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Method of Moments I: Isolated Dipole Antenna

1 Introduction

This paper, part 1 of 2, gives a detailed description on how to solve for the exact current

distribution on an isolated thin wire dipole antenna, and to then plot the total radiated fields. A

detailed derivation of the equations will be given along with a description of how the equations

can be calculated numerically; specific examples using MATLAB are given.

We need computational electromagnetics because we cannot measure every quantity but we

can formulate our problems in very rigorous ways that allow us to obtain solutions; these

solutions can be found by applying different methods or taking different approaches by using a

computer. The solution of boundary problems can be solved numerically and in practical

applications these are large scale problems that take a considerable amount of time for even a

computer to solve. The method of breaking these boundary value problems into matrix

equations that can be solved on a computer is called the method of moments; the approach taken

here uses Pocklingtons integral equation to find the current on the surface of a dipole antenna.

2 Derivation of Pocklingtons Integral Equation

Figure 1 shows a wire antenna, which is considered a perfectly conducting scatterer, centered

about the z -axis with a current I flowing in the positive z -direction. i E is the incident electric

field on the cylindrical surface, s E is the scattered electric field from the surface, and t E is the

total field and is given by sit E E E += . The boundary conditions between a conductive mediaand free-space can be applied to give the following equations which are valid on the surface of

the wire.

0=n E t

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(2)

(3)

i E

s E

z

J

i E

s E

J

2/ L

2/ L

z

0 I

0 I

),,( r P

x

z

y

r R

x

y

2/ L

2/ L

r R

Figure 1: From Left to right: 1) Induced current on a PEC cylindrical surface due to an incident field, 2)

Representation of the sinusoidal current fed to a thin wire and an induced current due to a incident field; the actual

current on the antenna is the superposition of the two currents, 3) Center fed cylindrical dipole antenna, and its 4)

Thin wire approximation with a z -directed current located on the z -axis.

From here, a relationship between the incident and scattered electric fields on the surface of

the wire can be derived.

( )

0=

+=

+=

n E n E

n E E n E

si

sit

n E n E si =

On the surface of the wire we assume that:

1) Only the z-component is the relevant tangential component

2) The radius of the wire is very small, a

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(4)(5)

(6)

(7)

(8)

(9)

(10)

(11)

H j E =

E j J H +=

E

=v

g

0 H =g

Equation (7) implies that the vector field given by H only has rotational aspects (solenoidal)

meaning that it can be completely described by the curl of another vector field; this other field is

denoted by A and is called the vector potential.

0 H =g A H =

Now place (8) into (4), and this shows that the vector field given by the sum of two fields in

A j E + does not have any rotational aspects (irrotational) which implies that it can be

expressed in terms of the gradient of a scalar field called the scalar potential and denoted by .

( ) A j E = ( )0=+ A j E

E j A + = E j A = vv

Now take (8) again but this time place it and the result from (10) into (5). This result can

then be equated with the identity for the vector triple product as ( ) ( ) 2 A A A = g .Finally, by applying the Lorentz gauge condition, A j = g , an expression for the vector wave equation can be obtained where is the wave number.

( ) ( ) A J j j A = +

( )2 J A j = +

( )2 J A A = + + g

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(12)

(13)

(14)

(15)

( ) 2 A A= g

J A A =+ 22

The general solution of the vector wave equation for the vector potential at r due to the

source distribution over a volume V is given by (13) where each differential volume element is

given by z d yd xd d = . The vector R is called the separation vector and it spans the

length from the source point at r to the field point at r , and it has a magnitude of

( ) ( ) ( )222 z z y y x x R ++= .

( ) ( )

=

d Re

r J r A R j

V 4

We now make the thin wire approximation ( a

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(16)

(17)

(18)

(19)

(20)

( ) ( ) ( ) ( ) ( )

==

2

2

4

4

L

L

R j R j

z d R

e z I z z d yd xd

Re

z y x z I r A

We now seek to find an expression for the scalar potential and to do this requires going back

to Maxwells equations. This derivation begins with the divergence of the electric field given in

(10), and the divergence of the vector potential is obtained by the Lorentz condition. This

expression can be related to the charge density using the Maxwell equation in (6) which gives the

wave equation for the scalar potential. The solution to this equation is given by (19).

