ZIP with MATLAB scripts and note:
exercise 6.5
exercise 6.5 notes:
c0=2998792586;
LengthTL=.03 % 3cm stub length
f0=6e9;
lambda0=c0/f0 % air fllled transmission line
ZLoad_1=0
Z0=100
D=L/lambda0
ZinTL=1j*Z0*tan(2*pi*D)
% if capacitor C were for instance 10e-12
C1=10e-12 % Farad
Zin=1./(1j*2*pi*f0*C1+1./ZinTL)
% Now calculating capacitor that brings resonance at f0:
% Resonance takes place when
% ZinTL=1j*Z0*tan(2*pi*D); % short-circuit stub
syms C0
C0=double(solve(2*pi*f0*C0-1./(Z0*tan(2*pi/lambda0*LengthTL))==0,C0))
Corrections on the solutions manual:
1. air-filled TL beta is 12.57m^-1, not 125.7m^-1
2. beta*L is not 216degree, but 21.6
3. Zin is half the shown value, it's not 1j*72.6 ohm, but 1j*39.61
4. therefore the sought capacitor value is not 0.36pF but 0.66pF
2*pi*f0/c0
2*pi*f0/c0*LengthTL % betal*L [rad]
2*pi*f0/c0*LengthTL*180/pi % betal*L [degree]
100*tand(2*pi*6e9/c0*.03*180/pi)
1/(2*pi*f0*100*tand(2*pi*6e9/c0*.03*180/pi))
To achieve the erroneous beta of 125.7 a material with the following relative permittivity would be needed:
1/(2*pi*f0/(c0*125.7))^2
in Appendix G, only Titania has a similar relative permittivity
lambda0 =
0.499798764333333
D =
0.060024158002904
ZinTL =
0.000000000000000 +39.610360288354627i
Zin =
0.000000000000000 - 2.842967026681587i
C0 =
6.696688355154338e-13
= 12.571430254655670
= 99.977247246073190
= 21.608696881045308
= 39.610360288354620
= 6.696688355154338e-13
= 99.977
f1=0.5e9;f2=12e9;f0=6e9
Nf=1e6 % amount freq points between f1 f2
df=abs(f1-f2)/(Nf+1) % frequency resolution
f=[f1:df:f2];
ZinTL=1j*Z0*tan(2*pi/c0*f*LengthTL);
% short-circuit stub, ideal short
Zin=1./(1j*2*pi*f*C0+1./ZinTL);
figure(1);plot(f,abs(Zin));grid on
title('|Zin| ideal short');xlabel('f')
Gamma=(Zin-Z0)./(Zin+Z0);
absGamma=abs(Gamma);
figure(2);plot(f,absGamma);title('|\Gamma|');grid on
title('|\Gamma| ideal shortcut');xlabel('f')
ZL=.01
ZinTL=...
Z0*(ZL+1j*Z0*tan(2*pi/c0*f*LengthTL))./...
(Z0+1j*ZL*tan(2*pi/c0*f*LengthTL));
% short-circuit stub, real short
Zin=1./(1j*2*pi*f*C0+1./ZinTL);
figure(3);plot(f,abs(Zin));grid on
title('|Zin| real short');xlabel('f')
Gamma=(Zin-Z0)./(Zin+Z0);
absGamma=abs(Gamma);
figure(4);plot(f,absGamma);title('|\Gamma|');grid on
title('|Zin| real short');xlabel('f')
b)
R=1e4
equivalent circuit
Q0=2*pi*f0*RL*C0
at resonance, an equivalent inductance can be defined
L0=(1/(C0^.5*2*pi*f0) )^2
Q0 =
2.524592032791982e+02
L0 =
1.050697439590425e-09
