reference/Asset_Management/Audio/db_to_lin.htm
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db_to_lin
db_to_lin
Click here to see this page in full context
db_to_lin
This function converts a gain expressed in decibels (dB) to a linear gain, which can be used with functions such asaudio_sound_gain.
Changes in gain can be expressed in dB. This corresponds better to how we perceive changes in amplitude (human perception of sound is not linear).
The conversion from a gain expressed in dB to a linear gain is done using the following formula: gain_linear = power(10, gain_db/20);
TIP 0dB means no change in gain (i.e. a linear gain of 1). An increase of 6dB roughly corresponds to a doubling of the linear gain, a decrease of 6dB (i.e. an increase of -6dB) to a halving of it.
Also see:lin_to_db
Syntax:
db_to_lin(x);
| Argument | Type | Description |
---
| x | Real | The value to convert |
Returns:
Real
Example:
var _lin =db_to_lin(-3);
The above code calculates the linear gain reduction corresponding to a reduction of 3dB. The result of the conversion is stored in a temporary variable _lin.
Back:Audio
Next:audio_create_stream
© Copyright YoYo Games Ltd. 2025 All Rights Reserved
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https://sim.ku.edu/strategic-math-addition-regrouping
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Strategic Math: Addition with Regrouping
For some students, addition difficulties begin with single-digit basic facts, but, for others, problems emerge when the curriculum progresses to multi-digit addition, especially if regrouping is involved. When students struggle with computation and word problems that require regrouping, specialized instruction is needed.
The specialized instruction in the Addition With Regrouping program includes concrete and representational lessons that involve the use of base-ten blocks and drawings to represent the regrouping process. These three- and two-dimensional models ensure that students conceptually understand WHY regrouping is needed. This understanding enables students to determine when regrouping is needed, remember the process involved in regrouping 10 ones into a ten or 10 tens into a hundred, and retain this important information over time.
Once conceptual knowledge is gained, students learn the procedural steps that must be followed to obtain accurate answers with problems involving numbers only. Finally, fluency is developed through motivating minute timings and exciting games using the “pig dice” that accompany the manual. Who knew computation and word problems that involve addition and regrouping could be so much fun?
Author(s):Susan Peterson Miller, Bradley J. Kaffar, and Cecil D. Mercer
Publication and Purchasing Information: University of Kansas, Center for Research on Learning / KU CRL Online Store
Resources:
Research Articles:
This product is available through the KUCRL Shop.
Please note that professional development, coaching, and infrastructure support are essential components to effective implementation of SIM instructional tools and interventions. It is highly recommended that you work with a SIM professional developer. See the SIM Event list for sessions or email simpd@ku.edu to learn more.
More in the Strategic Math Series
An accessible version of the documents on this site will be made available upon request. Please contact the KU CRL Professional Development Research Institute, at simpd@ku.edu to request the document be made available in an accessible format.
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https://teachy.ai/en/project/high-school-en-US/us-12th-grade/math/exploring-surface-area-of-cones-through-the-great-cone-challenge
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Activities of Exploring Surface Area of Cones through "The Great Cone Challenge"
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Spatial Geometry: Surface Area of the Cone
Project: Exploring Surface Area of Cones through "The Great Cone Challenge"
Lara from Teachy
Subject Math
Math
Source Teachy Original
Teachy Original
Topic Spatial Geometry: Surface Area of the Cone
Spatial Geometry: Surface Area of the Cone
Contextualization
Introduction to Spatial Geometry: Surface Area of the Cone
Spatial geometry is a branch of mathematics that deals with the properties and measurement of figures in space. One of the fundamental figures in spatial geometry is the cone. The cone is a three-dimensional geometric shape with a circular base and a pointed top, much like an ice cream cone. It's an important concept to understand because it's a figure that we see frequently in our world.
The surface area of a cone is the total area that the surface of the cone occupies. We can break the surface area of a cone into two components: the base and the lateral surface. The base is simply a circle, and we know that the area of a circle is calculated by the formula A = πr², where r is the radius of the circle. The lateral surface is a sector of a circle that has been rolled into a cone. The formula for the lateral surface area of a cone is A = πrl, where r is the radius of the base, and l is the slant height of the cone.
Why is Spatial Geometry Important?
Spatial geometry plays an integral role in the study of mathematics and also has many applications in the real world. For example, in architecture, spatial geometry is used to design and construct buildings. In engineering, it is used to design structures like bridges and tunnels.
Understanding spatial geometry also helps us develop skills in problem-solving, logical thinking, and spatial visualization. These skills are not only valuable in the field of mathematics but also in various other fields like computer science, physics, and even art and design.
Resources
To delve deeper into the topic, you can consult the following resources:
Khan Academy: Surface area of cones
Math is Fun: Surface Area of a Cone
Geometry: Surface Area of Cones
Book: "Geometry for Enjoyment and Challenge" by Richard Rhoad and George Milauskas.
These resources are a great starting point for understanding the surface area of cones. They provide clear explanations, examples, and even interactive exercises to help solidify your understanding of the topic. Happy learning!
Practical Activity
Activity Title: "The Great Cone Challenge"
Objective of the Project:
The objective of this project is to understand the concept of surface area of a cone and its real-world applications through a hands-on, engaging and collaborative activity.
Detailed Description of the Project:
In this project, students will work in groups of 3 to 5 to design and construct a three-dimensional model of a cone using paper, and then calculate the surface area of the cone. Each student will play a specific role in the group: a 'Designer' who will sketch the model, a 'Builder' who will construct the model, and a 'Mathematician' who will calculate the surface area. The roles can be swapped to ensure all students get a chance to experience each role.
Necessary Materials:
Large sheets of paper
Pencils, erasers, rulers, compasses
Scissors, glue
Calculators
String (to measure the slant height)
Detailed Step-by-Step for Carrying Out the Activity:
Preparation and Planning (1 hour): The group will discuss and plan out their cone design, sketching it on paper. They will decide on the dimensions of the cone (the base radius and the height) and the slant height of the cone.
Construction (1 hour): Using the planned dimensions, the group will create a cone model with paper. They will cut out a sector of a circle and form a cone by joining the straight edges.
Measurement and Calculation (1 hour): The group will measure the slant height of the cone using a piece of string and a ruler. They will then calculate the surface area of the cone using the formulas A = πr² (for the base) and A = πrl (for the lateral surface), where r is the radius and l is the slant height.
Report Writing (2 hours): The group will write a detailed report of their project, including a description of their cone model, the calculations of the surface area, and a reflection on the project.
Project Deliverables:
Cone Model: The group will present their cone model, which should be accurately constructed with the planned dimensions.
Surface Area Calculation: The group will present their calculations for the surface area of the cone, showing their understanding of the formulas and how to apply them.
Written Report: The group will submit a written report, which should include the following sections:
Introduction: A brief overview of the project, its relevance, and the objective.
Development: Detailed information about the theory behind the surface area of a cone, a step-by-step explanation of the activity, the methodology used, and the results obtained.
Conclusion: A summary of the project, the learnings obtained, and the conclusions drawn about the surface area of a cone.
Bibliography: A list of the resources used to work on the project, including books, web pages, videos, etc.
Project Duration:
The project is expected to be completed in one week, with each student committing approximately 8-10 hours to the project. The first three days will be dedicated to planning, construction, and calculation. The remaining two days will be for report writing and preparation for the presentation.
Group Size:
The project should be completed in groups of 3 to 5 students. This will encourage teamwork, collaboration, and problem-solving skills. Each group should ensure that all members actively participate and contribute to each phase of the project.
Project Assessment:
The assessment will be based on the accuracy of the cone model, the correctness of the surface area calculations, the depth of understanding demonstrated in the written report, and the quality of the group's presentation. The students should be able to clearly articulate their understanding of the surface area of a cone, the process they followed to construct their model, and the results of their calculations. They should also reflect on their collaboration and problem-solving skills in the report.
In this way, the project will not only assess their understanding of the mathematical concepts but also their ability to work in a team, manage their time effectively, and communicate their ideas clearly and effectively.
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How Do You Find the nth Term? A Comprehensive Guide
06.01.2024
Finding the nth term of a sequence is a fundamental skill in mathematics, essential for solving problems in algebra, calculus, and beyond. Whether you're tackling homework, preparing for exams, or diving into advanced math, understanding how to find the nth term can simplify complex problems and reveal hidden patterns. In this post, we’ll break down the process step by step, explore different types of sequences, and provide real-world examples to illustrate the concepts.
What Is the nth Term?
The "nth term" of a sequence is a formula that lets you calculate any term in the sequence without listing all the previous ones. In other words, if you have a sequence like 3, 5, 7, 9, … you can find the 50th term directly using a formula. The value "n" represents the position of the term in the sequence.
Types of Sequences and Their nth Term Formulas
Different types of sequences use different formulas to determine the nth term. Let’s explore the most common ones.
1. Arithmetic Sequences
An arithmetic sequence is one in which the difference between consecutive terms is constant. This constant is known as the common difference (d).
General Formula:
an is the nth term.
a1 is the first term.
d is the common difference.
n is the term number.
Example:
Consider the sequence: 4, 7, 10, 13, …
The first term, a1, is 4.
The common difference, d, is 7−4=3.
Plug these into the formula:
So, the 10th term is:
2. Geometric Sequences
A geometric sequence multiplies each term by a constant ratio to get the next term. This constant is called the common ratio (r).
General Formula:
an is the nth term.
a1 is the first term.
r is the common ratio.
n is the term number.
Example:
Consider the sequence: 2, 6, 18, 54, …
The first term, a1, is 2.
The common ratio, r, is 6/2=3.
Using the formula:
Thus, the 5th term is:
3. Other Sequences
While arithmetic and geometric sequences are the most common, some sequences follow quadratic, cubic, or even more complex patterns. For example, a quadratic sequence might have a general form like:
Finding the coefficients aaa, bbb, and ccc usually involves solving a system of equations based on the first few terms of the sequence.
Step-by-Step Guide: How to Find the nth Term
No matter what type of sequence you’re dealing with, the following steps will help you determine the nth term:
Identify the First Term:
Determine the first term (a1) of the sequence. This is your starting point.
Determine the Pattern:
For Arithmetic Sequences: Calculate the common difference (d) by subtracting any term from the subsequent term.
For Geometric Sequences: Calculate the common ratio (r) by dividing any term by the previous term.
Choose the Appropriate Formula:
Use the arithmetic or geometric nth term formula based on the type of sequence.
Substitute and Simplify:
Plug your values into the formula and simplify to get your nth term formula.
Apply the Formula:
Once you have your formula, substitute the value of n to find any term in the sequence.
Practical Examples
Example 1: Arithmetic Sequence
Problem:
Find the 12th term of the sequence: 5, 8, 11, 14, …
Solution:
First term a1=5
Common difference d=8−5=3
Use the arithmetic formula:
For n=12:
Example 2: Geometric Sequence
Problem:
Find the 6th term of the sequence: 3, 6, 12, 24, …
Solution:
First term a1=3
Common ratio r=6/3=2
Use the geometric formula:
For n=6:
Why Finding the nth Term Matters
Understanding how to find the nth term is not just a classroom exercise. It has real-world applications such as:
Predicting Future Values: In finance, determining the future value of an investment can sometimes be modelled using arithmetic or geometric sequences.
Pattern Recognition: In computer science, recognizing patterns is crucial for algorithms and data structures.
Engineering Applications: Engineers often use sequences to model loads, stresses, or other regularly changing variables.
By mastering this technique, you can solve a wide range of problems more efficiently.
Tips for Success
Practice Regularly: The more you practice finding the nth term, the more intuitive it becomes.
Double-Check Your Calculations: Small errors in identifying a1, d, or r can lead to incorrect results.
Understand the Underlying Concepts: Rather than memorizing formulas, focus on understanding why they work. This deep comprehension will help you adapt to various types of sequences.
Conclusion
Finding the nth term of a sequence is a powerful tool in mathematics that opens up a world of problem-solving strategies. Whether you’re working with an arithmetic sequence by adding a common difference or a geometric sequence by multiplying by a common ratio, the process is systematic and logical. By following the step-by-step guide outlined in this post, you can quickly determine any term in a sequence and apply these skills to both academic and real-world problems.
Ready to put your new skills to the test? Start by identifying sequences around you and practice calculating their nth terms. Happy calculating!
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https://ocw.mit.edu/courses/res-18-012-algebra-ii-student-notes-spring-2022/mit18_702s22_lect13.pdf
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Lecture 13: More Factorization 13 More Factorization 13.1 Factoring Integer Polynomials We’ve seen that Z, F[x], and Z[i] are unique factorization domains — in fact, we’ll later look into a neat application of unique factorization in Z[i] to a problem in number theory. First we’ll discuss factorization in Z[x]. We previously stated the following theorem: Theorem 13.1 If R is a unique factorization domain, then R[x] is also a unique factorization domain.
We’ll prove this for the case R = Z. The proof of the general case is very similar to the proof for Z — we essentially just have to replace the familiar construction of the gcd over integers with the more abstract notion of gcd in a general UFD, as discussed last time.
Guiding Question Given a polynomial P ∈Z[x], there’s two natural questions we can ask about its factorization: 1. Is it possible to factor P where the factors lie in Q[x]?
2. What about in Z[x]?
There are more tools available for approaching the second question, factoring in Z[x], which make it possible to reduce the potential factorizations to a finite number of possibilities. One such tool is reducing mod a prime p.
Example 13.2 Consider the polynomial P(x) = 3x2 + 2x + 2. We can show it’s impossible to factor P in Z[x] by reducing mod 2 — we have P(x) ≡x3 (mod 2).
But the only way x3 factors in F2[x] is as x3 ·1 or x2 ·x. If P factored as P1P2 where P1 and P2 had positive degree, then since their leading coefficients must be odd (as these leading coefficients multiply to 3), they must still have positive degree in F2[x]. So P1 and P2 must be congruent to x and x2 mod 2; in particular, both their free terms are divisible by 2. But the free term of P would then be divisible by 4, which is a contradiction.
This example illustrates one possible trick that can be used to show that a polynomial is irreducible in Z[x].
There are various other tricks as well. For example, the product of the free terms of the factors must equal the free term of the original polynomial; so we can look at all possible factorizations of the free term. On the other hand, factoring polynomials in Q[x] seems much more difficult, since these tricks no longer work — there’s infinitely many ways to factor the free term of the original polynomial as a product of rationals, so this argument can’t be used to reduce our search to finitely many possibilities.
Fortunately, it turns out that factoring over Z[x] and Q[x] are actually equivalent. To show this, recall the definition of a primitive polynomial from last class, here applied specifically to Z[x]: Definition 13.3 A polynomial P ∈Z[x] is primitive if the gcd of all its coefficients is 1.
Evidently, any nonzero P ∈Z[x] can be written as a product P = nQ, where Q is primitive and n is the gcd of the coefficients of P. In fact, any P ∈Q[x] can be scaled to a primitive polynomial, by clearing denominators and factoring out the gcd of its coefficients.
The key point in relating factorization in Z[x] and Q[x] is Gauss’s Lemma, which we proved last class: Lemma 13.4 (Gauss’s Lemma) If P and Q are primitive, then PQ is as well.
63 Lecture 13: More Factorization Proof Sketch. If PQ is not primitive, then there is some prime p which divides all coefficients of PQ. Now consider P and Q mod p; since both are nonzero mod p, it’s clear that their product is nonzero mod p as well — the integers mod p are a field, and for polynomials over a field, it’s impossible to multiply two nonzero polynomials and get the zero polynomial.
Using Gauss’s Lemma, we can reduce questions about divisibility in Z[x] to ones about divisibility in Q[x], via the following corollary: Corollary 13.5 If P, Q ∈Z[x] are such that P divides Q in Q[x] and P is primitive, then P divides Q in Z[x].
Proof. We have Q = P · S for some S ∈Q[x]. Now write S = aT/b where T ∈ZZ[x] is primitive, and a and b are integers with b ̸= 0. Then the equation can be rewritten as bQ = aPT.
By Gauss’s Lemma, PT is primitive, so the gcd of all coefficients of aPT is exactly a. Meanwhile, b certainly divides all coefficients of bQ, so it divides all coefficients of aPT as well, which means b | a. As a result, a/b ∈Z, so S ∈Z[x] and P divides Q in Z[x] as well.
Note 13.6 There’s a different way to phrase this proof — for polynomials P ∈Z[x], we can define the content of P, denoted c(P), as the gcd of the coefficients of P. It’s possible to extend this to polynomials in Q[x] as well, such that for any T ∈Q[x] and a ∈Q, we have c(aT) = a · c(T). Then Gauss’s Lemma states that c(PQ) = c(P)c(Q) for any P, Q ∈Z[x], and therefore for any P, Q ∈Q[x] as well. Now in this proof we have Q = PS, which means c(Q) = c(P)c(S). But c(Q) is an integer and c(P) = 1, so c(S) must be an integer as well; therefore S has integer coefficients.
As a result, factoring in Z[x] and Q[x] are in fact equivalent.
Example 13.7 The polynomial 3x2 + 2x + 2 is irreducible in Z[x], so it cannot be factored in Q[x] either.
Corollary 13.8 The irreducible elements in Z[x] fall into two categories: ±p for prime integers p, and primitive polynomials which are irreducible in Q[x].
It’s not necessarily easy to tell whether a polynomial is irreducible; but this does mean that answering the question of whether a polynomial is irreducible in Q[x] is as easy as answering it in Z[x].
Proof. It’s clear that both categories of elements are irreducible — the primes are clearly irreducible in Z[x], since the only way to factor a constant polynomial is as a product of constants. Meanwhile, if a polynomial is irreducible in Q[x], then the only way it can be factored is by pulling out constant factors; but this is impossible for a primitive polynomial, so all such polynomials must be irreducible in Z[x] as well.
On the other hand, if P is not of either form, then we’ll show that P can be factored and therefore is not irreducible. First, if deg(P) = 0 (meaning P is an integer), then it’s clear that it must be prime in order to be irreducible.
Now assume that deg(P) ≥1. If P is not primitive, then we can pull out the greatest common divisor of its coefficients. Meanwhile, if P is primitive but factors in Q[x], then P = Q1Q2. We can rescale Q1 by taking integers a and b such that aQ1/b is in Z[x] and primitive. Then by Lemma 13.5, aQ1/b must divide P in Z[x] as well, giving a nontrivial factorization of P.
Theorem 13.1 for Z follows as a corollary.
64 Lecture 13: More Factorization Corollary 13.9 The polynomials with integer coefficients, Z[x], form a unique factorization domain.
Proof. As usual, we need to prove that a factorization exists and is unique.
We’ll first prove existence. First, by factoring out constants we can write P = p1 · · · pℓP1 such that P1 is primitive and the pi are primes. If P1 is irreducible in Q[x], we are done, as it is also irreducible in Z[x].
Otherwise, P1 factors in Q[x], and we can rescale both factors so that they are primitive elements of ZZ[x], so then P1 factors in Z[x]. We can continue to attempt to factor the two resulting factors of P1. As we keep on factoring, the degrees of our polynomials decrease at every step, so the factorization process must terminate — which means that eventually, all our polynomials become irreducible.
Now we’ll prove uniqueness. As in all the other cases where we proved uniqueness, it is enough to show that if an irreducible polynomial P ∈Z[x] divides Q1Q2, then P divides either Q1 or Q2. (Then similarly to in the proof that every PID is a UFD, given two factorizations P1 · · · Pn and Q1 · · · Qm, we can show that P1 must appear in the second factorization as well, cancel it out from both, and repeat with the remaining factorizations until we’ve matched up all the factors.) In order to show this result, we have two cases. First, if P is an integer prime p ∈Z, then this follows directly from the fact that the product of two nonzero polynomials in (Z/pZ)[x] is nonzero, as Z/pZ is a field.
Otherwise, P is primitive and irreducible in Q[x]. We have that if P | Q1Q2, then P | Q1 or P | Q2 in Q[x] (since Q is a field, so Q[x] is a PID and therefore a UFD). By Lemma 13.5, since P is primitive, then P must divide Q1 or Q2 in Z[x].
So this shows that for any irreducible P ∈Z[x], if P | Q1Q2 then P | Q1 or P | Q2, as desired.
As mentioned earlier, the same proof used to show that Z[x] is a UFD would work if we replaced Z with any UFD R. For example, we can even take R to be Z[x], now that we know it’s a UFD; this shows that Z[x, y] is also a UFD.
13.2 Gaussian Primes Unique factorization in Z[i] has an interesting application — it can be used to solve a problem in number theory.
Guiding Question Which integers can be written as n = a2 + b2 for integers a and b?
Example 13.10 We can write 5 = 22 + 12, while 6 and 21 cannot be written as a sum of squares.
On the way to proving the answer, we’ll classify irreducible elements in Z[i]. (Since Z[i] is a UFD, the irreducible elements are exactly the primes, so we will use “prime” and “irreducible” interchangeably here.) This is an example of how the abstract property of unique factorization can lead to concrete results.
First, note that n = a2 + b2 can be rewritten as n = (a + bi)(a −bi). This makes it clear that if n and m can be written in the form a2 + b2, then so can mn — if n = αα and m = ββ, then (αβ)(αβ). So the property is multiplicative, which motivates considering the special case where n is prime.
Lemma 13.11 Let p ∈Z be a prime number. Then p = a2 + b2 if and only if p is not a prime in Z[i].
We refer to primes in Z[i] as Gaussian primes.
Proof. First, if p were a Gaussian prime and we could write p as a sum of squares, then we would have p = (a + bi)(a −bi), 65 Lecture 13: More Factorization which would mean p must divide either a + bi or a −bi. In either case, p would need to divide both a and b, which is impossible.
Meanwhile, if p is not a Gaussian prime, since it’s real and doesn’t factor in Z, it must factor as p = αα for some α ∈Z[i] which is not in Z. So then α = a + bi for some integers a and b, which means p = a2 + b2.
So answering our initial question for primes is equivalent to figuring out which integer primes are also Gaussian primes.
Lemma 13.12 Let p ∈Z be a prime number. Then p is not prime in Z[i] if and only if p = 2 or p ≡1 (mod 4).
Proof. First, 2 factors as 2 = (1 + i)(1 −i). Now suppose p > 2.
Claim. p is not a prime in Z[i] if and only if there exists α ∈Z[i] such that p ∤α, but p | αα.
Proof. By definition, p is not a prime in Z[i] if and only if there exist α and β such that p divides αβ, but not α or β. It immediately follows that if there exists an α ∈Z[i] with the described properties, then p is not prime.
On the other hand, if p is not prime, then take α and β such that neither is divisible by p but αβ is; then p | αβαβ = (αα)(ββ).
Since p is an integer prime, and both αα and ββ are integers, then p must divide one of them, and either α or β has the described properties.
This turns the question into one over Fp — then p is not a prime in Z[i] if and only if there exist a, b ∈Fp, which are not both 0, such that a2 + b2 = 0.
Since Fp is a field, we can divide by b2 and rewrite the equation as −1 = (ab−1)2, so this is true if and only if −1 is a square in Fp.
Now consider the abelian group F× p (the multiplicative group of Fp) which has order p −1. The only element of order 2 is −1, since x2 −1 = (x −1)(x + 1) has no roots other than ±1, and 1 has order 1. This gives a homomorphism φ : F× p →F× p sending α →α2.
Then ker(φ) = {±1}, so by the homomorphism theorem, im(φ) has (p −1)/2 elements.
But −1 is a square in Fp if and only if it is in the image of φ. Since im(φ) is a subgroup of F× p , this occurs if and only if im(φ) contains an element of order 2 (since the only possible element of order 2 is −1).
But im(φ) contains an element of order 2 if and only if |im(φ)| = (p −1)/2 is divisible by 2 — one direction follows from the fact that the order of every element divides the order of the group, and the other follows from the Sylow Theorems. So −1 is a square if and only if (p −1)/2 is even, or equivalently p ≡1 (mod 4).
So an odd integer prime p is not a Gaussian prime if and only if p ≡1 (mod 4).
This proof can be used to classify the Gaussian primes up to association (multiplying by units, here ±1 and ±i).
Theorem 13.13 The full list of primes in Z[i], up to association, can be constructed as follows: consider all integer primes p.
• If p ≡3 (mod 4), then p itself is a Gaussian prime.
• If p ≡1 (mod 4), then it factors as (a −bi)(a + bi), and both factors a ± bi are Gaussian primes.
• If p = 2, then it factors as (1 + i)(1 −i), and since 1 + i and 1 −i are associate, they correspond to the same Gaussian prime.
66 MIT OpenCourseWare Resource: Algebra II Student Notes Spring 2022 Instructor: Roman Bezrukavnikov Notes taken by Sanjana Das and Jakin Ng For information about citing these materials or our Terms of Use, visit:
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FINRA Rules
11000. UNIFORM PRACTICE CODE
11800. CLOSE-OUT PROCEDURES
11890. Clearly Erroneous Transactions
11892. Clearly Erroneous Transactions in Exchange-Listed Securities
The Rule Notices Guidance News Releases FAQs
(a) Procedures for Reviewing Transactions
(1) An Executive Vice President of FINRA's Market Regulation Department or Transparency Services Department, or any officer designated by such Executive Vice President (FINRA officer), may, on his or her own motion, review any over-the-counter transaction involving an exchange-listed security occurring outside of Normal Market Hours (9:30 a.m. Eastern Time to 4:00 p.m. Eastern Time) or eligible for review pursuant to paragraph (b)(1) of this Rule arising out of or reported through a trade reporting system owned or operated by FINRA or FINRA Regulation and authorized by the Commission, provided that the transaction meets the guidelines set forth in this Rule. A FINRA officer acting pursuant to this subparagraph may declare any such transaction null and void if the officer determines that (A) the transaction is clearly erroneous, or (B) such actions are necessary for the maintenance of a fair and orderly market or the protection of investors and the public interest, consistent with the guidelines set forth in this Rule. Absent extraordinary circumstances, the officer shall take action pursuant to this paragraph generally within 30 minutes after becoming aware of the transaction. When extraordinary circumstances exist, any such action of the officer must be taken no later than the start of trading on the day following the date of execution(s) under review.
(2) If a FINRA officer acting pursuant to this Rule declares any transaction null and void, each party involved in the transaction shall be notified as soon as practicable by FINRA, and the party aggrieved by the action may appeal such action in accordance with Rule 11894, unless the decision is made by a FINRA officer under Supplementary Material .02 of this Rule regarding transactions that occurred outside of the applicable Price Bands disseminated pursuant to the LULD Plan, and further provided that rulings made by FINRA in conjunction with one or more other self-regulatory organizations are not appealable.
(b) Clearly Erroneous Review
(1) Review of Transactions Occurring During Normal Market Hours
If the execution time of the transaction(s) under review is during Normal Market Hours, the transaction will not be reviewable as clearly erroneous unless the transaction:
(A) is in an NMS Stock that is not subject to the Plan to Address Extraordinary Market Volatility Pursuant to Rule 608 of Regulation NMS under the Act (the “Limit Up-Limit Down Plan” or “LULD Plan”). In such case, the Numerical Guidelines set forth in paragraph (b)(2) of this Rule will be applicable to such NMS Stock;
(B) was executed at a time when Price Bands under the LULD Plan were not available, or is the result of a member’s technology or systems issue that results in the transaction occurring outside of the applicable LULD Price Bands pursuant to Supplementary Material .02 of this Rule, or is executed after the primary listing market for the security declares a regulatory trading halt, suspension, or pause pursuant to paragraph (e) of this Rule. A transaction subject to review pursuant to this paragraph shall be found to be clearly erroneous if the price of the subject transaction to buy (sell) is greater than (less than) the Reference Price, described in paragraph (c) of this Rule, by an amount that equals or exceeds the applicable Percentage Parameter defined in Appendix A to the LULD Plan (“Percentage Parameters”); or
(C) involved, in the case of (i) a corporate action or new issue or (ii) a security that enters a Trading Pause pursuant to the LULD Plan and resumes trading without an auction, a Reference Price that is determined to be erroneous by a FINRA officer because it clearly deviated from the theoretical value of the security. In such circumstances, FINRA may use a different Reference Price pursuant to paragraph (c)(2) of this Rule. A transaction subject to review pursuant to this paragraph shall be found to be clearly erroneous if the price of the subject transaction to buy (sell) is greater than (less than) the new Reference Price, described in paragraph (c)(2) of this Rule, by an amount that equals or exceeds the Numerical Guidelines or Percentage Parameters, as applicable depending on whether the security is subject to the LULD Plan.
(2) Review of Transactions Occurring Outside of Normal Market Hours or Eligible for Review Pursuant to Paragraph (b)(1)(A)
(A) Subject to the additional factors described in paragraph (b)(2)(C) of this Rule, a transaction executed outside of Normal Market Hours, or eligible for review pursuant to paragraph (b)(1)(A) of this Rule, shall be found to be clearly erroneous if the price of the transaction is away from the Reference Price by an amount that equals or exceeds the Numerical Guidelines set forth below.
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| Reference Price: Circumstance or Product | Normal Market Hours Numerical Guidelines for Transactions Eligible for Review Pursuant to Paragraph (b)(1)(A) (Subject transaction's % difference from the Reference Price): | Outside Normal Market Hours Numerical Guidelines (Subject transaction's % difference from the Reference Price): |
| Greater than $0.00 up to and including $25.00 | 10% | 20% |
| Greater than $25.00 up to and including $50.00 | 5% | 10% |
| Greater than $50.00 | 3% | 6% |
| Multi-Stock Event — Events involving five or more, but less than twenty, securities whose executions occurred within a period of five minutes or less | 10% | 10% |
| Multi-Stock Event — Events involving twenty or more securities whose executions occurred within a period of five minutes or less | 30%, subject to the terms of paragraph (b)(2)(B) of this Rule | 30%, subject to the terms of paragraph (b)(2)(B) of this Rule |
| Leveraged ETF/ETN securities | N/A | Normal Market Hours Numerical Guidelines multiplied by the leverage multiplier (i.e. 2x) |
(B) Multi-Stock Events Involving Twenty or More Securities
Multi-Stock Events involving twenty or more securities may be reviewable as clearly erroneous if they occur outside of Normal Market Hours or are eligible for review pursuant to paragraph (b)(1)(A) of this Rule. During Multi-Stock Events, the number of affected transactions may be such that immediate finality is necessary to maintain a fair and orderly market and to protect investors and the public interest. In such circumstances, FINRA may use a Reference Price other than the consolidated last sale. To ensure consistent application across the markets when this paragraph is invoked, FINRA will promptly coordinate with other self-regulatory organizations to determine the appropriate review period, which may be greater than the period of five minutes or less that triggered application of this paragraph, as well as select one or more specific points in time prior to the transactions in question and use transaction prices at or immediately prior to the one or more specific points in time selected as the Reference Price. FINRA will nullify as clearly erroneous all transactions that are at prices equal to or greater than 30% away from the Reference Price in each affected security during the review period selected by FINRA and the other self-regulatory organizations consistent with this paragraph.
(C) Additional Factors
Except in the context of a Multi-Stock Event involving five or more securities, a FINRA officer may also consider additional factors to determine whether a transaction is clearly erroneous, provided the execution occurred outside of Normal Market Hours or is eligible for review pursuant to paragraph (b)(1)(A) of this Rule. Such additional factors include but are not limited to: system malfunctions or disruptions; volume and volatility for the security; derivative securities products that correspond to greater than 100% in the direction of a tracking index; news released for the security; whether trading in the security was recently halted/resumed; whether the security is an IPO; whether the security was subject to a stock-split, reorganization, or other corporate action; overall market conditions; Opening and Late Session executions; validity of the consolidated tapes' trades and quotes; consideration of primary market indications; and executions inconsistent with the trading pattern in the stock. Each additional factor shall be considered with a view toward maintaining a fair and orderly market and the protection of investors and the public interest.
(c) Reference Price
The Reference Price referred to in paragraphs (b)(1) and (b)(2) of this Rule will be equal to the consolidated last sale immediately prior to the execution(s) under review except for:
(1) in the case of Multi-Stock Events involving twenty or more securities, as described in paragraph (b)(2)(B) of this Rule;
(2) in the case of an erroneous Reference Price, as described in paragraph (b)(1)(C) of this Rule. In the case of paragraph (b)(1)(C)(i), FINRA would consider a number of factors to determine a new Reference Price that is based on the theoretical value of the security, including but not limited to, the offering price of the new issue, the ratio of the stock split applied to the prior day’s closing price, the theoretical price derived from the numerical terms of the corporate action transaction such as the exchange ratio and spin-off terms, and for an OTC up-listing, the price of the security as provided in the prior day’s FINRA Trade Dissemination Service final closing report. In the case of paragraph (b)(1)(C)(ii), the Reference Price will be the last effective Price Band that was in a limit state before the Trading Pause; or
(3) in other circumstances, such as, for example, relevant news impacting a security or securities, periods of extreme market volatility, sustained illiquidity, or widespread system issues, where use of a different Reference Price is necessary for the maintenance of a fair and orderly market and the protection of investors and the public interest, provided that such circumstances occurred outside of Normal Market Hours, or are eligible for review pursuant to paragraph (b)(1)(A) of this Rule.
(d) Multi-day Events
A series of transactions in a particular security on one or more trading days may be viewed as one event if all such transactions were effected based on the same fundamentally incorrect or grossly misinterpreted issuance information resulting in a severe valuation error for all such transactions (the "Event"). A FINRA officer, acting on his or her own motion, shall take action to declare all transactions that occurred during the Event null and void not later than the start of trading on the day following the last transaction in the Event. If trading in the security is halted before the valuation error is corrected, a FINRA officer shall take action to declare all transactions that occurred during the Event null and void prior to the resumption of trading. Notwithstanding the foregoing, no action can be taken pursuant to this paragraph with respect to any transactions that have reached settlement date or that result from an initial public offering of a security. To the extent transactions related to an Event occur on one or more other self-regulatory organization, FINRA will promptly coordinate with such other self-regulatory organization(s) to ensure consistent treatment of the transactions related to the Event, if practicable. Any action taken in connection with this paragraph will be taken without regard to the Percentage Parameters or Numerical Guidelines set forth in this Rule. FINRA will notify each member involved in a transaction subject to this paragraph as soon as practicable of a determination to declare such transaction null and void, and the party aggrieved by the action may appeal such action in accordance with Rule 11894.
(e) Transactions Occurring During Trading Halts
In the event of any disruption or malfunction in the operation of the electronic communications and trading facilities of a self-regulatory organization or responsible single plan processor in connection with the transmittal or receipt of a regulatory trading halt, suspension or pause, a FINRA officer, acting on his or her own motion, shall declare as null and void any transaction in a security that occurs after the primary listing market for such security declares a regulatory trading halt, suspension or pause with respect to such security and before such regulatory trading halt, suspension or pause with respect to such security has officially ended according to the primary listing market. In addition, in the event a regulatory trading halt, suspension or pause is declared, then prematurely lifted in error and is then re-instituted, a FINRA officer also shall declare as null and void transactions that occur before the official, final end of the regulatory halt, suspension or pause according to the primary listing market. Any action taken in connection with this paragraph shall be taken in a timely fashion, generally within thirty (30) minutes of the detection of the erroneous transaction and in no circumstances later than the start of normal market hours on the trading day following the date of the execution(s) under review. Any action taken in connection with this paragraph will be taken without regard to the Percentage Parameters or Numerical Guidelines set forth in this Rule. FINRA will notify each member involved in a transaction subject to this paragraph as soon as practicable of a determination to declare such transaction(s) null and void, and the party aggrieved by the action may appeal such action in accordance with Rule 11894.
• • • Supplementary Material: ------------------
.01 Determinations by a National Securities Exchange to Nullify and Void the Terms of One or More Transactions in an Exchange-Listed Security When There Are Corresponding or Related Transactions Reported Through a FINRA System.
FINRA believes that coordinating with other self-regulatory organizations with the goal of having consistency and transparency regarding the clearly erroneous process is important to the marketplace and to investors. Consequently, for OTC transactions in exchange-listed securities that are reported to a FINRA system, such as a FINRA Trade Reporting Facility (“TRF”) or Alternative Display Facility (“ADF”), FINRA will generally follow the determination of a national securities exchange to break a trade(s) when that national securities exchange has broken a trade(s) at or near the price range in question at or near the time in question (in FINRA staff's sole discretion) such that FINRA breaking such trade(s) would be consistent with market integrity and investor protection. In such a case where multiple national securities exchanges have related trades, FINRA will leave a trade(s) unbroken when any of those national securities exchanges has left a trade(s) unbroken at or near the price range in question at or near the time in question (in FINRA staff's sole discretion) such that FINRA breaking such trade(s) would be inconsistent with market integrity and investor protection.
.02 Transactions Occurring Outside of LULD Plan Price Bands
(a) If as a result of a member's technology or systems issue any transaction reported to a FINRA system, such as a FINRA TRF or ADF, occurs outside of the applicable Price Bands disseminated pursuant to the LULD Plan, a FINRA officer, acting on his or her own motion or at the request of a member, shall review and deem such transaction clearly erroneous, subject to the certification requirement of paragraph (b) of this Supplementary Material. Absent extraordinary circumstances, any such action of the FINRA officer shall be taken in a timely fashion, generally within thirty (30) minutes of the detection of the erroneous transaction. When extraordinary circumstances exist, any such action of the FINRA officer must be taken by no later than the start of normal market hours on the trading day following the date on which the execution(s) under review occurred. Each member involved in the transaction shall be notified as soon as practicable by FINRA, and a member aggrieved by the action may appeal such action in accordance with Rule 11894. In the event that a single plan processor experiences a technology or systems issue that prevents the dissemination of Price Bands, FINRA will make the determination of whether to deem transactions clearly erroneous based on Rule 11892(b)(1)(B).
(b) A member requesting review of a transaction pursuant to paragraph (a) of this Supplementary Material must certify, in the manner and form prescribed by FINRA, that the subject transaction(s) occurring outside of the applicable price bands disseminated pursuant to the Plan is the result of the member's bona fide technological or systems issue.
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| Amended by SR-FINRA-2022-027 eff. Oct. 1, 2022. Amended by SR-FINRA-2022-020 eff. July 19, 2022. Amended by SR-FINRA-2022-008 eff. April 6, 2022. Amended by SR-FINRA-2021-026 eff. Oct. 5, 2021. Amended by SR-FINRA-2021-004 eff. April 14, 2021. Amended by SR-FINRA-2020-036 eff. Oct. 16, 2020. Amended by SR-FINRA-2020-008 eff. April 17, 2020. Amended by SR-FINRA-2019-025 eff. Oct. 10, 2019. Amended by SR-FINRA-2019-011 eff. April 9, 2019. Amended by SR-FINRA-2015-034 eff. Dec. 20, 2015. Amended by SR-FINRA-2014-021 eff. June 19, 2014. Amended by SR-FINRA-2014-013 eff. Mar. 19, 2014. Amended by SR-FINRA-2013-041 eff. Sept. 24, 2013. Amended by SR-FINRA-2013-012 eff. Jan. 30, 2013. Amended by SR-FINRA-2012-038 eff. July 23, 2012. Amended by SR-FINRA-2012-005 eff. Jan. 24, 2012. Amended by SR-FINRA-2011-039 eff. Aug. 10, 2011. Amended by SR-FINRA-2011-037 eff. Aug. 5, 2011. Amended by SR-FINRA-2011-014 eff. Mar. 30, 2011. Amended by SR-FINRA-2010-065 eff. Dec. 8, 2010. Amended by SR-FINRA-2010-032 eff. Sept. 10, 2010. Amended by SR-FINRA-2009-068 eff. Feb. 15, 2010. Amended by SR-FINRA-2008-037 eff. July 8, 2008. Amended by SR-NASD-2006-104 eff. March 5, 2007. Amended by SR-NASD-2006-121 eff. Oct. 30, 2006. Amended by SR-NASD-2005-087 eff. Aug. 1, 2006 Amended by SR-NASD-2006-033 eff. Mar. 1, 2006. Amended by SR-NASD-2005-089 eff. Oct. 1, 2005. Amended by SR-NASD-2005-115 eff. Sep. 22, 2005. Amended by SR-NASD-2004-009 eff. July 27, 2005. Amended by SR-NASD-2003-125 eff. Aug. 8, 2003. Amended by SR-NASD-2002-127 eff. Jan. 22, 2003. Amended by SR-NASD-98-85 eff. Oct. 11, 1999. Amended by SR-NASD-98-94 eff. April 26, 1999. Amended by SR-NASD-96-51 eff. Feb. 23, 1998. Amended June 21, 1991; May 21, 1993. Adopted eff. Apr. 2, 1990. Selected Notices: 98-21, 99-29, 00-10, 03-11, 10-04, 10-43, 16-04. |
Regulatory Notice 16-19
FINRA Issues Guidance Regarding the Use of Stop Orders During Volatile Market Conditions
May 26, 2016
Regulatory Notice 13-12
FINRA Adopts Amendments Relating to Regulation NMS Plan to Address Extraordinary Market Volatility
March 06, 2013
Regulatory Notice 10-43
Amendments to FINRA Rules on Trading Pauses Due to Extraordinary Market Volatility and Clearly Erroneous Transactions in Exchange-Listed Securities
September 22, 2010
Regulatory Notice 10-04
SEC Approves Consolidated FINRA Rules Governing Clearly Erroneous Transactions
January 15, 2010
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6.1200J | Spring 2024 | Undergraduate
Mathematics for Computer Science
Lecture 17: More Counting Techniques
We continue the counting techniques by looking at the pigeonhole principle to solve counting problems.
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Prof. Erik Demaine
Dr. Zachary Abel
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Mathematics
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Computer Science
Algorithms and Data Structures
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Computation
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Population policy of Bangladesh
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The presentation provides an overview of population policies in Bangladesh. It discusses the country's high population growth rate and need for policies to manage resources. The objectives of policies are to lower fertility rates, reduce mortality, and achieve targets like maintaining the natural replacement rate of 1. Major strategies include advocacy campaigns, promoting small families, and ensuring access to family planning services. The roles of different ministries and organizations in implementing population programs are also outlined.
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In this document
Powered by AI
Slides 1 - 2 Introduction to Population Policy Presentation
Welcome audience to the presentation about the Population Policy of Bangladesh, introducing the group.View
Slides 3 - 4 Scope of Presentation
Presentation discusses the methodology, aims, objectives, and outline related to population policies in Bangladesh.View
Slide 5 Methodology and Data Sources
Data sourced from books and the internet, presented visually through pictures and graphs.View
Slide 6 Aim and Objectives
Focuses on understanding population issues, policy formulation, and implementation strategies.View
Slide 7 Historical Background
First population policy in Bangladesh post-independence in 1976 aimed at population control and health reforms.View
Slides 8 - 10 Defining Policy and Population Policy
Explains what policy is, the basis for policy, specifically focusing on what constitutes population policy.View
Slides 11 - 13 Components & Process of Population Policy
Discusses common elements of population policy and the process involved in policy-making.View
Slide 14 Demographic Information of Bangladesh
Presents demographic data of Bangladesh including population size, growth, birth and death rates from 2016.View
Slides 15 - 16 Need for Population Policy
Rationale behind needing a population policy due to resource limitations and socio-economic challenges.View
Slides 17 - 18 Population Policy Goals in Bangladesh
Describes the socio-economic development focus of Bangladesh's population policy and the need for continual updates.View
Slides 19 - 20 Vision and Objectives of Policy
Vision for population policy aiming for a healthier, wealthier Bangladesh along with specific objectives.View
Slide 21 Strategies for Implementation
Major strategies proposed to achieve population policy objectives, including advocacy, education, and service provision.View
Slides 22 - 28 Evolution of Population Policies
Chronological overview of various population policies in Bangladesh from 1973 to 2015 and their impacts.View
Slides 29 - 31 Role of Various Sectors in Population Policy
Explains the involvement of different ministries and NGOs in the implementation and development of population policies.View
Slide 33 Obstacles to Population Policy
Identifies social and political challenges that hinder the effective implementation of population policies.View
Slide 34 Conclusion and Future Directions
Encourages focus on the importance of population policy for national development and overcoming existing challenges.View
Slide 35 References and Acknowledgments
Bibliography listing used references and sources for the presentation.View
Slides 36 - 38 Closing Thank You
Thanking the audience for their attention and inviting questions regarding the presentation.View
Population policy of Bangladesh
Welcome To Our Presentation
Group No :( 11 ) Id No Name 1311003 Jannatul Ferdousi 1311034 Mst. Nasrin Nahar 1311044 Md. Omar Faruk 1311046 Sohanur Rakib 1311053 Rumi Aktar
Our Presentation Topic Population Policy Of Bangladesh
Outline Methodology Aim and Objective Introduction Whatis policy? Basis for policy What is population policy? Common Elements of Population Policy Process of population policy making Types of population policies Why we need population policy in Bangladesh Population Policy of Bangladesh Vision Objectives Major Strategies for achieve the vision Population Policies of Bangladesh (Different Years) Population Policy in Fourth five years (1990-1995) plan Population Policy in five years (1995-2000) plan Role of Different Ministries, Non-Government Organizations and Private Sector in Population Program Conclusion
Methodology Source of data Books Internet Representationof data By Picture By Graph
Aim and Objectiveof this presentation Know about population and development issues in Bangladesh Understand how population policies are formulated To know about the strategies for implementation of those policies
Introduction Population Policy inBangladesh was first articulated after the war of independence, when the economic imperative facing the government of containing population growth was paramount. Besides Bangladesh, faced many significant population and development problems. The first population policy of Bangladesh was formulated in 1976 when the rate of population growth was approximately 3% per year. In the population policy outline, population control and family planning activities were considered integral elements of social reform and national development with a view to reducing family size for ensuring sound maternal and child health, family welfare and higher standard of living.
What is policy? Setof ideas or plans that is used as a basis for decision making. Attitude an actions of an organization regarding a particular issue. General statement of understanding which guide decision making.
Basis for policy Setof values Commitments Assessment of current situation Image of a desired future situation
What is populationpolicy? Population policy may be defined as deliberately constructed or modified institutional arrangements and/or specific programs through which governments influence, directly or indirectly, demographic change. Measures formulated by a range of social institution including government which may influence the size, distribution or composition of human population (Driver, 1972) A set of coordinated laws aimed at reaching some demographic goals (Bourgeois – Piechat, 1994)
Common Elements ofPopulation Policy Rationale : demographic analysis. Objectives and goals. Targets : time bound level of fertility. Program Measures.
Process of populationpolicy making o Developing the constituency in favor of population policy. o Identifying the arguments favoring population policy. o Addressing the issue to a right place. o Visualizing the form a policy should take. o Recognizing the most advantageous time.
Types of populationpolicies Explicit Document by a national government announcing its intention to affect the population growth and composition. Implicit Directives not necessarily issued to influenced the population growth and composition but may have the effect of doing so. Example: Migration laws Example: Women education
Demography of Bangladesh Population162,951,560 (2016 est.) Growth rate 1.6% (2015 est.) Birth rate 21.14 births/1,000 population (2015 est.) Death rate 5.61 deaths/1,000 population (2015 est.) Life expectancy 70.94 years (2015 est.) • male 69.02 years • female 72.94 years Fertility rate 2.4 children born/woman (2015 est.) Infant mortality rate 44.09 deaths/1,000 live births (2015 est.) World population prospect: Revision, 2017
Why we needpopulation policy in Bangladesh
Bangladesh is aover populated country with limited resources. For the well distribution and the development of those resources we need population policy. So we need a policy (long and short time) for reduced problems and make a developed country. Besides, population policy also need for…… Standard Population growth Poverty Elevation Gender equality Good health & well being Quality education
Population Policy ofBangladesh Socio -economic development for all citizen is a cornerstone of Bangladesh constitution. The state has the responsibility to ensure to it’s citizens certain basis need such as ,
In order totranslate the constitutional aim to reality government need to peruse several population policies. Bangladesh follow both explicit and implicit types of population policy. It is essential to update the population policies and strategies in order to keep the population of the country within tolerable limit.
Vision Develop a healthier Happier Wealthier Bangladesh
Objectives of PopulationPolicy of Bangladesh Lower the total fertility rate (TFR) Increase family planning user Achieve NRR=1 Ensure safe motherhood Reduce maternal and infant mortality Improve mother health care Ensure the availability of family planning method. Ensure gender equity and women`s empowerment plans for developing the population into human resources with concerned Ministries Ensure easy access to information on reproductive health
Major Strategies forachieve the vision 1. Develop and launch advocacy campaigns to address special groups, such as, policy makers, opinion leaders. 2. Promote the small family norm through innovative schemes. 3. Reduce unmet need for family planning. 4. Ensure the provision of quality services especially to the poor to the rural areas and urban slums. 5.Coordinate and monitor a comprehensive network of family planning and reproductive health services. 6. Build strong partnership with concerned line Ministries, Non- Governmental Organizations and the private sector. 7. Ensure population and family life education for school and college students.
Population Policies ofBangladesh (Different Years) Bangladesh is an over populated country with small land area. So a policy is necessary for the development of our country The Government betake a policy each five year on the basis of previous population policies result. Planning Year Major steps Aim Population Policy in First two-years (1973- 1978) plan and five years ( 1980 -1985) plan Children and mother health Nutrient Family planning Training Below Population 10 Crore Population Policy in third five years plan (1985-1990) Children and Mother health Increase conscious of family planning Population growth rate decrease from 2.5% to 1.8%
Population Policy infive years(1997-2002) plan Poverty reduction strategy Decrease the discrimination of education between male and female Insure pure drinking water Elevation poverty Take strategy for achieving MDG Population Policy in sixth five years(2011-2015) plan Client-oriented service Youth friendly services Human resources development Environment friendly planning NRR=1by 2010 Stable population in 2060 Source: জনসংখ্যা ভূ গ াল । Population Policy in forth five years(1990-1995) plan Cannot achieve the aim of Population Policy in third five years plan (1980- 1985). Growth rate increase 2.11% from 2.5%. So taken some immediate action for reduce the NRR rate. Decrease the population growth rate 1.8% CBR 30.1%from 34.5% CDR11.9% from 13.4%
Population Policy inFourth five years(1990-1995) plan Major steps Reduce gender discrimination Compulsory women primary education Development of human resources Aim Birth rate=20.8% Death rate=7.7% Growth rate=1.32%
Bangladesh Census report,2001 0 20000000 40000000 60000000 80000000 100000000 120000000 140000000 1974 1981 1991 2001 76398000 89912000 111455185 130029749 2.5 2.33 2.15 1.54 Population Growth rate (1974-2001) Total population Growth rate (%)
Population Policy infive years(1995-2000) plan In 2000 UN declared the Millennium Development Goal (MDG) for develop and under developing country Bangladesh is a singing member of this program for implementation of MDG. The aim of MDG and population policy of Bangladesh are same Sustainable development. So the government publish population policy under the title of complete poverty reduction strategy. Identify the poverty on two different basis.
Identify of poverty level 1.Bytaken calorie (each person) 2.Land ownership National Urban Rural 0 10 20 30 40 50 1999 2004 46.2 40.9 49.9 43.6 45.6 40.1 1. By taken calorie each person National Urban Rural 45.53 38.88 21.53 9.01 37.9 0 10 20 30 40 50 60 70 Landless Small Medium Big Total 2. Land ownershipLand ownership National Urban Rural Bangladesh Census bureau, poverty survey, 2004 Bangladesh Census bureau, poverty survey, 2004
Population in sixthfive years (2011-2015) In 2015 the time limitation of MDG is over. So the government publish 6th five years plan with some aim and strategy for implementation and achieve NRR=1 and stable population by 2060. Strategies for implementation ( 2011-2015 ) Urban and Rural Health Care Empowerment of Women and Equal Partnership of Men and Women Adolescent Welfare Program Human Resources Developmen
Role of DifferentMinistries, Non-Government Organizations and Private Sector in Population Program Role of Different Ministries in Population Program Role of Non-Government and Private Organizations in Population Program Role of Different Ministries, Non-Government Organizations and Private Sector in Population Program The ministries and institutions as partners in population planning and development programs whose target groups are heavily affected by growth of population. In this regard the concerned ministries and institutions can play fruitful roles within the scope of their own activities.
Name of MinistryMajor Role Health and Family Welfare family planning maternal and child health and reproductive health services program. Public Administration take necessary initiatives to implement family planning and maternal and child health activities at the field level through division, district and upazilla administrations. Finance allocate necessary funds. Education include updated issues relating to over-population and its serious impact, health, education and life skills education in different text books and other learning materials. Primary and Mass Education equality between male and female population in education sector.
Name of MinistryMajor Role Information broadcasting information on health education, family planning, maternal and child health, reproductive health, equality of men and women, sexually transmitted diseases and HIV/AIDS through different public and private radio and television channels, and other different media. Women and Children Affairs Arrangement of loans for trained women, institutional training, and the rights and responsibilities of women. Social Welfare play a role in encouraging the communities to receive family planning services. Planning assist the government in its policy making and planning with serious deliberations on demography, population projection and development. Agriculture Use of modern agricultural technologies and to promote two-child families. Discouraging urban migration.
•Production and distributionof family planning commodities by non- government and private sectors •Employment-generating industries and other institutions for playing complementary roles in •Import, distribution and marketing of family planning commodities Role of Non-Government and Private Organizations in Population Program
Social Problem Poverty Religion Illiteracy Political problem unstable political condition Delaye the publishcation of policies Obstacleof population policy
Conclusion Population isa resource of any country unless its more than country`s natural resource and area. Population policy help to develop and solved many problems . Our country take many policies and try to control over population .There are some lacking ,.so the Government should more emphasis on the policies to achieve the goals.
References রউফ আ. ক.(২০০৭),জনসংখ্যাভূ গ াল, সুজগনষু প্রকাশনী. পৃ:২১১-২১৫। https:// en.wikipedia,org/Demographics of Bangladesh.com https:// en.wikipedia,org/ Population policy of Bangladesh,2012 3174 6a22_df81_4a82_b70c_24125dec56d7/Bangladesh-Population-Policy-2012.pdf
Thank You AllFor Stay With Us
HAVE YOU ANYQUESTIONS ?
Presented by Sohanur Rakib Session:2013-2014 Dept of Geograpgy and Environmental Science Email: sohanurrakib1995@gmail.com Begum Rokeya University, Rangpur Bangladesh……
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Geometric sequence
In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence.
For example, is a geometric sequence with common ratio and is a geometric sequence with common ratio ; however, and are not geometric sequences, as the ratio between consecutive terms varies.
More formally, the sequence is a geometric progression if and only if . A similar definition holds for infinite geometric sequences. It appears most frequently in its three-term form: namely, that constants , , and are in geometric progression if and only if .
Contents
[hide]
1 Properties
2 Sum
2.1 Finite
2.2 Infinite
3 Problems
3.1 Introductory
3.2 Intermediate
4 See also
Properties
Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio. Let be the first term, be the th term, and be the common ratio of any geometric sequence; then, .
A common lemma is that a sequence is in geometric progression if and only if is the geometric mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of geometric sequences.
Sum
A geometric series is the sum of all the terms of a geometric sequence. They come in two varieties, both of which have their own formulas: finitely or infinitely many terms.
Finite
A finite geometric series with first term , common ratio not equal to one, and total terms has a value equal to .
Proof: Let the geometric series have value . Then Factoring out , mulltiplying both sides by , and using the difference of powers factorization yields Dividing both sides by yields , as desired.
Infinite
An infinite geometric series converges if and only if ; if this condition is satisfied, the series converges to .
Proof: The proof that the series convergence if and only if is an easy application of the ratio test from calculus; thus, such a proof is beyond the scope of this article. If one assumes convergence, there is an elementary proof of the formula that uses telescoping. Using the terms defined above, Multiplying both sides by and adding , we find that Thus, , and so .
Problems
Here are some problems with solutions that utilize geometric sequences and series.
Introductory
2025 AMC 8 Problem 20
Intermediate
1965 AHSME Problem 36
2005 AIME II Problem 3
2007 AIME II Problem 12
See also
Arithmetic sequence
Harmonic sequence
Sequence
Series
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Hyperbola Formula
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HYPERBOLA FORMULA
In a simple sense, hyperbolas resemble mirrored parabolas. The branches are the two halves. A parabola is formed when the plane intersects the halves of a right circular cone at an angle parallel to the cone’s axis.
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HYPERBOLA FORMULA
What is Hyperbola?
Hyperbolas have two foci and two vertices. A hyperbola’s foci are away from its centre and vertices. A Hyperbola Formula is a conic section formed by the intersection of a double cone and a plane surface, but not necessarily at its centre. Like the ellipse, a Hyperbola Formula is symmetric along the conjugate axis. Hyperbolas have foci, directrix, latus rectum, and eccentricity. Examples of hyperbolas include the path taken by the shadow of a sundial, the scattering trajectory of subatomic particles, etc.
The purpose of this chapter is to provide an understanding of the hyperbola definition, formula, derivation of the formula, and standard forms of a Hyperbola Formula through the solving of examples.
What is Hyperbola?
Hyperbolas are smooth curves that lie on a plane and have two components or branches that are mirror images of each other and resemble infinite bows. Hyperbolas are sets of points whose distances from two foci are constant. It is calculated by subtracting the distance from the farther focus from the distance from the closer focus. The locus of the Hyperbola Formula is PF – PF’ = 2a for a point P(x,y) on the hyperbola and two foci F, F’.
Hyperbola Definition
In analytic geometry, a Hyperbola Formula is formed when a plane intersects a double right circular cone at an angle such that both halves are intersected. A hyperbola is created when the plane and cone intersect, resulting in two unbounded curves that are mirror images of one another.
Parts of a Hyperbola
The Hyperbola Formula has two foci with coordinates F(c, o) and F'(-c, 0).
Centre of hyperbola is the centre point of the line connecting the two foci.
A hyperbola’s major axis measures 2a units in length.
Any hyperbola’s minor axis measures 2b units length wise.
Hyperbola Formula vertices are the points where the axis intersects the hyperbola. (a, 0), (-a, 0) are the vertices of a hyperbola.
The latus rectum of a hyperbola passes through the foci of the hyperbola and is drawn perpendicular to its transverse axis. The length of the latus rectum of Hyperbola Formula is 2b2/a.
The transverse axis of the hyperbola passes through the two foci and the centre.
Hyperbola Formula conjugate axis: This line perpendicular to the transverse axis passes through the centre of the hyperbola.
Eccentricity of Hyperbola: (e > 1) Eccentricity is the ratio between the distance of the focus from the centre of the Hyperbola Formula, and the distance of the vertex from the centre. Focus distance is ‘c’ units, and vertex distance is ‘a’ units, so the eccentricity equals e = c/a.
Hyperbola Equation
A hyperbola can be represented by the following equation. The x-axis represents the transverse axis of the Hyperbola Formula, and the y-axis represents its conjugate axis.
x2/a2-y2/b2=1
Standard Equation of Hyperbola
The Hyperbola has two standard equations. Each hyperbola has a transverse axis and conjugate axis. The standard equation of hyperbola is x2/a2-y2/b2=1 has the transverse axis as the x-axis and the conjugate axis as y-axis. Further, another equation of Hyperbola Formula is y2/a2-x2/b2=1, and it has the transverse axis as y-axis and its conjugate axis is x-axis.
Derivation of Hyperbola Equation
P(x, y) should be on the hyperbola such that PF1 – PF2 = 2a
As a result of the distance formula, we have:
√ {(x + c)2 + y2} – √ {(x – c)2 + y2} = 2a
Or, √ {(x + c)2 + y2} = 2a + √ {(x – c)2 + y2}
On squaring both sides as well. As a result,
(x + c)2 + y2 = 4a2 + 4a√ {(x – c)2 + y2} + (x – c)2 + y2
As a result of simplifying the equation,
√ {(x – c)2 + y2} = x(c/a) – a
Squaring both sides again and simplifying further,
x2/a2 – y2/(c2 – a2) = 1
It is known that c2 – a2 = b2. As a result,
x2/a2 – y2/b2 = 1
As a result, any point on the Hyperbola Formula satisfies the equation:
x2/a2 – y2/b2 = 1
Hyperbola Formula
Graph of Hyperbola
Axis minor
Perpendicular to the major axis, it crosses the center of the hyperbola.
The minor axis has a length of 2b. Therefore, the equation is as follows:
x = x0
Axis major
A major axis is the line that crosses by the middle, the focus of the Hyperbola Formula, and the vertices. 2a is the length of the major axis. Here is the equation:
y = y0
Asymptotes
Asymptotes are two intersecting line segments that cross through the centre of the Hyperbola Formula without touching the curve. Asymptotes can be expressed as follows:
y = y0 + b/ax – b/ax0
y = y0 − b/ax + b/ax0
Each hyperbola consists of two curves, each with a vertex and a focus. A hyperbola’s transverse axis crosses both its vertices and foci, and its conjugate axis is perpendicular to it. If the equation of Hyperbola Formula is not in standard form, then one needs to complete the square to get standard form.
The Extramarks website provides more information about Hyperbola graphs.
Properties of a Hyperbola
Hyperbola can be better understood by considering the following properties related to different concepts.
Asymptotes are straight lines drawn parallel to the Hyperbola Formula and assumed to touch the hyperbola at infinity. Asymptotes of the hyperbola have equations y = bx/a and y = -bx/a.
A hyperbola with the same transverse and conjugate axes is called a rectangular hyperbola. In this case, 2a = 2b, or a = b. Hence, equation of rectangular Hyperbola Formula is equal to x2 – y2 = a2
Hyperbola points can be represented by the parametric coordinates (x, y) = (asecθ, btanθ). The parametric coordinates of these points on the hyperbola satisfy its equation.
The auxiliary circle is drawn using the endpoints of the transverse axis of the hyperbola as its diameter. The equation of the auxiliary circle of Hyperbola Formula is x2 + y2 = a2.
The director circle is the point of intersection of perpendicular tangents to the hyperbola. The equation of the director circle is x2 + y2 = a2 – b2.
Examples on Hyperbola
1: Find the hyperbola’s equation if e1 is the eccentricity of the ellipse, x2/16 + y2/25 = 1, and e2 is the eccentricity of the Hyperbola Formula passing through the foci of the ellipse, e1e2 = 1.
Solution: The eccentricity of x2/16 + y2/25 = 1 is
e1 = √(1-16/25) = 3/5
e2 = 5/3
This is obtained using the relation e1e2 = 1.
Hence, the foci of the ellipse are (0, ± 3)
Hence, the equation of the Hyperbola Formula is x2/16 – y2/9 = -1.
2: A hyperbola, having transverse axis of length 2 sin θ, is confocal with ellipse 3×2 + 4y2 = 12. Then its equation is
x2cosec2 θ – y2sec2 θ = 1 2. x2sec2 θ – y2cosec2θ = 1
x2sin2 θ – y2cos2 θ = 1 4. x2cos2 θ – y2sin2 θ = 1
Solution: Given ellipse is x2/4 + y2/3 = 1
Hence, a = 2 and b = √3
Hence, 3 = 4(1-e2) which gives e = 1/2
So, ae = 2.1/2 = 1
Thus, the eccentricity e1 for the Hyperbola Formula is
1 = e1 sin θ which means e1 = cosec θ.
So, b2 = sin2 θ(cosec2 θ – 1) = cos2 θ
Hence, equation of hyperbola is x2/sin2 θ – y2/cos2θ = 1
Or x2cosec2 θ – y2sec2 θ = 1
3: The circle whose equation is x2 + y2 – 8x = 0 and Hyperbola Formula x2/9 – y2/4 = 1 intersect at the points A and B.
An equation for a common tangent to the circle and hyperbola with a positive slope is…?
Find the equation of the circle which has a diameter AB.
Solution: (1) Equation of the tangent to the Hyperbola Formula having slope m is y = mx + √9m2-4
The equation of the tangent to the circle is y = m(x-4) + √16m2 +16
The equations will be identical when m = 2/√5
Therefore, the equation of common tangent is 2x – √5y + 4 = 0
(2) The equation of the Hyperbola Formula is x2/9 – y2/4 = 1 and that of circle is x2 + y2 – 8x = 0.
For their points of intersection, x2/9 + (x2– 8x)/ 4 = 1
So, this gives 4×2 + 9×2 -72x = 36
So, 13×2 -72x -36 = 0
This gives x = 6 and -13/6
But x = -13/6 is not acceptable
Now, x = 6, y = ± 2√3
Required equation is (x-6)2 + (y+2√3)(y-2√3) = 0
This gives x2 + y2 -12x + 24 = 0
4: Find equation of the tangent to the Hyperbola Formula having equation x2/9−y2=1 whose slope is 5
Solution:
Slope of tangent m = 5, a2 = 9, b2 = 1
Tangent equation in slope form is
y = mx ± √(a2m2 – b2 )
y = 5x ± √(9.52 – 1)
y = 5x ± 4√14
[Note: For the ellipse, director circle is x2 + y2 = a2 + b2, x2/a2 + y2/b2 = 1]
Normal:
Equation of the normal of x2/a2 – y2/b2 = 1 at (x1, y1)
a2x/x1 + b2y/y1 = a2 + b2
5: Find the normal at point (6, 3) to the Hyperbola Formula having equation x2/18 − y2/9 = 1
Solution:
Equation of the Normal at point (x1, y1) is a2 = 18, b2 = 9
a2x/x1 + b2y/y1 = a2 + b2
At point (6, 3), the normal’s equation is
18x/6 + 9y/3 = 18 + 9
x + y = 9
6: Find the equation of the chord of Contact of the point (2, 3) to the Hyperbola Formula x2/16 − y2/9 = 1
Solution:
T = 0 is the equation of the chord of contact
i.e. (xx1)/a2 – (yy1)/b2 – 1 = 0
Or, 2x/16 – 3y/9 = 1
Or, x/8 – y/3 = 1
A chord’s equation when its mid-point is known
T = (xx1)/a2 – (yy1)/b2 – 1 = x12/a2 – y12/b2 − 1.
7: Find equation of the chord of the hyperbola x2/9 – y2/4 = 1 whose midpoint is (5, 1).
Solution: Equation of chord of the Hyperbola Formula whose mid-points is (x1, y1)
T = (xx1)/a2 – (yy1)/b2 – 1 = x12/a2 – y12/b2 − 1
Given, midpoint is (5, 1).
=>5x/9 – y/4 – 1= 25/9 – ¼ – 1
=>5x/9 – y/4 = 91/36
Practice Questions on Hyperbola
Find the equation of parabola with focus (− √2, 0) and directrix x = √2.
Solution
The parabola is open left and has the x-axis and vertex (0,0) as its axis of symmetry.
Therefore, the equation of the parabola which is required is
(y – 0)2 = – 4√2 ( x – 0)
⇔ y 2 = – 4√2 x.
This parabola has a vertex of (5, -2) and a focus of (2, -2). Find the equation of this parabola.
Solution
The focal distance AS = a = 3 is measured from vertex A (5, -2), focus S (2, -2), and focal distance AS = b = 5.
The parabola is open on the left and symmetric about the x-axis.
Therefore, the equation of the parabola which is required is
(y + 2)2 = -4 (3)(x – 5)
⇔ y2 + 4 y + 4 = -12x + 60
⇔ y2 + 4 y +12x – 56 = 0.
Determine the equation of the parabola with vertex (-1, -2) and axis parallel to the y-axis and passing through (3, 6).
Solution
Since the axis is parallel to the y-axis, the parabola equation is:
(x +1)2 = 4 a (y + 2).
As this passes through (3,6), we get
(3 +1)2 = 4 a (6 + 2)
⇔ a = 1/2.
Then the equation of parabola is (x +1)2 = 2 (y + 2) which on simplifying yields,
x2 + 2x – 2y – 3 = 0.
In the parabola x2 – 4x – 5 y -1 = 0, find the vertex, focus, directrix, and length of the latus rectum.
Solution
For the parabola,
x 2 – 4 x – 5 y -1 = 0
⇔ x2 – 4x = 5 y +1
⇔ x2 – 4 x + 4 = 5 y +1+ 4.
⇒ (x – 2)2 = 5(y +1) is in standard form. Therefore, 4 a = 5 the focus is (2, 1/4) and the vertex is (2, -1).
Directrix equation is y + k + a = 0
y -1+ 5/4 = 0
4 y +1 = 0.
The length of the latus rectum is five units.
Find the equation of ellipse with foci (± 2, 0), vertices (± 3, 0).
Solution
SS′ = 2 c and 2 c= 4; A′A= 2 a= 6
⇒ a = 3 and c= 2,
⇒ b 2 = a 2 − c 2 = 9 − 4 = 5.
The major axis runs along the x-axis, since a > b.
Centre is (0, 0) and Foci are (±2, 0).
Thus, x 2/9 + y 2/5 = 1 is the equation of the ellipse
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FAQs (Frequently Asked Questions)
1. In conic sections, what is hyperbola?
The locus of a hyperbola is a point whose distance from two fixed points is constant. Hyperbolas have two fixed points called foci.
2. Rectangular hyperbolas: what are they?
Hyperbolas with equal major and minor axes are called rectangular hyperbolas. Thus, we have 2a = 2b, or a = b. The equation of rectangular Hyperbola Formula is x2 – y2 = a2.
3. Hyperbola's eccentricity: what is it?
There is a greater eccentricity than 1 in the hyperbola (e > 1). The eccentricity is the ratio between the distance between the focus and the vertex from the centre of the ellipse. The focus is ‘c’ units away, and the vertex is ‘a’ units away, so the eccentricity is e = c/a. Also, here c2 = a2 + b2.
4. Hyperbolas have foci, what are they?
There are two foci on either side of the hyperbola’s centre, and one on its transverse axis. In the Hyperbola Formula, there are two foci (c, 0), and (-c, 0).
5. A hyperbola has a conjugate axis. What is it?
Hyperbolas have conjugate axes that are perpendicular to their transverse axes and pass through their centres. A hyperbola’s conjugate axis is its y-axis.
6. Asymptotes of Hyperbola: What are they?
Asymptotes are the parallel lines that meet the Hyperbola Formula at infinity. Y = bx/a and y = -bx/a are the equations of asymptotes of the hyperbola.
7. In a hyperbola, what are the vertices?
A hyperbola’s vertex is the point where its transverse axis is cut. Hyperbolas have only two vertices, (a, 0), and (-a, 0).
8. What is the Transverse Axis of a Hyperbola?
A hyperbola’s transverse axis passes through the centre and both foci. A hyperbola’s transverse axis is its x-axis.
9. What is the equation of tangents and normal to the Hyperbola?
Using tangents and normals to the Hyperbola Formula, one can solve the equations given below: In point form, the equation of a tangent is written as: x sec θ a − y tan θ b = 1.
10. Which equations describe hyperbola asymptotes?
Asymptotes have the equation y = ± b a x. In the case of y 2 a 2 − x2 b2 = 1. Asymptotes have equations y = ± a b x. The rectangular Hyperbola Formula has an equal length transverse axis and conjugate axis.
11. What is the method for finding the vertices of a hyperbola?
There is an equation of the form y 2 a 2 + x 2 b 2 = 1 y 2 a 2 + x 2 b 2 = 1, which implies that the transverse axis lies on the y-axis. Since the Hyperbola Formula is centred at the origin, the vertices serve as the graph’s y-intercepts. To find the vertices, set x = 0, and find the value of y.
12. Hyperbola equation: what is it?
A second standard Hyperbola Formula equation is y 2 a2 + x 2 b2 = 1, where y is the transverse axis, and x is the conjugate axis.
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Tools & Reference>Urology
Hydronephrosis and Hydroureter Workup
Updated: Oct 30, 2024
Author: Dennis G Lusaya, MD; Chief Editor: Bradley Fields Schwartz, DO, FACS more...;)
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Hydronephrosis and Hydroureter
Sections Hydronephrosis and Hydroureter
Overview
Practice Essentials
Pathophysiology
Etiology
Epidemiology
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References;)
Workup
Laboratory Studies
Urinalysis is used to assess for signs of infection. Pyuria suggests the presence of infection. Microscopic hematuria may indicate the presence of a stone or tumor.
Complete blood cell count may reveal leukocytosis, which may indicate acute infection.
Serum chemistry studies can reveal an elevation of blood urea nitrogen (BUN) and creatinine levels, which may be the result of bilateral hydronephrosis and hydroureter. In addition, hyperkalemia can be a life-threatening condition.
Next:
Imaging Studies
Imaging in adults
Early diagnosis of urinary tract obstruction is important because most cases can be corrected and a delay in therapy can lead to irreversible renal injury.
Bladder catheterization should be performed initially if there is reason to suspect that bladder neck obstruction leading to acute or chronic urinary retention may be present. Possible clues to this diagnosis include suprapubic pain, a palpable bladder, or unexplained renal failure in an older man.
Radiologic tests are generally used to exclude obstruction at the level of the ureters or above by detecting dilatation of the collecting system. It is important to remember, therefore, that obstruction can occur without dilatation in the following three settings:
Within the first 1-3 days, when the collecting system is relatively noncompliant and less likely to dilate: In this setting, unilateral obstruction can usually be diagnosed by duplex Doppler ultrasonography, which detects an increased resistive index (a reflection of increased renal vascular resistance) in the affected compared with the contralateral kidney. This test is of no value with bilateral involvement because it cannot distinguish obstruction from intrinsic renal disease.
When the collecting systems are encased by retroperitoneal tumor or fibrosis: In this setting, hydronephrosis may be present in the absence of ureteral dilatation. Retroperitoneal fibrosis can occur in a number of settings, including retroperitoneal fibrosis (most commonly idiopathic or associated with beta-blocker or methysergide use, malignancy, or a connective-tissue disorder) and with the fibrotic reaction that surrounds a renal transplant. Thus, the diagnosis of renal insufficiency due to asymptomatic obstruction in a transplanted kidney may be made based on renal biopsy findings, which show diffuse tubular dilatation, rather than the signs of rejection or cyclosporine nephrotoxicity.
When the obstruction is mild, a setting in which renal function is not usually impaired
Renal ultrasonography is the test of choice to exclude urinary tract obstruction, avoiding the potential allergic and toxic complications of radiocontrast media. It can, in the majority of affected patients, help diagnose hydronephrosis and establish its cause; it can also detect other causes of renal disease such as polycystic kidney disease.
Although ultrasound can accurately identify hydronephrosis, it is less sensitive than computed tomography (CT) for detecting stones in the ureters. In one study, hydronephrosis on ultrasound had a positive predictive value of 0.77 for the presence of a ureteral stone and a negative predictive value of 0.71 for the absence of a ureteral stone.
Indications for CT scanning include the following:
Ultrasonography results are equivocal
The kidneys cannot be well visualized
The cause of the obstruction cannot be identified
The combination of a plain film of the abdomen (including tomographic cuts to detect radiopaque calculi), ultrasonography, and, if necessary, CT scanning is adequate for diagnostic purposes in over 90% of cases. It should be noted, however, that the false-positive rate for ultrasonography may be as high as 25% if only minimal criteria (any visualization of the collecting systems) are used to diagnose obstruction.
The advantages of intravenous pyelography (IVP) in relation to ultrasonography are that IVP has a very low false-positive rate, it can identify the site of obstruction, and it can help detect associated conditions such as papillary necrosis or calyceal blunting from previous infection. Disadvantages are that IVP is more cumbersome to perform and requires the administration of a radiocontrast agent. On balance, IVP can be used to screen for urinary tract obstruction in the following settings:
In patients with staghorn calculi or multiple renal or parapelvic cysts, since hydronephrosis is usually not distinguishable from cysts or stones by ultrasonography or CT
When CT scanning cannot identify the level of obstruction
With suspected acute obstruction due to kidney stones (or less frequently, to other problems, such as a sloughed papilla or blood clot): Dilatation of the collecting system may not be seen at this time, but the presence and location of the obstructing stone can be identified
Diffusion-weighted magnetic resonance imaging (MRI) may allow noninvasive detection of changes in renal perfusion and diffusion that occur during acute ureteral obstruction. The advantage of this technique is that it does not require the use of contrast agents. However, the clinical utility of diffusion-weighted MRI has not been adequately tested.
Hydronephrosis without apparent obstruction or with asymptomatic obstruction
In some cases, one of the above radiologic tests demonstrates hydronephrosis without evidence of obstruction. This is a normal finding in pregnant women. Megaureter due to previous marked vesicoureteral reflux or a dilated but nonobstructed extrarenal pelvis is the most common example of this problem. These patients are often being evaluated for back or flank pain, and the following two questions need to be answered:
Is obstruction present?
Is the obstruction responsible for the pain?
In this setting, 3 different tests have been used: diuretic renography, IVP (less often), and perfusion pressure flow studies.
Diuretic renography involves the administration of a loop diuretic (eg, 0.5 mg/kg of furosemide) prior to radionuclide renal scanning or during IVP, while the latter involves percutaneous insertion of a catheter into the dilated renal pelvis, followed by fluid perfusion into the pelvis at a rate of 10 mL/min. The marked increase in urine flow should, if obstruction is present, slow the rate of washout of the radioisotope during renal scanning, further increase the size of the collecting system on IVP, or elevate the renal pelvic pressure to above 22 mm Hg during a perfusion study. Furthermore, any of these procedures may precipitate pain similar to the patient's initial complaint.
Noninvasive diuretic renography is generally preferred. However, optimal interpretation of any of these test results is uncertain, because both false-positive and false-negative results may be seen. Nevertheless, the following general recommendations have been made:
Surgical correction should be considered in a patient with pain and positive diuretic renography findings.
No therapy is necessary in an asymptomatic patient with positive diuretic renography findings but normal renal function. These patients often are adults and have therefore had the partial obstruction for many years without apparent damage to the kidney. Hydronephrosis may first be noted after a radiologic study is performed for some other reason, or, as noted above, obstruction may be suspected when pain is induced after a period of high fluid intake leads to a diuresis that exceeds the rate at which urine can flow through the obstructed area. Similarly, the decreased washout observed on renography occurs only at a urine flow rate much higher than the patient is likely to achieve on his or her own. Periodic monitoring of renal function and renal parenchymal size (by ultrasonography) is indicated in these patients to exclude progressive renal injury.
No therapy is indicated in an asymptomatic patient with negative renography findings. Long-term follow-up has demonstrated stable renal function in most of these patients.
A perfusion pressure flow study should be performed in a symptomatic patient with negative or equivocal diuretic renography findings. Some nonrenal cause for the pain is probably present if the perfusion study is negative. On the other hand, a positive study is suggestive of obstruction and the need for surgical correction.
A perfusion pressure flow study may also be performed in patients with hydronephrosis and poor renal function. The diuretic renogram may be falsely negative in this setting, because the diuretic may not sufficiently raise the urine flow.
In general, approximately 50% of patients with positive diuretic renography findings eventually require surgery, either for pain or progressive parenchymal loss.
Imaging in children
The following grading systems have been developed for evaluating hydronephrosis severity in infants :
Anterior-posterior (AP) diameter of renal pelvis
Society for Fetal Urology (SFU)
Radiology
Urinary Tract Dilation (UTD)
Onen
The Onen grading system applies to both prenatal and postnatal ureteropelvic junctiontype hydronephrosis and includes two categories of findings: dilation of the pelvicalyceal system and, most important, the quality of the renal parenchyma (thickness and appearance). AP diameter is not important.
The Onen system has four grades. Grade 1 criterion is renal pelvic dilatation. Grade 2 criteria are as follows:
Pelvis and caliceal dilatation
Renal parenchyma (medulla and cortex) are normal (> 7 mm)
Grade 3 criteria are as follows:
Pelvis and caliceal dilatation
Medulla is short and thin
Cortex is normal
Total parenchymal thickness 2-5 mm second trimester, 2.5-6 mm third trimester, 3-7 mm postnatal
Corticomedullary differentiation is normal
Grade 4 criteria are as follows:
Pelvis and caliceal dilatation
No medulla is present (total loss)
Cortex is thin (second trimester, < 2 mm; third trimester, < 2.5 mm; postnatal, < 3 mm)
No corticomedullary differentiation is visible
Recesses between calyxes are significantly short and thin
Detection of antenatal hydronephrosis by ultrasound usually occurs in the second trimester. In this setting, hydronephrosis is defined as renal pelvic dilation (RPD) of 4 mm or more. Mild hydronephrosis (RPD of 4-10 mm or SFU grade 1 or 2) can be associated with Down syndrome or other chromosome anomalies. More severe dilatation increases the risk of renal and/or urinary tract disorders.
During the ultrasonography, the appearance of the fetal renal system can vary in both healthy fetuses without hydronephrosis and those with hydronephrosis. Therefore, this diagnosis should not be based on a single measurement. An increase of maternal hydration can also increase the RPD in both healthy fetuses and those with hydronephrosis.
If fetal hydronephrosis is detected, the following parameters need to be evaluated using ultrasonography, as they guide further need for evaluation and are helpful in determining the cause of hydronephrosis:
Severity of hydronephrosis: The likelihood of a congenital kidney or urinary tract anomaly increases with the severity of RPD.
Unilateral versus bilateral involvement: Bilateral involvement increases the risk of a significant renal abnormality and the risk of impaired postnatal renal function.
Ureter: Dilatation of the ureter can be consistent with vesicoureteral reflux (VUR) or obstructive uropathy distal to the ureteropelvic junction.
Postnatal radiologic studies
Postnatal radiologic evaluation of a newborn with antenatal hydronephrosis begins with an ultrasonography examination. The timing of ultrasonography and the need for other studies depend on the severity of postnatal hydronephrosis and whether there is bilateral involvement or an affected solitary kidney.
Ultrasonography of kidneys and bladder should be performed in the postnatal period on affected infants. The timing of the study depends on the severity of the antenatal hydronephrosis. In general, examination should be avoided in the first 2 days after birth because hydronephrosis may not be detected because of extracellular fluid shifts that underestimate the degree of hydronephrosis. However, infants with bilateral hydronephrosis and those with a severe hydronephrotic solitary kidney require urgent evaluation on the first postnatal day because of the increased likelihood of significant disease and a possible need for early intervention. For unilateral hydronephrosis without antenatal bladder pathology, performing postnatal sonography 1-4 weeks after birth is recommended.
A voiding cystourethrography (VCUG) is performed to detect VUR and, in boys, to evaluate the posterior urethra. For this procedure, a urinary catheter is inserted into the bladder and contrast material is instilled. Fluoroscopic monitoring is performed while the bladder is filling and during voiding. Infants usually tolerate this procedure well. Although the duration of fluoroscopy is minimized, the gonads, especially the ovaries, are exposed to radiation.
Diuretic renography is used to diagnose urinary tract obstruction in infants with persistent hydronephrosis and is usually ordered after a VCUG has demonstrated no vesicoureteral reflux. It measures the drainage time from the renal pelvis and assesses total and individual kidney renal function. The test requires insertion of a bladder catheter to relieve any pressure that can be transmitted to the ureters and kidneys. Intravenous access is needed for hydration and the administration of the radioisotope and diuretic. The preferred radioisotope is technetium Tc 99m-mercaptoacetyltriglycine (Tc99mMAG3), which is taken up by the renal cortex, filtered across the glomerular basement membrane to the renal tubules, and excreted into the renal pelvis and urinary tract.
Diuretic renography includes two phases. First, radioisotope is injected intravenously and renal parenchymal (cortical) uptake is measured during the first 2-3 minutes. The relative contribution of each kidney to overall renal function (called the split renal function) is assessed quantitatively and is useful as a baseline study. Subsequent studies can be compared to assess whether kidney function remains stable or has deteriorated, suggesting true obstruction.
Second, at peak renal uptake, intravenous furosemide is administered and the excretion of isotope from the kidney is measured, referred to as the washout curve. This phase indicates the extent of obstruction, if present. In a healthy kidney, furosemide administration results in a prompt washout. In a dilated system, if washout occurs rapidly after diuretic administration (< 15 min), the system is not obstructed. If washout is delayed beyond 20 minutes, the pattern is consistent with obstructive uropathy. However, a delayed washout must be interpreted with caution. [25, 26]
In a series of 39 infants with antenatal unilateral hydronephrosis followed without surgery, diuretic renography indicated obstruction in 24 patients whose renal function never decreased and who thus did not have obstruction. These results may be due, in part, to the normally low neonatal GFR that can be refractory to diuretic therapy. If washout is from 15-20 minutes, the study is indeterminate.
Gravity-assisted drainage imaging may assist in the assessment of pediatric hydronephrosis. Unlike customary diuretic renography, in which the patient remains supine, with gravity-assisted renography a single, static image is obtained after positioning the child in the upright position for 5 minutes to promote additional drainage of tracer from the collecting system. Notable improvement in drainage with this maneuver suggests that slowness in drainage is not due to urinary tract obstruction.
Split kidney function results are the most useful criteria to evaluate a decrease in kidney function. In patients with unilateral hydronephrosis (which is the most common clinical scenario), if the normal nonhydronephrotic kidney and hydronephrotic kidney both have equal function, conservative management without surgery is a safe option. In a cohort of 831 cases of antenatal hydronephrosis, renal scanning performed in 229 newborns demonstrated that 16% of patients had a significant decrease in function of one kidney, defined as 35% or less differential kidney function. A decrease in differential kidney function was associated with severe antenatal hydronephrosis (ie, renal pelvic diameter > 10 mm at 20-24 wk gestation and >16 mm at 33 wk gestation).
Magnetic resonance urography (MRU) in children is becoming more commonly used in the diagnosis and management of congenital uropathies such as UPJ obstruction. [29, 30] MRU is especially useful in the management of obstructed kidneys that have rotation or ascent anomalies, or are solitary. MRU can more clearly define the anatomy and delineate the proper surgical approach (ie, retroperitoneal vs transperitoneal). Newer MRU technology may even define obstruction, eliminating the need for diuretic renal scanning.
The disadvantage of MRU is that the study often requires general anesthesia or heavy conscious sedation in children. Furthermore, the contrast agent gadolinium can only be used if kidney function is normal (requiring a preprocedure serum creatinine test) because of reports of irreversible renal fibrosis in patients with kidney insufficiency.
Previous
Next:
Procedures
Antegrade or retrograde pyelography is usually used to relieve, rather than diagnose, urinary tract obstruction. However, these procedures can also be performed for diagnosis when the history is highly suggestive (eg, unexplained acute kidney injury in a patient with known pelvic malignancy), even though hydronephrosis may be absent (due to possible ureteral encasement) on ultrasonography and CT scanning.
Previous
Treatment & Management
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Holmes N, Harrison MR, Baskin LS. Fetal surgery for posterior urethral valves: long-term postnatal outcomes. Pediatrics. 2001 Jul. 108 (1):E7. [QxMD MEDLINE Link].
Morris RK, Malin GL, Quinlan-Jones E, Middleton LJ, Hemming K, Burke D, et al. Percutaneous vesicoamniotic shunting versus conservative management for fetal lower urinary tract obstruction (PLUTO): a randomised trial. Lancet. 2013 Nov 2. 382(9903):1496-506. [QxMD MEDLINE Link].[Full Text].
de Bessa J Jr, Rodrigues CM, Chammas MC, Miranda EP, Gomes CM, Moscardi PR, et al. Diagnostic accuracy of Onen's Alternative Grading System combined with Doppler evaluation of ureteral jets as an alternative in the diagnosis of obstructive hydronephrosis in children. PeerJ. 2018. 6:e4791. [QxMD MEDLINE Link].[Full Text].
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Dennis G Lusaya, MD Associate Professor II, Department of Surgery (Urology), University of Santo Tomas Faculty of Medicine and Surgery; Chairman, Institute of Urology, St Luke's Medical Center; Head of Urology Unit, Benavides Cancer Institute, University of Santo Tomas Hospital, PhilippinesDennis G Lusaya, MD is a member of the following medical societies: American Urological Association, Philippine College of Surgeons, Philippine Medical Association, Philippine Society of Urological Oncology, Philippine Urological AssociationDisclosure: Nothing to disclose.
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Edgar V Lerma, MD, FACP, FASN, FAHA, FASH, FNLA, FNKF Clinical Professor of Medicine, Section of Nephrology, Department of Medicine, University of Illinois at Chicago College of Medicine; Research Director, Internal Medicine Training Program, Advocate Christ Medical Center; Consulting Staff, Associates in Nephrology, SCEdgar V Lerma, MD, FACP, FASN, FAHA, FASH, FNLA, FNKF is a member of the following medical societies: American Heart Association, American Medical Association, American Society of Hypertension, American Society of Nephrology, Chicago Medical Society, Illinois State Medical Society, National Kidney Foundation, Society of General Internal MedicineDisclosure: Serve(d) as a speaker or a member of a speakers bureau for: Astra Zeneca
Author for: UpToDate, ACP Smart Medicine, Elsevier, McGraw-Hill, Wolters Kluwer.
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Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment.
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Bradley Fields Schwartz, DO, FACS Professor of Urology, Frank and Linda Vala Endowed Chair of Urology, Director, Center for Laparoscopy and Endourology, Department of Surgery, Division of Urology, Southern Illinois University School of MedicineBradley Fields Schwartz, DO, FACS is a member of the following medical societies: American College of Surgeons, American Urological Association, Association of Military Osteopathic Physicians and Surgeons, Endourological Society, Society of Laparoscopic and Robotic Surgeons, Society of University UrologistsDisclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: Endourological Society Board of Directors
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UNCLASSIFIED AD 408 661 DEFENSE DOCUMENTATION CENTER FOR SCIENTIFIC AND TECHNICAL INFORMATION CAMERON STATION. ALEXANDRIA, VIRGINIA UNCLASSIFIED NOTICE: When government or other drawings, speci-fications or other data are used for any purpose other than in connection with a definitely related government procurement operation, the U. S.
Government thereby incurs no responsibility, nor any obligation whatsoever; and the fact that the Govern-ment may have formulated, furnished, or in any way supplied the said drawings, specifications, or other data is not to be regarded by implication or other-wise as in any manner licensing the holder or any other person or corporation, or conveying any rights or permission to manufacture, use or sell any patented invention that my in any way be related thereto.
DEPARTMENT OF THE NAVY OFFICE OF THE CHIEF OF NAVAL OPERATIONS IASHINGTON 25, D. C.
Nl (OEG)485-63 C.., om: Director, Operations Evaluation Group qqTOT): Distribution List " Sibj: Operations Evaluation Group IRM-37; forwarding of _ Enc: (1) OEG IRM-37 "Estimation of Takeoff Ground-run C) Distances for Jet-Propelled Conventional and -.STOL Aircraft" Unclassified of 19 Apr 1963 C'o -"1. Enclosure (1) is forwarded for your information and ' - retention.
2. This memorandum presents methods for estimating the ~C-takeoff ground-run distances for two types of aircraft C.D "ihich use turbojet or turbofan propulsion systems: con-ventional takeoff and landing (CTOL) and short takeoff and landing (STOL). Such aircraft are defined as having fixed direction thrust. Only the ground-run phase is analyzed here. Later reports will consider the transition pullup and initial climb phases of the takeoff.
3. Attention is invited to the fact that this type of memorandum does not necessarily represent the opinion of the Operations Evaluation Group or the U. S. Navy.
4. Additional copies of the IRM may be obtained from OEG.
..
ri 1// ,, -HOWARD W. KREINER For the Director Operations Evaluation Group DISTRIBUTION: Attached List Distribution List for (OEG)485-63: DDR&E COM DDC (20) Dir, WSEG (2) S MDL J60 USNA ANNA (2) W6B BUWEPSREP APL/JHU J84 SUPTNAVPGSCOL (2) W7E CONAVAIRDEVCEN J95 PRESNAVWARCOL (2) W7C COMNAVAIRTESTCEN BUWEPS ONR R- 12 405 R- 14 R- 55 RA RAAD-33 RAPP MARINE CORPS ACTIVITIES COMDTMARCORPS Dir, MCLFDC, Quantico Dir, MCEC DEPT OF AIR FORCE ACTIVITIES Dept of AF (8) (Attn: PDO 4008) COMDT, Air University USAF Liaison Office, Rand Corp. AFSC C/S, U.S. Air Force (Attn: AFX-PDB) DEPT OF ARMY ACTIVITIES Dept of Arm~y (Attn: C/R&D) (for RAC) Dept of Army (Attn: Adj Gen) (6) CONDT, Army War College MISCELLANEOUS ACTIVITIES SRI British Navy Staff U. of Chicago Canadian Joint Staff DISTRIBUTED PREVIOUSLY TO: Op09B9 Op9O OpO5W OPO9E Op93 0p52 OpO3B 0p34 OpO7T r'-OES This is an interim report of continuing research. It does not necessarily represent the opinion of OEG or the U. S. Navy. It may be modified or withdrawn at any time. D D C JULI j 1 TISIA B Interim Research Memorandum OPERATIONS EVALUATION GROUP, WASHINGTON 25, D. C.
19 April 1963 INTERIM RESEARCH MEMORANDUM OPERATIONS EVALUATION GROUP ESTIMATION OF TAKEOFF GROUND-RUN DISTANCES FOR JET-PROPELLED CONVENTIONAL AND STOL AIRCRAFT by R.D. Linnell IRM-37 This is an interim report of continuing research. It does not necessarily rep-resent the opinion of OEG or the U. S.
Navy. It may be modified or withdrawn at any time.
INTERIM RESEARCH MEMORANDUM IRM-37 A BSTRACT This memorandum presents methods for estimating the takeoff ground-run distances for two types of aircraft which use turbojet or turbofan propulsion systems: conventional takeoff and landing (CTOL) and short takeoff and landing (STOL). Such aircraft are defined here as having fixed-direction thrust. The maximum lift coefficient for the landing and takeoff configuration is moderate (around 1. 5) for CTOL aircraft but is relatively large (around 3.0) for STOL aircraft. Both types of aircraft can be studied at one time by use of the maxi-mum lift coefficient as a parameter for analysis of takeoff and landing distances.
Only the ground-run phase of the takeoff is analyzed here. Later reports will consider the transition pullup and initial climb phases of the takeoff. Correla-tion of the estimation methods discussed here with experimental data for current and past aircraft will be presented later.
1 (REVERSE BLANK) INTERIM RESEARCH MEMORANDUM IRM-37 INTRODUCTION Takeoff is the operation of an aircraft from a condition of rest to controllhd flight, usually a climb, at a small altitude of the order of 50 feet. There arc three phases to this operation, i. e., the accelerated ground run, an upward-turning transition, and a climb phase, as sketched in figure i. The takeoff distance is defined as the distance along the surface of the ground to a specified altitude on the flight path. The specified altitude, z0 , is usually 50 feet, but other values are also used.
The ground surface on which the aircraft takes off, indicated in figure 1, is called the runway. It is not absolutely necessary that the ground under the transi-tion and climb phases of the takeoff be a prepared surface, but there must at least be no high obstacles there. The runway is usually very close to horizontal, but some runways may be inclined upward or downward (positive 13 or negative 9) in the direction of takeoff, as indicated in figure 1.
Besides f, three important environmental conditions for takeoff are the character of the runway location, runway surface, and the nature of winds along the runway. The average altitude of the runway can be specified by giving the atmospheric density, whose value also reflects the nature of the climate, i.e., hot, normal, or cold. The runway surface hardness can be given in terms of the coefficient of rolling friction, p , for wheeled landing gear. It is defined as the ratio of resistive horizontal force to the normal force supported by the wheels.
Its value depends on the runway material, the tire pressure, and the speed of the wheel relative to the runway. Approximate, constant values are as follows: Dry concrete, ice, snow, dry steel .02 Wet concrete, hard ground .05 Wet grass, soft sod .10 Mud, soft sand .30 The wind speed, Vw, is defined as the runway-component of the speed of the air near the surface of the runway, positive when its direction is opposite to the direction of takeoff. There is usually a wind-gradient, i.e., an increase of wind speed with increasing altitude. Neglect of the gradient is conservative (safe) when there is a positive Vw (headwind), but unconservative for negative Vw (tailwind). The gradient will, however, be neglected here.
The gross characteristics of takeoff that are of greatest interest are the takeoff distance, xTOz , and takeoff time, tTOz. The former gives the runway length and cleared length needed. The time is needed to obtain an estimate of 3 INTERIM RESEARCH MEMORANDUM IRM-37 uz z z 00 [' 00 z zJ 0 [' 4 INTERIM RESEARCH MEMORANDUM IRM-37 the fuel consumed during takeoff. It is desirable to analyze the takeoff in three phases which correspond to the piloting phases. The ground-run distance, xgrI and time, tgr , are obtained first, followed by Xtr and ttr for the transition, and x and t for the climb. These distances are indicated in figure 1, where the c c sum (xtr + Xc) is designated as the air distance, Xair' for convenience of nomenclature. Experimental values for the takeoff characteristics indicate that the exact techniques used by the pilot can cause considerable variation. At best, an accuracy of ±5 percent in the experimental data can be expected. Thus, an analysis of great refinement would not be consistent with the accuracy of the in-put information (pilot technique) or the possibility of verification (experimental data). On this basis, several approximations will be made in the analysis that follows.
There are several important critical points during the takeoff. Most of them are associated with the operation of mult-engine aircraft and are defined by critical airspeeds, i.e., critical speeds relative to the air. Each such speed can be associated with a distance along the runway, at which the speed occurs during takeoff. These critical distances are indicated in figure 2. If one engine fails, and the remaining engines induce a yawing moment, sufficient aerodynamic and braking moment to maintain control is available only at speeds greater than the Minimum Ground Control speed. If an engine fails at lower airspeeds, the remaining engines must be throttled to avoid an uncontrollable change of ground-path direction.
(ONE ENG OUT) (ONE ENG OUT) MIN GRD MIN AIR CONTROL STALL CONTROL START CRITICAL "TAKEOFF" (EQUAL DIST TO ONE-ENG-OUT TAKEOFF OR NO-ENG BRAKED STOP) FIG. 2: CRITICAL DISTANCES DURING TAKEOFF 5 INTERIM RESEARCH MEMORANDUM IRM-37 The Critical speed is the speed at which, if an engine fails, the aircraft can takeoff with the remaining engines, to the specified altitude, in the same additional distar -, that would be required to stop with full braking after immediately throttling back all the remaining engines. A little thought indicates that for airspeeds less that the Critical speed all engines should be cut when one fails, if the shortest distance to takeoff or stop is desired. However, for airspeeds greater than the critical speed the takeoff should be continued.
The Stall speed is physically well defined, but is subject to some question for estimation purposes. It will be defined here as the speed at which the lift is equal to the takeoff weight, and the maximum lift coefficient in the takeoff configuration, but ignoring ground effect, exists. Thus ground effect and any contribution of engine thrust to lift-direction forces are neglected.
As for ground control, there exists a minimum speed at which aerodynamic moment alone can just overcome yawing moment caused by the failure of one engine. This speed is the Minimum Air Control speed indicated in figure 2.
Until this speed is exceeded, the pilot should not terminate the ground run. A somewhat greater speed should be attained before lift-off and the initiation of the transition, to provide a margin of safety with greater controllability and maneuver-ability. This greater speed at lift-off is called the Takeoff speed.
Hereafter, it will be assumed that all of the critical speeds and distances are less than the stall speed, except the Takeoff speed. The latter is actually the lift-off speed, as described above, but it is conventionally called the Takeoff speed, and denoted by the symbol VTO.
A Reference Ground-Run Distance and Time An elementary analysis of the takeoff ground-run will provide very simple expressions for the ground-run distance and time. The phenomena that occur are elucidated by this analysis, and the resultant expressions are convenient reference values, which will be denoted by Zgr and tr, for later use. The approximations used to obtain these references are to neglect runway slope, runway friction drag, aircraft lift and aircraft drag during the ground run.
Then the accelerating force is simply the thrust, T, produced by the propulsion system, and this thrust is approximated as being constant.
With W denoting aircraft takeoff weight and g denoting the local acceleration of gravity, the acceleration equation becomes W d2x (1 g dt 'I INTERIM RESEARCH MEMORANDUM IRM-37 Direct integration or use of the identity (d 2x/dt 2) = (1/2XdV 2/dx), where V denotes the flight speed (airspeed), gives 2 2 x2 -x1 = (V 2 -Vl)/2g(T/W) (2) t 2 -t1 = (V 2 -V1)/g(T/W) For the reference parameters with zero wind, the initial conditions are selected as zero, and V2 becomes the takeoff speed. To provide a margin of safety explicitly, the takeoff speed is taken to be times the stall speed, where is greater than unity. Thus, VTO = Vt = 4'12(W/A)/pQCma (3) The asterisks are used to emphasize that takeoff conditions are to be used when evaluating the stall speed, aircraft weight and maximum lift coefficient.
The safety factor is often assumed to have the value of 1.2 for conventional takeoff and landing (CTOL) aircraft. The effect of variation of its value will be investigated later.
The reference values can now be written as V 22 2 TO -2(W/A) 13 42(W/A)psf ft gr 2g- = pg(T/W)C Lmax /W)C Lmax (4) V TO _ 4 2(W/A) _9 1 (W/A)psf t gr Fr7W) PC T/W) Csec gr g(T/W) (T/'W) p C Lmax Lmax where 7 denotes the density ratio, P/pSL' and pSL is the standard sea level air density. The first form is basic, but substitution for VTO shows how the aircraft parameters W/A and T/W affect the ground run distance and time.
The final forms are convenient for numerical evaluation. The reference values will be used hereafter as a standard, indicating the basic variation of takeoff distance and time. For example, 5r is linear in wing loading, W/A, quadratic gr in the safety factor 4, and inversely proportional to a, thrust-weight ratio and maximum lift coefficient. For variations of aircraft weight only, xg r is pro-portional to W2 .
If the thrust is proportional to air density, then increases of altitude lead to increases of 5r proportional to p-2 gr 7 INTERIM RESEARCH MEMORANDUM IRM-37 If the wind speed, V w , is not zero, the limits for the airspeed are V9 = VTO - Vw and V, = Vw .
In terms of the reference values, approximate values for the ground run distance and time, based on constant force by using equation 2, are 2 X = 5"[1 Xgr where = Vw/VTo =Vw/Vst The square brackets are correction factors needed to modify the reference values for the effect of wind speed.
Corrections for the Ground-Run Distance and Time A more complete analysis of the takeoff ground run is needed for new uses, such as analysis of STOL aircraft, analysis of overloaded VTOL aircraft, and operation from unprepared fields in undeveloped countries or in warfare. For modern conventional aircraft, the reference ground-run distance and time pro-vide quite accurate estimates, but a new analytical method is developed below for the newly interesting cases that require consideration of the effects of p, CDO' CL and Vw .
The forces acting are sketched in figure 3. The thrust will be considered constant during the ground run, i.e., independent of flight speed, but it can vary with air density. Actually, the direction of action of the thrust depends on its orientation with respect to a zero-lift reference line in the aircraft and the angle of attack of the wing of the aircraft. For the present analysis, the thrust direction is approximated as being parallel to the runway, along the direction of the x-axis, at all times. For small angles this still is a good approximation for the x-component, but there could be a significant z-component even for small angles. Such cases will be considered in later reports, in connection with vertical takeoff and landing aircraft.
8 INTERIM RESEARCH MEMORANDUM IRM-37 LIFT, L RUNWAY DRAG, D FRICTION HORIZONTAL I:(W cos /3 - L) FIG. 3: FORCES DURING TAKEOFF GROUND RUN For the lift and drag, a symmetrical parabolic polar is used. With asterisks to emphasize that takeoff conditions prevail, L=(1/2) pV 2AC LL (6) D=(1/2)pV 2A CDO +KC L2 1 Here C is the zero-lift drag coefficient, K is the drag-due-to-lift parameter, DO and V is the flight speed, the sum of the ground speed, V , and the runway-component, V w , of wind speed. g VV g + Vw (7) The air density, P, is essentially constant during the ground run. The lift coefficient, CL, during the ground-run is subject to control manipulation by the pilot. With nose-wheel landing gear the angle of attack is automatically close to zero, and the pilot must exert control forces to obtain a moderate or large angle of attack. For tail-wheel landing-gear the angle of attack is initially large, near the stall angle of attack, and pilot control is required to lift the tail to obtain small angles of attack. An optimum constant lift co-efficient exists, as shown on the following page, and will be assumed here.
9 INTERIM RESEARCH MEMORANDUM IRM-37 The flight speeds during takeoff are sufficiently small that C 0 is very ac-curately constant, but it is much larger than C DO for clean-configuration flight, because extended landing gear, partial flaps, and possibly external fuel tanks make major contributions. The value of K may be affected slightly by extended landing gear and ground effect, but it is quite accurately constant.
The acceleration equation for takeoff can be written 2 dV dV 2 2g[ T - pcosP-- sin--= dt dx (8) p [CD + KC 2 -PC (V +V )2 Inspection shows that a small positive lift coefficient increases the acceleration, but that a sufficiently large C will decrease it. By differentiation of the C" L L terms it is found that the optimum C for maximum acceleration is C opt =ii/2K (9a) KC L2 -C L opt= - 2/4K (9b) The drag-due-to-lift parameter, K, for takeoff is usually small, say approxi-mately. 05 for an aspect ratio of about 6. Taking ii equal to. 02 for a hard-surfaced runway, the optimum lift coefficient is about . 2 or quite moderate.
However, for a takeoff from a surface of soft sand, the friction coefficient becomes about .3 and the optimum lift coefficient rises to 3.0, which may not be obtainable.
For such cases the correction methods presented below for the takeoff ground run can still -he used as described in the next paragraph, even though they are based on operation at the optimum lift coefficient.
10 INTERIM RESEARCH MEMORANDUM IRM-37 For integration, equation 8 can be rewritten using equation 9b and two parameters, C 2 2 DO (H2/4K) _2 (T/W) - 4cosP - sinP 3 C Lmax (10) C=V V/Vo= Vw/vt ~w TO wV/ sVt The parameter g can be positive, negative or zero If the optimum lift coefficient is greater than CLmax , or if the optimum is not used for any reason, the term 2 4 /4K in the definition of should be replaced with the actual value of K C 2 -C I If a zero lift coefficient is assumed, then the term should be set equal to zero.
Using the definitions of . and C, the differential equation becomes dV = d 2g FT (11) dt dx W T The integration must include consideration of the sign of . which can be negative or positive. The result can be written in terms of the reference ground run characteristics, with correction factors, as follows.
g~ = (T/W) x F(., C) (12) Xg =(T/W) - Pcos3 - sin 3 gr !5 0, thenF -l --2C [tan -7- tan-l -1 1-C2 /T INTiRIM RESEARCH MEMORANDUM IRM- 37 ->0, then F = -1 In 1 -g -2C tanh-',/.- tanh-'C,/" - _ (T/W) t G(, ) (13) gr =(T/W) -iicos0- sin gr F0, thenG; I tani-' - tan- Cl7r g0, thenG= 1 tanh tanh C/ /6 For specific cases the equations can be evaluated, but, to show the nature and magnitude of the corrections, figures 4 and 5 present contours of constant magnitude of F and G in the g, C plane. It is seen that the correction factor is within ±10 percent of unity when C is close to zero and is in the normal range of I I less than about .25. Thus, in many cases the F and G correction factors can be neglected with fair accuracy. For difficult operating conditions, such as in undeveloped or new countries and for technical guerilla warfare, the corrections can be quite important, and are easily estimated by the use of the thrust-ratio factors of equations 12 and 13 and the functions F and G.
The preceding procedures assume a continuous thrust throughout a contin-uous takeoff. Discontinuous thrust such as short-duration Jet Assisted Takeoff (JATO) units may be used deliberately, or engine failure may cause a discon-tinuity of thrust. It is possible for other parameters to vary discontinuously, e.g., the ground-slope angle, 1, or the aerodynamic CDO. For such cases the preceding methods could be applied in steps of the ground run. This procedure is fairly simple when drag, friction and lift are ignored, because equations 2 are valid for each step. Use of the method of this section is more complicated.
A numerical integration by electronic computer may be simpler in such cases, even though approximations for the variation of C are required, thus degrading L the over-all accuracy of the result. For possible use in such cases the equations for the change of x and t that are obtained by integrating equation 11 between arbitrary limits are given below, Where the symbol r denotes the ratio VgVTOP and and C are as defined in equations 10.
12 INTERIM RESEARCH MEMORANDUM IRM-37 x 2 X, TOTW) F , +~ F( , C+r 2)] (14) t 2 - ti V TO/ _ [G ,C+r,) -G(F,,C+r 2 )j (15) (T/W) - 4±cosI3 - sinf3 The curves of figures 4 and 5 do not extend to large enough values of C for general use with these equations, but F and G can be obtained by evaluation of equations 12 and 13.
13 INTERIM RESEARCH MEMORANDUM IRM-37 cO 0.0 F- -. Q cc 0 CC 0 '00 C~ -0 -~ e'~ ~ 00 I I U 14r1 INTE RIM RESEARCH MEMORANDUM IRM-37 'Cz 0 cq 0 0 I-, 0 coo A UlAA -15 (REVRSE LANK
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Scheduled maintenance is planned for September 26–29. You may experience brief interruptions during this time.
LAWRENCE LEEMAN, MD, MPH, AND PATRICIA FONTAINE, MD, MS
A more recent article on hypertensive disorders of pregnancy is available.
Am Fam Physician. 2008;78(1):93-100
Patient information: See a related handout on high blood pressure during pregnancy.
Author disclosure: Nothing to disclose.
Article Sections
The National High Blood Pressure Education Program Working Group on High Blood Pressure in Pregnancy has defined four categories of hypertension in pregnancy: chronic hypertension, gestational hypertension, preeclampsia, and preeclampsia superimposed on chronic hypertension. A maternal blood pressure measurement of 140/90 mm Hg or greater on two occasions before 20 weeks of gestation indicates chronic hypertension. Pharmacologic treatment is needed to prevent maternal end-organ damage from severely elevated blood pressure (150 to 180/100 to 110 mm Hg); treatment of mild to moderate chronic hypertension does not improve neonatal outcomes or prevent superimposed preeclampsia. Gestational hypertension is a provisional diagnosis for women with new-onset, nonproteinuric hypertension after 20 weeks of gestation; many of these women are eventually diagnosed with preeclampsia or chronic hypertension. Preeclampsia is the development of new-onset hypertension with proteinuria after 20 weeks of gestation. Adverse pregnancy outcomes related to severe preeclampsia are caused primarily by the need for preterm delivery. HELLP (i.e., hemolysis, elevated liver enzymes, and low platelet count) syndrome is a form of severe preeclampsia with high rates of neonatal and maternal morbidity. Magnesium sulfate is the drug of choice to prevent and treat eclampsia. The use of magnesium sulfate for seizure prophylaxis in women with mild preeclampsia is controversial because of the low incidence of seizures in this population.
The National High Blood Pressure Education Program Working Group on High Blood Pressure in Pregnancy has defined four categories of hypertension in pregnancy: chronic hypertension, gestational hypertension, preeclampsia, and preeclampsia superimposed on chronic hypertension. A maternal blood pressure measurement of 140/90 mm Hg or greater on two occasions before 20 weeks of gestation indicates chronic hypertension. Pharmacologic treatment is needed to prevent maternal end-organ damage from severely elevated blood pressure (150 to 180/100 to 110 mm Hg); treatment of mild to moderate chronic hypertension does not improve neonatal outcomes or prevent superimposed preeclampsia. Gestational hypertension is a provisional diagnosis for women with new-onset, nonproteinuric hypertension after 20 weeks of gestation; many of these women are eventually diagnosed with preeclampsia or chronic hypertension. Preeclampsia is the development of new-onset hypertension with proteinuria after 20 weeks of gestation. Adverse pregnancy outcomes related to severe preeclampsia are caused primarily by the need for preterm delivery. HELLP (i.e., hemolysis, elevated liver enzymes, and low platelet count) syndrome is a form of severe preeclampsia with high rates of neonatal and maternal morbidity. Magnesium sulfate is the drug of choice to prevent and treat eclampsia. The use of magnesium sulfate for seizure prophylaxis in women with mild preeclampsia is controversial because of the low incidence of seizures in this population.
Hypertensive disorders represent the most common medical complication of pregnancy, affecting 6 to 8 percent of gestations in the United States.1 In 2000, the National High Blood Pressure Education Program Working Group on High Blood Pressure in Pregnancy defined four categories of hypertension in pregnancy: chronic hypertension, gestational hypertension, preeclampsia, and preeclampsia superimposed on chronic hypertension.1
| Clinical recommendation | Evidence rating | References |
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| In women without end-organ damage, chronic hypertension in pregnancy does not require treatment unless the patient's blood pressure is persistently greater than 150 to 180/100 to 110 mm Hg. | C | 1, 2, 4, 6 |
| Calcium supplementation decreases the incidence of hypertension and preeclampsia, respectively, among all women (NNT = 11 and NNT = 20), women at high risk of hypertensive disorders (NNT = 2 and NNT = 6), and women with low calcium intake (NNT = 6 and NNT = 13). | A | 26 |
| Low-dose aspirin (75 to 81 mg daily) has small to moderate benefits for the prevention of preeclampsia (NNT = 72), preterm delivery (NNT = 74), and fetal death (NNT = 243). The benefit of aspirin is greatest (NNT = 19) for prevention of preeclampsia in women at highest risk (previous severe preeclampsia, diabetes, chronic hypertension, renal disease, or autoimmune disease). | B | 27 |
| For women with mild preeclampsia, delivery is generally not indicated until 37 to 38 weeks of gestation and should occur by 40 weeks. | C | 7 |
| Magnesium sulfate is the treatment of choice for women with preeclampsia to prevent eclamptic seizures (NNT = 100) and placental abruption (NNT = 100). | A | 42 |
| Intravenous labetalol or hydralazine may be used to treat severe hypertension in pregnancy because neither agent has demonstrated superior effectiveness. | B | 1, 46 |
| For managing severe preeclampsia between 24 and 34 weeks of gestation, the data are insufficient to determine whether an “interventionist” approach (i.e., induction or cesarean delivery 12 to 24 hours after corticosteroid administration) is superior to expectant management. Expectant management, with close monitoring of the mother and fetus, reduces neonatal complications and stay in the newborn intensive care nursery. | B | 47–49 |
| Magnesium sulfate is more effective than diazepam (Valium; NNT = 8) or phenytoin (Dilantin; NNT = 8) in preventing recurrent eclamptic seizures. | A | 39, 54–56 |
Chronic Hypertension
Chronic hypertension is defined as a blood pressure measurement of 140/90 mm Hg or more on two occasions before 20 weeks of gestation or persisting beyond 12 weeks postpartum.1 Treatment of mild to moderate chronic hypertension neither benefits the fetus nor prevents preeclampsia.2–4 Excessively lowering blood pressure may result in decreased placental perfusion and adverse perinatal outcomes.5 When a patient's blood pressure is persistently greater than 150 to 180/100 to 110 mm Hg, pharmacologic treatment is needed to prevent maternal end-organ damage.1,2,4,6
Methyldopa (Aldomet; brand no longer available in the United States), labetalol, and nifedipine (Procardia) are oral agents commonly used to treat chronic hypertension in pregnancy. Angiotensin-converting enzyme inhibitors and angiotensin-II receptor antagonists are not used because of teratogenicity, intrauterine growth restriction (IUGR), and neonatal renal failure.4 The beta blocker atenolol (Tenormin) has been associated with IUGR,3 and thiazide diuretics can exacerbate intravascular fluid depletion if superimposed preeclampsia develops. Women in active labor with uncontrolled severe chronic hypertension require treatment with intravenous labetalol or hydralazine.7
Morbidity occurs primarily from superimposed preeclampsia or IUGR.4 A sudden increase in blood pressure, new proteinuria, or signs and symptoms of severe preeclampsia indicate superimposed preeclampsia. Fetal growth may be assessed by serial fundal height measurements supplemented by ultrasonography every four weeks starting at 28 weeks of gestation.4
Gestational Hypertension
Gestational hypertension has replaced the term pregnancy-induced hypertension to describe women who develop hypertension without proteinuria after 20 weeks of gestation.1 Gestational hypertension is a provisional diagnosis that includes women eventually diagnosed with preeclampsia or chronic hypertension, as well as women retrospectively diagnosed with transient hypertension of pregnancy. Fifty percent of women diagnosed with gestational hypertension between 24 and 35 weeks develop preeclampsia.8 Expectant management of mild gestational hypertension can reduce the increased rate of cesarean delivery associated with the induction of nulliparous women who have an unripe cervix.9 Women who progress to severe gestational hypertension based on the degree of blood pressure elevation have worse perinatal outcomes than do women with mild preeclampsia, and require management similar to those with severe preeclampsia.10
Preeclampsia
Preeclampsia is a multiorgan disease process of unknown etiology11 characterized by the development of hypertension and proteinuria after 20 weeks of gestation. Table 1 lists proposed etiologies and risk factors for preeclampsia.7,12–21 Prevention through routine supplementation with calcium, magnesium, omega-3 fatty acids, or antioxidant vitamins is ineffective.22–25 Calcium supplementation reduces the risk of developing preeclampsia in high-risk women and those with low dietary calcium intakes.26
| |
| Theories of pathogenesis |
| Abnormal placental implantation (defects in trophoblasts and spiral arterioles)13,14 |
| Angiogenic factors (increased sFlt-1, decreased placental growth factor levels)15,16 |
| Cardiovascular maladaptation and vasoconstriction |
| Genetic predisposition (maternal, paternal, thrombophilias)17–20 |
| Immunologic intolerance between fetoplacental and maternal tissue 7 |
| Platelet activation |
| Vascular endothelial damage or dysfunction7 |
| Risk factors7,12 |
| Antiphospholipid antibody syndrome |
| Chronic hypertension |
| Chronic renal disease |
| Elevated body mass index |
| Maternal age older than 40 years |
| Multiple gestation |
| Nulliparity |
| Preeclampsia in a previous pregnancy (particularly if severe or before 32 weeks of gestation) |
| Pregestational diabetes mellitus |
Low-dose aspirin (75 to 81 mg per day) is effective for women at increased risk of preeclampsia. Treating 69 women prevents one case of preeclampsia; treating 227 women prevents one fetal death.27 For women at highest risk from previous severe preeclampsia, diabetes, chronic hypertension, or renal or autoimmune disease, only 18 need to be treated with low-dose aspirin to prevent one case of preeclampsia.27
DIAGNOSIS
Blood pressure should be measured at each prenatal visit with an appropriately sized cuff and the patient in a seated position.28,29 Diagnostic criteria for preeclampsia are systolic blood pressure of 140 mm Hg or more or a diastolic blood pressure of 90 mm Hg or more on two occasions at least six hours apart.12,28,29 An increase of 30 mm Hg systolic or 15 mm Hg diastolic from baseline is no longer diagnostic for preeclampsia12 because similar increases are common in uncomplicated pregnancies.
The diagnostic threshold for proteinuria is 300 mg in a 24-hour urine specimen. A 24-hour determination is most accurate because urine dipsticks can be affected by variable excretion, maternal dehydration, and bacteriuria.7 A random urine protein/creatinine ratio of less than 0.21 indicates that significant proteinuria is unlikely with a negative predictive value of 83 percent; however, confirmatory 24-hour urine protein determination is recommended.30 Generalized edema (affecting the face and hands) is often present in patients with preeclampsia but is not a diagnostic criterion.1
Severe Preeclampsia. Preeclampsia is characterized as mild or severe based on the degree of hypertension and proteinuria, and the presence of symptoms resulting from involvement of the kidneys, brain, liver, and cardiovascular system (Table 2).12 Severe headache, visual disturbances, and hyperreflexia may signal impending eclampsia. Increased peripheral vascular resistance and pulmonary edema may occur. A decreased glomerular filtration rate may progress to oliguria and acute renal failure. The increased glomerular filtration rate of pregnancy lowers serum creatinine, and levels greater than 0.9 mg per dL (80 μmol per L) are abnormal in pregnancy. Liver manifestations include elevated transaminase levels, subcapsular hemorrhage with right upper quadrant pain, and capsular rupture with life-threatening intraabdominal bleeding. Obstetric complications include IUGR, placental abruption, and fetal demise.12
| | |
--- |
| Blood pressure ≥ 160 mm Hg systolic or 110 mm Hg diastolic on two occasions at least six hours apart during bed rest | |
| Proteinuria ≥ 5 g in a 24-hour urine specimen or 3+ or greater on two random urine specimens collected at least four hours apart | |
| Any of the following associated signs and symptoms: | |
| | Cerebral or visual disturbances |
| | Epigastric or right upper quadrant pain |
| | Fetal growth restriction |
| | Impaired liver function |
| | Oliguria < 500 mL in 24 hours |
| | Pulmonary edema |
| | Thrombocytopenia |
HELLP Syndrome. The acronym HELLP describes a variant of severe preeclampsia characterized by hemolysis, elevated liver enzymes, and low platelet count.31 HELLP syndrome occurs in up to 20 percent of pregnancies complicated by severe preeclampsia.32 The clinical presentation of HELLP syndrome is variable; 12 to 18 percent of affected women are normotensive and 13 percent do not have proteinuria.33 At diagnosis, 30 percent of women are postpartum, 18 percent are term, and 52 percent are preterm.32 Common presenting complaints are right upper quadrant or epigastric pain, nausea, and vomiting. Many patients have a history of malaise or nonspecific symptoms suggesting an acute viral syndrome.33 Any patient with these symptoms or signs of preeclampsia should be evaluated with complete blood count, platelet count, and liver enzyme determinations.34
Laboratory tests are used to diagnose HELLP syndrome (Table 333–35); a decreasing platelet count and an increasing l-lactate dehydrogenase level (indicative of both hemolysis and liver dysfunction) reflect disease severity.33,35 When the platelet count is less than 50,000 per mm3 (50 × 109 per L) or active bleeding occurs, coagulation studies (i.e., prothrombin time, partial thromboplastin time, and fibrinogen level) should be performed to rule out superimposed disseminated intravascular coagulation.
| | |
--- |
| Hemolysis | Abnormal peripheral blood smear (evidence of damaged erythrocytes, such as schistocytes and burr cells) |
| Serum bilirubin ≥ 1.2 mg per dL (21 μmol per L) |
| LDH > 600 U per L (10.02 μkat per L) |
| Elevated liver enzymes | AST (SGOT) elevated |
| ALT (SGPT) elevated |
| Low platelet count | < 100,000 per mm3 (100 × 109 per L) |
| or |
| Class 1: ≤ 50,000 per mm3 (50 × 109 per L) |
| Class 2: > 50,000 but ≤ 100,000 per mm3 |
| Class 3: > 100,000 but < 150,000 per mm3 (150 × 109 per L) |
MANAGEMENT OF PREECLAMPSIA
A common regimen for expectant management of mild preeclampsia is outlined in Table 4.1,7 Nonstress tests, amniotic fluid index measurements, and biophysical profiles are used to monitor patients for uteroplacental insufficiency.1,7 Umbilical artery systolic/diastolic ratios measured by Doppler ultrasonography may detect early uteroplacental insufficiency.36,37 The decision to deliver involves balancing the risks of worsening preeclampsia against those of prematurity. Delivery is generally not indicated for women with mild preeclampsia until 37 to 38 weeks of gestation and should occur by 40 weeks1,7 (Figure 17). Patients with severe preeclampsia are admitted to the hospital, placed on bed rest, and carefully monitored (Figure 27 and Table 51,7,12 ). The goals of treatment are to prevent seizures, lower blood pressure to avoid maternal end-organ damage, and expedite delivery.
| |
| Maternal monitoring |
| Measure blood pressure twice weekly |
| Obtain laboratory tests weekly: CBC, platelet count, ALT, AST, LDH, uric acid, creatinine |
| Assess for proteinuria: screen with dipstick or spot protein/creatinine ratio and obtain periodic 24-hour urine collections |
| Fetal monitoring |
| Obtain nonstress test twice weekly |
| Measure amniotic fluid index once or twice weekly |
| Biophysical profile may be done weekly in place of one of the twice-weekly nonstress tests and amniotic fluid index |
| Perform ultrasonography for fetal growth every three to four weeks |
Magnesium Sulfate. The use of magnesium sulfate helps prevent seizures in women with preeclampsia.38–40 One eclamptic seizure is prevented for every 100 women treated.38 The use of magnesium sulfate is controversial in women with mild preeclampsia because the incidence of eclamptic seizures is only 0.5 percent in these patients. Assuming one half of seizures are preventable with magnesium sulfate,38 400 women with mild preeclampsia would need to be treated to prevent one seizure.41 Magnesium sulfate has the additional benefit of reducing the incidence of placental abruption.42
Magnesium sulfate slows neuromuscular conduction and depresses central nervous system irritability without significant effects on blood pressure. One fourth of women will experience adverse effects, especially flushing.42 Table 5 outlines the standard dosing regimen.1,7,12 Serum magnesium levels should be monitored in women with elevated serum creatinine levels, decreased urine output, or absent deep tendon reflexes.43 Magnesium toxicity can lead to respiratory paralysis, central nervous system depression, and cardiac arrest. The antidote is calcium gluconate, 1 g infused intravenously over two minutes.44
| | | | |
--- --- |
| Bed rest with seizure precautions Vital signs (blood pressure, pulse, respiration); deep tendon reflexes; and mental status every 15 to 60 minutes until stable, then every 60 minutes while on magnesium sulfate Accurate intake and output; Foley catheter if needed Administer lactated Ringer's solution at 75 mL per hour IV to maintain urine output of 30 to 40 mL per hour; total intake (IV and oral) should not exceed 125 mL per hour or 3,000 mL per day Continuous fetal heart rate monitoring7 Laboratory tests Dipstick urine collection for protein level on admission 24-hour urine collection for total protein level CBC with platelets, peripheral blood smear BUN, creatinine, uric acid AST, ALT, LDH Fetal evaluation: nonstress test on admission; obstetric ultrasonography for estimated fetal weight, amniotic fluid volume, and umbilical artery Doppler measurements Medications Magnesium sulfate Loading dose of 4 to 6 g diluted in 100 mL of normal saline, given IV over 15 to 20 minutes, followed by a continuous infusion of 2 g per hour12 Assess serum magnesium level if urine output is < 30 mL per hour or there is a loss of deep tendon reflexes, decreased respiratory rate, or altered mental status Therapeutic range for serum magnesium is 4 to 7 mg per dL Corticosteroids (if between 24 and 34 weeks of gestation and not previously administered) Betamethasone (Celestone), 12 mg IM initially, then repeat in 24 hours or Dexamethasone, 6 mg IM initially, then repeat every 12 hours for three additional doses For systolic blood pressure > 160 mm Hg or diastolic > 110 mm Hg, one of the following should be given to achieve a systolic measurement of 140 to 155 mm Hg and/or a diastolic measurement of 90 to 105 mm Hg7: Hydralazine, 5 to 10 mg IV every 15 to 30 minutes (maximal dose: 30 mg)7 or Labetalol, 20 mg IV initially; if the initial dose is not effective, double the dose to 40 mg and then 80 mg at 10-minute intervals until target blood pressure is reached or a total of 220 mg has been administered1,7; the maximal dose of IV labetalol is 220 mg in a 24-hour period7,12 Calcium gluconate, 1 g IV; keep at bedside in case of respiratory depression from magnesium sulfate use | | | |
Bed rest with seizure precautions
Vital signs (blood pressure, pulse, respiration); deep tendon reflexes; and mental status every 15 to 60 minutes until stable, then every 60 minutes while on magnesium sulfate
Accurate intake and output; Foley catheter if needed
Administer lactated Ringer's solution at 75 mL per hour IV to maintain urine output of 30 to 40 mL per hour; total intake (IV and oral) should not exceed 125 mL per hour or 3,000 mL per day
Continuous fetal heart rate monitoring7
Laboratory tests
Dipstick urine collection for protein level on admission
24-hour urine collection for total protein level
CBC with platelets, peripheral blood smear
BUN, creatinine, uric acid
AST, ALT, LDH
Fetal evaluation: nonstress test on admission; obstetric ultrasonography for estimated fetal weight, amniotic fluid volume, and umbilical artery Doppler measurements
Medications
Magnesium sulfate
Loading dose of 4 to 6 g diluted in 100 mL of normal saline, given IV over 15 to 20 minutes, followed by a continuous infusion of 2 g per hour12
Assess serum magnesium level if urine output is < 30 mL per hour or there is a loss of deep tendon reflexes, decreased respiratory rate, or altered mental status
Therapeutic range for serum magnesium is 4 to 7 mg per dL
Corticosteroids (if between 24 and 34 weeks of gestation and not previously administered)
Betamethasone (Celestone), 12 mg IM initially, then repeat in 24 hours
or
Dexamethasone, 6 mg IM initially, then repeat every 12 hours for three additional doses
For systolic blood pressure > 160 mm Hg or diastolic > 110 mm Hg, one of the following should be given to achieve a systolic measurement of 140 to 155 mm Hg and/or a diastolic measurement of 90 to 105 mm Hg7:
Hydralazine, 5 to 10 mg IV every 15 to 30 minutes (maximal dose: 30 mg)7
or
Labetalol, 20 mg IV initially; if the initial dose is not effective, double the dose to 40 mg and then 80 mg at 10-minute intervals until target blood pressure is reached or a total of 220 mg has been administered1,7; the maximal dose of IV labetalol is 220 mg in a 24-hour period7,12
Calcium gluconate, 1 g IV; keep at bedside in case of respiratory depression from magnesium sulfate use
Antihypertensive Medications. The optimal level of blood pressure control in pregnancies complicated by hypertension is unknown.2,6 Less tight control may decrease the risk that the infant will be small for gestational age, but it may increase the risk of respiratory distress syndrome of the newborn, severe hypertension, and antenatal hospitalization.2,5 Although traditional recommendations are based on diastolic blood pressure, a retrospective review of 28 women with severe preeclampsia who experienced a cerebrovascular accident demonstrated that more than 90 percent had systolic blood pressure over 160 mm Hg, but only 12.5 percent had diastolic blood pressure over 110 mm Hg.45
Intravenous labetalol and hydralazine are commonly used for the acute management of preeclampsia.1,46 A Cochrane review showed no evidence that one parenteral agent had superior effectiveness.46 For women with severe preeclampsia undergoing expectant management remote from term, oral labetalol and nifedipine are acceptable options.7
Fluid Management. Excessive fluid administration can result in pulmonary edema, ascites, and cardiopulmonary overload, whereas too little fluid exacerbates an already constricted intravascular volume and leads to further end-organ ischemia. Urine output should be greater than 30 mL per hour44 and intravenous fluids limited to 100 mL per hour.35,44
Delivery Decisions in Severe Preeclampsia. Delivery is the only cure for preeclampsia. Decisions regarding the timing and mode of delivery are based on a combination of maternal and fetal factors. Fetal factors include gestational age, evidence of lung maturity, and signs of fetal compromise on antenatal assessment. Patients with treatment-resistant severe hypertension or other signs of maternal or fetal deterioration should be delivered within 24 hours, irrespective of gestational age or fetal lung maturity. Fetuses older than 34 weeks, or those with documented lung maturity, are also delivered without delay.7
For patients with severe preeclampsia between 24 and 34 weeks of gestation, the data are insufficient to recommend “interventionist” versus expectant management.47 Subspecialty consultation is indicated.48,49 Corticosteroids are administered to accelerate fetal lung maturity.7 Interventionist management advocates induction or cesarean delivery 12 to 24 hours after corticosteroid administration. Expectant management, with close monitoring of the mother and fetus, delays delivery when possible and reduces neonatal complications and length of stay in the newborn intensive care nursery.47–49 Contraindications to expectant management include persistent severe symptoms, multiorgan dysfunction, severe IUGR (i.e., estimated fetal weight below the 5th percentile), suspected placental abruption, or nonreassuring fetal testing.49
In women with HELLP syndrome, the fetus is delivered at an earlier gestation; specifically, fetuses older than 28 weeks are routinely delivered 24 to 48 hours after the first maternal dose of corticosteroids is administered.34 Conservative management of HELLP syndrome remains experimental and, for most women, the clinical course is too rapid to complete the steroid regimen before initiating delivery.33
Vaginal delivery is recommended for women with severe preeclampsia if there is no evidence of maternal or fetal compromise or other obstetric contraindication.1 Some experts recommend cesarean delivery for fetuses younger than 30 weeks when the cervix is not ripe, but a trial of induction may be considered.1,7 In patients with HELLP syndrome, cesarean delivery carries special risks, such as bleeding from thrombocytopenia and difficulty controlling blood pressure because of depleted intravascular volume.33,34
Postpartum Management. Most patients with preeclampsia respond promptly to delivery with decreased blood pressure, diuresis, and clinical improvement. Eclampsia may occur postpartum; the greatest risk of postpartum eclampsia is within the first 48 hours.43 Magnesium sulfate is continued for 12 to 24 hours, or occasionally longer if the clinical situation warrants. There are no reliable data on postpartum hypertensive management50; however, oral nifedipine is commonly used.7
Eclampsia
An eclamptic seizure may be preceded by increasingly severe preeclampsia, or it may appear unexpectedly in a patient with minimally elevated blood pressure and no proteinuria. Blood pressure is only mildly elevated in 30 to 60 percent of women who develop eclampsia.43 An eclamptic seizure usually lasts from 60 to 90 seconds, during which time the patient is without respiratory effort. A postictal phase may follow with confusion, agitation, and combativeness. The timing of an eclamptic seizure can be antepartum (53 percent), intrapartum (19 percent), or postpartum (28 percent).51 Late postpartum (more than 48 hours after delivery) onset of eclampsia was traditionally thought to be rare; however, a study of 29 cases of postpartum eclampsia demonstrated that 79 percent occurred in the late post-partum period.43,52
MANAGEMENT OF ECLAMPSIA
Initial management of an eclamptic seizure includes protecting the airway and minimizing the risk of aspiration by placing the woman on her left side, suctioning her mouth, and administering oxygen. A medical professional skilled in performing intubations should be immediately available.53 Close observation, soft padding, and use of side rails on the bed may help prevent trauma from falls or violent seizure activity. After the convulsion has ended and the patient is stabilized, plans should be made for prompt delivery. In rural or remote areas, physicians need to consider the risk of transfer versus the benefits of tertiary maternal and neonatal care.
It is important to avoid unnecessary interventions and iatrogenic complications.43,53 Magnesium sulfate is the drug of choice because it is more effective in preventing recurrent seizures than phenytoin (Dilantin) or diazepam (Valium).39,54–56 If a patient has already received a prophylactic loading dose of magnesium sulfate and is receiving a continuous infusion, an additional 2 g should be given intravenously. Otherwise, a 6-g loading dose is given intravenously over 15 to 20 minutes, followed by maintenance infusion of 2 g per hour. A total of 8 g of magnesium sulfate should not be exceeded over a short period of time.43,53
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The residues modulo m of products of random integers Citation for published version (APA): Baryshnikov, Y., & Stadje, W. (1999). The residues modulo m of products of random integers. (Report Eurandom; Vol. 99046). Eurandom.
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Download date: 29. Sep. 2025 Report 99-046 The Residues modulo m of Products of Random Integers Yuliy Baryshnikov, Wolfgang Stadje ISSN 1389-2355 The Residues modulo m of Products of Random Integers Yuliy Baryshnikov and Wolfgang Stadje University of Osnabriick Abstract For two (possibly stochastically dependent) random variables X and Y taking values in {O, ... ,m - 1} we study the distribution of the random residue U = XY mod m.
In the case of independent and uniformly distributed X and Y we provide an exact solution in terms of generating functions that are computed via p-adic analysis.
We show also that in the uniform case it is stochastically smaller than (and very close to) the uniform distribution. For general dependent X and Y we prove an inequality for the distance sUPxE[O,l] lFu(x) - xl.
1 Introduction Let X and Y be two (possibly dependent) random variables taking values in {O, 1, ... ,m - I}, where m 2:: 2 is some fixed integer. In this note we study the distribution of the random residue of the product U = XYmodm.
We consider first the case when X and Y are independent and uniformly distributed, i.e.
P(X = i, Y = j) = m-2 for i, j E {O, ... ,m - I}.
In Section 2 it is shown that the problem for general m can be reduced to that for m = pn, where p is some prime number and n E N, and that in this case it is sufficient to determine the cardinalities 1 We prove that for every prime number p the generating function Hp(T, Z) = ~ Np(l, n)TnZl of the double sequence Np(l, n) is given by n,l R (T Z) = (1 - pT)2(1 - p-lZ) - p2(1 - p-1 T)T(1 - Z) .
(1.1) p , (1 - Z)(1 - p-1Z)(1 - pT)2(1 - p2T) In the case p = 2 we derive a neat explicit formula for the distribution function of U. It is given by n-l P(U::; k) = (k + 1)2-n + 2-n+lL(1- 6i) i=O (1.2) for k = 0, ... ,2n - 1 , where 60,." ,6n-l E {O, 1} are the binary digits of k, defined by k = 60 + 261 + 462 +... + 2n - 18n _ 1.
It follows from (1.2) that the random 'fractional residue' 2-nU is stochastically smaller than a uniform random variable on [0,1), i.e. P(U/2n < u) ~ u for all u E [0, 1] and that the maximal deviation is given by sup (P(2-nU < u) - u) = (n + 2)2-(n+l) , 0<u9 (1.3) so that the distribution of 2-nU tends to the uniform distribution on [0,1] at an exponential rate (given by (1.3)), as n -+ 00. In fact, these stochastic dominance and convergence remain valid for arbitrary m.
The rest of the paper is devoted to an extension of this asymptotic equidistribution result to general m and dependent, non-uniform random variables X and Y.
We will show that ( IOgm) 1/2 sup IP(U/m < u) - ul ::; C --O<u<1 m (1.4) if the distribution of Y and the conditional distribution of X given Y do not deviate too much from uniformity and if the latter distribution satisfies a certain Lipschitz condition. Specifically, we assume that P(Y = k) ::; Co/m p(jlk) = P(X = j I Y = k) ::; Cdm I p(j 1I k) 11 C '" . 1/ p(j2Ik) -::; 2 JI - J2 m 2 for some constants Co, C1, C2 • Then (1.4) holds for a certain constant C which depends only on Co, C1 and C2 • From (1.4) we can conclude that U1m is for a large class of joint distributions of X and Y 'almost' uniformly distributed on [0,1] in the sense of weak convergence.
Deterministic sequences of integers whose residues are uniformly distributed are treated in Narkiewicz and Kuipers and Niederreiter . They play an important role in random number generation (Ripley ). In the realm of stochastic sequences already Dvoretzky and Wolfowitz studied weak convergence ofresidues for sums ofindependent, Z+-valued random variables; more recent papers on related questions are Brown , Barbour and Grubel , and Grubel .
The distribution of the fractional part of continuous random variables, in particular its closeness or convergence to the uniform distribution on [0,1), has been studied by many authors (e.g. Schatte , Stadje [14, 15], Qi and Wilms ).
2 The uniform case We start by deriving the exact probability distribution of U in the case m = 2n , n E N. For x E ll4 let frac(x) be the fractional part of x.
Proposition 1 We have n-1 P(U:::; k) = (k + 1)2-n+ 2-(n+1) L)l - 6i ), i=O (2.1) if m < n if m = n.
for every k E {a, 1, ... ,2n - I}, where 60 , ,6n- 1 E {O, ... ,n - I} are the binary digits of k, i.e. k = 6 0 + 261 + 462 + + 2n - 16n _ 1.
Proof. Obviously, 2n -1 P(U = k) = L 2-2n card{j E In I frac(ij2- n ) = k2-n}.
(2.2) i=O Let A _ { {i E In I i2-m is odd}, m -{O}, It is easily seen that d A { 2n-m-1, if mE {O, ... ,n -I} car m = 1, }'f m=n.
3 Consider i E Am and k E A! for some m, l E {O, ... ,n-1}, say i = (2p+1)2m and k = (2q + 1)2!. Then for any j E In, (2.3) is equivalent to (2p + 1)j - (2q + 1)2!-m = N2n- m for some integer N.
(2.4) For l < m the lefthand side of (2.4) is not integer, so there is no solution j of (2.3). Now let l ~ m. Since 2p + 1 and 2n are relatively prime, a simple result on residues implies that the numbers (2p + 1)j - (2q + 1)2!-m run through a complete set of residues mod 2n if j runs through (the complete set of residues) 0, 1, ... ,2n - 1. But N2n- m gives different residues mod 2n for N = 0, ... ,2m -1, while for larger values of N one only gets replications of these residues. Thus, the number of solutions j of (2.3) is 2n if l ~ m.
The same result also holds for m E As, Le. m = 0.
From (2.2) it now follows that if k E A! for some l < n we obtain n-l P(U = k2-n) = L 2-2n L card{j E In I int(ij2-n ) = k2-n} + 2-n8 ok m=O iEAm !
L 2-2n card(Am )2n m=O I L2-n2n-m-l m=O (l + 1)2-(n+l), (2.5) while if k E An, n-l P(U = 0) =L 2-2n card(Am )2n + 2-n m=O = (n + 2)2-(n+l).
(2.6) In particular, k I--t P(U = k) is constant on A! for every l. Therefore, the probability P(U E (2ma, 2ma+2m - 1]) is the same for every a E {O, ... ,2n- m_ 4 I}. It follows that P(U S k) -P(U = 0) + P(O < U < <5n_ 12n ) +~P (~O'2' <U ~ '~,0,2') n-l P(U = 0) +L P(O < Us <512').
1=0 (2.7) To compute the righthand side of (2.7), note that the number of integers i E Am satisfying 0 < i S 2' is equal to 21- m- 1 for m = 0, ... ,l-1 and equal to 1 for m = l. Hence, by (2.5), I P(O < U S 21) = L P(U E Am n {O, ." ,21}) m=O 1-1 L(l + 1)2-(n+l)21- m- 1 + (l + 1)2-(n+l) m=O _ 2-(n+l)(21+1 - 1).
Inserting (2.8) and (2.6) in (2.7) now yields (2.1).
(2.8) Proposition 2 1) For arbitrary m U is stochastically smaller than a uniform random variable on [0, 1]; 2) For arbitrary m sup (P(U < u) - u) = O(m-l+ f ), O 0; and 3) For m = 2n , sup (P(U < u) - u) = (n + 2)2-(n+l).
0<u9 Proof. We start with 1). It is clear that (2.9) (2.10) #{O s j < m: ijmodm S k} = gcd(i,m) (l t )J + 1).
(2.11) gcd ?',m 5 This implies 1 m-l (k ) P(U:::; k) = 2 Lgcd(i,m) L C )J + 1 > kim m i=O gcd z,m for all 0 :::; k < m, and hence proves 1).
Further, estimating (2.12) in an obvious way from above, we obtain (2.12) P(U:::; k) < ~2 E~~l gcd(i, m) (gCd~,m) + 1) < kim + -4 E~~l gcd(i, m) kim + ~2Ellm #{O :::; i < m : gcd(i, m) = l} (2.13) < kim + ~2 El1m l7 kim + d(m)/m, where d(m) denotes the number of divisors of m. It is known that d(m) = O(mE) for all € > 0, which implies 2).
To prove 3) define for 0 < u:::; 1 the integer k(u) by k(u)2-n < u :::; (k(u) + 1)2-n and let 00, ... ,On-l be its binary digits. By (2.1) we can write n-l P(U < u) - u = (k(u)2-n + 2-n - u) + 2-(n+l) L(1 - Oi), (2.14) i=O which is nonnegative by the definition of k(u). Further it is clear from (2.14) that sUPO<u9(P(U < u) - u) is approached as u 4- 0, yielding (2.10).
Now we derive the exact formulae for P(U = k) in the case of general mEN.
Let X and Y be independent and uniform on the set {a, ... ,m - 1}, which we identify with Z/mZ. Then P(U = a) is equal to m-2 times the number of solutions (x, y) E (Z/mZ) x (Z/mZ) of the equation xy= amodm.
Let m = Ilp~i be the prime factorization of m (pi primes, ni EN).
For a E Z/mZ we define a(i) E Z/p~iZ as the (unique) solution of a(i) = a modp~i.
Then as Z/mZ = Il(Z/p~iZ) (the Chinese remainder theorem), we have the following decomposition.
6 Lemma 1 The number of pairs (x, y) E (Z/mZ) x (Z/mZ) satisfying xy= amodm (2.15) is equal to the product of the numbers of solutions (x, y) E (Z/p~iZ) X (Z/p~iZ) of (2.16) By the Lemma, we only have to determine the number of solutions of (2.15) for m of the form m = pn.
Fix a prime number p and a natural number n. Observe first that the number of solutions (x, y) E (Z/pnz) x (ZpnZ) of xy = amodpn depends on a only through the p-adic norm of a, that is, through the exponent of the maximal power of p that divides a. Indeed, if there exists an invertible b in Z/pnz satisfying then #{(x,y) E (Z/pnZ) x (Z/pnZ) I xy= amodpn} =#((x, y) I xyb= pn-l modpn} =#((x, z) E (Z/pZ) x (Z/pZ) I xz= pn-l modpn} =Np(l,n).
To compute Np(l, n), we use the following well-known formula from the theory ofp-adic integration (Christol [4, Sect. 7.2.2, p. 466]). Let f(Xl,'" ,xr) be a polynomial with coefficients in Zp, the ring of p-adic integers, and let I. Ip denote the p-adic norm. Then for any real s > 0, !
If(Xl"" ,xr)I~Jl(dxl)'" Jl(dxr) = pS - (pS _l)Q(p-r-s), (2.17) (Zp)T where Jl is the Haar measure on Zp and Q(T) is a Poincare series: 00 Q(T) = LTk#{(Xl"" ,xr) E (Z/pkzy I f(xl"" ,xr)= Omodpk}.
k=O Theorem 1 The generating functions 00 00 n Gp,l(T) = LNp(Z,n)Tn, Hp(T,Z) = LLNp(Z,n)TnZl n=O n=O l=O 7 are given by H (T Z) = (1 - pT)2(1- p-1Z) - p2(1 - p-1 T)(1 - Z)T p , (1 - Z)(1 - p-1Z)(1 - pT)2(1 _ p2T) (2.19) Proof. We use formula (2.17) for r = 2 and f(x, y) = f,(X, y) = pixy. For the lefthand side of (2.17) we obtain f If,(X,y)l;tt(dx)tt(dy) = f p-1Ixl; Iyl; tt(dx)tt(dy) (Zp)2 (Zp)2 = p-l ([ Ixl; I'(dx)r By (2.17), f I I s (d) S (S 1) 1 1 - p-1 X ptt x = P -P -1 _ p-1-s = 1 _ p-1-s' Zp (Note that here Q(T) = 1/(1 - T), since #{x E 'llpnI'll I x 0 modpn} = 1 for all n). Furthermore, Thus, the coefficients on the righthand side of (2.17) are just the Np(l, n). It follows that Setting T = p-2-s, so that p-s = p2T we get 1 (1) (1- -1)2 p2T -p2T - 1 Gp,I(T) = p-l 1 _ ~T (2.20) and (2.18) follows from (2.20) by a short calculation. Similarly, multiplying (2.20) by Zl and summing over 1yields (2.19).
8 For example, if p = 2 the numbers Np(O, n) of solutions (x, y) of (x, y) 0 mod 2n is (n + 2)2n-t, as 00 (1 - 2T)2 - T G2,o(T) = ~ Np(O, n)T n = (1 _ 2T)2(1 _ 4T) 00 1 - T "'"'( ) n-ITn = (1 - 2T)2 = ~ n + 2 2 .
3 The inequality for dependent random variables We will now prove (1.4). For this we need some basic theory of continued fractions (see e.g. Hardy and Wright , Billingsley ) and a probability estimate due to Levy ).
Any x E [0, 1] has a continued fraction expansion x = [al (x), a2(x), ... ] providing a sequence of fractions usually denoted by For two positive numbers Po < PI let B(po, PI) = {x E [0,1] I Po < qk(X) < PI for some kEN}.
2po 1 Lemma 2 >'(B(po, PI)) ~ 1 -(1 + 2log2 Po) - PI .
PI - Po Proof. Let Q be the set of all finite sequences q= (qI,'" ,qk), kEN, of denominators ofpossible continued fraction expansions satisfying qk :::; Po. We set x(q) = Pk/qk, where Pk is the kth numerator corresponding to qI, ... ,qk, and I(q) = {x E [0,1] I (qI(X), ... ,qk(X)) = q} J(q) = I(q) n {x E [0,1] I qk+I(X) ~ PI or x = x(q)} J(O) = {x E [0,1] I qI(X) ~ pr}o The sets J(q), q E Q, and J(O) are pairwise disjoint intervals and B(po, PI) = [0,1]\ (J(O) u UJ(q)).
qEQ 9 Thus, A([O, l]\B(po, P1)) A(J(O)) + L A(J(q)) qEQ ko A(J(O)) +L L A(J(q)), k=l qEQ Iql=k (3.1) where Iql denotes the length of the sequence qand kois the maximum length of sequences in Q. Since > 2(k-1)/2 £ ( ) Q Po > qk _ or every q1,···, qk E , it follows that ko< 1+ 2log2 po.
(3.2) Now let U be a random variable that is uniformly distributed on [0,1]. Then if q E Q, Iql = k, it follows that A(J(q)) P(qk+l(U) ~ PI, U E I(q)) P(U E I(q))P(qk+l(U) ~ P11 U E I(q)) < P(U E I(q))P(ak+l(U) > P1 - po I U E I(q)) (3.3) -1 Po < P(U E I(q))2 (P1 ~ po) For the first inequality in (3.3) we have used the recursion qk+l = qkak+l+qk-1 which for q E Q,lql = k, implies that ak+l > (P1 - Po)/Po. The second inequality follows from a result of Levy [9, p. 296].
To estimate A(J(O)), note that q1(X) ~ Po implies that x ~ P1(X)/q1(X) = 1/Pl. Thus, by (3.1), (3.2) and (3.3).
A([O, 1]IB(Po,P1)) ~ pi1+ ko 2po L P(U E I(q)) P1 - Po --+ qEQ 1 2po ~ PI + (1 + 2log2 po) .
P1 - Po The Lemma is proved.
10 (3.5) Lemma 3 Let X be uniformly distributed on {O, 1, ... ,m - 1}. Then P(X/m (j. B(po, Pi)) :::; 2po(1 + 2log2Po) ( 1 + po) + pi 1 + m-1• Pi - Po m (3.4) Proof. For every half-open or open interval I in [0,1] we have IP(X/m E I) - .(1)1:::; m- 1• As J(O) and J(q) are half-open intervals, (3.1) and (3.4) yield P(X/m (j. B(po, pd) :::; .(J(O)) + L .(J(q)) qEQ (3.6) +m-1(1 + card Q).
It remains to find an upper bound for card Q. Let Q be the set of sequences in Q having maximal length, i.e., the set of those (ql(X), ... ,qk(X)) E Q for which qk+l(X) ~ Po. Since 1 1 1 .(I(ql" .. ,qk)) = qk(qk + qk-l) > 2q~ ~ 2p~ for (ql, . .. ,qk) E Q, we clearly have card Q < 2p~.
Inequality (3.4) now follows from (3.6), Lemma 2 and card Q :::; kocard Q< (1 + log2 Po) (2P5)' Lemma 4 Let p(j,k) = P(X =j, Y = k), j,k E {O, ... ,m-1} be the joint distribution of X and Y.
Assume that there are constants C1 and C2 such that p(jlk) = P(X = jlY = k) :::; Cdm I p(jllk) I ., 1/ p(j2Ik) - 1 :::; C21Jl - J2 m for allj,k,jl,h E {O, ... ,m-1}. Then IP(U/m < ulY = k) - ul :::; 3C2 + inf f (qn(~)) m n~l m 11 (3.7) (3.8) for all k E {O, ... ,m -I}, where f( ) - 3 (C1 + C2)q ~T q -- + , q E 1'1.
q m Proof. Let piq be an arbitrary fraction from the continued fraction expansion of kim. Let Ji = {(i - l)q, (i - l)q + 1, ... ,iq - I} Ji(u) = {j E Ji I frac (jklm) < u}, where frac(x) denotes the fractional part of x ~ O.
Then [m/q] P(Ulm < u) I Y = k) = L L P(X = j I Y = k) i=1 jEJi(U) + L P(X = j I Y = k) (3.9) kEJ[m/q]+l k<m 1 +11.
Clearly, (3.7) yields (3.10) Regarding the sum 1, we can write (3.11) where Ai = ~axp(jlk) and ai = rp.inp(jlk). l,From (3.8) we can conclude JEJi JEJi that (3.12) Obviously, card Ji = q. We need an upper bound for card Ji(u). Note that 12 For arbitrary j E Ji(u) write j = (i - l)q + h, where h E J1 ; we obtain ( k hk) frac(jk/m) = frac (i - l)qm + m = frac ((i -l)q~ +frac(~)) and frac ( ~) = frac (h (~ - ~) + h:) = frac ( a + h:) where lal < q-l. Recall that p and q are relatively prime. Thus, as h runs through J1 , frac( ':::) runs through the set of all values ~ + a, l E J 1• Let f3i = (i - l)qk/m.
Let Ji(U) be the number of values frac(f3i + (l/q)) in [0, u) for which l E J1.
Clearly, we have Ji(U) E {[qu], [qu] + I}. Since lal < q-l, it now follows easily that so that By (3.12) and (3.13), Ai card Ji(u) (1 C1q) qu + 3 C1q 3 3C2 ---~< +-<u+-+-+-.
ai card Ji -m q -m q m Inserting (3.14) and (3.10) in (3.9) we find that C2q 3 3C2 C1q P(U/m <u):::; u+ -+ - + -+-m q m m 3C2 =u+-+f(q)· m (3.13) (3.14) Minimizing with respect to all possible denominators q = qn(k/m) we arrive at 3C2 • (( k)) P(U/ m < u) - u :::; -+ mf f qn -.
m n2':1 m The analogous lower bound P(U/m < u) 2:: u - (3C2/m) - f(q) is derived along the same lines.
13 Theorem 2 Assume that the joint distribution ofX and Y satisfies conditions (3.7) and (3.8) and that P(Y = k) ~ Co/m, k = 0, ... ,m-1.
(3.15) for same constant co. Then there is a constant C depending only on Co, C1, C2 such that ( 10gm) 1/2 sup IP(U/m < u) - ul ~ C --.
o::;u::;I m Proof. By the formula of total probability and Lemma 4, we obtain (3.16) m-l P(U/m < u) = L P(Y = k)P(U/m < ulY = k) k=O < u + 3C,m-1 +~ P{Y = k) min [l,~~f (qn(~))] u+3C2m-1+E (min [l,~~rf (qn(:))]).
(3.17) Note that the right side of (3.17) is equal to Jo1(1 - G(x))dx, where Let C3 = C1 + C2. The function f(t) = 3t-1+ C3m-1t, t > 0, is strictly convex, has the unique minimum to = (3m/C3)1/2 and xo = f(to) = 2to 1.
Thus the equati on f(t) = x has no solution for x < xo and exactly two solutions t 1(x) < t2(x) for x > xo. If x > xo, a short calculation yields x 6C3 f(6/x) = f(mx/2C3 ) = -2 + -< x, mx and consequently t1(x) < 6/x < mx/2C3 < t2(X). These observations show that G(x) P(t1(x) < qn(Y/m) < t2(x) for some n E N} > P(6/x < qn(Y/m) < mx/2C3 for some n E N} (3.18) P(Y/m E B(6/x, mx/2C3 ).
l,From (3.15) and Lemma 3 it now follows that 1 - G(x) ~ H(x) + m-1, x E (0,1] 14 where the function H is defined by 2C ( 12C ) H(x) = _3 + 2Co (6/X)2m -1 + 2 3 (1 + 2logt(6/x)), x> xo.
mx mx -12C3 Thus, for any y E (xo, 1] we have the following estimate: E(min[1, f(qn(Y/m))]) =[(1-G(x)) dx:o Y+l H(x) dx.
(3.19) On (xo, (0) the function H(x) is positive and strictly decreasing from infinity at zero. FUrther, ( 36 12C3) 48 H(x) ~ 2 -2 +-2 (1 + 2log2(6/x)) ~ 12· -2' x E (xo, 1] (3.20) mx mx mx as Co ~ 1 and C3 ~ 1. Let Xl be the solution of H(x) = 1 in (xo, (0). For sufficiently large m we have Xl < 1 and then, by (3.20), Xl ~ max[12(C3/m)1/2, (576/m) 1/2].
Hence if Xl ~ X ~ 1, H(x) can be bounded as follows: 2C3 ( 36 12C3 ) / 2)) H(x) ~ -+ 2Co -2 + 2( (C / 2)) (1 + log2(36 Xl mx mx mx 1 -12 3 mX1 2C3 2Co ( 144) ~ -+ -2 36 + -C3 (1 + log2(36m/576)) mx mx 11 2C3 2Co ~ -+ -2(36 + 14C3)(10g2 m - 3).
mx mx For any y E [Xl, 1] we now find that 1 1 H( ) d 2C3 2Co(36 + 14C3)(log2 m - 3) y+ X x~y+-+ .
y my my Over y E (0, (0) the right-hand side of (3.21) is minimized for Yo = [2C3 + 2Co {36 + 14C3)(log2m - 3)jl/2m-1/2, (3.21) the corresponding minimum being equal to 2yo. A short calculation shows that H(yo) ~ (9 + 3C3)/(9 + 4C3) < 1, as m ~ 00.
Thus, Yo > Xl for sufficiently large m. Hence we may insert the value Yo in (3.21) for all but finitely many m. To summarize, it is now proved that JlOgm P(U/m < u) ~ u + C ----:;;;-15 for some constant C depending only on Co, Cl, and C2 . Similarly it can be shown that P(U/m < u) ~ u - C((1ogm)/m)1/2.
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Contents
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1 Properties
1.1 Proof
2 Sylvester's construction
2.1 Alternative construction
3 Hadamard conjecture
4 Equivalence and uniqueness
5 Special cases
5.1 Skew Hadamard matrices
5.2 Regular Hadamard matrices
5.3 Circulant Hadamard matrices
6 Generalizations
7 Practical applications
8 See also
9 Notes
10 Further reading
11 External links
Hadamard matrix
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From Wikipedia, the free encyclopedia
Mathematics concept
In mathematics, an Hadamard matrix, named after the French mathematician Jacques Hadamard, is a square matrix whose entries are either +1 or −1 and whose rows are mutually orthogonal. In geometric terms, this means that each pair of rows in a Hadamard matrix represents two perpendicular vectors, while in combinatorial terms, it means that each pair of rows has matching entries in exactly half of their columns and mismatched entries in the remaining columns. It is a consequence of this definition that the corresponding properties hold for columns as well as rows.
The n-dimensional parallelotope spanned by the rows of an n × n Hadamard matrix has the maximum possible n-dimensional volume among parallelotopes spanned by vectors whose entries are bounded in absolute value by 1. Equivalently, a Hadamard matrix has maximal determinant among matrices with entries of absolute value less than or equal to 1 and so is an extremal solution of Hadamard's maximal determinant problem.
Certain Hadamard matrices can almost directly be used as an error-correcting code using a Hadamard code (generalized in Reed–Muller codes), and are also used in balanced repeated replication (BRR), used by statisticians to estimate the variance of a parameter estimator.
Properties
[edit]
Let H be a Hadamard matrix of order n. The transpose of H is closely related to its inverse. In fact:
where In is the n × n identity matrix and HT is the transpose of H. To see that this is true, notice that the rows of H are all orthogonal vectors over the field of real numbers and each have length Dividing H through by this length gives an orthogonal matrix whose transpose is thus its inverse:
Multiplying by the length again gives the equality above. As a result,
where det(H) is the determinant of H.
Suppose that M is a complex matrix of order n, whose entries are bounded by |Mij| ≤ 1, for each i, j between 1 and n. Then Hadamard's determinant bound states that
Equality in this bound is attained for a real matrix M if and only if M is a Hadamard matrix.
The order of a Hadamard matrix must be 1, 2, or a multiple of 4.
Proof
[edit]
The proof of the nonexistence of Hadamard matrices with dimensions other than 1, 2, or a multiple of 4 follows:
If , then there is at least one scalar product of 2 rows which has to be 0. The scalar product is a sum of n values each of which is either 1 or −1, therefore the sum is odd for odd n, so n must be even.
If with , and there exists an Hadamard matrix , then it has the property that for any :
Now we define the matrix by setting . Note that has all 1s in row 0. We check that is also a Hadamard matrix:
Row 1 and row 2, like all other rows except row 0, must have entries of 1 and entries of −1 each. ()
Let denote the number of 1s of row 2 beneath 1s in row 1. Let denote the number of −1s of row 2 beneath 1s in row 1. Let denote the number of 1s of row 2 beneath −1s in row 1. Let denote the number of −1s of row 2 beneath −1s in row 1.
Row 2 has to be orthogonal to row 1, so the number of products of entries of the rows resulting in 1, , has to match those resulting in −1, . Due to (), we also have , from which we can express and and substitute:
But we have as the number of 1s in row 1 the odd number , contradiction.
Sylvester's construction
[edit]
Examples of Hadamard matrices were actually first constructed by James Joseph Sylvester in 1867. Let H be a Hadamard matrix of order n. Then the partitioned matrix
is a Hadamard matrix of order 2n. This observation can be applied repeatedly and leads to the following sequence of matrices, also called Walsh matrices.
and
for , where denotes the Kronecker product.
In this manner, Sylvester constructed Hadamard matrices of order 2k for every non-negative integer k.
Sylvester's matrices have a number of special properties. They are symmetric and, when k ≥ 1 (2k > 1), have trace zero. The elements in the first column and the first row are all positive. The elements in all the other rows and columns are evenly divided between positive and negative. Sylvester matrices are closely connected with Walsh functions.
Alternative construction
[edit]
If we map the elements of the Hadamard matrix using the group homomorphism , where is the additive group of the field with two elements, we can describe an alternative construction of Sylvester's Hadamard matrix. First consider the matrix , the matrix whose columns consist of all n-bit numbers arranged in ascending counting order. We may define recursively by
It can be shown by induction that the image of the Hadamard matrix under the above homomorphism is given by
where the matrix arithmetic is done over .
This construction demonstrates that the rows of the Hadamard matrix can be viewed as a length linear error-correcting code of rank n, and minimum distance with generating matrix
This code is also referred to as a Walsh code. The Hadamard code, by contrast, is constructed from the Hadamard matrix by a slightly different procedure.
Hadamard conjecture
[edit]
Unsolved problem in mathematics
Is there a Hadamard matrix of order 4k for every positive integer k?
More unsolved problems in mathematics
The most important open question in the theory of Hadamard matrices is one of existence. Specifically, the Hadamard conjecture proposes that a Hadamard matrix of order 4k exists for every positive integer k. The Hadamard conjecture has also been attributed to Paley, although it was considered implicitly by others prior to Paley's work.
A generalization of Sylvester's construction proves that if and are Hadamard matrices of orders n and m respectively, then is a Hadamard matrix of order nm. This result is used to produce Hadamard matrices of higher order once those of smaller orders are known.
Sylvester's 1867 construction yields Hadamard matrices of order 1, 2, 4, 8, 16, 32, etc. Hadamard matrices of orders 12 and 20 were subsequently constructed by Hadamard (in 1893). In 1933, Raymond Paley discovered the Paley construction, which produces a Hadamard matrix of order q + 1 when q is any prime power that is congruent to 3 modulo 4 and that produces a Hadamard matrix of order 2(q + 1) when q is a prime power that is congruent to 1 modulo 4. His method uses finite fields.
The smallest order that cannot be constructed by a combination of Sylvester's and Paley's methods is 92. A Hadamard matrix of this order was found using a computer by Baumert, Golomb, and Hall in 1962 at JPL. They used a construction, due to Williamson, that has yielded many additional orders. Many other methods for constructing Hadamard matrices are now known.
In 2005, Hadi Kharaghani and Behruz Tayfeh-Rezaie published their construction of a Hadamard matrix of order 428. As a result, the smallest order for which no Hadamard matrix is presently known is 668.
By 2014, there were 12 multiples of 4 less than 2000 for which no Hadamard matrix of that order was known. They are: 668, 716, 892, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948, and 1964.
Equivalence and uniqueness
[edit]
Two Hadamard matrices are considered equivalent if one can be obtained from the other by negating rows or columns, or by interchanging rows or columns. Up to equivalence, there is a unique Hadamard matrix of orders 1, 2, 4, 8, and 12. There are 5 inequivalent matrices of order 16, 3 of order 20, 60 of order 24, and 487 of order 28. Millions of inequivalent matrices are known for orders 32, 36, and 40. Using a coarser notion of equivalence that also allows transposition, there are 4 inequivalent matrices of order 16, 3 of order 20, 36 of order 24, and 294 of order 28.
Hadamard matrices are also uniquely recoverable, in the following sense: If an Hadamard matrix of order has entries randomly deleted, then with overwhelming likelihood, one can perfectly recover the original matrix from the damaged one. The algorithm of recovery has the same computational cost as matrix inversion.
Special cases
[edit]
Many special cases of Hadamard matrices have been investigated in the mathematical literature.
Skew Hadamard matrices
[edit]
A Hadamard matrix H is skew if A skew Hadamard matrix remains a skew Hadamard matrix after multiplication of any row and its corresponding column by −1. This makes it possible, for example, to normalize a skew Hadamard matrix so that all elements in the first row equal 1.
Reid and Brown in 1972 showed that there exists a doubly regular tournament of order n if and only if there exists a skew Hadamard matrix of order n + 1. In a mathematical tournament of order n, each of n players plays one match against each of the other players, each match resulting in a win for one of the players and a loss for the other. A tournament is regular if each player wins the same number of matches. A regular tournament is doubly regular if the number of opponents beaten by both of two distinct players is the same for all pairs of distinct players. Since each of the n(n − 1)/2 matches played results in a win for one of the players, each player wins (n − 1)/2 matches (and loses the same number). Since each of the (n − 1)/2 players defeated by a given player also loses to (n − 3)/2 other players, the number of player pairs (i, j) such that j loses both to i and to the given player is (n − 1)(n − 3)/4. The same result should be obtained if the pairs are counted differently: the given player and any of the n − 1 other players together defeat the same number of common opponents. This common number of defeated opponents must therefore be (n − 3)/4. A skew Hadamard matrix is obtained by introducing an additional player who defeats all of the original players and then forming a matrix with rows and columns labeled by players according to the rule that row i, column j contains 1 if i = j or i defeats j and −1 if j defeats i. This correspondence in reverse produces a doubly regular tournament from a skew Hadamard matrix, assuming the skew Hadamard matrix is normalized so that all elements of the first row equal 1.
Regular Hadamard matrices
[edit]
Regular Hadamard matrices are real Hadamard matrices whose row and column sums are all equal. A necessary condition on the existence of a regular n × n Hadamard matrix is that n be a square number. A circulant matrix is manifestly regular, and therefore a circulant Hadamard matrix would have to be of square order. Moreover, if an n × n circulant Hadamard matrix existed with n > 1 then n would necessarily have to be of the form 4u2 with u odd.
Circulant Hadamard matrices
[edit]
The circulant Hadamard matrix conjecture, however, asserts that, apart from the known 1 × 1 and 4 × 4 examples, no such matrices exist. This was verified for all but 26 values of u less than 104.
Generalizations
[edit]
One basic generalization is a weighing matrix. A weighing matrix is a square matrix in which entries may also be zero and which satisfies for some w, its weight. A weighing matrix with its weight equal to its order is a Hadamard matrix.
Another generalization defines a complex Hadamard matrix to be a matrix in which the entries are complex numbers of unit modulus and which satisfies H H = n In where H is the conjugate transpose of H. Complex Hadamard matrices arise in the study of operator algebras and the theory of quantum computation. Butson-type Hadamard matrices are complex Hadamard matrices in which the entries are taken to be qth roots of unity. The term complex Hadamard matrix has been used by some authors to refer specifically to the case q = 4.
Practical applications
[edit]
Olivia MFSK – an amateur-radio digital protocol designed to work in difficult (low signal-to-noise ratio plus multipath propagation) conditions on shortwave bands.
Balanced repeated replication (BRR) – a technique used by statisticians to estimate the variance of a statistical estimator.
Coded aperture spectrometry – an instrument for measuring the spectrum of light. The mask element used in coded aperture spectrometers is often a variant of a Hadamard matrix.
Feedback delay networks – Digital reverberation devices which use Hadamard matrices to blend sample values
Plackett–Burman design of experiments for investigating the dependence of some measured quantity on a number of independent variables.
Robust parameter designs for investigating noise factor impacts on responses
Compressed sensing for signal processing and under-determined linear systems (inverse problems)
Quantum Hadamard gate for quantum computing and the Hadamard transform for quantum algorithms.
See also
[edit]
Combinatorial design
Hadamard transform
Quincunx matrix
Walsh matrix
Weighing matrix
Quantum logic gate
Notes
[edit]
^ "Hadamard Matrices and Designs" (PDF). UC Denver. Retrieved 11 February 2023.
^ J.J. Sylvester. Thoughts on inverse orthogonal matrices, simultaneous sign successions, and tessellated pavements in two or more colours, with applications to Newton's rule, ornamental tile-work, and the theory of numbers. Philosophical Magazine, 34:461–475, 1867
^ Hedayat, A.; Wallis, W. D. (1978). "Hadamard matrices and their applications". Annals of Statistics. 6 (6): 1184–1238. doi:10.1214/aos/1176344370. JSTOR 2958712. MR 0523759..
^ Hadamard, J. (1893). "Résolution d'une question relative aux déterminants". Bulletin des Sciences Mathématiques. 17: 240–246.
^ Paley, R. E. A. C. (1933). "On orthogonal matrices". Journal of Mathematics and Physics. 12 (1–4): 311–320. doi:10.1002/sapm1933121311.
^ Baumert, L.; Golomb, S. W.; Hall, M. Jr. (1962). "Discovery of an Hadamard Matrix of Order 92". Bulletin of the American Mathematical Society. 68 (3): 237–238. doi:10.1090/S0002-9904-1962-10761-7. MR 0148686.
^ Williamson, J. (1944). "Hadamard's determinant theorem and the sum of four squares". Duke Mathematical Journal. 11 (1): 65–81. doi:10.1215/S0012-7094-44-01108-7. MR 0009590.
^ Kharaghani, H.; Tayfeh-Rezaie, B. (2005). "A Hadamard matrix of order 428". Journal of Combinatorial Designs. 13 (6): 435–440. doi:10.1002/jcd.20043. S2CID 17206302.
^ Đoković, Dragomir Ž; Golubitsky, Oleg; Kotsireas, Ilias S. (2014). "Some new orders of Hadamard and Skew-Hadamard matrices". Journal of Combinatorial Designs. 22 (6): 270–277. arXiv:1301.3671. doi:10.1002/jcd.21358. S2CID 26598685.
^ Wanless, I.M. (2005). "Permanents of matrices of signed ones". Linear and Multilinear Algebra. 53 (6): 427–433. doi:10.1080/03081080500093990. S2CID 121547091.
^ Kline, J. (2019). "Geometric search for Hadamard matrices". Theoretical Computer Science. 778: 33–46. doi:10.1016/j.tcs.2019.01.025. S2CID 126730552.
^ Reid, K.B.; Brown, Ezra (1972). "Doubly regular tournaments are equivalent to skew hadamard matrices". Journal of Combinatorial Theory, Series A. 12 (3): 332–338. doi:10.1016/0097-3165(72)90098-2.
^ Turyn, R. J. (1965). "Character sums and difference sets". Pacific Journal of Mathematics. 15 (1): 319–346. doi:10.2140/pjm.1965.15.319. MR 0179098.
^ Turyn, R. J. (1969). "Sequences with small correlation". In Mann, H. B. (ed.). Error Correcting Codes. New York: Wiley. pp. 195–228.
^ Schmidt, B. (1999). "Cyclotomic integers and finite geometry". Journal of the American Mathematical Society. 12 (4): 929–952. doi:10.1090/S0894-0347-99-00298-2. hdl:10356/92085. JSTOR 2646093.
^ Geramita, Anthony V.; Pullman, Norman J.; Wallis, Jennifer S. (1974). "Families of weighing matrices". Bulletin of the Australian Mathematical Society. 10 (1). Cambridge University Press (CUP): 119–122. doi:10.1017/s0004972700040703. ISSN 0004-9727. S2CID 122560830.
Further reading
[edit]
Baumert, L. D.; Hall, Marshall (1965). "Hadamard matrices of the Williamson type". Math. Comp. 19 (91): 442–447. doi:10.1090/S0025-5718-1965-0179093-2. MR 0179093.
Georgiou, S.; Koukouvinos, C.; Seberry, J. (2003). "Hadamard matrices, orthogonal designs and construction algorithms". Designs 2002: Further computational and constructive design theory. Boston: Kluwer. pp. 133–205. ISBN 978-1-4020-7599-5.
Goethals, J. M.; Seidel, J. J. (1970). "A skew Hadamard matrix of order 36". J. Austral. Math. Soc. 11 (3): 343–344. doi:10.1017/S144678870000673X. S2CID 14193297.
Kimura, Hiroshi (1989). "New Hadamard matrix of order 24". Graphs and Combinatorics. 5 (1): 235–242. doi:10.1007/BF01788676. S2CID 39169723.
Mood, Alexander M. (1964). "On Hotelling's Weighing Problem". Annals of Mathematical Statistics. 17 (4): 432–446. doi:10.1214/aoms/1177730883.
Reid, K. B.; Brown, E. (1972). "Doubly regular tournaments are equivalent to skew Hadamard matrices". J. Combin. Theory Ser. A. 12 (3): 332–338. doi:10.1016/0097-3165(72)90098-2.
Seberry Wallis, Jennifer (1976). "On the existence of Hadamard matrices". J. Comb. Theory A. 21 (2): 188–195. doi:10.1016/0097-3165(76)90062-5.
Seberry, Jennifer (1980). "A construction for generalized hadamard matrices". J. Statist. Plann. Infer. 4 (4): 365–368. doi:10.1016/0378-3758(80)90021-X.
Seberry, J.; Wysocki, B.; Wysocki, T. (2005). "On some applications of Hadamard matrices". Metrika. 62 (2–3): 221–239. doi:10.1007/s00184-005-0415-y. S2CID 40646.
Spence, Edward (1995). "Classification of hadamard matrices of order 24 and 28". Discrete Math. 140 (1–3): 185–242. doi:10.1016/0012-365X(93)E0169-5.
Yarlagadda, R. K.; Hershey, J. E. (1997). Hadamard Matrix Analysis and Synthesis. Boston: Kluwer. ISBN 978-0-7923-9826-4.
External links
[edit]
Skew Hadamard matrices of all orders up to 100, including every type with order up to 28;
"Hadamard Matrix". in OEIS
N. J. A. Sloane. "Library of Hadamard Matrices".
On-line utility to obtain all orders up to 1000, except 668, 716, 876 & 892.
R-Package to generate Hadamard Matrices using R
JPL: In 1961, mathematicians from NASA’s Jet Propulsion Laboratory and Caltech worked together to construct a Hadamard Matrix containing 92 rows and columns
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C Program to Convert Binary Number to Decimal and vice-versa
Updated on 13 December, 2024Associated Doc
Introduction
The conversion between binary and decimal number systems is a fundamental concept in computer science, crucial for understanding how computers process and store data. Binary numbers, consisting solely of 0s and 1s, are native to machine-level operations, while decimal numbers are used by humans for more intuitive understanding and interaction.
In this article, you will learn how to implement conversions between binary and decimal numbers in C programming. By examining detailed code examples, gain insights into the logic and methods for performing these conversions efficiently.
Converting Binary to Decimal
Understand the Conversion Process
Binary to decimal conversion involves reading a binary number (base-2) and transforming it into its decimal (base-10) equivalent. This process relies on understanding the weight (power of 2) of each binary digit based on its position.
Example: Binary to Decimal Conversion
Declare variables to store the binary number, decimal number, and the base exponent.
Use a loop to process each digit of the binary number from the least significant digit (rightmost) to the most significant digit (leftmost).
Extract each binary digit and multiply it by the appropriate power of two.
c
#include #include int binaryToDecimal(long long binaryNumber) { int decimalNumber = 0, i = 0, remainder; while (binaryNumber != 0) { remainder = binaryNumber % 10; binaryNumber /= 10; decimalNumber += remainder pow(2, i); ++ i; } return decimalNumber;} int main() { long long binaryNumber; printf("Enter a binary number: "); scanf("%lld", & binaryNumber); printf("Decimal number is: %d \n ", binaryToDecimal(binaryNumber)); return 0;}
In this code, binaryToDecimal function calculates the decimal value by adding the products of binary digits and their corresponding powers of two. The while loop continues until the binary number diminishes to zero.
Converting Decimal to Binary
Understand the Conversion Process
Converting a decimal number to binary involves dividing the decimal number by 2 and recording the remainder for each division operation until the number is reduced to 0. The remainders constitute the binary representation in reverse order.
Example: Decimal to Binary Conversion
Declare variables to store the decimal number and the binary equivalent.
Implement a loop to perform division by 2 and collect remainders.
Store or display the binary number in the correct order.
c
#include void decimalToBinary(int decimalNumber) { int binaryNumber, i = 0; while (decimalNumber > 0) { binaryNumber[i] = decimalNumber % 2; decimalNumber /= 2; i ++; } for (int j = i - 1; j >= 0; j --) { printf("%d", binaryNumber[j]); }} int main() { int decimalNumber; printf("Enter a decimal number: "); scanf("%d", & decimalNumber); printf("Binary number is: "); decimalToBinary(decimalNumber); printf(" \n "); return 0;}
This snippet utilizes an array binaryNumber to store binary digits. The conversion loop generates binary digits, stored in reverse, which are then printed in the correct sequence starting from the highest index.
Conclusion
Through clear examples, mastering the conversion of binary to decimal and vice versa in C becomes attainable. These fundamental operations are not only essential for various computing tasks but also serve as a great exercise for enhancing problem-solving and programming skills in C. Incorporate these conversions in projects where binary data manipulation is required, ensuring efficient data processing and manipulation.
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Algebraic Manipulation A. C. Norman January 1994 1 Introduction These notes cover the Lent Term lecture course on Computer Algebra. There are now a number of plausible textbooks that cover the relevant material too, but some of them have quite ridiculous prices while others key themselves very strongly to one particular (perhaps commercial) algebra system, so I find it hard to produce a clear-cut and simple recommendation on that front. I will mention some of the options in the lectures but have tried to collect the topics that need most careful documentation here. In parts of the course where I think that the lectures them-selves can give adequate coverage these printed notes may degenerate to section headings or cryptic comments.
There are a number of different algebra systems available these days, and it can be useful to try examples out on one of them. This can both give you an un-derstanding of why the various algebraic algorithms discussed here are important, and how the application of computer algebra feels. It should not matter much which system you use, but it is useful to have one single concrete syntax for ex-amples included here. I have used REDUCE, since that is both a system that I know well and it is also available to all of you both on Phoenix and CUS. To use REDUCE on CUS you will need to set up some Unix environment variables be-fore starting the system itself — the “man reduce” entry explains what has to be done. Phoenix users should try “help reduce” to ensure that they have up to date instructions for loading the system. REDUCE can also run on almost any other style or brand of computer you can think of, and there is a free demonstration ver-sion for MSDOS which is not capable of solving huge problems but which may be helpful for the small-scale investigations needed in support of this course. A full REDUCE manual is available on-line, but you almost certainly do not want to print it all out, so to help with experimentation and getting a feel for the system I include a glossary of the operators, keywords and switches that are provided.
People who have their own copies of Maple, Axiom, Macsyma, Mathematica or Derive will find that the same sorts of fundamental issues come up with those system as do with REDUCE, and although the details of how things are expressed will differ most of this course should still be relevant. Just as numerical analysis does not depend (much) on whether the processes are coded in Fortran, C or ADA, there is much useful that can be said about computer algebra that transcends the differences between particular packages.
This course covers Computer Algebra from two rather different perspectives.
The first views it as an area in which there are many interesting and unexpected algorithms to be described. It can thus be viewed as a specialised extension of 1 the earlier “Data Structures and Algorithms” course illustrating the challenges of one particular application domain. In this context the nature of Computer Algebra means that it can involve looking at algorithmic interpretations or uses of what had previously seemed to be inapplicable pure mathematical results. A signifi-cant amount of essential material on the algorithms needed in computer algebra is included in Knuth’s “Art of Computer Programming” (mainly Volume 2 on semi-numerical algorithms). Despite the age of that work it is still very readily available and its explanation of (for example) the polynomial GCD problem is quite sufficient for a first course.
The second thread in this course is in applications. As algebra systems be-come more widely available1 many calculations which would previously have been attacked by pure numerical methods will be treated symbolically. When this succeeds the results will frequently be more general or more reliable then the results of numerical analysis. With this in view this course includes some basic ideas about the proper use of algebra systems, analogous to the hints that will be incorporated in a first course on numerical analysis. Perhaps the one overview remark in this context is that algebra systems are generally expected to produce exactly correct values — and the big problem with numerical analysis is the con-trol or (rounding) error. So how can anything possibly go wrong if you work algebraically? The answer is that clumsy application of an algebra system can use all the memory and all the CPU time you have an still not produce a result, even in cases where alternative approaches could turn out results in very short order.
2 Computing with Polynomials 2.1 Introduction A common exercise in the use of programming languages that support dynamic storage allocation and records is that of representing and differentiating algebraic formulae. Formulae are represented directly as parse trees. Addition, subtraction, multiplication and so on just create new tree nodes, and substitution, differen-tiation and some other obviously useful operations are rather easy to code up.
Getting a user’s input into parse tree form is just a direct application of material from a course on compiler construction. Displaying a formula to typeset quality can be harder, but is fundamentally a problem in text handling or graphics. Parse 1For at least the last 20 years algebra systems have been (as reported by their designers) on the verge of massive wide-scale acceptance.
2 trees can represent all sorts of formulae almost equally well — nodes can stand for logarithms and trigonometric functions, or summations or integrals. But it is very difficult to answer questions about formulae (for instance to detect if they are zero!) or guarantee to simplify them properly.
A very good start towards a proper development of computer algebra is to con-centrate on a very limited class of expressions — polynomials. For these (made up out of coefficients 2 and indeterminates using the operations of addition and multiplication) everything can be made well-behaved. And it even turns out that many real-world applications of computer algebra require nothing more!
2.2 Canonical and Normal forms A Canonical Form for a class of expressions is one that formulae in the relevant class can be converted into such that any two equivalent formulae get mapped into exactly the same structure. Note that this may not be exactly the same as a naive interpretation of the word “simplificaton”. For instance a canonical form for polynomials can be obtained by multiplying out all brackets, collecting terms and then arranging the remaining terms in some standard order. But for a formula such as (a + b + c)10 performing these steps causes a small and well-structured expression to expand into something large and rather ugly. Despite this a good basis for a polynomial algebra system will involve a canonical form.
A weaker idea than that of a canonical form is a normal form. Here the only requirement is that when transformed any expression that is equivalent to zero gets mapped onto the object “0”. Note that with a normal form it is possible to make reliable comparisons (for equality) between pairs of expressions: to compare u and v just form u −v and reduce it to its normal form — if the result is visibly “0” the two original objects were equal. Given such an equality test theoreticians will assert that a canonical form can be produced. One enumerates all valid ex-pressions (is is not hard to show that there are only a countable number) and maps each expression onto the first item in this enumeration that it is equal to. Such a recipe is not very practical.
2usually just integers, but it is also reasonable, and sometimes very useful to support rational numbers.
3 2.3 Distributed and Recursive representations If you multiply (a + b + c)2 out you can write the result as 1 1a2 + 2 1ab + 2 1ac + 1 1b2 + 2 1bc + 1 1c2 where I have sorted the terms lexographically. Each term has a coefficient (which will be shown here as a rational number) and may mention several variables each raised to some integer power. This representation is known as distributed. With distributed representations it is often useful to suppose that there are just a few indeterminates and that their names are known in advance — then a term just has to record the exponents associated with each indeterminate. An exponent of zero allows for variables not visibly present. So for instance the first two terms of the above expression could be represented as records: [1, 1 : 2, 0, 0] % (1/1) aˆ2 bˆ0 cˆ0 [2, 1 : 1, 1, 0] % (2/1) aˆ1 bˆ1 cˆ0 and complete polynomials are now just lists of terms. The representation is clearly easy to implement and work with.
An alternative scheme would view (a + b + c)2 firstly as a polynomial in just a, which expands out to 1a2 + (2(b + c))a1 + (b + c)2a0.
Concentrating just on the top level of this structure it is made up out of terms each of which has a coefficient, a main variable and a degree. The trick is that the coefficients are now not just integers but they can be polynomials in variables that are “less important” that a. The benefit of this recursive representation is that many operations can be coded as if they were just working on univariate polynomials (an easier case), allowing the power of recursion into the coefficients deal with all the subsidiary variables.
If the terms in a polynomial are kept sorted then (in either case) adding poly-nomials is (almost) the same as merging lists as used in merge-sort.
Integration and differentiation of polynomials are basically easy term-by-term operations.
2.4 Polynomial multiplication: unsorted representations When one looks at polynomial multiplication it is slightly harder. Multiplication by a single term is very easy. So multiplication by a general polynomial can be 4 achieved by multiplying by each of the terms in it and adding the sub-results so generated. The bad news is that (unexpectedly, perhaps) this turns out to have a bad worst case cost. If the two input polynomials have n and m terms in them then naive multiplication can have cost proportional to m2n. To show this is unreasonable note that there are at worst mn terms in the product, so all the terms can be generated in mn stages, then ensuring that these are collected into the correct order can be done in mn log(mn) steps.
The bottleneck is ensuring that the product polynomial is created with its terms properly sorted. One way to respond to this challenge is to accept that the terms that make up a polynomial will not be kept in any special order. But now com-bining like terms on addition in a reasonably fast way becomes a non-obvious process. The lectures will explain an unordered merge algorithm (based on the use of hash tables) that can leave addition of unsorted polynomials fast while re-ducing the cost of multiplication to around mn steps. The main penalty of using it is that polynomials when printed out will be presented with their terms all jumbled up, or maybe extra sorting effort will be required in the print routines.
3 Computing with Truncated Power Series 3.1 Introduction Many mathematical and physical problems do not have neat solutions in closed form, but do have solutions that can be expanded as series in terms of one of the variables present in the problem. Even when closed form solutions do exist, these series solutions will still usually be easier to find and almost as useful: this sec-tion is concerned with the application of algebra systems to this task. There are many techniques available for deriving series solutions to equations, and it should be expected that whichever of these is used the same results will be produced.
Examples given here will show, however, that there can be dramatic differences in the amounts of computer time and memory that will be consumed when differ-ent approaches are taken to the solution of any given problem. There can also be large differences in the complexity of the mathematical and computational conse-quences of following different routes towards a solution, so a user has to develop an awareness of the various possibilities for compromise between simple but slow algorithms and elaborate and delicate but fast ones. When it is only necessary to produce the first few terms in the series expansion of some fairly simple function it will usually be appropriate to use the most direct method available, but even quite 5 modest looking problems can consume quite unreasonable amounts of computer time if attacked in too clumsy a way, and for large problems it will be essential to consider efficiency when planning solution techniques.
Even though proper power series are infinite in extent all useful computation with them will work with just their first few terms. The treatment of power series given here will treat the expansions as formal ones, that is to say it will not concern itself with the question as to whether any given series converges, or for what range of its independent variable it gives how accurate an approximation to some true result. Indeed there are some calculations that can quite properly be performed using series without any concern about convergence: for instance when the values of individual coefficients in the series are more interesting than the behaviour of the series as a whole 3. Although truncated series are only approximations to the function that they represent, the coefficients in them are expected to be calculated exactly using exact integer and fractional arithmetic: in this way the algebraic approximation of a series has a different character to numerical schemes that use floating point arithmetic and where it is hard to produce formal descriptions of the introduced error.
The first example problem given here is somewhat artificial, but will make it possible to introduce a number of the facilities of the Reduce algebra system. It is to find the coefficient of x20 in the series expansion of (1 + x)100. There are a number of ways of attacking this problem. With access to an algebra system the easiest is probably just to display (1 + x)100 with brackets multiplied out and pick out the term in x20 by eye. The request to Reduce is just: (1 + x)ˆ100; and the displayed result covers several pages. The next refinement would be to get Reduce to select out the required coefficient from the full formula. This can be done using the built-in selector function coeffn, which takes three arguments: an expression to inspect, the name of a variable and the degree in that variable of the coefficient required. Thus the desired value can be displayed by: coeffn((1 + x)ˆ100, x, 20); A rather different technique for solving our problem makes recourse to the binomial theorem and an assertion that the desired value is just the binomial co-efficient 100C20. To compute this we can define two Reduce procedures, one for factorials and the second for binomial coefficients: 3eg. when the series is thought of as the expansion of some generating function.
6 procedure fact n; for i := 1:n product i; procedure binom(n, r); fact n / (fact r fact(n-r)); binom(100, 20); This produces results noticeably more cheaply than the direct selection of the x20 term out of the fully expanded formula, but at the cost of requiring much more mathematical knowledge and of being a method that is not obviously adaptable to solving related problems, such as finding the coefficient of x20 in (1 + x + x2)70.
Even this scheme provides for further variations. The definition of binom(n,r) given above is computationally wasteful and can be rewritten as procedure binom1(n, r); if (2r > n) then (for i := r+1:n product i) / fact(n-r) else (for i := n-r+1:n product i) / fact r; which makes an attempt to avoid multiplying factors into the result if they will subsequently be divided out. The definition binom1 is slightly faster than binom, but it will only make enough difference to be noticed in cases where very heavy use is made of it. Conversely the same values can be defined in terms of a basic identity satisfied by binomial coefficients: procedure binom2(n, r); if r=0 or r=n then 1 else binom2(n - 1, r - 1) + binom2(n - 1, r); As n increases the cost of using procedure binom2(n,r) grows very rapidly.
Calculating binom2(100,20) would certainly require that the binom2 was called with either r=0 or r=n a number of times equal to the value eventually to be returned, i.e. 535983370403809682970. On the fastest Reduce implementa-tions on large mainframes it could be that a million such exits could be taken per second, in which case the entire calculation would complete in about 17 million years. Even binom2(20,10) a very substantial calculation.
As a final attack on the original problem it is worth considering another math-ematical result that states that the terms in a series expansion are related to the derivatives of the function being expanded. In Reduce the derivative of y with 7 respect to x is obtained by writing df(y,x), and the nth derivative can be re-quested as df(y,x,n). Evaluating an expression at a value v for the variable x is performed using a construct of the form sub(x=v,). Us-ing these we can obtain the coefficient we require by writing just: sub(x=0, df((1 + x)ˆ100, x, 20)) / fact 20; The method shown earlier using coeffn was only applicable if the expres-sion being decomposed was built up using just the operators +, −and ∗. This scheme using differentiation is a little more expensive but can be applied to a much larger range of functions.
3.2 Simple iterative methods When series expansions are produced it is almost invariably the case that inter-mediate calculations generate terms of much higher degree than will be required in the final result. It makes sense to discard such unwanted terms as early in the course of a computation as possible. In Reduce if terms in x with degree n or more are to be ignored a directive of the form let xˆn = 0; should be issued. This indicates that until further notice any terms involving xn or higher powers of x should be suppressed. If at some later stage in a calculation this action is not required the rule introduced by the “LET” statement can be removed by clear xˆn; In the code given earlier to find the coefficient of x20 in a large formula it would have been sensible to issue let xˆ21 = 0; before calculating (1 + x)100 thereby avoiding a large proportion of the work that want into finding high degree terms in the full expansion. When Reduce is used to compute with power series in some variable, x say, the readability of the results it produces is often enhanced by setting some output control flags by saying: factor x; on div,revpri; 8 These do not have any effect on Reduce’s internal workings or the values of the algebraic results that it produces: they just control the format in which results are displayed. Their effects can be cancelled to return to Reduce’s default print style by remfac x; off div,revpri; With these flags set and a truncation established so that powers of x higher than (say) the 10th will not appear it is possible to consider some techniques for exploiting power series arithmetic.
The first of these to be considered is Picard’s method for solving ordinary dif-ferential equations, which will be illustrated using the equation t′ = 1 + t2 (which of course has the exact solution t = tan(x) provided t(0) = 0). Picard’s method develops a solution to the equation starting from some initial approximation by repeatedly substituting its approximation into the right hand side of the equation and integrating to get a new approximation: let xˆ11 = 0; factor x; on div; tt := 0; for i := 1 : 6 do write tt := int(1 + ttˆ2, x); Running this program4 will cause Reduce to display a succession of values for the power series, and it will be observed that the initial terms in the series stabilise rapidly, and that in this particular case each integration leads to one more correct term in the series. Note that the above code relied on the Reduce integration oper-ator int handing back a result with a zero constant of integration. It is also worth noting how for even rather simple calculations the rational number coefficients that get generated can become quite complicated.
For an example differential equation where the required solution has a nonzero value at x = 0 consider r′ = −r2g′ for some known power series g (here we will use g = 1 + x). By choosing r(0) = 1/g(0) this has an exact solution r = 1/g.
Assuming that a suitable let statement is in force, the Reduce code is g := 1 + x; 4The code uses tt rather than just t as the variable since REDUCE reserves plain t and will get upset if you try to use it here.
9 r := r0 := 1 / sub(x=0, g); for i := 1:10 do r := r0 + int(-rˆ2df(g, x), x); where it should be remarked that sub(x=,) makes the specified substitution for x in the given expression (used here to evaluate g at x=0), and df(g,x) stands for the derivative of g with respect to x. With g set up as shown this program computes the series 1 −x + x2 −x3 + . . . which is perhaps not very exciting. However if its initial line is altered it can compute an expansion of the reciprocal of any power series despite the fact that Reduce does not have a built in capability for series division, and this is worthy of note. As before this Picard iteration exhibits first order convergence: to obtain n correct terms in the results it is necessary to perform the integration step n times.
In cases where the function defined by a differential equation is known, as in the two examples given so far, the terms in a series expansion can be found by differentiation, as in tt := for i := 0:10 sum (sub(x=0, df(tan x, x, i)) / fact i); which should generate the same series for tt as was produced earlier. In many cases it turns out that the Picard iteration produces results faster even though it at first seems to be doing more work. For instance a test on the tan(x) expan-sion finding terms up to degree 20 in x showed Picard’s method to be about 50% faster than repeated differentiation. Obviously the cost difference will vary greatly from problem to problem, and in some cases it will involve order of magnitude differences.
The important thing in the above iterations is not the presence of an integration operator, but the fact that if an expansion correct to order n in x is substituted into the right hand side result will be an expansion correct to some higher order. This can be achieved by multiplication by x as well as by integration, as in r := 1; for i := 1:10 do write r := 1 + xr; which provides another way to compute 1/(1 −x), and which can easily be ad-justed to expand other quotients in series form. Iterations of this form can often be derived by just separating an equation into a set of leading terms that do not depend on x and a correction factor that does. Even in cases where this does not at first seem possible a little rearrangement can help: consider the problem of ex-panding √1 + x as a power series. To find an iterative formula for deriving the 10 expansion it is necessary to put in a leading term for the expansion, and so express the square root in the form 1 + xq. Then the fact that this is the square root of 1 + x amounts to the identity (1 + q)2 = 1 + x and after a very small amount of rearrangement this leads to the iteration q := 1; for i := 1:10 do q := (1 + xqˆ2) / 2; which is certainly easier to program than a formula based on use of the binomial expansions.
A somewhat similar form of rearrangement can be needed if a differential equation has terms in it which are multiplied by powers of x, as in Legendre’s equation (1 −x2)y′′ −2xy′ + n(n + 1)y = 0 where the need to divide by (1 −x2) can be avoided by rearranging the equation to give a recurrence rule dy := dy0 + int(xˆ2df(dy, x) + 2xdy - n(n+1)y, x); y := y0 + int(dy, x); with y0 and dy0 providing initial conditions. Observe that in this case it is rea-sonable to use the derivative of dy in the right hand side of the first assignment because it is immediately multiplied by x2, which compensates for the shift in order produced in the differentiation.
The general idea behind Picard’s iteration is that evaluating the right hand side of the recurrence formula must produce a result correct to a higher degree in x than the previously best known approximation to the solution y. In many cases each iteration will just increase the order of accuracy of the solution by one, but in some cases (e.g. the one just given) each iteration may increase accuracy by two (or even more) terms.
3.3 Newton’s method and second order convergence Probably the best known iterative technique in numerical analysis is Newton’s method. To find a solution to the equation f(z) = 0 it starts with some initial approximation z0 and defines a sequence of further approximations by zn+1 = zn −f(zn)/f ′(zn).
11 Except when the solution for z that is being found is a repeated root of f(z) = 0 this iteration exhibits second order convergence, ie. the number of correct digits in the approximations zn roughly doubles each time n increases by one. This very general iteration can be applied in the context of power series calculations. The initial approximation can almost always be taken to be just the leading (constant) term of the desired series, and this is usually trivial to find. The general form of the iteration involves a power series division, and although some algebra systems support this directly Reduce does not. However the required quotient can be com-puted using a special case of the Newton iteration itself. Consider the function f(z) = y −1/z, then f(z) = 0 will be solved when z = 1/y. f ′(z) = 1/z2, and so Newton’s iteration simplifies to something in which no division is present. A suitable starting value for z0 will be obtained by substituting x = 0 into the power series y to obtain the leading term, which being numeric can be divided by. Thus the program for computing a power series a for 1/y becomes z := 1 / sub(x=0, y); for i := 1:n do z := z(2 - yz); together with a suitable let statement that will avoid terms with too high a degree in x from accumulating. Second order convergence in this context means that each step in the iteration doubles the number of correct terms in the series, and so if for example an expansion correct to terms in x16 is required it will be neces-sary to make the iteration count (n in the program fragment) four. This iterative scheme for computing the reciprocal of a power series can be packaged to provide a general series division capability for Reduce by encapsulating it in a procedure: let xˆ16 = 0; procedure tpsquotient(a, b, x); begin scalar z, z1; z := 1 / sub(x=0, b); repeat << z1 := z; z := z(2 - zb) >> until z = z1; return az end; The procedure tpsquotient will always perform one unnecessary iteration at the end of computing z, the reciprocal of b, since it decides when to stop by 12 observing when Newton’s formula does not lead to a change in the value of z, but in almost all circumstances this small inefficiency will be unimportant. Some of the work done by the above procedure is unnecessary for another reason: during early stages in the iteration it is known that the intermediate results will only be accurate to low order in terms of powers of x, but the global LET Xˆ16 = 0 reduction does not reflect this. A more refined version of the algorithm would arrange to truncate all intermediate results to keep the smallest possible number of terms in them. With Reduce the heavy use of LET and CLEAR that this would involve is clumsy, and for second order (where only a very few cycles of the iteration are needed) the extra complication is usually not worthwhile.
Having synthesised a power series division procedure it is now easy to use Newton’s method to more elaborate equations. For instance finding y = √a where a is some formula such as 1 + x can easily be achieved by starting with y := sqrt(sub(x=0, a)); and using the iteration y := (y + tpsquotient(a, y, x)) / 2; and for simple formulae a the results will match those that could have been pre-dicted using binomial expansions. The scheme is, however, equally easily used on problems where the solution can not obviously be obtained otherwise. Consider the equation 2y3 −y = 1 + x and the problem of expanding its solution y as a power series in x. It is first necessary to consider an initial approximation, and this can be found by discarding all instances of x from the original problem to leave 2y3 −y = 1 which has a solution y = 1. For this particular equation the other potential initial values for y are complex, and it might be that the originator of the problem can indicate that only a real solution is required: otherwise it would also be necessary to consider the other possible starting values for y, viz the complex numbers (−1+ i)/2 and (−1 −i)/2.
Newton’s formula then dictates the iteration to be used: yn+1 = yn −(2y3 n −yn −1 −x)/(6y2 n −1) and this translates directly into a program 13 y := 1; for i := 1:4 do y := y - tpsquotient(2yˆ3-y-1-x, 6yˆ2-1, x); As for the reciprocal program the number of correct terms in the result double each time the iteration is used, and so a very small number of cycles of the loop will usually be adequate. By changing the first line of the above from Y:=1; to Y:=(1+I)/2; the same program would find an expansion for one of the other solutions to the original equation. In a similar way the program that finds √1 + x using Newton’s method will naturally find the other solution to y2 = 1 + x, i.e.
−√1 + x, if started with −1 as an initial value for y.
Applying the iteration zn+1 = zn −f(zn)/f ′(zn) involves repeated evaluation of the derivative f ′(zn) and division by it, both of which may be expensive operations. A simple modification of the Newton Raph-son iteration keeps using the initial approximation to f ′ (which is often just a number), so that each step in the iteration is faster: zn+1 = zn −f(zn)/f ′(z0).
This modified Newton’s method generally exhibits first order convergence, but in cases where the function f ′ is more complicated than f the reduced cost per step can result in it be more efficient overall.
3.4 The use of undetermined coefficients The discussion so far has concentrated on the order of accuracy of series expan-sions. In the following section an alternative view will be presented. This consid-ers the leading error term in an approximation, and attempts to eliminate it. The result will generally be a new approximation where the leading is of higher degree, and so this new leading error term will be eliminated next. When expressions are dense (ie. almost all possible terms in the expression are present) there is no prac-tical difference between repeated approximation methods expressed as iterations over the degrees of formulae and those thought of as the successive elimination of leading error terms. For sparse expressions, however, term by term methods can lead to useful saving.
14 3.4.1 Series reversion Consider the problem where a function y(x) is defined by a power series y = y0 + y1x + y2x2 + . . .
and it is desired to express x as a function of y, also in series form: x = x0 + x1y + x2y2 + . . .
The coefficients xi can be derived using a repeated approximation algorithm. It is first necessary to exhibit the initial terms of the expansion. By drawing a graph of y against x it can be seen that the problem is only a sensible one if y0 = x0 = 0, and in that case x1 = 1/y1. The dependence of the remaining xi on the known coefficients yi can be derived in a step by step manner. Thus to find x2 a provisional expansion for x is set up with a new indeterminate acting as an undetermined value for x2. Here the symbol z will be used, and so to order 2 in y, x = (1/y1)y + zy2.
This can now be substituted into the original identity, keeping only those terms with degree no higher than 2 in y. The two sides of the identity should be in agreement in their constant and linear terms (because the those terms in the ex-pansion for x were supposed to be correct already), and so the y2 term can be used to give an equation to be solved for z. This establishes that x2 = −y2/y3 1. If higher order terms are required the same sequence of steps can be repeated. Re-using the symbol z, to order 3 in y, x = (1/y1)y −y2/y3 1y2 + zy3.
and substituting this into the equation for y in terms of x will again lead to an equation which can be solved to find x3, with similar calculations leading to as many more terms in the expansion for x as are required.
3.4.2 Series techniques with other than simple polynomials around Imagine a polynomial in the variables a, b, c and u, v and w. Now imagine that u stands for eix, v for eiy and w for eiz. Basic arithmetic on the polynomials is not altered by their interpretation as complex exponentials. The rules for integration and differentiation need altering, but not in very dramatic ways. The effect is that 15 for very little extra cost rather broader class of expressions can be handled. But who wants to worry with complex exponentials? Well the magic step is to write sin x = (eix −e−ix)/(2i) and similarly for cos(x) — suddenly a system capable of dealing with complex exponentials can deal with formula involving a bunch of polynomial style vari-ables and a collection of trig functions. The restricted sort of series with sines and cosines in are knows as Poisson Series.
As an example of repeated approximation applied to Poisson Series it is almost easy to generate an expansion of the solution to the Kepler equation, e = u + x sin(e) by starting with a first approximation e0 = 0 and repeatedly substituting into the right hand side of the equation. The calculation sin(e) can be performed by substituting the Poisson series approximation for e into the power series for the sin function.
3.5 History, state of the art, future prospects 3.5.1 Delaunay’s analytic lunar theory A major classical use of Poisson Series was in producing an analytic Lunar The-ory. One of the early triumphs of computer algebra was the reproduction of a series of massive calculations that had been performed by hand by the French-man Ch. Delaunay. Since then the same technology has been applied to artificial satellite theory.
3.5.2 Selection done during multiplication rather than afterwards In repeated approximation methods many operations are performed in circum-stances where it is known that high order terms in the result will not be mean-ingful. Major savings in both time and space can be achieved if the polynomial multiplication procedures are adjusted so that they respect the cut-off (known as a selection) and avoid generating parts of the result that would be of too high an order.
16 3.5.3 Specialised power series packages Various special purpose computer algebra systems have been written at various times with the specific aim of making polynomial, power series and Poisson series working as fast as possible. By fixing the names of variables and limiting the magnitude of exponents it becomes possible to use very tightly packed and neat datastructures, which again helps speed and space efficiency. In a Cambridge context the main system to note is CAMAL. As computers have become larger and faster the special purpose algebra systems have somewhat fallen from favour.
3.6 Case studies 3.6.1 Legendre polynomials done lots of different ways This section tries to illustrate that even for polynomial and series calculations there can be many very varied ways of calculating the same values. These may differ radically in terms of programming convenience or the demands that they place upon an algebra system. You might like to try the following out to see which are easier to get working and which run fastest.
1. Using pure polynomial arithmetic the Legendre Polynomials can be com-puted using a recurrence formula.
p0(x) = 1 p1(x) = x pn(x) = ((2n −1)xpn−1(x) −(n −1)pn−2(x))/n 2. The formula dn/dxn(1−x2)n/n! (Roderigue’s formula) computes the same polynomials using differentiation of polynomials.
3. If you write the power series expansion 1 √ 1 −2x2t + t2 = ∞ X i=0 pi(x)ti and work out explicit values for the coefficients pi(x) then once again you will have found the Legendre polynomials.
4. The differential equation (1−x2)y′′−2xy′+n(n+1)y = 0 has a polynomial solution that is the nth Legendre polynomial.
17 3.6.2 Van der Pol equation Consider the equation y′′ + y = ey′(1 −y2) subject to constraint y′(0) = 0. If e is small this is nearly just a simple harmonic oscillator. For small oscillations (ie. the average value of y is small) if e is positive the amplitude of the oscillations will tend to increase. For large amplitudes on average 1 −y2 will be negative and this will damp things down. Somewhere in between there is a stable state — a limit cycle. I will sketch (but not give full details of) how Poisson Series can be used to find the limit cycle. The technique is applicable to a wide range of weakly nonlinear periodic and and almost periodic systems, and can be thought of as a very simplified model of what was involved in lunar and satellite theory analysis.
3.7 Exercises 1. Compare performance for lots of ways of computing √1 + x, 1/(1 −x), tan(x) series.
2. Tchebychev polys: check them out in a suitable book and see how many different ways of evaluating them you can invent. For instance Tn(x) = cos(n arccos(x)).
3. Show how to interpolate a polynomial through (x1, y1), (x2, y2), . ..(xn, yn) 4. The Duffing equation is y′′ + y = ey3. Compare with Van Der Pol.
4 Rational Functions 4.1 Introduction In section two all expressions were put in the form of series. This made it possible to generate expansions of the solution to various algebraic and differential equa-tions, but meant that a simple fraction such as 1/(1 −x) had to be represented by some initial segment of the infinite series 1 + x + x2 + . . .
This transformation can be avoided if quotients are kept as such. The main technical problem that this raises for algebra system implementors is one of keep-ing the resulting fractions reduced to their lowest terms. Even when high powered 18 algorithms are used this process can be unexpectedly costly when the formulae be-ing processed are of high degree or involve many different indeterminates, and so the first part of this section discusses ways of reducing the need for the calculation of greatest common divisors (GCDs) in calculations. Of course for sufficiently small problems the techniques described will represent irrelevant complication, but as progressively larger computations are attempted it will often be the case that GCD calculation limits what can be done on a given computer in a reasonable amount of time.
The procedures discussed here are applicable to formulae built up out of in-tegers and indeterminates using addition, subtraction, multiplication and division.
The word “polynomial” will be taken to apply to those formulae with no division in them (not even fractional coefficients), and all other cases will be treated as quotients of pairs of polynomials. Such quotients will be called rational func-tions. Thus (1/2)x+(1/3) will be thought of as a rational function with the poly-nomial 3x + 2 as its numerator and the constant polynomial 6 as its denominator.
This is in fact the way that the formula is represented inside the Reduce algebra system, and the ON DIV flag used in 3.2 just changes the style in which results are printed, not the internal organisation of the system. A consequence of this is that in Reduce some built-in functions that require polynomials as arguments (e.g. COEFF, REDUCE) take a strict view and expect to be given expressions with whole number (and not fractional) coefficients.
The remainder of this section investigates the algorithms used for rational function manipulation and various closely related processes (e.g. factorisation).
Some understanding of these can help a user appreciate why large GCD calcula-tions can be so extraordinarily costly, while others that seem at first sight to be as complicated are completed very rapidly. The algorithms also provide an illustra-tion of the way in which techniques derived from abstract “modern” algebra find a direct applicability in the solution of apparently elementary problems. Some of the algorithms will be illustrated by code fragments showing how they could be implemented by a Reduce user: these may form prototypes for user written packages for other mathematical procedures.
4.2 Reducing the need for GCD calculations In many cases when a fraction p/q occurs in the course of an algebraic calculation, p and q will have no common factors. If they do have a common factor, say g = gcd(p, q), then the quotient should be represented as (p/g)/(q/g), where both divisions by g will be exact. For uniformity it will be normal to ensure that 19 the leading coefficient in the denominator q is positive, if necessary by multiplying top and bottom of the fraction by −1.
Throughout a calculation the repeated checking of fractions for factors that need cancelling represents an overhead. This is particularly so if it happens that no significant gcds are ever found. Unfortunately simple gcd algorithms exhibit their worst behaviour (ie. they consume most time and store) precisely when they are eventually going to report that their inputs are coprime, and so with early algebra systems the size of problem that could be solved was strongly limited by the form of the rational function arithmetic involved. With the larger memories and better algorithms now available many more algebraic problems can be handled without gcd calculation giving trouble, but it is still the case that for large calculations it can still be necessary to take steps to reduce the costs associated with reducing fractions to their lowest terms. Thus most problems should initially be presented to the algebra system as if rational function manipulation were not a problem: if resource limits prevent the required results from being obtained it may be worth considering some of the transformations discussed below.
The first principal to be applied when performing large scale algebra is to avoid computing a result that contains more information than is actually required.
Apparently equivalent ways of writing even quite simple fragments of code can lead to very different computational behaviours. For instance given four polyno-mials p, q, r and s it may be necessary to decide if p/q = r/s. The obvious test to apply would be if (p/q) = (r/s) then ...
but this involves reducing the fractions p/q and r/s to their lowest terms, which will turn out to be unnecessary. A second attempt, which although mathematically equivalent to the above will be computationally different is if (p/q) - (r/s) = 0 then ...
where not only are p/q and r/s reduced to their lowest terms, but if nonzero their difference also has to be so reduced. Since the only information that is required is whether the difference is zero, it will probably be better to concentrate on its numerator, avoiding all gcd calculations and writing if ps = rq then ...
Another example in the same vein involves a test to see if the polynomial q divides exactly into p. The test 20 r := p/q; if den r = 1 then ...
forms the rational function (p/q) and tests if its denominator is one. If so it is clear that q divided exactly into p, and in this case the quotient r will often be required for further processing. If q is not a factor of p the above code will reduce the fraction p/q to its lowest terms, and if the reduced fraction is not needed computing it will be a waste of time: the test if remainder(p, q) = 0 then ...
does not suffer from this, but does have the disadvantage that it does not auto-matically provide a value for the quotient p/q in the exact division case, even though the process of producing the remainder will have had to go through the steps needed to obtain it. Later on there will be an explanation of how users can obtain access to the function that Reduce uses internally for test division5, thus providing a way of obtaining further improved performance at the cost of having to understand more about the fine details of the algebra system being used.
In general if calculations are performed on rational functions and fractions are not reduced to lowest terms promptly the size of expressions grows explosively.
In special cases, however, it can be predicted in advance that this will not be so: for instance it may be known that some series of calculations will necessarily lead to an answer expressed in its lowest terms even without any gcd calculations being performed to ensure this.
In such cases it may be advantageous to instruct the algebra system not to attempt to reduce fractions to lowest terms. In Reduce this is done with a directive off gcd; the effect of which can be cancelled by ON GCD. Since calculations which benefit from this treatment are fairly rare it seems possible that the best use of OFF GCD mode is to allow the user to see how dramatically expressions can grow if not kept in reduced form, and hence how important a good gcd algorithm is in an algebra system!
Having adjusted a program so that it does not compute more than the minimal interesting part of an answer it can be becomes useful to seek ways of simplifying the problem formulation. Any transformation that reduces the degrees of polyno-mials found in the problem or which reduces the number of variables present will 5See “testdivide” defined shortly.
21 be useful, and to a lesser extent transformations which remove large numeric val-ues can help. It is often possible to make changes of variables so that factors that appear in the denominators of intermediate expressions become simply powers of some indeterminate — this renders greatest common divisor extraction trivial.
Thus if in some calculation there are two variables u and v, and the denominator of many fractions will involve powers of u + v it will be useful to introduce a new symbol (w say) to represent u + v and use a substitution such as ... sub(v = w - u, ) ...
to express original expressions in terms of it. A simple substitution at the end of the calculation will then make it possible to present results in terms of the original variables. It can make sense to introduce a new indeterminate to rep-resent the reciprocal of some complicated denominator even when it is not then possible to eliminate any of the original variables. For instance in calculations involving a matrix M and its inverse, the introduction of a symbol that stands for 1/determinant(M) can often simplify things substantially.
In some cases it is possible to reformulate calculations so that rational function arithmetic is not needed at all. If, for instance, there is a way of predicting in ad-vance some common denominator for all the expressions that will arise throughout a calculation then everything can be reduced to polynomial arithmetic, which is generally well behaved. Later in this section (in the discussion of the subresultant PRS method for computing gcds) there will be an example of a case where the mathematical structure of a calculation makes it possible to predict factors that are guaranteed to divide exactly into some of the intermediate results, and this dividing out of predicted exact factors saves significant amounts of time.
The costs of computing greatest common divisors can grow rapidly with the size of the expressions involved, and this means that it is generally better to per-form several small gcd calculations rather than allowing common factors to build up and finally do a single large reduction. An application of this principle can be found inside the implementation of Reduce where two rational functions are to be added. The natural way of forming the sum of (a/b) and (c/d) would be to com-pute the numerator p = ad + cb and the denominator q = bd of the result and then form them into a fraction (p/q) cancelling any common factors. The scheme that is in fact used tries to find common factors as early as possible so as to reduce the size of intermediate results and hence the cost of the final gcd calculation. First the greatest common divisor, g, of b and d is computed. Now the numerator of the final result can be evaluated as p = a(d/g) + c(b/g) and the denominator as q = b(d/g) where the divisions indicated are known to be ones that will be exact.
22 If it happens that g is large (for instance b and d might be the same) this will result in significantly smaller values for p and q. In forming the final quotient (p/q) it is still necessary to look for further common factors. If b and d are coprime the new method is no help, but in many realistic cases it speeds calculations up by a useful amount. Analogous optimizations are possible when forming products and quo-tients of rational functions, and potentially within any user-written code that ever generates the numerator of a rational function separately from its denominator.
A scheme that has been proposed (but not widely adopted) that subsumes sev-eral of the above ideas is that of representing expressions as partially factored forms. Rather than keeping all brackets multiplied out and the numerators and denominators of expressions separate, a partially factored representation holds values as products of powers of items. Negative powers are used to indicated factors that belong in the denominator of the expression. During multiplication and division common factors are merged, and in some cases gcd calculations will reveal further factorisations of some of the terms. For instance if the product of x2 −1 and x + 1 were to be required the process of merging lead to the result (x −1)(x + 1)2. During addition and subtraction and factors common to the ex-pressions being combined will remain separated out, but other terms will have to be expanded out thus destroying knowledge of their factor structure.
The final suggestion here for reducing the demands on a gcd procedure in-volves breaking away from the closed form representation of quotients and drop-ping back to the use of power series. Section 4.4.2 illustrates how it is possible to recover an exact rational result from its power series representation subject only to the need to have a bound on the degrees of the numerator and denominator of the expected rational function. For univariate calculations where it is known a priori that the result will be fairly simple but where intermediate stages in the working would lead to rational functions with very high degree denominators this radical transformation can perform wonders.
To end this section here is the REDUCE code for test division mentioned ear-lier. As defined here it returns the quotient of its arguments if the division involved is exact. Otherwise it returns 0. It expects to be given polynomial arguments, but does not check. The explanation of the incantation used to provide this interface to a system-level function within Reduce is outside the scope of this section.
symbolic procedure testdivide(p, q); mk!sq (quotf(numr simp! p, numr simp! q) ./ 1); flag(’(testdivide), ’opfn); 23 4.3 Polynomial Remainder Sequences 4.3.1 Integer GCD Computing the greatest common divisor of a pair of positive integers can be done simply and efficiently using Euclid’s algorithm: PROCEDURE numeric_gcd(a, b); IF b = 0 THEN a ELSE numeric_gcd(b, remainder(a, b)); 4.3.2 Polynomial Remainder and pseudo-remainder The same idea can be applied to find polynomials once we understand what is meant by the “remainder” operation in that case. Dividing one univariate polyno-mial by another is a pretty straightforward process and what gets left at the end is clearly the remainder. But observe the case when x2 + 1 is divided by 2x −1: the quotient comes out as 1 2x + 1 4 and the remainder as 3 4. The fractional coefficients here are not a help at all, and turn out not to contribute usefully to GCD calcula-tions, so it is normal to get rid of them. A few moment’s though will show that the only denominators that can be introduced are powers of the leading coefficient of the polynomial that is being divided by. If, before the remaindering step, the other polynomial is multiplied by a suitable power of same then the unwanted fractions will not appear. This variation on computing a remainder is known as the pseudo-remainder of the two polynomials.
If the recursive representation is used for polynomials then a scheme the com-putes univariate pseudo-remainders can be applied directly (to the top level of the recursive datastructure) to obtain multi-variate remainders. But note now that the result obtained will depend on the ordering of variables in the structure. You may like to try out a very simple case — evaluate the pseudo-remainder when a2 + b2 is divided by 2a + b first when each polynomial is treated as being in terms of a main variable a then in terms of b.
4.3.3 a PRS to compute a GCD If a polynomial a(x) is written anxn + . . . + a1x + a0 then we defined the content of a to be gcd(an, . . . , a0). For instance the contents of 2yx2+4(y3−1)x−8y+4 (when viewed as a polynomial in x) is just 2. The content of a polynomial does not have to be just number if all the coefficients have some common factor that 24 involves subordinate variables, as in (y −1)x+(y −1) which clearly has contents y −1. If a polynomial has a content of 1 it is referred to as being primitive and a dividing a polynomial by its content gives its primitive part.
Now suppose we have two primitive polynomials and want their GCD. The GCD will also be primitive. Is this obvious to you? Use the Euclidean GCD algorithm starting with the two inputs, but doing pseudo-remainders rather than true remainders all the way. At the end you will have a polynomial that is almost the desired GCD. But the pseudo-remaindering operation can have left unwanted extra coefficient-domain factors multiplied through, so the true answer wanted is just the primitive part of what comes out of the Euclidean method. You might suspect that the need for this final calculation of a primitive part was just because we used pseudo- rather than real remainders, but in fact use of real remainders would have left in equally nasty stray multipliers, but ones that could be fractional and hence even more painful to handle. If you work through a small example (I suggest x2 −1 and x2 + 2x + 1) you will see what happens quite clearly — it is not very difficult. The series of results computed during the gcd process is a polynomial remainder sequence or PRS.
Now I can take the GCD of pairs of primitive polynomials: what about the general case. Well that can now be achieved by splitting each of my real input polynomials into contents and primitive parts. Taking the gcds of the contents is a calculation in the coefficient domain so I suppose (by recursion) that it is easy.
Taking the gcd of the primitive parts is what I have just covered. I finally multiply the two sub-results together. Easy!
4.3.4 Euclidean, Primitive, Reduced, Subresultant, Trial division This section is not written out in detail here because Knuth covers it pretty well — but the main messages are: 1. If you compute a PRS of even medium-degree polynomials the coefficients in the intermediate results can become huge, and working with them domi-nates the computing time of the entire process. A simple GCD procedure is not useful for serious applications.
2. The coefficient growth can be cured somewhat by taking primitive parts at each step in the generation of the PRS. But then the cost of all the coeffi-cient GCDs involved in working out the contents that are to be divided by becomes an embarrassment.
25 3. Some clever mathematics (not covered in this course) makes it possible to predict factors that can be divided out. These factors are in fact just certain powers of the leading coefficients of polynomials that arose earlier in the PRS. The effect is that at least almost always the bad growth in coefficients can be cured at quite modest cost.
4. Since the values divided out by the Reduced and Sub-resultant methods are just powers of previous leading coefficients, an easier scheme to program and understand just keeps a list of these previous coefficients and uses test division to remove unwanted factors of them. The test division turns out to be much cheaper than the gcds that were at first needed to keep everything primitive.
4.3.5 Eliminating main variable between two polys: Resultant This is maybe an aside: if you have two equations each in two unknowns you may want to eliminate one of the variables between them. If you look at the steps that you take you will find that you are computing a PRS, and that the univariate equation you end up with is just the final polynomial in that sequence. Moral: PRSs arise in GCD calculations but have other uses too.
4.3.6 PRS as Gaussian elimination on a certain matrix Imagine two polynomials a and b with degrees n and m and coefficients ai and bi.
Consider the matrix an an−1 an−2 . . .
0 an an−1 . . .
0 0 an . . .
.
.
.
. . .
a1 a0 0 . . .
a2 a1 a0 bm bm−1 bm−1 . . .
0 bm bm−2 . . .
.
.
.
. . .
b1 b0 0 . . .
b2 b1 b0 26 If you imagine doing Gaussian Elimination on this matrix you will rapidly dis-cover that the steps you take bear an uncanny resemblance to those that need to be used when computing a PRS. The determinant of the above matrix is known as the resultant of the two polynomials a and b with respect to the variable x. The importance of resultants (here) is that they show a perhaps unexpected link be-tween PRS calculation and simple linear algebra (ie. solving simultaneous linear equations and computing determinants). This link is at the root of the mathemat-ics that led to the Reduced and Sub-resultant GCD algorithms. This course does not provide me with sufficient time to explore it further!
4.4 Evaluation/Interpolation methods The PRS methods for computing GCDs work, but a serious problem with them is that their worst case is when the GCD that is to be found is just 1. In that case the remainder sequence will be of maximum length and despite all attempts to keep them small intermediate coefficients still grow pretty alarmingly. But if you take two polynomials at random their GCD is probably 1. It would be nice to have a method which behaved best in this especially common case.
4.4.1 Modular Probabilistic Checking Consider first the univariate case. The only big cost in computing a PRS is because the coefficients tend to get big. Try approximate coefficient arithmetic to prevent this. In particular, choose a prime p and perform all arithmetic modulo p. If p is a single precision integer then this avoids all the painful overhead of allowing for multiple-precision intermediate results. But is the result computed by a modular PRS any use at all? Yes! If the modular calculation indicated that two primitive polynomials are coprime then they really are. The only possible problem is when the modular calculation says there is a real non-trivial GCD, since then it could be that the polynomials are really coprime and the apparent GCD is an artefact of the approximate arithmetic, and anyway the modular GCD does not instantly allow us to recover the real one.
For multivariate polynomials it can be useful to reduce to the univariate case by substituting random (well, arbitrary, or pseudo-random or something) integer values for all bar one indeterminant. If this is done mod p it can not lead to over-large numbers. Then a modular PRS is computed for the resulting univariate polynomials. This provides a cheap and in fact rather reliable test that identified most cases where the full GCD will be 1.
27 4.4.2 The Hensel Construction Given the above cheap check it would be very nice if the true GCD could be recon-structed from a modular image in non-trivial cases. For multi-variate polynomials the way of doing this is known as the Hensel Construction, and we will see it in action again when we come to factorisation. The presentation here is at first going to gloss over a few problem areas to try to give as clear an overview as possible.
So imagine a bi-variate case where polynomials in x and y are being used. We are computing the gcd of a(x, y) and b(x, y). First view a and b as power series in y, eg.
a(x, y) = a0(x) + a1(x)y + a2(x)y2 + . . .
Of course this will only be a finite series, stopping at the highest degree term in y present in a. Now suppose that the required gcd of a and b will be g(x, y), and write h = a/g. This division of course goes exactly since g, being the gcd of a and b, must divide neatly into a.
The key step is then to observe the (obvious) identity a = gh but to write it out in power series form. Looking at the various powers of y in this gives a series of identities (I have written just ai instead of ai(x) etc. here to keep things concise, but everything mentioned will be dependent on x).
a0 = g0h0 a1 = g0h1 + h0g1 −(g1h1) + a2 = g0h2 + h0g2 −(g1h2 + g2g1) + a3 = g0h3 + h0g3 . . .
I have collected terms on the two sides of the equals sign in a slightly curious way, but otherwise the equations are pretty straightforward. Next consider gcd(a0, b0).
If we are lucky6 this will be exactly g0, the leading term in the power series repre-sentation of the required gcd. Then we can compute h0 as a0/g0. The first of our equations has just been satisfied. Now suppose we can solve the next equation to find g1 and h1, then all the subsequent equations would be of the same form, ie.
g0hi + h0gi = and thus it will be possible to compute as many of the coefficients gi as are re-quired. The largest one that can possibly be nonzero is the one corresponding to 6The term “lucky” is treated as a technical term here!
28 the highest power of y present in the gcd g, and this is bounded by the degree in y of a and b.
To show how to solve the odd equations we have come up with I need to impose two technical constraints. The first is that g0 and h0 must be co-prime.
For now let me just observe that this can only possible fail to be so if the original polynomials a and b had repeated factors, and that it is possible to reduce the general problem one where no repeated factors are present. The second condition is that all the polynomials that we are working with have their coefficients reduced modulo a suitable7 prime p. The reason for this is that when working mod p it is possible to divide any coefficient by another, while when working with integer coefficients some such divisions may not go exactly.
Now before showing how to solve the given equation for polynomials let’s look at the analogous one for integers, and for given u and v try to find α and β such that αu + βv = 1. This can be achieved by looking at a remainder sequence that computes the gcd of a and b. Start by writing a = 1a+0b and b = 0a+1b, and note that if q = ⌊a/b⌋then remainder(a, b) = a −qb, or otherwise 1a + (−q)b.
Following through each value in the remainder sequence can be expressed as a combination of the two previous ones, and hence as a linear combination of the original values a and b. Eventually if gcd(a, b) = 1 this will lead to the final 1 in the remainder sequence being expressible in the form αa + βb as required. By multiplying these values α and β by any value c that value c can be expressed as a linear combination of a and b. The same basic process can be used on coefficients (with modular coefficients), and hence completes the method for extending a uni-variate gcd to a bivariate one. It can clearly be used to cope with polynomials in any number of variables.
4.4.3 Hensel and integer coefficients The Hensel construction as sketched above makes it possible to reconstruct a mul-tivariate gcd, but only with coefficients reduced mod p. To get back to integer coefficients a little more is needed. In fact this is remarkably easy and similar to the ideas we have already seen. Observe that if we write integers in radix p an integer n can be written out as a combination of digits8 n = n0 + n1p + n2p2 + . . .
7The rules about what counts as “suitable” are not especially complicated but are not discussed in depth here.
8Negative numbers might need a little more effort, but that is not a real difficulty and will be glossed over here.
29 where each digits is restricted to the range 0 to p −1. This looks sufficiently like a power series that the previous presentation of the Hensel construction can be applied to it. The methods even work cheerfully if instead of just numbers we work with polynomials written out as a(x, y) = a0(x, y) + a1(x, y)p + a2(x, y)p2 + . . .
where now the correct normalisation constraint is that all the coefficients in each ai(x, y) must be bounded by p. Use of the Hensel construction in this way makes it easy to extend a gcd that has been computed correctly modulo p to one that is correct modulo p2, p3, and so on until a value has been produced correct modulo some power of p that is larger than any possible integer coefficient in the true answer. At this stage the computed result must in fact be the true gcd.
4.4.4 Unluckiness Sometimes the modular-number based gcd processes will fail because the prime used interacts badly with the particular polynomials used. For instance if one worked modulo 3 and while trying to compute gcd(x+1, x+4) something would be bound to go wrong because modulo 3 the two inputs alias together. A com-plete program has to cope with this by being prepared to accept that the Hensel extension process may fail to converge. In that case trying other primes and and other small adjustments9 make it possible to proceed. Fortunately it can be shown in this case that almost all primes will be “lucky”.
4.5 Factorisation and algebraic number arithmetic Factorising polynomials follows on from gcd calculation as the next major algorithm-cluster. One first restricts attention to primitive polynomials. Then removes the frivolous complication of repeated factors.
4.5.1 Square-free decomposition If a polynomial u(x) has a repeated factor then gcd(u, du/dx) will be non-trivial.
By following up on this observation it is possible to do a number of gcd operations that will express u unambiguously in the form u = Ku1u2 2u3 3 . . .
9which will be mentioned briefly in the lectures.
30 where each ui is now free of repeated factors.
4.5.2 Kroneker’s algorithm A classical way of finding factors of a polynomial can be illustrated by looking for possible quadratic factors of u(x). Find all integer factors of each of u(0), u(1) and u(2). For each possible combination of one factor from u(0), one from u(1) and one from u(2) it is possible to interpolate a quadratic. If this quadratic has integer coefficients and divides exactly into u then we have found a factor as desired. If there is any quadratic factor we will come across it is this search.
This method can clearly be used to see factors of any desired degree, but in be-comes grotesquely expensive for non-trivial cases and is not obviously adaptable to the needs of multivariate cases.
4.5.3 Berlekamp/Hensel A better way to factorise polynomials will start by reducing an arbitrary input to a univariate polynomial modulo a prime p. If this can be factorised then the Hensel construction will make it possible to reconstruct the multivariate factors over the integers.
Berlekamp’s algorithm factorises a univariate polynomial modulo some given prime. The inputs are: A polynomial u(x), which should be primitive and square-free.
A prime p.
We will imagine that the (irreducible) factors of u(x) are w1(x), w2(x),. .., wj(x). All arithmetic mentioned here will be done modulo the selected prime p, and so all coefficients will be in the range 0 to p −1.
The factorisation process works by considering the class of polynomials v(x) that have the properties: degree(v) < degree(u) u divides exactly into (v −0)(v −1)(v −2)...(v −[p −1]).
It will turn out that these polynomials v can be found quite easily, and that from them it is possible to derive the factors wi that will be our final results. I will demonstrate the first of these facts first.
31 Finding the v polynomials: Because we are working modulo a prime p the formula (v −0)(v −1)...(v −[p −1]) can be simplified to v(x)p −v(x). A further result relating to polynomials with modular coefficients asserts that this in turn is equal to v(xp) −v(x). The justifi-cation of these two steps is not included in this note, but the results are standard mathematical ones. The condition that u(x) divides v(xp)−v(x) can be expressed as v(xp) −v(x) = 0 (mod u(x)).
A polynomial v can be characterised by a vector giving its coefficients v0, v1, . .., vn−1 (we know that the degree of v is to be less than that of u, and so can establish the number of coefficients to be considered). The equation we are solving now turns into v0(x0 −x0 mod u) + v1(xp −x1 mod u) + v2(x2p −x2 mod u) + vn−1(x(n−1)p −xn−1 mod u) = 0 where all the polynomials involved now have degrees less than that of u(x). Each of the polynomials xkp −xk can be written out as a row of coefficients, and the rows (one for each value of k) stacked to form a matrix, generally known as Q.
The equation that was to be solved now takes the form of a matrix multiplication: (v0, v1, . . .)Q = 0.
Solving such a set of equations is a standard piece of linear algebra. The com-plete set of solutions for v form a vector space and a set of independent vectors spanning this space is known as a null space basis for the matrix Q. To save space I will not fill in the full details here, but finding a null space basis proceeds by go-ing through the steps of Gaussian elimination as if the matrix involved contained coefficients for a set of linear equations. For equations it is unusual and generally an error for the matrix to be found to be singular: for a null-space finding algo-rithm this is just what is expected and the null vectors amount to nothing more than a record of how and when singularity was found.
If a basic set of j independent null vectors are found then all solutions to our original equation can be expressed as linear combinations of these. Since we are working (mod p) these linear combinations can only have numbers in the range 0 to p −1 as coefficients, and so a grand total of pj solutions for v(x) exist.
32 Deriving factors: If u(x) divides exactly into (v−0)(v−1)...(v−[p−1]) then necessarily all the factors wi(x) also divide into it. But except for the dull case that v is a constant, (v −0), (v −1) and so on are all pairwise coprime, and so this means that each wi must divide exactly into some particular one of the (v −s).
This leads to the observation that given some particular factor w1 (say) and one of our v polynomials, there is a constant s such that: w1 divides exactly into u(x) [of course it does!], w1 divides exactly into (v(x) −s).
Since w was a proper factor of u(x) its degree is greater than zero, and since it divides into v(x) −s its degree is at most the same as that of v, which is in turn strictly less than that of u. So u(x) and v(x)−s have a w as a common factor. We compute g as their greatest common divisor. The polynomial g will be either the factor w1 itself or some multiple of it, but since the degree of g is less than that of u the identity u(x) = g(x)(u(x)/g(x)) has exhibited a non-trivial split of u(x). There is no way of telling what value s will work in the above, but provided the prime p is fairly small it is quite cheap enough to try all the possible values 0, 1, ... , [p −1] until one is found to lead to a factorisation of u(x).
In fact the number of independent null vectors that are found is exactly equal to the number of irreducible factor that u(x) will have, and by performing gcd operations between u(x) and terms of the form (v −s) letting v range over the null space basis and s over the numbers from 0 to [p −1] it is always possible to complete the factorisation of u(x).
Example: Factorise u(x) = x7 + x5 + x3 + x2 + 1 modulo the prime 2.
Start by writing down x0 −x0 = 0 x2 −x1 = x2 + x (mod u(x)), (mod 2) x4 −x2 = x4 + x2 x6 −x3 = x6 + x3 x8 −x4 = x6 + x3 + x x10 −x5 = x4 + x x12 −x6 = x5 + x2 + 1.
33 The right hand sides are just the remainders left when the left hand sides are divided by u(x), and so in the first 4 cases where the degree of the left hand side is less than that of u(x) all that has happened is that mod 2 arithmetic has allowed us to write + instead of −.
This makes the Q matrix: • • • • • • • • X X • • • • • • X • X • • • • • X • • X • X • X • • X • X • • X • • X • X • • X • and I can exhibit two independent null vectors: [1, 0, 0, 0, 0, 0, 0] and [0, 1, 1, 0, 0, 1, 0].
You can check that if you multiply either of these row vectors by the matrix (mod 2) you get zero. Once again I am not going to go into the fine details of the (simple) calculation that leads to this null space basis - I will just go on from the fact that I have v-polynomials v1 = 1 and v2 = x5 + x2 + x to work from. The fact that there are two independent null vectors indicates that my polynomial will have 2 factors. I now compute gcd(u, v1 −0), gcd(u, v1 −1), gcd(u, v2 −0), gcd(u, v2 −1) and expect one of these to be non-trivial. In this case it turns out that gcd(u, v2 − 0) = x4 + x + 1, and so this is one of the factors of u. The other is easily found (by dividing the known one into u(x)) to be x3 + x + 1, and since we knew that u(x) had just two factors we are finished.
Observations: For large primes p this algorithm becomes costly because of the need to try all possible values of s. It is therefore usual to use it with primes in the range (say) up to 100. The case p = 2 can be handled particularly neatly using bit-vectors to represent all of the polynomials and exclusive-OR and shift instructions for some of the operations on them. For large primes (e.g. about the size of a machine word) there are slightly more elaborate variants on the algorithm - see Knuth vol. 2 for some pointers. A coded version of the complete algorithm is no more than a couple of pages long.
34 4.5.4 Algebraic number arithmetic The natural technical follow-on to gcd and factorisation algorithms involve alge-braic numbers. These are values that arise as the roots of polynomial equations.
Thus values such as √ 2 and 3 √ 5 are covered, as will be the imaginary i and the roots of any higher degree equation such as α5 + α + 1 = 0.
Three main approaches are taken, and these correspond to different interpre-tations of what a symbol such as √ 2 really stands for: 1. Root isolation: Given a polynomial u(x) set up a sequence of polynomials si with s0 = u, s1 = u′ and si+2 = −remainder(si, si+1). Note that apart from the negative signs this is just a PRS. Now look at the curious function S(z) which counts the number of sign changes in the sequence s0(z), s1(z), ... . For any rational number z with u(z) ̸= 0 it is clearly possible (if perhaps expensive) to compute the integer S(z) quite unambiguously. Now, as I hope I will be able to make slightly plausible by drawing pictures on the blackboard, |S(z1) −S(z2)| gives the number of real roots u(x) has between z1 and z2! Given this tool (A Sturm Sequence, but note it is just another sort of PRS)) the roots of u can be counted and isolated to utterly arbitrary accuracy.
2. Minimal Polynomial Arithmetic: Represent any algebraic number by a symbol and a polynomial equation that it satisfies. Thus for √ 2 write (ω : ω2 −2 = 0) and for 3 √ 5 use (η : η3 −5 = 0). Note that this representation simultaneously allows for both the √ 2 that is about 1.414 and the one that is -1.414, ie. what is being represented is really ± √ 2.
To combine values that are represented this way is actually rather straight-forward. Consider the addition of the the values mentioned above. Write out a set of equations: ω2 −2 = 0 η3 −5 = 0 ν −(ω + η) = 0 where the first two equations are the definitions of ω and η and the final one expresses the arithmetic operation being performed. Now just eliminate ω and η from these equations to leave one in just ν and a representation for the algebraic number sum has been derived as required. Hah! Eliminating 35 variables between polynomial equations was what resultants were about, and again we are really using a method based on the PRS.
3. Reduction rewrites: Set up rewrite rules like ( √ 2)2 →2 and i2 →−1.
Apply these throughout your calculations whenever possible. Hope or ex-pect that this will lead to consistent and well-simplified results. A major problem that can arise with the use of rewrites is that sometimes some com-plicated rewrites can interfere with each other so that the order in which reductions are performed influences the final result produced. For polyno-mial rewrites it is possible to extend any initial set of rules into a so-called Groebner base which has the good property that whatever strategy is used for selecting rewrites to apply the same end result will always be obtained.
Algorithms for the construction of Groebner bases have been a central part of research into Computer Algebra over the last about 15 years, and you should inspect the specialist literature if you want to know more about them or about more of their applications.
4.6 Exercises 4.6.1 Find an example that illustrates that when the sum (a/b) + (c/d) is evaluated as (a(d/g)+c(b/g))/(b(d/g)) [where g = gcd(b, d)] the final quotient can still need a common factor removed from the numerator and denominator.
4.6.2 Show how to form the product and quotient of (a/b) and (c/d) using sequences of small gcd calculations rather than a single large one. Note that it is possible to exploit the fact that the incoming expressions should already be in their lowest terms and so gcd(a,b)=gcd(c,d)=1 can be assumed.
4.6.3 Use Berlekamp’s algorithm to find a polynomial of the form x16 + xn + 1 which is irreducible mod 2. Clearly n will be in the range 1 to 15, and an argument of symmetry shows that it is only necessary to consider n from 1 to 7. A value of n that gives an irreducible polynomial corresponds to the position of a tap in a feedback shift-register (as shown below) that will cycle with largest possible 36 period: such shift registers can be used to generate checksums or pseudo-random bitstreams.
XOR stage 15 stage n-1 6 ?
0 1 2 14 15 ---- ...
--Each stage in the shift-register is a D type flip-flop (all driven by a common clock signal), with the shift register’s input being derived by forming the exclusive OR of the output from the final and some intermediate stage. Possible lengths of cycles in the bit patterns generated by this arrangement are related to the factor structure (mod 2) of a polynomial that has 1 coefficients at the places where taps are taken from the shift register. Textbooks on error correcting codes (e.g. Peter-son, 19xx) provide further explanation and background: it was for the analysis of this sort of circuit that Berlekamp’s algorithm was first introduced.
5 Transcendental Functions 5.1 Introduction Consistent algebraic calculation with formula involving transcendental functions is difficult. In particular, for many of the classes of algebraic formulae that seem natural it is known to be impossible to produce a systematic way of determining if an expression is zero or not. If E is an expressions which might possibly be zero, but there again might not be, the simplification of formulae such as AE+B can not proceed. Other questions become hard to answer: the expressions Eexp(x2) + x has an integral in closed (elementary) form if and only of E = 0, and so the inability to decide if an expression vanishes means that it is impossible to exhibit a reliable and complete integration algorithm.
A frustrating aspect of the main undecidability result for algebraic simplifica-tion is that key technical aspects of the proof suggest strongly that current under-standing in the area is incomplete, and that the pessimism expressed above may not be entirely justified. To explain this, a sketch of a proof (due to Richardson) 37 that identifying real-valued functions that are in fact zero will be given. Many details of the proof will be omitted or glossed over, but the line of argument given can be made watertight. The purpose of showing this proof is so that a certain key oddity in it can be exposed and not with the intent that the proof as a whole be followed in detail.
Richardson’s result shows that there is no algorithm that will reliably deter-mine if a real function of a single real variable is in fact identically zero. The class of functions it allows are built up from the variable x, the constant π and the integers, using the four normal arithmetic operations, together with uses of the normal elementary functions: sin, cos, exp, log and sqrt. It is supposed that any expression that can be shown to have a constant value can be tested to see if it is zero. If this were not true then deciding if expressions were zero would certainly be undecidable! Richardson restricts his analysis to real values. Since the log and sqrt functions are (in this context) undefined for negative arguments, a function from the class considered may be partial. The result proved will show that it is impossible to tell if the function is zero at such places as it has a value, thereby allowing for formulae such as ((x/x) −1) which are certainly zero everywhere except at x = 0, but are undefined there.
The proof proceeds by assuming that there was an algorithm to decide if func-tions from the given class were zero, and shows how this would imply an algorith-mic solution to a different problem that has already been shown to be algorithmi-cally undecidable. The known undecidable problem is that of solving Diophantine equations: that is finding sets of integer values that result in given multivariate polynomials (with integer coefficients) vanishing. Of course given any particular Diophantine equation it may be possible to find the solutions: the thing which can not be done is to produce a single algorithmic method that guarantees to solve any such equation in a finite amount of time. (In this context solving also includes the case of deciding that an equation does not have any solutions in terms of whole numbers).
Richardson shows how, for any given Diophantine equation, it is possible to derive an analytic function of one variable such that that function being identically zero corresponds means that the Diophantine equation has no solutions. Thus if it were possible to test the function for zero it would be possible to tell that the Diophantine equation could not be solved, and that is known to be impossible.
The first step in the derivation of this result looks like the first cheat! Con-sider a polynomial equation P(x,y,z)=0 where we are only interested interested in solutions where x, y and z take integer values. Then for any such solution it is 38 certainly true that P(x, y, z)2 + sin(πx)2 + sin(πy)2 + sin(πz)2 = 0 and indeed a similar statement can be made however many variables were present in the original formula. Furthermore if we restrict our attention to real values the new function can only possibly be zero at places that are integer solutions to P = 0 even if x, y and z are now thought of as variables that can range over all possible real values.
The next step shows that from any multivariate function (for instance the one just written down) it is possible to derive a function of one variable which takes the value zero at places related to the zeros of the original function. The essence of this is the construction of packing functions p1, p2,. ..(details are not given here) and the replacement of f(x, y, z) by f(t) = f(p1(t), p2(t), p3(t)) where as t varies the functions p1, p2, ... arrange that all possible integer values of x, y and z are visited eventually. Despite the fact that this may seem peculiar, this sort of packing transformation is a standard tool in the analysis of what is computable and what is not.
The univariate function f(t) is now multiplied by a large positive value so that any nonzero minima of f(t) are scaled to have value greater than 1. Inventing a scale value (which will in fact be a function of t derived from the expression that represented f(t)) is not a particularly obvious step, but can be done. If the scaled univariate function is F(t), then we now have F(t) > 1 except in the immediate vicinity of a place where F(t) = 0, and these places correspond to integer solutions of the Diophantine equation we started from.
Finally, Richardson shows how to switch from consideration of whether a function takes the value zero anywhere to a test on whether it takes a non-zero value anywhere. He observes that since we are working in real numbers, he can express a version of the absolute value function as, for instance abs(x) = (√x)2.
If the use of a square root is objected to he can achieve the same effect using exponentials and logarithms. He then defines h(x) = (abs(x −1) −(x −1))/2 and a quick sketch of the graph of h(x) shows that if its argument is less than 1 it is nonzero, while if x > 1 then h(x) vanishes. This is enough to show that h(F(t)) can only have nonzero values if F(t) takes values less than 1 somewhere, 39 and hence if f(t) has roots. So any procedure that can find out if h(F(t)) is ever nonzero can be used to solve arbitrary Diophantine equations!
In the above proof the most objectionable step is the synthesis of the absolute value function as (√x)2, which amounts to the synthesis of a non-analytic func-tion out of several analytic ones, but it has been found amazingly difficult to find any general way of prohibiting that while still allowing for calculations with an in-terestingly rich collection of functions. One obvious possibility would be to work with complex numbers rather than reals. Doing so reveals that the oddity relat-ing to the absolute value function is just the tip of a huge iceberg of oddities and problems that arise when complex-valued functions can have multiple branches.
For instance when working with algebraic formulae over complex numbers it is not obvious that √x = √x or log(x) = 2iπ + log(x) until somebody has decided if the characters we use to stand for a function should represent one branch or all possible ones, or what!
5.2 Reducing the number of distinct functions A good pragmatic approach to calculating with higher functions is to try to avoid it, or at least reduce the number of potentially interacting functions that are present in the formulae to be worked with.
5.3 Algebraic independence If we have a formula that involves x, log(x) and sin(1/x) it might make sense to try replacing the two nasty functions by new symbols, y and z say. That will reduce the formula to just a rational function which we already know how to cope with. The only extra effort needed will be to remember that y and z need special treatment when differentiated, for example dy/dx = 1/x. The case above will in fact survive this re-interpretation, but if the original formula included various closely related functions (eg. exp(x) and exp(2x + 1) or log(x) and log(5x2), or even just √ x2) the rational function version of the formula would fail to capture important aspects of the original.
5.4 Structure Theorems and their uses A Structure Theorem is a procedure for testing if a collection of exponentials and logarithms (or whatever) are sufficiently independent that renaming them into rational function variables is safe. The lectures will sketch the process applied in 40 either the exponential or logarithmic case. If the functions present in an original formula do not pass the test of a structure theorem the technology provides no guidance about what can or should be done — the idea is just to certify various good cases where reliable calculations can be performed and disallow the sorts of things that allowed Richardson’s pessimistic result to be proved.
5.4.1 Risch’s Integration algorithm The main technical triumph built on top of the idea of structure theorems is algo-rithmic indefinite integration. The idea behind this is that in well behaved cases it is possible to predict the form that an integral will have (if one exists). A general expression (with a load of undetermined coefficients in it) of that form is then written down and identified with the original integral. After differentiating both sides the use of a structure theorem makes it valid to consider everything as ra-tional function calculation. In particular it is permissible to compare coefficients, and doing so leads to a bunch of linear equations that can be solved to find values for the unknown coefficients that had been introduced. If the linear equations can be solved an integral has been found, otherwise the lack of a solution amounts to a proof that no integral in closed form exists. There are special (and extreme) complications that arise when integrands start off with square or higher roots, or when higher transcendental functions (error functions etc˙ ) are to be handled, and not all combinations can be coped with, but the vast majority of the integrals re-quired in engineering applications can now be done by computer much faster and more reliably than either hand work or inspection of books of tables allows.
6 REDUCE keywords and switches Note that Reduce10 is an evolving system — new releases come out every couple of years and these always add a number of new operators and capabilities. In some cases awkward old syntax is replaced by a neater way of expressing things.
The list will therefor not always contain a complete list of the functions that are available, but it does include the ones most important for any examples you might want to try for this course.
ABS Computes the absolute value of an expression 10And of course all the other major algebra systems.
41 ACOS Arc-cosine ACOSH Arc-cosh ADD Alternative for + ALGEBRAIC Used when switching between algebraic and symbolic modes.
ALLBRANCH Switch. Used with SOLVE. (on) ALLFAC Switch. If on, expressions are displayed with common factors first. (on) AND Logical operator ANTISYMMETRIC Declares operators to be antisymmetric in their arguments.
ARGLENGTH Number of arguments of top level operator in expression.
ARRAY For declaring arrays.
ASIN Arc-sine ASINH Arc-sinh ATAN Arc-tan ATANH Arc-tanh BEGIN END Compound statement.
BYE Finishes Reduce job, clears it from memory.
CARDNO! Fortran output option.
CLEAR For removing assignments and substitutions.
COEFF partitions polynomial expression into coefficients COMMENT Text between COMMENT and ; or $ is ignored.
COMP Switch. Used to invoke Lisp compiler. (off) CONS Alternative for dot operator (more usual to use .).
CONT Used to continue file input which has been PAUSEd.
CONVERT Switch. If on, integral real coefficients are replaced by integers (on) COS Cosine COSH Hyperbolic cosine COT Cotangent CREF Switch. If on, does a cross-reference analysis. (off) DEFINE To define synonyms for Reduce keywords and indentifiers.
DEFN Used in symbolic mode.
DEG Leading degree of polynomial in given variable.
DEMO Switch. If on, press return to execute next command from file input. (off) DEN Denominator of a rational expression.
DEPEND Sets up dependencies between variables/kernels.
DET Determinant of matrix.
42 DF Partial differentiation of expression.
DIFFERENCE Take difference between two arguments (more usual to use -).
DILOG dilogarithm.
DISPLAY For displaying previous inputs.
DIV Switch. If on, displays have rational fractions, negative powers. (off) DO Used in FOR loops and with WHILE E The base of natural logarithms (2.71828...).
ECHO Switch. If on, file input is echoed in display. (on) EDITDEF Allows interactive editing of user defined procedure.
ED Allows interactive editing of any previous command.
END Terminates a BEGIN END block or a file.
EPS High energy physics: antisymmetric tensor of order 4.
EQ Used in symbolic mode EQUAL Alternative to = ERF Error function.
EXP Exponential function.
EXP Switch. If on, expressions are expanded during evaluation. (on) EXPINT exponential integral.
EXPR Used in symbolic mode.
EXPT Alternative for or ˆ (raising to a power).
EZGCD Switch. If on, with GCD on, uses EZ-GCD algorithm to compute gcds. (off) FACTOR Switch. If on, expressions are displayed in factored form. (off) FACTOR Declares expressions as factors for displays.
FACTORIZE Factorizes polynomial expression.
FAILHARD Switch. If on, impossible integration returns error. (off) FIXP Returns true if expression is integer, else false.
FLOAT Switch. If on, allows use of floating point numbers. (off) FOR Start of program loop.
FORALL LET Declares new substitution rule(s).
FOREACH Used in symbolic mode.
FORT Switch. If on, display is in a Fortran notation. (off) FORTWIDTH! Fortran output option.
FREEOF True if first argument does not contain second argument.
G High energy physics: a Dirac gamma matrix expression.
GCD Switch. If on, greatest common divisors are cancelled. (off) GCD Returns the greatest common divisor of two polynomials.
43 GEQ Alternative for >= GO (TO) For use with a labelled statement within BEGIN END.
GREATERP Alternative for > HIPOW! Set to highest non-zero power when COEFF is used.
I Square root (-1).
IF THEN ELSE Conditional statement.
IN Takes input from external Reduce file(s).
INDEX High energy physics INFIX Declares new infix operators.
INPUT Used to reference previous inputs in new computations.
INT Switch. Controls whether file input is batch or not.
(default depends on implementation) INT Integration INTEGER Declares local integer variables in BEGIN END block.
KORDER Declares internal ordering for variables.
LAMBDA Used in symbolic mode.
LCM Switch. If on, least common multiple of denominators is used. (on) LCOF Leading coefficient of polynomial.
LEQ Alternative for <= LESSP Alternative for < LET Declares substitutions.
LINEAR Declares operators to be linear in their arguments.
LINELENGTH For setting output linelength - see Manual.
LISP Synonym for SYMBOLIC.
LIST Switch. If on expressions are displayed one term to a line. (off) LOAD Used in symbolic mode.
LOG Natural logarithm LOWPOW! Set to lowest non-zero power when COEFF is used.
LTERM Leading term of expression.
MACRO Used in symbolic mode.
MAINVAR Main variable of polynomial.
MASS High energy physics: assign masses to vectors.
MAT Used to assign values to matrices.
MATCH Declares substitutions (less flexible than LET) MATRIX Declares matrix variables.
MAX Returns maximum of any number of numerical expressions.
MCD Switch. If on, makes common denominators when adding 44 expressions. (on) MEMBER Used in symbolic mode.
MEMQ Used in symbolic mode.
MIN Returns minimum of any number of numerical expressions.
MINUS Alternative for -MODULAR Switch. If on, does arithmetic modulo SETMOD. (off) MSG Switch. If off, warning messages are not displayed. (on) MSHELL High energy physics: puts variables “on the mass shell”.
MULT Alternative for NAT Switch. If off, display is in form that could be used for input. (on) NEQ Not equal to.
NERO Switch. If on, zero assignments are not displayed. (off) NIL Synonym for zero.
NODEPEND Removes dependencies created by DEPEND.
NOLNR Switch. Integration: may be useful if no closed form solution. (off) NONCOM Declares operators to be non-commutative for multiplication.
NOSPUR High energy physics: (traces and Dirac matrix calculations).
NOT Logical operator.
NUM Numerator of a rational expression.
NUMBERP True if argument is a number, else false.
NUMVAL Switch. If on, expressions are evaluated numerically. (off) OFF Turns off the named mode switches.
ON Turns on the named mode switches.
OPERATOR Declares new prefix operators.
OR Logical operator.
ORDER Declares an ordering for variables in displays.
ORDP True if first argument is ordered ahead of second argument.
OUT Directs output to named file or to terminal (OUT T;).
OUTPUT Switch. If off, there is no printing at the top level. (on) OVERVIEW Switch. Factorization: connected with TRFAC. (off) PART Extracting parts of expressions PAUSE In file input, offers option of continuing from terminal.
PERIOD Switch. Fortran output option (re inclusion of decimal points). (on) PFACTORIZE Factorizes univariate polynomial, modulo given prime.
PGWD Switch. Used in symbolic mode. (off) 45 PI Circular constant.
PLAP Switch. Used in symbolic mode. (off) PLUS Alternative for + PRECEDENCE Sets precedence of new operators declared by INFIX.
PRECISION Sets precision for real arithmetic, used with ON ROUNDED.
PRET Used in symbolic mode. (off) PRI Switch. If off, all output declarations and switches are ignored. (on) PROCEDURE Names statement(s).
PRODUCT Used with FOR to find products.
PUT To define synonyms for ALGEBRAIC or SYMBOLIC.
PWRDS Switch. Used in symbolic mode. (on) QUIT Exit from Reduce.
QUOTIENT Take ratio of two arguments (more usual to use /).
RAISE Switch. If on case of letters is ignored in keywords, expressions. (on) RAT Switch. Used with FACTOR for displaying expressions. (off) RATIONAL Switch. If on, polynomials use rational numbers. (off) REAL Declares local real variables in BEGIN END block.
RECIP Take reciprocal of argument (more usual to use 1/argument).
REDERR Print error message.
REDUCT Reductum of expression with respect to variable.
REMAINDER Remainder when first polynomial is divided by second.
REMFAC Clears the effect of FACTOR.
REMIND High energy physics: removes the effect of INDEX.
REPEAT UNTIL Provides repetition RESUBS Switch. If off, no resubstitutions are made after the first. (on) RESULTANT Resultant of two polynomials with respect to given variable.
RETRY Tries to do the command in which the last error occurred.
RETURN For transfer out of BEGIN END.
ROUNDED Switch. If on, gives real number evaluation (see PRECISION). (off) SAVEAS Alternative for x:=ws$ SAVESTRUCTR Switch. If on, causes STRUCTR to store results. (off) SCALAR Declares local variables in BEGIN END block.
SETQ Alternative to := SETMOD Sets modular base, used with mode switch MODULAR.
46 SHARE Used in symbolic mode.
SHOWTIME Displays the elapsed time since last SHOWTIME.
SHUT Closes output file(s).
SIN Sine SINH Hyperbolic sine SMACRO Used in symbolic mode.
SOLVE Solves one or more simultaneous equations.
SOLVEINTERVAL Switch. If on, inexact roots are represented by intervals. (off) SOLVESINGULAR Switch. If on, solutions may include arbitrary constants. (on) SOLVEWRITE Switch. If on, solutions are displayed. (on) SPUR High energy physics: traces in Dirac matrix calculations.
SQRT square root STEP UNTIL Used in FOR loops STRUCTR Displays the structure of an expression.
SUB Replace variable by expression in an expression.
SUCH THAT Used in FORALL LET SUM Used with FOR to find sums SYMBOLIC Used when switching between algebraic and symbolic modes.
SYMMETRIC Declares operators to be symmetric in their arguments.
T Cannot be formal parameter or local variable in procedure.
TAN Tangent TANH Hyperbolic tangent TERMS Number of top level terms in numerator of argument.
TIME Switch. If on, cpu time used by each command is displayed. (off) TIMES Alternative for TIMINGS Factorization: connected with TRFAC. (off) TP Transposes a matrix.
TRACE Trace of a matrix.
TRFAC Switch. Factorization: if on, traces operation of the algorithm. (off) TRINT Switch. Integration: if on, traces operation of the algorithm. (off) UNTIL Used with FOR and WHILE VARNAME Fortran output option, for naming expressions.
VECDIM High energy physics: setting dimensions 47 VECTOR High energy physics: declaring vectors.
WEIGHT Asymptotic commnad for assigning weights.
WHILE DO Provides repetition WRITE Displays expressions, strings.
WS Used to reference previous outputs for new computations.
WTLEVEL Asymptotic command to reset weight level (default 2).
!MODE Displays the current mode (algebraic or symbolic).
; Terminator for a command, result is displayed.
$ Terminator for a command, result is not displayed.
:= Assignment symbol .
Dot operator, and decimal point.
<< >> Indicates a group statement.
( ) Simple brackets.
: May replace STEP 1 UNTIL in FOR loops, also use with labels.
, Used as separator in lists.
" " Delimits text in WRITE statement.
% Text between % and end of line is ignored.
!
Is an escape character for special symbols in an identifier.
’ Used in symbolic mode.
{ } Used to write lists.
The following arithmetic operators have their usual meaning on algebraic ex-pressions ( and ˆ are synonyms): + - / ˆ The following relational operators have their usual meaning for comparing numbers: = >= > <= < 48
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5
Under what condition does an exponential graph have an x intercept?
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An exponential graph of the form y=a⋅bx (where a is a constant and b is the base of the exponential) has an x-intercept when the graph crosses the x-axis. This occurs when y=0.
For the function y=a⋅bx:
If a>0 and b>1: The graph approaches the x-axis but never touches it, so there is no x-intercept.
If a>0 and 0<b<1: Similarly, the graph approaches the x-axis but never touches it, so there is no x-intercept.
If a<0: The graph will be below the x-axis, and depending on the base b:
For b>1, the graph will rise towards the x-axis but will not touch it, so no x-in
An exponential graph of the form y=a⋅bx (where a is a constant and b is the base of the exponential) has an x-intercept when the graph crosses the x-axis. This occurs when y=0.
For the function y=a⋅bx:
If a>0 and b>1: The graph approaches the x-axis but never touches it, so there is no x-intercept.
If a>0 and 0<b<1: Similarly, the graph approaches the x-axis but never touches it, so there is no x-intercept.
If a<0: The graph will be below the x-axis, and depending on the base b:
For b>1, the graph will rise towards the x-axis but will not touch it, so no x-intercept.
For 0<b<1, the graph will fall towards the x-axis but will not touch it, so no x-intercept.
If a=0: The graph is simply the x-axis, which means it has infinitely many x-intercepts.
In summary, an exponential graph will never have an x-intercept unless the constant a is zero; otherwise, it approaches the x-axis but does not intersect it.
Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
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Author has 3.9K answers and 4M answer views
·
Updated 4y
How do you write an exponential function from a graph?
Assuming that you have a plot of a function in the form:
y=abx
where the horizontal asymptote is the x-axis, then the plot will look like this:
To find a, look for the ordinate where the curve crosses the y-axis. In the plot, we see that happens at 4. Therefore, a=4. To find b, look for the ordinate where the curve crosses the vertical line at x=1. In the plot, we see that happens at 8. Then, b is this number divided by a, or b=8a=84=2. This plot is the curve:
y=4(2)x
Using this equation, notice that:
when x=0, the equation is y=4(2)0=4
and when x=1, the equation is y=4(2)1=8
Assuming that you have a plot of a function in the form:
y=abx
where the horizontal asymptote is the x-axis, then the plot will look like this:
To find a, look for the ordinate where the curve crosses the y-axis. In the plot, we see that happens at 4. Therefore, a=4. To find b, look for the ordinate where the curve crosses the vertical line at x=1. In the plot, we see that happens at 8. Then, b is this number divided by a, or b=8a=84=2. This plot is the curve:
y=4(2)x
Using this equation, notice that:
when x=0, the equation is y=4(2)0=4
and when x=1, the equation is y=4(2)1=8
which is exactly what the plot shows.
Sometimes, the horizontal asymptote is not the x-axis where y=0. In this case, the equation is shifted either up or down, and has the form:
y=abx+c
where y=c is the horizontal asymptote. Then, the plot looks like this:
where the green dashed line is the horizontal asymptote. In this case, notice that the horizontal asymptote is y=1. This means that c=1 and the entire curve has been translated or shifted up by 1. To find a and b, we do what we did before, but need to subtract 1 from the ordinates that we read off of the graph to reverse the shift. So, a=5–1=4 and b=9–1a=9–14=2 and we find that this is a plot of the exponential function:
y=4(2)x+1
which is the previous function that has been shifted up by 1 unit.
Here is the plot of a famous exponential function:
First, we notice that its horizontal asymptote is the x-axis, y=0. Therefore, there is no shift up or down and c=0. Since the curve intersects the y-axis at (0,1), we know that a=1. Finally, we see that the curve intersects the vertical line x=1 at (1,2.718...). The number 2.718... is also called e. Therefore, b=ea=e1=e. We conclude that this is a plot of:
y=ex
This is the plot of a related exponential function:
Notice that this curve is decreasing instead of increasing. Similar to the previous function, c=0 and a=1 but, in this case, b=1e=e−1≈0.3679
This function is y=(e(−1))x=e−x
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Johnny M
Master's Degree from Harvard University (Graduated 2011)
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Updated Sep 9
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6y
For an exponential function of form f(x)=ae^kx+c, the addative constant c would be the sole determinant as to whether or not the function has an x intercept. (This is assuming that a is not equal to zero and that a, e, k, and c are all constants)
Related questions
What is the Y intercept in an exponential equation?
For the exponential functions y=a^x and y= a^(-x), where a is any positive constant, the y-intercept is always equal to what?
What is the reason why an exponential function graph does not have an x-intercept?
How do you find the x intercept of a graph?
What does a positive x-intercept mean on a graph?
Reynaldo Loya
B.S. in Mathematics & Computer Science, Texas A&M University - Kingsville (Graduated 2021)
·
Author has 335 answers and 535.6K answer views
·
6y
For an exponential of the form f(x)=a∗bkx, there is no possible way for there to be an x-intercept so long as a>0 and b>0.
For an exponential of the form f(x)=a∗bkx+c, there is an x-intercept at x=logb(−ca)k for c<0.
Alavi Ullah
I love math.
·
6y
If its of the form y=ab^t - c where c is the same sign as a, it should have an x intercept.
0=ab^t - c
c/a=b^t
t=log(b, c/a)
Since c has the same sign as a c/a>0
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present)
·
Author has 6.2K answers and 4.3M answer views
·
1y
Related
How do we solve the x-intercept of an exponential function?
The parent function (such as e^x) has no x-intercept, because it is asymptotic on the x-axis. Generally these functions will not have an x-intercept unless they are shifted downward by some specified amount, or they are reflected about the x-axis, then shifted upward by some specified amount.
What follows is the graph of a child exponential function that has been reflected as above, then shifted upward by 4 units: f(x) = -e^x + 4:
Solving for the x-intercept is straightforward, and is really much the same as for other kinds of functions. Just set f(x) equal to 0, and solve for x! Easy-peasy.
But
The parent function (such as e^x) has no x-intercept, because it is asymptotic on the x-axis. Generally these functions will not have an x-intercept unless they are shifted downward by some specified amount, or they are reflected about the x-axis, then shifted upward by some specified amount.
What follows is the graph of a child exponential function that has been reflected as above, then shifted upward by 4 units: f(x) = -e^x + 4:
Solving for the x-intercept is straightforward, and is really much the same as for other kinds of functions. Just set f(x) equal to 0, and solve for x! Easy-peasy.
But what about an exponential function that is more general than this? Consider the following:
Related questions
What is the y-intercept of the graph of the exponential function y=24(2) ^x? Can someone explain it as well?
What is the y-intercept in an exponential function graph?
What does it mean when a graph has a negative Y intercept or positive X intercept?
What is the y-intercept of an exponential function y bx?
What is the formula for x intercept?
B. S. Thomson
Lived in Vancouver, BC
·
Upvoted by
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 1.2K answers and 2.9M answer views
·
Mar 5
Related
How many x-intercepts and y-intercepts can exist in the graph of a function?
A few [so far] well-meaning answers to a peculiar question. Let us, however, make it into a genuine question.
There is nothing to be said, even in jest, about f(0), but the other part of the question, properly thought out, is of some interest.
Definition. Let Define the zero set of [math]f[/math] to be
[math]Z_f ={x\in \R: f(x)=0} .\tag1[/math]
As a warm-up consider what sets can appear as the zero set of some polynomial. Then, of course, one needs to get more ambitious depending on your background.
Q1. What are the necessary and sufficient conditions on a set [math]E\subset \R[/math] in order that there can exists a functi
A few [so far] well-meaning answers to a peculiar question. Let us, however, make it into a genuine question.
There is nothing to be said, even in jest, about [math]f(0)[/math], but the other part of the question, properly thought out, is of some interest.
Definition. Let [math]f:\R\to\R.[/math] Define the zero set of [math]f[/math] to be
[math]Z_f ={x\in \R: f(x)=0} .\tag1[/math]
As a warm-up consider what sets can appear as the zero set of some polynomial. Then, of course, one needs to get more ambitious depending on your background.
Q1. What are the necessary and sufficient conditions on a set [math]E\subset \R[/math] in order that there can exists a function [math]f:\R\to\R[/math] so that [math]Z_f=E[/math]?
Q2. What are the necessary and sufficient conditions on a set [math]E\subset \R[/math] in order that there can exists a continuous function [math]f:\R\to\R[/math] so that [math] Z_f=E[/math]?
Q3. What are the necessary and sufficient conditions on a set [math]E\subset \R[/math] in order that there can exists an infinitely differentiable function [math]f:\R\to\R [/math] so that [math] Z_f=E[/math]?
Q4. What are the necessary and sufficient conditions on a set [math]E\subset \R[/math] in order that there can exists a nondecreasing function [math]f:\R\to\R[/math] so that [math]Z_f=E[/math]?
Q5. What are the necessary and sufficient conditions on a set [math]E\subset \R [/math] in order that there can exists a function of bounded variation [math] f:\R\to\R[/math] so that [math]Z_f=E[/math]?
Q6. What are the necessary and sufficient conditions on a set [math]E\subset \R[/math] in order that there can exists a differentiable function [math] F:\R\to\R[/math] with [math]F'(x)=f(x) [/math] everywhere so that [math]Z_f=E[/math]?
These are the kind of questions that naturally occur in real analysis. Perhaps curiousity about the cardinality of [math]Z_f[/math] as this post raised can lead to more serious musings, even serious research.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
4y
Related
How many x intercepts can a parabola have?
This is probably not quite the answer you expected…
But if we include some complex x values but just have real y values, the basic type of parabolas always cross the x plane where y = 0 TWICE!!!
THIS SHORT VIDEO IS WELL WORTH A LOOK!!!
2017-10-01 1432 Intro To Phantom Graphs-1.m4v
World's leading screen capture + recorder from Snagit + Screencast by Techsmith. Capture, edit and share professional-quality content seamlessly.
This is probably not quite the answer you expected…
But if we include some complex x values but just have real y values, the basic type of parabolas always cross the x plane where y = 0 TWICE!!!
THIS SHORT VIDEO IS WELL WORTH A LOOK!!!
2017-10-01 1432 Intro To Phantom Graphs-1.m4v
World's leading screen capture + recorder from Snagit + Screencast by Techsmith. Capture, edit and share professional-quality content seamlessly.
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987)
·
Author has 6.2K answers and 1.7M answer views
·
3y
Related
How can you write the equation of exponential graph if given on the graph is horizontal asymptote of 5 and an x-intercept of 4?
Let’s assume the following exponential function f(x) = ae^x + b
b is the horizontal asymptote → b = 5
0 = a•e^4 + 5 → a = -5/e^4 ≈ -0.091
Let’s assume the following exponential function f(x) = ae^x + b
b is the horizontal asymptote → b = 5
0 = a•e^4 + 5 → a = -5/e^4 ≈ -0.091
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Making smart financial decisions doesn't have to be complicated. In 2025, there are several simple yet highly effective money hacks that can make a huge difference in your financial health. These aren't complicated investment strategies; they are practical, everyday habits that help you keep more of your hard-earned money.
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Making smart financial decisions doesn't have to be complicated. In 2025, there are several simple yet highly effective money hacks that can make a huge difference in your financial health. These aren't complicated investment strategies; they are practical, everyday habits that help you keep more of your hard-earned money.
Here are 5 easy ways to boost your savings and make your income work for you:
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John Fryer
Author has 890 answers and 885.4K answer views
·
3y
Related
How do you find the base of an exponential function when given a point on the graph?
With difficulty
More information needed.
One point on a graph will not give us much information.
It would be helpful to know the form of the exponential equation.
Has it got a constant multiplier other than unity.
Is the power simply x or again a constant times x?
Is the graph displaced vertically.
Even with many points on a graph it is not easy to find the equation or even the base unless we have more information.
The one common pattern is the steep rise in values for the function when b is positive.
If the value is ( 0 , 1 ) then even for the simple y = a^x then a can take on an infinite number of va
With difficulty
More information needed.
One point on a graph will not give us much information.
It would be helpful to know the form of the exponential equation.
Has it got a constant multiplier other than unity.
Is the power simply x or again a constant times x?
Is the graph displaced vertically.
Even with many points on a graph it is not easy to find the equation or even the base unless we have more information.
The one common pattern is the steep rise in values for the function when b is positive.
If the value is ( 0 , 1 ) then even for the simple y = a^x then a can take on an infinite number of values
If we can take the equation as y = a^x we can get a result which may or may not be correct ( see first graph )
If the equation was y = a^2x then using the same point gives a completely different result
The graphs are not just similar but the same and so y = a^cx can be reduced to another base to the power x
In this example 1.6475² = 2.714
If the equation was y = b a ^x then instead of a y intercept of ( 0 , 1 ) we have an intercept of ( 0 , b )
Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
·
Author has 3.9K answers and 4M answer views
·
3y
Related
How do you write exponential function for a graph that includes given points (2,2) (3,16)?
An exponential function has the form:
From the first point math[/math], we get:
From the second point math[/math], we get:
If we divide the second equation by first equation, we get:
Then, substitute for [math]b[/math] into the first equation:
The required equation is:
A plot of this exponential function looks like this:
Note that you can use the properties of exponents to write this in the equivalent form:
An exponential function has the form:
[math]y=ab^x[/math]
From the first point math[/math], we get:
[math]2=ab^2[/math]
From the second point math[/math], we get:
[math]16=ab^3[/math]
If we divide the second equation by first equation, we get:
[math]\frac{16}{2}=\frac{ab^3}{ab^2}[/math]
[math]b=8[/math]
Then, substitute for [math]b[/math] into the first equation:
[math]2=a(8)^2[/math]
[math]a=\frac{1}{32}[/math]
The required equation is:
[math]y=\frac{1}{32}(8)^x[/math]
A plot of this exponential function looks like this:
Note that you can use the properties of exponents to write this in the equivalent form:
[math]y=2^{-5}(2^3)^x[/math]
[math]y=2^{-5}(2^{3x})[/math]
[math]y=2^{(3x-5)}[/math]
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
5y
Related
How do I find the equation of an exponential function from a graph?
For straightforward exponential graphs like y = b^x it is reasonably easy for whole number values of b.
Since b^0 = 1 they all go through (0, 1)
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Upvoted by
Dwight House
, Ph. D. Mathematics, Duke University (1972) · Author has 6.8K answers and 52.8M answer views
·
3y
Related
Can a graph have no y-intercept?
A graph does not have to cross the y axis but most graphs do.
Here are a few that do not…
And a really special one!!!
A graph does not have to cross the y axis but most graphs do.
Here are a few that do not…
And a really special one!!!
.
Related questions
What is the Y intercept in an exponential equation?
For the exponential functions y=a^x and y= a^(-x), where a is any positive constant, the y-intercept is always equal to what?
What is the reason why an exponential function graph does not have an x-intercept?
How do you find the x intercept of a graph?
What does a positive x-intercept mean on a graph?
What is the y-intercept of the graph of the exponential function y=24(2) ^x? Can someone explain it as well?
What is the y-intercept in an exponential function graph?
What does it mean when a graph has a negative Y intercept or positive X intercept?
What is the y-intercept of an exponential function y bx?
What is the formula for x intercept?
How do you find the intercepts of an exponential function graph?
How do I find the x and y intercepts of a graph?
What is the name of a function whose graph has an x intercept and a y intercept?
What is the significance of an x-intercept in a graph?
How do I graph the linear equation x = 4y by finding and plotting its intercepts?
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Compound interest is the concept of adding accumulated interest back to the principal, so that interest is earned on interest from that moment on.
Formula[]
Amount = Principal + Interest
The amount A from a certain principal P after appying compound interest at the rate of 100r% per year in t years n times is
If A = amount, P = principal, r = rate percent yearly (or every fixed period) and n is the number of years (or terms of the fixed period) the interest rates for the successive fixed periods are r1%, r2%, r3% ..., then A (amount) is given by
See also[]
Euler's number, where compounding continuously yields e.
Detailed Explanation on Compounding Interest[]
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Contract curve
The contract curve is a concept in economics that represents all efficient allocations of resources between two parties in a given market. It is derived from the points of tangency between the indifference curves of the parties involved.
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Arti Arti
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Updated on Jun 6, 2024
Reading time 3 minutes
In this guide
1. Contract curve
2. Key Takeaways
3. What is the Contract Curve?
4. Importance of the Contract Curve
5. How the Contract Curve Works
6. Examples of the Contract Curve
7. Real-World Application
Key Takeaways
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The contract curve shows the set of mutually beneficial agreements in an exchange economy.
It is derived from the points where the indifference curves of two parties are tangent to each other.
The contract curve is crucial for understanding the efficiency of resource allocation.
What is the Contract Curve?
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The contract curve is a graphical representation in the Edgeworth box, used in microeconomics to depict all the efficient allocations of resources between two individuals or entities. These allocations are efficient because they lie on the points of tangency between the individuals’ indifference curves, where no further mutual gains from trade are possible. Essentially, the contract curve represents the set of Pareto-efficient outcomes where both parties maximize their utility given their initial endowments.
Importance of the Contract Curve
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Efficiency: Highlights the most efficient distribution of resources between two parties.
Resource Allocation: Helps in understanding how resources can be optimally allocated to maximize overall welfare.
Negotiation: Provides a basis for negotiations and agreements between parties in an economy.
How the Contract Curve Works
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Derivation
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The contract curve is derived within the Edgeworth box, which is a graphical tool used to show the distribution of resources in a two-person, two-good economy. Each point on the contract curve corresponds to a combination of goods where the indifference curves of both parties are tangent, indicating that the marginal rate of substitution (MRS) is equal for both.
Properties
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Pareto Efficiency: Every point on the contract curve is Pareto efficient, meaning that no one can be made better off without making someone else worse off.
Mutual Benefit: The allocations on the curve are mutually beneficial, as they lie on higher indifference curves for both parties compared to their initial endowment points.
Practical Use
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In practical scenarios, the contract curve helps in understanding and predicting the outcomes of negotiations and trade between two parties. It is also used in welfare economics to assess the efficiency of resource allocations and in policy-making to design mechanisms that can achieve efficient outcomes.
Examples of the Contract Curve
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Labor and Leisure: In a labor-leisure model, the contract curve can show the efficient trade-offs between work hours and leisure time for two individuals.
Trade between Countries: For two countries trading two goods, the contract curve indicates the combinations of goods that can be traded efficiently to maximize the welfare of both countries.
Real-World Application
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Market Negotiations: Firms and labor unions can use the concept of the contract curve to reach efficient agreements on wages and working conditions.
Policy Making: Governments can design policies that aim to move the economy towards the contract curve, ensuring efficient resource allocation.
Trade Agreements: International trade agreements can be analyzed using the contract curve to find mutually beneficial terms of trade.
The contract curve is an essential concept in economics that illustrates the efficient allocations of resources between two parties. By understanding the contract curve, economists and policymakers can better analyze and design mechanisms for optimal resource distribution, ensuring that all parties involved reach mutually beneficial and efficient outcomes.
More definitions
Contingent asset
Contingent market
Contingent valuation
Continuous compounding
Continuous time
Contract for difference (CFD)
Contract note
Contribution margin
Contributory
Contributory pension scheme
Sources & references
Our editors fact-check all content to ensure compliance with our strict editorial policy. The information in this article is supported by the following reliable sources.
Arti
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Arti
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Arti is a specialized AI Financial Assistant at Invezz, created to support the editorial team. He leverages both AI and the Invezz.com knowledge base, understands over 100,000 Invezz related data points, has read every piece of research, news and guidance we\'ve ever produced, and is trained to never make up new...
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Question: 2. Consider the function f(x) = { x2 sin(1), x0 x = 0 (a) Show that f is differentiable everywhere (of course you'll have to rely on the limit definition to find the derivative at 0). (b) Show that the derivative f' is not continuous at 0. Thus f' is not differentiable at 0, so the original function f was differentiable, but not smooth (or even twice-
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Deriving Laws (Equations) of Motion Using Calculus
Functions & Calculus by Professor Calculish
2080 subscribers
25 likes
Description
2092 views
Posted: 8 Jul 2022
See how you can easily derive two fundamental laws of motion using integrals of rates of change. The concepts are simplified in a manner to help derive such equations and simultaneously allow the student to remember the technique.
The purpose of the channel is to learn, familiarize, and review the necessary precalculus and trigonometry/geometry topics that form a basis for calculus (with a focus on functions). Calculus topics include limits, derivatives, integrals, formula derivations, derivations/proofs, area/volumes of curves and calculus applications.
Utilize simple and easy techniques to get the material across to you.
Disclaimer: I am not a professor by profession.
Thank you.
CalculusBasedPhysicsofMotion #derivingLawsofMotion #IntegralCalculus #position #velocity #acceleration #kinematics #professorCalculish #PhysicsAndCalculus #IntegralsOfDerivatives #DerivativesofMotion #physics #motion #derivatives #integrals #ratesofchange #lawsofmotion #definiteintegration #physicseasytounderstand #linearmotion #physics #engineeringmaths #calculus #calculusfundamentals #derivatives #functions #linearmotion #motioninonedimension
4 comments
Transcript:
good evening everyone I am Mr ish thank you for joining me for what will be an interesting digression into one of these topics of physics we're not going to get too heavily into it we're just showing you some integral calculus applications with regards to the laws of motion we don't need to get too heavy with this because this is something you'll study in physics but my intent here is to show you how integral calculus can make some of these physics topics quite easy and I can say through experience never having known calculus when I was doing physics because I did the General Physics not the calculus based physics had I known these calculus Concepts those physics Concepts would have been much easier we'll look at two important laws of motions one is to do with acceleration and the other one is to do with velocity again all of this Falls within the Realms of integrals of rates of change we're looking at something which is something of this sort you have an integral but within this integral you have a derivative or you can do F derivative of t D of T but we're looking at integrals rate of change and I've stressed this concept quite heavily recently that when you have an acceleration you do it integration you lead to the velocity function and when you do its integral you end up at the position function we don't have to do too much more but you can always reverse this procedure back by means of derivative we've talked about this we are looking at two important laws of motion derivations two specific formula in this video and when you see these formula come about you'll know exactly what they are but you will see something here within the lens of integral calculus and everything I'm showing you here can be derive without having to use calculus but when you look at this procedure here you'll see and realize that calculus makes the derivation all the more easy the first one has to do with acceleration we all know acceleration is change in velocity divid by change in time if you are to consider this as a function and you were to integrate this by means of integrating the rates of change this is how it would come up about you would be looking at something with regards to this acceleration is dv/ DT and how do we proceed from here it's not hard take the DT on the other side DV is equal to acceleration DT you can think of acceleration here as a constant not as a variable and then you will integrate this when you integrate it you're adding something like this and you've seen something like this develop in the previous two videos where we've talked about velocity position and acceleration now we have to do this the variable here we can bring in our variable intervals we can look everything here from initial velocity to final velocity these are terms you should be familiar with time at zero and time at time T whatever that might be and now you have to integrate something which is basically a polom integral which is easy you know when nothing exists over here with regards to your variable the variable comes into place because here we're looking at something at V ^ 0 then you do + 1 n + 1/ n +1 right here you're looking at nothing with regards to the time variable you do that divid by that when you do this you end up coming to this you have the integral anti- dtive come through and the integral goes with but you end up coming with over here is the V the velocity variable from VF and v0 and then here you have an acceleration sitting outside as a constant and then the time variable comes here from time T to Time Zero you do the upper and the lower and the difference of the two you get a VF minus V 0 is equal to a t the T this right here T coming in in the Zer the zero is meaningless but when you solve for your final velocity you get the all important law of motion equation which you are all familiar with initial velocity plus acceleration time time is equal to your final velocity and this right here is one of the laws of motion we don't have to do much with this in terms of calculation because I'm not here to do calculations of this type of equation because this is not a physics video I'm just showing you how integral calculus can derive you one of these laws of motion when you think of everything with regards to integrals of rates of change or integrating a derivative you can easily come up to this formula now the previous derivation which you just saw related acceleration to Velocity in this next one which is the next law of motion we will relate velocity to position and you know velocity you can think about velocity as change in distance over change in time and if you want to be more better with our writing we can write change in position divid by change in time now if you think about everything here with regards to derivatives you can say DS over d t you can cross multiply this T upwards and hit it with that velocity velocity with respect to DT is equal to DS but what did we just derive we derived an important equation where the final velocity is equal to initial velocity plus a this right here is equal to VF which is is essentially equal to the V we're looking at over here you can substitute this right over here initial velocity plus a with respect to this DT is equal to DS now you have everything which can be integrated you have two rates of change which can be integrated and let's do that when you integrate them your equations will become this V 0 or initial velocity plus a withs to DT is equal to DS now we have to do is add our inter our intervals are this with respect to what we're looking here in terms of variable here the variable is time so it be zero here the variable is position so we can say initial position and we can just say final position with an SF or you can just write a capital S you don't even have to put that F designation but I will when you integrate this it's not hard you're integrating a polinomial with regards to the time here you're integrating something which looks like this zero when you integrate you do + one and 0 + one that we know here we're looking at everything with regards to time know how to do it instead of looking at x we're looking at time T so it's not hard when you integrate this you get the t pop up over here and here you get a t^2 / 2 because here we're integrating with respect to time you know you're looking at everything here from T and then zero upper limit lower limit here you just get the S come about SF and s0 when you do the definite integration over here the putting the t's and the zeros will lead you to exactly what you see initial velocity time plus a t^2 / 2 here you'll actually get final position minus initial position but we're looking at everything here with regards to the final position and let's bring it here the final position of an object is equal to initial position I'm taking this on the other side initial position plus all of this initial velocity time time plus half a t² and this right here is the all important law of motion which we all know from our study of physics derived Solly here through integral calculus by understanding that we are doing an integral of a rate of change or you can say we are doing integral of derivatives and you can see it over here you can either look at this SF as just a plain capital s or you can just say SF it's all the same some people use this some people use that basically this law of motion says that the final position of an object is equal to its initial position plus this portion over here which is the sum of the product of initial velocity and time and the product of .5 acceleration and Time Square and I've shown you how we've derived this using integral of rate of change and how we've shown you to derive this so all of this was just the intent of this video to show you the perspective of Laws of Motion through this integral calculus lens and you can probably realize that integrating this these two equations using integral calculus is not hard you can also think of this as a quadratic equation because it indeed is a quadratic think about it you have your constant you have your first order you have your second order you're looking at something which looks like f ofx is = to a x² + b x + C here's my C term here's my BX term here's my ax squ term order does not matter around these positives but this is exactly what we're looking at doing the derivation of this equation using quadratics is quite difficult we doing the derivation of the same equation using this framework is quite easy all you always have to remember is acceleration leads to Velocity by means of integral and velocity leads to position by means of and you can reverse all of this by means of derivatives and this framework right here is what I've repeatedly talked about in my previous two videos now keep talking about it because this is how I wish I had been taught these Topics by means of this framework it would have been so much easier for me any know I'm sharing it with you I hope you enjoyed this video thank you for watching bye
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Periodic Table of the Elements, in Pictures and Words
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The Periodic Table of the Elements,
in Pictures and Words
These colorful, fun, and informative periodic tables are great for elementary, middle, and high school students, as well as adults.
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© 2005-2016 Keith Enevoldsen. See Terms of Use.
The Periodic Table of the Elements, in Pictures
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Hi-res PDF for viewing/printing (2 pages, Pictures and Words)
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This pictorial periodic table is colorful, fun, and packed with information. In addition to the element's name, symbol, and atomic number, each element box has a drawing of one of the element's main human uses or natural occurrences. The table is color-coded to show the chemical groupings. Small symbols pack in additional information: solid/liquid/gas, color of element, common in the human body, common in the earth's crust, magnetic metals, noble metals, radioactive, and rare or never found in nature. It does not overload kids with a lot of detailed numbers, like atomic weights and valence numbers.
Note: Elements in Pictures and Elements in Words are a set. Either may stand alone, but they work best together.
See also: Interactive Periodic Table of the Elements, in Pictures and Words.
This contains most of the same information on an interactive web page.
See also: Links to translations in other languages.
The Periodic Table of the Elements, in Words
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This textual periodic table is packed with even more information. In addition to the element's name, symbol, and atomic number, each element box contains a textual description of the element's physical properties and a list of several of its human uses and/or natural occurrences. The table is color-coded to show the chemical groups, and each group is described in a panel of the same color. Other info panels describe atomic structure, chemical bonding, and radioactivity. It does not overload kids with a lot of detailed numbers, but it does provide some simple rules-of-thumb about atomic weights and valence numbers.
See also: The Elements — Descriptions, Uses and Occurrences.
This contains most of the same information, but in a less graphical, more textual html format. It also contains handy links to search for more info about each use.
The Periodic Table of the Elements, in Pictures and Words
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This has both Elements in Pictures and Elements in Words combined on a single page.
See also: Interactive Periodic Table of the Elements, in Pictures and Words.
This contains most of the same information on an interactive web page.
The Periodic Table of the Elements, in Pictures (Simplified)
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This is exactly the same as Elements in Pictures above, but with less information. The fine print — the legend boxes and the small symbols — have been removed. Use this if you want a cleaner image with less visual clutter, or you want less information, or you want to print smaller or at lower resolution.
See also: Plain Periodic Table
Elements Cards
Download zip file containing Hi-res PDF for printing cards (32 pages, 16 sheets)
Print at letter size (8.5x11 in) double-sided on card stock.
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Print-your-own elements cards. Use these however you want. It's fun to simply lay them out to make the whole periodic table. You can use them as flash cards to help you memorize the facts on the front and back of each card. If you want to play it as a game, you can invent your own game rules.
Print double-sided on card stock. Cut out cards with paper cutter or scissors. Nine cards per sheet. There's a card for every element, with a picture on the front and words on the back. Also included are key cards (symbol key and color key), info cards (atom info and bonding info), and group info cards. The zip file also includes a template for making a card box. It also includes printing instructions.
Sorry, card decks are not for sale at the Wlonk Shop. You need to print them yourself. See Links for cards at other shops.
Elements Sheets
Download zip file containing Hi-res PDFs for Printing Sheets (one sheet per element)
Print at letter size (8.5x11 in) single-sided.
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Print each element on a separate sheet. You can make a very large periodic table to cover a large wall. Or you can use the sheets separately, for example, each student chooses one element.
Print single-sided. Elements 1-98 have pictures. Also included are key sheets (symbol key and color key), info sheets (atom info and bonding info), and group info sheets.
Sorry, elements sheets are not for sale at the Wlonk Shop. You need to print them yourself.
Student Worksheets
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Print at letter size (8.5x11 in) single-sided.
Printable student worksheet. Print one or many per student. This worksheet has spaces for the student to write the element's symbol, name, atomic number, description, uses and/or occurrences, and a space to draw a picture.
Are you homeschooling? This activity can occupy your student for one or many days. If you want a big activity, print out 98 of these worksheets, and assign your student to research every element from 1 to 98 (the elements with uses) and fill out a worksheet for each element, one or a few elements per day. This not just busywork; knowing some facts about each element is useful basic knowledge for many careers: engineering, manufacturing, biology, medicine, ecology, agriculture, geology, physics, and, of course, chemistry!
Alternative Periodic Tables
Alternative periodic tables: left-step, spiral, cone, and cylinder
Students will have fun builiding paper models of all these alternative periodic tables. Then they can discuss how the models are different and yet the same.
Left-step Periodic Table
This left-step periodic table is the basis for the spiral, cone, and cylinder models.
Left-step periodic table
Left-step table, paper model
Left-step table, folded paper model
Left-step Periodic Table, PDF for viewing/printing (1 page)
This left-step table is a slightly modified version of the left-step table by Charles Janet (1928). This compact version places hydrogen, helium, lithium, and beryllium all in a single row.
Note that hydrogen is unique — it may be grouped with the alkali metals (like lithium) because it has one electron available to donate, grouped with the halogens (like fluorine) because gaining one electron will fill its valence shell, or it may be put in a group of its own.
Spiral Periodic Table
In this periodic table, the elements are arranged in a single flat spiral (not concentric circles).
Spiral periodic table
Spiral periodic table, paper model
Spiral Periodic Table, PDF for viewing/printing (1 page)
Conical Spiral Periodic Table
In this model, the elements are arranged in a single conical spiral (not stacked rings).
Conical spiral, paper model (front)
Conical spiral, paper model (back)
Conical Spiral Periodic Table, paper pattern, PDF for printing (1 page)
The paper model above is very easy to make. The more elaborate paper-and-plastic model below is more difficult to make.
Conical spiral, paper-and-plastic model (front)
Conical spiral, paper-and-plastic model (back)
Conical Spiral Periodic Table, paper-and-plastic pattern, PDF for printing (1 page)
Cylindrical Helix Periodic Table
In this model, the elements are arranged in a single cylindrical helix (not stacked rings).
Cylindrical helix, paper model (front)
Cylindrical helix, paper model (back)
Cylindrical Helix Periodic Table, paper pattern, PDF for printing (1 page)
The paper model above is very easy to make. The more elaborate paper-and-plastic model below is more difficult to make.
Cylindrical Helix, paper-and-plastic model (front)
Cylindrical helix, paper-and-plastic model (back)
Cylindrical Helix Periodic Table, paper-and-plastic pattern, PDF for printing (1 page)
Atomic Orbitals
Atomic Orbitals (Color)
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This color-coded chart shows what atoms look like. This chart shows all the fundamental atomic electron orbitals as electron probability density distributions (fuzzy clouds), which is close as you can get to visualizing what an atom really looks like. The orbitals are labeled. It describes other ways to visualize atoms, namely, electron orbits (like planets) and surfaces of constant probability (bulgy blobs). It has a small periodic table showing in which order the electron shells are filled.
Atomic Orbitals (Black)
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This white-on-black chart shows what atoms look like. This chart shows all the fundamental atomic electron orbitals as electron probability density distributions (fuzzy clouds), which is as close as you can get to visualizing what an atom really looks like. The orbitals are labeled. This elegant chart has little visual clutter.
See other web sites for 3D visualizations. See Links for 3D products.
See also: Plain atomic orbitals chart (color) and Plain atomic orbitals chart (black)
Particles
Particles (Standard Model)
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This chart shows what the universe is made of. This chart shows all the elementary particles in the standard model (SM) of particle physics, and many non-elementary particles too. It starts with the basics: an atom contains a nucleus of protons and neutrons, which are made of quarks. The chart organizes all the important particles and classes of particles: elementary fermions (quarks, leptons, electrons, neutrinos), elementary bosons (gluons, photons, W and Z bosons, Higgs, and gravitons [predicted]), composite particles (hadrons, baryons, protons, neutrons, mesons), anti-particles.
See also: Plain list of particles.
Reference: IUPAC is the world authority on the naming of elements (see IUPAC periodic table).
CC BY-SA Creative Commons Attribution-ShareAlike 4.0
Periodic Table of the Elements, in Pictures and Words© 2005-2016 Keith Enevoldsen elements.wlonk.com
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Also, visit my other website, thinkzone.wlonk.com
Keith Enevoldsen, Seattle, USA, Earth
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Quadratic systems: a line and a parabola | Equations | Algebra 2 | Khan Academy
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Posted: 19 Jul 2019
Keep going! Check out the next lesson and practice what you’re learning:
A system of equations that contains one linear equation and one quadratic equations can be solved both graphically and algebraically. Each method has its pros and cons. See an example using both methods.
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Transcript:
we're told the parabola given by y is equal to three x squared minus six x plus one and the line given by y minus x plus one equals zero our graph so you can see the parabola here in red and we can see the line here in blue and the first thing they ask us is one intersection point is clearly identifiable from the graph what is it and they want us to put it in here this is actually a screenshot from the exercise on khan academy but i'm just going to write on it if you're doing it on khan academy you would type it in but pause this video and see if you can answer this first part all right so one intersection point is clearly identifiable from the graph i see two intersection points i see that one and i see that one there this second one seems clearly identifiable because when i look at the grid it looks clearly to be at a value of x equals 2 and y equals 1. it seems to be the point 2 comma 1. so it's 2 comma 1 there and what's interesting about these intersection points is because it's a point that sits on the graph of both of these curves that means that it satisfies both of these equations that it's a solution to both of these equations so the other one is find the other intersection point your answer must be exact so they want us to figure out this intersection point right over here well to do that we're gonna we're gonna have to try to solve this system of equations and this is interesting because this is a system of equations where one of the equations is not linear where it is a quadratic so let's see how we could go about doing that so let me write down the equations i have y is equal to 3x squared minus 6x plus 1. and our next equation right over here y y minus x plus 1 is equal to 0. well one way to tackle and this is one way to tackle any system of equations is through substitution so if i can rewrite this linear equation as in terms of y if i can solve for y then i can substitute what y equals back into my first equation into my quadratic one and then hopefully i can solve for x so let's solve for y here actually let me color code it because this one is in red and this one is the line in that blue color so let's just solve for y the easiest way to solve for y is to add x to both sides and subtract 1 from both sides that was hard to see so and subtract 1 from both sides and so we are going to get y and then all the rest of the stuff cancels out is equal to x minus 1. and so now we can substitute x minus 1 back in for y and so we get x minus 1 is equal to 3 x squared minus 6x plus 1. now we can we want to get a 0 on one side of this equation so let's subtract x i'll do this in a neutral color now let's subtract x from both sides and let's add 1 to both sides and then what do we get on the left hand side we just get 0 and on the right hand side we get 3x squared minus seven x plus two so this is equal to zero now we could try to factor this let's see is there an obvious way to factor it can i think of two numbers a times b that's equal to the product of three and two three times two and if this looks unfamiliar you can review factoring by grouping and can i think of those same two a plus b where it's going to be equal to negative 7 and actually negative 6 and negative 1 work so i can what i can do is i can rewrite this whole thing as 0 is equal to 3x squared and then instead of negative 7x i can write negative 6x and minus x and then i have my plus 2. i'm just factoring by grouping for those of you who are not familiar with this technique you could also use a quadratic formula so then 0 is equal to so if i group these first two i can factor out a 3x so i'm going to get 3x times x minus 2 and in these second two i can factor out in these second two i can factor out a negative one so i have negative one times x minus two and then i can factor out a negative two i'll scroll down a little bit so i have some space so i have zero is equal to if i factor out an x minus 2 i'm going to get x minus 2 times 3 x minus 1. and so a solution would be a situation where either of these is equal to zero or i'll scroll down a little bit more so x minus two could be equal to zero or three x minus one is equal to zero the point which x minus two equals zero is when x is equal to two and for 3x minus 1 equals 0 add 1 to both sides you get 3x is equal to 1 or x is equal to 1 3. so we figured out the we already saw the solution where x is equals equals 2. that's this point right over here we already typed that in but now we figured out the x value of the other solution so this is x is equal to 1 3 right over here so our x value is 1 3 but we still have to figure out the y value well the y value is going to be the corresponding y we get for that x in either equation and i like to focus on the simpler of the two equations so we can figure out what is x when or what is y when x is equal to 1 3 using this equation we could have used the original one but this is even simpler it's already solved for y so y is equal to 1 3 minus 1. i'm just substituting that one-third back into this and so you get y is equal to negative two-thirds and it looks like that as well y is equal to negative two-thirds right over there so this is the point one-third comma negative two-thirds and we're done
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The History and Clinical Utility of Eponymous Triads – Clinical Correlations
Clinical Correlations
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The History and Clinical Utility of Eponymous Triads
May 10, 2021
Medical History 6 min read| Twitter| Facebook| Email
By Joshua Novack
Peer Reviewed
Beginning in medical school, in between lectures on biochemistry, anatomy, and physiology, students begin to learn the physical exam and its role in the diagnosis of medical conditions. Some of these findings are grouped into eponymous triads, named after the physicians of yore who first identified these findings and the diagnoses they suggest. How useful, though, are these triads in modern medical practice today? In this article, I will explore the recent evidence evaluating a dyad of triads and provide some history on the physicians whose names live on through these groupings.
Beck’s Triad
In 1935, thoracic surgeon Dr. Claude Schaeffer Beck published an article in the Journal of the American Medical Association describing how to diagnose cardiac tamponade in the setting of a pericardial effusion. In Dr. Beck’s paper, two triads were described, one for “acute compression” and the second for “chronic compression.” Beck’s triad of acute compression is the one we commonly learn today, describing pericardial tamponade as presenting with “(1) a falling arterial pressure, (2) a rising venous pressure and (3) a small quiet heart” . Or, as we would describe: hypotension, elevated jugular venous pressure, and muffled heart sounds . Beck’s triad of chronic compression consisted of ascites, elevated jugular venous pressure, and muffled heart sounds . I would recommend going here to watch a video of jugular venous pulsation in cardiac tamponade.
For the purposes of this article, I will focus on Beck’s triad of acute compression as a way to diagnose acute cardiac tamponade. A 2007 systematic review published in JAMA evaluated eight studies to determine the group’s sensitivity for the diagnosis of tamponade. The sensitivities of the individual components were as follows: hypotension 26 ± 10%, diminished heart sounds 28 ± 7%, and elevated jugular venous pressure 76 ± 14% . There was no estimation of the sensitivity for the entire triad of symptoms, but it is likely poor given the low individual sensitivities of hypotension and muffled heart sounds.
A more recent 2017 retrospective chart review of patients presenting to two emergency departments examined the sensitivity of Beck’s triad in 153 patients. While not a systematic review, the study did assess for the presence of Beck’s triad in each patient. The authors found that Beck’s triad was 0% (95% CI 0-19%) sensitive for the diagnosis of pericardial effusion, and that the sensitivity of any one of the triad’s components for identifying tamponade was 50 ± 22% .
Given the poor sensitivities of these findings, what is the most useful way to rule in or rule out tamponade at the bedside? In the 2007 systematic review, the most important physical exam finding to diagnose cardiac tamponade was a pulsus paradoxus greater than 10 mmHg with a sensitivity of 82 ± 10%. This finding had a positive likelihood ratio of 3.3 (95% CI 1.8-6.3), and a negative likelihood ratio of 0.03 (0.01-0.24). A higher cutoff for pulsus paradoxus provides even greater ability to detect tamponade. A pulsus paradoxus greater than 12 had a positive likelihood ratio of 5.9 (95% CI 2.4-14) . Pulsus paradoxus less than 10 argues strongly against the diagnosis of pericardial tamponade.
Claude Schaeffer Beck was born in 1894 and trained in neurosurgery at Harvard and Johns Hopkins Hospital before obtaining training in cardiovascular surgery at the University Hospitals of Cleveland. Besides his eponymous triad, he is also famous for performing the first successful defibrillation in 1947 and becoming the first professor of cardiovascular surgery in the United States in 1952. He passed away in 1971 .
Charcot’s Cholangitis Triad
In 1877, Jean-Martin Charcot, a French physician practicing at the Salpêtrière in Paris, presented to a group of physicians his criteria to diagnose acute cholangitis, in a talk titled “Leçons sur maladies du foie, des voies biliaires et des reins, faires a la faculte de medecine” [5,6]. Or, in English, as per Google Translate: “Lectures on Diseases of the Liver, Bile Ducts, and Kidneys, made at the Faculty of Medicine.” These criteria were fever, jaundice, and right upper quadrant pain [5,6]. Charcot’s criteria came to be known as Charcot’s cholangitis triad, to separate them from his neurological triad for multiple sclerosis: nystagmus, intention tremor, and scanning speech .
How useful is Charcot’s cholangitis triad in modern practice? Multiple studies have been done looking at the sensitivity and specificity. This is of vital importance, given the need for urgent treatment of this disease . A 2017 systematic review found 16 suitable studies in the literature. Sensitivity was calculated from all 16 studies, while specificity could only be calculated from three of them. The triad had a poor sensitivity of only 36.3%, with a stronger specificity of 93.2% . This leads to a positive likelihood ratio of 5.3, and a negative likelihood ratio of 0.68. Of note, there was great heterogeneity in the individually reported sensitivities, ranging from 7.7% in a cohort of patients with mild disease, to 72% in a sicker cohort . Overall, Charcot’s cholangitis triad appears more useful for ruling in a diagnosis of acute cholangitis, but is likely to miss many other cases.
In 1959, Drs. Benedict M. Reynolds and Everett L. Dargan of the Bronx Municipal Hospital Center reported a case series of patients with acute cholangitis who benefited from emergent surgical biliary decompression. They diagnosed these patients with what they termed acute obstructive cholangitis. These patients all had “the well known triad of chills and fever, jaundice, and right upper quadrant abdominal pain or tenderness,” but noted that “the addition of lethargy or mental confusion and shock to this triad” represented acute obstructive cholangitis as opposed to acute cholangitis . Today, we refer to this as Reynolds’ pentad, which we associate with suppurative cholangitis. However, in the systematic review above, the sensitivity of Reynolds’ pentad across nine of the studies was 4.82% .
Dr. Charcot was born in 1825 and received his medical doctorate from the University of Paris in 1853, before working at the Salpêtrière [7,10]. While many in medicine are familiar with Dr. Charcot for his cholangitis triad, he is better known for his contributions to neurology, and is often called “the father of neurology” . Charcot was a pathologist and gifted teacher who influenced many now-famous physicians, including Sigmund Freud and Charles Babinski [7,10]. Among the many diseases he helped characterize are multiple sclerosis, Charcot-Marie-Tooth Disease, amyotrophic lateral sclerosis, and Charcot’s joint. Dr. Charcot passed away in 1893 .
Dr. Reynolds graduated from New York University School of Medicine in 1948 and went on to practice surgery. Among other titles throughout his career, he was chairman of Surgery at Fordham Hospital from 1976 to 1982 . At the time his paper was published, Dr. Reynolds practiced at The Bronx Municipal Hospital Center, which consisted of Jacobi Hospital and Van Etten Hospital, a 500-bed sanitarium for the chronic care of patients with tuberculosis . Dr. Reynolds died in 2019 at the age of 94 .
Dr. Joshua Novack is a first-year resident in the NYU Langone Internal Medicine Residency
Peer Reviewed by David Kudlowitz, MD, assistant professor, Department of Medicine (Division of General Internal Medicine) and associate editor for Clinical Correlations
Image courtesy Wikimedia Commons
References
Beck CS. TWO CARDIAC COMPRESSION TRIADS. JAMA J Am Med Assoc. 1935;104(9):714. doi:10.1001/jama.1935.02760090018005
Stolz L, Valenzuela J, Situ-LaCasse E, et al. Clinical and historical features of emergency department patients with pericardial effusions. World J Emerg Med. 2017;8(1):29-33. doi:10.5847/wjem.j.1920-8642.2017.01.005. Accessed 10/26/2020.
Roy CL, Minor MA, Brookhart MA, Choudhry NK. Does This Patient With a Pericardial Effusion Have Cardiac Tamponade? JAMA. 2007;297(16):1810. doi:10.1001/jama.297.16.1810. Accessed 10/26/2020.
Dittrick Medical History Center. Claude Beck, defibrillation and CPR. Case Western Reserve University College of Arts and Sciences Dittrick Medical History Center. Accessed October 26, 2020.
Jean-Martin Charcot. Lecons sur les maladies du foie, des voies biliaires et des reins, faites a la faculte de medecine – – Jean-Martin Charcot – Hachette Bnf – Grand format – Le Hall du Livre NANCY.; 1877. Accessed October 29, 2020.
Lipsett PA, Pitt HA. Acute cholangitis. Front Biosci J Virtual Libr. 2003;8:s1229-1239. doi:10.2741/881. Accessed October 29, 2020.
Kumar DR, Aslinia F, Yale SH, Mazza JJ. Jean-Martin Charcot: The Father of Neurology. Clin Med Res. 2011;9(1):46-49. doi:10.3121/cmr.2009.883. Accessed October 29, 2020.
Rumsey S, Winders J, MacCormick AD. Diagnostic accuracy of Charcot’s triad: a systematic review. ANZ J Surg. 2017;87(4):232-238. doi:10.1111/ans.13907. Accessed October 29, 2020.
Reynolds BM, Dargan EL. Acute obstructive cholangitis; a distinct clinical syndrome. Ann Surg. 1959;150(2):299-303. doi:10.1097/00000658-195908000-00013. Accessed October 29, 2020.
Jean-Martin Charcot | French neurologist. Jean-Martin Charcot. Published August 12, 2020. Accessed October 29, 2020.
Benedict Reynolds Obituary (1925 – 2019) – The Journal News. The Journal News. Published April 16, 2019. Accessed October 29, 2020.
NYC Health + Hospitals. Jacobi Hospital History. Jacobi Hospital History. Accessed October 29, 2020.
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Yes, a parabola can have no x-intercepts. This occurs when the parabola opens upward (with a positive leading coefficient) and its vertex is above the x-axis, or when it opens downward (with a negative leading coefficient) and its vertex is below the x-axis.
In mathematical terms, for a quadratic function y=a x 2+b x+c y=a x 2+b x+c:
The x-intercepts are found by solving the equation a x 2+b x+c=0 a x 2+b x+c=0.
The discriminant D D of the quadratic equation, given by D=b 2−4 a c D=b 2−4 a c, determines the nature of the roots:
If D>0 D>0, there are two distinct x-intercepts.
If D=0 D=0, there is one x-intercept (a repea
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Yes, a parabola can have no x-intercepts. This occurs when the parabola opens upward (with a positive leading coefficient) and its vertex is above the x-axis, or when it opens downward (with a negative leading coefficient) and its vertex is below the x-axis.
In mathematical terms, for a quadratic function y=a x 2+b x+c y=a x 2+b x+c:
The x-intercepts are found by solving the equation a x 2+b x+c=0 a x 2+b x+c=0.
The discriminant D D of the quadratic equation, given by D=b 2−4 a c D=b 2−4 a c, determines the nature of the roots:
If D>0 D>0, there are two distinct x-intercepts.
If D=0 D=0, there is one x-intercept (a repeated root).
If D<0 D<0, there are no x-intercepts.
Thus, if the discriminant is negative, the parabola does not intersect the x-axis.
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Related questions
More answers below
Can a parabola have no Y intercept?
How many x intercepts can a parabola have?
Is it possible for a parabola to not cross with y intercept?
What does a parabola with just one x intercept look like? Also, how do you find the vertex?
How do you find the y-intercept when the parabola is not touching it?
Gary Russell
Former Professor at University of Iowa (1996–2025) · Author has 6K answers and 3.1M answer views
·4y
Sure! A quadratic function is a parabola. If f(x) = ax^2 + bx + c has no real roots, then there exists no real-valued number x such that f(x) = 0. That means that f(x) > 0 or f(x) < 0 for all values of x.
Draw a curve like y = x^2 + 1 and take a look:
Continue Reading
Sure! A quadratic function is a parabola. If f(x) = ax^2 + bx + c has no real roots, then there exists no real-valued number x such that f(x) = 0. That means that f(x) > 0 or f(x) < 0 for all values of x.
Draw a curve like y = x^2 + 1 and take a look:
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Johnny M
Master's Degree from Harvard University (Graduated 2011)
·Updated Sep 9
Does switching car insurance really save you money, or is that just marketing hype?
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·4y
Of course! If the discriminant b^2 - 4ac is a negative number, then the parabola has no x-intercepts—i.e., its equation has no real roots (albeit it does have two complex conjugate roots). If the discriminant is zero, then the parabola has one x-intercept (but its equation yields two identical real roots). If the discriminant is positive, then the parabola has two x-intercepts (while its equation has two nonidentical real roots).
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Luqman Khan
Author has 771 answers and 784.3K answer views
·4y
Related
Can a parabola have no Y intercept?
y^2=(x-1)
As you can see this parabola has no y intercept.
Therefore it is possible to have a parabola with no Y intercept
Continue Reading
y^2=(x-1)
As you can see this parabola has no y intercept.
Therefore it is possible to have a parabola with no Y intercept
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Related questions
More answers below
How many y-intercepts can a parabola have?
How do you know how many X intercepts a parabola has?
How do you find the x and y intercepts of a parabola?
How do you graph a parabola with no y-intercepts?
What are the x-intercepts of the parabola (-3.5,0) and (4.1,0)?
Navneet Teja
teacher · Author has 699 answers and 412.6K answer views
·4y
A parabola will have no x intercept of it’s vertex is above x axis and it opens upward, or if vertex is below x axis and it opens down ward.
y=a(x-b)^2+c. a ,c both greater than zero
y=a(x-b)^2+c. a,c both are less than zero
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Sripad Sambrani
Knows Sanskrit · Author has 6.8K answers and 2.9M answer views
·4y
For a general parabola, y=ax²+b+c, has X-intercepts based on the value of D=b²-4ac
No intercepts when D<0, parabola lies above(below) the X-axis based on a is positive (negative).
Single intercept when D=0 => y=(px+q)², parabola touches X-axis at (-q/p,0)
Two intercepts when D>0 parabola intercepts X-axis at points ([-b-√(b²-4ac)]/2a, 0) & ([-b+√(b²-4ac)]/2a, 0)
Trust this helps.
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Elaine Dawe
BMath, in Mathematics&Computer Science, University of Waterloo (Graduated 1985) · Author has 5.4K answers and 6.6M answer views
·2y
Related
What does a parabola with just one x intercept look like? Also, how do you find the vertex?
I’ll assume that parabolas are of the form: y=a x 2+b x+c y=a x 2+b x+c, which open up or down and have a vertical axis of symmetry.
A parabola with just one x x-intercept has vertex on the x x-axis.
Such parabolas can be written in the form y=a(x−h)2 y=a(x−h)2, where vertex =(h,0)=(h,0)
Examples:
y=−1 2 x 2⟹V=(0,0)y=−1 2 x 2⟹V=(0,0)
y=1.5(x−4)2⟹V=(4,0)y=1.5(x−4)2⟹V=(4,0)
y=2(x+3)2⟹V=(−3,0)y=2(x+3)2⟹V=(−3,0)
For general parabolas, we can find vertex by completing the square.
Example:
y=−2 x 2−20 x−47 y=−2 x 2−20 x−47
y=−2(x 2+10 x+25−25)−47 y=−2(x 2+10 x+25−25)−47
y=−2(x 2+10 x+25)+y=−2(x 2+10 x+25)+
Continue Reading
I’ll assume that parabolas are of the form: y=a x 2+b x+c y=a x 2+b x+c, which open up or down and have a vertical axis of symmetry.
A parabola with just one x x-intercept has vertex on the x x-axis.
Such parabolas can be written in the form y=a(x−h)2 y=a(x−h)2, where vertex =(h,0)=(h,0)
Examples:
y=−1 2 x 2⟹V=(0,0)y=−1 2 x 2⟹V=(0,0)
y=1.5(x−4)2⟹V=(4,0)y=1.5(x−4)2⟹V=(4,0)
y=2(x+3)2⟹V=(−3,0)y=2(x+3)2⟹V=(−3,0)
For general parabolas, we can find vertex by completing the square.
Example:
y=−2 x 2−20 x−47 y=−2 x 2−20 x−47
y=−2(x 2+10 x+25−25)−47 y=−2(x 2+10 x+25−25)−47
y=−2(x 2+10 x+25)+50−47 y=−2(x 2+10 x+25)+50−47
y=−2(x+5)2+3 y=−2(x+5)2+3
Vertex=(−5,3)Vertex=(−5,3)
For general conics of the form:
A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0
we get a parabola with horizontal axis when A=B=0 A=B=0 and C,D≠0 C,D≠0
Otherwise (when A≠0 A≠0), the nature of the conic will depend on its discriminant B 2−4 A C B 2−4 A C
B 2–4 A C<0:ellipse B 2–4 A C<0:ellipse
B 2–4 A C=0:parabola B 2–4 A C=0:parabola
B 2–4 A C>0:hyperbola B 2–4 A C>0:hyperbola
x x-intercepts occur where y=0 y=0, giving:
A x 2+D x+F=0 A x 2+D x+F=0
So we get exactly one x x-intercept to original equation when this equation has exactly one solution, which occurs when discriminant D 2−4 A F=0 D 2−4 A F=0.
Therefore, the general equation for a conic
A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0
must meet the following criteria so that it is a parabola with one x x-intercept:
A=0,C≠0 B=0,D≠0 or B 2−4 A C=0 D 2−4 A F=0 A=0,C≠0 B=0,D≠0 or B 2−4 A C=0 D 2−4 A F=0
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Dip Bhattacharya
M.S + Ed D in Mathematics&Secondary Math Education, Clarion University Of Pennsylvania (Graduated 1982) · Author has 3.1K answers and 2M answer views
·4y
Sure !
y = x^2 + 1 { it is above the x-axis)
Or y = -X^2 - 1 { it id below the x-axis)
b^2- 4ac < 0
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Richard Goldstone
PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995) · Upvoted by
Michael Jørgensen
, PhD in mathematics · Author has 1.8K answers and 3.9M answer views
·10mo
Related
What can you change in making a parabola that intersects the x-axis two times?
For y=a x 2+b x+c,y=a x 2+b x+c, the quadratic formula for the roots tells us that we must have b 2–4 a c>0.b 2–4 a c>0. But that only accounts for parabolas with axis of symmetry perpendicular to the x x-axis.
The general conic section equation is
a x 2+b x y+c y 2+d x+e y+f=0,a x 2+b x y+c y 2+d x+e y+f=0,
and this is a parabola if and only if b 2–4 a c=0,b 2–4 a c=0, so for a parabola we have
a x 2±2 x y√a c+c y 2+d x+e y+f=0.a x 2±2 x y a c+c y 2+d x+e y+f=0.
If this intersects the x x-axis, then there must be points of the form (t,0)(t,0) that satisfy the equation, and for those points we must have
a t 2+d t+f=0,or t=−d±√d 2−4 a f 2 a.a t 2+d t+f=0,or t=−d±d 2−4 a f 2 a.
In order to have tw
Continue Reading
For y=a x 2+b x+c,y=a x 2+b x+c, the quadratic formula for the roots tells us that we must have b 2–4 a c>0.b 2–4 a c>0. But that only accounts for parabolas with axis of symmetry perpendicular to the x x-axis.
The general conic section equation is
a x 2+b x y+c y 2+d x+e y+f=0,a x 2+b x y+c y 2+d x+e y+f=0,
and this is a parabola if and only if b 2–4 a c=0,b 2–4 a c=0, so for a parabola we have
a x 2±2 x y√a c+c y 2+d x+e y+f=0.a x 2±2 x y a c+c y 2+d x+e y+f=0.
If this intersects the x x-axis, then there must be points of the form (t,0)(t,0) that satisfy the equation, and for those points we must have
a t 2+d t+f=0,or t=−d±√d 2−4 a f 2 a.a t 2+d t+f=0,or t=−d±d 2−4 a f 2 a.
In order to have two values for t,t, we must have d 2–4 a f>0.d 2–4 a f>0.
So that’s it: in the equation
a x 2±2 x y√a c+c y 2+d x+e y+f=0,a x 2±2 x y a c+c y 2+d x+e y+f=0,
choose a a and c c not both zero so that a c≥0,a c≥0, choose e e and f f randomly, and then choose d d so that d 2>4 a f.d 2>4 a f.
