Chapter 12: Modulation and Demodulation

Example 1, Page 285

In [3]:
#Varibale declaration
Vmax=8.#Vmax=maximum peak to peak value of an AM voltage
Vmin=2.#Vmin=minimum peak to peak value of an AM voltage

#Calculations&Results
ma=(Vmax-Vmin)/(Vmax+Vmin)#ma=percentage modulation 
print "Percentage modulation ma=%.f %%"%(ma*100)
#ma=(Vmax-Vmin)/(2*VC) where VC=amplitude of the unmodulated carrier
VC=(Vmax-Vmin)/(2*ma)
print "Amplitude of the unmodulated carrier is VC=%.f V"%VC
print "In the textbook answer given is incorrect as they have further divided by 2 which is not the part of given formula."
Percentage modulation ma=60 %
Amplitude of the unmodulated carrier is VC=5 V
In the textbook answer given is incorrect as they have further divided by 2 which is not the part of given formula.

Example 2, Page 285

In [5]:
#Varibale declaration
fc=1000*(10**3)#fc=frequency of the carrier wave in Hz(hertz)
fmin=400
fmax=1600.#fmin and fmax represent the frequency range of audio signals by which the carrier wave is amplitude modulated.

#Calculations&Results
fs=fmax-fmin#fs=frequency span of each sideband
print "1.Frequency span of each sideband is %.f Hz"%fs
fumax=(fc+fmax)/1000#fumax=maximum upper side frequency
print "2.The maximum upper side frequency is %.1f kHz"%fumax
flmin=(fc-fmax)/1000#flmin=minimum lower side frequency
print "3.The minimum lower side frequency is %.1f kHz "%flmin
Wc=fumax-flmin#Wc=channelwidth
print "4.The channelwidth is %.1f kHz"%Wc
1.Frequency span of each sideband is 1200 Hz
2.The maximum upper side frequency is 1001.6 kHz
3.The minimum lower side frequency is 998.4 kHz 
4.The channelwidth is 3.2 kHz

Example 3, Page 286

In [7]:
#Varibale declaration
R=100#R=load resistance in ohms
Vc=100.#Vc=peak voltage of the carrier in volts
ma=0.4#ma=modulation factor

#Calculations&Results
Pc=(Vc**2)/(2*R)#Pc=unmodulated carrier power developed by an AM wave
print "The unmodulated carrier power is Pc = %.f W"%Pc
Pt=Pc*(1+((ma**2)/2))#Pt=total power developed
print "The total power developed by the AM wave is Pt=%.f W"%Pt
The unmodulated carrier power is Pc = 50 W
The total power developed by the AM wave is Pt=54 W

Example 4, Page 286

In [10]:
#Varibale declaration
ma=0.5#ma=modulation factor
Pc=20#Pc=unmodulated carrier power in kilowatts(kW)

#Calculations&Results
Ps=(1./2)*(ma**2)*Pc#Ps=total sideband power
print "The total sideband power is Ps=%.1f kW"%Ps
#modulator system efficiency is given as 70 per cent
Pa=Ps/0.7#Pa=audio power necessary toamplitude modulate a given carrier wave
print "The required audio power is %.2f kW"%Pa
The total sideband power is Ps=2.5 kW
The required audio power is 3.57 kW

Example 5, Page 286

In [11]:
#Varibale declaration
df=30.#df=maximum frequency deviation in kilohertz(kHz)
fm=15#fm=modulation frequency of a sinusoidal audio signal in kilohertz(kHz)

#Calculations&Results
mf=df/fm#mf=frequency modulation index
print "1.The modulation index is mf=%.f"%mf
fc=100#fc=carrier wave frequency in megahertz(MHz)
print "2.The three significant pairs of side frequencies are 100MHz+-15kHz(fc+-fm);100MHz+-30kHz(fc+-2fm);100MHz+-45kHz(fc+-3fm)"
wc=mf*3*fm#wc=channelwidth required for 3 above mentioned side frequency pairs
print "3.The required channelwidth is %.f kHz"%wc
1.The modulation index is mf=2
2.The three significant pairs of side frequencies are 100MHz+-15kHz(fc+-fm);100MHz+-30kHz(fc+-2fm);100MHz+-45kHz(fc+-3fm)
3.The required channelwidth is 90 kHz

Example 6, Page 286

In [13]:
import math

#Varibale declaration
R=0.2*(10**6)#R=load resistance in ohms in a diode detector 
C=150*(10**-12)#C=capacitance in farad in a diode detector 

#Calculations
#fmh=wmh/(2*%pi)where fmh=highest modulation frequency that can be detected with tolerable distortion and wmh=corresponding angular frequency
ma=0.5#ma=modulation factor or depth of modulation
fmh=(1/(2*math.pi*ma*R*C))/1000

#Result
print "The required frequency is fmh=%.2f kHz"%fmh
The required frequency is fmh=10.61 kHz

Example 7, Page 287

In [15]:
import math

#Varibale declaration
Pc=10#Pc=unmodulated carrier power in kilowatts(kW)
Pt=12.5#Pt=total power in kilowatts(kW)

#Calculations&Results
#Pt=Pc*(1+((ma^2)/2)) 
ma=math.sqrt(2*((Pt/Pc)-1))#ma=depth of modulation of the first signal
print "The depth of modulation is ma=%.3f"%ma
mb=0.6#mb=depth of modulation of the second signal
PT=Pc*(1+((ma**2)/2)+((mb**2)/2))#PT=the total radiated power
print "The total radiated power is PT=%.1f kW"%PT
The depth of modulation is ma=0.707
The total radiated power is PT=14.3 kW