( ) E j A = g g

( )2 j A = g

( ) = 2 j j= 22

= 22

=

V

R j

d R

e

4

1

Now an expression for the charge density needs to be found in terms of the current

distribution. Using the curl of the magnetic field and the divergence of the electric field from (5)

and (6) we can derive the differential form of the continuity equation; this is done by taking the

divergence of H which by vector identity must be equal zero. The divergence is taken with

respect to the primed coordinates and is denoted by the operator .

z z

y y

x x

+

+=

=+ 22

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(21)

(22)

(23)

(24)

(25)

( ) ( )0i

i

H J j E H J j E

E

= + = + =

=v v v

g g gvg

( )i J j E j j

= = =

v vg g

( ) ( ) ( ) ( ) ( )( )

i

I z J I z x y x y

z z

= = v

g

Rearranging (22) and using (23) we have an expression for the charge density; equation (24)

is then substituted into (19) to obtain an expression for the scalar potential in terms of a singleintegral over the length of the line source.

( ) ( )( )1 I z

x y j z

=

=

V

R j

d R

e

4

1

( )1 14

j R

iV

e J d

j R

= v

g

( ) ( )[ ] ( )

=

V

R j

z d yd xd R

e z z I

y x j

1

41

( ) ( ) ( )

=

2

2

4

1 L

L

R j

z d R

e

z

z I yd y xd x

j

( )

=

2

2

4

1 L

L

R j

z d R

e z z I

j

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(26)

(27)

(28)

(29)

The gradient of (25) will only have a z -component and is then given by:

z

z

=

( )

=

2

2

41

L

L

R j

z d R

e z z I

z j

=2

2

41

L

L

R j

z d R

e z z

I j

We have now found expressions for the vector potential in (16) and the scalar potential in(25). The electric field radiated from the source can now be found from (10), with the use of the

relations in (27) and (28) as follows.

j j j ==

j

j j ==

z A j E z z

s z

=

( )

+=2

2

2

2

41

41

L

L

R j L

L

R j

z d R

e z z

I j

z d R

e z I j

( )

+=

2

2

2

2

41

L

L

R j

L

L

R j

z d Re

z z I

j z d Re

z I j

( )

+

=

2

2

2

2

2

4

L

L

R j L

L

R j

z d R

e z z

I z d

Re

z I j

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(30)

( )

+

=

2

2

2

4

L

L

R j

z d R

e z I

z z j

( ) ( )2

2

,4

L

L

I z K z z j

=

From the boundary condition on the surface of the wire, where a= , we know that the

tangential components are the z-components of the field and so we can write ( ) ( )a E a E s z i z = .

This gives us equation (30) where i z E is the known incident field on the surface of the wire and

( ) z I is the unknown current which appears in the integrand which gives us a Pocklingtons

integral equation that must be numerically solved. The kernel ( ), K z z is called the thin wirekernel or sometime the thin wire Greens function.

( ) ( )2

2

,4

L

i z

L

E I z K z j

=

We now need to find the current and so far we have formulated the entire problem as

expressing i z E over the integral of the unknown current and this is exactly what you get in

mutual coupling calculations. If you have two antennas in an array, one will emit a radiation

pattern which will in turn induce a current in the other array element which will cause a reaction.

We assume that these currents are strictly surface currents which is safe to assume if the

dipole can be considered a perfect conductor. From the geometry of a dipole it is easier to find amathematical expression for its radiated field in cylindrical coordinates because this will allow

the surface current to be described in terms of its and components.

3 Solution of the Thin Wire Kernel

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The thin kernel ( ), K z z can be reduced by straight forward differentiation. The kernel is given below along with some of the intermediate derivatives that are used.

