Chapter 5 Frequency Modulation : Transmission

Example 5.1 Page no 209

In [5]:
#given
v=25*10**-3
f=750                        #deviation constant
vg=10.0*10**-3                 #deviation constant

#calculation
pfd=v*(f/vg)                 #positive frequency deviation
nfd=-v*(f/vg)                #negative frequency deviation

#result
print"(a) positive frequency deviation = ",pfd,"Hz"
print"negative frequency deviation = ",nfd,"Hz"
print"The total deviation is written as +-2.25kHz for the given input signal level"
print"(b) The carrier wil deviate ",pfd,"Hz","&",nfd,"Hz","at 400 Hz"

(a) positive frequency deviation =  1875.0 Hz
negative frequency deviation =  -1875.0 Hz
The total deviation is written as +-2.25kHz for the given input signal level
(b) The carrier wil deviate  1875.0 Hz & -1875.0 Hz at 400 Hz

Example 5.3 Page no 214

In [6]:
#given
d=20*10**3     #maximum deviation
fi=10.0*10**3  #input frequency

#calculation
mf=d/fi
a=mf*40

#result
print"total required bandwidth is ",a,"KHz"

total required bandwidth is  80.0 KHz

Example 5.4 Page no 214

In [2]:
#given
d=20*10**3                    #maximum deviation
fi=5.0*10**3                  #input frequency

#calculation
mf=d/fi
a=2*35

#print
print"the required bandwidth is ",mf

the required bandwidth is  4.0

Example 5.5 Page no 215

In [6]:
#given
Vm=2000
R=50.0                          #resistance, ohm

#calcultion
import math
fc=(2*math.pi*(10**8))/2.0*math.pi
P=(2000/math.sqrt(2))**2/R
mf=2                                                     #by inspection of FM equation
fi=(math.pi*10**4)/(2.0*math.pi)
d=(mf*fi)
BW=mf*40
bw=2*(d+fi)
P1=((0.58*2000/math.sqrt(2))**2)/R
P2=((0.03*2000/math.sqrt(2))**2)/R

#result
print"(a) carrier frequency = ",round(fc,-9),"Hz"        #by inspection of FM equation
print"(b) the peak voltage is 2000V P thus= ",P,"W"
print"(c) mf = 2"                                        #by inspection of FM equation
print"(d) the intelligence frequency fi = ",fi,"Hz"
print"(e) BW = ",bw ,"Hz"                                #using carson's rule 
print"(f) The smallest sideband J4 is 0.03 times the carrier = ",P2,"W"

(a) carrier frequency =  1000000000.0 Hz
(b) the peak voltage is 2000V P thus=  40000.0 W
(c) mf = 2
(d) the intelligence frequency fi =  5000.0 Hz
(e) BW =  30000.0 Hz
(f) The smallest sideband J4 is 0.03 times the carrier =  36.0 W

Example 5.6 Page no 218

In [1]:
#given
d=75*10**3                       #maximum deviation
fi=30.0                          #modulating frequency, Hz
fi1=15.0*10**3                   
d1=1*10**3
fi2=100                           #Hz
fi3=2.0*10**3

#calculation
mf1=d/fi
mf2=d/fi1
mf3=d1/fi2
mf4=d1/fi3
DR=d1/fi3

#result
print"(a)maximum deviation at 30 Hz = ",mf1                           
print"maximum deviation at 15kHz= ",mf2
print"(b) maximum deviation at 100Hz = ",mf3
print"maximum deviation at 2KHz= ",mf4
print "(c)Deviation Ratio ",DR                  #deviation ratio

(a)maximum deviation at 30 Hz =  2500.0
maximum deviation at 15kHz=  5.0
(b) maximum deviation at 100Hz =  10
maximum deviation at 2KHz=  0.5
(c)Deviation Ratio  0.5

Example 5.7 Page no 218

In [31]:
#given
mf=0.25
a=0.98
b=0.12
x=10*10**3                      #power, watt

#calculation
P=(a**2)*x
P1=(b**2)*x
t=P+2*P1

#result
print"power of each sideband = ",P1,"W"
print"total power = ",round(t,-4),"W"

power of each sideband =  144.0 W
total power =  10000.0 W

Example 5.8 Page no 222

In [34]:
#given
phi=0.5                          #maximum intelligence frequency
fi=5.0*10**3
x=75*10**3

#calculation
d=phi*fi
y=x/d

#result
print"o/p S/N = ",y

o/p S/N =  30.0

Example 5.9 Page no 222

In [36]:
#given
dm=10*10**3
x=(1/3.0)                      #N/S input ratio

#calculation
import math
phi=math.asin(x)
phi1=math.asin(x)
fi=3*10**3
d=phi1*fi
a=dm/d

#result
print"The S/N output will be ",round(a,0)

The S/N output will be  10.0