For example, we could choose a,c,e,a,c,e, and f f all to be 1 and then choose d d so that d 2>4,d 2>4, say d=3.d=3. The result is two parabolas,
x 2±2 x y+y 2+3 x+y+1=0 x 2±2 x y+y 2+3 x+y+1=0
whose graphs are
Perhaps we want the parabola(s) to pass through two specified points on the x x-axis. Say we want them to go through (−1,0)(−1,0) and (1,0).(1,0). Then since a t 2+d t+f=0,a t 2+d t+f=0, we must have
a+d+f=0 a−d+f=0 a+d+f=0 a−d+f=0
which reduces to f=−a f=−a and d=0.d=0. The equation is thus
a x 2±2 x y√a c+c y 2+e y−a=0,a x 2±2 x y a c+c y 2+e y−a=0,
and, for example, we could take
x 2±2 x y+y 2+y−1=0 x 2±2 x y+y 2+y−1=0
to get
In general, suppose we want our parabolas to pass through (α,0)(α,0) and (β,0)(β,0) with α≠β.α≠β. Then we have to solve the simultaneous equations
α 2 a+α d+f=0 β 2 a+β d+f=0 α 2 a+α d+f=0 β 2 a+β d+f=0
for a,d,a,d, and f.f. We could of course use row reduction techniques, but subtracting one equation from the other quickly leads to
(α−β)((α+β)a+d)=0,(α−β)((α+β)a+d)=0,
and since α≠β,α≠β, we get d=−(α+β)a d=−(α+β)a and
f=−α 2 a−α d=−α 2 a+α(α+β)a=(α β)a.f=−α 2 a−α d=−α 2 a+α(α+β)a=(α β)a.
The result is equations
a x 2±2 x y√a c+c y 2−(α+β)a x+e y+(α β)a=0,a x 2±2 x y a c+c y 2−(α+β)a x+e y+(α β)a=0,
passing through (α,0)(α,0) and (β,0)(β,0) with α≠β.α≠β. The parameters left to choose are a,c,a,c, and e e with a a and c c not both zero and a c≥0.a c≥0.
If we fix a=1=c,a=1=c, then we get the one-parameter family of parabola pairs indexed by e,e,
(x±y)2−(α+β)x+e y+α β=0.(x±y)2−(α+β)x+e y+α β=0.
Here’s what we get for the points (2,0)(2,0) and (3,0)(3,0) with values of e e ranging from −4−4 to 4,4, giving the equations
(x±y)2−5 x+e y+6=0:(x±y)2−5 x+e y+6=0:
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·4y
Related
How many x intercepts can a parabola have?
This is probably not quite the answer you expected…
But if we include some complex x values but just have real y values, the basic type of parabolas always cross the x plane where y = 0 TWICE!!!
THIS SHORT VIDEO IS WELL WORTH A LOOK!!!
2017-10-01 1432 Intro To Phantom Graphs-1.m4v World's leading screen capture + recorder from Snagit + Screencast by Techsmith. Capture, edit and share professional-quality content seamlessly.
Continue Reading
This is probably not quite the answer you expected…
But if we include some complex x values but just have real y values, the basic type of parabolas always cross the x plane where y = 0 TWICE!!!
THIS SHORT VIDEO IS WELL WORTH A LOOK!!!
2017-10-01 1432 Intro To Phantom Graphs-1.m4v World's leading screen capture + recorder from Snagit + Screencast by Techsmith. Capture, edit and share professional-quality content seamlessly.
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Raymond Beck
Former Infantry Sergeant · Author has 42.4K answers and 10.3M answer views
·10mo
if it’s an upward opening parabola with vertex above the x axis, then it has no x intercepts
y=x^2 + 1 has vertex (0,1) and has no x intercepts
all horizontal or diagonal parabolas have x intercepts
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Rick Marcus
Former R&D Engineer at Stanford University (1983–1989) · Author has 823 answers and 121.3K answer views
·Aug 9
Related
How do you find the x and y intercepts of a parabola?
Here is the simple parabola y=x²-1.
The x-intercept of the parabola is the point at which the y-value is equal to zero. This is the point at which the function crosses (or meets) the x-axis. In the case of the parabola y=x²-1, set y=0 and solve for x. x²-1=0, therefore x=1 or x=-1. Thus the x-intercepts are at the points (...
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Silvio Capobianco
Senior Researcher at Tallinn University of Technology (2009–present) · Author has 3.4K answers and 634.1K answer views
·4y
Yes, for the same reason why a second degree polynomial with real coefficients can have no real roots.
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Related questions
Can a parabola have no Y intercept?
How many x intercepts can a parabola have?
Is it possible for a parabola to not cross with y intercept?
What does a parabola with just one x intercept look like? Also, how do you find the vertex?
How do you find the y-intercept when the parabola is not touching it?
How many y-intercepts can a parabola have?
How do you know how many X intercepts a parabola has?
How do you find the x and y intercepts of a parabola?
How do you graph a parabola with no y-intercepts?
What are the x-intercepts of the parabola (-3.5,0) and (4.1,0)?
What is the Y intercept of a parabola equation?
What is the definition of a parabola? Does a parabola have to have x intercepts? Can a parabola have only one y intercept?
Can a vertex of parabola be a x-intercept?
How can you find the x and y intercepts of a parabola without solving for them?
X -intercepts are 3 and 7 and vertex of the parabola is (-4,5) What will be the equation of the parabola?
Related questions
Can a parabola have no Y intercept?
How many x intercepts can a parabola have?
Is it possible for a parabola to not cross with y intercept?
What does a parabola with just one x intercept look like? Also, how do you find the vertex?
How do you find the y-intercept when the parabola is not touching it?
How many y-intercepts can a parabola have?
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https://www.wcpss.net/cms/lib/NC01911451/Centricity/domain/2407/4thgrademath/parent_overview_unit_6_multiplicative_comparisons.pdf
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Dear Parents, We will begin our next unit of study in math soon. The information below will serve as an overview of the unit as you work to support your child at home. If you have any questions, please feel free to contact me. I appreciate your on-going support. Sincerely, Your Child’s Teacher Wake County Public Schools, Unit Overview for Parents This document should not replace on-going communication between teachers & parents. Unit Name: Multiplicative Comparisons Common Core State Standards: 4.OA.1 Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations. 4.OA.2 Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison. Essential Vocabulary: multiplicative comparisons multiplication/multiply division/divide addition/add subtraction/subtract compare/comparison equation unknown remainders reasonableness mental computation estimation rounding Unit Overview: In this unit, students will apply their understanding of place value to help them explore multiple groups or division. Students will utilize their knowledge of base ten and groups of numbers within another number to help them compare quantities or numbers. Students will be asked to not only identify, but will need to verbalize which quantity is being multiplied and which quantity tells how many times it is multiplied. Students will need to illustrate comparative situations with various strategies listed below. In addition, students will need to translate the situation into an equation. Strategies/Skills: Students will build on their understanding of addition and subtraction by using place value strategies to make sense of the standard algorithms. They are expected to use a variety of models to support their reasoning about numbers. Comparison Bars Number Lines Dear Parents, We will begin our next unit of study in math soon. The information below will serve as an overview of the unit as you work to support your child at home. If you have any questions, please feel free to contact me. I appreciate your on-going support. Sincerely, Your Child’s Teacher Wake County Public Schools, Unit Overview for Parents This document should not replace on-going communication between teachers & parents. Video Support: Video support can be found on LearnZillion. o Comparing numbers using bar models o See multiplication as a comparison using number sentences o Compare numbers using additive and multiplicative comparisons o Represent unknown numbers using symbols or letters o Solve multiplicative comparison word problems by using bar models o Solve multiplicative comparison word problems by using bar models to represent divison o Solve multiplicative comparison word problems using multiplication or division Additional Resources: If you have limited/no internet access, please contact your child’s teacher for hard copies of the resources listed in this document. NCDPI Unpacking Document:
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A video of this demonstration is available at this link.
The photograph above shows two torsion pendulums, each of which has a disc whose mass is 4.57 kg. The torsion wire on the pendulum on the right has a greater diameter than the wire on the pendulum on the left. The torsion wire on the right thus has a greater force constant than that of the one on the left; it is stiffer than the one on the left.
Turn the disc of one of the pendulums in either direction, and then release it. The disc now oscillates back and forth. The pendulum on the right, whose torsion wire is stiffer than that of the one on the left, does so at a higher frequency than does the pendulum on the left. Setting the ring on the disc of either pendulum, and thus increasing the moment of inertia, lowers the frequency of oscillation. Both rings have the same mass, 4.19 kg, and for each pendulum, adding the ring increases the moment of inertia by a factor of about 2.7 (vide infra).
A torsion pendulum is analogous to a mass-spring oscillator. Instead of a mass at the end of a helical spring, which oscillates back and forth along a straight line, hower, it has a mass at the end of a torsion wire, which rotates back and forth. To set the mass-spring in motion, you displace the mass from its equilibrium position by moving it in a straight line and then releasing it. The helical spring (or gravity, depending on whether or not the system is oriented vertically, and in which direction you displace the mass) exerts a (linear) force to restore the mass to its equilibrium position. To set the torsion pendulum oscillating, you turn the mass (rotate it about its center), and then release it. To do this, you must exert a torque about the bottom of the torsion wire. The torsion wire, in turn, exerts a restoring torque to bring the mass back to its original position.
The torsion wires on the pendulums shown in the photograph are brass. The one on the pendulum on the left has a diameter of about 2.8 mm, and the one on the pendulum on the right has a diameter of about 3.9 mm. Both are approximately one meter long. The torsion wire on the pendulum on the right is therefore considerably stiffer than the one on the pendulum on the left.
The disc on each pendulum has a mass of 4.57 kg and a radius of 0.127 m, for a moment of inertia ((1/2)MR2) = 3.69 × 10-2 kg-m2. The rings have a mass of 4.19 kg, and an average radius of (0.122 m), for a moment of inertia (MR2) of 6.24 × 10-2 kg-m2. So adding a ring to a disc increases the moment of inertia by a factor of 2.7 relative to that of the disc alone. A white double arrow on each disc provides a reference by which to observe the oscillatory motion of the pendulum. A white stripe on each ring provides a similar reference. (Make sure that when you set the ring on the disc, the white stripe faces the class.)
When you rotate the disc hanging from the bottom of the torsion wire through some angle θ, the torsion wire exerts a restoring torque, τ = - kθ. This is Hooke’s law, and is analogous to the linear case, F = -kx. For a body subject to a torque, τ = Iα, where I is the body’s moment of inertia, and α is the angular acceleration of the body. From these two equations, we see that -kθ = I(d2θ/dt2). (Ω = dθ/dt, and α = dΩ/dt. Here I use Ω to denote speed of rotation, to avoid confusion with ω, used for frequency of oscillation.) Which we can write -(k/I)θ = (d2θ/dt2), or (d2θ/dt2) + (k/I)θ = 0. This is exactly the same form as the equation for the mass-spring in demonstration 40.12 -- Mass-springs with different spring constants and masses (except for one constant). It thus has the same solution. If we substitute ω = √(k/I), we have
(d2θ/dt2) + ω2θ = 0.
This has the characteristic equation r2 + ω2 = 0, whose roots are r = ±iω. This leads to the general solution
x = C1 cos ωt + C2 sin ωt.
As shown in the page for demonstration 40.12 -- Mass-springs with different spring constants and masses, appropriate substitution for the constants C1 and C2, and use of a trigonometric identity give the solution
θ = θm cos (ωt - φ),
where θm is the maximum displacement of the pendulum from its equilibrium position, and thus the amplitude of the oscillation.
The frequency of oscillation, in units of radians per second, is ω = √(k/I). The frequency in cycles per second, ν, equals ω/(2π). The period of oscillation, T, equals (2π)/ω, or 1/ν. φ is a phase factor, which depends on how the motion starts. We see that the frequency at which the torsional pendulum oscillates is proportional to √k, and inversely proportional to √I.
We also find an interesting relationship among θ, Ω and α, the position, velocity and acceleration of the disc. If we take θ = θm cos (ωt - φ), then Ω = dθ/dt = -ωθm sin (ωt - φ), and α = dθ2/dt2 = -ω2θm cos (ωt - φ). These equations show us that at the extrema (±θm), the angular velocity is zero, and the angular acceleration is at its maximum, and that at the equilibrium position (θ = 0), the angular acceleration is zero and the angular velocity is at its maximum.
The work to displace the disc through a particular angle, and thus the potential energy stored in the torsion wire when the disc is so displaced, is W = ∫τ·dθ, which equals ∫kθ dθ, which gives U = (1/2)kθ2. The rotational kinetic energy of the disc is K = (1/2)IΩ2 (where Ω = dθ/dt). The total energy must remain constant, except for frictional losses. The equations above, for θ and Ω, show that at the extrema, the potential energy is at its maximum, and the rotational kinetic energy equals zero. At the equilibrium position, the potential energy equals zero, and the rotational kinetic energy is at its maximum. As noted above, their sum is constant.
1) Thomas, George B., Jr. and Finney, Ross L. Calculus and Analytic Geometry (Reading, Massachusetts: Addison-Wesley Publishing Company, 1992), pp. 1080-1081.2) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 312.3)
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Step-by-step math courses covering Pre-Algebra through Calculus 3.
Definite integrals of even and odd functions
What to do if you think the function is even or odd
Sometimes we can simplify a definite integral if we recognize that the function we’re integrating is an even function or an odd function. If the function is neither even nor odd, then we proceed with integration like normal.
To find out whether the function is even or odd, we’ll substitute ???-x??? into the function for ???x???. If we get back the original function ???f(x)???, the function is even. If we get back the original function multiplied by ???-1???, the function is odd. In other words,
If ???f(-x)=f(x)???, the function is even
If ???f(-x)=-f(x)???, the function is odd
If we discover that the function is even or odd, the next step is to check the limits of integration (the interval over which we’re integrating). In order to use the special even or odd function rules for definite integrals, our interval must be in the form ???[-a,a]???. In other words, the limits of integration have the same number value but opposite signs, like ???[-1,1]??? or ???[-5,5]???.
The rules for integrating even and odd functions
If the function is even or odd and the interval is ???[-a,a]???, we can apply these rules:
When ???f(x)??? is even,
???\int^a_{-a}f(x)\ dx=2\int^a_0f(x)\ dx???
When ???f(x)??? is odd,
???\int^a_{-a}f(x)\ dx=0???
Here are two videos. The first is a walkthrough of a definite integral of an even function; the second is a walkthrough of a definite integral of an odd function.
Even functions:
Odd functions:
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Integrating even functions
Example
Integrate.
???\int^2_{-2}3x^2+2\ dx???
First we’ll check to see if the function meets the criteria for an even or odd function. To see if it’s even, we’ll substitute ???-x??? for ???x???.
???f(x)=3x^2+2???
???f(-x)=3(-x)^2+2???
???f(-x)=3x^2+2???
After substituting ???-x??? for ???x???, we were able to get back to the original function, which means we can say that
???f(x)=f(-x)???
and therefore that the function is even.
Looking at the given interval ???[-2,2]???, we see that it’s in the form ???[-a,a]???.
Since we know that our function is even and that our interval is symmetric about the ???y???-axis, we can calculate our answer using the formula
???\int^a_{-a}f(x)\ dx=2\int^a_0f(x)\ dx???
???\int^2_{-2}3x^2+2\ dx=2\int^2_03x^2+2\ dx???
Now, instead of integrating the left-hand side, we can instead integrate the right-hand side, and evaluating over the new interval will be a little easier.
???\int^2_{-2}3x^2+2\ dx=2\left(\frac{3}{3}x^3+2x\right)\bigg|^2_0???
???\int^2_{-2}3x^2+2\ dx=\left(2x^3+4x\right)\bigg|^2_0???
???\int^2_{-2}3x^2+2\ dx=\left[2(2)^3+4(2)\right]-\left[2(0)^3+4(0)\right]???
???\int^2_{-2}3x^2+2\ dx=24???
Integrating odd functions
Example
Integrate.
???\int^7_{-7}3x^7+4\sin{x}\ dx???
First we’ll check to see if the function meets the criteria for an even or odd function. Let’s start by testing it to see if it’s an even function by substituting ???-x??? for ???x???.
???f(x)=3x^7+4\sin{x}???
???f(-x)=3(-x)^7+4\sin{(-x)}???
???f(-x)=-3x^7-4\sin{x}???
???f(x)\neq f(-x)???
In order for the function to be even, ???f(-x)=f(x)???. Since ???f(x)\neq f(-x)???, this function is not even.
So we’ll check to see if the function is odd. Remember that an odd function requires ???f(-x)=-f(x)???. We can test this by substituting ???-x??? for ???x???.
???f(x)=3x^7+4\sin{x}???
???f(-x)=3(-x)^7+4\sin{(-x)}???
???f(-x)=-3x^7-4\sin{x}???
???f(-x)=-\left(3x^7+4\sin{x}\right)???
???f(-x)=-f(x)???
Because ???f(-x)??? becomes ???-f(x)???, we can say that the function is odd. Looking at the given interval ???[-7,7]???, we see that it’s in the form ???[-a,a]???.
Since we know that our function is odd and that our interval is symmetric about the ???y???-axis, we can calculate the answer using the formula
???\int^a_{-a}f(x)\ dx=0???
???\int^7_{-7}3x^7+4\sin{x}\ dx=0???
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https://pmc.ncbi.nlm.nih.gov/articles/PMC3810850/
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A general species delimitation method with applications to phylogenetic placements - PMC
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. 2013 Aug 29;29(22):2869–2876. doi: 10.1093/bioinformatics/btt499
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A general species delimitation method with applications to phylogenetic placements
Jiajie Zhang
Jiajie Zhang
1 The Exelixis Lab, Scientific Computing Group, Heidelberg Institute for Theoretical Studies, D-68159 Heidelberg, Germany, 2 Graduate School for Computing in Medicine and Life Sciences, University of Lübeck, 3 Institut für Neuro- und Bioinformatik, University of Lübeck, 23538 Lübeck, Germany, 4 Natural History Museum of Crete, University of Crete, GR-71409 Irakleio, Crete, Greece and 5 Institute of Molecular Biology and Biotechnology, Foundation for Research and Technology-Hellas-FORTH, GR-70013 Heraklion, Crete, Greece
Find articles by Jiajie Zhang
1,2,3, Paschalia Kapli
Paschalia Kapli
1 The Exelixis Lab, Scientific Computing Group, Heidelberg Institute for Theoretical Studies, D-68159 Heidelberg, Germany, 2 Graduate School for Computing in Medicine and Life Sciences, University of Lübeck, 3 Institut für Neuro- und Bioinformatik, University of Lübeck, 23538 Lübeck, Germany, 4 Natural History Museum of Crete, University of Crete, GR-71409 Irakleio, Crete, Greece and 5 Institute of Molecular Biology and Biotechnology, Foundation for Research and Technology-Hellas-FORTH, GR-70013 Heraklion, Crete, Greece
Find articles by Paschalia Kapli
1,4, Pavlos Pavlidis
Pavlos Pavlidis
1 The Exelixis Lab, Scientific Computing Group, Heidelberg Institute for Theoretical Studies, D-68159 Heidelberg, Germany, 2 Graduate School for Computing in Medicine and Life Sciences, University of Lübeck, 3 Institut für Neuro- und Bioinformatik, University of Lübeck, 23538 Lübeck, Germany, 4 Natural History Museum of Crete, University of Crete, GR-71409 Irakleio, Crete, Greece and 5 Institute of Molecular Biology and Biotechnology, Foundation for Research and Technology-Hellas-FORTH, GR-70013 Heraklion, Crete, Greece
Find articles by Pavlos Pavlidis
1,5, Alexandros Stamatakis
Alexandros Stamatakis
1 The Exelixis Lab, Scientific Computing Group, Heidelberg Institute for Theoretical Studies, D-68159 Heidelberg, Germany, 2 Graduate School for Computing in Medicine and Life Sciences, University of Lübeck, 3 Institut für Neuro- und Bioinformatik, University of Lübeck, 23538 Lübeck, Germany, 4 Natural History Museum of Crete, University of Crete, GR-71409 Irakleio, Crete, Greece and 5 Institute of Molecular Biology and Biotechnology, Foundation for Research and Technology-Hellas-FORTH, GR-70013 Heraklion, Crete, Greece
Find articles by Alexandros Stamatakis
1,
Author information
Article notes
Copyright and License information
1 The Exelixis Lab, Scientific Computing Group, Heidelberg Institute for Theoretical Studies, D-68159 Heidelberg, Germany, 2 Graduate School for Computing in Medicine and Life Sciences, University of Lübeck, 3 Institut für Neuro- und Bioinformatik, University of Lübeck, 23538 Lübeck, Germany, 4 Natural History Museum of Crete, University of Crete, GR-71409 Irakleio, Crete, Greece and 5 Institute of Molecular Biology and Biotechnology, Foundation for Research and Technology-Hellas-FORTH, GR-70013 Heraklion, Crete, Greece
✉
To whom correspondence should be addressed.
Associate Editor: David Posada
Received 2013 Apr 9; Revised 2013 Aug 20; Accepted 2013 Aug 21; Issue date 2013 Nov 15.
© The Author 2013. Published by Oxford University Press. All rights reserved.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC3810850 PMID: 23990417
Abstract
Motivation: Sequence-based methods to delimit species are central to DNA taxonomy, microbial community surveys and DNA metabarcoding studies. Current approaches either rely on simple sequence similarity thresholds (OTU-picking) or on complex and compute-intensive evolutionary models. The OTU-picking methods scale well on large datasets, but the results are highly sensitive to the similarity threshold. Coalescent-based species delimitation approaches often rely on Bayesian statistics and Markov Chain Monte Carlo sampling, and can therefore only be applied to small datasets.
Results: We introduce the Poisson tree processes (PTP) model to infer putative species boundaries on a given phylogenetic input tree. We also integrate PTP with our evolutionary placement algorithm (EPA-PTP) to count the number of species in phylogenetic placements. We compare our approaches with popular OTU-picking methods and the General Mixed Yule Coalescent (GMYC) model. For de novo species delimitation, the stand-alone PTP model generally outperforms GYMC as well as OTU-picking methods when evolutionary distances between species are small. PTP neither requires an ultrametric input tree nor a sequence similarity threshold as input. In the open reference species delimitation approach, EPA-PTP yields more accurate results than de novo species delimitation methods. Finally, EPA-PTP scales on large datasets because it relies on the parallel implementations of the EPA and RAxML, thereby allowing to delimit species in high-throughput sequencing data.
Availability and implementation: The code is freely available at www.exelixis-lab.org/software.html.
Contact:Alexandros.Stamatakis@h-its.org
Supplementary information:Supplementary data are available at Bioinformatics online.
1 INTRODUCTION
DNA barcoding studies mostly rely on a single marker gene and are widely used for DNA taxonomy (Goldstein and DeSalle, 2011; Vogler and Monaghan, 2007). More recently, high-throughput sequencing of barcoding genes has been deployed to disentangle the structure of microbial communities (Caporaso et al., 2011) and in metabarcoding biodiversity (Coissac et al., 2012) studies. A central analytical task in such studies is to classify molecular sequences into entities that correspond to species; this is commonly denoted as OTU-picking in metagenomic studies (Sun et al., 2012). The main goals of such methods are to identify known species and delimit new species (Vogler and Monaghan, 2007).
Numerous approaches exist for associating anonymous reads/query sequences with known species, for instance, nearest-neighbor BLAST (Liu et al., 2008) or the naïve Bayesian classifier (Wang et al., 2007). These methods use sequence similarity to associate reads with taxonomic ranks. Phylogeny-aware methods for identifying reads were introduced independently and simultaneously with the evolutionary placement algorithm (EPA; Berger et al., 2011) and pplacer (Matsen et al., 2010). Instead of sequence similarity, they use the phylogenetic signal in the reference and query sequences to attain higher classification accuracy. Note that obtaining a taxonomic classification from phylogenetic placements represents a difficult task because phylogenies and taxonomies are frequently incongruent (Cole et al., 2009). Placement methods are similar to closed-reference OTU-picking (Bik et al., 2012) or taxonomy-dependent methods (Schloss and Westcott, 2011). Their ability to associate query sequences with species depends on the completeness of the taxon sampling in the reference data (Meyer and Paulay, 2005). Closed-reference or taxonomy-dependent methods generally lack the ability to delimit new species; consequently, they may underestimate the number of species and hence the diversity in the query sequences (see example in Supplementary Fig. S1).
To identify new species, taxonomy-independent methods or de novo OTU-picking approaches are used to initially cluster sequences into so-called molecular operational taxonomic units (MOTUs) (Floyd et al., 2002; Goldstein and DeSalle, 2011; Vogler and Monaghan, 2007). Then, one can use a representative sequence from each MOTU cluster and assign a taxonomic rank via taxonomy-dependent methods. Although taxonomic assignments may still be inaccurate due to incomplete reference data, coarse-grain biodiversity estimates can be accurate when MOTUs are assigned to higher taxonomic ranks. De novo OTU-picking usually relies on unsupervised machine learning methods (Cai and Sun, 2011; Edgar, 2010; Fu et al., 2012) that cluster sequences based on, mostly arbitrary, sequence similarity thresholds (Puillandre et al., 2012; Schloss and Westcott, 2011). However, it is currently unclear how MOTUs correspond to species (Vogler and Monaghan, 2007).
To delimit species using molecular sequences, we initially need to define our species concept. The phylogenetic species concept (PSC) was initially introduced by Eldredge and Cracraft (1980) and subsequently refined by Baum and Donoghue (1995); Cracraft (1983); Davis and Nixon (1992); and Nixon and Wheeler (1990). For a review of PSCs definitions please refer to Baum and Shaw (1995). In general, phylogenetic species are the smallest units for which phylogenetic relationships can be reliably inferred. The PSC, in particular, from the genealogical point of view (Baum and Shaw, 1995), states that species reside at the transition point between evolutionary relationships that are best represented phylogenetically and relationships that are best reflected by reticulating genealogical connections (Goldstein and Desalle, 2000).
There already exist several PSC-based species delimitation approaches (e.g. see reviews in Fujita et al., 2012; Sites and Marshall, 2003, 2004). The General Mixed Yule Coalescent (GMYC) model (Fujisawa and Barraclough, 2013; Pons et al., 2006) for delimiting species on single genes is frequently used in empirical studies (Carstens and Dewey, 2010; Fontaneto et al., 2007; Monaghan et al., 2009; Powell, 2012; Vuataz et al., 2011).
The GMYC method models speciation (among-species branching events) via a pure birth process and within-species branching events as neutral coalescent processes. GMYC identifies the transition points between inter- and intra-species branching rates on a time-calibrated ultrametric tree by maximizing the likelihood score of the model. It assumes that all lineages leading from the root to the transition points are different species. GMYC has been shown to work well for comparatively small population sizes and low birth rates (Esselstyn et al., 2012). One drawback of GMYC is that it depends on the accuracy of the ultrametric input tree. Obtaining an ultrametric tree from a given phylogeny is a compute-intensive and potentially error-prone process. Most state-of-the-art likelihood-based tree calibration methods such as BEAST (Drummond and Rambaut, 2007) or DPPDIV (Heath et al., 2012) rely on Bayesian sampling using MCMC (Markov Chain Monte Carlo) methods. When delimiting species in phylogenetic placements, which requires calibrating thousands of trees, it is almost impossible to deploy these methods in an automated pipeline, given the difficulties to assess MCMC chain convergence, for instance.
Inspired by the PSC, we introduce the PTP model that can delimit species using non-ultrametric phylogenies. Ultrametricity is not required because we model speciation rate by directly using the number of substitutions. The PSC implies that deploying phylogenetic reconstruction methods within a species is inappropriate. A hierarchical relationship can nonetheless be inferred for intra-species sequences using phylogenetic methods. However, we expect to observe significant (in the statistical sense) differences between the relationships reconstructed among and within species. These differences are reflected by branch lengths that represent the mean expected number of substitutions per site between two branching events. Thus, our fundamental assumption is that the number of substitutions between species is significantly higher than the number of substitutions within species. In a sense, this is analogous to the GMYC approach that intends to identify significant changes in the pace of branching events on the tree. However, GMYC uses time to identify branching rate transition points, whereas, in contrast, PTP directly uses the number of substitutions.
PTP is simple, fast and robust. Thus, it can easily be integrated with the EPA to calculate the number of species in a set of query sequences that have been placed into a specific branch of the reference phylogeny. We implemented an open reference species delimitation pipeline by integrating PTP with the EPA to identify known and new species.
Initially, we assess the performance of GMYC and the PTP approach as general de novo species delimitation methods using real and simulated data. We then compare PTP and GMYC with two representative OTU-picking methods UCLUST (Edgar, 2010) and CROP (Hao et al., 2011). UCLUST represents a fixed threshold OTU-picking approach, whereas CROP is a soft threshold method that attempts to detect sequence clusters using a Gaussian mixture model. Finally, we evaluate the performance of our open reference approach EPA-PTP. For a fair comparison, we also integrated CROP with the EPA (EPA-CROP).
2 METHODS
2.1 The Poisson tree processes model
Classic speciation models such as the birth–death process (BDP) assume that new species will emerge and current species will become extinct at certain rates that are measured in unit time (Barraclough and Nee, 2001). Usually, a time-calibrated tree is required as an input. Thus, for molecular sequence data, a molecular clock model must be applied to calibrate the tree. Coalescent theory also relies on unit time to describe the relationships among ancestors and descendants in a population.
Instead, we may consider the number of substitutions between branching and/or speciation events, by modeling speciations using the number of substitutions instead of the time. The underlying assumption is that each substitution has a small probability of generating a speciation. Note that the substitutions are independent of each other. If we consider one substitution at a time in discrete steps, the probability of observing η speciations for κ substitutions is given by a binomial distribution. Because we assume that each substitution has a small probability ρ of generating a speciation, and the number of substitutions in a population of size η is large, the process follows a Poisson distribution in continuous time with rate . Therefore, the number of substitutions until the next speciation event follows an exponential distribution.
Comparing this with the assumptions of a BDP, we observe that each generation (e.g. with a generation time of 20 years) on a time-calibrated ultrametric tree has a small probability of speciation. The BDP does not model substitutions, thus, substitutions are superimposed onto the BDP, whereas PTP explicitly models substitutions. Substitution information can easily be obtained by using the branch lengths of the phylogenetic input tree. Thus, in our model, the underlying assumptions for observing a branching event are consistent with the assumptions made for phylogenetic tree inference.
We can now consider two independent classes of Poisson processes. One process class describes speciation such that the average number of substitutions until the next speciation event follows an exponential distribution. Given the species tree, we can estimate the rate of speciations per substitution in a straightforward way. The second Poisson process class describes within-species branching events that are analogous to coalescent events. We assume that the number of substitutions until the next within-species branching event also follows an exponential distribution. Thus, our model assumes that the branch lengths of the input tree have been generated by two independent Poisson process classes.
In the following step, we assign/fit the Poisson processes to the tree. Let T be a rooted tree, and P i be a path from the root to leaf i, where and l is the number of leaves. Let be the edge lengths of P i, representing the number of substitutions. We further assume that are independent exponentially distributed random variables with parameter λ. Let be the sum over the edge lengths for . We further define . B ik is the number of substitutions of the k th branching event, and is the number of branching events below B ik. Note that constitutes a Poisson process. Thereby, T and together form a tree of Poisson processes, which we denote as Poisson Tree Processes (PTP). To a rooted phylogeny with m species, we apply/fit one among-species PTP and at most m within-species PTPs. An example is shown in Figure 1.
Fig. 1.
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Illustration of the PTP. The example tree contains 6 speciation events: R, A, B, D, E, F, and 4 species: C, D, E, F. Species C consists of one individual; species D, E, F have two individuals each. The thick lines represent among-species PTP, and the thin lines represent within-species PTPs. The Newick representation of this tree is ((C:0.14, (d1:0.01, d2:0.02)D:0.1)A:0.15, ((e1:0.015, e2:0.014)E:0.1, (f1:0.03, f2:0.02)F:0.12)B:0.11)R. The tree has a total of 16 different possible species delimitations. The maximum likelihood search returned the depicted species delimitation with a log-likelihood score of 24.77, and = 8.33 and = 55.05
In analogy to BP&P (Yang and Rannala, 2010) and GMYC (Pons et al., 2006), we conduct a search for the transition points where the branching pattern changes from an among-species to a within-species branching pattern. The total number of possible delimitations on a rooted binary tree with m tips ranges between m (caterpillar tree) and , depending on the actual tree shape (Fujisawa and Barraclough, 2013). Because the search space is generally too large for an exhaustive search, we need to devise heuristic search strategies. Given a fixed species delimitation, we fit two exponential distributions to the respective two branch length classes (among- and within-species branching events). We calculate the log-likelihood as follows:
(1)
where x 1 to x k are the branch lengths generated by among-species PTPs, to x n are the branch lengths of within-species PTPs, is the speciation rate per substitution and is the rate of within-species branching events per substitution. The rates and can be obtained via the inverse of the average branch lengths that belong to the respective processes. Based on Equation (1), we search for the species delimitation that maximizes L. A standard likelihood-ratio test with one degree of freedom can be used to test if there are indeed two classes of branch lengths. Large P-value indicates that either all sequences are one species or that every sequence represents a single species.
We developed and assessed three heuristic search strategies for finding species delimitations with ‘good’ likelihood scores, which are described in the online supplement. For the experimental results presented here, we used the heuristic that performed best, based on our preliminary experiments.
2.2 Species delimitation using phylogenetic placements
In the following, we describe the open reference species delimitation pipeline that combines the EPA with the PTP (EPA-PTP). The EPA initially places a large number of query sequences (short reads) into the branches of a given reference phylogeny. Thereafter, we execute PTP separately and independently for the query sequences assigned to each branch. This allows to annotate the branches of the reference tree by the number of species induced by the query sequences that were placed into each branch. The input of our pipeline is a reference alignment where each sequence represents one species and a reference phylogeny for that alignment. The PTP method and the pipeline are implemented in Python and rely on the python Environment for Tree Exploration package (Huerta-Cepas et al., 2010) for tree manipulation and visualization.
Our pipeline executes the following steps:
Run UCHIME (Edgar et al., 2011) against the reference alignment to remove chimeric query sequences.
Use EPA to place the query sequences onto the reference tree. Sequences that have a maximum placement likelihood weight of <0.5 (i.e. an uncertain placement, see Berger et al., 2011 for details) are discarded.
For each branch in the reference tree, we extract the set of query sequences that have been placed into that branch and infer a tree on them using RAxML (Stamatakis, 2006). Because the PTP method requires a correctly rooted tree, we use the following two rooting strategies: if the branch leads to a tip, apart from the query sequences, we extend the alignment by including the reference tree tip sequence and that reference sequence that is furthest away from the current tip. The most distant sequence is used as outgroup. Keep in mind, that thereby the tree will be rooted at the longest branch (see the discussion below). To analyze query sequence placements at internal branches, we use the RAxML −g constraint tree option to obtain a rooted tree of the query sequences. The constraint tree consists of the bifurcating reference tree and a polytomy comprising the query sequences attached to the reference tree branch under consideration. The result of this constrained ML tree search is a resolved tree of query sequences that are attached to the reference tree branch. The attachment point is used as root.
Because we assume that the reference phylogeny is a species tree that reflects our knowledge about the speciation process and rate, we initially estimate only once on the reference phylogeny. Thereafter, we apply PTP to each query sequence (one for each branch of the reference phylogeny) tree to delimit species. Note that in this scenario we will only need to estimate , as remains fixed.
When PTP is applied to a placement of query sequences on a terminal branch, those queries that are delimited as one population with the reference sequence at the tip will be assigned taxonomically to the species represented by this reference sequence. Otherwise, they are identified as new species in the reference tree.
As mentioned previously, we also combined EPA with CROP (EPA-CROP). The method works as EPA-PTP, with the only difference that CROP is used instead of PTP to calculate the number of MOTUs for each placement.
3 EXPERIMENTAL SETUP
We initially tested stand-alone PTP for general species delimitation and compared it with the single-threshold GMYC model. The single-threshold GMYC model infers a single cutoff time T where all nodes above T represent species. Although the more advanced multiple-threshold GMYC allows for several threshold times T i, the single-threshold GMYC is usually more accurate than the multi-threshold version (see Fujita et al., 2012 for details). For simulated data, we used RAxML (Stamatakis, 2006) to infer phylogenetic trees, and then used them as input to PTP. Subsequently, these phylogenies were made ultrametric by r8s (Sanderson, 2003) to test GMYC. For UCLUST and CROP, only molecular sequences are needed as input. In both programs, we set the sequence dissimilarity threshold to 97%, a widely accepted threshold for bacterial sequences (Stackebrandt and Goebel, 1994). For real datasets, we used the phylogenetic tree and ultrametric tree from the original publications whenever possible, otherwise we used the same settings as described previously.
We then tested our open reference species delimitation approaches (EPA-PTP and EPA-CROP). We also assessed the impact of incomplete taxon sampling on the accuracy of these approaches, by removing up to 50% of the reference sequences.
3.1 Empirical datasets
3.1.1 Arthropod datasets
The Rivancidella dataset comprises three genes (cyt b, COI, 16S) and was originally used in Pons et al. (2006). The total number of sequences is 472, which represents 24 morphological species and 4 outgroup taxa. The estimated number of putative species for the genus as inferred by GMYC was 48 (with confidence limits ranging between 46 and 52 species). Alternative methods (see Pons et al., 2006 for details) used in this study yielded 46 and 47 putative species, respectively.
We also used COI marker datasets (Papadopoulou et al., 2010) of the genera Dendarus, Pimellia and Tentyria. The datasets comprise 51, 56 and 59 sequences, respectively. The number of species that were attributed to each taxon using morphological criteria was seven, one and one.
3.1.2 Gallotia dataset
The lizard genus Gallotia comprises seven species (based on genetic and morphological markers) that are endemic to the Canary islands. The taxonomic species tree and the molecular phylogeny for this dataset are fully congruent. The data (Cox et al., 2010) comprises four mitochondrial genes (cyt b, COI, 12S, 16S) and a total of 90 sequences (76 representing Gallotia and 14 outgroup sequences).
3.1.3 Arthropod metabarcoding dataset
This dataset contains 673 full-length COI arthropod sequences with a length of 658 bp. The sequence was obtained via polymerase chain reaction amplification and Sanger sequencing. Subsequently, these 673 sequences were resequenced with a 454 sequencer to generate a total of 133 057 short reads (Yu et al., 2012). Using the Sanger data as reference, Yu et al. developed metabarcoding protocols that use the 454 reads to unravel the diversity in the reference data. The authors use a multistep OTU-picking procedure with different similarity thresholds for clustering the 454 reads and the full-length reference sequences. The method clustered the 673 sequences into 547 MOTUs. The OTU-picking results for the 454 data are summarized in Table 4. Our PTP model finds 545 putative species in the 673 full-length sequences when directly applied to the phylogenetic reference tree. To ensure comparability of results, we used the 547 MOTUs identified in the original study to build a reference tree and reference alignment for testing the EPA-PTP and EPA-CROP pipelines. Initially, we aligned 454 sequences with a length exceeding 100 bp to these 547 reference sequences with HMMER (Eddy, 2009). Yu et al. initially blasted the 454-MOTU (obtained via three alternative clustering methods) to the Sanger-MOTUs using a threshold of 1e-10 and 97% minimum similarity. The Sanger-MOTUs that did not match any of the 454-MOTUs are called ‘dropouts’ by the authors. Inversely, 454-MOTUs that did not match Sanger-MOTUs are called ‘no-matches’.
Table 4.
Arthropod dataset: number of estimated MOTUs and species for the complete reference data and tree
| No. reads | OTU-picking | EPA-PTP |
---
| | Number of cluster | Drop-out (%) | No-match (%) | Number of cluster | Drop-out (%) | No-match (%) |
| reads | 973 | 19 | 42.8 | 587 | 7.3 | 13.6 |
| reads | 602 | 24 | 25.4 | 516 | 11.5 | 6.2 |
| reads | — | 36 | — | 441 | 21.9 | 3.2 |
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Note: Sanger data (the reference dataset) has a total of 547 MOTUs. The ‘—’ indicates that the number is not available in the original publication.
Analogously, in our pipelines, when the delimited species from 454 sequence placements contain one of the full-length reference sequences (see step 4 in 2.2), we consider this as a ‘match’. Further, we denote a full-length reference sequence that is not included in any short read placement delimitation as ‘dropout’. Finally, we call a short read placement that is delimited as a new species (i.e. does not contain a reference sequence) as ‘no-match’.
3.2 Simulations
We generated simulated datasets using a Yule coalescent model. We used ms (Hudson, 2002) and BioPerl (Stajich et al., 2002) in combination with INDELible (Fletcher and Yang, 2009) to simulate sequences. Using a modified version of the BioPerl module Bio::Phylo that allows to vary the birth rate value in the simulations, we initially generated a set of random species trees . The leaves of each tree T i () represent extant species. All 600 simulated datasets we generated contain 30 species. In the next step, we used ms to generate a structured coalescent gene tree. The node ages of the phylogenetic tree T i are interpreted as divergence times between populations. In other words, we treat species as diverged populations that were completely isolated from each other after they diverged from their common ancestor. Thus, using ms we simulated a multispecies coalescent gene tree with 30 species and 100 individuals per species. For each species, we randomly selected 10 individuals to generate evenly sampled (in terms of the number of individuals per species) datasets. We also generated unevenly sampled datasets containing 2 species with 100 individuals, 4 with 50 individuals, 8 with 10 individuals and 16 with 2 individuals. Finally, we used INDELible to simulate DNA alignments of 250, 500 and 1000 bp on the previously mentioned multispecies coalescent trees.
We generated datasets with a scaled birth rate (); small values generate large evolutionary distances between species. For details on the simulations and on the scaled rate , please refer to the online supplement.
We used the normalized mutual information (NMI) criterion (Vinh et al., 2010) to asses how the delimitation accuracy of the different algorithms agrees with the ground truth. The mutual information (MI) of two distinct partitions of the same dataset quantifies how much information is shared by these; NMI scales MI to values between 0.0 and 1.0. In our case, NMI = 1 means that the delimitation is identical to the ground truth, whereas NMI = 0 means that the delimited species are randomly partitioned compared with the ground truth.
Finally, we also tested the EPA-PTP and EPA-CROP pipelines on simulated data. In each simulated alignment, we randomly selected one individual sequence per species as reference sequence and treated the remaining sequences (of that species) as query sequences. To assess the impact of incomplete reference trees on species delimitations, we randomly removed up to 50% of the reference sequences. We deployed the same metrics as mentioned previously to quantify delimitation accuracy.
4 RESULTS
4.1 General species delimitation
The number of putative species delimited for Dendarus, Pimelia, Tentyria and Gallotia are comparable for all four methods (Table 1). For the Gallotia dataset, GMYC and PTP yield identical results. Three of the Gallotia species were split into two separate groups according to geographical isolation of the corresponding populations on different islands (see Supplementary Fig. S4).
Table 1.
Number of species delimitated on real data
| Taxon | Morphological | GMYC | PTP | CROP | UCLUST |
--- --- --- |
| Rivacindela | 24 | 48 | 27/44a | 6 | 82 |
| Dendarus | 7 | 10 | 9/11a | 7 | 11 |
| Pimelia | 1 | 10 | 9/15a | 7 | 10 |
| Tentyria | 1 | 2 | 2/2a | 1 | 3 |
| Gallotia | 7 | 10 | 10/10a | 9 | 15 |
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a Using the ultrametric tree as an input for PTP.
On the Rivacindela dataset PTP yields a more conservative delimitation than GMYC. PTP identifies 27 putative species (GMYC: 48), which is closer to the number of morphological species (24) and the number of independent networks (25) obtained via statistical parsimony (Pons et al., 2006). This pronounced difference may be associated with the construction of the ultrametric tree. According to the r8s manual, the presence of many short (close to zero) branches in the tree can yield inaccurate results. When applying PTP to the ultrametric tree, the resulting estimate is substantially closer to the GMYC estimate (see Table 1). Thus, we believe that the overestimation of the Rivacidela species by GMYC is most probably because of an erroneous ultrametric tree reconstruction. CROP and UCLUST yield dissimilar results; CROP only detects 6 clusters, whereas UCLUST detects 82 clusters.
The results on evenly sampled simulated data are summarized in Table 2 and Supplementary Table S1. On average, PTP shows the best performance and outperforms GMYC in all the test scenarios. OTU-picking methods work well on datasets with small values that is when the evolutionary distances between species are large. For , UCLUST generally outperforms PTP and yields the best overall results. However, with increasing the accuracy of OTU-picking methods decreases steeply. As expected, for shorter sequence lengths (250 and 500 bp), accuracy deteriorates for all methods and in a more pronounced way for PTP and GMYC. However, even with sequence lengths of 250 bp, PTP still yields best results on datasets with .
Table 2.
Species delimitation accuracy (measured in NMI) on simulated evenly sampled data
| NMI | b′ | Mean (variance) |
:---:
| | 5 | 10 | 20 | 40 | 80 | 160 | |
| 1000 bp | | | | | | | |
| UCLUST | 0.969 | 0.959 | 0.938 | 0.892 | 0.782 | 0.575 | 0.852 (0.023) |
| CROP | 0.964 | 0.930 | 0.848 | 0.646 | 0.232 | 0.038 | 0.609 (0.151) |
| GMYC | 0.924 | 0.914 | 0.907 | 0.886 | 0.834 | 0.697 | 0.860 (0.007) |
| PTP | 0.944 | 0.935 | 0.922 | 0.905 | 0.882 | 0.857 | 0.907 (0.001) |
| 250 bp | | | | | | | |
| UCLUST | 0.967 | 0.954 | 0.930 | 0.871 | 0.735 | 0.522 | 0.829 (0.029) |
| CROP | 0.961 | 0.917 | 0.800 | 0.545 | 0.152 | 0.024 | 0.566 (0.159) |
| GMYC | 0.892 | 0.620 | 0.484 | 0.464 | 0.550 | 0.503 | 0.585 (0.025) |
| PTP | 0.946 | 0.927 | 0.907 | 0.881 | 0.833 | 0.780 | 0.879 (0.003) |
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On the unevenly sampled simulated datasets (Supplementary Table S2), the delimitation accuracy decreases for UCLUST and PTP. CROP and GMYC yield higher NMI scores than on evenly sampled dataset. On average, PTP yields the best results over all (evenly and unevenly sampled) simulated datasets.
4.2 Species delimitation with phylogenetic placements
By combining EPA with PTP (or CROP) and applying it to simulated data as described in Section 3.2, we can substantially improve the delimitation accuracy on simulated data (Table 3 and Supplementary Tables S3 and S5).
Table 3.
Species delimitation accuracy (measured in NMI) on simulated evenly sampled data using the EPA-PTP pipeline
| NMI | b′ | Mean (variance) |
:---:
| | 5 | 10 | 20 | 40 | 80 | 160 | |
| 1000 bp | | | | | | | |
| Full ref. | 0.989 | 0.978 | 0.962 | 0.933 | 0.884 | 0.836 | 0.930 (0.003) |
| 90% ref. | 0.984 | 0.972 | 0.955 | 0.925 | 0.876 | 0.830 | 0.923 (0.003) |
| 80% ref. | 0.976 | 0.966 | 0.949 | 0.921 | 0.872 | 0.823 | 0.917 (0.003) |
| 70% ref. | 0.971 | 0.959 | 0.943 | 0.912 | 0.868 | 0.816 | 0.911 (0.003) |
| 60% ref. | 0.966 | 0.956 | 0.939 | 0.908 | 0.860 | 0.805 | 0.905 (0.003) |
| 50% ref. | 0.962 | 0.950 | 0.934 | 0.904 | 0.853 | 0.787 | 0.898 (0.004) |
| 250 bp | | | | | | | |
| Full ref. | 0.978 | 0.968 | 0.949 | 0.918 | 0.863 | 0.811 | 0.914 (0.004) |
| 90% ref. | 0.967 | 0.955 | 0.935 | 0.907 | 0.854 | 0.800 | 0.903 (0.004) |
| 80% ref. | 0.956 | 0.944 | 0.926 | 0.895 | 0.846 | 0.786 | 0.892 (0.004) |
| 70% ref. | 0.942 | 0.926 | 0.912 | 0.880 | 0.830 | 0.773 | 0.877 (0.004) |
| 60% ref. | 0.927 | 0.911 | 0.893 | 0.861 | 0.813 | 0.755 | 0.860 (0.004) |
| 50% ref. | 0.909 | 0.891 | 0.871 | 0.838 | 0.784 | 0.732 | 0.837 (0.004) |
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Note: ref. indicates reference sequences
When the reference phylogeny includes >70% of the reference data, EPA-PTP outperforms all competing approaches, including stand-alone PTP. EPA-PTP outperforms PTP even when the reference phylogeny contains only 50% of the simulated reference data for . With increasing , the reference data need to be more complete for EPA-PTP to outperform PTP. This is because with increasing , internal branch lengths tend to get shorter and the EPA placement accuracy decreases. Hence, more data are needed to obtain accurate placements. Note that under extremely high speciation rates, EPA-PTP performs worse than PTP. The estimation errors may also because of (i) discarding sequences with low likelihood weights (see Section 2.2) (ii) errors in phylogenetic inferences or (iii) PTP heuristics failing to find the maximum likelihood species delimitation.
The results for the EPA-CROP pipeline are shown in Supplementary Tables S4 and S6. EPA-CROP outperforms the stand-alone version of CROP, but the results are worse than for EPA-PTP.
On the Arthropod metabarcoding data, the EPA-PTP pipeline yields substantially better results than the multistep OTU-picking pipeline used in the original publication (Table 4). When the complete full-length reference sequence tree is used, the EPA-PTP pipeline shows substantially lower ‘dropout’ and ‘no-match’ rates. It recovers 12.5% more species with respect to the reference data that represent an improvement of over 50%. Here, we apply an analogous criterion as in the original study where at least two reads need to be contained in an OTU cluster for it to be considered. In our case, reads need to be contained in a species delimitation. If an OTU cluster or species delimitation only contains one read, it is highly likely that it represents a sequencing error. However, the availability of the complete reference dataset is not granted for most metabarcoding analyses. Thus, as for the simulated data, we randomly removed up to 50% of the reference sequences and reran our pipelines. We then calculated the ratios between the number of species estimated on the reduced reference data relative to the number of species estimated on the complete reference data. The results are shown in Supplementary Figure S5 in the online supplement. When species are delimited with taxonomy-dependent approaches, such as the EPA, the number of estimated species is expected to decrease with the number of species in the reference data. When combined with PTP (using reads per delimitation as cutoff), EPA-PTP yields stable diversity estimates, irrespective of the completeness of the reference phylogeny. EPA-CROP also yields better results than the multistep OTU-picking pipeline and stand-alone CROP. The results are slightly worse than for EPA-PTP (Supplementary Table S7).
5 DISCUSSION
We introduced, implemented and made available a new model (PTP) for species delimitation that is mainly intended for delimiting species in single-locus molecular phylogenies. PTP can propose species boundaries that are consistent with the PSC. An important advantage of our method is that it models speciation in terms of the number of substitutions. Thereby, it circumvents the potentially error-prone and compute-intensive process of generating time-calibrated ultrametric trees, which are required as an input for GMYC.
Using real datasets, we show that delimitations inferred with PTP are comparable with delimitations obtained via GMYC. Simulations suggest PTP outperforms GMYC.
In addition, it is more straightforward to use because it only requires a standard phylogenetic tree as input and because it also is substantially faster. On the 673-taxon metabarcoding dataset (using a modern Intel desktop processor), for instance, r8s requires 3 days to complete, whereas RAxML in combination with PTP only requires a total of about 20 min to return a species delimitation.
We also compared GMYC and PTP with two clustering algorithms: CROP and UCLUST. From our point of view, the problem of species delimitation needs to incoporate data from various sources (e.g. sequences and trees) and also depends heavily on the species definition used. Thus, GMYC and PTP yield comparable results on real data because they are based on the PSC. In contrast, by their very definition, CROP and UCLUST simply identify sequence clusters. The fact that there is a difference between sequence clusters and PSC-based species delimitation is underpinned by our simulations.
We simulate the data in accordance with the GMYC model that essentially adopts the PSC. To demonstrate the impact of the parameter on clustering-based delimitation accuracy, we plotted the pairwise sequence distances within species and between directly adjacent species in the simulated tree, for and in Supplementary Figures S2 and S3 of the online supplement. Lower values lead to larger evolutionary distances between species, that is, the so-called barcoding gap (Puillandre et al., 2012) is present. Increasing reduces the evolutionary distances between species and the barcoding gap disappears (see Puillandre et al., 2012 for examples of this phenomenon on real data). Therefore, our simulations show that clustering algorithms work on datasets with the barcoding gap because phylogenetic species are mostly consistent with sequence clusters in this case. However, clustering methods are prone to fail when the barcoding gap is not present because sequences cannot be told apart any more via sequence similarity alone. As we show, GMYC and PTP delimitation performance is more robust to the absence of barcoding gap. Thus, when no prior information (barcoding gap presence) about the dataset is available and the goal is to delimit phylogenetic species, GMYC and PTP should be preferred.
Apart from the stand-alone PTP code, we also introduced the EPA-PTP pipeline that combines the EPA with PTP.
The EPA-PTP pipeline represents the first integrated approach for analyzing metagenomic data that combines the phylogenetic placement approach with an explicit statistical criterion for species delimitation. On a representative empirical dataset, our pipeline yields a substantially more accurate diversity estimate than traditional OTU-picking methods. Using simulated data, we show that, open reference-based approaches can improve delimitation accuracy compared with de novo approaches. More importantly, the EPA-PTP pipeline allows for deploying a widely accepted species concept to metagenomic data, where millions of sequences need to be processed. EPA-CROP (with the default setting of 2000 MCMC generations) is approximately twice as fast as EPA-PTP on the metabarcoding dataset. Note that 2000 generations may not be sufficient, and that CROP does not offer a built-in MCMC convergence assessment criterion.
In the following, we discuss the current limitations of our approach.
Readers should keep in mind that entities delimited by PTP are putative species only. The phylogenetic trees inferred on single-gene molecular sequences are gene trees rather than species trees, albeit the hierarchical relationships above the species boundaries are expected to be mostly consistent with the species tree. However, the boundaries inferred by PTP are only approximate. Additional data need to be integrated to further validate the delimitations, such as morphological characters and multi-gene sequence data (Ence and Carstens, 2011) within an integrative taxonomy framework (Padial et al., 2010; Sauer and Hausdorf, 2012). The putative species delimited by PTP, can, for instance, be used as initial hypothesis that can be further scrutinized with multilocus coalescent-based methods such as BP&P (Yang and Rannala, 2010). BP&P requires prior knowledge of species boundaries, and it represents a validation method, rather than a delimitation method. Owing to its computational complexity, BP&P can currently only handle up to 20 species.
Compared with OTU-picking methods, PTP and EPA-PTP require substantially more CPU time because of the phylogenetic calculations. Although most OTU-picking methods can run on an off-the-shelf desktop computer, the EPA-PTP pipeline requires a multicore server for analyzing large metagenomic datasets.
Because PTP initiates the search for the maximum likelihood delimitation at the root of the input phylogeny, the tree has to be correctly rooted to obtain accurate estimates. Also, PTP should be used with caution on datasets where the number of individuals sampled per species is unbalanced and where the over-sampled species exhibit small within-species variation (see Supplementary Tables S1 through 4).
In such cases, the inferred phylogeny will comprise both, subtrees (comprising one species and many individuals) with a large number of extremely short branches, and subtrees (comprising one species but only few individuals) with short, but not extremely short branches. Such unbalanced samples may require the introduction of a third λ parameter class of branches to accommodate (i) over-sampled within-species branches, (ii) within-species branches and (iii) among-species branches. Otherwise, the species that are not over-sampled cannot be delimited properly, that is, each individual is likely to be identified as a separate species. Hence, we either need a criterion for removing over-sampled sequences, or a criterion to decide when and how many additional classes of PTP (λ parameters) need to be introduced.
However, a major drawback of introducing additional PTP classes is that the delimitation search space becomes significantly larger. Hence, finding the maximum likelihood delimitation or a best known delimitation represents a challenging task. Thus, before extending the number of classes, we feel that more work on the design and performance of heuristic search strategies for species delimitation is required to better characterize and understand the problem. This also applies to the heuristics used in GMYC, given that the underlying optimization problems are similar.
Supplementary Material
Supplementary Data
supp_29_22_2869__index.html (975B, html)
ACKNOWLEDGEMENTS
The authors thank T. Barraclough and T. Fujisawa for help on re-implementing GMYC in Python and T. Flouri for a bug fix in r8s. They also thank R.P. Brown, A. Papadopoulou and J. Pons for providing the empirical datasets. They acknowledge the support of Graduate School for Computing in Medicine and Life Sciences, University of Lübeck.
Funding: Jiajie Zhang is supported by a HITS scholarship.
Conflict of Interest: none declared.
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高中数学解析几何｜求轨迹方程方法最全总结 - 每日头条
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高中数学解析几何｜求轨迹方程方法最全总结
2017-09-16由 数学派 發表于科学
...
一、直接法
若动点运动的条件是一些较为明确的几何量的等量关系，而这些条件易于表达成关于x，y的等量关系式，可以较为容易地得到轨迹方程（即遵循求轨迹方程的一般程序），这种方法我们一般称之为直接法.用直接发求轨迹方程一般都要经过建系、设点、列式、化简、验证这五个环节.
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二、定义法
若动点轨迹的条件符合某一基本而常见轨迹的定义（如圆、椭圆、双曲线、抛物线等）已从定义来确定表示其几何特征的基本量而直接写出其轨迹方程，或从曲线定义来建立等量关系式从而求出轨迹方程．
...
三、代入法
若动点运动情况较为复杂，不易直接表述或求出，但是能够发现形成轨迹的动点P（x,y）随着另一动点Q （X,Y）的运动而有规律的运动，而且动点Q的运动轨迹方程已经给定或极为容易求出，故只要找出两动点P，Q之间的等量关系式，用x，y表示X,Y再代入Q的轨迹方程整理即得动点P的轨迹方程，称之为代入法，也叫相关点法.
深入瞭解
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四、参数法
若动点运动变化情况较为复杂，动点的纵坐标之间的等量关系式难以极快找到，可以适当引入参数，通过所设参数沟通动点横坐标之间的联系，从而得到轨迹的参数方程进而再消去所设参数得出轨迹的（普通）方程，称之为参数法.
............
点悟：注意落实好图形特征信息提供的解题方向，前提是自信，实力是运算过关.本题还可有一些较为简捷的解法，不妨试试！！！
五、交轨法
若所求轨迹可以看成是某两条曲线（包括直线）的交点轨迹时，可由方程直接消去参数，也可引入参数来建这两条动曲线之间的联系，再消参而得到轨迹方程，称之为交轨法.可以认为交轨法是参数法的一种特殊情况.
...
点悟：交轨是一种动态解题策略，注意特殊或极限情况处理.
六、几何法
认真分析动点运动变化规律，可以发现图形明显的几何特征，利用有关平面几何的知识将动点运动变化规律与动点满足的条件有机联系起来，再利用直接法得到动点的轨迹方程，称之为几何法.
...
七、点差法
涉及与圆锥曲线中点弦有关的轨迹问题时，常可以把两端点设为(x1,y1),(x2,y2)，代入圆锥曲线方程，然后作差法求出曲线的轨迹方程，此法称之为点差法，也叫平方差法.运用此法要注意限制轨迹方程中变量可能的取值范围.
...
点悟：上述方法是通过设直线AB的方程引入参数b得到动点M轨迹的参数方程再消去参数得到普通方程，注意参数的取值范围，因而轨迹是一条线段.本题较为简捷的求法还可考虑点差法：
...
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选用距离公式、斜率公式等将其转化为x，y的方程式，并化简；⑤证明,证明所求方程即为符合条件的动点轨迹方程.代入法：动点P依赖于另一动点Q的变化而变化，并且Q又在某已知曲线上，则可先用x，y的代数式表示x0，y0，再将x0，y0代入已知曲线得要求的轨迹方程；坐标之间的关系不易直接找
一个高难度集几何问题的解答
2017-07-29
题目如下图：成都赖老师给出的答案是：那么，怎么作出这个图形呢？江苏金老师给出了一个“交轨法”的作图方法：
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Exercises and Problems in Linear Algebra John M. Erdman Portland State University Version July 13, 2014 c ⃝2010 John M. Erdman E-mail address: erdman@pdx.edu Contents PREFACE vii Part 1.
MATRICES AND LINEAR EQUATIONS 1 Chapter 1.
SYSTEMS OF LINEAR EQUATIONS 3 1.1.
Background 3 1.2.
Exercises 4 1.3.
Problems 7 1.4.
Answers to Odd-Numbered Exercises 8 Chapter 2.
ARITHMETIC OF MATRICES 9 2.1.
Background 9 2.2.
Exercises 10 2.3.
Problems 12 2.4.
Answers to Odd-Numbered Exercises 14 Chapter 3.
ELEMENTARY MATRICES; DETERMINANTS 15 3.1.
Background 15 3.2.
Exercises 17 3.3.
Problems 22 3.4.
Answers to Odd-Numbered Exercises 23 Chapter 4.
VECTOR GEOMETRY IN Rn 25 4.1.
Background 25 4.2.
Exercises 26 4.3.
Problems 28 4.4.
Answers to Odd-Numbered Exercises 29 Part 2.
VECTOR SPACES 31 Chapter 5.
VECTOR SPACES 33 5.1.
Background 33 5.2.
Exercises 34 5.3.
Problems 37 5.4.
Answers to Odd-Numbered Exercises 38 Chapter 6.
SUBSPACES 39 6.1.
Background 39 6.2.
Exercises 40 6.3.
Problems 44 6.4.
Answers to Odd-Numbered Exercises 45 Chapter 7.
LINEAR INDEPENDENCE 47 7.1.
Background 47 7.2.
Exercises 49 iii iv CONTENTS 7.3.
Problems 51 7.4.
Answers to Odd-Numbered Exercises 53 Chapter 8.
BASIS FOR A VECTOR SPACE 55 8.1.
Background 55 8.2.
Exercises 56 8.3.
Problems 57 8.4.
Answers to Odd-Numbered Exercises 58 Part 3.
LINEAR MAPS BETWEEN VECTOR SPACES 59 Chapter 9.
LINEARITY 61 9.1.
Background 61 9.2.
Exercises 63 9.3.
Problems 67 9.4.
Answers to Odd-Numbered Exercises 70 Chapter 10.
LINEAR MAPS BETWEEN EUCLIDEAN SPACES 71 10.1.
Background 71 10.2.
Exercises 72 10.3.
Problems 74 10.4.
Answers to Odd-Numbered Exercises 75 Chapter 11.
PROJECTION OPERATORS 77 11.1.
Background 77 11.2.
Exercises 78 11.3.
Problems 79 11.4.
Answers to Odd-Numbered Exercises 80 Part 4.
SPECTRAL THEORY OF VECTOR SPACES 81 Chapter 12.
EIGENVALUES AND EIGENVECTORS 83 12.1.
Background 83 12.2.
Exercises 84 12.3.
Problems 85 12.4.
Answers to Odd-Numbered Exercises 86 Chapter 13.
DIAGONALIZATION OF MATRICES 87 13.1.
Background 87 13.2.
Exercises 89 13.3.
Problems 91 13.4.
Answers to Odd-Numbered Exercises 92 Chapter 14.
SPECTRAL THEOREM FOR VECTOR SPACES 93 14.1.
Background 93 14.2.
Exercises 94 14.3.
Answers to Odd-Numbered Exercises 96 Chapter 15.
SOME APPLICATIONS OF THE SPECTRAL THEOREM 97 15.1.
Background 97 15.2.
Exercises 98 15.3.
Problems 102 15.4.
Answers to Odd-Numbered Exercises 103 Chapter 16.
EVERY OPERATOR IS DIAGONALIZABLE PLUS NILPOTENT 105 CONTENTS v 16.1.
Background 105 16.2.
Exercises 106 16.3.
Problems 110 16.4.
Answers to Odd-Numbered Exercises 111 Part 5.
THE GEOMETRY OF INNER PRODUCT SPACES 113 Chapter 17.
COMPLEX ARITHMETIC 115 17.1.
Background 115 17.2.
Exercises 116 17.3.
Problems 118 17.4.
Answers to Odd-Numbered Exercises 119 Chapter 18.
REAL AND COMPLEX INNER PRODUCT SPACES 121 18.1.
Background 121 18.2.
Exercises 123 18.3.
Problems 125 18.4.
Answers to Odd-Numbered Exercises 126 Chapter 19.
ORTHONORMAL SETS OF VECTORS 127 19.1.
Background 127 19.2.
Exercises 128 19.3.
Problems 129 19.4.
Answers to Odd-Numbered Exercises 131 Chapter 20.
QUADRATIC FORMS 133 20.1.
Background 133 20.2.
Exercises 134 20.3.
Problems 136 20.4.
Answers to Odd-Numbered Exercises 137 Chapter 21.
OPTIMIZATION 139 21.1.
Background 139 21.2.
Exercises 140 21.3.
Problems 141 21.4.
Answers to Odd-Numbered Exercises 142 Part 6.
ADJOINT OPERATORS 143 Chapter 22.
ADJOINTS AND TRANSPOSES 145 22.1.
Background 145 22.2.
Exercises 146 22.3.
Problems 147 22.4.
Answers to Odd-Numbered Exercises 148 Chapter 23.
THE FOUR FUNDAMENTAL SUBSPACES 149 23.1.
Background 149 23.2.
Exercises 151 23.3.
Problems 155 23.4.
Answers to Odd-Numbered Exercises 157 Chapter 24.
ORTHOGONAL PROJECTIONS 159 24.1.
Background 159 24.2.
Exercises 160 vi CONTENTS 24.3.
Problems 163 24.4.
Answers to Odd-Numbered Exercises 164 Chapter 25.
LEAST SQUARES APPROXIMATION 165 25.1.
Background 165 25.2.
Exercises 166 25.3.
Problems 167 25.4.
Answers to Odd-Numbered Exercises 168 Part 7.
SPECTRAL THEORY OF INNER PRODUCT SPACES 169 Chapter 26.
SPECTRAL THEOREM FOR REAL INNER PRODUCT SPACES 171 26.1.
Background 171 26.2.
Exercises 172 26.3.
Problem 174 26.4.
Answers to the Odd-Numbered Exercise 175 Chapter 27.
SPECTRAL THEOREM FOR COMPLEX INNER PRODUCT SPACES 177 27.1.
Background 177 27.2.
Exercises 178 27.3.
Problems 181 27.4.
Answers to Odd-Numbered Exercises 182 Bibliography 183 Index 185 PREFACE This collection of exercises is designed to provide a framework for discussion in a junior level linear algebra class such as the one I have conducted fairly regularly at Portland State University.
There is no assigned text.
Students are free to choose their own sources of information.
Stu-dents are encouraged to find books, papers, and web sites whose writing style they find congenial, whose emphasis matches their interests, and whose price fits their budgets. The short introduc-tory background section in these exercises, which precede each assignment, are intended only to fix notation and provide “official” definitions and statements of important theorems for the exercises and problems which follow.
There are a number of excellent online texts which are available free of charge. Among the best are Linear Algebra by Jim Hefferon, and A First Course in Linear Algebra by Robert A. Beezer, Another very useful online resource is Przemyslaw Bogacki’s Linear Algebra Toolkit .
And, of course, many topics in linear algebra are discussed with varying degrees of thoroughness in the Wikipedia and Eric Weisstein’s Mathworld .
Among the dozens and dozens of linear algebra books that have appeared, two that were written before “dumbing down” of textbooks became fashionable are especially notable, in my opinion, for the clarity of their authors’ mathematical vision: Paul Halmos’s Finite-Dimensional Vector Spaces and Hoffman and Kunze’s Linear Algebra . Some students, especially mathematically inclined ones, love these books, but others find them hard to read. If you are trying seriously to learn the subject, give them a look when you have the chance. Another excellent traditional text is Linear Algebra: An Introductory Approach by Charles W. Curits. And for those more interested in applications both Elementary Linear Algebra: Applications Version by Howard Anton and Chris Rorres and Linear Algebra and its Applications by Gilbert Strang are loaded with applications.
If you are a student and find the level at which many of the current beginning linear algebra texts are written depressingly pedestrian and the endless routine computations irritating, you might examine some of the more advanced texts. Two excellent ones are Steven Roman’s Advanced Linear Algebra and William C. Brown’s A Second Course in Linear Algebra .
Concerning the material in these notes, I make no claims of originality. While I have dreamed up many of the items included here, there are many others which are standard linear algebra exercises that can be traced back, in one form or another, through generations of linear algebra texts, making any serious attempt at proper attribution quite futile. If anyone feels slighted, please contact me.
There will surely be errors. I will be delighted to receive corrections, suggestions, or criticism at vii viii PREFACE erdman@pdx.edu I have placed the the L AT EX source files on my web page so that those who wish to use these exer-cises for homework assignments, examinations, or any other noncommercial purpose can download the material and, without having to retype everything, edit it and supplement it as they wish.
Part 1 MATRICES AND LINEAR EQUATIONS CHAPTER 1 SYSTEMS OF LINEAR EQUATIONS 1.1. Background Topics: systems of linear equations; Gaussian elimination (Gauss’ method), elementary row op-erations, leading variables, free variables, echelon form, matrix, augmented matrix, Gauss-Jordan reduction, reduced echelon form.
1.1.1. Definition. We will say that an operation (sometimes called scaling) which multiplies a row of a matrix (or an equation) by a nonzero constant is a row operation of type I. An operation (sometimes called swapping) that interchanges two rows of a matrix (or two equations) is a row operation of type II. And an operation (sometimes called pivoting) that adds a multiple of one row of a matrix to another row (or adds a multiple of one equation to another) is a row operation of type III.
3 4 1. SYSTEMS OF LINEAR EQUATIONS 1.2. Exercises (1) Suppose that L1 and L2 are lines in the plane, that the x-intercepts of L1 and L2 are 5 and −1, respectively, and that the respective y-intercepts are 5 and 1. Then L1 and L2 intersect at the point ( , ) .
(2) Consider the following system of equations.
w + x + y + z = 6 w + y + z = 4 w + y = 2 (∗) (a) List the leading variables .
(b) List the free variables .
(c) The general solution of (∗) (expressed in terms of the free variables) is ( , , , ) .
(d) Suppose that a fourth equation −2w + y = 5 is included in the system (∗). What is the solution of the resulting system? Answer: ( , , , ).
(e) Suppose that instead of the equation in part (d), the equation −2w −2y = −3 is included in the system (∗).
Then what can you say about the solution(s) of the resulting system? Answer: .
(3) Consider the following system of equations: x + y + z = 2 x + 3y + 3z = 0 x+ 3y+ 6z = 3 (∗) (a) Use Gaussian elimination to put the augmented coefficient matrix into row echelon form. The result will be 1 1 1 a 0 1 1 b 0 0 1 c where a = , b = , and c = .
(b) Use Gauss-Jordan reduction to put the augmented coefficient matrix in reduced row echelon form. The result will be 1 0 0 d 0 1 0 e 0 0 1 f where d = , e = , and f = .
(c) The solutions of (∗) are x = , y = , and z = .
(4) Consider the following system of equations.
0.003000x + 59.14y = 59.17 5.291x −6.130y = 46.78.
(a) Using only row operation III and back substitution find the exact solution of the system. Answer: x = , y = .
(b) Same as (a), but after performing each arithmetic operation round offyour answer to four significant figures. Answer: x = , y = .
1.2. EXERCISES 5 (5) Find the values of k for which the system of equations x + ky = 1 kx + y = 1 has (a) no solution. Answer: .
(b) exactly one solution. Answer: .
(c) infinitely many solutions. Answer: .
(d) When there is exactly one solution, it is x = and y = .
(6) Consider the following two systems of equations.
x + y + z = 6 x + 2y + 2z = 11 2x + 3y −4z = 3 (1) and x + y + z = 7 x + 2y + 2z = 10 2x + 3y −4z = 3 (2) Solve both systems simultaneously by applying Gauss-Jordan reduction to an appro-priate 3 × 5 matrix.
(a) The resulting row echelon form of this 3 × 5 matrix is .
(b) The resulting reduced row echelon form is .
(c) The solution for (1) is ( , , ) and the solution for (2) is ( , , ) .
(7) Consider the following system of equations: x −y −3z = 3 2x + z = 0 2y + 7z = c (a) For what values of c does the system have a solution? Answer: c = .
(b) For the value of c you found in (a) describe the solution set geometrically as a subset of R3. Answer: .
(c) What does part (a) say about the planes x −y −3z = 3, 2x + z = 0, and 2y + 7z = 4 in R3? Answer: .
6 1. SYSTEMS OF LINEAR EQUATIONS (8) Consider the following system of linear equations ( where b1, . . . , b5 are constants).
u + 2v −w −2x + 3y = b1 x −y + 2z = b2 2u + 4v −2w −4x + 7y −4z = b3 −x + y −2z = b4 3u + 6v −3w −6x + 7y + 8z = b5 (a) In the process of Gaussian elimination the leading variables of this system are and the free variables are .
(b) What condition(s) must the constants b1, . . . , b5 satisfy so that the system is consis-tent? Answer: .
(c) Do the numbers b1 = 1, b2 = −3, b3 = 2, b4 = b5 = 3 satisfy the condition(s) you listed in (b)?
. If so, find the general solution to the system as a function of the free variables. Answer: u = v = w = x = y = z = .
(9) Consider the following homogeneous system of linear equations (where a and b are nonzero constants).
x + 2y = 0 ax + 8y + 3z = 0 by + 5z = 0 (a) Find a value for a which will make it necessary during Gaussian elimination to inter-change rows in the coefficient matrix. Answer: a = .
(b) Suppose that a does not have the value you found in part (a). Find a value for b so that the system has a nontrivial solution.
Answer: b = c 3 + d 3a where c = and d = .
(c) Suppose that a does not have the value you found in part (a) and that b = 100.
Suppose further that a is chosen so that the solution to the system is not unique.
The general solution to the system (in terms of the free variable) is 1 α z , −1 β z , z where α = and β = .
1.3. PROBLEMS 7 1.3. Problems (1) Give a geometric description of a single linear equation in three variables.
Then give a geometric description of the solution set of a system of 3 linear equations in 3 variables if the system (a) is inconsistent.
(b) is consistent and has no free variables.
(c) is consistent and has exactly one free variable.
(d) is consistent and has two free variables.
(2) Consider the following system of equations: −m1x + y = b1 −m2x + y = b2 (∗) (a) Prove that if m1 ̸= m2, then (∗) has exactly one solution. What is it?
(b) Suppose that m1 = m2. Then under what conditions will (∗) be consistent?
(c) Restate the results of (a) and (b) in geometrical language.
8 1. SYSTEMS OF LINEAR EQUATIONS 1.4. Answers to Odd-Numbered Exercises (1) 2, 3 (3) (a) 2, −1, 1 (b) 3, −2, 1 (c) 3, −2, 1 (5) (a) k = −1 (b) k ̸= −1, 1 (c) k = 1 (d) 1 k + 1, 1 k + 1 (7) (a) −6 (b) a line (c) They have no points in common.
(9) (a) 4 (b) 40, −10 (c) 10, 20 CHAPTER 2 ARITHMETIC OF MATRICES 2.1. Background Topics: addition, scalar multiplication, and multiplication of matrices, inverse of a nonsingular matrix.
2.1.1. Definition. Two square matrices A and B of the same size are said to commute if AB = BA.
2.1.2. Definition. If A and B are square matrices of the same size, then the commutator (or Lie bracket) of A and B, denoted by [A, B], is defined by [A, B] = AB −BA .
2.1.3. Notation. If A is an m×n matrix (that is, a matrix with m rows and n columns), then the element in the ith row and the jth column is denoted by aij. The matrix A itself may be denoted by aij m i=1 n j=1 or, more simply, by [aij]. In light of this notation it is reasonable to refer to the index i in the expression aij as the row index and to call j the column index. When we speak of the “value of a matrix A at (i, j),” we mean the entry in the ith row and jth column of A. Thus, for example, A = 1 4 3 −2 7 0 5 −1 is a 4 × 2 matrix and a31 = 7.
2.1.4. Definition. A matrix A = [aij] is upper triangular if aij = 0 whenever i > j.
2.1.5. Definition. The trace of a square matrix A, denoted by tr A, is the sum of the diagonal entries of the matrix. That is, if A = [aij] is an n × n matrix, then tr A := n X j=1 ajj.
2.1.6. Definition. The transpose of an n×n matrix A = aij is the matrix At = aji obtained by interchanging the rows and columns of A. The matrix A is symmetric if At = A.
2.1.7. Proposition. If A is an m × n matrix and B is an n × p matrix, then (AB)t = BtAt.
9 10 2. ARITHMETIC OF MATRICES 2.2. Exercises (1) Let A = 1 0 −1 2 0 3 1 −1 2 4 0 3 −3 1 −1 2 , B = 1 2 3 −1 0 −2 4 1 , and C = 3 −2 0 5 1 0 −3 4 .
(a) Does the matrix D = ABC exist?
If so, then d34 = .
(b) Does the matrix E = BAC exist?
If so, then e22 = .
(c) Does the matrix F = BCA exist?
If so, then f43 = .
(d) Does the matrix G = ACB exist?
If so, then g31 = .
(e) Does the matrix H = CAB exist?
If so, then h21 = .
(f) Does the matrix J = CBA exist?
If so, then j13 = .
(2) Let A = " 1 2 1 2 1 2 1 2 # , B = 1 0 0 −1 , and C = AB. Evaluate the following.
(a) A37 = (b) B63 = (c) B138 = (d) C42 = Note: If M is a matrix Mp is the product of p copies of M.
(3) Let A = 1 1/3 c d . Find numbers c and d such that A2 = −I.
Answer: c = and d = .
(4) Let A and B be symmetric n × n-matrices. Then [A, B] = [B, X], where X = .
(5) Let A, B, and C be n × n matrices. Then [A, B]C + B[A, C] = [X, Y ], where X = and Y = .
(6) Let A = 1 1/3 c d . Find numbers c and d such that A2 = 0. Answer: c = and d = .
(7) Consider the matrix 1 3 2 a 6 2 0 9 5 where a is a real number.
(a) For what value of a will a row interchange be required during Gaussian elimination?
Answer: a = .
(b) For what value of a is the matrix singular? Answer: a = .
(8) Let A = 1 0 −1 2 0 3 1 −1 2 4 0 3 −3 1 −1 2 , B = 1 2 3 −1 0 −2 4 1 , C = 3 −2 0 5 1 0 −3 4 , and M = 3A3 −5(BC)2. Then m14 = and m41 = .
(9) If A is an n × n matrix and it satisfies the equation A3 −4A2 + 3A −5In = 0, then A is nonsingular 2.2. EXERCISES 11 and its inverse is .
(10) Let A, B, and C be n × n matrices. Then [[A, B], C] + [[B, C], A] + [[C, A], B] = X, where X = .
(11) Let A, B, and C be n × n matrices. Then [A, C] + [B, C] = [X, Y ], where X = and Y = .
(12) Find the inverse of 1 0 0 0 1 4 1 0 0 1 3 1 3 1 0 1 2 1 2 1 2 1 . Answer: .
(13) The matrix H = 1 1 2 1 3 1 4 1 2 1 3 1 4 1 5 1 3 1 4 1 5 1 6 1 4 1 5 1 6 1 7 is the 4×4 Hilbert matrix. Use Gauss-Jordan elimination to compute K = H−1. Then K44 is (exactly) . Now, create a new matrix H′ by replacing each entry in H by its approximation to 3 decimal places. (For example, replace 1 6 by 0.167.) Use Gauss-Jordan elimination again to find the inverse K′ of H′. Then K′ 44 is .
(14) Suppose that A and B are symmetric n × n matrices. In this exercise we prove that AB is symmetric if and only if A commutes with B. Below are portions of the proof. Fill in the missing steps and the missing reasons. Choose reasons from the following list.
(H1) Hypothesis that A and B are symmetric.
(H2) Hypothesis that AB is symmetric.
(H3) Hypothesis that A commutes with B.
(D1) Definition of commutes.
(D2) Definition of symmetric.
(T) Proposition 2.1.7.
Proof. Suppose that AB is symmetric. Then AB = (reason: (H2) and ) = BtAt (reason: ) = (reason: (D2) and ) So A commutes with B (reason: ).
Conversely, suppose that A commutes with B. Then (AB)t = (reason: (T) ) = BA (reason: and ) = (reason: and ) Thus AB is symmetric (reason: ).
□ 12 2. ARITHMETIC OF MATRICES 2.3. Problems (1) Let A be a square matrix. Prove that if A2 is invertible, then so is A.
Hint. Our assumption is that there exists a matrix B such that A2B = BA2 = I .
We want to show that there exists a matrix C such that AC = CA = I .
Now to start with, you ought to find it fairly easy to show that there are matrices L and R such that LA = AR = I .
(∗) A matrix L is a left inverse of the matrix A if LA = I; and R is a right inverse of A if AR = I. Thus the problem boils down to determining whether A can have a left inverse and a right inverse that are different. (Clearly, if it turns out that they must be the same, then the C we are seeking is their common value.) So try to prove that if (∗) holds, then L = R.
(2) Anton speaks French and German; Geraldine speaks English, French and Italian; James speaks English, Italian, and Spanish; Lauren speaks all the languages the others speak except French; and no one speaks any other language. Make a matrix A = aij with rows representing the four people mentioned and columns representing the languages they speak. Put aij = 1 if person i speaks language j and aij = 0 otherwise. Explain the significance of the matrices AAt and AtA.
(3) Portland Fast Foods (PFF), which produces 138 food products all made from 87 basic ingredients, wants to set up a simple data structure from which they can quickly extract answers to the following questions: (a) How many ingredients does a given product contain?
(b) A given pair of ingredients are used together in how many products?
(c) How many ingredients do two given products have in common?
(d) In how many products is a given ingredient used?
In particular, PFF wants to set up a single table in such a way that: (i) the answer to any of the above questions can be extracted easily and quickly (matrix arithmetic permitted, of course); and (ii) if one of the 87 ingredients is added to or deleted from a product, only a single entry in the table needs to be changed.
Is this possible? Explain.
(4) Prove proposition 2.1.7.
(5) Let A and B be 2 × 2 matrices.
(a) Prove that if the trace of A is 0, then A2 is a scalar multiple of the identity matrix.
(b) Prove that the square of the commutator of A and B commutes with every 2 × 2 matrix C. Hint. What can you say about the trace of [A, B]?
(c) Prove that the commutator of A and B can never be a nonzero multiple of the identity matrix.
2.3. PROBLEMS 13 (6) The matrices that represent rotations of the xy-plane are A(θ) = cos θ −sin θ sin θ cos θ .
(a) Let x be the vector (−1, 1), θ = 3π/4, and y be A(θ) acting on x (that is, y = A(θ)xt).
Make a sketch showing x, y, and θ.
(b) Verify that A(θ1)A(θ2) = A(θ1 + θ2). Discuss what this means geometrically.
(c) What is the product of A(θ) times A(−θ)? Discuss what this means geometrically.
(d) Two sheets of graph paper are attached at the origin and rotated in such a way that the point (1, 0) on the upper sheet lies directly over the point (−5/13, 12/13) on the lower sheet. What point on the lower sheet lies directly below (6, 4) on the upper one?
(7) Let A = 0 a a2 a3 a4 0 0 a a2 a3 0 0 0 a a2 0 0 0 0 a 0 0 0 0 0 .
The goal of this problem is to develop a “calculus” for the matrix A. To start, recall (or look up) the power series expansion for 1 1 −x.
Now see if this formula works for the matrix A by first computing (I −A)−1 directly and then computing the power series expansion substituting A for x. (Explain why there are no convergence difficulties for the series when we use this particular matrix A.) Next try to define ln(I + A) and eA by means of appropriate series. Do you get what you expect when you compute eln(I+A)? Do formulas like eAeA = e2A hold? What about other familiar properties of the exponential and logarithmic functions?
Try some trigonometry with A. Use series to define sin, cos, tan, arctan, and so on. Do things like tan(arctan(A)) produce the expected results? Check some of the more obvious trigonometric identities. (What do you get for sin2 A + cos2 A −I? Is cos(2A) the same as cos2 A −sin2 A?) A relationship between the exponential and trigonometric functions is given by the famous formula eix = cos x + i sin x. Does this hold for A?
Do you think there are other matrices for which the same results might hold? Which ones?
(8) (a) Give an example of two symmetric matrices whose product is not symmetric.
Hint. Matrices containing only 0’s and 1’s will suffice.
(b) Now suppose that A and B are symmetric n×n matrices. Prove that AB is symmetric if and only if A commutes with B.
Hint. To prove that a statement P holds “if and only if” a statement Q holds you must first show that P implies Q and then show that Q implies P. In the current problem, there are 4 conditions to be considered: (i) At = A (A is symmetric), (ii) Bt = B (B is symmetric), (iii) (AB)t = AB (AB is symmetric), and (iv) AB = BA (A commutes with B).
Recall also the fact given in (v) theorem 2.1.7.
The first task is to derive (iv) from (i), (ii), (iii), and (v). Then try to derive (iii) from (i), (ii), (iv), and (v).
14 2. ARITHMETIC OF MATRICES 2.4. Answers to Odd-Numbered Exercises (1) (a) yes, 142 (b) no, – (c) yes, −45 (d) no, – (e) yes, −37 (f) no, – (3) −6, −1 (5) A, BC (7) (a) 2 (b) −4 (9) 1 5(A2 −4A + 3In) (11) A + B, C (13) 2800, −1329.909 CHAPTER 3 ELEMENTARY MATRICES; DETERMINANTS 3.1. Background Topics: elementary (reduction) matrices, determinants.
The following definition says that we often regard the effect of multiplying a matrix M on the left by another matrix A as the action of A on M.
3.1.1. Definition. We say that the matrix A acts on the matrix M to produce the matrix N if N = AM. For example the matrix 0 1 1 0 acts on any 2 × 2 matrix by interchanging (swapping) its rows because 0 1 1 0 a b c d = c d a b .
3.1.2. Notation. We adopt the following notation for elementary matrices which implement type I row operations. Let A be a matrix having n rows. For any real number r ̸= 0 denote by Mj(r) the n × n matrix which acts on A by multiplying its jth row by r. (See exercise 1.) 3.1.3. Notation. We use the following notation for elementary matrices which implement type II row operations. (See definition 1.1.1.) Let A be a matrix having n rows. Denote by Pij the n × n matrix which acts on A by interchanging its ith and jth rows. (See exercise 2.) 3.1.4. Notation. And we use the following notation for elementary matrices which implement type III row operations. (See definition 1.1.1.) Let A be a matrix having n rows. For any real number r denote by Eij(r) the n × n matrix which acts on A by adding r times the jth row of A to the ith row. (See exercise 3.) 3.1.5. Definition. If a matrix B can be produced from a matrix A by a sequence of elementary row operations, then A and B are row equivalent.
Some Facts about Determinants 3.1.6. Proposition. Let n ∈N and Mn×n be the collection of all n × n matrices. There is exactly one function det: Mn×n →R: A 7→det A which satisfies (a) det In = 1.
(b) If A ∈Mn×n and A′ is the matrix obtained by interchanging two rows of A, then det A′ = −det A.
(c) If A ∈Mn×n, c ∈R, and A′ is the matrix obtained by multiplying each element in one row of A by the number c, then det A′ = c det A.
(d) If A ∈Mn×n, c ∈R, and A′ is the matrix obtained from A by multiplying one row of A by c and adding it to another row of A (that is, choose i and j between 1 and n with i ̸= j and replace ajk by ajk + caik for 1 ≤k ≤n), then det A′ = det A.
15 16 3. ELEMENTARY MATRICES; DETERMINANTS 3.1.7. Definition. The unique function det: Mn×n →R described above is the n × n determi-nant function.
3.1.8. Proposition. If A = [a] for a ∈R (that is, if A ∈M1×1), then det A = a; if A ∈M2×2, then det A = a11a22 −a12a21 .
3.1.9. Proposition. If A, B ∈Mn×n, then det(AB) = (det A)(det B).
3.1.10. Proposition. If A ∈Mn×n, then det At = det A.
(An obvious corollary of this: in conditions (b), (c), and (d) of proposition 3.1.6 the word “columns” may be substituted for the word “rows”.) 3.1.11. Definition. Let A be an n × n matrix. The minor of the element ajk, denoted by Mjk, is the determinant of the (n −1) × (n −1) matrix which results from the deletion of the jth row and kth column of A. The cofactor of the element ajk, denoted by Cjk is defined by Cjk := (−1)j+kMjk.
3.1.12. Proposition. If A ∈Mn×n and 1 ≤j ≤n, then det A = n X k=1 ajkCjk.
This is the (Laplace) expansion of the determinant along the jth row.
In light of 3.1.10, it is clear that expansion along columns works as well as expansion along rows. That is, det A = n X j=1 ajkCjk for any k between 1 and n. This is the (Laplace) expansion of the determinant along the kth column.
3.1.13. Proposition. An n × n matrix A is invertible if and only if det A ̸= 0. If A is invertible, then A−1 = (det A)−1C t where C = Cjk is the matrix of cofactors of elements of A.
3.2. EXERCISES 17 3.2. Exercises (1) Let A be a matrix with 4 rows. The matrix M3(4) which multiplies the 3rd row of A by 4 is . (See 3.1.2.) (2) Let A be a matrix with 4 rows. The matrix P24 which interchanges the 2nd and 4th rows of A is . (See 3.1.3.) (3) Let A be a matrix with 4 rows. The matrix E23(−2) which adds −2 times the 3rd row of A to the 2nd row is . (See 3.1.4.) (4) Let A be the 4 × 4 elementary matrix E43(−6). Then A11 = and A−9 = .
(5) Let B be the elementary 4 × 4 matrix P24. Then B−9 = and B10 = .
(6) Let C be the elementary 4 × 4 matrix M3(−2). Then C4 = and C−3 = .
(7) Let A = 1 2 3 0 −1 1 −2 1 0 −1 2 −3 and B = P23E34(−2)M3(−2)E42(1)P14A. Then b23 = and b32 = .
(8) We apply Gaussian elimination (using type III elementary row operations only) to put a 4 × 4 matrix A into upper triangular form. The result is E43( 5 2)E42(2)E31(1)E21(−2)A = 1 2 −2 0 0 −1 0 1 0 0 −2 2 0 0 0 10 .
Then the determinant of A is .
18 3. ELEMENTARY MATRICES; DETERMINANTS (9) The system of equations: 2y+3z = 7 x+ y−z = −2 −x+ y−5z = 0 is solved by applying Gauss-Jordan reduction to the augmented coefficient matrix A = 0 2 3 7 1 1 −1 −2 −1 1 −5 0 . Give the names of the elementary 3 × 3 matrices X1, . . . , X8 which implement the following reduction.
A X1 − − − − → 1 1 −1 −2 0 2 3 7 −1 1 −5 0 X2 − − − − → 1 1 −1 −2 0 2 3 7 0 2 −6 −2 X3 − − − − → 1 1 −1 −2 0 2 3 7 0 0 −9 −9 X4 − − − − → 1 1 −1 −2 0 2 3 7 0 0 1 1 X5 − − − − → 1 1 −1 −2 0 2 0 4 0 0 1 1 X6 − − − − → 1 1 −1 −2 0 1 0 2 0 0 1 1 X7 − − − − → 1 1 0 −1 0 1 0 2 0 0 1 1 X8 − − − − → 1 0 0 −3 0 1 0 2 0 0 1 1 .
Answer: X1 = , X2 = , X3 = , X4 = , X5 = , X6 = , X7 = , X8 = .
(10) Solve the following equation for x: det 3 −4 7 0 6 −2 2 0 1 8 0 0 3 4 −8 3 1 2 27 6 5 0 0 3 3 x 0 2 1 −1 1 0 −1 3 4 0 = 0.
Answer: x = .
(11) Let A = 0 0 1 0 2 4 1 2 3 . Find A−1 using the technique of augmenting A by the identity matrix I and performing Gauss-Jordan reduction on the augmented matrix. The reduction can be accomplished by the application of five elementary 3 × 3 matrices. Find elementary matrices X1, X2, and X3 such that A−1 = X3E13(−3)X2M2(1/2)X1I.
(a) The required matrices are X1 = P1i where i = , X2 = Ejk(−2) where j = and k = , and X3 = E12(r) where r = .
(b) And then A−1 = .
(12) det 1 t t2 t3 t 1 t t2 t2 t 1 t t3 t2 t 1 = (1 −a(t))p where a(t) = and p = .
3.2. EXERCISES 19 (13) Evaluate each of the following determinants.
(a) det 6 9 39 49 5 7 32 37 3 4 4 4 1 1 1 1 = .
(b) det 1 0 1 1 1 −1 2 0 2 −1 3 1 4 17 0 −5 = .
(c) det 13 3 −8 6 0 0 −4 0 1 0 7 −2 3 0 2 0 = .
(14) Let M be the matrix 5 4 −2 3 5 7 −1 8 5 7 6 10 5 7 1 9 .
(a) The determinant of M can be expressed as the constant 5 times the determinant of the single 3 × 3 matrix 3 1 5 3 3 .
(b) The determinant of this 3 × 3 matrix can be expressed as the constant 3 times the determinant of the single 2 × 2 matrix 7 2 2 .
(c) The determinant of this 2 × 2 matrix is .
(d) Thus the determinant of M is .
(15) Find the determinant of the matrix 1 2 5 7 10 1 2 3 6 7 1 1 3 5 5 1 1 2 4 5 1 1 1 1 1 . Answer: .
(16) Find the determinants of the following matrices.
A = −73 78 24 92 66 25 −80 37 10 and B = −73 78 24 92 66 25 −80 37 10.01 .
Hint. Use a calculator (thoughtfully). Answer: det A = and det B = .
(17) Find the determinant of the following matrix.
283 5 π 347.86 × 101583 3136 56 5 cos(2.7402) 6776 121 11 5 2464 44 4 2 .
Hint. Do not use a calculator. Answer: .
20 3. ELEMENTARY MATRICES; DETERMINANTS (18) Let A = 0 −1 2 0 1 2 0 0 1 2 1 2 1 2 0 −1 2 0 1 0 1 2 1 2 . We find A−1 using elementary row operations to convert the 4 × 8 matrix h A .
.
. I4 i to the matrix h I4 .
.
. A−1 i .
Give the names of the elementary 4 × 4 matrices X1, . . . , X11 which implement the following Gauss-Jordan reduction and fill in the missing matrix entries.
0 −1 2 0 1 2 .
.
.
1 0 0 0 0 0 1 2 1 2 .
.
.
0 1 0 0 1 2 0 −1 2 0 .
.
.
0 0 1 0 1 0 1 2 1 2 .
.
.
0 0 0 1 X1 / 1 0 1 2 1 2 .
.
.
0 0 1 2 1 2 .
.
.
1 2 0 −1 2 0 .
.
.
0 −1 2 0 1 2 .
.
.
X2 / 1 0 1 2 1 2 .
.
.
0 0 1 2 1 2 .
.
.
0 0 −3 4 −1 4 .
.
.
0 −1 2 0 1 2 .
.
.
X3 / 1 0 1 2 1 2 .
.
.
0 −1 2 0 1 2 .
.
.
0 0 −3 4 −1 4 .
.
.
0 0 1 2 1 2 .
.
.
X4 / 1 0 1 2 1 2 .
.
.
0 1 0 −1 .
.
.
0 0 −3 4 −1 4 .
.
.
0 0 1 2 1 2 .
.
.
X5 / 1 0 1 2 1 2 .
.
.
0 1 0 −1 .
.
.
0 0 1 1 3 .
.
.
0 0 1 2 1 2 .
.
.
X6 / 1 0 1 2 1 2 .
.
.
0 1 0 −1 .
.
.
0 0 1 1 3 .
.
.
0 0 0 1 3 .
.
.
X7 / 1 0 1 2 1 2 .
.
.
0 1 0 −1 .
.
.
0 0 1 0 .
.
.
0 0 0 1 3 .
.
.
X8 / 1 0 1 2 1 2 .
.
.
0 1 0 −1 .
.
.
0 0 1 0 .
.
.
0 0 0 1 .
.
.
X9 / 1 0 1 2 1 2 .
.
.
0 1 0 0 .
.
.
0 0 1 0 .
.
.
0 0 0 1 .
.
.
X10 / 1 0 1 2 0 .
.
.
0 1 0 0 .
.
.
0 0 1 0 .
.
.
0 0 0 1 .
.
.
X11 / 1 0 0 0 .
.
.
0 1 0 0 .
.
.
0 0 1 0 .
.
.
0 0 0 1 .
.
.
Answer: X1 = , X2 = , X3 = , X4 = , X5 = , X6 = , X7 = , X8 = .
X9 = , X10 = , X11 = .
3.2. EXERCISES 21 (19) Suppose that A is a square matrix with determinant 7. Then (a) det(P24A) = .
(b) det(E23(−4)A) = .
(c) det(M3(2)A) = .
22 3. ELEMENTARY MATRICES; DETERMINANTS 3.3. Problems (1) For this problem assume that we know the following: If X is an m × m matrix, if Y is an m × n matrix and if 0 and I are zero and identity matrices of appropriate sizes, then det X Y 0 I = det X.
Let A be an m × n matrix and B be an n × m matrix. Prove carefully that det 0 A −B I = det AB .
Hint. Consider the product 0 A −B I I 0 B I .
(2) Let A and B be n × n-matrices. Your good friend Fred R. Dimm believes that det A B B A = det(A + B) det(A −B).
He offers the following argument to support this claim: det A B B A = det(A2 −B2) = det[(A + B)(A −B)] = det(A + B) det(A −B) .
(a) Comment (helpfully) on his “proof”.
In particular, explain carefully why each of the three steps in his “proof” is correct or incorrect. (That is, provide a proof or a counterexample to each step.) (b) Is the result he is trying to prove actually true?
Hint: Consider the product I B 0 A −B A + B 0 0 I .
(3) Let x be a fixed real number which is not an integer multiple of π. For each natural number n let An = ajk be the n × n-matrix defined by ajk = 0, for |j −k| > 1 1, for |j −k| = 1 2 cos x, for j = k.
Show that det An = sin(n + 1)x sin x . Hint. For each integer n let Dn = det An and prove that Dn+2 −2Dn+1 cos x + Dn = 0.
(Use mathematical induction.) 3.4. ANSWERS TO ODD-NUMBERED EXERCISES 23 3.4. Answers to Odd-Numbered Exercises (1) 1 0 0 0 0 1 0 0 0 0 4 0 0 0 0 1 (3) 1 0 0 0 0 1 −2 0 0 0 1 0 0 0 0 1 (5) 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 , 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 (7) −8, −1 (9) P12, E31(1), E32(−1), M3(−1 9), E23(−3), M2( 1 2), E13(1), E12(−1) (11) (a) 3, 2, 3, −2 (b) 1 −1 1 −2 1 2 0 1 0 0 (13) 100, 0, −72 (15) −10 (17) 6 (19) (a) −7 (b) 7 (c) 14 CHAPTER 4 VECTOR GEOMETRY IN Rn 4.1. Background Topics: inner (dot) products, cross products, lines and planes in 3-space, norm of a vector, angle between vectors.
4.1.1. Notation. There are many more or less standard notations for the inner product (or dot product) of two vectors x and y. The two that we will use interchangeably in these exercises are x · y and ⟨x, y⟩.
4.1.2. Definition. If x is a vector in Rn, then the norm (or length) of x is defined by ∥x∥= p ⟨x, x⟩.
4.1.3. Definition. Let x and y be nonzero vectors in Rn. Then ∡(x, y), the angle between x and y, is defined by ∡(x, y) = arccos ⟨x, y⟩ ∥x∥∥y∥ 4.1.4. Theorem (Cauchy-Schwarz inequality). If x and y are vectors in Rn, then |⟨x, y⟩| ≤∥x∥∥y∥.
(We will often refer to this just as the Schwarz inequality.) 4.1.5. Definition. If x = (x1, x2, x3) and y = (y1, y2, y3) are vectors in R3, then their cross product, denoted by x × y, is the vector (x2y3 −x3y2, x3y1 −x1y3, x1y2 −x2y1).
25 26 4. VECTOR GEOMETRY IN RN 4.2. Exercises (1) The angle between the vectors (1, 0, −1, 3) and (1, √ 3, 3, −3) in R4 is aπ where a = .
(2) Find the angle θ between the vectors x = (3, −1, 1, 0, 2, 1) and y = (2, −1, 0, √ 2, 2, 1) in R6. Answer: θ = .
(3) If a1, . . . , an > 0, then n X j=1 aj n X k=1 1 ak ≥n2.
The proof of this is obvious from the Cauchy-Schwarz inequality when we choose the vectors x and y as follows: x = and y = .
(4) Find all real numbers α such that the angle between the vectors 2i + 2j + (α −2)k and 2i + (α −2)j + 2k is π 3 . Answer: α = and .
(5) Which of the angles (if any) of triangle ABC, with A = (1, −2, 0), B = (2, 1, −2), and C = (6, −1, −3), is a right angle? Answer: the angle at vertex .
(6) The hydrogen atoms of a methane molecule (CH4) are located at (0, 0, 0), (1, 1, 0), (0, 1, 1), and (1, 0, 1) while the carbon atom is at (1 2, 1 2, 1 2). Find the cosine of the angle θ between two rays starting at the carbon atom and going to different hydrogen atoms.
Answer: cos θ = .
(7) If a, b, c, d, e, f ∈R, then |ad + be + cf| ≤ p a2 + b2 + c2p d2 + e2 + f2.
The proof of this inequality is obvious since this is just the Cauchy-Schwarz inequality where x = ( , , ) and y = ( , , ) .
(8) The volume of the parallelepiped generated by the three vectors i + 2j −k, j + k, and 3i −j + 2k is .
(9) The equations of the line containing the points (3, −1, 4) and (7, 9, 10) are x −3 2 = y −j b = z −k c where b = , c = , j = , and k = .
(10) The equations of the line containing the points (5, 2, −1) and (9, −4, 1) are x −h a = y −2 −3 = z −k c where a = , c = , h = , and k = .
(11) Find the equations of the line containing the point (1, 0, −1) which is parallel to the line x −4 2 = 2y −3 5 = 3z −7 6 .
Answer: x −h a = y −j b = z + 1 4 where a = , b = , h = , and j = .
(12) The equation of the plane containing the points (0, −1, 1), (1, 0, 2), and (3, 0, 1) is x+by + cz = d where b = , c = , and d = .
(13) The equation of the plane which passes through the points (0, −1, −1), (5, 0, 1), and (4, −1, 0) is ax + by + cz = 1 where a = , b = , and c = .
(14) The angle between the planes 4x + 4z −16 = 0 and −2x + 2y −13 = 0 is a b π where a = and b = .
4.2. EXERCISES 27 (15) Suppose that u ∈R3 is a vector which lies in the first quadrant of the xy-plane and has length 3 and that v ∈R3 is a vector that lies along the positive z-axis and has length 5.
Then (a) ∥u × v∥= ; (b) the x-coordinate of u × v is 0 (choose <, >, or =); (c) the y-coordinate of u × v is 0 (choose <, >, or =); and (d) the z-coordinate of u × v is 0 (choose <, >, or =).
(16) Suppose that u and v are vectors in R7 both of length 2 √ 2 and that the length of u −v is also 2 √ 2. Then ∥u + v∥= and the angle between u and v is .
28 4. VECTOR GEOMETRY IN RN 4.3. Problems (1) Show that if a, b, c > 0, then 1 2a + 1 3b + 1 6c 2 ≤1 2a2 + 1 3b2 + 1 6c2.
(2) Show that if a1, . . . , an, w1, . . . , wn > 0 and Pn k=1 wk = 1, then n X k=1 akwk 2 ≤ n X k=1 ak2wk.
(3) Prove that if (a1, a2, . . . ) is a sequence of real numbers such that the series ∞ X k=1 ak2 con-verges, then the series ∞ X k=1 1 k ak converges absolutely.
You may find the following steps helpful in organizing your solution.
(i) First of all, make sure that you recall the difference between a sequence of numbers (c1, c2, . . . ) and an infinite series ∞ X k=1 ck.
(ii) The key to this problem is an important theorem from third term Calculus: A nondecreasing sequence of real numbers converges if and only if it is bounded.
(∗) (Make sure that you know the meanings of all the terms used here.) (iii) The hypothesis of the result we are trying to prove is that the series ∞ X k=1 ak2 converges.
What, exactly, does this mean?
(iv) For each natural number n let bn = n X k=1 ak2.
Rephrase (iii) in terms of the se-quence (bn).
(v) Is the sequence (bn) nondecreasing?
(vi) What, then, does (∗) say about the sequence (bn)?
(vii) For each natural number n let cn = n X k=1 1 k2 . What do we know about the sequence (cn) from third term Calculus? What does (∗) say about the sequence (cn)?
(viii) The conclusion we are trying to prove is that the series ∞ X k=1 1 kak converges absolutely.
What does this mean?
(ix) For each natural number n let sn = n X k=1 1 k|ak|. Rephrase (viii) in terms of the se-quence (sn).
(x) Explain how for each n we may regard the number sn as the dot product of two vectors in Rn.
(xi) Apply the Cauchy-Schwarz inequality to the dot product in (x). Use (vi) and (vii) to establish that the sequence (sn) is bounded above.
(xii) Use (∗) one last time—keeping in mind what you said in (ix).
4.4. ANSWERS TO ODD-NUMBERED EXERCISES 29 4.4. Answers to Odd-Numbered Exercises (1) 3 4 (3) √a1, √a2, . . . , √an , 1 √a1 , 1 √a2 , . . . , 1 √an (5) B (7) a, b, c, d, e, f (9) 5, 3, −1, 4 (11) 4, 5, 1, 0 (13) 1, 3, −4 (15) (a) 15 (b) > (c) < (d) = Part 2 VECTOR SPACES CHAPTER 5 VECTOR SPACES 5.1. Background Topics: real and complex vector spaces, vectors, scalars.
In the following definition F may be taken to be an arbitrary field.
For this collection of exercises, however, we will be interested in only two cases, F = R (the field of real numbers) and F = C (the field of complex numbers).
5.1.1. Definition. A vector space is a set V together with operations of addition and scalar multiplication which satisfy the following axioms: (1) if x, y ∈V , then x + y ∈V ; (2) (x + y) + z = x + (y + z) for every x, y, z ∈V (associativity); (3) there exists 0 ∈V such that x + 0 = x for every x ∈V (existence of additive identity); (4) for every x ∈V there exists −x ∈V such that x + (−x) = 0 (existence of additive inverses); (5) x + y = y + x for every x, y ∈V (commutativity); (6) if α ∈F and x ∈V , then αx ∈V ; (7) α(x + y) = αx + αy for every α ∈F and every x, y ∈V ; (8) (α + β)x = αx + βx for every α, β ∈F and every x ∈V ; (9) (αβ)x = α(βx) for every α, β ∈F and every x ∈V ; and (10) 1 x = x for every x ∈V .
When F = R we speak of V as a real vector space and when F = C we call it a complex vector space.
5.1.2. Definition. An n × n-matrix is nonsingular if its determinant is not zero.
33 34 5. VECTOR SPACES 5.2. Exercises (1) Let V be the set of all real numbers. Define an operation of “addition” by x ⊞y = the maximum of x and y for all x, y ∈V . Define an operation of “scalar multiplication” by α ⊡x = αx for all α ∈R and x ∈V .
Under the operations ⊞and ⊡the set V is not a vector space. The vector space axioms (see 5.1.1 (1)–(10) ) which fail to hold are , , , and .
(2) Let V be the set of all real numbers x such that x ≥0. Define an operation of “addition” by x ⊞y = xy + 1 for all x, y ∈V . Define an operation of “scalar multiplication” by α ⊡x = α2x for all α ∈R and x ∈V .
Under the operations ⊞and ⊡the set V (is/is not) a vector space. If it is not, list all the vector space axioms (see 5.1.1 (1)–(10) ) which fail to hold. Answer: The axioms which are not satisfied are .
(3) Let V be R2, the set of all ordered pairs (x, y) of real numbers. Define an operation of “addition” by (u, v) ⊞(x, y) = (u + x + 1, v + y + 1) for all (u, v) and (x, y) in V . Define an operation of “scalar multiplication” by α ⊡(x, y) = (αx, αy) for all α ∈R and (x, y) ∈V .
Under the operations ⊞and ⊡the set V is not a vector space. The vector space axioms (see 5.1.1 (1)–(10) ) which fail to hold are and .
(4) Let V be R2, the set of all ordered pairs (x, y) of real numbers. Define an operation of “addition” by (u, v) ⊞(x, y) = (u + x, 0) for all (u, v) and (x, y) in V . Define an operation of “scalar multiplication” by α ⊡(x, y) = (αx, αy) for all α ∈R and (x, y) ∈V .
Under the operations ⊞and ⊡the set V is not a vector space. The vector space axioms (see 5.1.1 (1)–(10) ) which fail to hold are , , and .
(5) Let V be the set of all n × n matrices of real numbers. Define an operation of “addition” by A ⊞B = 1 2(AB + BA) for all A, B ∈V . Define an operation of “scalar multiplication” by α ⊡A = 0 for all α ∈R and A ∈V .
Under the operations ⊞and ⊡the set V is not a vector space. The vector space axioms (see 5.1.1 (1)–(10) ) which fail to hold are , , and .
5.2. EXERCISES 35 (6) Below are portions of proofs of four results about vector spaces which establish the fact that multiplying a vector x by the scalar −1 produces −x, the additive inverse of x. Fill in the missing steps and the missing reasons. Choose reasons from the following list.
(H) Hypothesis (1)–(10) Vector space axioms, see 5.1.1 (PA) Proposition A (PB) Proposition B (PC) Proposition C (RN) Property of the Real Numbers 5.2.1. Proposition (A). A vector x in a vector space V has at most one additive inverse.
That is, if y and z are vectors in V such that x + y = 0 and x + z = 0, then y = z.
Proof. Suppose that x + y = 0 and x + z = 0. Then y = (reason: ) = y + (x + z) (reason: ) = (reason: (2) ) = (x + y) + z (reason: ) = (reason: (H) ) = (reason: (5) ) = z (reason: ).
□ 5.2.2. Proposition (B). If x ∈V where V is a vector space and x + x = x, then x = 0.
Proof. If x ∈V and x + x = x, then x = x + 0 (reason: ) = (reason: (4) ) = (x + x) + (−x) (reason: ) = (reason: (H) ) = 0 (reason: ).
□ 5.2.3. Proposition (C). If x is a vector in a vector space V , then 0x = 0.
Proof. If x ∈V , then 0x = (0 + 0) x (reason: ) = (reason: (8) ) Thus 0x = 0 (reason: ).
□ 36 5. VECTOR SPACES 5.2.4. Proposition (D). If x is a vector in a vector space V , then (−1)x is −x, the additive inverse of x.
Proof. If x ∈V , then x + (−1) · x = (reason: (10) ) = 1 + (−1) · x (reason: ) = 0 · x (reason: ) = 0 (reason: ).
It then follows immediately from that (−1) · x = −x.
□ (7) In this exercise we prove that multiplying the zero vector by an arbitrary scalar produces the zero vector. For each step of the proof give the appropriate reason. Choose reasons from the following list.
(1)–(10) Vector space axioms 5.1.1.
(PB) Proposition 5.2.2 (RN) Property of the Real Numbers 5.2.5. Proposition (E). If 0 is the zero vector in a vector space and α is a scalar, then α · 0 = 0.
Proof. Let 0 be the zero vector of some vector space. Then for every scalar α α · 0 = α · (0 + 0) reason: = α · 0 + α · 0 reason: It then follows immediately from that α · 0 = 0.
□ (8) In this exercise we prove that the product of a scalar and a vector is zero if and only if either the scalar or the vector is zero. After each step of the proof give the appropriate reason. Choose reasons from the following list.
(H) Hypothesis.
(1)–(10) Vector space axioms 5.1.1.
(PC),(PE) Propositions 5.2.3 and 5.2.5, respectively.
(RN) Property of the Real Numbers.
5.2.6. Proposition. Suppose that x is a vector and α is a scalar. Then αx = 0 if and only if α = 0 or x = 0.
Proof. We have already shown in and that if α = 0 or x = 0, then αx = 0.
To prove the converse we suppose that αx = 0 and that α ̸= 0; and we prove that x = 0. This conclusion results from the following easy calculation: x = 1 · x reason: = 1 α · α · x reasons: and = 1 α · (α · x) reason: = 1 α · 0 reason: = 0 reason: .
□ 5.3. PROBLEMS 37 5.3. Problems (1) Prove that if V is a vector space, then its additive identity is unique. That is, show that if 0 and e 0 are vectors in V such that x + 0 = x for all x ∈V and x + e 0 = x for all x ∈V , then 0 = e 0.
(2) Let V be the set of all real numbers x such that x > 0. Define an operation of “addition” by x ⊞y = xy for all x, y ∈V . Define an operation of “scalar multiplication” by α ⊡x = xα for all α ∈R and x ∈V .
Prove that under the operations ⊞and ⊡the set V is a vector space.
(3) With the usual operations of addition and scalar multiplication the set of all n×n matrices of real numbers is a vector space: in particular, all the vector space axioms (see 5.1.1 (1)– (10) ) are satisfied. Explain clearly why the set of all nonsingular n × n matrices of real numbers is not a vector space under these same operations.
38 5. VECTOR SPACES 5.4. Answers to Odd-Numbered Exercises (1) 3, 4, 7, 8 (3) 7, 8 (5) 2, 4, 10 (7) 3, 7, PB CHAPTER 6 SUBSPACES 6.1. Background Topics: subspaces of a vector space 6.1.1. Definition. A nonempty subset of M of a vector space V is a subspace of V if it is closed under addition and scalar multiplication. (That is: if x and y belong to M, so does x + y; and if x belongs to M and α ∈R, then αx belongs to M.
6.1.2. Notation. We use the notation M ⪯V to indicate that M is a subspace of a vector space V .
6.1.3. Notation. Here are some frequently encountered families of functions: F = F[a, b] = {f : f is a real valued function on the interval [a, b]} (6.1.1) P = P[a, b] = {p: p is a polynomial function on [a, b]} (6.1.2) P4 = P4[a, b] = {p ∈P : the degree of p is less than 4} (6.1.3) Q4 = Q4[a, b] = {p ∈P : the degree of p is equal to 4} (6.1.4) C = C[a, b] = {f ∈F : f is continuous} (6.1.5) D = D[a, b] = {f ∈F : f is differentiable} (6.1.6) K = K[a, b] = {f ∈F : f is a constant function} (6.1.7) B = B[a, b] = {f ∈F : f is bounded} (6.1.8) J = J [a, b] = {f ∈F : f is integrable} (6.1.9) (A function f ∈F is bounded if there exists a number M ≥0 such that |f(x)| ≤M for all x in [a, b]. It is (Riemann) integrable if it is bounded and R b a f(x) dx exists.) 6.1.4. Definition. If A and B are subsets of a vector space then the sum of A and B, denoted by A + B, is defined by A + B := {a + b: a ∈A and b ∈B}.
6.1.5. Definition. Let M and N be subspaces of a vector space V . If M∩N = {0} and M+N = V , then V is the (internal) direct sum of M and N. In this case we write V = M ⊕N.
In this case the subspaces M and N are complementary and each is the complement of the other.
39 40 6. SUBSPACES 6.2. Exercises (1) One of the following is a subspace of R3. Which one?
The set of points (x, y, z) in R3 such that (a) x + 2y −3z = 4.
(b) x −1 2 = y + 2 3 = z 4.
(c) x + y + z = 0 and x −y + z = 1.
(d) x = −z and x = z.
(e) x2 + y2 = z.
(f) x 2 = y −3 5 .
Answer: ( ) is a subspace of R3.
(2) The smallest subspace of R3 containing the vectors (2, −3, −3) and (0, 3, 2) is the plane whose equation is ax + by + 6z = 0 where a = and b = .
(3) The smallest subspace of R3 containing the vectors (0, −3, 6) and (0, 1, −2) is the line whose equations are x = a and z = by where a = and b = .
(4) Let R∞denote the vector space of all sequences of real numbers. (Addition and scalar multiplication are defined coordinatewise.) In each of the following a subset of R∞is described. Write yes if the set is a subspace of R∞and no if it is not.
(a) Sequences that have infinitely many zeros (for example, (1, 1, 0, 1, 1, 0, 1, 1, 0, . . . )).
Answer: .
(b) Sequences which are eventually zero. (A sequence (xk) is eventually zero if there is an index n0 such that xn = 0 whenever n ≥n0.) Answer: .
(c) Sequences that are absolutely summable. (A sequence (xk) is absolutely summable if P∞ k=1|xk| < ∞.) Answer: .
(d) Bounded sequences. (A sequence (xk) is bounded if there is a positive number M such that |xk| ≤M for every k.) Answer: .
(e) Decreasing sequences. (A sequence (xk) is decreasing if xn+1 ≤xn for each n.) Answer: .
(f) Convergent sequences. Answer: .
(g) Arithmetic progressions. (A sequence (xk) is arithmetic if it is of the form (a, a + k, a + 2k, a + 3k, . . . ) for some constant k.) Answer: .
(h) Geometric progressions. (A sequence (xk) is geometric if it is of the form (a, ka, k2a, k3a, . . . ) for some constant k.) Answer: .
(5) Let M and N be subspaces of a vector space V . Consider the following subsets of V .
(a) M ∩N. (A vector v belongs to M ∩N if it belongs to both M and N.) (b) M ∪N. (A vector v belongs to M ∪N if it belongs to either M or N.) (c) M + N. (A vector v belongs to M + N if there are vectors m ∈M and n ∈N such that v = m + n.) (d) M −N. (A vector v belongs to M −N if there are vectors m ∈M and n ∈N such that v = m −n.) Which of (a)–(d) are subspaces of V ?
Answer: .
6.2. EXERCISES 41 (6) For a fixed interval [a, b], which sets of functions in the list 6.1.3 are vector subspaces of which?
Answer: ⪯ ⪯ ⪯ ⪯ ⪯ ⪯ ⪯ .
(7) Let M be the plane x + y + z = 0 and N be the line x = y = z in R3. The purpose of this exercise is to confirm that R3 = M ⊕N. This requires establishing three things: (i) M and N are subspaces of R3 (which is very easy and which we omit); (ii) R3 = M + N; and (iii) M ∩N = {0}.
(a) To show that R3 = M + N we need R3 ⊆M + N and M + N ⊆R3. Since M ⊆R3 and N ⊆R3, it is clear that M + N ⊆R3. So all that is required is to show that R3 ⊆M + N. That is, given a vector x = (x1, x2, x3) in R3 we must find vectors m = (m1, m2, m3) in M and n = (n1, n2, n3) in N such that x = m + n. Find two such vectors.
Answer: m = 1 3 ( , , ) and n = 1 3 ( , , ).
(b) The last thing to verify is that M ∩N = {0}; that is, that the only vector M and N have in common is the zero vector. Suppose that a vector x = (x1, x2, x3) belongs to both M and N. Since x ∈M it must satisfy the equation x1 + x2 + x3 = 0.
(1) since x ∈N it must satisfy the equations x1 = x2 and (2) x2 = x3.
(3) Solve the system of equations (1)–(3).
Answer: x = ( , , ) .
(8) Let C = C[−1, 1] be the vector space of all continuous real valued functions on the interval [−1, 1]. A function f in C is even if f(−x) = f(x) for all x ∈[−1, 1]; it is odd if f(−x) = −f(x) for all x ∈[−1, 1]. Let Co = {f ∈C : f is odd } and Ce = {f ∈C : f is even }. To show that C = Co ⊕Ce we need to show 3 things.
(i) Co and Ce are subspaces of C. This is quite simple: let’s do just one part of the proof.
We will show that Co is closed under addition. After each step of the following proof indicate the justification for that step. Make your choices from the following list.
(A) Arithmetic of real numbers.
(DA) Definition of addition of functions.
(DE) Definition of “even function.” (DO) Definition of “odd function.” (H) Hypothesis (that is, our assumptions or suppositions).
42 6. SUBSPACES Proof. Let f, g ∈Co. Then (f + g)(−x) = f(−x) + g(−x) reason: = −f(x) + (−g(x)) reason: and = −(f(x) + g(x)) reason: = −(f + g)(x).
reason: Thus f + g ∈Co. reason .
□ (ii) Co ∩Ce = {0} (where 0 is the constant function on [−1, 1] whose value is zero). Again choose from the reasons listed in part (i) to justify the given proof.
Proof. Suppose f ∈Co ∩Ce. Then for each x ∈[−1, 1] f(x) = f(−x) reason: = −f(x).
reason: Thus f(x) = 0 for every x ∈[−1, 1]; that is, f = 0.
reason: .
□ (iii) C = Co + Ce. To verify this we must show that every continuous function f on [−1, 1] can be written as the sum of an odd function j and an even function k. It turns out that the functions j and k can be written as linear combinations of the given function f and the function g defined by g(x) = f(−x) for −1 ≤x ≤1. What are the appropriate coefficients?
Answer: j = f + g k = f + g.
(9) Let M be the line x = y = z and N be the line x = 1 2y = 1 3z in R3.
(a) The line M is the set of all scalar multiples of the vector ( 1 , , ).
(b) The line N is the set of all scalar multiples of the vector ( 1 , , ).
(c) The set M + N is (geometrically speaking) a in R3; its equation is ax + by + z = 0 where a = and b = .
(10) Let M be the plane x −y + z = 0 and N be the plane x + 2y −z = 0 in R3. State in one short sentence how you know that R3 is not the direct sum of M and N.
Answer: .
(11) Let M be the plane 2x −3y + 4z + 1 = 0 and N be the line x 4 = y 2 = z 3 in R3. State in one short sentence how you know that R3 is not the direct sum of M and N.
Answer: .
(12) Let M be the plane x + y + z = 0 and N be the line x −1 = 1 2y = z + 2 in R3. State in one short sentence how you know that R3 is not the direct sum of M and N.
Answer: .
(13) Let M be the line x = y = z and N be the line x 4 = y 2 = z 3 in R3. State in one short sentence how you know that R3 is not the direct sum of M and N.
Answer: .
6.2. EXERCISES 43 (14) Let M be the plane x + y + z = 0 and N be the line x = −3 4y = 3z. The purpose of this exercise is to see (in two different ways) that R3 is not the direct sum of M and N.
(a) If R3 were equal to M ⊕N, then M ∩N would contain only the zero vector. Show that this is not the case by finding a nonzero vector x in R3 which belongs to M ∩N.
Answer: x = ( , , 1 ) .
(b) If R3 were equal to M ⊕N, then, in particular, we would have R3 = M + N. Since both M and N are subsets of R3, it is clear that M + N ⊆R3. Show that the reverse inclusion R3 ⊆M + N is not correct by finding a vector x ∈R3 which cannot be written in the form m + n where m ∈M and n ∈N.
Answer: x = (−6, 8, a) is such a vector provided that a ̸= .
(c) We have seen in part (b) that M + N ̸= R3. Then what is M + N?
Answer: M + N = .
44 6. SUBSPACES 6.3. Problems (1) Let M and N be subspaces of a vector space V . Consider the following subsets of V .
(a) M ∩N. (A vector v belongs to M ∩N if it belongs to both M and N.) (b) M ∪N. (A vector v belongs to M ∪N if it belongs to either M or N.) (c) M + N. (A vector v belongs to M + N if there are vectors m ∈M and n ∈N such that v = m + n.) (d) M −N. (A vector v belongs to M −N if there are vectors m ∈M and n ∈N such that v = m −n.) For each of the sets (a)–(d) above, either prove that it is a subspace of V or give a counterexample to show that it need not be a subspace of V .
(2) Let C = C[0, 1] be the family of continuous real valued functions on the interval [0, 1].
Define f1(t) = t and f2(t) = t4 for 0 ≤t ≤1. Let M be the set of all functions of the form αf1 + βf2 where α, β ∈R.
And let N be the set of all functions g in C which satisfy Z 1 0 tg(t) dt = 0 and Z 1 0 t4g(t) dt = 0.
Is C the direct sum of M and N? (Give a careful proof of your claim and illustrate it with an example. What does your result say, for instance, about the function h defined by h(t) = t2 for 0 ≤t ≤1.) (3) Let V be a vector space.
(a) Let M be a family of subspaces of V . Prove that the intersection T M of this family is itself a subspace of V .
(b) Let A be a set of vectors in V . Explain carefully why it makes sense to say that the intersection of the family of all subspaces containing A is “the smallest subspace of V which contains A”.
(c) Prove that the smallest subspace of V which contains A is in fact the span of A.
(4) In R3 let M be the line x = y = z, N be the line x = 1 2y = 1 3z, and L = M + N. Give a careful proof that L = M ⊕N.
(5) Let V be a vector space and suppose that V = M ⊕N. Show that for every v ∈V there exist unique vectors m ∈M and n ∈N such that v = m + n. Hint. It should be clear that the only thing you have to establish is the uniqueness of the vectors m and n. To this end, suppose that a vector v in V can be written as m1 + n1 and it can also be written as m2 + n2 where m1, m2 ∈M and n1, n2 ∈N. Prove that m1 = m2 and n1 = n2.
6.4. ANSWERS TO ODD-NUMBERED EXERCISES 45 6.4. Answers to Odd-Numbered Exercises (1) (d) (3) 0, −2 (5) (a), (c), and (d) (7) (a) 2x1−x2−x3 , −x1+2x2−x3 , −x1−x2+2x3 , x1+x2+x3 , x1+x2+x3 , x1+x2+x3 , (b) 0, 0, 0 (9) (a) 1, 1 (b) 2, 3 (c) plane, 1, −2 (11) M is not a subspace of R3.
(13) M + N is a plane, not all of R3.
CHAPTER 7 LINEAR INDEPENDENCE 7.1. Background Topics: linear combinations, span, linear dependence and independence.
7.1.1. Remark. Some authors of linear algebra texts make it appear as if the terms linear de-pendence and linear independence, span, and basis pertain only to finite sets of vectors. This is extremely misleading. The expressions should make sense for arbitrary sets. In particular, do not be misled into believing that a basis for a vector space must be a finite set of vectors (or a sequence of vectors). While it is true that in most elementary linear algebra courses the emphasis is on the study of finite dimensional vector spaces, bases for vector spaces may be very large indeed. I recommend the following definitions.
7.1.2. Definition. Recall that a vector y is a linear combination of distinct vectors x1, . . . , xn if there exist scalars α1, . . . αn such that y = Pn k=1 αkxk. Note: linear combinations are finite sums. The linear combination Pn k=1 αkxk is trivial if all the coefficients α1, . . . αn are zero. If at least one αk is different from zero, the linear combination is nontrivial.
7.1.3. Example. In R2 the vector (8, 2) is a linear combination of the vectors (1, 1) and (1, −1) because (8, 2) = 5(1, 1) + 3(1, −1).
7.1.4. Example. In R3 the vector (1, 2, 3) is not a linear combination of the vectors (1, 1, 0) and (1, −1, 0).
7.1.5. Definition. Suppose that A is a subset (finite or not) of a vector space V . The span of A is the set of all linear combinations of elements of A. Another way of saying the same thing: the span of A is the smallest subspace of V which contains A. (That these characterizations are equivalent is not completely obvious. Proof is required. See problem 3 in chapter 6. We denote the span of A by span A. If U = span A, we say that A spans U or that U is spanned by A.
7.1.6. Example. For each n = 0, 1, 2, . . . define a function pn on R by pn(x) = xn. Let P be the set of polynomial functions on R. It is a subspace of the vector space of continuous functions on R.
Then P = span{p0, p1, p2 . . . }. The exponential function exp, whose value at x is ex, is not in the span of the set {p0, p1, p2 . . . }.
7.1.7. Definition. A subset A (finite or not) of a vector space is linearly dependent if the zero vector 0 can be written as a nontrivial linear combination of elements of A; that is, if there exist vectors x1, . . . , xn ∈A and scalars α1, . . . , αn, not all zero, such that Pn k=1 αkxk = 0. A subset of a vector space is linearly independent if it is not linearly dependent.
Technically, it is a set of vectors that is linearly dependent or independent. Nevertheless, these terms are frequently used as if they were properties of the vectors themselves. For instance, if S = {x1, . . . , xn} is a finite set of vectors in a vector space, you may see the assertions “the set S is linearly independent” and “the vectors x1, . . . xn are linearly independent” used interchangeably.
7.1.8. Example. The (vectors going from the origin to) points on the unit circle in R2 are linearly dependent. Reason: If x = (1, 0), y = −1 2, √ 3 2 , and z = 1 2, √ 3 2 , then x + y + (−1)z = 0.
47 48 7. LINEAR INDEPENDENCE 7.1.9. Example. For each n = 0, 1, 2, . . . define a function pn on R by pn(x) = xn. Then the set {p0, p1, p2, . . . } is a linearly independent subset of the vector space of continuous functions on R.
7.2. EXERCISES 49 7.2. Exercises (1) Show that in the space R3 the vectors x = (1, 1, 0), y = (0, 1, 2), and z = (3, 1, −4) are linearly dependent by finding scalars α and β such that αx + βy + z = 0.
Answer: α = , β = .
(2) Let w = (1, 1, 0, 0), x = (1, 0, 1, 0), y = (0, 0, 1, 1), and z = (0, 1, 0, 1).
(a) We can show that {w, x, y, z} is not a spanning set for R4 by finding a vector u in R4 such that u / ∈span{w, x, y, z}. One such vector is u = (1, 2, 3, a) where a is any number except .
(b) Show that {w, x, y, z} is a linearly dependent set of vectors by finding scalars α, γ, and δ such that αw + x + γy + δz = 0.
Answer: α = , γ = , δ = .
(c) Show that {w, x, y, z} is a linearly dependent set by writing z as a linear combination of w, x, and y. Answer: z = w + x + y.
(3) Let p(x) = x2 + 2x −3, q(x) = 2x2 −3x + 4, and r(x) = ax2 −1. The set {p, q, r} is linearly dependent if a = .
(4) Show that in the vector space R3 the vectors x = (1, 2, −1), y = (3, 1, 1), and z = (5, −5, 7) are linearly dependent by finding scalars α and β such that αx + βy + z = 0.
Answer: α = , β = .
(5) Let f1(x) = sin x, f2(x) = cos(x+π/6), and f3(x) = sin(x−π/4) for 0 ≤x ≤2π. Show that {f1, f2, f3} is linearly dependent by finding constants α and β such that αf1−2f2−βf3 = 0.
Answer: α = and β = .
(6) In the space C[0, π] let f, g, h, and j be the vectors defined by f(x) = 1 g(x) = x h(x) = cos x j(x) = cos2 x 2 for 0 ≤x ≤π. Show that f, g, h, and j are linearly dependent by writing j as a linear combination of f, g, and h.
Answer: j = f + g + h.
(7) Let u = (λ, 1, 0), v = (1, λ, 1), and w = (0, 1, λ). Find all values of λ which make {u, v, w} a linearly dependent subset of R3. Answer: (8) Let u = (1, 0, −2), v = (1, 2, λ), and w = (2, 1, −1). Find all values of λ which make {u, v, w} a linearly dependent subset of R3. Answer: (9) Let p(x) = x3 −x2 + 2x + 3, q(x) = 3x3 + x2 −x −1, r(x) = x3 + 2x + 2, and s(x) = 7x3 + ax2 + 5. The set {p, q, r, s} is linearly dependent if a = .
50 7. LINEAR INDEPENDENCE (10) In the space C[0, π] define the vectors f, g, and h by f(x) = x g(x) = sin x h(x) = cos x for 0 ≤x ≤π. We show that f, g, and h are linearly independent. This is accomplished by showing that if αf + βg + γh = 0, then α = β = γ = 0. So we start by supposing that αf + βg + γh = 0; that is, αx + β sin x + γ cos x = 0 (1) for all x ∈[0, π].
(a) We see that γ must be zero by setting x = in equation (1).
Now differentiate (1) to obtain α + β cos x = 0 (2) for all x ∈[0, π].
(b) We see that α must be zero by setting x = in equation (2).
Differentiate (2) to obtain −β sin x = 0 (3) for all x ∈[0, π].
(c) We conclude that β = 0 by setting x = in (3).
7.3. PROBLEMS 51 7.3. Problems (1) In the space C[0, 1] define the vectors f, g, and h by f(x) = x g(x) = ex h(x) = e−x for 0 ≤x ≤1. Use the definition of linear independence to show that the functions f, g, and h are linearly independent.
(2) Let a, b, and c be distinct real numbers. Use the definition of linear independence to give a careful proof that the vectors (1, 1, 1), (a, b, c), and (a2, b2, c2) form a linearly independent subset of R3.
(3) Let {u, v, w} be a linearly independent set in a vector space V . Use the definition of linear independence to give a careful proof that the set {u + v, u + w, v + w} is linearly independent in V .
(4) You are the leader of an engineering group in the company you work for and have a routine computation that has to be done repeatedly. At your disposal is an intern, Kim, a beginning high school student, who is bright but has had no advanced mathematics. In particular, Kim knows nothing about vectors or matrices.
Here is the computation that is needed.
Three vectors, a, b, and c are specified in R5. (Denote their span by M.) Also specified is a (sometimes long) list of other vectors S = {v1, v2, . . . , vn} in R5. The problem is to (1) determine which of the vectors in S belong to M, and (2) for each vector vk ∈S which does belong to M find constants α, β, and γ such that vk = αa + βb + γc.
Kim has access to Computer Algebra System (Maple, or a similar program) with a Lin-ear Algebra package. Write a simple and efficient algorithm (that is, a set of instructions) which will allow Kim to carry out the desired computation repeatedly. The algorithm should be simple in the sense that it uses only the most basic linear algebra commands (for example, Matrix, Vector, Transpose, RowReducedEchelonForm, etc. in Maple). Re-member, you must tell Kim everything: how to set up the appropriate matrices, what operations to perform on them, and how to interpret the results. The algorithm should be as efficient as you can make it. For example, it would certainly not be efficient for Kim to retype the coordinates of a, b, and c for each new vk.
Include in your write-up an actual printout showing how your algorithm works in some special case of your own invention. (For this example, the set S need contain only 5 or 6 vectors, some in U, some not.) (5) The point of this problem is not just to get a correct answer to (a)–(c) below using tools you may have learned elsewhere, but to give a careful explanation of how to apply the linear algebra techniques you have already encountered to solve this problem in a systematic fashion. For background you may wish to read a bit about networks and Kirchhoff’s laws (see, for example, Topic: Analyzing Networks, pages 72–77 or Electrical Networks, pages 538–542).
Consider an electrical network having four nodes A, B, C, and D connected by six branches 1, . . . , 6. Branch 1 connects A and B; branch 2 connects B and D; branch 3 connects C and B; branch 4 connects C and D; branch 5 connects A and C; and branch 6 connects A and D.
52 7. LINEAR INDEPENDENCE The current in branch k is Ik, where k = 1, . . . , 6.
There is a 17 volt battery in branch 1 producing the current I1 which flows from A to B. In branches 2, 4, and 5 there are 0.5 ohm resistors; and in branches 1, 3, and 6 there are 1 ohm resistors.
(a) Find the current in each branch.
(Explain any minus signs which occur in your answer.) (b) Find the voltage drop across each branch.
(c) Let pn be the potential at node n = A, B, C, D. The voltage drop across the branch connecting node j to node k is the difference in the potentials at nodes j and k.
Suppose the network is grounded at D (so that pD = 0). Find the potential at the other nodes.
7.4. ANSWERS TO ODD-NUMBERED EXERCISES 53 7.4. Answers to Odd-Numbered Exercises (1) −3, 2 (3) 7 (5) √ 3 −1, √ 6 (7) − √ 2, 0, √ 2 (9) −3 CHAPTER 8 BASIS FOR A VECTOR SPACE 8.1. Background Topics: basis, dimension.
8.1.1. Definition. A set B (finite or not) of vectors in a vector space V is a basis for V if it is linearly independent and spans V .
8.1.2. Example. The vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1) constitute a basis for the vector space R3.
8.1.3. Example. More generally, consider the vector space Rn of all n-tuples of real numbers. For each natural number k between 1 and n let ek be the vector which is 1 in the kth-coordinate and 0 in all the others. Then the set {e1, e2, . . . , en} is a basis for Rn. It is called the standard basis for Rn.
8.1.4. Example. For each n = 0, 1, 2, . . . define a function pn on R by pn(x) = xn. Then the set {p0, p1, p2, . . . } is a basis for the vector space P of polynomial functions on R.
Two important facts of linear algebra are that regardless of the size of the space every vector space has a basis and that every subspace has a complement.
8.1.5. Theorem. Let B be a linearly independent set of vectors in a vector space V . Then there exists a set C of vectors in V such that B ∪C is a basis for V .
8.1.6. Corollary. Every vector space has a basis.
8.1.7. Corollary. Let V be a vector space.
If M ⪯V , then there exists N ⪯V such that M ⊕N = V .
The next theorem says that any two bases for a vector space are the same size.
8.1.8. Theorem. If B and C are bases for the same vector space, then there is a one-to-one correspondence from B onto C.
8.1.9. Definition. A vector space V is finite dimensional if it has a finite basis. Its dimension (denoted by dim V ) is the number of elements in the basis. If V does not have a finite basis it is infinite dimensional.
8.1.10. Theorem. If M and N are subspaces of a finite dimensional vector space, then dim(M + N) = dim M + dim N −dim(M ∩N) .
55 56 8. BASIS FOR A VECTOR SPACE 8.2. Exercises (1) Let u = (2, 0, −1), v = (3, 1, 0), and w = (1, −1, c) where c ∈R. The set {u, v, w} is a basis for R3 provided that c is not equal to .
(2) Let u = (1, −1, 3), v = (1, 0, 1), and w = (1, 2, c) where c ∈R. The set {u, v, w} is a basis for R3 provided that c is not equal to .
(3) The dimension of M2×2, the vector space of all 2 × 2 matrices of real numbers is .
(4) The dimension of T2, the vector space of all 2×2 matrices of real numbers with zero trace is .
(5) The dimension of the vector space of all real valued polynomial functions on R of degree 4 or less is .
(6) In R4 let M be the subspace spanned by the vectors (1, 1, 1, 0) and (0, −4, 1, 5) and let N be the subspace spanned by (0, −2, 1, 2) and (1, −1, 1, 3). One vector which belongs to both M and N is (1, , , ). The dimension of M ∩N is and the dimension of M + N is .
8.3. PROBLEMS 57 8.3. Problems (1) Exhibit a basis for M2×2, the vector space of all 2 × 2 matrices of real numbers.
(2) Exhibit a basis for T2, the vector space of all 2 × 2 matrices of real numbers with zero trace.
(3) Exhibit a basis for S3, the vector space of all symmetric 3 × 3 matrices of real numbers.
(4) Let U be the set of all matrices of real numbers of the form u −u −x 0 x and V be the set of all real matrices of the form v 0 w −v . Exhibit a basis for U, for V, for U + V, and for U ∩V.
(5) Prove that the vectors (1, 1, 0), (1, 2, 3), and (2, −1, 5) form a basis for R3.
(6) Let V be a vector space and A be a linearly independent subset of V . Prove that A is a basis for V if and only if it is a maximal linearly independent subset of V . (If A is a linearly independent subset of V we say that it is a maximal linearly independent set if the addition of any vector at all to A will result in a set which is not linearly independent.) (7) Let V be a vector space and A a subset of V which spans V . Prove that A is a basis for V if and only if it is a minimal spanning set. (If A is a set which spans V we say that it is a minimal spanning set if the removal of any vector at all from A will result in a set which does not span V .) 58 8. BASIS FOR A VECTOR SPACE 8.4. Answers to Odd-Numbered Exercises (1) −2 (3) 4 (5) 5 Part 3 LINEAR MAPS BETWEEN VECTOR SPACES CHAPTER 9 LINEARITY 9.1. Background Topics: linear maps between vector spaces, kernel, nullspace, nullity, range, rank, isomorphism.
9.1.1. Definition. A function f : A →B is one-to-one (or injective) if u = v in A whenever f(u) = f(v) in B.
9.1.2. Definition. A function f : A →B is onto (or surjective) if for every b ∈B there exists a ∈A such that b = f(a).
9.1.3. Definition. A function f : A →B is a one-to-one correspondence (or bijective) if it is both injective and surjective (one-to-one and onto).
9.1.4. Definition. A map T : V →W between vector spaces is linear if T(x + y) = Tx + Ty for all x, y ∈V (9.1.1) and T(αx) = αTx for all x ∈V and α ∈F.
(9.1.2) Here F = R if V and W are real vector spaces and F = C if they are complex vector spaces.
A scalar valued linear map on a vector space V is a linear functional.
A linear map is frequently called a linear transformation, and, in case the domain and codomain are the same, it is often called a (linear) operator. The family of all linear transfor-mations from V into W is denoted by L(V, W). We shorten L(V, V ) to L(V ).
Two oddities of notation concerning linear transformations deserve comment. First, the value of T at x is usually written Tx rather than T(x). Naturally the parentheses are used whenever their omission would create ambiguity. For example, in (9.1.1) above Tx + y is not an acceptable substitute for T(x + y).
Second, the symbol for composition of two linear transformations is ordinarily omitted. If S ∈L(U, V ) and T ∈L(V, W), then the composite of T and S is denoted by TS (rather than by T ◦S). As a consequence of this convention when T ∈L(V ) the linear operator T ◦T is written as T 2, T ◦T ◦T as T 3, and so on.
For future reference here are two obvious properties of a linear map.
9.1.5. Proposition. If T : V →W is a linear map between vector spaces, then T(0) = 0.
9.1.6. Proposition. If T : V →W is a linear map between vector spaces, then T(−x) = −Tx for every x ∈V .
You should prove these propositions if (and only if) it is not immediately obvious to you how to do so.
9.1.7. Definition. Let T : V →W be a linear transformation between vector spaces. Then ker T, the kernel of T, is defined to be the set of all x in V such that Tx = 0. The kernel of T is also called the nullspace of T. If V is finite dimensional, the dimension of the kernel of T is the nullity of T.
Also, ran T, the range of T, is the set of all y in W such that y = Tx for some x in V . If the range of T is finite dimensional, its dimension is the rank of T.
61 62 9. LINEARITY 9.1.8. Notation. Let V be a vector space. We denote the identity map on V (that is, the map x 7→x from V into itself) by IV , or just I.
The following fundamental result is proved in most linear algebra texts.
9.1.9. Theorem. If T : V →W is a linear map between finite dimensional vector spaces, then rank(T) + nullity(T) = dim V .
9.1.10. Definition. Let T : V →W and S : W →V be linear maps. If ST = IV , then T is a right inverse for S and S is a left inverse for T. The mapping T is invertible (or is an isomorphism) if there exists a linear transformation, which we denote by T −1 : W →V , such that TT −1 = IW and T −1T = IV .
The vector spaces V and W are isomorphic if there exists an isomorphism T from V to W.
9.1.11. Notation. Let V and W be vector spaces. We denote by L(V, W) the set of all linear maps from V into W and by L(V ) the set of all linear operators T : V →V .
9.2. EXERCISES 63 9.2. Exercises (1) Define T : R3 →R4 by Tx = (x1 −x3, x1 + x2, x3 −x2, x1 −2x2) for all x = (x1, x2, x3) in R3.
(a) Then T(1, −2, 3) = ( , , , ) .
(b) Find a vector x ∈R3 such that Tx = (8, 9, −5, 0).
Answer: x = ( , , ).
(2) Define T : R4 →R3 by Tx = (2x1 + x3 + x4, x1 −2x2 −x3, x2 −x3 + x4) for all x = (x1, x2, x3, x4) in R4.
(a) Then T(2, 1, −1, 3) = ( , , ) .
(b) Find a vector x ∈R4 such that Tx = (3, −1, −3).
Answer: x = ( , , , ).
(3) Let T be the linear map from R3 to R3 defined by T(x, y, z) = (x + 2y −z, 2x + 3y + z, 4x + 7y −z).
The kernel of T is (geometrically) a whose equation(s) is(are) ; and the range of T is geometrically a whose equation(s) is(are) .
(4) Let T : R3 →R3 be the linear transformation whose action on the standard basis vectors of R3 is T(1, 0, 0) = (1, −3 2, 2) T(0, 1, 0) = (−3, 9 2, −6) T(0, 0, 1) = (2, −3, 4).
Then T(5, 1, −1) = ( , , ) . The kernel of T is the whose equation is x + ay + bz = 0 where a = and b = . The range of T is the whose equations are x 2 = y c = z d where c = and where d = .
(5) Let P be the vector space of all polynomial functions on R with real coefficients. Define linear transformations T, D: P →P by (Dp)(x) = p′(x) and (Tp)(x) = xp(x) for all x ∈R.
(a) Let p(x) = x3−7x2+5x+6 for all x ∈R. Then ((D+T)(p))(x) = x4−ax3+bx2−bx+c where a = , b = , and c = .
(b) Let p be as in (a). Then (DTp)(x) = ax3−bx2+cx+6 where a = , b = , and c = .
(c) Let p be as in (a). Then (TDp)(x) = ax3 −bx2 + cx where a = , b = , and c = .
(d) Evaluate (and simplify) the commutator [D, T] := DT −TD.
Answer: [D, T] = .
(e) Find a number p such that (TD)p = T pDp + TD. Answer: p = .
64 9. LINEARITY (6) Let C = C[a, b] be the vector space of all continuous real valued functions on the interval [a, b] and C1 = C1[a, b] be the vector space of all continuously differentiable real valued functions on [a, b]. (Recall that a function is continuously differentiable if it has a derivative and the derivative is continuous.) Let D: C1 →C be the linear transformation defined by Df = f′ and let T : C →C1 be the linear transformation defined by (Tf)(x) = Z x a f(t) dt for all f ∈C and x ∈[a, b].
(a) Compute (and simplify) (DTf)(x). Answer: .
(b) Compute (and simplify) (TDf)(x). Answer: .
(c) The kernel of T is .
(d) The range of T is {g ∈C1 : } (7) In this exercise we prove that a linear transformation T : V →W between two vector spaces is one-to-one if and only if its kernel contains only the zero vector. After each step of the proof give the appropriate reason. Choose reasons from the following list.
(DK) Definition of “kernel”.
(DL) Definition of “linear”.
(DO) Definition of “one-to-one”.
(H) Hypothesis.
(Pa) Proposition 9.1.5.
(Pb) Proposition 9.1.6.
(VA) Vector space arithmetic (consequences of vector space axioms, definition of subtraction of vectors, etc.) Proof. Suppose that T is one-to-one. We show that ker T = {0V }. Since 0V ∈ker T (reason: and ), we need only show that ker T ⊆{0V }; that is, we show that if x ∈ker T, then x = 0V . So let x ∈ker T. Then Tx = 0W (reason: and ) and T0V = 0W (reason: ). From this we conclude that x = 0V (reason: and ).
Now we prove the converse. Suppose that ker T = {0V }. We wish to show that T is one-to-one. Let x, y ∈V and suppose that Tx = Ty. Then T(x −y) = T(x + (−y)) reason: = Tx + T(−y) reason: = Tx + (−Ty) reason: = Tx −Ty reason: = 0W reason: and Then x −y ∈ker T (reason: ). So x −y = 0V (reason: ); that is, x = y (reason: ). Thus T is one-to-one (reason: and ).
□ (8) Let C1(R) be the vector space of all functions defined on the real line R which have continuous derivatives at each point of R and C(R) be the vector space of continuous functions on R. Define the function T : C1(R) →C(R) by (Tf)(t) = f ′(t) + 3f(t) 9.2. EXERCISES 65 for every t ∈R. (Notice that T is a linear map.) The kernel of T is the set of all scalar multiples of the function g where g(t) = for each t. Thus the kernel of the linear map T is the solution space of the differential equation .
(9) Let C2(R) be the vector space of all functions defined on the real line R which have continuous second derivatives at each point of R and C(R) be the vector space of continuous functions on R. Define the function T : C2(R) →C(R) by (Tf)(t) = f ′′(t) + f(t) for every t ∈R. (Notice that T is a linear map.) Assume that the kernel of T is two dimensional. Then ker T = span{g, h} where g(t) = and h(t) = for all t. Thus the kernel of the linear map T is the solution space of the differential equa-tion .
(10) Define a function k on the unit square [0, 1] × [0, 1] by k(x, y) = ( x, for 0 ≤x ≤y ≤1 y, for 0 ≤y < x ≤1 .
Define an integral operator K on the vector space C[0, 1] of continuous real valued functions on [0, 1] by (Kf)(x) = Z 1 0 k(x, y)f(y) dy for 0 ≤x ≤1. Find the function Kf when f is the function defined by f(x) = x2 for 0 ≤x ≤1.
Answer: (Kf)(x) = .
(11) Let T : R3 →R3 : x 7→(x1 + 3x2 −2x3, x1 −4x3, x1 + 6x2).
(a) The kernel of T is a in R3 given by the equation(s) .
(b) The range of T is a in R3 given by the equation(s) .
(12) Let T : R2 →R3 : (x, y) 7→(2x −3y, x + 2y + 1, 5x −2y). State in one short sentence how you know that T is not a linear transformation.
Answer: .
(13) Let a = (1, 0, 0, 0), b = (1, 1, 0, 0), c = (1, 1, 1, 0), and d = (1, 1, 1, 1).
Suppose that T : R4 →R7 is a mapping such that T(a) = T(b) = T(c) = T(d) = 0 and that T(3, −19, 7, −8) = (1, 1, 1, −3, 6, 2, 5).
State in a short sentence or two how you know that T is not a linear transformation.
Answer: .
(14) Suppose that T : R3 →R3 is a mapping (not identically zero) whose range is contained in the paraboloid z = x2 + y2. State in a short sentence or two how you know that T is not a linear transformation.
Answer: .
(15) Let T : R2 →R4 : (x, y) 7→(2x−3y, x−7y, x+2y +1, 5x−2y). State in one short sentence how you know that T is not a linear transformation.
Answer: .
66 9. LINEARITY (16) Let a = (1, 1, 0) and b = (0, 1, 1), and c = (1, 2, 1).
Suppose that T : R3 →R5 is a mapping such that T(a) = T(b) = 0 and that T(c) = (1, −3, 6, 2, 5). State in a short sentence or two how you know that T is not a linear transformation.
Answer: .
(17) Suppose that T : R2 →R2 is a mapping (not identically zero) such that T(1, 1) = (3, −6) and T(−2, −2) = (−6, 3). State in a short sentence or two how you know that T is not a linear transformation.
Answer: .
9.3. PROBLEMS 67 9.3. Problems (1) Let T : V →W be a linear transformation between vector spaces and let N be a subspace of W. Define T ←(N) := {v ∈V : Tv ∈N}. Prove that T ←(N) is a subspace of V .
(2) Prove that a linear transformation T : R3 →R2 cannot be one-to-one and that a linear transformation S : R2 →R3 cannot be onto. Generalize these assertions.
(3) Prove that one-to-one linear transformations preserve linear independence. That is: Let T : V →W be a one-to-one linear transformation between vector spaces and {x1, x2, . . . , xn} be a linearly independent subset of V . Prove that {Tx1, Tx2, . . . , Txn} is a linearly in-dependent subset of W. Hint. To prove that the vectors Tx1, Tx2, . . . , Txn are linearly independent, it must be shown that the only linear combination of these vectors which equals zero is the trivial linear combination. So suppose that Pn k=1 αkTxk = 0 and prove that every αk must be zero. Use the result proved in exercise 7.
(4) The goal of this problem is to understand and write up an introduction to invertible linear transformations. Your write-up should explain with spectacular clarity the basic facts about invertible linear transformations. Include answers to the following questions— giving complete proofs or counterexamples as required. (But don’t number things in your report to correspond with the items that follow.) (a) If a linear transformation has a right inverse must it have a left inverse?
(b) If a linear transformation has a left inverse must it have a right inverse?
(c) If a linear transformation has both a left and a right inverse, must it be invertible?
(That is, must the left and right inverse be the same?) (d) If a linear transformation T has a unique right inverse is T necessarily invertible?
Hint. Consider ST + S −I, where S is a unique right inverse for T.
(e) What is the relationship between a linear transformation being one-to-one and onto and being invertible?
(f) Let {v1, . . . , vn} be a linearly independent set of vectors in V . What condition should a linear transformation T : V →W satisfy so that {Tv1, . . . , Tvn} is a linearly inde-pendent subset of W?
(g) Let {u1, . . . , un} be a basis for a subspace U of V . What conditions should a linear transformation T : V →W satisfy so that {Tu1, . . . , Tun} is a basis for the sub-space T(U)?
(h) Suppose the vectors v1, . . . , vn span the vector space V and T : V →W is a linear transformation. If {Tv1, . . . , Tvn} is a basis for W what can you conclude about the vectors v1, . . . , vn? What can you conclude about the linear transformation T?
(i) When are two finite dimensional vector spaces isomorphic?
(Give a simple—but nontrivial—necessary and sufficient condition on the spaces.) (j) Suppose S : V →V is linear and V has finite dimension. What is the relationship between the following properties of S?
(1) S is one-to-one.
(2) S is onto.
(3) S is an isomorphism.
(5) A sequence of vector spaces and linear maps · · · − →Vn−1 jn − − − − →Vn jn+1 − − − − →Vn+1 − →· · · is said to be exact at Vn if ran jn = ker jn+1. A sequence is exact if it is exact at each of its constituent vector spaces. A sequence of vector spaces and linear maps of the form 0 − →U j − − − − →V k − − − − →W − →0 (1) is a short exact sequence. (Here 0 denotes the trivial 0-dimensional vector space, and the unlabeled arrows are the obvious linear maps.) 68 9. LINEARITY (a) The sequence (1) is exact at U if and only if j is injective.
(b) The sequence (1) is exact at W if and only if k is surjective.
(c) Let U and V be vector spaces. Then the following sequence is short exact: 0 − →U ι1 − − − − →U × V π2 − − − − →V − →0.
The indicated linear maps are defined by ι1 : U →U × V : a 7→(a, 0) and π2 : U × V →V : (a, b) 7→b.
(d) Suppose a < b. Let K be the family of constant functions on the interval [a, b], C1 be the family of all continuously differentiable functions on [a, b], and C be the family of all continuous functions on [a, b]. Specify linear maps j and k so that the following sequence is short exact: 0 − →K j − − − − →C1 k − − − − →C − →0.
(e) Let C be the family of all continuous functions on the interval [0, 2].
Let E1 be the mapping from C into R defined by E1(f) = f(1). (The functional E1 is called “evaluation at 1”.) Find a subspace F of C such that the following sequence is short exact.
0 − →F ι − − − − →C E1 − − − − →R − →0.
(f) Suppose that the following sequence of finite dimensional vector spaces and linear maps is exact.
0 − →Vn fn − − − − →Vn−1 fn−1 − − − − →· · · f2 − − − − →V1 f1 − − − − →V0 − →0 Show that n X k=0 (−1)k dim(Vk) = 0.
9.3.1. Definition. It is frequently useful to think of functions as arrows in diagrams.
For example, the situation j : R →U, f : R →S, k: S →T, h: U →T may be represented by the following diagram.
W V j / U W h U X f / X V g The diagram is said to commute (or to be a commutative diagram) if j◦h = g◦f.
(g) Suppose that in the following diagram of vector spaces and linear maps 0 / U f j / V g k / W h / 0 0 / U ′ j′ / V ′ k′ / W ′ / 0 the rows are exact and the left square commutes. Then there exists a unique linear map h: W →W ′ which makes the right square commute.
9.3. PROBLEMS 69 In parts (h)–(k) consider the diagram 0 / U f j / V g k / W h / 0 0 / U ′ j′ / V ′ k′ / W ′ / 0 where the rows are exact and the squares commute.
(h) If g is surjective, so is h.
(i) If f is surjective and g is injective, then h is injective.
(j) If f and h are surjective, so is g.
(k) If f and h are injective, so is g.
(6) Let V and W be vector spaces.
Prove that (under the usual pointwise operations of addition and scalar multiplication) L(V, W) is a vector space.
70 9. LINEARITY 9.4. Answers to Odd-Numbered Exercises (1) (a) −2, −1, 5, 5 (b) 6, 3, −2 (3) line, −x 5 = y 3 = z, plane, 2x + y −z = 0 (5) (a) 7, 8, 5 (b) 4, 21, 10 (c) 3, 14, 5 (d) I (e) 2 (7) Pa, DK, H, DK, Pa, H, DO, VA, DL, Pb, VA, H, VA, DK, H, VA, DO, H (9) sin t, cos t, y′′ + y = 0 (11) (a) line, x 4 = −3y 2 = z (b) plane, 2x −y −z = 0 (13) R4 is the span of a, b, c, and d, all of which T takes to 0; so were T linear, its range would contain only 0.
(15) T does not map 0 to 0.
(17) If T were linear, then T(−2, −2) would be −2T(1, 1) = −2(3, −6) = (−6, 12).
CHAPTER 10 LINEAR MAPS BETWEEN EUCLIDEAN SPACES 10.1. Background Topics: linear mappings between finite dimensional spaces, a matrix as a linear map, the repre-sentation of a linear map as a matrix.
10.1.1. Proposition. Let T ∈L(V, W) where V is an n-dimensional vector space and W is an m-dimensional vector space and let {e1, e2, . . . , en} be a basis for V . Define an m × n-matrix [T] whose kth column (1 ≤k ≤n) is the column vector Tek. Then for each x ∈V we have Tx = [T]x.
The displayed equation above requires a little interpretation. The left side is T evaluated at x; the right side is an m × n matrix multiplied by an n × 1 matrix (that is, a column vector). Then the asserted equality can be thought of as identifying (1) two vectors in Rm, (2) two m-tuples of real numbers, or (3) two column vectors of length m (that is, m × 1 matrices).
If we wished to distinguish rigorously between column vectors and row vectors and also wished to identify m-tuples with row vectors, then the equation in the preceding proposition would have to read Tx = T t .
To avoid the extra notation in these notes we will not make this distinction. In an equation interpret a vector as a row vector or as a column vector in any way that makes sense.
10.1.2. Definition. If V and W are finite dimensional vector spaces with bases and T ∈L(V, W), then the matrix [T] in the preceding proposition is the matrix representation of T. It is also called the standard matrix for T 71 72 10. LINEAR MAPS BETWEEN EUCLIDEAN SPACES 10.2. Exercises (1) Let T : R4 →R3 be defined by Tx = (x1 −3x3 + x4, 2x1 + x2 + x3 + x4, 3x2 −4x3 + 7x4) for every x = (x1, x2, x3, x4) ∈R4. (The map T is linear, but you need not prove this.) (a) Find [T]. Answer: .
(b) Find T(1, −2, 1, 3). Answer: .
(c) Find T t. Answer: .
(d) Find ker T. Answer: ker T = span{ } .
(e) Find ran T. Answer: ran T = .
(2) Let T : R3 →R4 be defined by Tx = (x1 −3x3, x1 + x2 −6x3, x2 −3x3, x1 −3x3) for every x = (x1, x2, x3) ∈R3. (The map T is linear, but you need not prove this.) Then (a) [T] = .
(b) T(3, −2, 4) = .
(c) ker T = span{ } .
(d) ran T = span{ } .
(3) Let Pn be the vector space of all polynomial functions on R with degree strictly less than n.
The usual basis for Pn is the set of polynomials 1, t, t2, t3, . . . , tn−1. Define T : P3 →P5 by Tf(x) = Z x 0 Z u 0 p(t) dt for all x, u ∈R.
(a) Then the matrix representation of the linear map T (with respect to the usual bases for P3 and P5 is .
(b) The kernel of T is .
(c) The range of T is span{ } .
10.2. EXERCISES 73 (4) Let P4 be the vector space of polynomials of degree strictly less than 4. Consider the linear transformation D2 : P4 →P4 : f 7→f′′.
(a) Then the matrix representation D2 of D2 (with respect to the usual basis {1, t, t2, t3} for P4) is given by D2 = .
(b) ker D2 = span{ } .
(c) ran D2 = span{ } .
(5) Let P4 be the vector space of polynomials of degree strictly less than 4 and T : P4 →P5 be the linear transformation defined by (Tp)(t) = (2 + 3t)p(t) for every p ∈P4 and t ∈R.
Then the matrix representation of T (with respect to the usual basis {1, t, t2, t3} for P4) is given by [T] = .
(6) Let T : R3 →R3 be the linear transformation whose standard matrix is 1 1 0 0 1 1 1 0 −1 . We know that T is not onto because the only vectors (u, v, w) that are in the range of T are those that satisfy the relation u + av + bw = 0 where a = and b = .
(7) Let T be the linear map from R3 to R3 defined by T(x, y, z) = (3x + 2y + z , x + 3z , −y + 4z).
(a) The matrix representation of T is given by [T] = .
(b) The range of T is (geometrically speaking) a whose equation is .
74 10. LINEAR MAPS BETWEEN EUCLIDEAN SPACES 10.3. Problems (1) Define T : R3 →R2 by Tx = (x1 + 2x2 −x3, x2 + x3) for all x = (x1, x2, x3) in R3.
(a) Explain how to find [T], the matrix representation for T.
(b) Explain how to use [T] to find T(x) when x = (−1, 2, −1).
(c) Explain how to use [T] to find a vector x in R3 such that Tx = (0, 1).
Carry out the computations you describe.
(2) Let P be the vector space of all polynomial functions on R with real coefficients. Define linear transformations T, D: P →P by (Dp)(x) = p′(x) and (Tp)(x) = x2p(x) for all x ∈R.
Explain carefully how to find matrix representations for the linear transformations D +T, DT, and TD (with respect to the usual basis {1, t, t2} for the space of polynomials of degree two or less). Carry out the computations you describe. Use the resulting matrices to find ((D + T)(p))(x), (DTp)(x), and (TDp)(x) where p(x) = 3x2 + 4x −3 for all x ∈R.
(3) Define T : R2 →R3 by Tx = (x1 −x2, x2 −x1, x1 + x2) for all x = (x1, x2) in R2.
(a) Explain carefully how to find [T], the matrix representation for T.
(b) How do we use [T] to find T(1, −2)?
(c) Are there any nonzero vectors x in R2 such that Tx = 0? Explain.
(d) Under what conditions is a vector (u, v, w) in the range of T? Explain.
(4) Let C1([0, 1]) be the vector space of all functions defined on the interval [0, 1] which have continuous derivatives at each point and C([0, 1]) be the vector space of continuous func-tions on [0, 1]. Define a function T : C1([0, 1]) →C([0, 1]) by (Tf)(x) = Z x 0 f(t) dt + f ′(x) for every x ∈[0, 1].
(a) Prove that the function T is linear.
(b) Let f(x) = sin x and g(x) = cos x for all x ∈[0, 1]. Explain why one of these functions belongs to the kernel of T while the other does not.
(5) Let P4 be the vector space of polynomials of degree strictly less than 4. Consider the linear transformation D2 : P4 →P4 : f 7→f′′.
Explain carefully how to find [T], the matrix representation of D2 (with respect to the usual basis {1, t, t2, t3} for P4). Then explain how to use [T] to find ker D2 and ran D2.
10.4. ANSWERS TO ODD-NUMBERED EXERCISES 75 10.4. Answers to Odd-Numbered Exercises (1) (a) 1 0 −3 1 2 1 1 1 0 3 −4 7 (b) (1, 4, 11) (c) (1, 4, 11) (or [1 4 11]) (d) (1, −9, 2, 5) (e) R3 (3) (a) 0 0 0 0 0 0 1 2 0 0 0 1 6 0 0 0 1 12 (b) {0} (c) x2, x3, x4 (5) 2 0 0 0 3 2 0 0 0 3 2 0 0 0 3 2 0 0 0 3 (7) (a) 3 2 1 1 0 3 0 −1 4 (b) plane, u −3v + 2w = 0 CHAPTER 11 PROJECTION OPERATORS 11.1. Background Topics: projections along one subspace onto another.
11.1.1. Definition. Let V be a vector space and suppose that V = M ⊕N. We know that for each v ∈V there exist unique vectors m ∈M and n ∈N such that v = m + n (see problem 5 in chapter 6). Define a function EMN : V →V by EMNv = n. The function EMN is the projection of V along M onto N. (Frequently we write E for EMN. But keep in mind that E depends on both M and N.) 11.1.2. Theorem. Let V be a vector space and suppose that V = M ⊕N. If E is the projection of V along M onto N, then (i) E is linear; (ii) E2 = E (that is, E is idempotent); (iii) ran E = N; and (iv) ker E = M.
11.1.3. Theorem. Let V be a vector space and suppose that E : V →V is a function which satisfies (i) E is linear, and (ii) E2 = E.
Then V = ker E ⊕ran E and E is the projection of V along ker E onto ran E.
11.1.4. Theorem. Let V be a vector space and suppose that V = M ⊕N. If E is the projection of V along M onto N, then I −E is the projection of V along N onto M.
77 78 11. PROJECTION OPERATORS 11.2. Exercises (1) Let M be the line y = 2x and N be the y-axis in R2. Then [EMN] = a a −b c and [ENM] = c a b a where a = , b = , and c = .
(2) Let P be the plane in R3 whose equation is x −z = 0 and L be the line whose equations are y = 0 and x = −z. Let E be the projection of R3 along L onto P and F be the projection of R3 along P onto L. Then [E] = a b a b c b a b a and [F] = a b −a b b b −a b a where a = , b = , and c = .
(3) Let P be the plane in R3 whose equation is x + 2y −z = 0 and L be the line whose equations are x 3 = y = z 2. Let E be the projection of R3 along L onto P and F be the projection of R3 along P onto L. Then [E] = 1 3 a −b c −d d d a −2d −b + 2d c + 2d and [F] = 1 3 3d 3e −3d d e −d 2d 2e −2d where a = , b = , c = , d = , and e = .
(4) Let P be the plane in R3 whose equation is x −y −2z = 0 and L be the line whose equations are x = 0 and y = −z. Let E be the projection of R3 along L onto P and F be the projection of R3 along P onto L. Then [E] = a b b −a c c a −a −a and [F] = b b b a −a −c −a a c where a = , b = , and c = .
(5) Let E be the projection of R3 along the z-axis onto the plane 3x −y + 2z = 0 and let F be the projection of R3 along the plane 3x −y + 2z = 0 onto the z-axis.
(a) Then [E] = .
(b) Where does F take the point (4, 5, 1)? Answer: ( , , ) .
(6) Let M be the y-axis and N be the plane x + y −2z = 0 in R3.
(a) Then the projection EMN of R3 along M onto N is .
(b) The projection ENM takes the vector (3, 2, 1) to ( , , ).
11.3. PROBLEMS 79 11.3. Problems (1) Let E be a projection on a vector space. Show that a vector x belongs to the range of E if and only if Ex = x. Hint. Recall (from Theorems 11.1.2 and 11.1.3) that a projection is a linear map E such that E2 = E.
(2) Prove Theorem 11.1.2.
(3) Prove Theorem 11.1.3.
(4) Prove Theorem 11.1.4. Hint. Use Theorem 11.1.3.
(5) Let P be the plane in R3 whose equation is x −z = 0 and L be the line whose equations are y = 0 and x = 1 2z. Explain carefully how to find the matrix representation of the operator ELP , that is, the projection of R3 along L onto P. Carry out the computation you describe.
(6) Let L be the line in R3 whose equations are x = y and z = 0, and let P be the plane whose equation is x−z = 0. Explain carefully how to find the matrix representation of the operator ELP , that is, the projection of R3 along L onto P. Carry out the computation you describe.
(7) Let P be the plane in R3 whose equation is x −3y + z = 0 and L be the line whose equations are x = −2y = −z. Explain carefully how to find the matrix representation of the operator ELP of R3 along L onto P and the projection EPL of R3 along P onto L.
(8) Prove that a linear transformation between vector spaces has a left inverse if and only if it is injective.
(9) Prove that a linear transformation between vector spaces has a right inverse if and only if it is surjective.
80 11. PROJECTION OPERATORS 11.4. Answers to Odd-Numbered Exercises (1) 0, 2, 1 (3) 0, 6, 3, 1, 2 (5) (a) 1 0 0 0 1 0 −3 2 1 2 0 (b) 0, 0, 9 2 Part 4 SPECTRAL THEORY OF VECTOR SPACES CHAPTER 12 EIGENVALUES AND EIGENVECTORS 12.1. Background Topics: characteristic polynomial, eigenvalues, eigenvectors, eigenspaces, algebraic multiplicity and geometric multiplicity of an eigenvalue.
12.1.1. Definition. A number λ is an eigenvalue of an operator T on a vector space V if ker(λIV −T) contains a nonzero vector. Any such vector is an eigenvector of T associated with λ and ker(λIV −T) is the eigenspace of T associated with λ. The set of all eigenvalues of the operator T is its (point) spectrum and is denoted by σ(T).
If M is an n × n matrix, then det(λIn −M) (where In is the n × n identity matrix) is a polynomial in λ of degree n. This is the characteristic polynomial of M. A standard way of computing the eigenvalues of an operator T on a finite dimensional vector space is to find the zeros of the characteristic polynomial of its matrix representation. It is an easy consequence of the multiplicative property of the determinant function (see proposition 3.1.9) that the characteristic polynomial of an operator T on a vector space V is independent of the basis chosen for V and hence of the particular matrix representation of T that is used.
12.1.2. Theorem (Spectral Mapping Theorem). If T is an operator on a finite dimensional vector space and p is a polynomial, then σ(p(T)) = p(σ(T)).
That is, if σ(T) = {λ1, . . . , λk}, then σ(p(T)) = {p(λ1), . . . , p(λk)}.
83 84 12. EIGENVALUES AND EIGENVECTORS 12.2. Exercises (1) If A = 1 −1 4 3 2 −1 2 1 −1 , then the eigenvalue has corresponding eigenvector ( , 1 , 1 ) , the eigenvalue has corresponding eigenvector ( , 4 , 1 ) , and the eigenvalue has corresponding eigenvector ( , 2 , 1 ) .
(2) Let A = 0 0 2 0 2 0 2 0 0 .
(a) The eigenvalues of A are , , and .
(b) The matrix A has a one-dimensional eigenspace.
It is the span of the vector ( 1 , , ).
(c) The matrix A has a two-dimensional eigenspace.
It is the span of the vectors ( 1 , 0 , ) and ( 0 , 1 , ).
(3) Choose a, b and c in the matrix A = 0 1 0 0 0 1 a b c so that the characteristic polynomial of A is −λ3 + 4λ2 + 5λ + 6.
Answer: a = ; b = ; and c = .
(4) Suppose that it is known that the matrix A = 1 0 −1 √ 3 a 17 2 0 b has eigenvalues 2 and 3 and that the eigenvalue 2 has algebraic multiplicity 2. Then a = and b = .
(5) The matrices A = a 1 −2 d and B = 1 25 114 48 48 86 have the same eigenvalues.
Then a = and d = .
(6) Let A = 3 4 2 0 1 2 0 0 0 .
(a) The eigenvalues of A are , , and .
(b) The matrix A has three one-dimensional eigenspaces. They are spanned by the vectors ( , , ), ( , , ), and ( , , ), respectively.
(7) Let A = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 .
(a) The eigenvalues of A−I are (which has algebraic multiplicity ) and (which has algebraic multiplicity ).
(b) The determinant of A −I is .
(8) Let T be the operator on R3 whose matrix representation is 1 −1 0 0 0 0 −2 2 2 . Then the eigenvalues of the operator T 5 −3T 4 + T 3 −T 2 + T −3I are , , and .
12.3. PROBLEMS 85 12.3. Problems (1) Suppose that A and B are n × n matrices. Prove that σ(AB) = σ(BA). Hint. Show that if λ is an eigenvalue of AB, then it is also an eigenvalue of BA. Deal with the cases λ = 0 and λ ̸= 0 separately.
(2) Let c ∈R. Suppose that A is an n × n matrix and that the sum of the entries in each column of A is c. Prove that c is an eigenvalue of A. Hint. Consider the sum of the row vectors of the matrix A −cI.
(3) This is a problem in cryptography. Read about Hill ciphers, then decode the following Hill 3-cipher given that the first two words of the plaintext are known to be “My candle”.
(See for example , section 11.16.) OGWGCGWGKK.EWVD.XZJOHZWLNYH USTFAIOS.A.KBN JRCENYQZV,IE LTGCGWGKC YYBLSDWWODLBVFFOS.H In many discussions of Hill ciphers letters of the alphabet are assigned numbers from 0 to 25 and arithmetic is done modulo 26. The encoding here is done slightly differently.
Here each letter is assigned its numerical position in the alphabet (including Z which is assigned 26). Furthermore, a space between words is assigned 27, a comma is assigned 28, and a period is assigned zero. Thus, for this code, all arithmetic should be done modulo 29.
(One reason for this is that some computer algebra systems have problems calculating inverses mod 26.) Note: the ciphertext contains exactly three spaces.
86 12. EIGENVALUES AND EIGENVECTORS 12.4. Answers to Odd-Numbered Exercises (1) −2, −1, 1, −1, 3, 1 (3) 6, 5, 4 (5) 2, 6 (or 6, 2) (7) (a) −1, 3, 3, 1 (b) −3 CHAPTER 13 DIAGONALIZATION OF MATRICES 13.1. Background Topics: similarity of matrices, triangular and diagonal matrices, diagonalization, annihilating and minimal polynomials, algebraic and geometric multiplicity of an eigenvalue, the Cayley-Hamilton theorem.
13.1.1. Definition. Two operators R and T on a vector space V are similar if there exists an invertible operator S on V such that R = S−1TS.
13.1.2. Proposition. If V is a vector space, then similarity is an equivalence relation on L(V ).
13.1.3. Definition. Let V be a finite dimensional vector space and B = {e1, . . . , en} be a basis for V . An operator T on V is diagonal if there exist scalars α1, . . . , αn such that Tek = αkek for each k ∈Nn. Equivalently, T is diagonal if its matrix representation [T] = [tij] has the property that tij = 0 whenever i ̸= j.
Asking whether a particular operator on some finite dimensional vector space is diagonal is, strictly speaking, nonsense. As defined the operator property of being diagonal is definitely not a vector space concept. It makes sense only for a vector space for which a basis has been specified.
This important, if obvious, fact seems to go unnoticed in many beginning linear algebra texts, due, I suppose, to a rather obsessive fixation on Rn in such courses. Here is the relevant vector space property.
13.1.4. Definition. An operator T on a finite dimensional vector space V is diagonalizable if there exists a basis for V with respect to which T is diagonal. Equivalently, an operator on a finite dimensional vector space with basis is diagonalizable if it is similar to a diagonal operator. If a matrix D is diagonalizable and Λ = S−1DS is diagonal, we say that the matrix S diagonalizes D.
13.1.5. Theorem. Let A be an n × n matrix with n linear independent eigenvectors. If S is a matrix with these eigenvectors as columns, then S diagonalizes A. The entries along the diagonal of the resulting diagonal matrix are all eigenvalues of A.
13.1.6. Definition. A polynomial is monic if its leading coefficient is 1. Thus a polynomial of degree n is monic if it takes the form xn + an−1xn−1 + · · · + a1x + a0.
13.1.7. Definition. Let p be a polynomial of degree at least one and T be an operator on some vector space. We say that p is an annihilating polynomial for T (or that p annihilates T) if p(T) = 0. For example, if T 3 −4T 2 + T −7I = 0, then the polynomial p defined by p(x) = x3 −4x2 + x −7 is an annihilating polynomial for T.
13.1.8. Definition. Let T be an operator on a finite dimensional vector space. The minimal polynomial of T is the unique monic polynomial of smallest degree which annihilates T. (It is left as a problem to verify the existence and the uniqueness of such a polynomial: see problem 8.) 13.1.9. Theorem (Cayley-Hamilton Theorem). On a finite dimensional vector space the charac-teristic polynomial of an operator T annihilates T.
Paraphrase: Every matrix satisfies its characteristic equation.
87 88 13. DIAGONALIZATION OF MATRICES 13.1.10. Definition. A square matrix A = aij is upper triangular if aij = 0 whenever i > j.
A matrix is triangulable (or triangulizable) if it is similar to an upper triangular matrix.
13.1.11. Theorem. Let T be an operator on a finite dimensional vector space and let {λ1, . . . , λk} be its distinct eigenvalues. Then: (1) T is triangulable if and only if its minimal polynomial can be factored into a product of linear factors. That is, if and only if there are positive integers r1, . . . , rk such that mT (x) = (x −λ1)r1 . . . (x −λk)rk.
(2) T is diagonalizable if and only if its minimal polynomial has the form mT (x) = (x −λ1) . . . (x −λk).
13.1.12. Corollary. Every operator on a complex finite dimensional vector space is triangulable.
13.1.13. Definition. An operator is nilpotent if some power of the operator is 0.
13.2. EXERCISES 89 13.2. Exercises (1) Let A = 1 1 1 1 1 1 1 1 1 . The characteristic polynomial of A is λp(λ −3)q where p = and q = .
The minimal polynomial of A is λr(λ −3)s where r = and s = .
(2) Let T be the operator on R4 whose matrix representation is 0 1 0 −1 −2 3 0 −1 −2 1 2 −1 2 −1 0 3 .
The characteristic polynomial of T is (λ −2)p where p = .
The minimal polynomial of T is (λ −2)r where r = .
(3) Let T be the operator on R3 whose matrix representation is 3 1 −1 2 2 −1 2 2 0 .
(a) Find the characteristic polynomial of T.
Answer: cT (λ) = (λ −1)p(λ −2)q where p = and q = .
(b) Find the minimal polynomial of T.
Answer: mT (λ) = (λ −1)r(λ −2)s where r = and s = .
(c) Find the eigenspaces M1 and M2 of T.
Answer: M1 = span { } and M2 = span { }.
(4) Let T be the operator on R5 whose matrix representation is 1 0 0 1 −1 0 1 −2 3 −3 0 0 −1 2 −2 1 −1 1 0 1 1 −1 1 −1 2 .
(a) Find the characteristic polynomial of T.
Answer: cT (λ) = (λ + 1)p(λ −1)q where p = and q = .
(b) Find the minimal polynomial of T.
Answer: mT (λ) = (λ + 1)r(λ −1)s where r = and s = .
(c) Find the eigenspaces V1 and V2 of T.
Answer: V1 = span { } and V2 = span { }.
(5) Let T be an operator whose matrix representation is 0 0 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 1 2 0 0 0 −1 −1 .
(a) Regarding T as an operator on R5 find its characteristic polynomial and minimal polynomial.
Answer: cT (λ) = λp(λ2 + 1)q where p = and q = .
and mT (λ) = λr(λ2 + 1)s where r = and s = .
(b) Regarded as an operator on R5 is T diagonalizable?
. Is it triangulable?
.
90 13. DIAGONALIZATION OF MATRICES (c) Regarded as an operator on C5 is T diagonalizable?
. Is it triangulable?
.
(6) Let T be the operator on R3 whose matrix representation is 2 0 0 −1 3 2 1 −1 0 .
(a) Find the characteristic and minimal polynomials of T.
Answer: cT (λ) = (λ −1)p(λ −2)q where p = and q = .
and mT (λ) = (λ −1)r(λ −2)s where r = and s = .
(b) What can be concluded from the form of the minimal polynomial?
Answer: (c) Find a matrix S (if one exists) that diagonalizes [T]. What is the diagonal form Λ of [T] produced by this matrix? Answer: S = a b a b b −c −b a b where a = , b = , and c = ; and Λ = λ 0 0 0 µ 0 0 0 µ where λ = and µ = .
(7) Let T be the operator on R3 whose matrix representation is 8 −6 12 −18 11 18 −6 −3 26 .
(a) Find the characteristic and minimal polynomials of T.
Answer: cT (λ) = (λ −5)p(λ −20)q where p = and q = .
and mT (λ) = (λ −5)r(λ −20)s where r = and s = .
(b) What can be concluded from the form of the minimal polynomial?
Answer: (c) Find a matrix S (if one exists) that diagonalizes [T]. What is the diagonal form Λ of [T] produced by this matrix? Answer: S = a b c d −a a b c b where a = , b = , c = , and d = ; and Λ = λ 0 0 0 µ 0 0 0 µ where λ = and µ = .
(8) Let Pn be the space of polynomials of degree strictly less than n and D be the differenti-ation operator on Pn. Then (a) the only eigenvalue of D is λ = ; (b) the corresponding eigenspace is the span of ; (c) the algebraic multiplicity of λ is ; and (d) the geometric multiplicity of λ is .
13.3. PROBLEMS 91 13.3. Problems (1) Prove that the trace function is a similarity invariant on the family of n×n matrices; that is, prove that if A and B are similar n × n matrices, then tr A = tr B. Hint. Prove first that if M and N are n × n matrices, then MN and NM have the same trace.
(2) Prove that the determinant function is a similarity invariant on the family of n×n matrices; that is, prove that if A and B are similar n × n matrices, then det A = det B.
(3) Prove that if two matrices are diagonalized by the same matrix, then they commute.
(4) Prove that if a matrix A is diagonalizable, then so is every matrix similar to A.
(5) Show that if A is an n × n matrix of real (or complex) numbers, then tr A is the sum of the eigenvalues of A and det A is their product.
(6) Suppose that T is an operator on a finite dimensional complex vector space and that σ(T) = {0}. Show that T is nilpotent.
(7) Let T be an operator on a finite dimensional vector space.
(a) Show that if p is an annihilating polynomial for T, then the minimal polynomial mT divides p. Hint. Suppose that p annihilates T (so that the degree of p is at least as large as the degree of mT ). Divide p by mT . Then there exist polynomials q (the quotient) and r (the remainder) such that p = mT q + r and degree of r < degree of mT .
Conclude that r = 0.
(b) Show that the minimal polynomial mT and the characteristic polynomial cT have exactly the same roots (although not necessarily the same multiplicities). Hint. To show that every root of mT is also a root of cT , it is enough to know that mT divides cT . Why is that true?
To obtain the converse, suppose that λ is a root of cT : that is, suppose that λ is an eigenvalue of T. Use the spectral mapping theorem 12.1.2 to show that mT (λ) = 0.
(8) Let T be an operator on a finite dimensional vector space V . Show that there is a unique monic polynomial of smallest degree which annihilates T. Hint. This asks for a proof of the existence and the uniqueness of the minimal polynomial for the operator T. The existence part is easy: If there are any polynomials at all which annihilate T, surely there is one of smallest degree. (How do we know that there is at least one polynomial that annihilates T?) We want the annihilating polynomial of smallest degree to be monic—is this a problem?
To verify the uniqueness of the minimal polynomial, consider the case of degree one separately. That is, suppose that p and q are monic annihilating polynomials of degree one. How do we know that p = q? Then consider polynomials of higher degree. Suppose that p and q are monic annihilating polynomials of smallest degree k where k > 1. What can you say about p −q?
92 13. DIAGONALIZATION OF MATRICES 13.4. Answers to Odd-Numbered Exercises (1) 2, 1, 1, 1 (3) (a) 1, 2 (b) 1, 2 (c) (1, 0, 2), (1, 1, 2) (5) (a) 3, 1, 2, 1 (b) no, no (c) no, yes (7) (a) 1, 2, 1, 1 (b) T is diagonalizable (c) 2, 1, 0, 3, 5, 20 CHAPTER 14 SPECTRAL THEOREM FOR VECTOR SPACES 14.1. Background Topics: the spectral theorem for finite dimensional vector spaces (writing a diagonalizable operator as a linear combination of projections).
The central fact asserted by the finite dimensional vector space version of the spectral theorem is that every diagonalizable operator on such a space can be written as a linear combination of projection operators where the coefficients of the linear combination are the eigenvalues of the operator and the ranges of the projections are the corresponding eigenspaces. Here is a formal statement of the theorem.
14.1.1. Theorem (Spectral theorem: vector space version). Let T be a diagonalizable operator on a finite dimensional vector space V , and λ1, . . . , λk be the (distinct) eigenvalues of T.
For each j let Mj be the eigenspace associated with λj and Ej be the projection of V onto Mj along M1 + · · · + Mj−1 + Mj+1 + · · · + Mk. Then (i) T = λ1E1 + · · · + λkEk, (ii) I = E1 + · · · + Ek, and (iii) EiEj = 0 when i ̸= j.
For a proof of this result see, for example, , page 215, theorem 11.
93 94 14. SPECTRAL THEOREM FOR VECTOR SPACES 14.2. Exercises (1) Let T be the operator on R2 whose matrix representation is −7 8 −16 17 .
(a) Find the characteristic polynomial and minimal polynomial for T.
Answer: cT (λ) = and mT (λ) = .
(b) The eigenspace M1 associated with the smaller eigenvalue λ1 is the span of ( 1 , ).
(c) The eigenspace M2 associated with the larger eigenvalue λ2 is the span of ( 1 , ).
(d) We wish to write T as a linear combination of projection operators. Find the (matrix representations of the) appropriate projections E1 and E2 onto the eigenspaces M1 and M2, respectively.
Answer: E1 = a b a b , where a = and b = , and E2 = −c c −d d , where c = and d = .
(e) Compute the sum and product of E1 and E2.
Answer: E1 + E2 = ; and E1E2 = .
(f) Write T as a linear combination of the projections found in (d).
Answer: [T] = E1 + E2.
(g) Find a matrix S which diagonalizes [T]. What is the associated diagonal form Λ of [T]?
Answer: S = 1 1 a b , where a = and b = , and Λ = .
(2) Let T be the operator on R3 whose matrix representation is −2 −1 −6 −6 −1 −12 2 1 6 .
(a) Find the characteristic polynomial and minimal polynomial for T.
Answer: cT (λ) = and mT (λ) = .
(b) The eigenspace M1 associated with the smallest eigenvalue λ1 is the span of (3 , , ).
(c) The eigenspace M2 associated with the middle eigenvalue λ2 is the span of (1 , , ).
(d) The eigenspace M3 associated with the largest eigenvalue λ3 is the span of (1 , , ).
(e) We wish to write T as a linear combination of projection operators. Find the (matrix representations of the) appropriate projections E1, E2, and E3 onto the eigenspaces M1, M2, and M3, respectively.
Answer: E1 = a c a 2a c 2a −b c −b ; E2 = −b d c −2a a c b −d c ; and E3 = c −d −a c −b −2a c d a , where a = , b = , c = , and d = .
(f) Write T as a linear combination of the projections found in (e).
Answer: [T] = E1 + E2 + E3.
(g) Find a matrix S which diagonalizes [T]. What is the associated diagonal form Λ of [T]?
14.2. EXERCISES 95 Answer: S = a b b 2a a c −c −b −b , where a = , b = , and c = , and Λ = .
(3) Find a matrix A whose eigenvalues are 1 and 4, and whose eigenvectors are (3, 1) and (2, 1), respectively.
Answer: A = .
96 14. SPECTRAL THEOREM FOR VECTOR SPACES 14.3. Answers to Odd-Numbered Exercises (1) (a) λ2 −10λ + 9, λ2 −10λ + 9 (b) 1 (c) 2 (d) 2, −1, 1, 2 (e) 1 0 0 1 , 0 0 0 0 (f) 1, 9 (g) 1, 2, 1 0 0 9 (3) −5 18 −3 10 CHAPTER 15 SOME APPLICATIONS OF THE SPECTRAL THEOREM 15.1. Background Topics: systems of linear differential equations, initial conditions, steady state solutions, the func-tional calculus for operators on finite dimensional vector spaces, Markov processes.
15.1.1. Theorem. Let du dt = Au (15.1.1) be a vector differential equation (that is, a system of ordinary linear differential equations) where A is an n × n matrix and suppose that u0 = u(0) is an initial value of the system.
If A is a diagonalizable matrix (so that A = SΛS−1 for some diagonal matrix Λ and some invertible matrix S), then the equation (15.1.1) has the solution u(t) = eAtu0 = SeΛtS−1u0.
15.1.2. Definition. A Markov matrix is a square matrix with nonnegative entries and with each column adding to 1.
15.1.3. Proposition (Facts about Markov matrices.). Let A be a Markov matrix. Then (i) λ1 = 1 is an eigenvalue.
(ii) The eigenvector e1 corresponding to λ1 is nonnegative and it is a steady state.
(iii) The other eigenvalues satisfy |λk| ≤1.
(iv) If any power of A has all entries strictly positive, then |λk| < 1 for all k ̸= 1 and Aku0 → u∞where the steady state u∞is a multiple of e1.
97 98 15. SOME APPLICATIONS OF THE SPECTRAL THEOREM 15.2. Exercises (1) Let A = −1 1 1 −1 .
(a) The eigenvalues of A are λ1 = and λ2 = .
(b) The corresponding eigenvectors are e1 = (1, a) and e2 = (a, −a) where a = .
(c) Then eAt = a 1 + e−bt 1 −e−bt 1 −e−bt 1 + e−bt where a = and b = .
(d) Let u(t) = (x(t), y(t)). The general solution to the system of equations du dt = Au with initial conditions x0 = 3 and y0 = 1 is x(t) = a + be−ct and y(t) = a −be−ct where a = , b = , and c = .
(e) Find the steady state solution to the system du dt = Au under the initial condi-tions given in (d). That is, find u∞= x∞ y∞ where x∞= limt→∞x(t) and y∞= limt→∞y(t).
Answer: u∞= .
(2) Suppose that at time t the population y(t) of a predator and the population x(t) of its prey are governed by the equations dx dt = 4x −2y dy dt = x + y.
If at time t = 0 the populations are x = 300 and y = 200, then the populations at all future times t are x(t) = aebt + 200ect and y(t) = debt + aect where where a = , b = , c = , and d = . The long run ratio of populations of prey to predator approaches to .
(3) Use matrix methods to solve the initial value problem consisting of the system of differential equations du dt = 4u −v −w dv dt = u + 2v −w dw dt = u −v + 2w and the initial conditions u(0) = 2 v(0) = −2 w(0) = 7.
Answer: u(t) = aebt −eat; v(t) = aebt −ceat; and w(t) = aebt + deat where a = , b = , c = , and d = .
(4) Consider the initial value problem: y′′ −y′ −2y = 0 with the initial conditions y0 = 3 and y′ 0 = 3.
(a) Express the differential equation in the form du dt = Au where u = (y, z) and z = y′.
Then A is the matrix a b c b where a = , b = , and c = .
15.2. EXERCISES 99 (b) The smaller eigenvalue of A is and the larger is . The corresponding eigenvectors are (1, a) and (1, b) where a = and b = .
(c) The diagonal form of A is Λ = a 0 0 b where a = and b = .
(d) Find the diagonalizing matrix S for A. That is, find S so that Λ = S−1AS. Answer: S = 1 1 a b where a = and b = .
(e) The matrix eAt is 1 a bect + edt −ect + edt −bect + bedt ect + bedt where a = , b = , c = , and d = .
(f) The solution to the initial value problem is y(t) = .
(5) Use the spectral theorem to solve the initial value problem y′′′ −3y′′ + 2y′ = 0 where y(0) = 2, y′(0) = 0, and y′′(0) = 3.
Answer: y(t) = a+bet +cedt where a = , b = , c = , and d = .
(6) Let G0 = 0 and G1 = 1 2. For each k ≥0 let Gk+2 be the average of Gk and Gk+1.
(a) Find the transition matrix A which takes the vector (Gk+1, Gk) to the vector (Gk+2, Gk+1).
Answer: A = .
(b) Find a diagonal matrix Λ which is similar to A.
Answer: Λ = .
(c) Find a matrix S such that A = SΛS−1.
Answer: S = .
(d) Determine the long run behavior of the numbers Gk.
Answer: G∞:= limk→∞Gk = .
(7) Let T be the operator on R2 whose matrix representation is −7 8 −16 17 . Use the spectral theorem to find √ T. (A square root of T is an operator whose square is T.) Answer: √ T = −1 a b c where a = , b = , and c = .
(8) Let A = 4 3 1 2 . Find A100. (Write an exact answer—not a decimal approximation.) Answer: A100 = 1 4 .
100 15. SOME APPLICATIONS OF THE SPECTRAL THEOREM (9) Let T be the operator on R3 whose matrix representation is 2 −2 1 −1 1 1 −1 2 0 .
(a) Write T as a linear combination of projections.
Answer: T = c1E1 + c2E2 + c3E3 where c1 = , c2 = , c3 = , E1 = a b −b a b −b a −b b , E2 = b a b b a b b a b , and E3 = b −b a −b b a −b b a where a = and b = .
(b) Calculate the following: E1E2 = ; E1E3 = ; E2E3 = .
(c) E1 + E2 + E3 = .
(d) Write T 3 as a linear combination of E1, E2, and E3.
Answer: T 3 = E1+ E2+ E3.
(10) Let A be the matrix whose eigenvalues are λ1 = −1, λ2 = 1/2, and λ3 = 1/3, and whose corresponding eigenvectors are e1 = 1 0 1 , e2 = 1 −1 0 , and e3 = 0 −1 0 .
(a) Solve the difference equation xk+1 = Axk (where xk = uk vk wk ) subject to the initial condition x0 = 10 20 30 .
Answer: uk = a(−b)k −cdk, vk = cdk, and wk = a(−b)k where a = , b = , c = , and d = .
(b) Each xk can be written as a linear combination of the vectors ( , , ) and ( , , ).
(c) The value of x1000 is approximately ( , , ) .
(11) Let A be as in the preceding exercise. Solve the differential equation dx dt = Ax subject to the initial conditions x0 = 10 20 30 . Answer: x(t) = ae−t −bect, bect, ae−t where a = , b = , and c = .
(12) Suppose three cities A, B, and C are centers for trucks. Every month half of the trucks in A and half of those in B go to C. The other half of the trucks in A and the other half of the trucks in B stay where they are. Every month half of the trucks in C go to A and the other half go to B.
15.2. EXERCISES 101 (a) What is the (Markov) transition matrix which acts on the vector a0 b0 c0 (where a0 is the number of trucks initially in A, etc.)?
Answer: .
(b) If there are always 450 trucks, what is the long run distribution of trucks? Answer: a∞= , b∞= , c∞= .
102 15. SOME APPLICATIONS OF THE SPECTRAL THEOREM 15.3. Problems (1) Initially a 2100 gallon tank M is full of water and an 1800 gallon tank N is full of water in which 100 pounds of salt has been dissolved. Fresh water is pumped into tank M at a rate of 420 gallons per minute and salt water is released from tank N at the same rate.
Additionally, the contents of M are pumped into N at a rate of 490 gallons per minute and the contents of N are pumped into M at a rate sufficient to keep both tanks full.
How long does it take (to the nearest second) for the concentration of salt in tank M to reach a maximum? And how much salt is there (to three significant figures) in M at that time?
(2) Show that if A is a diagonalizable n × n matrix, then det(exp A) = exp(tr A) .
Hint. What would be a reasonable definition of exp A if A were a diagonal matrix?
(3) Explain carefully how to use matrix methods to solve the initial value problem y′′ −y′ −6y = 0 under the initial conditions y0 = −2 and y′ 0 = 14.
Carry out the computations you describe.
(4) Explain carefully how to use matrix methods to solve the initial value problem consisting of the system of differential equations dv dt = −v + w dw dt = v −w and the initial conditions ( v(0) = 5 w(0) = −3.
Carry out the computation you describe.
(5) Show how to use the spectral theorem to solve the initial value problem consisting of the system of differential equations du dt = −7u −5v + 5w dv dt = 2u + 3v −2w dw dt = −8u −5v + 6w and the initial conditions u(0) = 2 v(0) = 1 w(0) = −1.
(6) Explain carefully how to use the spectral theorem to find a square root of the matrix A = 2 1 1 2 . Illustrate your discussion by carrying out the computation.
(7) Let A = 0 1 0 0 0 0 0 0 0 .
(a) Does A have a cube root? Explain.
(b) Does A have a square root? Explain.
(8) Let A be a symmetric 2×2 matrix whose trace is 20 and whose determinant is 64. Suppose that the eigenspace associated with the smaller eigenvalue of A is the line x −y = 0. Find a matrix B such that B2 = A.
15.4. ANSWERS TO ODD-NUMBERED EXERCISES 103 15.4. Answers to Odd-Numbered Exercises (1) (a) 0, −2 (b) 1 (c) 1 2, 2 (d) 2, 1, 2 (e) 2 2 (3) 3, 2, 5, 4 (5) 7 2, −3, 3 2, 2 (7) 2, −4, 5 (9) (a) −1, 1, 3, 0, 1 2 (b) 0 0 0 0 0 0 0 0 0 , 0 0 0 0 0 0 0 0 0 , 0 0 0 0 0 0 0 0 0 (c) 1 0 0 0 1 0 0 0 1 (d) −1, 1, 27 (11) 30, 20, 1 2 CHAPTER 16 EVERY OPERATOR IS DIAGONALIZABLE PLUS NILPOTENT 16.1. Background Topics: generalized eigenspaces, nilpotent operators 16.1.1. Definition. An operator T on a vector space is nilpotent if T n = 0 for some n ∈N.
Similarly, a square matrix is nilpotent if some power of it is the zero matrix.
16.1.2. Theorem. Let T be an operator on a finite dimensional vector space V . Suppose that the minimal polynomial for T factors completely into linear factors mT (x) = (x −λ1)r1 . . . (x −λk)rk where λ1, . . . λk are the (distinct) eigenvalues of T. For each j let Wj = ker(T −λjI)rj and Ej be the projection of V onto Wj along W1 + · · · + Wj−1 + Wj+1 + · · · + Wk. Then V = W1 ⊕W2 ⊕· · · ⊕Wk, each Wj is invariant under T, and I = E1 + · · · + Ek. Furthermore, the operator D = λ1E1 + · · · + λkEk is diagonalizable, the operator N = T −D is nilpotent, and N commutes with D.
16.1.3. Corollary. Every operator on a finite dimensional complex vector space is the sum of a diagonal operator and a nilpotent one.
16.1.4. Definition. Since, in the preceding theorem, T = D +N where D is diagonalizable and N is nilpotent, we say that D is the diagonalizable part of T and N is the nilpotent part of T.
The subspace Wj = ker(T −λjI)rj is generalized eigenspace associated with the eigenvalue λj.
105 106 16. EVERY OPERATOR IS DIAGONALIZABLE PLUS NILPOTENT 16.2. Exercises (1) Let T be the operator on R2 whose matrix representation is 2 1 −1 4 .
(a) Explain briefly why T is not diagonalizable.
Answer: .
(b) Find the diagonalizable and nilpotent parts of T. Answer: D = a b b a and N = −c c −c c where a = , b = , and c = .
(2) Let T be the operator on R2 whose matrix representation is 4 −2 2 0 .
(a) Explain briefly why T is not diagonalizable.
Answer: .
(b) Find the diagonalizable and nilpotent parts of T. Answer: D = and N = (3) Let T be the operator on R3 whose matrix representation is 1 1 0 0 1 0 0 0 0 .
(a) The characteristic polynomial of T is (λ)p(λ −1)q where p = and q = .
(b) The minimal polynomial of T is (λ)r(λ −1)s where r = and s = .
(c) Explain briefly how we know from (b) that T is not diagonalizable.
Answer: .
(d) The eigenspace M1 (corresponding to the smaller eigenvalue of T) is span{( , , 1)} and the eigenspace M2 (corresponding to the larger eigenvalue) is span{(1 , , )}.
(e) Explain briefly how we know from (d) that T is not diagonalizable.
Answer: .
(f) The generalized eigenspace W1 (corresponding to the smaller eigenvalue) is W1 = span{( , , 1)} and the generalized eigenspace W2 (corresponding to the larger eigenvalue) is span{(1, a, a) , (a, 1, a)}, where a = .
(g) The (matrix representing the) projection E1 of R3 along W2 onto W1 is a a a a a a a a b where a = and b = .
(h) The (matrix representing the) projection E2 of R3 along W1 onto W2 is a b b b a b b b b where a = and b = .
(i) The diagonalizable part of T is D = a b b b a b b b b and the nilpotent part of T is N = b a b b b b b b b where a = and b = .
16.2. EXERCISES 107 (j) A matrix S which diagonalizes D is a b a a a b b a a where a = and b = .
(k) The diagonal form Λ of the diagonalizable part D of T is a a a a b a a a b where a = and b = .
(l) Show that D commutes with N by computing DN and ND.
Answer: DN = ND = a b a a a a a a a where a = and b = .
(4) Let T be the operator on R3 whose matrix representation is 1 1 0 0 1 1 0 0 0 .
(a) The characteristic polynomial of T is (λ)p(λ −1)q where p = and q = .
(b) The minimal polynomial of T is (λ)r(λ −1)s where r = and s = .
(c) Explain briefly how we know from (b) that T is not diagonalizable.
Answer: .
(d) The eigenspace M1 (corresponding to the smaller eigenvalue of T) is span{(1 , , )} and the eigenspace M2 (corresponding to the larger eigenvalue) is span{(1 , , )} (e) Explain briefly how we know from (d) that T is not diagonalizable.
Answer: .
(f) The generalized eigenspace W1 (corresponding to the smaller eigenvalue) is W1 = span{(1 , , )} and the generalized eigenspace W2 (corresponding to the larger eigenvalue) is span{(1, a, a) , (a, 1, a)}, where a = .
(g) The (matrix representing the) projection E1 of R3 along W2 onto W1 is a a b a a −b a a b where a = and b = .
(h) The (matrix representing the) projection E2 of R3 along W1 onto W2 is a b −a b a a b b b where a = and b = .
(i) The diagonalizable part of T is D = a b −a b a a b b b and the nilpotent part of T is N = b a a b b b b b b where a = and b = .
(j) A matrix S which diagonalizes D is a a b −a b a a b b where a = and b = .
(k) The diagonal form Λ of the diagonalizable part D of T is a a a a b a a a b where a = and b = .
(l) When comparing this exercise with the preceding one it may seem that the correct answer to part (i) should be that the diagonalizable part of T is D = 1 0 0 0 1 0 0 0 0 and 108 16. EVERY OPERATOR IS DIAGONALIZABLE PLUS NILPOTENT the nilpotent part of [T] is N = 0 1 0 0 0 1 0 0 0 , because D is diagonal, N is nilpotent, and [T] = D + N. Explain briefly why this is not correct.
Answer: .
(5) Let T be the operator on R3 whose matrix representation is 3 1 −1 2 2 −1 2 2 0 .
(a) The characteristic polynomial of T is (λ −1)p(λ −2)q where p = and q = .
(b) The minimal polynomial of T is (λ −1)r(λ −2)s where r = and s = .
(c) The eigenspace M1 (corresponding to the smaller eigenvalue of T) is span{(1 , , )} and the eigenspace M2 (corresponding to the larger eigenvalue) is span{(1 , , )}.
(d) The generalized eigenspace W1 (corresponding to the smaller eigenvalue) is span{(1 , , )} and the generalized eigenspace W2 (corresponding to the larger eigenvalue) is span{(1, a, b) , (0, b, a)}, where a = and b = .
(e) The diagonalizable part of T is D = a a b b c b −c c c and the nilpotent part of T is N = c b −a c b −a 2c b −c where a = , b = , and c = .
(f) A matrix S which diagonalizes D is a a b b a b c b a where a = , b = , and c = .
(g) The diagonal form Λ of the diagonalizable part D of T is λ 0 0 0 µ 0 0 0 µ where λ = and µ = .
(h) Show that D commutes with N by computing DN and ND.
Answer: DN = ND = a b −c a b −c 2a b −a where a = , b = , and c = .
(6) Let T be the operator on R4 whose matrix representation is 0 1 0 −1 −2 3 0 −1 −2 1 2 −1 2 −1 0 3 .
(a) The characteristic polynomial of T is (λ −2)p where p = .
(b) The minimal polynomial of T is (λ −2)r where r = .
(c) The diagonalizable part of T is D = a b b b b a b b b b a b b b b a where a = and b = .
(c) The nilpotent part of T is N = −a b c −b −a b c −b −a b c −b a −b c b where a = , b = , and c = .
16.2. EXERCISES 109 (7) Let T be the operator on R5 whose matrix representation is 1 0 0 1 −1 0 1 −2 3 −3 0 0 −1 2 −2 1 −1 1 0 1 1 −1 1 −1 2 .
(a) Find the characteristic polynomial of T.
Answer: cT (λ) = (λ + 1)p(λ −1)q where p = and q = .
(b) Find the minimal polynomial of T.
Answer: mT (λ) = (λ + 1)r(λ −1)s where r = and s = .
(c) Find the eigenspaces M1 and M2 of T.
Answer: M1 = span{(a, 1, b, a, a)} where a = and b = ; and M2 = span{(1, a, b, b, b), (b, b, b, 1, a)} where a = and b = .
(d) Find the diagonalizable part of T.
Answer: D = a b b b b b a −c c −c b b −a c −c b b b a b b b b b a where a = , b = , and c = .
(e) Find the nilpotent part of T.
Answer: N = a a a b −b a a a b −b a a a a a b −b b −b b b −b b −b b where a = and b = .
(f) Find a matrix S which diagonalizes the diagonalizable part D of T. What is the diagonal form Λ of D associated with this matrix?
Answer: S = a b a a a b a b a a b a a b a a a a b b a a a a b where a = and b = .
and Λ = −a 0 0 0 0 0 a 0 0 0 0 0 a 0 0 0 0 0 a 0 0 0 0 0 a where a = .
110 16. EVERY OPERATOR IS DIAGONALIZABLE PLUS NILPOTENT 16.3. Problems (1) Explain in detail how to find the diagonalizable and nilpotent parts of the matrix A = −3 −4 5 6 8 −6 −2 −1 4 . Carry out the computations you describe.
(2) Consider the matrix A = 2 0 −2 0 0 2 0 2 0 . In each part below explain carefully what you are doing.
(a) Find the characteristic polynomial for A.
(b) Find the minimal polynomial for A. What can you conclude from the form of the minimal polynomial?
(c) Find the eigenspace associated with each eigenvalue of A. Do the eigenvectors of A span R3? What can you conclude from this?
(d) Find the generalized eigenspaces W1 and W2 associated with the eigenvalues of A.
(e) Find the projection operators E1 of R3 onto W1 along W2 and E2 of R3 onto W2 along W1.
(f) Find the diagonalizable part D of A. Express D both as a single matrix and as a linear combination of projections.
(g) Find a matrix S which diagonalizes D. What is the resulting diagonal form Λ of D?
(h) Find the nilpotent part N of A. What is the smallest power of N which vanishes?
(3) Let T be the operator on R3 whose matrix representation is 1 −1 0 1 3 −1 0 0 1 .
(a) Explain how to find the characteristic polynomial for T. Then carry out the compu-tation.
(b) What is the minimal polynomial for a matrix? Find the minimal polynomial for T and explain how you know your answer is correct. What can you conclude from the form of this polynomial?
(c) Find the eigenspaces associated with each eigenvalue of T. Do the eigenvectors of T span R3? What can you conclude from this?
(d) Find the generalized eigenspaces W1 and W2 associated with the eigenvalues of T.
(e) Find the projection operators E1 of R3 onto W1 along W2 and E2 of R3 onto W2 along W1.
(f) Find the diagonalizable part D of T. Express D both as a single matrix and as a linear combination of projections.
(g) Find a matrix S which diagonalizes D. What is the resulting diagonal form Λ of D?
(h) Find the nilpotent part N of T. What is the smallest power of N which vanishes?
16.4. ANSWERS TO ODD-NUMBERED EXERCISES 111 16.4. Answers to Odd-Numbered Exercises (1) (a) The single one-dimensional eigenspace does not span R2. (OR: the minimal polyno-mial (λ −3)2 has a second degree factor–see theorem 13.1.11.) (b) 3, 0, 1 (3) (a) 1, 2 (b) 1, 2 (c) The minimal polynomial has a second degree factor (see theorem 13.1.11).
(d) 0, 0, 0, 0 (e) The eigenspaces do not span R3.
(f) 0, 0, 0 (g) 0, 1 (h) 1, 0 (i) 1, 0 (j) 0, 1 (k) 0, 1 (l) 0, 1 (5) (a) 1, 2 (b) 1, 2 (c) 0, 2, 1, 2 (d) 0, 2, 1, 0 (e) 1, 0, 2 (f) 1, 0, 2 (g) 1, 2 (h) 4, 0, 2 (7) (a) 1, 4 (b) 1, 2 (c) 0, 1, 1, 0 (d) 1, 0, 2 (e) 0, 1 (f) 0, 1, 1 Part 5 THE GEOMETRY OF INNER PRODUCT SPACES CHAPTER 17 COMPLEX ARITHMETIC 17.1. Background Topics: complex arithmetic, absolute value and argument of a complex number, De Moivre’s theorem.
115 116 17. COMPLEX ARITHMETIC 17.2. Exercises (1) Re 2 + 3i 3 −4i = .
(2) Im 2 −3i 2 + 3i = .
(3) 21 + 7i 1 −2i = a √ b where a = and b = .
(4) Arg(−2 √ 3 + 2i) = .
(5) Let z = 2 − √ 3 −(1 + 2 √ 3)i 1 + 2i . Then z10 = a(1 −bi) where a = and b = , |z10| = , and Arg z10 = .
(6) If z = 1 √ 2 (1 −i), then z365 = a + bi where a = and b = .
(7) The cube roots of −27 are a + bi, a −bi, and c + di where a = , b = , c = , and d = .
(8) The complex number w = 1 + 2i has 13 thirteenth roots. The sum of these roots is a + bi where a = and b = .
(9) Let z1 = (1 , i , 1 + i), z2 = (i , 0 , 1 −i), and z3 = (−1 , 1 + 2i , 7 + 3i). Show that the set {z1, z2, z3} is linearly dependent in C3 by finding scalars α and β such that αz1 + βz2 − z3 = 0.
Answer: α = and β = .
(10) Let A = 1 1 1 i 1 i 1 i 0 1 + i 0 0 2 0 2 2i . Then the rank of A is and the nullity of A is .
(11) Let A = 0 1 1 0 i 1 i i i . Then A−1 = 1 2 −2a c −2b a + b −a −b c a −b a + b c where a = , b = , and c = .
(12) If A = i 0 −i −2i 1 −1 −4i 2 −i i 3 , then A−1 = −a −ai b + ai −a + bi a −di −c + ai −a −di −a + bi b + ai −a + bi where a = , b = , c = , and d = .
(13) Consider the initial value problem: y′′ + 4y = 0 with the initial conditions y0 = 3 and y′ 0 = 2.
(a) Express the differential equation in the form du dt = Au where u = (y, z) and z = y′.
Then A is the matrix a b −c a where a = , b = , and c = .
(b) The eigenvalues of A are and ; and the corresponding eigenvectors are (1, a) and (1, −a) where a = .
(c) The diagonal form of A is Λ = a 0 0 −a where a = .
17.2. EXERCISES 117 (d) Find the diagonalizing matrix S for A. That is, find S so that Λ = S−1AS. Answer: S = 1 1 a −a where a = .
(e) Find the matrix eAt. Express your answer in terms of real trigonometric functions— not complex exponentials.
Answer: eAt = f(t) a g(t) −b g(t) f(t) where a = , b = , f(t) = , and g(t) = .
(f) The solution to the initial value problem is y(t) = .
(14) Let A = 5 −12 2 −5 .
(a) Find the eigenvalues of A. Answer: λ1 = ; λ2 = .
(b) Find a factorization of A in the form SΛS−1 where Λ is a diagonal matrix.
Answer: S = ; Λ = ; S−1 = .
(c) Find a square root of A. Answer: √ A = .
118 17. COMPLEX ARITHMETIC 17.3. Problems (1) Explain in detail how to find all the cube roots of i.
(2) Show that three points z1, z2, and z3 in the complex plane form an equilateral triangle if and only if z12 + z22 + z32 = z1z2 + z1z3 + z2z3.
17.4. ANSWERS TO ODD-NUMBERED EXERCISES 119 17.4. Answers to Odd-Numbered Exercises (1) −6 25 (3) 7, 2 (5) 512, √ 3, 1024, −π 3 (7) 3 2, 3 √ 3 2 , −3, 0 (9) 2 −i, 1 + 3i (11) 1, i, 0 (13) (a) 0, 1, 4 (b) 2i, −2i, 2i (c) 2i (d) 2i (e) 1 2, 2, cos 2t, sin 2t, (f) 3 cos 2t + sin 2t CHAPTER 18 REAL AND COMPLEX INNER PRODUCT SPACES 18.1. Background Topics: inner products in real and complex vector spaces, the Schwarz inequality, the parallelogram law, the Pythagorean theorem, the norm induced by an inner product, the metric induced by a norm, orthogonal (or perpendicular) vectors, the angle between two vectors, rowspace and columnspace of a matrix.
18.1.1. Definition. Let V be a (real or complex) vector space. A function which associates to each pair of vectors x and y in V a (real or complex) number ⟨x, y⟩(often written x·y) is an inner product (or a dot product) on V provided that the following four conditions are satisfied: (a) If x, y, z ∈V , then ⟨x + y, z⟩= ⟨x, z⟩+ ⟨y, z⟩.
(b) If x, y ∈V and α ∈C (or R), then ⟨αx, y⟩= α⟨x, y⟩.
(c) If x, y ∈V , then ⟨x, y⟩= ⟨y, x⟩.
(d) For every nonzero x in V we have ⟨x, x⟩> 0.
Conditions (a) and (b) show that an inner product is linear in its first variable. It is easy to see that an inner product is conjugate linear in its second variable. (A complex valued function f on V is conjugate linear if f(x + y) = f(x) + f(y) and f(αx) = αf(x) for all x, y ∈V and α ∈C.) When a mapping is linear in one variable and conjugate linear in the other, it is often called sesquilinear (the prefix “sesqui-” means “one and a half”). Taken together conditions (a)–(d) say that the inner product is a positive definite conjugate symmetric sesquilinear form. When a vector space has been equipped with an inner product we define the norm (or length) of a vector x by ∥x∥:= p ⟨x, x⟩; (Notice that this is same definition we used in 4.1.2 for vectors in Rn.) 18.1.2. Notation. There are many more or less standard notations for the inner product of two vectors x and y. The two that we will use interchangeably in these exercises are x · y and ⟨x, y⟩.
18.1.3. Example. For vectors x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) belonging to Rn define ⟨x, y⟩= n X k=1 xkyk .
Then Rn is an inner product space.
18.1.4. Example. For vectors x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) belonging to Cn define ⟨x, y⟩= n X k=1 xkyk .
Then Cn is an inner product space.
121 122 18. REAL AND COMPLEX INNER PRODUCT SPACES 18.1.5. Example. Let l2 be the set of all square summable sequences of complex numbers. (A sequence x = (xk)∞ k=1 is square summable if P∞ k=1|xk|2 < ∞.) (The vector space operations are defined pointwise.) For vectors x = (x1, x2, . . . ) and y = (y1, y2, . . . ) belonging to l2 define ⟨x, y⟩= ∞ X k=1 xkyk .
Then l2 is an inner product space. (It is important to recognize that in order for this definition to make sense, it must be verified that the infinite series actually converges.) 18.1.6. Example. For a < b let C([a, b]) be the family of all continuous complex valued functions on the interval [a, b]. For every f, g ∈C([a, b]) define ⟨f, g⟩= Z b a f(x)g(x) dx.
Then C([a, b]) is an inner product space.
18.1.7. Definition. Let x and y be nonzero vectors in a real vector space V . Then ∡(x, y), the angle between x and y, is defined by ∡(x, y) = arccos ⟨x, y⟩ ∥x∥∥y∥ (Notice that this is the same definition as the one given in 4.1.3 for vectors in Rn.) 18.1.8. Definition. Two vectors x and y in an inner product space V are orthogonal (or perpendicular) if ⟨x, y⟩= 0. In this case we write x ⊥y. Similarly if M and N are nonempty subsets of V we write M ⊥N if every vector in M is orthogonal to every vector in N. When one of the sets consists of a single vector we write x ⊥N instead of {x} ⊥N. When M is a nonempty subset of V we denote by M⊥the set of all vectors x such that x ⊥N. This is the orthogonal complement of M.
18.1.9. Definition. A real valued function f on an inner product space V is uniformly contin-uous if for every number ϵ > 0 there exists a number δ > 0 such that |f(x) −f(y)| < ϵ whenever ∥x −y∥< δ in V .
The following result is one we have already seen for Rn (see 4.1.4).
18.1.10. Theorem (Cauchy-Schwarz inequality). If x and y are vectors in a (real or complex) inner product space, then |⟨x, y⟩| ≤∥x∥∥y∥.
18.2. EXERCISES 123 18.2. Exercises (1) In C3 let x = (3 + 2i, 1 −i, i) and y = (1 −2i, 1 −i, 4 + i). Then ∥x∥= ; ∥y∥= ; and ⟨x, y⟩= .
(2) In C2 let x = (2 −4i, 4i) and y = (2 + 4i, 4). Then ∥x∥= ; ∥y∥= ; and ⟨x, y⟩= .
(3) In C2 let x = (3 −2 i, √ 3 i) and y = (1 + i, 1 −i). Then ∥x∥= ; ∥y∥= ; and ⟨x, y⟩= 1 −√a + (√a −b) where a = and b = .
(4) Make the family of 2 × 2 matrices of real numbers into an inner product space by defining ⟨A, B⟩:= tr(AtB) (see problem 7) for A, B ∈M2,2(R). Let U = 1 4 −3 5 and V = α2 α −1 α + 1 −1 . Find all values of α such that U ⊥V in the inner product space M2,2(R).
Answer: α = and .
(5) Let w = i √ 3 , i √ 3 , i √ 3 , x = −2i √ 6 , i √ 6 , i √ 6 , y = i √ 6 , i √ 6 , −2i √ 6 , and z = 0 , −i √ 2 , i √ 2 .
(a) Which three of these vectors form an orthonormal basis for C3?
Answer: , , and .
(b) Write (1, 0, 0) as a linear combination of the three basis vectors you chose in part (a).
(Use 0 as the coefficient of the vector which does not belong to the basis.) Answer: (1, 0, 0) = −i √a w + 2 i √ 2a x + by + cz where a = , b = , and c = .
(6) Find all real numbers α such that the angle between the vectors 2 i + 2 j + (α −2) k and 2 i + (α −2) j + 2 k is π 3 . Answer: α = and .
(7) Let f(x) = x and g(x) = x2 for 0 ≤x ≤1. Then the cosine of the angle between f and g in the inner product space C([0, 1]) of all continuous real valued functions on [0, 1] is a 4 where a = .
(8) Let f(x) = x and g(x) = cos πx for 0 ≤x ≤1. In the inner product space C([0, 1]) of all continuous real valued functions on [0, 1] the cosine of the angle between f and g is −a √ b c2 where a = , b = , and c = .
(9) Let f(x) = x2 and g(x) = 1 −cx where 0 ≤x ≤1 and c is a constant. If c = , then f ⊥g in the inner product space C([0, 1]) of continuous real valued function on [0, 1].
(10) In R4 the subspace perpendicular to both (1, 4, 4, 1) and (2, 9, 8, 2) is the span of the vectors (−4, a, b, a) and (−b, a, a, b) where a = and b = .
(11) A complex valued function f on the interval [−π, π] is said to be square integrable if R π −π|f(x)|2 dx < ∞. Let F([−π, π]) be the family of all square integrable complex valued functions on [−π, π]. This family is an inner product space under the usual pointwise operations of addition and scalar multiplication and the inner product defined by ⟨f, g⟩= 1 2π Z π −π f(x)g(x) dx for all f, g ∈F([−π, π]).
Actually the preceding sentence is a lie: to be correct we should identify any two square integrable functions which differ only on a set of Lebesgue 124 18. REAL AND COMPLEX INNER PRODUCT SPACES measure zero and work with the resulting equivalence classes. Then we have an inner product space. What we actually have here is a so-called semi-inner product. For the purposes of the current exercise, however, this correction turns out to be unimportant; ignore it.
For each integer n (positive, negative, or zero) define a function en by en(x) = einx for −π ≤x ≤π.
(a) Then each en belongs to F([−π, π]) and ∥en∥= for every integer n.
(b) If m ̸= n, then ⟨em, en⟩= .
Now let f(x) = ( 0 if −π ≤x < 0, 1 if 0 ≤x ≤π.
(c) Then ⟨f, e0⟩= .
(d) If n is odd, then ⟨f, en⟩= .
(e) If n is even but not zero, then ⟨f, en⟩= .
(f) Write the sum of the middle eleven terms of the Fourier series for f in simplified form.
Hint: Use problem 8 in this chapter.
Answer: 5 X n=−5 ⟨f, en⟩en = a+b sin x+c sin 3x+d sin 5x where a = , b = , c = , and d = .
18.3. PROBLEMS 125 18.3. Problems (1) Prove that if x ∈V and ⟨x, y⟩= 0 for all y ∈V , then x = 0.
(2) Let S, T : V →W be linear transformations between real inner product spaces. Prove that if ⟨Sv, w⟩= ⟨Tv, w⟩for all v ∈V and w ∈W, then S = T.
(3) Prove that if V is a complex inner product space and T ∈L(V ) satisfies ⟨Tz, z⟩= 0 for all z ∈V , then T = 0. Hint. In the hypothesis replace z first by x+y and then by x+iy.
(4) Show that the preceding result does not hold for real inner product spaces.
(5) Let V be a complex inner product space and S, T ∈L(V ). Prove that if ⟨Sx, x⟩= ⟨Tx, x⟩ for all x ∈V , then S = T.
(6) Prove the Schwarz inequality 18.1.10. Hint. Let α = ⟨y, y⟩and β = −⟨x, y⟩and expand ∥αx + βy∥2. This hint leads to a slick, easy, and totally unenlightening proof. Perhaps you can find a more perspicuous one.
he polarization identity If x and y are vectors in a complex inner product space, then ⟨x, y⟩= 1 4(∥x + y∥2 −∥x −y∥2 + i ∥x + iy∥2 −i ∥x −iy∥2) .
What is the correct identity for a real inner product space?
(7) Let M2,2(R) be the vector space of all 2 × 2 matrices of real numbers. Show that this space can be made into an inner product space by defining ⟨A, B⟩:= tr(AtB) for all A, B ∈M2,2(R).
(8) Prove that for every real number θ eiθ = cos θ + i sin θ .
Derive from this that cos θ = 1 2 eiθ + e−iθ and sin θ = 1 2i eiθ −e−iθ .
(9) Let x and y be vectors in an inner product space. Prove that ∥x + y∥2 + ∥x −y∥2 = 2∥x∥2 + 2∥y∥2.
Give a geometric interpretation of this result.
(10) Show that the norm function ∥· ∥: V →R on an inner product space is uniformly contin-uous.
126 18. REAL AND COMPLEX INNER PRODUCT SPACES 18.4. Answers to Odd-Numbered Exercises (1) 4, 2 √ 6, 2 + 12 i (3) 4, 2, 3, 5 (5) (a) w, x, y (b) 3, 0, 0 (7) √ 15 (9) 4 3 (11) (a) 1 (b) 0 (c) 1 2 (d) −i nπ (e) 0 (f) 1 2, 2 π, 2 3π, 2 5π CHAPTER 19 ORTHONORMAL SETS OF VECTORS 19.1. Background Topics: orthonormal sets of vectors, orthonormal bases, orthogonal complements, orthogonal direct sums, Gram-Schmidt orthonormalization, the QR-factorization of a matrix.
19.1.1. Definition. A set B of vectors in an inner product space is orthonormal if x ⊥y whenever x and y are distinct vectors in B and ∥x∥= 1 for every x ∈B.
The set B is a maximal!orthonormal set provided that it is orthonormal and the only orthonormal set which contains B is B itself.
19.1.2. Theorem. Let B = {e1, . . . , en} be an orthonormal set in an inner product space V . Then the following are equivalent.
(a) B is a maximal orthonormal set in V .
(b) If ⟨x, ek⟩= 0 for k = 1, . . . , n, then x = 0.
(c) The span of B is all of V .
(d) If x ∈V , then x = Pn k=1⟨x, ek⟩ek.
(the Fourier series for x.) (e) ⟨x, y⟩= Pn k=1⟨x, ek⟩⟨ek, y⟩for all x, y ∈V .
(f) ∥x∥2 = Pn k=1|⟨x, ek⟩|2 for every x ∈V .
(g) dim V = n.
19.1.3. Definition. An orthonormal set in an inner product space V which satisfies any (and hence all) of the conditions listed in the preceding theorem is called an orthonormal basis for V .
19.1.4. Definition. An n × n matrix of real numbers is an orthogonal matrix if its column vectors are an orthonormal basis for Rn. An n×n matrix of complex numbers is a unitary matrix if its column vectors are an orthonormal basis for Cn.
19.1.5. Theorem (QR-factorization). If A is an n × n matrix of real numbers, then there exist an orthogonal matrix Q and an upper triangular matrix R such that A = QR. If A is an n × n matrix of complex numbers, then there exist a unitary matrix Q and an upper triangular matrix R such that A = QR.
(For a proof of the preceding theorem see , pages 425–427.) 19.1.6. Definition. Let M and N be subspaces of an inner product space V . We say that the space V is the orthogonal direct sum of M and N if M + N = V and M ⊥N. In this case we write V = M ⊕N.
Caution: Since the same notation is used for direct sums of vector spaces and orthogonal direct sums of inner product spaces, close attention should be paid to the context in which these concepts arise. For example, if M is the x-axis and N is the line y = x in R2, is it true that R2 = M ⊕N?
Yes, if R2 is regarded as a vector space. No, if it is regarded as an inner product space.
127 128 19. ORTHONORMAL SETS OF VECTORS 19.2. Exercises (1) Use the Gram-Schmidt procedure to find an orthonormal basis for the subspace of R4 spanned by w1 = (1, 0, 0, 0), w2 = (1, 1, 1, 0), and w3 = (1, 2, 0, 1). The basis consists of the vectors e1 = ( 1 , 0 , 0 , 0 ); e2 = 1 a ( 0 , 1 , 1 , b ); and e3 = 1 c ( b , 1 , −1 , 1 ) where a = , b = , and c = .
(2) Let P4 = P4([0, 1]) be the vector space of polynomials of degree strictly less than 4 with an inner product defined by ⟨p, q⟩= Z 1 0 p(x)q(x) dx for all p, q ∈P4. Let w0(x) = 1, w1(x) = x, w2(x) = x2, and w3(x) = x3 for 0 ≤x ≤ 1. Use the Gram-Schmidt process to convert the ordered basis {w0, w1, w2, w3} to an orthonormal basis {e0, e1, e2, e3} for P4.
Answer: e0(x) = e1(x) = √a(bx −1) where a = and b = e2(x) = √a(bx2 −bx + 1) where a = and b = e3(x) = √a(bx3 −cx2 + dx −1) where a = , b = , c = , and d = (3) Find the QR factorization of A = 3 0 4 5 .
Answer: A = QR = 1 a b −c c b a c 0 b where a = , b = , and c = .
(4) Let A = 0 0 1 0 1 1 1 1 1 . The QR-factorization of A is A = QR where Q = and R = .
(5) Let A = 1 4 −2 1 3 −1 1 2 −1 . The QR-factorization of A is A = QR where Q = a b −ab a 0 2ab a −b −ab and R = 3a 9a −4a 0 2b −b 0 0 ab where a = and b = .
19.3. PROBLEMS 129 19.3. Problems (1) Let {e1, e2, . . . , en} be a finite orthonormal subset of an inner product space V and x ∈V .
Show that n X k=1 |⟨x, ek⟩|2 ≤∥x∥2.
Hint. Multiply out x −Pn k=1⟨x, ek⟩ek, x −Pn k=1⟨x, ek⟩ek .
(2) Let M be a subspace of an inner product space V .
(a) Show that M ⊆M⊥⊥.
(b) Prove that equality need not hold in (a).
(c) Show that if V is finite dimensional, then M = M⊥⊥.
(3) Let M and N be subspaces of an inner product space. Prove that (M + N)⊥= M⊥∩N⊥.
(4) Let M be a subspace of a finite dimensional inner product space V . Prove that V = M ⊕M⊥.
(5) Give an example to show that the conclusion of the preceding problem need not hold in an infinite dimensional space.
(6) Prove that if an inner product space V is the orthogonal direct sum M⊕N of two subspaces M and N, then N = M⊥.
(7) Prove that if f is a linear functional on a finite dimensional inner product space V , then there exists a unique vector a ∈V such that f(x) = ⟨x, a⟩ for every x ∈V .
(8) In beginning calculus you found (by making use of the p-test) that the series ∞ X k=1 1 k2 converges. But you were not given a means of discovering what the series converges to.
Now you have enough machinery to accomplish this.
We denote by L2[0, 2π] the vector space of all complex valued functions f defined on the interval [0, 2π] such that Z 2π 0 |f(t)|2 dt < ∞.
(As in exercise 11 of chapter 18 this isn’t quite correct: the members of L2 are technically equivalence classes of functions.
For the purposes of this problem, use the preceding not-quite-right definition.) On the space L2[0, 2π] we define the following inner product ⟨f, g⟩= 1 2π Z 2π 0 f(t)g(t) dt.
For each integer n (positive, negative, or zero) define the function en on [0, 2π] by en(x) = einx for 0 ≤x ≤2π.
(a) Show that {en : n is an integer} is an orthonormal set in L2[0, 2π].
130 19. ORTHONORMAL SETS OF VECTORS In part (b) you may use without proof the following fact: for every function f in the inner product space L2[0, 2π] ∥f∥2 = ∞ X k=−∞ |⟨f, ek⟩|2 (∗) That is, in L2[0, 2π] the square of the length of a vector is the sum of the squares of its Fourier coefficients with respect to the orthonormal family given in part (a). This is the infinite dimensional version of Parseval’s formula.
(b) Find the sum of the infinite series ∞ X k=1 1 k2 . Hint. Apply (∗) to the function f(x) = x.
19.4. ANSWERS TO ODD-NUMBERED EXERCISES 131 19.4. Answers to Odd-Numbered Exercises (1) √ 2, 0, √ 3 (3) 5, 3, 4 (5) 1 √ 3 , 1 √ 2 CHAPTER 20 QUADRATIC FORMS 20.1. Background Topics: quadratic forms, quadric surfaces, positive (and negative) definite, positive (and negative) semidefinite, indefinite.
20.1.1. Definition. A symmetric matrix A on a real inner product V is positive definite if ⟨Ax, x⟩> 0 for every x ̸= 0 in V . It is negative definite if ⟨Ax, x⟩< 0 for every x ̸= 0 in V . It is indefinite if there are vectors x and y in V such that ⟨Ax, x⟩> 0 and ⟨Ay, y⟩< 0. Of course an operator on a finite dimensional vector space is positive definite, negative definite, or indefinite of its matrix representation is positive definite, etc.
The following useful result (and its proof) can be found on page 250 of .
20.1.2. Theorem. Let A be an n × n matrix. Then the following conditions are equivalent: (a) A is positive definite; (b) xtAx > 0 for every x ̸= 0 in Rn; (c) every eigenvalue λ of A is strictly positive; (d) every leading principal submatrix Ak (k = 1, . . . , n) has strictly positive determinant; and (e) when A has been put in echelon form (without row exchanges) the pivots are all strictly positive.
In the preceding, the leading principal submatrix Ak is the k × k matrix which appears in the upper left corner of A.
133 134 20. QUADRATIC FORMS 20.2. Exercises (1) Suppose that A is a 3 × 3 matrix such that ⟨Ax, x⟩= x12 + 5x22 −3x32 + 6x1x2 −4x1x3 + 8x2x3 for all x ∈R3. Then A = a b c b d e c e f where a = , b = , c = , d = , e = , and f = .
(2) A curve C is given by the equation 2x2 −72xy + 23y2 = 50. What kind of curve is C?
Answer: It is a(n) .
(3) The equation 5x2 + 8xy + 5y2 = 1 describes an ellipse. The principal axes of the ellipse lie along the lines y = and y = .
(4) The graph of the equation 13x2 −8xy +7y2 = 45 is an ellipse. The length of its semimajor axis is and the length of its semiminor axis is .
(5) Consider the equation 2x2 + 2y2 −z2 −2xy + 4xz + 4yz = 3.
(a) The graph of the equation is what type of quadric surface?
Answer: .
(b) In standard form the equation for this surface is u2 + v2 + w2 = .
(c) Find three orthonormal vectors with the property that in the coordinate system they generate, the equation of the surface is in standard form.
Answer: 1 √ 6 ( 1 , , ), 1 √ 2 ( , , 0 ), and 1 √ 3 ( 1 , , ).
(6) Determine for each of the following matrices whether it is positive definite, positive semi-definite, negative definite, negative semidefinite, or indefinite.
(a) The matrix 2 −1 −1 −1 2 −1 −1 −1 2 is .
(b) The matrix −2 1 1 1 −2 −1 1 −1 −2 is .
(7) Determine for each of the following matrices whether it is positive definite, positive semi-definite, negative definite, negative semidefinite, or indefinite.
(a) The matrix 1 2 3 2 5 4 3 4 9 is .
(b) The matrix 1 2 0 0 2 6 −2 0 0 −2 5 −2 0 0 −2 3 is .
(c) The matrix 0 1 2 1 0 1 2 1 0 2 is .
(8) Let B = 2 2 4 2 b 8 4 8 7 . For what range of values of b is B positive definite?
Answer: .
20.2. EXERCISES 135 (9) Let A = a 1 1 1 a 1 1 1 a . For what range of values of a is A positive definite?
Answer: .
136 20. QUADRATIC FORMS 20.3. Problems (1) You are given a quadratic form q(x, y, z) = ax2 + by2 + cz2 + 2dxy + 2exz + 2fyz. Explain in detail how to determine whether the associated level surface q(x, y, z) = c encloses a region of finite volume in R3 and, if it does, how to find that volume. Justify carefully all claims you make. Among other things, explain how to use the change of variables theorem for multiple integrals to express the volume of an ellipsoid in terms of the lengths of the principal axes of the ellipsoid.
Apply the method you have developed to the equation 11x2 + 4y2 + 11z2 + 4xy −10xz + 4yz = 8.
20.4. ANSWERS TO ODD-NUMBERED EXERCISES 137 20.4. Answers to Odd-Numbered Exercises (1) 1, 3, −2, 5, 4, −3 (3) −x, x (5) (a) hyperboloid of one sheet (b) −1, 1 (c) 1, −2, 1, −1, 1, 1 (7) (a) indefinite (b) positive definite (c) positive definite (9) a > 1 CHAPTER 21 OPTIMIZATION 21.1. Background Topics: critical (stationary) points of a function of several variables; local (relative) maxima and minima; global (absolute) maxima and minima.
21.1.1. Definition. Let f : Rn →R be a smooth scalar field (that is, a real valued function on Rn with derivatives of all orders) and p ∈Rn. The Hessian matrix (or second derivative matrix) of f at p, denoted by H f(p), is the symmetric n × n matrix H f(p) = ∂2f ∂xi∂xj (p) n i=1 n j=1 = fij(p) .
21.1.2. Theorem (Second Derivative Test). Let p be a critical point of a smooth scalar field f (that is, a point where the gradient of f is zero). If the Hessian matrix H f is positive definite at p, then f has a local minimum there. If H f is negative definite at p, then f has a local maximum there. If H f is indefinite at p, then f has a saddle point there.
139 140 21. OPTIMIZATION 21.2. Exercises (1) Notice that the function f defined by f(x, y) = (x2 −2x) cos y has a critical point (sta-tionary point) at the point (1, π). The eigenvalues of the Hessian matrix of f are and ; so we conclude that the point (1, π) is a (local minimum, local maximum, saddle point).
(2) Use matrix methods to classify the critical point of the function f(x, y) = 2x2 + 2xy + 2x + y4 −4y3 + 7y2 −4y + 5 as a local maximum, local minimum, or saddle point.
(a) The only critical point is located at ( , ).
(b) It is a .
(3) Use matrix methods to classify the critical point of the function f(x, y, z) = 1 2x4 −xy + y2 −xz + z2 −x + 3 as a local maximum, local minimum, or saddle point.
(a) The only critical point is located at ( , , ).
(b) It is a .
(4) Notice that the function f defined by f(x, y) = −1+4(ex −x)−5x sin y+6y2 has a critical point (stationary point) at the origin. Since the eigenvalues of the Hessian matrix of f are (both positive, both negative, of different signs) we conclude that the origin is a (local minimum, local maximum, saddle point).
(5) Use matrix methods to classify each critical point of the function f(x, y) = y3 −4 3x3 −2y2 + 2x2 + y −7 as a local maximum, local minimum, or saddle point.
Answer: ( 0 , 1 3 ) is a .
( 0 , ) is a .
( , 1 3 ) is a .
( , ) is a .
(6) Use matrix methods to classify each critical point of the function f(x, y, z) = x2y −4x −y sin z for 0 < z < π as a local maximum, local minimum, or saddle point.
Answer: The critical points are ( −1 , , ), which is a ; and ( 1 , , ), which is a (local minimum, local max-imum, saddle point).
(7) The function f defined by f(x, y) = x2y2−2x−2y has a stationary point at ( , ).
At this stationary point f has a (local minimum, local maximum, saddle point).
21.3. PROBLEMS 141 21.3. Problems (1) Use matrix methods to classify each critical point of the function f(x, y, z) = x3y + z2 −3x −y + 4z + 5 as a local maximum, local minimum, or saddle point. Justify your conclusions carefully.
(2) Let f(x, y, z) = x2y −yez + 2x + z. The only critical point of f is located at (−1, 1, 0).
Use the second derivative test to classify this point as a local maximum, local minimum, or saddle point. State the reasons for your conclusion clearly.
(3) Notice that the function f defined by f(x, y, z) = x2y + 2xy + y −yez−1 + 2x + z + 7 has a critical point (stationary point) at (−2, 1, 1). Use the second derivative test to classify this point as a local maximum, local minimum, or saddle point. State the reasons for your conclusion clearly.
(4) Explain in detail how to use matrix methods to classify each critical point of the function f(x, y) = −1 2xy + 2 x −1 y as a local maximum, local minimum, or saddle point. Carry out the computations you describe.
142 21. OPTIMIZATION 21.4. Answers to Odd-Numbered Exercises (1) −2, −1, local maximum (3) (a) 1, 1 2, 1 2 (b) local minimum (5) saddle point, 1, local minimum, 1, local maximum, 1, 1, saddle point (7) 1, 1, saddle point Part 6 ADJOINT OPERATORS CHAPTER 22 ADJOINTS AND TRANSPOSES 22.1. Background Topics: adjoint of an operator, transpose of an operator, conjugate transpose.
22.1.1. Definition. Let T : V →W be a linear transformation between real inner product spaces.
If there exists a linear map T t : W →V which satisfies ⟨Tv, w⟩= ⟨v, T tw⟩ for all v ∈V and w ∈W, then T t is the transpose of T.
In connection with the definition above see problem 1.
22.1.2. Theorem. Let T : Rn →Rm be a linear transformation. Then the transpose linear trans-formation T t exists.
Furthermore, the matrix representation [T t] of this transformation is the transpose of the matrix representation of T.
22.1.3. Definition. Let V be a real inner product space and T be an operator on V whose transpose exists. If T = T t, then T is symmetric. If T commutes with its transpose (that is, if TT t = T tT) it is normal.
22.1.4. Definition. If aij is an m × n matrix, its conjugate transpose is the n × m matrix aji .
22.1.5. Definition. Let T : V →W be a linear transformation between complex inner product spaces. If there exists a linear map T ∗: W →V which satisfies ⟨Tv, w⟩= ⟨v, T ∗w⟩ for all v ∈V and w ∈W, then T ∗is the adjoint (or conjugate transpose, or Hermitian conjugate) of T. (In many places, T ∗is denoted by T H or by T †.) 22.1.6. Theorem. Let T : Cn →Cm be a linear transformation. Then the adjoint linear trans-formation T ∗exists.
Furthermore, the matrix representation [T ∗] of this transformation is the conjugate transpose of the matrix representation of T.
22.1.7. Definition. Let V be a complex inner product space and T be an operator on V whose adjoint exists. If T = T ∗, then T is self-adjoint (or Hermitian). If T commutes with its adjoint (that is, if TT ∗= T ∗T) it is normal. A matrix is normal if it is the representation of a normal operator.
22.1.8. Definition. Let V and W be inner product spaces. We make the vector space V ⊕W into an inner product space as follows. For v1, v2 ∈V and w1, w2 ∈W let ⟨(v1, w1), (v2, w2) ⟩:= ⟨v1, v2⟩+ ⟨w1, w2⟩.
(It is an easy exercise to verify that this is indeed an inner product on V ⊕W.) 145 146 22. ADJOINTS AND TRANSPOSES 22.2. Exercises (1) Let C([0, 1], C) be the family of all continuous complex valued functions on the inter-val [0, 1]. The usual inner product on this space is given by ⟨f, g⟩= Z 1 0 f(x)g(x) dx.
Let φ be a fixed continuous complex valued function on [0, 1]. Define the operator Mφ on the complex inner product space C([0, 1], C) by Mφ(f) = φf. Then Mφ∗= .
(2) Let A = "3 −i 2 + 2i 1 1 −i 3i # . Find Hermitian (that is, self-adjoint) matrices B and C such that A = B + iC. Hint. Consider A ± A∗.
Answer: B = 1 a 4c b + c i b −c i d and C = 1 a −a b −c i b + c i 4c , where a = , b = , c = , and d = .
(3) Let P3 be the space of polynomial functions of degree strictly less than 3 defined on the in-terval [0, 1]. Define the inner product of two polynomials p, q ∈P3 by ⟨p, q⟩= R 1 0 p(t)q(t) dt.
Then the matrix representation of the transpose of the differentiation operator D on the space P3 (with respect to its usual basis {1, t, t2}) is . Hint. The answer is not the transpose of the matrix representation of D.
(4) Let V be a complex inner product space. Define an operator T : V ⊕V →V ⊕V by T(x, y) = (y, −x).
Then T ∗(u, v) =( , ).
22.3. PROBLEMS 147 22.3. Problems (1) Let T : V →W be a linear map between real inner product spaces. If S : W →V is a function which satisfies ⟨Tv, w⟩= ⟨v, Sw⟩ for all v ∈V and all w ∈W, then S is linear (and is therefore the transpose of T).
(2) Prove theorem 22.1.2. Show that, in fact, every linear map between finite dimensional real inner product spaces has a transpose.
(3) Let T be a self-adjoint operator on a complex inner product space V . Prove that ⟨Tx, x⟩ is real for every x ∈V .
(4) Let T be an operator on a complex inner product space whose adjoint T ∗exists. Prove that T ∗T = 0 if and only if T = 0.
(5) Let V be a complex inner product space and let φ be defined on the set A(V ) of operators on V whose adjoint exists by φ(T) = T ∗.
Show that if S, T ∈A(V ) and α ∈C, then (S + T)∗= S∗+ T ∗and (αT)∗= α T ∗. Hint.
Use problem 5 in chapter18.
Note: Similarly, if V is a real inner product space, A(V ) is the set of operators whose transpose exists, and φ(T) := T t, then φ is linear.
(6) Let T be a linear operator on a complex inner product space V . Show if T has an adjoint, then so does T ∗and T ∗∗= T. Hint: Use problem 5 in chapter 18. (Here T ∗∗means T ∗∗.) Note: The real inner product space version of this result says that if T is an operator on a real inner product space whose transpose exists, then the transpose of T t exists and T tt = T.
(7) Let S and T be operators on a complex inner product space V . Show that if S and T have adjoints, then so does ST and (ST)∗= T ∗S∗. Hint. Use problem 5 in chapter 18.
Note: The real inner product space version of this says that if S and T are operators on a real inner product space and if S and T both have transposes, then so does ST and (ST)t = T tSt.
(8) Let A: V →V be an operator on a real inner product space. Suppose that At exists and that it commutes with A (that is, suppose AAt = AtA). Show that ker A = ker At.
(9) Let A and B be Hermitian operators on a complex inner product space. Prove that AB is Hermitian if and only if AB = BA.
(10) Show that if T : V →W is an invertible linear map between complex inner product spaces and both T and T −1 have adjoints, then T ∗is invertible and (T ∗)−1 = (T −1)∗.
Note: The real inner product space version of this says that if T : V →W is an invertible linear map between real inner product spaces and both T and T −1 have transposes, then T t is invertible and (T t)−1 = (T −1)t.
(11) Every eigenvalue of a self-adjoint operator on a complex inner product space is real. Hint.
Let x be an eigenvector associated with an eigenvalue λ of an operator A. Consider λ∥x∥2.
(12) Let A be a self-adjoint operator on a complex inner product space. Prove that eigenvectors associated with distinct eigenvalues of A are orthogonal. Hint. Use problem 11. Let x and y be eigenvectors associated with distinct eigenvalues λ and µ of A. Start your proof by showing that λ⟨x, y⟩= µ⟨x, y⟩.
148 22. ADJOINTS AND TRANSPOSES 22.4. Answers to Odd-Numbered Exercises (1) Mφ (3) −6 2 3 12 −24 −26 0 30 30 CHAPTER 23 THE FOUR FUNDAMENTAL SUBSPACES 23.1. Background Topics: column space; row space; nullspace; left nullspace, lead variables and free variables in a matrix, rank, row rank and column rank of as matrix.
23.1.1. Definition. A linear transformation T : Rn →Rm (and its associated standard matrix [T]) have four fundamental subspaces: the kernels and ranges of T and T t. Over the years a rather elaborate terminology has grown up around these basic notions.
The nullspace of the matrix [T] is the kernel of the linear map T.
The left nullspace of the matrix [T] is the kernel of the linear map T t.
The column space of [T] is the subspace of Rm spanned by the column vectors of the ma-trix [T]. This is just the range of the linear map T.
And finally, the row space of [T] is the subspace of Rn spanned by the row vectors of the matrix [T]. This is just the range of the linear map T t.
For a linear transformation T : Cn →Cm the terminology is the same.
EXCEPT: in the preceding five paragraphs each appearance of “ T t ” must be replaced by a “ T ∗” (and, of course, Rn by Cn and Rm by Cm).
23.1.2. Definition. The row rank of a matrix is the dimension of its row space and the column rank of a matrix is the dimension of its column space.
23.1.3. Proposition. The rank of a matrix A is the dimension of the largest square submatrix of A with nonzero determinant.
Two useful facts that you may wish to keep in mind are: (i) row equivalent matrices have the same row space (for a proof see , page 56); and (ii) the row rank of a matrix is the same as its column rank (for a proof see , page 72).
Note that according to the second assertion the rank of a linear map T, the row rank of its matrix representation [T], and the column rank of [T] are all equal.
23.1.4. Theorem (Fundamental Theorem of Linear Algebra). If T is an operator on a finite dimensional complex inner product space, then ker T = (ran T ∗)⊥.
23.1.5. Corollary. If T is an operator on a finite dimensional complex inner product space, then ker T ∗= (ran T)⊥.
23.1.6. Corollary. If T is an operator on a finite dimensional complex inner product space, then ran T = (ker T ∗)⊥.
23.1.7. Corollary. If T is an operator on a finite dimensional complex inner product space, then ran T ∗= (ker T)⊥.
149 150 23. THE FOUR FUNDAMENTAL SUBSPACES Note: With the obvious substitutions of T t for T ∗, the preceding theorem and its three corollaries remain true for finite dimensional real inner product spaces.
23.2. EXERCISES 151 23.2. Exercises (1) Let T : C3 →C3 be the operator whose matrix representation is [T] = 1 i −1 1 + i 3 −i −2 1 −2i −6 + 5i 1 .
(a) The kernel of T (the nullspace of [T]) is the span of {( , , 10 ) }.
(b) The range of T (the column space of [T]) is the span of { ( 1 , 0 , ) , ( 0 , 1 , ) }.
(c) The kernel of T ∗(the left nullspace of T ∗) is the span of { ( 3 , , ) }.
(d) The range of T ∗(the row space of [T])is the span of { ( 10 , 0 , ), ( 0 , 10 , ) }.
(2) Find a basis for each of the four fundamental subspaces associated with the matrix A = 1 2 0 1 0 1 1 0 1 2 0 1 .
(a) The column space of A is the plane in R3 whose equation is .
It is the span of { ( 1 , 0 , ) , ( 0 , 1 , ) }.
(b) The nullspace of A is the span of { ( , -1 , , 0 ) , ( , 0 , 0 , 1 ) }.
(c) The row space of A is the span of { ( 1 , 0 , , 1 ) , ( 0 , , 1 , 0 ) }.
(d) The left nullspace of A is the line in R3 whose equations are = = 0.
It is the span of { ( , , 1 ) }.
(3) Let A = 1 2 0 2 −1 1 3 6 1 1 −2 1 5 10 1 5 −4 3 .
(a) Find a basis for the column space of A.
Answer: { ( 1 , 0 , ) , ( 0 , , 1 ) }.
(b) The column space of A is a plane in R3. What is its equation?
Answer: .
(c) The dimension of the row space of A is .
(d) Fill in the missing coordinates of the following vector so that it lies in the row space of A.
(4, , 6 , , , ).
(e) The dimension of the nullspace of A is .
(f) Fill in the missing coordinates of the following vector so that it lies in the nullspace of A.
( , 1 , , 1 , 1, 1 ) .
152 23. THE FOUR FUNDAMENTAL SUBSPACES (4) Let T : R5 →R3 be defined by T(v, w, x, y, z) = (v −x + z, v + w −y, w + x −y −z).
(a) Find the matrix representation of T.
(b) The kernel of T (the nullspace of [T]) is the span of { ( , , 1 , 0 , 0) , ( 0 , 1 , 0 , 1 , 0 ) , ( -1 , 1 , 0 , , 1 ) }.
(c) The range of T (the column space of [T]) is the span of { ( 1 , , -1 ) , ( 0 , 1 , ) }.
Geometrically this is a .
(d) The range of T t (the row space of [T]) is the span of { ( 1 , 0 , , 0 , 1 ) , ( , , , -1 , -1 ) }.
(e) The kernel of T t (the left nullspace of [T]) is the span of { ( 1 , , ) }.
Geometrically this is a .
(5) Let A be the matrix 1 0 2 0 −1 1 2 4 −2 −1 0 1 1 −1 0 2 3 7 −3 −2 . Find the following subspaces associated with A.
(a) The column space of A is the span of { (1 , 0 , -1/2 , ) , (0 , 1 , , ) }.
(b) The row space of A is the span of { (1 , 0 , , 0 , ) , (0 , 1 , 1 , , 0 ) }.
(c) The nullspace of A is the span of { ( , -1 , 1 , 0 , 0) , ( 0 , 1 , 0 , , 0 ) , ( 1 , 0 , 0 , , ) }.
(d) The nullspace of At is the span of { ( , , 1 , 0 ) , (-1/2 , , 0 , 1 )}.
(6) Let A = 1 −2 −1 3 2 −2 4 2 −6 −4 5 −10 −1 15 0 3 −6 1 9 −4 3 −6 −1 9 1 0 0 2 0 −5 .
(a) The nullspace of A is the span of {(2 , 1 , 0 , 0 , 0) , ( , 0 , 0 , 1 , 0) , ( , 0 , , 0 , 1)}.
23.2. EXERCISES 153 (b) The row space of A is the span of {(1 , −2 , 0 , , −1/2 ) , (0 , 0 , 1 , , )}.
(c) The column space of A is the span of {(1 , −2 , 0 , , 1/2 , ) , (0 , 0 , 1 , , , 1/2)}.
(d) The left nullspace of A is the span of {(2, , 0, −1, 1, 0, 0) , ( , 1, 0, 0, 0, 0) , (−1/2, 0, , 0, 1, 0) , ( , 0, −1/2, 0, 0, 1)} (7) Let A = 1 2 1 2 4 3 3 6 4 .
(a) Fill in coordinates of the following vector x so that it is perpendicular to the rowspace of A. Answer: x = (10, , ) .
(b) Fill in coordinates of the following vector y so that it is perpendicular to the columnspace of A. Answer: y = (3, , ) .
(8) In this exercise we prove a slightly different version of the fundamental theorem of linear algebra than the one given in theorem 23.1.4.
Here we work with real inner product spaces and the scope of the result is not restricted to finite dimensional spaces, but we must assume that the linear map with which we are dealing has a transpose.
23.2.1. Theorem (Fundamental theorem of linear algebra). Suppose that V and W are arbitrary real inner product spaces and that the linear transformation T : V →W has a transpose. Then ker T = ran T t⊥.
We prove the preceding theorem.
For each step in the proof give the appropriate reason. Choose from the following list.
DK Definition of “Kernel” DO Definition of “Orthogonal” DOC Definition of “Orthogonal Complement” DR Definition of “Range” DT Definition of “Transpose” H Hypothesis PIP Elementary Property of Inner Products Proof. We must show two things: (i) ker T ⊆(ran T t)⊥and (ii) (ran T t)⊥⊆ker T.
To prove (i) we suppose that x ∈ker T and prove that x ∈(ran T t)⊥. Let v be a vector in ran T t. Then there exists a vector w in W such that v = T tw (reason: ). We compute the inner product of x and v.
⟨x, v⟩= ⟨x, T tw⟩ = ⟨Tx, w⟩ (reason: ) = ⟨0, w⟩ (reason: and ) = 0 (reason: ) From this we infer that x ⊥v (reason: ) and consequently that x ∈(ran T t)⊥ (reason: ).
154 23. THE FOUR FUNDAMENTAL SUBSPACES To prove the converse we suppose that x ∈(ran T t)⊥and show that x ∈ker T. We know that x ⊥ran T t (reason: and ). If w ∈W then the vector T tw belongs to ran T t (reason: ); so x ⊥T tw for all w ∈W. Thus for all w ∈W 0 = ⟨x, T tw⟩ (reason: ) = ⟨Tx, w⟩ (reason: ) It follows from this that Tx = 0 (reason: ). That is, x ∈ker T (reason: ).
□ (9) The matrix x y z y 1 x has rank one if and only if the point (x, y, z) lies on the parametrized curve r(t) = ( , t , ) in R3. Hint. Use proposition 23.1.3.
(10) Let A be the 3 × 4 matrix whose nullspace is the subspace of R4 spanned by the vectors (1, 0, 1, 0) and (0, 1, 1, 0). Then the vectors ( , , , 0 ) and ( 0 , , , ) form an orthonormal basis for the row space of A.
(11) Let T : R3 →R2 be the linear transformation whose matrix representation is 1 0 2 1 1 4 and let x = (5, 4, −9).
(a) Find u ∈ker T and v ∈ran T t such that x = u + v.
Answer: u = ( , , ) and v = ( , , ).
(b) Find y ∈ran T and z ∈ker T t such that Tx = y + z.
Answer: y = ( , ) and z = ( , ).
23.3. PROBLEMS 155 23.3. Problems (1) Let T : R5 →R4 be a linear transformation whose matrix representation is [T] = 1 2 0 −5 3 −2 −4 3 1 0 −1 −2 3 −4 3 1 2 3 −14 9 .
Gauss-Jordan reduction applied to [T] yields the matrix B = 1 2 0 −5 3 0 0 1 −3 2 0 0 0 0 0 0 0 0 0 0 and ap-plied to the transpose of [T] yields C = 1 0 1 3 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 .
(a) From the matrices above we can read offthe dimension of the range of T and write down a basis for it. Explain carefully.
(b) From the matrices above we can read offthe dimension of the range of the transpose of T and write down a basis for it. Explain carefully.
(c) From the matrices above we can write down two equations which a vector (v, w, x, y, z) must satisfy to be in the kernel of T. Explain carefully. What are the equations?
Also explain carefully how we obtain from these equations the dimension of the kernel of T and find a basis for it. Carry out the calculation you describe.
(d) From the matrices above we can write down two equations which a vector (w, x, y, z) must satisfy to be in the kernel of the transpose of T. Explain carefully. What are the equations? Also explain carefully how we obtain from these equations the dimension of the kernel of T t and find a basis for it. Carry out the calculation you describe.
(2) Let T : R6 →R3 be a linear transformation whose matrix representation is [T] = 1 2 0 2 −1 1 3 6 1 1 −2 1 5 10 1 5 −4 3 , Gauss-Jordan reduction applied to [T] yields the matrix B = 1 2 0 2 −1 1 0 0 1 −5 1 −2 0 0 0 0 0 0 , and applied to the transpose of [T] yields C = 1 0 2 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 .
(a) From the matrices above we can read offthe dimension of the range of T and write down a basis for it. Explain carefully.
(b) From the matrices above we can read offthe dimension of the range of the transpose of T and write down a basis for it. Explain carefully.
(c) From the matrices above we can write down two equations which a vector (u, v, w, x, y, z) in R6 must satisfy to be in the kernel of T. Explain carefully. What are the equa-tions? Also explain carefully how we obtain from these equations the dimension of the kernel of T and find a basis for it. Carry out the calculation you describe.
(d) From the matrices above we can write down two equations which a vector (x, y, z) in R3 must satisfy to be in the kernel of the transpose of T. Explain carefully. What are the equations? Also explain carefully how we obtain from these equations the 156 23. THE FOUR FUNDAMENTAL SUBSPACES dimension of the kernel of T t and find a basis for it. Carry out the calculation you describe.
(3) Let T : R6 →R5 be the linear transformation whose matrix representation is A = [T] = 1 2 −1 −2 3 0 0 0 0 1 −1 2 2 4 −2 −4 7 −4 0 0 0 −1 1 −2 3 6 −3 −6 7 8 .
You may use the following fact: the reduced row echelon forms of the augmented matrix [A | b ] and of At are B = 1 2 −1 0 0 8 3b1 + 2b2 −b3 0 0 0 1 0 −2 −2b1 + b2 + b3 0 0 0 0 1 −4 −2b1 + b3 0 0 0 0 0 0 b2 + b4 0 0 0 0 0 0 −7b1 + 2b3 + b5 and C = 1 0 0 0 7 0 1 0 −1 0 0 0 1 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , respectively.
Suppose that x = t u v w y z .
(a) What are the free variables of the system Ax = 0 and which are the lead variables?
How do you know?
(b) What is the rank of A? Why?
(c) Write a general solution to the homogeneous equation Ax = 0 as a linear combination of vectors in R6 using the free variables as coefficients. Explain.
(d) Explain how to find the dimension of and a basis for the kernel of T. Do so.
(e) Explain how to find the dimension of and a basis for the range of T. Do so.
(f) What conditions must the vector b = b1 b2 b3 b4 b5 satisfy in order that the nonhomogeneous equation Ax = b have solutions?
(g) Find, if possible, the general solution to the nonhomogeneous equation Ax = 1 −3 2 3 3 .
(Write your answer as a general solution to the homogeneous equation plus a partic-ular solution.) (h) Explain how to find the dimension of and a basis for the range of T t. Do so.
(i) Explain how to find the dimension of and a basis for the kernel of T t. Do so.
(4) Prove the three corollaries to the fundamental theorem of linear algebra 23.1.4 for complex inner product spaces.
23.4. ANSWERS TO ODD-NUMBERED EXERCISES 157 23.4. Answers to Odd-Numbered Exercises (1) (a) 9 −3 i, 3 −i (b) 3, −2 (c) −2, −1 (d) −9 −3 i, −3 −i (3) (a) 2, 1 (b) 2x + y −z = 0 (c) 2 (d) 8, −22, 2, −8 (e) 4 (f) −4, 6 (5) (a) 1 2, 1 2, 3 2 (b) 2, −1, −1 (c) −2, 1, 0, 1 (d) 1 2, −1 2, −3 2 (7) (a) −5, 0 (b) 3, −3 (9) t2, t3 (11) (a) 6, 6, −3, −1, −2, −6 (b) −13, −27, 0, 0 CHAPTER 24 ORTHOGONAL PROJECTIONS 24.1. Background Topics: orthogonal and unitary operators, orthogonal projections 24.1.1. Definition. A linear operator T : V →V on a real inner product space is orthogonal if it is invertible and T t = T −1. A matrix is orthogonal if it is the representation of an orthogonal operator. An operator T on a complex inner product space V is unitary if it is invertible and T ∗= T −1. A matrix is unitary if it is the representation of a unitary operator.
The definitions for orthogonal and unitary matrices given above differs from the ones offered in 19.1.4. In problem 8 you are asked to show that in both cases the definitions are equivalent.
24.1.2. Definition. An operator T on an inner product space V is an isometry if it preserves the distance between vectors. Equivalently, T is an isometry if ∥Tx∥= ∥x∥for every x ∈V .
24.1.3. Definition. Let V be an inner product space and suppose that it is the vector space direct sum of M and N. Then the projection EMN : V →V is an orthogonal projection if M ⊥N (that is if V is the orthogonal direct sum of M and N).
24.1.4. Proposition. A projection E on a complex inner product space V is an orthogonal pro-jection if and only if E is self-adjoint. On a real inner product space a projection is orthogonal if and only if it is symmetric.
For a proof of (the real case of) this result see exercise 4.
24.1.5. Definition. We say that a linear operator T on an inner product space V is positive (and write T ≥0) if ⟨Tx, x⟩≥0 for all x ∈V . If S and T are two linear operators on V , we say that Q dominates (or majorizes) P if Q −P ≥0. In this case we write P ≤Q.
159 160 24. ORTHOGONAL PROJECTIONS 24.2. Exercises (1) The matrix representation of the orthogonal projection operator taking R3 onto the plane x + y + z = 0 is .
(2) Find a vector u = (u1, u2, u3) in C3 such that the matrix 1 m −1 m u1 i m −i m u2 1−i m 1−i m u3 is unitary.
Answer: u = 1 √n (2 + ai, 3 −bi, c + di) where a = , b = , c = , d = , m = , and n = .
(3) The orthogonal projection of the vector (2, 0, −1, 3) on the plane spanned by (−1, 1, 0, 1) and (0, 1, 1, 1) in R4 is 1 5 (1, a, b, a) where a = and b = .
The matrix which implements this orthogonal projection is 1 5 c −d e −d −d e d e e d c d −d e d e where c = , d = , and e = .
(4) Let E be a projection operator on a real inner product space. Below we prove (the real case of) proposition 24.1.4: that E is an orthogonal projection if and only if E = Et. Fill in the missing reasons and steps. Choose reasons from the following list.
(DK) Definition of “kernel”.
(DL) Definition of “linear”.
(DO) Definition of “orthogonal”.
(DOP) Definition of “orthogonal projection”.
(DT) Definition of “transpose”.
(GPa) Problem 5 in chapter 6.
(GPb) Problem 1 in chapter 11.
(GPc) Problem 5 in chapter 18.
(GPd) Problem 6 in chapter 22.
(H1) Hypothesis that M ⊥N.
(H2) Hypothesis that E = Et.
(PIP) Elementary property of Inner Products.
(Ti) Theorem 11.1.2, part (i).
(Tiii) Theorem 11.1.2, part (iii).
(Tiv) Theorem 11.1.2, part (iv).
(VA) Vector space arithmetic (consequences of vector space axioms, etc.) Let E = EMN be a projection operator on a real inner product space V = M ⊕N.
Suppose first that E is an orthogonal projection. Then M ⊥N (reason: ) . If x and y are elements in V , then there exist unique vectors m, p ∈M and n, q ∈N such 24.2. EXERCISES 161 that x = m + n and y = p + q (reason: ) . Then ⟨Ex, y⟩= ⟨E(m + n), p + q⟩ = ⟨Em + En, p + q⟩ reason: and = ⟨0 + En, p + q⟩ reason: and = ⟨En, p + q⟩ reason: = ⟨n, p + q⟩ reason: and = ⟨n, p⟩+ ⟨n, q⟩ reason: = 0 + ⟨n, q⟩ reason: and = ⟨m, q⟩+ ⟨n, q⟩ reason: and = ⟨x, q⟩ reason: = ⟨x, Eq⟩ reason: and = ⟨x, 0 + Eq⟩ reason: = ⟨x, Ep + Eq⟩ reason: and = ⟨x, E(p + q)⟩ reason: and = ⟨x, Etty⟩ reason: = ⟨Etx, y⟩ reason: .
From this we conclude that E = Et (reason: ) .
Conversely, suppose that E = Et. To show that M ⊥N it is enough to show that m ⊥n for arbitrary elements m ∈M and n ∈N.
⟨n, m⟩= ⟨En, m⟩ reason: and = ⟨n, Etm⟩ reason: = ⟨n, Em⟩ reason: = ⟨n, 0⟩ reason: and = 0 reason: Thus m ⊥n (reason: ) .
Note: Of course, the complex inner product space version of the preceding result says that if E is a projection operator on a complex inner product space, then E is an orthogonal projection if and only if it is self-adjoint.
(5) Let P be the orthogonal projection of R3 onto the subspace spanned by the vectors (1, 0, 1) and (1, 1, −1). Then [P] = 1 6 a b c b b −b c −b a where a = , b = , and c = .
(6) Find the image of the vector b = (1, 2, 7) under the orthogonal projection of R3 onto the column space of the matrix A = 1 1 2 −1 −2 4 .
Answer: ( , , ).
(7) Let u = (3, −1, 1, 4, 2) and v = (1, 2, −1, 0, 1). Then the orthogonal projection of u onto v is ( , , , , ).
162 24. ORTHOGONAL PROJECTIONS (8) Let u = (8, √ 3, √ 7, −1, 1) and v = (1, −1, 0, 2, √ 3). Then the orthogonal projection of u onto v is a b v where a = and b = .
(9) Let u = (5, 4, 3, 1 2) and v = (1, 2, 0, −2). Then the orthogonal projection of u onto v is a b v where a = and b = .
(10) Find the point q in R3 on the ray connecting the origin to the point (2, 4, 8) which is closest to the point (1, 1, 1).
Answer: q = 1 3 ( , , ) .
(11) Let e1 = 2 3, 2 3, −1 3 and e2 = −1 3, 2 3, 2 3 be vectors in R3. Notice that {e1, e2} is an orthonormal set.
(a) Find a vector e3 whose first coordinate is positive such that B = {e1, e2, e3} is an orthonormal basis for R3. Answer: 1 3 ( , , ).
(b) Suppose that x is a vector in R3 whose Fourier coefficients with respect to the basis B are: ⟨x, e1⟩= −2; ⟨x, e2⟩= −1; and ⟨x, e3⟩= 3. Then x = ( , , ).
(c) Let y be a vector in R3 whose Fourier coefficients with respect to B are ⟨y, e1⟩= q 8 − √ 37; ⟨y, e2⟩= q 5 − √ 13; and ⟨y, e3⟩= q 3 + √ 13 + √ 37.
Then the length of the vector y is .
(d) The orthogonal projection of the vector b = (0, 3, 0) onto the plane spanned by e1 and e2 is 2 3 ( , , ).
(e) The orthogonal projection of the vector b = (0, 3, 0) onto the line spanned by e3 is 1 3 ( , , ).
(f) What vector do you get when you add the results of the projections you found in parts (d) and (e)? Answer: ( , , ).
24.3. PROBLEMS 163 24.3. Problems (1) Prove that an operator T : V →V on a finite dimensional real inner product space V is orthogonal if and only if it is an isometry. Similarly, on a finite dimensional complex inner product space an operator is unitary if and only if it is an isometry.
(2) Prove that an operator T : V →V on a finite dimensional real inner product space V is orthogonal if and only if T tT = I.
What is the corresponding necessary and suffi-cient condition on a finite dimensional complex inner product space for an operator to be unitary?
(3) Show that if an operator U on a complex inner product space is both Hermitian and unitary, then σ(U) ⊆{−1, 1}.
(4) Let P and Q be orthogonal projections on a real inner product space. Show that their sum P + Q is an orthogonal projection if and only if PQ = QP = 0. Hint. Use proposi-tion 24.1.4.
(5) Explain in detail how to find the matrix which represents the orthogonal projection of R3 onto the plane x + y −2z = 0. Carry out the computation you describe.
(6) Let P and Q be orthogonal projection operators on a real inner product space V .
(a) Show that the operator PQ is an orthogonal projection if and only if P commutes with Q.
(b) Show that if P commutes with Q, then ran(PQ) = ran P ∩ran Q .
Hint. To show that ran P ∩ran Q ⊆ran(PQ) start with a vector y in ran P ∩ran Q and examine PQy.
(7) Let P and Q be orthogonal projections on an inner product space V . Prove that the following are equivalent: (a) P ≤Q; (b) ∥Px∥≤∥Qx∥for all x ∈V ; (c) ran P ⊆ran Q; (d) QP = P; and (e) PQ = P.
Hint. First show that (d) and (e) are equivalent. Then show that (a) ⇒(b) ⇒(c) ⇒ (d) ⇒(a). To prove that (b) ⇒(c) take an arbitrary element x in the range of P; show that ∥Qx∥= ∥x∥and that consequently ∥(I −Q)x∥= 0. To prove that (d) ⇒(a) show that (I −P)Q is an orthogonal projection; then consider ∥(I −P)Q∥2.
(8) In 19.1.4 and 24.1.1 the definitions for unitary matrices differ. Show that they are, in fact, equivalent. Argue that the same is true for the definitions given for orthogonal matrices.
164 24. ORTHOGONAL PROJECTIONS 24.4. Answers to Odd-Numbered Exercises (1) 1 3 2 −1 −1 −1 2 −1 −1 −1 2 (3) 3, 4, 3, 1, 2 (5) 5, 2, 1 (7) 2 7, 4 7, −2 7, 0, 2 7 (9) 4, 3 (11) (a) 2, −1, 2 (b) 1, −3, 2 (c) 4 (d) 1, 4, 1 (e) −2, 1, −2 (f) 0, 3, 0 CHAPTER 25 LEAST SQUARES APPROXIMATION 25.1. Background Topics: least squares approximation.
165 166 25. LEAST SQUARES APPROXIMATION 25.2. Exercises (1) Let A = 1 1 2 −1 −2 4 .
(a) Find an orthonormal basis {e1, e2, e3} for R3 such that {e1, e2} spans the column space of A.
e1 = 1 n( a , b , −b ) e2 = 1 n( b , a , b ) e3 = 1 n( b , −b , −a ) where a = , b = , and n = .
(b) To which of the four fundamental subspaces of A does e3 belong?
Answer: e3 belongs to the of A.
(c) What is the least squares solution to Ax = b when b = (1, 2, 7)?
Answer: b x = ( , ) .
(2) Find the best least squares fit by a straight line to the following data: x = 1 when t = −1; x = 3 when t = 0; x = 2 when t = 1; and x = 3 when t = 2.
Answer: x = + t.
(3) At times t = −2, −1, 0, 1, and 2 the data y = 4, 2, −1, 0, and 0, respectively, are observed.
Find the best line to fit this data. Answer: y = Ct + D where C = and D = .
(4) The best (least squares) line fit to the data: y = 2 at t = −1; y = 0 at t = 0; y = −3 at t = 1; y = −5 at t = 2 is y = −a 10 −b 5 t where a = and b = .
(5) Consider the following data: y = 20 when t = −2; y = 6 when t = −1; y = 2 when t = 0; y = 8 when t = 1; y = 24 when t = 2. Find the parabola which best fits the data in the least squares sense.
Answer: y = C + Dt + Et2 where C = , D = , and E = .
(6) Consider the following data: y = 2 when t = −1; y = 0 when t = 0; y = −3 when t = 1; y = −5 when t = 2. Find the parabola which best fits the data in the least squares sense.
Answer: y = C + Dt + Et2 where C = , D = , and E = .
(7) Find the plane 50z = a + bu + cv which is the best least squares fit to the following data: z = 3 when u = 1 and v = 1; z = 6 when u = 0 and v = 3; z = 5 when u = 2 and v = 1; z = 0 when u = 0 and v = 0.
Answer: a = ; b = ; c = .
(8) Consider the following data: y = 4 at t = −1; y = 5 at t = 0; y = 9 at t = 1.
(a) Then the best (least squares) line which fits the data is y = c + dt where c = and d = .
(b) The orthogonal projection of b = (4, 5, 9) onto the column space of A = 1 −1 1 0 1 1 is ( , , ) .
(9) The best least squares solution to the following (inconsistent) system of equations u = 1 v = 1 u + v = 0 is u = and v = .
25.3. PROBLEMS 167 25.3. Problems (1) Explain in detail how to use matrix methods to find the best (least squares) solution to the following (inconsistent) system of equations u = 1 v = 1 u + v = 0 . Carry out the computation you describe.
(2) The following data y are observed at times t: y = 4 when t = −2; y = 3 when t = −1; y = 1 when t = 0; and y = 0 when t = 2.
(a) Explain how to use matrix methods to find the best (least squares) straight line approximation to the data. Carry out the computation you describe.
(b) Find the orthogonal projection of y = (4, 3, 1, 0) on the column space of the matrix A = 1 −2 1 −1 1 0 1 2 .
(c) Explain carefully what your answer in (b) has to do with part (a).
(d) At what time does the largest error occur? That is, when does the observed data differ most from the values your line predicts?
168 25. LEAST SQUARES APPROXIMATION 25.4. Answers to Odd-Numbered Exercises (1) (a) 1, 2, 3 (b) left nullspace (c) 1, 2 (3) −1, 1 (5) 2, 1, 5 (7) −6, 73, 101 (9) 1 3 , 1 3 Part 7 SPECTRAL THEORY OF INNER PRODUCT SPACES CHAPTER 26 SPECTRAL THEOREM FOR REAL INNER PRODUCT SPACES 26.1. Background Topics: the spectral theorem for finite dimensional real inner product spaces.
26.1.1. Definition. An operator T on a finite dimensional real inner product space with an orthonormal basis is orthogonally diagonalizable if there exists an orthogonal matrix which diagonalizes T.
The following theorem (together with its analog for complex spaces) is the fundamental struc-ture theorem for inner product spaces. It says that any symmetric operator on a finite dimensional real inner product space can be written as a linear combination of orthogonal projections. The coefficients are the eigenvalues of the operator and the ranges of the orthogonal projections are the eigenspaces of the operator.
26.1.2. Theorem (Spectral Theorem for Finite Dimensional Real Inner Product Spaces). Let T be a symmetric operator on a finite dimensional real inner product space V , and λ1, . . . , λk be the (distinct) eigenvalues of T.
For each j let Mj be the eigenspace associated with λj and Ej be the projection of V onto Mj along M1 + · · · + Mj−1 + Mj+1 + · · · + Mk. Then T is orthogonally diagonalizable, the eigenspaces of T are mutually orthogonal, each Ej is an orthogonal projection, and the following hold: (i) T = λ1E1 + · · · + λkEk, (ii) I = E1 + · · · + Ek, and (iii) EiEj = 0 when i ̸= j.
171 172 26. SPECTRAL THEOREM FOR REAL INNER PRODUCT SPACES 26.2. Exercises (1) Let T be the operator on R3 whose matrix representation is 1 3 −2 3 −2 3 −2 3 5 6 −7 6 −2 3 −7 6 5 6 .
(a) Find the characteristic polynomial and minimal polynomial for T.
Answer: cT (λ) = .
mT (λ) = .
(b) The eigenspace M1 associated with the smallest eigenvalue λ1 is the span of (1 , , ).
(c) The eigenspace M2 associated with the middle eigenvalue λ2 is the span of ( , , −1).
(d) The eigenspace M3 associated with the largest eigenvalue λ3 is the span of ( , 1 , ).
(e) Find the (matrix representations of the) orthogonal projections E1, E2, and E3 onto the eigenspaces M1, M2, and M3, respectively.
Answer: E1 = 1 m a a a a a a a a a ; E2 = 1 n b −c −c −c a a −c a a ; E3 = 1 2 d d d d a −a d −a a where a = , b = , c = , d = , m = , and n = .
(f) Write T as a linear combination of the projections found in (e).
Answer: [T] = E1 + E2 + E3.
(g) Find an orthogonal matrix Q (that is, a matrix such that Qt = Q−1) which diagonal-izes T. What is the associated diagonal form Λ of T?
Answer: Q = a √ b c √ bc 0 a √ b −a √ bc a √c a √ b −a √ bc −a √c and Λ = λ 0 0 0 µ 0 0 0 ν where a = , b = , c = , λ = , µ = , and ν = .
(2) Let T be the operator on R3 whose matrix representation is 2 2 1 2 2 −1 1 −1 −1 .
(a) The eigenspace M1 associated with the smallest eigenvalue λ1 is the span of ( 1 , , ).
(b) The eigenspace M2 associated with the middle eigenvalue λ2 is the span of ( 1 , , ).
(c) The eigenspace M3 associated with the largest eigenvalue λ3 is the span of ( 1 , , ).
(d) Find the (matrix representations of the) orthogonal projections E1, E2, and E3 onto the eigenspaces M1, M2, and M3, respectively.
Answer: E1 = 1 mn a −a −b −a a b −b b c ; E2 = 1 m a −a a −a a −a a −a a ; E3 = 1 n a a d a a d d d d where a = , b = , c = , d = , m = , and n = .
(e) Write T as a linear combination of the projections found in (d).
Answer: [T] = E1 + E2 + E3.
(f) Find an orthogonal matrix Q (that is, a matrix such that Qt = Q−1) which diagonal-izes T. What is the associated diagonal form Λ of T?
26.2. EXERCISES 173 Answer: Q = a √ bc a √ b a √c −a √ bc −a √ b a √c − c √ bc a √ b 0 and Λ = λ 0 0 0 µ 0 0 0 ν where a = , b = , c = , λ = , µ = , and ν = .
174 26. SPECTRAL THEOREM FOR REAL INNER PRODUCT SPACES 26.3. Problem (1) Let A = 1 −4 2 −4 1 −2 2 −2 −2 .
(a) Does A satisfy the hypotheses of the spectral theorem 26.1.2 for symmetric operators on a finite dimensional real inner product space? Explain.
(b) Explain how to find an orthogonal matrix which diagonalizes the matrix A. Carry out the computation you describe.
(c) Explain in careful detail how to write the matrix A in part (b) as a linear combination of orthogonal projections. Carry out the computations you describe.
26.4. ANSWERS TO THE ODD-NUMBERED EXERCISE 175 26.4. Answers to the Odd-Numbered Exercise (1) (a) λ3 −2λ2 −λ + 2 (or (λ + 1)(λ −1)(λ −2) ); λ3 −2λ2 −λ + 2 (or (λ + 1)(λ −1)(λ −2) ) (b) 1, 1 (c) 2, −1 (d) 0, −1 (e) 1, 4, 2, 0, 3, 6 (f) −1, 1, 2 (g) 1, 3, 2, −1, 1, 2 CHAPTER 27 SPECTRAL THEOREM FOR COMPLEX INNER PRODUCT SPACES 27.1. Background Topics: the spectral theorem for finite dimensional complex inner product spaces.
27.1.1. Definition. An operator T on a finite dimensional real inner product space with an orthonormal basis is unitarily diagonalizable if there exists an orthogonal matrix which diag-onalizes T.
27.1.2. Theorem (Spectral Theorem for Finite Dimensional Complex Inner Product Spaces). Let T be a normal operator on a finite dimensional complex inner product space V , and λ1, . . . , λk be the (distinct) eigenvalues of T. For each j let Mj be the eigenspace associated with λj and Ej be the projection of V onto Mj along M1 + · · · + Mj−1 + Mj+1 + · · · + Mk. Then T is unitarily diagonalizable, the eigenspaces of T are mutually orthogonal, each Ej is an orthogonal projection, and the following hold: (i) T = λ1E1 + · · · + λkEk, (ii) I = E1 + · · · + Ek, and (iii) EiEj = 0 when i ̸= j.
27.1.3. Theorem. Let T be an operator on a finite dimensional complex inner product space V .
Then the following are equivalent: (1) T is normal; (2) T is unitarily diagonalizable; and (3) V has an orthonormal basis consisting of eigenvectors of T.
177 178 27. SPECTRAL THEOREM FOR COMPLEX INNER PRODUCT SPACES 27.2. Exercises (1) Let A = 2 1 + i 1 −i 3 .
(a) Use the spectral theorem 27.1.2 to write A as a linear combination of orthogonal projections.
Answer: A = αE1 + βE2 where α = , β = , E1 = 1 3 2 −1 −i , and E2 = 1 3 1 1 + i .
(b) Find a square root of A.
Answer: √ A = 1 3 4 1 + i .
(2) Let T be the operator on C2 whose matrix representation is 0 1 −1 0 .
(a) The eigenspace V1 associated with the eigenvalue −i is the span of ( 1 , ).
(b) The eigenspace V2 associated with the eigenvalue i is the span of ( 1 , ).
(c) The (matrix representations of the) orthogonal projections E1 and E2 onto the eigenspaces V1 and V2, respectively, are E1 = a b −b a ; and E2 = a −b b a where a = and b = .
(d) Write T as a linear combination of the projections found in (c).
Answer: [T] = E1 + E2.
(e) A unitary matrix U which diagonalizes [T] is a a −b b where a = and b = .
The associated diagonal form Λ = U ∗[T]U of [T] is .
(3) Let N = 1 3 4 + 2i 1 −i 1 −i 1 −i 4 + 2i 1 −i 1 −i 1 −i 4 + 2i .
(a) The matrix N is normal because NN∗= N∗N = a b b b a b b b a where a = and b = .
(b) Thus according to the spectral theorem 27.1.2 N can be written as a linear combination of orthogonal projections. Written in this form N = λ1E1 + λ2E2 where λ1 = , λ2 = , E1 = a a a a a a a a a , and E2 = b −a −a −a b −a −a −a b where a = and b = .
(c) A unitary matrix U which diagonalizes N is a −b −c a b −c a d 2c where a = , b = , c = , and d = . The associated diagonal form Λ = U ∗NU of N is .
27.2. EXERCISES 179 (4) Let T be an operator whose matrix representation is 1 2 −1 −1 .
(a) Regarded as an operator on R2 is T triangulable?
. As an operator on R2 is it diagonalizable?
.
(b) Show that T regarded as an operator on C2 is diagonalizable by finding numbers c and d such that the matrix S = −2 −2 c d is invertible and S−1TS is diagonal.
Answer: c = and d = .
(c) Show that despite being diagonalizable (as an operator on C2) T is not normal.
Answer: TT ∗= ̸= = T ∗T.
(d) Explain briefly why the result of part (c) does not contradict Theorem 27.1.3.
(5) Let T be the operator on C3 whose matrix representation is 1 6 8 −i 5 −2i 2 + 4i −5 + 2i 8 −i −4 + 2i −2 −4i −4 + 2i 14 + 2i .
(a) Find the characteristic polynomial and minimal polynomial for T.
Answer: cT (λ) = .
mT (λ) = .
(b) The eigenspace M1 associated with the real eigenvalue λ1 is the span of (1 , , ).
(c) The eigenspace M2 associated with the complex eigenvalue λ2 with negative imaginary part is the span of ( 1 , , ).
(d) The eigenspace M3 associated with the remaining eigenvalue λ3 is the span of ( 1 , , ).
(e) Find the (matrix representations of the) orthogonal projections E1, E2, and E3 onto the eigenspaces M1, M2, and M3, respectively.
Answer: E1 = 1 m 1 −b bc b a −c −bc −c d ; E2 = 1 n 1 b e −b a e e e e ; E3 = 1 p 1 −b −b b a a b a a where a = , b = , c = , d = , e = , m = , n = , and p = .
(f) Write T as a linear combination of the projections found in (e).
Answer: [T] = E1 + E2 + E3.
(g) Find a unitary matrix U which diagonalizes T. What is the associated diagonal form Λ of T?
Answer: U = a √ bc a √ b a √c d √ bc −d √ b d √c −bd √ bc e √ b d √c and Λ = λ 0 0 0 µ 0 0 0 ν where a = , b = , c = , d = , e = , λ = , µ = , and ν = .
(h) The operator T is normal because TT ∗= T ∗T = 1 6 a −bc 2bc bc a −2b −2bc −2b d where a = , b = , c = , and d = .
180 27. SPECTRAL THEOREM FOR COMPLEX INNER PRODUCT SPACES (6) Let T be the operator on C3 whose matrix representation is 1 3 5 + 2i 2 −i 2 −i 2 −i 5 −i 2 + 2i 2 −i 2 + 2i 5 −i .
(a) Find the characteristic polynomial and minimal polynomial for T.
Answer: cT (λ) = .
mT (λ) = .
(b) The eigenspace M1 associated with the real eigenvalue λ1 is the span of (1 , , ).
(c) The eigenspace M2 associated with the complex eigenvalue λ2 with negative imaginary part is the span of ( , , −1).
(d) The eigenspace M3 associated with the remaining eigenvalue λ3 is the span of ( , −1 , ).
(e) Find the (matrix representations of the) orthogonal projections E1, E2, and E3 onto the eigenspaces M1, M2, and M3, respectively.
Answer: E1 = 1 m a a a a a a a a a ; E2 = 1 n b b b b c −c b −c c ; E3 = 1 6 d −e −e −e a a −e a a where a = , b = , c = , d = , e = , m = , and n = .
(f) Write T as a linear combination of the projections found in (e).
Answer: [T] = E1 + E2 + E3.
(g) Find an orthogonal matrix Q (that is, a matrix such that Qt = Q−1) which diagonal-izes T. What is the associated diagonal form Λ of T?
Answer: Q = a √ b 0 c √ bc a √ b a √c −a √ bc a √ b −a √c −a √ bc and Λ = λ 0 0 0 µ 0 0 0 ν where a = , b = , c = , λ = , µ = , and ν = .
27.3. PROBLEMS 181 27.3. Problems (1) Let N be a normal operator on a finite dimensional complex inner product space V. Show that ∥Nx∥= ∥N∗x∥for all x ∈V .
(2) Let N be a normal operator on a complex finite dimensional inner product space V. Show that if λ1, . . . , λk are the eigenvalues of N, then λ1, . . . , λk are the eigenvalues of N∗.
(3) Let T be as in exercise 4. Show by direct computation that there is no invertible 2 × 2 matrix S = a b c d of real numbers such that S−1TS is upper triangular.
182 27. SPECTRAL THEOREM FOR COMPLEX INNER PRODUCT SPACES 27.4. Answers to Odd-Numbered Exercises (1) (a) 1, 4, −1 + i, 1, 1 −i, 2 (b) 1 −i, 5 (3) (a) 8 3, 2 3 (b) 2, 1 + i, 1 3, 2 3 (c) 1 √ 3 , 1 √ 2 , 1 √ 6 , 0, 2 0 0 0 1 + i 0 0 0 1 + i (5) (a) λ3 −5λ2 +8λ−6 (or (λ2 −2λ+2)(λ−3) ); λ3 −5λ2 +8λ−6 (or (λ2 −2λ+2)(λ−3) ) (b) i, −2i (c) −i, 0 (d) i, i (e) 1, i, 2, 4, 0, 6, 2, 3 (f) 3, 1 −i, 1 + i (g) 1, 2, 3, i, 0, 3, 1 −i, 1 + i (h) 19, 7, i, 40 Bibliography 1. Howard Anton and Chris Rorres, Elementary Linear Algebra: Applications Version, eighth ed., John Wiley and Sons, New York, 2000. vii, 51, 85 2. Robert A. Beezer, A First Course in Linear Algebra, 2004, vii 3. Przemyslaw Bogacki, Linear Algebra Toolkit, 2005, vii 4. William C. Brown, A second Course in Linear Algebra, John Wiley, New York, 1988. vii 5. Charles W. Curtis, Linear Algebra: An Introductory Approach, Springer, New York, 1984. vii 6. Paul R. Halmos, Finite-Dimensional Vector Spaces, D. Van Nostrand, Princeton, 1958. vii 7. Jim Hefferon, Linear Algebra, 2006, vii, 51 8. Kenneth Hoffman and Ray Kunze, Linear Algebra, second ed., Prentice Hall, Englewood Cliffs,N.J., 1971. vii, 93, 149 9. Steven Roman, Advanced Linear Algebra, second ed., Springer-Verlag, New York, 2005. vii, 127 10. Gilbert Strang, Linear Algebra and its Applications, second ed., Academic Press, New York, 1980. vii, 133 11. Eric W. Weisstein, MathWorld, A Wolfram Web Resource, vii 12. Wikipedia, Wikipedia, The Free Encyclopedia, vii 183 Index ∡(x, y) (angle between x and y), 25, 122 x · y (inner product of x and y), 121 M ⊥N (two sets are orthogonal), 122 x ⊥y (two vectors are orthogonal), 122 M ⪯V (M is a subspace of V ), 39 ⟨x, y⟩(inner product), 121 [T] (matrix representation of T), 71 [A, B] (commutator of A and B), 9 TS (notation for composition of linear maps), 61 ⟨x, y⟩(inner product of x and y), 25, 121 x · y (inner product of x and y), 25, 121 M ⊕N (direct sum of M and N), 39 ∥x∥(norm of a vector x), 121 At (transpose of A), 9 M ⊥(the orthogonal complement of M), 122 action of a matrix, 15 additive inverses, 33 adjoint of a linear map, 145 angle, 25, 122 annihilating polynomial, 87 associative, 33 basis, 55 orthonormal, 127 standard, 55 bijective, 61 bounded function, 39 Cn as an inner product space, 121 C([a, b]) as an inner product space, 122 Cauchy-Schwarz inequality, 25, 122 Cayley-Hamilton theorem, 87 characteristic polynomial, 83 cofactor, 16 column rank, 149 column index, 9 column space, 149 combination linear, 47 commutative, 33 diagram, 68 commutator, 9 commute, 9, 68 complement orthogonal, 122 complementary subspaces, 39 conjugate Hermitian, 145 linear, 121 transpose, 145 of a linear map, 145 of a matrix, 145 continuous uniformly, 122 continuously differentiable, 64 cross product, 25 definite negative, 133 positive, 133 dependence linear, 47 determinant, 16 det A (determinant of A), 16 diagonal matrix, 87 diagonalizable, 87 conditions to be, 88 orthogonally, 171 part of an operator, 105 unitarily, 177 diagonalizing matrix, 87 diagram, 68 commutative, 68 differentiable continuously, 64 differential equations, 97 dimension, 55 dim V (dimension of V ), 55 direct sum, 39 orthogonal, 127 dominate, 159 EMN (projection along M onto N), 77 eigenspace, 83 generalized, 105 eigenvalue, 83 eigenvector, 83 electrical networks problem on, 51 185 186 INDEX equivalent row, 15 even function, 41 exact sequence, 67 expansion Laplace, 16 factorization QR-, 127 finite dimension, 55 Fourier series, 127 function even, 41 odd, 41 square integrable, 123 uniformly continuous, 122 functional linear, 61 fundamental theorem of linear algebra, 149, 153 generalized eigenspace, 105 Hermitian conjugate, 145 operator, 145 Hessian matrix, 139 IV (identity operator on V ), 62 identity additive, 33 identity map, 62 indefinite, 133 independence linear, 47 index column, 9 row, 9 inequality (Cauchy-)Schwarz, 122 infinite dimensional, 55 injective, 61 inner product, 25, 121 space Cn as a, 121 Rn as a, 121 C([a, b]) as a, 122 l2 as a, 122 integrable, 39 inverse additive, 33 left, 62 right, 62 invertible, 62 isometry, 159 isomorphic, 62 isomorphism, 62 ker T (the kernel of T), 61 kernel, 61 L(V ) linear operators on V , 61 L(V, W) linear maps between vector spaces, 61 l2 as an inner product space, 122 square summable sequences, 122 Laplace expansion, 16 leading principal submatrix, 133 left inverse, 12, 62 nullspace, 149 length, 25 Lie bracket, 9 linear, 61 combination, 47 trivial, 47 conjugate, 121 dependence, 47 functional, 61 independence, 47 map, 61 adjoint of a, 145 conjugate transpose of a, 145 Hermitian conjugate of a, 145 transpose of a, 145 operator, 61 sesqui-, 121 transformation, 61 majorize, 159 map linear, 61 Markov matrix, 97 matrix conjugate transpose of a, 145 diagonal, 87 diagonalizing, 87 Hessian, 139 Markov, 97 nilpotent, 105 nonsingular, 33 normal, 145 notation for, 9 orthogonal, 127, 159 representation of an linear map, 71 second derivative, 139 standard, 71 symmetric, 9 transpose of a, 9 unitary, 127, 159 upper triangular, 9 maximal linearly independent set, 57 orthonormal set, 127 minimal polynomial, 87 existence and uniqueness of, 91 minimal spanning set, 57 minor, 16 monic polynomial, 87 negative definite, 133 INDEX 187 networks problem on, 51 nilpotent, 88, 105 part of an operator, 105 nonsingular, 33 norm, 25, 121 normal matrix, 145 operator, 145 nullity, 61 nullspace, 61, 149 left, 149 odd function, 41 one-to-one, 61 correspondence, 61 onto, 61 operator adjoint of an, 145 diagonalizable part of an, 105 Hermitian, 145 linear, 61 nilpotent, 105 nilpotent part of an, 105 normal, 145 orthogonal, 159 positive, 159 self-adjoint, 145 symmetric, 145 transpose of an, 145 unitary, 159 orthogonal, 122 complement, 122 diagonalization, 171 direct sum, 127 matrix, 127, 159 operator, 159 projection, 159 orthonormal, 127 basis, 127 perpendicular, 122 pivoting, 3 point spectrum, 83 polarization identity, 125 polynomial annihilating, 87 characteristic, 83 minimal, 87 existence and uniqueness of, 91 monic, 87 positive definite, 133 operator, 159 principal submatrix, 133 product cross, 25 inner, 25, 121 projection along one subspace onto another, 77 orthogonal, 159 QR-factorization, 127 Rn as an inner product space, 121 ran T (the range of T), 61 range, 61 rank column, 149 of a linear map, 61 row, 149 representation matrix, 71 Riemann integrable function, 39 right inverse, 12, 62 row equivalent matrices, 15 operations, 3 rank, 149 row index, 9 row space, 149 scaling, 3 Schwarz inequality, 25, 122 second derivative matrix, 139 test, 139 self-adjoint operator, 145 sequence exact, 67 sesquilinear, 121 short exact sequence, 67 σ(T) (spectrum of T), 83 similar, 87 space vector, 33 span, 47 span A (the span of the set A), 47 spectral mapping theorem, 83 spectral theorem for finite dimensional complex inner product spaces, 177 for finite dimensional real inner product spaces, 171 for finite dimensional vector spaces, 93 spectrum, 83 square integrable function, 123 square summable, 122 standard basis for Rn, 55 matrix for a linear map, 71 submatrix leading principal, 133 subspace, 39 subspaces complementary, 39 sum direct, 39 summable square, 122 surjective, 61 188 INDEX swapping, 3 symmetric matrix, 9 operator, 145 system of differential equations, 97 trace, 9 tr A (trace of A), 9 transformation linear, 61 transpose conjugate, 145 of a matrix, 9 of an operator, 145 triangulable, 88 conditions to be, 88 triangular upper, 9 trivial linear combination, 47 uniformly continuous, 122 unitary diagonalization, 177 matrix, 127, 159 operator, 159 upper triangular, 9 upper triangular, 88 vector space, 33 complex, 33 real, 33
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(PDF) Protein Electrophoresis in Clinical Diagnosis
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Title
Abstract
Limitations of Iep
Selection of the Dilution of Patient's Serum
Limitations of Immunofixation
Overview of the Electrophoretic Strip
Interpretation of the Individual Patient's Sample
Electrophoretic Methods to Study CSF
Case Studies for Interpretation 305
Case Studies for Interpretation 323
References
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Protein Electrophoresis in Clinical Diagnosis
John Paul Pulido
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Abstract
AI
Protein electrophoresis is an essential tool in clinical diagnostics, particularly for analyzing serum, urine, and cerebrospinal fluid. This revised edition updates significant advancements in techniques since the previous versions, including automated methods and enhanced resolution systems like capillary zone electrophoresis. The book also discusses new reagents and their applications in monitoring monoclonal free light chains, aiding in the diagnosis of various plasma cell disorders.
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Screening and measurement of monoclonal (M) proteins are commonly performed using capillary zone electrophoresis (CZE). The identification of M-protein or monoclonal component (CM) is an essential requirement for diagnosis and monitoring of monoclonal gammopathies. The detection of CM has been largely improved by CZE. Capillary electrophoresis estimates CM more accurately, because absence of variation due to different dye binding affinities of proteins as instead seen with agarose gel electrophoresis. However, interferences can be present in CZE. This occurs because all substances absorbing at 200 nm can be identified. Recognition and handling of specimens exhibiting such interferences is essential to ensure accurate diagnostic and patient safety. We herein report on an unusual case of serum protein electrophoresis, to highlight that laboratory staff must be aware of and familiarise with the information provided by laboratory instruments. For example, in the case of serum indices, a...
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Warde Report 287 water, thermal stability of 188
Wegener's granulomatosis 128 Western blotting 58, 275, 275, 276
Wiskott-Aldrich syndrome 133, 300, 307
X-linked agammaglobulinemia (XLA) 10, 135-6, 146, 300 zone electrophoresis 5-8, 11 early clinical use 9 zwitterion 1, 1
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Written By Uma A V
Last Modified 25-01-2023
Section Formula in 3D: Internal and External Division Formulae, Applications
Section Formula in 3D: Points are the most basic characteristics of geometry, and it has no dimensions. Two points can be connected using exactly one straight line. A point on this straight line divides the line into two parts. In two-dimensional geometry, the coordinates of the point that divide a line segment internally or externally in a particular ratio can be calculated using the coordinates of the endpoints.
A similar formula can be extended into three-dimensional space. It has a lot of applications in three-dimensional geometry, such as the ratio in which a point in 3D space divides a line segment and to find the collinearity of points.
Internal Division Formula in 2D
Let (B(x,\,y)) divides (AC) in the ratio (m : n), where (A(x_1,\,y_1)) and (C(x_2,\,y_2)) internally.
Then, the section formula in (2D) for internal division ( \to B\left( {x,\,y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right))
External Division Formula in 2D
Let (B(x,\,y)) divides (AC) where (A(x_1,\,y_1)) and (C(x_2,\,y_2)) externally in the ratio (m ∶ n).
Then, the section formula in (2D) for external division ( \to B\left( {x,\,y} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\,\frac{{m{y_2} – n{y_1}}}{{m – n}}} \right)).
Derivation of Section Formula for Internal Division in 3D
In three-dimensional geometry, let the point (R(x,\,y,\,z)) divides the line segment (\overline {PQ} ) joining the points (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) in the ratio (m : n).
That is, (\frac {PR}{QR} = \frac {m}{n}).
Step 1: Draw perpendiculars to the (XY) plane from the points (P,\,Q)and (R)to intersect the (XY)plane at the points (L,\,M,)and (N), respectively, such that (PL\parallel RN\parallel QM).
Step 2: Join the points (L,\,N,)and (M). Draw a line parallel to (LM)through (R). Let this line intersect (QM)at (T). Extend the line (PL)to intersect the parallel line at (S).
Since (ST\parallel LM) and (PL\parallel RN\parallel QM), the quadrilaterals (LNRS)and (NMTR)are parallelograms.
Also, by (AA) Similarity Theorem, (\Delta PSR \sim \Delta QTR).
Corresponding sides of similar triangles are proportional.
Also, you already have (\frac{{PR}}{{QR}} = \frac{m}{n})
Thus, (\frac{{PR}}{{QR}} = \frac{{RS}}{{RT}} = \frac{{SP}}{{QT}} = \frac{m}{n})
By the construction of the line segments,
(SP = SL – PL)
( = RN – PL)
( = z – z_1)
and
(QT = QM – TM)
( = QM – RM)
( = z_2 – z)
Using these measures:
(\frac{{SP}}{{QT}} = \frac{m}{n} = \frac{{z – {z_1}}}{{{z_2} – z}} \to nz – n{z_1} = m{z_2} – mz)
This can be simplified as (z = \frac{{m{z_2} + n{z_1}}}{{m + n}}).
Now, if we start with the perpendiculars to the (XZ)plane from the point (P,\,Q,\,R)where (R)divides the line segment (\overline {PQ} )in the ratio (m : n),
and following the same arguments, we get the (y-)coordinate of (R)as (y = \frac{{m{y_2} + n{y_1}}}{{m + n}})
We can draw perpendiculars (PL,\,RN,) and (QM)to the (YZ-)plane to get the (x-)coordinate of the point (R)as (x = \frac{{m{x_2} + n{x_1}}}{{m + n}}).
Therefore, we have the coordinates of the point (R)as:
(R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right))
Since the point (R) internally divides the line segment (\overline {PQ} ) in the ratio (m : n), this is called the section formula for internal division.
Section Formula for External Division in 3D
Consider a line segment (\overline {PQ} ) and the point (R)externally divides the line segment in the ratio (m : n).
Then the coordinates of the point (R) can be calculated by replacing (n) with (-n).
That is, the coordinates of (R)can be written as:
(R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right))
Special Cases of Section Formula
There are some special cases of the formula for specific values of (m)and (n).
Midpoint Formula
A midpoint bisects a line segment. When (m = n), the point (R) divides the line segment (\overline {PQ} ) into two congruent segments. In other words, (R) would be the midpoint of (\overline {PQ} ).
So, the values of (m) and (n) are equal; the ratio (m : n) becomes (1 ∶ 1).
The formula to find the coordinates of (R) is written as:
(R\left( {x,\,y,\,z} \right) = \;\left( {\frac{{{x_1} + {x_2}}}{2},\;\,\frac{{{y_1} + {y_2}}}{2},\,\frac{{{z_1} + {z_2}}}{2}} \right))
Points of Trisection
This formula is used to find the coordinates of the points that trisect a line segment. Consider the line segment (\overline {PQ} )that joins the points (P\left( {x_1,\,y_1,\,z_1}\right))and (Q\left( {x_2,\,y_2,\,z_2}\right))and the points (R)and (S)that divide (\overline {PQ} )into three equal parts.
Point (R) divides the line segment (\overline {PQ} ) in the ratio (1 : 2), and point (S) divides the line segment (\overline {PQ} ) in the ratio (2 ∶ 1).
Applying the section formula for (m = 1) and (n = 2), the coordinates of (R) is calculated as,
(R\left( {x,\,y,\,z} \right) = \left( {\frac{{\left( 1 \right){x_2} + \left( 2 \right){x_1}}}{{1 + 2}},\,\frac{{\left( 1 \right){y_2} + \left( 2 \right){y_1}}}{{1 + 2}},\,\frac{{\left( 1 \right){z_2} + \left( 2 \right){z_1}}}{{1 + 2}}} \right)) ( = \left( {\frac{{{x_2} + 2{x_1}}}{3},\,\frac{{{y_2} + 2{y_1}}}{3},\,\frac{{{z_2} + 2{z_1}}}{3}} \right))
Similarly, the coordinates of the point (S) can be calculated by employing the section formula for (m = 2) and (n = 1) as,
(S\left( {x’,\,y’,\,z’} \right) = \left( {\frac{{\left( 2 \right){x_2} + \left( 1 \right){x_1}}}{{2 + 1}},\,\frac{{\left( 2 \right){y_2} + \left( 1 \right){y_1}}}{{2 + 1}},\,\frac{{\left( 2 \right){z_2} + \left( 1 \right){z_1}}}{{2 + 1}}} \right))
( = \left( {\frac{{2{x_2} + {x_1}}}{3},\,\frac{{2{y_2} + {y_1}}}{3},\,\frac{{2{z_2} + {z_1}}}{3}} \right))
In general, if a point (R(x,\,y,\,z)) divides the line joining the points (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) internally in the ratio (1 : k), then the coordinates of the point (R) are written as,
(R\left( {x,\,y,\,z} \right) = \left( {\frac{{\left( 1 \right){x_2} + \left( k \right){x_1}}}{{1 + k}},\frac{{\left( 1 \right){y_2} + \left( k \right){y_1}}}{{1 + k}},\frac{{\left( 1 \right){z_2} + \left( k \right){z_1}}}{{1 + k}}} \right))
( = \left( {\frac{{{x_2} + k{x_1}}}{{k + 1}},\,\frac{{{y_2} + k{y_1}}}{{k + 1}},\,\frac{{{z_2} + k{z_1}}}{{k + 1}}} \right))
Applications of Section Formula
The section formula in (3D) can be applied to derive many other useful results in three-dimensional geometry. For example, the coordinates of the centroid of a triangle can be derived from the section formula.
1. Coordinates of Centroid of a Triangle
The centroid of a triangle is the point of intersection of medians. Centroid divides each median in the ratio of (2 : 1).
Consider (\Delta ABC) with the coordinates (A(x_1,\,y_1,\,z_1),\;B(x_2,\,y_2,\,z_2),) and (C(x_3,\,y_3,\,z_3)) with the point (G(x,\,y,\,z)) as the centroid.
(D)is the midpoint of the side (AB), and the coordinates of (D)can be written using the midpoint formula as,
(D\left( {\frac{{{x_1} + {x_2}}}{2},\;\frac{{{y_1} + {y_2}}}{2},\;\frac{{{z_1} + {z_2}}}{2}} \right))
Now, point (G)divides the line segment (CD)in the ratio (2 : 1). So, the coordinates of the point (G)are given by the section formula as,
(\left( {\frac{{{x_3} + 2\left( {\frac{{{x_1} + {x_2}}}{2}} \right)}}{3},\frac{{{y_3} + 2\left( {\frac{{{y_1} + {y_2}}}{2}} \right)}}{3},\frac{{{z_3} + 2\left( {\frac{{{z_1} + {z_2}}}{2}} \right)}}{3}} \right) = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3},\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right))
Therefore, the coordinates of the centroid of a triangle with vertices
(A\left( {{x_1},\,{y_1},\,{z_1}} \right),\,B\left( {{x_2},\,{y_2},\,{z_2}} \right),) and (C\left( {{x_3},\,{y_3},\,{z_3}} \right)) are given by: (\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3},\,\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)).
2. Collinearity of Three Points
Three points are collinear if they lie on the same line. For three points, (L(x_1,\,y_1,\,z_1),\,M(x_2,\,y_2,\,z_2)), and (N(x_3,\,y_3,\,z_3)), it is enough to show that one of the points divide the line segment in a particular ratio, say (1 : k).
If a point (R(x,\,y,\,z)) divides the line joining the points (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) internally in the ratio (1 : k), then the coordinates of the point (R) are written as, (R\left( {x,\;y,\;z} \right) = \;\left( {\frac{{{x_2} + k{x_1}}}{{k + 1}},\,\frac{{{y_2} + k{y_1}}}{{k + 1}},\,\frac{{{z_2} + k{z_1}}}{{k + 1}}} \right)).
Now, if we can find a value of (k) such that (N\left( {{x_3},\;{y_3},\;{z_3}} \right) = \;\left( {\frac{{{x_2} + k{x_1}}}{{k + 1}},\;\frac{{{y_2} + k{y_1}}}{{k + 1}},\;\frac{{{z_2} + k{z_1}}}{{k + 1}}} \right)), then the points (L,\,M,) and (N) are collinear.
Solved Examples – Section Formula in 3D
_Q.1. What is the midpoint of the line joining the points (J(-3,\,4,\,7)) and (K(9,\,0,\,3))?
Ans:_ The coordinates of the midpoint of the line joining the points (\left( {{x_1},\;{y_1},\;{z_1}} \right)) and (\left( {{x_2},\;{y_2},\;{z_2}} \right)) is given by (\left( {\frac{{{x_1} + {x_2}}}{2},\;\,\frac{{{y_1} + {y_2}}}{2},\,\frac{{{z_1} + {z_2}}}{2}} \right)).
So, the coordinates of the line joining the points (J(-3,\,4,\,7)) and (K(9,\,0,\,3))are:
(\left( {\frac{{ – 3 + 9}}{2},\frac{{4 + 0}}{2},\frac{{7 + 3}}{2}\;} \right) = \left( {3,\,2,\,5} \right)).
_Q.2. Find the coordinates of the points that divide the line segment joining the points (X(2,\,3\,-2))and (Y(6,\,3\,2))internally in the ratio (3 : 1)?
Ans:_ If point (R \left( {{x},\;{y},\;{z}} \right))divides the line segment (\overline {PQ} ) joining the points (P \left( {{x_1},\;{y_1},\;{z_1}} \right)) and (Q\left( {{x_2},\;{y_2},\;{z_2}} \right)) internally in the ratio (m : n), then the coordinates of (R) are given by
(R\left( {x,\,y,\,z} \right) = \;\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right))
Here, (\left( {{x_1},{y_1},\;{z_1}} \right) = \left( {2,\;3,\; – 2} \right)) and (\left( {{x_2},{y_2},\;{z_2}} \right) = \;\left( {6,\;3,\;2} \right)).
The coordinates of the point that divides the line segment internally in the ratio (3 : 1)are:
(\left( {\frac{{3\left( 6 \right) + 1\left( 2 \right)}}{{3 + 1}},\frac{{3\left( 3 \right) + 1\left( 3 \right)}}{{3 + 1}},\frac{{3\left( 2 \right) + 1\left( { – 2} \right)}}{{3 + 1}}} \right) = \left( {5,\;3,\;1} \right)).
_Q.3. Find the ratio in which the point ((9,\,-11,\,1))externally divides the line segment joining the points (D(1,\,5,\,-3))and (E(3,\,1,\,-2)).
Ans:_ If point (R(x,\,y,\,z))divides the line segment (PQ) joining the points (P(x_1,\,y_1,\,z_1))and (Q(x_2,\,y_2,\,z_2))externally in the ratio (m : n), then the coordinates of (R) are given by
(R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right))
Here, ((x_1,\,y_1,\,z_1 ) = (1,\,5,\,-3),\;(x_2,\,y_2,\,z_2 ) = (3,\,1,\,-2),) and ((x,\,y,\,z) = (9,\,-11,\,1)). We need to find the ratio (m ∶ n).
The coordinates of the point that divides the line segment externally in the ratio (m : n)are
(\left( {\frac{{m\left( 3 \right) – n\left( 1 \right)}}{{m – n}},\frac{{m\left( 1 \right) – n\left( 5 \right)}}{{m – n}},\frac{{m\left( { – 2} \right) – n\left( { – 3} \right)}}{{m – n}}} \right) = \left( {9,\; – 11,\;1} \right))
Equating the coordinates:
(\frac{{3m – n}}{{m – n}} = 9)
(\frac{{m – 5n}}{{m – n}} = – 11)
(\frac{{-2m + 3n}}{{m – n}} = 1)
These equations can be simplified as:
(-3m + 4n = 0)
That is, (3m = 4n)
or
(\frac{m}{n} = \frac{4}{3})
Therefore the ratio in which point ((9,\,-11,\,1)) externally divides (\overline {DE} ) is (4 : 3).
_Q.4. Find the centroid of (\Delta SRT)whose vertices have the coordinates (S(-4,\,1,\,0),\;R(5,\,3,\,-3)) and (T(2,\,-10,\,3)).
Ans:_ The coordinates of the centroid of a triangle with vertices (A(x_1,\,y_1,\,z_1),\;B(x_2,\,y_2,\,z_2)), and (C(x_3,\,y_3,\,z_3)) are (\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3},\,\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)).
The coordinates of the centroid of the triangle with vertices (S(-4,\,1,\,0),\;R(5,\,3,\,-3)) and (T(2,\,-10,\,3)) are given by (\left( {\frac{{ – 4 + 5 + 2}}{3},\frac{{1 + 3 + \left( { – 10} \right)}}{3},\frac{{0 + \left( { – 3} \right) + 3}}{3}} \right) = \left( {1,\; – 2,\;0} \right)).
_Q.5. Use the section formula to show that the points (A(2,\,-3,\,4),\;B(-1,\,2,\,1)) and (C(0,\,\frac{1}{3},\,2)) are collinear.
Ans:_ Let point (P)divides the segment (\overline {AE} ) in the ratio ( k : 1).
Then, using the section formula, the coordinates of (P)are:
(\left( {\frac{{k\left( { – 1} \right) + 1\left( 2 \right)}}{{k + 1}},\,\frac{{k\left( 2 \right) + 1\left( { – 3} \right)}}{{k + 1}},\,\frac{{k\left( 1 \right) + 1\left( 4 \right)}}{{k + 1}}} \right) = \left( {\frac{{ – k + 2}}{{k + 1}},\,\frac{{2k – 3}}{{k + 1}},\,\frac{{k + 4}}{{k + 1}}} \right)).
Now, if we can find a value of (k) for which these coordinates coincide with (C), we can say that (C) divides the line segment (\overline {AB} ) internally or externally in the ratio (k : 1), and therefore the points (A,\,B,) and (C) are collinear.
The (x-)coordinate of point (C) is zero.
(\frac{{ – k + 2}}{{k + 1}} = 0) only when (k = 2)
When (k = 2):
(\frac{{2k – 3}}{{k + 1}} = \frac{1}{3}) and (\frac{{k + 4}}{{k + 1}} = 2)
That is, when (k = 2), the point (P) coincides with the point (C).
In other words, point (C)divides the line segment (\overline {AB} ) internally in the ratio of (2 : 1).
Therefore, the three points are collinear.
Summary
The article provides a foundation of the concept of section formula for two dimensions before moving into three dimensions. The derivation of the section formula for internal division is explained well, whereas the same for the external division is procured by applying the internal division formula. The special cases of the formula like points of trisection and midpoint formula are also dealt with clarity.
The article also discusses a couple of applications of section formula such as, finding the coordinates of the centroid of a triangle and checking the collinearity of three points. Further, the article concluded with a few solved examples to reinforce the concepts and calculations learnt.
Frequently Asked Questions(FAQs)
_Q.1. What is the section formula in 3D?
Ans:_ If a point (R(x,\,y,\,z)) divides (\overline {PQ} ) where (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) internally in the ratio (m : n), then the section formula for internal division is given by, (R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right))
If a point (R(x,\,y,\,z)) divides (\overline {PQ} ) where (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) externally in the ratio m:n, then the section formula for external division is given by, (R\left( {x,\,y,\,z} \right) = \;\left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)).
_Q.2. What is the section formula for external division?
Ans:_ If a point (R(x,\,y,\,z)) divides (\overline {PQ} ) where (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) externally in the ratio (m:n,) then the section formula for external division is given by (R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)).
_Q.3. What is the section formula and distance formula?
Ans:_ If a point (R(x,\,y,\,z)) divides (\overline {PQ} ) where (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) internally in the ratio (m : n), then the section formula for internal division is given by (R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)).
The distance formula for finding the distance (d) between the points (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) is given by (d = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} ).
_Q.4. How do you find the ratio in which a point divides a line?
Ans:_ If a point (R(x,\,y,\,z)) divides (\overline {PQ} ) where (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) internally in the ratio (m : n), then the section formula for internal division is given by (R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right))
If a point (R(x,\,y,\,z)) divides (\overline {PQ} ) where (P(x_1,\,y_1,\,z_1)) and (Q(x_2,\,y_2,\,z_2)) externally in the ratio (m : n), then the section formula for external division is given by,
(R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)).
_Q.5. Where is the section formula used?
Ans:_ In coordinate geometry, the section formula is used to find the ratio in which a point divides a line segment internally or externally. There are other applications like finding the coordinates of the centroid, incentre, etc. and checking the collinearity of three points.
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CDFW Marine News Informational Notice - CDPH Warns Public Not to Consume Sport-Harvested Bivalve Shellfish from Santa Cruz County
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CDFW Marine News Informational Notice - CDPH Warns Public Not to Consume Sport-Harvested Bivalve Shellfish from Santa Cruz County
This bulletin was sent at 06/06/2025 02:26 PM PDT
View as a webpage / share INFORMATIONAL NOTICE June 6, 2025 The following safety notification is from the California Department of Public Health: CDPH Warns Public Not to Consume Sport-Harvested Bivalve Shellfish from Santa Cruz County ------------------------------------------------------------------------------------------ The California Department of Public Health (CDPH) is advising consumers not to eat sport-harvested mussels, clams, scallops, or oysters from Santa Cruz County. PSP toxins affect the nervous system, producing a tingling around the mouth and fingertips within a few minutes to a few hours after eating toxic shellfish. These symptoms are typically followed by loss of balance, lack of muscular coordination, slurred speech and difficulty swallowing. In severe poisonings, complete muscular paralysis and death from asphyxiation can occur. This warning does not apply to commercially sold mussels, clams, scallops, and oysters from approved sources. State law permits only state-certified commercial shellfish harvesters or dealers to sell these products. Shellfish sold by certified harvesters and dealers are subject to frequent mandatory testing to monitor for toxins. The annual mussel quarantine remains in effect for the entire California coast. The annual quarantine prohibits the sport-harvest of mussels for human consumption and applies to all species of mussels harvested along the California coast, as well as all bays and estuaries. The purpose of the quarantine, which will continue through at least October 31, is to prevent paralytic shellfish poisoning (PSP) and domoic acid poisoning. Cooking does not destroy the toxins. You can get the most current information on shellfish advisories and quarantines by calling CDPH's toll-free Shellfish Information Line at (800) 553-4133 or viewing the recreational bivalve shellfish advisory interactive map. For additional information, please visit the CDPH Marine Biotoxin Monitoring web page. The original press release from CDPH is posted here. CDFW Marine Region News ServiceOcean-related news and information Learn more about CDFW's Marine Region online at wildlife.ca.gov/regions/marine Read the CDFW Marine Management News blog at cdfwmarine.wordpress.com
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Period of sinx is 2π | sine function is a periodic function | Proof of period of sine function
Shah Hussain Maths Online
5110 subscribers
164 likes
Description
9424 views
Posted: 18 Aug 2020
The period of the sine function is 2π,
which means that the value of the function is the same every 2π units.
