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JNTU B.Tech III ANALOG COMMUNICATIONS Previous Question Paper (September - 2021 - R16)

B.Tech | III | I | R16 | September - 2021

Answer PaperPaper Code: 135SAMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

With block diagram and related equations explain the coherent detection of DSBSC wave. What are its disadvantages?

8 marks

Answer

Coherent detection of a DSBSC wave involves multiplying the received signal with a locally generated carrier that is perfectly synchronized in both frequency and phase with the original carrier. The received DSBSC signal is s(t) = Acm(t)cos(2πfct). This signal is multiplied by a local oscillator output c(t) = cos(2πfct), resulting in v(t) = Acm(t)cos2(2πfct). Using the trigonometric identity cos2(θ) = (1 + cos(2θ))/2, this becomes v(t) = (Acm(t)/2) + (Acm(t)/2)cos(4πfct). Passing v(t) through a low-pass filter with cutoff frequency fm removes the high-frequency component, leaving the baseband signal m(t) scaled by a constant. The primary disadvantage is the requirement for a complex synchronization circuit (such as a Costas loop or PLL) to maintain phase and frequency coherence, and the system is highly sensitive to phase errors, which can cause significant distortion in the recovered message signal.

b

Explain the working of square Law modulator with neat diagram.

7 marks

Answer

A square-law modulator generates an AM signal using a non-linear device, such as a diode, operating in its square-law region. The input to the non-linear device is the sum of the carrier signal c(t) = Accos(2πfct) and the message signal m(t). The output of the non-linear device is expressed as vo(t) = a1vi(t) + a2vi2(t), where vi(t) = m(t) + Accos(2πfct). Expanding this, the squared term produces a component 2a2Acm(t)cos(2πfct), which represents the desired AM wave. A band-pass filter centered at the carrier frequency fc is then used to extract this term while suppressing the baseband message signal and the second harmonic of the carrier frequency.

215 marks
a

Discuss the drawbacks of envelope detector and explain its working.

8 marks

Answer

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b

A carrier wave with amplitude 12 V and frequency 10 MHz is amplitude modulated to 50 % level with a modulated frequency of 1 KHz. Write down the equation for the above wave and sketch the modulated signal in frequency domain.

7 marks

Answer

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315 marks
a

Derive the equation for output power in SSB wave and calculate its efficiency when modulation index is 100%.

6 marks

Answer

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b

With a neat diagram, explain how SSB wave is generated using phase shift method.

9 marks

Answer

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415 marks
a

Show that a VSB modulated wave s(t) containing a vestige of the lower sideband is defined by the following relation:

s(t) = (Ac/2)m(t)cos(2πfct) ± (Ac/2)m̂(t)sin(2πfct)
8 marks

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b

Discuss the frequency domain description of SSB-SC.

7 marks

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515 marks
a

In a FM system, the modulating frequency fm=1KHz, the modulating voltage Am=2V and the deviation is 6 KHz. If the modulating voltage is raised to 4 V then what is the new deviation? If the modulating voltage is further increased to 8 V and modulating frequency is reduced to 500Hz. What will be deviation? Calculate the modulation index in each case? Comment on result.

8 marks

Answer

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b

Explain the detection of FM using zero crossing detector.

7 marks

Answer

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615 marks
a

Compare and contrast NBFM with WBFM.

7 marks

Answer

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b

Explain generation of FM wave using direct method.

8 marks

Answer

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7

Discuss the following with necessary equations:

15 marks
a

Noise equivalent bandwidth

5 marks

Answer

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b

Thermal noise

5 marks

Answer

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c

Narrow band noise

5 marks

Answer

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815 marks
a

What is TDM? Explain the principle of TDM with a neat block diagram.

8 marks

Answer

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b

Explain the demodulation of the PPM wave.

7 marks

Answer

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This page contains the question paper of ANALOG COMMUNICATIONS from JNTU for B.Tech ECE III I, September - 2021, R16.

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