Communication system Communication is the process of exchanging information between two points. Fig: communication system

(1)

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

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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

𝑉_𝑐=peak voltage 𝑊_𝑐=2πfc 𝑓_𝑐=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× 10 5_=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)

(3)

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)=𝑉_𝑚 Sin𝑊_𝑚t---(1) 𝑉_𝑚=peak voltage of message signal 𝑊_𝑚=2πfm

𝑓_𝑚=modulating frequency Lt carrier signal C(t) is C(t)=𝑉_𝑐 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

(4)

V(t)=𝑉_𝑐sinWct +Vm sinWct Sin Wmt [using Sin A. SinB=1

2[Cos(A-B)-Cos(A+B)] V(t)=𝑉_𝑐sinWct+ Vm1 2[Cos(Wct-Wmt) - Cos(Wct +Wmt)] V(t)=𝑉𝑐sinWct+ Vm 2 [Cos(Wc-Wm)t-Cos(Wc+Wm)t] V(t)=𝑉_𝑐sinWct+ Vm 2 Cos(Wc-Wm)t-Vm 2 cos(Wc+Wm)t---(A) Modulation index m=Vm 𝑉_𝑐 =

peak voltage of msg signal1 peak voltage of carrier

m𝑉_𝑐=Vm Eqn (A) = > V(t)=𝑉𝑐sinWct+ 𝑚𝑉_𝑐 2 Cos(Wc-Wm)t

-

𝑚𝑉_𝑐 2 cos(Wc+m)t

Carrier lower sideband upper sideband Bandwidth=Wc+Wm-(Wc-Wm)

= Wc+Wm-Wc+Wm B.W =2Wm

(5)

Modulation index in terms of

𝑽

_𝒎𝒂𝒙

&

𝑽

_𝒎𝒊𝒏

:

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

m𝑉_𝑐 =Vm

Fig: Amplitude modulated signal

𝑉

_𝑚𝑎𝑥

=𝑉

_𝑐

+ Vm

𝑉

_𝑚𝑎𝑥

= 𝑉

_𝑐

+ mVc

𝑉

_𝑚𝑎𝑥

= 𝑉

_𝑐

[1+ m]---(1)

𝑉

_𝑚𝑖𝑛

= 𝑉

_𝑐

-Vm

𝑉

𝑚𝑖𝑛

= 𝑉

𝑐

-mVc

𝑉

_𝑚𝑖𝑛

= 𝑉

_𝑐

[1-m]--- (2)

(1) (2)

=

𝑉𝑚𝑎𝑥 𝑉_𝑚𝑖𝑛

=

𝑉𝑐[1+m] 𝑉_𝑐[1−m]

𝑉

_𝑚𝑎𝑥

[1-m]= 𝑉

_𝑚𝑖𝑛

[1+m]

𝑉

_𝑚𝑎𝑥

- m𝑉

_𝑚𝑎𝑥

=𝑉

_𝑚𝑖𝑛

+m 𝑉

_𝑚𝑖𝑛

𝑉

_𝑚𝑎𝑥

– 𝑉

_𝑚𝑖𝑛

= m𝑉

_𝑚𝑖𝑛

+m 𝑉

_𝑚𝑎𝑥

𝑉

𝑚𝑎𝑥

– 𝑉

𝑚𝑖𝑛

= m[𝑉

𝑚𝑎𝑥

+ 𝑉

𝑚𝑖𝑛

]

𝑉_𝑚𝑎𝑥– 𝑉_𝑚𝑖𝑛 𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛

= m

m = Vm 𝑉𝑐 𝑉_𝑚𝑎𝑥 𝑉𝑚𝑖𝑛

=

1+m 1−m

m =

𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛 𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛

(6)

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

𝑃

𝐿𝑆𝐵

=𝑃

𝑈𝑆𝐵

=

𝑚2𝑉𝑐2 8

&𝑃

𝑐

=

𝑉𝑐2 2

Ƞ =

𝑚2𝑉𝑐2 + 8 𝑚2𝑉𝑐2 8 𝑃_𝑐[1+𝑚2 2 ]

=

2𝑚2𝑉𝑐2 8 𝑃_𝑐[1+𝑚2 2 ]

=

𝑚2𝑉𝑐2 4 𝑃_𝑐[1+𝑚2 2 ]

Ƞ =

𝑚2𝑉𝑐2 4 𝑉𝑐2 2 [1+ 𝑚2 2 ]

=

𝑚2 2 [2+𝑚2 2 ]

=

𝑚 2 2+𝑚2

Ƞ =

𝑚2 𝑚2₊₂

(7)

Spectrum power :(Total transmitted in AM wave )

