1

Er. No.............................Semester - IV Academic Year 2011-12

JAYPEE UNIVERSITY OF ENGINEERING & TECHNOLOGY, GUNATest -II

10B11EC412- Analog Communication

Time : 1:30 Hours Maximum Marks : 25———————————————————————————————————————————Instructions and information for students

• Do not write anything on the question paper except your enrolment no.• All questions are compulsory.• Make suitable assumptions if necessary, write them with your answer.

1) a) Write a short note on stereophonic frequency modulation (FM) broadcasting system .b) What are the advantages and disadvantages of pulse modulation over continuous wave (CW)

modulation?. [3+2 =5 marks]

2) A periodic square wavem(t) of periodT0 (Figure2(a)) frequency modulates a carrier of frequencyfc = 10 kHz with ∆f = 1 kHz. The carrier amplitude is A. The resulting FM signal is demodulatedas shown in Figure2(b). Sketch the waveforms at pointsb, c, d ande. [2+1+1+2=6 marks]

Solution:We havefc = 10 kHz with ∆f = 1 kHz. We know that

∆f =kf mp

2π

wheremp is the maximum value ofm(t).

kf = 2000π

2

For FM :

fi = fc +kf

2πm(t) (1)

Becausem(t) switches from1 to −1 and vice versa, the FM wave frequency switches back andforth between9kHz to 11kHz. Let φb(t), φc(t), φd(t) and φe(t) the expressions of the signals atpointsb, c, d ande

φb(t) = A cos(

ωc t+ kf

∫ t

−∞

m(α)dα)

φc(t) =dφb(t)

dt= −A

(

ωc + kfm(t))

sin(

ωc t + kf

∫ t

−∞

m(α)dα)

φd(t) = −A(

ωc + kfm(t))

φe(t) = −Akfm(t)

3

3) Design (only the block diagram) an Armstrong indirect FM modulator to generate an FM carrier witha carrier frequency of96MHz and∆f = 20 kHz. A narrow band FM generator withfc = 200 kHzand adjustable in the range of9 to 10Hz is available. The stock room also has an oscillator withadjustable frequency in the range of9 to 10MHz. There is a bandpass filter with any center frequencyand only frequency doublers are available. [5 marks]

Solution:The block diagram of an Armstrong indirect FM modulator is shown in Figure3.

We have only frequency doublers.

M1 = 2n1

M2 = 2n2

M1 M2 = 2n1+n2

M1M2 =∆ f

∆ f1

We have∆ f = 20KHz and9Hz < ∆ f1 < 10Hz

2000 < M1M2 < 2222

2000 < 2n1+n2 < 2222

There is only one possibilityM1M2 = 2048 = 211.

∆ f1 =∆ f

2048= 9.76563Hz

4

We have

fc = 200M1M2KHz − Y M2MHz (2)

SubstitutingM1M2 = 2048 andfc = 96MHz in equation (2), we obtain

Y M2 = 313.6 (3)

SupposeM2 = 16

Y = 19.6 (4)

It is not acceptable. Now we takeM2 = 32

Y = 9.8 (5)

It is acceptable. SoM2 = 32 andM1 =2048

32= 64.

5

4) An angle modulated signal with carrier frequencyfc = 106 is described by the equation

ϕEM(t) =√5 cos

(

ωct + 30 sin 1000πt+ 15 sin 2000πt)

.

a) Find the power of the modulated signal.b) Find the frequency deviation∆f .c) Find the phase deviation∆φ.d) Estimate the bandwidth ofϕEM(t). [1+1+1+2=5 marks]

Solution:a) LetP be the power of the modulated signal.

The carrier amplitude is√5.

P =(√5)2

2= 2.5Watts

b) Let θ(t) and ωi(t) be the angle and the instantaneous frequency of the modulated signalrespectively.Given θ(t) = ωc t+ 30 sin 1000π t + 15 sin 2000π t

ωi(t) =dθ(t)

dt= ωc + 30, 000π cos 1000π t+ 30, 000π cos 2000π t

Let ∆ω be the maximum frequency deviation

∆ω = max{ωi(t)− ωc} = max{30, 000π cos 1000π t+ 30, 000π cos 2000π t}= 60, 000 π radian/sec.

∆f =∆ω

2 π= 30, 000Hz = 30KHz

c)

∆φ = max{θ(t)− ωc t} = max{30 sin 1000π t+ 15 sin 2000π t}= 45 radian

d) Let B andBEM be the bandwidth of the modulating signal and the modulated signal respec-tively.

B = bandwidth of30 sin 1000π t+ 15 sin 2000π t

= 1000Hz = 1KHz

We know that

BEM = 2(∆ f +B)

= 62KHz

6

5) Signalsx1(t) = 104 rect(104 t) and x2(t) = δ(t) are applied at the inputs of ideal low-pass filtersH1(ω) = rect

(

ω40,000 π

)

andH2(ω) = rect(

ω20,000π

)

. The outputsy1(t) and y2(t) of these filters aremultiplied to obtain the signaly(t) = y1(t)y2(t). Find the minimum sampling rates of they1(t),y2(t) andy(t). [1+1+2=4 marks]

Solution:Let B1, B2 andB be the bandwidths of they1(t), y2(t) andy(t) respectively.Let fs1 , fs2 andfs be the minimum sampling rates of they1(t), y2(t) andy(t) respectively.We have

g(t) = rect( t

τ

)

⇔ G(ω) = τ sinc(ω τ

2

)

Let X1(ω), X2(ω), Y1(ω) andY2(ω) be the Fourier transforms of thex1(t), x2(t), y1(t) andy2(t)respectively.

x1(t) = 104 rect(104 t) ⇔ X1(ω) = sinc(10−4 ω

2

)

7

x2(t) = δ(t) ⇔ X2(ω) = 1

Y1(ω) = H1(ω)X1(ω),

Y2(ω) = H2(ω)X2(ω) = H2(ω)

From Figures 5.3 and 5.2 (b), It is obviousB1 = 10, 000Hz = 10KHz andB2 = 5, 000Hz = 5KHz.

B = B1 +B2 = 15KHz

fs1 = 2B1 = 20KHz

fs2 = 2B2 = 10KHz

fs = 2B = 30KHz