How to solve Microwave Engineering Pozar chapter 05 exercise 05 with ADS and MATLAB

Question



Answer

the literature source [POZAR] is available here:
https://www.amazon.co.uk/Microwave-Engineering-David-M-Pozar/dp/0470631554

There's also a solutions manual available here:
https://www.scribd.com/doc/176505749/Microwave-engineering-pozar-4th-Ed-solutions-manual





this is the OC series stub impedance match circuit on single frequency.



The arcs of the transmission line lengths on the Smith chart are:






And the OC series stub arcs:






There's also an alternative graph to find the required transmission line and stub lengths; |s11| surface.

clc;clear all;close all
ZL=90+1j*60;Z0=75;
Dstep=.001;drange=[0: Dstep:1];
D1=drange;D2=D1;
[D1,D2]=meshgrid(drange);

Z1=Z0*(ZL+1j*Z0*tan(2*pi*D1))./(Z0+1j*ZL*tan(2*pi*D1));
Zin_stub=Z0./(1j*tan(2*pi*D2)); % series oc stub
Zin=Z1+Zin_stub;
s11=(Zin-Z0)./(Zin+Z0);

surf(abs(s11),'LineStyle','none')
ax=gca
xlabel('D1');ylabel('D2');
ax.XTickLabelMode='manual'; ax.YTickLabelMode='manual';
ax.XTickLabel=[0:0.1:1];ax.YTickLabel=[0:0.1:1];
ax.XTick=[0:100:1000];ax.YTick=[0:100:1000];
ax.PlotBoxAspectRatio=[1 1 1]




hold all;x0=find(drange==.5) % plotting corner to box D1<.5 D2<.5
plot3([x0 x0 0],[0 x0 x0],[5 5 5],'Color',[1 0 0],'LineWidth',3)

% moving camera birdeye view
ax.CameraPosition=[500 500 10]

camzoom(ax,1.5) % zoom in a bit, camzoom is cumulative
% zoom_factor within [0 1) zooms out zoom_factor>1 zooms in
ax.CameraUpVector = [0 1 0]; % camera attitude
ax.CameraTarget = [500 500 0]; % centring
ax.CameraViewAngle =8*pi; % focus



automating peaks capture: findpeaks with MinPeakProminence=2.5 returns the right amount of peaks. MinPeakProminence larger or smaller than 2.5 then either too few or too many peaks.

With MinPeakHeight=2 command findpeaks doesn't catch the right amount of peaks for any MinPeakDistance, going from 6 peaks only to too many peaks.

With Threshold=2 doesn't catch a single peak but Threshold=1 gets the right amount of peaks.



To get zeros exact locations, it's useful to invert |s11| surface just plotted, with the Laplacian of the surface, command del2

V=1e3*del2(abs(s11));
figure(2);ax=gca;surf(V,'Lines','none');
xlabel('D1');ylabel('D2');
ax.XTickLabelMode='manual';
ax.YTickLabelMode='manual'; ax.XTickLabel=[0:0.1:1];ax.YTickLabel=[0:0.1:1];
ax.XTick=[0:100:1000];ax.YTick=[0:100:1000];
[pks,locs]=findpeaks(V( : ),'Threshold',1);
[nd1,nd2]=ind2sub(size(V),locs);
hold all;figure(2);plot3(nd2,nd1,V(nd2,nd1)+2,'ro'); % plot peaks
ax.PlotBoxAspectRatio=[1 1 1]

x0=find(drange==.5) % plot corner to box D1<.5 D2<.5
figure(2);
plot3([x0 x0 0],[0 x0 x0],[.5 .5 .5],'Color',[1 0 0],'LineWidth',3)
ax2=gca
ax2.CameraPosition=[500 500 10] % moving camera bird-eye view
ax2.CameraUpVector = [0 1 0]; % camera attitude

camzoom(ax,1.5) % zoom in a bit, camzoom is cumulative
% zoom_factor within [0 1) zooms out zoom_factor>1 zooms in
ax2.CameraTarget = [500 500 0]; % centring

abs(s11(sub2ind(size(V),nd1,nd2)))

= 0.001390798787950
0.001390798787950
0.002883920448033
0.002883920448033
0.001390798787950
0.001390798787950
0.002883920448033
0.002883920448033

unique(sort(drange(nd1)))
= 0.1470 0.3530 0.6470 0.8530

numel(nd1)
= 8

among these stub lengths

D1=unique(drange(nd1))
= 0.1470 0.3530 0.6470 0.8530

D2=unique(drange(nd2))
= 0.1740 0.4820 0.6740 0.9820


the stub lengths inside D<.5 are the smallest ones

Dstub1= D1([1 2])
= 0.1470 0.3530

Dstub2= D2([1 2])
= 0.1740 0.4820

The resulting frequency response is the same as the one obtained with the MATLAB solution



So the problem is simple enough to find transmission line and stub length on a surface.


Quick check how short circuit and open circuit stubs behave over frequency:

Zin_sc_stub=1j*Z0*tan(beta*L)

Zin_oc_stub=-1j*Z0*cot(beta*L)

da=pi/2000;a=[-2*pi:da:2*pi];
y_OC=-cot(a); y_SC=tan(a);
figure;plot(a,y_OC,a,y_SC);
grid on;axis([-2*pi 2*pi -10 10])
legend('SC','OC')
ylabel('Stub Z_i_n');xlabel('\betaL')
xticks([-2*pi -2*pi*3/4 -pi -pi/2 0 pi/2 pi 3*pi/2 2*pi])
xticklabels({'-2\pi','-3\pi/4','-\pi','-\pi/2','0','\pi/2','\pi','3\pi/2','2\pi'})




John BG
jgb2012@sky.com