chapter 09 : Ground wave propagation

Exa 9.1 : page 9-23

In [4]:
from math import sqrt
#given data :
HT=50 #in meter
HR=10 #in meter
f=60 #in MHz
P=10 #in KW
D=10 #in Km
D=D*10**3 #in m
c=3*10**8 #speed of light in m/s
lamda=c/(f*10**6) #in meter
#Part (i) 
d=3.55*(sqrt(HT)+sqrt(HR)) #in Km
print "Maximum line of sight range = %0.2f km " %d 
#Part (ii)
Et=88*sqrt(P*1000)*HT*HR/(lamda*D**2)
print "The field strength at 10 km = %0.1e V/m" %Et 
#Part (iii)
#Formula : Et=88*sqrt(p)*HT*HR/(lambda*D**2)
Et=1 #in mV/m
D=sqrt(88*sqrt(P*1000)*HT*HR/(lamda*Et*10**-3)) #in m
print "Distance = %0.3f km " %(D/1000)

Maximum line of sight range = 36.33 km 
The field strength at 10 km = 8.8e-03 V/m
Distance = 29.665 km

Exa 9.2 : page 9-24

In [5]:
from math import sqrt
#given data :
P=200 #in KW
D=20 #in Km
D=D*10**3 #in m
E=300*sqrt(P)/D #in V/m
print "Field Strength at 20 km = %0.2f mV/m " %(E*10**3)

Field Strength at 20 km = 212.13 mV/m

Exa 9.3 : page 9-24

In [6]:
#given data :
HT=10 #in meter
HR=3 #in meter
P=200 #in W
D=50 #in Km
D=D*10**3 #in Km
f=150 #in MHz
c=3*10**8 #speed of light in m/s
lamda=c/(f*10**6) #in meter
E=88*sqrt(P)*HT*HR/(lamda*D**2) #in m
print "Field Strength at 20 km = %0.2f microV/m " %(E*10**6)

Field Strength at 20 km = 7.47 microV/m

Exa 9.4 : page 9-25

In [7]:
from math import sqrt
#given data :
HT=100 #in meter
d=60 #in Km
#Formula :  d=4.12*(sqrt(HT)+sqrt(HR)) #in Km
HR=(d/4.12-sqrt(HT))**2 #in meter
print "Height of receiving antenna = %0.2f m" %HR

Height of receiving antenna = 20.82 m

Exa 9.5 : page 9-25

In [8]:
from math import sqrt
#given data :
HT=3000 #in meter
HR=6000 #in meter
d=4.12*(sqrt(HT)+sqrt(HR)) #in Km
print "Maximum possible distance = %0.2f km" %d

Maximum possible distance = 544.80 km

Exa 9.6 : page 9-25

In [9]:
from math import log10
#given data :
f_MHz=3000 #in MHz
d_Km=384000 #in Km
PathLoss=32.45+20*log10(f_MHz)+20*log10(d_Km) #in dB
print "Path loss = %0.2f dB " %PathLoss

Path loss = 213.68 dB

Exa 9.7 : page 9-26

In [10]:
from math import pi
#given data :
#Part (i)
D=10 #in Km
lamda=10000 #in meter
LP=(4*pi*D*1000/lamda)**2 #in dB
print "Path loss = %0.2f dB" %LP
#Part (ii)
D=10**6 #in Km
lamda=0.3 #in cm
LP=(4*pi*D*1000/(lamda*10**-2))**2 #in dB
print "Path loss = %0.2e dB " %LP 
#Note : Answer in the book is wrong as value putted in the solution is differ from given in question.

Path loss = 157.91 dB
Path loss = 1.75e+25 dB

Exa 9.8 : page 9-26

In [11]:
from math import sqrt
#given data :
HT=50 #in meter
HR=5 #in meter
d=4.12*(sqrt(HT)+sqrt(HR)) #in Km
print "Range of LOS system = %0.2f km"%d

Range of LOS system = 38.35 km

Exa 9.9 : page 9-26

In [14]:
from math import pi
#given data :
PT=5.0 #in KW
PT=PT*1000 #in W
D=100.0 #in Km
D=D*10**3 #in m
f=300.0 #in MHz
GT=1.64 #Directivity of transmitter
GR=1.64 #Directivity of receiver
c=3*10**8 #speed of light in m/s
lamda=c/(f*10**6) #in meter
Pr=PT*GT*GR*(lamda/(4*pi*D))**2
print "Maximum power received = %0.3e W"% Pr
# Answer wrong in the textbook.

Maximum power received = 8.516e-09 W