Communication system Communication is the process of exchanging information between two points

Elements of communication system:

source

Transduce

r

Modulator

&

Transmitter

Receiver

&

Demodulator

Transducer Destinatio

n

channel

Audio electrical signal electrical signal audio

Fig: communication system

Source: The information which has to be transmitted is generated by source ex: audio video text

etc.

Transducer: Transducer is a device which converts one form of energy into another form of

energy here transducer convert audio signal to electrical signal and vice versa.

Modulator and Transmitter: Here signal is modulated and transmitted over long distance.

Channel: The channel is a medium through which signals are transmitted to receiver channel

may be wired or wireless.

Receiver and demodulator: Here signals are received and information signal is detected by

demodulator

Destination: signal is received in original form i.e. (Audio).

Frequency ranges with application:

Frequency range Application

Super high frequencies 3GHⱬ-30GHⱬ Radar.

Ultra high frequncies300MHⱬ-3GHⱬ Communication satellites cellular phones

personal communication system.

Very high frequncies30MHⱬ-300MHⱬ TV and FM broadcast.

High frequncies3MHⱬ-30MHⱬ Short-wave broadcast commercial.

Medium frequencies300KHⱬ-3MHⱬ AM broadcast.

Low frequencies30KHⱬ-300KHⱬ Navigation, submarine communication.

Very low frequncies3KHⱬ-30KHⱬ Sub marine communication

Voice frequencies300Hⱬ-3KHⱬ Audio, submarine communication

Extremely low frequencies30Hⱬ-300Hⱬ Power Transmission

Modulation: Modulation is the process of varying carrier in accordance with instantaneous value

of information signal or modulating signal.

The carrier signal, c(t)=VcSinWct

𝑉𝑐=peak voltage 𝑊𝑐=2πfc 𝑓𝑐=carrier frequency

Need of modulation :

1. Modulation process helps to transmit the signal to longer distance.

2. Reduce the height of antenna.

Height of antenna , ℓ = 𝜆 4⁄ =

𝐶

4𝑓 𝜆 =𝐶

𝑓⁄

Ex: 1 f = 1 KHz

ℓ = 𝜆 4⁄ =

𝐶

4𝑓 =

3×108

4×1𝑘 = 0.75× 105=75000m

Ex: 2 f=1MHz

ℓ = 𝜆 4⁄ =

𝐶

4𝑓 =

3×108

4×1×106 = 75m

By above example it is clear that transmitting frequency is increased height of antenna is

decreased.

3. We can send multiple numbers of signals through signal communication channel having

wider bandwidth

4. The designing and processing of signal becomes easier for transmitter and receiver

5. Modulation Process reduces the effect of noise which is added in communication channel

Three types of modulations:

1. Amplitude modulation (AM)

2. Frequency modulation(FM)

3. Phase modulation(PM)

Amplitude modulation: Amplitude modulation is a process of varying amplitude of carrier

signal in accordance with instantaneous value of message signal keeping frequency and phase

constant.

Let message signal m (t) is

m (t)=𝑉𝑚 Sin𝑊𝑚 t----------(1)

𝑉𝑚=peak voltage of message signal

𝑊𝑚=2πfm

𝑓𝑚=modulating frequency

Lt carrier signal C(t) is

C(t)=𝑉𝑐 sinWct-------------(2)

Vc=peak voltage of carrier signal

Wc=2πfc

𝑓𝑐=carrier frequency

\From eqn(1) & (2) amplitude modulated signal is given by,

V(t)=(Vc+Vm Sin Wmt)sinWct

V(t)=𝑉𝑐sinWct +Vm sinWct Sin Wmt

[using Sin A. SinB=1

2[Cos(A-B)-Cos(A+B)]

V(t)=𝑉𝑐sinWct+ Vm1

2[Cos(Wct-Wmt) - Cos(Wct +Wmt)]

V(t)=𝑉𝑐sinWct+ Vm

2[Cos(Wc-Wm)t-Cos(Wc+Wm)t]

V(t)=𝑉𝑐sinWct+ Vm

2Cos(Wc-Wm)t-

Vm

2cos(Wc+Wm)t-----------(A)

Modulation index m=Vm

𝑉𝑐 =

peak voltage of msg signal1

peak voltage of carrier

m𝑉𝑐=Vm

Eqn (A) = >

V(t)=𝑉𝑐sinWct+ 𝑚𝑉𝑐

2 Cos(Wc-Wm)t-

𝑚𝑉𝑐

2cos(Wc+m)t

Carrier lower sideband upper sideband

Bandwidth=Wc+Wm-(Wc-Wm)

= Wc+Wm-Wc+Wm

B.W =2Wm

Fig: AM spectrum

Modulation index in terms of 𝑽𝒎𝒂𝒙&𝑽𝒎𝒊𝒏:

Modulation index is defined as ratio of peak amplitude of message signal to peak amplitudes of

carrier signal

m𝑉𝑐 =Vm

Fig: Amplitude modulated signal

𝑉𝑚𝑎𝑥 =𝑉𝑐 + Vm

𝑉𝑚𝑎𝑥 = 𝑉𝑐 + mVc

𝑉𝑚𝑎𝑥 = 𝑉𝑐 [1+ m]-----------(1)

