Chapter 20 : Satellite Communication

Example 1 : pg 754

In [1]:
 
# page no 754
# prob no 20.1
# part A)
from math import pi, sqrt
#calculate the velocity and orbital period of satellite in both cases
#given
d=500.;
#calculations and results
#By using the equation for velocity of a satellite
v=sqrt(4*10**11/(d+6400));
print 'A) The velocity of a satellite is',round(v,2),'m/s'
# The radius of orbit is 
r=(6400+d)*10**3#in m
#The orbital period of satellite is
T=(2*pi*r)/v;
print 'The orbital period of satellite is',round(T,2),'sec'
#part B)
d=36000.;
#By using the equation for velocity of a satellite
v=sqrt(4*10**11/(d+6400));
print 'B) The velocity of a satellite is',round(v,2),'m/s'
#The radius of orbit is 
r=(6400+d)*10**3#in m
#The orbital period of satellite is
T=(2*pi*r)/v;
print 'The orbital period of satellite is',round(T,2),'sec'
A) The velocity of a satellite is 7613.87 m/s
The orbital period of satellite is 5694.08 sec
B) The velocity of a satellite is 3071.48 m/s
The orbital period of satellite is 86735.85 sec

Example 2 : pg 757

In [2]:
 
# page no 757
# prob no 20.2
#calculate the required angle
#given
from math import atan, cos, sin, pi
R = 6400.#Radius of earth
L = 45.#earth station lattitude
H = 36000.#Height of satellite above the earth;
#calculations
ang = atan((6400. * sin(L * pi / 180.)) / (36000 + (6400 * (1 - cos(L * pi / 180.))))) * 180 / pi
#results
print "required angle is ",ang, "degrees"
required angle is  6.8137529672 degrees

Example 3 : pg 758

In [3]:
 
# page no 758
# prob no 20.3
#calculate the length of the path
#given
from math import sqrt, sin, cos, pi
#Determination of lenght of geostationary satellite with angle of elavation=30
#degree
r = 64. * 10 ** 5#Radius of earth
h = 36. * 10 ** 6#height of satellite
theta = 30 * pi / 180.#angle of elevation
#calculations
d = sqrt(((r + h) ** 2) - ((r * cos(theta)) ** 2)) - (r * sin(theta))
#results
print 'The length of the path is',round(d / 1000,3),'km'
The length of the path is 38836.175 km

Example 4 : pg 759

In [4]:
 
# page no 759
# prob no 20.4
#calculate the value of signal strength
#given
from math import log10
#A satellite transmitter operates at 4GHz with 7W & antenna gain 40dBi
#Receiver antenna gain 30dBi & path length is 4*10**7
Gt_dBi = 40.
Gr_dBi = 30.
Pt = 7
d = 40000.#in km
f = 4000.#in MHz
#calculations
Pr_Pt_dB = Gt_dBi + Gr_dBi - (32.44 + (20 * log10(d)) + (20 * log10(f)))
#Signal strength at transmitter
Pt_dBm = 10 * log10(Pt / 10 ** -3)
Pr_dBm = (Pt_dBm) + (Pr_Pt_dB)
#results
print 'The value of signal strength at receiver',round(Pr_dBm,3),'dBm'
The value of signal strength at receiver -88.071 dBm

Example 5 : pg 760

In [5]:
 
# page no 760
# prob no 20.5
#calculate the receiver
#given
from math import log10
# In the given problem
G = 40# receiving antenna gain
T_sky = 15.# noise temp
L = 0.4#loss between antenna and LNA input
T_eq = 40.# noise temperature f LNA
#calculations
# Fir-st we have to find G in dB
G_dB = G - L
# For the calculation of T, we have to convert the feedhorn loss into a ratio
# as follows
L = 10 ** (0.4 / 10)
Ta = ((L - 1) * 290. + T_sky) / L
# The receiver noise temperature is given wrt the chosen reference
# point,theefore
Ratio = G - 10 * log10(Ta + T_eq)
#results
print 'The receiver noise temperature is',round(Ratio,3),'dB'
The receiver noise temperature is 21.013 dB

Example 6 : pg 761

In [6]:
 
# page no 761
# prob no 20.6
#calculate the equivalent noise temperature
#given
NF_dB=1.5;# noise fig of a receiver
#calculations
NF=10**(NF_dB/10);
# Equivalent noise temperature is giveb as
T_eq=290*(NF-1);
#results
print 'Equivalent noise temperature is',round(T_eq,2),'K'
Equivalent noise temperature is 119.64 K

Example 7 : pg 761

In [7]:
 
# page no 761
# prob no 20.7
#calculate the carrier to noise ratio
#given
from math import log10
# refer prob no 20.5
d=38000.;#distance of satellite from the Earth surface
P=50.;#transmitter power
G=30.;#antenna gain
f=12000.;#frequency in MHz
B=10**6;# Bandwidth in MHz
#from problem no 2.5
G_T=21;
L_misc=0;
k_dBW=-228.6;#Boltzmann's constant in dBW
#calculations
# There are no miscellaneous loss
#The stellite transmitting power in dBW is 
Pt_dBW = 10*log10(P);
# The EIPR in dBW 
EIRP_dBW=Pt_dBW + G;
#FSL in dB
FSL_dB= 32.44 + (20*log10(d)) + (20*log10(f));
# The carrier to noise ratio is
ratio=EIRP_dBW - FSL_dB - L_misc + G_T - k_dBW - 10*log10(B);
#results
print 'The carrier to noise ratio at the receiver is',round(ratio,2),'dB'
The carrier to noise ratio at the receiver is 30.97 dB

Example 8 : pg 762

In [8]:
 
# page no 762
# prob no 20.8
#calculate the total time delay
#given
D=40000.;# distance of satellite from the earth station
v=3*10**8;# velo of light
d=80000.;# distance between two earth stations
#calculations
# time delay is given as
t=d/v;
# total time delay will be twice that of calculated above
T=2*t;
#results
print 'The total time delay is ',round(T,6),'sec'
The total time delay is  0.000533 sec

Example 9 : pg 769

In [9]:
 
# page no 769
# prob no 20.9
#calculate the gain and beamwidth
#given
from math import pi, log10
f_down = 4*10**9;# downlink freq
D=3.;#diameter
n=0.55;#efficiency
c=3.*10**8;#velo of light
#calculations
# The gain of a parabolic antenna is given as G=(n*%pi**2*D**2)/wl**2. Therefore wavelength is given as
wl=c/f_down
G=(n*pi**2*D**2)/wl**2;
G_dB = 10*log10(G);
# The beamwidth is given as
bw= (70*wl)/D;
#results
print 'The gain of TVRO is ',round(G_dB,2),'dB'
print 'The beamwidth is',round(bw,2),'degree'
The gain of TVRO is  39.39 dB
The beamwidth is 1.75 degree