( ), K z z = ( ) ( )2

22, , z z z z

z +

( ) R

z z R

z =

2

2

( ) ( ) z z R z

=

2222

( ) ( ) z z R R z

=

3222

( ) ( ) R

e z z je

z

R j R j

=

2

2

( )( ) R

e z z R j

R

e

z

R j R j

+=

12

2

( ) ( )

=

22

2

41

4 R

R z

ee z

R

z R

e

z

R j R j R j

( ) ( )=

341

Re z z e z z R j

z

R j R j

( ) ( )+

=

R j R

e z z z

R j

14

13

( ) [ ]( ) ( ) +++

=

3311

4

1

Re z z

z R j R j

z Re z z R j R j

( ) ( )( ) ( ) ( )

+

++

=

3331

41

Re

z z z z z

z Re

R j R

z z j

Re z z R j R j R j

( ) ( ) ( )( ) ( )

+++=

6

33

34

2

141

R

R z

ee z

R z z

Re

R j R

e z z j

R j R j R j R j

( ) ( ) ( ) ( ) ( )

+++=

7

23

34

2 31

4

1

Re z z Re z z R j

z z Re

R j R

e z z j

R j R j R j R j

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( ) ( ) ( )

+++=

5

2

34

2 311

4 R

R j z z

R R j

R z z

je R j

( ) ( ) ( ) ( )[ ]{ } R j z z R R j z z R j R

e R j

+++=

314

222

5

The kernel is then:

( ) ( ) ( )+

= z z z

z z j

z z K ,,1

, 222

= ( ) ( ) ( ) ( )[ ]{ }++++ R

e R j z z R R j z z R j

Re

j

R j R j

431

41 2222

5

= ( ) ( ) ( ) ( )[ ]{ }422225 3141

R R j z z R R j z z R j R

e j

R j

++++

( ) ( ) ( ) ( ) ( ) ( )[ ]{ }z d R R j z z R R j z z R j Re z I

j z E

L

L

R j

z ++++=

2/

2/

42222

5 3141

Evaluating z E on the surface of the cylinder, and then applying the boundary conditions, we get

the following relationship between the electric field and the incident/scattered field. We do not

worry about the phi-component and we hold the rho-component constant. The spatial vector

22 z a R += and the term ( ) z z is now simply evaluated at z .

( ) ( ) ( ), 0 2 , i s z z z E a H z H E z E z = =

Completely multiply out the terms in the bracket { }... in the expression for the z-component of the electric field and applying the above condition for the incident/scattered field results in the

following:

( ) ( ) ( ) ( )[ ] 42222 31 R R j z z R R j z z R j +++= ( )( ) ( ) ( )22222222 31 z a R R j z z a R j Rz j +++++

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= ( ) ( ) 22222222222 311 z Ra R Rz j z R j z a R j Rz j +++++++

=222222222222222 33 z Ra R R j Rz j Rz j Rz j z Rz j Ra j z a Rz j ++++++

Now simplify:

222222222222222 33 z Ra R R j Rz j Rz j Rz j z Rz j Ra j z a Rz j ++++++

= 2222222222222 22 z Ra R z R Ra j z a Rz j +++

= ( ) ( ) 2222222 22 a R z a R j z a ++++

= ( )( ) 22222 21 a R z a R j +++

= ( ) 222222 3221 a Ra z a R j +++

= ( )( ) 22222 321 a Ra R R j ++

Finally the kernel and the field at the surface of the cylinder can be given by the following

expressions.

( ) ( )( ){ }222225 32141

, a Ra R R j R

e j

z z K R j

m

++=

( ) ( )( )[ ] z d a R Rz ja R R j R

e z I

j E E

L

L

R ji

z s

z +++

==

2/

2/

222225 23414

1

4 Matrix Formulation of the Moment Method

We now have solved the kernel but the current is still in the integrand, and because the

current distribution is unknown we cannot solve this integral. The unknown we wish to solve for is in the domain of the integral operator and thus to solve for the current distribution we must

solve this integral equation. The current is solved for by first breaking the length of the antenna

into N segments ( fig. 1, left) over which the current has an unknown constant value. The

unknown current over each segment is approximated by a set of known expansion functions

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where the total current over the n th segment is I n multiplied by the n th expansion function

)( z F n .

( ) z I ( ) ( ) ( ) z F I z F I z F I N N +++ 2211 = ( )= N

nnn z F I

1(31)

2/ L

z

1=n2=n3=n

n=

2/ L

M

2/ L

2/ L

z

b

a

Figure 3: (left) Segmented dipole, (right) magnetic frill currentaround the feed point of the dipole

The point matching technique enforces the integral at a particular point which is taken at the

center of each segment here and denoted by z m for the mth point match. The integral over the nth

segment can then be expressed in terms of a function of the point matches which lie on the z -axis

and the source locations which lie on a line source located on the surface ( =a ) of the actual

cylindrical dipole.

