# impedance matching single stub

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Moza il 29 Nov 2023
Risposto: David Goodmanson il 30 Nov 2023
I am trying to plot the reflection coefficient magnitude from 1 to 3 GHz for a single stub Short circuit to get the result below.
I worked on the code but I am not getting the same result.
clear all;
close all;
clc;
f = linspace(1e9, 3e9, 2000);
omega = 2 * pi * f; % Angular frequency
Z0 = 50;
Gamma1 = zeros(size(f));
Gamma2 = zeros(size(f));
c = 3e8;
lambda = c ./ f
l_stub1 = 0.110*(c./2e9);
l_stub2 = 0.260*(c./2e9);
for i = 1:length(f)
lambda = c / f(i); % Wavelength
beta = 2 * pi / lambda; % Phase constant
% Load impedance (complex due to capacitor)
% Stub impedances (short-circuited stub)
Z_stub1 = -1i * Z0 * tan(beta * l_stub1); % Solution #1
Z_stub2 = -1i * Z0 * tan(beta * l_stub2); % Solution #2
% Total impedance at the point where the stub connects
Z_total1 = 1 ./ (1./ZL + 1./Z_stub1);
Z_total2 = 1 ./ (1./ZL + 1./Z_stub2);
% Reflection coefficients
Gamma1(i) = (Z_total1 - Z0) / (Z_total1 + Z0);
Gamma2(i) = (Z_total2 - Z0) / (Z_total2 + Z0);
end
absGamma1 = abs(Gamma1);
absGamma2 = abs(Gamma2);
figure;
plot(f / 1e9, absGamma1, 'b', f / 1e9, absGamma2, 'r--');
xlabel('Frequency (GHz)');
ylabel('|Γ|');
title('Magnitude of Reflection Coefficient |Γ| versus Frequency');
legend('Solution #1', 'Solution #2');
grid on;
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### Risposta accettata

David Goodmanson il 30 Nov 2023
Hi Moza,
You have most of it, but you need to add a step and change a sign error. The section terminated by R and C, I called that section A of length la, and the shorted section is section S of length ls.
At the point where the two lines connect, the impedance of section A is not ZL, which is at the right end of the line, but rather the impedance seen at left end of the line, which is
Za = Z0*(ZL + i*Z0*tan(beta*la))./(Z0 + i*ZL.*tan(beta*la)); (1)
and your shorted stub impedance has a sign error and should be
Zs = (i*Z0*tan(beta*ls)); (2)
I defined (same as what you have)
lambda0 = c/2e9; so
case 1 la = .110*lambda0; ls = .095*lambda0;
case 2 la = .260*lambda0; ls = .405*lambda0;
If you add the new Za and Zs in parallel and keep going you will reproduce the plots.
Incidentally, with vector f a for loop is unnecessary if enough .* and ./ are used, which it seems like you already have quite a few of but not all.
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