JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II PULSE AND DIGITAL CIRCUITS Previous Question Paper (July/August - 2021 - R16)

B.Tech | II | II | R16 | July/August - 2021

Answer PaperPaper Code: 134CCMaximum Marks: 75Time: 3 Hours

Answer any Five Questions. All Questions Carry Equal Marks.

Main

115 marks
a

Derive an expression for the output of a high-pass circuit excited by step input.

9 marks

Answer

For a high-pass RC circuit, the input voltage Vi(t) is related to the output voltage Vo(t) by the differential equation:

Vi(t) = Vo(t) + RC(dVo/dt)

For a step input of magnitude V, Vi(t) = V for t > 0. The Laplace transform of the equation is:

V/s = Vo(s) + RC[sVo(s) - Vo(0+)]

Since the capacitor is initially uncharged, Vo(0+) = V. Solving for Vo(s):

Vo(s) = V / (s + 1/RC)

Taking the inverse Laplace transform, the output voltage is:

Vo(t) = V e-t/RC

This shows that the output of a high-pass circuit for a step input is an exponential decay starting from the magnitude of the input step.

b

A symmetrical square wave whose peak-to-peak amplitude is 8V and whose average value is zero is applied to an RC integrating circuit. The time constant is equal to half-period of the square wave. Find the peak to peak value of the output amplitude.

6 marks

Answer

Given a square wave with peak-to-peak amplitude Vpp = 8V, the signal swings between +4V and -4V. The time constant τ = RC is equal to half the period T/2. Let T/2 = τ.

During the first half-cycle (0 ≤ t ≤ τ), the input is +4V. The output of an integrator is:

Vo(t) = V1 + (Vin - V1)(1 - e-t/τ)

At steady state, the output at the end of the first half-cycle (V2) must be the negative of the output at the start (V1), so V2 = -V1. Setting t = τ and Vin = 4V:

-V1 = V1 + (4 - V1)(1 - e-1)

Using e-1 ≈ 0.368:

-V1 = V1 + (4 - V1)(0.632)
-V1 = V1 + 2.528 - 0.632V1
1.368V1 = 2.528 ⇒ V1 ≈ 1.848V

The output swings from V1 = 1.848V to V2 = -1.848V. The peak-to-peak output amplitude is:

Vo(pp) = V1 - V2 = 1.848 - (-1.848) = 3.696V
215 marks
a

Draw and explain the response of RLC circuit for a step input.

8 marks

Answer

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b

Derive an expression for the percentage tilt of the output of a high pass circuit with large time constant excited by a symmetrical square wave with zero average value.

7 marks

Answer

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3

Classify different types of clipper circuits. Give their circuits and explain their operation with the aid of transfer characteristics.

15 marks

Answer

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415 marks
a

Describe in detail about clipping at two independent levels.

7 marks

Answer

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b

Design a negative clamper with positive and negative biases and then, explain the same.

8 marks

Answer

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515 marks
a

Give a brief note on piece-wise linear diode characteristics.

7 marks

Answer

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b

Write short notes on diode switching times.

8 marks

Answer

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615 marks
a

Discuss in detail about breakdown voltages of a transistor.

6 marks

Answer

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b

Write a short note on switching characteristics of transistor.

9 marks

Answer

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715 marks
a

Derive the relation between slope error, displacement error and transmission error.

7 marks

Answer

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b

With the help of a neat circuit diagram, explain the working of a transistor constant current sweep circuit.

8 marks

Answer

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815 marks
a

Explain the working of four diode sampling gate with the help of neat circuit diagram.

8 marks

Answer

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b

Why a OR gate is called mixing circuit? Draw a diode OR circuit for positive logic and explain how it works.

7 marks

Answer

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About this paper

This page contains the question paper of PULSE AND DIGITAL CIRCUITS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech ECE II II, July/August - 2021, R16.

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