( ) ( ) ( ), ,n

n m n m z

f z z F z K z z dz

= (32)From thin wire kernel solution above, (31), and (32) the incident electric field can be

approximated by the summation over each of the N segments:

( ) z d z z f I

j E mn

N

nn

i z

=,

1

1 (33)

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The basis functions can be modeled by the following sub-domain

functions: piecewise constant (34), piecewise linear (35), piecewise

sinusoidal (36), or truncated cosine (37) [2]. Sometimes entire domain

functions are used, such as (38), in order to reduce computational time [2].

( )

= e l s e

z z z z F nnn

0

11

(34)

( )

= ++

e l s e w h

z z z z

z z

z z z z

z z

z F nn

n

nnn

n

0

11

11

(35)

( )

[ ]

[ ]

= ++

e ls e w h e r

z z z z

z z

z z z z

z z

z F nn

n

nnn

n

0

)s i n ()(s in

)s in ()(s in

11

11

(36)

( ) 1 1cos 2n nn n

z z F z z z z =

(37)

( ) (2 1)cos2n

n z F z H z

H

=

(38)

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In MATLAB you could write a function that the main body of code or

subroutine could call as follows. Put the code in the m-file editor and save to

a location that is in the current path or set a new path; this allows MATLAB to

be able to find your function when it is called. Name the file m-file the same

name as the function, in this case get_expansion_function.m .

function [w z F] = get_expansion_function(m,zn,w_n,H,N,f,QP) % - - - - - - - % b = 2*pi/(3*10^8/f); switch w_n case 'pulse' ,

[w z] = quadr(zn(m),zn(m+1),QP); % weights/evaluation pointsF = ones(1,QP); % 1XQP row vector

case 'liner' ,dz = 2*H/(N+1);[w z] = quadr(zn(m),zn(m+2),QP);

z1 = z(1:QP/2);z2 = z(QP/2+1:end);F1 = z1-zn(m);F2 = zn(m+2)-z2;F = [F1 F2]/dz;

case 'sinsd' ,dz = 2*H/(N+1);[w z] = quadr(zn(m),zn(m+2),QP);z1 = z(1:QP/2);z2 = z(QP/2+1:end);F1 = sin(b*(z1-zn(m)))/sin(b*dz);F2 = sin(b*(zn(m+2)-z2))/sin(b*dz);F = [F1 F2];

case 'cosin' ,[w z] = quadr(zn(m),zn(m+1),QP);F = cos(b*(z-(zn(m+1)-zn(m))/2));

case 'entir' ,

[w z] = quadr(zn(m),zn(m+1),QP);F = cos((2*m-1)*pi*z/(2*H));otherwise ,

display( 'not a valid expansion function' );exit;

end % - - - - - % - - - - - % - - - - - % - - - - - % - - - - - % - - - - - %

Next, a model for the generator needs to be developed. The magnetic frill method (figure 3,

right) of determining the excitation vector is based on the problem of a monopole above a ground

plane and fed by a coaxial line. It can be shown through the equivalence principle and the image

method that this is equivalently an antenna of twice the length of the monopole driven by a frill

of magnetic current that is -directed. This current distribution (39) can then be used to

determine the incident field on the thin wire due to this source as (40).

)/ln(21

abM

=

(39)

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==

21

21

)/ln(21

),0( R

e R

eab

z E R j R j

i z

(40)

221 a z R +=

222 b z R += (41)

By integrating both sides of (40) over each of the N segments of the wire,

using the point matching technique along with a testing function (34-38),

then the m th segment is given below.