The sine function, like cosine, tangent, cotangent, and many other trigonometric function, is a periodic function, which means it repeats its values on regular intervals, or "periods." In the case of the sine function, that interval is 2π.
For instance, sin(π) = 0. If you add 2π to the x-value, you get sin(π + 2π), which is sin(3π). Just like sin(π), sin(3π) = 0. Every time you add or subtract 2π from our x-value, the solution will be the same.
You can easily see the period on a graph, as the distance between "matching" points. Since the graph of y = sin(x) looks like a single pattern repeated over and over again, you can also think of it as the distance along the x-axis before the graph starts to repeat itself.
On the unit circle, 2π is a trip all the way around the circle. Any amount greater than 2π radians means that you keep looping around the circle -- that's the repeating nature of the sine function, and another way to illustrate that every 2π units, the function's value will be the same.
Trigonometric functions
The ratios sinx, cosx,
tanx,secx , cosecx, and cotx are called trigonometric functions.
Relation between trigonometric functions
sin x = 1/cosecx
or
cosec x =1/sin x
cos x = 1/sec x
or
sec x = 1/ cos x
tan x = 1/cot x
or
cot x = 1/tan x
Even function and odd function
To determine a function is even or odd, you replace x by -x in function, if the resulting function is the same as the original function, then the function is even. If the resulting function is the negative of the original function, then the function is odd.
Even function
f(-x)= f(x)
Example
cos(-x)= cos x
sec(-x)= sec x
f(x) = x^2
For odd function
f(-x)= -f(x)
Example
sin(-x)= -sinx
cosec(-x)= -cosec x
tan(-x)= - tan x
cot(-x)= -cot x
f(x) = x
Period of trigonometric functions
The smallest positive number , which when added to the original circular measure of the angle gives the same value of the function is called period of trigonometric functions.
Mathematically
If sin( x+p)= sinx
Then p is called the period of trigonometric functions.
Periodicity of trigonometric functions
All the six trigonometric functions repeat their values for each increase or decrease of 2pai in angle .This behavior of trigonometric functions is called periodicity of trigonometric functions.
sin(2pai+x)= sinx
sin(4pai+x)=sin x
cos(2pai+x)= cos x
cos(4pai+x)= cos x
Frequency of trigonometric functions
Reciprocal of period of trigonometric functions is called frequency of trigonometric functions.
That is
Frequency = 1/period
or
f = 1/p
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SHAH_HUSSAIN_MATHS_ONLINE
PERIOD_OF_FUNCTION #PERIOD
TRIGONOMETRIC_PERIODIC_FUNCTION #PERIODICITY
TRIGONOMETRIC_IDENDITIES #PERIODIC
FUNCTION #PERIODICITY #GRAPH #SINX #COSX
TANX #SECX #COSECX #COTX #DOMAIN #RANGE #GRAPH
LINEAR_FUNCTION #QUADRATIC_FUNCTION
PARABOLA #STRAIGHT_LINE
MATHS_TRICKS
PERIOD_OF_TRIGONOMETRIC_FUNCTIONS
SHORT_TRICKS_ON_PERIOD_OF_FUNCTIONS
DERIVATIVE #ECAT_MCQS
DERIVATIVE_OF_TRIGONOMETRIC_FUNCTION
FUNCTIONS_AND_LIMITS
ENTRY_TEST_MCQS #SHORTCUT_TRICKS
DERUVATIVE_BY_SHORT_TRICKS
ECAT #FAST #NUST #GIKI #PIEAS #NET #LUMS #PU #COMSAT #ETEA #GAT #SAT #ENTRY_TEST #CALCULUS
LIMITS_PROBLEMS #CONCEPT_OF_LIMITS_AND_CONTINUITY
LIMITS #CONTINUOUS_FUNCTION #LIMITS
BSC_MATHEMATICS #UNIT #ABSOLUTE_VALUED_PROBLEMS #BASIC_LIMITS_PROPERTIES #THOMAS_CALCULUS #CONTINUITY #CONTINUOUS_FUNCTION #INVERSE_OF_FUNCTION #GAT #NET #UET #TAXILA #PU #BZU #COMSAT #RATIONAL_FUNCTION #VERTICAL_LINE_TEST #HORIZONTAL_LINE_TEST #ONE_TWO_ONE_FUNCTION
FUNCTION #DOMAIN #RANGE #PERIOD
FREQUENCY #AMPLITUDE #PHASE_SHIFT
GRAPG #SYMMETRY #SMART_TRICKS
SHORT_CT#TIPS #AMPLITUDE #GRAPH#DOMAIN #RANGE #FUNCTION #PHASE_SHIFT #PERIOD
FREQUENCY #SYMMETRY #SMOOTH_CURVE
SECX #COSECX #SINX
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शो मोर जो अलार्म दिया स्टूडेंट सा एवं शुभ सेंट फ्रॉम स्पॉइलिंग दीपालपुर आज हम लेकर नंबर दो में प्रूफ करेंगे इस के प्रॉडक्ट साइंटिफिक फंक्शन एंड स्पिरिट उपाय यह हमारा एक क्वेश्चन हमें यह प्रूफ करना है कि साइन प्री वेडिंग फंक्शन है और साइन का जो पीरियड है वह क्या है टूटा यह वैरी सिंपली ऐसे भी क्वेश्चन आ सकता है सो लेट्स पीरियड आफ साइंसेज 25 तो सबसे पहले हम वेट करेंगे प्रूफ करने के लिए लेट किया लेट पीटी आज भी विद ए पीरियड आफ ए पीरियड आफ साइंस फंक्शन मैं साइन संयम ने लैटर कर लिया कि प्रिवेंट करता है ब्रेड को डेन्वर डेफिनेशन आफ कि डेफिनेशन क्या थे स्मालेस्ट पॉजिटिव नंबर है जिसको हम ऐड करें ओरिजिनल फंक्शन में और वह क्या हमेशा व्यस्त हैं तो जेंडर डेफिनेशन इन थाइरॉएड डेफिनेशन आफ पीरियड हमारे पास पीरियड की डेफिनेशन डेफिनेशन आफ हम इसको कह सकते हैं साइन आफ यहां पर ले सकते हैं इसको ले सकते हैं यह साइन थीटा प्लस पीठ पर टू साइन थीटा फॉर थिस ब्लॉग टो सेट ऑफ रियल नंबर अगर पीरियड है अब हम इसको ऐड करेंगे उस में तो हम यहां पर ए प्रूफ करना है कि साइन का जो पीरियड ओपी अध्यक्ष अप्रूव करने के लिए यहां पर हम कूटकर लेते हैं फुट किया थीटा इज इक्वल टू 0 गेट अब जब मैं 350 करूंगा तो डेफिनेटली लेफ्ट साइड पर बनेगा साइंस पेपर और राइट साइड पर बनेगा 600 अब अगर मैं इसे सिंपली टाइप करता हूं तो हमारे पास लेफ्ट साइड पर साइंस पेपर और राइट साइड पर 000 हमारे 50 ग्राम इसे सिंपली करता हूं तो पी लो मैं साइन वर्ष 2080 कहां है 0137 पर तो इसका मतलब है हमारे पास PC की वैल्यू के आएगी पेट की वैल्यू 10 प्लस माइनस 5 प्लस माइनस न धुंध यह हमारे पास प्लस माइनस टू सॉन्ग इसका मतलब है कि यह हमारे पास की वैल्यू आगे अब यहां पर नोटिस करें अगर पीरियड है डेफिनेशन सब्सक्राइब नंबर होगा तो एक्टिवेट को सब्सक्राइब नहीं करेंगे नहीं करेंगे तो यहां पर जस्ट अगर यह की वैल्यू प्लस माइनस प्लस माइनस प्लस माइनस टू नियुक्त कर रही है कि यहां पर होता है और हमारे पास सब्सक्राइब करना है अब यह आपके क्योंकि पॉजिटिव वैल्यू है तो यहां से हम इसे सबसे पहले लेते हैं इफिजी आफ 125 यहां पर इसे हम क्वेश्चन नंबर वन कहते हैं यह हमारे पास क्वेश्चन नंबर 10 अब हम इसे रिपेयर करते हैं तो पोस्ट की बात और यह टेस्ट पॉजिटिव वैल्यू है इसका मतलब इसलिए हम यहां पे कहां से स्टार्ट कर रहे हैं और अब मैं बेल्ट करूंगा ए पी के पाई तो यह क्या बनेगा हमारे पास साइन इन यहां पर डाटा प्लान अधिक साइन थीटा अब हमने यह ट्रू है यह तो यह हमारा घाघरा कैसे अगर मैं यहां पर लेफ्ट साइड पर से एक्सप्लेन कर दी फादर चैनल सब्सक्राइब कर दूं आपने यूनिट में यह हमला हुआ था गौतम साइन आफ प्लस सिक्यूरिटी फोर्स बीटा प्लज फास्ट मैं साइन थीटा अगर मैं लेफ्ट साइड पर फ्रॉम लॉर्ड एक्सप्लेंड कर दूं तो यह लेफ्ट साइड पर हमारे पास क्या बनेगा साइन आफ ए कॉलेज बीटा प्लस कौस अल्फा मैं साइन थीटा और राइट साइड SIM राइट कर दी साइन थीटा अब आपको पता होना चाहिए क्वांटिटी पर कायम होगा - हमारे पास साइन थीटा माइनस वन और यह ऐसे 1020 प्लस माइनस माइनस माइनस साइन थीटा त्यौहार तो इसका मतलब यह जो पॉइंट है यह नहीं है यह हमारे पास इस में पाई इस नोट आ आ आ ए पीरियड आफ ए मैं साइन एक्स अब नेक्स्ट वैल्यू लेते हैं सिर्फ जो पिज़्ज़ा इक्वल टू अब नेक्स्ट पॉजिटिव ये कौन सी होगी टू माय लाइफ इज इक्वल टू 50 क्वेश्चन विक्टिम्स तो फिर हमारे पास इक्वेशन वाह क्या बनेगी तो यह बनेगा हमारे पास साइन आफ साइंटिफिक टॉपलेस 2.5 इंच इक्वल टू साइन थीटा अब इसको मैसेज चेक कर सकते हैं यह जरूरी है और यहां पर यह पता चल रहा था गिफ्ट कैसे यह हमारे पास थर्ड थर्ड में नेगेटिव और मल्टी सब्सक्राइब नहीं अगर हम इसको लाइक सबस्क्राइब सब्सक्राइब थर्ड थर्ड में नेगेटिव और यह बेसिकली नियुक्ति नहीं था - हमारे पास राइट साइड पर क्या था है पे था प्लस साइन थीटा तो एडजस्ट यहां पीएम राइड कर सकते हैं कि यह क्या है फौज है यह आंवला मैनें एक्सपेक्ट जस्ट इसलिए आपको एक्सप्लेन करने के लिए समझाने के लिए अब यहां पर भी अब हम इसे डायरेक्टर लाएंगे से ले सकते हैं यह हमारे पास सब्सक्राइब फर्स्ट में पॉजिटिव मल्टीप्ल आफ 9 को चेंज नहीं होती है कि यहां पर भी हमें पता चल रहा है यह बाथरूम है लेकिन अगर आपने टू स्टेप फादर सॉल्व करने हैं तो हम इसे इस पेमेंट कर देते हैं बिल्कुल इसी फार्मूले से यह बनेगा साइन अल्फा हक वाश बेटा प्लस कौस अल्फा मैं साइन थीटा इज इक्वल टू साइन थीटा यह बनेगा हमारे पास साइन थीटा इक्वल टू पांच मतलब 360 डिग्री कोर्स 361 और यह बनेगा और साइन 360 आप इतना याद रखें सब्सक्राइब होती है यह हम प्रीवियस लेक्चर में डिस्कस कर चुके हैं तो यहां से अगेन प्ले करें यह तो यह साइन थीटा है इसके एक वाला रहा है साइन थीटा कि यह बात क्या है प्रूफ है कि साइन थीटा इज इक्वल टू साइन तो अब हम की वैल्यू ले रहे हैं तो हमारे पास साइन थीटा प्लस टू इक्वल टू इसका मतलब इसका मतलब यह है कि पॉजिटिव यह नंबर है जिसको हम साइन थीटा में ऐड करते हैं तो हमें सेम वही ओरिजिनल फंक्शन मिलता है तो हम कह सकते हैं यहां से है हाउ टो पाए हु इज कि द पीरियड आफ मैं साइन 102 या इस तरह लिख लेते हैं सोंठ उपाय इस द पीरियड ऑफ - फैंस साइन इज द मैं साइन इज आप रिकॉर्डिंग फंक्शन प्री वेडिंग फंक्शन एक पैन इस पीरियड्स एंड इट्स है पीरियड्स आज टू पॉइंट यह हमें में झाल झाल
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2024 AMC 12A Problems/Problem 21
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (lazy + quick)
5 Solution 4 (transform)
6 Solution 5 (transform)
7 See also
Problem
Suppose that and the sequence satisfies the recurrence relation for all What is the greatest integer less than or equal to
Solution 1
Multiply both sides of the recurrence to find that .
Let . Then the previous relation becomes
We can rewrite this relation for values of until and use telescoping to derive an explicit formula:
Summing the equations yields:
Now we can substitute back into our equation:
Thus the sum becomes
We know that , and we also know that , so the requested sum is equivalent to . All that remains is to calculate , and we know that this value lies between and (see the note below for a proof). Thus,
so
and thus the answer is .
~eevee9406
Note:. It is obvious that the sum is greater than 1 (since it contains as one of its terms).
If you forget this and have to derive this on the exam, here is how:
and it is clear that . ~eevee9406
Solution 2
According to the equation given,
Suppose , , then , where , then we obtain that
Hence,
Notice that,
so
and thus the answer is .
~reda_mandymath
Solution 3 (lazy + quick)
We'll first try to isolate in terms of .
Now, as with many, many of these large summation problems, if we just evaluate the first few values in the series, a pattern should emerge quickly. Here it works out well since our product on the LHS cancels out.
Here it becomes glaringly obvious that .
So, .
We proceed with the same summation strategy as Solution 1 and get our answer of .
Note: You only have find the answer's units digit from the answer choices; that's for each sum, giving choice B.
~nm1728
Solution 4 (transform)
Set
~luckuso
Solution 5 (transform)
According to the equation given,
The rest continues similar to Solution 1 or 2
~luckuso
See also
2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20Followed by
Problem 22
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 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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Isosceles triangle
This article is a stub. Help us out by expanding it.
An isosceles triangle is a triangle in which at least two edges have equal length. Equivalently, an isosceles triangle is one in which at least two angles have equal measure. Equilateral triangles are a special class of isosceles triangles in which all three sides and angles are equal. Because Isosceles triangles have two sides of equal length, the angles that don't connect them are equal, meaning that if you had the angle that connects them's measure, you would be able to figure out the other two. In more clear definitions: Say we have an isosceles triangle where . You also know that . You can then find out angles and because those angles are congruent (geometry), due to . , so angles and are each .
Every isosceles triangle has reflectional symmetry. The axis of symmetry passes through the vertex which is shared by the edges of equal length, and it is also the triangle's median, altitude, and angle bisector of that vertex.
An isosceles triangle.
See Also
Isosceles trapezoid
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How to plot functions with asymptotic behavior like tan(x), ln(x), and log(x) in JavaScript?
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Asked 1 year, 5 months ago
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I'm working on a project where I need to plot graphs of mathematical functions in JavaScript, specifically functions that exhibit asymptotic behavior or approach infinity at certain points, such as tan(x), ln(x), and log(x). I'm struggling with how to accurately represent these functions, especially near the points where they approach infinity (for example, near vertical asymptotes for tan(x)).
This is my Code to draw functions:
```javascript
export function drawFunction(latexStr) {
let isTan;
// Draw function
console.log(canvas.height);
console.log(canvas.width);
let prevY; // Stores the previous Y value
let startY = true; // To know when a new line is needed
ctx.beginPath();
ctx.strokeStyle = "blue";
ctx.lineWidth = 2;
// Draw the function in small steps
for (let x = -canvas.width / 2; x <= canvas.width / 2; x += stepSize) {
let y = parser(latexStr, x); // f(x)
isTan = parser(latexStr, x);
console.log(isTan);
// Skip drawing the graph if it's infinite or too large for the canvas
if (Math.abs(y) > canvas.height / scaleFactor && !isTan) {
startY = true;
y = (y < 0 ? -1 : 1) (canvas.height / scaleFactor - 1);
}
const adjustedX = canvas.width / 2 + x scaleFactor; // Calculation of the small X line
const adjustedY = canvas.height / 2 - y scaleFactor; // Calculation of the small Y line
/ if (Math.round(x 100) / 100 === stepSize)
ctx.moveTo(adjustedX, adjustedY); /
// Draw or move to position
if (startY) {
ctx.moveTo(adjustedX, adjustedY);
startY = false;
} else {
if (prevY !== undefined && Math.abs(y - prevY) > 1 scaleFactor) {
if (isTan) {
ctx.stroke(); // Draw the current line
ctx.beginPath(); // Start a new line
ctx.moveTo(adjustedX, y < 0 ? canvas.height : 0); // Move to the edge of the canvas
} else {
ctx.moveTo(adjustedX, adjustedY); // Move to the new starting position
}
} else {
ctx.lineTo(adjustedX, adjustedY);
}
}
prevY = y;
//console.log(Math.round(x 100) / 100);
}
ctx.stroke(); // Draw the function
}
```
Drawing a normal function like "y = x+2" is no problem, but issues arise with a function like "y = tan(x)" as seen in the image below:
The values are correct, so it seems not to be an issue with the calculations but rather with the way the drawing is handled. I've searched everywhere but couldn't find a solution. Does anyone have suggestions on how I could improve the drawing algorithm to better handle infinity points for functions like tan(x)?
Thank you in advance for your help!
javascript
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asked Apr 12, 2024 at 14:00
SimoneSimone
61 1 1 silver badge 7 7 bronze badges
1
2 This would probably be easier to debug with a minimal reproducible example, but typically what I do is map the x canvas coordinate to the graph's x coordinate, calculate, and translate back to canvas-y. As for when to skip drawing, I would say to do it when the absolute difference between previous and next y value is above a certain threshold.General Grievance –General Grievance 2024-04-12 14:18:07 +00:00 Commented Apr 12, 2024 at 14:18
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Your problem is not quite clear, and you have not provided full code ...
From your image I'm going to assume that what you are not happy with is some lines not making it all the way up to the edge of the canvas, like image below:
Problem
Here is some simple code reproducing your problem:
```javascript
var canvas = document.getElementById('plotCanvas');
var ctx = canvas.getContext('2d');
var startX = -20;
var endX = 20;
var scaleFactor = 25;
ctx.beginPath();
for (var x = startX + 0.01; x <= endX; x += 0.01) {
xP = (x - startX) (canvas.width / (endX - startX));
yP = canvas.height / 2 - Math.tan(x) scaleFactor
if (xP > 0 && yP > 0 && xP < canvas.width && yP < canvas.height) {
ctx.lineTo(xP, yP);
} else {
ctx.moveTo(xP, yP)
}
}
ctx.strokeStyle = 'blue';
ctx.stroke();
```
css
canvas {
border: 1px solid black;
}
```xml
```
Run code snippet Edit code snippet
Hide Results Copy Expand
Solution
We have a hard condition to only draw inside the canvas:
if (xP > 0 && yP > 0 && xP < canvas.width && yP < canvas.height) {
that is good to optimize what we draw and not waste time drawing things that are not visible, but...
The condition for the boundaries needs to be a bit more flexible, an easi fix is to let it draw a few lines outside the canvas boundaries, instead of stopping at 0 like I had, I'm letting it go to:
-canvas.width that is all
```javascript
var canvas = document.getElementById('plotCanvas');
var ctx = canvas.getContext('2d');
var startX = -20;
var endX = 20;
var scaleFactor = 25;
ctx.beginPath();
for (var x = startX + 0.01; x <= endX; x += 0.01) {
xP = (x - startX) (canvas.width / (endX - startX));
yP = canvas.height / 2 - Math.tan(x) scaleFactor
if (xP > -canvas.width && yP > -canvas.height && xP < canvas.width && yP < canvas.height) {
ctx.lineTo(xP, yP);
} else {
ctx.moveTo(xP, yP)
}
}
ctx.strokeStyle = 'blue';
ctx.stroke();
```
css
canvas {
border: 1px solid black;
}
```xml
```
Run code snippet Edit code snippet
Hide Results Copy Expand
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edited Apr 14, 2024 at 15:02
answered Apr 13, 2024 at 19:59
Helder SepulvedaHelder Sepulveda
17.7k 4 4 gold badges 34 34 silver badges 66 66 bronze badges
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https://www.quora.com/How-many-numbers-from-1-to-100-inclusive-are-divisible-by-3-or-4-or-both
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How many numbers from 1 to 100 (inclusive) are divisible by 3 or 4 (or both)? - Quora
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5
How many numbers from 1 to 100 (inclusive) are divisible by 3 or 4 (or both)?
All related (61)
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·
Sep 6
Count numbers divisible by 3 or 4 using inclusion–exclusion.
Numbers divisible by 3 between 1 and 100: floor(100/3) = 33.
Numbers divisible by 4 between 1 and 100: floor(100/4) = 25.
Numbers divisible by both 3 and 4 (i.e., by lcm(3,4)=12): floor(100/12) = 8.
By inclusion–exclusion: 33 + 25 − 8 = 50.
Answer: 50.
Upvote ·
Related questions
More answers below
What are all the numbers between 1 and 100 divisible by 3 and 4?
How many numbers between 0 and 100 are divisible by 4?
How many positive integers between 100, and 999 are inclusive divisible by 3 and 4?
How many numbers less than 100 are multiples of both 3 and 4?
How many whole numbers between 1 to 100 are not divisible by 3 or by 11?
Michael Nichols
Environmental Science (2000–present) · Author has 240 answers and 1M answer views
·7y
For the number 3 there are 33 numbers divisible by 3 from 1 to 100.
99/3 = 33
The number 4 has 25 numbers divisible by 4 from 1 to 100
100/4 = 25
If you want the number divisible by both 3 and 4 it is 58 .
If you want the numbers only divisible by 3 or 4 but not both subtract the multiples of 12 which is the LCM ( least common multiple )
100/12 = 8 .333 or approximately 8.
58 - 8 = 50 numbers by 3 or 4 .
Upvote ·
99 13
9 1
Joseph Lionarons
multi-instrumentalist, linguist, music teacher, pokemon trainer · Author has 54 answers and 210.8K answer views
·Updated 7y
Originally Answered: How many numbers between 1 and 100 (inclusive) are divisible by 3 or 2? ·
67.
100 divided by 3 is 33.333333. This means that there are 33 numbers between 1 and 100 that are divisible by 3.
100 divided by 2 is 50. This means that there are 50 numbers between 1 and 100 that are divisible by 2.
However, some of these numbers are counted twice (e.g. 6 is divisible by both 3 and 2).
If we divide 100 by the LCD, which in this case is 6, we get 16.666667, which means that 16 numbers are counted twice.As in they are divisible by both 3 and 2. So we discount these, because they are not unique numbers and exist in both sets.
50 + 33 = 83 - 16 = 67.
Upvote ·
99 35
9 1
Chris Nash
rapidly aging Cambridge math graduate · Author has 155 answers and 708.5K answer views
·9y
Originally Answered: How many numbers between 1 and 100 (inclusive) are divisible by 3 or 2? ·
Here’s a good opportunity to learn about inclusion-exclusion. It’s nothing sophisticated, but if you’re out to “count” something complicated, split it into easier problems you can count. The trick is to know whether you’ve counted everything you need - too many times, or not enough times?
Let’s start with a picture, yup, it’s a Venn diagram:
and put our numbers in it. All the numbers divisible by 2, we’ll put in the left circle. All the numbers divisible by 3, go in the right. And - no surprise - there will be some numbers in both.
What goes in the left circle? 2, 4, 6, … hopefully it will be no
Continue Reading
Here’s a good opportunity to learn about inclusion-exclusion. It’s nothing sophisticated, but if you’re out to “count” something complicated, split it into easier problems you can count. The trick is to know whether you’ve counted everything you need - too many times, or not enough times?
Let’s start with a picture, yup, it’s a Venn diagram:
and put our numbers in it. All the numbers divisible by 2, we’ll put in the left circle. All the numbers divisible by 3, go in the right. And - no surprise - there will be some numbers in both.
What goes in the left circle? 2, 4, 6, … hopefully it will be no surprise there are 50 numbers in the left circle.
What goes in the right circle? 3, 6, 9, … again, I hope it’s pretty evident there are 33 numbers in the right.
So, 50+33, right? Well, not quite. There will be some numbers that got counted twice, the ones in both circles, divisible by 2 and 3. Rather nicely, we’ve turned a problem about or into a problem about and. So which numbers go in the red area? Those divisible by 6. 6, 12, …. 16 of those. Our 50+33 counted these numbers twice. To get back to counting them once, we have to subtract those. 50 + 33 - 16 = 67.
The inclusion-exclusion principle can be used with any number of “properties” - suppose there are N N objects, each of which could have any (or all) of P P properties. Usually it’s a lot easier to count how many of the objects have some known combination (“and”), rather than several options (“or”) - but the second question can be answered by addition and subtraction of the right combinations of the first.
Upvote ·
99 21
9 2
Related questions
More answers below
How many three digit numbers are divisible by 4?
Which the numbres that are divisible by 3 and 5 till 100?
How many possible combinations are in 100 numbers, 1 to 100?
What is the total number of digits from 1 to 100?
How many two-digit numbers are divisible by both 2 and 3?
Parinit Agarwal
Works at Deloitte (company) · Author has 144 answers and 1M answer views
·9y
Originally Answered: How many numbers between 1 and 100 (inclusive) are divisible by 3 or 2? ·
A2A ed
First lets see the numbers between 1 and 100 divisible by 3
Clearly there are 33 such numbers (333=99) as 33 is the highest multiplier of 3 such that the product is less than or equal to 100.
Now lets see the numbers between 1 and 100 divisible by 2
Clearly there are 50 such numbers (250=100) as 50 is the highest multiplier of 2 such that the product is less than or equal to 100. As 100 is in
Continue Reading
A2A ed
First lets see the numbers between 1 and 100 divisible by 3
Clearly there are 33 such numbers (333=99) as 33 is the highest multiplier of 3 such that the product is less than or equal to 100.
Now lets see the numbers between 1 and 100 divisible by 2
Clearly there are 50 such numbers (250=100) as 50 is the highest multiplier of 2 such that the product is less than or equal to 100. As 100 is inclusive.
So, going by above, we have 33 + 50 = 83 numbers divisible by 3 or 2 between 1 and 100.
However, some numbers for ex, 6, 12 etc are divisible by both 3 and 2 and are definitely counted twice in our 83. So how do we account for them?
We fin...
Upvote ·
99 18
Anil Antony
Education in Mathematics, University of the West Indies (Graduated 1992) · Author has 87 answers and 174K answer views
·Updated 6y
Originally Answered: How many numbers between 1 and 100 (inclusive) are divisible by 3 or 2? ·
We are basically looking for multiples of 2 and 3 between 1 and 100 (inclusive).
There are 50 multiples of 2 between 1 and 100 (including 100) because 100 divided by 2 is 50.
n(multiples of 2) = 50
There are 33 multiples of 3 between 1 and 100 because 100 divided by 3 has a quotient 33 and a remainder 1.
n(multiples of 3) = 33
We know that 2 x 3 = 6. Hence all multiples of 6 are the common multiples.
There are 16 multiples of 6 between 1 and 100 because 100 divided by 6 has a quotient 16 and remainder 4.
n(multiples of 6) = 16
Using the set theory,
n(A∪B) = n (A) + n(B) – n(A∩B)
Therefore, the required
Continue Reading
We are basically looking for multiples of 2 and 3 between 1 and 100 (inclusive).
There are 50 multiples of 2 between 1 and 100 (including 100) because 100 divided by 2 is 50.
n(multiples of 2) = 50
There are 33 multiples of 3 between 1 and 100 because 100 divided by 3 has a quotient 33 and a remainder 1.
n(multiples of 3) = 33
We know that 2 x 3 = 6. Hence all multiples of 6 are the common multiples.
There are 16 multiples of 6 between 1 and 100 because 100 divided by 6 has a quotient 16 and remainder 4.
n(multiples of 6) = 16
Using the set theory,
n(A∪B) = n (A) + n(B) – n(A∩B)
Therefore, the required answer is:
n(multiples of 2) + n(multiples of 3) – n(multiples of 6)
That is:
50 + 33 – 16
Answer: 67
Upvote ·
9 5
Ram Kushwah
Up and coming Most viewed writer · Author has 6.8K answers and 15.7M answer views
·5y
Originally Answered: How many numbers between 1 and 100 (inclusive) are divisible by 3 or 2? ·
from 1 to 100
Number divisible by 2
S2=100/2=50
Numbers divisible by 3 are
3,6…………………………………96,.99
S3=(99–3)/3=96/3=32
In these 32 number numbers divisible by 2 or divisible by 6
S6=(96–6)/6=9/6=15
Thus total; such numbers=S2+S3–S6
=82–15=67
Thus 67 are numbers divisible by 2 or 3
Upvote ·
9 7
9 1
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·3y
Originally Answered: How many numbers from 1 to 100 inclusive are divisible by both 3 and 4? ·
Here’s a big hint. If any integer is evenly divisible by both 3 and 4, then it must be evenly divisible by 12. So pull out all the numbers from 1 to 100 that fit this criterion. It won’t take long.
Seriously, I may not address any further questions from you for quite some time yet. You need to take charge of your own thinking, sooner or later. That’s how education happens. It doesn’t happen when people sponge around for answers to easy questions on Quora!
Upvote ·
9 3
Orion Rhoades
Studied Mathematics&Chemistry at Johns Hopkins University (Graduated 2021)
·7y
There are 50 numbers that meet the criteria.
33 numbers divisible by 3, 25 numbers divisible by 4, and 8 divisible by both.
Therefore, the answer is 33+25=58 33+25=58, but because of the eight numbers divisible by both that were double-counted, the answer is 58−8=50 58−8=50. Hope I helped, and thanks for the A2A!
-Orion
Upvote ·
9 3
Srijay
studying Elementary Number Theory · Author has 97 answers and 194.8K answer views
·7y
Originally Answered: What are all the numbers between 1 and 100 divisible by 3 and 4? ·
Lemma : For a,b,c∈Z a,b,c∈Z with a,b≠0 a,b≠0, if a|c a|c and b|c b|c with g c d(a,b)=1 g c d(a,b)=1 then a b|c a b|c
Proof :
Since g c d(a,b)=1 g c d(a,b)=1 citing Bezout’s theorem ∃∃integers x,y x,y such that : a x+b y=1 a x+b y=1
Since a|c a|c and b|c,∃b|c,∃integers m,n m,n such that c=a m c=a m and c=b n c=b n
If c=0,c=0,the result is obviously true. Suppose c≠0 c≠0
a x+b y=1 a x+b y=1
⟺(a x)c+(b y)c=c⟺(a x)c+(b y)c=c
⟺(a x)(b n)+(b y)(a m)=c⟺(a x)(b n)+(b y)(a m)=c
⟺a b(x n+y m)=c⟺a b(x n+y m)=c
Thus a b|c■a b|c◼
— — — — — — — — — — — — — — — — — — — — — —
Since g c d(3,4)=1 g c d(3,4)=1
All mutiples of 3.4=12 3.4=12 and none other
Continue Reading
Lemma : For a,b,c∈Z a,b,c∈Z with a,b≠0 a,b≠0, if a|c a|c and b|c b|c with g c d(a,b)=1 g c d(a,b)=1 then a b|c a b|c
Proof :
Since g c d(a,b)=1 g c d(a,b)=1 citing Bezout’s theorem ∃∃integers x,y x,y such that : a x+b y=1 a x+b y=1
Since a|c a|c and b|c,∃b|c,∃integers m,n m,n such that c=a m c=a m and c=b n c=b n
If c=0,c=0,the result is obviously true. Suppose c≠0 c≠0
a x+b y=1 a x+b y=1
⟺(a x)c+(b y)c=c⟺(a x)c+(b y)c=c
⟺(a x)(b n)+(b y)(a m)=c⟺(a x)(b n)+(b y)(a m)=c
⟺a b(x n+y m)=c⟺a b(x n+y m)=c
Thus a b|c■a b|c◼
— — — — — — — — — — — — — — — — — — — — — —
Since g c d(3,4)=1 g c d(3,4)=1
All mutiples of 3.4=12 3.4=12 and none other , between 1 1 and 100 100 i.e.
{12,24,36,48,60,72,84,96}{12,24,36,48,60,72,84,96}are divisble by both 3 a n d 4.■.3 a n d 4.◼.The “none other” part can be easily proved by contradiction.
Upvote ·
9 3
Matias Pereira Miranda
No Way Home style · Author has 486 answers and 341.9K answer views
·9mo
There are 33 multiples of 3 between 1 and 100. There are 25 multiples of 4 between 1 and 100. There are 8 multiples of both 3 and 4 (or multiples of 12)
33 + 25 — 8 = 50
There are 50 numbers that are multiples of 3 and/or 4 between 1 and 100
Upvote ·
Dave Palamar
Electrician (1978–present) · Author has 2.3K answers and 2.7M answer views
·7y
As there are 25 numbers divisible by 4 in 4 to 100.
And 33 numbers from 3 to 99 Divisible by 3…
We add 25 + 33 and derive 58 total for each category, however when we exclude those divisible by 12 as Category 3, there are 8 repeats, in divisible by 3 and 4, thus there are 50 without Category 3 ...
Upvote ·
Parag Kalita
Interested in numbers. · Author has 175 answers and 770K answer views
·6y
Originally Answered: How many numbers between 1 and 100 (inclusive) are divisible by 3 or 2? ·
Here, let 3 and 2 be two events
So now n(3)=100/3=33.33=33(its greatest integer function)
And, n(2)=100/2=50
n(3 n 2)=100/6=16.66=16
So we need to find n(3 u 2)=n(3)+n(2)-n(3 n 2)
=33+50–16=67
So the required ans is 67
Upvote ·
9 5
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Power of a Point Theorem
The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two linesintersect a circle and each other.
Contents
[hide]
1 Statement
1.1 Case 1 (Inside the Circle):
1.2 Case 2 (Outside the Circle):
1.2.1 Classic Configuration
1.2.2 Tangent Line
1.3 Case 3 (On the Border/Useless Case):
1.4 Alternate Formulation
1.5 Hint for Proof
2 Notes
3 Proof
3.1 Case 1 (Inside the Circle)
3.2 Case 2 (Outside the Circle)
3.3 Case 3 (On the Circle Border)
4 Problems
4.1 Introductory
4.2 Intermediate
4.3 Olympiad
5 See Also
5.1 External Links
Statement
There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application.
Case 1 (Inside the Circle):
If two chords and intersect at a point within a circle, then
Case 2 (Outside the Circle):
Classic Configuration
Given lines and originate from two unique points on the circumference of a circle ( and ), intersect each other at point , outside the circle, and re-intersect the circle at points and respectively, then
Tangent Line
Given Lines and with tangent to the related circle at , lies outside the circle, and Line intersects the circle between and at ,
Case 3 (On the Border/Useless Case):
If two chords, and , have on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is so no matter what, the constant product is .
Alternate Formulation
This alternate formulation is much more compact, convenient, and general.
Consider a circle and a point in the plane where is not on the circle. Now draw a line through that intersects the circle in two places. The power of a point theorem says that the product of the length from to the first point of intersection and the length from to the second point of intersection is constant for any choice of a line through that intersects the circle. This constant is called the power of point . For example, in the figure below
Hint for Proof
Draw extra lines to create similar triangles (Draw on all three figures. Draw another line as well.)
Notice how this definition still works if and coincide (as is the case with ). Consider also when is inside the circle. The definition still holds in this case.
Notes
One important result of this theorem is that both tangents from any point outside of a circle to that circle are equal in length.
The theorem generalizes to higher dimensions, as follows.
Let be a point, and let be an -sphere. Let two arbitrary lines passing through intersect at , respectively. Then
Proof. We have already proven the theorem for a -sphere (a circle), so it only remains to prove the theorem for more dimensions. Consider the plane containing both of the lines passing through . The intersection of and must be a circle. If we consider the lines and with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
Proof
Case 1 (Inside the Circle)
Join and .
In
(Angles subtended by the same segment are equal)
(Vertically opposite angles)
(Corresponding sides of similar triangles are in the same ratio)
Case 2 (Outside the Circle)
Join and
(Why?)
Now, In
(shown above)
(common angle)
(Corresponding sides of similar triangles are in the same ratio)
Case 3 (On the Circle Border)
Length of a point is zero so no proof needed:)
Problems
Introductory
Find the value of in the following diagram:
Solution
Find the value of in the following diagram:
Solution
(ARML) In a circle, chords and intersect at . If and , find the ratio .
Solution
(ARML) Chords and of a given circle are perpendicular to each other and intersect at a right angle at point . Given that , , and , find .
Solution
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is (Source)
Intermediate
Two tangents from an external point are drawn to a circle and intersect it at and . A third tangent meets the circle at , and the tangents and at points and , respectively (this means that T is on the minor arc ). If , find the perimeter of . (Source)
Square of side length has a circle inscribed in it. Let be the midpoint of . Find the length of that portion of the segment that lies outside of the circle. (Source)
is a chord of a circle such that and Let be the center of the circle. Join and extend to cut the circle at Given find the radius of the circle. (Source)
Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find (Source)
Let be a triangle inscribed in circle . Let the tangents to at and intersect at point , and let intersect at . If , , and , can be written as the form , where and are relatively prime integers. Find . (Source)
Olympiad
Given circles and intersecting at points and , let be a line through the center of intersecting at points and and let be a line through the center of intersecting at points and . Prove that if and lie on a circle then the center of this circle lies on line .
(Source)
Let be a point interior to triangle (with ). The lines , and meet again its circumcircle at , , respectively . The tangent line at to meets the line at . Show that from follows .
(Source)
See Also
Geometry
Planar figures
External Links
Handout on AoPS Forums
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Obstetrics & Gynecology/
Jeffcoate's Principles of Gynaecology (Ninth Edition)
Jeffcoate's Principles of Gynaecology (Ninth Edition)
Narendra Malhotra, Jaideep Malhotra, Richa Saxena, Neharika Malhotra Bora
$150.00
The book Jeffcoate’s Principles of Gynaecology is considered to be the Bible for those who want to achieve excellence in the field of Gynaecology. By maintaining the style set by Sir Norman Jeffcoate, an effort has been taken into revising and updating the contents of the book for the Ninth edition with addition of new topics like Genetics in Gynaecology, Female Genital Mutilation, Adenomyosis, Assisted Reproductive Technology, Contraception, and Single Incision Laparoendoscopic Surgery, etc.
This latest edition will be a comprehensive textbook for postgraduate students, residents, surgeons, practitioners and consultants in Obstetrics and Gynaecology.
Table of Contents
Buy as you go
Buy by the chapter and never pay more than the price of the full book.
Chapter 1: A Clinical Approach to Gynaecology
Chapter 2: Anatomy
Chapter 3: Ovarian Functions
Chapter 4: Menstruation and Other Cyclical Phenomena
Chapter 5: Clinical Aspects of Menstruation and Ovulation
Chapter 6: Puberty and Adolescent Gynaecology
Chapter 7: Conception
Chapter 8: Spontaneous Abortions (Including Recurrent Loss)
Chapter 9: Ectopic Pregnancy and Pregnancy of Unknown Location
Chapter 10: Gestational Trophoblastic Disease
Chapter 11: Breast Function and its Disorders
Chapter 12: Development of the Urogenital System
Chapter 13: Malformations and Maldevelopments of the Genital Tract
Chapter 14: Disorders of Sexual Development, Asexuality and Intersexuality
Chapter 15: Injuries
Chapter 16: Female Genital Mutilation
Chapter 17: Pelvic Organ Prolapse
Chapter 18: Other Displacements of the Uterus
Chapter 19: Torsion of Pelvic Organs
Chapter 20: Infections Including Sexually Transmitted Diseases
Chapter 21: Infections as they Affect Individual Organs
Chapter 22: Genital Tuberculosis
Chapter 23: Endometriosis and Chronic Pelvic Pain
Chapter 24: Polycystic Ovary Syndrome
Chapter 25: Hirsutism
Chapter 26: Epithelial Abnormalities of the Genital Tract
Chapter 27: Genital Cancers
Chapter 28: Tumours of the Vulva
Chapter 29: Tumours of the Vagina
Chapter 30: Tumours of the Cervix Uteri
Chapter 31: Tumours of the Corpus Uteri
Chapter 32: Adenomyosis
Chapter 33: Tumours of the Fallopian Tubes
Chapter 34: Tumours of the Pelvic Ligaments
Chapter 35: Tumours of the Ovary
Chapter 36: Chemotherapy in Gynaecological Malignancies
Chapter 37: Radiotherapy in Gynaecological Malignancies
Chapter 38: Immunotherapy in Obstetrics and Gynaecology
Chapter 39: Amenorrhoea, Hypomenorrhoea, and Oligomenorrhoea
Chapter 40: Abnormal and Excessive Uterine Bleeding
Chapter 41: Dysmenorrhoea
Chapter 42: Premenstrual Syndrome and Other Menstrual Phenomena
Chapter 43: Hormone Therapy in Gynaecology
Chapter 44: Vaginal Discharge
Chapter 45: Pruritus Vulvae and Vulvodynia
Chapter 46: Problems of Sex and Marriage
Chapter 47: Male and Female Infertility Overview
Chapter 48: Assisted Reproductive Technology
Chapter 49: Instruments in Gynaecological Procedures
Chapter 50: Ultrasonography in Gynaecology
Chapter 51: Ultrasound in Infertility
Chapter 52: Ultrasound in Menopause
Chapter 53: Robotics Surgery
Chapter 54: Endoscopic Surgery in Gynaecology
Chapter 55: Contraception
Chapter 56: Sterilisation and Termination of Pregnancy
Chapter 57: Urinary Incontinence
Chapter 58: Urinary Tract Infection
Chapter 59: Menopause
Chapter 60: Hysterectomy and its Aftermath
Chapter 61: Conditions of the Lower Intestinal Tract
Chapter 62: Preoperative and Postoperative Management: Postoperative
Chapter 63: Nutrition and Exercise in Women
Chapter 64: Applications of Laser in Gynaecology
Chapter 65: Genetics in Gynaecology
Copyright © 2025 Jaypee Brothers Medical Publishers (P) Ltd. All rights reserved.
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Orders and Primitive Roots Markus Farneb¨ ack† , Isak Fleig, Sixten Georgsson and Oskar ˚ Adahl Minervagymnasium May 2022 Contents 0 Prerequisites 2 1 Introduction 2 2 Orders Modulo a Prime 2 3 Euler’s Phi Function 4 4 General Orders 5 5 Primitive Roots 7 6 Miscellaneous Examples 9 7 Exercises 12 8 Hints 13 9 Further Reading 13 †Berzeliusskolan 1 0 Prerequisites To fully understand the theory described in this document, we strongly recom-mend that the reader is familiar with fundamental concepts in number theory, such as divisibility, primes and basic modular arithmetic.
An explanation of the above mentioned concepts can be found in most books on elementary number theory, see for example .
1 Introduction In this document we introduce orders and primitive roots, two closely related concepts in number theory that are often useful in problems where one has integer powers modulo some integer.
Each section begins with some theory and discussion and is followed by one or more worked-through example problems that demonstrate how the theory can be applied. But remember that the best way to get better at problem solving is by doing. For that reason, we have tried to include numerous exercises for the reader, both at the end of each section and in the section Exercises at the end of this document. For some of the exercises there is a hint available in the section Hints. We recommend that one only uses a hint if one has not achieved any progress on a problem in more than 10 minutes.
In the section named Further Reading, the interested reader can find external material with more problems and information about orders, primitive roots, as well as other related concepts.
2 Orders Modulo a Prime As mentioned in the introduction, orders are related to integer powers modulo some integer. More specifically, we are interested in when some power of an integer is 1 modulo another integer.
Definition 2.1 (Orders Modulo a Prime). Given a prime p, the order of an integer a modulo p, p ∤a, is the smallest positive integer d, such that ad ≡1 (mod p). This is denoted ordp(a) = d.
Now, a relevant question is: Does the order always exist? One could imagine that we could keep multiplying a by itself and never get 1 modulo p. But, the expression ad ≡1 (mod p) should remind us of something, namely Fermat’s Little Theorem. (If the reader is not familiar with this theorem one can learn about it in .) It tells us that ap−1 ≡1 (mod p) if p ∤a. Thus, we know not only that the order always exists, but we also know that it is smaller than p. In fact, we can reduce the possible values of the order even further.
Consider the following table with the values of the orders of the integers 1 to 10 modulo 11.
2 a ord11(a) 1 1 2 10 3 5 4 5 5 5 6 10 7 10 8 10 9 5 10 2 Table 1: Orders modulo 11 All the orders in the table above have one thing in common – they are divisors of 10 = 11 −1. This is nothing unique for the prime 11. We have, in fact, the following useful theorem.
Theorem 2.1 (Fundamental Theorem of Orders). If p is a prime and a is an integer, p ∤a, we have an ≡1 (mod p) ⇐ ⇒ordp(a) | n The proof of this theorem is not very difficult and is left as an exercise for the reader. The importance of Theorem 2.1. cannot be stressed enough; it is used in almost all applications of orders. This should not come as a surprise since it has the word “fundamental” in its name.
But enough theory now. What can we even use this for? It turns out that orders are a very powerful tool in many problems. Consider the following classic problem.
Example 2.1. Determine all n such that n | 2n −1.
In general, most problems where one is asked to find all n that satisfy some condition, tend to only have small solutions that are easy to find, and the tricky part is to show that no other values of n work. After testing some small values, we observe that n = 1 is a solution since 1 | 21 −1 = 1, we also guess that this is the only solution. So, from now on we only care about n > 1.
Since 2n −1 is odd, it follows that n must be odd as well (Why?). We notice that 2n ≡1 (mod n) looks like something we could use orders to work with. But we do not necessarily know that n is prime. For that reason we instead consider a prime p that divides n. We have 2n ≡1 (mod p) and 2p−1 ≡1 (mod p) (Fermat’s Little Theorem) 3 This, together with the Fundamental Theorem of Orders, gives us that ordp(2) | n and ordp(2) | p −1 Since gcd(2, p) = 1 we know that the order must exist. This does not look like it will help us since we still have several possible candidates for ordp(2). Unless we choose p to be the smallest prime dividing n. Then, we cannot have any other number than 1 dividing both n and p −1 and we have ordp(2) = 1 ⇐ ⇒21 ≡1 (mod p) which is impossible. Hence, the only solution is n = 1.
□ The above solution illustrates some of the main ideas used in most problems involving orders, especially the idea of choosing the smallest prime factor, which is worth remembering.
Exercise 2.1. Given a prime p. Show that any prime factor q of pp −1 that is larger than p is congruent to 1 modulo p.
Exercise 2.2. Prove the Fundamental Theorem of Orders.
Exercise 2.3. Let p > 3 be a prime, prove that any prime divisor of 2p + 1 is either 3 or on the form 2kp + 1, for some non-negative integer k.
3 Euler’s Phi Function Prior to introducing general orders and primitive roots, we must briefly mention Euler’s phi function, denoted φ (for a more complete description, see for example ). If this section feels too technical, it is not essential to fully understand it.
Most of this document can be understood anyway. Though, it is important to know that for a prime p, φ(p) = p −1. The phi function is defined as follows.
Definition 3.1 (Euler’s Phi Function). φ(1) = 1. For a positive integer n > 1 φ(n) is the number of positive integers smaller than n that are relatively prime to n.
For the purpose of this document, only two properties regarding this function are necessary to know. These are Euler’s Theorem, the general case of Fermat’s Little Theorem, as well as Euler’s Product Formula, an efficient way to calculate the value of the function. For the sake of brevity, these results are stated without proof, but proofs can easily be found online.
Theorem 3.1 (Euler’s Theorem). If n and a are relatively prime, then aφ(n) ≡1 (mod n) 4 Theorem 3.2 (Euler’s Product Formula).
φ(n) = n Y p|n (1 −1 p) Where the product is over the distinct primes dividing n.
The above formula is easier understood through an example.
Example 3.1. Calculate φ(20).
We have 20 = 22 · 5, so, the distinct primes dividing 20 are 2 and 5. Now, by Euler’s Product Formula, φ(20) = 20(1 −1 2)(1 −1 5) = 20 · 1 2 · 4 5 = 8.
Going back to the definition of φ, we find that the 8 numbers smaller than and relatively prime to 20 are 1, 3, 7, 9, 11, 13, 17, and 19.
□ Here are some exercises to get more familiar with the phi function.
Exercise 3.1. Using Euler’s Theorem, calculate (a) 517 (mod 12) (b) 1382 (mod 60) (c) 2128 (mod 49) Exercise 3.2. Using Euler’s Product Formula, calculate (a) φ(30) (b) φ(100) (c) φ(pk), where p is a prime and k is a positive integer.
Exercise 3.3. Show that the function φ is multiplicative, i.e. if a and b are relatively prime then φ(ab) = φ(a)φ(b).
Exercise 3.4. Using the results from Exercise 3.2. part (c) and Exercise 3.3., find another version of Euler’s product formula for φ(n) based on the prime factorization of n.
4 General Orders Earlier, we discussed orders where the modulo base was a prime. But we can easily generalize the concept of orders to work for all modulo bases. General orders work almost the same as prime orders, with the obvious difference that the modulo base need not be a prime (to clarify, prime orders are a special case of general orders). This has implications for which numbers have an order for a given modulo base.
5 Definition 4.1 (General Orders). Given two relatively prime integers a and n > 0, the order of a modulo n is the smallest positive integer d such that ad ≡1 (mod n).
So, only integers that are relatively prime to the modulo base can have an order, and the existence of this order follows from Euler’s Theorem. Just as the existence of prime orders follows from Fermat’s Little Theorem, as described earlier. To explain this more intuitively, we consider the modulo base n which has a factor k ̸= 1, and the integer a which also has the factor k. If for some d > 0, ad ≡1 (mod n) then, n | ad −1. Since k | n and k | a, then k must divide 1 which is impossible.
It should be noted that most properties of prime orders also apply to general orders. Most importantly, the Fundamental Theorem of Orders holds for general orders as well. General orders are more seldom used in problem solving but do still appear occasionally, for example in the following problem from the Saint Petersburg Mathematical Olympiad.
Example 4.1 (Saint Petersburg Mathematical Olympiad). Prove that for all positive integers a > 1 and n we have n | φ(an −1) We have not mentioned any method for calculating φ of something like an−1, so this suggests that we might not need to calculate it. Maybe we could use Euler’s Theorem instead. For any x, we have xφ(an−1) ≡1 (mod an −1) Now, this looks like something where we could use orders. We would like to have some x such that ordan−1(x) = n, by the Fundamental Theorem of Orders, n = ordan−1(x) | φ(an −1) The question is what this x might be, if it even exists. To begin with, we must have (an −1) | (xn −1). This is obviously true for x = a. But if x = a should work we must, for all d < n, have ad ̸≡1 (mod an −1) which is equivalent to (an −1) ∤(ad −1) But we know this is true since an −1 is larger than ad −1. We also know that the order exists since gcd(a, an −1) = 1. Thus x = a works, and we are done.
□ Exercise 4.1. Determine whether the order exists in the following cases and calculate it if it does.
(a) ord10(5) 6 (b) ord12(7) (c) ord15(2) Exercise 4.2. Given positive integers n and a, and a prime p such that ap ≡−1 (mod n), n ∤a −1, and a + 1 and n are relatively prime. Find ordn(a).
5 Primitive Roots As it turns out, often, the most useful and interesting case is when the order of some integer g modulo n is φ(n). This case is so important that it even has a special name.
Definition 5.1 (Primitive Roots). Given a positive integer n. If ordn(g) = φ(n) then g is a primitive root modulo n. (For a prime p, φ(p) = p −1.) For example, looking at Table 1, we see that 2, 6, 7 and 8 have order 10 = φ(11). So, they are the primitive roots modulo 11.
The natural question now, as with orders, is: Does there always exist a primitive root modulo n? We encourage the reader to investigate for what n there exist primitive roots and come up with an own conjecture before we spoil the answer in the following theorem.
Theorem 5.1 (Existence of Primitive Roots). A primitive root exists modulo n if and only if n = 1, n = 2, n = 4 or if n is in the form pk or 2pk for some positive integer k and odd prime p.
The proof of this result is difficult and out of the scope for this document.
The main takeaway from this theorem is that primitive roots exist for all primes.
This case is by far the most used.
Now, one very important and useful fact regarding primitive roots is the following.
Theorem 5.2. Given a prime p and a primitive root g modulo p, the set {g1, g2, g3, . . . , gp−1} forms a complete set of residues modulo p. (This is equiv-alent to saying that all gi are different modulo p.) Proof. There are p −1 possible values of gi. So, if all are not different then, by the pigeonhole principle, there must exist some i, j, 0 < i < j < p, such that gi ≡gj (mod p) Then, gj−i ≡1 (mod p) But this is impossible, since j −i < p −1 and ordp(g) = p −1. Hence such i, j do not exist and all gi are different modulo p.
One should note that this theorem has an analog for general primitive roots.
Proving that analog theorem is Exercise 5.3. at the end of this section. Now, let us move on to some applications of primitive roots.
7 Example 5.1 (Wilson’s Theorem, one direction). Given a prime p, then (p −1)! ≡−1 (mod p) .
The case of p = 2 is obvious, so from now on we assume p is an odd prime.
Let g be a primitive root modulo p.
(p −1)!
= 1 · 2 · 3 · · · (p −1) ≡ g1 · g2 · g3 · · · gp−1 (mod p) (Theorem 5.2.) = g1+2+3+···+(p−1) = g p(p−1) 2 (Sum of arithmetic sequence) ≡ g p−1 2 (mod p) (Fermat’s Little Theorem, strong form) It is important to note that in the second step, the numbers 1, 2, 3, . . . , p −1 do not necessarily correspond to the powers of g in the specific order they are written. Now, gp−1 ≡1 (mod p) (Fermat’s Little Theorem) ⇐ ⇒p | gp−1 −1 = (g p−1 2 −1)(g p−1 2 + 1) However, from the definition of primitive roots, p ∤g p−1 2 −1. So, we must have p | g p−1 2 + 1 = ⇒(p −1)! ≡g p−1 2 ≡−1 (mod p) □ Why did we take care of the case p = 2 in the beginning?
Exercise 5.1. Prove the other direction of Wilson’s Theorem, i.e. if (p −1)! ≡ −1 (mod p), then p is a prime.
We continue with another example which is also a useful result in its own right.
Example 5.2 (Sum of Powers Modulo n). Let p be a prime and x be a positive integer. Find all residues the sum 1x + 2x + 3x + · · · + (p −1)x can give when divided by p.
8 If p −1 | x, then we have, for some k, x = k(p −1). Now, by Fermat’s Little Theorem, 1x + 2x + 3x + · · · + (p −1)x = 1k(p−1) + 2k(p−1) + 3k(p−1) + · · · + (p −1)k(p−1) ≡ 1k + 1k + 1k + · · · + 1k (mod p) (Fermat’s Little Theorem) = p −1 So, from now on, we assume that p−1 ∤x. Let g be a primitive root modulo p. Then we can write the given sum as 1x + 2x + 3x + · · · + (p −1)x ≡ gx + g2x + g3x + · · · + g(p−1)x (mod p) (Theorem 5.2.) = gx · g(p−1)x −1 gx −1 (Sum of geometric sequence) = gx · (gx)p−1 −1 gx −1 ≡ gx · 1 −1 gx −1 (mod p) (Fermat’s Little Theorem) = 0 Note that is only valid if the denominator gx −1 is not 0 modulo p. But, since g is a primitive root and p −1 ∤x, it follows from the Fundamental Theorem of Orders that we cannot have gx −1 ≡0 (mod p).
Thus, the sum can only give the residues 0 and p −1 when divided by p. □ Exercise 5.2. Find all primitive roots in the following modulo bases.
(a) 10 (b) 30 (c) 17 Exercise 5.3. Given an integer n for which there exists a primitive root g modulo n.
Prove that the set {g1, g2, . . . , gφ(n)}, when considered modulo n, contains all integers relatively prime to n.
Exercise 5.4. Let p be a prime and let g be a primitive root modulo p. Given that g2 ≡g + 1 (mod p) , show that g −1 is also a primitive root modulo p.
6 Miscellaneous Examples In this section, we solve two challenging problems using the theory presented in the previous sections.
Example 6.1 (Fermat’s Christmas Theorem). Let p be an odd prime. Then there exists an integer n such that p | n2 + 1 if and only if p ≡1 (mod 4) .
9 This is an interesting problem, because we can use orders for proving one direction and primitive roots for the other. We start by assuming that there exists an integer n such that p | n2 + 1. This immediately reminds us of orders, because this is n2 ≡−1 (mod p) And, as a result, n4 ≡1 (mod p) Thus, ordp(n) divides 4, but it does not divide 2, and we must have ordp(n) = 4.
And, by Fermat’s Little Theorem and the Fundamental Theorem of Orders, we have 4 = ordp(n) | p −1 Hence, p −1 = 4k, for some positive integer k and p ≡1 (mod 4) as desired.
Now, for the other direction. We assume that p ≡1 (mod 4) and want to find some n such that n2 ≡−1 (mod p) We know that primitive roots are our best tool to control residues of powers.
So, we let g be a primitive root modulo p. Note that in the solution of Wilson’s Theorem, we proved that g (p−1) 2 ≡−1 (mod p) Maybe we could use this. Then we want n2 = g p−1 2 = ⇒n = ±g p−1 4 And since p ≡1 (mod 4), we know that p−1 4 is an integer, and hence is n = ±g p−1 4 . So, this construction is indeed possible, and we are done.
□ We finish this section with one part of a difficult problem from the Swedish TST 2022 (the other part uses methods that are out of the scope for this doc-ument).
Example 6.2 (Sweden TST 2022, modified). Let p be a prime such that p2 | 2p−1 −1. Prove that p −1 has at least two distinct prime factors.
The cases we want to rule out are if p −1 has less than 2 distinct prime factors, i.e. 0 or 1 distinct prime factors. The only possibility for p −1 to have 0 distinct prime factor is if p = 2, but we note that p = 2 does not satisfy the divisibility condition given for p, hence we can assume p > 2. Then p is odd, so, if p−1 has only 1 distinct prime factor, that factor must be 2. Thus p−1 = 2k, for some non-negative integer k.
10 So, suppose there exists a prime p that satisfies p2 | 2p−1 −1 and p −1 = 2k for some positive integer k. If k has any odd factor a > 1, then, using the factorization of a sum of odd powers, we get p = 2k + 1 = (2 k a )a + 1 = (2 k a + 1)(2( k a )a−1 −2( k a )a−2 + · · · + 1) The factor 2 k a + 1 is obviously more than 1 and smaller than p, so, p cannot be prime. This is a contradiction, hence k cannot have any odd factors and must be a power of 2.
When we combine p2 | 2p−1 −1 and p −1 = 2k, we get p2 | 2p−1 −1 = 22k −1 Note that the expression 22k −1 = (22k−1)2 −1 is a difference of squares, so, we can factorize it as 22k −1 = (22k−1 −1)(22k−1 + 1) Now, we can apply the same trick to the left factor, and, by repeating this idea inductively, we get 22k −1 = (22k−1 −1)(22k−1 + 1) = (22k−2 −1)(22k−2 + 1)(22k−1 + 1) .
.
.
= (220 −1)(220 + 1)(221 + 1) · · · (22k−1 + 1) = (220 + 1)(221 + 1) · · · (22k−1 + 1) (1) Now, we know that p = 2k + 1 = 22m + 1 and, since 2k−1 ≥k for all positive integers k (Why?), one of the factors in (1) must be p. Since p2 divides the product and p is prime, there must be some other factor that is divisible by p as well.
Here we can use orders. Since p = 22m + 1, we have 22m ≡−1 (mod p) and thus, 22m+1 = (22m)2 ≡1 (mod p) So, ordp(2) ∤2m and ordp(2) | 2m+1, implying that ordp(2) = 2m+1. Now, any of the factors in (1) before 22m + 1 is smaller than p and cannot be divisible by p. Thus, some of the factors larger than p must be. We let one of those factors 11 be 22n + 1. Now we get 0 ≡ 22n + 1 (mod p) = 22m+1·2n−(m+1) + 1 = (22m+1)2n−(m+1) + 1 ≡ 12n−(m+1) + 1 (mod p) = 2 which is a contradiction, since p > 2. Hence, our assumption about the existence of p is false, and we are done.
□ 7 Exercises The problems in this section are based on the theory presented in this document and are ordered by (estimated) difficulty. One should not feel disappointed if one does not manage to solve all the problems as some are really difficult. Please give each problem a try before looking at the corresponding hints in the next section. Good luck!
Exercise 7.1. If n | 2n + 1, n > 1, find the smallest prime factor of 2n + 1.
Exercise 7.2. Find all n such that n2 | 3n + 1.
Exercise 7.3. Suppose that k ≥2 and let n1, n2, . . . , nk ≥1 be natural numbers having the property that n2 | 2n1 −1, n3 | 2n2 −1, . . . , nk | 2nk−1 −1, n1 | 2nk −1 Show that n1 = n2 = · · · = nk = 1.
Exercise 7.4. Prove that any prime factor of 22n + 1 is congruent to 1 modulo 2n+1.
Exercise 7.5 (Brazil Iberoamerican TST, Round 3 Problem 1). Let p > 10 be a prime. Prove that there exist positive integers m, n with m + n < p, such that p | 5m7n −1.
Exercise 7.6. Given that there exists a primitive root modulo n. Prove that there exist φ(φ(n)) primitive roots modulo n.
Exercise 7.7 (Swedish Correspondence Course 2021/2022, Round 3 Problem 2). The numbers m and n are positive integers, and the number p, p > n, is a prime. Given that pm + n divides pp + 1, prove that n divides m.
12 8 Hints 2.1. What is ordq(p)?
2.2. Use Euclid’s Division Lemma.
2.3. Set q to be any prime divisor of 2p + 1. Use the Fundamental Theorem of Orders.
3.3. Put n = ab into Euler’s Product Formula and separate their respective prime factors.
3.4. Split n into the product of its distinct prime factors.
4.2. What number must ordn(a) divide? Try then to rule out some candidate values of ordn(a) using the given conditions.
5.1. What would happen if p was not prime?
5.3. Use the definition of φ and look at the proof of Theorem 5.2.
5.4. Factorize using difference of squares and then assume g −1 is not a primi-tive root. What would that imply?
7.1. Look at the solution of Example 2.1.
7.2. Can n be even? Investigate what happens modulo 4. Then use a standard order argument.
7.3. Let p be the smallest prime factor of any ni. What is ordp(2)?
7.4. If p is a prime divisor of 22n + 1, what is ordp(2)? Look at the solution of Example 6.2. for more help.
7.5. Express 5 and 7 as powers of a primitive root g. Express m and n using the exponents of g in a clever way.
7.6. Use the result of Exercise 5.3.. How many powers in that set has order φ(n)?
7.7. Take care of the case p = 2. Then, let q be a prime dividing pp + 1. Factor pp + 1. Can q divide both the factors? Find ordq(−p).
9 Further Reading Aditya Khurmi. Modern Olympiad Number Theory.
Brilliant. Order of an Element.
Brilliant. Primitive Roots.
Evan Chen. Orders Modulo a Prime.
Jim Hefferon. Elementary Number Theory.
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https://www.bugfree.ai/data-question/seven-game-series-probability
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Data Interview Question
Frequently Asked QuestionsPress to expand
Frequently Asked Questions
Or Customize Question Press to expand
a 7-Game Series
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Solution & Explanation
Solution & Explanation
To determine the probability that a best-of-7 series between two evenly matched teams goes to a decisive 7th game, we need to consider the conditions required for the series to reach this point. Specifically, both teams must win exactly 3 games each in the first 6 games.
Step-by-Step Solution:
Understanding the Problem:
Mathematical Representation:
Binomial Probability Formula:
The probability of a specific sequence of wins can be calculated using the binomial probability formula:
P(x wins)=(nx)⋅px⋅(1−p)n−xP(\text{x wins}) = \binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}P(x wins)=(xn)⋅px⋅(1−p)n−x
Where:
Calculate Combinations:
(63)=6!3!(6−3)!=6×5×43×2×1=20\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20(36)=3!(6−3)!6!=3×2×16×5×4=20
Calculate Probability:
P(3 wins for each team)=(63)⋅(0.5)3⋅(0.5)3P(\text{3 wins for each team}) = \binom{6}{3} \cdot (0.5)^3 \cdot (0.5)^3P(3 wins for each team)=(36)⋅(0.5)3⋅(0.5)3
P(3 wins for each team)=20×(0.5)6=20×164=2064=516P(\text{3 wins for each team}) = 20 \times (0.5)^6 = 20 \times \frac{1}{64} = \frac{20}{64} = \frac{5}{16}P(3 wins for each team)=20×(0.5)6=20×641=6420=165
Final Result:
Explanation:
Why Binomial Distribution?
Why 516\frac{5}{16}165?
This approach provides a comprehensive understanding of how to calculate the probability of a 7-game series in a competitive environment with evenly matched teams.
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Acute Disseminated Intravascular Coagulation (DIC) & Hyperfibrinolysis - EMCrit Project
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You are here: Home / IBCC / Acute Disseminated Intravascular Coagulation (DIC) & Hyperfibrinolysis
Acute Disseminated Intravascular Coagulation (DIC) & Hyperfibrinolysis
September 18, 2024 by Josh Farkas
CONTENTS
DIC (disseminated intravascular coagulation)
Pathophysiology
Symptoms
Lab studies
Causes
Differential diagnosis
Diagnostic criteria
Sepsis-Induced Coagulopathy (SIC)
Management
purpura fulminans
Common causes
Clinical findings
Diagnostic tests
Diagnosis
Treatment
Antibiotics & source control
Heparin & heparin resistance
Protein C concentrates
Blood product replacement
Vitamin K supplementation
Vasodilators
Surgical debridement
hyperfibrinolysis
Pathophysiology
Causes
Clinical manifestations
Laboratory diagnosis
Treatment
pathophysiology
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the general concept of DIC
Normally, clots form locally at sites of vascular damage. These clots are subsequently degraded after tissue damage is repaired. The processes of clot formation and clot degradation (fibrinolysis) are localized and tightly regulated, based on a balance of counterregulatory proteins (procoagulants vs. anticoagulants regulate clot formation; fibrinolytics vs. antifibrinolytics regulate clot breakdown).
The fundamental pathophysiologic signature of DIC is widespread, uncontrolled clot formation. Clotting may be initiated by various factors (e.g., bacterial lipopolysaccharides, tissue factor released by monocytes or the placenta, or damage to the endothelial glycocalyx). The fine balance of procoagulation vs. anticoagulation is broken, leading to widespread disseminated clot formation. This clot burden places a massive strain on the fibrinolytic system, which is subsequently tasked with breaking down all of the clot. The fine balance of fibrinolysis vs. antifibrinolysis may subsequently become disrupted as numerous proteins are depleted.
DIC may cause a variety of life threats
Disseminated microvascular thrombosis may cause tissue hypoperfusion and tissue damage. This may be exacerbated by the depletion of fibrinolytic proteins, which prevents clots from undergoing thrombolysis (“fibrinolytic shutdown”).
Depletion of anticoagulant proteins may predispose patients to form macrovascular thromboses (e.g., deep vein thrombosis and pulmonary emboli).
Depletion of procoagulant proteins may lead to hemorrhage.
acute versus chronic DIC
Acute DIC results from an acute coagulation trigger (e.g., sepsis). This leads to abrupt and exuberant depletion of coagulation factors, leading to hemostatic imbalances.
This chapter is predominantly about acute DIC – which is more immediately relevant to critical care medicine.
Chronic DIC refers to chronic activation of coagulation, often due to disseminated adenocarcinomas. This causes gradual consumption of coagulation factors, which can be compensated by the production of additional clotting factors. Symptoms and laboratory abnormalities are consequently less notable (this is sometimes termed “chronic compensated DIC”).
symptoms
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DIC may be asymptomatic, or it may cause bleeding and clotting. Symptoms may be dominated by either bleeding or clotting.
microvascular thrombosis causes organ failure
Microvascular thromboses may be an occult cause of organ failures. With the exception of the skin, it is often difficult or impossible to tell that microvascular thrombosis is occurring. The most common manifestations are:
(1) Renal failure.
(2) ARDS (acute respiratory distress syndrome).
(3) Brain involvement may cause delirium, coma, or seizure.
(4) Adrenal failure (Waterhouse-Friedrichsen syndrome) may result from microvascular thromboses, followed by hemorrhagic transformation. The net result is adrenal failure, which can precipitate an adrenal crisis. 📖
(5) Skin: purpura fulminans (occlusion of microvasculature in the skin potentially causing gangrene). This signifies an extreme form of DIC which requires aggressive management (more on this below: ⚡️)
macrovascular thrombosis
DVT (deep vein thrombosis).
PE (pulmonary emboli).
bleeding
Petechiae and purpura may occur on the skin.
Oozing may originate from intravenous catheters or mucus membranes.
Life-threatening hemorrhage can result from gastrointestinal or intracranial hemorrhage. However, in most forms of DIC, severe bleeding is uncommon.
laboratory studies
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DIC screening lab panel
Complete blood count (CBC).
INR and PTT.
Fibrinogen.
D-dimer.
D-dimer
D-dimer is invariably elevated in DIC, often dramatically (e.g., >4,000 ng/mL).
A normal D-dimer essentially excludes DIC.
An elevated D-dimer is nonspecific. Serial measurement that reveals an acutely rising D-dimer might be more suggestive of active DIC. (38861325) However, advanced sepsis-induced DIC may impair fibrinolysis (“fibrinolytic shutdown”), which limits further increases in D-dimer.
thrombocytopenia
Moderate thrombocytopenia is usually seen in DIC. Platelet counts of <30,000/uL are uncommon (25413378, 22735856, 26308340, 23159146, 19581801)
Thrombocytopenia may be the most sensitive laboratory test for the detection of DIC. (30828800) A down-trending platelet count is often the first laboratory abnormality to be noted. (30634199) Of course, thrombocytopenia is not specific to DIC: thrombocytopenia is very common in the ICU due to various causes. 📖
INR and PTT abnormalities
Elevations of INR and PTT support a diagnosis of DIC. However, they can be normal in over half of patients with DIC. (30634199)
These abnormalities may not correspond to clinical hypocoagulability (they measure only clotting factors but do not reflect deficiencies that may be occurring in anticoagulant factors, such as protein C and protein S).
In sepsis-induced DIC, PTT prolongation may lag behind PT prolongation as DIC develops. This occurs because the PTT is reduced by elevated factor VIII levels. (27578502)
fibrinogen
Fibrinogen is the least sensitive marker of DIC, with a sensitivity of only 25%. (30634199) Low fibrinogen supports a diagnosis of DIC but usually isn't seen.
Markedly low fibrinogen levels might raise the possibility of hyperfibrinolysis. ⚡️
Sepsis increases fibrinogen levels so that sepsis-induced DIC may have elevated or normal levels of fibrinogen. Falling fibrinogen may suggest ongoing DIC with the consumption of fibrinogen.
antithrombin-III levels
Antithrombin-III is generally consumed in DIC. Reduced antithrombin-III levels may contribute to heparin resistance among patients with DIC.
Reduced levels of antithrombin-III might be useful to identify early (nonovert) DIC since these levels often fall prior to the development of an obvious consumptive coagulopathy. (38861325)
Other causes of reduced antithrombin levels include:
Exposure to devices (e.g., hemodialysis, ECMO, and possibly IABP).
Medications (especially unfractionated heparin infusion). (71399)
Cirrhosis.
Nephrotic syndrome.
microangiopathic hemolytic anemia (MAHA) can occur
Microangiopathic hemolytic anemia can occur due to microthrombi in the capillaries. This causes intravascular hemolysis with the generation of schistocytes (fragmented RBCs).
role of thromboelastography (TEG) in DIC
Patients with DIC often have reduced levels of both clotting factors and also endogenous anticoagulant proteins. This may create a situation where patients appear to be hypocoagulable based on traditional labs (e.g., platelet count and INR) – but they are actually hypercoagulable.
TEG is a whole-blood integrative test that may help clarify the overall balance of coagulation.
TEG often shows hypercoagulability early in the course of DIC (e.g., reduced R-time). Later on, during the course of fulminant DIC, TEG may show hypocoagulability.
Reduction in the maximal amplitude (MA) may be the most common abnormality seen in DIC. (33682140)
causes of DIC
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sepsis due to a wide variety of pathogens, most notably:
Bacteria (thrombosis predominates):
Gram-negative bacteria.
Toxigenic organisms (e.g., group A streptococci).
Viruses (e.g., viral hemorrhagic fevers).
Malaria.
direct tissue damage
Surgery.
Trauma (especially involving traumatic brain injury).
Thermal injury:
Burns or frostbite.
Heat stroke (thrombosis predominates).
Fat emboli syndrome.
Intravascular hemolysis:
Hemolytic transfusion reaction.
Malaria.
malignancy, especially:
Acute promyelocytic leukemia (AML-M3) and monocytic leukemia (AML-M5) – Tend to cause DIC and hyperfibrinolysis, with prominent bleeding. 📖
Adenocarcinoma (especially pancreas, prostate, lung, gastric, ovary). This is usually a chronic, compensated form of DIC with a tendency towards thrombosis.
Chemotherapy.
obstetric catastrophe (hemorrhage & hyperfibrinolysis often predominate)
Placental abruption.
Amniotic fluid embolism.
Preeclampsia, HELLP syndrome.
Retained dead fetus.
Maternal septic shock (e.g., chorioamnionitis, group A streptococcus infection).
Acute fatty liver of pregnancy.
other
Post-Cardiopulmonary Resuscitation (thrombosis predominates). (29255070)
Severe collagen vascular disease, vasculitis.
Drug-induced.
Heparin-induced thrombocytopenia 📖 can cause DIC in ~10% of cases (although the two processes are usually separate).
Severe hepatitis.
Vascular malformations (e.g., giant hemangiomas or large aortic aneurysms; hemorrhage predominates).
CAPS 📖 (catastrophic antiphospholipid antibody syndrome).
HLH 📖 (hemophagocytic lymphohistiocytosis).
common differential diagnostic considerations
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[1/5] DIC vs. sepsis-induced thrombocytopenia
Sepsis may cause thrombocytopenia via a variety of mechanisms: (29255070)
Impaired platelet production.
Increased platelet consumption (sometimes due to hemophagocytosis).
Splenic sequestration of platelets.
Septic patients often have thrombocytopenia without other abnormalities in their coagulation factors (e.g., normal INR and PTT). Such patients do not have DIC.
[2/5] DIC vs. HIT 📖
A useful feature to sort out HIT vs. DIC is chronicity. DIC usually occurs gradually, beginning around the time of admission. Alternatively, HIT tends to begin abruptly, more than five days after exposure to heparin. The 4-T score may be used to quantify the risk of HIT more precisely.
Most patients with HIT do not have DIC. However, severe HIT may trigger DIC in ~10% of cases.
HIT alone typically causes the following pattern of findings:
Platelet counts aren't profoundly reduced (e.g., platelets >20,000/uL).
INR and fibrinogen levels are normal.
HIT complicated by DIC may cause the following pattern:
Platelets may be profoundly reduced (e.g., platelets <20,000/uL)
INR prolongation and hypofibrinogenemia may occur
[3/5] cirrhosis vs. DIC vs. AICF (accelerated intravascular coagulation and fibrinolysis)
Cirrhosis is a risk factor for DIC because patients with cirrhosis often have reduced levels of coagulation factors and also anti-coagulant proteins (e.g., protein C and protein S). With lower levels of these proteins at baseline, patients with cirrhosis have a delicate balance of procoagulants vs. anticoagulants, which may be more easily disrupted by an acute insult. Thus, cirrhosis may often coexist with DIC. Patients with advanced cirrhosis are also at risk for primary hyperfibrinolysis (termed: accelerated intravascular coagulation and fibrinolysis).
D-dimer:
Stable cirrhosis patients often have a mildly elevated D-dimer.
Markedly elevated D-dimer suggests DIC or AICF.
Platelet count:
Cirrhosis can cause thrombocytopenia, but this is often stable.
A stable platelet count argues against DIC. (29255070)
INR:
Cirrhosis often causes elevation of INR.
INR that is stable or close to its baseline argues against DIC. (29255070)
Factor VIII:
In stable cirrhosis, factor VIII levels will remain normal or elevated (since this is produced by endothelial cells).
A low factor VIII level suggests DIC.
TEG with elevated LY-30: If seen this would support hyperfibrinolysis (but it is very insensitive for ACLF).
Antithrombin III: This may be reduced by cirrhosis and/or DIC.
[4/5] DIC vs. other thrombotic microangiopathies (e.g., TTP)
Both DIC and thrombotic microangiopathies may lead to anemia, schistocyte formation, and thrombocytopenia. However, schistocyte formation is considerably more exuberant in primary thrombotic microangiopathy compared to DIC.
Patients with TTP, HUS, or drug-induced thrombotic microangiopathy should have normal coagulation factors (e.g., INR, PTT, fibrinogen) because clots are composed of platelets and don't involve coagulation system activation. Similarly, the D-dimer should be normal or only slightly elevated. (29255070)
Further discussion on the approach to thrombotic microangiopathy: 📖
[5/5] DIC vs. CAPS (catastrophic antiphospholipid antibody syndrome) 📖
This can be challenging because CAPS causes DIC in ~25% of cases.
Features that may suggest CAPS include:
A history of connective tissue disease (especially antiphospholipid antibody syndrome).
A history of pregnancy loss or illness onset during pregnancy.
Recent withdrawal of anticoagulation.
Skin manifestations (e.g., livedo reticularis or cutaneous necrosis).
diagnostic criteria of DIC
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DIC is a clinical diagnosis
No single lab test is diagnostic of DIC. Rather, the diagnosis of DIC rests roughly on three components:
Constellation of laboratory abnormalities, which is consistent with DIC.
Presence of an underlying disorder known to cause DIC (e.g., trauma, sepsis).
Exclusion of an alternative explanation for coagulation abnormalities (see the differential diagnosis section above).
DIC scoring systems
Several DIC scoring systems exist. The ISTH (International Society of Thrombosis and Hemostasis) score below is the most widely utilized.
Scoring systems help provide objectivity but aren't perfect.
Early in the process of DIC, the patient may not quite reach a positive DIC score (sometimes called “non-overt DIC”). If there is ongoing concern regarding DIC, coagulation labs may be repeated in 12-24 hours to determine if the patient is developing DIC. Often, changes in serial labs may be more illuminating than a single set of labs (e.g., a fibrinogen that is actively falling from 300 mg/dL to 100 mg/dL is more worrisome than a fibrinogen that is stably low at 100 mg/dL).
sepsis induced coagulopathy (SIC)
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what is sepsis-induced coagulopathy (SIC)?
Sepsis-induced coagulopathy essentially refers to DIC caused by sepsis.
Sepsis tends to cause a particular form of DIC with the following characteristics:
Hypercoagulability and microvascular thrombosis are usually the predominant problems.
Fibrinogen levels are generally within the normal range.
Fibrinolysis is suppressed, which may lead to impairment of the breakdown of microthrombi.
The Sepsis-Induced Coagulopathy score is designed specifically to detect DIC in the context of sepsis (table below).
For patients with sepsis, the Sepsis-Induced Coagulopathy score is more sensitive than the ISTH DIC criteria. For example, patients will generally meet sepsis-induced coagulopathy criteria before they deteriorate and meet ISTH DIC criteria. (31099127) However, patients with Sepsis-Induced Coagulopathy seem to have the same mortality as patients meeting the more stringent ISTH DIC criteria – suggesting that the increased sensitivity of the Sepsis-Induced Coagulopathy score may not come at the cost of impaired specificity. (31410983) Unfortunately, since there is no gold standard to diagnose DIC, it's impossible to definitively tell which scoring system is ideal.
how should sepsis-induced coagulopathy be used clinically?
The utility of Sepsis-Induced Coagulopathy is unclear because, currently, there are no proven therapies for this. We are probably missing many septic patients with sepsis-induced coagulopathy, but does this actually matter?
One way that Sepsis-Induced Coagulopathy could be used is as a screening tool for DIC among patients with sepsis (figure below). (31099127)
It must be borne in mind that many disease processes can cause thrombocytopenia plus a mildly elevated INR. Therefore, patients meeting the criteria for sepsis-induced coagulopathy don't necessarily have DIC. Rather, these patients should be thoughtfully investigated with other differential diagnostic possibilities borne in mind (e.g., cirrhosis, heparin-induced thrombocytopenia).
management
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some specific types of DIC require specific treatment:
DIC due to APL (acute promyelocytic leukemia): see 📖
Purpura fulminans: see 📖
treat the underlying disorder
The most important principle of DIC management is to treat the underlying cause.
DIC due to septic shock is perhaps the most notable example of this – treatment should overall focus on the basic tenets of sepsis care.
[1/2] anticoagulation for treatment of microthrombi or overt thrombosis
DVT prophylaxis
DVT prophylaxis should be provided to all critically ill patients unless contraindicated. Contraindications may include:
Active bleeding.
Profound thrombocytopenia (e.g., platelet count <30,000/uL). (30828800)
Profound hypofibrinogenemia (e.g., fibrinogen <80 mg/dL).
Planned procedure with high concern regarding bleeding (e.g., neurosurgery, lumbar puncture).
Subgroup analysis of the SCARLET and KYBERSEPT trials investigating antithrombin III and thrombomodulin anticoagulation suggest that these anticoagulants were potentially beneficial – but only among patients not taking concomitant heparin. This indirectly suggests that heparin could provide benefits analogous to more expensive anticoagulants. (31988789)
therapeutic heparin anticoagulation
Rationale:
Theoretically, anticoagulation with heparin could impede ongoing thrombosis and thereby stop the primary abnormality of DIC (disseminated activation of coagulation).
Unfortunately, research into therapeutic heparin anticoagulation has generally been unimpressive. Thus, heparin anticoagulation isn't indicated in most cases of DIC.
Potential indications for heparin anticoagulation:
Clinical thrombosis (e.g., DVT or pulmonary embolism).
💡 Have a low threshold to obtain DVT studies in this patient population if there is any potential concern for DVT.
Purpura fulminans (this subset of DIC requires an entirely different treatment strategy: 📖).
Severe digital ischemia might be an indication for considering anticoagulation. (33555051)
If there is another indication for anticoagulation (e.g., atrial fibrillation), the presence of DIC could argue for anticoagulation.
Heparin infusion may be preferred:
Patients with DIC often have heparin resistance due to low levels of antithrombin III. Consequently, higher doses of heparin may be required than usual. If low molecular weight heparin is administered in a weight-based fashion without measurement of anti-Xa levels, that could lead to subtherapeutic anticoagulation.
Heparin infusions should ideally be titrated using an anti-Xa level. Patients with DIC will often have an elevated PTT at baseline, which may confuse the use of PTT to monitor heparin dosing.
Further discussion on heparin resistance: 📖
[2/2] blood product replacement (usually avoided)
platelet transfusion
Platelet transfusion should generally be avoided unless:
Platelets <10,000/uL.
There is active bleeding.
There is a planned procedure with a substantial risk of bleeding.
For patients with active bleeding or a pending procedure, transfusion to >30,000-50,000/uL may be reasonable. (29255070)
Due to ongoing platelet consumption, results from platelet transfusion may be disappointing.
fibrinogen supplementation
Profoundly low fibrinogen levels may increase the risk of bleeding, including intracranial hemorrhage.
It may be reasonable to attempt to maintain fibrinogen levels above >~50-100 mg/dL (a target that may be personalized to some extent, depending on the risk of bleeding vs. thrombosis).
For patients with active bleeding or a planned procedure, targeting a fibrinogen level >~150 mg/dL may be reasonable. Various sources recommend targets ranging from >100 mg/dL to >200 mg/dL, with little supporting evidence. The target may be individualized for a specific patient when considering the overall risks of hemorrhage and the state of coagulation (e.g., for patients with marked thrombocytopenia, a higher fibrinogen target may compensate for the thrombocytopenia to a certain degree).
Fibrinogen levels may be supported using cryoprecipitate or fibrinogen concentrates.
coagulation factor replacement (e.g., fresh frozen plasma)
This is confusing because traditional labs (e.g., INR) don't reflect the true coagulation tendency of the blood. INR measures only clotting factors (while ignoring endogenous anticoagulants such as protein C and protein S).
Thromboelastography is likely a better tool to determine the true balance of coagulation. The use of thromboelastography for patients with DIC has not been specifically studied. However, thromboelastography has a proven track record for the management of patients with enormously complex coagulopathies in the operating room (e.g., hepatic transplantation, multiple trauma patients, cardiopulmonary bypass operations) – many of whom doubtless had DIC. Thus, it is a logical extension that thromboelastography could be used to tailor blood product selection in the intensive care unit.
It may be better not to treat INR and PTT values when possible. However, in patients with active bleeding or planned procedures, factor replacement is indicated if thromboelastography reveals enzymatic coagulopathy (e.g., with prolonged R-time).
Fresh frozen plasma (FFP) has traditionally been used for coagulation factor repletion. However, prothrombin complex concentrates may be another option (especially in patients with volume overload).
vitamin K administration
It may be difficult to determine whether vitamin K deficiency exists (since INR prolongation results from DIC).
If vitamin K deficiency is suspected, then vitamin K should be administered empirically (e.g., 10 mg intravenously infused over 30-60 minutes).
purpura fulminans
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Basics of Purpura Fulminans
Pathophysiology
Common causes
Clinical findings
Diagnostic tests
Diagnosis
Treatment
Antibiotics & source control
Heparin & heparin resistance
Protein C concentrates
Blood product replacement
Vitamin K supplementation
Vasodilators
Surgical debridement
basics
Purpura fulminans is an extreme thrombotic subtype of disseminated intravascular coagulation, marked by microvascular thrombosis causing skin necrosis (most typically involving the extremities and digits).
In adults, purpura fulminans is most commonly caused by severe infection. However, the primary risk to life and limb is often the purpura fulminans (rather than the underlying infection). It's not uncommon for patients to lose digits or limbs to this disease.
types of purpura fulminans
Three types:
(1) Congenital form (manifests shortly after birth due to congenital deficiency of protein C, protein S, or antithrombin III).
(2) Postinfectious form (extremely rare, caused by autoantibodies which deplete protein S activity).
(3) Acute infectious form (the most common form in adults, a complication of severe disseminated intravascular coagulation)
The remainder of this section discusses the acute infectious form.
pathophysiology
acquired protein C deficiency
Purpura fulminans can be caused by either congenital or acquired protein C deficiency. In the context of septic shock, protein C deficiency is acquired.
Causes of protein C deficiency include the following:(16635072)
Consumption of protein C due to overwhelming thrombin generation. (30396911)
Reduced hepatic synthesis due to liver dysfunction (purpura fulminans may be associated with shock liver).
Degradation of protein C by elastase released by white blood cells.
Depletion of protein C leads to an imbalance in the coagulation systems, with inadequate safeguards to prevent excessive coagulation.
consequences of protein C deficiency
Normally, protein C is activated by thrombin (IIa), an interaction that is facilitated by thrombomodulin (panel A above). Thrombin is generally procoagulant, but by activating protein C, thrombin participates in negative feedback inhibition of the coagulation system.
Activated protein C has several effects to dampen coagulation:
Inactivation of factors Va and VIIIa (inhibiting enzymatic coagulation).
Inhibition of plasminogen-activator inhibitor-1 (PAI-1) – which overall will tend to favor fibrinolysis.
Cleavage of protease-activated receptor-1 (PAR-1), which sends anti-inflammatory and anti-apoptotic signals to the endothelial cell (thereby protecting it).
Deficiency of protein C may lead to excess fibrin generation, as well as inadequate fibrinolysis. The result can be widespread fibrin deposition within the microvasculature, causing ischemic tissue damage.
common causes of purpura fulminans
Bacterial infections:
[#1] Neisseria meningitides (~20% of patients develop purpura fulminans, especially following treatment with eculizumab).
[#2] Streptococcus pneumoniae (especially asplenic or hypo-splenic patients). (34846563)
Streptococcus pyogenes.
Haemophilus influenzae.
Capnocytophaga canimorsus.
Staphylococcus aureus.
Gram-negatives (Klebsiella pneumoniae, Escherichia coli, Pseudomonas aeruginosa, Proteus mirabilis).
(Less common: Enterococcus faecalis, Leptospira, Rickettsia rickettsii, Vibrio parahaemolyticus)
Viral infections:
Varicella zoster virus.
West Nile virus.
Rubeola virus.
Plasmodium falciparum.
Fungal infections:
Cryptococcus neoformans.
Aspergillus spp.
Fusarium spp.
clinical findings
initial skin findings
Non-blanching, red, patchy macules on the skin with thin, lacy, irregular borders.
Lesions are often most prominent on the extremities.
Initially, the skin may be painful and indurated.
These findings seem to represent the initial manifestation of microthrombi in small dermal vessels (29157918)
rapid progression to skin necrosis within 24-48 hours
Central necrosis occurs, leading to black necrotic lesions (with full-thickness necrosis of the skin). With necrosis, skin may become insensate.
Necrosis may begin in the extremities and extend proximally.
Hemorrhage into necrotic skin may lead to bullae formation.
Eventually, hard eschars may form.
involvement of other organs
Skin findings are most obvious, but thrombosis of small-medium vessels elsewhere in the body may cause failure of other organs. As such, purpura fulminans may be conceptualized as an occlusive, small-vessel vasculopathy.
Adrenal hemorrhage (Waterhouse-Friderichsen syndrome):
This begins with infarction of small vessels within the adrenal glands, which subsequently leads to hemorrhagic transformation, with bleeding into the adrenal glands.
This may lead to adrenal insufficiency which precipitates adrenal crisis 📖.
Renal failure due to glomerular thrombosis.
diagnostic tests
laboratory studies
Prolonged coagulation times:
INR may be normal or elevated (in a range of ~1-3). (32356378)
PTT is usually elevated (~1-3 times the upper limit of normal). (32356378)
Fibrinogen:
This may be low, normal, or high. Infection tends to increase fibrinogen, whereas DIC may consume fibrinogen. Consequently, only a minority of patients have low fibrinogen levels.(32015972) Contrary to popular belief, a normal fibrinogen level excludes neither DIC nor purpura fulminans.
D-dimer:
This is perhaps the most sensitive lab abnormality in purpura fulminans.
The D-dimer is generally very elevated (typically in the range of ~5,000-30,000 ng/mL).
Platelet count: Thrombocytopenia is seen in most patients (e.g. in the range of roughly 25,000-150,000/uL).
Reduced levels of protein C:
Levels will be reduced (e.g., below <40% of normal).
This is a send-out test at most hospitals that takes days to return. As such, it cannot contribute to acute patient management.
(For a patient with high likelihood of having purpura fulminans, consider ordering: fibrinogen, D-dimer, protein C level, protein S level, antithrombin-III level, clotting factor levels including factors 2, 5, 7, 8, 9, 10).
skin biopsy
Skin biopsy may be used to differentiate purpura fulminans from other conditions (e.g., vasculitis). In purpura fulminans, the biopsy will show dermal vessel thrombosis without vasculitis.
Culturing the biopsy may reveal the underlying bacterial pathogen in patients with purpura fulminans due to Neisseria meningitidis. (34846563)
Skin biopsy is usually unnecessary. In situations where the diagnosis of purpura fulminans is extremely likely based on clinical context, avoid delaying therapy while waiting for a skin biopsy.
diagnosis
Diagnosis is often evident based on the observation of characteristic skin changes within a consistent clinical context (septic shock) and a consistent laboratory pattern. However, in situations where the diagnosis isn't clear, the following differential diagnosis should be considered: (29157918)
differential diagnosis
Catastrophic antiphospholipid syndrome (CAPS) 📖
Heparin-induced thrombocytopenia 📖
Necrotizing fasciitis.
Warfarin-induced skin necrosis.
Vasculitis (e.g., Henoch-Schonlein purpura, cryoglobulinemic vasculitis).
Thrombotic thrombocytopenic purpura (TTP).
Cocaine/levamisole toxicity.
treatment: antibiotics and source control
antibiotics
Selection will depend on the clinical scenario.
Consider the use of meningeal dosing if meningitis is possible (e.g., ceftriaxone 2 grams IV Q12).
stress-dose steroid
Adrenal infarction may occur, which may require treatment with stress-dose steroids.
Compared to septic shock in general 📖, there should be a somewhat lower threshold for steroid initiation. (27583208)
vasopressors & inotropes
Avoid vasoconstrictors as much as possible. Vasoconstrictors aren't the cause of tissue necrosis in purpura fulminans, but excess vasoconstrictors may nonetheless aggravate matters.
Utilize agents that increase cardiac output, if possible (e.g., choosing epinephrine as a primary agent rather than norepinephrine).
Avoid vasopressin since it may promote digital ischemia more than other agents do.
treatment: heparin and heparin resistance
therapeutic unfractionated heparin infusion is generally recommended (if possible)
Several studies have evaluated the use of heparin in septic shock. These studies have generally been negative. However, studies have included a wide variety of “septic” patients (most of whom did not have purpura fulminans).
Given that purpura fulminans is a procoagulant form of DIC marked by clinical thrombus formation, guidelines, and expert opinion recommend the use of therapeutic anticoagulation with heparin.
Thrombocytopenia is common but not necessarily a contraindication to heparin infusion. One center reported using therapeutic heparin infusions in patients with platelet counts >30,000/uL, which may be reasonable. (16635072)
heparin administration: nuts and bolts
Therapeutic anticoagulation with heparin is a bit complicated for the following reasons:
PTT will usually be elevated at baseline (making these labs unreliable in determining the heparin effect). Note that an elevated PTT is not necessarily a contraindication to heparin, as this may not correlate with true coagulation tendency within the context of DIC.
Patients are usually resistant to heparin due to a deficiency of antithrombin III, as well as the upregulation of heparin-binding proteins.
⚠️ Weight-based dosing of low molecular weight heparin is likely to fail in patients with purpura fulminans due to heparin resistance. Ideally, heparin should be provided as a continuous infusion of unfractionated heparin, titrated based on anti-Xa levels.
Management of heparin resistance might include the following strategies:
Use of higher doses of heparin (noting that there is no well-defined “maximal dose” of heparin). (29157918)
Repletion of antithrombin III using antithrombin-III concentrates. Some centers will routinely administer antithrombin-III with the goal of maintaining levels >80% normal. (36485139)
Repletion of antithrombin III using blood products that contain antithrombin (e.g., Kcentra or fresh frozen plasma; see discussion of these products below).
Further discussion of heparin resistance: 📖
treatment: protein C replacement +/- FFP
Deficiency of endogenous anticoagulant proteins is central to the pathophysiology of purpura fulminans (especially Protein C). Although high-level evidence is lacking for this rare condition, attempting to replace endogenous anticoagulants to re-balance hemostasis is rational. There are roughly four ways to do this, as described below. The selection of therapy will depend heavily on which products are available at your hospital and local regulations regarding their utilization (e.g., high doses of Kcentra are required, which may not receive pharmacy approval). Please note that purpura fulminans is a life- and limb-threatening disease. Although these treatments are expensive, they may still be cost-saving as compared to surgical amputation and debridement (see section below on surgical management).
[1/3] Protein C concentrate (CEPROTIN) 💉
Recombinant Protein C is a rational therapy for purpura fulminans, given that deficient protein C is a cornerstone of the pathophysiology of this entity. (27583208)
Studies on protein C in purpura fulminans have not been adequately powered to prove any effect on mortality or such patient-centered outcomes as amputation. However, small studies in patients with purpura fulminans have demonstrated that protein C concentrates can improve hematologic endpoints, including (16635072)
Improved protein C activity (thus, there doesn't appear to be a need to use activated protein C).
Reduced D-dimer (implying successful braking of enzymatic coagulation).
Increasing levels of fibrinogen and antithrombin III.
Reduction of pathologically elevated plasminogen activator inhibitor-I (PAI-1). (10632475)
Postmarketing surveillance and case series suggest that protein C replacement is safe. (20723255) In particular, protein C lacks the hemorrhage risk that was observed with activated protein C (XIGRIS).
A reasonable dose might be a loading dose of ~100 units/kg, followed by ~50 units/kg q6hr until D-dimer shows a decreasing trend. (31449632, 12794428) If protein C levels can be measured with rapid turnaround time, the dose may also be titrated against protein C level.
[2/3] Kcentra (four-factor PCC)
Kcentra contains 840-1640 units of Protein C per 1,000 units of KCentra (roughly 1:1).
Kcentra dosed at 100 U/kg would provide the following:
~100 units/kg of protein C.
~0.8-6 units/kg of Antithrombin III. For a 70-kg patient, this would be ~49-420 U of antithrombin III. A typical loading dose of antithrombin III for the management of antithrombin deficiency is 500 mg. So, this might provide a meaningful dose of antithrombin III.
Kcentra could be utilized if protein C concentrate isn't available. The optimal dose is unknown. To provide a standard replacement of protein C (100 U/kg) would require 1000 U/kg of Kcentra, which is an extremely large dose. It might be reasonable to cap the loading dose of Kcentra at 5,000 for this indication. Unfortunately, based on the short half-life of protein C, numerous doses of Kcentra would be required.
[3/3] FFP (fresh frozen plasma)
FFP contains protein C and antithrombin III at roughly physiological concentrations (e.g., ~1 U/ml). It also contains additional anticoagulants, which might provide a more robust strategy for re-establishing hemostatic balance.
The optimal dosing of FFP is not clear. Protein C has a very short half-life in the plasma, so repeated administration of FFP is needed (e.g., 2-4 units q6hrs).
In the absence of more advanced blood products (Protein C concentrate or Kcentra), large volumes of FFP would be needed to re-establish a normal balance of coagulation. Targeting 100 U/kg replacement of protein C would probably be impossible with FFP since this would require a massive dose (100 cc/kg, which for a 70-kg patient is 7 liters of FFP). However, it's probable that a lower dose could provide some benefit. This will provide a substantial fluid load, so diuresis or CRRT might be required to avoid volume overload.
In combination with protein C concentrates and antithrombin, less FFP would be required. Goshua G et al. describe an empiric regimen of 2 units FFP followed by 1 unit every four hours for the first three days, followed by 1 unit every six hours on days #4-5, and then 1 unit every 8 hours on days #6-7 in this context. (36485139)
treatment: replacement of other blood products
general considerations in blood product replacement:
Balance of hemostasis vs. thrombosis:
Overzealous replacement could theoretically aggravate thrombus formation.
However, allowing coagulation levels to fall extremely low could also be dangerous (particularly in patients who are receiving heparin infusions).
Concurrent use of heparin: For patients on therapeutic heparin, higher targets are often reasonable.
If FFP is utilized, note that one 250-ml unit of FFP contains roughly 750 mg of fibrinogen (~3 units/ml).
targets for product administration?
Consider maintaining a fibrinogen level of >100 mg/dL.
Consider maintaining a platelet level >~30,000/uL among patients on a therapeutic heparin infusion. For patients who aren't receiving heparin, a usual platelet transfusion threshold might be appropriate (e.g., targeting >~10,000/uL). (16635072)
treatment: vitamin K supplementation
administration of vitamin K
A key pathophysiological problem in purpura fulminans is a deficiency of protein C. Synthesis of this protein depends on vitamin K.
Replacement should be considered in patients with suspected or possible vitamin K deficiency. Some authors have suggested that vitamin K deficiency could contribute to a tendency towards developing purpura fulminans. (15186640)
Detection of vitamin K deficiency may be challenging because most patients will have an elevated INR due to disseminated intravascular coagulation.
avoidance of warfarin
One of the first effects of warfarin is the depletion of protein C and protein S. This may lead to a transient hypercoagulable state (which is the pathophysiology underlying warfarin-induced skin necrosis).
Initiation of warfarin in a patient with purpura fulminans would be illogical and potentially dangerous.
treatment: vasodilators
For patients with digital ischemia and the threat of digital amputation, the following treatments may be attempted:
topical vasodilators
Topical nitroglycerine for local vasodilation of hypoperfused extremities.
intravenous epoprostenol infusion
Low-dose epoprostenol infusion has been described in case reports as improving distal perfusion and potentially reducing tissue loss (e.g., starting at a dose of 2 ng/kg/min). (21396502) Other case reports have used iloprost, a similar agent unavailable intravenously in the United States. (30871501)
Evidence supporting epoprostenol consists solely of a few case reports. It is indirectly buttressed by evidence regarding epoprostenol in digital ischemia due to refractory Raynaud's phenomenon. (27465880)
In patients with threatened perfusion of their digits who are at risk of amputation, a clinical trial of epoprostenol might be considered under close supervision.
Epoprostenol can cause hypotension so that this therapy would be contraindicated in patients with severe hemodynamic instability. Other side effects are less serious (e.g., nausea, vomiting, abdominal pain, diarrhea).
treatment: surgical debridement
general conceptualization as a burn wound
Extensive skin necrosis due to purpura fulminans may mimic burns.
Some authors have suggested that patients with extensive involvement should be treated in burn centers, but this is rarely done in practice. (32513908)
extremity compartment syndrome
Pathophysiology:
(1) Aggressive fluid administration leads to tissue edema.
(2) Skin necrosis prevents the skin from stretching (especially if the skin is involved circumferentially with purpura fulminans).
Extremity compartment syndrome may be a major driver for the need to amputate limbs. Although purpura fulminans involves the dermis, it usually spares the underlying tissue.
Early surgical decompression with fasciotomy may potentially decrease the need for amputation. (31524150)
Compartment syndrome often is not obvious (and in many cases, it may go entirely unrecognized). When in doubt, surgical consultants should be involved early.
hyperfibrinolysis
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Similar to coagulation, there is a delicate balance between pro-fibrinolytic proteins (e.g., tPA, uPA, plasminogen) and anti-fibrinolytic proteins (e.g., TAFI, PAI-1, ⍺2AP). A relative deficiency of anti-fibrinolytic proteins will cause excessive fibrinolysis, which promotes clinical bleeding.
Hyperfibrinolysis may result from a primary abnormality of fibrinolysis (e.g., abnormality of one or more of the above proteins). Hyperfibrinolysis may also result as a secondary process due to DIC (i.e., the generation of large amounts of fibrin leads to an increase in fibrin breakdown). The following sections and discussion are directed at primary abnormalities of fibrinolysis.
hyperfibrinolysis: causes
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cirrhosis (termed AICF: accelerated intravascular coagulation and hyperfibrinolysis)
Precise rates of AICF are unclear, but this is not uncommon in advanced cirrhosis. Correlates with AICF include:
AICF may be more common among alcoholic cirrhosis, as opposed to nonalcoholic steatohepatitis. (33772480)
AICF is more common among acute-on-chronic liver failure (ACLF). (33772480)
⚠️ AICF is often missed. Failure to diagnose AICF may lead to intractable hemorrhage that doesn't respond to usual treatment.
Pathophysiology of AICF may involve:
Normally, the endothelium releases tiny amounts of tissue plasminogen activator (tPA), which is subsequently cleared by the liver. Failure of the liver to metabolize endogenous tPA leads to increasing tPA levels.
Decreased hepatic production of anti-fibrinolytic proteins (TAFI, PAI-1, and ⍺2AP).
APM (acute promyelocytic leukemia)
Discussed here: 📖
other exposures
Postpartum hemorrhage and placental disorders (release of plasminogen activators from the uterus and placenta).
Trauma.
Cardiopulmonary bypass.
Thrombolytic therapy.
congenital
Alpha2-antiplasmin deficiency.
Plasminogen activator inhibitor-1 deficiency.
Quebec platelet disorder (overexpression of uPA in platelets).
Hemophilia (enhanced lytic susceptibility of fibrin structure, impaired TAFI activation)
FXIII deficiency (enhanced lytic susceptibility of fibrin structure, impaired alpha2-plasmin crosslinking to fibrin).
Dysfibrinogenemias (abnormal fibrin structure increases susceptibility to lysis).
hyperfibrinolysis: clinical manifestations
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Hyperfibrinolysis is difficult to diagnose due to its rarity. Some unusual clinical features may suggest the possibility of hyperfibrinolysis:
Delayed bleeding (clots form following trauma/surgery but subsequently break down).
Intractable bleeding that fails to respond to usual therapy.
Mucosal bleeding (areas with high fibrinolytic activity):
Menorrhagia.
Epistaxis.
Gastrointestinal hemorrhage (especially hyperfibrinolysis related to cirrhosis).
Gingival bleeding.
Urological.
Continuous venous oozing from venipuncture or surgical sites.
Spontaneous bleeding (e.g., intramuscular hematoma).
hyperfibrinolysis: laboratory diagnosis
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general diagnosis of hyperfibrinolysis
D-dimer elevation.
D-dimer is generally profoundly elevated.
Elevated D-dimer is the most sensitive finding for hyperfibrinolysis.
Fibrinogen depletion.
Fibrinogen levels may respond initially to supplementation, but subsequently, fibrinogen levels continually drift downwards and require frequent repletion.
Thromboelastography.
Thromboelastography is insensitive to hyperfibrinolysis because tPA rapidly becomes inactivated in vitro. (26049070) Therefore, a normal LY-30 doesn't exclude hyperfibrinolysis.
If seen, LY30 >3% suggests hyperfibrinolysis. (29363269)
(Euglobulin clot lysis time is an assay for hyperfibrinolysis, but this usually isn't available.)
features that may support a diagnosis of hyperfibrinolysis rather than DIC
Platelet count may be relatively stable over time.
INR and PTT may be normal or close to baseline.
Patients with hyperfibrinolysis may have normal/elevated levels of factor VIII (whereas factor VIII would be reduced in DIC). (21475134)
Elevated LY-30 could be helpful (but this is rarely seen in practice, as discussed above).
hyperfibrinolysis: treatment
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restoration of fibrinogen levels
For patients with hemorrhage, fibrinogen levels must be restored by transfusion of cryoprecipitate or fibrinogen concentrate.
For intractable bleeding, targeting a relatively higher fibrinogen level (e.g., >150-200 mg/dL) could be helpful. Unfortunately, there is no high-level evidence regarding the optimal fibrinogen target.
A combination of fibrinogen replacement with fibrinolysis inhibition is often required to maintain adequate fibrinogen levels over time.
tranexamic acid or aminocaproic acid
A fibrinolysis inhibitor is often required to maintain adequate fibrinogen levels and prevent excessive clot degradation.
Tranexamic acid:
A starting dose may be 1 gram IV, followed by a continuous infusion of one gram over 8 hours repeatedly (i.e., 125 mg/hr infusion). (30986390) This is the standard dosing utilized in traumatology studies, such as the CRASH trials.
The optimal dose is unclear. In refractory cases of hyperfibrinolysis in cirrhosis, higher doses of tranexamic acid may be required (e.g., 200 mg/hour infusion). (31294331). The risk of seizure may increase at higher doses, so avoid using high-dose tranexamic acid in combination with other medications that reduce the seizure threshold.
After the bleeding has resolved, IV tranexamic acid may be converted to oral tranexamic acid (e.g., 1300 mg PO q6hr).
Aminocaproic acid is discussed further here: 📖
DDAVP is contraindicated
DDAVP increases plasminogen activator activity, which may enhance fibrinolysis. Consequently, DDAVP is contraindicated in patients with hyperfibrinolysis. (34209949)
By itself, DDAVP does not cause hyperfibrinolysis.
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When possible, try to avoid the use of blood products to “fix labs” in a patient with asymptomatic DIC (although this may be warranted for a profoundly low platelet count or fibrinogen level).
Sepsis-associated DIC generally causes patients to be prothrombotic. Therefore, holding DVT prophylaxis for patients with mild-moderate thrombocytopenia (e.g., platelet count >30,000) may be inadvisable.
Pitfalls on purpura fulminans:
Failure to involve surgical consultants early for patients who may require fasciotomy for compartment syndrome.
Failure to provide specific therapy for purpura fulminans (these patients will often respond poorly to usual sepsis care).
Excessive use of blood product replacement (e.g., fresh frozen plasma, platelets) in patients who are not bleeding.
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References
10632475 Ettingshausen CE, Veldmann A, Beeg T, Schneider W, Jäger G, Kreuz W. Replacement therapy with protein C concentrate in infants and adolescents with meningococcal sepsis and purpura fulminans. Semin Thromb Hemost. 1999;25(6):537-41. doi: 10.1055/s-2007-994962 [PubMed]
12794428 de Kleijn ED, de Groot R, Hack CE, Mulder PG, Engl W, Moritz B, Joosten KF, Hazelzet JA. Activation of protein C following infusion of protein C concentrate in children with severe meningococcal sepsis and purpura fulminans: a randomized, double-blinded, placebo-controlled, dose-finding study. Crit Care Med. 2003 Jun;31(6):1839-47. doi: 10.1097/01.CCM.0000072121.61120.D8 [PubMed]
15186640 Cone LA, B Waterbor R, Sofonio MV. Purpura fulminans due to Streptococcus pneumoniae sepsis following gastric bypass. Obes Surg. 2004 May;14(5):690-4. doi: 10.1381/096089204323093507 [PubMed]
16635072 Schellongowski P, Bauer E, Holzinger U, Staudinger T, Frass M, Laczika K, Locker GJ, Quehenberger P, Rabitsch W, Schenk P, Knöbl P. Treatment of adult patients with sepsis-induced coagulopathy and purpura fulminans using a plasma-derived protein C concentrate (Ceprotin). Vox Sang. 2006 May;90(4):294-301. doi: 10.1111/j.1423-0410.2006.00760.x [PubMed]
19581801 Sivula M, Pettilä V, Niemi TT, Varpula M, Kuitunen AH. Thromboelastometry in patients with severe sepsis and disseminated intravascular coagulation. Blood Coagul Fibrinolysis. 2009 Sep;20(6):419-26. doi: 10.1097/MBC.0b013e32832a76e1 [PubMed]
20723255 Veldman A, Fischer D, Wong FY, Kreuz W, Sasse M, Eberspächer B, Mansmann U, Schosser R. Human protein C concentrate in the treatment of purpura fulminans: a retrospective analysis of safety and outcome in 94 pediatric patients. Crit Care. 2010;14(4):R156. doi: 10.1186/cc9226 [PubMed]
21396502 Helviz Y, Hersch M, Shmuelevitz L, Atrash J, Einav S. Bad to worse. Am J Med. 2011 Mar;124(3):215-7. doi: 10.1016/j.amjmed.2010.11.007 [PubMed]
22735856 Massion PB, Peters P, Ledoux D, Zimermann V, Canivet JL, Massion PP, Damas P, Gothot A. Persistent hypocoagulability in patients with septic shock predicts greater hospital mortality: impact of impaired thrombin generation. Intensive Care Med. 2012 Aug;38(8):1326-35. doi: 10.1007/s00134-012-2620-2 [PubMed]
23159146 Ostrowski SR, Windeløv NA, Ibsen M, Haase N, Perner A, Johansson PI. Consecutive thrombelastography clot strength profiles in patients with severe sepsis and their association with 28-day mortality: a prospective study. J Crit Care. 2013 Jun;28(3):317.e1-11. doi: 10.1016/j.jcrc.2012.09.003 [PubMed]
25413378 Haase N, Ostrowski SR, Wetterslev J, Lange T, Møller MH, Tousi H, Steensen M, Pott F, Søe-Jensen P, Nielsen J, Hjortrup PB, Johansson PI, Perner A. Thromboelastography in patients with severe sepsis: a prospective cohort study. Intensive Care Med. 2015 Jan;41(1):77-85. doi: 10.1007/s00134-014-3552-9 [PubMed]
26308340 Panigada M, Zacchetti L, L'Acqua C, Cressoni M, Anzoletti MB, Bader R, Protti A, Consonni D, D'Angelo A, Gattinoni L. Assessment of Fibrinolysis in Sepsis Patients with Urokinase Modified Thromboelastography. PLoS One. 2015 Aug 26;10(8):e0136463. doi: 10.1371/journal.pone.0136463 [PubMed]
27477022 Kolev K, Longstaff C. Bleeding related to disturbed fibrinolysis. Br J Haematol. 2016 Oct;175(1):12-23. doi: 10.1111/bjh.14255. Epub 2016 Aug 1 [PubMed]
27465880 Cruz JE, Ward A, Anthony S, Chang S, Bae HB, Hermes-DeSantis ER. Evidence for the Use of Epoprostenol to Treat Raynaud's Phenomenon With or Without Digital Ulcers. Ann Pharmacother. 2016 Dec;50(12):1060-1067. doi: 10.1177/1060028016660324 [PubMed]
27578502 Toh CH, Alhamdi Y, Abrams ST. Current Pathological and Laboratory Considerations in the Diagnosis of Disseminated Intravascular Coagulation. Ann Lab Med. 2016 Nov;36(6):505-12. doi: 10.3343/alm.2016.36.6.505. Erratum in: Ann Lab Med. 2017 Jan;37(1):95 [PubMed]
27583208 Hale AJ, LaSalvia M, Kirby JE, Kimball A, Baden R. Fatal purpura fulminans and Waterhouse-Friderichsen syndrome from fulminant Streptococcus pneumoniae sepsis in an asplenic young adult. IDCases. 2016 Aug 16;6:1-4. doi: 10.1016/j.idcr.2016.08.004 [PubMed]
29157918 Colling ME, Bendapudi PK. Purpura Fulminans: Mechanism and Management of Dysregulated Hemostasis. Transfus Med Rev. 2018 Apr;32(2):69-76. doi: 10.1016/j.tmrv.2017.10.001 [PubMed]
29255070 Levi M, Scully M. How I treat disseminated intravascular coagulation. Blood. 2018 Feb 22;131(8):845-854. doi: 10.1182/blood-2017-10-804096 [PubMed]
29363269 Longstaff C. Measuring fibrinolysis: from research to routine diagnostic assays. J Thromb Haemost. 2018 Apr;16(4):652-662. doi: 10.1111/jth.13957 [PubMed]
30396911 Bendapudi PK, Robbins A, LeBoeuf N, Pozdnyakova O, Bhatt A, Duke F, Sells R, McQuiston J, Humrighouse B, Rouaisnel B, Colling M, Stephenson KE, Saavedra A, Losman JA. Persistence of endothelial thrombomodulin in a patient with infectious purpura fulminans treated with protein C concentrate. Blood Adv. 2018 Nov 13;2(21):2917-2921. doi: 10.1182/bloodadvances.2018024430 [PubMed]
30618150 Iba T, Levy JH, Wada H, Thachil J, Warkentin TE, Levi M; Subcommittee on Disseminated Intravascular Coagulation. Differential diagnoses for sepsis-induced disseminated intravascular coagulation: communication from the SSC of the ISTH. J Thromb Haemost. 2019 Feb;17(2):415-419. doi: 10.1111/jth.14354 [PubMed]
30634199 Thachil J. The Elusive Diagnosis of Disseminated Intravascular Coagulation: Does a Diagnosis of DIC Exist Anymore? Semin Thromb Hemost. 2019 Feb;45(1):100-107. doi: 10.1055/s-0038-1677042 [PubMed]
30828800 Scully M, Levi M. How we manage haemostasis during sepsis. Br J Haematol. 2019 Apr;185(2):209-218. doi: 10.1111/bjh.15821 [PubMed]
30871501 Hage-Sleiman M, Derre N, Verdet C, Pialoux G, Gaudin O, Senet P, Fartoukh M, Boissan M, Garnier M. Meningococcal purpura fulminans and severe myocarditis with clinical meningitis but no meningeal inflammation: a case report. BMC Infect Dis. 2019 Mar 12;19(1):252. doi: 10.1186/s12879-019-3866-x [PubMed]
30898966 Bhatti UF, Williams AM, Raghavendran K, Georgoff PE. Four-extremity amputation following disseminated intravascular coagulation and purpura fulminans. BMJ Case Rep. 2019 Mar 20;12(3):e228028. doi: 10.1136/bcr-2018-228028 [PubMed]
30991817 Patel P, Walborn A, Rondina M, Fareed J, Hoppensteadt D. Markers of Inflammation and Infection in Sepsis and Disseminated Intravascular Coagulation. Clin Appl Thromb Hemost. 2019 Jan-Dec;25:1076029619843338. doi: 10.1177/1076029619843338 [PubMed]
31099127 Iba T, Levy JH, Yamakawa K, Thachil J, Warkentin TE, Levi M; Scientific and Standardization Committee on DIC of the International Society on Thrombosis and Haemostasis. Proposal of a two-step process for the diagnosis of sepsis-induced disseminated intravascular coagulation. J Thromb Haemost. 2019 Aug;17(8):1265-1268. doi: 10.1111/jth.14482 [PubMed]
31410983 Iba T, Levy JH, Warkentin TE, Thachil J, van der Poll T, Levi M; Scientific and Standardization Committee on DIC, and the Scientific and Standardization Committee on Perioperative and Critical Care of the International Society on Thrombosis and Haemostasis. Diagnosis and management of sepsis-induced coagulopathy and disseminated intravascular coagulation. J Thromb Haemost. 2019 Nov;17(11):1989-1994. doi: 10.1111/jth.14578 [PubMed]
31443111 Iba T, Levi M, Levy JH. Sepsis-Induced Coagulopathy and Disseminated Intravascular Coagulation. Semin Thromb Hemost. 2020 Feb;46(1):89-95. doi: 10.1055/s-0039-1694995 [PubMed]
31449632 Kumar A, Shah NP, Menon V, Nissen SE. Purpura fulminans manifesting with Staphylococcus aureus endocarditis: a case report. Eur Heart J Case Rep. 2019 Jun 1;3(2):ytz077. doi: 10.1093/ehjcr/ytz077. PMID: 31449632 [PubMed]
31524150 Koch C, Taeger C, Geis S, Lonic D, Heidekrueger P, Dolderer J, Bitzinger D, Hirche C, Prantl L, Kehrer A. Early fasciotomies and plastic-surgical reconstruction may enhance preservation of functional extremity length in purpura fulminans. Clin Hemorheol Microcirc. 2020;75(3):267-278. doi: 10.3233/CH-190588 [PubMed]
31646137 Asif M, Quiroga L, Lagziel T, Ladd SB, Caffrey J. A Multidisciplinary Approach to the Management of Severe Purpura Fulminans in a Burn Center: A Case Series. Cureus. 2019 Aug 25;11(8):e5478. doi: 10.7759/cureus.5478 [PubMed]
31988789 Wiedermann CJ. Anticoagulant therapy for septic coagulopathy and disseminated intravascular coagulation: where do KyberSept and SCARLET leave us? Acute Med Surg. 2020 Jan 16;7(1):e477. doi: 10.1002/ams2.477 [PubMed]
32015972 Nasrullah A, Javed A, Tariq U, Young M, Moeen Z, Balaan M. Purpura Fulminans: a Rare but Fierce Presentation of Pneumococcal Sepsis. Eur J Case Rep Intern Med. 2019 Dec 30;7(1):001373. doi: 10.12890/2019_001373 [PubMed]
32513908 Djurdjevic N, Taweesedt PT, Paulson M, LaNou A, Radovanovic M, Patel JN, Veselinovic M, McDermott WR, Dumic I. Septic Shock and Purpura Fulminans Due to Streptococcus pneumoniae Bacteremia in an Unvaccinated Immunocompetent Adult: Case Report and Review. Am J Case Rep. 2020 Jun 9;21:e923266. doi: 10.12659/AJCR.923266 [PubMed]
32356378 Brozyna JR, Sardiña LA, Sharma A, Theil KS, Bergfeld WF. Acute purpura fulminans-a rare cause of skin necrosis: A single-institution clinicopathological experience. J Cutan Pathol. 2020 May 1. doi: 10.1111/cup.13732 [PubMed]
32641317 Beechar VB, de la Flor C, Medford RJ. Non-typeable Haemophilus influenzae and purpura fulminans. BMJ Case Rep. 2020 Jul 8;13(7):e234880. doi: 10.1136/bcr-2020-234880 [PubMed]
33555051 Adelborg K, Larsen JB, Hvas AM. Disseminated intravascular coagulation: epidemiology, biomarkers, and management. Br J Haematol. 2021 Mar;192(5):803-818. doi: 10.1111/bjh.17172 [PubMed]
33682140 Hu YL, McRae HL, Refaai MA. Efficacy of viscoelastic hemostatic assay testing in patients with sepsis-induced disseminated intravascular coagulation. Eur J Haematol. 2021 Jun;106(6):873-875. doi: 10.1111/ejh.13617 [PubMed]
33772480 von Meijenfeldt FA, Lisman T. Fibrinolysis in Patients with Liver Disease. Semin Thromb Hemost. 2021 Jul;47(5):601-609. doi: 10.1055/s-0040-1718924 [PubMed]
34846563 Contou D, Urbina T, de Prost N. Understanding purpura fulminans in adult patients. Intensive Care Med. 2022 Jan;48(1):106-110. doi: 10.1007/s00134-021-06580-2 [PubMed]
36485139 Goshua G, Bendapudi PK, Lee AI. Thrombosis questions from the inpatient wards. Hematology Am Soc Hematol Educ Program. 2022 Dec 9;2022(1):481-490. doi: 10.1182/hematology.2022000384 [PubMed]
34209949 Franchini M, Zaffanello M, Mannucci PM. Bleeding Disorders in Primary Fibrinolysis. Int J Mol Sci. 2021 Jun 29;22(13):7027. doi: 10.3390/ijms22137027 [PubMed]
37221630 Iba T, Helms J, Connors JM, Levy JH. The pathophysiology, diagnosis, and management of sepsis-associated disseminated intravascular coagulation. J Intensive Care. 2023 May 23;11(1):24. doi: 10.1186/s40560-023-00672-5 [PubMed]
38861325 Scarlatescu E, Iba T, Maier CL, Moore H, Othman M, Connors JM, Levy JH. Deranged Balance of Hemostasis and Fibrinolysis in Disseminated Intravascular Coagulation: Assessment and Relevance in Different Clinical Settings. Anesthesiology. 2024 Sep 1;141(3):570-583. doi: 10.1097/ALN.0000000000005023 [PubMed]
The Internet Book of Critical Care is an online textbook written by Josh Farkas (@PulmCrit), an associate professor of Pulmonary and Critical Care Medicine at the University of Vermont.
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Entomologist
Are you always pointing out cool bugs you see when you’re outside? If so, you’d probably fit right in as an entomologist, but what is an entomologist exactly?
Entomologists play a role in more industries than most people realize. We can thank them for many advancements in agriculture, public health, and even engineering. Think you have what it takes to join their ranks?
An entomologist is a person who studies insects. Entomology is sometimes considered a subset of wildlife biology, so an entomologist is like a bug biologist. This guide covers everything you need to know about entomology careers, including answers to the questions what does an entomologist study and how much do entomologists make.
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| Career Path Overview for Entomologists | |
| Education Requirements | Bachelor’s Degree |
| Recommended Degree Program | B.S. in Wildlife Conservation |
| Average Salary (2021) | $64,650 |
| Workers Employed in U.S. (2021) | 17,100 |
| Projected Job Openings (2021-31) | 100 |
| Projected Growth Rate | 1% (Little or no change) |
| Other Job Titles | Insect Zoologist, Bug Biologist, Bug Scientist |
| Related Careers | Zoologist, Environmental Scientist, Wildlife Conservationist |
Career Path Overview for Entomologists
Education Requirements
Bachelor’s Degree
Recommended Degree Program
B.S. in Wildlife Conservation
Average Salary (2021)
$64,650
Workers Employed in U.S. (2021)
17,100
Projected Job Openings (2021-31)
100
Projected Growth Rate
1% (Little or no change)
Other Job Titles
Insect Zoologist, Bug Biologist, Bug Scientist
Related Careers
Zoologist, Environmental Scientist, Wildlife Conservationist
Source: Bureau of Labor Statistics
Table of Contents
What Is An Entomologist?
Entomologists study insects, scorpions, spiders, and more. If it’s an arthropod (an animal with a hard exoskeleton), then an entomologist has likely studied it. And there is a lot to research since insects are one of the most prolific and diverse groups of animals on Earth. They may be small in size, but insects have giant impacts on the world. Entomologists look at the greater relationship between the earth’s smallest animals and the rest of the ecosystem.
Humans rely on a healthy relationship with insects for almost all aspects of life, from food production to medicine. Climate change is shifting our relationship to all life, but especially bugs. Some keystone insect species are going extinct, while others are wreaking havoc in the form of invasive species and overpopulation. Therefore, people who study bugs will continue to be vital to the health and sustainability of our planet in the future.
Junior Entomologist Job Description
Now that you can answer the question, “What is a bug scientist called?” let’s explore what the average bug scientist does at work. If you want to become a person who studies bugs, you will likely have to start in an entry-level position after college. What do entomologists do in entry-level positions? The duties and responsibilities of a junior entomologist include:
Advanced Entomologist Job Description
After you gain some years of experience in the field, you may have opportunities to advance your entomology career. Here are some examples of the job duties of a senior-level entomologist:
Where Do Entomologists Work?
Governmental agencies, agriculture organizations, and university departments employ many entomologists. Others work in food processing, pest control, health services, and more. While the growth of the field is slower than average for most careers, the possibilities for innovation are vast due to an increasing understanding of the insect world. From studying how insects have evolved physically and chemically, engineers and medical researchers are taking inspiration from the insect world to solve human problems.
In the Office
Depending on the role and specialty, most entomologists will spend at least some time in the office. Those further in their career might have more managerial tasks. Others might be analyzing data from the field season. Some might be advising other professionals on the insect impact of their field. As with most professional careers, skills in communication, organization, and administration will come in handy to any person who studies bugs.
In the Field
Being someone who studies insects can be exciting due to the number of field opportunities available, especially at the junior level. Conditions for fieldwork can be physically demanding or uncomfortable. You may be staying up all night conducting a moth survey by lamplight or traveling to tropical climates to study dengue fever in mosquito populations.
The key to fieldwork: be comfortable getting dirty. There are always risks when out in the field including insect stings/bites, exposure to allergens, potential for harsh weather conditions, or odd working hours. You will likely be required to use specific protective clothing and equipment.
Average Entomologist Salary
While the Bureau of Labor Statistics (BLS) does not have specific data on the average salary for entomologists, we will look at the data for wildlife biology and zoology because entomology is often considered a branch of zoology. The median annual salary for wildlife biologists in 2021 was $64,650. The top 10% earned more than $103,900, while entry-level workers earned less than $42,420. You can expect an entomologist salary to be somewhere in that range. The highest-paying states for this field are:
What is the Job Outlook for Entomologists?
As we continue to see the impacts of climate change and the loss of biodiversity across the planet, the need for entomologists may increase. As of 2021, the BLS data of zoologists and wildlife biologists shows an expected increase of 100 new jobs in the field in the next decade.
Entomologist Education Recommendations
Ready to learn the details of how to become an entomologist? To become an entomologist, you must like bugs. This is key! Then, you must meet certain educational requirements. Most entomologists have at least a bachelor’s degree, and many go on to earn a master’s and/or Ph.D. Aspiring entomologists will need a strong background in life sciences, laboratory experience, and critical-thinking skills.
Entomologist High School Recommendations
A high school student who is interested in becoming a studier of insects should take science courses including biology and environmental science (if your school offers them). Students serious about bug studies can practice their IDing skills in their own backyard or neighborhood. However, be careful handling unknown or poisonous insects.
Other ways to prepare for your future entomology career while still in high school include:
Entomologist College Education Recommendations
A college degree is essential for someone who studies bugs. While you can learn a lot through self-teaching and exploring your surroundings, you will need technical skills and research experience to start your entomology career. You should earn at least a bachelor’s degree, which takes about four years of full-time college. Look for programs that offer internship opportunities to help jumpstart your career before you even graduate.
Undergraduate
A person who studies bugs can start their higher education journey with a biology degree. If available, it is recommended to take courses on insect biology or get hands-on involvement through research or volunteering. Unity Environmental University offers a course called Humans, Parasites, and Wildlife: Understanding the Impact of Insects on Wildlife as part of our wildlife conservation degree.
Postgraduate
Many successful entomology careers are honed through post-graduate programs focused specifically on the study of wildlife including insects. Consider a master’s degree in wildlife conservation and management from Unity to help you gain leadership experience, more research opportunities, and networking opportunities with professionals in the field to boost your career.
Additional Certifications and Licenses
While a license is not required to be an entomologist, joining a professional organization or getting certified through one can make you stand out amongst the competition when applying for jobs. Furthermore, you can attend networking events to meet other insect lovers and potential employers. The main professional organization for bug biologists in the US is the Entomological Society of America.
Once you have met the education and experience requirements, you can become a Board Certified Entomologist or an Associate Certified Entomologist. The latter certification is specifically for pest management.
Recommended Unity Degrees and Courses
B.S. in Wildlife Conservation
This completely online degree program includes fieldwork and access to our career services to help guide you to the next step in your entomology career after graduation. The program includes coursework in biology, animal identification, and natural resource law and policy.
Explore More Career Paths Related to Entomologists
Zoologist
If you love all animals, you may enjoy a career as a zoologist. Zoologists don’t just work in zoos; they also work in national parks, aquariums, and wildlife refuges. Some zoologists help rehabilitate injured wildlife while others study animals in laboratories.
Environmental Scientist
Being an environmental scientist gives you the opportunity to apply for many different jobs including environmental technician, researcher, or consultant. While environmental scientists may work with insects, they take a more holistic approach to understanding and solving environmental problems.
Wildlife Conservationist
Do you want to work outside? Wildlife conservationists often spend a lot of time in the field studying wildlife (sometimes that includes insects) and collecting samples. Their goals are to understand how humans, animals, and the environment interact and help protect and conserve all living creatures.
You may also find Interesting
Environmental Consultant
Energy Engineer
Adventure Therapist
Unity Environmental University
70 Farm View Drive, Suite 200
New Gloucester, ME 04260
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Illustrative Mathematics
Loading [MathJax]/jax/input/TeX/config.js
Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Introducing the Distributive Property
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Alignments to Content Standards:3.MD.C.7.c
Student View
Task
Part I.
a. How many circles are there in all? Write down a number sentence that shows how you thought about it.
Â
b. Alonso said he figured out how many shaded circles there were first and then how many unshaded circles there were second. Once he knew how many of each, he added them together to find the total. Write a number sentence Alonso could have used that shows his reasoning.
c. Jennifer said, “I just saw 3 rows of 8 circles.â€Â Write a number sentence that Jennifer could have used that shows her reasoning.
​Part II.
a. The area model below shows the floor plan for a storage closet. The storage closet will have a tiled floor with grey tiles on the left and white tiles on the right. How many tiles are needed for the storage closet in all? Write down a number sentence that shows how you thought about it.
b. Think back to how Alonso figured out how many circles there were in Part I. Use Alonso’s same strategy here to find out how many tiles are needed. Write a number sentence with Alonso’s reasoning.
c. Think back to how Jennifer figured out how many circles there were in Part I. Use Jennifer’s same strategy here to find out how many tiles are needed. Write a number sentence with Jennifer's reasoning.
IM Commentary
This is an instructional task, best used when students are first working with the distributive property. Â The standard asks students to apply the distributive property to area models, though this task intentionally begins with array models. Â Students may need to become comfortable with breaking apart discrete sets before breaking apart continuous area models (just as students usually begin understanding multiplication through arrays before transitioning to area models.)
The teacher should facilitate each step of this task. Â When asking students how many circles they see, push students to think beyond counting. Â Some students may see 3 rows of 8 while others may see 4 sets of 6! Â Helping students recognize the equal groups that different students see in the same array is a crucial part of building fluency and flexibility with multiplication.
This task does not explicitly ask students to formally connect Alonso and Jennifer's reasoning strategies as evidence of the distributive property- the teacher will need to make this link explicit after students have had time to think through how Alonso and Jennifer's ways of thinking about the same problem are different. Â Both ways of thinking yield the same answer.
Â
Solutions
Solution: Part I
part a.
Solutions may vary. Â Here are several possibilities:
3 x 8 = 24
4 x 6 = 24 (the six come from every 2 columns of 3 grouped together)
3 x 6 + 3 x 2 = 24
8 + 8 + 8 = 24
part b.
Alonso found that there are 18 shaded dots and 6 unshaded dots, so there are 24 dots in all.
His number sentence would be: 3 x 6 + 3 x 2 = 24
part c.
Jennifer found that there are 24 dots in all.
Her number sentence would be: 3 x 8 = 24
Solution: Part II
part a.Â
Solutions may vary. Â Here are some examples:
5 x 7 = 35
10 + 10 + 10 + 5 = 35
5 + 5 + 5 + 5 + 5 + 5 + 5 = 35
5 x 5 + 10 = 35
part b.
Alonso found 25 shaded tiles and 10 white tiles, so 35 tiles in all.
His number sentence would be: 5 x 5 + 5 x 2 = 35Â
part c.
Jennifer found 35 tiles in total.
Her number sentence would be: 5 x 7 = 35
Introducing the Distributive Property
Part I.
a. How many circles are there in all? Write down a number sentence that shows how you thought about it.
Â
b. Alonso said he figured out how many shaded circles there were first and then how many unshaded circles there were second. Once he knew how many of each, he added them together to find the total. Write a number sentence Alonso could have used that shows his reasoning.
c. Jennifer said, “I just saw 3 rows of 8 circles.â€Â Write a number sentence that Jennifer could have used that shows her reasoning.
​Part II.
a. The area model below shows the floor plan for a storage closet. The storage closet will have a tiled floor with grey tiles on the left and white tiles on the right. How many tiles are needed for the storage closet in all? Write down a number sentence that shows how you thought about it.
b. Think back to how Alonso figured out how many circles there were in Part I. Use Alonso’s same strategy here to find out how many tiles are needed. Write a number sentence with Alonso’s reasoning.
c. Think back to how Jennifer figured out how many circles there were in Part I. Use Jennifer’s same strategy here to find out how many tiles are needed. Write a number sentence with Jennifer's reasoning.
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Diaphysis
Last revised by Andrew Murphy
on 23 Jun 2025
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Citation, DOI, disclosures and article data
Citation:
Jones J, Walizai T, Knipe H, et al. Diaphysis. Reference article, Radiopaedia.org (Accessed on 11 Aug 2025)
DOI:
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rID:
42632
Article created:
1 Feb 2016,
Jeremy Jones
Disclosures:
At the time the article was created Jeremy Jones had no recorded disclosures.
View Jeremy Jones's current disclosures
Last revised:
23 Jun 2025,
Andrew Murphy
Disclosures:
At the time the article was last revised Andrew Murphy had
no financial relationships to ineligible companies to disclose.
View Andrew Murphy's current disclosures
Revisions:
8 times, by
8 contributors -
see full revision history and disclosures
Systems:
Musculoskeletal
Sections:
Anatomy
Tags:
bone
Synonyms:
Diaphyses
Shaft
The diaphyses (singular: diaphysis), sometimes colloquially called the shafts, are the main portions of a long bone (a bone that is longer than it is wide) and provide most of their length.
The diaphysis has a tubular composition with a hard outer section of hard cortical bone and a central portion with cancellous bone and bone marrow cavity.
It is formed by endochondral ossification before longitudinal growth continues with secondary ossification at the physis.
See also
metaphysis
epiphysis
apophysis
physis
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Venn Diagrams
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Venn Diagrams
Resource overview
The first three sheets of this interactive excel file show Venn Diagrams of two sets and allow the student to explore other sets based on their union, intersection and complements. By clicking on the Boolean Algebra expression the region is shown on the Venn Diagram.
The next interactive sheets illustrate sets based upon three sets and various combinations of unions, intersections and complements.
The next sheet lists the elements of a Universal set and the elements of two other sets. The elements are numerical values which can be generated to produce a new example. The elements in associated sets including complements and intersections can be revealed as is the Venn Diagram.
The final interactive sheet is similar to the previous one except that it deals with a Universal set and three other sets.
Note:
This program was designed by be viewed on a screen with a resolution of 1024 x 768. Users may have to adjust the resolution of their screen for the pages to display as was originally intended. The program uses macros which need to be enabled on users' machines.
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Venn diagrams
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| Subject | Mathematics |
| Age group | 11-14, 14-16, 16-19 |
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Half-life (physics)
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From Simple English Wikipedia, the free encyclopedia
This article is about the property of radioactive elements. For the 1998 video game, see Half-Life (video game). For the video game series, see Half-Life (series).
| Number ofhalf-lives over | Partsremaining | As power of 2 |
---
| 0 | 1/1 | |
| 1 | 1/2 | |
| 2 | 1/4 | |
| 3 | 1/8 | |
| 4 | 1/16 | |
| 5 | 1/32 | |
| 6 | 1/64 | |
| 7 | 1/128 | |
| 8 | 1/256 | |
| 9 | 1/512 | |
| 10 | 1/1024 | |
| ... | ... |
| | | |
The half-life of a substance is the time it takes for half of the substance to decay. The word "half-life" was first used when talking about radioactive elements where the number of atoms get smaller over time by changing into different atoms. It is now used in other situations, such as the time it takes for a drug in the body to be half gone. A Geiger-Muller detector can be used to measure the radioactive half-life; it is the time when the activity is half the original.
Half-life depends on probability because the atoms decay at a random time. Half-life is the expected time when half the number of atoms have decayed, on average.
Radioactive isotopes are atoms that have unstable nuclei, meaning that the nucleus of each atom will decay after enough time has passed. Their nuclei are unstable because the arrangement of protons and neutrons in them change. This is known as radioactive decay. When they decay, they release particles such as alpha particles, beta particles, gamma rays. Sometimes they decay by fission, which means to break into pieces, to make smaller nuclei. For example, a radioactive carbon-14 atom releases a beta particle to become nitrogen-14. As an example of fission decay, a fermium-256 atom can split into xenon-140 and palladium-112 atoms, releasing four neutrons in the process.
Uranium-232 has a half-life of about 69 years. Plutonium-238 has a half-life of 88 years. Carbon-14, which is used to find the age of recent fossils, has a half-life of 5,730 years.
After ten half-lives, about 99.9% of the atoms have decayed into different atoms, so only 0.1% of the original atoms are left, and 99.9% of the radioactivity from the original kind of atom is gone. Some atoms decay into other atoms that are also radioactive, so the remaining radioactivity depends on the type of atom.
An example of a short half-life: the isotope, 260Sg, has a half-life of approximately 4 ms (or 4 thousandths of a second).
Related pages
[change | change source]
Biological half-life
Period 8 element
List of radioactive nuclides by half-life
References
[change | change source]
↑ Retrieved 2023-04-03
Other websites
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Online chart of isotopes by half-life from the National Nuclear Data Center
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Half-life (physics)
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Collinear points
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Alexander Katz
and
Hua Zhi Vee
contributed
In Geometry, a set of points are said to be collinear if they all lie on a single line. Because there is a line between any two points, every pair of points is collinear. Demonstrating that certain points are collinear is a particularly common problem in olympiads, owing to the vast number of proof methods.
Contents
Slope-based collinearity test
Linear algebra-based collinearity test
Other collinearity tests
See Also
Slope-based collinearity test
Collinearity tests are primarily focused on determining whether a given 3 points A,B, and C are collinear. This is because it is easily extensible: e.g. showing 4 points A,B,C,D are collinear can be done by first showing A,B,C are collinear then showing B,C,D are collinear as well.
When the coordinates of the points are given, this problem is relatively simple, if sometimes computationally challenging. One simple test simply finds the equation of the line between A and B using standard methods, then finding the equation of the line between B and C, and comparing.
Let A=(1,2),B=(2,4),C=(3,6). Are A,B,C collinear?
The slope of the line between A and B is 2−14−2=2, making the equation of the line between A and B y=2x+b, and setting x=1,y=2 we determine b=0. Thus, the equation of the line becomes y=2x. Similarly, the slope between B and C is 3−26−4=2, making the equation of the line between B and C also y=2x. This is the same line, so A,B,C are collinear.
This can be slightly improved:
Yes
No
Suppose we have points A,B and C, such that the slope between A and B is equal to the slope between B and C. Are A,B and C necessarily collinear points?
The correct answer is: Yes
Linear algebra-based collinearity test
Another, more advanced, method utilizes the fact that the polygon formed by A, B, and C is necessarily degenerate, and therefore has area 0. The shoelace formula, which uses some linear algebra, may thus be used to find the area of "triangle" ABC, and if it is 0 the points are necessarily collinear. This is a popular technique when using barycentric coordinates, where explicitly finding equations of lines is computationally difficult.
Let A=(1,2),B=(2,4),C=(3,6). Are A,B,C collinear?
We find that the determinant of the matrix
111123246
is 0. Thus, by the shoelace formula, the area of △ABC is 0 and so A,B,C are collinear.
Other collinearity tests
Collinearity is a particularly nice property when not explicitly constructed; in particular, there are a number of sets of interesting points that turn out to be collinear, and thus one method of proving collinearity is to demonstrate that the conditions for these theorems hold. For example, to show A,B,C are collinear, one might show that one can construct a configuration so that A,B,C are the intersections specified by Pascal's theorem (which states these intersections are collinear).
For a given triangle, orthocenter, circumcenter, and the center of the nine point circle are always collinear (regardless of the triangle). Together, they form the Euler line.
Menelaus' theorem gives a criterion for points on the sides of a triangle to be collinear.
Given 6 points on a conic, typically a circle, Pascal's theorem states that certain intersections are collinear.
Monge's theorem gives a criterion for three circles to induce a set of collinear points. This is a major win for projective geometry, and Desargues' theorem in particular.
See Also
Concurrency and collinearity - problem solving
Menelaus' Theorem
Simson Line Theorem
Radical Axis of 2 Circles
Cite as:
Collinear points.
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A box and whiskers plot (in the style of Tukey)
Source: R/geom-boxplot.R, R/stat-boxplot.R
The boxplot compactly displays the distribution of a continuous variable. It visualises five summary statistics (the median, two hinges and two whiskers), and all "outlying" points individually.
Usage
geom_boxplot(geom_boxplot( mapping = NULL, = NULL data = NULL, = NULL stat = "boxplot", = "boxplot" position = "dodge2", = "dodge2" ...,... outliers = TRUE, = TRUE outlier.colour = NULL, = NULL outlier.color = NULL, = NULL outlier.fill = NULL, = NULL outlier.shape = NULL, = NULL outlier.size = NULL, = NULL outlier.stroke = 0.5, =0.5 outlier.alpha = NULL, = NULL whisker.colour = NULL, = NULL whisker.color = NULL, = NULL whisker.linetype = NULL, = NULL whisker.linewidth = NULL, = NULL staple.colour = NULL, = NULL staple.color = NULL, = NULL staple.linetype = NULL, = NULL staple.linewidth = NULL, = NULL median.colour = NULL, = NULL median.color = NULL, = NULL median.linetype = NULL, = NULL median.linewidth = NULL, = NULL box.colour = NULL, = NULL box.color = NULL, = NULL box.linetype = NULL, = NULL box.linewidth = NULL, = NULL notch = FALSE, = FALSE notchwidth = 0.5, =0.5 staplewidth = 0, = 0 varwidth = FALSE, = FALSE na.rm = FALSE, = FALSE orientation = NA, = NA show.legend = NA, = NA inherit.aes = TRUE = TRUE))stat_boxplot(stat_boxplot( mapping = NULL, = NULL data = NULL, = NULL geom = "boxplot", = "boxplot" position = "dodge2", = "dodge2" ...,... orientation = NA, = NA coef = 1.5, =1.5 na.rm = FALSE, = FALSE show.legend = NA, = NA inherit.aes = TRUE = TRUE))
Arguments
mapping
: Set of aesthetic mappings created by aes(). If specified and inherit.aes = TRUE (the default), it is combined with the default mapping at the top level of the plot. You must supply mapping if there is no plot mapping.
data
: The data to be displayed in this layer. There are three options:
If `NULL`, the default, the data is inherited from the plot data as specified in the call to `ggplot()`.
A `data.frame`, or other object, will override the plot data. All objects will be fortified to produce a data frame. See `fortify()` for which variables will be created.
A `function` will be called with a single argument, the plot data. The return value must be a `data.frame`, and will be used as the layer data. A `function` can be created from a `formula` (e.g. `~ head(.x, 10)`).
position
: A position adjustment to use on the data for this layer. This can be used in various ways, including to prevent overplotting and improving the display. The position argument accepts the following:
The result of calling a position function, such as `position_jitter()`. This method allows for passing extra arguments to the position.
A string naming the position adjustment. To give the position as a string, strip the function name of the `position_` prefix. For example, to use `position_jitter()`, give the position as `"jitter"`.
For more information and other ways to specify the position, see the layer position documentation.
...
: Other arguments passed on to layer()'s params argument. These arguments broadly fall into one of 4 categories below. Notably, further arguments to the position argument, or aesthetics that are required can not be passed through .... Unknown arguments that are not part of the 4 categories below are ignored.
Static aesthetics that are not mapped to a scale, but are at a fixed value and apply to the layer as a whole. For example, `colour = "red"` or `linewidth = 3`. The geom's documentation has an Aesthetics section that lists the available options. The 'required' aesthetics cannot be passed on to the `params`. Please note that while passing unmapped aesthetics as vectors is technically possible, the order and required length is not guaranteed to be parallel to the input data.
When constructing a layer using a `stat_()` function, the `...` argument can be used to pass on parameters to the `geom` part of the layer. An example of this is `stat_density(geom = "area", outline.type = "both")`. The geom's documentation lists which parameters it can accept.
Inversely, when constructing a layer using a `geom_()` function, the `...` argument can be used to pass on parameters to the `stat` part of the layer. An example of this is `geom_area(stat = "density", adjust = 0.5)`. The stat's documentation lists which parameters it can accept.
The `key_glyph` argument of `layer()` may also be passed on through `...`. This can be one of the functions described as key glyphs, to change the display of the layer in the legend.
outliers
: Whether to display (TRUE) or discard (FALSE) outliers from the plot. Hiding or discarding outliers can be useful when, for example, raw data points need to be displayed on top of the boxplot. By discarding outliers, the axis limits will adapt to the box and whiskers only, not the full data range. If outliers need to be hidden and the axes needs to show the full data range, please use outlier.shape = NA instead.
outlier.colour, outlier.color, outlier.fill, outlier.shape, outlier.size, outlier.stroke, outlier.alpha
: Default aesthetics for outliers. Set to NULL to inherit from the data's aesthetics.
whisker.colour, whisker.color, whisker.linetype, whisker.linewidth
: Default aesthetics for the whiskers. Set to NULL to inherit from the data's aesthetics.
staple.colour, staple.color, staple.linetype, staple.linewidth
: Default aesthetics for the staples. Set to NULL to inherit from the data's aesthetics. Note that staples don't appear unless the staplewidth argument is set to a non-zero size.
median.colour, median.color, median.linetype, median.linewidth
: Default aesthetics for the median line. Set to NULL to inherit from the data's aesthetics.
box.colour, box.color, box.linetype, box.linewidth
: Default aesthetics for the boxes. Set to NULL to inherit from the data's aesthetics.
notch
: If FALSE (default) make a standard box plot. If TRUE, make a notched box plot. Notches are used to compare groups; if the notches of two boxes do not overlap, this suggests that the medians are significantly different.
notchwidth
: For a notched box plot, width of the notch relative to the body (defaults to notchwidth = 0.5).
staplewidth
: The relative width of staples to the width of the box. Staples mark the ends of the whiskers with a line.
varwidth
: If FALSE (default) make a standard box plot. If TRUE, boxes are drawn with widths proportional to the square-roots of the number of observations in the groups (possibly weighted, using the weight aesthetic).
na.rm
: If FALSE, the default, missing values are removed with a warning. If TRUE, missing values are silently removed.
orientation
: The orientation of the layer. The default (NA) automatically determines the orientation from the aesthetic mapping. In the rare event that this fails it can be given explicitly by setting orientation to either "x" or "y". See the Orientation section for more detail.
show.legend
: logical. Should this layer be included in the legends? NA, the default, includes if any aesthetics are mapped. FALSE never includes, and TRUE always includes. It can also be a named logical vector to finely select the aesthetics to display. To include legend keys for all levels, even when no data exists, use TRUE. If NA, all levels are shown in legend, but unobserved levels are omitted.
inherit.aes
: If FALSE, overrides the default aesthetics, rather than combining with them. This is most useful for helper functions that define both data and aesthetics and shouldn't inherit behaviour from the default plot specification, e.g. annotation_borders().
geom, stat
: Use to override the default connection between geom_boxplot() and stat_boxplot(). For more information about overriding these connections, see how the stat and geom arguments work.
coef
: Length of the whiskers as multiple of IQR. Defaults to 1.5.
Note
In the unlikely event you specify both US and UK spellings of colour, the US spelling will take precedence.
Orientation
This geom treats each axis differently and, thus, can thus have two orientations. Often the orientation is easy to deduce from a combination of the given mappings and the types of positional scales in use. Thus, ggplot2 will by default try to guess which orientation the layer should have. Under rare circumstances, the orientation is ambiguous and guessing may fail. In that case the orientation can be specified directly using the orientation parameter, which can be either "x" or "y". The value gives the axis that the geom should run along, "x" being the default orientation you would expect for the geom.
Summary statistics
The lower and upper hinges correspond to the first and third quartiles (the 25th and 75th percentiles). This differs slightly from the method used by the boxplot() function, and may be apparent with small samples. See boxplot.stats() for more information on how hinge positions are calculated for boxplot().
The upper whisker extends from the hinge to the largest value no further than 1.5 IQR from the hinge (where IQR is the inter-quartile range, or distance between the first and third quartiles). The lower whisker extends from the hinge to the smallest value at most 1.5 IQR of the hinge. Data beyond the end of the whiskers are called "outlying" points and are plotted individually.
In a notched box plot, the notches extend 1.58 IQR / sqrt(n). This gives a roughly 95% confidence interval for comparing medians. See McGill et al. (1978) for more details.
Computed variables
These are calculated by the 'stat' part of layers and can be accessed with delayed evaluation. stat_boxplot() provides the following variables, some of which depend on the orientation:
after_stat(width) width of boxplot.
after_stat(ymin) or after_stat(xmin) lower whisker = smallest observation greater than or equal to lower hinger - 1.5 IQR.
after_stat(lower) or after_stat(xlower) lower hinge, 25% quantile.
after_stat(notchlower) lower edge of notch = median - 1.58 IQR / sqrt(n).
after_stat(middle) or after_stat(xmiddle) median, 50% quantile.
after_stat(notchupper) upper edge of notch = median + 1.58 IQR / sqrt(n).
after_stat(upper) or after_stat(xupper) upper hinge, 75% quantile.
after_stat(ymax) or after_stat(xmax) upper whisker = largest observation less than or equal to upper hinger + 1.5 IQR.
References
McGill, R., Tukey, J. W. and Larsen, W. A. (1978) Variations of box plots. The American Statistician 32, 12-16.
See also
geom_quantile() for continuous x, geom_violin() for a richer display of the distribution, and geom_jitter() for a useful technique for small data.
Aesthetics
geom_boxplot() understands the following aesthetics. Required aesthetics are displayed in bold and defaults are displayed for optional aesthetics:
| | |
--- |
| • | x or y |
| • | lower or xlower |
| • | upper or xupper |
| • | middle or xmiddle |
| • | ymin or xmin |
| • | ymax or xmax |
| • | alpha | → NA |
| • | colour | → via theme() |
| • | fill | → via theme() |
| • | group | → inferred |
| • | linetype | → via theme() |
| • | linewidth | → via theme() |
| • | shape | → via theme() |
| • | size | → via theme() |
| • | weight | → 1 |
| • | width | → 0.9 |
Learn more about setting these aesthetics in vignette("ggplot2-specs").
Examples
p <- ggplot(mpg, aes(class, hwy))p <- ggplot(mpg, aes(class, hwy)) p<- ggplot(mpg aes(class hwy))p + geom_boxplot()p + geom_boxplot() p + geom_boxplot() # Orientation follows the discrete axis # Orientation follows the discrete axis # Orientation follows the discrete axisggplot(mpg, aes(hwy, class)) + geom_boxplot()ggplot(mpg, aes(hwy, class)) + geom_boxplot() ggplot(mpg aes(hwy class)) + geom_boxplot()p + geom_boxplot(notch = TRUE)p + geom_boxplot(notch = TRUE) p + geom_boxplot(= TRUE)#> Notch went outside hinges#>#> ℹ Do you want `notch = FALSE`?#> ℹ#> Notch went outside hinges#>#> ℹ Do you want `notch = FALSE`?#> ℹp + geom_boxplot(varwidth = TRUE)p + geom_boxplot(varwidth = TRUE) p + geom_boxplot(= TRUE)p + geom_boxplot(fill = "white", colour = "#3366FF")p + geom_boxplot(fill = "white", colour = "#3366FF") p + geom_boxplot(= "white" = "#3366FF")# By default, outlier points match the colour of the box. Use# By default, outlier points match the colour of the box. Use# By default, outlier points match the colour of the box. Use# outlier.colour to override# outlier.colour to override# outlier.colour to overridep + geom_boxplot(outlier.colour = "red", outlier.shape = 1)p + geom_boxplot(outlier.colour = "red", outlier.shape = 1) p + geom_boxplot(= "red" = 1) # Remove outliers when overlaying boxplot with original data points # Remove outliers when overlaying boxplot with original data points # Remove outliers when overlaying boxplot with original data pointsp + geom_boxplot(outlier.shape = NA) + geom_jitter(width = 0.2)p + geom_boxplot(outlier.shape = NA) + geom_jitter(width = 0.2) p + geom_boxplot(= NA) + geom_jitter(=0.2) # Boxplots are automatically dodged when any aesthetic is a factor # Boxplots are automatically dodged when any aesthetic is a factor # Boxplots are automatically dodged when any aesthetic is a factorp + geom_boxplot(aes(colour = drv))p + geom_boxplot(aes(colour = drv)) p + geom_boxplot(aes(= drv))# You can also use boxplots with continuous x, as long as you supply# You can also use boxplots with continuous x, as long as you supply# You can also use boxplots with continuous x, as long as you supply# a grouping variable. cut_width is particularly useful# a grouping variable. cut_width is particularly useful# a grouping variable. cut_width is particularly usefulggplot(diamonds, aes(carat, price)) +ggplot(diamonds, aes(carat, price)) + ggplot(diamonds aes(carat price)) + geom_boxplot() geom_boxplot() geom_boxplot()#> Warning: Continuous x aesthetic#>Warning: x#> ℹ did you forget `aes(group = ...)`?#> ℹggplot(diamonds, aes(carat, price)) +ggplot(diamonds, aes(carat, price)) + ggplot(diamonds aes(carat price)) + geom_boxplot(aes(group = cut_width(carat, 0.25))) geom_boxplot(aes(group = cut_width(carat, 0.25))) geom_boxplot(aes(= cut_width(carat0.25)))# Adjust the transparency of outliers using outlier.alpha# Adjust the transparency of outliers using outlier.alpha# Adjust the transparency of outliers using outlier.alphaggplot(diamonds, aes(carat, price)) +ggplot(diamonds, aes(carat, price)) + ggplot(diamonds aes(carat price)) + geom_boxplot(aes(group = cut_width(carat, 0.25)), outlier.alpha = 0.1) geom_boxplot(aes(group = cut_width(carat, 0.25)), outlier.alpha = 0.1) geom_boxplot(aes(= cut_width(carat0.25)) =0.1)# \donttest{# \donttest{# \donttest{# It's possible to draw a boxplot with your own computations if you # It's possible to draw a boxplot with your own computations if you # It's possible to draw a boxplot with your own computations if you# use stat = "identity":# use stat = "identity":# use stat = "identity":set.seed(1)set.seed(1)set.seed(1)y <- rnorm(100)y <- rnorm(100) y<- rnorm(100)df <- data.frame(df <- data.frame(df<-data.frame( x = 1, x = 1, = 1 y0 = min(y), y0 = min(y), = min(y) y25 = quantile(y, 0.25), y25 = quantile(y, 0.25), = quantile(y0.25) y50 = median(y), y50 = median(y), = median(y) y75 = quantile(y, 0.75), y75 = quantile(y, 0.75), = quantile(y0.75) y100 = max(y) y100 = max(y) = max(y))))ggplot(df, aes(x)) +ggplot(df, aes(x)) + ggplot(df aes(x)) + geom_boxplot( geom_boxplot(geom_boxplot( aes(ymin = y0, lower = y25, middle = y50, upper = y75, ymax = y100), aes(ymin = y0, lower = y25, middle = y50, upper = y75, ymax = y100), aes(= y0 = y25 = y50 = y75 = y100) stat = "identity" stat = "identity" = "identity" ) ))# }# }# }
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https://www.imomath.com/imocomp/sl06_0627.pdf
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Duˇ san Djuki´ c Vladimir Jankovi´ c Ivan Mati´ c Nikola Petrovi´ c IMO Shortlist 2006 From the book “The IMO Compendium” Springer c ⃝2007 Springer Science+Business Media, Inc.
All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, Inc. 233, Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholary analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden.
The use in this publication of trade names, trademarks, service marks, and similar items, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.
1 Problems 1.1 The Forty-Seventh IMO Ljubljana, Slovenia, July 6–18, 2006 1.1.1 Contest Problems First Day (July 12) 1. Let ABC be a triangle with incenter I. A point P in the interior of the triangle satisfies ∠PBA+∠PCA = ∠PBC +∠PCB.
Show that AP ≥AI, and that equality holds if and only if P = I.
2. Let P be a regular 2006-gon. A diagonal of P is called good if its endpoints divide the boundary of P into two parts, each composed of an odd number of sides of P. The sides of P are also called good.
Suppose P has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of P. Find the maximum number of isosce-les triangles having two good sides that could appear in such a configuration.
3. Determine the least real number M such that the inequality ab(a2 −b2)+bc(b2 −c2)+ca(c2 −a2) ≤M(a2 +b2 +c2)2 holds for all real numbers a, b and c.
Second Day (July 13) 4. Determine all pairs (x,y) of integers such that 1 +2x +22x+1 = y2.
5. Let P(x) be a polynomial of degree n > 1 with integer coefficients and let k be a positive integer. Consider the polynomial 2 1 Problems Q(x) = P(P(...P(P(x))...)), where P occurs k times. Prove that there are at most n integers t such that Q(t) = t.
6. Assign to each side b of a convex polygon P the maximum area of a triangle that has b as a side and is contained in P. Show that the sum of the areas assigned to the sides of P is at least twice the area of P.
1.1.2 Shortlisted Problems 1. A1 (EST) A sequence of real numbers a0, a1, a2, ... is defined by the formula ai+1 = [ai]·{ai}, for i ≥0; here a0 is an arbitrary number, [ai] denotes the greatest integer not exceeding ai, and {ai} = ai −[ai]. Prove that ai = ai+2 for i sufficiently large.
2. A2 (POL) The sequence of real numbers a0, a1, a2, ... is defined recursively by a0 = −1, n ∑ k=0 an−k k +1 = 0 for n ≥1.
Show that an > 0 for n ≥1.
3. A3 (RUS) The sequence c0, c1, ..., cn, ... is defined by c0 = 1, c1 = 0, and cn+2 = cn+1 + cn for n ≥0. Consider the set S of ordered pairs (x,y) for which there is a finite set J of positive integers such that x = ∑j∈J cj, y = ∑j∈J cj−1.
Prove that there exist real numbers α,β, and M with the following property: An ordered pair of nonnegative integers (x,y) satisfies the inequality m < αx+βy < M if and only if (x,y) ∈S.
Remark: A sum over the elements of the empty set is assumed to be 0.
4. A4 (SER) Prove the inequality ∑ i<j aia j ai +a j ≤ n 2(a1 +a2 +···+an) ∑ i CD. Points K and L lie on the line segments AB and CD, respectively, so that AK/KB = DL/LC.
Suppose that there are points P and Q on the line segment KL satisfying ∠APB = ∠BCD and ∠CQD = ∠ABC. Prove that the points P, Q, B, and C are concyclic.
16. G3 (USA) Let ABCDE be a convex pentagon such that ∠BAC = ∠CAD = ∠DAE and ∠ABC = ∠ACD = ∠ADE. The diagonals BD and CE meet at P.
Prove that the line AP bisects the side CD.
17. G4 (RUS) A point D is chosen on the side AC of a triangle ABC with ∠C < ∠A < 90◦in such a way that BD = BA. The incircle of ABC is tangent to AB and AC at points K and L, respectively. Let J be the incenter of triangle BCD. Prove that the line KL intersects the line segment AJ at its midpoint.
18. G5 (GRE) In triangle ABC, let J be the center of the excircle tangent to side BC at A1 and to the extensions of sides AC and AB at B1 and C1, respectively.
Suppose that the lines A1B1 and AB are perpendicular and intersect at D. Let E be the foot of the perpendicular fromC1 to line DJ. Determine the angles ∠BEA1 and ∠AEB1.
19. G6 (BRA) Circles ω1 and ω2 with centers O1 and O2 are externally tangent at point D and internally tangent to a circle ω at points E and F, repsectively.
Line t is the common tangent of ω1 and ω2 at D. Let AB be the diameter of ω perpendicular to t, so that A, E, and O1 are on the same side of t. Prove that the lines AO1, BO2, EF, and t are concurrent.
20. G7 (SVK) In an triangle ABC, let Ma, Mb, Mc, be resepctively the midpoints of the sides BC, CA, AB, and Ta, Tb, Tc be the midpoints of the arcs BC, CA, AB of the circumcircle of ABC, not couning the opposite vertices. For i ∈{a,b,c} let ωi be the circle with MiTi as diameter. Let pi be the common external tangent to 1.1 Copyright c ⃝: The Authors and Springer 5 ωj, ωk ({i, j,k} = {a,b,c}) such that ωi lies on the opposite side of pi than ωj, ωk do. Prove that the lines pa, pb, pc form a triangle similar to ABC and find the ratio of similitude.
21. G8 (POL) Let ABCD be a convex quadrilateral. A circle passing through the points A and D and a circle passing through the points B and C are externally tangent at a point P inside the quadrilateral. Suppose that ∠PAB+∠PDC ≤90◦ and ∠PBA+∠PCD ≤90◦. Prove that AB+CD ≥BC +AD.
22. G9 (RUS) Points A1, B1, C1 are chosen on the sides BC, CA, AB of a triangle ABC respectively. The circumcircles of triangles AB1C1, BC1A1, CA1B1 intersect the circumcircle of triangle ABC again at points A2, B2, C2 respectively (A2 ̸= A, B2 ̸= B, C2 ̸= C). Points A3, B3, C3 are symmetric to A1, B1, C1 with respect to the midpoints of the sides BC, CA, AB, respectively. Prove that the triangles A2B2C2 and A3B3C3 are similar.
23. G10 (SER)IMO6 Assign to each side b of a convex polygon P the maximum area of a triangle that has b as a side and is contained in P. Show that the sum of the areas assigned to the sides of P is at least twice the area of P.
24. N1 (USA)IMO4 Determine all pairs (x,y) of integers satisfying the equation 1 + 2x +22x+1 = y2.
25. N2 (CAN) For x ∈(0,1) let y ∈(0,1) be the number whose nth digit after the decimal point is the 2nth digit after the decimal point of x. Show that if x is rational then so is y.
26. N3 (SAF) The sequence f(1), f(2), f(3), ... is defined by f(n) = 1 n hn 1 i + hn 2 i +···+ hn n i , where [x] denotes the integral part of x.
(a) Prove that f(n +1) > f(n) infinitely often.
(b) Prove that f(n +1) < f(n) infinitely often.
27. N4 (ROM)IMO5 Let P(x) be a polynomial of degree n > 1 with integer co-efficients and let k be a positive integer. Consider the polynomial Q(x) = P(P(...P(P(x))...)), where P occurs k times. Prove that there are at most n integers t such that Q(t) = t.
28. N5 (RUS) Find all integer solutions of the equation x7 −1 x−1 = y5 −1.
29. N6 (USA) Let a > b > 1 be relatively prime positive integers. Define the weight of an integer c, denoted by w(c) to be the minimal possible value of |x| + |y| taken over all pairs of integers x and y such that ax + by = c. An integer c is called a local champion if w(c) ≥w(c ± a) and w(c) ≥w(c ± b). Find all local champions and determine their number.
6 1 Problems 30. N7 (EST) Prove that for every positive integer n there exists an integer m such that 2m +m is divisible by n.
2 Solutions 8 2 Solutions 2.1 Solutions to the Shortlisted Problems of IMO 2006 1. If a0 ≥0 then ai ≥0 for each i and [ai+1] ≤ai+1 = [ai]{ai} < [ai] unless [ai] = 0.
Eventually 0 appears in the sequence [ai] and all subsequent ak’s are 0.
Now suppose that a0 < 0; then all ai ≤0. Suppose that the sequence never reaches 0. Then [ai] ≤−1 and so 1 + [ai+1] > ai+1 = [ai]{ai} > [ai], so the se-quence [ai] is nondecreasing and hence must be constant from some term on: [ai] = c < 0 for i ≥n. The defining formula becomes ai+1 = c{ai} = c(ai −c) which is equivalent to bi+1 = cbi, where bi = ai −c2 c−1. Since (bi) is bounded, we must have either c = −1, in which case ai+1 = −ai −1 and hence ai+2 = ai, or bi = 0 and thus ai = c2 c−1 for all i ≥n.
2. We use induction on n. We have a1 = 1/2; assume that n ≥1 and a1,...,an > 0.
The formula gives us (n + 1)∑m k=1 ak m−k+1 = 1. Writing this equation for n and n +1 and subtracting yields (n +2)an+1 = n ∑ k=1 n +1 n −k +1 − n +2 n −k +2 ak which is positive as so is the coefficient at each ak.
Remark. By using techniques from complex analysis such as contour integrals one can obtain the following formula for n ≥1: an = Z ∞ 1 dx xn(π2 +ln2(x−1)) > 0.
3. We know that cn = φn−1−ψn−1 φ−ψ , where φ = 1+ √ 5 2 and ψ = 1− √ 5 2 are the roots of t2 −t −1. Since cn−1/cn →−ψ, taking α = ψ and β = 1 is a natural choice. For every finite set J ⊆N we have −1 = ∞ ∑ n=0 ψ2n+1 < ψx+y = ∑ j∈J ψ j−1 < ∞ ∑ n=0 ψ2n = φ.
Thus m = −1 and M = φ is an appropriate choice. We now prove that this choice has the desired properties by showing that, for any x,y ∈N with −1 < K = xψ +y < φ, there is a finite set J ⊂N such that K = ∑j∈J ψ j.
Given such K, there are sequences i1 ≤··· ≤ik with ψi1 + ··· + ψik = K (one such sequence consists of y zeros and x ones). Consider all such sequences of minimum length n. Since ψm +ψm+1 = ψm+2, these sequences contain no two consecutive integers. Order such sequences as follows: If ik = jk for 1 ≤k ≤t and it < jt, then (ir) ≺(jr). Consider the smallest sequence (ir)n r=1 in this ordering.
We claim that its terms are distinct. Since 2ψ2 = 1 + ψ3, replacing two equal terms m,m by m−2,m+1 for m ≥2 would yield a smaller sequence, so only 0 or 1 can repeat among the ir. But it = it+1 = 0 implies ∑r ψir > 2+∑∞ k=0 ψ2k+3 = φ, while it = it+1 = 1 similarly implies ∑r ψir < −1, so both cases are impossible, proving our claim. Thus J = {i1,...,in} is a required set.
2.1 Copyright c ⃝: The Authors and Springer 9 4. Since ab a+b = 1 4 a +b −(a−b)2 a+b , the left hand side of the desired inequality equals A = ∑ i<j aia j ai +a j = n −1 4 ∑ k ak −1 4 ∑ i<j (ai −a j)2 ai +a j .
The right hand side of the inequality is equal to B = n 2 ∑aia j ∑ak = n −1 4 ∑ k ak −1 4 ∑ i 0 is odd. A side of the 2006-gon is said to belong to triangle AiA jAk if it lies on the polygonal line AiAi+1 ...Ak. At least one of the odd number of sides AiAi+1,...,A j−1A j and at least one of the sides A jA j+1,...,Ak−1Ak do not belong to any other odd isosceles triangle; assign those two sides to △AiA jAk.
This ensures that every two assigned pairs are disjoint; therefore there are at most 1003 odd isosceles triangles.
An example with 1003 odd isosceles triangles can be attained when the diagonals A2kA2k+2 are drawn for k = 0,...,1002, where A0 = A2006.
9. The number c(P) of points inside P is equal to n −a(P) −b(P), where n = |S|.
Writing y = 1 −x the considered sum becomes 2.1 Copyright c ⃝: The Authors and Springer 11 ∑ P xa(P)yb(P)(x+y)c(P) = ∑ P c(P) ∑ i=0 c(P) i xa(P)+iyb(P)+c(P)−i = ∑ P a(P)+c(P) ∑ k=a(P) c(P) k −a(P) xkyn−k.
Here the coefficient at xkyn−k is the sum ∑P c(P) k−a(P) which equals the number of pairs (P,Z) of a convex polygon P and a k-element subset Z of S whose con-vex hull is P, and is thus equal to n k . Now the required statement immediately follows.
10. Denote by SA (R) the number of strawberries of arrangement A inside rectangle R. We write A ≤B if for every rectangle Q containing the top left corner O we have SB(Q) ≥SA (Q). In this ordering, every switch transforms an arrangement to a larger one. Since the number of arrangements is finite, it is enough to prove that whenever A < B there is a switch taking A to C with C ≤B. Consider the highest row t of the cake which differs in A and B; let X and Y be the positions of the strawberries in t in A and B respectively. Clearly Y is to the left from X and the strawberry of A in the column of Y is below Y. Now consider the highest strawberry X′ of A below t whose column is between X andY (including Y). Let s be the row of X′. Now switch X,X′ to the other two vertices Z,Z′ of the corresponding rectangle, obtain-ing an arrangement C . We claim that C ≤B. It is enough to ver-ify that SC (Q) ≤SB(Q) for those rectangles Q = OMNP with N ly-ing inside XZX′Z′. Let Q′ = OMN1P 1 be the smallest rectangle contain-ing X. Our choice of s ensures that SC (Q) = SA (Q′) ≥SB(Q′) ≥SB(Q), as claimed.
X′ O M N N1 P P1 t s X Y Z Z′ 11. Let q be the largest integer such that 2q | n. We prove that an (n,k)-tournament exists if and only if k < 2q.
The first l rounds of an (n,k)-tournament form an (n,l)-tournament. Thus it is enough to show that a (n,2q −1)-tournament exists and a (n,2q)-tournament does not.
If n = 2q, we can label the contestants and rounds by elements of the additive group Zq 2. If contestants x and x + j meet in the round labelled j, it is easy to verify the conditions. If n = 2qp, we can divide the contestants into p disjoint groups of 2q and perform a (2q,2q −1)-tournament in each, thus obtaining an (n,2q −1)-tournament.
For the other direction let Gi be the graph of players with edges between any two players who met in the first i rounds. We claim that the size of each connected component of Gi is a power of 2. For i = 1 this is obvious; assume it holds for i.
Suppose that the components C and D merge in the (i+1)-th round. Then some 12 2 Solutions c ∈C and d ∈D meet in this round. Moreover, each player in C meets a player in D. Indeed, for every c′ ∈C there is a path c = c0,c1,...,ck = c′ with cjcj+1 ∈Gi; then if d j is the opponent of cj in the (i+1)-th round, condition (ii) shows that each d jd j+1 ∈Gi, so dk ∈D. Analogously, all players in D meet players in C, so |C| = |D|, proving our claim. Now if there are 2q rounds, every component has size at least 2q +1 and is thus divisible by 2q+1, which is impossible if 2q+1 ∤n.
12. Let U and D be the sets of upward and downward unit triangles, respectively.
Two triangles are neighbors if they form a diamond. For A ⊆D, denote by F(A) the set of neighbors of he elements of A.
If a holey triangle can be tiled with diamonds, in every upward triangle of side l there are l2 elements of D, so there must be at least as many elements of U and at most l holes.
Now we pass to the other direction. It is enough to show the condition (ii) of the marriage theorem: For every set X ⊂D we have |F(X)| ≥|X|. Indeed, the theorem claims that then we can ”marry” the elements of D with the elements of U, which means exactly covering T by diamonds. So, assume to the contrary that |F(X)| < |X| for some set X. Note that two elements of D having a common neighbor must share a vertex; this means that we can focus on connected sets X.
Consider an upward triangle of side 3. It contains three elements of D; if two of them are in X, adding the third one to X increases F(X) by at most 1, so |F(X)| < |X| still holds. Continuing this procedure we will end up with a set X forming an upward sub-triangle of T and satisfying |F(X)| < |X|, which contradicts the conditions of the problem. This contradiction proves that |F(X)| ≥|X| for every set X.
13. Consider a polyhedron P with v vertices, e edges and f faces. Consider the map σ to the unit sphere S taking each vertex, edge or face x of P to the set of outward unit normal vectors (i.e. points on S) to the support planes of P containing x. Thus σ maps faces to points on S, edges to shorter arcs of big circles connecting some pairs of these points, and vertices to spherical regions formed by these arcs. These points, arcs and regions on S form a ”spherical polyhedron” G .
We now translate the conditions of the problem into the language of G . Denote by x the image of x in reflection in the center of S. No edge of P being parallel to another edge or face means that the big circle of any edge e of G does not contain any vertex V non-incident to e. Also note that vertices A and B of P are antipodal if and only if σ(A) and σ(B) intersect, and that the midpoints of edges a and b are antipodal if and only if σ(a) and σ(b) intersect.
Consider the union F of G and G . The faces of F are the intersections of faces of G and G , so their number equals 2A. Similarly, the edges of G and G have 2B intersections, so F has 2e+4B edges and 2 f +2B vertices. Now Euler’s theorem for F gives us 2e+4B+2 = 2A+2 f +2B, and therefore A−B = e−f +1.
14. The condition of the problem implies that ∠PBC + ∠PCB = 90◦−α/2, i.e.
∠BPC = 90◦+ α/2 = ∠BIC. Thus P lies on the circumcircle ω of △BCI. It 2.1 Copyright c ⃝: The Authors and Springer 13 is well known that the center M of ω is the second intersection of AI with the circumcircle of △ABC. Therefore AP ≥AM−MP = AM−MI = AI, with equal-ity if and only if P ≡I.
15. The relation AK/KB = DL/LC implies that AD,BC and KL have a common point O. Moreover, since ∠APB = 180◦−∠ABC and ∠DQC = 180◦−∠BCD, line BC is tangent to the circles APB and CQD. These two circles are homothetic with respect to O, so if OP meets circle APB again at P′, we have ∠PQC = ∠PP′B = ∠PBC, showing that P,Q,B,C lie on a circle.
16. Let the diagonals AC and BD meet at Q and AD and CE meet at R. The quadri-laterals ABCD and ACDE are similar, so AQ/QC = AR/RD. Now if AP meets CD at M, the Ceva theorem gives us CM MD = CQ QA · AR RD = 1.
17. Let M be the point on AC such that JM ∥KL. It is enough to prove that AM = 2AL.
From ∠BDA = α we obtain that ∠JDM = 90◦−α 2 = ∠KLA = ∠JMD; hence JM = JD and the tangency point of the incircle of △BCD with CD is the mid-point T of segment MD. Therefore, DM = 2DT = BD+CD−BC = AB−BC+ CD, which gives us AM = AD+DM = AC +AB−BC = 2AL.
18. Let A1B1 and CJ intersect at K.
Then JK is parallel and equal to C1D and DC1/C1J = JK/JB1 = JB1/JC = C1J/JC, so the right triangles DC1J and C1JC are similar; hence C1C ⊥ DJ. Thus E belongs to CC1. Now the points A1,B1 and E lie on the circle with diameter CJ. Therefore ∠DBA1 = ∠A1CJ = ∠A1ED, implying A J C1 B1 A1 B C D K E that BEA1D is cyclic; hence ∠A1EB = 90◦. Likewise, ADEB1 is cyclic because ∠EB1A = ∠EJC = ∠EDC1, so ∠AEB1 = 90◦.
Second solution. The segments JA1,JB1,JC1 are tangent to the circles with di-ameters A1B,AB1,C1D. Since JA2 1 = JB2 1 = JC2 1 = JD·JE, E lies on the first two circles (with diameters A1B and AB1), so ∠AEB1 = ∠A1EB = 90◦.
19. The homothety with center E mapping ω1 to ω maps D to B, so D lies on BE; analogously, D lies on AF. Let AE and BF meet at point C. The lines BE and AF are the altitudes of triangle ABC, so D is the orthocenter and C lies on t. Let the line through D parallel to AB meet AC at M. The centers O1 and O2 are the midpoints of DM and DN respectively.
A B C O K E F D O1 M O2 P 14 2 Solutions We have thus reduced the problem to a classical triangle geometry problem: If CD and EF intersect at P, we should prove that points A, O1 and P are collinear (analogously, so are B,O2,P). By the Menelaus theorem for triangle CDM, this is equivalent to CA AM = CP PD, which is again equivalent to CK KD = CP PD (because DM ∥AB), where K is the foot of the altitude from C to AB. The last equality immediately follows from the fact that the pairs C,D; P,K are harmoni-cally adjoint.
20. Let I be the incenter of △ABC. It is well known that TaTc and TaTb are the perpendicular bisectors of the segments BI and CI respectively. Let TaTb meet AC at P and ωb at U, and let TaTc meet AB at Q and ωc at V. We have ∠BIQ = ∠IBQ = ∠IBC, so IQ ∥BC; similarly IP ∥BC. Hence PQ is the line through I parallel to BC.
The homothety from Tb mapping ωb to the circumcircle ω of ABC maps the tangent t to ωb at U to the tangent to ω at Ta which is parallel to BC. It follows that t ∥BC. Let t meet AC at X. Since XU = XMb and ∠PUMb = 90◦, X is the midpoint of PMb. Similarly, the tangent to ωc at V meets QMc at its midpoint Y.
But since XY ∥PQ ∥MbMc, points U,X,Y,V are collinear, so t coincides with the common tangent pa. Thus pa runs midway between I and MbMc. Analogous conclusions hold for pb and pc, so these three lines form a triangle homothetic to the triangle MaMbMc from center I in ratio 1 2 which is therefore similar to the triangle ABC in ratio 1 4.
21. The following proposition is easy to prove: Lemma. For an arbitrary point X inside a convex quadrilateral ABCD, circles ADX and BCX are tangent at X if and only if ∠ADX +∠BCX = ∠AXB.
Let Q be the second intersection point of the circles ABP and CDP (we assume Q ̸≡P; the opposite case is similarly handled). It follows from the conditions of the problem that Q lies inside quadrilateral ABCD (since ∠BCP + ∠BAP < 180◦, C is outside the circumcircle of APB; the same holds for D). If Q is inside △APD (the other case is similar), we have ∠BQC = ∠BQP+∠PQC = ∠BAP+ ∠CDP ≤90◦. Similarly ∠AQD ≤90◦. Moreover, ∠ADQ + ∠BCQ = ∠ADP + ∠BCP = ∠APB = ∠AQB implies that circles ADQ and BCQ are tangent at Q.
Therefore the interiors of the semicircles with diameters AD and BC are disjoint and if M, N are the midpoints of AD and BC respectively, we have 2MN ≥ AD + BC. On the other hand, 2MN ≤AB +CD because − → BA+ − → CD = 2− − → MN, and the statement of the problem immediately follows.
22. We work with oriented angles modulo 180◦. For two lines a,b we denote by ∠(l,m) the angle of counterclockwise rotation transforming a to b; also, by ∠ABC we mean ∠(BA,BC).
It is well-known that the circles AB1C1, BC1A1 and CA1B1 have a common point, say P. Let O be the circumcenter of ABC. Denote ∠PB1C = ∠PC1A = ∠PA1B = ϕ. Let A2P,B2P,C2P meet the circle ABC again at A4,B4,C4, respectively. Since ∠A4A2A = ∠PA2A = ∠PC1A = ϕ and thus ∠A4OA = 2ϕ etc, △ABC is the image of △A4B4C4 under rotation R about O by 2ϕ.
2.1 Copyright c ⃝: The Authors and Springer 15 Therefore ∠(AB4,PC1) = ∠B4AB + ∠AC1P = ϕ −ϕ = 0, so AB4 ∥PC1.
Let PC1 intersect A4B4 at C5; define A5,B5 analogously. Then ∠B4C5P = ∠A4B4A = ϕ, so AB4C5C1 is an isosce-les trapezoid with BC3 = AC1 = B4C5.
Similarly, AC3 = A4C5, so C3 is the im-age of C5 under R; similar statements hold for A5,B5. Thus △A3B3C3 ∼ = △A5B5C5. It remains to show that △A5B5C5 ∼△A2B2C2.
We have seen that ∠A4B5P = ∠B4C5P, A B C P A1 B1 C1 A2 B2 C2 A4 B4 C4 A5 B5 C5 which implies that P lies on the circle A4B5C5. Analogously, P lies on the circle C4A5B5. Therefore ∠A2B2C2 = ∠A2B2B4 +∠B4B2C2 = ∠A2A4B4 +∠B4C4C2 = ∠PA4C5 +∠A5C4P = ∠PB5C5 +∠A5B5P = ∠A5B5C5, and similarly for the other angles, which is what we wanted.
23. Let Si be the area assigned to side AiAi+1 of polygon P = A1 ...An of area S.
We start with the following auxiliary statement.
Lemma. At least one of the areas S1,...,Sn is not smaller than 2S/n.
Proof. It suffices to show the statement for even n. The case of an odd n will then follow immediately from this case applied on the degenerated 2n-gon A1A′ 1 ...AnA′ n, where A′ i is the midpoint of AiAi+1.
Let n = 2m. For i = 1,2,...,m, denote by Ti the area of the region Pi inside the polygon bounded by the diagonals AiAm+i, Ai+1Am+i+1 and the sides AiAi+1, Am+iAm+i+1. We observe that the regions Pi cover the entire poly-gon. Indeed, let X be an arbitrary point inside the polygon, to the left (with-out loss of generality) of the ray A1Am+1.
Then X is to the right of the ray Am+1A1, so there is a k such that X is to the left of ray AkAk+m and to the right of ray Ak+1Ak+m+1, i.e. X ∈ Pk. It follows that T1 +···+Tm ≥S; hence at least one Ti is not smaller than 2S/n, say T1 ≥2S/n.
Let O be the intersection point of A1Am+1 and A2Am+2, and let us assume without loss of generality A1 A2 A3 Am+1 Am+2 Am+3 O X P1 P1 that SA1A2O ≥SAm+1Am+2O and A1O ≥OAm+1. Then we have S1 ≥SA1A2Am+2 = SA1A2O + SA1Am+2O ≥SA1A2O + SAm+1Am+2O = T1 ≥2S/n, which proves the lemma.
16 2 Solutions If, contrary to the assertion, ∑Si S < 2, we can choose rational numbers qi = 2mi/N with N = m1 + ··· + mn such that qi > Si/S. However, considering the given polygon as a degenerated N-gon obtained by division of side AiAi+1 into mi equal parts for each i and applying the lemma we obtain Si/mi ≥2S/N, i.e.
Si/S ≥qi for some i, a contradiction.
Equality holds if and only if P is centrally symmetric.
Second solution. We say that vertex V is assigned to side a of a convex (possibly degenerate) polygon P if the triangle determined by a and V has the maximum area Sa among the triangles with side a contained in P. Denote σ(P) = ∑a Sa and δ(P) = σ(P)−2[P]. We use induction on the number n of pairwise non-parallel sides of P to show that δ(P) ≥0 for every polygon P. This is obvi-ously true for n = 2, so let n ≥3.
There exist two adjacent sides AB and BC whose respective assigned vertices U and V are distinct. Let the lines through U and V parallel to AB and BC respec-tively intersect at point X. Assume w.l.o.g. that there are no sides of P lying on UX and VX. Call the sides and vertices of P lying within the triangle UVX passive (excluding vertices U and V). It is easy to see that no passive vertex is assigned to any side of P and that vertex B is assigned to every passive side.
Now replace all passive vertices of P by X, obtaining a polygon P′. Vertex B is assigned to sides UX and VX or P′, so the sum of areas assigned to passive sides increases by the area S of the part of quadrilateral BUXV lying outside P; the other assigned areas do not change. Thus σ increases by S. On the other hand, the area of the polygon also increases by S, so δ decreases by S.
Note that the change from P to P′ decreases the number of nonparallel sides.
Thus by the inductive hypothesis we have δ(P) ≥δ(P′) ≥0.
Third solution. To each convex n-gon P = A1A2 ...An we assign a centrally symmetric 2n-gon Q, called the associate of P, as follows. Attach the 2n vec-tors ±− − − − → AiAi+1 at a common origin and label them b1,··· ,b2n counterclockwise so that bn+i = −bi for 1 ≤i ≤n. Then take Q to be the polygon B1B2 ...B2n with − − − − → BiBi+1 = bi. Denote by ai the side of P corresponding to bi (i = 1,...,n).
The distance between the parallel sides BiBi+1 and Bn+iBn+i+1 of Q equals twice the maximum height of P to the side ai. Thus, if O is the center of Q, the area of △BiBi+1O (i = 1,...,n) is exactly the area Si assigned to side ai of P; therefore [Q] = 2∑Si. It remains to show that d(P) = [Q]−4[P] ≥0.
(i) Suppose that P has two parallel sides ai and a j, where a j ≥ai and remove from it the parallelogram D determined by ai and a part of side a j. We obtain a polygon P′ with a smaller number of nonparallel sides. Then the associate of P′ is obtained from Q by removing a parallelogram similar to D in ratio 2 (and with area four times that of D); thus d(P′) = d(P).
(ii) Suppose that there is a side bi (i ≤n) of Q such that the sum of the angles at its endpoints is greater than 180◦. Extend the pairs of sides adjacent to bi and bn+i to their intersections U and V, thus enlarging Q by two congruent triangles to a polygon Q′. Then Q′ is the associate of the polygon P′ ob-2.1 Copyright c ⃝: The Authors and Springer 17 tained from P by attaching a triangle congruent to BiBi+1U to the side ai.
Therefore d(P′) equals d(P) minus twice the area of the attached triangle.
By repeatedly performing the operations (i) and (ii) to polygon P we will even-tually reduce it to a parallelogram E, thereby decreasing the value of d. Since d(E) = 0, it follows that d(P) ≥0.
Remark. Polygon Q is the Minkowski sum of P and a polygon centrally sym-metric to P. Thus the inequality [Q] ≥4[P] is a direct consequence of the Brunn-Minkowski inequality.
24. Obviously x ≥0. For x = 0 the only solutions are (0,±2). Now let (x,y) be a solution with x > 0. Assume w.l.o.g. that y > 0. The equation rewritten as 2x(1 + 2x+1) = (y −1)(y + 1) shows that one of the factors y ± 1 is divisible by 2 but not by 4 and the other by 2x−1 but not by 2x; hence x ≥3. Thus y = 2x−1m + ε, where m is odd and ε = ±1. Plugging this in the original equation and simplifying yields 2x−2(m2 −8) = 1 −εm.
(∗) As m = 1 is obviously impossible, we have m ≥3 and hence ε = −1. Now (∗) gives us 2(m2 −8) ≤1 + m, implying m = 3 which leads to x = 4 and y = 23.
Thus all solutions are (0,±2) and (4,±23).
25. If x is rational, its digits repeat periodically starting at some point. If n is the length of the period of x, the sequence 2,22,23,... is eventually periodic modulo n, so the corresponding digits of x (i.e. the digits of y) also make an eventually periodic sequence, implying that y is rational.
26. Consider g(n) = [ n 1]+[ n 2]+···+[ n n] = n f(n) and define g(0) = 0. Since for any k the difference [ n k] −[ n−1 k ] equals 1 if k divides n and 0 otherwise, we obtain that g(n) = g(n −1) + d(n), where d(n) is the number of positive divisors of n. Thus g(n) = d(1) + d(2) + ···+ d(n) and f(n) is the arithmetic mean of the numbers d(1),...,d(n). Therefore, (a) and (b) will follow if we show that each of d(n+1) > f(n) and d(n+1) < f(n) holds infinitely often. But d(n+1) < f(n) holds whenever n +1 is prime, and d(n +1) > f(n) holds whenever d(n +1) > d(1),...,d(n) (which clearly holds for infinitely many n).
27. We first show that every fixed point x of Q is in fact a fixed point of P ◦P.
Consider the sequence given by x0 = x and xi+1 = P(xi) for i ≥0. Assume xk = x0. We know that u −v divides P(u) −P(v) for every two distinct integers u,v.
In particular, di = xi+1 −xi | P(xi+1)−P(xi) = xi+2 −xi+1 = di+1 for all i, which together with dk = d0 implies |d0| = |d1| = ··· = |dk|. Suppose that d1 = d0 = d ̸= 0. Then d2 = d (otherwise x3 = x1 and x0 will never occur in the sequence again). Similarly, d3 = d etc, and hence xi = x0 +id ̸= x0 for all i, a contradiction. It follows that d1 = −d0, so x2 = x0 as claimed. Thus we can assume that Q = P◦P.
If every integer t with P(P(t)) = t also satisfies P(t) = t, the number of solutions is clearly at most degP = n. Suppose that P(t1) = t2, P(t2) = t1, P(t3) = t4 i 18 2 Solutions P(t4) = t3, where t1 ̸= t2,3,4 (but not necessarily t3 ̸= t4). Since t1 −t3 divides t2 −t4 and vice versa, we conclude that t1 −t3 = ±(t2 −t4). Assume that t1 −t3 = t2−t4, i.e. t1−t2 = t3−t4 = u ̸= 0. Since the relation t1−t4 = ±(t2−t3) similarly holds, we obtain t1 −t3 + u = ±(t1 −t3 −u) which is impossible. Therefore, we must have t1 −t3 = t4 −t2, which gives us P(t1) +t1 = P(t3) + t3 = c for some c. It follows that all integral solutions t of the equation P(P(t)) = t satisfy P(t)+t = c, and hence their number does not exceed n.
28. Every prime divisor p of x7−1 x−1 = x6 + ··· + x + 1 is congruent to 0 or 1 modulo 7. Indeed, If p | x−1, then x7−1 x−1 ≡1 +···+1 ≡7 (mod p), so p = 7; otherwise the order of x modulo p is 7 and hence p ≡1 (mod 7). Therefore every positive divisor d of x7−1 x−1 satisfies d ≡0 or 1 (mod 7).
Now suppose (x,y) is a solution of the given equation. Since y−1 and y4 +y3 + y2 +y+1 divide x7−1 x−1 = y5 −1, we have y ≡1 or 2 and y4 +y3 +y2 +y+1 ≡0 or 1 (mod 7). However, y ≡1 or 2 implies that y4 +y3 +y2 +y+1 ≡5 or 3 (mod 7), which is impossible.
29. All representations of n in the form ax + by (x,y ∈Z) are given by (x,y) = (x0 + bt,y0 −at), where x0,y0 are fixed and t ∈Z is arbitrary. The following lemma enables us to determine w(n).
Lemma. The equality w(ax+by) = |x|+|y| holds if and only if: (i) a−b 2 < y ≤a+b 2 and x ≥y−a+b 2 ; or (ii) −a−b 2 ≤y ≤a−b 2 and x ∈Z; or (iii) −a+b 2 ≤y < −a−b 2 and x ≤y+ a+b 2 .
Proof. Assume w.l.o.g. that y ≥0. We have w(ax + by) = |x| + y if and only if |x+b|+|y−a|≥|x|+y and |x−b|+(y+a)≥|x|+y, where the latter is ob-viously true and the former clearly implies y < a. Then the former inequality becomes |x + b| −|x| ≥2y −a. We distinguish three cases: if y ≤a−b 2 then 2y −a ≤b and the previous inequality always holds; for a−b 2 < y ≤a+b 2 it holds if and only if x ≥y−a+b 2 ; and for y > a+b 2 it never holds.
Now let n = ax+by be a local champion with w(n) = |x|+|y|. As in lemma, we distinguish three cases: (i) a−b 2 < y ≤a+b 2 . Then x+1 ≥y−a+b 2 by the lemma, so w(n+a) = |x+1|+y (because n+a = a(x+1)+by). Since w(n+a) ≤w(n), we must have x < 0.
Likewise, w(n−a) equals either |x−1|+y = w(n)+1 or |x+b−1|+a−y.
The condition w(n −a) ≤w(n) leads to x ≤y−a+b−1 2 ; hence x = y−[ a+b 2 ] and w(n) = [ a+b 2 ]. Now w(n −b) = −x + y −1 = w(n) −1 and w(n + b) = (x + b) + (a −1 −y) = a + b −1 −[ a+b 2 ] ≤w(n), so n is a local champion.
Conversely, every n = ax+by with a−b 2 < y ≤a+b 2 and x = y−[ a+b 2 ] is a local champion. Thus we obtain b −1 local champions which are all distinct.
(ii) |y| ≤a−b 2 . Now we conclude from the lemma that w(n −a) = |x −1| + |y| and w(n + a) = |x + 1| + |y|, and at least one of these two values exceeds w(n) = |x|+|y|. Thus n is not a local champion.
2.1 Copyright c ⃝: The Authors and Springer 19 (iii) −a+b 2 ≤y < −a−b 2 . By taking x,y to −x,−y this case is reduced to case (i), so we again have b −1 local champions n = ax+by with x = y+[ a+b 2 ].
It is easy to check that the sets of local champions from cases (i) and (iii) coincide if a and b are both odd (so we have b −1 local champions in total), and are otherwise disjoint (then we have 2(b −1) local champions).
30. We shall show by induction on n that there exists an arbitrarily large m satisfying 2m ≡−m (mod n). The case n = 1 is trivial; assume that n > 1.
Recall that the sequence of powers of 2 modulo n is eventually periodic with the period dividing ϕ(n); thus 2x ≡2y whenever x ≡y (mod ϕ(n)) and x and y are large enough. Let us consider m of the form m ≡−2k (mod nϕ(n)). Then the congruence 2m ≡−m (mod n) is equivalent to 2m ≡2k (mod n), and this holds whenever −2k ≡m ≡k (mod ϕ(n)) and m,k are large enough. But the existence of m and k is guartanteed by the inductive hypothesis for ϕ(n), so the induction is complete.
A Notation and Abbreviations A.1 Notation We assume familiarity with standard elementary notation of set theory, algebra, logic, geometry (including vectors), analysis, number theory (including divisibility and congruences), and combinatorics. We use this notation liberally.
We assume familiarity with the basic elements of the game of chess (the movement of pieces and the coloring of the board).
The following is notation that deserves additional clarification.
◦B(A,B,C), A −B −C: indicates the relation of betweenness, i.e., that B is be-tween A and C (this automatically means that A,B,C are different collinear points).
◦A = l1 ∩l2: indicates that A is the intersection point of the lines l1 and l2.
◦AB: line through A and B, segment AB, length of segment AB (depending on context).
◦[AB: ray starting in A and containing B.
◦(AB: ray starting in A and containing B, but without the point A.
◦(AB): open interval AB, set of points between A and B.
◦[AB]: closed interval AB, segment AB, (AB)∪{A,B}.
◦(AB]: semiopen interval AB, closed at B and open at A, (AB)∪{B}.
The same bracket notation is applied to real numbers, e.g., [a,b) = {x | a ≤x < b}.
◦ABC: plane determined by points A,B,C, triangle ABC (△ABC) (depending on context).
◦[AB,C: half-plane consisting of line AB and all points in the plane on the same side of AB as C.
◦(AB,C: [AB,C without the line AB.
22 A Notation and Abbreviations ◦⟨− → a ,− → b ⟩, − → a ·− → b : scalar product of − → a and − → b .
◦a,b,c,α,β,γ: the respective sides and angles of triangle ABC (unless otherwise indicated).
◦k(O,r): circle k with center O and radius r.
◦d(A, p): distance from point A to line p.
◦SA1A2...An, [A1A2...An]: area of n-gon A1A2...An (special case for n = 3, SABC: area of △ABC).
◦N, Z, Q, R, C: the sets of natural, integer, rational, real, complex numbers (re-spectively).
◦Zn: the ring of residues modulo n, n ∈N.
◦Zp: the field of residues modulo p, p being prime.
◦Z[x], R[x]: the rings of polynomials in x with integer and real coefficients respec-tively.
◦R∗: the set of nonzero elements of a ring R.
◦R[α], R(α), where α is a root of a quadratic polynomial in R[x]: {a+bα | a,b ∈ R}.
◦X0: X ∪{0} for X such that 0 / ∈X.
◦X+, X−, aX +b, aX +bY: {x | x ∈X,x > 0}, {x | x ∈X,x < 0}, {ax+b | x ∈X}, {ax+by | x ∈X,y ∈Y} (respectively) for X,Y ⊆R, a,b ∈R.
◦[x], ⌊x⌋: the greatest integer smaller than or equal to x.
◦⌈x⌉: the smallest integer greater than or equal to x.
The following is notation simultaneously used in different concepts (depending on context).
◦|AB|, |x|, |S|: the distance between two points AB, the absolute value of the num-ber x, the number of elements of the set S (respectively).
◦(x,y), (m,n), (a,b): (ordered) pair x and y, the greatest common divisor of inte-gers m and n, the open interval between real numbers a and b (respectively).
A.2 Abbreviations We tried to avoid using nonstandard notation and abbreviations as much as possible.
However, one nonstandard abbreviation stood out as particularly convenient: ◦w.l.o.g.: without loss of generality.
Other abbreviations include: ◦RHS: right-hand side (of a given equation).
A.2 Abbreviations 23 ◦LHS: left-hand side (of a given equation).
◦QM, AM, GM, HM: the quadratic mean, the arithmetic mean, the geometric mean, the harmonic mean (respectively).
◦gcd, lcm: greatest common divisor, least common multiple (respectively).
◦i.e.: in other words.
◦e.g.: for example.
B Codes of the Countries of Origin ARG Argentina ARM Armenia AUS Australia AUT Austria BEL Belgium BLR Belarus BRA Brazil BUL Bulgaria CAN Canada CHN China COL Colombia CRO Croatia CUB Cuba CYP Cyprus CZE Czech Republic CZS Czechoslovakia EST Estonia FIN Finland FRA France FRG Germany, FR GBR United Kingdom GDR Germany, DR GEO Georgia GER Germany GRE Greece HKG Hong Kong HUN Hungary ICE Iceland INA Indonesia IND India IRE Ireland IRN Iran ISR Israel ITA Italy JAP Japan KAZ Kazakhstan KOR Korea, South KUW Kuwait LAT Latvia LIT Lithuania LUX Luxembourg MCD Macedonia MEX Mexico MON Mongolia MOR Morocco NET Netherlands NOR Norway NZL New Zealand PER Peru PHI Philippines POL Poland POR Portugal PRK Korea, North PUR Puerto Rico ROM Romania RUS Russia SAF South Africa SER Serbia SIN Singapore SLO Slovenia SMN Serbia and Montenegro SPA Spain SVK Slovakia SWE Sweden THA Thailand TUN Tunisia TUR Turkey TWN Taiwan UKR Ukraine USA United States USS Soviet Union UZB Uzbekistan VIE Vietnam YUG Yugoslavia
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Equation of tangents from external point to a circle
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I have point (p,q) and circle x2+y2+2gx+2fy+c=0.
I'm aware you could do substitute in y=mx+c and solve a quadratic or you could use y−(−f)=m(x−(−g))1+m2−−−−−−√ where m is the slope of the line and (−g,−f) is the coordinates of the centre. But are there other ways? perhaps using calculus/vectors/complex numbers
maybe take point (4,−5) and circle x2+y2−6x+−4y+4=0. as example
calculus
analytic-geometry
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edited Jan 22, 2021 at 10:41
Raffaele
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asked Nov 6, 2017 at 16:46
VriskVrisk
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The for y=mx+b misses vertical tangents, but you can find those for a circle by inspection.
– amd
Commented
Nov 6, 2017 at 16:55
@amd I'm sorry, i don't follow?
– Vrisk
Commented
Nov 6, 2017 at 17:02
Vertical lines can’t be described by equations of the form y=mx+b, so your method will not discover those. E.g., if you try to compute the tangents to x2+y2=1 from (1,1) in the way you describe, you’ll only get one value for m and miss the line x=1. Using this form of line equation when there’s a possibility of vertical lines and not checking for the latter is a common error.
– amd
Commented
Nov 6, 2017 at 17:10
Do you know gradients for implicitly defined functions?
– user2222
Commented
Nov 6, 2017 at 17:19
@Rookie, I can do a total differtiation of this yeah. not sure what to do with that
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Here are a few other ways to find the tangent lines via (more or less) direct computation. They might be overkill when working with circles, but these methods generalize to other types of conics.
Solve a simpler problem and transform:
It’s a fairly simple matter to find the tangents to the unit circle through the point (d,0): You can use either the Pythagoran theorem and similar triangles, or use the fact that the polar of this point is the line x=1d to find that the points of tangency are (1d,±d2−1√d), and the corresponding lines are x∓yd2−1−−−−−√=d. The general problem can be transformed into this one via translation, scaling and rotation, so the solution to the general problem can be obtained by transforming these lines back into the original coordinate system.
If we represent points as homogeneous column vectors, we can also represent lines as homogeneous row vectors, so that the equation ax+by+c=0 of a line can be written as lx=[a:b:c][x:y:1]T=0. Given the nonsingular point transformation x′=Mx, we have
lx=l(M−1Mx)=(lM−1)(Mx)=0,
which shows that lines are covariant and transform as l′=lM−1. So, to convert the unit circle tangent lines into the original coordinate system, apply the three transformations in reverse order, i.e., if M=RST is the transformation that maps points into the unit circle’s coordinate system, then the tangents are lM=[1:∓d2−1−−−−−√:d]RST. For the circle (x−h)2+(y−k)2=r2 and point p=[x0:y0:1]T, this expands into
[1∓d2−1−−−−−√−d]⎡⎣⎢⎢⎢⎢⎢ξξ2+η2√−ηξ2+η2√0ηξ2+η2√ξξ2+η2√0001⎤⎦⎥⎥⎥⎥⎥⎡⎣⎢⎢1r0001r0001⎤⎦⎥⎥⎡⎣⎢100010−h−k1⎤⎦⎥,
where (ξ,η)=(x−h,y−k) and d=1rξ2+η2−−−−−−√, i.e., the x-coordinate of RSTp. Since these are homogeneous matrices, you can simplify your computations a bit by multiplying R through by ξ2+η2−−−−−−√ and using S=diag(1,1,r) instead. This gives us
M=⎡⎣⎢ξ−η0ηξ0−hξ−kηhη−kξrξ2+η2−−−−−−√⎤⎦⎥.
Similarly, you can multiply the lines by r, producing the equivalent lines [r:∓ξ2+η2−r2−−−−−−−−−−√:−ξ2+η2−−−−−−√].
For your example, we have h=3, k=2, r=3 and p=[4:−5:1], giving [3:±41−−√:−52–√] for the unit circle tangents and finally
[3±41−−√−52–√]⎡⎣⎢170−71011−23152–√⎤⎦⎥=[3±741−−√−21±41−−√−117∓2341−−√]
as the solution to the problem. These lines and the circle are plotted below:
This method works for ellipses as well if you use a non-uniform scaling and modify ξ and η accordingly. If the ellipse isn’t axis-aligned, then you might need to add another rotation before scaling, but the basic method is the same.
Degenerate conic:
If you don’t need the equations of the individual tangent lines, then you can produce a degenerate conic that consists of this pair of lines quite directly. This works for any conic. Let C be its homogeneous matrix and p be the point through which the tangents are to be drawn. The dual conic C∗ consists of all of the lines tangent to C, and [p]T×C∗[p]× consists of only the tangents through p. Here, [p]× is the “cross product matrix:” if p=[x:y:z], then
[p]×=⎡⎣⎢0w−y−z0xy−x0⎤⎦⎥.
The dual conic matrix C∗ is the adjugate of C, or, if C is nonsingular, C−1. For your example,
[p]×=⎡⎣⎢015−104−5−40⎤⎦⎥
and
C=⎡⎣⎢10−301−2−3−24⎤⎦⎥,
so, using the adjugate,
[p]T×C∗[p]×=⎡⎣⎢407−1257−8−68−125−68160⎤⎦⎥.
After multiplying it out and pulling out a factor of two, this gives the equation 20x2+7xy−4y2−125x−68y+80=0. If you do need the individual lines, you can try to factor this equation or use the method described here to split the degenerate conic. The latter involves computing the adjugate of [p]T×C∗[p]×, among other things, but this for this problem this is extra work because you can instead compute the...
Dual to the intersection of line and conic:
Raffaele’s answer includes a method that computes the intersection of the polar line of the point through which tangents are to be drawn with the circle to find the points of tangency, from which you can compute the equations of the tangent lines. If you don’t need those points, however, you can compute the lines directly by working with the dual conic: the intersections of this conic with the given point are the tangent lines through that point. For the circle
C=⎡⎣⎢10−h01−k−h−kh2+k2−r2⎤⎦⎥
the dual conic is given by its adjugate
C∗=⎡⎣⎢h2−r2hkhhkk2−r2khk1⎤⎦⎥.
Finding the tangent lines is then a matter of solving the system
[λ:μ:τ]p=0[λ:μ:τ]C∗[λ:μ:τ]T=0.
You can also compute it directly using the method outlined here.
From above, we have
[p]T×C∗[p]×=⎡⎣⎢407−1257−8−68−125−68160⎤⎦⎥.
We need to find a coefficient α such that [p]T×C∗[p]×+α[p]× has rank 1. There are a couple of ways to do this, and one of them produces α=−∣∣∣4077−8∣∣∣−−−−−−−√=341−−√ for
[p]T×C∗[p]×+α[p]×=⎡⎣⎢407+341−−√−125+1541−−√7−341−−√−8−68+1241−−√−125−1541−−√−68−1241−−√160⎤⎦⎥.
We can take for our pair of tangent lines any row and column of this matrix.
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edited Nov 8, 2017 at 19:21
answered Nov 7, 2017 at 10:35
amdamd
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(ξ,η)=(x−h,y−k) should it be (ξ,η)=(x0−h,y0−k)? or how do you get numerical values in the example?
– Federico
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x2+y2−6x+4y+4=0
centre C(3;−2) radius r=3
P(4;−5)
Write the equation of the generic line passing through P
y+5=m(x−4)
adjust it in normal form
F:mx−y−4m−5=0
Calculate the distance d(m) from F to the centre C of the circle
d(m)=|3m+2−4m−5|m2+1−−−−−−√
The line F is tangent to the circle if the distance d(m) is equal to the radius
d(m)=3→|−3−m|m2+1−−−−−−√=3
square both sides
9+6m+m2=9m2+9→m1=0;m2=34
the two tangents are
y=−5;y=34x−8
Second method
Works with any conic
Find the equation of the polar line of P wrt the circle
Substitute
x2→xpx;y2→ypy;x→x+xp2;y→y+yp2;xy→xyp+yxp2
4x−5y−6x+42+4y−52+4=0
p:x−3y=18
The intersection points of this line with the circle are the tangent points of the tangents from P to the circle, namely K(3,−5);H(4.8,−4.4)
Hope this helps
...
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edited Nov 29, 2023 at 18:17
Prince Nagar
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answered Nov 6, 2017 at 18:29
RaffaeleRaffaele
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By Joachimsthal
s21=s⋅s11
defines the tangents from (p,q) as a line pair with
s=x2+y2+2gx+2fy+c=0,
s11=p2+q2+2gp+2fq+c,
s1=xp+yq+g(x+p)+f(y+q)+c.
I.e.
(xp+yq+g(x+p)+f(y+q)+c)2−(x2+y2+2gx+2fy+c)(p2+q2+2gp+2fq+c)=0
or
(−g2+2fq+q2+c)(x−p)2+(−2fg−2fp−2gq−2pq)(x−p)(y−q)+(−f2+2gp+p2+c)(y−q)2=0
which factorises as:
((−g2+2fq+q2+c)(x−p)−(y−q)(pq+gq+fp+fg−D√))((−g2+2fq+q2+c)(x−p)−(y−q)(pq+gq+fp+fg+D√))−g2+2fq+q2+c,
where
D=(g2+f2−c)(p2+q2+2gp+2fq+c)=(g2+f2−c)s11,
or
((p2+2gp−f2+c)(y−q)−(x−p)(pq+gq+fp+fg−D√))((p2+2gp−f2+c)(y−q)−(x−p)(pq+gq+fp+fg+D√))p2+2gp−f2+c,
making the actual lines
(y−q)=(x−p)pq+gq+fp+fg±D−−√p2+2gp−f2+c.
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answered Jan 22, 2021 at 13:46
Jan-Magnus ØklandJan-Magnus Økland
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My teacher said that a line is parallel to itself. : r/askmath
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•5 yr. ago
khufiie
My teacher said that a line is parallel to itself.
Geometry
But as far as I know parallel lines never meet but a line meets itself every time, at every point wtf.
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https://hellothinkster.com/math-tutor/exponents/16-to-power-of-5
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These coaching plans come with a learning guarantee and two tutors - a dedicated math coach for 1:1 live tutoring & an expert AI Learning Lab coach). Get access to world-class curriculum, homework help, and continuous personalization. Designed to make your child math confident for life!
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Math Tutor Explains: What is 16 to the Power of 5?
A math tutor would explain that to solve for 16 to the power of 5, we must first understand the structure. The number 16 is called the base, and the number 5 is called the exponent. In short, the base must be multiplied by itself the number of times as the exponent. Let’s look at this problem in more detail.
Math Tutor's Solution: 16 to the Power of 5 is equal to 1048576
Step-by-step from math tutor: finding 16 to the power of 5
The first step a math tutor would suggest is to understand what it means when a number has an exponent. The “power” of a number indicates how many times the base would be multiplied by itself to reach the correct value.
The second step is to write the number in the base-exponent form, and lastly calculate what the final result would be. Consider the example of 2 to the power of 4: in exponent form that would be
242^424
. To solve this, we need to multiply the base, 2 by itself, 4 times -
2⋅2⋅2⋅22\cdot2\cdot2\cdot22⋅2⋅2⋅2
= 16. So
24=162^4 = 1624=16
.
So re-applying these steps to our particular problem, we first convert our word problem to a base-exponent form of:
16516^{5}165
To simplify this, all that is needed is to multiply it out:
16 x 16 x 16 x 16 x 16 = 1048576
Therefore, 16 to the power of 5 is 1048576.
Math Tutor Suggests: Try related exponent problems
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https://byjus.com/maths/value-of-root-2/
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Published Time: 2019-04-04T16:50:47+05:30
The square root or root of a number “n” is a number that returns the value equal to “n” when multiplied by itself. For example, the value of the square root of 4 is ±2. Because 2 x 2 = 4. Finding the square root of a number is the inverse process of squaring the number. When we multiply the number to itself, we get the square of a number. Finding the square root of the perfect square number is easier than the non-perfect square numbers.
Examples for perfect square numbers:
4 = square of a number 2 = 2 x 2
9 = square of a number 3 = 3 x 3
16 = square of a number 4 = 4 x 4
25 = square of a number 5 = 5 x 5
It is a bit complicated to find the square root value for the non-perfect square numbers say 10. The symbol used to denote the square root is ‘√’. The square root symbol is also known as a radical symbol or radix. The number underneath the radical symbol is called as radicand. If the given number is not a square number, then the value should take either radical form or decimal form. Finding the value of root 2 is quite difficult since the number 2 is not a square number and there is no such easy method to find the value. The only possible method to find the square root values of the non-perfect square is a long division method.
Division Method to Find the Value of Root 2
By using the long division method, it is possible to find the value of square roots of any number. The steps for finding the value of root 2 is given below:
Step 1: The number 2 can be written as 2.000000, i.e., 2 = 2.000000
Step 2: Take the perfect square which is below 2. So the perfect square below 2 is 1.
Step 3: Write the number 1 in both divisor and quotient place and subtract 1 from 2, you will get the remainder 1.
Step 4: Next write down two zeros and write down after 1 and take the decimal point after 1 in the quotient.
Step 5: Now add the divisor part with the same number that we have taken already, .i.e. 1, we will get 2 and write down the number in the divisor place opposite to 100. So, after number 2, we have to take the number and the same number has to be taken in the quotient place also. So, when you take the number 4 after 2 and the same number has to be written in the quotient. Because when you multiply 24 with 4, you get 96 which is closer to the number 100. Finally, subtract 96 from 100 and write down the remainder.
Step 6: Now again write down two zero’s and repeat step 5.
Step 7: Finally, you get the quotient value as 1.41421 which is approximately equal to 1.414.
So, the value of root 2 is equal to 1.414.
√2 = 1.414
With the help of the long division method, you will find the values of non-perfect square values like root 3, root 5 etc.
Also Check:Value of root 3
Register with BYJU’S – The Learning App to know the root values of other numbers and also watch interactive videos to clarify the doubts.
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563
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https://stackoverflow.com/questions/1311049/how-to-map-atan2-to-degrees-0-360
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3
Note: The posted update method will not return zero degrees, but values from just above 0 to 360.0.
chux
–
chux
2013-09-27 02:37:59 +00:00
Commented
Sep 27, 2013 at 2:37
2
[How to Get angle from 2 positions] : stackoverflow.com/questions/9457988/…
MAnoj Sarnaik
–
MAnoj Sarnaik
2015-04-08 08:45:23 +00:00
Commented
Apr 8, 2015 at 8:45
This function works great, however the angle of "bearingDegrees" calculation is flipped. for example, 45 degrees would typically by the 1st quadrant, however this in the 4th quadrant. 135 degrees would typically be in the 2nd quadrant but this function returns it to be in the 3rd quadrant. i can simply take the function return value x and negate that from 360 to get the correct angle value however I'm curious to know why this is happening in the first place?
goelv
–
goelv
2015-05-18 10:16:32 +00:00
Commented
May 18, 2015 at 10:16
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Solution using Modulo
A simple solution that catches all cases.
```
degrees = (degrees + 360) % 360; // +360 for implementations where mod returns negative numbers
```
Explanation
Positive: 1 to 180
If you mod any positive number between 1 and 180 by 360, you will get the exact same number you put in. Mod here just ensures these positive numbers are returned as the same value.
Negative: -180 to -1
Using mod here will return values in the range of 180 and 359 degrees.
Special cases: 0 and 360
Using mod means that 0 is returned, making this a safe 0-359 degrees solution.
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edited Feb 14, 2017 at 11:00
answered Sep 8, 2014 at 13:06
Liam George BetsworthLiam George Betsworth
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I don't believe it's necessary to add 360. -1 % 360 is still 359 :)
yobiscus
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yobiscus
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I don't think this is correct in all languages. In Javascript -1 % 360 = -1
Startec
–
Startec
2016-09-14 09:36:54 +00:00
Commented
Sep 14, 2016 at 9:36
1
Also not a viable approach in Java
Hulk
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Hulk
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@pleasemorebacon Incorrect. In some languages -1 % 360 is -1.
Pharap
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```
(x > 0 ? x : (2PI + x)) 360 / (2PI)
```
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edited Oct 19, 2018 at 11:55
BartoszKP
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answered Aug 21, 2009 at 10:20
erikkallenerikkallen
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Probably also want x >= 0 for the x = 0 case.
bpw1621
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bpw1621
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For those not comfortable with this notation, and without the conversion to degrees built in: if(x>0) {radians = x;} else {radians = 2PI + x;} so we are just adding 2PI to the result if it is less than 0.
David Doria
–
David Doria
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Or (x >= 0 ? x : (2PI + x)) 180/PI as in (x < 0 ? 2PI + x : x) 180/PI
user3342816
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user3342816
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Add 360° if the answer from atan2 is less than 0°.
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edited Oct 12, 2022 at 20:23
Dave Jarvis
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answered Aug 21, 2009 at 10:17
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Which is the same as "just add 2 PI" if you're having one of those days.
Chris O
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Chris O
2014-10-08 14:43:36 +00:00
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Oct 8, 2014 at 14:43
Just adding the code to what dave4420 said: x = (x < 0) ? (x + 360) : x;
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javaEntu
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Or if you don't like branching, negate the two parameters and add 180° to the answer.
(Adding 180° to the return value puts it nicely in the 0-360 range, but flips the angle. Negating both input parameters flips it back.)
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edited Oct 12, 2022 at 20:24
Dave Jarvis
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answered Aug 21, 2009 at 12:08
aibaib
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I'd rather modify my code to use denormalized angles (<0, >=360) but there always seems to be someone aiming for that fake "optimized" feel; that's why I wanted to add this. (Or was it because this was the quicker way around some temporary debug code I used? hmm)
aib
–
aib
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Definitely not straightforward to grok, as I can concur after 2+ years. So: Adding 180° to the return value puts it nicely in the 0-360 range, but flips the angle. Negating both input parameters flips it back.
aib
–
aib
2014-11-19 10:57:10 +00:00
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Nov 19, 2014 at 10:57
This can have some issues when $x = 0$ and $y > 0$ iirc
Trinidad
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Trinidad
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@erikkallen is close but not quite right.
```
theta_rad = atan2(y,x);
theta_deg = (theta_rad/M_PI180) + (theta_rad > 0 ? 0 : 360);
```
This should work in C++: (depending on how fmod is implemented, it may be faster or slower than the conditional expression)
```
theta_deg = fmod(atan2(y,x)/M_PI180,360);
```
Alternatively you could do this:
```
theta_deg = atan2(-y,-x)/M_PI180 + 180;
```
since (x,y) and (-x,-y) differ in angles by 180 degrees.
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answered Aug 21, 2009 at 19:14
Jason SJason S
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if I understood you correctly in Fortran it is atan2(-y,-x) 180/PI + 180. Is that correct ?
gansub
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gansub
2016-07-19 16:09:02 +00:00
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sorry I don't know FORTRAN. But your math looks right.
Jason S
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Jason S
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I have 2 solutions that seem to work for all combinations of positive and negative x and y.
1) Abuse atan2()
According to the docs atan2 takes parameters y and x in that order. However if you reverse them you can do the following:
```
double radians = std::atan2(x, y);
double degrees = radians 180 / M_PI;
if (radians < 0)
{
degrees += 360;
}
```
2) Use atan2() correctly and convert afterwards
```
double degrees = std::atan2(y, x) 180 / M_PI;
if (degrees > 90)
{
degrees = 450 - degrees;
}
else
{
degrees = 90 - degrees;
}
```
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answered Aug 20, 2014 at 6:35
FibbsFibbs
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You sir, are a life savior. Just used this approach in Unity and works like a charm.
porfiriopartida
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porfiriopartida
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@Jason S: your "fmod" variant will not work on a standards-compliant implementation. The C standard is explicit and clear (7.12.10.1, "the fmod functions"):
if y is nonzero, the result has the same sign as x
thus,
```
fmod(atan2(y,x)/M_PI180,360)
```
is actually just a verbose rewriting of:
```
atan2(y,x)/M_PI180
```
Your third suggestion, however, is spot on.
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answered Aug 21, 2009 at 20:37
Stephen CanonStephen Canon
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Here's some javascript. Just input x and y values.
```
var angle = (Math.atan2(x,y) (180/Math.PI) + 360) % 360;
```
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answered Apr 21, 2020 at 2:24
PerspectiveInfinityPerspectiveInfinity
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This is what I normally do:
```
float rads = atan2(y, x);
if (y < 0) rads = M_PI2.f + rads;
float degrees = rads180.f/M_PI;
```
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answered Feb 6, 2017 at 14:59
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```
angle = Math.atan2(x,y)180/Math.PI;
```
I have made a Formula for orienting angle into 0 to 360
```
angle + Math.ceil( -angle / 360 ) 360;
```
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answered Feb 4, 2015 at 7:58
Saad AhmedSaad Ahmed
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An alternative solution is to use the mod () function defined as:
function mod(a, b) {return a - Math.floor (a / b) b;}
Then, with the following function, the angle between ini(x,y) and end(x,y) points is obtained. The angle is expressed in degrees normalized to [0, 360] deg. and North referencing 360 deg.
```
function angleInDegrees(ini, end) {
var radian = Math.atan2((end.y - ini.y), (end.x - ini.x));//radian [-PI,PI]
return mod(radian 180 / Math.PI + 90, 360);
}
```
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edited Feb 3, 2016 at 22:01
answered Feb 3, 2016 at 21:47
A. Torrez GarayA. Torrez Garay
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```
double degree = fmodf((atan2(x, y) (180.0 / M_PI)) + 360, 360);
```
This will return degree from 0°-360° counter-clockwise, 0° is at 3 o'clock.
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edited Feb 7, 2017 at 0:18
Pang
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answered Oct 9, 2016 at 6:49
FinallzFinallz
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A formula to have the range of values from 0 to 360 degrees.
f(x,y)=180-90(1+sign(x)) (1-sign(y^2))-45(2+sign(x))sign(y)
```
-(180/pi())sign(xy)atan((abs(x)-abs(y))/(abs(x)+abs(y)))
```
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answered Nov 25, 2018 at 13:18
theodore panagostheodore panagos
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Can you please explain how this relates to the question?
Klaus Gütter
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Klaus Gütter
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The R packages geosphere will calculate bearingRhumb, which is a constant bearing line given an origin point and easting/northing. The easting and northing must be in a matrix or vector. The origin point for a wind rose is 0,0. The following code seems to readily resolve the issue:
```
windE<-wind$uasE
windN<-wind$vasN
wind_matrix<-cbind(windE, windN)
wind$wind_dir<-bearingRhumb(c(0,0), wind_matrix)
wind$wind_dir<-round(wind$wind_dir, 0)
```
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answered May 19, 2016 at 14:32
ArianeAriane
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```
theta_rad = Math.Atan2(y,x);
if(theta_rad < 0)
theta_rad = theta_rad + 2 Math.PI; //if neg., add 2 PI to it
theta_deg = (theta_rad/M_PI180) ; //convert from radian to degree
//or
theta_rad = Math.Atan2(y,x);
theta_rad = (theta_rad < 0) ? theta_rad + 2 Math.PI : theta_rad;
theta_deg = (theta_rad/M_PI180) ;
```
-1 deg becomes (-1 + 360) = 359 deg
-179 deg becomes (-179 + 360) = 181 deg
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answered Feb 3, 2017 at 3:56
PhilCPhilC
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What's Math.PI? Is it the same as M_PI?
Pang
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Pang
2017-02-03 04:13:27 +00:00
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Feb 3, 2017 at 4:13
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For your application I suspect you don't need exact degrees and would prefer a more approximate compass angle, eg 1 of 16 directions? If so then this code avoids atan issues and indeed avoids floating point altogether. It was written for a video game so uses 8 bit and 16 bit integers:
```
/
349.75d 11.25d, tan=0.2034523
\ /
\ Sector /
\ 0 / 22.5d tan = ?2 - 1
15 | 1 33.75
| / 45d, tan = 1
14 | 2 _56.25
| / 67.5d, tan = 1 + ?2
13 | 3
| __ 78.75
|
12---------------+----------------4 90d tan = infty
| __ 101.25
|
11 | 5
|
10 | 6
|
9 | 7
8
/
// use signs to map sectors:
static const int8_t map = { / +n means n >= 0, -n means n < 0 /
/ 0: +x +y / {0, 1, 2, 3, 4},
/ 1: +x -y / {8, 7, 6, 5, 4},
/ 2: -x +y / {0, 15, 14, 13, 12},
/ 3: -x -y / {8, 9, 10, 11, 12}
};
int8_t sector(int8_t x, int8_t y) { // x,y signed in range -128:127, result 0:15 from north, clockwise.
int16_t tangent; // 16 bits
int8_t quadrant = 0;
if (x > 0) x = -x; else quadrant |= 2; // make both negative avoids issue with negating -128
if (y > 0) y = -y; else quadrant |= 1;
if (y != 0) {
// The primary cost of this algorithm is five 16-bit multiplies.
tangent = (int16_t)x32; // worst case y = 1, tangent = 25532 so fits in 2 bytes.
/
determine base sector using abs(x)/abs(y).
in segment:
0 if 0 <= x/y < tan 11.25 -- centered around 0 N
1 if tan 11.25 <= x/y < tan 33.75 -- 22.5 NxNE
2 if tan 33.75 <= x/y < tan 56.25 -- 45 NE
3 if tan 56.25 <= x/y < tan 78.75 -- 67.5 ExNE
4 if tan 78.75 <= x/y < tan 90 -- 90 E
/
if (tangent > y6 ) return map[quadrant]; // tan(11.25)32
if (tangent > y21 ) return map[quadrant]; // tan(33.75)32
if (tangent > y47 ) return map[quadrant]; // tan(56.25)32
if (tangent > y160) return map[quadrant]; // tan(78.75)32
// last case is the potentially infinite tan(90) but we don't need to check that limit.
}
return map[quadrant];
}
```
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answered Aug 24, 2021 at 19:49
Graham ToalGraham Toal
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Everything You Need to Know About Exponents (18 new tricks!)
The Tested Tutor
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Update: I am not available for tutoring! But still alive, :) Meanwhile, so many students have personally recommended Target Test Prep for GRE / GMAT that I asked them for a unique discount code, TestedTutor10, for this link: All you need to know to tackle exponents, all in one video: plus dozens of practice questions included. Featuring stacked exponents, factorising exponents, algebraic exponents, multiplying and dividing powers, picking numbers, negative exponents, exponent cancellation, exponent equations, the essential exponent rules, the exponents you should memorise and much much more...
Ideal for anyone studying the GRE or GMAT or any exam in which exponents / powers / indices are tested. Starts at beginner level all the way up to expert level.
I offer private GRE / GMAT tutoring online at a fixed rate of $140/hr. Please get in touch via the email below, or through my tutoring website:
Enquiries: philip@gretutorlondon.com
The GRE and GMAT are two of most used graduate exams for those seeking to study a Masters Degree or Masters in Business Administration (MBA) degree. The Graduate Record Exam and Graduate Management Admissions Test have been taken for over 50 years and are taken by approximately 600,000 and 300,000 people worldwide annually, respectively. These exams are typically taken between the ages of 18-40, by those who have completed, or are about to complete, their Bachelors Degree.
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Grateful for any support:
0:00 Introduction
2:53 Exponents To Memorise
7:21 Negative Numbers
11:23 Powers of Zero/One
13:11 Essential Exponent Rules
20:21 Negative Exponents
26:42 Exponents and Picking Numbers
33:37 Prime Factorising the Base
42:01 Base Cancellation (Warning)
43:59 Practice Interval
47:48 Joining Exponents (and when not!)
51:24 Exponents with Additions/Subtractions
54:40 Exponent Fractions
56:47 Simplifying Situations If you’re now ready to get into your dream MBA program, my former student, Angel, is offering virtual private Admissions coaching. She was admitted into Harvard Business School, The Wharton School of Business, and Columbia Business School (accepted into every school she interviewed for) after graduating from UCLA with a degree in Communication. She also got 3 perfect scores on the GRE’s Analytical Writing Assessment (99th percentile).
Here is her link - - and you can reach her at angellinzhu@gmail.com with the subject line, “PHILIP SENT ME.”
340 comments
Transcript:
Introduction i know that there are lots of materials out there on youtube and the internet about exponents otherwise known as powers for the gre and for the gmat but what i decided to do with this video is just give you one video packed full of real questions plus questions i've made up official questions and unofficial questions about exponents including everything you need to know for the test could have created 10 separate videos or created a playlist with 10 videos in it but i thought let's try an experiment let's just have one video with everything you need to know about exponents so this took weeks of preparation but i've compiled this document and this video for your benefit so please do let me know in the comments if you find it at all helpful quick point to make at the beginning is that it's going to start with a basic explanation of exponents and gradually cover the harder and harder topics as we go through so if you're just new to the gre gmat start at the beginning work your way through and don't get too stressed if the final few points on the video are quite hard to understand equally if you're a really advanced student who knows the basics don't feel bad about skipping to 10 minutes 20 minutes into the video to cover the areas that you're not sure on obviously i'd love it if everyone watches the whole video but really it comes down to what level you're at now and what you need to know with that said and a coffee in hand i'm now going to teach you everything you need to know about exponents starting with what one is have you seen those numbers where it says 3 to the power of 4 and the power is written in a small font at the top well the top is the exponent also known as the power the index or just that little number at the top that's what i'm referring to when i talk about exponents the bigger number is the base number so if we have 3 to the power 4 3 is the base four is the exponent i want to be clear this does not mean three times four one of the most common mistakes at the beginner level when dealing with exponents is to think the little number is something maybe that we need to multiply by no it doesn't mean three times four so what does it mean it means three times itself four times three times three times three times three there are four threes which is why it's three to the power of four it does not mean three times four it means three times itself four times big difference okay i hope that's really clear what an exponent is so let's get into some of the things that you are tested on with exponents the very first thing i wanted you to do Exponents To Memorise for all of my gre gmat students and even anyone learning about exponents is i think it's a really good idea to memorize certain exponents and on screen are the exponents that i want you to memorize so the box at the top represents all these square numbers from 1 to 15. now most people know 1 squared is 1 2 squared is 4 3 squared is 9. but people start hesitating usually when you get to 4 squared or 7 squared or especially 12 squared or 13 squared but these are very common numbers and very common calculations for gre gma and many other tests so rather than take the 10 15 seconds however quick you are with the calculator to work these out like 14 squared or 9 squared much better just to have them memorized for example if i talk about the number 37 i would want you to instantly realize that's very close to 6 squared 36 whereas if you're that type of person who just works things out in the spur of the moment oh eight times eight what is that eight times seven fifty six eight times eight sixty four you wouldn't spot the same thing you wouldn't spot the 37 is close to 36 because it's not memorized so strongly recommend that you memorize the square numbers from 1 to 15. by the way the numbers in blue in that top box are the square numbers obviously 8 isn't the square 7 isn't the square it's the result is the square number next i want you to remember the cube numbers in red in the second box 1 cubed equals 1 2 cubed equals 8 3 cubed equals 27 etc remember cubed doesn't mean for example four times three it means four times itself three times four times four times four is sixty-four five times five times five is 125. why memorize these well i could say they come up quite a lot in the grand jima and other tests but also they come up in volume situations that you may deal with in real life because cubes are often used when we talk about volumes and the final box the 2 to the 6 equals 64 and the 2 to the 10 equals 1024 you might be thinking that's weird why did he only pick those two for the powers of two to memorize well because they are easy to memorize first of all why are the powers of two important because they represent doubling two to the power of five means doubling five times think about it two times two times two times two times two or 2 to the power of 10 means doubling 10 times and you can just imagine the number of questions the gre gman other tests can come up with involving things doubling animals plants objects people whatever very common question also comes up in programming a lot in computing so great to know but again why did i pick those two they're easy to memorize because 2 to the power of 6 begins with a 6. it's 64. 2 to the power of 10 begins with a 10 it's 10 24 1024 so these two numbers are the easier one to memorize and you might be thinking well what if the question's 2 to the power of 7 well if you've memorized off by heart that 2 to the power of 6 is 64 then 2 to the power of 7 is just a doubling of that 128 or if the question's asking two to the power of nine instead of getting our calculator and pressing two nine times which might take a while or especially in your head doing two times two times two etcetera nine times if you've memorized that two to the power of ten is 1024 because it begins with the 10. you can just halve that that would be 512 to find out what 2 to the power of 9 is saves a huge amount of time in short i've picked these numbers out for a reason and i strongly recommend that you memorize them i know what a lot of you are going to do now because i've worked with thousands of students you're going to note down in your diary yes must memorize these and you're going to turn the page and you're never going to come back to it and you're never going to memorize it but for those students who do put in the extra work to memorize it i can almost guarantee you it will pay off and now what about negative numbers what happens when we have Negative Numbers a negative number to a certain exponent this is tested frequently i could almost say that about most of the topics we're going to cover in this video but what would happen for example as you can see on the top left if we have negative 2 squared or negative 2 squared means negative 2 times by itself twice minus 2 times -2 the two negatives would cancel out giving us plus four as you can see on the right but what about a negative cubed that would be negative two times negative two times negative two the first two negatives would cancel out but then we'll be left with the third negative so we get negative eight the point i'm trying to make is that whenever there's an even power notice we get a positive result when we have a negative number to a certain exponent but whenever we have an odd exponent it ends up being a negative result this observation will be tested in the gre gmat and other tests look at quantity a there is no time to work out the actual number either manually or even with a calculator but what can we do we can say that negative 5 to the power of 4 is going to be positive y because it's a negative number but to an even power so that first bracket is going to be positive the next bracket will be negative because we have a negative number negative seven to the power of an odd number three so so far we have a positive and a negative can you finish that off and tell me the sign of the remaining three brackets you should have gotten that negative 2 squared is going to be positive negative 11 to the 6 is going to be positive and negative 3 to the 7 is going to be negative because it's an odd power in total we had a positive negative positive positive and negative and if you notice that was three positives and two negatives and any time you have a pair of negative numbers they will cancel out when you're multiplying them so those two negatives that we were left with would cancel out to give us a positive if we were left with three negatives two of them would cancel out and the whole thing would be left as a negative if we were left with six negatives it would cancel out to be positive if we were left with seven negatives it would cancel out to be negative so that rule still applies either way we know here we're left with two negatives the negative seven cubed and the negative three to the power of 7 those would cancel out giving us a positive result that's a massive number obviously because all those numbers are big numbers and that's going to be a lot greater than 1. one final thing to note however is negative 3 in bracket squared and all these questions so far have been in brackets is not the same thing as negative 3 squared where there's no brackets notice with the bracket version we're squaring everything the negative and the three and the negative squared cancels out to be positive and three squared is nine so you get plus nine but on the right hand side the negative is not being squared because the negative on the right is not in a bracket it's not being squared so it stays on its own it stays negative so on the right the answer would be negative 9 whereas on the left the answer will be positive nine so you really need to pay attention to whether the negative number is in a bracket or not if it's being square then it's in a bracket it will end up being a positive if it's being squared and it's not in a bracket just with a negative sign on the outside it will stay being a negative and they can trick you like that on the test so do watch out for that moving on from negative numbers let's talk about two special cases powers of zero and powers of one here Powers of Zero/One things are fairly simple anything to the power of zero is one as you can see on the left it doesn't matter if it's a number to the power of zero or letters variable to the power of zero anything to the power of zero is one 17 to the power of zero is one xyz to the power of zero is one great to know anything to the power of one is itself so two to the power of one is just two x to the power of one is just x you might think well what's the point of the power of one well we'll get on to that later on in the video but regardless you need to know that if it's to the power of one it just stays as itself i see all too many students forgetting this basic fact as i've written down below even though the power of one does nothing doesn't change the number it's still worth writing and i'm going to test you on this later in the video so don't forget that even though a power of one does nothing it's still worth writing about so as i've given an example here if you see y multiplied by y cubed i want you to rewrite that as y to the power of 1 times y cubed because that's a lot easier to work out as you'll see later the y is on its own so it's y to the power of 1 because y to the power of 1 just means y to summarize if you're dealing with exponents rewrite individual letters like x and y as x to the power of 1 and y to the power of one now we're getting to the really really fun stuff i hope you're excited Essential Exponent Rules what about the essential exponent rules let's walk through the three main essential exponent rules first what happens when we multiply exponents as i've written in the top left x cubed multiplied by x to the power of four what happens we add the little numbers aka we add the exponents when you multiply two numbers two base numbers that are the same what happens is the exponents are added so x to the power of three times x to the power of four equals x to the power of seven or in my example on the right three to the power of minus five multiplied by three to the power of eight equals three to the power three why because negative five plus eight equals positive three we add the little numbers we add the exponents what about when we're dividing for example as you can see on the left two to the power of seven divided by two to the power of twelve well when we're dividing as you can probably predict we subtract the exponents notice the base number stays the same 2 divided by 2 is 1 but we don't call it 1 to the power of negative 5 the base number stays the same i could have said that with the example at the top 3 to the negative 5 times 3 to the 8 doesn't become 9 to the power of 3 no the base number stays the same the base stays as 3 just the power gets added giving us power 3. super common mistake that i don't want you to fall for and one further example of dividing at the bottom left x to the power of 8 divided by x to the power of 2 doesn't become x to the power of 4 because 8 divided by 2 is 4 no we subtract the powers remember so 8 take away 2 is 6. that's enough about multiplying and dividing what about when there's a bracket in between two exponents like 2 to the power 4 all in brackets to the power of 3. well what we do then is we multiply the little numbers giving us 2 to the power of 12 or if we had y squared in brackets to the power of five the two times five becomes ten so we have y to the power of ten just quickly for those who are curious the reason is two to the power four in brackets cubed means because it says cubed times by itself three times right so it's 2 to the power 4 times by 2 to the power 4 times by 2 to the power 4 bringing us back to rule number one what happens when we multiply numbers with exponents we add the powers so 2 to the power of 4 times 2 to the power 4 times 2 to the power 4 adding up the little fours 4 plus 4 plus 4 equals 12. obviously we don't do this each time because we now know the rule which saves us the time of adding three times we just do four times three but that's just quickly why we multiply bringing this all together can you now answer the practice question that you can see on the right what is the value of 2 to the power of 4 multiplied by 2 to the power of 3 over 2 to the power of 4 brackets cubed well using these different rules we have the numerator 2 to the power 4 times 2 to the power 3 becomes 2 to the power of 7 because we add the little numbers the denominator is 2 to the power 4 in brackets cubed and we multiply the little numbers giving us 2 to the power of 12. finally we have 2 to the power of 7 divided by 2 to the power of 12 and we subtract the little numbers giving us 2 to the power of negative 5 answer b that was a nice and gentle introduction but it's time for a practice interval the first one in this video where i'm going to show you some real gmat and gre questions i want you to answer in this case two gre questions these both came up in real ets tests so yes they're lower level but still real questions can you attempt to answer both of these questions the challenge is to say whether quantity a is bigger quantity b is bigger both quantities are equal or we can't tell feel free to pause the video and try these out yourself before you hear my explanation so the first question what do we do well 0.82 squared is being multiplied by 0.82 cubed and when we multiply we add the little powers giving us 0.82 to the power of 5 because 2 plus 3 is 5. so what's bigger 0.82 to the power of 5 or 0.82 to the power of 6. many of you would have said quantity b because it's a bigger power right it's to the power of 6 instead of to the power of 5. but don't forget when you multiply numbers that are between 0 and 1 like decimal like 0.82 when you multiply those numbers by themselves when you power them up they actually get smaller think about it 0.5 squared is smaller than 0.5 or if you do it on the calculator 0.5 cubed which is 0.5 times 0.5 times 0.5 is actually smaller than 0.5 squared so the fact that we have a bigger power for quantity b we have 0.82 to the power of 6 means that you've multiplied by an extra 0.82 meaning you've made the number smaller in summary when we're dealing with numbers between 0 and 1 the bigger the power the smaller the number so 0.82 to the power of 5 quantity a is actually bigger than 0.82 to the power of 6 even though there's a small exponent i hope that's clear what about the next question do you want to try that yourself again what do we do well as i said i would test you on earlier we have an x on its own so we're going to rewrite that as x to the power of 1 as i've done for quantity a we have a brackets so we multiply the two little powers giving us x to the power of eight many of you if i hadn't said that tip about writing x as x to the power of one would then be stuck like what do we do x times x to the power of a no idea is it x squared that have gotten it wrong but hopefully those students who listen to my earlier advice and rewrote the x as x to the power of one will now get the question right because it's just simply x to the power of 1 times x to the power of a adding the little numbers that's x to the power of 9. it's so much easier because we wrote x as x to the power of 1. so it's a great bit of advice there for you quantity b was a bit easier just one single bracket three times three is nine so we get x to the power of nine both quantities are equal right enough of the practice interval let's get to new learning what about negative exponents very Negative Exponents common mistake here as i've written down in blue is to think that a negative power means a negative answer or a negative number let me say again that a negative power does not mean a negative number or a negative answer just because it says 2 to the power of negative 1 does not mean the answer will be negative in fact i can tell you an extra secret that i haven't written down here which is if you have a positive number like 2 you can raise it to any power you want and you will never get a negative answer you could do 2 to the power of a billion or 2 to the power of 0 or 2 to the power of negative a billion or 2 to the power of 0.5 or 2 to the power of negative 0.5 it doesn't matter there's no power that you can raise a positive number two to make it negative so once you have a positive base the answer will always be positive no matter what the power is obviously that's different if you have a negative base so what does it mean if it doesn't give us a negative result what does a negative exponent actually do what is 2 to the power of negative 1 what the negative does is simply flip the fraction that's the only role of the negative i know what you're thinking there is no fraction but every number is a fraction if you think about it two is the same thing as two over one so if i just write a number three on the page that's the same thing as writing the number three over one right so the negative in the exponent don't worry about the number we'll deal with the 1 later but the negative in the exponent it flips a fraction meaning 2 over 1 becomes 1 over 2 as you can see and once you've flipped it then deal with the power what's two to the one because the number was one two to the one is just itself just two so the answer is one over two i know that's a lot to take in so let's do another example what about 3 to the negative 2 well what does the negative do it flips the fraction so instead of it being a 3 it's 1 over 3. then we bring in the power which is power of 2 and 3 to the power of 2 is 9 so the answer is 1 over 9. next example what about 1 over 5 to the negative 2. well this time something weird happens right just think of it like this a negative in the denominator just promotes the number to be in the numerator it's like a lift if you see a negative power in the denominator of a fraction it just goes up one level so now it's in the numerator if it was in the numerator it will become in the denominator as we saw with the first two examples if it was in the denominator a negative exponent means it would be in the numerator so now the 5 to the negative 2 gets promoted up to the numerator giving us 5 squared with no fraction let me illustrate that with a harder example if we have 3 to the minus 3 over 6 to the minus 2 what happens well the 6 to the minus 2 because of the negative in the exponent gets promoted up to the numerator to become 6 squared in the numerator and the 3 to the negative 3 because of the negative as we saw earlier it gets demoted down into the denominator so now the 3 cubed is in the denominator notice once we've moved those two numbers we've moved the six squared to the top and the three cubed down to the bottom then we don't write the negative anymore because the negative and the power has done its job is promoted and demoted the two numbers so then you just focus on calculating it 6 squared is 36 3 cubed is 27 simplifying that by dividing by 9 you get 4 over 3. so i know to many of you this will be new territory but look over the different examples on the screen and see how it works so before we move on from this topic let me give you a real practice question that came up which was this which two of the following numbers have a product that is between negative one and zero in case you didn't know product means multiplying them together let's quickly work out what two to the negative four is and three to the negative two can you work those two out for me 2 to the negative 4 well the negative exponent would demote the 2 to the 4 to a fraction where it's in the denominator and 2 to the 4 is 16. so this is 1 over 16. likewise the 3 to the negative 2 it gets demoted so it's one over three squared which is one over nine now we've calculated those two it's simply a matter of picking the two numbers multiplied to get an answer between negative one and zero which i think would be negative 10 and 1 over 16. that would multiply to get negative 10 over 16 which is around about negative 0.6 you can try the other combinations but it won't work for example negative 20 times 1 over 16 will be negative 20 over 16 which is like negative 1.1 outside of the range so it would be b and c being multiplied but of course we needed to know how to deal with negative exponents i've got one further question to test you on this topic and again it's an official question what's bigger quantity a or quantity b as we saw before when there's a fraction and negative exponents we apply the promotion and demotion theory so the 3 to the power of negative 1 gets demoted down to the denominator and the 4 to the negative 1 gets promoted up to the numerator the 4 to the power of 1 in the top is just 4 because any power of 1 is just itself and the 3 to the power of 1 in the denominator is just 3. so quantity a is in fact just the same as quantity b answer c again a lot to take in so do feel free to re-watch this section but let's now move on to exponents and Exponents and Picking Numbers picking numbers because i can't get through a video on exponents without talking about picking numbers and the importance of picking numbers here are two medium level exponent questions from the gre and if we only relied on theory they'd be fairly difficult but with picking numbers they become fairly easy now i could have just skipped this part of the video and just put it into a different video on picky numbers but i just wanted to show you that not all exponent questions come down to some advanced theory of how to solve exponents sometimes if you can't simplify something look at the question on the left 3 to the power of x plus 1 you can't simplify that or quantity b 4 to the power of x you can't simplify that you can't use any of the rules they're giving you to work anything out there's nothing to do and therefore what do we do we pick numbers so i wanted to make that point that as i've written down below when simplification just won't cut it sometimes you have to give up and just pick some numbers remember also to pick extreme values doing the question on the left it says x is an integer greater than one so yes you'd pick two but make sure that your next example isn't three for example go for six or ten or something much larger than two if we try the integer two we get three to the power of two plus one which is three cubed which is 27 compared to four to the power of 2 which is 16 so quantity a would be bigger but if we tried a massive number like 6 for example then obviously using a calculator 3 to the power of 6 plus 1 3 to the power of seven is two thousand one hundred and eighty seven whereas four to the power of six is four thousand and ninety six meaning quantity b is bigger yes we could have got that without picking numbers using logic but it would be a lot harder you'd have to realize that the 3 to the power of x has got a head start because of the plus 1 but that the 4 to the power of x because every time x increases by 1 that's another times by 4 but for quantity a it's another times by three slowly quantity b is catching up and eventually will overtake now the logic behind when there's an addition or subtraction in the exponent will be covered later in the video but for now i'll just say that you could do it logically but i would almost prefer that students realize they can just pick numbers and get the right answer that way so the answer will be quantity d because it depends on what your x is what about the question on the right i'll explain that and then i've got one further question for you to try on your own x and m are positive numbers and m is a multiple of three what's bigger quantity a or quantity b now again yes you could do it logically you could say quantity a we're subtracting so it's x to the power of n minus three whereas quantity b is x to the power of m over three and we don't know whether m minus three is bigger than m over three it depends on what m is yes if you're comfortable with that advanced logic great you get the answer right however for me and for many of you out there it might be just easier to pick numbers if m is a multiple of 3 i would start off with a small multiple like just the number three and then a big one like 12. let's try that if m is three quantity a becomes x to the power of three over x to the power of three and whenever you have the same number top and bottom of a fraction the answer is just one so quantity a is one quantity b is x to the power of three over three which is x to the power of one x to the power of one is just itself which is x so quantity a was one quantity b was x and so we don't know which one's bigger because we don't know what x is it said x is a positive number not a positive integer so it could be greater than one or less than one so that's already kind of proof of d but we could finish off with picking 12. if we pick 12 x to the power of 12 divided by x to the power of 3 is x to the power of 9 whereas in quantity b x to the power of 12 over 3 is x to the power of 4. what's bigger x to the power of 4 or x to the power of 9 well many of you would say x to the power of 9 must be bigger but remember what we covered earlier on in this video if we're dealing with numbers between 0 and 1 like 0.82 a bigger power makes the number smaller and the question never said x was a positive integer just that it's positive so be careful just because we have a bigger exponent doesn't mean one thing is bigger than the other thing so again that's proof of answer d we don't know which quantity is bigger it depends on what x is now we've shown some different ways of picking numbers let me give you a question to handle again another official question which quantity is bigger quantity a or quantity b here they said x is a negative integer so i would try out negative one and maybe negative five with negative one quantity a becomes two to the power of minus one which is one over two a half quantity b would become three to the power of negative one plus one which is zero three to the power of zero is remind me one anything to the power of zero is one and if quantity a is a half and quantity b is one quantity b is bigger let's try the next negative integer that i thought of which is negative five quantity a would be two to the negative five which is one over two to the 5 1 over 32 remember we talked about memorizing exponents in the second part of this video in contrast quantity b would be 3 to the power of negative 5 plus 1 which is 3 to the power of negative 4 which is 1 over 81. and in that case quantity b is smaller 1 over 81 is less than 1 over 32. because it's closer to 0 it's a smaller number so in that case quantity b is bigger so again the answer is d we don't know which quantity is bigger and it's much simpler just to realize if we pick extreme values we can see that it's d yes again you can do it logically but i find that several exponent questions that you might come across in the test can be answered quickest just by picking numbers okay i think i've covered enough about picking numbers time to move on to some new learning Prime Factorising the Base one of the most important things to cover in this video one of the most common questions in all of exponents which is prime factoring the base here are two more quantity comparison questions and we're going to talk about the importance of prime factorizing the base i don't know about you but quantity a and quantity b on the left they look so different from one another we've got a 20 to the power of seven whereas on the right we've got fours and fives so different and if we use the calculator it would take forever to work out and write down well i'll say forever it'll take three four minutes but how can we do it if we prime factorize 20 what we're going to do is we are going to prime factorize 20 on the left which means essentially we're going to break it down to its ingredients now we're not going to go all the way down necessarily to the base prime we're just going to get it to look a lot more like quantity b so we can make a fair comparison let me illustrate 20 to the power of 7 is the equivalent of 4 times 5 to the power of 7. yes i could have picked 10 times 2 so why did i pick 4 times 5 well because quantity b involves the numbers 4 and 5. so it makes it easier to compare if we break down the numbers using the ingredients that we can see elsewhere in the question so 20 is 4 times 5 to the power of 7 simply because 20 is 4 times 5. what happens when there's a multiplication in the brackets and then outside there's a power as we have here four times five all to the power of seven well both numbers inherit the power on the outside so they both become power of seven four to the seven five to the seven and now look how much easier it is to compare the left-hand side to the right-hand side we've changed 20 to the seven into four to the seven times five to the seven starting to look a lot more like quantity b is now but there is some work we have to do with quantity b first we have a division 4 to the 13 divided by 4 to the 6. what happens with a division we subtract the powers thirteen take away six is seven so that's four to the seven and what happens with the multiplication five to the four times five to the three we add the powers so that would be five to the power of seven 7. now of course we notice that they're the same thing 20 to the 7 was 4 to the 7 times 5 to the 7 and quantity b was also 4 to the 7 times 5 plus 7. so the two quantities equal answer c can you try and use that same logic for the other question on the right can you compare those two quantities well we have the number 420 and we could factorize down all the way to its core components how many twos it's got how many sevens how many fives but given that we're comparing it to quantity b let's just break it down until we get to the number 105. so notice i've changed the 420 to 4 times 105 why did i pick out 4 times 105 i could have done 210 times 2 i could have done 60 times 7 or seventy times six i did four times 105 because the number 105 appeared on the right look at quantity b it's got that number 105. just like i did with the 20 in the first question i changed it to 4 times 5 rather than 10 times 2 because 4 and 5 is what i want to compare with for quantity b same thing you can break down 420 in any way in fact you can break it all the way down to its primes if you like but better to break it down to 105 because that's what's inside quantity that's what we're comparing to anyway 420 is 4 times 105. what's 4 4 is 2 squared so essentially we have 2 squared on the outside and a 2 squared on the inside this is further prime factorization and what happens when you multiply the 2 squared times the 2 squared you get 2 to the power of 4. so this is quantity a if we rewrite it as a multiple of 105 we get 2 to the power of 4 times 105. and look how easy that is to compare to quantity b which is 2 to the 5 times 105. we can clearly see that quantity b is bigger because the power of 2 is just 1 higher is 2 to the power 5 rather than 2 to the power 4. so quantity b is bigger and yes we use the technique of prime factoring but we didn't go all the way down to the prime numbers we went down to the ingredients that were relevant for making the comparison now don't worry we are going to do further practice of this topic so this isn't the last you'll see of prime factorization not by a long shot but now it's time to move on to our next topic which is base cancellation again an extremely common topic on the gre and gmat let's take this question on the left what is y if nine to the power of three equals three to the power of two y plus five well hopefully based on the last lesson you would prime factorize the 9 to make the comparison easier 9 is 3 squared so the 9 to the 3 on the left becomes 3 squared to the power of 3. we have a brackets which means we multiply giving us three to the power of six equals three to the power of two y plus five so just like in the last part of the video we prime factorize the base into the ingredients that made the comparison easier now we have threes on the left and 3 is on the right but this time there's an equation involved so what do we do we essentially cancel out the base so we've got a 3 to the power of 6 on the left and a 3 to the power of two y plus five so we just cancel out those two threes we'll cancel out the base on both sides giving us the simple equation six which is the power on the left equals two y plus five which was the power on the right some of you who didn't zoom in enough in the video might have thought the plus 5 was just a number plus 5 at the end that was part of the power either way we cancelled out the base here and we can quickly solve that equation take away 5 from both sides so 1 equals 2y so a half equals y and that's the answer can you now apply that trick to the question on the right you can use all the bits of learning that you've gathered so far to answer this okay so we're comparing 9 to the power of x plus 1 to 27 to the power of x minus 1. to make the comparison easier we're going to prime factorize the base you probably get bored of me saying this 9 is the same thing as 3 squared while 27 is the same thing as 3 cubed and now we're dealing with 3s on both sides so it's an easier comparison when there's brackets we multiply the exponents so it becomes 3 to the power of 2x plus 2 because the plus 1 also gets multiplied by 2 on the left and 3 to the power of 3x minus 3 because the minus 1 also gets multiplied by 3. now you guessed it we cancel out the base giving us the equation 2x plus 2 equals 3x minus 3. solving that quickly take away 2x from both sides and add 3 to both sides gives us 5 equals x it's not quite the end of the question because technically the question said what is the value of x squared minus 1 and 5 squared minus 1 is 24. so don't rush and click the answer 5 thinking you've done all the hard work the actual question was asking for x squared minus one kind of a random question and this was a demonstration of the technique called base cancellation which is quite common with exponent questions time for one quick warning however Base Cancellation (Warning) there is a warning i must give you about base cancellation it works 99 of the time but there's a slight exception imagine we have a variable like x instead of a number x to the power of 7 equals x to the power of y i know what you're thinking just cancel out the base so 7 equals y there is a slight problem if x is one of the following three numbers minus one zero or one we can't cancel the base let me demonstrate if x was the number one one to the power of seven is the same thing as one to the power of ten but that doesn't mean that 7 equals 10 let me give you another example 0 to the power of a billion is the same thing as 0 to the power of minus 5. both of them equal 0. that doesn't mean that you can cancel the base and say a billion equals minus five so there's the warning if the base is minus one zero or one you can't cancel out the base which means if the base is a variable like x then you can only cancel out the base if they tell you that x isn't those three things for example they might say x is greater than one and then you're allowed to cancel out the base because it can't be negative one zero or one or they might say x is a prime number in which case you can cancel out the base because you know that x isn't minus one zero or one as you can see we're getting some more advanced topics here but that was just a little warning you can't cancel out the base if it's negative one zero or one or a variable like x where you're not told about it being greater than one for example so as the title of the video talked about everything you need to know for exponents i thought i would have to say that as well Practice Interval time for the practice interval this time you definitely should be pausing the video trying them out yourself because you're getting to the most common questions of all involving exponents we've covered the low level stuff this is the bulk of the medium level stuff this is what you really need to know and the topics that come after this are getting more and more advanced so it's imperative that you practice and learn and get good at these medium level questions pause the video and have a go at both of these questions okay let's tackle the first question s and t are positive integers and 32 to the power of s equals 2 to the power of t to make this comparison fair we need to change the 32 into a power of 2 to make the comparison easier essentially we're prime factoring the base as we saw earlier 32 is 2 to the power of 5 so it becomes 2 to the power of 5 to the power of s equals 2 to the t no point changing it to 4 times 8 or something or 16 times 2 because we're comparing it to the right hand side which is only talking about twos when there's a bracket between some exponents we multiply so it becomes 2 to the power of 5s equals 2 to the t and let's cancel out the base giving us 5s equals t finally what's the question what is s over t for quantity a well let's divide both sides by t so we get the s over t on the left giving us five s over t equals one why is equal one by the way because when you divide both sides by t t over t equals one whenever you have the same number on the top and bottom of a fraction it equals one finally to get the s over t on its own we divide both sides by 5. so s over t equals 1 over 5 meaning both quantities are equal great let's move on to the next question if 5 to the power of 5x times 25 equals 5 to the power of n where n and x are integers what is the value of n in terms of x now i know some of you are going to look at that and go oh there's so many steps involved or what does it mean n in terms of x don't get intimidated by all the different steps you think are going to be involved in the question simply get your head down work things out step by step and then hopefully the answer will just be easy at the end don't worry about all the steps to come don't worry about the entire journey just get your foot out the front door so the 25 is a perfect candidate for prime factoring 25 is 5 squared so now we have 5 to the power of 5x multiplied by 5 squared equals 5 to the n what happens when we multiply 5 to the 5x times 5 to the 2 well the numbers get added remember the 5 squared is not power it's just a long alongside so the numbers get added giving us 5 to the power of 5x plus 2 equals 5 to the n now what though we cancel out the base giving us 5x plus 2 equals n and finally the actual answer the question becomes a bit more obvious now all they were asking was what is n what is the value of n in terms of x like using the letter x well we've worked out without really intending to that n equals five x plus two answer b so you didn't necessarily have to see that ending coming to get the answer right just follow the steps that we've outlined so far in the video one by one and you'll get the right answer that's the beauty of mathematics right that was the practice interval so let's get to some amazing new learning what about joining exponents Joining Exponents (and when not!) when and when not well this question looks a bit different can you tell the difference with quantity a why it's so weird well with quantity a it's the exponents that are the same the 9 and the 9 not the base can you see that's a really important difference in all the questions we've dealt with so far in the video the base number has been the same whether it's x to the power 3 x to the power 4 or 2 to the power of 5 times 2 to the power of 8 the base number has always been the same here the base number is different and the exponents are the same so what happens when the base number is different but the exponents are the same well this is the only example of when you combine and change the base so the three to the nine times two to the nine becomes six to the power of nine normally changing the base is absolutely wrong and something you should never do for example 2 to the 3 times 2 to the 5 is 2 to the 8 not 4 to the 8. changing the base would be wrong the only exception is when the exponent is the same as here for quantity a three to the nine times two to the nine because they're the same exponent we can combine them and the 3 times 2 is 6 so that's 6 to the power of 9. that's the only time you can do that can you calculate quantity b for me 49 to the 5 i don't mean use a calculator i mean simplify it well 49 if you've memorized your squares from earlier you would know that's 7 squared so we have 7 squared to the power of 5. without knowing that the question becomes a lot harder if you didn't know 49 off by heart was 7 squared this would take a lot longer in fact you probably wouldn't even be able to answer the question multiplying the exponents gives us 7 to the power of 10. and we can clearly see that 7 to the power of 10 is going to be bigger than 6 to the power of 9. but none of this was obvious immediately we had to combine the bases for quantity a and we had to prime factorize the base in quantity b two different techniques seven to the ten is going to be bigger than six to the nine finally i've thrown in a cheeky question there at the bottom what about three to the nine times two to the eighth does that become six to the power of 17 maybe or 6 to the power of 8 or what does that become it becomes nothing if you've listened to the lesson so far you will know that the only time you can combine two different bases like a three and a two is if the exponent is the same as with quantity a if the exponent is not the same you can't combine two different bases so there's no way to work out three to the nine times 2 to the 8 they won't ask you that question unless of course they're trying to trick you there's nothing to be done it just stays as 3 to the 9 times 2 to the 8 at any point in this video obviously feel free to take a break because you're taking in so much information here and as i've written at the bottom you can't always save the world sometimes there's nothing to be done it's a little bit grim but there it is anyway for those ready for more learning let's move on to the next topic now if you thought those questions were interesting get ready for these what about as i promised earlier Exponents with Additions/Subtractions exponents that have additions in them or subtractions for example we have this question 5 to the power of x plus 2 equals 50. what's bigger 5 to the power of x or 2.5 the key thing is to think about how do we end up with a plus 2 in the power how do we end up with additions in an exponent well as we saw earlier we end up with additions in the exponent when we multiply numbers with exponents together in other words we got a plus two because 5 to the x was multiplied by 5 the 2. if i showed it to you that way round you guys would love it i know it if i said what's 5 to the x times 5 to the 2 everyone would put their hands up and go oh philip that's 5 to the x plus 2 we add the exponents but if i ask you the other way around and say what is 5 to the x plus 2 most students get really confused and say i don't know what is 5x to the x plus 2 it's the same thing but in reverse to undo the plus 2 in the exponent you have to realize that the way we got plus 2 was because we multiplied by 5 squared so 5 to the x plus 2 is just 5 to the x times 5 to the 2. once we realize that we can just simplify both sides 5 squared is 25 and dividing both sides of the equation by 25 gives us 5 to the x equals 2. and if 5 to the x equals 2 quantity a is smaller than 2.5 so quantity b is bigger what about exponents with subtractions now you can try this question yourself if you like because it's based on the exact same logic as the first question was based on the question is if 3 to the x minus 2 equals 7 is 3 to the x bigger than 8 squared well how did we end up with the minus 2 in the exponent because it was 3 to the x multiplied by 3 to the minus 2. that's how we ended up with the minus 2 in the exponent some of you may have also thought about that as 3 to the x divided by 3 to the 2 and that is also correct both ways are correct and get you the right answer either way let's work out 3 to the minus 2. that becomes not a negative number as we've learned that's just 1 over 3 squared which is 1 over 9. multiplying both sides by 9 we get 3 to the x equals 63 and if 3 to the x is 63 is that bigger than 8 squared no because 8 squared is 64. so the answer would be no but in both cases we realize that the plus 2 and the minus 2 in the exponents were achieved simply by multiplying by 5 to the 2 and multiplying by 3 to the minus 2 respectively so that's how you simplify exponents that have additions or subtractions if that's what the question demands right i think that's covered fairly Exponent Fractions thoroughly let's move on to exponent fractions a relatively rarer topic in the gre but it does come up occasionally slightly more common in the gmat and other exams quite simple to understand though whenever you have a base like 4 to the power of a fraction like 3 over 2 the denominator is the root meaning like cube root if it's a 3 in the denominator square root if it's a 2 in the denominator 4th root if it's a 4 in the denominator the numerator is the power like cubed if it's power of 3 in the top squared if it's power of 2 to the power of 4 if the numerator is 4 etc so just remember this with fractional exponents the denominator is the root the numerator is the power final thing as i've written down below always do the root first it's always easier to work out the root doing it manually in your head or on a calculator so here four to the power of three over two let's do the over 2 bit first the root means the square root of 4. square root 4 is 2. then we do the power the numerator is 3 so it's to the power of 3 2 to the power 3 is 8. so 4 to the 3 over 2 equals 8. now can you work out this next one on your own what about 8 to the power of 2 over 3 well the over 3 the root is going to be cube root because it's the denominator of 3. the cube root of 8 is 2. remember i asked you to memorize your cubes in the second slide if the cube root of a is two we then raise it to the power of two because the numerator was two two squared is four so the answer is four and that in a nutshell is how you deal with exponent fractions you might have been expecting something more glamorous but that's all it is so we can move on now to simplifying Simplifying Situations If you’re now ready to get into your dream MBA program, my former student, Angel, is offering virtual private Admissions coaching. She was admitted into Harvard Business School, The Wharton School of Business, and Columbia Business School (accepted into every school she interviewed for) after graduating from UCLA with a degree in Communication. She also got 3 perfect scores on the GRE’s Analytical Writing Assessment (99th percentile). situations and this is a nice relaxing topic because even though the questions look really hard really high level they really are as long as you follow some basic principles of simplification and i've written that principle basically in the bottom we don't like fractions and decimals so if you can get rid of fractions get rid of decimals let's take the first question on the left all of these are official questions what is the least integer n such that 1 over 2 to the power of n is less than 0.001 super confusing unless we get rid of that decimal 0.001 we can think of that as 1 over 1000 why because there are three zeros so it's going to be one over a thousand and now we've at least gotten rid of the decimal giving us one over two to the n is less than one over a thousand i know this is not answering the question about least integer it's just simplifying the situation i know what you're thinking you're thinking oh but philip you said you don't like fractions and now you just created a fraction well the very next step after having created these fractions is to cross multiply and remove the fractions leaving nice simple integers if you cross multiply this inequality we get a thousand is less than two to the power of n i've done other videos on cross multiplication essentially the bottom left goes up to the top right and the bottom right goes up to the top left now when i explain this to one of my diligent students last week she said to me i thought you couldn't cross multiply with inequalities you can only cross multiply with inequalities when you know the sign for example you know it's positive in which case the inequality doesn't flip or you know it's negative and the inequality flips i've covered this in my inequalities video here we know both 2 to the n and 1 000 are both positive because of what i talked about earlier any power keeps it being positive so we are allowed to cross multiply finally a little bit of a test of the memorization i said right at the beginning of the video what power of two in terms of integers gets us just over a thousand remember at the beginning of the video it was two to the power of ten two to the power ten was ten 10 24 remember it begins with a 10 and that will be the least integer 10 where 2 to the power of 10 is bigger than a thousand in other words a thousand is less than 2 to the power of 10. so we simplify the situation made it a lot easier and used a bit of memorization to get the question right obviously we could also use a calculator let's apply that same simplifying logic to the second question the one on the right because i know many of you would maybe call it two to the minus k and do some advanced factorization but there's a much simpler approach we don't like fractions so what could we multiply the left-hand side of this equation by to get rid of all of those fractions 2 to the k we have 2 to the k in the denominator of the first and second fraction so if we multiply everything by 2 to the k we immediately get rid of those fractions and remember we don't like decimals we don't like fractions multiplying it all by 2 to the k we get 1 plus 1 equals 2 to the k over 2 to the x how so well when we multiply 2 to the k by the first fraction it cancels out leaving just one same thing for the second fraction and for the fraction on the right if we multiply by 2 to the k we multiply the numerator the denominator doesn't change giving us 2 to the k over 2 to the x now 1 plus 1 is 2 of course and the 2 to the k over 2 to the x what happens when we're dividing exponents we subtract the powers giving us 2 to the power of k minus x finally what was that lesson i taught you about when you have the number 2 for example on its own we write it as 2 to the power of 1. any number in an exponent question at least which is on its own we rewrite with a power of one very important to remember that now let's use the base cancellation technique we covered a little while ago in the video cancelling out the twos to give us one equals k minus x not quite the final answer because they wanted to know x expressed in terms of k meaning we have to get the x on its own let's add x to both sides and take away one from both sides and you'll notice the x becomes on its own so the x jumps across to the left if we add x then the one jumps across to the right when we take away one giving us x equals k minus one answer b both of these questions illustrate that sometimes you don't need to do anything particularly advanced with excellent questions you just need to simplify the situation okay let's move on to a more advanced topic now we are getting to the harder and harder stuff so things get particularly intimidating at the moment don't stress out too much what would you say is bigger 50 to the power of 9 or 7 to the power of 18. now you're probably getting bored of me referring back to those excellents i want you to memorize but again they come in handy i would want you to spot that 50 is very close to 49 which is 7 squared if you don't spot that there's almost nothing you can do to answer this question it'd be very very hard to get it right there's no way you've got time to work out you have to spot at the higher level that 50 is very close to 49 so we're gonna estimate and change the 50 into a 49 even though obviously 50 is not the same thing as 49 why 49 partly because it's the nearest square number and second of all because 49 relates to 7 which is the key ingredient on the right hand side so if we change 50 to 49 we have 49 to the power of 9. 49 is 7 squared so we have 7 squared to the power of 9 and the exponents multiply to get 7 to the power of 18 which is the same as quantity b so is the answer c both quantities equal no because obviously with quantity a we rounded down a little bit we rounded from 50 down to 49 so quantity a is bigger than 7 to the power of 18 by a little bit meaning that quantity a is bigger and it all came down to noticing that 50 is very close to 49 which is a square number in fact the square of 7 which appears in quantity b let me give you another example of this what's bigger quantity a or quantity b as ever feel free to pause the video and work it out hopefully you notice that 950 is very close to a thousand and a thousand is a power of 10 which appears in quantity b a thousand is in fact 10 to the power of 3 because there are three zeros giving us 10 to the power of 3 to the power of 2000 multiplying that you get 10 to the power of 6000 which is the same right it's quantity b except this time we rounded up quantity a was actually 950 and we had to round it up to get to a thousand so really quantity a is smaller than a thousand to the power of 2000 it's smaller than 10 to the power of 6000 so quantity b is bigger in this instance by the way in case you didn't know the squiggly lines means roughly equal to so that is how you estimate exponents and it comes down to knowing your powers relatively well but don't worry they won't make it too hard it will be fairly close to what you need to get to now let's get on to the last of the really common exponent topics and for this i'm going to turn to a video i made earlier that you can also check out on my channel what i'm about to show you is almost guaranteed to come up in your test either for the gre or for the gmat it's one of my favorite topic areas and virtually every student i meet doesn't know how to do it or doesn't know how to do it properly this is one of those topics where i bet only about 10 20 of you could just pause the video and answer the question on the screen and get it right or certainly if i gave you the very last question in this set the boss level one even fewer could get it right so let's get straight to how to do it the question is what is the greatest prior factor of x if x equals 2 to the power of 8 plus 2 to the power 7 plus 2 to the power 6. now there's one thing i must get off my chest immediately and that is we can't literally just add up the little numbers many students would have gone ah i can outsmart philip i'm just going to add the little numbers and get 2 to the power 21 but it doesn't work like that when you add together numbers with exponents you can't simply add the little numbers what do we do if there's a plus sign or a minus sign between numbers with exponents that's the rule that most students don't know and it always amazes me because obviously that's the one the test often tries to catch you out on so what do we do step one and you might want to write this down is we factor out the smallest exponent so you look at the little powers see the smallest power that's 6 in this case so we're going to factor out 2 to the power of 6. so we write 2 to the power of 6 and then a bracket and i've demonstrated that here with three spaces in the bracket because we're going to have to think of the three different numbers that will go in the bracket but first i just want to be clear we're factoring out two to the 6 with brackets because 6 was the smallest exponent now we have to work out after we factor out the 2 to the 6 what will go inside the brackets but here's the real question what do we have to multiply 2 to the 6 by to get back to the original 2 to the 8. this is really important you pay attention at this point because many students get stuck here and don't realize what i'm asking but see that first slot in the bracket where i've put the underscore that's where you have to think of a number where 2 to the 6 times by that number gets us back to 2 to the 8. that's how you factor out you have to think in reverse if you want i recommend pausing the video and seeing if you can think of what that number would be less than half of students actually get it right the answer would be 2 to the power of 2. why is it 2 to the power 2 because think about it if we multiplied out that bracket 2 to the 6 times 2 to the 2 because it's multiplied which is because there's a bracket you add the little numbers 6 plus 2 gets us back to the a so in a sense the trick here is to think of what do i need to add to the little six and in this case that would be a two the other one should be a bit more obvious now two to the sixth times two to the one gets us back to the original two to the seven the last one's a bit more interesting it would be two to the six times one well done for those who got that but it's two to the six times one because two to the power of 6 times 1 gets you 2 2 to the power of 6. what's the next step well let me be clear that first step of factoring out is by far the hardest step and that's the step most students forget this next step is much easier but sometimes students forget this step as well what you do step two and you can write this down is you calculate the value of the bracket you see those numbers inside the bracket two to the two plus two to the one plus one you need to calculate what they actually are as numbers and add them up two to the two is four two to the one is two and one is of course one let's make something else clear here have you noticed how far into the question we've gotten without even wondering about greatest prime factor and that's something that's really important for me to emphasize these habits these skills these methods apply regardless of what the question is you see numbers with exponents being added or subtracted we're going to apply this factorization method regardless because when we're done our answer is going to be simpler than it was before and so we should be able to answer any question that they give us it's a bit like if i give you the fraction two over four you don't hesitate to call that one over two you're simplifying because no matter what the task is with two over four it will be easier to accomplish if we just call it one over two same thing with exponents we're going to factorize and simplify regardless of what the question is and of course here just to finish 4 plus 2 plus 1 is 7. so it becomes 2 to the 6 times 7. and now we can answer the question much more easily what is the greatest prime factor of x if x is all of that well what is the greatest prime factor of that answer we've got at the end it'll be seven the two prime factors would be two and seven and the biggest one the greatest one is seven so seven is the answer here notice how you wouldn't have spotted that just by looking at the numbers if you look at two to the eighth plus two to the seven plus two to the sixth it's not at all obvious that seven would be a factor of that answer so it's really important we follow this method now this is such an important method that i'm going to give a follow-up example which is much the same just to test out if you've understood what i was saying so with this question feel free to just pause the video and work out yourself or for those of you who want to hear my explanation let's get to it i'm going to ignore for now the 19 to the x 3 to the y bit i'm just going to focus on my core method of exponent factorization which number am i going to factorize you guessed it 3 to the power 3. that's the smallest exponent so that's the one that gets factorized this time i'll just write 3 to the power of 3 with a bracket what are the three items that go inside the bracket this is probably the hardest step that many students get confused at but what would you say the first one would be 3 to the 3 because 3 to the 3 times 3 to the 3 is 3 to the 6. remember you're adding up the little numbers when you multiply and once you get that one right the others are kind of obvious it will be take away because there's a take away sign 3 to the 2 because 3 to the 3 times 3 to the 2 is 3 to the 5 plus 1. there's always a 1 in there plus 1 minus 1. it gets a bit simple after a while next step you guessed it you calculate the value of the bracket 3 to the power of 3 is 27 minus 9 plus 1 is quick mental maths 19 so we have 3 to the power 3 times 19. and now let's emphasize something else that's important as i've written down below a number on its own should always be written as that number to the power of one when we're dealing with exponents and factorization so you see that 19 i didn't just write 19 did i i wrote 19 to the power of one and that's a good professional habit that i want you to get into all of the students watching this and i'm very grateful for the hundreds or thousands of people who will end up watching this and all of them i know are going to like and leave a comment i'm watching you but anyway i want all of my students to write to the power of one when there's a number that's on its own like 19 hair you can just trust me or see for yourself why that's so useful let's take this question by comparing it to the right hand side of the equation 19 to the power of x times 3 to the power of y we can quickly see that x would have to be the 1 and y would have to be 3. where it's not quite as obvious what the x is if you don't write the power of one by the way that's definitely not the only reason we write to the power of one but it's a great reason and of course if x is the one and y is three then x plus y is four but now we get to another core aspect of the video that i wanted to touch on which is the boss level question and its extra complications so with this question feel free again to pause the video and try it yourself and the question i'm talking about is the one that should now be appearing on the screen if 27 to the power of 2 plus 9 to the power of 3 over 3 to the power of x 2 to the power of y is an integer what is the greatest possible value of x y so tuffy i wouldn't say it's the absolute hardest they could possibly ask but it's right up there so try your best if you can or wait for my explanation the first thing that many of you will be confused by is the fact that it's 27 squared 9 cubed not a prime number like 2 or 3. and that's why i've written at the bottom pf or prime factorization is a prerequisite you have to prime factorize if you're going to deal with exponents or factorize exponents so we hate the fact that there's a 27 there and there's a 9 there because they're not primes we can't really apply our method unless there's primes so we're going to prime factorize those numbers what is 27 written in prime factored form it would be 3 cubed so i've replaced the 27 with the 3 cubed in brackets what would 9 be in prime factorized form it would be 3 squared so i just remove the 9 and write 3 squared in brackets in its place the next step some of you will know based on your exponent rule knowledge which is what to do with the little numbers with the exponents when there's a bracket in between what we do is we multiply the little numbers we multiply the three and the two in this case when there's a bracket in between two exponents you multiply so the numerator actually becomes 3 to the power of 6 plus 3 to the power 6 because both 3 times 2 and 2 times 3 becomes 6. and here's another step that i know some of you would have gotten really confused by how do we deal with 3 to the power 6 plus 3 to the power 6. some of the wrong answers here would be 3 to the power of 12 or 6 to the power 6. no you can't change the base like that and you don't add the little numbers when exponents added what you could do is simply follow our method which is we factor out the smallest one and i know they're identical but we can still do it and what would be in the brackets one plus one which is two which makes sense we've got one three to the sixth plus another one three to the sixth giving us two three to the power of 6 and if we're being professional how do we write that 2 to the power of 1 times 3 to the power 6. remember a number on its own gets that power of 1. now we can deal with the actual question they said if that fraction is an integer what does that mean it means the denominator must fit into must be a factor of the numerator if the denominator is not a factor of the numerator if it doesn't go in then it wouldn't be an integer it would still stay as a fraction so when they give you a fraction and tell you it ends up being an integer it just means the bottom line fits into neatly the top line now some of you are thinking but there's many numbers that x could be and y could be that would fit in if x was a three and y was a one that would fit in if x was a four and y was a zero that would also cancel out and fit in but the question was asking for the greatest possible value of x y and the biggest thing that x could be is six because then the three to the sixes would cancel and the biggest thing that y could be is a one because then it would cancel if x was any bigger than six or y was any bigger than one then it wouldn't cancel out and you wouldn't get an integer so x is six at the biggest y is one of the biggest which if you multiply them that gets us to be six at the most of course what's written there saying that x and y are positive integers is also essential because if they could be negatives then of course two negatives can multiply to be even bigger than six but as positives the biggest thing they could be would be six when you multiply them so six is the answer how do we factorize algebraic exponents can we use that same technique absolutely we can step one factor out the smaller of the two exponents so in fact you can pause and try this one the smaller of the two exponents on the left is two to the power of x so we factor out giving us two squared plus one calculate the value of the bracket 2 squared plus 1 is 5. cancel out the 5's on both sides so 2 to the x equals 2 to the 2y base cancellation so x equals 2y and that is in fact the answer to the question so i wanted to demonstrate that when we factorize using algebraic exponents it's the exact same technique as what we just saw the exact same principle step one factor out the smaller of the two exponents step two calculate the value of the bracket okay now we're getting to the end of my epic video on exponents with some truly fancy topics in fact two more topics first one pretty rare in fact both of them pretty rare but if you're gunning for that 170 or that 800 in the gmat this is what we need to know stacked exponents if n equals 2 cubed what is n to the power of n my key bit of advice here is to calculate the exponent separately otherwise you'll get confused so n to the power of n basically means 2 cubed to the power of 2 cubed right because n is 2 cubed now don't get mixed up in do we add do we multiply just calculate that exponent on the outside of the bracket separately just work out what is 2 cubed 2 cubed is 8. so we have 2 to the power of 3 so keep that bracket separate don't do anything with that to the power of 8. and now we can apply our techniques once we've sorted out that exponent what happens when we have 2 cubed to the power of 8 it multiplies to get 2 to the power of 24 and that's answer d following that exact same principle of working out the exponent first can you answer the next official question on the right if x equals 3 squared what is the value of x to the power of x well notice again we're never going to work out the bracket we're not going to work out the 3 squared is 9. that doesn't help because we want to prime factorize keep things in primes so instead we just write it out as 3 squared to the power of 3 squared step 1 most important of all calculate the exponent what is three squared three squared is nine so we don't calculate the three squared that's in the brackets we calculate the three squared that's outside the brackets we calculate the exponent first three squared is nine so we have three squared in brackets that's untouched we don't do anything with that to the power of nine maybe it'll be most clear if i phrase my advice is just saying start with the exponent calculate that first now what do we do we multiply the exponents when we have a bracket in between them so we get 3 to the power of 18 which is answer e and that's how you deal with stacked exponents you calculate the exponent outside the bracket first notice by the way i should have also said we have a bracket for the base so we make three squared the base and have it in brackets to the power of three squared i probably should have mentioned that earlier without that bracket things get really confusing so notice with the first question on the left the two cubes i put that in a bracket and then i had the other end the other two cubed outside the bracket i calculated the exponent outside the bracket first that's 2 cubed equals 8 and then i multiplied very important the order in which you do things and now for the hardest topic in this video we're getting to the super rare questions again my title was everything you need to know so i'm obliged to teach you this next topic which is joining twos and fives to make tens technically this is more of a number properties kind of question than an excellent question but let's cover it anyway if 1 over 2 to the 11 times 5 to the 17 is expressed as a terminating decimal how many non-zero digits will the decimal have virtually every student can't do this question because it is really quite difficult we have to realize is that we can only join two numbers together if they have the same exponent and 2 to the 11 and 5 to the 17 don't have the same exponent but we can still join them together if we just separate off 11 of the fives what do i mean by that we can rewrite the denominator as 2 to the 11 times 5 to the 11 times 5 to the 6. of course the 5 to the 11 and 5 to the 6 would make up the 5 to the 17. but why do we split up the 5 to the 17 like that well mostly so that has the same exponent as the two but partly because twos and fives join together to make tens and tens are really easy to calculate unlike twos or fives i hope that's really clear we split up the 5 to the 17 to become 5 to the 11 times 5 to the 6 because once we have a 5 to the 11 we could combine it with the 2 to the 11 to make 10 to the 11. if you're not sure how we combine exponents i covered this earlier in the video if we have a different base we can combine them as long as they have the same power so 2 to the 11 times 5 to the 11 becomes 10 to the 11. now why is this a real improvement we still have two numbers in the denominator with two different powers so how is this an improvement because 10 is much easier to calculate when we're doing decimals or really i should say 1 over 10 is much easier to calculate let me show you the next step because it's arguably the hardest we essentially split up the fraction to make it 1 over 10 to the 11 and 1 over 5 to the 6. this is really important for those advanced students who think this is unusual technique the reason we can do this is because we have a 1 in the numerator if that was anything else other than a 1 we couldn't split the fraction like this if it was a 2 in the numerator we couldn't call it 2 over 10 to the power of 11 and 2 over 5 to the power of 6 only because it's a 1 because 1 to the power of 11 doesn't make any difference i know this seems really strange but it's something that you can do splitting up a fraction only when you have one in the numerator and again many of you are still thinking but is that really any easier like how are we making progress the reason this is progress the reason we join the twos and fives together to make 10 is because 10 to any power is super easy to calculate 1 10 is simply 0.1 and one-fifth is 0.2 they're both really easy to calculate as decimals even when we apply really large powers to them watch this 0.1 to the power of 11 is simply 0.00000 however many zeros in fact 11 zeros in total one that's so much easier to calculate than 0.5 to the 11. if we just start with twos and fives the one over two will become zero point five and zero point five to the eleven would be a huge number because we combined the twos and the fives we got one tenth which is zero point one which is incredibly easy to calculate how many zeros will there be there will be exactly eleven zeros if we include the zero before the decimal point so it'll be zero point and then an additional ten zeros and then a one what about the one fifth which as a decimal is 0.2 what about that 0.2 to the power of 6. that's not that hard to calculate that brings us all the way back to that memorization sheet that we saw right at the beginning 2 to the power of 6 is 64 as you've probably memorized by now how many zeros well the way i like to remember it is you count how many numbers are for the decimal place which in the case of 0.2 is just one number after the decimal place and then multiply that by the power so one number after the decimal place times six means there will be six numbers after the decimal place that's why there are four zeros plus the six and the four to make 6 numbers after the decimal place so to recap you work out 2 to the power of 6 which is 64 and then just count the number of zeros to make up a total of six numbers off the decimal place i think you'll agree we're straying quite far away from exponents here this is moving into decimals and calculations i thought i'd throw this in as a bonus lesson we don't really care about the number of zeros though because the question was how many non-zero digits will the decimal have when we multiply these two brackets together we're going to get a massive number of zeros at the front it's going to be a tiny answer but then with 1 times 64 at the end which is 64. so there'll just be those two numbers right at the end after loads of zeros a six and a four meaning the answer is b there will be two non-zero digits at the end of the decimal i know many of you will think this doesn't have much to do with exponents but the one bit that it does have to do with exponents is the beginning which is that you can split up an exponent like 5 to the 17 into 5 to the 11 times 5 to the 6 for the reason of combining the twos and the 5 together to make tens which for most calculations will be a lot easier definitely this is a 170 level question and that brings us to the end of this video and as it says i want to thank you for getting this far and please do leave a like and a comment if you have found this helpful in any way even though i've taught you now everything you need to know about exponents from the really simple stuff all the way to the crazy hard stuff the gre and gmat can still come up with an exponent question that's difficult because it will bring in another topic maybe it will bring in geometry with exponents or bring in statistics and exponents like an average question or a median question or standard deviation just because you now know everything to do with exponents doesn't guarantee yet that you'll get every single exponent question right because sometimes the test is allowed to mix two topics together and if you don't know the other topic you might still get the question wrong so i don't want to pretend like you're guaranteed to get every single excellent question right now it depends on your knowledge when they merge two topics together but if it's a purely exponent based question which most of them are you now have a very very good chance of getting it right finally please please do let me know in the comments if you like this style of one big video covering everything you need to know or if you'd rather prefer a playlist or just prefer individual tiny videos on individual tricks and tips it really helps to let me know what you think thank you again for reaching the end well done take a break and see you in the next video
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What are Directed Angles?
Essentials Of Math
4840 subscribers
148 likes
Description
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Posted: 9 Jul 2018
Follow me on twitter @abourquemath
Enjoy another random upload. New series starts tomorrow!
Document on directed angles:
My proof:
15 comments
Transcript:
Introduction hello everybody in this video I'm going to be talking about what directed angles are directed angles basically allow geometric proofs to be more concise and especially get rid of casework if there's a problem and you have some edge cases our different configurations where some things some angles may be equal or something goals might be supplementary it just allows it allows something to be stated with just one line one configuration tragic directed angles basically will allow us to extend the flexibility of a geometric proof and I want to cover this video now not because I'm definitely not doing the inverse of geometry for my next series I think I have an idea of what it is but I'll leave that a secret but I'm sort of working on an upload schedule for the next school year and I think when I cover some harder geometry especially Olimpia geometry problems I'm going to be using direct and angles probably quite often and so I just want to make this video now just sort of get that out there so I don't think this is a very well documented topic I will link a a handout on directed angles which I'm pretty much basing this video on and so anyway if we have two lines here let's call them M Directed Angles and L then what we'll call the directed angle which is written written like this the directed angle of L and M is essentially the angle from L to M in the counterclockwise direction so it is this angle right here and so essentially since we have a set direction if if we looked at instead the directed angle between N and L these would not be equal in most cases these would not be equal because the director angle between m and L is actually this angle and so we see that these two are supplementary angles and so they have the property that they always add up to 180 degrees but that's sort of not very nice because sometimes where when we're looking at these edge cases and we're sort of one thing could be two angles could be equal or they could be supplementary it was sort of one a nice way to state that sort of those cases are the same and so what we can what we can do is take our angles mod 180 and basically we look at the remainder when we divide by 180 and that allows us to say if we had the angle of 150 degrees instead of 150 degrees we could say that is equal to negative 30 degrees mod 180 so all directed angles are taken mod 180 and so in fact we have the more general statement that if we switch the order that we're taking the angle then we get the negative of that angle measure and so we sort of have this this line like two lines but we usually see an angle like a traditional geometric angle would be written as say angle AOB where you would have a Oh B this is a o and B and so we define the directed angle of a OB to be the directed angle of instead of L and M we have a o and Bo Bo and these are lines you know there's a segment notation but whatever and and so it would be this angle in the counterclockwise direction it would be negative if a OB the order was clockwise and so let's see a particular example where this helps us so let's look at Cyclic Quadrilateral cyclic quadrilateral we know the basic definition of cyclic quadrilateral or at least one of the properties of cyclic quadrilateral is that if we have a B X and y on a circle or points on a circle then here we have that X a Y the angle X a Y is congruent to the angle X B Y but here we see that X a Y and X B Y are supplementary and all we think okay that's no good but here we see that X a Y and X B Y or so the the same orientation right there they're both yeah I think that's that's counterclockwise order so XA y is counterclockwise X B Y is counterclockwise and these are equal cases right here X a Y and X B Y are in opposite orientations so in fact the in in directed angles they would be negatives of each other yeah I don't know David they're going to be negative is each other but basically what I'm saying is that here we have equality and here we have supplementary condition but X B Y and X a Y go in opposite directions and so since they're supplementary we knew that X a Y is equal to this angle out here but this angle since it is 180 minus this angle it's basically the negative of X B Y I'm not the negative but well I I guess I should just write so here here we have that the norm that normally the angles are equal and if two angles are congruent then they're they're directed well I guess they have to be in the same orientation anyway what I'm trying to say here is that here X a Y is congruent to and xby and here we see that the directed angle xby is equal to since it's in the counterclockwise since xby is in the clockwise direction then xby the directed angle is equal to sort of 180 minus the normal XB y or if we have a point out here on this array then we're looking at this is angle Y BC but we know that this is also equal to the directed angle of X a Y so if we have two points or yeah so if we have two points X and y then a and B lie on a circle with x and y if the directed angles X B y equals the directed angle X a Y it took a while for me to sort of phrase that but we sort of get the same we get two different configurations one where angles are equal and one where angles are supplementary for one for one condition that the directed angles are in fact equal so yeah that was a mouthful but I'm not gonna do any problems with this I'm just sort of showing this for sort of documentation and because I will definitely be using directed angles in future geometric proofs so I will leave the link for the document that this comes from so you can read up more about it and then there's some problems at the end and I'll actually link a problem a solution to a problem that I wrote up that uses directed angles so you can sort of see how how they would be used in an approach because I know that the problems in the document get the solutions get a little involved I think the the one I have linked is the one I wrote up it's a little a little bit of an easier problem and you sort of take care of cases I don't talk about the cases in the problem but there's and there's two cases and you sort of get them for free if he's directed angles and same here so basically directed angles make the definition of cyclic watch much nicer than than what it is and there's a lot of problems especially on Olympiads the little sort of throw-in trick configurations where someone's argument might go wrong but if you use direct angles you're more more likely to guarantee that that won't happen as long as easy to use them correctly I guess some other notes would be to say that using directed angles with trigonometry is a bit of a nightmare especially if you're doing sort of law of sines chasing and the inscribed angle theorem I believe is the name where you go from sort of like say X B Y to you if I drew in the center X zero Y we know that these are twice that that XO y would be twice of X B Y but this is problematic because it doesn't make sense to take twice of an angle in mod 180 or for that matter to take half of an angle because mod mods will mess that up let me see that here that if we take half of 75 we get or sorry half of 150 we get 75 but if we take half of negative 30 we get negative 15 and these are not congruent mod 180 and so sort of in we define sort of directed arcs to deal with this case because the angle xly is really representing sort of them the measure of the arc XY and so we take directed arcs in some sense to be mod 360 because there's 360 our degrees in a circle anyway I've been rambling I'm a little bit tired at the time of recording this but yeah I just want to sort of show this topic and I promise but starting Tuesday we'll have more videos coming out so thank you thank you and I appreciate each and every one of you your support and I hope you enjoy the new series thank you
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Home > Theses, Dissertations and Culminating Projects > 714
Theses, Dissertations and Culminating Projects
Enumeration of Independent Sets in Graphs
Author
James Alexander, Montclair State University
Date of Award
5-2012
Document Type
Thesis
Degree Name
Master of Science (MS)
College/School
College of Science and Mathematics
Department/Program
Mathematical Sciences
Thesis Sponsor/Dissertation Chair/Project Chair
Jonathan Cutler
Committee Member
Aihua Li
Committee Member
Diana Thomas
Abstract
An independent set is one of the most natural structures in a graph to focus on, from both a pure and applied perspective. In the realm of graph theory, and any concept it can represent, an independent set is the mathematical way of capturing a set of objects, none of which are related to each other. As graph theory grows, many questions about independent sets are being asked and answered, many of which are concerned with the enumeration of independent sets in graphs. We provide a detailed introduction to general graph theory for those who are not familiar with the subject, and then develop the basic language and notation of independent set theory before cataloging some of the history and major results of the field. We focus particularly on the enumeration of independent sets in various classes of graphs, with the heaviest focus on those defined by maximum and minimum degree restrictions. We provide a brief, specific history of this topic, and present some original results in this area. We then speak about some questions which remain open, and end the work with a conjecture for which we provide strong, original evidence. In the appendices, we cover all other necessary prerequisites for those without a mathematical background.
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Alexander, James, "Enumeration of Independent Sets in Graphs" (2012). Theses, Dissertations and Culminating Projects. 714.
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Conjugating a Third Conjugation Verb
Nov 18, 2014
• 260 likes • 640 Views
Conjugating a Third Conjugation Verb. duco ducere duxi ductus. All About Verbs Each verb has 4 principal parts. duco. ducere. duxi. ductus. First Person Singular Present Active I lead. Present active Infinitive To lead. First Person Singular Perfect Active I have led.
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Conjugating a Third Conjugation Verb duco ducere duxi ductus
All About Verbs Each verb has 4 principal parts duco ducere duxi ductus First Person Singular Present ActiveI lead Present active Infinitive To lead First Person SingularPerfect ActiveI have led Perfect Passive ParticipleHaving been led
Finding the Present Stem Find the second principal part of the verb. duco ducere duxi ductus The present stem is the second principal part – re. ducere - re = duce
Present Active Tense In the present active tense change e on the present stem to i and add the present active endings.Present stem = Ducere – re = duce duc e i Singular Plural we o I mus First Person Second Person tis you s you he, she,it Third Person t nt they Present Active Endings
Present Active I leadI am leadingI do lead First Person We leadWe are leadingWe do lead duc o duci mus Second Person You leadYou are leadingYou do leading You leadYou are leadingYou do lead duci tis duci s He leadsHe is leadingHe does lead Third Person They leadThey are leadingThey do lead duci t ducu nt Nota Bene: There is no “i” in the first person singular and the e changes to “u” in the third person plural.
Present Passive Tense In the present passive tense change “e” on the present stem to “i” and add the present active endings.Present stem = Ducere – re = duce Singular Plural First Person mur we or I mini you Second Person you ris he, she,it ntur they Third Person tur Present Passive Endings
Present Passive I am led We are led First Person duc or duci mur Second Person You are led duci mini duce ris You are led He is led They are led Third Person duci tur ducu ntur Nota Bene: There is no “i” in the first person singular,the “e” is retained in the second person singular and the “e” changes to “u” in the third person plural.
Imperfect Tense Present Stem + BA + Present Endings Imperfect Active Imperfect Passive duce ba m I led duce ba r I was led duce ba s You led duce ba ris You were led He led duce ba tur He was led duce ba t We led duce ba mur We were led duce ba mus You led duce ba mini You were led duce ba tis They were led They led duce ba ntur duce ba nt
Future Tense The sign of the future in the third declension is e. Future Active Future Passive duc a m I will lead duc a r I will be led duc e s You will lead duc e ris You will be led duc e tur He will be led duc e t He will lead duc e mur We will be led duc e mus We will lead duc e mini You will be led duc e tis You will lead They will be led duc e ntur duc e nt They will lead Nota Bene: There is an “a” in the first person singular
Finding the Perfect Stem Find the third principal part of the verb. duco ducere duxi doctus The perfect stem is the third principal part – i. duxi - i = dux
Perfect Active Tense The Perfect Active Tense = Perfect Stem + Perfect Active Endings Singular Plural First Person imus we i I istis you Second Person you isti he, she,it erunt they Third Person it Perfect Active Endings
First Person dux i I ledI have ledI did lead dux imus We ledWe have ledWe did lead You ledYou have ledYou did lead Second Person dux istis dux isti You ledYou have ledYou did lead Third Person dux it dux erunt They ledThey have ledThey did lead He ledHe has ledHe did lead Perfect Active
Pluperfect Active Tense Perfect Stem + era + Present Endings Pluperfect Active dux era m I had led dux era s You had led dux era t He had led dux era mus We had led You had led dux era tis dux era nt They had led
Future Perfect Active Tense Perfect Stem + eri + Present Endings Future Perfect Active dux er o I will have led dux eri s You will have led dux eri t He will have led dux eri mus We will have led You will have led dux eri tis They will have led dux eri nt
Perfect Passive System 1. Uses the fourth principal part which is called the perfect passive participle (PPP). duco, ducere, duxi, ductus. -a. -um 2. The perfect passive participle (PPP) is a verbal adjective and must agree with the subject of the sentence. If the subject is puella (feminine singular), the PPP will be ducta. If the subject is puellae (feminine plural), the PPP will be ductae. If the subject is vir (masculine singular), the PPP will be ductus. If the subject is viri (masculine plural), the PPP will be ducti. If the subject is oppidum (neuter singular), the PPP will be ductum. If the subject is oppida (neuter plural), the PPP will be ducta.
Perfect Passive Tense PPP + Present Tense of Sum Perfect Passive ductus – a - um sum I have been led ductus – a - um es You have been led ductus –a -um est He has been led ducti –ae -a sumus We have been led You have been led ducti –ae –a estis They have been led ducti – ae -a sunt
Pluperfect Passive Tense PPP + imperfect Tense of Sum Pluperfect Passive ductus – a - um eram I had been led ductus – a - um eras You had been led ductus –a -um erat He had been led ducti –ae -a eramus We had been led You had been led ducti –ae –a eratis They had been led ducti – ae -a erant
Future Perfect Passive Tense PPP + Future Tense of Sum Future Perfect PassiveTense ductus – a - um ero I will have been led ductus – a - um eris You will have been led ductus –a -um erit He will have been led ducti –ae -a erimus We will have been led You will have been led ducti –ae –a eritis They will have been led ducti – ae -a erunt
The __ tense does not use the present stem. a. Future b. Past Perfect c. Imperfect d. Present next
Legere is the __ principal part of the word. a. First b. Second c. Third d. Fourth next
legisti can be translated _______ a. You have chosen b. You were chosen c. You are chosen d. You will choose next
Acta est can be translated _______ a. She had been driven b. She is driven c. She has been driven d. She will have been driven
Finis
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ANTENNAS JOHN D. KRAUS Second Edition TATA McGflAW-HILL EDITION SYMBOLS, PREFIXES AND ABBREVIATIONS See inside back cover for Constants and Conversions A ampere A angstrom = UK 10 m A vector potential, Wb m“ A , a area, m 2 Ae collecting aperture A€ effective aperture Am maximum effective aperture A„ effective aperture, receiving effective aperture, transmitting A t geometric aperture Ap physical aperture At scattering aperture AR axial ratio AU astronomical unit a atto — 10“ (prefix) a unit vector B, B magnetic flux density, T = Wb m -2 B susceptance, O B susceptance/unit Length, U m “ 1 BWFN beam width, first nulls C coulomb C capacitance, F C capacitance/unit lengthy F m“ 1 C, c a constant, c = velocity of light cc cubic centimeter CC degree Celsius D, D electric flux density, Cm -2 D directivity d distance, m deg degree, angle dB decibel - 10 log {PJP V ) dBi decibels over isotropic dt element of length (sqilar), m d\ element of length (vector), m ds element of surface (scalar), nr element of surface (vector), m 2 dv dement of volume (scalar), m 3 E, E electric field intensity, V m“ 1 E exa = 10 1 (prefix) emf electromotive force, V e electric charge, C F farad F, F force, N F femto — KT 15 (prefix) / frequency, Hz G giga = 10 (prefix) G conductance, U G conductance/unit length, Um“ G gain g gram H henry H, H magnetic field. Am" 1 HPBW half-power beam width Hz hertz = 1 cycle per second K effective height I, IJ current, A J joule J, J current density, A m -2 Jy jansky, 10‘ 2ft W m“ 2 Hz“ l K kelvin K, K sheet-current density, Am -1 K, k a constant k kilo = 103 (prefix) fcg kilogram L inductance, H L inductance/unit length, H m" l Liter l, L Length (scalar), m \ length (vector), m LCP Left circularly polarized LEP left elliptically polarized In natural logarithm (base e) log common logarithm (base 10) M mega = 106 (prefix) M, Af magnetization. Am' 1 Af polarization state of wave Ma polarization state of antenna m meter m milli = 10“ 3 (prefix) min minute N newton N, n number (integer) Np neper n nano = 10“ (prefix) n unit vector normal to a surface P, P polarization of dielectric, C m” P peta — 10' 3 (prefix) P polarization state = P(y, ^) P power, W P, normalized power pattern, dimensionless P pico = 10 12 (prefix) G.
charge, C R resistance, ft Rr radiation resistance RCP right-circular polarization REP right -elliptical polarization r revolution r radius, m; also coordinate direction r unit vector in r direction rad radian rad 2 square radian — steradian ^ sr S, S 5 5, s $ sr T T l U V V ir v W Wb w X X i x V r y y z z Z' z< Zl Z0 Z t Poynting vector, W m 2 flux density, W m~ 2 Hz 1 distance, m; also surface area, m 2 second (or time) steradian = square radian = rad 2 tesla - Wb m “ 2 tera - 10 12 (prefix) time, s radiation intensity, W sr -1 volt voltage (also emf), V emf (electromotive force), V velocity, m s _1 watt weber energy density, J m~ 3 reactance, ft reactance/unit length, ft m _I unit vector in x direction coordinate direction admittance, U admittance/unit length, U m“ 1 unit vector in y direction coordinate direction impedance, ft impedance/unit length, ft m ‘ 1 intrinsic impedance, conductor,’ ft per square intrinsic impedance, dielectric, ft per square load impedance, Q transverse impedance, rectangular waveguide, ft transverse impedance, cylindrical waveguide, ft intrinsic impedance, space, ft per square characteristic impedance, transmission line, ft unit vector in z direction coordinate direction, also red shift (alpha) angle, deg or rad attenuation constant, hep m 1 (beta) angle, deg or rad; also phase constant - 2n/X (gamma) angle, deg or rad 3 Co n e # K X p Pr P X ft ft U to (delta) angle, deg or rad (epsilon) permittivity (dielectric constant), FmM aperture efficiency beam efficiency stray factor relative permittivity permittivity of vacuum, Fm' 1 (eta) (theta) angle, deg or rad (theta) unit vector in 0 direction (kappa) constant (lambda) wavelength, m free-space wavelength (mu) permeability, H m “ 1 relative permeability permeability of vacuum, H m “ 1 (nu) (xi) (pi) = 3.1416 (rho) electric charge density, Cm -3 : also mass density, kg m 3 reflection coefficient, dimensionless surface charge density. Cm -2 linear charge density, Cm -1 (sigma) conductivity, U m“ 1 radar cross section (tau) tilt angle, polarization ellipse, deg or rad transmission coefficient (phi) angle, deg or rad (phi) unit vector in ^ direction (chi) susceptibility, dimensionless (psi) angle, deg or rad magnetic flux, Wb (capital omega) ohm (capital omega) solid angle, sr or deg2 beam area main beam area minor lobe area (upsidedown capital omega) mho (U = 1/ft = S, siemens) (omega) angular frequency ( = 2itf\ rad s“ 1 To Heinrich Hertz T who invented thefirst antennas .
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. and Guglielmo Marconi , iv//o ptonseiW jVi rfctfir practical application ANTENNAS Second Edition John D. Kraus Director, Radio Observatory Taine G, McDougal Professor Emeritus of Electrical Engineering and Astronomy The Ohio State University with sections on Frequency-Sensitive Surfaces by Benedikt A. Munk Radar Scattering by Robert G. Kouyoumjian and Moment Method by Edward H. Newman all of the Ohio State University Tata McGraw-Hill Publishing Company Limited — NEW DELHI McGraw-Hiif Offices New Delhi New \brk St Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal San Juan Singapore Sydney Tokyo Toronto Tata McGraw-Hill ^ A Division of The McGrmv IUU Companies ANTENNAS Copyright© 1988 by McGraw-Hill, Inc.
All rights reserved. No part of this publication may be reproduced, stored in a data base or retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher Tata McGraw-Hill edition 1997 Sixth reprint 2001 RCLYCRCLRACBB Reprinted in India by arrangement with The McGraw-Hill Companies, Inc,, New York For sale in India Only Library of Congress Cataloging-in-Publication Data Kraus, John Daniel, (date).
Antennas.
(McGraw-Hill series in electrical engineering.
Electronics and electronic circuits) Includes index.
1. Antennas (Electronics) I. Title.
TK7871.6.K74 1988 621.38 , 028 1 3 87-15913 ISBN 0-07-035422-7 When ordering this title use ISBN 0-07-463219-1 Published by Tata McGraw-Hill Publishing Company Limited, 7 West Patel Nagar, New Delhi 110 008, and printed at A P Offset, Shahdara, Delhi 110 032 L ABOUT THE AUTHOR John D. Kraus was born in Ann Arbor, Michigan, in 1910 and received his Ph.D.
degree in physics from the University of Michigan in 1933. He then did research in nuclear physics with Michigan's newly completed 100-ton cyclotron until World War II when he worked on the degaussing of ships for the U.S. Navy and on radar countermeasures at Harvard University. After the War he came to the Ohio State University where he is now Director of the Radio Observatory and McDougal Professor (Emeritus) of Electrical Engineering and Astronomy.
Dr. Kraus is the inventor of the helical antenna, the workhorse of space communication, the corner reflector, used by the millions for television reception, and many other types of antennas. He designed and built the giant Ohio radio telescope known as “Big Ear” He is the holder of many patents and has published hundreds of scientific and technical articles. He is also the author of the widely used classic textbooks Amermas (McGraw-Hill, 1950), considered to be the “Antenna Bible,” Electromagnetics (McGraw-Hill, 1953, second edition 1973, third edition, 1984), and Radio Astronomy {McGraw-Hill, 1966, second edition Cygnus Quasar, 1986). In addition, Dr. Kraus has written two popular books Big Ear (1976) and Our Cosmic Universe (1980).
Dr. Kraus received the U.S, Navy Meritorious Civilian Service Award in 1946. He was made a Fellow of the Institute of Electrical and Electronic Engi-neers (IEEE) in 1954 and was elected to the National Academy of Engineering in 1972. He received the Sullivant Medal, Ohio State University's top award, in 1970; the Outstanding Achievement Award of the University of Michigan in 1981 ; the prestigious Edison Medal of the IEEE in 1985; and the Distinguished Achievement Award of the Antennas and Propagation Society of the IEEE in the iame year.
Currently, Dr. Kraus is serving as antenna consultant to government and industry.
CONTENTS Symbols, Prefixes and Abbreviations Inside front cover .and facing inside front cover Constants and Conversions Facing inside back cover Gradient, Divergence and Curl in Rectangular, Cylindrical and Spherical Coordinates Inside back cover Preface xxiii 1 Introduction 1 1-1 Introduction 1 1-2 The Origins of Electromagnetic Theory and the First Antennas 1 1-3 Electromagnetic Spectrum ' 8 1-4 Dimensions and Units 11 1-5 Fundamental and Secondary Units 12 1-6 How to Read the Symbols and Notation 13 1-7 Equation Numbering 14 1-8 Dimensional Analysis 15 References 15 2 Basic Antenna Concepts 17 2-1 Introduction 17 2-2 Definitions 17 2-3 Basic Antenna Parameters 19 2-4 Patterns 20 2-5 Beam Area (or Beam Solid Angle) 23 2-6 Radiation Intensity 25 2-7 Beam Efficiency 25 2-8 Directivity 26 2-9 Examples of Directivity 26 2-10 Directivity and Gain 27 xf XU CONTENTS 2- 1 1 Directivity and Resolution 2-12 Aperture Concept 2-13 Effective Aperture 2-14 Scattering Aperture 2-15 Loss Aperture 2-16 Collecting Aperture 2-17 Physical Aperture and Aperture Efficiency 2-18 Scattering by Large Apertures 2-19 Effective Height 2-20 Maximum Effective Aperture of a Short Dipole 2-21 Maximum Effective Aperture of a Linear A/2 Antenna 2-22 Effective Aperture and Directivity 2-23 Beam Solid Angle as a Fraction of a Sphere 2-24 Table of Effective Aperture, Directivity, Effective Height and Other Parameters for Dipoles and Loops 2-2 5 F ri is Transmission Formula 2-26 Duality of Antennas 2-27 Sources of Radiation: Radiation Results from Accelerated Charges 2-28 Pulsed Opened-Out Twin-Line Antennas 2-29 Fields from Oscillating Dipole 2-30 Radiation from Pulsed Center-Fed Dipole Antennas 2-31 Antenna Field Zones 2-32 Shape-Impedance Considerations 2-33 Antennas and Transmission Lines Compared 2-34 Wave Polarization 2-35 Wave Polarization and the Poynting Vector 2-36 Wave Polarization and the Poincare Sphere 2-37 Cross-field 2-38 Table Summarizing Important Relations of Chapter 2 Problems 3 Point Sources 3-1 Introduction, Point Source Defined 3-2 Power Patterns 3-3 A Power Theorem and Its Application to an Isotropic Source 3-4 Radiation Intensity 3-5 Source with Hemispheric Power Pattern.
3-6 Source with Unidirectional Cosine Power Pattern 3-7 Source with Bidirectional Cosine Power Pattern 3-8 Source with Sine (Doughnut) Power Pattern 3-9 Source with Sine-Squared (Doughnut) Power Pattern 3-10 Source with Unidirectional Cosine-Squared Power Pattern 3-11 Source with Unidirectional Cosine" Power Pattern 3-12 Source with Unidirectional Power Pattern That Is Not Symmetrical 3-13 Directivity 3-14 Source with Pattern of Arbitrary Shape 3-L5 Gain 3-16 Field Patterns CONTENTS Xlll 3-17 Phase Patterns 108 3-18 General Equation for the Field of a Point Source 115 problems 116 4 Arrays of Point Sources 118 4-1 Introduction 118 4-2 Arrays of Two Isotropic Point Sources 118 4- 2a Case L Two Isotropic Point Sources of Same Amplitude and Phase 1 1 8 4-2b Case 2. Two Isotropic Point Sources of Same Amplitude but Opposite Phase 121 4-2c Case 3. Two Isotropic Point Sources of the Same Amplitude and in Phase Quadrature 122 4-2d Case 4, General Case of Two Isotropic Point Sources of Equal Amplitude and Any Phase Difference 125 4-2e Case 5, Most General Case of Two Isotropic Point Sources of Unequal Amplitude and Any Phase Difference 126 4-3 Nonisotropic but Similar Point Sources and the Principle of Pattern Multiplication 127 4-4 Example of Pattern Synthesis by Pattern Multiplication 130 4-5 Nonisotropic and Dissimilar Point Sources 134 4-6 Linear Arrays of n Isotropic Point Sources of Equal Amplitude and Spacing 1 37 4-6a Introduction 137 4-6b Case 1. Broadside Array (Sources in Phase) 140 4-6c Case 2. Ordinary End-Fire Array 141 4-6d Case 3. End-Fire Array with Increased Directivity 141 4-6e Case 4. Array with Maximum Field in an Arbitrary Direction.
Scanning Array 145 4-7 Null Directions for Arrays of n Isotropic Point Sources oF Equal Amplitude and Spacing 145 4-8 Broadside versus End-Fire Arrays. Turns versus Dipoles and 3-Dimensional Arrays 150 4-9 Directions of Maxima for Arrays of n Isotropic Point Sources of Equal Amplitude and Spacing 156 4-10 Linear Broadside Arrays with Nonuniform Amplitude Distributions General Considerations 159 4-1 1 Linear Arrays with Nonuniform Amplitude Distributions The Dolph-Tchebyscheff Optimum Distribution 162 4-12 Example of Dolph-Tchebyscheff Distribution for an Array of 8 Sources 171 4-13 Comparison of Amplitude Distributions for 8-Source Arrays 173 4-14 Continuous Arrays 175 4-15 Huygens' Principle 179 4-16 Huygens 7 Principle Applied to the Diffraction of a Plane Wave Incident on a Rat Sheet Physical Optics 183 4-17 Rectangular-Area Broadside Arrays 186 4-18 Arrays with Missing Sources and Random Arrays 189 Problems 190 XIV CONTENTS 5 The Electric Dipole and Thin Linear Antennas 200 5-1 The Short Electric Dipole,, 20Q 5-2 The Fields of a Short Dipole 201 5-3 Radiation Resistance of Short Electric Dipole 21 3 5-4 The Fields of Short Dipole by the Hertz Vector Method 217 5-5 The Thin Linear Antenna 218 5- 5a Case L kj2 Antenna 221 5-5b Case 2, Full-Wave Anterma 222 5-5c Case 3.
32/2 Antenna 222 5-5d Field at Any Distance from Center-Fed Dipole 223 5-6 Radiation Resistance of 2/2 Antenna 224 5-7 Radiation Resistance at a Point Which Is Not a Current Maximum 227 5-8 Fields of a Thin Linear Antenna with a Uniform Traveling Wave 228 5-8a Case L Linear a/2 Antenna 233 5-8b Case 2, Linear Antenna 5A Long 234 5-8c Case 3.
Linear Antennas A/2 to 25A Long 234 Problems 235 .6 The Loop Antenna 238 6-1 The Small Loop 238 6-2 The Short Magnetic Dipole. Equivalence to a Loop 241 6-3 The Short Magnetic Dipole. Far Fields 242 6-4 Comparison of Far Fields of Small Loop and Short Dipole 244 6-5 The Loop Antenna. General Case 244 6-6 Far-Field Patterns of Circular Loop Antennas with Uniform Current 247 6-7 The Small Loop as a Special Case 249 6-8 Radiation Resistance of Loops 250 6-9 Directivity of Circular Loop Antennas with Uniform Current 253 6-10 Table of Loop Formulas 254 6-il Square Loops 254 6-12 Radiation Efficiency, Q, Bandwidth and Signal-to-Notse Ratio 256 Problems 263 7 Tbe Helical Antenna 265 7-1 Introduction 265 7-2 Helical Geometry 271 7-3 Transmission and Radiation Modes of Monofilar Helices 274 7-4 Practical Design Considerations for the Monofilar Axial-Mode Helical Antenna 276 7-5 Axial-Mode Patterns and the Phase Velocity of Wave Propagation on Monofilar Helices 287 7-6 Monofilar Axial-Mode Single-Turn Patterns 295 7-7 Complete Axial-Mode Patterns of Monofilar Helices 300 7-8 Axial Ratio and Conditions for Circular Polarization of Monofilar Axial-Mode Helical Antennas 301 7-9 Wideband Characteristics of Monofilar Helical Antennas Radiating in tbe Axial Mode 307 7-10 Table of Pattern, Beam Width, Gain, Impedance and Axial Ratio Formulas 309 CONTENTS XV 741 Radiation from Linear Periodic Structures with Traveling Waves with Particular Reference to the Helix as a Periodic Structure Antenna 309 7-12 Arrays of Monofilar Axial-Mode Helical Antennas 319 7-12a Array of 4 Monofilar Axial-Mode Helical Antennas 321 742b Array oF96 Monofilar Axial-Mode Helical Antennas 323 7 13 The Monofilar Axial-Mode Helix as a Parasitic Element 323 Helix-Helix 323 Polyrod-Helix 323 Horn- Helix 323 Corner- Helix 324 The 2-Wi re- Line- Helix 324 Helix-Helix 325 Helix Lens 325 744 The Monofilar Axial-Mode Helical Antenna as a Phase and Frequency Shifter 325 7-15 Linear Polarization with Monofilar Axial-Mode Helical Antennas 326 746 Monofilar Axial-Mode Helical Antennas as Feeds 327 7-17 Tapered and Other Forms of Axial-Mode Helical Antennas 329 7-18 Multifilar Axial-Mode (Kilgus Coil and Patton Coil) Helical Antennas 332 749 Monofilar and Multifilar Normal-Mode Helical Antennas, The Wheeler Coil 333 Problems 338 8 The Biconical Antenna and Its Impedance 340 8 -1 Introduction 340 8-2 The Characteristic Impedance of the Infinite Biconical Antenna 341 8-3 Input Impedance of the Infinite Biconical Antenna 346 8-4 Input Impedance of the Finite Biconical Antenna 347 8-5 Pattern of Biconical Antenna 353 8-6 fnput Impedance of Antennas of Arbitrary Shape 354 8-7 Measurements of Conical and Triangular Antennas.
