Chapter 4 : Angle Modulation

Example 1 : pg 139

In [1]:
#page no 139
#prob no. 4.1
#Calculate the o/p frequency
#An FM modulator is given with kf=30kHz/V operate at carrier freq 175MHz
#given
fc=175.*10**6;kf=30.*10**3;
#a)Determination of o/p freq for modulating signal value em1=150mV 
em1=150*10**-3;
#calculations and results
fsig1=fc+(kf*em1);
print 'a)The value of o/p freq is ',fsig1/(10**6),'MHz'
#b)Determination of o/p freq for modulating signal value em2=-2V 
em2=-2;
fsig2=fc+(kf*em2);
print 'b)The value of o/p freq is ',fsig2/(10**6),'MHz'

a)The value of o/p freq is  175.0045 MHz
b)The value of o/p freq is  174.94 MHz

Example 2 : pg 140

In [2]:
#page no 140
#prob no. 4.2
#calculate the value of deviation
from math import sqrt
#An FM modulator is given which is modulated by sine wave 3V
#given
v=3.;
kf=30.*10**3;
#calculations
#Determination of peak value 
Em=v*sqrt(2);
#Determination of deviation delta
delta=kf*Em;
#results
print 'The value of deviation is ',round(delta/1000.,3),'kHz'

The value of deviation is  127.279 kHz

Example 3 : pg 140

In [3]:
#page no 140
#prob no. 4.3
#calculate the value of modulation index in both cases
#An FM broadcaster transmitter operate at max deviatn of 75kHz
#given
delta=75.*10**3;
#a)Determination of modulation index with modulating freq of signal =15kHz
fm1=15.*10**3;
#calculations and results
mf1=delta/fm1;
print 'a)The value of modulation index for fm=15kHz is ',mf1
#b)Determination of modulation index with modulating freq of signal =50Hz
fm2=50;
mf2=delta/fm2;
print 'b)The value of modulation index for fm=50Hz is ',mf2

a)The value of modulation index for fm=15kHz is  5.0
b)The value of modulation index for fm=50Hz is  1500.0

Example 4 : pg 141

In [4]:
 
#page no 141
#prob no. 4.4
from math import pi, sqrt
#calculate the rms voltage
#A phase modulator is given with kp=2rad/V 
#given
kp=2;
#Peak phase deviation of 60 degree
#calculations
#Converting degree in radian 
phi=(2*pi*60)/360;
#Determination of peak voltage that cause that deviation 
Vp=phi/kp;
#Determination of rms voltage
Vrms=Vp/(sqrt(2));
#results
print 'The rms voltage that cause deviation is ',round(Vrms,2),'V'

The rms voltage that cause deviation is  0.37 V

Example 6 : pg 145

In [5]:
 
#page no 145
#prob no. 4.6
#calculate the freq deviation 
#given
#Phase modulator with sensitivity kp=3rad/V & sine wave i/p 2 V peak at 1kHz
kp=3.;Vp=2.;f=1*10**3;
#calculations
#As max value of sine functn is 1, hence max value of phi is kp*Vp
phi_max=kp*Vp;
#phi_max is nothing but mp
mp=phi_max;
#value of mf is same as mp if signal is considered as freq modulation
#Determination of freq deviation
dev=mp*f;
#results
print 'The freq deviation produce is',dev/1000,'kHz'

The freq deviation produce is 6.0 kHz

Example 7 : pg 149

In [9]:
 
#page no 149
#prob no. 4.7
#calculate the rms voltage in all cases 
%matplotlib inline
from math import sqrt, log10
import numpy
import matplotlib
from matplotlib import pyplot
#given
#An FM signal has deviation 3kHz & modulating freq 1kHz with total power Pt=5W
#developed across 50 ohm with fc=160 MHz
dev = 3. * 10 ** 3
fm = 10 ** 3
Pt = 5.
Rl = 50.
fc = 160. * 10 ** 6
#calculations and results
#a)Determination of RMS signal voltage
Vt = sqrt(Pt * Rl)
print 'a)The rms signal voltage is',Vt,'V'
######/b)Determination of rms voltage at carrier freq
#for that modulation index needs to be found out
mf = dev / fm
#From bessel function table, the coeff for the carrier first 3 side bands
J = ([0.26,0.34,0.49,0.31])
V = numpy.zeros(4)
print 'b)The rms voltage of side bands are'
for i in range(0,3):
    V[i] = J[i] * Vt

print 'Vc=',round(V[0],2)
print 'V1=',round(V[1],2)
print 'V2=',round(V[2],2)
print 'V3=',round(V[3],2)
#####/c)Determination of freq of each side bands########
print 'c)The 3 side bands at different freq. are '
f_usb = numpy.zeros(3)
for j in range(0,2):
    f_usb[j] = fc / 10 ** 6 + (fm * j / 10 ** 6)

print 'f_usb1=',round(f_usb[0],2)
print 'f_usb2=',round(f_usb[1],2)
print 'f_usb3=',round(f_usb[2],2)

f_lsb = numpy.zeros(3)
for j in range(0,2):
    f_lsb[j] = fc / 10 ** 6 - (fm * j / 10 ** 6)

print 'f_lsb1=',round(f_lsb[0],2)
print 'f_lsb2=',round(f_lsb[1],2)
print 'f_lsb3=',round(f_lsb[2],2)

