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Impedance on thin wire dipole using method of moments

  • Thread starter Cram
  • Start date
  • #1
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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 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).

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) {​

zn=n*dz-(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:

Answers and Replies

  • #2
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Bumping as thread has changed categories.
 

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