Chapter 5 Fundamentals of frequency modulation

Example 5.1 Page no 153

#Given
f = 915*10**6
fm_deviation =12.5*10**3

#Calculation
max_deviation = f + fm_deviation
min_deviation = f - fm_deviation

#Result
print"Maximum frequency occur during modulation is ",max_deviation/1000.0,"KHz"
print"Minimum frequency occur during modulation is ",min_deviation/1000.0,"KHz"

Maximum frequency occur during modulation is  915012.5 KHz
Minimum frequency occur during modulation is  914987.5 KHz

Example 5.2 Page no 160

#Given
max_deviation = 25*10**3
fm =15.0*10**3

#Calculation
mf =max_deviation/fm

#Result
print"The deviation ratio of the TV sound is ",round(mf,3)

The deviation ratio of the TV sound is  1.667

Example 5.3 Page no 162

#Given
mf = 2.2
fd = 7.48*10**3

#Calculation
fm = fd/mf

#Result
print"The maximum modulating frequency is ",fm/1000.0,"KHz"

The maximum modulating frequency is  3.4 KHz

Example 5.4 Page no 164

#Given
J0 = -0.4
J1 = -0.07
J2 = 0.36
J3 = 0.43
J4 = 0.28

#Result
print"The amplitude of the carrier is ",J0
print"Amplitudes of the first four sidebands are "," \n ", J1,"\n ",J2,"\n ",J3,"\n ",J4

The amplitude of the carrier is  -0.4
Amplitudes of the first four sidebands are   
  -0.07 
  0.36 
  0.43 
  0.28

Example 5.5 Page no 165

#Given
fd = 30*10**3
fm = 5*10**3
N=9

#Calculation
bw1 = 2*fm*N
bw2 = 2*(fd+fm)

#Result
print"The maximum bandwidth of the fm signal is ",bw1/10**3,"KHz"
print"Bandwidth using carson's rule ",bw2/10**3,"KHz"

The maximum bandwidth of the fm signal is  90 KHz
Bandwidth using carson's rule  70 KHz

Example 5.6 Page no 167

#Given
S_N = 2.8
fm = 1.5*10**3
fd =4*10**3

#Calculation
import math
fi= math.asin(1/S_N)
delta = fi*fm
SN =fd/delta

#Result
print"(a) The frequency deviation caused by the noise ",round(delta,1),"Hz"
print"(b) The improved output signal to noise ratio is ",round(SN,1)

(a) The frequency deviation caused by the noise  547.8 Hz
(b) The improved output signal to noise ratio is  7.3