( ) ( ) ( ) ( )1

,m m

N i

m z m n n mn z z

j F z E z dz F z I f z z dz =

= ( ) ( ) ( ) ( ) ( )1

1 1 , ,m N

m m N N m F z I F z K z z dz I F z K z z dz dz

+ +

L

The summation above can be rearranged to get the following for 1 nN :

( ) ( ) ( ) ( ) ( ) ( ) E z F jdz z z f z F I dz z z f z F I dz z z f z F I mmmm z

i z m

z m N m N

z mm

z mm

=+++ ,,, 2211

For each of the m axial segments we get a set of linear equations 1 mN :

( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1

1 1 1 1 2 1 2 1 1 1 1, , ,i

N N z z z z z

I F z f z z dz I F z f z z dz I F z f z z dz j F z E z d

+ + + = ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2

1 2 1 2 2 2 2 2 2 2 2, , ,i

N N z z z z z

I F z f z z dz I F z f z z dz I F z f z z dz j F z E

+ + + =

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( ) ( ) ( ) ( ) ( ) ( ) E z F jdz z z f z F I dz z z f z F I dz z z f z F I N N N N z

N z

N N N N z

N N z

N N

=+++ ,,, 2211

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )( )

( )

=

E z F

E z F

E z F

j

I

I

I

dz z z f z F dz z z f z F dz z z f z F

dz z z f z F dz z z f z F dz z z f z F

dz z z f z F dz z z f z F dz z z f z F

N N N N z N

z

z

N z

N N N z

N N z

N N

z N

z z

z N

z m

z

1

1

222

111

1

1

2

1

21

22222212

1121111

,,,

,,,

,,,

This can be turned into a system of linear equations where each term in the generalized

impedance matrix and the excitation matrix are computed via numerical integration.

][]][[ VIZ =

( ) ( )

=m z

mnmmn dz z z f z F Z ,

( ) ( )dz z E z F V

m z

i

z mm =

5 Numerical Solution of Integrals

The numerical solution of the impedance matrix can be found by a Q-point quadrature which

is a numerical method of solving a definite integral. The weights can simply be computed as

shown below using the Legendre function call in MATLAB, where Q is the number of

quadrature points. P = LEGENDRE(N,X) computes the associated Legendre functions of

degree N and order M = 0, 1, ..., N , evaluated for each element of X . N must be a scalar integer and X must contain real values between -1

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The unscaled weights are wm = 2/ [(1-z m 2 )*P' N ( z m )2 ], m=0 ,1,..., N , where P' N ( z m ) are the

derivatives of the order-N Legendre polynomial, evaluated at the Legendre zeros. The second

row of LEGENDRE( N , z ) contains these derivatives. A very fast, and more accurate, way of

calculating these weights has been given by S. J. Orfanidis; here is an edited version:

% GQuad.m - Gauss-Legendre quadrature weights and evaluation points%% Usage: [w,x] = GQuad(a,b,Q)%% a,b = integration interval [a,b]% Q = number of weights in quadrature formula (even Q avoids x=0)%% w = length-Q column vector of (symmetric) weights% x = length-Q column vector of shifted/scaled Legendre evaluation points%% S. J. Orfanidis - 1999 - www.ece.rutgers.edu/~orfanidi/ewafunction [w,x] = GQuad(a,b,Q) P = legendre(Q,0); % evaluate Legendre functions at x=0m = (0:Q)'; % coefficient indexP = (-1).^m .* P ./ gamma(m+1); % order-Q Legendre polynomial coefficients z = sort(roots(flip(P))); % sort roots in increasing magnitudex = (z*(b-a) + a + b)/2; % shifted Legendre rootsk = (0:N-1)';c = (1 + (-1).^k) ./ (k+1); % this is \int_{-1}^1 t^k dtA = []; % coefficient matrix of the system A*w = cfor m=1:Q

A = [A, z(m).^k]; % build the columns of Aendw = A\c * (b-a)/2;w = w'; x = x'; % return w and x as row vectors (1xQ)

For the integral over the space m z , which covers a single axial segment, we get the Q-point

Gaussian quadrature weights and evaluation points by the following:

1 2[ ] ( , , ) p pw z GQuad z z Q=

Similarly, for the integral over a single source segment at a = we get:

1 2[ ] ( , , )q qw z GQuad z z Q =

With the above values we can approximate each integral in the impedance matrix and excitation

vector by a summation over each of their corresponding weights and evaluation points, and for a

particular segment the integral approximations will have the following form:

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1

1

( ) ( )

