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JNTU B.Tech III ANALOG COMMUNICATIONS Previous Question Paper (August - 2022 - R13)

B.Tech | III | I | R13 | August - 2022

Answer PaperPaper Code: 115AKMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Draw the square law Modulator and explain the process of generation of AM waves.

7 marks

Answer

A square law modulator consists of a non-linear device, such as a diode, operating in its square-law region, followed by a bandpass filter. The input to the non-linear device is the sum of the carrier signal c(t) = Ac cos(2πfct) and the message signal m(t). The output of the non-linear device is represented by the power series vo(t) = a1vi(t) + a2vi(t)2, where vi(t) = m(t) + Ac cos(2πfct).

Expanding the squared term produces components including m(t)2, Ac2 cos2(2πfct), and 2a2Acm(t) cos(2πfct). The bandpass filter is tuned to the carrier frequency fc with a bandwidth sufficient to pass the modulated signal while rejecting the low-frequency m(t)2 and high-frequency harmonic components. The resulting output is the AM wave: s(t) = a1Ac[1 + (2a2/a1)m(t)] cos(2πfct), which represents the amplitude-modulated signal.

b

Derive the expression for total power in DSB-SC AM system. Find the total power ratio of DSB and DSB-SC AM.

8 marks

Answer

In a DSB-SC system, the modulated signal is s(t) = Acm(t) cos(2πfct). Assuming a sinusoidal message m(t) = Am cos(2πfmt), the signal becomes s(t) = AcAm cos(2πfmt) cos(2πfct). Using the trigonometric identity, s(t) = (AcAm/2) [cos(2π(fc+fm)t) + cos(2π(fc-fm)t)]. The total power PDSB-SC is the sum of the powers of the two sidebands: PDSB-SC = (Ac2Am2/8) + (Ac2Am2/8) = Ac2Am2/4.

For standard AM, the total power is PAM = Pc(1 + μ2/2), where Pc = Ac2/2 is the carrier power and μ is the modulation index. The power ratio of DSB (standard AM) to DSB-SC is given by the ratio of PAM to PDSB-SC. Since DSB-SC transmits only the sidebands, it is more power-efficient than standard AM, which wastes power in the carrier component.

215 marks
a

Comparison and contrast different AM Techniques.

7 marks

Answer

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b

The peak amplitude of an amplitude modulated signal varies from 2V to 10V. Calculate the modulation index, modulation efficiency and total power.

8 marks

Answer

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315 marks
a

List out the methods for generation of SSB-SC signal and explain any one of the method in detail.

7 marks

Answer

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b

Find the various frequency components and their amplitudes in the voltage given by v(t) = 50 (1+ 0.7 Cos 5000t - 0.3 Cos 1000t) sin 5×106t. Draw the single sided spectrum. Also evaluate the modulated and sideband power.

8 marks

Answer

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415 marks
a

Explain the process of generation of VSB waves.

8 marks

Answer

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b

Write the applications of different AM Systems in detail.

7 marks

Answer

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515 marks
a

Explain in detail about NBFM and WBFM. Derive the expression for bandwidth of wideband FM.

8 marks

Answer

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b

Distinguish PM and FM.

7 marks

Answer

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615 marks
a

Explain the threshold effects in angle modulation system with suitable diagrams.

8 marks

Answer

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b

Draw and explain the working of Zero crossing detector.

7 marks

Answer

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715 marks
a

What is noise? Explain the difference between thermal noise and shot noise.

6 marks

Answer

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b

Write a short note on following: i) Average Noise Bandwidth ii) Effective Noise Temperature iii) Average Noise Figures.

9 marks

Answer

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815 marks
a

Describe the synchronization procedure for PAM, PWM and PPM signals.

8 marks

Answer

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b

List out the advantages and disadvantages of TRF receivers.

7 marks

Answer

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