P = numpy.zeros(4)
a = numpy.zeros(4)
######d)Determination of power of each side band########/
for i in range(0,3):
    P[i] = ((V[i]) ** 2) / Rl
    a[i] = (P[i]) / (10 ** -3)

print 'd)The power of each side band is'
print 'Pc=',round(P[0],2)
print 'P1=',round(P[1],2)
print 'P2=',round(P[2],2)
print 'P3=',round(P[3],2)

#####e)Determination of power that is uncounted
P = P[0] + 2 * (P[2] + P[3] + P[1])
#As total power is 5 W
P_x = Pt - P
#Percentage of total power uncounted
Px = (P_x / P) * 100
print 'e)Percentage total power which is uncounted is',Px,'%',
#####f)Ploting the signal in freq domain##########/
#Converting power in dBm
P_dBm = numpy.zeros(4)
for i in range(0,3):
    #a(k)=(P(k))/(10**-3);
   P_dBm[i] = 10 * log10(a[i]) 

print 'f)Power of each side bands in dBm is'
print 'Pc(dBm)=',round(P_dBm[0],2)
print 'P1(dBm)=',round(P_dBm[1],2)
print 'P2(dBm)=',round(P_dBm[2],2)
print 'P3(dBm)=',round(P_dBm[3],2)

x = ([159.997,159.998,159.999,160.0,160.001,160.002,160.003])
y = ([26.8,30.8,27.6,25.3,27.6,30.8,26.8])
pyplot.plot(x,y);
pyplot.show();

a)The rms signal voltage is 15.8113883008 V
b)The rms voltage of side bands are
Vc= 4.11
V1= 5.38
V2= 7.75
V3= 0.0
c)The 3 side bands at different freq. are 
f_usb1= 160.0
f_usb2= 160.0
f_usb3= 0.0
f_lsb1= 160.0
f_lsb2= 160.0
f_lsb3= 0.0
d)The power of each side band is
Pc= 0.34
P1= 0.58
P2= 1.2
P3= 0.0
e)Percentage total power which is uncounted is 28.3697047497 % f)Power of each side bands in dBm is
Pc(dBm)= 25.29
P1(dBm)= 27.62
P2(dBm)= 30.79
P3(dBm)= 0.0

Example 9 : pg 157

In [10]:
 
#page no 157
#prob no. 4.9
#calculate the SNR at detector o/p
from math import log10
#given
#An FM signal has freq deviation of 5kHz modulating freq fm=1kHz with SNR at i/p is 20 dB
#Converting dB in voltage ratio
fm=1.*10**3;dev_s=5.*10**3;snr=20.;
#calculations
Es_En=10**(snr/20);
#Since Es>>En then 
phi=1/(Es_En);
m_fn=phi;#modulation index equal to phi_n
dev_n=(m_fn)*fm;#Equivalent freq deviation due to noise
#SNR as a voltage ratio is given as
SNR=(dev_s)/(dev_n);
#Converting this voltage ration in dB
SNR_dB=20*(log10(SNR));
#results
print 'The SNR at detector o/p is',round(SNR_dB,3),'dB'

The SNR at detector o/p is 33.979 dB

Example 10 : pg 163

In [12]:
 
#page no 163
#prob no. 4.10
#calculate whether the freq is with in the acceptable range
#Refer the fig. 4.19
#given
# We know this transmitter is designed for voice frequencies,so we have to use trial 
#and error method to produce a carrier null for a deviation of 5kHz
mf=2.4;# starting with the first null for mf=2.4
dev=5;#in kHz
#calculations and results
fm=dev/mf;
if (0.3 <= fm and 3>=fm):
    print 'The freq is with in the acceptable range',round(fm,3),'kHz'
else:
    mf=5.5;
    fm=dev/mf;
    print 'The freq is with in the acceptable range',round(fm,3),'kHz'
# for this calculated fm, set the function generator to the value of fm so that the deviation is 5kHz

The freq is with in the acceptable range 2.083 kHz