Chapter 14 : Transmission lines

Example 1 : pg 461

In [1]:
 
#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'
The inductance of 1m length is 225.0 nH/m

Example 2 : pg 462

In [2]:
 
#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'
a)The characteristic impedance of conductor is 227.399 ohm
b)The characteristic impedance of coaxial cable is 54.784 ohm

Example 3 : pg 463

In [3]:
 
#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'
The value of velocity factor is 0.69
The value of propagation velo. is 207019667.803 m/s

Example 4 : pg 468

In [4]:
 
#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'
The total voltage at the load is 0.333 V
The voltage across the line 0.333 V

Example 5 : pg 472

In [5]:
 
#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'
The length reqd for phase shift is 0.12375 m

Example 6 : pg 476

In [6]:
 
#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
The value of SWR is 2.0

Example 7 : pg 477

In [7]:
 
#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'
The amount of power reflected is 12.5 mW
The power dissipated in load is 37.5 mW

Example 8 : pg 478

In [8]:
 
#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'
The power absorbed by load is 44.444 W

Example 9 : pg 480

In [9]:
 
# 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'
The impedance looking toward the load (40-30j) ohm

Example 10 : pg 481

In [10]:
 
#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'
The length should be 0.1425 m

Example 11 : pg 481

In [11]:
 
#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'
The transmitter power must be 151.356 W

Example 13 : pg 490

In [13]:
 
#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'
Input impedance is (19.36+5.44j) ohm

Example 14 : pg 492

In [14]:
 
#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
The required turn ratio is (0.752156137344+0j)

Example 15 : pg 494

In [15]:
 
#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)
Zo =  66.476

Example 16 : pg 494

In [16]:
 
#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'
Capacitance is 2.12206590789e-11 F

Example 17 : pg 497

In [17]:
 
#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'
The value of stude is 0.014 S

Example 18 : pg 501

In [18]:
 
#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'
The distance is 168.0 m

Example 19 : pg 503

In [19]:
 
#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'
The wavelength is 46.0 cm
The freq is 619.565 MHz

Example 20 : pg 504

In [20]:
 
#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)
The value of SWR is 2.15