# Chapter 10 ACTIVE MICROWAVE CIRCUITS¶

## Example:10.1 page.no:554¶

In :
# program to determine the equivalent noise temperature of the amplifier .
from math import log10

T1=290;P1=-62;G=100;B=10**9;k=1.38*10**-23;
T2 =77; P2 = -64.7; Ts =450;
Y=P1-P2; # Yfactor in db.
Y=10**0.27;
Te=(T1-Y*T2)/(Y-1);
Po=G*k*B*(Ts+Te);
Po=10*log10(Po/0.001); #/ converting in to dBm.
print "the equivalent noise temperature in kelwin = ",Te
print "the total noise power out of the amplifier in dBm will be = ",Po

the equivalent noise temperature in kelwin =  170.074792027
the total noise power out of the amplifier in dBm will be =  -60.6767683732


## Example:10.2 page.no:557¶

In :
# program to find the dynamic range of the amplifier .
from math import log10

G=20;F=3.5; # in db.
k=1.38*10**-23;To=290;B=2*10**9;
# output noise power => No=G⇤F⇤k⇤To⇤B.so in dbm it will be
No=20+3.5+10*log10((k*To*B)/0.001);
DR=10-No;
print "the dynamic range in dB = %.3f"%DR

the dynamic range in dB = 67.467


## Example:10.3 page.no:558¶

In :
# program to calculate the noise figure ig anteena is replaced by amplifier .
from math import log10

L=10**0.2;T=300;To=290;Te=150;
Fl=1+(L-1.)*(T/To);
Fld=10.*log10(Fl); # converting in to dBm.
Fa=1.+(Te/To)
Fad=10.*log10(Fa);# converting in to dBm.
Fcas=Fl+L*(Fa-1.);
Fcasd=10*log10(Fcas);# converting in to dBm.
print "the noise figure of the cascade in dB = ",Fcasd
print "the noise figure of the amplifier in dB = ",Fad
print "the noise figure of the line in dB = ",Fld

the noise figure of the cascade in dB =  2.0
the noise figure of the amplifier in dB =  0.0
the noise figure of the line in dB =  2.0


## Example:10.4 page.no:562¶

In :
#program to calculate the impedence of the diode. 70

Cp=0.1e-12;Lp=2e-9;Cj=0.15e-12;Rs=10.;Is =0.1e-6;
Io1=0.;Io2=60e-6;alpha=0.04e3;
R1j=1./(alpha*(Io1+Is)); # for Io=0.
R2j=1./(alpha*(Io2+Is)); # for Io=60 mA.
print "junction resistance for Io=0, in ohm = ",R1j
print "junction resistance for Io=0, in ohm = %.3f"%R2j

junction resistance for Io=0, in ohm =  250000.0
junction resistance for Io=0, in ohm = 415.973


## Example:10.5 page.no:579¶

In :
# program to obtain the greatest ratio of off to on attenuation .
from math import pi,log10
from sympy import I

Cj=0.1*10**-12;Rr=1;Rf=5;
Li=0.4*10**-9;f=5*10**9;Zo=50;
w=2*pi*f;
Zr=Rr+I*((w*Li)-(1/(w*Cj)));
Zf=Rf+(I*w*Li);
# for series circuit .
ILon=-20*log10(abs((2*Zo)/(2*Zo+Zf)));
ILoff=-20*log10(abs((2*Zo)/(2*Zo+Zr)));
# for shunt circuit .
ILon1=-20*log10(abs((2*Zr)/(2*Zr+Zo)));
ILoff1=-20*log10(abs((2*Zf)/(2*Zf+Zo)));
print "for series circuit = %.2f"%ILon
print "for series circuit = %.2f"%ILoff
print "for shunt circuit = %.2f"%ILon1
print "for shunt circuit = %.2f"%ILoff1

for series circuit = 0.49
for series circuit = 10.16
for shunt circuit = 0.03
for shunt circuit = 7.62