( ) ( )

m

n

Q

p p p z

Q

q qq z

f z dz w f z

f z dz w f z

=

=

Impedance Matrix:

mn Z ( ) ( ),m

m n m z

F z f z z dz

=

( ) ( ) ( ),m n

m n z z

F z F z K z z dz dz

=

( ) ( ) ( )1

,m

Q

m q n q qq z

F z w F z K z z dz =

( ) ( ) ( )1 1

,Q Q

p m p q n q p q p q

w F z w F z K z z = =

( ) ( ) ( )1 1

,Q Q

p m p q n q p q p q

w F z w F z K z z = =

=

Excitation Vector:

mV ( ) ( )0,

m

im z

z

F z E z dz

= =

( ) ( )1

0,Q

i p m p z p

p

w F z E z =

=( ) 1 2

1 1 22 ln( / )

p p j R j RQ p m p

p p p

w F z e e

b a R R

=

=

Where 2 21 p p R z a= + and 2 22 p p R z b= + .

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A strait forward, but computational expensive and unadvisable, way writing the above in

MATLAB is as follows:% for the mnth element of the impedance matrix Z(m,n)for m = 1:N % for each axial segment

[wp zp] = GQuad(z_n(m),z_n(m+1),quad_N); % axis segmentFm = get_mth_basis_function(m,zp,);

for n = 1:N % for each surface segment (rho=a)

[wq zq] = GQuad(z_n(n),z_n(n+1),quad_N); % source segmentFn = get_nth_testing_function(n,zq,);

% calculates the mutual impedance between the n th segment% along the surface of the dipole and the m th axial segment% by performing numerical integration (results in evaluating% the double integral on p.440)for p = 1:Q % for each quadrature point

Z(m,n)= Z(m,n)- wq(p)*Fn(p)*sum( wp.*Fm.*K(a,zq,zp(p)) );end

end % numerical integration over the mth axial segment to obtain% the excitation vector components using magnetic frill modelb = 2.3*a;R1p = sqrt(zp.^2+a^2);R2p = sqrt(zp.^2+b^2);V(m) = sum(wm .* Fm .* 1/(2*log(b/a)) .* (exp(-j*b_o*R1p)./R1p-exp(-j*b_o*R2p)./R2p));

end

6 Minimizing the Computational Requirements

The current of a symmetrically driven dipole will have current coefficients following the relation

1 2 1 3 2, , ... N N N I I I I I I = = = . This can also be seen by looking at the generalized impedance

matrix whose elements can also be referred to as the self-coupling coefficients. This matrix has atoeplitz symmetry which is due to the nature of the thin-wire kernel; that is the input to the kernel

only depends on the difference between z and z

. For this reason, the kernel is sometimes

written as (| |) K z z , and this means that 12 21 23 32... Z Z Z Z = = = , 13 31 24 42... Z Z Z Z = = = , and soon. Computation of the generalized impedance matrix can thus be done in MATLAB by only

finding the elements in the first row and then filling the rest of the matrix by using the following

code:% basis function for nth (n=1) segment - - - - - - - - - - - - - - - - - - -[wn zn Fn] = get_expansion_function(n,nodes,w_n,H,N,f,QP);

% for each axial segment - - - - - - - - - - - - - - - - - - - - - - -for m = 1:N [wm zm Fm] = get_expansion_function(m,nodes,w_n,H,N,f,QP);

[Z(m) V(m)] = solve(a,zn,wn,Fn,zm,wm,Fm,QP,f);end

% Fill Impedance Matrix - - - - - - - - - - - - - - - - - - - - - - - - - -Z_matrix = zeros(N,N);for m = 1:N

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for n = 1:NZ_matrix(m,n) = Z(abs(m-n)+1);

endend% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

function [Z V] = solve(a,zn,wn,Fn,zm,wm,Fm,QP,f) % - - - - - - - - - - - - - Z = 0;lam = (3*10^8)/f; % wave lengthb_o = 2*pi/lam; % free-space propagation constant % impedance matrixfor q = 1:QP

Z = Z - (wn(q)*Fn(q)) * sum( wm.*Fm.*K(a,zm,zn(q)) );end % voltage vector (Magnetic Frill Excitation)b = 2.3*a;R1 = sqrt(zm.^2+a^2);R2 = sqrt(zm.^2+b^2);V_m = 1/(2*log(b/a)) * (exp(-j*b_o*R1)./R1-exp(-j*b_o*R2)./R2);V = sum( wm.*Fm.*V_m );

% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Note, the above for-loop fills the impedance matrix with toeplitz symmetry, MATLAB also has a

command to do this but you should not use it. The values of each of the coefficients will be

complex, and when there are complex numbers entered in to toeplitz() function, it will fill the

matrix with their complex conjugates.