The Brown-Woodward (Bow-Tie) Antenna 354 8-8 The Stacked Biconical Antenna and the Phantom Biconical Antenna 356 Problems 358 9 The Cylindrical Antenna. id Moment Method (MM) 359 9-1 Introduction 359 9-2 Outline of the Integral-Equation Method 360 9-3 The Wave Equation in the Vector Potential A 361 9-4 H alien's Integral Equation 363 9-5 First-Order Solution of Hallen’s Equation 365 9-6 Length -Thick ness Parameter IT 368 9-7 Equivalent Radius of Antennas with Noncircular Cross Section 368 9-8 Current Distributions 369 9-9 Input Impedance 371 9-10 Patterns of Cylindrical Antennas 376 9-11 The Thin Cylindrical Antenna 377 9-12 Cylindrical Antennas with Conical Input Sections 379 XVi CONTENTS 9-13 Antennas of Other Shapes. The Spheroidal Antenna 9-14 Current Distributions on Long Cylindrical Antennas 9-15 Integral Equations and the Moment Method (MM) in Electrostatics 9-16 The Moment Method (MM) and Its Application to a Wire Antenna 9-17 Self-Imped a nee. Radar Cross Section and Mutual impedance of Short Dipoles by the Method of Moments by Edward H, Newman Additional References for Chap. 9 Problems 10 Self and Mutual Impedances 10-1 Introduction 10-2 Reciprocity Theorem for Antennas 10-3 Self- Impedance of a Thin Linear Antenna 10-4 Mutual Impedance of Two Parallel Linear Aniennas 10-5 Mutual Impedance of Parallel Antennas Side-hy-Side 10-6 Mutual Impedance of Parallel Collinear Antennas 10-7 Mutual impedance of Parallel Antennas in Echelon 10-8 Mutual Impedance of Other Conliguraiions 10-9 Mutual Impedance in Terms of Directivity and Radiation Resistance Additional References for Chap. 10 Problems 1 1 Arrays of Dipoles and of Apertures 1 1 - 1 Introduction 1 1 -2 Array of 2 Driven kj 2 Elements, Broadside Case I l-2a Field Patterns II -2b Drivingr Point Impedance 1 1-2c Gain in Field Intensity 1 1-3 Array of 2 Driven 2/2 Elements. End-Fire Case, 11-3a Field Patterns 1 L3b Driving-Point Impedance lL3e Gain in Field Intensity 1 1-4 Array of 2 Driven // 2 Elements, Gener|M7ase with Equal Currents of Any Phase Relation 1 1-5 Closely Spaced Elements I l-5a introduction I I -5b Closely Spaced Elements and Radiating Efficiency. The W8JK Array 11-6 Array of n Driven Elements 11-7 Horizontal Antennas above a Plane Ground I L7a Horizontal 2/2 Antenna above Ground I I -7b W8JK Antenna above Ground 1 1-7c Slacked Horizontal 2/2 Antennas above Ground 1 1-8 Vertical Antennas above a Ground Plane 1 1 -9 Arrays with Parasitic Elements II -9a Introduction 1 1 -9b The Yagi-Uda Array ll-9c Square-Corner-Yagi-Uda Hybrid 380 380 384 389 397 407 408 409 409 410 413 422 424 428 428 430 432 433 433 435 435 436 436 439 440 445 445 447 448 449 453 453 454 459 461 461 468 470 472 476 476 481 483 CONTENTS XVII 1 1-9d Circular Polarization with a Yagi-Uda Antenna ! l-9e The Landsdorfer Shaped-Dipole Array 1 1- 10 Phased Arrays 1 l-10a Introduction I l-10b Phased Array Designs 1 1-lOc Rotatable Helix Phased Array 11-11 Frequency-Scanning Arrays 1 l-lla Frequency-Scanning Line- Fed Array 1 1-Mb Frequency-Scanning Backward Angle-Fire Grid and Chain Arrays 11-12 Retro-Arrays. The Van Atta Array ! 1-13 Adaptive Arrays and Smart Antennas 1 LI 3a Literature on Adaptive Arrays 11-14 Microstrip Arrays 11-15 Low-Sidelobc Arrays 1 L 1 6 Long-Wire Antennas 1 1- 16a V Antennas |] -16b Rhombic Antennas N-16c Beverage Antennas 11-17 Curtain Arrays 11-18 Location and Method of heeding Antennas 11-19 Folded Dipole Antennas 1 1-20 Modifications of Folded Dipoles 1 L21 Continuous Aperture Distribution 11-22 Fourier Transform Relations between the Ear-Field Pattern and the Aperture Distribution 11-23 Spatial Frequency Response and Pattern Smoothing 11-24 The Simple {Adding) Interferometer 1 L25 Aperture Synthesis and Multi-aperture Arrays 1 1-26 t jrating Lobes Additional References Problems 484 484 485 485 486 489 490 490 491 496 496 499 501 501 502 502 503 508 509 510 511 514 515 517 520 522 533 535 536 537 12 Reflector Antennas and Their Feed Systems 543 12-1 Introduction 12-2 Plane Sheet Reflectors and Diffraction 12-3 Corner Reflectors !2-3a Active (Kraus) Corner Reflector l2-3b Passive (Retro) Corner Reflector 12-4 The Parabola. General Properties 12-5 A Comparison between Parabolic and Corner Reflectors 12-6 The Paraboloidal Reflector 12-7 Patterns of Large Circular Apertures with Uniform Illumination 12-8 The Cylindrical Parabolic Reflector 12-9 Aperture Distributions and Efficiencies 12-10 Surface Irregularities and Gain Loss 12-11 OfLAxis Operation of Parabolic Reflectors 12-12 Cassegrain Feed, Shaped Reflectors, Spherical Reflectors and Offset Feed 543 545 549 549 561 561 563 564 569 572 573 587 592 594 CONTENTS XVIII CONTENTS 12-13 Frequency-Sensitive (or -Selective) Surfaces (FSS) by Benedikt A. Munk Effect of Element Spacings dx and dw Effect of Angle of Incidence 0 Control of Bandwidth Cascading or Stacking More FSS Element Types 12-14 Some Examples of Reflector Antennas Bonn Anecibo Bell Telephone Laboratories Nobeyama Ohio State University Gorki Five College Observatory Offsat 12-15 Low-Sidelobe Considerations References Problems 600 600 602 603 603 604 605 605 605 605 609 610 614 615 61 6 616 619 621 13 Slot, Horn and Complementary Antennas 624 13-1 Introduction 624 13-2 Slot Antennas 624 13-3 Patterns of Slot Antennas in Flat Sheets. Edge Diffraction 628 13-4 Babinet’s Principle and Complementary Antennas 632 135 The Impedance of Complementary Screens 635 1^-6 The Impedance of Slot Antennas 638 13-7 Slotted Cylinder Antennas 642 13-8 Horn Antennas 644 13-9 The Rectangular Horn Antenna 648 13-10 Beam-Width Comparison 653 13-11 Conical Horn Antennas 653 13-12 Ridge Horns 654 13-13 Septum Horns 6 55 13-14 Corrugated Horns 657 13-15 Aperture-Matched Horn 659 References 659 Problems 660 14 Lens Antennas 661 14-1 Introduction 661 14-2 Nonmetallic Dielectric Lens Antennas. Fermat’s Principle 663 143 Artificial Dielectric Lens Antennas 670 14-4 E-Plane Metal-Plate Lens Antennas 673 1±5 Tolerances on Lens Antennas 680 14-6 tf- Plane Metal-Plate Lens Antennas 683 14-7 Reflector-Lens Antenna 684 14-8 Polyrods 685 xix 14-9 Multiple- Helix Lenses 14-10 Luneburg and Einstein Lenses Additional Reference Problems 687 683 690 690 15 Broadband and Frequency-Independent Antennas 15-1 Broadband Antennas 15-2 The Frequency-Independent Concept: Rumsey's Principle 15-3 The Frequency-Independent Planar Log-Spiral Antenna 15-4 The Frequency- Independent Conical-Spiral Antenna 15-5 The Log-Periodic Antenna 1 5-6 The Yagi-Uda-Corner-Log-Periodic (YUCOLP) Array Problems 692 692 696 697 701 703 708 710 16 Antennas for Special Applications: Feeding Considerations 711 j 16-1 Introduction 16-2 Electrically Small Antennas 16-3 Physically Small Antennas 16-4 Antenna Siting and the Effect of Typical (Imperfect) Ground 16-5 Ground-Plane Antennas 16-6 Sleeve Antennas 16-7 Turnstile Antenna 16-8 Superturnstile Antenna 16-9 Other Omnidirectional Antennas 16-10 Circularly Polarized Antennas 16-11 Matching Arrangements, Baiuns and Traps 16-12 Patch or Microstrip Antennas 16-13 The High-Gain Omni 16-14 Submerged Antennas 16-15 Surface-Wave and Leaky-Wave Antennas 16-16 Antenna Design Considerations for Satellite Communication 16-17 Recei ving versus Transmitting Considerations 16-18 Bandwidth Considerations 16-19 Gravity-Wave Antennas, Rotating Boom for Transmitting and Weber Bar for Receiving Problems 711 711 714 716 723 725 726 729 731 732 "734 745 749 749 754 762 766 767 768 770 17 Antenna Temperature, Remote Sensing, Radar and Scattering 774 17-1 Introduction 17-2 Antenna Temperature, Incremental and Total 17-3 System Temperature and Signal -to-Noise Ratio 17-4 Passive Remote Sensing 17-5 Radar, Scattering and Active Remote Sensing by Robert G< Kouyoumjian Additional References Problems 774 774 782 787 791 797 799 XX CONTENTS 18 Antenna Measurements 805 IK- 1 Introduction IK-2 Patterns 18-3 Pattern Measurement Arrangements IS- 3a Distance Requirement for Uniform Phase 18-3b Uniform Field Amplitude Requirement 1 8-3c Absorbing Materials lS-3d The Anechoic Chamber Compact Range 18-3e Pattern and Squint Measurements Using Celestial and Satellite Radio Sources 18-4 Phase Measurements 18-5 Directivity 18-6 Gain 18-6a Gain by Comparison 1 8-6b Absolute Gain of Identical Antennas l8-6c Absolute Gain of Single Antenna 18-6d Gain by Near- Field Measurements 18-6e Gain and Aperture Efficiency from Celestial Source Measurements 18-7 Terminal Impedance Measurements 18-8 Current Distribution Measurements 18-9 Polarization Measurements l8-9a Polarization- Pattern Method 18-9b Linear-Component Method 18-9c Circular-Component Method 18-10 Antenna Rotation Experiments 18-11 Model Measurements 18-12 Measurement Error Additional References References on Radiation Hazards Problems 805 805 807 809 811 814 818 822 823 824 824 824 826 828 829 830 832 834 835 836 838 838 840 840 841 842 K43 843 Appendix A Tables for Reference A-l Table of Antenna and Antenna System Relations A- 2 Formu las for I nput Im ped ance of Te rmi nated Transmission L ines A-3 Reflection and Transmission Coefficients and VSWR A-4 Characteristic Impedance of Coaxial, 2-Wire and Microstrip Transmission Lines j A-5 Characteristic Impedance of Transmission Lines in Terms of Distributed Parameters A-6 Material Constants (Permittivity, Conductivity and Dielectric Strength) A-7 Permittivity Relations A-8 Celestial Radio Sources for Pattern, Squint, Gain and Aperture Efficiency Measurements A-9 Maxwelfs Equations A-1Q Beam Width and Sidelobe Level for Rectangular and Circular Aperture Distributions 845 845 848 849 849 850 851 852 853 854 856 contents XXI Appendix B Computer Programs (Codes) 857 B-l Additional Computer Program References S59 B-2 BASIC Phased-Array Antenna Pattern Programs Appendix C Books and Video Tapes C-l Books g65 C-2 Video Tapes Appendix D Answers to Starred Problems 866 870 Appendix E Problem Supplement 873 Index Although there has been an explosion in antenna technology in the years since Antennas was published, the basic principles and theory remain unchanged My aim in this new edition is to blend a central core of basics from the first edition with a representative selection of important new developments and advances resulting in a much enlarged, updated book. It is appropriate that .1 ^appearing just 100 years from the date on which the first antennas were invented by Hein-rich Hertz to whom, along with Guglielmo Marconi, this new edit,on is dedi-CatCd As with the first edition, physical concepts are emphasized which aid in the visualization and understanding of the radiation phenomenon. More worked examples are given to illustrate the steps and thought processes required in going from a fundamental equation to a useful answer. The new edition stresses practi-cal approaches to real-world situations and much information of value is made available in the form of many simple drawings, graphs and equations.
As with the first edition my purpose is to give a unified treatment ol antennas from the electromagnetic theory point of view while paying attention to important applications. Following a brief history of antennas in the first chapter to set the stage, the next three chapters deal with basic concepts and the theory of point sources. These are followed by chapters on the linear, loop, helical, bicon-ical and cylindrical antennas.
Then come chapters on antenna arrays, reflectors, slot, horn, complemen-tary and lens antennas. The last four chapters discuss broadband and frequency-independent antennas, antennas for special applications including electrically small and physically small antennas, temperature, remote sensing, radar, scat-tering and measurements. The Appendix has many useful tables and references.
The book has over 1000 drawings and illustrations, many of which are unique, providing physical insights into the process of radiation from antennas.
The book is an outgrowth of lectures for antenna courses I have given at xxiil Xxiv PREFACE Ohio State University and at Ohio University. The material is suitable for use at late undergraduate or early graduate level and is more than adequate for a one-semester course. The problem sets at the end of each chapter illustrate and extend the material covered in the text. In many cases they include important results on topics listed in the index. There are over 500 problems and worked examples.
Antennas has been written to serve not only as a textbook but also as a reference tjpok for the practicing engineer and scientist. As an aid to those seeking additional information on a particular subject, the book is well docu-mented with references both in footnotes and at the ends of chapters.
A few years ago it was customary to devote many pages of a textbook to computer programs, some with hundreds of steps. Now with many conveniently packaged programs and codes readily available this is no longer necessary.
Extensive listings of such programs and codes, particularly those using moment methods, are given in Chapter 9 and in the Appendix. Nevertheless, some rela-tively short programs are included with the problem sets and in the Appendix.
From my IEEE Antennas and Propagation Society Centennial address (1984) I quote.
With mankind's activities expanding into space, the need for antennas will grow to an unprecedented degree. Antennas wil] provide the vital links to and from every-thing out there. The future of antennas reaches to the stars, Robert G, Kouyoumjian, Bencdikt A. Munk and Edward H. Newman of the Ohio State University have contributed sections on scattering, frequency-sensitive surfaces and moment method respectively, I have edited these contribu-tions to make symbols and terminology consistent with the rest of the book and any errors are my responsibility.
In addition, I gratefully acknowledge the assistance, comments and data from many others on the topics listed : Walter D. Burnside , Ohio State University, compact ranges Robert S Dixon , Ohio State University, phased-arrays Von R , Eshleman, Stanford University, gravity lenses Paul E. Mayes, University of Illinois, frequency-independent antennas Robert E. Munson Ball Aerospace, microstrip antennas Leon Peters , Jr„ Ohio State University, dipole antennas David M. Pozar , University of Massachusetts, moment method Jack H Richmond , Ohio State University, moment method Helmut E. Schrank, Westinghouse, low-sidelobe antennas Chen-To Tai , University of Michigan, dipole antennas Throughout the preparation of this edition, I have had the expert editorial assistance of Dr + Erich Pacht.
Illustration and manuscript preparation have been handled by Robert Davis, Kristine Hall and William Taylor.
McGraw-Hill editors were Sanjeev Rao, Alar Etken and John Morriss.
PREFACE XXV Although great care has been exercised, some errors or omissions in the text tables, lists or figures will inevitably occur. Anyone finding them will do me by waiting .o me so that .hey can be eortected .a .ubsequen.
Pr"“7L appreciate the very helpful comment, of Ronald N. Bmcewell. Stan-ford University, who reviewed the manuscript for McGraw-Hill.
Finally, I thank my wife, Alice, for her patience, encouragement and dedica-tion through’all the years of work it has taken.
John D. Kraus Ohio State University CHAPTER 1 INTRODUCTION 1-1 INTRODUCTION. Since Hertz and Marconi, antennas have become increasingly important to our society until now they are indispensable. They are everywhere: at our homes and workplaces, on our cars and aircraft, while our ships, satellites and spacecraft bristle with them. Even as pedestrians, we carry them.
Although antennas may seem to have a bewildering, almost infinite variety, they all operate according to the same basic principles of electromagnetics. The aim of this book is to explain these principles in the simplest possible terms and illustrate them with many practical examples.
In some situations intuitive approaches will suffice while in others complete rigor is needed The book pro-vides a blend of both with selected examples illustrating when to use one or the other.
This chapter provides an historical background while Chap. 2 gives an introduction to basic concepts. The chapters that follow develop the subject in more detail, 1-2 THE ORIGINS OF ELECTROMAGNETIC THEORY AND THE FIRST ANTENNAS 1 Six hundred years before Christ, a Greek mathe-matician, astronomer and philosopher, Thales of Miletus, noted that when amber is rubbed with silk it produces sparks and has a seemingly magical power to ] y D. Kraus, "Antennas Since Hertz and Marconi,” IEEE Trans. Anrs. Prop^ AP-33, 131-137, 1985.
See also references at end of chapter.
1 2 1 introduction attract particles of fluff and straw. The Greek word for amber is elektron and from this we get our words electricity, electron and electronics .
Thales also noted the attractive power between pieces of a natural magnetic rock called loadstone, found at a place called Magnesia , from which is derived the words magnet and magnetism, Thales was a pioneer in both electricity and magnetism but his inter-est, like that of others of his time, was philosophical rather than practical, and it was 22 centuries before these phenomena were investigated in a serious experi-mental way.
It remained for William Gilbert of England in about A.D. 1600 to perform the first systematic experiments of electric and magnetic phenomena, describing his experiments in his celebrated book, De Magnete . Gilbert invented the electro-scope for measuring electrostatic effects. He was also the first to recognize that the earth itself is a huge magnet, thus providing new insights into the principles of the compass and dip needle.
In experiments with electricity made about 1750 that led to his invention of the lightning rod, Benjamin Franklin, the American scientist-statesman, estab-lished the law of conservation of charge and determined that there are both posi-tive and negative charges.
Later, Charles Augustin de Coulomb of France measured electric and magnetic forces with a delicate torsion balance he invent-ed. During this period Karl Friedrich Gauss, a German mathematician and astronomer, formulated his famous divergence theorem relating a volume and its surface.
By 1800 Alessandro Volta of Italy had invented the voltaic cell and, con-necting several in series, the electric battery. With batteries, electric currents could be produced, and in 1819 the Danish professor of physics Hans Christian Oersted found that a current-carrying wire caused a nearby compass needle to deflect, thus discovering that electricity could produce magnetism . Before Oersted, electricity and magnetism were considered as entirely independent phenomena.
The following year, Andre Marie Ampere, a French physicist, extended Oersted's observations. He invented the solenoidal coil for producing magnetic fields and theorized correctly that the atoms in a magnet are magnetized by tiny electric currents circulating in them. About this time Georg Simon Ohm of Germany published his now-famous law relating current, voltage and resistance.
However, it initially met with ridicule and a decade passed before scientists began to recognize its truth and importance.
Then in 1831, Michael Faraday of London demonstrated that a changing magnetic field could produce an electric current. Whereas Oersted found that electricity could produce magnetism, Faraday discovered that magnetism could produce electricity , At about the same time, Joseph Henry of Albany, New York, observed the effect independently. Henry also invented the electric telegraph and relay.
Faraday’s extensive experimental investigations enabled James Clerk Maxwell, a professor at Cambridge University, England, to establish in a pro-found and elegant manner the interdependence of electricity and magnetism. In his classic treatise of 1873, he published the first unified theory of electricity and 12 THE ORIGINS OF ELECTROMAGNETIC THEORY AND THE FIRST ANTENNAS 3 magnetism and founded the science of electromagnetics. He postulated that light vt'as electromagnetic in nature and that electromagnetic radiation of other wave-lengths should be possible.
Maxwell unified electromagnetics in the same way that Isaac Newton unified mechanics two centuries earlier with his famous Law of Universal Gravi-tation governing the motion of all bodies both terrestrial and celestial.
Although Maxwell’s equations are of great importance and, with boundary, continuity and other auxiliary relations, form the basic tenets of modern electro-magnetics, many scientists of Maxwell’s time were skeptical of his theories. It was more than a decade before his theories were vindicated by Heinrich Rudolph Hertz.
Early in the 1880s the Berlin Academy of Science had offered a prize for research on the relation between electromagnetic forces and dielectric polariz-ation. Heinrich Hertz considered whether the problem could be solved with oscil-lations using Leyden jars or open induction coils. Although he did not pursue this problem, his interest in oscillations had been kindled and in 1886 as pro-fessor at the Technical Institute in Karlsruhe he assembled apparatus we would now describe as a complete radio system with an end-loaded dipole as transmit-ting antenna and a resonant square loop antenna as receiver.
1 When sparks were produced at a gap at the center of the dipole, sparking also occurred at a gap in the nearby loop. During the next 2 years, Hertz extended his experiments and demonstrated reflection, refraction and polarization, showing that except for their much greater length, radio waves were one with light.
Hertz turned the tide against Maxwell around.
Hertz’s initial experiments were conducted at wavelengths of about 8 meters while his later work was at shorter wavelengths, around 30 centimeters. Figure 1-1 shows Hertz’s earliest 8-meter system and Fig. 1-2 a display of his apparatus, including the cylindrical parabolic reflector he used at 30 centimeters.
Although Hertz was the father of radio, his invention remained a labora-tory curiosity for nearly a decade until 20-year-old Gugltelmo Marconi, on a summer vacation in the Alps, chanced upon a magazine which described Hertz’s experiments. Young Guglielmo wondered if these Hertzian waves could be used to send messages. He became obsessed with the idea, cut short his vacation and rushed home to test it.
In spacious rooms on an upper floor of the Marconi mansion in Bologna, Marconi repeated Hertz’s experiments. His first success late one night so elated him he could not wait until morning to break the news, so he woke his mother and demonstrated his radio system to her.
Marconi quickly went on to add tuning, big antenna and ground systems for longer wavelengths and was able to signal over large distances.
In mid-December 1901, he startled the world by announcing that he had received radio 1 His dipole was called a Hertzian dipole and the radio waves Hertzian waves.
4 ] rNTROUl.'t'TlON' a b Figu'e l-I Heinrich Hertz's complete radio system of 1886 with end-loaded dipole transmitting antenna ICC') and resonant loop receiving antenna \abcd) for - S m. Wtih induction coil \A) turned on T sparks at gap B induced sparks at M in the loop receiving antenna. [From Heinrich Hertz's hook Electric Metres, Macmillan, IH93: redrawn hi’rJi dinM'nsions added.) Figure 1-2 Hertzs sphere-loaded A.2 dipole and spark gap (resting on floor in foreground) and cylindrical parabolic reflector for 30 centimeters (standing at left). Dipole with spark gap is on the parabola focal axis. {Photograph by Edward C Jordan.) 1-2 THE ORIGINS OF ELECTROMAGNETIC THEORY AND THE FIRST ANTENNAS 5 signals at St John's, Newfoundland, which had been sept across the Atlantic from a station he had built at Poldhu in Cornwall, England, The scientific estab-lishment did not believe his claim because in its view radio waves, like light, should travel in straight lines and could not bend around the earth from England to Newfoundland, However, the Cable Company believed Marconi and served him with a writ to cease and desist because it had a monopoly on transatlantic communication.
The Cable Company's stock had plummeted following Marconi's announcement and it threatened to sue him for any loss of revenue if he persisted. However, persist he did, and a legal battle developed that continued for 27 years until finally the cable and wireless groups merged.
One month after Marconi's announcement, the American Institute of Elec-Irical Engineers (AIEE) held a banquet at New York's Waldorf-Astoria to cele-brate the event. Charles Protius Steinmetz, President of the AIEE, was there, as was Alexander Graham Bell, but many prominent scientists boycotted the banquet. Their theories had been challenged and they wanted no part of it.
Not long after the banquet, Marconi provided irrefutable evidence that radio waves could bend around the earth. He recorded Morse signals, inked automatically on tape, as received from England across almost all of the Atlantic while steaming aboard the SS Philadelphia from Cherbourg to New York. The ship’s captain, the first officer and many passengers were witnesses.
A year later, in 1903, Marconi began a regular transatlantic message service between Poldhu, England, and stations he built near Glace Bay, Nova Scotia, and South Wellfleet on Cape Cod, In 1901, the Poldhu station had a fan aerial supported by two 60-meter guyed wooden poles and as receiving antenna for his first transatlantic signals at St. John's, Marconi pulled up a 200-meter wire with a kite, working it against an array of wires on the ground. A later antenna at Poldhu, typical of antennas at other Marconi stations, consisted of a conical wire cage. This was held up by four massive self-supporting 70-meter wooden towers (Fig.
1-3). With inputs of 50 kilowatts, antenna wires crackled and glowed with corona at night.
Local residents were sure that such fireworks in the sky would alter the weather Rarely has an invention captured the public imagination like Marconi's wireless did at the turn of the century. We now call it radio but then it was wireless: Marconi's wireless. After its value at sea had been dramatized by the SS Republic and SS Titanic disasters, Marconi was regarded with a universal awe and admiration seldom matched. Before wireless, complete isolation enshrouded a ship at sea. Disaster could strike without anyone on the shore or nearby ships being aware that anything had happened. Marconi changed all that. Marconi became the Wizard of Wireless.
Although Hertz had used 30-centimeter wavelengths and Jagadis Chandra Bose and others even shorter wavelengths involving horns and hollow wave-guides, the distance these waves could be detected was limited by the technology of the period so these centimeter waves found little use until much later. Radio developed at long wavelengths with very long waves favored for long distances. A popular “rule-of-thumb” of the period was that the range which could be 6 1 INTRODUCTION Fign-e 13 Square-cone antenna at Marconi's Poldhu, England, station in 1905. The 70-meter wooden towers support a network of wires which converge to a point just above the transmitting and receiving buildings between the towers.
achieved with adequate power was equal to 500 times the wavelength. Thus, for a range of 5000 kilometers, one required a wavelength of 10000 meters.
At typical wavelengths of 2000 to 20000 meters, the antennas were a small fraction of a wavelength in height and their radiation resistances only an ohm or less. Losses in heat and corona reduced efficiencies but with the brute power of many kilowatts, significant amounts were radiated. Although many authorities favored very long wavelengths, Marconi may have appreciated the importance of radiation resistance and was in the vanguard of those advocating shorter wave-lengths, such as 600 meters. At this wavelength an antenna could have 100 times its radiation resistance at 6000 meters.
In 1912 the Wireless Institute and the Society of Radio Engineers merged to form the Institute of Radio Engineers, 1 In the first issue of the Institute's Pro-ceedings, which appeared in January 1913, it is interesting that the first article was on antennas and in particular on radiation resistance. Another Proceedings article noted the youthfulness of commercial wireless operators. Most were in their late teens with practically none over the age of 25. Wireless was definitely a young man's profession.
The era before World War I was one of long waves, of spark, arc and alternators for transmission; and of coherers Fleming valves and De Forest 1 In 1963, the Institute of Radio Engineers and the American Institute of Electrical Engineers merged to form the Institute of Electrical and Electronic Engineers {IEEE}.
1-2 THE ORIGINS OF ELECTROMAGNETIC THEORY AND THE FIRST ANTENNAS 7 audions for reception. Following the war, vacuum lubes became available for transmission; continuous waves replaced spark and radio broadcasting began in the 200 to 600-meter range.
Wavelengths less than 200 meters were considered of little value and were relegated to the amateurs. In 1921, the American Radio Relay League sent Paul Godley to Europe to try and receive a Greenwich, Connecticut, amateur station operating on 200 meters. Major Edwin H, Armstrong, inventor of the super-heterodyne receiver and later of FM, constructed the transmitter with the help of several other amateurs. Godley set up his receiving station near the Firth of Clyde in Scotland. He had two receivers, one a 10-tube superheterodyne, and a Beverage antenna. On December 12, 1921, just 20 years to the day after Marconi received his first transatlantic signals on a very long wavelength, Godley received messages from the Connecticut station and went on to log over 30 -other LIS.
amateurs.
It was a breakthrough, and in the years that followed, wavelengths from 200 meters down began to be used for long-distance communication.
Atmospherics were the bane of the long waves, especially in the summer.
They were less on the short waves but still enough of a problem in 1930 for the Bell Telephone Laboratories to have Karl G. Jansky study whether they came from certain predominant directions. Antennas for telephone service with Europe might then be designed with nulls in these directions, Jansky constructed a rotating 8-element Bruce curtain with a reflector oper-ating at 14 meters (Fig.
1-4). Although he obtained the desired data on atmo-spherics from thunderstorms, he noted that in the absence of all such static there was always present a very faint hisslike noise or static which moved completely around the compass in 24 hours.
After many months of observations, Jansky Figure 14 Karl Guthe Jansky and his rotating Bruce curtain antenna with which he discovered jadio emission from our galaxy. {Courtesy Beit Telephone Laboratories: Jansky inset courtesy Mary Jansky Striffler.) 8 I INTRODUCTION concluded that it was coming from beyond the earth and beyond the sun.
It was a cosmic static coming from our galaxy with the maximum from the galactic center. Jansky's serendipitous discovery of extraterrestrial radio waves opened a new window on the universe. Jansky became the father of radio astronomy.
Jansky recognized that this cosmic noise from our galaxy set a limit to the sensitivity that could be achieved with a short-wave receiving system. At 14 meters this sky noise has an equivalent temperature of 20000 kelvins. At centi-meter wavelengths it is less, but never less than 3 kelvins. This is the residual sky background Level of the primordial fireball that created the universe as measured four decades later by radio astronomers Arno Penzias and Robert Wilson of the Bell Telephone Laboratories at a site not far from the one used by Jansky.
For many years, or until after World War II, only one person, Grote Reber, followed up Jansky's discovery in a significant way. Reber constructed a 9-meter parabolic reflector antenna (Fig.
1-5) operating at a wavelength of about 2 meters which is the prototype of the modern parabolic dish antenna. With it he made the first radio maps of the sky. Reber also recognized that his antenna -receiver constituted a radiometer, i.e., a temperature-measuring device in which his recei-ver response was related to the temperature of distant regions of space coupled to his antenna via its radiation resistance.
With the advent of radar during World War II, centimeter waves, which had been abandoned at the turn of the century, finally came into their own and the entire radio spectrum opened up to wide usage. Hundreds of stationary com-munication satellites operating at centimeter wavelengths now ring the earth as though mounted on towers 36000 kilometers high. Our probes are exploring the solar system to Uranus and beyond, responding to our commands and sending back pictures and data at centimeter wavelengths even though it takes more than an hour for the radio waves to travel the distance one way. Our radio telescopes operating at millimeter to kilometer wavelengths receive signals from objects so distant that the waves have been traveling for more than tO billion years.
With mankind's activities expanding into space, the need for antennas will grow to an unprecedented degree. Antennas will provide the vital links to and from everything out there. The future of antennas reaches to the stars.
1-3 ELECTROMAGNETIC SPECTRUM. Continuous wave energy radi-ated by antennas oscillates at radio frequencies. The associated free-space waves range in length from thousands of meters at the long-wave extreme to fractions of a millimeter at the short-wave extreme. The relation of radio waves to the entire electromagnetic spectrum is presented in Fig.
1-6. Short radio waves and long infrared waves overlap into a twilight zone that may be regarded as belonging to both.
The wavelength X of a wave is related to the frequency / and velocity v of the wave by U) Figure 1-5 Grote Reber and his parabolic reflector anterma with which he made the first radio maps of the sky. This antenna, which he built in 1938, is the prototype of the modem dish antenna. (Reber courtesy Arthur C. Clarke.) 10 I INTRODUCTION Atmosphere opaque Optical window Infra-red windows it Abscrpirtn by intarEttllir gas III Radio window Ionosphere opaque Objects different 0l -] ' Si26^ Atomic nuctau a = 3 > « -5 £'3 Is E MoH&u lr linoj I I TV T_ 1AM longyvityt Gamma rays Atom \ + X-rays Ultra -9 r'Vt a n«ar-Jf t ^ / Radar Snortuvai/f _ Hydrogen ^ Redwood | line traa Infra-red VI Radio/ J_ L -I I L Wavelength _l l__j 1 101001 101 00 1 101001 101001 10100 1 10100 1 101001 meters 10 100 attometers 10“ 16 femtometers 1 Q~ 15m pico meters 10 -nanometers 1Q- mjpro meters 10 millimeters 10“ 3m kilometers 10 3 m Figure lfS The electromagnetic spectrum with wavelength on a logarithmic scale from the shortest gamma rays to the longest radio waves. The atmospheric-ionospheric opacity is shown at the top with the optical and radio windows in evidence.
Thus, the wavelength depends on the velocity v which depends on the medium. In this sense, frequency is a more fundamental quantity since it is independent of the medium When the medium is free space (vacuum) u = c = h 10 s m s" 1 (2) Figure 1-7 shows the relation of wavelength to frequency for v = c (free space). Many of the uses of the spectrum are indicated along the right-hand edge of the figure A more detailed frequency use listing is given in Table 1-1.
Table 1-1 Radio-frequency band designations Frequency Wavelength Baud designation 30-300 Hz 10-1 Mm ELF (extremely low frequency) 300-3000 Hz 1 Mm-100 km 3-30 kHz 100-10 km VLF (very low frequency) 30-300 kHz 10-1 km LF (low frequency) 300- 3000 kHz 1 km-100 m MF (medium frequency) 3-30 MHz 100-10 m HF (high frequency) 30-300 MHz 10-1 m VHF (very high frequency) 300-3000 MHz 1 m-10 cm UHF (ultra high frequency) 3-30 GHz KM cm SHF (super high frequency) 30-300 GHz 1 cm-1 mm EHF (extremely high frequency) 300-3000 GHz 1 mm-100 /mi Frequency Wavelength IEEE Radar Baud designation 1-2 GHz 30-15 cm L 2-4 GHz 15-7.5 cm $ 4—8 GHz 7.5-3.75 cm C 8-12 GHz 3.75-2.50 cm X 12-18 GHz 2,50-1.67 cm Ku 18-27 GHz 1.67-1.11 cm K 27—40 GHz 1,11 cm- 7. 5 mm Ka 40-300 GHz 7.5-10 mm mm AnugntiH H M DIMENSIONS AND UNITS 11 Band Wavelength (for u = c) Figure 1-7 Wavelength versus frequency for v = c.
Example of wavelength for a given frequency. For a frequency of 300 MHz the cor-responding wavelength is given by c 3 x 10 8 m s _1 A ~f~ 300 x 10 Hz “ m (3) In a lossless nonmagnetic dielectric medium with relative permittivity cr = 2, the same wave has a velocity and v k c _ 3 x 1Q B v 2.12 x 10" / ~ 300 x !0‘ = 2.12 x 10® m s“’ = 0.707 m — 707 mm (4) (5) M DIMENSIONS AND UNITS. Lord Kelvin is reported to have said : When you can measure what you are speaking about and express it in numbers you know something about it; but when you cannot measure it, when you cannot express it in numbers your knowledge is of a meagre and unsatisfactory kind; it may 12 ] INTRODUCTION be the beginning ofknowledge but you have scarcely progressed in your thoughts to the stage of science whatever the matter may be.
To this it might be added that before we can measure something, we must define its dimensions and provide some standard, or reference unit, in terms of which the quantity can be expressed numerically, A dimension defines some physical characteristic. For example, .length, mass, time, velocity and force are dimensions. The dimensions of length , mass , time , electric current , temperature and luminous intensity are considered as the funda-mental dimensions since other dimensions can be defined in terms of these six.
This choice is arbitrary but convenient. Let the letters L, Af, T, /, and J represent the dimensions of length, mass, time, electric current, temperature and luminous intensity.
Other dimensions are then secondary dimensions.
For example, area is a secondary dimension which can be expressed in terms of the fundamental dimension of length squared (L 2 ). As other examples, the fundamen-tal dimensions of velocity are L/T and of force are Af L/T 2 .
A unit is a standard or reference by which a dimension can be expressed numerically. Thus, the meter is a unit in terms of which the dimension of length can be expressed and the kilogram is a unit in terms of which the dimension of mass can be expressed. For example, the length (dimension) of a steel rod might be 2 meters and its mass (dimension) 5 kilograms.
1-5 FUNDAMENTAL AND SECONDARY UNITS, The units for the fundamental dimensions are called the fundamental or base units. In this book the International System of Units, abbreviated SI, is used.
1 In this system the werer.
kilogram second , ampere , kelvin and candela are the base units for the six funda-mental dimensions of length, mass, time, electric current, temperature and lumin-ous intensity. The definitions for these fundamental units are: Meter (m). Length equal to 1 650763.73 wavelengths in vacuum corresponding to the 2p 10-5d^ transition of krypton-86.
Kilogram (kg). Equal to mass of international prototype kilogram, a platinum-iridium mass preserved at Sevres, France. This standard kilogram is the only artifact among the SI base units.
Second (s), Equal to time duration of 9 192631 770 periods of radiation correspond-ing to the transition between two hyperfine levels of the ground state of cesium-133.
The second was formerly defined as 1/86400 part of a mean solar day. The earth’s rotation rate is gradually slowing down, but the atomic (cesium-133) transition is 1 The International System of Units is the modernized version of the metric system. The abbreviation SI is from the French name Sy&tkme Internationale if Unites, For the complete official description of the system see U.S, Nall. Bur. Stand. Spec. Pub. 330, 1971.
1-6 HOW TO READ THE SYMBOLS AND NOTATION 13 much more constant and is now the standard. The two standards differ by about 1 second per year.
Ampere (A).
Electric current which if flowing in two infinitely long parallel wires in vacuum separated by 1 meter produces a force of 200 nanonewtons per meter of length (200 nN m 1 — 2 x 10" 7 Nm _I ).
Kelvin (K), Temperature equal to 1/273.16 of the triple point of water (or triple point of water equals 273.16 kelvins).
1 Candela (cd). Luminous intensity equal to that oF 1/600000 square meter of a perfect radiator at the temperature of freezing platinum.
The units for other dimensions are called secondary or derived units and are based on these fundamental units.
The material in this book deals principally with the four fundamental dimensions length , mass, time and electric current (dimensional symbols L, Af, T and /). The four fundamental units for these dimensions are the basis of what was formerly called the meter-kilogram-second-ampere (mksa) system, now a sub-system of the SL The book also includes discussions of temperature but no refer-ences to luminous intensity.
The complete SI involves not only units but also other recommendations, one of which is that multiples and submultiples of the SI units be stated in steps of 10 3 or 10“ 3 .
Thus, the kilometer (1 km — m) and the millimeter (1 mm = 10“ 3 m) are preferred units of length, but the centimeter {= 10“ 2 m) is not.
For example, the proper SI designation for the width of motion -picture film is 35 mm, not 3.5 cm.
In this book rationalized SI units are used. The rationalized system has the advantage that the factor An does not appear in Maxwell's equations (App. A), although it does appear in certain other relations. A complete table of units in this system is given in the Appendix of Electromagnetics, 3rd ed,, by J. D. Kraus (McGraw-Hill, 1984).
1-6 HOW TO READ THE SYMBOLS AND NOTATION, In this book quantifies, or dimensions, which are scalars, like charge Q, mass M or resis-tance R, are always in italics. Quantities which may be vectors or scalars are boldface as vectors and italics as scalars, e.g., electric field E (vector) or E (scalar).
Unit vectors are always boldface with a hat (circumflex) over the letter, e.g., x or r .
2 ' Note that the symbol for degrees is not used with kelvins. Thus, the boiling temperature of water i!00"Q is 373 kelvins (373 KX not 373K. However, the degree sign is retained with degrees Celsius.
1 In longhand notation a vector may be indicated by a bar over the letter and hat ( ') over the unit lector.
14 1 INTRODUCTION Units are in roman type, i.e., not italic; for example, H for henry, s for second, or A for ampere.
1 The abbreviation for a unit is capitalized if the unit is derived from a proper name; otherwise it is lowercase (small letter). Thus, we have C for coulomb but m for meter. Note that when the unit is written out, it is : always lowercase even though derived from a proper name. Prefixes for units are : also roman, like n in nC for nanocoulomb or M in MW for megawatt.
Example 1.
D = x 200 pCm" 3 means that the electric flu density D is a vector in the positive x direction with a magnitude of 200 picocoulombs per square meter ( = 2 x 10“ 10 coulomb per square, meter).
1 Example 2 .
V = to V means that the voltage V equals 10 volts. Distinguish carefully between V (italics) for voltage, V (roman) for volts, v (lowercase, boldface) for velocity and v (lowercase, italics) for volume.
Example 3 S = 4 W m" means that the flux density S (a scalar) equals 4 watts per square meter per hertz.
This can also be written S — 4 W/m 3/Hz or 4 W/(m 2 Hz), but the form W m“ 2 Hz -1 is more direct and less ambiguous.
Note that for conciseness, prefixes are used where appropriate instead of exponents. Thus, a velocity would be expressed in prefix form as v = 215 Mm s l ; (215 megameters per second) not in the exponential form 2.15 x 1G 8 m s _1 ,i However, in solving a problem the exponential would be used although the final answer might be put in the prefix form (215 Mm s“ } The modernized metric (SI) units and the conventions used herein combine to give a concise, exact and unambiguous notation, and if one is attentive to the details, it will be seen to possess both elegance and beauty.
1-7 EQUATION NUMBERING, Important equations and those referred to in the text are numbered consecutively beginning with each section. When reference is made to an equation in a different section, its number is preceded by the chapter and section number.
Thus, (14-15-3) refers to Chap.
14, Sec, 15 Eq, (3). A reference to this same equation within Sec.
15 of Chap. 14 would read simply (3). Note that chapter and section numbers are printed at the top of each page.
1 In longhand notation no distinction is usually made between quantities (italics) and units (roman).
However, it can be done by placing a bar under the letter to indicate italics or writing the letter with a d i si i net slant.
REFERENCES 15 1-8 DIMENSIONAL ANALYSIS, It is a necessary condition for correct-ness that every equation be balanced dimensionally. For example, consider the hypothetical formula AY — = DA L where AY = mass L = length D = density (mass per unit volume) A = area The dimensional symbols for the left side are AY/L, the same as those used. The dimensional symbols for the right side are AY , Af -r L 2 = — L?
L Therefore, both sides of this equation have the dimensions of mass per length, and the equation is balanced dimensionally. This is not a guarantee that the equation is correct; i.e,, it is not a sufficient condition for correctness.
It is, however, a necessary condition for correctness, and it is frequently helpful to analyze equations in this way to determine whether or not they are dimensionally balanced.
Such dimensional analysis is also useful for determining what the dimensions of a quantity are. For example, to find the dimensions of force, we make use of Newton’s second law that Force = mass x acceleration Since acceleration has the dimensions oflength per time squared, the dimensions of force are Mass x length Time 2 ir in dimensional symbols Force — references flose, Jagadis Chandra : Collected Physical Papers, Longmans, Green, 1927.
Rose, Jagadis Chandra' “On a Complete Apparatus for the Study of the Properties of Electric Waves,” Elect. Ertgr. [Land,), October 1896.
Brown, George H.: 11 Marconi,” Cosmic Search, 2, 5—8, Spring 1980, Dunlap, Orrin E.
r Marconi— The Man and His Wireless, Macmillan, 1937.
faraday, Michael: Experimental Researches in Electricity, B. Quaritch, London, 1855.
Gundlach Friedrich Wilhelm: 11 Die Technit der kiirzeslen elecktromagnetischen Wellen seix Hein-rich Hertz,” Eiektrotech. Zeii.
(ETZ ), 7, 246, 1957.
16 I JNTRODUCrrON 1 Hertz, Heinrich Rudolph: “Ober StrahJen elecktrischer Kraft,'’ Wiedemanns Ann . Phys., 36, 769-783, 1889.
Hertz; Heinrich Rudolph: Electric Waves, Macmillan, London, 1893; Dover, 1962.
Hertz, Heinrich Rudolph: Collected Works, Barth Verlag, 1895.
Hertz, Heinrich Rudolph: “The Work of Hertz and His Successors—Signalling through Space without Wires,” Electrician Publications, 1894, 1898, 1900, 1908, Hertz, Johanna: Heinrich Hertz , San Francisco Press, 1977 (memoirs, letters and diaries of Hertz).
Kraus, John D.: Big Ear, Cygnus-Quasar, 1976.
Kraus, John D.: “ Karl Jansky and His Discovery of Radio Waves from Our Galaxy," Cosmic Search, 3, no. 4, 8-12, 1981, Kraus, John, D.: "Grote Reber and Ihe First Radio Maps of the Sky” Cosmic Search, 4, no.
1, 14-18, 1982 Kraus, John D.: “Karl Guthe Jansky’s Serendipity, Its Impact on Astronomy and Its Lessons for the Future,” in K. Kellermann and B. Sheets (eds.), Serendipitous Discoveries in Radio Astronomy, National Radio Astronomy Observatory, 1983.
Kraus, John D.: Electromagnetics , 3rd ed,, McGraw-Hill, 1984.
Kraus, John D.: “Antennas Since Hertz and Marconi,” IEEE Trans. Ants . Prop.. AP-33, 131-137, February 1985 (Centennial Plenary Session Paper), Kraus, John D.: Radio Astronomy h 2nd ed., Cygnus-Quasar, 1986; Sec, 1-2 on Jansky, Reber and early history.
Kraus, John D.: “Heinrich Hertz—Theorist and Experimenter,”' /£££ Trans. Microwave Theory Tech. Hertz Centennial Issue, MTT-36, May 1988.
Lodge, Oliver J.: “Signalling through Space without Wires,” Electrician Publications, 1898, Marconi, Degna: My Father Marconi, McGraw-Hill, 1962.
Maxwell, James Clerk: A Treatise on Electricity and Magnetism , Oxford 1873, 1904, Newton, Isaac: Principle , Cambridge, 1687, 1 Poincare, Henri, and F.
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Ramsey, John F.: “Microwave Antenna and Waveguide Techniques before 1900” Proc. IRE , 46, 405-415, February 1958.
Rayleigh, Lord: "On the Passage of Electric Waves through Tubes or the Vibrations of Dielectric Cylinders,” Phil. Mag., 43, 125-132, February 1897, Righi, A.
: “ L’Ouica della Qscillaztoni Elellriche,” Zanichelli, Bologna, 1897, Rothe, Horst: "Heinrich Hertz, der Enldecker der elektromagnetischen Wellen,” Elektrotech. Zeit , (ETZ), 7, 247 251, 1957.
Wolf, Franz: l Heinrich Hertz, Leben and Werk," Elektrotech. Zeit (ETZ), 7, 242-246, 1957.
L CHAPTER 2 BASIC ANTENNA CONCEPTS 2-1 INTRODUCTION The purpose of this chapter is to provide intro-ductory insights into antennas and their characteristics. Following a section on definitions, the basic parameters of radiation resistance, temperature, pattern , directivity t gain , beam area and aperture are introduced. From the aperture concept it is only a few steps to the important Friis transmission formula. This is followed by a discussion of sources of radiation, field zones around an antenna and the effect of shape on impedance. The sources of radiation are illustrated for both transient (pulse) and continuous waves. The chapter concludes with a dis-cussion of polarization and cross- field.
2-2 DEFINITIONS. A radio antenna 1 may be defined as the structure associated with the region of transition between a guided wave and a free-space wave, or vice versa.
In connection with this definition it is also useful to consider what is meant by the terms transmission iine and resonator.
A transmission line is a device for transmitting or guiding radio-frequency energy from one point to another. Usually it is desirable to transmit the energy 1 In Us zoological sense, an antenna is the feeler, or organ of touch, of an insect. According to usage in the United States the plural of “insect antenna" is “antennae," but the plural of "radio antenna" is “ antennas.” 17 18 : BASIC ANTENNA CONCEPTS with a minimum of attenuation, heat and radiation losses being as small as pos-sible, This means that while the energy is being conveyed from one point to another it is confined to the transmission line or is bound closely to it. Thus, the wave transmitted along the line is 1-dimensional in that it does not spread out into space but follows along the line. From this general point of view one may extend the term transmission line (or transmission system) to include not only coaxial and 2-wire transmission lines but also hollow pipes, or waveguides.
A generator connected to an infinite, lossless transmission line produces a uniform traveling wave along the line.
If the line is short-circuited, the outgoing traveling wave is reflected, producing a standing wave on the line due to the interference between the outgoing and reflected waves, A standing wave has associated with it local concentrations of energy, if the reflected wave is equal to the outgoing wave, we have a pure standing wave. The energy concentrations in such a wave oscillate from entirely electric to entirely magnetic and back twice per cycle. Such energy behavior is characteristic of a resonant circuit, or reson-ator. Although the term resonator, in its most general sense, may be applied to any device with standing waves, the term is usually reserved for devices with stored energy concentrations that are large compared with the net flow of energy per cycle, 1 Where there is only an outer conductor, as in a short-circuited section of waveguide, the device is called a cavity resonator.
Thus, antennas radiate (or receive) energy , transmission lines guide energy , while resonators store energy.
A guided w rave traveling along a transmission line which opens out, as in Fig. 2-1, will radiate as a free-space wave. The guided wave is a plane wave while the free-space wave is a spherically expanding wave. Along the uniform part of the line, energy is guided as a plane wave with little loss, provided the spacing between the wires is a small fraction of a wavelength. At the right, as the trans-mission line separation approaches a wavelength or more, the wave tends to be radiated so that the opened-out line acts like an antenna which launches a free-space wave. The currents on the transmission line flow out on the transmission line and end there, but the fields associated with them keep on going. To be more explicit, the region of transition between the guided wave and the frce-space wave may be defined as an antenna.
We have described the antenna as a transmitting device. As a receiving device the definition is turned around, and an antenna is the region of transition between a free-space wave and a guided wave. Thus, an antenna is a transition device , or transducer, between a guided wore and afree-space wave, or vice versa 1 While transmission lines (or waveguides) are usually made so as to mini-1 The ratio of the energy stored to that lost per cycle is proportional to the Q , or sharpness of resonance of the resonalor (see Sec. 6-12}.
2 We note that antenna parameters, such as impedance or gain, require that the antenna terminals be specified.
M BASIC ANTENNA PARAMETERS 19 Figure 2-1 The antenna is a region of transition between a wave guided by a transmission line and a free-space wave. The transmission line conductor separation is a small fraction of a wavelength while the separation at the open end of the transition region or antenna may be many wavelengths. More generally, an antenna interfaces between electrons on conductors and photons in space. The eye is another such device mize radiation, antennas are designed to radiate (or receive) energy as effectively as possible.
The antenna, like the eye, is a transformation device converting electromag-netic photons into circuit currents; but, unlike the eyeT the antenna can also convert energy from a circuit into photons radiated into space.
1 In simplest terms an antenna converts photons to currents or vice versa > Consider a transmission line connected to a dipole 2 antenna as in Fig. 2-2.
The dipole acts as an antenna because it launches a free-space wave. However, it may also be regarded as a section of an open-ended transmission line. In addi-tion, it exhibits many of the characteristics of a resonator, since energy reflected from the ends of the dipole gives rise to a standing wave and energy storage near the antenna. Thus, a single device, in this case the dipole, exhibits simultaneously properties characteristic of an antenna , a transmission line and a resonator.
2-3 BASIC ANTENNA PARAMETERS.
Referring to Fig.
2-2, the antenna appears from the transmission line as a 2-terminal circuit element having an impedance Z with a resistive component called the radiation resistance R r , 1 A photon is the quantum unit of electromagnetic energy equal to tif where h - Planck? constant ( -6.63 x I0' J J s) and / = frequency (Hz), : A positive electric charge q separated a distance from an equal but negative charge constitutes an electric dipole.
if the separation is l, then ql is the dipole moment. A linear conductor which, at a given instant, has a positive charge at one end and an equal but negative charge at the other end may act as a dipole antenna.
[A loop may be considered to be a magnetic dipole antenna of moment IA, where l — loop current and A = loop area.) 20 2 BASIC ANTENNA CONCEPTS mSL ThC““ ' aUnCheS 3 free-Spa“— but ™impedance lo the ,ran S -is charac,erized by its or pa,,ems an.,nL he „ radiat K°?
reSiStanCe ' iS n0t aSSOC,ated with resistance in the “T J[ OPe \ K ^ TIT C°Upled fr°m the antenna and its environment indtSTn CtoT?
"0" '"iS, n“ “ ^ 2' 13 2' 14 Associated with the radiation resistance is also an antenna temperature T.
For a lossless antenna this temperature has nothing to do with the physical tern-perature of the antenna proper but is related to the temperature of distant regions of space (and nearer surroundings) coupled to the antenna via its radi-atton resistance. Actually, the antenna temperature is not so much an inherent property of the antenna as it is a parameter that depends on the temperature of the regions the antenna is “looking at.” In this sense, a receiving antenna may be regarded as a remote -sen sing, temperature-measuring device (see Chap.
1 7).
Both the radiation resistance Rr and the antenna temperature T. are single-valued scalar quantities. The radiation patterns, on the other hand, involve the variation of field or power (proportional to the field squared) as a function of the two spherical coordinates 0 and .
PATTERNS.
Figure 2-3a shows a field pattern where r is proportional to the field intensity at a certain distance from the antenna in the direction B, d> The pattern has its main-lobe maximum in the z direction (0 = 0) with minor lobes (side and back) in other directions. Between the lobes are nulls in the directions of zero or minimum radiation.
anJ rad,ac J°" , An electromagnet,c wave consists of electric and magnetic fields propacan™ c i SP3<» ajield being a region where electric or magnetic forces act. The electric and maimetir radiation iBC"Spacc Wav^ traveling outward at a large distance from an antenna convey energy called 2-4 PATTERNS 21 Field pattern 1 I Main lobe axis Power pattern Main lobe axis Main lobe or main beam Field ^ components 2 Side \ Field in f lobes \ \ 8r $ direction ^ \ \ 7 / j / Half-power beam / Jrf width (HPBW) / Back lobes Beam width between first nufls {BWFNf dB pattern Minor or side lobes i \ First side lobe Null / Al A Figure 2-3 (a) Antenna fidd pattern with coordinate system, (fr) Antenna power pattern in polar coordinates (linear scale), (c) Antenna pattern in rectangular coordinates and decibel (logarithmic) scale Patterns (h) and (c) are the same To completely specify the radiation pattern with respect to field intensity and polarization requires three patterns: 1. The 6 component of the electric field as a function of the angles 0 and tj> or «t)(Vn") 2. The tj> component of the electric field as a function of the angles 0 and tj> or £J0,^)(Vait' 1 ) 22 2 BASIC ANTENNA CONCEPTS 3. The phases of these fields as a function of the angles 8 and 0 or <5^ 0) and <5(0, 0) (rad or deg) Dividing a field component by its maximum value, we obtain a normalized field pattern which is a dimensionless number with a maximum value of unity.
Thus, the normalized field pattern for the 6 component of the electric field is given by )H = ^ " (dimensionless) ( 1 ) At distances that are large compared to the size of the antenna and large com-pared to the wavelength, the shape of the field pattern is independent of distance.
Usually the patterns of interest are for this far-field condition (see Chap.
18).
Patterns may also be expressed in terms of the power per unit area [or Poynting vector S(0, 0)] at a certain distance from the antenna.
1 Normalizing this power with respect to its maximum value yields a normalized power pattern as a function of angle which is a dimensionless number with a maximum value of unity. Thus, the normalized power pattern is given by 0) = cvi -(dimensionless) 1 2) Wmax V } where 5(0, 0) = Poynting vector = [£ 2 (0, 0) + £$0, 0)]/Zo , W m“ 2 = r 2 dft (1) where = solid angle subtended by the area dA The area of the strip of width r dO extending around the sphere at a con-stant angle B is given by (2?ir sin 9) (r d8). Integrating this for 8 values from 0 to n yields the area of the sphere. Thus, Area of sphere = 2;rr 2 j sin 8 dd = 2rcr2 [ — cos 0]S = 4nr 2 (2) where 4 n = solid angle subtended by a sphere, sr Thus, 1 steradian = 1 sr = (solid angle of sphere)/(47t) = 1 rad 2 = C—j (deg2 ) = 3282.8064 square degrees (3) Therefore, 4n steradians = 3282.8064 x 4n = 41 252.96 = 41 253- square degrees r = solid angle in a sphere (4) 24 2 BASIC ANTENNA CONCEPTS This strip has area = 2?rr sfn 8 r d8 r sin 0 d$ Area dA -r 1 sin 8 d6 d$ \ = r d i1 , where dQ - solid angle \ = sin 8 d& d$ s Azimuth angle Fienre 2-5 Spherical coordinates in relation to the area dA of solid angle dU = sin 8 d& Now the beam area (or beam solid angle) for an antenna is given by the integral of the normalized power pattern over a sphere (4n sr) or = r f Vjto, Jo Jo ) dQ.
(sr) where dCl = sin B dB d Referring to Fig, 2-6, the beam area QA of an actual pattern is equivalent to the same solid angle subtended by the spherical cap of the cone-shaped (triangular cross-section) pattern.
Equivalent solid angle liA Actual pattern of beam area li,, Half-power beam width 0HP Fifite 2-6 Cross section of symmetrical power pattern of antenna showing equivalent solid angle for a cone-shaped {triangular) pattern.
2-1 BEAM EFFICIENCY 25 This solid angle can often be described approximately in terms of the angles subtended by the half-power points of the main lobe in the two principal planes as given by — ^hp ^hp (sr) (6) where 0 H p and 0 HP are the half-power beam widths (HPBW) in the two principal planes, minor lobes being neglected 2-6 RADIATION INTENSITY The power radiated from an antenna per unit solid angle is called the radiation intensity U (watts per steradian or per square degree). The normalized power pattern of the previous section can also be expressed in terms of this parameter as the ratio of the radiation intensity U(6, $), as a function of angle, to its maximum value Thus, PM U(fl.
V(6 t U. m 0) ( 1 ) Whereas the Poynting vector S depends on the distance from the antenna (varying inversely as the square of the distance), the radiation intensity U is inde-pendent of the distance, assuming in both cases that we are in the far field of the antenna (see Sec 2-35).
2-7 BEAM EFFICIENCY- The (total) beam area (or beam solid angle) consists of the main beam area (or solid angle) plus the minor-lobe area (or solid angle) Q^,, 1 Thus, = (1) The ratio of the main beam area to the (total) beam area is called the (main) beam efficiency zM Thus, — beam efficiency A The ratio of the minor-lobe area to the (total) beam area is called the stray factor Thus, Qm t t £ — — = stray factor 0.
It follows that (4) If the main beam is not bounded by a deep null its extent becomes an arbitrary act of judgment.
26 1 BASIC ANTENNA CONCEPTS 2-8 DIRECTIVITY. The directivity D of an antenna is given by the ratio of the maximum radiation intensity (power per unit solid angle) U(d , to the average radiation intensity Ulr (averaged over a sphere). Or, at a certain^ istance from the antenna the directivity may be expressed as the ratio of the maximum to !
the average Poynting vector. Thus, (dimensionless) Both radiation intensity and Poynting vector values should be measured in the far field of the antenna (see Sec. 2-35).
Now the average Poynting vector over a sphere is given by S(0, L -±rr« 4>) d£i (W m - 2 ) Thus, the directivity wriz dn s If The smaller the beam solid angle, the greater the directivity.
2-9 EXAMPLES OF DIRECTIVITY.
If an antenna could be isotropic (radiate the same in all directions) (j > ) = 1 (for all 0 and (1) ^en = 4ft (2) This is the smallest directivity an antenna can have- Thus, QA must always be equal to or less than 4te, while the directivity D must always be equal to or greater than unity.
Neglecting the effect of minor lobes, we have from (2-8-3) and (2-5-6) the simple approximation 1 D 4te 41 OOP ^HP^HP 0HP0HP ^ ^ 1 4irsr- 41 253 square degrees. Since (4) is an approximation 41 253 is rounded off to 41 000, Ml DIRECTIVITY AND RESOLUTION 27 where 0HP — half-power beam width in 0 plane, rad HP = half-power beam width in plane, rad 0f, P = half-power beam width in 8 plane, deg <£hp = half-power beam width in 0 plane, deg Equation (4) is an approximation and should be used in this context. To avoid inappropriate usage, see the discussion following Eq. (17) of Sec. 3-13.
If an antenna has a main lobe with both half-power beam widths (HPBWs) = 20°, its directivity from (2-8-4) and (2-5-6) is approximately 4rc(sr) 41 000 (deg 2 ) = 41000(deg ; ) ^(sr)^ 20° x 20° ^ 103 = 20 dBi (dB above isotropic) (5) which means that the antenna radiates a power in the direction of the main-lobe maximum which is about 100 times as much as would be radiated by a nondirec-tional (isotropic) antenna for the same power input.
2-10 DIRECTIVITY AND GAIN The gain of an antenna (referred to a lossless isotropic source) depends on both its directivity and its efficiency.
If the efficiency is not 100 percent, the gain is less than the directivity. Thus, the gain G = kD (dimensionless) (1) where k = efficiency factor of antenna (0 < k < 1), dimensionless This efficiency has to do only with ohmic losses in the antenna. In transmitting, these losses involve power fed to the antenna which is not radiated but heats the antenna structure.
2-11 DIRECTIVITY AND RESOLUTION. The resolution of an antenna may be defined as equal to half the beam width between first nulls (BWFN/2), 2 For example, an antenna whose pattern BWFN = 2° has a resolution of 1° and, accordingly, should be able to distinguish between transmit-ters on two adjacent satellites in the Clarke geostationary orbit separated by T.
Thus, when the antenna beam maximum is aligned with one satellite, the first null coincides with the other satellite.
] When gain is used as a single-valued quantily (like directivity) its maximum nose-on main-beam value is implied in the same way that the power rating of an engine implies its maximum value.
Multiplying the gain <3 by the normalized power pattern PH{& $) gives the gain as a function of angle.
1 Often called the Rayleigh resolution . See Sec, 11-23 and also J. D, Kraus, Radio Astronomy, 2nd ed., Cygnus-Quasar, 1986, pp. 6-19.
28 2 BASIC ANTENNA CONCtPTS Half the beam width between first nulls is approximately equal to the half-power beam width (HPBW) or BWFN ^ HPBW so from (2-5-6) the product of the BWFN/2 in the two principal planes of the antenna pattern is a measure of the antenna beam area.
1 Thus, /BWFN\ /BWFN\ It then follows that the number N of radio transmitters or point sources of radi-ation distributed uniformly over the sky which an antenna can resolve is given approximately by where — beam area, sr However, from (2-8-4), and we may conclude that ideally the number of point sources an antenna can resolve is numerically equal to the directivity of the antenna or D = N (5) Equation (4) states that the directivity is equal to the number of beam areas into which the antenna pattern can subdivide the sky and (5) gives the added signify cance that the directivity is equal to the number of point sources in the sky that the antenna can resolve under the assumed ideal conditions of a uniform source dis-tribution. 2 2-12 APERTURE CONCEPT, The concept of aperture is most simply introduced by considering a receiving antenna.
Suppose that the receiving antenna is an electromagnetic horn immersed in the field of a uniform plane wave as suggested in Fig, 2-7. Let the Poynting vector, or power density, of the plane wave be 5 watts per square meter and the area of the mouth of the horn be A 1 Usually BWFN/2 is slightly greater than HPBW and from (3-13-18) we may conclude that (2) is actually a better approximation to QA than QA - 0HP as given by (2-5-b).
2 A strictly regular distribution of points on a sphere is only possible for 4, 6, 8 P 12 and 20 points corresponding to the vertices of a tetrahedron, cube, octahedron, isoahedron and dodecahedron.
2-11 RFFRCT1VR APRUTCKF.
29 Figure 2-7 Mane wave inddeni on electromagnetic horn of mouth aperture A.
square meters.
If the horn extracts all the power from the wave over its entire area A, then the total power P absorbed from the wave is P = SA (W) (H Thus, the electromagnetic horn may be regarded as an aperture, the total power it extracts from a passing wave being proportional to the aperture or area of its mouth.
It will be convenient to distinguish between several types of apertures, namely, effective aperture , scattering aperture, loss aperture , collecting aperture and p/iysicfff aperture. These different types of apertures are defined and discussed in the following sections.
In the following discussion it is assumed, unless otherwise stated, that the antenna has the same polarization as the incident wave and is oriented for maximum response.
2-13 EFFECTIVE APERTURE. Consider a dipole receiving antenna f//2 or less) situated in the field of a passing electromagnetic wave as suggested in Fig. 2 -8a. The antenna collects power from the wave and delivers it to the termi-nating or load impedance Z T connected to its terminals. The Poynting vector, or power density of the wave, is S watts per square meter. Referring to the equiva-lent circuit of Fig. 2-8b y the antenna may be replaced by an equivalent or Theve-nin generator having an equivalent voltage V and internal or equivalent antenna mpedance ZA . The voltage V is induced by the passing wave and produces a current / through the terminating impedance Z T given by Z T + ZA (I) w here / and V are rms or effective values.
30 2 BASIC ANTENNA CONCEPTS Dipole antenna (fl) (M Figure 2-8 Schematic diagram of dipole antenna terminated in impedance Z T with plane wave incident on antenna (a) and equiv-alent circuit {/>).
where P = power in termination, W S = power density of incident wave, W m ~ 2 A = area, m 2 If 5 is in watts per square wavelength (W k~ 2 ) then A is in square wavelengths (X 2 \ which is often a convenient unit of measurement for areas.
m scattering apertcrf.
31 Substituting (7) into (8} we have A V 2 R T S[( r + Rl + Rj) 1 + {XA + ,9 ’ Unless otherwise specified, it is assumed that V is the induced voltage when the antenna is oriented for maximum response and the incident wave has the same polarization as the antenna. The value of A as indicated by (9) takes into account any antenna losses as given by R L and any mismatch between the antenna and its terminating impedance.
Let us consider now the situation where the terminating impedance is the complex conjugate of the antenna impedance (terminal or load impedance matched to antenna) so that maximum power is transferred. Thus, -X a ( 10 ) and = + (11) Introducing (10) and (11) in (9) yields the effective aperture A e of the antenna.
Thus, + R l ) (m 2 or ;/) If the antenna is lossless (ft Yj = 0) we obtain the maximum effective aperture A em of the antenna. Thus, a~=4sr; lm2 °r '-Z) The aperture A em given by (13) represents the area over which power is extracted from the incident wave ancl delivered to the load.
Sometimes the terminating impedance is not located physically at the antenna terminals as suggested in Fig. 2-8.
Rather, it is in a receiver which is connected to the antenna by a length of transmission line. In this ease Z T is the equivalent impedance which appears across the antenna terminals.
If the trans-mission line is lossless, the power delivered to the receiver is the same as that delivered to the equivalent terminating impedance Z 7 If the transmission line has attenuation, the power delivered to the receiver is less than that delivered to the equivalent terminating impedance by the amount lost in the line.
2-14 SCATTERING APERTURE.
In the preceding section we discussed the effective area from which power is absorbed.
Referring to Fig. 2-8fr, the voltage induced in the antenna produces a current through both the antenna impedance Z A and the terminal or load impedance Z r , The power P absorbed by the terminal impedance is, as we have seen, the square of this current times the real part of the load impedance. Thus, as given in (2-13-5), P = I 2 R T . Let us now inquire into the power appearing in the antenna impedance Z A . The real part of 32 2 BASIC ANTENNA CONCEPTS this impedance has two parts, the radiation resistance Rr and the loss resist-ance R l (Ra = R r + Rl) Therefore, some of the power that is received will be < dissipated as heat in the antenna, as given by ' P = I 2R l (1) < The remainder is “dissipated"' in the radiation resistance, in other words, is reradiated from the antenna. This reradiated power is P M = I 2 Rr (2) ; This reradiated or scattered power is analogous to the power that is dissipated in a generator in order that power be delivered to a load. Under conditions of maximum power transfer, as much power is dissipated in the generator as is i delivered to the load.
This reradiated power may be related to a scattering aperture or scattering ' cross section. This aperture As may be defined as the ratio of the reradiated power to the power density of the incident wave. Thus, j pff A, = scattering aperture = — (3) j S t Thus, under conditions for which maximum power is delivered to the terminal impedance, an equal power is reradiatedfrom the receiving antenna.
Now suppose that the load resistance is zero and X T = — X A (antenna resonant). This zero- load-resistance condition may be referred to as a resonant short-circuit (RSC) condition. Then for RSC the reradiated power is 1 Antenna matched.
3 Referring to Fig, 2-8a, note that if the direction of the incident wave changes, the scattered power could increase while V decreases. However, ZA remains the same.
L 2 14 SCATTERING APERTURE 33 Figure 2-9 a/2 dipole antenna receiving (and re radiating) power from A/2 dipole transmitting antenna.
and the scattering aperture becomes or A.-4a (9) Thus, for the RSC condition, the scattering aperture of the antenna is 4 times as great as its maximum effective aperture .
Figure 2-9 sho_ws two 2/2 dipoles, one transmitting and the other receiving.
Let the receiving antenna be lossless {RL = 0). Consider now three conditions of the receiving antenna: 1. Antenna matched 2. Resonant short circuit 3. Antenna-open-circuited (Z r = oc) For condition 1 (antenna matched). A, = Am , but for condition 2 (resonant short circuit), As = 4/4 t„ and 4 times as much power is scattered or rcradiated as under condition L Under condition 2 (resonant short circuit), the “ receiving ” antenna acts like a scatterer and, if close to the transmitting antenna, may absorb and reradiate sufficient power to significantly alter the transmitting antenna radiation pattern.
Under these conditions one may refer to the “receiving” antenna as a parasitic element . Depending on the phase of the current in the parasitic element, it may act either as a directgr or a reflector (see Sec.
1 l-9u). To control its phase, it may be operated off-resonance (X T ^ — AT^), although this also reduces its scattering aperture.
For condition 3 (antenna open-circuited), / = 0, A e = 0 and A, = 0.
1 ] This is an idealization. Although the scattering may be small it is not zero. See Table 17-2 for scattering from short wires.
34 2 BASIC ANTF-NNA l.O^CEPTS Figure MO Variation of effective j aperture A t , scattering aperture Aa j and collecting aperture A t as a j function of the relative terminal j resistance R T/flr of a small j antenna.
It is assumed that R L = Jf, = X r = 0.
To summarize: Condition 1, antenna matched: Ar “ j4m Condition 2, resonant short circuit: A t = 4Am Condition 3, antenna open-circuited: Aa = At = 0 The ratio AJAtm as a function of the relative terminal resistance R T/Rr is!
shown in Fig. 2-10. For RT/Rr = 0, AJA em = 4, while as R r/iC approaches intin- j ity (open circuit), AJAm approaches zero j The ratio of the scattering aperture to the effective aperture may be called] the scattering ratio that is, \ x I Scattering ratio = — = (dimensionless) (10).!
1 The scattering ratio may assume values between zero and infinity (0 < /? < oo).
'> For conditions of maximum power transfer and zero antenna losses, the!
scattering ratio is unity. If the terminal resistance is increased, both the scattering ' aperture and the effective aperture decrease, but the scattering aperture decreases j more rapidly so that the scattering ratio becomes smaller. By increasing the ter-] minal resistance, the ratio of the scattered power to power in the load can bej made as small as we please, although by so doing the power in the load is also] reduced.
\ The reradiated or scattered field of an absorbing antenna may be con-; sidered as interfering with the incident field so that a shadow may be cast behind; the antenna as illustrated in Fig. 2-11-1.
| Although the above discussion of scattering aperture is applicable to a< single dipole (A/2 or shorter), it does not apply in general. (See Sec. 2-18. See also; Sec.
17 - 5 .) / M?
PHYSICAL APERTURE AND APERTURE EFFICIENCY 35 245 LOSS APERTURE, If R L is not zero [k ^ 1 in (2-10-1)], some power is dissipated as heat in the antenna. This may be related to a loss aperture A L which is given by l 2R L V 2 Rl S SWr + L + Rt) Z + tX A + t ) 2 ] dl 2-16 COLLECTING APERTURE, Three types of apertures have now been discussed: effective, scattering and loss. These three apertures are related to three ways in which power collected by the antenna may be divided: into power in the terminal resistance (effective aperture); into heat in the antenna (loss aperture); or into reradiated power (scattering aperture). By conservation of energy the total power collected is the sum of these three powers. Thus, adding these three apertures together yields what may be called the collecting aperture as given by V 2(Rf +R l + R t ) St{Rr + R L + R T ) 2 + iX A + X r) 2 ] —3- = A, + A l + Ar The variation of A c with RjjRr for the case of A L — 0 is shown in Fig. 2-10 2-17 PHYSICAL APERTURE AND APERTURE EFFICIENCY, It is often convenient to speak of a fifth type of aperture called the physical aperture A p . This aperture is a measure of the physical size of the antenna. The manner in which it is defined is entirely arbitrary. For example, it may be defined as the physical cross section (in square meters or square wavelengths) perpendicular to the direction of propagation of the incident wave with the antenna oriented for maximum response. This is a practical definition in the case of many antennas For example, the physical aperture of an electromagnetic horn is the area of its mouth, while the physical aperture of a linear cylindrical dipole is the cross-36 2 BASIC ANTENNA CONCEPTS sectional area of the dipole. However, in the case of a short stub antenna mounted on an airplane, the physical aperture could be taken as the cross-: sectional area of the stub or, since currents associated with the antenna may flowj over the entire surface of the airplane, the physical aperture could be taken as the cross-sectional area of the airplane. Thus, the physical aperture has a simple, definite meaning only for some antennas. On the other hand, the effective aper-ture has a definite, simply defined valuefor all antennas.
The ratio of the effective aperture to the physical aperture is the aperture^ efficiency eJpl that is, £ap = ~f (dimensionless) ( 1) Although aperture efficiency may assume values between zero and infinity, it cannot exceed unity for large (in terms of wavelength) broadside apertures.
j 2-18 SCATTERING BY LARGE APERTURES, In Sec 2-14 it wa shown that the scattering aperture of a single dipole was equal to the (maximum effective aperture for the condition of a (conjugate) match and 4 times as muc for a resonant short circuit. For a large broadside aperture A (dimensions k matched to a uniform wave, all power incident on the aperture can be absor' over the area A , while an equal power is forward-scattered. Thus, the total co lecting aperture is 2A. If the large aperture is a nonabsorbing perfectly conductin flat sheet the power incident on the area A is backscattered while an equal pow is forward-scattered, yielding a scattering (and collecting) aperture 2A In thi case the scattering aperture may be appropriately called a total scattering cro section ( + H0 3 = - V m where R0 = intrinsic resistance of space = 377 O Zj — load impedance = space cloth in parallel with space behind it " = Ro/2 t = transmission coefficient = j Sheet of space cloth ncident Reflected Transmitted wave wave ] wave (a) Figure 2-11-3 (a) A plane wave traveling to the right incident nor-mally on an infinite sheet of space (b) cloth is partially reflected, partially absorbed and partially transmit-ted, (ft) Analogous transmission-line arrangement.
38 1 BASIC ANTENNA CONCEPTS The electric field intensity of the reflected wave traveling to the left of th& sheet is F _ ^ ~ _ (o/2> ~ 0 | v -j " p ' ZL + Z0 (R0/2) + R 0 3 V m 2> where p = reflection coefficient = It is apparent that a sheet of space doth by itself is insufficient to terminate an incident wave without reflection. This may also be seen by considering the analogous lossless transmission line arrangement shown in Fig. 2-11-3&, where the load resistance R0 is in parallel with the tine to the right with characteristic resistance ^ For both space wave and transmission line, ^ [ = (i) 2 )] of the incident power is reflected or scattered back, f [ = (3) 2 ] of the incident power is transmit; ted or forward-scattered and the remaining f absorbed in the space cloth or loadi If the area of the space cloth equals A , then the effective aperture A e = and the scattering aperture As = f A.
In order to completely absorb the incident wave without reflection or trans^ mission, let hn infinite petfectly conducting sheet or reflector be placed parallel t<j the space cloth and A/4 behind it, as portrayed in Fig. 2-lMu. Now the imped-i ance presented to the incident wave at the sheet of space cloth is 377 Cl, being thl impedance of the sheet in parallel with an infinite impedance. As a consequence, this arrangement results in the total absorption of the wave by the space cloth. ^ There is, however, a standing wave and energy circulation between the cloth and the conducting sheet and a shadow behind the reflector.
The analogous transmission -line arrangement is illustrated in Fig. 2-11-4b the .A/4 section (stub) presenting an infinite impedance across the load R 0 .
In the case of the plane wave, the perfectly conducting sheet or reflected effectively isolates the region of space behind it from the effects of the wave. In ait analogous manner the shorting bar on the transmission line reduces the wavei beyond it to a small value.
When the space cloth is backed by the reflector the wave is matched. In a similar way, the fine is matched by the load /? 0 with A/4 stub. 2 A transmission line may also be terminated by placing a resistance across /the line which is equal to the characteristic resistance of the line, as in Fig.j 2-1 l-3h, and disconnecting the line beyond it. Although this provides a practical method of terminating a transmission line, there is no analogous counterpart irij the case of a space wave because it is not possible to ‘‘disconnect” the space to] the right of the termination. A region of space may only be isolated or shielded] as by a perfectly conducting sheet. 3 1 J. D. Kraus, Electromagnetics, 3rd ed, T McGraw-Hill, 1984, pp. 461—462.
1 The stub Length can differ from 4/4 provided the load presents a conjugate match.
5 The spacing of the transmission line is assumed to be small (<i)and radiation negligible.
18 SCATTY R INC BY LAROH APERTURES 39 Figure 2-11-4 (a) A plane wave leveling to the right incident normally on an infinite sheet of spaa do"h backed hi L infinite perfectly conducting reflecting sheet, as shown, .. comply absorbed without reflection lb) Analogous transmission-line arrangement in whrch a wave traveling to the ngh 35 completely absorbed in the load without reflection.
If the space cloth reflector area A is large (dimensions /.) but not infinite in extent the power incident on A is absorbed (as in the infinite case) but there is now scattering of an equal power so that the total collecting aperture A, is twice A or A c - A e + A s - 2A where A e - effective aperture = A , m 2 A s = scattering aperture = A , m 2 Thus, as much power is scattered as is absorbed (maximum power transfer condition) = A e ).
If only the flat perfectly conducting reflector of area A is present (no space cloth) the wave incident on the reflector is backscattered instead of absorbed and the wave is totally scattered (half back, half forward) so that the collecting aper-ture is all scattering aperture and equal to 2 A (>1, = 2A = o„ see Table 17-1, las row column 3). In both cases (with and without space cloth) the incident wave front is disturbed and the energyflow redirected over an area twice the area A.
40 2 BASIC ANTENNA CONCEPTS 2-10 EFFECTIVE HEIGHT 41 Absorption is also possible by methods other than the single space cloth technique as, for example, using thick {multiple space cloth) or other absorbing structures as discussed in Sec.
18-3c. These structures, as welt as a single space cloth, constitute a distributed toad. The above conclusions regarding large, but not infinite, apertures also apply to a large uniform broadside array of area A connected to a lumped load or a uniformly illuminated parabolic reflector of area A with power brought to a focus and delivered to a lumped load .
In all cases (distributed load, broadside array and parabolic reflector), the effective aperture Af — A (= physical aperture A p) and the scattering aperture A s also equal A (= A p). The aperture efficiency in these cases is given by which is the maximum possible value (100 percent efficiency) for large broadside antennas. In theory, the 100 percent limit might be exceeded slightly by using supergain techniques. However, as shown by Rhodes, 1 the practical obstacles are enormous. In practice, less than 100 percent efficiency may be necessary in order to reduce the sidelobe level by using tapered (nonuniform) aperture distributions.
Accordingly, targe aperture antennas are commonly operated at 50 to 70 percent aperture efficiency.
The single dipole and the large-area antenna may be considered to rep-resent two extremes as regards scattering, with other antenna types intermediate.
Table 2-1 summarizes the scattering parameters for large space cloth or array apertures, for transmission lines and for a single dipole (A/2 or shorter).
2-19 EFFECTIVE HEIGHT. The effective height h (meters) of an antenna is another parameter related to the aperture. Multiplying the effective height by the incident field E (volts per meter) of the same polarization gives the voltage V induced. Thus, V = HE (1) Accordingly, the effective height may be defined as the ratio of the induced voltage to the incident field or A = f ' Thus (fot | lneal polarization) we can write V = ht E = ht BvoQ' where hr — effective height and polarization angle of anteop. ® _ i E = field intensity and polarization angle of incident V m ft = angle between polarization angles of antenna and wave tfeU In a still more general expression (for any polarization ftat^ i> the angle between polarization states on the Poincare sphere (see Sec, 2-36).
2 20 MAXIMUM EFFECTIVE APERTURE OK A SHORT DIPOLE 43 Direction of incident vvave Figure 2-13 Short dipole with uniform current induced by incident wave.
origin as shown in Fig. 2-13, with a plane wave traveling in the negative x direc-tion incident on the dipole. The wave is assumed to be linearly polarized with E in the y direction. The current on the dipole is assumed constant and in the same phase over its entire length, and the terminating resistance is assumed equal to the dipole radiation resistance Rr . The antenna loss resistance RL is assumed equal to zero.
The maximum effective aperture of an antenna is obtained from (2-13-13) as V 2 Aem ~ 4SR r {1) where the effective value of the induced voltage V is here given by the product of the effective electric field intensity at the dipole and its length, that is, V = El (2) The radiation resistance Rr of a short dipole of length J with uniform current will be shown later (in Sec. 5-3) to be 1 where X = wavelength /av = average current / 0 = terminal current The power density, or Poynting vector, of the incident wave at the dipole is related to the field intensity by S "F I"' where Z = intrinsic impedance of the medium 1 This relation for the radiation resistance of a short dipole was worked out by Max Abraham in 1^04 and R. Rudcnberg in 1908.
It is very clearly set forth in Jonathan Zenneck's textbook editions of 1905 and 1908 and its English translation, Wireless Technology, McGraw-Hill, 1915.
44 2 BASIC ANTENNA CONCEPTS In the present case, the medium is free space so that Z ~ 12Orc Q. Now substitut-ing (2), (3) and (4) into (1), we obtain for the maximum effective aperture of a short dipole (for / av = / 0) A 120n£ 2 f2 A 2 320n 2E 2 l 2 (5) Equation (5) indicates that the maximum effective aperture of a short dipole is somewhat more than ^ of the square wavelength and is independent of the length of the dipole provided only that it is small (I ^ A). The maximum effective aperture neglects the effect of any losses, which probably would be considerable for an actual short dipole antenna. If we assume that the terminating impedance is matched to the antenna impedance but that the antenna has a loss resistance equal to its radiation resistance, the effective aperture from (2-13-12) is \ the maximum effective aperture obtained in (5).
2-21 MAXIMUM EFFECTIVE APERTURE OF A LINEAR k/2 ANTENNA. As a further illustration, the maximum effective aperture of a linear A/2 antenna will be calculated.
It is assumed that the current has a sinu-soidal distribution and is in phase along the entire length of the antenna.
It isj further assumed that RL — 0. Referring to Fig, 2-14^, the current l at any point y 1 is then J = /O cos^p (1) A plane wave incident on the antenna is traveling in the negative x direction. The wave is linearly polarized with E in the y direction. The equivalent circuit is shown in Fig. 2- 14b. The antenna has been replaced by an equivalent or Theve-nin generator. The infinitesimal voltage dV of this generator due to the voltage L 2,21 MAXIMUM EFFECTIVE APERTURE OF A LINEAR a ;2 ANTENNA 45 „ Linear }2 antenna Figure 2-15 (a) Maximum effective aperture of Linear a/2 antenna is approximately represented by rectangle j by on a side, (fc) Maximum effective aperture of Linear a/2 antenna represented by elliptical area of 0.1 3i 3 .
induced by the incident wave in an infinitesimal element of length dy of the antenna is 2tiv dV — E dy cos -j-(2) It is assumed that the infinitesimal induced voltage is proportional to the current at the infinitesimal element as given by the current distribution (1).
The total induced voltage V is given by integrating (2) over the length of the antenna. This may be written as .
£ cos — r - dy A Performing the integration in (3) we have 7T The value of the radiation resistance Rr of the linear A/2 antenna will be taken as 73 ft .
1 The terminating resistance R T is assumed equal to R r . The power density at the antenna is as given by (2-20-4). Substituting (4), (2-20-4) and Rr = 73 into (2-13-13), we obtain, for the maximum effective aperture of a linear A/2 antenna, 120tt£^ = j0_ 2 2 ' ro 4k2£ x 73 73h °' 13 A Comparing (5) with (2-20-5), the maximum effective aperture of the linear A/2 antenna is about 10 percent greater than that of the short dipole.
The maximum effective aperture of the A/2 antenna is approximately the same as an area \ by JA on a side, as illustrated in Fig. 2- 15a. This area is JA a .
An elliptically shaped aperture of 0,13a 2 is shown in Fig. 2-15b. The physical ' The derivation of this value is given in Sec. 5-6, 46 2 BASTO ANTENNA CONCEPTS significance of these apertures is that power from the incident plane wave is absorbed over an area of this sizeby the antenna and is delivered to the terming tmg resistance.
A typical thin a/2 antenna may have a conductor diameter of so that its physical aperture is only soqA 2 , For such an antenna the maximum effective aperture of OA3X 2 is about 100 times larger.
2-22 EFFECTIVE APERTURE AND DIRECTIVITY. There is an important relation between effective aperture and directivity of all antennas as will now be shown.
Consider the electric field Er at a large distance in a direction broadside to a radiating aperture as in Fig. 2-16. If the field intensity in the aperture is constant and equal to E„ (volts per meter), the radiated power is given by p -~r A (i) where A = antenna aperture, nr Z = intrinsic impedance of the medium, O The power radiated may also be expressed in terms of the field intensity E, (volts per meter) at a distance r by ^ = L|L riQ A (2) where = beam solid angle of antenna, sr It may be shown {Sec. 11-21) that the field intensities Er and Ea are related by i£j=--r p) where X = wavelength, m Substituting (3) in (2) and equating (1) and (2) yields X2 = AQj where X = wavelength, m A — antenna aperture, m 2 = beam solid angle, sr 2-23 BEAM SOLID ANGLE AS A FRACTION OF A SPHERE 47 In (4) the aperture A is the physical aperture A p if the field is uniform over the aperture, as assumed, but in general A is the maximum effective aperture Am itsses equal zero). Thus, We note that A em is determined entirely by the antenna pattern of beam area QA .
According to this important relation, the product of the maximum effective aper-ture of the antenna and the antenna beam solid angle is equal to the wavelength squared. Equation (5) applies to all antennas. From (5) and (2-8-4) we have that 4te D = JI (6) When, for simplicity, A e is substituted for A em in (5) or (6), zero losses are assumed.
Three expressions have now been given for the directivity D of an antenna.
They are „ Vie, <Mma, s<0, 4>u% I 4n D =o; 4it D = Atm From (2-10-1), the gain C = kD, where k = AJAtm , so 4tt 2-23 BEAM SOLID ANGLE AS A FRACTION OF A SPHERE. A short dipole with directivity D = \ has a beam solid angle Putting fisph = 4it = solid angle of a sphere, the dipole beam solid angle aA = insph (2) Thus, the dipole radiation pattern may be said to fill § of a sphere. The larger the directivity of an antenna, the smaller is the fraction of a sphere filled by its radi-ation pattern. At the other extreme, a non directional (isotropic) antenna with D = l completely fills a sphere. This concept, emphasized by Harold A, Wheeler (1964), provides an interesting way' of looking at directivity and beam area.
48 2 BASIC ANTENNA CONCEPTS 2-24 TABLE OF EFFECTIVE APERTURE, DIRECTIVITY, EFFECTIVE HEIGHT AND OTHER PARAMETERS FOR DIPOLES AND LOOPS Arnault Radiation resistnce§ Maximum effective aperture A„ , X 1 Effective height, maximum value, A, m Sphere filling .
factor Directivity D ZHdBi) Isotropic — = 0.079 4ir 1 1 0 Short dipole.t length f Kt)' 3 - = 0.119 8te (/. h 2 i 1 2 L76 Short dipole,t 1 = 2/ 10 (/» = /o) 7.9 0.119 a/10 i 2 2 1.76 Short dipole,t / = 2/10 ('. = Vo) 1.98 0-1 19 a/20 i 1 2 1,76 Linear, a/2 dipole (sinusoidal current distribution) 73 30 — ^0,13 73n a 21 rt rt 0.61 1.64 2.15 Small loop! (single turn), any shape 31 — = 0.119 8rt 2n — i i 1.76 Small square loop? (single turn), side length = f Area A — t2 — (2/10) 1 3,12 3 — = 0.119 Sir Ink Too i i 1.76 § See Chaps 5 and 6.
t Length / £ a/ 10.
J Area A £ aj/100, see Sec, 6-8. For FMum loop, mluhiply R w by n2 and h by n.
Although the radiation resistance, effective aperture, effective height and directivity are the same for both receiving and transmitting, the current distribu-tion is, in general, not the same. Thus, a plane wave incident on a receiving antenna excites a different current distribution than a localized voltage applied to a pair of terminals for transmitting.
2-25 FRIIS TRANSMISSION FORMULA, The usefulness of the aper-ture concept will now be illustrated by using it to derive the important Frits transmission formula published in 1946 by Harald T. Friis of the Bell Telephone Laboratories.
1 Referring to Fig. 217, this formula gives the power received over a radio j communication circuit. Let the transmitter T feed a power P t to a transmitting ; 1 H. T. Friis, “A Note on a Simple Transmission Formula, 1 Proc. IKE, 34, 254—256, 1946.
| L 2-25 FRIIS TRANSMISSION FORMULA 49 figure 217 Communication circuit with waves from transmitting antenna arriving at the receiving antenna by a direct path of length r.
antenna of effective aperture A tl . At a distance t a receiving antenna of effective aperture intercepts some of the power radiated by the transmitting antenna and delivers it to the receiver R. Assuming for the moment that the transmitting antenna is isotropic, the power per unit area at the receiving antenna is If the antenna has gain G IS the power per unit area at the receiving antenna will be increased in proportion as given by S, =^ (W) (2) Now the power cd&ctad by the receiving antenna of effective aperture is /-r = S,A„=^£^ (W) (3) From (2-22-10) the gain of the transmitting antenna can be expressed as Substituting this in (3) yields the Friis transmission formula p _ p r r r t 2 1 2 where Pr = received power (antenna matched), W P, = power into transmitting antenna, W A tl = effective aperture of transmitting antenna, m 2 A tt = effective aperture of receiving antenna, m2 r = distance between antennas, m X — wavelength, m It is assumed that each antenna is in the far field of the other.
50 2 BASIC ANTENNA CONCEPTS Space quantities Circuit quantities Radiation resistance, Rr Antenna temperature, TA Current distribution Transition region Field j patterns £{0 t>) Power patterns 1 pje.
Beam Solid angle, Directivity, D Gain, G Effective aperture, A f .Scattering aperture,^ Figure 2-18 Schematic diagram of basic antenna parameters, illustrating the duality of an antenna:!
a circuit device (with a resistance and temperature) on the one hand and a space device (with radi-ation patterns, beam angles, directivity, gain and aperture) on the other.
2-26 DUALITY OF ANTENNAS The duality of an antenna, as a circuit j device on the one hand and a space device on the other, is illustrated schemati-] cally in Fig. 2-18, ] 2-27 SOURCES OF RADIATION: RADIATION RESULTS FROM ACCELERATED CHARGES A stationary electric charge does not radiate, (Fig. 2-19 a) and neither does an electric charge moving at uniform velocity along a straight wire (Fig. 2-196).
1 However, if the charge is accelerated, i.e., its velocity!
changes with time, it radiates. Thus, as in Fig. 2-19e, a charge reversing direction!
on reflection from the end of a wire radiates. The shorter the pulse for a given charge, the greater the acceleration and the greater the power radiated, or, as in' Fig. 2-19d, a charge moving at uniform velocity along a curved or bent wire is accelerated and radiates.
Consider a pulse of electric charge moving along a straight conductor in the!
x direction, as in Fig. 2-20. This moving charge constitutes a momentary electric!
current / as given by I — Ql where qL = charge per unit length, C m 1 This can be seen from relativistic considerations, since, for an observer m a reference frame moving with the charge, it will appear stationary, 1 SOURCES OF RADIATION: RADIATION RESULTS FROM ACCELERATED C HA ROES Static electric charge does not radiate.
\ X =h i —^ + + / x X / ++ Electric charge moving with uniform velocity along a straight wire does not radiate.
However, when charge reaches end of wire and reverses direction it undergoes acceleration and radiates.
The shorter or more compact the pulse of charge, the stronger the radiation.
Electric charge moving at uniform velocity l' along a curved or bent wire is accelerated and radiates.
Electric charge oscillating beck and forth in simple harmonic motion along a wire undergoes periodic acceleration and radiates.
Figure 2-19 A static electric charge or a charge moving with uniform velocity in a straight line does not radiate. An accelerated charge, however^ does radlate.
Multiplying by the length ( of the pulse as measured along the conductor Ii ^ qLl !i = qv bubble with velocity v = c in free space. The charges are assumed to move with' ( With radiation from the curved section, the energy of th® P11^ decreases as energy is j ost t0 radi-ation according to (2-27-5). Thus, stated another way, it is assumed that due to pr j0r rac|j a|i(m negligible charge reaches the open end, being absorbed in radiation resistance. Energy | ost lR at ion resistance is energy radiated.
2-30 RADIATION FROM PULSED CENTER-FED DIPOLE ANTENNAS 55 Figure 2-12 Oscillating electric dipole consisting of two electric charges in simple harmonic motion, showing propagation of an electric held line and its detachment (radiation) from the dipole. Arrows next to the dipole indicate current (/) direction-velocity v = c along the dipole. At the next stage [(a) middle] the charges reach the ends of the dipole, are reflected (bounce back) and move inward toward the generator [(a) bottom].
If the generator is an impedance match, the pulses are absorbed at the generator but the field lines join, initiating a new pulse from the center of the dipole with the pulse fields somewhat later, as shown in (ft).
Maximum radiation is broadside to the dipole and zero on axis as with a harmonically excited dipole. Broadside to the dipole (0 = 90°) there is a sym-metrical pulse triplet, but, at an angle such as 30° from broadside (0 = 60°), the middle pulse of the triplet splits into two pulses so that the triplet becomes a quadruplet as shown in (ft). Thus, the pulse pattern is a function of angle. The 2-30 RADIATION FROM PULSED CENTER-FED DIPOLE ANTENNAS 59 time T between pulses 1 and 3 is the time required for the charges to travel out and back from the generator. This is the same as the travel time over the length L of the dipole, or r = -(l) C Thus, the dipole length determines the pulse spacing T while the pulse length t determines the much shorter wavelength of the pulse radiation.
If the generator is not an impedance match, then the charges will continue to bounce back and forth on the dipole, resulting in a longer pulse train as suggested in (c) for the case of a poor match (some pulse energy is absorbed in the generator but more is reflected). The first 5 pulses broadside (7 at 30° from broadside) of an indefinitely long (gradually damping) pulse train are indicated.
It is evident from Fig, 2-24 that radiation occurs from the points where charge is accelerated, i,e., at the center or feed point and at the ends of the dipole but not along the dipole itself.
1 ] G. Franceschetti and C. H. Papas, “Pulsed Antennas," Sensor and Simulation Note 203, Cal. Tech,, 1973 .
60 2 BASFC ANTFVNA CONCEPTS 2-31 ANTENNA FIELD ZONES The fields around an antenna may be divided into two principal regions, one near the antenna called the near field or Fresnel zone and one at a large distance called the far field or Fraunhofer zone.
Referring to Fig. 2-25, the boundary between the two may be arbitrarily taken to j be at a radius D 21} R =~ (m) 0) where L = maximum dimension of the antenna, m X = wavelength, m In the far or Fraunhofer region, the measurable field components are trans-, verse to the radial direction from the antenna and all power flow is directed radially outward. In the far field the shape of the field pattern is independent of the distance.
In the near or Fresnel region, the longitudinal component of the I electric field may be significant and power flow is not entirely radial. In the near field, the shape of the field pattern depends, in general, on the distance.
Enclosing the antenna in an imaginary boundary sphere as in Fig. 2-26a, it ; is as though the region near the poles of the sphere acts as a reflector. On the ; other hand, the waves expanding perpendicular to the dipole in the equatorial region of the sphere result in power leakage through the sphere as if partially ^ transparent in this region.
J This results in reciprocating (oscillating) energy flow near the antenna accompanied by outward flow in the equatorial region. The outflow accounts for the power radiated from the antenna, while the reciprocating energy represents reactive power that is trapped near the antenna like in a resonator. This over-simplified discussion accounts in a qualitative way for the field pattern of the X/2 \ Flgm 2-25 Antenna region, Fresnel region and Fraunhofer region.
dipole antenna as shown in Fig, 2-26b. The energy picture is discussed in more detail in Sec. 5-2 and displayed in Fig.
5-7.
For a Xjl dipole antenna, the energy is stored at one instant of time in the electric field, mainly near the ends of the antenna or maximum charge regions, while a ^-period later the energy is stored in the magnetic field mainly near the center of the antenna or maximum current region.
Note that although the term power flow is sometimes used, it is actually energy which flows, power being the time rate of energy flow, A similar loose usage occurs when we say we pay a power bill, when, in fact, we are actually paying for electric energy.
2-32 SHAPE-IMPEDANCE CONSIDERATIONS.
It is possible in many cases to deduce the qualitative behavior of an antenna from its shape. This may be illustrated with the aid of Fig, 2-27. Starting with the opened-out two-conductor transmission line of Fig. 2- 27a, we find that, if extended far enough, a nearly constant impedance will be provided at the input (left) end for d ^ X and D > X.
62 2 BASIC ANTENNA CONCEPTS Flgire 2-27 Evolution of a thin cylindrical antenna (d) from an opened-out twin line (a). Curving the conductors as in (?) results in the spiral antenna.
In Fig. 2-27fc, the curved conductors are straightened into regular cones and in Fig. 2-27c the cones are aligned colinearly forming a biconical antenna.
In Fig. 2-27d the cones degenerate into straight wires. In going from Fig. 2~27u to d> the bandwidth of relatively constant impedance tends to decrease. Another differ-ence is that the antennas of Fig, 2-27a and b are unidirectional with beams to the right, while the antennas of Fig. 2-27c and d are omnidirectional in the horizon-tal plane (perpendicular to the wire or cone axes), A different modification is shown in Fig. 2-27e. Here the two conductors are curved more sharply and in opposite directions, resulting in a spiral antenna with maximum radiation broadside (perpendicular to the page) and with polariz-2 32 SHAPE-IMPEDANCE CONSIDERATIONS 63 Figm 2-28 Evolution of stub (monopole) antenna (e) from volcano-smoke antenna (a).
ation which rotates clockwise. This antenna, like the one in Fig. 2-27o, exhibits very broadband characteristics (see Chap.
15).
The dipole antennas of Fig. 2-27 are balanced, i.e., they are fed by two-conductor (balanced) transmission lines. Figure 2-28 illustrates a similar evolu-tion of monopole antennas, i.e., antennas fed from coaxial (unbalanced) transmission lines.
By gradually tapering the inner and outer conductors of a coaxial transmis-sion line, a very wide band antenna with an appearance reminiscent of a volcanic crater and puff of smoke is obtained, as suggested in the cutaway view of Fig. 2-2 8a.
In Fig.
2-28f?
the volcano form is modified into a double dish and in Fig. 2-28c into two wide-angle cones. All of these antennas are omnidirectional m 64 2 BASJC ANTENNA CONCEPTS a plane perpendicular to their axes and all have a wide bandwidth. For example an actual biconical antenna as in Fig. 2-28c, with a full cone angle of 120' has an omnidirectional pattern and nearly constant 50-ft input impedance (power reflec-tion less than 1 percent or VSWR < 1.2) over a 6 to 1 bandwidth with cone diameter D — 2 at the lowest frequency.
1 Increasing the lower cone angle to ISO 11 or into a flat ground plane while reducing the upper cone angle results in the antenna of Fig, 2-28d. Collapsing the upper cone into a thin stub we arrive at the extreme modification of Fig. 2-28e. If the antenna of Fig. 2-28a is regarded as the most basic form, the stub type of Fig. 2-2%e is the most degenerate form with a relatively narrow bandwidth.
As we depart further from the basic type the discontinuity in the Eransmis-sion line becomes more abrupt at what eventually becomes the junction of the ground plane and the coaxial line. This discontinuity results in some energy being reflected back into the tine. The reflection at the end of the antenna also increases for thinner antennas. At some frequency the two reflections may compensate but the bandwidth of compensation is narrow.
Antennas with large and abrupt discontinuities have large reflections and act as reflectionless transducers only over narrow frequency bands where the reflections cancel. Antennas with discontinuities that are small and gradual have small reflections and are in general relatively reflectionless transducers over wide frequency bands.
2-33 ANTENNAS AND TRANSMISSION LINES COMPARED. A uniform transmission line has a constant characteristic impedance determined by the geometry of its cross section. Thus, the space between the conductors of the ; coaxial line of Fig, 2-29a can be mapped into curvilinear squares with each !
square having the characteristic resistance of space ( = 376.7 377 Cl).
2 Therefore this line has a characteristic resistance of 1 ‘ 377 Z0 =— = 68.5 11 (1) : If the line is cut and terminated in a load of 68.5 fi there would be, ideally, : no reflection (line matched). Put another way if the line is cut and terminated in : a load of 5 resistors of 377 ft, one for each full square, and two 377 Cl resistors in I D.
A.
McNamara D.
E, Baker and L Botha "Some Design Considerations for Biconical Antennas IEEE Ants . Prop . inf. Sjmjp, Digest 173, 1984, I I D. Kraus Electromagnetics , 3rd ed, 5 McGraw-Hill 1984, sec. 3-19.
The analytical (boundary-value) solution (medium air or vacuum) is ZQ = 138 log “ ffl) = 68.5 Q a where b = inner diameter of the outer conductor = 62.7 mm a — outer diameter of the inner conductor = 20.0 mm 2 H ANTENNAS AND TRANSMISSION LINES COMPARED 65 () Curvilinear square (b) Elgin 229 Coaxial transmission line with field map (aX and terminated with an array of 377 resistors one for each curvilinear square (6), The total resistance of the array is 68.5 a series for the ^ square, as in Fig. 2-29b> the line would be matched. A sheet of space cloth (resistance 377 ft per square) connected across the end of the line would also behave as a matched load.
Now consider an infinite biconical antenna, as in Fig. 230a. With a simple graphical field map drawn on a spherical surface, as in Fig. 2- 30b, the character-istic impedance of the biconical antenna can be obtained and also shown to be a constant, i.e., independent of the distance from the terminals.
Let spherical space be divided into 15° sectors in azimuth (^). Now con-sidering a half-cone angle 9 = 3G D , the map will require about 5 squares in series from equator to cone and 24 squares in parallel. The map below the equator (not shown) is a mirror image of the one above. Thus, the characteristic resistance of this infinite biconical antenna is z0 = 5 i 377 -^r 377 = 157 « (2) Np 24 where Nt = number of squares in series (from cone to cone) Np = number of squares in parallel SchelkunofTs formula (8-2-20) gives a characteristic resistance of 158.0 ft for a half-cone angle of 30° From spherical geometry it follows that the characteristic impedance of an infinite biconical antenna is a (radiation) resistance of constant value since the 66 7 BASIC ANTENNA CONCEPTS 30 / 377 0 resistor Figure 2-30 Infinite biconical antenna of half-cone angle 30“ (a) with field cells (squares) (ft).
ratio of the curvilinear squares in series to the number in parallel is constant, th ‘ solid angle subtended by each square being independent of radius.
Figure 2-31 illustrates some of the similarities and differences of antennal and transmission lines. In Fig. 2-31 a an infinite biconical antenna with 0h = and characteristic resistance 160 12 is compared with an infinite uniform two-conductor transmission line (characteristic resistance 300 12). Waves traveling oul are entire|y of the Transverse ElectroMagnetic (TEM) type and th< VSWR = l for both.
) In Fig. 2-3 1 b both antenna and line have load resistors equal to their char-^ vow?
0 1 reistance connected as shown.' Beyond the load resistors the VSWR = 1, but between the feed points and the loads the VSWR == 2.
1 In Fig. 2-3 le the cones and the line are truncated at the load. On the line!
the VSWR = 1 (line matched) but the biconical antenna is not matched {VSWR # 1 on the cones).
If d ^ k there are no significant waves (radiation negligible) beyond the load at the end of the transmission line, but the biconical antenna i£ an opened-out radiating system, and higher-order mode waves exist?
beyond the load in the outer region.
^ 1 Instead of connecting a single 160-0 resistance between the cones as in Fig, 2-31(1, a grid of distribJ uted resistors would provide a better arrangement. For example, a 10 x 24 grid of 240 resistors could] be employed, one for each curvilinear square of Fig. 2-306 with the resistor for each square equal to!
377 n for a tqtdf resistance of 157 flf. The spheric] surface could also be covered with a continuous!
sneei of space cloth (377 fl per square).
2-J3 ANTENNAS AND TRANSMISSION LINES COMPARED 67 Equivalent toad Reflection 1 ^^ J/ ^\ ^ ^flection ^ ] 4 Reflection w \ / {d\ _V//fo= 360 G X } . Equivalent circuit for biconical antenna Figure 2-31 Biconical antenna and transmission line compared for several conditions.
In Fig. 2-3 Id the load resistors are removed, resulting in an infinite VSWR 0n the transmission line but a finite VSWR on the cones of the antenna, because the radiation field of the antenna acts like an equivalent load ZL . This load lmpedance transforms to an input (or terminal) impedance Z t — Z0 -¥jZ$ tan Z0 + tan as though the cones were a transmission line of length /, For the truncated cones there is reflection at the ends of the cones but no reflection of waves radiating in the equatorial plane.
0 p l For character quency o departs s: versus fn extremes.
1 n AIMTtNNAS AND TRANSMISSION LINES COMPARED 69 Polar or antenna axis Figure 2-33 Field configuration near dipole antenna, cones there is an abrupt discontinuity, while at the equator there is none- Hence, there is a large reflection at the end of the cones, and little energy is radiated in this direction. On the other hand, at the equator the energy' continues into the outer region, and radiation is a maximum in this direction.
The E lines of principal-mode fields must end on conductors and, hence, cannot exist in free space. The waves which can exist and propagate in free space are higher mode forms in which the E lines form closed loops- The principal-mode wave is called a zero-order wave, and higher-order waves are of order 1 and greater. The configuration of the £ lines of a first-order wave in the outer region is illustrated in Fig. 2-33, This wave has been radiated from a short dipole antenna. The wave started on the antenna as a principal-mode wave, has passed through the boundary sphere and has been transformed.
1 The field has a radial component which is largest near the polar axis. At the equatorial plane the radial component is zero and the E lines at this plane travel through the boundary sphere without change. Since the radial components of the field attenuate more rapidly than the transverse components, the radial field becomes negligible in comparison with the transverse field at a large distance from the antenna.
Although the field at a large distance from the antenna is of a higher-order type, 1 Some first-order mode is also present inside the antenna boundary sphere as a reflected wave. This and higher-order modes may caist both inside and outside of the boundary sphere in such a way that there is continuity of the fields at the boundary sphere.
70 2 BASIC ANTENNA CONCEPTS the measurable components are only of the transverse type To suggest the fact that the radial field components are weak and become negligible at large dis-tances, the £ lines in the polar region in Fig, 2-33 are dashed 2-34 WAVE POLARIZATION An important property of an electromag-netic wave is its polarization , a quantity describing the orientation of the electric field E l Consider a plane wave traveling out of the page (positive z direction) as in Fig. 2-34a, with the electric field at all times in the y direction. This wave is said to be linearly polarized (in the y direction). As a function of time and position the electric field of a linearly polarized wave (as in Fig. 2-34a) traveling in the posi-tive z direction (out of the page) is given by Ej, - E 2 sin (art — fiz) (1) In general, the electric field of a wave traveling in the z direction may have both a y component and an x component, as suggested in Fig 2-346.
In this more general situation the wave is said to be elliptically polarized. At a fixed value of z the electric vector E rotates as a function of time, the tip of the vector describing an ellipse called the polarization ellipse The ratio of the major to minor axes of the polarization ellipse is called the axial ratio (AR). Thus, for the wave in Fig 2- 34b, AR = E 2/E v Two extreme cases of elliptical polarization cor-respond to circular polarization, as in Fig 2- 34c, and linear polarization, as in Fig. 2-34a.
For circular polarization = Ez and AR = 1, while for linear polarization E t =0 and AR = qo.
In the most general case of elliptical polarization the polarization ellipse may have any orientation, as suggested in Fig 2-35 This eUiptically polarized wave may be expressed in terms of two linearly polarized components, one in the x direction and one in the y direction. Thus, if the wave is travelling in the positive z direction (out of the page), the electric field components in the x and y directions are Ex -E x sin (cor — fiz) (2) Ey = E2 sin (art — fiz + &} (3) where = amplitude of wave linearly polarized in x direction E 2 = amplitude of wave linearly polarized in y direction S — time-phase angle by which Ey leads Ex Combining (2) and (3) gives the instantaneous total vector field E: E = x£ t sin (art — /Jz) + y£ 2 sin (^ — fiz + 6) (4) J Thus, a Imeariy polarized wave with E vertical is called a vertiegify polarized wave, the accompany-ing magnetic field H being horizontal.
2 WAVE POLARIZATION 71 Linear Elliptical Circular polarization polarization polarization At z = 0, Ex — £j sin art and Ey = E2 sin (cur + S\ Expanding Ey yields Ey = E 2 (sin art cos <5 + cos art sin 6) (5) From the relation for Ex we have sin cut = EJE V and cos cut = ^/l — {EJE^) 2 .
Introducing these in (5) eliminates cut, and on rearranging we obtain El 2Ex E y cos El E\ £,£2 + E\ — sin 2 S aEl — bEx Ey + cEl = 1 E\ sin 2 6 2 cos S E tE2 sin 2 S E\ sin 2 $ Equation (7) describes a (polarization) ellipse, as in Fig.
2-35 The line segment OA is the semimajor axis and the line segment OB is the semiminor axis. The tilt angle of the ellipse is t The axial ratio is AR = — (1 < AR < oo) OB For E t = 0, the wave is linearly polarized in the y direction For E2 = 0, the wave is linearly polarized in the x direction.
If <5 = 0 and E x — £ 2 , the wave is also linearly polarized but in a plane at an angle of 45° with respect to the x axis (i = 45°) For Ei = £2 and 6 — ±90°, the wave is circularly polarized. When <5 = +90°, the wave is left-circularly polarized, and when & = —90°, the wave is right-circularly polarized For the case 6 = +90° and for z = 0 and f = 0 we have from (2) and (3) that E = y£2 , as in Fig. 2-36o One-quarter cycle later (art = 90°) E = xE t , as in Fig 2-366 Thus, at a fixed position (z = 0) the electric field vector rotates clockwise (viewing the wave approaching) According to the IEEE 72 2 BASIC ANTENNA CONCEPTS Figure 2-35 Polarization ellipse at lilt angle t showing instantaneous components E^ and Er and amplitudes (or peak values) E 1 and E2 r definition, this corresponds to left-circular polarization.
1 The opposite rotation direction (<5 = —90°) corresponds to right-circular polarization.
Polarization ellipses, as a function of the ratio E 1jE 1 and phase angle & {wave approaching), are shown in Fig 2-37, In special cases, the ellipses become straight lines (linear polarization) or circles (circular polarization).
If the wave is viewed receding (from negative z axis in Fig. 2-36), the electric vector appears to rotate in the opposite direction. Hence clockwise rotation of E with the wave approaching is the same as counterclockwise rotation with the wave receding. Thus, unless the wave direction is specified there is a possibility of ambiguity as to whether the wave is left- or right-handed. This can be avoided by defining the polarization with the aid of helical antennas (see Chap, 7). Thus a right-handed monofilar axial-mode helical antenna radiates (or receives) right-circular (IEEE) polarization. 2 A right-handed helix, like a right-handed screw, is right-handed regardless or the position from which it is viewed. There is no possi-bility here of ambiguity.
The concept of polarization extends to antennas. Thus, an antenna which radiates a linearly polarized , wave can be described as a linearly polarized 1 This IEEE definition is opposite to the classical optics definition.
2 A left-handed monofilar axial-mode helical antenna radiates (or receives) left-circular (IEEE) polarization.
2-35 WAVE POLARIZATION AND THE POYNTING VECTOR 73 Wave approaching y\ it Ll3/= 0 4 x uit — 90 Figure 2-36 Instantaneous orientation of electric field vector £ at two instants of time for a left-circularly polarized wave which is approaching (out of page).
antenna, as, for example, a dipole antenna. Likewise an antenna which radiates a circularly polarized wave can be called a circularly polarized antenna as, for example a monofilar axial-mode helical antenna (see Chap.
7).
2-35 WAVE POLARIZATION AND THE POYNTING VECTOR, In complex notation the Poynting vector is S = ±E x H (I) The average Poynting vector is the real part of (I), or S„v = Re S — ±Re E x H (2) 0 0 T Ei s l S 1 D i a 0 i a 1 Q l B 1 s c Right-handed Left-handed 1 s s u lZI u D S N Counter-Clockwise clockwise 2 S3 O S3 2?, £3 53 0 — 1 80 s - 135' - 90' - 45' O' & + 45 + 90‘ + 135’ + 180 H Figure 2-37 Polarization ellipses as a function of the ratio E2/£ : £ro phase angle S with wave approaching. Clockwise rotation of the resultant E corresponds to left-handed polarization (IEEE definition) while counterclockwise corresponds to right-handed polarization.
74 2 BASIC ANTENNA CONCEPTS Referring to Fig.
2-35, let the elliptically polarized wave have x and y com-ponents with a phase difference & as given by \ EI = E L eKa- fz' (3) j 'J Ey = E2 (4) ] At 2 = 0 the total electric field (vector) is then !
E = xEx ± y E y = ftE^ + y£2 + « (5) where x = unit vector in x direction y = unit vector in y direction j Note that E has two components each involving both a space vector and a time-phase factor (phasor, with a>f explicit).
t The H-field component associated with Ex is j Hy = H^e--® (6) j where £ is the phase lag of Hf with respect to E. The H-field component associ-; ated with Er is Hx = (7) \ The total H field (vector) at z = 0 for a wave traveling in the positive z direction i is then \ H = yHy - \HX = y - xH2 e i{ait+fi ~ i) (8) | The complex conjugate of H is equal to (8) except for the sign of the exponents; j that is, j H = - HH2 e'™-® (9) ] Substituting (5) and (9) in (2) gives the average Poynting vector at i = 0 as S,T = iRe [(X x y)Ex H - tf x j = iz Re (£,// + £,«') (10) j where £ is the unit vector in the z direction (direction of propagation of the wave).
's It follows that the average power of the wave per unit area is j S1¥ = %HE XH X Re eji + E 2 H 2 Re \ = ^EiH x +£ 2 H 2)cos £ (W m -2 ) (11) •; It is to be noted that SIV is independent of S.
In a lossless medium, £ — 0 (electric and magnetic fields in the time phase) : and Ei/H t = E2/H 2 = Z0 (where Z0 , the intrinsic impedance of the medium, is j real), and (1 1) reduces to S„ -£,», +£ 2 H2) = iHH + Hl)Z0 = iif/% (Wm 2 ) (12) .-j 1-& WAVE POLARIZATION AND THE POINCARg SPHERE 75 where H — + H\ is the amplitude of the total H field. We can also write Sav = 2 — = jy-(Wm' 2 } (13) where E = ^Je\ + E\ is the amplitude of the total E field.
Example. An elliptically polarized wave travelling in the positive z direction in air has x and y components Ex = 3 sin (on - pz) (Vm“ l ) Er = 6 sin (cor - pz + 75°) (Vm -1 ) Find the average power per unit area conveyed by the wave.
Solution. The average power per unit area is equal to the average Poynting vector, which from (13) has a magnitude 1 £2 1 £2 + Ej ' 2 Z0 2 Z0 From the stated conditions, the amplitude = 3 V m 1 and the amplitude E 2 — f V m^ 1 . Also, for air, Z0 = 376.7 fl Hence, 1 3 2 + 6 2 376.7 2-36 WAVE POLARIZATION AND THE POINCARE SPHERE, In the Poincare sphere 1 representation of wave polarization, the polarization state is described by a point on a sphere where the longitude and latitude of the point are related to parameters of the polarization ellipse (see Fig. 2-38) as follows: Longitude = 2t Latitude = 2e (1) where t = tilt angle, 0° <; z £ 180° £ = cot "" 1 ( + AR), -45° <;e<; +45° The axial ratio (AR) and angle £ are negative for right-handed and positive for left-handed (IEEE) polarization.
The polarization state described by a point on a sphere can also be expressed in terms of the angle subtended by the great circle drawn from a 1 H. Poincart, Thtari Alathemattque de la Luminiire, G, Carre, Paris, 1892.
G. A. Deschamps, “Geometrical Representation of the Polarization of a Plane Electromagnetic Wave, Froc. IRE, 54a May 1951.
4 76 2 BASIC ANTENNA CONCEPTS reference point on the equator and the angle between the great circle and the equator (see Fig. 2-38) as follows : Great -circle angle = 2y Equator-to-great -circle angle = <5 (2) where y = tan -1 (E 1/E 1\ 0° ^ y < 90° 6 = phase difference between Er and EX7 - 180 < S < + 180° The geometric relation of t, c and yto the polarization ellipse is illustrated in Fig. 2-39. The spherical trigonometric interrelations of t, e, y and S are as follows: Knowing e and r, one can determine y and b or vice versa.
It is convenient to describe the polarization state by either of the two sets of angles (e, r) or (y, 5) which describe a point on the Poincare sphere (Fig. 2-38). Let the polarization state as a function of £ and t be designated by M(£> r) or simply M and the polarization state as a function of y and <5 be designated by P(y, £) or simply P, as in Fig. 2-38. Two special cases are of interest.
Case 1. For $ = 0 or S = ± 180% Ex and Er are exactly in phase or out of phase, so that any point on the equator represents a state of linear polarization. At the origin (e = i = 0) the polarization is linear and in the x direction (t -= 0), as suggested in Fig. 2~40n. On the equator 90° to the right, the polarization is linear with a tilt angle 2-3 WAVE POLARIZATION AND THE POINCA£ SPHERE 77 Flgm 1-39 Polarization ellipse showing relation of angles a, y and t.
z = 45% while 180° from the origin the polarization is linear and in the y direction (t = 90%. See Fig. 2-4Gu and 6. One octant of the Poincare sphere is shown in Fig. 2-40a, the full sphere being shown in Fig, 2-406 in rectangular projection.
Case 2. For <5 — ±90° and E 2 = E t [2y — 90 r and 2e = ±90°) and Ey have equal amplitudes but are in phase quadrature, which is the condition for circular polariza-tion. Thus, the poles represent a state of circular polarization, the upper pole rep-resenting left-circular polarization and the lower pole right-circular {IEEE) polarization, as suggested in Fig. 2-40a and b.
Cases 1 and 2 represent limiting conditions. In the general case, any point on the upper hemisphere describes a left-eUiptieally polarized wave ranging from pure left circular at the pole to linear at the equator. Likewise, any point on the lower hemisphere describes a right-eUiptically polarized wave ranging from pure right circular at the pole to linear at the equator.
Several elliptical states of polarization are shown by ellipses with appropriate tilt angles r and axial ratios AR at points on the Poincare sphere in Fig. 2-40a and b.
As an application of the Poincare sphere representation, it may be shown that the voltage response V of an antenna to a wave of arbitrary polarization is given by J 1 G. Sinclair, “The Transmission and Reception of Elliptically Polarized Waves” Proc. IRE, 38, 151, 1950.
78 2 BASIC ANTENNA CONCEPTS 2-37 CROSS-FIELD 79 Right circular polarization f h ) Figure 2-40 (a) One octant of Poincare sphere with polarization stales. (6) Rectangular projection of Poincare sphere showing full range of polarization states.
where MMa = angle subtended by great-circle line from polarization state M to Ma M — polarization state of wave Ma = polarization state of antenna k = constant The polarization state of the antenna is defined as the polarization state of the wave radiated by the antenna when it is transmitting. The factor in (4) involves the field strength of the wave and the size of the antenna. An important result to note is that if MM, = 0°, the antenna is matched to the wave (polarization state of wave same as for antenna) and the response is maximized.
However, if MM = 180°, the response is zero. This can occur, for example, if the wave is linearly polarized in the y direction while the antenna is linearly polarized in the x direction; or if the wave is left-circularly polarized while the antenna is right-circularly polarized. More generally we may say that an antenna is blind to a wave of opposite (or antipodal) polarization state.
A more complete discussion of polarization including unpolarized waves , partial polarization , Stokes parameters and the wave-to-antenna coupling factor is given by Kraus.
L 2-37 CROSS-FIELD For elliptical and circular polarization, the electric field vector E at a fixed point rotates with time in a plane perpendicular to the direction of wave propagation. There are situations, however, where E rotates in a plane parallel to the direction of wave propagation. This condition is called cross-field.
2 This situation can occur if there is a component of E in the direction of propagation. This condition never exists in the case of a single plane wave in free space since such a wave has no field component in the direction of propaga-tion. However, in the near field of an antenna there are field components in both the direction of propagation and normal to this direction so that cross-field is present. This is the case, for example, in the near field of a dipole antenna with field components E r and E9 as suggested in Fig. 2-41a.
/ iff) (t>) E-line Wave direction A,r E\ Conducting medium ic) Figure 241 Three situations in which cross-field is present: (t0 in the near field of a dipole antenna and fb) in the region exposed to radiation from two dipole antennas. At (c) cross-field is present near the surface of a conducting medium along which a plane wave is traveling.
1 J. D, Kraus. Radio Astronomy, 2nd cd,, Cygnus-Quasar, 198G, chap. 4; J. D. Kraus, Electromag-netics 3rd ed., McGraw-Hill, 1984, sec.
1 1-5.
1 A, Alford, J. D. Kraus and E. C Barkofsky, Very High Frequency Techniques, McGraw-Hill, 1947, chap. 9, p. 200.
80 2 BASIC ANTENNA CONCEPTS Perfect conductor Figure 2-42 Circular cross-field above a perfectly conducting plane at a point where the incident and reflected linearly polarized waves are in time-phase quadrature.
Cross-field may also be present where two waves of the same frequency and traveling in different directions cross. Thus, in the region exposed to radiation fr6m two dipole antennas, as in Fig, 2-416, there is cross-field- The region may be in the far fields of the antennas. Both antennas are linearly polarized in the plane of the page and both radiate at the same frequency (both connected to the same generator). In general, the tip of the E vector describes a locus that is an ellipse in a manner similar to that in elliptical polarization except that E is confined to the plane of the antennas (plane of the page).
Another situation in which cross-field is present is near the surface of a conducting medium along which a plane wave is traveling (Fig, 2-41c). Unless the medium is perfectly conducting, the E field is tilted forward near the surface of the medium so that E has components both normal to the surface (E„) and paral-lel or tangential to the surface (£,). Since, in general, these components are not in time-phase, elliptical cross-field is present, 1 Figure 2-42 illustrates the presence of cross-field at a point P above a perfect fiat conductor with a linearly polarized plane wave incident at an angle of 45°. At a distance 0.177 wavelength above the conductor the reflected and inci-dent waves cross at right angles and are in time-phase quadrature. This results in circular cross-field at P as suggested in the figure.
With pure cross-field, the tip of E describes, in general, an ellipse (elliptical cross-field) which in special cases may become a straight line (linear cross-field) or a circle (circular cross-field). This may be demonstrated analytically in a manner similar to that used in Sec. 2-34 for wave polarization. 2 Finally, consider the situation at a point where two circularly polarized waves cross at right angles as shown in Fig. 2-43, If the waves are of equal ampli-tude and the same frequency, the loci of E at P are ellipses (with AR = ^fl) which project in the plane of the page as a line AA (ellipse seen edge-on) or as a 1 J. D. Kraus Electromagnetics, 3rd ed., McGraw-Hill, 1984, p. 585.
1 J. D. Kraus, Electromagnetics, Sst ed., McGraw-Hill, 1953, p. 385.
2-3$ TABLE SUMMARIZING IMPORTANT RELATIONS OF CHAPTER 2 81 Direction of CP waves Figaro 2-43 Two orthogonal monofilar axial-mode helical antennas producing equal circu-larly polarized fields result in loci of E at P that are ellipses which project as line AA or as circle C in plane of page depending on phase.
circle C (ellipse seen at a slant angle) depending on the time phase. These loci represent neither pure cross-field nor pure polarization but a combination or hybrid situation; however, provided the two waves are of the same frequency, the locus of the tip of E always lies in a plane and, in general, describes an ellipse.
2-38 TABLE SUMMARIZING IMPORTANT RELATIONS OF CHAPTER 2.
Wavelenglh-freq uency V A = j (m) Beam area flj = $) dft (sr or deg 2 ) Beam area (approx.) - 0Hp ^hp (sr or deg 2 ) Beam efficiency =— (dimensionless) Directivity „ W tf>) $(0, A) & c (dimensionless) U m ^av Directivity 4te D — — (dimensionless) Directivity 4x4, 0 (dimensionless) Directivity (approx.) „ 4tt 41 000 u ~ a—T~~ ~ no jld" (dimensionless) &HP Directivity (better approx.) 41 OOPeju t (dimensionless) t See (3-13-18).
82 2 BASIC ANTENNA CONCEPTS Gain G — kD (dimensionless) Effective aperture and beam area A, 11, = k1 (mJ ) Scattering aperture (antenna matched) At = At (m2 or X2 ) Scattering aperture (antenna short -circuited) A, = 4A hT (m2 or X2 ) Scattering aperture (antenna open-circuited, small antenna) (m2 or a2 ) Collecting aperture A c = At + At + A l (m 2 or X 2 ) Aperture efficiency £ - — (dimensionless) Effective height V h‘ = T V Friis transmission formula <> Current-charge continuity relation ll = qb (A ms -1 ) Radiation power Near-field-far-field boundary Average power per unit area of elliptically polarized wave in air uzQ2 b2 m ojtZ 2L 1 J , k “-t (ra) S„ = ii^-=~i fW m I ) PROBLEMS 1 2-1 Directivity. Show that the directivity D of an antenna may be written m ri Z D= j ff 4lt JJx, z 2-2 Directivity.
Show that the directivity of an antenna may be expressed as 4 |j £<, y) dx dy || E(x, y) dx dy X2 || £{.x, y)£(x, >') dx dy where E{x , y) is the aperture field distribution.
2-3 Effective aperture. What is the maximum effective aperture (approximately) for a beam antenna having half-power widths of 30° and 35° in perpendicular planes intersecting in the beam axis? Minor lobes are small and may be neglected.
Answers to starred () problems are given in App. D.
problems 83 1-4 Effective aperture. What is the maximum effective aperture of a microwave antenna with a directivity of 900?
2-5 Received power. What is the maximum power received at a distance of ) phase quadrature (90 Q difference in phase) and (c) phase octature (45° difference in phase)?
2-11 Two LP waves. A wave traveling normally out of the page (toward the reader) has two linearly polarized components Ex — 2 cos cot Er = 3 cos (cut + 90°) {a) What is the axial ratio of the resultant wave?
(b) What is the tilt angle r of the major axis of the polarization ellipse?
(c) Does E rotate clockwise or counterclockwise?
2_|2 Two EP waves A wave traveling normally outward from the page (toward the reader) is the resultant of two elUptically polarized waves, one with components of E given by E' y = 2 cos E r x = 6 cos ^ 84 2 BASIC ANTENNA CONCEPTS and the other with components given by EJ — 1 cos cot El = 3 cos { tat — -V 2 (a) What is the axial ratio of the resultant wave?
(b) Does E rotate clockwise or counterclockwise?
2-13 Two LP components. An elliptical] y polarized plane wave traveling normally out of the page (toward the reader) has linearly polarized components Ex and Ey . Given that Ex — Et = IV m” 1 and that Ef leads Ex by 72, () Calculate and sketch the polarization ellipse.
() What is the axial ratio?
(c) What is the angle r between the major axis and the x axis?
214 Two LP components. Answer the same questions as in Prob. 2-13 for the case where Ey leads Ex by 72 as before but Ex = 2 V m 1 and Er — 1 V m 1 .
2-15 Two CP waves. Two circularly polarized waves intersect at the origin. One (y wave) is traveling in the positive y direction with E rotating clockwise as observed from a point on the positive y axis. The other ( x wave) is traveling in the positive x direc-tion with E rotating clockwise as observed from a point on the positive x axis. At the origin, E for the y wave is in the positive z direction at the same instant that E for the x wave is in the negative z direction. What is the locus of the resultant E vector at the origin?
2-16 Tilt angle. Show that the tilt angle t can be expressed as t — y tan 2£^£j cos ^ E\-E\ 2-17 Spaceship near moop. A spaceship at lunar distance from the earth transmits 2-GHz waves.
If a power of 10 W is radiated isotropically, find (a) the average Poynting vector at the earth, (6) the rms electric field E at the earth and (c) the time it takes for the radio waves to travel from the spaceship to the earth. (Take the earth-moon distance as 380 Mm.) (J) How many photons per unit area per second fall on the earth from the spaceship transmitter?
2-18 CP waves. A wave traveling normally out of the page is the resultant of two circu-larly polarized components £fi(k , — and Eleft = 2e -Awt + 9tn {V m" 1 ) t Find (a) the axial ratio AR, (6) the tilt angle t and (c) the hand of rotation (left or right).
2-19 EP wave. A wave traveling normally out of the page (toward the reader) is the resultant of two linearly polarized components E = 3 cos wt and Ep = 2 cos (out + 90°)l For the resultant wave find (a) the axial ratio AR, (£>} the tilt angle x and (c) the hand of rotation (left or right).
2-20 CP waves Two circularly polarized waves traveling normally out of the page have fields given by EJtf, - 2e“-M and E^, = 3c^(V m~ l ) (rms). For the resultant wave find (a) AR, (6) the hand of rotation and (c) the Poynting vector.
2-21 EP waves. A wave traveling normally out of the page is the resultant of two ellip-tically polarized (EP) waves, one with components Ex = 5 cos tot and E t = 3 sin cot and another with components Er — 3£Jw,r and E t = 4e For the resultant wave, find (a) AR, (6) t and (r) the hand of rotation PROBLEMS 85 2-22 CP waves. A wave traveling normally out of the page is the resultant of two circu-larly polarized components £, = 2^ and E t — + For the resultant wave find (a) AR, (6) x and (c) the hand of rotation.
2-23 More power with CP. Show that the average Poynting vector of a circularly pol-arized wave is twice that of a linearly polarized wave if the maximum electric field E is the same for both waves. This means that a medium can handle twice as much power before breakdown with circular polarization (CP) than with linear polariz-ation (LP).
2-24 PV constant for CP. Show that the instantaneous Poynting vector (PV) of a plane circularly polarized traveling wave is a constant.
2-25 EP wave power. An ellipticaliy polarized wave in a medium with constants a — 0, P, = 2, e, - 5 has H-field components (normal to the direction of propagation and normal to each other) of amplitudes 3 and 4 A m“ l . Find the average power con-veyed through an area of 5 m 2 normal to the direction of propagation.
2-26 Circular-depolarizatton ratio.
If the axial ratio oF a wave is AR, show that the circular-depolarization ratio of the wave is given by K = ™—1 AR + 1 Thus, for pure circular polarization AR = 1 and R = 0 (no depolarization) but for linear polarization AR = oo and R = l.
2-27 Super!mniiut phase velocity near dipole, (a) By measuring the distances between Pt F and P" determine the amount of superluminal {v > c) phase velocity of the waves near the dipole in Fig.
2-23.
(&) Under what other conditions are superluminal velocities encountered?
3-2 POWER PATTERNS 87 CHAPTER 3 POINT SOURCES 3-1 INTRODUCTION, POINT SOURCE DEFINED-Let us con-sider an antenna contained within a volume of radius b as in Fig. 3-la, Confining our attention only to the far field of the antenna, we may make observations of the fields along an observation circle of large radius /?. At this distance the mea-surable fields are entirely transverse, and the power flow, or Poynting vector, is entirely radial. It is convenient in many analyses to assume that the fields of the antenna are everywhere of this type. In fact, we may assume, by extrapolating inward along the radii of the circle, that the waves originate at a fictitious volumeless emitter, or point source, at the center 0 of the observation circle. The actual field variation near the antenna, or “near field,” is ignored, and we describe the source of the waves only in terms of the “far field' 1 it produces.
Provided that our observations are made at a sufficient distance, any antenna, regardless or its size or complexity, can be represented in this way by a single point source.
Instead of making field measurements around the observation circle with the antenna fixed, the equivalent effect may be obtained by making the measure-ments at a fixed point Q on the circle and rotating the antenna around the center O This is usually the more convenient procedure if the antenna is small.
In Fig.
3-1 a the center O of the antenna coincides with the center of the observation circle.
If the center of the antenna is displaced from 0, even to the extent that O lies outside the antenna as in Fig. 3-lh, the distance d between the two centers has a negligible effect on the field patterns at the observation circle 86 provided R > d, R > b, and R > h However, the phase patterns 1 will generally differ, depending on d.
If d = 0, the phase shift around the observation circle is usually a minimum. As d is increased, the observed phase shift becomes larger.
As discussed in Sec. 2-4, a complete description of the far field of a source requires three patterns : two patterns of orthogonal field components as a func-tion of angle [E(0, ) and EJfi, #)] and one pattern of the phase difference of these fields as a function of angle [<5(0, <£)]. For many purposes, however, such a complete knowledge is not necessary. It may suffice to specify only the variation with angle of the power density or Poynting vector magnitude (power per unit area) from the antenna [S^(0, <£)]. In this case the vector nature of the field is disregarded, and the radiation is treated as a scalar quantity. This is done in Sec. 3-2. The vector nature of the field is recognized later in the discussion on the magnitude of the field components in Sec. 3-16. Although the cases considered as examples in this chapter are hypothetical, they could be approximated by actual antennas.
3-2 POWER PATTERNS. Let a transmitting antenna in free space be rep-resented by a point-source radiator located at the origin of the coordinates in Fig. 3-2 (see also Fig. 2-5). The radiated energy streams from the source in radial lines. The time rate of energy flow per unit area is the Poynting vector, or power density (watts per square meter). For a point source (or in the far field of any antenna), the Poynting vector S has only a radial component Sr with no com-ponents in either the 0 or> directions (S = S+ — 0). Thus the magnitude of the Poynting vector, or power density, is equal to the radial component ( | S | = Sr).
1 Phase variation around the observation circle.
88 3 POINT SOURCES A source that radiates energy uniformly in all directions is an isotropic source .
For such a source the radial component Sr of the Poynting vector is independent of 8 and . A graph of Sf at a constant radius as a function of angle is a Poynting vector, or power-density, pattern, but is usually called a power pattern , The 3-dimensional power pattern for an isotropic source is a sphere- In 2 dimensions the pattern is a circle (a cross section through the sphere), as sug-gested in Fig. 3-3, 3-3 A POWER THEOREM 89 {a) {b) {£) (d) Figure 34 Power pattern (a), relative power pattern (bX radiation-intensity pattern (c) and relative radiation-intensity pattern fd) for the same directional or anisotropic source.
All patterns have the same shape. The relative power and radiation-intensity patterns {b and d) also have the same magni-tude and, hence, are identical.
Although the isotropic source is convenient in theory, it is not a physically realizable type. Even the simplest antennas have directional properties, i.e., they radiate more energy in some directions than in other?. In contrast to the isotropic source, they might be called anisotropic sources. As an example, the power pattern of such a source is shown in Fig, 3-4a where Srm is the maximum value of 5,-If St is expressed in watts per square meter, the graph is an absolute power pattern. On the other hand, if Sr is expressed in terms of its value in some refer-ence direction, the graph is a relative power pattern , It is customary to take the reference direction such that Sr is a maximum. Thus, the pattern radius for rela-tive power is SJS^ where is the maximum value of Sr . The maximum value of the relative power pattern is unity, as shown in Fig, 3-4b+ A pattern with a maximum of unity is also called a normalized pattern.
3-3 A POWER THEOREM 1 AND ITS APPLICATION TO AN ISOTROPIC SOURCE, If the Poynting vector is known at all points on a sphere of radius r from a point source in a lossless medium, the total power 1 This theorem is a special case of a more general relation for the complex power flow through any closed surface as given by i> = i x H») (1) where P is the total complex power flow and £ and H are complex vectors representing the electric and magnetic Adds, H being the complex conjugate of H. The average Poynting vector is $ = i Re (E x H) (2) Now the power flow in the far field is entirely real; hence, taking the real part of (1) and substituting (2X we obtain the special case of (3).
90 3 POINT SOURCES radiated by the source is the integral over the surface of the sphere of the radial component Sr of the average Poynting vector. Thus, P = = || V \ = element of solid angle, sr Thus, the power theorem may be restated as follows: The total power radiated is given by the integral of the radiation intensity U over a solid angle of 4n.
As already mentioned in Sec. 2-6, power patterns can be expressed in terms of either the Poynting vector (power density) or the radiation intensity. A power SOURCE WITH HEMISPHERIC POWER PATTERN 91 pattern in terms of U is shown in Fig. 3-4c. The maximum value Um is in the direction 8 = 0. A relative U/Um pattern has a maximum value of unity as shown in Fig. 3-4d. Relative power-density and radiation -intensity patterns are identical.
Applying (2) to an isotropic source gives P = 4nV 0 (W) (3) where U 0 = radiation intensity of isotropic source, W sr -1 .
3-5 SOURCE WITH HEMISPHERIC POWER PATTERN. As a further example let us consider a source with a power pattern which is a hemi-sphere; i.e., the radiation intensity equals a constant Um in the upper hemisphere and is zero in the lower hemisphere, as illustrated by the 3-dimensional diagram of Fig. 3-5a and its 2-dimensional cross section of Fig. 3-5b. Then, the total power radiated is the radiation intensity integrated over a hemisphere, or n /2 Um si I sin 8 d$ d\ “ 2nU„ Assuming that the total power P radiated by the hemispheric source is the same as the total power radiated by an isotropic source taken as a reference, (1) and (3-4-3) can be equated, yielding 2nUm = 4nU0 (2) — = 2 = directivity the ratio of Um to U0 in (3) equals ihe directivity of the hemispheric source.
The directivity of a source is equal to the ratip of its maximum radiation intensity to its average radiation intensity. The directivity of a source may also be stated as the ratio of its maximum radiation intensity^ to the radiation intensity of an iso-jtropic source radiating the same total power. By (3), the directivity of the hemi-spheric source is 2; that is to say, the power per unit solid angle Vm in one (> m w Figwt 3-5 Hemispheric power patterns, (a) and (/>), and comparison with isotropic pattern (c).
92 3 POINT SOURCES hemisphere from the hemispheric source is twice the power per unit solid angle U 0 from an isotropic source radiating the same total power. This we would expect, since a power P radiated uniformly over one hemisphere will give twice the power per unit solid angle as when radiated uniformly over both hemi-spheres, The power patterns of a hemispheric source and an isotropic source are compared in Fig. 3-5c for the same power radiated by both.
3-6 SOURCE WITH UNIDIRECTIONAL COSINE POWER PATTERN-Let us consider next a source with a cosine radiation-intensity pattern, that is, U = U m cos 9 (1) where U m = maximum radiation intensity The radiation intensity V has a value only in the upper hemisphere (0 < 8 ^ n/2 and 0 < < 2ji) and is zero in the lower hemisphere. The radiation intensity is a maximum at 9 - 0. The pattern is shown in Fig.
3-6. The space pattern is a figure of revolution of this circle around the polar axis.
To find the total power radiated by the cosine source, we apply (3-4-2) and integrate only over the upper hemisphere. Thus pa ri2 P = I Um cos 0 sin 9 dd = nUm (2) Jo jo If the power radiated by the unidirectional cosine source is the same as for an isotropic source, then (2) and (3-4-3) may be set equal, yielding U m = 4teU q or Directivity = — 4 (3) U0 Thus, the maximum radiation intensity Um of the unidirectional cosine source (in the direction 9 = 0) is 4 times the radiation intensity l/ 0 from an iso-tropic source radiating the same total power. The power patterns for the two sources are compared in Fig. 3-7 for the same total power radiated by each.
0 = 0 I Polar axis 1 JKL H alt-power point Figure 3-4 Unidirectional cosine power pattern.
i 3’B SOURCE WITH SJNE {DOUGHNUT} POWER PATTERN 93 Figure 3-7 Power patterns of unidirectional cosine source com-pared with isotropic source for same power radiated by both.
3-7 SOURCE WITH BIDIRECTIONAL COSINE POWER PATTERN.
Let us assume that the source has a cosine pattern as in the pre-ceding example but that the radiation intensity has a value in both hemispheres, instead of only in the upper one. The pattern is then as indicated by Fig, 3-8.
It follows that P is twice its value for the unidirectional cosine power pattern, and hence the directivity is 2 instead of 4.
3-8 SOURCE WITH SINE (DOUGHNUT) POWER PATTERN, Consider next a source having a radiation -in tensity pattern given by V=Um sm6 (1) The pattern is shown in Fig.
3-9, The space pattern is a figure of-revolution of this pattern around the polar axis and has the form of a doughnut. Applying (3-4-2), the total power radiated is P=Um \ [sxn 2 9 d9d~nU m (2) Jo Jo 0 = 0 Flgm 3-8 Bidirectional cosine power pattern.
Figure 3-9 Sine power pattern.
94 3 POINT SOURCES If the power radiated by this source is the same as for an isotropic source taken as reference, we have n2Va = 4nU 0 (3) U 4 and Directivity = ~ = - = 1.27 (4) U o 71 3-9 SOURCE WITH SINE-SQUARED (DOUGHNUT) POWER PATTERN. Next consider a source with a sine-squared radiation-intensity or power pattern. The radiation-intensity pattern is given by U - Um sin 2 0 (1) The power pattern is shown in Fig. 3-10. This type of pattern is of considerable interest because it is the pattern produced by a short dipole coincident with the polar (9 = 0) axis in Fig. 3-10. Applying (3-4-2), the total power radiated is p = P' psin 3 0Md+ = (2) Jo Jo If P is the same as for the isotropic source, 1/ 3 and Directivity =~ = - = 1,5 (3) t/Q 2 3-10 SOURCE WITH UNIDIRECTIONAL COSINE-SQUARED POWER PATTERN. Let us consider next the case of a source with a uni-directional cosine-squared radiation- intensity pattern as given by U = Um cos 2 9 (1) with the radiation intensity having a value only in the upper hemisphere. The pattern is shown in Fig. 3-11, The 3-dimensional or space pattern is a figure-of-0 = 0 9 = 0 oo Figure 3-10 Sine-squared power pattern Figure ^1 1 Unidirectional cosine-squared power pattern .
3-11 SOURCE WITH UNIDIRECTIONAL COSINE" POWER PATTERN 95 revolution of this pattern around the polar (0 = 0) axis and has the form of a prolate spheroid (football shape). The total power radiated is J 2w fn!2 cos 2 0 sin 0 d9 d = (2) 0 Jo If P is the same as radiated by an isotropic source, = 4l/0 and Directivity = — = 6 (3) Vo Thus, the maximum power per unit solid angle (at 0 = 0) from the source with the cosine-squared power pattern is six times the power per unit solid angle from an isotropic source radiating the same power.
3-11 SOURCE WITH UNIDIRECTIONAL COSINE" POWER PATTERN. A more general case for a unidirectional radiation-intensity pattern which is symmetrical around the polar (0 = 0) axis is given by U=UM cos" 0 (t) where n is any real number. In Fig.
3-12, relative radiation-intensity or power patterns plotted to the same maximum value are shown for the cases where n — 0, 1, 2, 3 and 4. The case for n = 0 is the same as the source with the hemispheric power pattern discussed in Sec. 3-5. The cases for n = 1 and n = 2 were treated in Secs. 3-6 and 3-10. When n ~ i, 3 and 4, the directivity is 3, 8 and 10, respectively, 1 These calculations are left to the reader as an exercise. A graph of the directivity of a unidirectional source as a function of n is presented in Fig. 3-13.
1 It may be shown that the directivity of sources with power patterns of the type given by (1) can be reduced to the simple expression, directivity = 2(n +-!). The proof is left to the reader as an exercise.
96 3 POINT SOURCES n Figure 5-13 Directivity versus n for unidirectional sources with cos 0 power patterns.
The half-power beam widths and exact directivity [f) = 2(n 4-1)] for cos" 0 patterns are listed in Table 3-1 for n values between 0 and 100. The directivity is also given in dBi (dB over isotropic) for both exact and approximate values, where the approximate value is based on the HPBW 2 from (2-9-4). The difference between the exact and approximate values is tabulated in the last column. The approximation is within ±02 dB of the exact value for n between 4 and 10, A further discussion of the (2-9-4) approximation is given following (17) of Sec. 3-13.
3-12 SOURCE WITH UNIDIRECTIONAL POWER PATTERN THAT IS NOT SYMMETRICAL.
All the patterns considered thus far have been symmetrical around the polar axis; i.e. n the space pattern could be constructed as a figure-of-revolution about the polar axis. Let us now consider a more general case in which the pattern is unidirectional but is unsymmetrical around its major axis. In discussing this type of pattern it will be convenient to shift the direction of the major axis or direction of maximum radiation from the polar (8 = 0) axis to a direction in the equatorial plane as shown in Fig. 3-14 (0 = 90°, (j> = 90°) The 8 = 90° plane coincides with the xy plane and the Table 3-1 Half-power beam widths and directivity of sources with unidirectional cos' 1 B power patterns.
rt HPBW, deg D Exact D, dBi Approx.
D, dBi Difference, dB 0.0 180.0 2 3.0 LI - 1.9 0.5 151.0 3 4.8 2.6 - 2.2 1 120.0 4 6.0 4.6 -1.4 2 90.0 6 7,8 7,1 - 0.7 3 74.9 8 9,0 8.7 — 0.3 4 65.5 10 10.0 9.8 - 0.2 5 59.0 12 10.8 10.8 0.0 6 54.0 14 11.5 11,5 0.0 8 47.0 18 L 2.6 12.7 0.1 10 42.2 22 13,4 13.6 0.2 20 30,0 42 16.2 16.6 0,4 50 19.0 102 20.1 20,6 0.5 too 13.5 202 23.1 23.5 0,4 i 1 M3 DIRECTIVITY 97 Figure 3-14 Unidirectional source radiating maximum power in the direction 0 — 9CT, ^ = 90" or along the y axis.
= 90 c plane with the yz plane. A rather general expression for the radiation intensity with its maximum at 8 = 90° and tp = 90° is then given by V = U m sin" 0 sin" 1 <p (1) where n = any real number m — any real number and the radiation intensity U has a value only in the right-hand hemisphere (Fig. 3-14) (0 < 8 < n; 0 < (p < tt). When m — n, (1) becomes the equation for a symmetrical power pattern of the same form as considered in Sec. 3-1 1 . When m and n are not the same, (1) represents the general case in which the pattern has different shapes in the 8 = 90° and and its maximum value by ,4) where U9 = a constant For the special case where m >— = i (5!
then 17,. = U9 and (3) can be written u = u„m 4>) w The average radiation intensity is _ p f f v.m ) (7i U°~4n~ te where P = total power radiated dft — sin 9 dO dip = element of solid angle The directivity D is then given by n _ Vjn _ V.m Ui 471/(9, 4>)„., «, Up ~ ff u.m ) dii Equation (8) can be reexpressed as 4n 4ti D= ffm 4>) dQ ~ nA M 4>L„ where tl, is defined as the beam area? or beam solid angle It is given by ifm hp t e half-power beam width in the plane. Then, neglecting the effect of minor lobes, we have approximately — 0HP $HP and, as already presented in (2-9-4), An An fljl 0HP ^HP 41 OOP ^HP ^HP where #h p = half-power beam width in 0 plane, rad = half-power beam width in plane, rad 0J P = half-power beam width in 9 plane, deg = half-power beam width in plane, deg - 4180/) square degrees - 41 253 square re.
the:re are 41 M square degrees in a sphere. For the approximate relation (16) the value is rounded off to 41 000.
100 3 POINT SOURCES For a pencil beam where 0HP = <£ KPl 4n 41000 ^hp (^hp) 2 (17) Equation (16) has been used very widely since appearing in the first edition, in most cases appropriately but sometimes inappropriately. Although accompa-nied by a three-paragraph footnote cautioning about its limitations, some authors have ignored these guidelines and argued that the numerator should be 26 358, 32 750 or 34250 instead of 41 253 square degrees, without realizing that the four cases involve four different classes of antennas.
My original rationale was that formulas tike (16) and (17) are useful in the context that if we know that an antenna has a l c pencil beam and small minor lobes, its gain is approximately 41 000 or 46 dB, give or take a decibel or two, but not that the result is accurate to of a decibel or better.
Some of the limitations of (16) are that: (1) the effect of minor lobes is neglected, (2) the angle product (0hp$hp) may not be rigorously related to the true solid angle of the main beam and (3) the angle product relation to the true solid angle varies according to the type of antenna pattern involved.
It is pointed out in the first edition footnote that by introducing a correc-tion factor the result can be improved. It is further pointed out that the value of this factor depends on the antenna type and may be relatively constant for a certain class of antennas.
Instead of introducing one correction factor, let us introduce two, 1 one involving the beam efficiency to correct for the minor lobes and the other to correct for pattern shape so that 41 kp&HP ^HP (18) where K = = beam efficiency - 0.75 ± 015 for most large antennas = pattern factor = 1.0 for a uniform field distribution across the antenna aperture For “ball-park” values (16) may suffice but (18) should be used for closer approximations. The effect of pattern shape is well illustrated by the comparison summarized in Table 3-1, Sec 3-11 For a Gaussian distribution (and beam) 2 1 J. D. Kraus, Radio Astronomy, 1st ed. s McGraw-Hill, 1966, p.
158; 2nd ed., Cygmis-Quasar, 1986, p. 6-6, 3 See Prob, 11-20, 3-14 SOURCE WITH PATTERN OF ARBITRARY SHAPE 101 3-14 SOURCE WITH PATTERN OF ARBITRARY SHAPE.
Exunple. Consider a pattern of arbitrary shape as shown in Fig. 3-16 in both polar and rectangular coordinates. The pattern has a pencil beam (symmetrical around the 6 = 0° axis) with a main-lobe HPBW of approximately 22° and four minor lobes. Find the directivity S&lution The directivity is given by where the denominator equals the total beam area , Since the pattern is symmetrical (no variation with 0), the integral with respect to $ yields 2 ji and (1) reduces to We have only the pattern graph available (no analytical expression) so let us divide the pattern (Fig 3- 1 6b) into 36 steps of 5 each. The approximate value of the into gral in the first (m = 1) 5 g section ( = ji/36 rad) is given by n .
, rt n 10 + 0.93 .
A -PJfl,)..
— sm2.5“ Figure 3-16b Power pattern with main lobe and several minor lobes Tor worked example calculation of directivity. The pattern is symmetri-cal around the 0 = 0 axis (vertical) with the 3-dimtnsionaJ pattern a figu'rc-of-revolution around this axis The same pattern PJfy is shown in rectangular coordinates in Fig. 3- 16b, Relative 102 J POINT SOURCES 1 lobe minor minor minor minor lobe lobe lobe lobe Iback lobe) Figure 3l6b Power pattern with main lobe and several minor lobes for worked example calculation of directivity. D equals the ratio of the dot-filled to cross-hatched areas. The same pattern in polar coordinates is shown in Fig. 3- 1 6a.
and the approximate directivity is then given by the summation of all 36 sections or by 4rc D ^ „ (4 ) 2^/36) £ PtfJ„ sin 0a Completing the summation we obtain D 4rt 4 jt _ 72 ^ ° " tT ~ 27t(7t/36X0.25 + 0.37 + 0.46 + 0.12 + 0,07) " 1.27 " (5) Main Firet Secoiul Third Fourth Lobe minor minor minor minor lobe Lobe tobe lobe (back lobe) or D 5= 12 6 dBi It is noteworthy that the second minor lobe contributes most to the total beam area, the first minor lobe almost as much and the main lobe less than either. Thus, the directivity is greatly affected by the minor lobes, which is a common situation with actual antennas. For this antenna pattern the beam efficiency is given by ^ = 0.20 " 1.27 If the second minor lobe were eliminated, the directivity would increase to 14.5 dBi (up 1.9 dB) and if both first and second minor lobes were eliminated the directivity would increase to 17.1 dBi (up 4,5 dB).
MS OATN 103 The directivity obtained in the above worked example is approximate By sufficiently reducing the step size (5° in the example), the summation can be made as precise as the available data will allow. Computation of this numerical integra-tion can be facilitated by using a computer.
The half-power beam width of the pattern in the example is about 22°.
Taking kp = 1 and as in (6), the approximate directivity is then 41 000 kp x HPBW 41000 x 0.2 ( 22°) 2 = 16.9 or 12.3 dBi (7) which is 0.3 dB less than obtained by the 36-step summation.
The beam area of an isotropic source equals 4n steradians, In Fig. 3- 16b this corresponds to the area A under the sin 0 curve. The beam area of the source in the worked example corresponds to the area a under the PJQ) sin 9 curve.
Thus, the directivity is simply Aja or the ratio of the area of the isotropic source to the area of the source being measured Hence, If the areas A and a are cut from a lead sheet of uniform thickness, the directivity equals the ratio of the weight of A to the weight of a.
GAIN, The definition of directivity in the preceding section is based entirely on the shape of the radiated power pattern. Antenna efficiency is not involved. The gain parameter does involve antenna efficiency. The gain 1 of an antenna is defined as q maximum radiation intensity maximum radiation intensity from a ^ reference antenna with same power input Any type of antenna may be taken as the reference. Often the reference is a linear a/2 antenna. Gain includes the effect of losses both in the antenna under con-sideration (subject antenna) and in the reference antenna.
In many situations it is convenient to assume that the reference antenna is an isotropic source of 100 percent efficiency. The gain so defined for the subject antenna is called the gain with respect to an isotropic source or maximum radiation intensity from subj ect antenna radiation intensity from (lossless) isotropic source with same power input The gain G as here defined is sometimes called power gain. This quantity is equal to the square of he gam in field intensity Gf . Thus, iff, is the maximum electric field intensity from the antenna at a large distance R and £0 is the maximum electric field intensity from the reference antenna with the same power inout at (he same distance it, then the power gain G is given by G = {EjE^f = Gj.
104 J POFNT SOURCES As given in (2-10-1), the gain with respect to the directivity is given by G = kD (3) where k = efficiency factor of antenna (0 < k < 1) D = directivity Thus, the gain of an antenna over a lossless isotropic source equals the directivity if the antenna is 100 percent efficient (fc = 1) but is less than the directivity if any Losses are present in the antenna (fc < 1). In decibels the gain over an isotropic source as in (2) is expressed as dBi. Directivity is always dBi.
3-16 FIELD PATTERNS, The discussion in the preceding sections is based on considerations of power. This has afforded a simplicity of analysis, since the power flow from a point source has only a radial component which can be con-sidered as a scalar quantity. To describe the field of a point source more com-pletely, we need to consider the electric field E and/or the magnetic field H (both vectors). For point sources we deal entirely with far fields so E and H are both entirely transverse to the wave direction, are perpendicular to each other, are in-phase and are related in magnitude by the intrinsic impedance of the medium (E/H = Z = 377 Q for free space). For our purposes it suffices to consider only one field vector and we arbitrarily choose the electric field E.
Since the Poynting vector around a point source is everywhere radial, it follows that the electric field is entirely transverse, having only E# and E com-ponents. The relation of the radial component Sr of the Poynting vector and the electric field components is illustrated by the spherical coordinate diagram of Fig, 3- 17a. The conditions characterizing the far field are then: 1 Poynting vector radial (5, component only) 2.
Electric field transverse (£ and components only) 3-lG F1E1D PATTERNS 105 The Poynting vector and the electric field at a point of the far field are related in the same manner as they are in a plane wave, since, if r is sufficiently large, a small section of the spherical wave front may be considered as a plane.
The relation between the average Poynting vector and the electric field at a point of the far field is 1 F 2 S'-3T (1 » where Z 0 = intrinsic impedance of medium and E = JeI + E% (2) where £ = amplitude of total electric field intensity E9 = amplitude of 0 component Eq — amplitude of component The field may be elliptically, linearly or circularly polarized.
If the field components are rms values, rather than amplitudes, the Poyn-ting vector is twice that given in (1).
A pattern showing the variation of the electric field intensity at a constant radius r as a function of angle (0, 4>) is called a field pattern. In presenting infor-mation concerning the far field of an antenna, it is customary to give the field patterns for the two components, E# and £^, of the electric field since the total electric field E can be obtained from the components by (2), but the components cannot be obtained from a knowledge of only £ When the field intensity is expressed in volts per meter, it is an absolute field pattern.
1 On the other hand, if the field intensity is expressed in units relative to its value in some reference direction, it is a relative field pattern . The reference direction is usually taken in the direction of maximum field intensity. The relative pattern of the Ee component is then given by and the relative pattern of the ^component is given by where = maximum value of E+m — maximum value of E+ 1 The magnitude depends on the radium varying inversely as the distance (E oc 1/r).
M7 PHASE PATTERNS 109 108 3 POINT SOURCES un (t» Figure 3-19 (a) Relative patterns of Ef and components of the electric field and the total field E for antenna of Example 3. (fr) Relative total power pattern.
For this antenna, we may speak of two types of power patterns. One type shows the power variation for one component of the electric field. Thus, the power in the E# component of the field is as shown by Fig, 3-18b and the power in the E component by Fig, 3-17h. The second type of power pattern shows the variation of the total power. This is proportional to the square of the total electric field intensity.
Accordingly, the relative total power pattern for the composite antenna is The relative pattern in the equatorial plane for the total power is, therefore, a circle of radius unity as illustrated by Fig. 3-19E.
We note in Fig. 3-1943 that at = 45 D the magnitudes of the two field com-ponents, E$ and E,, are equal. Depending on the time phase between Es and E, the field in this direction could be plane, elliptically or circularly polarized, but regard-less of phase the power is the same. To determine the type of polarization requires that the phase angle between Et and be known. This is discussed in the next section.
3-17 PHASE PATTERNS, Assuming that the field varies harmonically with time and that the frequency is known, the far field in all directions from a source may be completely specified by a knowledge of the following four quan-tities: 1 1 In general, for the near or far field, six quantities are required. These are E, , &+ and rf f each as a function of r, 9, and in addition the amplitude of the radial component of the electric field Er and its phase lag behind both as a function of r, 0. 4 Since Er ^ 0 in the far field, only the four quantities given are needed to describe completely the field in the Fraunhofer region.
1, Amplitude of the polar component E& of the electric field as a function of r, 0 and -2, Amplitude of the azimuthal component E of the electric field as a function of r, 8 and 3, Phase lag <5 of E# behind Ee as a function of 0 and 4 Phase lag r\ of either field component behind its value at a reference point as a function of r, 8 and Since we regard the field of a point source as a far field everywhere, the above four quantities can be considered as those required for a complete knowl-edge of the field of a point source.
If the amplitudes of the field components are known at a particular radius from a point source in free space, their amplitudes at all distances are known from the inverse-distance law. Thus, it is usually sufficient to specify Es and E+ as a function only of 8 and as, for example, by a set of field patterns.
As shown in the preceding sections, the amplitudes of the field components give us directly or indirectly a knowledge of the peak and effective values of the total field and Poynting vector. However, if both field components have a value, the polarization is indeterminate without a knowledge of the phase angle 5 between the field components. Focusing our attention on one field component, the phase angle r\ with respect to the phase at some reference point is a function of the radius and may also be a function of 8 and A knowledge of tj as a function of 8 and is essential when the fields of two or more point sources are to be added.
We now proceed to a discussion of the phase angles, <5 and /j, and of phase patterns for showing their variation. Let us consider three examples.
Example L Consider first a point source that radiates uniformly in the equatorial plane and has only an E + component of the electric field. Then at a distance r from the source, the instantaneous field in the equatorial plane is sin (tor - J?r) (1) r where E+ = rms value of ^ component of electric field intensity at unit radius from the source co = 2nf, where/ = frequency, Hz ff = 2nfk y where X = wavelength, m The relation given by (1) is the equation for the field of a spherical wave traveling radially outward from the source. The equation gives the instantaneous value of the field as a function of time and distance. The amplitude or peak value of the field is ^flEJr . The amplitude is independent of space angle (6 and ^>) but varies inversely with the distance r. The variation of the instantaneous field with distance for this example is illustrated by the upper graph in Fig. 3-20 in which the amplitude is taken as unity at a distance r. When r = 0, the variation of the instan-taneous field varies as sin cot. It is often convenient to take this variation as a refer-'no 3 POINT SOURCES Figure 3-20 Illustration for Ex-ample l.
Phase of Eq of point source radiating uniformly in plane is a function or r but is independent of tj>.
Phase lag ij increases linearly with distance r ence for the phase, designating it as the phase of the generator or source. The fact that the amplitude at r = 0 is infinite need not detract from using the phase at r = 0 as a reference. The phase at a distance r is then retarded behind that at the source by the angle pr. A phase retardation or lag of E+ with respect to a reference point will, in general, be designated as rj.
In the present case the reference point is the source; 1 hence, - pr =~ (rad) (2) A Thus, the phase lag tf increases linearly with the distance r from the source.
This is illustrated by the chart of phase lag versus distance in Fig. 3-20.
The phase lag r\ in this example is assumed to be independent of To demonstrate experimentally that tj depends on r but is independent of tf>, the arrangement shown at the lower left in Fig. 3-20 could be used. The outputs of two probes or small antennas are combined in a receiver. With both probes at or very near the same point, the receiver output is reduced to a minimum by adjusting the length of one of the probe cables. The voltages from the probes at the receiver are then in phase opposition. With one probe fixed in position, the other is then moved 1 in he phase is referred to some point at a distance rx from the source, then ft) becomes E# = (\/2 EJr) sin (cut — AJV where d — r — r.
, 3-17 PHASE PATTERNS 111 in such a way as to maintain a minimum output. The locus of points for minimum output constitutes a contour (or front) of constant phase. For the point source under consideration, each contour is a circle of constant radius with a separation of A between contours. The radius of the contours is then given by r j ± nX, where is the radius to the reference probe and n is any integer.
We may define a phase front as a (3 -dimensional) surface of constant or uniform phase. If our observation circle coincides with a phase front, then we have constant phase along it.
Example 2, Consider next the case of a point source that has only an E+ component and that radiates nonuniformly in the equatorial or ^ plane. Let the instantaneous value in the equatorial plane be given by flE E&i = — — cos $ sin (cut — pr) (3) where E^n = rms value of E component at unit radius in the direction of maximum field intensity Let a point at unit radius and in the direction tf> = 0 be taken as the reference for phase. Then at this radius, E# = cos sin tot (4) <= iacr Figure 3-21 Illustration for Example 2. Field pattern is shown at (a), the phase pattern ir> rectangular coordinates at (A) and in polar coordinates at fc).
112 3 POINT SOURCES ^17 PHASE PATTERNS 113 Figure 3-22 Const ant -phase contours for source of Example 2.
Setting sin cor = 1, the relative field pattern of the E + component as a function of is, therefore, E = cos 0 (5) as illustrated in Fig.
3-21a. A pattern of this type could be obtained by a short dipole coincident with the y axis at the origin. The phase lag rj as a function of $ is a step function, as shown in the rectangular graph of Fig. 3-21 b and in the polar graph of Fig. 3-21 c. The variation shown is at a constant radius with the phase in the direction tfi = 0 as a reference. We note that r\ has an apparent discontinuity of 180° as passes through 90° and 270°, since at these angles cos changes sign while passing through zero magnitude. The phase angle t\ is accordingly a contin-uous, linear function of r but a discontinuous, step function of & To demonstrate this variation experimentally, the two-probe arrangement described in Example 1 may be used. In practice, attenuators (not shown) would be desirable in the probe leads to equalize the probe outputs.
Referring to Fig 3-22, if both fixed and movable probes are in the lower quadrants (1 and 4), a set oF constant or equiphase circles is obtained with a radial separation of h If one probe is fixed in quadrant l while the upper quadrants are explored with the movable probe, a set of equiphase circles is obtained which have a radial separation of A but are displaced radially from the set in the lower quadrants by A/2. Thus, the constant-phase contours have an apparent discontinuity at the y axis, as shown in Fig. 3-22. The phase of the field of any linear antenna coincident with the y axis exhibits this discontinuity at the y axis.
1 1 It is to be noted that this phase change is actually a characteristic of the method of measurement, since by a second method no phase change may be observed between the upper and lower hemi-spheres.
In the second method the probe is moved from the upper to the lower hemisphere along a circular path in the xz plane at a constant radius from the source. However, for a linear antenna the second method is trivial since it is equivalent to totaling the antenna on its own axis with the probe at a fixed position.
Example 3, Consider lastly a point source which radiates a field with both £0 and E+ component in the equatorial plane, the instantaneous values being given by Et y/2 £.
r sin 4> sin (cur — fir) (6) and cos sin in - 0r -(7) Referring to Fig. 3-23, a field of the form of the Ee component in the equatorial plane could be produoed by a small loop at the origin oriented parallel to the yz plane. A field of the form of the E+ component in the equatorial plane could be produced by a short dipole at the origin coincident with the y axis. Let a point at unit radius in the first quadrant be taken as the reference for phase. Assuming that the loop and dipole radiate equal power.
Elm “ m Then at unit radius the relative patterns as a function of and t are given by and E0 , = sin sin tot E# = cos 0 sin = —cos (fi cos cur (9) (10) The relative field patterns in the equatorial plane are shown in Fig. 3-23. With the loop and dipole fed in-phase, their field components are in phase quadrature (& = te/2). In quadrants 1 and 3, E+ lags £ by 90°, while in quadrants 2 and 4, E+ leads E$ by 90°, The phase patterns in the equatorial plane for E0 and E^ are shown in polar form by Fig. 3-24 and in rectangular form by Fig. 3-25u.
Since E01 E4 and <5 are known, the polarization ellipses may be determined.
These polarization ellipses for different directions in the equatorial plane are shown in Fig. 3-25b.
It is to be noted that in quadrants t and 3, where lags E0 , the E Figure 323 Field patterns for source of Example 3.
114 3 POINT SOURCES vector rotates counterclockwise, while in quadrants 2 and 4, where E leads E ff , the rotation is clockwise.
At four angles the polarization is circular, E rotating counterclockwise at = 45° and 225° and rotating clockwise at tf> = 135 c and 315°. The polarization is linear at four angles, being horizontally polarized at 0° and 180° and vertically polarized at 90 s and 27(T. At all other angles the polarization is elliptical, but the power is constant as a function of (regardless of the polarization).
-oOO I OO o — o O 0 I 0 O o — O' 45' 90' 135' 180 225' 270' 315' 360" 22.5 67,5 4 Figure 3-25 Phase patterns in rectangular coordinates for source of Example 3 at (a) with polarization ellipses for every 22.5 th interval of 4 > at (b) r 3-1S GENERAL EQUATION FOR THE FIELD OF A POINT SOURCE 115 3-18 GENERAL EQUATION FOR THE FIELD OF A POINT SOURCE, Both components of the far field of a point source in free space vary inversely with the distance.
Therefore, in general, the two electric field com-ponents may be expressed as II II 0) r and (2) where E$m = rms value of £ fl component at unit radius in the direction of maximum field E$m = rms value of E component at unit radius in the direction of maximum field /i and f 2 are, in general, different functions of 9 and ) sin (o>r -tj) (3) and E# = ^ /2(& <£) sin (tut — tj — S) (4) where = fi{r - r t ) +/#, <£) <5 =U0, ) r = radius to field point {r, 0, ) = radius of point to which phase is referred / 3 and / 4 are, in general, different functions of 0 and The instantaneous value of the total electric field at a point (r, 9, ) due to a point source is the vector sum of the instantaneous values of the two com-ponents. That is, E| = ae E$i + a£^ (5) where a0 = unit vector in 6 direction a^ — unit vector in direction Substituting (3) and (4) into (5) then gives a general equation for the electric field 116 3 POINT SOURCES of a point source at any point (r, 0, ) as follows : Ej = ae^ £°" ) sin (tut -tj -(6) In this equation the instantaneous total electric field vector E ( is a function of both space and time; thus E« = /(r, 0, y t) (7) The far field is entirely specified by (6). When fx and f 2 are complicated expres-sions, it is often convenient to describe E by means of graphs for the four quan-tities E ff and S , as has been discussed.
It is assumed that the field varies harmonically with time and that the frequency is known.
PROBLEMS 1 Directivity.
(a) Calculate the exact directivity for three unidirectional sources having the fol-lowing power patterns: U = U m sin 9 sin 2 tj> U = U m sin 9 sin 3 $ U = V m sin 2 6 sin 3 U has a value only for 0 < 6 < n and 0 <, £ n and is zero elsewhere, (b) Calculate the approximate directivity from the product of the half-power beam widths for each of the sources.
(c) Tabulate the results for comparison.
3-2 Directivity Show that the directivity for a source with a unidirectional power pattern given by U = Um cos" 9 can be expressed as D = 2(n 4-1). U has a value only for 0 < 0 < njl and 0 < 4> <; and is zero elsewhere.
3-3 Solar power The earth receives from the sun 2.2 g cal min” 1 cm -2 .
What is the corresponding Poynting vector in watts per square meter?
(h) What is the power output of the sun, assuming that it is an isotropic soufcc?
(c) What is the rms field intensity at the earth due to the sun's radiation, assuming all the sun's energy is at a single frequency?
Note: 1 watt = 14.3 g cal min” 1 Distance earth to sun = 149 x 106 km 3-4 Directivity and minor lobes. Prove the following theorem: if the minor lobes of a radiation pattern remain constant as the beam width of the main lobe approaches zero, then the directivity of the antenna approaches a constant value as the beam width of the main lobe approaches zero.
1 Answers to starred () problems are given in App, D.
PROBLEMS 117 Directivity by integration.
r (a) Calculate by graphical integration or numerical methods the directivity of a source with a unidirectional power pattern given by U = cos 0. Compare th directivity value with the exact value. V has a value only for 0 < 0 5 a/2 and 0 £ ^ In and is zero elsewhere.
2 {b) Repeat for a unidirectional power pattern given by U = cos^ 0.
{c} Repeat for a unidirectional power pattern given by V = cos 9 .
M Directivity. Calculate the directivity for a source with relative field pattern E — cos 29 cos Or CHAPTER 4 ARRAYS OF POINT SOURCES 4-1 INTRODUCTION.
In Chap. 2 an antenna was treated as an aperture.
In Chap. 3 an antenna was considered as a single point source. In this chapter we continue with the point-source concept, but extend it to a consideration of arrays of point sources. This approach is of great value since the pattern of any antenna can be regarded as produced by an array of point sources. Much of the dis-cussion will concern arrays of isotropic point sources which may represent many different kinds of antennas. Arrays of nonisotropic but similar point sources are also treated, leading to the principle of pattern multiplication. From arrays of discrete point sources we proceed to continuous arrays of point sources and Huygens 1 principle.
4-2 ARRAYS OF TWO ISOTROPIC POINT SOURCES.
Let us intro uce the subject of arrays of point sources by considering the simplest situ-a ion, namely, that of two isotropic point sources. As illustrations, five cases involving two isotropic point sources will be discussed.
Ph ^5 Two IsolroP'c Point Sources of Same Amplitude and ase.
e first case we shall analyze is that of two isotropic point sources aving equal amplitudes and oscillating in the same phase. Let the two point urces, 1 and 2, be separated by a distance d and located symmetrically with spec to the origin of the coordinates as shown in Fig, 4-1 a. The angle is measured counterclockwise from the positive x axis. The origin of the coordi-118 U2 ARRAYS OF TWO ISOTROPIC POINT SOURCES 119 Figive 4-1 (a) Relation to coordinate system of 2 isotropic point sources separated by a distance d, (b) Vector addition of the fields from two isotropic point sources of equal amplitude and same phase located as in (a), {c) Field pattern of 2 isotropic point sources of equal amplitude and same phase located as in {a) for the case where the separation d is Xfi.
nates is taken as the reference for phase. Then at a distant point in the direction 4> the field from source 1 is retarded by %d T cos , while the field from source 2 is advanced by idr cos 0, where dr is the distance between the sources expressed in radians; that is, The total field at a large distance r in the direction (j> is then E = E0 e~M2 + E0 e +#> 2 (1) where ^ = dT cos ^ and the amplitude of the field components at the distance r is given by E0 1 120 4 ARRAYS OF POINT SOURCES The first term in (1) is the component of the field due to source 1 and the second term the component due to source 2 Equation (1) may be rewritten „ e+w + e-w E = 2E 0 .
— (2) which by a trigonometric identity is E - 2 £ 0 cos y = 2£ 0 cos ^y cos ^ This result may also be obtained with the aid of the vector diagram 1 shown in Fig. 4-1 fr, from which (3) follows directly. We note in Fig. 4-1 b that the phase of the total field E does not change as a function of \p, To normalize (3), that is, make its maximum value unity, set 2£0 = 1. Suppose further that d is >,/2. Then dr = n. Introducing these conditions into (3) gives E = cos cos (j^j (4) The field pattern of E versus <p as expressed by (4) is presented in Fig. 4-tc. The pattern is a bidirectional figure-of-eight with maxima along the y axis. The space pattern is doughnut-shaped, being a figure-of-revolution of this pattern around the x axis.
The same pattern can also be obtained by locating source I at the origin of the coordinates and source 2 at a distance d along the positive x axis as indicated in Fig. 4-2a. Taking now the field from source 1 as reference, the field from source 2 in the direction <p is advanced by dr cos <p, Thus, the total field £ at a large distance r is the vector sum of the fields from the two sources as given by E = E0 + E0 e +J ( 5) — dv COS ' The relation of these fields is indicated by the vector diagram of Fig. 4-2h. From the vector diagram the magnitude of the total field is £ = 2£ 0 cos y = 2£0 cos d T cos <p as obtained before in (3), The phase of the total field £ is, however, not constant in this case but is \jtf2, as also shown by rewriting (5) as £ = £0(l+) = 2£ 0 e/2 -) = 2E0 e»‘ 2 cos — 2 1 U is to be noted that the quantities represented here by vectors art not true space vectors but merely vector representations of the time phase (i.e. T phasors).
1 ARRAYS OF TWO ISOTROPIC POINT SOURCES 121 Figu 4-2 (a) Two isotropic point sources with the origin of the coordinate system coincident with one of the sources, (b) Vector addition of the fields from 2 isolropic point sources of equal amplitude and same phase located as in (#). (c) Phase of total field as a function of $ for 2 isotropic point sources of same amplitude and phase spaced Xf2 apart. The phase change is zero when referred to the center point of the array but is ^/2 as shown by the dashed curve when referred to source J.
Normalizing by setting 2£ n = 1, (7) becomes £ = e^ {1 cos y = cos y j\p/2 (8) In (8) the cosine factor gives the amplitude variation of E, and the exponential or angle factor gives the phase variation wrrii respect to source 1 as the reference.
The phase variation for the case of a/2 spacing (dT = n) is shown by the dashed line in Fig. 4-2c. Here the phase angle with respect to the phase of source 1 is given by ip/2 = (?i/2) cos <p, The magnitude variation for this case has already been presented in Fig 4-1 e. When the phase is referred to the point midway between the sources (Fig. 4-lu), there is no phase change around the array as shown by the solid line in Fig. 4-2e.
Thus, an observer at a fixed distance observes no phase change when the array is rotated {with respect to m of maximum field are obtained by setting the argument of (11) equal to ±(2k + \)n/2. Thus, ^ cos \m = ±(2k + 1) | (11a) where k — 0, 1, 2, 3 For k = 0, cos \„ = ± 1 and 4>m = 0° and 180°.
The null directions 0 are given by - cos c — ± kn For k = 0,^o = ±90°.
The half-power directions are given by ? cos = ±(2k + 1) 7 2 4 For = 0, <£ = ±60", ± 120°, The field pattern given by (11) is shown in Fig. 4-3. The pattern is a rela-tively broad figure-of-eight with the maximum field in the same direction as the line joining the sources {x axis). The space pattern is a figu re- of-revolution of this pattern around the x axis. The two sources, in this case, may be described as a simple type of “end-fire” array. In contrast to this pattern, the in-phase point sources produce a pattern with the maximum field normal to the line joining the sources, as shown in Fig. 4-lc. The two sources for this case may be described as a simple “ broadside ” type of array.
4-2c Case 3, Two Isotropic Point Sources of the Same Amplitude and in Phase Quadrature.
Let the two point sources be located as in Fig. 4- la. Taking the origin of the coordinates as the reference for phase, let source 1 be retarded 2 ARRAYS OF TWO ISOTROPIC POINT SOURCES 123 Figure 4-3 Relative field pattern for 2 isotropic point sources of the same amplitude but opposite phase, spaced X/2 apart.
by 45° and source 2 advanced by 45°. Then the total field in the direction ^ at a large distance r is given by E-E", [ +J(i^ + 1)] + exp + !)] (12) From (12) we obtain £ = 2£0 cos ^ + y cos ^ (13) Letting 2£ 0 = 1 and d = A/2, (13) becomes E = cos ^ ^ cos (14) The field pattern given by (14) is presented in Fig. 4-4. The space pattern is a figure-of-revolution of this pattern around the x axis. Most of the radiation is in the second and third quadrants.
It is interesting to note that the field in the direction 0 = 0° is the same as in the direction = 180°. The directions of maximum field are obtained by setting the argument of (14) equal to kn, where k = 0, 1, 2, 3 In this way we obtain n n .
.
- + - cos = kn (15) 124 4 ARRAYS OF POINT SOURCES Figure 4-4 Relative field pat-tern of 2 isotropic point sources of the same amplitude and in phase quadrature for a spacing of a/2 The source to the right leads that to the left by 90 c .
For k = 0, 71 .
71 -cos ^ = --(16) and = 120° and 240“ (17) If the spacing between the sources is reduced to 2/4, (13) becomes E = cos ^ ^ cos <t) j (18) The field pattern for this case is illustrated by Fig 4-5a.
It is a cardioid-shaped, unidirectional pattern with maximum field in the negative x direction. The space pattern is a figure-of-revolution of this pattern around the x axis, A simple method of determining the direction of maximum field is illus-trated by Fig. 4-5b. As indicated by the vectors, the phase of source 2 is 0° (vector to right) and the phase of source 1 is 270° (vector down) Thus, source 2 leads source 1 by 90°.
To find the field radiated to the left, imagine that we start at source 2 (phase 0 G ) and travel to the left, riding with the wave (phase 0°) like a surfer rides a breaker The phase of the wave we are riding is 0° and does not change but by the time we have traveled A/4 and arrived at source 1, a J-period has elapsed so the current in source 1 will have advanced 90° (vector rotated ccw) from 270° to 42 ARRAYS OF TWO ISOTROPIC POINT SOURCES 125 Figure 4-5 (a) Relative field pattern of 2 isotropic sources of same amplitude and in phase quadra-ture for a spacing of a/4. Source 2 leads source 1 by 90 c .
(£) Vector diagrams illustrating field reinforcement in the —x direction and field cancellation in the +x direction.
0°, making its phase the same as that of the wave we are riding, as in the middle diagram of Fig 4- 5b. Thus, the field of the wave from source 2 reinforces that of the field of source 1, and the two fields travel to the left together in phase produc-ing a maximum field to the left which is twice the field of either source alone.
Now imagine that we start at source 1 with phase 27G G (vector down) and travel to the right. By the time we arrive at source 2 the phase of its field has advanced from 0 to 90° so it is in phase opposition and cancels the field of the wave we are riding, as in the bottom diagram in Fig. 4-5h t resulting in zero radi-ation to the right.
4-2d Case 4. General Case of Two Isotropic Point Sources of Equal Amplitude and Any Phase Difference. Proceeding now to a more general situ-ation, let us consider the case of two isotropic point sources of equal amplitude but of any phase difference <5 The total phase difference ^ between the fields from 126 ARRAYS OF POINT SOURCES source 2 and source 1 at a distant point in the direction tfr {see Fig. 4- 2a) is then \j/ — dr cos 4> + <5 (19) Taking source 1 as the reference for phase, the positive sign in (19) indicates that source 2 is advanced in phase by the angle <5. A minus sign would be used to ; indicate a phase retardation.
If, instead of referring the phase to source 1, it is referred to the centerpoint of the array, the phase of the field from source 1 at a distant point is given by —ijf/ 2 and that from source 2 by + ^/2. The total field is then E = E f)(ei 11 + = 2 E g cos | (20) Normalizing (20), we have the general expression for the field pattern of two isotropic sources of equal amplitude and arbitrary phase, £ = cos ~ (21) where ^ is given by (19). The three cases we have discussed are obviously special cases of (21). Thus, Cases 1, 2 and 3 are obtained from (21) when <5 = 0 C \ 180 D and 90° respectively.
4-2e Case 5, Most General Case of Two Isotropic Point Sources of Unequal Amplitude and Any Phase Difference, A still more general situation, involving two isotropic point sources, exists when the amplitudes are unequal and the phase difference is arbitrary. Let the sources be situated as in Fig. 4-6a with source 1 at the origin. Assume that the source 1 has the larger amplitude and that its field at a large distance r has an amplitude of £ 0 , Let the field from source 2 be of amplitude aE 0 (0 ^ a 1) at the distance r. Then, referring to Fig. 4-6fr, the magnitude and phase angle of the total field £ is given by /(l + a cos ^) 2 /arctan [a sin where \jt = dr cos tp + S and the phase angle ( i _ ) is referred to source. 1. This is the phase angle £ shown in Fig. 4-6b.
Figure 4-6 (a) Two isotropic point sources of unequal amplitude and arbitrary phase with respect to the coordinate system, (i) Vector addition of fields from unequal sources arranged as in (a}. The amplitude of source 2 is assumed to be smaller than that of source 1 by the factor a.
4-3 NONISOTROPIC BUT SIMILAR POINT SOURCES AND PATTERN MULTIPLICATION 127 4-3 NONISOTROPIC BUT SIMILAR POINT SOURCES AND THE PRINCIPLE OF PATTERN MULTIPLICATION. The cases considered in the preceding section all involve isotropic point sources. These can readily be extended to a more general situation in which the sources are non-isotropic but similar .
The word similar is here used to indicate that the variation with absolute angle <p of both the amplitude and phase of the field is the same.
1 The maximum amplitudes of the individual sources may be unequal.
If, however, they are also equal, the sources are not only similar but are identical.
As an example, let us reconsider Case 4 of Sec, 4-2d in which the sources are identical, with the modification that both sources 1 and 2 have field patterns given by £0 = £o sin (1) Patterns of this type might be produced by short dipoles oriented parallel to the x axis as suggested by Fig. 4-7. Substituting (1) in (4-2-20) and normalizing by setting 2E' 0 = 1 gives the field pattern of the array as £ = sin 0 cos ^ (2) where ^ = d, cos ) by the pattern of two isotropic point sources (cos t/r/2).
If the similar but unequal point sources of Case 5 (Sec. 4-2e) have patterns as given by (1), the total normalized pattern is £ = sin + a cos $) 2 + a 2 sin2 (3) Here again, the result is the same as that obtained by multiplying the pattern of the individual source by the pattern of an array of isotropic point sources.
1 The patterns not only must be of the same shape but also must be oriented in the same direction to be called “similar." 128 4 ARRAYS OF POINT SOURCES These are examples illustrating the principle of pattern multiplication , which may be expressed as follows: The field pattern of an array of nonisotropic hut similar point sources is the product of the pattern of the individual source and the pattern of an array of isotrqpic point sources, having the same locations , relative amplitudes and phases as the nonisotropic point sources.
This principle may be applied to arrays of any number of sources provided only that they are similar The individual nonisotropic source or antenna may be of finite size but can be considered as a point source situated at the point in the antenna to which phase is referred. This point is said to be the “phase center, 1 ' The above discussion of pattern multiplication has been concerned only with the field pattern or magnitude of the field If the field of the nonisotropic source and the array of isotropic sources vary in phase with space angle, i.e„ have a phase pattern which is not a constant, the statement of the principle of pattern multiplication may be extended to include this more general case as follows: The total field pattern of an array of nonisotropic but similar sources is the product of the individual source pattern and the pattern of an array of isotropic point sources each located at ike phase center of the individual source and having the same relative ampli-tude and phase , wJtile the total phase pattern is the sum of the phase patterns of the individual source and the array of isotropic point sources.
The total phase pattern is referred to the phase center of the array In symbols, the total field E is then Field pattern Phase pattern where f(0, \) = field pattern of individual source f p(0, (j>) = phase pattern of individual source F[9 t ) = phase pattern of array of isotropic sources The patterns are expressed in (4) as a function of both polar angles to indi-cate that the principle of pattern multiplication applies to space patterns as well as to the two-dimensional cases we have been considering.
To illustrate the principle, let us apply to it two special modifications of Case 1 (Sec. 4-2a).
Example 1. Assume two identical point sources separated by a distance d, each source having the field pattern given by (1) as might be obtained by two short dipoles arranged as in Fig 4-7 Let d = 2/2 and the phase angle 5 = 0. Then the total field pattern is E = sin <p cos (5) This pattern is illustrated by Fig.
4-8c as the product of the individual source pattern (sin shown at (a) and the array pattern (cos [<ji/2) cos <£]} as shown at NONISOTROPIC BUT SIMILAR POINT SOURCES AND PATTERN MULTIPLICATION 129 Figure 4-8 Example of pattern multiplication. Two nonisotropic but identical point sources of the same amplitude and phase, spaced i/2 apart and arranged as in Fig. 4-7, produce the pattern shown at (c)i The individual source has the pattern shown at (a), which, when multi-plied by the pattern of an array of 2 isotropic point sources (of the same amplitude and phase) as shown at (b), yields the total array pattern of (c)i (fr) The pattern is sharper than it was in Case 1 (Sec. 4- 2a) for the isotropic sources.
In this instance, the maximum field of the individual source is in the direction — 90°, which coincides with the direction of the maximum field for the array of two isotropic sources.
Example 2. Let us consider next the situation in which d = 2/2 and 5 = 0 as in Example 1 but with individual source patterns given by E0 Eo cos (6) This type of pattern might be produced by short dipoles oriented parallel to the y axis as in Fig, 4-9 Here the maximum field of the individual source is in the direc-tion (0 = 0) of a null from the array, while the individual source has a null in the direction ( = 90°) of the pattern maximum of the array. By the principle of pattern multiplication the total normalized field is 130 4 ARRAYS OF POINT SOURCES Figure 4-10 Example of pattern multiplication. Total array pattern (c) as the product of pattern {a) of individual nonisotropic source and pattern (/s) of array of 2 isotropic sources. The pattern ( b ) for the array of 2 isotropic sources is identical with that of Fig. 4^86, but the individual source pattern (a) is rotated through 90° with respect to the one in Fig. 4-8a.
The total array pattern in the xy plane as given by (7) is illustrated in Fig.
4- 10c as the product oF the individual source pattern (cos ) shown at (a) and the array pattern {cos [(n/2) cos <£]} shown at (6). The total array pattern in the xy plane has four lobes with nulls at the x and y axes.
The above examples illustrate two applications of the principle of pattern multiplication to arrays in which the source has a simple pattern. However, in the more general case the individual source may represent an antenna of any com-plexity provided that the amplitude and phase of its field can be expressed as a function of angle, that is to say, provided that the field pattern and the phase pattern with respect to the phase center are known. If only the total field pattern is desired, phase patterns need not be known provided that the individual sources are identical.
If the arrays in the above examples are parts of still larger arrays, the smaller arrays may be regarded as non isotropic point sources in the larger array— another application of the principle of pattern multiplication yielding the complete pattern.
In this way the principle of pattern multiplication can be applied n times to find the patterns of arrays of arrays of arrays.
4-4 EXAMPLE OF PATTERN SYNTHESIS BY PATTERN MULTIPLICATION, The principle of pattern multiplication, discussed in the preceding section, is of great value in pattern synthesis. By pattern synthesis is meant the process of finding the source or array of sources that produces a desired pattern. Theoretically an array of isotropic point sources can be found that will produce any arbitrary pattern. This process is not always simple and may yield an array that is difficult or impossible to construct. A simpler, less elegant approach to the problem of antenna synthesis is by the application of pattern multiplication to combinations of practical arrays, the combination EXAMPLE OF PATTERN SYNTHESIS BY PATTERN MULTIPLICATION 131 S Figvc 4-11 (a) Requirements for pattern of broadcast station and (b) idealized pattern fulfilling ihttK whifch best approximates the desired pattern being arrived at by a trial-and-error process.
To illustrate this application of pattern multiplication, let us consider the following hypothetical problem. A broadcasting station (in the 500- to 150Q^kHz frequency band) requires a pattern in the horizontal plane fulfilling the conditions indicated in Fig. 4-1 In. The maximum field intensity, with as little variation as possible, is to be radiated in the 90° sector between northwest and northeast. No nulls in the pattern can occur in this sector. However, nulls may occur in any direction in the complementary 270° sector, but, as an additional requirement, nulls must be present in the due east and the due southwest directions in order to prevent interference with other stations in these directions. An idealized sector-shaped pattern fulfilling those requirements is illustrated in Fig. 4-1 it. The antenna producing this pattern is to consist of an array of four vertical towers.
The currents in all towers are to be equal in magnitude, but the phase may be adjusted to any relationship. There is also no restriction on the spacing or geo-metrical arrangement of the towers.
Since we are interested only in the horizontal plane pattern, each tower may be considered as an isotropic point source. The problem then becomes one of finding a space and phase relation of four isotropic point sources located in the horizontal plane which fulfills the above requirements.
The principle of pattern multiplication will be applied to the solution of this problem by seeking the patterns of two pairs of isotropic sources which yield the desired pattern when multiplied together. First let us find a pair of isotropic sources whose pattern fulfills the requirements of a broad lobe of radiation with maximum north and a null southwest. This will be called the “primary pattern.
Two isotropic sources phased as an end-fire array can produce a pattern with a broader major lobe than when phased as a broadside’ array (for example, compare Figs. 4-lc and 4-5). Since a broad lobe to the north is desired, an end-fire arrangement of two isotropic sources as shown in Fig. 4-12 will be tried.
132 A ARRAYS OF POINT SOURCES N E Flgprt 4-12 Arrangement of 2 isotropic point sources for S both primary and secondary arrays.
From a consideration of pattern shapes as a function of separation and phase/ a spacing between A/4 and 3A/8 appears suitable (see Fig.
11-11), Accordingly, let d — 0.3A. Then the field pattern for the array is & E — cos — (1) 2 where 0 = 0,6 cos $ + <5 (2) For there to be a null in the pattern of (1) at ^ = 135° it is necessary that 2 ^ = (2/c + l) (3) where k = 0, 1, 2, 3, . ..
Equating (2) and (3) then gives —0.6 -7= + S ~ (2k + 1) (4) yft-or $ ={2k + 1)jt + 0,425 (5) For k = 0 ? S — — 104°. The pattern for this case {d = 0,3A and <5 = —104°) is illustrated by Fig, 4-l3a, Next, let us find the array of two isotropic point sources that will produce a pattern that fulfills the requirements of a null at = 270° and that also has a broad lobe to the north. This will be called the secondary w pattern. This pattern 1 See, for example, G, H. Brown, “ Directional Antennas," Proc. IRE, 25, January 1937; F. E. Terman, Radio Engineers' Handbook, McGraw-Hill, New York, 1943, p. 804; C.
E.
Smith, ' Directional Antennas, Cleveland Institute of Radio Electronics, Cleveland, Ohio, 1946.
2 The azimuth angle 4> (Fig. 4-12) is measured counterclockwise (ccw) from the north. This is consis-tent with the engineering practice of measuring positive angles in a counterclockwise sense. However, it should be noted that the geodetic azimuth angle of a point is measured in the opposite, or clockwise (cw), sense from the reference direction, which is sometimes taken as south and sometimes as north.
44 EXAMPLE OF PATTERN SYNTHESIS BY PATTERN MULTIPLICATION 133 ^=0 = 0 = 0 Total array Figm 4-13 Field patterns of primary and secondary arrays of 2 isotropic sources which multiplied together give pattern of total array of 4 isotropic sources.
multiplied by the primary array pattern will then yield the total array pattern. If the secondary isotropic sources are also arranged as in Fig. 4-12 and have a phase difference of 180°, there is a null at - 270°. Let the spacing d - 0.6A.
Then the secondary pattern is given by (1) where iff ™1.2 cos tj> + (6) The pattern is illustrated by Fig, 4-13b. By the principle of pattern multiplication, the total array; pattern is the product of this pattern and the primary array pattern, or E = cos (54° cos 0 — 52°) cos (108° cos tf> + 90°) (7) This pattern, which is illustrated by Fig. 4-13c, satisfies the pattern requirements.
The complete array is obtained by replacing each of the isotropic sources of the secondary pattern by the two-source array producing the primary pattern. The midpoint of each primary array is its phase center, so this point is placed at the location of a secondary source. The complete antenna is then a linear array of four isotropic point sources as shown in the lower part of Fig. 413, where now each source represents a single vertical tower. All towers carry the same current.
The current of tower 2 leads tower 1 and the current of tower 4 leads tower 3 by 104% while the current in towers 1 and 3 and 2 and 4 are in phase opposition.
The relative phase of the current is illustrated by the vectors in the lower part of Fig, 4-13c.
134 4 ARRAYS OF POINT SOURCES {u) Primary pattern.
0) Secondary pattern.
Figure 4-14 Phase patterns of primary, secondary and total arrays having the field patterns shown in Fig. 4-13. Phase patterns are given for the phase center at the midpoint of the array and at the southernmost source, the arrangement of the arrays and the phase centers being shown at (d). The phase angle q is adjusted to zero at $ — 0 in ah cases.
The solution obtained is only one of an infinite number of possible solu-tions involving four towers.
It is, however, a satisfactory and practical solution to the problem The phase variation £ around the primary, secondary and total arrays is shown in Fig- 4- 14a, h and c with the phase center at the centerpoint of each array and also at the southernmost source. The arrangement of the arrays with their phase centers is illustrated in Fig. 4-14d for both cases.
4-5 NONISOTROPIC AND DISSIMILAR POINT SOURCES-In Sec 4-3 nonisotropic but similar point sources were discussed, and it was shown that the principle of pattern multiplication could be applied However, if the sources are dissimilar, this principle is no longer applicable and the fields of the sources must be added at each angle for which the total field is calculated.
Total phase angle.
4 5 NONISOTROPIC AND. DISSIMILAR POINT SOURCES 135 180 B 4 > Total pattern j Mid point as phase center Southern source as phase center Primary arrays o 1 Secondary array Phase center [x) at mid -point of arrays Phase center U) at southern-most Thus, for two dissimilar sources 1 and 2 situated on the x axis with source 1 at the origin and the sources separated by a distance d (same geometry as Fig. 4-6) the total field is in general E = £j + E 2 = Eoy/_f{) + aF(4>) cos ^] 2 + [aF(<£) sin </,] 2 /f p(4>) + arctan [oFQft) sin ^/(/(4r ) + aH4>) cos ft)] 0) where the field from source 1 is taken as E^Eo/miU® (2) E2 = aE0 h'(4>) /Ep(4>) + dr cos + 5 and from source 2 as 136 4 ARRAYS OF POINT SOURCES Figure 4-15 Relation of 2 nonisotropic dissimilar sources to coordinate system.
where E0 = constant a = ratio of maximum amplitude of source 2 to source 1 {0 < a < 1) \jt = dr cos + $ + Fffl, where S = relative phase of source 2 with respect to source 1 /(<£) = relative field pattern of source 1 = phase pattern of source 1 F((j>) = relative field pattern of source 2 = phase pattern of source 2 :In (1) the phase angle (Z_) is referred to the phase of the field from source 1 in some reference direction ( = 4>q\ In the special case where the field patterns are identical but the phase pat-terns are not, a = 1, and /W) = F(4>) (4) from which E = 2E 0 cos | where phase is again referred to source 1 in some reference direction .
As an illustration of non iso tropic, dissimilar point sources, let us consider an example in which the field from source 1 is given by and from source 2 by where = dr cos + 6 = cos (j> j0 E 2 = sin (j> jj/_ (6 ) 0) The relation of the two sources to the coordinate system and the individual field patterns is shown in Fig. 4-15. Source 1 is located at the origin. The total field E is then the vector sum of E t and E 2 , or E = cos ^ + sin /£ (8) Let us consider the case for kj4 spacing (d = k/4) and phase quadrature of the sources (5 = n/2). Then (cos 0 +1) (9) The calculation for this case is easily carried out by graphical vector addition.
The resulting field pattern for the total field £ of the array is presented in Fig.
4^S LINEAR ARRAYS OF ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING 137 90" Figure 4-16 Field pattern of array of 2 nonisotropic dissimilar sources of Fig. 4-15 for d = IfA 270 s and 5 = 90" $ Figure 4-17 Phase pattern of array having field pattern of Fig.
4-16. The phase angle £ is with respect to source 1 as phase center.
4-16, and the resulting phase pattern for the angle £ is given in Fig. 4-17. The angle £ is the phase angle between the total field and the field of source 1 in the direction = 0.
4-6 LINEAR ARRAYS OF n ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING, 1 4-6a Introduction, Let us now proceed to the case of n isotropic point sources of equal amplitude and spacing arranged as a linear array, as indicated in Fig.
1 S. A, Schelkunoff, Electromagnetic Waves, Van Nostr&nd, New York, 1943, p. 342.
J. A. Stratton, Electromagnetic Theory, McGraw-Hill, New York, 1941, p. 45 L 138 4 ARRAYS OF POINT SOURCES 0-90° 7 - 0 Figure 4-18 Arrangement of linear array 1 2 3 4 5 n of n isotropic point sources.
4-18, where n is any positive integer. The total field £ at a large distance in the direction 4 > is given by £ = l + e» + e 3 + + --- +e-1 » (0 where ^ is the total phase difference of the fields from adjacent sources as given by cos 4> + d = dr cos $ + 3 (2) where 5 is the phase difference of adjacent sources, i.e., source 2 with respect to 1, 3 with respect to 2, etc.
The amplitudes of the fields from the sources are all equal and taken as unity. Source 1 (Fig, 4-18) is the phase reference. Thus, at a distant point in the direction the field from source 2 is advanced in phase with respect to source 1 by the field from source 3 is advanced in phase with respect to source 1 by 2\j/ y etc.
Equation (1) is a geometric series. Each term represents a phasor, and the amplitude of the total field £ and its phase angle £ can be obtained by phasor (vector) addition as in Fig 4-19.
Analytically, £ can be expressed in a simple trigonometric form which we now develop as follows: Multiply (1) by e, giving Ee - e + e + <? + + eSn+ (3) Now subtract (3) from (1) and divide by 1 — yielding 1 _ e 1 E ~ -e» (4) Equation (4) may be rewritten as £ e!2 \ ( 5) 4-6 LINEAR ARRAYS OF id ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING 139 I ib) Figure 4-19 (a) Vector addition of fields at a large distance from the Linear array of $ isotropic point sources of equal amplitude with source 1 as the phase center (reference for phase). (6) Same, but with midpoint of array (source 3) as phase center.
In this case the phase pattern is a step function as given by the sign of (8), The phase of the field is constant wherever £ has a value but changes sign when £ goes through zero.
When $ = G, (6) or (8) is indeterminate so that for this- case £ must be obtained as the limit of (8) as $ approaches zero. Thus, for ^ = 0 we have the 0 10" 20 30 40 50 s 60 70 SO 90 100110120130140150160170180 36035034033032031 0300290280 270260250 240 230 2 202102001 90180 + $ Figure 4-20 Universal field-pattern chart for arrays of various numbers n of isotropic point sources of equal amplitude and spacing-I l relation that E = n (8a) This is the maximum value that E can attain. Hence, the normalized value of the total field for = n is E = l sin M/2) m ji sin (0/2) The field as given by (9) will be referred to as the “array factor,” Values of the array factor as obtained from (9) for various numbers of sources are presented in Fig, 4-20, If \jr is known as a function of 0, then the field pattern can be obtained directly from Fig, 4-20, We may conclude from the above discussion that the field from the array will be a maximum in any direction 0 for which 0=0. Stated in another way, the fields from the sources all arrive at a distant point in the same phase when 0 = 0, In special cases, 0 may not be zero for any value of 0, and in this case the field is usually a maximum at the minimum value of 0.
To illustrate some of the properties of linear arrays (9) wilt now be applied to several special cases. See BASIC programs in App. B for calculating patterns involving these different cases. See also Probs. 4-35 and 4-40.
4-6b Case 1. Broadside Array (Sources in Phase). The first case is a linear array of n isotropic sources of the same amplitude and phase. Therefore, <5 = 0 4-6 LINEAR ARRAYS OF 1SOTHOFIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING 141 and 0 = dr cos 0 (10) To make 0 = 0 requires that 0 = (2fc + IX/2), where k = 0, 1, 2, 3 The field is, therefore, a maximum when , n , 3?r 0 = 2 and y (10a) That is, the maximum field is in a direction normal to the array. Hence, this condition, which is characterized by in-phase sources {5 = 0), results in a “ broadside ” type of array.
As an example, the pattern of a broadside array of four in-phase isotropic point sources of equal amplitude is shown in Fig 4-2ia. The spacing between sources is A/2.
1 The field pattern in rectangular coordinates and the phase pat-terns for this array are presented in Fig 4-2 lh.
4-6c Case 2. Ordinary End-Fire Array.
Let us now find the phase angle between adjacent sources that is required to make the field a maximum in the direction of the array (0 = 0). An array of this type may be called an “end-fire” array. For this we substitute the conditions 0 = 0 and 0 = 0 into (2), from which $=-d r (11) Hence, for an end-fire array, the phase between sources is retarded progressively by the same amount as the spacing between sources in radians. Thus, if the spacing is A/4, source 2 in Fig 4-18 should lag source 1 by 90 a , source 3 should lag source 2 by 90°, etc.
As an example, the field pattern of an end-fire array of four isotropic point sources is presented in Fig 4-22u, The spacing between sources is A/2 and S = —n. The field pattern in rectangular coordinates and the phase patterns are shown in Fig 4-22h. The same shape of field pattern is obtained in this case if S = + n since, with d = A/2, the pattern is bidirectional. However, if the spacing is less than A/2, the maximum radiation is in the direction 0 = 0 when 6 = - dr and in the direction 0 = 180° when S = +dr .
4-6d Case 3. End-Fire Array with Increased Directivity. The situation dis-cussed in Case 2, namely, for S = — 4 r , produces a maximum field in the direc-tion 0 = 0 but does not give the maximum directivity.
It has been shown by 1 If the spacing between dements exceeds A, siddobes appear which are equal in amplitude to the main (center) kibe. These are called grating lobes (see Sec.
1 1-26).
142 4 ARRAYS OF POINT SOURCES 0^9 0 “ 0" 90° 1 80" £70 360" Figure 4-21 (u) Field pattern of broadside array of 4 isotropic point sources of the same amplitude and phase. The spacing between sources is a/2, (fr) Field pattern in rectangular coordinates and phase patterns of same array with phase center at midpoint and at source 1. The reference direction for phase is at = 90.
Hansen and Woodyard that a larger directivity is obtained by increasing the phase change between sources so that This condition will be referred to as the condition for “increased directivity” 1 W. W. Hansen and J. R. Woodyard, "A New Principle in Directional Antenna Design,” Proc. tRE y 26,333 345, March 1938.
4-6 LINEAR ARRAYS OF h ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING 143 Thus for the phase difference of the fields at a large distance we have ^ = dr(cos <fi -1) — ^ (13) As an example, the field pattern of an end-fire array of four isotropic point sources for this case is illustrated in Fig. 4-23. The spacing between sources is A/2, and therefore S = -(5rc/4). Hence, the conditions are the same as for the array with the pattern of Fig, 4-22, except that the phase difference between sources is increased by jt/4. Comparing the field patterns of Figs. 422a and 4-23, it is apparent that the additional phase difference yields a considerably sharper main 0 - 90 ' Figure 422 (a) Field pattern of ordinary end-fire array of 4 isotropic point sources of same ampli-tude. Spacing is kfl and the phase angle & = -k, (6) Field pattern in rectangular coordinates and phase patterns of same array with phase center at midpoint and at source 1. The reference direction for phase is at $ 0.
lobe in the direction ~ 0. However, the back lobes in this case are excessively large because the large value of spacing results in too great a range in \j/.
To realize the directivity increase afforded by the additional phase differ-ence requires that \j»\ be restricted in its range to a value of Ti n at = 0 and a value in the vicinity of at = 180°. This can be fulfilled if the spacing is reduced. For example, the field pattern of an end-fire array of 10 isotropic point sources of equal amplitude and spaced //4 apart is presented in Fig. 4-24a for the phase condition giving increased directivity (5 = -0.6w).
In contrast to this pattern, one is presented in Fig. 4-246 for the identical antenna with the phasing of an ordinary end-fire array (5 = — 0.5 ti). Both patterns are plotted to the same Ffenre 4-14 Field patterns of end-fire arrays of 10 isotropic point sources of equal amplitude spaced A/4 apart.
The pattern at (a) has the phase adjusted for increased - directivity (£ = -O.dnk while the pattern at (fr) has the phasing of an ordinary end-fire array (3 - - 0 .5).
E l -7 NULL DIRECTIONS FOR ARRAYS OF n ISOTROPIC POINT SOURCES 145 Table 4-1 Ordinary end-lire Cod-flre array with array increased directivity Beam width between half-power points 69 c 38" Beam width between first nulls 106= 74"' Directivity 11 19 maximum. The increased directivity is apparent from the greater sharpness of the upper pattern. Integrating the pattern, including the minor lobes, the directivity of the upper pattern is found to be about 19 and of the lower pattern about 11.
The beam widths and directivities for the two patterns are compared in Table 4-1.
The maximum of the field pattern of Fig.
4-24a occurs at = 0 and $ — —n/n.
In general, any increased directivity end-fire array, with maximum at ^ = — 70t, has a normalized field pattern given by E / n \ sin (nt/tf2) Sm [in) sin </2) (14) 46e Case 4, Array with Maximum Field in an Arbitrary Direction Scanning Array, Let us consider the case of an array with a field pattern having a maximum in some arbitrary direction ^ not equal to kn/2 where it = 0, 1, 2 or 3. Then (2) becomes 0 = dr cos k + S (15) By specifying the spacing the required phase difference $ is then determined by (1 5)+ Conversely, by changing 6 the beam direction t/> 1 can be shifted or scanned.
As an example, suppose that n = 4, d = 2/2 and that we wish to have a maximum field in the direction of $ - 60°. Then d = - nfl , yielding the field pattern shown in Fig. 425.
4-7 NULL DIRECTIONS FOR ARRAYS OF n ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING In this section simple methods are discussed for finding the directions of the pattern nulls of the arrays considered in Sec. 4-6.
Following the procedure given by Schelkiinoffl 1 the null directions for an array of n isotropic point sources of equal amplitude and spacing occur when 1 S. A. SchelkunofT, Electromagnetic Waves Van Nostrand, New York, 1943, p. 343.
&. A- Schdkunoff, “A Mathematical Theory of Arrays,” Bell System Tec)i.
/>, 22, 80-107, January 1943.
146 Ait hays of point sources Figure 4-15 Field pattern of array of 4 isotropic point sources of equal amplitude with phasing adjusted to give the maximum at = 60°. The spacing is E = 0 or, provided that the denominator of (4-6-4) is not zero, when e'" = 1 Equation (1) requires that n\j/ = ± 2Kn where K = 1, 2, 3, ..
.
Equating the value of ^ in (2) to its value in (4-6-2) gives .
2Kn t/f = dr cos <!>$ + <5 = ± = arccos where <p 0 gives the direction of the pattern nulls Note that values of K must be excluded for which K — where m = 1, 2, 3, ...
. Thus, if K = mn, (2) reduces to ^ = ± 2m/r and the denominator of (4-6-4) equals zero so that the null condition of (1), that the numerator of (46-4) be zero, is insufficient.
In a broadside array 6 - 0, so that for this case (4) becomes = arccos (±^jn) = arccos ( ±S) <5) As an example, the field pattern of Fig, 4-21 (n — 4, d = A/2, 6-0) has the null directions 0 — arccos (4) (6) 4-7 NULL DIRECTIONS FOR ARRAYS OF it ISOTROPIC POINT SOURCES 147 For K = 1, = ±60° and + 120% and for K = 2, = 0° and 180^. These are the six null directions for this array.
If <j} 0 in (3) is replaced by its complementary angle y 0 (see Fig. 4-18), then (5) becomes y 0 = a resin j±— If the array is long, so that nd Kk y KX y0 = ± — (8) nd The first nulls either side of the maximum occur for K = 1. These angles will be designated y01 . Thus, 7oi - ± ^ (9) and the total beam width of the main lobe between first nullsfor a long broadside array is then For the field pattern in Fig. 4-21 this width is exactly 60°, while as given by (10) it is 1 rad, or 57.3°. This pattern is for an array 2A long. The agreement would be better with longer arrays.
Turning next to end-fire arrays , the condition for an ordinary end -fire array is that S = —dr < Thus, for this case (3) becomes cos tp0 — 1 = + from which we obtain o .
/ Kx — = aresm ( ± — 1 V V ndr, 0 = 2 arcsin ± As an example, the field pattern of Fig. 4-22 (n = 4, d = A/2, S — — te) has the null directions o = 2 arcsin ± — For K = 1, 0 = ±60°; for K = 2, = ±90% etc.
148 ARRAYS OF POINT SOURCES If the array is long, so that nd P Kk. (13) becomes .
, l2Kl 0 ^ ± —T (15) V nd The first nulls either side of the main lobe occur for K = 1. These angles will be designated tp0i . Thus, oi ± H (16) V nd and the total beam width of the main lobe between first nulls a long ordinary end-fire array is then ^ oi yi (,?) For the field pattern in Fig. 4-22 this width is exactly 120°, while as given by (17) it is 2 rad, or 115°.
For end-fire arrays with increased directivity as proposed by Hansen and Woodyard, the condition is that S = ~{d, + x/n). Thus, for this case (3) becomes (/.(cos — 1) — - = ±2— (18) n n from which — = arcsin 2 {2K-I) or = 2 arcsin j^± ^j— (2K -1)J (20) If the array is long, so that nd i> Kk, (20) becomes ± ™ The first nulls either side of the main lobe, 0O1 , occur for K = L Thus, and the total beam width of the main lobe between first nulls for a long end-fire array with increased directivity is then 4-7 NULL DIRECTIONS KOft ARRAYS OF x ISOTROPIC POINT SOURCES 149 This width is 1/^/2, or 71 percent, of the width of the ordinary end-fire array. As an example, the ordinary end-fire array pattern of Fig. 4-24b has a beam width between first nulls of 106 c . The width of the pattern in Fig 4-24u for the array with increased directivity is 74°, or 70 percent as much.
Table 4-2 lists the formulas for null directions and beam widths for the different arrays considered above. The null directions in column 2 apply to arrays of any length. The formulas in the third and fourth columns are approximate and apply only to long arrays.
The formulas in Table 42 have been used to calculate the curves presented in Fig. 4-26. These curves show the beam width between first nulls as a function of ndx for three types of arrays: broadside, ordinary end-fire and end-fire with increased directivity. The quantity ndx ( — nd/?,) is approximately equal to the length of the array in wavelengths for long arrays. The exact value of the array length is (h — 1 )dk .
The beam width of long broadside arrays is inversely proportional to the array length, whereas the beam width of long end-fire types is inversely pro-portional to the square root of the array length. Hence, the beam width in the plane of a long linear broadside array is much smaller than for end-fire types of the same length as shown by Fig. 4-26.
It should be noted, however, that the broadside array has a disc-shaped pattern with a narrow beam width in a plane Table 4-2 Null directions and beam widths between first nulls for linear arrays of n isotropic point sources of equal amplitude and spacing.
{For n £ 2. The angles in columns 3 and 4 are expressed in radians. To convert to degrees, multiply by 57.1) Type of Ordinary end-fire End-fire with increased directivity (Hansen and Woodyard) Null directions (array any length) , 17 ±2" a n np 0 = arccos j — o \ — LV n / ^ 0 ^arcsm(± g) -—{M Null directions (long array) Beam width between first (long array) 2A 2y0 i^ 2oi = 2 H /-jPK-n - 2 J-, 150 4 ARRAYS OF POINT SOURCES (approx, array length} Figure 4-26 Beam width between first nulls as a function of ndx for arrays of n isotropic point sources of equal amplitude.
For Jong arrays, ndx is approximately equal to the array Length.
through the array axis but a circular pattern (360° beam width) in the plane normal to the array axis. On the other hand, the end-fire array has a cigarv shaped pattern with the same beam width in all planes through the array axis.
4-8 BROADSIDE VERSUS END-FIRE ARRAYS, TURNS VERSUS DIPOLES AND 3-DIMENSIONAL ARRAYS. Assuming that the half-power beam width (HPBW) is J the beam width between first nulls (BWFN) (an approximation), we have from (2-9-4) and (4-7-10) that the direc-tivity of a linear broadside array is given approximately by where n = number of sources d = spacing between sources, m A = wavelength, m L x = ndjX = L/A = length of array in wavelengths, dimensionless (1) The pattern is disc-shaped so 0 = 360°, or 2n radians.
It is assumed that <£hp = BWFN/2 = X/nd; also that Lx > 1 so that L 2= nd.
From (4-7-17), the directivity of an ordinary end-fire array is approximately D = An ™= 2tiL x T (2) 4-8 BROADSIDE VERSUS END-FIRE ARRAYS 151 while from (4-7-23) the directivity of an end-fire array with increased directivity is approximately D =; 4tt — = 47TL; (3) A A general expression for directivity, as given by (2-22-9), is where A e = effective aperture, m 2 For a square unidirectional broadside array or aperture with 100 percent aperture efficiency (uniform aperture distribution) we have At = A r = 1} (m J ) and D = 4jtLjf (5) For 100 percent aperture efficiency (uniform aperture distribution) the directivity of a circular unidirectional broadside array or aperture is D = n 2dl (6) where dk = d/X = diameter of array or aperture in wavelengths, dimensionless These directivity relations are summarized in Table 4-3, which also gives half-power beam widths and numerical directivities for array lengths (or diameters) of 1 to 1000 as measured in wavelengths.
For array dimensions of the order of A, the increased directivity end-fire array and square broadside array have comparable directivities. We note, however, that the end-fire directivity is proportional to the length L x while the directivity of the broadside square array is proportional to the square of the side length UA+ Hence, for a high directivity an end-fire array must have a much greater dimension than a square broadside array. For example, a square array with 1000A on a side has a directivity (see Table 4-3) of 12.6 million. To equal this directivity an increased-directivity end-fire array must be 1 million X long and an ordinary end-fire array 2 million X long. Even if all the sources or elements of such a long end-fire array could be fed with equal amplitude and in proper phase, the great length of the array is a severe disadvantage. Thus, it is apparent why broadside arrays or apertures are invariably used for high-directivity (high-gain) applications. The broadside aperture may consist of an array of X/2 dipoles or it may be the aperture of a parabolic reflector antenna with single feed point or, as discussed next, a broadside array of intermediate-length end-fire antennas.
With uniform amplitude of all X/2 dipole elements across a large aperture (and dipole spacing no more than A/2), there is nothing to be gained by replacing each A/2 dipole by a more directional end-fire antenna. However, if each end-fire 152 4 ARRAYS OF POINT SOURCES Table 4-3 Directivities and beamwnUhs of arrays and apertnresf Array {or aperture)^ Directivity Cdnndx Dh activity for Lj or dx equal to 1 10 100 1000 beam widths Linear broadside array of length La 2 2 20 200 2000 50.8 —— x 360" Li Ordinary end-fire array of Length Lx IxLi 6,3 63 630 6300 108" Ju I ncreased^directivity end-fire array of length Lx 4xLi 116 126 1260 12600 52" Square broadside aperture with side length L± 4Ll 126 1260 126000 1.26 x 107 50.8 50.8 Lx Li Circular broadside aperture with diameter dx Flat array (length L J of ordinary end-fire antennas (length L^)§ 9.9 990 99000 9.9 x 10 58 Same but square (L« = 2J§ Flat array (length L ,,) of increased-di rectivity end-fire antennas (length 1^)5 4jc£,i yJ~LxE r 8.9 281 8900 281000 Same but square 4 126 398 12600 398 000 t Tht directivities for the arrays (broadside and end-fire) aie approximate while for tbc apertures (square and circular) the directivities are exact. Note that if the directivity of the square or circular aperture is calculated using the approximation 41 000/0^ 4^ tbe result is larger than the directivity given in the table. See discussion regarding this approximation in connection with (31316} and (3-13-18).
Arrays or apertures are assumed to be large compared to the wavelength and to have uniform amplitude distribution. Specifically, the square and circular apertures are assumed to have 1 00 percent aperture efficiency. See text for other assumptions involved.
Tbe directivities for tbe square and circular apertures are identical in form to tbe more general exact relation given by (4), which applies to an aperture of any shape.
t » array length, wavelengths dj =; array diameter, wavelengths L1S = ordinary cad-fire array length, wavelengths “ increased-directivity end-fire array length, wavelengths 5 Beam off edge, nor perpendicular lo flat side.
I \ 4-K BROADSIDE VERSUS END-FIRE ARRAYS 153 antenna is used to replace several A/2 dipoles then some benefits may accrue as described in the following example involving a broadside array of end-fire antennas. An array of this type may be called a volwne or 3dimensional array .
Figure 4-27a is an edge view of a 12 x 12 broadside array of 144 point sources with A/2 spacing. The figure shows 12 dements as seen from one edge.
Let each source consist of a A/2 dipole element as suggested in partial broadside view in Fig. 4-27e. The array of Fig. 4-27rz is bidirectional but if backed at A/4 spacing by an identical array fed with equal amplitude currents and 90° phase difference, each pair of dipoles has a unidirectional pattern (as in Fig, 4-5) and Figure 4-27 Equivalence or broadside array of X/2 dipoles and 3-dimensional broadside-end-fire array. Parts a, b t c and d arc edge or side views while e is a Trout or broadside view.
154 4 ARRAYS OF POINT SOURCES the entire 12x12x2 array is unidirectional as suggested in Fig. 4-276. An alter-native, simpler arrangement is to replace the second array with a conducting flat-sheet reflector or ground plane at an appropriate spacing, as in Fig. 4-27c + A A/2 dipole has a directivity D = 1.64, and taking the directivity of dipole with reflector as twice this value or 3.28 ( = 2 x 1.64), the equivalent effective aperture is = 128 ^ = 0 26kl (7) An An or approximately A z /4, as suggested in Fig, 4-27e.
The 12 x 12 array has 144 feed points. By substituting an end-fire array of appropriate directivity and effective aperture for a group of A/2 dipole elements, the number of elements and feed points can be reduced. Thus, let us replace 9 A/2 dipoles by a single end-fire array. The 9 A/2 elements have an effective aperture of A t ~0V =\? = 2-25^ (8) For an increased-directivity end-fire array to provide an effective aperture of 2.25A2 requires an array directivity ^(required) = .2 e = An x 2.25 = 28.3 (9) The required length of the increased-directivity end-fire array is then Hrequired) 4 JL ,10) ^ n 4n An An effective end-fire array which meets the above requirements is a 9-turn monofilar axial-mode helical antenna (see Chap. 7) with A/tt diameter and A/4 spacing between turns making the length L = 9 x 0.25A = 2,25A + This end-fire antenna has the remarkable properties of increased directivity, wide bandwidth (over 2 to 1) and very small mutual coupling between adjacent helices.
It is fed from one end through the ground plane by a coaxial transmission line (as in Fig. 4-27d) which may be 50 or other convenient impedance.
With each 9-tum helix replacing 9 A/2 dipoles the broadside array of 144 A/2 dipoles is replaced by 16 (= 144/9) helices which reduces the complexity of the feed system and provides an array readily capable of wide bandwidth operation.
The effective aperture of the array is 6A x 6A = 36AZ for a directivity D = 4rc x 36 = 452 (or 26.6 dBi) (11) As a further step, the 16 9-turn helices could be replaced by 4 36-source end-fire arrays (36-tum helices), and as a final step, the 4 helices by a single 144-source end-fire array (144-tum helix) 36A long. Such a long helix is not prac-BROADSIDE VERSUS END- FIRE ARRAYS 155 tical but, even if it were, its great length is a disadvantage as compared to a more compact array with a number of shorter helices.
In the above example each turn of a helix (as an end-fire source) has a directivity equal to a A/2 dipole with reflector (as a broadside source), but polariz-ations differ (helix circularly polarized and dipole linearly polarized).
In summary, we considered the following cases: 1 12 x 12 array of reflector-backed A/2 dipoles: 12 x 12 = 144 sources 2.4x4 array of 9-turn helical end-fire antennas: 4 x 4 x 9 = 144 sources 3, 2 x 2 array of 36-tum helical end-fire antennas: 2 x 2 x 36 = 144 sources 4. Single 144-turn helical end-fire antenna: 1 x 144 = 144 sources Thus, for constant directivity and effective aperture, the number of sources is a constant whether the sources are arranged in a flat broadside array or in 3-dimensional broadside-end -fire configurations, all sources having uniform amplitude. Although the constant directivity and effective aperture in the above example may not apply fully for all 3-dimensional arrays, the example illustrates the principle that for a given number of sources, various configurations may produce (ideally) similar, if not identical, directivities and effective apertures.
Another broadside-end-fire combination is a flat (or planar) array consisting or a linear array with only a single row of end-fire antennas (helices) as in Fig.
4-27d (no other arrays stacked perpendicular to the page). The beam width in one plane is determined by the broadside length L x and the beam width in the other plane by the end-fire length Lkt (for ordinary end-fire) or length L XE > (for increased directivity).
The directivity for the ordinary end-fire case is D(ordinary) = nL x ^f%LXE (!2) and if L x = L XE (square flat array), D(ordinary) = ( 13 ) The directivity for the increased directivity case is Decreased directivity) = 4nLx JT^> (14) and if Lx = (square flat array), ^increased directivity) = Ati^/lI (15) These relations are summarized in Table 4-3.
An early application of end-fire antennas in a large 3-dimensional array, which I designed and built in 1951, is shown in Fig 7-4.
It has 96 helices of 11 turns each in a 4 x 24 configuration.
Equivalent to an array of 1056 (= 1 1 x 96)A/2 dipoles with reflectors, it has a wide bandwidth (over 2 to 1) and is simple to feed.
156 4 ARRAYS OF. POINT SOURCES 4-9 DIRECTIONS OF MAXIMA FOR ARRAYS OF n ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND SPACING, Let us now proceed to a discussion of the methods for locating the positions of the pattern maxima. The major-lobe maximum usually occurs when ^ = 0 This is the case for the broadside or ordinary end-fire array The main lobes of the broadside array are then at = 90° and 270° while for the ordinary end-fire array the main lobe is at 0° or 180° or both. For the end-fire array with increased directivity the main-lobe maximum occurs at a value of $ = ±nfn with the main lobe at 0° or 1 80° Referring to Fig. 4-24u, the main-lobe maximum (first maximum) for this case occurs at the first maximum of the numerator of (4-6-8).
The maxima of the minor lobes are situated between the first- and higher-order nulls.
It has been pointed out by Schelkunoff that these maxima occur approximately whenever the numerator of (4-6-8) is a maximum, ie, when mb sun y = 1 (1) Referring to Fig. 4-28, we note that the numerator of (4-6-8) varies as a function of 4f more rapidly than the denominator sin (^/2). This is especially true when n is large. Thus, although the nulls occur exactly where sin (n^/2) = 0, the maxima occur approximately where sin (n^/2) = 1. This condition requires that f-±PK+l)§ where X = 1, 2, 3, .
.
.
Substituting the value of iff from (2) into (4-6-2) gives , , .
e ±<2X + Dn dr cos 4>m + <5 = Therefore 4>m — arccos{[“Mj} where m — arccos ±<2X + 1) For X = l t 4>m = +41.4° and ± 1386°. These are the approximate directions for the maxima of the four minor lobes of this pattern.
4-9 DIRECTIONS OF MAXIMA FOR ARRAYS OF h ISOTROPIC POINT SOURCES 157 .
^ S(n “ 2 ’O a m >-t K'O £ 5 2 c £ -O = 1 SO" fll Or c c — + £ -f £ kl £ 5 Figure 4-28 Graphs of the numerator (sin mf//2) and denomi-nator (sin tf/2) of the array factor as functions of tfr, showing the values of $ corresponding to maxima and nulls of a fidd pattern for the case n = 8.
For an ordinary end-fire array, 6 — —dr so that (4) becomes 4>m ~ arccos ±(2X + 1)2H while, for an end-fire array with increased directivity 6 = — (dr + njn) and 4>m ^ arccos j— [1 ± (2K + 1)] + 1 j (8) The above formulas for the approximate location of the minor-lobe maxima are listed in Table 4-4 (K = 1 for first minor lobe, X = 2 for second minor lobe, etc.).
The amplitudes of the field at the minor-lobe maxima are also of interest.
It has been shown by Schelkunoff that since the numerator of (4-6-9) is approx-imately unity at the maximum of a minor lobe, the relative amplitude of a minor-lobe maximum is given by n sin (^/2) (9) 158 4 ARRAYS OF POINT SOURCES Table 4-4 Directions of minor-lobe maxima for linear arrays of n isotropic point sources of equal amplitude and spacing Type of array Broadside Ordinary end-fire End-fire with increased directivity (Hansen and Woodyard) Directions of minor-lobe maxima +(2x: + i)i r±(2 + in..~ "" arcH 2nd + _ = arccos — [l ± (2 K + 1)] + 1 Introducing the value of \jt from (2) into (9) yields E 1 ( 10) ML " n sin [(2K + l)n/2n] When n ^ K, that is, for the first few minor lobes of an array of a large number of sources, we have the further approximation £ml ~(2K + l)n (H Thus, for arrays of a large number of sources the relative amplitude of the first few minor lobes is given by (11) for K = 1, 2, 3, etc.
In a broadside or ordinary end-fire array, the major-lobe maximum is unity so that the relative amplitudes of the maximum and first five minor lobes for arrays of these types and many sources are 1, 0.21, 0.13, 0.09, 0.07 and 0.06. From the curve for n = 20 in Fig. 4-20 we have the corresponding relative amplitudes given by 1, 0.22, 0,13, 0.
09, 0.07 and 0.06. For an end-fire array with increased directivity the maximum for = 0 and n = 20 occurs at & = ji/20 = 9°. At this value of $ the array factor is 0.63. Putting the maximum equal to unity then makes the relative amplitudes 1, 0.35,0.21,0.14,0.11 and 0,09. It is interesting to note in (10) that the maximum amplitude of the smallest minor lobe occurs for2fC + l = n. Then .
r(2K + l)l_, ( 12) sin L^r^J-!
The condition 2K + 1 = n is exactly fulfilled when m is odd for the minor-lobe maximum at = 1 SO ’ (see Fig 4-20). When H is even, the condition is approx-imately fulfilled by the minor lobes nearest \fi = 180°. Thus, the maximum ampli-tude of the smallest minor lobe of the field pattern of any array of n isotropic 4-10 LINEAR BROADSIDE ARRAYS WITH NONUNIFORM AMPLITUDE DISTRIBUTIONS 159 point sources of equal amplitude and spacing will never be less than 1 jn of the major-lobe maximum. An exception to this is where the range of \p ends after a null in the array factor has been passed but before the next maximum has been reached. In this case the maximum of the smallest minor lobe may be arbitrarily small.
4-10 LINEAR BROADSIDE ARRAYS WITH NONUNIFORM AMPLITUDE DISTRIBUTIONS. GENERAL CONSIDERATIONS, In the preceding section, our discussion was limited to linear arrays of n isotropic sources of equal amplitude. This discussion will now be extended to the more general case where the amplitude distribution may be nonuniform. In introducing this subject, it is instructive to compare field patterns of four types of amplitude distributions, namely, uniform, binomial, edge and optimum. To be specific, let us consider a linear array of five isotropic point sources with a/2 spacing.
If the sources are in phase and all equal in amplitude, we may calculate the pattern as discussed in Sec 4-6, the result being as shown in Fig. 4-29 by the pattern desig-nated uniform. A uniform distribution yields the maximum directivity. The pattern has a half-power beam width of 23°, but the minor lobes are relatively large. The amplitude of the first minor lobe is 24 percent of the major-lobe maximum (see Fig. 4-20, n = 5) Tn some applications this minor-lobe amplitude may be undesirably large-To reduce the sidelobe level of linear in-phase broadside arrays, John Stone Stone 1 proposed that the sources have amplitudes proportional to the coefficients of a binomial series of the form (a + b)"-t = a -1 + (n -1^~ 2b + ~ ~ ^ a <~ i'b2 + • (1) where n is the number of sources. Thus, for arrays of three to six sources the relative amplitudes are given by n Relative amplitudes (Pascals triangle) 3 12 1 4 13 3 1 5 1 4 6 4 1 6 1 5 10 10 5 1 where the amplitudes are arranged as in Pascal's triangle (any inside number is equal to the sum of the adjacent numbers in the row above).
Applying the binomial distribution to the array of five sources spaced a/2 apart, the sources have the relative amplitudes 1, 4, 6, 4, 1. The resulting pattern, 1 John Stone Stone, U-S. Patents 1,643,323 and 1,715433.
160 A ARRAYS OF POINT SOURCES 11111 11111 .iii.
1 4 G 4 1 m mu 1 1.9 1 1.6 1.6 1...1 1 0 0 0 1 id) Figure 4-29 Normalized field patterns of broadside arrays of 5 isotropic point sources spaced i/2 apart. AH sources are in the same phase, but the relative amplitudes have four different distributions: uniform, binomial, optimum and edge. Only the upper half of the pattern is shown. The relative amplitudes of the 5 sources are indicated in each case by the array below the pattern, the height of the line at each source being proportional to its amplitude.
All patterns are adjusted to the same maximum amplitude.
designated binomial is shown in Fig, 4~29. Methods of calculating such patterns are discussed in the next section. The pattern has no minor lobes, but this has been achieved at the expense of an increased beam width (31°), For spacings of ^/2 or less between elements, the minor lobes are eliminated by Stone's binomial distribution. However, the increased beam width and the large ratio of current amplitudes required in large arrays are disadvantages.
A.t the other extreme from the binomial distribution, we might try an edge distribution in which only the end sources of the array are supplied with power, the three central sources being either omitted or inactive. The relative amplitudes of the five-source array are, accordingly, 1, 0, 0, 0, 1, The array has, therefore, 4^10 LINEAR BROADSIDE ARRAYS WITH NONUNIFORM AMPLITUDE DISTRIBUTIONS 161 degenerated to two sources 2k apart and has the field pattern designated as edge jn Fig.
4-29. The beam width between half-power points of the “main” lobe (normal to the array) is 15°, but “minor" lobes are the same amplitude as the “ main” lobe.
Comparing the binomial and edge distributions for the five-source array with 2/2 spacing, we have Half-power Minor-lobe amplitude Type of distribution beam width (% of major tube) Binomial 3T 0 Edge 15" 100 Although for most applications it would be desirable to combine the 15° beam width of the edge distribution with the zero minor-lobe level of the bino-mial distribution, this combination is not possible. However, if the distribution is between the binomial and the edge type, a compromise between the beam width and the sidelobe level can be made; i.e., the sidelobe level will not be zero, but the beam width will be less than for the binomial distribution. An amplitude distribu-tion of this nature for linear in-phase broadside arrays has been proposed by Dolph 1 which has the further property of optimizing the relation between beam width and sidelobe level; i.e,, if the sidelobe level is specified, the beam width between first nulls is minimized, or, conversely, if the beam width between first nulls is specified, the sidelobe level is minimized. Dolpffs distribution is based on the properties of the Tchebyscheff polynomials and accordingly will be referred to as the Dolph-Tchebyscheff or optimum distribution.
Applying the Dolph-Tchebyscheff distribution to our array of five sources with 2/2 spacing, let us specify a sidelobe level 20 dB below the main lobe, i.e., a minor-lobe amplitude 10 percent of the main lobe. The relative amplitude dis-tribution for this sidelobe level is 1, 1.6, 1.9, 1.6, 1 and yields the pattern desig-nated optimum in Fig. 4-29. Methods of calculating the distribution and pattern are discussed in the next section. The beam width between half-power points is 27°, which is less than for the binomial distribution. Smaller beam widths can be obtained only by raising the sidelobe level. The Dolph-Tchebyscheff distribution includes all distributions between the binomial and the edge. In fact, the binomial and edge distributions are special cases of the Dolph-Tchebyscheff distribution, the binomial distribution corresponding to an infinite ratio between main- and sidelobe levels and the edge distribution to a ratio of unity. The uniform distribu-tion is, however, not a special case of the Dolph-Tchebyscheff distribution.
1 C.
L. Dolph, MA Current Distribution for Broadside Arrays Which Optimizes the Relationship between Beam Width arid Side- Lobe Level," Proc. IRE, 34, no. 6, 335-348, June 1946.
H. J. Riblel, “Discussion on Dolph’s Paper," Froc. IRE, 35, no.
5, 489-492, May 1947.
162 4 ARRAYS OF POINT SOURCES.
Referring to Fig 4-29, we may draw a number of general conclusions regarding the relation between patterns and amplitude distributions. We note that if the amplitude tapers to a small value at the edge of the array (binomial distribution), minor lobes can be eliminated. On the other hand, if the distribu-tion has an inverse taper with maximum amplitude at the edges and none at the center of the array (edge distribution), the minor lobes are accentuated, being in fact equal to the "main” lobe. From this we may quite properly conclude that the minor-lobe level is closely related to the abruptness with which the amplitude distribution ends at the edge of the array. An abrupt discontinuity in the distribu-tion results in large minor lobes, while a gradually tapered distribution approach-ing zero at the edge minimizes the discontinuity and the minor lobe amplitude.
In the next section, we shall see that the abrupt discontinuity produces large higher “harmonic” terms in the Fourier series representing the pattern. On the other hand, these higher harmonic terms are small when the distribution tapers gradually to a small value at the edge. There is an analogy between this situation and the Fourier analysis of wave shapes. Thus, a square wave has relatively large higher harmonics, whereas a pure sine wave has none, the square wave being analogous to the uniform array distribution while the pure sine wave is analo-gous to the binomial distribution.
The preceding discussion has been concerned with arrays of discrete sources separated by finite distances. However, the general conclusions concerning ampli-tude distributions which we have drawn can be extended to large arrays of con-tinuous distributions of an infinite number of point sources, such as might exist in the case of a continuous current distribution on a metal sheet or in the case of a continuous field distribution across the mouth of an electromagnetic horn.
If the amplitude distribution follows a Gaussian error curve, which is similar to a binomial distribution for discrete sources, then minor lobes are absent but the beam width is relatively large. An increase of amplitude at the edge reduces the beam width but results in minor lobes, as we have seen. Thus, in the case of a high-gain parabolic reflector type of antenna, the illumination of the reflector by the primary antenna is usually arranged to taper toward the edge of the parab-ola. However, a compromise is generally made between beam width and side-lobe level so that the illumination is not zero at the edge but has an appreciable value as in a Dolph-Tchebyscheff distribution.
4-11 LINEAR ARRAYS WITH NONUNIFORM AMPLITUDE DISTRIBUTIONS.
THE DOLPH-TCHEBYSCHEFF OPTIMUM DISTRIBUTION. In this section linear in-phase arrays with nonuniform amplitude distributions are analyzed, and the development and application of the Dolph-Tchebyscheff distribution are discussed.
Let us consider a linear array of an even number n t of isotropic point sources of uniform spacing d arranged as in Fig. 4-30a.
All sources are in the same phase. The direction 6 = 0 is taken normal to the array with the origin at the center of the array as shown. The individual sources have the amplitudes A 0 > 4-11 LINEAR ARRAYS WITH NONUN]FORM AMPLITUDE DISTRIBUTIONS 163 At Aj A s Ai 2Aq A\ Ai A 3 At Figure 4-30 Linear broadside arrays of n isotropic sources with uniform spacing for n even (a) and n odd ibl A t , A 2 , etc., as indicated, the amplitude distribution being symmetrical about the center of the array. The total field Ent from the even number of sources at a large distance in a direction 0 is then the sum of the fields or the symmetrical pairs of sources, or 3^ fn. -1 Ent = 2A 0 cos | + 2A t cos — + -T 2A k cos I — ^ ) (!) where ^ — —j— sin 0 = dr sin 0 (2) Each term in (1) represents the field due to a symmetrically disposed pair of the sources.
Now let 2( + 1) = n.
n€ -1 2k + 1 2 “ 2 where it = 0, 1, 2, 3, ... so that 164 ARRAYS OF POINT SOURCES Then (1) becomes =n-i /?k -L-1 \ B^ = 2 At cos y—Y~ ^ J where N = nJ2 Next let us consider the case of a linear array of an odd number n0 of isotropic point sources of uniform spacing arranged as in Fig. 4-305. The ampli-tude distribution is symmetrical about the center source. The amplitude of the center source is taken as 2A 0 , the next as A it the next as A 2 etc. The total field E from the odd number of sources at a large distance in a direction 0 is then Nf E^ = 2A 0 + 2Ai cos t/r + 2A 2 cos 2\p + “ + 2A k cos ^ ^j {4} Now for this case let 2k + 1 = where k = 0, 1, 2, 3, — Then (4) becomes ^ cos (a|) (5) where N = -l)/2 The series expressed by (4) or by (5) may be recognized as a finite Fourier series of JV terms, 1 For k = 0 we have a constant term 2A 0 representing the contribution of the center source. For k = 1 we have the term 2 A { cos ^ rep-resenting the contribution of the first pair of sources on either side of the center source. For each higher value of k we have a higher harmonic term which in each case represents the contribution of a pair of symmetrically disposed sources.
Thus, the total field pattern is simply the sum of a series of terms of increasing order in the same way that the waveform of an alternating current can be rep-resented as a Fourier series involving, in general, a constant term, a fundamental term and higher harmonic terms. The field pattern of an even number of sources as given by (1) or (3) is also a finite Fourier series but one which has no constant term and only odd harmonics. The coefficients A 0 , A lt ...
in both series are arbitrary and express the amplitude distribution.
To illustrate the Fourier nature of the field-pattern expression, let us con-sider the simple example of an array of 9 isotropic point sources spaced >1/2 apart, having the same amplitude and phase. Hence, the coefficients are related as follows: 2A 0 - A x = A 2 = A$ — A± = The number of sources is odd; hence the expression for the field pattern is then given by (5) as E$ = i + cos tj/ + cos 2$ + cos 3 iff + cos 4^ (6) 1 Irving Wolff, "Determination of the Radiating System Which Will Produce a Specified Directional Characteristic," Proc> /K£. 25, 630-643, May 1937.
4-11 LINEAR ARRAYS WITH NONUNIFORM AMPLITUDE DISTRIBUTIONS 165 t/> Total pattern \ ' 2\ 3\ ' 4X k = 0 k = l k = 2 k - 3 k = A ia) ib) id) b?) Figure 4-31 Resolution of total pattern of array of 9 isotropic sources into Fourier components due to center source and pairs of symmetrically disposed sources. The relative field pattern of the entire array is shown by (/). The lower halves of patterns are not shown. (Note that the end-fine lobes are wider than the broadside lobes.) The first term (k = 0) is a constant so that the field pattern is a circle or amplitude { as shown in Fig. 4-31u. The second term (/c = 1) may be regarded as the funda-mental term of the Fourier series and gives the pattern of the two sources (4^ in Fig. 4-305) on either side of the center. This pattern has 4 lobes of maximum amplitude of unity, as illustrated in Fig. 4-315. The next term (k — 2) may be regarded as the second harmonic term and gives the pattern of the next pair of sources (A 2 in Fig. 4-305). This pattern has 8 lobes as shown by Fig. 4-31c. The last two terms represent the third and fourth harmonics, and the patterns have 12 and 16 lobes, respectively, as indicated by Fig. 4-3 li and e. The above relations may be summarized as in Table 4-5.
The algebraic sum of the patterns given by the five terms is the total far-field pattern of the array which is presented in Fig. 4-31/. If the middle source of the array has zero amplitude or is omitted, the total pattern is then the sum of the four terms for which k = 1, 2, 3 and 4. If in addition the pair of sources is omitted, the total pattern is the sum of three terms for which k = 2, 3 and 4.
Since these are higher harmonic terms, we may properly expect that in this case the minor lobes of the total pattern will be accentuated.
It is apparent from the above discussion that the field pattern of any symmetrical amplitude distribution can be expresed as a series of the form of (3) or (5).
Table 4-5 k Sources Spacing Fourier term Pattern 0 1 0 Constant Circle 1 2 u Fundamental 4 lobes 2 2 2X Second harmonic 8 Jobes 3 2 u Third harmonic 12 lobes 4 2 4X Fourth harmonic 16 Lobes 166 4 ARRAYS OF POINT SOURCES Proceeding now to the Dolph-Tchebyscheff amplitude distribution, it will be shown that the coefficients of the pattern series 1 can be uniquely determined so as to produce a pattern of minimum beam width for a specified sidelobe level.
The first step in the development of the Dolph-Tchebyscheff distribution is to show that (3) and (5) can be regarded as polynomials of degree — 1 and nd — 1, that is, polynomials of degree equal to the number of sources less 1. In the present discussion we shall consider only the case of the broadside type of array, i.e., where S = 0 + Thus, \j/ = dT sin 0 (7) Now by de Moivre’s theorem, J ^ ^ ^ = cos m—hjsinm — — | cos — + j sin -r 1 2 V 2 2 On taking real parts of (8) we have { ijf Ib cos m — = Re ( cos — + j sin -r 2 V 2 2 Expanding (9) as a binomial series gives $ m \p w(m - 1) cos m — — cos — — _ .
cos' 2 2 2 !
(10) i Putting sin 2 (^/2) = 1 - cos 2 (^/2), and substituting particular values of m, (IQ) then reduces to the following: m = 0, m= 1, m = 2, m — 3, m = 4, etc.
(ii) Equations (1XPX (4) and (5).
J 4-11 LINEAR ARRAYS WITH NONUNIFORM AMPLITUDE DISTRIBUTIONS 167 Now let X — cos — whereupon the equations of (1 1) become cos m — = 1, cos m — = x, cos m — — 2x 2 — T when m — 0 when m = 1 when m — 2 The polynomials of (13) are called Tchebyscheff polynomials, which may be designated in general by TJx) = cos m ^ (14) For particular values of m, the first eight Tchebyscheff polynomials are T 0(x) = 1 ' 7l(x) = x T 2(x) = 2x2 -1 T (jc) = 4x3 - 3x > (15) 7i(x) = 8x - 8x2 4-1 T s(x) = 16xJ - 20x3 + 5x T 6(x) = 32x - 48x + 18x2 -1 T 7(x) = 64x 7 -1 12xs + 56x3 - lx 1 We note in (15) that the degree of the polynomial is the same as the value of m.
The roots of the polynomials occur when cos m( ij//2) = 0 or when where k = 1, 2, 3, .
The roots of x, designated x\ are thus -— x = cos (2k -1) 2m (17) 168 4 ARRAYS OF POINT SOURCES 4-1 1 UNEAR ARRAYS WITH NONUN1FORM AMPLITUDE DISTRIBUTIONS 169 We have shown that cos 2) can be expressed as a polynomial of degree e Thus, (3) and (5) are expressible as polynomials of degree 2k + l and 2k respectively, since each is the sum of cosine polynomials of the form cos mfi^jl).
For an even number n e of sources 2k + 1 = — 1, while for an odd number noi 2k = na — 1. Therefore, (3) and (5), which express the field pattern of a symmetric in-phase equispaced linear array of rt isotropic point sources, are polynomials of degree equal to the number of sources less 1. If we now set the array polynomial as given by (3) or (5) equal to the Tchebyscheff polynomial of like degree (m = n — 1) and equate the array coefficients to the coefficients of the Tchebys-cheff polynomial, then the amplitude distribution given by these coefficients is a Tchebyscheff distribution and the field pattern of the array corresponds to the Tchebyscheff polynomial of degree n — 1.
The Tchebyscheff polynomials of degree m = 0 through m = 5 are presented in Fig. 4-32. Referring to Fig. 4-3 2 t the following properties of the polynomials are worthy of note: 1 All pass through the point {1, 1).
2. For . values of x in the range — 1 < x < +1, the polynomials all lie between ordinate values of +1 and —1. All roots occur between — 1 <> x <, + 1, and all maximum values in this range are + 1.
We may now describe Dolph's method of applying the Tchebyscheff poly-nomial to obtain an optimum pattern. Suppose that we have an array of 6 sources. The field pattern is -then a polynomial of degree 5. If this polynomial is equated to the Tchebyscheff polynomial of degree 5, shown in Fig. 4-33, then the optimum pattern may be derived as follows.
Let the ratio of the main-lobe Figure 4-33 Tchebyscheff polynomial of fifth degree with relation to coordinate scales.
maximum to the minor-lobe level be specified as R; that is, main -lobe maximum sidelobe level The point (x0 , R) on the 7 5 (x) polynomial curve then corresponds to the main-lobe maximum, while the minor lobes are confined to a maximum value of unity.
The roots of the polynomial correspond to the nulls of the field pattern. The important property of the Tchebyscheff polynomial is that if the ratio R is speci-fied , the beam width to the first null (x = x'J is minimized. The corollary also holds that if the beam width is specified, the ratio R is maximized (sidelobe level minimized).
The procedure will now be summarized. Let us write (3) and (5) again. It is to be noted that they are functions of ^/2. Thus, £ = 2 1 £ \ COS [(2 4-1) £1 (n even) (18) fc o L k ~ N ( tA and = 2 A k cos ( 2t 2) {n odd) (19) Since we are usually interested only in the relative field pattern, the factor 2 before the summation sign in (18) and (19) may be dropped.
For an array of n sources, the first step is to select the Tchebyscheff poly-nomial of the same degree as the array polynomial, (3) or (5). This is given by ^n-iW (20) 170 arrays of pojnt sources where n is the number of sources and m = n 1. Next we choose R and solve T m{x 0) = R (21) for x 0 . Referring to Fig 4-33, we note that, for R > I, x0 is also greater than unity. This presents a difficulty since, according to (12), x must be restricted to the range — 1 <, x < + 1.
If, however, a change of scale is made by introducing a new abscissa w (Fig. 4-33), where w = — (22) o then the restriction of (12) can be fulfilled by putting w — cos (23) where now the range of w is restricted to — 1 < w < + 1. The pattern polynomial, (18) or (19), may now be expressed as a polynomial in w. The final step is to equate the Tchebyscheff polynomial of (20) and the array polynomial obtained by substituting (23) into (18) or (19). Thus, T 9 - X(x)-Em (24) The coefficients of the array polynomial are then obtained from (24), yielding the Dolph-Tchebyscheff amplitude distribution which is an optimum for the side-lobe level specified.
As a proof of the optimum property of the Tchebyscheff polynomial, let us consider any other polynomial P(x) of degree 5 which passes through (x0 , R) in Fig. 4-33 and the highest root x\ and for all smaller values of x lies between + 1 and - 1.
If the range in ordinate of P(x) is less than ± 1, then this polynomial would give a smaller sidelobe level for this same beam width, and T 5(x) would not be optimum. Since F(.) lies between +1 in the range —x\ <x< + xi it must intersect the curve T 3 (x) in at least m + 1 = 6 points, including (x 0 , R). Two polynomials of the same degree m which intersect in m + 1 points must be the same polynomial, 1 so that p(x)=r 5 (x) and the T 5 (x) polynomial is, therefore, the optimum.
If the spacing between sources exceeds A/2, it should be noted that as the spacing approaches X a large lobe develops at 0 = +90° which equals the main lobe when d = X. However, if the individual sources of the array are nonisotropic, i.e, are directional with the maximum at & = 0 and with little or no radiation 1 This follows from the fact that a polynomial of degree m has w + I arbitrary constants. Further, if m -F 1 points on the polynomial's curve are specified, m + 1 independent equations with m + 1 unknowns can be written and the m + l constants thereby determined.
-Fl2 EXAMPLE OF DOLPH-TCHEBYSCHEFF DISTRIBUTION 171 at 0 = ±90°, then by pattern multiplication the lobes of the total pattern at 0 — +90" can be made small 4-12 EXAMPLE OF DOLPH-TCHEBYSCHEFF DISTRIBUTION FOR AN ARRAY OF 8 SOURCES- To illustrate the method for finding the Dolph-Tchebyscheff distribution, let us work the following problem.
An array of n = 8 in-phase isotropic sources, spaced k/2 apart, is to have a sidelobe level 26 dB below the main-lobe maximum. Find the amplitude distribu-tion fulfilling this requirement that produces the minimum beam width between first nulls, and plot the field pattern.
Since Sidelobe level in dB below main-lobe maximum = 20 log ]0 R (1) 1 it follows that R = 20 (2) The Tchebyscheff polynomial of degree n — 1 is r 7(x). Thus, we set T 7( 0) = 20 (3) The value of x0 may be determined by trial and error from the T 7(x) expansion as given in (4-11-15) or x0 may be calculated from 0 = + Jr 2 -() 1/m + (R - Jr 1 - i) I/m ] w Substituting R = 20 and m = 7 in (4) yields (5) Now substituting (4-1 1-23) in (4-1 1-18) and dropping the factor 2, we have E fl = A g w + Aj(4 w 3 - 3w) + A 2( 16w 5 - 20w + 5w) + A 3(64w 7 — I12w s + 56W — 7w) (6) But w = x/x0 , so making this substitution in (6) and grouping terms of like degree, 64 A 3 7 t 16A 2 - 112A l 5 .
4A x - 20 A 2 + 56 A 3 __ 3 Ed — , X 4" c X +" -i x A0 — 3A x + 5A 2 — 7A 3 l x Xn The Tchebyscheff polynomial of like degree is T ?(x) = 64x7 -1 1 2xs + 56x - lx (8) Now equating (7) and (8), E, = r 7(x) (9) 172 4 ARRAYS OF POINT SOURCES i he value of x corresponding to a given value of 0, as obtained from (13), is then introduced in the appropriate Tchebyscheff polynomial, in this case T 7(x), or scaled from a graph of this polynomial, as shown in Fig. 4-34. The value of the polynomial for this value of x is then the relative field strength in the direction 0 In general, as 0 ranges from —njl to +ir/2, the variables 2, w and x range as indicated by Table 4-6 Thus, in general, as 0 ranges from —n/2 to 0 to + tt/2, x ranges from some point, such as a in Fig. 4-34, to x D and back again to a , the ordinate value giving the relative field intensity.
In our problem, dr = n and = 1.15, so that the range of x is as shown in Table 4-7. Hence, at $ = —90° we start at the origin in Fig. 4-34 (point b) t and as 4-13 COMPARISON OF AMPLITUDE DISTRIBUTIONS FOR ^SOURCE ARRAYS 173 Table 4-6 0 approaches (T we proceed to the right along the polynomial curve, reaching the point (x0 , R — 1.15, 20) when 0 = 0°. As 0 continues to increase, we retrace the polynomial curve, reaching the origin when 0 = 90°, Thus, the pattern is sym-metrical about the 0 = (T direction.
As a preliminary step to plotting the field pattern, it is usually helpful io make a plot of x versus 0 from (13). Then, knowing the values of x for the nulls and maxima of the T m(jc) curve, the corresponding values of 0 may be determined.
As many intermediate points as are needed may also be obtained in the same manner.
Following this procedure, the field pattern for our problem of the 8-source array is presented in Fig. 4-35a in rectangular coordinates and in Fig. 4-356 in polar coordinates.
4-13 COMPARISON OF AMPLITUDE DISTRIBUTIONS FOR 8-SOURCE ARRAYS, In the problem worked in the preceding section, the sidelobe level was 26 dB below the maximum of the main beam ( R = 20). It is of interest to compare the amplitude for this case with the distributions for other sidelobe levels This is done in Fig. 4-36, in which the relative amplitude distribu-tions are shown for 8-source arrays with sidelobe levels ranging from 0 dB to an infinite number of decibels below the main beam maximum The infinite decibel case corresponds to Ji = oc (zero sidelobe level) and is identical with Stone’s binomial distribution. The relative amplitudes for this case are 1, 7, 21, 35, 35, 21, Table 4-7 174 4 ARRAYS OF POINT SOURCES Figure 4-55 Relative field pattern of broadside array of 8 isotropic sources spaced Xf2 apart. The amplitude distribution gives a minimum beam width for a sidelobe level ^ of the main lobe. The pattern is shown in rectangular coordinates at (a) and in polar at (b). Both diagrams show the pattern only from —90^ to + 9CT, the other half of the pattern being identical.
7, l.
1 The ratio of amplitudes of the center sources to the edge sources is 35 to 1.
Such a large ratio would be very difficult to achieve in practice. As the sidelobe level increases ( R decreases), the amplitude distribution becomes more uniform, the ratio of the center to edge amplitudes being only about 3 to 1 for the 26-dB (R = 20) case. The 20-dB case (R — 10) is more uniform, with an amplitude ratio of only 1.7 to L The 14-dB case {/? = 5) exhibits a still more uniform distribution but shows an inversion, the maximum amplitude having shifted to the outermost sources (1 and 8). The uniform distribution is not a special case of the Dolph-Tchebyscheff distribution, an inversion occurring before the uniform case is reached. As the sidelobe level is raised still further, the distribution tends more towards an edge type, the amplitude of the inner sources decreasing still further In the extreme case, where the sidelobes are equal to the main-lobe level (0 dB, or R — 1}, the amplitudes of all of the inner sources are zero, and the distribution is of the edge type discussed in connection with Fig. 4-29d. Thus, both the binomial and edge distributions are special cases of the Dolph-TchebyschefT distribution, but the uniform amplitude distribution is not The point of nearest approach to the uniform distribution is for an R value between 5 and 10. Referring to Fig.
4-20 and interpolating for n = 8 between the curves for n = 10 and n = 5, it is interesting to note that the ratio of the main-lobe maximum to the minor-lobe s may be noted by extending Pascal's triangle {Sec. 4-10) to « = 8.
4-14 CONTINUOUS ARRAYS 175 Linear array of 8 sources Figure 4-36 Comparison of Dolph-Tchebyscheff am plitude distribution envelopes for various sidelobe levels.
maxima ranges from about 4.3 to 8 for an array of eight sources of uniform amplitude.
The Dolph-Tchebyscheff optimum amplitude distribution, as discussed in the preceding sections, is optimum only if d > a/2, which covers the cases of most interest for broadside arrays. By a generalization of the method, however, cases with smaller spacings can also be optimized.
1 In conclusion, it should be pointed out that the properties of the Tchebys-cheff polynomials may be applied not only to antenna patterns as discussed above but also to other situations.
It is necessary, however, that the function to be optimized be expressible as a polynomial 4-14 CONTINUOUS ARRAYS, In the preceding sections, the discussion has been restricted to arrays of discrete point sources, i.e., to arrays of a finite number of sources separated by finite distances. We now proceed to a consider-ation of continuous arrays of point sources, i.e., arrays of an infinite number of sources separated by infinitesimal distances. By Huygens' principle, a continuous array of point sources is equivalent to a continuous field distribution. In this way, our discussion of continuous arrays can be extended to include the radiation patterns of field distributions across apertures, as, for example, the pattern of an electromagnetic horn where the field distribution across the mouth of the horn is known.
1 H. J. RiblcL Proc. IRE , 35, no, 5, 489-492, May 1947.
176 ARRAYS OF POINT SOURCES 2 Figure 4-37 Continuous broadside array of point sources of length a.
We shall now develop an expression for the far field of a continuous array of point sources of uniform amplitude and of the same phase. Let the array of length a be parallel to the v axis with its center at the origin as indicated in Fig. 4-37. Then the field dE at a distant point in the direction 6 due to the point sources in the infinitesimal length dy at a distance y from the origin is A A dE = — eMt-irtMi dy = e- pt dy m ri ?i w where P = (o/c = 2%fk and A is a constant involving amplitude The total field £ at the distant point is then the integrated value of (1) over the array of length a as given by ^ I' A J-a/2 'l eXut-firii dy Both A and the time factor may be taken outside the integral, and r, may also be if r x a. Thus, _ r r J However, referring to Fig. 4-37, r i = r — y sin 0 Substituting (4) in (3) and taking the constant factor e~ j0r outside the integral we have C12 \ = A> \ e J-t{2 e0 3 dy Ae- 0 where 4 J4 CONTINUOUS ARRAYS 177 Integrating (5) yields which may.be written as 2i4 ' ei0ui 4i n — e ~ P sin 9 2\ F 2A ' ($a a E = .
sin -- sin 0 p sin 0 \2 0' = pa sin 6 - ar sin 0 where ar = pa = Inajk = array length, rad However, from (9), so that (10) becomes Normalizing (II) gives finally p 2A' .
0' E-^ Sm 2 it' JS sin 6 = — a E = aA , sin Wi2) sin ('/2) Equation (12) expresses the far field, or Fraunhofer diffraction pattern, of a continuous broadside array of length a having uniform amplitude and phase. For n discrete, equally spaced sources, it was previously shown by (4-6-9) that the normalized value of the total field is £=^ > (13) n sin (0/2) 1 where 0 = d cos 0 + 5 For in-phase sources, 5 = 0.
Comparing Figs.
4-18 and 4-37 we note that 0 = 0 + tt/2, so that 0 = -dr sin 9 = - fid sin 9 (14) For small values of 0, which occur for small values of 0, d or both, (13) can be expressed as ^ _ sin (n0/2) _ sin _(findf2) sin ff] mft/2 (Pnd/2) sin 9 (15) 178 4 ARRAYS OF POINT SOURCES The length a of the array of discrete sources is a = d(n -1) where n = number of sources d = spacing If rt $> 1, a ^ nd and (15) becomes sin [(/to/2) sin 0] _ sin [(ar/2) sin 0] (/to/2) sin 9 {oJ2} sin 0 where ar = /to — 2nai?» By (9) this can now be expressed as sin W!2) which is identical with the value obtained in (12^ for the continuous array Thus, the field pattern for an array of many discrete sources (n P 1) and for small values of ^ is the same as the pattern of a continuous array of the same length.
If the array is long, that is, if nd ^ A, the main beam and the first minor lobes are confined to small values of 0 It therefore follows that the main features of the pattern of a large array are the same, whether the array has many discrete sources or is a continuous distribution of sources. Many of the conclusions derived in previous sections concerning amplitude distributions for arrays of dis-crete sources can also be applied to continuous arrays provided that the arrays are large.
The null directions 0O of the continuous array pattern are given by — = +Kn 2 where K = I, 2, 3, ...
/ = arcsin I ±— j For a long array (20) can be expressed as n K , ^ 57,3K , 0O =; ± — (rad) as ± -—~ (deg) &x a x where a ^ afX The beam width between first nulls (K = 1) for a long array is then 20O i - ~ frad) ^— (deg) (22) 4-15 HUYOENS L PRINCIPLE 179 level v & IO Figure 4-38 Main-lobe field patterns of continuous 0 uniform broadside arrays 5, 10 and 50a long.
It is to be noted that (20), (21) and (22) are identical with the expressions given for the broadside array of discrete sources, if nd is replaced by a (see Table 4-2, Sec. 4-7). Therefore, the null locations for long arrays of either discrete or contin-uous sources are the same provided only that n l.
The field patterns of the main beam of continuous arrays of point sources 5, 10 and 50A long are compared in Fig. 4-38, It may be noted that the beam width between half-power points, 0HP , of a long, uniform broadside array is given approximately by 0HP — O.90oi — ak (rad) (23) or 0hp — (deg) &x (24) 4-15 HUYGENS’ PRINCIPLE, 1 The principle proposed by Christian Huygens (1629-1695) has been of fundamental importance to the development of wave theory. Huygens 1 principle states that each point on a primary wavefront can be considered to be a new source of a secondary spherical wave and that a second-ary wave front can be constructed as the envelope of these secondary waves , as suggested in Fig. 439. Thus, a spherical wave from a single point source propa-gates as a spherical wave as indicated in Fig. 4-39a, while an infinite plane wave continues as a plane wave as suggested in Fig. 4-396. This principle of physical optics can be used to explain the apparent bending of electromagnetic waves around obstacles, i.e., the diffraction of waves, a diffracted ray being one that follows a path that cannot be interpreted as either reflection or refraction.
1 C Huygens, Traite de la Lumiere, Leyden, 1690.
Max Bom, Opfifc, Springer-Verlag, 1933.
Arnold Sommerfdd, “Theorie der Beugung," in Frank and Yon Mises (eds). Differential und Inte-gralgleichungen der Mechanik und Physik Vieweg, 1935, 180 4 arrays of poir-rr sources 4-13 HUYGENS’ PRINCIPLE 181 Figure 4-39 Spherical and plane wave fronts with secondary waves of Huygens.
Let us consider the situation shown in Fig 4-40a in which an infinite plane electromagnetic wave is incident on an infinite flat sheet which is opaque to the waves. The sheet has a slot of width a and of infinite length in the direction normal to the page. The field everywhere to the right of the sheet is the result of the section of the wave that passes through the slot. If a is many wavelengths, the field distribution across the slot may be assumed, in the first approximation, to be uniform, as shown in Fig. 4-40&. By Huygens' principle the field everywhere to r 'y T a L / Sheet Plane wave FlptftMO p[ane Relative amplitude ib) the right of the sheet is the same as though each point in the plane of the slot is the source of a new spherical wave. Each of these point sources is of equal ampli-tude and phase. Thus, by Huygens 1 principle the slotted sheet with a uniform field across the opening can be replaced by a continuous array of point sources which just fills the opening. The field pattern in the xy plane (Fig. 4-40— / (2) where a is the width or aperture of the slot, which is assumed to be large. Thus, the larger the aperture or the shorter the wavelength, the greater must be the distance at which the pattern is measured if we wish to avoid the effects of Fresnel diffraction. This is discussed further in Sec. l$-3a.
wave incident on opaque sheet with sJot of width a.
182 4 ARRAYS OF POINT SOURCES | id) Figure 4-41 Fresnel and Fraunhofer patterns of a slot of width a.
A nearly uniform type of field distribution across an aperture such as dis-cussed above in connection with Figs. 4-40 and 4-41 occurs in optics when a beam of light is incident on a slit.
It also may be realized by the field distribution across the mouth of a long electromagnetic horn antenna as in Fig. 4-42a. Since the pattern of a uniform field distribution is the same as the pattern of a uniform distribution of point sources of equal extent, another form of antenna equivalent to the optical slit or electromagnetic horn is a uniform curreni sheet. This can be approximated by a “billboard" type of array, as in Fig. 4-42b, having many dipole antennas carrying equal currents. The expressions which have been devel-oped can thus be applied to a calculation of the Fraunhofer diffraction pattern of an optical slit or the far field of a horn or uniform current sheet.
If the field or current distribution across the slit or antenna aperture is not uniform, the form factor for the distribution will appear in the integral for the field expression. If the aperture is large, the relations developed for amplitude distributions of arrays of discrete sources can be applied to the case of continuous arrays of sources.
4-1 HUYGENS’ PRINCIPLE APPLIED TO THE DIFFRACTION OF A PLANE WAVE 183 Array of dipoles with ref factor Figure 4-42 Electromagnetic horn antenna and array of dipoles with reflector.
Huygens' principle is not without its limitations. Thus, it neglects the vector nature of the electromagnetic field.
It also neglects the effect of currents which flow at the edge of the slot, as in Figs. 4-40 and 4-41, or at the edge of the horn, as in Fig. 4-42a. However, if the aperture is sufficiently large and we confine our attention to directions roughly normal to aperture, the scalar theory of Huygens' principle gives satisfactory results.
4-16 HUYGENS’ PRINCIPLE APPLIED TO THE DIFFRAC-TION OF A PLANE WAVE INCIDENT ON A FLAT SHEET PHYSICAL OPTICS, Consider a uniform plane wave incident on a perfectly conducting half-plane, as in Fig. 4-43.
1 We want to calculate the electric field at point P at a distance r behind the plane. By Huygens' principle, X »JU3 where dE is the electric field at P due to a point source at a distance x from the origin, as in Fig. 4-43b. Thus, J. D. Kraus, Radio Astrojtomy, 2nd ed rj Cygnus-Quasar Boots, 1986.
184 4 ARRAYS OF POINT SOURCES Conducting Wavefront of incident plane wave half-plane^ p— X— Secondary Geometric optics 5 Shadow side Illuminated side \physical optics kalk = constant!
^ Figure 4-43 Plane wave incident from above onto a conducting half-plane with resultant power-density variation below the plane as obtained by physical optics.
SO that If <5 < r y it follows that -?~f x S —r When we let 2jrk = k2 and kx = u f (3) becomes which can be rewritten as E = e ser j e du kr :n as e -'— 1 du -I c du The integrals in (6) have the form of Fresnel integrals so (6) can be written E = f?
e ~m ^ + ^~ LCika) +^(fca)] i rta nu2 where C(fca) = cos— du = Fresnel cosine integral Jo 2 f 1 TIU 2 £(fca) = ) sin —- du = Fresnel sine integral Jo 2 where ka = /—a, dimensionless-4-16 HUYGENS’ PRINCIPLE APPLIED TO THE DIFFRACTION OF A PLANE WAVE 185 0 Figure 444 Cornu spiral show-ing Qka) and Sika) as a function of ka values along the spiral.
For example when ka 1.0 C{ka) = D.7&D and S(fca) = 0.338.
When ka = cc C{ka) = S{ka) = A graph of C(ka) and S(ka) yields the Cornu spiral (Fig, 4-44) Since C(— Jt<a) = — C(fca) and S(— ka) = — S(ka) t the spiral for negative values of ka is in the third quadrant and is symmetrical with respect to the origin for the spiral in the first quadrant.
The power density as a function of ka is then EE s„ =— = So {[ - C(fca)] 2 + a - S{ka)] 2 } (W m “ 2 ) (10) F 2 1 where S0 = -^-(W m“ 2 ) (11) 2Zr The power density variation of (10) as a function of ka (with r, k and k constant) is shown in Fig. 4-43c, Assuming that the plane wave originates from a distant source we have 1. For no obscuration, ka = — ao and Sav = S0 .
2. For source, observer and edge of obscuring plane in line, ka = 0 and S1V = lc 4 J 0 3 For complete obscuration, ka = + oo and = 0 Thus, the power density does not go to zero abruptly as the point of observation goes from the illuminated side (ka < 0) to the shadow side (ka > 0); rather, there are fluctuations followed by a gradual decrease in power density.
From (10) and (1 1) the relative power density as a function of ka is S„(relative) = = ±{[i - C(ft«)] 2 + [i - Sfke)] 1 ?
( 12) 186 4 ARRAYS OF POINT SOURCES The relative power density (12) is equal to ^R 2 , where R is the distance from a ka value on the Cornu spiral to the point (see Fig. 4-44), For large posi-tive values of ka, R approaches l/nka, so that (12) reduces approximately to S av(relative) = £ rk 4tz 1 a 2 (13) where r — distance from obstacle (conducting haJLplane), m k = wavelength, m a = distance into shadow region, m 1 Equation (13) gives the relative power density for large ka ( > 3} (well into the shadow region). For this condition it is apparent that the power flux density (Poynting vector) due to diffraction increases with wavelength and with distance (from edge) but decreases as the square of the distance a into the shadow region.
Example. A vertical conducting wall 25 m high extends above a flat ground plane.
A / = 10-cm transmitter is situated 25 m above the ground plane at a large distance to one side of the vertical wall and a receiver is located on the ground plane 100 m to the other side of the wall. Find the signal level at the receiver due to diffraction over the wall as compared to the direct path signal without the wall.
Solution The constant k = s/2Irk = ^2/ 100 x 0.1 =0,44 and a = 25 m, so ka = 11 which is greater than 3. Thus, { 13) is applicable and Slv(relative) = TA 4nW 100 x 0.1 1 4ji 2 x 25 1 ~ 2500 °‘ — 34 dB Thus, the vertical wall causes 34 dB of attenuation as compared to a direct path signal.
If the half-plane in Fig. 4-43 is replaced by a strip of width D and length $>D , diffraction occurs from both edges, scattering radiation into the shadow region behind the strip. On the centerline of the strip, diffraction fields from both edges are equal in magnitude and of the same phase since the path lengths from both edges are equal Thus, the diffracted field has a maximum or central peak on the centerline.
If the strip is replaced by a disk of diameter D, there is diffraction around its entire edge and all diffracted fields arrive in phase on the centerline behind the disk producing a larger central peak. In optics this peak is called the axial bright spot . In a similar way, the diffracted fields from the feed system at the focus of a parabolic dish reflector can produce a back lobe on the axis of the parabola. See additional discussions on diffraction in Secs, 2-18, 12-2, 13-3, 17-5 and 18-3d.
4-17 RECTANGULAR-AREA BROADSIDE ARRAYS. The method of obtaining the field patterns of linear arrays discussed in the preceding sections 4-17 RECTANGULAR AREA BROADSIDE ARRAYS 187 Figure 4-4$ Rectangular broadside array of bright a and Length b with relation to coordinates.
can be easily extended to the case of rectangular broadside arrays, ie, arrays of sources which occupy a flat area of rectangular shape, as in Fig 4-45 For such a rectangular array, the field pattern in the xy plane (as a function of B) depends only on the y-dimension a of the array, white the field pattern in the xz plane (as a function of 0) depends only on the z-dimension b of the array. The assumption is made that the field or current distribution across the array in the y direction is the same for any values of z between ±bf2.
Likewise, it is assumed that the amplitude distribution across the array in the z direction is the same for all values of y between ±a/2 Therefore, the field pattern in the xy plane is calcu-lated as though the array consists only of a single linear array of height a coin-cident with the y axis (y array). In the same way, the pattern in the xz plane is obtained by calculating the pattern of a single linear array of length b coincident with the z axis (z array).
If the array also has depth in the x direction, ie. has end-fire directivity, then the pattern in the xy plane is the product of the patterns of the single linear x and y arrays, while the pattern in the xz plane is the product of the patterns of the jc and z arrays.
If the area occupied by the array is not rectangular in shape, the above principles do not hold. However, the approximate field patterns may be obtained in the case of an array of elliptical area, for example, by assuming that it is a rectangular area as in Fig. 4-46u or in the case of a circular area by assuming that it is square as in Fig 4-46f>.
From the field patterns in two planes (xy and xz) of a rectangular array the beam widths between half-power points can be obtained.
If the minor lobes are ]$# 4 ARRAYS OF POINT SOURCES Figure 4-46 Elliptical array wilh equiva-lent rectangular array (a) and circular array with equivalent square array [h'-not large, the directivity D is then given approximately by 41000 0\4>l ( 1 ) where and are the half-power beam widths in degrees in the xy and xz planes respectively.
The limitations of (1) are discussed following (3-13-16), leading to (3-13-18) which includes correction factors.
An expression for the directivity of a large rectangular broadside array of height a and width b (Fig, 4-45) and with a uniform amplitude distribution may also be derived rigorously as follows. By (3-13-8) the directivity of an antenna is given by jj /(0, (j > ) sin 0 dO d (2 ) where /(0, tj>) is the space power pattern, which varies as fhe square of the space held pattern. From (4-14-17) the space field pattern of a large rectangular array is m ) = sin [(ar sin 9)/ 2] sin [(hr sin <£)/2] { a r sin 9}/2 (br sin 0)/2 (3) where aT = 2i\ajk br = 2xbfk The main beam maximum is in the direction 9 = = 0 in Fig. 4-45. In (3), 0 = 0 at the equator, while in (2), 9 .= 0 at the zenith. For large arrays and relatively sharp beams we can therefore replace sin 9 and sin in (3) by the angles, while sin 9 in (2) can be set equal to unity. Assuming that the array is unidirectional (no field in the — x direction), the integral in the denominator of (2) then becomes V 2 |;2 sjn 2 s jn -!2 J- jc/2 InaBfk) 1 {n b/k)2 (4) Making the limits of integration — oc to +00 instead of — n/2 to + rc/2, (4) may be evaluated as k 2 fab> Therefore, the approximate directivity D of a large uni-directional rectangular broadside array with a uniform amplitude distribution is D = 4kab ^ ah IT ‘ 116 7 (5) 4-IS ARRAYS WITH MISSING SOURCES AND RANDOM ARRAYS 189 I As an example, the directivity of a broadside array of height a = 1CU and length b = 20k is, from (5), equal to 2520, or 34 dB.
4-18 ARRAYS WITH MISSING SOURCES AND RANDOM ARRAYS- A linear array of 5 isotropic point sources with k/2 spacing is dis-cussed in Sec. 4-10 for several amplitude distributions including uniform, bino-mial and Dolph-Tchebyscheff. Let us consider this 5-source array again with all sources of equal amplitude with pattern as in Fig. 4-47a (same as Fig. 4- 29a) and Figure 4«47 Field patterns of linear array of 5 isotropic point sources of equal amplitude and A/2 spacing: (a) all 5 sources on, (b) one source (next to the edge) off, (c ) one source (at the center) off and (4) one source (at the edge) off 190 4 ARRAYS OF POINT SOURCES note what happens to the pattern if one of the sources is turned off (amplitude reduced to zero), Tf the off source is next to the edge source, the beam width is essentially unchanged but the minor-lobe level is up and nulls are tilled as shown in Fig, 4-47h. When the center source is off, the beam width is reduced but the minor-lobe level is higher, as indicated in Fig. 4-47c. If an edge source is off, the array is identical with a uniform array of 4 sources and, as shown in Fig. 4-47d, has a larger beam width with the minor-lobe level slightly higher than for the uniform source array of Fig. 4-47a. (Compare curves for n - 4 and 5 in Fig. 4-20.) If the amplitude distribution of the array is tapered (binomial or Dolph-Tchebyscheff) as in Fig. 4-296 and c , turning off a source at the edge will have less effect than in the uniform-amplitude case.
In an array of a large number of sources, it is of interest to know what happens if one or more sources are turned off either intentionally or inadvertent-ly. Also to reduce cost, a designer would like to know how many (and which) sources can be omitted without appreciably affecting the performance character-istics.
It has been shown by Lo and Maher and Cheng 1 that if sources are posi-tioned with random instead of uniform spacing in a large array (L > 2) the number n of sources can be reduced without affecting the beam width appre-ciably. The gain, however, is proportional to n and to keep the largest siddobes below a certain level, a minimum n is required.
PROBLEMS 2 4-1 Two point sources.
(a) Show that the relative E(0) pattern of an array of 2 identical isotropic in-phase point sources arranged as in Fig, P4-1 is given by = cos [(^2) sin ^], where dr = 2nd/X, Figure P4-1 Two point sources.
1 Y. T. Lo, “A Probabilistic Approach to the Design of Large Antenna Arrays,’' IEEE Trans. Anti.
Prop., AP-11, 95-96, 1961 T. M. Maher and D. K. Cheng, “ Random Removal of Radiators from Large Linear Arrays,” IEEE Trans. Antst. Prop., AP-11, 106-112, 1963 2 Answers to starred {) problems are given in App. D.
PROBLEMS 191 (b) Show that the maxima, nulls and half-power points of the pattern are given by the following relations: Maxima: Nulls: Half-power points: = arcsin (2 lbf 2d where k = 0, 1, 2, 3, .. ., (c) For d = A find the maxima, nulls and half-power points, and from these points and any additional points that may be needed plot the pattern for 0° < £ 3W. There are four maxima, four nulls and eight half-power points.
(J) Repeat for d = 3 a/2.
(e) Repeat for d - 4A.
(/) Repeat for d = a/4. Note that this pattern has two maxima and two half-power points but no nulls. The half-power points are minima.
4-2 Four sources in square array, () Derive an expression for E(tf>) for an array of 4 identical isotropic point sources arranged as in Fig. P4-2, The spacing d between each source and the center point of the array is 32/8. Sources 1 and 2 are in phase, and sources 3 and 4 in opposite phase with respect to 1 and 2.
() Plot, approximately, the normalized field pattern.
d L Figure P4-2 Four sources in square array.
4-3 Two point sources.
(a) What is the expression for E(0) for an array of 2 point sources arranged as in the figure for Prob. 4-1. The spacing d is 32/8, The amplitude of source 1 in the plane is given by |cos and the phase by The amplitude of the field of source 2 is given by [cos (0 - 45°) | and the phase of the field by ^ - 45 c .
(i?) Plot the normalized amplitude and the phase of £(^) referring the phase to the centerpoint of the array.
4-4 Four sources in broadside array.
() Derive an expression for £(<£) for a linear in-phase broadside array of 4 identi-cal isotropic point sources. Take $ = 0 in the broadside direction. The spacing between sources is 52/8.
() Plot, approximately, the normalized field pattern ((F <. £ <, 360°).
192 4 ARRAYS OF POINT SOURCES (ci') Repeat parts (a) and (0) with the changed condition that the amplitudes of the 4 sources are proportional to the coefficients of the binomial series for {a + bf~ 1 4-5 Tchebyscheff r 3 (x)and 7 6(r) (a) Calculate and plot cos 0 as x and cos 30 as y 5 for — 1 < x < +1. Compare with the curve for (0) Calculate and plot cos & as x and cos 60 as y, for - 1 < x <. + 1. Compare with the curve for F 6(x).
4-6 Five source Dolph-Tchebyscheff (D-T) distribution.
fa\ Find the Dolph-Tchebyscheff current distribution for the minimum beam width of a linear in-phase broadside array of five isotropic point sources. The spacing between sources is a/2 and the sidelobe level is to be 20 dB down. Take $ — 0 in the broadside direction.
(0) Locate the nulls and maxima of the minor lobes (c) Plot, approximately, the normalized field pattern (0° <; 4> < 360°).
(if) What is the half-power beam width?
47 Eight-source D-T distribution.
(n) Find the Dolph-Tchebyscheff current distribution for the minimum beam width of a linear in-phase broadside array of 8 isotropic sources. The spacing between elements is 3 a/4 and the sidelobe level is to be 40 dB down. Take $ = 0 in the broadside direction.
(,h ) L ;ate the nulls and the maxima of the minor lobes.
{c} Plot, approximately, the normalized field pattern (O ' < # < 360''").
fd) What is the half-power beam width?
4-8 n -source array.
(a) Derive an expression for for an array of n identical isotropic point sources where = /(<£, d» £).
is the azimuthal position angle with — 0 in the direc-tion of the array.
<5 is the phase lag between sources as one moves along the array in the $ = 0° direction and d is the spacing.
(h) Plot the normalized field as ordinate and t/ as abscissa for n = 2, 4, 6, 8, 10 and 12 for 0° < 4f < 180 C .
4-9 Ten-source end-fire array.
(g) Plot E() for an end-fire array of n = 10 identical isotropic point sources spaced 3//8 apart with 6 = — 3?r/4.
(0) Repeat with d = — jc[(3/4) -I- (1/m)].
4-10 Two-source broadside array.
(u) Calculate the directivity of a broadside array of two identical isotropic in-phase point sources spaced a/2 apart along the polar axis, the field pattern being given by E = cos cos 0^ where 9 is the polar angle, (0) Show that the directivity for a broadside array of two identical isotropic in-phase point sources spaced a distance d is given by 2 D 1 + (A/2izd) sin (2nd/A) problems 193 4-11 Two-source end-fire array.
(a) Calculate the directivity of an end-fire array of two identical isotropic point sources in phase opposition, spaced i/2 apart along the polar axis, the relative field pattern being given by E = sin ^ cos 0^ where 0 is the polar angle.
(b) Show that the directivity of an ordinary end-fire array of two identical isotropic point sources spaced a distance d is given by D -I + {XjAnd) sin (4nd/X) 4-12 Four-tower BC array. A broadcasting station requires the horizontal plane pattern indicated by Fig. P4^12. The maximum field intensity is to be radiated northeast with as little decrease as possible in field intensity in the 90° sector between north and east. No nulls are permitted in this sector. Nulls may occur in any direction in the complementary 270" sector. However, it is required that nulls must be present for the directions of due west and due southwest, in order to prevent interference with other stations in these directions.
Design a four-vertical-tower array to fulfill these requirements. The currents are to be equal in magnitude in all towers, but the phase may be adjusted to any relationship. There is also no restriction on the spacing or geometrical arrange-ments of the towers. Plot the field pattern.
4-13 Two-source patterns. Calculate and plot the field and phase patterns for an array of two isotropic sources of the same amplitude and phase, for two cases-M d = \k {) d = U Plot the field pattern in polar coordinates and phase pattern in rectangular coordi-nates with: 1. Phase center at source 1 2. Phase center at midpoint 194 4 ARRAYS OF POINT SOURCES 4~14 Field and phase patterns. Calculate and plot the field and phase patterns of an array of 2 nonisotropic dissimilar sources for which the total field is given by E — cos $ + sin where ^ = d cos + <5 = — (cos $ + 1 ) Take source 1 as the reference for phase. See Fig, P4-14.
Figure P4-14 Field and phase patterns, + 4-15 DT 6-source array.
Calculate the Dolph-TchebyschefT distribution of a six-source broadside array for R = 5, 7 and 10. Explain the variation.
4-16 Two-unequal-source array. In Case 5 (Sec. 4-2e) for 2 isotropic point sources of unequal amplitude and any phase difference show that the phase angle of the total field with midpoint of the array as phase center is given by 4-17 Field and phase patterns. Calculate and plot the field and phase patterns for the cases of Figs. 4-21 and 4-22 and compare with the curves shown.
4-18 Five-source array.
(a) What is an expression for the field pattern of an array of 5 identical isotropic point sources arranged in line and spaced a distance d (<A/2) apart? The phase lead of source 2 over 1, 3 over 2, etc,, is 5.
(b) What value should 5 have to make the array a broadside type? For this broad-side case, what are the relative current magnitudes of the sources for: 1. Maximum directivity 2. No sidelobes 3. Sidelobes equal in magnitude to “ main ” lobe 4-19 Two-tower BC army. A broadcast array of 2 vertical towers with equal currents is to have a horizontal plane pattern with a broad maximum of field intensity to the norjfi and a null at an azimuth angle of 131 e measured counterclockwise from the north. Specify the arrangement of the towers, their spacing, and phasing. Calculate and plot the field pattern in the horizontal plane, 4-20 Three-lower BC array. A broadcast array with 3 vertical towers arranged in a straight horizontal line is to have a horizontal -plane pattern with a broad maximum of field intensity to the north and nulls at azimuth angles of 105% 147 c and 213° measured counterclockwise from the north. The towers need not have equal currents. For the purpose of analysis the center tower (2) may be regarded as PROBLEMS 195 2 towers, 2a and 2b t 2a belonging to an array of itself and tower 1 and 2b to an array of itself and tower 3. Specify the arrangement of towers, their spacing, cur-rents and phasing. Calculate and plot the field pattern in the horizontal plane.
4-21 Four-tower BC array. A broadcast array of 4 vertical towers with equal currents is to have a symmetrical 4-lobed pattern in the horizontal plane with maximum field intensity to the north, east, south and west arid a reduced field intensity to the northeast, southeast, southwest and northwest equal to -j the maximum. Specify the array arrangement, orientation, spacing and phasing. Calculate and plot the field pattern in the horizontal plane.
4-22 Eight-source end-fire array.
(a) Calculate and plot the field pattern of a linear array of 8 isotropic point sources of equal amplitude spaced 0.2A apart for the ordinary end-fire condition.
(b) Repeat, assuming that the phasing satisfies the Hansen and Woodyard increased-directivity condition.
(c) Calculate the directivity in both cases by graphical or numerical integration of the entire pattern.
4-23 Rectangular current sheet. Calculate and plot the patterns in both planes perpen-dicular to a rectangular sheet carrying a current of uniform density and everywhere of the same direction and phase if the sheet measures 10 by 202. What is the approximate directivity?
4-24 Twelve-source end-fire array.
(a) Calculate and plot the field pattern of a linear end-fire array of 12 isotropic point sources of equal amplitude spaced A/4 apart for the ordinary end-fire condition.
(b) Calculate the directivity by graphical or numerical integration of the entire pattern. Note that it is the power pattern (square of field pattern) which is to be integrated. It is most convenient to make the array axis coincide with the polar or z axis of Fig. 3-2 so that the pattern is a function only of 0 , {c) Calculate the directivity by the approximate half-power beam-width method and compare with that obtained in (b).
4-25 Twelve-source broadside array.
(a) Calculate and plot the pattern of a linear broadside array of 12 isotropic point sources of equal amplitude spaced A/4 apart with all sources in the same phase.
(b) Calculate the directivity by graphical or numerical integration of the entire pattern and compare with the directivity obtained in Prob. 4-24 for the same size array operating end-fire.
{c) Calculate the directivity by the approximate half-power beam-width method and compare with that obtained in (b).
4-26 Twelve-source end-fire array with increased directivity.
(a) Calculate and plot the pattern of a linear end-fire array of 12 isotropic point sources of equal amplitude spaced A/4 apart and phased to fulfill the Hansen and Woodyard increased-directivity condition.
(b) Calculate the directivity by graphical or numerical integration of the entire pattern and compare with the directivity obtained in Probs. 4-24 and 4-25.
(c) Calculate the directivity by the approximate half-power beam-width method and compare with that obtained in (b).
4-27 n-sonrce array. Variable phase velocity.
Referring to Fig. 40 S, assume that the uniform array of n isotropic point sources is connected by a transmission system 1% 4 ARRAYS OF POINT SOURCES extending along the array with the feed point at source 1 so that the phase of source 2 Jags l by wid/u, 3 Jags 1 by 2rW/i?, etc., where v is the phase velocity to the right along the transmission system. Show that the far field is given by (4-6-8) where $ = Jr[cos — 3L = Lf [cos $ — (1/p)], where p = vfc as in Prob.
4-27. Show also that for the four cases considered in Prob. 4-27 the p values are the same except for the increased-directivity end-fire case where p = 1/[1 + (1/2LJ], 4-29 Binomial distribution. Use the principle of pattern multiplication to show that a linear array with binomial amplitude distribution has a pattern with no minor lobes.
430 Two-source array. Show that for a 2-source array the field patterns sin (rt^/2) sin W2) and E ^ = 2 cos — 7 are equivalent, 4-31 Directivity of ordinary end-fire array. Show that the directivity of an ordinary end-fire array may be expressed as 1 + {Xflnnd) ^ [{rc — k)//t] sin (4tikd/X) k = I Note that [sin (n^/2f| 2 n ~ l & , , =n + X 2{n - k) cos 2k ^ L sm W2) J = i 2 4-32 Directivity of broadside array. Show that the directivity of a broadside array may be expressed as 1 + {Xjnruf) ^ [(n — k)fk] sin {InkdjX) = 1 4-33 Phase center. Show that the phase center of a uniform array is at its centerpoint.
4-34 Three-source array. The center source of a 3-source array has a (current) amplitude of unity. For a sidelobe level 0.1 of the main-lobe maximum field, find the Dolph-Tchebyscheff value for the amplitude of the end sources. The source spacing d = 2/2 .
6 4-35 End-fire arrays. 2/2 spacing. The following BASIC program provides antenna field-pattern graphs in polar and rectangular coordinates for an array of 4 sources as illustrated in Fig. P4-35. Using modifications of this program, produce graphs of PROBLEMS 197 the patterns for a larger number of sources spaced 2/2 apart with end- fire phasing, such as (a) 5 sources, (b) 6 sources, (c) 8 sources and (d ) 12 sources.
End fire 0 ) Figure P4-35 End- ft re array of 4 sources with 2 : 2 spacing. Field patterns in polar and rec-tangular coordinaies.
END-FIRE ARRAY N - 4 d = 2, 2 POLAR PLOT 10 HOME 20 HGR 30 HCOLOR = 3 40 FOR A - .02 TO 3.12 STEF 01 50 R = 15 SIN(6.28 (COS(A) - 1))/SIN(1.57 (COS(A) -1)) 60 HPLOT 138 + R C0S(A),79 + fl SfiM(A) 61 HPLOT 138 + R (-COS(A)) r 79 + R SIN(A) 70 NEXT A RECTANGULAR PLOT (POLAR PLOT STEPS 10 60 61 70 OMITTED) 60 HPLOT A 30, R + 75 61 HPLOT (A + 3.1 6) 30, -R + 75 70 NEXT A See also App, B r 4-36 End-fire arrays, a/4 spacing.
Repeat Prob. 4-35 for the case where the spacing is a/4 instead of a/2. The patterns in this case will be unidirectional instead of bidirec-tional as with a/2 spacing.
4-37 Two-element interferometer Using a computer as in the above problems, produce graphs of the field patterns of 2 isotropic in-phase sources with spacings of (a) 8a, (b) 16a and (c) 32a, 4-38 Two sources in phase. Two isotropic point sources of equal amplitude and same phase are spaced 2a apart, (a) Plot a graph of the field pattern, (b) Tabulate the angles for maxima and nulls, , 4-39 Two sources in opposite phase. Two isotropic sources of equal amplitude and opposite phase have 1.5a spacing. Find the angles for all maxima and nulls.
PROBLEMS 199 198 ARRAYS OF POINT SOURCES 4-40 Broadside arrays, A/2 spacing. The following BASIC program provides antenna field-pattern graphs in polar and rectangular coordinates for an array of 4 sources as illustrated in Fig. P4-40. Using modifications of this program, produce graphs of the patterns for a larger number of in-phase sources spaced A/2 apart, such as (a) 6 sources, (fr) 8 sources atid (c) 12 sources.
Figure P4-40 Broadside array of 4 sources with A/2 spacing. Field patterns in polar and rectangular coordinates.
BROADSIDE ARRAY N-4 6 = A/2 POLAR PLOT 10 HOME 20 HGR 30 H CO LOR = 3 40 FOR A = .02 TO 6,26 STEP .01 50 R = 16 SIN(6.28 C0S(A))/SIN(1 .57 COS(A)) 60 HPLOT 138 + R COS(A),79 + R SIN(A) 70 NEXT A RECTANGULAR PLOT (POLAR PLOT STEPS 10 60 70 OMITTED) 6JH4PL0T A 30, R + 75 "ib NEXT A See also App, B.
4-41 Three unequal sources. Three isotropic in-line sources have kj4 spacing. The middle source has 3 times the current of the end sources. If the phase of the middle source is 0°, the phase of one end source +90° and the phase of the other end source — 90°, make a graph of the normalized field pattern.
4-42 Long broadside array. Show that the HPBW of a long uniform broadside array is given (without approximation) by 50.87^, where Lx = Ljk = length of array in wavelengths.
4-43 Phase center of 2-source array. An array consists of 2 isotropic point sources, one at the origin and one at a distance of A/2 in the x direction. If the source at the origin has twice the amplitude (field) of the other source, find the position of the phase center of the array.
4-44 24-source end-ftre array. A uniform linear array has 24 isotropic point sources with a spacing of A/2.
If the phase difference 3 = —n/2 (ordinary end-fire condition), calculate exactly (a) the HPBW, (b) the first sidelobe level, (c) the beam solid angle, id) the beam efficiency, (e) the directivity and (/) the effective aperture.
4-45 Stray factor and directive gain- The ratio of the main beam solid angle to the (total) beam solid angle is called the main beam efficiency. The ratio of the minor-lobe solid angle to the (total) beam solid angle fl.
is called the stray factor.
It follows that QA. + « = 1- Show that the average directive gain over the minor lobes of a highly directive antenna is nearly equal to the stray factor. The directive gain is equal to the directivity multiplied by the normalized power pattern [=DPf8, )], making it a function of angle with the maximum value equal to D.
4-46 Power patterns. Write and run power-pattern programs for Probs. 4-35 and 4-40.
4-47 End-fire array with increased gain. Write and run normalized field- and power-pattern programs for end- fire arrays with d = A/4, 5 = — {2itdfk) — (ft/fl) for n = 4, 8, 12 and 16.
4-48 Grating lobe pattern. Write and run field-pattern programs for broadside arrays with d = k for n = 4, 8, 12 and 16. With d £ A, grating lobes appear.
CHAPTER 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS 5-1 THE SHORT ELECTRIC DIPOLE. Since any linear antenna may be considered as consisting of a large number of very short conductors connected in series, it is of interest to examine first the radiation properties of short conduc-tors. From a knowledge of the properties of short conductors, we can then proceed to a study of long linear conductors such as are commonly employed in practice, A short linear conductor is often called a short dipole.
In the following discussion, a short dipole is always of finite length even though it may be very short. If the dipole is vanishingly short, it is an infinitesimal dipole.
Let us consider a short dipofo such as shown in Fig. 5-la. The length L is very short compared to the wavelength (L <s A). Plates at the ends of the dipole provide capacitive loading. The short length and the presence of these plates result in a uniform current/ along the entire length L of the dipole. The dipole may be energized by a/balanced transmission line, as shown.
It is assumed that the transmission liite does not radiate and, therefore, its presence will be disre-garded. Radiation from the end plates is also considered to be negligible. The diameter d of. the dipole is small compared to its length (d < L), Thus, for pur-poses of analysis we may consider that the short dipole appears as in Fig. 5-16.
Here it consists simply of a thin conductor of length L with a uniform current / 200 V2 THE FIELDS OF A SHORT DIPOLE 201 Figure 5-1 A short dipole antenna (a) and its equivalent (6).
and point charges q at the ends. The current and charge are related by dq = ; dt 5-2 THE FIELDS OF A SHORT DIPOLE, Let us now proceed to find the fields everywhere around a short dipole. Let the dipole of length L be placed coincident with the z axis and with its center at the origin as in Fig. 52.
The relation of the electric field components^ £,, E9 and £4, is then as shown.
It is assumed that the medium surrounding the dipole is air or vacuum.
In dealing with antennas or radiating systems, the propagation time is a matter of great importance. Thus, if a current is flowing in the short dipole of Fig, 5-3, the effect of the current is not felt instantaneously at the point P, but only after an interval equal to the time required for the disturbance to propagate over the distance r. We have already recognized this in Chap, 4 in connection with the pattern of arrays of point sources, but here we are more explicit and describe it as a retardation effect.
Accordingly, instead of writing the current / as 1 1 = 1^ ( 1 ) which implies instantaneous propagation of the effect of the current, we introduce the propagation (or retardation) time as done by Lorentz and write = (2) where [/] is called the retarded current.
Specifically, the retardation time rjc results in a phase retardation torfc = Infrjc radians - 360° fr/c = 360° f/T, where T = 1//= time of one period or cycle (seconds) and / = frequency (hertz, Hz = cycles per second). The brackets may be added as in (2) to indicate explic-itly that the effect of the current is retarded.
Equation (2) is a statement of the fact that the disturbance at a time t and at a distance r from a current element is caused by a current [/] that occurred at an earlier time f — r/c. The time difference r/c is the interval required for the disturbance to travel the distance r, where c is the velocity of light ( = 300 Mm s-).
Electric and magnetic fields can be expressed in terms of vector and scalar potentials. Since we will be interested not only in the fields near the dipole but also at distances which are large compared to the wavelength, we must use retar-ded potentials , i.e., expressions involving t — r/c . For a dipole located as in Fig, 5-2 or Fig. 5- 3a, the retarded vector potential of the electric current has only one component, namely, A x . Its value is ? r 4 J-LI2 5 (3) H is assumed that wc Lake either the real (cos cut) or imaginary (sin oitj part of e.
5-2 THE FIELDS OF A SHORT DIPOLE 203 where [/] is the retarded current given by [/] = (3a) In (3) and (3 a), z = distance to a point on the conductor J0 = peak value in time of current (uniform along dipole) — permeability of free space = 4n x 10“ 7 H m“ J If the distance from the dipole is large compared to its length (r L) and if the wavelength is large compared to the length (J. ^ L ), wc can put s = r and neglect the phase differences of the field contributions from different parts of the wire.
The integrand in (3) can then be regarded as a constant, so that (3) becomes (4) 4nr The retarded scalar potential V of a charge distribution is 1 fM, = --(!
4KEq Jy where [p] is the retarded charge density given by |>] = p0 ea[t -wc ' ] (6) and dx = infinitesimal volume element e0 = permittivity or dielectric constant of free space = 8.85 x 10" 11 F m" 1 Since the region of charge in the case of the dipole being considered is confined to the points at the ends as in Fig. 5-1 ft, (5) reduces to 204 J THE ELECTRIC DIPOLE AND THJN LINEAR ANTENNAS From (5-1-1) and (3 a), Substituting (8) into (7), J 0 IV- AnE0jO) L Ji - (3/tH ^ _ [£] fci/01 ^»U -IJ2/C)]" Referring to Fig. 5-3b, when r p L, the lines connecting the ends of the dipole and the point P may be considered as parallel so that L s x =r--ca%0 (10) , £ and s2 = r + — cos 0 (11) Substituting (10) and (1 1) into (9) and clearing fractions, we have _ 4it£0jw gji&Ltlc) CM#[r + (L/2J cos QJ _ e -JW2^^^r _ (L/2 j Cos X ' ~ p (12) where the term 1} cos 2 0/4 in the denominator has been neglected in comparison with r 2 by assuming that r L. By de Moivre’s theorem (12) becomes l t - ir}£)] AnSojcar 2 ioL cos 0 + j sin / gjL cos 0 , c&L cos 0 — cos j sin V 2c 2c o>L cos Q\f L A n 2c + ~2 C°S °) ')(r-fcos ) If the wavelength is much greater than the length of the dipole (x ^ L) f then toL cos 0 nL cos 0 cos——— = cos ^ 1 (14) , (oL cos 9 oL cos 9 and sin ^ —-(15) 2c 2c Introducing (14) and (15) into (13), the expression for the scalar potential then reduces to I 0 L cos OeW-wn /\ ^j_\ 4jte0 c \r jo) r1) Equations (4) and (16) express the vector and scalar potentials everywhere due to a short dipole. The only restrictions are that r p L and \ L. These equations give the vector and scalar potentials at a point P in terms of the distance r to the 3-3 THE FIELDS OF A SHORT DIPOLE 205 {/ I / Figure 5-4 Resolution of vector potential into A r and V y Ae components.
point from the center of the dipole, the angle 0, the length of the dipole £, the current on the dipole and some constants.
Knowing the vector potential A and the scalar potential V , the electric and magnetic fields may then be obtained from the relations E = -jayA - \V (17) and H - - V x A (18) It will be desirable to obtain E and H in polar coordinates. The polar coordinate components for the vector potential are A = f.4 r + 6a, + (19) Since the vector potential for the dipole has only a z component, Af — 0, and A r and A e are given by (see Fig. 5-4) A r = A z cos e (20) Ag = ~A Z sin 8 (21) where A. is as given by (4). In polar coordinates the gradient of V is dv ISV 1 sv VK = aT + - "5^ + “ ~r— (22) dr r dd r sin 6 c Calculating now the electric field E from (17), let us first express E in its polar coordinate components. Thus, E = ,£, + £, + «,£, (23) From (17), (19) and (22) the three components of E are then 6V £ r = -jwA r - — (24) Eg = -jcoAj, - -— = -jr 3 / I0 L cos B /J_ 1 \ 27i£ ft \cr2 j&r/ f P L sin Q hjo_ J_ 1 \ 4tt£0 \c 2r + cr 2 + jajr3/ In obtaining (29) and (30) the relation was used that ^0 e0 = 1/c 2 , where c = velocity of light.
Turning our attention now to the magnetic field, this may be calculated by (18). In polar coordinates the curl of A is r pfsin Q)A+ d{A $) r sin 8 30 d dir sin Q)A > ~Jr <3i > Since A = 0, the first and fourth terms of (31) are zero. From (4) and (20) and (21) we note that A r and Ag are independent of <£, so that the second and third terms of (3t) are also zero Thus, only the last two terms in (31) contribute, so that V x A, and hence also H, have only a $ component Introducing (20) and (21) into (31), performing the indicated operations and substituting this result into (18), we have ]H| = //,= I 0 L sin fjw I Ht = H$ = 0 Thus, the fields from the dipole have only three components Ert EA and The components Hr and are everywhere zero.
When r is very large, the terms in 1/r2 and 1/r3 in (29), (30) and (32) can be neglected in favor of the terms in 1/r. Thus, in the far field Er is negligible, and we have effectively only two field components, £fl and given by >J 0 L sin 0 I0 pL t = =J :r~ S1 41t£0 c r 4jt£n cr sin 0 gMt-irf)] jml0 L me™-™ JspL .
„ = : = j —— sin 0 e™ 4ttct 4jtr and (35) 5-2 THE FIELDS OF A SHORT DIPOLE 207 Figure 5-5 Near- and far-fittld patterns of Et and components for short dipole (n) and near-field pattern of E t component (fi).
Taking the ratio of Eg to H+ as given by (34) and (35), we obtain = 1^ = 311 Q.
(36) H &0 c V £o This is the intrinsic impedance offree space (a pure resistance).
Comparing (34) and (35) we note that E and are in time phase in the far field. We note also that the field patterns of both are proportional to sin 0. The pattern is independent of , so that the space pattern is doughnut-shaped, being a figure-of-revolution of the pattern in Fig. 5-5a about the axis of the dipole Referring to the near-field expressions given by (29), (30) and (32), we note that for a small r the electric field has two components Er and E9 , which are both in time-phase quadrature with the magnetic field, as m a resonator At intermediate distances, E$ and Er can approach time-phase quadrature so that the total elec-tric field vector rotates in a plane parallel to the direction of propagation, thus exhibiting the phenomenon of crossfield For the E# and H components, the near-field patterns are the same as the far-field patterns, being proportional to sin 0 (Fig, 5- 5a). However, the near -field pattern for £r is proportional to cos 0 as indicated by Fig 5-5fr. The space pattern for Er is a figure-of-revolution of this pattern around the dipole axis Let us now consider the situation at very low frequencies. This will be referred to as the quasi-stationary, or dc, case Since from (8), [/] = = jwlq] (37) (29) and (30) can be rewritten as \jf\L cos 9 / jto 1 \ r 2rce0 \cr2 rl) and [g]L sin $ f a?2 1 ” 4tte0 \ c 2r cr2 r3 (38) (39) (40) 208 5 THE ELECTRIC DIPOLE AND THIN UNEAR ANTENNAS Table 5-1 Fields of a short electric dipole f t The restriction applies that r S> L and k L The quantities in the table are in SI units that is, E in volts per meter, H in amperes per meter, / in amperes, r in meters, etc. [f] is as given by (37) Three of the field components of an electric dipole are everywhere zero, that is, At low frequencies, <a approaches zero so that the terms with m in the numerator can be neglected. As co -» 0 , we also have M = = 90 (41) and = /„ (42) Thus, for the quasi -stationary, or dc, case, the field components become, from (38), (39) and (40), g0 L, cos 0 <43) F 1° Lsine (44> „ /o Lsin 0 "• - -P“ (45) The restriction that r L still applies.
The expressions for the electric field, (43) and (44), are identical to those obtained in electrostatics for the field of two point charges, +q0 and -q separated by a distance L. The relation for the magnetic field, (45), may be recog-nized as the Biot-Savart relation for the magnetic field of a short element carry-ing a steady or slowly varying current.
Since in the expressions for the quasi-stationary case the fields decrease as 1 /r3 or l/r\ the fields are confined to the vicinity of the dipole and there is negligible radiation. In the general expres-sions for the fields, (38), (39) and (40), it is the 1/r terms which are important in the far field and hence take into account the radiation.
The expressions for the fields from a short dipole developed above are sum-marized in Table 5-1 1 If we had been interested only in the far field, the development beginning with (5) could have been much simplified. The scalar potential V does not con-tribute to the far field, so that both E and H may be determined from A alone 3-7 THE FIELDS OF A SHORT DIPOLE 209 Thus, from (17), E and H of the far field may be obtained very simply from |E| = E= -ja>Ae (4 5a) where Z = v/V £o = 377 H+ may also be obtained as before from (18) and £# from H0 . Thus, = = — | V x A | and neglecting terms in 1 /r2 , E9 = ZH# = — | V x A | Vo Equation (30) for the 0 component of the electric field may be reexpressed as „ Joe^Zsinflr 1 , E» = j 1 — / — 360 r 3 + 90 +^ /— 360° r, + °° + ^4 / — 360° r, - 9<r ] (Vm" 1 ) (46) where Lk = Ljk r^rjl Z = 377 n The restrictions apply that If we let (46) becomes L (Lx < 1 ) r > L (rA > LJ A =— / — 360“ r, +90° i B =—; / — 360° r1 + 0“ Anri ‘ “ C = / — 360° r, - 90° /0 c'“L 1 Z sin 0 „ „ E» = 2 i (A + B + C) The magnitudes of the components A, B and C are shown in Fig. 5-6 as a function of rx (distance in wavelengths). For rK greater than the radian distance 210 3 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS 2t Figure 5-6 Variation of the magnitudes of the components of Et of a short etectnc dipole as a function of distance {rfX^ The magnitudes of all components equal n at the radian distance 1/(2). At larger distances energy is mostly radiated, at smaller distances mostly stored.
[1/(2ti)], component A of the electric field is dominant, for rx less than the radian distance component C of the electric field is dominant, while at the radian dis-tance only B contributes ( = rc) because although |4| = |i)| = |C| = n, A and C are in phase opposition and cancel.
Equation (32) for the 0 component (only component) of the magnetic field may be reexpressed as = I^'L> sin 8 |"1 2&)0 90 + _L / — 360° n + 0°1 (49) 0 A L 2ri 4nr\ — J The restrictions of (47) apply.
For greater than the radian distance the first component of the magnetic field (in brackets) is dominant, Tor rx less than the radian distance the second component of the magnetic field is dominant, while at the radian distance both components are equal ( = it) and in phase quadrature.
5-2 THE FIELDS OF A SHORT DIPOLE 211 Note that the ratio of E9 to as given by the ratio of (46) to (49), is an impedance whose value becomes exactly equal to Z, the intrinsic impedance of space ( = 377 ft), for rx > l/(2n).
For the special case where 9 = 90° (perpendicular to the dipole in the xy plane of Fig. 5-2) and at rx p 1/(2tc), =^ (Am-1 ) (50) while at rk ^ l/(2a).
L If , = M; 4reri (50a) which is identical to the relation for the magnetic field perpendicular to a short linear conductor carrying direct current as given by (45).
The magnetic field at any distance r from an infinite linear conductor with direct current is given by 2nr (50b) which is Ampere s law.
1 Remarkably, the magnitude of the magnetic field in the equatorial plane (i9 = 90°) in the far field of an oscillating Xj2 dipole is identical to (50b) (Ampere’s law).
It is assumed that the current distribution on the X/2 dipole is sinusoidal.
This is discussed in more detail in Sec, 5-5. The above magnetic field relations are summarized in Table 5-2.
Rearranging the three field components of Table 5-1 for a short electric dipole, we have [/lLiZcosfff 1 1 "I X [inrl ~ J 4mJ [/UiZsinOr 1 1 .
I 1 A L J > + «r!
J 8MJ {5D (52) We note that the constant factor in each of the terms in brackets differs from the factors of adjacent terms by factor of 2n, as do the constant factors in j4, B and C of (48), 1 Sec, for example, J. D. Kraus, Electromagnetics, 3rd ed., McGraw-Hill, 1984, p. 170.
212 5 the ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS Table 5-2 Magnetic fields from dipoles and linear conductors!
Short oscillating dipole [JJ^I = -at tx 2> 1/(2) r p L, X P L (A m l ) Short oscillating dipole | H+ \ = — at r ^ l/(2ir) Anr r > L, Xp L Short linear conductor with direct current r L _hL h~a^ (Am-') 1/2 oscillating dipole, far held \n+\ = ~ (A m 1 ) Infinite linear conductor with direct current at any r H4 = -— (Ampere’s law) (A m J ) 2itr t Magnetic field at distance r from dipoles and linear conductors fin direction perpendicular to dipole, in xy plane of Fig. 5-2) with current /„ .
At the radian distance [rA = 1/(27t)], (51), (52) and (53) reduce to ^ 2^<f]L,Zcosg / 45tl -7i[/]Lj Z sin 0 (56) The magnitude of the average power flux or Poynting vector in the d direc-tion is given by s$ = \ Re Er H = Re I/- 90° = jEr H+ cos (-90°) = 0 (57) indicating that no power is transmitted. However, the product E r //„ represents imaginary or reactive energy that oscillates back and forth from electric to mag-netic energy twice per cycle.
In like manner the magnitude of the power flux or Poynting vector in the r direction is given by Sr = \E0 cos ( - 45°) = indicating energy flow in the r direction.
/ S-J RADIATION RESISTANCE OF SHORT ELECTRIC DIPOLE 213 Much closer to the dipole [rA ^ 1/(2ti)], but with the restrictions of (47) still applying, (51), (52) and (53) reduce approximately to Ea = -j . [J]L a Z sin & 8n 2 kr\ [/]LA sin 0 From these equations it is apparent that Sr = Se = 0. However, the pro-ducts Er H 4f and E0H represent imaginary or reactive energy oscillating back and forth but not going anywhere. Thus, close to the dipole there is a region of almost complete energy storage.
Remote from the dipole [rA 1/(271)], (51), (52) and (53) reduce approx-imately to £ r = 0 (62) Since Er = 0, there is no energy flow in the 0 direction [S$ = 0). However, since. E0 and H# are in time phase, their product represents real power flow in the outward radial direction. This power is radiated.
The Poynting vector or power flux around a short dipole antenna is shown by means of vectors in Fig. 5-7a. The length of the vectors is proportional to the Poynting vector magnitude. Double-ended vectors indicate imaginary or reactive power (vars per square meter) while single-ended vectors represent real power flow (watts per square meter) in the direction indicated.
The region near the dipole is one of stored energy (reactive power) while regions remote from the dipole are ones of radiation. The radian sphere at r k = 1/(2ti) marks a zone of transition from one region to the other with a nearly equal division of the imaginary and real (radiated) power.
The region close to the dipole may be likened to a spherical resonator within which pulsating energy is trapped, but with some leakage which is radi-ated. There is no exact boundary to this resonator region, but if we arbitrarily put it at the radian distance a qualitative picture may be sketched as in Fig. 5-7b.
5-3 RADIATION RESISTANCE OF SHORT ELECTRIC DIPOLE. Let us now calculate the radiation resistance of the short dipole of Fig. 5-lb. This may be done as follows. The Poynting vector of the far field is integrated over a large sphere to obtain the total power radiated- This power is 214 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS Region of radiation Region of energy storage Divide lengths by 100 v, V“ 1 Multiply v J lengths by^ „ 10^000 ^rrh = \f{2r) 2 Y Short .^dipole Transition at radian sphere.
^ = 1/12 t) Distance, rf\ Figure 5-7 Power flu vectors at three distances near a short dipole antenna. Double-ended vectors indicate reactive power (vars per square meter) while single-ended vectors represent real power (watts per square meter). At the innermost distance [r^ = I/(2) a J, the power is almost entirely imaginary (reactive) with stored energy oscillating from electric to magnetic twice per cycle. At the outermost distance, the power is almost entirely real and flowing radially outward as radiation. At the radian sphere [rA = 1 A2ir)J, the condition is in transition with energy pulsating in the & direction and also radiating in the radial direction. Some stored energy (not shown) is also pulsating in the radial direc-tion, Note that for proper scale, vectors qt the innermost distance should be 10000 times larger while at the outermost distance they should be 100 limes smaller. The three radial distances are not to scale.
The other quadrants are mirror images.
^ Radian sphere resonator / s / Energy storage v Radiation / % \ ^ ^ 1 1 1 \ 'V ! ^ J Radiation Figure 5-7h Sketch suggesting that within the radian sphere the situation is like that inside a reson-ator with htgh^density pulsating energy accompanied by leakage which is radiated.
RADIATION RESISTANCE OF SHORT ELECTRIC DIPOLE 215 then equated to I 2R where J is the rms current on the dipole and R is a resist-ance, called the radiation resistance of the dipole.
The average Poynting vector is given by S = iRe(ExH) (1) The far-field components are E9 and so that the radial component of the Poynting vector is Sr = i Re E$ HJ (2) where £# and JJJ are complex.
The far-field components are related by the intrinsic impedance of the medium. Hence, Et = H+Z = H+ /-Thus, (2) becomes S, = \ Re = il //1 2 Re Z = il WJ The total power P radiated is then |/fJV sin 0 dS d<)> where the angles are as shown in Fig. 5-2 and i is the absolute value of the magnetic field, which from (5-235) is ,«> 4rccr Substituting this into (5) we have The double integral equals 8tt/ 3 and (7) becomes Ijipiie (8) V e 12ti This is the average power or rate at which energy is streaming out of a sphere surrounding the dipole. Hence, it is equal to the power radiated. Assuming no losses, it is also equal to the power delivered to the dipole. Therefore, P must be equal to the square of the rms current I flowing on the dipole times a resistance Rr called the radiation resistance of the dipole. Thus, 216 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS Solving for Rr For air or vacuum •-ft /fhMo = 377 = 120tt fl so that (10) becomes 1 , = 80tt2 %Qjt 2L 2 A = 790Lf As an example suppose that L = Then Rr = 7.9 £2.
If = 0.01, then Rr — 0.08 fl. Thus, the radiation resistance of a short dipole is small.
In developing the field expressions for the short dipole, which were used in obtaining (11), the restriction was made that X £> L, This made it possible to neglect the phase difference of field contributions from different parts of the dipole.
If \ we violate this assumption, but, as a matter of interest, let us find what the radiation resistance of a Xj2 dipole is, when calculated in this way.
Then for L A = ^ we obtain Rr = 197 fl. The correct value is 168 fl (see Prob.
5-3), which indicates the magnitude of the error introduced by violating the restriction that X L to the extent of taking L = X/2.
It has been assumed that with end-loading (see Fig. 5-1 a) the dipole current is uniform. However, with no end-loading the current must be zero at the ends and, if the dipole is short, the current tapers almost linearly from a maximum at the center to zero at the ends, as in Fig. 2-126, with an average value of ^ of the maximum. Modifying (8) for the general case where the current is not uniform on the dipole, the radiated power is fctlk V £ 1271 where / av = amplitude of average current on dipole (peak value in time) The power delivered to the dipole is, as before, P = £/£/?, (W) (13) where I Q = amplitude of terminal current of centerTed dipole (peak value in time) Equating the power radiated (12) to the power delivered (13) yields, for free space 0 = Po and £ = £o)> 3- radiation resistance -- 79 °(fc )’1 vVo/fo ^ 376.73 Q. 377 and 120ir arc convenient approximations.
As already given by (2-20-3). See also footnote accompanying (220-3).
THE FIELDS OF A SHORT DIPOLE BY THE HERTZ VECTOR METHOD 217 For a short dipole without end-loading, we have /av = ^/ 0 , as noted above, and (14) becomes Rr =\91Lj (fl) (15) 5-4 THE FIELDS OF A SHORT DIPOLE BY THE HERTZ VECTOR METHOD.
In Sec, 5-2 the fields of a short dipole were obtained by a method involving the use of vector and scalar potentials. Another equivalent method which is sometimes employed makes use of the Hertz vector. Since this method is frequently found in the literature, it will be of interest to use it to find the fields of a short electric dipole. The fields so obtained are identical with those found by the vector-scalar potential method, indicating the equivalence of the two procedures.
The retarded vector potential of any electric-current distribution is given by where the retarded current density [J] is given by = (2) Multiplying numerator and denominator by e, (1) may be written as <?n = j r dt 47!); jv r where r represents time and r volume. The quantity n is the retarded Hertz vector or retarded Hertzian potential. Since [J] is the only time-dependent quan-tity on the right-hand side of (4), we have for the retarded Hertz vector i fjui a i r [j] ""tej, .
-^JtT 151 we obtain from (3) n - n0 a — j(opten and (7) 218 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS If the retarded Hertz vector is known, both E and H everywhere can be calculated from the relations E = wVIT + V(V • n) (8) H = jeaeV x n (9) Thus, E and H are derivable from a single potential function, n. Substituting (7) into (8) and (9), these relations may be also reexpressed in terms of A alone. Thus, E = -jcoA-— V(VA) (10) 0)jX£ H = - V x A (11) ft Let us now find the retarded Hertz vector for a short electric dipole. The vector potential for the dipole has only a z component as given by (5-2-4). There-fore, from (7) the Hertz veetdr has only a z component given by n M 4nraie ( 12 ) In polar coordinates H has two components, obtained in the same way as the components of A in (5-2-20) and (52-21). Thus, II = rll r cos 0 -sin 0 (13) Substituting (12) into (13), and this in turn in (9) and performing the indicated operations, yields the result that _ [/]L_sinfl fjm 1 4ji \cr + r 2 This result is identical with that obtained previously in (5-2-32). We could have anticipated this result since substituting (7) into (9) gives (1 1), from which (5-2-32) was obtained.
Substituting (12) into (13) and this in turn in (8) then gives the electric field E everywhere. The expressions for the two components, E r and E8t so obtained are identical with those arrived at in (5-2-29) and (5-2-30) by the use of vector and scalar potentials.
5-5 THE THIN LINEAR ANTENNA, In this section expressions lor the far-held patterns of thin linear antennas will be developed.
It is assumed that the antennas are symmetrically fed at the center by a balanced two-wire transmission line. The antennas may be of any length, but it is assumed that the current dis-tribution is sinusoidal. Current -distribution measurements indicate that this is a good assumption provided that the antenna is thin, i.e., when the conductor diameter is less than, say, A/ 100. Thus, the sinusoidal current distribution approx-imates the natural distribution on thin antennas. Examples of the approximate 5-5 THE THIN LINEAR ANTENNA 219 v ' Figure 5-8 Approximate natural-current X A —X 2X distribution for thin, linear, center-fed 2 4 4 2 antennas of various lengths natural-current distributions on a number of thin, linear center-fed antennas of different length are illustrated in Fig.
58. The currents are in phase over each A/2 section and in opposite phase over the next.
Referring to Fig.
5-9, let us now proceed to develop the far-field equations for a symmetrical, thin, linear, center-fed antenna of length L. The retarded value of the current at any point z on the antenna referred to a point at a distance s is [/] = / 0 sin[?(Hh In (1) the function is the form factor for the current on the antenna. The expression (L/2) + z is used when z < 0 and (L/2) - z is used when z > 0. By regarding the antenna as made up of a series of infinitesimal dipoles of length dz , the field of the entire antenna may then be obtained by integrating the fields from all of the dipoles making up -\ To \ distant \ point Figure 5-9 Relations for symmetrical, 4 bin, linear, center-fed antenna of length L.
220 5 THF ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS the antenna. The far fields dE & and dH^ at a distance s from the infinitesimal dipole dz are (see Table 51) ;607i[/] sin 6 dz dE = ^ j[J] sin 8 dz Since Ee — ZH# ^ !2 it will suffice to calculate H CM «)z}c l The integrals are of the form sin (c + ^ La sin <c + bx'~ b «• + fca (9) 5^5 THE THIN LINEAR ANTENNA llY where for the first integral a = jfi cos 0 c = PL/2 For the second integral a and c are the same as in the first integral, but b = -p.