Power, p =

𝑉2𝑟𝑚𝑠 𝑅

=

(𝑉 √2) 2 𝑅

=

𝑉2 2 Assume R =1Ω

The total power of amplitude modulated wave ,

𝑃

_𝑡

= 𝑃

_𝑐

+𝑃

_𝑈𝑆𝐵

+𝑃

_𝐿𝑆𝐵

--- (1)

= (

𝑉𝑐 √2

)

2

+(

𝑚𝑉𝑐 2 √2

)

2

+(

𝑚𝑉𝑐 2 √2

)

2

𝑃

_𝑐

= (

𝑉𝑐 √2

)

2

=

𝑉𝑐 2 2

=

𝑉𝑐 2 2

+

𝑚2𝑉𝑐2 4 2

+

𝑚2𝑉𝑐2 4 2

𝑃

𝑈𝑆𝐵

=𝑃

𝐿𝑆𝐵

=(

𝑚𝑉𝑐 2 √2

)

2

=

𝑚 2_𝑉 𝑐2 8

=

𝑉𝑐 2 2

+

𝑚2_𝑉 𝑐2 4

×

1 2

+

𝑚2_𝑉 𝑐2 4

×

1

2

LSB=lower side band

USB=upper side band

=

𝑉𝑐2 2

+

𝑚2𝑉𝑐2 8

+

𝑚2𝑉𝑐2 8

=

𝑉𝑐 2 2

+

2𝑚2𝑉_𝑐2 8

=

𝑉𝑐 2 2

+

𝑚2𝑉_𝑐2 4

𝑃

_𝑡

=

𝑉𝑐 2 2

[1+

𝑚2 2

]

𝑃

_𝑡

= 𝑃

_𝑐

[1+

𝑚 2 2

]

If modulation index m=1,

𝑃

_𝑡

= 𝑃

_𝑐

[1+

1 2 2

] = 𝑃

𝑐

[

1+1 2

]

𝑃

_𝑡

= 𝑃

_𝑐

[

3 2

]

or 50% more than

𝑃

_𝑐

𝑃

_𝑡

=1.5 𝑃

_𝑐

(8)

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 1̸fc <<Rc<<1̸ fm

Tc = time period of carrier Fc = carrier frequency

Fm = modulating signal frequency Tm = Time period of modulating signal

(9)

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.

(10)

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 𝑃_𝑡 =2500 𝑃_𝑡= 𝑃_𝑐+𝑃_𝑈𝑆𝐵+𝑃_𝐿𝑆𝐵---(1) 𝑃_𝑡= 𝑃_𝑐[1+𝑚 2 2 ] 2500=𝑃_𝑐[1+(0.8) 2 2 ] 2500=𝑃𝑐[1.32] 𝑃𝑐= 2500 1.32 =1893.9W 𝑃_𝑡=𝑃_𝑐+𝑃_𝑈𝑆𝐵+𝑃_𝐿𝑆𝐵 𝑃_𝑡-𝑃_𝑐 = 𝑃_𝑈𝑆𝐵+𝑃_𝐿𝑆𝐵 𝑃_𝑈𝑆𝐵+𝑃_𝐿𝑆𝐵 =2500-1893.9 𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 =606.1W 𝑃_𝑈𝑆𝐵=𝑃_𝐿𝑆𝐵=606.1 2 W 𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=303.50W

(11)

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 𝑃_𝑡= 𝑃_𝑐[1+𝑚 2 2 ] 𝑃_𝑡=800[1+(0.9)2\2] = 1124W 𝑃_𝑡=𝑃_𝑐 + 𝑃_𝑈𝑆𝐵+𝑃_𝐿𝑆𝐵 𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 = 𝑃𝑡 –𝑃𝑐 = 1124-800 𝑃_𝑈𝑆𝐵+𝑃_𝐿𝑆𝐵 = 324W 𝑃_𝑈𝑆𝐵 = 𝑃_𝐿𝑆𝐵= 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

Given𝑃_𝑡=600W 𝑃_𝑈𝑆𝐵=𝑃_𝐿𝑆𝐵=75W m = ? 𝑃_𝑡 = 𝑃_𝑐 + 𝑃_𝑈𝑆𝐵 + 𝑃_𝐿𝑆𝐵 𝑃_𝑐 = 𝑃_𝑡− 𝑃_𝑈𝑆𝐵 − 𝑃_𝐿𝑆𝐵 𝑃_𝑐 = 600-75-75 𝑃_𝑐 =450W 𝑃𝑡= 𝑃𝑐[1+ 𝑚2 2 ] 600 =450[1+𝑚 2 2 ] 600 450= 1 + 𝑚2 2 1.33-1= 𝑚 2 2 0.33×2 =𝑚2 0.66 =𝑚2 m=√0.66 =0.816 m=0.816 % m = 0.816×100 =81.6%

(12)

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