𝑉𝑚𝑖𝑛 = 𝑉𝑐 -Vm

𝑉𝑚𝑖𝑛 = 𝑉𝑐 -mVc

𝑉𝑚𝑖𝑛 = 𝑉𝑐[1-m]------------- (2)

(1)

(2) =

𝑉𝑚𝑎𝑥

𝑉𝑚𝑖𝑛=

𝑉𝑐[1+m]

𝑉𝑐[1−m]

𝑉𝑚𝑎𝑥 [1-m]= 𝑉𝑚𝑖𝑛 [1+m]

𝑉𝑚𝑎𝑥 - m𝑉𝑚𝑎𝑥 =𝑉𝑚𝑖𝑛+m 𝑉𝑚𝑖𝑛

𝑉𝑚𝑎𝑥 – 𝑉𝑚𝑖𝑛 = m𝑉𝑚𝑖𝑛+m 𝑉𝑚𝑎𝑥

𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛 = m[𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛] 𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛

𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛 = m

m = Vm

𝑉𝑐

𝑉𝑚𝑎𝑥

𝑉𝑚𝑖𝑛=

1+m

1−m

m = 𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛

𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛

Spectrums of amplitude modulated signals are:

Fig: AM modulated signal spectrum

Fig: DSB-SC spectrum Fig: SSB Spectrum

DSB-SC: Double side band –suppressed carrier here carrier signal is suppressed only side hands

are transmitted

SSB: Single side band here carrier and lower side band are not transmitted only upper side band

is transmitted

Efficiency (Ƞ) :( % of power)

Ƞ = 𝑃𝐿𝑆𝐵 +𝑃𝑈𝑆𝐵

𝑃𝑡 𝑃𝐿𝑆𝐵=𝑃𝑈𝑆𝐵 =

𝑚2𝑉𝑐2

8&𝑃𝑐=

𝑉𝑐2

2

Ƞ =

𝑚2𝑉𝑐2 +

8

𝑚2𝑉𝑐2

8

𝑃𝑐[1+𝑚2

2]

=

2𝑚2𝑉𝑐2

8

𝑃𝑐[1+𝑚2

2]

=𝑚2𝑉𝑐2

4

𝑃𝑐[1+𝑚2

2]

Ƞ =

𝑚2𝑉𝑐2

4

𝑉𝑐2

2[1+

𝑚2

2] =

𝑚2

2

[2+𝑚2

2] =

𝑚2

2+𝑚2

Ƞ = 𝑚2

𝑚2+2

Spectrum power :(Total transmitted in AM wave )

Power, p = 𝑉2

𝑟𝑚𝑠

𝑅 =

(𝑉

√2)

2

𝑅 =

𝑉2

2

Assume R =1Ω

The total power of amplitude modulated wave ,

𝑃𝑡= 𝑃𝑐+𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵---------- (1)

= (𝑉𝑐

√2)

2+(

𝑚𝑉𝑐2

√2)

2

+(𝑚𝑉𝑐

2

√2)

2

𝑃𝑐 = (𝑉𝑐

√2)

2 =

𝑉𝑐2

2

= 𝑉𝑐2

2+

𝑚2𝑉𝑐2

4

2 +

𝑚2𝑉𝑐2

4

2 𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=(

𝑚𝑉𝑐2

√2)

2

=𝑚2𝑉𝑐

2

8

= 𝑉𝑐

2

2 +

𝑚2𝑉𝑐2

4×

1

2+

𝑚2𝑉𝑐2

4×

1

2 LSB=lower side band

USB=upper side band

= 𝑉𝑐

2

2 +

𝑚2𝑉𝑐2

8 +

𝑚2𝑉𝑐2

8

= 𝑉𝑐

2

2 +

2𝑚2𝑉𝑐2

8

= 𝑉𝑐

2

2 +

𝑚2𝑉𝑐2

4

𝑃𝑡= 𝑉𝑐

2

2[1+

𝑚2

2]

𝑃𝑡= 𝑃𝑐[1+𝑚2

2]

If modulation index m=1,

𝑃𝑡= 𝑃𝑐[1+12

2] = 𝑃𝑐[

1+1

2]

𝑃𝑡 = 𝑃𝑐[3

2]

or 50% more than 𝑃𝑐 𝑃𝑡=1.5 𝑃𝑐

AM Detection [ AM Demodulation]:

(a) Demodulation circuit

(b)demodulated waveform

The process of recovery of message signal from amplitude modulated signal is called am

demodulation or detection

Fig (a) shows demodulation circuit and fig(3) demodulated waveform

As v(t) rises to peak the diode conducts and capacitor charges through diode

When v(t) falls below peak the diode stops conducting and capacitor begins to discharge

through resistor R

This process repeat for all cycles

The time constant T = RC

The Conduction Tc<<Rc<<Tm

1fc <<Rc<<1 fm

Tc = time period of carrier

Fc = carrier frequency

Fm = modulating signal frequency

Tm = Time period of modulating signal

Frequency modulation :[FM]

The frequency modulation is the process of varying the frequency of carrier signal in

accordance with instantaneous value of message signal keeping amplitude and phase

constant.