Carrying through the two integrations, adding the results and simplifying yields „ _ Kip] [cos [(pL cos 0)/2] - cos (pL/2fl jUol r 2 nr Multiplying by Z - 120tt gives E # as _ j60CJ o ] ^ cos [{fiL cos fl)/2] - cos (/?L/2) sin 6 vhere [/ 0 ] = Equations (10) and (11) are the expressions for the far fields, H0 and ESl of a symmetrical, center-fed, thin linear antenna of length L. The shape of the far-field pattern is given by the factor in the brackets. The factors preceding the brackets in (10) and (11) give the instantaneous magnitude of the fields as functions of the antenna current and the distance r To obtain the rms value of the field, we let [/ 0 ] equal the rms current at the location of the current maximum. There is no factor involving phase in (10) or (It), since the center of the antenna is taken as the phase center. Hence any phase change of the fields as a function of 6 will be a jump of ISO"" when the pattern factor changes sign.
As examples of the far-field patterns of linear center-fed antennas, three antennas of different lengths wifi be considered.
Since the amplitude factor is independent of the length, only the relative field patterns as given by the pattern factor will be compared.
Cas Antenna When L = A/2, the pattern factor becomes __ cos [(te/2),cos fl] This pattern is shown in Fig. 5-I0a.
Tt is only slightly more directional than the pattern of an infinitesimal or short dipole which is given by sin 0. The beam width between half-power points of the A/2 antenna is 78° as compared to 90° for the short dipole.
222 5 THE ELECTRIC DIPOLE AND THIN LINtAX ANTENNAS FigiEf 5-10 Far-fidd patterns of a/2, full-wave and 3a/2 antennas. The antennas are center-fed and the current distribution is assumed to be sinusoidal 5-5b Case 2. Full-Wave Antenna. When L = X t the pattern factor becomes ^ cos (te cos 0 + 1 sin 6 This pattern is shown in Fig. 5-10b. The half-power beam width is 47° 5-5c Case 3. 3A/2 Antenna. When L = 3A/2, the pattern factor is £ _ cos (frc cos 8} sin 0 The pattern for this case is presented in Fig 5-10c With the midpoint of the antenna as phase center, the phase shifts 180° at each null, the relative phase of the lobes being indicated by the + and — signs. In all three cases, (a), (b) and (c), the space pattern is a figure-of-revolution of pattern shown around the axis of the antenna.
5 j THE THIN LINEAR ANTENNA 223 fj Figure 5-ii Symmetrical center-fed dipole with sinusoidal // current distribution. The field component £, at any distance // can be expressed as the sum of 3 components radiating from the ends and the center of the dipole.
5_5d Field at any Distance from Center-Fed Dipole. The geometry for the field at the point P from a symmetrical center-fed dipole of length L with sinu-soidal current distribution is presented in Fig 5-11. The maximum current is Jo-lt may be shown that the z component of the electric field at the point P is given by -jI 0 Z[e~^ e „ pL-1 E , = -+ 2 cos ^ (15) 4te L $ i s2 2 r J The component of the magnetic field at the point P (Fig. 5-1 1) is given by H — ———“ ( e _^a 4;rr sin B \ i + e ifii _ 2 cos Whereas the other field equations for oscillating dipoles given in this chapter apply only with the restrictions of (5-2-47), (15) and (16) apply without distance restrictions. Equations (15) and (16) are reminiscent of the pulsed center-fed dipole of Fig. 2-24 in that the field at P is made up of 3 field components, one from each end of the dipole and one from the center.
If P lies on the y axis (0 = 90®) and the dipole is 2/2 long, (15) becomes Ez — -° / — 360° i + d + rj - 90° (Vm“ l ) and (16) becomes a =A 2nr + r\ + 90° where rx = rjX Z = 377 a Jq = maximum current = terminal current 224 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS At a large distance the ratio of £ r as given by {17) to as given by (18) is E z ^ ~ ^ = i nt™sic impedance (resistance) of space (19) The magnitude of is as given in Table 5-2 t 5-6 RADIATION RESISTANCE OF A/2 ANTENNA. To find the radiation resistance, the Poynting vector is integrated over a large sphere yielding the power radiated, and this power is then equated to )/ 0/v '/ 2) 2 fi (l , where R 0 is the radiation resistance at a current maximum point and / 0 is the peak value in time of the current at this point. The total power P radiated was given in (5-3-5) m terms of H4 for a short dipole. In (5-3-5), |HJ is the absolute value !
hC correspondin8 va,ue of for a linear antenna is obtained from (5-5-10) by putting !;[/<>] t = /0 • Substituting this into (5-3-5), we obtain . _ 15Jg f 2 f f n -0 Jo cos [(/?L/2) cos fl] - cos (PL/2)} 2 {cos [(PL/2) cos fl] -sin 0 cos {fiLj2)} 2 Equating the radiated power as given by (2) to l\ R 0/2 we have R0 = 60 (cos [(jSL/2) cos 0] - cos {$Lj2)\ 2 where the radiation resistance R 0 is referred to the current maximum. In the case ot a A,-2 antenna this is at the center of the antenna or at the terminals of the transmission line (see Fig. 5-8).
Proceeding now to evaluate (4), let u = cos 0 by which (4) is transformed to du = — sin 8 dd [cos (pLv/2) - cos (fiL/2)] 2 'P = fJS-ds = (6 ) 5"6 RADIATION RESISTANCE OF a/2 ANTENNA 225 However 1 -U 2 (1 4 wXl - u) 2 Vl + W I - « Also putting k = f$L{2, (6) becomes R n = 30 f 1 (cos ku — cos k) 2 (cos ku - cos k}2 i 1 4 u 1 — u This integral gives the radiation resistance for a thin linear antenna of any length L. For the special case being considered where L = A/2, we have k = n/2. Thus, in the case of a thin A/2 antenna, (8) reduces to Ro = 30 1 fcos3 (tiu/2) cos 2 (tcu/2)~| — ; +— ; du i | 1 1 — u J Now in the first term let , v j dv 1 + u = -and du = — rc n and in the second term let v 1 dv 1 - u = — and du n n Noting also that (r — n)j 2 = (nr — i/)/2, Eq. (9) becomes Ji» V But cos 2 (x/2) = ^{1 4- cos x) so that 1 + cos (i? — tt) „ f 2 l — cos v — 30 I dv Jo » The last integral in (13) is often designated as Cin (x) t Thus, Cin (x) = P 1 - cc Jo v dv — In yx — Ci (x) = 0.577 4 In — Ci (x) where y = = 1,781 or In y = c = 0.577 = Euler’s constant The part of this integral given by Ci (x) = In yx — Cin (x) is called the cosine integral The value of this integral is given by ^ r v f cos v , , x 2 x x d C' W J.~ " - 573 + 414 - 6!6 + 226 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS 5-7 RADIATION RESISTANCE AT A POINT WHICH IS NOT A CURRENT MAXIMUM 227 When x is small (x < 0.2), Ci (x) s In yx = 0.577 H- In x (17) When x is large (x> 1), Ci (x) = sm x x (18) A curve of the cosine integral as a function of x is presented in Fig. 5- 12a. It is to be noted that Ci (x) converges around zero at large values of x. From (16) and (14) we obtain Cin (x) as an infinite series, Cin W = 2!2 x x 6 4! 4 + 6! 6 (19) While discussing Cin (x) and Ci (x), mention may be made of another inte-gral which commonly occurs in impedance calculations. This is the sine integral.
Si (x), given by Si (x) = r sin v .
dv = v x J x J X ~ 3! 3 + 515 ” When x is small (x < 0.5), (20) (21 ) Figjure 5-1 2b Sine integral.
When x is large (x ^ 1), Si (x) 7t 2 COS X X (22) A curve of the sine integral as a function of x is presented in Fig 5-12b.
It is to be noted that Si (x) converges around nf2 at large values of x.
Returning now to (13), this can be written as R 0 = 30 Cin (2w) = 30 x 2.44 = 73 n (23) This is the well-known value for the radiation resistance of a thin, linear, center-fed, 2/2 antenna with sinusoidal current distribution. The terminal impedance also includes some inductive reactance in series with R 0 (see Chap.
10). To make the reactance zero, t.e., to make the antenna resonant, requires that the antenna be a few per cent less than 2/2. This shortening also results in a reduction in the value of the radiation resistance.
5-7 RADIATION RESISTANCE AT A POINT WHICH IS NOT A CURRENT MAXIMUM, If we calculate, for example, the radiation resistance of a 32/4 antenna (see Fig. 5-8) by the above method, we obtain its value at a current maximum. This is not the point at which the transmission line is connected. Neglecting antenna losses, the value of radiation resistance so obtained is the resistance R 0 which would appear at the terminals of a transmis-sion line connected at a current maximum in the antenna, provided that the current distribution on the antenna is the same as when it is center-fed as in Fig.
58. Since a change of the feed point from the center of the antenna may change the current distribution, the radiation resistance R0 is not the value which would be measured on a 32/4 antenna or on any symmetrical antenna whose Si (x) — x 228 5 the electric dipole and thin linear antennas Figure 5-13 Relation of current /, at transmission -Line terminals to current /0 at current maximum.
length is not an odd number of A/2. However, J? 0 can be easily transformed to the value which would appear across the terminals of the transmission line connected at the center of the antenna.
This may be done by equating (5-6-3) to the power supplied by the trans-mission line, given by 1\RJ2, where l x is the current amplitude at the terminals and is the radiation resistance at this point (see Fig. 5-1 3). Thus, 0 ) where is the radiation resistance calculated at the current maximum. Thus, the radiation resistance appearing at the terminals is t (2 ) The current I x at a distance x from the nearest current maximum, as shown in Fig. 5-13, is given by / x = J 0 cos fix where / t = terminal current / 0 = maximum current Therefore, (2) can be expressed as cos 2 fix ( 3 ) (4) When x = 0, = -R 0 ; but when x = A/4, = oo if /i0 ^ 0, However, the radi-ation resistance measured at a current minimum (x = A/4) is not infinite as would be calculated from (4), since an actual antenna is not infinitesimally thin and the current at a minimum point is not zero. Nevertheless, the radiation resistance at a current minimum may in practice be very large, i.e., thousands of ohms.
5-8 FIELDS OF A THIN LINEAR ANTENNA WITH A UNIFORM TRAVELING WAVE. The foregoing discussion has been VH FIELDS OF A THIN LINEAR ANTENNA WITH A UNIFORM TRAVELING WAVE 229 t Relative current Distance along antenna Relative phase angle (lag!
Distance along antenna Figure 5-!4 Current amplitude and phase relations along an antenna carrying a single uniform traveling wave.
confined to the case of antennas with sinusoidal current distributions.
This current distribution may be regarded as the standing wave produced by two uniform (unattenuated) traveling waves of equal amplitude moving in opposite directions along the antenna.
If, however, only one such wave is present on the antenna, the current distribution is uniform. The amplitude is a constant along the antenna, and the phase changes linearly with distance as suggested by Fig, 5-14.
The condition of a uniform traveling wave on an antenna is one of con-siderable importance, as this condition may be approximated in a number of antenna systems. For example, a single-wire antenna terminated in its character-istic impedance, as in Fig. 5-15a, may have essentially a uniform traveling wave.
1 This type of antenna is often referred to as a Beverage or wave antenna, A ter-minated rhombic antenna (Fig. 5-156) may also have essentially a single traveling wave. The Beverage and rhombic antennas are discussed further in Chap.
16.
Other types of antennas that have, in the first approximation, a single outgoing traveling wave, are a long monofilar axial-mode helical antenna and a long, thick linear antenna as illustrated in Fig, 5-15c and d. These antennas have no termina-ting impedance but behave in a similar way to terminated antennas. Thus, the thick linear conductor has a current distribution similar to a thin terminated linear conductor, and the patterns are similar if the conductor diameter is not too large. The results for a traveling wave on a linear conductor can be applied to a helix, as shown in Chap.
7, by considering that the helix consists of a number of short linear segments. On the linear antennas, the phase velocity of the traveling wave is substantially equal to the velocity of light. However, the phase velocity along the conductor of a monofilar axial-mode helical antenna may differ appre-ciably from the velocity of light. Hence, to make the results applicable to any of the antenna types shown in Fig. 5-15, the fields from an antenna with a traveling 1 Since the fields of an antenna are nol confined to the immediate vicinity of the antenna, it is not possible to provide a nonreflecting termination with a lumped impedance. However, a lumped imped-ance may greatly reduce reflections at the termination.
230 5 THE electric dipole and thin linear antennas Wave Terminated rhombic antenna M>2MaQfl2fl2£L (c> l Long helical antenna Figure 5-15 Various antennas having essentially a single traveling wave.
wave will be developed for the general case where the phase velocity v of the wave along the conductor may have any arbitrary value.
1 Proceeding now to find the field radiated by a traveling wave on a thin linear conductor, let us consider a conductor of length b coincident with the z axis and with one end at the origin of a cylindrical coordinate system (p, £, z) as in Fig. 516.
It is assumed that a single, uniform traveling wave is moving to the right along the conductor.
Since the current is entirely in the z direction, the magnetic field has but one component The £ direction is normal to the page at P in Fig. 5-16, and its positive sense is outward from the page. The magnetic field can be obtained from the Hertz vector fl. Since the current is entirely in the 2 direction, the Hertz 1 A. Alford, WA Discussion of Methods Employed in Calculations of Electromagnetic Fields of Radi-aling Conductors," Elec. Commun., IS, 70-88, July 1936. Treats case where velocity is equal to Light.
J. D. Kraus and J, C. Williamson, “ Characteristics of Helical Antennas Radiating in the Axial Mode,'" J. App. Phys., 19, 87-96, January 1948^ Treats general case.
J. Grosstopf, “Ober die Verwendung zweicr Lbsungsansatze der Maxwellscben Gleiehungen bei der Bcrechnung der dectrotnagnetischen Felder strahlender Letter," Hockfreqenztechnik und Electro-akustik. 49, 205-211, June 1937. Treats case where velocity is equal to light.
5-B FIELDS OF A THIN LINEAR ANTENNA WITH A UNIFORM TRAVELING WAVE 231 vector has only a z component. Thus, = jcoeffi x II)4 = -jcuE where n r is the z component of the retarded Hertz vector at the point P> as given by where |TJ = sin cu^t - ^ (3) where z x — a point on the conductor v /a\ and v = pc or p = -W In (4), p is the ratio of the velocity along the conductor v to the velocity of light c.
This ratio will be called the relative phase velocity.
All the conditions required for calculating the magnetic field due to a single traveling wave on the linear conductor are contained in the relations (1) through (4). That is, if [/] in (3) is substituted into (2) and U z from this equation into (1) and the indicated operations performed, we obtain the field H Let us now proceed to carry through this calculation. To do this, let U = t-^ (5) C V Now since r = IV - Zi ) 2 + yVT we have du z — L dz x rc pc (6) (?) 232 5 THE ELECTRIC DIPOLE AND THIN LINEAR ANTENNAS Equation (2) now becomes nx = f^ — — du 4njo)e JKl z-z x - (r/p) where the new limits are r x r, i «]— — — and u2 = t —- — Introducing (8) into ft) we have >.
Iq c $ sin wu 5 4n dp J,,, 2 - z 2 - (r/p) “ Confining our attention now to the far field, i.e., at a large distance r, which is very much largerthan b T the quantity z v can be neglected and the denominator of the integrand considered to be a constant z - (r/p). Therefore (10) becomes fj _ _ J 0 C ^ — COS OJU2 + COS QJUt l " 4na> 8p | z - (r/p) J 11 ^ Performing the differentiation with respect to p, (1 1) becomes h _ { z ~ (r/p) where /„ = re-integrating (1) with respect to time, (2) JLOf-i v The magnetic moment of the loop is IA. Equating this to the moment of the magnetic dipole, we have Substituting (2) in (3), This may be reexpressed as = —1A /„/ = -j y as in Fig. 6-7, Then integrating over the loop, the total vector potential is obtained, and from this the far-field components are derived.
Since the current is confined to the loop, the only component of the vector potential having a value is ,4^. The other components are zero: A s = A r = 0. The infinitesimal value at the point P of the component of A from two diametrically 1 Donald Foster, "Loop Antennas with Uniform Current," Proc. IRE. 32, 603-607, October 1944. A discussion of circular loops of circumference less than a/2 < 4) with nonuniform current distribu-tion is given by G. Glinski, "Note cm Circular Loop Antennas with Nonuniform Current Distribu-tion," J. Appi Phya ., 18, 638-644, July 1947.
246 ft THE LOOP ANTENNA opposed infinitesimal dipoles is where dM is the current moment due to one pair of diametrically opposed infini-tesimal dipoles of length, a dtp. In the ^ — 0 plane (Fig. 6-7) the component of the retarded current moment due to one dipole is [f]n dtp cos <p (2) where [/] = I Q e™ [i “ and / 0 is the peak current in time on the loop Figure 6-8 is a cross section through the loop in the xz plane of Fig. 6-7.
Referring now to Fig, 6-8, the resultant moment dM at a large distance due to a pair of diametrically opposed dipoles is \p dM = 2j_Qa dtp cos tp sin — (3) where \p = ipa cos tp sin 0 radians Introducing this value for ip into (3) we have dM = 2;[/]a cos tp [sin (pa cos sin 0)] dtp (4) Now substituting (4) into (1) and integrating, /uf/ltf f A# = —t sin (Pa cos tp sin 0) cos tp dtp (5) 2nr J0 or J i(Pa sin 0 ) <6) 6-6 FAR-FIELD PATTERNS OF CIRCULAR LOOP ANTENNAS WITH UNIFORM CURRENT 247 where J l is a Bessel function of the first order and of argument (Pa sin 0). The integration of (5) is performed on equivalent dipoles which are all situated at the origin but have different ohentations with respect to tp, The retarded current [/] is referred to the origin and, hence, is constant in the integration.
The far electric field of the loop has only a tp component given by = -jaiA# 1 7 ) Substituting the value of A# from (6) into (7) yields 60npa[H J : (/ta sin 0) This expression gives the instantaneous electric field at a large distance r from a loop of any radius a. The peak value of E# is obtained by putting [/] = / 0 , where /0 is the peak value (in time) of the current on the loop. The magnetic field H# at a large distance is related to by the intrinsic impedance of the medium, in this case, free space. Thus, Jf ( 10 ) This expression gives the instantaneous magnetic field at a large distance r from a loop of any radius a.
6-6 FAR-FIELD PATTERNS OF CIRCULAR LOOP ANTENNAS WITH UNIFORM CURRENT- The far-field patterns for a loop of any size are given by (659) and (65-10). Fbr a loop of a given size. Pa is constant and the shape of the far-field pattern is given as a function of 9 by MC.srne) (1) where C k is the circumference of the loop in wavelengths. That is, c, = lj f = &a {2) The value of sin 9 as a function of 6 ranges in magnitude between zero and unity.
When 6 = 90°, the relative field is JJCJ, and as 9 decreases to zero, the values of the relative field vary in accordance with the J 1 curve from to zero. This is illustrated by Fig. 6-9 in which, a rectified first-order Bessel curve is shown as a function of Cx sin 0.
As an example, let us find the pattern for a loop 1A in diameter (C = tt = 3.14). The relative field in the direction 0 = 90 3 is then 0.285. As 9 decreases, the field intensity rises, reaching a maximum of 0.582 at angle 0 of about 36°. As 0 decreases further, the field intensity also decreases, reaching zero at 0 = (F. The 248 6 THE LOOP ANTENNA 6-7 THE SMALL LOOP AS A SPECIAL CASE 249 Figure, 6-9 Pattern chart for loops with uniform current as given b> first-order Bessel curve as a function of C ; sin 9.
pattern in the other four quadrants is symmetrical, the complete pattern being as presented in Fig. 6-106.
It is possible to obtain the pattern by a graphical construction. This is illustrated for the case we have just considered of C- = n by the auxiliary circle quadrant in Fig. 6-9. The angle 0 is laid off around the arc of the circle. The radius of the circle is equal to CA sin W = C At which in this case is n. The field in the direction 0 = 60°, for instance, is then given by drawing a perpendicular to the axis of the abscissa and continuing this perpendicular until it intersects the Ji curve, giving a value of relative field, in this case, of 0.443, as shown in Fig. 6-9.
Turning now to a consideration ofloops of other size, it is to be noted from Fig. 6-9 that the maximum field is in the direction 0 = 90° for all loops which are less than 1.84/ in circumference (less than 0.585>. in diameter). As an example, the pattern for a loop A/10 in diameter is presented in Fig. 6- 10a. The pattern is practically a sine pattern as would be obtained with a very small loop.
By way of contrast, the pattern for a loop 5/ in diameter is shown in Fig.
6-10d. In this case, which is typical for large circular loops with uniform current, the maximum field is in a direction nearly normal to the plane of the loop, while the field in the direction of the plane of the loop is small.
All patterns in Fig, 6-10 are adjusted to the same maximum. The space patterns for the five cases in Fig, 6-10 are fig ores -of- revolution of the patterns around the polar axis.
It is to be noted that the field exactly normal to the loop is always zero, regardless of the size of the loop.
6-7 THE SMALL LOOP AS A SPECIAL CASE, The relations of (6-5-9) and (6-5-10) apply to loops of any size.
It will now be shown that for the special case of a small loop, these expressions reduce to the ones obtained pre-viously.
For small arguments of the first-order Bessel function, the following approximate relation can be used: 1 AM = | (1) where x is any variable. When x = the approximation of (1) is about 1 percent in error The relation becomes exact as x approaches zero. Thus, if the perimeter of the loop is //3 or less (Cx < (1) may be applied to (6-5-9) and (6-5-10) with an error which is about 1 percent or less. Equations (6-5-9) and (6-5-10) then 1 For small arguments, the J 1 curve is nearly linear (see Fig, 6-9). The general relation for a Bessel function of any order n is ,/,() i /«! 2", where \x \ 1.
250 6 THE LOOP ANTENNA lo Figure 6-11 Loop and transmission line.
These far- fie Id equations for a small loop are identical with those obtained in earlier sections (see Table 6-1).
6-8 RADIATION RESISTANCE OF LOOPS, 1 To find the radiation resistance of a loop antenna, the Poynting vector is integrated over a large sphere yielding the total power P radiated. This power is then equated to the square of the effective current on the loop times the radiation resistance R r : P = yR, 0) where / 0 = peak current in time on the loop.
The radiation resistance so obtained is the value which would appear at the loop terminals connected to the transmission line, as shown in Fig. 6-1 1. The situation shown in Fig. 6-1 1 occurs naturally only on small loops. However, it will be assumed that the current is uniform and in phase for any radius a, this condition being obtained by means of phase shifters, multiple feeds or other devices (see Fig.
16-19). The average Poyn-ting vector of a far field is given by Sr = ±|H|ReZ (2) where | H | is the absolute value of the magnetic field and Z is the intrinsic imped-1 The procedure follows that given by Donald Foster, “Loop Antennas with Uniform Current," Proc.
IRE, 32, 603-607, October 1944, 6-s radiation resistance of loops 251 ance of the medium, which in this case is free space. Substituting the absolute value of from (6-5-10) for | H | in (2) yields ^ \5n{{ial(y) 2 5 , Sr = -2-J($a sin 0) (3) The total power radiated P is the integral of .S’, over a large sphere; that is.
P = JJ Sr ds= 1 5a(Pal0 ) 2 j j Jf(j8o sin 0) sin 0 dd d P = Wn2UiaI0) 2 f J 2(pa sin 0) sin 0 d9 In the case of a loop that is small in terms of wavelengths, the approx-imation of (6-7-1) can be applied. Thus (5) reduces to P = ^ n2ifiafl 2 0 sin3 6 dO = 10n 20aI2 (6) 2 Jo Since the area A = no 1 , (6) becomes P= 10^A 2 l% (7) Assuming no antenna losses, this power equals the power delivered to the loop terminals as given by (1). Therefore, A- = io/sM 2 /i; (8) Rr = 31 171 l) 1 = >97Ct or K,- 31 200^T) () (10) This is the radiation resistance of a small single-turn loop antenna, circular or square, with uniform in-phase current. The relation is about 2 percent in error when the loop perimeter is A/3. A circular loop of this perimeter has a diameter of about A/ 10. Its radiation resistance by (10) is nearly 2.5 fl.
The radiation resistance of a small loop consisting of one or more turns is given by 1 fi r = 31 20 r ) dy ( 12 ) I quality (12) and (1) and putting pa = C x yields fzcx R r = 60n 2 C x J 2(y) dy (ft) (13) Jo This is the radiation resistance as given by Foster for a single-turn circular loop with uniform in-phase current and of any circumference Cx , When the ]oop is large (CA > 5), we can use the approximation riv.x J 2(y) dy -1 (14) so that (13) reduces to Rr = 60n 2 C, = 592C, = 3720 -(15) & For a loop of 102 perimeter, the radiation resistance by (15) is nearly 6000 Q, For values of C. x between ^ and 5 the integral in (13) can be evaluated using the transformation rivx J 2 (y) dy = J n(y) dy - 2J i (2Cx) (16) Jo Jo where the expressions on the right of (16) are tabulated functions.
2 For perimeters of over 52 (C A > 5) one can also use the asymptotic develop-ment, I JM d> ' 1 “ [ ,in ( 2 - j) + it c°‘ i 2x -;)] 11 1 ' where x = pa = C x 1922^ ^ 7 realise on the Theory of Bessei Functions, Cambridge University Press, London, ^The integral involving J 0 for the interval 0 < x < 5 (where x = CJ is given by A, N. Lowan^and M.
amowitz, J. Math. Phys,, 22, 2-12, May 1943; and also by Natl. Bur Standards Tech. Memo 20, 6-<J nlRECTIVITY OF CJRrUMR LOOP ANTENNAS WITH UNIFORM CURRENT 253 Loop circumference, C\ Figure 6-12 Radiation resist-ance of single-turn circular loop with uniform, in-phase current as a function of the loop circum-ference in wavelengths, .
For small values of x, one can use a series obtained by integrating the ascending power series for J 2 Thus, 1 x ( My) dy = j f 1 x x x x - T + 56 “ T080 + 31 680 When x = C x = 2 (perimeter of 22), the result with four terms is about 2 percent in error. This same percentage error is obtained with one term when the perim-eter is about 2/3.
A graph showing the radiation resistance of single-turn loops with uniform current as a function of the circumference in wavelengths is presented in Fig.
6-12. The data for the curve are based on Foster's formulas as given above.
Curves for the approximate formulas of small and large loops are shown by the dashed lines.
6-9 DIRECTIVITY OF CIRCULAR LOOP ANTENNAS WITH UNIFORM CURRENT, The directivity D of an antenna was defined in (2-8- 1) as the ratio of maximum radiation intensity to the average radiation intensity. The maximum radiation intensity for a loop antenna is given by r z times (6-8-3). The average radiation intensity is given by (6-8-5) divided by 4tt.
Thus, tfie directivity of a loop is 2C x[J{Ca sin 0)]^ JS •/,(?) 6 w 254 6 THE LOOP ANTENNA Loop Circumference, C\ Figure 6-13 Directivity of circular loop antenna with uniform, in-phase current as a function ofloop circum-ference in wavelengths, C; . {After D.
Foster, “ Loop Antennas wirfr Uniform Current ' ' Proc.
IRE, 32, 603-607 October 1944,) This is Foster’s expression for the directivity of a circular loop with uniform in-phase current of any circumference C^. The angle 0 in (!) is the value for which the field is a maximum.
For a small loop (C^ < £), the directivity expression reduces to since the field is a maximum at 0 = 90°. The value of f is the same as for a short electric dipole. This is to be expected since the pattern of a short dipole is the same as for a small loop.
For a large loop (C A > 5), (1) reduces to V = 2C&KC, sin 0)!U (3) From Fig, 6-9 we note that for any loop with Cy ^ 1.84, the maximum value of JifC^sinfl) is 0.582.
Thus, the directivity expression of (3) for a large loop becomes D = 0.68 (4) The directivity of a loop antenna as a function of the loop circumference is presented in Fig. 6-13. Curves based on the approximate relations of (2) and (4) for small and large loops are indicated by dashed lines.
6-10 TABLE OF LOOP FORMULAS- The relations developed in the preceding sections are summarized in Table 6-2. The general and large loop for-mulas are based on Foster’s results.
6-U SQUARE LOOPS.
It was shown in Sec. 63 that the far-field patterns of square and circular loops of the same area are identical when the loops are small (A < A 2/10G). As a generalization, we may say that the properties depend (-I!
SQUARE LOOPS 255 Table 6-2 Formulas for circular loops with uniform current Small loopf General expression A < X 2 ! 100 Lirg# loop Quantity (any stzr loop) Cx < -j Ci >5 Far E+ -sin &) 1207t 2 [J] sin 9 A Same as general 7^ r r [/]Cj J,(Cj sin 8) ?t{7] sin 0 A Same as general Far H § —... — — — 2r r X 1 Radiation resistance, ft ric, j J:(y) Ay 31200^- 197Ct 3720 ^ = 592Ci Directivity sin 0)],^ dy 3 2 4.25 ^ — 0,68Ci A = area of loop; “ dreumferenct of circular Loop, wavelengths.
t The small loop formulas apply oof only to circular loops but also to square loops of area A and in fact to small loops of any shape having an area A. The formula involving C } applies, of course, only to a circular loop. See Sec. 16-9 about simulating uniform current on a large loop.
only on the area and that the shape of the loop has no effect when the loop is small. However, this is not the case when the loop is large. The pattern of a circular loop of any size is independent of the angle 4> but is a function of 0 (see Fig. 6-2). On the other hand, the pattern of a large square loop is a function of both 0 and Referring to Fig. 6-14, the pattern in a plane normal to the plane of the loop and~paral1el to two sides 1 and 3), as indicated by the line AA\ is simply the pattern of two point sources representing sides 2 and 4 of the loop. The pattern in a plane normal to the plane of the loop and passing through diagonal corners, as indicated by the line BB\ is different. The complete range in the 256 fi THE- LOOP ANTtNNA Figure 615 Pattern of square loop with uniform, in-phase current.
The loop ts 4.44a on a side. The pattern is in a plane normal to the plane of the loop and through the line AA' of Fig. 6-14.
pattern variation as a function of 0 is contained in this 45' interval between AA and BB ' in Fig r 6-14.
An additional difference of large circular and square loops is in the 8 pat-terns For instance, Fig, 6-10d shows the pattern as a function of 0 for a circular loop 5a in diameter. By way of comparison, the pattern for a square loop of the same area is presented in Fig r 6-15. The square loop is 4,44a on a side. The pattern is in a plane perpendicular to the plane of the loop and parallel to the sides (plane contains AA in Fig. 6-14), Comparing Figs, 6-lOd and 6-15, we note that the pattern lobes of the circular loop decrease in magnitude as 8 approaches 9(T while the lobes of the square loop are of equal magnitude. This illustrates the difference of the Bessel function pattern of the circular loop and the trigonomet-ric function pattern of the square loop. In the above discussion, uniform in-phase currents arc assumed.
612 RADIATION EFFICIENCY, Q, BANDWIDTH AND SIGNAL-TO-NOISE RATIO, in Sec. 2-10 we noted that the gain G of an antenna with respect to an isotropic source is identical with the antenna's direc-tivity D provided no losses other than radiation are present. For the more general case we write as in (2-iO-i) that G = kD (1) where k = efficiency factor (0 < k < 1), dimensionless For a lossless antenna, k = l, but with ohmic losses k is less than l h 6-]2 RADIATION EFFICIENCY, Q h BANDWIDTH AND SIGNA LTO- NOl 5E RATIO 257 If an antenna has a radiation resistance Rr and a loss resistance RLf then its (radiation) efficiency factor and the gain Rr + R l R r 4jiA em k R r + R l /} For antennas which are small compared to the wavelength, the radiation resistance R r is small and, if ohmic losses are significant radiation efficiency is reduced. Thus, short dipoles and small loops may be inefficient radiators when losses arc present.
For example, when R L = Rr the radiation efficiency is 50 percent; only half of the power input to the antenna is radiated, the other half being dissipated as heat in the antenna structure.
An rf wave entering a conductor attenuates to 1/e of its surface value in a distance S given by 1 V fnW where / = frequency, Hz p = permeability of medium, H m _I a = conductivity of medium, U m ' 1 It is assumed that a cue. The induced current density in the conductor also attenuates in the same way. This means that the current density associated with a wave traveling along a conductor is greatest close to the surface, the so-called skin effect. The quantity $ is referred to as the I/e depth of penetration.
It follows that the rf resistance of a round wire or solid cylindrical conductor is equivalent to the dc resistance of a hollow tube of the same material of wall thickness b It is assumed that the wire or conductor diameter is much larger than S.
Thus, assuming that the perimeter or circumference L is much smaller than the wave-length so that the current is essentially uniform around the loop, the ohmic (or loss resistance) of a small loop antenna is given by = = § (il) (5) tjTiao a \ ncr where L = loop length (perimeter or circumference), m d = wire or conductor diameter, m 1 J. D. Kraus, Electromagnetics „ 3rd ed., McGraw-Hill, 1984, pp r 447-451.
258 6 THE LOOP ANTENNA From (6-8-10) the radiation resistance of a small loop is K r = 31 200^ = 197C (6) where A = loop area (square or circular), m2 CA — C!
K where C = circumference of circular loop Assuming that the loop's inductive reactance is balanced by a capacitor, the ter-minal impedance will be resistive and equal to Rj = Rr + (?) and the radiation efficiency, or ratio of power radiated to input power, will be 1 + lRJRr) For a 1-turn copper-conductor circular loop (perimeter L = C) in air (a = 5.7 x 10 7 U m _1 , = 4n x 10 -7 H m“ ), l _ 3430 Rr C'fZiL d where C = circumference of loop, m /mh. = frequency, MH 2 d = wire (or conductor) diameter, m For small square loops of side length / (L — 4/), we may take C = 3.5/.
Example. Find the radiation efficiency of a 1-m diameter loop (C = n m) of 10-mm diameter copper wire at (a) 1 MHz and (ft) at 10 MHz.
(10) (10o) (ID The radiation efficiency as a Function of frequency for a small single-turn copper loop in air is shown in Fig. 6-16.
It is assumed that the loop is small compared to the wavelength (C a) and that the wire or conductor diameter is small compared to the loop circumference (d <1 C). Dielectric losses are neglected.
6-\2 RADIATION EFFICIENCY, Q. BANDWIDTH AND SIGNAL-TO-NOISE RATIO 259 In spite of the low efficiency of a small loop, there are many applications where such loops are useful in receiving applications provided the received signal-to-noise ratio is acceptable as discussed later in this section [see (19)], For loops with n turns, Rr increases in proportion to n 2 while R L increases in proportion to n + Hence, for multiturn loops (9) becomes R l 3430 ( 12) and the radiation efficiency k is increased by a factor which approaches n if Ri/R r is large. In (12) the effect of capacitance between turns has been neglected but if the turns are spaced sufficiently and are few in number, (12) can be a useful approximation.
The radiation efficiency of a multiturn loop or coil antenna can be increased by introducing a ferrite rod into the coil as in Fig. 6-17, Here the coil (horizontal to receive vertical polarization) serves the function of both an antenna and also (with a series capacitor) of the resonant circuit for the first (mixer) stage of a broadcast receiver (500 to 1600 kHz).
The radiation resistance of a ferrite loaded loop or coil is given by 260 6 THE LOOP ANTF.NNA and the loss resistance (due to the ferrite rod) by Rf = - to n 2 “ (Q) fr (14) where / = frequency, Hz li eT — effective relative permeability of ferrite rod, dimensionless = real part of relative permeability of ferrite material, dimensionless !i' = imaginary part of relative permeability of ferrite material, dimension-less p 0 = 4tt x 10“ 7 , H m“ 1 n = number of turns a — ferrite rod cross-sectional area, m 1 i = length of ferrite rod m Because of its open geometry {as contrasted to a closed core or ring) a ferrite rod with a relative permeability nr will have a smaller effective relative permeability (due to demagnetization effect). Typically for a rod with = 250 and a length-diameter ratio of 10, the effective relative permeability is about 50.
The ohmic loss resistance J? L of the coil is as given by (5), The radiation efficiency factor for the ferrite rod coil antenna is then Dielectric loss is neglected.
Knowing the total resistance of the ferrite rod antenna, one can calculate the Q and bandwidth of the tuned circuit of which it is a part. The U RADIATION tETKTENCY Q. BANDWIDTH AND SLGNAl-TO-NO($H RATIO 261 S' (ratio of energy stored to energy lost per cycle) is given by n 2nj0 L f 0 Q R, + R L + Rf A/hp where /o = center frequency, Hz L = p trr ft 2 afi 0jl ~ inductance, H 1 &fHP = bandwidth at half-power, Hz Example, The multi turn ferrite rod antenna of a broadcast receiver has 10 turns of 1 mm diameter enameled copper wire wound on a ferrite rod l cm in diameter and 10 cm long The ferrite rod p T — p r — p" = 250 — j'2,5. Take per = 50. At 1 MHz find (a) the radiation efficiency, {/?} the Q and (cj the half-power bandwidth.
Solution f rom (13|, R r = L91 x JO -4 ft. From (14), R f = 0.31 Q, From (4), d = 7 x 10 _i m. Thus, the ratio d/d -14.3 so we can use (5) (times n), which makes R L = 0.026 Q.
Accordingly, {R tr 4- R f )/R r = 1790 and k = 1/1790 = 5.6 x 10“ 4 (Dielectric losses are neglected.) (b) From (16), Q = 162.
(c) From (16), A/ HP = 6 170 kHz.
Although 6.17 kHz is adequate front-end selectivity for the 10-kHz channel spacing of the broadcast band, the low aperture efficiency of less than 0.06 percent makes it uncertain whether the sensitivity is adequate. To determine this, a calculation of the signat-ro-noise ratio for a typical application is required.
From the Friis transmission formula (2-25-5), the power received from a trans-mitter of power P t at a distance r is P,A A„ (W) ([7) r/r where A et — effective aperture of transmitting antenna, m 2 A er = effective aperture of receiving antenna, m 2 For a small loop receiving antenna, D = \ so l,5/.
2 fc , ^ where k = radiation efficiency factor The signal -to-noise ratio (S/N) is given by S P — = ” (dimensionless) 1 Distinguish between L for inductance in (16) and L for Length in (5).
262 6 THE LOOP ANTENNA where P r — received power, W N = fc7^A/[from (17-3-8)] At 1 MHz, 7^ s is dominated by the sky (antenna) temperature T 5 , so taking T SJi = T s we have for this case N — kT s A/ (20) where k = Boltzmann's constant = 1,38 x 10“ 33 J K -1 T s = sky background temperature, K A/ = bandwidth, Hz Distinguish between k in (20) for Boltzmann's constant and k in (18) for the radiation efficiency factor.
Example. Find the S/N ratio for a receiver with the ferrite rod antenna of the above example at a distance of 100 km from a 1-MHz 10-kW broadcast station with an omnidirectional antenna. Take the receiver band width as Af — 104 Hz.
Solution . Assuming that the radiation pattern of the transmitting antenna fills a half-sphere, its directivity is 2 and, hence, , 2 A '< = T" 6-9 Loop and dipole for circular polarization.
If a short electric dipole antenna is mounted inside a small loop antenna (on polar axis. Fig, 6-6) and both dipole and loop are fed in phase with equal power, show that the radiation is everywhere circularly polarized with a pattern as in Fig. 6-6.
CHAPTER 7 THE HELICAL ANTENNA 7-1 INTRODUCTION.
In 1946, a few months after joining he faculty at Ohio State University, I attended an afternoon lecture on traveling-wave tubes by a famous scientist who was visiting the campus. In these tubes an electron beam is fired down the inside of a long wire helix for amplification of waves traveling along the helix. The helix is only a small fraction of a wavelength in diameter and acts as a guiding structure. After the lecture I asked the visitor if he thought a helix could be used as an antenna, "No,” he replied, “Eve tried it and it doesn’t work,” The finality of his answer set me thinking.
If the helix were larger in diameter than in a l raveling-wave tube, I felt that it would have to radiate in some way, but how, I did not know.
I determined to find out.
That evening in the basement of my home I wound a 7-turn helical coil of wire 12 in circumference and fed it via coaxial line and ground plane from my 12-cm oscillator (Fig. 71).
I was thrilled to find that it produced a sharp beam of circularly polarized radiation off its open end.
Next I wound other helices with larger and smaller diameters, noting little change in behavior. Adding more turns, however, resulted in sharper beams.
Although my invention/discovery had come quickly, I realized then that much work would be required to understand this remarkable antenna. Actually it took years of extensive measurements and calculations, I published many articles, a few with students to whom I had assigned studies of specific properties of the 265 266 i THE HELICAL ASTENNA Figure 7-1 I he hrst axial-mode helical anlenna \ 19461. When 1 rotated the hand-held dipole there was no change in response, indicating circular polarization.
antenna.
1 1 also derived equations suitable for engineering design purposes and summarized them in Chap. 7 of the first edition of Antennas.
The steps taken to unravel the mystery of the helix went something like this. The input impedance was measured and found to he essentially resistive and constant over a wide bandwidth. This suggested that the helix behaved like a terminated (matched) transmission line. This was hard to understand because the open end of the helix was completely unterminaied. New insights came when we measured the current distribution along the helix. This we did by rotating a helix and its ground plane while holding a small loop (current probe) under the helical conductor (Fig.
7-2), At a low frequency (helix circumference about //' 2) there was an almost pure standing wave (VSWR- x) all along the helix (outgoing and reflected waves nearly equal) (Fig. 7-3ul, but as the frequency increased, the dis-1 J D Kraus.
LL Helical Beam Antenna," FJecirrmici. 20. 109 111. April 1947.
J D. K raus and J.
Williamson, TL Ch aract eristics of Helical Anicrmas Radiating in the Axial Mode," J Appi. Phys ..
19, 87-96, January 1948.
O J Glasser and J D. Kraus, “ Measured Impedances of Helical Beam Antennas," J. Appi. Pkys., 19, 393-197, February 3948 J D.
Kraus.
“Helical Beam Antennas for Wide-Band Applications" Proc. IRF r , 36, 1236-1242.
October 1948.
J. D Kraus, “The Helical Antenna." Proc. IRE, 37, 263-272. March 1949.
J. D. Kraus, “ Helical Beam Antenna Design Techniques,” Orntmunirdriurtt, 29. 6-9, 34—35, September 1949.
T.
E. Tice and J. D. Kraus, “The Influence of Conductor Size on the Properties of Helical Beam Amen n as," Pro c IRE. 37, 1296, November 1949 7-1 INTEtOUlTTION 267 Figure 7-2 Helix and ground plane mounted to rotate on the helix axis for current distribution measurements along the helical conductor using a loop probe.
{After Kraus arui WiliUtmsan.
" Characteristics of Heliaif KudiarrFty rn ihe Axial Mode," J Appl.
Phys., 19, 87-96, January 1948.) As the helix was rotated, the probe was moved horizontally to follow the hdica) conductor.
tribution changed dramatically.
For a helix circumference of about 1/ three regions appeared : near the input end the current decayed exponentially, near the open end there was a standing wave over a short distance, while between the ends there was a relatively uniform current amplitude (small VSWR) which extended over most of the helix (Fig, 7-3 /j). The decay at the input end could be understood as a transition between a helix-lo-ground-plane mode and a pure helix mode. The reflection of the outgoing wave at the open end also decayed exponentially to a much smaller reflected w fave, leaving the outgoing wave dominant over most of the helix (VSWR small). The small VSWR ripple was sufficient, however, to measure the relative phase velocity ( = Ah//0) along the helix, which was useful for an understanding of the radiation patterns. The current distribution resolved into outgoing and reflected waves is shown in Fig ; -7- 3c.
Our extensive pattern measurements showed that the end-fire beam mode persists over a frequency range of about 2 to 1 centered on the frequency for which the circumference is 1/. Thus, the diameter I had chosen for the first helix I tried was optimum!
Although the helix is continuous, it can also be regarded as a periodic structure. Thus, assuming that an n-turn helix is an end-fire array of n sources, I calculated the pattern using the formula (4-6-9) for the ordinary end-fire condi-tion.
Surprisingly, the measured helix patterns were much sharper. Could the helix be operating in the increased -directivity condition?
1 calculated patterns for this condition using the formula (4-6-14) and obtained good agreement with the 268 7 THE HELICAL ANTENNA Figure 7-3 Typical measured current distribution {a) at a frequency below the axial mode of oper-ation -^ (b } at a frequency near the center of the axial mode region, (c) Resolution of currents into outgoing and reflected waves, (From Kraus and Williamson, “Characteristics of Helical Antennas Radiating in the Axial Mode,” J. Appl. Phys,, 19, 87-96, January 1948.) Compare with distribution on the long, thick cylindrical conductor iti Fig 9-18, measured patterns. Furthermore, this condition persisted over a wide bandwidth, indicating that the phase velocity on the helix changes by just the right amount to maintain the increased-directivity condition. The phase velocity measurements we had made also confirmed this. Thus, the helix locks onto the increased-directivity condition and automatically stays locked over the full bandwidth. Not only does the helix have a nearly uniform resistive input over a wide bandwidth but it also operates as a supergain end-fire array over the same bandwidth! Fur-thermore, it is noncritical to an unprecedented degree and is easy to use in arrays because of an almost negligible mutual impedance.
The helix immediately found wide application.
I employed it in an array of 96 11-turn helices in a radio telescope I designed and built with my students in 1951 (Fig. 7-4). Operating at frequencies of 200 to 300 MHz, the array measured 50 m in length and had a gain of 35 dB, With it we produced some of the first and most extensive maps of the radio sky.
1 Others employed the helix over a wide range of frequencies, some at frequencies as low as 10 MHz (Fig, 7-5), 1 J. D. Kraus t Radio Astronomy „ 2nd ed, Cygnus-Quasar, 1986, p, 8-2.
Figure 7-5 Rotatable (in azimuth) 6-turn helical antenna about 45 m long for operation at fre-quencies around 10 MHz (a — 30 m). Note workmen on arch at far end for scale. (Coiuresy Electro-Physics Laboratory.) 270 ' mt hilkal anthnna Figure 7-6 FleeKaleom geoslationary relay satellite with monoHlar axial-mode helical antennas lor transmission and reception with one as the feed for a dish. These satellites provide global communica-tions for the US. military forces.
I Cowries y TRtV Corp.: H. E. King ami J , L. Wong, "Antenna System .for the EleetSatConi Satellites." IFUE Internationa! Symposium on Antennas and Propagation, pp 349-352, !9 7') Following Sputnik the helical antenna became the workhorse of space com-munications for telephone, television and data, being employed both on satellites and at ground stations. Many U.S.
satellites, including its weather satellites, Comsat, Fleetsatcom (Fig.
7-6}, GOES (global environmental satellites), Lcasat, Navstar-GPS (global position satellites) (Fig.
7-7), Wcstur and tracking and data-relay satellites, all have helical antennas, the latter with arrays of 30. Russiar satellites also have helical antennas, each of the Ekran class satellites being 1-2 HtUt.AL CiKOVTF.TRY 271 Apogee rocket motor Figure 7-7 Navstar GPS (global position] satellite with array of 10 monofilar axial-mode helical antennas.
Eighteen of these satellites are m elliptical orbits around the earth. From them one can determine one’s absolute position anywhere on the earth (latitude, longitude and altitude!, at any lime and in any weather, to a precision of a few centimeters and relative position to a few millimeters.
equipped with an array of 96 helicals. The helical antenna has been carried to the Moon and Mars.
It is also on many other probes of planets and comets, being used alone, in arrays or as feeds for parabolic reflectors its circular polarization high gain and simplicity making it especially attractive for space applications.
1 This short account provides a brief introduction to the helix in which som of the experimental and analytical steps taken to understand its behavior art outlined. Specifically the helix discussed can be described as a monofilar (one- wire) axial-mode helical antenna.
Its operation is explained in more detail in the follow-ing sections along with treatments of many variants and related forms of helical antennas including, later in the chapter, helices with 2 or more wires.
7-2 HELICAL GEOMETRY. The helix is a basic 3-dimensional geometric form. A helical wire on a uniform cylinder becomes a straight wire when unwound by rolling the cylinder on a flat surface. Viewed end-on, a helix projects as a circle. Thus, a helix combines the geometric forms of a straight line, a circle and a cylinder. In addition a helix has handedness; it can be either left- or right-handed.
1 A more detailed personal account of my early work on the helical antenna is given in my book Big Ear , Cygnus-Quasar, 1976.
272 n THE HELICAL AMENNA 7-2 HELICAL GEOMETSV 273 The following symbols will be used to describe a helix (Fig. 7-8): D = diameter of helix (center to center) C = circumference of helix = ttD S = spacing between turns (center to center) % = pitch angle = arctan S;tiD L = length of 1 turn n — number of turns A = axial length = nS d — diameter of helix conductor The diameter D and circumference C refer to the imaginary cylinder whose surface passes through the centerline of the helix conductor. A subscript /.
sig-nifies that the dimension is measured in free-space wavelengths. For example, D k is the helix diameter in free -space wavelengths.
If 1 turn of a circular helix is unrolled on a flat plane, the relation between the spacing 5. circumference C, turn length L and pitch angle 2 are as illustrated by the triangle in Fig, 7-9.
The dimensions of a helix are conveniently represented by a diameter-spacing chart or, as in Fig. 7-10, by a circumference-spacing chart. On this chart the dimensions of a helix may be expressed either in rectangular coordinates by the spacing S A and circumference C A or in polar coordinates by the length of 1 turn L k and the pitch angle 2 . When the spacing is zero, 2 = 0 and the helix becomes a loop. On the other hand, when the diameter is zero, 2 = 90 f and the helix becomes a linear conductor. Thus, in Fig. 7-10 the ordinate axis represents loops while the abscissa axis represents linear conductors.
The entire area between the two axes represents the general case of the helix.
Suppose that we have a 1-turn helix with a turn length of 1/. (L- = 1). When 2 = 0 , the helix is a loop of \k circumference or of diameter equal to l/.fn. As the pitch angle 2 increases, the circumference decreases and the dimensions of the helix move along the L k = 1 curve in Fig. 7-10 until, when 2 = 90\ the helix' is a straight conductor 1>. long.
Surface of imaginary helix cylinder Figure 7-8 Helix and associated dimensions.
Figure 7-9 Relation between circumference, spacing, turn length and pitch angle of a helix.
Pitch angle, tr Figure 7-10 Helix chart showing the location of different modes of operation as a function of the helix dimensions (diameter spacing and pitch angle). As a function of frequency the helix moves along a line of constant pitch angle. Along the vertical axis the helices become loops and along the horizon-tal axis they become linear conductors.
274 1 THE HELICAL. ANTENNA Figurt 7-1 1 ImUitoancLU^ cluigc urid ]k\iJ configuration'.
on Vk’IkV' for different ie-ji ns mission modes.
L he helix dimensions for various- operational modes arc indicated in Fig.
7-10. Reference to these will be made as wc proceed through the chapter.
7-3 TRANSMISSION AND RADIATION MOOTS OF MONO’ FILAR 1 HELICES.
In our discussions it is necessary to distinguish between irunsmissitm and rudiurmit modes.
The term transmission mode is used to describe the manner in which an electromagnetic wa\e is propagated along an infinite helix as though the helix constitutes an inlinile transmission line or waveguide. A variety of different trans-mission modes is possible.
The term nuliotion mode is used to describe the general form of the far-held pattern of a finite helical antenna. Although many patterns are possible, two kinds arc of particular interest. One is the oxiul (or beam) mode of radiation I R 1 mode, beam on axis) and the other is the mtrmot mode of radiation (K 0 mode, radiation maximum perpendicular to axis).
The lowest transmission mode for a helix has adjacent regions of positive and negative charge separated by many turns. This mode is designed as the T <t transmission mode and the instantaneous charge distribution is as suggested by Fig.
7-1 Id. The mode is important when the length of l turn is small com-pared to the wavelength [L /.! and is the mode occurring on low-frequency inductances.
It is also the important transmission mode in the traveling-wave tube. Since the adjacent regions of positive and negative charge arc separated by an appreciable axial distance a substantial axial component of the electric held is present, and in the traveling-wave tube this field interacts with the electron stream (Fig r 7-1 hi).
If the criterion L < y is arbitrarily selected as a boundary for the T 0 transmission mode, the region of the helix dimensions for which this mode is important is within the 7^ R 0 area in Fig, 7-10.
' Momtfilar - unifier - one wire - Singh wire = single conductor (terms used to distinguish the one-wire hdu from helices with 2 or more wiresl-to TRANSMISSION AMI JiAOIATlON MtJLJEi Uf Ml J !N<.) K I [_ A R HELICES 275 STRAIGHT CONDUCTORS (a = 90") Short Long fi-1 T = 1 T-l G = 0 0 i 2 -\ Figure 7-12 Patterns of straight conductor, loop und helix compared.
and ! 2 represent current magnitudes of waves traveling in opposite directions cm antennas.
If 1 2 ~ F tlt ert b a pure standing wave.
If I 2 = 0.
only n pure traveling wave is present.
U m = velocity of wave along antenna, c = velocity of light, C = circumference.) A helix excited in the T 0 transmission mode may radiate. Let us consider the case when the helix is very short (nL /) and the current is assumed to be of uniform magnitude and in phase along the entire helix.
It is theoretically possible to approximate this condition on a small, end-loaded helix. However, its radi-ation resistance is small. The maximum field from the helix is normal to the helix axis for all helix dimensions provided only that nL /, Thus, this condition is called a ^normal radiation mode' (R 0 )' Any component of the field has a sine variation with 8 as shown in Fig. 7-12 (lower left). The space pattern is a figure-of-revolution of the pattern shown around the vertical^ axis.
The field is, in general, elliptically polarized but for certain helix dimer^ions may be circularly 276 7 l'HF: HELICAL ANTENNA 7^ PRACTICAL DESIGN CONSIDERATIONS FOR THE MONOFILAR AXlAl MODE HELICAL ANTENNA 277 polarized and for other dimensions, linearly polarized (see Sec. 7-19), The trans-mission mode and radiation mode appropriate for very small helices can be described by combining the T 0 and designations as T 0 R 0 , This designation is applied in Fig.
7- 10 to the region of helix dimensions near the origin, A first-order transmission mode on the helix, designated Tu is possible when the helix circumference C in free-space wavelengths is of the order of t/« For small pitch angles, this mode has regions of adjacent positive and negative charge separated by approximately 4-turn (or near the opposite ends of a diameter) as shown in Fig.
7-1 Ih and also in end view by Fig. 7-1 lc.
It is found that radiation from helices with circumferences of the order of 1/ (C- — 1) and a number of turns > 1) is a well-defined beam with a maximum in the direction of the helix axis. Hence, this type of operation is called the “ axial (or beam) mode of radiation' with designation R x . The radiation from this mono filar axial mode helical antenna is circularly polarized or nearly so.
The monofilar axial mode of radiation occurs over a range of dimensions as indicated by the shaded area in Fig. 7-10; being associated with the 1\ transmission mode, the combined designation appropriate to this region of helix dimensions is T 1 R 1 .
Still higher-order transmission modes, T z , T 3 and so forth, become per-missible for larger values of C A . For small pitch angles, the approximate charge distribution around the helix for these modes is as suggested by Fig. 7-1 1c.
The axial ( 7^ i?
! > and normal radiation mode (T 0 /i 0 ) patterns of a helix are compared with the radiation patterns for straight conductors and loops in Fig.
7-12-It is Lo be noted that the patterns of a short linear conductor, a small loop and a small helix arc the same.
7-4 PRACTICAL DESIGN CONSIDERATIONS FOR THE MONOFILAR AXIAL-MODE HELICAL ANTENNA.
1 Before ana lyzing the many facets of the antenna individually, an overall picture will be given by describing the performance of some practical designs.
The monofilar axial-mode helical antenna is very noncritical and one of the easiest of all antennas to build. Nevertheless, attention to details can maximize its performance.
The important parameters arc: L Beam width 2. Gain 3. Impedance 4.
Axial ratio When T fust described the helical beam antenna in 1946, there was little chance of ambiguity but now there are many variations of the basic type making more explicit names desirable- -hence the term, monofilar axial-mode helical antenna. Uniform 1-wire end-fire helix or single-conductor helical beam antenna are also possible designations while some call it a Kraus-type helix or Kraus coil.
Gain and beam width, which are interdependent [G x (1/HPBW 2 )], and the other parameters are all functions of the number of turns, the turn spacing (or pitch angle) and the frequency. For a given number of turns the behavior of the beam width, gain, impedance and axial ratio determines the useful bandwidth .
The nominal center frequency of this bandwidth corresponds to a helix circum-ference of about 1/ {(T = i) t For a given bandwidth to be completely useful, all 4 parameters must be satisfactory over the entire bandwidth.
The parameters are also functions of the ground plane size and shape, the helical conductor diameter, the helix support structure and the feed arrangement.
The ground plane may be flat (either circular or square) with a diameter or side dimension of at least 3//4 or the ground plane (launching structure) may be cup-shaped forming a shallow cavity (Fig. 7 -I3h).
A 2-turn flush-mounted design described by Bystrom and Bernsten 1 for aircraft applications is shown in Fig. 7-1 3c. These authors found that 2 turns are required to obtain satisfactory pattern and impedance characteristics but that no significant improvement is obtained with a deeper cavity and a larger number of turns since the size of the aperture opening remains the same (like an open-ended cylindrical waveguide).
The deep conical arrangement of Fig. 7- 13d is effective in reducing the side-and back-lobe radiation.
2 Launching a wave on the helix may also be done without a ground plane, producing a back-fire beam (see Fig. 7-56). By slicing the outer conductor of the coaxial cable longitudinally and separating it gradually from the inner conductor over the first turn or two of the helix, Munk and Peters devised a ground-planeless feed that produces a forward end-fire beam. 3 Conductor size is not critical4 and may range from 0.005/. or less lo 0.05/ or more (Fig. 7-14). The helix may be supported by a few radial insulators mounted on an axial dielectric or metal rod or tube whose diameter is a few hundredths of a wavelength, by one or more longitudinal dielectric rods mounted peripherally (secured directly to the helical conductor) or by a thin-wall dielectric tube on which the helix is wound. With the latter arrangement the operating bandwidth is shifted to lower frequencies so that for a given frequency the antenna is smaller.
Several of these mounting arrangements are illustrated in Fig. 7-15.
! A. Bystrom, Jr., and D. G.
Bernsten, "An Experimental Investigation of Cavity-Mounted Helical Antennas/' IRE Trans. Ants. Prop,, AP-4, 53 58, January 1956-1 K.
R. Carver, “"The Helicons: A Circularly Polarized Antenna with Low Side Lobe Level," Proc.
IEEE, 55, 559, April 1967.
K. R. Carver and B. M. Potts, “Some Characteristics of the Helicone Antenna,' Antennas and Propa-gation International Symposium, 1970, pp. 142-150.
3 B. A. Munk and L, Peters, “A Helical Launcher for the Helical Antenna,'' IEEE Trans. Ants . Prop., AP-16, 362-363, May 1968.
4 T. E. Tice and J. D. Kraus, “The Influence of Conductor Size on the Properties of Helical Beam Antennas/' Prot\ IRE. 37, 12%, November 1949.
278 7 THE HELICAL ANTENNA 7-4 PRACTICAL DESKjN CONSIDERATIONS FOR THE MONOE1LAR AXIAL-MODE HELICAL ANTENNA 279 Figure 7-13 (n) Monofilar axial-mode helical antenna on flat ground plane and (ft) in shallow cupped ground plane {see also Fig 7-16c).
(r) General-purpose flush-mounted 2-turn monofilar axial-mode helical antenna wilh taper feed for matching to a 50-11 coaxial line {after Bystrom and Bernsten, ref. L p. 2? 7) (sec also Fig. 7-l6a and ft), (if) Deep conical ground-plane enclosure for reducing side and back lobes. {After K. R. Career, rtf 2, p r 277).
The helix may be fed axially, peripherally or from any convenient location on the ground-plane launching structure with the inner conductor of the coaxial line connected to the helix and the outer conductor bonded to the ground plane.
With axial feed the terminal impedance (resistive) is given within 20 percent by R = 140C, (Q) while with peripheral feed Baker 1 gives its value within 10 percent as 150 Tel u) (2 ) D.
E. Baker Design of a Broadband Impedance Matching Section for Peripherally Fed Helical Antennas, ' Antenna Applications Symposium, University of Illinois, September 1980.
i % i A Figure 7-14 Peripherally fed monofilar axial-mode helical antennas with helix conductors of 0.055.
0.017 and 0.00422 diameter at center frequency of 400 MHz for determining effect of conductor diameter on helix performance. Only minor differences were measured. [After 7, E. Tice and J. D.
Kraus, “ The Influence of Conductor Size on the Properties of Helical Beam Antennas.
' Proe. IRE. 37, 1296, November 1949 .) The 0.0552 diameter tubing (4.1 cm diameter) is about the largest size which could be bent to the radius of 1 1 cm { = 2/2jt).
These relations have the restrictions that 0,8 < C A < 1,2 12 < x < 14" and h > 4, With a suitable matching section the terminal impedance (resistive) can be made any desired value from less than 50 to more than 150 fi. Thus, by bring-ing the last i-turn of the helix parallel to the ground plane in a gradual manner, a tapered transition between the 140-or 150-0 helix impedance and a 50-Q coaxial line can be readily accomplished.
1 This can be done with either axially or peripherally fed helices but is more convenient with a peripheral feed. Details of a suitable arrangement are shown in Fig, 7-16u_and b , As the helix tubing is brought close to tfier ground plane, it is gradually flattened until it is completely flat at the termination, where it is spaced from the ground plane by a dielectric sheet (or slab). The appropriate height h (or thick-ness of the sheet) is given by 2 [377/(V£ZoH-2 where w = width of conductor at termination h = height of conductor above ground plane (or thickness of dielectric sheet) in same units as w c r = relative permittivity of dielectric sheet Zq = characteristic impedance of dielectric sheet l J, D. Kraus, “A 50-ohm Input Impedance for Helical Beam Antennas," IEEE Trans. 4ms. Prop,, AP-2S, 9 1 3 h November 1977, J I. D. Kraus. Electromagnetics, 3rd ed., McGraw-Hill, 1984, pp. 397-398.
280 7 THE HELICAL ANTENNA Figure 7-15 Monofilar axial-mode helical antenna supported by axial metal or dielectric rod (or tube) with radial insulators (a), by four, peripheral dielectric rods secured to the helix (&) and by a dielectric tube on which the helix is wound (c).
Example.
If the flattened tubing width is 5 mm, find the required thickness of a polystyrene sheet (e, = 2,7) for matching to a 50-fi coaxial transmission line.
Solution. From (3), h = — 1.9 mm [377/(v/Z7 x 50)] - 2 A typical peripherally fed monofilar axial-mode helical antenna with cup ground-plane launcher matched to a 50-ft line, as in Fig. 7-16o and b, is shown in Flattened .tubing Dielectric „ sheet """""" ' plane - 50 -ft coaxial A > connectpr Detailed cross-section through AA ' as seen from right.
Dielectric sheet , Flatten i y— 1 / tubing ZZZZjJ^ ^ Ground plane 50-ft coaxial connector Figure 7-16 {a) Gradually tapered transition from helix to coaxial line with detailed cross section at (6).
?- Pit ACTtCAt" DESIGN CONSIDERATIONS FOR THE MONOFILAR AXIAL-MODE HELICAL ANTENNA 281 Cupped ground plane Figure 7-l6c Typical peripherally fed monofilar axial-mode helical antenna with cupped ground plane matched to a 50-11 coaxial transmission line as in Fig. 7- 16a and b. The turn spacing S = 0.225/ and the circumference C = a at the center frequency. The relative phase velocity p changes automati-cally by just the right amount to lock onto end-fire ^ (F) with supergain over a frequency range of about one octave. Typical dimensions of the cupped ground plane are a = 0.75a and b = a/2 at the center frequency.
Fig.
7- 16c with dimensions given in wavelengths at the center frequency for which C ; = l. Support may be an axial rod with radial insulators or one or more peripheral rods (Fig, 7-1 Sa and b).
A monofilar axial-mode helical antenna with flat circular ground plane and supported by dielectric members is shown in Fig. 7-17 and one with cupped ground plane supported by a dielectric cylinder is illustrated in Fig. 7-18.
Measured patterns of a 6-turn helix as a function of frequency are presented in Fig. 7-19 and patterns at the center frequency (CA = l) as a function of length (number of turns) are shown in Fig. 7-20.
Based on a large number of such pattern measurements which we made during 1948 and 1949, the beam widths were found to be given by the following qua si-empirical relations.
52 HPBW (half-power beam width) ^ -j= (deg) (4) 115 BWFN (beam width between first nulls) ^ y= (deg) (5) CA y/fiSk The HPBW as given by (4) is shown graphically in Fig. 7-21, Dividing the square of (4) into the number of square degrees in a sphere (41253) yields an FigW 7-18 Thm-wall plastic-cylinder-suppo rted 6^-turn monofilar axial-mode helical antenna with solid metal cupped ground plane. Feeding is via a matching transition from a 50-fl cable connected through a fitting mounted on the back of the cup ground plane at a point between the plastic cylinder and the lip of the cup. The helix is a flat metal strip bonded to the plastic cylinder. The strip width is (equivalent to a 0.01 5A diameter round conductor). The pilch angle is 13.8 : . Built by the author for UHF TV band operation with YSWR < 2 from channel 25 to S3 (524 to 890 MHz) and less than 1.2 from channel 27 to 75 (548 to 842 MHz).
Figure 7-20 Hflect of number of turns on measured field patterns. Helices have 12-2 pitch angle and 2, 4, 6, R, 10 turns. Patterns shown are average of measured E4 and patterns. {After Kraus.) 284 7 THE HELICAL ANTENNA 7-4 PRACTICAL DESIGN CONSIDERATIONS FOR THE MONOFILAR AXIAL MODE HELICAL ANTENNA 285 i i i b i > 1 l | i > 1 1 1 1 1— J 3 4 5 6 7 B 910 12 14 161820 25 30 35 40 45 Number of turns tfll for t\ = TO and a = 12.5 Figure 7-21 Half-power beam width of monofilar axial-mode helical antenna as a function of the axial Length and circumference in free-space wavelengths and also as a function of the number of turns for C A - LQ and a = 12.5° (lower scale). {After JQ'dwsO approximate directivity relation: 1 D ^ 1 5C2 x nSx (6) This calculation disregards the effect of minor lobes and the details of the pattern shape. A more realistic relation is D^\2C\nSx (7) Restrictions are that (4) to (7) apply only for 0,8 <C X < 1.15, 12° < a < 14 and n > 1 Note that from (4-8-3) D = 4nnSx , The measured gains of King and Wong^ for 12.8° monofilar axial-mode helical antennas are presented in Fig, 7-22 as a function of helix length (L A = 1 It is assumed that the patterns of both field components are of the same shape and arc hgures-of-revolution around the helix axis.
H. E. King and J. L. Wong, "Characteristics of 1 lo S Wavelength Uniform Helical Antennas" /£££ Trans, Prop, AP-28, 291, March I960-\ Relative frequency (or circumference, CJ Figure 7-22 Measured (dashed) gain curves of monofilar axial-mode helical antennas as a function of relative frequency for different numbers of turns for a pitch angle of a = 12,B a . {After H .
E. King and J. L. Wong, " Characteristics cf t to S Wavelength Uniform Helical Antennas” IEEE Trans. Ants.
Prop,, AP-28, 291, March J980 ,) Calculated (solid) gain curves are also shown for different numbers of turns.
and frequency. Although higher gains are obtained by an increased number of turns, the bandwidth tends to become smaller. Highest gains occur at 10 to 20 percent above the center frequency for which C = 1. Although the gains in Fig. 7-22 tend to be less than calculated from (7), they were measured on helices with 0.08A diameter axial metal tubes.
Although pitch angles as small as 2°, as noted by MacLean and Kouyoum-jian, 1 and as large as 25°, as noted by Kraus, can be used, angles of 12° to 14 (corresponding to turn spacings of 0.21 to 0.252 at Cx — 1) are optimum. King and Wong found that on helices with metal axial tubes, smaller pitch angles (near 12°) resulted in a slightly higher (1 dB) gain but a narrower bandwidth than larger angles (near 14°).
Turning to other parameters, the pattern, axial ratio and impedance (VSWR) performance as a function of frequency for a 6-tum, 14° pitch angle monofilar axial-mode helical antenna are summarized in Fig. 7-23. This is the same antenna for which the patterns are shown in Fig, 7-19. The half-power beam width is taken between half-power points regardless of whether these occur 1 T. S. M. MacLean and R, G. Koujroumjian, "The Bandwidth of Helical Antennas," IRE Trans Ants . Prop., AP-7, S379-386, December 1959.
7-23 Summary of measured performance of 6-lurn, 14° monofilar axial-mode helical antenna The curves show the HPBW for both field components, the axial ratio and the VSWR on a 53-0 line as a function of the relative frequency (or circumference Trends of (relative) resistance R and reactance X are shown in the VSWR inset. Note the relatively constant R and small X for C, > 07.
' {After Itroiis.) i i on the major lobe or on minor lobes. This definition is arbitrary but is conve-nient to take into account a splitting up of the pattern into many lobes of large amplitude at frequencies outside the beam mode Beam widths of 180 or more are arbitrarily plotted as 180°. The axial ratio is the value measured in the direc-tion of the helix axis. The standing-wave ratio is the value measured on a 53-fl coaxial line. A transformer section A/4 long at the center frequency is located at the helix terminals to transform the terminal resistance of approximately 130 to 53 fl. Considered altogether, these pattern, polarization and impedance charac-M PRACTJCAL DESIGN CONSIDERATIONS EOR THE MONOFILAJt AXIAL MODE HELICAL ANTENNA 287 teristics represent remarkably good performance over a wide frequency range for a circularly polarized beam antenna The onset of the axial mode at a relative frequency of about 07 is very evident with axial-mode operation, extending from this frequency over at least an octave for VSWR and axial ratio and almost an octave for pattern Several arrangements have been proposed to reduce ,the axial ratio and VSWR to even lower values. These include a conical end-taper section by Wong and King, 1 Donn/ Angelakos and Kajfez and Jamwal and Vakil,4 and a flat Relative frequency (or circumference, Cl t Figire 7-24 Reflection coefficient and VSWR for an impedance-matched peripherally fed iG-tum, 13.8 monofilar axial-mode helical antenna as a function of relative frequency (or circumference <TJ without spiral termination (solid) and with it (dashed), [After D .
E. Baker , Design of a Broadband Impedance Matching Section for Peripherally Fed Helical Antennas'' Antenna Applications Sympo-sium, University of Illinois^ September 1980.) 1 J. L. Wong and H. E. King “ Broadband Quasi-Taper Helical Antennas," IEEE Trans. Ants: Prop., AP-27, 72-78, January 1979.
1 C. Donn, "A New Helical Antenna Design for Better On-and-OfT Boresighl Axial Ratio Per-formance' 1 /£££ Trans. Ants. Prop. AP-28, 264—267 March 1980.
3 D.
J. Angelakos and D. Raj fez, " Modifications on the AxiaJ-Mode Helical Antenna" Proc. /£££ 55558-559, April 1967.
4 K. K.
S. Jamwal and R. Vakil, “Design Analysis of Gain-Optimized Helix Antennas for X-band Frequencies” Microwave J, r 177-183, September 1985.
288 7 THE HELICAL ANTENNA 7-5 AXIAL-MODE PATTERNS AND THE PHASE VELOCITY OF WAVE PROPAGATION 289 spiral termination by Baker.
1 The flat spiral adds no axial length to the helix.
The reflection coefficient (or VSWR) as measured by Baker for a 10-turn periph-erally fed monofilar axial-mode helical antenna with and without the spiral ter-mination is presented in Fig. 7-24.
The improvement occurs at relative frequencies above IT which, however, is a region where the gain is decreasing, A dielectric tube supporting a helical conductor may significantly affect per-formance. The magnitude of the effect depends on the dielectric’s properties and its geometry, especially the thickness of the tubing wall. For a peripherally fed helix supported on a polyvinyl chloride (PVC) tube with zr — 2.70, Baker, 1 using the VSWR as a criterion, found that the relative frequency for the onset of the axial mode shifted from 0.72 without the tube to 0.625 with the tube for a ratio of 1T5.
Thus, the effective relative permittivity £eff (with tube) = 1.32 ( - 1.1 5 2 ), making the terminal resistance at the center frequency 130 Q (=150/^/1^32). A precision matching section designed by Baker converts this to 50 Q with mea-sured VSWR < 1.2 (p v < -20 dB) over a 1,7 to 1 bandwidth. The helix wire is wound in a groove of half the wall thickness machined with a computer-controlled lathe.
All dimensions of the helix, matching section and supporting structure are specified in Baker’s design.
7-5 AXIAL-MODE PATTERNS AND THE PHASE VELOCITY OF WAVE PROPAGATION ON MONOFILAR HELICES.
2 As a first approximation, a monofilar helical antenna radiating in the axial mode may be assumed to have a single traveling wave of uniform amplitude along its con-ductor. By the principle of pattern multiplication, the far-field pattern of a helix is the product of the pattern for 1 turn and the pattern for an array of n isotropic point sources as in Fig. 7-25. The number n equals the number of turns. The spacing S between sources is equal to the turn spacing. When the helix is long (say, nS^ > 1), the array pattern is much sharper than the single-turn pattern and hence largely determines the shape of the total far-field pattern.
Hence, the approximate far-field pattern of a long helix is given by the array pattern.
Assuming now that the far-field variation is given by the array pattern or factor and that the phase difference between sources of the array is equal to the phase shift over 1 turn length for a single traveling wave, it is possible to obtain a simple, approximate expression for the phase velocity required to produce axial-mode radiation. This value of phase velocity is then used in pattern calculations.
The array pattern or array factor E for an array of n isotropic point sources arranged as in Fig. 7-25 is given by (4-6-8). Thus, ^ sin (m^/2) E = ~ 1 (1) sin w/2) 1 D.
E. Baker, “Design of a Broadband Impedance Matching Section for Peripherally Fed Helical Antennas," Antenna Applications Symposium, University of Illinois, September 1980-2 J. D. Kraus, “The Helical Antenna," Proc. IRE+ 37, 263-272, March 1949, To distant point Figure 7-25 Array of isotropic sources, each source representing 1 turn of the helix.
where n = number of sources and \j/ = Sr cos + & (2) where Sr = 2 nS.}X In the present case, (2) becomes \j/ — 2^5^ cos tf> — (3) where p = v/c = relative phase velocity of wave propagation along the helical conductor, v being the phase velocity along the helical conductor and c being the velocity of light in free space.
If the fields from all sources are in phase at a point on the helix axis ( — 0 and equating (3) and (4), we have ^ = S + (5) P When m = 1 and p = 1, we have the relation L, - S = L or L - S = X (6) This is an approximate relation between the turn length and spacing required for a helix radiating in the axial mode. Since for a helix L 2 — n 20 2 + S 2 , (6) can be rewritten as D> = J2S, + I Ck = S, + 1 (7) 290 T THE HELICAL ANTENNA Equation (7) is shown graphically by the curve marked C A = p2S x 4 1 in Fig.
7-10, The curve defines approximately the upper limit of the axial- or beam-mode region.
When m = l s (5) is appropriate for a helix operating in the first-order (7L ) transmission mode. When m = 2, (5) is appropriate for the T 2 transmission mode , etc. A curve for m = 2 is shown in Fig. 7-10 by the line marked = 2^/Sx 4 1.
Hence, m corresponds to the order of the transmission mode on a helix radiating a maximum field in the axial direction. The case of particular interest here is where m — 1.
The case where m = 0 does not represent a realizable condition, unless p exceeds unity, since when m = 0 and p = 1 in (5) we have L = S. This is the condition for an end-fire array of isotropic sources excited by a straight wire connecting them (s = 90°). However, the field in the axial direction of a straight wire is zero so that there can be no axial mode of radiation in this case.
Returning now to a consideration of the case where m = 1 and solving (5) for p, we have ,8) From the triangle of Fig, 7-9, (8) can also be expressed as sin a 4 [(cos a)/CJ Equation (9) gives the required variation in the relative phase velocity p as a function of the circumference CA for in-phase fields in the axial direction. The variation for helices of different pitch angles is illustrated in Fig. 7-26. These curves indicate that when a helix is radiating in the axial mode (4 < Q < f) the value of p may be considerably less than unity. This is borne out by direct mea-surements of the phase velocity. In fact, the observed phase velocity is found to be slightly less than called for by (8) or (9), Calculating the array pattern for a 7-turn helix using values of p from (8) and (9) yields patterns much broader than observed. The p value of (8) or (9) corresponds to the ordinary end-lire condition discussed in Chap. 4.
If the increased directivity condition of Hansen and Woodyard is presumed to exist, (4) becomes = — ^2nm + ~) Now r equating (10) and (3), putting = 0 and solving for p we have S, 4 m 4 (l/2n) For the case of interest m = 1 and ' S x + i(2n 4 1 )/2u] ( 12) axial.mode patterns and the phase velocity of wave propagation 291 Figure 7-26 Relative phase velocity p for different pitch angles as a function of the heltx circum-ference Cx for the condition of in-phase fields in the axial direction.
For large values of n, (12) reduces to (8). Equation (12) can also be expressed as 1 D = -— — (13) sin a 4- [(2 n 4 l)/2n][(co$ oc)/CJ Using p as obtained from (12) or (13) to calculate the array factor yields patterns in good agreement with measured patterns. The p value from (12) or (13) also is in closer agreement with measured values of the relative phase velocity. Hence, it appears that the increased directivity condition is approximated as a natural con-dition on helices radiating in the axial mode,?
Another method of finding the relative phase velocity p on helical antennas radiating in the axial mode is by measuring the angle 0 O at which the first minimum or null occurs in the far-field pattern. This corresponds to the first null in the array factor, which is at (sec Fig 4-20). Then in this case (4) becomes ijf — — (2nm 4 0 O) U 4) Now equating (14) and (3) and putting m = 1 and solving for p , we have p =— ~—r U5) S x cos 00 4 1 + (0 o/2ji) 1 II is to be noted that, as n becomes targe, this relation (1.3) for increased directivity reduces to 19).
1 The axial mode region is shown by the shaded (T^R,) area in Fig. 7-10. Helices with dimensions in ibis region radiate in the axial mode, and (9)> or more properly (13), applies. Outside this region these equations generally do not apply.
292 ?
THE HELICAL ANTENNA Three relations for the relative phase velocity p have been discussed for helices radiating in the axial mode with transmission in the T x mode. These are given by (9), (13) and (15).
A fourth relation for p appropriate to the T, and higher-order transmission modes on infinite helices has been obtained by Bagby 1 by applying boundary conditions approximating a helical conductor to a solution of the general wave equation expressed in a new coordinate system he called “helicoidal cylindrical coordinates.’ Bagby's solution is obtained by applying boundary conditions to the two points c and d in Fig. 7-27. His value of the relative phase velocity is given by p = C \ D .
06) m cos a + hR sin or where Jt/i = tan 2 m J%kR) Jm -,{kR) Jm+1{kR) (17) 1 C. K. Bagby, “A Theoretical Investigation of Electro-magnetic Wave Propagation on the Helical Beam Antenna,” Master's thesis. Electrical Engineering Department, Ohio State University, 1948.
7-5 AXTAL-MODE PATTERNS ANfJ THE PHASE VELOCITY OF WAVE PROPAGATION 293 1.2 1.1 1.0 0.9 0.8 a I G 0.7 0.6 0.5 0.4 0.3 0.3 0.4 0.5 0.6 0.7 0.8 0,9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Helix circumference = Ci - 27f/X Figure 7-28 Relative phase velocity p.as a function of the helix circumference Cf for 1 V helices. The solid curve is measured on a 13'\ 7-tum helix. Curves A t and A l are as calculated by Bagby Tor T t and 7a transmission modes on an infinite 13° helix. Curves 8, and B2 are calculated Tor in-phase fields and curves C, and C2 for increased directivity for and T 2 transmission modes. Curve D is from data by Chu and Jackson as calculated for the T 0 transmission mode. {After JfraiH.) where m = order of transmission mode (=1,2, 3, .. ,) (m # 0) Ji = radius of helix cylinder kR = fC\ - (hR?
h = constant J - Bessel function of argument kR The variation of p as a function of C for a 13 n helix as calculated by (16) and (17) for the case m = 1 is illustrated by the curve A t in Fig. 7-28. A curve for the T] transmission mode (m = 1) as calculated for the in-phase condition from (9) is shown by Bj. A curve for the increased directivity condition on a 13°, 7-turn helix, with m = 1, is presented by C,.
Curves for the T 2 transmission mode for each of the three cases considered above are also presented in Fig. 7-28. In addition, a curve of the measured rela-tive phase velocity on a 7-tum helix is shown for circumferences between about 0.4 and 1.5/.
It is to be noted that in the circumference range where the helix is radiating in the axial mode (i < Ck < % the increased directivity curve, of the three calculated curves, lies closest to the measured curve.
1 The measured curve gives the value of the total or resultant phase velocity owing to all modes 1 The increased directivity curve is the only curve calculated for a helix of 7 turns. The in-phase field curve refers to no specific length while Bagby's curve is for an infinite helix.
7 MONOFILAR AXIAL-MODE SINGLE-TURN PATTERNS 295 294 7 THE HELICAL ANTENNA present (T 0 , 7^, etc.) as averaged over the region of the helix between the third and sixth turns from the feed end The vertical lines indicate the spread, if any, in values observed at one frequency. In general, each transmission mode propagates with a different velocity so that when waves of more than one transmission mode arc present the resultant phase velocity becomes a function of position along the helix and may vary over a considerable range of values.
1 When \ < C x < f the phase velocity as measured in the region between the third and sixth turns corre-sponds closely to that of the T, transmission mode. The T 0 mode is also present on the helix but is only important near the ends. When the circumference < y, the T 0 mode may be obtained almost alone over the entire helix and the mea-sured phase velocity approaches that for a pure T 0 mode indicated by curve D in Fig, 7-28, based on data given by Chu and Jackson. 2 This curve indicates that at small circumferences the relative velocity of a pure T 0 mode wave attains values considerably greater than that of light in free space. At C x = y, curve D has decreased to a value of nearly unity, and if no higher-order transmission mode were permissible, the phase velocity would approach that of light for large circumferences.
However, higher-order modes occur, and, when C 2 exceeds about the resultant velocity drops abruptly, as shown by the measured curve in Fig, 7-28. This change corresponds to a transition from the T 0 to the Ti transmission mode. For a circumference in the transition region, such as 0.72, both and T, modes are of about equal importance.
When C 2 is about \ or somewhat more, the measured phase velocity approaches a value associated with the Tj mode. As Cx increases further, the relative phase velocity increases in an approximately linear fashion, agreeing most closely with the theoretical curve for the increased directivity condition (curve Cj). When C k reaches about y, a still higher order transmission mode (T 2 ) appears to become partially effective, causing further dips in the measured curve.
However, the radiation may no longer be in the axial mode.
The formulas given for helical antennas operating in the first-order trans-mission mode{m = 1) are summarized in Table 7-1.
As mentioned above, the approximate far-field pattern of a monofilar helix radiating in the axial mode is given by the array factor for n isotropic point sources, each source replacing a single turn of the helix (see Fig, 7-25).
The normalized array factor is n sin (nij//2) In sin (t^/2) where tp = 27i[5 i cos — (L^/p)] (18) 1 I. A. Marsh “ Measured Current Distributions on Helical Antennas," Proc. IRE , 39, 66H-675, June 1951.
2 L. J. Chu and J. D. Jackson, “ Field Theory of Traveling Wave Tubes," Proi r . IRE , 36, 853-863, July 1948.
Table 7-1 Relative phase velocities for first-order transmission mode on helical antennas Condition Relative phase velocity In-phase fields (ordinary end-fire) Increased directivity From first null of measured field pattern Helicoid al cylindrical coordinate solution 5^ + 1 sin s + [(cos oO/Cj] Si + [(2n + l)/2n] 1 sin a + [(2n + l)/2n][(cos a)/CJ Lx ^ Si cos + (^a/2fr) + 1 C-p ™ cos a + hR sin a where hR is as given by (17) The normalizing factor is sin (njln) instead of \jn since the increased directivity end-fire condition is assumed to exist (see Sec. 4-6d, Case 3). For a given helix, S A and L x are known and p can be calculated from (12) or (13).
}t is then obtained as a function of From (18), these values of \js give the held pattern.
As an illustration, the calculated array factor patterns for a 7-turn, 12° helix with C k = 0,95 are shown in Fig, 7-29 for p values corresponding to increased directivity and also in-phase fields and for p = 1, 0.9 and 0.725. A measured curve (average of E and E $ ) is shown for comparison, ft is apparent that the pattern calculated for the increased-directivity condition (p = 0.76) agrees most closely with the measured pattern. The measured pattern was taken on a helix mounted on a ground plane 0,88/ in diameter. The calculated patterns neglect the effect of a ground plane. This effect is small if the back lobe is small compared to the front lobe, as it is for p = 0.802 and p = 0.76, The sensitivity of the pattern to the phase velocity is very apparent from Fig. 7-29. In particular, we note that as little as a 5 percent difference in phase velocity from that required for the increased directivity condition (p = 0,76) pro-duces marked changes as shown by the patterns for p — 0,802 (5 percent high) and p = 0.725 (5 percent low).
7-6 MONOFILAR AXIAL-MODE SINGLE-TURN PATTERNS. In this section expressions will be developed for the far-held patterns from a single turn of a monofilar helix radiating in the axial mode.
It is assumed that the single turn has a uniform traveling wave along its entire length. The product of the single-turn pattern and the array factor then gives the total helix pattern, A circular helix may be treated approximately by assuming that it is of square cross section. The total field from a single turn is then the resultant of the 5 percent less than ,76 Uivphase fields condition I Figure 7-29 Array factor patterns for 12% 7-tutn helix with Cx = 095. Patterns are shown for p = 1, 0,9, 0.802 (in-phase fields or ordinary end-fire condition) 0.76 (increased directivity) and 0.725. A measured curve is also presented. All patterns are adjusted to the same maximum. The sensitivity of the pattern to phase velocity is evident. A change of as little as 5 percent produces a drastic change in pattern, as may be_ noted by comparing the pattern for p = 0,802 (5 percent high) and the one for p = 0.725 (5 percent low) with the one for p = 0.76 which matches the measured pattern-fields of four short, linear antennas as shown in Fig. 7-30a, A helix of square cross section can, of course, be treated exactly by this method Measurements indicate that the difference between helices of circular and square cross section is small Referring to Fig. 7-31, the far electric field components, £^ r and £eT , in the xz plane will be calculated as a function of 4> for a single-turn helix.
Let the area of the square helix be equal to that of the circular helix so that U) where g is as shown in Fig, 7-30a.
The far magnetic field for a linear element with a uniform traveling wave is given by (5-8-15). Multiplying (5-8-15) by the intrinsic impedance Z of free space, putting y = (3ti/2) + a + 4>, t = 0 and h = £//cos x, w re obtain the expression for the 4> component £01 of the far field in the xz plane due to element 1 of the square helix as follows: £ A1 = k sin BA l ba\ 7-6 MONOFlLAR AXIAL-MODE SINGLE-TURN PATTERNS 297 yX /rr Hhx /' I j conductor t \ -/ 2l c V [^ \ Conductor To point P To point P Figure 7-30 Square helix used in calculating single-turn pattern.
where k = A = 1 — p cos y 2pc cos a The expressions for E+2 , £j , etc., due to elements 2, 3 and 4 of the square turn are obtained in a similar way. Since the elements are all dissimilar sources, the total component, £ T) from a single square turn is obtained by adding the 1 -turn helix x.
Figure 7-31 Field components with relation to single-turn hdix.
7 THE HELICAL ANTENNA fieids from the four elements at each angle for which the total field is calcu-lated. The sum of the fields from the four elements is then sin BA" sin a sin 0 /f Lo) w (S 1\ ~4^ + 7T sin y /r .
Lcn tu / S cos + k— Sin BA ' l-BA -2p-c + 7{-^-sin BA" sin ac sin 0 /f „ 3Lgj a ( + 1 A-/Ll^ ~^ + cV S COS + g sm + g sin 35 cos (p 4 “ ri )_ where y = — + & + / = ~ — oc + 7 — arccos (sin jx cos 0) ,4 = 1 — p cos y, /T — i _ p cos ^4" = 1 — p cos y" When a helix of circular cross section is being calculated, L = tiD/cos a in (3), while for a helix of square cross section L = 4b.
If the contributions of elements 2 and 4 are neglected, which is a good approximation when both a and are small, the expression for E+ T is consider-ably simplified. Making this approximation, letting k = 1 and = constant, -we obtain sin v Et =~ sin BA H-BA) + sin BA' /[ - BA' -B + n(S, cos 4> + -,/n D 1 sin <£)] (4) A Equation (4) applies specifically to helices of circular cross section, so that B in (4) 1S D 71 3/2 B = (5) 2p cos a Equation (4) gives the approximate pattern of the component of the far field in the xz plane from a single-turn helix of circular cross section In the case of the 8 component of the Tar field in the xz plane, only elements 2 and 4 of the square turn contribute.
Putting k = 1, the magnitude of the approximate 0 pattern of the far field of a single-turn helix of circular cross section can be shown to be sin y" sin BA" cos at 9T A tf {\ — sin 2 a cos 2 x sin cos s^n — 2^/ttB] where B is as given by (5) and y" and A" are as in (3) + (6) 7-6 MONOEILAR AXIAL-MODE SINGLE-TURN PATTERNS 299 a=T2 \ 1 Figure 7-32 Calculated patterns for t^r and £ar fields of single turn of a 12"' helix.
Qs Total pattern of one turn Wave direction Figure 7-3$ Individual E patterns of e.^ ments 1 and 3 and total pattern of single turn, E#r . The single turn is shown in plan view (in the xz plane of Fig. 7-30). The single turn and coordinate axes have been rotated around the y axis so that the z direction (6 = 0) is toward the top of the page.
3Q0 7 THE HELICAL ANTENNA As an example, the E+T and £#r patterns for a single-turn 12° helix with Cx — 107 have been calculated and are presented in Fig. 7-32. Although the two patterns are of different form, both are broad in the axial direction (# = 0).
The individual E4 patterns of elements 1 and 3 of the single turn are as suggested in Fig, 7-33. One lobe of each pattern is nearly in the axial direction, the tilt angle i being nearly equal to the pitch angle a. The individual patterns add to give the E+ r pattern for the single turn as shown (see also Fig, 7~32)l 7-7 COMPLETE AXIAL-MODE PATTERNS OF MONOFILAR HELICES, By the principle of pattern multiplication, the total far-field pattern of a helix radiating in the axial mode is the product of the single-turn pattern and the array factor E. Thus, the total 0 component E+ of the distant electric field of a helix of circular cross section is the product of (7-6-4) and (7-5-18) or E+ = E+t E (1) The total 6 component E is the product of (7-6^6) and (7-5-18) or E9 = E,tE (2) As examples, the approximate E+ and E# patterns, as calculated by the above procedure, for a 12°, 7-tum uniform helix of circular cross section with Ci = 1.07 are presented in Fig 7-34 at (a) and (c)l The helix is shown at (e), with in the plane of the page and E normal to the page. The array factor is shown at (b). The single-turn patterns are as presented in Fig. 7-32.. The value of p used in these calculations is approximately that for the increased directivity condition.
The product of the single-turn patterns (Fig. 7-32) and the array factor pattern at (b) yields the total patterns at (a) and (c). The agreement with the measured pat-terns shown at (d) and (/) is satisfactory.
Comparing the patterns of Figs. 7-32 and 7-34, it is to be noted that the array ifector is much sharper than the single-turn patterns. Thus, the total and En patterns (u) and (c) (Fig 7-34) are nearly the same, in spite of the difference in the single-turn patterns. Furthermore, the main lobes of the E+ and E$ patterns are very similar to the array factor pattern. For long helices (say, > 1) it is, therefore, apparent that a calculation of only the array factor suffices for an approximate pattern of any field component of the helix. Ordinarily the single-turn pattern need not be calculated except for short helices.
The far-fiHd patterns of a helix radiating in the axial mode can, thus, be calculated to a good approximation from a knowledge of the dimensions of the helix and the wavelength. The value of the relative phase velocity used in the calculations may be computed for the increased-directivity condition from the helix dimensions and number of turns.
The effect of the ground plane on the axial-mode patterns is small if there are at least a few turns, since the returning wave on the helix and also the back lobe of the outgoing wave are both small. Hence, the effect of the ground plane may be neglected unless the helix is very short (nSA < £).
7-S AXIAL RATIO AND CONDITIONS FOR CIRCULAR POLARIZATION 301 Flgpre 7-34 Comparison of complete calculated patterns (product of single-turn pattern and array factor) with measured patterns for a ]T\ 7-turn helix with C = 1.07 radiating in the axial mode.
Agreement is satisfactory.
The approximate pattern of an axial-mode helix can be calculated very simply, while including the approximate effect of the single-turn pattern, by assuming that the single-turn pattern is given by cos + Then the normalized total radiation pattern is expressed by / , 90&\ sir = sin — I —r V n / si sin (n^/2) sin(^/2) ' where n = number of turns and $ ~ 360°[S,i(l - cos i>) + (l/2n)] (4) The value of $ in (4) is for the increased-directivity condition and is obtained by substituting (7-5-12) in (7-5-3) and simplifying. The first factor in (3) is a normalizing factor, i.e., makes the maximum value of £ unity.
7-8 AXIAL RATIO AND CONDITIONS FOR CIRCULAR POLARIZATION OF MONOFILAR AXIAUMODE HELICAL ANTENNAS.
1 In this section the axial ratio in the direction of the helix axis 1 For a general discussion of elliptical and circular polarization see Secs. 2-34 through 2-37; also see i. D. Kraus, Radio Astronomy „ 2nd ed.» Cygnus-Quasar, 1986. chap. 4.
302 1 THE HELICAL ANTENNA Figure 705 Field components as viewed from the helix axis.
will be determined, and also the conditions necessary for circular polarization in this direction will be analyzed.
Consider the helix shown in Fig.
7-35, Let us calculate the electric field components and E6t as shown, at a large distance from the helix in the z direction. The helix is assumed to have a single uniform traveling wave as indi-cated. The relative phase velocity is p , The diameter of the helix is D and the spacing between turns is S. Unrolling the helix in the xz plane, the relations are as shown in Fig, 7-36, The helix as viewed from a point on the z axis is as indicated in Fig. 7-37. The angle Z is measured from the xz plane. The coordi-nates of a point Q on the helix can be specified as r, z The point Q is at a distance / from the terminal point T as measured along the helix. From the geometry of Figs. 7-36 and 7-37, we can write h — l sin a zp — h = zp — l sin x t 5 r£ a = arctan — = arccos — nD l rf = l cos a where z p is the distance from the origin to the distant point P on the z axis.
At the point P the component E+ of the electric field for a helix of an integral number of turns n is , f 2".
, L / z, Uma E# — Eq sin Z exp t -H Jo L \ c c pc/J where E0 is a constant involving the current magnitude on the helix.
(2) 7 8 AXIAL RATIO AND CONDITIONS FOR CIRCULAR POLARIZATION 303 From (1) the last two terms of the exponent in (2) may be rewritten. Thus, / sin a l rZ { 1 \ rffl — = — tan fx — = —“ \ c pc c \ p cos x/ c q = tan at — p cos x When a = 0, the helix becomes a loop and q = - i/p.
The relation being obtained is, thus, a general one, applying not only to helices but also to loops as a special case. Equation (2) now reduces to E^ = £ 0 c /r riitn -fisp } j sjn £ giks where quantities independent of Z have been taken outside the integral and where k = Prq = Lx ^sin 0 - ^ 304 7 THE HELICAL ANTENNA On integration (5) becomes £-zrrr (7, where E v — £0 In a similar way we have for the 9 component Ee of the electric field at the point f\ £# = E° I cos exp [ ;w ( r + d < (8) Making the same substitutions as in (2), we obtain from (8) e. -- n (91 The condition for circular polarization in the direction of the z axis is , .
Es = ±J dO) The ratio of (7) to (9) gives E t _ _L _ _ L E« jk k (ID Accordingly, for circular polarization in the axial direction of a helix of an inte-gral number of turns, k must equal + l t Equation (11) indicates that and Ee are in time-phase quadrature. There-tore, the axial ratio AR is given by the magnitude of (11) or The axia! ratio will be restricted here to values between unity and infinity. Hence, u t L z) is less than unity, its reciprocal is taken-Substituting the value of k from (6) into (12) yields IL^sin a - (I/p)]| (t3) or AR = L^sin a ~~j (14) Either (1 3) or (14) is used so that 1 < AR < ao.
From (13) and (14), it appears that the axial ratio can be calculated from the turn length L x and pitch angle tx of the helix, and the relative phase velocity p, If we introduce the value of p for the condition of in-phase fields (see Table 7-1), it is found that AR = I.
In other words, the in-phase field condition is also the condition for circular polarization in the axial direction.
7-8 AXIAL RATIO AND CONDITIONS FOR CIRCULAR POLARIZATION 305 This may also be shown by noting that (11) satisfies the condition for circu-lar polarization when k - — l t or LA sin a — -= — 1 Solving (15) for p, we obtain S,+ 1 06) which is identical with the relation for in-phase fields (ordinary end-fire condition).
Our previous discussion on phase velocity indicated that p followed more closely the relation for increased directivity than the relation for in-phase fields.
Thus, introducing p in (14) for the condition of increased directivity, we obtain AR (on axis) = 2n + 1 2n 07) where n is the number of turns of the helix. If n is large the axial ratio approaches unity and the polarization is nearly circular, 1 When I first derived (17) in 1947, it came as a pleasant surprise that the axial ratio could be given by such a simple expression.
As an example, let us consider the axial ratio in the direction of the helix axis for a 13°, 7-turn helix. The axial ratio is unity if the relative velocity for the condition of in-phase fields is used. By (17) the axial ratio for the condition of increased directivity is 15/14 = 1,07. This axial ratio is independent of the fre-quency or circumference Cx as shown by the dashed line in Fig. 7-38.
In this figure, the axial ratio is presented as a function of the helix circumference Cx in free-space wavelengths.
If the axial ratio is calculated from (13) or (14), using the measured value of p shown in Fig, 7-28, an axial ratio variation is obtained as indicated by the solid curve in Fig- 7-38. This type of axial ratio versus circumference curve is typical of ones measured on helical beam antennas. Usually, however, the measured axial ratio increases more sharply as CA decreases to values less than about This difference results from the fact that the calculation of axial ratio by (13) or (14) neglects the effect or the back wave on the helix. This is usually small when the helix is radiating in the axial mode but at lower frequencies or smaller circum-ferences (C^ < £) the back wave is important. The back wave on the helix pro-duces a wave reflected from the ground plane having the opposite direction of 1 With circularly polarized feed an AR = 1 can be obtained on the axis for a helix of any length according to R. G. Vaughan and J. B. Andersen (“Polarization Properties of the Axial Mode Helix Antenna." IEEE Trans. Ants. Prop. y AP-33, 10-20, January 1985). They also deduce the axial ratio as a function of the off-axis angle.
306 7 THE HELICAL ANTENNA Helix circumference, C Figure 7-38 Axial ratio as a function of helix circumference Cx for a 13^, 7-turn monofilar axial-mode helical antenna. The dashed curve is from (17). {After Kraus.) field rotation to that produced by the outgoing traveling wave on the helix. This causes the axial ratio to increase more rapidly than indicated in Fig. 7-38.
The foregoing discussion applies to helices of an integral number of turns Let us now consider a long helix where the number of turns may assume non-integral values Hence, the length of the helical conductor will be specified as ^ instead of 2?^. It is further assumed that k is nearly unity Thus, (5) becomes £ = :| i (18) J0 Since k ^ — 1, + I ^ 0T and it follows that t+lKl ^ 1 +j{k + l)£ x {19) Now integrating (18) and introducing the condition that k is nearly equal to -1 and the approximation of (19), we have Similarly the 0 component E ff of the electric field is (20) (21 ) When the helix is very long 7-9 WIDEBAND CHARACTERISTICS OF MONOFTLAR HELICAL ANTENNAS 307 and (20) and (21) reduce to E = -J and E» = + Taking the ratio of E to which fulfills the condition for circular polarization.
Still another condition resulting in circular polarization is obtained when {k ± 1)^! = 2nm f where m is an integer. This condition is satisfied when either the positive or negative sign in k + 1 is chosen but not for both.
To summarize, the important conditions for circular polarization are as follows: L The radiation in the axial direction from a helical antenna of any pitch angle and of an integral number of 1 or more turns will be circularly polarized if k = — 1 (in-phase fields or ordinary end-fire condition).
2. The radiation in the axial direction from a helical antenna of any pitch angle and a large number of turns, which are not necessarily an integral number, is nearly circularly polarized if k is nearly — 1.
7-9 WIDEBAND CHARACTERISTICS OF MONOFILAR HELICAL ANTENNAS RADIATING IN THE AXIAL MODE, The helical beam antenna 1 has inherent broadband properties, possessing desirable pattern, impedance and polarization characteristics over a relatively wide fre-quency range. The natural adjustment of the phase velocity so that the fields from each turn add nearly in phase in the axial direction accounts for the persist-ence of the axial mode of radiation over a nearly 2 to 1 range in frequency. If the phase velocity were constant as a function of frequency, the axial-mode patterns would be obtained only over a narrow frequency range The terminal impedance is relatively constant over the same frequency range because of the large attenu-ation of the wave reflected from the open end of the helix. The polarization is nearly circular over the same range in frequency because the condition of fields in phase is also the condition for circular polarization As shown in Fig, 7-39o, the dimensions of a helix in free-space wavelengths move along a constant pitch-angle line as a function of frequency.
If is the lower frequency limit of the axial mode of radiation and F2 the upper frequency limit of this mode, then the range in dimensions for a 10° helix would be as suggested by the heavy line on the diameter-spacing chart of Fig. 7-39a The center frequency F0 is arbitrarily defined as F0 = (Fj + F 2)/2 Or monofilar axial-mode helical antenna.
The properties of a helical beam antenna are a function of the pitch angle.
The angle resulting in a maximum frequency range F 2 — of the axial mode of radiation is said to be an “optimum" 4 pitch angle. To determine an optimum angle, the pattern, impedance and polarization characteristics of helical antennas may be compared on a diameter-spacing chart as in Fig. 7-3%. The three con-tours indicate the region of satisfactory pattern, impedance and polarization values as determined by measurements on helices of various pitch angle as a function of frequency The axial length of the helices tested is about 1.6/ at the center frequency. The pattern contour in Fig. 7-3% indicates the approximate region of satisfactory patterns. A satisfactory pattern is considered to be one with a major lobe in the axial direction and with relatively small minor lobes. Inside the pattern contour, the patterns are of this form and have half-power beam 4 , 3 !
7-11 RADIATION FROM LINEAR PERIODIC STRUCTURES 309 widths of less than 60° and as small as 30.
Inside the impedance contour in Fig. 7-3% the terminal impedance is relatively constant and is nearly a pure resistance of 100 to 150 £1 Inside the axial ratio contour, the axial ratio in the direction of the helix axis is less than 125 Note that all contours lie below the line for which D 2 = y/2Sx 4-1 fn. This line may be regarded as an upper limit for the beam mode.
It is apparent that the frequency range F2 ~ Fj is small if the pitch angle is either too small or too large. A pitch angle of about 12 or 14° would appear to be “optimum” for helices about 16A long at the center fre-quency Since the properties of the helix change slowly in the vicinity of the optimum angle, there is nothing critical about this value. The contours are arbi-trary but are suitable for a general-purpose beam antenna of moderate direc-tivity, The exact values of the frequency limits F t and F2 , are also arbitrary but are relatively well defined by the close bunching of the contours near the fre-quency limits.
Based on the above conclusions, f constructed a 14°, 6-turn helix and mea-sured its properties. The helix has a diameter of 0312 at the center frequency (400 MHz), The diameter of the conductor is about 0,022, Conductor diameters of 0.005 to 0,052 can be used with little difference in the properties of this helix in the frequency range of the beam mode.
1 The measured patterns between 275 and 560 MHz are presented in Fig 7-19.
It is apparent that the patterns are satisfactory over a frequency range from 300 MHz (C = 0.73) to 500 MHz {C2 = 1.22).
A summary of the characteristics of this antenna are given in Fig. 7-23 in which the half-power beam width, axial ratio and standing-wave ratio are shown as a function of the helix circumference.
h 7-10 TABLE OF PATTERN, BEAM WIDTH, GAIN, IMPEDANCE AND AXIAL RATIO FORMULAS, Expressions devel-oped in the preceding sections for calculating the pattern, beam width directivity, terminal resistance and axial ratio for axial-mode helical antennas are sum-marized in Table 7-2. These relations apply to helices for 12° < a < 15, ^ < f and n > 3 or to the more specific restrictions listed in the footnote to the table.
7-11 RADIATION FROM LINEAR PERIODIC STRUCTURES WITH TRAVELING WAVES WITH PARTICULAR REFERENCE TO THE HELIX AS A PERIODIC STRUCTURE ANTENNA Radiation from continuous linear antennas carrying a traveling wave was dis-cussed in Sec 5-8. Although the helical beam antenna consists of a continuous 1 Design data for a 12.5° helix are given by J. D. Kraus, “ Helical Beam Antenna Design Techniques,” Communications, 29, September 1949.
310 1 THE HELICAL ANTENNA Table 7-2 Formulas for monofilar axial-mode helical antennas / 90 : \ sin [rtif/-£ = f sin — — -\ n j sin (ip/.
re ^ --6oj\s.i.
-^ ;2) , cos <p m where $ - 360 c 5^(1 - cos ) + — it Beam width (half-power) HPBW = — - -See restrictions Beam width (first nulls) See restrictions Directivity (or gaini) See restrictions Terminal resistance See restrictions Axial ratio (on axis) Axial ratio (on axis) BWFN = tW«SA D = 12Cl n.S.
R = 140Ci fl (axial feed) R = JSO/y'G A (peripheral feed) AR = —+ - (increased directivity) 2n AR — 1 f./ sin Gt di (p unrestricted) n = number of turns of helix Cx = circumference in free-space wavelengths Si = spacing between turns in free-space wavelengths Lj = tmn length in tree-space wavelengths a = pilch angle p = rdalive phase velocity ip = angle with resided to helix axis Restriction for beam width and directivity; 0.8 < C2 < 1.15; 12 < a < 14 T n > y Restriction for terminal resistance: 0.8 < C x < 1-2; 12" < a c 14' ; n ^ 4.
t Assuming no losses.
conductor carrying a traveling wave, it is also a periodic structure with period equal to the turn spacing as considered in Sec. 7-5-Now let us develop the periodic structure approach in a more general way which illustrates the relation of helical antennas to other periodic- structure (dipole) antennas.
1 A linear array of n isotropic point sources of equal amplitude and spacing is shown in Fig. 7-40 representing a linear periodic structure carrying a traveling wave. As discussed previously the phase difference of the fields from adjacent sources as observed at a distance point is given by if/ = ^ S cos 4> — 3 (0 1 Although the helix is a periodic structure, it is continuous. The dipole arrays are discontinuous.
7-11 RADIATION FROM LINEAR PERIODIC STRUCTURES 311 Figure 7-40 Linear array of it isotropic point sources of equal amplitude and spacing, 5, representing a linear periodic structure carrying a traveling wave.
where S — spacing between sources, m a 0 = free-space wavelength, m 0 = angle between array axis and direction of distant point, rad or deg S — phase difference of source 2 with respect to source 1, 3 with respect to 2, etc., rad or deg We assume that the array is fed by a wave traveling along it from left to right via a guiding structure which may, for example, be an open-wire transmis-sion line, a waveguide or a helix.
1 The phase constant of the traveling wave is given by 0 = (rad m 1 ) = (deg ra" ) (2) /0 p AoP • where k0 = free-space wavelength, m p = v/c = relative phase velocity, dimensionless u — wave velocity, ms 1 c = velocity of light, ms -1 The phase difference between sources is given by 3 = S (rad) = “— S (deg) (3) / 0 p a0 P where 5 — spacing between sources, m In general, for the fields from the n sources to be in-phase at a distant point requires that ^ = 2nm (rad or deg) (4) where m = mode number = 0, ±1, ±2, etc.
1 We assume a uniform traveling wave. Although such a wave Is approximated with a monofilar axial-mode helix, it is not necessarily realized with the dipole arrays presented in this section without the addition of suitable impedance matching networks (not shown). See Sec. 11-10 on Phased Arrays and Sec. 16-21 on Leaky Wave Antennas.
312 7 THE HELICAL ANTENNA Introducing (3) and (4) into (1) yields „ 2n _ , 2/r 2 Jim = — o cos <p — -— j A 0 2?rm -/?0 S cos 0 — 0S where = y- = phase constant of free-space wave, rad m For mode number m = 0, the phase difference of the fields from adjacent sources at a distant point is zero; for m = 1, the phase difference is 2n; for m = 2, the phase difference is 4ti; etc, 2jt iff = = phase constant of guided wave, rad m' 1 4oP 0O 5 = electrical distance between sources for a free-space wave, rad /?5 — electrical distance between sources for the guided wave, rad /?0 S cos 0 = electrical distance between sources for a free-space wave in direction of distant point From (6) we have ^ rt 2nm h cos 0 = iff +— fi 2nm 1 m cos0= _ + __ = - +_ Let us now consider several examples.
Example 1 Mode number m — (X For different relative phase velocities p, the beam angles 0, as given by (3), are as tabulated: Relative phase velocity, p COS 0 0 Beam direction 1 (p = c) l o c End-fire 0 90" Broadside -1 (p = c with wave right-to-left) -1 180° Back-fire <1 >1 Imaginary No beam For the last entry (p < Ik 0 is imaginary. This implies that all of the wave energy is bound to the array (guided along it) and that there is no radiation (no beam).
Summary. For p values from +1 to + oo and — oo to — 1, the beam swings from end-fire (0 = 0°) through broadside (0 = 9(F) to back-fire (0 = 180°). For p values between — 1 and +1(— 1 < p < 1)0 is imaginary (no radiation). We note that these results are independent of the spacing S.
7-11 RADIATION FROM LINEAR PERIODIC STRUCTURES 313 Example 2 Mode umber m - — t. Relative phase velocity p -1 (v = c).
Fields from adjacent sources have 2n ( = 36(F) phase difference at a distant point in the direction of the beam maximum. For different spacings S f the beam angles 0, as given by (8), are as tabulated: Spacing S COS 0 0 Beam direction 0 Broadsidet Aq/2 -1 180° Back-fire} t For Ibis case, there are also equal beams end-lire (0 =<H and back-fire {0^ 180% X For ibis care, there is an equal end-fire Lobe (0«O n X Summary For spacings between XJ2 and a0 the beam swings between 90 and ISO.
Larger spacings are required to swing the beam to angles less than 90° but other lobes also appear.
Example 3 Mode number m = — L Relative phase velocity p = j (slow wave). For different spacings S t the beam angles 0, as given by (8), are as tabulated: h Sp«c5 cos 0 0 Beam direction I0 1 0 a , 90 D , ISO" End-fire, broadside and backfire J.q/2 0 90° Broadside V3 -1 180s Back-fire Summary. For spacings between XJ3 and Aq the beam swings from back-fire (18(F) through broadside (90°) to end-fire ((F). but for S = Aq broadside and back-fire lobes also appear. For spacings greater than or less than 0 is imaginary (no beam).
Example 4 Mode Bomber m = -1. Relathe phase velocity p = $ (slow wave). For different spacings S, the beam angles 0, as given by (8), are as tabulated : Spacing S cos 0 0 dfeectloa XJ4 1 0° End-fire 0 90° Broadside ktjb IKF Back-fire 314 7 THF HELICAL ANTENNA Beam anflle r </> Figure 7-41 Relation of spacing and beam angle for linear arrays with trav-eling waves of relative phase velocities p -l n i and \ (tramples 2, 3 and 4) all with mode number m - -1.
Summary. For spacing between 4/4 and 4/6 the beam swings from end -fire, through broadside to back-fire For spacing less than 4/6 or more than /. 0/4, 4> » imaginary lno beam).
The results of Examples 2, 3 and 4 are shown graphically in Fig. 7-41.
Another way of analyzing an array, and periodic structures in general, is to plot the electrical spacing (i0 S of the free-space wave (as ordinate) versus the electrical spacing /fS of the guided wave traveling along the array (as abscissa).
Dividing both coordinates by 2n, we obtain S/4 as ordinate and S/(p/. 0) - M as abscissa (where / = p4 = wavelength on the array). This type of S-S diagram is presented in Fig. 7-42, illustrating the three arrays of Examples 2, 3 and 4 (shown also in the S-(j) diagram of Fig, 7-41), For a relative phase velocity p = 1, S/2 = S/4, and the array operates along the p = 1 line {at 45" to the axes). Back-fire occurs at S// 0 -2 and broad-side at S/A0 = 1 (Example 2).
, r For a relative phase velocity p = i the array operates along the p - t ll"e with back-fire at S/X Q = i broadside at S/A 0 - i and end-fire at S/X 0 = 1 at right edge of diagram (Example 3).
' Some authors use k for P0 and refer lo the graph v a fc diagram, also called a Brilloum diagram after Leon Brillouiri, Wave Propagation in Periodic Structures, McGraw-Hill, 1946.
316 7 the helical antenna For a relative phase velocity p = L the array operates along the p = line with back-fire at S//0 = i broadside at S/k0 = i and end-fire at S/^, 4 (Example 4). For a higher mode number m = -2 the array again produce beams swinging from back-fire at S/A0 = i through broadside at S/,.0 - f to end-fire at S//. 0 = i (off diagram at right).
.
..
irf.
Let us consider several more examples of traveling-wave penodi c- structure arrays. On an S-S diagram each type of array occupies a unique location or niche which is characteristic of the antenna’s behavior. Differences between arrays are clearly evident from their locations on the diagram.
Example 5 Scanning array of dipoles with S -»« center frequency (Fig. 7-43).
Beam scanning is by shifting frequency. Physical element sparing is constant. Array is fed from the left end with a 2-wire transmission line ip = 1). The mode number m = -1. so (8) becomes 11,1 COS 0 = 7T7^“ -V - vt\ p O; t-K At the center frequency S = a, and from (9) cw - 0 and - > broadside). Halving the frequency makes S = a0/2 and rj> - 180 (beam back-fire .
Doubling the frequency makes S = 2a„ and - bff (only 30' beyond broadside) To swing the beam further toward end-fire requires a farther increase m frequency The position of this scanning array is shown in Fig. 7-42 (line labeled Examples 2 and 5) It is to be noted that there are other lobes present not given by (9).
Example 6 Seaming array of alternately reversed dipoles with S = a„/2 at ceuter frequency (Fig.
7-44). Beam scanning is by shifting frequency.
Physical element spacing is constant. Array is fed from the left end with a 2-wire transmission line (p ^ i y The mode number m = — so (8) becomes At the center frequency S = ^2 and from (10) a = 0 and = W (beam broadside). Halving the frequency makes S = V4 and 4> - 180 (beam back-fit Doubling the frequency makes 5 = swinging the beam to - 60 . The position of this scanning array is shown in Fig 7-42, Example 7 Scanning array of dipoles with S = 20/2 at center frequency with slow wave (p = 1) (Fig.
7-45). Beam scanning is by shifting frequency.
Physical element Dipoles 2 Example 5 3 4-r .. n Finn 7-43 Scanning array of dipoles with S = at center frequency, relative phase velocity p!T ( r = c) and mode number m-—i. The beam angle # (with respect to dipole t) outward from the page. (Example 5.) 7-n RADIATION FROM LINEAR PERIODIC STRUCTURES 317 Dipoles Figure 7-44 Scanning array of alternately reversed dipoles with S = at center fre-quency, relative phase velocity p = 1 and mode number m — The beam angle 4> (with respect to dipole 1) is outward from the page. (Example 6.) spacing is constant. Array is fed from the left end by a 2-wire transmission line with line length 2S between dipoles so that p - f The mode number m = - 1, so (8) becomes cos 0 = 2 ——-(n) oMo At the center frequency S = i0/2 and from (11) ^ = 90° {beam broadside). Reducing the frequency so S = A0/3 makes qh = ISO 1 (beam back-fire). Doubling the frequency makes S = a# and = (F (beam end-fire). The position of this scanning array is shown in Fig. 7-42 (line labeled Examples 3 and 7).
1 2 3 4 .
.
. n Example 6 Example 8 Scanning army of alternately reversed dipoles with S = V4 ce,rr frequency with slow wave (p = }> (Fig. 7-46). Beam scanning is by shifting frequency.
Physical element spacing is constant- Array is fed from the left end by a 2-wire transmission line with line length 2S between dipoles so that p = i The mode number m = — so (8) becomes At the center frequency S = X0/4 and from (12) 0 = 90 u (beam broadside). Decreas-ing the frequency to £ makes S = a0/6 and 0 = 180° (beam back-fire). Doubling the frequency makes 5 -'kJ2 and 0 = 0" (beam end-fire). The position of this scanning array is also shown in Fig. 7-42.
Dipoles Example 7 Figure 7-45 Scanning array of dipoles with S = i^/2 at center frequency, relative phase veloc-ily p = { (stow wave) and mode number is= 1 . (Example 7.) 318 7 THE HELICAL ANTENNA Dipoles 2 3 4 Example B Figure 7-46 Scanning array of alternately reversed dipoles with S = ;.0/4 at center frequency, relative phase velocity p = ± (slow waveUnd mode number m = -^.(Example 8 } Now lei us consider the monofilar helical antenna in comparison to the above examples of other periodic structure arrays with traveling waves.
Example 9 Monofilar axial-mode helical antenna with turn-spacing S - V and circumference C = ;.0 at center frequency (Fig. 7-47). As discussed in Sec.
-; monofilar ami-mode helical antenna operates in the mcreascd-directivity condition resulting in a supergam end-fire beam ( - 0 3 ). The mode number m- -1 and from (5) we have for (j> = 0 r that n 2n In S — 2n — - = —- S — n f-c p where n - number of turns .
p = (n/c) sin a - relative phase velocity of wave in direction of helix axis (not along the helical conductor as in Sec. 7-5) ot - pitch angle Rearranging ( 1 3) we obtain For large n.
2n + 1 1 2n Sjk o (14) 1 + [l/(5/A0 fl (15) Ground plane / M Coaxial S line Beam direction 6A=cn Example 9 Figure 7-47 Monofilar axial -mode helical antenna which locks on end-fire = G c ) with increased directivity over more than an octave frequency range by an automatic shift in relative phase velocity p (see Fig. 7-42).
The mode number m = — 1.
7-12 ARRAYS OF MONOFILAR AXIAL-MODE HELECAL ANTENNAS 319 Thus, end-fire (4> = 0 & ) with S = a0/4 requires that p = 0.20.
If the frequency decreases so that S - a„/5 (a 25 percent change in frequency) we have from (5) that 11 I r „ COS (b = — = 5 = U v p s/;.0 0.20 making 4> - 90 : (beam broadside). However, with frequency change the beam does not swing broadside but remains locked on end-fire {tj> = 0°) because the phase veloc-ity changes automatically by just the right amount to not only compensate for the frequency change but also to provide increased directivity and supergain. This is one of the very remarkable properties of the monofilar axial-mode helical antenna.
The in ere ased-di recti vity condition involves not only the turn spacing S but also the number of turns n. However, for large n the difference in p for the two end-fire conditions is small. Thus, for large rt we find from (15) that p values range from about 0.25 for 5 = 2^/3 through 0.20 for S = a0/4 to 0.167 for 5 = 2Q/5 y locking the beam on end-fire with supergain over a frequency range of about 2 to 1, which is in marked contrast to the beam-swinging of the scanning arrays discussed above. The position or the monofilar helical antenna over a 2 to 1 frequency range is shown in Fig. 7-42 for n = 4, 8 and 16 turns. For very large n f the position moves along the end-fire line which intersects S/a = 1 on the $// 0 = 0 axis. We note that while other arrays move along constant p lines as the frequency changes, the mono-filar axial-mode helical antenna moves along a constant beam-angle (end-fire) line, cutting across lines of constant p value .
Another remarkable property and great advantage of the monofilar helical antenna is that the input impedance is an almost constant resistance over an octave bandwidtff the resistance being easily set at any convenient value from 50 to 150 fl.
This is in contrast to the large impedance fluctuations of the above dipole arrays with change in frequency.
7-12 ARRAYS OF MONOFILAR AXIAL-MODE HELICAL ANTENNAS With arrays of monofilar axial-mode helical antennas the designer must strike a bala ice between the number and length of helices needed to achieve a desired gain. As discussed in Sec. 4-8 the choice is between more lower-gain antennas and fewer higher-gaip antennas appropriately spaced. As an illustration consider the following problem.
Example Design a circularly polarized antenna using one or more end-fire elements to produce a gain of 24 dB for operation at a given wavelength L Solution. The highest end-fire gain is obtained with the increased-directivity condi-tion which is automatic with monofilar axial-mode helical antennas. From (7-4-7) for ot = 12.7" and Cx = 1.05, the required length of a single helix is 252 L = nS = Ti7W =m requiring an 80-turn helix (Fig. 7-48n), 320 7 THE HELICAL ANTENNA Front view 1 helix (tf) 1 6 helices Figure 7-48 Single monofilar axial-mode helix with 80 turns (gJ compared with an array of four 20-tum helices (fc), an array of nine 9-tum helices (c) and sixteen 5-tum helices (<fi for worked example. All have 24 dB, gain. Note also that the product of the number of helices and number of turns for each array equals 80 (± 1). See Sec. 4-8, A more compact configuration results if four 20-turn helices are used in a broadside array. Assuming uniform aperture distribution, the effective aperture of each helix is Da 1 4ji ( 2 ) Assuming a square aperture, the side length is 2.24/ ( = s/50), With each helix placed at the center of its aperture area, the spacing between helices is 2.24/ (Fig.
7-48b).
A third configuration results if nine 9-turn helices are used in a broadside array (Fig. 7-4Sc), 7. 12 ARRAYS OF MONOFILAR AXIAL-MODE HELICAL ANTENNAS 321 Figure 7-49 Resistive (R) and reactive (Jf) components of the mutual impedance of a pair of same-handed 8-turn monofilar axial-mode helical antennas of 12° pitch angle as a function of the separation distance (center-to-oenter) in wavelengths.
Helix cir-cumference C x = 1.
Conductor diameter is 0.016/.
Self- imped ante Zu i 140 + jOft (After E.
A.
flfasi, “Theory and Application of the Radiation Mutual Coupling Factor,” MS.
thesis.
Electrical 0 Engineering Department , Ohio State University , 1952.) A fourth possible configuration results if sixteen Mum helices are used in a broadside array with a spacing of 1.12/ between helices (Fig. 7-4&f), Which of the above configurations should be used? The decision will depend on considerations of support structure an4 feed connections. The single helix has a single feed point and a small ground plane but is very long. The other configu-rations have larger ground planes but are more compact. The 4- and 16-helix con-figurations have the advantage that the helices can be fed by a symmetrical corporate structure. Also the 16-helix configuration can be operated as a phased array.
With the multiple-helix arrays the mutual impedance of adjacent helices is a consideration. Figure 7-49 shows the resistive and reactive components of the mutual impedance of a pair of same-handed 8-turn monofilar axial-mode helical antennas as a function of the separation distance in wavelengths (C = l s a = 12°) as measured by Blasi.
1 At spacings of a wavelength or more, as is typical in helix arrays, the mutual impedance is only a few percent or less of the helix self-impedance {140 n resistive). Thus, in designing the feed connections for a helix array the effect of mutual impedance can often be neglected without significant consequences. As examples, let us consider the feed systems for two helix arrays, one with 4 helices and the other with 96 helices.
7-12a Array of 4 Monofilar Axial-Mode Helical Antennas. This array, shown in Fig. 7-50a, which I constructed in 1947, has four 6-turh, 14° pitch angle helices mounted with 1.5A spacing on a 2.5 x 2.5X square ground plane. 2 The helices are fed axially through an insulated fitting in the ground plane. Each feed 1 E. A. Blast, “ Theory and Application of the Radiation Mutual Coupling Factor," M.S, thesis. Elec-trical Engineering Department, Ohio State University, 1952.
2 I. D. Kraus, “Helical Beam Antennas for Wide-Band Applications," Proc, /RE, 36, 1236-1242, October 1948-Separation, X 1 | 1 ^ J I 600 700 800 900 1000 Frequency, MHz 1 l J I I i 600 700 800 900 1 000 Frequency, MHz m Figure 7-50 \a) Conslruclional details for broadside array of Four 6-turn T 14 monofilar axiai-mode heii^Ei] antennas. Dimensions arc in wavelengths at the center frequency, (h} Measured performance of 4-helix array of tu) showing beam widths, axial ratio and VSWR on a 53-H line as a function of frequency. {After J.
D. Kraus y "Helical Bean Antennas for Wide- Band Applications," Proc, 1RF, 36.
1236—1242, October fV48_} 7-li Ihe monofilar axial-moof helix as a parasitic element 323 point is connected on the back side of the ground plane to a junction point at the center of the ground plane. Each conductor acts as a single-wire versus ground-plane transmission line with a spacing between wire and ground plane which tapers gradually so that the approximately 140 Q at each helix is transformed to 200 Q at the junction. The four 200-fi lines in parallel yield 50 O at the junction which is fed through an insulated connector to a 5G-fi coaxial fitting on the front (helix) side of the ground plane. Since the taper sections are about U long, the arrangement provides a low VSWR on the 50-0 line connected to the junction over a wide bandwidth. Beam width, axial ratio and VSWR performance of the array are presented in Fig 7-506. The array gain at the center frequency (800 MHz) is about 18.5 dB and at 1000 MHz about 21.5 dB.
7-l2b Array of 96 Monofilar Axial-Mode Helical Antennas. This array, shown in Fig, 7-4, which I designed and constructed in 1951, 1 has 96 11-turn 12. 5° pilch angle helices mounted on a tillable flat ground plane 40/ tong by 5/ wide for operation at a center frequency of 250 MHz. Each helix is fed by a 150-& coaxial cable. Equal-length cables from each helix of one bay or group of 12 helices are connected in parallel to one end of" a 2/ long tapered transition section which transforms the 12.5-fl (=150/12) resistive impedance to 50 Equal-length 50-0 coaxial cables then connect the transition section of each bay to a central location resulting in a uniform in-phase aperture distribution with low VSWR over a wide bandwidth. The array produces a gain of about 35 dB at the centerfrequency of 250 MHz (/ = 12 m) and increased gain at higher fre-quencies, At 250 MHz the beam is fan shaped with half-power beam widths of l by 8".
7-13 THE MONOFILAR AXIAL-MODE HELIX AS A PARA-SITIC ELEMENT (see Fig. 7-51) Helix-helix (Fig 7-51 a).
If the conductor of a 6-turn monofilar axial-mode helical antenna is cut at the end of the second turn, the antenna continues to operate with the first 2 turns launching the wave and the remaining 4 turns acting as a parasitic director.
Polyrod-helix (Fig. 7-516), By slipping a parasitic helix of several turns over a linearly polarized polyrod antenna, it becomes a circularly polarized antenna.
Hom-helix (Fig.
7-51c) By placing a parasitic helix of several turns in the throat of a linearly polarized pyramidal horn antenna without touching the horn walls, the horn radiation becomes circularly polarized.
1 J. D. Kraus, “The Ohio State Radio Telescope" Sky and TeL 12, 157- 159, April 1953.
324 T THE-.
HELJf.Al. ANTENNA - -=-rw ms Helix helix Figure 7-51 The monofihir axial-mode helical antenna in 7 applica-tions as a parasitic element.
Corner-helix (Fig.
1-S\d). A parasitic helix in front of a corner- reflector antenna results in a circularly polarized antenna.
The 2-wire-Jine-helix (Fig. 7-51 ef).
If a parasitic helix of many turns is slipped over a 2-wire transmission line without touching it (helix diameter slightly greater than line spacing), it is reported that the combination becomes a linearly po-larized end-fire antenna with E parallel to the plane of the 2-wirc line, 1 1 G. Broussaud and E. Spil£, Tl Fndtire Antennas/' PrtK. IRE, 49, 5 1 5-516, February 1961, 7-14 THE MONOFILAR AXIAL MODE HELICAL ANTENNA 325 Helix -helix (Fig.
7-51/).
If a parasitic helix is wound between the turns of a driven monofilar axial-mode helical antenna without touching it (diameters the same), Nakano el ai report that the combination gives an increased gain of about 1 dB without an increase in the axial length of the antenna. The increased gain occurs for helices of any number of turns between 8 and 20. The parasitic helix may be regarded as a director for the driven helix.
1 Helix lens (Fig.
A monofilar axial-mode helical antenna (or, for that matter, any end-fire antenna) acts as a lens. An array of parasitic helices of appropriate length arranged in a broadside configuration can operate as a large aperture-lens antenna (see Fig. 14-22).
7-14 THE MONOFILAR AXIAL-MODE HELICAL ANTENNA AS A PHASE AND FREQUENCY SHIFTER The monofilar axial mode helical antenna is a simple, beautiful device for changing phase or fre-quency-Thus, if the monofilar axial-mode helical antenna in Fig, 7-52 is transmitting at a frequency F , rotating the helix on its axis by 90 y will advance the phase of the radiated wave by 90° (or retard it, depending on the direction of rotation). Rotating the helix continuously / times per second results in radiation at a frequency F ±f depending on the direction of rotation (sec also Sec.
18-10).
As an application, consider the 3-helix lobe-sweeping antenna of Fig. 7-53.
All helices are of the same hand. By rotating helix 1 clockwise and helix 3 counterclockwise with helix 2 at the center stationary, a continuously swept lobe is obtained as suggested in the figure. In operation a small Jobe appears about 30° to the left, then grows in amplitude while sweeping to the right, reaching a maximum at 0® (at right angles to the array). Sweeping further to the right, the lobe decreases to a small amplitude at an angle of about 30° and simultaneously a small lobe appears at 30 u to the left and the process is repeated, giving a contin-uously sweeping lobe (left to right) which crosses the 0 C direction n times per 1 H. Nakano, T, Yamane, J. Yamauchi and Y, Samada, "+ Helical Antenna with Increased Power Gain" IEEE APS Int . Symp., 1, 4 1 7—420. 1984.
H. Nakano, Y. Samada and J. Yamauchi, “Axial Mode Helical Antennas,” IEEE Trans, Ants. Prop. AP-34, 1143-1148, September 1986.
326 7 THF' HELICAL ANTENNA 7-\6 MONOFILAR AXIAL-MODE HELICAL ANTENNAS AS FEEDS 327 30' ° 30° \ Figure 7-53 Array of 3 right-handed monofilar axial-mode helical antennas with outer 2 rotating in opposite directions to produce a continuously sweeping lobe.
minute for a helix rotation speed of n revolutions per minute. By using more helices, the beam width of the swept lobe can be made arbitrarily small.
1 designed, built and operated one of these 3 helix arrays in 1957 for observ-ing radio emissions from the planet Jupiter at frequencies of 25 to 35 MHz.
Each helix had 3 turns and was 3 m in diameter.
Another application of phase -shifting with a helix is discussed in the next section in connection with helices for linear polarization.
7-15 LINEAR POLARIZATION WITH MONOFILAR AXIAL-MODE HELICAL ANTENNAS.
If two monofilar axial-mode helical antennas are mounted side by side and fed equal power, the radiation on axis will be linearly polarized provided the helices are of opposite hand but otherwise identical (Fig. 7-54n). With switch 1 right, switch 2 left and switch 3 up, polariz-ation is linear. Rotating one of the helices on its axis 90 rotates the plane of linear polarization by 45°. Rotating one helix through 180°" rotates the plane of linear polarization 90'. With switches I and 3 left, as in the figure, the polariz-ation is LCP (left-handed circular polarization). With switches 2 and 3 right, the polarization is RCP (right-handed circular polarization).
Thus, the two helices can provide either left or right circular polarization or any plane of linear polarization.
1 J. D Kraus, “ Planeiary and Solar Radio Emission at 11 Meters Wavelength," Proc. IRE, 4. 266-274, January 1958.
LCP RCP or LP LP hand Right ?
e Left hand £ {<) (£> Figure 754 (a) Arrangement for producing left circular polarization (LCP). right circular polariz-ation (RCP) or any plane of linear polarization (LP), (b) Two helices of opposite hand in series for producing Linear polarization (LP).
Another method of obtaining linear polarization is to connect a left- and a right-handed helix in series as in Fig, 7-54b.
A third method has already been discussed in Sec. 7-13 (2-wire-line-helix).
Elliptical polarization approaching linear polarization can be obtained by flattening a helix so that its cross section is elliptical instead of circular.
7-16 MONOFILAR AXIAL-MODE HELICAL ANTENNAS AS FEEDS, Figure 7-55 shows a driven helix feeding an array of crossed dipoles acting as directors for producing circular polarization^ Although this arrange-ment has less gain and bandwidth than a full helix of the same length the crossed 328 7 THI IIKKAI ANTENNA Figure 7-55 Monofilar axial-mode helical amentia as feed for an end-fire arra> of crossed dipoles.
dipoles may be simpler to support than a tong helix. The feed connections for the helix are also simpler than for a pair of crossed Yagi-Uda antennas, which require equal power with voltages in phase quadrature.
Helices arc useful as feed elements for parabolic dish antennas. An example constructed by Johnson and Cotton 1 is shown in Fig, 7-56 in which a 3.5-turn monofilar axial-mode helix operates in the back-fire mode as a high-power unpressurized (200 kW) circularly polarized feed element for a parabolic dish reflector. Without a ground plane the helix naturally radiates in the backward axial direction. Another example of a back-fire helix feed is shown in Fig. 7-6." A monofilar back-fire helical antenna was also constructed by Patton 3 for compari-son with bifjlar back-fire helical antennas.
Short monofilar axial-mode end-fire helices of a fcw r turns with cupped ground plane are also useful as feeds for parabolic dish reflectors for producing sharp beams of circularly polarized radiation. Short conical helices (a constant, D and S increasing) are also useful because of their broad patterns for short focal-length dishes (see the helix in Fig, 7-59tf).
For dish feeds covering a frequency range greater than provided by a single helix, two or more helices can be mounted coaxially inside each other with phase centers coincident as shown in Fig, 7-57, This combination is superior to a log-pcriodic antenna as the feed since the phase center of a log-pcriodic antenna moves with frequency, resulting in defocusing of the parabolic reflector system.
Holland 4 has built a feed of this type with a larger helix for the L band and a smaller helix for the S band. The number of turns required for the helix feed antennas depends on the beam width desired. For the pattern to be 10 dB down at the edge of the parabolic dish reflector, the required number of turns is approximately given by 1 R.
C. Johnson and R.
B. Cullon.
'A Backfire Helical Feed" Georgia Inshlute of Technology, Engineering Experimental Station Report, 1982.
- H F. King and J L. Wong, “Antenna System for the FleetSatCom Satellites," IEEE International S’vwposjKm on Antennas and Propagation, pp. 349-352, 1977, W T. Patton, “The Backfire Bifilar Helical Anienna, ' Ph.D. thesis. University of Illinois, 1963.
M, Holland, “Multiple Feed Antenna Covers F. S and C-Band Segments, Microwave J ., 82-85, October 1981.
747 TAPERED AND OTHER FORMS OF AXIAL-MODE HEUCAL ANTENNAS 329 / / / \ \ \ Figive 7-56 Short back-fire monofilar axial-mode helix as high-power (200 kW) circularly polarized feed for a parabolic dish antenna, {After K, C, Johnson and R. B. Cotton, “A Backfire Helical Feed," Georgia Institute of Technology, Engineering Experimental Station Report, 1982,) where 0 = 10 dB beam width = turn spacing in wavelengths Thus, if 8^ = 0,21 (a = 12°) and the required value of — 1 15° we have from (1) that% = 3.
To reduce mutual coupling of the helices, Holland placed the peripheral feed points of the two helices on opposite sides of the axis, as suggested in Fig.
7-57, obtaining 2-port fixed-phase-center operation over a 5 to 1 bandwidth, 7-17 TAPERED AND OTHER FORMS OF AXIAL-MODE HELICAL ANTENNAS-In this section a number of variants of the uniform (constant diameter, constant pitch) monofilar axial-mode helical antenna are dis~ cussed. Some of these forms are shown in Figs. 7-58, 7-59 and 7-60 which are reproduced here without changes from the first edition of Antennas (1950). In .L-band helix Figure 7-57 Coaxially-mounted peripherally -fed monofilar axial-mode helical antennas of same hand as parabolic dish feeds with same stationary phase centers for covering a 5 to 1 fre-quency range.
( After J , Holland, ' Multiple Feed Antenna Covers L, S and C-Bamf Segments " Microwave J., S2-#5, October I9SE) S-band helix 330 7 THE HELICAL ANTENNA Thin die tact ric (d) \ <>?) (i) Figure 7-58 Axial-mode helices showing various constructional and feed arrangements.
{After Kraus.) Fig. 7-58 we recognize in (a) a uniform helix with ground plane and in {c) a uniform helix without ground plane (the same configuration as in Fig. 7-56 for a back-fire feed). The double winding in (i) is similar to the one of Nakano ef al.
(Fig. 7-51/) except that both windings in (i) are driven while in Nakano et al.
y one is parasitic.
Figure 7-59 shows 9 forms of tapered monofilar axial-mode helical antennas grouped into 3 classes: (1) pitch angle a constant but turn spacing S and diameter D variable, (2) diameter D constant but pitch angle a and turn spacing S Increasing Decreasing Envelope constant (C> constant ^RTT ^ constant ^ <• « Figure 7-59 Types of tapered monofilar axial-mode helical antennas, {After Kraus.) 7-17 TAPERED AND OTHER FORMS OF AXIAL-MODE HELICAL ANTENNAS 331 Figure 7-60 Types of tapered monofilar axial-mode helical antennas including ones in which con-ductor size is tapered, {After Kraus.) variable and (3) turn spacing S constant but pitch angle a and diameter D vari-able. Many of these forms have been investigated—the class (2) form by Day.
1 Day measured patterns of monofilar axial-mode helical antennas of 6 turns with the diameter constant but cc increasing or decreasing (D constant but a and S variable as in Fig. 7-59d, e and / or the middle row of Fig. 7-59), The helix conductor diameter was 0.O2A Pitch angles were varied on a given helix from 1 to 20°, ito 17° or 9 to 15°, both increasing and decreasing. These were compared with a constant pitch angle of 12.5° at helix circumferences Ck of 0.6, 0.8, 1.0, 1.2 and 1.4—a total of 35 cases. For pitch angle tapers between 5 and 17° and 0.8 < Ck <; 1.2, the pattern variations are minor. However, at Ck = 1.2 and with the pitch angle decreasing from 17° at the feed end to 5° at the open end, the gain is 1 dB more than for Ck = 1.2 and oc = 12.5° (constant pitch). This is a significant improvement since the highest gain for a uniform 6-turn helix occurs when ot is approximately 12.5° and the circumference Ck approximately 1.2.
Thus, the center helix (D constant, ot and S decreasing), of the 9 shown in Fig. 7-59 (i.e., Fig.
7-59e), is a useful variant of the uniform helix.
The conical helix in Fig. 7-59a for which a is constant and D (or C) and S are . increasing has been investigated by Chatterjee^ Nakano, Mikawa and Yamauchi 3 and others. With small pitch angles Chatteijee found that very broad 1 P.
C. Day “Some Characteristics of Tapered Helical Beam Antennas," M.S.
thesis. Ohio Stale University, 1950.
J J.
S. Chauerjce, “Radiation Field of a Conical Helix,” J.
Appi. Phys. t 24, 550-559, May 1953; 26, 331-335. March 1955.
3 H. Nakano, T, Mikawa and J. Yamauchi, “Numerical Analysis of Monofilar Conical Helix," IEEE AP-S /m. Symp. t 1 , 1 77- 1 SO. 1984.
332 7 THE HELICAL ANTENNA patterns can be obtained over a 5 to 1 bandwidth.
According to Nakano, Mikama and Yamauchi, the currents involved are those of the attenuating wave-launching region close to the feed point (see Fig. 7-3fr and c).
Additional tapered types are shown in Fig, 7-60, The one at (a) has a tapered and uniform section but reversed in order from the uniform-to-taper type of Wong and King 1 and others. The other designs shown in Fig. 7-60 involve variation in the diameter of the helical conductor d or the width w of a flat strip conductor. Thus, there are 4 quantities which can be varied, ot, D, S and d (or w).
Since the characteristics of the monofllar axial-mode helical antenna are rela-tively insensitive to moderate changes in dimensions, the effect of moderate departures from uniformity is, in general, not large. However, some changes may produce significant increases in gain, as discussed above, and significant decreases in axial ratio and VSWR (see Fig. 7-24), See also Sec. 15-4.
7-18 MULT1FILAR AXIAL-MODE (K1LGUS COIL AND PATTON COIL) HELICAL ANTENNAS. Four wires, each Xj2 long and forming ^-turn of a helix as in Fig, 7-61, produce a cardiod-shaped back-fire circularly polarized pattern (HPBW ^ 120°) when the two pairs are fed in phase quadrature. This Kilgus coil may be described as two ^-turn bifilar helices or one ^-tum quadrifilar helix, 2 The antenna is resonant and the bandwidth is narrow (about 4 percent). The 4 wires can also be A/4 or A long, ror these lengths the lower ends are open-circuited instead of short-circuited as for the A/2 wires of Fig. 7-61. Each bifilar helix can be balun-fed at the top from a coaxial line extending to the top along the central axis. By increasing the number of turns Kilgus reports that shaped-conical patterns can be obtained which may be more useful for some applications than a cardiod (heart-shaped) pattern, 3 A bifilar helix end-fed by a balanced 2-wire transmission line produces a back-fire beam when operated above the cutoff frequency of the principal mode of the helical waveguide. The maximum directivity of this Patton coil occurs slightly above the cutoff frequency.4 The pattern broadens with increasing fre-quency and at pitch angles of about 45° the back-fire beam splits and scans toward side-fire.
Below the cutoff frequency, there is a standing-wave current distribution along the helical conductor. Above cutoff, the standing wave gives way to a grad-ually decaying traveling wave. With a further increase in frequency the rate of decay increases and a low-level standing wave appears, indicating the existence of 1 J. L. Wong and H. E. King, Broadband Quasi-Taper Helical Antennas" IEEE Traits. Ants. Prop, AP-17, 72-78, January 1979.
2 C. C. Kilgus, Resonant Quadrifilar Helix," IEEE Trans, jtnfs. Prop„ AP-17, 349—351, May 1969.
C. C. Kilgus, " Shaped-Conical Radiation Pattern Performance of the Backfire Quadrifilar Helix," IEEE Trans. Ants. Prop., AP-23, 392-397, May 1975.
4 W, T, Patton, "The Backfire Helical Antenna," Ph.D. thesis. University of Illinois, 1963.
7' 1 9 MONOFILAR AND MULTIFILAR NORMAL-MODE HELICAL ANTENNAS 333 I Pattern maximum Figure 7-61 Resonant narrowband back-lire quadrifilar Kilgus coil for very broad circularly polarized pattern.
{After C.
C.
Kilgus, “Resonant QuadrifUar Helix f IEEE Trans. Ants. Prop., AP-17, 349-351, May 1969.) Wires are situated in space as though wrapped around a cylinder as suggested in the figure.
a higher-order helical waveguide mode. This establishes the upper frequency limit of the bifilar helix back-fire radiation, Quadrifilar and octofilar axial-mode forward end-fire circularly polarized L helical antarmas using large pitch angles (30 to 60°) have been investigated by Gerst and Worden 1 and Adams el al.
1 7-19 MONOFILAR AND MULTIFILAR NORMAL-MODE HELICAL ANTENNAS. THE WHEELER COIL- The previous sec tions deal with axial-mode helical antennas with maximum radiation in the direc-tion of the helix axis. The radiation may be (forward) end-fire or back-fire. In this section the normal mode of radiation is discussed, normal being used in the sense of perpendicular to or at right angles to the helix axis. This radiation with its maximum normal to the helix axis may also be described as side-fire or broadside , When the helix circumference is approximately a wavelength the axial mode of radiation is dominant, but when the circumference is much smaller the normal mode is dominant. Figure 7-62a and c shows helices radiating in both modes while Fig, 7-62b shows a 4-Iobed mode helix (Chireix coil) with the relative sizes for producing the modes being indicated.
1 C. W. Gerst and R, A, Worden, "Hdix Antennas Take a Turn for the Better." Electronics, 100-1 10, Aug. 22. 1966 2 A.
A, Adams, R. K. Gr«nough h R.
F. Wallenberg, A. Mendelovicz and C. Lumjiak, l'The Quadri-filar Helix Antenna IEEE Trans . Ants. Prop., A P-22, 173— 17E h March 1974.
334 7 THE-: helical antenna (a) (b) ^ 4 lobtid mode Figure 7-62 Field patterns of axial, 4-iobed and normal radiation modes of helical antennas with relative size indicated.
Now let us examine the requirements for normal-mode radiation in more detail. Consider a helix oriented with axis coincident with the polar or z axis as in Fig. 7-63a. // the dimensions ate small (nL <§ /), the maximum radiation is in the xy plane for a helix oriented as in Fig. 7~63a, with zero held in the z direction.
When the pitch angle is zero, the helix becomes a loop as in Fig. 7-63h.
When the pitch angle is 90°, the helix straightens out into a linear antenna as in Fig. 7-63c, the loop and straight antenna being limiting cases of the helix.
7.19 MONOFILAR AND MULTlFILAR NORMAL-MODE HELICAL ANTENNAS 335 d Figure 7-64 Modified helix for normal-mode calculations.
The far field of the helix may be described by two components of the elec-tric field, Eq and E as shown in Fig. 7 -63a. Let us now develop expressions for the far-held patterns of these components for a small short helix. The develop-ment is facilitated by assuming that the helix consists of a number of small loops and short dipoles connected in series as in Fig. 7-64a. The diameter D of the loops is the sameas the helix diameter, and the length of the dipoles S is the same as the spacing between turns of the helix. Provided-that the helix is small, the modified form of Fig. 7-64a is equivalent to the true helix of Fig. 7-63a. The current is assumed to be uniform in magnitude and in phase over the entire length of the helix. Since the helix is small, the far-field pattern is independent of the number of turns. Hence, it suffices to calculate the far-field patterns of a single small loop and one short dipole as illustrated in Fig. 7-64fr.
The far field of the small loop has only an component Its value is given in Table £-1 as r 120jt2[/] sin 0 A where the area of the loop A = nD 2 !
4 The far field of the short dipole has only an E s component.
Its value is given in the same table as . 60tc[/] sin 0 S where 5 has been substituted for L as the length of the dipole.
Comparing (1) and {2), the j operator in (2) and its absence in (1) indicates that E # and are in phase quadrature. The ratio of the magnitudes of (1) and (2) then gives the axial ratio of the polarization ellipse of the far field. Hence, divid-ing the magnitude of (2) by (1) we obtain for the axial ratio: Three special cases of the polarization ellipse are of interest. (1) When E+ = 0, the axial ratio is infinite and the polarization ellipse is a vertical fine indicating 336 7 THE HEUCAL ANTENNA 7 19 MONO FI LA ft AND MULTtHLAR NORMAL-MODE HELICAL ANTENNAS 337 Field pattern C,- 0 .14 oc = 4 a S^Q G1A Figure 7-65 Resonant narrowband circularly pol-arized monofilar normal-mode Wheeler coil.
Pattern is that of a short dipole.
linear vertical polarization. The helix in this case is a vertical dipole.
(2) When E9 = 0, the axial ratio is zero 1 and the polarization ellipse is a horizontal line indicating linear horizontal polarization. The helix in this case is a horizontal loop. (3) The third special case of interest occurs when | E# \ = | E^|. For this case the axial ratio is unity and the polarization ellipse is a circle, indicating circular polarization. Thus, setting (3) equal to unity yields nD = JlSi or Cj. = V'2S^ (4) This relation was first obtained by Wheeler in an equivalent form. 2 The radiation is circularly polarized in all directions in space but with zero field on axis (2 direction. Fig. 7-63a), A monofilar normal-mode helix or Wheeler coil fulfilling condition (4) is shown in Fig, 7-65, It is a resonant, narrowband antenna.
We have considered three special cases of the polarization ellipse involving linear and circular polarization. In the general case, the radiation is clliptically polarized. Therefore, the radiation from a helix of constant turn-length changes progressively through the following forms as the pitch angle is varied. When x = 0, we have a loop (Fig. 7-636) and the polarization is linear and horizontal.
As x increases, let us consider the helix dimensions as we move along a constant line (circle with center at origin, Fig.
7-10). As -j.
increases from zero, the polarization becomes elliptical with the major axis ofjhe polarization ellipse horizontal. When x reaches a value such that C — v 25 the polarization is cir-cular. With the aid of Fig. 7-9, this value of -a is given by .
-1 + V T + L] a = aresm ^ 1 (5) As a increases still further, the polarization again becomes elliptical but with the major axis of the polarization ellipse vertical.
Finally, when a reaches 90'\ we have a dipole (Fig. 7-63c) and the polarization is linear and vertical, Wheeler’s The axial ratio is here allowed to range from 0 lo x , instead of from 1 to x as customarily, in order to distinguish between linear vertical and linear horizontal polarization.
H A. Wheeler. "A Helicai Antenna for Circular Polarization/’ Proc. IRE, 35 f 1484^1488 December 1947, £K = 0,013 H Figure 7-66 Short resonant narrowband monofilar normal-mode helical antenna mounted over a ground plane as substitute for a A/4 stub.
relation for circular polarization from a helix radiating in the normal mode as given by (4) or (5) is shown in Fig. 7-10 by the curve marked Cx = In the preceding discussion on the normal mode of radiation, the assump-tion is made that the current is uniform in magnitude and in phase over the entire length of the helix. This condition could be approximated if the helix is very small (nL <£ /) and is end-loaded. However, the bandwidth of such a small helix is very narrow, and the radiation efficiency is low. The bandwidth and radiation efficiency could be increased by increasing the size of the helix, but to approximate the uniform, in-phase current distribution requires that some type of phase shifter be placed at intervals along the helix. This may be inconvenient or impractical. Hence, the production of the normal mode of radiation from a helix has practical limitations.
— -An antenna having four slanting dipoles that is suggestive of a fractional-turn quadrifilar helix radiating in the normal mode had been built by Brown and Woodward 1 (see Fig.
16-20/). Their arrangement is based on a design described by Lindenblad, 2 Resonant monofilar normal-mode helical antennas are useful as short, essentially vertically polarized, radiators.
Referring to Fig, 7-66, the helix mounted on a ground plane with axis vertical acts as a resonant narrowband substitute fora z/4 vertical stub or monopole above a ground plane. The helix in Fig. 7-66 is 0.06/. in height or about ^-height of a //4 stub. From (3) the axial ratio of the helix is given by AR 25- _ 2 x 0.Q1 Cl ~ (0.04) 2 12.5 (6} 1 G. H. Brown and O. M. Woodward, “Circularly Polarized Omnidirectional Antenna.” RCA Rev S 259- 269June 1947.
1 N. E, Lindenblad, “Antennas and Transmission Lines at the Empire State Television Station/ - Com-nunications, 21, 10-14, 24—26, April 1941.
338 7 THE HELICAL ANTENNA PROBLEMS 339 with the major axis of the polarization ellipse vertical. The polarization is, thus, essentially linear and vertical with an omnidirectional pattern in the horizontal plane (plane of ground plane). The radiation resistance is nearly the same as for a short monopole of height h x above the ground plane where h x = nS, which from (2-20-3) or (5-3-14) (for a short dipole) is given by Assuming a sinusoidal current distribution (maximum current at ground plane, zero at open end), Rs = 395 x (^j x 0.06 2 = 0.6 ft (8) This is the radiation resistance between the base of the helix and ground. Con-nection to a coaxial line would require an impedance transformer, but with the shunt feed of Fig, 7-66 the helix can be matched directly to a coaxial line by adjusting the tap point on the helix. With such a small radiation resistance, any loss resistance can reduce efficiency (see Secs. 2-15 and 6-12). The advantage of the helix over a straight stub or monopole is that its inductance can resonate the antenna.
A center-fed monofilar helix (a = 30°) with Sx = 1, L x = 2 and C x = has a 4-lobed pattern, 1 lobe each way on axis and 2 lobes normal to the axis. Its location is indicated on the m = 1 line of Fig. 710 where the 1-^ = 2 and a = 30° lines intersect, for which also C x — 1/73 and = 1 (Chireix coil; see also Fig, 7-62b).
Patton 1 has demonstrated that a bifilar helix end-fed by a balanced 2-wire transmission line can produce circularly polarized omnidirectional side-fire radi-ation when pitch angles of about 45° are used.
Some other monofilar and multifilar normal mode (side-fire) helices for omnidirectional FM and TV broadcasting are described by King and Wong and by DuHamel 2 PROBLEMS 3 7-1 An Saturn helix, A monofilar helical antenna has a = 12°, n = 8, D — 225 mm, (n) What is p at 400 MHz for (1) in-phase fields and (2) increased directivity? (b) Calcu-late and plot the field patterns for p = 1.0, 0.9 and 0.5 and also for p equal to the value for in-phase fields and increased directivity. Assume each turn is an isotropic point source, (c) Repeat (b) assuming each turn has a cosine pattern.
1 W. T. Patton, “The Backfire Helical Antenna," Ph.D. thesis, University of Illinois, 1965.
i H. E. King and J.
L. Wong, pp.
13-18, and R. H. DuHamel, pp, 28-35, in R. C. Johnson and H.
Jasik (eds.). Radio Engineering Handbook, McGraw-Hill, 1984, 3 Answers to starred () problems are given in App. D, 7-2 A 10-turn helix, A right-handed monofilar helical antenna has 10 turns, 100 mm diameter and 70 mm turn spacing. The frequency is 1 GHz. (a) Calculate and plot the far-field pattern. (6) What is the HPBW? (c) What is the gain? (d) What is the polarization state? (e) Repeat the problem for a frequency of 300 MH 2 r 7-3 A 30-turn helix, A right-handed monofilar axial-mode helical antenna has 30 turns, 2/3 diameter and 2/5 turn spacing. Find (a) HPBW, (b) gain and (c) polarization state.
Note regarding Probs, 7-1, 7-2 and 7-3: The patterns for monofilar axial-mode helices may be calculated using the BASIC program in App. B-2 where in line 10: N = number of turns D = spacing = 2xS x S = phase shift between turns = + {n/N) MF - multiplying or normalizing factor - 67 sin {njlN) and to account for the single-turn pattern, line 80 should read : R = MFABS(R)CA.
7-4 Helices, left and right. Two monofilar axial-mode helical antennas are mounted side-by-side with axes parallel (in the x direction). The antennas are identical except that one is wound left-handed and the other right-handed. What is the polarization state in the x direction if the two antennas are fed (a) in phase and (b) in opposite phase?
7-5 A 6-tum helix, A monofilar axial-mode helical antenna has 6 turns, 231 mm diam-eter and 181 mm turn spacing. Neglect the effect of the ground plane. Assume that the relative phase velocity p along the helical conductor satisfies the increased directivity condition. Calculate and plot the following patterns as a function of 4> (0 1<^36CT) in the 0 = 90 n plane at 400 MHz. Use the square helix approximation, (a) E^t for a single turn and E for the entire helix, (b) Repeat (n) neglecting the contribution of sides 2 and 4 of the square turn, (c) E$J for a single turn and E# for the entire helix.
1 7-6 Normal-mode helix, (a) What is the approximate relation required between the diameter D and height H of an antenna having the configuration shown in Fig. F7-6, in order to obtain a circularly polarized far field at all points at which the field is not zero.
The loop is circular and is horizontal, and the linear conductor of Length H is vertical. Assume D and H are small compared to the wavelength, and assume the current is of uniform magnitude and in phase over the system.
(b) What is the pattern of the far circularly polarized field?
Figure P7-6 Normal-mode helix.
CHAPTER 8 THE BICONICAL ANTENNA AND ITS IMPEDANCE 8-1 INTRODUCTION Sir Oliver Lodge constructed a biconical antenna in 1897, while the single cone working against ground was popularized by Marconi (Fig.
1-3). The fan (fiat triangular) antenna was also used by Marconi and others.
The broadband characteristics of monoconical (single-cone) and biconical (double-cone) antennas make them useful for many applications. In this chapter a fundamental analysis is given and both theoretical and experimental results are presented.
In the chapters preceding 7 it is usually assumed that the antenna conduc-tor is thin, in fact, infinitesimally thin. From known or assumed current distribu-tions, the far-field patterns are calculated. The effect of the conductor thickness on the pattern is negligible provided that the diameter of the conductor is a small fraction of a wavelength. Thus, the patterns calculated on the basis of an infini-tesimally thin conductor are applicable to conductors of moderate thickness, say for d < G.05A where d is the conductor diameter.
The radiation resistance of thin linear conductors and loops is calculated in Chaps. 5 and 6. This calculation is based on a knowledge of the pattern and a known or assumed current distribution. The values so obtained apply strictly to an infinitesimally thin conductor. The conductor thickness, up to moderate diam-340 the characteristic impedance of the infinite biconical antenna 341 \ t \ t \ t f \ f \ t \ Figure 8-1 An infinite biconical antenna tu) is analogous to an infinite uniform transmission Line {&).
eters, has only a small e^ffect on the resistance at or near a current loop but may have a large effect on the resistance at or near a current minimum.
1 In this chapter, we consider the problem of finding the input terminal resist-ance and also the reactance, taking into account the effect of conductor thickness.
This problem is most simply approached by SchelkunofTs treatment of the biconical antenna 2 which will be outlined in the following sections. Beginning with the infinite biconical antenna, the analysis proceeds to terminated biconical antennas, ie„ ones of finite length. This method of treatment bears a striking similarity to that usually employed with transmission lines in which the infinite transmission line is discussed first, followed by the terminated line of finite Length.
8-2 THE CHARACTERISTIC IMPEDANCE OF THE INFINITE BICONICAL ANTENNA, The infinite biconical antenna is analogous to an infinite uniform transmission line. The biconical antenna acts as a guide for a spherical wave in the same way that a uniform transmission line acts as a guide for a plane wave. The two situations are compared in Fig. 8-1.
The characteristic impedance of a biconical antenna will now be derived and will be shown to be uniform. Let a generator be connected to the terminals 1 This is discussed in more detail in Cbap, 9, 1 S. A. Schclkunoff, Electromagnetic (Vaues, Van Nostrand, New York, 1943, chap. II „ p. 441.
342 i THE BTCON1CAL ANTENNA AND ITS IMPEDANCE Flgtra 8-2 Infinite biconical antenna showing voltage V and current l at a dis-tance r from the terminals.
of an infinite biconical antenna as in Fig- 8-2, The generator causes waves with spherical phase fronts to travel radially outward from the terminals as suggested.
The waves produce currents on the cones and a voltage between them. Let V be the voltage between points on the upper and lower cones a distance r from the terminals as in Fig, 8-2. Let J be the total current on the surface of one of the cones at a distance r from the terminals. As on an ordinary transmission line, the ratio Vjl is the characteristic impedance of the antenna. For the characteristic impedance to be uniform, it is necessary that the ratio Vjl be independent of r.
Before V and / can be calculated, we must determine the nature of the electric and magnetic fields existing in the space between the conducting cones.
Although the biconical transmission line can support an infinite number of trans-mission modes, let us assume that only the TEM or principal transmission mode is present. For the TEM mode, both E and H are entirely transverse, i.e., they have no radial component. The E lines are along great circles passing through the polar axis as shown in Fig, 8-3. This satisfies the boundary conditions since E is normal to the surface of the cones, Jhe H lines are circles’ lying in planes normal to the polar axis.
Maxwell’s equation from Faraday’s law for harmonically varying fields is V x E = —jo)fM (1) The biconical antenna is most readily handled in spherical coordinates. Let the spherical coordinates r, 0, be related to the antenna as in Fig, 8-4, Expanding 2 the characteristic impedance of the infinite biconical antenna 343 the left side of (1) in spherical coordinates, we have v ^ e _ pfrsinflEJ d(rE9)1 r2 sin 0 [_ dd d<p J fl p£, d(r sin 6 £j ~| p(r£,) dE~I rsin0|_30 dr J + r[ B8 J W Since E has only a 0 component, which by symmetry is independent of , only the fifth term of (2) does not vanish. Thus, V x E 8(rE«) r dr (3) Expanding the right side of (1) in spherical coordinates, -joiftH = -jw)4iHr + §He + (4) Since H has only a component, (4) reduces to -jto/tfi = (5) Now equating (3) and (5) we have 1 djrEe ) r dr (6) This is Maxwell’s equation (1) reduced to a special form appropriate to a spher-ical wave.
344 & THE BICONICAL ANTENNA AND ITS IMPEDANCE Maxwell’s equation from Ampere’s law for harmonically varying fields in a nonconducting medium is V x H = jcusE (7) H has only a component and E only a 0 component. Since Et — 0 it follows that flsin 0 H) ^ Q (8) dG Equation (7) can be reduced by a similar procedure as used for (1) to the form = -joMrE,) (9) dr Now differentiating (9) with respect to r and introducing (6), we obtain a wave equation in (rH^), Thus,.
6-2 THE CHARACTERISTIC IMPEDANCE OF THE INFINITE BICONICAL ANTENNA 345 Figure 8-5 Et and field components at a distance r from the termi-nals of a biconical antenna.
Hence, a solution of (10) which also fulfills (1 1) is H =—H n e~^ where ft = <$yf}xz = 2njX This solution represents an outgoing traveling wave on the antenna. Since the biconical antenna is assumed to be infinitely long, only the outgoing wave need be considered The electric and magnetic fields of a TEM wave are related by the intrinsic impedance Z 0 of the medium. Thus, we have Ea — Zft H J. — r sin 0 Equations (12) and (13) give the variation of the magnetic and electric fields of a TEM outgoing wave in the space between the cones of a biconical antenna as a function of 0 and r. The fields are independent of .
The voltage V(r) between points 1 and 2 L on the cones at a distance r from the terminals (see Fig. 8-5) can now be obtained by taking the line integral of £„ along a great circle between the two points. Thus, r ^ V(r) = Ea r d6 (14) 346 I THE B[CON[CAL ANTENNA AND ITS IMPEDANCE where 0hi is the half-angle of the cone. Substituting (13) in (14) we have V(r) = Z0 H0 e;^ f' -y- = Z0 H0 e~» In ^ f (15) sin 0 tan (0hc/2) or V{r ) = 2Z0 H 0 e' jfr In cot y (16) The total current J(r) on the cone at a distance r from the terminals can be obtained by applying Ampere's law. Thus, /(r) = J H+r sin 0 d = InrH# sin 0 (17) Now substituting from (12) in (17) yields I(r) = 2nH Q e~^ (18) The characteristic impedance Zk of the biconical antenna is the ratio of K(r) to /(r) as given by (16) and (18) or Zt = — = — In cot y (19) /(r) Jt 2 For a medium of free space between the cones, Z0 = 120ft Q so that (19) becomes Zk = 120 In cot y <) (20) When 6hc is small (0hc < 20°), cot (Fht/2) ~ 2/0hc so that Zk = 120 In (fl) (21) Equations (20) and (21) are SchelkunofTs relations for the characteristic imped-ance of a biconical antenna.
Since these equations are independent of r, the biconical antenna has a uniform characteristic impedance.
8-3 INPUT IMPEDANCE OF THE INFINITE BICONICAL ANTENNA- The input impedance of a biconical antenna with TEM waves is given by the ratio K(r)//(r) as r approaches zero. For an infinite biconical antenna this ratio is independent of r, so that the input impedance of the infinite biconical antenna equals the characteristic impedance. The input impedance depends only on the TEM wave and is unaffected by higher-order waves. Thus Z, = Zt (1) where Z x is the input impedance of the biconical antenna and Zk is the character-istic impedance as given by (8-2-20) or for small cone angles by (8-2-21)- The characteristic and input impedances are pure resistances, a characteristic resist-INPUT IMPEDANCE OF THE FINITE BICONICAL ANTENNA 347 0 , 01 " 0 . 02 0 ,04 0 , 1 0 , 2 0 .4 “ 1 . 0 2 4 " 10 20 " 40 " 90 " HaJf'Cone angle H 0^ Figure 8-6 Characteristic resistance of biconical antenna and of single cone with ground plane (rnonocomcal antenna) as a function of the half-cone angle in degrees. If the antenna is infinitely long, the terminal impedance is equal to the characteristic resistance as given in the figure ” ance Rk and an input resistance /?. They are given by • = R, = 120 In cot ^ (fi) (2) The variation of this resistance as a function of the half-cone angle 0hc is presented by the solid curve in Fig. 8-6. An infinite biconical antenna of 2° total cone angle (0bc =1°) has a resistance of 568 fi, while one with a total cone angle of 100° (0hc = 50°) has a resistance of 91 fi.
If the lower cone is replaced by a large ground plane (see insert in Fig. 8-6), the resistance is ^ the value given by (2), as shown by the dashed line in Fig.
8-6.
8-4 INPUT IMPEDANCE OF THE FINITE BICONICAL ANTENNA-In this section we will consider the finite biconical antenna. This is analogous to a finite or terminated transmission line.
A TEM-mode wave can exist along the biconical conductors, but in the space beyond the cones transmission can be only in higher-order modes. Schelku-noff has defined the sphere coinciding with the ends of the cones as the boundary sphere, as indicated in Fig. 8-7. The radius of the sphere is K being equal to the length of the cones (r = f).
Inside this sphere TEM waves can exist, and also higher-order modes may be present, but outside only the higher-order modes can exist.
350 THE BICONICAL ANTENNA AND ITS IMPEDANCE S-4 INPUT IMPEDANCE OF THE FINITE BJCONICAL ANTENNA 351 () Figivt 8-10 Thin finite biconical antenna and transmission-line equivalent for finding ZL .
The real part Rm is the same as the radiation resistance at a current maximum of a very thin linear antenna. It has been calculated by SchelkunofT as 1 R m = 60 Cin 20/ 4- 30(0.577 + In 0/ - 2 Ci 20/ + Ci 40/) cos 20/ + 30(Si 40/ - 2 Si 20/) sin 20/ (G) (4) Provided only that the antenna is thin, the radiation resistance R m is inde-pendent of the shape of the antenna (i.e„ whether cylindrical or conical).
However, the radiation reactance depends on the shape and has been calculated by SchelkunofT for a thin cone as Xm = 60 Si 20/ + 30(Ci 40/ - In 0/ - 0.577) sin 20/ - 30(Si 40/) cos 20/ (G) (5) Now substituting (3) for ZL into (1), the input impedance is Zk +jZm tan 0/ +-/Z tan 0/ (6 ) where / = length of one cone Zk ~ value given by (8-221) Zm = Rm + jXnt> where Rm = value given by (4) and Xm = value given by (5) The value of Zm becomes independent of cone angle for thin cones. Thus, the real and imaginary parts of Zm , as given by (4) and (5), are independent of the cone angle, being functions only of the cone length f. However, the characteristic impedance Zk is a function of the cone angle. Hence, the input impedance Z { as calculated by (6) is a function of both the cone angle and the cone length. The 300 200 x, n too 0 0 100 200 300 400 R, n Figure 8-11 Resistance and reactance Xm of radiation impedance Z„ of a biconical antenna as a function of the cone length in wavelengths limitation in calculating Z„ (that the cone angle be small) also limits the use of (6) to small cone angles, say, half-cone angles of less than about 3V The radiation impedance Z„ at the current maximum of Schelkunoff’s biconical antenna as given by (4) and (5) is presented in Fig. 8-11. The impedance is given as a function of cone length, lit in wavelengths, where lA = IjX. This impedance applies to small cone angles.
Introducing Zm into (6), the input impedance can be obtained for cones of different characteristic impedance. As illustrations, the input impedance of a biconical antenna of 1000 G characteristic impedance (half-cone angle, 0hc -0.027°) and for one of 450 G characteristic impedance (half-cone angle, 0hc = 2J C ) are given in Fig. 8-1 2, 1 as functions of the cone length in wavelengths {/J.
If the lower cone is replaced by a large ground plane (see insert in Fig, 8-6), the input impedance is halved.
11 is significant that the terminal impedance of the thicker biconical amenna (lower characteristic impedance) is more constant as a function of cone length than the impedance of the thinner antenna. This difference in impedance behav-ior of thick and thin antennas is typical not only of conical antennas but also of antennas of other shapes, such as cylindrical antennas. We thus conclude that the impedance characteristics of a thick antenna are, in general, more suitable for wideband applications than those of a thin antenna.
The curve in Fig.
,8-12 Tor the 2.7° halfangle biconical antenna spirals inward and would eventually end at the point R = 450, X = 0, when the length lk becomes infinite. Likewise, the curve for the 0.027° antenna spirals into R = 1000, X = 0, when lk -co. The effect of the cone angle is greatest near the second, fourth, or even, resonances (/ ^ 1, etc.) and least near the first, third, or odd, resonances [lk ^ ±, etc,).
1 Approximate solutions for wide cone angles are discussed by C. T. Tai, “Application of a Varia-tional Principle to Biconical Antenna." J. Appi Phys., 29, 1076-1084. November 1949.
P, D. P. Smith, “The Conical Dipole of Wide Angle," J . Appl. Fhys 19, 1 1 —23, January 1948.
1 The curves in Figs. 8-11 and 8-12 are plotted from data given by S A SchelkunofT, Electromagnetic Waves y Van Nostrand, New York, 1943.
1 f equals half the total length of the antenna. In Chap.
5, L is twice this value, being equal to the total antenna length {thal is, L = 21).
352 8 THE BICONICAL ANTENNA AND ITS [MPRDANCR S-5 PATTERN OF BICONICAL ANTENNA 353 -3000 + 2000 -1000 X, - 1000 -2000 3000 -i V h = A / > zk = 100011 3.027“ jh\ Zi - 45011 K = 2.V \ \ \ 1 V .5 20 00 30,8 O00 50 1 f 00 :l,V \ \ IL V / / / — — ; A 'L ' V >-r Figure 8-12 Calculated input impedance of biconical antennas with IT half-cone angle (solid curve) and with 0.027 half-cone angle (dashed curve). The resistance K and reactance X of the input imped-ance Z, are presented as a function of the length of one cone in wavelengths {1J, the length being indicated in 0.1 k intervals.
We note in Fig, 8-12 that the geometric mean resistance R 13 of the resist-ance at the first and second resonances is about \ the characteristic resistance of the biconical antenna. We take R 12 — s!^A > where R ; is the resistance at the first resonance — 4} and R 3 is the resistance at the second resonance ^ 2 )-Thus, for the antenna with 17° half-cone angle, R 12 = 224 , which is about ± the characteristic resistance (R fc = 450). For the antenna with the 0.027 half-cone angle, R l2 = 500 or j the characteristic resistance (Rt = 1000 ), The geometric mean resistance R 33 of the resistance at the second and third resonances is closer to the characteristic resistance. We take R 23 — \/^2 Ra , where R 3 is the resist-ance at the third resonance (Jj. — 4), Thus, for the antenna with the 2.7 half-cone angle, R 3 3 = 317 (R = 450} while for the antenna with the 0.027° half-cone angle, R 2i = 710 (R fc = 1000). The geometric mean of successive higher resonant resist-ances would be expected to approach closer yet to the characteristic resistance around which the impedance spiral converges.
The impedance spirals in Fig. 8-12 are for a biconical antenna.
If the lower cone is replaced by a large ground plane, the impedance values are halved. Mea-sured impedances of single cones with ground plane are presented in Fig. 8-13 for cones with half-angles of 5, 10, 20 and 30° and characteristic resistances (Rk = Zk) Figu^f 8-13 Measured input impedance of single cones with top hat as a function of cone length in wavelengths (iA ). Impedance curves are presented for cones with half-angjcs of 5, 10, 20 and 3Q D , of 188, 146, 104 and 80 Q respectively .
1 The cones measured had a top hat con-sisting of an inverted cone of 90° total included angle (see insert in Fig. 8-13). It is to be noted that the trend toward reduced impedance variation with increasing cone angle, as predicted by the calculated curves of Fig. 8- 12 , is continued for the larger cone angles, 8-5 PATTERN OF BICONICAL ANTENNA, The far-field pattern of a biconical antenna will be nearly the same as for an infinitesimally thin linear antenna provided that the cone angle is small.
It is assumed that the current distribution is sinusoidal. Thus, Eqs. (5-5-10) and (5-5- 11 ) can be used for thin biconical antennas, the substitution being made that L = 2/, where / is the length of one cone.
1 The curves in Fig.
8 13. are plotted from data presented by A. Dome, in Very High Frequency Techniques, by Radio Research Laboratory Staff, McGraw-Hill, New YorIc, 1947, chap. 4.
354 8 THE BICONICAL ANTENNA AND ITS IMPEDANCE Figure 8-14 Cylindrical antenna and equivalent bicomcal antenna and transmission line.
8-6 INPUT IMPEDANCE OF ANTENNAS OF ARBITRARY SHAPE- Schelkunoff has extended his analysis for thin biconical antennas, as outlined above, to thin antennas of other shapes by considering the average char-acteristic impedance of the antenna. Whereas the characteristic impedance of a biconical antenna is uniform, the impedance of antennas of shape other than conical is nonuniform. Thus, as an approximation the input impedance of the cylindrical antenna in Fig. 8- 14a can be calculated as though it were a biconical antenna of characteristic impedance equal to the average characteristic imped-ance of the cylindrical antenna. The cylindrical antenna is replaced by the equiva-lent biconical antenna as suggested in Fig. 8-14a. The transmission-line circuit, equivalent to the antenna, is shown in Fig. 8-14h, it being assumed that the line oflength i has a uniform characteristic impedance equal to the average character-istic impedance of the cylindrical antenna.
This topic is discussed further in Sec. 9-11.
8-7 MEASUREMENTS OF CONICAL AND TRIANGULAR ANTENNAS THE BROWN-WOODWARD (BOW-TIE) ANTENNA, In 1945 Brown and Woodward made an extensive set of impedance measure-ments (published in 1952) 1 of both cones and triangles operating against a ground plane and also patterns of both biconical and triangular (bow-tie) dipoles.
The impedance data apply, of course, also to biconical and triangular dipoles by multiplying impedance values by 2.
Figure 8-15 shows Brown and Woodward’s results for conical and triangu-lar antennas operating against a ground plane as a function of the length (or height) 2 for flare angles 0 of 30, 60 and 90°. Although the conical measurements were made with open-ended cones. Brown and Woodward found no significant 1 G. H Brown and O. M. Woodward, 11 Experimentally Determined Radiation Characteristics of Conical and Triangular Antennas," RCA Rev., 13, 425-452, December 1952.
3 Note that here is measured perpendicular to the ground plane whereas in Fig. EM3 it is measured along the side of the cone.
8 7 MEASUREMENTS OF CONICAL AND TRIANGULAR ANTENNAS 355 a Length, h , Figure 8-15 Measured resistance (a) and reactance (M values for monoconical and monotriangular (flat sheet) antennas as a function of length 1 4 for flare angles 9 of 30, 60 and 90°, {After G, H. Brown ^ and O- M, Woodward “ Experimentatiy Determined Radiation Characteristics of Conical and Triangu-lar Antennas RCA Rev., 13, 425-452, December 1952.) Brown and Woodward give results in smaller flare angle increments between 0 and 90\ difference in impedance values for a 60° cone with and without end caps. The gain of conical dipoles of length 2lx with respect to a 2/2 dipole is shown in Fig. 8-16. The gains are calculated from measured patterns.
Although the conical antennas have a smaller resistance fluctuation with frequency than the triangular antennas, the flat geometry of the triangles is attractive. The measured performance of a Brown-Woodward (bow-tie) antenna 34 cm long connected to a 300-D twin line for frequencies between 480 and 900 MHz (UHF TV channels 15 to 83) is presented in Fig. 8-17.
A biconical antenna with a flare angle of 120° is shown in Fig. 2-28c, which has a VSWR < 2 over a 6 to 1 bandwidth with a cone diameter D — X at the lowest frequency.
356 8 THL BICONICAL ANTENNA AND ITS IMPEDANCE Length, 2fx Figure 8-16 Gain or biconical antennas with respect to a A/2 dipole as a function of length 21A for full cone (flare) angles B of 30, 60 and 90°.
{After G. H. Brown and 0.
M.
Woodward ' 1 Experimentally Determine d Radiation Characteristics of Conical and Triangular Antennas” RCA Rev., 13, 425-452„ December 1952.) M THE STACKED BICON1CAL ANTENNA AND THE PHANTOM BICONICAL ANTENNA. By stacking two biconical antennas, the vertical-plane beam width can be reduced with approximately a 3-dB increase in gain. A coaxially-fed stacked pair of 120° biconical antennas is shown in cross section in Fig. 8-18, Each biconical antenna has a 50-Zl resistive (a) Figure 8-17 Gain in dBi and V$WR or UHF Brown-Woodward (bow-tie) antenna with 6Q G flare angle as a Function of the length 2l x .
\After G- H. Brown and O. M , Woodward, ” Experimentally Determined Radiation Characteristics of Conical and Triangular Antennas,” RCA Rev r , 13, 425^452, December 1952.) The field patterns shown in (a) are actually those with the plane of the bow-tie perpendicular to the page (rotated 9QP on its axis) instead of with the plane of the bow-tie parallel to the page as drawn.
Figvt 8-19 Phantom 30° biconical antenna with 4 radial rods replacing each cone.
358 a THE BICONICAL ANTENNA AND ITS IMPEDANCE input impedance (see Fig, 2-28c) which is transformed by a tapered transition section of coaxial line to 100 ft at the junction with the main 50 ft coaxial line.
For reduced wind resistance each cone can be replaced by several conduct-ing radial rods resulting in a "phantom” biconical antenna. A view of a phantom arrangement for a 30° biconical antenna with 4 radials for each cone is shown in Fig. 8-19, By making the radial rods easily removable, the phantom biconical is well adapted for portable applications, A disadvantage of the phantom biconical antenna is that currents on the lower "cone” are not as well decoupled from the mast as with a solid cone. Currents induced on the mast may affect the pattern and gain adversely.
PROBLEMS 1 „ ^ u # .
8-1 Biconical antenna with unequal cone angles. Confirm SchelkunofFs result that the characteristic impedance of an unsymmetrical biconical antenna (with unequal cone angles) is Zk = 60 In ^cot Y cot y) where & hB = half the upper cone angle 0£c = half the lower cone angle 8-2 Single cone and ground plane. Prove that the characteristic impedance A for a single cone and ground plane is half Z ft for a biconical antenna.
8-3 The 2 a cone. Calculate the terminal impedance of a conical antenna of 2° total angle operating against a very large ground plane. The length l of the cone is 3A/8.
8-4 Bow-tie antenna. What is the gain (dBi) and input impedance of a U Brown-Woodward bow-tie antenna with 60° flare angle?
8-5 Monotriangular antenna. What is the input impedance of a monotriangular antenna 0.39/ long coaxially fed from a ground plane if the flare angle is 90 ?
8~6 Monoconkal antenna. What is the input impedance of a monoconical antenna 0.39/ long coaxially fed from a ground plane if the full-cone angle is 90 , Answers to starred (} problems are given in App. D.
CHAPTER 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD (MM) 9-1 INTRODUCTION.
1 In previous chapters, the assumption is made that the^current distribution on a finite antenna is sinusoidal. This assumption is a good one provided that the antenna is very thin. In this chapter, a method for calculating the current distribution of a cylindrical center-fed antenna will be discussed, taking into account the thickness of the antenna conductor.
This is a bound ary-value problem. The antenna as a bound ary-value problem was treated many years ago by Abraham, 2 who obtained an exact solu-tion for a freely oscillating elongated ellipsoid of revolution. However, the earliest treatments of the cylindrical center-driven antenna as a bound ary-value problem 1 In olher chapters sufficient steps are given in mosi analyses that the reader should be able to supply the intermediate ones without undue difficulty. However, this is not the case in this chapter since in most instances a large number of steps is omitted between those given in order to reduce the length of the development.
1 M. Abraham, “Die elect rischen Schwingungen um einen stabfbrmigen Leiter, behandelt nach der Maxwellschen Theoric,” Ann. Physik , 66,435-472, IB98.
359 360 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD (MMJ are those of Hallen 1 and King.
2 More recently the problem has been discussed by Synge and Albert.
3 Hallen’s method leads to an integral equation, approximate solutions of which yield the current distribution. Knowing the current distribu-tion and the voltage applied at the input terminals, the input impedance is then obtained as the ratio of the voltage to the current at the terminals.
Hallen T s integral-equation method will not be presented in detail, but the important steps and results will be discussed in the following sections. Later in the chapter the Moment Method (MM) is applied to the solution or the current distribution, impedance and radar cross section of a short dipole antenna.
9-2 OUTLINE OF THE INTEGRAL-EQUATION METHOD.
Since this method is a long one, an outline of the important steps is given in this section.
The objective of the method is twofold: L To obtain the current distribution of a cylindrical center-fed antenna in terms of its length and diameter.
2 To obtain the input impedance.
An outline of the procedure is given by the following steps. These are treated more fully in the sections which follow.
L The field E inside the conductor is expressed in terms of the current and skin effect resistance.
2. The field E outside the conductor is expressed in terms of the vector poten-tial.
3. The tangential components of E are equated, obtaining a wave equation in the vector potential A, Steps 1 through 3 are discussed in Sec. 9-3, 4. The wave equation in A is solved as the sum of a complementary function and a particular integral.
1 Erik Hallen, "Theoretical Investigations into the Transmitting and Receiving Qualities of Antennae," Nova Acta Regiae Soc. Sci , Vpsaliensisy ser. IV, II, no. 4, 1—44, 1938.
2 L. V, King, lOn the Radiation Field of a Perfectly Conducting Base-Insulated Cylindrical Antenna over a Perfectly Conducting Plane Earth, and the Calculation of the Radiation Resistance and Reac-tance;' Phil. Trans. R. Soc. [Lond.% 236, 381-422, 1937.
3 G. E. Albert and J.
L. Synge, “The General Problem of Antenna Radiation.
I, M Q . Appl. Math,, 6, 117-131, July 1948; and J.
L. Synge, “The General Problem of Antenna Radiation and the Funda-mental Integral Equation, With Application to an Antenna of Revolution.
II," Q. Appl Math 6, 133-156, July 1948.
9-3 THE WAVE EQUATION [N THE VECTOR POTENTIAL A 361 S. The constant C2 in the solution is evaluated in terns of the conditions at the input terminals, 6‘ The vector Potential A is expressed in terms of the antenna current /.
7. The value of C 3 from 5 and of A from 6 are inserted in the solution 4 obtaining Hallen’s integral equation. This is an integral equation in the current /.
Steps 4 through 7 are discussed in Sec. 9-4.
8. A partial solution for the current I is then obtained by evaluating one of the integrals so that the current is expressed as the sum of several terms, some of which also involve /.
9. Neglecting certain terms in /, an approximate (zero-order) solution is obtained for /.
10. This value off is substituted back in the current equation, obtaining a first-order approximation for the current. This process of iteration can be contin-ued, yielding second-order and higher-order solutions.
11. The constant C, is evaluated and an asymptotic expansion obtained for the current; that is, / _ Jv r fsin P(l-\y) + {bjit) + (fr2/fl'2 ) + --- 1 1 L cos pi + (dJO) + (dJST1 ) + • J where Q' = 2 In (2// a) and that the radius is very small compared to the wavelength (/fa ^ 1). The effect of the end face can then be neglected and the current l z taken equal to zero at z = ±t. Then E: = Zl z (3) where Z = conductor impedance in ohms per meter length of the conductor due to skin effect Iz — total current The electric field E outside the conductor is derivable entirely from the vector potential A, i.e., as given by (5-4-10), c2 E = -j — V(V A) -ja>A (4) CO Neglecting the end faces, the tangential E outside the conductor will have only an Ez component. Since the current is entirely in the z direction, A has only 9-4 HALLENTS INTEGRAL EQUATION 363 Figure 9-1 The tangential components of the electric field at the surface of the antenna are equal.
a z component. Hence, at the conductor surface (4) becomes Ei = ~j ^{i^ +plA) (5) Now equating (3) and (5) in accordance with the boundary condition of (1), we obtain a wave equation, J-PLzi 3z1 + P 1 a> 1 This completes the first three steps in the outline of Sec. 9-2.
9-4 HA I.I.EN’S INTEGRAL EQUATION. We next proceed to obtain a solution of (9-3-6), which is a one-dimensional wave equation in the vector poten-tial A z . The equation is of the second order and first degree.
If the antenna conductivity is infinite, Z = 0 and the equation becomes homogeneous. However, when Z is not zero, the equation is not homogeneous and its solution is given as the sum of a complementary function Ac and a particular integral A r ; that is, A z = A c + Ar (1) Introducing the values of Ae and Ari (1) becomes A x = - i (C t cos fk + C2 sin + — | J{s) sin J0(z - s) ds c c In (2 ) 364 9 THE CYLINDRICAL ANTENNA ; THE MOMENT METHOD |MM> Assume that the antenna is excited symmetrically by a pair of closely spaced terminals. Then l 1(z) = I z{-z) 0 ) and A 2(r) — A s { — z) The constant C2 in (2) may be evaluated as equal to £ the applied terminal voltage VT . Thus, C 2 = iV T (4) Let us now express the vector potential A z in terms of the current on the antenna. For a conductor of length z = — / to z — +J, as shown in Fig. 9-3, the vector potential A t at any point outside the conductor or at its surface is fie™ C +l 4ti r where r = [p 2 T (z - Zj) 2 ] 1/2 z x = a point on the conductor { — / < < +0 »-5 FIRST-ORDER SOLUTION OF HALLtN EQUATION 365 Inserting the values of C 2 and A z from (4) and (5) in (2) and rearranging yields Hallen’s integral equation, 1 jcfie™ f + lz V T f z ~ 4ff”“ J ^ ~L~ r &2 x = cos ftz + — sin z \ — Z J /(^) sin (z — s) ds (6) The absolute value sign on z in the second term of the right side of (6) has been introduced because of the symmetry condition of (3). Hallen’s equation {6) is an equation in the current Itl on the conductor.
If (6) could be solved for / the current distribution could be obtained as a function of the antenna dimensions and the conductor impedance.
The term with Z has a negligible effect provided that the antenna is a good conductor. Thus, assuming that Z = 0 (conductivity infinite), we can put Halten's integral equation in a simplified form as follows: C +l 1 e- T y J i — dZl = Cl C°S Pz + ^2 sin &\ z \ (7) In (7) we have put = 1 and written cfijAn — 30. This completes steps 4 through 7 in the outline of Sec. 9-2.
9-5 FIRST-ORDER SOLUTION OF HALLEN’S EQUATION,2 The problem now is to obtain a solution of (9-4-7) for the antenna current lz which can be evaluated. As a first step in the solution, let the integral in (9-4-7) be expanded by adding and subtracting i z \ that is, 1 An integral equation is an equation in which an unknown function appears under the integral sign.
In this case, the unknown function is the antenna current / In the integral equation approach to a boundary-value problem, the independent variable ranges over the boundary surface tin this case, the antenna) so that the boundary conditions are incorporated in the integral equation. This is in contrast to the situation with the differential equation approach, in which the independent variable ranges throughout space, with a solution being sought that satisfies the boundary conditions.
1 The development in this and following sections is similar to that given by Erik Hallen, “Theoretical Investigations into the Transmitting and Receiving Qualities of Antennae," Nova Acta Region Sac.
Sci Upsaliensis, scr. IV, 11, no.
4, 1-44, 1938; also by Ronold King and C. W. Harrison, Jr., “The Distribution of Current along a Symmetrical Center-Driven Antenna," Proc. IRE , 31, 548-5d7 October 1943.
366 9 THE CYLINDRICAL ANTENNA: THE MOMENT METHOD (MM) Integrating the first term in (2) and putting p = a we have r>—r)]-2i total antenna length Q' - 2 In — = 2 In —r a conductor radius “d -{i [7 I+ (rh)’ + + ]} Substituting (3) into (2), and this in turn into (9-4-7), yields h = ^ c i cos Pz + \ vt sin P 1 2 At the ends of the antenna the current is zero. Thus, when z = l f I z = 0 so that (6) reduces to -if 1 \ 1 f + '/ e~ Jpri 0 =w ( Cl cos pr 2 sin pi ) + a J_ f "V dZl (7) where r 1 = yf{l — z x ) 2 + a 2 Now subtracting (7) from (6) as done by Hallen, we have / 1 = ^Cj(cos Pz - cos pi) + ^ VV(sin p \ z \ - sin pi) Proceeding with Hallen s solution, the quantity in the braces in (8) is taken as zero so that the current J r , given by the terms in the brackets, becomes a zero-order approximation, designated F za > Thus, 1 Distinguish between the coefficient O' (with prime) as given by (4) and O (without prime) for ohms as given with impedances.
5 FIRST-ORDER SOLUTION OF HALLfcN’S EQUATION 367 where the following symbols have been introduced F0x = cos t?z — cos pi (10) G 0z = sin p\z\ - sin JffJ Subsisting /I0 , as given by (9), for Iz on the right side of (8), a first-order approximation Izl can be obtained for the current; that is, K"“- + « ) + 1 K '(G"' + If)] <"> where Flz = F,(z) - F L (l) FiU) = -F0z In [l - (jJ + F0z 5 - j' F°»c ~ F° dZl fi(0 = ~ G ll = G 1(z)-G 1 (0 Gj(z) is the same as Ff(z) with G substituted for F and G^/) is the same as /r 1 {/) with G substituted for F.
If (II) is now substituted for l z on the right side of (8), a second-order approximation for the current can be obtained. Continuing this process yields third-order and higher-order approximations, and the solution for the antenna current \ z takes the form /i=^[c{foi+^ +^ + '') + ^ Ki(Gor+ « ;£+^ + ")] (l2) Substituting l x as given by (12) into (7) yields .
1 r rc0(Q-K 1 /n j)G 1to + ---~ ' 2 tFo(0 + U/n')T 1W + --_ Inserting C t from (13) in (12) and rearranging, the current is given by the asymp-totic expansion, _ jVj rsm jg(f-|z|) + (^/n-)T(V^ J)+ -] ' 6Gfi'L cos pi + (dj/tt) + (d 2 iCl rl ) + J y where b : — F^z) sin pt — F^l) sin p\ z\ + G^O cos fiz — G x (z) cos pi = r»(D Neglecting b z t dz and higher-order terms, the first-order solution for the antenna current is _ jV T [~sin -j z | ) + (Vfl')l 1 6Qff[ cos pi + (dJO) J (15) 368 9 THE CYLINDRICAL ANTENNAl THE MOMENT METHOD (MM) The quantities b { and d x have been calculated in terms of real and imaginary functions' by King and Harrison for several values or i and curves given. 2 This completes steps 8 through 1 1 in the outline of Sec. 9-2.
<^6 LENGTH-THICKNESS PARAMETER . The above development is based on the assumptions that / P a and {Sa < /. The condition that l P a will be arbitrarily taken to mean that The ratio I/a equals the ratio of the total length of the cylindrical antenna to the diameter. Thus, Total length 21 l Diameter 2 a a When ifa - 60, the value of fi r from (9-5-4) is 21 0 H = 2 In — = 2 In 120 - 9.6 a A graph of Q' as a function of the ratio of the total length to the conductor diameter is presented in Fig. 9-4.
Another reason for restricting Ija to large values (f/a ^ 60) is that for asymptotic convergence of (95-14) Q' must exceed a certain value.
If W is too small, the series may diverge.
9-7 EQUIVALENT RADIUS OF ANTENNAS WITH NON-CIRCULAR CROSS SECTION. The above discussion in this and preced-ing sections deals with uniform cylindrical antennas, i.e. T antennas of circular cross section (radius = u). According to Hallen, 5 uniform antennas with non-circular cross section can also be treated by taking an equivalent radius. For squares cross sections of side length g (Fig. 9-5), the equivalent radius is a = 0,59g (1) while for thin flat strips of width w the equivalent radius is a = 0,25w (2) ] = M\ +jM" and d, - A\ + JA “ 2 Ronold King and C W. Hanison, Jr., "The Distribution of Current aJong a Symmetrical Center-Driven Antenna/' Proc . IRE, 31, 548-567, October 1943 Erik Hallen, Theoretical Investigations into the Transmitting and Receiving Qualities of Antennae/ 1 Nova Acta Regiae Soc. Sci. Upsaliensis, ser IV, 11, no. 4, 1938.
9-K CURRENT DISTRIBUTIONS 369 For 3ny shape of cross section there exists equivalent radius and hence a value of i Z .u E ! / aSCS U 15 assumed that the cross section is uniform over the entire length of the antenna, Wl CURRENT DISTRIBUTIONS. The ampiitude and phase of the current along cylindrical antennas of three lengths and two values or the total length-diameter ratio (i/a) are presented in Figs. 9-6, 9-7 and 9-8. Figure 9-6 is for c k \ an ,tenna/ 2/ = Fie' 9-7 for a fall-wavelength antenna (21 = k) and ’j r°r//^4 antcnna (2/ = 5; -/4)- For each length the relative amplitude and phase of the current are presented for fl' = 10 and O' = oo corresponding to total length-diameter ratios (l/a) of 75 and oo. The amplitude curves are adjusted to the same maximum value, and all phase curves are adjusted to the same value at the ends of the antenna.
It is generally assumed that the current distribution of an infinitesimally thin antenna (I/a = oo) is sinusoidal, and that the phase is constant over a a/2 conductor Flgive 9-5 Conductors of square and flat cross section with equivalent circular conductors of radius a.
370 v T^LE ^'YL| NDRLCAI- ANTENNA; I HE MOMENT METHOD iMMi Figure 9-6 Relative current amplitude and phase along a center-fed }J2 cylindrical antenna (2l = >.jl) for tola! length-diameter ratios (f/n) of 75 and infinity. (After R. King and C W. Harmon, Jr., “ The Distribution of Current along a Symmetrical Center-Driven Antennaf Proc. IRE, 31, 548-567+ October 1943.) Distance from the center of the antenna is expressed in wavelengths.
interval, changing abruptly by 1 8(F between intervals. This behavior is illustrated by the solid curves in Figs. 9-6, 9-7 and 9-8.
The dashed curves illustrate the current amplitude and phase variation for 1/a = 75 (fT = 10). The difference between these curves and the solid curves (If a = x) is not large but is appreciable. The dashed curves (If a = 75} are from the distributions given by King and Harrison 1 as calculated from (9-5-15), the current being expressed in terms of its amplitude and the phase angle relative to a reference point. Thus, = !A W The effect of the length -thickness ratio on the current amplitude is well illustrated by Fig. 9-7 for a full-wavelength antenna. When the antenna is infini-tesimally thin, the current is zero at the center. As the antenna becomes thicker, the current minimum increases and at the same time shifts slightly toward the end of the antenna. For still thicker antermas (i/a < 75), Eq. (9-5-15) is no longer a good approximation for the current, but it might be expected that the above trend would continue.
The effect of the length-thickncss ratio on the phase variation is well illus-trated by Fig. 9-8 for a 52/4 antenna. When the antenna is infinitesimally, thin, the phase varies as a step function, being constant over 2/2 and changing by 180 at the end of the a/2 interval (solid line, Fig. 9-8). This type of phase variation is observed in a pure standing wave. As the antenna becomes thicker, the phase shift at the end of the 2/2 int-rval tends to become less abrupt (dashed curve for Ifa = 75). For still thicker antennas (Ifa < 75), it might be expected that this trend would continue and for very thick antennas would tend to approach that of a pure traveling wave, as indicated by the straight dashed lines in Fig. 9-8.
1 Remold King and C. W. Harrison, Jr., "The Distribution of Current along a Symmetrical Center-Driven Antenna,' Proc. IRE, 31, 548-567, October 1943.
input impedance 371 Distance from center of antenna, X Figure 9-7 Relative current amplitude and phase along a center-fed full-wavelength cylin-drical antenna {21 = k) for total length-diameter ralios (Ifa) of 75 and infinity.
(After R r King and C.
W r Harrison, Jr., ‘'The Distribu-tion of Current along a Symmetrical Center-Driven Antennaf Proc.
IRE, 31, 548-567+ October 1943 .} Distance from the center of the antenna is expressed in wave-lengths.
9-9 INPUT IMPEDANCE, The input impedance ZT of a cenier-fed cylin-drical antenna is found by taking the ratio of the input or terminal voltage VT and the current I T at the input terminals; that is, Z T - y- - R t +JX T (1) 1 T where l T = L(0) R t = terminal resistance X T = terminal reactance Distunes from center of antenna, X Figure 9-8 Relative current amplitude and phase along a center-fed 5i/4 cylindrical antenna (21 = 5A/4) for total length-diameter ratios (Ifa) of 75 and infinity. (After R, King and C. W. Harr\son+ Jr.+ “The Distribution of Current along a Symmetrical Center-Driven Antennaf Proc. IRE, 31, 548-567, October 1943 .} Distance from the center of the antenna is expressed in wavelengths.
372 9 THF- CYLINDRICAL ANTENNA; THE MOMENT METHOD (MM) Therefore, setting z - 0 in (9-5-15) and inserting this value of current in (1) yields Hallen’s relation for the input impedance, Z r = ;6Ofi' cos pi + (di/fT) sin + (2 ) This is a first-order approximation for the input impedance.
If the second-order terms are included [see (9-5-14)], Hallen’s input-impedance expression has the form ZT -76QQ' cos pl + (d l/Q f ) + (d2/Q t2 ) ~ sin pi + (bjtt) + {b 1f^ 1 )_ (3) This relation has been evaluated by Hallen 1 who has also presented the results in chart form.
2 Impedance spirals based on Hallen’s data are presented in Fig, 9-9 for center-fed cylindrical antennas with ratios of total length to diameter {Ifa) of 60 and 2000, The half-length l of the antenna is given along the spirals in free-space wavelengths. The impedance variation is that which would be obtained as a function of frequency for an antenna of fixed physical dimensions. The differ-ence in the impedance behavior of the thinner antenna ( Ifa — 2000) and of the thicker antenna ( Ifa = 60) is striking, the variation in impedance with frequency of the thicker antenna being much less than that of the thinner antenna.
The impedance, given by (2) or (3), applies to center -fed cylindrical antennas of total length 21 and diameter 2a. To obtain the impedance of a cylindrical stub antenna of length l and diameter 2a operating against a very large perfectly con-ducting ground plane, (2) and (3) are divided by 2. The impedance curve based on H alien's calculations for a cylindrical stub antenna with an Ifa ratio of 60 is given by the solid spiral in Fig, 9-10, The length l of the stub is indicated in free-space wavelengths along the spiral. The measured impedance variation of the same type of antenna {Ifa = 60) as given by Dome 3 is also shown in Fig. 9-10 by the dashed spiral. The agreement is good considering the fact that the measured curve includes the effect of the shunt capacitance at the gap and the small but finite antenna terminals.
The measured input impedance of a cylindrical stub antenna with an Ifa ratio of 20 is shown in Fig, 9-11. Comparing this curve with the dashed curve of Fig r 9-10, it is apparent that the trend toward decreased impedance variation 1 Erik Hallen, “On Antenna Impedances," Trans. Roy, Inst. Technol Stockholm, no. 13, 1947.
3 Erik Hallen, “Admittance Diagrams for Antennas and the Relation between Antenna Theories," Cruft Laboratory Tech, Rept, 46, Harvard University, 1948.
3A, Dome, in Very High Frequency Techniques, by Radio Research Laboratory Staff, McGraw-Hill, New York, 1947, chap. 4, See also G. H. Brown and O, M. Woodward, “Experimentally Determined Impedance Char-acteristics of Cylindrical Antennas, 11 Proc, IRE t 33, 257-262, April 1945; D. D. King, “The Measured Impedance of Cylindrical Dipoles," J. Appi Phys., 17, 844-852, October 1946; and C-T. Tai, “ Dipoles and Monopoles," in Antenna Engineering Handbook, McGraw-Hill, 1984, chap. 4.
M INPUT IMPEDANCE 373 &-K :-4 Xfl (My-jiooo 1500 2000 12500 3000 A— / / Z(ave) = 454 / — 2000 a IT = 16.6 Zjt(ave) = 873 figure 9-9 Calculated input impedance (R + jX, O') for cylindrical center-fed antennas with ratios of total length to diameter (2//2n) of 60 and 2000 as a function of /j (along spiral).
( After E. Hallert.) with smaller Ifa ratio (increased thickness) suggested by Fig. 9-9 is continued to smaller Ifa ratios. A measured impedance curve for Ifa = 472 is also included in Fig. 9-1 1.
1 An antenna is said to be resonant when the input impedance is a pure resistance. On the impedance diagrams of Figs, 9-9, 9-10 and 9-11 resonance occurs where the spirals cross the X = 0 axis. At zero frequency all the imped-ance spirals start at R = 0 and X = — oo. As the frequency increases, the reac-tance decreases and the resistance also increases, although more slowly. The first resonance occurs when the length l of the antenna is about XfA. The resistance at the first resonance is designated /?j. As the frequency is increased, the length of the antenna becomes greater and the second resonance occurs when the length l is about X/2. The resistance at the second resonance is designated R 2 . At the third resonance (resistance = ft3 ), the antenna length l is about 3X/4 and at the 1 The curves in Fig. 9-1 1 are based on data presented by Dome (Ref. 3 on p. 372 )l 374 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD {MM| rigitft 5M0 Comparison of calculated (solid ^rve) and measured (dashed curve) input impedance (K + jX , O') for cylindrical stub antenna with ground plane for the length-radius ratio (I/a) of 60 as a function of ^ (along spiral).
fourth resonance (resistance = RJ l is about U, As the frequency is increased indefinitely, an infinite number of such resonances can be obtained except where the impedance stays reactive.
Since it is common practice to operate antennas at or near resonance, the values of the resonant resistances are of interest. Curves based on Hallen s calcu-lated graphs 1 are presented in Fig. 9-12 for the first four resonances of a cylin-drical stub antenna with large ground plane as a function of the length-radius ratio (l/a). The lowest value of l/a for which Hallen gives data is 60, since the accuracy of (3) tends to deteriorate for smaller t/a values. Thus, the solid part of the curves (l/a > 60) are according to Hallen’s calculated values. The dashed parts of the curves are extrapolations to smaller values of l/a . The extrapolation is without theoretical basis but is probably not much in error, A few measured values of resonant resistance from Dome's data 2 are shown as points in 1 Erik Hallen, “Admittance Diagrams for Amentias and the Relation between Antenna Theories,” Cruft Laboratory Tech. Rept. 46, Harvard University, 1948.
3 A. Dome, in Very High Frequency Techniques, by Radio Research Laboratory Staff, McGraw-Hill, New York, 1947, chap. 4.
INPUT IMPEDANCE 375 Fig. 9^12, the dotted lines indicating to which resonant resistance the points correspond.
Figure 9-12 illustrates the difference in the effect of antenna thickness on the resistance at odd and even resonances. The resistance at odd resonances (R^ /? 3 , etc.) is nearly independent of the antenna thickness. The first resonant resistance R l is about 35 O and the third resonant resistance is about 50 ft over a large range of t/a ratios. On the other hand, the antenna thickness has a large effect on the resistance at even resonances Z? 4 , etc,). The thicker the antenna, the smaller the resistance. For example, the second resonant resistance Zf 2 is about 200 ft when l/a = 10 and increases to about 1500 ft at l/a = 1000. The fourth resonant resistance behaves in a similar fashion, the values being somewhat less.
The difference in the resistance behavior at odd and even resonances is related to the current distribution. Thus, at odd resonances the antenna length l is an odd number of A/4 (approximately), and a current maximum appears at or near the input terminals At even resonances the antenna length Ms an even number of A/4 (approximately), and a current minimum appears at or near the input terminals. As indicated by the current distribution curves of Figs. 9-7 and 9-8, one of the most noticeable effects of an increase in antenna thickness is the 376 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD (MM) $-11 THE THIN CYLINDRICAL ANTENNA 377 Figure 9-12 Resonant resistance of cylindrical stub antenna with ground plane as a function of the length -radius ratio (//a). Curves are shown for the first four resonances. For cylindrical center-fed antennas (total length 20 multiply the resistance by 2.
increase of the current at current minima. Thus, when a current minimum is at or near the input terminals, an increase in the antenna thickness raises the input current IT for a constant input voltage V T so that the resonant resistance given by the ratio VT/l T is reduced.
9-10 PATTERNS OF CYLINDRICAL ANTENNAS. Formulas for calculating the far-field patterns of thin linear antennas were developed in Chap.
5. Although these relations apply strictly to infinitesimally thin conductors, they provide a first approximation to the pattern of even a relatively thick cylin-drical antenna. This is illustrated by the patterns in Fig 9-13 for center-fed linear cylindrical antennas of total length 21 equal to A/2, 1A, 3A/2 and 2A. The calcu-lated patterns for infinitesimally thin antennas are shown in the top row. Three of these, patterns were given previously in Fig. 5-10, In the next three rows patterns Figure 9-13 Field patterns of cylindrical center-fed linear antennas of total length 21 as a function of the total length-diameter ratio (//a) and also as a function of the total length (20 in wavelengths.
measured by Dome 1 are given for l/a ratios of 450, 50 and 87, The principal effect of increased antenna thickness appears to be that some of the pattern nulls are filled in and that some minor lobes are obliterated (note the patterns in the third column for 21 = 3A/2).
9-11 THE THIN CYLINDRICAL ANTENNA.
If the assumption is made that the cylindrical antenna is infinitesimally thin (IT -i oo), the current 1 A. Dome, in Very High Frequency Techniques, by Radio Research Laboratory Staff, McGraw-Hill New York, 1947 chap, 4, 378 9 THE CYLINDRICAL ANTENNA. THF. MOMENT METHOD |MMh distribution given by (9-5-14) or (9-5-15) reduces to _ jVT sin p(l - \z) z 60ft r cos f$l Although ft' approaches infinity, the ratio VT/ft may be maintained constant by also letting VT approach infinity. According to (1), the shape of the current dis-tribution is sinusoidal; that is T I z = k sin 0(1 -| z | ) (2) where k = a constant The input impedance Z T is the ratio VT/1 T where J r is the current at the terminals (z — 0), Thus from (1), Z T = j-= -J60Q' cot pi (3) In (3) we. may regard ft as large but finite. The terminal impedance Z T according to (3)’is a pure reactance X T . Equation (3) is identical to the relation for the input impedance of an open -circuited lossless transmission line of length (sec App. A, Sec. A -2) provided that 60ft' is taken equal to the characteristic impedance of the line.
If, by analogy, 600' is taken equal to the average characteristic impedance Z k (ave) of the center-fed cylindrical antenna then, from the value of ft given in Sec. 9-6, ( 21 Z k (ave) - 60ft = 120 In -(4) This relation is of the same form as SchelkunofTs expression for the characteristic impedance Z k of a thin biconical antenna given by (8-2-21) since for small cone angles 0hc = a/i so that (8-2-21) becomes Zk = 120 In -(5) a where a — end radius of the cone as shown in Fig. 9-14.
The average characteristic impedance of a center-fed cylindrical antenna as given by Schelkunoff is Zk (ave) = 120^ In — — 1^ (6) The average impedance of a cylindrical stub antenna with a large ground plane is \ the value of (6), As lla —> oo, (6) reduces to the form given in (4). However, for finite values of //a, the average characteristic impedance of a cylindrical antenna is the same as for a biconical antenna of the same length I but with an end radius a which is larger than the radius of the cylindrical conductor. This is suggested in Fig.
9-12 CYLINDRICAL ANTENNAS WITH CONICAL INPUT SECTIONS 379 Ui 8-14a. For example, a cylindrical antenna with an Ija ratio of 500 has an average characteristic impedance equal to that of a biconical antenna of the same length with an end radius 2.8 times larger than the radius of the cylindrical conductor.
In Fig.
9-9 the calculated input impedance is presented for cylindrical center-fed antennas with total length-diameter ratios {lljla = ija ) of 60 and 2000, The average characteristic impedance of these antennas by (6) is 454 and 873 ft respectively. The curve for the ija ratio of 60 [Zk (ave) = 454 ft] has approx-imately the same form as the calculated impedance spiral in Fig. 8-12 for a 2.7° half-angle biconical antenna (Zk = 450 ft).
In Fig. 9-11 the measured input impedance is shown for cylindrical stub antenrihs with Ija ratios of 20 and 472, The average characteristic impedance of these antennas as given by ^ of (6) is 161 and 350 ft respectively. The curve for ija = 20 [Zk (ave) = 161 ft] is of the same general form (although displaced downward), as would be anticipated from Fig. 8-13 since a spiral for Z te (ave) = 161ft should lie between those shown in Fig. 8-13 for Zk = 146 ft and Zk = 188 ft.
9-12 CYLINDRICAL ANTENNAS WITH CONICAL INPUT SECTIONS-It is common practice to construct cylindrical antennas with short conical sections at the input terminals as indicated at the bottom of Fig. 9-13.
If the cylinders are of large cross section, the conical sections are particularly desir-able in order to reduce the shunt capacitance at the gap.
Since the measured impedance of an antenna includes the effect of the gap capacitance and the small but finite terminals, the measured impedances will differ more or less from the theoretical values. It is to be expected that measured values will agree better with calculated ones when end cones are used rather than when the ends of the cylin-ders are butted close together.
380 V THE CYLINDRICAL ANTENNA; THE MOMENT METHOD |MM) Figure 9-15 Prolate spheroidal atuenna.
9-13 ANTENNAS OF OTHER SHAPES.
THE SPHEROIDAL ANTENNA. The solution of a boundary -value problem may be facilitated if the boundary can be specified by one coordinate of an appropriate coordinate system, A spherical antenna or one in the shape of an elongated ellipsoid of revolution (prolate spheroid), as in Fig. 9-15, is amenable to such treatment since the surface of the spheroid corresponds to a particular value of one coordinate of a spheroidal coordinate system. By varying the eccentricity of the ellipsoid, one may study the properties of the sphere at the one extreme of eccentricity and of a long thin conductor at the other extreme. This problem has been treated at length by Stratton and Chu 1 and by Page and Adams.
2 Stratton and Chu give admittance and impedance curves for various length-diameter (L/D) ratios (see Fig. 9-15). For long, thin ellipsoids the impedance characteristics are similar to those deduced by other methods. The current distribution for thin A/2 spheroids is also found to be nearly sinusoidal.
A point of interest is that for spheroids of the order of A/2 long, the imped-ance variation with frequency decreases with decreasing L/D ratios (thicker spheroids); that is to say, resonance with thick spheroids is broader than with thin ones. This is in agreement with the well-known fact that thick antennas have broader band impedance characteristics than thin ones.
9-14 CURRENT DISTRIBUTIONS ON LONG CYLINDRICAL ANTENNAS, On a matched lossless transmission line an outgoing wave has a uniform current magnitude and a linear phase change with distance (Fig. 9-16). If the line is mismatched and the reflected (returning) wave is ^ the magnitude of the outgoing wave, a standing wave appears on the line with VSWR given by 3 VSWR = Jq + Ji _ 1 + 1 ' 0 - /i 1 - i 1 J, A Stratton and L j. Chu, " Steady State Solutions of Electromagnetic Field Problems/' J , AppL Phys., 12, 230-248, March 1941.
L Page and N.
I. Adams. “The Electrical Oscillations of a Prolate Spheroid," Pkys, Rev., S3, 819-831 , 1938 .
3 Property this should be ISWR for the current standing-wave ratio. However, the VSWR =? ISWR, although their standing-wave patterns are displaced in position.
P-14 CURRENT DISTRIBUTIONS ON LONG CYLINDRICAL ANTENNAS 381 Distance, X Distance, X Figure 9^16 Current distribution and phase variation (1} of single uniform trav-eling wave (solid lines), (2) of two waves traveling in opposite directions with magni-tudes 1 and i (dash-dot lines) and (3) of 2 waves of equal magnitudes (dashed lines).
The last case represents a full (pure) stand-ing wave where I0 current magnitude of outgoing wave /j — current magnitude of returning wave The phase change is also nonuniform (fluctuating with distance), as indicated in Fig. 9-16.
When the line is completely mismatched (open- or short-circuited), so that the returning wave equals the outgoing wave, the VSWR = oo and the phase changes in a stepwise fashion (Fig. 9-16).
The phase velocity of a wave on the line is given by j3(x) d(f>/dx where {/3 +-/f + 2/ 0 /, cos [(0, + ft,) + y]}/|o/o-f dli + / o M0o - 0t)cos 00 1 -+ y] + sin Kfs, + h Jj where f$0 = 2rc/A0 0 ! = Inj^i A0 = wavelength for outgoing wave, m Xi = wavelength for returning wave, m y = phase difference of two waves, rad or deg For the case where y = 0, the velocity of both waves is the same (p l = /?0), / 0 and 7 , are constant with distance and (3) reduces to oi{Iq + ij + 2/ q 7 1 cos 2ffix) " = ‘ un - n) 1 J L A. Marsh, +lA Study of Phase Velocity on Long Cylindrical Conductors " Ph.D. thesis, Ohio State University, 1949-9-14 CURRENT DISTRIBUTIONS ON LONG CYLINDRICAL ANTENNAS 383 ioL X Outgoing wave f Returning wave • 0 1 2 3 4 Distance, \ Figure 9-18 Measured current distribution on a long (5/), thick (0.2a diameter) cylindrical conduc-tor with attendant phase and relative phase velocity (p) of the total wave. {After J. A. Marsh, klA Study of Phase Velocity on Long Cylindrical Conductors, 1 ' PU.D.
thesis , Ohio State University, 1949.) Resolution into outgoing and returning (reflected) waves is indicated.
Dividing by c ( = tu/jS0) yields the relative phase velocity p. The ratio of the maximum to minimum relative phase velocity is (Jq + 1,) 1 U0-/1) 1 Comparing (5) and (1) we note that ) and current magnitude (\I\) for 7 t = are presented in Fig, 9-17 over a distance of 2A. We note that the VSWR = 3 and the relative phase velocity ratio equals 9, and also that the relative phase velocity p is a maximum where 7 is a maximum.
The current and phase measured by Marsh along a 5A open-ended cylin-drical conductor 0.2A in diameter are shown in Fig. 9-18, as well as the deduced 384 9 THE CYLINDRICAL ANTENNA: THE MOMENT METHOD (MM) Table 9-1 Comparison of currents on long, thick cylindrical antenna and on helical antennas Relative phase Antenna Mode velocity, p{ = r/c)t Remarks Cylindrical conductor All T 0 1 Gradual attenuation of outgoing and returning T 0 waves Helix, Q = 0.6 All T 0 >1 Uniform equal outgoing and returning T 0 waves Helix, C. = 1.07 T d at ends.
1 T Q waves attenuate rapidly (axial mode) Ti over remainder A T p 0o T, waves uniform t For single traveling wave.
relative phase velocity variation and the magnitudes of the outgoing and reflected waves. The attenuation of the outgoing and reflected waves is evident.
It is interesting to compare the current distribution of Fig 9-18 for the long (5 A), thick (0.2X diameter open-ended) cylindrical conductor with the distributions of Fig. 7-3 for a 7-turn helix with much thinner conductor (0.02 k. diameter at Ci = 107). When Cx = 0.60, the helix has a nearly uniform standing wave, indi-cating outgoing and returning To mode waves of almost equal amplitude. When C x = 1 07 the outgoing r 0 mode wave attenuates rapidly with energy transferred to a nearly uniform 7"i mode wave over the rest of the helix. At the open end, a reflected or returning T 0 mode wave is excited which attenuates rapidly while transforming into a small nearly uniform returning 7\ mode wave.
On the cylindrical conductor (Fig. 9-18) a T 0 mode wave attenuates grad-ually over the length of the conductor and on reflection from the open end excites a gradually attenuating returning wave.
The behavior of the two antennas is summarized in Table 9-L 9-15 INTEGRAL EQUATIONS AND THE MOMENT METHOD (MM) IN ELECTROSTATICS, As an introduction to the moment method, let us consider its application to an electrostatics problem.
In calculus we deal with differential equations as, for example.
or the rate of change of distance (x) with time (t) equals the velocity (d).
It is implied that we know x as a function of z or how x varies with time [((}].
On the other hand, suppose we know how the velocity varies as a function of time Then the distance is given by the integral of the velocity with 9-li INTEGRAL EQUATIONS AND THE MOMENT METHOD (MM) TN ELECTROSTATICS 385 Charge Q Figmre 9-19 Electric potential V at point P is inversely proportional to the distance r from charge Q.
respect to time or x = 1 p(t) dt If v is constant, then (2) becomes or x is the distance traveled at velocity v in time Now suppose that x is known at t = 0 and f = but v is not known during this period of time. Then (2) is an integral equation with the problem being to obtain a solution for the velocity as a function of time Referring to Fig. 9-19, a basic relation of electrostatics is that the electric potential V at a point P due to a charge Q is given by where Q = charge, C r = distance from P to the charge Q, m e = permittivity of medium, F m“ 1 For a line of charge of density pL (C m -1 ) as in Fig. 9-20, then V at some observation point P is given by the integral of (4) over the length ( of the line or 1 fVrt 4E£ J0 r where pL(x) = charge per unit oflength of line as a function of x, C m" / Total charge Q Rod with charge densitv plU) x Figure 9-20 Electric potential V at point P due to a rod with charge density pLU) which is a function of position (x).
386 9 THE CYLINDRICAL ANTENNA- THE MOMENT METHOD (MM) If p L{x) is known as a function of x, then (5) can be integrated in a straight-forward manner. However, if pL(x) is not known, (5) represents an integral equa-tion with the problem being to find a solution for pL{x).
Example, Charge distribution on wire Let the line be an isolated conducting rod or wire of radius a and length / = 8a on which a total charge + Q has been placed.
Since like charges repel, it may be anticipated that the charge wilt tend to separate and pile up near the ends of the rod, making the charge distribution along the rod nonuniform. The problem is to determine this charge distribution pL(x) using an incremental numerical technique or moment method as an introduction to integral equations.
Solution.
First, let us divide the rod of length f into 4 segments or increments with each segment of length 2a as in Fig. 9-21 (; = ga). Let the total charge on segment 1 be Q x and on segment 2 be Q2 . By symmetry, the charge on segment 3 is the same as on segment 2, or g 2 . Likewise, the charge on segment 4 is equal to Q lm Specifi-cally, our problem is to find the ratio of Q L to Q% [a first step or approximation in solving for pjx)].
Let us assume that all of the charge on each segment is concentrated on a circle on the surface of the segment around its midpoint with the observation or test points on the wire axis. Since the distance r from an observation point to any point on the circle of charge is constant, we may consider that all of the charge of a segment is at one point (charge or source point), as in Fig. 9-21. The situation may now be regarded as one with 4 points of charge (source points) in empty space with the potential at observation or test points on the axis to be determined. Thus, from (4) the potential at point P 12 is given by Likewise at point P l3 the potential is given by LJo 2 + 9a' , 1 /o + 25a 2] -& 1 /a 1 + 9a 2J A boundary condition is that (even though the charge density varies along the rod) the potential is constant. Therefore, V(P i2) = F(/> ) so that equating (6) and (7) we find that Gi = 1 45Q 2 (8) Conducting Observation points Q Charge or source points Figure 9-21 Charged rod of radius a divided into 4 segments of equal length (2a) Sor calcu-lation of charge ratios.
Relative charge density 9-15 INTEGRAL EQUATIONS AND THE MOMENT METHOD (MM) IN ELECTROSTATICS 387 Figure 9-22 Relative charge density along straight conducting rod of radius a and Length 80 as calculated by the moment method using 4, 6 and 8 segments.
Thus, the charge (or average charge density) for the outer segments is 45 per cent greater than for the inner segments and we can write Q\'Qi = 1-45 : 1.00 (9) Dividing the rod into 6 segments and proceeding as above results in the ratios Ci 1-84: 1.03: LOO (10) The charge density distribution along the rod is shown by the step or pulse functions in Fig. 9-22 for the cases of 4, 6 and 8 segments. A smooth curve is also drawn through the centers of the pulse functions.
To simplify the above calculations we neglected the effect of the end surfaces of the rod. For 4 segments the cylindrical area of a segment is 4 times the end of the rod so that the effect of neglecting the ends is not large. However, with more seg-ments the effect becomes greater, especially for the charge on the end segments.
Let us now discuss the problem more formally. From (5), V ^r— 4ke J0 r Ui) Referring to Fig. 9-23, let the line be divided into N segments of equal length Ax with average charge density pL(x). for segment Ax„. Then the charge on segment tt is given by Gif = p = L X 3 N and the total charge on the wire by e=Ie.
H= l ( 13 ) 388 9 THE CYLINDRICAL ANTENNA- THE MOMENT METHOD (MM) Observation points Source points Figure 9-13 Charged rod for calculation of charge density distribution.
liquation (1 1) can now be written as I.LG.-'i ji = 1 where ” m = 1 , 2, 3, .
, , T M 47tErPM = vV + (x - x') 1 a = rod radius x = axial distance of observation or test point m x 1 = axial distance of source point at middle of segment n In matrix notation (14) [UCCJ = tKJ (U) (15) Thus, the integral equation (11) has been transformed into a set of N simulta-neous linear algebraic equations (17) where lm represents a known function (the inverse distance relation), V m represents potentials determined by the boundary con-ditions and Qn represents the N unknown charges whose values are sought.
m the charged rod example we have from symmetry that M = N/2 and from the boundary condition that Vl -v 2 = v>— -(IS) For 4 segments {m - 2, n = 4), (17) then reduces to Introducing (15) for lmi (19) is identical in form with (6) and (7) for increments of Ax = 2a, where m designates the test or observation points (P 12 for m = 1 and P 2 j 916 THE MOMENT METHOD {MM} AND ITS APPLICATION TO A WIRE ANTENNA 389 for m = 2 in Fig. 9-21) and rnn is equal to the distance between the test point m and the source point of segment n. Thus, r 2J is the distance between test point m = 2 in Fig, 9-21) and the center of segment 3.
In the above we have used a pulse or step-function approximation in which the boundary condition (V = constant) is not enforced everywhere along the rod but only at certain observation or test points. In between observation points the boundary condition may not be satisfied. However, as the number of segments and observation points increase, the boundary condition is enforced at more points (the solution converging) and the accuracy of the results should improve In the sense that the residual discrepancies or moments should vanish with a sufficient number of properly selected pulse functions, the procedure we have discussed may be called a moment method.
9-16 THE MOMENT METHOD (MM) AND ITS APPLICATION TO A WIRE ANTENNA, As discussed in the previous section, an integral equation can be transformed into a set of simultaneous linear algebraic equations (or matrix equation) which may then be solved by numerical techniques Roger Harrington 1 has unified the various procedures into a general moment method (MM) now widely used with powerful computers for solving electromagnetic field problems.
In this section the method will be developed for a wire antenna and.applied to an example for a short dipole.
Consider a cylindrical current-carrying conductor (or wire) of radius a iso-lated in free space (Fig. 924). Let its conductivity a = oc so that we can consider the radio-frequency current to be entirely on the surface (1/e depth = 0), The total current at point z' on the conductor is J(z') = K{z'\2na (1) where K(z ' ) = surface current density at z' (Am -1 ) All of the current is at a distance a from the conductor axis (z axis), and we will consider it as flowing in empty space along an infinitesimally thin filament parallel to the z axis at a distance a, as in Fig 9-25, with the conductor no longer present.
The electric field of charges and current is given by E = —jcofitQ A — VV (2) where A = vector potential V = scalar potential 1 R, F. Harrington, Field Computation by Moment Methods, Macmillan 196&.
3$0 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD (MM) Fign 9-24 Cylindrical conductor of radius Figure 9-15 Conductor replaced by current a with surface current density m l ).
filament / — 2xaK lA) at distance a from the z-axis (8) 16 THE MOMENT METHOD (MM) AND ITS APPLICATION TO A WIRE ANTENNA 391 TV™ ,or fm ^Observation point r Current .filament Figure 9-26 Source point on current filament with field dEt at dis-tance r on the z-axis.
For a current dement dzy the vector potential f(z>“ J dz where e ^/r = G„- = free space Green’s function r = J{z- z'f + a1 z — observation point z' = source point (see Fig. 9-26) The field from this current element is then <,o» For a conductor of length L, the total field is given by the integral of (10), known as Pocklington’s equation: 1 1 CL>1 ( d 2 \ £‘ =^J (U) Es is the radiated field due to the current I{z% resulting from an impressed or source field Ext from, for example, a voltage applied at the antenna terminals or from an incident plane wave (scattering case). On (and inside) the conductor the sum of these fields must vanish (er = ao) so £,= ( 12) H. C. Pocklington, "'Electrical Oscillations in Wires,” Comb, PkiL Soc Proc ., 9, 324—332, 1897.
392 9 THE CYLINDRICAL ANTENNA^ THE MOMENT METHOD (MM} Richmond 1 has differentiated and rearranged (11) in a more convenient form as follows: _£,=^ r e -f- [(1 + jprX2r 2 - 3a 2 ) + fi 2 a lr]M M (Vm' 1 ) (13) where r = distance between source and observation points = y/{z' - z) 2 + a1 , m Z0 - 377 Q With parameters in dimensionlessform , (13) becomes — V = — Az E(z f ) -[“ {“ +J2”+ -Xf)’] + 4”'4M 42 ‘ (VI where r2 = r/A, dimensionless V — voltage developed by E(z') over Azkt V For brevity let -E(z r ) in (14) be written as where G(r„J = fL{2 — E(z') = /(z')G(r„J dz' (V m' 1 ) J ~L(1 jg(^)|( | +J2«rj[2 -3(H)'] + (fl m 2 ) m = observation point n = source point Approximating the current with a series expansion we let m= £ I.FJiz’) (17) = l where F^z 1 ) is a pulse function (equal to zero or unity) for incremental segments Az' H , (Other functions are possible, e.g., overlapping segments, each with a tri 1 Jack H Richmond, “ Digital Computer Solutions of the Rigorous Equations for Scattering Prob-lems," Proc. IEEE, 53, 796-804, August 1965.
16 THE MOMENT METHOD (MM) AND ITS APPLICATION TO A WIRE ANTENNA 393 angular or piecewise sinusoidal current distribution.) 1 For the mth segment we have -EAzj = X /„ f G{rm)dz' (V, =1 Jazb ' (18) becomes approximately EAzm) — ^ i ^mi + Gm2 + ' + 4- K + + 1^ Gm4 (20) and the antenna equation now takes the form of a network equation. Writing (20) for each of the AT segments (m = 1, 2, 3, , N), we obtain a set of equations: /iG n + / 2 G 12 + + IN G lN = -EA x ) 11^21 F / 2 G22 + ' + — —EztZj) (21) + / 2 GN2 + ' ' ' + Jn Gnn = —Ez (zn) which may be expressed in matrix form as and in compact notation by -[GJ|7J= -[£J (V / -I or V m _1 ) where m =1, 2, 3, N = 1,2,3,,. „JV Multiplying both sides of the equation by the distance Azt JC/J = (V) (22) (23) (24) 1 The piecewise sinusoidal distribution (section of a sine curve) is assumed to be somewhat more appropriate than a strictly triangular distribution, A piecewise sinusoidal distribution is used in Sec. 9-17, The differences between distributions are most significant for a small number of segments.
With a large number of segments the different distribution functions should all give equivalent results 394 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD IMMf we have on the left the product of an impedance (Z) and current (I) and on the right a voltage {V) as in an electric circuit equation (Ohm's law): [2„n][/„] = -[K„] (V) (25) Example. Current distribution and impedance of short dipole. To illustrate the appli-cation of the moment method (circuit equation) technique, use (16), (19) and (2t) to calculate the current distribution and input impedance of a perfectly conducting center-fed cylindrical dipole 0.1a long with a radius of 0.001A. We presume no knowledge of what the impedance or current distribution should be except that the current distribution is symmetrical.
Solution.
Referring to Fig. 9-27, the dipole is divided into 3 segments Az r = 0.033A long and each assumed to have a uniform current over each segment (pulse function) given by i lT S 2 and I 3 . Thus, N — 3 and (21) becomes + — E(z i ) f 1 G2 i 4 1 2 Gj2 4 f 3 G 2 j — “ £{2 2) (26) fjGji 4 l 2 G i2 4 I 3 G 33 = — E(2j) The upper and lower halves of the dipole are symmetrical so that ru - rlt = r13 — r i2 and = r3i . Also rM = r22 - r 33 , Accordingly, we need to evaluate (16) for only 3 distances, r ll!h r l2 and r 13 , between the source and observation points (point matching).
Figure 9-27 Short center-fed dipole Q.U long divided into 3 segments Azj — Az a — &z3 — 0.033A long.
9 1 THt MOMENT METHOD (MM) AND m APPLICATION TO A WIRE ANTENNA 395 Figure !WS Half of one dipole segment divided into 5 subsegments for calculations of C u in the example.
Introducing numerical values for r 12 = 0.G66A, a = 0.0OU and Az into (16), = -J377 ^cos 2n x 0,066 sin 2tt x 0.066) 13 8te 7 x 0.066 3 x W 4 ;2,x 0.066)[2-3(^J] + 4^x0.001^ = -25.8 -/1184 (Or') Similarly, G, , = - 20.0 + /52 700 (fl 2“ ’) G„ = -25.6 — /12800 (« 2“‘) 0.033/ (27) (28) (29) Since the integrand of (16) is sensitive to small changes in r when r is small (particularly in the region where r — y'7-5 a or less), G, , was divided into 5 sub-segments, as indicated in Fig. 9-28. The segment with the smallest r contributes negligible resistance to the dipole impedance but a large negative reactance. For larger r's there is a contribution to the resistance and the reactance becomes posi-396 ?
THE CYLINDRJCAL ANTENNA- THE MOMENT METHOD (MM) Figure 9-29 Stepped (pulse function) approximation and actual (triangular) current distribution on short center-fed dipole antenna.
rive. The change in sign of the reactance occurs at r = ^/O a. To perform the inte-gration [by summation of (19)], the contributions of the subsegments are added to obtain the value in (28), Likewise, G l2 was subdivided into 2 segments and the contributions added to obtain (29).
Introducing (27), (28) and (29) in (26) and multiplying by As ( = 0,033J.) we obtain Amperes Ohms Volts M0.66 - jl 739) + 1 2 {0.85 + >422) + ^(0.85 + j39) = V v (30) 9-17 SELF-IMPEDANCE, RADAR CROSS SECTION AND MUTUAL IMPEDANCE OF SHORT DIPOLES 397 solved with a small pocket calculator. It took the author an hour or so using one.
However, such calculations if done frequently are most appropriately accom-plished with a computer, especially one programmed to accommodate more seg-ments and a range of dipole parameters such as length and diameter The literature describing programs for doing this is extensive. The book Antenna Design Using Personal Computers by David M. Pozar 1 includes a number of programs on wire antennas. One of these called DIPOLE uses a standard moment method solution in a 254-step BASIC program which can handle a wide range of dipole dimensions.
The use of personal computers for antenna problems has been presented by Miller and Burke3 with discussions of the Numerical Electromagnetics Code (NEC) and its subset MININEC for wire antennas. Many additional references on computer programs are listed at the end of this chapter and in App, B.
The next section includes 3 moment method examples which can be done on a pocket calculator.
J 1(0,85+;422) + - jll39) + 1 (0.85 +>422) = V 2 (31) / ± (0.S5 + ;39) + J 2(0.85 + J422) + J 5(0.66 -;1739) = (32) By symmetry l 2 = / 3 ; also for a center-fed dipole V 1 = - 0 so (30) and (32) are identical. Introducing the condition that Fj = 0 in (30) yields a current ratio of j- - 0.25 + /0.0002 (arts.) f 2 Putting this ratio in (31), dividing by / z and setting V 2 — 1 V yields an input imped-ance of y Z = -5-K+jX = l,l-jl528 Il (arts.) (33) 1 For a short center-fed dipole the current distribution is nearly triangular so that the pulse function current ratio fJJ 2 should be ] as suggested in Fig. 9-29.
Ideally, the moment method should yield this value. If this ratio (-J) is substituted in (3 1 ) we obtain Z = 1,2 -;1458Q (34) For a dipole of the same length (0,1 X) but only the radius (0.000 U), Richmond’s convergence value of Z = 1.852 — j 1895 O (see Prob. 9-6).
It is not expected with the few segments used and the assumptions made that high accuracy would be obtained. The purpose of the example is to illustrate the moment method with actual numerical values for a simple case which can be 9-17 SELF-IMPEDANCE, RADAR CROSS SECTION AND MUTUAL IMPEDANCE OF SHORT DIPOLES BY THE METHOD OF MOMENTS By Edward H. Newman.3 In this section the moment method (MM) with piecewise sinusoidal current modes is used to calculate the current distribution, input impedance and radar cross section of a short dipole The mutual coupling of two adjacent short dipoles is also calculated. Using a simplification of the mutual impedance equations of Howard E. King4 the calculations can be done on a scientific hand calculator in an hour or so. Thus, the MM can be illustrated without resorting to a digital computer and taking the time and effort to write an appropriate computer program.
As in (9-16-17), the dipole current can be approximated by the series expansion iv) = i (i) = i where FH(z') is a piecewise sinusoidal mode. For example, the dipole may be divided into 4 equal segments of length d = 1/4 as shown in Fig. 9-30. Segment n 1 Artech House, 1985.
1 E. K. Miller and G. J. Burke, “ Personal Computer Applications in Electromagnetics,” /£££ Antj, Prop. Soc „ Newsletter, 5-9, August 1983.
J EJectroScicnce Laboratory, Ohio State University.
4 H.
E. Ring, “Mutual Impedance of Unequal Length Antennas in Echelon,” /£££ Trans.
Ants.
Prop., AP-5, 306-3 1 3, July 1957.
5 We assume that the dipole is perfectly conducting and that the surrounding medium is free space.
398 D THE CYLINDRICAL ANTENNA: THE MOMENT METHOD (MM] extends from zn to zw+ ^ The piecewise sinusoidal modes are placed on the dipole in an overlapping fashion, with mode n existing on segment rt and n + 1. Mode n has endpoints and zH + 2 > and center or terminals at zH + l . FK is a filament of electric current, located a radius a from the wire centerline (i,e., on the surface of the wire) and with current sin ft(d — \ z — zn+ !
| ) where p — Injh Fn(z) is zero at its endpoints and rises sinusoidally to a maximum at its center with terminal current of /b0 = Fj(z n + 1 ) ^ 1 A. Note that the piecewise sinusoidal modes produce a current which is continuous and also zero at the dipole end’ points.
Except at the dipole endpoints, the dipole current of (1) at zH+l is fB amperes. Equation (1) produces a sinusoidal interpolation of the current values at the JV + 2 points.
We require that the radiated and impressed fields satisfy (9-16-12). Substi-tuting (1) into (9-16-12) we have - £ /„ E zll ~ Ez .
(3) where E zn — free-space z component of the electric field of FK j E z , = z component of the incident field \ \ I E zn is available in terms of simple functions given by Schelkunoff and Friis and i King.
2 j The weighting functions in the MM solution are chosen identical to the j expansion functions, except that they are located along the centerline of the dipole. This is because we enforce (9-16-12) on the centerline. Then multiplying j both sides of (3) by the sequence of N weighting functions, Fm (m = 1, 2, . .., N), : (3) becomes an N x JV system of simultaneous linear algebraic equations which l can be written compactly in matrix form as j [Z]I = V (4) ; Here / is the current column vector whose N components contain the /„ of (1), [Z] is the N x N impedance matrix whose typical term is \ zm„= -E„Fm d ( 5) ; Jm In general [Z] is dependent on the geometry and material composition of the scattered but not on the incident fields, A typical element of the right-hand -side 1 S. A. Schelkunoff and H. T. Friis, Antennas Theory and Practice, Wiley, 1952, 2 H.
E. King, “Mutual Impedance of Unequal Length Antennas in Echelon,” IEEE Trans , Ants.
Prop,. A P-5, 306-313, July 1957.
9 17 SELF-IMPEDANCE RADAR CROSS SECTION AND MUTUAL IMPEDANCE OF SHORT DIPOLES 399 Figure 9-30 Throe piecewise sinusoidal dipole modes on a dipole divided into 4 equal segments for Example 1.
or voltage vector V is given by V„ = I Ez . F„ dz (6) Jm The integration in (5) and (6) is on the dipole centerline, and over the extent of Fn> that is, from z = zm to zm + 7 , The dimensions of the elements of [Z] and V are volt-amperes (VA), while the elements of I are dimensionless. If the ZM were divided by /w0 /B0 , then the Z^ would have dimensions of ohms (Q). Since in our case the modal terminal currents are In0 = 1 A, Zmn can be considered to have the dimensions of ohms. In any case, the [Z] matrix is usually referred to as an impedance matrix and V as a voltage vector since the matrix equation (4) resem-bles an /V-port generalization of Ohm’s law.
The major problem in an MM solution is usually the evaluation of the elements in the impedance matrix. Typically this involves numerical integrations and/or the evaluation of special functions. As a result, most MM solutions are done on a digital computer and require a great deal of programming time and effort. For this rtason, most MM solutions are not suitable as a simple example problem which can be accomplished in about an hour using only a hand calcu-lator. Relatively simple expressions for the elements in the dipole MM impedance matrix are presented here, thus eliminating the need for a digital computer to carry out the MM solution to the examples given.
For the dipole antenna, the elements in the impedance matrix, as given by (5), are the mutual impedances between parallel piecewise sinusoidal dipole modes.
Figure 9-31 shows two parallel piecewise sinusoidal dipole modes of length Id. The bottom of weighting mode m is located a distance h above the center of expansion mode n, and the modes are staggered by the distance r. For convenience, the expansion mode has its center at z = 0. Exact expressions for 400 9 THE CYLINDRICAL ANTENNA: THE MOMENT METHOD (MM) Figure 9^31 Geometry for the mutual impedance of 2 parallel piecewise sinusoidal dipoles.
the mutual impedance between these modes has been given by King.
1 King's expressions are very lengthy and also require the evaluation of sine and cosine integrals. In order to simplify King's expressions we assume that the modes are electrically small and electrically close. Specifically, if we assume that fid, fih and fir are all < 1, it may be shown that King's expressions for the mutual impedance between modes m and n reduce to 2 (7) R mK = 20(fid) 2 X™= -pd [-4/4 + f>B ~ AC + D + £ + 4/iln(2v4 + 2h) -6(d+h) In (2fi + 2h + 2d) + 4(2<i + h) In (2C + 2ft + 4d) + (d-h) In (2D + 2h-2d)~ (3d + h) In (2E + 2h + &/}] where A = ,/ r 2 + h1 B = yV2 + (rf + ft) 2 C = + (2<f + ft) 2 D = yV + (< ~ A) 2 £ = xA + (3rf + h) 1 H, E. King, "Mutual Impedance of Unequal Length Antennas in Echelon," IEEE Trans. Ants.
Prop., AP-S, 306-313, July 19S7.
I appreciate the assistance of Linda Bingham in obtaining these simplified expressions.
9-1 T SELF-IMPEDANCE, RADAR CROSS SECTION AND MUTUAL IMPEDANCE OF SHORT DIPOLES 401 Equations (8) and (9) are suitable for a hand calculator, since they involve no operations more complicated than a logarithm and square root. Note that R is the well-known formula for the radiation resistance or a short dipole, and" is independent of mode separation. Equation (9) can be further simplified if we assume r = a < d (wire radius much less than the segment length) and also con-sider certain special values of ft. For self-impedance terms, m = n, and = -4 = - ^ [~ 4 + 4 lnQ ( 10) For adjacent modes with one overlapping segment, | m - n \ = 1 and ---sb” e?)] If | m — n | — 2, then the modes share a single point and = <0 = - y d [- 0.68] ( 12) If | m — n | > 3, then the modes are not touching and Y < “I, »4(2d + ft)4 , (2d + ft)(ft - d) 1 Pd Id " (d + hf(h - d%3d + ft) + " (d + hf(3d + ft) 3J ( ' 3) Now, consider the evaluation of the right-hand-side vector V, As seen in (6), V is dependent upon the z component of the incident electric field.
First, consider the case where the dipole is excited by a voltage generator , The simplest, and probably the most commonly used, model for a voltage gener-ator is the so-called delta-gap model.
1 A delta-gap generator is one that creates an extremely large, but highly localized, electric field polarized parallel to the wire centerline. A delta-gap generator located at z = z' has the incident field Ez , = v S{z - z f ) (14) where v = generator voltage d(z) - unit area Dirac delta function Normally the generators are placed at the center or terminals of the piecewise sinusoidal mode£ Thus, referring to Fig. 9-30 for a dipole with N + 3 modes, the generator could be placed at z — z 2 , z3 or z4 , which would be at the terminals of modes 1, 2 or 3 respectively Inserting the incident field from (14) into (6) shows that, if a delta-gap generator of vm volts is placed at the terminals of mode m, then K, = v„ U5> 1 W. L. SUitzman and G. A Thiele, Antenna Theory and Design, Wiley, 1981, chap, 7.
402 9 THE CYLINDRICAL ANTENNA, THE MOMENT METHOD (MM) Element m of V is nonzero only if a nonzero generator is placed at the terminals of mode m.
, .
Now consider the effect of placing a lumped load m the wire, A lumped load of ZIm ohms, placed at the terminals of mode m, will produce a voltage of — Im Zim volts at these terminals.
If we treat this voltage as a dependent delta-gap generator, then according to (15) we should add — Im Ztm to V m . However, this is an unknown voltage, since initially Im is unknown. Since it is conventional to write all unknowns on the left-hand side of the matrix equation, we add i m Zlm to both sides of row m of the matrix equation. Thus, it can be seen that a lumped load of Z Jm ohms placed at the terminals of mode m simply results in being replaced by Zmm + Zim .
There is no physical break or gap in the wire where a generator or load is placed. Thus, the current is continuous through generators and loads. However, there is a slope discontinuity, or jump in the derivative of the current, at the generator or load. Note that the piecewise sinusoidal modes account for this behavior by enforcing continuity of current on the wire and by allowing a slope discontinuity at their terminals.
Next consider the situation where the wire is excited by a normally incident plane wave.
If a z-polarized plane wave is incident from the +x axis with magni-tude E0 , then Ex , = E 0 e x ( 16> Inserting (16) into (6) and integrating yields Example 1 Dipole current distribution and input impedance. Compute the current distribution and input impedance of a center-fed dipole antenna. The dipole length l = A/10 and radius a = A/10000.
Solution As illustrated in Fig. 9-30 wc use N = 3 piecewise sinusoidal inodes on the dipole and segment the dipole into N + 1 = 4 equal segments of length d — f/4 = 0.025A. In this case the 3 x 3 MM matrix (4) can be explicitly written as Since the current on the center-fed dipole is symmetric, /] = / 3 ^ In this case, we can add column 3 of the matrix equation to column 1 and reduce the order 3 matrix equation to the order 2 matrix equation L1576 0,4885 + /132.2 Reducing the order of the matrix equation from 3 to 2 greatly reduces the effort in the hand calculations required to solve the matrix equation.
Although [Z] in (18) and (19) contain 9 elements, only 3 are distinct, since from the symmetry of the dipole, Zii — — Z 33 (20) Zli — %2l — Z2 3 — (21 ) Z, 3 = Z 31 (22) The real part of each ZmK is given by (8) as = 0.4935 VA (23) The imaginary part of the can be computed from (9); however, here we choose to use the simpler forms of (10), (1 1) and (12). Table 9-2 shows the etements in the first row of the [Z] matrix of (18) computed by (7), (8), (10), (11) and (12) and by King’s exact expressions. Note that the approximate values of [Z] are within 11 percent of the exact values.
If the approximate values of | /.~\ from Table 9-2 are substituted into (19) we obtain 0.9869 -;3324 0 0.9869 +j‘3506 0 1.4935 +jl753Y/ 1 "l TO" 1.4935 — j3454j|_/ aJ ~ [l where we have set V 2 = 1 VA since there is a 1-V generator at the terminals of mode 2. Equation (24) can be easily solved using Cramer’s rule. The results for the ele^ ments in the dimensionless current vector are / 1 = 0.0003286 /89.892° (25) l 2 = 0.000 623 0 /89-926° (26) Dividing (26) by (25), the current ratio l 2jl v is very nearly equal to 1.9, indicating a nearly triangular current distribution on the dipole.
The dipole current in amperes can now be obtained by inserting these coeffi-cients into (1) with N — 3. Thus, the dipole input or terminal current is I 2 amperes.
The input impedance is given by the ratio of the input voltage to the input current; that is, zjn = — = 1083 - j'1605 Q fans.) (27) 404 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD (MM) By contrast, if we were to use the exact values of [Z] from Table 9-2, then the results for the current distribution and input impedance would be = J 3 = 0.0002498 /90.0 (28) I 2 = 0.000521 9 /S9.9" (29) Zin = 1.892 -j\ 916 ft (ians .) (30) Example 2 Scattering from a short dipole. Compute the radar, cross section 1 of the same dipole considered in Example 1 for a wave at normal (broadside) incidence with the dipole terminated in a (conjugate) matched load.
Solution. To terminate the dipole in its conjugate matched load, we place a lumped load of Zf n at the center of the dipole, i.e., at the terminals of mode 2. Using the value of Zin from (27), Zi2 = = 2.083 + >1605 ft (31) The impedance matrix for the loaded dipole is identical to that of the unloaded dipole except that we add Zf2 to the self-impedance of mode 2 to obtain Z 22 = (0.4935 — >3454) + (2,083 + >1605) = 2.576 ->1849 VA -(32) If the incident electric field is a unit amplitude z-polarized plane wave incident from the + x axis, then the elements of the right- hand-side vector are identical and given by (17) with £0 = 1: V m = 0.02505 VA, m = 1, 2, 3 (33) Since the excitation and loading of the antenna are symmetric with respect to the center of the dipole, the current on the dipole remains symmetric. Thus, the current vector can still be computed from the order 2 matrix (19): p0.9869 — j3324 0.4935 +;1753T/, i.0.9869 +J3506 2.576 — jlS49 M / 3 I 7, (“0.025 05] /J~Lo02505j Equation (34) can be solved using Cramer’s rule. The results for the elements in the dimensionless current vector are — 1$ — 0.006515 /assr /, = 0.01235 /0.548 c The scattered field is given by % - Z where Etm is the free-space electric field of expansion mode Fn as in (3). For a field point on the +x axis (i.e., in the backscatter direction) and in the far zone of the See Sec. 17,5, 17 SELF-IMPEDANCE, RADAR CROSS SECTION AND MUTUAL 1MPEDANCE - OF SHORT DIPOLES 405 dipole the electric field of Fn will be z polarized and given by — >60 tan Using (38) and the above values for the f„ , the far-zone backscattered electric field is £, = 0.1199 /906" (Vm’ 1 ) The radar cross section of the dipole is |£ \ 2 ° ~ nxl TtV ” ians.) (40) !
I By contrast, if we were to repeat this example with the exact value of [Z] from Table 9-2, the results would be h =I 3 = 0.006199 /0W (41) l z = 0,01295 /OiF (42) a = 0.18012 2 (ans.) (43) From (2-20-5) the maximum effective aperture of a short matched lossless dipole is given by = —x2 = o.\m 2 provided that the dipole length l a. The total scattering aperture equals A^ and the radar cross section is this value times the short dipole directivity D( = 1.5)or 1.5 x 3 .
.
= DA^ = — a 2 = 0.1 79/ 2 oTT as compared to 0.1806 and 0.1801/, 2 above.
Example 3 Mutual impedance of 2 short dipoles. Compute the mutual impedance of 2 short side-by-side dipoles separated by A/100 as in Fig. 9-32. The dipoles are identical to the ones in Examples 1 and 2.
Solution, To simplify the computations, we place only one piecewise sinusoidal mode on each dipole. Thus, the order N = 2 matrix equation for this example is (46) Only Z n and Z l2 need be computed, since from the symmetry of the dipoles zn = Z2Z and Z l2 = Z 2i . Z iL is evaluated from (7), (8) and (10) with d = — h -1/2 — 0.05/, and r — a = 0,0001a. Z l2 is evaluated from the same equations with a replaced by s = Q.QU, since h = -rf. The results for Zu and Z 12 are shown in 406 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD {MM) Dipole Dipole 1 2 Figure 9-31 Geometry for 2 coupled dipoles of Example 3, The radius of each dipole is O.OU 0.000U, Table 9-3 where they are compared with the exact values of King 1 and of Rich-mond. 2 For this simple one mode per dipole solution, Z 12 is the mutual impedance between the two dipoles.
Inserting the approximate values from Table 9-3 into (46) we obtain [l.9739-;i992 15739 - /232.8TM Til (47) [l.9739 -j232.8 1.9739 -/1992 ±I 2] |a1 In (47) we set K, = 1, since to compute the input impedance of dipole 1, we place a 1-V generator at the terminals of mode 1. Equation {47} has the solution 1 1 = 0.0005090 /89,955 p 1 2 = 0.000059 49 / — 89.616° The input impedance is then zta = 3- = 1539 -j'1964n '1 The N = 1 mode solution for the input impedance of dipole 1 alone is 2^ = Z 1X = 1.974 -/1992 ft Table 9-3 Elements of the [Z] matrix (VA) for Example 3 Element Approximate Exact Z, t 1.9739 - /1992 2,0000 - /1921 Z l2 1.9739 - /232,8 1.9971 ->325-1 1 H. E, King, "Mutual Impedance of Unequal Length Antennas in Echelon," IEEE Trans.
Ants.
Prop., AP-S, 306-313, July 1957.
2 J. H, Richmond, “ Radiation and Scattering byThin-Wire Structures in a Homogeneous Conducting Medium," IEEE Trans. Ants. Prop., AP-2Z, 365, March 1974.
ADDITIONAL REFERENCES 407 Table 9-4 Summary of self-impedance, mutual impedance and radar cross section for short dipolest Modes Z t — 1 .974 — ;1992 ft N = 1 (approx. King) — 2,000 — > 1 92 1 Q N = l (exact King) = 2,083 — }1605 ft N = 3 (approx. King) = 1.892 — >1916 ft N = 3 (exact King) = 1.864 —>1905 ft A1 = 5 (exact King) = 1.856 — > 1 899 ft /V = 7 (exact King) Z m = 1.974 - j'233 ft N = 1 (approx. King) .= 1,997 ~}325 ft N = l (exact King) = Zi — Z m = 0.000 — jl 759 ft N = 1 (approx. King) = 0.003 — >1 596 ft /V = 1 (exact King) ff = 0.1806a 2 (approx. King) = O.lSOl^ 2 (exact King) = Q.179^ 2 (short dipole theory. Chap. 2) Z t -self' impedance = input impedance of isolated dipole, fl Zm - mutual-impedance, 11 a = radar cross section, X1 Dipole length, t = a/ 10 Dipole radius, a = a/ 10000 Dipole separation, s = a/ 100 (for ZJ Note that Z, should approach the Irge value as Af increases provided the solution converges, also that the self' mutual difference ZA + 0 as s - 0 and Z4 - Z r as s — oo {Zm -> 0).
If the exact values of Table 9-3 are used, the impedance of dipole 1 in the presence of dipole 2 is ZiB = 1.382 -/1822 ft (52) Table 9-4 gives a summary of self-impedance, mutual impedance and radar cross section for short dipoles.
ADDITIONAL REFERENCES FOR CHAP. 9 Andrcasen, M. G.: "Scattering from Parallel Metallic Cylinders with Arbitrary Cross Sections," IEEE Trans. Ants. Prop ., AP-12, 746-754, November 1964.
Andreasen, M. G.: "Scattering from Bodies of Revolution by an Exact Method" IEEE Trans , Ants.
Prop., AP-13, 303-310, March 1965.
Balanis, C. A.: Antenna Theory : Analysis and Design, Harper and Row, 1982: Burke, G. J.: The Numerical Electromagnetics Code (NEC), The SCEEE Press, 1981.
Elliott, R, $.: Antenna Theory and Design, Prentice-Hall, 1981.
Kennaugh, E. M.; M Multipole Field Expansions and Their Use in Approximate Solutions of Electro-magnetic Scattering Problems,". Ph.D. dissertation, Ohio State University, 1959.
Mei, K. K., and J. G, Van Bladel: M Scattering by Perfectly Conducting Rectangular Cylinders,” /£££ Trans. Ants. Prop^, AP-1I, 185-192, March 1963.
Newman, E, H.: Simple Examples of the Method ofMoments in Electromagnetics, pending publication.
Equation (51) Z i; (exact). Table 9-3 (27) (30) T H, Richmond, FORTRAN IV program (see References below) Z x 2 (approx.), Table 9-3 Zi2 (exact). Table 9-3 (51) — Z 12 (approx.) Zu (exact) — Z i2 (exact) (40) (43) (45) 408 9 THE CYLINDRICAL ANTENNA; THE MOMENT METHOD (MM) Richmond, J. H + ; “Scattering by a Dielectric Cylinder of Arbitrary Cross Section Shape,” IEEE Trans. Ants. Prop AP-13, 334-341, May 1965, Richmond, J, H.: “Scattering by an Arbitrary Array of Parallel Wires/ 1 IEEE Trans. Microwave Theory and Tectu MTT-13, 408-412, July 1965.
Richmond, J. H.: “Digital Computer Solutions of the Rigorous Equations for Scattering Problems, 1 Proc. IEEE, 53, 796-804, August 1965.
Richmond, J. H,: “ Radiation and Scattering by Thin-Wire Structures in a Homogeneous Conducting Medium " FORTRAN IV program giving current distribution, impedance, gain, patterns, scat-tering cross section and other parameters for dipoles, loops, plates, spheres, cones, and wire grid models of aircraft and ships, AS1S-NAPS Document NAPS-02223, 1 IEEE Trans. Anw.
Prop-, AP-22, 365, March 1974, Stutzman, W, L,, and G- A. Thiele: Antenna Theory and Design, John Wiley, 1981.
Tsai, L, S„ and C. E Smith: “Moment Methods in Electromagnetics for Undergraduates,” IEEE Trans. Edac. t E-ll, 14—22, February 1978, Additional references are in App. B.
PROBLEMS 9-1 HaHen’s equation.
( then an equal current (in both amplitude and phase) will he obtained at the terminals of antenna A if the same emf is applied to the termi-nals of antenna B> It is assumed that the emfs are of the same frequency and that the media are linear, passive and also isotropic. An important consequence of this theorem is the fact that under these conditions the transmitting and receiving patterns of an antenna are the same. Also, for matched impedances, the power flow is the same either way.
As an illustration of the reciprocity theorem for antennas, consider the fol-lowing two cases.
Case 1. Let an emf V 9 be applied to the terminals or antenna A as in Fig, 10-2a. This antenna acts as a transmitting antenna, and energy flows from it to antenna B, which may be considered as a receiving antenna, producing a current J at its ter-minals.
1 It is assumed that the generator supplying the emf and the ammeter for measuring the current have zero impedance or, if not zero, that the generator and ammeter impedances are equal.
Case 2.
If an emf V b i# applied to the terminals of antenna 5, then it acts as a 1 Although the emf V m and the current Jh are scalar space quantities, they are complex or vector quantities with respect to time phase. The term “phasor” is sometimes used to distinguish such a quantity from a true space vector.
412 10 SELF AND MUTUAL IMPEDANCES transmitting antenna and energy flows from it to antenna A as in Fig. 10-2b, producing a current J fl at its terminals.
Now if V b = V m> then by the reciprocity theorem Ia = Ib .
The ratio of an emf to a current is an impedance. In Case 1, the ratio of V a to I b may be called the transfer impedance Zabt and in Case 2 the ratio V b to Ia may be called the transfer impedance Zba . Then, by the reciprocity theorem it follows that these impedances are equal Thus, In order to prove the reciprocity theorem for antennas, let the antennas and the space between them be replaced by a network of linear, passive, bilateral impedances. Since any 4-terminal network can be reduced to an equivalent T section, 1 the antenna arrangement of Case 1 (see Fig. 10-2n) can be replaced by the network of Fig.
10-3 a.
The current through the meter is where h = I i Z 2 + Z 3 (2 ) K , _ K(z2 + z3 ) Zj T[Z 2 Z^Z 2 + Z 3)] Z 1Z 2 + Z 2 Z i ^Zi Z 1 (3) 1 This is true insofar as the amplitude and phase of the input voltage and output current art con-cerned.
10 3 SELF-IMPEDANCE OF A THIN LINEAR ANTENNA 413 Introducing (3) into (2) yields the current through the meter in terms of the emf V a and the network impedances. Thus, z 1z 2 + z 3 z 3 + z 3 z 1 ' ' If the locations of the emf and current meter are interchanged, as in Fig.
10-36, we obtain fl ZfZ 2 + Z 2 Z 3 + Z 3 Zj v ' Comparing (4) and (5), it follows that if V a = V h then l 0 -I b , proving the theorem.
10-3 SELF-IMPEDANCE OF A THIN LINEAR ANTENNA, In Sec. 9-16 we discussed how the current distribution and self-impedance of a dipole antenna can be obtained using the moment method . In this section we consider an induced emf method used by Carter 1 in 1932, decades before the general availability of powerful computers. Only the self-impedance can be deter-mined with the induced emf method, the current distribution being assumed at the outset.
Since measurements indicate that the current distribution on thin dipoles is nearly sinusoidal, except near current minima, this form of distribution is assumed. It results in satisfactory values for dipoles with length-diameter ratios as small as 100, provided the terminals are at a current maximum.
Consider a thin center-fed dipole antenna with lower end located at the origin of the coordinates as shown in Fig.
10-4. The antenna is situated in air or vacuum and is remote from other objects. Since the antenna is thin, a sinusoidal current distribution will be assumed with the maximum current I L at the termi-nals. Only lengths L which are an odd multiple of A/2 will be considered so that the current distribution is symmetrical, with a current maximum at the terminals.
The current distribution shown in Fig. 10-4 is for the case where L = A/2. The current at a distance z from the origin is designated I2 . Then / r = I t sin fiz ( 1 ) Suppose that an emf V Li applied to the terminals of the antenna of Fig, 10-4 produces a current I z at a distance z from the lower end. The ratio of Vll to Iz 1 P.
S. Carter, 11 Circuit Relations in Radiating Systems and Applications to Antenna Problems,” Proc. IRE, 20, 1QG4-1Q4£ June 1932 A. A.
Pistolkors, "The Radiation Resistance of Beam Antennas,” Proc. IRE , 17, 562-579, March 1929.
R. Bechmann, “Calculation of Electric and Magnetic Field Strengths of Any Oscillating Straight Conductors,” Proc, IRE, 19, 461-466, March 193 1-R r Bechmann, “On the Calculation of Radiation Resistance of Antennas and Antenna Com-binations " Proc. IRE, 19, 1471-1480, August 1931.
414 10' SELF AND MUTUAL IMPEDANCES dt 1 Figpf MM Center-fed linear X/2 antenna.
may be designated as the transfer impedance Z tl .Thus, Next let the applied field at the antenna and parallel to it be Ex . This is the field produced by the antenna's own current as though it were flowing in empty space. This field induces a field Ezi at the conductor such that the boundary conditions are satisfied. For a perfect conductor these conditions are that the total field Ezt is zero or that Est = Ex + Eti = 0 and therefore Exi = —Er . The emf dV z produced by the induced field over a length dz is —Ez dz or dV^.-E.dz (3) If the antenna is short-circuited, this emf will produce a current dl t at the termi-nals. Then the transfer impedance Zxl is given by Since the reciprocity theorem {Sec.
10-2) holds not only for 2 separate antennas but also for 2 points on the same antenna, it follows that the transfer impedances of (2) and (4) are equal. Therefore, h± = z =z =^ =^£ /, Zu Zxt dl , dl , (5) and V„ dl, = 1,EI dz (6) The terminal impedance Z,, of the antenna is given by the ratio of V t , to the total terminal current Thus, Z“‘ I, (7) 10-3 SELF-IMPEDANCE OF A THIN LINEAR ANTENNA 415 The impedance Z 1: is a constant and is independent of the current amplitude.
This follows from the fact that the system is linear. Therefore, Zn can also be expressed as the ratio of an infinitesimal emf dV^ 2 at the terminals to an infinitesi-mal current dl j at the terminals, or from which Substituting (9) into (6), Integrating (10) over the length of the antenna, we obtain v„ dV, i m /l dl.
dl, = 1, dV„ m -/ - z Ez dz (10) = - f m Jo where is the emf which must be applied at the terminals to produce the current at the terminals. The terminal impedance Zn is then y 1 ft Zn = _ii = IiEidz 02) 1 1 t Jo Since the antenna is isolated, this impedance is called the self-impedance. In (12) Ez is the z component of the electric field at the antenna caused by its own current.
It wilt be convenient to indicate explicitly this type of field by the symbol £n in place of E t . Introducing also the value J x from (1) into (12), we obtain for the self-impedance E.i = -f [\i J 1 Jo sin ftz dz To evaluate (13), it is first necessary to derive an expression for the field £ lt along the antenna produced by its own current. Substituting this into (13) and integrating, it is possible to obtain an expression which can be evaluated numeri-cally. The steps in this development are given in the following paragraphs.
If expressions can be written for the retarded scalar potential V due to charges on the antenna and for the retarded vector potential A due to currents on the antenna, then the electric field everywhere is derivable from the relation E = -VV^jaA (14) More particularly, the z component of E is given by dV ( 15 ) 416 10 SELF AND MUTUAL IMPEDANCES Referring to Fig. 10-5, let the antenna be coincident with the z axis. A point on the antenna is designated by its distance z x from the origin. A point P in space is given in cylindrical coordinates by p, z. Other distances are as shown. Only lengths L which are an odd multiple of k/2 will be considered. Thus, where n = 1, 3, 5, .
..
The scalar potential V at any point is given by where p is the volume charge density, r the distance from the charge element to the point and dr is a volume element. From Fig, 10-5, r = Jp 1 + (z - Zi) 2 In the case of a thin wire of length L, (16) reduces to where pL = linear charge density on the wire The vector potential A at any point is given by where J = the current density 10-3 SELF-IMPEDANCE OF A THIN LINEAR ANTENNA 417 In the case of a thin wire (18) reduces to a _ f L !l± r 4ie Jo r where / ri = the current on the wire By the continuity relation between current and linear charge density The current on the antenna is assumed to have a sinusoidal distribution as given by (1).
Introducing the retarded time factor, we have for the retarded current I ZI = it sin (21) Substituting (21) into (20) and performing the indicated operations, the retarded linear charge density is Pl = COS jSz , e “ Wr,J (22) cu Introducing (22) into (17) and noting that p/w = 1 fc t the retarded scalar potential is (23) 4ns0 c J0 r Likewise, introducing (21) into (19), the z component of the retarded vector potential is J f sin Pz x e iPr j A =— — — dz 1 4 Jo r By de Moivre’s theorem.
cos fizt = He 11 +e^ Jfizt ) sin Pz l = ™(e^11 — Making these substitutions in (23) and (24), V = } -^~\ i ±£_ iz x (27) &7Z£0 C Jo r~ it, I fjvl fL --jAlu+r} _ JRn-r) and (28) 8« 1 r Equations (27) and (28) give the retarded scaflar and vector potentials caused by current on the antenna with the assumed sinusoidal distribution. Sub-418 10 SELF AND MUTUAL IMPEDANCES stituting these equations into (15) yields an expression for the z component of the electric field everywhere. Thus, jl ! e1™1 CL d /e~mil+r) + eif,I, ~ r}\ E~ 8?tE0 C jo dz \ T ) dZl a)Ho he 1 fL + 8 1 { ; ) b ' E _ _ jl,e^ te-"-1 + r 4ke0 c\ r 2 r 2 ) r i = \fp 2 + z 1 {31) and r 2 = p 1 + (L - z) 1 (32) The factor l/4n&0 c s; 1 20tt/4jc = 30. Also putting the time factor equal to its absolute value = 1, Eq. (30) becomes— +—> At the antenna (31) and (32) become and r2 = L - z (35) Substituting these into (33) yields the value of the z component of the electric field £u at the antenna due to its own current. Thus, ~z + -r^r) Introducing (36) into (13) we obtain the self-impedance Z n of a thin linear antenna an odd number of 2/2 long. Hence, rt / -jpt -jPiL - xl\ Z tl =j’30 J ^ — 1——— J sin Pz dz Applying de Moivre’s theorem to sin 0z, , _ 15 fT^" 1 ^ ifL{eJ2,z -1) ~ 11 Jo L r L - z J For L = nkjl where n = 1, 3, 5, ...,e' ln = c“ J’" - -1, so that (38) becomes r i 1 - e ~ i2tl fL 1 - eilf ‘ Iu = 15 dz + 15 I V^— Jo 2 Jo £ z 10-3 SELF- IMPEDANCE OF A THTN LINEAR ANTENNA 419 In the first integral let u = 20z or du — 20 dz The upper limit z = L becomes u = 20L = 2nn t while the lower limit is unchanged. The first integral then transforms to In the second integral let v — 2P(L — z) or dv — — 20 dz The upper limit becomes zero while the lower limit becomes 2 nn, The second integral then transforms to ro t _ i?) ri h i _ — 15 I -- - - --dv-15 i—L hn » Jo » Equations (41) and (42) are definite integrals of identical form. Since their limits are the same, they are equal Therefore (40) becomes If we now put w = ju , (43) transforms to J 2 i _ €— du o u ransforms to J }2 xn l P ~ w dw O » The integral in (44) is an exponential integral with imaginary argument It is designated by Ein (jy). Thus, 1 CJj 1 - e“ w Em O'y) = < Jo w In our case y = 2nn. This integral can be expressed in terms of the sine and cosine integrals discussed in Sec.
5-6 Thus, Ein {jy) = Cin (y) + j Si (y) (46) or Ein O'y) = 0-577 + In y — Ci (y) + j Si (y) (47) Hence, the self-impedance is Zn = Ru 4- jX^ = 30[Cin (Inn) + j Si (2tm)] (48) or Zn = 30[0.577 + In (Inn) - Ci (21) +; Si (2nnj] (ft) (49) 1 See, for example, S. A, Schdkunoff, Applied Mathematics for Engineers and Scientists, Van Nostrand.
New York, 1948, p. 377.
420 10 SELF AND MUTUAL IMPEDANCES SELF-IMPEDANCE OF A THIN LINEAR ANTENNA 421 (ff) (b) Figure 10-6 a/2 and 3 X/2 antennas.
The self- resistance is R n = 30 Cin [2nn] = 30[0.577 + In {Inn) - Ci {InnJ] (ft) (50) and the self-reactance is X lt = 30 Si ilnn) (Q) (51) These equations give the impedance values for a thin linear center-fed antenna that is an odd number (n) of a/2 long. The current distribution is assumed to be sinusoidal (Fig, 10-6a). The values arc those appearing at the ter-minals at the center of the antenna.
In the case of a 2/2 antenna as shown in Fig, 10-6a, n = 1, and we have for the self- resistance and self-reactance Rn =30 Cin (2ir) (fi) (52) and X lx = 30 Si {In) (ft) (53) The value of (52) is identical with that given for the radiation resistance of a 2/2 antenna, in Sec. 5-6, Eq, (5-6-23), Evaluating (52) and (53), we obtain for the self-impedance Z n = R vv + /X n = 73 T/42.5 Q (54) Since X n is not zero, an antenna exactly 2/2 long is not resonant. To obtain a resonant antenna, it is common practice to shorten the antenna a few percent to make X , t = 0. In this case the self-resistance is somewhat less than 73 ft.
For a 32/2 antenna as shown in Fig, 10-66, n = 3, and the self-impedance is Z M = 30[Cin (6rc) + / Si (6tt)] or ZM = 105.5 + /45.5 ft (55) It is interesting that the self-reactance of center-fed antennas, an exact odd number of 2/2 long, is always positive since the sine integral Si {Inn) is always positive. For large n the sine integral converges around a value of nil (see Fig, 5-1 2b) which corresponds to a reactance of 47,1 ft.
It should be noted that for antenna lengths not an exact odd number of 2/2 the reactance may be positive or negative. However, the foregoing analysis of this section is limited to antennas that are an exact odd number of 2/2 long.
For large n, the self- resistance expression (50) approaches the value R Xi = 30[0.577 + In (2™}] l-L Ground 2 plane ^Li W) Figure ID-7 Stub antenna of length t at (a) and center-fed antenna oflength L at (b).
since Ci (2nn) approaches zero. Thus, the self-resistance continues to increase indefinitely with increasing n but at a logarithmic rate.
The more general situation, where the antenna length L is not restricted to an odd number of 2/2, has also been treated.
1 The antenna is center-fed, and the current distribution is assumed to be sinusoidal (see Fig. 5-8), The self-resistance for this case is R i i = 30^1 - cot 2 ^0 Cin 2fiL + 4 cot2 ^ Cin f}L + 2 cot ^ (Si 2pL - 2 Si £L)J (ft) (57) When the length L is small, (57) reduces very nearly to =5(pL) 1 (ft) (58) which is the same as (2-20-3) when I av = (1/2) / 0 , For the special case of L = n2/2, where n= 1, 3, 5, ...
, (57) reduces to the relation given previously by (50), The above discussion of this section applies to balanced center-fed antennas. For a thin linear stub antenna of height / perpendicular to an infinite, perfectly conducting ground plane as in Fig, 10-7n, the self-impedance is \ that for the corresponding balanced type (Fig.
10-76), The general formula (57) for self-resistance can be converted for a stub antenna above a ground plane by changing the factor 30 to 15 and making the substitution L = 21 The formulas (50) and (51) can be converted for a stub antenna with ground plane where the antenna is an odd number n of 2/4 long by changing the factor 30 to 15, Thus, for 1 G. H. Brown and R, King, “High Frequency Models in Antenna Investigations,” Proc, //EE, 22, 457-480, April 1934, # J. Labus, “Recherischc Ermittlung der Impcdanz von Antennen," Hockfrequet\ztechnik und Electrty-akustik y 17, January 1933.
(56) 422 10 SELF AND MUTUAL IMPEDANCES a 2/4 antenna perpendicular to an infinite, perfectly conducting ground plane, the | seif-impedance is ZM = 36,5 + J2\ £1 10-4 MUTUAL IMPEDANCE OF TWO PARALLEL LINEAR ANTENNAS. The mutual impedance of 2 coupled circuits is defined in circuit theory as the negative of the ratio of the emf V 2 i induced in circuit 2 to the current / x flowing in circuit 1 with circuit 2 open. Consider, for example, the coupled circuit of Fig. 10-8 consisting of the primary and secondary coils of a transformer. The mutual impedance Z 2l is then ( 1 ) where V 1X is the emf induced across the terminals of the open-circuited secondary by the current J x in the primary. The mutual impedance, so defined, is not the same as a transfer impedance such as discussed in connection with the reciprocity theorem in Sec. 10-2. In general, a tranter impedance is the ratio of an emf impressed in one circuit to the resulting current in another with all circuits dosed.
For example, if the generator in Fig. 10-8 is removed from the primary and is connected to the secondary terminals, the ratio of the emf V applied by this generator to the current I t in the closed primary circuit is a transfer impedance Z T . Thus, - = Zr (2) h This impedance is not the same as the mutual impedance Z 2l given in (1).
Instead of the coupled drcuit of Fig.
10-8, let us consider now the case of 2 coupled antennas 1 and 2 as shown in Fig. 109. Suppose a current I x in antenna | 1 induces an emf V 21 at the open terminals of antenna 2. Then the ratio of - V 21 to I !
is the mutual impedance Z1V Thus, If the generator is moved to the terminals of antenna 2, then by reciprocity the mutual impedance Z 12 or ratio of ” F12 to J 2 is the same as before, where V 12 l-) ± j Si (v'l (5) Thus, the mutual resistance is J? 21 = 30{2 Ci {M ~ Ci Wy/d 2 + 1} + Lfl - Ci mjd 2 + L 2 - Lfl} (11) (6) sin $z dz (3) m (4) Colhnear Staggered or in echelon [«) (b) U) Figure l(M0 Three arrangements of two parallel antennas.
105 MUTUAL IMPEDANCE OF PARALLEL ANTENNAS SJDE^BY-SIDE 425 and the mutual reactance is X2l = — 30{2 Sim - Si + L)] - Si [JS{Jd 2 + li - L)]} (ft) (7) where ^ 2 i = Z12 = Rn + JX.ii (8) and where L = tix/2 for n odd.
The mutual resistance and reactance calculated by (6) and (7) for the case of kjl antennas (L = A/2) are presented by the solid curves in Fig. HM2 as a func-tion of the spacing d. The mutual resistance R 2l is also listed in Table 10-1.
An integral-equation method for the calculation of the mutual impedance of linear antennas has been presented by King and Harrison 1 and by Tai. 2 The method is related to that discussed in Chap. 9. In this method the diameter of the antenfia conductor is a factor. By way of comparison, curves for the mutual resistance and reactance given by Tai are also shown in Fig. 10-12. The dashed curves are for a total length-diameter ratio (L/D) of 11 000 (very thin antenna) and the dotted curves for a ratio of 73.
In Table 10- 1 the quantity — R 3lJ which is important in array calcu-lations, is also tabulated. When d is small, it has been shown by Brown 3 that this quantity is given approximately by the simple relation where A = the free-space wavelength 1 R. King and C W Harrison.
Jr. p “Mutual and Self Impedance for Coupled Antennas," J , Appi Phys. y 15, 481-495, June 1944.
1 C T. Tai, “ Coupled Antennas," Pr&c. IRE, 36, 487^500. April 1948.
3 G. H. Brown, personal communication to the author, June 16, 19TB.
426 10 SELF AND MUTUAL IMPEDANCES Distance between antennas, d y Figure UM 2 Curves of mutual resistance (R it ) and reactance (£ ;| ) of two parallel side-by-side Linear //2 antennas as a function of distance between them, Solid curves are for infinitesimally thin antennas as calculated from Cartel formulas. Dashed and dotted curves between 0 and 1.0/ spacing are from Tai's data for antennas with L/D ratios of 1 1 000 and 73 respectively.
This relation is accurate to within \ percent when d <, 0.05/ and to within about 5 percent when d < 0.1a.
In the more general situation where the antenna length L is not restricted to an odd number of A/2, the mutual resistance and reactance are given by Brown and King 1 as - 4 cos 2 fy j^Ci ^ ( s/4drTH ~l) + Ci ^ (v"4 (10) 1 G. H, Brown and R. King, “High Frequency Models in Antenna Investigations," Proc. IRE, 22, 457-480, April 1934.
1 0-5 MUTUAL IMPEDANCE OF PARALLEL ANTENNAS SlDE BY-SlDE 427 Table 10-1 Mutual resistance versus spacing for thin center-fed side-by-side A/2 antennas (fiL = 180 n ), with sinusoidal current distribution Mutual Self minus resistance mutual resistance Spacing d r21,q 0.00 73.13 0.00 0,01 73.07 0.06 0.05 71.65 1.48 0,10 67,5 5,63 0.125 64.4 8.7 0.15 60.6 12.5 0.20 51.6 21.5 0.25 40.9 32.2 0.3 29,4 43.7 0.4 + 6.3 66.8 0.5 -12,7 85.8 0.6 -23.4 96.5 0.7 -24,8 97.9 0.8 -18.6 91.7 0.9 -7.2 80.3 JO + 3.8 69.3 1.1 + 12,1 61.0 1.2 + 15,8 57.3 1.3 + 12,4 60.7 1.4 + 5.8 67.3 1.5 -2,4 75.5 1.6 -8,3 Si.
4 1.7 -10.7 83.8 1.8 -9.4 82.5 1.9 -4.8 77.9 2.0 + 1.1 72.0 and ,ri'- 305^){-2(2 + cosWSi '''' + 4 cos 2 fy ^Si £ {^4d2 + L 2 - L) + Si | (J4d 2 + I? + L)J - cos pL[Si Ky/tf + I - L) + Si ppjd2 + + L)] + sin pL^Ci pi^/d 2 + I2 + L) - Ci p{Jd 2 + ii - L) - 2 Ci | {J4d 2 ^l3 + L) + 2 Ci ^ ^4d2 + L 2 -L)JJ (U) 01) 428 10 SELF AND MUTUAL IMPEDANCES 1 2 Ground ptane Figure IQ-13 Two coupled linear parallel stub antennas.
In the special case of L = nA/2, where n is odd, (10) and (1 1) reduce to the relations given previously by (6) and (7).
The above relations of this section apply to balanced center-fed antennas.
The mutual impedance of two stub antennas of height f = L/2 above an infinite, , perfectly conducting ground plane as in Fig. 10-13 is } that given by (6) and (7) or (10) and (11), These relations are converted to the ground-plane case by chang-ing the factor 30 to 15 and making the substitution L = 21.
10-6 MUTUAL IMPEDANCE OF PARALLEL COLL1NEAR ANTENNAS, Let each antenna be an odd number of A/2 long and arranged as in Fig, 10-10h. For the case where h is greater than L, Carter 1 gives the mutual resistance and reactance as R 2l = - 15 cos 0/^-2 Ci 2ph + Ci 20(h - L) + Ci 20(A + L) - In + 15 sin fih [2 Si 20ft - Si 20(ft — L) — Si 20(ft + L}] (£1) {1) and X lx - - 15 cos ph[2 Si 2fih - Si 2p(h - l) - Si 2 + L)] - L 2Y 2 Ci 2ph - Ci 2fi{k - L) - Ci 2p(h + L) — In ( —-— 1 (ft) (2) Curves for R 2L and X 1L of parallel collinear A/2 antennas (L = A/2) are presented in Fig. 10-14 as a function of the spacing s where s = h — L (see Fig. 10-10b), 10-7 MUTUAL IMPEDANCE OF PARALLEL ANTENNAS IN ECHELON- For this case the antennas are staggered or in echelon as in Fig. 10-10c, Each antenna is an odd number of A/2 long. The mutual resistance 1 P.
S, Carter, 11 Circuit Relations in Radiating Systems and Applications to Antenna ProbleniS," Proc. IRE, 20, 1004-1041, June 1932.
Spacing, s h Figure JG-M Curves of mutual resistance and reactance (A 21 ) of two parallel collinear infini-tesimally thin a/2 antennas as a function of the spacing s between adjacent ends.
and reactance of two such antennas are given by Carter 1 as H 21 = - 15 cos /}fK-2 Ci A - 2 Ci A + Ci B + Ci B' + Ci C + Ci C) + 15 sin fih(2 Si A — 2 Si A - Si B + Si B' - Si C + Si C) (H) (1) and X2l = - 15 cos 0h(2 Si A + 2 Si A — Si B — Si B' — Si C — Si C') + 15 sin 0h(2 Ci A - 2 Ci A — Ci B + Ci « — Ci C + Ci O) (U) (2) where A = 0^/d1 + h1 + h) A’ = P(^/d 2 + h2 - h) B = ptjd2 + (ft - L)2 + (ft - t)] « = JlVd2 + (ft - L? -(ft - Z,)] C = Pljd 1 + (ft + L) 2 + (ft + L)] C = frjd 2 + (ft + L) 2 ~(h + L)] Values of the mutual resistance in ohms as calculated from (1) are listed in Table 10-2^ as a function of d and h Tor the case where the antennas are A/2 long (L — A/2), as indicated in Fig, 10-15.
The staggered or echelon arrangement is the more general situation of which the side-by-side position (Sec, 10-5) and the collinear position (Sec. 10-6) are special cases.
1 P.
$. Carter, "Circuit Relations in Radiating Systems and Applications to Antenna Problems," Proc. IRE, 2Q, 1004—1041, June 1931 See also H. £. King, "Mutual Impedance of Unequal Length Antennas in Echelon," IEEE Trans. Ants. Prop., AP-5, 306-313, July 1957.
2 All but a few values are from a table by A. A.
Pistolkors, “The Radiation Resistance of ft™ Antennas" Proc. IRE , 17, 562-579, March 1929.
430 10 SELF AND MUTUAL IMPEDANCES 10-8 MUTUAL IMPEDANCE OF OTHER CONFIGURATIONS, There are many other antenna configurations for which the mutual impedance may be of interest. The variety is enormous but two will be mentioned and references given which the reader may consult for further information.
1 Parallel antennas of unequal height. This case has been treated by Cox.
1 His data apply specifically to stub antennas perpendicular to an infinite perfectly conducting ground, but can be used with symmetrical center-fed antennas of twice the length by multiplying the resistance and reactance values by 2 (see also Howard E. King 2 ), 2 V or skew antennas. Some antenna systems involve nonparallel linear radi-ators. The mutual impedance of such inclined antennas are readily calculated by the moment method as for example, by J, Richmond’s FORTRAN IV program ASIS-NAPS Document NAPS-G2223 (see References at the end of Chap.
9), For two short dipoles however a simple useful relation can be derived as follows.
Referring to Fig. 10-16, consider the 2 short center-fed dipoles i and 2 of length Lk and L\ separated by a distance rk with orientation angles 0 and O' as indicated. The mutual impedance Z21 is given by the ratio of the voltage F 21 induced in dipole 2 by the current / j flowing in dipole 1 . Then for Lk 1, L\ 1 and r A P I we have from (5-2-34), Z 21 = sin 6 e~^ fL\ sin (1) 1 C.
R. Cox T “Mutual Impedance between Vertical Antennas of Unequal Heights" Prof. IRE. 35 1367-1370, November 1947, 2 H. E, King, “Mutual Impedance of Unequal Length Antennas in Echelon" IEEE Trans.
Artts.
Protf. T AP-5 306-313, July 1957.
10-9 MUTUAL IMPEDANCE: OF OTHER CONFIGURATIONS 43 !
Table IQ-2 Mutual resistance as a function of d and h (Fig 10-15) for thin A/2 antennas in echelon Spacing h Spacing d o,o; 0.5/ 1.0/ 1.5/ 2.0/ 2.5/ 3.0/ 0.0/ + 73.1 + 26.4 -4.1 + 1.8 - 1.0 + 0.6 -0.4 0.5/ -12,7 -11.8 -0,8 + 0 8 -1,0 + 0.5 -0.3 1.0/ + 3.8 + 8.8 + 3.6 — 2.9 + 11 -0.4 + 0,1 1.5a -2A -5.8 -6,3 + 2.0 + 0 6 -1.0 + 0.9 2M + 1.1 + 3.8 + 6.1 + 0.2 -2.6 + 1.6 -0.5 2,5J — 0,8 -2.8 — 5.7 -2,4 + 2.7 -0.3 -0.1 3.0/ +04 + 1.9 + 4,5 + 3.2 -2,1 -1.6 + 1,7 3.5/ -0.3 -1.5 -3.9 -3.8 +0.7 + 2.7 -1.0 4.0/ + 0.2 + 1.1 + 3.1 + 3.7 + 0.5 -2.5 -0.1 4.5/ -0.2 -0,9 -2.5 — 3,4 -1.3 + 2,0 + 1.1 5o; + 0.2 + 0.7 + 2,1 + 3.1 + 1,8 -1.4 -1,9 5.5/ -01 — 0.6 -1.8 -2.9 -2.2 + 0.5 + 1,8 6.0/ + 0.1 + 0.5 + 1.6 + 2.6 + 2.3 -0.1 -2.0 6.5/ -0.1 -0,5 -1,2 -2,3 -2.3 -0.5 + 1.7 7.0/ + 0.1 + 0.4 + 1.1 + 2.1 + 2.3 +0,9 -1.3 7.5/ 0,0 -0.3 -1,0 -1.9 -2,1 -1.0 + 0.7 Magnitude Orientation . fiQrcL; Lx f , a , Periodic function (sin 0 sin O') (sin 2nrx + j cos 2^) with maximum value .
60ttL x L\ Z 2i (max) = (Q) We note in (2) that there are 3 factors: the first is a magnitude factor involv-ing the lengths of the dipoles and their separation the second involves thetr mutual orientation ., while the third factor is a periodic or complex function of unit magnitude giving the phase as a function of the separation distance. The mutual impedance for antennas in general involves these 3 factors.
Flgirc 1\ d) in a horizontal plane (0 — 90° or xy plane in Fig.
1 l-2a) is £ 1 (0} = fc/ l ( 1 ) where It is a constant (fl m _1 ) involving the distance D and I L is the terminal current. Equation (1) is the absolute field pattern in the horizontal plane.
It is independent of so that the relative pattern is a circle as indicated in Fig. 1L26.
Next let the elements be replaced by isotropic point sources of equal ampli-tude. The pattern as a function of 0 in the horizontal plane for two such isotropic in-phase point sources is given by (4-2-6) as /d_ cos d>\ E iJ) = 2E 1J cos y 2 -j (2) 438 U ARRAYS OF DIPOLES AND OF APERTURES where dr is the distance between sources expressed in radians; that is, 4-^ (3) Applying the principle of pattern multiplication, we may consider that E0 is the field intensity from a single element at a distance D. Thus, Eo = £ 1 (0) = fc/ 1 t4) Introducing (4) into (2) yields the field intensity E() as a function of 0 in the horizontal plane at a large distance D from the array, or fdr cos \ fd r cos E() = £j(0)2 cos f J - 2W, cos (—-J (5) This expression may be called the absolute field pattern in the horizontal plane.
The electric field at points in this plane is everywhere vertically polarized. The shape of this pattern is illustrated in Fig. il-2c, and also partially in Fig.
1 t-2a, for the case where d = A/2. The maximum field intensity is at = 90 u or broad-side to the array.
The field intensity £^0) as a function of 9 from a single A/2 element at a distance D in the vertical plane (yz plane in Fig.
1 1 -2a> is, from (5-5-12), given by £ t (0) - kl v cos {(nil) cos 0] The shape of this pattern is shown in Fig.
1 l-2d.
It is independent of the angle <£.
The pattern £jso(0) in the vertical plane for two isotropic sources in place of the two elements is Elwm = 2Eo (la) Applying the principle of pattern multiplication, we put E0 = EM (lb) so that the field intensity E{0) in the vertical plane at a distance D from the array E(0) = 2kl cos \nj2) cos &] This may be called the absolute field pattern in the vertical plane. This pattern has the same shape as the pattern for a single element in the vertical plane and is independent of the spacing. The relative pattern is presented in Fig.
\ i-2e and also partially in Fig.
1 l-2a + The relative 3-dimensional field variation for the case vhere d = A/2 is suggested in Fig.
11-2a. This pattern is actually bidirectional, only half being shown.
11-2 ARRAY OF TWO DRIVEN i/3 ELEMENTS. BROADSIDE CASE 439 Figure 11-3 Broadside array of 2 linear a/2 ele-ments with arrangement for driving elements with equal in-phase currents.
II -2b Driving-Point Impedance. Suppose that the array is energized by the transmission-line arrangement shown in Fig 11-3. Two transmission lines of equal length l join at P to a third line extending to a transmitter. Let us find the driving-point impedance presented to the third line at the point P} This will be called the driving point for the array.
Let V v be the emf applied at the terminals of element 1, Then Vx = iZn + JiZu (9) where is the current in element 1, l 2 the current in element 2, Z It is the self-impedance of element 1 and Z l7 is the mutual impedance between the two elements. Likewise, if V 2 is the emf applied at the terminals of element 2, Vi=l 2 zxi + I xz l7 (10) where Z 22 = the self-impedance of element 2 The currents are equal and in phase so /i = /» (It) Therefore, (9) and (10) become V1 =I i (Z li + Z l2 ) (12) and v 7 = l 2(Z22 + Z 12 ) (13) The terminal impedance Z x of element 1 is Zi=-jr = z,.+z.i < 14) 1 G H, Brown, “A Critical Study of the Characteristics of Broadcast Antennas as Affected by Antenna Current Distribution/ 1 Proc „ 7RIi, 24, 48-81, January 1936.
G. H. Brown, “ Directional Antennas," Proc. IRE, 25, 78-145, January 1937.
440 M ARRAYS OF DIPOLES AND OF APERTURES and the terminal impedance Z 2 for element 2 is Z 2 = ^ = Z 21 + Z 12 (15) J 2 Since the elements are identical Z 22 = Z U (16) Therefore, the terminal impedances given by (14) and (15) are equal; that is Z l = Z 2 — Z n + Z ]2 (17) Since Z t — Z 2 and it is necessary that the emf Vx applied at the terminals of dement 1 be equal and in phase with respect to the emf V 2 applied at the terminals of element 2.
For the case where the spacing d is A/2, the terminal impedance Z t of each element is Zj = Z tl +Z 12 = R 1X + /?!2 + /(^u + X12) = 73 — 13 +X43 - 29) = 60 + j14 ft (18) Suppose that the reactance of 14 ft is tuned out at the terminals by a series capacitance.
1 The terminal impedance then becomes a pure resistance of 60 ft.
If the length / of each transmission line between the antenna terminals and P is A/2 the driving-point impedance of the array at P is a pure resistance of 30 ft. This value is independent of the characteristic impedance of the A/2 lines. However, a resistance of 30 ft is too low to be matched readily by an open-wire transmission line. Therefore, a more practical arrangement would be to make l equal to A/4.
Suppose that we wish to have a driving-point resistance of 600 ft . To do this, we let the characteristic impedance of each A/4 line be ^1200 x 60 = 269 ft, 2 Each line transforms the 60 ft to 1200 ft and since two such lines are connected in parallel at P, the driving-point impedance for the array is a pure resistance of 600 ft. This is the impedance presented to the line to the transmitter. For an impedance match this line should have a characteristic impedance of 600 ft.
ll-2c Gain ill Field Intensity, As the last part of the analysis of the array, let us determine the gain in field intensity for the array. This could be done by L h is often simpler to resonate the elements by shortening them slightly. This modifies the resistive component of the impedance and also alters the E(0) field pattern, but to a first approximation these effects can usually be neglected.
1 For the special case of a IjA line, the general transmission-line formula reduces to Z,n = Zj/ZL where Z lK - is the input impedance, Z 0 the characteristic impedance and ZL the load impedance. Thus, z a = jk~zL .
U-2 ARRAY OK TWO DRIVEN i/2 ELEMENTS BROADSIDE CASE 441 pattern integration as in Chap.
3 but with self- and mutual-impedance values available a shorter method is as follows.
Let the total power input (real power) to the array be P.
1 Assuming no heat losses, the power P x in element 1 is p i = + Rn) (19) and the power P 2 in dement 2 is P 7 = ^<22 + Rn ) (20) where l x and l 2 are rms currents. However, R 12 = R X1 and I 2 = I Making these substitutions and adding ( 1 9) and (20) to obtain the total power F, we have P = P l +P2 = 2I 2(R li +R l2) and / _ / P 1 V2<Pn + ;) Suppose that we express the gain with respect to a single A/2 element as the reference antenna.
Let the same power P be supplied to this antenna. Then assuming no heat losses, the current J 0 at its terminals is where R 00 is the self-resistance of the reference antenna ( — P X1 ).
In general, the gain in field intensity 2 of an array over a reference antenna is given by the ratio of the field intensity from the array to the field intensity from the reference antenna when both are supplied with the same power The com-parison 'is, of course, made in the same direction from both the array and the reference antenna. In the present case it will be convenient to obtain two gain expressions, one for the horizontal plane and the other for the vertical plane.
In the horizontal plane the field intensity EHW(<£), as a function of <p, at a distance D from a single vertical center-fed A/2 reference antenna is of the form of (1). Thus, £hw(0) = kl0 (24) 1 It is important that the antenna power P be considered constant. Most transmitters are essentially constant power devices which can be coupled to a wide raflge of antenna impedance.
Until the antenna power was considered constant by G. H. Brown (Free. IRE, January 1937) the advantages of closely spaced elements were not apparent. Prior to this time the antenna current had usually been considered constant.
1 The power gain discussed in Chap. 2 is equal to the square of the gain in Jieid intensity. The power gain is the ratio of the radiation intensities (power per unit solid angle) for the array and reference antennas, the radiation intensity being proportional to the square of the field intensity.
(21 ) (22) 442 1] ARRAYS OF DIPOLES AND OF APF-RTURES U 2 ARRAY OF TWO DRIVEN }.:1 ELEMENTS. BROADSIDE CASE 443 where f 0 is the terminal current and “HW” indicates 44 HalfWavelength (A/2) antenna ” Substituting the value of / 0 from (23), we obtain £hw(0) = ^ The field intensity E[tp) in the horizontal plane at a distance D from tne array is given by (5). Introducing the value of the terminal current / j from (22) into (5) yields E() = k 2P R tl + /? L2 cos (26) The ratio of (26) to (25) gives the gain in field intensity of the array (as a function of <p in the horizontal plane) with respect to a vertical A/2 reference antenna with the same power input. This gain will be designated by the symbol G^)[A/HW] where the expression in the brackets is by way of explanation that it is the gain in field of the array (A) with respect to a kaif-waveiength reference anzenna (HW) 1 in the same direction from both array and reference antenna. Thus, Gj{<P) m 2K n l + ^12 (27) The absolute value bars || are introduced so that the gain will be confined to positive values (or zero) regardless of the values of dr and <p+ A negative gain would merely indicate a phase difference between the fields of the array and the reference antenna.
If the gain is the ratio of the maximum field of the array to the maximum field of the reference antenna, it is designated by G f (not a function of angle).
The self-resistances R 00 = R lt = 73 £1 For the case where the spacing is A/2, dr = n and R j : = — 1 3 £1 so that (27) becomes GM A ~ HW 1.56 cos (28) In the broadside direction (2. The pattern of a single vertical a/2 reference antenna with the same power input is shown for comparison.
It is also of interest to find the angle 0 for which the gain is unity. For this condition (28) becomes or ±56" or ±124° These angles are shown in Fig.
1 1-4. The array has a gain of greater than unity in both broadside directions over an angle of 68 G .
The gain as a decibel ratio is given by the relation Gain — 20 log l0 G f (dB) where G f = gain in field intensity Thus, a field -in tensity gain of L56 is equal to 3.86 dB.
Turning our attention now to the gain in the vertical plane {yz plane of Fig.
l.l-2a), the field intensity £Hw(0) as a function of 6 in this vertical plane at a distance D from a single vertical A/2 reference antenna with the same power input is of the form of (6). Thus, EUM = kl 0 cos [(n/2) cos 0] sin 0 (30) where / 0 = the terminal current Substituting its value from (23), we get P cos [(n/2) cos fl] Roo Ehw(0) — fe sin 0 (31) 444 11 ARRAYS of dipoles and of apertures Figure 11-5 Vertical-plane pattern of broadside array of 2 vertical in-phase a/2 elements spaced a/2. The pattern of a single vertical A/2 refer-ence antenna with the same power input is shown for comparison.
The field intensity £(0) as a function of 9 in the vertical plane at a distance D from the array is given by (8). Introducing the value of the terminal current I Y from (22) into (8), we have The ratio of (32) to (31) gives the gain in field intensity, C^IIA/HW], of the array as a function of 0 in the vertical plane over a vertical 2/2 reference antenna with the same power input. Thus, C J A ~|_ m _ A IhwJ £hw(0) The gain is a constant, being independent of the angle 6. For the case where the spacing is 2/2, (33) becomes GfiO) = 1.56 (or 3.86 dB) (34) The shape of the pattern for the array and for the 2/2 reference antenna is the same as shown in Fig.
11-5, but the ratio of the radius vectors in a given direc-tion is a constant equal to T56.
If the reference antenna is an isotropic source instead of a 2/2 antenna, the gain in the. vertical plane is a function of the angle 0. The maximum gain in held intensity of the array over an isotropic source with the same power input is ^1.64 times greater than the voltage gain over a 2/2 reference antenna [D(2/2) = 1.64, see Sec. 2-24], Thus, when the spacing is 2/2, the maximum gain in E{9) = k j 2P cos [(tc/2) cos 0] fin + R 12 sin 9 11-3 ARRAY Of 2 DRIVEN A/3 ELEMENTS END FIRE CASE 44$ ^ d— \ 2 Figure 11-6 End-fire array of 2 linear a/2 elements with currents of equal magnitude but opposite phase.
field intensity of the array with respect to an isotropic source is = 156 = 20 (or 60 dBi)1 l35) This value is in the broadside direction (<£ = 9 = 90 5 ).
11-3 ARRAY OF 2 DRIVEN Xj2 ELEMENTS-END-FIRE CASE, Consider an array of 2 center-fed vertical 2/2 elements (dipoles) in free space arranged side by side with a spacing d and equal currents in opposite phase as in Fig.
11-6. The only difference between this case and the one discussed in Sec. 11-2 is that the currents in the elements are taken to be in the opposite phase instead of in the same phase. As in Sec.
11-2, the analysis will be divided into 3 subsections on the field patterns, driving-point impedance and gain in field inten-sity.
ll-3a Field Patterns. The field intensity £i($) as a function of ^ at a distance D in a horizontal plane (xy or tf> plane in Fig.
1 l-7a) from a single element is EM) = Wi where k = a constant involving the distance D j 1 1 = the terminal current Replacing the elements by isotropic point sources of equal amplitude, the pattern £„(<£) in the horizontal plane for two such isotropic out-of-phase sources is given by (4-2-10) as / cos 4>\ EJ&) = 2E0 sin 0) 1 Distinguish between "dB” for gain with respect to a rtference antenna {A/2 dipole in the present ose) and " dBi” for gain with respect to an isotropic source.
446 II ARRAYS OF DIPOLES AND OF APERTURES \ Z (ib ) (C) Figtrfe 11-7 Patterns for end-fire array of 2 linear qut-of-phase i/2 elements with spacing d = i/2.
Applying the principle of pattern multiplication, we may consider that E0 is the field intensity from a single element at a large distance D. Thus E0 - £, MO = Jfcl, (2) and the field intensity E{) as a function of in the horizontal plane at a large distance D from the array is /A cos <b\ Em = 2kl t sin (3) This is the absolute field pattern in the horizontal plane. The electric field at points in this plane is everywhere vertically polarized. The relative pattern for the case where the spacing d is 2/2 is shown in Fig.
11 -lb and also partially in Fig. 11 -7a. The maximum field intensity is at 0 = 0° and </> = 180° Hence, the array is commonly referred to as an “end-fire” type.
M 3 ARRAY OF 2 DRIVEN A. 2 ELEMENTS EN|>FIRE CASE 447 The field intensity £^0) as a function of 6 from a single 2/2 element at a distance D in the vertical plane (xz plane in Fig. ll-7a) is, from (5-5-12), given by ^ Jr cos [(n/2) cos 0] —r«— <41 The pattern EiM(0) as a function of 0 in the vertical plane, for fwo isotropic sources in place of the two elements is, from (4-2-10), d r sin Gf EiJfi) = 2£0 sin Note that 0 is complementary to <p in (4-2-10), so cos $ = sin 6.
Putting Eq = Ei(d) t the field intensity £(0) as a function of 0 in the vertical plane at a large distance D from the array is r r cos [(/2) cos 0] .
(dr sin 0\ Le ""H" <« This is the absolute field pattern in the vertical plane. The relative pattern is illustrated in Fig.
1 1 -7c, and also partially in Fig.
ll-7a, for the case where the spacing is 2/2. The relative 3dimensional field variation for this case (d = 2/2) is suggested in Fig. 11 -la. This pattern is actually bidirectional, only half being shown.
ll-3b Driving-Point Impedance-Let V x be the emf applied to the terminals of element 1. Then Ki — IiZ x j + I 2 Z l2 (7) Likewise, if V2 is the emf applied to the terminals of element 2, — / 2 + I ;Z 12 (8) The currents are equal in m agnitude but opposite in phase so (9) Therefore, (7) and (8) become V i — 1 1{Z 11 — Z iz) (10) and v 2 ^iaz 22 ^z 12 ) (H) The terminal impedance Z, of element 1 is Z, =~ = Zn 1 (12) and the terminal impedance Z 2 of element 2 is ( 13 ) 448 II ARRAYS OF DIPOLES AND OF APERTURES 1-4 ARRAY OF 2 DRIVEN ;/2 ELEMENTS 449 Figure 11-8 End-fire array of 2 linear a/2 dements with arrangement for driving elements with currents of equal magnitude but opposite phase.
Figure 11-9 Horizontal plane pattern (a) and vertical plane pattern (ft) of end-fire array of 2 vertical a/2 dements with a/2 spacing. The patterns of a vertical a/2 reference antenna with the same power input are shown for comparison Therefore, or Zj “ Z 2 — Zn Z i2 Vx = Ya h h (14) (15) Since J 2 = -I x it follows from (15) that V 2 = - F,. This means that the 2 ele-ments must be energized with emfs which are equal in magnitude and opposite in phase. This may be done by means of a crossover in the transmission line from the driving point P to one of the dements as shown in Fig.
1 1-8 The length l of each line is the same.
For the case where the spacing between elements is A/2, the terminal imped-ance of each element is Z l = R lx -R ll ^rj{X xx -X ll ) = 86 + J12 Q (16) Consider that the reactance of 72 Q is tuned out by a series capacitance at the terminals of each dement. The terminal impedance is then a pure resistance of 86 Q. To obtain a driving-point resistance of 600 let the length l of the line from P to each dement be A/4 and let the line impedance be yf1200 x 86 = 321 Q. For an impedance match, the line from the driving point P to the trans-mitter should have a characteristic impedance of 600 Q.
ll-3c Gain in Field Intensity, Using the same method as in Sec.
ll-2c, the current f v in each element for a power input P to the array is given by / (17) It is assumed that there are no heat losses The current / 0 in a single A/2 reference antenna is given by (11-2-23) The gain in field intensity G/^)[A/HW] as a func-tion of tj) in the horizontal plane with respect to a A/2 reference antenna is obtained by substituting (17) in {3} and taking the ratio of this result to (1 1-2-25).
This yields For a spacing of A/2, (18) reduces to (19) In the end-fire directions ($ = 0° and 180 c ) the pattern factor becomes unity, and the gain is 1 ,3 or 23 dB. This is the gain Gf (see Fig, 1 1 -9).
The gain in field intensity C^)[A/HW] as a function of 0 in the vertical plane (xz plane of Fig 1 1 -7^1) with respect to a A/2 reference antenna is found by substituting (17) in (6) and taking the ratio of this result to (1 1-2-31), obtaining which is of the same form as the gain expression (18) for the horizontal plane (note that maximum radiation is in a direction 0 = 90°, = 0°).
The gain in field intensity G f of the array over an isotropic source with the same power input is 1.3 x V/L64 = 1.66 (or 4.4 dBi).
11^4 ARRAY OF 2 DRIVEN L/2 ELEMENTS. GENERAL CASE WITH EQUAL CURRENTS OF ANY PHASE RELATION.
1 In the preceding sections 2 special cases of an array of two A/2 driven elements have 1 For a more detailed discussion of this case and also of the most general case where the current amplitudes are unequal see G. H. Brown, “Directional Antennas," Proc. IRE , 25 78-145, January 1937.
450 It ARRAYS OF DIPOLES AND OF APERTURES ^Element 2 i <5 = 0 Figure 11-10 Array of 2 side-by-side elements normal to plane of page.
been treated.
In one case the currents, in the elements are in phase (phase difference = 0°) and in the other the currents are in opposite phase (phase difference = 180 G ). In this section the more general case is considered where the phase difference may have any value. As in the preceding cases, the two A/2 elements are arranged side by _ide with a spacing d and are driven with currents of equal magnitude.
For the general-phase case the rad i ation -fie Id pattern in the horizontal plane (jcy plane of Fig. ll-7a) is, from (4-2-20), given by E() = 2kJ 1 cos ~ (1) where & is the total phase difference between the fields from element 1 and ele-ment 2 at a large distance in the direction (see Fig.
1 1-10). Thus, i = dT cos + & (2) where <5 — the phase difference of the currents in the elements A positive sign in (2) indicates that the current in element 2 of Fig. 11-10 is advanced in phase by an angle & with respect to the current in element 1 ; that is, h = /i !A or i=Iztz£ (3) The voltages applied at each element are — I + l 2 Z l2 - l Y [Z lx + Z 12 $ (4) and V 2 = I 2 Z 22 + l,Z l2 = 1 2(Z 22 + Zu (-S) (5) The driving-point impedances of the elements are then z l = %- = z 11 + Zuli (6) J 1 y and Z 2 = -- = Z 12 + Z 12 f — S (7) 2 The real parts of the driving-point resistances are JR 1 = + \Z l2 \ cos (t + <5) = Riz + \Ztz \ cos (t -<5) and (8) (9) 11-4 ARRAY OF 2 DRIVEN A/2 ELEMENTS 451 where r = the phase angle of the mutual impedance Z 12 (that is, t = arctan X i^/Ri 2 where Z 12 = -f- jX j^) Therefore, the power in element 1 is P, = 11,1^ = |/ 1 | 1[R 11 + |Zia |co«(T + tf)] (10) and the power P2 in element 2 is R2 I f 2 + I Z 12 | cos (t — <5)] (11) Since Jtn = R 22 the total power P is P = P^ + P2 = \I i \ 2 {2R 11 + \ Z 12 [[cos (t + S) + cos (t - <5)]} — 2 1 t 2CRn + \ Z X2 I cos t cos <5) = 2|/i t 2(R M + R 12 cos 3 ) (12) It follows that the gain in field intensity as a function of d> in the horizontal plane 1 of the array over a single A/2 element with the same power input is (13) A polar plot of (13) with respect to the azimuth angle gives the radiation-field pattern of the array in the horizontal plane, the ratio of the magnitude of the radius vector to a unit radius indicating the gain over a reference A/2 antenna.
Brown 2 has calculated such patterns as a function of phase difference 6 and spacing dF Examples of these are shown in Fig, 11-11.
The radiation-field pattern in the vertical plane containing the elements (in the plane of the page of Fig. 11-12) is E{9) = 2kl { cos dF sin 0 + 3\ cos [(rc/2) cos 0] Thus, the pattern in the vertical plane has the shape of the patterns of Fig, 1 1-1 1 multiplied by the pattern of a single A/2 antenna. The gain in the vertical plane over a vertical A/2 reference antenna with the same power input is then d F sin 8 + £ 2 (15) It is often convenient to refer the gain to an isotropic source with the same power input. Since the power gain of a A/2 antenna over an isotropic source is 1.64 [Z)(A/2) = 1.64, see Sec. 224], the gain in field intensity as a function of 8 in 1 This is the plane of the page in Fig 11-10.
3 G. H. Brown, “ Directional Antennas,” Proc. IRE, 25, 78-145, January 1937.
, — rf—H >>lk Figure 11-11 Horizontal-plane field patterns of 2 vertical dements as a function of the phase differ-ence & and spacing d. '{After G. H. Brown "Directional Antennas," Proc. IRE, 25, 78-145, January 193 7.) Both elements are the same length and have currents of equal magnitude. The circles indicate the field intensity of a single reference element of the same Length with the same power input.
M i CLOSELY SPACED ELEMEfVTS 453 the vertical plane of a vertical 1/2 antenna in free space over an isotropic source J cos [(rc/2) cos 0] I The gain in field intensity in the vertical plane of the array over the isotropic source is then the product of (15) and (16) or c4s]- g4hwM?
j 3.2SR n ftn -j- R 12 cos S dr sin 0 + d\ cos [(tt/2) cos 0] I 11-5 CLOSELY SPACED ELEMENTS ll-5a Introduction. In 1932 I joined the Institute of Radio Engineers (IRE, now IEEE). The following year I received my Ph.D. degree in physics from the University of Michigan, publishing my dissertation research in the Proceedings of the IRE .
I read every issue with interest On opening the January 1937 Pro-ceedings I delved into a monumental treatise on "Directional Antennas” by George H, Brown of RCA.
1 Buried deep in the article was, to me, an astonishing calculation which indicated that parallel linear dipoles with spacings of 2/8 or less had higher gains than the customary larger spacings Within one week of the time I received my Proceedings I had designed and built an array of 4 close-spaced 2/2 dipoles at my amateur station W8JK Oper-ating at a wavelength of 20 m, the array was phenomenally effective.
I published the design and in subsequent articles extended it to an entire family of close-spaced arrays. The antennas outperformed all others.
I called them "flat-top beams” but everyone else called them W8JK arrays 2 They were soon in use by the thousands world-wide.
1 G. H. Brown, “ Directional Antennas," Proc. IREt 25, 78-145, January 1937.
J J. D. Kraus, "A Small but Effective Flat Top Beam Antenna," Radio, no. 213, 56-58, March 1937 and no. 216, HM6, June 1937.
J. D. Kraus, " Rotary Flat Top Beam Antennas," Radio, no. 222, 1 1-16, December 1937.
J Kraus, “Directional Antennas with Closely-Spaced Elements," QST, 22, 21-23, January 1938.
J. D. Kraus, “Characteristics of Antennas with Closely-Spaced Elements" Radio, no. 236, 9-19, Feb-ruary 1939.
J. D. Kraus, “Antenna Arrays with Closely-Spaced Elements," Proc. IRE , 28, 76-84, February 1940, J. Kraus, “ W8JK 5-Band Rotary Beam Antenna," QST , 54, U-14, July 1970.
J, Kraus, “The W8JK Antenna," QST, 66, 11-14, June 1982.
454 n arrays of dipoles and of apertures L1- <-'U>S£LV SPACED ELEMENTS 455 In 1937 close spacing was a new and revolutionary concept.
In George Brown's autobiography 1 he states: Ironically, the particular portion of my paper which John Kraus used so effectively was a small paper which I submitted to the Proceedings in 1932 only to have it rejected by a reviewer who denied its validity. When I prepared “ Directional Antennas” I tucked this older material into the middle of this bulky manuscript on the assumption that the reviewer would not notice it.
George Brown’s ruse worked and the world finally learned of his idea but only after it had languished in obscurity for five years!
In the next part of this section, the benefits and limitations of close spacing are analyzed and its application to the W8JK array is outlined.
ll-5b Closely Spaced Elements and Radiating Efficiency. The WtfJK Array. The end-fire array of two side-by-side, out-of-phase A/2 dipole elements discussed in See. 11-3 can produce substantial gains when the spacing is decreased to small values. As indicated by the R L = 0 curve in the gain-versus-spacing graph of Fig, 11-131, the gain approaches 3.9 dB at small spacing. At A/2 spacing the gain is 2.3 dB. This curve is calculated from (11-3-18) for — 0° or (1 1-3-20) for 0 = 90°. As the spacing d approaches zero, the coupling factor becomes infinite, but at the same time the pattern factor approaches zero. The product of the two or gain stays finite, leveling off at a value of about 3.9 dB (6.0 dBi) for small spacing, as illustrated by Fig. 11-13&. The fact that increased gain is associated with small spacings makes this arrangement attractive for many applications.
Thus far it has been assumed that there are no heat losses in the antenna system. In many antennas such losses are small and can be neglected. However, in the W8JK antenna such losses may have considerable effect on the gain (Fig.
1 1-13). Therefore, the question of losses and of radiating efficiency will be treated in this section in connection with a discussion of arrays of 2 closely spaced, out-of-phase elements. The term “closely spaced” will be taken to mean that the elements are spaced A/4 or less.
A transmitting antenna is a device for radiating radio-frequency power. Let the radiating efficiency be defined as the ratio of the power radiated to the power input of the antenna. The real power delivered to the antenna that is not radiated is dissipated in the loss resistance and appears chiefly in the form of heat in the antenna conductor, in the insulators supporting the antenna, etc. An antenna with a total terminal resistance R iT may be considered to have a terminal radi-ation resistance Rj and an equivalent terminal loss resistance R 1L (see Sec. 2-13) 1 G. H. Brown, “And Part of Which I Was," Ang^ Cupar, 117 Hunt Drive, Princeton, NJ 08>40, 1982 [b) Figure 11-13 Gain of end-fire array of 2 out-of-phase A. '2 dements (WKJK array) with respect to a a 2 reference antenna as a function of the spacing for 5 values of the loss resistance R L . (M Gain curve for R L = 0 with variation of its component factors, the coupling factor and the pattern factor, for 0 = 0.
such that R IT = t + ^1L ( 1 ) It follows that Radiating efficiency (%) = —— x (2) Rt + ML Since many types of high-frequency antennas have radiation resistances that are large compared to any toss resistance, the efficiencies are high.
In an 456 n arrays of dipoles and of apertures Spacing d, \ Figure 11-14 Current / L and radiation resistance in each element of a W8JK antenna as a function of the spacing. The current is calculated for a constant input power of 100 W to the array.
array with closely spaced, out-of-phase elements, however, the radiation resist-ance may be relatively small and the antenna current very large, as illustrated by Fig.
11-14, Hence, a considerable reduction in radiating efficiency may result from (he presence of any loss resistance. The radiating efficiency may also be small for low-frequency antennas which are very short compared to the wave-length. Although the effect of loss resistance will be discussed specifically for an array of 2 closely spaced a/2 dements, the method is general and may be applied to any type of antenna.
Let the equivalent loss resistance at the terminals of each element be R iL .
The elements are center fed and are arranged side by side with a spacing d. The total terminal resistance R 1T is as given by (1)- The terminal radiation resistance R, is given by 1 =n -K| 2 (3) Substituting (3) in {1) the total terminal resistance for each element is then ^ 1 T ~ ^11 + “ ^12 (4) If a power P is supplied to the 2-element array, the current in each element is i, = / P (5) V 2(ftu + ~ ^12) The total terminal resistance R 0r of a single, center-fed 2/2 reference antenna is Rqt = Rq0 + (6 ) where R 00 is the seff-resistance and R0L the loss resistance of the reference antenna. The current I0 at the terminals of the reference antenna is then Rnn + Am (7) Ll-S CLOSELY SPACED ELEMENTS 457 With the array elements vertical, the gain in field intensity as a function of in the horizontal plane (xy plane in Fig.
1 l-7a) is obtained by substituting (5) in (1 13-3), (7) in ( 1 1-2-24) and taking the ratio which gives dr cos ) of a volume array as a function of 0 and is Eifi.
) — pattern of single element EJfi, \) = pattern of linear array of point sources in z direction The product of the last 3 terms in (1) is the pattern of a volume array of point sources.
If, for instance, we wish to obtain the pattern of the entire array E(4>) as a function of \ in the xy plane (# = 90°), we introduce the appropriate pattern expression in this plane for each component array in {1). For the example being considered the normalized pattern becomes sin (In sin 0) [n 1 £W>) =—— ~ m .—~ cos - (1 - cos 0) {2) 4 sin [(tt/2) sin <£] L4 J 1 } Only the broadside pattern and the Ej() end-fire pattern contribute to the array pattern in the xy plane, since in this plane the E^) pattern of a single element and the Ez( in the xy plane (0 = 90 ) for the array of Fig. 11-17 with respect to a single vertical A/2 dement with the same power input is Y 11 + U + 13 + 15 + 17 + i(12 + 16> + («,„ + 1„) sin (In sin $) sin [(tt/2) sin 0] n "1 i - (1 - cos ) where /in = self- resistance of one dement /£ 1L = loss resistance of one element R t2 - mutual resistance between element 1 and element 2 R 13 = mutual resistance between element 1 and dement 3, etc.
The numbering of the dements is as indicated in Fig. 1M7.
It is assumed that d ~ h = -12 and a — A/4 and that the current magnitudes are equal, the currents in the front 8 elements being alt in the same phase but retarded 90° with respect to the currents in the rear 8 elements.
11-7 HORIZONTAL ANTENNAS ABOVE A PLANE GROUND.
In the previous discussions it has been assumed that the antenna is in free space, i.e., infinitely remote from the ground. Although the fields near elevated micro-wave antennas may closely approximate this idealized situation, the fields of most antennas are affected by the presence of the ground. The change in the pattern from its free-space shape is of primary importance. The impedance relations may also be different than when the array is in free space, especially if the array is very close to the ground.
In this section the effect of the ground on horizontal antennas is discussed. In Sec. 11-8 the effect of the ground is analyzed for vertical antennas. A number of special cases are treated in each section, these being limited to single elements or to simple arrays of several elements. Perfectly con-ducting ground is assumed. The effect of imperfect (ordinary) ground is discussed in Sec. 16-4.
11 -7a Horizontal k/2 Antenna above Ground. Consider the horizontal //2 antenna shown in Fig.
1 1-18 at a height h above a plane ground of infinite extent.
Owing to the presence of the ground, the field at a distant point P is the resultant of a direct wave and a wave reflected from the ground as in Fig, 1 119, Assuming that the ground is perfectly conducting, the tangential component of the electric 462 II ARRAYS OK IMPOLES aND OF APERTURES H-7 HORIZONTAL ANTENNAS ABOVE A PLANE GROUND 463 A 2 T / Antenna mmmmym | ^Ground h k U-\ 2 Image Figure 11-18 Horizontal a/2 antenna at height h above ground with image at equal distance below ground.
field must vanish at its surface. To fulfill this boundary condition, the reflected wave must suffer a phase reversal of 1 80° at the point of reflection.
