# Chapter13 Transmission Lines And Cables¶

## Example 13.5.2,Pg.no.475¶

In [1]:
import math
a=0.0006  #The attenuation coeff in N/m
#a) Determinaion of the attenuation coeff in dB/m
a_dB=8.686*a
a_dB=round(a_dB,3)
print 'The attenuation coeff is',a_dB,'dB/m'
#b) Determination of attenuation coeff in dB/mile
k=1609   #conversion coeff for meter to mile
a_dB_mile=k*a_dB
a_dB_mile=round(a_dB_mile,2)
print 'The attenuation coeff is',a_dB_mile,'dB/mile'

The attenuation coeff is 0.005 dB/m
The attenuation coeff is 8.04 dB/mile


## Example 13.10.1,Pg.no.485¶

In [33]:
import math
Z0=50.0   #measured in ohm
VSWR=2.0
d=0.2     #distance from load to firstt minimum
T=(VSWR -1)/(VSWR+1)
pi=180.0
Ql=pi*(4*0.2-1)       #using Euler’s identity
e=complex(math.cos(Ql),math.sin(Ql))   #expansion for e ˆ( jQl )
a=T*e
ZL=Z0*(1+a)/(1-a)
Zlr=round(ZL.real,2)
Zli=round(ZL.imag,2)
print 'a)The equivalent series resistance is',Zlr,'ohm'
print '  The equivalent series reactance is',Zli,'ohm'
Yl=1/ZL
Yl1=round(1/Yl.real,2)
Yli=round(1/Yl.imag,2)
print 'b)The equivalent parallel resistance is',Yl1,'ohm'
print '  The equivalent parallel reactance is',Yli,'ohm'

a)The equivalent series resistance is 37.15 ohm
The equivalent series reactance is 27.63 ohm
b)The equivalent parallel resistance is 57.7 ohm
The equivalent parallel reactance is -77.57 ohm


## Example 13.11.1,Pg.no.488¶

In [34]:
import math
from math import pi
d=0.1   #length of 50ohm short−circuited line
Z0=50   #in ohm
f=500*10**6   #freq in Hz
Bl=2*pi*d
#a) Determination of equivalent inductive reactance
Z=complex(0,Z0*math.tan(Bl))
Z=round(Z.imag,2)
print 'The equivalent inductive reactance is',Z,'j','ohm'

The equivalent inductive reactance is 36.33 j ohm


## Example 13.17.1,Pg.no.513¶

In [35]:
import math
from math import pi
VSWR=2.0
l_min=0.2
Z0=50.0
Ql=((4*l_min )-1)*pi
tl=(VSWR -1)/(VSWR+1)
Tl=complex(0,tl*math.e**(Ql))
Zl=Z0*(1+Tl)/(1-Tl)
Zr=round(Zl.real,2)
Zm=round(Zl.imag,2)
print 'a) The equivalent series resistance is',Zr,'ohm'
print '   The equivalent series reactance is',Zm,'ohm'
Yl=1/Zl
Yl=1/Yl.real
Yl=round(Yl,2)
print 'b) The equivalent parallel resistance is',Yl,'ohm'

a) The equivalent series resistance is 46.93 ohm
The equivalent series reactance is 17.24 ohm
b) The equivalent parallel resistance is 53.27 ohm


## Example 13.17.2,Pg.no.514¶

In [20]:
import math
ZL=complex(30,0)-complex(0,23)
l=0.5     #length of line in m
Z0=50.0     #characteristic impedance in ohm
wl=0.45   #wavelength on the line in m
B=2*pi/wl
Tl=(ZL-Z0)/(ZL+Z0)
VI=1.0     #reference voltage in volt
VR=VI*Tl
z=complex(B*l)
Vi=VI*math.e**(z)
Vr=VR*math.e**(-z)
V=Vi+Vr
I=(Vi-Vr)/Z0
Z=V/I
Z1=round(Z.real,2)
Zi=Z.imag*10**5
Zi=round(Zi,2)*10**-5
print 'The input impedance is',Z1,'+',Zi,'j','ohm'

The input impedance is 50.0 + -2.87e-05 j ohm


## Example 13.17.3,Pg.no.515¶

In [24]:
import math
from math import pi,sqrt
Z0=600.0
Zl=73.0  #in ohm
F=0.9
QF=(2*pi*F)/4
#For matching , the effective load impedance on the main line must equal the characteristic impedance of the mail line
Zl1=Zl
Z01=sqrt(Zl1*Zl)
Tl=(Zl-Z01)/(Zl+Z01)
VI=1  #reference voltage
Vi=VI*math.e**(complex(0,QF));
Vr=Tl*VI*math.e**-(complex(0,QF))
V_in=Vi+Vr
I_in=(Vi-Vr)/Z01
Z_in=V_in/I_in
print'The input impedance is',Z_in,'ohm'
#the voltage reflection coeff is
TL_F=(Z_in -Z0)/(Z_in+Z0)
#the VSWr is given as
VSWR_F=(1+TL_F)/(1-TL_F)
VSWR_Fr=round(VSWR_F.real,2)
VSWR_Fi=VSWR_F.imag
print 'The VSWR is',VSWR_Fr,'+',VSWR_Fi,'j'

The input impedance is (73+0j) ohm
The VSWR is 0.12 + 0.0 j