Message signal m(t)

m(t)=VmSinWmt

The frequency modulated signal

f(t) = A Sin [Wc +Kf m(t)]t-----------(1)

modulation index : from eqn(1) f = fc + Kf m(t)

f = fc + KfVmsinWmt

Δf = kfvm =peak frequency deviation

Modulation index = peak frequncy deviation

modulating frequency

m = Δf

fm

Band width: ( BW)fm = 2 [Δf+fm]

Δf=frequency deviation.

fm=frequency of modulating signal.

Phase modulation[PM]: Phase modulation is process of varying phase of carrier signal in

accordance with instantaneous value of message signal keeping amplitude &phase constant

The phase modulated signal is given by

P(t) = A Sin[Wct + Kpm(t)]

Where m(t) = Vm Sin Wmt

NOTE :FM& PM are angle modulation

Information signal message signal modulating signal all are same

Comparison of amplitude and frequency modulation:

characteristics AM FM

1 Bandwidth B W =2 fm

Lesser bandwidth

B W =2 [Δf + fm]

larger bandwidth

2 Operating carrier frequency

Am uses lower carrier

frequency

Fm uses higher carrier

frequency

3 Transmission efficiency Less efficient More efficient

4 Area of reception Am covers more distance Fm covers limited distance

5 Noise performance poor better

6 channel Small channel is sufficient Wider channel is required

7 Common channel interference Distortion occurs Less Distortion

8 Tuning Tuning is not

required

Tuning is required

Problems :

1. Determine the power content of the carrier and each of sideband for an AM signal having a

percent modulation of 80% and totl power of 2500w

m= 80% m=80\100 =8\10=0.8 𝑃𝑡 =2500

𝑃𝑡= 𝑃𝑐+𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵----------(1)

𝑃𝑡= 𝑃𝑐[1+𝑚2

2]

2500=𝑃𝑐[1+(0.8)2

2]

2500=𝑃𝑐[1.32]

𝑃𝑐=2500

1.32 =1893.9W

𝑃𝑡=𝑃𝑐+𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵

𝑃𝑡-𝑃𝑐 = 𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵

𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 =2500-1893.9

𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 =606.1W

𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=606.1

2 W

𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=303.50W

2.The total power control of an AM Signal is 1000W determine the power being transmitted at

carrier frequency and at each of sidebands when percent modulation is 100%

𝑃𝑐=666.67W 𝑃𝑈𝑆𝐵 =𝑃𝐿𝑆𝐵 =166.66W

3.An amplitude modulated wave has a power content of 800W at its carrier frequency.

Determine the power content of each sidebands for a 90% modulation

Given :𝑃𝑐= 800W

m=90% =90\100 =0.9

𝑃𝑡= 𝑃𝑐[1+𝑚2

2]

𝑃𝑡=800[1+(0.9)2\2] = 1124W

𝑃𝑡=𝑃𝑐 + 𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵

𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 = 𝑃𝑡 –𝑃𝑐 = 1124-800

𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 = 324W

𝑃𝑈𝑆𝐵 = 𝑃𝐿𝑆𝐵= 324

2 =162W

4 .The total power content of an AM wave is 600W. Determine the percent modulation of the

signal if each of sidebands contains 75W

Given𝑃𝑡=600W

𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=75W

m = ?

𝑃𝑡 = 𝑃𝑐 + 𝑃𝑈𝑆𝐵 + 𝑃𝐿𝑆𝐵

𝑃𝑐 = 𝑃𝑡 − 𝑃𝑈𝑆𝐵 − 𝑃𝐿𝑆𝐵

𝑃𝑐 = 600-75-75

𝑃𝑐 =450W

𝑃𝑡= 𝑃𝑐[1+𝑚2

2]

600 =450[1+𝑚2

2]

600

450= 1 +

𝑚2

2

1.33-1= 𝑚2

2

0.33×2 =𝑚2

0.66 =𝑚2

m=√0.66 =0.816

m=0.816 % m = 0.816×100 =81.6%

Question paper questions :

1. Write the block diagram of communication system explain each element

2. What are commonly used frequency range in communication system? Mention application

of each range

3. What is modulation ?explain need of modulation

4. Define amplitude modulation?

Show that𝑉𝑚𝑎𝑥

𝑉𝑚𝑖𝑛=

1+𝑚

1−𝑚

5. Explain amplitude modulation with expression

6. Derive pt= 𝑝𝑐[1 +−2

𝑚2]

7. Derive total power transmitted in AM

8. Write short note on (a )frequency modulation

(b) phase modulation

9. Explain AM detection circuit

10. List the difference between AM & FM

11. what is modulation index for AM ?

Show that m =𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛

𝑉𝑚𝑎𝑥+𝑉𝑚𝑖𝑛