To obtain the field at a distant point P t it is convenient to transform the problem by the method of images. In this method the ground is replaced by an image of the antenna situated a distance h below the ground plane. By taking the current in the image equal in magnitude but reversed in phase by 18(T with respect to the antenna current, the condition of zero tangential electric field is met at all points along a plane everywhere equidistant from the antenna and the image. This is the plane of the ground which the image replaces. In this way, the problem of a horizontal antenna above a perfectly conducting ground 1 of infinite extent can be transformed into the problem already treated in Sec. 11-3 of a so-called end-fire array. One point of difference is that in developing the gain expression it is assumed that if a power P is delivered to the antenna, an equal Hgure ] 1-19 Antenna above ground with image showing- direct and reflected waves.
1 It is also possible to apply the method of images to the case of a ground of infinite extent but of finite conductivity a and of dielectric constant t by properly adjusting the relative magnitude and phase of the image current with respect to the antenna current.
Height above ground, X Figure 11-20 Driving-or feed point resistance R i at the center of a horizontal a/2 dipole antenna as a function of its height above a perfectly conducting ground.
power is also supplied to the image. Hence, a total power IP is furnished to the “ end-fire array ” consisting of the antenna and its image.
Owing to the presence of the ground, the driving-point impedance of the antenna is, in general, different from its free-space value.
Thus, the applied voltage at the antenna terminals is K.-i.Zu+'iZ., U) where = the antenna current I 2 = the image current Z y ! = the self-impedance of the anienna Zm = the mutual impedance of the antenna and its image at a distance of 2h Since l 1 = - ly, the driving- or feed-point impedance of the antenna is Z 1 =%- = ZU -Zm (2) 1 1 The real part of (2) or d.iving-point radiation resistance is The variation of this resistance at the center of the k/2 antenna is shown in Fig. 11-20 as a function of the antenna height h above the ground. As the height becomes very large, the effect of the image on the resistance decreases, the radi-ation resistance approaching its free-space value.
Since the antenna and image have currents of equal magnitude but opposite phase, there is zero radiation in the horizontal plane, i.e., in the direction for 46 4 II ARRAYS OF DIPOLES AND OF APERTURES FiEDre H-21 Vertical-plane field patterns of a horizontal A/2 dipole at various heights fi above a perfectly conducting ground as calculated from (4) for R 1L = 0, Patterns give gain in field intensity over a a/2 dipole with the same power input. Note that (he presence of the ground increases the field t>y approximately 6 dB or more in certain directions.
which the elevation angle ot is zero (see Fig.
11-19). If the height h is 2/4 or less, the maximum radiation is always in the vertical direction (a = 90°). For larger heights the maximum radiation is, in general, at some elevation angle between 0 and 90°.
It is convenient to compare the horizontal 2/2 antenna at a height h above ground with respect to a a/2 antenna in free space with the same power input. At a large distance the gain in field intensity of the “ Half-Wave length antenna Above Ground” (HWAG) with respect to the "Half-Wavelength antenna in Free Space ” (HWFS) is given by GM) -HWAG] rfitter LhwfsJ + 1 2 sin (hr sin a) [ (4) where hr = (2n/k)h R n ~ self-resistance of a/2 antenna ^il = loss resistance of kj 2 antenna R m = mutual resistance of 2/2 antenna and its image at a distance of 2h Equation (4) gives the gain in the vertical plane normal to the antenna as a function of a (see Fig.
11-21).
H-7 HORIZONTAL ANTENNAS ABOVE A PLANE GROUND 465 Figure 1 1-22 Horizontal antenna at height h above ground plane) showing azimuth angle ^ and elevation angle a for a distant point P.
The vertical-plane patterns of a horizontal 2/2 antenna are shown in Fig.
1 1-21 for heights h = 0.1, 0.25, 0.5 and 1.02. The circular pattern is for a 2/2 antenna in free space (i.e., with the ground removed) with the same power input.
It is assumed that loss resistances are zero.
It is also of interest to calculate the field pattern as a function of the azimuth angle for a constant elevation angle a. The radius vector to the distant point P then sweeps out a cone as suggested in Fig.
1 1-22. To find this field pattern, let us first consider the field pattern of a horizontal antenna in free space as in Fig, 1 1-23. The xy plane is horizontal. The field intensity at a large distance Figure 11-23 Geometrical construction for finding the field intensity at a constant elevation ancle a in direction of line OA .
6 466 ]l ARRAYS OF DIPOLES AND OF APERTURES in the direction of a and 0 is then given by the length OA between the origin and the point of intersection of a cone of elevation angle x and the surface of the 3-dimcnsional doughnut field pattern of the antenna as suggested in Fig.
i 1-23.
This length is obtained from the field-pattern formula of the antenna in free space by expressing the polar angle 0' from the antenna axis in terms of a and 0. For the spherical right triangle in Fig.
1 1-23 we have cos 4> = cos cos a (5) or sin = v ' 1 — cos 2 cos 2 a (6) Substituting these relations in the pattern formula, we get the field intensity in the direction a, 0< For example, by substituting (5) and (6) into (5-5-12), noting that 0' in (5) and (6) equals 0 in (5-5-12), we obtain for the field of a A/2 horizontal dipole antenna cos [(jt/2) cos 0 cos a] 0) = y (') sf\ — cos' 0 cos at Then the relative field pattern of the horizontal 111 dipole antenna in free space as a function of 0 at a fixed elevation angle x0 is given by „ „ cos [(it/2) cos 0 cos oc0] yj\ - cos 0 cos^ a0 To obtain the field pattern of the antenna when situated at a height h above a perfectly conducting ground, we multiply the above free-space relations by the pattern of 2 isotropic point sources of equal amplitude but opposite phase. The sources are separated by a distance 2h along the z axis. From (4-2-tO) the pattern of the isotropic sources becomes in the present case E ii0 = sin (hr sin a) (9) "'here K the height of the antenna above ground in radians; that is, The pattern is independent of the azimuth angle 0. Multiplying the free-space field pattern of any horizontal antenna by (9) yields the field pattern for the antenna above a perfectly conducting ground. Thus, for a horizontal A/2 dipole antenna above a perfectly conducting ground, the 3-dimensional field pattern as a function of both a and 0 is obtained by multiplying (7) and (9) which gives cos Un/2) cos 0 cos a] .
.
E =— sin (/i, sin a) — cos 2 0 cos 2 a where hr = the height of the antenna above ground, rad As an example, the field patterns as a function of the azimuth angle 0 at ele-vation angles For this case (1) reduces to 1.18(1 - cos ff) >/j?i l + (Vm-1 ) (2) where E is the field intensity along the ground at a distance of 1.61 km for a power input of 1 kW. The variation of E as given by (1) is presented in Fig, 11-35 as a function of antenna length; 1, 2 The vertical-plane patterns calculated by (1) as a function of the elevation angle a for vertical antennas of various heights are presented in Fig.
1 1-36. 1, 2 A length of about 0,642. yields the greatest field inten-sity along the ground, but as pointed out by Brown 1 the large high-angle radi-1 G.
H. Brown, “A Critical Study of the Characteristics of Broadcast Antennas as Affected by Antenna Current Distribution, Proc. JRE, 25, 7&-145, January 1937.
3 C E Smith, Directional Cleveland Institute of Radio Electronics, Cleveland, Ohio, 1946.
11-8 VERTICAL ANTENNAS ABOVE A PLANE GROUND 475 Height / of antenna, X Figure 1 1-35 Field intensity at the ground (zero elevation angle) at a distance of 1,61 km from a ver-tical antenna with 1 kW input as a function of its height i Perfectly conducting ground is assumed.
The solid curve is for an assumed Joss resistance Rl = 0 and the dashed curve for RL = 1 fl.
ation (at a = 60°) for this length reduces the nonfading range at broadcast frequencies (500 to 1500 kHz) as compared, for example, with an antenna about a/2 long. The nonfading range is largest for an antenna height of 0.5282.
It is assumed that the loss resistance = 0, that is, the entire input to the antenna is radiated. The small amount of high-angle radiation, which is an important factor in reducing fading, is apparent for the / = 0.5282 antenna as compared to other lengths (see Fig.
1 1-366 and c).
The analysis of arrays of several vertical stub antennas can be reduced in a similar fashion to arrays of symmetrical center-fed antennas. Many of these have been treated in previous sections. In this case it is often convenient to compare ib) W Figure 1 1-36 Vert ical -plane field patterns of vertical antennas for several values of antenna neight /.
The field intensity is expressed in millivolts per meter at a distance of 1.61 km for 1 kW input.
Perfectly conducting ground and zero loss resistance are assumed.
the pattern and refer the gain to a single vertical stub antenna with the same power input. The situation of a symmetrical center-fed vertical antenna with its lower end some distance above the ground can also be treated by the method of images. In this case the antenna is reduced to a collinear array.
For the case of a linear array of vertical elements of equal height and of the same current distribution, the pattern £(0) as a function of the azimuth angle at a constant elevation angle <x is given by m = EaP) X Et ( 3) where = relative pattern of array of isotropic point sources used to replace elements = relative field intensity of a single vertical element at the elevation angle a The angle 0' in the pattern formula of the array of isotropic sources is the angle with respect to the array axis or x axis in Fig. ll-37a.
Before inserting this formula into (3), it is necessary to express 0' in terms of the azimuth angle 0 and elevation angle <x (Fig.
1 l-37a). This is done by the substitutions cos 0' = cos 0 cos a (4^ sin 0' = J\ — cos 2 0 cos 2 If the relative field intensity formula £ x of a single vertical element is given in terms of the polar angle 8, the elevation angle a is introduced by means of the substitution 0 = 90° — a, since, as indicated in Fig. 11-37&, 6 and ot are comple-mentary angles.
11-9 ARRAYS WITH PARASITIC ELEMENTS ll'9a Introduction, In the above sections it has been assumed that all the array elements are driven, i,e„ all are supplied with power by means of a trans-it ARRAYS WITH PARASITIC ELEMENTS 477 h J, 2 -l' 1 2 ’'V Parasitic element Driven element \ 0 Figare 11-38 Array with one driven dipole element and one parasitic element.
mission line. Directional arrays can also be constructed with the aid of elements in which currents are induced by the fields of a driven element Such elements have no transmission-line connection to the transmitter and are usually referred to as “parasitic elements." Let us consider the case of an array in free space consisting of one driven k/2 dipole element (element 1) and one parasitic element (element 2), as in Fig.
1 1-38, The procedure follows that used by Brown.
1 Suppose that both ele-ments are vertical so that the azimuth angle 0 is as indicated. The circuit rela-tions for the elements are Vi = f i^i 1 + 12 ^12 0=i IiZn + liZ12 From (2 ) the current in element 2 is / 2 = IZ.il/l Z 1 1 , /T — Tt 7 1 ^22 IZ22I& 7 - -11 ^22 or 1 2 -Z 1 2 i — /< 1 7 t-L &21 where £ = n + - t 2t in which 12 = arctan 12 z 2 = arctan G. H. Brown, “Directional Antennas," free. IRE, 25, 78-145, January 1937.
478 11 ARRAYS OF WPOLES AND OF APERTURES where R JZ + j^n = Z l2 ~ mutual impedance of elements 1 and 2, Q R 22 + jX 21 = Z 12 = self-impedance of the parasitic element, fl The electric field intensity at a large distance from the array as a function of 0 is £(0) = W i +JW4 cr where dr = fid = -r d Substituting (4) for 1 2 in (5), £(<£) = fc/,^1 + li + rf, cos d j Solving (1) and (2) for the driving-point impedance Z x of the driven element, we get v Hi-? JUAB Z L — Z n rr ,7 I t— The real part of Z : is Z 22 I hi fii = fin - ^ cos (2tm - z 2 ) ill Adding a term for the effective loss resistance if any is present, we have fi,=fi u + fi 1L -cos (2 t„ - t 2 ) ^22 For a power input P to the driven element, 1 V^i V^n + r il ~ \ Z 2 mIz zi I cos {2rm - r ; ) and substituting (10) for l t in (6) yields the electric field intensity at a large dis-tance from the array as a function of 0. Thus, £(0) = All + - \Zi 2/Z22 \ cos (2r (I1 - t 2 ) i' +k j f cos For a power input P to a single vertical 2/2 element the electric field intensity at the same distance is £hw(0) — 0 — k j -+- flr (Vm-1 ) ]|-9 ARRAYS WITH PARASITIC ELEMENTS 479 where /? 00 = seif- resist anee of single A/2 element, = loss resistance of single 2/2 element, Q The gain in field intensity (as a function of 0) of the array with respect to a single 2/2 antenna with the same power input is the ratio of (1 1) to (12). Since R 00 = R lt and letting R$ L = R 1L , we have If Z 22 is made very large by detuning the parasitic element (ie., by making X 22 large), (13) reduces to unity, that is to say, the field of the array becomes the same as the single 2/2 dipole comparison antenna.
By means of a relation equivalent to (13), Brown 1 analyzed arrays with a single parasitic element for various values of parasitic element reactance (X 2 z) and was the first to point out that spacings of less than 2/4 were desirable.
The magnitude of the current in the parasitic element and its phase relation to the current in the driven element depends on its tuning. The parasitic element may have a fixed length of 2/2, the tuning being accomplished by inserting a lumped reactance in series with the antenna at its center point. Alternatively, the parasitic element may be continuous and the tuning accomplished by adjusting the length.
This method is often simpler in practice but is more difficult of analysis. When the 2/2 parasitic dipole element is inductive (longer than its resonant length) it acts as a reflector , When it is capacitive (shorter than its resonant length) it acts as a director } Arrays may be constructed with both a reflector and a director, A 3-element array of this type is shown in Fig.
11-39, one parasitic element acting as a reflec-tor and the other as a director. The analysis for the 3-element array is more complex than for the 2-elemenl type treated above.
Experimentally measured field patterns of a horizontal 3-element array situ-ated 12 above a square horizontal ground plane about 132 on a side are present-ed in Fig.
1 1-40. The element lengths and spacings are as indicated. The gain at a =15" for this array at a height of 12 is about 5 dB with respect to a single 2/2 1 G. H. Brown, il Directional Antennas," Proc IRE, 25, 7S- 145. January 1937.
2 From Fig. 9-9, we note that the reactance of a thin linear element varies rapidly as a function of frequency when its length is about 2/2, going irom positive (inductive) reactance through zero reac-tance (resonance) to negative (capacitive) reactance values as the length is reduced.
48U II ARRAYS OF DIPOLES AND OF APERTURES Maximum^ radiation Reflector Driven Director element Figure 11-39 Three-element array.
= 90 " CD ^Reflector -T — ^ Director 10 10 j Ground plane Side view Relative field intensity for Driven element Reflectory j .Director Plan view Figure 11-40 Measured vertical plane pattern (a) and horizontal plane patterns (i) at 3 elevation angles for a 3-dcmem array located li above a large ground plane. (Patterns by D, C. Cleckner, Ohio Slate University,} II 9 ARRAYS WITH PARASITIC ELEMENTS 481 Open-wire Coupler line / jj/ ... If C C C 1 C C Boom Figure ll-41a Shintaro Uda’s experimental antenna with 1 reflector and 7 directors on the roof of his laboratory at Tohoku University Tor vertically polarized transmission tests during 1927 and 1928 over land and sea paths up to 135 km using a wavelength A = 4.4 m. The horizontal wooden boom supporting the array elements is 1 5 m long.
dipole antenna at the same height.
1 The vertical plane pattern is shown in Fig.
ll-40a.
It is interesting to note that because of the finite size of the ground plane there is radiation at negative elevation angles. This phenomenon is charac-teristic of antennas with finite ground planes, the radiation at negative angles being largely the result of currents on the edges of the ground plane or beneath it.
The azimuthal patterns at elevation angles a = 10, 15 and 20" are shown in Fig. 11-406. A parasitic array of this type with closely spaced elements has a small driving-point radiation resistance and a relatively narrow bandwidth.
Il-9b The Yagi-Uda Array.
Shintaro Uda, an assistant professor at Tohoku University, had not turned 30 when he conducted experiments on the use of parasitic reflector and director elements in 1926, This led to his publication of a series'of 1 1 articles (from March 1926 to July 1929) in the Journal of the Institute of Electrical Engineers of Japan titled “On the Wireless Beam of Short Electric Waves.” 2 He measured patterns and gains with a single parasitic reflector, a single parasitic director and with a reflector and as many as 30 directors. One of his many experimental arrays is showm in Fig.
1 1-41 a. He found the highest gain with the reflector about A/2 in length and spaced about A/4 from the driven element, while the best director lengths were about 10 percent less than A/2 with optimum spacings about A/3, Even though many patterns were measured in the 1 Note that ii is necessary to specify both the height and elevation angle at which the comparison is made.
In comparing one antenna with another, the gain as a function of elevation angle at a given height (or as a function of height at a given elevation angle) may, in general, range from zero to infinity.
2 S. Uda, “On the Wireless Beam of Short Electric Waves/' JIEE [Japan), March 1926, pp. 273-282 (I); April 1926 (II), July 1926 (111); January 1927 (IV); January 1927 (V); April 1927 (VI); June 1927 I VII); October 1927 (VIII); November 1927, pp. 1209-1219 (IX); April 1928 (X); July 1929 (XI).
482 IE ARRAYS OF DIPOLES AND OF APERTURES near field, these lengths and spacings agree remarkably well with optimum values determined since then by further experimental and computer techniques. After George H, Brown demonstrated the advantages of close spacing, the reflector-to-driven-element spacings were reduced.
Hidetsugu Vagi, professor of electrical engineering at Tohoku University and 10 years Uda's senior, presented a paper with Uda at the Imperial Academy on the Projector of the Sharpest Beam of Electric Waves” in 1926, and in the same year they both presented a paper before the Third Pan-Pacific Congress in Tokyo titled “ On the Feasibility of Power Transmission by Electric Waves.” The narrow beams of short waves produced by the guiding action of the multidirector periodic structure, which they called a “wave canal,” had encouraged them to suggest using it for short-wave power transmission, an idea now being considered for beaming solar power to the earth from a space station or from earth to a satellite {but not necessarily with their antennas).
It is reported that Professor Yagi had received a substantial grant from Sendai businessman Saito Zenuemon which supported the antenna research done by Uda with Yagi’s collaboration. Then in 1928 Yagi toured -the United States presenting talks before Institute of Radio Engineers sections in New York, Wash-ington and Hartford, and in the same year Yagi published his now famous article on “Beam Transmission of Ultra Short Waves" in the Proceeding of the IRE} Although Yagi noted that Uda had already published 9 papers on the antenna and acknowledged that Uda’s ingenuity was mainly responsible for its successful development, the antenna soon came to be called “a Yagi.” In deference to Udas contributions, I refer to the array as a Yagi-Uda antenna, a practice now becom-ing common. Uda has summarized his researches on the antenna in two data packed books. 2 A typical modern-version 6-element Yagi-Uda antenna is shown in Fig.
ll-4Ib. It consists of a driven dement (folded 2/2 dipole) fed by a 300-ft 2-wire transmission line (twin line), a reflector and 4 directors. Dimensions (lengths and spacings) are indicated on the figure. The antenna provides a gain of about 12 dBi (maximum) with a bandwidth at half-power of 10 percent. By adjusting lengths and spacings appropriately (tweeking), the dimensions can be optimized, producing an increase in gain of another decibel. 3 However, the dimensions are critical.
H- ^agi, Beam Transmission of Ultra Shorl Waves," Proc. IRE T 16, 715-740, June 1928, 2 S. Uda and Y. Mushiake, Yagi-Uda Antenna, Maruzden, Tokyo, 1954, S^Uda, Short Wave Projector; Historical Records cf My Studies in the Early Days, published by Uda, C, A. Chen and D. K. Cheng, “Optimum Spacings for Yagi-Uda Arrays," IEEE Trans , Ants. Prop AP-21, 615-623, September 1973.
C. A. Chen and D. K. Cheng, “Optimum Element Lengths for Yagi-Uda Arrays," IEEE Trans Ants rrop., AP-23, 8-15, 1975, P P. Viezbicke, “ Yagi Antenna Design," NBS Technical Note 688, December 1968.
11-9 ARRAYS WITH PARASITIC ELEMENTS 483 1 . b \ / = 0.475X 0.46A 0.44X 0.44A 0.43X 0.40X Figure l!-41b Modern -version 6-elemcnl Yagi-Uda antenna with dimensions.
It has a maximum directivity of about ! 2 dBi at the center of a bandwidth of 10 percent at half-power.
The inherently narrow bandwidth of the Yagi-Uda antenna can be broadened to 1.5 to 1 by lengthening the reflector to improve operation at low frequencies and shortening the directors to improve high-frequency operation.
However, this is accomplished at a sacrifice in gain of as much as 5 dB.
U-9c Square-Corner-Yagi-Uda Hybrid. The corner reflector (Sec.
12-3) is an inherently wideband antenna producing substantial gains. Thus, a square corner reflector with maximum dimension equal to the 1.5/ length of the Yagi-Uda antenna of Fig, ll-41b produces about 1 dB more gain. Furthermore, the corner reflector can operate over a 2 to 1 bandwidth with 5 or 6 dB more gain than a 1.5 to 1 bandwidth Yagi-Uda antenna of the same size.
Although reducing the length of the reflector elements and reducing their number (by increasing the spacing between them) decreases the gain of a corner reflector antenna, the gain may still be more than that of a wideband Yagi-Uda antenna. However, if the increased spacing is as much as //6 or //4 at the low-frequency end of a 2 to 1 band, the spacing becomes //3 to 2/2 at the high-frequency end, reducing the effectiveness of the reflector. To compensate for this effect at the high-frequency end a number of directors can be added in front of the driven element of the corner reflector, resulting in a corner reflector with directors or corner- Yagi-Uda hybrid antenna as shown in Fig.
1142. The length and spacing of the directors are made appropriate for the high-frequency end of the band. The thinned corner reflector is effective at the low and mid frequencies while the directors are effective at the high frequencies, resulting in an average gain of 7 or 8 dBi over a bandwidth of nearly 2 to 1, which is adequate for the UHF U.S. TV band. Although less than the gain of a full-element corner reflector antenna, the hybrid has less wind resistance.
In the competitive high-volume UHF TV antenna market, in which mil-lions of corner reflector and Yagi-Uda antennas have been sold, the hybrid offers an alternative. At least half of the reflector elements can be removed from the 484 ]| ARRAYS OF DIPOl-tS AND OF APERTURES Figure 11 -42 Square-corner-YagMJda hybrid anten-na.
It has an average directivity of about 8 dBi over a 2 to I bandwidth.
corner reflector. The directors which are added are much shorter and may be fewer in number, resulting in a net savings in cost.
Compare the wide bandwidth comer reflector of Fig.
12-14 and also the Yagi-Uda corner-log-periodic array described in Sec.
15-6, 1 1-9(1 Circular polarization with a Yagi-Uda Antenna. To produce circular polarization, 2 Yagi-Uda antennas can be crossed (elements at right angles on the same boom) with the driven elements fed in phase quadrature, or both driven elements can be fed in phase but with one array displaced A/4 along the boom with respect to the other. Another alternative is to feed the crossed director pairs with a monofilar axial-mode helical antenna (see Fig, 7-55). An advantage of this arrangement ts that it can be fed by a single coaxial transmission line.
11 -9e The Landsdorfer Shaped-Dipole Array, The gain of a Yagi-Uda antenna can be increased by adding more directors and increasing the length of the array. As with all end-fire arrays, -twice the gain (3 dB improvement) requires a 4-fold increase in length. Another method of obtaining a 3 dB improvement is to stack 2 arrays. As yet another alternative, Landsdorfer 1 has demonstrated that higher gain can be obtained by extending and shaping the conductors of a 3-element close-spaced Yagi-Uda antenna.
Consider the center-fed 3 a/2 dipole shown in Fig.
1 l-43a. Assuming a sinus-oidal current distribution, the field pattern is as indicated in {h) (see also Fig.
5-9).
There are small broadside lobes and also large lobes at an angle. If the center A/2 1 F. M. Landsdorfer "Zur Optimalen Form von Linearantennen/’ Frequertz, 30 344-349, 1976.
F M. Landsdorfer, “A New Type of Directional Antenna," Ant, Prop. Soc.
Jut. Symp. Digest, 169-172, 1976.
F. M. Landsdorfer and R. R. Sacher, Optimization of Wire Amentias, Wiley. 1985.
n-10 PHASED ARRAYS 485 Field patterns Figure 11-43 fti) A 3/ 2 center-fed dipole and its field pattern.
(<-) Center section folded into a A/4 stub reducing length to 1/ and (d) the field pattern, b?} Further shaping and addition of shaped director and reflector forms a Landsdorfer array with the field pattern [fy section is folded into a A/4 stub as in (c) the antenna reduces to an in-phase 1/ dipole with a bidirectional broadside field pattern as in (d). Now pulling the stub apart and shaping it and the A/2 sections, it can be arranged with a similarly shaped director and reflector as done by Landsdorfer and shown in (e) with the unidirectional pattern at (/). The overall length is 1.3A with gain reported to be about 11,5 dBi. This compares to about 8.5 dBi for a close-spaced 3-eiement array of A/2 dipoles.
11-10 PHASED ARRAYS 11-IOa Introduction. Although the elements of any antenna array must be phased in some manner, the term phased array has come to mean an array of many elements with the phase (and also, in general, the amplitude) of each ele-ment being a variable, providing control of the beam direction and pattern shape including sidelpbes. These arrays are discussed in the second part of this section.
Specialized phased arrays given different names are the frequency scanning array, the retro-array and the adaptive array.
In the scanning array, phase change is accomplished by varying the fre-quency. These frequency scanning arrays are among the simplest phased arrays 486 11 ARRAYS OP DIPOLES AMD OF APERTURES since no phase control is required at each element. Several of these arrays were discussed in Sec, 7-1 1. Additional ones arc described in the next section {l M 1).
A retro-array is one which automatically reflects an incoming signal back toward its source. This type of array is considered in Sec.
11-12.
Adaptive arrays have an awareness of their environment and adjust to it in a desired fashion.
Thus, an adaptive array can automatically steer its beam toward a desired signal while steering a null toward an undesired or interfering signal. In a more versatile adaptive array the output of each element is sampled, digitized and processed by a computer which can be programmed to accomplish tasks limited mainly by the sophistication of the computer program and the available computer power. Such an array may be called a smart antenna. These arrays are described in Sec. 11-13.
11 -10b Phased Array Designs. An objective of a phased array is to accom-plish beam steering without the mechanical and inertial problems of rotating the entire array.
In principle, the beam steering of a phased array can be instanta-neous and, with suitable networks, all beams can be formed simultaneously.
However, the look angle or field of view of a planar phased array may be more restricted than for a steerable array (although a phased array on a curved surface may cover as much solid angle). Also the beam of a rotatable array maintains its shape with change in direction whereas a phased array beam may not.
Another objective of the phased array is to provide beam control at a fixed frequency or at any number of frequencies within a certain bandwidth in a frequency-independent manner.
In its most simplistic form, beam steering of a phased array can be done by mechanical switching. Thus, consider the case of the rudimentary 3-element array of Fig.
I l-44a. Let each element be a X/2 dipole (seen end-on in the figure). An incoming wave arriving broadside as in (a ) will induce voltages in the transmis-sion lines (or cables) in the same phase so that if all cables are of the same length — ^2 = ^ 3 ) the voltages will be in phase at the (dashed) in-phase line. By bring-ing all 3 cables to a common point as in (6), the 3-element array will operate as a broadside array.
For an impedance match, the cable to the receiver (or transmitter) should be \ the impedance of the 3 cables, or a 3 to 1 impedance transformer can be inserted at the common junction point with all cables of the same impedance.
Now consider a wave arriving at an angle of 45" from broadside as in Fig, 11-44.
If the wave velocity v = c on the cables, the (dashed) in-phase line is parallel to the wave front of the incoming wave, as suggested in (c). However, if v < c t the lengths l 2 and /3 must be increased as suggested in order for all phases to be the same (the in-phase condition). Then, if cables of these lengths are joined as in (d) the 3-element array will have its beam 45° from broadside.
By installing a switch at each antenna element and one at the common feed point as in Fig. 11-44 and mechanically ganging all switches together, the beam can be shifted from broadside to 45° by operating the ganged switch.
mo phased arrays 487 Figure 11-44 [a) Array of three 1/2 dipoles (seen end-on) with incoming wave broadside.
\b) Equal length cables joined, (c) Incoming wave at 45" from broadside.
(</) Appropriate lengths of cable jqined.
(e) Switches for shifting from broadside 10 45' reception.
By adding more switch points and more cables of appropriate length the beam can be steered to an arbitrarily large number of directions. With more elements, narrower beams can be formed. With diodes (PIN type 1 ) in place of mechanical switches, control can be electronic. However, even with these modifi-1 PIN (Positive Intrinsic Negative): high open-circuit impedance, low short-circuit impedance.
488 IL ARRAYS OF DIPOLES AND OF APERTURES cations, it is obvious that for precision beam steering with many antenna ele-ments, the required number of interconnecting cables can become astronomic Many schemes have been proposed to reduce the required number of cables Qne of these, called a Butler matrix, l is a cable-connected matrix which is the hard-wire equivalent of a discrete fast Fourier transform. For an JV element, N output-port matrix, forming N simultaneous beams, the number of required cables is reduced from N 1 to /V In N, resulting in a significant economy for large values of JV. Computers can do the same thing by appropriate programming of sampled signals.
Instead of controlling the beam by switching cables, a phase shifter can be installed at each element. Phase shifting may be accomplished by a ferrite device.
The same effect may be produced by the insertion of sections of cable (delay line) by electronic switching Thus, insertion of cables of A/4, A/2, 3A/4 (and no cable) provides phase increments of 90°, For more precise phasing, cables with smaller incremental differences are used.
Figure ll-45n is the schematic of a phased array with a phase shifter and attenuator at each element. The feed cables are all of equal length in a corporate structure 2 arrangement. Figure 1 1-456 shows an end-fed phased array, also with individual element phase shifter and attenuator Since a progressive phase shift is introduced between elements with a frequency change, the phase shifters must introduce opposing phase changes to compensate, in addition to making the desired phase changes.
Figure 1 l-45c shows a 4-element end-fed phased receiving array with each element fed from a transmission line via a directional coupler The transmission line has a matched termination (zero reflection) so that (ideally) there is a pure traveling wave on the line Phasing is accomplished by physically sliding the directional couplers along the line Element amplitude is controlled by changing the closeness of coupling. Reduction of amplitude can control or eliminate minor lobes as with a 1 : 3 : 3 : 1 (binomial) amplitude distribution.
The literature on phased arrays is extensive. Robert Maiiloux gives a very comprehensive overview of the subject, updating an earlier review article by L.
Stark/ 'J.
Butler and R.
Lowe, " Beamforming Matrix Simplifies Design of Eleccronically Scanned Antennas," Elect. Des ., 9, 170-173 1961, J.
L.
Butler, in R. C. Hansen (ed-X Microwave Scanning Hnrenmu, vol, 3, Academic Press, 1966 chap. 3 (includes 80 references}.
1 Named after the organizational structure of a corporation with president over 2 vice presidents each over 2 subordinates etc. the diagram in Fig. lt-45a being an upside-down version.
3 For explanation of directional coupler action, see J. D. Kraus flecircwwpneUts, 3rd ed., McGraw Hill, 1984, pp, 419—420 R.
J. Maiiloux, "Phased Array Theory and Technology” Proc. IEEE , 70, 246-291 March 1982 (includes 1 74 references}, L. Stark, “Microwave Theory of Phased Array Antennas—A Review," Proc. IEEE, 62, 1661-1701, 1974 11-10 PHASED ARRAYS 489 Directional Mai chad couplers terminations Figure 11-45 (a) Schematic of phased array fed by corporate structure and (6} end-fed.
(fj Phased array with each element fed from a matched transmission line via a directional coupler.
11-lOc Rotatable Helix Phased Array. With monofilar axial-mode helices as dements of the array, phasing can be accomplished by rotating the helices on their axes, a rotation of 90" providing a 90° shift in phase of the (circularly polarized) wave. For example, with 3 helices of the same hand connected as in Fig. 11-46 the beam direction can be steered by rotation of the outer helices (l and 3) with helix 2 stationary. Thus, continuously rotating helix J clockwise and helix 3 counterclockwise will result in a continuously sweeping beam between two angular extremes, as suggested in the figure. In this type of operation a lobe appears at the left extreme of the sweep angle, grows in amplitude as it sweeps to the right, reaching maximum amplitude at broadside It then becomes smaller as it sweeps further to the right. After reaching the extreme right of the sweep angle, a new lobe appears at the left extreme and the process repeats. The angle of sweep is determined by the pattern of a single helix.
I designed and built a 3-helix beam-sweeping array of this type in 1958 for operation at 25 to 35 MHz at the Ohio State University Radio Observatory for 490 H ARRAYS OF DIPOLES AND OF APERTURES MU FREQUENCY -SCANNING ARRAYS 49 1 Figure 11-46 Three-helix array with outer 2 helices rotating in opposite directions to produce continuously sweeping beam or lobe.
planetary (Jupiter) and solar radio observations.
1 The helices were 3 m in diam-eter, the outer 2 rotating in opposite directions. Each helix had 3 turns so that the beam width between first nulls (and sweep angle) was about 130°. With helix rotation at 3 revolutions per hour, the equatorial zone of the sky was swept or scanned from east to west every 20 min.
11-11 FREQUENCY-SCANNING ARRAYS 11-1 la Frequency-Scanning Line-Fed Array, Consider a lineTed array of uniformly spaced elements (dipoles) with a receiver connected at the right end of the line as suggested by the schematic of Fig.
1 147. Each element is fed from the transmission line via a directional coupler. This arrangement is similar to that in Fig. ll-45c but in the present case the coupler positions are fixed, with beam sweeping or scanning done by changing the frequency. The transmission line is matched to eliminate reflections and insure an essentially pure traveling wave on the line. From (7-1 1-8), cos = I _5L P + d/xo ( 1 ) where = beam angle from array axis, rad or deg p = phase velocity on transmission line = v/c, dimensionless m — mode number, dimensionless d — element spacing, m A0 = free-space wavelength at center frequency of array operation, m Figure 1 1-47 Frequency-scanning line-fed array of uniformly spaced elemertls with tunable receiver •M left end. Beam angle is a function of the frequency.
For p = 1 and m — 0, - 0 (beam fixed at end-fire independent of / 0 ), Consider now the situation for p — 1, m — — 1, d = 1 m and A 0 = 1 m, or cos —! — 1 = 0 (2) and tp = 90 (beam broadside).
Suppose next that the frequency is increased so that the wavelength is 0,9A o or 0.9 m; then cos (ft = 1 - 0.9 = 0.1 (3) and (p = 84.3 , or 5,7 right of broadside. Shifting to a lower frequency so that the wavelength is 1,l/.<yor 1.1 m.
cos =.
1 -1.1 - -0 1 and — 95.7\ or 5,7 left of broadside.
Thus, a + 10 percent shift in wavelength (or frequency) swings the beam ±5.7-' from broadside (total scan 11.4 ) with larger frequency shifts resulting in larger scan angles. To eliminate or reduce beams at = —90 (mirror image), and also end-fire and back-fire beams, the a 2 dipoles can be replaced by identi-cal unidirectional elements, the scan angle then being restricted to the beam width of the individual element.
This frequency-scanning array has no moving parts, no phase shifters and no switches, making it one of the simplest types of phased arrays.
11-1 lb Frequency-Scanning Backward Angle-Fire Grid and Chain Arrays, 1 The current on folded-wire antennas usually assumes a sinusoidal (standing-wave) distribution. In the 1930s 1 had constructed and used folded-wire 1 Some cull ihe backward angle-fire grid array a Kraus grid and the frequency-scanning backward tingle-fire chain array a T\uri chain l J. D. Kraus, “Planetary and Solar Radio Emission at 11 m Wavelength, Froc. IRE, 46, 266-274, January 1958, J, D. Kraus, Radio Astronomy, 2nd ed., Cygnus-Quasar, 1986, p. 6-40 (1st ed„ J966, p. 192) Figure 11-48 Frequency -scanning Kraus grid array. When fed a l point 2 (with terminals at point 1 short-circuited and a matched load connected at point 3) the beam is at an angle which is a function of the frequency, A frequency shift of ± 14 percent swings the beam angle $ through about 75 Switching the feed point to 3 and load to 2 puts the beam in the right-hand quadrant, mating the total scan angle J 5CT. Typically, l ^ a, s < 2/2, h < a/4.
antennas of this kind, such as the Bruce curtain (Fig 1 1-586), and became fam-iliar with their operation. A broadside array of these antennas (for increased gain) may require many interconnecting transmission lines to feed them In thinking about these arrays during the winter of 1961-1962, I wondered if it would be possible to construct a continuous wire grid as a broadside array and feed it at a single centrally located point.
The basic arrangement is illustrated in Fig.
1 1-48. The dimensions of each mesh of the grid are / ss A by s 2= A/2. Assuming standing waves, the instantane-ous current distribution would be as indicated by the arrows, one located at each current maximum point. Currents on all of the short sides (A/2 long, horizontal in the figure) would be in phase, while on the long sides of the meshes (A long, vertical in the figure) there are as many current maxima in one direction as in the opposite so that radiation broadside from the long sides should (ideally) be zero.
Thus, I reasoned, the array should produce a linearly polarized (horizontal in the figure) beam broadside to the array with a gain proportional to the number of A/2 sides (or elements) (31 in the figure).
LM1 F REQCENCY-5CANN I N(J ARRAYS 493 I constructed an array similar to the one in Fig 11-48, mounted it approx-imately A/4 from a flat conducting ground plane and fed it with a balanced trans-mission line at the central point (1 in the figure). To my surprise, I found that the radiation was not in a single broadside beam but was split into 2 equal lobes, one left and one right of broadside.
It was apparent that the current distribution was not that of a resonant standing wave but rather of two traveling waves, one to the left and one to the right from point 1 Accordingly, in order to have only one (left-to-right) traveling wave across the entire grid, T short-circuited the terminals at point l and fed the grid with a coaxial line at the left edge (point 2). This resulted in a single beam in the back-fire direction (opposite to the traveling wave) as indicated in Fig.
1 1-48, with the beam angle — 2.37 — —77 s/A (5) (6 ) D. Kraus, “A Backward Angle-Fire Array Antenna," IEEE Trans, Ants, Prop., AP-12, 48-50, January 1964.
J, D. Kraus, "Backward Angle Traveling Wave Wire Mesh Antenna Array/ 1 LLS, Patent 3,290,688, applied for June 1 1, 1962, granted Dec. 6, 1966.
494 II ARRAYS Of-' DIPOLES AND OF APERTURES Figure 11-49 Beam angle ft as a function of frequency expressed in terms of s/7. The dashed curves are calculated for different values of relative phase velocity p. The solid curve is measured, suggesting that p is a (weak) function of the fre-quency.
The beam direction 0 H (complement to (f> in Fig. 11-48} varies from about 15 to 9 4- (A/2)]. Typically, f ~ 2,75s so that the effective phase velocity is about 0.4c, making the grid a slow-wave structure. The average gain of a grid array as in Fig.
1 1-48 is about 17 dBi.
A microstrip version of the array is described by Conti et ai l The Kraus grid array principle has been extended by Tiuri, Tallqvist and Urpo 3 and by Tallqvist 3 to an array in which the meshes are divided into parallel matched chains, with a typical configuration as in Fig. 11-50. This Tiuri chain 1 R. Conti, J. Toth, T. Dowling and J. Weiss, "The Wire Grid Microstrip Antenna,'’ IEEE Trans.
Tnrs. Prop., AP-29, 157-166, January 19S1.
2 . M. Tiuri. S Tallqvist and S. Urpo, “Chain Antenna," IEEE Ant . Prop. Soe.
fn/. Symp. Digest , June 10, 1974, 3 S.
Tallqvist, "Theory of the Meander Type Chain and Grid Antenna” Lie. thesis, Helsinki Uni-versity of Technology, 1977.
496 IL ARRAVS OK DlPQI-ES AND OK APERTURES 1L 13 ADAPTIVE ARRAVS AND SMART ANTENNAS 497 design is also well adapted to microstrip or printed circuit construction. The array shown has an endpoint input impedance of 50 ft for an impedance of 300 ft for the individual chains. The current attenuation from input to matched output is about 10 dB, which is considered optimum. Higher attenuation reduces the gain due to the larger taper in current distribution while lower attenuation lowers the gain because more power is lost in the matched load. Average aperture efficiencies arc typically about 50 percent.
By bending the chain elements, Hcndriksson, Markus and Tiuri have devel-oped a circularly polarized chain array, 1 11-12 RETRO-ARRAYS, THE VAN ATTA ARRAY, If a wave inci-dent on an array is received and transmitted back in the same direction, the array acts as a retro-reflector or retro-array.
The passive square-corner reflector (Sec, 12-36) does the same thing.
In general, each element of a retro-array reradiates a signal which is the conjugate of the received signal. The Van Atta array of Fig, 1 1-51 is an example.
2 The 8 identical elements may be 2/2 dipoles, shown in end view in the figure.
With element pairs (1 and 8, 2 and 7, 3 and 6, and 4 and 5) connected by identical equal-length cables, as indicated, a wave arriving at any angle 4> is transmitted back in the same direction. The array shown in Fig.
11-51, like the square-corner reflector, is passive. An adaptive (active) array (Sec.
11-13) can also be made rctrodirective by using a mixer to produce a conjugate phase shift for each dementi An advantage of an active array is that the elements need not be arranged in a line or, in a 2-dimensional case, in a plane. Active retro-arrays can also incorporate amplifiers, 4 11-13 ADAPTIVE ARRAYS AND SMART ANTENNAS- The antenna elements and their transmission-line interconnections discussed so far produce a beam or beams in predetermined directions. Thus, when receiving, these arrays look in a given direction regardless of whether any signals are arriv-ing from that direction or not. However, by processing the signals from the indi-vidual elements, an array can become active and react intelligently to its environment, steering its beam toward a desired signal while simultaneously steering a null toward an undesired, interfering signal and thereby maximizing the signal- to-noise ratio of the desired signal. The term adaptive array is applied to this kind of antenna.
1 J. Hendrik sson, K. Markus and M. Tiuri, "A Circularly Polarized Traveling-Wave Chain Antenna, 1" European Microwave Conf. y Brighton, September 1979.
1 L. C. Van Atta, “ Electromagnetic Reflector," U.S, Patent 2,909,002, Oct. 6, 1959, 1 C. Y. Pon. “ Retrodireclive Array Using the Heterodyne Method " IEEE Trans. Ants. Prop. y AP-12, 176- ISO, March 1964.
4 S. N. Andre and D. J r Leonard, “An Active Retrodirective Array for Satellite Communications," IEEE Trans. Anis. Prop. y AP-12, 181 -186, March 1964.
Figure 1 1-51 Eight-element Van Atta reiro-array Element pairs are connected by equal length lines.
Also, by suitable signal processing, performance may be further enhanced, giving simulated patterns' of higher resolution and lower sidelobes.
In addition, by appropriate sampling and digitizing the signals at the terminals of each dement and processing them with a computer, a very intelligent or smart antenna can, in principle, be built. For a given number of elements, such an antenna’s capabilities are limited, mainly by the ingenuity of the programmer and the avail-able computer power. Thus, for example, multiple beams may be simultaneously directed toward many signals arriving from different directions within the field of view of the antenna (ideally =2n sr for a planar array). These antennas are some-times called Digital Beam Forming (DBF) antennas, 2 As a rudimentary example of an adaptive array, a simple 2-element system is shown in Fig. 11-52 with 2/2 spacing between the elements at the signal fre-quency^. Let each element be a 2/2 dipole seen end-on in Fig. 11-52 so that the patterns of the elements are uniform in the plane of the page. With elements operating in phase, the beam is broadside (up in the figure).
Consider now the case of a signal at 30° from broadside as suggested in Fig. 11-52 so that the wave arriving at element 2 travels 2/4 farther than to element 1, thus retarding the phase of the signal by 90° at element 2, Each element is equipped with its own mixer. Voltage-Con trolled Oscillator (VCO), intermediate frequency amplifier and phase detector. An oscillator at the interme-1 Simulated patterns are ones that exist only in the sign a!- processing domain.
2 H. Slcyskal, 'Digital Beamforming Antennas," Microwave J,, 30. 107-124, January 1987.
498 n ARRAYS OF DIPOLES AND OF APERTURES Figure 1 1-52 Two-element adaptive array with signal -processing circuitry.
diate frequency / 0 is connected to each phase detector as reference. The phase detector compares the phase of the downshifted signal with the phase oj the refer-ence oscillator and produces a voltage proportional to the phase difference. This voltage, in turn, advances or retards the phase of the VCO output so as to reduce the phase difference to zero (phase locking). The voltage for the VCO of element 1 would ideally be equal in magnitude but of opposite sign to the voltage for the VCO of dement 2 so that the downshifted signals from both elements are locked in phase, making \ = 2 = o where y = phase of downshifted signal from element 1 <p 2 = phase of downshifted signal from element 2 0 rj = phase of reference oscillator With equal gain from both IF amplifiers the voltages V y and V2 from both elements should be equal so that Vi&i = Vi&i & making the voltage from the summing amplifier proportional to 2Vy { — 2V2 ) and maximizing the response of the array to the incoming signal by steering the beam onto the incoming signal. In our example, 45° phase corrections of opposite sign would be required by the VCOs{ + for element 1, —for dement 2), ll-n ADAPTIVE ARRAYS AND SMART ANTENNAS 499 Figure 11-53 Patterns of 2-elemern adaptive array for signals from 0 and 30 : directions. For the 0 signal, nulls are at 90 and 270 while for the 30' signal, nulls are at 210 and 330'. These patterns are identical with those of Figs. 4-1 and 4-4.
In our rudimentary 2-element example, the beam will be in the (F direction for a signal from the 0° direction and at 3CP for a signal from that direction, as shown by the patterns in Fig. 1F53.
If interfering signals are arriving from the 210 and 330° directions when the main signal is at 30°, the nulls at 210 and 33(T will suppress the interference. However, an interfering signal at 150“ would be at a pattern maximum, the same as the desired signal at 30°. To provide more effective adaptation to its environment, an array with more elements and more sophisticated signal processing is required For example, the main beam may be steered toward the desired signal by changing the progressive phase difference between elements, while, independently, one or more nulls are steered toward interfering signals by modifying the array element amplitudes with digitally con-trolled attenuators.
ll-13a Literature on Adaptive Arrays. There is an extensive literature on adaptive arrays. Three special issues of the IEEE Transactions on Antennas and Propagation have been devoted to adaptive arrays. They are; vol. AP-12, March 1964; vol AP-24, September 1976; and vol. AP-34, March 1986. Additional refer-ences are as follows: Bickmore, R. W.: "Time Versus Space in Antenna Theory,' 1 in R. C Hansen (edA Microwave Scanning Antennas , vol. 3, Academic Press, 1966, pp. 289-339.
Blank, S.: "An Algorithm for the Empirical Optimization of Antenna Arrays," IEEE Trans. Ants , Prop., AP-31, 685^689, July 1983.
Butler, J.
L.: "Digital, Matrix, and Intermediate Frequency Scanning," in R. C.
Hansen (ed.). Microwave Scanning Antennas , vol.
3, Academic Press, 1966, pp. 217-288.
Dinger, R.
J.
: "A Computer Study of Interference Nulling. by Reactively Steered Adaptive Arrays,” Ant . Prop . Soc. Int. Symp , Proc., 2, 807-810, 1984.
Einarsson, O, : Optimization of Planar Arrays, IEEE Trans. Ants.
Prop., AP-27, 86-92, January 1979.
500 11 ARRAYS OF DIPOLES AND OF APFRTUfttS Ersoy, O.: "Real Discrete Fourier Transform,” /£££ Trans.
Acoustics , Speech ’ and Signal Processing , ASSP-33, 880-882, August 1985.
Fan, H., E. L El-Masry and W t K. Jenkins: 44 Resolution Enhancement of Digital Beamformers," /£££ Trans, Acoustics , Speech and Signal Processing, ASS P-32, 1041 1052, October 1984.
Griffiths, L.
J., and C, W, Jim: "An Alternative Approach to Linearly Con-strained Adaptive Beamform i n g,” IEEE Trans.
Ants , Prop A P-30, January 1982.
Gupta, L J., and A. A. Ksienski: "Effect of Mutual Coupling on the Performance of Adaptive Arrays,” /£££ Trans. Ants. Prop ., AP-31, 785-791, September 1983.
Hansen, R. C: "Gain Limitations of Large Antennas,” JRE Trans. Ants. Prop., AP-8, 491-495, September I960.
Hatcher, B. R.
: "Granularity of Beam Positions in Digital Phased Arrays,” Proc.
/£££, 56, November 1968.
Johnson, H. W., and C S.
Burrus: "The Design of Optimal DFT Algorithms Lf s ing Dynamic Programm ingJ 1 /£££ Trans.
A co u sr ic s Speech and Signal Processing, ASSP-31, 378—387, April 1983.
Laxpati, S.
R.: 44 Planar Array Synthesis with Prescribed Pattern Nulls,” J£££ Tnm.s. /Infs. Prop., AP-30, 1 176-1183, November 1982.
Mucci, R.
A.: “A Comparison of Efficient Beamforming Algorithms,” /£££ Trans , Acoustics Speech , and Signal Processing , ASSP-32, 548-558, June 1984.
Ricardi, L.
J,; "Adaptive Antennas,” in R. C.
Johnson and H.
Jasik (edsjj.
Antenna Engineering Handbook , McGraw-Hill, 1984, chap. 22, Shelton, J.
P, T and K.
S. Kelleher: “Multiple Beams from Linear Arrays,” IRE Trans. Ants. Prop,, A P-9, 154-161, March 196 L Sorenson, H. V„ M, T. Heideman and C.
S, Burrus: "On Computing the Split-Radix FFT,” /£££ Trans. Acoustics, Speech , and Signal Processing , ASSP-34, 152-156, February 1986.
Steinberg, B + D t : "Design Approach for a High-Resolution Microwave Imaging Radio Camera,” J . Franklin Inst., 296, 41 5—432, December 1973, Volakis, J. L + , and J. D. Young: "Phase Linearization of a Broad-Band Antenna Response in Time Domain,” /£££ Trans, Ants.
Prop., AP-30, 309-313, March 1982.
Vu, T.
B.: “Null Steering by Controlling Current Amplitudes Only," Ant, Prop.
Soc. Int. Sytnp. Proc., 2, 811-814, 1984.
Waldron, T. P,, S. K, Chin and R. J. Naster: “Distributed Beamsteering Control of Phased Array Radars,” Microwave J ., 29, 133-146, September 1986.
Weber, M, E., and R, Heisler: 44A Frequency-Domain Beamforming Algorithm for Wideband Coherent Signal Processing,” J.
Acoustic Society of America , 76, October 1984.
Wheeler, H.
A.: “The Grating-Lobe Series for the Impedance Variation in a Planar Phased-Array Antenna," /£££ Trans. Ants Prop ., AP-14, 707-714, November 1966.
L I -1 S low-sidelobf ARRAYS 501 Ctiioradti.) 11-14 MICROSIRIP ARRAYS.
Printed circuit and microstrip techniques facilitate the construction of multielement arrays for microwave frequencies, 1 The Kraus grid and Tiuri chain arrays (Sec.
11-116) are examples. The 896-element microstrip antenna for space research shown in Fig. M-54 is another.
All ele-ments are photoctched from one side of a printed circuit board. The corporate structure feed has amplitude taper in the narrow dimension. The 9,5 x 2.4 m array has a 34-dBi gain at /. ^ 30 cm.
Although the microstrip element is inherently narrowband, log-periodic patch arrays have achieved bandwidths of 4 to 1 , 2 Patch (or microstrip) antennas are discussed further in Chap, 16, 11-15 LOW-SIDELOBE A R RAYS.
Ideally it may be desirable for an antenna to have a narrow, well-defined beam with no sidelobes, or at least none above a certain prescribed level, A prime factor affecting the sidelobe level is the aperture distribution (Sec, 1 1-22), Although some distributions yield (theoretically) zero sidelobes, they may be more difficult of realization on reflector antennas than on phased arrays. The typical large parabolic reflector antenna (Chap, 12} may also have significant sidelobes due to other causes such as aper-ture blocking and diffraction from its prime focus feed or Cassegrain reflector, from struts of the supporting structure, from feed spillover, from surface irregu-larities and from the edge of the parabola. The edge diffraction may be reduced by using a rolled edge (sec Sec.
18-3J). Although offsel feeds or integral horn reflectors may have significantly lower minor lobes, accurate construction tech-niques providing precise amplitude and phase control of all dements of a phased 1 H. G. Oilman and D. A- Huebner, “ Electromagnelically Coupled Microstrip Dipoles,” tE.EE Trans.
4n< 5 , Prop, T AP-29, 1 51 J 57, January 198].
P.
S. Hall, “Microstrip Antenna Array wiih Mulfi-Gciave Bandwidth7 Microwave 7., 29, 131-139 March 1986.
502 11 ARRAYS OF DIPOLES AND OF APERTL'RFS array have resulted in phased array designs wiui sidelobe levels down 50 dB or rru>re. This is as good or better than has been achieved to date with reflector antennas.
1 However, phased arrays are inherently narrower band than reflector antennas, U-16 LONG-WIRE ANTENNAS Most of the preceding parts of this chapter deal with arrays of individual, discrete elements (usually //2 long) inter-connected by transmission lines. A linear wire antenna, many wavelengths long, may also be regarded as an array of kfl elements but connected in a continuous linear fashion with each dement serving as both a radiator and a transmission line. The long-wire antennas discussed in this section are the V, rhombic and Beverage types. The V antenna may be. either unterminated {with standing wave) or terminated {with traveling wave). The rhombic and Beverage antennas are almost always terminated (with traveling wave) 11 -16a V Antennas, 3 By assuming a sinusoidal (standing-wave) current dis-tribution, the pattern of a long thin wire antenna can be calculated as described in Chap.
5. A typical pattern is shown in Fig.
I l-55u for a wire 2k long. The main lobes are at an angle - W with respect to the wire. By arranging two such wires in a V with an included angle y = 72° as in Fig. 11-556, a bidirectional pattern can be obtained. This pattern is the sum of the patterns of the individual wires or legs. Although an included angle y = results in the alignment of the major lobes at zero elevation angle (wires horizontal) and in free space, it is necessary to make y somewhat less than 2/? in order to obtain alignment at ele-vation angles greater than zero.’ This is because the space pattern of a single wire is conical, being obtained by revolving the pattern of Fig.
1 1 -55a, for example, with the wire as the axis.
If the legs of the thin-wire V antenna are terminated in their characteristic impedance, as in Fig.
11 -55c, so that the wires carry only an outgoing traveling wave, the back-radiation is greatly reduced. The patterns of the individual wires can be calculated, assuming a single traveling wave as done in Chap.
5.
A similar effect may be produced without terminations by the use of V conductors of considerable thickness. The reflected, wave on such a conductor may be small compared to the outgoing wave, and a condition approaching that of a single traveling (outgoing) wave may result. For example, a V antenna con-] H. E Schrank, 11 Low Sidelobe Phased Array Antennas," IEEE Ant. Prop. Soc , Newsletter, 25, 5-9, April 1983.
' P.
S.
Carter, C. W.
Hansell and N.
E.
Lindenblad, Development of Directive Transmitting Antennas by R. C. A. Communications, Inc.." Proc. IRE , 19, 1773-1842, October 1931.
P.
S. Carter, "Circuit Relations in Radiating Systems and Applications to Antenna Problems" Proc.
IRE, 20, 1004 1041, June 1932.
The A R R.L. Antenna Book, American Radio Relay League, West Hartford, Conn,, 1984, p. 7-4.
11-16 LONG-WIRE ANTENNAS 503 Terminated V- antenna Figure 1 1-55 (a) Calculated pattern of 22 wire with standing wave, (fr) V antenna of two such wires, (d terminated V antenna with legs 22 Long and (d) V antenna of cylindrical conductors 1.25a long with measured pattern.
sisting of two cylindrical conductors 125k long and kj20 diameter with an included angle y = 90° has the highly unidirectional pattern 1 of Fig 1 l-55d.
11-16b Rhombic .Antennas2 A rhombic antenna may be regarded as a double-V type. The wires at the end remote from the feed end are in dose prox-imity, as in Fig. ll-56n. A terminating resistance, usually 600 to 800 0, can be l A. Dome, in Very High Frequency Techniques, Radio Research Laboratory Staff, McGraw-Hill, New York, 1947, chap. 4, p.
1 15.
1 E.
Bruce, Development in Short-wave Directive Antennas," Proc. IRE, 19, 1406-1433, August 1931.
E.
Bruce, A. C Beck and L. R. Lowry, ^Horizontal Rhombic Antennas,” Proc. IRE , 23, 24—46, January 1935, A. E. Harper, Rhombic Antenna Design, Van Nostrand, -New York, 194L Donald Foster, “ Radiation from Rhombic Antennas," Proc. IRE, 25, 1327-1353, October 1937, 504 1] ARRAYS OF DIPOLES AND OF APERTURES Figure 11-56 Terminated rhombic antenna (u) with azimuthal pattern [b ) and vertical plane pattern U'l f° r a rhombic 6/ long on each leg, ^ — 70', and al a height of 1.1a above a perfectly conducting ground. [After A. E. Harper.
Rhombic Antenna Design, Van j\os-trand , New York 1941.) conveniently connected at this location so that there is substantially a single outgoing traveling wave on the wires. The length of each leg is L , and half of the included side angle is $. The calculated patterns 1 of a terminated rhombic with legs 6/ long are shown in Fig. 11-56/j and c. The rhombic is assumed to be 1.1/ above a perfectly conducting ground, and = 70°, In designing a rhombic antenna, the angle <£, the leg length and the height above ground may be so chosen that (1) the maximum of the main lobe coincides with the desired elevation angle a (alignment design) or (2) the maximum relative field intensity E for a constant antenna current is obtained at the desired ele-vation angle a (maximum E design). 2 If the height above ground is less than that required for these designs, align-ment may be obtained by increasing the leg length.
If the height is maintained but the leg length is reduced, alignment may be obtained by changing the angle As a third possibility, if both the height and the leg length are reduced, the angle $ can be changed to produce alignment. Any of these 3 modifications results in a so-called compromise design 2 having reduced gain. If moderate depar-tures from optimum performance are acceptable, a rhombic antenna can be oper-ated without adjustment over a frequency band of the order of 2 to 1.
1 From A. E. Harper, Rhombic Antenna Design Van Noslrand, New York, 1941.
11-16 LONG-WIRE ANTENNAS 505 The pattern of a rhombic antenna may be calculated as the sum of the patterns of four tilted wires each with a single outgoing traveling wave. The effect of a perfectly conducting ground may be introduced by the method of images.
For a horizontal rhombic of perfectly conducting wire above a perfectly conduct-ing plane ground, Bruce, Beck and Lowry 2 give the relative field intensity E in the vertical plane coincident with the rhombic axis 3 as a function of a, $ , L x and H x as (cos = half included side angle of rhombic antenna Hi = Hia = height of rhombic antenna above ground L x = Lia = leg length Hf = 2 tiH k — 2k{H/a) Lr = 2nL A = InL/k $ = (1 — sin cos a)/2 A uniform antenna current is assumed and mutual coupling is neglected.
Following the procedure of Bruce, Beck and Lowry, the various parameters may be determined as follows. For the maximum E condition, £ is maximized with respect to that is, w re make which yields which is satisfied when where n = 1, 3, 5, .
, , cos (2k sin a) = 0 2nH i sin a = n -2 For the lowest practical height, n — L Therefore, 4 sin a 2 E. Bruce, A. C- Beck anti L.
R. Lowry, "Horizontal Rhombic Antennas," Pror. IRE.
23, 24—26, January 1935.
s The radiation in this plane is horizontally polarized. However, in other planes the polarization h not, in general, horizontal.
506 M ARRAYS Of-' DIPOLES AND OK APERTURES Equation (3) gives the height H A for the antenna. To find the leg length, E is maximized with respect to y obtaining 1 2(1 — sin 0 cos a) Finally, by maximizing E with respect to and the leg length in wavelengths L A , for maximum E at the desired elevation angle a. This is for a uniform antenna current It does not follow that the field intensity at the desired elevation angle is a maximum for a given power input to the antenna. However, it is probably very close to this condition.
It is also of interest that for the maximum E condition the maximum point of the main lobe of radiation is not, in general, aligned with the desired elevation angle.
In the alignment design the maximum point of the main lobe of radiation is aligned with the desired elevation angle cl. For this condition, E at 7 is slightly less than for the maximum E condition. Alignment is accomplished by maximiz-ing E with respect to a. and introducing the condition of (3), This gives = -:— 7 w 1 - sin ^ cos 7 Substituting (7) in (1) and maximizing the resulting relation for the field with respect to tp gives \ and for alignment of the maximum point of the main lobe of radiation with the desired elevation angle a.
Only the length is different in the alignment design, being 0.37 1/05 = 0.74 of the value for the maximum E design.
The above design relations are summarized in Table ll-l together with design formulas for 3 kinds of compromise designs.
An end-to-end receiving array of a number of rhombics may be so con-nected as to provide an electrically controllable vertical plane pattern which can u-i LONG-WIRE antennas 507 Table 11-1 Design formulas for terminated rhombic antennas!
Type or rhombic antenna Formulas Maximum E at elevation angle a ^ = 90 - a Alignment of major lobe with elevation angle 1 \ - W — a Reduced height H Compromise design for alignment at elevation angle 2 $ = 90 - i tan [(xrLi) sin 2 3 sin 2 a] 1 H'.
in a [ 2 ji sin a tan (//; sin 2 ) IV IV where If\— — and //; = 2rr — Reduced length L Compromise design for alignment at elevation angle a Reduced height IV and length L Compromise design for alignment at elevation angle a = arcsi .
~L\ - 0.371" esm — L\ cos 1 where L\ = L/X Solve this equation for rfr.
\ = 1 l\ sin tan a tan (Hi sin a) 4rr0 tan (ijfU r ) 1 - sin 0 cos a L where \j> = and L =2n-+ After E. Bruce, A- C- Beck, and L.
R. la>wry. "’Horizontal Rhombic Anlentiai" Frw.
7R£,23. 24-26, January m$.
be adjusted to coincide with the optimum elevation angle of downcoming waves.
This Multiple Unit Steerable Antenna, 1 or MUSA, is a vertically steerable system of this kind for long-distance short-wave reception of horizontally polarized downcoming waves.
1 H. T.
Friis and C.
B. Feldman, "A Multiple Unit Steerable Antenna for Short-Wave Reception," Proc. IRE, 25, 841 9 1 7 + July 1937.
SOS II ARRAYS f)V DIPOLLS AND OF APERTURES Ground Hgurr 11-57 Ut\ Wave front over a perfect conductor.
[h] Wave front over imperfect conductor.
ic\ Beverage antenna.
11 -16c Beverage Antennas. The electric field of a wave traveling along a per-fectly conducting surface is perpendicular to the surface as in Fig, 1 1 -57a.
However, if the surface is an imperfect conductor, such as the earth's surface or ground, the electric field lines have a forward tilt near the surface as in Fig, 1 1-576. Hence, the field at the surface has a vertical component E y and a horizontal component Ex .
] The component Ex is associated with that part of the wave that enters the surface and is dissipated as heat. The Ey component con-tinues to travel along the surface.
The fact that a horizontal component E x exists is applied in the wave antenna of Beverage, Rice and Kellogg for receiving vertically polarized waves, 2 This antenna consists of a long horizontal wire terminated in its characteristic impedance at the end toward the transmitting station as in Fig, 11 -57c. The ground acts as the imperfect conductor. The emfs induced along the antenna by the E x component, as the wave travels toward the receiver, all add up in the same phase at the receiver. Energy from a wave arriving from the opposite direction is 1 Actually the wave exhibits elliptical cross-field, ie., the electric vector describes an ellipse whose plane is parallel to the direction of propagation. However, the axial ratio of this ellipse is usually very large 2 H, H.
Beverage, C. W. Rice and B. W, Kellogg, "The Wave Antenna, a New Type of Highly Directive Antenna," Truna. A IEE , 42, 215, 1923.
U-17 CURTAIN ARRAYS 509 Figure 11-58 (a) Array of X}2 dipoles with reflectors, (b) symmetrical Brut antenna, FI Sierba curtain array and Chireix-Mesny array. Arrows indicate instantaneous current directions and dots indicate current minimum points.
largely absorbed in the termination. Hence, the antenna exhibits a directional pattern in the horizontal plane with maximum response in the direction of the termination (to the left in Fig.
ll-57c). The Beverage antenna finds application as a receiving antenna in the low- and medium-frequency range, 11-17 CURTAIN ARRAYS.
In short-wave communications the curtain type of array finds many applications. As an example, a curtain type is illustrated in Fig, ll-58u that consists of an array of A/2 dipoles with a similar curtain at a distance of about A/4 acting as a reflector, 1 If the array is large in terms of wave-lengths, the reflector curtain is nearly equivalent to a large sheet reflector.
H. Briickmann, Antenmn, [hr? Theorie und Technik, S. Hirzel, Leipzig, 1939, p. 300, 510 li ARRAYS or DIPOLES AND OF APERTURES im folded dipole antennas 511 Figure 11-59 (u| Loop with 2-wire feed for horizontal polarization, {b) loop with 1-wire feed from L-oaml line for vertical polarization.
It) center-fed broadside array of two 2/2 dipoles, (ef> end-fed end -tire array of two / 2 dipoles and fo) end- fed broadside array of two 2/2 dipoles. Arrows indicate instantaneous current directions and dots indicate current minimum points.
Several other examples of curtain arrays are the Bruce type of Fig.
11-586, the Sterha type 1 of Fig. 11-5&? and the Chireix-Mesny type 2 of Fig r ll-58d. The arrows are located at or near current maxima and indicate the instantaneous current direction. The small dots indicate the locations of current minima.
1 1-18 LOCATION AND METHOD OF FEEDING ANTENNAS.
It is interesting to note the effect that the method and location of feeding has on the characteristics of an antenna. As illustrations, let us consider the following cases.
If an antenna is fed with a balanced 2-wire line, equal out-of-phase currents must flow at the feed point. Thus, a square loop 1A in perimeter and fed at the bottom as in Fig.
1 1 -59a must have the current distribution indicated. The 1 El.
J. Sterba.
‘ l Theoretical and Practical Aspects of Directional Transmitting Systems,” Proc. IRE, 19.
L 1 84-121 5, July 1931.
K. Chireix, ,l French System of Directional Aerials for Transmission on Short Waves,” Exp. Wire-less and H ireless Efi^r.. 6, 235, May 1929.
arrows indicate the instantaneous current directions and the dots the locations of current minima. The radiation normal to this loop is horizontally polarized.
Consider now the situation shown in Fig.
It -596. Here the loop is fed at the same location. However, the loop is continuous and is fed at a point by an unbal-anced line. In this case, the antenna currents flowing to the feed point are equal and in phase, so that the current distribution on the antenna must be as indi-cated. The radiation normal to this loop is vertically polarized.
The location at which an antenna is energized also may be important For example, two A/2 elements have in-phase currents when symmetrically fed as in Fig.
1 1 -59c but out-of-phase currents when fed from one end as in Fig.
t1-59<f.
For the currents to be in phase when the array is fed from one end requires that the line between the elements be transposed as in Fig, 1 -59e.
11-19 FOLDED DIPOLE ANTENNAS, A simple A/2 dipole has a ter-minal resistance of about 70 fi so that an impedance transformer is required to match this antenna to a 2-wire line of 300 to 600 Q characteristic impedance.
However, the terminal resistance of the modified A/2 dipole shown in Fig.
i l-GOa is nearly 300 Cl so that it can be directly connected to a 2-wire line having a characteristic impedance of the same value.
This 44 ultra close-spaced type of array” is called a folded dipole. More specifically the one in Fig.
11 -60a is a 2-wire folded A/2 dipole. The antenna consists of 2 closely spaced A/2 elements connected together at the outer ends. The currents in the elements are substan-tially equal and in phase.
Assuming that both conductors of the dipole have the same diameter, the approximate value of the terminal impedance may be deduced very simply as t~ C ± T d K 2 2 wire folded dipole 2 V Uf) 3-wire folded dipole Figure 11-40 Folded dipoles.
512 n ARRAVS OF DIPOLES AMD OF APERTURES follows.
1 Let the emf V applied to the antenna terminals be divided between the 2 dipoles as in Fig. 11-606. Then y = /,z rl +/ 2 z 12 (1) where l x = current at terminals of dipole 1 } 1 ~ current at terminals of dipole 2 Z 1X = self-impedance of dipole 1 Z l2 = mutual impedance of dipoles ! and 2 Since / t — I 2 , (1) becomes F = 2/ 1 (Z 11 + Z 12 ) (2) Further, since the 2 dipoles are dose together, usually d is of the order of 2/100, Zj 2 — Zn - Thus, the terminal impedance Z of the antenna is given by Z = f = 4Z„ 0) 1 Taking Z n = 70 + ;0 0 for a 2/2 dipole, the terminal impedance of the 2-wire i folded dipole becomes Z ^ 280 ft h For a 3-wire folded 2/2 dipole as in Fig.
11 -60c the terminal resistance calculated in this way is 9 x 70 = 630 ft. In general, for a folded 2/2 dipole of N wires, the terminal resistance is 10N 1 ft. Equal currents in all wires are assumed.
Several other types of folded-wire antennas 2 are shown in Fig.
11-61. The one at (a) is a 3-wire type which differs from the one in Fig, 11 -60c in that there arc no closed loops. The measured terminal resistance of this antenna is about 1200 ft. The antenna at (6) is a 4-wire type with a measured terminal resistance of about 1400 ft. Thus far, all the folded dipoles discussed have been 2./ 2 types. The total current distribution for these types is nearly sinusoidal, the same as for a simple 2/2 dipole.
Folded dipoles of length other than 2/2 are illustrated in Fig, 1 1-61 c and d. The one at (<?) is a 2-wire type 32/4 long and that at (.d > is a 4-wire type 32/8 long. The instantaneous current directions, the current distribu-tion on the individual conductors and the total current distribution are also indi-cated. Half of the 2-wire 32/4 dipole can be operated with a ground plane as in Fig.
I l-61c, yielding the 32/8 stub antenna with total current distribution shown.
' R. W.
P. King, H R. Vtimno and A. H. Wing, Transmission Lines, and Ware Guides, McGraw-Hill, New York, 1945, p. 224.
W. V.
B. Roberts, "Input Impedance of a Folded Dipole." RC4 Re t., 8, 289- 300, June 1947, which treats folded dipoles with conductors of equal diameter and also unequal diameter.
' J- D. Kraus, " Multi-wire Dipole Antennas," Electronic s f 13, 26-27, January 1940.
IL-19 FOLDED DIPOLE ANTENNas 513 Figure 11-61 (a) Three- wire folded 2 dipole, 4- wire folded k/2 dipole, {c) 2-wire 3A/4 antenna, (d) 4-wire 3A.8 antenna and e) 2-wire 3y./8 stub antenna. Arrows indicate instantaneous current direc-tions and dots indicate current minimum points.
| After Kraus.) The measured terminal resistance of the 2-wire 32/4 dipole is about 450 ft, of the 4-wire 32/8 dipole about 225 ft and of the 2-wire 32/8 stub antenna about 225 ft.
An application of the 3-wire folded 2/2 dipole of Fig.
1 1 -6 1 a to a W8JK array with 2/5 spacing is shown in Fig. U-62, 1 The impedance of each folded dipole in free space is about 1200 ft (resistive) but in the array is reduced to 300 ft, which transforms via a 2/4 600-ft line to 1200 ft. At the junction of the two transformers the impedance is half 1200 ft, or 600 ft, matching a 600-ft line to the transmitter or receiver. Thus, the W8JK array is fed entirely by lines of constant impedance (600 ft) with no resonant stubs or tuning adjustments required.
1 J. D. Kraus, "Twin-Three Flat-Top Beam Anlenna " Radio, no, 243, 10-16, November 1939 r 514 11 AAKAYS of dipoles and of apertures Figure 11-62 WBJK array, with 3-wire folded dipole dements fed by transmission Lines of constant impedance. The dipoles are separ-ated by wooden or plastic spreaders and supported by nylon rope.
11 20 MODIFICATIONS OF FOLDED DIPOLES.
Consider a 2-wire folded dipole shown in Fig.
11 -63 a. The terminal resistance is approx-imately 300 ft. By modifying the dipole to the general form shown in Fig. 11-636, a wide range of terminal resistances can be obtained, depending on the value of 0 > This arrangement is called a T-match antenna.
1 Dimensions in wavelengths for providing an impedance match to a 600 ft line are shown in Fig. ll-63c.
11-21 CONTINUOUS APERTURE DISTRIBUTION 515 Figure 11-64 la) Two wire folded dipole and (6) as modified to form singloturn loop. |r| Fourwire folded dipole and Id) as modified to form 2-turn loop.
A 2-wire folded a/2 dipole is also shown in Fig.
1 1-64a. The arrows indicate the instantaneous current direction and the small dots indicate the Locations of current minima. By pulling the dipole wires apart at the center, the single-turn loop antenna of Fig. 11-646 is obtained The length of each side is /./4. The loop has aJower terminal resistance than the folded dipole.
A 4-wire folded A/2 dipole is shown in Fig. ll-64c. This dipole is the same type as shown in Fig.
1 1-616.
It is, however, sketched in a different manner. By pulling this dipole apart at the center the 2-turn loop or quad antenna of Fig, 1 1-64d results.
The directivity of all the types shown in Fig.
1 1-64 is nearly the same as for a simple A/2 dipole, although the types at (6) and {d) are equivalent to 2 horizon-tal dipoles stacked ^O.ISA giving a small increase. With the loop types vertical and the terminals at the lowest corner, the radiation normal to the plane of the loops is horizontally polarized.
11-21 CONTINUOUS APERTURE DISTRIBUTION.
1 Extending our discussion of Sec. 4-14, consider now a continuous-current sheet or field dis-1 The following sections (1 1-21 through 11-25) are from J.
D. Kraus, Radio Astronomy. 2nd ed„ Cygnus-Quasar, 19B6 516 II ARRAYS OF DIPOLES AND OF APERTURES Aperture | Figure U-65 Aperture of width a and amplitude distribution £(xj.
tribution overman aperture as in Fig, 11-65, Assuming a current or field perpen-dicular to the page (y direction) that is uniform with respect to y, the electric field at a distance r from an elemental aperture dx dy is 1 dE = —jw ds# v — — e jfir dx dy (1) > 4 nr Z where = vector potential — jJJ -1 di\ in general J, Vsm 1 J y = current density. Am -2 = aperture electric-field distribution, V m -1 Z = intrinsic impedance of medium, Q square” 1 (jj = 2nf{f— frequency), rad s” 1 /i — permeability of medium, H m ” 1 For an aperture with a uniform dimension perpendicular to the page and with the field distribution over the aperture a function only of the electric field as a function of 0 at a large distance from the aperture (r > a) is, from (1) E() = E(x yfi dx The magnitude of E() is then |£(0)l = jil_ r° i: ^r0 ^ J - 0/2 Eix)^' dx 1 Noie that E\x)jZ — E f{x)fZ = H r = J where z tup in Fig.
M-G5| is the thickness of the current sheet. Also dE = dEF , FOURIER TRANSFORM RELATIONS 51 ?
where (1 - 2ni/.. For a uniform aperture distribution [£() = £J.( 3 ) reduces to |£(0)l = ^ / r pjfix ain 4t ^ and on axis i = 0) we have I £<0)1 = ~r where A = aperture area ( = ay,) £. = electric field in aperture plane For unidirectional radiation from the aperture (in direction <}> = 0 but not in direction 0 - I80 l ), | E(4>)| is twice the value given in (5).
Integration of (4) gives |£(0)l = k { sin _{fiaj2) sin 0] lffti/2) sin 0 From (4-14-17) the field of a long array of n discrete sources of spacing d is sin Ufia’jl) sin (M/2) sin 0 “ W where the length of the long array is «' = („- l)d ~ „d and £0 = field of one source. It is also assumed m (8) that (j> is restricted to small angles. This is not an undue restriction if the array is large and only the main lobe and first sidelobes are or interest. Under these conditions it is dear that the field pattern (8) of the long array of discrete sources is the same as the pattern (6) for the continuous array of the same length (a = a f ) (as already noted in Sec. 4-14).
transform relations between the FAR-FIELD pattern and the aperture distribution.
According to Booker and Clemmuw 1 a 1 -dimensional aperture distribution £(,) and its far-field distribution £(sin 0) are reciprocal Fourier transforms as given H, G. Booker and P, C. Clemmow, "The Concept of an Angular Spectrum of Plane Waves and Its c anon to That of Polar Diagram and Aperture Distribution/ Proc, instn. Ele t\ Engrs London, ser 397, 11-17, January 1950 v ^ wr f or the more general 2-dimensional case see R. N, Bracewell, “ Radio Astronomy Techniques" in S lugge (ed.J, Handbuch der Physik, vol. 54, Springer-Verlag OHG, Berlin, 1962, pp r 42-129 518 M ARRAYS OF DIROLE5 AND OF APERTURES E($in ) = 1 Etx^ 2'-^" dx x — X (1) and E(xi) = j 1 £(sin ) (2) where , - = x/A. For real values of , |sin | < 1, the field distribution represents rad iated^power, while for | sin 1 it represents reactive or stored power-The field distribution E(sin # or angular spectrum refers to an angular^d.stnbri ion of plane waves. Except for I sin J >\ > 1 the angular spectrum for a a^ rt“ is the same as the far-field pattern £(</») (the far-field condition r S> n does not hold for an infinite aperture, i.e., where a = >- Thus, for a finite aperture Fourier integral representation of (1) may be written E(4>) = r+ekf E{x x . -ai;l This is identical with (11-21-2) except for constant factors Equation >t' 2I -2 J “ an absolute relation, whereas (3) is relative. Examples of the far-field patterns Eft) for several aperture distributions E(x,) of the same extent are presented m F '6 ' Taking the uniform distribution as reference, the more tapered distributions (triangular and cosine) have larger beam widths and smaller minor lobes, w i e the most gradually tapered distributions (cosine squared and Gaussian) have still larger beam widths but no minor lobes. On the other hand, an averse taper (ess amplitude at the center than at the edge), such as shown in Fig.
11-66/, yields a smaller beam width but larger minor lobes than for the uniform distribution Such an inverse taper might inadvertently result from aperture blocking due o feed structure in front of the aperture. Carrying the inverse taper to its extreme limit results in the edge distribution of Fig.
1 1-66g. Th^dislnbutiomsequiva^m to that of a 2-element interferometer and has a beam width i that of the uniform distribution but sidelobes equal in amplitude to the main lobe.
, .
A uniform line source or rectangular aperture distribution (Fig.
-a) produces the highest directivity. However, the first sidelobe is only about 13 dB down. Thus, aperture distributions used in practice are a trade-off or compromise between desired directivity (or gain) and sidelobe level, as already discussed in Sec. 4-10 and Sec. 4-11 on the Dolph-Tchebyscheff distribution. A table of beam-width and sidelobe level for various rectangular and circular aperture distnbu-tions is given in App. A n Sec. A-10.
. d. R. Rhodes, “The Opt.nu.rn Lme Source tor the Best Mean Square Approximation to a Given Radiation Pattern/- IEEE Trans. A ms. Prop., AP-ll, +#M46, July 1963.
FOURIER TRANSFORM RELATIONS 519 11-23 field pattern is) Figure 11-66 Seven different anten-na aperture distributions with assoc-iated far-field patterns.
Theoretically, the directivity of a uniform aperture distribution can be exceeded (supergain condition) by large field fluctuations near the edges of the aperture. However, according to Rhodes, 1 to obtain a ^ dB increase in directivity from a 5k aperture requires that the field near the edges be,at least 20 times the field for a uniform aperture. The currents to produce these fields would have I 2R losses which offset any gain unless all conductors were perfectly conducting and 1 (k R. Rhodes, “On an Optimum Line Source for Maximum Directivity," IEEE Trans , Xnfs Prop 485-492, July 197J.
520 II ARRAYS OF DIPOLES AND OF APERTURES the surrounding media were completely lossless. Furthermore, the antenna would have an enormous Q and an extremely narrow bandwidth, making supergain attempts impractical.
The situation here is reminiscent of the W8JK. array which theoretically has a 4-dB gain over a single dipole even when the spacing between elements approaches zero; however, enormous currents would be required. With appre-ciable losses and for usable bandwidths a minimum spacing of /./8 is a practical limit, as indicated in Fig. ll-13n.
A useful property of (3) is that the distribution may be taken as the sum of 2 or more component distributions, E,(xJ. E 2(x,), etc., the resulting pattern being the sum of the transforms of these distributions. Thus, e,(4>) + e 2 (4>) + -F + axil C ^ a> - : ^ = £ 1U ) > a x .
If a source moves through the beam of an antenna (or if the source is fixed and the antenna is rotated) the observed response of this scanning process is pro-portional to the convolution of the antenna power pattern and the source bright-ness distribution. Thus, 00 $()P„(4>a -4>) dtp J -OC' (2) M U SPATIAL FREQUENCY RESPONSE AND PATTERN SMOOTHING 521 where S(0 O) = observed power distribution, W m -2 Hz -L = true source brightness distribution, W m “ 2 Hz -1 sr -1 = mirror image of normalized antenna power pattern 0 O = displacement angle (scan angle) Tt follows that S(xJ = BiXjjPixJ (3) where the bars mean the Fourier transform Since varies as the autocorre-lation function of the aperture distribution, it follows that 5(xJ and S(0 O) are zero where P{xxJ = 0 This means that there is a cutoff Tor all values of greater than ax .
1 The quantity x Ao is called the spatial frequenc y (wavelengths per aperture) and a x its cutoff value. Thus, = a i (4) where xXt = spatial-frequency cutoff R. N. BraccweLl and J. A. Roberts, “Aerial Smoothing in Radio Astronomy,” Australian J Phys 7, 615-640, 1954.
522 11 ARRAYS or depou:s and of apertures True distribution. B (a) Figure ll-6Ha Smoothed distribution 5 observed with in ten n it pattern P.
The reciprocal of x Xt gives an angle 1 57.3 = — rad = deg (5) It follows that this (cutoff) angle 4>c is equal to \ the Beam Width between First Nulls (BWFN) for a uniform aperture distribution ( t = BWFN/2), and is 12 percent greater than the beam width at half-power = 1.12 HPBWf The significance of is that structure in the source distribution having a period of less than BWFN/2 will not appear in the observed response. Thus, the antenna tends to smooth the true brightness distribution: 1 This is illustrated in Fig.
11 -68a. Half of the beam width between first nulls (BWFN/2) is equal to the Raylei&h resolution. 1 Thus, 2 point sources separated by this distance will be just resolved as indicated at the right in Fig.
1 1 -68a.
The observed half-power width as a function of the source width in half-power beam widths for a large uniform linear-aperture antenna and a uniform 1 -dimensional source is shown in Fig.
1 1 -68b. A source of half-power width equal to the antenna half-power beam width produces about 20 percent beam broaden-ing, or an observed width of 1.2 beam widths.
For larger source widths the observed width approaches the actual source width. Thus, from the amount of broadening an estimate may be made of the equivalent source extent.
1 1-24 THE SIMPLE (ADDING) INTERFEROMETER, The resolution of an antenna or of a radio telescope can be improved, for example, by increasing the aperture a , However, this may not be economically feasible. A 1 R. N. Bracewell and J. A. Roberts, Aerial Smoothing in Radio Astronomy," Australian J. Phys., 7, 615-640, 1954 The Rayleigh resolution can, in principle, be improved upon in the signal-processing domain of an adaptive array I see Sec.
11-13).
xn 2 x for — l < m < + 1 and .= 0 for m < — l and m > +1.
Suppose that the visibility function is like the one in Fig.
M-74 with the visibility going through zero at an interferometer spacing of 57.3a = 57.3) or x (source width) = 1 rad 57,3" with the resulting source brightness B0 = y - 57JS0 (Wm“ 2 Hz -1 rad -1 ) where = source power density The source has this brightness in the range — ^ < & < +t°( — I <m< +1) and is zero outside, as indicated in Fig, 1 1-75.
By scanning a source with an interferometer over a range of spacings the visibility function was obtained. Then by taking the Fourier transform of the visibil-ity a source brightness distribution was reconstructed (Fig, 1 1-75) which is the same 532 11 ARRAYS Of DIPOLES AND OF A F FUTURES Figure 11-76 (a; Visibility functions for a source of uniform brightness with holes of various widths (fr) visibility function for a uniform source with a bright center.
one discussed earlier in our analysts. Thus, in this hypothetical example we have gone full circle.
Examples of visibility functions for other source distributions arc given in Fig 1 1-76, The case of a uniform source of extent ?. with a symmetrical hole of extent (i is shown in Fig.
1 1 -76a for several cases of hole width. When & approaches the distribution approaches that of 2 point sources with a separa-tion x. When fi = 0, the source distribution is uniform. In Fig. 11-766 the visibil-ity function is presented for a uniform source with a bright center of 4 times the side brightness.
It is to be noted that the visibility functions of Fig.
I l-76a can be obtained as the visibility for a uniform source distribution of width a minus the visibility of a uniform source of width /?, while in Fig. 11-766 the result can be obtained as the sum of 2 uniform distributions of widths x and /J.
Comparing (17) with (1 1-22-1) and (1 1-22-3), it is apparent that the complex visibility function is related to the source brightness distribution in the same manner that the far-field pattern of an antenna is related to the antenna aperture distribution. Accordingly, the graphs of Fig. 11-66 may also be interpreted as IL-25 APERTURE SYNTHESIS AND MULTI APERTURE ARRAYS 533 giving the visibility functions for various source distributions and the graphs of Fig.
11-76 the field patterns for various aperture distributions. The restriction holds that the source extent is small and the aperture extent large.
For simplicity only 1-dimensional distributions have been considered. This case is of considerable practical importance. The principles may be extended also to the more general 2-dimensional case. A thorough treatment of the 2-dimensional problem is given by Bracewell.
1 In the above discussion monochromatic radiation at a single frequency is assumed.
If the antennas and receiver respond uniformly over a bandwidth / 0 + Af/2 with the radiation considered to be made up of mutually incoherent monochromatic components, the result for a point source is similar in form to the one above for a uniform source of width a, but with a/2 replaced by Af/2. A result of too wide a bandwidth is that the higher-order fringes may be obliter-ated.
11-25 APERTURE SYNTHESIS AND MULTI-APERTURE ARRAYS- The example of the preceding section is illustrative of 1-dimensional aperture synthesis. To obtain resolution of the l c uniform source of the example to a Rayleigh resolution ( = BWFN/2) of 0.P requires a single aperture 573/ in length [= 114.67(2 x 0.1°)]. An interferometer, on the other hand, can produce the same result with two small (say, 5/) apertures operated at various spacings provided that the source is strong enough to give a satisfactory signal-to-noise (S/N) ratio with the small apertures. Thus, the interferometer can synthesize a large continuous aperture—hence the term aperture synthesis.
Instead of using only 2 apertures and moving one or both of them, a number of apertures can be employed with unequal spacings in order to provide data points on the visibility curve. Many types of interferometers including phase-switched (multiplying), multi-element, grating, compound and cross arrays and the use of phase-closure and self-calibration techniques are discussed by Kraus.
; In radio astronomy observations, the baseline separation of the apertures and the angle of the baseline changes with the earth's rotation providing more visibility data, making it possible to do 2-dimensional aperture synthesis and produce 2-dimensional maps of the source distribution. Spreading the apertures over a plane (instead of all in-line) also improves the data. Typically, a celestial object is tracked (as the earth rotates) by a number of radio telescope antennas with each antenna-pair combination producing a complex visibility function (amplitude and phase) as a function of displacement (hour) angle, A map or image 1 R. N. Bracewell “Radio Interferometry of Discrete Sources." Proc , IRE. 46,97-105, January 1958.
See also A, R, Thompson, J. M. Moran and G. W Swenson, Jr., Interferometry and Synthesis in Radio Astronomy, Wiley- In terscience, 1986, 1 J, D, Kraus. Radio Astronomy 2nd ed., Cygn us- Quasar, 1986, chap.
6, 534 II ARRAYS OF DIPOLES AND OF APERTU R KS Figure JU77 Antennas of the Very Large Array (VLA) of the National Radio Astronomy Observa-tory in compact configuration The array, on the Plains of San Augustin, New Mexico, has 27 steer-able Cassegrain-type reflect or antennas 25 m in diameter mounted on 3 rail tracks (radials at 120^ each 21 km long for deployment in a variety of aperture spaemgs. (Courtesy NRAOjAUL) of the object or sky region is then constructed as a Fourier transform of the complex visibility.
The largest and most elaborate aperture synthesis array is presently the VLA (Very Large Array) of the National Radio Astronomy Observatory built at a cost of $7BM (1975 $) on the Plains of San Augustin near Socorro, New Mexico. This array consists of 27 Cassegrain-type reflector antennas (apertures), shown in Fig, 11-77, each with a diameter of 25 m. Each dish can be moved along one of three radial railroad tracks 21 km long so that the dishes can be deployed in a Y-shaped configuration with spacings of 0.6 to 36 km in order to maximize the visibility data with the earth’s rotation. In the photograph 9 dishes are shown in their most compact, close-in configuration along one track. Synthe-sized sky maps of 107 pixels (picture dements) can be produced with sensitivities of l mJy or better. At A — 6 cm the resolution is about ^ aresecond.
1 Radio telescope antennas separated by intercontinental distances have also been operated as interferometers at even higher resolution. With antenna aper-tures separated by 10000 km, milliarcsecond resolution is possible at ). = 6 cm.
In this Very Long Baseline Interferometry (VLBT), signals are downcon verted P J. Napier, A. R. Thompson and R. D. Ekers, "The Very Large Array: Design and Performance of a Modern Synthesis Radio Telescope;' Froc. IEEE, 71, 1295-1320, November 19S3 (includes 155 11-26 GRATING LOBES 535 and the IF signals taped and sent to a central location. By synchronizing the tapes using atomic standards, a real-time comparison can be simulated. Such observations require a high degree of coordination between participating obser-vatories and being time-consuming have only been performed on an intermittent basis. A dedicated full-time Very Long Baseline Array (VLBA) has been proposed which has at least 10 antennas (apertures) with locations coast-to-coast in the continental United States and in Puerto Rico and Hawaii.
1 The cost is estimated at $75M (1985 S) with completion scheduled for 1995.
A logical next step is to add an orbiting antenna to the array, increasing the resolution and visibility data resulting in maps with better detail and dynamic range. The rapid change of baseline distances for an orbiting antenna also reduces the mapping time. Ultimately with 2 or more orbiting antennas, all of the interferometry could be done above the earth’s atmosphere obtaining higher phase stability. High elliptical or circular orbits of 10000 to 60000 km apogee or radius are contemplated. 2 Aperture-synthesis arrays were pioneered by Martin Ryle of Cambridge University, England, and used by him and his students for mapping celestial radio sources.
3 Currently, the most complete and definitive work on aperture synthesis is the book by Thompson, Moran and Swenson.4 11-26 GRATING LOBES.
Tf a uniform linear array of n elements has a spacing d x between elements exceeding unity, sidelobes appear which are equal in amplitude to the main (center) lobe. These so-called grating lobes have a spacing from the main lobe of .
_ , m = sm — (rad), where m = 1, 2, 3, ...
(1) If P 1, this reduces approximately to G = j~ (rad) (2) 1 K. L Ke Nermann and A.
R. Thompson, “The Very Long Baseline Array;' Science, 229, 123-130, 1985, 1 B, F. Burke, " Orbititig^VLBI: A Survey,” in R. Fanti, It. KeUermann and G. Setti (eds.), VLSI and Compact Radio Sources, Reidel, 1984.
R, A. Preston, B. F. Burke, R, Doxsey, J. F. Jordan, S. H. Morgan, D. H, Roberts and I, I. Shapiro, “The Future of VLSI Observations in Space, 7 in F. Biraud (ed,), Very Long Baseline Interferometry Techniques , Cepadues, 1983.
3 See, for example, M. Ryle, A. Hewteh and J. R. Shakeshaft, “The Synthesis of Large Radio Tele-scopes by the Use of Radio Interferometers,” IRE Trans, Ants, Prop., AP-7, Si 20- 124, December 1959.
4 A, R. Thompson, J, M. Moran and G. W. Swenson, Jr„ Interferometry aid Synthesis in Radio Astronomy, Wiley-Interscience, 1986.
536 II ARRAYS OF D I POLKS AND OF APERTURES PROBLEMS 537 Figure 11-78 Grating Jobes with array of n widely spaced dements. Solid line: pattern with isolropic elements (array factor). Dashed line: total pattern with directional elements, each with pattern of dotled line.
with the first sidelobe (m = 1) at \jd x as given by the solid line (array factor) in Fig, 1 1-78 [individual array elements are nondirectional (isotropic)]. To suppress all grating lobes including the first requires directional elements, each with an aperture of approximately dk , This puts the first null of the individual element pattern on the first grating lobe, but the array is now equivalent to a continuous aperture. With less directivity (smaller aperture) dements, with pattern as sug-gested by the upper dashed line in Fig.
11-78, the resultant (total) pattern is as suggested by the lower dashed line with some grating lobe suppression but not elimination.
ADDITIONAL REFERENCES Allen, J. L.; "Array Antennas: New Applications for an Old Technique” /£££ Spectrum L 1 15-MO, November 1964. Excellent introductory article with 50' references.
Barton, P : "Digital Beamforming for Radar," /££ Proc., 127, August 1980.
Fcnn. A.
J.: “Maximizing Jammer Effectiveness for Evaluating the Performance of Adaptive Nulling Array Antennas" IEEE Trans. Ants. Prop ,, AP-33, 1 Mi-1142. October 1985.
Gabriel, W. K: "Spectral Analysis and Adaptive Array Superresolution," Proc. IEEE, 68, June 1980.
Howell, J. M,: " Phased Arrays for Microwave Landing Systems," Microwave 7,, 30, 129 137, January 1987.
Hudson, J. E.: Adaptive Array Principles, Peregrinus, 1981.
Kinzd. j A : “GaAs Technology for Millimeter-Wave Phased Arrays," IEEE Prop. Sac. News-letter, 29, 12-14, February 1987.
Kominami. M., D. M. Pozar and D, H. Schaubert: “Dipole and Slot Elements and Arrays on Semi-infinite Substrates," IEEE Trans. Ants. Prop., AP-33, 600-607, June 1985.
Lindell. L V.. E. Akmen and K. Mannersaio: "Exact Image Method for Impedance Compulation of Antennas above the Ground," IEEE Trans. Ants, Prop. AP-33. 937-945, September 1985.
Mailloux, R.
J.: "Antennas and Radar, 11 Microwave J.. 30, 26 30, January 1987 Moynihan. R. L.: " Phased Arrays for Airborne ECM," Microwave J,, 30, 34—53, January 1987.
Pozar, D.
M.: "General Relations for a Phased Array of Printed Antennas Derived from Infinite Current Sheets," IEEE Trans.
Prop, AP-33, 498-504, May 1985.
Rhodes, D. R .
: "On a Fundamental Principle in the Theory of Planar Antennas," Proc IEEE , 52, 1013-1021, September 1964.
Rhodes. D R.: "On the Stored Energy of Planar Apertures, 11 IEEE Trans. Tnrs. Prop., AP-14, 676-683, November 1966.
Rhodes. D. R.
: "On the Aperture and Pattern Space Factors for Rectangular and Circular Aper-tures," IEEE Trans Ants . Prop,, AP-19, 76W70, November 1971.
Rhodes, D R.: "Observable Stored Energies of Electromagnetic Systems" J.
Franklin Inst. 302, 225-237, September 1976-Sato, M.. M. Iguchi and R. Sato: "Transient Response of Coupled Linear Dipole Antennas," IEEE Trans. Ants . Prop„ AP-32, 133^140, February 1984.
Tai, C.-T : “The Optimum Directivity of Uniformly Spaced Broadside Arrays of Dipoles," IEEE Trans. Ants. Prop., AP-12, 447—454, 1964.
Taylor. T. T.: "Design of Line-Source Antennas for Narrow Beamwidth and Low Side Lobes.
/RE Trans. Ants. Prop., AP-3, 16-28. 1955-Woodward, P. M-. and J. D. Lawson: "The Theoretical Precision with W rhich an Arbitrary Radiation Pattern May Be Obtained from a Source of Finite Size," 7, Irtsi. Elec. Engng„ pi.
3, 95. 36>-370. September 1948.
problems 1 11-1 Two k/2-ekment broadside array.
{a) Calculate and plot the gain of a broadside array of 2 side-by-side a/2 elements in free space as a function of the spacing d for values of d from 0 to 2. Express the gain with respect lo a single A/2 element. Assume all elements are 1 00 percent efficient.
ib) What spacing results in the largest gain?
(c) Calculate and plot the radiation field patterns for a/2 spacing. Show also the patterns of the 7/2 reference antenna to the proper relative scale, 11-2 Two k/2-element end-fire array. A 2-element end- fire array in free space consists of 2 vertical side-by-side a/2 elements with equal out-of-phase currents.
At what angles in the horizontal plane is the gain equal to unity : (a) When the spacing is a/2?
(b) When the spacing is a/4?
11-3 Two-element VP array. Calculate and plot the field and phase patterns of the far field for an array of 2 vertical side-by-side A/2 elements in free space with A/4 spacing when the elements are : (g) In phase and (6) 180 ' out of phase.
For the in-phase case also include on the graph the patterns in both the yz or vertical plane and xy or horizontal plane of Fig. \-2a. For the out-of-phase case do the same for the patterns in both the 2 or vertical plane and xy or horizontal plane of Fig.
1 l-7a.
11-4 W8JK array. Calculate the vertical and horizontal plane free-space held patterns of a W8JK antenna consisting of two horizontal out-of-phase z/2 elements spaced ;./8. Assume a loss resistance of 1 fl and show the relative patterns of a z/2 refer-ence antenna with the same power input.
1 1-5 Sixteen-element and W8JK array gain equations. Confirm ( 11-6-6) and (11-7-11).
1 1 -6 Two-element array with unequal currents.
(a) Consider two A/2 side-by-side vertical elements spaced a distance d with cur-rents related by - al, {6. Develop the gain expression in a plane parallel to the elements and the gain expression in a plane normal to the elements, taking 1 Answers to starred () problems are given in App. D, il 538 <1 ARRAYS Or DIPOLES AND OF APERTURES a vertical 2/2 element with the same power irfput as reference (0 < a < 1).
Check that these reduce -to (11-4-15) and II 1-4- 13) when a = 1 ih) Plot the field patterns in both planes and also show the field pattern of the reference antenna in proper relative proportion for the case where d = 2/4 0 — \ and £ — 120 c , 1 1-7 Impedance of D-T array ia) Calculate the driving-point impedance at the center of each element of an in-phase broadside array of 6 side-by-side 2/2 elements spaced 2/2 apart. The currents have a Dotph-TchebyschcfF distribution such that the minor lobes have ^ the field intensity of the major lobe, (b) Design a feed system for the array.
11-8 Stacked X/2 elements and W8JK array above ground.
{a) Develop (lt-7-12), (b) Calculate and^ plot from (II -7- 12) the gain in field intensity for an array of 2 in-phase horizontal 2/2 elements stacked 2/2 apart (as in Fig, M -3 1 ) over a 2/2 antenna in free space with the same power input as a Function of h up to h = 1.5 A for an elevation angle = 10 Also calculate and plot for comparison on the same graph the gains at % = 10 for a 2-element horizontal W8JK antenna over a single horizontal 2/2 antenna as a function of the height above ground from h = 0 to h = 1.52. Note differences of these curves and those for 1 = 20 in Fig, 1 1-32.
11-9 Two-tower BC array, A broadcast-station antenna array consists of two vertical z4 towers spaced 2/4 apart. The currents are equal in magnitude and in phase quadrature. Assume a perfectly conducting ground and zero loss resistance. Calcu-late and plot the azimuthal field pattern in millivolts (rms) per meter at 1.6 km with 1 kW input for vertical elevation angles = 0, 20, 40, 60 and 80° The towers are series fed at the base. Assume that the towers are infinitesimally thin.
H-10 Two-lower BC array. Calculate and plot the relative field pattern in the vertical plane through the axis of the 2-tower broadcast array fulfilling the requirements of Prob. 4-19 if the towers are a/4 high and are series fed at the base. Assume that the towers are infinitesimally thin and that the ground is perfectly conducting:' U-U Three-tower BC array Calculate and plot the relative field pattern in the vertical plane through the axis of the 3-tower broadcast array fulfilling the requirements of Prob. 4-20 if the towers are 32/8 high and are series fed at the base. Assume tha the towers are infinitesimally thin and that the ground is perfectly conducting.
I M2 BC array with null at at = 30°, Design a broadcast -station antenna array of 2 vex tical base-fed towers a/4 high and spaced 32/8 which produces a broad maximun of field intensity to the north in the horizontal plane and a null at an elevatio angle = 3CP and azimuth angle — 135 measured ccw from north to reduce interference via ionospheric bounce. Assume that the towers are infinitesimally thin that the ground is perfectly conducting and that the base currents of the two towers are equal. Specify the orientation and phasing of the towers. Calculate and plot the azimuthal field pattern at z = Q° and a = 30" and also the pattern in the vertical piane through <£ = 135". The suggested procedure is as follows. Solve fl 1-8-4) for & at the null. Then set in the pattern factor in (11-4-13) equal to tp r and solve for the value of b which makes the pattern factor zero. The relative field intensity at any angle (<£ ) is then given by (1 M- 14) where sin 0 = cos $ = cos cos in the first pattern factor and 0 = 90 J - a in the second pattern factor.
PROBLEMS 539 H-13 BC array with null to west at all ou Design a broadcast-station array of 2 vertical base-fed towers 2/4 high that produces a broad maximum of field intensity to the north in the horizontal plane and a null at all vertical angles to the west. Assume that the towers are infinitesimally thin and that the ground is perfectly conducting.
Specify the spacing orientation and phasing of the towers. Calculate and plot the azimuthal relative field patterns at elevation angles of ot = 0 30 and 60 g .
11-14 Impedance and gain of 2-element array. Two thin center-fed 2/2 antennas are driven in phase opposition. Assume that the current distributions are sinusoidal. If the antennas are parallel and spaced 0.22 {a} Calculate the mutual impedance of the antennas.
(b) Calculate the gain of the array in free space over one of the antennas alone, 11-15 Triactgle array. Three isotropic point sources of equal amplitude are arranged at the corners of an equilateral triangle, as in Fig. Pit- 15. If all sources are in phase, determine and plot the far-ficld pattern.
Figure Pi 1-15 Triangle array.
11-16 Square array. Four isotropic point sources of equal amplitude are arranged at the corners of a square, as in Fig. PI 1-16.
if the phases are as indicated by the arrows determine and plot the far -field pattern.
Figure Pll-16 Square array.
1 1-17 Terminated V. Traveling wave (a) Calculate and plot the far-field pattern of a terminated-V antenna with 5/. legs and 45 s included angle.
(h) What is the HPBW?
11-18 Seven short dipoles. 4-dB angle. A linear broadside (in-pfiase) array of 7 short dipoles has a separation of 0.35>.
between dipoles.
Find the angle from the maximum field for which the field is 4 dB down (to nearest 0 1').
540 U ARRAYS OF DIPOLES AND OF APERTURES PROBLEMS 541 11-19 Square array Four identical short dipoles (perpendicular to page) are arranged at the corners of a square a/2 on a side. The upper left and lower right dipoles are in the same phase while the 2 dipoles at the other corners are in the opposite phase.
If the direction to the right (x direction) corresponds to 0 = 0\ hnd the angles 0 for ail maxima and minima of the field pattern in the plane of the page.
11-20 Pencil-beam patterns. For symmetrical circular aperture pencil-beam patterns (function only of 0) show that the main beam solid angle Qw is given by 1.13 0yiF for a Gaussian pattern 0,988 Tor a (sin x}/x pattern 1,008 P for a Bessel pattern where 0HP is the half-power beam width. Also show that Om is given by 1.036 0Hp. Find Un d and (6) d so that there is a maximum field at 0 = 45' (northeast) and a null at 0 = 90° (north).
There can be other nulls and maxima, but no maximum can exceed the one at 45\ The distance d must he less than a/2.
11-23 Eight-source scanning array. A linear broadside array has 8 sources of equal amplitude and a/2 spacing. Find the progressive phase shift required to swing the beam (a) 5 ib) KT and (c) 15 from the broadside direction, (4) Find BWFN when all sources are in phase.
11-24 A 24-dipole scanning array. A linear array consists of an in-line configuration of 24 2 dipoles spaced /. ? The dipoles are fed with equal currents but with an arbi-trary progressive phase shift 6 between dipoles. What value of .5 is required to put the mam-lobe maximum (a) perpendicular to the line of the array (broadside condition), (ft) 25? from broadside, (t| 50 ; from broadside and (if) 75 from broad-side? H\i Calculate and plot the four field patterns in polar coordinates. (/) Discuss the feasibility of this arrangement for a scanning array by changing feed-line lengths to change (5 or by keeping the array physically fixed but changing the frequency. What practical limits occur in both cases?
11-25 Three-helix scanning array. Three 4-turn right-handed monofilar axial mode helical antennas spaced 1.5/ apart are arranged in a broadside array as in Fig.
11-46. The pitch angle a -12.5 ; and the circumference C = /, (a) If the outer two helices rotate on their axes in opposite directions while the center helix is fixed, determine the angle ij> of the main lobe with respect to the broadside direction and describe how vanes as the helices rotate.
(6) What is the maximum scan angle 4>-(c)V< hat is the main-lobe HPBW as a function of </)?
11-26 E-type rhombic. Design a maximum E-type rhombic antenna for an elevation angle a - 17.5°.
11-27 Alignment rhombic. Design an alignment -type rhombic antenna for an elevation angle a = 175°, 11-28 Compromise rhombic. Design a compromise-type rhombic antenna for an ele-vation angle a = 17,5° at a height above ground of a/2.
11-29 Compromise rhombic. Design a compromise-type rhombic antenna for an ele-vation angle a = 1 7,5° with a leg length of 3a, 11-30 Compromise rhombic. Design a com promise-type rhombic antenna for an ele-vation angle <x = 17.5 at a height above ground of a/2 and a leg length of 3/..
11-31 Rhombic patterns. Calculate the relative vertical plane patterns in the axial direc-tion for the rhombics of Probs, 1 1-26, 11-27, 11-28, 11-29 and 11-30. Compare the patterns with the main lobes adjusted to the same maximum value, 11-32 Rhombic equation. Derive (11-16-1) for the relative field intensity of a horizontal rhombic antenna above a perfectly conducting ground, 11-33 Alignment rhombic equation.
Verify (1 1-16-3), (11-16-8) and (i 1-16-9) for the align-ment design rhombic antenna.
11-34 Sixteen source broadside array. A uniform linear array has 16 isotropic in-phase point sources with a spacing of A/2. Calculate exactly (a) the half-power beam width, (fr) the level of the first sidelobe, (c) the beam solid angle, (4) the beam efficiency, (e) the directivity and (/) the effective aperture.
11-35 Four aperture distributions. For the following aperture distributions show that the far-field patterns are as given: (a) (b) (c) (d) Stepped Circular "jriangula^ TriaAguiar asymmetric ^ 2 sin 0 1 sin w/2 ^ = 3 f + 3 V/2 2JJ») £#) = sin 2 0 “T 1 sin 0 ( cos 0 sin 0\ -) where 0 = sin 0 where 0 = sin 0 where 0 = — Lx sin 0 where 0 = nLk sin 0 11-36 Fourier transform. Apply the Fourier transform method to obtain the far-field pattern of an array of 2 equal in-phase isotropic point sources with a separation d.
Reduce the expression to its simplest trigonometric form.
11-37 Interferometer. Pattern multiplication. An interferometer antenna consists of 2 square broadside in-phase apertures with uniform field distribution.
() if the apertures are 10a square and are separated 6QA on centers, calculate and plot the Tar-field pattern to the first null of the single aperture pattern, () How many lobes are contained between first nulls of the aperture pattern?
(c) What is the effective aperture?
(4) What is the HPBW of the central interferometer lobe?
542 11 ARRAYS G DIPOLtS AND OF APERTURES {e) How does this HPBW compare with the HPBW for the central lobe of two isotropic in-phase point sources separated 60a?
11-38 Visibility function. Show that the visibility function observed with a simple inter-ferometer of spacing s for a uniform source of width cl with a symmetrical uniform bright center of width ff - a/6 is ViA -sin (rcsj a) 4- 3 sin (jt.^ a/6) 3^ a/2 if the center brightness is 4 times the side brightness {see Fig.
11 -76b).
11-39 Visibility function. Show that the visibility function observed with a simple inter-ferometer of spacing s for 2 equal uniform sources of width a/6 spaced between centers by 5a/6 is sin 2 sin (2ns; a/3) rts^a 3 2^ a/3 Pattern smoothing. An idealized antenna pattern-brightness distribution is illus-trated by the 1-dimensional diagram in Fig. Pll-40. The brightness distribution consists of a point source of flux density S and a uniform source T wide, also of flux density S. The point source is T from the center of the 2" source. The antenna pattern is triangular (symmetrical) with a 2" beam width between zero points and with zero response beyond.
(a) Draw an accurate graph of the observed flux density as a function of angle from the center of the 2 U source.
(b) What is the maximum ratio of the observed to the actual total flux density (2S)?
Figure PM-40 Pattern smoothing.
11-41 Interferoqwter output. Show that the output of a simple interferometer of spacing s and bandwidth / 0 ± A//2 for a point source is given by I Insi/ \ J Sm 7T 1 SoA/l+ ““tH where c — velocity of light.
11-42 Interferometer bandwidth. Show that for an interferometer with bandwidth / 0 ± A//2 (Prob.
1 1-41), the condition A/// 0 < l/n> where n = fringe order, must hold in order that the fringe amplitude not be decreased.
11-43 Number of elements. In Fig.
1 1-78 how many elements n have been assumed?
CHAPTER 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS 121 INTRODUCTION.
Reflectors are widely used to modify the radiation pattern of a radiating element. For example, the backward radiation from an antenna may be eliminated with a plane sheet reflector of large enough dimen-sions, In the more general case, a beam of predetermined characteristics may be produced by means of a large, suitably shaped, and illuminated reflector surface.
The characteristics of antennas with sheet reflectors or their equivalent are con-sidered in this chapter.
Several reflector types are illustrated in Fig, 1 2-1. The arrangement in Fig. 12-la has a large, flat sheet reflector near a linear dipole antenna to reduce the backward radiation (to the left in the figure). With small spacings between the antenna and sheet this arrangement also yields a substantia! gain in the forward radiation. This case was discussed in Sec.
11 -7a with the ground acting as the fiat sheet reflector. The desirable properties of the sheet reflector may be largely pre-served with the reflector reduced in size as in Fig. 12-lb and even in the limiting case of Fig.
12-lc. Here the sheet has degenerated into a thin reflector dement.
Whereas the properties of the large sheet are relatively insensitive to small fre-quency changes, the thin reflector element is highly sensitive to frequency changes. The case of a kjl antenna with parasitic reflector element was treated in Sec, 11 -9a, 543 544 12 REFLECTOR ANTENNAS AND TKF.1R FEED SYSTEMS Active Passive Parabolic corner corner reflector id) if) (s) 0>) (') Figure 11-1 Reflectors of various shapes.
With two flat sheets intersecting at an angle y {<180°) as in Fig.
12-1 d, a sharper radiation pattern than from a flat sheet reflector (ot = 180 ) can be obtained This arrangement called an active corner reflector antenna , is most practical where apertures of 1 or 2/ are of convenient size. A corner reflector without an exciting antenna can be used as a passive reflector or target for radar waves. In this application the aperture may be many wavelengths, and the comer angle is always 9QP. Reflectors with this angle have the property that an incident wave is reflected back toward its source as in Fig.
12-le, the corner acting as a retrorefteaor , When it is feasible to build antennas with apertures of many wavelengths, parabolic reflectors can be used to provide highly directional antennas. A para-bolic reflector antenna is shown in Fig. 12-1/ The parabola reflects the waves originating from a source at the focus into a parallel beam, the parabola trans-forming the curved wave front from the feed antenna ^at the focus into a plane i Z-2 PLANE SHEET REFLECTORS AND DIFFRACTION 545 wave front. Many other shapes of reflectors can be employed for special applications. For instance, with an antenna at one focus, the elliptical reflector (Fig, 12-lg) produces a diverging beam with all reflected waves passing through the second focus of the ellipse. Examples of reflectors of other shapes are the hyperbolic 1 and the circular reflectors 2 shown in Fig.
12-lft and i .
The plane sheet reflector, the corner reflector, the parabolic reflector and other reflectors are discussed in more detail in the following sections. In addition, feed systems, aperture blockage, aperture efficiency, diffraction, surface irregu-larities, gain and frequency -selective surfaces are considered, 12-2 PLANE SHEET REFLECTORS AND DIFFRACTION. The problem of an antenna at a distance S from a perfectly conducting plane sheet reflector of infinite extent is readily handled by the method of images.
3 In this method the reflector is replaced by an image of the antenna at a distance 2S from i he antenna, as in Fig.
122, This situation is identical with the one considered in Sec.
11-7, of a horizontal antenna above ground.
If the antenna is a lj2 dipole ibis in turn reduces to the problem of the W8JK antenna discussed in Sec.
1 1-5.
Assuming zero reflector losses, the gain in field intensity of a A/2 dipole antenna at a distance S from an infinite plane reflector is, from (1 1-5-8), - 2Jr+ 1f isin ' (l) Flat sheet reflector Image 1 Antenna O— 5—4—5-+ Figure 11-2 Antenna with flat sheet reflector.
1 G, Stavis and A. Dome, in Very High Frequency Techniques, Radio Research Laboratory Staff, McGraw-Hill, New York, 1947, chap. 6.
J. Ashmead and A. B. Pippard, ‘The Use of Spherical Reflectors as Microwave’ Scanning Aerials" JIEE (Lond-X 93, pt. 1IIA, no. 4, 627-632, 1946.
3 See, for example, G. H. Brown, " Directional Antannas Proc. IRE, 25, 122, January 1937, 546 32 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Figure 12-3 Field patterns of a 2/2 antenna at spacings of ±, { and -fax from an infinite flat sheet reflector Patterns give gain in field intensity over a X/2 antenna in free space with same power input.
For 2/S spacing, the gain is l.li, = 6.7 dB = 8.9 dBi).
where Sr — 2tt -The gain in (1) is expressed relative to a 2/2 antenna in free space with the same power input. The field patterns of 2/2 antennas at distances S = //4, 2/8 and 2/16 from the flat sheet reflector are shown in Fig.
12-3, These patterns are calculated from (1) for the case where RL = 0, Figure 12^ Gain in field intensity of 2 2 dipole antenna at distance 5 from flat sheet reflector. Gain is relative to /. 2 dipole antenna in free space with the same power input. Gain in dBi is also shown.
Gain is in direction tp — Garni is shown for assumed loss resistances R y = 0, 1 and 5 fl.
\ 2-2 PLANE SHEET REFLECTORS AND DIFFRACTION 547 Figure 12-5 Array of 2/2 elements with flat sheet reflector (billboard antenna).
The gain as a function of the spacing S is presented in Fig. 12-4 for assumed antenna loss resistances R L = 0, 1 and 5 Q. These curves are calculated from (1) for 0 = 0.
It is apparent that very small spacings can be used effectively provided that losses are small. However, the bandwidth is narrow for small spacings, as discussed in Sec.
11-5. With wide spacings the gain is less, but the bandwidth is larger. Assuming an antenna loss resistance of 1 Q, a spacing of 0. 1 25/ yields the maximum gain.
A large flat sheet reflector can convert a bidirectional antenna array into a unidirectional system. An example is shown in Fig.
12-5. Here a broadside array of 16 in-phase 2/2 elements spaced 2/2 apart is backed up by a large sheet reflec-tor so that a unidirectional beam is produced. The feed system for the array is indicated, equal in-phase voltages being applied at the 2 pairs of terminals F-F.
If the edges of the sheet extend some distance beyond the array, the assumption that the flat sheet is infinite in extent is a good first approximation. The choice of the spacing S between the array and the sheet usually involves a compromise between gain and bandwidth, [f a spacing of 2/8 is used, the radiation resistance of the elements of a large array remains about the same as with no reflector present.
1 This spacing also has the advantage over wider spacings of reduced interaction between elements. On the other hand, a spacing such as 2/4 provides a greater bandwidth, and the precise value of S is less critical in its effect on the element impedance.
When the reflecting sheet is reduced in size, the analysis is less simple. The situation is shown in Fig. 12-6g. There are 3 principal angular regions: V H. A, Wheeler. “The Radiation Resistance of an Antenna in an Infinite Array or Waveguide" Proc IRE, 36. 478-487, April 1948.
548 REFUECTOR ANTENNAS AND THEIR FEED SYSTEMS 12 3 CORNER REFT ECTOR S 549 Region I {above or in front of the sheet). In this region the radiated field is given by the resultant of the direel field of the dipole and the reflected field from the sheet.
Region 2 (above and below at the sides of the sheet). In this region there is only the direct field from the dipole. This region is in the shadow of the reflected field.
Region 3 (below or behind the sheet).
In this region the sheet acts as a shield, producing a full shadow (no direct or reflected fields, only diffracted fields).
If the sheet is 1 or 2k in width and the dipole is close to it, image theory accounts adequately for the radiation pattern in region 1. In region 2, the distant field is dominated by the direct ray from the dipole. In the fulfshadow behind the sheet (region 3) the Geometrical Theory of Diffraction (GTD) must be used.
1 The pattern in this region is effectively that of 2 weak line sources, one along each edge. The fields in the 3 regions are shown in Fig.
12-6i> for the case of the sheet width dimension D = 2.25k and the dipole spacing d = 32/8, It is assumed that the sheet is very long perpendicular to the page (^D).
Narrower reflecting sheets result in more radiation into region 3 but this diffracted radiation can be minimized by using a rolled edge (radius of curvature > 2/4) and absorbing material, as suggested in Fig, 12-6c, 12-3 CORNER REFLECTORS 12-3a Active (Kraus) Corner Reflector In 1938 while analyzing the radi-ation from a dipole parallel to and closely spaced from a flat reflecting sheet, I realized that when the sheet is replaced by its image, the dipole and its image form a W8JK array. When the flat sheet (180° included angle) is folded into a square (90°) corner the theory calls for 3 images, and my calculations showed correspondingly higher gain. Thus,1he corner reflector developed as an extension of my analysis of the W8JK array.
I immediately constructed several corner reflectors to obtain experimental confirmation. I tried parallel-wire grid reflectors, modifying both the spacing and length of the reflector wires to determine the limiting dimensions required. Figure 12-7 shows my first comer reflector, a 90° corner for 2 = 5 m operation built in 1938. Subsequent tests were done with smaller comers of various angles at X — 1 m with patterns measured by rotating the antennas on a turntable.
1 Diffraction is also discussed in Secs. 2-1S, 4-15, 4-16, 13-3, 17-5 and IS- 3d- For more details on the principles of physical and geometrical diffraction theories see J. D. Kraus, Electromagnetic^ 3rd ed^ McGraw-Hill, 1984, pp r 524—531 (2nd cd., 1972, pp. 464-179), and J. D. Kraus, Radio Astronomy, 2nd ed., Cygnus-Quasar, 1986, pp. 642-646.
550 L2 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS 1^ CORNER REFLECTORS 55] Figure 1 1—7 The author with the first corner reflector which he built and tested in 1938 for L 5 m operation.
The following presentation is based mainly on my publications on the corner reflector in the Proceedings of the Institute of Radio Engineers (November 1940), in Radio (March 1939) and in a, patent application filed Jan. 31, 1940 1 Two flat reflecting sheets intersecting at an angle or corner as in Fig. 12-8 form an effective directional antenna. When the corner angle a = 90°, the sheets intersect at right angles, forming a square-corner reflector. Corner angles both greater or less than 90° can be used although there are practical disadvantages to angles much less than 9(T A corner reflector with 2 = 180° is equivalent to a flat sheet reflector and may be considered as a limiting case of the corner reflector This case has been treated in Sec.
12-2.
Assuming perfectly conducting reflecting sheets of infinite extent, the method of images can be applied to analyze the corner reflector antenna for angles 2 = lB07n, where n is any positive integer This method of handling corners is well known in electrostatics. 2 Corner angles of 180° (flat sheet), 90'% 60 , etc., can be treated in this way. Corner reflectors of intermediate angle cannot be deter-mined by this method but can be interpolated approximately from the others.
In the analysts of the 90" corner reflector there arc 3 image elements, 2, 3 and 4, located as shown in Fig. 12-9 a. The driven antenna 1 and the 3 images have currents of equal magnitude. The phase of the currents in 1 and 4 is the same. The phase of the currents in 2 and 3 is the same but 180 out of phase with respect to the currents in 1 and 4. All elements are assumed to be k/2 long.
At the point P at a large distance D from the antenna, the field intensity is E(4>) = 2kl { \ [cos {Sr cos ) - cos (Sr sin <£)] | (1) where /, — current in each element Sr = spacing of each element from the corner, rad = 2 n(S/k) k = constant involving the distance D, etc.
Figure 12-8 Comer reflector antenna.
] I D, Kraus, “The Corner Reflector Anienna," Prat. IRE , 28, 513-519, November 1940, l D. Kraus, “The Square-Corner Reflector" Radio , no 237, 19^24, March 1939.
I D. Kraus, “Corner Reflector Antenna," US. Patent 2,270,314, granted Jan. 20, 1942.
See, for example, Sir Janies Jeans, Mathematical Theory of Electricity and Magnetism, 5lh ed., Cam-bridge University Press, London, p. 188 For arbitrary corner angles, analysis involves integrations of cylindrical functions which can be approximated by infinite sums as shown by R. W. Klopfenstein, Corner Reflector Antennas with Arbitrary Dipole Orientation and Apex Angle" IRE Trans Ants -Prop,, AP-S, 297-305, July 1957.
552 i: REFLECTOR ANTENNAS AND THEIR FTFD SYSTEMS The emf Vj at the terminals at the center of the driven element is = / 1 Z 1I + I 1 Z 14 -2I 1Z 12 (2) where Z n - self-impedance of driven element J?i L = equivalent loss resistance of driven element ,Z 12 — mutual impedance of elements 1 and 2 Z 14 = mutual impedance of elements 1 and 4 Similar expressions can be written for the emf s at the terminals of each of the images. Then if P is the power delivered to the driven element (power to each image element is also P), we have from symmetry that Vll + 1L + ^14 -1 Substituting (3) in (1) yields £() - cos (Sr sin 0)] | (4) The field intensity at the point P at a distance D from the driven A/2 element with the reflector removed is 11 + 1L where k = the same constant as in.(l) and (4) This is the relation for field intensity of a A/2 dipole antenna in free space with a power input P and provides a convenient reference for the comer reflector antenna.
Thus, dividing (4) by (5), we obtain the gain in field intensity of a Corner reflector \ i To point P / ^Driven Figure 12-9 Square-comer reflector with images used in analysis (a) and Globed pattern of driven element and images (b).
CORNER REFLECTORS 553 square-corner reflector antenna over a single A/2 antenna in free space with the same power input, or Gtf) m ' 2 V 11 +^+^2^,, 1 [COS {S' COS - C0s <5' Sin Ml I W where the expression in brackets is the pattern factor and the expression included under the radical sign is the coupling factor The pattern shape is a function of both the angle and the antenna-to-corner spacing S. The pattern calculated by (6) has 4 lobes as shown in Fig.
12-9h. However, only one of the lobes is real Expressions for the gain in field intensity of corner reflectors with corner angles of 60, 45°, etc., can be obtained in a similar manner. The driven element is a A/2 dipole. For the 60° corner the analysis requires a total of 6 elements, 1 actual antenna and 5 images as in Fig.
12-10, Gain-pattern expressions for comer reflectors of 90 and 60° are listed in Table 12-L The expression for a 180 s “ corner ” or flat sheet is also included.
Table 12-1 Gain-pattern formulas for corner reflector antennas Number of Corner dement# in Gain in field intensity oier X/2 uitcnm in free space with ogle, deg analysis seme power input 554 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Antenna -to -corner spacing, X Figure 12-H Gain of corner reflector antennas over a a(2 dipole antenna m free space with the same power input as a function of the antenna-to-corner spacing. Gain is in the direction ^ = 0 and is shown for zero Loss resistance {solid curves) and for an assumed loss resistance of 1 U (ft lL -1 Si) (dashed curves), {After Kraus, ) In the formulas of Table 12-1 it is assumed that the reflector sheets are perfectly conducting and of infinite extent. Curves of gain versus spacing calcu-lated from these relations are presented in Fig. 12-11. The gain given is in the direction 0 = 0 Two curves are shown for each comer angle. The solid curve in each case is computed for zero losses (ft 1L = 0), while the dashed curve is for an assumed loss resistance R 1L = 1 0.
It is apparent that for efficient operation too small a spacing should be avoided. A small spacing is also objectionable because of narrow bandwidth. On the other hand, too large a spacing results in less gam.
The calculated pattern or a 90° corner reflector with antenna-to-corner spacing S — 0,52 is shown in Fig. 12-1 2a. The gain is nearly 10 dB over a refer-ence 2/2 antenna or 12 dBi. This pattern is typical if the spacing S is not too large.
If S exceeds a certain value, a multilobed pattern may be obtained. For 1 Displacing the driven dipole from the bisector of the corner angle shifts (squints) the beam direction to the other side of the bisector {see Prob.
16-20).
12-2 CORNER REFLECTORS 555 Figure 12-12 Calculated pat-terns of square-corner reflector antennas with antenna-to-corner spacings of {a) 0,52.
(b) J,0A and (c| 1.5>..
Patterns give gain relative to the x/2 dipole anienna in free space with the same power input example, a square-corner reflector with S = 1.0/ has a 2-lobed pattern as in Fig.
12- 12b. If the spacing is increased to 1.5, the pattern shown in Fig. 12-1 2c is obtained with the major lobe in the 0 = 0 direction but with minor lobes present.
This pattern may be considered as belonging to a higher-order radiation mode of the antenna. The gain over a single //2 dipole antenna is 12.9 dB (= 15 dBi).
Restricting patterns to the lower-order radiation mode (no minor lobes), it is generally desirable that S lie between the following limits: a S 90 3 0.25-0.72 180° (flat sheet) 0.1-0.32 The terminal resistance R T of the driven antenna is obtained by dividing (2) by / j and taking the real parts of the impedances. Thus Rp = R tl + /? 1L + R l4 — 2R l2 (7 ) Terminal radiation resistance, £1 556 12 reflector antennas and their feed systems Figtra 12-l3a Terminal radiation resistance of driven A/2 dipole antenna as a function of the antenna-to-comer spacing for corner reflectors of 3 comer angles. {After Kraus) If R 1L = 0, the terminal resistance is all radiation resistance. The variation of the terminal radiation resistance of the driven element is presented in Fig.
1 2- 1 3a as a function of the spacing 5 for corner angles a = 180, 90 and 60° We note that for a = 90° and S = 0.352, the resistance of the driven 2/2 dipole is the same as for a 2/2 dipole in free space.
In the above analysis it is assumed that the reflectors are perfectly conduct-ing and of infinite extent, with the exception that the gains with a finitely con-ducting reflector may be approximated with a proper choice of . The analysis provides a good first approximation to the gain-pattern characteristics of actual corner reflectors with finite sides provided that the sides are not too small.
Neglecting edge effects, a suitable value for the length of sides may be arrived at by the following line of reasoning. An essential region of the reflector is that near the point at which a wave from the driven antenna is reflected parallel to the axis.
For example, this is the point A of the square-corner reflector of Fig, 12- 13b, This point is at a distance of 1.4 IS from the corner C, where S is the antenna-to-corner spacing. If the reflector ends at the point B at a distance L = 25 from the 120 CORNER REFLECTORS 557 Figure 12-1 3b Square-earner reflec-tor with sides of length L equal to twice the antenna-to-comer spacing $.
corner, as in Fig.
12-13b, the reflector extends approximately 0.65 beyond A .
With the reflector ending at 0, it is to be noted that the only waves reflected from infinite sides, but not from finite sides, are those radiated in the sectors Fur-thermore, these waves are reflected with infinite sides into a direction that is at a considerable angle 0 with respect to the axis. Hence, the absence of the reflector beyond B should not have a large effect. It should also have relatively little effect on the driving-point impedance. The most noticeable effect with finite sides is that the measured pattern is appreciably broader than that calculated for infinite sides and a null does not occur at = 45 3 but at a somewhat larger angle. If this is not objectionable, a side length of twice the antenna-to-corner spacing (L = 25) is a practical minimum value for square-corner reflectors.
Although the gain of a corner reflector with infinite sides can be increased by reducing the corner angle, it does not follow that the gain of a corner reflector with finite sides of fixed length will increase as the corner angle is decreased. To maintain a given efficiency with a smaller corner angle requires that S be increased. Also on a 60 c reflector, for example, the point at which a wave is reflected parallel to the axis is at a distance of 1,735 from the corner as compared to 1,415 for the square-corner type. Hence, to realize the increase in gain requires that the length of the reflector sides be much larger than for a scfuare-comer reflector designed for the same frequency. Usually this is a practical disadvantage in view of the relatively small increase to be expected in gain.
To reduce the wind resistance offered by a solid reflector, a grid of parallel wires or conductors can be used as in Fig.
12-14. The supporting member joining S58 Li REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Figure 12-14 Square-corner (grid) reflector with bow-tie dipole for wideband operation (see Table 12-2).
the midpoints of the reflector conductors may be either a conductor or an insula-tor.
\n general the spacing s between reflector conductors should be equal to or less than A/8. With a driven element the length R of the reflector conductors should be equal to or greater than 0.7a. If the length R is reduced to values of less than 0.6A, radiation to the sides and rear tends to increase and the gain decreases.
When K is decreased to as little as 0.3A, the strongest radiation is no longer forward and the reflector” acts as a director.
Table 12-2 Design data for wideband 91/2 to 32/4 monopole spaced 0.92 from the corner with d = 22, a beam is obtained in a direc-tion making approximately equal angles with the 3 coordinate axes, and Inagaki 2 reports a gain of 17 dBi with higher gains for smaller corner angles. By placing a cylindrical surface of constant radius from the apex of a corner reflector (of 2 or 3 dimensions). Elkamchouchi 3 reports improved performance.
To distinguish the corner reflector with dipole (driven element) discussed in this section from corners without dipoles (or monopoles), the ones with the driven element may be called active (Kraus) corners and the others passive (retro) corners. While the active corners may have any included angle, with 90° the most common, the passive corners are always square (90°). These corners are discussed further in the next section.
Figure 12-16 Retroreflector of 8 square comers for reflecting back waves from any direction. Path of ray returning via triple bounce is shown.
1 J. L. Wong and H. E, King, “A Wide-Band Corner-Reflector Antenna for 240 to 400 MHz," IEEE Trans. Ants. Prop., AP-33, 891-892, August 19S5.
2 N. lnagaki, “Three Dimensional Corner Reflector Antennas/' /£££ Traits. /4ms. Prop., AP-22, 580, July 1974.
3 H. M.
Elkamchouchi, “Cylindrical and Three-Dimensional Corner Reflector Antennas,” IEEE Trans. Ants Prop., AP-31, 451^455, May 1983.
12-4 THE PARABOLA. GENERAL PROPERTIES 561 12-3b Passive (Retro) Corner Reflector. A passive (retro) corner consisting of 2 flat reflecting sheets is shown in Fig.
12-le.
With 3 mutually perpendicular reflecting sheets, as in Fig.
12-16, each sheet extending a distance d( = + x x = ±y, = + Zj) from the origin, we have a duster of eight 3-dimensional square-corner retroreflectors. Each square corner occupies one octant (4tt/8 sr = ji/2 sr = 5157 square deg). Any ray or wave incident within this solid angle is reflected back in the same direction, as suggested in the figure.
Together the cluster of 8 square corners form a retroreflector for waves from any direction within a full sphere solid angle { = 4rc sr = 41 253 square deg) with the equivalent (normal incidence) flat-sheet reflecting area being a function of the angle of incidence.
Its maximum value is yfld 1 except that in the 6 directions, exactly on the 3 axes (x , y and z\ the area is Ad 1 . Just off these directions the area approaches zero.
The enhanced reflection of radar signals from such passive corner clusters makes them useful for many applications. For example, small water-craft com-monly carry one (usually with reflecting surfaces of wire mesh) on a tall mast to make the craft s presence more visible on radar screens. To be most effective, the reflector dimensions should be many 2 and the periphery of the mesh hole <2/2.
The surface should also be flat to better than 2/16 and, to increase the probability that the radar echo will be noticed, the reflector can be rotated to avoid a persis-tent null condition.
Truncating the sides of the reflecting sheets along the diagonal lines (dashed lines in Fig. 12-16) results in 8 truncated corners, each bounded by an equilateral triangle (included angle 60 5 ) with an aperture area of \d2 .
If the surface of a sphere is divided into triangles (in the manner of a Buckminster Fuller geodesic dome) and a truncated corner inserted in each triangular area, it is possible te-array dozens of corner reflectors over the sphere and obtain a more uniform echo area as a function of angle of incidence over 47t sr, but at the expense of a smaller maximum value.
Retroreflectors can also be constructed in other ways. Thus, a Luneburg lens (Sec. 14-10) with a reflecting cap over, say it sr, acts as a retroreflector over this angle. The Van Atta array (Sec.
1 1-12) is another example.
12-4 THE PARABOLA- GENERAL PROPERTIES Suppose that we have a point source and that we wish to produce a plane-wave front over a large aperture by means of a sheet reflector. Referring to Fig.
12-1 la, it is then required that the distance from the source to the plane-wave front via path 1 and 2 be equal or 1 2L = R(1 + cos 0) (1) and R = 2L 1 + cos 6 ( 2 ) This is an application of the principle of equality of path length | Fermat 's principle) to the special ease where all palhs are in the same medium. For the more general situation involving more than one tedium see Chap.
14.
562 \2 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS This is the equation for the required surface contour.
It is the equation of a parabola with the focus at F, Referring to Fig. 12-176, the parabolic curve may be defined as follows. The distance from any point P on a parabolic curve to a fixed point F, called the focus, is equal to the perpendicular distance to a fixed line called the directrix.
Thus, in Fig. 12-176, PF = PQ. Referring now to Fig.
12- 17c, let AA be a line normal to the axis at an arbitrary distance QS from the directrix.
Since PS = QS - PQ and PF = PQ, it follows that the distance from the focus to S is PF + PS = PF + QS - PQ = QS (3) Thus, a property of a parabolic reflector is that all waves from an isotropic source at the focus that are reflected from the parabola arrive at a line AA ' with equal phase. The "image” of the focus is the directrix, and the reflected field along the line AA appears as though it originated at the directrix as a plane wave. The plane BB (Fig.
12-17c) at which a reflector is cut off is called the aperture plane.
{a) Paraboloid Figure 12-18.
Line source and cylindrical parabolic reflector (a) and point source and paraboloi-dal reflector (fr).
]:o A COMPARISON Hi I }.l N FAR A UOL.lt AN I? t'ORNER REFLECTORS ' 563 A cylindrical parabola converts a cylindrical wave radiated by an in-phase line source at the focus, as in Fig.
12-18u. into a plane wave at the aperture, or a paraboloid-of-revolution converts a spherical wave from an isotropic source at the focus, as in Fig 12-186, into a uniform plane wave at the aperture. Confining our attention to a single ray or wave path, the paraboloid has the property of directing or collimating radiation from the focus into a beam parallel to the axis (see Fig.
12-17).
12-5 A COMPARISON BETWEEN PARABOLIC AND CORNER REFLECTORS.
Referring to Fig, 12-17, any radiation from the primary source or feed antenna at the focus of the parabola which is not directed into the parabola is not collimated but is distributed by direct paths over a large solid angle. This is not only inefficient but the distributed radiation can degrade the pattern of the radiation from the parabola. Thus, it is essential that a parabolic reflector have a directional feed which radiates alt or most of the energy into the parabola. A corner reflector, on the other hand, docs not require a directional feed since the direct and reflected waves are properly combined (image theory).
Furthermore, a corner reflector has no specific focal point.
Practical aperture dimensions for a square-conifer reflector arc 1 to 2L For larger apertures parabolic reflectors should be used, and for a large parabola of many k aperture, a practical choice for the feed can be a corner reflector with a corner angle of 90 to 180 depending on the F./D ratio of the parabola (see Sec.
12-6). Corner angles of about 120° have the advantage that the beam widths are approximately equal in both principal planes.
Although the corner reflector differs in principle from the parabolic reflec-tor, there are situations in which the two may be nearly equivalent. This may be illustrated with the aid of Fig.
12-19. Let a linear antenna be located at the focus Figure 12-19 Cylindrical parabolic reflector compared wilh square-corner reflector.
564 .
12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS i i-fr THE PARABOLOIDAL REFLECTOR 565 F of a cylindrical parabolic reflector, and let this arrangement be compared with a square-corner reflector of the same aperture and with an antenna- to-corner spacing AF. The parabolic and corner reflectors are superimposed for compari-son in Fig.
12-19. A wave radiated in the positive y direction from F is reflected at 0 by the corner reflector and at O' by the cylindrical parabolic reflector Hence, this wave travels a shorter distance in the corner reflector by an amount 00'.
If AF = 2A, the electrical length of Otf is about 1 80^ so that a marked difference would be expected in the field patterns of the two reflectors.
However, if AF = 035a the electrical length of 00 is only about 30% and this will cause only a slight difference in the field patterns, ft follows that if AF is small in terms of the wavelength the exact shape of the reflector is not of great importance. The practical advantage of the corner reflector is the simplicity and ease of construc-tion of the flat sides.
12-6 THE PARABOLOIDAL REFLECTOR.
1 The surface generated by the revolution of a parabola around its axis is called a paraboloid or a parabola of revolution. If an isotropic source is placed at the focus of a paraboloidal reflector as in Fig.
1 2-20a t the portion A of the source radiation that is intercepted by the paraboloid is reflected as a plane wave of circular cross section provided that the reflector surface deviates from a true parabolic surface by no more than a small fraction of a wavelength.
If the distance L between the focus and vertex of the paraboloid -is an even number of x/4, the direct radiation in the axial direction from the source will be in opposite phase and will tend to cancel the central region of the reflected wave.
However, if where n — 1, 3, 5, .
the direct radiation in the axial direction from the source will be in the same phase and will tend to reinforce the central region of the reflected wave.
Direct radiation from the source can be eliminated by means of a direc-tional source or primary antenna 2 as in Fig.
12-20b and c. A primary antenna with the idealized hemispherical pattern shown in Fig. 12-20b (solid curve) results in a wave of uniform phase over the reflector aperture. However, the amplitude is 1 S. Silver {ed.). Microwave Antenna Theory and Design , McGraw-Hill, New York, 1949.
H. T. Friis and W L D, Lewis, “Radar Antennas;' Bell System Tech. J., 26, 219-317, April, 1947 C C.
Cutler, "Parabolic Antenna Design for Microwaves; 1 Proc. IRE, 37 1 1284^1294 November 1947.
J C. Slater, Micro-wave Transmission, McGraw-Hill, New York, 1942, pp. 272-276.
It is convenient to refer to the pattern of the source or primary antenna as the primarv pattern and the pattern of the entire antenna as the secondary pattern.
vr Tapered illumination with dashed primary pattern Nearly uniform illumination with solid primary pattern Relative field intensity Figure 12-20 Parabolic reflectors of difFere.it focal lengths (L) and with sources of different pallerns.
tapered as indicated. To obtain a more uniform aperture field distribution or illumination, it is necessary to make small, as suggested in Fig. 12-20c, by increasing the focal length L while keeping the reflector diameter D constant.
1 If the source pattern is uniform between the angles ±0!
(solid curve), the aperture illumination is more nearly uniform (solid curve) but not entirely so. The path length from F to the edges (at and P2 ) is greater than from F to V (at the vertex). Although ray paths via the edges (at P and P2 ) and via the vertex V are of equal total length to the plane wave front (see Fig. 12-17u), giving phase equal-ity, there is more path length to the edges in a spherical (1 /r) attenuating wave.
1 Thai is, by increasing the ratio of L to D. This ratio is called the F ratio, the F/D ratio or the focal ratio(^L/D = FV/D).
Thus, the field at the edges is weaker, as will be calculated. To make the field completely uniform across the aperture would require a feed pattern with inverse taper.
A typical pattern for a directional source as indicated by the dashed curve at (c) (left) gives a more tapered aperture distribution as shown by the dashed curve at (c) (right). The greater amount of taper with resultant reduction in edge illumination may be desirable in order to reduce the minor-lobe level, this being achieved, however, at some sacrifice in directivity.
The arrangement of Fig 12-20b illustrates the case of a small focal ratio.
The arrangement at (c) illustrates the case of a larger ratio.
Suitable directional patterns may be obtained with various types of primary antennas. As examples, a 2/2 antenna with a small ground plane is shown in Fig, 12-2la, and a small horn antenna in Fig. 12-216.
The presence of the primary antenna in the path of the reflected wave, as in the above examples, has 2 principal disadvantages. These are, first, that waves reflected from the parabola back to the primary antenna produce interaction and mismatching.
1 Second, the primary antenna acts as an obstruction, blocking out the central portion of the aperture and increasing the minor lobes. To avoid both effects, a portion of the paraboloid can be used and the primary antenna dis-placed as in Fig. 12-2 lc, This is called an offset feed.
Let us next develop. an expression for the field distribution across the aper-ture of a parabolic reflector. Since the development is simpler for a cylindrical parabola, this case is treated first, as an introduction to the case for a paraboloid.
Consider a cylindrical parabolic reflector with line source as in Fig. 12-22a. The line source is isotropic in a plane perpendicular to its axis (plane of page). For a [ This may be greatly reduced by using a circularly polarized primary antenna, such as an axial-mode helix.
If the primary antenna radiation is right-circularly polarized, the wave reflected from the para-bola is mostly left -circularly polarized and the primary antenna is insensitive to this polarization.
12-6 THE PARABOLOIDAL REFLECTOR 567 (8) 568 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS The field-intensity ratio in the aperture plane is equal to the square root of the power ratio or where £9/E0 is the relative field intensity at a distance y from the axis as given by y = R sin 9 Turning now to the case of a paraboloid-of-revolution with an isotropic point source as in Fig. 12-22fi, the total power P through the annular section of radius p and width dp is P = 2npdpSp (10) where Sp = the power density at a distance p from the axis, W m~ 2 This power must be equal to the power radiated by the isotropic source over the solid angle 2n sin 0 d9. Thus, P = 2b sin BdSU (11) where V = the radiation intensity, W sr“ 1 Then or p dp Sp — sin 9 d9 V Sg _ sin 9 V ~ p(dp/d6f) where p — R sin 0 — 2L sin 6 1 + cos 0 (12) (13) This yields S, (1 + cos 9) 2 4L 2 (14) The ratio of the power density S$ at the angle 9 to the power density S0 at 9 = 0 is then S9 (1 + cos 9) 2 S~ 0 4 (15) The field-intensity ratio in the aperture plane is equal to the square root of the power ratio or E9 _ 1 + cos 9 _ = (16) where E^jE^ is the relative field intensity at a radius p from the axis as given by p = R sin 9, 12-1 PATTERNS OF LARGE CIRCULAR APERTURES WITH UNIFORM ILLUMINATION 569 12-7 PATTERNS OF LARGE CIRCULAR APERTURES WITH UNIFORM ILLUMINATION, The radiation from a large paraboloid with uniformly illuminated aperture is essentially equivalent to that from a circular aperture of the same diameter D in an infinite metal plate with a uniform plane wave incident on the plate as in Fig, 12-23. The radiation -fie Id pattern for such a uniformly illuminated aperture can be calculated 1 by applying Huygens' principle in a similar way to that for a rectangular aperture in Chap.
4. The normalized field pattern Eft) as a function of 0 and D is Eft) 2k Ji_{nD/k) sin ] nD sin where D = diameter of aperture, m / = free-space wavelength, m - angle with respect to the normal to the aperture (Fig. 12-23) Jj = first-order Bessel function The angle 0 O to the first nulls of the radiation pattern are given by nD — sin (j> Q = 3,83 0 ) (a) [b > Figure 12-23 Large paraboloid with uniformly illuminated aperture (a) and equivalent uniformly illuminated aperture of same diameter D in infinite flat sheet (h).
1 See, for example, L C. Slater and N, H, Frank, Introduction to Theoretical Physics, McGraw-Hill, New York, 1933, p. 325. Also see S. Silver (ed.), Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, p, 194.
570 L2 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS since Jj(x) = 0 when x = 3.83. Thus, 3.83a .
1 ,222 ...
Q — arcsin —— = arcsin ^ Pi When 4>0 is very small {aperture large) 0 ^ (rad) = ^ (deg) (4) where D x = D/2 = diameter of aperture, 2 The beam width between first nulls is twice this. Hence for large circular aper-tures, the beam width between first nulls 140 BWFN =— {deg) (5) By way of comparison, the beam width between first nulls for a large uniformly illuminated rectangular aperture or a long linear array is 115 J , ,,, BWFN -— (deg) (6) L x where L K = length of aperture, 2 The beam width between half-power points for a large circular aperture is 1 58 HPBW = — (deg) (7) &JL The directivity D of a large uniformly illuminated aperture is given by , area /0 , D = An -7T-( 8 ) For a circular aperture tiD 2 D = 4tt —=- = 9£lDl 4/r where D^ = the diameter of the aperture, 2 The power gain G of a circular aperture over a 2/2 dipole antenna is G = 6D\ (10) For example, an antenna with a uniformly illuminated circular aperture 10/. in diameter has a gain of 600 or nearly 28 dB with respect to a 2/2 dipole antenna ( =s 30 dBi).
1 S. Silver (ed.). Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, p. 194.
\2-f PATTERNS OK LARGE CIRCULAR APERTURES WITH UNIFORM ILLUMINATION 571 For a square aperture , the directivity is D = 4s ~ = 12.6Ll i 2 and the power gain over a 7J2 dipole is G = 1.1Ll where Lx = the length of a side, 2 (H) (12) For example, an antenna with a square aperture 102 on a side has a gain of 770 or nearly 29 dB over a 2/2 dipole ( = 31 dBi).
The above directivity and gain relations are for uniformly illuminated aper-tures at least several wavelengths across. Tf the illumination is tapered, the direc-tivity and gain are less.
The patterns for a square aperture 102 on a side and for a circular aperture 102 in diameter are compared in Fig.
12-24. In both cases the field is assumed to be uniform in both magnitude and phase across the aperture. The patterns are given as a function of in the xy plane. The patterns in the xz plane are identical to those in the xy plane. Although the beam width for the circular aperture is greater than for the square aperture, the sidelobe level for the circular aperture is smaller, A similar effect could be produced with the square aperture by tapering the illumination.
Beam widths, directivities and gains are summarized in Table 12-3. Beam widths are compared with horn antennas in Table 131.
Figure 12-24 Relative radiation patterns of circular aperture of diameter D — UU and of square aperture of side length L = 10a, 572 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Table 12-3 Beam widths, directivities and gains of cir-cular and rectangular apertures with uniform aperture distributions Aperture Circular Rectangular 58 L 51 = Half-power beam width D-~L-140° 115" Beam width between first nulls D a T~ Directivity (gain over isotropic source) 12.6L;/-, Gam over X/2 dipole 6D\ 7.7L} L\ where 0- = diameter, A L x = side length, a L\ = length of other side (if aperture is square, L\ = t Apertures are assumed 10 be large compared to With tapered distributions beam widths are larger, and directivities, gains and minor lobes are less 12-8 THE CYLINDRICAL PARABOLIC REFLECTOR, The cylin-drical parabolic reflector is used with a line-source type of primary antenna. Two types are illustrated in Fig.
12-25. Both produce fan beams, i,e., a held pattern that is w ride in one plane and narrow in the other. The antenna in Fig, 12-25a has a line source of 8 in-phase A/2 antennas and produces a beam that is narrow in the £ plane (xz plane) and wide in the H plane (xy plane). The antenna in Fig, 12-25fr produces a beam that is wide in the £ plane (xz plane) and narrow in the H plane (xy plane). The primary antenna consists of a driven stub element Figure 11-25 Parabolic reflector with linear anay of 8 in-phase a/2 dipole antennas (a) and “ pillbox " or " cheese " antenna {by 11-9 APERTURE DISTRIBUTIONS AND EFFICIENCIES 573 with a reflector element. The driven element is fed by a coaxial line. The side plates act as a parallel -plane waveguide. They guide the radiation from the primary antenna to the parabolic reflector. This type of antenna is called a “pillbox ” or “cheese” antenna.
If L < A/2, propagation between the planes is restricted to the principal or TEM mode. In this case the source may be a stub antenna of length less than A/2 (as in Fig.
12-256), or the Source may be an open-ended waveguide or small horn. Neglecting edge effects, the patterns of the antennas of Fig. 12-25 are those of rectangular apertures of side dimensions L by H.
If the illumination is substantially uniform over the aperture, the relations developed for rectangular apertures in Chap. 4 can be applied to calculate the patterns, provided that the side length is large compared to the wavelength.
12-9 APERTURE DISTRIBUTIONS AND EFFICIENCIES.
1 As an introduction to aperture distributions and efficiencies, a few basic concepts dis-cussed in earlier chapters are reviewed briefly. Then a number of criteria useful in antenna design arc developed.
Let a plane wave of power density 5 (W m -2 ) be incident on an antenna, as in Fig, 1226. The power P delivered by the antenna to the receiver is then P = SA t (1) where A6 = effective aperture of the antenna or If ohmic losses are not negligible, as assumed above, we may distinguish R ~P Figure 12-26 Wive of flux density 5 incident on antenna delivers a power P to the receiver li.
1 This section (12-9) and the following sections (12-10 and 12-11) are from J. D. Kraus, Radio Astronomy, 2nd ed., Cygnus-Quasar, 1986, 574 i; REFLECTOR ANTENNAS AND WlR FEED SYSTEMS between the actual effective aperture {including the effect of ohmic losses) and an effective aperture based entirely on the pattern (losses neglected). Thus, we may write A e = k 0 A", ( 3 ) where A e = actual effective aperture, m : k0 — ohmic -loss factor, dimensionless (0 < k0 < 1) A ep = effective aperture as determined entirely by pattern, tn 2 Using these symbols, the directivity of an antenna is given by 4tt 4it (4) D = ci A = T lA" from which 1 (5) and A p Q a = kv /} where Zl A = antenna beam solid angle, sr (6) A = wavelength, m The aperture efficiency is defined as the ratio of the effective aperture to the physical aperture, or E = — (7) p A p so that the ratio of the aperture and beam efficiencies is ep k0 A ^ A p Q.M A p CL^ where - beam efficiency, dimensionless nM = main beam area, sr £lA = total beam area, sr Although the physical aperture A p may not have a unique meaning on some apertures, its value tends to be readily defined on apertures that are large in terms of wavelength.
The directivity of an antenna depends only on the radiation pattern, so that the directivity D is given by D = ^Aep (9) 1 In (2-22-5) this is given as = a 1 where Am as defined in Chap. 2 is equal to Atp as defined here.
i2 9 APERTURE DISTRIBUTIONS AND EFFICIENCIES 575 The gain G is then 4n G = DkQ = k0 jy A tp (10) I'he maximum directivity Dm will be defined as the directivity obtainable from an antenna (assumed to be large in terms of A) if the field is uniform over the aper-ture, i.e., the physical aperture. Hence, _ 4n j D = ff A (ID In designing an antenna, one may have a certain design directivity D4 one wishes to achieve. In general, this will be less than Dm , since, to reduce sidelobes, some taper will probably be introduced into the aperture distribution. Thus we may write Dd = ^A p k, = Dm ku (12) where Dd = design directivity The factor ku is called the utilization factor.
It is the ratio of the directivity chosen by design to that obtainable with a uniform aperture distribution, or . = £ 0<k„<l (13) After designing and building the antenna and measuring its performance, it will probably be found, however, that the actual directivity D is less than the design directivity Dd .
It is possible, but unlikely, that D exceeds Dd , The actual direc-tivity may then be expressed as D = Dd ka = Dm ku k a (14) where ka is the achievement factor, which is a measure of how well the objective has been achieved. Thus, kA = — 0 < kj < 1 (usually) The gain G of the antenna can now be written as 1 2 — 2 ^p £tp where A p = physical aperture, m2 = ohmic efficiency factor, dimensionless kM = utilization factor, dimensionless ka = achievement factor, dimensionless = aperture efficiency, dimensionless 576 12 REFLECTOR ANTFNNAS AND THEIR FEED SYSTEMS We may also write A p Af = k0 kM ka A p f — — — k k k tap — £ — where A fP = effective aperture (as determined entirely by pattern) A e — actual effective aperture A basic definition Tor the directivity of an antenna is that the directivity is equal to the ratio of the maximum to the average radiation intensity from the antenna (assumed transmitting). By reciprocity the directivity will be the same in the receiving case. Hence, 0 = — (20) u v sv where Um = maximum radiation intensity, W sr“ 1 U„ = average radiation intensity, W sr“ l The average value may be expressed as the integral of the radiation intensity [/(0, 0) over a solid angle of 4tt divided by 4;r. Thus m 1 (T -me, .
4n J.
where P is the total power radiated. From (21) the effective power (power which would need to be radiated if the antenna were isotropic) is DP = 4nVm (22) Let us now consider an aperture under two conditions. Under Condition 1 the field distribution is assumed to be uniform, so that the directivity is Dm . The power radiated is P'. Under Condition 2 the field has its actual distribution, and the directivity and power radiated are D and P respectively.
If the effective powers are equal in the two conditions.
DP = D-F n n F 471 ; D - D m p -A t E„EJ, II-y)£(x, y) 12'9 APERTURE DISTRIBUTIONS AND EFFICIENCIES 577-In (24) the surface of integration has been collapsed over the antenna, with part of the surface coinciding with the aperture.
Further, it is assumed that all the power radiated flows out through the aperture. In (24) Z is the intrinsic imped-ance of the medium (O per square) and Eav is the average field across the aper-ture, as given by = ™jj >') dx dy where E(x, y) is the field at any point (x, y) of the aperture. Rearranging (24), we have for the actual directivity D = -^ A„ dJh^rJh^rJ This relation was developed originally by Ronald N.
BracewelL 1 Following BracewelPs discussion and also elaborating on it by introducing the utilization factor, we can write for the design directivity ^ — ji where the primes indicate the design field values. The right-hand factor in (27) may be recognized as the utilization factor ku , Multiplying and dividing (26) by k K , as given in (27), yields 4tt 1 'i !
'i rrr£ , (Jc,y)T£'(x,y)i „ ± We also have, from (14), D = D„ k u k.
1 R. N. BracewdK “Tolerance Theory of Large Antennas,' IRE Trans.
Anfs.
Prop., AP-9, 49-58, January 1961.
578 IZ REFLECTOR AKTENNAS AND THETR FEED SYSTEMS Thus, the last factor in (28) is the achievement factor, a result given by Braceweil.
Further, as done by Braceweil, let E(x> y) = Eav + (30) £(, y) = 1 + $ where 3 is the complex deviation {factor} of the field from its average value. Thus, the denominator of the last factor in (28) may be written as — ff u + ^xi + <>) dy <32 > Since the average of 3 or <5 + over the aperture is zero, (32) simplifies to ill lH S3 dx dy = \ + var <5 where var S is used to signify the variance of S or average of S3 { = 1 <5 !
2 ) over the aperture. Thus, (28) may be written more concisely as 4n 1 1 + var 3 1 + var S' 1 + var (5 where = design value <5 = actual value We also have that the utilization factor and the achievement factor k“ 1 + var d’ 1 4- var 3 1 + var i5 Turning now to the beam efficiency , or the ratio of the solid angle of the main beam fiM to the total beam solid angle , w re have IIm mam i Ifg Itjbc p{8, ) dn 0 < em < 1 where P{9, ) = antenna power pattern ( = ££ = 1 Ej 1 ) Let us consider next the effect of the aperture field distribution on the beam and aperture efficiencies. We take first the simplest case of a 1-dimensional dis-tribution, i.e., a rectangular aperture (L k by L k) with a uniform distribution in the y direction (aperture L k ) and a distribution in the x direction (aperture LJ as in Fig. 12-27, given by «(jc) = K I + K^1 ] (38) where E(x) = field distribution K i = constant (see Fig. 12-27) K 2 = constant (see Fig. 12-27) L k = Lja = aperture width, / n = integer (=1, 2, 3, ..
.) If K 2 = 0, the distribution is uniform.
If K-, = 0, the distribution is parabolic for n = 1 and more severely tapered toward the edges for larger values of n , as indi-cated in Fig 12-27. The beam and aperture efficiency of a 1-dimensional aperture with n = 1 has been calculated by Nash 1 as a function of the ratio K l /[K i with the results shown in Fig.
12-28, For K t =0, the distribution tapers to zero at the edge (maximum taper). As the abscissa value in Fig, 32-28 [ratio K 1 f(K x + X 2 )] increases, the taper decreases until at an abscissa value of 1 (K 2 — 0), there is no taper; i.e., the distribution is uniform.
2 The curves of Fig. 12-28 show that the beam efficiency tends to increase with an increase in taper but the aperture efficiency decreases. Maximum aperture efficiency occurs for 1 R. T. Nash, “Beam Efficiency Limitations of Large Antennas,” iEEE Trans. Ants.
Prop., A P-12, 91S-923, December 1964.
2 The step function or pedestal is an idealization which is not physically realizable. The actual field cutoff must be more gradual. Rhodes shows that the electric field component perpendicular to the edge of a planar structure must vanish as the first power of the distance from the edge and that the component parallel to the edge must vanish as the second power. See D. R Rhodes, “On a New Condition for Physical Realizability of Planar Antennas," IEEE Trans. Ants. Prop AP-19, 162-166.
March 1971, 580 |2 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS e:-9 aperture distributions and efficiencies 581 taper A^+A'z tape r Figure 12-28 Beam and aperture efficiencies for a 1-dimensional aperture as a function of taper. The aperture efficiency is a maximum with no taper, while the beam efficiency is a maximum with full taper A parabolic dislribution for K 2 is assumed I see n = 1 in Fig 12-271 [After R , T, Nash, “Beam Efficiency Limitations for Large Antennasf IEEE Trans. Ants Prop. A P-1 2, 918-923, December 1964.) a uniform aperture distribution^ but maximum beam efficiency occurs for a highly tapered distribution. In most cases a taper is used that is intermediate between the two extremes of Fig. 12-28 [K t/(Ki + K z ) = 0 or 1], and a compromise is reached between large beam and aperture efficiencies.
For a 2-dimensional aperture distribution, i.e., a rectangular aperture (T 2 by Lf) with the same type of distribution in both the x and y directions, the beam and aperture efficiencies as a function of taper have been calculated by Nash, 1 A field distribution as given by (38) with n = l is assumed (K 2 part of distribution parabolic). Thus, for this case E<x, y) = {, + , 1 -+ K r 1 - (jQ 2 | (39) 1 R, T. Nash +l Beam Efficiency Limitations of Large Antennas," IEEE Trans. Ams. Prop., AP-12, 918-923, December 1964.
Figure 12-29u Aperture efficiency (solid) and beam efficiency Idashedl of a rectangular aperture as a function of taper and phase error A parabolic distribution for K 2 is assumed (see n -l in Fig. 12-27).
\ Alter R T Nash, Beam Efficiency Limitations of Large Antennas:' IEEE Trans. Ants, Prop.
AP-12.
918-923, December 1964), Nashs data, which also show the effect of random phase errors across the aper-ture, are presented in Fig. 12-29a. There is a family of 4 curves for the aperture efficiency (solid) and another family of 4 cufves for the beam efficiency (dashed) for random displacements (or errors) of 0, 0.04, 0.06 and 0.082 rms deviation. To obtain these curves the aperture and beam efficiencies calculated for the smooth distribution (39) were multiplied by the gain-degrad at ion (or gain-loss) factor of Ruze, J given by k = e -wP {40) where S is the rms phase front displacement from planar over the aperture.
It is assumed that the correlation intervals of the deviations are greater than the wavelength. The curves of Fig. 12-29a indicate that the controlling effect of the taper on the efficiencies (beam and aperture) tends to decrease as the phase error increases. The efficiencies are also reduced by the presence of the phase error, ] J, Ruze, " Physical Limitations of Antennas," MIT Res. Lab. Electron Tech, Kept. 248, 1952.
582 J2 RFKl.F:C'TOR ANTFNNA AND THEIR FEED SYSTEMS Figure 1219b Aperture efficiency tsol.d) and beam efficiency Washed) of a eircular ap^Wre as a function of taper and phase error. A parabohc distribul.on for K, ^ ^umed (,« n g-{ After R T. Nash: Beam Eff^ncy UmUationM of Urge Aniens lEEh Trans. Ants. Erop , 9iti 921, December 1964).
since such errors tend to scatter radiation into the side!obe regions. Thus, the nhase errors constitute a primary limitation on the antenna efficiency.
The gain-toss formula (40) of Ruze assumes specifically that the deviations from the test-fit paraboloid are random and distnbuted m a Gaussian that the errors are uniformly distributed, that the region over are substantially constant is large compared to the wavelength and that the number of such uncorrelated regions is large.
.
.
.
na , A circular aperture of diameter D l with a distribution as given by (38J where x is replaced by r and L, by D k , has also teen mvest.Bated by Nas^wUh the results shown in Fig. 12-29b. Two families of 4 curves each are g.ven for the beam and aperture efficiencies for 4 condi turns of random assumed that n = 1 (K 2 part of distribution parabolic). The ^ (circular aperture) are seen to be very similar to those o lg.
rectangular aperture. For reflector antennas with an rms surface deviation 6 should be noted that the phase front deviation & in (40) will be approximate y twice as large; that is, 6 = 2<5'.
.
Another problem to be considered with reflector antennas .s the efficiency with which the primary, or feed, antenna illuminates the reflector. This may be ]’-9 APERTURE DISTRIBUTIONS AND EFFICIENCIES 583 defined as thefeed efficiency ef , where PM, ) dO PM, 4>) dQ where P^B, $) — power pattern of feed fl = solid angle subtended by reflector as viewed from feed point If the first null of the feed antenna pattern coincides with the edge of the reflector, the feed efficiency given by (41) is identical with the beam efficiency of the feed.
1 In the general situation with a reflector antenna, the beam efficiency of the system may be expressed as FAB, ) dQ 9, \ (in 4>) dQ where S’ = rms surface error of reflector P(tf, = power pattern of reflector due to aperture distribution produced by the feed assuming no phase error Py{6 , ) = power pattern of feed Surface leakage is neglected. If it is appreciable, another factor is required.
An experimental procedure for determining the beam efficiency of a large antenna using a ceiestial source is as follows. The main-beam solid-angle is evaluated from where kp = factor between about 1.0 for a uniform aperture distribution and 1.13 for a Gaussian power pattern 0HP = half-power beam width in 0 plane, sr <pHP = half-power beam width in 0 plane, sr The half-power beam widths are measured, while kp may be calculated or esti-mated from the pattern shape. The half-power beam width in right ascension in degrees is given by - the observed half-power beam width of a drift profile in 1 Common practice is to taper the feed illumination by 10 dB or so at the edges of the reflector.
584 U REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS minutes of time multiplied by cos S/4, where S is the declination of the observed point source; that is, v HPBW(min) cos S HPBW(deg) = (44) The total beam solid angle is obtained from the relation k j 2 <45 > A w here the effective aperture A e is determined by observing a celestial point source of known flux density ; 1 that is, Combining relations, the beam efficiency is then given by _ _ kp $hp 0hf 2k7A where k = Boltzmann's constant (=138 x 10 23 J K “ ') T a = antenna temperature due to radio source (measured value corrected for cable loss), K S = flux density of radio source, W m 2 Hz” 1 A — wavelength, m k a = antenna ohmic-loss factor, dimensionless and where k p> 0HP and tf>HP are as defined in (43).
A procedure for determining the aperture efficiency eap of a large antenna is to observe a celestial source of known flux density to find A e by (46), from which (48) where A p — physical aperture It is often of interest to the antenna designer, however, to breast £ap up into a number of factors, as in (19), in order to account as completely as possible for all the causes of efficiency degradation. Thus, from knowledge of the antenna structure and its conductivity the ohmic-loss factor ka may be determined. The utilization factor itu can be calculated from design considerations. The achieve-ment factor ka could be calculated from the relation in (28) if the actual fields across the aperture were accurately known; however, these are rarely measured .
2 As an alternative the achievement factor can be separated into many factors See the table of celestial soirees, App. A, Sec. A-8. See also Sec.
18-6e.
See the discussion on p. 615 regarding a holographic measurement of the apeTture distribution.
I2- APERTURE. DISTRJBUTJONS AND EFFICIENCIES 'S85 involving the random surface error, feed efficiency:, aperture blocking, feed dis-placement normal to axis (squint or coma), fedj displacement parallel to axis (astigmatism), etc., of the (parabolic) reflector antenna, [t is assumed that each of these subfactors can be independently calculated or estimated.
The aperture efficiency £ap may then be expressed as A £p ^ ~ = K ku ka — k0 fcM k { k 2 k N = KK FI kn (49) ft = i where kn = ohmic-loss factor k u = utilization factor (by design) k a = achievement factor k l = e {4ltJ ,jl)1 = random reflector-surface-error (gain-loss) factor = k k 2 = Cf = feed -efficiency factor 9 k 3 = aperture-blocking factor k+ = squint factor = astigmatism factor k 6 = surface-leakage factor etc.
A dose agreement between sap as calculated by (49) and as measured by (48) does not necessarily mean that the designer has taken all factors properly into account (some could have been overestimated and others underestimated), but it does provide some confidence in understanding the factors involved. However, if there is significant disagreement between the two methods, the designer knows the analysis is incorrect or incomplete in one or more respects. This comparative method has been used by Nash J in analyzing the performance characteristics of a 110-m radio telescope (Big Ear). A discussion of tolerances in large antennas is given by Bracewell 2 For in-phase fields over a lossless aperture, the aperture efficiency is given from (26) by (50) where E = field at any point in aperture £av — average of E over aperture (E^)»v = average of E 2 over aperture R T. Nash, "A Mulii-relleclor Meridian Transit Radio Telescope Amen n a for the Observation of Waves of Extra-terrestrial Origin," Ph.D. thesis, Ohio Stale University, 196 l r 2 R. N. Bracewell, "Tolerance Theory of Large Antennas," IRE Trans.
Ants.
Prop.
A P-9 49-5R January 1961, 586 1-REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS tMO SLR FACT IR&EGLLASITIES AND GAIN LOSS 587 Example. Determine the aperture efficiency and the beam efficiency for a 50/ diameter circular aperture with aperture distribution of the form (1 — r ) , where r = 0 at the center and r = I at the rim for various values of m. Show the results graphically.
Solution. The aperture efficiency is obtained from (50). The beam efficiency is obtained by integrating the pattern over An for Q A and over the main beam for QM , using numerical methods. The results are shown in Fig. 12-30 with efficiency as ordinate and sidelobe level as abscissa, with aperture distribution shape as given by n also indicated. For an n = 2 distribution the aperture efficiency is about 56 percent and the first sidelobe 31 dB down.
Figure 12-30 Aperture and beam efficiencies for various aperture distributions and sidelobe levels from worked example. Uniform phase is assumed. With phase variation, aperture effi-ciency decreases and sidelobes increase.
For continuous apertures which are large (£>/) some conclusions are: 1. A uniform amplitude distribution yields the maximum directivity (nonuniform edge-enhanced distributions for supergain being considered impractical).
2. Tapering the amplitude from a maximum at the center to a smaller value at the edges reduces the sidelobe level but results in larger (main-lobe) beam width and less directivity.
3. A distribution with an inverse taper (amplitude depression at center) results in smaller (main-lobe) beam width but in increased sidelobe level and less direc-tivity. (The amplitude depression at the center might be produced inadver-tently by a primary (feed) antenna blocking the center of the aperture.) 4. Maximum aperture efficiency occurs for a uniform aperture distribution, but maximum beam efficiency occurs for a highly tapered distribution.
5. Aperture phase errors are a primary limitation on antenna efficiency.
6. Depending on aperture size (in /.) and phase error, there is a frequency (or /) for which the gain peaks, rolling off to smaller values as the frequency is raised (see Fig.
12-34.) 12 10 SURFACE IRREGULARITIES AND GAIN LOSS.
1 Referring to Fig.
12-31, consider the idealized case of a reflecting surface with irregularities which depart a distance & above and below the ideal surface. Plane waves reflec-ted from the irregular surface will be advanced or retarded with respect to waves reflected from the uniform surface by 2n — x 2 = 2n -(rad) (1) where S' = surface deviation (surface error) S = twice surface deviation = 2<5' Waves I deaf uniform surface Figure 12-31 Geometry for deter-mining the effect of surface irregu-larities.
1 W. N. Christiansen and J. A. Hogbom {Radioteiescopes, Cambridge University Press, 1985) give an excellent, lucid discussion or the effect of surface irregularities on gain' which served as an inspiration in preparing parts of this section.
588 12 REFLECTOR AIVTENNAS AND THEIR FEED SYSTEMS Referring to the phase diagram of Fig. 12-32, the resultant field E = E 0 COS A0 ® ; where &.
) of the irregularities is at least as large as the wavelength and that there are as many positive as negative irregularities, the (normalized) surface gain-loss factor is kg = ^ E° c°s = cos 2 &<p = cos 2 ^720“ £) (3) where kt = gain-loss factor (0 < k9 ^ 1), dimensionless The variation of the gain-loss factor as a function of rms surface deviation in wavelengths and degrees as calculated from (3) is given in Fig.
12-33. The gam-loss factor as calculated from the Ruze relation (12-9-40) (=exp [-(4a6'jk) ]) is also shown. Correlation distances of at least a wavelength are assumed.
Example. A surface has an rms deviation 5 r = a/20. Find the reduction in gain.
Solution, From (3), A = An ^ (rad) = 720° - = 720° ^ = 36° (4) and fc, = cos1 36° = 0.65 or a gain reduction of 1.8 dB (as may be noted in Fig. 12-33).
Assuming 50 percent aperture efficiency (A e = %A p) and a circular dish diameter D, 1 /DV A ‘ 2 \ 2 / Introducing the gain-loss factor for surface irregularities from (3), where S r = rms surface deviation E=Eq cos A4 E=Eo sin A0 FIf 12-32 Phase variation due to surface irregu-larities.
12 10 SURFACE IRREGULARITIES AND GAIN LOSS 589 Figure 12-33 Gain-loss factor of reflector antenna as a function of twice the rms surface deviation expressed in degrees of phase angle (of the reflected wave) as calculated from (3) and (12-9-40). The vertical dashed lines correspond to phase angles for rms surface deviations expressed in fractions of a wavelength. See text for assumptions involved.
Values of the gain calculated from (6) are shown in Fig. 12-34 for various surface deviations and circular dish diameters D = 4, 20 and 100 m. Measured gains of three radio telescope dishes (dashed curves) are shown for comparison. It may be inferred from Fig, 12-34 that the equivalent rms surface deviations of the Onsala and Nobeyama dishes are about 150 and of the Bonn dish about 500 jim, Referring to Fig. 12-32, the quadrature field components are given by E4 - E0 sin A0 (7) 590 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS ]2-]0 SURFACE IRREGULARITIES AND GAIN LOSS 591 Figure 12-34 Gain peaking and roll-off with decreasing wavelength. The curves show antenna gain as a function of operating wavelength for various rms surface deviations (fl'J and antenna diameters (£) equal to 4, 20 and 100 m as calculated from (6)> Since scales arc logarithmic; other values of D and 3 H can be readily interpolated. Curves for Onsala, Nobeyama and Bonn dishes are shown dashed. See text for assumptions involved. Gain is in dBi.
For equal positive and negative irregularities these total zero. Suppose, however, that in some other direction they ail add in phase , The power level of this minor lobe relative to the main lobe is then 'wY = (NE0 sin A^V FMjoJ \tt£0 cos A) = tan2 A^ where N = number of irregularities (8 ) Equation (8) gives the maximum level a minor lobe could have due to these irregularities. Consider, on the other hand, that the quadrature fields add random-ly. Their total is then y/N E0 sin Atp and the power level of the resulting minor lobe relative to the main lobe is fminarY = /y/jv £q Sill Atj >\ 2 £m.j.r/ \ NE0 COS A(j> } 1 N tan 2 A(j> Example.
If the rms phase error is 3fr and the number of irregularities N — 100, find the maximum and random sidelobe levels.
Solution. From (8), Maximum sidelobe level = tan 2 36 = 0.527 Or down 2.8 dB from the main lobe. From (9), 1 0.527 Sidelobe level (random case) = — tan 2 36 = 0.00527 N 100 or down 22.8 dB from the main lobe.
Although the gain-loss factor (3) is not a function of the number N of irregularities, the sidelobe level {9) for the random case does depend on N and decreases as N increases.
Small-scale irregularities with correlation distances which are small compared to the wavelength result in fields which smooth out at large distances from the reflecting surface and, accordingly, these small-scale irregularities do not affect the gain or sidelobe level as discussed above.
For a reflector antenna with perfect surface the main lobe and near side-lobes are determined principally by the aperture distribution and taper, while the minor lobes at larger angles to 90° or so are influenced by scatter from the feed support structure. Back -radiation is a function of spillover and diffraction around the edge of the reflector For a circular dish a substantial back lobe may occur on axis (180° from the main lobe) due to all diffracted waves adding in phase (see Fig.
12-35). To reduce diffraction the reflector should have a rolled edge with radius of curvature at least a/4 at the longest wavelength of operation.
1 Due to the force of gravity, a ground-based steerable parabolic reflector deforms as the reflector is tilted so that for a given wavelength there is a maximum diameter which cannot be exceeded by adding metal to the backup structure. However, this limit can be exceeded by a homologous design in which one paraboloid deforms into another. The next limit is then imposed by thermal deformation.
2 1 W. D.
Burnside, M. C Gilreath B. M, Kent and G. C.
Clerici, “Curved Edge Modification of Compact Range Reflector," IEEE Trans. Anls, Prop., AP-35, 176-182, Feburary 1987, 2 S. Von Hoerner, “The Design and Improvement of Tiltable Radio Telescopes,' Fisfos in Astron., 20, pi. 4,411-444, 1977.
592 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Front half of field pattern diffraction BB Back half of V field pattern |l 80 Figure 12-35 Typical parabolic reflector with feed (or subreflector). Front part of field pattern (main lobe and sidelobes) is determined by aperture distribution, surface irregularities and scattering from feed and struts. Rack half of pattern (axial back lobe and near lobes) is determined by spillover and diffraction around dish.
12-11 OFF-AXIS OPERATION OF PARABOLIC REFLECTORS.
When the feed of a parabolic reflector antenna is displaced laterally from the focal point the beam is shifted (squinted) off-axis in the opposite direction to the feed displacement. Such squinting is accompanied by gain toss, beam broadening and the appearance of a coma sidelobe. The amount of such degradation in per-formance is a function of the F/D ratio and the feed displacement, as illustrated in Fig.
12-36.
1 1 D. E. Baker, “The Focal Regions of Paraboloidal Reflectors of Arbitrary F/D Ratio/ 1 Ph.D. thesis, Ohio Slate University, 1974.
Loss, dB IJ-ll OFF-AXIS OPERATION OF PARAROL1C REFLECTORS ®3 Squint in HPBWs Figure 12-36 Universal squint dia-gram for parabolic antennas of any size showing the gain Joss as a func-tion of the squint displacement in HPBWs for 3 values of F/D ratio.
The loss for other F/D ratios may be interpolated.
(After D. E.
Baker, "The Focal Regions of Paraboloidal Reflectors of Arbitrary F/D Ratio/ 1 Ph.D.
thesis Ohio State University 1974).
HPRw1k t F/D .rJ;?: the gain l0SS is 03 dB for a recd displacement or 10 HPBWs but with F/D = 0.25 the loss is 10 dB for the same displacement. Thus a S “ more of“ a-in Fi« C 1217 T P s atT Sf0 F 3 53°; ‘ diame,er Parabola with F/D = 1.2 are shown Mg. 12-37 for 5 values of squint angle as calculated by Baker.
1 The aperture 594 1 1 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS distribution is uniform, resulting in symmetrical first sidelobes 17,6 dB down as shown for squint. As the squint angle increases, the gain decreases, the beam width increases, the pattern becomes asymmetrical and a first sidelobe increases (coma lobe). Thus, at a squint of 4" the gain is down 1 dB, the HPBW has increased from 8' to 8.5' and the coma lobe is only 10.5 dB down (up 7.1 dB from the 0 squint condition). Squinting with multiple feeds deployed near the prime focus (or by a movable feed for tracking) is frequently useful in spite of some degradation in performance. More detailed discussions of off-axis operation are given by Baker 1 , Lo, 2 Rudge and Withers, 3 Rusch and Ludwig,4 Sandler 5 and von Gniss and Ries.
6 12-12 CASSEGRAIN FEED, SHAPED REFLECTORS, SPHER-ICAL REFLECTORS AND OFFSET FEED.
Sub-refiectors offer flex-ibility of design for reflecting telescopes. The classical arrangement introduced by N, Cassegrain of France over 300 years ago uses a sub-reflector of hyperbolic shape which surrounds the prime focus point of the main parabolic reflector.
Referring to Fig.
12-38, we require that all rays from the focal point F form a spherical wave front (circle of radius CF') on reflection from the (hyperbolic) sub-reflector (as though radiating isotropically from the parabola focus F') or by Fermat’s principle of equality of path length that CA' + FA' — CA + FA (U Noting that CA = CF' - AF and that FA — AF = 2 OA we obtain FA' - A'F' = 2 OA = BA (2 > which is the relation for an hyperbola with standard form 1 D. E. Baker, "The Focal Regions of Paraboloidal Reflectors of Arbitrary F/D Ratio," Ph D. thesis, Ohio Slate University, 1974.
1 Y. T. Lo, “On the Beam Deviation Factor of a Paraboloidal Reflector,” IRE Trans. Ants. Prop ., AP-S, 347, May 1960.
J A. W. Rudge and M.
J. Withers, “New Technique for Beam Steering with Fixed Parabolic Reflec-tors,” Proc. IEE, 1 IS, 857, July 1971.
W, V.
T. Rusch and A. C. Ludwig, “Determination of the Maximum Scan-Gain Contours of a Beam-Scanning Paraboloid and Their Relation lo the Petzval Surface,” IEEE Trans. Ants.
Prop., AP-2L 141, March 1973.
3 S. S. Sandler, “Paraboloidal Reflector Patterns for Off-Axts Feed” IRE Trarts.Ants. Prop., AP-8, 368, July I960.
H. von Gniss and G. Ries, “ Feld Bild um den Brennpunkt von Parabolreflektoren mit ICleinen F/D Verhaknis,” Archiv der Elek. Ubertrag 23,481, October 1969.
Vertex of parabola $96 1= REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS correct the primary pattern of the feed and produce a more uniform field dis-tribution across the parabola aperture, 1 Referring to Fig, 12-39, the surface at AA is deformed to enlarge or restrict the incremental ray bundle thereby decreasing or increasing the watts per steradian in the bundle and finally the watts per square meter in the aperture plane of the parabola. This shaping technique may be extended over the entire sub-reflector and often both sub-reflector and parabola are shaped. As a result a more uniform aperture distribution and higher aperture efficiency can be 1 Iflhe rays are allowed to pass through the focus of the parabola instead of being intercepted by the hyperbolic (convex) Cassegrain surface, as in Fig. 12-38, they can be reflected by a concave ellipsoi surface beyond (to the right of) the focus. This type of sub-reflector is called Gregorian after James Gregory of England who devised it about 1660-Vertex of parabola Focus of parabola Center of circle Figure 12-40 Circle and parabola compared, with radius of circle equal to twice the focal length of the parabola.
achieved but with higher first sidelobes and also more rapid gain loss as the feed is moved off-axis to squint the beam.
A constraint on the Cassegrain arrangement is that to minimize blockage the sub-reflector should be small compared to the parabola, yet the sub-reflector must be large compared to the wavelength. Details on Cassegrain reflectors are given by Love.
1 In Fig. 12-40 a parabola is given by y 2 = 4/t (5) where/= focal distance = VF This parabola is compared with a eircte of radius R = VC.
It may be shown that for small values of x, the circle is of nearly the same form as the parabola when R = 2/ (6) Over an angle 0 and aperture radius r = R sin 6 (7) the circle differs from the parabola by less than AR.
If AR < k (or specifically When the elements resonate and = Re ZL - 0 the array can provide complete reflec-tion. Distinguish between R L for load resistance as used here and RL for loss resistance as used in some earlier sections.
Alternatively, an array of slots in a conducting surface, as in Fig. 12-436, can make the surface completely transparent . 3 Frequency-sensitive or -selective surfaces are used extensively for hybrid radomes (a mixture of FSS and dielectric slabs), dichroic (dual-frequency) surfaces, analog absorbers, etc.
Effect of Element Spacing^ dx and d% .
Let us consider the reflection or trans-mission characteristics with frequency of the FSS array in Fig. 12-43a for an incident wave traveling in the ? direction making an angle 0 with the y axis. The reflection coefficient p0 = -rr (dimensionless) where E t = electric field of incident wave Er = electric field of wave reflected in specular direction ( 1 ) If the spacings dx and dx between elements are small enough that no grating lobes occur,4 then, for 2f < 0.62, Pv= -(R^ + KJ+jPO + XJ (dimensionless) ElectroScience Laboratory, Ohio State University.
2 References to the pioneering work of ICicburtz and Ishimaru, Twersky and Ott er ai. are listed at the end of the chapter.
1 The dipoles and slots constitute complementary surfaces as discussed in Secs. 13-4, 13-5 and 13-6.
4 No grating lobes occur if dt < kf{ 1 + sin 0) for scan in the xy plane or dx < A/fl + sin 0) for scan in the yz plane.
12-n FREQU ENCY -SENSITIVE (OR -SELECTIVE} SURFACES [FSS] 601 Figarc 12-43 (a) Reflecting surface of dipole-like elements acts as a band-rejection filter. The response curve is shown at right, (i) Transparent surface of slots acts as a bandpass filter. The response curve is shown at right. The transmission coefficient = 1 + p v r Z p 2 where = 0 , fl (for scan in xy plane) Zax ux COS v z 2 Ra = 2d d ' ^ ^or 50311 plane) ZQ = 377 Q i r P — 7” f{) exp {—jflf •!) dz = pattern function, 1 m J-t /0 = element terminal current, A (3fl) (3) (4) For short elements, p varies little with angle of incidence and frequency.
602 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Figure 12-44 (a) Reflection coefficient of FSS for 2 values pf dement spacing. The smaller spacing results in the broader bandwidth. (h) Reflection coefficient of F^S for different angles of incidence 0.
If we consider p v in (2) as a function of frequency, we find that RA as given by (3) varies relatively little. However, X A + A\, which is much more compli-cated to determine than R A , changes dramatically.
Typical curves of pv for normal incidence (0 = 0°) are shown in Fig. 12-44a. Changing dx and/or dt , X A + X L remains essentially constant, provided dx and dx < A/2, However, is inversely proportional to both of these, i.e., becomes larger for smaller.^ and dx Inspection of (2) shows that smaller spacings make pv more broadbanded, as illustrated in Fig, 12-44a. However, the maximum magnitude of will always be unity provided R L = 0 and no grating lobes are present Effect of Angle of Incidence 0 Let us next consider the case where the spao ings dx and dz are fixed (< ^ 0.45/) and determine how the reflection coefficient p v varies with angle of incidence 0, We again find that XA + X L changes rela-tively little with angle of incidence. However, by inspection of (3), we note that R 4 increases with angle of incidence for scan in the xy plane and decreases for scan in the yz plane. Further, we observe that the bandwidth of p„ becomes larger and smaller respectively, as indicated in Fig. 12-44b. This change of bandwidth with angle of incidence is quite large in all FSS regardless of the types of elements used.
It can be compensated for in a number of ways. The most promising 12-13 FfttOUENCY^SENSITIVE (OR SELECTIVE) SURFACES fF5S| 603 Loaded wire arrays t Figure 12-45-1 (a) Single loaded^dipole reflecting surface for vertical polarization.
(f>) Double loaded-dipole reflecting surface for arbitrary polarization.
This form is generally preferable because of the cross-polarization component in the single loaded-dipole lype at high angle of inci-dence in the vertical (yz) plane (Fig. 12-43).
appears to be by using dielectric slabs mounted on both sides of the FSS as done by Munk 1 and by Munk and Kornbau, 2 Control of Bandwidth It was shown above that the bandwidth could be broadened by decreasing dx and/or dtt or vice versa, for smaller element spacings (this feature is quite universally independent of element type). Another way to affect the bandwidth is to make the load reactance vary faster with frequency.
This makes pv more narrowbanded as shown by Munk 3 and by Munk, Kouy-oumjian and Peters. 4 A practical type of load may typically consist of a 2-wire transmission line stub as shown in Fig, 12-45-la. A variation using two transmis-sion lines in parallel is shown in Fig. 12-45-lb, It has the advantage of being capable of handling arbitrary polarizations.
Finally, the diameter of the wire elements can be reduced, leading to an even narrower bandwidth of pv . However, compared to the other two methods above, this is much less effective since it changes only logarithmically with wire diameter.
Cascading or Stacking More FSS A single FSS, as shown above, has only one frequency where |p„| = 1. However, by cascading or stacking two or more surfaces the reflection coefficient curve can be flattened at the top and can roll off faster as indicated in Fig.
12-45-2. This approach may be used for the reflecting (dipole) as well as the transparent (slot) surfaces. However, while a single dipole 1 B. A. Munk, “Space Filler,” U.S. Patent 4,125,841, November 1978.
2 B, A. Munk and T. W. Kombau, “On Stabilization of the Bandwidth of a Dichroic Surface by Use of Dielectric Slabs,” Electromag., 5, 349-373, 1985, J B. A. Munk, ” Periodic Surface for Large Scan Angle,” U.S. Patent 3,789,404, Jan. 29, 1974.
4 B, A. Munk, R. G. Kouyoumjian and L.
Peters, Jr, “Reflection Properties of Periodic Surfaces of Loaded Dipoles,” IEEE Trans. Ants. Prop., A P-19; 612-617, September 197L 604 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Front Back Figflrt 12-45-2 Cascading or stacking 2 or more FSS {in y direction) as at left can produce a flatter-topped reflection coefficient curve with Taster roil-off as shown at right. Note that the figure shows two 3 x 3 ( = 9) element arrays stacked one in front of {or to (he left of) the other, surface has the same reflection curve as its complementary surface’s transmission curve (Babinet's principle), this is not the case when surfaces are stacked or when dielectric is added to a single surface.
Element Types. Nearly any type of element will resonate at some frequency.
However, some types are much better than others. First, the element should be sufficiently small that the element spacing dx and dx can remain small, meeting the grating lobe conditions.
Consequently, a straight element as shown in Fig. 12-45-3a is rarely used for high-performance FSS, being too long (^A/2), Much better performance can be obtained with inductively loaded elements as shown in Fig, 12-45- 3b, c and d (also Fig, 12-45- la and b), The latter can also be constructed in 3-legged loaded form as shown in Fig. 12-45-3i and in the unloaded 3-legged form of Fig, 12-45-3/ (Pelton and Munk 1 ).
Four-legged unloaded elements of 2 crossed wires (or slots) are used but are not recommended since they will, in general, exhibit a double resonance in one scan plane but not in the other, a feature considered undesirable for most applications, as pointed out 1 E. L. Pdton and B, A. Munk, "Periodic Antenna Surface of Tripole Slot Elements,” AF invention l(U7a Aug. 17, 1976.
12-14 SOME EXAMPLES OF REFLECTOR ANTENNAS 605 Wire elements </) Of) (ft) Figure 12-45-3 Various types of wire dements for use with FSS by Pelton and Munk, 1 All of the elements described are characterized by having medium or narrow bandwidth (4 to 25 percent).
In contrast, the two loop ele-ments shown in Fig.
12-45-33 and & can be made to operate over more than an octave bandwidth.
12-14 SOME EXAMPLES OF REFLECTOR ANTENNAS. A number of reflector antennas illustrative of different designs are given in this section.
Bonn. The world’s largest fully steerable reflector telescope is shown in Fig: 12- 46. Standing 40 stories tall in a valley of the Eifel mountains near Bonn, West Germany, the 3200-tonne 100-m diameter parabolic reflector antenna is usedTor radio astronomy and space research. The dish can be elevated 20° per miiute and rotated in azimuth on a circular track 4fT per minute.
It can operate at wavelengths as short as 1 cm. Feeds may be either prime focus or Cassegrain. See the gain versus wavelength curve in Fig. 12-34.
Arecibo. The fixed 305-m diameter spherical reflector at Arecibo, discussed in the preceding section, is shown in cross section in Fig. 12-47 and from above in Fig. 12-48.
^ Bell Telephone Laboratories The 7-m diameter parabolic reflector with offset feed of the Bell Telephone Laboratories is shown in Fig. 12-49. This antenna, 1 E. L. Pella n and B. A. Munk, “Scattering from Periodic Arrays of Crossed Dipoles," IEEE Trans Ants. Prop AP-27, 323-330, May 1979.
Figure 12-46 Fully steerable lOG-m diameter antenna near Bonn West Germany standing nearly 40 stories tall. The moving parts weigh 3200 tonnes. Note the massive backup structure required to provide rigidity to the dish. fCouriesy Dr. R, Wielebinski, Max Planck Institute for Radioastronomy\ In -coming wave Figure 12-47 Elevation cross section of 305-m diameter fixed spherical reflector suspended in moun-tain valley at Arecibo Puerto Rico, 606 1M4 SOME EXAMPLES OF REFLECTOR antennas 607 Figure 12-48 Air view of 305-tn Arecbo dish. Feed structure ls supported by cable, from 3 towers.
fro™iheendh y oni° Vln8 | ,h f b™^ all°WS observa,lorls at an 8<es »P <° 20” from the zenith. Away y Par ‘ °f the reflect° r ““ ta wi,h ™- 1«-« dUctaJ located at Crawford Hill, New Jersey, has a 100-/rm rms surface accuracy permit-ting operation at wavelengths as short as 3 mm (100 GHz) with 59 percent aper-ture efficiency, a gam of 77 dBi and HPBW of 2 areminutes.
1 At 1 cm (30 GHz) Mm,W ter'w,vW A T lISO ":-Lr Eneland ' D A Gray and W- E Le®’ “ The Crswford Hill 7-m Millimeter Wave Antenna.
BSTJ, 57.
1 257-1 288. May-June 1978.
Figure 12-49 The 7-m offset Cassegrain-fed low-siddobe millimeter- wave antenna of the Bdl Tele-phone Laboratories at Crawford Hill, New Jersey. Key structural parts are covered with insulation to reduce thermal effects. (Courtesy Dr. T-S Oiw, Bell Telephone Laboratories.) the aperture efficiency is 69 percent and the gain 66.5 dBi, With a 15-dB feed taper and no aperture blockage the sidelobe level is over 40 dB down l y off-axis at 30 GHz. The antenna also has very low cross- polarization (over 40 dB down across the main beam). These low levels are achieved using a quasi-optical feed system and conical corrugated horn with hybrid-mode launcher (see Fig.
1331).
1214 SOME EXAMPLES OF REFLECTOR ANTENNAS 609 Whereas the atmosphere contributes a temperature of only 2.3 K at 4 GHz (at the zenith), the dry atmosphere temperature at 100 GHz is about 100 K so that the temperature contribution due to spillover (12 K) and mylar waveguide window loss (1 K) add little of significance to the system temperature at 100 GHz.
1 In radio astronomy observations, receiver noise reflected back into prime focus or Cassegrain feeds can produce spurious spectral lines. This effect is elimi-nated with an offset feed as in the Crawford Hill antenna.
Nobeyama. The world's largest mil lime ter-wave dish, shown in Fig.
12-50, is figure 12-50 The 45-m fully steerable parabolic reflector antenna with gam of nearly 90 dBi at wavelengths of 2 to 3 mm al the Nobeyama Radio Observatory in the Yatsugaiake Mountains.
Japan. IjYRO photo.) T-S Chu. personal communication. t985.
610 12 REFLECTOR ANTENNA AND THHTR FEED SYSTEMS ITU $OVE EXAMPLES Ob REFLECTOR ANTENNAS' 6t t located at the Nobeyama Radio Observatory at a high elevation in the central mountains of Honshu Island, Japan.
It has a 200-^m rms overall equivalent surface deviation and a pointing accuracy of 2 arcseconds.
1 Maximum gain is almost 90 dBi at 2 to 3 mm wavelengths.
See the gain curve in Fig.
12-34.
Thermal effects are minimized by covering the dish structure with insulated panels and circulating air inside the enclosure. Each of the 600 rigid panels of the dish have a surface accuracy of 60 /;m rm^, Ohio State University. The first radio telescope antenna I designed and built at the Ohio State University in 1951 consisted of an array of 96 helical antennas mounted on a tiltable flat panel 50 m long for operation at 200 to^ 300 MHz, The array is shown in Fig. 7-4.
A few years later I began designing a larger dual -reflector antenna on which construction began in 1956.
2 Work was done mostly by part-time university stu-dents and took over 10 years to complete. The antenna, often called a Kraus-type reflector, consists of a fixed-standing-curved reflector and a tiltable-flat reflector.
The one we built, called “Big Ear,” is shown in Fig.
12-51. The standing-curved reflector is a rectangular section of a sphere or paraboloid- of-revolution with dimensions of 1 10 by 21 m. The til table-flat reflector dimensions are 104 by 31 m and the two reflectors are joined by a flat conducting ground plane. The flat reflector is shown in Fig.
12-52.
The basic design consideration of the antenna was that of obtaining the maximum aperture per unit of cost. It also has other advantages such as less than t of 1 percent effective aperture blocking, reduced susceptibility to terrestrial interference because of the low profile of the Teed (right at ground level) and shielding by the large reflectors, and a spacious underground receiver laboratory directly below the prime focus where weight restrictions are not a consideration.
Tilting the flat reflector allows observations between declinations of —36 and +64 (a range of KXU) which gives a coverage of 90 percent of the sky observable from the site. Movement of the feed car, shown in Fig, 12-53, permits beam steering of 15" in azimuth (or tracking of sources for an hour or more in 1 K. Akabane, M. Morimolo, N. Kaifu and M. Ishigro. "The Nobeyama Radio Observatory," Sky W 7VL 495, December 1983 3 J D Kraus, “The Ohio State Radio Telescope,' 1 Sky and Tel., 12, 157-159, April 1955.
J. D. Kraus.
LL Radio Telescopes." Sci. Am., 192, 33-45, March 1955.
J. D. Kraus.
LL Radio Telescopes of Large Aperture," Pnx. IRE, 46. 92-97, January 1958.
J. D. Kraus, R. T. Nash and H.
t’. Ko. “Some Characteristics of the OSU 360-fl Radio Telescope," IRb 7 rans. .4ms. Prop., AP-9. 4 8, January 1961 .
J. D. Kraus, "The Large Radio Telescope of the Ohio State University," Skv and Tel., 26, 12-16. July 1963 J. D. Kraus, Hiy Ear. Cy gnus- Quasar, 1976-Kiswe 12-51 "Big Ear,” the 1 10-m Kraus-type telescope antenna at the Ohio Slate University. The Low- profile design features a large aperture per unii cost, negligible aperture blocking, reduced suscep-tibility lo lerrestria!
interference, convenient feed location, long focal length and extended tracking capability. This telescope was used for the Ohio Sky Survey, in which 20000 radio sources were cataloged and mapped at 1415 MHz and many unique sources discovered, including the most distant known objects in the universe, i Tom Root photo.) Figure 12-52 The liltable flat reflector of Big Far For scale, note the man halfway up )he ladder on the far side.
612 1 2 REFLECTOR ANTENNAS AND THFIR FEED SYSTEMS right ascension). The telescope has a very long focal length with F/D = 1.17 in azimuth (right ascension) and F/D = 6.0 in elevation (declination) so that it is possible to deploy many feed systems efficiently in the focal region for simulta-neous operation.
It is noteworthy that because of the long focal length the curved reflector can be described as either a parabola or a sphere since both arc almost identical except at the extreme E-W edges where they differ by only a few milli-meters. The antenna has been operated routinely at frequencies as low as 20 MHz (15 m) and as high as 3 GHz (10 cm). With installation of a finer mesh screen, efficient operation could extend to even higher frequencies.
The principle of operation is indicated in the elevation cross section of Tig.
12-54, Incoming waves are deflected by the flat reflector into the parabola, which brings the waves to a focus at ground level near the base of the flat reflec-tor. By moving the flat reflector through 50 c the antenna beam is tilted through a lQtr range in elevation.
The antenna may be operated in two modes.
In one mode, illustrated in Fig.
12-5 5a, the feed horn axis is aligned with the center of the parabola and the ground plane is incidental.
In the second mode (Fig. 12-55ck the horn axis is coincident with the conducting ground plane, which joins the parabola and flat reflector. The ground plane serves as a guiding boundary surface.
In this mode polarization must be vertical, and the feed horn required is £ the height and ^ the length of the horn required in the first mode. This difference in horn size may be 12-14 SOME EXAMPLES OF REFLECTOR ANTENNAS 613 Figu-e 12-54 Elevalion cross section through Big Ear.
inferred with the aid of the 3 diagrams in Fig.
12-55.
If a horn of the first mode (Fig. 12-55a) is placed with its axis coincident with the ground plane, the lower half is an image and may be discarded, as in Fig.
12-55b. However, the beam width of the horn is too narrow (by a factor of 2), so that its dimensions must be halved, as in Fig. 12-55c Although the ground plane serves no primary function in the first mode of operation, its presence reduces the antenna temperature sig-nificantly by shielding minor lobes from direct ground pickup.
Another application of Big Ear is as a compact measuring range for fre-quencies below 3 GHz, as discussed in See. 18-3d + TiJ table hat Figure 12-55 Arrangement for feeding antenna without ground plane (n) requires 4 times horn height of arrangement at (c} using ground plane. Diagram {b) shows that half the horn in (a) produces too sharp a pattern when used with the ground plane and must be reduced in size, as at (r).
614 \2 REFLECTOR ANTENNAS ANO THEIR FEED SYSTEMS Parabolic reflector Flat reflector vertical Figure 1256 Arrangement of standing-parabolic and tillable-flat reflectors of 25-m millimeter-wave antenna at Gorki, USSR, used by Albert Kislyakov at a wavelength of 1 mm, with flat reflector turned vertically, for self-calibration.
Gorki Albert Kislyakov has constructed a standing-parabola tillable -flat-reflector antenna near Gorki, USSR, for operation at millimeter wavelengths.
Although the basic principle is the same as for Big Ear, K'slyakov has added a unique feature of self-calibration by providing that the flat reflector can be set vertically.
1 With flat reflector vertical and with feed displaced laterally off-axis (and transmitting), radiation is brought to a focus on the opposite side of the axis where the signal can be received as suggested by Fig.
12-56. By moving the recei-ving horn sideways or up and down, the far-field pattern of the antenna can be measured with an arbitrarily high S/N ratio.
With the same configuration (flat reflector vertical), panels of the reflectors can be adjusted to maximize the gain. For example, with panels 1, 2, 3, 5, 6 and 7 covered with absorbing material, only panels 4> 8 and 8 are operational, and.
'A. 0 Kislyakov, M Radioastronomical Investigations at Millimetric and Submillimetnc Wave’ lengths,” UFH, 101, 607, 1970.
Figure 12-57 From and back views of the Five College Radio Astronomy Observatory 14-m radome-enclosed radio telescope.
with 4 and 4' as reference, panels 8 and 8' can he adjusted for maximum signal or gain. The same procedure is then repeated with other pairs of panels until all panels are adjusted.
Five College Observatory. The 14-m dish of the Five College Radio Astronomy Observatory is an example of a millimeter-wave antenna enclosed in a Buckminster Fuller triangular-panel geodesic radome. Front and back views of the dish are shown in Fig, 12-57, The reflector is constructed of 72 panels con-toured to a section of a paraboloid with a 65-^m rms surface accuracy, or /i/20 at 1.3 mm. To set the panels accurately a holographic technique was used in which the amplitude and phase of a 38-GHz satellite beacon signal was measured over an angular extent much larger than the main beam of the antenna. The ampli-tude and phase of the antenna aperture field distribution was then obtained by a 2-dimensional Fourier transform.
1 This indicated a 90-jun rms residual panel 1 A complete description of the procedure is given by P. F. Goldsmith and N. R. Frick son in see. 6-22 ofi. D. Kraus, Radio Asirrtrtrwny, 2nd ed,. Cygnus-Quasar. 19K6. For additional information on holo-graphic techniques see Bennet es aL, Godwin, Whitaker and Anderson, Mayer ei aL and R ah mat-Samii in the references listed at the end of this chapter.
jlti 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS rigare 12-58 The 5.5 by 2-4 m Ojfsat uplink-downlink antenna with full offset feed for C and lIu bands. {Comtech Antenna Corporation.) position error with the antenna at a 45° elevation angle. At 38 GHz (A — 7.9 mm) the gain-loss factor from (12-10-3) is kg = cos2 1 720° 0,09 mm\ 7.9 mm / = 0.98 or only 009 dB loss from surface error At 230 GHz (A = 1.3 mm) the loss is larger and irregular lens behavior of the radome fabric becomes significant.
OfFsat. A parabolic one-piece offset-feed fiberglass 55 by 2.4 m antenna is shown in Fig. 12-58, This Offsai antenna was developed by Comtech Antenna Corporation to meet the FCC-1TU 2° spacing requirement for satellites in the Clarke orbit. The full-offset prime-focus feed eliminates aperture blocking The gain at 4 GHz is 40 dBi, at 6 GHz is 46 dBi, at 12 GHz is 51dBi and at 14 GHz is 52 dBi. The dish weighs 1 tonne, has a noise temperature of 17 K at 40° ele-vation angle with the first sidelobe 24 dB down (See Sec. 16-16.) 12-15 LOW-SIDELOBE CONSIDERATIONS.
Referring to Fig. 12-35, it is to be noted that the sidelobes in the forward direction (0 to 90°) of a large I2'I5 LQW-SIDF.LOBE CONSIDERATIONS 617 Figure 12-59 Main beam and sidelobes with respect to the isotropic level.
parabolic dish reflector are determined (1) by the aperture distribution, (2) by the irregularities of the dish surface, (3) by scattering or diffraction from the feed structure and support struts and (4) by diffraction from the edge of the dish Effect (3) is absent with offset feeds or with the horn-reflector antenna. The side-lobes to the rear (90 to 180°) are determined (1) by the spillover and (2) by diffrac-tion around the edges of the dish Schrank 1 gives a thorough summary of low-sidelobe reflector antennas.
For an isotropic source the directivity D = 1 (0 dBi) so the isotropic level and the relative power level (uniform in all directions) are the same. As D increases, the main lobe rises above the isotropic level in proportion to D. In the hypothetical (but typical) pattern shown in Fig.
12-59, the directivity D = 22.5 dBi ( = lossless gain) and the highest sidelobe is 17 dB below (the main-lobe maximum) or 5.5 dB above the isotropic level (5.5 dBi). Sidelobe levels are usually referred to the main lobe but sometimes to the isotropic level As shown in Fig.
1 1-66, a triangular or cosine tapered aperture distribution drops to zero field at the edge of the aperture, yet it results in front sidelobes.
Back sidelobes should, in principle, be absent However, the cosine squared or Gaussian distributions have no sidelobes but the HPBWs are greater.
1 H.
E. Schrank, + Low Sidelobe Reflector Antennas;' IEEE Ant. Prop .
Soc.
Newsletter 27, 5-16 April 1985.
618 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Sharp edge Serrated edge Figure 12-60 (tf) Sharp edge and (6) serrated or sawtooth edge for reduced siddobe levd.
Dish surface irregularities with a regular (periodic) spacing of a wavelength or more are apt to result in sidelobes called grating lobes.
1 Diffraction from the sharp edge of a dish also contributes to the sidelobes both front and back. To randomize the phase of the diffracted rays, the edge may be serrated (saw-tooth edge) as in Fig.
1260. The tooth dimensions should be of the order of a wave-length or more. The straight -versus-diagonal or sawtooth effect may also be noted with a square dish or ground plane. Thus, as suggested in Fig, 12-61, the sidelobes in the plane of the diagonal tend to be less than in the plane of the sides.
Edge diffraction may be reduced by means of a rolled edge with (or without) absorbing material as illustrated in Fig.
12-6c. An oversize parabolic dish with reduced-edge illumination might also be used to reduce edge diffraction but a problem with this approach is that the underilluminated edge and outer regions of the parabola (with its irregularities) still contribute to a diffracted field.
However, if the outer region of the dish is curved away from a parabolic contour Grating lobes are typical for arrays with interdement spacings of /. or more (see Sec.
1 1-26), ADDITIONAL REFERENCES 619 and blended into a rolled or curved edge, diffraction effects are much reduced 1 (see Sec. 18-3d).
Although the horn-reflector antenna (Fig. 12-42) has very low wide-angle sidelobes, a sidelobe tends to appear at 0 = 90°, which may be objectionable.
Thomas 2 has used a serrated edge (blinder) to reduce it.
Another side and back lobe suppression technique involves the addition of a cylindrical absorbing shroud attached to the edge of the dish as in Fig. 12-62: The outer surface of the shroud may be metal or dielectric. Dybdal and King found that with a shroud twice as long as the dish diameter the far-out sidelobes were 60 to 75 dB down, although the on-axis back lobe (0 = 1 80®) was onlv 50 dB down.
ADDITIONAL REFERENCES Bach, H., and H-H. Viskum: “The SNFGTD (Spherical Near-Field Geometrical Theory of Diffraction) Method and Its Accuracy," IEEE Trans. Ants. Prop., AP-35, 169-175, February 1987, Bennet, J, C, A.
P. Anderson, P. A, Mclnnes and A.
J. T, Whitaker; “ Microwave Holo-graphic Metrology of Large Reflector Antennas," IEEE Trans. Ants. Prop AP-24 295-303, 3976.
Burnside, W, D,, M, C. Gilreath and B. M. Kent; “Rolled Edge Modification of Compact Range Reflector," AMTA Symp., San Diego, 1984.
Chu, T-S: “An Imaging Beam Waveguide Feed,” IEEE Trans . Ants. Prop., AP-31 614-619, July 1983.
Chu, T-S, and R. H. Turrin: “Depolarization Properties of Offset Reflector Antennas,” IEEE Trans . Ants. Prop„ AP-21, 339-345, May 1973.
1 W, D.
Burnside, M, C. Gilreaih, B. M. Kent and G. C.
Clerici, “Curved Edge Modification of Compact Range Reflector,” IEEE Trans. Ants. Prop., AP-35, 176-182, Feburary 1987.
2 D. T. Thomas, “Design of Multiple-Edge Blinders for Large Horn Reflector Antennas,' IEEE Trans. Ants, Prop., AP-21, 153-158, March 3973, 1 R. B. Dybdal an^ H, E. King, “ Performance of Reflector Antennas with Absorber- Lined Tunnels” IEEE Symp. Aitfs. Prop., Seattle, June 1979.
620 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS Chuane C. W , and W. D, Burnside: “A Diffraction Coefficient for a Cylindrical^ Trun-cated Planar Surface (Rolled Edge Effect)," IEEE Trans. Ants. Prop., AP-28, 177-182, March 1980, Crawford, A.
D. C. Hogg and L. E. Hunt: “A Horn-Reflector Antenna for Space Com-munication," BSTJ, 40, 1045-1116, July 196L Cross, D, C, D. D. Howard and J, W. Titus: “ Mireor-Antenna Radar Concept, Mjcw- wave 29, 323-335, May 1986.
„ , , Dragone, C:“ Offset Multireflector Antennas with Perfect Pattern Symmetry and Polariz-ation Discrimination,” BSTJ , 57, 2663-2684, September 1978.
Ehrenspeck, H. W.: “The Short Backfire Antenna,” Proc. IEEE , 53, 1138-1140, August 1965.
A Godwin, M, F, A.
J. T. Whitaker and A.
P. Anderson: “Microwave Diagnostics ol the Chilbolton 25 m Antenna Using the OTS Satellite ” Proc , Inst. Elec. Eng « Int. Conf York, England, pp+ 232-236, 1981 James, G.
L.: Geometrical Theory of Diffraction for Electromagnetic Waves, Peregrmus, ' 1980.
Kaplan, P.
D.: “Predicting Antenna Sidelobe Performance (Sidelobe Pop-Up Probability),” Microwave J., 29, 201-206, September 1986.
Keller, J. B.: “ Geometrical Theory of Diffraction,” J. Opt. Soc. Am., 52, 116-130, February ' t962.
.
-.
_ .
Kieburtz, R.
B., and A. Ishimaru: “Scattering by a Periodically Apertured Conducting Screen," IRE Trans. Ants. Prop., AP-9, 506-514, November 1961.
Kieburtz, R. B , and A. Ishimaru: “Aperture Fields of an Array of Rectangular Apertures, IRE Trans. Ants. Prop, AP-10, 663-671, November 1962.
Koffman, I " Feed Polarization for Parallel Currents in Reflectors Generated by Come Sections," IEEE Trans. Ants. Prop., AP-14, 37^0, January 1966.
Kouyoumjian, R. G-, and P. H. Pathak: “A Uniform Geometrical Theory of Diffraction for an Edge in a Perfectly Conducting Surface,” Proc.
IEEE, 62, 1448-1461, November 1974.
.
' Lamb, J.
W; “Analysis of a Cassegrain Antenna Having a Gaussian Illumination Pattern” Helsinki Univ. Tech. Kept. S147, 1983.
Ludwig, A C “ Low Sidelobe Aperture Distribution for Blocked and Unblocked Circular Apertures,” IEEE Trans. Ants. Prop., 30, 933-946, September 1982.
Mayer C E J H Davis, W. L. Peters and W, J. Vogel: “A Holographic Surface Mea-surement of the Texas 4.9-metcr Antenna at 86 GHz," IEEE Trans. Instrum. Meas., IM-32, 102-109, 1983.
Nash, R. T.: “Stepped Amplitude Distributions,” IEEE Trans. Ants. Prop., AP-12, July 1964 Ott, R. H., R. G. Kouyoumjian and L. Peters, Jr.: “The Scattering by a Two-Dimensional Periodic Narrow Array of Plates,” Radio Sci., November 1967.
_ Ott, R. H., and J. H. Richmond: “A Linear Equation Solution for Scattering by a Periodic Plane Array of Thin Conducting Plates” ElectroSd. Lab. Rept. 1774-9, 1965.
Peters, L., Jr, and C. E. Ryan, Jr,: “ Empirical Formulas for E-Plane Creeping Waves on General Smooth Conducting Bodies,” JEEE Trans. Ants. Prop., AP-18, 432-43 May 1970.
Tt .
Rahmat-Samii.
Y.: “Surface Diagnosis of Large Reflector Antennas Using Microwave Holographic Metrology—An Iterative Approach," Radio Sci, 13, 1205-1217, September-October 1984.
Rahmat-Samii, Y: “Microwave Holography of Large Reflector Antennas: Simulation Algorithms," IEEE Trans. Ants. Prop., AP-33, 1194-1203, November 1985.
PROBLEMS 621 Ratna&iri, P. A. J., R, G. Kouyoumjian and P. H. Pathak: "The Wide Angle Side Lobes of Reflector Antennas,” OSU Elect. Sci. Lab. Rept. 5635-4, 1970.
Rudge, A. W., and N. A, Adatia: “Offset-Parabolic-Antennas: A Review," Proc IEEE, 66, 1592-1618, December 3978.
Rudge, A. W., and M. J. Withers: “Design of Flared-Horn Primary Feeds for Parabolic Reflector Antennas," Proc. IEE, 117, 1741-1747, September 1970.
Rudge, A, W., and M. J. Withers: “New Technique for Beam Steering with Fixed Parabol-ic Reflectors ” Proc. IEE , 118, 857-863, July 1971.
Rusch, W, V, T., Y. Rahmat-Samii and R. A. Shore: “The Equivalent Paraboloid of an Optimized Off-Set Cassegrain Antenna ” Dig . IEEE Int , Symp. Ants. Prop., 1986.
Scott, P. F, and M. Ryle: "A Rapid Method for Measuring the Figure of a Radio Tele-scope Reflector," Royal Asl Soc. Mon. Notices, 178, 539-545, 1977.
Shore, R. A.: “A Simple Derivation of the Basic Design Equation for Offset Dual Reflector Antennas with Rotational Symmetry and Zero Cross Polarization,” IEEE Trans.
Ants . Prop AP-33, 114-116, January 1985.
Sletten, C.
J., W. Payne and W, Shillue: “Offset Dual Reflector Antennas for Very Low Sidelobes," Microwave J., 29, 221-240, May 1986.
Tanaka, H., and M. Mizuasawa: “Elimination of Cross- Polarization in Offset Dual-Reflector Antennas,” Elec. Commun. (Japan), 58-B, 71-78, 1975.
Twersky, V r : “Multiple Scattering of Waves and Optical Phenomena,” J. Opt. Soc. Am., 52, 145-171, 1962, PROBLEMS 1 12-1 Flat sheet reflector Calculate and plot the radiation pattern of a 2/2 dipole antenna spaced 0,1 5 a from an infinite flat sheet for assumed antenna loss resist-ances Rl = 0 and 10 ft, Express the patterns in gain over a 2/2 dipole antenna in free space with the same power input (and zero loss resistance).
12-2 Square-corner reflector A square-corner reflector has a driven 2/2 dipole antenna spaced a/2 from ihe corner. Assume perfectly conducting sheet reflectors of infinite extent (ideal reflector). Calculate and plot the radiation pattern in a plane at right angles to the driven element.
12-3 Square-comer reflector. Calculate and plot the pattern of an ideal square-corner reflector with 2/2 driven antenna spaced A/2 from the comer but with the antenna displaced 20° from the bisector of the comer angle. The pattern to be calculated is in a plane perpendicular to the antenna and to the reflecting sides.
12-4 Paraboloidal reflector.
Calculate and plot the radiation patterns of a paraboloidal reflector with uniformly illuminated aperture when the diameter is 8A and when the diameter is 16A.
12-5 Cylindrical parabolic reflector. Calculate the radiation pattern of a cylindrical parabolic reflector of square aperture 16A on a side when the illumination is uniform over the aperture and when the field intensity across the aperture follows a cosine variation with maximum intensity at the center and zero intensity at the edges. Compare the two cases by plotting the normalized curves on the same graph.
1 Answers to starred () problems are given in App. D.
622 12 REFLECTOR ANTENNAS AND THEIR FEED SYSTEMS + 12-6 Square-comer reflector, {a) Calculate and plot the pattern of a 90° corner reflector with a thin center-fed a/2 driven antenna spaced 0.35a from the corner. Assume that the corner reflector is of infinite extent.
(M Calculate the radiation resistance of the driven antenna.
(c) Calculate the gain of the antenna and corner reflector over the antenna alone.
Assume that losses are negligible.
12-7 Square-corner reflector versus array of its image elements. Assume that the corner reflector of Prob, 12-6 is removed and that in its place the three images used in the analysis are present physically, resulting in a 4-element driven array.
(0) Calculate and plot the pattern of this array, {h) Calculate the radiation resistance at the center of one of the antennas, fc) Calculate the gain of the array over one of the antennas alone.
+ 12-8 Square-corner reflector array. Four 90 c corner-reflector antennas are arranged in line as a broadside array. The corner edges are parallel and side-by-side as in Fig. P12-8- The spacing between corners is IA, The driven antenna in each corner is a A/2 dement spaced 0.4A from the corner. All antennas are energized in phase and have equal current amplitude. Assuming that the properties of each corner are the same as if its sides were of infinite extent, what is (u) the gain of the array over a single A/2 antenna and (h) the half-power beam width in the H plane?
Figure PI2-8 Square-corner reflector array.
129 Paraboloidal reflector aperture distribution.
(a) Show that the variation of field across the aperture of paraboloidal reflector with an isotropic source is proportional to 1/'[1 + (p/2L) 2 ] where p is the radial distance from the axis of the paraboloid. Show that this relation is equivalent to (1 + cos 0)/2.
[b) If the parabola extends to the focal plane and the feed is isotropic over the hemisphere subtended by the parabola, calculate the aperture efficiency.
12-10 Square-comer reflector.
(a} Show that the relative field pattern in the plane of the driven A/2 element of a square-corner reflector is given by E = [1 — cos ( S r sin 0)] cos (90° cos 0) sin 0 PROBLEMS 623 where 0 is the angle with respect to the element axis. Assume that the corner-reflector sheets are perfectly conducting and of infinite extent.
(6) Calculate and plot the field pattern in the plane of the driven dement for a spacing of A/2 to the corner. Compare, with the pattern at right angles (Prob.
12-2).
12-11 Corner reflector. X}4 to the driven element. A square-corner reflector has a spadng of A/4 between the driven A/2 element and the corner. Show that the directivity D = 12.8 dBi.
12-12 Corner reflector 1/1 to the driven element. A square-corner reflector has a driven A/2 element A/2 from the corner.
() Calculate and plot the far-fidd pattern in both principal planes.
() What are the HPBWs in the two principal planes?
(c) What is the terminal impedance of the driven element?
(d) Calculate the directivity in two ways: (i) from impedances of driven and image dipoles and (2) from HPBWs, and compare. Assume perfectly conducting sheet reflectors of infinite extent, 12-13 Parabolic reflector with missing sector. A circular parabolic dish antenna has an effective aperture of 100 m 2 . If one 45“ sector of the parabola is removed, find the new effective aperture. The rest of the antenna, including the feed, is unchanged.
12-14 Efficiency of rectangular aperture with partial taper. Calculate the aperture effi-ciency and directivity of an antenna with rectangular aperture x ty t with a uniform field distribution in the y direction and a cosine field distribution in the x direction (zero at edges, maximum at center) if x t = 20A and y1 = 10A.
12-15 Efficiency of rectangular aperture with full taper. Repeat Prob. 12-14 for the case where the aperture field has a cosine distribution in both the x and y directions.
12-16 Efficiency of aperture with phase ripple. A square unidirectional aperture (x^i) is 10A on a side and has a design distribution for the electric field which is uniform in the x direction but triangular in the y direction with maximum at the center and zero at the edges. Design phase is constant across the aperture. However, in the actual aperture distribution there is a plus-and-minus-30° sinusoidal phase varia-tion in the x direction with a phase cycle per wavelength. Calculate (a) the design directivity, (b) the utilization factor, (c) the actual directivity, (d) the achievement factor, (e) the effective aperture and (/) the aperture efficiency.
12-17 Rectangular aperture. Cosine taper. An antenna with rectangular aperture x t y, has a uniform field in the y direction and a cosine field distribution in the x direction (zero at edges, maximum at center). If x t = 16A and y L - 8A, calculate (a) the aperture efficiency and (6) the directivity.
12-18 Rectangular aperture. Cosine tapers. Repeat Prob.
12-17 for the case where the aperture field has a cosine distribution in both the x and y directions.
12-19 A 20X line source. Cosine-squared taper (a) Calculate and plot the far-field pattern of a continuous in-phase line source 20A long with cosine-squared field distribution.
(b) What is Ihe HPBW?
SLOT ANTENNAS 625 CHAPTER 13 SLOT, HORN AND COMPLEMENTARY ANTENNAS 13-1 INTRODUCTION.
Sections 13-2 and 13-3 of this chapter deal with slot antennas and their patterns. These antennas are useful in many applications, especially where low-profile or flush installations are required as, for example, on high-speed aircraft Sections 13-4, 13-5 and 13-6 discuss the relation of slots to their complementary dipole forms. Any slot has its complementary form in wires or strips, s6 that pattern and impedance data of these forms can be used to predict the patterns and impedances of the corresponding slots. The discussion is based largely on a generalization and extension of flabinet’s (Ba-bi-nay’s) prin-ciple by Henry Booker. A slotted cylinder antenna is discussed in Sec, 13-7.
1 The next four sections of the chapter describe horn antennas, both rectangular and conical. The remaining sections involve ridge, septum, corrugated and matched horns.
13-2 SLOT ANTENNAS. The antenna shown in Fig.
13-la, consisting of two resonant A/4 stubs connected to a 2-wire transmission line, forms an ineffi-cient radiator. The long wires are closely spaced (w < A) and carry currents of opposite phase so that their fields tend to cancel. The end wires carry currents in 1 Patch or microstrip art term as, which mu) 1 be regarded as derived from slot antennas, are discussed in Sec.
16-12.
624 Metal sheet Figure 13-1 Parallel connected /JA stubs (u) and simple slot antenna [bl the same phase, but they are too short to radiate efficiently. Hence, enormous currents may be required to radiate appreciable amounts of power The antenna in Fig.
13-1 /j, on the other hand, is a very efficient radiator, [n this arrangement a A/2 slot is cut in a flat metal sheet. Although the width of the slot is small {w 4 A), the currents are not confined to the edges of the slot but spread out over the sheet. This is a simple type of slot antenna. Radiation occurs equally from both sides of the sheet If the slot is horizontal, as shown, the radi-ation normal to the sheet is vertically polarized.
A slot antenna may be conveniently energized with a coaxial transmission line as in Fig.
13-2a. The outer conductor of the cable is bonded to the metal sheet. Since the terminal resistance at the center of a resonant A/2 slot in a large sheet is about 500 Q and the characteristic impedance of coaxial transmission lines is usually much less, an off-center feed such as shown in Fig.
13-2 h may be Used to provide a better impedance match. For a 50-Q coaxial cable the distance s should be about z.2Q.
Slot antennas fed by a coaxial line in this manner are illustrated in Fig.
l3-2c and d. The radiation normal to the sheet with the hori-zontal slot (Fig.
13-2c) is vertically polarized while radiation normal to the sheet with the vertical slot (Fig.
13-2^) is horizontally polarized. The slot may be A 2 long, as shown, or more.
A flat sheet with a /. 2 slot radiates equally on both sides of the sheet.
However, if the sheet is very large {ideally infinite) and boxed in as in Fig. 13-3u, radiation occurs only from one side.
If the depth d of the box is appropriate {d - A 4 for a thin slot), no appreciable shunt susceptancc appears across the terminals. With such a zero susceptancc box, the terminal impedance of the res-onant A; 2 slot is nonreactive and approximately twice its value without the box, or about 1000 Q, The boxed-in slot antenna might be applied even at relatively long wave-lengths by using the ground as the flat conducting sheet and excavating a trench A, 2 long by A/4 deep as suggested in Fig, 13-3h. The absence of any structure above ground level might make this type of antenna attractive, for example, in applications near airports. To improve the ground conductivity, the walls of the trench and the ground surrounding the slot can be covered with copper sheet or 626 Li SLO-L.
AND f OM Pl.WMY VI A K> ANTI-MVAS screen. Radiation is maximum in all directions at right angles to the slot and is zero along the ground in the directions of the ends of the slot, as suggested in Fig.
1 3-3b, The radiation along the ground is vertically polarized.
Radiation from only one side of a large flat sheet may also be achieved by a slot fed with a waveguide as in Fig.
13-4u. With transmission in the guide in the TE, 0 mode, the direction of the electric field E is as shown. The width L of the guide must be more than y,/2 to transmit energy, but it should be less than 1 k to suppress higher-order transmission modes. WiLh the slot horizontal, as show r n, Figure 13-3 Boxed-in slot antenna {«) and application to provide flush radiator (M.
13 2 SLOT ANTKNNAS 627 Figure 0-4 Waveguide-fed slot (a) and T-fed slot \h) the radiation normal to the sheet is vertically polarized. The slot opening consti-tutes an abrupt termination to the waveguide.
It has been found 1 that the resulting impedance mismatch is least over a wide frequency band if the ratio L/w is less than 3.
A compact wideband method for feeding a boxed-m slot is illustrated in Fig.
1 3-4b. In this T-fed arrangement 1 the bar compensates the impedance char-acteristics so as to provide a VSWR on a 50-Q feed line of less than 2 over a frequency range of nearly 2 to 1. The ratio L/w of the length to width of the slot is about 3, Dispensing with the flat sheet altogether, an array of slots may be cut in the waveguide as in Fig.
13-5 so as to produce a directional radiation pattern. 2 With transmission in the guide in the TE I0 mode, the instantaneous direction of the electric field E inside the guide is as indicated by the dashed arrows. By cutting inclined slots as shown at intervals of kJ2 (where is the wavelength in the guide), the slots are energized in phase and produce a directional pattern wilh maximum radiation broadside to the guide.
If the guide is horizontal and E inside the guide is vertical, the radiated field is horizontally polarized as sug-gested in Fig.
13-5.
3 A. Dgrne and D. Lazarus, in Very High Frequency Techniques, Radio Research Laboratory Sltiff McGraw-Hill, New York, l947 T t;hap.
7.
1 W. H. Watson, The Physical Principles of Wave Guide Transmission and Antenna Systems, Oxford University Press, London, 1947.
628 13 SLOT. HORN AND t OU PU Alt NTAft'.
\NTKNNAS Figure 13-5 Broadside array of slots in waveguide.
13-3 PATTERNS OF SLOT ANTENNAS IN FLAT SHEETS EDGE DIFFRACTION. Consider the horizontal X/2 slot antenna of width w in a perfectly conducting flat sheet of infinite extent, as in Fig, 13-6a. The sheet is energized at the terminals FF.
It has been postulated by Booker 1 that the radiation pattern of the slot is the same as that of the complementary horizontal /J2 dipole consisting of a perfectly conducting flat strip of width w and energized at the terminals FF, as indicated in Fig.
13-66, but with two differences. These arc (1) that the electric and magnetic fields are interchanged and (2) that the component of the electric field of the slot normal to the sheet is discontinuous from one side of the sheet to the other, the direction of the field reversing. The tangential component of the magnetic field is, likewise, discontinuous.
The patterns of the X/2 slot and the complementary dipole are compared in Fig.
13-7 The infinite flat sheet is coincident with the xz plane, and the long dimension of the slot is in the x direction (Fig 13-7a), The complementary dipole is coincident with the x axis (Fig, 13-76). The radiation-field patterns have the same doughnut shape, as indicated, but the directions of E and H are inter-changed, The solid arrows indicate the direction of the electric field E and the dashed arrows the direction of the magnetic field H, If the xv plane is horizontal and the z axis vertical as in Fig, 13-7 a, the radiation from the horizontal slot is vertically polarized everywhere in the xy plane. Turning the slot to a vertical position (coincident with the i axis) rotates Figure 13-6 A x/2 sloi in an infinite flat sheet (a) and a complementary x/2 dipole antenna (fr).
1 H. G. Booker, h+ Slot Aerials and Their Relation to Complementary Wire Aerials," JIEE \Lond, I, 93, pt. IIIA, no, 4, 1946.
O-3 PATTERNS OK SLOT ANTENNAS IN FLAT SHEETS EDGE DIFFRACTION 629 Figure 13-7 Radiation- field patterns of slot in -an infinite sheet and of complementary dipole antenna [h) The patterns have the same shape but with E and H interchanged.
the radiation pattern through 90 ' to the position shown in Fig.
13-8. The radi-ation in this case is everywhere horizontally polarized: i.e., the electric field has only an E component.
If the slot is very thin (vv < X) and X/2 long (L = x/2), the variation of E as a function of 6 is, from (5-5-12), given by im cos [(te/2) cos 0] EJO) = : : Figure 13-8 Radiation pattern of vertical slot in an infinite flat sheet.
630 1-1 SLOT, hlORN and L’t >M PLtm lntar y antennas (a) ^ Figure 13-9 Solid curves show patterns in xy plane of big, 13-8 for slot in finite sheet of length L Slot is open on both sides in (a) and closed on left side in (£>), Dashed curves show pattern for infinite sheet. All patterns idealized.
Assuming that the sheet is perfectly conducting and infinite in extent, the magnitude of the field component E# remains constant as a function of 0 for any value of Thus, E0(0) == constant (2) Consider now the situation where the slot is cut in a sheet of finite extent as suggested by the dashed lines in Fig.
13-8, This change produces relatively little effect on the E^(0) pattern given by (1). However, there must be a drastic change in the £^{0) pattern since in the x direction, for example, the fields radiated from the two sides of the sheet are equal in magnitude but opposite in phase so that they cancel. Hence, there is a null in all directions in the plane of the sheet. For a sheet of given length L in the x direction, the field pattern in the xy plane might then be as indicated by the solid curve in Fig.
13-9tf. The dashed curve is for an infinite sheet (L = oo). If one side of the slot is boxed in, there is radiation in the plane of the sheet as suggested by the pattern in Fig. 13-%.' With a finite sheet the pattern usually exhibits a scalloped or undulating characteristic, as suggested in Fig.
13-9. As the length L of the sheet is increased, the pattern undulations become more numerous but the magnitude of the undulations decreases, so that for a very large sheet the pattern conforms closely to a circular shape. Measured patterns 2 illustrating this effect are shown in Fig. 13-10 for 3 values of L. A method due to Andrew Alford for locating the angular positions of the maxima and minima is described by Dome and Lazarus.
2 In this method the assumption ] According to H. G Booker. “Slot Aerials and Their Relation to Complementary Wire Aerials," JILL [Lond.y 93. pi.
Ill A, no. 4, 1946, the energy density m the ^ 0 or 180' directions is \ that for an infinite sheet, or the field intensity is 0.707 that for an infinite sheet.
1 A Dome and D. Lazarus, in Very Hi&h Frequency Techniques, Radio Research Laboratory Staff, McGraw-Hill, New York, 1947, chap. 7 (see sec.
7-3).
130 patterns of slot antennas in flat sheets edge diffraction 631 Figure 13-10 Measured tf>-piane patterns of k/2 boxed-in slot antennas in finite sheets of three lengths L ~ 0.5, 2.75 and 5.3/. The width of the slots is 01/.
( After A. Dome and 0, Lazarus, in Very High Frequency Techniques, Radio Research Laboratory Staff, McGraw-Hill^ New York , 194Z) is made that the far field is produced by three sources (see Fig.
13-11), one (1) at the slot of strength 1 sin on and two (2 and 3) at the edges of the sheet (edge diffraction effect) with a strength k sin (tor -<5), where It <£ 1 and 3 gives the phase difference of the edge sources with respect to the source (1) at the slot. At the point P at a large distance in the direction 0, the relative field intensity is then E = sin on + k sin (cur - £ - c) + k sin (cuf -<5 + s) (3) where t = (jt/'/JL cos 0 X X X^^ To point P Figra-e 13-11 Construction for locating maxima and minima of pattern for slot in a finite sheet.
632 li SLOT HORN AND COM FLfcMENTAfO ANTENNAS By trigonometric expansion and rearrangement, E - (i + 2k cos 6 cos c) sin tot - (2 k sin ^ cos e) cos tot (4) and the modulus of E is | £| = v/(l + 2A cos cos t:) 2 + (2A sin 6 cos i:) 2 (5) Squaring and neglecting terms with A 2 , since A 1, (5) reduces to 1 E | = r \ -b 4A cos 6 cos t: (h) The maximum and minimum values oT | £ | as a function of £ occur when c — nn, so that - L cos — nn (7) a where n is an integer. Thus n/ , , n/ cos = — and tp = arceos y-(a) L L The values of for maxima and minima in the \ 2 ) As Case 2 let the first screen be replaced by its complementary screen and the field behind it be given by Fa =f1(x , y, z) (2) 1 See, for example, Max Born, Optik Verlag Julius Springer, Berlin, 1933, p.
1 55.
].M (iA HINHT'S PRINCIPLE AND COMPLEMENTARY ANTENNAS 633 Case 3 Figure 13-12 Illustration of Babinet's principle.
As Case 3 with no screen present the field is =/3 t> z) (3) Then Babinet’s principle asserts that at the same point x„ y l5 z u Fs + F„ = F0 (4) The source may be a point as in the above example or a distribution of sources.
The principle applies not only to points in the plane of observation B as sug-gested in Fig, 13-12 but also to any point behind screen A. Although the prin-ciple is obvious enough for the simple shadow case above, it also applies where diffraction is considered.
634 U SLOT NOR N AND COMPLEMENTARY ANTENNAS Babi net's principle has been extended and generalized by Booker 1 to take into account the vector nature of the electromagnetic field. In this extension it is assumed that the screen is plane, perfectly conducting and infinitesimally thm.
Furthermore, if one screen is perfectly conducting (<r = x), the complementary screen must have infinite permeability (p = < I- Thus, if one screen is a perfect conductor of electricity, the complementary screen is a perfect “conductor" of magnetism. No infinitely permeable material exists, but the equivalent effect may bJ obtained by making both the original and complementary screens of perfectly conducting material and interchanging electric and magnetic quantities every-where. The only perfect conductors are superconductors which soon may be avail-able at ordinary temperatures for antenna applications. However, many metals, sucb as silver and copper, have such high conductivity that we may assume the conductivity is infinite with a negligible error in most applications.
As an illustration of Booker's extension of Babinet's principle, consider the cases in Fig.
13-13. The source in all cases is a short dipole, theoretically an infinitesimal dipole.
In Case 1 the dipole is horizontal and the original screen is an infinite, perfectly conducting, plane, infinitesimally thin sheet with a vertical slot cut out as indicated. At a point P behind the screen the field is E^. In Case 2 the original screen is replaced by the complementary screen consisting of a per-fectly conducting, plane, infinitesimally thin strip of the same dimensions as the slot in the original screen. In addition, the dipole source is turned vertical so as to interchange E and H. At the same point P behind the screen the field is E 2 . As an alternative situation for Case 2 the dipole source is horizontal and the strip is also turned horizontal.
Finally, in Case 3 no screen is present and the field at point P is E v . Then, by Babinet’s principle, E i + E 2 — E 0 Babinet's principle may also be applied to points in front of the screens. In the situation of Case 1 (Fig. 13-13) a large amount of energy may be transmitted through the slot so that the field E, may be about equal to the field E0 with no intermediate screen (Case 3). In such a situation the complementary dipole acts like a reflector and E t is very small. (Recall Sec.
12-13 on frequency-sensitive surfaces.) Since a metal sheet with a A/2 slot or, in general, an orifice of at least U perimeter may transmit considerable energy, slots or orifices of this size should be assiduously avoided in sheet reflectors such as described in Chap. 12 when E is not parallel to the slot 1 H.G. Booker, “Slot Aerials and Their Relation to Complementary Wire Aerials," JIEE {Lond\ 93, pi, 111A, no. 4, 1946.
U S THE IMPEDANCE OF COMPLEMENTARY SCREENS 635 " /-/ !
Figure 13-13 Illustration of Babinet's principle applied to a slot in an infinite metal sheet and the complementary metal strip.
13-5 THE IMPEDANCE OF COMPLEMENTARY SCREENS.
In this section Babinet's principle is applied with the aid of a transmission- line analogy to finding the relation between the surface impedance Z x of a screen and the surface impedance Z2 of the complementary metal screen.
1 Consider the infinite transmission line shown in Fig. 13-14a of character-istic impedance Z0 or characteristic admittance Y 0 = 1/Z0 . Let a shunt admit-tance be placed across the line. An incident wave traveling to the right of voltage Vj is partly reflected at Yx as a wave of voltage V r and partly transmitted beyond Y as a wave of voltage K The voltages are measured across the line.
1 The treatment follows that given by H. G, Booker. See "Slot Aerials and Their Relation to Comple-mentary Wire Aerials,” JIEE (LotuI.X 93, pt. II1A + no. 4, 1946.
636 L3 slot, horn and complementary antennas Screen Figure 13-14 Shunt admittance across transmission line lei) is analogous to infin-ite screen in path of plane wave {fr) Method of measuring surface admittance of screen is suggested in (c).
This situation is analogous to a plane wave of field intensity E { incident normally on a plane screen of infinite extent with a surface admittance, or admit-tance per square, of Y1 \ that is, the admittance measured between the opposite edges of any square section of the sheet as in f ig.
13-1 4c-is Yt . Neglecting the impedance of the leads the admittance = -p (U per square) (1) The value of Y is the same for any square section of the sheet. Thus, the section may be l cm square or 1 meter square. Hence, (1) has the dimensions of admit-| tance rather than of admittance per length squared and is called a surface admit-, tance , or admittance per square. The field intensities of the waves reflected and transmitted normally to the screen are E r and £,. Let the medium surrounding the screen be free space.
It has a characteristic admittance Y 0 which is a pure conductance G 0 , Thus, ( 2 ) ii-S THE IMPEDANCE OF COMPLEMENTARY SCREENS 637 The ratio of the magnetic to the electric field intensity of any plane traveling wave in free space has this value. Hence, ’•Hr-S-t <°> e> where H c , H r and H t are the magnetic field intensities of the incident, reflected and transmitted waves respectively.
The transmission coefficient for volatge i c , of the transmission line 1 (Fig 13~l4a) is 2 Y 0 r K 2Y 0 +Y 1 By analogy the transmission coefficient for the electric field (Fig.
13-14/?) is E t 2Y 0 Ze ~E~ 2Y 0 + Y 1 (5) If now the original screen is replaced by its complementary screen with an admit-tance per square of Yj, the new transmission coefficient is the ratio of the new transmitted field E\ to the incident field. Thus, , E\ 2V0 Applying Babinet T s principle, we have from (13-4-6) that E t e; ^ + e = 1 (7> + T r = Therefore, 2Y 0 2^+5^ 2Y 0 + Y 2 and we obtain Booker’s result that Y,Y,=4ri Since Y, = 1/Z„ Y 2 = 1 /Z 2 and Y 0 = 1/Z0 , 7 7 Z« Z > Z = T or -z z —— V — -See. for example. S. A. Schelkunoff, Electromagnetic Waves Van Nostrand, New York, 1943, p. 212.
638 11 SLOT, HtJlN AND rOMPLPMTNIARY ANlhNNAS Thus, the geometric mean of the impedances of the two screens equals ± the intrinsic impedance of the surrounding medium.
Since, for free space, Z 0 -376.7 £1, If screen 1 is an infinite grating of narrow parallel metal strips as in Fig. 13-15a, then the complementary screen (screen 2) is an infinite grating of narrow slots as shown in Fig. 13-156. Suppose that a low-frequency plane wave is incident normally on screen 1 with the electric field parallel to the strips. Then the grating acts as a perfectly reflecting screen and zero field penetrates to the rear. Thus Z, = 0 and, from (12), Z 2 = x, so that the complementary screen of slots (screen 2) offers no impediment to the passage of the wave.
If the frequency is increased sufficiently, screen 1 begins to transmit part of the incident wave.
If at the frequency F 0 screen 1 has a surface impedance Z, = J 1 88 £1 per square, the impedance Z 2 of screen 2 is -j 188 £2 per square, so that both screens transmit equally well. If screen 5 becomes more transparent {'/ .
larger) as the frequency is further increased, screen 2 will become more opaque (Z 2 smaller). At any fre-quency the sum of the fields transmitted through screen 1 and through screen 2 is a constant and equal to the field without any screen present.
The dipoles and slots of Sec.
12-13 are more specialized examples of com-plementary surfaces or screens.
13-6 THE IMPEDANCE OF SLOT ANTENNAS, In this section a relation is developed for the impedance Z s of a slot antenna in terms of the impedance Zd of the complementary dipole antennal Knowing Z d for the dipole, the impedance ZA of the slot can then be determined.
1 The irearnient follows lhal given by H. ti. Booker, “ Slot Aerials and Their Relation to Complemen-tary Wire Aerials,” JIEE (Lend.), 93, pt. 111A, no. 4, 1946, with minor embellishments suggested by V H. Rumsey. See also Sec.
15-2.
iv the impedance of slot antennas 639 V////////A FA I Figure 13-16 Slot antenna (a) and complementary dipole antenna (6).
Considcr the slot antenna shown in Fig.
13-16a and the complementary dipole antenna shown in Fig.
13-16 /j. The terminals of each antenna are indicated by f and it is assumed that they are separated by an infinitesimal distance. It is assumed that the dipole and slot are cut from an infinitesimally thin, plane, per-fectly conducting sheet.
Let a generator be connected to the terminals of the slot. The driving-point impedance Z s at the terminals is the ratio of the terminal voltage F s to the ter-minal current Let Es and H s be the electric and magnetic fields of the slot at any point P, Then the voltage V 5 at the terminals FF of the slot is given by the line integral of E T over the path C i (Fig, 13-16a) as approaches zero. Thus, K s = lim E s d\ U) C] -0 Jc, where d\ = infinitesimal vector element of length d\ along the contour or path The current /, at the terminals of the slot is 1=2 lim H, d\ The path C 2 is just outside the metal sheet and parallel to its surface. The factor 2 enters because only j of the closed line integral is taken, the line integral over the other side of the sheet being equal by symmetry.
Turning our attention to the complementary dipole antenna, let a generator be connected to the terminals of the dipole. The driving-point impedance Zd at the terminals is the ratio of the terminal voltage V a to the terminal current Let Ed and Hd be the electric and magnetic fields of the dipole at any point P, Then the voltage at the dipole terminals is Vj = lim [" E„ • d\ (3) 640 L3 si.o-i.morn anu complementary antennas and the current is 2 li lim H d c i-o Jc L However, lim Ej d\ = A lim - d\ (5) Jc 1 Cj-O Jc 2 lim \ Hj d\ 1 = Z lim j E s d\ (61 fi -o Jci Ci - n fC I where Z 0 is the intrinsic impedance of the surrounding medium, Subsisting { 3 1 and (2) in n) yields Substituting (4) and (l) in (6l gives Multiplying (71 and (8) we have w ^ — /, KKt^zl Ll/~ 4 yl y2 ZiZtf_ 4 Zl “4Z„ Thus, wc obtain Booker s result that the terminal impedance Zs of a slot antenna is equal to \ of the square of the intrinsic impedance of the surrounding medium divided by ihe terminal impedance Zd of the complementary dipole antenna. For free space = 376,7 £L so 1 _ Zl 35 476 4Z, Z, 1 If the intrinsic impedance Z 0 of free space were unknown, (II) provides a means of deiermming it t>y mta sure merits of the impedance Zs of a slot antenna and the impedance Zd of the complementary dipole antenna. The impedance Z„ is twice the geometric means of Z.and Z^or Z0 = 2v '/ZJZd B2) O-fi THE IMPHIJANCF. OK SLOT ANTtNNAS 641 X/2 dipole X/2 slot Resonant X/2 dipole Resonant X/2 slot Full X dipole Full X slot Figure 13- 1 7 Comparison of impedances of cylindrical dipole antennas with complementary slot antennas. The slot in [c] matches directly to the 50 1] coaxial line.
The impedance of the slot is proportional to the admittance of the dipole, or vice versa. Since, in general, Zd may be complex, we may write 35476 +-]X d Rj + Xj <Rj jXd) (13) where and X d are the resistive and reactive components of the dipole terminal impedance Zd , Thus, if the dipole antenna is inductive, the slot is capacitative, and vice versa.
Lengthening a A/2 dipole makes it more inductive, but lengthening a A/2 slot makes it more capacitalive.
Let us now consider some numerical examples, proceeding from known dipole types to the complementary slot types. The impedance of an infinitesimally thin A/2 antenna {L — 0.5A and LjD = oo) is 73 + /42,5 Q (see Chap.
10). There-fore, the terminal impedance of an infinitesimally thin A/2 slot antenna (L — 0.5A and Ljw — x ) is Zi 35476 73 T J42.S = 363 -;211 Q (14) See Fig. 13-1 7a.
642 13 SLOT, HORN AND COMPLEMENTARY ANTENNAS As another more practical example, a cylindrical antenna with a length-diameter ratio of 100 (L/D = 100) is resonant when the length is about 0.475/ (/ = 0.475/). The terminal impedance is resistive and equal to about 67 fi. The terminal resistance of the complementary slot antenna is then Z, =^^ = 530 +/0 fl (15) 67 Sec Fig. 13-175.
The complementary slot has a length L = 0.475/, the same as for the dipole, but the width of the slot should be twice the diameter of the cylindrical dipole. As indicated in Sec. 9-7, a flat strip of width w is equivalent to a cylindrical conduc-tor of diameter D provided that w = 2D. Thus, in this example, the width of the complementary slot is 2L 2 x 0.475/ ^ .
w = 2D — = — ^ 0.01/ 100 100 As a third example, a cylindrical dipole with an L/D ratio of 28 and length of about 0.925/ has a terminal resistance of about 7lO+/0fi. The terminal resistance of the complementary slot is then about 50 + /0 fi so that an imped-ance match will be provided to a 50 fi coaxial line. See Fig.
13- 17c.
If the slots in these examples are enclosed on one side of the sheet with a box of such size that zero susceptance is shunted across the slot terminals, due to the box, the impedances are doubled.
The bandwidth or selectivity characteristics of a slot antenna are the same as for the complementary dipole.
Thus, widening a slot (smaller L/w ratio) increases the bandwidth of the slot antenna, the same as increasing the thickness of a dipole antenna (smaller L/D ratio) increases its bandwidth.
The above discussion of this section applies to slots in sheets of infinite extent If the sheet is finite, the impedance values are substantially the same pro-vided that the edge of the sheet is at least a wavelength from the slot. However, the measured slot impedance is sensitive to the nature of the terminal connec-tions.
13-7 SLOTTED CYLINDER ANTENNAS.
1 A slotted sheet antenna is shown in Fig, 13-18a + By bending the sheet into a U-shape as in (5) and finally into a cylinder as in (c), we arrive at a slotted cylinder antenna. The impedance of the path around the circumference of the cylinder may be sufficiently low so that 1 A. Alford, ' Long Slot Antennas/ Prvc. Natl. Electronics Con/., 1946, p. 143 A. Alford, '‘Antenna for F-M Station WGHF," Common icuti'ens, 26, 22, February 1946.
E. C. Jordan and W. E. Miller, "SloLlcd Cylinder Antennas/ Electronics, 20,90-93, February 1947.
George Sinclair, “ The Patterns of Slotted Cylinder Antennas,” Proc. IRE , 36, 1487- 1492, December 1948 DOTTED CYLINDER ANTENNAS 643 <°> () (c) Figure 13-1 Evolution of dotted cylinder from dotted sheet.
most of the current tends to flow in horizontal loops around the cylinder as suggested.
If the diameter D of the cylinder is a sufficiently small fraction of a wavelength, say, less than A/8, the vertical slotted cylinder radiates a horizontally polarized field with a pattern in the horizontal plane which is nearly circular. As the diameter of the cylinder is increased, the pattern in the horizontal plane tends to become more unidirectional with the maximum radiation from the side of the cylinder with the slot. For resonance, the length L of the slot is greater than A/2 This may be explained as follows. Referring to Fig. 13-19«, the 2-wire transmis-sion line is resonant when it is A/2 long. However, if this line is loaded with a senes of loops of diameter D as at (fo), the phase velocity of wave transmission on the line can be increased, so that the resonant frequency is raised. With a suffi-cient number of shunt loops the arrangement of (b) becomes equivalent to a slotted cylinder of diameter D t Typical slotted cylinder dimensions for resonance are D = 0.125/, L = 0.75/ and the slot width about 0.02/, This type of antenna, pioneered by Andrew Alford, has found considerable application for broadcasting a horizontally polarized wave with an omnidirec-tb) Figure 13-19 Slotted cylinder as a loop-loaded transmission Line.
644 [J SLOT HORN AND COMPLEMENTARY ANTENNAS Cion.1 or circular pattern in the horizontal plane. Vertical-plane directivity may be increased by using a long cylinder with stacked, i.e„ colhnear, slots.
,, S HORN ANTENNAS. A horn antenna may be regarded as a flared-t lor onened-outl waveguide. The function of the horn is to produce a uniform » larger «P«rtu.c Oran lha, of Che .a.egcnde and Kence grea.er directivity. Horn antennas are not new. Jagadm Chandra Bose constructed a antennas are ill!,rated in Fig. 13-20. Those in the lefl column ^ Angular horns All are energised from mctangular wavegu.des ,he right column arc circular types. To minimise refections of .he gullied wave, the transilion region or horn between the waveguide at the throat rectangular horns circular horns (a) Exponentially tapered pyramidal ie) Exponentially tapered (,/) Pyramidal ^ TE oi bicomcal Figure 13-10 Types of rectangular and circular horn antennas. Arrows indicate E-field directions.
Li-W HORN ANTENNAS 645 and free space at the aperture could be, given a gradual exponential taper as in pig. 13-20a or e. However, it is the general practice to make horns with straight flares as suggested by the other types in Fig.
1 3-20.
1 The types in Fig. 13-20b and c are sectoral horns. They are rectangular types with a dare in only one dimen-sion. Assuming that the rectangular waveguide is energized with a TE 10 mode wave electric field {E in the v direction), the horn in Fig, 13-20b is flared out in a plane perpendicular to E, This is the plane of the magnetic field H. Hence, this type of horn is called a sectoral horn flared in the H plane or simply an H-plane sectoral horn .
The horn in Fig. 13-20c is flared out in the plane of the electric field E, and, hence, is called an E-plane sectoral horn. A rectangular horn with flare in both planes, as in Fig. 13-20J, is called a pyramidal horn. With a TEj 0 wave in the waveguide the magnitude of the electric field is quite uniform in the y direc-tion across the apertures of the horns of Fig.
1 3-206, c and d but tapers to zero in the x direction across the apertures. This variation is suggested by the arrows at the apertures in Fig. 13-206, c and ti. The arrows indicate the direction of the electric field E, and their length gives an approximate indication of the magnitude of the field intensity. For small flare angles the field variation across the aperture of the rectangular horns is similar to the sinusoidal distribution of the TE 10 mode across the waveguide.
The horn shown in Fig, 1 3-20/" is a conical type , When excited with a circu-lar guide carrying a TE n mode wave, the electric field distribution at the aper-ture is as shown by the arrows. The horns in Fig. 13-20# and h are biconicat types. The one in Fig. 13-20# is excited in the TEM mode by a vertical radiator while the one in Fig. 13-206 is excited in the TE01 mode by a small horizontal loop antenna. These biconical horns are nondircctional in the horizontal plane.
The biconical horn of Fig. 13-20# is like the one shown in Fig. 2- 28c.
Neglecting edge effects, the radiation pattern of a horn antenna can be determined if the aperture dimensions and aperture field distribution are known.
For a given aperture the directivity is maximum for a uniform distribution.
Variations in the magnitude or phase of the field across the aperture decrease the directivity. Since the H-plane sectoral horn (Fig. 13-206) has a field distribution over the x dimension which tapers to zero at the edges of the aperture, one would expect a pattern in the xz plane relatively free of minor lobes as compared to the yz plane pattern of an £ -pi ahe sectoral horn (Fig, 13- 20c) for which the magni-tude of E is quite constant over the y dimension of the aperture. This is borne out experimentally.
The principle of equality of path length (Fermat's principle) is applicable to the horn design but with a different emphasis. Instead of requiring a constant phase across the horn mouth, the requirement is relaxed to one where the phase may deviate, but by less than a specified amount <5, equal to the path length difference between a ray traveling along the side and along the axis of the horn.
1 Homs with a straight flare tend to have a constant phase center while those wiih a taper do not £46 L3 SLOT. HORN AND COMPLEMENTARY ANTENNAS <> llllllmiMim i, 1 \ I Horn mouth iEM \ , 1 H'S| Mil Hi.il Rectangular waveguide Plane of horn mouth Figure 15-21 frd Pyramidal horn antenna, (fc) Cross section with dimensions used in analysis. The diagram can be for either E-plane or JF-plane cross sections. For the E plane the flare angle is 0£ and aperture u t ; . For the H plane the flare angle is 0H and the aperture a fl ,See Fig. 13-22.
From Fig.
13-21, 9 L cos - = 2 E b .
0 a Sm 2 _ 2(L + b) 0 a tan - = — 2 2L where 9 — flare angle (tfL for £ plane, 0H for H plane) a = aperture (u E for £ plane, aH for H plane) L = horn length Frotn the geometry we have also that L = ~ (ML) 8 = 2 tan 1 — = 2 cos 1 2L L + <5 (5) 13-S HORN ANTENNAS 647 In the E plane of the horn, b is usually held to 0.251 or less However, in the H plane, S can be larger, or about G,4/, since E goes to zero at the horn edges (boundary condition, E t — 0 satisfied).
To obtain as uniform an aperture distribution as possible, a very long horn with .a small flare angle is required. However, from the standpoint of practical convenience the horn -should be as short as possible. An optimum horn is between these extremes and has the minimum beam width without excessive sidelobe level (or most gain) for a given length.
If i5 is a sufficiently small fraction of a wavelength, the field has nearly uniform phase over the entire aperture. For a constant length £, the directivity of the horn increases (beam width decreases) as the aperture-a and flare angle 9 are increased. However, if the aperture and flare angle become so large that <5 is equivalent to 180 electrical degrees, the field at the edge of the aperture is in phase opposition to the field on the axis For all but very large flare angles the ratio L/(L + <5) is so nearly unity that the effect of the additional path length b on the distribution of the field magnitude can be neglected. However, when <5 — 180°, the phase reversal at the edges of the aperture reduces the directivity (increases stdelobes) It follows that the maximum directivity occurs at the largest flare angle for which b does not exceed a certain value (<5 0 ). Thus, from (1) the optimum horn dimensions can be related by cos (Oil) ^ (6) L b 0 cos (0/2 ) (7) or " 1 - cos (0/2) It turns out that the value of b$ must usually be in the range of 0.1 to 0.4 free-space wavelength.
1 Suppose that for an optimum horn <5 0 = 0.25/ and that the axial length L = 10A. Then from (5), 6 = 25°. This flare angle then results in the maximum directivity for a 10/ horn.
The path length, or £ effect, discussed above is an inherent limitation of all horn antennas of the conventional type.
2 The relations of (1) through (7) can be applied to all the horns of Fig. 13-20 to determine the optimum dimensions.
However, the appropriate value of <5 0 may differ as discussed in the following sections. Another limitation of horn antennas is that for the most uniform aper-ture illumination higher modes of transmission in the horn must be suppressed.
It 1 At a given frequency the wavelength in the horn kk is always equal to or greater than I he Free-space wavelength /. Since depends on the horn dimensions, it is more convenient to express in free-space wavelengths 2, 2 In the lens-compensated type^of horn antenna (see Fig.
14-1 8b and Fig. 13-30) the velocity of the wave is increased near the edge of the horn with respect to the velocity at the axis in order to equalize the phase over the aperture.
£48 |J SLOT. HORN AND COMPLEMENTARY ANTENNAS follows that the width of the waveguide at the throat of the horn must be between A/2 and I A, of if the excitation system is symmetrica], so that even modes are not energized, the width must be between A; 2 and 3A/_.
13-9 THE RECTANGULAR HORN ANTENNA.
1 Provided that the aperture in both planes of a rectangular horn exceeds I/., the pattern in one plane is substantially independent of the aperture in the other plane. Hence, in general, the H-plane pattern of an H-plane sectoral horn is the same as the H-plane pattern of a pyramidal horn with the same H-plane cross section.
Likewise, the E-plane pattern of an E-plane sectoral horn is the same as the E'-plane pattern of a pyramidal horn with the same E'-plane cross section. Referring to Fig, 13-22, the total flare angle in the £ plane is 0 F and the total flare angle in the H plane is 1 W.
L. Barrow and F. D Lewis, “The Sectoral Electromagnetic Horn” Proc. IRE , 27, 41-50, January 1939.
W. L. Barrow and L.
J Chu, "Theory or the Electromagnetic Horn," Proc. IRE , 27, 51-64, January 1939.
L J Chu and W. L. Barrow/ 1 Electromagnetic Horn Design " Truns. AIEE , 58, 333-337 T July 1939 F, E. Term an, Engineer/ Handbfjok, McGraw-Hill, New York, 1943, pp, 824-837 (this refer-ence includes a summary of design data on horns).
' J, R. Risseu m S, Silver fed ), Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, chap.
10, pp. 349-365.
G.
Slavic and A. Dome, in Very High frequency Techniques, Radio Research Laboratory Stan, McGraw-Hill, New York, 1947 t chap. 6 13-9 THE RECTANGULAR HORN ANTENNA 649 Figure 13-23 Measured E- and //-plane held patterns of rectangular horns as a function of flare angle. and hom length, (After D. R . Rhodes, ‘An Experimental investigation of the Radiation Patterns of Electromagnetic Horn Antennas/ Proc, IRE, 36, 11G1-JIQ5 , September 1948.) 0H . The axial length of the horn from throat to aperture is L and the radial length is /?. Patterns measured by Donald Rhodes 1 are shown in Fig. 13-23, In (a) the patterns in the E plane and H plane are compared as a function of R. Both sets are for a flare angle of 20°, The E-plane patterns have minor lobes whereas the //-plane patterns have practically none. In (h) measured patterns for horns with R = 8/ are compared as a function of flare angle. In the upper row E-plane patterns are given as a function of the £-plane flare angle 0E and in the lower row //-plane patterns are shown as a function of the //-plane flare angle 0H . For a flare angle 6E = 50° the E-plane pattern is split, whereas for 0H = 50° the //-plane pattern is not. This is because a given phase shift at the aperture in the E-plane horn has more effect on the pattern than the same phase shift in the //-plane horn. In the //-plane horn the held goes to zero at the edges of the aperture, so L D, R Rhodes, “An Experimental Investigation of the Radiaiiori Patterns of Electromagnetic Horn Antennas/ Proc. IRE, 1101-1 105, September 1948.
650 U SUIT. HORN AMI COMPLEMENTARY ANTENNAS fi„ = 0.4X-Horn length. L y Fieure 13-24 Experimentally determined optimum dimensions for rectangular horn antennas Solid curves give relation of flare angle fl t in E plane and flare angle 0H in H plane to horn length (see Fig.
13-22). The corresponding half-power beam widths and apertures in wavelengths are indicated along the curves. Dashed curves show calculated dimensions for \ -= 0,25 and 0.4a.
the phase near the edge is relatively less important.
Accordingly we should expect the value of S 0 for the H plane to be larger than for the £ plane. This is illustrated in Fig. 13-24 and discussed in the next paragraph.
From Rhodes’s experimental patterns, optimum dimensions were selected for both E- and H-plane flare as a function of flare angle and horn '™gt!i £ These optimum dimensions are indicated by the solid lines in Fig.
13-24. The corresponding half-power beam widths and apertures in wavelengths are also indicated. The dashed curves show the calculated dimensions for a path length ,j 0 = 0.25/ and = 0.4/. The value of 0.25/ gives a curve close to the experimen-tal curve for £-plane flare, while the value of 0.4/ gives a curve close to the experimental one for H -plane flare over a considerable range of horn length.
Thus, the tolerance in path length is greater for H-plane flare than for £-p!ane flare, as indicated above.
Suppose we wish to construct an optimum horn with length L = 10/, rrom Fig.
13-24 we note that for this length the HPBW (E plane) = ll a and the FIPBW (H plane) = 13°, the £-plane aperture o E = 4.5/ and the //-plane aperture a H = 5,8/, Thus, although the E-plane aperture is not so large as the H-plane aperture, the beam width is less (but minor lobes larger) because the E-plane aperture distribution is more uniform. For horn operation over a frequency ban it is desirable to determine the optimum dimensions for the highest frequency to be used, since S as measured in wavelengths is largest at this highest frequency.
13 9 THE RECTANGULAR HORN ANTENNA 651 The directivity (or gain, assuming no loss) of a horn antenna can be expressed in terms of its effective aperture. Thus, 471/4, A „ where A m = effective aperture, m 2 A p = physical aperture, m 2 Eap — aperture efficiency = AJAP X — wavelength, m For a rectangular horn A p = aE aH and for a conical horn A p = nr 2 , where r = aperture radius.
It is assumed that aEt aH or r are all at least \X. Taking eap 0.6, (1) becomes or D 10 log (dBi) (3) For a pyramidal (rectangular) horn (3) can also be expressed as 10log(T5u£,uH ,) (4) where aEk = £-plane aperture in X aH ± = //-plane aperture in X Example, (a) Determine the length L, //-plane aperture and flare angles and 0M (in the E and H planes respectively) of a pyramidal horn as in Fig. 13-204 for which the E-plane aperture % = 10T The horn is fed by a rectangular waveguide with TE l£> mode. Let <5 = 0,2-A in the E plane and 0.375/ in the M plane, (fr) What are the beam widths? (c) What is the directivity?
Solution, Talcing 6 = Xj$ in the E plane, we have from (13-8-4) that the required horn length 151 85 8/5 and from (13-8-5) that the flare angle in the £ plane flE = 2tan _1 ^-= 2 tan -1 -^ = 9.1° (6) Taking S - 3a/8 in the H plane we have from (13-8-5) that the flare angle in the H plane Wp jo Wp ' 652 13 SLOT HORN AND COMPLEMENTARY ANTENNAS Figure 13-25 Dimensions of rectangular (pyramidal) horns (in wavelengths) versus directivity (or gain, if no loss). Thus, noting the dashed lines, a gain of 19 dBi requires a horn length LA = 4.25, an ff-plane aperture aul — 3.7 and an E-plane aperture a£i = 2.9. These are inside dimensions.
It is assumed that 6 (E plane) = 0.25A and £ (H plane) ~ 0.42, making the dimensions dose to optimum.
It is also assumed that s4p = 0.6.
and from (13-8-5) that the H -plane aperture aM = 2 L tan — = 2 x 62.5/ tan 6.26 s = 13.7/ (8) H 2 From Table 13-1 56 G 56 & HPBW (E plane) =— = 5.6° (9a) “n.
10 67 fl 67° HPBW (H plane) = — =— « 4.9" (96) aHk 13.7 From (3), = 10 log (7.5 x 10 x 13.7) = 30.1 dBi (10) The <5 values used in this example are conservative. For an optimum horn, the 5 values are larger, resulting in a considerably shorter horn but at the expense of slightly less gain (because fields are less uniform across the aperture of an optimum horn).
Figure 13-25a shows the optimum dimensions for pyramidal (rectangular) horns versus gain (or directivity, if no loss).
1 For a given desired gain, the graph gives the dimensions for the length £-plane aperture aEJL and H-plane aper-ture all in wavelengths. For a given length, the graph also gives the appro-priate apertures and the gain. The dimensions are close to optimum.
Such dimensions are only of importance on large horns (many wavelengths long) 1 A similar but somewhat different version has been given by H. Schrank (“Optfmum Horns, 1 IEEE Ant . Prop Soc. Newsletter, 27, 13-14, February 1985), who credits T. A. Milligan for sending him the data.
13-tl CONICAL HORN ANTENNAS 653 Table 13-1 f Type of apertret Beam width, 4tg Between Between fot mdb haif-power points Uniformly illuminated rectangular aperture or linear array Uniformly illuminated circular aperture Optimum £-plane rectangular horn Optimum tf-plane rectangular horn 115 140 fiT 115 aEl 172 aRl La 58 >a 56 aEl 67 am t L, - length of rectangular aperture or linear array in free-spare wavelengths Dk - diameter of circular aperture in free-spare wavelengths - aperture in E plane in free-spare wavelengths .
“ aperture in H plane hj free-spare wavelengths where it is desired that the length be a minimum. For small (short) horns, opti-nuzation is usually unwarranted, Y 13-10 BEAM-WIDTH COMPARISON.
It is interesting to compare the beam width between first nulls and between half-power points for uniformly illu-mutated rectangular and circular apertures obtained in previous chapters with those for optimum rectangular horn antennas (sectoral or pyramidal). This is done in Table 13-1. In general, the relations apply to apertures that are at least several wavelengths long. The beam widths between nulls for the horns are calcu-lated and the half-power beam widths are empirical.
1 I«.
CONICAL HORN ANTENNAS. The conical horn (Fig. 13-20/) can directly excited from a circular waveguide. Dimensions can be determined from (13-8-5), (13-8-6) and (13-8-7) by taking S0 = 032L and A in Very High Frequency Techniques, Radio Research Laboratory Staff McGraw-Hill, New York, 1947, chap. 6.
_ P SiSouthwonh and A. P. King, “ Metal Homs as Directive Receivers of Uitrashort Waves.” Ptac 27, 95-102, February 1939.
A. P. King, The Radiation Characteristics of Conical Horn Antennas,” Proc. lREt 3%, 249-251 Mrr 95°- F™0plimUm conlcal ho™Ki" 8ivc half-power beam widths of fO/aEI in the £ plane and 70/awl in the H plane. These are about 6 per cent more than the values for a rectaneular horn a gjven in Table 13-1.
6 654 L} SLOT. HORN AND COMpLeM HNTARY ANTENNAS Figure l3-25b Dimensions of conical horn (in wavelengths) versus directivity (or gain, if no loss).
Thus, noting the dashed lines, a gain of 20 dBi requires a horn length L A = 6.0 and a diameter 0. - 4.3.
These {inside} dimensions are dose to optimum.
The biconical horns 1 (Fig. 13-20# and ft) have patterns that are nondirec-tional in the horizontal plane (axis of horns vertical). These horns may be regard-ed as modified pyramidal horns with a 360° flare angle in the horizontal plane.
The optimum vertical-plane flare angle is about the same as for a sectoral horn of the same cross section excited in the same mode.
Figure 13-25b shows optimum dimensions for conical horns versus direc-tivity (or gain if no loss) as adapted from King.
2 For a given desired gain, the graph gives the length Lx and diameter , or for a given length, the graph gives the appropriate aperture and gain.
13-12 RIDGE HORNS.' A central ridge loads a waveguide and increases its useful bandwidth by lowering the cutoff frequency of the dominant mode.
3 A DQ Double ridge (b) Fin t Diode Fin line with diode (c) Figure 13-26 Single- and double-ridge rectangular waveguide and fin- line with diode.
E W. L. Barrow, L. J. Chu and J.
J. Jansen. “ Biconical Electromagnetic Horns Pnx. IRE , 27, 769-779, December 1939.
2 A.
P. King.
11 The Radiation Characteristics of Conical Horn Antennas," Proc. IRE , 38, 249 25 U 1950.
J S B Cohn, “ Properties of the Ridge Wave Guide " Proc. IRE, 35, 783-789, August 1947.
T-S Chen, “Calculation of the Parameters of Ridge Waveguides, IRE Trans. Microwave Theory Tech ., MTT-26, 726 732, October 1978.
ITI3 SEPTUM HORNS 655 Figure LV27 Double-ridge horn with coaxial feed. The view at \a) is a cross section at the feed point.
rectangular guide with single ridge is shown in Fig, 13-26a and with a double ridge in Fig.
1 3-266. A very thin ridge or fin is also effective in producing the loading of a central ridge.
It may consist of a metal-clad ceramic sheet which facilitates the installation of shunt circuit dements as suggested in Fig.
13- 26c, 1 Of course, the cutoff frequency can be lowered by placing dielectric material in the waveguide, but this does not increase the bandwidth and it may increase losses.
By continuing a double-ridge structure from a waveguide into a pyramidal horn as suggested in Fig.
13-27, the useful bandwidth of the horn can be increased manyfold 2 (see also Fig.
15-1). A quadruple-ridge horn connected to a dual-fed quadruple-ridge square waveguide can provide dual orthogonal linear polarizations over bandwidths of more than 6 to l.
13-13 SEPTUM HORNS, Although the electric, field in the H plane of a pyramidal horn tends to zero at fhe edges, resulting in a tapered distribution and reduced sidelobcs, the electric field in the £ plane may be close to uniform in amplitude to the edges, resulting in significant sidelobcs. By introducing septum plates bonded to the horn walls, a stepped-amplitude distribution can be achieved in the £ plane with a reduction in E-plane sidelobes. Typically, the first sidelobes of a uniform amplitude distribution are down about 13 dB. With a 2-septum horn Peace and Swartz 3 were able to achieve a sidelobe level more than 30 dB down, which is lower than the sidelobe level in the H plane.
A cosine field distribution is approximated with a 1:2:1 stepped ampli-tude distribution with apertures also in the ratio 1:2:1 as suggested in Fig.
13-28, To achieve this distribution the septums must be appropriately spaced at the throat of the horn.
Figure 13-29 shows a stepped-amplitude septum horn with metal-plate lens for feeding “ Big Ear/ 1 the Ohio State University 1 10-m radio telescope. The dual-1 P.
J. Meier, “Integrated Fin-Line Millimeter Components, ' IEEE Trans. Microwave Theory Tech., MTT-22, 1209 1 2 1 6, December 1974 1 K.
L. Walton and V. C. Sundberg. “Broadband Ridged Horn Design," Mimm’drc J., 7. 96-101, March 1964.
1 G. M. Peace and L.
E. Swart/, ‘'Amplitude Compensated Horn Antenna." AfrcrtJh-m.-p J..
7. 66-68.
February 1964.
656 13 SLOT, HORN AND COMPLEMENTARY ANTENNAS Top aperture Central aperture Bottom aperture Septum plates Cosine distribution B at horn mouth Figure 13-28 Two-septum horn with 1:2:1 stepped amplitude distribution in field intensity at mouth of horn | approximating a cosine distribution)-feed stacked twin horn with metal-piate £-plane lens, as in Fig. 13-28, provides a compact arrangement \ the length of a single horn. This twin horn is one of a pair used for radio astronomy observations at 1 to 2 GHz.
1 Both the aperture and field distributions have the ratios 1:2:1 (binomial series ratio).
IM4 CORRUGATED HORNS 657 13-14 CORRUGATED HORNS- Corrugated horns can provide reduced edge diffraction, improved pattern symmetry and reduced cross- polarization (less E field in the H plane).
Corrugations on the horn walls acting as A/4 chokes are used to reduce E to very low values at all horn edges for all polarizations. These prevent waves from diffracting around the edges of the horn (or surface currents flowing around the edge and over the outside).
1 Consider the corrugations of width w and depth d shown in Fig.
13-30. A square cross section (w by w), as indicated in the figure, constitutes the open end of a short-circuited field-cell transmission line of Length d with a characteristic impedance Z = 377 ft. The reactance at the open end is given approximately by A: = 377 tan fey j (ft) where d = depth, A The reactance for any square area of the corrugations (such as 3w x 3w) is also as given by (1). Thus, this is the surface reactance in ohms per square.
It is assumed that the corrugations are air-filled and that the wall thickness is small enough to be neglected.
When d = A/4, X becomes infinite, while when d = A/2, X — 0 and, assuming no radiation or loss, /? = 0 and, hence, Z = 0.
As an example, a circular waveguide-fed corrugated horn with a corrugated transition is shown in Fig. 13-3T In the transition section the corrugation depth changes from A/2, where the corrugations act like a conducting surface, to A/4, where the corrugations present a high impedance. The corrugation spacing or 1 A. F. Kay, “The Scalar Feed,” AFCRL Rept. 64-347, March 1964, 1 J. D, Kraus, Electromagnetics, 3rd ed., McGraw-Hill, 1984, see, 10-5, 658 1.1 SLOT H Oft IN AND CO MPt.KMHNTARV ANTENNAS ADDITIONAL REFERENCES 659 Figure 13^31 Cross section of circular waveguide-fed corrugated horn with corrugated transition.
Corrugations with depth of /„ 2 at waveguide act like a conducting surface while corrugations with 4 depth in horn present a high impedance. {After T. S. Chu el aL "Crawford Hill 7-m Millimeter Ware Antenna:' Bell Sys, Tech. J, 57, 1257-1288, May- June 1978.) Figure 13-32 Cross section of circular waveguide with flange and 4 chokes for wide-beam-width high -efficiency feed of low F/D parabolic reflectors. (After R. Wohlleben , //, Mattes and O. Lochner, Simple. Small, Primary Feed for Large Opening Angles and High Aperture Efficiency:' Electronics Letters, 8, 19, September 1972,) width w = XI 10. The corrugations are air-filled. This corrugated nom was devel-oped by Chu et ai 1 for feeding a 7-m mil lime ter-wave reflector antenna of the Bell Telephone Laboratories.
A simpler feed with choke corrugations was developed for deep dishes with F/D ratios of less than 0.35 by Wohlleben, Mattes and Lochner2 for use on the Bonn 100-m radiotelescope.
It consists of a circular waveguide equipped with a disc (or flange) projecting U beyond the guide and 4 chokes X/5 deep as shown in Fig.
13-32. The location of the disc with chokes 3A/8 behind the waveguide opening gives a broad 130° 10-dB beam width with a steep edge taper. This results in high aperture efficiency.
13-15 APERTURE-MATCHED HORN.
By attaching a smooth curved (or rolled) surface section to the outside of the aperture edge of a horn, Burnside and Chuang 3 have achieved a significant improvement in the pattern, impedance and bandwidth characteristics.
This arrangement, shown in Fig. 13-33 is an attractive alternative to a corrugated horn. The shape of the rolled edge is not critical but its radius of curvature should be at least Jt/4 Curved edge section Figure 13-33 Cross section of Burnside and Chuang's aperture-matched horn. The radius of curvature r of the rolled edge should be ai least yl/4.
ADDITIONAL REFERENCES Ando, M., K.. Sakurai and N. Goto: “ Characteristics of a RadiaJ Line Slot Antenna for 12 GHz Band Satellite TV Reception," IEEE Trans. Ants. Prop^ AP-34, 1269-1272, October 1986.
Oamcoats, P.
J.
B., and A, D. Qlver: Corrugated Horns for Microwave Antennas, IEE Electromag-netic Wave Series, 1984.
Dragone, C: “A Rectangular Horn of Four Corrugated Plates," IEEE Trans.
Ants.
Prop., AP-33, 160-164, February 1985.
1 T.
S. Chu, R. W, Wilson, R. W. England, D.
A. Gray and W, E. Legg, “Crawford Hill 7-m Milli-meter Wave Antenna," Bell Sys. Tech . J ., 57, 1257-1288, May-June 1978.
1 R, Wohlleben, H, Mattes and O, Lochner, “Simple, Small, Primary Feed for Large Opening Angles and High Aperture Efficiency," Electronics Letters, 8y 19, September 1972.
1 W. D, Burnside and C. W. Chuang, “An Aperture-Matched Horn Design,” IEEE Trans.
Ants.
Prop., A P-30, 790-796, July 1982.
.
660 13 SLOT, HORN AND COMPLEMENTARY ANTENNAS Geyer, H.: " R under Hornstrahler mil Ringformigen Sperrtopfen zur GLeichzritigen Uberiragen zweier Polarization sen tkoppelter Wellen" Frequenz, 20 22-28 1966, Kay, A. F.:"The Scalar Feed" AFCRL Rept. 64-347, 1964.
Kisliuk, M-, and A. Axelrod: “"A Broadband Double-Slot Waveguide Antenna," Microwave J, 30 223-226, September 1987.
Lagrone, A. H„ and G.
F, Roberis: "Minor Lobe Suppression in a Rectangular Horn Antenna through Ihe Utilization of a High Impedance Choke Flange," IEEE Trans, ,4ms. Prop., AP-14 102-104. 1966 Love, A. W.: Electromagnetic Horn Antennas IEEE Press New York, 1976, Menlzer C.
A., and L, Peters Jr.: "Pattern Analysis of Corrugated Horn Antennas," IEEE Trans .
,4 nrs. Prop . AP-24, 304^309, May 1976.
Rumsey, V. H.: "Horn Antennas with Uniform Power Patterns around Their Axes" IEEE Trans, A ms. Prop ,. AP-14, 656-658, September 1966.
Yngvesson, K.
S. J.
J. Johansson and E. K. Kollberg: "A New Integrated Slot Element Feed Array for Multibeam Systems," IEEE Trans. Ants. Prop., AP-34, 1372-1376 November 1986.
PROBLEMS 1 13-1 Boxed-slot impedance. What is the terminal impedance of a slot antenna boxed to radiate only in one half-space whose complementary dipole antenna has a driving-point impedance of Z — 100 + jO O? The box adds no shunt susceptance across the terminals.
13-2 Open-slot impedance What dimensions are required of a slot antenna in order that its terminal impedance be 75 4- j'O fi? The slot is open on both sides.
13-3 Optimum horn gain. What is the approximate maximum power gain of an optimum horn antenna with a square aperture 102 on a side?
13-4 Horn pattern la) Calculate and plot the E-plane pattern of the horn of Prob.
13-3 assuming uniform illumination over the aperture.
(6) What is the half-power beamwidth and the angle between first nulls?
13-5 Two Xy'2 slots. Two /,/2-slot antennas are arranged end-to-end in a large conduct-ing sheet with a spacing of l/. between centers.
If the slots are Ted with equal in-phase voltages, calculate and plot the far-held pattern in the 2 principal planes.
Note that the H plane coincides with the line of the slots.
13-6 Boxed slot The complementary dipole of a slot antenna has a terminal impedance Z - 90 + yiO Q, If the slot antenna is boxed so that it radiates only in one half-space. what is the terminal impedance of the slot antenna? The box adds no shunt susceptance at the terminals.
13-7 Pyramidal horn.
(a) Determine the length L, aperture ai{ and half-angles in E and H planes for a pyramidal electromagnetic horn for which the aperiure a F = 8/.. The horn is fed with a rectangular waveguide wilh TE 1(i mode. Take S = /.. to in the E plane and <5 = >.. 4 in the H plane, tb) What are the HPBWs in both E and H planes?
\c) What is the directivity?
id} What is the aperture efficiency?
1 Answers to starred () problems are given in App. D.
CHAPTER 14 I FISTS ANTENNAS 14-1 INTRODUCTION. Lens antennas may be divided into two distinct types: (1) delay lenses in which the electrical path length is increased by the lens medium and (2) fast lenses in which the electrical path length is decreased by the lens medium. In delay lenses the wave is retarded by the lens medium. Dielectric lenses and H-plane metal-plate lenses are of the delay type £-plane metal-plate lenses are of the fast type. The actions of a dielectric tens and an E-plane metal-plate lens are compared in Fig.
14-1.
The dielectric lenses may be divided into two groups: I. Lenses constructed of nonmetallic dielectrics, such as lucite or polystyrene X Lenses constructed of metallic or artificial dielectrics These types are considered in the next two sections (14-2 and 14-3), E-plane metal-plate lenses are discussed in Sec, 14-4 tolerances in Sec. 14-5 and the H-plane metal-plate lens in Sec.
14-6, A reflector lens is presented in Sec.
14-7.
All lens antennas of the delay type may be regarded basically as end-fire antennas with the polyrod and monofilar axial-mode helical antennas as rudi-mentary forms as suggested in Fig. 14-2.
Likewise, the director structure of a many-element Yagi-Uda antenna is a rudimentary lens. Polyrods are covered in Sec.
14-8, monofilar axial-mode helical antennas having already been discussed in Chap. 7 r Yagi-Uda antennas are considered in Chap 11.
Lenses of multiple helices are also described in Sec, 14-9. The last section of this chapter (14-10) discusses two spherically symmetric lenses of special type, the Luneburg and Ein-stein lenses the latter utilizing a large mass as the focusing device.
661 662 14 LENS ANTENNAS primary antenna ))) Dielectric lens Wave fronts ''Wave retarded £-plane metal plate lens Source or primary antenna Wave fronts vWave accelerated Figure 14-1 Comparison of dielectric (delay) lens and E- plane metahplale (fast) lens actions.
Wave fronts -\ \ \ XT i i i Wave direction | Monofilar axial'mode helix Wave fronts Wave direction Yagi Uda 1 :i : V; ; Wave direction Figyre 14-2 Three forms of rudimentary lens antennas.
14-2 NONMETALLlC DIELECTRIC LENS ANTENNAS, FERMAT'S PRINCIPLE 663 At millimeter wavelengths low-loss dielectric lens antennas are competitive in weight and performance with reflector antennas.
1 14-2 NONMETALLlC DIELECTRIC LENS ANTENNAS. 2 FERMAT’S PRINCIPLE. This type is similar to the optical lens-It may be designed by the ray analysis methods of geometrical optics As an example, let us determine the shape of the plano-convex lens of Fig. 14- la for transforming the spherical wave front from an isotropic point source or primary antenna into a plane wave front.
3 The field over the plane surface can be made everywhere in phase by shaping the lens so that all paths from the source to the plane are of equal electrical length. This is the principle of equality of electrical (or optical) path length (Fermat’s principle). Thus, in Fig.
143, the electrical length of the path OPP' must equal the electrical length of the path OQQ Q f \ or more simply OP must equal OQ. Let OQ — L and OP = R t and let the medium surrounding the lens be air or vacuum. Then R L R cos 0 — L Figure 14-3 Path lengths in dielectric ^ lens.
1 P. F. Goldsmith and E. L. Moore, ‘"Gaussian Optics Lens Antennas (GOLAs)," Microwave J. 26, 153-156, July 1984.
P.
F, Goldsmith, w Quasioptical Techniques at Millimeter and Suhmillimeter Wavelengths,” in It.
Button fed.), Infrared and Millimeter Waves voL 6, Academic Press, 1982, pp. 277-343.
P. F. Goldsmith and G. J. Gill, "Dielectric Wedge Conical Scanned Gaussian Optics Lens Antenna,” Microwave 29, 207-212, September 1986.
1 A detailed discussion is given by J.
R. Kisser, in S. Silver (ed.), Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, chap.
11.
i A wave front is defined as a surface on all points of which the field is in the same phase.
664 14 LENS ANTENNAS where A0 = wavelength in free space (air or vacuum) = wavelength in the lens Multiplying(l) by /0> R = L + n(R cos 0 — L) where n = JL 0/Ad = index of refraction In general, n _ >// Vi vW% where /= frequency, Hz Vq — velocity in free space, m s _1 vd = velocity in dielectric, ms -1 /i = permeability of the dielectric medium, H m -1 e = permittivity of the dielectric medium, F m“ 1 p0 = permeability of free space — 4^ x 10“ 7 , H m _1 e0 = permittivity of free space = 8.85 x 10“ 12 , F m _1 However, / = Vo & and e = Eq Er where ^ ™= relative permeability of dielectric medium (2) ( 3) (4) ( 5 ) £r = — = relative permittivity of dielectric medium eo Thus, from (3), n -(6) For nonmagnetic materials nr is very nearly unity so that n = y/% The index of refraction of dielectric substances is always greater than unity. For vacuum, er = 1 by definition. For air at atmospheric pressure, Er = 1,0006, but in most applications it is sufficiently accurate to take £,, = 1 for air. The relative permittivity (or dielectric constant), index of refraction and power factor for a number of lens materials are listed in Table 14-1 in order of increasing e,, Although the permittivity of materials may vary with frequency (e, for water is 81 at radio frequencies and about 1.8 at optical frequencies), the table values are 4.>.
£ f i!
1+-2 NON METALLIC DIELECTRIC LENS ANTENNAS. FERMATS PRINCIPLE 665 Table 14-1 Relative Index of permittivity refraction Power Material factor^ Paraffin 2.1 1.4 0.0003 Polyethylene 2.2 1.5 0.00Q3 Lucile or Plexiglas (methacrylic resin) 2.6 1.6 0,01 Polystyrene 2.5 1,6 0,0004 Flint glass 7 2.5 0.004 Polyglas (Ti02 or Ulanate fillers) 4-16t 2-4 0.003 Rutile (TiO = ) 85-170 9-13 0.0006 t Depends on composition.
t Depends on orientation of crystal with respect to field.
§ Equals cosine of angle between conduction and total currents.
appropriate at radio wavelengths down to the order of 1 cm. The power factor is also a function of frequency. The values listed merely indicate the order of magni-tude at radio frequencies.
Returning now to Eq. (2) and solving for R, we have (» ~ 1)1 n cos 0 — 1 (7) This equation gives the required shape of the lens. It is the equation of a hyper-bola. Referring to Fig.
14-3, the distance L is the focal length of the lens.
1 The asymptotes of the hyperbola are at an angle fl0 with respect to the axis. The angle 0„ may be determined from (7) by letting R = no. Thus, 0o — arccos -n (8) The point 0 is at one focus of the hyperbola. The other focus is at O'. For a point source at the focus, the 3-dimensional lens surface is a spherical hyperbola obtained by rotating the hyperbola on its axis. For an in-phase line source normal to the page (Fig 14-3) as the primary antenna, the lens surface is a cylin-drical hyperbola obtained by translating the hyperbola parallel to the line source.
1 The F number of a lens is (he ratio of the focal distance to the diameter A of (he lens aperture.
Thus, F = LJA.
666 1 LEMS ANTENNAS Although Eq. (7) for the lens surface was derived without using Snell’s laws of refraction , 1 these laws are satisfied by the lens boundary as given by (7), The plane wave emerging from the right side of the lens produces a second-ary pattern with maximum radiation in the direction of the axis. The shape of the secondary pattern is a function of both the aperture A and the type of illumi-nation. This aperture-pattern relation has been discussed in previous chapters.
For an isotropic point-source primary antenna and a given focal distance L , the field at the edge of the lens (8 = 8 is less than at the center {0 = 0), the effects of reflections at the lens surfaces and losses in the lens material being neglected. The variation of field intensity in the aperture plane of the spherical lens can be determined by calculating the power per unit area passing through an annular section of the aperture as a function of t^e radius p} Referring to Fig.
14-4, the total power P through the annular section of radius p and width dp is given by P = 2np dp Sp (9) where Sp = power density or Poynting vector (power per unit area) at radius p This power must be equal to that radiated by the isotropic source over the solid angle In sin 0 d8. Thus, P = 2k sin SdSV 00) where U = radiation intensity of the isotropic source (power per unit solid angle) Equating (9) and (10), p dp S fi = sin 8 d8 U Sg __ sin 0 V ~ p(dpfd8) (ID (12) 1 Snell's laws of refraction are (1) that the incident ray, the refracted ray and the normal to the surface lie in a plane and (2) that the ratio of the sine of the angle of incidence to the sine of the angle of refraction equals a constant for any two media. If the medium of the incident wave is air, the constant is the index of refraction n of the medium with the refracted ray. Thus, sin a/sin ft - n, where a is the angle between the incident ray in air and the normal to the surface and ft is the angle between the refracted ray in the dielectric medium and the normal to the surface.
1 J. R.
Risser, in S.
Silver (ed.fc Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, chap.
11.
14-2 NON METALLIC DIELECTRIC LENS ANTENNAS FERMAT’S PRINCIPLE 667 However, p = R sin 9 f and introducing the value of R from (7), -( n cos 0 - If p («- D^n-cosUU, 3 (13) The ratio of the power density S6 at the angle 8 to the power density S0 at the axis (8 = 0) is given by the ratio of (13) when 8 -0, to (13) when 0 = 0. Thus, Sg _ (n cos 8 — l) 3 S 0 (n -l) 2 (tt - cos 8) In the aperture plane the field-intensity ratio is equal to the square root of (14), or E± = [S# _ 1 /(a cos 8 - fp Eg \ S0 n — 1 y n — cos 8 The ratio EJEG is the relative field intensity at a radius p given by p = R sin 8.
For n — 1.5, -r = 0.7 at 8 = 20° £ and —- = 0.14 at 8 - 40° Eo Hence, for nearly uniform aperture illumination an angle 8 1 to the edge of the lens even less than 20 is essential unless the pattern of the primary antenna is an inverted type, i.e., one with less intensity in the axial direction (0 = 0) than in directions off the axis.
Instead of uniform aperture illumination, a tapered illumination may be desired in order to suppress minor lobes. Thus, in the above example with 8 t = 40°, the field at the edge of the lens is 0,14 its value at the center. The disadvan-tage of this method of producing a taper is (hat the lens is bulky (Fig. 14-5a). An alternative arrangement, shown in Fig, 14-5h, has a lens of smaller 8 t value with the desired taper obtained with a directional primary antenna at a larger focal distance (relative to the aperture). The lens in this case is less bulky, but the focal distance is larger (F number = L/A larger).
For compactness and mechanical lightness it would be desirable to combine the short focal distance of the lens at (a) with the light weight of the lens at (6), This combination may be largely achieved with the short focal distance zoned 1 Equation (15a) is for a spherical lens, Allenualkm in the lens is neglected. For a cylindrical lens the field-intensity ratio is E± n cos 8 — 1 =— „ (15b) ^/[n — !Xn “ 0) where EJE0 is the relative field intensity at a distance y from the axis given by y = R sin 8.
668 L4 LENS ANTENNAS Primary antenna <r) Figure 14-5 Short-focus lens (a), long-focus Jens (b) and zoned lens (4 lens of Fig.
14- 5c, The weight of this lens is reduced by the removal of sections or zones, the geometry of the zones being such that the lens performance is substan-tially unaffected at the design frequency. Whereas the unzoned lens is not fre-quency sensitive, the zoned lens is, and this may be a disadvantage. The thickness z of a zone step is such that the electrical length of z in the dielectric is an integral number of wavelengths longer (usually unity) than the electrical length of z in air.
Thus, for a 1 X difference.
(16) / 0 or Z= J-o n — 1 (17) For a dielectric with index of refraction n = t.5, z - 2/0 that is, each zone step is twice the free-space wavelength. Since n = X0/Xdt 2 = 3A„ L4-2 NONMETALLIC DIELECTRIC LENS ANTENNAS. FERMAT'S PRINCIPLE 669 Figure 14-6 Reflected waves entering primary antenna (a) and refocused to one side of primary antenna (b).
Thus, in this case, the electrical length of z in the dielectric is 3 A, while the electrical length of z in air is 2 X (see Fig. 14-5c).
In lens antennas the primary antenna does not interfere with the plane wave leaving the aperture as it does in a symmetrical parabolic reflector antenna (with prime focus feed) (see Chap.
12). However, the energy reflected from the tens surfaces may be sufficient to cause a mismatch of the primary antenna to its feed line or guide. In the lens of Fig. 14-6a reflections from the convex surface of the lens do not return to the source except from points at or near the axis. This is not serious, but the wave reflected internally from the plane lens surface is refocused at the primary antenna and may be disturbing. In this case, the wave is reflected at normal incidence, and the reflection coefficient is Z0 -Z Z0 + Z (18) where Z0 -intrinsic impedance of free space = Z = intrinsic impedance of dielectric lens material = v//ji/£ Thus, (ZJZ) -1 n — 1 P {ZJZ) +1 n + 1 (19) where n = the index of refraction of the dielectric lens material For n = L5, p = 0.2, while for n = 4, p = 0.6. Hence, for a small reflection a low index of refraction is desirable. The reflection can also be minimized by other methods. For example, a A/4 plate can be applied to the plane lens surface with the refractive index of the plate made equal to yfn, where n is the refractive index 670 14 LENS ANTENNAS of the lens proper 1 Another method is to tilt the icns slightly as indicated in Fig. 14-66 so that the reflected wave refocuses to one side of the primary antenna.
Even though the lens is tilted, the antenna beam remains on axis, 14-3 ARTIFICIAL DIELECTRIC LENS ANTENNAS.
Instead of using ordinary, nonmetaflic dielectrics for the lens. Rock 3 has demonstrated that artificial or metallic dielectrics can be substituted, generally with a saving in weight. Whereas the ordinary dielectric consists of molecular particles of micro-scopic size, the artificial dielectric consists of discrete metal particles of macro-scopic size. The size of the metal particles should be small compared to the design wavelength to avoid resonance effects.
It is found that this requirement is satisfied if the maximum particle dimension (parallel to the electric field) is less than //4. A second requirement is that the spacing between the particles be less than X to avoid diffraction effects.
The particles may be metal spheres, discs, strips or rods. For example, a plano-convex lens constructed of metal spheres is illustrated in Fig, 14-7. The spheres are arranged in a 3-dimcnsional array or lattice structure. ^Such an arrangement simulates the crystalline lattice of an ordinary dielectric substance but on a much larger scale. The radio waves from the source or primary antenna cause oscillating currents to flow on the spheres. The spheres are, thus, analogous to the oscillating molecular dipoles of an ordinary dielectric.
An artificial dielectric lens can be designed in the same manner as an ordi-nary dielectric lens (Sec.
14-2). To do this, it is necessary to know the effective index of refraction of ihe artificial dielectric. This can be measured experimentally wjth a stab of the material, or it can be calculated approximately by the following method of analysis. 2 Metal discs or strips are generally preferable to spheres because they are lighter in weight. The strips may be continuous in a direction Figure 14-7 Artificial dielectric lens of metal sphere^ 1 In general the refractive index of a 2/4 matching plate between two media should be equal to the geometric mean of the indices of the two media. This is equivalent to saying that the intrinsic imped-ance Zp of the plate material is made equal to the geometric mean of the intrinsic impedances Z 1 and Zj of the two media. Thus, Z, = s/Z lZ 1 , \ 1 W. E. Koclt, “ Metallic Delay Lens,” Bell System Tech. J 27, 5&-B2, January 1948.
.
ARTIFICIAL DIELECTRIC LENS ANTENNA' 671 perpendicular to the electric field as indicated in Fig.
14-8. Since, however, the sphere is more readily analyzed, the method will be illustrated for the case of the sphere.
Let an uncharged conducting sphere be placed in an electric field E as in Fig.
14-9a, The field induces positive and negative charges as indicated. At a distance the effect of these charges may be represented by point charges + q and — q separated by a distance l as in Fig.
14-96. Such a configuration is an electric dipole of dipole moment ql At a distance r t the potential due to the dipole is given by ql cos 0 4k£0 r2 The polarization P of the artificial dielectric is given by P = Nq\ where N = number of spheres per cubic meter I = vector of length / joining the charges q ( 1 ) (2) tiEure 14-9 Charged sphere and equivalent dipole.
672 14 LENS ANTENNAS The electric displacement D, the electric held intensity fe and the polarization P are related by D = eE = c0 E + P (3) where £ 0 = dielectric constant of free space Thus, the effective dielectric constant e of the artificial dielectric medium is P q\ £ = eo + ^ — Eo + N ^ (4) Hence, if the number of spheres per unit volume and the dipole moment of one sphere per unit applied field are known, the effective dielectric constant can be determined. Let us now determine the dipole moment per unit applied field.
We have E = — VF. Then in a uniform field the potential m r V = — E cos 0 dr — —Er cos 0 (5) Jo where 0 is the angle between the radius vector and the field (see Fig.
14-9fr). The potential F 0 outside the sphere placed in an originally uniform field is the sum of (1) and (5), or V a = —Er cos S + ql cos B At the sphere (radius a) (6) becomes 1 0 = — Ea cos B + ql cos 0 4ti£0 a 2 and solving for ql/E we obtain = 4rc£0 a 3 Introducing this value for the dipole moment per unit applied field in (4) £ = e0 + 4ti£0 Na 2 or £, = 1 + 4nNa (8) where £, = effective relative permittivity of the artificial dielectric If the effective relative permeability of the artificial dielectric is unity, the index of refraction is given by the square root of (8). However, the lines of mag-netic field of a radio wave are deformed around the sphere since high-frequency fields penetrate to only a very small distance in good conductors. The effective The potential of the sphere is zero since there is as much positive as negative charge on its surface.
4 E- PLANE METAL -PLATE LENS ANTENNAS 673 Table 14-2 Artificial dielectric materialst Retime Type of permittivity particle cr Sphere J + 4rcWu 3 Disc ] + 5.33jVa J Strip 1 + 7 8 5JVh-+ :V = number of spheres or discs per cubic meter u = radius of sphere or disc, rti A - number of strips per square meter in lens cross section (see Fig. N’SaJ w = width of strips, m isec Fig. I4-8| relative permeability or an artificial dielectric of conducting spheres is l±r —\— InNa 2 ^ The effective index of refraction of the artificial dielectric of conducting spheres is then given by = \fwr = VO + 47iNfl 3 Xl - 2nNa l ) (10) Equation (10) gives a smaller n than obtained by the square root of (8) alone.
According to (10) the index of refraction of an artificial dielectric of conducting spheres can be calculated if the radius a of the sphere (in meters) and the number A/ of spheres per cubic meter are known. The relative permeability of disc or strip-type artificial dielectrics is more nearly unity so that one can take ^fzr as their index of refraction. Theoretical values of £„ ^ and n for artificial dielectrics made of conducting spheres, discs and strips are listed in Table 14-2.
1 According to Kock the table values are reliable only for e, < 1,5, and only approximate for larger £,. For >1.5, N becomes sufficiently large that the particles interact because of their close spacing. This effect is neglected by the formulas.
14-4 f-PLANE METAL-PLATE LENS ANTENNAS. 2 Whereas the ordinary and artificial dielectric lens depend for their action on a retardation of the wave in the lens, the £-plane metal-plate type of lens depends for its action on an acceleration of the wave by the lens. In this type of lens the metal plates are parallel to the E plane (or plane of the electric field). Referring to Fig.
14-10, the velocity v of propagation of a TE 10 wave (E as indicated) in the x direction 1 Erom W. E, Kook, “Metallic Delay Lens” ile/J System Tech. J > 27, 58—82, January 1948.
2 W, E. Kock, " Metal Lens Antennas " Proc. 1RET 34, 828-836, November 1946, Relative permeability Index of Mr refraction n I - InNa J{\ + 4tiNa^ } - 2nNgr ) -1 Jl + 5.33fra ^ I V/1 + 7.85NV 674 14 LENS ANTENNAS Figure 14-10 Wave between plates in £-p1ane type of metal plate lens.
between two parallel conducting plates of large extent is given by v = Viy U) v'l ~(V2b) 2 where u 0 = velocity in free space A 0 = wavelength in free space b = spacing of plates or sheets The plates act as a guide, transmitting the wave for values of b > /.0/2. The spacing b = A„/2 is the critical spacing since for smaller values of b the guide is op.qu‘ and ,h« wav. is no. transmittal Th. variation of the v.ta.y or a fo.d wavelength as a function of the plate spacing b is illustrated in Fig.
14-11. The velocity of the wave between the plates is always greater than the free-space velocity v0 . It approaches infinity as b approaches 0.5/ o , and it approaches r 0 as b becomes infinite.
0 o?5 TTo 20 b in free-space wavelengths, A 0 Figure 14-11 Velocity u of wave between parallel plates and equivalent index of refraction n as a func-tion of spacing b between plates.
1 L J Chu and W. L Barrow, " Electromagnetic Waves in Hollow Metal Tubes of Rectangular Cross Section; 1 Proc. IRE , 16, 1 520- 1555 , December 1938, However, for typical spacing of b ' 3A/4 at normal incidence the transmission coeirtcient is nearly Unity.
J4-4 E-PLANE METAL-PLATE LENS ANTENNAS 675 The equivalent index of refraction of a medium constructed of many such parallel plates with a spacing b is The index is always less than unity, as shown in Fig. 14-1 L The acceleration of waves between plates has been applied 1 in a metal-plate lens for focusing radio waves. For instance, a metal lens equivalent to the plano-convex dielectric lens of Fig. 14-la or Fig. 14-3 is a plano-concave type as illus-trated in Fig. 14-12. The plates are cut from flat sheets, the thickness t at any point being such as to transform the spherical wave from the source into a plane wave on the plane side of the lens. The electric field is parallel to the plates.
The lens plate on the axis of the lens in Fig. 14-12 is shown in Fig.
14-13.
The shape of the plate can be determined by the principle of equality of electrical path length (Fermat’s principle). Thus, in Fig 14-13 OPF must be equal to OQQ' in electrical length, or L R L — R cos 0 Z = T 0 + T t (3) where A0 = wavelength in free space = wavelength in lens W, E, Kock, w Metal Lens Antennas," Proc. IRE, 34 , S25-836, November 1946, 676 L4 LENS ANTENNAS <) Figiw 14-14 Cross sections of constrained types of £-plane metal-plate lenses.
Then (1 ~ n)L 1 — n cos 8 (4) This relation is identical with (14-2-7). However, to keep both numerator and denominator positive (since n < 1 in the present case), the numerator and denominator of (14-2-7) should be multiplied by minus 1. With n < 1, (4) is the equation of an ellipse. The 3-dimensional concave surface of the lens in Fig. 14-12 would be generated by rotating the contour for the center plate, as given by (4), on the axis.
If the primary antenna were a line source perpendicular to the page in Fig.
.4-13, all the plates would be identical and the lens surface would be in the form of an elliptical cylinder.
Waves entering the lens of Fig, 14-12 at the point P obey Snell's laws of refraction. However, this is not necessarily the case for waves entering at P where the metal plates constrain the wave to travel between them. E-plane metal-plate lenses may be constructed that have only such constrained refraction. Two types are illustrated in Cross section in Fig.
14-14. Both have a line source normal to the page. The electric field E is parallel to the source.
All lens cross sections perpendicular to the line sources are the same as the ones shown in the figure. In the lens at (a) the spacing between plates is uniform, but the width varies from plate to plate. In the lens at (b) all plates have the same width, but the spacing varies.
A disadvantage of the E-plane metal-plate lens as compared to the dielectric type is that it is frequency -sensitive, i.e., the lens has a relatively small bandwidth.
To determine the bandwidth, 1 consider the geometry of Fig.
14-15. At the design frequency /we have from (3) that 1 I.
R.
Risser, in S.
Silver (ed). Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, chap, 1 L 14-4 E-PLANE METAL-PLATE LENS ANTENNAS 677 or L = R + nt (6) where n = index of refraction at the design frequency / At some other frequency f\ E 4- & = R -r ti r t (7) where <5 = the difference in electrical path length of OQ and OPF n f = index of refraction at the frequency /' Subtracting (6) from (7), $ = An t (8) where An — n — n 678 i LESS ANTKNSAS If we arbitrarily take <5 = 0.25a, (%l (,3) _ ^ 05 and t = 6/, the bandwidth ror rt B = 5.5% Tu the usable frequency band for this antenna is 5.5 percent of the design U ^enc y } Although zoning a dielectric lens introduces frequency sensitivity, the A 1 of zoning an E-plane metal-plate lens is to decrease the frequency sensi-C Heoce, zoning is desirable with E-plane metal-plate lens, both to save tW1 and to increase the bandwidth. An E-plane metal-plate lens 40/ square with zones dlustrated in Fig.
14-16. The patterns of this lens, fed with a hort primary horn antenna, are shown in rig. 14-17.
The bandwidth of a zoned E-planc metal-plate lens is given approximately by B = -^~ (%) (14) 1 + Kn where = index of refraction at the design frequency = number of zones, the zone on the axis of the lens being counted as the first zone \ ^iied ^ns comparable to the unzoned lens of n = 0.5, t = 6/ 0 and & = 5.5 has rt = 0.5 and K = 3 since with n = 0.5 K - tJL The bandwidth B of this z°ne^ lens is 10 percent, or nearly double the bandwidth of the unzoned lens.
The aperture efficiency to be expected of large lens antennas is about 0.6 so that directivity is about the same as for optimum horns of the same size fUlre (see Chap, 12).
^ Referring to Fig. 14-1 8«, the thickness 2 of a zone step is given by — - — = 1 A0 kq 26)-A i _ 0//a) ~0^i) /(/1 ~/a) ^ fi h. = —L (foT A/ 1 J n < I-be tilted a considerable angle i with respect to the axis through the pr.mary antenna and center of the lens (see Fig. 14-0.) without serious effects.
The above-mentioned tolerances are summarized m Table 14-3.
[olerances for zoned lenses are also listed. These are derived from the unzoned lens toler-ances by taking the dielectric lens thickness as nearly equal to a0l (n -1) and the metal-plate lens thickness as nearly equal to z0/(l - «) AH tolerances in the ta e assume A0/32 rms for the individual lens variations and a0/64 rms for the reflector variation resulting, in each case, in a gain-loss factor fce = 0.16 dB as calculated from (12-10-3), provided the correlation distance >A and other conditions o (12-10-3) are met.
For a combination of random variations (as index a } Little difference in radiation-field patterns of an E-plane metal-plate lens antenna is revealed for a tilt angle t as large as 30" according to patterns presented by Fnis and Lewis. See H. T. Frns and W. D. Lewis, Radar Antennae" BfU System Tech. J., 26, 270, April 1947, H PLANE MF.TAl.-Pl.ATF LI NS ANTENNAS 683 thickness) the net effective variation is given by the quadrature sum of the indi-vidual variations.
14-6 //-PLANE METAL-PLATE LENS ANTENNAS, 1 A wave enter-ing a stack of metal plates oriented parallel to the H plane (perpendicular to the E plane) as in Fig.
14- 19a is little affected in its velocity. However, the wave is constrained to pass between the plates so that, once inside, the path length can be increased if the plates are deformed, as suggested in Fig. 14-196. An increase in path length can also be produced by slanting the plates as at (c). The increase of path length is S — 7\ Using the slant-plate method of increasing the path length, an //-plane metal-plate lens can be designed by applying the principle of equality of electrical path length. This type of lens is called an //-plane type since the plates are parallel to the magnetic field (perpendicular to the E plane).
Referring to Fig. 14-19J, the condition for equality of electrical path length requires that Figure 14-19 (a) //-plane stack of flat metal plates. (6) //-plane stack with increased path length, (c) [ Slanted //-plane plates. {d\ Biplane metal-plate lens using slanted plate construction.
1 W. E. Kock, 41 Path Length Microwave Lenses, 4 /Vue, IRE , 37, 852-855, August 1949.
684 14 LENS ANTENNAS or (n ~ 1)L n cos 8 — 1 m where n = 1 /cos £ = effective index of refraction of the slant: plate lens medium In this case the index of refraction is equal to or greater than unity so that (2) is identical with (14-2-7) for a dielectric lens. The index n depends only on the plate slant angle £ and is not a function of the frequency as in the E-plane type of metal-plate lens. The most critical dimension is the path length S in the lens. This may be affected by a change in T or in £ Assuming a maximum allowable varia-tion S = V8 in electrical path length, the tolerance in S is given by ±0.06ao (3) A disadvantage of the //-plane metal-plate lens is that this type of construction tends to produce unsymme.trical aperture illumination in the £ plane.
14-7 REFLECTOR-LENS ANTENNA About 1960 I filled a notebook with many unique reflector designs and at long last in 1982 published several of them.
1 One of these combines a dielectric lens with a flat reflecting sheet as sug-gested in Fig. 14-20. An incoming ray traverses the lens twice and is brought to a focus at E. For thin lenses the distance R is given approximately by (in - 1)2L (2n -1) cos 0 -1 0 ) where n = index of refraction of lens L = focal length (in same units as /l) This reflector lens is approximately half the thickness and half the weight of a simple dielectric lens (Figs. 14-la or 14-3) and is an alternative to a parabolic 1 Jr D. Kraus. “Some Unique Reflector Antennas." IEEE Ant. Prop. Soc. Newsletter^ 24, 10, April 1982.
-9 POLYRODS 685 reflector. An interesting, specialized application would be to use a pool of con-ducting liquid for the flat surface with a long focal length reflector lens situated above as in Fig. 14-20. Beam squinting by horizontal displacements of the feed would allow observations of a region near the zenith. Only the thickness of the lens is critical since the flatness of the reflecting surface is maintained automati-cally, thanks to gravity. The lens need not be in contact with the flat surface, A meridian- transit millimeter-wave radio astronomy application at low cost is envisioned.
14-8 POLYRODS- A dielectric rod or wire can act as a guide for electro-magnetic waves, 1 The guiding action, however, is imperfect since considerable power may escape through the wall of the rod and be radiated. This tendency to radiate is turned to advantage in the polyrod antenna , 2 so called because the dielectric rod is usually made of polystyrene. A 62 long polyrod antenna is shown in cross section in Fig.
14-21 a. The rod is fed by a short section of cylindrical FigHTe 14-21 (at Cross section of cylindrical polystyrene antenna 6k long.
\b) Radiation pattern, fAfter G. E.
Mueller and W. A .
TyrreW, “ Polyrod Antennas" Bell System Tech.
J., 26, #37-45A October 1947.) 1 D. Hondros and P. Debye, "Elektromagnelische Wellen an dieleklrischen Drahten," Ann. Pkysik^ 32,465-476, 1910, S. A. Schelkunoff, Electromagnetic Waves, Van Nostrand, New York, 1943, pp. 425-428.
ft. M. Whitmer, 44 Fields in Non-metaLlic Guides," Proc. /HE, 36, 1 105-1109, September 1948, J G.
E, Mueller and W. A. Tyrrell, “Polyrod Antennas," Bell System Tech . A, 26, 837-851, October 1947.
686 L4 LENS ANTENNAS waveguide which, in turn, is energized by a coaxial transmission line. This type of polyrod acts as an end-fire. antenna.
1 The phase velocity of wave propagation in the rod and also the ratio of the power guided outside the rod to the power guided inside are both functions of the rod diameter D in wavelengths and the dielectric constant of the rod material 3 For polystyrene rods with D < a/4, the rod possesses little gutdmg effect on the wave, and only a small fraction of the power is confined to the inside of the rod. The phase velocity in the rod is also close to that for the surrounding medium (free space). For diameters of the order of a wavelength, however, most of the power is confined to the rod, and the phase velocity in the rod is nearly the same as in an unbounded medium of polystyrene. For increased directivity oper-ation the diameter D x in free-space wavelengths of a uniform rod (length L t > 2 and 2 < cr < 5) is In practice, polystyrene rod diameters in the range 0.5 to 0.3/. are used, 4 The rod may be uniform or to reduce minor lobes can be tapered as in Fig. I4-21«. This polyrod is tapered halfway and is uniform in cross section the remainder of its length. The diameter D is 0.5/ at the butt end and 0.32 at the far end. The radiation-field pattern for this polyrod as given by Mueller and Tyrrell is shown in Fig.
14-21/i. The gain is about 16 dBi.
To a first approximation the radiation pattern of a polyrod antenna excited uniformly along its length may be calculated by assuming that it is a continuous array of isotropic point sources with a phase shift of about 360 (1 + 1/2LJ deg'wavelength of antenna, where is the total length of the antenna in free-space wavelengths.
5 The relative field pattern as a function of the angle 9 from the axis is then given by £( 0 ) sin (»72 ) f/2 (2 ) where v = 2tiLj cos 9 — ^ ~ j — 2.“j F.lcos 0 1) J ‘ An end-fire poly rod Vienna may be regarded as a degenerate or rudimentary form of ton antenna with an effective lens cross section of the order of a wavelength. See Gilbert Wilkes.
Wavelength Lens," Pror. IRE, 36, 206-212, February 1948.
3 The relative permittivity e, -2.5 for polystyrene. See Table 14-1.
! G.
E. Mueller and W, A. Tyrrell, "Polyrod Antennas," Bell System Tech. J„ 26, 837-851, October 1947.
4 To transmit the lowest (TE, ,) mode in a circular waveguide, the diameter D of the guide_musl be at least aHi/VS, where i is the free-space wavelength and a, is the relative perrn.mvUy of the guide.
Thus, for a rod of polystyrene (Sufonrte'o 5 haS the pr°Perty that an incident plane wave is broughiT a s™„if h °PP° SIte side of the sphere as suggested for wave 2 in Fig 14-23 Simultaneously, waves from other directions will be brought to a focus at a not.
on e opposite side of the sphere, as suggested in Fig. 14-23 for waves 1 and 3 Thus stgnaJs can be recaved simultaneously with a Luneburg lens from as many dn-ections as there is space available on the sphere to place feed horns or otJr F°r Steenng a sin®le beam thc receiver (or transmitter) can be t0 H d,ffcrent feed homs’ « a movable feed horn can T u^Tt^ ,j “e 'ndEX reared can be obtained with an artificial dielectric material f a Lunlbn C°T triC ShCUS °f 'dieleC‘ric different indices of refraction, side a " “ “ half and 3 refleCti"e Shect P‘aced « ^ flat ncidenL Str^^!
; “T” reSU 'tS With ,nCOmin« wave at an angle of Tte fufi? h , r ?
31 C corresP°nding angle of reflection 0, = ff,.
dinates W and S F ?
pr0V,deS beam steenng in both polar coor-Son louth^t, T l "f °nly °ne Coordmate a plane (parallel-sided) section through the center of the sphere can be used. However the beam is no rive? ^ ?
me i” b°th coordinatcs due t0 vingetting in the 6 direction Keileher gives a good review on the Luneburg lens and its variants.
1 in Sc,W^flQl6 idea th ad fi been mcntioned earlier. Albert Einstein’s brief note of a star such as ?! ana 'ys,s of lens action by the gravitational field are deflected tb ,, SUn' In‘:ldeilt electromagnetic waves passing around a star hlonebt t ?
r0Ug an anfile which is proportional to the mass of the star and S ™thC f? f dC’ ^ SUBgeSted by the ^ fens of rniuimtm r Th ? a IS "° foCal point’ rather a f°cai line extending from a minimum foca! distance to infinity. The gain of the lens is proportional to the mass of the star and inversely proportional to the wavelength. At X = 1 mm a so ar lens can, in principle, give a gain of more than 80 dB. If the spacecraft on 2/ K ^ IICher-’ Electromechanical Scanning Antennas," in R C. Johnson and H Jasik (edsl Antennas Engineering Handbook, 2nd ed t McGraw-Hill, 1984, chap; 18.
SrtnSjRMtST 1 AC,i0'1 °f 3 S,Sr b> 'he DeTia,l°n °f Ligh' " ,he ' G™»>"«I Field" 3 Actually any large mass—the sun, Jupiter, the earth, a neutron star or a black hol^would do 690 i4 LENS ANTENNAS PROBLEMS 691 I: I Incident & Figure 14-14 Einstein gravity lens. A plane wave incident on a large mass (such as the sun) is brought to a focus along a line extending from a minimum distance to infinity. Large gams are possible.
14-3 Unzoned metal-plate lens. Design an unzoned plano-concave £-p(ane type of metal-plate lens of the unconstrained type with an aperture KU square for use with a 3-GHz line source 1(M long. The source is to be 20a from the lens (F = 2) Make the index of refraction 0.6.
h (a) What should the spacing between the plates be?
(b) Draw the shape of the lens and give dimensions.
(c) What is the bandwidth of the lens if the maximum tolerable path difference is a/4 ?
14-4 Helix lens. Confirm (14-5-1).
14-5 Cylindrical lens.
Prove {14-2-15Jbf the focal line has an antenna with 80 dB gain, the total system gain is 160 dB equivalent to the gain of an array of 100 million 80 dB-gain antennas.
The minimum focal distance in the case of the sun is about a dozen times the distance of Pluto so that wc must wait until it is possible to send a properly equipped spacecraft to that distance before the sun can be put to use as a gravity lens.
1 give more details on the Einstein lens with a worked example in my book Radio Astronomy, 1 based on the very extensive treatment of Von R. Eshleman.
ADDITIONAL REFERENCE Goldsmith, P, F., and G.
J, Gill: “Dielectric Wedge Conical Scanned Gaussian Optics Lens Antenna/' Microwave J. T 29, 202-212, September 1986.
PROBLEMS 3 14-1 Dielectric lens .
(a) Design a plano-convex dielectric lens for 5 GHz. with a diameter of 10/, The lens material is to be paraffin and the F number is to be unity. Draw the lens cross section.
(b) What type of primary antenna pattern is required to produce a uniform aper-ture distribution?
14-2 Artificial dielectric. Design an artificial dielectric with relative permittivity of 1.4 for use at 3 GHz when the artificial dielectric consists of {a) copper spheres, (&} copper discs, (c) copper strips.
1 J. D. Kraus, Radio Astronomy, 2nd ed, t Cygnus-Quasar, 1986.
1 V, R. Eshleman, “Gravitational Lens of the Sun: Its Potential for Observations and Communica-tions over Interstellar Distances," 205, 1133, 1979, -1 Answers to starred () problems are given in App. D.
CHAPTER 15 BROADBAND AND FREQUENCY-INDEPENDENT ANTENNAS i 15-1 BROADBAND ANTENNAS. Many antennas are highly resonant, operating over bandwidths of only a few percent.
Such “tuned, narrow-bandwidth antennas may be entirely satisfactory or even desirable for single-frequency or narrowband applications.
In many situations, however, wider bandwidths may be required.
The volcano smoke unipole antenna of Fig- 15-la and the twin Alpine horn antenna of Fig. 15-l/> are examples of basic wide-bandwidth antennas 1 (see earlier discussion of the antennas in Sec, 2-32). The gradual, smooth transition from coaxial or twin line to a radiating structure can provide an almost constant input impedance over very wide bandwidths The high-frequency limit of the Alpine horn antenna may be said to occur when the transmission -line spacing d > A/10 and the low-frequency limit when the open end spacing D < kfL Thus, if D = lOOOd, the antenna has a theoretical 200 to 1 bandwidth-A compact version of the twin Alpine horn, shown in Fig. 15-lc and d , has a double-ridge waveguide as the launcher on an exponentially flaring 2-conductor balanced transmission line^ The design in Fig. 15-lc and d incorporates features 1 l built, and tested a volcano smoke antenna at the Harvard University Radio Research Laboratory in 1945 at the suggestion of Andrew Alford. The impedance bandwidth was very broad as anticipated.
Drawings of both volcano smoke and twin Alpine horn antennas appeared in the first edition of this book (1950).
692 151 BROADBAND antennas 693 used by Kerr 1 and by Baker and Van der Neut. 2 The exponential taper is of the form y = (i) where k t and k2 are constants. The exact curvature ts not critical provided it is gradual.