7 Solving for the Currents and the Input Impedance

Once the generalized impedance matrix and voltage matrix have been found, the currents can be

solved for by matrix inversion is shown below. The input impedance is easily obtained once the

currents have been solved by taking the center current element, which is the input current, as

given below:

in

inin

I V

Z =

For a 1-volt gap voltage this reduces to 1 /in in Z I = . In MATLAB:

% compute current and input impedance - - - - - - - - - - - - - - - - - - -I = pinv(Z_matrix)*V.';Zin = 1/I(round(N/2));I = abs(I);% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

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The transpose command in MATLAB can be taken by using the symbol ,, and by using . this

will take the non-conjugate transpose

8 Total Radiated Field of an Isolated Dipole

Given moment method solution for the currents, the radiated field for each individual segment

can be derived for a constant current from (16). The total radiated field (44) is given by the

superposition of each of the contributions where (42) and (43) give expressions for the and z

components of the nth segment. In order to get the total field expressions you need to start out

with the expression for the vector potential for the nth segment.

Solving for the magnetic and then the electric field in cylindrical coordinates:

( ) ( )1 1 1 1 z z n n A A A A A H A A z H z z

= = + + =

rr

14

n

j Rn

n z

I e H dz R

=

r4

n

j Rn

z

I e dz R

= 2 314 n

j Rn

z

I j e dz R R

= +

( )

( )

1 1 1 1

1 1

n n n zn znn n n

nn

n zn

H H H H H E H H z

j j z z

H H z

j z j

E E z

= = + + = +

= +

r r

2

3 4 5

3 3( )

4n

j Rnn

z

I j E z z e dz

j R R R

= + (42)

2 2 2 2

2 3 4 5

2 2 3 34

n

j Rn zn

z

I j j E e dz

j R R R R

+ = + (43)

( )4 4

n n

j R j Rn

zn nV z

I e e A r J dv dz R R

= = r rr

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2 2

1

N

total n znn

E E E =

= + r r (44)In MATLAB you can integrate this numerically to calculate the total field, another option would

be to use the far field approximations (magnitude/phase terms) to evaluate these integrals exactly

or to use the near field equations derived from the sinusoidal assumption.

% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -for m = 1:length(I)

[wp rp] = GQuad(zn(m),zn(m+1),QN); % returns column vectors for n = 1:QN

E1 = GET_E_rho(r, rp(n), theta);E2 = GET_E_z(r, rp(n), theta);

E_r(n,:) = I(m) * ( wp(n) * (E1.*sin(theta) + E2.*cos(theta)) );E_t(n,:) = I(m) * ( wp(n) * (E1.*cos(theta) - E2.*sin(theta)) );

endend E = sqrt(sum(E_r).^2 + sum(E_t).^2);% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

9 Verification of MoM Code

In order to verify results you can use measured data to show that the code developed agrees

with experimental results. The cylindrical dipole used by Mack * has a radius a=7.02210 -3 and

a length of L=0.47 .

% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -% Mack's Data% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -bh = [1.2 1.35 1.432 1.501 1.570 1.652 1.796 1.997 ...

2.355 2.751 3.141 3.398 3.649 3.8 3.926]; G0 = [4.12999 12.28333 15.51296 13.56922 10.1704 7.09591 4.24183 ...

2.63298 1.63816 1.23747 1.02096 0.91726 0.87181 0.91248 1.03854]; B0 = [ 9.59504 9.12585 2.93684 -2.22898 -4.43037 -4.77136 -3.92426 -2.72756 ...

-1.33809 0.18730 1.00032 1.91899 2.99706 3.80520 4.65317];

Y0 = G0 + j*B0;

ZA = 1./(G0 + j*B0);RA = real(ZA);XA = imag(ZA);% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

* R. W. P. King, G. J. Fikioris, and R. B. Mack, Cylindrical Antennas and Arrays (Rev. Sub. edition). CambridgeUniversity Press, 2002.