exercise 5.23
ZIP with MATLAB scripts and note:
exercise 5.23 notes:
|gamma| over beta*L
ZL=100;Z0=50;a=1/L*log(Z0/ZL)
gamma(L)=.5*log(ZL/Z0)*exp(-1j*beta*L)*sin(b*L)/(b*L) same as
gamma(D)=.5*log(ZL/Z0)*exp(-1j*2*pi*D)*sin(2*pi*D)/(2*pi*D) with D=1 same as b*L=2*pi
gamma_mod=abs(.5*log(ZL/Z0).*exp(-1j*2*pi*D).*sin(2*pi*D)./(2*pi*D));
plot(D,gamma_mod);grid on;
Exponential impedance tapers should be used with length multiples of lambda/2, not lambda/4.
Finding L that obtains |gamma|=<0.05 over a 100% BW
nD1=find(gamma_mod<=.05)
hold all
plot(D(nD1),0.05,'ro')
one could certainly consider the first burst of consecutive D meeting |gamma|=<0.05
min(D(nD1))
nD1_HF=nD1(find(D(nD1)<.75));
% all D around D=0.5 gamma_mod<.05
% gamma_mod(nD1(find(D(nD1)<.75))) % check
max(nD1(find(D(nD1)<.75)))
D_HF=[min(D(nD1)) max(nD1(find(D(nD1)<.75)))]
nD1_LF=nD1(find(D(nD1)>.75)); % low frequency band
D1=D(nD1_LF(1)) % LF starts here, highest frequency of LPF
plot(D(nD1_LF),0.05,'bo')
the term is the way to ask for as much band stuck to the highest
frequency of the low frequency band or base band.
When no actual frequencies given, it's useful to think in terms of octaves: What values of L would fit octave up octave down around an undefined central frequency within baseband.
Understanding D1 is the highest frequency of the Low Pass or base band, then the central frequency D0=(D1+D2)/2; lower frequency limit D2=2*D0; and upper frequency limit D1=.5*D0 In optics people use lambdas or Ds rather than frequencies, which is a bit of upside down for the 'baseband' people (UHF, microwave, K, ..) from their point of view.
Note that the solutions manuals plots over betal*L but this plot is over D: beta*L = 2*pi*D
D0=D1*2
D2=D0*2
so the result is:
D_LF=[D1 D2]
centred at
D0
octave up, octave down is a 150% (fractional) bandwidth (2*f0-.5*f0)/f0=1.5
100% BW is half-octave up, half octave down: (1.5*f0-.5*f0)/f0=1, therefore
D0=D1*2
D2=D0*1.5
D_LF=[D1 D2]
same centre
D0
D0 = 1.718000000000000
D2 = 3.436000000000000
D_LF= 0.859000000000000 3.436000000000000
D0 = 1.718000000000000
D0= 1.718000000000000
D2 = 2.577000000000000
D_LF=1.718000000000000 2.577000000000000
D0= 1.718000000000000
How many Chebyshev sections would be required for the base band of previous point
BLm is beta*L, for 100% fractional bandwidth
BLm100=(1-2)*pi/4 % =pi/4
gamma_L=(ZL-Z0)/(ZL+Z0);
gamma_m=.05;
% reversing sec(BLm)
N=acosh(abs(gamma_L/gamma_m))/acosh(sec(BLm100))
N_df100=ceil(N)
BLm150=(1.5-2)*pi/4 % broader freq band means narrower BL
N=acosh(abs(gamma_L/gamma_m))/acosh(sec(BLm150))
N_df150=ceil(N)
BLm100 = -0.785398163397448
N_df100 = 3
BLm150 = -0.392699081698724
N_df150 = 7
N_df100 =3 means the Chebyshev transformer would be 3*lambda/4 long while the exponential transformer would have length D0=1.71
The Cheby impedance transformer would be a bit shorter, D=0.21, 0.21*lambda0 saved.
It’s worth mentioning that despite the gamma=(ZL-Z0)/(ZL+Z0) ~ ln(ZL/Z0) broadly used throughout this chapter, when using it here to calculate the amount of Chebyshev impedance transformer sections, then
N_df100_approx=ceil(cosh(abs(log(ZL/Z0)/(2*gamma_L)))/acosh(sec(BLm100)))
N_df150_approx=ceil(cosh(abs(log(ZL/Z0)/(2*gamma_L)))/acosh(sec(BLm150)))
N_df100_approx = 2
N_df150_approx = 4