The fields are bound sufficiently close to the ridges that the horn beyond the launcher may be omitted. The version shown is a compromise with the top and bottom of the horn present but solid sides replaced by a grid of conductors with a spacing of about A/10 at the lowest frequency. The grid reduces the pattern width in the //-plane, increasing the low-frequency gain. The Chuang-Burnside 3 cylindrical end sections on the ridges reduce the back radiation and VSWR Absorber on the top and bottom of the ridges (or horn) also reduces back radiation and VSWR.
Depending on the ratio of the open end dimension D to the spacing d of the ridges at the feed end, almost arbitrarily large bandwidths are possible with gain increasing with frequency. Thus, for the antenna of Fig. 15-lc and dT the feed dimension d = 1.5 mm, so that the shortest wavelength A = 15 mm (A/10 = 1.5 mm) and the open-end dimension D = 128 mm so that the longest wavelength A = 256 mm (A/2 = 128 mm) for a bandwidth of 17 to 1 4 = 256/15).
For a similar antenna without end cylinders, Kerr reports bandwidths of 17 to 1 with gains up to 14 dBi.
The design of Fig.
15-lc and d is linearly polarized (vertical) With two orthogonal sets of ridges forming a quadruply ridged waveguide-fed horn, either vertical or horizontal or circular polarization can be obtained.
The 120° wide-cone-angle biconical antenna of Fig. 15-lc may be regarded as an evolved form of the volcano smoke antenna of Fig 15-la The properties of the biconical antenna have been discussed in Chap. 8. An example of a 120° wide-cone biconical antenna is shown in Fig, 2-28c and, even though not fitted with end caps or absorber, has a nearly constant 50-Q input impedance over a 6 to 1 bandwidth.
Although the parabolic reflector is also a broadband device, its useful band-width is often restricted by the bandwidth of its feed. Other examples of broad-band antennas are the monofilar axial-mode helical antenna (Chap. 7) and the corner reflector (Chap.
12), both with bandwidths of 2 to 1 or more. Tapered helices or clustered helices have reported bandwidths of 5 to 1 or more (see 1 J.
L. Kerr, "Short Axial Length Broad Band Horns, 1 IEEE Trans. Ants. Prop. T AP-21, 710-714, September 1973-1 D.
E.
Baker and C A. Van der Neut, "A Compact, Broadband, Balanced Transmission Line Antenna from Double- Ridged Waveguide," IEEE- Ant. Prop, Soc, Syrup,, Albuquerque, 1982, pp. 568-570.
3 C. W. Chuang and W. D. Burnside, “A Diffraction Coefficient for a CyLindricalLy Truncated Planar Surface," IEEE Trans , Ants. Prop„ AP-28, 177-182, March 1980, 694 15 BROADBAND AND FREQUENCY^DEPENDENT ANTENNAS Side view £ Mi 15-1 BROADBAND ANTENNAS 695 Absorber Figure 15-1 Wideband antennas. Volcano smoke (a) in cutaway view and twin Alpine horn (b) are basic types. Volcano smoke is omnidirectional in azimuth. The twin Alpine horn is unidirectional.
Practical design derived from the twin Alpine horn with 17 to l design bandwidth is shown in side view (r) and end view (</). The 1 20 c wide-cone-angle biconical antenna in (c) has end caps with absorber.
Secs. 7-16, 7-17 and 15-4). The rhombic antenna (Sea 11-1 6b) and the discone (Sec.
16-5) are also broadband types.
All of the above antennas may have relatively constant input impedance and satisfactory pattern and gain over wide bandwidths with beam widths tending to become smaller and gains larger with increasing frequency. Although the increased gain may be highly desirable and useful, these antennas are not frequency independent in the sense that all parameters (impedance, pattern, polarization, gain) are constant or nearly so as a function of frequency.
Consider, for example, the arrangement of Fig.
15-2, which consists of an adjustable X/2 dipole made of two drum-type (roll-up) pocket rulers.
If L is adjusted to approximately A/2 at the frequency of operation, the impedance and pattern remain the same. Strictly speaking, the element thickness or width w and the size of the drum housing should also be adjusted, but if these dimensions remain small compared to A, this effect is small and for many purposes may be Sliding contact Figure 15-1 Adjustable a/2 dipole of 2 drum-type rulers illustrates the require-ment that to be frequency independent an antenna must expand or contract in pro-portion to the wavelength.
150 THE FREQUENCY-INDEPENDENT PLANAR LOG-SPIRAL ANTENNA 697 696 (5 BROADBAND AND FREQUENCY -INDEPENDENT ANTENNAS negligible.
This simple antenna illustrates the requirement that the antenna should expand or contract in proportion to the wavelength in order to be fre-quency independent, or if the antenna structure is not mechanically adjustable as above, the size of the active or radiating region should be proportional to the wavelength. Although the roll-up pocket ruler of Fig. 152 can be adjusted to different frequencies, it does not provide frequency independence on an instanta-neous basis. A dipole with tuned traps can provide instantaneous resonant oper-ation at a number of widely separated frequencies but not over a continuous wide bandwidth. A true frequency-independent antenna is physically fixed in size and operates on an instantaneous basis over a wide bandwidth with relatively constant impedance, pattern, polarization and gain. These kinds of antennas are discussed in subsequent sections.
15-2 THE FREQUENCY-INDEPENDENT CONCEPT: RUMSEY’S PRINCIPLE-Beginning while at the Ohio State University in the early 1950s, continuing from 1954 to 1957 at the University of Illinois, and later at the Uni-versity of California, first at Berkeley and subsequently at San Diego, Victor H.
Rumsey developed and introduced a new way of looking at antennas and their operation as a function of the frequency.
1 Rumsey was intrigued with Mushiake's observation 2 in 1949 that self-complementary antennas 3 have a constant impedance of Z0/2, or half the intrinsic impedance of space, at all frequencies. This is remarkable since there is an infinity of self-complementary shapes. A self-complementary planar antenna has a metal area congruent to the open area, he,, the two areas can be brought into coin-cidence by a rigid motion. Three examples of self-complementary antennas are shown in Fig.
15-3. The metal and open areas are congruent since a rotation of either brings both into coincidence.
The slot and complementary dipole antennas of Chap. 13 are similarly related but usually require a translation for coincidence, Mushiake’s Z0/2 result comes directly from Booker's relation for complementary slots and dipoles as given by (13-6-10).
Rumsey's principle is that the impedance and pattern properties of an antenna 1 V. H. Rumsey, Frequency Independent Antennas, Academic Press, 1966.
E-C- Jordan, G.
A. Deschamps, J.
D. Dyson and P.
E.
Mayes, “ Developments in Broadband Antennas," IEEE Spectrum, I, 58-71, April 1964.
P.
E. Mayes, “Frequency-Independent Antennas: Birth and Growth of an Idea," /£££ Prop.
Soc, Newsletter, 14, 5-S, August 1982.
2 Y. Mushiake, J. IEE Japan, 69, 1949.
S. Uda and Y, Mushiake, " Input Impedance of Slit Antennas," Tech. Rept. Tohoku University 14, pp.
46-59, 1949.
1 The concept of complementary antennas applies strictly only to infinitesimally thin planar, perfectly conducting shapes of infinite extent.
s £ s f I# -S.
', : F Figure 15-3 Three self-complementary planar antennas. Theoretical terminal impedance U 188 fl will be frequency independent if the antenna shape is specified only in terms of angles. Thus, an infinite logarithmic spiral should meet the requirement The biconical antenna of Chap. 8 is an example of an antenna that can be specified only in terms of the included cone angle, but it is frequency independent only if it is infinitely long. When truncated (without a matched termination) there is a reflected wave from the ends of the cones which results in modified imped-ance and pattern characteristics.
To meet the frequency-independent requirement in a finite structure requires that the current attenuate along the structure and be negligible at the point of truncation. For radiation and attenuation to occur, as stated in Sec.
2-37, charge must be accelerated (or decelerated) and this happens when a con-ductor is curved or bent normally to the direction in which the charge is trav-eling. Thus, the curvature of a spiral results in radiation and attenuation so that, even when truncated, the spiral provides frequency-independent operation over a wide bandwidth.
Rumsey's principle was implemented experimentally by John D. Dyson at the Uni versity of III inoi s, who cons t ructed the first practical frequency-independent spiral antennas in 1958, first the bidirectional planar spiral and then the unidirectional conical spiral. These two types are described in the next two sections.
15-3 THE FREQUENCY-INDEPENDENT PLANAR LOG-SPIRAL ANTENNA The equation for a logarithmic (or log) spiral is given by r = (1) or In r = 0 In a (2) where, referring to Fig. 15-4, r — radial distance to point P on spiral 0 = angle with respect to x axis a = a constant 698 I broadband AND frequency-independent antennas Figure 15-4 Logarithmic or log spiral.
From (t)> the rate of change of radius with angle is — = aB In a - r In a Aft The constant a in (3) is related to the angle 0 between the spiral and a radial line from the origin as given by , dr 1 u\ Thus, from (4) and (2), r dO tan 8 = ian /3 In r The log spiral in Fig.
15-4 was constructed so as to make r -1 at 0 - 0 and r = 2 at 6 = Jr These conditions determine the value of the constants a and p.
Thus, from (4) and (5), fi = 77.6° and a = 1.247. Thus, the shape of the spiral is determined by the angle p which is the same for all points on the spiral.
’ Although it is a broadband anltnna, the Archimedes spiral is not regarded as completely frequency independent. The angle ft for an Archimedes spiral as given by r = aB K not a ^ “ function of position along the spiral. However, remote from the origin on a tight Archimedes spiral 9 approaches a nearly constant angle and the Archimedes spiral becomes a close approsima ion o a tightly wound log spiral. See R. Bawer and J.
J. Wolfe. “A Printed Craut Baiun hrU =w> Spiral Antennas," IRE Trans. Microwm Th. Tech, MTT-8, 319-325, May I960; andIP. E. Mayes and tt Dyson, "A Note on the Difference between Equiangular and Archimedes Spiral Antennas, IRt Trans. Microwave Th. Tech,, MTT-9. 203^205, March 1961-15-3 THE FREQUENCY -INDEPENDENT PLANAR LOG-SPIRAL ANTENNA 699 Let a second log spiral, identical in form to the one in Fig.
15-4, be gener-ated by an angular rotation <5 so that (1) becomes ^ = a~ b (6) and a third arid fourth spiral given by r = <-(7) and r4 = a 9 J (8) Then, for a rotation 3 = itf2 we have 4 spirals at 90° angles. Metalizing the areas between spirals 1 and 4 and 2 and 3, with the other areas open, self-complementary and congruence conditions are satisfied. Connecting a generator or receiver across the inner terminals, we obtain Dyson's frequency-independent planar spiral antenna of Fig.
1 5-5.
1 The arrows indicate the direction of the outgoing waves traveling along the conductors resulting in right-circular I y polarized (RCP) radiation {IEEE definition) outward from the page and teft-circul a rly polarized radiation into the 1 I. D. Dyson, “The Equiangular Spiral Antenna,” IRE Trans.
Prop.^ AP-7, 181-187, April 1959.
700 15 BROADBAND AND FREQUENCY-INDEPENDENT ANTENNAS Coaxial cable bonded Ground plane Figure 1 5- Frequency-independent planar spiral antenna cut from large ground plane, page. The high-frequency limit of operation is determined by the spacing d of the input terminal and the low-frequency limit by the overall diameter D. The ratio Did 'for the antenna of Fig 15-5 is about 25 to 1, If we take d — kj 10 at the high-frequency limit and D — k/2 at the low-frequency limit, the antenna band-width is 5 to 1 The spiral should be continued to a smaller radius but, for clarity, the terminal separation shown in Fig 15-5 is larger than it should be. Halving it doubles the bandwidth.
In practice, it is more convenient to cut the slots for the antenna from a large ground plane, as done by Dyson, and feed the antenna with a coaxial cable bonded to one of the spiral arms as in Fig.
15-6, the spiral acting as a balun 1 A dummy cable may be bonded to the other arm for symmetry but is not shown.
Radiation for the antennas of Figs. 15-5 and 15-6 is bidirectional broadside to the plane of the spiral. The patterns in both directions have a single broad lobe so that the gain is only a few dBi The input impedance depends on the parameters 6 and a and the terminal separation. According to Dyson, typical values are in the range 50 to 100 fi, or considerably less than the theoretical 1 Balance-to- unbalance transformer.
15-4 THE FREQUENCY]NDEPENDENT CONICALSPIRAL ANTENNA 701 188 D (=Z0/2), The smaller measured values are apparently due to the finite thickness of spirals.
Referring to Fig.
15-5, the- ratio K of the radii across any arm, such as between spirals 2 and 3, is given by the ratio of (7) to (6), or = ^ = 0 -+' (9) 2 For the antenna of Fig. 15-5, S = n/2 so K = ^ = a ~« = 0.707 ( = 1/72) (10) r2 This is seen to be the ratio of the radial distances to the spiral of Fig. 15-4 at successive 90° intervals.
15-4 THE FREQUENCY-INDEPENDENT CONICAL-SPIRAL ANTENNA. A tapered helix is a conical- spiral antenna and these were described and investigated extensively in the years following 1947, In my first article on the helical antenna published in 1947, 1 I describe a tapered helix which 1 constructed and measured Figure 15-7a and b shows tapered helical or conical spiral antennas in which the pitch angle is constant with diameter and turn spacing variable. These figures appeared in the first edition of this book (1950) and also appear as Fig 7-59a and b of the present edition. These tapered helix or conical spirals were investigated by Springer (1950), 2 by Chatterjee (1953, 1955) 3 (b) Figure 15-7 Tapered helical or conical-spiral (forward-fire) CP antennas.
1 J. D. Kraus, “Helical Beam Antenna" Electronics, 20, 109-111, April 1947, 1 P.
S. Springer, “End-Loaded and Expanding Helices as Broad-Band Circularly Polarized Radi-ators," Electronic Subdiv. Tech, Rept 6104, Wrighl-Patterson AFB, 1950.
3 J.
S. Chatteqee, “Radiation Field of a Conical Helix," J.
Appl.
Phys., 24, 550-559, May 1953; “Radiation Characteristics of a Conical Helix of Low Pitch Angle," J. Appi Phys., 26, 331-335 March 1955.
7U2 15 BROADBAND AND FREQUENCY-INDEPENDENT ANTENNAS 15-5 THE LOG-PERIODIC ANTENNA 703 Figure 15-8 Dyson 2-arm balanced conical-spiral (backward-fire) antenna.
Polarization is RCP. Inner conductor of coax connects to dummy at apex.
and others, and more recently by Nakano, Mikawa and Yamauchi, 1 and found capable of bandwidths of 5 to 1 or more (see Sec. 7-17). Chatterjee also ^escribed a planar spiral antenna.
However, it was not until 1958 that John D, Dyson 2 at the University of Illinois made the tapered helix or conical spiral fully frequency independent by wrapping or projecting multiple planar spirals onto a conical surface.
A typical balanced 2-arm Dyson conical spiral is shown in Fig.
1 5-8 The conical spiral retains the frequency- independent properties of the planar spiral while providing broad-lobed unidirectional circularly polarized radiation off the small end or apex of the cone As with the planar spiral, the two arms of the conical spiral are fed at the centerpoint or apex from a coaxial cable bonded to one of the arms, the spiral acting as a balun. For symmetry a dummy cable may be bonded to the other arm, as suggested in Fig.
15-8 In some models the metal straps are dispensed with and the cables alone used as the spiral conductors.
According to Dyson, the input impedance is between 100 and 150 for a pitch angle a = IT and full cone angles of 20 to 60°. The smaller cone angles (30° or less) have higher frotit-to-back ratios of radiation. The bandwidth, as with the 1 H. Nakano, T, Mikawa and 1. Yamauchi, “Numerical Analysis of Monofilar Conical Helix" IEEE AF-S Jni. Symp., 1, 177-lSO, 1984.
1 j. D. Dyson, “The Unidirectional Spiral Antenna," IRE Trans. Ants. Prop., AP-7, 329^334, October 1959, planar spiral, depends on the ratio of the base diameter (—i/2 at the lowest frequency) to the truncated apex diameter ( ~ A/4 at the highest frequency). This ratio may be made arbitrarily large.
Conical and planar spirals with more than 2 arms are also possible and have been investigated by Dyson and Mayes 1 and by Deschamps, 2 all at the University oflllinois, and also by Alia and Mei. 3 15-5 THE LOG-PERIODIC ANTENNA. While the planar and conical spirals were being developed, Raymond DuHamel and Dwight Isbell,4 also at the University of Illinois, created a new type of frequency-independent antenna with a self-complementary toothed structure as suggested in Fig.
15-9. In an alterna-tive version, the metal and slot areas of Fig. 15-9 are interchanged.
Since Figure 15-9 Self-complementary toothed log-periodic antenna of DuHamel and IsbelL 1 J. D. Dyson and P. F Mayes, "New Circularly- Polarized Frequency -Independent Antennas with Conical Beam or Omnidirectional Patterns," IRE Trans r Ants. Prop., AP-9, 334—342, July 1961.
3 G.
A. Deschamps. "Impedance Properties of Complementary Multiterminal Planar Structures, IRE Trans Ants, Prop,, AP-7, S37I 378, December 1959 3 A. E. Atia and K. K. Mei. "Analysis of Multiple Arm Conical Log-Spiral Antennas," IEEE Trans Arus. Prop , APl9t 320-331, May 1971.
4 R. H. DuHamel and D. F- Isbell, “Broadband Logarithmically Periodic Antenna Structures," IRE jVmf. Conr. Rec.. pt.
1 T 119-128, 1957.
704 15 BROADBAND AND FREQUENCY -INDEPENDENT ANTENNAS Inactive {stop) region |f>X/2i Figure 15-10 Isbell log-periodic frequency -independent type ol dipole array of 7 dBi gain with 11 dipoles showing active central region and inactive regions (left and right ends).
0 + fi 2 — 90° the self-complementary condition is fulfilled.
The expansion parameter and the tooth-width parameter K. + .1 Further work at the University of Illinois showed that the self-complementary condition was not required, and by 1960 Dwight Isbell 1 had demonstrated the first log+periodic dipole array The basic concept is that a grad-ually expanding periodic structure array radiates most effectively when the array elements (dipoles) are near resonance so that with change in frequency the active (radiating) region moves along the array. This expanding structure array differs from the uniform arrays considered in Sec. 7-11 The log-periodic dipole array is a popular design. Referring to Fig.
15-10, the dipole lengths increase along the antenna so that the included angle oc is a constant, and the lengths / and spacings s of adjacent elements are scaled so that l±± - ^±1 = k (3) L su 1 D. E. Isbell, “Log Periodic Dipole Arrays," tRE Trans. Anis. Prop., AP-8, 260^-267, May 1960.
1 So called because the structure repeats periodically with the Logarithm of the frequency. Put another way, the structure doubles for each doubling of the wavelength [see (8)].
15-5 THE LOG' PERIODIC ANTENNA 70S where k is a constant. At a wavelength near the middle of the operating range, radiation occurs primarily from the central region- of the antenna, as suggested in Fig.
15-10. The elements in this active region are about A/2 long.
Elements 9, 10 and 11 are in the neighborhood of 1/ long and carry only small currents (they present a large inductive reactance to the line). The small currents in elements 9, 10 and 1 1 mean that the antenna is effectively truncated at the right of the active region. Any small fields from elements 9, 10 and 11 also tend to cancel in both forward and backward directions. However, some radi-ation may occur broadside since the currents are approximately in phase. The elements at the left (1, 2, 3, etc.) are less than A/2 long and present a large capac-.
itative reactance to the line.
Hence, currents in these elements are small and radiation is small.
Thus, at a wavelength A, radiation occurs from the middle portion where the dipole elements are —A/2 long. When the wavelength is increased the radi-ation zone moves to the right and when the wavelength is decreased it moves to the left with maximum radiation toward the apex or feed point of the array.
At any given frequency only a fraction of the antenna is used (where the dipoles are about A/2 long). At the short-wavelength limit of the bandwidth only 15 percent of the length may be used, while at the long-wavelength limit a larger fraction is used but still less than 50 percent.
From the geometry of Fig. 15-11 for a section of the array, we have tan a = (K+i-O/2 or from (3), tan s = s (5) Taking /„+. , = /J2 (when active) we have I - d/t) “ST -where a = apex angle k = scale factor 5 , = spacing in wavelengths shortward of A/2 element (6 ) 706 BROADBAND AND f REQUENCY-INDEPENDENT ANTENNAS Figure 15-12 Relation of log-periodic array parameters of apex angle x, scale factor k and P“«"8 [from (6)] with optimum design line and gain values according to Carrel and others (see fo details).
Specifying any 2 of the 3 parameters st, k and sx determines the third. The relationship of the 3 parameters is displayed in Fig. 15-12 with the optimum design line (maximum gain for a given value of scale factor k ) and gam along this line from calculations of Carrel, 1 Cheong and Ring, 2 De Vito and Stracca and Butson and Thomson, 4 f The length J (and spacing s) for any element n + 1 is k greater than lor element 1, or inli = fc" = F h where F = frequency ratio or bandwidth Thus, if k = 1.19 and n = 4, F = kA = 1.194 = 2 and element 5 ( = n + 1) is twice the length (, of element 1. Thus, with 5 elements and k = 1.19, the frequency ratio is 2 to i.
Example. Design a log-periodic dipole array with 7 dBi gain and a 4 to 1 band-width Specify apex angle scale constant k and number of elements, 1 R r L Carrel, “The Design of Log- Periodic Dipole Antennas,” IRE Int. Corn, Rec. t 1 , 61-75, 1961.
2 W. M. Cheong and R. W.
P. King, "Log Periodic Dipole Antenna,” Radio Set., 1 , 1315-1325, November 1967, 3 G. De Vito and G. B, Stracca, "Comments on the Design of Log- Periodic Dipole Antennas, IEEE Trans. Ants. Prop., AP-21, 303-308, May 1973; AP-12, 7l-718. September 1974.
4 P. C. Butson and G. T, Thomson, “A Note on the Calculation of the Gain of Log-Periodic Dipole ^ Antennas.” IEEE Trans . Ants. Prop., AP-24. 105-106. January 1976.
15' THE LOG-PERIODIC ANTENNA 707 Solution, From Fig.
15-12. the 7 dBi point on the maximum gain line corresponds to the apex angle or = 15 and k = L2. (We also note that sA — 0.15) From (7).
(t“ = F or n In k = In F <> and In F In 4 1.386 n = — = = 7.6 (9) In k In 1.2 0.182 Taking n = 8. n + 1 = 9, Adding 2 more elements for a conservative design brings the total to 1 L l The array in Fig.
15-10 corresponds to the parameters of the above example. The E-plane HPBW = 60 The H-plane beam width is a function of the gain as given by HPBW (H plane) (deg) (10) For the antenna of the example D - 5, since log 10 5 = 7dBj, so HPBW <H plane) <; = 137° (1 1) Details of construction and feeding are shown in Fig.
15-13. The arrange ment in (a) is fed with coaxial cable, the one at (6) with twin line.
To obtain more gain than with a single log- periodic dipole array, 2 arrays may be stacked. However, for frequency-independent operation, Ruinsey’s prin-ciple requires that the locations of all elements be specified by angles rather than distances. This means that both log-periodic arrays must have a common apex, and, accordingly, the beams of the 2 arrays point in different directions For a stacking angle of 60°, the situation is as suggested in Fig. 15-14a for dipole arrays of the type shown in Figs. 15-10 and 15-13. The array in Fig 15-146 is a skeleton-tooth or edge-fed trapezoidal type. Wires supported by a central boom replace the teeth of the antenna of Fig, 15-9.
For very wide bandwidths the log-periodic array must be correspondingly long. To shorten the structure, Paul Mayes and Robert Carrel, 2 of the University of Illinois, developed a more compact V-dipole array which can operate in several modes. In the lowest mode, with the central region dipoles 2 long, operation is as already described. However, as the frequency is increased to the point where the shortest elements are too long to give 2/2 resonance, the longest elements become active at 32/2 resonance. As the frequency is increased further, . The number or extra elements needed depends, for example, on the design gain, 'thus, for a high-gain design the active region requires more elements than for a Low-gain design so that the bandwidth is less than given by fc", 1 P. E, Mayes and R, L Carrel, “Log Periodic Resonant-V Arrays," San Francisco Wescon Conf., August 1961.
708 15 BROADBAND AND FREQUENCYINDEPENDENT ANTENNAS m Fioure 1513 Construe!,™. and feed details of log-periodic dipole array Arrangement aUn) has a £for coaaial feed. The one a, (b) has criss-crossed open-wtre hne for 30M1 tw.n-hne feed.
the active region moves to the small end in the 3A/2 mode. With still further increase in frequency the large end becomes active in still higher-order modes.
The forward tilt of the V-dipoles has tittle effect on the A/2 mode but in the higher modes provides essential forward beaming. An example of a V-dipole array is shown in Fig. 15-15.
15-6 THE YAGI-UDA^CORNER-LOG-PERIODIC (YUCOLP) ARRAY. For ultimate compactness and gain, to cover the 54 to 890 MHz U.S.
TV and FM bands, a hybrid YUCOLP (Yagi-Uda-Corner-Log- Periodic) array li-6 THE YACjI-UDA CORNER LOG -PERIODIC (YUCOLP) ARRAY 709 is a popular design. A typical model, shown in Fig.
15-15, has an a — 43 n , k — 1.3 LP array of 5 V-dipoles to cover the 54 to 108 MHz TV and FM bands with a 6 dBi gain in the A/2 mode, the 174 to 216 MHz band with 8 or 9 dBi gain in the 3A/2 mode and a square-corner-YU array to cover the 470 to 890 UHF TV band with a 7 to 10 dBi gain. The total included angle of the V-dipoles is 120°. The corner-YU array is similar in design to the one in Fig. 11-42. As frequency increases, the active region moves from the large to the small end of the LP array in the A/2 mode, then from the large to the small end in the 3A/2 mode, next to the corner reflector and finally to the YU array. The corner-YU array provides 710 15 BROADBAND AND mmiENCY -INDEPENDENT AN1KNNAS Square corner Hgve 15-15 YUCOLP (Yagi-Uda-Comer-Log-Periodic) hybrid array for covering U S.
VHP TV and FM bands and UHF TV band.
The YU-comer combination provides higher gain for the UHF TV band than an extension of the LP dipole array.
more gain for the UHF band than possible with a high-frequency extension of the LP array.
PROBLEMS 15-1 Log spiral. Design a planar log-spiral antenna or the type shown in Fig, 15-5 to operate a!
frequencies from 1 to l O' GHz, Make a drawing with dimensions in millimeters.
I 15-2 Log periodic.
Design an ^optimum' log-periodic antenna of ihe type shown in Fig. 15-10 to operate at frequencies Ifrom 50 to 250 MHz, Make a drawing with dimensions in meters.
15-3 Stacked l.Ps, Two LP arrays like in the worked example of Sec. 15-5 are stacked as in Fig, 15-I4d.
la) Calculate and plot the vertical plane field pattern. Note that pattern multiplica-tion cannot be applied.
^/0) and the effective height more nearly equal to the physical height. This will not quadruple R, but could increase it by some factor less than 4, The currents on the flat top are in phase opposition and, being separated by a small fraction of a wavelength, have only a small effect on the radiation: What steps, if any, are taken to increase the radiation efficiency involve trade-offs with cost considerations a determining factor.
According to Harold Wheeler, 1 a figure-of-merit for an electrically small antenna is its radiation power factor: PF radiated power reactive power X where R r = radiation resistance X = reactance This is not the same as I /Q since e = energy stored per unit time energy lost per unit time X _ center frequency Rf + Rl ~ bandwidth (5) Thus, increasing either or R Li or both, reduces Q and broadens the band-width, However, only an increase in Rr increases Wheeler’s radiation power factor, Harold Wheeler defines an electrically small antenna as one which occupies a volume of less than a radian sphere , i.e., a sphere of radius r — k/2n (=0,16 k).
The significance of this definition is that stored energy dominates inside the radian sphere while radiated energy is important outside, Wheeler shows further that the radiation power factor for a small antenna is equal to the ratio of the antenna volume to a radian sphere or _ antenna volume _ jnr3 _ /2rcry radian sphere fji(A/2Jt) 3 \ k ) 1 H. A, Wheeler, “Fundamental Limitations of Small Antennas," Ptoc, IRE, 35, 1479-1484, Decem-ber 1947.
H. A. Wheeler, “ Small Antennas IEEE Trans . Aras, Prop., AP-23, 462-469. July 1975, See also R. C. Hansen, “Fundamental Limitations of Antennas," Proc. IEEE, 69, 17{M&2, February 1981, It is assumed in (2) and (3) that the average antenna current is j the terminal current.
714 16 ANTENNAS FOR SPECIAL. APPLICATIONS: FFEPtNO CONSIDERATIONS where r = radius of antenna volume, m 2 = wavelength, tn According to (6), the PF or figurc-of-merit or a small antenna varies as the cube of its dimension. If R L — 0 in (5), we have from (4) and (6) that (7) Thus, an electrically small lossless antenna (2nrx 1} is inherently high Q and narrow bandwidth. We note from (5) that losses decrease Q and increase band-width.
To reduce the attenuation through salt water, very low frequencies (10 to 50 kHz) are used for transmitting to submerged submarines. At a frequency of 10 kHz (2 = 30 km) antennas may have flat tops covering many square kilo-meters and rank as the world’s largest antennas (physically), yet they are electri-cally small {h e < 0.01/). (See Prob. 16-21.) 16-3 PHYSICALLY SMALL ANTENNAS Heinrich Hertz and other radio pioneers used meter and centimeter wavelengths, but the demonstration of long-distance communication with kilometer wavelengths by Guglielmo Marconi and others soon moved interest from the short to the long wavelengths. However, over the intervening years this trend has been reversed and now active radio technology extends down to millimeter and submillimeter wavelengths. One of the bonuses of the short wavelengths is the spectrum space available for many wide-bandwidth video and high-data-rate channels.
Antennas for these short wavelengths include printed and patch antennas (Sec. 16-12) and ones with built-in (integrated) active elements (amplifiers and detectors) {Sec.
16-13), These active elements can compensate for the increased transmission-line losses at millimeter wavelengths.
The twin Alpine exponential horn described in Secs, 2-2, 2-28 and 15-1 is well adapted for printed -circuit fabrication, as suggested in Fig, 16-2. Here the 16-3 PHYSICALLY SMALL ANTENNAS 715 horn takes the form of an exponential notch in the conducting surface of a circuit board with coupling from a 50-ft strip line on the other surface of the board.
Bandwidths of 5 to 1 are possible but with only small gain, 1 However many such printed notch radiators can be stacked to form highly directional phased arrays. 2 Visible wavelengths (blue to red) extend from 400 to 700 nm. Infrared wave-lengths extend from 700 nm into the micrometer range where they merge with radio. Whether these micrometer wavelengths are called infrared or radio is arbi-trary. With printed -circuit technologies now able to fabricate structures in the micrometer and nanometer range 3 it may be possible to construct antennas for visible wavelengths.
An array of 2/2 dipoles connected to detector (rectifier) elements has been described by Marks4 for the direct high-efficiency conversion of light to dc power. An important application is power generation from solar radiation.
In the design proposed by Marks, a broadside array of 2/2 dipoles is used to collect the incident radiation and convert it to direct current via a 2-conductor RF transmission line which terminates in a full-wave bridge rectifier. The dc output ofthc rectifier feeds the conductors of a dc bus. However my variant, shown in Fig. 16-3 consists of an array of 2 dipoles (2 cobnear 2/2 dipoles), reducing the number of dipoles required to From the table of Sec. 2-24 the effective aperture of a resonant 2/2 dipole is 0.132 2 .
If the array is backed by a suitable reflector (solid or grid) the effective aperture of each dipole can be doubled (approximately) to 0.262 2 , which is about the same as an area 2/2 square, as shown shaded for one 2/2 dipole in Fig. 16-3.
It is apparent that the aperture efficiency of the array should be about 100 percent 5 This efficiency implies linear polarization. However, solar radiation is unpo-larized (equivalent to 2 orthogonal polarizations in random phase), so that the aperture efficiency for sunlight is only about 50 percent. To capture the cross-polarized radiation requires an additional orthogonal array (with grid reflector) placed above the other array with all dc outputs combined. Phasing between the 1 S, N. Prasad and S. Mahapatra “A New MIC Slot-Line Aerial," IEEE Trans, Artis, Prop., AP-31, 525-527 May 983.
2 J.
L. Armitage, “ Electronic Warfare Solid-State Phased Arrays' Microwave 29, 109-122, Feb-ruary 1986.
J A. L. Robinson, “ Cornell Submicron Facility Dedicated," Science, 214 777-778, 1981.
4 A. M. Marti, “Device for Conversion of Light Power lo Electric Power," U,S, Patent 4445050, Apr. 24, 1984.
Note that proper phasing Tor broadside operation requires that adjacent dipoles (whether A/2 tong as in Marks 1 design or A Long as in Fig.
16-3) be cross-connected. An alternative is to space the dipoles by 1/ and have 2 such identical arrays each with its own transmission line and rectifiers one array above (or below) and staggered i/2 with respect to the other array. The reflector could be spaced i/4 and 3i/4 respectively from the 2 arrays.
6 Two orthogonal linear polarizations or left- and right-circular polarizations. Note that an unpo-larized wave is not the same as a single circularly polarized wave.
716 [6 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS Figure 16-3 Dipole array for direct conversion of light to dc power. The wavelength k — 600 nm.
arrays is of no concern. With such a combination array, 100 percent aperture efficiency is possible, in principle. Although transmission line and rectifier loss could reduce the overall )ight-to-dc efficiency to 75 or 80 percent, this is several times the efficiency of present photocell devices. Since the array may be imbedded in a dielectric, the dipole dimensions for resonance will be less than, in air or less than about 500 nm for the center-fed full-/ dipole.
Although not operating in a strict sense as an antenna, the rods and cones of the retina of the human eye are very similar in design to a frolyrod antenna (see Sec.
14-8), with conversion of photon to dc impulses transmitted via bipolar cells, dendrites and axons (transmission lines) to the brain for signal processing and image recognition, 1 The retina contains an array of over 100 million rods and cones. A typical cone is 50 /mi long with diameter tapering from about 3 pm at one end to 1 /im at the other. At a light wavelength of 500 nm (i /un) these dimensions correspond to a length of 100A and diameters of 6 to 2x. It is inter-esting that the index of refraction of the cone medium is nearly the same as employed in typical commercial optical fibers (n =; 1-45).
16-4 ANTENNA SITING AND THE EFFECT OF TYPICAL (IMPERFECT) GROUND.
In Secs.
11-7 and 11-8 the vertical plane pat-terns for horizontal and vertical antennas were calculated assuming that the 1 J. D. Kraus, Electromagnetics 3rd eL McGraw-Hill, 1984, p. 596, ,J. M. Enoch and F. L. Tobey feds.). Vertebrate Photoreceptor Optics, Springer Verlag, 1981, IH ANTENNA SITING AND THE EFFECT OF TYPICAL (IMPERFECT) GROUND 717 Direct ray Figure IM Geometry for horizontal dipole at height h above a flat earth or ground.
ground was perfectly conducting ( 1 ), px =-l (5) and the patterns for the two cases are the same.
Consider now the situation for vertical polarization. Referring to Fig.
16-5, the electric field from the short vertical dipole is parallel (||) to the page or plane of incidence. Assuming that the ground is nonmagnetic (p = it may be shown that the reflection coefficient (py) is given by 1 _ Er sin g — Jet - cos 2 g ^ tr sin a 4- y/Er — cos2 a If £r 1, (6) reduces to n sin a — 1 rt sin a + l where n - J7 r — index of refraction 1 See, for example, J. D, Kraus, Electromagnetics, 3rdjpd., McGraw-Hill, 1984, p. 518.
MU ANTENNA SrnNG AND THE EFFECT of TYPICAL (IMPERFECT) GltOUNO 719 As with the horizontal dipole, the field at a large distance is the resultant of the direct ray from the vertical dipole and the ray reflected from the ground, as Suggested in Fig. 16-5, as given by £,! = cos a[ 1 + P]J2f$h sin oe] (8) where = 2k/X h = height of center of short vertical dipole above ground a = elevation angle pH = reflection coefficient 2fih sin a = path length difference of direct and reflected rays, rad From (7) we note that if the ground is: (1) perfectly conducting (cr = oo) or (2) a lossless dielectric (a = 0) of large relative permittivity (4 £> 1), P„ = +l W and the patterns for the two cases are similar, except at small values of sin a.
To illustrate the significance of the above relations let us consider several important cases.
1 Example 1. A horizontal a/2 dipole is situated 1/2 above a homogeneous flat ground of rich, clay soil typical of many midwestern U S.
states (e, — 16, cr = 10’ 2 U m” 1 ). Calculate and plot the vertical -plane electrical field pattern broadside to the dipole at (a) 1 MHz and (h) 100 MHz, Solution. From (4)^ the loss term of the relative permittivity at i MHz is equal to ^ _ ™~ 2 1£0 ' wen In x 106 x 8.85 x 10“ 12 which is large compared to the dielectric term c' — 16, However, at 100 MHz, e" = 1.8 and is small compared to E^ = 16.
[a) I MHz case. Putting Er = 16-/180 in (2) for p± as a function of the elevation angle a and then in (3) with k = 1/2, the electric field pattern as a function of a is given by the solid curve in Fig. 16-6. The loss permittivity e, is sufficiently large at 1 MHz that the pattern is essentially the same as for a perfectly conducting ground.
fb) 100 MHz case Putting er »= 16 - j'1.8 in (2) for p L as a function or a and then in (3) with h = 1/2, the pattern is as shown by the dashed curve in Fig.
16-6.
Comparing the 2 cases, we note that at 1 MHz the pattern has a vertical null (at a = 90°) which is filled in at 100 MHz and also that the gain is greater at the lower frequency (up 1 dB at a = 30°).
1 For table of ground and water constants, see App. A s Sec. A-6.
720 16 ANTENNAS FOR SPECIAL APPLICATIONS : FEEDING CONSIDERATIONS ANTENNA SITING AND THE EFFECT OF TYPICAL (IMPERFECT! GROUND 721 Figure 16-6 Vertical plane patterns for Example l of horizontal a/2 dipole a/2 above a Hat earih with < = 16 and e = 1CT 1 O m“‘ at 1 MHz (solid curve) and at 100 MHz (dashed curve). The pattern for perfectly conducting ground {a = &) is essentially the same as the solid curve. Patterns to left (90" < a < 180") are mirror images.
Example 2. A short vertical dipole (J < A/10) [£(0) = cos a] is located A/2 above a ground with the same constants as for Example t. Calculate and plot the electric field pattern at (a) 1 MHz and (ft) 100 MHz.
{a) I MHz case. Introducing < - 180 into (7) for as a function of a and then in (8) with ft = XjX the pattern is given by the dashed curve in Fig.
16-7.
(£) 100 MHz case. Introducing £ = 16 into (7) for p H and then in (8) with ft = lj\ the pattern is as shown by the dotted curve in Fig. 16-7. The solid pattern is for perfectly conducting ground (c7 — ao) (same at all frequencies).
Figure 16-7 Vertical plane patterns for Example 2 of short vertical dipole i/2 above a flat earth with < - 16 and = 10“ 3 U m" 1 at 1 MHz (dashed curve) and at 100 MHz (dotted curve). The pattern for perfectly conducting ground (ff = oo) 5 as shown by the solid curve, is the same at all frequencies. Patterns to left (90° < i < ISO") are mirror images.
Ftyve 1641 Vertical plane patterns for Example 3 of short vertical dipok at the surface Jf a flat earth with c; = 16 and ir = at 1 MHz (dashed curve) and at 100 MHz (dotted curve). The pattern for perfectly conducting ground (same at alt frequencies) is shown by the solid curve. Patterns to left (90 < a < 180") are mirror images.
Example 3. Repeat Example 2 for the case where the vertical dipole is at the ground (vertical monopole) so that we may set ft — 0.
Solution Since ft = 0, (8) reduces to E = cos «(1 + p ||) ( 10) and introducing (7), _ 2rt sin a E = cos a n sm a + 1 UD Evaluating (11) for the 2 frequencies results in the curves shown in Fig. 16-8. The pattern for 1 MHz is given by the dashed curve and at 100 MHz by the dotted curve. The solid pattern is for perfectly conducting ground ( 1, More rigorous, and necessarily more complex, studies have been con-722 , (1 VNfHSNAS KOR SPK'HL APPLll’ VHONS M ' t -'D I Nt i CONSIDERATIONS i ducted beginning with Arnold Sommcrfelds classic solution of 1909 Nc«rth,-le SS our examples illustrate some of the principle changes caused by typical around as compared to perfectly conducting ground.
Let us summarise some of the principal differences of nonperfectly versus perfect! v conducting ground. With a perfectly conducting ground the rcflec ed-ray amplitude is equal to the direct-ray amplitude (Fig.
16-4). which means that jn some directions the 2 fields may add in phase, doubling the field (quadrup mg the power) for a 6-dB gain. On the other hand, in some other directions the ^ ravs may he out of phase or cancel, resulting in zero field (zero power) for an infinite dB loss. Thus, with a perfectly conducting ground, there is the possibi i y of anything from a +6dBloa - x dB change from the free-space condition 'With nonperfectly conducting ground the reflectcd-ray amplitude tend, be less than the dirccl-ray amplitude. Thus, in directions for which !he fieldsi add in phase, the maximum gain is less than 6dB but ,n directions foi' wh he fields are out of phase there tends not to be complete signal cancellation la ) except at low frequences. Referring to Figs.
16-6. 16-7 and 16-8, the ahovc noted trends are apparent, with less gain and filled nulls m Figs.
16-6 and 16-7 and also much reduced field strength along the ground (3 = 0} for the vertical antennas i FlgS C^mpanng Figs.
16-6 and 16-7, we note that for perfectly conducting ground the maximum for horizontal polarization is at an elevation angle of 0 with a null at 0\ while for vertical polarization the null is at 30 with he maximum at 0 .
If both the vertical and horizontal antennas are short dipoles, z > above perfectly conducting ground, and are connected together as a George Brown turnstile to transmit circular polarization, a circularly polarizedl antenna will receive a constant signal as a function of elevation between 0 and 30 (no nulls, no maxima) but 6 dB below the vertical or horizontal maxima.
in the above discussion it is assumed that the direct and reflected rays a parallel (distance very large). However, if the receiving antenna is c,oser “ the direct and reflected rays are no< parallel, the received signal can fluctuate many times between maxima and nulls as a function of height, linear polarization being assumed.
In general, avoiding a null with linear polarization may require either raising or lowing the receiving antenna (see Prob.
16-14, the accompany mg figure and solution).
Thus, although the siting procedure of Hg. 16-9 unorthodox, the result has some credibility.
I 1 A. Sommerfeld. “Cber die Ausbrritiing der Wdlen in der drahllosen Telegraphic" Am. Phys, 28, t 6 E. Terman. Radio Engineering Handbook.
1st ed., McGraw-Hill, 1943, sec.
10. 28 refs.
R W P Kina, Theory of Linear Antennae Harvard, 1956. chap. 7. 75 refs I V.
L. Lindell, E. AbiJn and K Mannersalo, " Exwi Image Method for Lmpal^e Computation of Antennas above the Ground, 1 ' Helsinki University Radio Lab, RepL S 161, 1984, 23 rets R W. P King, "Electromagnetic Surface Waves; New Formulas and Applications, 1EEL Frans.
,4nfs. Prop, A P-33. 1204-1212, November !985.
1&-5 GROUND'PLANE antennas 723 BEETLE HAILEY ^ MOHT WA IJ( ER Figure 16-9 Siting the antenna. {Reprinted with special permission of King Features Syndicate, /nr.) Another siting effect occurs when, for example, a Yagi-Uda array is located less than A/2 above ground, the proximity detuning the elements and reducing the gain.
16-5 GROUND-PLANE ANTENNAS-Several types of ground-plane or related antennas are shown in Fig, 16-10. The type at (a) has a vertical A/4 stub with a circular-sheet ground plane about A/2 in diameter. The anienna is fed by a coaxial transmission line with the inner conductor connected to the A/4 stub and the outer conductor terminating in the ground plane. In (h) the ground plane has been modified to a skirt or cone. By replacing the A/4 stub with a disc as in (c), a Kandoion discone antenna 1 is obtained. The dimensions given are appropriate for the center frequency of operation. In Fig. 16-10d the solid-sheet ground plane is replaced by 4 radial conductors. A modification of this antenna is shown at (el in which a short-circuited A/4 section of coaxial line is connected in parallel with the antenna terminals.
2 This widens the impedance band width and also places the stub antenna at dc ground potential. This is desirable to protect (he transmission line from lightning surges.
With reference to solid-sheet ground-plane antennas, it should be noted that the radiation pattern of a vertical A/4 stub on a finite ground sheet differs appreciably from the pattern with an infinite sheet. This is illustrated by Fig.
16-11. The solid curve is the calculated pattern with a ground sheet of infinite extent The dashed curve is for a sheet several wavelengths in diameter and the 1 A, G, Kandoian, “Three New Antenna Types and Their Applications,” Proc. 1RE< 34, 70W 75W, February 1946.
A. G, kandoian, W. Sichak and G, A. Fdsenheld, " High Gain with Discone Antennas" Proc. Natl.
Electronics Conf^ 3, 318-328, 1947.
1 These radial conductor ground-plane antennas were invented in 1938 by George H. Brown, Jess Epstein and Robert Lewis.
724 l& ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS 16-6 SLEEVE ANTENNAS 725 Figure 16-10 (a) Stub antenna with flat circular ground plane, (b) same antenna with ground plane modified to skirt or cone, (c) Kandoian discone antenna, (d) stub antenna with 4 radial conductors to simulate ground plane and (e) a method of feeding ground-plane antenna.
I •f dotted curve for a sheet or the order of U in diameter. With finite solid-sheet ground planes the maximum radiation is generally not in the direction of the ground plane but at an angle . At any instant of time the pattern is a figure-of-eight of the same 16-7 TURNSTILE ANTENNA 727 Figure 16-14 George Brown turnstile with short (infini-tesimal) dipoles.
shape as for a single infinitesimal dipole. An instantaneous pattern is shown in Fig. 16-14h for wr = 135'. As a function of time this pattern rotates, completing 1 revolution per cycle.
In the case being considered in Fig.
16-14, the pattern rotates clockwise. Thus, the phase of the field as a function of 0 is given by 0 + <oi = constant and, if the constant is zero, by cot = -0 (3) If the field is a maximum in the direction 0 = 0 at a given instant, then according to (3) the field is a maximum in the 0 = — 45' direction ^-period later.
The above discussion concerns the field in the 0 plane (plane of the crossed dipoles). The field in the axial direction (normal to the crossed infinitesimal dipoles) has a constant magnitude given by | E\ = v/cos 2 + sin 2 wf = 1 (4) Thus, the field normal to the infinitesimal dipoles is circularly polarized.
In the case being considered in Fig. 16-14 the field rotates in a clockwise direction.
Replacing the infinitesimal dipoles by Xj2 dipoles results jn a practical type of antenna with approximately the same pattern characteristics. This kind of antenna is a George Brown turnstile antenna.
1 Since the pattern of a Xj2 element is slightly sharper than for an infinitesimal dipole, the 0-plane pattern of the turn-stile with A/2 elements is not quite circular but departs from a circle by about ± 5 percent. The relative pattern is shown in Fig. 16-15u. The relative field as a func-tion of 0 and time is expressed by cos (90 J cos 0) cos (90 sin 0) r E = ;— cos cot H sin cot sin 0 cos 0 Although the 0-plane pattern with X/2 elements differs from the pattern with infinitesimal dipoles, the radiation is circularly polarized in the axial direc-tion from the X/2 elements provided that the currents are equal in magnitude and in phase quadrature.
A turnstile antenna may be conveniently mounted on a vertical mast. The mast is coincident with the axis of the turnstile. To increase the vertical plane 1 G. H. Brown “The Turnstile Anienna" Electronics, 9, 15 April 1936, 728 I ANTENNAS FOft SP'ECIAL APPLICATIONS- FEEDING CONSIDERATIONS (a) <> fright 16-15 George Brown turnstile with kjl dipoles.
directivity, several turnstile units can be stacked at about 2/2 intervals as in Fig. 16-15b. The arrangement at (i>) is called a “4-bay” turnstile.
It requires 2 bays to obtain a field intensity approximately equal to the maximum field from a single 2/2 dipole with the same power input.
In order that the currents on the A/2 dipoles be in phase quadrature, the dipoles may be connected to separate nonresonant lines of unequal length.
Suppose, for example, that the terminal impedance of each dipole in a single-bay turnstile antenna is 70 + jO Q. Then by connecting 70-0 lines (dual coaxial type), as in the schematic diagram of Fig. 16- 1 6a, with the length of one line 90 electri-cal degrees longer than the other, the dipoles will be driven with currents of equal magnitude and in phase quadrature. By connecting a 35-11 line between the junc-tion point P of the two 70-fl lines and the transmitter, the entire transmission-line system is matched.
Another method of obtaining quadrature currents is by introducing reac-tance in series with one of the dipoles.
1 Suppose, for example, that the length and diameter of the dipoles in Fig 16-1 6b result in a terminal impedance of 70 _ j7Q By introducing a scries reactance (inductive) of +/70 at each ter-Ftgwe 16-16 Arrangements for feeding turnstile antennas.
1 G. H Brown and J, Epstein, "A Pretuned Turnstile Antenna," Electronics, 18, 102-107, June 1945.
ita superturnstile antenna 729 minal of dipole l as in Fig. 16-166, the terminal impedance of this dipole becomes 70 + /70 O. With the 2 dipoles connected in parallel, the currents are 1 70 + j70 and j ^ 70-/70 where V = impressed emf / ! = current at terminals of dipole 1 l 2 - current at terminals of dipole 2 and ‘‘-nl+tE \ so that /, and 1 2 are equal in magnitude but I 2 leads I t by 90“ The 2 imped-ances in parallel yield y_l_ 1 Y Cl/(70 + j70>] + [1/(70 — j70)] “ 70 +J° (n} (8} so that a 10-fl (dual coaxial) line will be properly matched when connected to the terminals FF\ 16-8 SUPERTURNSTILE ANTENNA, In order to obtain a very low VSWR over a considerable bandwidth, the turnstile described above has been modified by Masters 1 to the form shown in the photograph of Fig/16-17. In this arrangement, or Masters superturnstile, the simple dipole elements are replaced by flat sheets or their equivalent.
Each "dipole 11 is equivalent to a slotted sheet about 0.7 by 0,5/ as in Fig.
16-18 a, The terminals are at FF. As in the slotted cylinder antenna, the length of the slot for resonance is more than k(2 (about 0,7A), The dipole can be mounted on a mast as in Fig. 16-186. To reduce wind resistance, the solid sheet is replaced by a grid of conductors. Typical dimensions for the center frequency of operation are shown. This arrangement gives a VSWR of about 1.1 or less over about a 30 percent bandwidth, which makes it convenient as a mast-mounted television transmitting antenna for frequencies as low as about 50 MHz. Unlike the simple turnstile there is relatively little radiation in the axial direction (along the mastk and only one bay is required to obtain a field intensity approximately R. W. Masters, "The Super-turnstile Antenna, 1 Broadcast News, 42, January 1946.
730 It ANTENNAS FOE SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS i Figure 16-17 Six-bay Masters superwrnstite antenna.
((Townesy KC/4 J Figure 16-18 Single dipole dement of Masters superturnstile antenna, (a) Solid sheet const (b) tubing construction showing method of mounting on mast.
tfr-9 OTHER OMNIDIRECTIONAL ANTENNAS 731 equal to the maximum field from a single k/2 dipole with the same power input For decreased beam width in the vertical plane the superturnstile bays are slacked at intervals of about \k between centers. Impedance matching is dis-cussed by Sato ei aL 16-9 OTHER OMNIDIRECTIONAL ANTENNAS, The radiation pat-terns of the slot ted -cylinder and turnstile antennas are nearly circular in the hori-zontal plane. Such antennas are sometimes referred to as omnidirectional types, it being understood that omnidirectional ” refers only to the horizontal plane.
As shown in Chap.
6, a circular loop with a uniform current radiates a maximum in the plane of the loop provided that the diameter D is less than about 0.58/., The pattern is doughnut shaped with a null in the axial direction as suggested by the vertical plane cross section in Fig r 16-19a, Figure 16-19 (a) Circular loop antenna and (b) approximately equivalent arrangements of “clover-leaf 1 type, fc) “ triangular-loop" type and frf) square or Alford loop.
1 G. Sato, H. Kawakami, H. Sato and R, W, Masters, “ Design Method for Fine Impedance Matching of Supertumstile Antenna and Characteristics of the Modified Batwing Antenna," Trans, IECE (Japanl E6S, 271-278, May 1982.
732 I ft ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS 16-10 CIRCULARLY POLARIZED ANTENNAS 733 One method of simulating the uniform loop is illustrated in Fig. 16-196.
Here 4 smaller loops are connected in parallel across a coaxial line. This arrange-ment is called a LL doverleaf ” antenna.
1 Another method is shown in Fig. 16-19c, 3 folded dipoles being connected in parallel across a coaxial line.
2 A third method utilizing a square loop is illustrated in Fig. 16-19d. 3 The terminals are at FF. The side length L may be of the order of A/4. A single equivalent loop or bay of any of these types produces approximately the same field intensity as the maximum field from a single A/2 dipole with the same power input. For increased directivity in the vertical plane, several loops may be stacked, forming a multibay arrangement.
16-10 CIRCULARLY POLARIZED ANTENNAS.
Circularly polarized radiation may be produced with various antennas. The monofilar axial-mode helical antenna (Fig.
16-20a) is a simple, effective type of antenna for generating circular polarization.
Circular polarization may also be produced in the axial (o) X/2 dipoles Axis tb) ic) \ Axis ZJ w Coaxial lines id) Figm 16-20 Antenna types for circular polarization.
J P. H.
Smith, “ Cloverleaf Antenna for FM Broadcasting," Proc. IRE, 35, 1556-1563, December 1947.
2 A, G. Kandoian and R. A. Felstnhekl, "Triangular High-Band TV Loop Antenna System," Commu-jiications, 29, 16-18, August 1949.
3 A. Alford and A. G. KLandoiati, “ Ultra-high Frequency Loop Antennas," Trans. AIEE t 59, 843-848, 1940.
direction from a pair of crossed A/2 dipoles with equal currents in phase quadra-ture (Fig. 16-206) as in a turnstile antenna.
If radiation in one axial direction is right-circularly polarized, it is left-circularly polarized in the opposite axial direc-tion.
A third type of circularly polarized antenna consists of 2 in-phase crossed dipoles separated in space by 2/4 as in Fig.
16-20c, With this arrangement the type of circular polarization is the same in both axial directions.
Any of these 3 arrangements can serve as a primary antenna that illumi-nates a parabolic reflector or they can be placed within a circular waveguide so as to generate a circularly polarized TE n mode wave. By flaring the guide out into a conical horn, a circularly polarized beam can be produced.
Another technique by which a circularly polarized beam may be obtained with a parabolic reflector of large focal length with respect to the diameter is with the aid of a metal grid or grating of parallel wires spaced A/8 from the reflector and oriented at 45° with respect to the plane of polarization of the wave from a linearly polarized primary antenna.
Three arrangements for producing an omnidirectional pattern of circularly polarized radiation are illustrated by Fig, 16-20d, e and / At (/) 4 short axial-mode helices of the same type are disposed around a meta! cylinder with axis vertical and fed in phase from a central coaxial line.
1 In the system at (e) verti-cally polarized omnidirectional radiation is obtained from two vertical A/2 cylin-ders when fed at FF and horizontally polarized omnidirectional radiation is obtained from the slots fed at F'F. By adjusting the power and phasing to the 2 sets of terminals so that the vertically polarized and horizontally polarized fields are equal in magnitude and in phase quadrature, a circularly polarized omnidi-rectional pattern is produced. 2 At (/) 4 in-phase A/2 dipoles are mounted around the circumference of an imaginary circle about A/3 in diameter. 3 Each dipole is inclined to the horizontal plane as suggested in the figure.
In general, any linearly polarized wave can be transformed to an eliiptically or circularly polarized wave, or vice versa, by means of a wave polarizer. 4 For example, assume that a linearly polarized wave is traveling in the negative z direction in Fig.
16-21 and that the plane of polarization is at a 45° angle with respect to the positive x axis. Suppose that this wave is incident on a large grating of many dielectric slabs of depth L with air spaces between. A section of 1 i.
D.
K.raus h "Helical Beam Antenna for Wide-Band Applications," Proc. IRE, 36, 1236-1242, October 1948.
2 C.
E. Smith and R. A. Fouty, “Circular Polarization in F-M Broadcasting," E/ecrronics, 21, 103— 107, September 1948.
1 G.
H. Brown and Q. M, Woodward, Jr., w Circularly- Polarized Omnidirectional Antenna," RCA Rev.. 8, 259-269, June 1947.
4 F. Braun, “ Elektrische Schwingungen und drahtlose Telegraphic," Jahrbuch der drahtlosen Tele-graphie und Telephonic, 4, no.
1, 17, 1910.
734 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS this grating is shown in Fig.
16-21, The slab spacing (in the x direction) is assumed to be a small part of a wavelength.
The incident electric field E can be resolved into two components, one parallel to the x axis (£) and the other parallel to the y axis (Ey); that is, E = \E X + yEy . The x component (£) will be relatively unaffected by the slabs. However, £ will be retarded (velocity reduced). If the depth L of the slabs is just sufficient to retard Ey by 90° in time phase behind £, the wave emerging from the back side of the slabs will be circularly polarized if l £ I = I £y I Viewing the approach-ing wave from a point on the negative z axis, the E vector rotates clockwise.
If the depth of the slabs is increased to 2L, the wave emerging from the back side will again be linearly polarized since E x and Er are in opposite phase, but E is at a negative angle of 45° with respect to the positive x axis. Increasing the slab depth to 3L makes the emerging wave circularly polarized but this time with a counterclockwise rotation direction for E (as viewed from a point on the negative z axis). Finally, if the slab depth is increased to 4E, the emerging wave is linearly polarized at a slant angle of 45°, the same as the incident wave. The dielectric grating in this example behaves in a similar way to the atomic planes of a uniaxial crystal, such as calcite or rutile, to the propagation of light. For such crystals the velocity of propagation of light, linearly polarized parallel to the optic axis, is different from the velocity for light, linearly polarized perpendicular to the optic axis.
16-11 MATCHING ARRANGEMENTS, BALUNS AND TRAPS, Impedance matching between a transmission line and antenna may be accom-plished in various ways.
1 As illustrations, several methods for matching a trans-1 Only arrangements with transmission-line elements will be described. These are convenient at high frequencies. However, at low or medium frequencies the length of the required transmission-line sec-tions may be inconveniently large so that it is the usual practice to use matching circuits with lumped elements. Radio-frequency transformers and rc, T and L sections are employed in this application.
16-U MATCHING ARRANGEMENTS, BALUNS AND TRAPS 735 mission line to a simple X/2 dipole will be considered. Suppose that the antenna is a cylindrical dipole with a length-diameter ratio of 60 {LjO — 60) and that the measured terminal impedances at 5 frequencies are as follows: Frequency Antenna length, Terminal impedance, 1-1 5Fa L = a 53 1 10 + j90 107Fa L = 0.49 80 + j40 F0 = center frequency £ = 0.46 65 + jO 0.93F0 L = 0,43 52 -;40 0.85F0 L = 0.39 40 — jlOO The center frequency F0 corresponds to the resonant frequency of the antenna. At this frequency the terminal impedance is 65 + jO fl.
The most direct arrangement for obtaining an impedance match is to feed the dipole with a dual coaxial transmission Line of 65 Q characteristic impedance as in Fig. 16-22a. The variation of the antenna impedance referred to 65 Q is shown by the solid curve in the Smith chart 1 of Fig.
16-23. The normalized impedances plotted on the chart are obtained by dividing the antenna terminal impedances by 65. The VSWR on the 65-tJ line as a function of frequency and antenna length is shown by the solid curve in Fig.
16-24.
The dipole antenna may also be energized with a 2-wire open type of trans-mission line. Since the characteristic impedance of convenient sizes of open 2-wire line is in the range of 200 to 600 D, an impedance transformer is required between the line and the antenna. A suitable transformer design may be deduced as follows. Referring to Fig. 16-22i>, the impedance Z B at the terminals of a loss-less transmission line terminated in an impedance ZA is .
a + jZ0 tan fix 0 Z0 +jZA tan jB where Jffx = (2tt/2)x = length of line, rad Z0 = characteristic impedance of the transmission line (since the line is assumed to be lossless, Z0 is a pure resistance), O Equation (1) may bz reexpressed as jZJXan fix) +jZ0 " -0 (ZJtan fix) +)ZA 1 P. H. Smith, “An Improved Transmission Line Calculator," EUctronicz, 17, 130, January 1944. See J, D. Kraus, “Electromagnetics’ 3rd ed+, McGraw-Hill, 1984, pp. 408^411, for Smith chart solutions of single and double stub matching \/2 antenna and 65-f? line """ V2 antenna and 500-12 line with one A/4 transformer “ A'2 antenna and 500-12 line with two A/4 transformers Figure 15-23 Normalized impedance variation for cylindrical A/2 dipole antenna {L/D = 60) fed directly by a 65-fl line (solid), by a 50G-f2 line with one A/4 transformer (dashed) and by a 500-fl line with two A 4 transformers in series (dash-dot).
Figure 16-24 VSWR as a function of antenna Length L in wavelengths and us a function or the fre-quency (the resonant frequency Fa is taken as unity).
The VSWR curves are for the same 3 cases of Fig. 16-21 738 1 fi ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS shown in Fig.
I6-22c. At the center frequency, Z, = 65 Q. Supposing that the characteristic impedance of the line we wish to use is 50° Q (Z„ = 500 0), we have from (5) that the characteristic impedance of the z/4 section should be Z 0 -This type of transformer gives a perfect match (zero reflection coefficient) at only the center frequency. At a higher frequency the antenna impedance is differ-ent, and the line length is also greater than A/4. The resultant impedance varia-tion with frequency on the 500-fi line for the L/D = 60 dipole antenna and 180-0 transformer (A/4 at F0) is shown by the dashed curve in Fig.
1 6-23 and the VSWR on the 500-0 line is indicated by the dashed curve in Fig.
16-24.
is apparent that this arrangement is more frequency sensitive than the arrangement with the dual coaxial 65-0 line.
Instead of making the transformation from the 500-12 line to the antenna in a single step with a single -section transformer, two sections may be connected in series as in Fig. 16-224 Each is //4 long at the center frequency F 0 - At Fo the first section (Z0 = 108 01 transforms the antenna resistance of 65 il to 180 II.
The second section (Z0 = 300 O) transforms this to 500 fl. The antenna and line are perfectly matched at only the center frequency, as before. However, this 2-section arrangement is less frequency sensitive than the single section. The nor-malized impedance variation with the 2-section transformer is indicated by the dash-dot curve in Fig. 16-23, and the VSWR on the 500-Q line is shown by the dash-dot curve in Fig. 16-24.
, .
, If the number of sections in the transformer is increased further, it should be possible to approach closer to the frequency sensitivity with the direct con-nected 65-fi line.
1 As the number of sections is increased indefinitely, we approach in the limit a transmission line tapered gradually in characteristic impedance over a distance of many wavelengths/ At one end, the line has a characteristic impedance equal to the antenna resistance (65 II in the example) and at the other end has a characteristic impedance equal to that of the transmis-sion line we wish to use (500 fl in the example).
' The logarithms of the impedance ratios may be made to correspond to «. "nom£ (See J.C.
Slater, m-romwe Transmission.
McGraw-Hill, New York, 1942 P ) „ ' logarithms of the impedance ratios for 2-, 3- and 4-Section transformers would be as in Pascals triangle: 2-sections: 1:2:1 3-$eciions; l : 3 : 3 : 1 4-sections: 1 : 4 : 6 : 4 : 1 In the 2-seciion transformer of Fig. 16-22d these ratios are followed since 108 300 500 to, 6F : "Tw :loe 355 3,1:2 1 C.
R. Barrows, “The Exponential Transmission Line," Bell System Tech, J .
17, 555 573, October 1938 H. A. Wheeler. “Transmission Lines with Exponential Taper," Fn. IRE , 27, 65-71, January 1939 611 MATCHING ARRANGEMENTS. BALUNS AND TRAPS 739 Another more frequency -sensitive method of matching a 500-fi line to a 2/2 dipole is with a stub, 1 as shown in Fig. 1622^ The line between the stub and the transmitter may be nonresonant or perfectly matched to the antenna at one fre-quency with the stub as shown. The stub may also be placed A/2 farther from the antenna, as shown by the dashed lines.
2 In this case, however, the resonant line between the stub and antenna is longer, and this arrangement is more frequency sensitive than with the stub closer to the antenna. In general, it is desirable to place matching or compensating networks as close to the antenna as possible if frequency sensitivity is to be a minimum.
With the single stub as in Fig.
16-22? both the length of the stub and its distance S from the antenna are adjustable. The stub may be open or short-circuited at the end remote from the line, the stub length being a/4 different for the 2 cases. To adapt this arrangement to a coaxial line requires that a line stretcher be inserted between the stub and the antenna. An alternative arrange ment is a double-stub tuner which has 2 stubs at fixed distances from the antenna but with the lengths of both stubs adjustable.
The frequency sensitivity 3 of a dipole antenna may be made less than for the L/D = 60 dipole with a direct-connected 65-fi line, as above, in several ways.
A large-diameter dipole can be used (smaller L/D ratio) since, as shown in Chap. 9, the impedance, variation with frequency is inherently less for thick dipoles as compared to thin dipoles. The thick dipole is desirable for very wide-band applications. If such a dipole is inconvenient, the impedance variation can often be reduced over a moderate bandwidth by means of a compensating network. For example, the frequency sensitivity of the L/D = 60 dipole with a direct-connected 65-fi line can be reduced over a considerable bandwidth by connecting a compensating line in parallel with the antenna terminals as in Fig. 16-22/ If this line or stub has an electrical length of X/2 at the center fre-quency and has a 65-fi characteristic impedance, the same as the transmission line, the variation of normalized antenna terminal impedance with frequency, as referred to 65 fi, is shown by the dash-dot curve in Fig. 16-25u. The variation without compensation (antenna of Fig. \6-22a) is given by the solid curve (same curve as in Fig.
16-23). The VSWRs on a 65-fi line are compared in Fig. 16-25& for the antenna without compensation (solid curve) and with the compensating stub (dash-dot curve). The frequency sensitivity of the compensated arrangement is appreciably less over the frequency range shown. For instance, the, bandwidth for VSWR < 2 is about 14 percent for the uncompensated dipole but is about 18 percent for the compensated dipole.
The action of the parallel-connected compensating line or stub is as follows.
1 F.
E, Terman, Radio Engineers' Handbook, McGraw-HUl, New York, 1943, pp.
187- L9L Gives design charts for open stub, closed stub and reentrant matching arrangements, 2 In general, the distance of the stub from the antenna can be increased by nk/2 where n is an integer 3 Frequency sensitivity as used here refers only to impedance, not antenna pattern.
740 | -6 ANTENNAS fob special applications- heeding considerations — X/2 antenna with 65-12 line —- X/2 antenna and 65-12 line with 65-0 compensating stub — A/2 antenna and 120-12 line with 65-tl compensating stub (a) Figifre 16-25 Normalized impedance (a) and VSWR (b) for cylindrical a/2 dipole (L/D — 60) fed directly with a 65-0 line as in Fig. 16-22 (solid curves); with a 65-0 line and 65-0 2/2 compensating stub as in Fig 1 (>-22f (dash-dot curves); and with a 120-0 line and 65-0 a/2 compensating stub (dashed curves).
At the center frequency F0 it is 180" in Length Since it is open ended it places an infinite impedance across the antenna terminals and has no effect. At a frequency slightly above F0 the line bepomes capacitative Hence it places a positive sus-ceptance in parallel with the antenna admittance which at this frequency has a negative susceptance.
1 Admittances in parallel are additive so this tends to reduce the total susceptance at the antenna terminals and therefore the VSWR on the line. At a frequency slightly below F 0 the result is similar, but in this case the stub is inductive and the antenna has capacitative reactance.
The above matching arrangements provide for a perfect impedance match (VSWR = 1) at the resonant frequency of the antenna. Sometimes a perfect impedance match is not required at any frequency and it is sufficient to make the 1 The antenna impedance at this frequency has a positive reactance. Hence, Y = -~.
G —jB Z R + jX where G is the conductance component and B the susceptance component of the admittance Y, L ft-L 1 MATCHING ARRANGEMENTS, BALUNS AND TRAPS 741 VSWR less than a certain value over as wide a frequency band as possible. For example, the VSWR for the /.?2 dipole [L/D = 60) may be made less than 2 over nearly the entire frequency band under consideration if the antenna with a 65-Q compensating stub is fed with a 120-Q line instead of a 65-Q line. The impedance and VSWR curves for this case are shown by the dashed lines in Fig, 16-25u and h r Although the above discussion deals specifically with matching arrange-ments between a /./2 dipole and a 2-conductor transmission line, the principles are general and can be applied to other types of antennas and to coaxial lines Antenna impedance characteristics may also be compensated by scries reac-tances or by combinations of series and parallel reactances.
1 Many of the tech-niques of impedance compensation are discussed with examples by J. A. Nelson and G. Stavis.
2 It is often convenient to use a single coaxial cable to feed a balanced antenna. This may be accomplished with the aid of a bal ancc-to-unbalance trans-former or balun.
3 One type of balun suitable for operation over a wide frequency band is illustrated in Fig.
16-I3t\ Another more compact type is shown in Fig. l6-26a. The gap spacing at the center of the dipole is made small to mini-mize unbalance. The length L may be about z.,/4 at the center frequency with operation over a frequency range of 2 to 1 or more. With this arrangement a reactive impedance Z = jZn tan flL appears in parallel with the antenna imped-ance at the gap, Z0 being the characteristic impedance of the 2-conductor line of length L. Yet another form of balun is shown in Fig. 16-266. This form provides a balanced transformation only when L is //4 and accordingly is suitable only for operation over a few percent bandwidth.
If a /J2 dipole antenna is fed by a single coaxial line as in Fig. 16-27u, current can flow back along the outside of the coaxial cable making the current distribution on the dipole unbalanced, as suggested in the figure. To feed the dipole with a single coaxial line in a balanced manner, a dummy x/4 cable can be placed parallel to the active coaxial line as in Fig, !6-28) Balanced 2/2 dipole balun-fed from coaxial line.
16-11 MATCHING ARRANGEMENTS, BaLUNS AND TRAPS 743 Dipole antenna Dipole antenna Figure 16-28 {g) 2/4 balance- to-unbalance transformer fbalun). \b) More compact balun with cables wound on a ferrite core.
antennas with high efficiency since the 2-conductor aluminum line has low losses.
By sliding the trombones, the antenna may be resonated and matched at any frequency over a range of several octaves, with the bandwidth at a particular setting dependent on the Q of the antenna.
Drawing not to scale Figure 16-29 Trombone tuners for resonant antenna system with balun.
744 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS /at Fy Figm 16-30 {a) Dipole with traps for operation at 2 frequencies separated by an octave {Fz = 2F, ).
fh> Four in-phase 2 elements with phase-reversing coils.
|c) Vertical omnidirectional in-phase 3//4 monopole.
In many wide-bandwidth applications it is not necessary to have a frequency-independent antenna for continuous spectrum coverage but rather an antenna which can operate at spot frequencies. This is possible, for example, with a center-fed dipole by means of tuned traps, as shown in Fig.
16- 30a, each trap consisting of a parallel-tuned LC circuit. At frequency F u for which the dipole is A/2 long, the traps introduce some inductance so that the resonant length of the dipole is reduced.
At twice the frequency F 2 ( = 2 f) the traps are resonant [wL = 1/ojC) and the high impedance they introduce effectively isolates the outer sections of the dipole, the inner part becoming a resonant A/2 dipole at frequency Fz , Thus, in this simple example, the antenna can perform simultaneously as a matched A/2 dipole at 2 frequencies, F 1 and F 2 , separated by an octave. With more traps, operation may be extended to other frequencies. Although in our example the 2 frequencies are harmonically related, this is not a requirement.
Note, however, that the end segment must present a low impedance to the trap to be isolated (i.e., mismatched). In Fig.
16-30a the end segment is ^A/4 long at F 2 so this requirement is met, A coil (or trap) can also act as a 180° phase, shifter as in the collinear array of 4 in-phase A/2 elements in Fig.
1 6- 30b, Here the elements present a high impe-dance to the coil which may be resonated without an external capacitance due to its distributed capacitance. The coil may also be thought of as a coiled-up A/2 element. This 4-element array has a gain of 6.4 dBi as compared to 3.S and 5.3 dBi gain for 2- and 3-element collinear arrays of A/2 elements.
i-13 patch or microstrip antennas 745 Cutting the antenna of Fig, 16-306 at point A and turning the section AB vertical above a ground plane results in the in-phase 3A/4 omnidirectional mono-pole antenna of Fig. 16-30c with 8.3 dBi gain. To match the approximately 150-ft terminal resistance to a 50-to 75-0 coaxial line, a capacitance-inductance L network can be used.
16-12 PATCH OR MICROSTRIP ANTENNAS, These antennas are popular for low-profile applications at frequencies above 100 MHz (A0 < 3 m).
They commonly consist of a rectangular metal patch on a dielectric-coated ground plane (circuit board). A microstrip or patch antenna is shown in Fig.
16-31 with the dielectric substrate material having a typical relative permittivity Er ^2 and thickness r ^ A 0/100. They differ in size and method of feeding from the flush-disc antenna of Fig, 16-12J.
Typical patch dimensions of length F, width W and thickness t are indi-cated in Fig. 16-31 with feed from a coaxial line at the center of the left edge. The horizontal components of the electric fields at the left and right edges are in the same direction, giving in-phase linearly polarized radiation with a maximum broadside to the patch fup.m Fig.
16-316).
The patch acts as a resonant A/2 parallel-plate microstrip transmission line with characteristic impedance equal to the reciprocal of the number n of parallel 746 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS F ringing field Figmc 16-32 View from coax feed side of patch of Fig.
16-31 with gap (or slot) divided into (square) field cells.
field-cell transmission lines.
1 Each field-cell transmission line has a characteristic impedance equal to the intrinsic impedance of the medium where For air, pr — e — 1 and Z, = Z0 = 377 Cl.
An end view of the patch (from the left side) is shown in Fig.
16-32. The cross section (as drawn) has 10 parallel field-cell transmission lines so that for er = 2 the microstrip characteristic impedance Zc is given by riy/s, 10^2 Since n = Wft, a more general relation is - 26.7 Cl = 7= Vt Wyfc This relation neglects fringing of the field at the edges. Since W is typically even much larger than the t value in this example, this effect is small for a patch.
However, for a microstrip transmission line where the ratio Wft is smaller, the fringing effect can be accounted for approximately by adding 2 cells, giving a more accurate formula for microstrip line impedance as ‘ VaRW/0 + 2] The resonant length L of the patch is critical and typically is a couple of percent less than 2/2, where 2 is the wavelength in the dielectric (2 = k0l^J7 f).
Radiation from the patch occurs as though from 2 slots, at the left and right in Fig. 16-31. Let us calculate their impedance for the case where the dielectric substrate is air {iT = 1).
1 J. D, Kraus, Electromagnetic^ 3rd ed., McGraw-Hill, 1984, pp 388, 397.
]fi-J2 PATCH OR MICROSTRIP ANTENNAS 747 From (13-6-10) the impedance Zs of a slot antenna is given by 7 Z^ 35476 where Z0 = intrinsic impedance of empty space = 377 Q = impedance of complementary dipole, O For a slot A long and a few 2/100 wide, the complementary dipole impedance is approximately 700 +jQCl. Thus, the approximate slot impedance is = 50 Cl This is the impedance of a slot in a conducting sheet open on both sides.
The 2/2 microstrip is equivalent to a 2/2 section of transmission line shunted with a radiation (and loss) resistance at each end. The centerpoint of the line is at low potential and c^h be grounded with little effect on operation. Thus, the slot at each end is effectively boxed-in, doubling its impedance. However, the 2 patch slots act in parallel, which halves the impedance resulting in an input resistance R in ^ 50 Q for W 20 , which is a typical value even with er > L if W is smaller, R in increases proportionately The radiation pattern of the patch is broad Typically, the beam area C1A is ^ of a half-space or about tt sr. The resulting patch directivity D is then given by ^ 4tt 4ti = — = — = 4(or6 dBi) (7) n The impedance bandwidth of a patch antenna is usually much narrower than its pattern bandwidth The impedance bandwidth is proportional to the thickness r of the dielectric substrate. Since r is small, the bandwidth is small, typically a few percent.
Although the concept of effective height he may not be appropriate when applied to a patch, it is interesting to evaluate h tf for a typical patch antenna From (2-19-6) the effective height of an antenna is given by K = where Rr = radiation resistance, Cl Ae = effective aperture, 23 Z0 = intrinsic impedance of space, Cl 748 16 ntenns for SPECIAL APPLICATIONS; feeding considerations If we take D = 4 and R. = 50 n for a typical patch we have as its effective aperture D/.l k% A e -= — 4ji n and as its effective height (9) It is interesting that an antenna which extends only V 100 above a flat ground plane has an “ effective height ” 30 times as much.
The dimensions of a patch are not electrically small and at, say, 2 CjHz n = Ujl = 106 mm) a patch is also not physically very small.
Although the above discussion is considerably simplified, it outlines some of the important properties of patch antennas. Many other shapes and configu-rations are possible. For example, for matching purposes the feed point can be moved in from the edge. Patches also lend themselves to microstnp arrays in which the patches are fed by microstrip transmission lines. The configuration ot a 4-patch array with e, = 4 is shown in Fig.
16-33. See also Fig.
1 1-54 showing an array of 896 microstrip elements.
.
The literature on patch or microstrip antennas is extensive. Some^ basic references are those of Munson, 1 Derneryd 1 Carver and Mink and Pozar, Monolithic Microwave Integrated Circuits (MMICs) combine microstrip antennas and associated circuitry in very compact form with applications at fre-quencies from 50 MHz to 100 GHz. 5 Printed -circuit technology is suitable for printing a variety of antenna elements as, for example, dipoles.6 To feed these balanced center-fed elements i R E Munson, “Conformal Microstrip Anlennas and Microslnp Phased Arrays," IEEE Trans.
tt - , 1 A Derneryd, A Theoretical Investigation of the Rectangular Microstnp Antenna Element, IEEE Tram, Ants. Prop., API6, 532-535, July 1978.
5 K R. Carver and J. W. Mink, “ Microstrip Antenna Technology," IEEE Tram. Ants .
Prop., AP-I9, 25-38, January 1981, .
X1 , 4 D. M, Pozar, “An Update on Microstrip Antenna Theory and Design Including Some Novel Feeding Techniques," IEEE Ants. Prop. Sac. Newsletter^ IS, 5-9, October 1986.
3 E, Brookner, ^ Array Radars: An Update," Microwave 30, 134^138, February 1987, and 167-17 , March 1987.
6 D, M. Po/ar, -Considerations for Millimeter Wave Printed Antennas," IEEE Trans. Ants. Prop., 31 , 740^747, September 1983.
„ ...
r E, Levine, J. Ashenasy and D. Treves. “Printed Dipole Arrays on a Cylinder, Microbe J..
30, 85-92, March 1987.
J6-M SUBMERGED ANTENNAS 749 Figure lfc-33 Microstrip array of 4 patches with corporate feed by microstrip transmission lines.
requires 2-conductor microstrip transmission lines, which may be a disadvantage.
However, a dipole element occupies less area than a patch.
It is interesting that the input resistance variation with dielectric thickness f of a center-fedM l dipole is similar to that of a patch whose slot is complementary to the dipole.
16-13 THE HIGH-GAIN OMNI, A high-gain (high -directivity) antenna which is also omnidirectional involves a contradiction. The directivity D of an antenna is given by where Q A = beam area A hypothetical isotropic point source has a beam area = 4ji, making it completely omnidirectional. From (1), its directivity D = 1, which is the lowest possible directivity. For the directivity (or gain of a lossless radiator) to be more than unity requires that the beam area be less than 4a so that the antenna is no longer omnidirectional. Thus, for a simple antenna, high directivity is incompat-ible with an omnidirectional (4a sr) pattern. The combination of both is a theo-retical and a practical impossibility. However, in the digital signal-processing domain of a large phased array the combination is theoretically possible, giving an arbitrarily large number of simultaneous high-directivity beams but, due to inherent losses, not necessarily ones of high gain (see Sec. 11-13).
16-14 SUBMERGED ANTENNAS. In (16-4-1) for the reflection coeffi-cient it is assumed that the wave originates in the less-dense medium (air) and is incident on the earth or ground of relative permittivity sr .
If the wave orig-inates in the denser medium and travels from it into air (16-4-1) becomes where £r = relative permittivity of denser medium 750 IA ANTF.NNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS If i/e r < cos 2 a, p L is complex and |p ± | = 1. Under this condition, the incident wave is totally reflected back into the more-dense medium, 1 When the radical in (1) is zero, p K = I/O'', which defines the critical angle (see Fig. 16-34) ( 2 ) For all angles less than |/>_J = l, and the wave originating in the denser medium is reflected back from the interface.
It may be shown that for a < the electric field in the less-dense medium decays exponentially away from the inter-face (evanescent wave) and propagates without loss along the^interface (surface wave), 2 Figure 16-34 Rays from submerged antenna at angles a below S3. 6 are totally internally reflected.
Rays between S3. 6 and 9CT are transmitted through the surface into air above and spread oul over almost 180'.
1 This is the case for either perpendicular or parallel polarization, 2 J, D. Kraus, Electromagnetics, 3rd ed., McGraw-Hill, 1984, p. 515. For a detailed exposition of antennas in material media, set R. W. P. King and G.
S. Smith (with M. Owens and T. T. Wu), Antennas in Matter,. MIT Press, 1980.
See also P. Delogne, Leaky Feeders and Subsurface Radio Comma meat ion, I EE, London, 1982.
16-M SUBMERGED ANTENNAS 751 Example. A 60-MHz {XQ = 5 m) radio transmitter feeding a monofilar axial-mode helical antenna is situated 20 m below the surface of a freshwater lake with con-stants fir = 1, C r = 80 and a = 1,33 x 10“ 2 U m” 1 The helix axis is vertical. See Fig, 16-34. (a) For a given transmitter power how many turns should the helix have to maximize the signal at an elevation angle of 10° above the surface? (b) What is the relative power density radiated above the surface as a function of the elevation angle above the surface? (c) If the transmitter power is 100 W, what is the signal level at an elevation angle of 10 at a distance of 1 km from the submerged antenna site? The receiving antenna is a George Brown turnstile with reflector elements.
Solution (a) At 60 MHz the loss term of the relative permittivity g 1.33x10" 5 £ = — 4 (Tl r 8.85 x 1(T 12 x 2n 60 x 10 fi and the complex relative permittivity r = r“X = 80-j4 (4) The water is not lossless but c' r p e"\ so neglecting e" we have from (2) that the critical angle is given by "7s-8)581 ISI Thus, all rays from the helix at elevation angles less than 83.58 are reflected back into the water, and only rays at elevation angles greater than 83,58 emerge into the air, as suggested in Fig. 16-34. The hole through which the rays emerge is 1 2.84 1 wide [ = 2 (90 - 83.58)] centered on the zenith (a = 90}.
From Snell's law we have approximately that cos % t a cos % - 8.94 cos y (6) where ac f - elevation angle of ray transmitted above the surface For a T = 10°, we have from (6} that % = 83.68 or just 0,F ( = 83.68 - 83.58) above the critical angle.
For a monofilar axial-mode helical antenna with number of turns n > 3, pitch angle 12° < % < 14 U and circumference 0,8 < C < 1.15, the beam width and directivity from (7-4-4) and (7-4-7) are 52 HPBW ^ = (7) and (8) Taking Cx = 1, a = 12,5 and n = 12, the HPBW = 32.0 and D = 31.7.
1 The half-power angle off axis is given by HPBW/2 = 16.0, as compared to 6,32 1 Do not confuse a here for helix pitch angle with a in (6) for elevation angle.
752 J6 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS ( = 90° — 83,68°) for the ray which emerges at z, = 10°. Thus, with 12 turns the beam is broad enough that at or close to the critical angle the signal level is down from the maximum (on axis) only about 0,3 dB. Doubling the length of the helix (n = 24) increases the directivity from (8) by about 3 dB, but narrows the beam width sufficiently (by l/,y2) that ai the critical angle the signal Level is down about another 0.5 dB, Thus, doubling n results in a net increase in the level of the radiation emerging at an elevation angle a, = 10° above the surface.
However, the narrower beam requires that the helix axis be set within 1 of vertical. Thus, a helix With about 20 turns is a reasonable compromise.
(h) The relative power density 5 of a ray reflected back into the water is given by « -(1Q> The relative power density St transmitted through the surface and emerging above it is then given by S r = 1 -S r -(H) Evaluating (10) and (11) for er = 80 as a function of the elevation angle oc r yields the curve of Fig, 16-35a and the pattern of Fig. 16-356. Both graphs give the power radiated relative to its value at the zenith (a ( = 90°).
(c) Note that the dimensions of the helix are reduced to 1 — 0. 1 1 or 11 percent of the dimensions in air. Thus, the submerged helix diameter D = = 5 m/(v/807t) = 0.178 m.
The loss component of the permittivity ^ = 4 and since e," <£ e the attenu-ation constant for the water is given by 1 It.
e" 71 4 5 Jm and the attenuation in the water path d by Attenuation (water) = 20 Log 0.281 Np m” 1 20 log e" 0 - 2 1 10 = 49 dB The loss at the interface is given from (LI) for elevation angle ot (water) = 83.68° as 10 log 0.51 = 3 dB.
1 Do not confuse a here as the attenuation constant ( — real part of propagation constant y) with a for helix pitch angle or a for elevation angle. Unfortunately, the English and Greek alphabets do not have enough letters to go around-tfi-l SUBMERGED ANTENNAS 753 Figure 16-35 (a) Ratio of power density at elevation angle or, to power density at zenith (90°) for waves transmitted above surface of lake from submerged transmitter (/>) Field pattern or radiation transmitted above surface in polar coordinates.
The directivity of the 20-turn helix is given by D- 12ClnSA 12 x 20 x 0.22 = 52.8 and its effective aperture by , D/& 52.8 x 5 2 A, = —— — — — = 105 m 2 4te An A George Brown turnstile (Sec.
16-7) with reflector elements has a directivity D ^ 3 and effective aperture at = 5 m of 6 m2 . Taking the distance between transmitter and receiver as 1 km we have from the Frits transmission formula that the received power less water attenuation and surface (interface) Loss is P tA rt Aer _ 100 x 105 x3 r lOOO2 x 5 2 125 The water attenuation and surface (interface) Loss total 52 dB ( = 49 -I- 3) so the net (actual) received power is given by Pr = 1,25 x t0“ 3/antilog 5.2 = 8 x 10“ 9 W = 8 nW or a bit less than 1 mV on a 100-ti transmission line.
The transmitted energy-carrying field emerging above the surface goes to zero at the surface (elevation angle i f = 0°), as indicated in Fig. 16-35u and 6. There is 754 I ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS also a reactive wave traveling out along (above) the surface that accompanies the totally internally reflected wave. However, it is a reactive wave and carries no enercv (E and H are in time-phase quadrature).
Its fields decay exponentially with height above the surface (being simply the matching fields at the surface of the reflected-wave fields below) and are called evanescent fields. Due to the difference in transmission of the parallel and perpendicular field components through the surface, the received wave is not necessarily circularly polarized (AR 1). However, the receiving antenna should be circularly polarized and of the same hand as the trans-mitting antenna, an implicit assumption made in solving the example problem.
It is also assumed in the example that the water surface is smooth. If it is no smooth, as under windy conditions, the situation is different and fluctuations (noise) will occur in the received signal.
The problem of electromagnetic wave transmission from water into air is analogous to that of transmission through the earth’s ionosphere to an extrater-restrial point above it.
In each case, refraction is from a medium with lower velocity to one with higher velocity of wave propagation with ray-bending a total internal reflection at small enough elevation angles. (See Probs. 17-18 and l7 '19 iowever there are some fundamental differences in that the ionosphere is inhomogeneous and anisotropic (a magnetized plasma) with polanzatton changes by Faraday rotation. These effects make circularly polarized antennas essential.
We note that the radio situation differs from the optica one in that a _ for water at radio wavelengths but er - 1.75 for water for light waves the critical angle «c = 41° as compared to 84 for radio waves.
16-15 SURFACE-WAVE AND LEAKY-WAVE ANTENNAS.
Traveling-wave antennas discussed in previous chapters include the monofilar axial-mode helix, long-wire and polyrod antennas. Surface-wave and leaky-wave antennas are also traveling-wave antennas but are ones which are adapted to flush or low-profile installations as on the skin of high-speed a.rcraff Typically, their bandwidth is narrow (10 percent) and their gam is moderated 15 dBifi Consider the plane boundary between air and a perfect conductor as shown in Fig, 16-36a with a vertically polarized plane TEM wave traveling io the right along the boundary. From the boundary condition that the tangential com-ponent of the electric field vanishes at the surface of a perfect conductor, the electric field of a TEM wave traveling parallel to the boundary must be exact y normal to the boundary, or vertical, as in Fig. 16-36o. However ff the conduct.v-ity ff of the conductor is not infinite, there will be a tangential electric field Ex a the boundary, as well as the vertical component ET , so that the total field will have a forward tilt as in Fig. 16-36f>.
The direction and magnitude of the power flow per unit- area are given by the Poynting vector with average value S if = | Re E x H (W m“ 2 ) (1) I6-J5 SURFACE-WAVE AND LEARY-J^AYE ANTENNAS 755 Medium 1 TE Direction of _ energy How Medium 1 lair) Direction of energy flow Medium 2 (perfect conductor) cr= Medium 2 (conductor witH finite conductivity a) Figure 16-36 Vertically polarized wave traveling to right (a) along surface of a perfectly conducting medium and (A) along surface of medium with finite conductivity.
At the conducting surface the average power into the conductor ( — y direction) is Sy = Re EX H (2) This is power dissipated as heat in the conductor.
The space relation of Ex , Hz (or H) and Sy is shown in Fig. 16-37a, Since H,' Z-01 where Zc = intrinsic impedance of the conductor then from (2) the power flow into the conductor can be written as Re Zc = Re Zc (4) where Hz = H z0 e~ j H = H z0 et +yx = complex conjugate of H z £ = phase lag of Hx with respect to Ex y = propagation constant = a + jp The power flow parallel to the surface { x direction) is Sx = fRe£,J/f (Wm‘ 2 ) (5) Medium 1 (air) Medium 2 Figwe 16-37 Fields and Poynting vector at the surface of a conducting medium with a vertically polarized wave traveling parallel to the surface.
si 756 Iti ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS The space relation of Ey , H z and S, is shown in Fig.
16-37f>. Since ^ = Z, U where Z s = intrinsic impedance of the dielectric medium (air) above the conduc-tor | .
then from (5) the power flow along the conductor (x-ditection) can be written S, = & Re Z, (W m _I ) < 7> The total average Poynting vector is then Sav = S, + yS, = (i Re z, - y Re ZJ (8) Figure 16-37c shows the relation of Sav to its x and y components. It is apparent that the average power flow is not parallel to the surface but downward at an angle t. This angle is the same as the forward tilt angle of the electric field (see Fig.
16-36i>).
If the conductivity of the conductor is infinite (a = a>), the tilt is zero (t = 0).
From (8) the tilt angle _i Re Zc 1 “ Re Zd Example 1. Find the tilt angle t for a vertically polarized 3 GHz wave traveling in air along a flat copper sheet.
Solution. At 3 GHz, we have for copper that Re Z c = 14.4 mil The intrinsic imped-ance of air is real and independent of frequency ( = 377 Q) From (9) we have 144 9— = 0 .0022° ( 10) Although t is very small it is not zero, indicating some power flow into the copper sheet.
If o is small or if the conductor is replaced by a dielectric, t may amount to a few degrees. In the Beverage antenna (Sec. Il-I6c) the honzonta field component (Ex> is parallel to the antenna wire and induces emf s in it.
Example % Find the forward tilt angle t for a vertically polarized 3-GHz wave traveling in air along the smooth surface of a freshwater lalte.
Solution. At 3 GHz the conduction current of fresh water is negligible compared with the displacement current « 4 e(l so the water may be regarded as a dielectric medium with e, = e, = 80, Thus, from (9), r = tan 1 —•== — 6,4° v'SS 16-15 SURFACE-WAVE AND LEAKY-WAVE ANTENNAS 757 Figure 16-38 (a) Magnitude variation with time of Ey and Ex components of E in air at the surface of a copper region for a 3-GHz TEM wave traveling parallel to the surface, (b) Resultant values of E (space vector) at 22.5' intervals over one cycle, illustrating elliptical cross-field at the surface of the copper region. The wave is traveling to the right. Note that although E has x and y components (cross-field) it is linearly polarized as seen from the .v direction. (From J. D. Kraus Electromagnetics, 3rd ttt. McGw-Hili 1984.) The tilt angle i calculated above is an average value. In general, its instan-taneous direction varies with time. For the wave traveling along the copper sheet, £y and Ex are in phase octature (45° phase difference), so that at one instant the total field E may be in the x direction and ^-period later in the y direction. For the 3-GHz wave of Example 1, Ex and Ey vary with time as in Fig, 16-38a. The locus of the tip of E describes a cross-field ellipse as in Fig, 16-386, with positions shown as a function of time (cuf). The ellipse is not to scale, the Ex values being magnified by 5000, The variation of the instantaneous Poynting vector is shown in Fig, 16-39 with ordinate values magnified by 5000, It is of note that the tip of the Poynting vector travels around the ellipse twice per cycle.
Whereas copper has a complex intrinsic impedance, fresh water at 3 GHz (as in Example 2) has a real intrinsic impedance so that Ex and are essentially in phase and the cross-field ellipse collapses to a straight line with a forward tilt of 6.4°.
The forward tilt of E or downward tilt of S in the above examples tend to make the wave energy hug the surface, resulting in a bound ware or surface wave.
The phase velocity v of this wave is less than the velocity c of light, i.e„ it is a slow wave (r < c). To initiate a wave along the surface, a launching device, such as a 758 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS Figure 16-39 Poynting vector in air at a point on the surface of a copper region for a 3-GHz TEM wave traveling along the surface (to right). The Poynting vector is shown at 22.5'’ intervals over a 1 -cycle. The ordinate values are magnified 5000 times compared with the abscissa values. The tip ot the Poynting vector travels around the ellipse twice per cycle. {From J. D. Kraus, Electromagnetics.
,3rd ed., McGraw-Hill y f9$4.) horn, with a height of several wavelengths can be used as in Figs.
16-40 and 16-41.
.
, . In 1899 Arnold Sommerfeld 1 showed that a wave could be guided by a round wire of finite conductivity. Jonathan Zenneck 1 pointed out that for similar reasons a wave traveling along the earth's surface would tend to be guided by it.
The guiding action of a flat conducting surface can be enhanced by adding metal corrugations. These and the horn launcher in Fig. 16-40 form a surface-wave antenna. The corrugations have many teeth per wavelength (s < A/5). The slots between the teeth support a TEM wave involving E x , which travels up and down the depth of the slot (a standing wave). Each slot acts like a short-circuited section of a parallel-plane 2-conductor transmission line of depth d. Assuming lossless materials, the impedance Z presented to a wave traveling down the slots is a pure reactance given by (ID C2) Conductor Figure 16-40 Corrugated surface-wave antenna with horn wave-launcher.
1 A. Sommerfeld, “ Fortpflanzung elektrodynamischer Wellen an einem zylmdrischen Leiter, Ann.
Phys u. Chem„ 67, 233, December 1899, 3 J. Zenneck, “ Ober die Fortpflanzung ebener elektromagnetischeT Wellen langs einer ebenen Leiter-flache und ihre Beziehung zur drahtlosen Telegraphic," Annt Phys. t 23, 846-866, 1907.
16-15 SURFACE-WAVE AND LEAKY-WAVE ANTENNAS 759 i Wave Conductor Figure 16-41 Dielectric-slab surface-wave antenna with horn wave-launcher.
where Zd = intrinsic impedance of medium filling the slots, Q £r = relative permittivity of the medium, dimensionless x0 = free-space wavelength, m d = depth of slots, m For air-filled slots (Zd = 377 and er = 1), {12) reduces to Z jllOn tan (fi) 03) The slots store energy from the passing wave. When d < 20/4, the plane along the top of the teeth is inductively reactive. When d = a 0/4, Z = co and the plane along the top is like an open circuit (nothing below), while when d = a0/2, Z — 0 and the plane appears tike the conducting sheet below it (a short circuit).
The guiding action of a flat conducting surface can also be enhanced by adding a dielectric coating or slab of thickness d With a launcher, as in Fig.
16-41, the combination forms another type of surface-wave antenna. The electric field is vertically polarized but has a small forward tilt at the dielectric surface.
For a sufficient thickness d y the fields attenuate perpendicular to the surface (y direction) as e -a,, T where (Np m x ) (14) For Er = 2, the attenuation is over 50 dB Thus, the fields are confined close to the surface.
Corrugated and dielectric slab surface-wave antennas, as in’ Figs. 16-40 and 16-41, with a length of several k and a width of \k or more (into page), produce end-fire beams with gain proportional to their length and width.
It is assumed that the conducting surface (ground plane) extends beyond the end of the corru-gations or slab. If not, the beam direction tends to be off end-fire (elevated). For optimum patterns, the depth of the slots or thickness of the slab may be tapered at both ends.
Although surface-wave antennas take many forms, the ones described above are typical. They are traveling-wave antennas carrying a bound wave with the energy flowing above the guiding surface and with velocity v < c (slow wave). E would be perpendicular to the surface except for its forward tilt.
760 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS Beam Figur 16-42 Open-top waveguide antenna with continuous energy leakage Leaky-wave antennas are also traveling-wave types but ones in which radi-ating energy leaks continuously or periodically from along the length of the guiding structure, with most of the energy flow wtt/tm the structure. Typically, but not necessarily, the structure carries a fast mwe {v > c). A hollow metal wave-guide is an example. With one wall removed, energy can leak out continuously all along the guide.
.
A leaky-waveguide antenna of this type is shown in Fig.
16-42. bince tne wave velocity v in the guide is faster than the velocity c of light, the radiation forms a beam inclined at an angle tj> with the guide as given by — cos ” 1 -<>5) V For t> = 1.5c, 4> = 48°. Since r is a function of the frequency, the beam angle (p may be scanned by a change in frequency.
A leaky-wave antenna may also be constructed by cutting holes or slots at a regular spacing along the waveguide wall as in Fig. 16-43 (see also the slotted waveguide of Fig. 13-5). Leakage may be controlled by the slot or hole size. For a slot or hole perimeter of -1/, leakage is large but decreases rapidly with a decrease in perimeter. This periodic structure radiates at a beam angle tj> given from (7-11-8) by where s — hole or slot spacing, m A0 = free-space wavelength, m X9 = wavelength in guide, m m = mode number, 0, ± ± 1, Example. Find the beam angle $ for = 1-5J.0 , s = and m = — 1.
Solution.
FrorrUlb), $ = cos -1 jj "5 ” 109 513 back-fire direction) 16 IS SURFACE-WAVE AND LEAKY-WAVE ANTENNAS 761 Figure 16-43 Slotted-waveguide antenna with leakage at periodic (discrete) intervals.
If the slots are spaced Ijl instead of 1/, proper phasing requires that alter-nate ones be placed on opposite sides of the centerline. For this case m = — \ in (16). The beam angle 4> may be scanned by a change in frequency. Although described here as a leaky-wave antenna, the antenna of Fig. 16-43 may also be considered as simply an array of waveguide slots.
Consider now a dielectric slab waveguide of thickness d with wave injected at small enough e (below the critical angle) so that the wave is totally internally reflected and propagates by multiple reflections inside the slab. The situation is similar to that for the submerged radio transmitter in Sec.
16-14, except that here the denser medium has upper and lower boundaries and a thickness of the order of 1A0 , Although the energy is transmitted inside the slab, fields exist above and below. These, however, are evanescent and decay exponentially away from the slab. They convey no energy. (Recall the evanescent wave above the water surface in Sec, 16-14.) It might be supposed that any wave injected into the slab at an angle below the critical value will propagate, but, because of multiple -re flection interference, waves will only propagate at certain angles (eigenvalues).
1 Figure 16-44 is an end view of a dielectric slab waveguide carrying a wave with electric field parallel to Evanescent Figure 16-44 End view of dielectric slab waveguide with E parallel to the slab surfaces. Propagation is into the page. Arrows and graph (at left) indicate variation in magnitude of E across the thickness d of the slab. The field outside the slab is evanescent and transmits no energy.
1 J. D. Kraus. Electromagnetics 3rd ed., McGraw-Hill, 1984, p, 590.
D, Marcuse, Theory of Dielectric Optical Waveguides, Academic Press, 1974.
762 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS the surfaces (perpendicular to the plane of incidence).
In the notation of Sec.
16-14 this is an E ± field. Propagation is into the page. The evanescent field above and below the slab matches the field inside but carries no energy. However, dis-continuities on the slab surface can cause energy leakage from inside and radi-ation. Thus, a dielectric slab with surface discontinuities can act as a leaky-wave antenna.
A dielectric cylinder can also serve as a guide in the same manner as a siao.
At or near optical wavelengths the cylinder diameter can be physically small or threadlike. Such guides, or optical fibers, consist typically of a transparent glass core surrounded by a glass sheath or cladding of slightly lower index of refraction, with the sheath enclosed in an opaque protective jacket. A typical core fiber 25 pm in diameter is as fine as a human hair and can carry one thousand 2-way voice communication channels with an attenuation as small as 1 dB km at light or infrared wavelengths (^ to 1 pm).
The literature on surface- and leaky-wave antennas is extensive. Summary treatments are given by Zucker and by Mittra.
1 16-16 ANTENNA DESIGN CONSIDERATIONS FOR SATELLITE COMMUNICATION In 1945, while a radar officer in the Royal Air Force, Arthur C. Clarke 2 published an article in Wireless World in which he proposed the use of artificial satellites and, in particular, those in a geostationary orbit, as a solution to the world's communication problem. The idea then seemed far fetched to many but 12 years later Sputnik went up and only 6 years after that, in 1963, the first successful geostationary satellite, Syncom 2, was put into orbit and Clarke's proposal became a reality. Now there are hundreds of satellites 3 in the geostationary or Clarke orbit with more being added at frequent intervals. The satellites move with the earth as though attached to it so that from the earth each appears to remain stationary above a fixed point on the equator. These satellites form a ring around the earth at a height of 36000 km above the equator, putting the earth in the company of other ringed planets, Saturn and Uranus, with the difference that the earth's ring is man-made and not natural. Arthur Clarke has observed that the ancient superstition that our destinies are controlled by celes-tial bodies has at last come true, except that the bodies are ones we have put up there ourselves.
Having a transmitting (transponder) antenna on a Clarke-orbit satellite is like having it on an invisible tower 36000 km high. Since earth-station dishes look upward at the sky, ground reflections are eliminated but the ground is still 1 F.
J Zucker (chap.
12) and R. Mittra (chap.
10), in Amelina Engineerinfl Handbook, McGraw-Hill, 1984.
1 A. C. Clarke, +l Extra-terrestrial Relays," Ifiretess World, 51, 305-308. October 1945, 3 Nearly 500 at the end of 1986.
IfMS ANTENNA DESIGN CONSIDERATIONS FOR SATELLITE COMMUNICATION 763 involved through its effect on the antenna noise temperature via side and back lobes.
The spacing of satellites in the Clarke orbit is closely connected with the design of both satellite and earth-station antennas. Thus, if beam widths and sideiobe levels are reduced, the spacing can be reduced and more satellites accommodated in orbit.
The full orbit utilization problem involves not only satellite placement and spacing but also the available spectrum. Thus, satellites may be parked closer together than beam widths would allow if they operate in different frequency bands. However, both orbit space and spectrum space are limited resources which must be allocated in an optimum manner to meet the tremendous international demand for satellite communications. The problem of maximizing access to the Clarke orbit is under study and planning by the satellite orbit-use (ORB) sessions of the World Administrative Radio Conference (WARC), with its first session (ORB-1) in 1985 and a second session (ORB-2) scheduled for 1988.
1 To permit closer spacings than otherwise, North American satellites sharing the same frequencies use an alternate linear-polarization technique. For example, the even-numbered channels of a satellite may be polarized parallel to the orbit, with odd-numbered channels perpendicular to it, while the adjacent satellites on either side have the polarizations reversed (odd-numbered parallel and even-numbered perpendicular to the orbit). Some satellites serving other parts of the world use circular polarizations of opposite hand to do the same thing. Thus, the response level of the earth-station antenna to the opposite state of polarization is a factor.
It is necessary that the earth-station antenna have gain and pattern proper-ties capable of providing a satisfactory S/N ratio with respect to noise and inter-ference sources (as from adjacent satellites). The satellite transponder power, antenna gain and pattern are also factors.
At the Clarke-orbit height of 36000 km, a satellite spacing of V amounts to a physical separation of 630 km.
Although each satellite wanders about its assigned orbital location (usually in a systematic manner), this “ station-keeping motion is required by regulation to be less than 60 km (0.1°) so that spacing even closer than 1° is possible without danger of satellites colliding. 2 Some of the antenna design aspects may be illustrated by an example.
Example, Determine the required parabolic dish diameter of a 4-GHz (C-band) earth-station antenna if its system temperature = 100 K for an S/N ratio of 20 dB J H. J. Weiss, w Maximizing Access to the Geostationary-Satellite Orbit," ITU J., 53, 469-477, August 1986, 1 The satellite antenna pointing accuracy in roll, pitch and yaw are also usually of the order of 0,1 c or less.
Roll, pitch and yaw are of primary concern in the satellite antenna design but of secondary importance in the earth-station antenna design. The Astra satellites have specifications of 0,06“ roll, 0.07° pitch, 0,22“ yaw and ±0.05° station keeping 764 IS ANTENNAS FOR SPECIAL APPLICATIONS; FEEDING CONSIDERATIONS and bandwidth = 30 MHz with satellite transponder power = 5W, satellite para-bolic dish diameter = 2m and spacing between satellites = T .
Solution. From the Friis transmission formula and the Nyquist noise-power equa-tion, we have from (17-3-9) that the S/N (actually C/N or carrier- to-noise) ratio for isotropic antennas and a transmitter power of 1 W is given by N \6n 2 r2kT %ys B where k = wavelength r = downlink distance, 36000 km k — Boltzmann’s constant = 1,38 x 10“ 23 J K 7^ = system temperature, K B = bandwidth, Hz Introducing the indicated values in (1) we obtain —61.8 dB (2) N for the downlink at 1 W isotropic.
For the D = 2 m diameter satellite dish at 50 percent aperture efficiency, the antenna gain C, is given by G, = - 35.5 dB (3) where Ac -^ — 1-6 m 2 Jl = 0.075 m The transponder power of 5 W gives an additional 7 dB ( = log 5) gain. The required earth-station antenna gain G E must then be sufficient to make the system S/N ratio ^20dB,or Thus, the required earth-station effective aperture - 3.8 m 2 At an assumed aperture efficiency of 50 percent the required physical aperture A p is twice this, or 7,6 m 2 , making the required diameter of the earth-station antenna D e — 2 j— = 3,1 m This diameter meets the S/N ratio requirement. To determine if it also meets the T spacing requirement we should know the illumination taper across the dish aper-16-16 ANTENNA DESIGN CONSIDERATIONS FOR SATELLITE COMMUNICATION 765 r 0‘ 1 UL-j “l “ Variation Sate Elite positions (nominaf) Figure 16-45 Pattern of C-band earth-station dish for worked example.
Pattern is shown with respect to satellite positions at 2 D spacing in the Clarke orbit.
ture. This is not given. However, we assumed 50 percent aperture efficiency which implies substantial taper. Accordingly, the beam widths may be estimated as HPBW ~ ^ = 1-6“ Dx 3.1 m/0.075 m and 145° BWFN i 3.5° Dt (8) Thus, first nulls or minima are approximately 1.75“ either side of the satellite or 0.25° less than the T spacing of the adjacent satellites, as indicated in Fig.
1645.
766 l ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS The first sidelobe level of a tapered circular aperture should be at least 20 dB down and since the adjacent satellites are closer to the first nulls or minima than the first sidelobe maxima, the adjacent satellite levels should be at least 25 dB down, which could make the total noise level about 18 dB down. Although this may not be satisfactory, the use of an opposite state of polarization on the adjacent satellites should reduce their interfering signal level to an acceptable level provided the earth-station response to cross- polarization is small.
The station-keeping accuracy of the satellites and the pointing error of the earth-station antenna are also factors.
Since these may vary independently, the effects will be random and must be assessed by statistical methods giving upper and lower limits to the overall S/N ratio.
Assuming an earth-station antenna pointing error of ±0.1°, the same as the required satellite station -keeping tolerance, means that the satellite position may vary a maximum of ±0.2 C with respect to the first null. This is indicated in Fig.
16-45. We note that at one extreme the satellite is almost at the null but at the other extreme is only about 4 dB below the first sidelobe maximum. These and other factors are discussed by Jansky and Jeruchim.
1 Although we have made a number of assumptions, the example illustrates many of the important factors involved in determining suitable dimensions and pattern characteristics of an earth-station antenna. (See Fig. 12-58.) 16-17 RECEIVING VERSUS TRANSMITTING CONSIDER-ATIONS According to the principle of reciprocity, the field pattern of an antenna is the same for reception as for transmission However, it does not always follow that because a particular antenna is desirable for a given transmit-ting application it is also desirable for reception. In transmission the main objec-tive is usually to obtain the largest field intensity possible at the point or points of reception. To this end, high efficiency and gain are desirable. In reception, on the other hand, the primary requirement is usually a large signal-to-noise ratio Thus, although high efficiency and also gain may be desirable, they are important only insofar as they improve the signal-to-noise ratio. As an example, a receiving antenna with the pattern of Fig, 16-46n may be preferable to a higher-gain antenna with the more directional pattern of Fig. 16-46b, if there is an interfering signal or noise arriving from the back direction as indicated Although the gain of the antenna with the pattern at (a) is less, it may provide a higher signal-to-noise ratio since its pattern has a null directed toward the source of the noise or inter-ference (see Sec.
1 1 - 1 3 on adaptive arrays).
However, by way of contrast, suppose that circuit noise in the receiver is the 1 D. M. Jansky and M. C. Jeruchim, L'om/nurttcuN'on Satellites in rhe Geosfaftomio' Orbjt y Artech House 1983.
bandwidth considerations 767 Figure 16-46 Low-gain antenna with high front-to-back ratio at (a) is better than higher-gain antenna with lower froni-lo^back ratio at (hj for reception in the presence of a back-side interfering signal.
However for transmission the higher-gain antenna at (b) is superior.
limiting factor. Then high antenna gain and efficiency would be important in order to raise the signal-to-noise ratio.
In direction-finding antennas the directional characteristic of the antenna is employed to determine the direction of arrival of the radio wave If the signal-to-noise ratio is high, a null in the field pattern may be used to find the direction of arrival. With a low signal-to-noise ratio, however, the maximum of the main lobe may provide a more satisfactory indication 1 16-18 BANDWIDTH CONSIDERATIONS. The useful bandwidth of an antenna depends, in general, on both its pattern and impedance characteristics 2 In thin dipole antennas the bandwidth is usually determined by the impedance variation since the pattern changes less rapidly.
3 However, with very thick cylin-drical antennas or biconical antennas of considerable cone an^le, the impedance characteristics may be satisfactory over so wide a bandwidth that the pattern variation determines one or both of the frequency limits. The pattern may also determine the useful bandwidth of horn antennas, metal-plate lens antennas or zoned lens antennas If the acceptable bandwidth for pattern exceeds that for impedance, the bandwidth can be arbitrarily specified by the frequency limits F t and F2 at which the VSWR on the transmission line exceeds an acceptable value What is accept-able varies widely depending on the application. In some cases the VSWR must be close to unity. In others it may be as high as 10 to 1 or higher. The bandwidth 1 A. Alford J. D. Kraus and E. C. Barkofsky, in Very Rigk Frequency Techniques, Radio Research Laboratory Staff, McGraw-Hill, New York 1947, chap. 9.
1 The pattern is understood to include polarization characteristics.
3 A dipole a/2 long has a half-power beam width of 78°. If the frequency is reduced so that the dipole length approaches an infinitesimal fraction of a wavelength, the beam width only increases from 78 to 90', while if the frequency is doubled so that the dipole is U long the beam width decreases from 78 to about 47^ 768 16 ANTENNAS FOR SPECIAL APPLICATIONS: FEEDING CONSIDERATIONS can be specified as the ratio of F 2 - f\ to F0 (the center or design frequency) or in percent as x 100 Another definition is the simple ratio FJF X (or FJF t to 1) where F 2 > F v The bandwidth due to the impedance can also be specified (if the bandwidth is small) in terms of its reciprocal or Q at F0 , where Q = 2n total energy stored by antenna energy dissipated or radiated per cycle 16-19 GRAVITY-WAVE ANTENNAS. ROTATING BOOM FOR TRANSMITTING AND WEBER BAR FOR RECEIVING.
Gravity, which we frequently think of as a force which holds us to our seat in a chair, is a manifestation of Albert Einstein's theory of general relativity in which space, time and matter are related in a geometric concept.
1 Although accelerating masses may, in theory, radiate gravity waves (analogous to the electromagnetic waves radiated by accelerated charges), gravitational radiation is a smaller-order effect and tends to be weak. 2 Consider, for example, the gravitational radiation from a uniform solid steel bar rotating end-over-end as a gravitywave transmitting antenna. According to Misner, Thorne and Wheeler 3 in their classic book Gravitation, the bar will radiate a power 1.5 x 10“ 54M 2 Loj6 (W) (1) where M = mass of bar, kg L = length of bar, m to = angular velocity, rad s' 1 If the bar’s mass is 500 tonnes (M = 5 x 10 s kg), is 20 m long (L = 20 m) and rotates end-over-end 270 times per minute, which is about as fast as the tensile strength of the bar will permit (a> = 2n x 270/60 rad s 1 ). we have, intro-ducing these values in (1), that the very small power p = 2.2 x 10 -21 W (2) is radiated in gravity waves.
1 A. Einstein, "Zut allgemeinen ReLativilatstheorie/' Preuss. Akod, 47, 778-7B6, 799-801. 1915, 1 Electric charges, being of two signs (positive and negative), can form dipoles but masses, being of only one sign, cannot. Thus, gravitational dipole radiation is not possible but gravitational quadru-polar radiation is.
1 C. W, Misner, K, S, Thome and J, A. Wheder, Gravitation, Freeman, 1973, 16-19 GRAVITY-WAVE ANTF.NNAS 769 Transmitting Receiving system system and Weber-bar receiving antenna. The Weber bar should be suspended in a cushioned, evacuated.
refrigerated tank (not shown).
Although the above example shows that gravity-wave radiation is very inef-ficient and weak, let us design a hypothetical gravity-wave communication link.
Instead of a uniform solid bar as the transmitting antenna let us use a pair of moveable weights on a rotating boom as in Fig.
16-47. The rotating boom idles with the weights at the ends. With weights drawn in, the boom speeds up and sends out waves of a higher frequency to which a Weber-bar receiving antenna is tuned. With the position of the weights under the control of a telegraph key, we have a frequency- shift CW system with the transmitted gravity-wave frequency equal to twice the boom rotation rate. The velocity of propagation of the gravity waves is presumed to be equal to the velocity of light.
Turning next to the receiving system, let us use an antenna similar to the one developed by the pioneer gravity-wave scientist Professor Joseph Weber of the University of Maryland.
Tt consists of a several-tonne aluminum bar, or Weber-bar antenna, suspended at its midpoint and isolated from sound and vibration by enclosing it in a refrigerated, cushioned, evacuated tank (Fig, 16-47).
A passing gravity wave should make the bar vibrate as though tapped by a tiny hammer. The vibration of the bar generates electrical signals in strain sensors attached to the bar. These signals are then amplified and recorded.
For optimum performance the bar's natural frequency should be the same as the frequency of the gravity waves from the rotating transmitting antenna. The Weber-bar antenna should also be broadside to the wave direction. The sensi-770 Iti ANTENNAS FOR SPECIAL APPLICATIONS: feeding considerations tivity of the Weber-bar antenna is proportional to its mass and its stiffness (or Q) and inversely proportional to its temperature.
As of the present date (1987) no gravity waves have been detected for certain although a number of very sensitive systems are in operation.
Indirect evidence of their existence comes from measurements of rotating double (binary) star systems where the measured rate of decay or slow-down of the orbital period is found to agree closely with the energy loss and decay expected from gravity-wave radiation.
Gravity waves are now (1987) in a status between their theoretical postu-lation by Einstein and their detection and demonstration, similar to the status of radio waves in the years after Maxwell published his theory but before Hertz produced and measured them.
Additional ^cussions of gravity-wave detection are given by Kraus.
1 PROBLEMS 2 16-1 VSWR for dlpok antenna.
Calculate the VSWR on a 65-D line connected to the L/D = 60 dipole antenna or Fig. 16-22/ over a 30 percent bandwidth if an open-ended line of 40 ft characteristic impedance is connected in parallel with the antenna terminals. The line is 180" long at the center frequency F0 1 6-2 Stub impedance.
(a) What is the terminal impedance of a ground-plane mounted stub antenna fed with a 50-0 air-filled coaxial line if the VSWR on the line is 2.5 and the first voltage minimum is 0.17a from the terminals?
(b) Design a transformer so that the VSWR = 1.
16-3 Square loop. Calculate and plot the far-field pattern in the plane of a loop antenna consisting of four A/2 center-fed dipoles with sinusoidal current distribution arranged to form a square A/2 on a side. The dipoles are all in phase around the square.
16-4 Triangular loop. Calculate and plot the far-field pattern in the plane of a loop antenna consisting of three A/2 center-fed dipoles with sinusoidal current distribu-tion arranged to form an equilateral triangle A/2 on a side. The dipoles are all in phase around the triangle.
16-5 Microstrip line. For a polystyrene substrate (s, = 2.7) what width-substrate thick-ness ratio results in a 50-0 microstrip transmission line?
16-6 Surface-wave powers. A 100-MHz wave is traveling parallel to a copper sheet (| Zc | = 3.7 x 10“ ) with E ( = 100 V m“ 1 rms) perpendicular to the sheet Find { a ) the Poynting vector (watts per square meter) parallel to sheet and (b) the Poynting vector into the sheet.
1 J. D. Kraus, Radio Astronomy, 2nd ed, T Cygnus-Quasar, 1986.
pp, 9-29, 9-33; J. D. Kraus, Our Cosmic Universe Cygnus-Quasar, 1980, pp. 211-214, 225-3 Answers to starred () problems are given in App. D.
problems 77 1 16-7 Surface-wave powers. A 100-MHz wave is traveling parallel to a conducting sheet for which \ Z e \ — 0.02 O. If E is perpendicular to the sheet and equal to 150 V m“ 1 (rms) 5 find (a) watts per square meter traveling parallel to the sheet and (b) watts per square meter into the sheet.
16-8 Surface-wave power. A plane 3-GHz wave in air is traveling parallel to the bound-ary of a conducting medium with H parallel to the boundary. The constants for the conducting medium are a = K/ U m“ l and = Ef -1.
If the traveling-wave rms electric field E = 75 mV m“ \ find the average power per unit area lost in the conducting medium.
16-9 Surface-wave current sheet. A TEM wave is traveling in air parallel to the plane boundary of a conducting medium. Show that if K = p s v t where K is the sheet-current density in amperes per meter, p s is the surface charge density in coulombs per square meter and v the velocity of the wave in meters per second, it follows that K = H, where H is the magnitude of the H field of the wave.
16-10 Coated-surface wave power. Show that for a dielectric-coated conductor, as in Fig. 16-41, the ratio of the power transmitted in the dielectric P4 to the power transmitted in the air Pa is given by where d is the thickness of the dielectric coating.
16-11 Coated-surface wave cutoff. A perfectly conducting flat sheet oflarge extent has a dielectric coating (e, = 3) of thickness d = 5 mm. Find the cutoff frequency for the TM 0 (dominant) mode and its attenuation per unit distance.
16-12 Lunar communication by surface wave. Discuss the possibilities of using dielectric-slab surface-wave modes for radio communication around the moon over long distances (1000 km or more). Note that the moon has no ionosphere.
See, for example, W. W, Salisbury and D, L. Fernald, " Postocculation Reception of Lunar Ship Endeavour Radio Transmission/ Nature, 234, 95, Nov.
12, 1971; also A. F.
Wickersham, Jr., Generation, Detection and Propagation on the Earth of HF and VHF Radio Surface Waves/ Nature, 230, 125-130, Apr 5, I97L 16-13 Horizontal dipole above ground. A thin A/2 dipole is parallel to a flat, perfectly conducting ground at a height h above it.
(a) Calculate and plot the gain of the dipole in the zenith direction as a function of height h for heights from zero to A.
Express the gain with respect to a A/2 dipole in free space. Assume zero losses, (b) Repeat (o) for dipole loss resistance R L = 1 13.
16-14 Overbad TV for HP, VP and CP.
(a) A typical overland microwave communication circuit for AM, FM or TV between a transmitter on a tall building and a distant receiver involves 2 paths of transmission, one a direct path (length r0) and one an indirect path with ground reflection (length r l + r2 ), as suggested in Fig. P16-14. Let = 300 m and d = 5 km. For a frequency of 100 MHz calculate the ratio of the power received per unit area to the transmitted power as a function of the height h 2 of the receiving antenna.
Plot these results in decibels as abscissa versus h 2 as ordinate for 3 cases with transmitting and receiving antennas both ( 1 ) vertically polarized, (2) horizontally polarized and (3) right-circular] y polarized for h 7 values from 0 to 100 m. Assume that the transmitting antenna is isotropic and 772 I ANTENNAS FOR SPECIAL A PPLICATJDNS: FEEDINfi CONSIDERATIONS Transmjttrog antenna Figure P16-14 Overland microwave communication circuit that the receiving antennas are also isotropic (all have the same effective aperture). Consider that the ground is flat and perfectly conducting, {/?) Compare the results for the 3 types of polarization and show that circular polarization is best from the standpoint of both the noncriticalness of the height h l and or the absence or echo or ghost signals. Thus, for horizontal or vertical polarization the direct and ground-reflected waves may cancel at certain heights while at other heights, where they reinforce, the images on the TV screen may be objectionable because the time difference via the 2 paths produces a double image {a direct image and its ghost).
(c) Extend the comparison of (6) to consider the effect of other buildings or struc-tures which may produce additional paths of transmission.
Note that direct satellite-to-earth TV downlinks are substantially free of these reflection and ghost image effects.
16-15 Horizontal dipole above imperfect ground.
Calculate the vertical plane field pattern broadside to a horizontal A/2 dipole antenna A/4 above actual homogeneous ground with constants £' = 12 and a — 2 x 10 ^ U m 1 at (o) 100 kHz and (6) 100 MHz.
16-16 Short vertical dipole above imperfect ground. The center of a short vertical dipole (l < A/ 10) is Located A/4 above actual homogeneous ground with the same con-stants as in Prob.
16-15. Calculate the vertical plane field pattern at (a) 1 MHz and (b) 100 MHz.
16-17 DF and monopulse. Many direction-finder (DF) antennas consist of small (in terms of A) loops giving a figure-of-eight pattern as in Fig, P!6-17a. Although the null is sharp the bearing (direction of transmitter signal) may have considerable uncertainty unless the S/N ratio is large. To resolve the 180 ambiguity of the loop pattern, an auxiliary antenna may be used with the loop to give a cardiod pattern with broad maximum in the signal direction and null in the opposite direction.
The maximum of a beam antenna pattern, as in Fig, PI6-176, can be employed to obtain a bearing with the advantage of a higher S/N ratio but with reduced pattern change per unit angle. However, if 2 receivers and 2 displaced beams are used, as in Fig. P16-17c, a large power-pattern change can be combined with a high S/N ratio. An arrangement of this kind for receiving radar echo signals problems 773 0=0 0 = 0 <) (b) ( c ) Figure P16-17 Direction finding: (a) with loop null, (h ) with beam maximum and (c) with double beam (mono pulse).
can give bearing information on a single pulse (monoputee radar).
If the power received on beam l is P t and on beam 2 is P 2 , then if P 2 > P v the bearing is to the right.
If Pj > P 2 the bearing is to the left and if = P 2 the bearing is on axis (boresight). (With 4 antennas, bearing information left-right and up-down can be obtained.) () If the power pattern is proportional to cos 0, as in Fig. P16-17c, determine P2/P t if the interbeam (squint) angle 2 - 40° for A0 = 5 and 10.
() Repeat for 2 = 50'\ (tj Determine the PJP A of the single power pattern of Fig. PI 6-1 76 for Ad ~ 5 and 10 if the power pattern is also proportional to cos ft.
(d) Tabulate the results for comparison and indicate any improvement of the double over the single beam.
16-18 Path difference on overland radio link.
If h, = h z and d h v in Prob. 16-14 (Fig, P16-14), show that the path difference of direct and reflected rays is 2hf/d.
16-15 Antennas over imperfect ground. Write and run computer programs for Probs.
16-15 and 16-16 with a menu for height, frequency and ground constants e' and er.
16-20 Square-corner monopuke radar antenna. Design a square-corner reflector antenna with 2 off-axis feeds to operate as a monopulse radar. See footnote preceding equation (12-3-7).
16-21 Signalling to submerged submarines.
Calculate the depths at which a 1 /jV m^ 1 field will be obtained with E at the surface equal to 1 V m _1 at frequencies of 1, 10, 100 and 1000 kHz. What combination of frequency and antennas is most suit-able?
CHAPTER 17 ANTENNA TEMPERATURE, REMOTE SENSING, RADAR AND SCATTERING 17-1 INTRODUCTION. The concept! of antenna temperature and sytum nerature concept to remote sensing is then discussed. The next sectio Tadar and the radar equation , leading to considerations of scattering radar or scattering cross section.
17 2 ANTENNA TEMPERATURE, INCREMENTAL AND TOTAL Referring to Fig. 17-la, the noise power per unit handwidth available at if ” of rasUlanca R and l=n.p«ia.u,e T is ev.n b, .ha Nyquisl" relation as p = kT (W Hz -) (1) , ~ Kraus, Radio Astronomy, 2nd ed Cyg„ us-Quasar m6.
H. NyqUi.C-Ttarin.1 AUUh ofMeCM - Conduct Pi.,, Kr,, 32, 1 10^1 13 , 192S.
774 17-2 ANTENNA TEMPERATURE, INCREMENTAL AND TOTAL 775 (a) w t (C l figure 17-1 {a) Resistor at temperature T. {h) Antenna in an anechoic chamber at temperature T\ ic) Antenna observing sky at temperature T. The same noise power per unit bandwidth is available at the terminals in all 3 cases.
where p = power per unit bandwidth, W Hz -1 k = Boltzmann’s constant = 1,38 x 10 23 J K 1 T = absolute or Kelvin temperature, K If the resistor R is replaced by a lossless antenna of radiation resistance J? in an anechoic chamber, as in Fig, 17-16, at temperature T the noise power per unit bandwidth available at the terminals is unchanged.
Now if the antenna is removed from the chamber and pointed at a sky of temperature 7\ the noise power at the terminals : s the same as for the two pre-vious cases.
1 .
If p is independent of frequency, the total power P is given by p multiplied by the bandwidth Af or P = kT Af (W) (2 ) where P — power, W Af = bandwidth, Hz Suppose that the antenna of Fig. 17- It has an effective aperture A and that its beam is directed at a source of radiation which produces a power density per unit bandwidth or flux density S at the antenna. The power received from the source is then given by P = SA e Af (W) P) where S = power density per unit bandwidth, W m Hz A e = effective aperture, m 2 Af = bandwidth, Hz It is assumed that the entire antenna pattern sees " the sky of temperature T, 776 17 ANTENNA TEMPERATURE, REMOTE SENSING. RADAR AND SCATTERING Equating (3) and {2), the power density per unit bandwidth or flux density from the source at the antenna is k AT S = (W m “ 2 Hz " ) (4) A where AT^ = incremental antenna temperature due to the source, K To separate the power received from the source from other sources of radi-ation, AT^ is measured as the difference in antenna temperature with the antenna beam on and off the source, or A r = -jf ( 5) This incremental antenna temperature A T A is equal to the change in tem-perature of the resistor R (substituted for the antenna) required to produce an equal power P as given by (2).
In practice it is not necessary to make the resistor temperature change equal to AT in order to measure AT^. Thus, the receiver may be switched alter-nately to antenna and resistor Assuming that the receiver output indicator has a linear power scale, let the deflection with the antenna be dA and with the resistor dR , If the resistor temperature change is AT R , then the incremental antenna tem-perature AT^ is given by AT a = ^AT r (6) where AT^ = incremental antenna temperature, K dA = deflection with antenna, arbitrary units dR = deflection with resistor, same units as for dA AT B = resistor temperature change, K It is assumed that the antenna and resistor are both matched to the trans-mission line and receiver.
If the antenna has no side or back lobes and its beam is narrower than the source, the incremental antenna temperature A T A is equal to the source tem-perature 7^. Thus, the antenna and receiver act as a passive remote sensing device which can determine the temperature of near or distant regions within the antenna beam.
It is as though the antenna's radiation resistance R is coupled by the beam to the remote regions and acquires a temperature AT^ equal to the remote region (or source) temperature T s .
Let us consider now a different situation, i.e., one in which the antenna beam is much wider than the source extent, as in Fig, 1 7-2 Then, obviously, A T A will be less than T s . However, if the source solid angle Q/and the antenna beam solid angle are known, the source temperature is given very simply by (7 ) 17-2 ANTENNA TEMPERATURE, INCREMENTAL AND TOTAL 777 Figure 17-2 Situation where source extent is smaller than the beam area £1^ where T^ = source temperature, K A 7^ = incremental antenna (noise) temperature, K Os = source solid angle (see Fig.
17-2), sr = antenna beam solid angle (see Fig.
17-2), sr It is important to note that the antenna temperature has nothing to do with the physical temperature of the antenna provided the antenna is lossless.
Example 1 Mats temperature. The incremental antenna temperature for the planet Mars measured with the US. Naval Research Laboratory 1 15-m radio tele-scope antenna at 31,5 mm wavelength was 0.24 K, Mars subtended an angle of 0.005° at the time of the measurement. The antenna HPBW = 0.tl6°.
Find the average temperature of Mars at 31,5 mm wavelength.
Solution. Assuming that UA is given by the solid angle within the HPBW, we have from (7) that the Mars temperature AT, Q.ilfi 2 ct(0.005 2 /4) 0.24 = 164 K This temperature is less than the infrared temperature measured for the sunlit side (250 K), implying that the 31,5-mm radiation may originate further below the Martian surface than the infrared radiation. This is an example of remote-sensing the surface of another planet from the earth.
The source temperature in the above discussion and example is an eqviva-lent temperature.
It may represent the physical temperature of a planetary surface, as in the example, but, on the other hand, a celestial plasma cloud with oscil-lating electrons which is at a physical temperature close to absolute zero may generate radiation with an equivalent temperature of thousands of kelvins. The temperatures we are discussing are thermal (noise) temperatures like those of a perfect emitting-absorbing object called a blackbody. A hot object filling the beam of a receiving antenna will ideally produce an antenna temperature equal to its thermometer-measured temperature. 2 However, the oscillating currents of a transmitting antenna can produce an equivalent temperature of millions of degrees (K) even though the antenna structure is at normal outdoor temperature.
1 C. H. Mayer, T.
P. McCullough and R. M, Sloanaker, “Observations of Mars and Jupiter at a Wavelength of 3,15 0117 Astropkys. J. T 127, 11-16, January 1958, 1 Assuming the object's intrinsic impedance = 377 IT 778 17 ANTENNA TEMPERATURE, REMOTE SENSING. RADAR AND SCATTERING "Emory" sky 3 k Distant quasar > 1G S K •Mars, 164 K Groundr 290 K pjgure J 7_3 Horn antenna directed at various objects senses different temperatures as suggested.
It may be said that the antenna (and its currents) have an equivalent blackbody (or noise) temperature of millions of degrees.
All objects not at absolute zero produce radiation which, in principle, may be detected with a radio antenna- receiver. A few objects are shown in Fig. 17-3 with the equivalent temperatures measured when the horn antenna is pointed at them. Thus, the temperature of a distant quasar is over 10 6 K, of Mars 164 K, of a transmitter on the earth 106 K, of a man 310 K, of the ground 290 K, 1 while the empty sky at the zenith is 3 K. This temperature, called the 3 K sky back-ground, is the residual temperature of the primordial fireball which created the universe and is the minimum possible temperature of any antenna looking at the sky.
An assumption was made in the above discussion which requires comment.
It was assumed that the antenna and source polarizations were matched (same polarization states on the Poincare sphere; see Sec. 2-36). Although this is pos-sible for the transmitting antenna in Fig. 17-3, it is not possible for the other sources because their radiation is unpolarized 2 and any antenna, whether linearly or circularly polarized, receives only half of the available power. Hence, for such sources the flux density of the source at the antenna is given by twice (4) or s = _ ATk (W m' 1 Hz' ‘) (8) A e 1 Due to reflection, more realistic values for a man and the ground might be less.
2 See J. D. Kraus, Radio Astronomy^ 2nd ed, Cygmis-Quasar, 1986, Secs. 4-4, 4-5 and 4-6, 17-2 ANTENNA TEMPERATURE. INCREMENTAL AND TOTAL 779 We have considered two extreme cases, one where the source extent is much broader than the antenna beam width arid one where the source extent is much less than the beam width.
Lei us consider now the general situation for any source beam width-size relation. For this general situation the total antenria t&n-perenure is T a = -L f ] m ) P„(0, 4>) d£i ) = brightness temperature of source or sources as a function of angle, K Pn{0, fi) = normalized antenna power pattern, dimensionless = antenna beam solid angle, sr dD = sin 0 df) d(j} = infinitesimal element of solid angle, sr Note that T A in (9) is the total antenna temperature including not only con-tributions from a particular source in the main beam but from sources of radi-ation in all directions in proportion to the pattern response. Note also that the temperatures are in kelvins, K { = Celsius degrees above absolute zero).
Example 2 Antenna temperature. A circular reflector antenna of 500 vn 1 effective aperture operating at k = 20 cm is directed at the zenith. What is the total antenna temperature assuming the sky temperature is uniform and equal to 10 K? Take the ground temperature equal to 300 K and assume that half the minor-lobe beam area is in the back direction (toward the ground).
1 The beam efficiency is 0.7 ( = 0^/0^), Solution. Assuming that the antenna aperture efficiency is 50 percent, its physical aperture is 1000 m 2 and its diameter 35.7 m ( = 2v 1000, 7t m). At 7. = 0.2 m the diameter is 1797, implying that the HPBW = 0.4 n ( = 707179). Thus, the antenna is highly directional with the main beam directed entirely at the sky (close to the zenith).
Since the (main) beam efficiency is 0.7, 70 percent of the beam area Z1A is directed at the 10 K sky, half of the remainder or 15 percent at the sky and the other half of the remainder or 15 percent at the 300 K ground. Thus, integrating (9) in 3 steps, we have Sky contribution = ~ (10 x 0.7 = 7 K Sidelobe contribution = pj- (10 x \ x 0.3 Q^,) = 1.5 K Back -lobe contribution = (300 x \ x 0.3 Q^) = 45 It 1 Since much sky radiation reaches the antenna via reflection from the ground, a more realistic ground temperature might be less than 300 K, 780 17 ANTENNA TEMPERATURE, REMOTE SENSING, RADAR AND SCATTERING and = 7 4- 1,5 + 45 — 53.5 K.
Note that 45 of the 53.5 K, or 84 percent, of the total antenna temperature results from the back-lobe pickup from the ground. With no back lobes the antenna temperature could ideally be only 10 K, so that in this example the back Lobes are very detrimental to the system sensitivity {see Sec, 17-3).
It is for this reason that radio telescope and space communication antennas are usually designed to reduce back- and si delobe response to a minimum.
The information given regarding aperture and wavelength is relevant to the problem only to the extent that it indicates that the main beam is directed entirely at the sky.
In contrast to the above example, let us recall the temperature measure-ments made by Arno Penzias and Robert Wilson 1 in 1965 at 4 GHz on their 6.2-m horn-reflector antenna which resulted in their discovery of the 3-K sky background. When directed at regions of L1 empty “ sky near the zenith, Penzias and Wilson measured a total antenna temperature T A = 6.7 K..
Contributions to this temperature were measured as 2 2,3 ± 0.3 K. due to the atmosphere 0.8 ± 0.4 K due to ohmic tosses <0,1 K due to back lobes into the ground 3.2 + 0.5 The difference, 6.7 - 3.2 = 3.5 K, they attributed to the sky background. Theirs was the first measurement of the residual temperature of the primordial (Big Bang) fireball which created the universe, and sets a lower limit to the tem-perature of any antenna looking at the sky.
The 0.1-K ground pickup by the antenna of Penzias and Wilson is one of the smallest values ever measured for an antenna. Note also that, in their analysis, they attributed 0.8 K to ohmic losses in the antenna and rotary joint.
The antenna noise temperature from the sky as a function of frequency (and wavelength) is presented in Fig. 17-4. A beam angle (HPBW) of less than a few degrees and 100 percent (main) beam efficiency are assumed. Curves are given for beam angles from the zenith (complementary to elevation angles). At lower fre-quencies the temperature is dominated by radiation from the galaxy. At higher frequencies the atmosphere introduces noise due to absorption. Above the earth s atmosphere (in space) this noise is avoided, but there is a universal photon or quantum noise temperature limit at still higher frequencies given by the photon energy hfdivided by Boltzmann's constant, or 1 A. A. Penzias and R. W. Wilson, “A Measurement of Excess Antenna Temperature at 4080 MHz, Astrephys. J., 14X 419-421, 1965.
1 A discussion of the errors is given in Sec, 18-12.
l7 ‘3 antenna temperature, incremental and total 781 Figure 17-4 Antenna (noise) temperature from the sky as a function of frequency. See text for expla-nation, (From J.
Z>. Kraus, Radio Astronomy, 2nd Cygnus-Quosar, 19&6.) hf T = i (K) (10) where h = Planck's constant = 6,63 x 10“ 34 i s / = frequency, Hz k = Boltzmann’s constant = 1.38 x 10“ 23 J K _1 Across the spectrum between these sources of noise there is the noise background or floor of 3 K (or more precisely 2.7 K) due to radiation from the primordial fireball.
The low-noise region between galactic radiation and atmospheric absorption defines an earth-bosed radio window, while the region between galactic radiation and quantum limit establishes a cosmic radio window.
782 17 ANTENNA TEMPERATURE. REMOTE SENSING, RAlMft AND SCATTERING figure 17-5 Antenna, transmission line and receiver for system temperature determination.
17-3 SYSTEM TEMPERATURE AND SIGNAL-TO-NOISE RATIO. An antenna is part of a receiving system consisting, in general, of an antenna, a receiver and a transmission line which connects them. The tem-perature of the system, or system temperature is a critical factor in determining the sensitivity and signal -to -noise ratio of a receiving system.
Let us consider a receiving system as shown schematically in Fig.
17-5, with an antenna, a receiver and a transmission line (or waveguide) connecting them.
The system temperature depends on the noise temperature of the sky, the ground and antenna environs, the antenna pattern, the antenna thermal effi-ciency the receiver noise temperature and the efficiency of the transmission line (or waveguide) between the antenna and receiver. The system temperature at the antenna terminals is given by T„-T J + 7-^J- - l) + l) + ^ (') where T A = antenna noise temperature [as given by (17-2-9)], K TAp = antenna physical temperature, K e, = antenna (thermal) efficiency (0 < e x < 1), dimensionless T lp = line physical temperature, K z 2 = line efficiency (0 ^ < l), 1 dimensionless 7^ = receiver noise temperature (see next paragraph), K The receiver noise temperature is given by T 2 r 3 n .
T = T ' + t + ^l + " where T, = noise temperature of first stage of receiver, K T 2 — noise temperature of second stage, K T 3 = noise temperature of third stage, K G x = power gain of first stage G 2 = power gain of second stage ' e 2 = f where a = attenuation constant (Np m and l = length of line (m).
Antenna terminals 17-3 SYSTEM TEMPERATURE AND SIGN AUTO.NOISE RATIO 783 Terms for additional stages may be required if the temperatures are sufficiently high and the gains sufficiently low.
Example 1 System temperature. A receiving system has an antenna with a total noise temperature of 50 K, a physical temperature of 300 K and an efficiency of 99 percent, a transmission line at a physical temperature 300 K and an efficiency of 90 percent, and a receiver with the first 3 stages all of 80 K. noise temperature and 13 dB gain. Find the system temperature.
Solution. From (2) the receiver noise temperature is T k = 80 + — + —r = 80 + 4 + 0.2 = 84.2 K 20 20 ^ From (1) the system temperature is T- - » + KoS -') + HiS -') 55 “'2 = 50 + 3 + 33.3 + 93.6 m 180 K Note that due to losses in the antenna, its physical temperature contributes 3 K.
The line contributes about 33 K and the receiver about 94 K..
The sensitivity, or minimum detectable temperature A7^,in , of a receiving system is equal to the rms noise temperature AT rmi of the system, as given by Ar mi„ = "Trr = A7™“ (3) VAff where k f = system constant (order of unity), dimensionless 7^ a = system temperature [sum of antenna, line and receiver temperatures as given by (1)], K = rms noise temperature = A7^j„, K Af= predetection bandwidth of receiver, Hz t = postdetection time constant, s The criterion of detectability is that the incremental antenna temperature A T a due to a radio source be equal to or exceed A 7^ ifl , that is.
ATa 3= Ar rai (4) 784 17 ANTENNA TEMPERATURE. REMOTE SENSING. RADAR AND SCATTERING and the signal-to-noise (S/N) ratio is then 1 S N <5) 2 Many space communication systems, radio telescopes and remote sensing systems operate at such high sensitivity (low signal levels) that a low system temperature is essential Example 2 MiBiimim detectable flux density. The Ohio State University 110 by 21 m radio telescope antenna (see Fig. 12-5t) has a physical aperture of 2208 m 2 , and at 1415 MHz an aperture efficiency of 54 percent and a system temperature of 50 K. The rf bandwidth is 100 MHz, the output time constant is 10 s and the system constant is 2.2. Find the minimum detectable flux density.
Stflmtioit. From (3) the minimum detectable temperature is AF mi 2.2 x 50 /too X 106 X 10 : - 0.003 S K.
The effective aperture A t = Ap etp — 2208 x 0.545 1203 m J . From (17-2-8) the minimum detectable flux density is ASt 2k AT min 2 x 1.38 x 10" 23 x 0.0035 At 1203 = 8d x 10 “ 29 W m -2 Hz” 1 - 8mJy 3 By repeating observations apd averaging, the minimum can be further reduced { <x Ji/re, where n = number 6f observations).
In a large sky survey at 1415 MHz, about 20000 radio sources were detected and cataloged at flux densities above 180 mJy Thus, the signal-to-noise ratio for these cataloged sources is S; _ 180 mJy = AT A ^ 0.0785 K ^ ^ N ~ 8.0 mJy ~ AT min ~ 0.0035 K ~ Let us consider next the signal-to-noise ratio for a receiving system which is part of a communication link.
If a transmitter radiates a power P t isotropically and uniformly over a bandwidth A£, it produces a flux density at a distance r of PJ(4xr2 A£). A receiving antenna of effective aperture A tr at a distance r can 1 Distinguish between 5 here for signal and 5 elsewhere for flux density or Poyntittg vector.
1 The minimum detectable signal is given by S = S or S/S = 1. The ratio is . sometimes expressed {5 + S)/N and for the minimum detectable signal case this ratio equals 2.
1 8millijanslcy,wheie 1 Jy = 10“M Wm“ J Hz -1 , I i7-3 SYSTEM TEMPERATURE AND SIGNA L-TO-NOISE RATIO 785 collect a power ?,A~4fr 4w 2 Af where P, = radiated transmitter power, W A er = effective aperture of receiving antenna, m 1 Af r = receiver bandwidth, Hz Aft — transmitter bandwidth, Hz r = distance between transmitter and receiver, m ft is assumed that Af r < A/ fT With a transmitting antenna of directivity D = 4jiAJa\ the received power becomes p = p ‘ A" M (W) (71 r 2 ;.1 af t twj (7) where l = wavelength, m A et = transmitter effective aperture, m 2 For Af = A/ f (bandwidths matched), (7) is the Friis transmission formula (2-25-5).
Whether this amount of received power is useful for communication depends on the signal-to-noise ratio SjN, where the signal power is given by (7) and the noise power by the Nyquist relation (17-2-1) as Pa = kT syt Af r = N ( 8 ) where k = Boltzmann's constant = 1.38 x 10 23 J K“ 1 T ays = system temperature, K For matched bandwidths (A/ r = A/ f), the ratio of (7) to (8) gives the signal-to-noise ratio as S P r _ P t A„A et N PH kT^)?Af r (dimensionless) Example 3 Down-link signal-to-noise ratio.
A Gar k e -orbit -sate 11 it e-to-earth FM-TV downlink, as in Fig.
17-6, has a transmitter (transponder! power P t = 5 W, transmitter antenna physical aperture A pi = 6 irv 2 , earth-station receiving antenna physical aperture A^ = 12 m 2 , path length r as 40000 km and frequency 4 GHz r The earth-station receiving antenna has an antenna temperature contribution from the main beam directed at the sky region of the Clarke orbit when the galactic plane is crossing (worst condition) (see Fig.
17-4) of 7 K. Beam efficiency is 80 percent with half of the minor lobes above the horizon “seeing” an average of 10 K while the remaining minor lobes "see” the earth at 290 K. The receiver has a 3-stage front-end amplifier with the same characteristics as the receiver in Example 1. The receiving antenna has a 99 percent thermal efficiency and short transmission line (amplifier close to antenna terminals) with 98 percent efficiency. The transmitting antenna on the Clarke-orbit satellite and the earth-station receiving antenna both 786 17 ANTENNA TEMPERATURE. REMOTE SENSING, RADAR AND SCATTERING Figure 17-6 Clarke-orbh downlink.
have an aperture efficiency £ap of 60 percent. Find the signal-to-noise ratio for a 30-MHz bandwidth. Both receiver and transmission line are at a physical (ambient) temperature of 300 K.
It is assumed that beams are aligned and that polarizations and bandwidths are matched.
Solution, At 60 percent aperture efficiency, the effective apertures of the two antennas are A rl — ZtpAy, = 0.6 x 6 = 3.6 m 2 ^ = X„ = 0.6x 12 = 7.2 m1 The receiving antenna temperature from (17-2-9) is T a = 0.8 x 7 + 0.1 x 10 + 0.1 x 290 ~ 36 K.
The receiving system temperature from (1) is T %yt i 36 + 300 x 0.01 + 300 x 0.02 + 1,02 x 84 = 36 + 3 + 6 + 86 = 131 K At 4 GHz, X = 0.075 m. The signal-to-noise ratio from (9) is S 5 x 7,2 x 3.6 N = 08 x 10“ 23 x 131 x4J x 10 14 x 0.075 1 x 3 x 10 7 = 265 or 24 dB which would be satisfactory for most applications.
17 4 PASSIVE REMOTE SENSING 787 Examples 2 and 3 dealt with receiving systems operating at 1.4 and 4 GHz, Referring to Fig.
17-4, we note that at these frequencies and at large elevation angles (small zenith angles) the sky temperature is less than 10 K, However, at angles near the horizon (zenith angle >80°) the temperature may approach 100 K due to atmospheric absorption. At frequencies below 1 GHz the noise from our galaxy becomes important. At 50 MHz the galactic noise ranges from about 2000K when the antenna is looking at the galactic poles to 20000 K when it is looking at the galactic center. Under these conditions reducing the noise temperature of a receiver from, say, 200 to 100 K would make but a small difference on the system temperature and the signal-to-noise ratio. Thus, if T A = 5000 K, T r = 200 K and neglecting other contributions, the system temper-ature T sys = 5200 K as compared to 5100 K for T R = 100 K, The improvement in T sys and minimum noise level is only 0,08 dB [=10 log (52/51)].
Sometimes the parameter noise figure is used instead of the noise tem-perature . They are related as follows: T = (F - 1)T 0 (10) where T = noise temperature, K T 0 = 290 K F = noise figure, dimensionless Thus, or F(dB) = 10 log F (12) where F(dB) = noise figure in decibels. The relation of the noise figure F and its value in decibels to the noise temperature T are shown in Fig. 17-7 17-4 PASSIVE REMOTE SENSING, A radio telescope is a remote sensing device whether it is earth-based and pointed at the sky for observing celestial objects 1 or on an aircraft or satellite and pointed at the earth. In this section we consider the case where the radiation detected or sensed by the tele-scope originates in the objects being observed, making for a passive remote sen-sing system in distinction to radar or active remote sensing where signals are transmitted and their reflections observed and analyzed. The active case is dis-cussed in the next section.
1 See the Mars temperature example, Example 1 in Sec.
17-2, and the minimum delectable flux density example. Example 2 in Sec. 17-3.
T (noise temperature, kelvin) 788 1?
ANTfcJSNA TEMPERATURE, REMOTE SENSING, RADAR AND SCATTERING PdR- Noise figure (dfl) Figure 17-7 Noise-temperaturc-noise-figure chart.
Consider the situation of Fig 17-8nT in which the earth-based radio tele-scope antenna beam is completely subtended by a celestial source of temperature T, with an intervening absorbing-emitting cloud of temperature 7; . With no cloud present y the incremental antenna temperature AT a = T sy but with the cloud it may be shown that the observed antenna temperature AT a - r f ( 1 - e~'<) + ( 1 ) 17-1 PASSIVE REMOTE SENSING 789 Earth -based radio telescope Radio telescope on satellite Receiver Antenna / terminals terminals^ Transmission ime T r, P Figure 17-8 (a) Earth-based radio telescope remote-sensing celestial source through inter-vening interstellar cloud (f?) Radio telescope on satellite remote-sensing the earth through forest, (t ) Receiver detecting antenna output through transmission line. The doud T forest and transmission line have analogous emit ting-absorbing properties.
where = absorption coefficient of the cloud 1 ( = 0 for no absorption and = oo for infinite absorption), rhus, knowing 7^ and rf , the cloud’s equivalent black-body temperature T c can be determined.
Now, referring to Fig 17-86, let us reverse the situation and put the radio telescope on an orbiting satellite for observing the surface of the earth at tem-perature 7^ with the antenna beam completely subtended by a large forest at a temperature 7]r , The incremental satellite antenna temperature is then AT a = 7}(1 - e-'f) + T t e~ x (K) (2) where if = absorption coefficient of the forest. Knowing T e and Ty, the tem-perature of the forest can be determined, or knowing T t and T f y the absorption coefficient can be deduced.
It is by such a technique that the whole earth can be surveyed and much information obtained about the temperatures of land and water areas, and from absorption coefficients about the nature of the surface Example Forest temperature by remote sensing. The remote-sensing antenna of 3-GHz orbiting satellite measures a temperature AT^ - 300 K when directed at tropical forest region having an absorption coefficient = 0693 at vertical im dence. If the earth temperature 7^ = 305 K, find the temperature of the forest 1 Astronomers call the “optical depth." The quantity e is equivalent to the efficiency factor £ ir 790 17 antenna temperature, remote sensing, radar and scattering Solution.
Since t,- = 0.693, e r' = 0.5, so from (2), _ _ AT4 - T a e-'-< 300 - 305 x 0.5 _ MJ R (-i _<>-'/ 1 -0.5 If the antenna-transmission Sine-receiver system is viewed from the receiver terminals as in Fig. 17-8c (instead of from the antenna terminals as in Fig. 17-5), we note that the analogy between the remote-sensing situations discussed above extends here to the transmission line. Thus, the emitting-absorbing transmission line of Fig 17-8c is like the emitting-absorbing cloud of Fig. 17-8a and like the emitting-absorbing forest of Fig.
17-85. The analogy may be emphasized by com-paring (1) and (2) with the temperature as seen from the receiver terminals, so that the equations for the 3 situations have identical form as follows: Antenna looking at celestial source A T A = T c (l - f ") + T.e-(K) (J| (see Fig.
17-8a): Antenna looking at earth from satellite AT^ = T/(l — e 0 + (see Fig, 17-8h); Receiver looking at antenna T = 7^1 - + T A e~« (K) ( 5) (see Fig, 17-8c): where AT = incremental antenna temperature, K T € = cloud temperature, K .
, rc = cloud absorption coefficient (optical depth), dimensionless 7; = celestial source temperature, K T f = forest temperature, K if = forest absorption coefficient, dimensionless T e = temperature of earth, K T lp = transmission line physical temperature, K 'At = transmission line attenuation constant, Np m / = length of transmission line, m Note that the system temperature should be referred to the antenna termin-als as in (17-3-1) and not to the receiver terminals as in (5\ Thus, if the line is completely lossy (e - ' = e 2 = 0\ (17-3-1) gives an infinite system temperature which is correct, meaning that the system has no sensitivity whatever. However, with this condition, a “system temperature” viewed from the receiver terminals, as in (5), would equal the temperature T LF of the line plus the receiver tem-perature, a completely misleading result since it indicates that the system still has sensitivity.
17-5 RADAR. SCATTERING AND ACTIVE REMOTE SENSING BY ROBERT G kOUYOUMJlAN 791 17-5 RADAR/ SCATTERING AND ACTIVE REMOTE SENSING By Robert G, Kouyoumjian 2 Consider an object in the far zone of a radar antenna as shown in Fig.
17-9.
Typically, in pulse radar the antenna is connected to the transmitter for the pulse transmission and then switched to the receiver in time to receive the pulse echo.
In practice, the antenna may be connected continuously to both the transmitter and receiver but with the receiver blanked during transmission of the pulse.
During transmission the power density incident at the object (radar target) is \G 4nr2 (W m rl ) (1) where P t = transmitted power, W G = antenna gain, dimensionless r — distance between antenna and radar target, m The radar target scatters the incident power in all directions. The total amount of this scattered power is P ra as given by Pa = S^t r t (W) (2) where c t = total scattering cross section, m 2 The total scattering cross section o t may be regarded as an effective aperture which “collects” a power Pts from the incident wave and reradiates it in all directions (4tt sr).
Figure 17-9 Radar system and scattering object (radar target).
I 1 RADAR is an acronym for RAdio Direction And Range.
2 Department of Electrical Engineering, The Ohio State University.
792 IT antenna temperature, remote sensing, radar and scattering The scattered power incident back on the antenna (back scattered power) Sr is given by s = £±1ji = Di£l (W m 2 ) (3) r 4 nr 2 4nr1 where D s = radar target (backsca tiering) directivity and the antenna is in the far zone of the radar target.
The power Pr reaching the radar receiver is then = tW ) (4) f 4nr 1 where a = Ds v t = radar cross section , m 2 ^ 2 A = A'tPq -effective aperture of the radar antenna, m 2, ^ A f = — G = maximum effective aperture of radar antenna, m An p = polarization mismatch factor = cos 2 (MM fl/2), dimensionless MM fl = angular distance between polarization states M and JVf a M = polarization state of wave (see Sec. 2-36 on Poincare sphere).
M = polarization state of antenna d r a n n 4 j.
1 'p q = impedance mismatch factor = ^ + ft ) 2 + (X + X ) 2 dimensionless R a = antenna resistance, ft ft f receiver resistance, ft X a = antenna reactance, ft X r = receiver reactance, ft The value of the product pq in in (4) can range between 0 and T Finally, introducing (1) in (4) we obtain the radar equation P t (4ic) 3 r4 (dimensionless) We note that the received power Pr varies as the inverse fourth power of the distance r since r~ 2 factors are involved in both transmission from the radar to target and again from the target to radar, A simpler, frequently used form of (5) is G 2 ?
2 /“' TsrKct 0 = dimensionless o/Z)„jn dHnensmnte^ — ” — „ ;a 2A naiatflA, Object bounded by a smooth Tt^a tw^ 2i a 9 » curved surface with principal radii a 1 and a 2 at the specular point _ rt 2 _ a i 2na 2 = 2A S i Sphere, radius a na — ...
.
„ r i tf ; 4aL (rc/2)Li Cylinder of radius u and Znfli L KL ' length L. normal incidence =Ck L ^ Flat disc of radius a, (2mj) tw normal incidence =^ SV =C] - 4^„a ' Flat plate of area A .
1,.:.
4ji4 2 /a1 ‘'jiT'a 2-4 normal incidence = ^ - w. C - circumference - 2-t. C. - O -- » A " "™e,ri,: “^ cross section = A p -Targcls arc assumed to be perfectly conducting-thinner loop resonating closer to this value. Also it is worth loops at first resonance is larger than « the geometric or physical cross section A 9 (=V The effective aperture producing scattering equals A„ while the forward (small angle) scattering also has an effec-tive aperture equal to A s , making ff, = 2A g .
In Table 17-1 it is seen that the frequency behavior of a is very shape depen-dent whereas in Table 17-2 it is evident that the low-frequency behavior of a 1 H. C. Van de Holst, Light Scattering by Small Particles, Wiley, 1951, p, 107.
ADDITIONAL REFERENCES 797 varies as l//4 for all targets, which is characteristic of scattering in the Rayleigh region. Additional information on radar cross-section calculations and measure-ments is given in the references at the end of the chapter. See especially W.
E.
Blore and the /£££ Proceedings special issues for August 1965 and February 1985.
Whereas passive remote sensing can provide information on temperature, polarization and absorption and also on velocity (for line emitting objects), active (radar) remote sensing has the capability of providing much additional informa-tion such as distance, shape and composition. A returning pulse carries a charac-teristic signature by which different objects may be recognized, the pulse response being the inverse Fourier transform of the object's frequency response. Both passive and active remote -sen sing techniques can be used for mapping or imaging of extended objects.
ADDITIONAL REFERENCES Arvas, E.„ R.
F. Harrington and. J. R. Mautz: “Radiation and Scattering from Electrically Small Conducting Bodies of Arbitrary Shape," IEEE Trans. Ants. Prop., AP-34, 66^77, January 1986.
Blore, W.
E.: "The Radar Cross-Section of Ogives, Double-Backed Cones, Double-Rounded Cones and Cone Spheres," IEEE Trans. Ants . Prop., AP-12, 582-590, September 1964.
798 k 7 ANTtNNA TEMPERATURE REMOTE SENSING. RADAR AND SCATTERING PROBLEMS 799 Bowman, I J T B.
A. Senior and P.
L.
E.
Uslenghi: Electromagnetic ami Acoustic Scattering by Simple Shapes. North Holland. Amsterdam, 1969.
Chuang. C W. and D. L Moffati: 'Natural Resonances of Radar Targets via Prony's Method and Target Disc rim mat ion." /£££ Trans. Aerospace ariJ Licet. Sps., AES-12, September 1976.
Clarke, J (etl.): in Radar Techniques. Peregrinus.
1985.
Dalle Mees, F.. M. Mancianti, L Verra^ani and A, Cantoni: "Target Identification by Means of Radar," M i crowd re J-, 27, 85-102, December 1984.
Davidovii 7 vM., and W r-M. Boerner: " Extension of Kennaugh s Optimal Polarization Concept to the Asymmetric Scattering Matrix Case," IEEE Trans. Ants. Prop.. A P-34. 569-574, April 1986.
E>eschamps, G. A.: “High Frequency Diffraction by Wedges,' IEEE Trans , Ants. Prop.. A P-33, 357, April 1985(56 references).
Fftimiu, C.: Scattering by Rough Surfaces: A Simple Model," IEEE Trans Ant s.
Prop.. AP-34.
626-630. May 1986.
Evans, J. V.. and T. Hagfors: Radar Astronomy. McGraw-Hill, 1968.
Hall, R. C.. and R. Mittra: “Scattering from a Periodic Array of Resislive Strips,' - IEEE Trans. Ants Prop.. A P-33, 1009-1011, September 1985.
Hansen. R. C (ed.): Geometric Theory of Diffraction, IEEE Press, 198!
Keller. J B.
“ Back sea tiering from a Finite Cone," IRE Trans. Anrs. Prop., 9. 41 1-4! 2, 1961 Kennaugh, E M.:"The K -Pulse Concept," IEEE Trans. Ants. Prop , AP-29, 327-331, March 1981.
Kcnnaugh, E. M, and D, L.
Moffati: “Transient and Impulse Response Approximation," Proe.
IEEE. 53. August 1965.
Kim, H.
T.. N Wang and D. L MofTatt: “K-Pulse for a Thin Circular Loop," IEEE Trans , Ants.
Prop., A P-33, 1403-1407, December 1985.
Kouyoumjian, ft. G,. and P H. Pathak : "A Uniform Geometrical Theory of Diffraction for an Edge in a Perfectly Conducting Surface,’’ Proc. IEEE.. 62, 1448-1461. November 1974, Medgyesi-Mitschang, L. N, and J. M Putnam: "Electromagnetic Scattering from Extended Wires and 2- and 3- Dimensional Surfaces,” IEEE Trans.
Ants.
Prop.. AP-33, 1090-1100, October 1985.
Nakano. H , A, Yoshizawa and J. Yamauchi: "Characteristics of a Crossed-Wire Scatlerer without a Junction Point for an Incident Wave of Circular Polarization," IEEE Trans.
Ants.
Prop..
A P-33. 409-415, April 1985.
Newman.
E. H.: “TM and TF. Scattering by a Dielectric Ferrite Cylinder in the Presence of a Half Plane/’ IEEE Trans Ants . Prop AP-34, 804^813, June 1986.
Rao.
S. M . T. K. Sakar and S. A. Dianat: “A Novel Technique to the Solution of Transient Electro-magnetic Scattering from Thin Wires," IEEE Trans. Ants. Prop., AP-34, 630 634, May 1986.
Richmond, J. H.: “Digital Computer Solutions of the Rigorous Equations ior Scattering Problems," Proc. IEEE .
53. 796-804, August 1965.
Richmond, J H.: "On the Edge Mode in the Theory of TM Scattering by a Strip or Strip GratingC IEEE Trans Ants Prop., AP-2S, 883 887, November 1980.
Richmond. J. H, Isee also other Richmond references at the end of Chap. 9).
Rusch, W. V T., and R, J.
Poyorzelski: “A Mixed- Field Solution for Scattering from Composite Bodies," IEEE Trans Anfs. Prop., AP-34, 955-958, July 1986.
Schuerman, D W.: Light Scattering by Irregularly Shaped Particles , Plenum Press, 1980.
Senior, T. B.
A., and J, L. Votakis: "Scattering by an Imperfect Right-Angled Wedge," IEEE Trans.
Ants Prop.. AP-34. 681-689, May 1986 Shaffer, J F.: “EM Scattering from Bodies of Revolution with Attached Wires," IEEE Trans. Ants.
Prop., A P-30, 426 431, May 1982.
Siegel, K, M.: " Far Field Scattering from Bodies of Revolution," Appi Sci. Res., 7B, 293-328, 1959.
SkoJnik, M. I. fed ): Radar Handbook , McGraw-Hill 1980.
Ulaby, F. T„ R. K. Moore and A. K. Fung: Microwave Remote Sensing, Active and Passive „ Addison-Wesley, 1981-Umashankar, K„ A. Taflove and S. M. Rao: "Electromagnetic Scattering by Arbitrary Shaped 3-Dimensional Homogeneous Lossy Dielectric Objects," /£££ Trans.
Ants, Prop ., AP-J4, 758-766, June 1986.
Voiakis, J, L-, W. D, Burnside and L. Peters, Jr: "Electromagnetic Scattering from Appendages on a Smooth Surface," /£££ Trans. Ants , Prop., AP-J3, 736-743, July 1985, Wang, D.
and L. N. Medgyesi-Mitschang: "Electromagnetic Scattering from Finite Circular and Elliptic Cones," J£££ Trans, Anns. Prop., A P-33, 488-497, May 1985.
Wang, N.: "Electromagnetic Scattering from a Dielectric-Coated Circular Cylinder" /£££ Trans.
Ants. Prop,, AP-33, 960-963, September 1985.
Yaghjian, A. D-, and R. V, McGahan: “Broadside Radar Cross Section of the Perfectly Conducting Cube," /£££ Trans. Ants. Prop,, AP-33, 321-329, March 1985.
Proceedings of the /£££, Special Issue on Radar Reflectivity, 53. August 1965.
Proceedings of the IEEE , Special Issue on Radar, 73, February 1985.
See also back scat ter references at the end of Chap.
9.
PROBLEMS 1 17-1 Satellite TV downlink. A transmitter (transponder) on a Clarke orbit satellite pro-duces an effective radiated power (£RP) at an earth station of 35 dB over 1 W isotropic, () Determine the S/N ratio (dB) if the earth station antenna diameter is 3 m, the antenna temperature 25 K, the receiver temperature 75 K and the bandwidth 30 MHz, Take the satellite distance as 36000 km, Assume the antenna is a parabolic reflector (dish- type) of 50 percent efficiency.
() If a 10-dB S/N ratio is acceptable, what is the required diameter of the earth station antenna?
Note : For FM -modulated video signals, as employed by the Clarke-orbit satel-lites, the S/N ratio as used above is actually a carrier-to- noise (C/N) ratio, the ultimate video signal -to- noise ratio for typical North American domestic system satellites being almost 40 dB higher This is an advantage of FM modulation.
If the C/N exceeds a few decibels, then, in principle, a perfect picture results.
However, a C/N > 10 dB is desirable to allow for misalignment of the earth-staiion antenna, attenuation due to water or snow in the dish, a decrease in trans-ponder power, etc, 17-2 Antenna temperature. An end- fire array is directed at the zenith. The array is located over flat nonreflecting ground. If 0.9 £2^ is within 45° of the zenith and 0.08 between 45^ and the horizon calculate the antenna temperature. The sky brightness temperature is 5 K between the zenith and 45° from the zenith, 50 K between 45^ from the zenith and the horizon and 300 K for the ground (below the horizon). The antenna is 99 percent efficient and is at a physical temperature of 300 K 17-3 Earth-station antenna temperature. An earth-station dish of 100 m 2 effective aper-ture is directed at the zenith. Calculate the antenna temperature assuming that the sky temperature is uniform and equal lo 6 K. Take the ground temperature equal to 300 K and assume that ^ of the minor-lobe beam area is in the back direction.
The wavelength is 75 mm and the beam efficiency is 0.8.
1 Answers to starred () problems are given in App. D.
800 17 ANTENNA TEMPERATURE. REMOTE SENSING. RADAR AND Sf'ATTFRING 17-4 System temperature. The digital output of a E4-GHz radio telescope gives the following values (arbitrary units) as a function of the sidereal time while scanning a uniform brightness region. The integration time is 14 s, with 1 s idle time for print-out. The output units are proportional to power.
Time Output Time Output 31 m 30 v 234 32m45 E 229 31 45 235 33 00 236 32 00 224 33 15 233 32 15 226 33 30 230 32 30 239 33 45 226 If the temperature calibration gives 170 units for 2.9 K applied, find (a) the rms noise at the receiver, ( b ) the minimum detectable temperature, (c) the system tem-perature and (d) the minimum detectable flux density. The calibration signal is introduced at the receiver. The transmission line from the antenna to the receiver has 0.5 dB attenuation. The antenna effective aperture is 500 m 2 The receiver bandwidth is 7 MHz, The receiver constant k' = 2.
17-5 System temperature, Find the system temperature of a receiving system with 15 K antenna temperature, 0,95 transmission-line efficiency, 300 K transmission- line temperature, 75 K receiver first-stage temperature, 100 K receiver second-stage temperature and 200 K receiver third-stage temperature. Each receiver stage has 16 dB gain, 17-6 Minimum detectable temperature, A radio telescope has the following character-istics: antenna noise temperature 50 1C, receiver noise temperature 50 K, transmission -line between antenna and receiver 1 dB loss and 270 K physical tem-perature, receiver bandwidth 5 MHz, receiver integration time 5 s, receiver (system) constant k r = ir/v/2 and antenna effective aperture 500 m 2 . If two records are averaged, find (a) the minimum detectable temperature and (b) the minimum detectable flux density.
17-7 Minimum detectable temperature, A radio telescope operates at 2650 MHz with the following parameters: system temperature 150 K., predetection bandwidth 100 MHz, postdetection time constant 5 s, system constant k 1 = 2.2 and effective aperture of antenna 800 m 2 . Find (a) the minimum detectable temperature and (b) the minimum detectable flux density, (c) If four records are averaged, what change results in (a) and (b)?
17-8 Antenna temperature with absorbing cloud, A radio source is occulted by an inter-vening emitting and absorbing cloud of unity optical depth and brightness tem-perature 100 K. The source has a uniform brightness distribution of 200 K and a solid angle of 1 square degree. The radio telescope has an effective aperture of 50 m 2 .
If the wavelength is 50 cm, find the antenna temperature when the radio telescope is directed at the source. The cloud is of uniform thickness and has an angular extent of 5 square degrees. Assume that the antenna has uniform response over the source and cloud.
17-9 Passive remote-sensing antenna- Design a 30-GHz antenna for an earth-resource passive remote-sensing satellite to measure earth-surface temperatures with 1 km 2 resolution from a 300-km orbital height.
problems 801 17-10 Forest absorption. An earth- resource satellite passive remote- sensing antenna directed at the Amazon River Basin measures a night-time temperature T 4 ^ 21 n C, If the earth temperature T e = 2TC and the Amazon forest temperature 7} = 1 5 r'C t find the forest absorption coefficient 17-11 Solar interference to earth station, (a) Twice a year the sun passes through the apparent declination of the geosta-tionary Garke-orbit satellites, causing solar-noise interference to earth sta-tions.
If the equivalent temperature of the sun at 4 GHz is 50000 K, find the sun's signal-to-noise ratio (in decibels) for an earth station with a 3-m parabol-ic dish antenna at 4 GHz. Take the sun’s diameter as 0,5' and the earth -station system temperature as 100 K..
(b) Compare this result with that for the carrier -to- noise ratio calculated in Prob.
17-1 for a typical Clarke-orbit TV transponder, (c) How long does the interference last?
Note that the relation / 2/A tf gives the solid beam angle m steradians and hot in square degrees.
17-12 Radar defection, A radar receiver has a sensitivity of 10“ 12 W.
If the radar antenna effective aperture is 1 m 2 and the wavelength is 10 cm, find the transmit-ter power required to detect an object with a 5-m 2 radar cross section at a distance of 1 km, 17-13 Venus and moon radar.
(a) Design an earth-based radar system capable of delivering 10“ 15 W of peak echo power from Venus to a receiver. The radar is to operate at 2 GHz and the same antenna is to be used for both transmitting and receiving. Specify the effective aperture of the antenna and the peak transmitter power. Take the earth-Venus distance as 3 light -minutes, the diameter of Venus as 12,6 Mm and the radar cross section of Venus as 10 percent of the physical cross section.
(b) If the system of (u) is used to observe the moon, what will the received power be? Take the moon diameter as 3.5 Mm and the moon radar cross section as 10 percent of the physical cross section, 17-14 Thompson scatter. The alternating electric field of a passing electromagnetic wave causes an electron (initially at rest) to oscillate. This oscillation of the electron makes it equivalent to a dipole radiator. Show that the ratio of the power scat-tered per steradian to the incident Poynting vector is given by (p^e 2 sin fi/Anm) 1 , where e and tn are the charge and mass of the electron and 0 is the angle of the scattered radiation with respect to the direction of the electric field E of the inci-dent wave. This ratio times 4 jt is the radar cross section of the electron. Such reradiation is called Thompson scatter.
17-15 Thompson-scatter radar, A ground-based vertical-looking radar can be used to determine electron densities in the earth's ionosphere by means of Thompson scatter (see Prob.
17-14). The scattered-power radar return is proportional to the electron density.
If a short pulse is transmitted by the radar, the backscattered power as a function of time is a measure of the electron density as a function of height. Design a Thompson- scat ter radar operating at 430 MHz capable of mea-suring ionospheric electron densities with 1 km resolution in height and horizontal position to heights of 1 Mm. The radar should also be capable of detecting a minimum of 100 electrons at a height of 1 Mm. The design should specify radar peak power, pulse length, antenna size and receiver sensitivity. See W.
E. Gordon, 802 n antenna TEMPERATURE, remote sensing, radar and scattering “ ftadar Backscatter from the Earth’s Ionosphere, IEEE Trans .
/Infs, Prop ., AP-12, &73-S76, December 1964.
17-16 Jupiter signals. Flux densities or IQ" 20 W m' 2 Hz^.are commonly received from Jupiter at 20 MHz, What is the power per unit bandwidth radiated at the source?
Take the earth-Jupiter distance as 40 light -minutes and assume that the source radiates isotropically.
17-17 Red shifts. Powers, Some radio sources have been identified with optical objects and the Doppler or red shift z ( = AA/a) measured from an optical spectrum.
Assume that the objects with larger red shift are more distant, according to the Hubble relation v m — 1 c ^ m + 1 JT0 where R = distance in megaparsecs (1 megaparsec = 1 Mpc = 3.26 106 light-years), v = velocity of recession of object in m s“ \ m = {z + l) 2 c - velocity of light and Hq = Hubble's constant = 75 km s' 1 Mpc” 1 , Determine the distance R in light-years to the following radio sources: (a) Cygnus A (prototype radio galaxy), z = 0.06; (ft) 3C273 (quasistellar radio source, or quasar), z = 0.16; and (c) OQ172 (distant quasar), z = 3.53. The above sources have flux densities as follows at 3 GHz: Cygnus A, 600 Jy; 3C273, 30 Jy; OQ172, 2 Jy (1 Jy = 10" 26 W m" Hz" }. (d) Determine the radio power per unit bandwidth radiated by each source.
Assume that the source radiates isotropically.
17-18 Critical frequency, MUF. Layers may be said to exist in the earth’s ionosphere where the ionization gradient is sufficient to refract radio waves back to the earth, [Although the wave actually may be bent gradually along a curved path in an ionized region of considerable thickness, a useful simplification for some situations is to assume that the wave is reflected as though from a horizontal perfectly con-ducting surface situated at a (iftrtuuf) height The highest frequency at which this layer reflects a vertically incident wave back to the earth is called the critical frequency f 0 . Higher frequencies at vertical incidence pass through. For waves at oblique incidence (4 > 0 in Fig, PI 7-18) the maximum usable frequency (MUF) for point-to-point communication on the earth is given by MUF =fjcos 0, where £ wangle of incidence. The critical frequency / 0 = 9^/N, where N = electron density (number m“ 3 ). N is a function of solar irradiation and other factors. Both f 0 and h vary with time of day, season, latitude and phase of the 11 -year sunspot cycle. Find the MUF for {a) a distance d = 1.3 Mm by fyiayer (ft = 325 km) Earth's surface Figure PI7-I& Communication path via reflection from ionospheric layer.
problems 803 reflection with F 2 -layer electron density N = 6 x 10 tl m" J ;(6)a distance d — L5 Mm by fylayer (ft = 275 km) reflection with N = 10 12 m -3 ; and (c) a distance d = 1 Mm by sporadic £-!ayer (ft = 100 km) reflection with N = 8 x 10 11 m _i Neglect earth curvature.
17-19 mUF for Clarke-orbit satellites.
Stationary communication (relay) satellites are placed in the Clarke orbit at heights of about 36 Mm. This is far above the iono-sphere, so that the transmission path passes completely through the ionosphere twice, as in Fig. P17-19, Since frequencies of 2 GHz and above are usually used, the ionosphere has little effect. The high frequency also permits wide bandwidths.
If the ionosphere consists of a layer 200 km thick between heights of 200 and 400 km with a uniform electron density N = 10 2 m‘ find the lowest frequency (or minimum usable frequency, mUF) which can be used with a communication satellite (a) for vertical incidence and (b ) for paths 30° from the zenith, (c) For an earth station on the equator, what is the mUF for a satellite 15 n above the eastern or western horizon?
To communication satellite Figure P17-19 Communication path via geostationary Garke-orbit relay satellite.
17-10 S/M ratio. Show that the S/N ratio for a radio link with t W transmitter and isotropic antennas is S X1 N \6n 2 r 2kT svt where symbols are as given in (17-3-6) and (17-3-8).
17-21 Disc backscatter. Show that the back scattering efficiency of a large broadside flat disc is twice its back sea tiering directivity or Cf - 2Dabj .
17-22 Large sphere. Show that the total scattering cross section of a large sphere (radius r X) is twice its geometric or physical cross section.
7-23 Effect of resonance on radar cross section of short dipoles-(a) Calculate the radar cross section of a lossless resonant dipole (ZL = —jX A ) with length = X/ 10 and diameter = A/100. (See Secs. 2-14, 2-20 and 9-17).
(ft) Calculate the radar cross section of the same dipole from Table 1 7-2.
(c) Compare both values with the maximum radar cross section of Kouyoumjian shown in Fig.
17-11. Comment on the results.
4 „^ ^ 17-24 Loop, wire, SP>« »« disc radr "^^f^onanceTmc^rtha^he radar cross S^pS“oT^« .1.™m resonances. Why is the loop's , the highest'?
CHAPTER 18 ANTENNA MEASUREMENTS 18-1 INTRODUCTION. The understanding of physical phenomena involves a balance of theory and experiment. Since theoretical analyses usually treat idealizations or simplifications of actual situations, theory may only approx-imate the real world. So while theory is essential to our understanding, experi-mental measurements determine the actual performance, but only if the measurements are done properly.
In this chapter methods and techniques are discussed for experimental mea-surements on antennas. There are sections on the measurement of pattern, gain, current distribution, impedance and polarization. According to the reciprocity relation, the same pattern will be measured whether the antenna is transmitting or receiving. Reciprocity also applies to certain other characteristics, 1 so that it is convenient in some cases to regard the antenna as a radiator and in other situ-ations as a receiver.
18-2 PATTERNS- The far-held pattern of an antenna is one of its most important characteristics. The complete field pattern is a 3-dimensional or space pattern and its complete description requires field intensity measurements in all directions (over 4n sr).
L , Consider that the antenna under test is situated at the origin of the coordi-nates of Fig.
18-1 with the 2 axis vertical. Then patterns of the Q and tj> com-1 But not to current distributions, 805 906 IB ANTENNA MEASUREMENTS Figure IB-1 Antenna and coordinates for pattern measurements.
ponents of the electric field <£, and £,) are measured as a function of along constant 9 circles, where d> is the longitude or azimuth angle and 9 the zenith angle (complement of the latitude or elevation angle). These patterns may be determined by moving the measuring antenna with the antenna under test fixed or by rotating the antenna under test on its vertical (2) axis with the measuring antenna fixed. For complete information, the phase angle 5 between £9 and E is also required to establish the polarization, although the polarization ellipse may also be determined by rotation of a linearly polarized measuring antenna (see Sec.
1 8-9). With sufficient data, 3-dimensional intensity diagrams of Ee , £# d can be produced. An alternative is to make a 3-dimensional contour map of the power (proportional to E\ + £j) and superimpose polarization ellipse axes as suggested in Fig.
18-2.
Although detailed pattern measurements as above are sometimes required, fewer patterns are frequently sufficient. Thus, suppose that the antenna is a direc-tional type with a main beam in the x direction, as suggested in Fig.
18-J. then two patterns, called principal plane patterns , bisecting the mam beam may suffice.
If the antenna is horizontally polarized, then and xy plane patterns of £, as indicated in Fig. 18-3u, are measured.
If the antenna is vertically polarized t en xz and JC>- plane patterns of £„, as indicated in Fig. 18-3h, are measured.
If t e antenna is elliptically or circularly polarized, both sets of measurements (4 patterns) plus axial ratio data are required. Even if the antenna is believed to be linearly polarized, measurement of the 4 patterns plus axial ratios may be desir-able to establish polarization purity.
IS- 3 PATTERN MEASUREMENT ARRANGEMENTS 807 1 80° Figure 18-2 Three-dimensional power pattern in 5-dB increments for circularly polarized antenna with main beam vertical (in direction of z axis or out of the page!- Near sidelobes are 10 dB down, others 15 and 20 dB down. The polarization ellipses (for circular and elliptical polarization) are shown dotted with the solid orthogonal lines indicating major and minor axes. For essentially pure linear polarization (vertical, horizontal or slant) the ellipse collapses to the major axis line. The solid contours represent signal strength. The main beam is circularly polarized (CP), while the near side-lobes are elliptically polarized with the major axis vertical (VHP), Other minor lobes are linearly polarized vertically (VLP) S horizontally (HLP) or at a slant angle (SLP).
To summarize, the 4 patterns are: E^(d = 90°, 0) — pattern of component of electric field.^as a function of in xy plane = 9(T) ^ E^d = 0°) — pattern of component as a function of 0 in xz plane ( — 0 ) eJo = 90°, 4>) = pattern of 0 component as a function of in xy plane (0 = 90°) tf> = 0°) = pattern of 0 component as a function of 0 in xz plane { = 0°) 18-3 PATTERN MEASUREMENT ARRANGEMENTS, Consider the arrangement in Fig. 18-4. with the antenna under test acting as a receiving 18-3 PATTERN MEASUREMENT ARRANGEMENTS 809 antenna situated under suitable “illumination” from a transmitting antenna as suggested. The transmitting antenna is fixed in position and the antenna under test is rotated on a vertical axis by the antenna support shaft. Assuming that both antennas are linearly polarized, the Et (6 = 90°, 0) pattern is measured by rotat-ing the antenna support shaft with both antennas horizontal as in Fig. 18-4. To measure the (0, 0 = 0) pattern, the antenna support shaft is rotated with both antennas vertical.
Indication may be on a direct-reading meter calibrated in field intensity or the meter may always be adjusted to a constant value by means of a calibrated attenuator. Where many pattern measurements are involved, work is facilitated with an automatic pattern recorder.
18-3a Distance Requirement for Uniform Phase. For an accurate far-field or Fraunhofer pattern of an antenna a first requirement is that the measurements be made at a sufficiently large distance that the field at the antenna under test approximates a uniform plane wave. Suppose that the antenna to be measured is a broadside array consisting of a number of in-phase linear elements as suggested in Fig 18-5. The width or physical size of the array is a. At an infinite distance normal to the center of the array, the fields from all parts of the array will arrive in the same phase. However, at any finite distance r, as in Fig. 18-5, the field from the edge of the array must travel a distance r + & and, hence, is retarded in phase by 360“^ with respect to the field from the center of the array.
If & is a large enough fraction of a wavelength, the measured pattern will depart appreciably from the true far-field pattern.
1 Referring to Fig. 18-5, r 2 + 2rd + d2 = r 2 + W If S a and <5 r t T a 1 -~ j • /"Broadside array Figure 18-5 Geometry for distance requirement.
If ihe tetance is insufficient the near field or Fresnel pattern . measure^ In general, tins pattern a function of the distance at which it is measured. However, srncc the far-field pattern .s the Fourvr transform of the aperture field distribution (see Sec. 11-22), » . posstWe to deduce afar-fieU pattern Jrom near-field measurements.
See, for example, R^CJohnson. », A p ““ "d ' “Determination of Far Fidd Pattern, from Near Field Measurements.
Proc. IEEE. 61, 1668-1694, December 1973.
810 LH ANTENNA MEASUREMENTS Table 18-1 Tolerable phase difference data Maximum tolerable phase difference, deg d k Thus, the minimum distance r depends on the maximum value of <5 which can be tolerated. Some workers 1 recommended that <5 be equal to or less than Xj 16.
Then In general^ the constant factor [equal to 2 in (3)] may be represented by k Thus, r > k -(4) The phase difference equals 360° 5//, which for 6 = a/16 is 22. 5°. In some special cases phase differences of more than 22.5° can be tolerated and in other cases less. Hacker and Schrank 2 indicate that on very low sidelobe antennas (-30 to — 40 dB), phase differences of 5° or less may be required to resolve near-in sidelobes but for measurements of the far-out sidelobes 22,5 is satisfac-tory. Table 18-1 gives the constant factor k in (4) for 5 values of tolerable phase difference.
According to (4) the minimum distance of measurement is a function of both the antenna aperture a and the wavelength A. In the case of antennas of large physical aperture and small wavelength, large distances may be required For 1 C. C Cutler, A, P. King and W, E. Kock, “Microwave Antenna Measurements," Proc. IRE , 35, 1462-1471, December 1947.
2 P.
S. Hacker and H, E, Schrank, Range Distance Requirements for Measuring Low and Ullralow Sidelobe Antenna Patterns," IEEE Trans. Ants. Prop., AP-30, 956-966, September 1982.
18-3 PATTERN MEASUREMENT ARRANGEMENTS 811 example, consider a 3-GHz broadside beam antenna with a physical size of 20 m.
Taking k = 2, we obtain for (he minimum distance r = 4 km.
The constant phase front of a wave radiated by a point source is a sphere.
At a distance r the phase front departs by distances S r and 3" from the ends of a nominally perpendicular line of length a as indicated in Fig. 18-6. By comparing phases at the ends of the line with the phase at the center, 3 f and 3" can be determined and knowing the length a , the distance r of the source is given by .a 4{S' + <5") (5) Thus, changing the distance requirement of (3) around, we use a phase mea-surement to determine the distance. This is a passim distance measuring method.
It has been proposed that with radio telescopes in space on a very long baseline, the distance to all radio-emitting objects in the universe can be measured by this method.
1 18-3b Uniform Field Amplitude Requirement.
Further requirements are that the field in the test or target zone (volume containing antenna or scattering object under test) have small amplitude taper, small amplitude ripple and small cross-polarization (or high polarization purity). The polarization requirement means that for linear vertical or horizontal polarization the polarization state be on or dose to the equator of the Poincare sphere and for circular polarization on or close to one of the poles.
On outdoor measuring ranges, field variations can be produced by inter-ference of the direct wave with waves reflected from the ground, as in Fig.
18-7, or from other objects. The effect of the ground reflection may be reduced by using a directional transmitting antenna and placing both antennas on towers as in Fig. 18-8fl or near the edges of adjacent buildings as in Fig.
18-86. With such arrangements the amplitude of the reflected wave is reduced since the ground-ward radiation from the transmitting antenna is less and also since the path N. Kardashev it at., Acad, Sri, USSR, Space Res. Inst, Repl. PR-373, 1977, 812 IS ANTENNA MEASUREMENTS TACt Transmitting ' esi antenna Direct location Specular reflection point ^ lcs ^ati°n Transmitting Antenna (a) <> Figure 18-8 Antenna test arrangements.
length of the reflected wave is considerably greater than the path length of the direct wave.
In a typical case, the variation in field ntensUy as a function of height at the test location may be as indicated by the solid curve in Fig.
18-transmitting antenna is directional and is at a fixed height h. There is a consider-able target zone region near the height h with but small ripple. If the transmuting antenna is nondirectional the ripple is much greater, as suggested by the dashed curve in Fig.
18-9.
Relative field intensity at test location Figure 18-9 Variation or ripple in field intensity with height at the test location with a directional transmitting antenna relatively dose (solid curve). If the transmitting antenna is nondirectional in the vertical plane, the ripple is greater (dashed curve).
PATTERN MEASUREMENT ARRANGEMENTS 813 Transmitting antenna «o Ground Relative field intensity Figure 18-10 Variation of field intensity with height at the test location with a directional transmit-ting antenna at a large distance.
Sometimes the distance requirement of (4) is so large that the tower height needed may be impractical. In this case, the test antenna can be situated in a region of maximum field intensity such as at the heights h x or h 2 in Fig.
18-10.
(See a]so Prob.
16-14.) This arrangement has the limitation that the height of the test antenna may need to be adjusted for each change in frequency.
To reduce ground reflection, the ground can be covered with absorbing material around the specular reflection point or one or more conducting fences can be installed. However, the edge of a fence can diffract Waves into the test area and generate ripple.
Ranges constructed indoors require appropriately placed wave-absorbing material. In the simple, inexpensive range shown in Fig. 18-11, absorbing panels are placed at the specular reflection points on the walls, ceiling and floor as suggested. With small aperture antennas the distance requirement may be satis-fied while with a directional transmitting antenna and the absorbing panels the field uniformity may be adequate.
PLAN VIEW Back wall Figure 18-11 Simple low-cost indoor absorber rangC in plan view with wave-absorbing panels only at specular reflection points.
For minimum field taper across the test volume, the transmitting antenna beam width should be considerably greater than the test volume Jangle (HPBW > 3} but considerably smaller than the angle of specular- reflect ion (HPBW ^ ffy.
SI4 IS ANTENNA MEASUREMENTS 18-3c Absorbing Materials. The wave-absorbing materials mentioned above, and more extensively in the following sections on anechoic chambers and compact ranges, are now an integral part of antenna technology. They are used both in measurement ranges and also as antenna components for reducing side-and back-lobe radiation.
The use of a sheet of space cloth (Z = 377 Q per square) placed /0/4 from a reflecting plate to completely absorb a normally incident wave was discussed in Sec 2-18 This technique was invented by Winfield Salisbury 1 at the Harvard Radio Research Laboratory during World War II and the resistive (carbon-impregnated) cloth sheets he used are called Salisbury screens.
At normal incidence the arrangement gives a 1 .3 to 1 bandwidth for a reflected wave at least 20 dB down.
2 The transmission line equivalent is shown in Fig. 18-1 2a with the charac-teristic impedance of the transmission line equal to 377 CT For simplicity let us (d) Salisbury sheets .Reflecting plate Figure 18-12 (a) Transmission line and single load with 2„/4 stub as m&tched termination.
(b ) Same configuration in terms of normalized impedances, (c ) Line with 6 loads distributed over U0 as wideband matched termination, (d) Space equivalent with stack of Salisbury sheets.
1 W, W. Salisbury, "Absorbent Body for Electromagnetic Waves," U.S, Patent 2,599,944, June 10, 1952.
.
, .
1 The wave absorption resonates also for sheet-to-plate distances of 3 A/4, 5A/4, etc,, but the band-width is narrower.
]6-i PATTERN MEASUREMENT ARRANGEMENTS 815 Flgm 18-13 Path of normalized admit-tance Y from short via incremental steps to match point at center of Smith chart for the line with 6 distributed loads of Fig. t8-12r <5 to l bandwidth for reflected wave 20 dB down).
divide by 377 ft, obtaining normalized (dimensionless) impedances as in Fig. 18-12.
Consider now the situation shown in Fig.
18-12c with a number of resist-ances shunting the line over a distance of about 1A + The (normalized) resistances range from small to large values with distance from the short. The spacings also increase with distance from the short.
1 As shown in Fig. 18-13, the path of the normalized line admittance Y moves from the Y — oc position at the short via incremental steps of distance and conductance to the center of the chart (match point). The advantage of this incrementally tapered termination is that it provides a low reflection coefficibnt over wider bandwidths than the single- resistor termi-nation of Fig.
1 8-12a and b. The space equivalent of Fig. 18-12e is shown in Fig.
18-1 2d consisting of a stack of Salisbury sheets with impedances per square as indicated and backed by a reflecting plate. Stacks of thiskind whh sheets sandwiched between layers of plastic (dielectric) were developed in Germany during World War II by J. Jaumanm An historical summary of the development of wave-absorbing material and its application is given by Emerson.
2 1 Both resistance and spacing increase in an approximately exponential manner.
2 W. H. Emerson, “Electromagnetic Wave Absorbers and Anechoic Chambers through the Years," IEEE Trans, Ants. Prop ., AP-21, 484-489, July 1973 (49 references).
816 IB antenna measurements Increasing both the resistance and number of sheets (and decreasing their spacing) results in the limit in a continuously tapered medium.
If both the permeability p and permittivity e of a medium include a loss term but with = ^ ,6) the medium, although lossy, will have a real impedance equal to that of free space as given by = /£ = [K ° [K = 377 IK = 377 q V £ V % V £r V £r In principle, a wave incident on a uniform medium of such material can enter it without reflection and, if the medium is thick enough, be completely absorbed. Such a medium, being uniform, is in contrast to the tapered media discussed above.
Example 1. Find the reflection coefficient \p„\ for a 3-mm thick absorbing sheet backed by a flat perfectly conducting metal plate at 3 GHz if the constants of the sheet are a = 0, \ir = iT - 10 - jlO, Solution. The propagation constant "/ = ®+j0=J^ (10-/10) (8) -0 c 310‘ and ^ = 7 = J7705 ^ a ' m Therefore, a = 2n x 100 = 628 Np m 1 and the relative field intensity of the wave emerging from the sheet after reflection from the metal plate is given by — = ] p | = e' J“ = (-“ 3 ',, = 0.023 Eo or down 33 dB from the incident wave.
Although lossy media with pr = £r appear attractive m principle, the param-eters of the more popular types of absorber are typically: pr = 1 and zT = 2 -j L Popular shapes are in the form of pyramids and wedges as illustrated ifn Fig.
18- 1 4a and b. The pyramids behave like a tapered transition (as discussed above) for normal (nose-on) incidence.
However, DeWitt and Burnside 1 find that pyramid absorbers tend to scatter as a random rough surface with large reflec-1 B T.
DeWiit and W r D.
Burnside, “Electromagnetic Scattering by Pyramidal and Wedge Absorber,” IEEE Trans. Ants , Prop., 1988.
18-3 PATTERN MEASUREMENT ARRANGEMENTS 817 Angle of incidence Wedges 1B-I4 (a) Pyramid and (b) wedge forms of wave absorbers.
tion coefficient at large angles of incidence, wedges being far superior at these angles provided the wave direction is nearly parallel to the ridge of the wedge. At large incidence angles the wave direction is almost broadside to the side faces of the pyramids, resulting in reflection due to media mismatch.
Example 1 (a) Find the normal (nose-on) reflection coefficient at 3 GHz (4 = 100 mm) for an array of pyramids 30 cm from tip 1 to base with a = 0,^ = 1 and ^ — 2 - jt. (b) Find the reflection coefficient at 10 GHz (2 = 30 nun) assuming that J Q J and the attenuation constant a = ^ x -0.35 = 22 Np m“' The reflection coefficient is then IpJ = e-2 22 ' 1 =0.0123 (a) At 3 GHz this is -38 dB.
(b| At 10 GHz it is about - 125 dB.
In practice, the reflection coefficient is unlikely to be as small as this at 10 GHz although it may be substantially smaller than at 3 GHz, The inhomoge-neity of some commercial absorbers can also increase the reflection coefficient and backscatter.
1 18-3d The Anechoic Chamber Compact Range. The transition from the simple indoor range of Fig.
18-1 1 to what might be called an anechoic [no echo) chamber is accomplished by completely covering all room surfaces with absorb-ing material Thus, the side walls, ceiling and floor are covered by wedge absorbers with a ridge direction parallel to the path from the transmitter to the test site, while the back wall and wall behind the transmitter are covered with pyramids. The philosophy is to provide a nonreflecting environment like in outer space except that the walls are at ambient ( ~ 300 K) temperature instead of 3 K (or somewhat more), but with the distinct advantage that the room provides shielding from all of the external electromagnetic noise and interference (natural 1 B.
T.
DeWitt and W.
D.
Burnside, " Electromagnetic Scattering by Pyramidal and Wedge Absorber," IEEE Trans. Anis . Prop. y 1988.
18 3 PATTERN MEASUREMENT ARRANGEMENTS 819 Edge scattered for near-point (feed wave-absorber field point for far-condition field condition) Figure 18-15 Compact range configuration-Radiation scat-tered from edges of the parabol-ic reflector degrades desired plane (far-held) wavefront at test site into one with ripples.
and man-made). However, simple indoor ranges are usually limited by the dis-tance requirement, A parabolic reflector is a spherical -to-plane-wave transformer and may be used to produce a plane wave in the test zone of an indoor range as suggested in Fig, 18-15.
In effect, the parabolic reflector moves the far-field region in very close, making this foreshortened or compact range equivalent to a much larger conventional range.
1 A conventional parabolic dish reflector will diffract significant radiation from its edges into the test zone of a compact range, limiting the usable test volume as suggested in Fig. 18-15.
Serrated edges and absorber material are sometimes used to reduce this scattering.
Burnside et at.
2 have shown that a rolled edge is effective in reducing the edge scattering effect. A minimum radius of curvature r > /L/4 at the lowest frequency is required. Furthermore, it is shown by Pistorius and Burnside 3 that diffraction effects can be made still smaller with a very smooth or blended transition of the reflector surface from the parabola to the rolled edge, as suggested in Fig.
18-16. The compact range with 5.8 m high elliptic rolled-edge parabolic reflector at the Electro-Science Laboratory at the Ohio State University is shown in Fig.
18-17.4 1 R C. Johnsoir, H. A, Ecker and R. A. Moore, "Compact Range Techniques and Measurements," tEEE Trans. dnts. Prop AP-17, 568-576 September 1969, 1 W D Burnside, M. C Gilrealh, B, M. Kent and G. C Clerici, “Curved Edge Modification of Compact Range Reflector," /£££ Trans. Ants. Prop., AP-35, 176-182, February 1987.
W. D.
Burnside, M. C Gilreath and B. M. Kent.
11 Rolled- Edge Modification of Compact Range Reflector," A\1 T A Synip., San Diego, 1984.
J C. W I. Pisiorius and W. D. Burnside, "An Improved Main Reflector Design for Compact Range Applications,” /£££ Trans. 4ms. Prop., AP-35, 342-346, March 1987.
4 E.
K.. Walton and J D. Young, "The Ohio State University Compact Radar Cross-Section Mea-surement Range." /£££ Trans, Ants. Prop., AP-32, 1218-1223, November 1984.
g20 I ANTENNA M H ASI JRKMUNTS -Figure IK-16 Rolled edge wilh smooth blended transition to parabola.
Pistorius and Burnside also indicate that combining a 4-point parabola as in Fig 18-18 with a blended rolled edge loads to even greater improvement, the 4-point design being superior to circular or rectangular shapes.
P The ultimate in compact range design, as proposed by Pistonus, Clcnci and Figure 18-17 Compact range a, the Flec.ro- Science Laboratory of the Ohm Stale 5.8-m high blended- rolled-edge parabolic dish reflector. The lest object pedestal » m the ground and the ttansmiller feed close to the floor between ,1 and the dish Wedge .ta°rkn me he Boon ceiling and side walls with pyram.d absorbers on the wall behind the dish and the back wall (behind our point -of- view in the photo). {Courtesy W . D, fiurnsnte, iifcrfro-Stttfmf a oral or>-]^3 PATTERN MEASUREMENT ARRANGEMENTS 821 4 -point Figure 18-18 Four-point parabola compand with circular and rectangular designs. {After C W.
I.
Pis-torius and W< D. Burnside, “An Improved Main Reflec-tor Design for Compact Range Applications,” IEEE Trans. Ants. Prop, AP-35, 342-346 p March 1987.) Burnside, 1 is a dual-chamber Gregorian fed system as shown in Fig. 18-19. This arrangement can provide very small taper, ripple and cross-polarization in the test zone.
Gating (or turning the receiver on and off) in the interval of pulse return from the target can further improve performance and the ability to measure the echo from objects with very small radar cross section.
fine 18-19 Dual-chamber Gregonan-fed Compact Range (After C. W. I. Piswrius, G CUrici and n i _ _ iCtj. Cmdiivirt B/in/tf A nniiratinn.-IEEE Trans. Ant. Prop., AP-36, 1988.) . C W Pistorius, G Oeriti and W. D. Burnside, “A Dual-Chamber Gregorian Subreflector System for Compact Range Applications.- IEEE Trans. Ana. Prop AP-. 1988.
822 18 ANTENNA M EASU JIEM ENTS To Andromeda Figure 18-20 Big Ear radio telescope of the Ohio State University as a compact range for scattering measurements of large objects. (Sec Fig. 12-51 for photograph.) Actual full-size vehicles, aircraft and in-flight helicopters can be accommodated. The flat reflector deflects waves into the empty sky, replacing absorbing materials used in standard compact ranges.
For lower-frequency operation, even large indoor compact ranges may be too small. Although not initially designed as one, the Ohio State University radio telescope (photo in Fig. 12-51) is almost ideally suited as a compact range for frequencies from 3 GHz to 100 MHz or even lower, and is used for this purpose.
Figure 18-20 shows the arrangement for measuring the echo area of full size vehi-cles, aircraft and in-flight helicopters.
1 Although a prime objective of a compact range is usually to provide a far-held environment in the test zone, there are situations where near-field pat-terns are desired, such as where the test object may actually be located in the near field.
It is shown by Rudduck et al.
2 that by defocusing a compact range reflector it is possible to produce a spherical-wave environment in the test zone with an adjustable radius of curvature to suit a wide range of near-field condi-tions. The spherical wave is produced by moving the feed to points along a line from the focal point to the center of the parabola, such as the point indicated in Fig. 18-15.
18-3e Pattern and Squint Measurements Using Celestial and Satellite Radio Sources, Celestial and satellite radio sources can be used to measure the 1 J. D.
Kraus, “Ohio States Big Ear Detects Edge of the Universe and Doubles as a Compact Ranged IEEE /ims. Prop. Soc. Newsletter, 27. 5-10, January 1985.
3 R. C Ruddock, M, C Liang, W. D, Burnside and J.
S. Yu, "Feasibility of Compact Ranges for Near-Zone Measurements,” IEEE Trans : Ants. Prop., AP-3S, 280-286, March 1987.
18 -4 PHASE MEASUREMENTS 823 I far-field patterns and squint of antennas, especially ones with large apertures in terms of wavelengths where the distance to the far field is large (10s or 100s of kilometers). For pattern measurements, the radio source should have a small angular extent (much less than the antenna HPBW), be strong and isolated from nearby sources. For squint measurements, the position of the source should be accurately known. The table in Sec. A-8 (App. A) lists a few celestial radio sources which meet most or all of the above requirements. Most of these sources are at distances of more than l billion light-years (1 light-year = 10 13 kilometers).
For more details see Kraus.
1 Celestial radio sources can also be used for antenna gain, beam efficiency and aperture efficiency measurements (see Sec. 18-6e).
18-4 PHASE MEASUREMENTS The preceding sections on pattern mea-surements deal only with the magnitude of the field intensity. To measure the phase variation of the field, an arrangement such as shown in Fig. 18-21 can be used. The antenna under test is operated as a transmitting antenna. The output of a receiving antenna is combined with the signal conveyed by cable from the oscillator. The receiving antenna is then moved so as to maintain either a minimum or a maximum indication. The path traced out in this way is a line of constant phase.
In another type of measurement the receiving antenna is moved along a reference line, A calibrated line stretcher or phase shifter is then adjusted to Tran smitt i ng Probe antenna y path 1 J. D : Kraus, Radio Astronomy, 2nd ed., Cygnus-Quasar, 1986, chap.
6.
824 Lft ANTENNA MEASUREMENTS maintain a maximum or minimum indication. The measured phase shift can then be plotted as a function of position along the reference line.
1 18-5 DIRECTIVITY. The directivity of an antenna can be determined from the measured field pattern Thus, as defined in Chap. 2, the directivity of an antenna is 4/T Ml n = “— (1) || f{0, 4>) sin 0 dB d where f{6 , ) = relative radiation intensity (power per square radian) as a func-tion of the space angles 0 and $ (see Fig.
18-1) Since the radiation intensity is proportional to the square of the field intensity, the directivity expression (l) can be written as || Pn(8, <fi) sin 8 dO where. P„(0, ) = normalized power pattern - {Ei + EjMEj + Eft™ The directivity is determined from the shape of the field pattern by integra-tion and is independent of antenna loss or mismatch 18-6 GAIN. The gain of an_antenna over an isotropic source is defined in Chap. 2 as C 0 = k0 D (1) where G 0 -- gain with respect to an isotropic source (G without a subscript indi-cates the gain with reference to some antenna other than an iso-tropic source) D = directivity k 0 = ohmic loss factor (0 < k0 < ) The factor k 0 = l if the antenna is lossless.
It is assumed that the antenna is matched.
18-6a Gain by Comparison. Gain may be measured with respect to a com-parison or reference antenna whose gain has been determined by other means. A ' C. C. Cutler, A.
P. King and W. E. Kock, Microwave Antenna Measurements,' Proc. IRE , 35, 1462-1471, December 1947.
H. Knitter, in S. Silver (ed), Microwave Antenna Theory and Design, McGraw-Hill, New York, 1949, chap, 1 5, p. 543.
Harley lams, “Phase Plotter for Centimeter Waves," RC/I Kec., 8, 270-275, June 1947. Describes automatic device for plotting phase fronts near antennas.
1 18-6 GAIN 825 Transmitting antenna Calibrated attenuator Oscillator .Antenna under reference antenna Figure 18-22 Gain measurement by comparison: A/2 dipole antenna or a horn antenna are commonly used as references. The gain G is then given by where P x = power received with antenna under test, W p 2 = power received with reference antenna, W V l = voltage received with antenna under test V y z = voltage received with reference antenna, V It is assumed that both antennas arc properly matched.
If both are also lossless and the reference is a A/2 dipole, the gain G0 over a lossless isotropic source is G0 = 1.64 G = 10 log (1.64 c; (dBi) (3) The comparison should be made with both antennas in a suitable location where the wave from a distant source is substantially plane and of constant amplitude. The requirements of Secs 18-3a and 18-3b should be fulfilled.
Both antennas may be mounted side by side as in Fig. 18-22 and the com-parison made by switching the receiver from one antenna to the other. The ratio VJV 2 is observed on an output indicator calibrated in relative voltage. An alter-native method is to adjust the power radiated by the transmitting antenna with a calibrated attenuator so that the received indication is the same for both antennas. The ratio P VIP 2 is then obtained from the attenuator settings Mounting both antennas side by side as in Fig.
18-22 but in too close proximity may affect the measurements because of coupling between the antennas. To avoid such coupling, a direct substitution may be made with the idle antenna removed to some distance.
If the antennas are of unequal gain, it is more important that the high-gain antenna be thus removed.
If the gain of the antenna under test is large, it is often more convenient to use a reference antenna of higher gain than that of a A/2 dipole. At microwave frequencies electromagnetic horns are frequently employed for this purpose.
1 H. Knitter, in S. Silver (ed.), Microwave Antennas, McGraw-Hill, New York, 1949, chap. 15, p. 543.
826 IS ANTENNA MEASUREMENTS Short-wave directional antenna arrays, such as used in transoceanic com-munication, are situated al a fixed height above the ground. The gain of such antennas is customarily referred to either a vertical or a horizontal a/2 antenna placed at a height equal to the average height of the array. This gain comparison is at the elevation angle a of the downcoming wave. If the directional antenna is a high-gain type and any mutual coupling exists between it and the A/2 antenna, the directional antenna can be rendered completely inoperative by lowering it to the ground or sectionalizidg its elements when receiving with the A/2 antenna.
In the above discussion it has been assumed that the antennas are perfectly matched. It is not always practical to provide such matching. This is particularly true with wideband receiving antennas that are only approximately matched to the transmission line. In general, another mismatch may occur between the trans-mission line and the receiver. In such cases the measured gain is a function of the receiver input impedance and the length of the transmission line.
1 To determine the range of fluctuation of gain of such wideband antennas with a given receiver as a function of the frequency and line length, the length of the line can be adjusted at each frequency to a length giving maximum gain and then to a length giving minimum gain. The average of this maximum and minimum may be called the average gain.
HWib Absolute Gain of Identical Antennas. The gain can also be measured by a so-called absolute method2 in which two identical antennas are arranged in free space as. in Fig. 18-23 a. One antenna acts as a transmitter and the other as a receiver. By the Friis transmission formula — = — A 1 (dimensionless) H) P t AV where P r = received power, W P t — transmitted power, W = effective aperture of receiving antenna, m 2 A — effective aperture of transmitting antenna, m 2 A = wavelength, m r = distance between antennas, m The distance requirement of Sec. 18-3a should be fulfilled.
If r is large compared 1 J r D. Kraus, H. K. Clark, E. C Barkofsky and G. Stavis, in Very High Frequency Techniques, Radio Research Laboratory Staff, M<^jraw-Hill, New York, 1947, chap. 10, pp. 232 and 271, i C C. Cutler, A. F, King and W, E. Kock, “Microwave Antenna Measurements,” Proc. /RE, 35, 1462-1471, December 1947; also H. Knitter, m S.
Silver (ed.k Microwave Antennas, McGraw-Hill, New York, 1949, chap. 15, p. 543. The gain is absolute in the sense that it depends on distance and power measurements which are independent of the antenna itself or the gain of other antennas.
IS-6 CAIN 827 828 IS ANTENNA MEASUREMENTS and since it is assumed that A er = A eI , (4) becomes tor IP, (7) and G°~ l }p, Thus by measuring the ratio of the received to transmitted power, the distance r and the wavelength A, the gain of either antenna can be determined. Although it may have been intended that the antennas be identical, they may actually differ in gain by an appreciable amount. The gain measured in this case is Go = GoiGo^ ^ where G 0) - gain of antenna 1 of the “identical” pair Gq 2 = gan °f antenna 2 of the “ identical ” pair with both gains referred to an isotropic source. To find G 0 i and G n2 , the above measurement is supplemented by a comparison of each of the antennas with a third reference antenna whose gain need not be known. This gives a gam ratio between “ identical ” antennas of C’.Sx (9) where G x = gain of antenna 1 over reference antenna G 2 = gain of antenna 2 over reference antenna Then since we have r = = Got g 2 g B2 g01 — Go .
_ Go The U.S. National Bureau of Standards uses a modified 3-antenna tech-nique for accurate antenna gain measurements, 18-6c Absolute Gain of Single Antenna. Using radar techniques the method of the preceding section involving 2 or 3 antennas can be extended in several ways to measuring the absolute gain of a single antenna.
1 A, C. Newell, C F. Suibenrauch and R. C Baird, “ Calibration of Microwave Antenna Gam Stan-dards," JVoc, IEEE, 74 129-132, January 1986.
18-6 GAIN 829 1. By flat sheet reflector By replacing the second antenna of Fig. 18-23a with a sufficiently large, flat, perfectly reflecting sheet, as in Fig.
18-23b, the gain of the single (transmit ting-receiving) antenna is given by (7) where r now equals the distance from the antenna to its image behind the reflector. This distance must meet the far-field requirement and this may require a very large flat sheet reflector.
Z By reflecting sphere. As discussed in Sec.
17-5, the radar cross section tr of a perfectly reflecting sphere is equal to its physical cross section (tuj 2 ) when its radius a > /. With a sphere as the radar target, as in Fig, 18-23c, we have from the radar equation (17-5-6) that the antenna gain where r = distance from antenna to sphere, m a — radius of sphere, m The distance r must meet the far-field requirement while the sphere radius requirement is that a ^ L 3. By parabolic reflector. A more compact configuration involves the use of a parabolic reflector as in Fig. l8-23d with the antenna at the focus of the parabola. For this configuration the gain g = 4w^p, where rx = focal distance of parabola in wavelengths, dimensionless Equation (14) is identical to (7).
18-6d Gain by Near-FieW Measurements, Referring to Fig, 18-23?, mea-surements of the near field of a large antenna with a probe can be used to obtain the gain from BracewelFs relation (12-9-26) as <,5) a where E(x, y) = electric field at any point x, y in the aperture, V m E„ = -~ || £(jc, y) dx dy = average electric field over -I the aperture V m A p = area of (aperture) plane over which measurements are made, m 2 = complex conjugate Antenna pattern, dB g30 19 ANTENNA MEASUREMENTS 18-6 GAIN 831 Figiire 18-24 Far-field pattern in dB of high-gain paraboloidal reflector antenna over a 64 square-degree area centered on the beam axis as obtained from near-field measurements by Newell, Stuben-rauch and Baird of the U.S, National Bureau of Standards, {From A C. Newell, C. F. Sftrtenrauch and R. C, Baird, ” Calibration of Microwave Antenna Gain Standards , Proc. IEEE, 74, 129-132, January 1986 .
) It is assumed that all of the radiated power flows through A p .
This general method 1 is employed by the U,S. National Bureau of Stan-dards (Boulder, Colorado) for gain measurements to an overall accuracy of the order of ±0.2 dB. In addition, far-field patterns are obtained using the Fourier transform. An example of the pattern of a high-gain antenna obtained in this way is shown in Fig.
18-24. 2 18-6e Gain and Aperture Efficiency from Celestial Source Measure ments.
3 For gain measurements using a celestial radio source, an accurate flux density of the source is required and, generally, the source should be essentially unpolarized (less than 1 or 2 percent). Most of the sources in the table of Sec. A-8 meet these requirements, but since flux densities are given at only discrete fre-quencies it may be necessary to interpolate the fluxes at other frequencies.
1 W. H. Rummer and E, S. Gillespie, “Antenna Measurements," Proc. /EEE,66, 483-507, April 1978, 2 A. C, Newell, C, F. Stubenrauch and R. C Baird, “ Calibration of Midowave Antenna Gain Stan-dards," Proc. FEEE t 74, 129—132, January 1986.
3 See also Secs. 18- 3e and 12-9.
I From (17-2-8) the effective aperture A e of an antenna is related to the known flux density 5 and measured incremental antenna temperature A7^ as given by 2k A T a (16) from which the gain is G = Sirk A T a S k2 (17) where k = Boltzmann's constant = 1.38 x 10 -23 J K A 7^ = measured source temperature, K S = source flux density, W m 2 Hz -1 k = wavelength, m Thus, knowing S (from the table of Sec. A-8) and /, a measurement of AT^ deter-mines the gain. This measurement includes the effect of any (ohmic) loss in the antenna and any mismatch.
Example, Find the gain and aperture efficiency of the Ohio State University 110-m radio telescope antenna at 1.4 GHz if the measured increase in antenna temperature from Cygnus A = 687 K. The physical aperture is 2208 m 2 .
Solution. From the table of Sec. A-8, the flux density of Cygnus A at 1.4 GHz is 1590 Jy. From (16) the effective aperture A, 2k &T a 2 x 1.38 x 10“ 23 x 687 S ~ 1590 x 10“ 26 1 193 m 2 and from (17) the pin G 4nA € 4tt x 1193 0.214 2 = 3.27 x 10 5 = 55 dBi Since the physical aperture A p of the antenna is 2208 m± the aperture efficiency £ = dr = 11?? = 0.54 or 54 percent At 2208 The U.S.
National Bureau of Standards offers a service for measuring antenna gain using celestial radio sources.
1 See discussion of (12-9-43) to (12-9-47) for beam efficiency measurements with a celestial source.
1 R. C Baird “Microwave Antenna Measurement Services at the National Bureau of Standards." A nr, Meas . Symp. (>lnr. Mens. Tech. Xssor.), October 1981, 832 18 ANTENNA MEASUREMENTS 18-7 TERMINAL IMPEDANCE MEASUREMENTS.
In general, any antenna impedance Z A terminating a transmission line will produce a reflected wave with reflection coefficient pv and a voltage standing wave ratio (VSWR) related as follows: Ireflected voltage] | K\ VSWR — 1 ~ | incident voltage] | FJ VSWR 4-\ where ttuTvSWR is the ratio of the maximum to minimum voltage on the line.
The reflection coefficient is a complex quantity with magnitude | p„ I and phase angle 0V . Thus, P„ = lftl/0|i ^ The VSWR can be measured by moving a voltage probe along a slotted measuring line (Fig. .18-25). The value of |p„| is then given from (!) Using the probe to locate the voltage minimum point on the line, the reflection coefficient phase angle 0^ is found from ev = -£) (3 ) where = distance of voltage minimum from antenna terminals, m A = wavelength along line, m Knowing both the magnitude and phase angle of the reflection coefficient, the antenna impedance ZA is given by where Z { = transmission line impedance, fl In a slotted (coaxial) transmission line a probe is introduced through a longitudinal slot in the outer conductor as indicated in Fig.
18-25. Since currents Figure 18^25 Slotted coaxial Line for antenna terminal impedance measurements. The line is shown in longitudinal and transverse cross sections with movable probe for measuring voltage as a function of distance along the linr L9 7 TERMINAL IMPEDANCE MEASUREMENTS 833 flow parallel to the slot, little field escapes the line. With the probe connected to a voltage indicator, the voltage variation along the line can be determined, giving both VSWR and x vllt . Replacing the antenna with a short circuit, two successive minima V mip (short) are located on the line. Their separation is equal to A/2, With the antenna connected, the distance between a voltage minimum point V min (am) and F mjn (short) is equal to x vmt as shown in Fig.
1825. It is usually preferable to measure xvm with respect to the first F mj(l (short) rather than to the terminals, due to uncertainties from end effects, as from an insulator, which modify the electrical distance.
Instead of a slotted line, directional couplers can be used which give outputs proportional to the reflected and incident voltages V t and V { from which [p tf | can be determined as in (1) and the VSWR from VSWR = -—ft 1 (5) 1 -I ft !
With directional couplers and a sweep frequency generator, the VSWR can be monitored over a bandwidth and displayed continuously on a cathode ray tube (CRT) while adjustments are made on the antenna. If the phase difference of V r and Vi is also monitored, the antenna impedance can be displayed over a bandwidth on a CRT Smith chart/ Although the impedance measuring arrangement shown in Fig.
18-25 is appropriate for monopoles (or other unbalanced antennas), it can also be used to measure the terminal impedance of a center-fed dipole (or other balanced antenna) by measuring \ of the antenna and multiplying the measured impedance by 2. Thus, instead of measuring a balanced A/2 dipole with a 2-wire transmission line, measurements are made on ^ of the dipole or A/4 monopole (or stub) antenna with a large ground plane (Fig.
18-25). Ideally the ground plane should be perfectly conducting and infinite in extent to produce a perfect image of the stub antenna. The ground plane of finite extent used in practice should, therefore, be as large as possible. Even though the ground plane is several wavelengths in diameter, the measured impedance of a stub antenna varies appreciably as a function of the ground-plane diameter. 2 This variation is reduced as the ground-plane diameter is increased.
Meier and Summers 2 found that a large square ground plane results in about half the variation of impedance observed with a circular ground plane of approximately the same size.
The antennas were mounted symmetrically on both ground planes. The reduced variation with the square ground plane is presumably due to the partial cancellation of waves reflec-ted to the antenna terminals from the edge of the ground plane. These waves 1 For more details and procedures for using Smith charts see, for example, J. D. Kraus. Electromag-netic.^ 3rd ed. t McGraw-Hill, 1984, secs. JfV7 and 10-8.
1 A.
S. Meier and W. P. Summers, “Measured impedance of Vertical Antennas over Finite Ground Planes/' Proc. IRE, 37, 609 616, June 1949.
834 I AKTENNA MEASUREMENTS travel different distances on a square ground plane and, hence, all camota in the same phase. The ratio of the longest to the shortest distance is the ratio tf the diagonal of a square to the length of one side (1.41). With a circular ground plane and symmetrically located antenna, all the waves reflected from the edge return in the same phase.
, The ground (image) plane technique can also be used to advantage in mea-suring the terminal impedance of slot antennas. A sheet with a half-slot (equal in length to the full-slot but of the width) is butted against an image plane placed perpendicular to the slot plane. The half-slot is energized by a coaxial hue with the inner conductor connected to the terminal of the slot and the outer conduc terminated in the image plane. The termina! impedance of the u1 -lotjs mice the impedance of the half-slot. The impedance Zs/2 of the ha f- slot s related to the impedance Zda of the complementary stub antenna or half-dipole by sil 8869/ Whh horn or slot antennas that are fed with a waveguide, measurements of the field in the guide can be made with a slotted waveguide and probe arrange-mlntln this wly mf.urtmtn.s of .ho VSWR, reflection coeftan. end =,«».-lent load impedance may be obtained in a manner analogous to that used with a coaxial line, 18-8 CURRENT DISTRIBUTION MEASUREMENTS. In many cases it is important to know the current distribution along an antenna^ For example, if both the magnitude and phase of the current is known at all po along an antenna, the far field of the antenna can be calculated The current can be sampled by a small pickup loop placed close to t antenna conductor. Loop and indicator may be combmed into a smgk un t However, at short wavelengths the indicating instrument may be too large to place near the antenna without disturbing the field. To remove the ^icatorfro the antenna field the arrangement of F,g. 18-26 can be ^^. Here the loop pro-iects through a longitudinal slot in the hollow antenna conductor. The outpu cable from the loop is confined within the antenna conductor and is brought om through the end of a grounded stub as shown. The arrangement m Fig. 18-26 permits both amplitude and phase measurements. The phase is measured by comparison with a reference current as suggested by the dashed con ^s in the figure. The signal picked up by the loop is mixed with a ^^fwtthtte imately equal amplitude extracted by a probe on a matched slotted ‘'ne. With antenna sampling loop fixed, the line probe is moved to give a minimum ind ca-tion When the antenna sampling loop is displaced to a new location, the line probe is moved so as to maintain a minimum indication. The phase shift between the line-probe positions then equals the phase shift between the two antenna sampling loop locations. The phase shift is a linear unction of di®£notcm » me with matched termination. Assuming the phase velocity equa s that of light free space, the phase shift 6 along the line in degrees per unit length '^givenby 360W where is the free‘ sPace wavelength of the applied signal. T p 18-9 POLARIZATION MEASUREMENTS 835 Oscillator .Matched termination ^Slotted antenna ^Samphog .oop^ Receiver S 4 1 1 Small coaxial I line from ^Ground plane sampling loop Fignrc 18-26 Slotted antenna and sampling loop arranged for measurement of both current ampli-tude and phase.
change between two points on the line is then the distance between the points multiplied by 9.
In Chap. 9 on cylindrical antennas Fig, 9-18 shows the current distribution on a 52 cylindrical monopole antenna 0.22 in diameter measured using an arrangement similar to that in Fig. 18-26.
18-9 POLARIZATION MEASUREMENTS. Four methods for polariza-tion measurements are : 1. Polarization-pattern method. A linearly polarized Antenna is used to measure a polarization pattern and two circularly polarized antennas are used to deter-mine the hand of rotation.
2. Linear-component method. Two perpendicular linearly polarized antennas are used to measure the linearly polarized components of the wave and also their phase difference.
3. Circular-component method. Two circularly polarized antennas are used to measure the circularly polarized components of the wave of opposite hand and the phase angle between them, 4 Power measurement (without phase) method. Some waves may consist of the superposition of a large number of statistically independent waves of a variety of polarizations. The resultant wave is said to be randomly polarized or unpo-larized. Thus, in general, waves may be partially polarized and partially unpo-larized.
In ordinary communications the waves are usually completely polarized but in radio astronomy the waves from celestial sources are m general, partially polarized and in many cases completely unpolanzed. To deal with the most general situation it is convenient to use Stokes parameters. A detailed discussion of polarization parameters, Stokes parameters and 836 IS ANTENNA M E A 5L' R F.MENTS Table 18-2 Wave characteristics determined by power mea-surements of 6 antennas (no phase measurements required) VP HP + 45° -45° RCP LCP Wave dipole dipole dipole dipole helix belix VP l 0 1 2 i i i i 4 HP 0 l 1 2 1 2 4 i +45" LP 4 £ l 0 4 -45° LP 1 2 1 2 0 l i i i RCP 1 2 1 2 1 1 i 0 LCP 1 2 1 2 i i 0 i U npolarized 1 2 \ i 1 2 1 2 1 2 polarization measurements is given by Kraus.
1 Of interest here is that the polarization characteristics of a wave (including any unpolarized components) may be completely determined without any phase measurements by noting the power response or 6 antennas: 1. vertically polarized (VP), 1 horizontally po-larized (HP), 1 linearly polarized (LP) at a slant angle of +45", 1 linearly polarized (LP) at a slant angle of -45“ and 2 circularly polarized (CP) antennas, one right-circularly polarized (RCP) and the other left-circular ly polarized (LCP). The linearly polarized antennas may be dipoles and the cir-cularly polarized antennas monofilar axial-mode helices, one wound right-handed and the other left-handed. For a completely polarized wave only 3 independent measurements are necessary so there is some redundancy.
An example of the responses of the 6 antennas to a wave of unit incident power density is shown in Table 18-2. The power response of all 6 antennas is normalized to unity for- a wave of unit incident power density of the same polarization. We note that each type of wave polarization produces a different set of power responses.
18-9& PokmzationrPatterii Method.
In this method a rotatable linearly po-larized antenna, such as the 2/2 dipole antenna in Fig. 18-27a, is connected to a receiver calibrated to read relative voltage.
1 Let the wave be approaching (out of page). Then as the antenna is rotated in the plane of the page, the voltage observed at each position is proportional to the maximum component of E in the direction of the antenna. Such measurements of the incident wave with a rotat-able linearly polarized antenna do not yield the polarization ellipse of the wave but rather its polarization pattern (Fig. 18- 27b). Thus, if the tip of the electric vector E describes the polarization ellipse shown in Fig. 18-27b (dashed curve).
1 J. D, Kraus, Radio Astronomy, 2nd ed., CygiHis-Qua&ai\ 1986, chap, 4.
. In practice a linearly polarized antenna of considerable directivity is preferable to a simple A/2 dipole.
19-« POLARIZATION MEASUREMENTS 837 Figure 18-27 (a) Schematic arrangement for measuring wave polarization by the polarization+pattern method, (b) Measured polarization pattern and polarization ellipse for elliptical polarization, (c) Mea-sured polarization pattern for linear polarization with polarization ellipse collapsed to a Line, the variation measured with a linearly polarized receiving antenna is given by the polarization pattern in Fig. 18-275 (solid line). For a given orientation OP of the linearly polarized antenna, the response is proportional to the greatest ellipse dimension measured normally to OP . As shown in Fig, 18-275, this is the length OF.
If the linearly polarized antenna orientation is OQ, the response is pro-portional to the length OQ H . For the case or linear polarization, the polarization ellipse degenerates to a straight line and the corresponding polarization pattern is a figure-of-eight, as indicated in Fig, 18- 27c. By graphical construction as in Fig, 18-275 and c, the polarization ellipse can be constructed if the polarization pattern is known, or vice versa. To determine the direction of rotation of E an auxiliary measurement is necessary. For example, the output of 2 circularly po-larized antennas could be compared, one responsive to right- and the other to left-circular polarization. The rotation direction of E then corresponds to the polarization of the antenna with the larger response.
Thus, by this method the polarization ellipse can be drawn and the rotation direction indicated. Although such a diagram completely describes the polariza-tion characteristics of a wave, it is simpler to measure merely the maximum amplitude A/2 and the minimum amplitude B/2 and take the ratio of the two amplitudes which is the axial ratio of the polarization ellipse or simply the axial ratio (AR). The axial ratio is expressed so that it is equal to or greater than unity.
838 I ANTENNA MEASUREMENTS (ff) ^ Figure 18-28 Schematic arrangement for measuring polarization by the linear-component method with vertical and horizontal components given by (a) and phase by (b).
The axial ratio of the polarization ellipse of Fig. 18-27b is Thus, by specifying AR, the tilt angle and the rotation direction of E the polariza-tion characteristics are completely described. (See Secs. 2-34, 2-35 and 2-36.) 18_9b Linear-Component Method.
In this method 2 fixed linearly polarized antennas can be mounted at right angles, like the two 2/2 antennas in Fig.
18- 28a. The wave is approaching normally out of the page. By connecting the receiver first to the terminals of one antenna and then the other, the ratio £ 2/£, can be measured. Then, by connecting both antennas to a phase comparator, the angle <5 can be measured. This may be done as in Fig. 18-286, using a matched slotted line. From a knowledge of E lt E, and S the polarization ellipse can be calculated and the direction of rotation E determined.
18-9c Circular-Component Method.
In this method 2 circularly polarized antennas of opposite hand are connected successively to the receiver and the amplitudes EL and ER of the circularly polarized component waves measured.
The antennas can very conveniently consist of 2 long monofilar axial-mode helical antennas, one wound left-handed and the other wound right-handed as in Fig. 18-29. The left-handed helix responds to left-circular polarization and the right-handed helix to right-circular polarization (IEEE definition). The left-circular component E L of the wave is measured with the switch to the left as in Fig. 18-29 so that the receiver is connected to the left-handed helix. The right-circular component E R of the wave is measured with the switch thrown to the 18-9 POLARIZATION MEASUREMENTS 839 Wave direction Figure 18-29 Arrangement for measuring polariza-tion by the circular-component method. Left- and right-handed components are measured by individual helices and phase angle by rotating one helix with both connected.
right so that the receiver is connected to the right-handed helix. The axial ratio (AR) of the received wave is then given by According to (1) the axial ratio may have values between +1 and + oo and between -1 and - oo. For positive values of AR the wave is right-elliptical and for negative values it is left-elliptical The tilt angle i of the polarization ellipse may be measured by finding the direction of maximum E with a rotatable lin-early polarized antenna, or t may be determined with the helical antennas of Fig.
18-29 by rotating one helix on its axis with both helices connected in parallel to the receiver (switch segment up in Fig. 18-29). Assuming that the axes of the helices are in a horizontal plane, let the helix rotation angle be & and let its reference point (S' = 0) be taken when the receiver output is a minimum for a horizontally polarized incident wave. Then for any type of polarization with the polarization ellipse at a tilt angle t to the horizontal, t = $'fZ Thus, 3 measure-ments, El , E k and with the helical antennas determine the polarization char-acteristics of the received wave completely.
The circular-component method using helical antennas is suitable for mea-surements over a considerable frequency range. The accuracy depends on the circularity of polarization of the helices. This is improved (AR nearer unity) by making the helices long since In + 1 where n = number of turns of the helix 840 |g ANTENNA MEASUREMENTS 18-10 ANTENNA ROTATION EXPERIMENTS. Consider the radio circuit shown in Fig.
18-30a in which both the transmitting and receiving antennas are linearly polarized.
If either of the antennas is rotated about its axis at a frequency /(r/s), the received signal is amplitude modulated at this frequency.
The direction of rotation is immaterial.
Consider next the radio circuit shown in Fig.
18-30b in which one antenna is circularly polarized and the other is linearly polarized.
If one of the antennas is rotated about its axis at a frequency / (r/s), the received signal is shifted to F ±/, where F is the transmitter frequency. This experiment may also be conducted with 2 circularly polarized antennas of the same type. The frequency /is added or subtracted from F depending on the direction of antenna rotation relative to the rotation direction of E (or hand of circular polarization).
18-11 MODEL MEASUREMENTS.
Pattern and impedance measure-ments of actual antennas are often difficult or impractical because of the large size of the antenna system. In such cases a scale mode! of the antenna system may be built to a convenient size and then measurements made on the properties of the model.
1 This technique is especially useful in measuring patterns of antennas mounted on aircraft. Although the antenna proper may be small, it may excite currents over much of the airplane surface so that the entire airplane becomes part of the antenna system, and, hence, the measurements must be made of the antenna with airplane. Another advantage is that the patterns of antennas on aircraft in flight (remote from the ground) can be easily simulated by the model technique by placing the model on a suitable tower. To measure such patterns on actual aircraft is both tedious and expensive.
Let the scale factor for the model be p. Then any length dimension Lm on the model is related to the corresponding dimension L on the actual antenna by L„ = -U) P Then the frequency fm used to measure the model must be related to the fre-quency/used with the actual antenna by /. = pf <2> A further requirement of an accurate model for pattern and impedance measure-ments is that the conductivity of the antenna metal be scaled according to the relation o = p<t < 3> 1 George Sinclair. “Theory of Models of Electromagnetic Systems," Proc.. IRE, 36, 1364-1370, November 1948.
„ D JDr G. H. Brown and Remold King, " High-Frequency Model m Antenna Investigation, Proc. IRL ZZ, 457-480, April 1914, 18- Ll MEASUREMENT ERROR 841 Antenna axis Flgjvt lfr-30 Arrangements for antenna rotation experiments, (a) Antenna rotation produces ampli-tude modulation, (b) Rotating the monofilar axial-mode helix increases or decreases the signal fre-quency by the rotation rate.
where = conductivity of metal in model (j = conductivity of metal in actual antenna However, if a is large enough, the metal can be considered to be a “perfect conductor" (<r = oo) and the conductivity need not be modeled. Thus, actual antennas of copper can usually be modeled in copper.
It is assumed that ferro-magnetic materials are excluded from both actual antenna and model and that the model is measured in air.
18-12 MEASUREMENT ERROR All measured quantities involve error.
Thus, the measured area of a sphere might be given as 276 + 003 m , indicating an error or uncertainty of ±0.03 m 2 . If this is the root mean square (rms) or stan-dard deviation?
it means that the chances are roughly 2 to 1 that the true value is between the limits, i.e., greater than 2.73 and less than 2.79 m. For a finite number n of readings the rms deviation = jd\ + d+-f » -1 where. d u d2 , etc, are the measured deviations from the mean of a set of n obser-842 IS ANTENNA MEASUREMENTS Sometimes the error given is the probable error , which is 0,6745 times the rms error. A probable error indicates that the chances are even (or 1 to 1) that the true value is between the limits cited.
When an error is given it usually implies that a set of measurements has been made. A single measurement is anomalous, and any error associated with it must be an estimate.
Wherever measured values are cited in this book it is understood that they are subject to error whether explicitly stated or not. For example, the gain of an antenna may be quoted as 36.5 dBi. However, to be explicit, an appropriate error should be included Thus, if the error is ±0.5 dBi the gain should be given as 36.5 ± 0.5 dBi, However, for brevity, errors have usually not been included. An exception involves the antenna temperature measurements of Penzias and Wilson reported in Sec 17-2. Their results are 2.3 ± 0.3 K due to the atmosphere 0.8 ± 0.4 K due to ohmic losses <0.1 K due to back lobes into the ground 3.2 ±0.5 K total 1 Their measured sky temperature was 6.7 ± 0,8 K which, less 3.2 + 0,5 K, gave a residua] of 3.5 ± 1.0 K.
!
Penzias and Wilson's attention to the errors led to their discovery of the 3-K sky background for which they subsequently received a Nobel prize.
ADDITIONAL REFERENCES Burnside, W. D,, and R. W. Burgener "High Frequency Scattering by a Thin Lossless Dielectric Slab," IEEE Trans , Ants. Prop, AP-31, 104^1 10, January 1983.
Collinglon, G., Y, Michael, F. Robin and J. C Bolomey: “Quiet Microwave Field Mapping for Large Antennas," Microwave J., 15, 129—1 32, December 1982.
IEEE Standard Test Procedures for Antennas, IEEE Standard 149, 1979.
Johnson, R, C.: "Some Design Parameters for Point-Source Compact Ranges" IEEE Trans.
Ants.
Prop., AP-34, 845-847, June 1986.
Kummer, W. H., and E. S, Gillespie: “Antenna Measurements," Proc. /£££, 66, 483-507, April 1978.
Mostafavi, M., J. C Bolomey and D, Picard: "Far- Field Accuracy Investigation Using Modulated Scattering Technique for Fast Near-Field Measurements," IEEE Trans . Ants.
Prop., AP-33, 279-285, March 1985.
Nortkr, J. R., C. A, Van der Neut and D. E. Baker: "Tables for the Design of Jaumann Microwave Absorber,” Microwar# J., 30, 219-222, September 1987, Rhodes, D. R,: “On Minimum Range for Radiation Patterns" Proc. IRE, 42, 1409-1410, September 1954.
1 Note that these total errors are rss (root sum square) (quantities unrelated) not rms (root mean square) (single quantity error). Thus, 0 3 and 0.4 (in above tabulation) are rms but the total 0.5 is rss.
PROBLEMS 843 References on Radiation Hazards Mumford, W. W.: “Some Technical Aspects of Microwave Radiation Hazards," Proc. IRE, 49, 427-447, February 1961.
Tell, R. A,, and F. Harlen: "A Review of Selected Biological Effects and Dosimetric Data Useful for Development of Radio Frequency Safety Standards for Human Exposure," J. Microwave Power, 14,405-424, 1979.
PROBLEMS 1 18-1 Absorbing materia), 1/e depths. A medium has constants a = 102 U m~ l , pr = 2 and e, = 3.
If the constants do not change with frequency, find the 1 fe and 1 percent depths of penetration at (n) 60 Hz, (6) 2 MHz and (c) 3 GHz.
18-2 Lossy medium.
At 200 MHz a solid fern te-titan ate medium has constants o = 0, p r = 15(1 ->3) and et = 50(1 -j 1). Find (a)Z/Z0) (h) kik0 ,{c) vjv0 , (d\ l/e depth, (e) the dB attenuation for a 5-mm thickness and (/) the reflection coefficient pv for a wave in air incident normally on the flat surface of the medium. The zero sub-scripts refer (o parameters for air (or vacuum), 18-3 Lossy medium. Complex constants. A medium has constants <r = 3.34 U m 1 and p r = Rr = 5 + j2. Find the l/e depth of penetration at 30 GHz.
18^4 Attenuation by lossy slab. A nonconducting slab 200 mm thick has constants /r = £r = 2 — }2. Find the dB attenuation of the slab to a 600-MHz wave.
18-5 Attenuation of lossy sheet. A nonconducting sheet 6 mm thick has constants — £ r = 5 — jS. Find the dB attenuation of the sheet to a 300- MHz wave.
18-6 Attenuation of conducting sheet. A nonmagnetic conducting sheet 2 mm thick has a conductivity g = 10 J Um“\ Find the dB attenuation of the sheet to an 800-MHz wave.
18-7 Attenuation in lossy medium. A medium has constants «r = L112 x 10“ 2 U m -1 , = 5 — j4 and cr = 5 — j2.
At 100 MHz find (a) the impedance of the medium and (b) the distance required to attenuate a wave by 20 dB after entering the medium.
18-8 Absorbing sheet. A large flat sheet oFnonconducting material is backed by alumin-um foil. At 500 MHz the constants of the ferrous dielectric medium are = e.r = 6 — y6. How thick must the sheet be for a 500-MHz wave (in air) incident on the sheet to be reduced upon reflection by 30 dB if the wave is incident normally?
18-9 Reflection from dielectric medium. A plane 3-GHz wave is incident normally from air onto a half-space of dielectric with constants a = 0 S pr = 1 and er = 2 — j2.
Find the dB value of the reflected power.
18-10 CP wave reflection and transmission. A circularly polarized 200-MHz wave in air with E = 2 V m -1 (rms) is incident normally on a half-space of nonconducting medium with p r = er = 3 — j3. Find the Poynting vector (a) for the reflected wave and (b) for the transmitted wave at a depth of 200 mm in the medium.
18-11 Radar cross section. The radar return from an object on a compact range is 8 dB more than for a 10A diameter sphere when substituted for the object. What is the object's radar cross section?
1 Answers to starred () problems are given in App. D. Permittivity relations are given in Sec, A-7.
944 i< antenna measurements .
„ y, i 1 , 17 hut he increase in antenna tern-iwa jsr£?spr — » « k ^ - ,he aperture 18-13 ^-i^'^bsorbe,. T^rce absorber sheet.°j‘250 .
£ square are stacked at A/4 intervals in an JVC ^ ^ the bandwidth for which the reflected wave is at least 20 dB down. (See non Van der Neut and Baker reference.) APPENDIX A TABLES FOR REFERENCE A-l TABLE OF ANTENNA AND ANTENNA SYSTEM RELATIONS 1 A E2 Aperture efficiency, «tp = —1 = (dimensionless) (£ )mv X 2 Aperture, effective, At = — (m 2 ) Array factor (n sources of equal amplitude and spacing), En - — 7 (dimensionless) n sin ()/fj2) where ^ = f$ d cos 0 4 & (rad or deg) Beam efficiency, ^ (dimensionless) Beam solid angle, = (T PJiO t tf>) ^ (ST) 1 See index for page references giving more details on these relations. Also sec index for tables (list of) for other relations.
845 846 APPENDIX A -TABLES FOR REFERENCE Beam solid angle (approx ), C1A = 0HP $ HP (sr) - 0hp$hp (deg 2 ) Charge-current continuity, // = qb xJO Circular aperture, HPBW = — (uniform distribution) (Di = diameter in A) itv Circular aperture, BWFN — (uniform distribution) Circular aperture, directivity = 9.9 D 2 (uniform distribution) Circular aperture, gain over A/2 dipole = 6Oj (uniform distribution) n 4kA e 4k P( 0 , Directivity, D = —r^~ = 7T = S" (c A r1T (dimensionless) Directivity (approx.), D ^ 4;i(sr) 41 OOP (deg2 ) HP tf'Hptsr) 0«P ^HP _ 41 000 Ei# +1 uuy Directivity (better approx,), D ^—-—— p ^hp <Php Dipole (short), directivity, D = 1.5 ( = 1.76 dBi) Dipole (short), radiation resistance, R = 80rm Dipole (A/2), directivity, D = 1.64 ( = 2.15 dBi) Dipole (A/2), self-impedance Z = Kr + jX = 73 + j'42.5 O « Pr A„ A# Friis formula, — = —ttv p r/ (dimensionless) Flux density, S = — A (W m 2 Hz ) ^ <1 2k AT m :„ Flux density, minimum detectable, ASmiA = — Gain, G = kD (dimensionless) VI if' Height, effective, he = — = — hp = — I(z) dz t 1 q f o Jo (W m 2 Hz 1 ) A’ I TABLE OF ANTENNA AND ANTENNA SYSTEM RELATIONS 847 fC\ 2 nS Helical antenna, monofilar axial-mode, directivity, D = 12( — J — Linear array (long, uniform, in phase), HPBW = — (rad) = Loop (single turn) radiation resistance, Rr = 197CJ (D) Near-field-far-field boundary, R = —r (m) Noise power, receiver, N = kT syi Af (W) Nyquist power, w — kT (W Hz~ l ) P A 2& Radar equation, — = —m.
(dimensionless) Pt 4ns/ r Radiation power, P = ti 2 q 2 v 1 Radiation resistance, ^ Rectangular aperture, HPBW = -j— (uniform distribution) 1 Rectangular aperture, BWFN = Rectangular aperture, directivity = ]2.6L^L\ Rectangular aperture, gain over A/2 dipole = ULy L\ BWFN Resolution angle ^—-S Pt P t A et A„ Signal-to-noise rat.o, - = - = ^ Signal -to-noise ratio, S _ A 2 (above 1 W isotropic) N 167iV/cT fiys Af System temperature, = T A + r LPl - — 1 ) H (K) Temperature, antenna (through emitting-absorbing cloud), T A = T £(-e-") + T a e^ (K) #4$ APPENDIX A TABLES FOR REFERENCE Temperature, minimum detectable, A7^ in = — AT lms V' A f Wave power (average, elliptically polarized).
1 „ E\ + E\ {W m~ 2 ) Wavelength and frequency, k — y (m) (air or vacuum) A-2 FORMULAS FOR INPUT IMPEDANCE OF TERMINATED TRANSMISSION LINES. Formulas for the input impedance Zx appearing at a distance ,\ from a load or terminating impedance ZL on a transmission line of characteristic impedance Z0 as shown in Fig. A-l are listed in the table for 3 load conditions; (1) any value of impedance Z L , (2) ZL = 0 or short-circuited line and (3) ZL = co or open-circuited line. For each load condition there are columns for 2 cases: (1) the general case in which attenuation is present on the line (a / 0) and (2) the lossless case where the line losses are negligible (a = 0).
Figure A-l Transmission line of characteristic impedance Z0 , length x and load ZL .
Load condition Any value ZL General era Z L + Z0 lanh yx ZQ + Z L lanh yx z _ z ZL + JZ0 lan P 0 ZD + jZL tan fix z , = zyzs ZL = 0 Short-circuited line Zx — Z0 tanh yx ^ tanh ax + j lan fix 0 1 +j lanh aot lan fix Z T = Zn colh xvt Zr =jZ0 tan px ZL ^ x Open-circuited Line Zx = Z0 coth yx 1 + j tanh ctx tan px 0 lanh xx A j lan px Zx — Zq lanh axt Zx ~ —jZQ cot px + When fix = im, 2 where n = L 3, 5 In the table y = Ajfi where a - atttnuattoa constant aad fi 2/a.
A -4 CHARACTERISTIC IMPEDANCE OF COAXIAL 2-WIRE AND MLCROSTRIP TRANSMISSION LINES 849 A-3 REFLECTION AND TRANSMISSION COEFFICIENTS AND VSWR. For a transmission line of characteristic impedance Z0 terminated in a load impedance ZL , the reflection coefficient for voltage pv1 the reflection coeffi-cient for current p it the transmission coefficient for voltage or relative voltage at the load rpl the transmission coefficient for current or relative current at the load t, and the VSWR are given by Reflection coefficient for voltage ZL -Z0 ZL + Z0 Reflection coefficient for current Zp — ZL Zo + ZL “P, Transmission coefficient for voltage 2Zl Z0 + ZL = 1 + Pv Transmission coefficient for current 2Z0 Z0 + ZL = i + Pi Voltage standing-wave ratio (VSWR) Magnitude of reflection coefficient 1+Ip,I _ 1+1piI 1-IpJ 1-tPil Pv \ = Ipfl “ VSWR -1 VSWR + 1 A-4 CHARACTERISTIC IMPEDANCE OF COAXIAL, 2-WIRE AND MICROSTRIP TRANSMISSION LINES Type Mine Coaxial (filled with medium of relative permittivity s r ) Coaxial (air-filled) Twowire (in medium of relative permittivity £ r) (D a) Two-wire (in air) (f>> a) Microstrip (w ^ 2ft) CfcfinKtcmtk impedira, O where h = inside radius of outer conductor a = radius of inner conductor or wire D = spacing between centers of wires w — width of strip h — height or thickness of substrate V g50 APPENDIX A TABLES FOR REFERENCE It is assumed the lines are lossless (or R < taL and G < cuC) and also that the currents are confined io the conductor surfaces to which the radii refer. This , condition is approximated at high frequencies owing to the small depth of pen-etration.
It is also assumed that the lines are operating in the TEM mode.
The microstrip relation approaches exactness as the ratio w/h becomes very large. For strips of width w less than 2h , the formula for a single wire above a ground plane can be used. Thus, z0 138 log ^=138 log^ (£1) where D = spacing between wire and its image = 2h 1 a = wire radius = wj 4 The flat strip is considered equivalent to a circular conductor of radius 4; of the !
strip width.
a e CHARACTERISTIC IMPEDANCE OF TRANSMISSION LINES IN TERMS OF DISTRIBUTED PARAMETERS In the following table the characteristic impedance Z 0 of a transmission line is given for 3 cases: (1) general case where losses are present, (2) special case where losses are small and (3) lossless case.
In the table Z 0 = characteristic impedance, ft K 0 = characteristic resistance, ft Z = series impedance, ft m~ 1 R = series resistance, ft m 1 L — series inductance, H m _l V = shunt admittance, U m _1 G = shunt conductance, O m“ 1 C = shunt capacitance, F m“ 1 Z = R + joiL Y = G 4- jwC General case Small losses Lossless caset R = o n G = 0 I if.
| t Also holds approx imaidy for the case where losses' are not zero but tuL R and ojC 3> G-A- MATERIAL CONSTANTS (PERMITTIVITY, CONDUCTIVITY AND DIELECTRIC STRENGTH) 851 A-6 MATERIAL CONSTANTS (PERMITTIVITY, CONDUCTIVITY AND DIELECTRIC STRENGTH) Relative permittivity Conductivity Dielectric strength, Material £ e; 1 MV m" Air (atmospheric 1.0006 0 0 3 pressure) Aluminum 1 0 3.5 x 10 7 Bafcelite 5 0.05 10“ 14 25 Carbon 3 x 10 Copper 1 0 5,8 x 10" Glass (plate) 6 0.03 10 13 30 Graphite 10 5 Mica 6 0.2 10“' 5 200 Oil, mineral 2.2 0.0002 10“ 14 15 Paper (impregnated) 3 0.1 50 Paraffin 2.1 0.0004 ~nr 15 20 Plexiglas 3.4 Polyfoam -1.05 Polystyrene 2,7 0.0002 10“ 20 Polyvinyl chloride (PYC) 2,7 Porcelain 5 0,004 PVC (expanded) -1.1 Rubber, neoprene 5 0.02 10“ 13 25 Quartz 5 0.00 1 10 1 ’ 35 Rutile 100 0.02 (titanium dioxide) Snow, fresh 1.5 0,5-0.0003 Soil, day 14 5 x 10“ 3 Soil, sandy 10 2 x 10“ 3 Stone (limestone) 10“ 1 Stone, slate 7 Styrofoam 1.03 Urban ground 4 2 x 10' 4 Vacuum It 0 0 Vaseline 2.2 0.0003 Teflon 2.1 0,005 I0' ,s 60 Water, distilled 80 10“ 4 Water, fresh 80 10‘ 2 to 10“ 3 Water, sea SO 4 to 5 Wood, fir plywood 2 0.04 t By definition.
Note: Both £ r and e" are, in general, a function of frequency. Values given are typical of the kilohertz to gigahertz range- The permittivity is also a function of the temperature. Values given are typical of temperatures near 25°C except for snow.
£52 appendix a tables for reference A-7 permittivity relations Er = «r-X' t i tr — £r “ i(£rh £k) er ~ e; -XPF for sma11 PF Power loss = cE 2 + <de vE 2 (W m 3 ) where er = relative permittivity = e/eq, dimensionless e = permittivity, F m " 1 £0 = permittivity of vacuum — 8.85 x 10 12 F m ^ = relative permittivity related to displacement current E" = relative permittivity related to equivalent conduction current c ~ dc conductivity, Urn -1 o' = equivalent conductivity, 13 m 1 = relative permittivity related to hysteresis effects £" = relative permittivity related to dc conductivity PF = power factor = a f/(oe for a' < toe E = electric field, V m" 1 V B = 0 !
I: I t; II 1 1 The first table gives Maxwell’s equations in differential form and the second table in integral form.
The equations are stated for the general case, free-space case, harmonic-variation case, steady care (static fields but with conduction currents) and static case (static fields with no currents). In the table giving the integral form, the equivalence is. also indicated between the various equations and the electric potential or emf V t the magnetic potential or mmf U, the electric current I , the electric flux ^ and the magnetic flux $m .
o II II « II II E t O li T3 II N II Q II ^ ^ ^ 'S.
o II w w ki tx] '-©-i II II II II II i Li ^ II II 5 <5 X 3 II II 3 CjO f 856 APPENDIX A TABLES FOB BEFEBENCE A-10 BEAM WIDTH AND SIDELOBE LEVEL FOR RECTANGULAR AND CIRCULAR APERTURE DISTRIBUTIONS'!
1 Aperture fleM 4htributi« Recliiigular or linttr apertures £() Half-power Level of first beam width sideftobe, dB Tapered lo j at edge 10 dB down) £)x) = 1 — 2x J/3 - 1 0 x + 1 Tapered to zero at edge E(x) = 1 - x 2 = cos (Jtuc/2) Tapered to zero at edg£ £(x) = cos 1 (nx/2) Circubr apertures Tapered to \ at edge 10 dB down) = 1 — 2r 2/3 1 0 r 1 Tapered to zero at edge E(r) = 1 -r !
Tapered to zero at edg£ t L. - Lfi, 0, - D/a.
It is »auoed lh»l L, • I and D, > 1. For a uniform rectanjular or linear aperture HPBW .= Sr/L^ with tint eiddobe - 13 dB, See abo Tables 123 and 13-1, I I APPENDIX B COMPUTER PROGRAMS (CODES) 1. Horizontal dipole arrays over imperfect ground. Computer programs for pat-tern and gain calculations of HF horizontal dipole arrays over imperfect ground were developed in 1986 by the International Radio Consultive Com-mittee (CCIR). These programs in BASIC are available on disks from the International Telecommunication Union, General Secretariat (Sales Section), Place de Nations, CH-1211 Geneva 20, Switzerland, The programs have the code name HFMULSLW-HFDUASLW, 2. Three-/ pattern plots. A computer program that plots antenna patterns in a 3-dimensional form (with hidden lines not plotted) was written in 1982 by W. A, Sandrin in FORTRAN IV.
It is available as ASIS-NAPS Document NAPS-04053 in photocopy or microfiche from NAPS c/o Microfiche Pub-lications, PO Box 3513, Grand Central Station, New York, NY 10163.
10163, 3b The Numerical/Electromagnetics Code (NEC).
1 This program or code has been under development for many years.
It is a hybrid code which uses an Electric Field Integral Equation (EFIE) to model wire-like objects and a Mag-netic Field Integral Equation (MFIE) to model surface-like objects with time-1 J. X. Break)], G.
J, Burke and E X, Miller, “The Numerical/Electromagneiics Code," Lawrence Livermore Natl. Lab., Document UCRL-90560, 19R4, 857 858 APPENDIX 0 COMPUTER PROGRAMS (CODES) .
i sinusoidal spline basis s used for the wire harmonic excitation -surface current. Antennas and scattered can current and a pit sc asi Users of this program and also of almost all be modeled ,n their environments. Users ofth> PJ ^ ^ computer mode s should be a,are^ ^gment. or surface patches ade-vuhdity. Therefore, are e d may actually be ambiguous if con-quate? Adding more can be expensive ana ni vergence is not monotonic.
po and R R Mi[!er of An 8000-line .
G^ (^ is available for radiation or scat-Lawrence L.vermore National Uboratory d 19g0).
tering from a wire structure in free space or o e o 8 „ ennn , lnP FORTRAN IV program by R.
J- Mar-4- Wire Antenna Program. A SOW-hn > FOR ^ w D Burnside of the hefka of the ElectroSciencc Lab tOSU ^a ^^ ^ conductjng laboratory for radiation, gain and scattering structures. Incorporates NEC (^e 1978^ by R H . Newman 5. Wire and Plate Program. A «0Wme FOR^ P } avajlable from E H and D. M poza° 1 e radiation and scattering of 3-dimensional Newman for impedan , , s E H . Newman and D. M. Pozar, objects consisting oEwires _ and/or e Wl[e and Surface Geometries,” “Electromagnetic Modeling of Compost IEEE Trans. Ants , Prop., AP-26, November Other codes (programs) are listed at the end of Chap. 9.
_ nrtiTin]Si A i COMPUTER PROGRAM REFERENCES B-l ADDITIONAL Transient and Broadband Analysis and Synthesis of Baum C E.: “Emerging Technology for Transient ^ Antennas and ScaUerers," Ptoc. !EEE,bA, November l^fo.
„ .
T x Wi i| nur,. -p0 ie Extraction from Real-Frequency Informa-Briuingham, J. N, E. X. Milter and J, L. Willows.
Koie lion " Proc. IEEE, 68. 263-273, February 1980, , Burke G. V. aud E. K Miller: -Meddle, Antennas Near to aud Penelralmg a Lossy Surface, Lawrence Livermore Lab. Rep.. 89838. >983.
r j _ Lytfc and , T okada: .. Compuler Burke. G.
J., E. k Miller, J N. Bnllingh .
, -u,,, .
j, 29-49, 1981. Includes . ..re:il dis-Modeling of Antennas near the Ground.
Eleciru^-tributions and elevation patterns of Beverage anienaas „ Harrington, R. F., and J. R. Manta: “ Theory of Charactensbc Modes for Conducing Bod.es, IEEE Truns. dm. Prop Solution of Unbounded Field Problems,” f£££ McDonald, B. H„ and A. Wexler: f Dumber ,972 Trans. Microwave Theory Teth^ MTT-20, S41 .
, w flwrence Eivcrmorc Lab. Rept. syblJ, Miller E K.; “Numerical Modeling Techniques, Lawrcip .
_ , _ Miller’ E K., G.
J. Burke and E. S. Selden: '‘Accuracy Modetmg Cmde nes or Imegral-Equa.ion Evaluauon of Th.n-Wire Structures," IEEE Tram dm. Prop., AM. »7L _ -v Aspects of Thm-Wire Modeling, in R.
Miller, E. K, and F, J. D«dnck: Ei'Ctromagnetie. Springer-Verlag. 1975, Mittra (ed.), Numerical and Asymptotic J ^chmque Miller t 3 K 'f J Deadriek and G.
J. Burke: "Compute- Graphics Applications in Electromagnetic Computer ^^ . f£££ Rautio, J. G: “Reflection Coefficient Analysis ol tne pm Ants. Prop. Soc. Newsletter, 29, 5—1 1> February B-2 BASIC PHASED-ARRAY ANTENNA PATTERN PROGRAMS 859 Wilton, D. R. 3 S-S- M. Rao and A. W, Glisson, “Triangular Patch Modeling of Arbitrary Bodies,” fEEE Ants. Prop. Soc. Symp ., University of Washington, 1979, See also references at the end of Chap. 9.
B-2 BASIC PHASED-ARRAY ANTENNA PATTERN PRO-GRAMS 1 These BASIC programs supplement those of Chap, 4 for linear arrays of isotropic sources of equal amplitude and spacing. These programs, based on (4-6-9) and (4-6-14), provide normalized field and power patterns in polar and rectangular coordinates for a variety of conditions including broadside, end-fire, angle-fire and interferometer cases for AT sources with spacing D (A), phasing S (rad) and multiplying (or scale) factor MF. These programs are Apple-soft BASIC. To change to IBM GW-BASIC see note on page 862.
Genera) program (field pattern , polar plot): ION = ?: D = ?: S = ?: MF = ?
20 HOME: CD = 6.28D 30 HGR 40 HCOLOR = 3 50 FOR A = .01 TO 6.27 STEP .01 60 CA = COS(A): PF = CDCA + S 70 R = SIN{NPF/2) / SIN(PF/2) 80 R = MFABS(R) 90HPLOT 138 4 RCA, 79 + RSIN(A) 100 NEXT A Program L Broadside array of 4 sources with A/2 spacing.
First or menu line should read : 10N=4:D = .5:S=0: MF = 16.8 Program 2. Ordinary end-fire array of 8 sources with A/4 spacing.
First line should read : ION = 8:D = .25: S = -1.57: MF = 8.4 Program 3.
End-fire array with increased-directivity of 8 sources with A/4 spacing. First line should read: 10 N = 8: D = ,25: S = -1.96: MF = 13.1 Program 4> End-fire array with increased-directivity of 24 sources with A/4 spacing. First line should read: ION = 24: D = .25: S - -1.70: MF = 438 1 The assistance of Marc Abel in preparing these Apple compatible programs is gratefully acknowl-edged-860 APPENDIX B COMPUTER PROGRAMS (CODES) Figure B-l Polar field patterns for the 8 computer program examples of linear phased arrays of N isotropic sources along the horizontal axis. The 3-dimensional patterns are figures-of-revolution around the horizontal axis. Thns T in the broadside patterns 1 and 8 the mainlobe is a disc and in the end-fire patterns 2, 3 and 4 the main lobe is like a balloon or zeppdin. The main lobe in 5 is conical as are the lobes between broadside and end-fire in 6 and 7. The inner circle is at half-power Program 5, Angle-fire array (beam at 30° from broadside) of 16 sources with a/2 spacing. First line should read: ION = 16: D = ,5: S = -1:57: MF = 4.2 Program 6 Interferometer 2-source array with 4k spacing.
First line should read: ION = 2: D = 4:S =0: MF = 33.5 Program 7.
Broadside array of 8 sources with 2k spacing and grating lobes.
First line should read: 10 N = 8: D = 2: S = 0: MF = 8.4 Program 8-Broadside array 12 sources with A/2 spacing. First line should read: 10N = 12: D = .5: S = 0: MF = 5.6 Polar field patterns for the above 8 programs are shown in Fig.
B-l.
Compare the grating lobes of (7) with Fig.
1 1-78.
For different numbers of sources (N), spacings (D) and phasings (S) an unlimited variety of patterns are possible. Combinations of broadside and end-fire arrays are left as an exercise. For example, a broadside array (as in Program 1 but with D = L5) of 4 end-fire arrays (as in Program 3) results in a 32-source area array of high gain and small sidelobes in the plane of the array.
Although the magnification factor MF is arbitrary, the product of N and MF should be a constant for all pattern maxima to be equal (or normalized) when sources add in phase.
Thus, in the above examples NMF = 67 or B 2 BASIC PHASED-ARRAY ANTENNA PATTERN PROGRAMS 861 MF = 67/N. For increased directivity, however, all sources do not add in phase at the pattern maximum so make MF = 67SIN(1.57/N).
For power pattern, polar plot change line 80 to: 80 R = MFABS(R)ABS(R)/N For adding polar plot coordinate lines to programs 3 through 8 continue from line 80 as follows: 90HPLOT 138 + R.83CA, 79 + RSIN(A) 100 NEXT A 1 10 FOR X = — 58 TO 58 STEP 1 1 20 Y = 0 130HPLOT 138 + X, 79 + Y 140 NEXT X 150 FOR Y = -67 TO 67 STEP 1 160 X = 0 170 HPLOT 138 + X, 79 + Y 180 NEXT Y 190 FOR X = -50 TO 50 STEP 1 200 Y = L2 577X 210 HPLOT 138 + X, 79 + Y 220HPLOT 138 -X, 79-Y 230 HPLOT 138 + X, 79-Y 240 HPLOT 138 -X, 79 + Y 250 NEXT X 260 FOR X = - 29 TO 29 STEP 1 270 Y = 2.08 X 280 HPLOT 138 + X, 79 + Y 290 HPLOT 138-X, 79-Y 300 HPLOT 138 + X, 79-Y 310 HPLOT 138-X, 79 + Y 320 NEXT X 330 FOR A = .01 TO 6.27 STEP .01 340 HPLOT 1 38 + 56COS(A), 79 + 67SIN(A) 350 HPLOT 138 + 4QCOS(A), 79+47SlN(A) 360 NEXT A RUN By entering the above 36-line program and storing it on a disc, an infinity of array patterns can be calculated and drawn by simply entering a new line 10 with the desired parameters as in the 8 example programs.
The factor .83 in line 90 equalizes the X and Y scales on the- printer used.
Equivalent equalizing factors are written into the X and Y instructions for the coordinates. These factors may be modified or omitted.
862 appendix b computer programs (COOES) For field pattern, rectangular plot change lines 80 and 90 to : 80 R = MFR 90HPLOT A30, 75 — R For power pattern, rectangular plot change lines 80 and 90 to : 80 R = MFABS(R)ABS(R)/N 90 HPLOT A 30, 75-R To run the above programs as IBM GW-BASIC programs: Change HOME to CLS.
Change HGR to SCREEN 2.
Delete line 40.
Change HPLOT to PSET with parentheses enclosing the rest ot the line.
APPENDIX c BOOKS AND VIDEO TAPES C-l BOOKS Abraham, M., and R, Becker: Electricity and Magnetism , Stechert, 1932.
Aharoni, J.: Antennae, Oxford, 1946, Bah), L J., and P. Bhartia: Microstrip Antennas, Artech House, 1980.
Balanis, C. A,: Antenna Theory ; Analysis and Design, Harper and Row, 1982.
Biraud, F,{ed.): Very Long Baseline Interferometry Techniques, Cepadues, 1983.
Blake, L. V.: Antennas , Artech House, 1984.
Born, M.: Optik, Springer, 1933-Bowman, J.
J,, T, B. A. Senior and P.
L.
E. Uslenghi: Electromagnetic and Acoustic Scattering by Simple Shapes, North Holland, Amsterdam, 1969, Brace well, R N.: The Fourier Transform and Its Applications, McGraw-Hill, 1965.
Bracewell, R. N.: The Hartley Transform , Clarendon Press, Oxford, 1986.
Brillouin. L,: Wave Propagation in Periodic Structures, McGraw-Hill, 1946.
Brown, George H.: And Pari of Which I Was , Angus Cupar (117 Hunt Drive, Princeton, Ni 08540), 1982.
Briickmann, H.: Antennert, ihre Theorie und Technik, Hirzel, 1939-Burrows, M, L,: ELF Communications Antennas, Peregrines, 1978.
Cady, W, M., M. B. karelitz and L. A. Turner: Radar Scanners and Radomes, McGraw-Hill, 1948.
Christiansen, W N,, and J. A, Hogbom: Radio Telescopes, Cambridge, 1969, 1985.
Clarke, J. (ed,); Advances in Radar Techniques , Peregrinus, 1985-Clarri coats, P. J. B,> and A. D. Olver: Corrugated Horns for Microwave Peregrinus, 1984.
Collin, R, E.: Antennas and Radiowave Propagation, McGraw-Hill, 1985.
Collin, R. E„ and F. J Zucker (eds.): Antenna Theory, McGraw-Hill, 1969.
Cornbleet, S.: Microwave Optics ; The Optics of Microwave Antenna Design , Academic Press, 19H4.
Delogne, P.
: Leaky Feeders and Subsurface Radio Communication, 1EE, London, 1982 F-lliott, R. S.
: Antenna Theory and Design, Prenlice-Hal!, 1981.
Evans. J. V., and T. Hagfors (eds.): Radar Astronomy„ McG raw -Hill „ 1968-863 864 APPENDIX C BOOKS AND VIDEO TAPES C-2 VIDEO TAPES 865 Fanli, R,, K, Kellermann and G. Settifeds.): VLBl and Compact Radio Sources, Reidel, 1984.
[ Faraday, M.: Experimental Researches in Electricity., Quarjtch, 1839, 1855.
^ Galejs, J.: Antennas in Inhomogeneous Media, Pergamon, 1969.
Hall, G, L. (ed.): ARRL Antenna Book, American Radio Relay League, 1984.
H alien, E.: Tcoretisk Electricitetsl'dra, Skrivbyran Standard, 1945.
Hansen, R. C,: Microwave Scanning Antennas, vols.
1, 2, 3, Academic Press, 1966-Harper, A. E.: Rhombic Antenna Design, Van Nostrand, 1941.
Harrington? R. F.: Field Computation by Moment Methods, Macmillan, 1968-Hertz, Heinrich, R.: E/ectric Waves, Macmillan, 1893; Dover, L962.
Hertz, Heinrich : Memoirs, Letters, Diaries, San Francisco Press, 1977.
Hoird, R. M.: Remote Sensing, Methods and Applications , Wiley, 1986.
Hudson^ J. E.; Adaptive Array Principles, Peregrinus, 1981.
] Huygens, C.
; Trait e de la Luminiere, Leyden, 1690.
James, G, L,: Geometrical Theory of Diffraction for Electromagnetic IVat'fj, Peregrinus, 1980.
James, J. R., P. S, Hall and C. Wood: Microstrip Antenna Theory and Design , Peregrinus, 1982.
Jansky, D.
M., and M. C. Jeruchim: Communication Satellites in the Geostationary Orbit , Artech House, 1983.
1 Johnson, R. C., and H. Jasik (eds.l: Antenna Engineering Handbook, McGraw-Hill, 1984.
Johnson, R. C, and H.
Jasik (eds): Antenna Applications Reference Guide , McGraw-Hill, 1987.
(Selected chapters from Antenna Engineering Handbook.) J Jordan, E. C., and K, G. Balmain: Electromagnetic Waves and Radiating Systems, Prentice- Hall, 1968.
^ Jull, E. V,: Antennas and Diffraction Theory, Peregrinus, 1981.
| Kiely, D. G.: Dielectric Aerials , Methuen, 1953.
J King, R. W. P.: The Theory of Linear Antennas, Harvard, 1956.
J King, R. W.
P., R.
B. Mack and S, S. Sandler: Arrays of Cylindrical Dipole Antennas , Cambridge, | 1968.
1 King, R. W, P., H.
R. Mimno and A.
H. Wing: Transmission Lines, Antennas and Wave Guides, McGraw-Hill, 1945.
: | King, R. W. P. 3 and G. S. Smith: Antennas in Matter, MIT Press, 1980.
] Kraus, J, D.: Electromagnetics „ McGraw-Hill, 1953, 1973, 1984, Kraus, J. D.: Radio Astronomy, Cygnus-Quasar (PO Box 85, Powell, OH 43065), L966, 1986.
.
Kraus J. D.: Big Ear, Cygnus-Quasar, 1976.
Kuzmin, A, D., and A.
S.
Solomonovich: Radio Astronomical Methods for the Measurement of Antenna Parameters , Soviet Radio, 1974.
Landau, L., and E. Lifshitz; The Classical Theory of Fields, Addison-Wesley. 1951.
La port, E. A.: Radio Antenna Engineering, McGraw-Hill, 1952-Law, P. E,. Jr.: Shipboard Antennas, Artech House, 1986.
' Leonov, A.
I., and K, 1, Fomichev: Monopulse Radar , Artech House, 1986.
Lo, Y. T. (ed.): Handbook of Antenna Theory and Design, Van Nostrand Reinhold, 1987.
Love, A. W.
: Electromagnetic Horn Antennas, IEEE Press, 1976.
Love, A. W.; Reflector Antennas, IEEE Press, 1978.
Luneburg, R. K.: Mathematical Theory of Optics, Brown University Press, 1944.
Ma, M. T.: Theory and Application of Antenna Arrays , Wiley, 1974.
Mar, J.
W., and H. Liebowitz feds.): Structures Technology for Large Radar and Radio Telescope Systems, MIT, 196-9.
Marconi, Degna: My Father Marconi, McGraw-Hill, 1962.
Marcuse, D.
: Theory of Dielectric Optical Waveguides , Academic Press, 1974.
Maxwell, J. C.: A Treatise on Electricity and Magnetism, Oxford, 1873.
Monzingo, R, A,, and T, W. Miller: Introduction to Adaptive Arrays , Wiley, 1980.
Moullin, E. B.
: Radio Aerials , Oxford, 1949.
Poincare, H.
: Theorie Mathematique de la Luminiere, Carre, 1 892, Popovic, B. D., M. B. Dragovic and A. R. Djordjevic: Analysis and Synthesis of Wire Antennas, Wiley, 1982, } Pozar, D-: Antenna Design Using Personal Computers , Artech House, 1985.
.
Rayleigh] Lord; The Theory of Sound, Macmillan, 1877, 1878, L929, 1937, Reich, H. J, fed.): Very High Frequency Techniques, McGraw-Hill, 1947.
Reinljes, J. F. (ed.): Principles of Radar, McGraw-Hill, 1946, Rhodes D. R.: Introduction to Monopulse, McGraw-Hill, 1959.
Rhodes, D. R.: Synthesis of Planar Antenna Sources, Oarendon Press, Oxford, 1974.
Rudge» A. W., K. Milne, A. D. Olver and P. Knight (eds.): Handbook of Antenna Design, Peregrinus, 1983.
Rumsey, V, H.: Frequency Independent Antennas, Academic Press, 1966.
Rusch, W. V, T., and P. D. Potter: Analysis of Reflector Antennas , Academic Press, 1970.
Schelkunoff, S. A,: Electromagnetic Waves , Van Nostrand, 1948, Schelkunoff, S. A.: Advanced Antenna Theory, Wiley, 1952, Schelkunoff, S. A., and H. T, Friis: Antennas: Theory and Practice , Wiley, 1952.
Sherman, S. M,: Monoputse Principles and Techniques , Artech Houses 1984.
Silver, S.: Microwave Antenna Theory and Design , McGraw-Hill, 1949, Skolnik, M. 1.: introduction fo Radar Systems, McGraw-Hill, 1980.
Slater, J. G: Microwave Transmission , McGraw-Hill, 1942.
Slater, J, G,, and N. H. Frank: Introduction to Theoretical Physics, McGraw-Hill, 1933.
Smith, C. E.: Direefionol Antennas, Cleveland Institute of Radio Electronics, 1946.
Smith, C.
E.: Theory and Design of Directional Antenna Systems, National Association of Broad-casters, 1949.
Steinberg, B. D.: Principles of Aperture and Array System Design, Wiley, 1976.
Stutzman; W. L., and G, A. Thiele: Antertna Theory and Design, Wiley, 1981.
Tai, C-T: Dyadic Green's Functions in Electromagnetic Theory , Intext, 1971.
Terman, F, E.: Radio Engineers' Handbook, McGraw-Hill, 1943, Thompson, A.
R,, J, M. Moran and G, W. Swenson, Jr: Interferometry and Synthesis in Radio Astronomy, Wiley-Intersdcncc, 1986, Tseitlin, N, M.: Practical Methods of Radioastronomiced and Antenna Techniques t Soviet Radio, 1966.
Uchida, H.: Fundamentals of Coupled Lines and Multiwirc Antennas, Sasaki (Sendai), 1967.
Uda, Shintaro: On rhe Wireless Beam of Short Electric Waves. Series of 1 1 articles on his wave canal .(Yagj-Uda) antenna published in J. IEE Japan, between March 1926 and July 1929, plus earlier and later articles on meter wavelength experiments. Privately published bound volume.
Uda, S.: Short Wave Projector, Tohoku University, 1974.
Uda, S., and Y. Mushiake: Yagi-Uda Antenna, Tohoku University, 1954.
Ulaby, F. T., R. K. Moore and A. K. Fung: Microwatt Remote Sensing, Active and Passive, Addison-Wesley, 198 L Wail, J, R,: Antennas and Propagation, Peregrinus, 1986.
Walter, C. H.: Traveling Wave Antennas, Dover, 1972, Watson, W r H.: The Physical Principals of Wave Guide Transmission and Antenna Systems, Oxford, 1947.
Weeks, W. L.: Antenna Engineering, McGraw-Hill, 1968, Wood, P. J.: Reflector Antenna Analysis and Design, Peregrinus, 1980.
Zenneck, J.: JYirete Telegraphy, McGraw-Hill, 1915, C-2 VIDEO TAPES Kraus, J. D.: “Antennas and Radiation," Lecture-demonstration, excellent teaching supplement, 70 min. color, VHS. Cygnus-Quasar, P,(X Box 85, Powell OH 43065.
Landt, J. A., and E. K. Miller; “Computer Graphics of Transient Radiation and Scattering Pheno-mena on Antennas and Wire Structures" Fields and currents in slo^ motion, 15 min, color, VHS. Cygnus-Quasar, P.O, Box 85, Powell, OH 43065.
APPENDIX D ANSWERS TO STARRED PROBLEMS CHAPTER 2 2-4, 7l.6A J 2-6.
-11 kW 2-9.
152 m1 RCP ML fa) AR = 1.5; (> t = 90% fa) CW M3. 0) AR = 1.38, fa) t = 45° M5w Straight line with r = 45° MS. fa) AR - -2.33(RH),0)i - -45%fa)RH 2-20. fa) AR - - 5, 0) RH, fa) 34 mW m " Ml fa) AR = 3.0, 0) t = -22,5%fa)CW,fa)LH 2-25.
14.9 kW CHAPTER 3 3-1 .
(ii> 5,1, 6, 7.07; (6> 3 S, 4.6, 6,1 3-3, fa) 1539 Wm“% 0)4.29 x 10“ W, fa) 762 V m" 1 CHAPTER 4 4-1.
fa) Max. at 0% + 90%180° Nulls al ±30°, ± ISO6 Half-power at ± 14.5 s , ±165.5°, ±48.6°, ±131.4° 866 answers to starred problems 867 ic) 4-6, fa) m (dl 4-10.
fa) 4-15, ft = R ft 4-18 (a) Man. at 0%180% ±41.8°, ±138.2° Nulls at ±19.4, ±90, ±160.6° Half-power at ±9.6°, ±170.4°, ±30°, ±150°, ±56.5°, ±123.5° Max. at 0°, 180°, ±14,5% ±165,5% ±30°, ±150% ±49% ±131% ±90° Nulls at +7 ± 173% ±22°, ±158% ±39°, ±141°, ±61% ± 119 Half-power at ±3.6", ± 176.5% ±11°, ±l®°, + 18.5°, ± 161 -5". ± 26°’ ±154° ±345 ° ±145.5", ±43.5", ±136.5", +54.5", ±125.5°, ±70°. ±110° ) Max. at 0", 180° Half-power at ±90° 1 sin tfn sin 4>) j 15 , ) E(^) = cosi (i sin ) or \ cos (Jn sin <£) + i cos sin 0.52, 0.82, 1.00,0.82,0.52 Max, at ± 39% ± 141°, ±90° Nulls at ±30°, ±54°, ±126% ±150° 24° i 2 = 5 : 0.93, 0.84, 1 .00, 1 ,00, 0.84, 0.93 ^ 7 1 0,69, 0.80, 1.00, 1.00,0.80,0.69 = 10; 0.53, 0,78, 1,00, 1.00,0.78,0.53 ) f = where $ = d r cos $ + $ sin /2 ) ^ — 0, 0): 1, L 1, L % (2): % 4, 6, 4, 1, (3): 1 T 0, 0, 0, 1 /m i tnninr and .5 minor lobes, fa) ordinary D — 7; inc. dir.. D — 12 4-23 2500 4-25, 0)6.6, fa) 6.3 4-34.
0.61 4-38 Max. 0°, 180% ±60% ±90°, ±120° Min. ±41.4% ±75.5% ±104.5, +138.6 444, fa)44%0) - 13.3 dB,fa)0.17n sr, fa) 0.89, fa) 24,(/) 19X1 CHAPTER 5 Ql cos 9 _ G' sin 0 _ «• £ _0 5-3.
(n) E - Ian 9 sin0 cos (6) 168 Q, (< > 168. 73; 197 Q ^ (fl) £ -[sin Ji-cos ^l;4-lobcd patterns 1-pcosO (b) 40-lobed pattern[-•(J-—)] ; 5-7 3.3311 5-9, fa) sr, 0) 4 5-1 L 354 Q 5-13 fa) 8.16,0)6530 868 APPENDIX D CHAPTER 6 6-2.
4-lobed pattern 64.
1890 Cl 6-7 .
(1)180 0,15; (2) 1550 ft 1-2; (3) 4100 [2, 3.6 CHAPTER 7 7-1.
(1)0.802,(2)0.763 7-6.
ia)D t = CHAPTER 8 8-3 270 + j L350 ft CHAPTER 10 10-1.
121 +;46 tl 10-6 «. = 50ft K,=.20n,K t = 500 CHAPTER 11 11-1.
) -0.67L 11-7, (a) 1 and 6: 63 + j'29; 2 and 5: 46 -j‘2; 3 and 4; 53 + jlOfl 11-16 (a) SZ-jiinUbJG/Onax)^ 1.55 11-17. (6) ~ 17° 11-16 11.7° 11-22, (0)0.3542,(6) -/2 11-16 H1 = 0.83. 4> = 72.5 s , ^ = 5.5 11-28. = 72.5°, Li = 5.14 11-30. = 60?
11-34. (a) 6"22 ', (6) - 13.15 dB, (r) rr/4 sr, (d) 0.89, (e) 16,(/) 1.27a 11-46 i CHAPTER 12 12-6 (6) 73 ft (c) 10 dB 12-6 (a) !6dBi.(6) 12.6 s 12-13. 76.6 m 1 12-16 0.81, 2036 12 15. 0,66, 1651 12-17. 81,1%, 1304 or 31,2 dBi 12-19. (6)4,2° answers to starred problems 869 CHAPTER 13 1>1.
710 n 13-3.
750 13-6.
779- y87 Q CHAPTER 14 14-3.
fa) 62.5 mm, fa) 28% CHAPTER 16 161 56 — j50 £1 16-6.
fa) 26,5 W m -2 , (6) 184 m~ 2 16-8, 1.35 nW m“ 2 16-11. f € = 0, 8,89/a0 Npm" 1 16-14. See App, £ and Fig. E-I, 16-17.
fa) 5\ l.l dB; 10M2 dB (ib ) 5\ 1.4 dB; 10 a,19 dB fa) 5\ 0.06 dB; 10”, 0.26 dB 16-21.
106,35, 11 and 3.5 m CHAPTER 17 17-1.
(0)12,2 dB, (6) 2.3 m 17-2.
14.5 K 174.
fa) 0.08 K, (b) 0.09 It, fa) 445 K, fa) 500 mJy 174k (a) 0,06 K, (6) 320 mJy 17-8.
23.6 It 17-11. fa) 12.7 dB, (6) 0.5 dB, fa) -13 min 17-16, 65.1 tW Hz" 1 1718. (0)15,6 MHz, ib) 26.1 MHz, (c) 4U0 MHz CHAPTER 18 18-1, (a) 4.60 m, 11,2 m; (6) 25.2 mm, 116 mm; fa) 650 3,00 mm 18-3.
398 fan 18-5.
)64dB 187.
fa) 377 ft fa) 275 mm 18-8, 27.5 mm 18-9.
-9.9 dB 18-12. 57% 18-13, —3:1 Receiving antenna height.
APPENDIX E PROBLEM SUPPLEMENT 16-14 (ft) Note that the level with the CP antennas is the same as would be obtained with either the VP or HP antennas if the signal was received by the direct path only (no ground reflection). However, with CP antennas the signal level is essentially independent of the -1 5 -1 0' - 5 0 +6 Signal leva I, dB Figm E-l Solution to part (a) of Prob. 16-14 showing variation of vertical, horizontal and circular polarization signals with height above ground r 870 PROBLEM SUPPLEMENT 871 height, while with the HP and VP antennas the level varies from 6 dB more to no signal at all Another important factor for TV reception is that with CP antennas the signal is received (ideally) over only the direct path while with the HP and VP antennas the signal is received via both direct and reflected paths. At maximum, the direct and reflected path signals are essentially equal in level but arrive at different times. If the time difference is of the order of a microsecond, objectionable ghost images will occur, degrading the picture quality.
NAME INDEXt Abel, Marc, 859n Abraham. M-, 43n, 359, 863 Abramowitz, M., 252n Adams, A. A., 333n Adams, N.
I., 380n Adatia, N. A., 621 Aharoni, J., 863 Ajioka, T. S., 688n Akabane, K., 609n Alanen, E,, 536, 722n Albert, G, E-, 360n Alford, Andrew, 79n, 230n, 251n, 630, 642n, 643, 692n, 732n, 767n, Allen, J. L., 536 Ampire, Andre M., 2 Andersen, J, B., 305n Anderson, A, P., 619, 620 Ando, M., 659 Andre, S. N.,496n Andreasen, M. G., 408 Angelakos, D. J,, 287n Armilage, J. L., 715n Armstrong, Major Edwin FT, 7 Arvas, E., 797 Ashenasy, J , 74Sn Ashmead, J., 545n, 598 Atia, A. E,, 703 t n after a page number signifies footnote.
Bach, H„ 619 Bagby, C. K., 292 BaKL, 1, J., 863 Bailey, Beetle, 723 Baird, ft. C , 828n, 830, 831n Baker, D. E., 64n, 278n, 287n, 593, 594, 693, 842 Balanis, C. A., 408, 863 Balmain, K. G„ 864 Barkofsky, E, C, 79n, 767n, 827n Barrow, W. L,, 648n, 654n, 674n Barton, P-, 536 Baum, C E., 858 Bawer, ft., 698 n Bechmann, R,,4l3n Beck, A. C, 503n, 505 Becker, R„ 863 Bell, Alejcander G„ 5 Bennel, J. C, 619 Bennett, F. D„741n Bemsien, D. G„ 277n Beverage, H. H., 508 Bhartia, P-, 863 Bickmore, R. W., 499 Bingham, Linda, 400n Biraud, F., 863 Blake, L. V , 863 Blank, S., 499 Blasi, E A., 32 In Blore, W. E , 797 Bock, E. L., 72 5n Boemer, W.M.;798 «73 g74 NAME INDEX Bolomey, J, C. 842 Booker, H, G., 628, 630n, 634, 638n,517 Bom, Man, I79n, 63 2n, 863 Bose, Jagadis C, 5, 15. 644 Botha, L., 64n Bowman, J. J,, 798, 863 Bracewell, R N., xxv, 5l7n, 521n, 522n, 528, 533, 577n, 578. 585, 863 Braun, F. h 733n Breakail, J, K., 857n Brillouin, L, 3l4n, 863 Brittingham, J, N-, 858 Brookner, E., 748n Broussaud, G r , 324n Brown, G, H„ 15, I32n, 337n, 354n, 372n, 42ln, 425 n, 426n, 439n, 449n, 45 In, 452, 453, 454, 473, 474, 479, 482, 503n, 545n, 723n, 727, 728n, 733n, 840n, 863 Bruce, E,, 505 Biuckmann, H r , 509n, 863 Burgener, R. W r , 842 Burke, B. F., 535n Burke, G. J„ 397n, 408, 8^7n, 858 Burnside, W, D„ xxiv, 59ln, 6I9 + 620, 659, 693, 799, 816, 6l8n, 819, 820, 82ln, 822n, 842 Burrows, C, R-, 738n Burrows, M. L., 863 Burras, C S., 500 Butler, J, L., 488n, 499 Butson, P. C., 706 Byslrom, A;, Jr,, 277n Cady, W. M., 863 Cantoni, A., 798 Carrel, R, L„ 706, 707 Carson, J. R., 410n Carter, P. S., 4l3n,428n, 429n, 429n, 460n, 502n Carver. X R., 277n, 278, 494n, 748 Chaiteijee, J. S., 331n, 701 Chen, C A.,482n Chen, T-S, 654n Cheng, D. K , 190n,482n Cheong, W.
706 Chin, S, K., 500 Chireix, H., 5l0n Christ, J,, 1 Christiansen, W, N„ 587n, 863 Chu, L. J., 294n, 380n, 648n, 654n, 674n Chu,T-S, 605n, 609n, 6!9, 658 Chuang, C. W„ 620,659, 693, 798 Clark., H, X., 827n Clarke, A, C, 9, 762 Clarke, J., 798, 863 Clarricoats, P. J. B., 659, 863 Cleckner, D. C,, 480 Clemmow, P. C, 517 Clerici, G. C, 59 In!, 6l9n, 8l9n, 820, S21n Cohn, S. B., 654n Coleman, P, D., 74ln Collin, R. E., 863 Collmgton, G., 842 Conti, R., 495 Cornbleet, S,, 863 Cotton, R r B-, 328n Coulomb, Charles A. de, 2 Cowan, John D r , Jr. 57 Cox, C R., 4E0n Crawford, A. B„ 620 Cross, D. C., 620 Cutler, C. C, 564n, 8l0n, 824n, 827n Dalle Mees, E., 798 Davidovitz, M-, 798 Davis, J. H,, 620 Davis, R„ xxiv Day, P> C, 33ln DeVito, G., 706 Deadrick, F, J,, 858 Debye, P., 685n Delogne, P., 863 Demeryd, A,, 748 Deschamps, G. A., 75n, 696n, 703, 798 Dewitt, B.T.,816,Sl8n Dtanat, S. A,, 798 Dinger, R, J r , 499 Dixon, R. S,, xxiv Djordjevic, A, R., 864 Dolph, C L., 161n Donn, C r , 287n Dome, A„ 353n, 372n, 374n, 377n, 503n, 545n, 627n, 630, 648n, 653n, 725n Dowling, T., 495n Doxsey, R., 535n Dragone, C., 620, 659 Dragovic, M, B., 864 DuHamel, X H., 338n, 703 Dunlap,' Orrin E., 15 Dybdal, R. B., 619 Dyson, J r D„ 696n, 697, 698n, 699, 702, 703 Eeker, H. A., 809n, 819n Eflimiu, C., 798 Bhrenspeck, H. W., 620 Einarsson, O., 499 Einstein, A., 689, 768 EkervR, D, 534n El-Masry, E.
1., 500 ElectroSciencc Lab, 858 Elkamchouchi, H. M., 560 Elliott, R. S , 408, 863 Emerson, W. H„ 815 England, R, W. h 658n Enoch, J, M , 71 6n Epstein, J,, 723 n, 728n Erickson, N. R., 615n Ersoy, O., 500 Eshleman,V R,, xxiv, 690 Evans, J. V., 798, 863 Fan, H„ 500 Fanti, R„ 864 Faraday, M., 2, 15, 864 Feldman, C. B,, 507n Felsenheld, G, A., 723n Felsenheld, R, A., 732n Fenn, A. J., 536 Flugge, S., 5l7n Fomichev, K. L, 864 Foster, Donald, 245n, 250n, 254, 503n Fouty, R. A., 733n Franceschetti, G., 59n Frank, N. H„ 569n, 865 Franklin, Benjamin, 2 Friis, H, T , 48, 398n, 507n, 564n, 682n, Fung, A. K-, 798, 865 Gabriel, W. F., 536 GaLejs, J„ 864 Gauss, Karl F., 2 Gersl, C W., 333n Geyer, H. 659 Gilbert, William, 2 Gill, G, J„ 663 n, 690 Gillespie, E. S,, 830n, 842 Gitieath, M. C, 591n, 619, 819n Glasser, O, J„ 266n Glinski, G., 245n Glisson, A. W,, 859 Godwin, M, P,, 620 Goldsmith, P. F,, 61 Jn, 663n, 690 Gordon, W. E., 801 Goto, N., 659 Gray, D. A., 605 n, 658n Greenough, R. X., 333n Gregory, James, 596n Griffiths, L. J., 500 Grosskopf, J., 230n Gundlach, Friedrich W„ 15 Gupta, 1. 7, 500 Hacker, P. S.,810 Hagfors, T., 79i, 863 name index 875 Hall, G. L., 864 Hall, K, h xxiv HaO, P S., 501n, 864 Hall, R, C„ 798 H allin, E r , 360n, 365n, 368n, 372n, 374n, 864 Hansell, C. W., 502n Hansen, R. C., 488n, 499, 500, 713n, 798, 864 Hansen, W. W., 142n Harlen, F., 843 Harper, A, E., 503 n, 504, 864 Harrington, X F., 389n, 797, 658, 864 Hamson, C W., Jrv 36ln, 368n, 370n, 425n Hatcher, B. R., 500 Heideman, M, T r , 500 Heister, R., 500 Hendriksson, J r , 496 Henry, Joseph, 2 Hertz, Heinrich R., frontispiece, xxiit, I, 3, 16, 864 Hertz, Johanna, 16 Hcwish, A., 535n Hogbom, L A., 587n, 863 Hogg, D, C„ 620 Holland, L, 328n Hollis, J. S., 809n 855 Hondros, D., 685n Hord, X M„ 864 Horton, C W., 687n Howard, T>. D,, 620 Howei), J M , 536 Hudson, J. E., 536, 864 Huebner, D, A,, 50 In Hunt, L. E,, 620 Huygens, C, 179n,864 tarns, Harley, S24n Iguchi, M,, 537 Enagaki, N,, 560 Isbell, D. E., 703, 704 Eshigro, M,, 609n Ishimaru, A,, 599, 620 Jackson, J, D-, 294n James, G r L., 620, 864 James, J. R., 864 Jamieson, H. W., 74 tn Jamwal, K K S. 287n Jansen, J. J., 654n Jansky, D. M , 766, 864 Jansky, Karl G„ 7 Jasik,H„ 500,864 Jaumann, J., 815, 842 Jenkins, W, K , 500 Jeruchim, M. C,, 766, 864 NAME INDEX NAME INDEX 877 876 Jim, C. W., 500 Johansson J.
J-, 660 Johnson H. W. 500 Johnson, R. C, 328n, 809n, 819n 842 864 Jordan, E. C,, 4 642n, 696n 864 Jordan J. F., 535n Jull, E. V., 864 Kaifu, N.,609n Kajfez D., 2B7n Kandoian, A, G., 251 n, 723n 732n Kaplan, P. D. 620 Kardashev N., 8Un Karelitz M. B, 863 Kawakami, H, 73ln Kay, A. F., 657n 659 Kelleher K, S 500 6$9 Keller, J.B., 620 798 Kellermann, K.
I. 535n 864 Kellogg E. W., ’508 Kennaugh E. M,, 57408 798 Kent, B, M. 59ln 619 819n Ken J. L., 693 Kieburtz R, B- 599n, 620 Kiely D. G., 687n, 864 Kildal, P-S 598 Kilgus, C. C. 332n Kim, H. T., 798 King A, P. 653n 654n 8l0n 824n 827n King D. D„ 372a King, H, E. 270, 284 287n, 328n 332n 338n, 397n, 398n 400n 4Q6n, 429n 430n 560, 619 King, L. V., 360n King R. W. P 36 Ln, 365n 36Bn 370n, 421n 425n426n 512n 706 722n, 750n 840n, 864 Kinzd, J A, 536 Kislyakov A. G, 614 Klopfenslein R, W: 55 In Knight P., 865 Ko, H. C, 6l0n Kock W, E. 670 673 675n, 679, 683n 810n, 824n, 827n Koffman L, 620 Kollberg, E. K., 660 Kominami, M., 536 Kombau, T. W, 603 Kouyoumjian R G. xxiv 285n 603 620 621 791794, 798 Kroner, H„ 824n 825n, 827n Ksienski A, A, 500 Kummer, W. FL 830n, 842 Kuzmin A, D., 864 Lager, D, L. 858 Lagrone A. H, 659 Lamb J. W. 620 Landau L., 52n, 864 Landsdorfer, F. M., 484 Landt J. A., 865 Laport E A. 864 Law, P. E, Jr. 864 Lawrence Livermore National Laboratory, 858 Lawson, J. D. 537 Laxpati S. R., 500 Lazarus D., 627n 630 Legg W. E.,605n 658n Leonard D. J., 496n Leonov, A. L 864 Levine E 748n Lewis, F, D. 648n Lewis, Robert, 723n Lewis, W D., 564n 682n Liang M. C, 822n Uebowitz, H 864 Ufshilz E, 52n 864 Lindell, I. V. L, 536 722n Lindenblad N, E 337n, 502n Lo Y.T. 190m 593 864 Lochner 0, t 659 Lodge Sir Oliver 1 6 340 Love A, W. 597 598 659 864 Lowan A N. 252n Lowe R. 488n Lowry, L. R. 503n 505 Ludwig A. C 593 620 Lumjiak, C, 333n Luncburg R. K, T 6B8n 864 Lytle R. J. 858 Ma M. T. 864 Mack R B , 864 MacLean T S. M, 285n Magnesia 2 Mahapaira 5 7l5n Maher, T. M 190n MaiLloux R. J. 488 536 Mandanti M. 798 Mannersalo K> 536, 722n Mantz J, R, 858 Mar, J, W 864 Marconi Degna l£ 864 Marconi, Gugbelmo, frontispiece xxiii 1 3, 340 Marcuse D., 761n, 864 Marks A. M715 Markus, K-, 496 Marsh J. A., 294n 3S2n Masters, R W. 72973ln Mattes H., 659 Mautz J. R. 797 Maxwell, J. C. 2, 16 864 Mayer, C. E., 620 Mayer C. H., 777n Mayes P. E. n xxiv, 696n, 698n, 703, 707 McCullough, T. P. 777 n McDonald B. H., 858 McGahan, R. V., 799 Mclnnes, P. A„ 619 McNamara D. A^64n Medgyesi-Mitschang L. N. 798 799 Mei K. K 408, 703 Meier, A. S, 74ln 833 Meier P. J, 655n Mendelovicz A. 333n Mentzer, C. A., 660 Michael Y, 842 Mie, G., 794 Mikawa T„ 33 In 702 Miller, E. K., 397n, 858 865 Miller T W. 864 Miller, W. E., 642n Milligan, T. A., 652n Milne, K, 865 Mimno H. R.> 51 2n 864 Mink J- W/748 Misner, C. W , 768 Mima, R., 762 798 Mizuasawa, M., 621 Moffau D. L., 798 Monzingo, R. A 864 Moore, E. L., 663n Moore R. A., 819n Moore, R. K. 798, 865 Moran, J. M., 533n, 535n, 865 Morgan S. H-, 535n Morimoto M. t 609n Mostafavi M, 842 Moullin, E. B, 864 Moynihan, R, L., 536 Mjucci R. A., 5tX) Mueller, G, E. 685, 687n Mumford, W r W r , 843 Munk, B. A., xxiv 277n, 599 603, 604, 605 Munson RE xxiv, 501 748 Mushiake Y., 433 482n696 865 Nakano H., 325n 331n, 702, 798 Nakatami, D. T., 688n Napier P. J. 534n Nash R. T, 579, 580 582 585, 6l0n, 620 656n Naster, R.
J., 500 Nelson J. A., 725n 741 Newell, A. C 828n 830 Newman E.H , xxiv 397, 408, 798 858 Newton Isaac 3 16 Nortier, J. R. 842 L Nyquist H. 770 Oersted, Hans C, 2 Ohm, Georg S. 2 Okada, J. T. 858 Oilman, H. G-, 501n Olver, A. D„ 659 863 865 Olt, R. H-, 599, 620 Owens M: ( 750n Pacht E xxiv Page, L., 380n Papa 1 ;, C, H, 59n Paihak P. H, 620, 621798 Patton W. T., 328n, 332n 338n Payne, W,, 621 Peace M., 655 Pel ton E. L- 604, 605 Penzias A. A, 8 599 780, 842 Peters, L- Jr., xxiv 277n, 603, 620 660, 799 Picard, D, 842 Pippard A. B- 54 5n, 598 Pistolkors, A. A., 4l3n, 429n 725n Pislorius, C. W. L 819, 820, 82ln Pocklington, H, C 39 In Poggjo, A. J., 858 Poincare, Henri 16, 75n 864 Pon, C. Y„ 496n Popovic, B. D,, 864 Potter, P. D., 865 Polls, B. M., 277n Poyorzdski R. J., 798 Pozar D, M., xxiv, 397, 536 748 858 864 Prasad, S. N, 715n Preston, R A, 535n Putnam J- M 798 Rahmat-Samii Y., 620 621 Ramsey John F. 16 Rao, $. M. 798 Rao S. S. M. 859 Ratnasiri P. A., 621 Rautio J. C. 859 Rayleigh Lord, 16, 41Gn, 794 865 Reber Grote 8 Regier F.472 Rdch H. J.. 865 Rjeintjes J. F., 865 Rhodes D, R 40n> 433 5l8n 5l9n 536 537 579n 649 72 Sn 842 865 Labus J., 421 n 473 £7$ NAME INDEX Riblel, H. J., 161n, 175n Ricardi, L. J., 500 Rice, C. W, 508 Richmond, J. H., xxiv, 392n, 406n, 408, 430, 620,798 Ries, G., 593 Rjghi, A., 16 Risser, J. R. P 648n. 663n, 676n, 6S0n Robbins, T. E., 741n Roberts, D. H,, 535n Roberts, G, F-, 659 Roberts, J.
A", 521n, 522n Roberts, W. V. B., 5l2n Robin, F., 842 Robinson, A. L,, 7l5n Rolhe, Horst, 16 Rowland, H J., 74 In Rudduck, R. C., 822 Rudenbcrg, R., 43n.
Rudge, A. W., 593, 621,865 Rumsey, V. H., 598, 660, 696, 865 Rusch, W. V r T., 593, 621, 798, 865 Ruze, 1, 58 1 Ryan, C. E.,Jr, 620 Ryle, Martin, 535, 621 Sacher, R. R., 484n Sakar, T. K„ 798 Sakurai, K., 659 Salisbury, W. W., 813 Samada. Y., 32 5n Sandler, S. S,, 593, 864 Sato, G., 731 Sato, M., 537 Sato, R>, 537 Schaubert, D. H„ 536 Schelkunoff, S. A,, I45n, 157, 34ln, 35ln, 398n, 419n, 637n, 68 Sn, 687, 865 Schell, A, C, 598 Schrank, H, E., xxiv, 502n, 617, 652n, 810 Selden, E. S., 858 Senior, T, B. A., 798, 863 Setti, G„ 864 Shaffer, J. F., 798 Shakeshaft, J. R,, 535n Shapiro, I. I„ 535n Shelton, J, P„ 500 Sherman, S. M., 865 Shillue, W , 621 Shore, R. A„ 621 Sichak, W., 723 n Siegpl, ft, M., 798 Silver, S., 569n, 648n, 663n, 676n, 680n, 824n.
825n, 827n, 865 Sinclair, G., 77n, 642, 840n Skolnik, M.
I., 798, 865 Slater, J, C, S64n, 569n, 738n, 865 Sletten, C, J.,.598,621 Sloanaker, R, M,, 777n Smith, C. E„ I32n, 40«,474h, 733 n, 865 Smith, G. S., 750n, 864 Smith, P, D. P., 351 Smith, R H.,732n, 735n Solomonovich, A. S., 864 SommerfeLd, A., 179n, 722, 758 Sorenson, H, V., 500 Southworth, G. C, 653n Spencer, R. C, 598 Spitz, E,, 324n Springer, P. S., 701 S$ Republic, 5 SS Titanic, 5 Stark, L., 488n Stavis, G., 545n, 648n, 653n, 741, 827n Steinberg, B. D., 500, 865 Steinmeiz, Charles P., 5 Sterba, E, J„5l0n Steyskal, H,, 497n Stone, John S., 159 Stracca, G. B., 706 Stratton, J. A., 380n Striffler, Mary J., 7 Stubenrauch, C F., 828n, 830 Sturgeon, S. S., 5l4n Stutzman, W. L., 401n, 408, 865 Summers, W. P., 833 Sundberg, V, C, 655n Swartz, E. E., 655 Swenson, G, W L| Jr., 533n, 535n, 865 Synge, J. L-, 360n Taflove, A., 798 Tai, C-T, xxiv, 35 In, 372n, 425n, 537, 792n, 865 Tallqvist, S., 495 Tanaka, H„ 621 Taylor, T.T., 537 Taylor, W., xxiv Tell, R. A., 843 Terman, F. E., I32n, 648n, 722n, 739n, 865 Thales of Miletus, l Thiele, G, A., 401n, 408, 865 Thomas, D. T., 619 Thompson, A, R., 533n, 534n, 535n, 865 Thomson, G T., 706 Thorne, K, S., 768 Tkx, T. E,266n, 277n, 279 Titus, J. W., 620 Tiuri, M,, 495, 496 Tobey, F. L., 716n Toth, J„ 495n Treves, D., 748n Tsai, L, S„ 408 Tseitlin, N. M., 865 Turner, L. A., 863 Turrin, R. H„ 619 Twersky, V., 599, 621 Tyrrell, W. A„ 685, 6S7n LIS. National Bureau of Standards, 831 Uchida, H., 865 Uda, S , 433,481,482, 696n, 865 Ulaby, F. T„ 798, 865 Utnashankar, K., 798 Urpo, S., 495 Uslenghi, R L. E„ 798, 863 Vakil, R., 287m Van Atta, L. C-, 496n Van Blade], J, G., 408 Van de Hulst, H. C-, 794n, 796n Vander Neul, C. A., 693, 842 Vaughan, R. G„ 305n Verrazzani, L., 798 Viezbicke, P. P., 482n Viskum, H-H, 619 Vogel, W r J,, 620 Volakis, J. L., 500, 798,799 Volta, Alessandro, 2 von Gniss, H-, 593 Von Hoemer, S., 591n Vreeland, F. K., 16 Vu, T, B., 500 Wait, J, R., 865 Waldron, T. P„ 500 Wallenberg, R. F., 333n Walsh, J. E., 598 Walter, C H , 865 Walton, E. K-, 8l9n Walton, K.. L. p 655n Wang, D. S., 799 Wang, N., 798, 799 Watson, G. K. 252n NAME INDEX 87 Watson, R. B., 687n Watson, W.H„627n, 865 Weber, Joseph, 769 Weber, M. E., 500 Weeks, W. L., 865 Weiss, H. J., 763n Weiss, J. 495n Wexler, A., 858 Wheeler, H. A., 47, 336n, 500, 547n, 713, 738n Wheeler, J. A., 768 828, 830, Whinnery, J, R., 74ln Whitaker, A. J. T„ 619, 620 Whitmer, R, 685n Williamson, J, C., 230n, 266n Willows, J. L,, 858 Wilson, R. W., 8, 599, 605n, 65Sn, 780, 842 Wilton, D. R., 859 Wing, A. H., 512n, 864 Withers, M. J., 593, 621 Wohlleben, R., 659 Wolf, Franz, 16 Wolfe, J. J., 698n Wolff, Irving, I64n Wong, J. L,, 270, 284, 287n, 328n, 332n, 338n, 560 Wood, C, 864 Wood, P. J., 865 Woodward, O, M., 337n, 354n, 372n Woodward, P. M., 537 Woodward, O. M., Jr,, 733n Woodyard, J. R., 142n Worden, R. A., 333n Wu, T. T., 750n Yaghjian, A. D., 799 Yagi, H., 482 Yamane, T,, 325n Yamauchi, J., 325n, 33 In, 702, 798 Yngvesson, K, $., 660 Yoshizawa, A„ 798 Young, J. D., 500, 819n Yu, J. S„ 822n Zenneck, J„ 43n, 758. 865 Zenuemon, Saito, 482 Zucker, F. J^, 762, 863 SUBJECT INDEX Absorbing materials, 813-818 1 /e depths, 843 (Prob. 18-1) medium, example, 8 1 6 sheet, 843 (Prob. 18-8) shroud, 619 Jaumann sandwich, 843 (Prob, 18-13) Salisbury sheets, 36, 813 wedges, 816 Accelerated charges, 50-52 Achievement factor, 575 Adaptive arrays, 496-500 Adjustable A/2 dipole, 695 Alford loop, 731 Alpine hom antenna, 52, 692, 694 Ampere, definition of, 13 Anechoic chamber, 818-822 Annular zone, 666 Answers to problems, 866-869 Antenna : concepts, basic, 17-82 definwl, 17 early history, 1 parameters, 19-20 relations, table, 845-848 (compared to) transmission line, 64-70 Aperture: collecting, 35 concept, 28-29 t n after a page number signifies footnote.
SUBJECT INDEXt distributions and efficiencies, 573-587 distributions compared, 519 effective, 29-31, 46-47, 845 efficiency, 574, 845 with celestial source, 584, 830-831 from Cygnus A, 843 (Prob, 18-12) ioss, 35 physical, 35-36 scattering, 31-34 synthesis, 533-535 Aperture-matched horn, 659 Apertures, circular and rectangular, data on, 572 Archimedes spiral, 698 Arecibo, 605 Arrays : (see also specific type) adaptive, 496 billboard, 182, 547 broadside, 140-141, 439, 191 (Prob. 4-4), 192 (Prob. 4-10X 195 (Prob, 4-25), 196 (Prob, 4-32), 198 (Prob. 4-42) non-uniform distribution, 159-162, versus end-fire, 150-155 circular, 188 collinear, 540 (Prob. 11-21) of dipoles and of apertures 435-542 elliptical, 188 end-fire, increased directivity, 141-145 end-fire, ordinary, 141 factor, 140, 845 grid type, 491-496 of helices, 319-323 i linear, 1 37-145, 314 lobe sweeping, 326 minor-lobe maxima, 1 58 with missing sources, 189 190 of r driven dements, 459-461 null directions, 145-150 with parasitic elements, 476-485 pattern maxima, 156 of point sources, 1 18-199 phased, 485-496 primary pattern, 131-133 rectangular, 186-189 secondary pattern. 132-133 smart, 496 three-dimensional, 150 155 with traveling waves, 314 of turns versus dipoles. L50 of two driven A/ 2 elements broadside ease, 436 end-fire case, 445 general case, 449-453 Attenuation : of conducting sheet, 843 (Prob.
18-6) of lossy sheet, 843 (Prob. 18-5) by lossy slab, 843 (Prob. 18-4) in lossy medium, 843 (Prob.
1 B-7) Autocorrelation, 523 Axial bright spot, 186 Axial mode, 274 Axial ratio, 71 Axons, 716 Babinet's principle, 632 Backfire helix. 328 Backward angle-fire grid array, 491-496 Baiuns, 734- 745 Bandwidth, 256-263, 767-768 Ba twin g antenna, 731 BC arrays : 4 tower, 193 (Prob. 4-12) 4 towers, 195 (Prob. 4-21) with null at a = 30, 538 (Prob. 11-12) with null to west at all st, 539 (Prob.
I M3) 3 towers, 194 (Prob. 4-20) 2 towers, 194 (Prob. 4-19) Beam : area, 23-25 efficiency, 25, 578, 845 efficiency with celestial source, 583 mode, 274 solid angle, 23—25, 845 solid angle (approx.), 846 width and sidelobe level, table, 856 Bell Aerospace antenna, 501 Bell Telephone Laboratories, 605 Bessel curve, 248 Beverage antennas, 229, 508-509 Biconical antenna, 340-358 boundary sphere, 348 characteristic impedance, 346 finite length, 347-353 impedance of, 352 infinite, impedance of, 341-346 input impedance, 346-347 patterns, 353 phantom, 356-358 slacked, 356- 358 ' transmission line equivalent, 350 wilh unequal cone angles, 358 (Prob. 8-1) with wide-cone angle, 695 Big Far, 27ln, 585,610-613 Big Ear compact range, 822 Billboard array, see Arrays Binomial distribution, 160, 196(Prob. 4-29) Bipolar cells, 716 Black body, 777 Bonn, 589. 605 Books and tapes, 863-865 Bound wave, 757 Bow-tie antenna, 354, 358 (Prob. 8-4) Brillouin diagram, 314n Broadband antennas, 692-696 Broadcast station example, 131 Brown's equation, 433 (Prob. 10-4) Brown-Woodward antenna, 354—356 Brown-Wood ward (bow-tie) dipole, 559 Bruce antenna, 509 Bruce curtain, 492 Buckminster Fuller geodesic dome, 561 Butler matrix, 4S8 Candela, definition of, 13 Carrier-to-noise (C/N) ratio, 764 Cassegrain : feed, 594-600 reflector, 501 Cavity resonator, 18 Celestial: radio sources, table, 853 source example, 83 1 source measurements, 583—584, 822^-823, 830-83 1 Chain array, 491-496 Charge-current continuity, 846 Charge distribution, 407 (Prob. 9-2) Cheese antenna, 573 Chireix coil, 333 Chireix-Mesny array, 509 882 St' EJECT INDEX SUBJECT INDEX 883 Choke feed h 658 Circular aperture: BWFK846 directivity, 846 gain, 846 HPaW, 846 Circular apertures, 569-572 Circular array I 88 Circular-component method, S3 8 S39 Ci rcu la r-de polarization ratio, 85 (Prob, 2-26) Circular polarization, 70, see also Polarization Circularly polarized: antennas, 732-734 wave reflection, 843 (Prob. 18-10) waves, S4 (Prob. 2-20), 85 (Prob 2-22) Cladding, 761 Clarke (geostationary) orbit/27 (see also Satellite) downlink, 786 satellite, 762 Closely spaced elements, 453-459 Ooverleaf antenna, 732 Codes computer, see Programs computer Collecting aperture, 35 Compact range, 818 822 Complementary antennas 624—660 Complementary dipole antenna, 639 Complementary screens, 633 impedance of, 635-638 Complex deviation (factor) 578 Complex visibility function, 527 Computer codes or programs see Programs, computer Comsat ’satellite, 270 Comtech antenna, 616 Conductivity, table 851 Cone, 2'\ 358 (Prob, 8-3} Cones of retina, 716 Conical : antenna 354 horn, 645 651-654 pattern, 236 (Probs, 5-9 5-10) spiral antenna 701-703 Constants and conversions, inside back cover Constrained lens 676 Continuous aperture distribution 515-517 Continuous arrays 175-179 Corner-helix 324 Comer reflector : active 548-560 with bow-lie dipole design data 558 2/2 to the driven element 623 (Prob, 12-12) 2/4 to the driven element 623 (Prob. 12-1 1) gain- patient formulas, 553 passive (retro) 560-561 resistance of, 556 square. 548-561 Cornu spiral, 185 Corrugated horns 657-659 Corrugated surface-wave antenna, 758 Cosine integral, 226 Cosine" power pattern, 95-96 Cosmic radio window 781 Creeping wave 548 Critical angle, 750 Critical frequency MUF, 802-803 (Prob. 17-18) Cross-field, 79-81,207757 Cross section, radar, 792 Cross section, scattering 791 Crossed dipoles, 83 (Prob, 2-10) Current, retarded, 202 Current distribution: on cylindrical antennas, 369-371 on helix, 267-268 measurements, 834-835 Curtain arrays, 509-510 Cygnus A, 802 (Prob. 17-17) Cylindrical antennas, 359-408 with conical input sections, 379 current distribution, 369-371, 380-383 impedance, 371-376 pattern, 376- 377 Cylindrical lens, 691 (Prob. 14-5) Cylindrical parabolic reflector 563, 572-573, 621 (Prob, 12-5) De Forest audions, 7 Delay lens 663 Delta-gap generator, 40 1 model, 401 Dendrites 7 1 6 Depth of penetration 257 Design directivity 575 Dielectric: artificial 670-673, 690 (Prob. 14-2) lens, 663-670, 690 (Prob. 14-1) slab 759 762 strength 851 Diffraction (see also scattering) by compact range reflector 818-822 by flat sheets (physical optics), 183-186 by flat sheets (GTD), 548-549 from sheet with slot 628-632 of rays or waves 179-183 Digital Beam Forming (DBF) antennas 49?
Dimensional analysis, \5 Dimensions and units 11-12 f $ I f s DIPOLE BASIC program, 397 Dipoles: 200 234 adjustable, 695 center-fed 223, 414 horizontal, above typical ground, 719 and loops compared, 244 2/2 directivity 846 2/2 impedance, 846, 407 (Prob. 9-3) power flux from. 214 pulsed center-fed, 54-59 radiation resistance, 213 217 short, directivity, 846 short, radiation resistance 846 vertical, above typical ground 720 Dirac della function 401 Direction-finder (DF) antenna, 772 -773 (Prob.
16-17) Directional couplers, 488—490, 833 Directive gain 198 (Prob. 4-45) Directivity: 26 46-47, 574 r 824, 846 actual 575 approximation, 99 calculation, example, 101 design, 575 by integration 1 16 (Prob. 3-5) of loop, 253-254 and minor lobes 116 (Prob. 3-4) Disc back sea tier, 794-796, 803 (Prob. 17-21) Distance requirement for uniform phase, 809-811 Distributions: binomial, 160 compared 173-175 Dolph-Tchebyscheff, 161518 edge 160 Gaussian 162 optimum 160 tapered 162 uniform, 160 Dolph-Tchebyscheff (D-T) distribution 161-171, 518 8 sources, example of, 171-173, 192 (Prob, 4-7) 5 sources 192 (Prob. 4-6) 6 sources, 194 (Prob, 4-15) Double-beam DF 772-773 (Prob. 16-17) Double-ridge horn, 655 Down-Links (see Satellite) Driving-point impedance 409 439 447r448 Dual-chamber Gregorian-fed compact range 821 Duality of antennas 50 Dyson conical spiral, 702 Earth-station (see Satellite) Edge diffraction 628-632 Edge distribution J60 Effect of typical (imperfect ground) 7 1 6—722 Effective aperture, 29-31, 46-47, 82 (Prob. 2-3), 83 (Prob. 2-4) Effective height 40-42 Efficiency (see also Aperture efficiency): of aperture with phase ripple, 623 (Prob.
12-16) of rectangular aperture with full taper, 623 (Prob. 12-15} of rectangular aperture with partial taper 623 (Prob. 12-14) Einstein (gravity) lens, 688-690 Einstein's theory of general relativity, 768 Ekran class satellites 270 Electrically small antennas 711-714 Electro Science Laboratory, Ohio State University 819 Electromagnetic spectrum, 8 II Electromagnetic theory origins 1- 8 Emit ting-absorbing object, 777 End-fire arrays 141-159, 445 (see also Helical antennas) Equation numbering 14 Equivalence of pattern factors, 237 (Prob 5- 1 5) Equivalent radius of antennas 368-369 Exponential integral, 419 Exponential notch antenna 714 F number 665n F ratio, 565n F/O ratio 565n Far field 206 Farfields of small electric dipoles and loops table of 244 Fast wave 760 Feed systems, 749, 543-619 Feeding considerations, 71 1-773 Fermat's principle 561 645 663 Ferrite loaded loop, 259 Ferrite rod antenna 260 Field zones 60, 206 Field and phase patterns 194(Probs. 4-14, 4-17) Field-cell transmission line, 746 Field components 209 Field patterns, 10 z-r r = d_ _d_ d^ dr 7j> 7z A r rA# A z Spherical Coordinates v/= f ^ + e i f fL + $ —L_ dr r d9 ^ r sin 0 £0 V A ^ “7 f r A r + — 1— 7, ^ {A « sitl ^ r f r r sin 0 d9 r sin 6 d v - A - f rirs [I 1'v“ w - ^f] + • Ksrs - i M) CONSTANTS AND CONVERSIONS Quantity Symbol or ibbreviition Astronomical unit Boltzmann's constant Earth mass Earth radius {average) Electron charge Electron rest mass ^ Electron chargers em ratio .
Flux density (power) Jy Hydrogen atom (mass) Hydrogen line rest frequency Light-second Light, velocity of c Light -ycar .
.
