Chapter 18 : Terrestrial Microwave system

Example 1 : pg 676

In [1]:
 
# page no 676
#calculate the height of each tower
# prob no 18_1
#given
#Transmitter and receiver have same height at dist 40km
d=40;#dist is 40 km
#calculations
h=(d**2)/68.;# As d=sqrt(17h)+sqrt(17h)
#results
print 'The height of each tower must be at least',round(h,3),'m'

The height of each tower must be at least 23.529 m

Example 2 : pg 678

In [2]:
 
# page no 678
# prob no 18_2
from math import sqrt
#calculate the distance required
#A line of sight radio link at freq 6GHz with seperation 40 km betn antennas
#given
f=6.;d1=10.;d2=30;#obstacle located at 10 km
#calculations
#Determination of dist R to clear obstacle 
R=10.4*sqrt((d1*d2)/(f*(d1+d2)));
#results
print 'The distance by which beam must clear the obstacle is',round(R,2),'m'

The distance by which beam must clear the obstacle is 11.63 m

Example 3 : pg 679

In [3]:
 
# page no 679
# prob no 18_3
#calculate the power delivered
from math import log10
#given
#A transmitter and receiver at 6GHz seperated by 40km with o/p power 2 W
f = 6 * 10 ** 9
d = 40.
Pt = 2# power in watt
#transmitting antenna gain Gt=20dBi,receiving antenna Gr=25dBi
Gt = 20
Gr = 25
f_mhz = 6000#f=6000 MHz
#calculations
Pr_Pt_dB = (Gt + Gr) - (32.44 + (20 * log10(d)) + (20 * log10(f_mhz)))
Pt_dBm = 10 * log10(Pt / 10 ** -3)
Pr_dBm = Pt_dBm + Pr_Pt_dB
#results
print 'The power delivered to the Rx is',round(Pr_dBm,2),'dBm'

The power delivered to the Rx is -62.03 dBm

Example 4 : pg 680

In [4]:
 
# page no 680
#calculate the noise temperature
# prob no 18_4
#given
T_sky = 120.# Sky temp expressed in K
L_dB = 2# antenna feedline loss
#calculations
L = 10 ** (L_dB / 10)
# the noise temp is given as
Ta = ((L - 1) * 290 + T_sky) / L
#results
print 'Noise temperature is',Ta,'K'

Noise temperature is 120.0 K

Example 5 : pg 681

In [5]:
 
# page no 681
# prob no 18.5
#calculate the equivalent noise temperature
#given
NF_dB = 2
#calculations
NF_power = 10 ** (NF_dB / 10.)
T_eq = 290. * (NF_power - 1)
#results
print 'The equivalent noise temperature',round(T_eq,3),'K'

The equivalent noise temperature 169.619 K

Example 6 : pg 681

In [6]:
 
# page no 681
# prob no 18.6
# refer example no 18.4 and 18.5
#calculate the noise power in both cases
from math import log10
#given
# The antenna and feedline combination from ex 18.4 is used with the Rx from ex
# 18.5
Ta = 182.# noise temp of the antenna and feedline combination expressed in K
Teq = 169.# noise temperature of the Rx
B = 20. * 10 ** 6# BW of the receiver
#calculations
Tn_sys = Ta + Teq#Noise temp for the system
k = 1.38 * 10 ** -23#Boltzmann constant
# Noise power of the system is given as
Pn = k * Tn_sys * B# where k is Boltzmann constant
Pn_dBm = 10 * log10(Pn / 10 ** -3)
#results
print 'The noise power is',Pn,'W'
print 'The thermal noise power is',round(Pn_dBm,3),'dBm'

The noise power is 9.6876e-14 W
The thermal noise power is -100.138 dBm

Example 7 : pg 682

In [7]:
 
# page no 682
# prob no 18.7
#calculate the carrier to noise ratio
# refer ex no 18.3 and 18.6
#given
Pr_dBm = -62.#power at the receiver in dBm
Pn_dBm = -100.#thermal noise power in dBm
#calculations
# carrier to noise ratio in dB is given as
C_N = Pr_dBm - Pn_dBm
#results
print 'Carrier to noise ratio is',C_N,'dB'

Carrier to noise ratio is 38.0 dB

Example 8 : pg 683

In [8]:
 
# page no 683
# prob no 18.8
#calculate the energy per bit to noise density ratio
# refer ex 18.7
from math import log10
#given
fb = 40 * 10 ** 6# bit rate in bps
Pr_dBm = -62#power at the receiver in dBm
Pr = 10 ** (Pr_dBm / 10) * 10 ** -3# power at the receiver in W
Eb = Pr / fb# the energy per bit in J
k = 1.38 * 10 ** -23#Boltzmann constant
T = 350.
#calculations
# the noise power density is
No = k * T
# Energy per bit to noise density ratio in dB is
Eb_No = 10 * log10(Eb / No)
#results
print 'Energy per bit to noise density ratio is ',round(Eb_No,3),'dB'

Energy per bit to noise density ratio is  27.14 dB

Example 9 : pg 686

In [9]:
 
# page no 686
# prob no 18.9
# refer fig 18.7(b)
#calculate the local oscillator frequency
#given
#This is the standard superheterodyne receiver
fc=6870.;# the received carrier freq in MHz
IF=70.;# IF in MHz
#calculations
# The local oscillator freq is given as
f_lo=fc-IF;
#results
print 'The local oscillator freq is ',f_lo,'MHz'

The local oscillator freq is  6800.0 MHz

Example 10 : pg 688

In [10]:
 
# page no 688
# prob no 18.10
# refer fig 18.9a)
#calculate the freq shift in all cases
#given
fc_r = 6870.# carrier freq of received signal in MHz
fc_t = 6710.#carrier freq of transmitted signal in MHz
IF = 70.#in MHz
#calculations and results
# the freq of shift oscillator is
fso = fc_r - fc_t
print 'The freq of shift oscillator is ',fso,'MHz'
#the local oscillator freq is given as
flo = fc_t - IF
print 'The local oscillator freq is ',flo,'MHz'
#from fig, mixer 3 will produce an o/p as
op_mix3 = flo + fso
print 'O/P of Mixer 3 is',op_mix3,'MHz'

The freq of shift oscillator is  160.0 MHz
The local oscillator freq is  6640.0 MHz
O/P of Mixer 3 is 6800.0 MHz

Example 11 : pg 690

In [11]:
 
# page no 690
# prob no 18.11
#calculate the baud rate and bits per symbol
#given
from math import log
# A typical microwave digital radio system uses 16-QAM.
fb = 90.524#bit rate expressesd in Mbps
n = 16# for 16-QAM system
#calculations and results
#part a) calculation of no of bits per symbol
m = log(n) / log(2)
print 'The number of bits per symbol are',m,'bits'
# part b) calclation of baud rate
# baud rate is 1/4th of the bit rate
baud = fb / 4
print 'The baud rate is',baud,'Mbaud'

The number of bits per symbol are 4.0 bits
The baud rate is 22.631 Mbaud