Chapter 16 : Communication link analysis and Design

Example 16.1 Page No : 518

In [1]:
import math

# Variables
c = 3.*10**8;  #speed of light
Pt = 5.    #W
GtdB = 13.;  #dB
GrdB = 17.;  #dB
d = 80.*10**3;  #metre
f = 3.*10**9;   #frequency

# Calculations and Results
lembda = c/f;   #wavelength
print 'The wavelength is,  =  %.1f m'%(lembda);

Gt = 10**(GtdB/10);
Gr = 10**(GrdB/10);
print ' Gt = %.2f  '%(Gt);
print ' Gr = %.2f  '%(Gr);
Pr = lembda**2*Gt*Gr*Pt/((4*math.pi)**2*d**2);
print ' Pr = %.1f pW '%(Pr*10**12);

The wavelength is,  =  0.1 m
 Gt = 19.95  
 Gr = 50.12  
 Pr = 49.5 pW

Example 16.2 Page No : 520

In [2]:
import math 
# Variables
c = 3.*10**8;  #speed of light
Pt = 5.    #W
GtdB = 13.;  #dB
GrdB = 17.;  #dB
d = 80.;  #in km
f = 3.;   #frequency in GHz

# Calculations and Results
PtdBW = 10*math.log10(Pt);
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #dB
print 'The path loss is, (alfa_1dB)  =  %.2f dB'%(alfa1_dB);

PrdBW = PtdBW+GtdB+GrdB-alfa1_dB;   #calculation of recieved power in dB
print ' PrdBW) = %.2f  dBW'%(PrdBW)
 
Pr = 10**(PrdBW/10);   #recieved power in Watts
print ' Pr = %.1f pW '%(Pr*10**12);

The path loss is, (alfa_1dB)  =  140.04 dB
 PrdBW) = -103.05  dBW
 Pr = 49.5 pW

Example 16.3 Page No : 521

In [3]:
import math 

# Variables
d = 240000.*1.609;  #in km
#part a
f = 100.;   #frequency in MHz

# Calculations and Results
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+32.44;  #dB
print 'a) The path loss is %.2f dB'%(alfa1_dB);
#part b
f = 1;   #frequency in GHz
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #dB
print ' b) The path loss is %.2f dB'%(alfa1_dB);
#part c
f = 10;   #frequency in GHz
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #dB
print ' c) The path loss is %.2f dB'%(alfa1_dB);

a) The path loss is 184.18 dB
 b) The path loss is 204.18 dB
 c) The path loss is 224.18 dB

Example 16.4 Page No : 522

In [4]:
import math 

# Variables
f = 1.;  #in GHz
#part a
d = 1.;   #in Km

# Calculations and Results
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #dB
print 'a) The path loss is %.2f dB'%(alfa1_dB);
#part b
d = 10;   #in km
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #dB
print ' b) The path loss is %.2f dB'%(alfa1_dB);
#part c
d = 100;   #in km
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #dB
print ' c) The path loss is %.2f dB'%(alfa1_dB);

a) The path loss is 92.44 dB
 b) The path loss is 112.44 dB
 c) The path loss is 132.44 dB

Example 16.5 Page No : 522

In [5]:
import math 

# Variables
Pr = 50*10**-12;  #in Watts
GtdB = 3;  #dB
GrdB = 4;  #dB
d = 80;  #kilo-metre
f = 500;   #frequency in MHz

# Calculations and Results
PrdBW = 10*math.log10(Pr);   #in dB conversion
print 'PrdBW) = %.2f  dBW'%(PrdBW)
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+32.44;  #path loss in dB
print ' The path loss is, %.2f dB'%(alfa1_dB);
PtdBW = PrdBW+alfa1_dB-GtdB-GrdB;   #calculation of transmitted  power in dB
print ' PtdBW) = %.2f  dBW'%(PtdBW)
Pt = 10**(PtdBW/10);   #transmitted power in Watts
print ' Pt = %.1f W '%(Pt);

PrdBW) = -103.01  dBW
 The path loss is, 124.48 dB
 PtdBW) = 14.47  dBW
 Pt = 28.0 W

Example 16.6 Page No : 523

In [7]:
import math 

# Variables
Pr = 200.;  #in f-Watts
GtdB = 30.;  #dB
GrdB = 20.;  #dB
d = 40000.;  #kilo-metre
f = 4.;   #frequency in GHz