, log x = logo, x (common logarithm) In x = log, x (natural logarithm} Logarithm, base e reciprocal of base e Logarithm conversion Moon distance (average) Moon mass Moon radius (average) ~ pc Parsec r „ pc Parsec r „ pc Parsec Permeability of vacuumt ^ Permittivity of vacuum+ c° Pl l Planck’s constant Proton re 4 mass Radian ^ Space, impedance oft z Sphere, solid angle Sphere, solid angle Square degree d Stefan- Boltzmann constant Steradian ( = square radian) s Sun, distance ^ Sun mass Sun radius (average) Year (tropical) Nominal value 1.5 x 10s km 1.38 X 10“ 21 J K" 1 6.0 x 1034 kg 6.37 Mm — 1.60 x 10 " 19 C 9,11 x 10“ 11 kg 1.76 x 10 11 C kg' 1 10“ 26 W m“ 2 Hz" 1.673 x 10“ 12 kg 1420.405 MHz 300 Mm 300 Mm s“ l 9 46 x 1012 km More accurate vilnct 1,496 x 10s 1.38062 x 10" 25 5.98 x 1034 -1.602 x 10“^ 1.758803 x 10 11 10 16 (by definition) 299.7925 9.4«>5 x 10 11 2.72 0.368 In x - 2.3 log x log x - 0.43 In x 380 Mm 6.7 x 10 12 kg 1,738 Mm 3 1 x IQ 1 km 3.26 Ly 2.06 x 1G 5 MJ 1260 nH m " 1 8.85 pF m“‘ 3.14 6.63 < 10 34 J s 1.67 x 10' 37 kg 57.3 r -376.7 (a 120) ft 116 sr 41253 deg 2 3 05 x 10‘ 4 sr 5.67 x 10 s Wm“ 2 K 4 3283 deg 3 1.5 x )0 8 km 2 0 x 10JO kg 700 Mm 365.24 days = 3.1556925 x 10" s 2.718282 0 36788 In x - 2.3026 log log x = 0 4343 In x 3,0856 x 10 15 3.2615 2.06265 400n (exact value!
8.854185 - 1/PoC 2 3.1415927 6 62620 x 10" i4 1 67261 x 10" iT 57,2958° 376.7304 = 4ic = 12.5664 41252 96 304617 x 10“ 4 5,6692 x 10“ 0 080/tt) 2 - 3282.806 1.496 x IO 0 1.99 x \0 ia 6953 " , 7 tioie that the values tor these
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Direct Variation and Proportional Change
Resource ID: A1M4L12
Grade Range: 9 - 12
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Lung Adenocarcinoma in Never Smokers: Problems of Primary Prevention from Aspects of Susceptible Genes and Carcinogens | Anticancer Research
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Review Article Reviews R
Lung Adenocarcinoma in Never Smokers: Problems of Primary Prevention from Aspects of Susceptible Genes and Carcinogens
ISAO OKAZAKI, SHIGEMI ISHIKAWA, WATARU ANDO and YASUNORI SOHARA
Anticancer Research December 2016, 36 (12) 6207-6224;
ISAO OKAZAKI
1 Department of Internal Medicine, Sanno Hospital, International University of Health and Welfare, Tokyo, Japan
2 Department of Internal Medicine, International University of Health and Welfare Hospital, Nasushiobara, Japan
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For correspondence: iokazaki@iuhw.ac.jp
SHIGEMI ISHIKAWA
3 Department of Chest Surgery, International University of Health and Welfare Hospital, Nasushiobara, Japan
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WATARU ANDO
4 Department of Clinical Pharmacy, Center for Clinical Pharmacy and Sciences, School of Pharmacy, Kitasato University, Tokyo, Japan
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YASUNORI SOHARA
5 Department of Chest Surgery,Saitama Prefecture Central Hospital, Okegawa, Japan
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Abstract
Global statistics estimate that approximately 25% of patients with lung cancer are never smokers. We suggest that genes related to susceptibility to metabolic syndrome were present among those related to susceptibility to lung adenocarcinoma (AC) in never smokers. There are many questions concerning lung AC in never smokers, which is increasing in incidence, with female predominance, good prognosis, unique genes related to susceptibility and good response to treatment with specific agents. The purpose of this review was to investigate the carcinogenesis of lung AC in never smokers focusing on genes related to susceptibility to lung AC and carcinogens, including environmental factors. In order to clarify the carcinogenesis of lung AC in never smokers, the definition of never smokers, survey of environmental tobacco smoke, the presence of the physical characteristics of metabolic syndrome, and other carcinogens should be investigated for primary prevention of lung AC.
Never smoker
lung adenocarcinoma
environmental tobacco smoke (ETS)
susceptible gene
metabolic syndrome
review
Lung cancer was the leading cause of death in developed countries in 2012 in both males and females, and the leading cause in less developed countries in males and the second cause in females (1). Lung cancer is classified into two morphological groups, small cell lung cancer (SCLC) and non-small cell lung cancer (NSCLC). NSCLC includes squamous cell lung cancer (SC) and adenocarcinoma (AC). Lung AC is on the rise in both sexes, especially women, although the number is not as high as among men, in both developed and developing countries (2). Lung AC is associated with life-style throughout the world (1, 3).
Islami et al. reported that global incidence and mortality rates of lung cancer are still closely associated with smoking prevalence (4). However, high-income countries have shown remarkable decrease in smoking in both males and females (4). Parkin et al. reported that 15% of lung cancer cases in men and 53% in women are not attributable to smoking, and approximately 25% of patients with lung cancer were never smokers, according to global statistics in 2002 (5). This was a very shocking report. A review article by Sun, et al. (6) and others (7-10) reported that never-smoker patients with lung AC were predominantly female (6, 10), were significantly younger (7, 8) and had better prognosis (6, 7, 10-12), with especially good response to treatment with epidermal growth factor receptor-tyrosine kinase inhibitors (EGFR-TKIs) (7, 13-15). Such patients were frequently Asian females (4, 16).
We previously wrote a review on genes related to susceptibility to lung AC in never smokers and demonstrated that genes associated with metabolic syndrome were present among these (17). Many questions arose, e.g. whether lung AC may be a different disease (6, 7), whether it may involve different genes (8), why female predisposition is present, whether the number of patients continues to increase (16), and why there are so many patients with AC among never smokers with NSCLC (6, 16, 18). For primary prevention, it is most important to clarify the mechanism of carcinogenesis, especially to identify carcinogens related to genes susceptibility to lung cancer. Moreover, it should also be elucidated whether and how metabolic syndrome plays a role in these patients (17).
Concerning lung carcinogenesis in never smokers, environmental tobacco smoke (ETS) at home, work and other places have been supposed to be the cause of lung cancer in never smokers (4, 19-24). If never smokers have been exposed to ETS, it may act as a trigger, converting stem cells into cancer cells in the lung (25) with involvement of metabolic syndrome (26). Besides ETS, reported carcinogens include radon (27-29), cooking oil vapor (9, 30), indoor coal burning, hormonal factors (after hormonal replacement therapy) (6, 8, 9), occupational chemical exposure including asbestos/heavy metals (8), infectious factors (31, 32), and air pollution (8).
The purpose of the present study was to clarify the carcinogenesis of lung AC in never smokers from the viewpoints of both genes related to susceptibility, environmental factors and lifestyle. It is necessary to clarify why lung AC is increasing in Asian women never smokers (15, 29). Finally this review shows the interrelation between genes related to susceptibililty and suspected carcinogens from original articles on lung AC in never smokers, and discusses a primary prevention strategy for lung cancer in never smokers.
Selection of Bibliography
The articles cited in this review dated before 2000 were selected from review articles. Original articles after 2000 sought from PubMed were screened by us. The key words used in the search were as follows: lung cancer, lung AC, never smoker, NSCLC, incidence rate of global lung cancer, mortality rate of lung cancer, smoking prevalence, genes susceptible to lung cancer, and genes susceptible to lung AC. Of the articles published after 2000 we selected those we considered to be most important and appropriate for this study. The purpose of this review was to investigate the carcinogenesis of lung AC in never smokers focusing on genes related to susceptibility and suspected carcinogens, including environmental factors, in order to improve primary prevention of lung AC.
Original Articles on Genes Related to Susceptibility to Lung AC in Never Smokers
Thirty-one original articles listed in our previous review (17) were examined once again from the following standpoints: the hypothesis or purpose stated by the authors, the genes they had investigated, the study style used, the definition of never smokers, and the etiology of lung AC they analyzed such as ETS. Among these, the study hypothesis was considered the most important. If the authors had found genes related to susceptibility to lung AC and these genes were analyzed by the grade of heavy smoking but did not include discussion on never smokers, we deleted these articles from the list of originals in the present study. We focused more precisely on genes related to susceptibility to lung AC in never smokers (Table I).
1. Classical Genes
The hypothesis stated in the original articles on classical genes (20, 33-43) is as follows: exposure of never smokers to ETS resulted in lung AC, hence the genes affected in these cases are the same as those seen in the tobacco-related cases (33-36). The relative risk of the genes suspected in patients with lung AC exposed to ETS was higher than the frequency of suspected genes in patients with lung AC without exposure to ETS (Table I). The slow and the fast genotypes among the polymorphisms of N-acetyltransferase 2 (NAT2) were shown to be related to the risk of lung cancer among women never smokers (37). NAT2 participates in the detoxification of aromatic amines. The authors took into consideration exposure to cooking oil fumes among never smoker women who developed lung cancer.
The tumor suppressor genes, TP53 and, TP63 may be involved in carcinogenesis of lung AC in female never smokers (38, 39), although TP53 mutation has been reported in smokers with lung SC (17). These two reports (38, 39) did not describe the survey methods regarding ETS. Of course the authors understood the possibility of ETS, or other environmental risk factors, such as exposure to combustion products of indoor heating and cooking solid fuel and cooking oil fumes, as possible etiologies (38). Zhang et al. (39) introduced studies by Cianchi et al. (68) and Yao and Rahman (69), reporting that exposure to external carcinogens, including cigarette smoking, infectious agents, and dietary carcinogens, can result in inflammation and play a role in tumor development, but ETS in their cases was not analyzed because it was not their purpose. The purpose of their studies was to discover novel genes. DNA-repair genes were investigated in never smokers (40-43) and the authors strongly suggested the possibility of ETS (41, 42) or cooking oil fumes (40) as suspected carcinogens.
2. Susceptible Genes Found by Genome-wide Association Study (GWAS)
Nicotine acetylcholine esterase [cholinergic receptor nicotinic alpha (CHRNA)] genes found by GWAS for association with lung cancer reported by Amos et al. (70), Hung et al. (71) and Thorgeirsson et al. (72) indicated the association of chromosome 15q variants for CHRNA (rs16969968 and rs8034191) and risk of lung cancer with odds ratios (ORs) between 1.30 and 1.32. This was not found in patients with lung AC in Asia by Wu et al. (44), who identified four novel single-nucleotide polymorphisms (SNPs) for CHRNA3 (rs2036534c>T, rs667282C>T, rs12916984G>A, and rs6495309T>C) associated with significantly increased lung cancer risk and smoking behavior. Shiraishi et al. reported CHRNA SNPs to be associated with lung cancer susceptibility in a small subset of the Japanese population in a smoking-independent manner (45). These two reports from Asia did not focus on never smokers, and the results regarding smoking habits were very different between the reports, stressing the vital importance of the survey method used when analyzing smoking habits in patients as well as in controls (Table I). Wang et al. found no evidence of association between 6p21.33 or 15q25.1 variation and risk of lung AC in never smokers (47), and this finding was widely accepted in an article written by authors of 69 Institutions (73).
Telomerase reverse transpeptidase (TERT)–Cleft lip and palate transmembrane 1-like protein (CLPTM1L) variants may be involved in lung AC as a gene related to lung AC in never smokers (46-51) (Table I). The survey method regarding ETS was not described in this excellent and important article (Table I). The 5p15.33 region was associated significantly with lung AC in Asian female never smokers (48). Landi et al. revealed that 5p15.33 rs2736100 (TERT) was associated with risk of AC [odds ratio (OR)=1.23] (49). Jin et al. confirmed that 5p15.33, especially in the TERT gene, may also predispose to susceptibility to lung AC in Chinese female never smokers (50). Fine-mapping analysis of genetic variants in the 5p15.33 region conducted by Pande et al. revealed four SNPs associated with lung cancer risk (46): rs4975538, which is an intronic SNP in TERT, rs451360 and rs370348, which are intronic SNPs in CLPTM1L, and rs4975615, which is in the intergenic region between the two genes. However, these excellent articles did not discuss any association with ETS among never smokers.
Lan et al. made the extremely important finding that the strongest association signal, rs7086803 at 10q25.2, located at intron 7 of the vesicle transport through interaction with t-SNAREs homologue 1A gene (VTl1A) gene was implicated in lung carcinogenesis (52). VTl1A is involved in ACRP30-containing vesicles in adipocytes, and lower amounts of VTl1A in cultured adipocytes can inhibit adiponectin secretion (74). However, it is not clear how this international team defined the term never smokers or how several thousand cases and controls were surveyed for smoking habits.
Li et al. demonstrated a strong correlation between the transcription level of the gene glypican-5 (GPC5) and genotypes of the replicated SNP (rs2352028 at 13q31.3) in 77 non-tumor lung tissue samples, and the expression levels of GPC5 in the matched lung AC tissue were lower by half than in normal tissues (54). The main function of membrane-attached glypican is to regulate the signaling pathway of wingless transformation, hedgehog, fibroblast growth factors, and bone morphogenetic proteins (75). The authors showed that down-regulation of GPC5 contributes to the development of lung cancer in never smokers. Their results were discussed from the viewpoints of former smokers and never smokers who had quit smoking (54) in relation to ETS exposure in never smokers (76). Landi et al. did not agree with this observation (55). However, the study by Li et al. (54) opened the way to a new epidemiological approach to clarify the interrelationships between the susceptible genes found by GWAS and carcinogens in ETS.
Tessema et al. reported that never smokers with primary AC had a significantly higher prevalence of methylation of tumor necrosis factor receptor superfamily member 10C (TNFRSF10C), basic helix-loop-helix (bHLH) transcription factor 5 (BHLHB5) and boule-like RNA-binding protein (BOLL) (regulating meiotic G 2/M transition) than current and former smokers (56). Genotypes of C3ORF21, which plays an important role in the formation of NOTCH EGF repeats, were nominally associated with a reduced risk of lung AC among never smokers (39). Another locus, 18p11.22, near the adenomatous polyposis coli down-regulated 1 (APCDD1), N-ethylmaleimide-sensitive factor attachment protein gamma (NAPG) and family with sequence similarity 38, member B (FAM38B) genes was reported in Korean never smokers with NSCLC by Ahn et al. (53). These findings for never smokers are very important, but the survey method of smoking was not sufficiently described.
3. Driver Genes
Driver genes are of interest as genes related to susceptibility to lung AC in Asian women never smokers from the standpoints of signal transduction for cell proliferation, survival migration and angiogenesis, as well as good treatment response to EGFR-TKIs if EGFR mutation was present (57-64) (Table I). Driver genes include mutations of EGFR, Kirsten rat sarcoma viral oncogene (KRAS), protooncogene B-Raf (BRAF), phosphatidylinositol-4,5-bis phosphate 3-kinase, catalytic subunit alpha (PIK3CA), and fusion of echinoderm microtubule-associated protein-like 4 (EML4) and the intracellular signaling portion of the anaplastic lymphoma kinase (ALK) protein (EML4–ALK).
Among never smokers with AC, the frequency of EGFR mutation was reported to be 42 ~50% in Asia (57, 58, 60), and 55% in those of European descent (63), while the frequencies of those with fusion of EML4–ALK, PTEN, PIK3CA, c-MET, KRAS, STK11, and BRAF were 9.3%, 9.1%, 5.2%, 4.8%, 4.5%, 2.7%, and 1.9%, respectively (57). Among smoker patients with AC, the frequency of EGFR mutation was reported to be 22.0% (57) or 10% to 20% (60), while the mutation frequency of STK11 was 19.0%, and of KRAS was 12.0% (57). That is, never smokers with lung AC showed increased EGFR mutation and EML4–ALK fusion protein. These mutations were at low frequency in smokers, but other mutations were seen in smokers with lung cancer (57). Although some cases showed overlap of EGFR and KRAS mutations, these were generally few in AC (57). The EML4–ALK fusion gene has been identified in a small subset of patients with lung AC (58). Ren et al. reported that 82.7% of lung AC in Asian female non-smokers showed well-known oncogenic mutations in EGFR, KRAS, HER2, BRAF and PIK3CA, and a majority of the mutations were mutually exclusive as noted above, except two with EGFR mutation and BRAF mutation, one with EML4–ALK fusion and PIK3CA mutation (59). Li et al. identified seven (among 208 cases; 3.37%) patients with the EML4–ALK fusion gene, of whom four had variant 3, two had variant 1, and one case had variant 1. Of these cases, six were non-smokers, and five were found among 33 cases of female non-smokers. EML4–ALK translocation was predominant in never smokers with AC (58).
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Table I.
List of publications selected for the present review on genes related to lung adenocarcinoma (AC) in never smokers.
Lung AC with EGFR mutation in never smokers showed somewhat good prognosis (57). The effect of ETS on EGFR mutation was reported in lung cancer in never smokers, and the rate of EGFR mutation among never smokers exposed to ETS was significantly low, as discussed below (60). Moreover, EGFR-TKIs, e.g. gefitinib and erlotinib, have been reported to be associated with high sensitivity to AC in Asian women non-smokers (57-61, 77, 78). Activating EGFR mutation, including 19 in-frame deletions and exon 21 L858R substitutions, have been shown to be the most potent biological predictors of sensitivity to EGFR-TKIs (60). Jou et al. found significant frequency of EGFR intron 1 SNP in Asian female never smoker patients with lung AC, as shown in Table I (61).
Other driver genes, human epidermal growth factor receptor type 2 (HER2) and BRAF, have been reported in female never smokers with AC (62, 63). Jo et al. found that three SNPs of HER2 (-3444C>T, -1985G>T and P1170A C>G) were significantly frequent in Asian female non-smokers and nondrinker patients with lung AC (62). Suzuki et al. examined HER2 gene mutation in 1,275 patients (1,055 ACs, 146 SCs, and 74 others) and detected HER2 gene mutation in 46 (3.6%) of all cases. Mutation-positive cases were all ACs, comprising 4.3% of ACs (79). These positive cases were younger never smokers, with smaller tumor size (79).
Jang et al. found novel fusion transcript formed between exon 1-9 of staphylococcal nuclease domain-containing protein 1 (SND1) and exons 2 to 3’ end of BRAF (63). This was observed in 3/89 tested tumors and 2/64 never-smoker lung ACs (63). BRAF mutations and their fusion transcripts lead to constitutive activation of Ser/Thr kinase activity and are located downstream of the RAF/MEK/ERK pathway in lung, melanoma, thyroid and colon cancer (63). SND1 is a component of the RNA-induced silencing complex and plays a role as a regulator for transcription of specific mRNAs by mediating RNA interference reported in colon, prostate and liver cancer (63).
4. Genes Related to Inflammation and Natural Immunity
Kang et al. used an Affymetrix custom-made GeneChip for finding novel genes related to susceptibility to lung cancer in female never smokers, and found three SNPs involved: colony stimulating factor 1 receptor (CSF1R) rs10079250 A>G, TP63 rs7631358G> A and core-pressor interacting with RBPJ 1 (C1R1) rs130090791 (65). CSF1R rs10079250 A>G exhibited an increased level of phosphorylated c-Jun NH(2)-terminal kinase (JNK), a downstream molecule of the CSF1-CSF1R signaling pathway. They did not analyze genes from the viewpoint of ETS. Kiyohara et al. reported that an association of IL1B (rs1143634, 3954 C>T) polymorphism with lung cancer was seen in never smokers (OR=1.11); higher risk was of course observed for smokers (OR=2.48) (66). Olivo-Marston, et al. reported that ETS exposure during childhood is associated with increased lung cancer risk among never smokers, especially never smokers with the haplotype mannose-binding lectin-2 (MBL2) (OR=2.52), suggesting alteration in inherited genetic variants of innate immunity genes (67). In these reports, the authors described in detail their surveys of smoking habits.
Definition of Never Smoker and Survey Methods for Smoking Habits
We sent e-mails with a questionnaire regarding the ETS exposure of never smokers to authors who had published original articles on genes and lung AC among never smokers in December of 2015. Very few responses were received. The reason for this is easily understandable because the hypothesis or purpose of the original articles was not to clarify carcinogenesis but focus on novel genes. Then, we examined the original articles listed in Table I again.
It is globally accepted that the definition of never smoker is as follows: a never smoker is an individual who has had a lifetime exposure of fewer than 100 cigarettes (University of California at Los Angeles, Harvard, Mayo, and International Agency for Research on Cancer studies) (24). This definition is the same as that of World Health Organization nomenclature (80).
Other definitions such as the Hawaii study (those who smoked fewer than 180 cigarettes in their lifetime), the Seoul study (those who smoked fewer than 200 cigarettes in their lifetime), the Liverpool study (those who never smoked more than 10 cigarettes per week regularly), the CREST study (those who either smoked less than 400 cigarettes in their lifetime or less than one cigarette per day for one year), the Aichi and GenAir studies (those who reported they had never smoked) have been reported (24).
We checked the definitions of never smoker described in original articles on novel susceptibility genes for lung AC in never smokers as shown in Table I, and found that most described never smokers as individuals with a lifetime exposure of fewer than 100 cigarettes.
Regarding the method used to survey the smoking habits of the patients, most authors described that a trained interviewer administered the questions regarding smoking. ETS at home, work or other places has been reported to be the cause of lung cancer in never smokers (19-24; Figure 1). De Andrade et al. showed that from 1997 to 2001, 810 women with lung cancer were interviewed to obtain data including the source, intensity, and duration of ETS exposure (76). In this descriptive study, relationships between smoking history, ETS exposure, and lung cancer histological subtypes were analyzed. Among the 810 patients, 773 (95.4%) reported personal smoking or ETS exposure, including 170 out of 207 (82%) never smokers. Among the never smokers with a history of ETS exposure, the mean years of exposure were 27 from smoking spouses, 19 from parents, and 15 from co-workers. For each major subtype of lung cancer (AC, SC, unclassified NSCLC, small cell, or carcinoids) among never smokers, 75-100% of patients had ETS exposure. Trends for AC, SC and small cell carcinoma were found to be statistically significant using the Cochran-Armitage Test for Trend I (p<0.001) among never smokers without ETS exposure, never smokers with ETS exposure, former smokers, and current smokers.
Kurahashi et al. reported a population-based prospective study (Cohort I in 1990 and Cohort II in 1993) including 28,414 nonsmoking women aged 40-69 years old, 28,414 nonsmoking women with exposure to ETS from their husband, at the workplace and during childhood (21). Over 13 years of follow-up, 109 cases with lung cancer were diagnosed. Among them 82 had lung AC. The hazard ratio (HR) in women with smoker husbands compared to those with never-smoker husbands was 1.34, and an association with lung AC had an HR of 2.03, with a dose–response relationship with the husbands' smoking. Tse et al. demonstrated an association between ETS on lung cancer in nonsmokers using a population-based, case-referent study in Hong Kong during 2004-2006, including 132 Chinese male nonsmoker cases with lung cancer and 536 nonsmoking community referents (22). They found a weak association between lung cancer and ETS exposure from household/workplace with OR of 1.11, and an increased risk of lung AC with OR of 1.68.
A recent report by the International Lung Cancer Consortium (ILCCO) in 2014 (24) is valuable in understanding the situation of ETS exposure among never smokers; it includes eight studies in North America, four studies in Europe and six studies in Asia/Oceania. The data comprise 12,688 lung cancer cases and 14,452 controls of which 2,504 cases and 7,276 controls were never smokers. This study showed that exposure to ETS increased risk of lung cancer among both ever smokers and never smokers. The association between ETS exposure and lung cancer development among never smokers was follows: OR=1.35 in males, OR= 1.27 in females; OR=1.56 in those ≥65 years old, OR=1.10 in those<65 years old. The adjusted OR comparing SCLC with NSCLC was 1.28 in the overall population and 2.11 in never smokers.
Clinical Features of Lung AC in Never Smokers
As noted above, global statistics estimate that 15% of lung cancer in men and 53% in women are not attributable to smoking, indicating that worldwide approximately 25% of patients with lung cancer are never smokers (5). Moreover, never smoker patients with lung cancer with AC were female, younger and had a better prognosis (6, 7, 10-12), although Pallis et al. noted that a correlation between younger age in never smoker lung cancer and the severity of cases was doubtful (9). NSCLC in never smokers is currently on the rise (16, 18, 22) and there is a question as to whether it may not be a different disease altogether (23, 24). Moreover, lung AC harboring an EGFR mutation responds to EGFR-TKIs, while that without EGFR mutation does not (85, 86). HER2 amplification was an unfavorable prognostic factor, but HER2 phosphorylation was a favorable prognostic factor (79).
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Figure 1.
Several scenarios showing sources of exposure to environmental tobacco smoke (ETS). Upper right image shows the possibility of ETS at the workplace. Lower image shows ETS at unidentified places. Upper left shows ETS at the home.
Histopathology of Lung Cancer in Never Smokers
Incidence of lung AC in men surpassed the incidence of lung SC in the 1960s and 1970s in the USA and Europe, and in the 1980s and 1990s in Japan (3). The report of the International Association for the Study of Lung Cancer (IASLC) noted imbalance with respect to gender and histology: among 2,341 female patients, 55% had AC, 25% SC, while among 6,796 male patients, 30% had AC and 57% SC (1990~2000) (87).
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Table II.
Genes related to susceptibility to metabolic syndrome resembling genes related to susceptibility to lung adenocarcinoma (AC) in never smokers.
Yang et al. showed a strong association between cigarette smoking history and lung AC in a prospective cohort of 41,836 Iowa women aged 55-69 years with 13 years of follow-up (84). Two-thirds of the enrolled population were never smokers; lung AC was seen in 25%, SC in 8% and SCLC in 4% of all lung cancer. Moreover, they found that women who developed SC consumed more alcohol but less fruit than women with other cancer types, and women who developed SCLC had higher waist circumferences than women with other types of cancer. Women who developed lung cancer were compared with smoker cancer cases, and the differential factors seen in never-smoker cases were higher education, consumption of more fruit, less alcohol and less physical activity.
Seki et al. reported the effects of ETS of spouses on lung cancer in 1,670 cases and 5,855 controls and noted a marginal association of ETS with female lung cancer risk (OR=1.31), whereas no significant association was observed for lung cancer in men (23). Moreover, a recent report by ILCCO noted that ETS exposure in never smokers or former and current smokers showed a strong association with SCLC (24).
Possibility of Metabolic Syndrome Participating in Lung AC in Never Smokers
Yang et al. surveyed body mass index (BMI) and waist circumferences in a prospective cohort of 41,836 Iowa women aged 55-69 years with 13 years of follow-up as described above (84). They showed patients with lung cancer had significantly higher waist circumference (p<0.15). Zhang et al. reported that a comprehensive analysis of adiponectin quantitative trait loci (QTLs) associated with gene expression correlation identified genes related to metabolic syndrome with a potential role in carcinogenesis (88). As discussed above, genes related to susceptibility to lung AC in never smokers include classical genes (participating in detoxifying or metabolizing carcinogenic agents derived mainly from tobacco smoke, tumor-suppressor genes such as TP53 and TP63, and DNA damage-repair genes), genes found by GWAS (nicotine acetylcholine esterase genes, TERT and CLPTM1L, VTL1A, GPC5, TNFRSF10C, C3ORF21, hypermethylation of TNFSF10C, BHLHB5, and BOLL), and driver genes (EGFR, KRAS, BRAF, PIK3CA, EML4-ALK) have been reported (17). Although these genes are not related to metabolic syndrome, EGFR, VTL1A, TNFRSF10C, C3ORF21 and hyper-methylation of TNFSF10C, BHLHB5, and BOLL are involved in the metabolic pathways of metabolic syndrome (Table II). Mazieres et al. reported a close relationship between lung AC in never smokers and metabolic syndrome (89). They examined 140 women with AC (63 never smokers and 77 former/current smokers) and found that never smokers were characterized by a higher frequency of lipidic features (60.3% vs. 37.7%) compared with smokers (89). Obesity, lack of physical activity, heavy alcohol consumption and a diet of food with high fat content all lead to metabolic syndrome and a high percentage of cancer (90, 91). Metabolic syndrome is frequently complicated with type-2 diabetes mellitus, which is associated with increased risk of lung cancer, especially among female diabetic patients with RR=1.14 (92).
Predominance of Women Among Never Smokers
Among the never-smoker patients with lung AC, there were fewer female patients than male patients. However, the percentage of never smokers among total females with lung AC was over 50%. It has been reported that these female patients were younger and had better prognoses, as mentioned above (6-12).
It remains unknown why lung AC is seen in Asian women never smokers (4, 16). Association of EGFR mutation and estrogen receptor (ER)-α and –β with lung carcinogenesis has been reported (93, 94). Mazieres et al. observed increased frequency of EGFR mutation and ERα expression in never smokers with AC with higher frequency of lipidic features (89). Li et al. also revealed that EGFR mutations were seen in 24.5% (51/208 cases) of patients with lung cancer. These mutations were identified with higher frequency in females (47.5% vs. 15.0% in males), never smokers (42.3% vs. 13.9% in smokers), and patients with AC (44.2% vs. 8.0% in patients with non-AC) (58). Moreover, human papillomavirus (HPV) infection may be involved in Asian non-smoking lung AC, which responds to EGFR-TKIs (95). Women with early menarche or late menopause showed significantly increased risk of lung cancer (96). Further epidemiological research should be conducted to clarify these points.
Interaction of Susceptibility Genes and Suspected Carcinogens
There is much information on genes related to susceptibility to lung AC in never smokers as detailed above. Comparing original articles of the 1990s with those published after 2000 by meta-analysis, Okazaki et al. reported that the role of classical genes associated with lung cancer is decreasing and novel genes are emerging (2010), a fact that may reflect changes in lifestyle in Japan (3). Although the carcinogenesis of lung AC in smokers is simple, that in never smokers is complicated. However, clarification is necessary for primary prevention of lung AC. Moreover, the numbers of patients with lung AC in never smokers is increasing for both sexes. Generally the frequency among female never smokers was predominant in lung AC. In developed countries, even though the frequency in men is less than in women, that in never smoker male patients increases after quitting smoking (4).
Different genes associated with susceptibility to cancer of course relate to different metabolic pathways to lung AC. For example, never smokers exposed to ETS exhibited the changes related to classical genes as Bennett et al. suggested (34). Members of the same group, Olivo-Marston et al. showed that ETS exposure during childhood might result in more susceptibility to ETS in never smokers with a haplotype of MBL2, suggesting alteration in inherited genetic variants of innate immunity genes (67). Li et al. found glypican-5 to be a novel gene related to susceptibility to lung AC in never smokers, and showed the possibility of a role for ETS in carcinogenesis in never smokers (54). Although the role of carcinogens in never smokers with most defined genes related to susceptibility remains unknown, the reported interrelations are shown in Table III.
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Table III.
Suspected carcinogens and genes involved in susceptibility to lung AC reported among never smokers.
Lee et al. clearly demonstrated that EGFR mutation rate was 56.6% in lung cancer in never smokers without ETS exposure, 44.0% in lung cancer in never smokers exposed for <45 ETS smoker years and 25.7% in lung cancer in never smokers exposed for ≥45 ETS smoker years (60). The same trend was seen for childhood ETS exposure, and household vs. workplace ETS, i.e. EGFR mutation rate was higher in never smokers without ETS, but a high level of ETS was associated with a low EGFR mutation rate; the EGFR mutation rate was lowest in smokers with lung cancer. According to the review of Choi et al., radon is recognized as the second leading carcinogen of lung cancer, and they listed the genes related to susceptibility as shown in Table III (29).
Subramanian and Govindan reviewed pre-existing lung diseases, oncogenic viruses and human papillomavirus in carcinogenesis of lung cancer in never smokers (97), and they also referred to Schabath et al.'s study (98), in which estrogen replacement therapy was used to exert a protective effect on women against developing lung cancer. Hormone or hormone replacement therapy should be investigated from the viewpoint of carcinogenesis between carcinogen and susceptible genes; further research is necessary.
We reported the possibility of genes associated with metabolic syndrome being related to lung AC in never smokers (17). However, while ETS and the other agents shown in Table III are suspected carcinogens, it is not yet clear whether they actually are. The references shown in Table II did not reveal the physical characteristics of metabolic syndrome such as body weight, body length, BMI, waist circumferences, hypertension, hyperlipidemia, blood sugar levels and hemoglobin A1c (HbA1c) levels (99). Even if the study subjects of these reports had the physical characteristics of metabolic syndrome, ETS or other carcinogens, drugs, infectious agents might be responsible for causing lung cancer because metabolic syndrome, obesity and type 2 diabetes mellitus are known to be complicating factors in malignant disease. The carcinogen(s) involved in never-smoker patients with lung cancer and metabolic syndrome requires further elucidation.
Future Directions
For primary prevention of lung AC in never smokers, of course, it is necessary to educate the general population that they should quit smoking and reduce ETS. We should promote the conduct of epidemiological surveys of suspected carcinogens in individual districts in order to clarify the carcinogenesis of lung AC in each case with genes related to susceptibility to cancer.
The relationship between AC and metabolic syndrome is especially important, because in general Asians are increasingly suffering from increased obesity and type-2 diabetes mellitus (92) as their life-styles change to include more Western-type foods and they engage in less physical activity. A recent report by Zanetti et al. showed the importance of ethnic groups with respect to susceptibility genes (100). The interaction of these genes and carcinogens, even if they are only suspected, should be investigated. The well-known review by Sun et al. (6) pointed out that EGFR mutation in never smokers vs. KRAS mutation and TP53 mutation in never smokers are frequently seen, and these differences in never smokers support the idea that different carcinogens are involved for different groups of never smokers. Moreover, the study of different oncogenic mutation spectra of lung AC in never smokers can lead to better selection of effective treatments and improve prognosis. We could establish strategies for primary prevention of lung AC in never smokers if we start to conduct surveys in individual districts.
Acknowledgements
The Authors thank Ms. Yoshie Muroya for her systematic research of the literature and Ms. Cecilia Hamagami for her assistance in the English revision of the manuscript.
The Authors are grateful to the Smoking Research Foundation (Tokyo, Japan) for support of this research in 2015~2016 (IO. SI, YS).
Footnotes
This article is freely accessible online.
Conflicts of Interest
The Authors confirm that there are no conflicts of interest regarding the contents of this article. This study was approved by the Ethics Committee of the International University of Health and Welfare (13-B-130; October 10, 2015).
Received October 8, 2016.
Revision received October 30, 2016.
Accepted November 3, 2016.
Copyright© 2016 International Institute of Anticancer Research (Dr. John G. Delinassios), All rights reserved
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Vol. 36, Issue 12
December 2016
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Lung Adenocarcinoma in Never Smokers: Problems of Primary Prevention from Aspects of Susceptible Genes and Carcinogens
ISAO OKAZAKI, SHIGEMI ISHIKAWA, WATARU ANDO, YASUNORI SOHARA
Anticancer Research Dec 2016, 36 (12) 6207-6224;
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Abstract
Selection of Bibliography
Original Articles on Genes Related to Susceptibility to Lung AC in Never Smokers
1. Classical Genes
2. Susceptible Genes Found by Genome-wide Association Study (GWAS)
3. Driver Genes
4. Genes Related to Inflammation and Natural Immunity
Definition of Never Smoker and Survey Methods for Smoking Habits
Clinical Features of Lung AC in Never Smokers
Histopathology of Lung Cancer in Never Smokers
Possibility of Metabolic Syndrome Participating in Lung AC in Never Smokers
Predominance of Women Among Never Smokers
Interaction of Susceptibility Genes and Suspected Carcinogens
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Lung Cancer in Non-smokers in Czech Republic: Data from LUCAS Lung Cancer Clinical Registry
Molecular characterization of non-small cell lung cancer tumors in Latin American patients from Brazil, Chile and Peru uncovers novel potentially driver mutations
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Proteolytic Enzyme Therapy in Complementary Oncology: A Systematic Review
Multimodal Treatment of Primary Advanced Ovarian Cancer
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Lung Adenocarcinoma in Never Smokers: Problems of Primary Prevention from Aspects of Susceptible Genes and Carcinogens
ISAO OKAZAKI, SHIGEMI ISHIKAWA, WATARU ANDO, YASUNORI SOHARA
Anticancer Research Dec 2016, 36 (12) 6207-6224;
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How to Solve Radical Inequalities?
An inequality that contains a variable within a radical is called a "radical inequality". To solve a radical inequality, we can use a graph or algebra.
A step-by-step guide tosolving radical inequalities
In mathematics, a radical is the opposite of an exponent, denoted by the symbol √, also known as the root.
An inequality that has variables inside the radicand is called a radical inequality. In other words, a radical inequality is an inequality that has a variable or variables inside the radical symbol.
We can solve radical inequalities using algebra. Some radical inequalities also have variables outside the radical, and we can use algebra to calculate them as well. The following steps can be used to solve radical inequalities:
Step 1:Check the index of the radical.
If the index is even, the final calculated value of the radicand cannot be negative and must be positive. This is called domain restriction.
Step 2: If the index is even, consider the value of the radicand as positive. Solve for the variable x in radicands.
Therefore, we solve for the variable 𝑥 x for this radicand when it is greater than or equal to zero. That is, we consider the radicand as 𝑥≥0 x≥0 from the radical inequality 𝑥⎯⎯√𝑛<𝑑 x n<d and calculate the variable 𝑥 x. If the index is odd, however, then consider the radicand as 𝑥<𝑑 x<d.
Step 3: Solve the original inequality expression algebraically and also remove the radical symbol from the expression.
We eliminate the radical by taking the index and using it as the exponent in terms of both sides of the inequality. (i.e., 𝑥⎯⎯√𝑛<𝑑→(𝑥⎯⎯√𝑛)𝑛<𝑑 𝑛)x n<d→(x n)n<d n). Note here that when using the index as an exponent on the radical expression, it nullifies the radical symbol, thus removing it.
Step 4: Test the values to check the solution.
To test the values of 𝑥 x, we consider a random value that satisfies the inequality. And we will also consider values outside the equality so that we can confirm the correctness of our solution.
Solving Radical Inequalities– Example 1:
solve 3+4 𝑥−4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√≤7 3+4 x−4≤7.
Solution:
To solve this radical inequality, first, we check the index of the given radical inequality. Since the index value is not given, the index value is 2 2. Since the index is even, the radicand of the square root will be greater than or equal to zero.
4 𝑥−4≥0 4 x−4≥0
4 𝑥≥4 4 x≥4
𝑥≥1 x≥1………….. (1)(1)
We now solve the radical inequality algebraically and also remove the radical symbol to simplify it. First, we isolate the radical.
3+4 𝑥−4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√≤7→4 𝑥−4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√≤4 3+4 x−4≤7→4 x−4≤4
Now, we remove the radical symbol by taking the index as an exponent on both sides of the inequality.
(4 𝑥−4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√)2≤4 2(4 x−4)2≤4 2
4 𝑥−4≤16 4 x−4≤16
4 𝑥≤20 4 x≤20
𝑥≤5 x≤5………….. (2)(2)
Here, we got two inequalities for the value of 𝑥 x from equations 1 1 and 2 2. So we combine them both and write it as a compound inequality. Then our final answer is:
1≤𝑥≤5 1≤x≤5
Exercises forSolving Radical Inequalities
Solve.
𝑥+3⎯⎯⎯⎯⎯⎯⎯⎯⎯√3≥2 x+3 3≥2
−2 𝑥+1⎯⎯⎯⎯⎯⎯⎯⎯⎯√≤−6−2 x+1≤−6
4 𝑥+1⎯⎯⎯⎯⎯⎯⎯⎯⎯√3≥12 4 x+1 3≥12
𝑥≥5 x≥5
𝑥≥8 x≥8
𝑥≥26 x≥26
by: Effortless Math Team about 3 years ago (category: Articles)
Effortless Math Team
3 weeks ago
Effortless Math Team
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Substitution
Substitution is when one replaces all instances of a variable (or expression) with another equivalent variable (or expression).
Contents
1 Uses
1.1 System of Equations
1.2 Observing Common Parts
2 Problems
2.1 Introductory
2.2 Intermediate
Uses
System of Equations
Main article: System of equations
Substitution is a relatively universal method to solve simultaneous equations. It is generally introduced in a first year high school algebra class. A solution generally exists when the number of equations is exactly equal to the number of unknowns. The method of solving by substitution includes:
Isolation of a variable
Substitution of variable into another equation to reduce the number of variables by one
Repeat until there is a single equation in one variable, which can be solved by means of other methods.
For example, consider the below system. An example of solving the system by substitution is when we start by isolating in the top equation to get . Then, we can replace all instances of with in the second equation. Doing so results in an equation with one variable, and solving it results in After solving for , we can "plug in" the value for to get , so the solution to the system is . As usual, we can check by substituting for and for .
This same method is used for simultaneous equations with more than two equations.
Observing Common Parts
Substitution can also be used when an expression has multiple common parts.
For instance, consider the equation . Note that with exponent properties, we can rewrite the equation as . Because there are multiple instances of in the equation, we can let and substitute for to make the equation easier to solve.
Doing so results in Finally, we can substitute for to get or , resulting in or . As usual, we can check by substituting and for separately.
Problems
Introductory
1961 AHSME Problems/Problem 5
1969 AHSME Problems/Problem 17
2007 iTest Problems/Problem 2
Intermediate
1983 AIME Problems/Problem 3
Retrieved from "
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[Solved] The range of ab if |a|≤ 1 and a + b = 1, (a, b &isi
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The range of ab if |a|≤ 1 and a + b = 1, (a, b ∈ R) is:
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[1 4,2]
[−2,1 4]
[0,1 4]
[0, 2]
Answer (Detailed Solution Below)
Option 2 : [−2,1 4]
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Concept:
Range of a function:The range of a function may refer to either of two closely related concepts: The codomain of the function, The image of the function Given two sets X and Y, a binary relation f between X and Y is a function if for every x in X there is exactly one y in Y such that f relates x to y.
(or)The range of a function is the set of all output values (Y-values).
AM ≥ GM ≥ HM where,AM - Arithmetic mean, GM - Geometric mean, HM - harmonic mean
Calculations:
Given,|a|≤ 1 and a + b = 1, (a, b ∈ R)
We know that AM≥ GM
⇒a+b 2≥(a b)1 2
Since a + b = 1 and squaring on both sides we get
⇒1 4≥a b( is the maximum value)
And also we have|a|≤ 1
⇒ -1≤ a≤ 1
If a = 1, then b = 0 |∵ a + b = 1
If a = - 1, then b = 1 - a
b = 2 |∵ a + b = 1
and in this case ab = -1 × 2 = - 2 ( is the minimum value)
Hence the range of ab is[−2,1 4]
Hence, the correct answer is option 3)
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4.1.1 Solving Linear Equations in One Variable | Introductory Algebra
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Introductory Algebra
Chapter 4: Linear Equations and Inequalities in One Variable
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4.1.1 Solving Linear Equations in One Variable
Solving Linear Equations
Learning Objectives
Determine whether or not an equation in one variable is defined as a linear equation
Use the addition and subtraction property to solve a linear equation
Use the multiplication and division property to solve a linear equation
KEY WORDS
Equality: The state of two or more entities having the same value.
Equation: a statement that two expressions are equal.
Solution: a number or numbers that make an equation true.
Equivalent equations: two or more equations that have identical solutions.
Linear Equations
An equation with just one variable is said to be linear when the highest power on the variable is 1 1. Remember that x 1 x 1 is equivalent to x x, so any equation that can be simplified to a x+b=c a x+b=c (where a,b,c a,b,c are real numbers) is a linear equation in one variable. For example, the equation 2 x+3=7,5 x=7,−3 4 x+7 3=6 5 2 x+3=7,5 x=7,−3 4 x+7 3=6 5 are all linear equations because the highest power of x x is 1 1. On the other hand, the equation x 3+2 x 2=4 x−2 x 3+2 x 2=4 x−2 is NOT linear because the highest power of x x is 3 3, not 1 1.
EXAMPLE
Determine whether the equation is linear:
a)3 x−7=6 x+2 3 x−7=6 x+2
b)x 3−5 x 2=5 x+6 x 3−5 x 2=5 x+6
c)x(x+6)=8 x(x+6)=8
d)x−3 8=1 2 x−3 8=1 2
Solution:
a) 3 x−7=6 x+2 3 x−7=6 x+2 Yes, it is linear since the highest power of x x is 1 1
b) x 3−5 x 2=5 x+6 x 3−5 x 2=5 x+6 No, it is NOT linear since the highest power of x x is 3 3, not 1 1
c) x(x+6)=8 x(x+6)=8 Use the distributive property to simplify the left side of the equation:x 2+6 x=8 x 2+6 x=8. Now we can see that the equation is NOT linear since the highest power of x x is 2 2, not 1 1
d) x−3 8=1 2 x−3 8=1 2 Yes, it is linear since the highest power of x x is 1 1
TRY IT
Determine whether the equation 6 x−7=8 x 2+5 6 x−7=8 x 2+5 is linear.
Show Solution
Solution:
6 x−7=8 x 2+5 6 x−7=8 x 2+5 is NOT linear since the highest power of x x is 2 2, not 1 1.
TRY IT
Determine whether the equation 4 5 x+6=−7 3 x 4 5 x+6=−7 3 x is linear.
Show Solution
Solution:
4 5 x+6=−7 3 x 4 5 x+6=−7 3 x is linear since the highest power of x x is 1 1.
The Addition and Subtraction Property of Equality
When an equation contains a variable such as x x, this variable is considered an unknown value. In many cases, we can find the possible values for x x that would make the equation true. We can solve the equation for x x.
For some equations like x+3=5 x+3=5 it is easy to guess the solution: the only possible value of x x is 2 2, because 2+3=5 2+3=5.However, it becomes useful to have a process for finding solutions for unknowns as problems become more complex.
In order to solve an equation, we need to isolate the variable. Isolating the variable means rewriting an equivalent equation in which the variable is on one side of the equation and everything else is on the other side of the equation. The variable is an unknown quantity whose value we are trying to find.We have a solution when we reorganize the equation into the form x x = a constant.
An equation is two expressions that are set equal to each other. If we perform an operation (+, –, ·, ÷, etc) to one side of the equation, it will upset the balance of the equation. Our goal is to keep the balance of the equation intact. This can be likened to a balance scale where the equation x+3=5 x+3=5 is illustrated:
The second we add or subtract a “weight” to one side of the scale, it will become unbalanced. In order to right that balance, we must add or subtract the same “weight” to the other side of the sale.
To solve this equation for x x, we need to isolate x x by removing the +3+3 from the left side of the equation. We do this this by “undoing” the addition of+3+3 by subtracting 3 3. In other words we use the inverse operation of addition, which is subtraction. But if we subtract 3 3 from the left side of the equation, we must also subtract 3 3 from the right side to keep the equation balanced.
x+3=5 x+3−3=5−3 x=2 x+3=5 x+3−3=5−3 x=2
We can verify that x=2 x=2 is indeed the solution by substituting 2 2 in for x x in the original equation: x+3=5 2+3=5 5=5 x+3=5 2+3=5 5=5 Since this is true,x=2 x=2 is the solution of x+3=5 x+3=5.
If the equation is x−5=9 x−5=9, we isolate the variable by adding 5 5 to both sides of the equation. This is because, to isolate x x we must “undo” subtract 5 5 by adding 5 5:
x−5=9 x−5+5=9+5 x=14 x−5=9 x−5+5=9+5 x=14
Again, we can verify the solution:x−5=9 14−5=9 9=9 x−5=9 14−5=9 9=9 True.
When we add or subtract the same term to both sides of an equation, we get an equivalent equation. Equivalent equations are two or more equations that have identical solutions. This leads us to the the Addition and Subtraction Property of Equality.
Addition & subtraction Property of Equality
For all real numbers a,b a,b, and c c, if a=b a=b, then a+c=b+c a+c=b+c and a−c=b−c a−c=b−c.
Adding or subtracting the same term to both sides of an equation will result in an equivalent equation.
EXAMPLE
Solve: x+11=−3 x+11=−3
Solution:
To isolate x x, we undo the addition of 11 11 by using the Subtraction Property of Equality.
x+11=−3 x+11=−3
Subtract 11 from each side to “undo” the addition.x+11−11=−3−11 x+11−11=−3−11
Simplify.x=−14 x=−14
Check:x+11=−3 x+11=−3
Substitute x=−14 x=−14 .−14+11?=−3−14+11=?−3
−3=−3✓−3=−3✓
Since x=−14 x=−14 makes x+11=−3 x+11=−3 a true statement, we know that it is a solution to the equation.
TRY IT
EXAMPLE
Solve: m−4=−5 m−4=−5
Solution:
m−4=−5 m−4=−5
Add 4 to each side to “undo” the subtraction.m−4+4=−5+4 m−4+4=−5+4
Simplify.m=−1 m=−1
Check:m−4=−5 m−4=−5
Substitute m=−1 m=−1 .−1+4?=−5−1+4=?−5
−5=−5✓−5=−5✓
The solution to m−4=−5 m−4=−5 is m=−1 m=−1
TRY IT
Linear Equations Containing Fractions
It is not uncommon to encounter equations that contain fractions; therefore, in the following examples, we will demonstrate how to use the addition property of equality to solve an equation with fractions.
EXAMPLE
Solve: n−3 8=1 2 n−3 8=1 2
Solution
n−3 8=1 2 n−3 8=1 2
Use the Addition Property of Equality.n−3 8+3 8=1 2+3 8 n−3 8+3 8=1 2+3 8
Find the LCD to add the fractions on the right.n 3 8+3 8=4 8+3 8 n 3 8+3 8=4 8+3 8
Simplify.n=7 8 n=7 8
Check:n−3 8=1 2 n−3 8=1 2
Substitute n=7 8 n=7 8 7 8−3 8?=1 2 7 8−3 8=?1 2
Subtract.4 8?=1 2 4 8=?1 2
Simplify.1 2=1 2✓1 2=1 2✓
The solution checks.
TRY IT
Watch this video for more examples of solving equations that include fractions and require addition or subtraction.
Linear Equations Containing Decimals
Decimals will be encountered any time money or metric measurements are used.
eXAMPLE
Solve a−3.7=4.3 a−3.7=4.3
Solution
a−3.7=4.3 a−3.7=4.3
Use the Addition Property of Equality.a−3.7+3.7=4.3+3.7 a−3.7+3.7=4.3+3.7
Add.a=8 a=8
Check:a−3.7=4.3 a−3.7=4.3
Substitute a=8 a=8 .8−3.7?=4.3 8−3.7=?4.3
Simplify.4.3=4.3✓4.3=4.3✓
The solution checks.
TRY IT
The Multiplication and Division Properties of Equality
Just as we can add or subtract the same exact quantity on both sides of an equation, we can also multiply or divide both sides of an equation by the same quantity to write an equivalent equation. This makes sense because multiplication is just repeated addition and division is multiplication by a reciprocal.
For example, to solve the equation 3 x=15 3 x=15 we need to isolate the variable x x which is multiplied by 3 3. To “undo” this multiplication, we divide both sides of the equation by 3 3:\frac{3x}\color{blue}{{3}}=\frac{15}\color{blue}{{3}}\frac{3x}\color{blue}{{3}}=\frac{15}\color{blue}{{3}}. Since 3 3=1 3 3=1 and 1 x=x 1 x=x, this simplifies to x=5 x=5 and we have our solution.
On the other hand, in the equation 1 2 x=−5 1 2 x=−5 the variable x x is multiplied by 1 2 1 2. To isolate the variable we need to turn 1 2 1 2 into 1 1 by multiplying by the reciprocal 2 2: 2(1 2 x)=2(−5)2(1 2 x)=2(−5). This simplifies to x=−10 x=−10 and we have our solution.
This characteristic of equations is generalized in the Multiplication & Division Property of Equality.
multiplication & Division Property of Equality
For all real numbers a,b,c a,b,c, and c≠0 c≠0, if a=b a=b, then a c=b c a c=b c and a c=b c a c=b c.
Multiplying or dividing the same non-zero term to both sides of an equation will result in an equivalent equation.
Technically we could multiply both sides of an equation by zero but that would wipe out our entire equation to 0=0 0=0. However, we can never divide by zero as that is undefined.
When the equation involves multiplication or division, we can “undo” these operations by using the inverse operation to isolate the variable.
Example
Solve 3 x=24 3 x=24. When you are done, check your solution.
Solution
Divide both sides of the equation by 3 3 to isolate the variable (this is will give you a coefficient of 1 1).Dividing by 3 3 is the same as multiplying by 1 3 1 3.
3 x–––=24–––3 3 x=8 3 x _=24 _ 3 3 x=8
Check by substituting your solution, 8 8, for the variable in the original equation.
3 x=24 3⋅8=24 24=24 3 x=24 3⋅8=24 24=24
The solution is correct!
Answer
x=8 x=8
example
Solve: a−7=−42 a−7=−42
Solution
Here a a is divided by −7−7. We can multiply both sides by −7−7 to isolate a a.
a−7=−42 a−7=−42
Multiply both sides by −7−7 .−7(a−7)=−7(−42)−7(a−7)=−7(−42)
[latex]\frac{-7a}{-7}=294[/latex]
Simplify.a=294 a=294
Check your answer.a−7=−42 a−7=−42
Let a=294 a=294 .294−7?=−42 294−7=?−42
−42=−42✓−42=−42✓
try it
Another way to think about solving an equation when the operation is multiplication or division is that we want to multiply the coefficient by the multiplicative inverse (reciprocal) in order to change the coefficient to 1 1.
In the following example, we change the coefficient to 1 1 by multiplying by the multiplicative inverse of 1 2 1 2.
Example
Solve 1 2 x=8 1 2 x=8 for x x.
Solution
We can multiply both sides by the reciprocal of 1 2 1 2, which is 2 1 2 1.
(2 1)1 2 x=(2 1)8(1)x=16 x=16(2 1)1 2 x=(2 1)8(1)x=16 x=16
The video shows examples of how to use the Multiplication and Division Property of Equality to solve one-step equations with integers and fractions.
example
Solve: 4 x=−28 4 x=−28
Solution:
To solve this equation, we use the Division Property of Equality to divide both sides by 4 4.
4 x=−28 4 x=−28
Divide both sides by 4 to undo the multiplication.4 x 4=−28 4 4 x 4=−28 4
Simplify.x=−7 x=−7
Check your answer.4 x=−28 4 x=−28
Let x=−7 x=−7. Substitute −7−7 for x.4(−7)?=−28 4(−7)=?−28
−28=−28−28=−28
Since this is a true statement, x=−7 x=−7 is a solution to 4 x=−28 4 x=−28.
try it
As we solve equations that require several steps, it is not unusual to end up with an equation that looks like the one in the next example, with a negative sign in front of the variable.
example
Solve: −r=2−r=2
Solution:
Remember −r−r is equivalent to −1 r−1 r.
−r=2−r=2
Rewrite −r−r as −1 r−1 r .−1 r=2−1 r=2
Divide both sides by −1−1 .−1 r−1=2−1−1 r−1=2−1
Simplify.r=−2 r=−2
Check.−r=2−r=2
Substitute r=−2 r=−2−(−2)?=2−(−2)=?2
Simplify.2=2✓2=2✓
try it
The equations x=4 x=4 and 4=x 4=x are equivalent. Both say that x x is equal to 4 4. This is an example of The Reflection Property of Equality.
The Reflection Property of Equality
If a=b a=b, then b=a b=a, for all real numbers a a and b b.
This implies that it does not matter which side of the equation the variable term ends up on.
The next video includes examples of using the division and multiplication properties to solve equations with the variable on the right side of the equal sign.
Two-Step Linear Equations
If the equation is in the form a x+b=c a x+b=c, where x x is the variable, we can solve the equation as before. First we must isolate the x−x−term by “undoing” the addition or subtraction. Then we isolate the variable by “undoing” the multiplication or division.
Example
Solve: 4 x+6=−14 4 x+6=−14
Solution:
In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side.
Since the left side is the variable side, the 6 is out of place. We must “undo” adding 6 6 by subtracting 6 6, and to keep the equality we must subtract 6 6 from both sides. Use the Subtraction Property of Equality.4 x+6−6=−14−6 4 x+6−6=−14−6
Simplify.4 x=−20 4 x=−20
Now all the x x s are on the left and the constant on the right.
Use the Division Property of Equality.4 x 4=−20 4 4 x 4=−20 4
Simplify.x=−5 x=−5
Check:4 x+6=−14 4 x+6=−14
Let x=−5 x=−5 .4(−5)+6=−14 4(−5)+6=−14
−20+6=−14−20+6=−14
−14=−14✓−14=−14✓
Solve: 2 y−7=15 2 y−7=15
Solution:
Notice that the variable is only on the left side of the equation, so this will be the variable side, and the right side will be the constant side. Since the left side is the variable side, the 7 7 is out of place. It is subtracted from the 2 y 2 y, so to “undo” subtraction, add 7 7 to both sides.
2 y−7 2 y−7 is the side containing a variable.15 15 is the side containing only a constant.
Add 7 7 to both sides.2 y−7+7=15+7 2 y−7+7=15+7
Simplify.2 y=22 2 y=22
The variables are now on one side and the constants on the other.
Divide both sides by 2 2.2 y 2=22 2 2 y 2=22 2
Simplify.y=11 y=11
Check:2 y−7=15 2 y−7=15
Let y=11 y=11 .2⋅11−7?=15 2⋅11−7=?15
22−7?=15 22−7=?15
15=15✓15=15✓
Example
Solve 3 y+2=11 3 y+2=11
Solution
Subtract 2 2 from both sides of the equation to get the term with the variable by itself.
3 y+2=11−2−2–––––––––––––3 y=9 3 y+2=11−2−2 _ 3 y=9
Divide both sides of the equation by 3 3 to get a coefficient of 1 1 for the variable.
3 y–––=9–3 9 y=3 3 y _=9 _ 3 9 y=3
Answer
y=3 y=3
Try It
In the following video, we show examples of solving two step linear equations.
Remember to check the solution of an algebraic equation by substituting the value of the variable into the original equation.
Candela Citations
CC licensed content, Original
Revision and Adaptation. Provided by: Lumen Learning. . Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Linear Equations with No Solutions of Infinite Solutions.. Authored by: James Sousa . Provided by: (Mathispower4u.com) for Lumen Learning.. Located at: License: CC BY: Attribution
Definition of a Linear Equation. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
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Beginning and Intermediate Algebra.. Authored by: Tyler Wallace.. Located at: License: CC BY: Attribution
Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. . Provided by: Monterey Institute of Technology and Education.. Located at: License: CC BY: Attribution
Licenses and Attributions
CC licensed content, Original
Revision and Adaptation. Provided by: Lumen Learning. . Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Linear Equations with No Solutions of Infinite Solutions.. Authored by: James Sousa . Provided by: (Mathispower4u.com) for Lumen Learning.. Located at: License: CC BY: Attribution
Definition of a Linear Equation. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
CC licensed content, Shared previously
Beginning and Intermediate Algebra.. Authored by: Tyler Wallace.. Located at: License: CC BY: Attribution
Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. . Provided by: Monterey Institute of Technology and Education.. Located at: License: CC BY: Attribution
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14.4 Basic Solutions and Hydrolysis of Salt Solutions
Learning Objectives
By the end of this module, you will be able to:
Predict whether a salt solution will be acidic, basic, or neutral
Calculate the concentrations of the various species in a salt solution
Describe the process that causes solutions of certain metal ions to be acidic
As we have seen in the section on chemical reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word “neutralization” seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall now see.
Acid-Base Neutralization
A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:
A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the conjugate partners produced are of negligible strength: HCl(aq)+NaOH(aq)⇌NaCl(aq)+H2O(l)
A strong acid and a weak base yield a weakly acidic solution, not because of the strong acid involved, but because of the conjugate acid of the weak base.
A weak acid and a strong base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of the weak acid and the conjugate acid of the strong base. The conjugate acid of the strong base is a weaker acid than water and has no effect on the acidity of the resulting solution. However, the conjugate base of the weak acid is a weak base and ionizes slightly in water. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.
A weak acid plus a weak base can yield either an acidic, basic, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acid and the weak base will have the same strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral. To predict whether a particular combination will be acidic, basic or neutral, tabulated K values of the conjugates must be compared.
Stomach Antacids
Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO3. The reaction,
CaCO3(s)+2HCl(aq)⇌aCl2(aq)+H2O(l)+CO2(g)
not only neutralizes stomach acid, it also produces CO2(g), which may result in a satisfying belch.
Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)2. It works according to the reaction:
Mg(OH)2(s)⇌Mg2+(aq)+2OH-(aq)
The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that
H3O++OH-⇌2H2O(l)
This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect.
Several antacids have aluminum hydroxide, Al(OH)3, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.
Culinary Aspects of Chemistry
Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO3 is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.
Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 1). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.
Figure 1. A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.
Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.
Salts of Weak Bases and Strong Acids
When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH4Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:
NH3(aq)+HCl(aq)⟶NH4Cl(aq)
A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:
NH4+(aq)+H2O(l)⇌H3O+(aq)+NH3(aq)
The equilibrium equation for this reaction is simply the ionization constant. Ka, for the acid NH4+:
[H3O+][NH3][NH4+]=Ka
We will not find a value of Ka for the ammonium ion in Ionization Constants of Weak Acids. However, it is not difficult to determine Ka for NH4+ from the value of the ionization constant of water, Kw, and Kb, the ionization constant of its conjugate base, NH3, using the following relationship:
Kw=Ka×Kb
This relation holds for any base and its conjugate acid or for any acid and its conjugate base.
Example 1: The pH of a Solution of a Salt of a Weak Base and a Strong Acid
Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [C6H5NH3+]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?
C6H5NH3+(aq)+H2O(l)⇌H3O+(aq)+C6H5NH2(aq)
Show Answer
The new step in this example is to determine Ka for the C6H5NH3+ ion. The C6H5NH3+ ion is the conjugate acid of a weak base. The value of Ka for this acid is not listed in Ionization Constants of Weak Acids, but we can determine it from the value of Kb for aniline, C6H5NH2, which is given as 4.6 × 10−10 (Relative Strengths of Acids and Bases and Ionization Constants of Weak Bases):
Ka( for C6H5NH3+)×Kb( for C6H5NH2)=Kw=1.0×10−14
Ka(forC6H5NH3+)=KwKb( for C6H5NH2)=1.0×10−144.6×10−10=2.2×10−5
Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H3O+, and the pH:
With these steps we find [H3O+] = 2.3 × 10−3M and pH = 2.64
Check Your Learning
Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of C6H5NH3+ is 2.3 × 10−3 and the pH is 2.64.
What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions NH4+ and NO3-. Use the data in Table 2 on Relative Strengths of Acids and Bases to determine Kb for the ammonium ion. Which is the stronger acid C6H5NH3+ or NH4+?
Show Answer
Ka(forNH4+)=5.6×10−10, [H3O+] = 7.5 × 10−6M
C6H5NH3+ is the stronger acid.
Salts of Weak Acids and Strong Bases
When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH3CO2, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:
CH3CO2H(aq)+NaOH(aq)⟶NaCH3CO2(aq)+H2O(aq)
A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:
CH3CO2-(aq)+H2O(l)⇌CH3CO2H(aq)+OH-(aq)
The equilibrium equation for this reaction is the ionization constant, Kb, for the base CH3CO2-. The value of Kb can be calculated from the value of the ionization constant of water, Kw, and Ka, the ionization constant of the conjugate acid of the anion using the equation:
Kw=Ka×Kb
For the acetate ion and its conjugate acid we have:
Kb(forCH3CO2-)=KwKa(forCH3CO2H)=1.0×10−141.8×10−5=5.6×10−10
Some handbooks do not report values of Kb. They only report ionization constants for acids. If we want to determine a Kb value using one of these handbooks, we must look up the value of Ka for the conjugate acid and convert it to a Kb value.
Example 2: Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with [CH3CO2-]=0.050M and [OH−] = 2.5 × 10−6M at equilibrium. The reaction is:
CH3CO2-(aq)+H2O(l)⇌CH3CO2H(aq)+OH-(aq)
Show Answer
We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.
The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine Kb as follows:
Kb( for CH3CO2-)=KwKa( for CH3CO2H)=1.0×10−141.8×10−5=5.6×10−10
Now find the missing concentration:
Kb=[CH3CO2H][OH-][CH3CO2-]=5.6×10−10
=CH3CO2H(0.050)=5.6×10−10
Solving this equation we get [CH3CO2H] = 1.1 × 10−5M.
Check Your Learning
What is the pH of a 0.083-M solution of CN−? Use 4.0 × 10−10 as Ka for HCN. Hint: We will probably need to convert pOH to pH or find [H3O+] using [OH−] in the final stages of this problem.
Show Answer
11.16
Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base
In a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the Ka of the weak acid and the Kb of the weak base. If Ka > Kb, the solution is acidic, and if Kb > Ka, the solution is basic.
Example 3: Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
KBr
NaHCO3
NH4Cl
Na2HPO4
NH4F
Show Answer
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:
The K+ cation and the Br− anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.
The Na+ cation is a spectator, and will not affect the pH of the solution; while the HCO3- anion is amphiprotic, it could either behave as an acid or a base. The Ka of HCO3- is 4.7 × 10−11, so the Kb of its conjugate base is 1.0×10−144.7×10−11=2.1×10−4.
Since Kb >> Ka, the solution is basic.
The NH4+ ion is acidic and the Cl− ion is a spectator. The solution will be acidic.
The Na+ ion is a spectator, while the HPO4− ion is amphiprotic, with a Ka of 3.6 × 10−13
so that the Kb of its conjugate base is 1.0×10−143.6×10−13=2.8×10−2. Because Kb >> Ka, the solution is basic.
The NH4+ ion is listed as being acidic, and the F− ion is listed as a base, so we must directly compare the Ka and the Kb of the two ions. Ka of NH4+ is 5.6 × 10−10, which seems very small, yet the Kb of F− is 1.4 × 10−11, so the solution is acidic, since Ka > Kb.
Check Your Learning
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
K2CO3
CaCl2
KH2PO4
(NH4)2CO3
AlBr3
Show Answer
(1) basic; (2) neutral; (3) basic; (4) basic; (5) acidic
The Ionization of Hydrated Metal Ions
If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, Al(H2O)63+, dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this Al(H2O)63+ cluster as well:
Al(NO3)3(s)+6H2O(l)⟶Al(H2O)63+(aq)+3NO3-(aq)
We frequently see the formula of this ion simply as “Al3+(aq)”, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, H3O+ to H+. However, in this case, the hydrated aluminum ion is a weak acid (Figure 2) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:
Al(H2O)63+(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)5(OH)2+(aq)Ka=1.4×10−5
As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:
Al(H2O)63+(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)5(OH)2+(aq)
Al(H2O)5(OH)2+(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)4(OH)2+(aq)
Al(H2O)4(OH)2+(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)3(OH)3(aq)
Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the right side of the equilibrium expressions and as bases when they appear on the left side.
Figure 3. When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.
However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:
Fe(H2O)63+(aq)+H2O(l)⇌H3O+(aq)+Fe(H2O)5(OH)2+(aq)Ka=2.74
Cu(H2O)62+(aq)+H2O(l)⇌H3O+(aq)+Cu(H2O)5(OH)+(aq)Ka= 6.3
Zn(H2O)42+(aq)+H2O(l)⇌H3O+(aq)+Zn(H2O)3(OH)+(aq)Ka=9.6
Example 4: Hydrolysis of [Al(H2O)6]3+
Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [Al(H2O)6]3+ in solution.
Show Answer
In spite of the unusual appearance of the acid, this is a typical acid ionization problem.
Determine the direction of change. The equation for the reaction and Ka are:Al(H2O)63+(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)5(OH)2+(aq)Ka=1.4×10−5The reaction shifts to the right to reach equilibrium.
Determine x and equilibrium concentrations. Use the table:
Solve for x and the equilibrium concentrations. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:Ka=[H3O+][Al(H2O)5(OH)2+][Al(H2O)63+]=(x)(x)0.10−x=1.4×10−5Solving this equation gives:x=1.2×10−3MFrom this we find:[H3O+]=0+x=1.2×10−3MpH=-log[H3O+]=2.92(an acidic solution)
Check the work. The arithmetic checks; when 1.2 × 10−3M is substituted for x, the result = Ka.
Check Your Learning
What is [Al(H2O)5(OH)2+] in a 0.15-M solution of Al(NO3)3 that contains enough of the strong acid HNO3 to bring [H3O+] to 0.10 M?
Show Answer
2.1 × 10−5M
The constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.
Key Concepts and Summary
The characteristic properties of aqueous solutions of Brønsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Brønsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic.
Solutions that contain salts or hydrated metal ions have a pH that is determined by the extent of the hydrolysis of the ions in the solution. The pH of the solutions may be calculated using familiar equilibrium techniques, or it may be qualitatively determined to be acidic, basic, or neutral depending on the relative Ka and Kb of the ions involved.
Exercises
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
Al(NO3)3
RbI
KHCO2
CH3NH3Br
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
FeCl3
K2CO3
NH4Br
KClO4
Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 × 10−6. Is a solution of novocaine acidic or basic? What are [H3O+], [OH−], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.
Show Selected Answers
2. The solutions can be categorized as follows:
FeCl3 dissociates into Fe3+ ions (acidic metal cation) and Cl− ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.
K2CO3 dissociates into K+ ions (neutral metal cation) and CO32− ions (the conjugate base of a weak acid and therefore basic). The aqueous solution is therefore basic.
NH4Br dissociates into NH4+ ions (a weak acid) and Br− ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.
KClO4 dissociates into K+ ions (neutral metal cation) and ClO4− ions (the conjugate base of a strong acid and therefore neutral). The aqueous solution is therefore neutral.
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https://www.youtube.com/watch?v=BV2jd4r070U
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Finding Percentage Change | Applied Mathematics | Odia Grade 7 | Math | Khan Academy
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Posted: 30 Jun 2025
In this video, we break down the concept of percentage change into clear, easy steps using everyday examples. We start by understanding what change means in a quantity—whether it's an increase or a decrease—and how to measure that shift. Then we quickly recall the idea of percentage as a way to express parts of a whole out of 100. With that foundation, we dive into the formula to calculate percentage change using the difference between the final and original values. To bring it all together, we walk through a practical example—calculating the percentage discount on a sale item—so you can see exactly how the math works in real life.
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now!
Timestamps:
00:00 What is Change?
00:45 Recalling Percentage
01:45 Percentage Change
03:39 Solved Example on Percentage Change
04:25 Discount Percentage
Khan Academy India is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We have videos and exercises that have been translated into multiple Indian languages, and 15 million people around the globe learn on Khan Academy every month.
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Created by Mahak Wasan
Transcript:
What is Change? This is Nirj. He checked his weight in November and it was 70 kg. By the time summers arrived, he checked his weight again in June and this time it was 63 kes. Now, do we see a change in the weight? Yes, the weight decreased by 7 kg. Right? His friend Sheila, she bought some pencils for 20 rupees. A day later she sold them for 25 rupees. Again, do we see a change in the price? Yes, the price increased by 5 rupees. So whenever a certain value increases or decreases, we say that it has changed. Now let us recall something that we have Recalling Percentage learned earlier which is percentage. Percentage means out of 100. Let's say if you have a chocolate cut into 100 pieces and then you eat 25 out of those. So the percentage of chocolate that you've eaten is 25%. You have eaten 25% of the chocolate. Right? Let's take one more example. Let us say that you uh gave a test and you scored 18 out of 20 in the test. Now what will be your marks percentage? It will be 18 divided by 20 into 100. We calculate percentage by multiplying it by 100. Right? So let's do it. 20 into 5 is 100. 18 into 5 is 90. So you scored 90% in your test. Now we know what is change. And we know what is percentage. How about we combine these two. So how will we find the percentage change? What is percentage change? Percentage change is uh like when something has changed and you want Percentage Change to know by how much percentage it has changed. Okay. So let's find this. Let us use the same example of the weight change that we just saw earlier. The weight in November is 70 kg. The weight in June is 63 kg. And the change in weight is 70 - 63 that is 7 kg. Now think about this. For initial weight of 70 kg, the change in weight is 7. The weight initially was 70 and change was of 7 kg. Now let's use the unit three method here. For initial weight of 1 kg, the change in weight would be 7 by 70. Right? Since we need to find percentage change, that means out of 100. So for initial weight of 100 kgs, the change in weight would be 7 by 70 into 100. 7 into 10 is 70. So that is 100 by 10 which is 10 kg. So for a weight of 100 kgs the change is 10 kg. That means the weight decreases by 10 kgs per 100 kg that is 10 by 100 or 10%. Now we do not need to do this unitary method every time. We have a quick formula to find out the percentage change which of course follows the same logic. So percentage change is actually equal to the change in value upon original value into 100. And that's what we did right. We use change in value which was 7 upon original value which was 70 into 100. That's what we did right and this is equal to 10%. So the formula for percentage change is percent change is equal to change in value upon original value into 100. Let us take one more example. We have 1 kg of apples which costed 40 rupees. But Solved Example on Percentage Change over time their price changed from 40 to 42 rupees. Now what is the change here? Changes of 2 rupees. I need to find this percentage of this price change. So let's recall the formula. Percentage changes change in value upon additional value into 100. The change here is change here is of 2 rupees original value is 40 into 100. So that becomes 200 divided by 42 which is 5%. So the price has increased by 5%. Let us look into one last example. A girl runs a boutique and she would usually sell this bag for 500 rupees but now she's giving Discount Percentage a discount. We often see discounts in shopping malls and shops, right? So after the discount, she's selling this bag for 400 rupees. Again, there is a change in the value of the price. The price has decreased by 100 rupees. Now I need to find the discount percentage. You know the formula for percentage change, right? It is the change which is 100 upon original value which is 500 into 100. So these two zeros would cancel out with these two it becomes 100 by 5 which is 20%. So the so we have received a discount of 20%. Now using this can we generate a formula for discount percentage? Let's try. So discount percentage is equal to this change which is nothing but the discount discount upon original price into 100. And that's the formula for discount percentage.
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https://math.stackexchange.com/questions/394387/solving-the-domain-and-range-of-a-region-satisfying-two-inequalities
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linear algebra - Solving the domain and range of a region satisfying two inequalities? - Mathematics Stack Exchange
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Solving the domain and range of a region satisfying two inequalities?
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The question I was provided was:
"Find the domain and range of the region satisfied by the following inequalities: i) y≥(x−1)2 y≥(x−1)2 ii)y≤2 x+1 y≤2 x+1
Any help would be greatly appreciated. Would you recommend graphing or solving algebraically?
linear-algebra
functions
inequality
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edited May 17, 2013 at 11:03
missiledragonmissiledragon
asked May 17, 2013 at 10:54
missiledragonmissiledragon
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hint: solved the equations,you will have some points and then check the max and min.chenbai –chenbai 2013-05-17 10:59:36 +00:00 Commented May 17, 2013 at 10:59
I seriously have no clue what to do. I tried expanding at all. Could you please provide a full explanation in the answers?r missiledragon –missiledragon 2013-05-17 11:06:19 +00:00 Commented May 17, 2013 at 11:06
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I am giving you a very basic way to find out the regions graphically. For
y≥(x−1)2 y≥(x−1)2
Note that y=(x−1)2 y=(x−1)2 is a polynomial which has R R as domain and obviously R≥0 R≥0 as its range. To find out what that inequality tells you, you can pick two points in and out of the region the parabola made in R 2 R 2. As you see P P is in and Q Q is out with respect the parabola. Now satisfy the coordinates of P P into the inequality. You see
(1/2−1)2=1/4(1/2−1)2=1/4
and it is smaller than 2 2. The same for Q Q tells us
(3−1)2=4(3−1)2=4
is greater than 1 1. So, the desired region satisfying
y≥(x−1)2 y≥(x−1)2
is the parabola an the region which is enclosed by it.
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answered May 17, 2013 at 11:14
MikasaMikasa
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Thanks, I got that :D, but how would you solve the domain and range?missiledragon –missiledragon 2013-05-17 11:20:55 +00:00 Commented May 17, 2013 at 11:20
@missiledragon: The domain doesn't change but the range will be restricted. I cannot say that it is all R≥0 R≥0.Mikasa –Mikasa 2013-05-17 11:25:12 +00:00 Commented May 17, 2013 at 11:25
I'm lost. I've only tried analysing the graph but I got that both the domain and range are infinite. Can you please show me how to do it?missiledragon –missiledragon 2013-05-17 11:35:50 +00:00 Commented May 17, 2013 at 11:35
@missiledragon: Both are infinite. I don't think we can show the range in any understandable form but {y∣y≥(x−1)2}{y∣y≥(x−1)2}.Mikasa –Mikasa 2013-05-17 11:40:55 +00:00 Commented May 17, 2013 at 11:40
1 @BabakS.:+➊Ⓝⓘⓒⓔ+➊Ⓝⓘⓒⓔ ⓦⓞⓡⓚSoftware –Software 2013-05-17 11:52:06 +00:00 Commented May 17, 2013 at 11:52
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I explain more detail:
first ,let y=2 x+1,y=(x−1)2 y=2 x+1,y=(x−1)2, you will have two points(0,1)(0,1) and (4,9)(4,9)
y≤2 x+1 y≤2 x+1 means all points below the red line, for example, (0,0),(1,0),(1,1)(0,0),(1,0),(1,1)
y≥(x−1)2 y≥(x−1)2 means all points within the Parabola，for example (1,1)(2,2)(1,1)(2,2)
so two curves will have a closed area. the domain and range are: [0,4][0,4] and [0,9][0,9]
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answered May 17, 2013 at 12:38
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https://saylordotorg.github.io/text_international-economics-theory-and-policy/s22-08-the-keynesian-cross-diagram.html
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19.8 The Keynesian Cross Diagram
Learning Objective
Learn how to use the Keynesian cross diagram to describe equilibrium in the G&S market.
The Keynesian cross diagramDepicts the equilibrium level of national income in the G&S market model. depicts the equilibrium level of national income in the G&S market model. We begin with a plot of the aggregate demand function with respect to real GNP (Y) in Figure 19.1 "Aggregate Demand Function". Real GNP (Y) is plotted along the horizontal axis, and aggregate demand is measured along the vertical axis. The aggregate demand function is shown as the upward sloping line labeled AD(Y, …). The (…) is meant to indicate that AD is a function of many other variables not listed. There are several important assumptions about the form of the AD function that are needed to assure an equilibrium. We discuss each assumption in turn.
Figure 19.1 Aggregate Demand Function
First, the AD function is positively sloped with respect to changes in Y, ceteris paribus. Recall that ceteris paribus means that all other variables affecting aggregate demand are assumed to remain constant as GNP changes. The positive slope arises from the rationale given previously that an increase in disposable income should naturally lead to an increase in consumption demand and a smaller decrease in CA demand, resulting in a net increase in aggregate demand. Next, if GNP rises, ceteris paribus, it means that taxes and transfer payments remain fixed and disposable income must increase. Thus an increase in GNP leads to an increase in AD.
Second, the AD function has a positive vertical intercept term. In other words, the AD function crosses the vertical axis at a level greater than zero. For reasons that are not too important, this feature is critical for generating the equilibrium later. The reason it arises is because autonomous consumption, investment, and government demand are all assumed to be independent of income and positive in value. These assumptions guarantee a positive vertical intercept.
Third, the AD function has a slope that is less than one. This assumption means that for every $1 increase in GNP (Y), there is a less than $1 increase in aggregate demand. This arises because the marginal propensity to consume domestic GNP is less than one for two reasons. First, some of the additional income will be spent on imported goods, and second, some of the additional income will be saved. Thus the AD function will have a slope less than one.
Also plotted in the diagram is a line labeled AD = Y. This line is also sometimes called the forty-five-degree line since it sits at a forty-five-degree angle to the horizontal axis. This line represents all the points on the diagram where AD equals GNP. Since GNP can be thought of as aggregate supply, the forty-five-degree line contains all the points where AD equals aggregate supply.
Because of the assumptions about the shape and position of the AD function, AD will cross the forty-five-degree line, only once, from above. The intersection determines the equilibrium value of GNP, labeled Y′ in the diagram.
Key Takeaways
The Keynesian cross diagram plots the aggregate demand function versus GNP together with a forty-five-degree line representing the set of points where AD = GNP. The intersection of these two lines represents equilibrium GNP in the economy.
An equilibrium exists if the AD function crosses the forty-five-degree line from above. This occurs if three conditions hold:
The AD function has a positive slope. (It does.)
The AD function has a slope less than one. (It does.)
The AD function intersects the vertical axis in the positive range. (It does.)
Exercise
Jeopardy Questions. As in the popular television game show, you are given an answer to a question and you must respond with the question. For example, if the answer is “a tax on imports,” then the correct question is “What is a tariff?”
Of positive, negative, or zero, the slope of an aggregate demand function with respect to changes in real GNP.
Of positive, negative, or zero, the value of the vertical intercept of an aggregate demand function.
Of greater than one, less than one, or equal to one, the value of the slope of an aggregate demand function with respect to changes in real GNP.
The equality that is satisfied on the forty-five-degree line in a Keynesian cross diagram.
The value of this variable is determined at the intersection of the aggregate demand function and the forty-five-degree line in a Keynesian cross diagram.
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Depicts the equilibrium level of national income in the G&S market model.
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What does it mean when a function has its derivative undefined at a point, and how would you find/show this if it were true (in calculus)?
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Matthew Bond
Ph.D., harmonic analysis, Michigan State, 2011 · Upvoted by
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Pepito Moropo
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What is happening to a function when the derivative is not defined?
It’s a little hard to classify something by the absence of a structure, but differentiability is a very common property of our most popular and easy-to-use functions. “Nice” functions “look straight when you zoom in”, and “bad” functions don’t. And there’s more than one way to be “bad”.
I’ll start with some common examples that are closer to basic functions before giving a rough idea of what more advanced cases look like, as these are not usually mentioned in a Calc 1 class.
Any place where a function is not continuous is the simplest case. This can be at a jump, or a “compressed spring” shape s
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It’s a little hard to classify something by the absence of a structure, but differentiability is a very common property of our most popular and easy-to-use functions. “Nice” functions “look straight when you zoom in”, and “bad” functions don’t. And there’s more than one way to be “bad”.
I’ll start with some common examples that are closer to basic functions before giving a rough idea of what more advanced cases look like, as these are not usually mentioned in a Calc 1 class.
Any place where a function is not continuous is the simplest case. This can be at a jump, or a “compressed spring” shape such as sin(1/x)sin(1/x), or the function can be all over the place and nowhere continuous, like the function f(rational)=1, f(irrational)=0.
(Above: f(x)=sin(1/x),f(0)=0)f(x)=sin(1/x),f(0)=0). Limit at 0 DNE.)
Continuous-but-not-differentiable-at-a-point is somewhat more interesting, because then we actually get to say something about the derivative. The simplest case is a bounce, such as |x||x| at the origin - the slope wants to be two different things. Other common cases are cusps (a bounce that approaches pure vertical, such as √|x||x| or x 2/3 x 2/3 at the origin), or vertical tangents (such as x 1/3 x 1/3).
There are actually many more cases, and some functions are known to be continuous everywhere, but nowhere-differentiable. They were not discovered until the late 1800s, and can be hard to draw or imagine. These are functions that are “jittery” but still continuous, like a stock market graph. Brownian motion(let’s do the one-variable case, a particle moving up and down over time) is a famous example that moves randomly but continuously, preferring neither up or down. It changes direction too much to stay close to any tangent line, no matter how small you slice it. It always has secant lines sloping both upward and downward.
The first example of a function proven to be everywhere continuous but nowhere differentiable is credited to Weierstrass: Weierstrass function. This construction is not random, so you can get a specific example and plug numbers into the function. The idea is to add together a lot of functions that get smaller, but rapidly more “jittery”, so that you throw the derivative into chaos without the function itself blowing up.
For example, something like:
f(x)=cos(x)+1 2 cos(10 x)+1 4 cos(100 x)+1 8 cos(1000 x)+...+1 2 n cos(10 n x)+...f(x)=cos(x)+1 2 cos(10 x)+1 4 cos(100 x)+1 8 cos(1000 x)+...+1 2 n cos(10 n x)+...
(The following graph is a Weierstrass function, but not necessarily same one given above)
It used to be thought that continuous functions were differentiable-with-maybe-a-few-exceptions, but now we know better.
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What is narcissistic abuse?
Narcissistic abuse is a funnel which leads to a state of mental and emotional claustrophobia.
At first, it appears to be a wide-open space filled with opportunity. But as you pour more and more energy into it, you feel the space around you decrease and the psychological walls converging.
First impressions last, so you find it difficult to forget the feelings of hope, warmth and possibility which dominated the beginning of the relationship. Because narcissistic abuse is enforced during everyday situations with a ‘loved one’, it is not consciously felt. Instead, you are shamed, ridiculed, disregar
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Narcissistic abuse is a funnel which leads to a state of mental and emotional claustrophobia.
At first, it appears to be a wide-open space filled with opportunity. But as you pour more and more energy into it, you feel the space around you decrease and the psychological walls converging.
First impressions last, so you find it difficult to forget the feelings of hope, warmth and possibility which dominated the beginning of the relationship. Because narcissistic abuse is enforced during everyday situations with a ‘loved one’, it is not consciously felt. Instead, you are shamed, ridiculed, disregarded, manipulated and frightened in ways that seem justified due to the narcissist’s shamelessness. Nobody would purposely do such things, would they? you ask.
Someone with healthy shame senses when they have wounded another person’s emotions, and therefore adapts their behaviour to restore pride to the other. A shamed person exhibits signs of it, including in their facial expression, their posture, and their subsequent behaviour. Anyone with a healthy sense of shame knows when someone is experiencing shame, because they themselves have experienced it. It is as unpleasant to see as it is to feel it. A narcissist instead views shame as a magic potion which can cripple their target’s self-esteem and give them the upper hand.
The narcissist strings together a never-ending series of attacks on their target which cause the target to feel more and more shame, fear and guilt. As a result, the target experiences a psychological pressure which grows over time and results in a psychological cage. The target’s self-esteem shrinks, their imagination loses its potency, and they can no longer see beyond their constant suffering. This is the funnelling process which occurs over the course of a relationship with a narcissist. Before you know it, you are attached to the narcissist and trapped in your role, overwhelmed by the mental and emotional heaviness which was spoon fed to you within the context of what you believed to be a fair relationship.
In short, narcissistic abuse is the ritualistic herding of a target from a place of total freedom to a dystopian cage of suffering with little room to move or breathe.
If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist.
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What does it mean when a derivative is undefined?
What does it mean to integrate a function in calculus?
What is the result of taking the derivative of a non-differentiable (but still integrable) function?
What does it mean to find the derivative of a function of its original function?
What is an undefined derivative in calculus?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
Actually there are very few graphs which are not differentiable anywhere.
The question usually refers to the case where a graph is not differentiable at a particular point but it is differentiable everywhere else.
I would like to explain this concept in very simple terms so that it is clearly understood. Too many explanations involve very precise higher mathematical terms with unintelligible definit
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Actually there are very few graphs which are not differentiable anywhere.
The question usually refers to the case where a graph is not differentiable at a particular point but it is differentiable everywhere else.
I would like to explain this concept in very simple terms so that it is clearly understood. Too many explanations involve very precise higher mathematical terms with unintelligible definitions involving limits.
“CONTINUOUS” at a point simply means “JOINED” at that point.
“DIFFERENTIABLE” at a point simply means “SMOOTHLY JOINED” at that point.
(i.e. at the point, the gradient on the left hand side has to equal the gradient on the right hand side.)
(ie left hand...
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Peter Shea
B. Sc in Mathematics&Computer Science, Monash University (Graduated 1972) · Author has 5.2K answers and 1.2M answer views
·2y
Originally Answered: What does it mean when a derivative is undefined? ·
For a variety of possible reasons, either the function is undefined, or there is no clear limit.
Examples:
f(x) = 1/(x-5)^2 is undefined at x = 5, and (f(x+h)-f(x)) / h does not converge near x = 5.
f(x) = |x| has positive values everywhere. But its slope is -1 for -ve x and +1 for +ve x. There is no obvious slope definition for x = 0.
Also, consider the Heaviside step function:
U(t) = 0 for x < 0,
= 1 for x ≥ 1.
What is its slope at x = 1?
Or the Dirac delta function:
δ(x) = +∞ for x=0,
=0 for x≠0
and, on the interval [-∞,+∞], ∫ δ(x) dx = 1.
What is its slope at x=0?
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·10mo
Originally Answered: Can you explain what it means for a derivative to be undefined at certain points? ·
See my answer here…
What does it mean when a function has its derivative undefined at a point, and how would you find/show this if it were true (in calculus)?
Just scroll down to my name.
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Is the derivative of a function at a point always unique?
What does it mean for a function to be undefined at some point and its derivative to be zero at that point?
How do you know, graphically, if the second derivative is undefined from the original function?
What does taking the derivative of a function actually do?
What does it mean if the derivative of a function is undefined at a particular value?
Ilkka Vuorio
PhD in Physics, Massachusetts Institute of Technology (Graduated 1987) · Author has 17.7K answers and 2.2M answer views
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Originally Answered: What does it mean if the derivative of a function is not defined at a particular value? ·
Let’s call the “particular value” by x0. The function could be discontinuous at x0. Or the graph (“curve”) could have a “sharp turn” at x0.
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When does a function not have a derivative and why?
The derivative of a function at a point is the slope of the function at that point. Hence, if at a certain point a function does not have a clearly defined slope, then it does not have a derivative at that point. For example, f(x)=|x|f(x)=|x| at x=0 x=0 is not differentiable. From the left, the slope is −1−1, from the right the slope is 1 1. Hence at x=0 x=0, f(x)=|x|f(x)=|x| does not have a derivative.
If a function is not defined at a point, then it is not differentiable. For example f(x)=1 x f(x)=1 x is not differentiable at x=0 x=0 since f(0)f(0) is not defined.
If a function is not continuous at a point, then also
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The derivative of a function at a point is the slope of the function at that point. Hence, if at a certain point a function does not have a clearly defined slope, then it does not have a derivative at that point. For example, f(x)=|x|f(x)=|x| at x=0 x=0 is not differentiable. From the left, the slope is −1−1, from the right the slope is 1 1. Hence at x=0 x=0, f(x)=|x|f(x)=|x| does not have a derivative.
If a function is not defined at a point, then it is not differentiable. For example f(x)=1 x f(x)=1 x is not differentiable at x=0 x=0 since f(0)f(0) is not defined.
If a function is not continuous at a point, then also it is not differentiable there. f(x)=s g n(x)f(x)=s g n(x) is not differentiable at x=0 x=0 because its left and right limits are different.
But the most fun functions are those that look like fractals, such as the Weierstrass function or the Takagi curve. These are continous everywhere, but differentiable nowhere since every point has protusions and kinks.
Weierstrass Function
Takagi Curve
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When does a function not have a derivative and why?
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Eric Dietrich
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What is happening to a function when the derivative is not defined?
Thanks for the A2A!
Three things could be happening:
The function is discontinuous at the point, undefined at the point, or it could be a sharp edge. The last case is like x=0 x=0 on |x||x|, or the cusp of a cardioid.
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What is happening to a function when the derivative is not defined?
Its graph will then be a never-ending straight vertical line. The derivative of a graph is similar to the tangent of an angle a line makes with the x-axis. But when a derivative is undefined, the line will be perpendicular to the x-axis, where the tangent of ninety degrees is also undefined.
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Mark Regan
Calculus I & II in Calculus, The Ohio State University · Author has 1.5K answers and 2.9M answer views
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What is happening to a function when the derivative is not defined?
The function is typically taking on a vertical slope, or it doesn’t exist for a particular range of inputs because the function doesn’t exist on that range, or the limit of f(x) as h approaches x goes to infinity. If the value of the function is not defined then the derivative is not defined.
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Terry Moore
Calculus fan since 1958. · Author has 16.6K answers and 29.4M answer views
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Can the derivative of a function be equal to zero at points where it doesn't exist?
Can the derivative of a function be equal to zero at points where it doesn't exist?
As Luboš Motl pointed out, if the derivative is zero then the derivative does exist.
However, I think the question might mean, Can the derivative of a function be equal to zero at points where the function doesn't exist?
Consider the function f(x)=(x−1)3 x−1 f(x)=(x−1)3 x−1. When x=1 x=1 this function reduces to 0 0 0 0 and therefore doesn’t exist. However, except when x=1 x=1, f(x)=(x−a)2 f(x)=(x−a)2 whose derivative is 2(x−1)2(x−1). But, although that is 0 0 when x=1 x=1 that is not the derivative of f(x)f(x) at x=1 x=1 because neither the function nor the
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Can the derivative of a function be equal to zero at points where it doesn't exist?
As Luboš Motl pointed out, if the derivative is zero then the derivative does exist.
However, I think the question might mean, Can the derivative of a function be equal to zero at points where the function doesn't exist?
Consider the function f(x)=(x−1)3 x−1 f(x)=(x−1)3 x−1. When x=1 x=1 this function reduces to 0 0 0 0 and therefore doesn’t exist. However, except when x=1 x=1, f(x)=(x−a)2 f(x)=(x−a)2 whose derivative is 2(x−1)2(x−1). But, although that is 0 0 when x=1 x=1 that is not the derivative of f(x)f(x) at x=1 x=1 because neither the function nor the derivative exist there.
So the answer is “No, the derivative cannot exist where the function doesn’t exist.” However, sometimes it is possible to extend the function so that the derivative does exist. In this case just declare that g(x)=f(x)g(x)=f(x) when x≠1 x≠1 and g(1)=0 g(1)=0. This new function is differentiable everywhere.
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B. S. Thomson
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If a derivative does not exist at a point, does that mean the function does not exist at that point?
Well Uta, you just need to read definitions exactly and get used to the language.
If a function f f of the kind you are studying in the calculus “does not exist at that point” then that point is not in the domain of the function. That means loosely that point is of no concern to the function and the point is irrelevent to many definitions of the calculus. For example
To say that a function is continuous at a point is only meaningful if that point is in the domain of the function.
To say that a function is differentiable at a point is only meaningful if that point is in the domain of the function.
H
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Well Uta, you just need to read definitions exactly and get used to the language.
If a function f f of the kind you are studying in the calculus “does not exist at that point” then that point is not in the domain of the function. That means loosely that point is of no concern to the function and the point is irrelevent to many definitions of the calculus. For example
To say that a function is continuous at a point is only meaningful if that point is in the domain of the function.
To say that a function is differentiable at a point is only meaningful if that point is in the domain of the function.
However … in discussions of limits lim x→x 0 f(x)lim x→x 0 f(x) the point x 0 x 0 does not have to be in the domain of the function. This is the reverse of the first two: for limits the value of the function at a point is completely irrelevant.
So the answer to your question? You have it all backwards as stated. If the function “does not exist at that point” then continuity and differentiability are not available. If the point is in the domain of the function maybe the derivative exists and maybe not. You will have to check. But, certainly if the derivative does not exist at a point x 0 x 0 then you may conclude …only that the derivative does not exist there.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
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Can a function have no derivative at all points? If so, what does it look like?
A function does not have to be a curve. It does not need to have an actual equation either!
I will just make up a function to show you.
(However, if I were to add another point like (3, 2) it would not be a function anymore because a fu...
Continue Reading
A function does not have to be a curve. It does not need to have an actual equation either!
I will just make up a function to show you.
(However, if I were to add another point like (3, 2) it would not be a function anymore because a fu...
Upvote ·
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Is the speed of sound 330m/s or 343m/s? In what condition is it any of the two?
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The speed of sound in air is commonly cited as approximately 343 m/s at 20°C (68°F) and at standard atmospheric pressure. However, the speed of sound can vary based on several factors, including temperature, humidity, and altitude.
Speed of Sound in Air:
At 0°C (32°F): Approximately 331 m/s.
At 20°C (68°F): Approximately 343 m/s.
At 0°C (32°F) with 100% humidity: Approximately 334 m/s.
Conditions Affecting Speed:
Temperature: The speed of sound increases with the temperature of the air. A rough rule of thumb is that the speed of sound increases by about 0.6 m/s for every degree Celsius increase in t
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The speed of sound in air is commonly cited as approximately 343 m/s at 20°C (68°F) and at standard atmospheric pressure. However, the speed of sound can vary based on several factors, including temperature, humidity, and altitude.
Speed of Sound in Air:
At 0°C (32°F): Approximately 331 m/s.
At 20°C (68°F): Approximately 343 m/s.
At 0°C (32°F) with 100% humidity: Approximately 334 m/s.
Conditions Affecting Speed:
Temperature: The speed of sound increases with the temperature of the air. A rough rule of thumb is that the speed of sound increases by about 0.6 m/s for every degree Celsius increase in temperature.
Humidity: The speed of sound is higher in humid air than in dry air because water vapor is less dense than the nitrogen and oxygen it replaces.
Altitude: At higher altitudes, the air is less dense and colder, which can reduce the speed of sound.
In summary, 343 m/s is a standard value for the speed of sound at room temperature (20°C), while 330 m/s might refer to lower temperatures or specific conditions.
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Is 340 m/s the speed of sound?
The speed of sound in air is 332m/s at 0°c. What is its speed at 20°c?
The speed of sound in the air is 343 MS to the power of negative one.What is the unit in km per day?
What is the speed of sound?
Why is ultrasound (1540m/s) faster than sound (343m/S)?
Alan MacRobbie
Former Analog Design Engineer · Upvoted by
Bert Willke
, Ph.D. Physics, Massachusetts Institute of Technology (1963)
·4y
Hi Abiona — the speed of sound in air is dependent upon the air temperature, which affects the density and the characteristic impedance. At 0 C, it is 331.6 m/s and at 20 C it is 343 m/s. It is useful to note that the density of air is somewhat greater at 0 C than it is at 20 C (1.293 kg/m^3 vs. 1.21 kg/m^3, respectively). The characteristic impedance of air is higher at 0 C than at 20 C - 428 Pa . s/m vs. 415 Pa . s/m, respectively. In these examples the air pressure is the same and is 1.013 x 10^5 Pa. or 1 ATM, or approximately 760 mm Hg. Caution - this relationship does not hold true for al
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Hi Abiona — the speed of sound in air is dependent upon the air temperature, which affects the density and the characteristic impedance. At 0 C, it is 331.6 m/s and at 20 C it is 343 m/s. It is useful to note that the density of air is somewhat greater at 0 C than it is at 20 C (1.293 kg/m^3 vs. 1.21 kg/m^3, respectively). The characteristic impedance of air is higher at 0 C than at 20 C - 428 Pa . s/m vs. 415 Pa . s/m, respectively. In these examples the air pressure is the same and is 1.013 x 10^5 Pa. or 1 ATM, or approximately 760 mm Hg. Caution - this relationship does not hold true for all gases, fluids, or solids. You should consult the tables that are available to compare the speed of sound in different materials.
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Sam Greene
BSEE in Electrical and Electronics Engineering, Michigan Technological University (Graduated 1992) · Author has 65 answers and 12.2K answer views
·3y
Originally Answered: Is the speed of sound 330 m/s or 332 m/s? Can anyone explain? ·
The speed of sound is 334 m/s at 20C at sea level. The speed of sound through any media is dependent on the coefficient of stiffness (Ks) and the density of the media (p) where c is the speed of sound below.
This image has been removed for violating Quora's policy.
Sound will travel faster through steel than through air. If the density of air changes, the speed of sound in air changes. It is not absolute.
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The speed of sound is 334 m/s at 20C at sea level. The speed of sound through any media is dependent on the coefficient of stiffness (Ks) and the density of the media (p) where c is the speed of sound below.
This image has been removed for violating Quora's policy.
Sound will travel faster through steel than through air. If the density of air changes, the speed of sound in air changes. It is not absolute.
Upvote ·
Jim.Moore
Physics, Math, Systems Engineer, Educator · Author has 22.3K answers and 17.5M answer views
·3y
Originally Answered: Is the speed of sound 330 m/s or 332 m/s? Can anyone explain? ·
It’s “about” both of those numbers. If you are going to be more precise, you need a chart!
You need to specify temperature, pressure, and even humidity!
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It’s “about” both of those numbers. If you are going to be more precise, you need a chart!
You need to specify temperature, pressure, and even humidity!
Upvote ·
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Is the speed of sound 330 m/s or 332 m/s? Can anyone explain?
What is the temperature of air when the speed of sound is 343 m/s?
What is the speed of sound at 35,000 feet?
How does the speed of air molecules (500 m/s) affect the sound barrier (343 m/s)?
How is the speed of sound in the air 340m/s?
Evert De Ruiter
PhD in Urban Physiology&Acoustic and Noise Control Engineering, Delft University of Technology (Graduated 2005) · Author has 5.2K answers and 2.1M answer views
·4y
It is not one or the other, see other answers. But…what do you care? It's variation is limited, as is its importance.
I don't consider waves in fluids, solids and other gases than air, for two reasons. 1. Hearing takes place while in the Earth's atmosphere; exceptions are irrelevant. 2. In solids several types of waves occur, with different speeds, sometimes dependent on frequency.
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Charlie Chicklis
M.S. in Physics&Mathematics, Northeastern U. · Author has 629 answers and 499.5K answer views
·4y
I believe 330 is at 0 degrees and 343 is at room temperature.
T=330+.6(delta T)
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Andrew McGregor
Studied for a PhD in Physics · Author has 14.4K answers and 98.8M answer views
·3y
Originally Answered: Is the speed of sound 330 m/s or 332 m/s? Can anyone explain? ·
It’s not constant, the speed of sound depends a bit on air pressure, temperature, and humidity (although at ‘normal’ air pressures, it hardly depends on pressure).
So, any particular number you see is an approximation unless it’s stated to be for very specific conditions.
You can find data for whatever actual conditions you want to use.
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David J Ford
Former Manager 'In Computers' Retired Years Ago (1959–2003) · Author has 2.1K answers and 700.9K answer views
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Here’s a list of the speed of sound in various materials and under various conditions - hope you find it useful.
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Gordon Taylor
Property Developer. · Author has 1.4K answers and 663.5K answer views
·3y
Originally Answered: Is the speed of sound 330 m/s or 332 m/s? Can anyone explain? ·
The speed of sound isn't constant. It depends on the density of the medium it is travelling through. So, as the pressure and humidity of the atmosphere change slightly, the speed of sound also changes slightly. In addition, the atmosphere becomes less dense with increasing altitude which will also affect the speed of sound.
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Hannibal
40+years of study in Physics, Aviation, Meteorology, Chemistry · Author has 10.1K answers and 1.1M answer views
·3y
Originally Answered: Is the speed of sound 330 m/s or 332 m/s? Can anyone explain? ·
The speed of sound is 331.5 meters/second at 0°C and goes up .6m per second with every 1°C rise in temperature. The warmer the air, the faster the speed of sound. At 80,000 feet mach 3 is 2100mph. Much slower than at sea level on a warm, windless day on the beach. 👍😎
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Julian Danzer
Knows German · Author has 27.8K answers and 4.5M answer views
·3y
Originally Answered: Is the speed of sound 330 m/s or 332 m/s? Can anyone explain? ·
it depends on the medium
in air it mostly depends on temperature
whcih can vary
hwich is why you will find “normal” refernece values for the speed of sound ranging from 300 to 341 m/s depending on what conditions they’re given for
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Brent Meeker
Studied Physical Sciences&Computer Science at The University of Texas at Austin (Graduated 1974) · Author has 22.4K answers and 17.4M answer views
·4y
The speed of sound in a gas is directly proportional to the average speed of the molecules, which is proportional to the square root of the temperature. So there’s only 5.6 degK difference to produce those two sound speeds. It’s 330m/s at 271degK.
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John
Physics Teacher · Author has 347 answers and 274.7K answer views
·Mar 6
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Is 340 m/s the speed of sound?
That’s the speed of sound in air at 59°F (15°C).
The speed of sound in air increases by about 0.60 m/s for every 1°C temperature increase. So, at 104°F (40°C), the speed of sound is about 355 m/s.
The speed of sound in seawater is about 1500 m/s, 4x faster.
Sound is a wave that needs a medium to travel in (to exist). Sound can’t travel in a vacuum, like space. Sound is a series of (mechanical) energy waves moving through SOMETHING (SOMETHING is the medium that supports the waves, like air, water or the metal tines of a tuning fork, etc.).
So, to correctly ask the question, we need to know what the
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That’s the speed of sound in air at 59°F (15°C).
The speed of sound in air increases by about 0.60 m/s for every 1°C temperature increase. So, at 104°F (40°C), the speed of sound is about 355 m/s.
The speed of sound in seawater is about 1500 m/s, 4x faster.
Sound is a wave that needs a medium to travel in (to exist). Sound can’t travel in a vacuum, like space. Sound is a series of (mechanical) energy waves moving through SOMETHING (SOMETHING is the medium that supports the waves, like air, water or the metal tines of a tuning fork, etc.).
So, to correctly ask the question, we need to know what the medium is (air) and to get kinda picky, the temperature of the air makes a difference too.
Upvote ·
Related questions
Is 340 m/s the speed of sound?
The speed of sound in air is 332m/s at 0°c. What is its speed at 20°c?
The speed of sound in the air is 343 MS to the power of negative one.What is the unit in km per day?
What is the speed of sound?
Why is ultrasound (1540m/s) faster than sound (343m/S)?
Is the speed of sound 330 m/s or 332 m/s? Can anyone explain?
What is the temperature of air when the speed of sound is 343 m/s?
What is the speed of sound at 35,000 feet?
How does the speed of air molecules (500 m/s) affect the sound barrier (343 m/s)?
How is the speed of sound in the air 340m/s?
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Related questions
Is 340 m/s the speed of sound?
The speed of sound in air is 332m/s at 0°c. What is its speed at 20°c?
The speed of sound in the air is 343 MS to the power of negative one.What is the unit in km per day?
What is the speed of sound?
Why is ultrasound (1540m/s) faster than sound (343m/S)?
Is the speed of sound 330 m/s or 332 m/s? Can anyone explain?
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Ramsey Theory on subsets of ${1, 2, \dots, n}$, $k$ colored.
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Prove that there exists a $n$ natural for each $k > 0$ natural such that, by coloring all the subsets of ${1, 2, \dots, n}$ with $k$ colors, there exist two disjoint subsets among them, $X$ and $Y$, such that $X, Y, X \cup Y$ are identically colored.
I am aware of Schur's theorem, which has a relatively similar statement, but it is actually for numbers in a set with their sum, while this problem is for sets in a 'family' with their union. I'm thinking to create a bijective indicator, that transforms each of the elements of $S \in \mathcal{P(n)}$ into a natural number $\omega(S)$, indicator function that also have the property that for distinct sets: $$\omega(A) + \omega(B) = \omega(A \cup B)$$
Actually, If I proved this, I would have solved the problem since Schur's theorem may be used. However, I couldn't find a proof for this fact on a general case.
For example, a valid (satisfies the definition conditions, but does not form a correct Schur-type set) function for $n = 3$ would be: $\omega(\emptyset) = 0$, $\omega({1}) = 1$, $\omega({2}) = 3$, $\omega({3}) = 5$, $\omega({1, 2}) = 4$, $\omega({1, 3}) = 6$, $\omega({2, 3}) = 8$ and $\omega({1, 2, 3}) = 9$.
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This theorem does have a proof based on Schur's theorem, but we would also have to invoke a version of the hypergraph Ramsey's theorem.
Let $S(k)$ be the least number such that if ${1, 2, \dots, S(k)}$ are $k$-colored, there is a monochromatic sum $a+b=c$ (with $a,b$ not necessarily distinct).
Let $n$ be large enough that in any coloring of $\mathcal P(n)$, there is a subset $D$ of size $S(k)$ such that the color of every $i$-element subset of $D$ is the same (possibly depending on $i$). This gives us a coloring of ${1, 2, \dots, S(k)}$ by giving each value $i$ the color of an $i$-element subset of $D$. Now within that coloring, find a monochromatic sum $a+b=c$; we are done by choosing disjoint $A,B \subseteq D$ with $|A|=a$ and $|B|=b$.
This is massive overkill, but has the advantage that it generalizes to using Folkman's theorem (an analog of Schur's theorem which finds a larger set such that the sum of any nonempty subset is the same color) to prove the finite unions theorem (the same, but for a collection of disjoint sets, with $\cup$ instead of $+$).
It does not seem that there is an easy direct argument that uses Schur's theorem to find a monochromatic disjoint union. (There sure is one in reverse, though!)
However, there is a shorter proof using Ramsey's theorem for triangles, which is completely analogous to the proof of Schur's theorem using the same result.
Starting from a coloring of $\mathcal P(n)$, take a complete graph with vertices $v_0, v_1, v_2, \dots, v_n$, and give each edge $v_i v_j$ (assuming $i) the color of the set ${i+1, i+2, \dots, j}$. Then the three edges of a monochromatic triangle form a triple $A, B, A\cup B$ in which the two smaller sets are disjoint.
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answered Feb 11, 2022 at 20:20
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Rock Bass
SCIENTIFIC NAME: Ambloplites rupestris
CHARACTERISTICS:Rock bass, also called goggle-eye, can be easily distinguished from shadow bass by the well-defined rows of dark spots along the sides, particularly below the lateral line, which usually contains 37 to 46 scales. The dorsal fin has 10 to 13 spines and 10 to 12 rays. The anal fin has five to seven spines and nine to 11 rays. The mouth is large, and the upper jaw extends to below the middle of the eye. The back and sides of this species are olive to light brown with distinct horizontal rows of black spots. The vertical fins have a light yellowish background. The basal areas are mottled with brown or gray, and the distal portion has two to four dark bands. The anal fin on adult males has a distinct black margin. The iris is red to reddish orange. Venter color can vary from white to light brown or charcoal. See Rafinesque (1817b) for original description.
ADULT SIZE:8 to 10 in (203 to 254 mm).
STATE RECORD: a list of the State Record Freshwater Fish.
DISTRIBUTION:Rock bass are limited to the Tennessee River system in Alabama, but they could eventually enter the upper Tombigbee River system through the Tennessee-Tombigbee Waterway.
HABITAT AND BIOLOGY:Most of our collections have come from small to medium streams with rubble and gravel substrates and slow to moderate current. Individuals linger around submerged vegetation or tree roots in pools. Pflieger (1975) reports that spawning occurs from April through June in Missouri. Although we have not observed spawning in Alabama, it probably occurs in March and April, when the shadow bass spawn. Primary food items include crayfishes, large aquatic insect larvae, and small fishes (Probst et al., 1984). Redmon and Krumholz (1978) report that rock bass have a maximum life span of eight years.
REMARKS: Rock bass are an excellent small-stream game fish, particularly when caught with ultralight spinning gear or a fly rod.
ORIGINAL DESCRIPTION: The rock bass was described by Rafinesque in 1817.
ETYMOLOGY: Ambloplites means blunt armature. Rupestris means among the rocks.
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MIT OpenCourseWare 3.091SC Introduction to Solid State Chemistry, Fall 2010 Transcript – Session 4 The following content is provided under a Creative Commons License. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit mitopencourseware@ocw.mit.edu. PROFESSOR: OK, settle down. Let's get started. One announcement, yesterday we had our first weekly I would say minor celebration. And by and large, it went well. Please make sure that you go to your assigned recitation. If you miss your recitation, you need to get down to see Hillary so that we make sure that we have enough copies of the weekly quizzes on hand. If you've joined the class, you have to check in with her. She's down the hall here in room 8-201. And you'll be assigned to a section. What else do I have by way of announcements? Oh yes. Just reminding you, this was from 2003. See it doesn't change. It's the same s-block, p-block, and d-block elements. So that's coming up a week from tomorrow, two celebrations next. And of course the contests, the contests with hot prizes. All right. Let's get down to business. Last day we looked at the Rutherford-Geiger-Marsden experiment. Oh there's one other one. If you look at the readings, there's this one section called the archives. My predecessor, Professor Wit, wrote a set of lecture notes. And they look something like this. And some students have said that they find these a little more expository in certain sections on certain topics than the book to be. And I have no preference. But if you take a look at this it'll say LN1, lecture notes 1. If you go to this, read this. If you find that's helpful, good. If you don't find it helpful, than stick with the book. Just letting you know what that is. All right, so last day we looked at the Rutherford-Geiger-Marsden experiment in which a high energy beam of alpha particles bombarded a thin, gold foil. And on the basis of the scattering results, namely most of the particles went through with minor scattering. And a tiny fraction of them were scattered through large angles. The Thomson plum pudding model was rejected in favor of Rutherford's nuclear model of the atom. And then subsequently, Bohr came up with the quantitative representation off the Rutherford nuclear model. And we were partway through the treatment of Bohr last day when we adjourned. So let's right pick up the thread from where we left off. And so just to remind you, the Bohr model is for a 1-electron atom gas phase. So this is either atomic hydrogen, it could be helium plus lithium 2 plus roentgenium 110 plus Its doesn't matter how many protons. there's always only want electron. And it's a planetary model. The positive charge concentrated in the nucleus, Z is the proton number. And at a distance r from the nucleus is a circular orbit in which resides 1 electron. It has a charge of minus E. And I just designated them q1 and q2. You could have done it the other way. But I had to choose something. So we went through and looked at the constitutive equations here. So first of all, the energy of the system-- and this is only going to be the energy the electron. Because we assume that the nucleus is far more massive. And so we don't have to get into things like reduced mass or anything like that. So just measure the energy of the electron. 1/2 mv squared is the newtonian component. And then coulombic energy that's stored is z times e squared over 4 pi epsilon zero r, where epsilon zero is the permittivity of vacuum. And it's the factor, the 4 pi epsilon zero, is the factor that allows us to take electrostatic energies and put them on the same plane as mechanical energies. When we run through this, we always end up in joules. Then there's a force balanced to make sure that the electron neither falls into the nucleus nor flees and breaks free of the atom. And the force balance is if you put a ball at the end of a string, and you whip it around on a tether, you have a centrifugal force that's trying to get the ball to break away. And then the string is pulling in. So the pull in, in this case, is the coulombic force. And the force that makes the ball want to flee is this mv squared over r. And that must be net zero. Otherwise we're going to have a shift in orbit. And then lastly we have the quantum condition. And this was the breakthrough of Bohr where he enunciated that the quantum condition is going to give us this energy level quantization. And this was a big departure from what had been in the past. The only antecedent idea of this nature was the work by Planck who said that light is quantized. But as I told you last day, who knows what light really is. The Newtonian notion of a ball orbiting was very compelling here. And the notion that the movement of the electron could in some way be discontinuous was quite a major departure. So I left you last day with three equations and three unknowns. And what I'm not going to do right now is solve the equations. Because I've been lecturing long enough to know that's the way to kill interest, quench a lecture. And so if you really want to go through the algebra, be my guest. You're smart enough to do that. Instead I'm going to show you the results. But those are the three equations that you need. So we have an equation in r. We have an equation at v, and an equation in e. So let's go after them. If you first look at the solution for r. This is the radius of the orbit of the electron, the orbit of the electron. If you go through and solve, you'll end up with this, Epsilon zero times the square of the Planck constant divided by pi times m. It's always the electron. So this is the radius of the electron orbit. This is the mass of the electron times the square of the elementary charge-- that whole thing I'm going to group-- times the square of the quantum number divided by z, the proton number. So what we see here? Well, everything inside the parentheses is constant. These are all constant. Pi obviously is geometric constant. And the rest of these are constants you could look up in your table of constants. And we noticed that there is a set of solutions to this. The radius of the electron can occupy various discrete values defined by n. So we say that the radius takes on a plurality of values a function of n. And furthermore, the functionality goes as the square of n. It's n squared times a constant, where that constant is inside those parentheses. And we notice that because the r goes as n squared it's nonlinear. It's nonlinear. This is so important I'm going to write it down one more time. So the radius of the electron orbit takes multiple values. It takes multiple values. And they're discreet. The physicists like to use a different term. When something is discretized, the physicists say it is quantized. So these values are quantized. You cannot continuously vary the radius and nonlinear values, multiple values. So let's plug in. Because I want to get a sense of scale. So let's look at the simplest one. The most primitive 1-electron atom would be hydrogen. In which case, Z equals 1. So I've just got a proton orbited by an electron. So look at atomic hydrogen. So in that case, Z equals 1. And I'm going to look at n equals 1, which is the lowest number here, right? R scales as n squared. So the lowest value or r is obtained when n equals 1. And this is termed the ground state. The ground state. So I want to ask what is the radius of the electron orbit ground state in atomic hydrogen? And if I plug in these values, I'll call this r sub 1. It turns out to be 5.29 times 10 to the minus 11 meters, or 0.529 angstroms. I love the angstrom. It's a great unit. It's a great unit. It's not an SI unit. But I like the angstrom. I'll show you why. If you try to express this in SI units, well there's 10 to the minus 11 meters. The Si units go in units of clusters of 1,000. So for example, you've got the meter. You've got the kilometer. You've got the micrometer. You've got 10 to the minus 9 meters, which is the nanometer. This is a Goldilocks problem. This one's too big. And then the next one down here is 10 to the minus 12 meters, which is the picometer. So this is either 52.9 picometers, or a 0.0529 nanometers. And that's no good. I want numbers like 3, 7, simple to remember. So 0.529, this is about 1/2 angstrom. It's good to know. But you try to publish, you know what happens in a scientific literature today? The Literary Lions that control the journals, they'll circle that and say you have to convert to SI units. And so they have to right some goofy nanometer thing or something. I know you think I'm crazy, but I love the angstrom. So there. Anyway, so here it is. Once you know that this is 0.529, this is on your table of constants. It's right on your table of constants. So you don't have to go and calculate all this stuff. Which means if you do your homework with your table of constants, you will know where those numbers lie, as opposed to opening this thing up for the first time on the first celebration of learning on October the 7th, and with 47 entries and they're tiny, tiny font. And you're wondering where is that thing. Just a word to wise. So now we know what this is. We know this is 0.529 angstroms. So now I can write an equation for the radius of a 1-electron atom anywhere, anytime. r of n is going to be equal to a naught which is this value here. And it is termed the Bohr radius. So you can write it as a function of the Bohr radius, times the square of n divided by Z. So that's for all 1-electron atoms, gas phase. And you can see that as Z goes up, the r goes down, which makes sense. So suppose instead of hydrogen, we talk helium plus What's the only difference? Helium plus has 2 protons in the nucleus. Which means that the coulombic force of attraction between the same 1-electron and now a doubly charged nucleus is going to be stronger. So the first orbit is going to get pulled in. And all the other orbits are going to get pulled in. By how much are they going to get pulled in? By that much. So this is the functional representation of all of that physics. All right, there's three equations, three unknowns. Let's look at energy. So if you go through and solve for energy, you get this one. Minus this big monstrosity, mass of the electron to the fourth power of the elementary charge times 8 times the square of the permittivity of vacuum times the square of the Planck constant, all times the square of the Proton number divided by the square of the quantum number. And I just to make sure everybody is with me here. I always want to write n here. n equals 1, 2, 3, takes on integer values. And we'll say it again here. N equals 1, 2, 3, et cetera, et cetera. So again we say we see that e is a function of n. It's discretized. It's quantized. Why? Because once you impose the quantum condition here on angular momentum, it propagates through the entire model. So radius this quantized, energy is quantized, you're going to see velocity is quantized. Because the quantum condition is pervasive. So e to the n, and I'm going to take this whole quantity in parentheses and just call it giant K. These are all positive quantities. Mass is positive. And squares and fourth powers of numbers must be positive. So this is K times Z squared over n squared That's good. And we can go and evaluate K. And when we evaluate L in SI units, we get 2.18 times 10 to the minus 18 joules. This is joules per atom. Or you can multiply this by Avogadro's number. If you multiply it by Avogadro's number, then that will give you 1.312 megajoules per mol. So that's the energy of the electron in the ground state of atomic hydrogen. And then we can mediate that with Z and n, and go to electrons that are outside the ground state, above the ground state, or ground state electrons in atoms that have more than 1 proton, or both. And so let's take a look at the graphical representation of that. So instead of Cartesian coordinates, because this is spatial distribution, I'm going to go to energy coordinates and give you an energy level diagram. So again, not to scale. Because this thing goes is 1 over the square. So that's going to be messy. So let's start here. And on the left side I'm going to designate the energy. And on the right side I'm going to designate the quantum number. I'm going to start down here. That's the lowest energy. You see these are all negative values, first of all. They're all negative values. Because Z is a square, n is a square, and K is a positive quantity. So n equals 1 is the lowest state. It's the ground state. It has a value of minus K. So I'm going to do this one just for atomic hydrogen. So I'm going to write atomic H. So now Z equals 1. You can do it later for Z equals 2, 3, whatever. So this is atomic hydrogen. So ground state energy is minus K. What happens if we go to n equals 2? n equals 2 it becomes K divided by 2 squared 4. So it should be 3/4 of the way up. I'm not going to go quite 3/4 of the way. Because I want to leave room for some fine structure. That's why it's not to scale. All right so this is minus L over 4. What if we go to n equals 3? Well it's not symmetric here. It's nonlinear. But this should be really what? Minus K over 3 squared is 9. You get the picture. 3, you can go 4, 5, and so on until n equals infinity. What happens when n equals infinity? I've got minus K over infinity, which is vanishingly small, zero. Where is the electron when n equals infinity? r is n squared times the Bohr radius. That's a great, great distance away. What does it mean? Physically it means that the electron is so far away that it is no longer bound. It's no longer part of the atom. And when it's no longer part of the atom, and the potential energy that's stored is a result of the charges coming together from infinity. I'm starting to talk like someone out of 802. What's the energy if I take 2 charged particles at infinite separation? I bring them into some finite separation. Voila. There it is. So when they're in infinite separation there's no energy stored. Hence, you are at that point. So n equals infinity means r equals infinity, which means E equal zero. There's no stored energy. So this means the electron is no longer bound. And therefore, if it's no longer bound, we have a term for it. It's called free. It's a free electron. And if the electron is free, then the atom is electron deficient. So if the electron is free, that means the atom is now an ion. Because it's lost an electron. It's no longer net neutral. Or we say an atom hasn't turned into an ion. Or the electron has been ionized. What's the energy for that? We can calculate what that energy is. It would be called the ionization energy. So if I started with an electron down in here, and I sent it all the way to infinity. See that's an energy space. Which is the equivalent in Cartesian space to go from here to infinity, same idea. Do you see the models? This is Cartesian. It's like a. Map This is energy coordinates. It's different. And you're going to be able to think from one model to another. What's the energy consequences? What are the Cartesian consequences? Until we get to the point where there is no Cartesian representation. Because the abstraction level is too high, we'll have to content ourselves with this. So get comfortable moving from there to there. And then some day we're going to say it's too complicated. There's no Cartesian thing. We'll be comfortable by then with this. OK so now we're taking an electron from here up to here. While we ask, what is the ionization energy? The ionization energy must equal the delta E of the transition. So what's that? The delta E of the transition is going to equal always E final minus E initial. E final minus E initial. Which is equal to E at infinity minus E1, the ground state. Well E infinity, we just said, is zero. And the ground state energy is equal to minus K. So minus minus K is K. So you also get the energy. From infinity down to ground state is the ionization energy. So we can define the ionization energy in terms of this transition. Define ionization energy as the minimum energy to remove an electron from the ground state of an atom in a gas phase. So that means there's no solids, no liquids. There's no work function here. There's no lattice energy, and so on. So there's a definition of the ionization energy. And we can be a little bit more elaborate. Even though right now I'm just going to do a little break here. I don't want to mislead people. But just an aside. I'm nonlinear. I can have multiple conversations at once. And you are capable of stacking. So we're going to break now. We're not going to talk about 1-electron atom. We're going to follow the thread of ionization energy. I'm going to take lithium. Lithium in its normal state has 3 protons, 3 electrons. So I'm going to take lithium gas. And I'm going to ionize it and make lithium plus. So this is a lithium plus ion in the gas phase. It's still got 2 electrons. So this isn't Bohr model stuff. But anyway, here we are. So the energy for this action would be called the ionization energy. Because I took a neutral atom, and I pulled an electron out of the ground state, and so on. So this is an ionization energy. But now I can continue this process. And I can take Lithium plus. And I can lose an electron from that, which will than give me lithium 2 plus. And this is called also an ionization energy. This is called the second ionization energy. So this is the first ionization energy. But just as when you write an equation, when the coefficient is 1, you don't write the 1. I don't write 1 lithium here. I know it's 1. This is the second ionization energy. And then I can keep stripping away electrons. And I can take lithium 2 plus in the gas phase, and take away that electron leaving me with just the lithium nucleus, lithium 3 plus, plus electron. And that's called the third ionization energy. OK, now what can we say? What's the relationship here between any of this except the definition and the Bohr model? well? The Bohr model applies only to 1-electron atoms. Are there any 1-electron atoms on this board? So we can calculate the energy, the third ionization energy. We can get that from the Bohr model. You can do it in your head. You can do it in your head right? It's just K times Z squared right here. It's going from 1. So if this is 2.18, it's going to be 9 times that trivially. OK, so this is just 3 squared times K. And this when you have to get from the literature. So you have to go to primary sources. Which is why you're going to learn how to use the proper database. And this one here also you get from the literature. But this is on your periodic table. Your periodic table, one of the data points it gives is the first ionization energy of all of the elements. And so even though lithium normally is a solid at room temperature, the ionization energy for lithium as given on your periodic table is for this reaction. It's for the gas. Anyway. So that's little aside. All right, the last quantity that we could get from the Bohr model is v, the velocity. And I solve for the velocity. We don't talk about this very much. But we're going to do so once today. And here it is. So I went through the algebra. And you get nh over 2 pi mr, where r is the radius. There's the quantum number. This is quantized as well. So I regrouped this. And I already have a nice, cool expression for r in terms of the Bohr radius. So I use that because that's on the table. So this is 2 pi times the mass of the electron times the Bohr radius, 1/2 angstrom, times Z proton number divided by n, where n equals 1, 2, 3, and so on. So let's again get a sense of scale. So let's try for sense of scale. Let's do velocity of the ground state electron in atomic hydrogen. So that means Z equals 1, n equals 1. So I plug in the numbers. And I get v1 for hydrogen, atomic hydrogen, gives me 2.18 times 10 to the 6 meters for second. I don't know. Is that fast? Is that slow? I don't know. But I do know this much. I know that the speed of light is equal to 3 times 10 to the 8 meters per second. So 10 to the 8 divided by 10 to the 6. 2 and 3, that's roughly 1, speaking as an engineer. Who cares? So this is about 1% of the speed of light. That's pretty good. That gives me something I can hang on to. I would say that if this thing is zipping around at 1% of the speed of light, I would say that's relatively fast. One more time, 1% of the speed of light. That's relatively fast. Remember last day I told you that Bohr simply dismissed the concept of the use of classical electrodynamics down to atomic dimensions. Well here's another example of why a lot of these assumptions aren't going to work so well. We're talking about the ground state electron velocity in this putative planetary model of a 1-electron atom. You're already getting into relativistic effects. So, just another example. By the way, why do we use the letter c for speed of light? It comes from the latin word celeritas, which means swiftness. And we get the modern word acceleration, deceleration from that. OK. All right. So the Bohr model, we've now rolled it all out. We have the energy portrait. We have the radii, discrete. We have quantization. And we have velocities if we ever want to look at those again. Now what's the next thing we do in science? We compare the predictions of the Bohr model with data. Are there any data to support this? Because remember, all Rutherford said was plum pudding doesn't make sense. Instead I'm going to concentrate the positive mass in the center. And then Bohr came along and said, not only is it going to be a planetary model, I'm going to have circular orbits. So now we've gone a long way from Geiger-Marsden. So is there any data? Well there were data in 1853. Remember, Bohr published this in 1913. In 1853 there was a spectroscopist by the name-- I'm going to tell you his name-- in Uppsala, Sweden. And his name was Angstrom. Angstrom was doing experiments on hydrogen in gas discharge tubes. So he measured emissions from gas discharge tube. And it was filled with various gases including atomic hydrogen. And in order to take his data, what he used was this device here-- there's the Bohr radius, just an example. He used the prism spectrograph. So here's a gas discharge tube. And I'm going to show you the physics of that in a second. Basically you've got a pair of electrodes. You've got gas in the tube. And as this cartoon shows, you apply a potential across the electrodes. And beyond a certain threshold potential, the tube begins to glow. And the glow goes in all directions. And a blinds you when you're in the lab. So what you do, is you cover this up a bit. You have a narrow slit. And then you force the light to come through in a thin ribbon, and then expose it to a prism. What the prism does, is it takes the light and breaks it into its components, sort of rainbow-like, and magnifies the difference. As refraction goes, different wavelengths will refract different amounts. And then you shoot this across the room. There's two counters. One is a scintillation screen and an army of graduate students who sit there in the dark. But they're no good because you can't stick them into the publication. You need to have data that people will rely upon. So instead you use a photographic plate. And if you put even a tiny, tiny angle of separation across a great enough distance, you start to get enough line splitting that you can see. And then you go backwards. And you know the geometry here. And you can figure out what the wavelength must have been to go this distance, et cetera, et cetera. And these are all color coded, not because they had color film in those days, but just to let you know that 656 nanometers, if you were the graduate student sitting there, you'd see a red line, a green line, a blue line, and a violent line. So that's how he made the measurements. And he published those measurements. And so they lay. And then the story gets a little thicker. In 1885, there's a Swiss high school math teacher. I mean, I can't make this stuff. This is true story. There's a Swiss high school math teacher by the name of J. J. Balmer. We had J. J. Thomson. Now we've got J. J. Balmer. And J. J. Balmer, he loved to play with numbers. And he was studying this set of lines. And he was trying to come up with a pattern. Can you see a pattern there? 410 434 486, 656, do they go squares? Are they primes? What's the pattern there? So Balmer puzzled over this for awhile. And he finally came up with the equation to represent those lines. So he studied Angstrom's data found the pattern. And here's the pattern that he found. He said that those are wavelengths. If I take instead wave number, nu bar is called wave number, which is the reciprocal of the wavelength. So if I take the reciprocal of the wavelength, I end up with those 4 lines conforming to a series that goes like this. 1 over 2 squared minus 1 over n squared, where n equals 3, 4, 5, 6. And there's a constant here which we're going to designate r. And the value of R-- I'm going to put that on the next board-- the value of R as expressed in SI units today, would be 1.1 times 10 to the 7 reciprocal meters. Wavelength is a meter. Reciprocal wavelength or wave number must be reciprocal meters. So how this all of this support the Bohr model? Well in order to explain it, I've got a first tell you what the physics of the gas discharge tube are. So let's go inside the gas discharge tube and understand those physics. So here's the gas discharge tube. It's made are borosilicate glass. And we fill it with gas. And in this case, the gas is going to contain among other things hydrogen. So this is a hydrogen gas phase atom. And this is probably at low pressure. And then I said I need electrode. I'll get the electrodes inside. So I've got to have a really good glass blower who can make a glass to metal seal, and have a feedthrough to an electrode. So this is still a vacuum seal. How do they get the gas in the first place? We don't show you this in the books. I'll tell you because I did this in my Ph.D. What you do is you have a little side tube here. You evacuate. This goes to a vacuum pump. And then over here you have a gas source. You evacuate. Put in the gas to whatever pressure you want. And then the glass blower disconnects. This pulls this down to a reduced pressure, and seals it. But the books don't show you that. That's a secret. So we've got a gas at low pressure. And I've got an electrode over here and an electrode over here. And they're connected to a variable voltage power supply. So I'm going to put an arrow with the V meaning it's variable voltage. I can change the voltage. And this convention, this is the negative side. So this means the electrons leave the power supply and go like this. Which means this electrode will be negative. And this electrode will be positive. And thanks to Michael Faraday, we will call this one the cathode. And this one we will call the anode. Now what happens? We start turning up the pressure, turning up the voltage rather. Low pressure here, but electrical pressure is voltage. The voltage gets high enough, eventually the electrons will boil off the cathode. And this is a gas at low pressure. And they will accelerate from rest and go all the way across the tube and crash into the anode. And we complete the circuit. But if there's a gas in here, some of these electrons are going to hit the gas molecules. And when they hit the gas molecules, if they have enough energy to do so, they will cause electrons inside the gas molecules to be excited. And if they're excited enough, the electrons will jump up to a higher energy level. But they can't be sustained. Because this is a ballistic collision. It's a one of. It's like a bowling alley. One ball, one pin, one impact. Now the pin is in the air. What happens to the pin? It falls back down. Why? Because gravity pulls it down. In this case, you've got the energetics pulling the electron back down. Now the electron goes from high energy to low energy. And when that happens, the energy difference is given off in the form of a photon. So I get photon emission when this falls back down. And this photon has a wavelength. What I'm going to show you is that the set of lines that you get from exactly this configuration using this equation give you the Balmer series. So now you've got a model that Bohr postulated for atomic hydrogen on 1-electron atom that exactly predicts that set of 4 lines. Which were measured 50 years before. So let's go. So first of all let's get the energy here. And I'm going to get the energy of this electron. This electron I'm going to call a ballistic electron. Why do I call it a ballistic electron? Because it's not bound. It's free. It boils off the cathode, flies through free space, and crashes into an anode. Clearly it's not part of an atom. But there's a second electron in this story. And it's the ground state electron in hydrogen. And it lives here. So what's the energy of the ballistic electron? Well that's just 1/2 mv squared And where did it get its energy from? It got its energy from the power supply. And what's the electrostatic energy? It's the product of the charge on the species times the voltage through which it was accelerated. So away we go. I know the charge on the electron is minus E. Whatever the voltage is there, 1 volt, 10 volts, 100 volts, whatever. Away we go. By the way, I'm going to show you just one other thing in terms of order of magnitude. The kinds of voltages you see along here are 1 volt, 10 volts, that sort of thing. So suppose, to get an order of magnitude, suppose we had a species of charge E. So in other words, it's only 1 times the elementary charge. A species of charge E influenced by voltage of 1 volt. So this is 1 in the voltage units, and 1 in the elementary charge units. How much energy would that be? That's equivalent to making 1 volt and accelerate an electron from rest across this gap. And the result would be, the energy then would simply equal 1.6 times 10 to the minus 19 coulombs times 1 volt. And what's the energy going to be? Well I've got coulombs times volts. And I don't know how I convert one to the other. I don't have to. Why not? Because that's an SI unit. And that's an SI unit. This is an energy. So with impunity, I write 1.6 times 10 to the minus 19 joules. That's the beauty of SI units. So that's a good news. I know it's joules. The bad news is I hate this number. It's a stupid number, 1.6 times 10 to the minus 19. it's crazy. Why don't I come up with a number like 3, 7? So what I could do, is I could define. I could define a unit such that when the elementary charge is accelerated across the unit voltage, I would call that unit 1 electron volt. And so somebody thought of this before me. And hence, this is the unit of the electron volt. It takes these crazy things that we've been spewing here up until now, and rationalizes them into numbers that people can carry around in their heads. So what's K now? K is 2.18 times 10 to the minus 18 joules. Yuck! Let's convert that to electron volts. So I divide by 1.6 times 10 to the minus 19. And I got 13.6 electron volts. You'll remember that on your death bed, ionization energy of atomic hydrogen. Maybe we don't have to give out tables of constants. You just know this stuff. It's OK. All right. Now one last thing about this. So this has got gas in it. This is the cathode. And this beam of electrons, back in the 1800s, there was a popular term, it was called the ray. So instead of a beam of light, people refer to the ray of light. So then when they got to particle beams, they talk to them as rays. So this is now not an electron beam, it's an electron ray. And it comes off the cathode. And it's in a vacuum tube. So this could be called a cathode ray tube, a CRT. Now see, I could flatten this. And I could spray it with phosphors. And then I could put some charge plates here. The electrons have a negative charge. So if I charge these plates, and I was clever about how I charge them and varied the charge, I could raster the electron beam like this about 30 times a second all up and down the screen. And then I could put some program signal in there. And I could sit here. And I could watch TV. It all started with the gas discharge tube. It says nothing about the content unfortunately. Very nice physics, but no content. All right. So now we've got the electron. The electron is moving, the ballistic electron. Now I want to look at what happens when the ballistic electron smashes into one of those hydrogen atoms. So let's go back over here. So here's the incident particle. And it's going to be an electron in this case. So I'm going to designate this electron. This is my ballistic electron. So here's the incident electron. And this is ballistic, just to be clear. It's the ballistic incident electron. Now this is a mixed metaphor here. Because I'm representing this in Cartesian space. But I've moved into energy space. So some people are going to get really upset. Because they're going to say, well this is Cartesian, but this isn't. It doesn't matter. It's my lecture. It's my model. It works. I'm the professor. So we've got this mixed metaphor here. But anyways, it helps a lot. So what happens when this thing comes in? It depends on how much energy it has. Now if the incident energy, if E incident is tiny, nothing happens. This thing just zooms right on through. But if the incident energy, if E of the incident ballistic electron is greater than delta E for any transition that's feasible-- and in this case I'm going to assume I don't have thermal distribution of electrons. If I gave you Avogadro's number of hydrogen atoms because of the thermal distribution of energies-- and we'll come back to this later-- there might actually be, at any moment, some electrons that are thermally excited above the ground state. We're going to forget about that for now. We're going to spiral up the learning curve here. So first time we're going to assume all the electrons are in the ground state. If I don't enough energy to go from n equals 1 to n equals 2, nothing happens. If I have more than enough energy to go for n equals 1 to n equals 2, I will take that energy. And the electron will jump, steal that amount of energy, and then this thing moves on-- and I'm purposely making this vector shorter than the incident vector-- with that amount of energy raw. And this is called the scattered electron. Go back to the bowling ball analogy. The bowling ball comes in with a certain energy, hits the pin, continues to roll. But you know that there's a loss of kinetic energy in the bowling ball. That's what we're seeing here. It's purely ballistic. Now let's say we do have more than enough energy. Suppose I have enough energy to go from n equals 1 halfway between n equals 2 and n equals 3. There's only n equals 1, n equals 2. I can't take the electron up to n equals 2.3. It's unallowed. These are the only allowed states. So that differential amount of energy then resides with the electron that's ballistic. And it moves on here. So we've got conservation of energy here. We can say that E incident will then equal the energy that's lost in the transition plus the energy that's still left with the scattered electron. And we can calculate what that transitional energy is. That transitional energy is going to equal 1/2 mv squared incident. This is the velocity, the incident electron. What's the energy to go from n equals 1 to n equals n? Whatever it is, its minus K times Z squared If it's hydrogen, Z is 1. It's going to be 1 over nf. The final quantum number squared minus 1 over the square of the initial quantum number. And then, what's left over after this has been robbed from the incident ballistic energy is 1/2 mv squared of the scattered ballistic electron. And the quantization dictates that only if E incident is greater than delta E going 1 up to n-- here I'm assuming everything is in its ground state. Later on we're going to be more sophisticated. But for first time through, all are ground state electrons. I have to have enough energy to go at least to n equals 2. I can go to n equals 3, m equals 4. In principle, if this thing had more than 13.6 electron volts, what would happen? It would kick this electron out. Gone! And you'd have 2 free electrons. So if it's greater than this, then the consequence is electron promotion. So we're moving along. But this excited state is unstable. This excited state is unstable because it's like the bowling pin that got thrown up. So what happens? The electron standing up there on n equals 2, for example, looking down to n equals 1. And it falls. When it falls it gives off radiation. And that radiation is conservation of energy there. And what we know is when it gives off an energy of the emitted photon, the energy of the emitted photon must equal delta E of the transition falling from 2 to 1. And we know how to calculate that. That's just that thing flipped around. And this thing is equal to what? This is equal to h nu. According to Planck it's hc over lambda is equal to hc nu bar. You know what this one is. This is minus KZ squared 1 over-- in this case it's going to be-- 1 over nf squared 1 over 1 squared minus 1 over, in this case, 1 over 2 squared And you can generalize this 1 over nf. So where am I going with this? Well I'm going to flip all of this around. And when I flip it all around, what I'm going to end up with is this equation here. And better than that, I'm going to end up with this equation. And I'm going to end up with this as the constant. And when I get that, we're going to say Bohr has done it. The data support the theory. So that's what we're going to do. But I think we're going to stop at this point today. So let me just jump here. I mentioned to you that if you go here on your periodic table, there's the 13.6 electron volts. In the case of lithium, this is 5.4 electron volts. So you can see all the various values. This by the way, people no noise. No noise. It's 11:52. I'm holding court until 11:55. I'm simply changing topics. It's not, oh this is the part where I can talk to my neighbor. What are the rules? No talking. No food. No horseplay until 11:55. Then, still no horseplay, gentle talking, no food, no drink. This is a sacred space. I'm not kidding you. Do you know why? In this secular America, this is sacred space because this is where people learn. The lecture hall is sacred space. Now, here's the 13.6 electron volts. And there's the 1.6 times 10 to the minus 19 joules. And if you multiply those 2, you'll get the 2.18 over here. All right. Here's a cartoon showing the photon, higher energy orbit, lower energy orbit, electron emission transition, right out of your book. And there's a postulate 6. And we're going to finish this up at the beginning of the Friday lecture. So here's the whole series from n equals 3 to 2, n equals 4 to 2, and so on. See how that works. OK. One of the things that we've learned here, is that it doesn't matter what the incident energy is here. The emission is characteristic of the energy levels inside the gas. Instead of using an incident electron, I could use an incident proton. I could use an incident alpha particle. I could use an incident neutron. Anything that has enough energy to kick this up from n equals 1 to n equals 2 will result in photon emission of this frequency. That means the set of those lines is unique to the target. And this is the beginning of chemical analysis. I can use this to characterize species, to it, stars. How do we analyze the composition a stars? Well, do we send a NASA spaceship out 25 light-years and grab some gas and bring it back to the lab? No. All we've got is the spectrograph. The star is hot. That means thermal excitation and cascading down with photon emission. And the lines we get are related to the energy levels within the stars. So from a distance of a 100 light-years, I can tell you what the composition is. And if there's two gases there, what if there's hydrogen and helium? the helium lines will be there. And they'll be superimposed on the hydrogen lines unless they lie directly on top of one another. I'm going to be able to figure out what's there. That's how it works. Now here's a story about an astronomer, Cecilia Payne. She's the first woman graduate student in astronomy at Harvard. She went on to chair the Faculty of Arts and Sciences, awarded tenure, but denied a professorship for 18 years because she was a woman. And in her thesis, she was the first person to figure out to the sun is dominantly hydrogen, not iron, which is what most astronomers thought. Why? Well, because the earth is made of iron. Meteorites are made of iron. The whole universe must be made of iron. Never mind the fact that the sun is glowing. So here's how spectroscopy works. Look at this. See, what does this mean? This is an analogy. What could that mean? Well, they said iron again. I know this is misspelled. But maybe it's just a glitch in the instrumentation. So most people would look at that and say, the message is iron. But it's not about the word. It's about the pattern. It's about the pattern. And that's what's spectroscopy is. By the way, if you somehow didn't catch this lecture. And you walked in, and all you saw was those four lines. You know those four lines. That set of lines is characteristic of atomic hydrogen, and nothing else. By the way, those lines are very faint. This is not to scale. And they're so faint, they're ghostlike. And what is the latin word for ghost? Specter. So what is a spectrum? It is a set of ghostlike lines. To this day, the term spectroscopy refers to the ability to study data that are so faint they're ghostlike. All right, we'll see you on Friday. MIT OpenCourseWare 3.091SC Introduction to Solid State Chemistry, Fall 2010 Please use the following citation format: Donald Sadoway, 3.091SC Introduction to Solid State Chemistry, Fall 2010. (Massachusetts Institute of Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit:
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TrianglesPerpendicular
Caroline P.
asked • 09/16/15
Please help me find the solution to this question! Thank you!
The y-interepts, P and Q, of two perpendicular lines intersecting at the point A(6,8)have a sum of zero. What is the area of ΔAPQ?
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Jordan K.answered • 09/16/15
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Hi Caroline,
Sounds like a challenging problem for sure, but we'll step through it and all will be clear.
Let's begin by determining what are the base and the altitude of triangle APQ, since these are the two items needed for calculating the area of a triangle:
The base will be side PQ since we are told that points P and Q both lie on the y-axis (these points being y-intercepts). We are further told that the sum of these y-intercepts is zero. This means that each y-intercept is the additive inverse of the other and can be represented as "b" and "-b", respectively. Therefore, the length of the base of the triangle will be the difference of these two y-intercepts (b - (-b)) or 2b units.
The altitude of the triangle will be the horizontal distance from point A to the portion of the y-axis bounded by points P and Q (the vertical base of the triangle). This distance is the difference between the x coordinate of point A (6) and the x coordinate of the y-axis (0): 6 - 0 (6 units).
All that is needed now is to determine the values of our two y-intercepts (b and -b).
Let's begin by putting the equations of the lines representing sides AP and AQ into slope-intercept form (noting that we are told these lines are perpendicular to each other and, therefore, their slopes are the negative reciprocals of each other):
Line AP will descend slightly from point P (0,b) to point A (6,8), so it will have a small negative slope and a positive y-intercept:
y = -(1/m)x + b
Line AQ will ascend steeply from point Q (0,-b) to point A (6,8), so it will large positive slope and a negative y-intercept:
y = mx - b
Now let's plug in the coordinates point A (common intersection point of these two lines) into the slope-intercept equations for both lines and then add the equations together to eliminate b and solve for m:
8 = -(1/m)(6) + b [equation of line AP]
8 = m(6) - b [equation for line AQ]
16 = 6m - (6/m) [equations added together]
(m)(16) = (m)(6m) - m(6/m)
16m = 6m 2 -6
6m 2 - 16m - 6 = 0
2(3m 2 - 8m - 3) = 0
3m 2 - 8m - 3 = 0
3m 2 + (-9 + 1)m - 3 =0
3m 2 - 9m + m - 3 = 0
3m(m - 3) + 1(m - 3) = 0
(m - 3)(3m + 1) = 0
m - 3 = 0
m = 3
3m + 1 = 0
3m = -1
m = -1/3
We can now see that the roots of the quadratic equation formed from the addition of our two line equations, gives us the two negative reciprocal slopes (-1/3 and 3) of our two perpendicular lines forming sides AP and AQ, respectively.
We are now in position to calculate the absolute value of y-intercept (b) by plugging into either of our two line equations: our slope and the coordinates of common point A:
8 = (6)(-1/3) + b [equation of line AP]
8 = -2 + b
b = 8 + 2
b = 10 (absolute value)
8 = (3)(6) - b [equation of line AQ]
8 = 18 - b
b = 18 - 8
b = 10 (absolute value)
So now we know that since the absolute value of our y-intercept (b) is 10, we also know that our positive and negative y-intercepts are 10 and -10, respectively. Therefore, the length of base P is
(10 - (-10)) = 10 + 10 = 20
Putting it all together:
Area of triangle APQ = 1/2(base x height)
Area of triangle APQ = 1/2(PQ)(altitude)
Area of triangle APQ = 1/2(20)(6)
Area of triangle APQ = 1/2(120)
Area of triangle APQ = 60 square units
Below is the link to our diagram for this problem showing the vertices of triangle APQ with altitude AB drawn to base PQ, so you can have a visual aid supporting our solution:
Thanks for submitting this problem and glad to help.
God bless, Jordan.
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http://dorothydingzx.github.io/2018/01/11/Bolted-connections-in-Eurocode-3-1/
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Bolted connections in Eurocode 3(1) | Hello, life
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Bolted connections in Eurocode 3(1)
发表于 2018-01-11 |分类于Connections|阅读次数
Eurocode 3: Design of steel structures - Part 1-8: Design of joints
Design assumptions
(1) Joints shall be designed on the basis of a realistic assumption of the distribution of internal forces and moments. The following assumptions shall be used to determine the distribution of forces:
(a) the internal forces and moments assumed in the analysis are in equilibrium with the forces and moments applied to the joints,
(b) each element in the joint is capable of resisting the internal forces and moments,
(c) the deformations implied by this distribution do not exceed the deformation capacity of the fasteners or welds and the connected parts,
(d) the assumed distribution of internal forces shall be realistic with regard to relative stiffnesses within the joint,
(e) the deformations assumed in any design model based on elastic-plastic analysis are based on rigid body rotations and/or in-plane deformations which are physically possible,
(f) any model used is in compliance with the evaluation of test results (see EN 1990).
Connections made with bolts, rivets or pins
Bolts, nuts and washers
General
(1) All bolts, nuts and washers should comply with 1.2.4 Reference Standards: Group 4.
(2) The rules in this Standard are valid for the bolt classes given in Table 3.1.
(3) The yield strength f y b f y b and the ultimate tensile strength f u b f u b for bolt classes 4.6, 5.6, 5.8, 6.8, 8.8 and 10.9 are given in Table 3.1. These values should be adopted as characteristic values in design calculations.
Preloaded bolts
(1) Only bolt assemblies of classes 8.8 and 10.9 conforming to the requirements given in 1.2.4 Reference Standards: Group 4 for High Strength Structural Bolting for preloading with controlled tightening in accordance with the requirements in 1.2.7 Reference Standards: Group 7 may be used as preloaded bolts.
Categories of bolted connections
Shear connections
(1) Bolted connections loaded in shear should be designed as one of the following:
Category A: Bearing type
In this category bolts from class 4.6 up to and including class 10.9 should be used. No preloading and special provisions for contact surfaces are required. The design ultimate shear load should not exceed the design shear resistance, obtained from 3.6, nor the design bearing resistance, obtained from 3.6 and 3.7.
Category B: Slip-resistant at serviceability limit state
In this category preloaded bolts in accordance with 3.1.2(1) should be used. Slip should not occur at the serviceability limit state. The design serviceability shear load should not exceed the design slip resistance, obtained from 3.9. The design ultimate shear load should not exceed the design shear resistance, obtained from 3.6, nor the design bearing resistance, obtained from 3.6 and 3.7.
Category C: Slip-resistant at ultimate limit state
In this category preloaded bolts in accordance with 3.1.2(1) should be used. Slip should not occur at the ultimate limit state. The design ultimate shear load should not exceed the design slip resistance, obtained from 3.9, nor the design bearing resistance, obtained from 3.6 and 3.7. In addition for a connection in tension, the design plastic resistance of the net cross-section at bolt holes N n e t,R d N n e t,R d, (see 6.2 of EN 1993-1-1), should be checked, at the ultimate limit state.
The design checks for these connections are summarized in Table 3.2.
Tension connections
(1) Bolted connection loaded in tension should be designed as one of the following:
Category D: non-preloaded
In this category bolts from class 4.6 up to and including class 10.9 should be used. No preloading is required. This category should not be used where the connections are frequently subjected to variations of tensile loading. However, they may be used in connections designed to resist normal wind loads.
Category E: preloaded
In this category preloaded 8.8 and 10.9 bolts with controlled tightening in conformity with 1.2.7 Reference Standards: Group 7 should be used.
The design checks for these connections are summarized in Table 3.2.
Positioning of holes for bolts and rivets
(1) Minimum and maximum spacing and end and edge distances for bolts and rivets are given in Table 3.3.
(2) Minimum and maximum spacing, end and edge distances for structures subjected to fatigue, see EN 1993-1-9.
Design resistance of individual fasteners
Bolts and rivets
(1) The design resistance for an individual fastener subjected to shear and/or tension is given in Table 3.4.
(2) For preloaded bolts in accordance with 3.1.2(1) the design preload, F p,C d F p,C d ,to be used in design calculations should be taken as:
F p,C d=0.7 f u b A s/γ M 7 F p,C d=0.7 f u b A s/γ M 7
NOTE: Where the preload is not used in design calculations see not to Table 3.2.
(3) The design resistances for tension and for shear through the threaded portion of a bolt given in Table 3.4 should only be used for bolts manufactured in conformity with 1.2.4 Reference Standard: Group 4.
For bolts with cut threads, such as anchor bolts or tie rods fabricated from round steel bars where the threads comply with EN 1090, the relevant values from Table 3.4 should be used.
For bolts with cut threads where the threads do not comply with EN 1090 the relevant values from Table 3.4 should be multiplied by a factor of 0.85.
(4) The design shear resistance F v,R d F v,R d given in Table 3.4 should only be used where the bolts are used in holes with nominal clearances not exceeding those for normal holes as specified in 1.2.7 Reference Standards: Group 7.
(5) M12 and M14 bolts may also be used in 2 mm clearance holes provided that the design resistance of the bolt group based on bearing is less than or equal to the design resistance of the bolt group based on bolt shear. In addition for class 4.8, 5.8, 6.8, 8.8 and 10.9 bolts the design shear resistance F v,R d F v,R d should be taken as 0.85 times the value given in Table 3.4.
(6) Fit bolts should be designed using the method for bolts in normal holes.
(7) The thread of a fit bolt should not be included in the shear plane.
(8) The length of the threaded portion of a fit bolt included in the bearing length should not exceed 1/3 of the thickness of the plate, see Figure 3.2.
(9) The hole tolerance used for fit bolts should be in accordance with 1.2.7 Reference Standards: Group 7.
(10) In single lap joints with only one bolt row, see Figure 3.3, the bolts should be provided with washers under both the head and the nut. The design bearing resistance F b,R d F b,R d for each bolt should be limited to:
F b,R d=1.5 f u d t/γ M 2 F b,R d=1.5 f u d t/γ M 2
NOTE: Single rivets should not be used in single lap joints.
(11) In the case of class 8.8 or 10.9 bolts, hardened washers should be used for single lap joints with only one bolt or one row of bolts.
(12) Where bolts or rivets transmitting load in shear and bearing pass through packing of total thickness t p t p greater than one-third of the nominal diameter d d, see Figure 3.4, the design shear resistance F v,R d F v,R d calculated as specified in Table 3.4, should be multiplying by a reduction factor β p β p given by:
β p=9 8 d+3 t p β p=9 8 d+3 t p
but β p≤1 β p≤1
(13) For double shear connections with packing on both sides of the splice, t p t p should be taken as the thickness of the thicker packing.
(15) For grade S 235 steel the “as driven” value of f u r f u r may be taken as 400 N/mm2.
Group of fasteners
(1) The design resistance of a group of fasteners may be taken as the sum of the design bearing resistances F b,R d F b,R d of the individual fasteners provided that the design shear resistance F v,R d F v,R d of each individual fastener is greater than or equal to the design bearing resistance F b,R d F b,R d . Otherwise the design resistance of a group of fasteners should be taken as the number of fasteners multiplied by the smallest design resistance of any of the individual fasteners.
Long joints
(1) Where the distance L j L j between the centres of the end fasteners in a joint, measured in the direction of force transfer (see Figure 3.7), is more than 15 d 15 d, the design shear resistance F v,R d F v,R d of all the fasteners calculated according to Table 3.4 should be reduced by multiplying it by a reduction factor β L f β L f, given by:
β L f=1−L j−15 d 200 d β L f=1−L j−15 d 200 d
β L f≤1.0 β L f≤1.0 but β L f≥0.75 β L f≥0.75
(2) The provision in 3.8(1) does not apply where there is a uniform distribution of force transfer over the length of the joint, e.g. the transfer of shear force between the web and the flange of a section.
Slip-resistant connections using 8.8 or 10.9 bolts
Design Slip resistance
(1) The design slip resistance of a preloaded class 8.8 or 10.9 bolt should be taken as:
F s,R d=k s n μ γ M 3 F p,C F s,R d=k s n μ γ M 3 F p,C
F s,R d,s e r=k s n μ γ M 3,s e r F p,C F s,R d,s e r=k s n μ γ M 3,s e r F p,C
where:
k s k s is given in Table 3.6
n n is the number of the friction planes
μ μ is the slip factor obtained either by specific tests for the friction surface in accordance with 1.2.7 Reference Standards: Group 7 or when relevant as given in Table 3.7.
(2) For class 8.8 and 10.9 bolts conforming with 1.2.4 Reference Standards: Group 4, with controlled tightening in conformity with 1.2.7 Reference Standards: Group 7, the preloading force F p,C F p,C to be used in equation (3.6) should be taken as:
F p,C=0.7 f u b A s F p,C=0.7 f u b A s
Combined tension and shear
(1) If a slip-resistant connection is subjected to an applied tensile force, F t,E d F t,E d or F t,E d F t,E d,ser, in addition to the shear force, F v,E d F v,E d or F v,E d,s e r F v,E d,s e r, tending to produce slip, the design slip resistance per bolt should be taken as follows:
for a category B connection:
F v,E d,s e r=k s n μ(F p,C−0.8 F t,E d,s e r)γ M 3,s e r F v,E d,s e r=k s n μ(F p,C−0.8 F t,E d,s e r)γ M 3,s e r
for a category C connection:
F v,E d=k s n μ(F p,C−0.8 F t,E d)γ M 3 F v,E d=k s n μ(F p,C−0.8 F t,E d)γ M 3
(2) If, in a moment connection, a contact force on the compression side counterbalances the applied tensile force, no reduction in slip resistance is required.
Deductions for fastener holes
General
(1) Deduction for holes in the member design should be made according to EN 1993-1-1.
Design for block tearing
(1) Block tearing consists of failure in shear at the row of bolts along the shear face of the hole group accompanied by tensile rupture along the line of bolt holes on the tension face of the bolt group.
(2) For a symmetric bolt group subject to concentric loading the design block tearing resistance, V e f f,1,R d V e f f,1,R d is given by:
V e f f,1,R d=f u A n t/γ M 2+(1/3–√)f y A n v/γ M 0 V e f f,1,R d=f u A n t/γ M 2+(1/3)f y A n v/γ M 0
A n t A n t is net area subjected to tension;
A n v A n v is net area subjected to shear.
(3) For a bolt group subject to eccentric loading the design block shear tearing resistance V e f f,2,R d V e f f,2,R d is given by:
V e f f,2,R d=0.5 f u A n t/γ M 2+(1/3–√)f y A n v/γ M 0 V e f f,2,R d=0.5 f u A n t/γ M 2+(1/3)f y A n v/γ M 0
Prying forces
(1) Where fasteners are required to carry an applied tensile force, they should be designed to resist the additional force due to prying action, where this can occur.
NOTE: The rules given in 6.2.4 implicitly account for prying forces.
Distribution of forces between fasteners at the ultimate limit state
(1) When a moment is applied to a joint, the distribution of internal forces may be either linear (i.e. proportional to the distance from the centre of rotation) or plastic, (i.e. any distribution that is in equilibrium is acceptable provided that the resistances of the components are not exceeded and the ductility of the components is sufficient).
(2) The elastic linear distribution of internal forces should be used for the following:
– when bolts are used creating a category C slip-resistant connection,
– in shear connections where the design shear resistance F v,R d F v,R d of a fastener is less than the design bearing resistance F b,R d F b,R d,
– where connections are subjected to impact, vibration or load reversal (except wind loads).
(3) When a joint is loaded by a concentric shear only, the load may be assumed to be uniformly distributed amongst the fasteners, provided that the size and the class of fasteners is the same.
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# Bolted connections# Design Codes
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Bolted connections in Eurocode 3(2)
文章目录
站点概览
Dorothy
Just Do It
15 日志
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1.Connections made with bolts, rivets or pins
2.Bolts, nuts and washers
2.1.General
2.2.Preloaded bolts
3.Categories of bolted connections
3.1.Shear connections
3.2.Tension connections
4.Positioning of holes for bolts and rivets
5.Design resistance of individual fasteners
5.1.Bolts and rivets
6.Group of fasteners
7.Long joints
8.Slip-resistant connections using 8.8 or 10.9 bolts
8.1.Design Slip resistance
8.2.Combined tension and shear
9.Deductions for fastener holes
9.1.General
9.2.Design for block tearing
10.Prying forces
11.Distribution of forces between fasteners at the ultimate limit state
© 2018 Dorothy
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Miles to Meters (m) Converter
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Miles to Meters Converter
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Meters to miles ►
How to convert miles to meters
1 mile is equal to 1609.344 meters:
1mi = 1609.344m
The distance d in meters (m) is equal to the distance d in miles (mi) times 1609.344:
d(m) = d(mi) × 1609.344
Example
Convert 20 mi to meters:
d(m) = 20mi × 1609.344 = 32186.88m
How many meters in a mile
One mile is equal to 1609.344 meters:
1mi = 1mi × 1609.344 = 1609.344m
How many miles in a meter
One meter is equal to 1/1609.344 miles:
1m = 1m/1609.344 = 6.213712e-4mi
How to convert 10mi to meters
Multiply 10 miles by 1609.344 to get meters:
10mi = 10mi × 1609.344 = 16093.44m
Miles to meters conversion table
| Miles (mi) | Meters (m) |
--- |
| 0.01 mi | 16.09344 m |
| 0.1 mi | 160.9344 m |
| 1 mi | 1609.344 m |
| 2 mi | 3218.688 m |
| 3 mi | 4828.032 m |
| 4 mi | 6437.376 m |
| 5 mi | 8046.720 m |
| 6 mi | 9656.064 m |
| 7 mi | 11265.408 m |
| 8 mi | 12874.752 m |
| 9 mi | 14484.096 m |
| 10 mi | 16093.440 m |
| 20 mi | 32186.880 m |
| 30 mi | 48280.320 m |
| 40 mi | 64373.760 m |
| 50 mi | 80467.200 m |
| 60 mi | 96560.640 m |
| 70 mi | 112654.080 m |
| 80 mi | 128747.520 m |
| 90 mi | 144840.960 m |
| 100 mi | 160934.400 m |
Meters to miles ►
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概率统计笔记（五）：正态分布 - 知乎
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概率统计笔记（五）：正态分布
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概率统计笔记（五）：正态分布
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参考资料：概率论与数理统计（浙大第四版）,A First Course in Probability 8th Edition，
an introduction to mathematical statistics and its applications
本期关键词：正态分布
前置知识：概率分布函数，期望，方差，矩母函数，微积分，排列组合
正态分布（Normal Distribution）又叫高斯分布，对，就是那个画十七边形的高斯他发现的一种分布类型。之所以名字中带着个“normal”，是因为正态分布实在是太常见了，各种数据都能沾上一丝正态分布的气息。
分布函数
概率密度函数
f X(x)=1 2 π σ e−(x−μ)2 2 σ 2 f_{X}(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}
记作 X\sim N(\mu,\sigma ^{2})
\mu 被称作位置参数，代表分布的均值；
\sigma ^{2} 被称为形状参数，代表分布的方差。
然后验证一下 \int_{-\infty}^{\infty}f_{X}(x)dx=1 （因为正态分布的PDF不是初等积分，这里需要用到重积分证明）
令 t=\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx
则 t^{2}=\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx\int_{-\infty}^{\infty}e^{-\frac{y^{2}}{2}}dy
=\int_{R^{2}}^{}e^{-\frac{(x^{2}+y^{2})}{2}}dxdy
用极坐标换元：
\left{\begin{matrix} x=rcos\theta \y=rsin\theta \end{matrix}\right.
原式 =\int_{0}^{2\pi}d\theta\int_{0}^{\infty}e^{-\frac{r^{2}}{2}}rdr
=2\pi(-e^{-\frac{r^{2}}{2}})|_{0}^{\infty}
=2\pi
t=\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx=\sqrt{2\pi} （这样最重要的证明部分就完成了）
则 \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx=1
令 z=\mu +\sigma x
\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(z-\mu)^{2}}{2\sigma^{2}}}dz=1
累积分布函数
F_{X}(x)=\int_{-\infty}^{x}f_{X}(x)dx （这里没有具体的表达式...）
有时也记作 \phi(x)
性质（PDF）：
PDF
对称性
正态分布关于 x=\mu 对称
f_{X}(x)=f_{X}(-x)
F_{X}(-x)=1-F_{X}(x)
极值
正态分布在 x=\mu 取得最大值（峰值）
图像变换
\mu 变化使图像平移，对称中心落在 x=\mu
\sigma^{2} 变化使图像上下伸缩：
\sigma^{2} 越小图像越尖锐，可以理解为数据越集中。
\sigma^{2} 越大图像越扁平，可以理解为数据越分散。
对于 X\sim N(0,1) 的随机变量，我们称其服从标准正态分布。
正态分布也具有连续型随机变量一般的性质：
我们设 X\sim N(0,1)
则 f_{X}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}
Y=\mu +\sigma X
期望
E(X)=\int_{-\infty}^{\infty}xf_{X}(x)dx
=\int_{-\infty}^{\infty}x\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx=0 （奇函数在对称区间的积分为0）
E(Y)=\sigma E(X)+\mu
=\mu
方差
Var(X)=E(X^{2})-E(X)^{2}
=E(X^{2})
=\int_{-\infty}^{\infty}x^{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx
=-\frac{1}{\sqrt{2\pi}}[xe^{-\frac{x^{2}}{2}}|{-\infty}^{\infty}-\int{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx]
=-\frac{1}{\sqrt{2\pi}}[0-\sqrt{2\pi}]=1
E(Y)=\sigma ^{2}E(X)=\sigma^{2}
矩母函数
M_{X}(t)=\int_{-\infty}^{\infty}e^{tx}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx
=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-t)^{2}}{2}+\frac{t^{2}}{2}}dx
=e^{\frac{t^{2}}{2}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-t)^{2}}{2}}d(x-t)
=e^{\frac{t^{2}}{2}}1=e^{\frac{t^{2}}{2}}
M_{Y}(t)=e^{\mu t}M_{X}(\sigma t)
=e^{\mu t+\frac{\sigma^{2}t^{2}}{2}}
应用
正态分布可以应用于许多现实中的场景，比如：
学生身高、体重的分布
鸡蛋大小的分布
工厂中零件尺寸的分布
这些分布虽然不能完全匹配正态分布，但我们可以利用正态分布的一些性质对它进行一定的估计，以应对生产生活需要。
3\sigma原则
对于 X\sim N(\mu,\sigma ^{2})
P(\mu-\sigma\leq X\leq \mu+\sigma)=0.6827
P(\mu-2\sigma\leq X\leq \mu+2\sigma)=0.9545
P(\mu-3\sigma\leq X\leq \mu+3\sigma)=0.9973
假如一个工厂生产的零件尺寸的均值为90，方差为4
那么我们可以近似估计零件尺寸 X\sim N(90,4)
那么零件尺寸处于 (86,94) 的概率近乎为95%
当然，这只是一种粗略的估计，当正态分布和中心极限定理结合起来的时候，才是它真正发挥威力的时刻。
这就是后面数理统计的内容了，这里先挖个坑。
发布于 2021-03-27 07:39
正态分布
概率论与数理统计
统计学
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©U U2b0D1S2Z PKPu6tRaT bSToAfSt1wLaRrceE 2LWLICs.c m WAKlWlP Yrnilgahhtls4 LrSe2sTe5rDv6eRdx.o T NMuacdKeM OwBiEtyhW 7IonBfziCnAiLtZeD nAyligUeebwr1aN e2H.h Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period_ Date___ Vertex Form of Parabolas Use the information provided to write the vertex form equation of each parabola.
1) y = x2 + 16 x + 71 2) y = x2 − 2 x − 5 3) y = − x2 − 14 x − 59 4) y = 2 x2 + 36 x + 170 5) y = x2 − 12 x + 46 6) y = x2 + 4 x 7) y = x2 − 6 x + 5 8) y = ( x + 5)( x + 4) 9) 1 2( y + 4) = ( x − 7)2 10) 6 x2 + 12 x + y + 13 = 0 11) 162 x + 731 = − y − 9 x2 12) x2 − 12 x + y + 40 = 0 13) y = x2 + 10 x + 33 14) y + 6 = ( x + 3)2 -1-©u k2F0W1c2K aKduOtEaO jSsoRfntbwxa0rBeG yLvL7CW.F h uAdlLlB yrriOguhztfsj srUeOs0evrYv3eTdt.1 r DMJaldseo BwPiEtihm PIenxfJilnoiVtkeI oALlLgReVbNrza4 b2w.2 Worksheet by Kuta Software LLC Identify the vertex and axis of symmetry of each. Then sketch the graph.
15) f (x) = −3 ( x − 2)2 − 4 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 16) f (x) = −1 4 ( x − 1)2 + 4 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 17) f (x) = 1 4 ( x + 4)2 + 3 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 18) f (x) = 1 4 ( x + 5)2 + 2 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 19) f (x) = −2 ( x + 5)2 − 3 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 20) f (x) = ( x + 2)2 − 1 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 -2-©C O2B071W2v gKAuXtEaj 2S4oxfNtNwhaarMe9 RLKLrCF.M 6 DAxlHlE 5rvitgrhDtoso urRewsBePrav9eid6.P 1 iMzaHd5eK HwSiItBh8 UIrnnfnirnoibtcee 3AelYgverbBria9 n2y.i Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period Date__ Vertex Form of Parabolas Use the information provided to write the vertex form equation of each parabola.
1) y = x2 + 16 x + 71 y = ( x + 8)2 + 7 2) y = x2 − 2 x − 5 y = ( x − 1)2 − 6 3) y = − x2 − 14 x − 59 y = − ( x + 7)2 − 10 4) y = 2 x2 + 36 x + 170 y = 2 ( x + 9)2 + 8 5) y = x2 − 12 x + 46 y = ( x − 6)2 + 10 6) y = x2 + 4 x y = ( x + 2)2 − 4 7) y = x2 − 6 x + 5 y = ( x − 3)2 − 4 8) y = ( x + 5)( x + 4) y = ( x + 9 2) 2 − 1 4 9) 1 2( y + 4) = ( x − 7)2 y = 2 ( x − 7)2 − 4 10) 6 x2 + 12 x + y + 13 = 0 y = −6 ( x + 1)2 − 7 11) 162 x + 731 = − y − 9 x2 y = −9 ( x + 9)2 − 2 12) x2 − 12 x + y + 40 = 0 y = − ( x − 6)2 − 4 13) y = x2 + 10 x + 33 y = ( x + 5)2 + 8 14) y + 6 = ( x + 3)2 y = ( x + 3)2 − 6 -1-©5 H27031q24 EK4uIt1aw sSSo5fJtIwAaNr5e9 OL1LYCv.I y yAllulK trtiDgdhetgsx UrPe8sheQryvJebdc.k v bMnaOdzeI tw9iCt4hA lIvngfAiDnWiItweT 6AAlGg6eBbZrUat i2n.K Worksheet by Kuta Software LLC Identify the vertex and axis of symmetry of each. Then sketch the graph.
15) f (x) = −3 ( x − 2)2 − 4 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 Vertex: (2, −4) Axis of Sym.: x = 2 16) f (x) = −1 4 ( x − 1)2 + 4 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 Vertex: (1, 4) Axis of Sym.: x = 1 17) f (x) = 1 4 ( x + 4)2 + 3 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 Vertex: (−4, 3) Axis of Sym.: x = −4 18) f (x) = 1 4 ( x + 5)2 + 2 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 Vertex: (−5, 2) Axis of Sym.: x = −5 19) f (x) = −2 ( x + 5)2 − 3 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 Vertex: (−5, −3) Axis of Sym.: x = −5 20) f (x) = ( x + 2)2 − 1 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 Vertex: (−2, −1) Axis of Sym.: x = −2 -2-Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com
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Determining Limits using Algebraic Manipulation - GeeksforGeeks
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Determining Limits using Algebraic Manipulation
Last Updated : 23 Jul, 2025
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Limits give us the power to approximate functions and see the values they are approaching. Limit is not the value of the function at a particular point. It is the value which the function is approaching as one moves towards the given point. There are many ways to solve the limits, often limits are easy to solve, but sometimes they evaluate into indeterminate forms which are not defined and are difficult to estimate. The goal is to avoid these forms and solve the limits of the function. Let's look at some algebraic ways to solve such limits.
Table of Content
Limits
Indeterminate Forms
Limits using Algebraic Manipulation
Limits by Factoring
Limits by Rationalization
Trigonometric limits using Pythagorean Identities
Double angle Identities
Sample Problems
Limits
A limit is a value the function or a sequence approach as the input or the index approaches some value. These concepts are essential in calculus and real analysis as they help us define continuity,differentiability, and integrals. In the formula, the limit for a function f(x) at a point x = c is usually denoted as,
limx→c f(x)\lim_{x \to c} f(x)lim x→cf(x)
Often only the substitution method is enough for calculating the limit. But sometimes some limits might be evaluated to indeterminate forms.
Indeterminate Forms
An indeterminate form is usually encountered where the limit involves more than one function. It is defined as an expressioninvolving two or more whose limit cannot be determined solely from the individual functions.
The most common indeterminate forms are generated from the limits of ratio of functions,f(x)g(x)\frac{f(x)}{g(x)}g(x)f(x), when both functions evaluate either to 0 or ∞ generating the limits of the form 0 0\frac{0}{0}0 0or∞∞\frac{\infty}{\infty}∞∞. Other indeterminate forms include 0 x ∞, ∞ - ∞, 0∞ etc.
Certain algebraic methods can help us avoid these forms.
Limits using Algebraic Manipulation
These techniques from algebra can help in avoiding the indeterminate forms in the limits. Some of these forms include evaluating limits by factoring and sometimes rationalizing. In the case of trigonometric functions, some other tricks such as using thePythagorean Identity or trigonometric limit using double angle identity can help us solve these limits.
Limits by Factoring
Usually, in the ratio functions consisting of polynomials, the indeterminate form stems from one of the factors occurring in the expression. For example, in the function f(x) given below, the indeterminate form is due to the factor (x - 1).
limx→1 x 2−1 x−1\lim_{x \to 1}\frac{x^2 - 1}{x - 1}lim x→1x−1 x 2−1
In such cases, we factorize both the polynomials such that the common factor cancels out.
limx→1 x 2−1 x−1\lim_{x \to 1}\frac{x^2 - 1}{x - 1}lim x→1x−1 x 2−1
⇒limx→1(x−1)(x+1)x−1\lim_{x \to 1}\frac{(x - 1)(x + 1)}{x - 1}lim x→1x−1(x−1)(x+1)
⇒limx→1(x+1)\lim_{x \to 1}(x + 1)lim x→1(x+1)
⇒limx→1(1+1)\lim_{x \to 1}(1 + 1)lim x→1(1+1)
⇒2
Limits by Rationalization
In this method, indeterminate form is dealt with by rationalizing the function.
limx→∞x 2+x+1−x 2+1\lim_{x \to \infty} \sqrt{x^2 + x + 1} - \sqrt{x^2 + 1}lim x→∞x 2+x+1−x 2+1
Notice that this generates the indeterminate form of ∞ - ∞.
In such cases, we rationalize the expression.
limx→∞x 2+x+1−x 2+1\lim_{x \to \infty} \sqrt{x^2 + x + 1} - \sqrt{x^2 + 1}lim x→∞x 2+x+1−x 2+1
⇒limx→∞(x 2+x+1−x 2+1)(x 2+x+1+x 2+1)(x 2+x+1+x 2+1)\lim_{x \to \infty} (\sqrt{x^2 + x + 1} - \sqrt{x^2 + 1})\frac{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}lim x→∞(x 2+x+1−x 2+1)(x 2+x+1+x 2+1)(x 2+x+1+x 2+1)
⇒limx→∞(x 2+x+1−(x 2+1))(x 2+x+1+x 2+1)\lim_{x \to \infty} \frac{(x^2 + x + 1 - (x^2 + 1))}{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}lim x→∞(x 2+x+1+x 2+1)(x 2+x+1−(x 2+1))
⇒limx→∞x(x 2+x+1+x 2+1)\lim_{x \to \infty} \frac{x}{(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}lim x→∞(x 2+x+1+x 2+1)x
⇒limx→∞x(x 1+1 x+1 x 2+1+1 x 2)\lim_{x \to \infty} \frac{x}{(x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}})}lim x→∞(x 1+x 1+x 2 1+1+x 2 1)x
⇒limx→∞1(1+1 x+1 x 2+1+1 x 2)\lim_{x \to \infty} \frac{1}{(\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}})}lim x→∞(1+x 1+x 2 1+1+x 2 1)1
⇒1(1+1∞+1∞2+1+1∞2)\frac{1}{(\sqrt{1 + \frac{1}{\infty} + \frac{1}{\infty^2}} + \sqrt{1 + \frac{1}{\infty^2}})}(1+∞1+∞2 1+1+∞2 1)1
⇒1 2\frac{1}{2}2 1
Trigonometric limits using Pythagorean Identities
Pythagorean Identities consists of following three identities:
sin 2 x + cos 2 x = 1
1 + tan 2 x = sec 2 x
1 + cot 2 x = cosec 2 x
These identities can be used to substitute in the functions when trigonometric functions show indeterminate forms.
Double angle Identities
Following double angle identities can also be used for substitution in trigonometric functions:
cos2x = 2cos 2 x - 1 = 1 - 2sin 2 x
sin2x = 2sinxcosx
Let's see some examples of these concepts.
Sample Problems
Question 1: Find out the following limit.
limx→1 x 2−3 x+2 x 2−1\lim_{x \to 1}\frac{x^2 -3x + 2}{x^2 - 1}lim x→1x 2−1 x 2−3 x+2
Solution:
limx→1 x 2−3 x+2 x 2−1\lim_{x \to 1}\frac{x^2 -3x + 2}{x^2 - 1}lim x→1x 2−1 x 2−3 x+2
This limit is of the 0/0 form. Factorization can be used here.
limx→1 x 2−3 x+2 x 2−1\lim_{x \to 1}\frac{x^2 -3x + 2}{x^2 - 1}lim x→1x 2−1 x 2−3 x+2
⇒limx→1 x 2−2 x−x+2 x 2−1 2\lim_{x \to 1}\frac{x^2 -2x -x + 2}{x^2 - 1^2}lim x→1x 2−1 2 x 2−2 x−x+2
⇒limx→1 x(x−2)−1(x−2)(x−1)(x+1)\lim_{x \to 1}\frac{x(x -2) -1(x - 2)}{(x- 1)(x + 1)}lim x→1(x−1)(x+1)x(x−2)−1(x−2)
⇒limx→1(x−1)(x−2)(x−1)(x+1)\lim_{x \to 1}\frac{(x - 1)(x -2)}{(x- 1)(x + 1)}lim x→1(x−1)(x+1)(x−1)(x−2)
⇒limx→1(x−2)(x+1)\lim_{x \to 1}\frac{(x -2)}{(x + 1)}lim x→1(x+1)(x−2)
⇒limx→1(1−2)(1+1)\lim_{x \to 1}\frac{(1 -2)}{(1 + 1)}lim x→1(1+1)(1−2)
⇒−1 2\frac{-1}{2}2−1
Question 2: Find out the following limit.
limx→2 x 2−5 x+6 x 2−3 x+2\lim_{x \to 2}\frac{x^2 -5x + 6}{x^2 -3x +2}lim x→2x 2−3 x+2 x 2−5 x+6
Solution:
limx→2 x 2−5 x+6 x 2−3 x+2\lim_{x \to 2}\frac{x^2 -5x + 6}{x^2 -3x +2}lim x→2x 2−3 x+2 x 2−5 x+6
This limit is of the 0/0 form. Factorization can be used here.
limx→2 x 2−5 x+6 x 2−3 x+2\lim_{x \to 2}\frac{x^2 -5x + 6}{x^2 -3x +2}lim x→2x 2−3 x+2 x 2−5 x+6
⇒limx→2 x 2−3 x−2 x+6 x 2−x−2 x+2\lim_{x \to 2}\frac{x^2 -3x -2x + 6}{x^2 -x -2x +2}lim x→2x 2−x−2 x+2 x 2−3 x−2 x+6
⇒limx→2(x−3)(x−2)(x−1)(x−2)\lim_{x \to 2}\frac{(x -3)(x -2)}{(x -1)(x-2)}lim x→2(x−1)(x−2)(x−3)(x−2)
⇒limx→2(x−3)(x−1)\lim_{x \to 2}\frac{(x -3)}{(x -1)}lim x→2(x−1)(x−3)
⇒limx→2(2−3)(2−1)\lim_{x \to 2}\frac{(2 -3)}{(2 -1)}lim x→2(2−1)(2−3)
⇒ -1
Question 3: Find out the following limit.
limx→9 x−3 x−9\lim_{x \to 9}\frac{\sqrt{x} - 3}{x -9}lim x→9x−9 x−3
Solution:
Substitute x = 9:
9−3 9−9=3−3 0=0 0\frac{\sqrt{9} - 3}{9 - 9} = \frac{3 - 3}{0} = \frac{0}{0}9−9 9−3=0 3−3=0 0
This is an indeterminate form, so we need to simplify.
Simplify using conjugate:
Multiply numerator and denominator by the conjugate of the numerator, √x + 3:
x−3 x−9×x+3 x+3=(x−3)(x+3)(x−9)(x+3)=x−9(x−9)(x+3)=1 x+3\dfrac{\sqrt{x} - 3}{x - 9} \times \dfrac{\sqrt{x} + 3}{\sqrt{x} + 3} \= \dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} \= \dfrac{x - 9}{(x - 9)(\sqrt{x} + 3)} \ = \dfrac{1}{\sqrt{x} + 3}x−9 x−3×x+3 x+3=(x−9)(x+3)(x−3)(x+3)=(x−9)(x+3)x−9=x+3 1
Evaluate the simplified limit:
limx→9 1 x+3=1 9+3=1 3+3=1 6\lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}lim x→9x+3 1=9+3 1=3+3 1=6 1
Question 4: Find out the limit of,
l i m x→2 x−2(x+2−2)lim_{x\to2}\frac{x-2}{(\sqrt{x+2}-2)}l i m x→2(x+2−2)x−2
Solution:
Putting the limit x tends to 2 to see the value obtained,
l i m x→2 x−2(x+2−2)=0 0 lim_{x\to2}\frac{x-2}{(\sqrt{x+2}-2)}\=\frac{0}{0}\l i m x→2(x+2−2)x−2=0 0
= Undefined.
As, it is clear that the answer is undefined, rationalize the denominator,
l i m x→2 x−2(x+2−2)=l i m x→2(x−2)(x+2+2)(x+2−2)(x+2+2)=l i m x→2(x−2)(x+2+2)(x−2)=l i m x→2 x+2+2=2+2=4 lim_{x\to2}\frac{x-2}{(\sqrt{x+2}-2)}\=lim_{x\to2}\frac{(x-2)(\sqrt{x+2}+2)}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}\=lim_{x\to2}\frac{(x-2)(\sqrt{x+2}+2)}{(x-2)}\=lim_{x\to2}\sqrt{x+2}+2\=2+2=4 l i m x→2(x+2−2)x−2=l i m x→2(x+2−2)(x+2+2)(x−2)(x+2+2)=l i m x→2(x−2)(x−2)(x+2+2)=l i m x→2x+2+2=2+2=4
Practice Problems
Problem 1: Find the limit:limx→3 x 2−9 x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}lim x→3x−3 x 2−9
Problem 2: Find the limit:limx→2 x 2−4 x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}lim x→2x−2 x 2−4
Problem 3: Find the limit:limx→0 x 2+2 x x\lim_{x \to 0} \frac{x^2 + 2x}{x}lim x→0x x 2+2 x
Problem 4: Find the limit:limx→1 x 3−1 x−1\lim_{x \to 1} \frac{x^3 - 1}{x - 1}lim x→1x−1 x 3−1
Problem 5: Find the limit:limx→−1 x 2−1 x+1\lim_{x \to -1} \frac{x^2 - 1}{x + 1}lim x→−1x+1 x 2−1
Problem 6: Find the limit:limx→0 x 3+8 x+2\lim_{x \to 0} \frac{x^3 + 8}{x + 2}lim x→0x+2 x 3+8
Problem 7: Find the limit:limx→4 2 x 2−8 x x−4\lim_{x \to 4} \frac{2x^2 - 8x}{x - 4}lim x→4x−4 2 x 2−8 x
Problem 8: Find the limit:limx→5 x 2−25 x−5\lim_{x \to 5} \frac{x^2 - 25}{x - 5}lim x→5x−5 x 2−25
Problem 9: Find the limit:limx→1 x 3−x x−1\lim_{x \to 1} \frac{x^3 - x}{x - 1}lim x→1x−1 x 3−x
Problem 10: Find the limit:limx→−2 x 2+4 x+4 x+2\lim_{x \to -2} \frac{x^2 + 4x + 4}{x + 2}lim x→−2x+2 x 2+4 x+4
Summary
Determining limits using algebraic manipulation is a crucial technique in calculus, allowing for the simplification of expressions to find the value a function approaches as the input approaches a specific point. This method is particularly useful for resolving indeterminate forms and simplifying complex expressions. By mastering the techniques of algebraic manipulation for finding limits, students can handle a wide variety of problems in calculus, providing a deeper understanding of the behavior of functions and laying the groundwork for more advanced topics.
Explore Related Articles:
| Limits in Calculus |
| Strategy in Finding Limits |
| Limits by Direct Substitution |
| Calculus | Differential and Integral Calculus |
| Estimating Limits from Graphs |
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quantum mechanics - Eigenstates for $\vec{L}^2, L_z, L_x$ and $L_y$? - Physics Stack Exchange
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Eigenstates for L⃗2,L z,L x L→2,L z,L x and L y L y? [closed]
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Homework-like questions and check-my-work questions are considered off-topic here, particularly when asking about specific computations instead of underlying physics concepts. Homework questions can be on-topic when they are useful to a broader audience. Note that answers to homework questions with complete solutions may be deleted!
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I am asked to find states |j,m⟩|j,m⟩ that are simultaneously eigenstates for L⃗2,L z,L x L→2,L z,L x and L y L y. I know that the L i L i operators do not commute and hence you cannot have a state |ϕ⟩|ϕ⟩ that is common to L i L i and L j L j because that would imply that
[L i,L j]|ϕ⟩=0⃗[L i,L j]|ϕ⟩=0→
Since 0⃗0→ is not an eigenvector by definition, I believe that you cannot have any common eigenstates between L i L i and L j L j. What am I missing?
quantum-mechanics
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edited Apr 8, 2021 at 11:46
Tobias Fünke
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asked Apr 8, 2021 at 11:16
Y2HY2H
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Do you know what is the requirement for operators to have common eigenstates? I don't give the answer because this would invalidate this home work question.my2cts –my2cts 2021-04-08 11:21:23 +00:00 Commented Apr 8, 2021 at 11:21
3 I suspect that you have misunderstood what is being asked.mike stone –mike stone 2021-04-08 11:28:29 +00:00 Commented Apr 8, 2021 at 11:28
@my2cts it’s not actually a HW question. I’m self-studying a course.Y2H –Y2H 2021-04-08 11:35:47 +00:00 Commented Apr 8, 2021 at 11:35
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Note that if two observables do not commute, then there exists no common eigenbasis. However, this in general does not mean that there are no states which are eigenstates of both operators.
For your specific example: You could try to express L^x L^x and L^y L^y in terms of the ladder operators L^+L^+ and L^−L^−. Applying L^x L^x and L^y L^y to |l,m⟩|l,m⟩ and require them to be eigenstates then yields conditions on l l and m m.
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edited Apr 8, 2021 at 11:54
answered Apr 8, 2021 at 11:34
Tobias FünkeTobias Fünke
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I believe the only state that is a simultaneous eigenstate of L 2,L z,L x L 2,L z,L x and L y L y is |l=0,m=0⟩|l=0,m=0⟩ (thus the ground state of the Hydrogen atom) with eigenvalue 0 0 for all the concerned observables. As expected, this doesn't form an orthonormal basis by itself.
It is easy enough to see why this is the only such state in existence. The states |l,m⟩|l,m⟩ are by definition the simultaneous eigenstates of L 2 and L z.L 2 and L z. Now, we need to find specific values of l l and m m for which these states are also eigenstates of both the L x L x and L y L y operator. At first site, this looks impossible as L x L x or L y L y are linear combinations of the two ladder operators L+L+ and L−L− which change the value of m m by one and keeps the value of j j the same; thereby producing a completely different state. The only exception is when the state is such that both the ladder operators annihilate it. This is only possible for the state with |l=0,m=0⟩|l=0,m=0⟩.
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edited Apr 9, 2021 at 5:45
answered Apr 9, 2021 at 4:25
ArnabArnab
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Uniform Distribution
Uniform Distribution
5 min readSep 14, 2022
Uniform Distribution is defined by having a constant probability within given domain.
The uniform distribution is defined by two parameters, a and b,
a is the minimum
b is the maximum
The uniform distribution is written as U(a, b)
Area
Since it is a nice an simple rectangle and we can find the area by using the formula for the area of a rectangle.
Like all probability distributions for continuous random variables, the area under the graph of a random variable is always equal to 1.
From the graph above.
So,
Putting all together we get.
There are 2 types of uniform distribution, continuous and discrete.
Continuous Uniform Distribution
In continuous uniform distribution the expected output takes a value in a specified range. Ex: In a range 0 to 1 it can take any value such as 0.1, 0.2, 0.22, etc.
Probability Density Function (PDF)
The Probability Density Function (PDF) for a continuous uniform distribution taking values in the range a to b is:
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) for a continuous uniform distribution taking values in the range a to b is:
Probability
From any continuous probability density function we can calculate probabilities by using integration.
Mean or Expected Value
The mean of a continuous uniform distribution is.
Variance
The variance of a continuous uniform distribution is given by.
Standard Deviation
The standard deviation of a continuous uniform distribution is given by.
Example
You arrive into a building and are about to take an elevator to the your floor. Once you call the elevator, it will take between 0 and 40 seconds to arrive to you. We will assume that the elevator arrives uniformly between 0 and 40 seconds after you press the button. In this case a = 0 and b = 40.
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Join Medium for free to get updates from this writer.
To calculate the probability that elevator takes less than 15 seconds to arrive we set d = 15 and c = 0.
Discrete Uniform Distribution
In discrete uniform distribution the expected output takes a finite set of values. Ex: 1, 2, 3, 4, etc.
Probability Mass Function (PMF)
A discrete uniform random variable X with parameters a and b has Probability Mass Function (PMF).
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) for a uniform distribution is given by.
Probability
From any discrete probability mass function we can calculate probabilities by using a summation.
Mean or Expected Value
The mean of a discrete uniform distribution is.
Variance
The variance of a discrete uniform distribution is given by.
Standard Deviation
The standard deviation of a discrete uniform distribution is given by.
Example
Lets take an example of throwing a Dice.
To calculate the probability that the dice lands on 2 or 3 we set d = 3 and c = 2.
Standard Uniform Distribution
The standard uniform distribution is where a = 0 and b = 1 and is common in statistics, especially for random number generation. Its expected value is 1/2 and variance is 1/12.
Mass vs Density Function
A probability mass function (PMF) differs from a probability density function (PDF) in that the latter is associated with continuous rather than discrete random variables.
Probability mass functions are used for discrete distributions. It assigns a probability to each point in the sample space. Whereas the integral of a probability density function gives the probability that a random variable falls within some interval.
Takeaways
Uniform distributions are probability distributions with equally likely outcomes.
In a discrete uniform distribution, outcomes are discrete and have the same probability.
In a continuous uniform distribution, outcomes are continuous and infinite.
I hope this article provides you with a good understanding of Uniform Distribution.
If you have any questions or if you find anything misrepresented please let me know.
Thanks!
Uniform Distribution
Machine Learning
Artificial Intelligence
Data Science
Statistics
## Written by Gajendra
232 followers
·94 following
| AWS MLS, SAA, CLF | MIT - ADSP | Software Engineer | Data Scientist | Machine Learning | Artificial Intelligence | Hobby Blogger |
Responses (1)
Write a response
What are your thoughts?
Shouvik Pal
Jun 26, 2023
```
Hi Gajendra - Thanks for the write-up.
Could you pls correct a typo:
The mean of a continuous uniform distribution is mentioned as (b-a)/2. It should be (a+b)/2.
```
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4th Grade Common Core: 4.NBT.4
Common Core Identifier: 4.NBT.4 / Grade: 4
Curriculum: Number And Operations In Base Ten: Use Place Value Understanding And Properties Of Operations To Perform Multi-Digit Arithmetic.
Detail: Fluently add and subtract multi-digit whole numbers using the standard algorithm.
213 Common Core State Standards (CCSS) aligned worksheets found:
Subtraction Task Cards (4-Digits)
Students solve to find the answers to the 4-digit subtraction problems on the task cards. Then check by scanning the barcode.
Level:
View PDF
4-Digit Subtraction #2
Here's another 4-digit subtraction worksheet, similar to the one above. This printable has ten vertical problems and two word problems.
Level:
View PDF
Subtraction Crossword (4 and 5-Digit)
Subtract to find th four and five digit differences. The plug the numbers into the puzzle.
Level:
View PDF
Addition Crossword (4 and 5-Digit)
Find the sums and use the answers to fill in the puzzle.
Level:
View PDF
Addition Crossword (Three Addends)
Add the three digit numbers and plug them into the math crossword puzzle.
Level:
View PDF
Addition Practice (4-digit Numbers)
These word problems require students to find the sums of four-digit addends.
Level:
View PDF
Balloon Counting (4-Digit)
Complete the number lines by filling in the missing numbers on each set of balloons.
Level:
View PDF
Addition: 3 AddendsFREE
Four-digit column addition problems with three addends; Includes one word problem
Level:
View PDF
Subtraction Puzzle: Earth Day
Subtract the four-digit minus three digit subtraction problems. The complete the puzzle based on the answers.
Level:
View PDF
Addition Crossword (4-Digit)FREE
Add the numbers and write the answers in the crossword puzzle.
Level:
View PDF
Dino Picture PuzzleFREE
Solve the addition problems. Then match the sums with the picture puzzle pieces. Put the puzzle together to reveal a dinosaur.
Level:
View PDF
Subtraction (4-Digit Numbers)
This QR scan worksheet has word problems with 4-digit numbers that students must subtract.
Level:
View PDF
4-Digit SubtractionFREE
Solve the four-digit subtraction problems to find the differences. (example 6,397 - 2,976)
Level:
View PDF
Addition: 4 Addends
Four-digit column addition problems with four addends; Includes one word problem
Level:
View PDF
Subtracting Word Problems: Making Change (Medium)
Subtract the amount each person gives the cashier from the amount due. Requires subtraction across zero.
Level:
View PDF
Addition (4-Digit Numbers)
This worksheet contains four QR codes. Students scan each code to reveal a word problem. They solve the problem in the space provided on the worksheet.
Level:
View PDF
4-Digit Subtraction (Across Zero)
Solve the subtraction problems to find the differences. Requires subtraction across zero with 4-digit numbers. (example 4,000 - 1,374)
Level:
View PDF
3-Digit Numbers, 4 Addends
Three-digit column addition problems with three or four addends; Includes one word problem
Level:
View PDF
Graph Paper Subtraction Across Zero
Graph paper subtraction will ensure kids can line up their place value columns perfectly.
Level:
View PDF
Subtracting Money (Across Zero)
Subtract the money amounts on this page. Includes word problems and standard problems.
Level:
View PDF
4-Digit: Astronaut Picture Puzzle
Solve the 4-digit minus 3-digit subtraction problems. Glue the answers on the grid to see an astronaut picture.
Level:
View PDF
M.L.K. Math - Adding (3-Digit Numbers)
Find the sums of three-digit addends. These problems require knowledge of regrouping. Includes an MLK-themed word problem and a picture of King.
Level:
View PDF
Subtraction Practice (3-Digits/ Regrouping)FREE
Practice subtracting 3-digit numbers with borrowing / regrouping.
Level:
View PDF
Subtraction: 3-Digits
Subtract the three-digit numbers to find the differences. All problems require some borrowing. There are also two word problems.
Level:
View PDF
Subtraction Worksheet (3-digits / Regrouping)
Practice 3-digit Subtracting with borrowing / regrouping. Check by adding.
Level:
View PDF
More & Less (4-Digit)
Students are given a number. They must add 1, 10, and 100 to the number. They must also subtract 1, 10, and 100.
Level:
View PDF
3-Digit Numbers, 3 Addends
Three-digit column addition problems with three addends; Includes one word problem
Level:
View PDF
Word Problems: Addition and Subtraction
Students must determine whether they need to add or subtract to solve each problem.
Level:
View PDF
Graph Paper Subtraction: Money Across Zero
Subtracting money with zero on graph paper. One digit per box will help kids line up columns. (example: $9.00 - 3.48)
Level:
View PDF
Subtraction Crossword (4-Digit)
Find the answers to the subtraction problems and complete the number crossword puzzle.
Level:
View PDF
M.L.K. Math - Subtraction (3-Digit Numbers)
Subtract three-digit numbers. These problems require borrowing/regrouping. There is an MLK word problem and picture.
Level:
View PDF
Math Story - Camping (Grades 4-5)
Read the story of Wendy's Camping Adventure, then answer the math questions the follow.
Level:
View PDF
Monster Math (Addition:3-Digits w/ Regrouping)
This monster-themed practice worksheet has three-digit adding problems. These problems require students to regroup/carry.
Level:
View PDF
Subtraction Bingo (3-digits / Regrouping)
Solve the subtraction problems and color the answers on the bingo boards. Can you get a bingo?
Level:
View PDF
Subtraction Across Zero Worksheet
Practice 3-digit Subtracting problems that require borrowing across zero
Level:
View PDF
Dragon Subtraction (3-digits / Regrouping)
This subtraction worksheet has pictures of cute dragons. 3-digits with borrowing.
Level:
View PDF
Word Problems (2 & 3 digits)
These subtraction story problems involve borrowing. Students work with 2 and 3-digit numbers on this activity.
Level:
View PDF
3-Digit Numbers, 3 Addends Word Problems #1
Four word problems that require students to use column addition to solve; Includes space for students to show their work
Level:
View PDF
Addition (3-digit)
Read and solve these 3-digit addition word problems. There is plenty of space for students to show their work.
Level:
View PDF
Graph Paper Subtraction: Money
Subtracting money on grid paper. One digit per box ensures kids line up numbers correctly. (example: $9.26 - 2.88)
Level:
View PDF
Grid Puzzle Subtraction - Across Zero
Use the grid puzzle to solve the subtraction problems; 4-digit subtraction with zeros. (example 3,400 - 454)
Level:
View PDF
3-Digit Subtraction - Hanukkah
Students will need to regroup to solve the 3-digit subtraction problems on this worksheet. This page is has Hanukkah-related illustrations.
Level:
View PDF
Toy Shop: Subtraction Across Zero
Add to find the total cost of two toys pictured on the shelf. Then subtract across zero to make change.
Level:
View PDF
Subtraction: Easter Eggs (3-Digit Subtraction)
This page has 3-digit subtraction problems, written on Easter eggs. After students solve all of the problems, they can color the eggs.
Level:
View PDF
Sea Animal Subtraction Picture ( 3-Digit / Regrouping)
These subtraction problems have 3-digit quantities and require borrowing. The worksheet features pictures of sea animals.
Level:
View PDF
Snowmen Subtraction Picture (Money)
Subtraction problems with money amounts; Winter snowmen pictures
Level:
View PDF
Graph Paper Subtraction (3-digits / Regrouping)
Subtracting on graph paper gives kids plenty of room to write their numbers and illustrates place value
Level:
View PDF
Subtracting Money (No Zeros)
Subtract the money amounts; Requires regrouping, but does not subtract across zero
Level:
View PDF
3-Digit Numbers, 3 Addends Word Problems #2
More story problems in which students practice column adding
Level:
View PDF
Addition Puzzle: Earth DayFREE
Add to find the sums, then use the answers to glue the picture puzzle together.
Level:
View PDF
3-Digit Subtraction with Place Value Blocks
Use base 10 blocks (place value blocks) to solve the subtraction problems on this worksheet.
Level:
View PDF
Subtraction Arrows (2-Digits / Regrouping)
Subtraction arrow chain problems; 2-digits minus 1-digit; Regrouping
Level:
View PDF
Turkey Addition (Regrouping)
Adding problems with 2-digits; Requires regrouping; Thanksgiving turkey pictures.
Level:
View PDF
Michael's Birthday Money (Mini-Book)
Michael receives $9.99 for his 9th birthday. Subtract each time Michael spends money to show how much is left.
Level:
View PDF
Subtraction Mixed Review
Mixture of different types of subtraction problems, including a couple of word problems. (3 and 4-digit)
Level:
View PDF
C9 - Seven Thousand, One Hundred Thirty-Seven
Lucky sevens - there's a pair of them in the mystery number. There's also a pair of unlucky thirteens in this four-digit number.
Level:
View PDF
Missing Addends Activity
Use subtraction to find the missing addend in each equation.
Level:
View PDF
Subtract and Color: Castle
Color the picture of the knight, dragon, and castle. Then solve the subtraction problems.
Level:
View PDF
Subtraction (3 digits - 2 digits)
Rewrite each problem vertically, then solve. Students required to "line up the digits" correctly.
Level:
View PDF
Addition (2-digit with regrouping)
Solve these two-digit addition word problems. All problems require carrying (regrouping).
Level:
View PDF
Addition Bar Models With Work Space
These seven addition bar models (or tape diagrams) have space for students to show their work on the right-hand side.
Level:
View PDF
Variety Worksheet (5-Digit Subtraction)
Practice several different types of problems with this 5-digit subtraction variety worksheet.
Level:
View PDF
Addition Boxes (5-Digit Addends)
Solve the addition problems in the boxes. This worksheet has space for students to rewrite problems vertically (if needed) and show their work.
Level:
View PDF
Find the Mistakes
These math problems were already solved, but there are mistakes. Explain how the problems were incorrectly solved. Then calculate the correct sums.
Level:
View PDF
Mixed Math: C1FREE
The mixed review problems on this page include adding, subtracting, counting money
Level:
View PDF
ABC Order - Word Sets (B-Thanksgiving)
Write each group of Thanksgiving spelling words in ABC order. Includes review and challenge words.
Level:
View PDF
Daily Word Problems D-141 through D-145
Assigning one of these word problem worksheets every day is a great way to make sure your students are keeping their math skills fresh!
Level:
View PDF
5-Digit SubtractionFREE
Solve the 5-digit subtraction problems. This worksheet includes 10 vertical problems and 2 word problems.
Level:
View PDF
3-Digit Addition: Match Puzzle Picture (Gingerbread)
Find the sum for each 3-digit addition problem. Then cut out the puzzle pieces. Match the number (sum) on each puzzle piece with the corresponding addition problem. The completed puzzle shows a lovely gingerbread house.
Level:
View PDF
Task Cards
This file has 30 task cards with vertical and horizontal money addition problems.
Level:
View PDF
2-Digit Numbers, 4 Addends
Double-digit column addition problems with three or four addends; Includes one word problem
Level:
View PDF
SCM Subtraction: 3 Digits
After decoding the numbers using the key, students will subtract to find the differences.
Level:
View PDF
Daily Word Problems D-111 through D-115
Students will review lines of symmetry, fill in a chart, and more with this set of daily word problems.
Level:
View PDF
Variety Worksheet: 3-Digit Addition
Use this worksheet to help students practice adding with three digits.
Level:
View PDF
Daily Word Problems D-11 through D-15
This set includes a problem related to number patterns, solving 4-digit operations, and reading & interpreting data on a table.
Level:
View PDF
Even & Odd Sums (4 Digits)
Solve each 4-digit addition problem. Then cut out the cards and sort them into two different categories: Odd Sums and Even Sums.
Level:
View PDF
Math Puzzle Picture (3-Digit Addition)FREE
Find the sums for these 3-digit addition problems. Then use the answers to put the puzzle together and see what this Day of the Dead mystery picture looks like.
Level:
View PDF
Daily Word Problems E-1 through E-5FREE
This file has 5 worksheets, one for each day of the week. Topics include adding with fractions, finding the area of a rectangle, and more.
Level:
View PDF
Find the Mistakes (Multiple Addends)
Study the completed math problems and look for errors. Explain the mistakes and solve correctly.
Level:
View PDF
Daily Word Problems D-81 through D-85
This week we'll practice comparing fractions, reading and interpreting a table, and solving multistep problems with various operations.
Level:
View PDF
Daily Word Problems E-41 through E-45
This set of daily word problems features skills such as place value, division with remainders, and addition with decimals.
Level:
View PDF
Learning About Leap Years
This worksheet has questions about the number of days in a Leap Year and non-Leap Year. Students are also asked to circle the LYs in a list. They are also asked to explain how you calculate which years are LYs and which are not.
Level:
View PDF
Magic Digits Game (3-Digit Subtraction)
In this game, students must figure out the correct locations for the missing digits in each subtraction problem. Includes 12 task cars and a student answer page.
Level:
View PDF
Task Cards: 4-Digit Subtraction
These thirty task cards have 4-digit subtraction problems on them. You can laminate the cards, and have students solve the problems with a dry-erase marker. Or use them to build a learning center. Or post a card each morning for your students to solve.
Level:
View PDF
Addition-Subtraction Bar Models (5-Digit)
This sheet has six bar models. Students add or subtract to find the missing numbers. This worksheet has mostly 5-digit numbers.
Level:
View PDF
D6 - Five Hundred Eighty Seven
How many years are in a century? How many degrees are in a right angle? How many days in March? How many days in a leap year? Add the answers to find the number.
Level:
View PDF
Secret Code Math: 5-Digit Subtraction Across Zero
Use the key to decode the symbols and find the 5-digit numbers. Then solve the subtraction problems and write the answers in the secret code.
Level:
View PDF
Graph Paper Addition (5-Digit Addends)
These addition problems are written on graph paper squares to help students better see the place value column alignment.
Level:
View PDF
Find the Mistakes (Version 1)
On this worksheet, students will look carefully at the completed subtraction problems for mistakes. They will explain the mistakes made and solve the problems correctly.
Level:
View PDF
The Late Broom (Subtracting Money Across Zero)
Why was the broom late? To discover the answer to this riddle, kids must subtract pairs of money amounts.
Level:
View PDF
Dollar Money Bonds #2 (Subtraction)
Subtract find the missing amount for each number bond. This is an alternate version of the worksheet above.
Level:
View PDF
Find the Mistakes
Look for the mistakes in the way two subtraction problems were solved and then solve them correctly.
Level:
View PDF
Bees Going to School (Add/Subtract; 3 Digit)
How do bees get to school in the morning? To find out, figure out the answers to the 3-digit addition and subtraction problems.
Level:
View PDF
Math Riddle: Duck (4 Digit Add & Subtract)
What did the duck eat for lunch? Solve the 4-digit addition and subtraction problems and match the letters with the answers to find out.
Level:
View PDF
Word Problems: 2-Digit Addition (No Regrouping)
Solve these double-digit addition word problems. These problems do NOT require regrouping.
Level:
View PDF
Secret Code Math: Subtraction 4-Digit Numbers
First use the cypher key to decode the mystery symbols into regular numbers. Then subtract.
(example: 6,526 - 2,818)
Level:
View PDF
Daily Word Problems D-126 through D-130
This week students will use their geometry, fraction, and other math skills to complete their daily word problems.
Level:
View PDF
Variety Worksheet: Adding Money
Practice adding money with this worksheet. It includes several styles of math problems, such as word problems and fill-in-the-blanks.
Level:
View PDF
Daily Word Problems D-26 through D-30
Answer math questions about real-world scenarios related to money and making change, line plots, and division.
Level:
View PDF
Subtracting Numbers: Even/Odd Sort
Cut apart the math cards and solve the 4-digit subtraction problems. Then sort the differences into two categories on the t-chart: "Even Differences" and "Odd Differences."
Level:
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Daily Word Problems D-66 through D-70
Continue practicing 4th grade math skills with this next set of daily word problems. It covers skills like reading and interpreting data from a table, adding fractions, and more.
Level:
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Daily Word Problems E-16 through E-20
Here are five more worksheets with daily word problems for 5th graders. Topics include finding volume, calculating time, and comparing decimals.
Level:
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Secret Code Math: 4-Digit Addition
Use the key to decode the addends in the secret symbol numbers. Then solve the 4-digit addition problems.
Level:
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Daily Word Problems D-96 through D-100
Print out this file to continue practicing grade-appropriate word problems with your students each day.
Level:
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Secret Santa Addition (Money)
Add the money amounts to determine how much you'll spend on your family members at the Secret Santa Market.
Level:
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Math Buzz: Week 5 Worksheets 21 through 25
Within this set you'll find a variety of word problems, geometry, place value up to 7 places, fractions, advanced addition and subtraction.
Level:
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Balance Scales
On this worksheet, students must balance the scale by writing in missing addends or sums. Note: This may be a tricky activity for younger students.
Level:
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Shape Addition (4-Digit)
At the top of this page are numbers written inside of polygons. Students add numbers in similar shapes. For example: Find the sum of the numbers in the octagons.
Level:
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Subtraction Circles
Choose a number from each wheel, and subtract the number from the second wheel from the number from the first wheel.
Level:
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Easter Bunnies Math Picture Puzzle (3-Digit Subtraction)
The answers to these subtraction problems are the key to unscrambling this mystery picture! Solve the equations, and then cut and clue the pictures onto the squares with the corresponding answers.
Level:
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Math Riddle: Rabbits on Vacation
Riddle: How does a rabbit travel? To find the answer, solve three-digit subtraction problems. Then match the answers with the letters to decode the solution to the riddle.
Level:
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Adding and Subtracting with Bar Models
In this math activity, students will find the missing value for each bar model, then they will write an addition and subtraction number sentence for each one.
Level:
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Even and Odd Subtraction Sort
Students will solve the subtraction problems on the first page of this file, and then cut each card out. Then they will sort each one into either the "Odd Answers" and "Even Answers" section on the second page.
Level:
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Find the Mistakes
These two addition problems have been solved incorrectly. Can you find the errors? Tell how each problem could be solved correctly.
Level:
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Find the Mistakes (3 Addends)FREE
These addition problems were solved incorrectly. Examine them carefully and explain the mistakes.
Level:
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Math Riddle: Captain's Choice
What do boats eat in the morning for breakfast? To solve the riddle, students will need to complete column addition problems. Each one has three, 3-digit addends.
Level:
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Addition-Subtraction Bar Models (Tape Diagrams)
This page contains tape diagrams (also known as bar models). Students will need to add or subtract to find the missing numbers.
Level:
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Graph Paper Subtraction (5-Digit)
These subtraction problems are written on graph paper squares so students can better see the alignment of each column.
Level:
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Math Buzz: Week 3 Worksheets 11 through 15
On these printable worksheets, your fifth graders will review parallel/intersecting/perpendicular lines, long division (without remainders), estimating basic operations, factors, area models, and word problems.
Level:
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Find the Mistakes
Look carefully at the completed addition problems. There are mistakes. Re-solve each problem to find the correct sum. Then explain the mistakes in the original solutions.
Level:
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Addition Circles: Practice Adding
Choose numbers from each circle. Add them together to find the sums. Do this six times.
Level:
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Daily Word Problems D-136 through D-140
Keep working on those math skills with these engaging daily word problem worksheets!
Level:
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Subtraction Puzzle Picture: Groundhog Day (3-Digit Subtraction)
Complete the subtraction problems and use your answers to put together the puzzle pieces and reveal a fun Groundhog Day image!
Level:
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3-Digit Subtraction: Match Puzzle Picture (Santa)
Solve the 3-digit subtraction problems. (Most problems require regrouping.) Then cut out the puzzle parts. Match each difference with the correct problem to complete the puzzle. The completed puzzle shows Santa placing gifts below a tree.
Level:
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Task Cards Subtracting Money
This file has 30 task cards with money subtraction on them.
Level:
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Daily Word Problems D-76 through D-80
Review a variety of operations and math skills with these next 5 word problems.
Level:
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Daily Word Problems E-31 through E-35
These word problems reinforce key math skills, such as finding area and performing operations with decimals and fractions.
Level:
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SCM: Multiple Addends Up to 4 Addends; 4 Digits
These coded problems have up to 4 addends, with 2, 3, or 4 digits in each.
Level:
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Daily Word Problems D-106 through D-110
Skills in this batch of worksheets include finding area and perimeter, as well as more geometry skills, reading a table, and operations with decimals and fractions.
Level:
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Variety Worksheet: Subtracting Money
This page includes word problems, fill-in-the-blanks, and simple subtraction problems.
Level:
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Daily Word Problems D-6 through D-10
Four out of five of these problems are multi-step. Skills include basic operations with whole numbers, counting money, and interpreting data from a table.
Level:
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Addition Task Cards
These cards have addition problems with 4-digit addends. Use these cars in your math learning center, with your classroom document camera, or as cards for math games.
Level:
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Shape Subtraction: 3 Digits
On the top of this page, there are pictures of shapes with 3-digit numbers in them. Students find differences between pairs of shapes. For example: _Subtract the number in the pentagon from the number
in the octagon._
Level:
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5-Digit Subtraction (Across Zero)
Solve these subtraction problems to find the differences. Each problem requires subtracting across zero.
Level:
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Subtraction Variety (3-digit)
Solve a variety of subtraction problems, including word problems, vertical and horizontal equations, and missing digit questions.
Level:
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Magic Digits Game (2-Digit Subtraction)
Students are given four digits. They must place them in the squares correctly to make the given difference. This game has 12 task cards and a student answer page.
Level:
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Addition-Subtraction Bar Models With Work Space (5-Digit)
This page has 5-digit addition-subtraction tape diagrams (aka bar models). There are seven problems on this file.
Level:
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Secret Code Math: 5-Digit Subtraction with Borrowing
Decode the picture symbols into 5-digit numbers. Then subtract and write the answers in the secret code too.
Level:
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Task Cards: 5-Digit Addends
This file contains 30 task cards. Each card has a vertical addition problem with 5-digit addends.
Level:
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Find the Mistakes
This worksheet has pictures of subtraction problems that somebody has already tried to solve. But there are mistakes in their work. Find the errors and explain where they went wrong.
Level:
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Math Crossword - Add, Subt, Mult, Div
Add, subtract, divide, and multiply to solve this math crossword. (Grade 4 and up)
Level:
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A Skunk Joke (Adding Money)
Did you hear the joke about the skunk? On this worksheet, students calculate the sums of two money amounts. Then they can use the sums to find the answer to the riddle.
Level:
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Dollar Money Bonds #1 (Subtraction)
Use subtractions to find the missing amount for each number bond. Each bond adds up to $1, so this skill will require subtracting across zero.
Level:
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Task Cards: Subtraction (with Regrouping)
This PDF file has a class set of task cards. Use them to play math games, in your math learning center, or as part of your lesson with your document camera. You could also laminate these cards and kids can solve problems with dry-erase markers.
Level:
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5-Digit Subtraction: Astronaut Puzzle Picture
Solve the 5-digit subtraction problems to unscramble the puzzle picture.
Level:
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Addition Picture Puzzle: Hot Air Balloon
Add the money amounts to find the sums. Then cut out the picture puzzle pieces. Match each numbered (sum) piece to the correct problem. The completed picture shows a hot air balloon.
Level:
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Math Riddle: Blue Elephant (5 Digit Add & Subtract)
Add and subtract 5-digit numbers. Find the answer to all problems to solve the riddle: What should you do when you see a blue elephant?
Level:
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SCM Subtraction: 4 Digits (Across Zero)
Translate the picto-code numbers into "regular numbers" and then solve. All problems require students to subtract 4-digit numbers across zero. (example: 5,000-2,614)
Level:
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Daily Word Problems D-121 through D-125
Continue reviewing important math skills with more fourth grade word problems this week.
Level:
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Variety Worksheet: 5-Digit Addition
Practice a variety of 5-digit addition skills with this worksheet.
Level:
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Daily Word Problems D-21 through D-25
Calculate elapsed time for a movie. Count to find the value of dollar bills and coins. Solve two multi-step problems using the four basic operations.
Level:
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Subtraction: Even/Odd Sort Activity
Cut out the math cards. Solve the subtraction problems. Then sort the answers into "even differences" and "odd differences." After that, students glue the math cards in the correct columns on the t-chart.
Level:
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4-Digit Subtraction - Figure Skater Picture Puzzle
Solving this puzzle requires subtracting 3-digit numbers from 4-digit numbers.
Level:
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Daily Word Problems E-11 through E-15
These five worksheets reinforce key math skills for 5th grade. Each worksheet includes a daily word problem, most of which involve multiple steps. Topics include subtracting fractions, dividing money, and interpreting a chart.
Level:
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Secret Code Math: 3-Digit Addition
Decode the secret numbers using the cypher key at the top of the page. Then add them together.
Level:
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Daily Word Problems D-91 through D-95
Compare fractions, identify a number pattern, perform operations with money, and more in this week's set of daily word problems.
Level:
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Daily Word Problems E-51 through E-55
Skills covered in these daily worksheets include reading and interpreting frequency charts, reducing fractions, and adding fractions with unlike denominators.
Level:
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Math Buzz: Week 3 Worksheets 11 through 15
The third week of Math Buzz for 4th grade features skills like: division, place value, advanced subtraction, rounding, fractions, area, and perimeter.
Level:
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Magic Digits Game
In this game, students will work with task cards that have digits. They must arrange the digits to complete the addends in the addition problem.
Level:
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Cut-and-Glue Addition Puzzle: Rocket
First solve the 4-digit plus 4-digit problems. Then glue the picture puzzle pieces over the answers to make a picture of a rocket ship blasting through outer space.
Level:
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SCM Subtraction: 3 Digits (Across Zero)
These 3-digit subtraction problems require students to borrow across zero. (example: 400-126)
Level:
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Daily Word Problems D-116 through D-120
Practice finding perimeter and area, solve for a missing fraction in an equation, review obtuse angles, and more with this set of word problems.
Level:
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Variety Worksheet: 4-Digit Addition
This worksheet includes a variety of 4-digit addition problems, including two word problems, horizontal and vertical equations, and more.
Level:
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Daily Word Problems D-16 through D-20
Use addition, subtraction, multiplication, and division to solve these five single and multi-step problems.
Level:
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Shape Math: Subtracting Money
At the top of the page, students are presented with a series of shapes. They subtract numbers from pairs of shapes and find the differences. For example: Subtract the number in the rhombus from the number in the trapezoid.
Level:
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Addition Arrows
Addition arrow chains; Repeated addition.
Level:
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Math Puzzle Picture (3-Digit Subtraction with Regrouping)
Solve the 3-digit subtraction problems and use the answers to find out what the puzzle picture looks like.
Level:
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Daily Word Problems E-6 through E-10FREE
Five worksheets are included in this file. Each worksheet features a daily math word problem. Topics include frequency charts, place value, and more.
Level:
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Find the Mistakes (2 Digits)
Look for mistakes in these completed addition problems. Explain the errors. Then solve.
Level:
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Daily Word Problems D-86 through D-90
These word problems use realistic scenarios to help students practice important math skills, such as adding and subtracting fractions, reading a table, and more.
Level:
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Daily Word Problems E-46 through E-50
These worksheets cover topics such as rounding decimals to the nearest whole number, converting improper fractions to mixed numbers, and more.
Level:
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3-Digit Addition with Regrouping
Add the 3-digit numbers starting with the ones, then tens, then hundreds. Boxes are available to show regrouping/carrying.
Level:
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Magic Digits Game (3-Digit Addition)
Figure out where to place the digits for each addition problem.
Level:
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Task Cards: 4-Digit Subtraction Across Zero
Use these task cards for learning centers, math lessons, math games, or review work. Most subtraction problems require borrowing across zero. Includes 30 task cards and an answer key.
Level:
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Money Operations with Bar Models
Find the missing value for each bar model. Then write an addition and subtraction number sentence for each.
Level:
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Task Cards: 5-Digit Subtraction
Each of these task cards features a unique 5-digit subtraction problem. You can use the cards in learning centers, post a card each morning for your students to solve, or use them any other way you like.
Level:
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Odd and Even Sums
Add each pair of addends (up to 5 digits). Then sort them into two categories: odd sums and even sums.
Level:
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Find the Mistakes (Version 2)
Here's another "Find the Mistakes" worksheet. This one has different problems than the worksheet above.
Level:
View PDF
Math Riddle: Amazing Talking Dog
Find the sum for each problem. All addition requires carrying or regrouping. Use the answers to solve the riddle: What's more amazing than a talking dog?
Level:
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Addition Bar Models (Tape Diagrams)
Write the missing number for each bar model (aka tape diagram or part-part-whole diagram). Then write an addition and a subtraction problem to go along with each diagram.
Level:
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Subtraction Boxes
This worksheet has 12 subtraction problems with room for students to write them vertically and solve.
Level:
View PDF
Math Buzz: Week 1 Worksheets 1 through 5FREE
This is the first week of the 5th grade math buzz series. This file contains 5 worksheets, reviewing basic skills from the fourth grade. Skills covered include: adding and subtracting 5-digit whole numbers, factor pairs, lines of symmetry, rounding, and equivalent fractions.
Level:
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Thanksgiving Grocery Math
Students can cut out pictures of Thanksgiving foods from a grocery store ad paper. Then add up the prices to find the total cost of the items.
Level:
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Addition Circles Activity
Add six pairs of numbers. Choose numbers from the circles at the top of the page.
Level:
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Daily Word Problems D-131 through D-135
Print out these pages and test your students' skills in changing fractions to decimals, finding area and perimeter of irregular shapes, and more!
Level:
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Adding 5-Digit NumbersFREE
This page has vertical problems with 5-digit addends, followed by two word problems.
Level:
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Math Puzzle Picture (Cut and Glue)
First solve 16 3-digit subtraction problems. Then cut out the numbered puzzle pieces. Then attach each numbered piece onto the correct problem. This reveals a mystery picture of a school bus with students.
Level:
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Shape Subtraction
Subtract numbers in the shapes shown. For instance: Subtract the number in the hexagon from the number in the pentagon.
Level:
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Daily Word Problems D-71 through D-75
Solve multistep word problems relating to realistic scenarios with this week's daily word problems.
Level:
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Daily Word Problems E-26 through E-30
These worksheets reinforce key 5th grade math skills. Topics include reading and interpreting a bar graph and converting measurements.
Level:
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Secret Code Math: 3 Addends, 3 Digits
Use the cypher key at the top of the worksheet to decode the addends in these column addition problems.
Level:
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Daily Word Problems D-101 through D-105
Practice fourth grade math skills involving geometry, fractions, and more this week!
Level:
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Variety Worksheet: 4-Digit Subtraction
This worksheet includes 2 word problems, vertical and horizontal equations, and fill-in-the-missing-number questions.
Level:
View PDF
Daily Word Problems D-1 through D-5FREE
This file has 5 pages, with a word problem for each day of the week. There are 3 multi-step problems and 2 single-step. Add, subtract, multiply, and/or divide to solve.
Level:
View PDF
Addition Practice: Task Cards
This file has 30 addition task cards. Each one has a problem with 3-digit addends. There's also a student answer worksheet. Makes a great learning center.
Level:
View PDF
Shape Addition (Money)
At the top of this page are money amounts with shapes around them. Add the numbers in similar shapes together. For example: Find the sum of the numbers in the triangles.
Level:
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Task Cards: 5-Digit Subtraction Across Zero
Each task card includes a problem that deals with subtraction across zero. You can use these task cards in many different ways depending on what you're doing with your students or class.
Level:
View PDF
Easter Chick Math Picture Puzzle (3-Digit Addition)
Students can use the solutions to these equations to unscramble an adorable Easter picture.
Level:
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Lazy Skeleton (4-Digit Column Addition)
What do you call a skeleton who won't work? This worksheet has a set of tricky 4-digit column addition problems.
Level:
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Addition-Subtraction Bar Models Including Work Space
Here is another tape diagram (bar model) worksheet with seven problems. This version has empty space for students to show their work on the right-hand side.
Level:
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5-Digit Subtraction: Mixed Review
This mixed-review worksheet instructs students to subtract the 5-digit numbers and then check their work with addition.
Level:
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Addition Circles
Choose a number from each of the two circles and add them together. Choose six different pairs of numbers to add.
Level:
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Find the Mistakes
These completed subtraction problems have mistakes in them. Figure out what's wrong with each one and explain.
Level:
View PDF
Math Crossword - Add, Subt, Mult, Div
Another division, multiplication, addition, and subtraction puzzle. (Grade 4 and up)
Level:
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Cookie Went to the Doctor (4-Digit Addition)
Why did the cookie visit his doctor? The answer will make your students laugh out loud. Solve 4-digit addition problems to decode the punchline.
Level:
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Adding Activity: Addition Circles
Choose a money amount from each wheel. Then calculate the sum.
Level:
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Daily Word Problems D-146 through D-150
You're setting your students up for success by having them complete these daily word problems. If your students have completed the whole level D series, they may be ready to move on to level E!
Level:
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5-Digit Subtraction Crossword
Solve these subtraction problems and use the differences to complete the math crossword.
Level:
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Cut-and-Glue Addition Puzzle: Parrot
Find the sums. Then cut out the numbered puzzle pieces. Match each puzzle piece (sum) with the correct addition problem.
Level:
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Math Riddle: Graveyard Fences
Why are there fences around graveyards? To discover the answer to this math riddle, students will answer a series of mental math addition problems. Addends are multiples of 10, 100, or 1,000. (example: 4,000+2,000)
Level:
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Assoc. & Comm. Properties - Intermediate
This intermediate-level worksheet features larger numbers into the thousands and money amounts.
Level:
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2-Digit Numbers, 3 Addends
Double-digit column addition problems with three addends; Includes one word problem
Level:
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D9 -Ninety-Seven Thousand, Four Hundred One
Start with the product of a dozen and a dozen, then double it. Then add ten thousand and then another eighty thousand.
Level:
View PDF
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https://www.savemyexams.com/dp/physics/ib/23/sl/revision-notes/nuclear-and-quantum-physics/radioactive-decay/binding-energy-per-nucleon-curve/
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IBPhysicsDPSLRevision NotesNuclear & Quantum PhysicsRadioactive DecayBinding Energy per Nucleon Curve
Binding Energy per Nucleon Curve (DP IB Physics): Revision Note
Author
Katie M
Last updated
DP Physics: SLDP Physics: HL
Binding Energy per Nucleon Curve
In order to compare nuclear stability, it is useful to look at the binding energy per nucleon
The binding energy per nucleon is defined as:
The binding energy of a nucleus divided by the number of nucleons in the nucleus
A higher binding energy per nucleon indicates a higher stability
In other words, more energy is required to separate the nucleons contained within a nucleus
By plotting a graph of binding energy per nucleon against nucleon number, the stability of elements can be inferred
Key Features of the Graph
At low values of A:
Nuclei have lower binding energies per nucleon than at large values of A, but they tend to be stable when N = Z
This means light nuclei have weaker electrostatic forces and will undergo fusion
The gradient is much steeper compared to the gradient at large values of A
This means that fusion reactions release a greater binding energy than fission reactions
At high values of A:
Nuclei have generally higher binding energies per nucleon, but this gradually decreases with A
This means the heaviest elements are the most unstable and will undergo fission
The gradient is less steep compared to the gradient at low values of A
This means that fission reactions release less binding energy than fusion reactions
Iron (A = 56) has the highest binding energy per nucleon, which makes it the most stable of all the elements
Helium (4He), carbon (12C) and oxygen (16O) do not fit the trend
Helium-4 is a particularly stable nucleus hence it has a high binding energy per nucleon
Carbon-12 and oxygen-16 can be considered to be three and four helium nuclei, respectively, bound together
Comparing Fusion & Fission
Similarities
In both fusion and fission, the total mass of the products is slightly less than the total mass of the reactants
The mass defect is equivalent to the binding energy that is released
As a result, both fusion and fission reactions release energy
Differences
In fusion, two smaller nuclei combine into a larger nucleus
In fission, an unstable nucleus splits into two smaller nuclei
Fusion occurs between light nuclei (A < 56)
Fission occurs in heavy nuclei (A > 56)
In light nuclei, attractive nuclear forces dominate over repulsive electrostatic forces between protons, and this contributes to nuclear stability
In heavy nuclei, repulsive electrostatic forces between protons begin to dominate over attractive nuclear forces, and this contributes to nuclear instability
Fusion releases much more energy per kg than fission
Fusion requires a greater initial input of energy than fission
Worked Example
The equation below represents one possible decay of the induced fission of a nucleus of uranium-235.
The graph shows the binding energy per nucleon plotted against nucleon number A.
Calculate the energy released
(a) by the fission process represented by the equation
(b) when 1.0 kg of uranium, containing 3% by mass of U-235, undergoes fission
Answer:
Part (a)
Step 1: Use the graph to identify each isotope’s binding energy per nucleon
Binding energy per nucleon (U-235) = 7.5 MeV
Binding energy per nucleon (Sr-91) = 8.2 MeV
Binding energy per nucleon (Xe-142) = 8.7 MeV
Step 2: Determine the binding energy of each isotope
Binding energy = Binding Energy per Nucleon × Mass Number
Binding energy of U-235 nucleus = (235 × 7.5) = 1763 MeV
Binding energy of Sr-91 = (91 × 8.2) = 746 MeV
Binding energy of Xe-142 = (142 × 8.7) = 1235 MeV
Step 3: Calculate the energy released
Energy released = Binding energy after (Sr + Xe) – Binding energy before (U)
Energy released = (1235 + 746) – 1763 = 218 MeV
Part (b)
Step 1: Calculate the energy released by 1 mol of uranium-235
There are NA (Avogadro’s number) atoms in 1 mol of U-235, which is equal to a mass of 235 g
Energy released by 235 g of U-235 = (6 × 1023) × 218 MeV
Step 2: Convert the energy released from MeV to J
1 MeV = 1.6 × 10–13 J
Energy released = (6 × 1023) × 218 × (1.6 × 10–13) = 2.09 × 1013 J
Step 3: Work out the proportion of uranium-235 in the sample
1 kg of uranium which is 3% U-235 contains 0.03 kg or 30 g of U-235
Step 4: Calculate the energy released by the sample
Energy released from 1 kg of Uranium =
Examiner Tips and Tricks
Checklist on what to include (and what not to include) in an exam question asking you to draw a graph of binding energy per nucleon against nucleon number:
Do not begin your curve at A = 0, this is not a nucleus!
Make sure to correctly label both axes AND units for binding energy per nucleon
You will be expected to include numbers on the axes, mainly at the peak to show the position of iron (56Fe)
Strong Nuclear Force
In a nucleus, there are
Repulsive electric forces between protons due to their positive charge
Attractive gravitational forces due to the mass of the nucleons
Gravity is the weakest of the fundamental forces, so it has a negligible effect compared to electric repulsion between protons
If these were the only forces acting, the nucleus would not hold together
Therefore, there must be an attractive force acting between all nucleons which is stronger than the electric repulsive force
This is known as the strong nuclear force
The strong nuclear force acts between particles called quarks
Protons and neutrons are made up of quarks, so the interaction between the quarks in the nucleons keeps them bound within a nucleus
Whilst the electrostatic force is a repulsive force in the nucleus, the strong nuclear force holds the nucleus together
Properties of the Strong Nuclear Force
The strength of the strong nuclear force between two nucleons varies with the separation between them
This can be plotted on a graph which shows how the force changes with separation
The strong nuclear force is repulsive below a separation of 0.5 fm and attractive up to 3.0 fm
The key features of the graph are:
The strong force is highly repulsive at separations below 0.5 fm
The strong force is very attractive up to a nuclear separation of 3.0 fm
The maximum attractive value occurs at around 1.0 fm, which is a typical value for nucleon separation
The equilibrium position, where the resultant force is zero, occurs at a separation of about 0.5 fm
In comparison to other fundamental forces, the strong nuclear force has a very small range (from 0.5 to 3.0 fm)
Comparison of Electrostatic and Strong Forces
The graph below shows how the strength of the electrostatic and strong forces between two nucleons vary with the separation between them
The red curve represents the strong nuclear force between nucleons
The blue curve represents the electrostatic repulsion between protons
At separations between 0.5 and 3.0 fm, the attraction of the strong force is far more powerful than the repulsion of the electrostatic force
The repulsive electrostatic force between protons has a much larger range than the strong nuclear force
However, it only becomes significant when the proton separation is more than around 2.5 fm
The electrostatic force is influenced by charge, whereas the strong nuclear force is not
This means the strength of the strong nuclear force is roughly the same between all types of nucleon (i.e. proton-proton, neutron-neutron and proton-neutron)
This only applies for separations between 0.5 and 3.0 fm (where the electrostatic force between protons is insignificant)
The equilibrium position for protons, where the electrostatic repulsive and strong attractive forces are equal, occurs at a separation of around 0.7 fm
Examiner Tips and Tricks
You may see the strong nuclear force also referred to as the strong interaction
Make sure you can describe how the strong nuclear force varies with the separation of nucleons - make sure you remember the key values: range = 0.5 to 3.0 fm and typical nuclear separation ≈ 1.0 fm.
Remember to write that after 3 fm, the strong force becomes 'zero' or 'has no effect' rather than it is ‘negligible’.
Recall that 1 fm, or 1 femtometre, is 1 × 10–15 m
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Previous:Mass Defect & Nuclear Binding EnergyNext:Spontaneous & Induced Fission
Author:Katie M
Expertise: Physics Content Creator
Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.
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Chapter 1 Dimensional analysis Mathematical models are equations that express relationships between given quanti-ties of interest. The equations may be of any type, and the quantities may be of any type, either variable or constant. In this chapter, we outline various results about the units and dimensions of quantities, which can lead to insights, and point the way towards simpler, more concise forms of any mathematical model.
1.1. Units and dimensions Throughout our developments we consider equations involving real-valued quantities expressed in given units of some given dimension. By a unit for a quantity we mean a scale for its measurement, such as a foot, hour, or gram. By the dimension of a quantity we mean its intrinsic type, such as length, time, or mass. Whereas a unit for a quantity can be chosen arbitrarily, the dimension of a quantity is a characteristic property that is fixed.
Not all quantities have a dimension of their own. Indeed, by virtue of their def-inition, the dimensions of some quantities can be expressed as combinations of oth-ers. Thus only a basic set or basis of dimensions is required to describe a collection of quantities. For example, a standard dimensional basis for quantities arising in simple physical systems is (1.1) {length (𝐿), time (𝑇), mass (𝑀), temperature (𝛩)}.
Different bases could be considered depending on the context. In systems for which forces are important but not masses, the basis could include the dimension of force instead of mass. A similar change could be made if energies were important but not masses. In systems that include electrical quantities, the basis would be enlarged to include the dimension of electric current. As a different example, to describe quantities arising in a simple ecological system, a dimensional basis might consist of (1.2) {carnivore (𝐶), herbivore (𝐻), plant (𝑃), insect (𝐼), time (𝑇)}.
1 2 1. Dimensional analysis To any dimensional basis we associate a corresponding choice of units. These units may have some standard size, or any other arbitrary, nontrivial size, and they may have some standard name, or any other arbitrary name for convenience. For exam-ple, for the dimensional basis in (1.1), one choice of units is {meter, second, kilogram, kelvin}. For the dimensional basis in (1.2), one choice of units could be {herd, flock, field, swarm, month}, where, for example, 1 herd may be defined as 20 carnivores, 1 flock may be defined as 12 herbivores, and so on. Thus we will consider real-valued quantities, with values specified in a given choice of units, in a given dimensional basis.
The following notation will be used throughout.
Definition 1.1.1. Let 𝑞∈ℝbe a quantity specified in units {𝑈1, ... , 𝑈𝑚} in a dimen-sional basis {𝐷1, ... , 𝐷𝑚} for some 𝑚≥1. By [𝑞] we mean the dimension of 𝑞expressed as a product of powers of the basis elements, namely (1.3) [𝑞] = 𝐷𝑎1 1 𝐷𝑎2 2 ⋯𝐷𝑎𝑚 𝑚.
The numbers 𝑎1, ... , 𝑎𝑚are called the dimensional exponents of 𝑞in the given basis.
The array of exponents is denoted by Δ𝑞= (𝑎1, ... , 𝑎𝑚) ∈ℝ𝑚.
Example 1.1.1. Let 𝑝= 3 kg⋅m2 s2 , 𝑔= 9.8 m s2 , and 𝑞= 100 kelvin s . These quantities are expressed in units {meter, second, kilogram, kelvin} in the dimensional basis {𝐿, 𝑇, 𝑀, 𝛩}. The dimensions and corresponding dimensional exponents for 𝑝, 𝑔and 𝑞 in this basis are (1.4) [𝑝] = 𝑀𝐿 2/𝑇2 = 𝐿 2𝑇−2𝑀𝛩0, Δ𝑝= (2, −2, 1, 0), [𝑔] = 𝐿/𝑇2 = 𝐿𝑇−2𝑀0𝛩0, Δ𝑔= (1, −2, 0, 0), [𝑞] = 𝛩/𝑇 = 𝐿 0𝑇−1𝑀0𝛩, Δ𝑞= (0, −1, 0, 1).
Recall that, in the notation 𝑔= 9.8 m s2 , the number 9.8 is the numerical value of the quantity, and the tag m s2 is an explicit reminder of the units for the quantity. When we say that one quantity is a function of another, we mean that a relation exists between their numerical values, with respect to a given choice of units, in a given dimensional basis. Thus when we write 𝑝= 𝑓(𝑞), we mean that the numerical value of 𝑝is com-pletely determined by the numerical value of 𝑞. The function 𝑓is simply a map from one real value to another, and may be defined by a formula or graph in the usual way.
1.2. Axioms of dimensions We adopt the basic axioms that addition and subtraction are dimensionally meaning-ful only for quantities of the same dimension, whereas multiplication and division are meaningful for quantities of arbitrary dimension. To state these axioms in a more pre-cise way, let 𝑝, 𝑞, 𝑟, 𝑠∈ℝbe quantities with given units in a given dimensional basis.
The basic axiom on addition and subtraction reflects the idea that only quantities of the same dimension can be added and subtracted in a dimensionally meaningful way. Thus the statement 𝑟= 𝑝±𝑞has a dimensional meaning only when 𝑝and 𝑞, and hence 𝑟, have the same dimension. For instance, “1 meter + 2 meter” is a meaningful statement, whereas “1 meter + 2 second” is not.
1.3. Dimensionless quantities 3 The basic axiom on multiplication and division reflects the idea that quantities of any dimension can be multiplied and divided; indeed, this is how more complicated dimensions are derived from elementary ones. Thus the statements 𝑟= 𝑝𝑞and 𝑠= 𝑝/𝑞 (𝑞≠0) have a dimensional meaning for all 𝑝and 𝑞, and the dimensions of the results 𝑟 and 𝑠are well defined in each case. Moreover, this axiom can be extended to arbitrary powers, integration, and differentiation.
Axiom 1.2.1. Let 𝑝, 𝑞∈ℝbe quantities specified in units {𝑈1, ... , 𝑈𝑚}, in a dimensional basis {𝐷1, ... , 𝐷𝑚}, with dimensions [𝑝], [𝑞]. Then (1) [𝑝± 𝑞] is defined if and only if [𝑝] = [𝑞], (2) [𝑝𝑞] = [𝑝][𝑞] for all 𝑝, 𝑞, (3) [𝑝/𝑞] = [𝑝]/[𝑞] for all 𝑝, 𝑞with 𝑞≠0, (4) [𝑞𝛼] = [𝑞]𝛼for all 𝑞> 0 and real 𝛼, (5) [∫𝑝𝑑𝑞] = [𝑝][𝑞] for any integrable function 𝑝= 𝑓(𝑞), (6) [𝑑𝑝/𝑑𝑞] = [𝑝]/[𝑞] for any differentiable function 𝑝= 𝑓(𝑞).
In property (4) the condition 𝑞> 0 ensures that 𝑞𝛼is defined for any power 𝛼.
While it would suffice to only consider rational powers, we assume that the property holds for all real powers. The content of properties (2)–(4) can be translated to the dimensional exponents Δ𝑝and Δ𝑞in a straightforward way, namely (1.5) Δ𝑝𝑞= Δ𝑝+ Δ𝑞, Δ𝑝/𝑞= Δ𝑝−Δ𝑞, Δ𝑞𝛼= 𝛼Δ𝑞.
1.3. Dimensionless quantities The concept of a quantity with no dimension as defined next will play an important role throughout our developments. We note that such quantities can arise when con-sidering combinations of other quantities, and can also arise naturally in other ways.
Definition 1.3.1. A quantity 𝑞∈ℝis called dimensionless if its dimensional expres-sion is [𝑞] = 1, or equivalently its array of dimensional exponents is Δ𝑞= 0, in any units in any dimensional basis.
Example 1.3.1. (1) Let 𝑞= 𝑎𝑏/𝑐, where 𝑎= 4 ft hour, 𝑏= 3 1 hour, 𝑐= 2 ft hour2 . Consid-ering dimensions we have [𝑎] = 𝐿𝑇−1, [𝑏] = 𝑇−1, and [𝑐] = 𝐿𝑇−2, and we find that [𝑞] = [𝑎][𝑏]/[𝑐] = 1. Thus 𝑞is a dimensionless quantity; its value is 𝑞= 6.
(2) Let 𝑎be an arbitrary quantity with dimension [𝑎], and let 𝑏= 𝑎+𝑎and 𝑐= 𝑎⋅𝑎⋅𝑎.
Then it is natural to rewrite these quantities as 𝑏= 2𝑎and 𝑐= 𝑎3. In these latter expressions, we note that the coefficient 2 and exponent 3 are dimensionless; they are purely mathematical entities called pure numbers. The dimensions of 𝑏and 𝑐are [𝑏] = [2𝑎] = [𝑎] and [𝑐] = [𝑎3] = [𝑎]3.
(3) Let 𝜃be an arbitrary angle, which when inscribed in a circle of radius 𝑟subtends an arc of length ℓ. Then, in the radian unit of measurement, we have 𝜃= ℓ 𝑟and we find [𝜃] = 1. Hence angles and the radian unit of measurement are dimensionless.
4 1. Dimensional analysis Similarly, since it only differs in size, the degree unit of measurement is dimensionless.
(4) Any ratio of two quantities of the same dimension is dimensionless. The value of such a ratio can be expressed as a pure number, or in terms of any arbitrary dimension-less unit such as a percentage or parts-per-hundred.
1.4. Change of units Here we outline the effect of a change of units on an arbitrary quantity. For our pur-poses it will be sufficient to only consider changes in the dimensional units associated with a given dimensional basis, with any dimensionless units held fixed. We assume that any two units of the same dimensional type are related by a multiplicative conver-sion factor as introduced below.
To state the result, we consider an arbitrary quantity 𝑞∈ℝ, expressed in units {𝑈1, ... , 𝑈𝑚}, in a dimensional basis {𝐷1, ... , 𝐷𝑚}, with dimensional exponents Δ𝑞= (𝑎1, ... , 𝑎𝑚) ∈ℝ𝑚.
Result 1.4.1. If units {𝑈1, ... , 𝑈𝑚} are changed to { ˜ 𝑈1, ... , ˜ 𝑈𝑚}, then the quantity 𝑞is changed to ̃ 𝑞, where (1.6) ̃ 𝑞= 𝑞𝜆𝑎1 1 𝜆𝑎2 2 ⋯𝜆𝑎𝑚 𝑚.
Here 𝜆𝑖> 0 (𝑖= 1, ... , 𝑚) are unit-conversion factors; each factor 𝜆𝑖quantifies the number of units of ˜ 𝑈𝑖per unit of 𝑈𝑖.
The above result follows from straightforward algebra and the axioms on dimen-sions regarding multiplication and division. Note that if 𝑞is dimensionless, then Δ𝑞= (0, ... , 0), and we obtain ̃ 𝑞= 𝑞. Thus dimensionless quantities are not affected by a change of dimensional units.
Example 1.4.1. Let 𝑔= 9.8 m s2 . This quantity is expressed in units {m, s} in the dimen-sional basis {𝐿, 𝑇}. Since [𝑔] = 𝐿𝑇−2, its dimensional exponents are Δ𝑔= (𝑎1, 𝑎2) = (1, −2). If the units are changed to {km, min}, then the unit-conversion factors are (1.7) 𝜆1 = 1 1000 km m , 𝜆2 = 1 60 min s .
In the new units we have (1.8) ̃ 𝑔= 𝑔𝜆𝑎1 1 𝜆𝑎2 2 = (9.8 m s2 )( 1 1000 km m )( 1 60 min s ) −2 = 35.28 km min2 .
1.5. Unit-free equations In the modeling of various types of systems, we will usually consider a set of real-valued quantities 𝑞1, ... , 𝑞𝑛, specified in units {𝑈1, ... , 𝑈𝑚}, in a dimensional basis {𝐷1, ... , 𝐷𝑚}, for some 𝑛≥2 and 𝑚≥1. We will often seek to construct and study equations of the form (1.9) 𝑞1 = 𝑓(𝑞2, ... , 𝑞𝑛), 1.5. Unit-free equations 5 where 𝑓∶ℝ𝑛−1 →ℝis some function. The function notation above indicates that the numerical value of 𝑞1 is completely determined by the numerical values of 𝑞2, ... , 𝑞𝑛 in the given units. In our pursuits, we will only consider equations that are unit-free as defined next.
Definition 1.5.1. An equation 𝑞1 = 𝑓(𝑞2, ... , 𝑞𝑛) is called unit-free if it transforms into (1.10) ̃ 𝑞1 = 𝑓( ̃ 𝑞2, ... , ̃ 𝑞𝑛) under an arbitrary change of units on arbitrary values of 𝑞1, ... , 𝑞𝑛.
The key point of a unit-free equation is that the function 𝑓is unaffected by the choice of units. All the equations that we consider will be unit-free in this sense. Note that, without this property, the function 𝑓may change whenever the units are changed, and the equation would have limited value as a model. Indeed, it would be tedious to document each different version of the equation for each different choice of units. Thus a unit-free equation can be viewed as a well designed equation. Model equations de-rived from fundamental physical laws are naturally unit-free; they inherit this property from the laws on which they are based. In contrast, empirical equations derived from curve fitting procedures are not naturally unit-free, but can always be re-designed to have this property by introducing appropriate dimensional constants. In the most ba-sic sense, a unit-free equation can be viewed as a dimensionally meaningful equation, consistent with the axioms of dimensions, and this property can always be achieved.
Example 1.5.1. Let 𝑥, 𝑡and 𝑔be specified in units {m, s}, in the dimensional basis {𝐿, 𝑇}, with dimensions [𝑥] = 𝐿, [𝑡] = 𝑇and [𝑔] = 𝐿𝑇−2, and exponents Δ𝑥= (1, 0), Δ𝑡= (0, 1) and Δ𝑔= (1, −2). Suppose that the value of 𝑥is determined by the values of 𝑡and 𝑔through the equation (1.11) 𝑥= 𝑓(𝑡, 𝑔) = 1 2𝑔𝑡2.
(Unless mentioned otherwise, unnamed quantities such as the factor 1 2 and exponent 2 can be interpreted as pure numbers.) To determine if the above equation is unit-free, we consider a change of units from {m, s} to arbitrary units { ˜ 𝑈1, ˜ 𝑈2}, defined by arbitrary conversion factors 𝜆1, 𝜆2. In the new units, the values of 𝑥, 𝑡and 𝑔become (1.12) ̃ 𝑥= 𝑥𝜆1, ̃ 𝑡= 𝑡𝜆2, ̃ 𝑔= 𝑔𝜆1𝜆−2 2 .
Substitution of these expressions into 𝑥= 1 2𝑔𝑡2 gives (1.13) ( ̃ 𝑥𝜆−1 1 ) = 1 2( ̃ 𝑔𝜆−1 1 𝜆2 2)( ̃ 𝑡𝜆−1 2 ) 2 .
In the above, all factors with 𝜆1 and 𝜆2 cancel, and we get (1.14) ̃ 𝑥= 1 2 ̃ 𝑔̃ 𝑡2.
Thus (1.11) is unit-free since it has exactly the same form in any choice of units. The original equation 𝑥= 𝑓(𝑡, 𝑔) is transformed into ̃ 𝑥= 𝑓( ̃ 𝑡, ̃ 𝑔), with the same function 𝑓.
6 1. Dimensional analysis Example 1.5.2. Let 𝑥, 𝑡and 𝑔be as before, and let 𝑐be an additional quantity, say a constant, with [𝑐] = 𝑇and Δ𝑐= (0, 1). For purposes of comparison, consider the two different equations (1.15) 𝑥= 1 2𝑔𝑡2𝑒−𝑡, 𝑥= 1 2𝑔𝑡2𝑒−𝑡/𝑐.
(Here 𝑒𝑞= exp(𝑞) is the natural exponential function; the base 𝑒can be interpreted as a pure number.) Considering an arbitrary change of units as above, we get (1.16) ̃ 𝑥= 1 2 ̃ 𝑔̃ 𝑡2𝑒−̃ 𝑡𝜆−1 2 , ̃ 𝑥= 1 2 ̃ 𝑔̃ 𝑡2𝑒−̃ 𝑡/ ̃ 𝑐.
The first equation is not unit-free since it changes form: a unit-conversion factor re-mains in the equation and does not cancel out. In contrast, the second equation is unit-free since all the unit-conversion factors cancel. Note how the first equation be-comes unit-free by introduction of the constant 𝑐. The equations in (1.15), written in units {m, s}, would be numerically the same when 𝑐= 1s. However, the second equa-tion is advantageous since it would have exactly the same form in any units.
Example 1.5.3. Let 𝑣and 𝑡be quantities specified in units {lb, hr}, in the dimensional basis {𝑀, 𝑇}, with dimensions [𝑣] = 𝑀/𝑇and [𝑡] = 𝑇. Suppose that the value of 𝑣is determined by the value of 𝑡through an empirical equation (1.17) 𝑣= 3.7𝑡2 −sin(5.4𝑡).
Here we rewrite this equation in a unit-free form. To begin, we introduce constants 𝑎, 𝑏, 𝑐with values 3.7, 1, 5.4 in units {lb, hr} and consider (1.18) 𝑣= 𝑎𝑡2 −𝑏sin(𝑐𝑡).
We next determine the dimensions of these constants to make the equation unit-free.
Accordingly, let Δ𝑎= (𝛼1, 𝛼2), Δ𝑏= (𝛽1, 𝛽2) and Δ𝑐= (𝛾1, 𝛾2) be the unknown dimen-sional exponents. Under an arbitrary change of units with conversion factors 𝜆1, 𝜆2, using the fact that Δ𝑣= (1, −1) and Δ𝑡= (0, 1), we get, after dividing out the conver-sion factors from the left side of the equation, (1.19) ̃ 𝑣= 𝜆1−𝛼1 1 𝜆−3−𝛼2 2 ̃ 𝑎̃ 𝑡2 −𝜆1−𝛽1 1 𝜆−1−𝛽2 2 ̃ 𝑏sin(𝜆−𝛾1 1 𝜆−1−𝛾2 2 ̃ 𝑐̃ 𝑡).
Note that the unit-free condition will be satisfied when the exponents of all the conver-sion factors in the above expression are zero, which requires Δ𝑎= (1, −3), Δ𝑏= (1, −1) and Δ𝑐= (0, −1). Thus the dimensions of the constants are completely determined, and in units {lb, hr} we have (1.20) 𝑎= 3.7 lb hr3 , 𝑏= 1 lb hr, 𝑐= 5.4 1 hr.
In any other units, the equation would be ̃ 𝑣= ̃ 𝑎̃ 𝑡2 − ̃ 𝑏sin( ̃ 𝑐̃ 𝑡), where ̃ 𝑎, ̃ 𝑏, ̃ 𝑐are the values of the constants in the new units. In our function notation, the equation in (1.18) would be written as 𝑣= 𝑓(𝑡, 𝑎, 𝑏, 𝑐).
1.6. Buckingham 𝜋-theorem 7 1.6. Buckingham 𝜋-theorem Here we outline a classic result known as the Buckingham 𝜋-theorem. It states that, for any unit-free equation 𝑞1 = 𝑓(𝑞2, ... , 𝑞𝑛), the function 𝑓cannot depend on 𝑞2, ... , 𝑞𝑛in a completely arbitrary way; it can only depend on certain dimensionless combinations.
For simplicity we state the result only for positive values of 𝑞1, ... , 𝑞𝑛. Similar results hold for nonpositive values, but at the expense of more complicated statements.
Definition 1.6.1. By a power product of 𝑞1, ... , 𝑞𝑛> 0 we mean a quantity 𝜋> 0 of the form (1.21) 𝜋= 𝑞𝑏1 1 ⋯𝑞𝑏𝑛 𝑛, for some powers 𝑏1, ... , 𝑏𝑛∈ℝ. We say that 𝜋includes 𝑞𝑖if 𝑏𝑖≠0.
The condition that each 𝑞𝑖be positive ensures that 𝜋is well defined for arbitrary powers. In any dimensional basis {𝐷1, ... , 𝐷𝑚}, we note that each quantity 𝑞𝑖will have dimensional exponents Δ𝑞𝑖∈ℝ𝑚, and the power product 𝜋will have dimensional exponents Δ𝜋∈ℝ𝑚. From the definition in (1.21), together with the properties that Δ𝑝𝑞= Δ𝑝+ Δ𝑞and Δ𝑞𝛼= 𝛼Δ𝑞given in (1.5), we deduce that (1.22) Δ𝜋= 𝑏1Δ𝑞1 + ⋯+ 𝑏𝑛Δ𝑞𝑛= 𝐴𝑣, where 𝐴= (Δ𝑞1, Δ𝑞2, ... , Δ𝑞𝑛) ∈ℝ𝑚×𝑛and 𝑣= (𝑏1, ... , 𝑏𝑛) ∈ℝ𝑛. Here all one-dimensional arrays are considered as columns, and we assume 𝑛≥2 and 𝑚≥1 with 𝑛≥𝑚.
Given 𝑞1, ... , 𝑞𝑛we will be interested in forming power products 𝜋that are dimen-sionless. In this respect, we note that (1.23) 𝜋dimensionless ⇔ Δ𝜋= 0 ⇔ 𝐴𝑣= 0.
Furthermore, we will only be interested in nontrivial power products 𝜋≢1, which cor-respond to 𝑣≠0. The following result, which essentially is a definition, characterizes the dimensionless power products that we seek.
Result 1.6.1. If 𝐴𝑣= 0 has a total of 𝑘independent solutions 𝑣1, ... , 𝑣𝑘, then a total of 𝑘 independent dimensionless power products 𝜋1, ... , 𝜋𝑘can be formed. Any such 𝜋1, ... , 𝜋𝑘 is called a full set. This set is further called normalized if 𝜋1 includes 𝑞1 (with power 𝑏1 = 1), and 𝜋2, ... , 𝜋𝑘do not include 𝑞1.
Recall that, through the usual process of row reduction, any nontrivial solution of 𝐴𝑣= 0 will be expressed in terms of certain free variables. If there are 𝑘free vari-ables, then there are 𝑘independent solutions 𝑣, and hence 𝑘independent dimension-less power products 𝜋. While any independent choices of the free variables can be made to form a full set of solutions, a deliberate choice of these variables is required to form a normalized set. Specifically, the normalization condition requires that the first solution have 𝑏1 = 1, and any other solutions have 𝑏1 = 0.
Example 1.6.1. Let 𝑥, 𝑡, 𝑔, ℎ, 𝑚> 0 be quantities with dimensions [𝑥] = 𝐿, [𝑡] = 𝑇, [𝑔] = 𝐿𝑇−2, [ℎ] = 𝐿𝑇−3 and [𝑚] = 𝑀. A dimensional basis is {𝐿, 𝑇, 𝑀}, and the 8 1. Dimensional analysis dimensional exponent matrix in this basis is (1.24) 𝐴= (Δ𝑥, Δ𝑡, Δ𝑔, Δℎ, Δ𝑚) = ( 1 0 1 1 0 0 1 −2 −3 0 0 0 0 0 1 ) .
An arbitrary power product has the form 𝜋= 𝑥𝑏1𝑡𝑏2𝑔𝑏3ℎ𝑏4𝑚𝑏5. The equation 𝐴𝑣= 0, where 𝑣= (𝑏1, ... , 𝑏5), has two free variables, and the general solution is (1.25) 𝑏1 = −𝑏3 −𝑏4, 𝑏2 = 2𝑏3 + 3𝑏4, 𝑏5 = 0, 𝑏3, 𝑏4 free.
Since there are two free variables, there are two independent solutions. For one so-lution we choose 𝑏3 = −1, 𝑏4 = 0, which gives 𝑣1 = (1, −2, −1, 0, 0), and hence 𝜋1 = 𝑥/(𝑔𝑡2). For a second solution we choose 𝑏3 = −1, 𝑏4 = 1, which gives 𝑣2 = (0, 1, −1, 1, 0), and hence 𝜋2 = 𝑡ℎ/𝑔. By choice, we arranged for 𝜋1 to include 𝑥with an exponent of unity, and for 𝜋2 to exclude 𝑥. Hence 𝜋1, 𝜋2 is a full set of dimensionless power products for 𝑥, 𝑡, 𝑔, ℎ, 𝑚, and this set is normalized with respect to 𝑥. Note that 𝑚will not be included in any dimensionless power product.
The next result shows that any unit-free equation can only depend on dimension-less power products. No assumption on the form or continuity properties of the func-tion 𝑓are required. For simplicity, the results are stated only for positive quantities; similar results can be derived to account for negative and zero quantities.
Result 1.6.2. [𝜋-theorem] Let 𝜋1, ... , 𝜋𝑘> 0 be a full set of dimensionless power prod-ucts for 𝑞1, ... , 𝑞𝑛> 0 where 𝑘≥1 and 𝑛≥2. If the set 𝜋1, ... , 𝜋𝑘is normalized, then any unit-free equation 𝑞1 = 𝑓(𝑞2, ... , 𝑞𝑛) for some function 𝑓, is equivalent to an equation (1.26) 𝜋1 = 𝜙(𝜋2, ... , 𝜋𝑘) for some function 𝜙. In the case that 𝑘= 1, the function 𝜙reduces to some constant 𝐶.
The normalization condition ensures that the reduced equation (1.26) is explicit, just as the original equation, in terms of 𝑞1. We remark that a more general form of the theorem states that any unit-free equation in the general implicit form 𝐹(𝑞1, ... , 𝑞𝑛) = 0 is equivalent to 𝛷(𝜋1, ... , 𝜋𝑘) = 0, without any normalization condition on the power products. Also, if the only dimensionless power product for quantities 𝑞1, ... , 𝑞𝑛is trivial, then the only unit-free relation among these quantities is trivial; in this case, the set 𝑞1, ... , 𝑞𝑛would need to be enlarged in order for a nontrivial unit-free relation to exist. The proof of the theorem is based on a change of variable argument that exploits the unit-free condition and the definition of the power products, which will be outlined after some examples.
Example 1.6.2. Let 𝑥, 𝑡, 𝑔, ℎ, 𝑚> 0 be quantities as in the previous example. A nor-malized set of power products for these quantities is 𝜋1 = 𝑥/(𝑔𝑡2) and 𝜋2 = 𝑡ℎ/𝑔. Thus any unit-free equation of the form 𝑥= 𝑓(𝑡, 𝑔, ℎ, 𝑚) must be equivalent to an equation of the form (1.27) 𝜋1 = 𝜙(𝜋2) or 𝑥 𝑔𝑡2 = 𝜙(𝑡ℎ 𝑔), 1.6. Buckingham 𝜋-theorem 9 which can be rearranged to yield (1.28) 𝑥= 𝑔𝑡2 ⋅𝜙(𝑡ℎ 𝑔).
Thus 𝑥cannot depend on 𝑚, and must depend on 𝑡, 𝑔and ℎin a specific way. If 𝜙(0) is defined, then the special case when ℎhas a fixed value of zero can be considered, and the relation becomes 𝑥= 𝛽𝑔𝑡2, where 𝛽= 𝜙(0) is a dimensionless constant.
Example 1.6.3. Here we explicitly find the reduced form of the unit-free equation (1.29) 𝑢= 𝑓(𝑥, 𝑡, 𝛼, 𝛽) = 𝛼𝑥 𝑡𝑒−𝑥2/(𝛽𝑡2), where [𝑢] = 𝛩, [𝑥] = 𝐿, [𝑡] = 𝑇, [𝛼] = 𝛩𝑇/𝐿, and [𝛽] = 𝐿 2/𝑇2. A dimensional basis is {𝛩, 𝐿, 𝑇}, and the dimensional exponent matrix is (1.30) 𝐴= (Δᵆ, Δ𝑥, Δ𝑡, Δ𝛼, Δ𝛽) = ( 1 0 0 1 0 0 1 0 −1 2 0 0 1 1 −2 ) .
An arbitrary power product is 𝜋= 𝑢𝑏1𝑥𝑏2𝑡𝑏3𝛼𝑏4𝛽𝑏5. The equation 𝐴𝑣= 0, where 𝑣= (𝑏1, ... , 𝑏5), has two free variables. By choosing these variables in a similar way as before, we obtain the full, normalized set 𝜋1 = 𝑢/(𝛼√𝛽) and 𝜋2 = 𝑥/(𝑡√𝛽). By the 𝜋-theorem, the original equation 𝑢= 𝑓(𝑥, 𝑡, 𝛼, 𝛽) must be equivalent to 𝜋1 = 𝜙(𝜋2) for some function 𝜙. Here this result can be verified directly due to the explicit form of the original equation. Specifically, dividing the equation by 𝛼√𝛽, and then substituting, we obtain (1.31) 𝑢= 𝛼𝑥 𝑡𝑒−𝑥2/(𝛽𝑡2) ⇔ 𝑢 𝛼√𝛽 = 𝑥 𝑡√𝛽 𝑒−𝑥2/(𝛽𝑡2) ⇔ 𝜋1 = 𝜋2𝑒−𝜋2 2.
Example 1.6.4. A simple theory of sound waves in a gas proposes that the speed of sound 𝑣> 0 should depend on only the mass density 𝜌> 0, pressure 𝑝> 0, and viscosity 𝜇> 0 so that (1.32) 𝑣= 𝑓(𝜌, 𝑝, 𝜇), for some function 𝑓. Here we use the 𝜋-theorem to find an equivalent and possibly simpler form of (1.32) assuming that it is unit-free. This can be viewed as an important first step in exploring any new or proposed relation of interest.
The quantities 𝑣, 𝜌, 𝑝, 𝜇have dimensions [𝑣] = 𝐿/𝑇, [𝜌] = 𝑀/𝐿 3, [𝑝] = 𝑀/(𝐿𝑇2) and [𝜇] = 𝑀/(𝐿𝑇). A dimensional basis is {𝐿, 𝑇, 𝑀}, and the dimensional exponent matrix in this basis is (1.33) 𝐴= (Δ𝑣, Δ𝜌, Δ𝑝, Δ𝜇) = ( 1 −3 −1 −1 −1 0 −2 −1 0 1 1 1 ) .
An arbitrary power product has the form 𝜋= 𝑣𝑏1𝜌𝑏2𝑝𝑏3𝜇𝑏4. The equation 𝐴𝑣= 0, where 𝑣= (𝑏1, ... , 𝑏4), has one free variable, and the general solution is (1.34) 𝑏1 = −2𝑏3, 𝑏2 = −𝑏3, 𝑏4 = 0, 𝑏3 free.
10 1. Dimensional analysis Since there is only one free variable, there is only one independent solution. For this so-lution we choose 𝑏3 = −1/2, which gives 𝑣1 = (1, 1/2, −1/2, 0), and hence 𝜋1 = 𝑣√𝜌/𝑝.
Hence 𝜋1 is a full set of independent dimensionless power products for 𝑣, 𝜌, 𝑝, 𝜇, and this set is normalized with respect to 𝑣.
By the 𝜋-theorem, the equation in (1.32) must be equivalent to (1.35) 𝜋1 = 𝐶 or 𝑣= 𝐶√ 𝑝 𝜌, where 𝐶> 0 is some dimensionless constant. Thus any experimental investigation of sound waves under the given hypothesis should be aimed at (1.35), and the determina-tion of the unknown constant 𝐶.
Note that, even though an unknown constant is involved, there is valuable, direct information implied by (1.35). For instance, it implies that the speed of sound must be independent of the viscosity, and would increase with pressure at fixed density, and decrease with density at fixed pressure. Moreover, the speed of sound would remain unchanged when pressure and density are both increased or decreased in a simultane-ous way.
Sketch of proof: Result 1.6.2. Let 𝑞1, ... , 𝑞𝑛> 0 and 𝜋1, ... , 𝜋𝑘> 0 be given, and let 𝐴∈ℝ𝑚×𝑛be the dimensional exponent matrix whose columns are Δ𝑞𝜍 = (𝑎1,𝜍, ... , 𝑎𝑚,𝜍) ∈ℝ𝑚, where 𝜎= 1, ... , 𝑛. Note that, to each dimensionless power product 𝜋𝜌, there is an independent solution 𝑣𝜌= (𝑏1,𝜌, ... , 𝑏𝑛,𝜌) ∈ℝ𝑛of 𝐴𝑣= 0, where 𝜌= 1, ... , 𝑘. Thus the row-reduced form of 𝐴𝑣= 0 has 𝑘columns without pivots, which correspond to the free variables, and 𝑛−𝑘columns with pivots.
Due to the normalization condition, we have 𝜋1 = 𝑞1𝑞 𝑏2,1 2 ⋯𝑞 𝑏𝑛,1 𝑛 , and any re-maining power products 𝜋2, ... , 𝜋𝑘involve only 𝑞2, ... , 𝑞𝑛. In view of this, we consider 𝐴′ ∈ℝ𝑚×(𝑛−1), defined to be the submatrix of 𝐴obtained by omitting the first column, and 𝑣′ ∈ℝ𝑛−1, defined to be the subvector of 𝑣obtained by omitting the first entry.
The assumption that 𝐴𝑣= 0 has a full set of 𝑘independent solutions that satisfy the normalization condition implies that 𝐴′𝑣′ = 0 has precisely 𝑘−1 independent solu-tions, and hence precisely as many free variables. Consequently, the row-reduced form of 𝐴′𝑣′ = 0 has 𝑛−𝑘columns with pivots, so that 𝐴′ has rank 𝑛−𝑘.
Let 𝑞1 = 𝑓(𝑞2, ... , 𝑞𝑛) be given and consider an arbitrary change of units that changes 𝑞1, ... , 𝑞𝑛into ̃ 𝑞1, ... , ̃ 𝑞𝑛, and note that ̃ 𝑞1 = 𝑓( ̃ 𝑞2, ... , ̃ 𝑞𝑛) by the unit-free as-sumption. In view of the above expression for 𝜋1, we introduce the function 𝐹(𝑞2, ... , 𝑞𝑛) = 𝑓(𝑞2, ... , 𝑞𝑛)𝑞 𝑏2,1 2 ⋯𝑞 𝑏𝑛,1 𝑛 so that 𝜋1 = 𝐹(𝑞2, ... , 𝑞𝑛). Similarly, beginning from the analogous expression for ˜ 𝜋1, we find ˜ 𝜋1 = 𝐹( ̃ 𝑞2, ... , ̃ 𝑞𝑛). Because it is dimensionless, we have 𝜋1 = ˜ 𝜋1, which im-plies 𝐹(𝑞2, ... , 𝑞𝑛) = 𝐹( ̃ 𝑞2, ... , ̃ 𝑞𝑛). Thus the function 𝐹is invariant under an arbitrary change of units.
To establish the result of the theorem, we consider different cases depending on the number 𝑘of power products. In the case when 𝑘= 1, we consider the change of unit relations ̃ 𝑞𝜍= 𝑞𝜍𝜆 𝑎1,𝜍 1 ⋯𝜆 𝑎𝑚,𝜍 𝑚 for 𝜎= 2, ... , 𝑛, where 𝜆1, ... , 𝜆𝑚are the conversion fac-tors. From this we obtain the log-linear system ln( ̃ 𝑞𝜍/𝑞𝜍) = 𝑎1,𝜍ln 𝜆1+⋯+𝑎𝑚,𝜍ln 𝜆𝑚.
1.7. Case study 11 In matrix form, we have 𝐴′𝑇𝑢= 𝑔′, where 𝑢= (ln 𝜆1, ... , ln 𝜆𝑚) ∈ℝ𝑚and 𝑔′ = (ln( ̃ 𝑞2/𝑞2), ... , ln( ̃ 𝑞𝑛/𝑞𝑛)) ∈ℝ𝑛−1. When 𝑘= 1, the rank of 𝐴′ and 𝐴′𝑇is 𝑛−1, and the columns of 𝐴′𝑇span ℝ𝑛−1. Thus, for arbitrary old values 𝑞2, ... , 𝑞𝑛, we can always find a change of units to obtain any specified new values ̃ 𝑞2, ... , ̃ 𝑞𝑛, say a value of one for each. The required change of units can be found by setting ̃ 𝑞2 = 1, ... , ̃ 𝑞𝑛= 1 in this log-linear system and solving for the conversion factors 𝜆1, ... , 𝜆𝑚. Due to the in-variance property of 𝐹, for arbitrary 𝑞2, ... , 𝑞𝑛we get 𝜋1 = 𝐹(1, ... , 1) = 𝐶, where 𝐶is some fixed constant, which establishes the result for this case.
In the case when 𝑘= 𝑛, the system 𝐴′𝑣′ = 0 has 𝑛−1 independent solutions 𝑣′ 𝜌= (𝑏2,𝜌, ... , 𝑏𝑛,𝜌) ∈ℝ𝑛−1 for 𝜌= 2, ... , 𝑛. Let 𝐵′ ∈ℝ(𝑛−1)×(𝑛−1) be the matrix whose columns are these solutions, and note that it is square and has full rank, and hence is invertible. For this case, we consider the power products 𝜋𝜌= 𝑞 𝑏2,𝜌 2 ⋯𝑞 𝑏𝑛,𝜌 𝑛 for 𝜌= 2, ... , 𝑛, and obtain the log-linear system ln 𝜋𝜌= 𝑏2,𝜌ln 𝑞2 + ⋯+ 𝑏𝑛,𝜌ln 𝑞𝑛.
In matrix form, we have 𝐵′𝑇𝑤′ = ℎ′, where 𝑤′ = (ln 𝑞2, ... , ln 𝑞𝑛) ∈ℝ𝑛−1 and ℎ′ = (ln 𝜋2, ... , ln 𝜋𝑛) ∈ℝ𝑛−1. Since 𝐵′𝑇is invertible, we find that 𝑞2, ... , 𝑞𝑛can be uniquely expressed in terms of 𝜋2, ... , 𝜋𝑛. This implies that 𝜋1 = 𝐹(𝑞2, ... , 𝑞𝑛) = 𝜙(𝜋2, ... , 𝜋𝑛), for some function 𝜙, which establishes the result for this case.
In the case when 1 < 𝑘< 𝑛, the system 𝐴′𝑣′ = 0 has 𝑘−1 independent solutions 𝑣′ 𝜌= (𝑏2,𝜌, ... , 𝑏𝑛,𝜌) ∈ℝ𝑛−1 for 𝜌= 2, ... , 𝑘, and has rank 𝑛−𝑘as noted earlier.
Without loss of generality, up to a reordering of 𝑞2, ... , 𝑞𝑛, we may suppose that the pivots in the system 𝐴′𝑣′ = 0 all occur in the leading 𝑛−𝑘columns, whereas the free variables all occur in the latter 𝑘−1 columns. We now consider the 𝑛−𝑘change of unit relations ̃ 𝑞𝜍= 𝑞𝜍𝜆 𝑎1,𝜍 1 ⋯𝜆 𝑎𝑚,𝜍 𝑚 for 𝜎= 2, ... , 𝑛−𝑘+ 1, and again consider the system ln( ̃ 𝑞𝜍/𝑞𝜍) = 𝑎1,𝜍ln 𝜆1+⋯+𝑎𝑚,𝜍ln 𝜆𝑚. Since the dimensional exponent vectors (𝑎1,𝜍, ... , 𝑎𝑚,𝜍) are the leading 𝑛−𝑘columns of 𝐴′, they are independent. Thus for arbitrary 𝑞2, ... , 𝑞𝑛−𝑘+1 we can find 𝜆1, ... , 𝜆𝑚to achieve ̃ 𝑞2 = 1, ... , ̃ 𝑞𝑛−𝑘+1 = 1. We next consider the 𝑘−1 power products 𝜋𝜌= 𝑞 𝑏2,𝜌 2 ⋯𝑞 𝑏𝑛,𝜌 𝑛 for 𝜌= 2, ... , 𝑘. Since 𝜋𝜌= ˜ 𝜋𝜌and ̃ 𝑞2 = 1, ... , ̃ 𝑞𝑛−𝑘+1 = 1, we get the reduced expressions 𝜋𝜌= ̃ 𝑞 𝑏𝑛−𝑘+2,𝜌 𝑛−𝑘+2 ⋯̃ 𝑞 𝑏𝑛,𝜌 𝑛 , which leads to the system ln 𝜋𝜌= 𝑏𝑛−𝑘+2,𝜌ln ̃ 𝑞𝑛−𝑘+2 + ⋯+ 𝑏𝑛,𝜌ln ̃ 𝑞𝑛. For each 𝜌, we note that (𝑏𝑛−𝑘+2,𝜌, ... , 𝑏𝑛,𝜌) are the 𝑘−1 free variables from the system 𝐴′𝑣′ = 0, which were independently chosen to generate the solution set. Thus this log-linear system is square and has full rank, and we find that ̃ 𝑞𝑛−𝑘+2, ... , ̃ 𝑞𝑛can be uniquely expressed in terms of 𝜋2, ... , 𝜋𝑘. This implies 𝜋1 = 𝐹( ̃ 𝑞2, ... , ̃ 𝑞𝑛) = 𝐹(1, ... , 1, ̃ 𝑞𝑛−𝑘+2, ... , ̃ 𝑞𝑛) = 𝜙(𝜋2, ... , 𝜋𝑘), for some function 𝜙, which establishes the result.
1.7 . Case study Setup. To illustrate the preceding results on dimensional methods, and the process of modelling a simple mechanical system, we study the motion of a pendulum released from rest. Figure 1.1 illustrates the system, which consists of a string of length ℓ, with one end attached to a fixed support point, and the other end attached to a ball of mass 𝑚. We assume the string is always in tension and hence straight, and we let 𝜃denote the angle between the string and a vertical line through the support point, and arbitrarily take the positive direction to be counter-clockwise. We assume that gravitational ac-celeration 𝑔is directed in the downward, vertical direction. When the ball is raised 12 1. Dimensional analysis g m θ θ j y x i o e e r θ r Fstring Fgravity Figure 1.1.
and released from the rest conditions 𝜃= 𝜃0 and 𝑑𝜃 𝑑𝑡= 0 at time 𝑡= 0, the ball will swing back-and-forth in a periodic motion. We seek to understand various aspects of this motion; for example, how the period depends on the parameters 𝑚, 𝑔, ℓ, and 𝜃0.
Outline of model. We assume that the motion occurs in a plane and introduce an origin and 𝑥, 𝑦coordinates as shown. The standard unit vectors in the positive 𝑥 and 𝑦directions are denoted by ⃗ 𝑖and ⃗ 𝑗, and the position vector for the ball is denoted by ⃗ 𝑟. It will be convenient to introduce unit vectors ⃗ 𝑒𝑟and ⃗ 𝑒𝜃that are parallel and perpendicular to ⃗ 𝑟. For any angle 𝜃, the components of these vectors are ⃗ 𝑟= ℓsin 𝜃⃗ 𝑖+ ℓcos 𝜃 ⃗ 𝑗, ⃗ 𝑒𝑟= sin 𝜃⃗ 𝑖+cos 𝜃 ⃗ 𝑗, and ⃗ 𝑒𝜃= cos 𝜃⃗ 𝑖−sin 𝜃 ⃗ 𝑗. By differentiating the position with respect to time, we obtain the velocity and acceleration vectors (1.36) 𝑑⃗ 𝑟 𝑑𝑡= ℓcos 𝜃𝑑𝜃 𝑑𝑡 ⃗ 𝑖−ℓsin 𝜃𝑑𝜃 𝑑𝑡 ⃗ 𝑗, 𝑑2 ⃗ 𝑟 𝑑𝑡2 = [ℓcos 𝜃𝑑2𝜃 𝑑𝑡2 −ℓsin 𝜃(𝑑𝜃 𝑑𝑡) 2 ] ⃗ 𝑖−[ℓsin 𝜃𝑑2𝜃 𝑑𝑡2 + ℓcos 𝜃(𝑑𝜃 𝑑𝑡) 2 ] ⃗ 𝑗.
We assume that only two forces act on the ball: one due to gravity, and another due to the pull of the string. Thus we neglect any other forces, such as that due to air resistance. The force of gravity has the form ⃗ 𝐹 gravity = 𝑚𝑔⃗ 𝑗, and the force in the string has the form ⃗ 𝐹 string = −𝜆⃗ 𝑒𝑟, where 𝜆is an unknown tension, which is nonconstant in general. Note that, although the magnitude of this force is unknown, its direction is known: it is always parallel to ⃗ 𝑒𝑟. Newton’s law of motion for the ball requires that the product of its mass and acceleration be equal to the sum of the applied forces, or equivalently, (1.37) 𝑚𝑑2 ⃗ 𝑟 𝑑𝑡2 = ⃗ 𝐹 gravity + ⃗ 𝐹 string.
To put the above equation in a concise form, and eliminate the unknown magni-tude of ⃗ 𝐹 string, we consider the vector dot-product of the above with the unit vector ⃗ 𝑒𝜃, namely (1.38) 𝑚𝑑2 ⃗ 𝑟 𝑑𝑡2 ⋅⃗ 𝑒𝜃= ⃗ 𝐹 gravity ⋅⃗ 𝑒𝜃+ ⃗ 𝐹 string ⋅⃗ 𝑒𝜃.
By direct calculation, using the component expressions for all vectors involved, and the facts that ⃗ 𝑖⋅⃗ 𝑖= 1, ⃗ 𝑗⋅ ⃗ 𝑗= 1 and ⃗ 𝑖⋅ ⃗ 𝑗= 0, and noting that ⃗ 𝑒𝑟⋅⃗ 𝑒𝜃= 0 because they are perpendicular, we obtain (1.39) 𝑑2 ⃗ 𝑟 𝑑𝑡2 ⋅⃗ 𝑒𝜃= ℓ𝑑2𝜃 𝑑𝑡2 , ⃗ 𝐹 gravity ⋅⃗ 𝑒𝜃= −𝑚𝑔sin 𝜃, ⃗ 𝐹 string ⋅⃗ 𝑒𝜃= 0.
1.7. Case study 13 By substituting (1.39) into (1.38), and dividing out the mass and rearranging, we arrive at a differential equation for the pendulum motion. When the release conditions at time 𝑡= 0 are included, we obtain (1.40) ℓ𝑑2𝜃 𝑑𝑡2 + 𝑔sin 𝜃= 0, 𝑑𝜃 𝑑𝑡|𝑡=0 = 0, 𝜃|𝑡=0 = 𝜃0, 𝑡≥0.
The equations in (1.40) form a second-order, nonlinear, initial-value problem for the pendulum angle 𝜃as a function of time 𝑡. This function also naturally depends on the parameters 𝑔, ℓ, and 𝜃0 that appear in the equations, and we note that the mass 𝑚 was eliminated along the way. The theory of ordinary differential equations guarantees that there exists a unique solution 𝜃= 𝑓(𝑡, 𝑔, ℓ, 𝜃0), for some function 𝑓. Moreover, provided that the initial velocity is zero and the initial angle satisfies 𝜃0 ∈(0, 𝜋), this solution will be periodic in time with a period 𝑃= 𝐹(𝑔, ℓ, 𝜃0), for some function 𝐹.
Although they can be written in terms of certain special (elliptic) functions, there are no elementary expressions for 𝑓or 𝐹. Here we use dimensional methods to find a reduced form of the period relation and examine some implications.
Reduced equation for period. The quantities 𝑃, 𝑔, ℓ, 𝜃0 have dimensions [𝑃] = 𝑇, [𝑔] = 𝐿/𝑇2, [ℓ] = 𝐿and [𝜃0] = 1. A dimensional basis is {𝑇, 𝐿}, and the dimensional exponent matrix in this basis is (1.41) 𝐴= (Δ𝑃, Δ𝑔, Δℓ, Δ𝜃0) = ( 1 −2 0 0 0 1 1 0 ) .
An arbitrary power product has the form 𝜋= 𝑃𝑏1𝑔𝑏2ℓ𝑏3𝜃𝑏4 0 . The equation 𝐴𝑣= 0, where 𝑣= (𝑏1, ... , 𝑏4), has two free variables, and the general solution is (1.42) 𝑏1 = −2𝑏3, 𝑏2 = −𝑏3, 𝑏3 and 𝑏4 free.
Since there are two free variables, there are two independent solutions. For the first solution, we choose 𝑏3 = −1/2 and 𝑏4 = 0, which gives 𝜋1 = 𝑃√𝑔/ℓ. For the second solution, we choose 𝑏3 = 0 and 𝑏4 = 1, which gives 𝜋2 = 𝜃0. This is a full set of independent dimensionless power products, and is normalized with respect to 𝑃.
By the 𝜋-theorem, the period equation 𝑃= 𝐹(𝑔, ℓ, 𝜃0) must be equivalent to (1.43) 𝜋1 = 𝜙(𝜋2) or 𝑃= √ ℓ 𝑔𝜙(𝜃0), for some function 𝜙. Thus the relation between the quantities 𝑃, 𝑔, ℓ, 𝜃0 is not charac-terized by an unknown function of three quantities 𝐹(𝑔, ℓ, 𝜃0), but is instead charac-terized by an unknown function of one quantity 𝜙(𝜃0). Equivalently, the dependence of 𝐹(𝑔, ℓ, 𝜃0) on the quantities 𝑔and ℓis completely dictated by dimensional consider-ations.
Some implications. The reduced form of the period relation given in (1.43) has some interesting implications, as summarized next.
(1) A single curve of 𝜋1 versus 𝜋2 completely determines the function 𝜙, and hence the relation between the quantities 𝑃, 𝑔, ℓ, and 𝜃0. Thus the goal of any experi-ment or further analysis should be aimed at determining this curve.
14 1. Dimensional analysis (2) Consider any two pendula released from the same initial angle 𝜃0. Let {𝑔1, ℓ1, 𝜃0} be the parameters of the first pendulum, and {𝑔2, ℓ2, 𝜃0} be the parameters of the second. In view of (1.43), the periods of the two pendula are given by 𝑃 1 = √ℓ1/𝑔1 𝜙(𝜃0) and 𝑃 2 = √ℓ2/𝑔2 𝜙(𝜃0). By dividing these two expressions, we ob-tain a fundamental period law for pendula, namely (1.44) 𝑃 1 𝑃 2 = √ ℓ1𝑔2 ℓ2𝑔1 .
(3) The dependence of the period 𝑃on each of the quantities 𝑔, ℓ, and 𝜃0 can be char-acterized using (1.43). Specifically, for fixed 𝑔and 𝜃0, the period 𝑃is an increasing function of ℓ; for fixed ℓand 𝜃0, the period 𝑃is a decreasing function of 𝑔; and for fixed ℓand 𝑔, the period 𝑃increases or decreases with 𝜃0 depending on the function 𝜙.
A detailed analysis shows that the function 𝜙(𝜃0), 𝜃0 ∈(0, 𝜋) is positive, mono-tone, increasing, and has the limits (1.45) lim 𝜃0→0+ 𝜙(𝜃0) = 2𝜋, lim 𝜃0→𝜋−𝜙(𝜃0) = ∞.
Thus the period satisfies 𝑃≈2𝜋√ℓ/𝑔for a pendulum released from rest with initial angle 𝜃0 ≈0, which corresponds to a nearly vertical, downward position. Interestingly, the period 𝑃is arbitrarily large for initial angles 𝜃0 ≈𝜋, which corresponds to a nearly vertical, upward position. The special cases of 𝜃0 = 0 and 𝜃0 = 𝜋correspond to rest or equilibrium positions of the system in which no motion occurs. Such states and further properties of dynamical systems will be considered in later chapters.
Reference notes Classic references for the material presented here are the books by Birkhoff (2015) and Bridgman (1963). A recent treatment with a wealth of details and examples is given in Szirtes (2007), and a concise guide that illustrates various diverse applications is given in Lemons (2017).
Exercises 1. Let 𝑥, 𝑡, 𝑎, 𝑏be quantities in units {m, s}, and let ̃ 𝑥, ̃ 𝑡, ̃ 𝑎, ̃ 𝑏be the corresponding quantities in units {cm, hr}, with dimensions [𝑥] = 𝐿, [𝑡] = 𝑇, [𝑎] = 𝐿/𝑇and [𝑏] = 𝑇. Change the equation from 𝑥, 𝑡, 𝑎, 𝑏to ̃ 𝑥, ̃ 𝑡, ̃ 𝑎, ̃ 𝑏. Is the equation unit-free?
𝑥= 𝑎𝑡3 arctan(𝑡) (𝑏+ 4𝑡)2 .
(a) 𝑥= 𝑎𝑡2 arctan(𝑡/𝑏) 𝑏+ 4𝑡 .
(b) 2. Let 𝑥, 𝑦and 𝑝, 𝑞, 𝑟, 𝑠be quantities in units {ft, lb} in the basis {𝐿, 𝑀}. Also, let 𝑣= 𝑑𝑦 𝑑𝑥. Assuming [𝑥] = 𝐿and [𝑦] = 𝑀, find the dimensions [𝑝], [𝑞], [𝑟], [𝑠] as needed to make the given equation unit-free.
Exercises 15 𝑦= 𝑝𝑥2 + 𝑞sin(𝑟𝑥).
(a) 𝑦= 𝑝ln(𝑠𝑥) + 𝑞 𝑟+ 𝑥2 .
(b) 𝑦= (𝑝𝑥+ 𝑞𝑥2)𝑒𝑥/𝑟.
(c) 𝑦= 𝑝+ 𝑞𝑒𝑟𝑥2 𝑠+ 𝑥 .
(d) 𝑣= 𝑝𝑥−𝑞𝑥3 −𝑟.
(e) 𝑣= 𝑝𝑦−𝑞𝑥2 −𝑟𝑥𝑦2.
(f) 3. Let 𝑃, 𝑄, 𝑅, 𝑆be quantities in given units in a basis {𝐷1, 𝐷2}. Show that 𝑃= 𝑄+ 𝑅+ 𝑆is unit-free if and only if [𝑃] = [𝑄] = [𝑅] = [𝑆].
4. Let 𝑥, 𝑦, 𝑧> 0 and 𝑝, 𝑞, 𝑟> 0 be quantities with the following dimensions in the basis (1.1): [𝑥] = 𝐿, [𝑦] = 𝑀𝐿/𝑇, [𝑧] = 𝛩𝑀/(𝐿𝑇), [𝑝] = 𝐿 2, [𝑞] = 𝑀/𝑇, [𝑟] = 𝑀/𝛩.
Find a reduced form of the given equation assuming it is unit-free. If not possible, explain why.
𝑥= 𝑓(𝑦, 𝑞, 𝑟).
(a) 𝑦= 𝑓(𝑥, 𝑧, 𝑝).
(b) 𝑧= 𝑓(𝑥, 𝑦, 𝑟).
(c) 𝑦= 𝑓(𝑥, 𝑝, 𝑞).
(d) 5. Let 𝑢, 𝑣, 𝑤> 0 and 𝛼, 𝛽, 𝛾, 𝛿> 0 be quantities with the following dimensions in the basis (1.2): [𝑢] = 𝐶/𝑇, [𝑣] = 𝐻/𝑇, [𝑤] = 𝑃/𝑇, [𝛼] = 𝐶/𝐻, [𝛽] = 𝑃/𝐻, [𝛾] = 𝐻/(𝐶𝑇), [𝛿] = 1/𝑇.
Find a reduced form of the given equation assuming it is unit-free. If not possible, explain why.
𝑢= 𝑓(𝑣, 𝑤, 𝛼, 𝛽).
(a) 𝑣= 𝑓(𝑤, 𝛽, 𝛾, 𝛿).
(b) 𝑤= 𝑓(𝛼, 𝛽, 𝛾).
(c) 𝑤= 𝑓(𝑢, 𝑣, 𝛼, 𝛽, 𝛿).
(d) 6. An experiment to measure the temperature 𝑢in a furnace at time 𝑡is performed.
A curve fitting procedure applied to the 𝑢, 𝑡data yields the empirical relation 𝑢= 3.7𝑡1.5 + 4.2𝑡+ 293.2 in units {kelvin, minutes}. Here we explore different ways to make this relation unit-free.
Consider 𝑢= 𝑎𝑡1.5 + 𝑏𝑡+ 𝑐, where 𝑎, 𝑏, 𝑐are dimensional constants with values 3.7, 4.2, 293.2 in units {kelvin, minutes}. What must be the dimensions of 𝑎, 𝑏, 𝑐so that the equation is unit-free? What would be the values of 𝑎, 𝑏, 𝑐 in units {kelvin, hours}?
(a) Alternatively, consider 𝑢= 𝛽[3.7( 𝑡 𝛼)1.5 + 4.2 𝑡 𝛼+ 293.2], where 𝛼and 𝛽are dimensional constants, with values 𝛼= 1 minute and 𝛽= 1 kelvin, and 3.7, 4.2 and 293.2 are dimensionless (pure) numbers. Show that this form of the equation is also unit-free.
(b) 16 1. Dimensional analysis 7. Data is collected on the height 𝑦of certain trees at time 𝑡during their lifetimes.
A curve fitting procedure gives the empirical relation 𝑦= 52.4 −52.4𝑒−0.1𝑡− 3.3𝑡𝑒−0.2𝑡in units {foot, year}. Rewriting as 𝑦= 𝑎−𝑏𝑒−𝑐𝑡−𝑝𝑡𝑒−𝑞𝑡, find the di-mensions of 𝑎, 𝑏, 𝑐, 𝑝, 𝑞so that the equation is unit-free. Convert the equation to units {yard, decade}.
8. According to a simple theory of growth, the ultimate radius 𝑟> 0 of a single-celled organism is determined by the nutrient absorption rate 𝑎> 0 through its surface, and nutrient consumption rate 𝑐> 0 throughout its volume, via a unit-free equation 𝑟= 𝑓(𝑎, 𝑐). Find a reduced form of the relation using [𝑎] = 𝑀/(𝐿 2𝑇) and [𝑐] = 𝑀/(𝐿 3𝑇).
9. A metal forming process involves a pressure 𝑃> 0, length 𝑥> 0, time 𝑡> 0, mass 𝑚> 0 and density 𝜌> 0, and is described by a unit-free equation 𝑃= 𝑓(𝑥, 𝑡, 𝑚, 𝜌).
Find a reduced form of this relation and express the result explicitly in terms of 𝑃. Recall that [𝑃] = 𝑀/(𝐿𝑇2) and [𝜌] = 𝑀/𝐿 3.
10. A sphere of radius 𝑟> 0 is immersed in a fluid of density 𝜌> 0 and viscosity 𝜇> 0. When subject to a force 𝑞> 0, the sphere attains a terminal velocity 𝑣> 0.
We suppose there is a unit-free equation 𝑣= 𝑓(𝑟, 𝑞, 𝜌, 𝜇), where [𝑞] = 𝑀𝐿𝑇−2, [𝜌] = 𝑀𝐿 −3, [𝜇] = 𝑀𝐿 −1𝑇−1.
Find a reduced form of 𝑣= 𝑓(𝑟, 𝑞, 𝜌, 𝜇).
(a) For fixed 𝑞, 𝜌, 𝜇, show that 𝑣is proportional to a power of 𝑟.
(b) 11. A model for the digestion process in animals states that the absorption rate 𝑢> 0 of a given nutrient is determined by the concentration 𝑐> 0, residence time 𝜏> 0, and breakdown rate 𝑟> 0 of the nutrient in the gut, along with the volume 𝑣> 0 of the gut. We suppose there is a unit-free equation 𝑢= 𝑓(𝑐, 𝜏, 𝑟, 𝑣), where [𝑢] = 𝑀/𝑇, [𝑐] = 𝑀/𝐿 3 and [𝑟] = 𝑀/(𝐿 3𝑇).
Find a reduced form of 𝑢= 𝑓(𝑐, 𝜏, 𝑟, 𝑣).
(a) For fixed 𝑐, 𝜏, 𝑟, show that 𝑢is proportional to 𝑣.
(b) 12. In an explosion, a circular blast wave of intense pressure expands from the point of explosion into the surrounding air. A simple theory asserts that the radius 𝑟> 0 of the wave is determined by the elapsed time 𝑡> 0, the energy 𝐸> 0 released in the explosion, and the density 𝜌> 0 of the surrounding air, via a unit-free equation 𝑟= 𝑓(𝑡, 𝐸, 𝜌). Find a reduced form of this relation and show that the radius must increase with time in a nonlinear way; specifically, 𝑟increases as 𝑡2/5.
Here [𝐸] = 𝑀𝐿 2/𝑇2 and [𝜌] = 𝑀/𝐿 3.
13. In a domino toppling show, a long line of dominoes topple over, one by one, in a chain reaction. It is hypothesized that the speed 𝑣> 0 of the toppling wave depends on the spacing 𝑑> 0 and height ℎ> 0 of each domino, and gravitational acceleration 𝑔> 0, via a unit-free equation 𝑣= 𝑓(𝑑, ℎ, 𝑔). (The speed is assumed to be insensitive to the thickness and width of each domino.) Exercises 17 Show that a reduced form of the speed equation is 𝑣= √𝑔ℎ𝜙(𝑑/ℎ) for some function 𝜙.
(a) The figure below shows a plot of 𝜋1 = 𝜙(𝜋2), where 𝜋1 = 𝑣/√𝑔ℎand 𝜋2 = 𝑑/ℎ, made with data from different domino experiments. Note that the graph is approximately linear on the interval 0.1 ≤𝜋2 ≤0.7. Find a linear expres-sion for 𝜋1 = 𝜙(𝜋2) valid on this interval. Use this expression to write 𝑣in terms of 𝑑, ℎ, 𝑔.
π 0.1 0.7 1.5 1.2 π 2 1 (b) 14. Drone airplanes of surface area 𝑠use fuel of energy content 𝑒to fly at velocity 𝑣 through air of viscosity 𝜇. A theory proposes that the fuel consumption rate 𝑘 is determined by a unit-free equation 𝑘= 𝑓(𝑠, 𝑒, 𝑣, 𝜇), where [𝑘] = 𝐿 3𝑇−1, [𝑒] = 𝑀𝐿 −1𝑇−2 and [𝜇]=𝑀𝐿 −1𝑇−1.
Find a reduced form of 𝑘= 𝑓(𝑠, 𝑒, 𝑣, 𝜇).
(a) Using data from two experiments, and linear interpolation in a plot of the reduced form, find 𝑘when (𝑠, 𝑒, 𝑣, 𝜇) = (6, 3, 20, 1).
Data in some appropriate units 𝑠 𝑒 𝑣 𝜇 𝑘 experiment 1: 5 2 10 1 1 experiment 2: 7 3 15 2 5 (b) Mini-project. As developed in Section 1.7, a model for an ideal pendulum released from rest is g θ ℓ𝑑2𝜃 𝑑𝑡2 + 𝑔sin 𝜃= 0, 𝑑𝜃 𝑑𝑡|𝑡=0 = 0, 𝜃|𝑡=0 = 𝜃0, 𝑡≥0.
Here 𝜃is the pendulum angle, ℓis the length, 𝑔is gravitational acceleration, and 𝑡is time. The above system has a unique solution 𝜃= 𝑓(𝑡, 𝑔, ℓ, 𝜃0), for some function 𝑓, and provided that the initial velocity is zero and the initial angle satisfies 𝜃0 ∈(0, 𝜋), this solution will be periodic in time with a period 𝑃= 𝐹(𝑔, ℓ, 𝜃0), for some function 𝐹.
Here we study the period relation and construct an approximate formula for it using some data. All quantities are in units of meters and seconds.
(a) As outlined in the text, show that the reduced form of the period relation is 𝑃= √ℓ/𝑔𝜙(𝜃0), for some function 𝜙. Given that 𝜙and 𝐹are both unknown, what is 18 1. Dimensional analysis the conceptual advantage of the form 𝑃= √ℓ/𝑔𝜙(𝜃0) compared to the form 𝑃= 𝐹(𝑔, ℓ, 𝜃0)?
(b) The table below shows experimental measurements of 𝑃for five pendula with dif-ferent values of 𝑔, ℓ, and 𝜃0. Compute the value of 𝜙for each case; make a table or plot of 𝜙versus 𝜃0. Over the given interval of 𝜃0, what are the qualitative features of 𝜙? Does the function appear to be increasing or decreasing? Concave up or concave down?
𝑔, m/s2 ℓ, m 𝜃0, rad 𝑃, s case 1: 9.80 0.20 𝜋/12 0.9015 case 2: 9.80 0.10 𝜋/6 0.6457 case 3: 9.80 0.50 𝜋/4 1.4760 case 4: 9.80 0.30 𝜋/3 1.1798 case 5: 9.80 0.25 𝜋/2 1.1845 (c) Using your 𝜙versus 𝜃0 table, and linear interpolation between entries, predict the period 𝑃for given parameter values {𝑔, ℓ, 𝜃0} = {9.8, 0.5, 5𝜋 12 }, {9.8, 0.7, 5𝜋 12 }, {4.9, 0.6, 𝜋 3 }.
More generally, what would be an approximate formula for 𝑃, valid for any ℓ> 0, 𝑔> 0 and 𝜃0 ∈[ 𝜋 3 , 𝜋 2 ]?
(d) Use Matlab or other similar software to numerically solve the pendulum differen-tial equation, along with the initial conditions, to produce a plot of 𝜃versus 𝑡. Run a simulation with each set of parameters {𝑔, ℓ, 𝜃0} from part (c) and directly estimate the period from the plot for each case. Do the estimates agree with the predictions from (c)?
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## Cantor’s Paradise
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Fun with Number Theory: Primitive Roots
A gentle introduction using just multiplication and remainders
Russell Lim
8 min read
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Oct 27, 2022
Number theory questions such as the one below often appear on junior high school mathematics competition and aptitude tests:
Find the last digit of 2²⁰²².
If you’ve never come across something like this before, it seems like a computational nightmare. Until you dive in, start investigating the first few powers of 2 and a pattern emerges:
The last digits repeat in cycles of length 4: {2, 4, 8, 6, …}.
So the last digit of 2²⁰²² will be the same as the last digit of 2².
Problem solved!
If you find this cyclical pattern to be strangely satisfying, then you have been moved by the magic ✨ of number theory.
Investigating the cycles a bit further, they will lead us to a concept in number theory called “primitive roots”. Stick around, it won’t take too long… 🧚♀️
Why powers repeat in cycles
The fact that the last digits repeat in cycles is not specific to the base number 2. Any base at all would also give a repeating pattern, although the cycle length may be shorter or longer than 4.
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## Published in Cantor’s Paradise
38K followers
·Last published 19 hours ago
Medium’s #1 Math Publication
## Written by Russell Lim
3.3K followers
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I teach high school mathematics in Melbourne, Australia. I like thinking about interesting problems and learning new things.
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rezwits
Oct 29, 2022
```
0.5, 0.25, 0.125, 0.0625, 0.03125, 0.015625, 0.0078125, 0.00390625...
This is what pisses me off about those who believe in Infinity with Calculus. Because no matter what, if you take a summation of 1/2^n, n = 1 -> ∞
You will see that no matter what…
```
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Fresnel integrals
Version 1.1.0.0 (5.04 KB) by Christos Saragiotis
Calculates FresnelC, FresnelS integrals and their variations (C_1, C_2 and S_1 and S_2)
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FRESNEL(X) calculates the values of the Fresnel integrals for real values of vector X,
i.e.
C = \int_0^x cos(pit^2/2) dt, (0a)
S = \int_0^x sin(pit^2/2) dt (0b)
Also, it evaluates the following variations of the Fresnel integrals
C1 = \sqrt(2/pi) \int_0^x cos(t^2) dt, (1a)
S1 = \sqrt(2/pi) \int_0^x sin(t^2) dt (1b)
and
C2 = \sqrt(1/2/pi) \int_0^x cos(t) / \sqrt(t) dt, (2a)
S2 = \sqrt(1/2/pi) \int_0^x sin(t) / \sqrt(t) dt (2b)
The integrals are calculated as follows:
Values of X in the interval [-5,5] are looked up in Table 7.7 in and interpolated ('cubic'), if necessary. This table has values of the Fresnel integrals with 7 significant digits and even linear interpolation gives an error of no more than 3e-004.
Values outside the interval [-5,5] are evaluated using the approximations under Table 7.7 in . The error is less than
3e-007.
NOTE: The Tables were OCR'ed and although thoroughly checked, there might be some mistakes. Please let me know if you find any.
REFERENCE
Abramowitz, M. and Stegun, I. A. (Eds.). "Error Function and Fresnel Integrals." Ch. 7 in "Handbook of mathematical functions with formulas, Graphs, and mathematical tables", 9th printing. New York: Dover, pp. 295-329, 1970.
EXAMPLES OF USAGE
Y = fresnel(X,'c');
returns the Fresnel C integral according to eq. (0a)
Y = fresnel(X,'s',2);
returns the Fresnel S integral values according to eq. (2b)
[C,S] = fresnel(X,[],1)
returns the both the Fresnel C and S integral values according to eqs. (1a), (1b)
[C,S] = fresnel(X,[])
returns the both the Fresnel C and S integral values according to eqs. (0a), (0b)
Cite As
Christos Saragiotis (2025). Fresnel integrals ( MATLAB Central File Exchange. Retrieved September 29, 2025.
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Acknowledgements
Inspired:FresnelS and FresnelC
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fresnel(x,cs,var)
| Version | Published | Release Notes | |
--- --- |
| 1.1.0.0 | 29 Oct 2008 | Corrected the OCR typos mentioned by Peter Volegov (thanks, Peter, for taking the time) and three or four others as well. I checked it with some other relevant functions and it seems to be working OK now (differences are of the order of 1e-8). | Download |
| 1.0.0.0 | 27 May 2008 | | Download |
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https://www.youtube.com/watch?v=d7VPJ4DYwQQ
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Density (conceptual) | Middle school chemistry | Khan Academy
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Posted: 26 Mar 2025
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now:
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Density reveals how much mass is packed into a given volume. Different substances have different densities. Water’s density is 1.0 g/cm³, meaning any material with a density higher than this will sink in water, while lower densities float. Since each substance has a consistent density, this property helps identify materials and explain why some objects are “light” or “heavy” for their size.
00:00 A ton of bricks vs a ton of feathers
01:32 What is density?
04:18 Density values of common materials
04:45 Check your understanding of density
05:42 Water, floatation, & neutron stars
07:00 Summary
Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help!
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Transcript:
A ton of bricks vs a ton of feathers here's an old joke what weighs more a ton of bricks or a ton of feathers when I first heard this I was like of course bricks weigh more than the feathers come on that's obvious but wait we asked what weighs more a ton of bricks or a ton of feathers since they're both one ton they must weigh exactly the same but this feels weird my brain is telling me bricks must weigh more than the feather why is that well because when I hear this uh question I can completely ignore the one ton part and I consider them having equal volumes I consider equal volumes of bricks and feather and if you do consider equal volumes of bricks and feather then of course bricks will weigh more than the feather but the question asked one ton of bricks and feather let's see what would that look like if it took a ton of bricks it'll occupy a space which is almost the same as a household refrigerator but if you take a ton of feathers you need as as much you know Lo space of a small house okay only then they will have the same mass so clearly this is very weird and we usually don't think like that so we completely ignore the ton part and we assume instinctively that they have the same volume equal volume and therefore we conclude that bricks will weigh more than the feathers or we conclude that hey metals are heavier than plastic for example we're instinctively thinking that they have the same or equal volume and this is interesting because what we are instinct thinking about is something called the What is density? density what exactly is density you ask well density is a measure of how much mass is packed in a given volume now this will make more sense if we take some concrete example to do that let's first think about the units in the metric system for Mass we can use the units of kilograms or grams um for volume we can use cubic cm or cubic meters let's use cubic centim over here so let's now consider exactly 1 cm cube of bricks and 1 cm cube of feathers how big is a ctim cube you ask well just consider a box a cube of length and width and height all 1 cm that will have a volume of 1 cm cub and this is a very small volume it's same size as a small Dy so CM cube is pretty small it's actually the same as a milliliter cenim cube is the same as a milliliter as well but anyways if you consider 1 cm cube of bricks you know how much will be its mass it will have a mass of 1 6 G and if you take 1 cm cube of feather it will have a mass of 0.0025 G very little Mass so now over here look 1.6 G of bricks is packed per CM Cube so we say the density of the bricks is 1.6 G per CM Cube similarly over here look 025 G of feather is packed per CM Cube so its density is 025 G per CM CU that's what density is all about and look density of the bricks is much higher than density of the feathers and that's what we've been instinctively thinking about when we think that bricks are heavier than feathers we're not thinking about their Mass we're thinking about their density pretty cool right okay let's test our understanding before moving forward what if I had 10 cm cube of the same bricks same density what would be its mass can you pause and think about it all right since each CM Cube has a mass of 1.6 G if I have 10 10 cm cubes 10 times as much then the mass will also be 10 times as much so it'll be 10 1.6 which is 16 G does that makes sense and if I had for example 100 cm cube of the bricks then its mass would be 100 times more60 G this also helps us understand that density is an intrinsic property meaning density does not depend upon the amount of stuff it doesn't matter how many bricks or I have or how big the bricks are the density will be the same mass on the other hand is an extrinsic property it does depend upon the amount of stuff look more stuff more mass volume also depends upon the amount of stuff more stuff more volume but density which is a mass per volume that does not depend upon how much stuff you have that is an intrinsic property of the material itself so now let's look at the density Density values of common materials values of some common things you can see gases have very low density metals tend to have very high density osmium for example is one of the densest metals that we know but compare that with aluminium it's also a metal but it has a very low density and that's why it is used in construction material because you can build lightweight stuff that's pretty cool right and before we talk about more interesting things about density let's again check our understanding here's a pop quiz if I Check your understanding of density take the same volume of copper and gold which one do you think will weigh more why don't you pause and think about it all right I know that gold has a higher density compared to Copper I can see that over here meaning more mass is packed in a given volume compared to Copper which means for equal volumes gold must be heavier than copper okay another question a slightly harder one I have three materials aluminium copper and gold but this time I have the same mass which of the three will have the biggest size why don't you pause and try all right let's see let's say the mass is 100 G okay now of the three aluminium has the lowest density meaning it has the least amount of mass packed in a given volume so it'll take a lot of aluminium to get this up to 100 G and therefore if they need to have the same mass aluminium needs to be the biggest next would be copper and the last would be gold makes sense right okay another Water, floatation, & neutron stars interesting thing to note is water has a density of exactly 1.0 G per CM cube is that some kind of a neat coincidence no that's how we Define what a gram is we Define a gram as the mass of 1 cm cube of water that's our definition of a gram and that's why water ends up having a density of 1.0 G per CM Cube and this is pretty useful in predicting with the stuff will sink or float stuff that has density less than that of water will float and stuff that has density more than that of water will sink this is why for example if you take a big piece of wood let's say oak wood since it has a density less than that of water it will float and even if you take a very tiny piece of an iron nail because it has a density more than that of water it's going to sink but finally guess what's the densest thing in the universe well it happens to be something called the neutron stars their density is this big that's right if you take a CM cube of a neutron star which is like again of one die worth of neutron star its mass would be about 900 billion kilog can you imagine that that is about the mass of a Mount Everest a tiny die worth of neutron star will have that much mass wow so If You Now cal down and summarize Summary density is the amount of mass that is packed in a given volume mass per volume it can be measured in grams per CM Cube or other units like kilog per meter cube and so on but what's important is that it's an intrinsic property of a material it does not depend upon how much stuff you have
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Role of Puborectalis Muscle in the Genesis of Urethral Pressure | Journal of Urology
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No AccessJournal of UrologyInvestigative Urology 1 Oct 2012
Role of Puborectalis Muscle in the Genesis of Urethral Pressure
M. Raj Rajasekaran,
Dongwan Sohn,
Mitra Salehi,
Valmik Bhargava,
Helga Fritsch,and
Ravinder K. Mittal
M. Raj Rajasekaran
M. Raj Rajasekaran
Division of Gastroenterology, Department of Medicine, University of California-San Diego and Department of Medicine, San Diego Veterans Affairs Healthcare System, San Diego, California
More articles by this author
,
Dongwan Sohn
Dongwan Sohn
Department of Urology, Catholic University, Seoul, Korea
More articles by this author
,
Mitra Salehi
Mitra Salehi
Division of Gastroenterology, Department of Medicine, University of California-San Diego and Department of Medicine, San Diego Veterans Affairs Healthcare System, San Diego, California
More articles by this author
,
Valmik Bhargava
Valmik Bhargava
Division of Gastroenterology, Department of Medicine, University of California-San Diego and Department of Medicine, San Diego Veterans Affairs Healthcare System, San Diego, California
More articles by this author
,
Helga Fritsch
Helga Fritsch
Department of Anatomy, Medical University of Innsbruck, Innsbruck, Austria
More articles by this author
,and
Ravinder K. Mittal
Ravinder K. Mittal
Division of Gastroenterology, Department of Medicine, University of California-San Diego and Department of Medicine, San Diego Veterans Affairs Healthcare System, San Diego, California
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Abstract
Purpose:
The internal (smooth muscle) and the external (rhabdosphincter striated muscle) urethral sphincters have important roles in the genesis of urethral closure pressure. The U-shaped pelvic floor puborectalis muscle is important in the closure of anal and vaginal orifices in humans. We defined the contribution of the puborectalis to urethral pressure.
Materials and Methods:
A total of 11 female rabbits were anesthetized and prepared to measure urethral, vaginal and anal canal pressure using manometric methods. Pressure was recorded at rest, after administration of pharmacological agents and during electrical stimulation of the puborectalis and rhabdosphincter sphincter muscles. Phenylephrine, sodium nitroprusside (Sigma-Aldrich®) and rocuronium bromide (PharMEDium, Lake Forest, Illinois) were used to define the relative contribution of smooth and striated muscles to urethral pressure. Histology of the pelvic floor hiatus was also studied.
Results:
At rest mean ± SEM maximum urethral pressure was 13 ± 6 mm Hg. Sodium nitroprusside (50 μg/kg) infusion resulted in a 30% to 40% decrease in resting urethral pressure (mean 7.2 ± 0.2 mm Hg). Phenylephrine produced a dose dependent increase in urethral pressure (mean 17 ± 6, 25 ± 5 and 29 ± 6 for 5, 10 and 50 μg/kg intravenously, respectively). Electrical stimulation of the puborectalis muscle induced a stimulus dependent increase in urethral, vaginal and anal canal pressure. On the other hand, rhabdosphincter stimulation induced a stimulus intensity dependent increase in urethral pressure only. The increase in urethral pressure after puborectalis muscle stimulation was more than twofold higher than after rhabdosphincter stimulation.
Conclusions:
Our data prove that the puborectalis, a component of the pelvic floor muscles, is an important contributor to urethral pressure in the rabbit.
© 2012 by American Urological Association Education and Research, Inc.
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References
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Cited byHernández-Bonilla C, Zacapa D, Zempoalteca R, Corona-Quintanilla D, Castelán F and Martínez-Gómez M(2023)Multiparity Reduces Urethral and Vaginal Pressures Following the Bulboglandularis Muscle Stimulation in RabbitsReproductive Sciences, 10.1007/s43032-023-01263-3, VOL. 30, NO. 11, (3379-3387), Online publication date: 1-Nov-2023. Hernandez-Reynoso A, Rahman F, Hedden B, Castelán F, Martínez-Gómez M, Zimmern P and Romero-Ortega M(2023)Secondary urethral sphincter function of the rabbit pelvic and perineal musclesFrontiers in Neuroscience, 10.3389/fnins.2023.1111884, VOL. 17 Sorkhi S, Seo Y, Bhargava V and Rajasekaran M(2022)Preclinical applications of high-definition manometry system to investigate pelvic floor muscle contribution to continence mechanisms in a rabbit modelAmerican Journal of Physiology-Gastrointestinal and Liver Physiology, 10.1152/ajpgi.00295.2021, VOL. 322, NO. 1, (G134-G141), Online publication date: 1-Jan-2022. Tuttle L, Zifan A, Swartz J and Mittal R(2021)Do resistance exercises during biofeedback therapy enhance the anal sphincter and pelvic floor muscles in anal incontinence?Neurogastroenterology & Motility, 10.1111/nmo.14212, VOL. 34, NO. 1, Online publication date: 1-Jan-2022. Sánchez-García O, López-Juárez R, Corona-Quintanilla D, Ruiz Á, Martínez-Gómez M, Cuevas-Romero E and Castelán F(2021)(2021)Estrogens influence differentially on the pelvic floor muscles activation at somatovisceral reflexes involved in micturition of rabbitsMenopause, 10.1097/GME.0000000000001838, VOL. 28, NO. 11, (1287-1295) Tai J, Sorkhi S, Trivedi I, Sakamoto K, Albo M, Bhargava V and Rajasekaran M(2021)Evaluation of Age- and Radical-Prostatectomy Related Changes in Male Pelvic Floor Anatomy Based on Magnetic Resonance Imaging and 3-Dimensional ReconstructionThe World Journal of Men's Health, 10.5534/wjmh.200021, VOL. 39, NO. 3, (566), . Castelán F, Cuevas-Romero E and Martínez-Gómez MThe Expression of Hormone Receptors as a Gateway toward Understanding Endocrine Actions in Female Pelvic Floor MusclesEndocrine, Metabolic & Immune Disorders - Drug Targets, 10.2174/1871530319666191009154751, VOL. 20, NO. 3, (305-320) Rajasekaran M, Fu J, Nguyen M, Wang Y, Albo M and Bhargava V(2018)Age and multiparity related urethral sphincter muscle dysfunction in a rabbit model: Potential roles of TGF‐β and Wnt‐β catenin signaling pathwaysNeurourology and Urodynamics, 10.1002/nau.23889, VOL. 38, NO. 2, (607-614), Online publication date: 1-Feb-2019. Mittal R(2019)Anorectal Anatomy and FunctionAnorectal Disorders, 10.1016/B978-0-12-815346-8.00002-3, (9-28), . Tuttle L, Zifan A, Sun C, Swartz J, Roalkvam S and Mittal R(2018)Measuring length-tension function of the anal sphincters and puborectalis muscle using the functional luminal imaging probeAmerican Journal of Physiology-Gastrointestinal and Liver Physiology, 10.1152/ajpgi.00414.2017, VOL. 315, NO. 5, (G781-G787), Online publication date: 1-Nov-2018. Sánchez‐García O, López‐Juárez R, Rodríguez‐Castelán J, Corona‐Quintanilla D, Martínez‐Gómez M, Cuevas‐Romero E and Castelán F(2018)Hypothyroidism impairs somatovisceral reflexes involved in micturition of female rabbitsNeurourology and Urodynamics, 10.1002/nau.23594, VOL. 37, NO. 8, (2406-2413), Online publication date: 1-Nov-2018. Corona‐Quintanilla D, López‐Juárez R, Zempoalteca R, Cuevas E, Castelán F and Martínez‐Gómez M(2015)Anatomic and functional properties of bulboglandularis striated muscle support its contribution as sphincter in female rabbit micturitionNeurourology and Urodynamics, 10.1002/nau.22788, VOL. 35, NO. 6, (689-695), Online publication date: 1-Aug-2016. Oversand S, Atan I, Shek K and Dietz H(2016)Association of urinary and anal incontinence with measures of pelvic floor muscle contractilityUltrasound in Obstetrics & Gynecology, 10.1002/uog.14902, VOL. 47, NO. 5, (642-645), Online publication date: 1-May-2016. Dobberfuhl A, Spettel S, Schuler C, Levin R, Dubin A and De E(2015)Noxious electrical stimulation of the pelvic floor and vagina induces transient voiding dysfunction in a rabbit survival model of pelvic floor dystoniaKorean Journal of Urology, 10.4111/kju.2015.56.12.837, VOL. 56, NO. 12, (837), . Mittal R, Sheean G, Padda B and Rajasekaran M(2014)Length Tension Function of Puborectalis Muscle: Implications for the Treatment of Fecal Incontinence and Pelvic Floor DisordersJournal of Neurogastroenterology and Motility, 10.5056/jnm14033, VOL. 20, NO. 4, (539-546), Online publication date: 1-Oct-2014. Sohn D, Hong C, Chung D, Kim S, Kim S, Chung J, Joung J, Lee K and Seo H(2013)Pelvic floor musculature and bladder neck changes before and after continence recovery after radical prostatectomy in pelvic MRIJournal of Magnetic Resonance Imaging, 10.1002/jmri.24299, VOL. 39, NO. 6, (1431-1435), Online publication date: 1-Jun-2014. López-García K, Mariscal-Tovar S, Martínez-Gómez M, Jiménez-Estrada I and Castelán F(2014)Fiber type characterization of striated muscles related to micturition in female rabbitsActa Histochemica, 10.1016/j.acthis.2013.10.004, VOL. 116, NO. 3, (481-486), Online publication date: 1-Apr-2014. Corona‐Quintanilla D, Zempoalteca R, Arteaga L, Castelán F and Martínez‐Gómez M(2013)The role of pelvic and perineal striated muscles in urethral function during micturition in female rabbitsNeurourology and Urodynamics, 10.1002/nau.22416, VOL. 33, NO. 4, (455-460), Online publication date: 1-Apr-2014. Mittal R, Bhargava V, Sheean G, Ledgerwood M and Sinha S(2014)Purse-string morphology of external anal sphincter revealed by novel imaging techniquesAmerican Journal of Physiology-Gastrointestinal and Liver Physiology, 10.1152/ajpgi.00338.2013, VOL. 306, NO. 6, (G505-G514), Online publication date: 15-Mar-2014. Andersson K(2012)This Month in Investigative UrologyJournal of Urology, VOL. 188, NO. 4, (1065-1066), Online publication date: 1-Oct-2012.
#### Volume 188 #### Issue 4 #### October 2012
Page: 1382-1388
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© 2012 by American Urological Association Education and Research, Inc.
Keywords
urethra
pelvic floor
female
muscle
striated
urinary incontinence
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M. Raj RajasekaranDivision of Gastroenterology, Department of Medicine, University of California-San Diego and Department of Medicine, San Diego Veterans Affairs Healthcare System, San Diego, California More articles by this author
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