#page no 461
#prob no. 14.1
#calculate the inductance
#A coaxial cable with capacitance=90pF/m & characteristic impedance=50 ohm
#given
C=90.*10**-12;Zo=50.;
#Determination of inductance of 1m length
#calculations
L=(Zo**2)*C;
#results
print 'The inductance of 1m length is',L*10**9,'nH/m'
#page no 462
#prob no. 14.2
#calculate the imepdance
from math import log10,sqrt
#a)Determination of impedance of open wire with diameter 3mm & r=10mm
#given
D=3./2.;r=10.;#All values are in mm
#calculations and results
Zo1=276*log10(r/D);
print 'a)The characteristic impedance of conductor is',round(Zo1,3),'ohm'
#b)Determination of impedance of coaxial with er=2.3,inner diameter=2mm & outer diameter=8mm
er=2.3;D=8.;d=2;#All diameter values in mm
Zo2=(138/sqrt(er))*log10(D/d);
print 'b)The characteristic impedance of coaxial cable is',round(Zo2,3),'ohm'
#page no 463
#prob no. 14.3
#calculate the velocity factor and propagation velocity
#Cable with teflon dielectric er=2.1
from math import sqrt
#given
er=2.1;c=3.*10**8;#Velocity of light
#calculations and results
#Determination of velocity factor
Vf=1/sqrt(er);
print 'The value of velocity factor is',round(Vf,3)
#Determination of propagation velocity
Vp=Vf*c;
print 'The value of propagation velo. is',Vp,'m/s'
#page no 468
#prob no. 14.4
#calculate the voltage
#Refer fig. 14.13(a)
vs=1;#source voltage
Rs=50.;#source resistance
Zo=50.;#line impedance
RL=25.;#load resistance
l=10.;#length of line
vf=0.7;#velocity factor
Vi=0.5;
c=3.*10**8;#velo of light
#calculations
#Vs will divide between Rs and Zo of the line.Since two resistors are equal,the voltage will divide equally.
#Therefore at t=0,the voltage at the source end of the line will rise from zero to 0.5V.
#The voltage at the load will remain zero untill the surge reaches it.The time for this is
T=l/(vf*c);
# After T sec, the voltage at the load will rise.The reflection coefficient is given as
refl_coeff=(RL-Zo)/(RL+Zo)
#Now reflection voltage is
Vr=refl_coeff * Vi;
#The total voltage at the load is
Vt=Vr+Vi;
# The reflected voltage will propogate back along the line,reaching
#the source at time 2T.After this the voltage will be 0.3335V all along the line
#The voltage across the line, and the load will be
VL=vs*(RL/(RL+Zo));
#results
print 'The total voltage at the load is',round(Vt,3),'V'
print 'The voltage across the line',round(VL,3),'V'
#page no 472
#prob no. 14.5
#calculate the length reqd
#Standard coaxial cable RG-8/U with 45 degree phase shift at 200MHz
#given
p=45.;f=200.*10**6;c=3.*10**8;#Speed of light in m/s
vf=0.66;#velo. factor for this line
#calculations
vp=vf*c;#Determination of propagation velo.
wav=vp/f;#Determination of wavelength of signal
#Determination of reqd length for 45 degree phase shift
L=wav*(p/360.);
#results
print 'The length reqd for phase shift is',L,'m'
#page no 476
#prob no. 14.6
#calculate the value of SWR
#A 50ohm line terminated in 25ohm resistance
Zo=50.;Zl=25.;
#calculations
#Determination of SWR
SWR=Zo/Zl;#In this case Zo>Zl
#results
print 'The value of SWR is',SWR
#page no 477
#prob no. 14.7
#calculate the power
#A generator sends 50mW at 50ohm line & reflection coeff I=0.5
#given
Pi=50.;I=0.5;
#calculations
#Determination of amount of power reflected
Pr=(I**2)*Pi;
#Determination of remainder power that reaches load
Pl=Pi-Pr;
#results
print 'The amount of power reflected is',Pr,'mW'
print 'The power dissipated in load is',Pl,'mW'
#page no 478
#prob no. 14.8
#A transmitter supplies 50W with SWR 2:1
#calculate the power
#given
Pi=50.;SWR=2.;
#calculations
#Determination of power absorbed by load
Pl=(4*SWR*Pi)/(1+SWR)**2;
#results
print 'The power absorbed by load is',round(Pl,3),'W'
# page no 545
# prob no 14.9
#calculate the impedance
from math import pi,tan
#given
Zo=50.;# line impedence in ohm
ZL=100.;# load impedance in ohm
vf=0.8;#velocity factor
l=1.;#length of line
f=30.*10**6;# freq of operation
c=3.*10**8;#velo of light
#calculations
# we have to find the length of line in degree
wl=vf*c/f#wavelength
# Then the length of line in degree is
ang=l/wl*360
# calculation of impedance
Z=Zo*(ZL+(1j*Zo*tan(ang*pi/180)))/(Zo+(1j*ZL*tan(ang*pi/180.)));
print 'The impedance looking toward the load',Z,'ohm'
#page no 481
#prob no. 14.10
#calculate the length required
#A series tuned ckt tuned at 1GHz
#given
vf=0.95;c=3.*10**8;f=10**9;
#calculations
vp=vf*c;#determination of propagation velo.
