exercise 6.6

Z0=100;

ZLoad=.1+1j*50

RL=real(ZLoad)

XL=imag(ZLoad)

syms t                               % t=tand(beta*Length)

ZR=Z0*(RL+1j*(XL+Z0*t))/(Z0-XL*t+1j*RL*t)   

% to have ZRight=conj(ZLeft) right in the middle of the circuit t=tand(beta*LengthTL/2)

t0=double(solve(imag(ZR)==0,t))

imThreshold=.01

t0=t0(imag(t0)<imThreshold)     % removing non-null Im() root

t0=real(t0)

betaL=atand(t0)*2                  % betaL=beta*Length

% replacing negative length with positive values

betaL(betaL<0)=betaL(betaL<0)+180                     

% although betaL(1)==betaL(2) returns 0,

% betaL(1) is quite the same as betaL(2).

if numel(betaL>1) betaL=betaL(1); end

% by definition betaL=beta*Length=2*pi/lambda*D*lambda=2*pi*D

Length_TL_resonance=betaL/360

% alternatively:

dD=.0001

D=[0:dD:1];

syms D

Zin=Z0*(ZLoad+1j*Z0*tan(2*pi*D))./(Z0+1j*ZLoad*tan(2*pi*D))

reZin=real(Zin)

imZin=imag(Zin)

D1=double(solve(reZin==RL,D))

D2=double(solve(imZin==-XL,D))

D1=real(D1)     % ignore small imaginary residue

D2=real(D2)

% D is symmetric, so D<0 where resonance takes place,

% circuit also resonates at -D

D1=-D1;

D2=-D2;

betaL1=tan(2*pi*D1)

betaL2=tan(2*pi*D2)

Length1_TL_resonance=betaL1/360

Length2_TL_resonance=betaL2/360

% Q=2*pi*f0*L/R=XL/R with total resistance R=2*RL

R=2*RL;Q=XL/R