Chapter 15 : Radio Wave Propagation

Example 1 : pg 517

In [1]:
#page no 517
#calculate the characteristic impedance of polyethylene
#prob no. 15.1
from math import sqrt
#Dielectric constt=2.3
#given
er=2.3;
#calculations
#Determination of characteristic impedance
Z=377./sqrt(er);
#results
print 'The charasteristic impedance of polyethylene is',round(Z,3),'ohm'
The charasteristic impedance of polyethylene is 248.586 ohm

Example 2 : pg 518

In [2]:
#calculate the max power density 
#page no 518
#prob no. 15.2
#given
#Dielelectric strength of air=3MV/m
e=3*10**6;#electric field strength
Z=377.;#impedance of air
#calculations
Pd=(e**2)/Z;#Determination of power density
#results
print 'The max power density is',round(Pd/10**9,3),'GW/m2'
The max power density is 23.873 GW/m2

Example 3 : pg 520

In [3]:
#calculate the power density 
#page no 520
#prob no. 15.3
from  math import pi
#given
#An isotropic radiator with power 100W & dist given is 10km
Pt=100;r=10*10**3;
#calculations
#Determination of power density at r=10km
Pd=Pt/(4*pi*(r**2));
#results
print 'Power density at a point 10km',Pd*10**9,'nW/m2'
Power density at a point 10km 79.5774715459 nW/m2

Example 4 : pg 521

In [4]:
#calculate the electric field strength 
#page no 521
#prob no. 15.4
from math import sqrt
#given
#An isotropic radiator radiates power=100W at point 10km
Pt=100.;r=10.*10**3;
#calculations
#Determination of electric field strength
e=sqrt(30*Pt)/r;
#results
print 'The electric field strength is',e*1000,'mW/m'
The electric field strength is 5.47722557505 mW/m

Example 5 : pg 525

In [5]:
#calculate the power delivered 
#page no 525
#prob no. 15.5
from math import log10
#A transmitter with power o/p=150W at fc=325MHz.antenna gain=12dBi receiver antenna gain=5dBi at 10km away
#given
#considering no loss in the system
d=10.;Gt_dBi=12.;Gr_dBi=5.;fc=325.;Pt=150.;
#calculations
#Determination of power delivered
Lfs=32.44+(20*log10(d))+(20*log10(fc))-(Gt_dBi)-(Gr_dBi);
Pr=Pt/(10**(Lfs/10));
#results
print 'The power delivered to receiver is',round(Pr*10**9,3),'nW'
The power delivered to receiver is 405.812 nW

Example 6 : pg 525

In [7]:
#calculate the power delivered to receiver 
#page no 525
#prob no. 15.6
#given
from math import log10
#A transmitter with o/p power=10W at fc=250MHz,connected to Tx 10m line with loss=3dB/100m t0 antenna with gain=6dBi.
#Rx antenna 20km away with gain=4dBi 
#Refer fig.15.6,assuming free space propagation
d=20;fc=250.;Gt_dBi=6.;Gr_dBi=4.;loss=3./100;Zl=75.;Zo=50.;L=10.;Pt=10.;
#calculations
Lfs=32.44+(20*log10(d))+(20*log10(fc))-Gt_dBi-Gr_dBi;#path loss

L_tx=L*loss;#Determination of loss
ref_coe=(Zl-Zo)/(Zl+Zo);#Reflection coefficient
L_rx=1-(ref_coe**2);#Proportion of incident power that reaches load
L_rx_dB=-10*log10(L_rx);#Converting that proportion in dB
#Determination of total loss Lt
Lt=(Lfs)+(L_tx)+(L_rx_dB);
#Determination of power delivered to receiver
Pt_Pr=10**(Lt/10);#Power ratio
Pr=Pt/Pt_Pr;
#results
print 'The power delivered to receiver is',Pr,'W'
The power delivered to receiver is 2.04329571558e-09 W

Example 7 : pg 530

In [8]:
#calculate the angle of refraction 
#page no 530
#prob no. 15.7
#given
from math import sqrt, sin, asin,pi
#A radio wave moves from air(er=1) to glass(er=7.8).angle of incidence=30 deg
theta_i=30.;er1=1;er2=7.8;
#calculations
#determination of angle of refraction
theta_r=asin((sin(theta_i*pi/180.))/(sqrt(er2/er1)));
#results
print 'The angle of refraction is',round(theta_r*180/pi,3),'degree'
The angle of refraction is 10.313 degree

Example 8 : pg 537

In [9]:
#calculate the max usable freq 
#page no 537
#prob no. 15.8
#given
from math import cos,pi
#A Tx statn with fc=11.6MHz & angle of incidence=70 degree
theta_i=70.;fc=11.6;#in MHz
#calculations
#determination of max usable freq(MUF)
MUF=fc/(cos(theta_i*pi/180.));
#results
print 'The max usable freq MUF is',round(MUF,2),'MHz'
The max usable freq MUF is 33.92 MHz

Example 9 : pg 539

In [12]:
#calculate the max common distance in both cases 
#page no 539
#prob no. 15.9
#given
from math import sqrt
#A taxi compony using central dispatcher with antenna height=15m & taxi antenna height=1.5m
ht=15.;hr=1.5;
#calculations and results
#a)Determination of max commn dist betn dispatcher and taxi
d1=sqrt(17*ht)+sqrt(17*hr);
print 'a)The max common distance between dispatcher & taxi',round(d1,3),'km'
#b)Determination of max ommn dist betn 2 taxis
d2=sqrt(17*hr)+sqrt(17*hr);#ht=hr=height of antenna of taxi cab
print 'The max common distance between two taxi is',round(d2,3),'km'
a)The max common distance between dispatcher & taxi 21.018 km
The max common distance between two taxi is 10.1 km

Example 11 : pg 545

In [13]:
#calculate the fading period in both cases 
# page no 545
# prob no 15.11
#given
# An automobile travels at 60km/hr
v=60.*10**3/(60*60);#conversion of car's speedto m/s
c=3.*10**8;#speed of light
#part a) calculation of time between fades if car uses a cell phone at 800*10**6Hz
f=800.*10**6;
#calculations and results
T=c/(2*f*v);
print 'The fading period is',T,'sec'
#part b) calculation of time between fades if car uses a PCS phone at 1900*10**6Hz
f=1900.*10**6;
T=c/(2*f*v);
print 'The fading period is',round(T,5),'sec'
# Note that the rapidity of the fading increases with both the frequency of the transmissions and the speed of the vehicle
The fading period is 0.01125 sec
The fading period is 0.00474 sec

Example 12 : pg 550

In [14]:
#calculate the no. of cell sites 
#page no 550
#given
# problem no 15.12
A=1000.;#metropolitian area expressed in sq. km
r=2;#radius of cell in km
#calculations
# Number of cell sites given as
N=A/(3.464*r**2);
#results
print 'Number of cell sites are',round(N)
Number of cell sites are 72.0