wav=vp/f;#Determination of wavelength
#Determination of length
L=wav/2;#Since half wavelength section wiil be series resonant
#results
print 'The length should be',L,'m'
#page no 481
#prob no. 14.10
#calculate the transmitter power
#A Tx deliver 100W to antenna through 45m coaxial cable with loss=4dB/100m
#given
loss=4./100;L=45.;Pout=100.;
#calculations
loss_dB=L*loss;#Determination of loss in dB
Pin_Pout=10**(loss_dB/10);
#Determination of Tx power
Pin=Pout*Pin_Pout;
#results
print 'The transmitter power must be',round(Pin,3),'W'
#page no 490
#prob no. 14.13
#calculate the input impedance
Zo=50.;#line impedance in ohm
f=100.*10**6;#operating freq
vf=0.7;#velocity factor
L=6.;#length in m
c=3.*10**8;#velo of light
ZL=50+1j*50;#load impedance in ohm
#calculations
# we have to calculate length in degree,so for this first find wl
wl=vf*c/f;#wavength in m
ang=360*L/wl;
# now from the graph input impedance is 19.36+%i5.44;
Zi=19.36+1j*5.44;
#results
print 'Input impedance is',Zi,'ohm'
#page no 492
#prob no. 14.14
from cmath import sqrt
#given
Zo=50.;#line impedance in ohm
ZL=75.+1j*25;
# the requirment of this is simply to match the 50ohm line to the impedsnce at this point on the line,which is 88.38 ohm,resistive.
Z2=88.38;#in ohm
#calculations
#The required turn ratio is
N1_N2=sqrt(Zo/Z2);
#results
print 'The required turn ratio is',N1_N2
#page no 494
#prob no. 14.15
#calculate the impedance
# refer prob no 14.14
from math import sqrt
#given
Zo=50.;#line impedance in ohm
Z2=88.38;#in ohm
#calculations
Zo_=sqrt(Zo*Z2);
#results
print 'Zo = ',round(Zo_,3)
#page no 494
#prob no. 14.16
#calculate the capacitance
from math import pi
#given
Zo=50.;#line impedance in ohm
f=100.*10**6;#operating freq in Hz
ZL1=50+1j*75;# load impedance with Xc=75
Xc=75;
#calculations
# Capacitance in farads is given as
C=1/(2*pi*f*Xc);
#results
print 'Capacitance is',C,'F'
#page no 497
#prob no. 14.17
#calculate the value of stude
#given
Zo=72.;#line impedance in ohm
ZL=120.-1j*100;#load impedance
#calculations
#The stub must be inserted at a point on the line where the real part of the load admittance is correct. This value is
s=1/Zo;
#results
print 'The value of stude is',round(s,3),'S'
#page no 501
#prob no. 14.18
#calculate the distance
#given
#A TDR display shows dscontinuity at 1.4us & vf=0.8
t=1.4*10**-6;vf=0.8;c=3.*10**8;#Speed of light
#Determination of distance of fault
#calculations
d=(vf*c*t)/2;#One-half time is used to calculate
#results
print 'The distance is',d,'m'
#page no 503
#prob no. 14.19
#calculate the wavelength and freq
#given
#2 adjacent minima on slotted are 23cm apart with velo factor=95%
L=23*10**-2;vf=0.95;c=3*10**8;#Velo. of light in m/s
#calculations and results
#Determination of wavelength
wav=2*L;#Minima are seperated by one-half wavelength
print 'The wavelength is',wav*100,'cm'
#Determination of freq.
f=(vf*c)/wav;#vp=vf*c
print 'The freq is' ,round(f/10**6,3),'MHz'
#page no 504
#prob no. 14.20
#calculate the value of SWR
from math import sqrt
#given
#Frwd power in Tx line is 150W,Reverse power=20W
Pi=150.;Pr=20.;#All power in watt
#calculations
#Determination of SWR
SWR=(1+sqrt(Pr/Pi))/(1-sqrt(Pr/Pi));
#results
print 'The value of SWR is',round(SWR,3)