# Calculations and Results
PrdBf = 10*math.log10(Pr);   #in dBf conversion
print 'PrdBf) = %.2f  dBf'%(PrdBf)
alfa1_dB = 20*math.log10(f)+20*math.log10(d)+92.44;  #path loss in dB
print ' The path loss is, %.2f dB'%(alfa1_dB);
PtdBf = PrdBf+alfa1_dB-GtdB-GrdB;   #calculation of transmitted  power in dBf
PtdBW = PtdBf-150;   #calculation of transmitted  power in dBW
print ' PtdBf) = %.2f  dBf OR %.2f dBW'%(PtdBf,PtdBW)
Pt = 10**(PtdBW/10);   #transmitted power in Watts
print ' Pt = %.2f W '%(Pt);

PrdBf) = 23.01  dBf
 The path loss is, 196.52 dB
 PtdBf) = 169.53  dBf OR 19.53 dBW
 Pt = 89.80 W

Example 16.7 Page No : 525

In [8]:
import math 

# Variables
hT = 50.;  #m
hR = 5.;   #m

# Calculations
d_km = math.sqrt(17*hT)+math.sqrt(17*hR);   #in km

# Results
print ' dkm) = %.2f Km '%(d_km);

 dkm) = 38.37 Km

Example 16.8 Page No : 528

In [9]:
import math 

# Variables
Pt = 10000.;  #Watts
Gt = 25.;     #dB
f = 3;        #GHz
d = 50;     #km
sigma = 20   #radar cross section in m**2

# Calculations and Results
alfa2_dB = 20*math.log10(f)+40*math.log10(d)+163.43-10*math.log10(sigma);   #alfa2(dB) calculation
print ' The two way path loss is (alfa2dB) =  %.2f dB'%(alfa2_dB);
PtdBW = 10*math.log10(Pt);   #transmitted  power in dB
print ' PtdBW) = %i  dBW'%(PtdBW)
PrdBW = PtdBW+2*Gt-alfa2_dB   #dBW
print ' PrdBW) = %.2f dBW '%(PrdBW);
Pr = 10**(PrdBW/10);
print ' Pr = %.2f fW'%(Pr*10**15);

 The two way path loss is (alfa2dB) =  227.92 dB
 PtdBW) = 40  dBW
 PrdBW) = -137.92 dBW 
 Pr = 16.14 fW

Example 16.9 Page No : 530

In [10]:
# Variables
c = 3*10**8;    #speed of light in m/s
Td = 400*10**-6     #s

# Calculations
d = c*Td/2.       #in m

# Results
print ' d = %.0f Km '%(d*10**-3);

 d = 60 Km

Example 16.10 Page No : 530

In [11]:
# Variables
c = 3.*10**8;    #speed of light in m/s
fp = 2.*10**3;   #Hz

# Calculations and Results
T = 1./fp     #s
dmax = c*T/2       #in m
print 'a) d max = %.0f Km '%(dmax*10**-3);
tau = 6.*10**-6;   #s
dmin = c*tau/2   #m
print 'b) d min = %.0f m '%(dmin);

a) d max = 75 Km 
b) d min = 900 m

Example 16.11 Page No : 532

In [13]:
# Variables
c = 3.*10**8;    #speed of light in m/s
fc = 15.*10**9;   #Hz
v = 25.     # speed in m/s

# Calculations
fD = 2*v/c*fc;    #Hz

# Results
print 'Doppler shift , fD = %.0f Hz '%(fD);

Doppler shift , fD = 2500 Hz

Example 16.12 Page No : 532

In [14]:
# Variables
c = 186000.;    #speed of light in mi/s
fc = 10.*10**9;   #Hz
fD = 2.*10**3;    #frequency shift in Hz

# Calculations and Results
v = c*fD/(2*fc);  #speed in mi/s
print 'Speed of automobile , v = %.2f*10**-3 mi/s '%(v*10**3);
v = 3600*v;
print ' v = %.1f mi/hr '%(v);

Speed of automobile , v = 18.60*10**-3 mi/s 
 v = 67.0 mi/hr

Example 16.13 Page No : 535

In [15]:
import math 

# Variables
n1 = 1.;    #refraction index of air
E2 = 4.     #material dielectric consmath.tant
theta_i = 50.   #angle of incidence in degree  (misprinted in the solution)

# Calculations
n2 = math.sqrt(E2);
theta_r = math.asin(n1/n2*math.sin(theta_i*math.pi/180));

# Results
print ' The angle of refraction is %.2f  using angle of incidence  = 50)'%(theta_r*180/math.pi);
#misprinted angle 

 The angle of refraction is 22.52  using angle of incidence  = 50)