# Chapter 12: Modulation and Demodulation¶

## Example 1, Page 285¶

In :
#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 :
#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 :
#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 :
#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 :
#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 :
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 :
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