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Analog communication

Solution scheme for 2nd test

1.

S (t) - modulated waveCos 2fl (t) - oscillator freq to which modulate wave has to be

shifted

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Spectrum of DSB-SC wave translated downwards in freq

S (t) = m (t) cos 2fct

V1 (t) = m (t) cos 2fct. cos2flt= 1/2 m (t) cos 2 (fc-fl) t + 1/2m (t) cos 2 (fc+fl) tV2 (t) = 1/2 m (t) cos 2 (fc-fl) t

= 1/2 m(t) cos (2f0t)

The device used for fre translation of a modulated wave is called mixer.This is constructed by using non-linear or switching devices.

Mixing is linear operation & it preserves the relation of the S.B's of theincoming modulated wave to the carrier.

When number of independent message signals are Txed over a commonchannel such that the signals are kept apart so that they do not interfere witheach other. If this separation is based on fre division, then the process is saidto be as fre division multiplexing.

2.

Basic Elements of an AM receiver of the super heterodyneType

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Signal received by an antenna is in the form ofelectromagnetic waves. RF section selects wantedsignal and reject others. It reduces the effect ofnoise. Mixer mixes fRF and fLO and differencefrequency components fRF-fLO is selected andothers are rejected. To maintain constant difference, ganged tuningis used. IF signal is then amplified by one or more

stages. Sensitivity and selectivity of receiver dontchange with incoming frequency. IF signal is detected by detector to recover themodulating signal.

2b.

Aspect ratio:-It is the ratio of width of the picture to height of the

picture.

Kell Factor:-

Gives the effective number of horizontal scanning

lines those can be scanned because of practicallimitations.

Band width:-

B= K/2 (b/a) [N-2Nvr/ T-Thr]

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B- video bandwidthK- Kell Factor 0.6 to 0.7b/a - Aspect Ratio 4:3N - Total lines/ Frames 525Nvr Vertical RetraceT - Total line timeThr Horizontal retrace time

3a.

If either phase or frequency of the carrier is

varied in accordance with the message signal, keepingamplitude of carrier as constant, then it is called anglemodulation.

S (t) = Ac cos [ (t)]

Ac- Amplitude of the carrier maintained constant. (t) - Angular argument varied by a message signal m(t).

Fi (t)= fc + Kfm(t)

t

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(t)= 2fc (t) + 2kf 0 m(t) dtt

S (t) = Ac cos [2fc (t) + 2Kfo m(t) dt ]

S(t)= Ac cos [(t)]

(t)= 2fc(t) + Kpm(t)

S(t)= Ac cos [ 2fc(t) + Kpm(t) ]

Ac- Amplitude of the carrier maintained constant(t)- Angular argument varied by a message signalm (t) - Message signal

3b.

Envelop of FM wave is constant; so that averagepower of such a wave dissipated in 1 resistor is alsoconstant.

P= Ac2

P= Ac2 Jn2 ()

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Jn2 () 1

There fore

4.

S1 (t) = A1 cos [2f1 (t) + 1 (t)]t

1 (t) = 2K1o m (t) dt

If angle 1 (t) is small compared to 1 radian

Cos [1 (t)] = 1 and sin [1 (t)] = 1 (t)

S1 (t) = A1 cos 2f1 (t) cos1 (t) A1 sin 2f1 (t). sin1 (t)t

= A1 cos 2f1 (t) - A1 sin 2f1 (t) [2K10 m(t) dt]

Importance of selection of 0.3 has to be written

S2(t) = a1S1(t) + a2S12(t) + .

P= Ac2

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t

S (t) = Ac cos [2fc (t) + 2Kf0 m(t) dt]Kf= nK1

5. Explain the detection process of FM signals using balanced frequency discriminator

with relevant

H1(f)={j2 c+B/2) , fc B/2

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S(t)=Ac cos( m(t) dt)

(t) =Ac m(t) dt)

1(f)={ j2 B/2) (f) B/2

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PART 36. Explain FM detection using PLL for non linearised model

VCO has 900 Phase shift w.r.t. unmodulated carrierS(t)= AC sin[2fct +1(t)]

1(t)=2kf m(t) dt

r(t) =Av cos[2fct +2(t)]

s(t) r(t)= higher and lower frequency componentHigher frequency component km Ac Av sin[4fct +1(t) +2(t)]Lower frequency component km Ac Av sin[1(t)- 2(t)]Km------multiplier gain/ volte(t)= km Ac Av sin[ e(t)]e(t)= 1(t)- 2(t)V(t)= e() h(t-) dh(t) impulse response of filterde(t)/dt = d 1(t)/dt - 2k0 sin[e()] h(t-) dk0 = km kv Ac Av

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7. Derive an epression for spectrum of FM wave with sinusoidal modulationS(t) = Ac cos [2fct + sin 2fmt]

=Ac{cos 2fct cos(Sin2fmt) sinfct sin(sin 2fmt)}SI(t)=AC Cos[ Sin(2fmt)]

SQ(t) = ACSin[ sin(2fmt)]

(t) = AC ej sin 2fmtS(t) = Re[ (t) ej2fmt](t) = Cn ej2nfmtCn =fm +1/2fm-1/2fm (t) e-j2nfmt dt

= AC fm +1/2fm-1/2fm ejSin2fmt j2nfm t dtX= 2fmt

Cn =AC/2 - ej( sinx-nx) dxJn () = 1/2 - ej(sin-nx)dxCn =AC Jn ()S(t) = Ac Jn() ej2nfmt

S(t) =ACJn() cos[2(fc+nfm)t]

S(f)=AC/2 Jn() [(f-fc-nfm) + (f+fc+nfm)]J-n()={ jn () n even

-jn() n oddFig. Of plot of Bessel function of first kind*********************************************************************************8. A modulating signal 5 cos 2(5*10^3 t) angle modulates a carrier Acos wct.Find the modulation index and the bandwidth of the FM system . Determine thechange in the band width and the modulation index if FM is reduced to 5KHZ.what is the conclusion of these two results?Sol. m(t)= 5 cos 2 15*103 t

C(t)=A coswctS(t)=AC cos[2fct + 2kf0t m(t) dt]

=Ac cos [2fCt + 2kf 0t 5 cos 215*103 t]= Ac cos [2fCt + 2kf(5 sin 2( 15*103)t55)/15*103]= Ac cos [wCt + 2(5kf) ( sin 2( 15*103)t)/15*103]

f=Am kf=5 kf Assume kf=1

f=5fm=15*(10)3

= f/fm

=5/(15*103

)=1/(3*103)

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B.w=2f+ 2fm=2(5)+2(15*103)=30.010 KHZ

When m(t) is (5*103) KHZ

S(t)= Ac cos [wCt + 2(kf) ( sin 2( 5*103

)

t)/5*103

]f=Am kff=1 Assume kf=1

= f/fm=1/(5*103)

B.W.= B.w=2f+ 2fm=2(1)+2(5*103)=10.002KHZ

As the modulating frequency and amplitude of modulating wave is reduced ,its modulation index and also Band width reduces.

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