- #1

Cram

- 3

- 1

## Homework Statement

Write code to calculate the input impedance on a thin wire antenna using the below form of Pocklington's integral equation. Use Method of moments.

Use a pulse as the basis function, point matching and delta-gap generator as the 'source'.

Assume the wire is broken up into 7 segments (N=7), the wire is 0.47λ long with radius 0.005λ.

## Homework Equations

Pocklington's integral equation was given as

[itex]E^{i}_{z}=\frac{j}{ωε}∫^{L/2}_{-L/2}I_{z}(z')\frac{e^{-jkR}}{4 \pi R^{5}}[(1+jkR)(2R^{2}-3a^{2})+(kaR)^{2}]dz'[/itex] where [itex] R=\sqrt{a^{2}+(z-z')^{2}}[/itex]

Note that the source point (vertical z) is on the outside of the wire and the field/observation point (z') is on the axis of the wire.

## The Attempt at a Solution

I have broken down pocklington's equation to try and fit it into the matrix expression [itex]

*= [Z]^{-1}[E] [/itex]. I ignored the pulse function as it was a constant 1 over the integral for each segment. [itex]*

*[/itex] is the set of weights representing the current on each segment which ends up out the front of the integral since they don't have dependence on z'.*

For the E column, I used all zeroes except for the 'gap' (relating to middle segment N=4) where it is V/Δz.

In this case V=1. I multipiled [E] by constant -j4∏ωε as I moved them from the RHS to the LHS in the above equation.

For the Z matrix, I am using the integral of the Green's function, which is the everything under the integral to the right of [itex]I_{z}(z')[/itex]. I am numerically integrating it over each segment using quadgk() in Matlab.

To getFor the E column, I used all zeroes except for the 'gap' (relating to middle segment N=4) where it is V/Δz.

In this case V=1. I multipiled [E] by constant -j4∏ωε as I moved them from the RHS to the LHS in the above equation.

For the Z matrix, I am using the integral of the Green's function, which is the everything under the integral to the right of [itex]I_{z}(z')[/itex]. I am numerically integrating it over each segment using quadgk() in Matlab.

To get

*, multiply inverse of [Z] by [E].*

Then finally to get the input impedance, I divide V (=1) by the centre segment current N=4.

The result I am getting (1.5541e+01 + 4.8658e+02j) doesn't match up with what I am expecting (closer to 100 + j100).

N=7; freq=3e8; omega=2*pi*freq; c=3e8; lambda=c/freq; k=2*pi/lambda; a=0.005*lambda

epsilon=8.85e-12; dz=L/N;

<<

for(m=1:N) {

};

E=zeros(N,1);

E(4,1) = -j*4*pi*omega*epsilon/dz;

I=inv(Z)*E;

Zin=1/I(4,1);

>>

Any assistance on the general approach would be greatly appreciated. I have tried different integrators which has yielded different but silly results.

I believe the integration of the Green's function is the problem.

Thanks,

CramThen finally to get the input impedance, I divide V (=1) by the centre segment current N=4.

The result I am getting (1.5541e+01 + 4.8658e+02j) doesn't match up with what I am expecting (closer to 100 + j100).

*The gist of my program is shown by pseudocode:*N=7; freq=3e8; omega=2*pi*freq; c=3e8; lambda=c/freq; k=2*pi/lambda; a=0.005*lambda

epsilon=8.85e-12; dz=L/N;

<<

for(m=1:N) {

zm=m*dz-(dz/2);

for(n=1:N) {

for(n=1:N) {

zn=n*dz-(dz/2);

//This integrates over nth segment length

Z(m,n)=quadgk(greens,zn-dz/2, zn+dz/2);

};//This integrates over nth segment length

Z(m,n)=quadgk(greens,zn-dz/2, zn+dz/2);

E=zeros(N,1);

E(4,1) = -j*4*pi*omega*epsilon/dz;

I=inv(Z)*E;

Zin=1/I(4,1);

>>

Any assistance on the general approach would be greatly appreciated. I have tried different integrators which has yielded different but silly results.

I believe the integration of the Green's function is the problem.

Thanks,

Cram

Last edited: