GATE Electronics and Communication Engineering Chapter-wise Solved Papers (2000-2020) 9788126571956, 9788126595372

Table of contents :
Front Cover
TCY's Unique Success Mantra
Title Page
Copyright
Note to the Aspirants
Contents
Chapter 1: Engineering Mathematics
Important Formulas
Questions
Answers with Explanation
Chapter 2: Networks, Signals and Systems
Questions
Important Formulas
Important Tables
Answers with Explanation
Chapter 3: Electronic Devices
Important Formulas
Questions
Answers with Explanation
Chapter 4: Analog Circuits
Important Tables
Important Formulas
Questions
Answers with Explanation
Chapter 5: Digital Circuits
Important Formulas
Questions
Answers with Explanation
Chapter 6: Control Systems
Important Formulas
Questions
Answers with Explanation
Chapter 7: Communications
Important Formulas
Questions
Answers with Explanation
Chapter 8: Electromagnetics
Important Formulas
Questions
Answers with Explanation
APPENDIX Solved GATE (ECE) 2019
APPENDIX Solved GATE (ECE) 2020
Back Cover

Citation preview

The book comprises previous years’ (2000 to 2020) GATE questions in Electronics and Communication Engineering covered chapter-wise with detailed solutions. Each chapter begins with a detailed year-wise analysis of topics on which questions are based and is followed by listing of important formulas and concepts for that chapter. The book is designed to make the students well-versed with the pattern of examination, level of questions asked and the concept distribution of questions and thus bring greater focus to their preparation. It aims to be a must-have resource for an essential step in their preparation, that is, solving and practicing previous years’ papers.

Salient Features l

Chapter-wise coverage of GATE Electronics and Communication Engineering previous years’ questions (2000 to 2018). Chapter-wise 2019 and 2020 GATE papers available as appendix

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Chapter on Engineering Mathematics included for complete coverage of the technical section of the GATE paper

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Detailed topic-wise analysis of questions for each chapter

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Important Formulas and Concepts summarized for easy recall in each chapter

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All questions marked for level of difficulty (i.e. 1 Mark or 2 Marks)

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Detailed solutions for all questions, tagged topic-wise

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Packaged with unique scratch code for free 7-day subscription for online GATE tests

Customer Care +91 120 6291100 [email protected] www.wileyindia.com www.wiley.com

eISBN 978-81-265-9537-2

9 788126 595372

Maini | Agrawal | Maini

Wiley India Pvt. Ltd.

GATE

ELECTRONICS AND COMMUNICATION ENGINEERING

CHAPTER-WISE SOLVED PAPERS

(2000-2020)

Inside

The Graduate Aptitude Test in Engineering (GATE) is an All-India level examination for engineering graduates aspiring to pursue Master’s or Ph.D. programmes in India. The Public Sector Undertakings (PSU’s) also use GATE used as a recruiting examination. The examination is highly competitive, and the GATE score plays an important role in fulfilling the academic and professional aspirations of the students. This book is aimed at supporting the efforts of GATE aspirants in achieving a high GATE score.

GATE ELECTRONICS AND COMMUNICATION ENGINEERING CHAPTER-WISE SOLVED PAPERS (2000-2020)

About the Book

Dr Anil Kumar Maini Varsha Agrawal Nakul Maini

l

Detailed Exam Analysis Chapter-wise and Topic-wise.

l

Questions from previous years’ (2000 – 2020) papers.

l

Unique scratch code that provides access to w

Free online test with analytics.

w

7-Day free subscription for topic-wise GATE tests.

w

Instant correction report with remedial action.

To get access code for the free tests, please write to us at [email protected], with a copy of the Invoice to verify purchase.

GATE

ELECTRONICS AND COMMUNICATION ENGINEERING

CHAPTER-WISE SOLVED PAPERS

(2000-2020)

Dr Anil Kumar Maini Outstanding Scientist and former Director of Laser Science and Technology Centre DRDO, New Delhi

Varsha Agrawal Senior Scientist, Laser Science and Technology Centre DRDO, New Delhi

Nakul Maini Analyst Ericsson (India)

GATE Electronics and Communication Engineering Chapter-wise Solved Papers (2000—2020) Copyright © 2020 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 Edition: 2020 ISBN: 978-81-265-7195-6 ISBN: 978-81-265-9537-2 (ebk) www.wileyindia.com Printed at:

Note to the Aspirants About the Examination Graduate Aptitude Test in Engineering (GATE) is an examination conducted jointly by the Indian Institute of Science (IISc), Bangalore and the seven Indian Institutes of Technology (at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee) on behalf of the National Coordination Board (NCB)GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD), and Government of India. Qualifying in GATE is a mandatory requirement for seeking admission and/or financial assistance to:



• M  aster’s programs and direct Doctoral programs in Engineering/Technology/Architecture. • Doctoral programs in relevant branches of Science, in the institutions supported by the MHRD and other Government agencies. Even in some colleges and institutions, which admit students without MHRD scholarship/assistantship, the GATE qualification is mandatory. • Many Public Sector Undertakings (PSUs) have been using the GATE score in their recruitment process.

Graduate Aptitude Test in Engineering (GATE) is basically an examination on the comprehensive understanding of the candidates in various undergraduate subjects in Engineering/Technology/ Architecture and post-graduate level subjects in Science. GATE is conducted for 24 subjects (also referred to as “papers”) are generally held at the end of January or early February of the year. The GATE examination centres are spread in different cities across India, as well as, in six cities outside India. The examination is purely a Computer Based Test (CBT). The GATE score reflects the relative performance level of the candidate in a particular subject, which is quantified based on the several years of examination data.

Eligibility for GATE Electronics and Communication Engineering

Bachelor’s degree holders in Electronics and Communication Engineering (4 years after 10+2 or 3 years after B.Sc./Diploma in Electronics and Communication Engineering).

About the Book This book GATE Chapter-wise Electronics and Communication Engineering Solved Papers is designed as a must have resource for the students preparing for the M.E./M.Tech./M.S./Ph.D. in Electronics and Communication Engineering. It offers Chapter-wise solved previous years’ GATE Electronics and Communication Engineering questions for the years 2000– 2020. This book will help students become well-versed with the pattern of examination, level of questions asked and concept distribution in questions. Key Features of the Book



• C  omplete solutions provided for every question, tagged for the topic on which the question is based. • Chapter-wise analysis of GATE questions provided at the beginning of the book to make students familiar with chapter-wise marks distribution and weightage of each. • Topic-wise analysis of GATE Questions provided at the beginning of each chapter to make students familiar with important topics on which questions are commonly asked. • Unique scratch code in the book that provides access to Free online test with analytics. 7-Day free subscription for topic-wise GATE tests. Instant correction report with remedial action. These features will help students develop problem-solving skills and focus in their preparation on important chapters and topics. ■

■

■

Validity

The GATE score is valid for THREE YEARS from the date of announcement of the results.

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Analog Circuits

Control Systems

Communications

Electromagnetics

6

7

8

2

3

3

3

4

2

Electronic Devices

3

Digital Circuits

5

Networks, Signals and Systems

2

4

2

Engineering Mathematics

5

GATE 2011

GATE 2012

GATE 2013

GATE 2014

GATE 2015

GATE 2016

GATE 2017

GATE 2018

GATE 2019

GATE 2020

2

4

3

3

3

4

5

4

3

2

2

3

3

3

6

1

3

4

3

3

3

4

6

3

4

4

1

3

2

1

6

4

5

3

3

1

2

3

7

1

3

1

2

2

2

2

9

3

2

5

5

1

5

1

8

2

10

8

7

14

14

11

23

20

12

17

15

10

14

14

24

15

5

6

8

8

9

7

19

20

11

9

10

10

14

5

18

4

9

8

6

7

12

9

17

14

15

14

8

9

10

8

14

12

5

4

6

3

11

8

9

6

6

9

4

10

10

8

11

8

3

3

1

3

3

2

4

7

3

3

3

4

2

5

6

4

3

3

1

4

0

4

3

7

4

4

5

3

2

5

4

3

2

4

2

3

3

2

5

4

2

3

4

4

4

4

5

4

1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks

1

S.No. Chapter Name

GATE 2010

The table given below depicts the chapter-wise question and marks distribution of previous years’ (2010–2020) GATE Electronics and Communication Engineering papers. This will help students understand the relative weightage of each chapter in terms of number of questions asked and thus bring focus in their preparation.

Gate Electronics and Communication Engineering: Chapter-Wise Question Distribution Analysis 2010–2020

Contents

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Note to the Aspirants

iii

Chapter 1: Engineering Mathematics Important Formulas Questions Answers with Explanation

1 1 26 43

Chapter 2: Networks, Signals and Systems Important Formulas Important Tables Questions Answers with Explanation

71 71 75 84 130

Chapter 3: Electronic Devices Important Formulas Questions Answers with Explanation

197 197 199 220

Chapter 4: Analog Circuits Important Formulas Important Tables Questions Answers with Explanation

243 243 249 250 287

Chapter 5: Digital Circuits Important Formulas Questions Answers with Explanation

323 323 325 356

Chapter 6: Control Systems Important Formulas Questions Answers with Explanation

385 385 389 413

Chapter 7: Communications Important Formulas Questions Answers with Explanation

445 445 450 475

Chapter 8: Electromagnetics Important Formulas Questions Answers with Explanation

513 513 518 539

Solved GATE (ECE) 2019

A1

Solved GATE (ECE) 2020

B1

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Engineering Mathematics

CHAPTER1

Syllabus Linear Algebra: Vector space, basis, linear dependence and independence, matrix algebra, eigen values and eigen vectors, rank, solution of linear equations – existence and uniqueness. Calculus: Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partial ­derivatives, maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series. Differential Equations: First order equations (linear and nonlinear), higher order linear differential equations, Cauchy’s and Euler’s equations, methods of solution using variation of parameters, complementary function and particular integral, partial differential equations, variable separable method, initial and boundary value problems. Vector Analysis: Vectors in plane and space, vector operations, gradient, divergence and curl, Gauss’s, Green’s and Stoke’s theorems. Complex Analysis: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula; Taylor’s and Laurent’s series, residue theorem. Numerical Methods: Solution of nonlinear equations, single and multi-step methods for differential equations, convergence criteria. Probability and Statistics: Mean, median, mode and standard deviation; combinatorial probability, probability ­distribution functions - binomial, Poisson, exponential and normal; Joint and conditional probability; Correlation and regression ­analysis.

Chapter Analysis Topic

GATE 2009

GATE 2010

GATE 2011

Linear Algebra

1

1

1

Calculus

1

1

Differential Equations

2

1

1

1

1

GATE 2012

1

Vector Analysis

GATE 2013

GATE 2014

GATE 2015

GATE 2016

GATE 2017

GATE 2018

2

7

6

4

4

2

1

10

5

7

6

2

1

5

5

2

2

3 2

1

Complex Analysis Numerical Methods Probability and Statistics

1

2

1

1

1

1

2

3

2

4

1

2

3

1

1

1

2

1

8

3

3

1

Important Formulas Linear Algebra 1.





Types of Matrices (a) Row matrix: A matrix having only one row is called a row matrix or a row vector. Therefore, for a row matrix, m = 1. (b) Column matrix: A matrix having only one column is called a column matrix or a column vector. Therefore, for a column matrix, n = 1. (c) Square matrix: A matrix in which the number of rows is equal to the number of columns, say n, is called a square matrix of order n.

Ch wise GATE_ECE_CH01_Imp formula.indd 1



(d) Diagonal matrix: A square matrix is called a diagonal matrix if all the elements except those in the leading diagonal are zero, i.e. aij = 0 for all i ≠ j.



(e) Scalar matrix: A matrix A = [aij ]n × n is called a ­scalar matrix if •  aij = 0, for all i ≠ j. •  aii = c, for all i, where c ≠ 0.



(f) Identity or unit matrix: A square matrix A = [aij]n × n is called an identity or unit matrix if •  aij = 0, for all i ≠ j. •  aij = 1, for all i = j.

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GATE ECE Chapter-wise Solved papers



(g) Null matrix: A matrix whose all the elements are zero is called a null matrix or a zero matrix.



(h) Upper triangular matrix: A square matrix A = [aij] is called an upper triangular matrix if aij = 0 for i > j. (i) Lower triangular matrix: A square matrix A = [aij]



is called a lower triangular matrix if aij = 0 for i < j. 2.

Types of a Square Matrix (a) Nilpotent matrix: A square matrix A is called a nilpotent matrix if there exists a positive integer n such that An = 0. If n is least positive integer such that An = 0, then n is called index of the nilpotent matrix A.



(e) Cancellation laws: If A, B and C are matrices of the same order, then A+B=A+C⇒B=C B+A=C+A⇒B=C

5.

Some important properties of matrix multiplication are: (a) Matrix multiplication is not commutative.



(b) Matrix multiplication is associative, i.e. (AB)C = A(BC).



(c)  Matrix multiplication is distributive over matrix addition, i.e. A(B + C) = AB + AC.



(d) If A is an m × n matrix, then ImA = A = AIn. (e) The product of two matrices can be the null matrix while neither of them is the null matrix.



(b) Symmetrical matrix: It is a square matrix in which aij = aji for all i and j. A symmetrical matrix is necessarily a square one. If A is symmetric, then AT = A.



(c) Skew-symmetrical matrix: It is a square matrix in which aij = −aji for all i and j. In a skew-symmetrical matrix, all elements along the diagonal are zero.

6.



(d) Hermitian matrix: It is a square matrix A in which (i, j)th element is equal to complex conjugate of the (j, i)th element, i.e. aij = a ji for all i and j.



(b) (k + l) ⋅ A = kA + lA



(c) (kl) ⋅ A = k(lA) = l(kA)



(d) (−k) ⋅ A = −(kA) = k(−A)



(e) Skew-Hermitian matrix: It is a square matrix A = [aij] in which aij = − aij for all i and j.



(f) Orthogonal matrix: A square matrix A is called orthogonal matrix if AAT = ATA = I.

3.

Equality of a Matrix Two matrices A = [aij]m × n and B = [bij]x × y are equal if (a)  m = x, i.e. the number of rows in A equals the number of rows in B.



(b) n = y, i.e. the number of columns in A equals the number of columns in B. (c)  aij = bij for i = 1, 2, 3, …, m and j = 1, 2, 3, …, n.

4.

Some of the important properties of matrix addition are: (a) Commutativity: If A and B are two m × n matrices, then A + B = B + A, i.e. matrix addition is commutative.



(b) Associativity: If A, B and C are three matrices of the same order, then





(A + B) + C = A + (B + C)

i.e. matrix addition is associative.

(c) Existence of identity: The null matrix is the identity element for matrix addition. Thus, A + O = A = O+A (d) Existence of inverse: For every matrix A = [aij]m × n, there exists a matrix [aij]m × n, denoted by −A, such that A + (−A) = O = ( −A) + A

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Some of the important properties of scalar multiplication are: (a)  k(A + B) = kA + kB

(e) 1 ⋅ A = A (f)  −1 ⋅ A = −A Here A and B are two matrices of same order and k and l are constants. If A is a matrix and A2 = A, then A is called idempotent matrix. If A is a matrix and satisfies A2 = I, then A is called involuntary matrix. 7. Some of the important properties of transpose of a matrix are: (a)  For any matrix A, (AT)T = A (b)  For any two matrices A and B of the same order (A + B)T = AT + BT (c)  If A is a matrix and k is a scalar, then (kA)T = k(AT) (d) If A and B are two matrices such that AB is defined, then

(AB)T = BTAT

8.

Some of the important properties of inverse of a matrix are: (a)  A−1 exists only when A is non-singular, i.e. |A| ≠ 0.

(b)  The inverse of a matrix is unique.



(c) Reversal laws: If A and B are invertible matrices of the same order, then (AB)−1 = B−1A−1

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3

Chapter 1  • Engineering Mathematics



(d)  If A is an invertible square matrix, then (AT)−1 = (A−1)T



(e) The inverse of an invertible symmetric matrix is a symmetric matrix.



(f) Let A be a non-singular square matrix of order n. Then



|adj A| = |A|n −1

adj (AB) = (adj B)(adj A)

(h) If A is an invertible square matrix, then adj AT = (adj A)T



(i)  If A is a non-singular square matrix, then adj (adj A) = |A|n − 2 A



(j)  If A is a non-singular matrix, then



1 |A | = |A| , i.e. | A | = | A| −1

−1

−1

(k) Let A, B and C be three square matrices of same type and A be a non-singular matrix. Then AB = AC ⇒ B = C

and 9.



(m) The rank of transpose of a matrix is equal to rank of the original matrix. rank (A) = rank (AT)

(n) The rank of a matrix does not change by pre-multiplication or post-multiplication with a ­ non-singular matrix.



(o) If A − B, then rank (A) = rank (B).

(g) If A and B are non-singular square matrices of the same order, then



(l)  If A is any n-rowed square matrix of rank, n − 1, then adj A ≠ 0

BA = CA ⇒ B = C



The rank of a matrix A is commonly denoted by rank (A). Some of the important properties of rank of a matrix are: (a)  The rank of a matrix is unique.



(b)  The rank of a null matrix is zero.



(c)  Every matrix has a rank.



(d) If A is a matrix of order m × n, then rank (A) ≤ m × n (smaller of the two)



(e) If rank (A) = n, then every minor of order n + 1, n + 2, etc., is zero.



(f) If A is a matrix of order n × n, then A is non-singular and rank (A) = n.



(g)  Rank of IA = n.

(h) A is a matrix of order m × n. If every kth order minor (k < m, k < n) is zero, then rank (A) < k



(i) A is a matrix of order m × n. If there is a minor of order (k < m, k < n) which is not zero, then rank (A) ≥ k



(j) If A is a non-zero column matrix and B is a non-zero row matrix, then rank (AB) = 1.



(k) The rank of a matrix is greater than or equal to the rank of every sub-matrix.

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(p) The rank of a product of two matrices cannot exceed rank of either matrix. rank (A × B) ≤ rank A or rank ( A × B) ≤ rank B

(q) The rank of sum of two matrices cannot exceed sum of their ranks. (r) Elementary transformations do not change the rank of a matrix.

10. Determinants Every square matrix can be associated to an expression or a number which is known as its determinant. If A = [aij] is a square matrix of order n, then the determinant of A is denoted by det A or |A|. If A = [a11] is a square matrix of order 1, then determinant of A is defined as |A| = a11 a12 ⎤ ⎡a If A = ⎢ 11 ⎥ is a square matrix of order 2, then ⎣ a21 a23 ⎦ determinant of A is defined as |A| = a11a23 − a12a21 ⎡ a11 If A = ⎢⎢ a21 ⎢⎣ a31

a12 a22 a32

a13 ⎤ a23 ⎥⎥ is a square matrix of order 3, a33 ⎥⎦

then determinant of A is defined as |A| = a11 (a22a33 − a23a32) − a21 (a12a33 − a13a32) + a31 (a12a23 − a13a22) or  |A| = a11 (a22a33 − a23a32) − a12 (a21a33 − a23a31) + a13 (a21a32 − a22a31) 11. Minors The minor Mij of A = [aij] is the determinant of the square sub-matrix of order (n − 1) obtained by removing i th row and j th column of the matrix A. 12. Cofactors The cofactor Cij of A = [aij] is equal to (−1)i + j times the determinant of the sub-matrix of order (n − 1) obtained by leaving i th row and j th column of A.

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4

GATE ECE Chapter-wise Solved papers

13. Some of the important properties of determinants are: (a) Sum of the product of elements of any row or column of a square matrix A = [aij] of order n with their cofactors is always equal to its determinant. n

∑a c i =1



ij ij

n

= | A| = ∑ aij cij j =1

(b) Sum of the product of elements of any row or column of a square matrix A = [aij] of order n with the cofactors of the corresponding elements of other row or column is zero. n

∑a c i =1

ij ik

(d) Consider a square matrix A = [aij] of order n ≥ 2 and B obtained from A by interchanging any two rows or columns of A, then |B| = −A.



(e) For a square matrix A = [aij] of order (n ≥ 2), if any two rows or columns are identical, then its determinant is zero, i.e. |A| = 0.



(f) If all the elements of a square matrix A = [aij] of order n are multiplied by a scalar k, then the determinant of new matrix is equal to k|A|. (g) Let A be a square matrix such that each element of a row or column of A is expressed as the sum of two or more terms. Then |A| can be expressed as the sum of the determinants of two or more matrices of the same order.

•  If A ≠ 0, system is consistent and has a unique solution given by X = A−1 B. •  If A = 0 and (adj A)B = 0, system is consistent and has infinite solutions. •  If A = 0 and (adj A)B ≠ 0, system is inconsistent. 16. Cramer’s Rule Suppose we have the following system of linear equations: a1x + b1 y + c1z = k1

a2x + b2 y + c2z = k2



a3x + b3 y + c3z = k3



(h) Let A be a square matrix and B be a matrix obtained from A by adding to a row or column of A a scalar multiple of another row or column of A, then |B| = |A|. (i) Let A be a square matrix of order n (≥ 2) and also a null matrix, then |A| = 0.

(j) Consider A = [aij] as a diagonal matrix of order n (≥ 2), then |A| = a11 × a22 × a33 × … × ann



(k) Suppose A and B are square matrices of same order, then |AB| = |A| ⋅ |B| 14. There are two cases that arise for homogeneous systems: (a) Matrix A is non-singular or |A| ≠ 0.  The solution of the homogeneous system in the above equation has a unique solution, X = 0, i.e. x1 = x2 = … = xj = 0.

(b) Matrix A is singular or |A| = 0, then it has infinite many solutions. To find the solution when |A| = 0, put z = k (where k is any real number) and solve any

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(b) If AX = B is a system with linear equations equal to the number of unknowns, then three cases arise:

j =1







n

(c) For a square matrix A = [aij] of order n, |A| = |AT|.



15. For the method to solve a non-homogeneous system of simultaneous linear equations, please note the number of unknowns and the number of equations. (a) Given that A is a non-singular matrix, then a system of equations represented by AX = B has the unique solution which can be calculated by X = A−1 B.

= 0 = ∑ aij ckj





two equations for x and y using the matrix method. The values obtained for x and y with z = k is the solution of the system.



Now, if ⎡ a1 Δ = ⎢⎢ a2 ⎢⎣ a3

b1 b2 b3

c1 ⎤ c2 ⎥⎥ ≠ 0 c3 ⎥⎦

⎡ k1 Δ1 = ⎢⎢ k2 ⎢⎣ k3

b1 b2 b3

c1 ⎤ c2 ⎥⎥ ≠ 0 c3 ⎥⎦

⎡ a1 Δ 2 = ⎢⎢ a2 ⎢⎣ a3

k1 k2 k3

c1 ⎤ c2 ⎥⎥ ≠ 0 c3 ⎥⎦

⎡ a1 Δ 3 = ⎢⎢ a2 ⎢⎣ a3

b1 b2 b3

k1 ⎤ k2 ⎥⎥ ≠ 0 k3 ⎥⎦

Thus, the solution of the system of equations is given by Δ1 Δ Δ y= 2 Δ Δ3 z= Δ x=

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5

Chapter 1  • Engineering Mathematics

17. Augmented Matrix Consider the following system of equations:

problem. To solve the problem, we need to determine the value of X’s and l’s to satisfy the above-mentioned vector. Note that the zero vector (i.e. X = 0) is not of our interest. A value of l for which the above equation has a solution X ≠ 0 is called an eigenvalue or ­characteristic value of the matrix A. The corresponding solutions X ≠ 0 of the equation are called the eigenvectors or characteristic vectors of A corresponding to that eigenvalue, l. The set of all the eigenvalues of A is called the spectrum of A. The largest of the absolute values of the eigenvalues of A is called the spectral radius of A. The sum of the elements of the principal diagonal of a matrix A is called the trace of A.

a11 x1 + a12 x2 +  + a1n xn = b1 a21 x1 + a22 x2 +  + a2 n xn = b2 





am1 x1 + am 2 x2 +  + amn xn = bm

This system can be represented as AX = B.

⎡ a11 ⎢a where A = ⎢ 21 ⎢  ⎢ ⎣ am1

a12 a22  am 2

 a1n ⎤ ⎡ x1 ⎤ ⎢ ⎥ ⎥ x  a2 n ⎥ , X = ⎢ 2 ⎥ and ⎢   ⎥  ⎥ ⎢ ⎥ ⎥  amn ⎦ ⎢⎣ xn ⎥⎦

⎡b1 ⎤ ⎢ ⎥ b B = ⎢ 2 ⎥. ⎢ ⎥ ⎢ ⎥ ⎢⎣bm ⎥⎦



⎡ a11 ⎢ a The matrix ⎡⎣ A B ⎤⎦ = ⎢ 21 ⎢  ⎢ ⎢⎣ am1

a12 a22  am 2

 a1n b1 ⎤ ⎥  a2 n b2 ⎥ is called    ⎥ ⎥  amn bm ⎥⎦

augmented matrix. 18. Cayley–Hamilton Theorem According to the Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equations. Hence, if A − l I = ( −1) n ( l n + a1l n −1 + a2 l n − 2 +  + an )

is the characteristic polynomial of a matrix A of order n, then the matrix equation X n + a1 X n −1 + a2 X n − 2 +  + an I = 0



is satisfied by X = A.

19. Eigenvalues and Eigenvectors If A = [aij]n × n is a square matrix of order n, then the vector ⎡ x1 ⎤ ⎢x ⎥ ⎢ 2⎥ equation AX = lX, where X = ⎢⎢. ⎥⎥ is an unknown vector ⎢. ⎥ ⎢. ⎥ ⎢ ⎥ ⎢⎣ xn ⎥⎦ and l is an unknown scalar value, is called an eigenvalue

Ch wise GATE_ECE_CH01_Imp formula.indd 5

20. Properties of Eigenvalues and Eigenvectors Some of the main characteristics of eigenvalues and eigenvectors are discussed in the following points: (a) If l1, l2, l3, …, ln are the eigenvalues of A, then kl1, kl2, kl3, …, kln are eigenvalues of kA, where k is a constant scalar quantity.

(b) If l1, l2, l3, …, ln are the eigenvalues of A, then 1 1 1 1 , , , ..., are the eigenvalues of A−1. l1 l 2 l3 ln



(c) If l1, l2, l3, …, ln are the eigenvalues of A, then

l1k , l 2k , l3k , ..., l nk are the eigenvalues of Ak.

(d) If l1, l2, l3, …, ln are the eigenvalues of A, then A

,

A

,

A

l1 l 2 l3

, ...,

A

ln

are the eigenvalues of adj A.



(e) The eigenvalues of a matrix A are equal to the eigenvalues of AT.



(f) The maximum number of distinct eigenvalues is n, where n is the size of the matrix A.



(g) The trace of a matrix is equal to the sum of the eigenvalues of a matrix.



(h) The product of the eigenvalues of a matrix is equal to the determinant of that matrix.



(i) If A and B are similar matrices, i.e. A = IB, then A and B have the same eigenvalues.



(j) If A and B are two matrices of same order, then the matrices AB and BA have the same eigenvalues. (k) The eigenvalues of a triangular matrix are equal to the diagonal elements of the matrix.



Calculus 21. Rolle’s Mean Value Theorem Consider a real-valued function defined in the closed interval [a, b], such that

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(a)  It is continuous on the closed interval [a, b].

(b)  It is differentiable on the open interval (a, b). (c)  f(a) = f(b). Then, according to Rolle’s theorem, there exists a real number c ∈( a, b) such that f ′(c) = 0. 22. Lagrange’s Mean Value Theorem Consider a function f(x) defined in the closed interval [a, b], such that (a)  It is continuous on the closed interval [a, b]. (b)  It is differentiable on the closed interval (a, b). Then, according to Lagrange’s mean value theorem, there exists a real number c ∈( a, b), such that f ′( c ) =

f ( b) − f ( a) b−a

23. Cauchy’s Mean Value Theorem Consider two functions f(x) and g(x), such that (a)  f (x) and g(x) both are continuous in [a, b]. (b)  f ′(x) and g ′(x) both exist in (a, b). Then, there exists a point c ∈( a, b) such that f ′( c ) f ( b) − f ( a) = g ′( c ) g ( b) − g ( a) 24. Taylor’s Theorem If f(x) is a continuous function such that f ′( x ), f ′′( x ), …, f n −1 ( x ) are all continuous in [a, a + h] and f n ( x ) exists in (a, a + h) where h = b − a, then according to Taylor’s theorem, h2 f ′′( a) +  2! hn −1 hn n f (a) + f n −1 ( a) + n! (n − 1)!

Suppose f(x) is a real-valued function defined at the interval (a, b). Then f(x) is said to have minimum value, if there exists a point y in (a, b) such that



h2 f ′′(0) +  2! hn −1 hn n + f n −1 (0) + f (0 ) n! (n − 1)!



f (x) = f (y) for all x ∈ (a, b)

Ch wise GATE_ECE_CH01_Imp formula.indd 6

(a) If f ′( y ) = 0 and f ′( x ) changes sign from positive to negative as ‘x’ increases through ‘y’, then x = y is a point of local maximum value of f(x). (b) If f ′( y ) = 0 and f ′( x ) changes sign from negative to positive as ‘x’ increases through ‘y’, then x = y is a point of local minimum value of f(x).

28. Maximum and minimum values in a closed interval [a, b] can be calculated using the following steps:

(a) Calculate f ′( x ).



(b) Put f ′( x ) = 0 and find value(s) of x. Let c1, c2, ..., cn be values of x. (c) Take the maximum and minimum values out of the values f(a), f(c1), f(c2), …, f(cn), f(b). The ­maximum and minimum values obtained are the absolute maximum and absolute minimum values of the function, respectively.



29. Partial Derivatives Partial differentiation is used to find the partial derivative of a function of more than one independent variable. The partial derivatives of f(x, y) with respect to x and y are defined by ∂f f ( x + a, y ) − f ( x, y ) = lim ∂x a → 0 a ∂f f ( x , y + b) − f ( x , y ) = lim b → 0 ∂y b

f ( h) = f (0) + hf ′(0) +

26. Maxima and Minima Suppose f(x) is a real-valued function defined at an ­internal (a, b). Then f(x) is said to have maximum value, if there exists a point y in (a, b) such that

Local maxima and local minima of any function can be calculated as: Consider that f(x) be defined in (a, b) and y ∈ (a, b). Now,

27. Some important properties of maxima and minima are given as follows: (a) If f(x) is continuous in its domain, then at least one maxima and minima lie between two equal values of x. (b) Maxima and minima occur alternately, i.e. no two maxima or minima can occur together.

f ( a + h) = f ( a) + hf ′( a) +

25. Maclaurin’s Theorem If the Taylor’s series obtained in 24 is centered at 0, then the series we obtain is called the Maclaurin’s series. According to Maclaurin’s theorem,

f (x) ≥ f (y) for all x ∈ (a, b)



and the above limits exist. ∂f ∂x is simply the ordinary derivative of f with respect to x keeping y constant, while ∂f ∂y is the ordinary derivative of f with respect to y keeping x constant. Similarly, second-order partial derivatives can be calculated by

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Chapter 1  • Engineering Mathematics



∂ ⎛ ∂ f ⎞ ∂ ⎛ ∂ f ⎞ ∂ ⎛ ∂f ⎞ ∂ ⎛ ∂f ⎞ and is, respec, ⎜ ⎟, ⎜ ⎟, ∂x ⎝ ∂x ⎠ ∂x ⎜⎝ ∂y ⎟⎠ ∂y ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂2 f ∂2 f ∂2 f ∂2 f tively, denoted by , , , . ∂x 2 ∂x ∂y ∂y ∂x ∂y 2

30. Some of the important results from the above relation are given as follows: (a) If u = f(x, y) and y = f(x), then ∂u ∂u ∂u ∂y = + ⋅ ∂x ∂x ∂y ∂x

(b) If u = f(x, y) and x = f1(t1, t2) and y = f2(t1, t2), then ∂u ∂u ∂ x ∂ u ∂ y = ⋅ + ⋅ ∂t1 ∂x ∂t1 ∂y ∂t1

31. Integration Using Table Some of the common integration problems can be solved by directly referring the tables and computing the results. Table 1 shows the result of some of the common integrals we use. Table 1 |  Table of common integrals

1 ∫ ax + b dx 1

∫ (ax + b)

2

1

∫ (ax + b)

n

dx

f ′ (x)

∫ f ( x ) dx 2

∫ sin

3

1 ln ax + b + C where C is a a constant −

dx

1 ∫ a2 + x 2 dx

∫ sin

Result

−

1 +C a ( ax + b ) 1

a ( n − 1) ( ax + b )

n −1

2

xdx

x 1 + sin x cos x + C 2 2

∫ cos

3

xdx

1 sin x − sin 3 x + C 3

∫ cos

n

1 n −1 cos n −1 x sin x + n n

xdx

∫ cos

2

ln x + x 2 ± a 2 + C

x ± a2 2

∫ x sin nxdx

1 (sin nx − nx cos nx ) + C n2

∫ x cos nxdx

1 (cos nx + nx sin nx ) + C n2

∫e

ax

e ax ( a sin bx − b cos bx )

sin bxdx

a2 + b2 e ax ( a cos bx + b sin bx )

ax ∫ e cos bxdx

a2 + b2

ln f ( x ) + C x 1 − sin x cos x + C 2 2

xdx

1 − cos x + cos3 x + C 3 1 n −1 − sin n −1 x cos x + n n n−2 ∫ sin xdx + C

∫x

2

+C +C

1 ( − n2 x 2 cos nx + 2 cos nx n3 + 2nx sin nx ) + C

+C

1 ⎛ x⎞ tan −1 ⎜ ⎟ + C ⎝ a⎠ a

xdx + C

1 x−a ln +C 2a x + a

dx − a2 dx

∫

n−2

tan n −1 x − ∫ tan n − 2 xdx + C n −1

n ∫ tan xdx

2 ∫ x sin nxdx

xdx

n ∫ sin xdx

Result

∫ cos

∫x

∂u ∂u ∂x ∂u ∂y = ⋅ + ⋅ and ∂t 2 ∂x ∂t 2 ∂y ∂t 2

Integration

Integration

7

1 2 2 ( n x sin nx − 2 sin nx n3 + 2nx cos nx ) + C

cos nxdx

32. Integration by Partial Fraction The formula which come handy while working with ­partial fractions are given as follows: 1

∫ x − a dx = ln ( x − a) + C ∫a

2

∫a

2

1 1 ⎛ x⎞ dx = tan −1 ⎜ ⎟ + C 2 ⎝ a⎠ a +x

(

)

x 1 dx = ln a 2 + x 2 + C 2 2 +x

(Continued)

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33. Integration Using Trigonometric Substitution Trigonometric substitution is used to simplify certain integrals containing radical expressions. Depending on the function we need to integrate, we substitute one of the following expressions to simplify the integration:

(a) For

a 2 − x 2 , use x = a sin q .



(b) For

a 2 + x 2 , use x = a tan q .

(c) For

x − a , use x = a sec q .



2

keeping x constant and between the limits y1, y2 and the resulting expression is integrated with respect to x within the limits x1, x2. x2 y2

∫∫

f ( x, y ) dxdy =



y2 x2

f ( x, y ) dxdy =

Q

34. Some of the important properties of definite integrals are given as follows: (a) The value of definite integrals remains the same with change of variables of integration provided the limits of integration remain the same. b

b

a

a

∫ f ( x) ⋅ dx = ∫ f ( y) ⋅ dy b



(b) 

∫ a

b



(c) 

∫ a

2a



(d) 

∫ 0



(e)  (f) 

y2 x2

∫∫



f ( x, y ) dydx

b

a

c

37. The length of the arc of the polar curve r = f (q ) between

a

a

f ( x ) ⋅ dx = ∫ f ( x ) ⋅ dx + ∫ f ( 2a − x ) ⋅ dx 0

0

the points where q = α , β is

0

β

∫ α



⎡ 2 ⎛ dr ⎞ 2 ⎤ ⎢ r + ⎜⎝ ⎟⎠ ⎥ dq . dq ⎦ ⎣

The length of the arc of the curve x = f (t ) , y = g (t ) between the points where t = a and t = b is

a

f ( x ) ⋅ dx = 2∫ f ( x ) ⋅ dx , if the function is even.

⎡⎛ dx ⎞ 2 ⎛ dy ⎞ 2 ⎤ ⎢⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ ⎥ dt . dt ⎦ ⎣ dt

b

∫

0

a

f ( x ) ⋅ dx = 0, if the function is odd.

−a

na

a

0

0

∫ f ( x) ⋅ dx = n∫ f ( x) ⋅ dx

if f(x) = f(x + a)

35. Some of the important properties of double integrals are: (a) When x1, x2 are functions of y and y1, y2 are constants, then f(x, y) is integrated with respect to x keeping y constant within the limits x1, x2 and the resulting expression is integrated with respect to y between the limits y1, y2.

38. Line integrals Let us say that points (ak, bk) are chosen so that they lie on the curve between points (xk−1, yk−1) and (xk, yk). Now, consider the sum, n

∑ {P (α k =1

y2 x2

Q

Ch wise GATE_ECE_CH01_Imp formula.indd 8

, βk ) Δxk + Q(α k , βk ) Δyk }

∫ [ P ( x, y)dx + Q( x, y)dy] c

y1 x1

(b) When y1, y2 are functions of x and x1, x2 are constants, f(x, y) is first integrated with respect to y,

k

The limit of the sum as n → ∞ is taken in such a way that all the quantities Δ xk and Δ yk approach zero, and if such limit exists, it is called a line integral along the curve C and is denoted by

∫∫ f ( x, y) dxdy = ∫ ∫ f ( x, y) dxdy

∫∫

x1 y1

a

(g) 

f ( x, y ) dydx

x1 y1

c

f ( x ) ⋅ dx = ∫ f ( x ) ⋅ dx + ∫ f ( x ) ⋅ dx  if a < c < b

0

∫

∫∫

x2 y 2

f ( x, y ) dxdy =

y1 x1



b

∫ f ( x) ⋅ dx = ∫ f (a − x) ⋅ dx ∫

y1 x1

36. Change of Order of Integration As already discussed, if limits are constant

a

f ( x ) ⋅ dx = − ∫ f ( x ) ⋅ dx

a

−a

∫∫

x2 y2

f ( x, y ) dxdy =

Hence, in a double integral, the order of integration does not change the final result provided the limits are changed accordingly. However, if the limits are variable, the change of order of integration changes the limits of integration.

a

a



f ( x, y ) dydx

(c) When x1, x2, y1 and y2 are constants, then

∫∫

2

∫∫

x1 y1

Q



The limit exists if P and Q are continuous at all points of C. To understand better, refer to Fig. 1.

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Chapter 1  • Engineering Mathematics

9

Differential Equations

y (xk + 1, yk + 1) B(a2, b2) (αk + 1, βk + 1) (xk, yk)

C

A(α1, β1)

40. Variable Separable If all functions of x and dx can be arranged on one side and y and dy on the other side, then the variables are separable. The solution of this equation is found by integrating the functions of x and y.

∫ f ( x)dx = ∫ g ( y)dy + C

(x1, y1)

41. Homogeneous Equation Homogeneous equations are of the form

a1

a2

x

dy f ( x, y ) = dx g ( x, y )

Figure 1 |  Line integral.



In the same way, a line integral along a curve C in the three-dimensional space is given by



where f(x, y) and g(x, y) are homogeneous functions of the same degree in x and y. Homogeneous functions are those in which all the terms are of nth degree. To solve homogeneous equation,



(a) Put y = vx , then



(b) Separate v and x and then integrate.

∫ [ A dx + A dy + A dz ] 1

2

3

c



where A1, A2 and A3 are functions of x, y and z, respectively.

39. Surface Integrals Suppose g(x, y, z) is single valued and continuous for all values of C. Now, consider the sum n

∑ g (α k =1





k

where (ak, bk, gk) is any arbitrary point of ΔCk. If the limit of this sum, n→∞ is in such a way that each ΔCk → 0 exists, the resulting limit is called the surface integral of g(x, y, z) over C. The surface integral is denoted by

∫∫ g ( x, y, z ) ⋅ ds C



Figure 2 graphically represents surface integrals.

dy = f ( y/ x ) dx y/ x = v dv dx = f (v) − v x

⇒ ⇒

, βk , γ k ) Δc p

dy dv = v+x . dx dx

dv

∫ f (v) − v = log x + C

⇒

42. Linear Equation of First Order If a differential equation has its dependent variables and its derivatives occur in the first degree and are not multiplied together, then the equation is said to be linear. The standard equation of a linear equation of first order is given as dy + Py = Q dx

∆Ck



where P and Q are functions of x.

C



Integrating factor = (I.F.) = e ∫ y ⋅ e∫

P ⋅ dx

= ∫ Q ⋅ e∫

P ⋅ dx

P ⋅ dx

dx + C

⇒ y(I.F.) = ∫ Q(I.F.) dx + C

C′

∆Ak = ∆xk ∆yk

Figure 2 |  Surface integrals.

Ch wise GATE_ECE_CH01_Imp formula.indd 9

43. Exact Differential Equation A differential equation of the form M(x, y) dx + N(x, y) dy = 0, if df = M dx + N dy where f(x, y) is a function, is called an exact differential equation.

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Solution of differential equation is f(x, y) = C. The necessary condition for the differential equation to be exact is



Differentiating above equation w.r.t. x, we get dp dp + f ′( p) dx dx dp ⇒ [ x + f ′( p)] = 0 dx p = p+ x

∂M ∂N = ∂y ∂x

The exact differential equation can be solved using the following steps: (a) First, integrate M with respect to x keeping y constant. (b) Next, integrate the terms of N with respect to y which do not involve x. (c) Then, the sum of the two expressions is equated to an arbitrary constant to give the solution.

∫ M ⋅ dx + ∫ N ⋅ dy = C



44. Integrating Factor A differential equation which is not exact can be converted into an exact differential equation by multiplication of a suitable function. This function is called integrating factor. Some of the important formulae are: (a) If the given equation of the form M dx + N dy = 0 is homogeneous and Mx + Ny ≠ 0, then I.F. =



1 Mx + Ny



Therefore, [ x + f ′( p)] = 0  or 

dp =0 dx

dp = 0, then p = c. dx



Now, if



Thus, eliminating p from above Eqs., we get



y = cx + f (c) as the general solution of the given equation. Hence, the solution of Clairaut’s equation is obtained on replacing p by c.

46. Linear Differential Equation The general form of a differential equation of nth order is denoted by

dn y d n −1 y d n − 2 y + p p2 n − 2 +  + pn y = X 1 dx n dx n −1 dx where p1, p2, …, pn and X are functions of x.

d If operator is denoted by D, then substituting it in dx (b) If the given equation is of the form f ( x, y ) y∂x + g ( x, y ) x∂y = 0 above equation, we get f ( x, y ) y∂x + g ( x, y ) x∂y = 0 and Mx − Ny ≠ 0, then I.F. =

1 Mx − Ny

(c) If the given equation is of the form Mdx + Ndy = 0 1 ⎛ ∂M ∂N ⎞ − = f ( x ) where f ( x ) is a funcN ⎜⎝ ∂y ∂x ⎟⎠ tion of x, then and

I.F. = e ∫



f ( x )⋅∂x



(d) If the given equation is of the form Mdx + Ndy = 0 1 ⎛ ∂N ∂M ⎞ − = f ( y ) where f ( y ) in a funcand M ⎜⎝ ∂x ∂y ⎟⎠ tion of y, then I .F . = e ∫



f ( y )⋅∂y



45. Clairaut’s Equation

dy An equation of the form y = px + f ( p), where p = dx is called a Clairaut’s equation. We know that



Ch wise GATE_ECE_CH01_Imp formula.indd 10

y = px + f ( p)

D n y + p1 D n −1 y + p2 D n − 2 y +  + pn y = X

⇒ f ( D) y = X



where f(D) = Dn + p1Dn - 1 +  + pn = 0. As already mentioned before, the equation can be generalized as f ( D) y = X



The general solution of the above equation can be given as   y = (Complementary Function) + (Particular Integral)



The general form of the linear differential equation with constant coefficients is given as



dn y d n −1 y d n−2 + k + k +  + kn y = X 2 1 dx n dx n −1 ∂x n − 2 where k1, k2, …, kn are constants and X is a function of x only. The equation dn y d n −1 y d n−2 + k1 n −1 + k2 n − 2 +  + kn y = 0 n dx dx dx

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where k1, k2, …, kn are constants can also be denoted as ( D + k1 D n





+ k2 D

n−2

+  + kn ) y = 0 n −1

n−2

and the equation D + k1 D + k2 D +  + kn = 0 is called the auxiliary equation (A.E.). Now, if m1, m2, …, mn are the roots, then the complementary functions can be calculated using the following cases: Case I: If the roots of A.E. are distinct and real, i.e. m1 ≠ m2 ≠ m3 ≠  ≠ mn, then n

y = c1e

m1 x

+ c2 e

m2 x

+ c3 e

m3 x

+

Case II: If some roots of A.E. are real and some are equal, i.e. m1 = m2 ≠ m3, then



n −1

y = (c1 + c2 x )e m1 x + c3 c m3 x +  Case III: If the roots of A.E. are complex, i.e. α + iβ , α − iβ , m3 ,..., mn , then

=

(c) When X = cos ax, then

where C1 = c1 + c2 and C2 = i(c1 − c2 ).



If the complex roots are equal, then y = eα x [(c1 x + c2 ) cos β x + (c3 x + c4 ) sin β x ] +  + cn e mn x

=

P.I. =



P.I. =

Expansion of [ f ( D )]−1 is to be carried up to the term Dm because (m + 1)th and higher derivatives of xm are zero.



(e) When X = e ax v( x ), then P.I. =

P.I. =



1 ax e , if f ( a) ≠ 0  f ( a)

P.I. =



x e ax , if f ′( a) ≠ 0  f ′( a)



P.I. =

48. Homogeneous Linear Equation Cauchy’s homogeneous linear equation is given by xn

x2 e ax , if f ′′( a) ≠ 0 f ′′( a)

(b) When X = sin ax, then P.I. =

Ch wise GATE_ECE_CH01_Imp formula.indd 11

Thus, the following steps should be used to calculate the complete solution:   (i) Find the auxiliary equation (A.E.) from the differential equation.  (ii) Calculate the complementary function using the cases given before. (iii)  Calculate particular integral (P.I.) using the cases explained before.   (iv) To find the complete solution, use the following expression: y = C.F. + P.I.

If f ′( a) = 0, then

1 sin ax f (D2 )

1 x v( x ) f ( D)

⎡ f ′( D ) ⎤ 1 = ⎢x − ⋅ v( x ) f ( D ) ⎥⎦ f ( D ) ⎣

1 e ax f ( D)

If f ( a) = 0, then

1 v( x )  f ( D + a)

(f) When X = x v(x), then

The following cases arise for particular integrals: (a) When X = eax, then

=

1 e ax v( x ) f ( D)

P.I. = e ax

1 X D n + k1 D n −1 + k2 D n − 2 +  + kn

P.I. =

1 x m = [ f ( D )]−1 x m  f ( D)



( D n + k1 D n −1 + k2 D n − 2 +  + kn ) y = X The particular integral of the equation is given by

1 cos ax , if f ( − a 2 ) ≠ 0 f ( −a2 )

(d) When X = xm, then

47. Particular Integrals Consider the equation given in the previous section

1 cos ax f (D2 )

P.I. =

y = eα x (C1 cos β x + C2 sin β x ) + c3 e m3 x +  + cn e mn x

1 2 sin ax , if f ( − a ) ≠ 0  f ( −a2 )



n−2 n −1 dn y y n−2 ∂ n −1 d + k x + k x +  + kn y = X 2 1 dx n dx n −1 ∂x n − 2

where k1, k2, …, kn are constants and X is a function of x.

49. Bernoulli’s Equation

dy + Py = Qy n, where P and Q are funcdx tions of x and can be reduced to Leibnitz’s linear equation and is called the Bernoulli’s equation. The equation

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To solve the Bernoulli’s equation, divide both sides by y n , so that dy y−n + Py1− n = Q dx

Here, a, b and c are constants. The method of variation of parameters to find solution to non-homogeneous Euler-Cauchy equation is given. As a first step, divide equation by ax2 so as to make coefficient of y ′′ as unity.

Putting y1− n = z , we get



y ′′ +

or

y ′′ +

(1− n) y − n

dy dz = dx dx

Therefore, 1 dz + Pz = Q 1− n dx dz ⇒ + P (1 − n) z = Q (1 − n) dx

which is Leibnitz’s linear equation in z and can be solved using the methods mentioned under Integrating Factor.

50. Homogeneous Euler–Cauchy Equation An ordinary differential equation is called homogeneous Euler-Cauchy equation, if it is in the form

ax 2 y ′′ + bxy ′ + cy = 0  where a, b and c are constants. This equation has two linearly independent solutions, which depend on the following quadratic equation:

am2 + (b - a) m + c = 0



This quadratic equation is known as characteristic equation. Let us consider for x > 0 as this equation is singular for x = 0. (a) If the roots of the characteristic equation are real and distinct, say m1 and m2, two linearly independent solutions of the differential equations are xm1 and xm2.



(b) If the roots of characteristic equation are real and equal, that is, m1 = m2 = m; the linearly independent solutions of the differential equation are xm and xm ln x. The general solution for the first case is given by y = c1xm1 + c2xm2 The general solution for the second case is given by y = (c1 + c2 ln x) xm

(c) If the roots of characteristic equation are complex conjugate, that is, m1 = a + jb and m2 = a - jb, Linearly independent solutions of differential are xa cos (b ln x) and xa sin (b ln x).

 The general solution is given by    y = xa [C1 cos (b ln x) + C2 sin (b ln x)] 51. Non-Homogeneous Euler–Cauchy Equation The non-homogeneous Euler-Cauchy equation is of the form

ax y ′′ + bxy ′ + cy = r( x ) 

Ch wise GATE_ECE_CH01_Imp formula.indd 12

2

b c r( x )  y′ + 2 y = ax ax ax 2 b c y ′ + 2 y = r( x)  ax ax r( x ) r( x) = ax 2



Particular solution is given by y ( x )r( x ) y ( x )r( x ) y p ( x ) = − y1 ( x )∫ 2 dx + y2 ( x )∫ 1 dx W ( y1 , y2 ) W ( y1 , y2 )



General solution to non-homogeneous Euler-Cauchy equation is therefore given by the following equation:



y(x) = C1   y1(x) + C2   y2(x) + yp(x)

52. Variation of Parameters Method The method of variation of parameters is a general method that can be used to solve non-homogeneous linear ordinary differential equations of the form y ′′ + py ′ + qy = x 

where p, q and x are functions of x. The particular integral in this case is given by y X yX P.I. = − y1 ∫ 2 dx + y2 ∫ 1 dx W W  Here, y1 and y2 are the solutions of y ′′ + py ′ + qy = 0 and W =

y1 y1′

y2 is called the Wronskian of y1 and y2. y2′

53. Separation of Variables Method (a) Partial differential equations representing engineering problems include the one-dimensional heat flow equation given by ∂u ∂2u = c2 2 dt dx (b) wave equation representing vibrations of a stretched string given by



∂2 y ∂2 y = c2 2 2 ∂t ∂x (c)  two-dimensional heat flow equation that becomes two-dimensional Laplace equation in steady state and given by ∂2u ∂2u + =0 ∂x 2 ∂y 2

transmission line equations, two-dimensional wave equation and so on.

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Vector Analysis 54. Vectors Vector is any quantity that has magnitude as well as direction. If we have two points A and B, then vector between A and B is denoted by AB. Position vector is a vector of any points, A, with respect to the origin, O. If A is given by the coordinates x, y and z.  OA = x 2 + y 2 + z 2 55. Zero vector is a vector whose initial and  final points are same. Zero vectors are denoted by 0 . They are also called null vectors. 56. Unit vector is a vector whose magnitude  is unity or one. It is in the direction of given vector A and is denoted by A . 57. Equal vectors are those which have the same magnitude and direction regardless of their initial points. 58. Addition of Vectors According to triangle law of vector addition, as shown in Fig. 3, C c b

A

B a

Figure 3 |  Triangle law of vector addition.

   c = a+b



According to parallelogram law of vector addition, as shown in Fig. 4, B

C

59. Some important properties of vector addition are:   (a) If we have two vectors a and b     a+b = b +a    (b) If we have three vectors a , b and c       a+b +c = a+ b +c

(

)

(

)

60. Multiplication of Vectors (a) Multiplication of a vector with scalar: Consider a  vector a and a scalar quantity v. Then   ka = k a (b)  Multiplication of a vector with another vector using dot product: Dot product or scalar product of two vectors is given by     a ⋅ b = a b cos q    where a = magnitude of vector a , b = magnitude of    vector b and q = angle between a and b, 0 ≤ q ≤ p. 61.

Some important laws of dot product are: (a)  A ⋅B = B ⋅A (b)  A ⋅(B + C) = A ⋅B + A ⋅C (c)  k(A ⋅B) = (kA)⋅B = A ⋅(kB) (d)  i  ⋅ i = j  ⋅ j = k  ⋅ k = 1, i  ⋅ j = j  ⋅ k = k  ⋅ i = 0

62. Multiplication of Vectors Using Cross Product The cross or vector product of two vectors is given by     a × b = a b sin q nˆ,   where a = magnitude of vector a   b = magnitude of vector b   q = angle between a and b , 0 ≤ q ≤ p   n = Unit vector perpendicular to both a and b.   The result of a × b is always a vector. iˆ

c

b

63. Some important laws of cross product are: (a)  A × B = −B × A O

a

A

Figure 4 |  Parallelogram law of vector addition.

   c = a+b



If we have two vectors represented by adjacent sides of parallelogram, then the sum of the two vectors in magnitude and direction is given by the diagonal of the parallelogram. This is known as parallelogram law of vector addition.

Ch wise GATE_ECE_CH01_Imp formula.indd 13



(b)  A × (B + C) = A × B + A × C



(c)  m(A × B) = (mA) × B = A × (mB)



(d)  i × i = j × j = k × k = 0, i × j = k, j × k = i, k × i = j



(e) If A = A1i + A2    j + A3k and B = B1i + B2   j + B3k, then i A × B = A1 B1

j A2 B2

k A3 B3

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64. Derivatives of Vector Functions The derivative of vector A(x) is defined as dA A( x + Δx ) − A( x ) = lim Δ x → 0 dx Δx if the above limits exists. If A(x) = A1(x)i + A2(x)j + A3(x)k, then dA dA dA dA1 = i+ 2 j+ 3 k dx dx dx dx If A(x, y, z) = A1(x, y, z)i + A2(x, y, z)j + A3(x, y, z)k, then dA =

dA dA dA dx + dy + dz dx dy dz

d dB dA ( A ⋅ B) = A + B dy dy dy d dB dA ( A × B) = A × + ×B dz dz dz

 A unit vector perpendicular to two given vectors a and b is given by   a×b  c=   |a×b |

65. Gradient of a Scalar Field If we have a scalar function a(x, y, z), then the gradient of this scalar function is a vector function which is defined by ∂f ∂f ∂f  i+ j+ k grad a = ∇a = ∂x ∂y ∂z

⎛ ∂Ay ∂Ax ⎞ ⎛ ∂A ∂Ay ⎞ ⎛ ∂Ax ∂Az ⎞ j+⎜ k i+⎜ − − =⎜ z − ⎟ ⎟ ∂y ⎟⎠ ∂x ⎠ ∂z ⎠ ⎝ ∂z ⎝ ∂x ⎝ ∂y  where Ax, Ay and Az are the components of vector A. 68. Some important points of divergence and curl are:     ∂2 A ∂2 A ∂2 A (a)  ∇ ⋅ ∇ A = ∇2A = + + ∂x 2 ∂y 2 ∂z 2  (b)  ∇ × ∇ A = 0  (c)  ∇ ⋅ ∇ × A = 0    (d)  ∇ × (∇ × A ) = ∇(∇ ⋅  A ) - ∇2 A    (e)  ∇ (∇ ⋅  A ) = ∇ × (∇ × A ) + ∇2 A     (f)  ∇ ( A + B ) = ∇ ⋅  A + ∇ ⋅  B     (g)  ∇ × ( A + B ) = ∇ × A + ∇ × B       (h)  ∇ ⋅ ( A × B ) = B  ⋅ (∇ × A ) − A  ⋅ (∇ × B )   (i) ∇ × ( A × B ) = (B  ⋅ ∇) A − B (∇⋅A) − (A∇) B + A (∇B) 69. Directional Derivative The directional derivative of a scalar function, f ( x ) = f ( x1 , x2 , …, xn ) along a vector v = ( v1 , …, vn ) is a function defined by the limit ∇ v f ( x ) = lim h→ 0



66. Divergence of a Vector  If we have a differentiable vector (x, y, z), then diverA  gence of vector A is given by

 where Ax, Ay and Az are the components of vector A.

i ∂ = ∂x Ax

Ch wise GATE_ECE_CH01_Imp formula.indd 14

j ∂ ∂y Ay

k ∂ ∂z Az

h

If the function f is differentiable at x, then the directional derivative exists alone any vector v,

where ∇ on the right-hand side of the equation is the v gradient and v is the unit vector given by v = . v

67. Curl of a Vector  The curl of a continuously differentiable vector A is given by   ⎛ ∂ ∂ ∂⎞ curl A = ∇ × A = ⎜ i + j + k ⎟ × ( Ax i + Ay j + Az k ) ⎝ ∂x ∂y ∂z ⎠

)

∇ v f ( x ) = ∇f ( x ) ⋅ v

  ∂A ∂Ay ∂Az + div A = ∇ ⋅ A = x + ∂x ∂y ∂z



(

f x + hv − f ( x )

The directional derivative is also often written as follows: d  ∂ ∂ ∂ = v ⋅ ∇ = vx + vy + vz dv ∂x ∂y ∂z

70. Scalar Triple Product The scalar product of three vectors a, b and c is defined as

[a, b, c] = a ⋅ (b × c ) = b ⋅ (c × a) = c ⋅ (a × b)

The volume of a parallelepiped whose sides are given by the vectors a, b and c, as shown in Fig. 5, can be calculated by the absolute value of the scalar triple product.

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Chapter 1  • Engineering Mathematics

Vparallelepiped = a ⋅ (b × c ) a×b

cos z =

e iz + e − iz 2

tan z =

sin z cos z

with cosec z, sec z and cot z as respective reciprocals of above equation.

Now, cos z + i sin z = c

z = x + iy. b

15

e iz + e − iz e iz − e − iz +i = e iz , where 2 2i

Also, e iy = cos y + i sin y, where y is real.

a Figure 5 |  Parallelepiped with sides given by the vectors a, b and c.

71. Vector Triple Product If we have three vectors a, b and c, then the vector triple product is given by a × (b × c ) = b ( a ⋅ c ) − c ( a ⋅ b ) 72. Stokes’ Theorem



This is called Euler’s theorem. Now, according to De Moivre’s theorem for complex number,

(cos q + i sin q )n = (eiq )

73. Green’s Theorem ∂f 2 ⎞

∫∫ ⎜⎝ ∂x − ∂y ⎟⎠ dxdy = ∫ ( f dx + f dy) 2

s

sin hx =

ex − e− x 2

1

cos h x =

ex + e− x 2

tan h x =

ex − e− x ex + e− x

cot h x =

ex + e− x ex − e− x

sec h x =

2 ex + e− x

c

74. Gauss Divergence Theorem  ∫∫∫ ∇ ⋅ AdV = ∫∫ A ⋅ ndS v

s

Complex Variables 75. A number of the form x + iy, where x and y are real numbers and i = ( −1) , is called a complex number.

x is called the real part of x + iy and is written as R(x + iy), whereas y is called the imaginary part and is written as I(x + iy).

76. Exponential Function of Complex Variables iq

z = re . 77. Circular Function of Complex Variables The circular functions of the complex variable z can be defined by the equations: sin z =

Ch wise GATE_ECE_CH01_Imp formula.indd 15

e iz − e − iz 2i

= cos nq + i sin nq ,

whether q is real or complex.

s

⎛ ∂f1

n

78. Hyperbolic Functions of Complex Variables Hyperbolic functions can be given as

∫ A ⋅ dr = ∫∫ (∇ × A) ⋅ nds c

Hence, e i q = cos q + i sin q , where q is real or complex.

cosec h x =

2 ex − e− x

Hyperbolic and circular functions are related as follows: sin ix = i sin h x cos ix = cos h x tan ix = i tan h x

79. Logarithmic Function of Complex Variables If z = x + iy and w = u + iy are related such that e w = z , then w is said to be a logarithm of x to the base e and is written as

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w = log e z Also, e w + 2inp = e w ⋅ e 2inp = z [∵ e 2inp = 1 ] ⇒ log z = w + 2 in p 80. Cauchy–Riemann Equations A necessary condition for w = u ( x, y ) + iv ( x, y ) to be analytic in a region R is that u and v satisfy the following equations:



81. Cauchy’s Theorem If f ( z ) is an analytic function and f ′ ( z ) is continuous at each point within and on a closed curve c, then according to Cauchy’s theorem,

∫ f ( z ) ⋅ dz = 0 

c 82. Some of the important results that can be concluded are: (a) The line integral of a function f ( z ) , which is analytic in the region R, is independent of the path joining any two points of R.



83. Cauchy’s Integral Formula If f(x) is analytic within and on a simple closed curve c and a is any point interior to c, then



f (α ) =

(b) Extension of Cauchy’s theorem: If f(z) is analytic in the region R between the two simple closed curves c and c1, then

∫ f ( z ) ⋅ dz = ∫ f ( z ) ⋅ dz c

c1

(c) If c1 , c2 , c3 , … be any number of closed curves within c, then

Ch wise GATE_ECE_CH01_Imp formula.indd 16

f ( z ) dz 1 2p i ∫c z − α

f (z)

in above Eq., which is analytic at all z −α points within c except at z = α . Now, with a as center and r as radius, draw a small circle c1 lying entirely within c. Generally, we can write that

The derivative of f ( z ) is then given by

The real and imaginary parts of an analytic function are called conjugate functions.



c3

Consider

1 ∂u ∂v ∂v ∂u + = −i = v y − iu y i ∂y ∂y ∂y ∂y

c2

+ ∫ f ( z ) ⋅ dz + 



Above equations are called Cauchy–Riemann equations. If the partial derivatives in above equation are continuous in R, the equations are sufficient conditions that f ( z ) be analytic in R.

f ′ (z) =



c1

∂u ∂v = ∂y ∂x

⎛ ∂u ∂v ⎞ or  f ′ ( z ) = lim ⎜ +i Δx → 0 ⎝ i∂y i∂y ⎟⎠



c

∂u ∂v = ∂z ∂y

⎛ ∂u ∂v ⎞ ∂u ∂v f ′ ( z ) = lim ⎜ + i ⎟ = +i = ux + iv x Δy → 0 ⎝ ∂x ∂x ⎠ ∂x ∂x





∫ f ( z ) ⋅ dz = ∫ f ( z ) ⋅ dz + ∫ f ( z ) ⋅ dz

f (z) n! dz ∫ 2p i c ( z − α )n +1

f n (α ) =

84. Taylor’s Series of Complex Variables If f(z) is analytic inside a circle C with centre at a, then for z inside C, f ( z ) = f (α ) + f ′ (α ) ( z − α ) + + +

f n (α ) n!

f ′′ (α ) 2!

( z − α )2

( z − α )n

85. Laurent’s Series of Complex Variables If f(z) is analytic in the ring-shaped region R bounded by the two concentric circles C and C1 of radii r and r1 (such that r > r1 ) and with center at a, then for all z in R ∞

f ( z ) = ∑ α ( z − α ) = a0 + a1 ( z − α ) + a2 ( z − α ) +  2

n

−∞

+ a−1( z − α ) + a−2 ( z − α ) −1

where an =

−2

f (t ) 1 ∫ (t − α )n+1 dt . 2p i 

86. Zeros and Poles of an Analytic Function (a) A zero of an analytic function f(z) is the value of z for which f(z) = 0. (b) A singular point of a function f(z) is a value of z at which f(z) fails to be analytic.

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Chapter 1  • Engineering Mathematics

87. Residues



The coefficient of ( z − α ) in the expansion of f ( z ) around an isolated singularity is called the residue of f ( z ) at the point. −1

It can be found from the formula 1 d n −1 ⎡ ( z − α )n f ( z )⎤⎦  z → α ( n − 1)! dz z −1 ⎣

a−1 = lim



where n is the order of the pole.



The residue of f(z) at z = a can also be found by



Res f (α ) =

1 ∫c f ( z ) 2p i 

x2 = ( x1 + x0 )/ 2



If f ( x2 ) = 0 , then x2 itself is the root. If not, the root lies either between x0 and x2 or between x1 and x2 . That is, if f ( x0 ) and f ( x1 ) are of opposite signs, then root lies between x0 and x2; otherwise if f (x1) and f(x2) are of opposite signs, the root lies between x1 and x2 . The new interval for searching the root is therefore ( x0 , x1 ) in the first case and ( x1 , x2 ) in the second case, which is much less compared to the first interval ( x0 , x1 ). This process is illustrated in Fig. 6. This new interval will be the beginning interval for the next iteration, and the processes of bisection and finding another such x2 are repeated. y or f(x)

88. Residue Theorem If f(z) is analytic in a region R except for a pole of order n at z = a and let C be a simple closed curve in R containing z = a, then



∫ f ( z ) dz = 2p i ×

(sum of the residue at the singular

c

points within C )

89. Calculation of Residues (a) If f(z) has a simple pole at z = a, then Res f (α ) = lim ⎡⎣( z − α ) f ( z )⎤⎦ z →α



(b) If f ( z ) = ϕ ( z ) /ψ ( z ) , where

ψ ( z ) = ( z − α ) f ( z ) , f (α ) ≠ 0 , then Res f (α ) =

ϕ (α ) ψ ′ (α )

(c) If f(z) has a pole of order n at z = ¥∞, then   Res f (α ) =

⎫ 1 ⎧ d n −1 ⎡ n ⎨ n −1 ⎣( z − α ) f ( x )⎤⎦ ⎬ 1 ! n − dz ( )⎩ ⎭ z =α

Numerical Methods 90. Bisection Method We begin the iterative cycle by choosing two trial points x0 and x1 , which enclose the actual root. Then  f ( x0 ) and f ( x1 ) are of opposite signs. The interval  ( x0 , x1 ) is bisected and its midpoint x2 is obtained as

Ch wise GATE_ECE_CH01_Imp formula.indd 17

f(x)

Actual root

x0 x′

x =(x0+x1) x1 2

x

Third iteration Second iteration Starting interval Figure 6 |  Illustration of bisection method.

91. Regula–Falsi Method (Method of False Position Method) The iterative procedure is started by choosing two values x0 and x1 such that f ( x0 ) and f ( x1 ) are of opposite signs. Then the two points [ x0 , f ( x0 )] and [ x1 , f ( x1 )] are joined by a straight line. The intersection of this line with the x-axis gives x2. If f ( x2 ) and f ( x0 ) are of opposite signs, then replace x1 by x2; otherwise, replace x0 by x2. This yields a new set of values for x0 and x1. The present range is much smaller than the range or interval between the first chosen set of x0 and x1. The convergence is thus established, and the iterations are carried over with the new set of x0 and x1. Another x2 is found by the intersection of the straight line joining the new f ( x0 ) and f ( x1 ) points with x-axis. Each new or successive interval is smaller than the previous interval, and it is guaranteed to converge to the root. The procedure is illustrated in Fig. 7.

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GATE ECE Chapter-wise Solved papers

y

y

A (x0, f(x0))

y =f(x)

f(x0)

x3 0

x0

x2 x1

b x

P(x)

x2

Figure 7 |  Illustration of Regula–Falsi method.

From Fig. 7, it can be seen (from the equation of the straight line) that y − f ( x0 ) =



93. Secant Method The secant method can be thought of as a finite difference approximation of Newton–Raphson method.

f ( x1 ) − f ( x0 ) ( x − x0 ) x1 − x0 x0 − x1 f ( x0 ) f ( x1 ) − f ( x0 )



where    x2 = x0 −



which is an approximation to the root. x3

92. Newton–Raphson Method Newton’s method, also known as Newton–Raphson method, is another iteration method for solving e­ quations f ( x ) = 0, where f is assumed to have a continuous derivative f  ′. The underlying idea is that we approximate the graph of f by suitable tangents. The Newton–Raphson method has second-order convergence.

Using an approximate value x0 obtained from the graph of f, we let x1 be the point of intersection of the x-axis and the tangent to the curve of f at x0 (Fig. 8). Then tan β = f ′( x0 ) =

Hence, x1 = x0 −



x1 − f ( x1 ) f ′( x1 )

in the third step x3 from x2 again by the same formula, and so on.

Ch wise GATE_ECE_CH01_Imp formula.indd 18

x2

x1

Figure 9 |  Secant method.



f ( x0 ) x0 − x1

f ( x0 ) f ′( x0 )

x0

f(x)

Consider Fig. 9. The slope of line from points (x0, f(x0)) to (x1, f(x1)) can be given by the following equation. y=



In the second step, we compute x2 =



x

x0

Figure 8 |  Newton–Raphson method.

B (x1, f(x1))



x1

f ( x1 ) − f ( x0 ) x1 − x0

We now calculate the solution of x by putting y = 0. x = x1 − f ( x1 )



( x − x1 ) + f ( x1 )

x1 − x0 f ( x1 ) − f ( x0 )

We use the new value of x as x2 and repeat the same process using x1 and x2 instead of x0 and x1. This process is continued in the same manner unless we obtain xn = xn −1 .

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Chapter 1  • Engineering Mathematics



x2 = x1 − f ( x1 )

x1 − x0 f ( x1 ) − f ( x0 )

x3 = x2 − f ( x2 )

x2 − x1 f ( x2 ) − f ( x1 )

The generalized solution for the secant method as xn = xn −1 − f ( xn −1 )



xn −1 − xn − 2 f ( xn −1 ) − f ( xn − 2 )

94. Jacobian If u and v are functions of two independent variables x ∂ u /∂ x ∂ u /∂ y and y, then the determinant is called the ∂ v /∂ x ∂ v /∂ y Jacobian of u, v with respect to x, y and is written as ∂ (u, v )

⎛ u, v ⎞ . or J ⎜ ⎝ x, y ⎟⎠ ∂ ( x, y )

Probability and Statistics 95. Types of Events (a) Each outcome of a random experiment is called an elementary event. (b)  An event associated with a random experiment that always occurs whenever the experiment is performed is called a certain event. (c) An event associated with a random experiment that never occurs whenever the experiment is performed is called an impossible event. (d) If the occurrence of any one of two or more events, associated with a random experiment, presents the occurrence of all others, then the events are called mutually exclusive events. (e) If the union of two or more events associated with a random experiment includes all possible outcomes, then the events are called exhaustive events. (f) If the occurrence or non-occurrence of one event does not affect the probability of the occurrence or non-occurrence of the other, then the events are independent. (g) Two events are equally likely events if the ­probability of their occurrence is same. (h) An event which has a probability of occurrence equal to 1− P, where P is the probability of ­occurrence of an event A, is called the complementary event of A. 96. Axioms of Probability (a) The numerical value of probability lies between 0 and 1.

Ch wise GATE_ECE_CH01_Imp formula.indd 19

19

Hence, for any event A of S, 0 ≤ P ( A) ≤ 1 . (b) The sum of probabilities of all sample events is unity. Hence, P(S) = 1. (c) Probability of an event made of two or more sample events is the sum of their probabilities. 97. Conditional Probability Let A and B be two events of a random experiment. The probability of occurrence of A if B has already occurred and P ( B ) ≠ 0 is known as conditional probability. This is denoted by P ( A/B). Also, conditional probability can be defined as the probability of occurrence of B if A has already occurred and P ( A) ≠ 0. This is denoted by P ( B /A). 98. Geometric Probability Due to the nature of the problem or the solution or both, random events that take place in continuous sample space may invoke geometric imagery. Hence. geometric probabilities can be considered as non-negative quantities with maximum value of 1 being assigned to subregions of a given domain subject to certain rules. If P is an expression of this assignment defined on a domain S, then 0 < P ( A) ≤ 1, A ⊂ S and P ( S ) = 1

The subsets of S for which P is defined are the random events that form a particular sample spaces. P is defined by the ratio of the areas so that if σ ( A) is defined as the area of set A, then σ ( A) P ( A) = σ ( s)

99. Rules of Probability Some of the important rules of probability are given as follows: (a)  Inclusion–Exclusion principle of probability: P ( A ∪ B) = P ( A) + P ( B) − P ( A ∩ B)

I f A and B are mutually exclusive events, P ( A ∩ B) = 0 and then formula reduces to P ( A ∪ B) = P ( A) + P ( B)

(b)  Complementary probability: P ( A) = 1 − P ( Ac )     where P ( Ac ) is the complementary probability of A. (c)  P ( A ∩ B) = P ( A) ∗ P ( B /A) = P ( B) ∗ P ( A/B)   where P ( A/B) represents the conditional prob­ ability of A given B and P ( B /A) represents the conditional probability of B given A.

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GATE ECE Chapter-wise Solved papers

  If B1 , B2 ,..., Bn are pairwise disjoint events of positive probability, then ⎛ A⎞ ⎛ A⎞ P ( A) = P ⎜ ⎟ P ( B1 ) + P ⎜ ⎟ P ( B2 ) ⎝ B1 ⎠ ⎝ B2 ⎠ ⎛ A⎞ +  + P ⎜ ⎟ P ( Bn ) ⎝ Bn ⎠

where

P ( A ∩ B ) = P ( B) ∗ P ( A/B) ⇒ P ( A/B ) =

    Or

P ( A ∩ B) P ( B)



th



⎡ ( n + 1) ⎤ Now, if n is odd, then median = ⎢ ⎥ value ⎣ 2 ⎦

If n is even, then median th ⎡⎛ n ⎞ th ⎤ ⎛n ⎞ = ⎢⎜ ⎟ value + ⎜ + 1⎟ value ⎥ ⎝ ⎠ ⎝ ⎠ 2 ⎢⎣ 2 ⎥⎦ 2

P ( A ∩ B) P ( B /A) = P ( A)

(e) Bayes’ theorem: Suppose we have an event A corresponding to a number of exhaustive events B1 , B2 ,..., Bn .



If P ( Bi ) and P ( A/Bi ) are given, then P ( Bi /A) =

P ( Bi ) P ( A/Bi ) ∑ P ( Bi ) P ( A/Bi )



(f) Rule of total probability: Consider an event E which occurs via two different values A and B. Also, let A and B be mutually exhaustive and c­ollectively exhaustive events. Now, the probability of E is given as P( E ) = P( A ∩ E ) + P( B ∩ E ) = P ( A) ∗ P ( E /A) + P ( B) ∗ P ( E /B)

i =1

101. Median Median for Raw Data Suppose we have n numbers of ungrouped/raw values and let values be x1 , x2 ,..., xn . To calculate median, arrange all the values in ascending or descending order.

(d)  Conditional probability rule:

n

N = ∑ fi

Median for Grouped Data To calculate median of grouped values, identify the class containing the middle observation.

⎤ ⎡ ( N + 1) − ( F + 1) ⎥ ⎢ 2 Median = L + ⎢ ⎥×h fm ⎥ ⎢ ⎥⎦ ⎢⎣ where L = lower limit of median class

N = total number of data items = ∑ f



F = cumulative frequency of class immediately preceding the median class

   This is called the rule of total probability.

 = frequency of median class f m h = width of class

100. Arithmetic Mean Arithmetic Mean for Raw data Suppose we have values x1 , x2 ,..., xn and n are the total number of values, then arithmetic mean is given by

102. Mode Mode of raw values of data is the value with the highest frequency or simply the value which occurs the most number of times.

x=

1 n ∑ xi n i =1

where x = arithmetic mean.

Arithmetic Mean for Grouped Data (Frequency Distribution) Suppose f i is the frequency of xi , then the arithmetic mean from frequency distribution can be calculated as x=

Ch wise GATE_ECE_CH01_Imp formula.indd 20

1 n ∑ f i xi N i =1



Mode of Raw Data Mode of raw data is calculated by simply checking which value is repeated the most number of times. Mode of Grouped Data Mode of grouped values of data is calculated by first identifying the modal class, i.e. the class which has the target frequency. The mode can then be calculated using the following formula:



Mode = L +

f m − f1 ×h 2 f m − f1 − f 2

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Chapter 1  • Engineering Mathematics

where L = lower limit of modal class

104. Geometric Mean Geometric Mean of Raw Data

f m = frequency of modal class f1 = frequency of class preceding modal class



f 2 = frequency of class following modal class h = width of model class 103. Relation Between Mean, Median and Mode When an approximate value of mode is required, the given empirical formula for mode may be used. There are three types of frequency distribution: (a) Symmetric distribution: It has lower half equal to upper half of distribution. For symmetric distribution, Mean = Median = Mode

n

x1 ⋅ x2 ⋅ x3 ⋅⋅⋅⋅⋅ xn

Geometric Mean of Grouped Data Geometric mean for a frequency distribution is given by 1 N

n

∑f i =1

i

log( xi )

n

i =1

105. Harmonic Mean

Harmonic Mean of Raw Data Harmonic mean of n numbers x1 , x2 , x3 ,..., xn is calculated as n H.M. = n 1 ∑ x i =1 i



Harmonic Mean of Grouped Data Harmonic mean for a frequency distribution is calculated as

(b) Positively skewed distribution: It has a longer tail on the right than on the left. Mode ≤ Median ≤ Mean



=

where N = ∑ f i .



(c)  Negatively skewed distribution: It has a long tail on the left than on the right. Mean ≤ Median ≤ Mode

1/ n

log G.M. =

Empirical mode = 3 Median − 2 Mean

Geometric mean of n numbers x1 , x2 ,..., xn is given by ⎛ n⎞ ⎜⎝ ip=1⎟⎠



21

Figure 10 shows the three types of frequency distribution.

H.M. =

N n

∑ ( f /x ) i =1

i

i

n

where N = ∑ f i . i =1

(a) (a)

(b) (b)

106. Mean Deviation Mean Deviation of Raw Data Suppose we have a given set of n values x1 , x2 , x3 ,..., xn , then the mean deviation is given by M.D. =

(c) (c) Figure 10 |  (a) Symmetrical frequency distribution, (b) Positively skewed frequency distribution and (c) Negatively skewed frequency distribution.

Ch wise GATE_ECE_CH01_Imp formula.indd 21

1 n ∑ xi − x n i =1



where x = mean.



The following steps should be followed to calculate mean deviation of raw data:



(a)  Compute the central value or average ‘A’ about which mean deviation is to be calculated.



(b)  Take mod of the deviations of the observations about the central value ‘A’, i.e. xi − x .



(c) Obtain the total of these deviations, i.e. ∑ x2 − x .

n

i =1

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GATE ECE Chapter-wise Solved papers

(d) Divide the total obtained in step 3 by the number of observations. Mean Deviation of Discrete Frequency Distribution For a frequency deviation, the mean deviation is given by M.D. =

1 N

n

∑f i =1

i

xi − x



(d) Divide the sum by n to obtain the value of variance, i.e.

σ2 =

(e) Take out the square root of variance to obtain standard deviation,

n

where  N = ∑ fi .





The following steps should be followed to calculate mean deviation of discrete frequency deviation: (a) Calculate the central value or average ‘A’ of the given frequency distribution about which mean deviation is to be calculated. (b) Take mod of the deviations of the observations from the central value, i.e. xi − x .



(c) Multiply these deviations by respective frequencies and obtain the total ∑ f i xi − x .



(d) Divide the total obtained by the number of observan

tions, i.e. N = ∑ f i to obtain the mean deviation. i =1



Mean Deviation of Grouped Frequency Distribution For calculating the mean deviation of grouped frequency distribution, the procedure is same as for a discrete frequency distribution. However, the only difference is that we have to obtain the mid-points of the various classes and take the deviations of these mid-points from the given central value.



If we have n values x1 , x2 , ..., xn of X, then 1 n σ = ∑ ( xi − X ) 2 n i =1

Standard Deviation of Discrete Frequency Distribution If we have a discrete frequency distribution of X, then

σ2 =

⎤ 1 ⎡ n ∑ ( xi − X )2 ⎥ N ⎢⎣ i =1 ⎦

σ=

⎤ 1 ⎡ n ( xi − X ) 2 ⎥ ∑ ⎢ N ⎣ i =1 ⎦

n

where N = ∑ f . i =1



The following steps should be followed to calculate variance if discrete distribution of data is given: (a)  Obtain the given frequency distribution.



(b) Calculate mean ( X ) for given frequency distribution.



(c) Take deviations of observations from the mean, i.e. ( xi − X ).



(d) Square the deviations obtained in the previous step and multiply the squared deviations by respective frequencies, i.e. f i ( xi − X ).



(e)  Calculate the total, i.e. ∑ f i ( xi − X ) 2 .



(f) Divide the sum ∑ f i ( xi − X ) 2 by N, where N = ∑ f i ,

107. Standard Deviation Standard Deviation of Raw Data

1 n ∑ ( xi − X )2 n i =1

σ2 =

i =1



1 n ∑ ( xi − X )2 n i =1

n

i =1

2

⇒ σ=

1 n ∑ ( xi − X )2 n i =1

The following steps should be followed to calculate standard deviation for discrete data:



(a)  Calculate mean ( X ) for given observations.



(b) Take deviations of observations from the mean, i.e. ( xi − X ) .



(c) Square the deviations obtained in the previous step and find n

∑ (x i =1

Ch wise GATE_ECE_CH01_Imp formula.indd 22

i

− X )2

n

i =1

to obtain the variance, σ 2.

(g) Take out the square root of the variance to obtain standard deviation, σ =

1⎡n ⎤ ∑ f i ( xi − X ) 2 ⎥ . ⎢ i = 1 n⎣ ⎦



Standard Deviation of Grouped Frequency Distribution



For calculating standard deviation of a grouped frequency distribution, the procedure is same as for a discrete frequency distribution. However, the only difference is that we have to obtain the mid-point of the various classes and take the deviations of these midpoints from the given central point.

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Chapter 1  • Engineering Mathematics

108. Coefficient of Variation Coefficient of variation (C.V.) is a measure of variability which is independent of units and hence can be used to compare two data sets with different units. C.V. =

112. General Discrete Distribution Suppose a discrete variable X is the outcome of some random experiment and the probability of X taking the value xi is Pi , then

σ × 100 X

P ( X = xi ) = Pi for i = 1, 2,...

where σ represents standard deviation and X represents mean.



where, P ( xi ) ≥ 0 for all values of i and ∑ P ( xi ) = 1.



The set of values xi with their probabilities Pi of a discrete variable X is called a discrete probability distribution. For example, the discrete probability distribution of X, the number which is selected by picking a card from a well-shuffled deck is given by the following table:

109. Random Variable

Random variable can be discrete and continuous.



Discrete random variable is a variable that can take a value from a continuous range of values.



Continuous random variable is a variable that can take a value from a discrete range of values.



If a random variable X takes x1 , x2 ,..., xn with respective probabilities P1 , P2 ,..., Pn , then X : x1 P ( X ) : P1



x2 P2

X = xi : Pi :

... xn ... Pn

x3 P3

X = xi : Ace 2 3 4 5 6 7 1 1 1 1 1 1 1 Pi : 13 13 13 13 13 13 13



is known as the probability distribution of X.

110. Properties of Discrete Distribution

(a)  ∑ P ( x ) = 1



(b)  E ( x ) = ∑ xP ( x )

10 Jack Queen King

1 1 1 13 13 13

1 13

1 13

1 13

The distribution function F(x) of discrete variable X is defined by n

i =1

2

2



where E(x) denotes the expected value or average value of a random variable x and V(x) denotes the variance of a random variable x.

where x is any integer. The mean value ( x ) of the probability distribution of a discrete variable X is known as its expectation and is denoted by E(x). If f(x) is the probability density function of X, then n

E ( x ) = ∑ xi f ( xi )

111. Properties of Continuous Distribution

9

F ( x ) = P ( X ≤ x ) = ∑ Pi

(c)  V ( x ) = E ( x 2 ) − [ E ( x )] = ∑ x 2 P ( x ) − [ ∑ xP ( x )]

8

i =1

(a)  Cumulative distribution function is given by



Variable of a distribution is given by

x

F ( x) =

∫

f ( x ) dx

n

σ 2 = ∑ ( xi − x ) 2 f ( xi )

−∞



(b)  E ( x ) =

i =1

∞

∫

xf ( x ) dx

−∞

(c)  V ( x ) = E ( x 2 ) − [ E ( x )] = 2

∞

∫

x 2 f ( x )dx

−∞

⎤ ⎡∞ − ⎢ ∫ xf ( x )dx ⎥ ⎣ −∞ ⎦

2

P ( a < x < b) = P ( a ≤ x ≤ b) = P ( a < x ≤ b) (d)  b

= P ( a ≤ x ≤ b) = ∫ f ( x ) dx a

Ch wise GATE_ECE_CH01_Imp formula.indd 23

113. Binomial Distribution Binomial distribution is concerned with trails of a respective nature whose outcome can be classified as either a success or a failure. Suppose we have to perform n independent trails, each of which results in a success with probability P and in a failure with probability X which is equal to 1 − P. If X represents the number of successes that occur in the n trails, then X is said to be binomial random variable with parameters (n, p).

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GATE ECE Chapter-wise Solved papers





The binomial distribution occurs when the experiment performed satisfies the following four assumptions of Bernoulli’s trails. (a)  They are finite in number.



(b)  There are exactly two outcomes: success or failure.



(c) The probability of success or failure remains same in each trail.



(d)  They are independent of each other.



The probability of obtaining x successes from n trails is given by the binomial distribution formula,

116. Geometric Distribution Consider repeated trails of a Bernoulli experiment E with probability of success, p, and probability of failure, q = 1 − p. Let x denote the number of times E must be repeated until finally obtaining a success. The distribution is



A random variable X, taking on one of the values 0, 1, 2, …, n, is said to be a Poisson random variable with parameters m if for some m > 0, P( x) =



e−m mx x!

Mean = E ( x ) = m



Variance = V ( x ) = m



m

P( x) =

C nx C y − x m+ n

Cx

Ch wise GATE_ECE_CH01_Imp formula.indd 24

C xnC y − x = m + nC y

m

Also,   p( x ) = q x p, x = 0, 1, 2, ..., q = 1 − p ∞

x

= p

x=0

1 =1 1− q



The mean of geometric distribution = q /p.



The variance of geometric distribution = q /p 2 .



Types of Continuous Distribution

117. General Continuous Distribution When a random variable X takes all possible values in an interval, then the distribution is called continuous distribution of X. A continuous distribution of X can be defined by a probability density function f(x) which is given by p( −∞ ≤ x ≤ ∞) =

∞

∫

f ( x )dx = 1

−∞



The expectation for general continuous distribution is given by E (x) =



∞

∫

−∞

x f ( x ) dx

The variance for general continuous distribution is given by V (x) = σ 2 =

∞

∫ ( x − X ) f ( x ) dx −∞ 2

118. Uniform Distribution If density of a random variable X over the interval −∞ < a < b < ∞ is given by f (x) =

, x = 0, 1, ..., y, y ≤ m, n

This distribution is known as hypergeometric distribution.

i =1

x=0

Therefore, the expected value and the variance of a Poisson random variable are both equal to its parameters m.

115. Hypergeometric Distribution If the probability changes from trail to trail, one of the assumptions of binomial distribution gets violated; hence, binomial distribution cannot be used. In such cases, hypergeometric distribution is used; say a bag contains m white and n black balls. If y balls are drawn one at a time (with replacement), then the probability that x of them will be white is

i =1

∞

For Poisson distribution,



n

∑ P( x) = P ∑ q

where P is the probability of success in any trail and (1 − p) is the probability of failure.

114. Poisson Distribution Poisson distribution is a distribution related to the probabilities of events which are extremely rare but which have a large number of independent opportunities for occurrence.

n

∑ p( x) = 1, since ∑

P ( X ) = nC x p x (1 − p) n − x

For hypergeometric distribution,

1 , a< x 0 is given by

∑ XY

where X is deviation from the mean ( x − x ) , Y is deviation from the mean ( y − y ) , σ x is standard deviation



In uniform distribution, x takes the values with the same probability. The variance of uniform distribution is given by

25

σy σx

.

σx . σy

Line of regression of x on y x = a′+b′y Line of regression of y on x y = a +bx

x′

°

x

y′ Figure 11 |  Line of regression.

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GATE ECE Chapter-wise Solved papers

QUESTIONS Linear Algebra 1.

⎡2 − 1 0 ⎢0 3 0 The eigen values of the matrix ⎢ ⎢0 0 − 2 ⎢ ⎣0 0 − 1 (a) (b) (c) (d)

4.

0⎤ 0 ⎥⎥ are 0⎥ ⎥ 4⎦

⎡ −1⎤ ⎡ 2⎤ (c) ⎢ ⎥ (d) ⎢ 2⎥ ⎣ ⎦ ⎣ −1⎦

2, −2, 1, −1 2, 3, −2, 4 2, 3, 1, 4 None of the above

(GATE 2005, 2 Marks)

(GATE 2000: 2 Marks)

5.

⎡ 1 1 1 1⎤ ⎢ 1 1 −1 −1⎥ ⎥. 2. Given an orthogonal matrix A = ⎢ ⎢ 1 −1 0 0 ⎥ ⎥ ⎢ 1 −1⎦ ⎣0 0 −1 6. What is the value of ⎡⎣ AAT ⎤⎦ ? ⎡1 ⎢4 ⎢ ⎢0 ⎢ (a) ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎣

⎡1 ⎢0 (c) ⎢ ⎢0 ⎢ ⎣0

0

0

1 4

0

0

1 2

0

0

0 1 0 0

0 0 1 0

⎡ − 4 2⎤ Given the matrix ⎢ ⎥ , the eigenvector is ⎣ 4 3⎦ ⎡ 3⎤ ⎡4⎤ (a) ⎢ ⎥ (b) ⎢ ⎥ ⎣ 2⎦ ⎣ 3⎦

⎤ ⎡1 0⎥ ⎢2 0 ⎥ ⎢ ⎢0 1 0⎥ ⎥ ⎢ 2 ⎥ (b) ⎢ ⎢0 0 0⎥ ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢0 0 2⎦ ⎣ ⎡1 ⎢4 0 ⎢ 0⎤ ⎢0 1 ⎥ 0⎥ ⎢ 4 (d) ⎢ 0⎥ ⎢0 0 ⎥ ⎢ 1⎦ ⎢ ⎢0 0 ⎣

0 0 1 2 0 0 0 1 4 0

⎤ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 1⎥ ⎥ 2⎦

7.

⎤ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 1⎥ ⎥ 4⎦

(a) 0 (c) 2

(b) 1 (d) 3 (GATE 2006, 1 Mark)

⎡ 4 2⎤ For the matrix ⎢ ⎥ , the eigenvalue corresponding to ⎣ 2 4⎦ ⎡101⎤ the eigenvector ⎢ ⎥ is ⎣101⎦ (a) 2 (b) 4 (c) 6 (d) 8 (GATE 2006, 2 Marks) The eigenvalues and the corresponding eigenvectors or a 2 × 2 matrix are given by eigenvalues l1 = 8 and l2 = 4 and eigenvectors ⎡1⎤ ⎡1 ⎤ v1 = ⎢ ⎥ , v2 = ⎢ ⎥ 1 ⎣⎦ ⎣ −1⎦ The matrix is ⎡4 6⎤ ⎡6 2⎤ (a) ⎢ (b) ⎢6 4⎥ ⎥ ⎣ ⎦ ⎣2 6⎦ ⎡ 4 8⎤ ⎡ 2 4⎤ (c) ⎢ (d) ⎢ 8 4⎥ ⎥ ⎣ ⎦ ⎣ 4 2⎦

(GATE 2005, 2 Marks) ⎤ ⎡1 ⎡ 2 −0.1⎤ −1 ⎢ 2 a ⎥ , then (a + b) is 3. Let A = ⎢ and A = ⎥ ⎢ 3⎥⎦ ⎣0 ⎣ 0 b⎦

⎡1 1 1⎤ ⎢ ⎥ The rank of the matrix ⎢1 −1 0 ⎥ is ⎢⎣1 1 1⎥⎦

(GATE 2006, 2 Marks) 8.

(a)

7 3 (b) 20 20

(c)

11 19 (d) 20 60

It is given that X 1 , X 2 ,..., X M are M non-zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X 1 , X 2 ,...., X M , − X 1 , − X 2 ,..., − X M is (a) 2M (b) M + 1 (c) M (d) dependent on the choice of X 1 , X 2 ,..., X M

(GATE 2005, 2 Marks)

(GATE 2007, 2 Marks)

Ch wise GATE_ECE_CH01_Part 1.indd 26

12/4/2018 10:38:19 AM

Chapter 1  • Engineering Mathematics

9.

p12 ⎤ ⎡p All the four entries of the 2 × 2 matrix P = ⎢ 11 ⎥ ⎣ p21 p22 ⎦ are non-zero, and one of its eigenvalues is zero. Which one of the following statements is true? (a)

14. The minimum eigenvalue of the following matrix is ⎡ 3 5 2⎤ ⎢ 5 12 7⎥ ⎢ ⎥ ⎢⎣ 2 7 5⎥⎦

p11 p22 − p12 p21 = 1

(a) 0 (c) 2

(b) p11 p22 − p12 p21 = −1 (c)

(b) 1 (d) 3 (GATE 2013, 1 Mark)

p11 p22 − p12 p21 = 0

(d) p11 p22 + p12 p21 = 0 (GATE 2008, 1 Mark) 10. The system of linear equations 4x + 2 y = 7 2x + y = 6 has (a) a unique solution (b) no solution (c) infinite number of solutions (d) exactly two distinct solutions (GATE 2008, 1 Mark) ⎡ −1 3 5⎤ ⎢ ⎥ 11. The eigenvalues of the matrix ⎢ −3 −1 6 ⎥ are ⎢⎣ 0 0 3⎥⎦ (a) 3, 3 + 5j, 6 − j (b) −6 + 5j, 3 + j, 3 − j (c) 3 + j, 3 − j, 5 + j (d) 3, −1 + 3j, −1 − 3j (GATE 2009, 2 Marks) 12. The eigenvalues of a skew-symmetric matrix are (a) always zero (b) always pure imaginary (c) either zero or pure imaginary (d) always real (GATE 2010, 1 Mark)

15. Let A be an m × n matrix and B be an n × m matrix. It is given that determinant ( I m + AB) = determinant ( I n + BA), where I K is the k × k identity matrix. Using the above property, the determinant of the matrix given below is ⎡2 ⎢1 ⎢ ⎢1 ⎢ ⎣1

18. The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is . (GATE 2014, 1 Mark) 19. Consider the matrix ⎡0 ⎢0 ⎢ ⎢0 J6 = ⎢ ⎢0 ⎢0 ⎢ ⎢⎣1

(a) λ = 6, μ = 20

(GATE 2011, 2 Marks)

Ch wise GATE_ECE_CH01_Part 1.indd 27

(b) 5 (d) 16 (GATE 2013, 2 Marks)

17. A real (4 × 4) matrix A satisfies the equation A2 = I, where I is the (4 × 4) identity matrix. The positive eigenvalue of A is . (GATE 2014, 1 Mark)

has no solution for values of l and m given by

(d) λ ≠ 6, μ ≠ 20

1⎤ 1⎥⎥ 1⎥ ⎥ 2⎦

1 1 2 1

16. For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold? (a) (M T)T = M (b) (cM)T = c(M)T T T T (c) (M + N) = M  + N  (d) MN = NM (GATE 2014, 1 Mark)

x+ y+z =6 x + 4 y + 6 z = 20 x + 4y + λz = μ

(c) λ ≠ 6, μ = 20

1 2 1 1

(a) 2 (c) 8

13. The system of equations

(b) λ = 6, μ ≠ 20

27

0 0 0 0 1 0

0 0 0 1 0 0

0 0 1 0 0 0

0 1 0 0 0 0

1⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 0 ⎥⎦

which is obtained by reversing the order of the columns of the identity matrix I6. Let P = I6 + a J6, where a is a non-negative real number. The value of a for which det(P) = 0 is . (GATE 2014, 2 Marks)

12/4/2018 10:38:21 AM

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GATE ECE Chapter-wise Solved papers

20. The system of linear equations



⎛ 2 1 3⎞ ⎛ a⎞ ⎛ 5 ⎞ ⎜ 3 0 1 ⎟ ⎜ b ⎟ = ⎜ − 4⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝1 2 5⎠ ⎝ c ⎠ ⎝14 ⎠ has (a) (b) (c) (d)

⎡ p ⎤ ⎡1 1 ⎢ q ⎥ = ⎢1 w1 3 ⎢ ⎥ ⎢ ⎢⎣ r ⎥⎦ ⎢⎣1 w32

a unique solution infinitely many solutions no solution exactly two solutions

If another sequence [p, q, r] is derived as 1 ⎤ ⎡1 0 w32 ⎥⎥ ⎢⎢0 w32 w34 ⎥⎦ ⎢⎣0 0

22. Which one of the following statements is NOT true for a square matrix A? (a) If A is upper triangular, the eigenvalues of A are the diagonal elements of it (b) If A is real symmetric, the eigenvalues of A are always real and positive (c) If A is real, the eigenvalues of A and AT are always the same (d) If all the principal minors of A are positive, all the eigenvalues of A are also positive (GATE 2014, 2 Marks) ⎡1 ⎤ ⎢ 23. The value of P such that the vector ⎢ 2⎥⎥ is an eigenvector ⎢⎣ 3⎥⎦ ⎡4 1 2⎤ . of the matrix ⎢⎢ P 2 1 ⎥⎥ is ⎢⎣14 −4 10 ⎦⎥ (GATE 2015, 1 Mark) 24. Consider the system of linear equations: x - 2y + 3z = -1 x - 3y + 4z = 1 and -2x + 4y - 6z = k, The value of k for which the system has infinitely many solutions is . (GATE 2015, 1 Mark) 25. Two sequences [a, b, c] and [A, B, C] are related as ⎡ A⎤ ⎡1 1 ⎢ B ⎥ = ⎢1 w −1 3 ⎢ ⎥ ⎢ ⎢⎣C ⎥⎦ ⎢⎣1 w3−2 where w3 = e

Ch wise GATE_ECE_CH01_Part 1.indd 28

j

2π 3

1 ⎤ ⎡a⎤ w3−2 ⎥⎥ ⎢⎢ b ⎥⎥ w3−4 ⎥⎦ ⎣⎢ c ⎦⎥

⎤ ⎥ ⎥ ⎥⎦

then the relationship between the sequences [p, q, r] and [a, b, c] is (a) [p, q, r] = [b, a, c] (b) [p, q, r] = [b, c, a] (c) [p, q, r] = [c, a, b] (d) [p, q, r] = [c, b, a] (GATE 2015, 1 Mark)

(GATE 2014, 2 Marks) 21. The maximum value of the determinant among all 2 × 2 . real symmetric matrices with trace 14 is (GATE 2014, 2 Marks)

0 ⎤ ⎡ A/ 3 0 ⎥⎥ ⎢⎢ B / 3 w34 ⎥⎦ ⎢⎣ C / 3

az + b . If f(z1) = f(z2) for all z1 ≠ z2, a = 2, cz + d b = 4 and c = 5, then d should be equal to .

26. Let f ( z ) =

(GATE 2015, 1 Mark) 27. The value of x for which all the eigenvalues of the matrix given below are real is ⎡10 5 + j ⎢x 20 ⎢ ⎢⎣ 4 2 (a) 5 + j (c) 1 - 5j

4 ⎤ 2 ⎥⎥ −10 ⎥⎦ (b) 5 - j (d) 1 + 5j (GATE 2015, 1 Mark)

tan x ⎤ ⎡ 1 , the determinant of ATA-1 is 28. For A = ⎢ 1 ⎥⎦ ⎣ − tan x (a) sec2 x (c) 1

(b) cos 4x (d) 0 (GATE 2015, 1 Mark)

29. Let M 4 = I, (where I denotes the identity matrix) and M ≠ I, M 2 ≠ I and M 3 ≠ I. Then, for any natural number k, M − 1 equals: (a) M 4k + 1 (b) M 4k + 2 (c) M 4k + 3 (d) M 4k (GATE 2016, 1 Mark) 2 4 ⎞ ⎛ 3 ⎜ 30. The value of x for which the matrix A = 9 7 13 ⎟ ⎜ ⎟ ⎝ −6 −4 −9 + x ⎠ has zero as an eigenvalue is __________. (GATE 2016, 1 Mark) ⎡σ x ⎤ 31. Consider a 2 × 2 square matrix: A = ⎢ ⎥ , where x is ⎣ω σ ⎦ unknown. If the eigenvalues of the matrix A are (σ + jω) and (σ − jω), then x is equal to

12/4/2018 10:38:23 AM

Chapter 1  • Engineering Mathematics

(a) +jω (b) −jω (c) +ω (d) −ω (GATE 2016, 1 Mark) ⎛a ⎜2 32. The matrix A = ⎜ ⎜0 ⎜⎝ 0

0 3 7⎞ 5 1 3⎟ ⎟ has det(A) = 100 and 0 2 4⎟ 0 0 b⎟⎠

trace(A) = 14. The value of |a − b| is __________. (GATE 2016, 2 Marks) 33. Consider the 5 × 5 matrix ⎡1 ⎢5 ⎢ A = ⎢4 ⎢ ⎢3 ⎢⎣ 2

2 1 5 4 3

3 2 1 5 4

4 3 2 1 5

5⎤ 4 ⎥⎥ 3⎥ ⎥ 2⎥ 1 ⎥⎦

It is given that A has only one real eigenvalue. Then the real eigenvalue of A is (a) -2.5 (c) 15

(b) 0 (d) 25 (GATE 2017, 1 Mark)

⎡5 10 10 ⎤ 34. The rank of the matrix M = ⎢1 0 2 ⎥ is ⎢ ⎥ ⎢⎣3 6 6 ⎥⎦ (a) 0 (c) 2

(b) 1 (d) 3 (GATE 2017, 1 Mark)

35. Consider the following statements about the linear dependence of the real-valued functions y1 = 1, y2 = x and y3 = x2, over the field of real numbers. I. y1, y2 and y3 are linearly independent on -1 ≤ x ≤ 0.

29

36. The rank of the matrix ⎡ 1 −1 0 0 0 ⎤ ⎢ 0 0 1 −1 0 ⎥ ⎥ ⎢ ⎢ 0 1 −1 0 0 ⎥ is ____________. ⎥ ⎢ ⎢ −1 0 0 0 1 ⎥ ⎢⎣ 0 0 0 1 −1⎥⎦ (GATE 2017, 1 Mark)



37. Let M be a real 4 × 4 matrix. Consider the following statements: S1: M has 4 linearly independent eigenvectors S2: M has 4 distinct eigenvalues S3: M is non-singular (invertible) Which one among the following is TRUE? (a) S1 implies S2 (b) S1 implies S3 (c) S2 implies S1 (d) S3 implies S2 (GATE 2018, 1 Mark) 2k ⎤ ⎡k ⎡ x1 ⎤ 38. Consider matrix A = ⎢ 2 and vector x = ⎢ ⎥. 2⎥ ⎣k − k k ⎦ ⎣ x2 ⎦ The number of distinct real values of k for which the equation Ax = 0 has infinitely many solutions is . (GATE 2018, 1 Mark)

Calculus 39. Let δ (t) denotes the delta function. The value of the inte∞ ⎛ 3t ⎞ gral ∫ δ (t ) cos ⎜ ⎟ dt is ⎝ 2⎠ −∞ (a) 1 (c) 0

(b) −1 π (d) 2 (GATE 2001: 1 Mark)

40. The derivative of the symmetric function drawn in the given figure will look like

II. y1, y2 and y3 are linearly dependent on 0 ≤ x ≤ 1. III.   y1, y2 and y3 are linearly independent on 0 ≤ x ≤ 1.

IV.  y1, y2 and y3 are linearly dependent on -1 ≤ x ≤ 0.



Which one among the following is correct? (a) Both I and II are true.

(a)

(b) Both I and III are true. (c) Both II and IV are true. (d) Both III and IV are true. (GATE 2017, 1 Mark)

Ch wise GATE_ECE_CH01_Part 1.indd 29

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30

GATE ECE Chapter-wise Solved papers

(b)

y

3 2

(c)

1 x −1 (a) 1.0 (c) 4.0 (d)

1

2

3 (b) 2.5 (d) 5.0 (GATE 2007, 1 Mark)

46. Which of the following functions would have only odd powers of x in its Taylor’s series expansion about the point x = 0? (a) sin (x3) (b) sin (x2) (c) cos (x3) (GATE 2005, 2 Marks) 41. As x increased from −∞ to ∞, the function f ( x ) =

ex 1+ ex

(a) monotonically increases (b) monotonically decreases (c) increases to a maximum value and then decreases (d) decreases to a minimum value and then increases (GATE 2006, 2 Marks) 42. The integral

∫

π

0

sin 3 θ dθ is given by

(a) 1/2 (c) 4/3

(b) 2/3 (d) 8/3 (GATE 2006, 2 Marks)

sin(θ / 2) 43. lim is θ →0 θ (a) 0.5 (c) 2

(d) cos (x2) (GATE 2008, 1 Mark)

47. In the Taylor’s series expansion of exp(x) + sin(x) about the point x = p, the coefficient of (x − p)2 is (a) exp(p) (b) 0.5 exp(p) (c) exp(p) + 1 (d) exp(p) − 1 (GATE 2008, 2 Marks) 48. The value of the integral of the function g ( x, y ) = 4 x 3 + 10 y 4 along the straight line segment from point (0, 0) to point (1, 2) in the xy plane is (a) 33 (b) 35 (c) 40 (d) 56 (GATE 2008, 2 Marks) 49. Consider points P and Q in the xy plane, with P = (1, 0) Q

and Q = (0, 1). The line integral 2∫ ( xdx + ydy ) along (b) 1 (d) not defined (GATE 2007, 1 Mark)

44. For the function e−x, the linear approximation around x = 2 is (a) (3 − x )e −2

(b) 1 − x

(c) ⎡⎣3 + 2 2 − (1 + 2 )x ⎤⎦ e (d) e −2 (GATE 2007, 1 Mark) −2

45. The following plot shows a function y which varies line2

arly with x. The value of the integral I = ∫ ydx is

P

the semicircle with the line segment PQ as its diameter (a) is −1 (b) is 0 (c) is 1 (d)  depends on the direction (clockwise or anticlockwise) of the semicircle. (GATE 2008, 2 Marks) sin x 50. The Taylor’s series expansion of at x = p  is given x −π by (a) 1 +

( x − π )2 ( x − π )2 +  (b) −1 − + 3! 3!

1

Ch wise GATE_ECE_CH01_Part 1.indd 30

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Chapter 1  • Engineering Mathematics

(c) 1 −

( x − π )2 ( x − π )2 +  (d) −1 + + 3! 3! (GATE 2009, 2 Marks)

 51. If A = xya x + x 2 a y ,

  ∫ A ⋅ dl over the path shown in the

56. The maximum value of the function f(x) = ln(1 + x) − x (where x > − 1) occurs at x = . (GATE 2014, 1 Mark) 57. The series

∑

(a) 2 ln 2 (c) 2

C

figure is

31

y

∞ n= 0

1 converges to n! (b) 2 (d) e (GATE 2014, 1 Mark)

58. The Taylor’s series expansion of 3 sin x + 2 cos x is (a) 2 + 3 x − x 2 −

3

x3 + 2

x3 + 2 x3 (c) 2 + 3 x + x 2 − +  2 (b) 2 − 3 x + x 2 −

C 1

(d) 2 − 3 x − x 2 +

x 0

1/√3

2/√3

x3 + 2 (GATE 2014, 2 Marks)

(a) 0

(b)

2

59. For a function g(t), it is given that

3

∫

(d) 2 3

(c) 1

53. C is a closed path in the z-plane given by |Z| = 3. The ⎛ z2 − z + 4⎞ value of the integral  ∫ C ⎜⎝ z + 2 j ⎟⎠ dz is (a) −p (1 + j2) (c) −4p (3 + j2)

(b) 4p (3 − j2) (d) 4p (1 − j2) (GATE 2014, 1 Mark)

54. For 0 ≤ t < ∞, the maximum value of the f­unction f(t) = e−t − 2e−2t occurs at (a) t = loge4 (b) t = loge2 (c) t = 0 (d) t = loge8 (GATE 2014, 1 Mark) 55. The value of

t

−∞

(a) 0 (c) −

Ch wise GATE_ECE_CH01_Part 1.indd 31

∫

+∞ −∞

y(t ) dt is (b) −j

j j (d) 2 2 (GATE 2014, 2 Marks)

60. The volume under the surface z(x, y) = x + y and above the triangle in the xy plane defined by {0 ≤ y ≤ x and 0 ≤ x ≤ 12} is . (GATE 2014, 2 Marks) 61. The maximum value of f(x) = 2x3 − 9x2 + 12x − 3 in the interval 0 ≤ x ≤ 3 is . (GATE 2014, 2 Marks) 62. For a right-angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is (a) 12° (b) 36° (c) 60° (d) 45° (GATE 2014, 2 Marks)

63. ⎛ 1⎞ lim ⎜1 + ⎟ is x →∞ ⎝ x⎠ (a) ln 2 (b) 1.0 (c) e (d) ∞ (GATE 2014, 1 Mark) x

2

g (t )e − jωt dt = ω e −2ω for any real value ω If

y(t ) = ∫ g (t )dt , then

(GATE 2010, 2 Marks) 52. The maximum value of q until which the approximation sin q = q holds to within 10% error is (a) 10° (b) 18° (c) 50° (d) 90° (GATE 2013, 1 Mark)

+∞ −∞

A function f(x) = 1 - x2 + x3 is defined in the closed interval [-1, 1]. The value of x, in the open interval (-1, 1) for which the mean value theorem is satisfied, is (a) −

1 2

(b) −

1 3

12/4/2018 10:38:28 AM

32

GATE ECE Chapter-wise Solved papers

(c)

1 3

(d)

1 2

(GATE 2015, 1 Mark) 64. The maximum area (in square units) of a rectangle whose vertices lie on the ellipse x2 + 4y2 = 1 is . (GATE 2015, 1 Mark) 65. Which one of the following graphs describes the function f ( x ) = e − x ( x 2 + x + 1)? (a)

f(x) 1

x (b)



P: If f(x) is continuous at 𝑥 = 𝑥0, then it is also differentiable at 𝑥 = 𝑥0.

Q: If f(x) is continuous at 𝑥 = 𝑥0, then it may not be differentiable at 𝑥 = 𝑥0.

R: If f(x) is differentiable at 𝑥 = 𝑥0, then it is also continuous at 𝑥 = 𝑥0.

(a) P is true, Q is false, R is false (b) P is false, Q is true, R is true (c) P is false, Q is true, R is false (d) P is true, Q is false, R is true (GATE 2016, 1 Mark)

69. As x varies from −1 to + 3, which one of the following describes the behaviour of the function f(x) = x3 − 3x2 + 1? (a) f(x) increases monotonically. (b) f(x) increases, then decreases and increases again. (c) f(x) decreases, then increases and decreases again. (d) f(x) increases and then decreases. (GATE 2016, 1 Mark)

f(x) 1

x (c)

68. Given the following statements about a function 𝑓: ℝ → ℝ, select the right option:

f(x) 1

70. The integral

∫

1

0

dx 1− x

is equal to __________. (GATE 2016, 1 Mark)

1 ( x + y + 10)dx dy, where D denotes 2p ∫∫D the disc: 𝑥2 + 𝑦2 ≤ 4, evaluates to __________. (GATE 2016, 2 Marks)

x

71. The integral (d)

f(x) 1

72. In the following integral, the contour C encloses the 1 sin z points 2πj and −2πj: − ∫ C ( z − 2π j )3 dz. 2π 

x (GATE 2015, 2 Marks) ∞

66. The value of the integral .

∫ 12 cos( 2π t )

−∞

sin( 4π t ) dt is 4π t

(GATE 2015, 2 Marks) 67. A vector field D = 2r2ar + z az exists inside a cylindrical region enclosed by the surfaces r = 1, z = 0 and z = 5. Let S be the surface bounding this cylindrical region. The surface integral of this field on S ⎛  ∫∫ D ⋅ ds⎞⎠ is ⎝ S . (GATE 2015, 2 Marks)

Ch wise GATE_ECE_CH01_Part 1.indd 32

The value of the integral is __________. (GATE 2016, 2 Marks)

π ⎧ ⎨( ρ, φ , z ) : 3 ≤ ρ ≤ 5, ≤ 8 ⎩ π ⎫ φ ≤ , 3 ≤ z ≤ 4.5⎬ in cylindrical coordinates has 4 ⎭ volume of __________. (GATE 2016, 2 Marks)

73. The region specified by

74. A triangle in the xy-plane is bounded by the straight lines 2x = 3y, y = 0 and x = 3. The volume above the triangle and under the plane x + y + z = 6 is __________. (GATE 2016, 2 Marks)

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Chapter 1  • Engineering Mathematics 2

75. Let F ( x ) = e x + x for real x. From among the following, choose the Taylor series approximation of F(x) around x = 0, which includes all powers of x less than or equal to 3. 3 (a) 1 + x + x2 + x3 (b) 1 + x + x 2 + x 3 2 7 3 (c) 1 + x + x 2 + x 3 (d) 1 + x + 3 x 2 + 7 x 3 2 6 (GATE 2017, 2 Marks) 76. A three-dimensional region R of finite volume is described by x2 + y2 ≤ z3;  0 ≤ z ≤ 1

where x, y, z are real. The volume of R (up to two decimal places) is ____________. (GATE 2017, 2 Marks)

81. Consider p(s) = s3 + a2s2 + a1s + a0 with all real coefficients. It is known that its derivative p′(s) has no real roots. The number of real roots of p(s) is (a) 0 (b) 1 (c) 2 (d) 3 (GATE 2018, 1 Mark) 82. Let r = x2 + y - z and z3 - xy + yz + y3 = 1. Assume that x and y are independent variables. At (x, y, z) = (2, -1, 1), ∂r the value (correct to two decimal places) of is . ∂x (GATE 2018, 2 Marks)

Differential Equations 83. The following differential equation has

77. Let I = ∫ ( 2 z dx + 2 y dy + 2 x dz ) where x, y, z are real,

3

⎛ d2 y⎞ ⎛ dy ⎞ 3 ⎜ 2 ⎟ + 4 ⎜ ⎟ + y 2 + 2 = x. ⎝ dt ⎠ ⎝ dt ⎠

C

and let C be the straight line segment from point A: (0, 2, 1) to point B: (4, 1, -1). The value of I is ___________. (GATE 2017, 2 Marks) 78. The values of the integrals ⎛1 x− y ⎞ ∫0 ⎜⎝ ∫0 ( x + y)3 dy⎟⎠ dx 1

and 1

⎛1 x− y

∫ ⎜⎝ ∫ ( x + y) 0

(a) (b) (c) (d)

3

0

79. An integral I over a counter-clockwise circle C is given by

C

z2 −1 z e dz z2 + 1

If C is defined as |z| = 3, then the value of I is (a) -p i sin(1) (b) -2p i sin(1) (c) -3p i sin(1) (d) -4p i sin(1) (GATE 2017, 2 Marks) 1 x( x 2 − 3) in 3 the interval -100 ≤ x ≤ 100 occurs at x = __________.

80. The minimum value of the function f ( x ) =

(GATE 2017, 2 Marks)

Ch wise GATE_ECE_CH01_Part 1.indd 33

(GATE 2005, 1 Mark) 84. The solution of the following differential equation is given by d2 y dy −5 + 6y = 0 2 dx dx (a)  y = e 2 x + e −3 x

(b)  y = e 2 x + e3 x

(c) y = e −2 x + e 3 x

(d)  y = e −2 x + e −3 x (GATE 2005, 1 Mark)

(GATE 2017, 2 Marks)

I =∫

(a) Degree = 2, order = 1 (b) Degree = 1, order = 2 (c) Degree = 4, order = 3 (d) Degree = 2, order = 3

⎞ dx ⎟ dy are ⎠

same and equal to 0.5 same and equal to -0.5 0.5 and -0.5, respectively -0.5 and 0.5, respectively

33

85. Which of the following is a solution to the differential dx(t ) equation + 3 x (t ) = 0 ? dt (a)  x(t ) = 3e − t

(b)  x(t ) = 2e −3t

3 (c)  x (t ) = − t 2 2

(d)  x(t ) = 3t 2 (GATE 2008, 1 Mark)

86. The order of the differential equation 3

    (a) 1 (c) 3

d 2 y ⎛ dy ⎞ + ⎜ ⎟ + y 4 = e − t is dt 2 ⎝ dt ⎠ (b) 2 (d) 4 (GATE 2009, 1 Mark)

12/4/2018 10:38:33 AM

34

GATE ECE Chapter-wise Solved papers

87. Match List I with List II and select the correct answer using the codes given below the lists: List I List II A. 

dy y = dx x

1. Circles

B. 

dy y =− dx x

2.  Straight lines

C. 

dy x = dx y

3. Hyperbola

D. 

dy x =− dx y

91. A system described by a linear constant coefficient and ordinary, first-order differential equation has an exact solution given by y(t ) for t > 0, when the forcing function is x(t ) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes −2 y(t ) for t > 0, we need to (a) change the initial condition to − y(0) and the forcing function to 2x(t ) (b) change the initial condition to 2 y(0) and the forcing function to −x(t ) (c) change the initial condition to j 2 y(0) and the forcing function to j 2 x(t ) (d) change the initial condition to −2 y(0) and the forcing function to −2x(t ) (GATE 2013, 2 Marks)

Codes:

A  B  C  D (a) 2    3    3    1 3     2    1 (b) 1    (c) 2    1     3    3 (d) 3    2     1    2



92. If the characteristic equation of the differential equation

(GATE 2009, 2 Marks)

88. A function n( x ) satisfies the differential equation d 2 n( x ) n( x ) − 2 = 0 , where L is a constant. The boundary dx 2 L conditions are n(0) = k and n(∞) = 0. The solution to this equation is (a)  n( x ) = k exp( x /L)

d2 y dy + 2α +y=0 2 dx dx has two equal roots, then the values of a are (a) ± 1 (b) 0, 0 (c) ± j (d) ±1/2 (GATE 2014, 1 Mark)

93. Which one of the following is a linear non-­homogeneous differential equation, where x and y are the independent and dependent variables, respectively? (a)

dy + xy = e − x dx

(b)

dy + xy = 0 dx

(c)

dy + xy = e − y dx

(d)

dy + e− y = 0 dx

(b)  n( x ) = k exp( − x / L ) (c)

n( x ) = k 2 exp( − x / L)

(d)  n( x ) = k exp( − x / L)

(GATE 2014, 1 Mark) (GATE 2010, 1 Mark)

dy 89. The solution of the differential equation = ky, y(0) = c dx is (a)  x = ce − ky (b)  x = ke cy (c)  y = ce kx

(d)  y = ce − kx

94. If z = xy ln(xy), then (a) x

∂z ∂z ∂z ∂z +y = 0 (b) y =x ∂x ∂y ∂x ∂y

(c) x

∂z ∂z ∂z ∂z =y (d) y +x =0 ∂x ∂y ∂x ∂y

(GATE 2011, 1 Mark)

(GATE 2014, 1 Mark)

90. With initial condition x(1) = 0.5, the solution of the difdx + x = t is ferential equation, t dt 1 1 (a)  x=t− (b)  x = t2 − 2 2 2 t t (c)  x= (d)  x= 2 2 (GATE 2012, 1 Mark)

95. If a and b are constants, the most general solution of the d2x dx differential equation 2 + 2 + x = 0 is dt dt −t (a) ae (b) ae−t + bte−t (c) aet + bte−t (d) ae−2t (GATE 2014, 1 Mark)

Ch wise GATE_ECE_CH01_Part 1.indd 34

96. With initial values y(0) = y'¢(0) = 1, the solution of the differential equation

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Chapter 1  • Engineering Mathematics

103. The particular solution of the initial value problem given below is

d2 y dy + 4 + 4y = 0 dx dx 2 at x = 1 is

. (GATE 2014, 2 Marks) 2

97. The solution of the differential equation with y(0) = y ′(0) = 1 is (a) (2 - t)et (c) (2 + t)e-t

d y 2dy + + y=0 dt dt 2

(b) (1 + 2t) e-t (d) (1 - 2t)et (GATE 2015, 1 Mark)

98. The general solution of the differential equation dy 1 + cos 2 y = is dx 1 − cos 2 x (a) tan y - cot x = c (c is a constant) (b) tan x - cot y = c (c is a constant) (c) tan y + cot x = c (c is a constant) (d) tan x + cot y = c (c is a constant) (GATE 2015, 1 Mark) dx = 10 − 0.2 x with dt initial condition x(0) = 1. The response x(t) for t > 0 is (a) 2 - e-0.2t (b) 2 - e0.2t (c) 50 - 49e-0.2t (d) 50 - 49e0.2t (GATE 2015, 1 Mark)

99. Consider the differential equation

100. The contour on the x-y plane, where the partial derivative of x2 + y2 with respect to y is equal to the partial derivative of 6y + 4x with respect to x, is (a) y = 2 (b) x = 2 (c) x + y = 4 (d) x - y = 0 (GATE 2015, 1 Mark) 101. Consider the differential equation d 2 x (t ) dx(t ) +3 + 2 x(t ) = 0. 2 dt dt Given x(0) = 20 and x(1) = 10/e, where e = 2.718, the value of x(2) is . (GATE 2015, 1 Mark) dx 102. The ordinary differential equation = −3 x + 2, dt with x(0) = 1 is to be solved using the forward Euler method. The largest time step that can be used to solve the equation without making the numerical solution unstable is __________. (GATE 2016, 2 Marks)

Ch wise GATE_ECE_CH01_Part 1.indd 35

35

d 2 y 12dy dy + + 36 y = 0 with y(0) = 3 and 2 dx dx dx (a) (3 − 18x)e−6x (c) (3 + 20x)e−6x

= −36 x=0

(b) (3 + 25x)e−6x (d) (3 − 12x)e−6x (GATE 2016, 2 Marks)

104. The general solution of the differential equation d2 y dy + 2 − 5y = 0 2 dx dx in terms of arbitrary constants K1 and K2 is (a) K1e ( −1+

6)x

+ K 2 e ( −1−

6)x

(b) K1e ( −1+

8)x

+ K 2 e ( −1−

8)x

(c) K1e ( −2 +

6)x

+ K 2 e ( −2 −

6)x

(d) K1e ( −2 +

8)x

+ K 2 e ( −2 −

8)x

(GATE 2017, 1 Mark) 105. Which one of the following is the general solution of the first-order differential equation dy = ( x + y − 1) 2 dx where x, y are real? (a) y = 1 + x + tan-1 (x + c), where c is a constant. (b) y = 1 + x + tan (x + c), where c is a constant. (c) y = 1 - x + tan-1 (x + c), where c is a constant. (d) y = 1 - x + tan (x + c), where c is a constant. (GATE 2017, 2 Marks) 106. Let f ( x, y ) =

ax 2 + by 2 , where a and b are constants. xy

∂f ∂f at x = 1 and y = 2, then the relation between = ∂x ∂y a and b is

If

(a) a =

b b (b) a= 4 2

(c) a = 2b

(d) a = 4b (GATE 2018, 1 Mark)

107. A curve passes through the point (x = 1, y = 0) and satisfies the differential equation dy x 2 + y 2 y = + dx 2y x

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36

GATE ECE Chapter-wise Solved papers

The equation that describes the curve is

113. The directional derivative of f ( x, y ) =

(b)

1 ⎛ y2 ⎞ ln ⎜1 + 2 ⎟ = x − 1 2 ⎝ x ⎠

  114. If r = xa x + ya y + za z and r = r , then div( r 2 ∇(ln r ))

y⎞ ⎛ (c) ln ⎜1 + ⎟ = x − 1 ⎝ x⎠

=

(GATE 2018, 2 Marks) 108. The position of a particle y(t) is described by the differential equation: d2 y dy 5 y =− − 2 dt 4 dt The initial conditions are y(0) = 1 and

dy = 0.The dt t = 0

position (accurate to two decimal places) of the particle at t = p is . (GATE 2018, 2 Marks)

Vector Analysis 109. ∇ × ∇ × P, where P is a vector, is equal to (a) P × ∇ × P − ∇ 2 P (b) ∇ 2 P + ∇(∇ × P ) (c) ∇ 2 P + ∇ × P (d) ∇(∇ ⋅ P ) − ∇ 2 P (GATE 2006, 1 Mark) 110.

∫∫ (∇ × P ) ⋅ ds, where P is a vector, is equal to (a)

∫ P ⋅ dl (b) ∫ ∇ × ∇ × P ⋅ dl

(c)

∫∫∫ ∇ ⋅ Pdv ∫ ∇ × P ⋅ dl (d) (GATE 2006, 1 Mark)

 111. The divergence of the vector field A = xa x + ya y + za z is (a) 0 (b) 1/3 (c) 1 (d) 3 (GATE 2013, 1 Mark) 112. The magnitude of the gradient for the function f(x,y,z) = x2 + 3y2 + z3 at the point (1,1,1) is . (GATE 2014, 1 Mark)

Ch wise GATE_ECE_CH01_Part 1.indd 36

. (GATE 2014, 2 Marks)

1 ⎛ y⎞ (d) ln ⎜1 + ⎟ = x − 1 2 ⎝ x⎠



xy

( x + y ) at 2 π (1, 1) in the direction of the unit vector at an angle of 4 with y-axis is given by . (GATE 2014, 1 Mark)

⎛ y ⎞ (a) ln ⎜1 + 2 ⎟ = x − 1 x ⎠ ⎝ 2

     115. A vector P is given by P = x 3 yax − x 2 y 2 a y − x 2 yzaz . Which one of the following statements is TRUE?  (a) P is solenoidal, but not irrotational  (b) P is irrotational, but not solenoidal  (c) P is neither solenoidal nor irrotational  (d) P is both solenoidal and irrotational (GATE 2015, 1 Mark) 116. If C denotes the counterclockwise unit circle, the value 1 of the contour integral . ∫c Re( z ) dz is 2π j  (GATE 2015, 1 Mark) 117. Which one of the following is a property of the solutions to the Laplace equation: ∇2𝑓 = 0? (a) The solutions have neither maxima nor minima anywhere except at the boundaries. (b) The solutions are not separable in the coordinates. (c) The solutions are not continuous. (d) The solutions are not dependent on the boundary conditions. (GATE 2016, 1 Mark) 118. Consider the time-varying vector I = x 15cos (ω t ) + y 5sin (ω t ) in Cartesian coordinates, where ω > 0 is a constant. When the vector magnitude | I | is at its minimum value, the angle θ that I makes with the x-axis (in degrees, such that 0 ≤ θ ≤ 180) is __________. (GATE 2016, 1 Mark) 119. Suppose C is the closed curve defined as the circle x2 + y2  = 1 with C oriented anticlockwise. The value of 2 2 ∫ ( xy dx + x ydy) over the curve C equals __________. (GATE 2016, 2 Marks)

120. If the vectors e1 = (1, 0, 2), e2 = (0, 1, 0) and e3 = (−2, 0, 1) form an orthogonal basis of the three-dimensional real space ℝ3, then the vector u = (4, 3, −3) ∈ ℝ3 can be expressed as

12/4/2018 10:38:41 AM

Chapter 1  • Engineering Mathematics

2 11 (a) u = −  e1 − 3e2 −  e3 5 5

37

(a) 2π c1 (b) 2π (1 + c0 ) (c) 2π jc1 (d) 2π j (1 + c0 )

2 11 (b) u = − e1 − 3e2 +  e3 5 5 2 11 (c) u = −  e1 + 3e2 +  e3 5 5

(GATE 2009, 1 Mark) 1 − 2z 125. The residue of a complex function x ( z ) = 1) ( z − 2) z z − ( at its poles are

2 11 (d) u = −  e1 + 3e2 −  e3 5 5 (GATE 2016, 2 Marks)

Complex Analysis 121. The value of the contour integral

1 dz in posz −i = 2 z + 4

∫

(a)

1 1 , and 1 2 2

(c)

1 1 3 3 , 1 and − (d) , −1 and 2 2 2 2

2

itive sense is (a) ip/2 (b) −p/2 (c) −ip/2 (d) p/2 (GATE 2006, 2 Marks) 122. If the semicircle D of radius 2 is as shown in the figure, 1 then the value of the integral  ∫D ( s2 − 1) ds is

jw

(b)

1 1 , and −1 2 2

(GATE 2010, 2 Marks) 126. If x = −1 , then the value of xx is (a) e − π / 2 (b) eπ / 2 (c) x (d) 1 (GATE 2012, 1 Mark) 127. Given f ( z ) =

1 2 − . If C is a counterclockwise z +1 z + 3

path in the z-plane such that z + 1 = 1, the value of

j2 D

1 ∫ C f ( z ) dz is 2π j 

0 2

σ

−j2

(a) −2 (b) −1 (c) 1 (d) 2 (GATE 2012, 1 Mark)

128. The real part of an analytic function f(z) where z = x + jy (a) jp (b) −jp is given by e−ycos(x). The imaginary part of f(z) is (c) −p (d) p (a) ey cos(x) (b) e−y sin(x) (GATE 2007, 2 Marks) (c) −ey sin(x) (d) −e−y sin(x) 1 (GATE 2014, 2 Marks) 123. The residue of the function f ( z ) = at 2 ( z + 2 ) ( z − 2 )2 z = 2 is 129. Let Z = x + iy be a complex variable, consider continuous integration is performed along the unit circle in 1 1 (a) − (b) − anticlockwise direction. Which one of the following 32 16 statements is NOT TRUE? 1 1 1 z (c) (d) (a) The residue of 2 at z = 1 is 16 32 2 z −1 (GATE 2008, 2 Marks) 2 (b)  ∫ z dz = 0 1+ f ( z ) 124. If f ( z ) = c0 + c1 z , then  dz is given by ∫ z unit circle

C

−1

Ch wise GATE_ECE_CH01_Part 1.indd 37

(c)

1 1 dz = 1 ∫  2π i C z

12/4/2018 10:38:44 AM

38

GATE ECE Chapter-wise Solved papers

(d)  z (complex conjugate of z) is an analytical function (GATE 2015, 1 Mark) 130. If C is a circle of radius r with centre z0, in the complex dz z-plane and if n is a non-zero integer, then  ∫C ( z − z0 )n+1 equals (a) 2p nj nj (c) 2π

(b) 0

y C

-1

The value of the integral

dz 1 is ∫ 2  Πj c z − 1

(d) 2p n

131. Consider the complex valued function f (z) = 2z3 + b |z|3, where z is a complex variable. The value of b for which the function f(z) is analytic is __________.

Numerical Methods 137. Match List I with List II and select the correct answer using the codes given below the lists:

(GATE 2016, 1 Mark) 132. For f (z) = (sin (z)/z ), the residue of the pole at z = 0 is __________. 2

(GATE 2016, 1 Mark) ⎛ ez ⎞ 1 133. The values of the integral ∫c ⎜⎝ z − 2 ⎟⎠ dz along a 2π j  closed contour c in anti-clockwise direction for (i) the point z0 = 2 inside the contour c, and (ii) the point z0 = 2 outside the contour c, respectively, are (a) (i) 2.72, (ii) 0 (b) (i) 7.39, (ii) 0 (c) (i) 0, (ii) 2.72 (d) (i) 0, (ii) 7.39

List I

134. The residues of a function 1 f ( z) = are ( z − 4)( z + 1)3 (a)

−1 −1 1 −1 and (b) and 27 125 125 125

(c)

−1 1 1 −1 and (d) and 27 5 125 5 (GATE 2017, 1 Mark)

0

2

/ 2)

dt around

x = 0 has the form F(x) = a0 + a1x + a2x2 +  The coefficient a2 (correct to two decimal places) is equal to ______. (GATE 2018, 1 Mark) 136. The contour C given below is on the complex plane z = x + jy, where j = −1.

Ch wise GATE_ECE_CH01_Part 1.indd 38

List II

A. Newton–Raphson method

1. Solving non-linear equations

B. Runge–Kutta method equations

2. Solving simultaneous linear equations

C. Simpson’s rule equations

3. Solving ordinary differential

D.  Gauss elimination

4.  Numerical integration 5. Interpolation 6. Calculation of Eigenvalues

(GATE 2016, 2 Marks)

x

.

(GATE 2018, 2 Marks)

(GATE 2015, 1 Mark)

135. Taylor series expansion of f ( x ) = ∫ e − ( t

x

1

Codes: A  B  C  D (a) 6   1   5   3 (b) 1   6   4   3 (c) 1   3   4   2 (d) 5   3   4   1 (GATE 2005, 2 Marks) 138. The equation x 3 − x 2 + 4 x − 4 = 0 is to be solved using the Newton–Raphson method. If x = 2 is taken as the initial approximation of the solution, then the next ­ approximation using this method will be (a)

2 4 (b) 3 3

(c) 1

(d)

3 2

(GATE 2007, 2 Marks) 139. The recursion relation to solve x = e − x using Newton– Raphson method is

12/4/2018 10:38:46 AM

39

Chapter 1  • Engineering Mathematics

144. How many distinct values of x satisfy the equation sin(x) = x/2, where x is in radians?

(a) xn +1 = e − xn (b) xn +1 = xn − e − xn (c) xn +1

(a) 1 (c) 3

e − xn = (1 + xn ) 1 + e − xn xn2 − e − xn (1 + xn ) − 1

(b) 2 (d) 4 or more (GATE 2016, 1 Mark)

dy( x ) − y( x ) = x with dx the initial conditions y(0) = 0 . Using Euler’s first-order

145. Consider the first-order initial value problem y′ = y + 2x − x2, y(0) = 1, (0 ≤ x  0

(minima)

For x = 1, −2 1 f (1) = (1)3 − 1 = 3 3

Therefore, option (c) is the correct answer.

79. (d)

Now, f  (-100) = -333433.33

Im(Z) +3i

|Z| = 3

× +i −3

× −i

Re 3

−3i

I= ∫

z2 −1 z e dz z2 + 1

= 2p i(Res ( z = i ) + Res ( z = −i ))

Ch wise GATE_ECE_CH01_Part 2.indd 55

and f  (100) = 333233.33 (Since x = 100 and -100 are end points of interval) Therefore, minimum value occurs at x = -100. 81. (b) p(s) will have three roots in total. If derivative p′(s) has no real roots, this means it will have two complex roots. Complex roots always come in pair. This means p(s) has one real root. 82. (4.5)  Given r = x2 + y - z 3 z - xy + yz + y3 = 1

(1) (2)

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GATE ECE Chapter-wise Solved papers

Differentiating Eq. (1) we have (since x and y are independent)

∂r ∂z = 2x − ∂x   ∂x

(3)

Differentiating Eq. (2) we have



y ∂z =  ∂x 3 z 2 + y

x = e −3t + e c

⇒

x = c1e −3t

⇒

x = c1e −3t

(4)

87. (a)  dy y = dx x dy dx dy dx = ⇒∫ =∫ y x y x log y = log x + log c = log cx

A.   

From Eqs. (3) and (4) we have ∂r y = 2x − 2 ∂x 3z + y

So, ∂r 1 1 = 2×2+ = 4 + = 4.5 ∂x ( 2, −1,1) 3 −1 2

y = cx , which is the equation of a straight line. dy − y = dx x

B.  

dy − dx dy dx = ⇒∫ = −∫ y x y x log y = − log x + log c log y + log x = log c log yx = log c yx = c

Differential Equations 83. (b)  The highest derivative term of the equation is 2, hence order = 2. The power of highest derivative term is 1, hence degree = 1. 84. (b)  We have d2 y dy −5 + 6y = 0 2 dx dx

y = c/ x , which is the equation of a hyperbola. dy x = , ydy = xdx ⇒ ∫ ydy = ∫ xdx dx y

C.  

The equation can be written as

y 2 x 2 c2 − = → const 2 2 2 y 2 − x 2 = c2

( D 2 − 5 D + 6) y = 0 Auxiliary equation is given by D 2 − 5D + 6 = 0 ⇒ ( D − 2)( D − 3) = 0 ⇒ D = 2, 3 Therefore, y = e 2 x + e 3 x . 85. (b)  We have dx(t ) + 3 x (t ) = 0 dt ⇒ ⇒ ⇒

Ch wise GATE_ECE_CH01_Part 2.indd 56

dx = −3 x dt dx = −3dt x dx ∫ x = −3dt

(c1 = e c )

86. (b)  The highest derivative of differential is 2. Hence, the order of the equation is 2.

∂z {3 z 2 + y} = y ∂x

    

⇒

Hence, from the given options, x(t ) = 2e −3t is the correct answer.

∂z ∂z 3z − y+ y =0 ∂x ∂x 2

⇒

ln x = −3t + c



y2 x2 − = 1 , which is the equation of a hyperbola. c2 c2

D.  

dy − x = ⇒ ∫ ydy = − ∫ xdx dx y y2 x 2 c2 =− + 2 2 2 2 2 2 y x c + = 2 2 2

x 2 + y 2 = c 2 , which is the equation of a circle. 88. (d)  We know that d 2 n( x ) n( x ) − 2 =0 dx 2 L

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Chapter 1  • Engineering Mathematics

57

The equation can be written as 1 Also, we are given the initial condition x(1) = . 2 The equation is of the form,

⎛ 2 1⎞ ⎜⎝ D − 2 ⎟⎠ n( x ) = 0 L ⇒ D2 − ⇒ D2 =

dx + Px = Q dt

1 =0 L2

1 1 ⇒D=± L L2

where P =

Therefore, the solution is n( x ) = c1e −1/ Lx + c2 e1/ Lx (1)



1 and Q = 1 t

Integrating factor = e ∫

Now, at x = 0, we have

c2 = 0

t2 +C 2 t C x= + 2 t

xt =

Hence, c1 = k . Now substituting values of c1 and c2, we get the required solution. Thus, the solution is n( x ) = ke − x / L . 89. (c)  We have dy = ky dx dy = k dx y

⇒

Integrating on both sides, we get ln y = kx + A

Now putting x(1) =

1 , we get 2

⇒

1 C 1 = + 2 1 2

⇒

C=0

So, x =

t is the required solution. 2

91. (d)  We have

Putting x = 0, we get ln y(0) = A \ A = ln c

dy(t ) + ky(t ) = x(t ) dt

[∵ y(0) = c]

Taking Laplace transformation of both sides, we have

Hence,

sY ( s) − y(0) + kY ( s) = X ( s) ln y = kx + ln c

Y ( s)[ s + k ] = X ( s) + y(0)

⇒            ln y − ln c = kx ⇒

⎛ y⎞ ⇒              ln ⎜ ⎟ = kx ⎝ c⎠ ⇒           

X ( s) y ( 0) + s+k s+k

y(t ) = e − kt x(t ) + y(0) e − kt

y = e kx ⇒ y = ce kx c

dx +x=t dt dx x ⇒ + =1 dt t

y ( s) =

Taking inverse Laplace transform, we have

So if we want −2 y(t ) as a solution, both x(t ) and y(0) have to be multiplied by −2 and hence x(t ) and y(0) be changed to −2x(t ) and −2 y(0) , respectively.

90. (d)  The given differential equation is t

Ch wise GATE_ECE_CH01_Part 2.indd 57

= e loge t = t

x ⋅ t = ∫ 1⋅ t ⋅ dt + C

n(∞) = c1e −∞ + c2 e ∞ = 0

⇒

1

∫ t dt

x ⋅ (I.F.) = ∫ Q ⋅ (I.F.) dt + C

and at x = ∞, we have

c2 e ∞ = 0

=e

The solution is given by

n(0) = c1 + c2 = k

⇒

Pdt

92. (a)  For equal roots, discriminant b 2 − 4 ac = 0.

Thus,   

4α 2 − 4 = 0 ⇒ α = ±1

12/4/2018 10:39:55 AM

58

dy 1 + cos 2 y = dx 1 − cos 2 x dy dx ⇒ = 1 − cos 2 y 1 + cos 2 x dy dx ⇒ = (vvariable-separable) 2 2 sin y 2 cos 2 x

GATE ECE Chapter-wise Solved papers

dy 93. (a)  Option (a), + xy = e − x is a first-order linear equadx dy tion (non-homogeneous). Option (b), + xy = 0 is a dx first-order linear equation (homogeneous). Options (c) and (d) are non-linear equations.

⇒ ∫ cos ec 2 ydy = ∫ sec 2 xdx ⇒ − cot y = tan x + k ⇒ − tan x − cot y = k

⇒ tan x + cot y = c where c = − k 99. (c)  Given differential equation is,

94. (c)  We are given z = xy ln(xy)

dx = 10 − 0.2 x, x(0) = 1 dt

⎤ ⎡ ∂z 1 = y ⎢ x × × y + ln xy ⎥ = y (1 + ln xy ) ∂x xy ⎣ ⎦ and

∂z ∂z ∂z = x [1 + ln xy ] ⇒ x =y ∂y ∂x ∂y



m 2 + 2m + 1 = 0 ⇒ m = −1, −1 Thus, general solution is x = ( a + bt ) e



xp =

y = ( a + bx ) e

Given, −2 x

(1)

x(0) = 1 ⇒ C + 50 = 1 ⇒ C = 49 x = 50 − 49e ( −0.2 )t

y = ( a + bx )( −2e −2 x ) + e −2 x (b) (2)

Using y(0) = 1 and y'¢(0) = 1, Eqs. (1) and (2) give a = 1 and b = 3. Therefore y = (1 + 3 x ) e −2 x



At x = 1, y = 4e −2 = 0.541

2

97. (b)  Differential equation is d y + 2dy + y = 0 dt dt 2 Thus,

D 2 + 2 D + 1 = 0 ⇒ ( D + 1) = 0 ⇒ D = −1, −1



Thus, solution is y(t ) = (c1 + c2 t )e − t → C.F.





2



1 10e 0 t 10e 0 t = = 50e 0 t = 50 0.2 D + (0.2)

x = xc + x p = Ce( −0.2 ) t + 50

Thus, solution is

Complementary solution, xc = Ce ( −0.2 )t

m 2 + 4 m + 4 = 0 ⇒ m = −2, −2

The auxiliary equation is given as

−t

96. (0.541)  We can write the auxiliary equation as

dx + (0.2) x = 10 dt

m + 0.2 = 0 ⇒ m = −0.2

95. (b)  The auxiliary equation is given by



⇒

⇒ y ′(t ) = c2 e − t + (c1 + c2 t )( −e − t ) y(0) = 1; y ′(0) = 1 gives c1 = 1 and c2 + c1 ( −1) = 1 ⇒ c2 = 2

Therefore, y(t ) = (1 + 2t )e − t 98. (d)  We are given that dy 1 + cos 2 y = dx 1 − cos 2 x dy dx ⇒ = 1 − cos 2 y 1 + cos 2 x dy dx ⇒ = (vvariable-separable) 2 2 sin y 2 cos 2 x

100. (a)  The partial derivative of x 2 + y 2 with respect to y is 0 + 2 y ⇒ 2 y.



The partial derivative of 6 y + 4 x with respect to x is 0 + 4 = 4. Given that both are equal. ⇒ 2y = 4 ⇒ y = 2

101. (0.8556) Given d 2 x (t ) dx(t ) +3 + 2 x (t ) = 0 2 dt dt x(0) = 20 10 x(1) = e Auxiliary equation is given by

m 2 + 3m + 2 = 0 ⇒ m = −1, −2 Complementary solution, xe = c1e − t + c2 e −2t

Particular solution, xp = 0 Finally, x = xc + x p = c1e − t + c2 e −2t x (0 ) = 20 ⇒ 20 = c1 + c2

(1)

10 10 ⇒ = c1e −1 + c2 e −2 e e ⇒ 10 = c1 + c2 e −1

( 2)

x (1) =

⇒ ∫ cos ec ydy = ∫ sec xdx 2

⇒ − cot y = tan x + k Ch wise GATE_ECE_CH01_Part 2.indd 58 tan x − cot y = k ⇒−

2

12/4/2018 10:39:58 AM

Chapter 1  • Engineering Mathematics



From Eq. (1), c1 = 20 − c2

104. (a)    

10 = ( 20 − c2 ) + c2 e −1 −1 10 = ( 20 ⇒ 10 = c−(ce2−)1 +− c12)e+ 20

d2 y dy + 2 − 5y = 0 dx dx 2

⇒ (D2 + 2D - 5)y = 0 ⇒ D2 + 2D - 5 = 0

2

1 ⇒ 10 = 10 c2 (−e −20 − 1) + 2010 10 10e ⇒ c2 = −1 = − −1 = = −1 − 10 10 10e 10 20 ⇒ c2 = e −1 − 1 = − e −1 − 1 = 1 − e −1 = e − 1 e e−−1 20 e −10 e e−−120 1−−10ee 10 1 e 20 ⇒ c1 = 20 − = = 10 e 20 e 20 10 e 10 e − 20 − − e −1 = e −1 ⇒ c1 = 20 − e − 1 = e −1 e −1 e −1

10e − 20 − t 10e −2t e + e e −1 e −1 ⎛ 10e − 20 ⎞ −2 ⎛ 10e ⎞ −4 x ( 2) = ⎜ e +⎜ e = 0.8556 ⎝ e − 1 ⎟⎠ ⎝ e − 1⎟⎠

x (t ) =

⇒ D=

−2 ± 4 + 20 1 = −1 ± 24 2 2

⇒ D = −1 ± 6 Therefore, m1 = −1 + 6 , m2 = −1 − 6 We know y(t ) = K1e m1 x + K 2 e m2 x ⇒ y(t ) = K1e ( −1+

102. (0.67)  We have dx = −3 x + 2, with dt

x (0 ) = 1

If |1 − 3h| < 1, the solution is stable. −1 < 1 − 3h < 1 −2 < − 3h < 0 0 < h < 2/3

Let d/dx = D. Therefore, (D2 + 12D + 36)y = 0

1+



dp dy −1 = dx dx dp − 1 = p2 dx



dp = dx 1 + p2

⇒ c1 = 3 Now, dy = (c1 + c2 x )( −6 × e −6 x ) + e −6 x × c2 dx dy dx

= −36 = c1 ( −6) + c2 ⇒ c2 = −36 + 18 = −18 x=0

Therefore, y = (c1 + c2x)e − 6x = (3 − 18x)e− 6x

Ch wise GATE_ECE_CH01_Part 2.indd 59

(4)

Integrating both sides of Eq. (4), we get tan −1 p = x + c x + y − 1 = tan( x + c) y = 1 − x + tan( x + c)

(m + 6)2 = 0 ⇒ m = − 6 and −6 The complementary solution is

y(0) = 3 = (c1 + 0)e− 0

dy dp = (3) dx dx



m2 + 12m + 36 = 0

It is given that

6)x

Differentiating Eq. (2), we get

The auxiliary equation is

y = (c1 + c2x)e− 6x

+ K 2 e ( −1−

dy = ( x + y − 1) 2 (1) dx Let x + y -1 = p(2)

2 = = 0.67 3

103. (a)  It is given that d 2 y 12dy + + 36 y = 0 dx dx 2

6)x

105. (d)     

Therefore, hmax

59

106. (d)    f ( x, y ) =

ax 2 + by 2 ax by = + xy y x

∂f a by       ∂x = y − x 2 (1)       ∂f − ax b  = 2 + ∂y x y Since

(2)

∂f ∂f a by − ax b = ⇒ − 2 = 2 + ∂x ∂y y x x y ⇒

a ax by b + = + y y2 x2 x

12/4/2018 10:40:00 AM

60

GATE ECE Chapter-wise Solved papers

At x = 1, y = 2, we get



a a 2b 3 + = + b ⇒ a = 3b ⇒ a = 4b 2 4 1 4 2 107. (a)    dy = x + y + y  dx 2 y 2 x

Let

dy dx



dt +t  dx

(2)

dt x tx +t = + +t dx 2t 2

⇒

dt 1 t = + dx 2t 2

⇒

dt 1 + t 2 2t = ⇒ dx = dt dx 2t (1 + t 2 )

∫∫ ( ∇ × P ) ⋅ ds = ∫ P ⋅ d l 111. (d)  We know that  ∂A ∂A ∂A ∇⋅ A = x + x + z ∂x ∂y ∂z  ∂ ∂ ∂ ∇⋅ A = ( x) + ( y) + ( z) = 1 + 1 + 1 ∂x ∂y ∂z  ∴ ∇⋅ A = 3

⎛ y2 ⎞ Therefore, ln ⎜1 + 2 ⎟ = x − 1 x ⎠ ⎝

   112. (7)    (∇f ) P (1,1,1) = [i ( 2 x ) + j (6 y ) + k (3 z 2 )]P (1,1,1)    = ( 2i + 6 j + 3k )

108. (-0.21) Given



( ∇f ) P

(1) 113. (3)   

⎡

5⎤

⇒ ⎢ s 2 + s + ⎥ Y ( s) − ( s + 1) y(0) = 0 4⎦ ⎣

Ch wise GATE_ECE_CH01_Part 2.indd 60

f =

= 4 + 36 + 9 = 7

(x 2

1

2

y + xy 2

)

 ⎡ 2 xy + y 2 ⎤  ⎡ x 2 + 2 xy ⎤ ⇒ ∇f = i ⎢ ⎥+ j⎢ ⎥ 2 ⎦ 2 ⎦ ⎣ ⎣ at (1, 1), ∇f =

Using Laplace transform: 5 ⎡⎣ s 2Y ( s) − sy(0) ⎤⎦ + sY ( s) − y(0) + Y ( s) = 0 4

2

1⎞ 1⎞ ⎛ ⎛ ⎜⎝ s + ⎟⎠ + 1 ⎜⎝ s + ⎠⎟ + 1 2 2

110. (a)  According to Stokes’ theorem,

⇒ C = −1

d 2 y dy 5 y + + =0 dt 2 dt 4

1/ 2

∇ × ∇ × P = (∇ ⋅ P )∇ − (∇ ⋅ ∇) P = ∇(∇ ⋅ P ) − ∇ 2 P

⇒ 1 + C = log(1)

Rewriting Eq. (1) we have

+

We get

2 2 ⇒ x + C = log(1 + y /x ) (∵ x = 1; y = 0)



2

Now putting, A = ∇, B = ∇ and C = P

x + C = log ⎣⎡1 + ( y /x ) 2 ⎤⎦



1⎞ ⎛ ⎜⎝ s + ⎟⎠ + 1 2

( s + 1/ 2)

=

109. (d)  From the property of vector triple product, we have A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B)C

Now t = y/ x, so we have

dy =0 y(0) = 1 and dt t = 0

2

as y(0) = 1

1 ⎤ ⎡ y(p ) = e − p / 2 ⎢cos p + sin p ⎥ = −e − p / 2 = -0.2078 2 ⎣ ⎦

x + C = log(1 + t2)

d 2 y − dy 5 y = −  dt 4 dt 2

( s + 1)

5 4

Vector Analysis

Integrating both the sides, we get



s +s+

1 ⎡ ⎤ L−1 [Y ( s)] = y(t ) = e − t / 2 ⎢cos t + sin t ⎥ 2 ⎣ ⎦ At t = p

x xt dt ⇒x = + dx 2t 2



⇒ Y ( s) =

(1)

Putting Eq. (2) in Eq. (1) we get x

s +1 2

y = t ⇒ y = xt . So x

= x

⇒ Y ( s) =



3  3  i+ j 2 2

e = unit vector in the direction making an angle of p /4 with y-axis. So

p⎞  ⎛ p⎞ ⎛ e = ⎜ sin ⎟ i + ⎜ cos ⎟ j ⎝ 4⎠ ⎝ 4⎠

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Chapter 1  • Engineering Mathematics





Thus, directional derivative is

3 3 e ⋅ ∇f = + = 3 2 2   114. (3)  We are given r = xa x + ya y + za z and r = r. Then,

 r  ⇒ div r 2 ∇ (ln r ) = div ( r ) = 3 2 r ⎡ ∂ ⎛ 1⎞ ⎛ x ⎞ ⎤ ⎢∇ (ln r ) = Σ a x ∂x (ln r ) = Σ a x ⎜⎝ r ⎟⎠ ⎜⎝ r ⎟⎠ ⎥ ⎢ ⎥  ⎢ ⎥ 1 r = 2 Σ a x x = 2 ⎢ ⎥ r r ⎣ ⎦

(

∇ (ln r ) =

)

115. (a)  We are given,     P = x 3 yax − x 2 y 2 a y − x 2 yza2   ∇ ⋅ P = 3x 2 y − 2 x 2 y − x 2 y = 0

It is solenoidal,

1. f(x, y, z) is the average of f values over a spherical surface of radius, R, which is centered at (x, y, z). ⎛ 1 ⎞ f ( x, y, z ) = ⎜ ∫ fda ⎝ 4p R 2 ⎟⎠ Sphere

This is for the case of three-dimensional solution. In case of one-dimensional solution, f (x) is the average of f(x + a) and f(x − a). For any a, f ( x) =

a y

∂ ∂ ∇× P = ∂x ∂y 3 x y − x2 y2

Similarly, for two-dimensional solution, the value f (x, y) is the average of values of f on any circle of radius R centered at the point (x, y).

2.  Laplace’s equation does not endure local minima or maxima. In general, the extreme values of f must occur at endpoints or boundary. Thus, we conclude that only the property in option (a) is true.

a z

I = x 15 cos (ω t ) + y 5 sin (ω t )

∂ ∂z − x 2 yz

Therefore, I = (15 cos ω t ) 2 + (5 sin ω t ) 2 = 25 + 200 cos 2 ω t Here, | I | is minimum when cos(ω t) = 0. Hence, q = ωt = 90°.

= a x ( − x 2 y ) − a y ( −2 xyz ) + a z ( −2 xy 2 − x 3 ) ≠ 0 So, P is solenoidal but not irrotational. 116. (1/2)  We are given,

119. (0)  By Green’s theorem, we have

∫ xy dx + x 2

1 ∫C Re( z )dz where C is z = 1 2p j 

2

⎡d ⎤ d y dy = ∫ ∫ ⎢ ( x 2 y ) − ( xy 2 ) ⎥ dx dy dy ⎣ dx ⎦ = ∫ ∫ [2 xy − 2 xy ]dx dy = 0

Put

z = e jθ ⇒ dθ = je jθ dθ 2p

∫ Re(e

120. (d)  It is given that e = (1, 0, 2) = i + 2k 1

jθ

) je jθ dθ

0

1 = 2p j

2p

∫ cos θ j(cos θ + j sin θ )ddθ

e2 = (0, 1, 0) = j e3 = ( −2, 0, 1) = −2i + k u = ( 4, 3, −3) = 4i + 3 j − 3k (1)

0

Let us consider

2p 2p ⎤ j ⎡ 2 θ θ = d + cos cos θ sin θ dθ ⎥ ⎢∫ ∫ 2p j ⎣ 0 0 ⎦

=

11 ⎞ ⎛ 2 u = ⎜ − e1 + 3e2 − e3 ⎟ ⎝ 5 5 ⎠

1 j (p + 0) = 2p j 2

117. (a)  The solution of Laplace function is called harmonic function and has the following properties. These properties are true irrespective of the number of dimensions (one-, two- or three-­dimensions) in which we solve the equation for ∇2 f = 0.

Ch wise GATE_ECE_CH01_Part 2.indd 61

1 [ f ( x + a) + f ( x − a)] 2

118. (90)  We have

a x

1 2p j

61



2 11 = − (i + 2k ) + 3( j ) − ( −2i + k ) 5 5 ⎛ 2  22 ⎞ ⎛ 4 11⎞ = ⎜ − i + i⎟ + 3 j + k ⎜ − − ⎟ ⎝ 5 ⎠ ⎝ 5 5⎠ 5 = 4i + 3 j − 3k (2) As we see, Eqs. (1) and (2) are same. Therefore, option (d) is correct.

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Complex Analysis

Here, n = 2 (pole of order 2) and a = 2. 1⎧d ⎡ 2 ⎫ ⎨ ⎣( z − 2) f ( z )⎤⎦ ⎬ 1! ⎩ dz ⎭z=2

∴ Res f ( 2) =

121. (d)  We have 1 1 = z + 4 ( z + 2i ) ( z − 2i ) 2

⎧⎪ d ⎡ ⎤ ⎫⎪ 1 2 = ⎨ ⎢( z − 2 ) ⎥⎬ dz ( z + 2)2 ( z − 2)2 ⎥⎦ ⎭⎪ z = 2 ⎩⎪ ⎢⎣

Pole (0, 2i) lies inside the circle z − i = 2, whereas pole (0, −2i) lies outside it, as can be seen from the following figure: (0,3)

⎧⎪ d ⎡ 1 ⎤ ⎫⎪ −3 =⎨ ⎢ = ⎡ −2 ( z + 2) ⎤ ⎥ 2 ⎬ ⎣ ⎦z=2 dz ⎩⎪ ⎢⎣ ( z + 2) ⎥⎦ ⎭⎪ z = 2

(0,2)

=

(0,1) (0,0)

124. (d)  We have

(0,−1) (0,−2)

∫

1+ f ( z )

unit circle

z

−2

( 2 + 2)

3

=−

1 32

f ( z ) = c0 + c1 z −1 dz has one pole at origin, which is inside

the unit circle.

∫ f ( z ) dz = 2p i C

[ Residue at these poles which are inside C ]

= 2p i Res f ( 2i ) 1 p = 2p i = (2i + 2i ) 2

So,

∫

⎡⎣1 + f ( z )⎤⎦ dz = 2p j  [Residue f ( z ) at z = 0] z

Since f ( z ) = c0 + c1 z −1 ⇒ f (0 ) = c0

122. (a)  We have I= ∫

1 1 ds =  ∫ ( s + 1)( s − 1) ds ( s 2 − 1)

= 2p j × (Sum of residues) Pole s = −1 is not inside the contour D, but s = 1 is inside D. Residue at pole s = 1 is z = lim s →1

⇒

∫ ( s

2

1 ( s − 1) = ( s − 1)( s + 1) 2

1 1 ds = 2p j × = jp 2 − 1)

2 123. (a) Since lim ⎡( z − 2) f ( z )⎤ is finite and non-zero, ⎦ z→2 ⎣ f ( z ) has a pole of order two at z = 2. The residue at z = a is given for a pole of order n as

Res f ( a ) =

Ch wise GATE_ECE_CH01_Part 2.indd 62

⎫ 1 ⎧d n ⎨ n −1 ⎡⎣( z − a ) f ( z )⎤⎦ ⎬ (n − 1)! ⎩ dz ⎭z=2 n −1

= 2p j ⎡⎣1 + f (0 )⎤⎦



∴   

∫

1+ f ( z ) z

unit circle

dz

= 2p j (1 + c0 )

125. (c)  We have     x ( z ) =

1 − 2z z ( z − 1) ( z − 2)

Poles are z = 0, z = 1 and z = 2. Residue at z = 0 Residue = Value of =

1 − 2z at z = 0 ( z − 1) ( z − 2)

1− 2 × 0 1 = (0 − 1) (0 − 2) 2

Residue at z = 1 Residue = Value of =

1 − 2z at z = 1 z ( z − 2)

1− 2 ×1 =1 (1) (1 − 2)

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Chapter 1  • Engineering Mathematics

Residue at z = 2 1 − 2z at z = 2 Residue = Value of z ( z − 1) =

x = cos

p p + i sin = e 2 2

p i 2

2

=

On integrating, we get

u = x, v = -y ⇒ ux = 1 and v x = 0 u y = 0 and v y = −1 ⇒ ux ≠ v y that is, C-R equations are not satisfied. Thus, z is not analytic. (a)  z = 1 is a simple pole.

127. (c)  We are given that





129. (d)    f ( z ) = z = x − iy

Now, x x = i i = (ep i / 2 )i = e i p / 2 = e − p / 2

f (z) =

= e − y cos xdx − e − y sin xdy = d ⎣⎡e − y sin x ⎤⎦ v = e − y sin x

1− 2 × 2 3 =− 2 ( 2 − 1) 2

1 3 Therefore, the residue of x(z) at its poles are , 1 and − . 2 2 126. (a)  We have x=i Then in polar coordinates, we have

63

Therefore, residue

( z + 3) − 2 ( z + 1) 1 2 − = z +1 z + 3 ( z + 1) ( z + 3)

−z +1 ( z + 1) ( z + 3)

z

( z − 1) at z = 1 is 2

z 1 z = lim = z 2 − 1 z →1 z + 1 2

lim( z − 1) ⋅ z →1

(b) Since z2 is analytic everywhere

Hence, we have poles at −1 and −3, i.e. at (−1, 0) and (−3, 0). From the figure of z + 1 = 1, we see that (−1, 0) is inside the circle and (−3, 0) is outside the circle.

Therefore, using Cauchy’s integral theorem,

∫ z dz = 0 2

C

130. (b)  By Cauchy’s integral formula, f ( z)

∫ ( z − z )

n +1

dz =

0

C

2p if n ( z0 ) n!

∵ f ( z) = 1 ⎛ ⎞ dz 2p i n = × = 0 0 ⎜⎝ f ( z ) = 0 at any z0 ⎟⎠ ∫C ( z − z0 )n+1 n!

(−1,0)

131. (0)  Here, |z|3 is differentiable at the origin, but it is not analytic. However, 2z3 is analytic everywhere. Therefore, b = 0 for f(z) to be analytic. Residue theorem states that 1 esidue of those poles which are ∫C f ( z ) dz = Rinside 2p j  C. 1 ∫C f ( z ) dz is given by the 2p j  residue of function at pole (−1, 0) (which is inside the circle). − ( −1) + 1 2 This residue is = = 1. ( −1 + 3) 2 So the required integral

128. (b)  We have the real part u = e − y cos x dv =

∂v ∂v ∂u ∂u dx + dy = − dx + dy ∂x ∂y ∂y ∂x [Using C-R equations]

Ch wise GATE_ECE_CH01_Part 2.indd 63

132. (1) 

f ( z) =

⎤ sin z ⎛ 1 ⎞ ⎡ ⎛ z 3 ⎞ ⎛ z 5 ⎞ ⎛ z 7 ⎞ = ⎜ 2 ⎟ ⎢ z − ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ + ⎥ 2 ⎝ ⎠ z z ⎣ ⎝ 3! ⎠ ⎝ 5! ⎠ ⎝ 7 ! ⎠ ⎦

3 ⎛ 1⎞ ⎛ z ⎞ ⎛ z ⎞ = ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − ⎝ z ⎠ ⎝ 3 !⎠ ⎝ 5 ! ⎠

1 Reside at z = 0 is the coefficient of ⎛⎜ ⎞⎟ in the expansion ⎝ z⎠ of f (z), that is residue R = 1. 133. (b)  It is given that f ( z) =

⎛ ez ⎞ 1 dz ∫  2p j c ⎜⎝ z − 2 ⎟⎠

The singular point is at z = 2. If z = 2 lies inside the contour, then the value of f (z) is the residue of f (z) at z = 2.

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Therefore,



Coefficient of x2 around x = 0 is a2 =

⎡ e z ( z − 2) ⎤ 2 f ( z ) = lim ⎢ ⎥ = e = 7.39 z →2 ⎣ z−2 ⎦

a2 =

If z = 2 is lies outside of the contour it means the given function is analytic, then the value of f(z) = 0. 134. (b)     f ( z ) =

y

Residue of f (z) at z = a is f(z) (z - a)|z = a 1 at z = 4 is ( z − 4)( z + 1)3

1 ⎛ 1 ⎞ ( z − 4) z = 4 = ⎜ 3 ⎝ z + 1⎟⎠ ( z − 4)( z + 1)

3

z=4

1 = 125

 

f ( z) =

1 with poles at z = -1 is ( z − 4)( z + 1)3



1 ⎛ d2     = ⎜ 2 2 ⎝ dz     =

    =

⎛ 1 ⎞⎞ ⎜⎝ ⎟ z − 4 ⎠ ⎟⎠

z = −1

⎛ t2 ⎞ −⎜ ⎟ ⎝ 2⎠

dt

0

f ′′( x ) =

2

/2

−1 − x 2 e × 2( x ) 2

f ′′( x ) = − xe − x ⇒ f ′′(0) = 0 By Taylor’s series, f (x) at x = 0 is f ( x ) = f (0) + xf ′(0) +

Ch wise GATE_ECE_CH01_Part 2.indd 64

1 1 2 1 . = . p j ( z 2 − 1) 2p j ( z − 1)( z + 1)

Both the poles are encircled by the contour, z = -1 in clockwise and z = 1 in anticlockwise. Using Cauchy’s Integral Theorem we have

∫

c

⎡ 2 ⎡ ⎛ 1 ⎞ f ( z ) = 2p i ⎢ ⎢− ⎜ ⎟ ⎢⎣ 2p i ⎣ ⎝ z − 1⎠

⎛ 1 ⎞ +⎜ ⎝ z + 1⎟⎠ z = −1

⎤⎤ ⎥⎥ z =1 ⎦ ⎥ ⎦

137. (c)  Option (c) gives the correct matching between List I and List II. 138. (b) Here, x0 = 2 We are given

f ( x ) = x3 − x 2 + 4 x − 4

Then,

f ′ ( x ) = 3x 2 − 2 x + 4

Also,

f ( x0 ) = f ( 2) = 8



f ′ ( x0 ) = f ′ ( 2) = 12

⇒

2



x

Numerical Methods

1⎡ 2 ⎤ 1     = ⎢ =− 3⎥ 2 ⎣ ( z − 4) ⎦ z = −1 125

f ′( x ) = e − x

1

z = −1

1 ⎡ ( z − 4) 2 .0 − ( −1)2( z − 4) ⎤ ⎥ ⎢ 2⎣ ( z − 4) 4 ⎦ z = −1

135. (0)     f ( x ) = ∫ e

-1

⎡ ⎛ −1⎞ ⎛ 1 ⎞ ⎤ = 2 ⎢− ⎜ ⎟ + ⎜ ⎟ ⎥ = 2 ⎣ ⎝ 2 ⎠ ⎝ 2⎠ ⎦

1 ⎛ d ⎛ −1 ⎞ ⎞ 2 ⎜⎝ dz ⎜⎝ ( z − 4) 2 ⎟⎠ ⎟⎠

x

C2

C2 → anticlockwise contour (+ve)

1 ⎛ d n −1 ⎞ [( z − a) n f ( z )]z = a ( n − 1)! ⎜⎝ dz n −1 ⎟⎠

⎞⎤ 1 ⎡ d2 ⎛ 1 × ( z + 1)3 ⎟ ⎥ ⎢ 2⎜ 3 2 ! ⎣ dz ⎝ ( z − 4)( z + 1) ⎠ ⎦ z = −1

C1

Let C1 → clockwise contour (-ve)

Residue of f (z) with multiple pole at z = a with order n is

Residue of

0 =0 2!

136. (2)

1 ( z − 4)( z + 1)3

So, residue of f ( z ) =

f ′′(0) . So 2!

2

x f ′′(0) +  2!

x1 = x0 −

f ( x0 )

f ′ ( x0 )

= 2−

8 4 = 12 3

139. (c)  The given equation is x = e− x The equation can be rewritten as f ( x) = x − e− x = 0 f ′ ( x) = 1 + e− x

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Chapter 1  • Engineering Mathematics

The Newton–Raphson iterative formula is x n +1 = x n −



Here

f ′ ( xn ) = 1 + e







x n +1 = x n −

f ′ ( xn )

xn − e − xn e − xn xn + e − xn = 1 + e − xn 1 + e − xn

e − xn = (1 + xn ) 1 + e − xn

Here, x0 = 0, y0 = y( x0 ) = y(0) = 0 x1 = x0 + h = 0 + 0.1 = 0.1



dy f ( x, y ) = = y+x dx ⇒

y1 = y0 + h f ( x0 , y0 ) = 0 + 0.1 × f (0, 0 ) = 0 + 0.1 × (0 + 0 ) = 0

Now,  x1 = 0.1, y1 = 0

Now f ′ ( x ) = 1 + ⇒ f ′ ( x0 ) = 1 +

1 2 x 1 2 2

Then, x1 = x0 −

f ( x0 )

f ′ ( x0 )

⇒

= 2−

2 −1 1 1+ 2 2

⇒      x1 = 1.694 142. (b) 143. (2.4903)  We are given f ( x) = x3 − 5x 2 + 6 x − 8 x0 = 5 f ′( x ) = 3 x 2 − 10 x + 6 By Newton–Raphson method, x1 = x0 −

f ( x0 ) f (5) 22 = 5− = 5− f ′( x0 ) f ′(5) 31

= 5 − 0.7097 = 2.4903 144. (c)  Let y = sinx and y = x/2 be two curves. The solutions of sinx = x/2 are the points ­intersected these two curves.

    x2 = x0 + 2h = 0 + 2 × 0.1 = 0.2 ⇒ y2 = y1 + h f ( x1 , y1 ) = 0 + 0.1 × f (0.1, 0 ) = 0 + 0.1(0.1 + 0 ) = 0.01        Now, x2 = 0.2, y2 = 0.01

f ′ ( xn )

   f ( x0 ) = 2 + 2 − 3 = 2 − 1

140. (b)  We have dy − y = x , y ( 0) = 0 dx Step size = h = 0.1 Euler’s first-order formula is yi +1 = yi + h f ( xi , yi ) y1 = y0 + h f ( x0 , y0 )

f ( xn )

Putting x = 2, we get

− xn

∴ The Newton–Raphson iterative formula is given by x n +1 = x n −

Using Newton–Raphson method, we have

f ( xn )

f ( xn ) = xn − e − xn

y=

−2π −π

0

π

x 2

2π

x3 = x0 + 3h = 0 + 3 × 0.1 = 0.3 y3 = y2 + h f ( x2 , y2 )

65

y = −1 y = sin x

From the graph shown here, we see that the intersection = 0.01 + 0.1 × f (0.2, 0.01) = 0.01 + 0.1(0.2 + 0.01) = 0.031       is at three points. = 0.01 + 0.1 × f (0      .2, 0.01) = 0.01 + 0.1(0.2 + 0.01) = 0.031 145. (0.0628)  It is given that Therefore, at x3 = 0.3, y3 = 0.031. y1 = y + 2x − x2 ⇒ f(x, y) = y + 2x − x2 Hence, the correct answer is option (b). From Runge−Kutta second-order method, we have 141. (c)  We have 1 y1 = y0 + ( k1 + k2 ) f (x) = x + x − 3 = 0 2

Ch wise GATE_ECE_CH01_Part 2.indd 65

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where k1 = h f (x0, y0) and k2 = h f (x0 + h, y0 + k1). It is given that x0 = 0, y0 = 1, h = 0.1. Then, k1 = (0.1)(y0 + 2x0 − x20)  = (0.1)(1) = 0.1

147. (c) Px (x) is a continuous random variable, then the probability that X takes on any particular event is zero. 148. (a)  d ( n) = x( n) − x( n −1)

k2 = 0.1f(0 + 0.1, 1 + 0.1)

Squaring both sides and taking expectation both sides, we get

= (0.1)f(0.1, 1.1) = (0.1)(1.1 + 2(0.1) − 0.01) = 0.129

E[d ( n)]2 = E[ x( n) − x( n − 1)]2

Therefore, 1 y1 = y(0.1) = y0 + (0.1 + 0.129) = 1.1145  2 However, the exact solution is

Probability and Statistics

(1)

E[d ( n)]2 = E[ x( n)]2 + E[ x( n − 1)]2 − 2 E[ x( n) ⋅ x( n − 1)]

σ d2 = σ x2 + σ x2 − 2 Rxx (1) Rxx (τ ) = E [ x( n) ⋅ x( n − τ )]

y(x) = x2 + ex

y(0.1) = (0.1)2 + e0.1 = 1.1152

(2)

The percentage difference is 1 ⎛ 1.1152 −1.1145 ⎞ = 0.0628 ⎜⎝ ⎟⎠ % = 6.28 × 1.1152 100

Given σ d2 =

σ x2 10

σ x2 = σ x2 + σ x2 − 2 Rxx (1) 10 2 Rxx (1) =

146. (0.68)  F ( x ) = x 3 + x − 1 = 0 (1) By Newton–Raphson’s iterative method, we have

      xn +1

F ( xn ) (2) = xn − F ′ ( xn )

Given x0 = 1 ⇒ F ( x0 ) = 1 + 1 − 1 = 1

19 2 σx 10

Rxx (1) 19 = = 0.95 10 σ x2 149. (d)  Probability of getting an odd number Po = 3/6 =1/2 Probability of getting an even number Pe = 3/6 = 1/2 As both events are independent of each other,

F ′( x ) = 3 x 2 + 1

P( o / e ) =

F ′( x0 ) = 3 + 1 = 4 So, from Eq. (2), we have

150. (a)  We have

3

⎛ 3⎞ ⎛ 3⎞ F ( x1 ) = ⎜ ⎟ + ⎜ ⎟ − 1 = 0.1718 ⎝ 4⎠ ⎝ 4⎠

Comparing with

∞

x2 = x1 −

F ( x1 ) F ′( x1 )

3 0.1718 x2 = − = 0.686 4 2.6875

Ch wise GATE_ECE_CH01_Part 2.indd 66

2

∞

1

σ 2p

2∫ 0

1

σ 2p

∫e

( x − μ )2 2σ

2

dx = 1

−∞

e

−

x2 2σ

2

dx = 1

Given, −

Therefore,

(− x / 8) dx

We can provide m and s any value. Here, putting m = 0

2

⎛ 3⎞ ⇒ F ′( x1 ) = 3 ⎜ ⎟ + 1 = 2.6875 ⎝ 4⎠

∫e 2p 0

1 3 = 4 4

Therefore, from Eq. (1),

∞

1

I=

F ( x0 ) x1 = x0 − F ′( x0 ) x1 = 1 −

1 1 1 × = 2 2 4

x2 x2 =− 2 8 2σ

⇒         s = 2, Now putting s = 2 in the above equation, we get ∞

∫ 0

1 2p

e

−

x2 8

=1

12/4/2018 10:40:21 AM

Chapter 1  • Engineering Mathematics ∞

151. (c)      

∫

p( x ) dx = 1

−∞ ∞

∫ Ke

−α x

dx = 1

−∞ 0

∫

−∞

⇒

∞

K eα x dx + ∫ K e − α x dx = 1 0

( )

K αx e α

0 −∞

⇒

+

(

K −α x e −α

)

∞

0

156. (c)  In the first toss, results can be 1, 2, 3, 4, 5. For 1, the second toss results can be 2, 3, 4, 5, 6. For 2, the second toss results can be 3, 4, 5, 6. For 3, the second toss results can be 4, 5, 6. For 4, the second toss results can be 5, 6. For 5, the second toss results can be 6. The required probability =

=1

K K + =1 α α 2K = α

67

1 5 1 4 1 3 1 2 1 1 5 ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = 6 6 6 6 6 6 6 6 6 6 12

157. (b)  −1< x < 1 and −1 ≤ y ≤ 1 is the entire rectangle. 1 The region in which maximum of (x, y) is less than is 2 shown below as the shaded region inside the rectangle.

⇒ K = 0.5α 152. (d)  S → supply by Y, d → defective Probability that the computer was supplied by Y, if the product is defective P (S ∩ d ) P ( S /d ) = P (d )

y 1

(−1, 1)

1/2 −1

1/2

x

P ( d ) = 0.6 × 0.1 + 0.3 × 0.02 + 0.1 × 0.03 = 0.015 0.006 = 0.4 0.015

153. (c) Let A denote the event of failing in Paper 1 and B denote the event of failing in Paper 2. Then we are given that P ( A) = 0.3

−1

(−1, −1)

Probability of failing in both, P ( A ∩ B ) = P ( B ) ∗ P ( A/B ) = 0.2 ∗ 0.6 = 0.12    154. (c)  Probably of only the first two tosses being heads = P(H, H, T, T, T, …, T) As each toss is independent, the required ­probability = P ( H ) × P ( H ) × [ P (T )]8 2

2

⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ =⎜ ⎟ ⎜ ⎟ =⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠

10

155. (d)  Coin is tssed 4 times. Hence, the total outcomes = 24 = 16. Favorable outcomes (the number of times heads show up is more than the number of times tails show up) = {HHHH, HHTH, HHHT, HTHH, THHH} = 5 5 Hence, the required probability = . 16

Ch wise GATE_ECE_CH01_Part 2.indd 67

(1, −1)

1⎞ Area of shaded region ⎛ P ⎜ max x, y < ⎟ = ⎝ 2 ⎠ Area of entire rectangle 3 3 × 9 = 2 2= 2 × 2 16

P ( B ) = 0.2

P ( A/B ) = 0.6

1

0

P ( S ∩ d ) = 0.3 × 0.02 = 0.006 P ( S /d ) =

(1, 1)

158. (c)  Probability that the number of toss is odd = Probability of the number of toss is 1, 3, 5, 7, … Probability that the number of toss is 1 = Probability of getting head in first toss = 1/2 Probability that the number of toss is 3 = Probability of getting tail in first toss, tail in second toss and head in third toss =

1 1 1 1 × × = 2 2 2 8

Probability that the number of toss is 5 = P(T, T, T, T, H) 5

1 ⎛ 1⎞ = ⎜ ⎟ = ⎝ 2⎠ 32 So, probability that the number of tosses are odd =

1 1 1 + + + 2 8 32

12/4/2018 10:40:23 AM

68

GATE ECE Chapter-wise Solved papers

Sum of infinite geometric series with 1 1 1 and r is = 2 = a= 1 2 4 1− 4

1 2 =2 3 3 4

159. (0.667) Let E1 = one child family, E          2 = two children family and A = picking a child. Then by Bayes’ theorem, required probability is given by 1 x 2 2 P ( E2 / A) = = = 0.667 1.x 1 3 + x 2 2 2 where x is number of families. 160. (0.33)  Given that three random variables X1, X2, and X3 are uniformly distributed on [0, 1], we have the following possible values: X1 > X2 > X3 X1 > X3 > X2 X2 > X1 > X3 X2 > X3 > X1 X3 > X1 > X2 X3 > X2 > X1 Since all the three variables are identical, the probabilities for all the above inequalities are same. Hence, the probability that X1 is the largest is P{X1 is the largest} = 2/6 = 1/3 = 0.33 161. (50)  X = 1, 3, 5,..., 99 ⇒ n = 50 (number of observations) Therefore 1 n 1 E ( X ) = ∑ X i = [1 + 3 + 5 +  + 99] 50 n i= j =

1 2 ⋅ (50 ) = 50 50

162. (c)  P[fourth head appears at the tenth toss] = P[getting 3 heads in the first 9 tosses and one head at tenth toss]

∞

E ( x ) = ∑ xp ( x ) = 1 × x =1

Let, S = 1×

⇒



Event (After getting the first H)

x

Probability, p(x)

T

1

1/2

HT

2

HHT and so on …

Ch wise GATE_ECE_CH01_Part 2.indd 68

3

1/2 × 1/2 = 1/4 1/8

1 1 1 + 2 × + 3 × +  (1) 2 4 8

1 1 1 1 S = + 2 × + 3 × +  (2) 2 4 8 16

Subtracting Eq. (2) from Eq. (1) we get 1 1 1 1 ⎛ 1⎞ ⎜⎝1 − ⎟⎠ S = + + + +  2 2 4 8 16 1 ⇒ S= 2





1 2 1−

1 2

= 1 ⇒ S = 2 ⇒ E (x) = 2

Hence, the expected number of tosses (after first head) to get first tail is 2 and same can be applicable if first toss results in tail. Thus, the average number of tosses is 1 + 2 = 3.

164. (0.16) Given x1, x2 and x3 be independent and identically distributed with uniform distribution on [0, 1]. Let z = x1 + x2 - x3. Then P { x1 + x2 ≤ x3 } = P { x1 + x2 − x3 ≤ 0} = P { z ≤ 0}



Let us find probability density function of random variable z. Since z is summation of three random variables x1 , x2 and −x3 , overall PDF of z is convolution of the PDF of x1 , x2 and −x3 .

PDF of {x1 + x2} is

9 ⎡ ⎛ 1 ⎞ ⎤ ⎡ 1 ⎤ 21 = ⎢ 9C3 ⋅ ⎜ ⎟ ⎥ ⎢ ⎥ = = 0.082 ⎝ 2 ⎠ ⎥⎦ ⎣ 2 ⎦ 256 ⎢⎣

163. (3)  Let the first toss be head. Let x denotes the number of tosses (after getting first head) to get first tail. We can summarize the event as:

1 1 1 + 2 × + 3 × + 2 4 8

1

0

1

2

PDF of {−x3} is 1

−1 P { z ≤ 0} =

0

∫

−1

( z + 1)2 dz = ( z + 1)3 2

6

0

= −1

1 = 0.16 6

12/4/2018 10:40:25 AM

Chapter 1  • Engineering Mathematics

171. (0.25)  We are given,

165. (0.8) 1

f (x) =

2p

e− x

2

/2

, −∞ < x < ∞

y (t ) =

E{ x } =

∞

−∞

= =

1 2p

∞

x 2 ∫ xe − x / 2 dx = 2

0

2 2p

∞

−u ∫ e du





Mean = Variance = 2

2 = 0.797 = 0.8 p

(a)  It is lost by the first post office (b) It is passed by first post office but lost by the second post office 1 4 1 9 P(parcel is lost) = + × = 5 5 5 25 P(parcel lost by second post office if it passes first post office) = P(Parcel passed by first post office) × P(Parcel 4 1 4 lost by second post office) = × = 5 5 25 P(parcel lost by second post office | parcel lost) = 4/25 4 = = 0.44 9/25 9

However, we know that variance is σ2 = m2 − m21 (1) where m2 is the second central moment and m1 is the mean. From Eq. (1), we get

λ = 2 − λ2 λ2 + λ − 2 = 0 ⇒ (λ + 2) (λ − 1) = 0 Therefore,

λ = −2 or λ = 1 Since, variance (λ) is positive, λ = 1. 173. (0.33)  We have 1

P ( x + y ≤ 1) = ⇒



P(Tail)⋅P(Tail)⋅P(Tail)⋅P(Tail)⋅P(Head) = (0.7)4 (0.3) = 0.072

V ( x ) = E ( x 2 ) − {E ( x )}2

∫

∞

1 x ∫−∞ 2 x e xdx = 0

( ∵ the function is odd) E( x2 ) =

∞

∫

x 2 f ( x )dx =

−∞

=

∞

∫x

−∞

= 0.33 0

The probability of getting ‘head’ for the first time in the fifth toss:

1 x e − x is probability density 2 function of random variable X.

−∞

1

P(Head) = 0.3 = p and P(Tail) = 0.7 = q

168. (6)  Given that f ( x ) =

xf ( x )dx =

∫ ∫

1− x

⎛ y2 ⎞ ( x + y )dx dy = ∫ ⎜ xy + ⎟ dx 2 ⎠0 x=0 ⎝ 1

174. (0.072)  We have

Thus, (a), (b) and (d) are correct. Hence, (c) is false.

∞

1 1− x

x x3 = − 2 6

Also, if A and B are independent then A and B are also independent, that is,

E ( x) =

f xy ( x, y ) dx dy

1 ⎛ (1 − x ) 2 ⎞ dx = ∫ ⎜ x(1 − x ) + 2 ⎟⎠ 0⎝

A and B are independent then P ( A/B) = P ( A) and P ( B / A) = P ( B ).

P ( A ∩ B ) = P ( A) ⋅ P ( B )

∫ ∫

x=0 0

P ( A ∩ B) = P ( A) ⋅ P ( B)



1

x=0 y=0

167. (c)  We know that A and B are independent, then

n

172. (1)  For this Poisson distribution,

0

166. (0.44)  Parcel will be lost if



∑X

⎛ 3T ⎞ ⎡ 3p / 4 ⎤ 1 = = 0.25 Ryy ⎜ ⎟ = ⎢1 − ⎝ 4⎠ ⎣ p ⎥⎦ 4

x f (x)

∫

⎡ τ⎤ p(t − nT − ϕ ) Ryy ( z ) = ⎢1 − ⎥ ⎣ T⎦

∞

n = −∞

Therefore



69

2

1 x e − x dx 2

175. (0.25) Since X1, X2, X3 and X4 are independent normal random variables with zero mean and unit variance, we have

P(X1 = smallest) = P(X2 = smallest) = P(X3 = smallest) = 1 P(X4 = smallest) = = 0.25. 4

∞

2 3 −x x e dx (∵ function is even) 3 ∫2

= 3! = 6

Ch wise GATE_ECE_CH01_Part 2.indd 69

12/4/2018 10:40:27 AM

Ch wise GATE_ECE_CH01_Part 2.indd 70

12/4/2018 10:40:27 AM

Networks, Signals and Systems

CHAPTER2

Syllabus Network solution methods: nodal and mesh analysis; Network theorems: superposition, Thevenin and Norton’s, maximum power transfer; Wye-Delta transformations; Steady state sinusoidal analysis using phasors; Time domain analysis of simple linear circuits; Solution of network equations using Laplace transform; Frequency domain analysis of RLC circuits; Linear 2-port network parameters: driving point and transfer functions; State equations for networks.   Continuous-time signals: Fourier series and Fourier transform representations, sampling theorem and applications; Discrete-time signals: discrete-time Fourier transform (DTFT), DFT, FFT, z-transform, interpolation of discrete-time signals; LTI systems: definition and properties, causality, stability, impulse response, convolution, poles and zeros, parallel and cascade structure, frequency response, group delay, phase delay, digital filter design techniques.

Chapter Analysis Topic

GATE 2009

GATE 2010

Network solution methods Network theorems

1 2

GATE 2011

GATE 2012

GATE 2013

1

Time domain analysis of simple linear circuits

1

Solution of network equations using Laplace transform

3

2

1

Continuous-time signals

2

Discrete-time signals

1

1

LTI Systems

3

2

2

GATE 2017

GATE 2018 1

2

4

6

5

3

1

2

2

3

3

1

1

1

2

1

1

2

6

2

5

3

3

2

1

4

1 2

2

3

Frequency domain analysis of RLC circuits Linear 2-port network parameters

GATE 2016

1

1 2

GATE 2015

2

Wye-Delta transformation Steady state sinusoidal analysis using phasors

GATE 2014

1

1

1

1

3

1

1

2

1

5

2

3

1

1

1

1

2

3

2

3

3

3

5

7

5

1

9

6

1 3

2

3

2 1

1

Important Formulas 1.

The equivalent resistance Req of n resistors in series is

3.

Req = R1 + R2 +  + Rn 2.

The equivalent resistance Req of n resistors in parallel is 1 1 1 1 = + ++ Req R1 R2 Rn

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 71

The equivalent capacitance Ceq of n capacitors in series is 1 1 1 1 = + ++ Ceq C1 C2 Cn

4.

The equivalent capacitance Ceq of n capacitors in parallel is Ceq = C1 + C2 +  + Cn

12/4/2018 11:05:01 AM

72 5.

GATE ECE Chapter-wise Solved papers

The equivalent inductance Leq of n inductors in series is Leq = L1 + L2 +  + Ln

6.

The equivalent inductance Leq of n inductors in parallel is 1 1 1 1 = + ++ Leq L1 L2 Ln

7.

T to p -Transformations: ZA =

Z1 Z 2 + Z 2 Z3 + Z3 Z1 Z2

Z Z + Z 2 Z3 + Z3 Z1 ZB = 1 2 Z3 ZC = 8.

Z1 Z 2 + Z 2 Z3 + Z3 Z1 Z1

p to T-Transformations: ZAZB Z1 = Z A + Z B + ZC

9.

Z2 =

Z B ZC Z A + Z B + ZC

Z3 =

Z A ZC Z A + Z B + ZC

if the load impedance ZL is a complex conjugate of the impedance Z of the network. 13. Thevenin’s theorem: A linear two-terminal circuit containing energy sources and impedances can be replaced by an equivalent circuit consisting of a voltage source VTH in series with an impedance ZTH, where VTH is referred to as the Thevenin’s equivalent voltage and is the open-circuit voltage at the terminals and ZTH is referred to as the Thevenin’s equivalent impedance and is the input or equivalent impedance at the terminals when the independent sources are made inactive. 14. Norton’s theorem: It states that a linear two-terminal network can be replaced by an equivalent circuit consisting of a current source iN in parallel with impedance ZN, where iN is the short-circuit current through the terminals and ZN is the input or equivalent impedance at the terminals when the independent sources are turned off. 15. Reciprocity theorem: It states that in a linear bilateral single source circuit, the ratio of excitation to response is constant when the positions of excitation and response are interchanged. 16. According to Millman’s theorem n

E=

Mesh analysis: ⎡ Z11 ⎢Z ⎢ 21 ⎢  ⎢ ⎣ Z m1

Z12 Z 22  Zm2

E1Y1 + E2Y2 +  + EnYn = Y1 + Y2 +  + Yn

∑EY i =1 n

∑Y i =1

Z13 Z 23  Zm3

.... Z1m ⎤ ⎡ I1 ⎤ ⎡ V1 ⎤ .... Z 2 m ⎥⎥ ⎢⎢ I 2 ⎥⎥ ⎢⎢V2 ⎥⎥ = or   ⎥⎢  ⎥ ⎢  ⎥ ⎥⎢ ⎥ ⎢ ⎥ .... Z mm ⎦ ⎣ I m ⎦ ⎣Vm ⎦

[ Z ][ I ] = [V ] 10. Nodal analysis: ⎡ Y11 Y12 Y13 ⎢Y ⎢ 21 Y22 Y23 ⎢    ⎢ Y Y Y m3 ⎣ m1 m 2

.... Y1m ⎤ ⎡ V1 ⎤ ⎡ I1 ⎤ .... Y2 m ⎥⎥ ⎢⎢V2 ⎥⎥ ⎢⎢ I 2 ⎥⎥ =   ⎥ ⎢  ⎥ ⎢  ⎥ or ⎥⎢ ⎥ ⎢ ⎥ .... Ymm ⎦ ⎣Vm ⎦ ⎣ I m ⎦

[Y ][V ] = [ I ] 11. Superposition theorem: In any linear circuit containing multiple independent sources, the current or voltage at any point in the network may be calculated as algebraic sum of the individual contributions of each source acting alone. 12. Maximum power transfer theorem: The maximum power will be delivered by a network to a load impedance ZL,

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 72

and    Z =

1 = Y1 + Y2 +  + Yn

i i

i

1 n

∑Y i =1

i

17. Delta-to-Wye transformation and Wye-to-delta transformations: (i)  Delta-to-Wye transformation R1 =

Rb Rc Ra + Rb + Rc

R2 =

Rc Ra Ra + Rb + Rc

R3 =

Ra Rb Ra + Rb + Rc

(ii)  Wye-to-delta transformation Ra = R2 + R3 +

R2 R3 R1

Rb = R1 + R3 +

R1 R3 R2

Rc = R1 + R2 +

R1 R2 R3

12/4/2018 11:05:03 AM

Chapter 2  •  Networks, Signals and Systems

18. Series RC steady-state sinusoidal response:

28. The time constant of an RC circuit is

τ = RC

2

⎛ 1 ⎞ υ (t ) = I R + ⎜ sin (ω t − θ ) or ⎝ ω C ⎟⎠

73

2

i (t ) =

V R 2 + (1/ω C ) 2

29. The time constant of an RL is

τ=

sin(ω t + θ )

19. Series RL steady-state sinusoidal response:

30. The resonant frequency for a series and parallel RLC ­circuits is

υ (t ) = I R 2 + (ω L) 2 sin(ω t + θ ) or i (t ) =

V R 2 + (ω L) 2

sin(ω t − θ )

20. Impedance:      V V ∠ϕ V =  ∠(ϕ − ψ ) Z =  and Z ∠θ =  I ∠ψ I I    21. If n impedances Z1 , Z 2 , … , Z n are connected in series, then     Z eq = Z1 + Z 2 +  + Z n    22. If n impedances Z 1 , Z 2 , …, Z n are connected in parallel, then 1 1 1 1  =  +  +  +  Z eq Z1 Z 2 Zn

L R

ωo =

1 LC

31. For a series RLC circuit, the quality factor is Qo =

ωo L 1 1 L = = R ω oCR R C

32. For a parallel RLC circuit, the quality factor is Qo =

C R = ω oCR = R ωo L L

33. Laplace transform X(s) of a signal x(t) is X ( s) =

∞

∫ x (t )e

− st

dt

−∞

34. For the parallel interconnection of two systems,

  23. Admittance (Y ) is the reciprocal of the impedance ( Z ).

h(t) = h1(t) + h2(t) and H(s) = H1(s) + H2(s)

 1 1 Y =  =  ∠ - q Z Z    24. If n admittances Y 1, Y 2, …, Y n are connected in series, then

35. For the series interconnection of two systems,

1 1 1 1  =  +  +  +  Yeq Y1 Y 2 Yn    25. If n admittances Y 1, Y 2, …, Y n are connected in ­parallel, then     Y eq = Y1 + Y1 +  + Yn 26. The voltage across a capacitor in an RC circuit is Vo ⎧ υ (t ) = ⎨ − t / RC ⎩Vs + (Vo − Vs )e

for t < 0 for t ≥ 0

27. The current through an inductor in an RL circuit is Io ⎧ ⎪ i(t ) = ⎨Vs ⎛ Vs ⎞ − tR / L ⎪ R + ⎜⎝ I o − R ⎟⎠ e ⎩

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 73

for t < 0 for t ≥ 0

h(t) = h1(t)*h2(t) and H(s) = H1(s)H2(s) 36. For the feedback interconnection of two systems, H ( s) =

H1 ( s ) 1 + H1 ( s ) H 2 ( s )

37. The inverse Laplace transform of X(s) is σ + jω

x (t ) =

1 X ( s)e st ds 2π j σ −∫jω

, z12 =

V1 I2

, z22 =

V2 I2

I1 = 0

I1 I , y = 1 V1 V2 = 0 12 V2

V1 = 0

38. z11 =

V1 I1

z21 =

V2 I1

39. y11 =

I2 = 0

I2 = 0

I1 = 0

12/4/2018 11:05:06 AM

74

GATE ECE Chapter-wise Solved papers

I y21 = 2 V1 40. h11 =

h21 =

41. g11 =

V2 = 0

I1 V1

I2 = 0

, h22 =

, g12 =

50. [IL] is [Ib] = [B]T[IL]

V1 = 0

51. Bt = [At-1Al]T

V1 V , h12 = 1 I1 V2 = 0 V2 I2 I1

49. [Vb] = [Q]T[Vn] and [Vb] = [A]T[Vn]

I , y22 = 2 V2 V2 = 0

52. Ql = At-1Al

I1 = 0

53. BQT = QBT = 0

I2 V2

I1 = 0

I1 I2

V1 = 0

54. The Fourier series representation of a continuous-time signal x(t) is x (t ) =

∞

∑ce

k = −∞

V g 21 = 2 V1 42. A =

V1 V2

I C= 1 V2 V 43. A′ = 2 V1 C′ =

I2 V1

V , g 22 = 2 I2 I2 = 0

I2 = 0

, B=−

V1 I2

∞

∑ce

k = −∞

jk ( 2π / T ) t

k

56. The continuous-time Fourier transform X( j w) of a signal x(t) is

V2 = 0

∞

∫ x (t )e

X ( jω ) = V1 = 0

I2 I1

V1 = 0

I1 = 0

=

1 x(t ) rms = a02 + ( a12 + a22 + a32 +  + b12 + b22 + b32 + ) 2

V , B′ = − 2 I1 I1 = 0 , D′ = −

jk ω 0 t

55. The rms value of a signal x(t) in terms of Fourier series coefficients is

V1 = 0

V2 = 0

I , D=− 1 I2 I2 = 0

k

− jω t

dt

−∞

57. The inverse Fourier transform of X( j w) is x (t ) =

44. The voltage transfer function is G21 ( s) =

V2 ( s) V1 ( s)

∞

1 2π

∫ X ( jω )e

M

H ( jω ) =

∑ b ( jω ) k =0 N

k =0

V2 ( s) I1 ( s )

48. The number of trees of a graph are T = det{[A][A] � T

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 74

k

k

59. For band-limited signals, f S ≥ 2 f M . 60. For band-pass signals, f S ≥ integer not exceeding

47. The impedance transfer function is Z 21 (s) =

k

k

∑ a ( jω )

46. The admittance transfer function is I ( s) Y21 (s) = 2 V1 ( s)

dω

−∞

58. The transfer function H( jw) of a continuous-time LTI system

45. The current transfer function is I ( s) α 21 (s) = 2 I1 ( s )

jω t

fu . fB

2 fu , where k is the largest k

61. The Fourier series representation of a discrete-time periodic signal x[n], having fundamental period, N, is x[n] =

∑ae

k= N

k

jk ω 0 n

=

∑ae

k= N

jk ( 2π / N ) n

k

12/4/2018 11:05:09 AM

Chapter 2  •  Networks, Signals and Systems

62. Discrete-time Fourier transform X(ejw) of a signal x[n] is +∞

∑ x[n]e

X ( e jω ) =

69. For a continuous-time LTI system,

− jω n

y (t ) =

n = −∞

∫ X (e

jω

)e

jω n

dω

2π

70. The commutative property of a discrete-time LTI system is +∞

∑ h[k ]x[n − k ]

k = −∞

71. The commutative property of a continuous-time LTI system is

N −1

X ( k ) = ∑ x[n]WN kn n= 0

65. The z-transform of a discrete-time signal x[n] is +∞

∑ x[n]z

x(t ) * h(t ) = h(t ) * x(t ) =

+∞

∫ h(τ ) x(t − τ )dτ

−∞

−n

72. The distributive property of a discrete-time LTI systems is

n = −∞

66. The inverse z-transform of X(z) is given by 1 n −1 x[n] = ∫ X ( z ) z dz. 2π j 

x[n] * ( h1 [n] + h2 [n]) = x[n] * h1 [n] + x[n] * h2 [n] 73. The distributive property of a continuous-time LTI systems is

67. The unilateral z-transform of a discrete-time signal x[n] is +∞

X ( z ) = ∑ x[n]z − n .

x(t ) * [ h1 (t ) + h2 (t )] = x(t ) * h1 (t ) + x(t ) * h2 (t ) 74. The associative property of a discrete-time LTI systems is

n= 0

68. For a discrete-time LTI system, y[n] =

∫ x(τ )h(t − τ )dτ = x(t ) * h(t )

x[n] * h[n] = h[n] * x[n] =

64. DFT of a sequence x[n]

X ( z) =

+∞

−∞

63. The inverse of discrete-time Fourier transform 1 x[n] = 2π

75

x[n] * ( h1 [n] * h2 [n]) = ( x[n] * h1 [n]) * h2 [n]

+∞

∑ x[k ]h[n − k ] = x[n] * h[n]

k = −∞

IMPORTANT tABLES Table 1 |  Common Laplace transform pairs along with their region of convergence. S. No.

Signal

Transform

ROC

 1.

δ(t)

1

All s

 2.

u(t)

1 s

Re{s� > 0

 3.

−u(−t)

1 s

Re{s� < 0

 4.

t n−1 u(t ) (n −1)!

1 sn

Re{s� > 0

t n−1 u( − t ) (n −1)!

1 sn

Re{s� < 0

e − α t u (t )

1 s+α

Re{s� > −α

 5.   6.

−

(Continued )

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 75

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S. No.

Signal

Transform

ROC

 7.

− e − α t u( −t )

1 s+α

Re{s� < −α

 8.

t n −1 − α t e u (t ) ( n − 1)!

1 ( s + α )n

Re{s� > −α

t n −1 − α t e u( −t ) ( n − 1)!

1 ( s + α )n

Re{s� < −α

−

 9. 10.

δ (t − T )

e − sT

All s

11.

[cos w0t]u(t)

1 s + ω 02

Re{s� > 0

12.

[sin w0t]u(t)

ω0 s 2 + ω 02

Re{s� > 0

13.

[ e − α t cos w0t]u(t)

s+α ( s + α ) 2 + ω 02

Re{s� > −α

14.

[ e − α t sin w0t]u(t)

ω0 ( s + α ) 2 + ω 02

Re{s� > −α

sn

All s

1 sn

Re{s� > 0

15. 16.

un (t ) =

2

d n δ (t ) dt n

u− n ( t ) = u ( t ) *  * u ( t )    n times

Table 2 |   Properties of Laplace transform. Property

Signal**

Laplace Transform

ROC

Linearity

ax1(t) + bx2(t)

aX1(s) + bX2(s)

At least R1 ∩ R2

Time shifting

x(t − t0)

e − st0 X ( s)

R

− s0 t

x (t )

X(s − s0)

Shifting in the s-domain

e

Time scaling

x(at)

1 ⎛ s⎞ X⎜ ⎟ a ⎝ a⎠

Conjugation

x *(t )

X *( s*)

R

Convolution

x1 (t ) * x2 (t )

X 1 ( s) X 2 ( s)

At least R1 ∩ R2

Differentiation in the time domain

d x (t ) dt

sX ( s)

At least R

Differentiation in s-domain

−tx(t )

d X ( s) ds

R

1 X ( s) s

At least R1 ∩ {Re(s) > 0�

t

Integration in the time domain

∫ x(τ )dτ

−∞

Shifted version of R (i.e., s is in the ROC if s − s0 is in R) Scaled ROC (i.e., s is in the ROC if s/a is in R)

(Continued )

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Chapter 2  •  Networks, Signals and Systems

Property

Signal**

Laplace Transform

Initial value theorem

x(0+) = lim sX ( s)

Final value theorem

lim x(t ) = lim sX ( s)

77

ROC

s →∞

t →∞

s→0

** Let x1(t), x2(t) and x(t) be with Laplace transforms X1(s), X2(s) and X(s) with ROC R1, R2 and R, respectively. Table 3 |  Time domain and s-domain representation of basic network elements. Time Domain

s-Domain Voltage/ Current Expression

s-Domain

i

i R

R i

V ( s) = RI ( s)

I(s)

i(0+)

L

−+

sL

i

I ( s) =

Vo ( s) i(0 + ) + sL s

I(s)

i(0+)

L

Li(0+)

I(s) + + Vo(s) i(0 )/s −

sL

V ( s) = sL I ( s) − Li(0 + )

+−

sL I ( s) + Li(0 + )

sL

Li(0+) I(s)

sL

i

C +

I(s) −

+ Vo−

i

C +

+ Vo(s) −

−

+

1/sC − +−

Vo/s I(s) CV(0+) + I/sC Vo(s) −

I(s) 1/sC + − −+

− Vo +

Vo/s I(s) CV(0+) I/sC

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 77

i(0+)/s

I ( s) =

Vo ( s) i(0 + ) − sL s

V ( s) =

I ( s) Vo + sC s

I ( s) = SCVo ( s) − CV (0 + )

V ( s) =

I ( s) Vo − sC s

I ( s) = SCVo ( s) + CV (0 + )

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Table 4 |  Laplace transform of common waveforms. F ( s) = Lf ( t ) =

f (t)

∞

∫ f ( t )e

− st

Table 5 |  Laplace transform of different operations.

dt

0

u(t)

1 s

eat

1 s−a

sinw t, cosw t

sinhat, coshat

ω s , 2 2 2 s + ω s + ω2 a s , s2 − a2 s2 − a2 F(s + a)

e−atf (t)

f (t)

F ( s) = L f ( t )

f(t   - t0)u(t - t0)

e − t0 s F ( s )

d f (t ) dt

sF(s) - f(0+)

t

∫ f (t ) dt

F ( s) s

t f (t)

−

1 f (t ) t

∫ F ( s) ds

0

d F ( s) ds

∞

s

n! s n +1

tn

Table 6|   Properties of continuous-time Fourier series. Property

Periodic Signal

Fourier Series Coefficients

x(t ) ⎫ Periodic with period T and ⎬ y(t )⎭ fundamental frequency ω 0 = 2π /T

ak bk

Linearity

Ax(t ) + By(t )

Aak + Bbk

Time shifting

x (t − t 0 )

ak e − jkω0 t0 = ak e − jk ( 2π /T )t 0

Frequency shifting

e jM ω0 t x(t ) = e jM ( 2π /T )t x(t )

Conjugation

x*(t )

a*− k

Time reversal

x( − t )

a− k

Time scaling

x(α t ), α > 0 ( periodic with period T /α )

ak

Periodic convolution

∫ x(τ )y(t − τ )dτ

Takbk

ak − M

T

Multiplication Differentiation

x (t ) y (t )

∞

∑ ab

l = −∞

d x (t ) dt

l k −l

⎛ 2π ⎞ jk ω 0 ak = jk ⎜ ⎟ ak ⎝T ⎠ (Continued )

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Chapter 2  •  Networks, Signals and Systems

Property

Periodic Signal t

∫ x(t )dt

Integration

Fourier Series Coefficients

(finite valued and periodic only if

−∞

a0 = 0)

⎛ 1 ⎞ ⎛ ⎞ 1 ⎜⎝ jk ω ⎟⎠ ak = ⎜⎝ jk ( 2π /T ) ⎟⎠ ak 0

Conjugate symmetry for real signals

x(t) real

ak = a*− k ⎧ ⎪Re{a } = Re{a } k −k ⎪⎪ Im{ a } = Im{ a ⎨ k −k } ⎪ ak = a− k ⎪ ⎪⎩ ∠ak = −∠a− k

Real and even signals

x(t) real and even

ak real and even

Real and odd signals

x(t) real and odd

ak purely imaginary and odd

Even–odd decomposition of real signals

⎧ xe (t ) = Ev{x(t )} ⎨ ⎩ xo (t ) = Od{x(t )}

Parseval’s relation for periodic signals

79

Re {ak } j Im {ak }

[ x(t ) real] [ x(t ) real]

∞ 1 2 x(t ) dt = ∑ ak ∫ T T k = −∞

2

Table 7|   Common Fourier transform pairs. Signal +∞

∑ae

k = −∞

k

Fourier Transform jk ω 0t

+∞

2π

∑

k = −∞

ak δ (ω − k ω 0 )

e jω 0 t

2πδ (ω − ω 0 )

cos ω 0 t

π [δ (ω − ω 0 ) + δ (ω + ω 0 )]

sin ω 0 t

∑

δ (t − nT )

n = −∞

⎪⎧1, t ≤ T1 x (t ) = ⎨ ⎩⎪0, t > T1

sinWT πt

⎪⎧1, ω < W X ( jω ) = ⎨ ⎪⎩0, ω > W

δ (t )

1 1 + πδ (ω ) jω

δ (t − t 0 )

e jω t 0

e − at u(t ), Re{a} > 0 2 sin k ω 0T1 δ (ω − k ω 0 ) k k = −∞

1 a + jω

te − at u(t ), Re{a} > 0

1 ( a + jω ) 2

2π T

t n −1 − at e u (t ) ( n − 1)! Re{a} > 0

π [δ (ω − ω 0 ) − δ (ω + ω 0 )] j

+∞

∑

x (t + T ) = x (t ) +∞

Fourier Transform

u(t)

Periodic square wave ⎧1, t ≤ T1 ⎪ x (t ) = ⎨ T ⎪0, T1 < t ≤ ⎩ 2 and

Signal

2π k ⎞ ⎛ δ ⎜ω − ∑ ⎟ ⎝ T ⎠ k = −∞ +∞

1 ( a + jω ) n

2 sin ω T1 ω

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Table 8|   Properties of continuous-time Fourier transform. Property

Aperiodic Signal

Fourier Transform

x (t ) y (t )

X ( jω ) Y ( jω )

Linearity

ax(t ) + by(t )

aX ( jω ) + bY ( jω )

Time shifting

x (t − t 0 )

e − jω t0 X ( jω )

Frequency shifting

e jω 0 t x ( t )

X ( j (ω − ω 0 ))

Conjugation

x*(t )

X *( − jω )

Time reversal

x ( −t )

X ( − jω )

x( at )

1 ⎛ jω ⎞ X⎜ ⎟ a ⎝ a ⎠

Time and frequency scaling Convolution

x(t )*y(t )

X(j w)Y(j w)

Multiplication

x (t ) y (t )

1 X ( jω )*Y ( jω ) 2π

Differentiation in time

d x (t ) dt

jω X ( jω )

Integration

t

⎛ 1 ⎞ ⎜⎝ jω ⎟⎠ X ( jω ) + π X (0)d (ω )

∫ x(t )dt

−∞

tx(t)

⎛ d ⎞ j⎜ X ( jω ) ⎝ dω ⎟⎠

x(t) real

⎧ X ( jω ) = X *( − jω ) ⎪ ⎪Re{ X ( jω )} = Re{ X ( − jω )} ⎪ ⎨Im{ X ( jω )} = − Im{ X ( − jω )} ⎪ X ( jω ) = X ( jω ) ⎪ ⎪⎩ ∠X ( jω ) = −∠X ( − jω )

Symmetry for real and even signals

x(t) real and even

X ( jω ) real and even

Symmetry for real and odd signals

x(t) real and odd

X ( jω ) purely imaginary and odd

Even–odd decomposition for real signals

⎧ xe (t ) = Ev{x(t )} ⎨ ⎩ xo (t ) = Od{x(t )}

Differentiation in frequency

Conjugate symmetry for real signals

+∞

Parseval’s relation for aperiodic signals

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 80

∫

−∞

2

x(t ) dt =

1 2π

+∞

∫

[ x(t ) real] [ x(t ) real]

Re { X ( jω )} j Im { X ( jω )}

2

X ( jω ) dω

−∞

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Chapter 2  •  Networks, Signals and Systems

81

Table 9 |   Properties of discrete-time Fourier series. Property

Periodic Signal

Fourier Series Coefficients

x[n]⎫ Periodic with period N and ⎬ y[n]⎭ fundamental frequency ω0 = 2π /N

ak ⎫ ⎬ Periodic with period N bk ⎭

Linearity

Ax[n] + By[n]

Aak + Bbk

Time shifting

x[n − n0 ]

ak e

Frequency shifting

e jM ( 2π/N ) n x[n]

ak − M

Conjugation

x*[n]

a*− k

Time reversal

x [ − n]

a− k

if n is a multiple of m ⎧ x[n/m] x m [n] = ⎨ 0 if ple of m n is not a multip ⎩ (periodic with period mN)

1 ak m (viewed as periodic with period mN)

Time scaling

Periodic convolution

∑

x[r ] y[n − r ]

− jk ( 2π / N ) n0

Nakbk

r= N

Multiplication

∑ ab

x[n] y[n]

l= N

First difference

x[n] − x[n −1]

Running sum

∑ x[ k ]

l k −l

(1 − e − jk ( 2π / N))ak 1 ⎛ ⎞ a ⎜⎝ − jk ( 2π / N ) ⎟ k ⎠ 1− e

n

k = −∞

(finite valued and periodic only if a0 = 0) Conjugate symmetry for real signals

x[n] real

⎧ ak = a *− k ⎪ ⎪Re{ak } = Re{a− k } ⎪ ⎨Im{ak } = − Im{a− k } ⎪ ak = a− k ⎪ ⎪ ⎩ ∠ak = −∠a− k

Real and even signals

x[n] real and even

ak real and even

Real and odd signals

x[n] real and odd

ak purely imaginary and odd

Even−odd decomposition of real signals

⎧ xe [n] = Ev{x[n]} ⎨ ⎩ xo [n] = Od{x[n]}

[ x[n] real] [ x[n] real]

Parseval’s relation for periodic signals

1 N

ak

∑

n= N

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 81

2

x[n] =

∑

k= N

Re {ak } j Im {ak }

2

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Table 10 |   Discrete-time Fourier transform pairs. Signal +∞

∑ae

Fourier Transform 2π

jk ( 2 n /N ) n

k

k= N

+∞

⎛

∑ a δ ⎜⎝ ω −

k = −∞

k

2π k ⎞ ⎟ N ⎠

+∞

2π ∑ δ (ω − ω 0 − 2π l )

e jω 0 n

l = −∞

+∞

cos ω 0 n

π ∑ {d (ω − ω 0 − 2π l ) l = −∞

+ d (ω + ω 0 − 2π l )}

π +∞ ∑ {δ (ω − ω 0 − 2π l ) j l = −∞ − δ (ω + ω 0 − 2π l )}

sin ω 0 n

+∞

2π ∑ δ (ω − 2π l )

x[n] - 1

2π

⎧1, n ≤ N1 ⎪ x[n] = ⎨ N ⎪0, N1 < n ≤ ⎩ 2 and

+∞

⎛

∑ a δ ⎜⎝ ω −

k = −∞

k

2π k ⎞ ⎟ N ⎠

a 1

[r n cos ω 0 n]u[n]

1 − [r cos ω 0 ]z −1 1 − [2r cos ω 0 ]z −1 + r 2 z −2

z >r

[r n sin ω 0 n]u[n]

[r sin ω 0 ]z −1 1 − [2r cos ω 0 ]z −1 + r 2 z −2

z >r

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Table 13 |  Properties of z-transform. Property

Signal

z -Transform

Linearity

ax1[n] + bx2[n]

aX1(z) + bX2(z)

At least the intersection of R1 and R2

Time shifting

x[n − n0]

z − n0 X ( z )

R, except for the possible addition or deletion of the origin

Scaling in the z-domain

e jω0 n x[n]

X ( e − jω 0 z ) ⎛ z⎞ X⎜ ⎟ ⎝z ⎠

R

X ( a −1 z )

Scaled version of R (i.e., a R

z0n x[n]

z0R

0

a n x[n]

= the set of points

{ a z} for

z in R) Time reversal

X(z )

Inverted R (i.e., R−1 = the set of points z−1, where z is in R)

X(zk)

R1/k (i.e., the set of points z1/k, where z is in R)

X*(z*)

R

−1

x[−n]

Conjugation

⎧ x[r ], n = rk x( k ) [n] = ⎨ n ≠ rk ⎩0, x*[n]

Convolution

x1 [n]* x2 [n]

X1(z)X2(z)

At least the intersection of R1 and R2

First difference

x[n] − x[n − 1]

(1 − z −1 ) X ( z )

At least the intersection of R and z > 0

Accumulation

∑

1 X ( z) 1 − z −1

Differentiation in the z-domain

nx[n]

Initial value theorem

If x[n] = 0 for n < 0, then x[0] → lim X ( z )

Final value theorem

lim x [ n] = lim (1− z −1 ) X ( z )

Time expansion

n k = −∞

for some integer r

x[k ]

−z

dX ( z ) dz

At least the intersection of R and z > 1 R

z →∞

n →∞

z →1

QUESTIONS 1.

(a) eat – ebt (c) aeat – bebt

In the circuit shown in the following figure, the voltage V(t) is 1Ω

1Ω

2. eat

+ V (t) 1 H −

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 84

(b) eat + ebt (d) aeat + bebt (GATE 2000: 1 Mark)

In the circuit shown in the following figure, the value of the voltage source E is

ebt

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Chapter 2  •  Networks, Signals and Systems

7.

0V

V2 + 2 V−

−

−

+ 1V

85

One period (0, T) each of two periodic waveforms, W1 and W2 are shown in the following figure. The magnitudes of the nth Fourier series coefficients of W1 and W2, for n ≥1, n odd, are respectively proportional to 1

1

+ −

W2

W1

E=?

T/2

T T

0

0

T/2

+ + 4V

−1

5V

V1

10 V

(a) –16 V (c) –6 V 3.

(b) 4 V (d) 16 V (GATE 2000: 1 Mark) −3t 2

The Fourier Transform of the signal x(t ) = e following form, where A and B are constants: (a) (c)

Ae − B f A+ B f

is of the

(b) Ae − Bf 2

(d) Ae − Bf



−1

  

(a)

n −2 and n −3 n −3 and n −2 (b)

(c)

n −1 and n −2 (d) n −4 and n −2 (GATE 2000: 2 Marks)

8. Let u(t) be the step function. Which of the waveforms shown in the options (a)–(d) corresponds to the convolution of u(t) – u(t − 1) with u(t) – u(t – 2)? (a) 1

2

(GATE 2000: 1 Mark) 4.

5.

A system with an input x(t) and output y(t) is described by the relation: y(t) = t x(t). This system is (a) linear and time-invariant (b) linear and time varying (c) non-linear and time-invariant (d) non-linear and time-varying (GATE 2000: 1 Mark)

0 (b)

2

1

Use the data of (a). The current i in the circuit of (b) is R2 R1 + 10 V

R2 R3

R1

0

1

2

(c)

R3

3 1.5

20 V + R4

2A

I=?

R4 t

(a)            (b) (a) −2 A (b) 2 A (c) −4 A (d) +4 A (GATE 2000: 2 Marks) 6.

1

A linear time invariant system has an impulse response e2t, t > 0. If the initial conditions are zero and the input is e3t, the output for t > 0 is (a) e3t − e2t (b) e5t 3t 2t (c) e + e (d) None of the above (GATE 2000: 2 Marks)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 85

0 (d)

1.5

2

3

1

t 0

1

3 (GATE 2000: 2 Marks)

9.

In the following figure, the steady state output voltage corresponding to the input voltage 3 + 4 sin 100t V is

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13. If each branch of a Delta circuit has impedance 3Z, then each branch of the equivalent Wye circuit has impedance

1 kΩ 10 µF

Input

Output

Z

(a)

π⎞ ⎛ (a) 3 + sin ⎜100t − ⎟ V ⎝ 4⎠ 2

(c) 3 3Z

4

(d)

Z 3

(GATE 2001: 1 Mark)

π⎞ ⎛ (b) 3 + 4 2 sin ⎜100t − ⎟ V ⎝ 4⎠ (c)

(b) 3Z

3

14. The admittance parameter Y12 in the 2-port network in the following figure is (a) −0.2 mho (b) 0.1 mho (c) −0.05 mho (d) 0.05 mho

3 4 π⎞ ⎛ + sin ⎜100t + ⎟ V ⎝ 2 4⎠ 2

π⎞ ⎛ (d) 3 + 4 sin ⎜100t + ⎟ V ⎝ 4⎠

20 Ω

I1

I2

(GATE 2000: 2 Marks) 10. The Hilbert transform of cos ω1t + sin ω2 t is

E1

(a) sin ω1t – cos ω2t (b) sin ω1t + cos ω2t (c) cos ω1t – sin ω2t (d) sin ω1t + sin ω2t

(GATE 2001: 1 Mark)

11. A system has a phase response given by ϕ (ω), where ω is the angular frequency. The phase delay and group delay at ω = ω0 are respectively given by

ϕ ( ωo ) d ϕ ( ω) (a) − ω = ωo ,− ωo dω

(c)

ωo d ϕ ( ω) ,− ω = ωo ϕ ( ωo ) dω

(d) ωo ϕ ( ωo ), ∫

−∞

15. The region of convergence of the z-transform of a unit step function is (a) |z| > 1 (b) |z| < 1 (c) (Real part of z) > 0 (d) (Real part of z) < 0 (GATE 2001: 1 Mark) 16. If a signal f(t) has energy E, the energy of the signal f(2t) is equal to E (a) E (b) 2 (c) 2E (d) 4E (GATE 2001: 1 Mark)

d 2 ϕ ( ωo ) ω = ωo d ω2

ωo

10 Ω E2

−

(GATE 2000: 2 Marks)

(b) ϕ ( ωo ), −

5Ω

17. The voltage e0 in the following figure is

ϕ ( λ)d λ

2Ω

(GATE 2000: 2 Marks)

16 Ω

12. The voltage e0 in the following figure is 4Ω

2Ω

8A

10 Ω

6Ω

12 Ω

+ eo −

+ 12 V

(a) 2 V (c) 4 V

4Ω

eo −

2Ω

4 V 3 (d) 8 V (GATE 2001: 1 Mark)

(b)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 86

(a) 48 V (c) 36 V

(b) 24 V (d) 28 V (GATE 2001: 2 Marks)

18. In the following figure, the value of the load resistor R which maximizes the power delivered to it is

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Chapter 2  •  Networks, Signals and Systems

10 Ω

20. The Z parameters Z11 and Z21 for the 2-port network in the following figure are

1H

2Ω

I1 Em cos 10t

I2 4Ω

R E1

10 E1 (b) 10 Ω (d) 28.28 Ω (GATE 2001: 2 Marks)

(a) 14.14 Ω (c) 200 Ω

19. When the angular frequency ω in the following figure is varied from 0 to ∞, the locus of the current phasor I2 is given by i(t) Em cos ωt

i1(t)

i2(t)

(a) Z11 = −

6 16 Ω; Z 21 = Ω; 11 11

(b) Z11 =

6 4 Ω; Z 21 = Ω; 11 11

(c) Z11 =

6 16 Ω; Z 21 = − Ω; 11 11

(d) Z11 =

4 4 Ω; Z 21 = Ω; 11 11 (GATE 2001: 2 Marks)

R2 R1 C

(a)

21. The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by h1(t) = 1 h2(t) = u(t)

w=0 I2

h3 (t ) =

w=∞ E = Em ∠0°

Em

Em

2R2

2R2

(b)

w = 0 E = Em ∠0°

Em 2R2 Em 2R2

w=∞ I2

Em

Em

2R2 2R2

2R2

w=∞

Em 2R2 Em 2R2

where u(t) is the unit step function. Which of these ­systems is time invariant, causal, and stable? (a) S1 (b) S2 (c) S3 (d) S4 (GATE 2001: 2 Marks)

w=0

π sec 350 π (d) sec 175 (GATE 2001: 2 Marks)

1 sec 350 1 (c) sec 700

(b)

E = Em ∠0°

(d) Em 2R2

h4 (t ) = e −3t u(t )

(a)

(c)

Em

u (t ) t +1

22. The Nyquist sampling interval, for the signal sinc (700t ) + sinc (500t ) is

I2

w=0

E2

E = Em ∠0°

23. The PSD and the power of a signal g(t) are, respectively, Sg(w) and Pg. The PSD and the power of the signal ag(t) are, respectively (a) a 2 S g ( ω) and a 2 Pg (c) aS g ( ω) and a 2 Pg





(b) a 2 S g ( ω) and aPg (d) aS g ( ω) and aPg (GATE 2001: 2 Marks)

I2 w=∞

24. The dependent current source shown in the following figure (GATE 2001: 2 Marks)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 87

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GATE ECE Chapter-wise Solved papers

(a) TP = Tg = constant

5Ω

(b) TP ∝ f and Tg ∝ f (c) TP = constant and Tg ∝ f

+ I −

5Ω

V1 = 20 V

(a) delivers 80 W (c) delivers 40 W

V1 A 5

(b) absorbs 80 W (d) absorbs 40 W (GATE 2002: 1 Mark)

(d) TP ∝ f and Tg = constant (GATE 2002: 1 Mark) 30. In the network of the following figure, the maximum power is delivered to RL if its value is I1

25. In the following figure, the switch was closed for a long time before opening at t = 0, the voltage Vx at t = 0+ is

40 Ω 0.511

t=0

20 Ω

RL 50 V

20 Ω

2.5 A 5H 20 Ω Vx

− (a) 25 V (c) −50 V

+

(b) 50 V (d) 0 V (GATE 2002: 1 Mark)



26. Convolution of x (t + 5) with impulse function δ (t – 7) is equal to (a) x (t – 12) (b) x (t + 12) (c) x (t – 2) (d) x (t + 2) (GATE 2002: 1 Mark) 27. Which of the following cannot be the Fourier series expansion of a periodic signal? (a) x(t) = 2cos t + 3 cos 3t (b) x(t) = 2cos π t + 7 cos t (c) x(t) = cos t + 0.5 (d) x(t) = 2 cos 1.5 π t + sin 3.5 π t (GATE 2002: 1 Mark) 1 . 28. The Fourier transform F {e u(t )} is equal to 1 + j 2π f ⎧ 1 ⎫ Therefore, F ⎨ ⎬ is ⎩1 + j 2 π t ⎭ (a) e fu(f ) (b) e−fu(f ) (c) e fu(−f ) (d) e−fu(−f ) (GATE 2002: 1 Mark) −t

29. A linear phase channel with phase delay TP and group delay Tg must have

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 88

40 (a) 16 Ω (b) Ω 3 (c) 60 Ω (d) 20 Ω (GATE 2002: 2 Marks) 31. If the 3-phase balanced source in the following figure delivers 1500 W at a leading power factor of 0.844, then the value of ZL (in ohm) is approximately ZL 3-Phase balanced source

(a) 90 ∠32.44° (c) 80 ∠ − 32.44°

ZL

400 V ZL

(b) 80 ∠32.44° (d) 90 ∠ − 32.44° (GATE 2002: 2 Marks)

32. The Laplace transform of a continuous-time signal x(t) 5−s . If the Fourier transform of this sigis X ( s) = 2 s −s−2 nal exists, then x(t) is (a) e2t u(t) – 2e−t u(t) (b) –  e2tu(−t) + 2e−tu(t) (c) –  e2tu(−t) – 2e−tu(t) (d) e2tu(−t) – 2e−tu(t) (GATE 2002: 2 Marks) 33. The minimum number of equations required to analyze the circuit shown in the following figure is

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Chapter 2  •  Networks, Signals and Systems

C

i(t)

C

2W

2H

+ sin t R R

V

R

C

(a) 3 (c) 6

(b) 4 (d) 7 (GATE 2003: 1 Mark)

34. A source of angular frequency 1 rad/s has a source impedance consisting of 1 Ω resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is (a) 1 Ω resistance. (b) 1 Ω resistance in parallel with 1 H inductance. (c) 1 Ω resistance in series with 1 F capacitor. (d) 1 Ω resistance in parallel with 1 F capacitor. (GATE 2003: 1 Mark) 35. The maximum power that can be transferred to the load resistor RL from the voltage source in the circuit shown in the following figure is

d 2 i (t ) di(t ) +2 + i(t ) = sin t 2 dt dt

(a)

2

(b)

d 2 i (t ) di(t ) +2 + 2i(t ) = cos t 2 dt dt

(c)

2

(d)

d 2 i (t ) di(t ) +2 + 2i(t ) = sin t dt dt 2

d 2 i (t ) di(t ) +2 + i(t ) = cos t dt dt 2

(GATE 2003: 1 Mark) 38. The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by g p (t ) =

∞

∑ce

n = −∞

n

j 2π nf 0 t

It is given that c3 = 3 + j5. Then c−3 is (a) 5 + j3 (b) −3 − j5 (c) −5 + j3 (d) 3 − j5 (GATE 2003: 1 Mark) 39. A sequence x(n) with the z-transform X ( z ) = z 4 + z 2 − 2 z + 2 − 3 z −4 is applied as an input to a linear, time-invariant system with the impulse response h( n) = 2δ ( n − 3) where

100

10 V

1F

R

+ −

⎧1, n = 0 δ ( n) = ⎨ ⎩0, otherwise

RL

The output at n = 4 is

(a) 1 W (c) 0.25 W

(b) 10 W (d) 0.5 W (GATE 2003: 1 Mark)

36. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is (a) 25 (b) 50 (c) 100 (d) 200 (GATE 2003: 1 Mark) 37. The differential equation for the current i(t) in the circuit of the figure is

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 89

(a) −6 (c) 2

(b) zero (d) −4 (GATE 2003: 1 Mark)

40. Let x(t) be the input to a linear, time-invariant system. The required output is 4 x(t − 2). The transfer function of the system should be (a) 4 e j 4π f (b) 2 e − j 8π f (c) 4 e − j 4π f (d) 2 e j 8π f (GATE 2003: 1 Mark) 41. Twelve 1 Ω resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is

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GATE ECE Chapter-wise Solved papers

⎛ 5⎞ (a) ⎜ ⎟ Ω (b) 1Ω ⎝ 6⎠ ⎛ 6⎞ ⎛ 3⎞ (c) ⎜ ⎟ Ω (d) ⎜⎝ ⎟⎠ Ω ⎝ 5⎠ 2 (GATE 2003: 2 Marks) 42. The current flowing through the resistance R in the circuit shown in the following figure has the form P cos 4t where P is M = 0.75 H 1/10.24 F

R = 3.92 Ω

3Ω

+ −

V = 2 cos 4t

(a) (0.18 + j 0.72) (c) -(0.18 + j 1.90) 43. An

input

20 V

(b) (0.46 + j 1.90) (d) -(0.192 + j 0.144) (GATE 2003: 2 Marks)

υ (t ) = 10 2 cos(t + 10°) + 10 5

voltage

cos( 2t + 10°)V is applied to a series combination of resistance R = 1Ω and an inductance L = 1 H. The resulting steady-state current i(t) in ampere is (a) 10cos(t + 55°) + 10cos(2t + 10°+ tan-1 2) (b) 10 cos (t + 55°) + 10

3 cos( 2t + 55°) 2

45. I1(s) and I2(s) are the Laplace transforms of i (t) and i (t), 1 2 respectively. The equations for the loop currents I1(s) and I2(s) for the circuit shown in the given figure, after the switch is brought from position 1 to position 2 at t = 0, are 1 ⎡ ⎢ R + Ls + Cs (a) ⎢ ⎢ − Ls ⎢⎣

⎤ − Ls ⎥ ⎡V ⎤ ⎡ I1 ( s ) ⎤ ⎢ ⎥ = ⎥⎢ ⎥ ⎢s⎥ 1 I ( s) R + ⎥ ⎣ 2 ⎦ ⎣0 ⎦ Cs ⎥⎦

1 ⎡ ⎢ R + Ls + Cs (b) ⎢ ⎢ − Ls ⎢⎣

⎤ − Ls ⎥ ⎡V ⎤ ⎡ I1 ( s ) ⎤ ⎢ ⎥ = ⎥⎢ ⎥ ⎢s⎥ 1 I ( s) R + ⎥ ⎣ 2 ⎦ ⎣0 ⎦ Cs ⎥⎦

1 ⎡ ⎢ R + Ls + Cs (c) ⎢ ⎢ − Ls ⎢⎣

⎤ ⎥ ⎡ I1 ( s ) ⎤ ⎡ − V ⎤ ⎢ s⎥ ⎥⎢ ⎥= ⎥ 1 ⎥ ⎣ I 2 ( s) ⎦ ⎢ R + Ls + ⎣0 ⎦ ⎥ Cs ⎦

1 ⎡ ⎢ R + Ls + Cs (d) ⎢ ⎢ − Ls ⎢⎣

⎤ ⎥ ⎡ I1 ( s ) ⎤ ⎡ V ⎤ ⎢ ⎥ ⎥⎢ ⎥= s 1 ⎥ ⎣ I 2 ( s) ⎦ ⎢ ⎥ R + Ls + ⎣0 ⎦ Cs ⎥⎦

(c) 10cos(t - 35°) + 10cos(2t + 10° - tan 2) 3 cos( 2t − 35°) 2

1 1Ω

C

i1(t)

2

1´ (a) (b) (c) (d)

1

i2(t) L

R

2Ω

2Ω

1Ω

2´

Z11 = 2.75 Ω and Z12 = 0.25 Ω Z11 = 3 Ω and Z12 = 0.5 Ω Z11 = 3 Ω and Z12 = 0.25 Ω Z11 = 2.25 Ω and Z12 = 0.5 Ω (GATE 2003: 2 Marks)

47. The driving-point impedance Z(s) of a network has the

R C

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 90

2 2Ω

Common Data for Questions 44 and 45: The circuit for questions 44 and 45 is shown in the following figure. Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0.

V

− Ls

46. The impedance parameters Z11 and Z12 of the two-port network shown in the following figure are

(GATE 2003: 2 Marks)

S

− Ls

(GATE 2003: 2 Marks) -1

(d) 10 cos (t − 35°) + 10

44. At t = 0+, the current i (t) is 1 −V −V (a) (b) 2R R −V (c) (d) zero 4R (GATE 2003: 2 Marks)

pole-zero locations as shown in the following figure. If Z(0) = 3, then Z(s) is

12/4/2018 11:05:56 AM

Chapter 2  •  Networks, Signals and Systems

M

Im s-plane

1

1

−3

2

Re

−1

denotes zero denotes pole

(c)

3( s + 3) 2( s + 3) (b) s 2 + 2s + 3 s 2 + 2s + 2 3( s − 3) 2( s − 3) (d) 2 2 s − 2s − 2 s − 2s − 3 (GATE 2003: 2 Marks)

(a) L1 + L2 + M (b) L1 + L2 - M (c) L1 + L2 + 2M (d) L1 + L2 - 2M (GATE 2004: 1 Mark) 52. The circuit shown in the following figure, with R = 1/ 3 Ω, L = 1/ 4 H, C = 3 F , has the input voltage v(t) = sin 2t. The resulting current i(t) is i(t)

v(t) Common Data for Questions 48 and 49: The system under consideration is an RC low-pass filter (RC–LPF) with R = 1.0 KΩ and C = 1.0 mF. 48. Let H(  f  ) denote the frequency response of the RC-LPF. Let f1 be the highest frequency such that 0 ≤ f ≤ f1 ;

H ( f1 ) H (0 )

(a) 327.8 (c) 52.2

L2

L1

−1

(a)

91

≥ 0.95. Then f1 (in Hz) is (b) 163.9 (d) 104.4

R

(a) 5 sin(2t + 53.1°) (c) 25 sin(2t + 53.1°)

(a) 0.717  (b)  7.17  (c)  71.7  (d)  4.505 (GATE 2003: 2 Marks) 50. Let P be the linearity, Q be the time-invariance, R be the causality and S be the stability. A discrete-time system has the input-output relationship, n ≥1 ⎧ x( n), ⎪ y( n) = ⎨ 0, n=0 ⎪ x( n + 1), n ≤ −1 ⎩ where x(n) is the input and y(n) is the output. The above system has the properties (a) P, S but not Q, R (b) P, Q, S but not R (c) P, Q, R, S (d) Q, R, S but not P (GATE 2003: 2 Marks) 51. The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the following figure is

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 91

(b) 5 sin(2t - 53.1°) (d) 25 sin(2t - 53.1°) (GATE 2004: 1 Mark)

53. For the circuit shown in the following figure, the time constant RC = 1 ms. The input voltage is vi (t ) = 2 sin 103 t . The output voltage vo(t) is equal to

(GATE 2003: 2 Marks) 49. Let tg(f ) be the group delay function of the given RC-LPF and f 2 = 100 Hz. Then t g ( f 2 ), in ms, is

C

L

R +

+ C

vi(t) −

vo(t) −

(a) sin(103t - 45°) (b) sin(103t + 45°) (c) sin(103t - 53°) (d) sin(103t + 53°) (GATE 2004: 1 Mark) 54. For the RL circuit shown in the following figure, the input voltage v (t) = u(t). The current i(t) is 1

1H

v1(t)

i(t)

2

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GATE ECE Chapter-wise Solved papers

i(t)

(a)

0.5 0.31

t(s)

2 i(t) 1 (b) 0.63

1 2

t(s)

i(t)

56. The Fourier transform of a conjugate symmetric function is always (a) imaginary (b) conjugate antisymmetric (c) real (d) conjugate symmetric (GATE 2004: 1 Mark) 57. The impulse response h[n] of a linear time invariant system is given by h[n] = u[n + 3] + u[n − 2] − 2u[n − 7] where u[n] is the unit-step sequence. The above system is (a) stable but not causal (b) stable and causal (c) causal but unstable (d) unstable and not causal (GATE 2004: 1 Mark) z . If the z − 0.2 ROC is z < 0.2, then the impulse response of the ­system is (a) (0.2) n u [ n] (b) (0.2) n u [ − n − 1]

58. The z-transform of a system is H ( z ) =

(c) 0.5 0.31

1 2

t(s)

(c) −(0.2) n u [ n] (d) −(0.2) n u [ − n − 1] (GATE 2004: 1 Mark)

i(t)

59. The transfer function H ( s) =

1 (d) 0.63

Vo ( s) of an RLC circuit Vi ( s)

106 . The quality factor s 2 + 20 s + 106 (Q-factor) of this circuit is (a) 25 (b) 50 (c) 100 (d) 5000 (GATE 2004: 2 Marks)

is given by H ( s) = 2

t(s)

(GATE 2004: 1 Mark)

55. Consider the network graph shown in the following ­figure.

60. For the circuit shown in the following figure, the initial V ( s) conditions are zero. Its transfer function H ( s) = o Vi ( s) is +

Which one of the following is NOT a ‘tree’ of this graph?

vi(t) (a)

(c)

(b)

−

(d) (GATE 2004: 1 Mark)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 92

10 kΩ

10 mH 100 µF

+

vo(t) −

(a)

106 1 (b) 2 6 6 s + 10 s + 10 s + 103 s + 106

(c)

103 106 (d) s 2 + 103 s + 106 s 2 + 106 s + 106

2

(GATE 2004: 2 Marks)

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Chapter 2  •  Networks, Signals and Systems

61. Consider the following statements S1 and S2: S1: At the resonant frequency, the impedance of a series RLC circuit is zero. S2: In a parallel GLC circuit, increasing the conductance G results in increase in Q factor. Which one of the following is correct? (a) S1 is False and S2 is True (b) Both S1 and S2 are True (c) S1 is True and S2 is False (d) Both S1 and S2 are False

64. A rectangular pulse train s(t) as shown in the following figure is convolved with the signal cos 2 ( 4π × 103 t ). The convolved signal will be a s(t) 1

t

(GATE 2004: 2 Marks) 62. The circuit shown in the following figure has initial current iL(0-) = 1 A through the inductor and an initial voltage vC(0-) = -1 V across the capacitor. For input v(t) = u(t), the Laplace transform of the current i(t) for t ≥ 0 is 1Ω

0 0.1ms (a) DC (b) 12 kHz sinusoid (c)  8 kHz sinusoid (d) 14 kHz sinusoid

1H

(GATE 2004: 2 Marks)

+ i(t)

+ 1F −

v(t)

65. Let x(t) and y(t) [with Fourier transforms X(f ) and Y(f ), respectively] be related as shown in the ­following figure. Then Y(f ) is x(t)

−

y(t)

1 (a) (c)

s s+2 (b) s2 + s + 1 s2 + s + 1 s−2 s−2 (d) s2 + s + 1 s2 + s − 1 (GATE 2004: 2 Marks)

63. For the lattice circuit shown in the following figure, Za = j2 Ω and Zb = 2 Ω. The values of the open circuit impedZ12 ⎤ ⎡Z ance parameters Z = ⎢ 11 ⎥ are ⎣ Z 21 Z 22 ⎦ 1

t

2 −1

(a) −

1 ⎛ f⎞ 1 ⎛ f ⎞ − j 2π f (b) − X ⎜ ⎟ e j 2π f X⎜ ⎟e 2 ⎝ 2⎠ 2 ⎝ 2⎠

⎛ (c) − X ⎜ ⎝

f ⎞ j 2π f ⎛ f⎞ (d) − X ⎜ ⎟ e − j 2π f ⎟⎠ e ⎝ 2⎠ 2

4

66. A 1 kHz sinusoidal signal is ideally sampled at 1500 samples/s and the sampled signal is passed through an ideal low-pass filter with cut-off frequency 800 Hz. The output signal has the frequency (a) zero Hz (b) 0.75 kHz (c) 0.5 kHz (d) 0.25 kHz (GATE 2004: 2 Marks)

Za Zb

j⎤ ⎡ 1− j 1+ (b) ⎥ ⎢ −1 + j 1 − j⎦ ⎣

j⎤ j ⎥⎦

⎡ 1 + j −1 + j ⎤ ⎡1 + j 1 + j ⎤ (c) ⎢ (d) ⎢ ⎥ ⎥ ⎣ −1 + j 1 + j ⎦ ⎣1 − j 1 − j ⎦ (GATE 2004: 2 Marks)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 93

0

(GATE 2004: 2 Marks)

Za

⎡1 − j 1 + (a) ⎢ ⎣1 + j 1 +

−2

0

3 Zb

2

−2 −1 t

67. Consider the sequence x[n] = [ − 4 − j 5 1 + j 2 4] The conjugate anti-symmetric part of the sequence is (a) (b)

[ − 4 − j 2.5 [ − j 2.5 1

j 2 4 − j 2.5] j 2.5]

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GATE ECE Chapter-wise Solved papers

(c) (d)

[ − j 2.5 j 2 [ − 4 1 4]

0] (GATE 2004: 2 Marks)

72. The ABCD parameters of an ideal n : 1 transformer shown ⎡n 0 ⎤ in the following figure are ⎢ ⎥ . The value of X will be ⎣0 X ⎦ I1

68. The impulse response h[n] of a linear time invariant system is given as ⎧ −2 2 , n = 1, − 1 ⎪⎪ h[n] = ⎨ 4 2 , n = 2, − 2 ⎪ 0, otherwise ⎪⎩

V1

If the input to the above system is the sequence e jpn/4, then the output is 4 2e − j π n / 4 (a) 4 2e jπ n / 4 (b) (c) 4e jπ n / 4 (d) −4e jpn/4 (GATE 2004: 2 Marks)

69. A causal LTI system is described by the difference equation 2 y [ n] = α y [ n − 2] − 2 x [ n] + β x [ n − 1] The system is stable only if (a) α = 2, β < 2 (b) α > 2, β > 2 (c) α < 2, any value of β (d) β < 2, any value of α (GATE 2004: 2 Marks) 70. The condition on R, L and C such that the step response y(t) in the following figure has no oscillations, is L

R +

+ y(t)

− − (a) R ≥

1 L L (b) R≥ 2 C C

(c) R ≥ 2

L L (d) R= C C (GATE 2005: 1 Mark)

71. In a series RLC circuit, R = 2 kΩ, L = 1 H, and C = 1/ 400 mF. The resonant frequency is 1 (a) 2 ×´ 104 Hz (b) × 10 4 Hz π (c) 104 Hz (d) 2p ×´ 104 Hz (GATE 2005: 1 Mark)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 94

V2

1 n 1 (c) n2 (d) n2 (GATE 2005: 1 Mark) (a) n

(b)

73. The first and the last critical frequency of an RC-driving point impedance function must, respectively, be (a) a zero and a pole (b) a zero and a zero (c) a pole and a pole (d) a pole and a zero (GATE 2005: 1 Mark) 74. Choose the function f(t), −∞ < t < ∞, for which a Fourier series cannot be defined. (a) 3 sin (25t) (b) 4cos (20t + 3) + 2 sin(710t) (c) exp( − t ) sin( 25t ) (d) 1 (GATE 2005: 1 Mark) 75. The region of convergence of z-transform of the n n ⎛ 5⎞ ⎛ 6⎞ sequence ⎜ ⎟ u( n) − ⎜ ⎟ u( − n − 1) is ⎝ 6⎠ ⎝ 5⎠ 5 5 (b) z > 6 6 5 6 6 (c) < z < (d) < z a + 2 (c) Re{s} < 2

(b) Re{s} > a + 7 (d) Re{s} > a + 5 (GATE 2005: 2 Marks)

1

1

85. In what range should Re{s} remain so that the Laplace transform of the function e(a + 2)t + 5 exists?

1/2

1/2

(a) —2

0

2

86. The h-parameters of the circuit shown in the following figure are

20 Ω

1/2

−

1/2

—3

—1

1

1 ⎤ ⎡30 20 ⎤ ⎡10 (c) ⎢ (d) ⎥ ⎢ ⎥ ⎣ 20 20 ⎦ ⎣ −1 0.05⎦ (GATE 2005: 2 Marks) 87. For a signal x(t) the Fourier transform is X( f  ). Then the inverse Fourier transform of X (3 f + 2) is given by

5

n

1

1 1/2

1/2

(c) —6

—4

—2

1 ⎛ t ⎞ j 3π t 1 ⎛ t ⎞ − j 4π t / 3 x ⎜ ⎟ e (b) x⎜ ⎟ e 2 ⎝ 2⎠ 3 ⎝ 3⎠

(c) 3 x(3t )e

3

2

⎡10 −1 ⎤ ⎡ 0.1 0.1⎤ (a) ⎢ ⎥ (b) ⎢ 1 0.05⎥ − 0 . 1 0 . 3 ⎣ ⎦ ⎣ ⎦

− j 4π t

n

(b)

V2

−

6

1

1 +

V1

4

2

I2

10 Ω

+

(a)

for n odd

will be

(a)  3 V   (b)  - 3 V  (c)  4 V  (d)  - 4 V (GATE 2005: 2 Marks)

I1

for n even

0

2

n

2

(d) x(3t + 2) (GATE 2005: 2 Marks)

Linked Answer Questions 88 and 89: A sequence x (n) has non-zero values as shown in the following figure. 2

1

1 1/2

1/2

(d)

x(n)

—5

—3

—1

1

3

n

(GATE 2005: 2 Marks) 1

1

89. The Fourier transform of y(2n) will be 1/2

—2

1/2

—1

0

1

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 96

2

(a) e − j 2ω [cos 4ω + 2 cos 2ω + 2] n

(b)

[cos 2ω + 2 cos ω + 2]

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Chapter 2  •  Networks, Signals and Systems

(c) e − jω [cos 2ω + 2 cos ω + 2]

94. The Dirac delta function δ (t ) is defined as

(d) e − jω / 2 [cos 2ω + 2 cos ω + 2]

⎧1, t = 0 (a) δ (t ) = ⎨ ⎩0, otherwise

(GATE 2005: 2 Marks)

⎧ ∞, t = 0 (b) δ (t ) = ⎨ ⎩0, otherwise

90. A signal x( n) = sin (ω 0 n + ϕ ) is the input to a linear time-invariant system having a frequency response H (e jω ). If the output of the system is Ax ( n − n0 ), then the most general form of ∠H (e jω ) will be (a) − n0ω 0 + β for any arbitrary real b

⎧1, t = 0 (c) δ (t ) = ⎨ and ⎩0, otherwise ⎧ ∞, t = 0 (d) δ (t ) = ⎨ and ⎩0, otherwise

(b) − n0ω 0 + 2π k for any arbitrary integer k (c) n0ω 0 + 2π k for any arbitrary integer k (d) −n0ω 0ϕ 91. The output y(t) of a linear time invariant system is related to its input x(t) by the following equation:

(a)

The filter transfer function H (ω ) of such a system is given by

(c)

A(ω ) = Cω , ϕ (ω ) = k ω 2

96. Consider the function f (t) having Laplace transform

(d) (1 − 0.5 cos ω T )e − jω td

F ( s) =

(GATE 2005: 2 Marks) 92. Let x(t ) ↔ X ( jω ) be Fourier transform pair. The Fourier transform of the signal x(5t − 3) in terms of X ( jω ) is given as 1 − ( j 3ω / 5) ⎛ jω ⎞ e X⎜ ⎝ 5 ⎟⎠ 5



ω0 s 2 + ω 02

Re{s} > 0

The final value of f (t) would be (a) 0 (b) 1 (c) −1 ≤ f (∞) ≤ 1 (d) ∞ (GATE 2006: 2 Marks)

97. A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is

1 ( j 3ω / 5) ⎛ jω ⎞ (b) e X⎜ ⎝ 5 ⎟⎠ 5

1 j 3ω ⎛ jω ⎞ e X⎜ ⎝ 5 ⎟⎠ 5

A(ω ) = Cω 2 , ϕ (ω ) = k ω 3

(GATE 2006: 1 Mark)

(c) (1 − cos ω T )e − jω td

(d)

∫ δ (t )dt = 1

−∞

(d) A(ω ) = C , ϕ (ω ) = k ω −1

(b) (1 + 0.5 cos ω T )e − jω td

1 − j 3ω ⎛ jω ⎞ e X⎜ ⎝ 5 ⎟⎠ 5

∞

(b) A(ω ) = Cω 2 , ϕ (ω ) = k ω

(a) (1+ cos ω T )e − jω td

(c)

∫ δ (t )dt = 1

−∞

95. A low-pass filter having a frequency response H ( jω ) = aA(ω )e j ϕ (ω ) does not produce any phase distortion, if

y(t ) = 0.5 x(t − t d + T ) + x(t − t d ) + 0.5 x(t − t d − T ).

(a)

∞

(GATE 2006: 1 Mark) (GATE 2005: 2 Marks)



97

I(s) 0.002 s (GATE 2006: 1 Mark)

1 2 93. If the ROC of x1 [n] + x2 [n] is < z < , then the ROC 3 3 of x1 [n] − x2 [n] includes 1 2 (a) < z < 3 (b) < z 0, the voltage across the resistor is 1 3

[e − (

3 / 2)t

123. The h-parameter matrix for this network is 3 ⎤ ⎡ −3 −1 ⎤ ⎡ −3 (b) (a) ⎢ ⎥ ⎢ 3 0.67⎥ ⎦ ⎣ ⎣ −1 0.67⎦ 1 ⎤ 3  ⎡3 3 (c)  (d) ⎥ ⎢  ⎣ −3 −0.67⎦ 1 0.67 

− e − (1/ 2 )t ]

(GATE 2008: 2 Marks) 124. The driving point impedance of the following network 0.2 s is given by Z ( s) = 2 . The component values s + 0.1s + 2 are

Z(s) L

(d)

−

(GATE 2008: 2 Marks)

⎡ ⎛ 3t ⎞ 1 ⎛ 3t ⎞ ⎤ − (b) e − (1/ 2 )t ⎢cos ⎜ sin ⎜ ⎟⎥ ⎟ 3 ⎝ 2 ⎠ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ (c)

+ 1.5 V

2´

(GATE 2008: 2 Marks)

(a)

+ V2 −

⎡1.5 1.5⎤ 1.5 4.5 (b) (a) ⎢ ⎥ 1.5 4.5 ⎣ 4.5 1.5⎦  

120. For t > 0, the output voltage VC(t) is − 2 ⎛ − 12 t ⎜e − e 3⎝

6V

− A + 2

2

122. The z-parameter matrix for this network is +

δ(t) +

−

1 + V1 −

+ (GATE 2008: 2 Marks)

1H

S2

+ A − 1

C

R

⎛ 3t ⎞ 2 − (1/ 2 )t e sin ⎜ ⎟ 3 ⎝ 2 ⎠ ⎛ 3t ⎞ e − (1/ 2 )t cos ⎜ ⎟ 3 ⎝ 2 ⎠

2

(GATE 2008: 2 Marks) Linked Answer Questions 122 and 123: A two-port network shown below is excited by external DC sources. The voltages and currents are measured with voltmeters V1 and V2 and ammeters A1 and A2 (all assumed to be ideal) as indicated. Under following switch conditions, the readings obtained are listed as follows:

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 101

 (a) (b)   (c) (d)

L = 5 H, R = 0.5 Ω, C = 0.1 F L = 0.1 H, R = 0.5 Ω, C = 5 F L = 5 H, R = 2 Ω, C = 0.1 F L = 0.1 H, R = 2 Ω, C = 5F (GATE 2008: 2 Marks)

125. The signal x(t) is described by ⎧1 for − 1 ≤ t ≤ +1 x (t ) = ⎨ ⎩0 otherwise

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Two of the angular frequencies at which its Fourier transform becomes zero are (a) π , 2π (b) 0.5π , 1.5π

(a) ( P1 , R1 ), ( P2 , R3 ), ( P3 , R4 ) (b) ( P1 , R2 ), ( P2 , R3 ), ( P3 , R4 ) (c) ( P1 , R3 ), ( P2 , R1 ), ( P3 , R2 )

(c) 0, π (d) 2π , 2.5π

(d) ( P1 , R1 ), ( P2 , R2 ), ( P3 , R3 )

(GATE 2008: 2 Marks) Linked Answer Questions 126 and 127: In the network shown in the following figure, the switch is closed at t = 0 − and the sampling starts from t = 0. The sampling frequency is 10 Hz. 10 µF

S

X[z] x(n)

+ —

200 kΩ

Sampler (fs=10 Hz)

z-Transform

5V

126. The samples x( n) ( n = 0, 1, 2, …) are given by 5e −0.05 n (a) 5(1 − e −0.05 n ) (b) 5e −5 n (c) 5(1 − e −5 n ) (d) (GATE 2008: 2 Marks) 127. The expression and the ROC of the z-transform of the sampled signal are 5z , z < e −5 −5 z−e 5z (b) , z < e −0.05 z − e −0.05 5z (c) , z > e −0.05 z − e −0.05 5z (d) , z > e −5 z − e −5

(GATE 2008: 2 Marks) 129. A discrete-time linear shift-invariant system has an impulse response h[n] with h[0] = 1, h[1] = −1, h[2] = 2, and zero otherwise. The system is given an input sequence x[n] with x[0] = x[2] = 1 and zero otherwise. The number of non-zero samples in the output sequence y[n], and the value of y[2], respectively, are (a) 5, 2 (b) 6, 2 (c) 6, 1 (d) 5, 3 (GATE 2008: 2 Marks) Linked Answer Questions 130 and 131:  The impulse response h(t) of a linear time-invariant continuous-time system is given by h(t ) = exp( −2t )u(t ), where u(t) denotes the unit step function. 130. The frequency response H (ω ) of this system in terms of angular frequency ω , is given by H (ω ) =

(a)

(a)

1 sin (ω ) (b) 1 + j 2ω ω

(c)

1 jω (d) 2 + jω 2 + jω (GATE 2008: 2 Marks)

131. The output of this system, to the sinusoidal input x(t ) = 2 cos ( 2t ) for all time t, is (a) 0 (b) 2 − 0.25 cos ( 2t − 0.125π ) (c) 2 − 0.5 cos ( 2t − 0.125π )

(GATE 2008: 2 Marks) 128. Let x(t) be the input and y(t) be the output of a ­continuous-time system. Match the system properties P1 , P2 and P3 with system relations R1 , R2 , R3 and R4 . Properties P1: Linear, but NOT time-invariant.

(d) 2 − 0.5 cos ( 2t − 0.25π ) (GATE 2008: 2 Marks) 132. In the interconnection of ideal sources shown in the ­following figure, it is known that the 60 V source is absorbing power.

P2:Time-invariant, but NOT linear.

+−

20 V

P3: Linear and time-invariant. Relations R1: y(t ) = t 2 x(t )

I

+ −

60 V 12 A

R2: y(t ) = t x(t ) R3: y(t ) = x(t ) R4: y(t ) = x( t − 5)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 102



Which of the following can be the value of the c­ urrent source I ?

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Chapter 2  •  Networks, Signals and Systems

(a) 10 A (c) 15 A

(b) 13 A (d) 18 A (GATE 2009: 1 Mark)

133. A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that during the talk-time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the following figure. How much energy does the battery deliver during this talk-time? (a) 220 J (b) 12 kJ (c) 13.2 kJ (d) 14.4 J v(t) 12 V

136. A function is given by f (t ) = sin 2 t + cos 2t . Which of the following is true? (a) f  has frequency components at 0 and (1/ 2π ) Hz. (b) f  has frequency components at 0 and (1/π ) Hz. (c) f has frequency components at (1/ 2π ) Hz and (1/π ) Hz. (d) f has frequency components at 0, (1/ 2π ) and (1/π ) Hz. (GATE 2009: 1 Mark) 137. The ROC of z-transform of the discrete time sequence n n x( n) = (1/ 3) u( n) − (1/ 2) u( − n − 1) is (a)

1 1 1 z > < z < (b) 2 3 2

(c)

z
2, but non-zero for ws < 2 (GATE 2009: 2 Marks)

140. The switch in the circuit shown was on position a for a long time, and is moved to position b at time t = 0. The current i(t) for t > 0 is given by 10 kΩ

a

144. Consider a system whose input x(t) and output y(t) are

b

∞

related by the equation y(t ) =

i(t)

∫ x(t − τ ) h (2τ ) dτ

where

−∞

100 V

0.2 µF

+ −

h(t) is depicted in the graph shown in the following figure. 5 kΩ h(t)

0.5 µF

0.3 µF t 0

(a) 0.2e-125tu(t) mA (c) 0.2e-1250tu(t) mA

(b) 20e-1250tu(t) mA (d) 20e-1000tu(t) mA (GATE 2009: 2 Marks)

141. The time-domain behavior of an RL circuit is repredi sented by L + Ri = V0 (1 + Be − Rt / L sin t )u(t ) . For an inidt V tial current of i(0) = 0 , the steady-state value of the R current is given by (a) i(t ) →

V0 R

V0 (1 + B) R



BIBO: Bounded input gives a bounded output.



Causal: The system is causal.



LP: The system is low pass.



LTI: The system is linear and time-invariant. (b) BIBO, LTI (a) Causal, LP (d) LP, LTI (c) BIBO, Causal, LTI (GATE 2009: 2 Marks)

S1 : H ( z ) is a low-pass filter S2 : H ( z ) is an FIR filter.

2V (d) i(t ) → 0 (1 + B) R (GATE 2009: 2 Marks) 142. Given that F(s) is the one-sided Laplace transform of t

f(t), the Laplace transform of

Which of the following four properties are possessed by the system?

145. A system with transfer function H(z) has the impulse response h( n) defined as h( 2) = 1, h(3) = −1 and h( k ) = 0 otherwise. Consider the following statements.

2V (b) i(t ) → 0 R (c) i(t ) →



∫

f (τ )dτ is

0

1 (a) sF(s) − f(0) (b) F ( s) s s 1 (c) ∫ F (τ ) dτ (d) [ F ( s) − f (0)] s 0 (GATE 2009: 2 Marks) s2 + 1 and s 2 + 2s + 1 input x(t) = sin(t +1) is in steady state. The output is sampled at a rate ws rad/s to obtain the final output {y(k)}. Which of the following is true?

143. An LTI system having transfer function

(a) y(x) is zero for all sampling frequencies ws (b) y(x) is non-zero for all sampling frequencies ws

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 104

Which of the following is correct? (a) Only S2 is true. (b) Both S1 and S2 are false. (c) Both S1 and S2 are true, and S2 is a reason for S1 . (d) Both S1 and S2 are true, but S2 is not a reason for S1 . (GATE 2009: 2 Marks) 146. For a parallel RLC circuit, which one of the following statements is NOT correct? (a)  The bandwidth of the circuit decreases if R is increased (b) The bandwidth of the circuit remains same if L is increased (c) At resonance, input impedance is a real quantity (d) At resonance, the magnitude of input impedance attains its minimum value (GATE 2010: 1 Mark)

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Chapter 2  •  Networks, Signals and Systems

147. For the two-port network shown in the following figure, the short-circuit admittance parameter matrix is

1Ω

0.5 Ω

1Ω

2

1 0.5 Ω

105

1Ω

0.5 Ω

+ −

10 V

1A 1´

1Ω

2´

1Ω

− 0.5⎤ ⎡ 4 − 2⎤ ⎡ 1 S (b) S (a) ⎢ ⎥ ⎢ 1 ⎥⎦ ⎣−2 4 ⎦ ⎣ − 0.5

(a) 0 W (c) 10 W

⎡ 1 0.5⎤ ⎡ 4 2⎤ (c) ⎢ S (d) ⎥ ⎢ 2 4⎥ S ⎣0.5 1 ⎦ ⎦ ⎣ (GATE 2010: 1 Mark)

2A

(b) 5 W (d) 100 W (GATE 2010: 2 Marks)

152. The current I in the circuit shown in the following figure is

148. If the scattering matrix [S ] of a two port network is 20 mH

⎡ 0.2∠0° 0.9∠90°⎤ [S ] = ⎢ ⎥ ⎣0.9∠90° 0.1∠90° ⎦

20∠0° V ω =103 rad/s

then the network is (a) loss less and reciprocal (b) loss less but not reciprocal (c) not loss less but reciprocal (d) neither loss less nor reciprocal (GATE 2010: 1 Mark)

I

(a) -j1 A (c) 0 A

1Ω

50 µF

(b) j1 A (d) 20 A (GATE 2010: 2 Marks)

153. In the circuit shown, the switch S is open for a long time + and 149. Consider the z-transform X ( z ) = 5 z + 4 z + 3; 0 < z < ∞ . is closed at t = 0. The current i(t) for t ≥ 0 is X ( z ) = 5 z 2 + 4 z −1 + 3; 0 < z < ∞. The inverse z-transform x[n] is t=0 10 Ω (a) 5δ [ n + 2] + 3δ [ n] + 4δ [ n − 1] S 15 mH 1.5 A 10 Ω (b) 5δ [ n − 2] + 3δ [ n] + 4δ [ n + 1] i(t) 10 Ω (c) 5u [ n + 2] + 3u [ n] + 4u [ n − 1] 2

−1

(d) 5u [ n − 2] + 3u [ n] + 4u [ n + 1]

(a) i(t) = 0.5 - 0.125 e-1000t A

(GATE 2010: 1 Mark) 150. Two discrete time systems with impulse responses h1 [n] = δ [n − 1] and h2 [n] = δ [n − 2] are connected in cascade. The overall impulse response of the cascaded system is (a) δ [n − 1] + δ [n − 2] (b) δ [n − 4]

(b) i(t) = 1.5 - 0.125 e-1000t A (c) i(t) = 0.5 - 0.5 e-1000t A (d) i(t) = 0.5 - 0.375 e-1000t A (GATE 2010: 2 Marks) 154. A continuous time LTI system is described by d 2 y (t ) dy(t ) dx(t ) +4 + 3 y (t ) = 2 + 4 x (t ) 2 dt dt dt

(c) δ [n − 3] (d) δ [n − 1] δ [n − 2]

(GATE 2010: 1 Mark)

151. In the circuit shown, the power supplied by the voltage source is

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 105

Assuming zero initial conditions, the response y(t) of the above system for the input x(t) = e−2tu(t) is given by (a) (et − e3t)u(t) (b) (e−t − e−3t)u(t)

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GATE ECE Chapter-wise Solved papers

(a) 5 Ω (c) 15 Ω

(c) (e−t + e−3t)u(t) (d) (e + e )u(t) t

3t

(b) 10 Ω (d) 20 Ω (GATE 2011: 1 Mark)

(GATE 2010: 2 Marks) 155. The transfer function of a discrete time LTI system is given by 2 − (3/ 4) z −1 H ( z) = 1 − (3/ 4) z −1 + (1/8) z −2

158. The circuit shown in the following figure is driven by a sinusoidal input Vi = Vpcos(t/RC). The steady output Vo is R

C

Consider the following statements:

+

1 S1:  The system is stable and causal for ROC : z > . 2 1 S2:  The system is stable but not causal for ROC : z < . 4 S3:  The system is neither stable nor causal for 1 1 ROC : < z < . 4 2 Which one of the following statements is valid? (a) Both S1 and S2 are true (b) Both S2 and S3 true (c) Both S1 and S3 are true (d) S1 , S2 and S2 are all true (GATE 2010: 2 Marks) 156. In the circuit shown in the following figure, the Norton equivalent current in amperes with respect to the terminals P and Q is j30 Ω P

16∠0°A

−j50 Ω

25 Ω

Q

15 Ω

(a) 6.4 - j4.8 (c) 10 + j0

(b) 6.56 - j 7.87 (d) 16 + j 0 (GATE 2011: 1 Mark)

157. In the circuit shown below, the value of RL such that the power transferred to RL is maximum is 10 Ω

+ Vi − Vp

159. If the unit step response of a network is (1 − e-at), then its unit impulse response is (a) ae−at (b) a −1e−at −1 −at (c) (1 − a )e (d) (1 − a)e−at (GATE 2011: 1 Mark) 160. The trigonometric Fourier series of an even function does not have the (a) DC term (b) cosine terms (c) sine terms (d) odd harmonic terms (GATE 2011: 1 Mark) 161. A system is defined by its impulse response h(n) = 2nu(n − 2). The system is (a) stable and causal (b) causal but not stable (c) stable but not causal (d) unstable and non-causal (GATE 2011: 1 Mark) 162. In the circuit shown below, the current I is equal to

j4 Ω

−j4 Ω

6Ω 14∠0° V 1A

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 106

Vo −

I

10 Ω + − 2V

C

Vp  t   t  (b) cos  sin    RC 3 3    RC  Vp Vp  t   t  (d) (c) cos  sin    RC 2 2    RC  (GATE 2011: 1 Mark) (a)

10 Ω

+ − 5V

R

RL

6Ω

6Ω

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Chapter 2  •  Networks, Signals and Systems

(a) 1.4 ∠ 0° A (c) 2.8 ∠ 0° A

(b) 2.0 ∠ 0° A (d) 3.2 ∠ 0° A (GATE 2011: 2 Marks)

163. In the following circuit shown, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t = 0. The current i(t) at a time t after the switch is closed is

The voltage gain

V2 is V1

1 1 (b) − 90 90 1 1 (c) − (d) − 11 99

(a)

(GATE 2011: 2 Marks)

i(t)

167. Two systems H1 ( z ) and H 2 ( z ) are connected in cascade as shown in the following figure. The overall output y[n] is the same as the input x[n] with a one unit delay. The transfer function of the second system H 2 ( z ) is

10 Ω 100 V

+ − − 50 µF +

(a) (b) (c) (d)

i(t) = 15 exp(-2 × 103t) A i(t) = 5 exp(-2 × 103t) A i(t) = 10 exp(-2 × 103t) A i(t) = -5 exp(-2 × 103t) A

x[n]

y[n]

H2(z)

(a)

z −1 (1 − 0.6 z −1 ) (1 − 0.6 z −1 ) (b) (1 − 0.4 z −1 ) z −1 (1 − 0.4 z −1 )

(c)

(1 − 0.4 z −1 ) z −1 (1 − 0.4 z −1 ) (d) −1 −1 z (1 − 0.6 z −1 ) (1 − 0.6 z )

(GATE 2011: 2 Marks) 164. An input x(t ) = exp( −2t ) u(t ) + δ (t − 6) is applied to an LTI system with impulse response h(t) = u(t). The output is (a) [1 − exp( −2t )]u(t ) + u(t + 6)

(1—0.4z —1) (1—0.6z —1)

H1(z)=

(GATE 2011: 2 Marks) 168. In the circuit shown below, the current through the inductor is

(b) [1 − exp( −2t )]u(t ) + u(t − 6) (c) 0.5 [1 − exp( −2t )]u(t ) + u(t + 6)

j1 Ω

1Ω

(d) 0.5 [1 − exp( −2t )]u(t ) + u(t − 6) (GATE 2011: 2 Marks)

1∠0 V −

2( s + 1) , then the initial and final s + 4s + 7 values of f(t), respectively, are (a) 0, 2 (b) 2, 0 (c) 0, 2/7 (d) 2/7, 0 (GATE 2011: 2 Marks)

165. If F ( s) = L[ f (t )] =

2

166. In the circuit shown in the following figure, the network N is described by the following Y matrix: ⎡ 0.1S −0.01S⎤ Y =⎢ ⎥ ⎣0.01S 0.1S ⎦ 25 W I1 I2 +

+ 100 V

+ −

V1

N

−

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 107

V2 −

100 W

−j1 Ω

+ 1∠0 A

1∠0 A 1∠0 V −

+

1Ω

(a)

2 −1 A (b) A 1+ j 1+ j

(c)

1 A 1+ j

(d) 0 A (GATE 2012: 1 Mark)

169. The average power delivered to an impedance (4 - j3) Ω by a current 5 cos (100pt + 100) A is (a) 44.2 W (b) 50 W (c) 62.5 W (d) 125 W (GATE 2012: 1 Mark)

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GATE ECE Chapter-wise Solved papers

170. The impedance looking into nodes 1 and 2 in the circuit shown in the following figure is

173. If VA - VB = 6 V, then VC - VD is VA

R

2Ω

VB

R

ib

R

99 ib

1 kΩ

−

R

R

1Ω

R

R

10 V + R

9Ω

+ − 5V

1 100 Ω

(b) 100 Ω (d) 10.1 kΩ (GATE 2012: 1 Mark)

2Ω + 10 V −

t=0

Circuit A (a) 0.8 Ω (c) 2 Ω

C1

C2 i(t)

(a) zero (b) a step function (c) an exponentially decaying function (d) an impulse function (GATE 2012: 1 Mark) n

n

⎛ 1⎞ ⎛ 1 ⎞ 172. If x [ n] = ⎜ ⎟ − ⎜ ⎟ u [ n] , then the ROC of its z⎝ 3⎠ ⎝ 2 ⎠ transform in the two-plane will be 1 1 1 < z < < z < 3 (b) 3 3 2 1 1 (c) < z < 3 (d) < z 2 3 (GATE 2012: 1 Mark)

(a)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 108

(b) 2 V (d) 6 V (GATE 2012: 2 Marks)

174. Assuming that both voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B (see the following figure) is

171. In the circuit shown in the following figure, C1 and C2 are ideal capacitors, C1 had been charged to 12 V before the ideal switch S is closed at t = 0, The current i(t) for all t is

S

VD 2A

(a) -5 V (c) 3 V

2

(a) 50 Ω (c) 5 kΩ

VC

R −j1 Ω

+ 3V −

Circuit B (b) 1.4 Ω (d) 2.8 Ω (GATE 2012: 2 Marks)

Common Data for Questions 175 and 176: With 10 V DC connected at port A in the linear non-reciprocal two-port network shown in the following figure, the following were observed: (i)  1 Ω connected at port B draws a current of 3 A. (ii) 2.5 Ω connected at port B draws a current of 2 A. + A

B −

175. For the same network, with 6 V DC connected at port A, 1 Ω connected at port B draws (7/3) A. If 8 V DC is connected to port A, the open circuit voltage at port B is (a) 6 V (b) 7 V (c) 8 V (d) 9 V (GATE 2012: 2 Marks) 176. With 10 V DC connected at port A of the network shown, the current drawn by 7 Ω connected at port B is

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Chapter 2  •  Networks, Signals and Systems

3 A 7

(b)

5 A 7

(c) 1 A

(d)

9 A 7

(a)

(a) k2 (b) k 1 (c) k (d) k (GATE 2013: 1 Mark)

(GATE 2012: 2 Marks) 177. The Fourier transform of a signal h(t) H ( jω ) = ( 2 cos ω )(sin 2ω ) /ω . The value of h(0) is (a) 1/4 (c) 1

is

(b) 1/2 (d) 2 (GATE 2012: 2 Marks)

178. Let y[n] denote the convolution of h[n] and g[n], where h[n] = (1/2)nu[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals 1 (a) 0 (b) 2 3 (c) 1 (d) 2

182. The impulse response of a system is h(t) = tu(t). For an input u(t −1), the output is t2 t (t − 1) u(t ) (b) u(t − 1) 2 2 (t − 1) 2 t2 −1 u(t − 1) (d) u(t − 1) (c) 2 2 (GATE 2013: 1 Mark) (a)

183. Assuming zero initial condition, the response y(t) of the system (shown in the following figure) to a unit step input u(t) is

U(s)

Y(s) 1/s

(GATE 2012: 2 Marks) 179. The input x(t) and output y(t) of a system are related as t

y(t ) =

∫ x(τ ) cos (3τ )dτ. The system is

(c)

−∞

(a) (b) (c) (d)

(a) u(t) (b) tu(t)

time-invariant and stable stable and not time-invariant time-invariant and not stable not time-invariant and not stable (GATE 2012: 2 Marks)

184. The transfer function ­follows is

180. A source vs (t ) = V cos 100π t has an internal impedance of (4 + j3) Ω. If a purely resistive load c­ onnected to this source has to extract the maximum power out of the source, its value (in Ω) should be (a) 3 (b) 4 (c) 5 (d) 7 (GATE 2013: 1 Mark) 181. Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k > 0, the elements of the corresponding star equivalent will be scaled by a factor of

t2 u(t ) (d) e−tu(t) 2 (GATE 2013: 1 Mark) V2 ( s) of the circuit shown as V1 ( s)

100 µF +

+ 10 kΩ

V1(s)

V2(s) 100 µF

−

−

(a)

0.5s + 1 3s + 6 (b) s +1 s+2

(c)

s+2 s +1 (d) s +1 s+2 (GATE 2013: 1 Mark)

RC

Ra Rb

Rc

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 109

RA

RB

2

185. Let g (t ) = e − π t , and h(t) is a filter matched to g(t). If g(t) is applied as input to h(t), then the Fourier transform of the output is

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GATE ECE Chapter-wise Solved papers 2

2

(a) e − π f (b) e −π f / 2 2

(c) e − π f (d) e −2π f (GATE 2013: 1 Mark) 186. A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is (a) 5 kHz (b) 12 kHz (c) 15 kHz (d) 20 kHz (GATE 2013: 1 Mark) 187. Two systems with impulse responses h1(t) and h2(t) are connected in cascade. Then the overall impulse response of the cascaded system is given by (a) product of h1(t) and h2(t) (b) sum of h1(t) and h2(t) (c) convolution of h1(t) and h2(t) (d) subtraction of h2(t) from h1(t) (GATE 2013: 1 Mark) 188. Which one of the following statement is NOT TRUE for a continuous time causal and stable LTI system? (a) All poles of the system must lie on the left side of the jw -axis. (b) Zeros of the system can lie anywhere in the s-plane. (c) All poles must lie within s = 1.

C2

189. The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An AC voltage VWX1 = 100 V is applied across WX to get an open circuit voltage VYZl across YZ. Next, an AC voltage VYZ2 = 100 V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1/VWX1, Vwx2/VYZ2 are, respectively, W 1 : 1.25 Y

Z

80 100 125 80 and (a) (b) and 100 100 100 100 100 100 80 80 (c) and (d) and 100 100 100 100

C3

C1 (a) 2.8 and 36 (c) 2.8 and 32

(b) 7 and 119 (d) 7 and 80 (GATE 2013: 2 Marks)

Common Data for Questions 191 and 192: Consider the ­following figure: I1

5Ω Is 10 V

+ Vs

(d) A  ll roots of the characteristic equation must be located on the left side of the jw -axis. (GATE 2013: 1 Mark)

X

190. Three capacitors C1, C2 and C3 whose values are 10 µF, 5 µF, and 2 µF, respectively, have breakdown voltages of 10 V, 5 V and 2 V, respectively. For the interconnection shown in the following figure, the maximum safe voltage in volts that can be applied across the combination, and the corresponding total charge (in µC) stored in the effective capacitance across the terminals are, respectively,

−

1Ω

2Ω

2A

191. The current Is in amperes in the voltage source, and voltage Vs in volts across the current source respectively, are (a) 13, -20 (b) 8, -10 (c) -8, 20 (d) -13, 20 (GATE 2013: 2 Marks) 192. The current in the 1 Ω resistor (in amperes) is (a) 2 (b) 3.33 (c) 10 (d) 12 (GATE 2013: 2 Marks) 193. In the circuit shown in the following figure, if the source voltage Vs = 100 ∠53.13°V , then the Thevenin’s equivalent voltage (in volts) as seen by the load resistance RL is 3Ω

j4 Ω +

Vs

i1

− VL1 + j40i2 −

j6 Ω

3Ω

RL =10 Ω + 10VL1 i2 −

(GATE 2013: 2 Marks)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 110

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Chapter 2  •  Networks, Signals and Systems

(a) 100 ∠90° (b) 800 ∠0° (c) 800 ∠90° (d) 100∠60° (GATE 2013: 2 Marks) 194. Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor q at the same operating frequency is 1 1 + (a) q1 + q2 (b) q1 q2 (c)

q1 R2 + q2 R1 q1 R1 + q2 R2 (d) R1 + R2 R1 + R2

For R = 1 Ω and currents i1 = 2 A, i4 = -1 A, i5 = - 4 A, which one of the following is TRUE? (a)  i6 = 5 A (b)  i3 = -4 A (c) Data is sufficient to conclude that the supposed currents are impossible. (d) Data is insufficient to identify the currents i2, i3 and i6 (GATE 2014: 1 Mark) 198. The magnitude of current (in mA) through the resistor R2 in the figure shown is .

R2

(GATE 2013: 2 Marks)

1kΩ

195. A system is described by the differential equation d 2 y (t ) dy(t ) +5 + 6 y(t ) = x(t ). Let x(t) be a rectangular 2 dt dt pulse given by

R1 2kΩ

10 mA

⎧1 0 < t < 2 x (t ) = ⎨ ⎩0 otherwise

(GATE 2014: 1 Mark)

1 − e −2 s e −2 s (b) s( s + 2)( s + 3) s( s + 2)( s + 3)

e −2 s 1 − e −2 s (c) (d) ( s + 2)( s + 3) ( s + 2)( s + 3) (GATE 2013: 2 Marks) 196. The impulse response of a continuous time system is given by h(t ) = δ ( t − 1) + δ ( t − 3). The value of the step response at t = 2 is (a) 0 (b) 1 (c) 2 (d) 3 (GATE 2013: 2 Marks) 197. Consider the configuration shown in the figure which is a portion of a larger electrical network i5 i2

R

R i3

i1

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 111

199. For the maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z1 of the first section to the input impedance Z2 of the second section is (a)  Z2 = Z1 (b)  Z2 = −Z1 Z 2 = Z1* (c) 

Z 2 = − Z1* (d)  (GATE 2014: 1 Mark)

200. In the figure shown, the value of the current I (in Amperes) is . 5W

5W I

+ 5V −

1A

10 W

(GATE 2014: 1 Mark) 201. Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance (a) in series with a current source (b) in parallel with a voltage source (c) in series with a voltage source (d) in parallel with a current source (GATE 2014: 1 Mark)

R

i4

2 mA

R4 3kΩ

dy(t ) = 0 at t = 0, the LapAssuming that y(0) = 0 and dt lace transform of y(t) is (a)

R3 4kΩ

i6

202. A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8

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GATE ECE Chapter-wise Solved papers

leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is (a) (18 + j1.5) kVA (b) (18 – j1.5) kVA (c) (20 + j1.5) kVA (d) (20 – j1.5) kVA

(d) i(t)

0

t (GATE 2014: 1 Mark)

(GATE 2014: 1 Mark) 203. In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in mA) through the 4 kΩ resistor at t = 0+ is . 4 kW

5 kW 10 V + −

1 kW

i 10 mF

1 mH t=0 (GATE 2014: 1 Mark)

204. A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage Vs, the resistance R, the capacitance C, and the current i(t) is given below: Vs = Ri(t ) +

1 t i(u )du. C ∫0

Which one of the following represents the current i(t)?

205. A system is described by the following differential equation, where u(t) is the input to the system and y(t) is the output of the system. . y(t) + 5y(t) = u(t) when y(0) = 1 and u(t) is a unit step function, y(t) is (a) 0.2 + 0.8e−5t (b) 0.2 - 0.2e−5t −5t (c) 0.8 + 0.2e (d) 0.8 - 0.8e−5t (GATE 2014: 1 Mark) 206. A two-port network has scattering parameters given by s ⎤ ⎡s [ s] = ⎢ 11 12 ⎥ . If the port-2 of the two-port network is ⎣ s21 s22 ⎦ short circuited, the s11 parameter for the resultant oneport network is s −s s +s s (a) 11 11 22 12 21 1 + s22 s −s s −s s (b) 11 11 22 12 21 1 + s22 s +s s +s s (c) 11 11 22 12 21 1 − s22

(a) i(t)

s −s s +s s (d) 11 11 22 12 21 1 − s22 (GATE 2014: 1 Mark) 0

t

(b) i(t)

207. The state transition matrix ⎡ x1 ⎤ ⎡0 1 ⎤ ⎡ x1 ⎤ ⎢ x ⎥ = ⎢0 0 ⎥ ⎢ x ⎥ is ⎦⎣ 2⎦ ⎣ 2⎦ ⎣

ϕ (t )

of a system

⎡t 1 ⎤ 1 0  (a) ⎢ (b) ⎥ t 1  ⎣1 0 ⎦   0

t

(c)

(GATE 2014: 1 Mark)

i(t)

0

0 1 1 t  (d) (c)   0 1 1 t   

t

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 112

208. Consider two real valued signals, x(t) band-limited to [−500 Hz, 500 Hz] and y(t) band-limited to [−1 kHz, 1 kHz]. For z(t) = x(t) . y(t), the Nyquist sampling frequency (in kHz) is . (GATE 2014: 1 Mark)

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Chapter 2  •  Networks, Signals and Systems

209. Let x(t) = cos(10p t) + cos(30p t) be sampled at 20 Hz and reconstructed using an ideal low-pass filter with cut-off frequency of 20 Hz. The frequency/frequencies present in the reconstructed signal is/are (a) 5 Hz and 15 Hz only (b) 10 Hz and 15 Hz only (c) 5 Hz, 10 Hz and 15 Hz only (d) 5 Hz only (GATE 2014: 1 Mark) 210. A modulated signal is y(t) = m(t)cos(40000pt), where the baseband signal m(t) has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which y(t) should be sampled to recover m(t) is . (GATE 2014: 1 Mark) ⎛ π n⎞ 211. Consider a discrete time periodic signal x[n] = sin ⎜ ⎟ . ⎝ 5⎠ Let ak be the complex Fourier series coefficients of x[n]. The coefficients {ak } are non-zero when k = Bm ± 1, where m is any integer. The value of B is . (GATE 2014: 1 Mark) 212. Let x[n] = x[–n]. Let X(z) be the z-transform of x[n]. If 0.5 + j 0.25 is a zero of X(z), which one of the following must also be a zero of X(z)? (a) 0.5 – j0.25 (b) 1/(0.5 + j 0.25) (c) 1/(0.5 – j0.25) (d) 2 + j4 (GATE 2014: 1 Mark) 213. A Fourier transform pair is given by n

Ae − j 6π f ⎛ 2 ⎞ − j 2π f 1− ⎜ ⎟ e ⎝ 3⎠ where u[n] denotes the unit step sequence. The value of A is . (GATE 2014: 1 Mark) ⎛ 2⎞ FT ⎜⎝ ⎟⎠ u[n + 3] ←⎯→ 3

214. The sequence x[n] = 0.5n u[n], where u[n] is the unit step sequence, is convolved with itself to obtain y[n]. Then +∞

∑ y[n]

is

.

n = −∞

(GATE 2014: 1 Mark) 215. A discrete-time signal x[n] = sin(p2n), n being an integer, is (a) periodic with period p (b) periodic with period p2 (c) periodic with period p/2 (d) not periodic (GATE 2014: 1 Mark)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 113

113

216. A continuous, linear time-invariant filter has an impulse response h(t) described by ⎧3 for 0 ≤ t ≤ 3 h(t ) = ⎨ ⎩0 otherwise

When a constant input of value 5 is applied to this filter, the steady state output is . (GATE 2014: 1 Mark)

217. An FIR system is described by the system function 3 7 H ( z ) = 1 + z −1 + z −2 2 2 The system is (a) maximum phase (b) minimum phase (c) mixed phase (d) zero phase (GATE 2014: 1 Mark) 218. The phase response of a passband waveform at the receiver is given by f (f) = –2pa (f – fc) – 2pb fc where fc is the centre frequency, and a and b are positive constants. The actual signal propagation delay from the transmitter to receiver is αβ (a) α − β (b) α +β α+β (c) a (d) b (GATE 2014: 1 Mark) 219. A real-valued signal x(t) limited to the frequency band f ≤ W / 2 is passed through a linear time invariant system whose frequency response is W ⎧ − j 4π f f ≤ ⎪⎪e 2 H( f ) = ⎨ W ⎪ 0, f > ⎪⎩ 2 The output of the system is (a) x(t + 4) (b) x(t – 4) (c) x(t + 2) (d) x(t – 2) (GATE 2014: 1 Mark) 220. In the magnetically coupled circuit shown in the figure, 56% of the total flux emanating from one coil links the other coil. The value of the mutual inductance (in H) is .

10 Ω

4H

M

5H

(1/16) F

60 cos(4t + 30°)V (GATE 2014: 2 Marks)

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GATE ECE Chapter-wise Solved papers

221. In the circuit shown in the figure, the value of node ­voltage V2 is −

+

The value of Re/R is

(GATE 2014: 2 Marks)

V2 4W

1W j6 W

6W

−j3 W

4Ð0° A

(a) 22 + j  2 V (c) 22 − j  2 V

.

225. In the circuit shown in the figure, the angular frequency w (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b′ is purely resistive, is .

10Ð0° V

V1



(b) 2 + j  22 V (d) 2 − j  22 V (GATE 2014: 2 Marks)

222. The steady state output of the circuit shown in the figure is given by y(t ) = A(ω ) sin(ω t + ϕ (ω )). If the amplitude A(ω ) = 0.25, then the frequency w is

R

C y(t)

1F b

10 cos w t (Volts)

+ 0.5H − b¢ (GATE 2014: 2 Marks)

226. For the Y-network shown in the figure, the value of R1 (in Ω) in the equivalent Δ-network is . R1

5W

3W

7.5 W

C

+ sin(ω t) −

C

(a)

1

2 (b) 3RC 3RC

(c)

1 2 (d) RC RC

(GATE 2014: 2 Marks)

(GATE 2014: 2 Marks)

227. In the circuit shown in the figure, the value of capacitor C (in mF) needed to have critically damped response i(t) is .

223. In the circuit shown in the figure, the value of vo(t) (in Volts) for t → ∞ is .

ix + 2ix -

5W

5W

+ vo(t) -

(GATE 2014: 2 Marks) 224. The equivalent resistance in the infinite ladder network shown in the figure, is Re.

2R

R

R

R

R

Re

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 114

R

R

C 4H i(t)

2H

10 u(t) A

40 W

R

+ − Vo

(GATE 2014: 2 Marks) 228. A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L, C and R are 1 H, 1 F and 1 Ω, respectively, the operating angular frequency (in rad/s) is . (GATE 2014: 2 Marks) 229. In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t > 0?

12/4/2018 11:06:53 AM

Chapter 2  •  Networks, Signals and Systems

I

1 kW 5V

C

R2

+ −

1 mF

2 kW

(GATE 2014: 2 Marks) 1 . s2 + s + 1 Which one of the following is the unilateral Laplace transform of g(t) = t ⋅ f(t)?

232. The unilateral Laplace transform of f(t) is

5 2 (a) I (t ) = (1 − e − t / τ ), τ = ms 3 3 (b) I (t ) =

5 2 (1 − e − t / τ ), τ = ms 2 3

5 (c) I (t ) = (1 − e − t / τ ), τ = 3 ms 3 (d) I (t ) =

x1(t) = −1, x2(t) = 2 x1(t) = −e−t, x2(t) = 2e−t x1(t) = e−t, x2(t) = −e−2t x1(t) = −e−t, x2(t) = −2e−t

(a) (b) (c) (d)

R1

115

5 (1 − e − t / τ ), τ = 3 ms 2 (GATE 2014: 2 Marks)

230. Consider the building block called ‘Network N’ shown in the figure. Let C = 100 mF and R = 10 kΩ. Network N +

+

(a)

−s −( 2 s + 1) (b) 2 ( s 2 + s + 1) 2 ( s + s + 1)

(c)

s 2s + 1 (d) 2 2 ( s + s + 1) ( s + s + 1) 2

2

2

(GATE 2014: 2 Marks) 233. In the h-parameter model of the 2-port network given in the figure shown, the value of h22 (in S) is . 3W

3W

1

3W

2

C V1(s)

R

V2(s)

−

−

2W 2W 2W

1¢

2¢

Two such blocks are connected in cascade, as shown in the figure. (GATE 2014: 2 Marks) + V1(s)

+ Network N

−

+ Network N

−

V3(s)

234. For the two-port network shown in the figure, the impedance (Z) matrix (in Ω) is

30 W

−

1

2 +



The transfer function

V3 ( s) of the cascaded network is V1 ( s)

s (b) s2 (a) 1+ s 1 + 3s + s 2 2

⎛ s ⎞ s (c) ⎜ (d) ⎝ 1 + s ⎟⎠ 2+ s (GATE 2014: 2 Marks) 231. An unforced linear time invariant (LTI) system is represented by ⎡ x1 ⎤ ⎡ −1 0 ⎤ ⎡ x1 ⎤ ⎢ x ⎥ = ⎢ 0 −2⎥ ⎢ x ⎥ ⎦⎣ 2⎦ ⎣ 2⎦ ⎣

If the initial conditions are x1(0) = 1 and x2(0) = −1, the solution of the state equation is

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 115

+ 10 W

− 1¢

60 W − 2¢

⎡6 24 ⎤ ⎡9 8 ⎤ (b) (a) ⎢ ⎥ ⎢8 24 ⎥ ⎦ ⎣ 42 9 ⎦ ⎣ ⎡9 6 ⎤ ⎡ 42 6 ⎤ (c) ⎢ (d) ⎥ ⎢ ⎥ ⎣6 24 ⎦ ⎣6 60 ⎦ (GATE 2014: 2 Marks) 235. A Y-network has resistances of 10 Ω each in two of its arms, while the third arm has a resistance of 11 Ω. In the equivalent Δ-network, the lowest value (in Ω) among the three resistances is . (GATE 2014: 2 Marks)

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GATE ECE Chapter-wise Solved papers

236. A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x(t) is . x(t) 1

0

t T/2

T/2 (GATE 2014: 2 Marks)

237. The N-point DFT X of a sequence x[n], 0 ≤ n ≤ N − 1 is given by X [k ] =

1 N

N −1

∑ x[n]e

−j

2π nk N

, 0 ≤ k ≤ N − 1.

n= 0

Denote this relation as X = DFT(x). For N = 4, which one of the following sequences satisfies DFT (DFT(x)) = x? (a) x = [1 2 3 4] (b) x = [1 2 3 2] (c) x = [1 3 2 2] (d) x = [1 2 2 3] (GATE 2014: 2 Marks)

238. The z-transform of the sequence x[n] is given by 1 X ( z) = , with the region of convergence (1 − 2 z −1 ) 2 |z | > 2. Then, x[2] is

. (GATE 2014: 2 Marks)

239. Consider a discrete-time signal ⎧n for 0 ≤ n ≤ 10 x[n] = ⎨ ⎩0 otherwise If y[n] is the convolution of x[n] with itself, the value of y[4] is . (GATE 2014: 2 Marks)



242. A stable linear time invariant (LTI) system has a transfer 1 function H ( s) = 2 . To make this system causal s + s−6 it needs to be cascaded with another LTI system having a transfer function H1(s). A correct choice for H1(s) among the following options is (a) s + 3 (b) s – 2 (c) s – 6 (d) s + 1 (GATE 2014: 2 Marks) 243. A causal LTI system has zero initial conditions and impulse response h(t). Its input x(t) and output y(t) are related through the linear constant-­coefficient differential equation d 2 y (t ) dy(t ) +α + α 2 y(t ) = x(t ). 2 dt dt Let another signal g(t) be defined as t

dh(t ) + α h(t ). dt 0 If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is . (GATE 2014: 2 Marks) g (t ) = α 2 ∫ h(τ )dτ +

244. In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in Volts) across the capacitor is . 4Ω 0.1 mH + 10 cos wt (Volts)

240. The input-output relationship of a causal stable LTI ­system is given as y[n] = ay[n - 1] + bx[n]

⎡ 1 ⎞⎤ ⎛ The value of the expectation E ⎢π x(t ) x ⎜ t − ⎟ is ⎝ 4 w ⎠ ⎥⎦ ⎣ . (GATE 2014: 2 Marks)

1 µF −

If the impulse response h[n] of this system satisfies the condition ∑ n = 0 h[n] = 2, the relationship between a and

(GATE 2015: 1 Mark)

∞

b is (a) a = 1 - b/2 (b) a = 1 + b/2 (c) a = 2b (d) a = –2b (GATE 2014: 2 Marks)

245. In the network shown in the figure, all resistors are identical with R = 300 Ω. The resistance Rab (in Ω) of the network is . a R

R

R

R

R 241. The power spectral density of a real stationary random process x(t) is given by

R R

R

R

R ⎧1 ⎪ , SX ( f ) = ⎨ w ⎪0, ⎩

Rab

R = 300 W R

R

R

R

R

f ≤w f >w

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 116

b (GATE 2015: 1 Mark)

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Chapter 2  •  Networks, Signals and Systems

246. In the given circuit, the values of V1 and V2 respectively are

250. For the circuit shown in the figure, the Thevenin equivalent voltage (in Volts) across terminals a – b is .

4Ω

a

I

+ V2

4Ω

5A

117

2I

+

4Ω

12 V

1A

V1 −

−

b (GATE 2015: 1 Mark)

(a) 5V, 25V (c) 15V, 35V

(b) 10V, 30V (d) 0V, 20V (GATE 2015: 1 Mark)

247. In the circuit shown, the voltage Vx (in volts) is

251. The voltage (VC) across the capacitor (in Volts) in the network shown is . 80 V

VC

40 V

.

0.5 Vx

10 W

100 V, 50 Hz (GATE 2015: 1 Mark)

+ 5A

Vx

20 W

+ −

8W

−

0.25 Vx

252. In the circuit shown, the average value of the voltage Vab (in Volts) in steady state condition is .

(GATE 2015: 1 Mark) 248. At very high frequencies, the peak output voltage Vo (in Volts) is .

1 kW 5π sin (5000t)

1 µF b −

2 kW

a 1 mH +

5V

Vab 100 mF

(GATE 2015: 1 Mark) 1 kW

1 kW 100 mF

Vo

+ 1.0 sin(w t) V − 1 kW

253. In the circuit shown, the switch SW is thrown from position A to position B at time t = 0. The energy (in mJ) taken from the 3 V source to charge the 0.1 mF capacitor from 0 V to 3 V is

1 kW

100 mF

3V

(GATE 2015: 1 Mark)

4 Ð0 Vrms

RL

(GATE 2015: 1 Mark)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 117

A

0.1 mF

(a) 0.3 (c) 0.9 j2 W

SW B

t=0

249. In the given circuit, the maximum power (in Watts) that can be transferred to the load RL is . 2W

120 Ω

(b) 0.45 (d) 3 (GATE 2015: 1 Mark)

254. Let the signal f(t) = 0 outside the interval [T1, T2], where T1 and T2 are finite. Furthermore, f (t ) < ∞. The region of convergence (RoC) of the signal’s bilateral Laplace transform F(s) is

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118

GATE ECE Chapter-wise Solved papers

(a) a parallel strip containing the j Ω axis (b) a parallel strip not containing the j Ω axis (c)  the entire s-plane (d) a half plane containing the j Ω axis (GATE 2015: 1 Mark)

258. Two causal discrete-time signals x[n] and y[n] are

255. The 2-port admittance matrix of the circuit shown is given by

259. The magnitude and phase of the complex Fourier series coefficients ak of a periodic signal x(t) are shown in the figure. Choose the correct statement from the four choices given. Notation: C is the set of complex numbers, R is the set of purely real numbers, and P is the set purely imaginary numbers.

5W 10 W

10 W

n

related as y[n] = ∑ x[m]. If the z-transform of y[n] is m=0

2 , the value of x[2] is z ( z - 1) 2

2 1 k

−5 −4 −3 −2 −1 0

256. The waveform of a periodic signal x(t) is shown in the figure.

1

2

3

4

1

2

3

4

∠ak

(GATE 2015: 1 Mark) −5 −4 −3 −2 −1

k 0 −π

x(t) (a) x(t ) ∈ R

3

(b) x(t ) ∈ P

−2

1

4

−1

2

t

3

−3

3

2

⎡0.3 0.4 ⎤ ⎡3.33 5 ⎤ (c) ⎢ ⎥ (d) ⎢0.4 0.3 ⎥ 5 3 . 33 ⎣ ⎦ ⎣ ⎦

−4 −3

(GATE 2015: 1 Mark)

ak|

3 ⎡0.3 0.2⎤ ⎡15 5 ⎤ (b) (a) ⎢ ⎥ ⎢ ⎥ ⎣0.2 0.3⎦ ⎣5 15⎦

.

⎛ t − 1⎞ A signal g(t) is defined by g (t ) = x ⎜ . The average ⎝ 2 ⎟⎠ power of g(t) is . (GATE 2015: 1 Mark)

257. For the discrete-time system shown in the figure, the poles of the system transfer function are located at Y [n]

X[n] −1 6

5 6 z−1

z−1

(c) x(t ) ∈ (C - R) (d) The information given is not sufficient to draw any conclusion about x(t). (GATE 2015: 1 Mark) 260. The result of the convolution x(−t)*d (−t − t0) is (a) x(t + t0) (b) x(t − t0) (c) x(−t + t0) (d) x(−t − t0) (GATE 2015: 1 Mark) 261. The impulse response of an LTI system can be obtained by (a) differentiating the unit ramp response (b) differentiating the unit step response (c) integrating the unit ramp response (d) integrating the unit step response (GATE 2015: 1 Mark) 262. Consider a four-point moving average filter defined 3

1 1 1 1 (a) 2, 3  (b)  , 3   (c)  ,   (d)  2, 3 2 2 3 (GATE 2015: 1 Mark)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 118

by the equation y[n] = ∑ α i x[n - i ] . The condition i =0

on the filter coefficients that results in a null at zero frequency is.

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Chapter 2  •  Networks, Signals and Systems

(a) (b) (c) (d)

a1 = a2 = 0; a0 = −a3 a1 = a2 = 1; a0 = −a3 a0 = a3 = 0; a1 = a2 a1 = a2 = 0; a0 = a3

(c) (d) (GATE 2015: 1 Mark)

263. In the circuit shown, the Norton equivalent resistance (in Ω) across terminals a - b is . 2W

1 2π LC 1 2π LC

1-

119

L R 2C

1 - R2

C L

(GATE 2015: 2 Marks)

268. In the circuit shown, the current I flowing through the 50 Ω resistor will be zero if the value of capacitor C (in  mF) is .

a 50 W + −

4I

2W

4W

1 mH

1 mH

I 5p sin (5000t)

1 mH

C

I b (GATE 2015: 2 Marks)

(GATE 2015: 2 Marks)

264. A coaxial capacitor of inner radius 1 mm and outer radius 5 mm has a capacitance per unit length of 172 pF/m. If the ratio of outer radius to inner is doubled, the capacitance per unit length (in pF/m) is . (GATE 2015: 2 Marks)

269. Let x(t) = as(t) + s(−t) with s(t) = be−4tu(t), where u(t) is unit step function. If the bilateral Laplace transform of x(t) is

265. The damping ratio of a series RLC circuit can be expressed as R 2C 2L (a) (b) 2L R 2C (c)

t=0

3W

SW 2W

5 F 6

+ −

vC(t)

(GATE 2015: 2 Marks) 267. An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is (a) (b)



1

2π LC

Then the value of b is

270. A network is described by the state model as x1 = 2 x1 − x2 + 3u x2 = −4 x2 − u y = 3 x1 − 2 x2

⎛ Y ( s) ⎞ is The transfer function H ( s) = ⎜ ⎝ U ( s) ⎟⎠ (a)

11s − 35 11s + 35 (b) ( s − 2)( s + 4) ( s − 2)( s + 4)

(c)

11s + 38 11s − 38 (d) ( s − 2)( s + 4) ( s − 2)( s + 4) (GATE 2015: 2 Marks)

271. The ABCD parameters of the following 2-port network are (5 + j4) W

(5 − j4) W

(2 + j0) W

2π LC 1

16 , −4 < Re {s} < 4; s − 16 2

(GATE 2015: 2 Marks)

2 L R C (d) R C 2 L (GATE 2015: 2 Marks)

266. In the circuit shown, switch SW is closed at t = 0. Assuming zero initial conditions, the value of vC(t) (in  Volts) at t = 1 second is .

10 V + −

X ( s) =

1 − R2

C L

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 119

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120

GATE ECE Chapter-wise Solved papers

20.5 ⎤ ⎡3.5 + j 2 (a) ⎢ 20 . 5 3 . 5 − j 2⎥⎦ ⎣

x[n]

+ z−1

30.5 ⎤ ⎡3.5 + j 2 (b) ⎢ 0 . 5 3 . 5 − j 2⎥⎦ ⎣ 2 + j 0⎤ ⎡ 10 (c) ⎢ 10 ⎥⎦ ⎣2 + j0

+

y[n]

z−1 −2/9

0.5 ⎤ ⎡7 + j 4 (d) ⎢ . 30 5 7 − j 4 ⎥⎦ ⎣ (GATE 2015: 2 Marks) 272. Consider a continuous-time signal defined as ⎛ sin(π t / 2) ⎞ ∞ x (t ) = ⎜ ⎟ * ∑ δ (t − 10 n) ⎝ (π t / 2) ⎠ n =−∞

−5/3

1 +

where ‘*’ denotes the convolution operation and t is in seconds. The Nyquist sampling rate (in samples/s) for . x(t) is (GATE 2015: 2 Marks)

273. Consider two real sequences with time-origin marked by the bold value x1[n] = {1, 2, 3, 0}, x2[n] = {1, 3, 2, 1} Let X1(k) and X2(k) be 4-point DFTs of x1[n] and x2[n], respectively. Another sequence x3[n] is derived by taking 4-point inverse DFT of X3(k) = X1(k)X2(k). The value of x3[2] is . (GATE 2015: 2 Marks)

5/3

n

n

n

n

⎛ 2⎞ ⎛ 1⎞ (a) 4 ⎜ − ⎟ u[n] − 5 ⎜ − ⎟ u[n] ⎝ 3⎠ ⎝ 3⎠ ⎛ 1⎞ ⎛ 2⎞ (b) 5 ⎜ − ⎟ u[n] − 3 ⎜ − ⎟ u[n] 3 ⎝ ⎠ ⎝ 3⎠ (c)

n

n

n

n

⎛2⎞ ⎛1⎞ 5 ⎜ ⎟ u[n] − 5 ⎜ ⎟ u[n] 3 ⎝ ⎠ ⎝3⎠

⎛1⎞ ⎛2⎞ (d) 5 ⎜ ⎟ u[n] − 5 ⎜ ⎟ u[n] 3 ⎝ ⎠ ⎝3⎠ (GATE 2015: 2 Marks) 277. The pole-zero diagram of a causal and stable discretetime system is shown in the figure. The zero at the origin has multiplicity 4. The impulse response of the system is h[n]. If h[0] = 1, we can conclude Im(z)

⎛πn⎞ x[n] = 1 + cos ⎜ ⎟ be periodic signal with ⎝ 8 ⎠ period 16. Its DFS coefficients are defined by

274. Let

´

1 15 ⎛ π ⎞ ak = ∑ x[n]exp ⎜ − j kn ⎟ for all k. The value of the 16 n = 0 ⎝ 8 ⎠ coefficient a31 is . (GATE 2015: 2 Marks) 275. Suppose x[n] is an absolutely summable discrete-time signal. Its z-transform is a rational function with two poles and two zeros. The poles are at z = ±2j. Which one of the following statements is TRUE for the signal x[n]? (a) It is a finite duration signal (b) It is a causal signal (c) It is a non-causal signal (d) It is a periodic signal (GATE 2015: 2 Marks) 276. A realization of a stable discrete time system is shown in figure. If the system is excited by a unit step sequence input x[n], the response y[n] is

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 120

0.5

´

4 −0.5 ´

(a) (b) (c) (d)

Re(z) 0.5

−0.5 ´

h[n] is real for all n h[n] is purely imaginary for all n h[n] is real for only even n h[n] is purely imaginary for only odd n (GATE 2015: 2 Marks)

278. Input x(t) and output y(t) of an LTI system are related by the differential equation y ′′(t ) − y ′(t ) − 6 y(t ) = x(t ) . If the system is neither causal nor stable, the impulse response h(t) of the system is

12/4/2018 11:07:07 AM

Chapter 2  •  Networks, Signals and Systems

(a)



1 3t 1 e u( −t ) + e −2t u( −t ) 5 5

1 1 (b) − e 3t u( −t ) + e −2t u( −t ) 5 5 (c)

At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects? (a) 2E  (b)   2E   (c)  E  (d)  E/2 (GATE 2016: 1 Mark)

282. In the figure shown, the current i (in ampere) is

1 3t 1 e u( −t ) − e −2t u( −t ) 5 5

121

1 1 (d) − e 3t u( −t ) − e −2t u( −t ) 5 5

1A

(GATE 2015: 2 Marks)

5Ω 1Ω

8V+ − + 8V 1Ω −

1Ω

279. A zero mean white Gaussian noise having power spectral density N 0 / 2 is passed through an LTI filter whose impulse response h(t) is shown in the figure. The variance of the filtered noise at t = 4 is

.

i 1Ω

(GATE 2016: 1 Mark) h(t)

283. In the circuit shown below, VS is a constant voltage source and IL is a constant current load

A

+

t 0

1

2

3

R VS

−A

3 2 3 2 A N 0 (b) A N0 2 4 1 2 (c) A2N0 (d) A N0 2 (GATE 2015: 2 Marks)

(a)

280. Two sequences x1[n] and x2[n] have the same energy. Suppose x1[n] = a 0.5nu[n], where a is a positive real number and u[n] is the unit step sequence. Assume ⎪⎧ 1.5 for n = 0,1 x2 [n] = ⎨ ⎩⎪ 0 otherwise Then the value of a is



IL

−

The value of IL that maximizes the power absorbed by the constant current load is (a) (c)

VS VS (b) 4R 2R VS (d) ∞ R (GATE 2016: 1 Mark)

284. The switch has been in position 1 for a long time and abruptly changes to position 2 at t = 0. 3Ω

.

1

(GATE 2015: 2 Marks) 281. The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2d. d

+ − 10 V 2 Ω 0.1 F



2

4Ω

t=0 + VC 2 Ω −

2Ω

5A

If time t is in seconds, the capacitor voltage VC (in volts) for t > 0 is given by (a) 4[1 - exp(-t/0.5)] (b) 10 - 6 exp(-t/0.5) (c) 4[1 - exp(t/0.6)] (d) 10 - 6 exp(-t/0.6) (GATE 2016: 1 Mark)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 121

12/4/2018 11:07:09 AM

122

GATE ECE Chapter-wise Solved papers

285. The figure shows an RLC circuit with a sinusoidal current source.

Zb(jw) =Rb + jw Zb

Za 1 Imsinw t

R 10 Ω



IC

IL

IR

1′

L

C 10 mH 10 µF



At resonance, the ratio I L / I R , i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is . (GATE 2016: 1 Mark)

286. In the RLC circuit shown in the figure, the input voltage is given by vi(t) = 2 cos (200t) + 4 sin (500t). The output voltage vo(t) is 100 µF

0.25 H +

+ 2Ω

10 µF

0.4 H vi(t)

vo(t) 2Ω

−

−

(a) cos(200t) + 2 sin(500t) (b) 2 cos(200t) + 4 sin(500t) (c) sin(200t) + 2 cos(500t) (d) 2 sin(200t) + 4 cos(500t)

2′

Then the value of Rb (in Ω) equals . (GATE 2016: 1 Mark)

289. A continuous-time function x(t) is periodic with period T. The function is sampled uniformly with a sampling period Ts. In which of the following cases is the simplified signal periodic? (a) T = 2TS (b) T = 1.2 Ts (c) Always

(d) Never (GATE 2016: 1 Mark)

290. A continuous-time sinusoid of frequency 33 Hz is multiplied with a periodic Dirac impulse train of frequency 46 Hz. The resulting signal is passed through an ideal analog low-pass filter with a cutoff frequency of 23 Hz. The fundamental frequency (in Hz) of the output is . (GATE 2016: 1 Mark) 291. Consider the signal x(t) = cos(6π t) + sin(8π t), where t is in seconds. The Nyquist sampling rate (in samples/ second) for the signal y(t) = x(2t + 5) is (a) 8

(b) 12

(c) 16

(d) 32 (GATE 2016: 1 Mark)

(GATE 2016: 1 Mark) 287. Consider a two-port network with the transmission ⎛ A B⎞ . If the network is reciprocal, then matrix: T = ⎜ ⎝ C D ⎟⎠

sin(t ) sin(t ) * with ∗ denoting the πt πt convolution operation, then x(t) is equal to

292. If the signal x(t ) =

(a)

sin( 2t ) sin(t ) (b) 2π t πt

(c)

⎛ sin(t ) ⎞ 2sin(t ) (d) ⎜ πt ⎟ πt ⎝ ⎠

(a) T −1 = T

2

(b) T   = T (c) Determinant (T ) = 0 (d) Determinant (T) = 1 2

(GATE 2016: 1 Mark) (GATE 2016: 1 Mark)

288. The Z-parameter matrix for the two-port network shown is jω ⎞ ⎛ 2 jω ⎜ ⎟ ⎝ jω 3 + 2 jω ⎠

2 Zc

where the entries are in Ω. Suppose

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Qns_Part 1.indd 122

293. Consider the sequence x[n] = anu[n] + bnu[n], where u[n] denotes the unit-step sequence and 0  0. Therefore, Re{s} >(a + 2) 10 Ω

5Ω

b

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 142

86. Topic: Linear 2- Port Network Parameters: Driving Point and Transfer Functions (d) The h-parameters can be calculated using the formula

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143

Chapter 2  •  Networks, Signals and Systems

h11 =

V1 I1 V2 = 0

h12 =

V1 V2

h21 =

I2 I1

h22 =

I2 V2

88. Topic: Discrete Time Siganals: Discrete Time Fourier Transform (DTFT) (a)  It is given that ⎛n ⎞ y( n) = x ⎜ - 1⎟ , n even ⎝2 ⎠

I1 = 0

= 0 for n odd Therefore, for

V2 = 0

I1 = 0

When V2 = 0, the circuit can be redrawn as shown in the following figure: + 10 Ω

I1

y( n) = x( -1) = 1 y ( n) = x ( 0 ) = 2 y( n) = x(1) = 1 y( n) = x( 2) = 1 / 2

Hence, y(n) given in option (a) is the correct answer. 89. Topic: Discrete Time Fourier Transform (DTFT) (c)  y( 2n) = x( n - 1) Therefore,

I2

V1

1 δ ( n + 1) + δ ( n) + 2δ ( n - 1) 2 1 + δ ( n - 2) + δ ( n - 3) 2 which is from the figure shown in the given data. Let f(n) = y(2n). Taking z-transform, we get y( 2n) =

− From this figure, we can see that I1 = - I 2 V1 = 10 I1 . Therefore, h21 = and

n = 0, n = 2, n = 4, n = 6,

and

I2 = -1 I1

V h11 = 1 = 10 I1

When I1 = 0, then V1 = V2 since there is no drop in 10 Ω resistance. Therefore, h12 =

F( z) =

Substituting z = e jω in the above equation, we get 1 jω 1 e + 1 + 2e - j ω + e -2 j ω + e -3 j ω 2 2 1 1 ⎛ ⎞ = e - j ω ⎜ e 2 j ω + e j ω + 2 + e - j ω + e -2 j ω ⎟ ⎝2 ⎠ 2

F ( e jω ) =

V1 =1 V2

Also,

⎤ ⎡ e +2 j ω + e -2 j ω = e - jω ⎢ + e j ω + e - j ω + 2⎥ 2 ⎦ ⎣

V2 = 20 I 2 Therefore, h22 =

I2 1 = = 0.05 V2 20

The h-parameter matrix is given by 1 ⎤ ⎡10 ⎢ -1 0.05⎥ ⎣ ⎦ 87. Topic: Contiunous- Time Signals: Fourier Series and Fourier Transform Representations (b)  Applying frequency scaling and frequency shifting properties, we get ⎡ ⎛ 2⎞ ⎤ X [3 f + 2] = X ⎢3 ⎜ f + ⎟ ⎥ ⎝ 3⎠ ⎦ ⎣ 1 ⎛t⎞ FT ←⎯ → x ⎜ ⎟ e - j 4p t / 3 3 ⎝ 3⎠

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 143

1 1 z + 1 + 2 z -1 + z -2 + z -3 2 2

Therefore, f ( n) = y( 2n) = e - jω [cos 2ω + 2 cos ω + 2] 90. Topic: LTI Systems: Frequency Response (b)  Given that output is y( n) = Ax ( n - n0 ) Taking Fourier transform, we get Y (e jω ) = Ae - jω0 n0 X (e jω ) Therefore, H ( e jω ) =

Y ( e jω ) = Ae - jω0 n0 X ( e jω )

Therefore, ∠H (e jω ) = -ω 0 n0

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144

GATE ECE Chapter-wise Solved papers

For the discrete-time LTI system, the phase and frequency of H(ejw) are periodic with period 2p . Therefore, in general form, it is given as

θ (ω ) = - n0ω 0 + 2p k 91. Topic: LTI Systems: Digital Filter Design Techniques (a)  Given that y(t ) = 0.5 x(t - t d + T ) + 0.5 x(t - t d - T ) + x(t - t d ) Taking the Fourier transform, we get Y ( jω ) = ⎣⎡0.5e jω ( - td +T ) + 0.5e jω ( - td -T ) + e - jωtd ⎤⎦ X ( jω ) The filter transfer function is H ( jω ) =

Y ( jω ) X ( jω )

= e - jω td ⎡⎣0.5e jω T + 0.5e - jω T + 1⎤⎦ = (1 + cos ω T )e - jω td 92. Topic: Contiunous- Time Signals: Fourier Series and Fourier Transform Representations (a)  Using time shifting and time scaling properties x(t - t0 ) ↔ X ( jω )e - jωt0 1 ⎛ jω ⎞ X⎜ and x ( at ) = ⎟ (here t0 = 3/5 and a = 5), we get a ⎝ a ⎠ ⎡ ⎛ 3⎞ ⎤ 1 ⎛ jω ⎞ - j 35ω x ⎢5 ⎜ t - ⎟ ⎥ ↔ X ⎜ ⎟e 5 ⎝ 5 ⎠ ⎣ ⎝ 5⎠ ⎦ 93. Topic: Discrete Time Signals: z-Transform (d)  The ROC remains the same after addition and sub1 2 traction in z-domain. Hence, the answer is < z < . 3 3 94. Topic: Continuous- Time Signal (d)  The Dirac delta function δ (t )  is ⎧ ∞, t = 0 and δ (t ) = ⎨ ⎩0, otherwise

∞

∫ δ (t )dt = 1

-∞

95. Topic: LTI Systems: Frequency Response (b)  For a transmission in which no phase distortion takes place, dϕ (ω ) = constant dω Therefore, the correct option is (b). 96. Topic: Solution of Network Equations Using Laplace Transform (c)  Inverse Laplace transform of the given equation is L [ F ( s)] = sin ω 0 t -1

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 144

Therefore, f(t) = sin w0t. Since the value of a sine function varies between −1 and +1, we get −1 ≤ f (∞) ≤ 1 97. Topic: Solution of Network Equations Using Laplace Transform (a)  The voltage across the inductor is given by Ldi V= dt Taking Laplace Transform on both sides, we get V(s) = sLI(s) - Li(0+) It is given that Li(0+) = 1 mV.  Therefore, 1 × 10 -3 i(0 + ) = = 0.5 A 2 × 10 -3 98. Topic: Solution of Network Equations Using Laplace Transform (c)  The impedance of a parallel combination of R and C is given by R × (1/Cs) R Z= = R + (1/Cs) RCs + 1 The output voltage vo(t) is given by vo (t ) ⎛ ⎞ ⎛ ⎞ 4 × 103 ⎜ ⎟ ⎜⎝ 4 × 103 × 10 -6 s + 1⎟⎠ ⎟ vi (t ) =⎜ ⎜⎛ ⎞⎟ ⎞ ⎛ 4 × 103 1 × 10 3 ⎜⎜ -6 -6 3 3 ⎟⎟ ⎟ +⎜ ⎝ ⎝ 4 × 10 × 10 s + 1⎠ ⎝ 1 × 10 × 4 × 10 s + 1⎠ ⎠ = 0.8vi ( t ) Therefore, vo(t) = 0.8 × 10u(t) = 8u(t) V 99. Topic: Linear 2- Port Network Parameters: Driving Point and Transfer Functions (d)  The following figure shows the two-port network terminated by resistance RL: I1

I2 +

+ Two-port network

V1 −

RL V2 −

The voltages and currents of the network can be expressed in terms of A, B, C and D parameters as V1 = AV2 - BI 2 and I1 = CV2 - DI 2 Now, V2 = - I 2 RL Therefore, V1 AV2 - BI 2 - AI 2 RL - BI 2 = = I1 CV2 - DI 2 -CI 2 RL - DI 2

12/4/2018 11:08:30 AM

Chapter 2  •  Networks, Signals and Systems

The input impedance is

The following figure (b) shows the resultant signal.

V1 ARL + B = I1 CRL + D

G(jf)

100. Topic: Linear 2- Port Network Parameters: Driving Point and Transfer Functions (b)  The Z12 and Z21 parameters can be calculated using the following formulas: Z12 =

V1 I2

145

I1 = 0

and Z 21 =

V2 I1

f

−20 kHz

20 kHz (a) M(jf)∗G(jf)

I2 = 0

When I1 = 0, then V1 = 0. Therefore, Z12 =

−20 kHz

V1 =0 I2

(b)

When I 2 = 0, then V2 = -b I1r0 . Therefore, Z 21 =

V2 = -b r0 I1

101. Topic: Driving Point and Transfer Functions (b)  RC impedance function has the first critical frequency due to pole and last critical frequency due to zero. 102. Topic: Linear 2- Port Network Parameters: State Equations for Networks (a)  For z2(s) to be postive real, |Rneg| < Re z1 ( jw),  ∀w 103. Topic: Contiunous- Time Signals: Fourier Series and Fourier Transform Representations (c)  It is given that the signal m(t) with a bandwidth of 500 Hz is multiplied by a signal g (t ) =

∞

∑ ( -1) δ (t - 0.5 × 10 k

-4

k)

This signal is passed through a low-pass filter with cutoff frequency of 1 kHz. After the low-pass filtering with f c = 1 kHz , the output is zero. 104. Topic: LTI Systems: Definition and Properties (c)  It is given that ⎛ 5 ⎞ y [ n] = ⎜ sin p n⎟ x [ n] ⎝ 6 ⎠ The sine value of a function varies between −1 and +1. Therefore, for bounded input x[n], the output is bounded. Hence, the system is stable. The system is linear and non-invertible. 105. Topic: Network Theorems: Maximum Power Transfer (d)  It is given that the source impedance is Z s = Rs + jX s From maximum power transfer theorem, the maximum power is transferred to the load impedance when it is a complex conjugate of the source impedance, that is, Z L = Zs* = Rs - jX s

k = -∞

The following figure shows the frequency domain representation of m(t)[M(jf )]. M(jf)

106. Topic: Network Theorems: Thevenin and Norton (d) Let VTH be the Thevenin’s equivalent voltage across X–Y. Applying Kirchhoff’s current law at node X, we get 2=

-500

f

Given that impulse train time period = 0.5 × 10-4 s. Therefore, sampling frequency 1 = Hz = 20 kHz 0.5 × 10 -4 The following figure (a) shows the frequency domain representation of g(t) [G(jf )]. The resultant signal is given by m(t ) g (t ) ↔ M ( jf ) * G ( jf )

VTH VTH VTH - 2i + + 2 1 1

where i=

500

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 145

20 kHz

VTH 1

Therefore, ⎛ 1 1 1 2⎞ 2 = VTH ⎜ + + - ⎟ ⎝ 2 1 1 1⎠ Hence, VTH = 4 V From the figure, the short-circuit current is I sc = 2 A

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Therefore, RTH

From this graph, we can see that the circuit is a bandpass filter.

V 4 = TH = = 2 Ω I sc 2

107. Topic: Solution of Network Equations Using Laplace Transform (d)  The final value theorem is applicable only when all the poles of system lie on the left half of the s-plane. Since s = 1 is a right s-plane pole. Therefore, the system is unbounded. 108. Topic: Frequency Domain Analysis of RLC Circuits (c) At w → ∞, the capacitor acts as a short circuit element and the circuit looks like as shown in the following figure. R +

+

Vi

R

10 V = 0.5 mA 20 × 103

3 3 -6 τ = ReqC = ( 20 × 10 20 × 10 ) × 4 × 10 = 40 ms

At w → 0, the capacitor acts as an open circuit and the circuit looks like as shown in the following figure. R

Using the formula i (t) = i (∞) - [i (∞) - i (0)]e-t/τ C C C C we get -3 i (t) = 0 - [(0 - 0.5)]e-t/40 × 10 = 0.5e-25t mA C

111. Topic: Frequency Domain Analysis of RLC Circuits (d)  The bandwidth of series RLC circuit is

+

R

110. Topic: Time Domain Analysis of Simple Linear Circuits (a)  At t = 0+, capacitor is short circuit and at t = ∞, capacitor is open circuit. Therefore,

The current through the capacitor at t = ∞ is iC(∞) = 0 The time constant of the circuit is

Vo =0 Vi

Vi

(5 - 3 j ) × (5 + 3 j ) 5-3j +5+3j

⎛ 25 + 9 ⎞ = 5∠30° × ⎜ = 17∠30° V ⎝ 10 ⎟⎠

iC (0 + ) =

Therefore,

+

= 5∠30° ×

Vo

−

−

109. Topic: Steady State Sinusoidal Analysis Using Phasors (d)  We know that VAB = Current ´× Impedance That is, VAB = 5∠30° × (5 - 3 j )  (5 + 3 j )

R L

Vo

The bandwidth of filter 1 is −

− B1 =

Therefore, Vo =0 Vi

R L1

The bandwidth of filter 2,

The frequency response of the circuit is shown in the following figure:

B2 =

R L2

L2 =

L1 4

It is given that Vo/Vi

Therefore, ω

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 146

B1 1 = B2 4

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112. Topic: Discrete- Time Signals: z-Transform (b)  It is given that X [ z] =

1

5

2

0.5 1 - 2 z -1

and the ROC includes a unit circle. Therefore, the signal is a left-handed signal. Hence, x[n] = -(0.5)( 2) - n u( - n - 1) 4 3 6 Therefore, the number of cut-sets is 6.

From the above equation, x [0] = 0. 113. Topic: LTI Systems: Digital Filter Design Techniques (a)  The Laplace transform of e - t u (t ) =

1 s +1

The magnitude at 3-dB frequency is 1

1 2

; therefore,

1 1 = s 1 + 2 1+ ω2 Solving the above equation, we get w = 1 rad. Therefore, 1 f = Hz 2p =

114. Topic: Time Domain Analysis of Simple Linear Circuits (b)  At t = 0+, the inductor behaves as an open circuit. Therefore, VL = IsRs Also, di VL = L (0 + ) dt Therefore, IR V di + (0 ) = L = s s dt L L 115. Topic: Network Solution Methods (c)  The different trees (P) are shown in the following figure:

116. Topic: LTI Systems: Causality (c)  A system is causal if the output at any time depends only on values of the input at the present time and in the past. Therefore, the system y(t ) = ( t + 4) x( t - 1) is causal. 117. Topic: LTI Systems: Stability (d)  It is given that h(t ) = ea t u(t ) + e b t u( -t ) For the system to be stable, ∞

∫ h(t )dt < ∞

-∞

For the above condition to be true, h(t) should be as shown in the following figure. h(t) e b tu(−t)

e a t u(t)

t

Therefore, a < 0 and b > 0. 118. Topic: Network Theorems: Thevenin and Norton’s (a)  The figure given below shows the equivalent circuit of the network given in the problem, with the circuit elements being replaced by their impedances and the independent sources being deactivated. P 1/s s 1Ω

1Ω

Q

Therefore, the number of trees is 4. The different cutsets (Q) are shown in the following figure:

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 147

Therefore, the Thevenin’s equivalent resistance between P and Q is given by

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Therefore,

⎛ 1 ⎞ ( s + 1)[1 + (1/ s)] Z TH = ( s + 1)  ⎜1 + ⎟ = ⎝ s ⎠ ( s + 1) + 1 + (1/ s) =

⎡ ⎛ 3 ⎞ 1 ⎛ 3 ⎞⎤ VR (t ) = e - t / 2 ⎢cos ⎜ t sin ⎜ t ⎟⎥ ⎜ 2 ⎟⎟ ⎜ ⎟ 3 ⎢⎣ ⎝ ⎠ ⎝ 2 ⎠ ⎥⎦

( s + 1) / s =1 ( s + 1)( s + 1)/ s 2

119. Topic: Time Domain Analysis of Simple Linear Circuits (c)  The waveform of voltage VC(t) is shown below.

122. Topic: Linear 2- Port Network Parameters: Driving Point and Transfer Functions (c)  It is given that when switch S1 is open and S2 is closed, then V1 = 4.5 V; V2 = 1.5 V; I 2 = 1 A; I1 = 0

VC(t)

Therefore,

1 2T t T

Z12 =

V1 I2

Z 22 =

V2 I2

−1

In mathematical form, we have VC(t) = tu(t) - 2(t - T)u(t - T) + 2(t - 2T)u(t - 2T)... ∞

= tu(t ) + 2∑ ( -1) n (t - nT )u(t - nT ) n =1

120. Topic: Solution of Network Equations Using Laplace Transform (d)  The Laplace transform of the voltage across the capacitor is 1 ⎛1⎞ ⋅⎜ ⎟ s + 1 + (1/s) ⎝ s ⎠ 1 1 = 2 = s + s + 1 [ s + (1/2)]2 + ( 3/2) 2

VC ( s) =

Taking inverse Laplace transform, we get VC (t ) =

⎛ 3 ⎞ e - ( t / 2 ) sin ⎜ t ⎜ 2 ⎟⎟ 3 ⎝ ⎠

2

121. Topic: Solution of Network Equations Using Laplace Transform (b)  The Laplace transform of the voltage across the resistor is s 1 VR ( s) = ⋅1 = 2 s + 1 + (1 / s) s + s +1 =

s + (1/ 2) [ s + (1/ 2)]2 + ( 3 / 2) 2

-

(1/ 2) ⋅ ( 3 / 2) ⋅ ( 2 / 3 ) [ s + (1/ 2)]2 + ( 3 / 2) 2

Taking inverse Laplace transform, we get ⎛ 3 ⎞ 1 -t / 2 ⎛ 3 ⎞ VR (t ) = e - t / 2 cos ⎜ t e sin ⎜ t ⎜ 2 ⎟⎟ ⎜ 2 ⎟⎟ 3 ⎝ ⎠ ⎝ ⎠

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 148

=

4.5 = 4.5 Ω 1

=

1.5 = 1.5 Ω 1

I1 = 0

I1 = 0

It is given that when switch S1 is closed and S2 is open, then I1 = 4 A; V1 = 6 V; V2 = 6 V; I 2 = 0 Therefore, Z11 =

V1 I1

I2 = 0

Z 21 =

V2 I1

I2 = 0

=

6 = 1.5 Ω 4

=

6 = 1.5 Ω 4

So, the z-parameter matrix is ⎡1.5 4.5⎤ ⎢1.5 1.5 ⎥ ⎣ ⎦ 123. Topic: Linear 2- Port Network Parameters: Driving Point and Transfer Functions (a)  The network equations for h-parameters are given by V1 = h11 I1 + h12V2 and I 2 = h21 I1 + h22V2 Therefore, h12 =

V1 V2

h22 =

I2 V2

h11 =

⎛ Z Z ⎞ V1 = Z11 - 12 21 ⎟ Z 22 ⎠ I1 V2 = 0 ⎜⎝

I1 = 0

I2 I1

4.5 =3 1.5

=

1 = 0.67 1.5

I1 = 0

= 1.5 h21 =

=

4.5 × 1.5 = -3 1.5 =

V2 = 0

- Z 21 -1.5 = = -1 Z 22 1.5

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So, the h-parameter matrix is 3 ⎤ ⎡ -3 ⎢ -1 0.67⎥ ⎣ ⎦

The voltage across the resistor in time-domain is given by VR (t ) = 5e -0.5t

124. Topic: Linear 2- Port Network Parametrs: State Equations for Networks (d)  It is given that the impedance is 0.2 s Z ( s) = 2 s + 0.1s + 2 Therefore the admittance is 2 s 2 + 0.1s + 2 s 1 = + + 0.2 s 0.2 2 0.2 s 10 = 5s + 0.5 + s The admittance Y(s) of a parallel RLC circuit is given by Y ( s) =

Y ( s) = Cs + Therefore,

1 1 + R Ls

C = 5F 1 = 2Ω 0.5 1 L= = 0.1 H 10

R=

125. Topic: Contiunous- Time Signals: Fourier Series and Fourier Transform Representations (a)  The Fourier transform of a signal x(t) is given by ∞

X ( jω ) =

∫ x (t )e

- jω t

dt

It is given that the samples are taken at 10 Hz. Therefore, the sampling time instants are n/10. Therefore, the samples are x( n) = 5e -0.5 n /10 = 5e -0.05 n 127. Topic: Discrete Time Signals: z-Transform (c)  Taking z-transform of signal x(n), we get

Therefore, the Fourier transform of the given signal is - jω t ∫ e dt =

-1

e - jω t - jω

1

= -1

1 jω (e - e - jω ) jω

X ( jω ) = 0, when e jω - e - jω = 0. Therefore, e jω -

1 =0 e jω

∞

∞

n= 0

n= 0

X [ z ] = ∑ 5e -0.05 n z - n = 5∑ (e -0.05 z -1 ) n Therefore, X [ z ] =

5z and ROC is z > e -0.05 z - e -0.05

128. Topic: LTI Systems: Definition and Properties (a)  y(t ) = x(t ) is not linear, but this function is time invariant. Options (a) and (b) may be correct. y(t ) = t x(t ) is not linear; thus, option (b) is wrong and (a) is correct. 129. Topic: LTI Systems: Impulse Response (d)  It is given that h[n] = [1, −1, 2] and x[n] = [1, 0, 1]. y[n] = x[n] ∗ h[n] Length of y[n] = Length of x[n] + Length of h[n] − 1  = 3 + 3 − 1 = 5 Now,

-∞

1

149

y[n] =

∞

∑ x[k ]h[n - k ]

k =-∞

Therefore, y[2] =

∞

∑ x[k ]h[2 - k ]

k =-∞

Hence, y[2] = x[0]h[2 − 0] + x[1]h[2 − 1] + x[2]h[2 − 2] = h[2] + 0 + h[0] = 2 + 1 = 3

Solving the above equation, we get e 2 jω = 1. Therefore, e jω = ±1. Hence, ω = p , 2p . 126. Topic: Discrete- Time Signals: z-Transform (b)  After closing the switch, the voltage across the resistor is given by

130. Topic: LTI Systems: Frequency Response (c)  It is given that the impulse response is h(t ) = exp( -2t )u(t ) Therefore, H ( jω ) =

∞

∫ h(t )e

- jω t

dt

-∞

⎛ ⎞5 200 × 103 VR ( s) = ⎜ 3 -6 ⎝ 200 × 10 + [1 / (10 + 10 s)] ⎟⎠ s =

-6

5 × 2 × 10 × 10 × 10 5 = s + 0.5 2 × 105 × 10 -5 s + 1 5

∞

= ∫ e -2t e - jωt dt 0

∞

= ∫ e - ( 2 + jω )t dt 0

1 = ⋅ e - ( 2 + jω )t 2 + jω

∞

0

1 = 2 + jω Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 149

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H ( jω ) =

∞

∫ h(t )e

- jω t

dt

-∞ ∞

= ∫ e -2t e - jωt dt 0

150

∞ GATE ECE Chapter-wise Solved papers

= ∫ e - ( 2 + jω )t dt 0

1 = ⋅ e - ( 2 + jω )t 2 + jω

∞

Now,

0

Therefore,

1 = 2 + jω

12 - I > 0  or  I < 12 A.

131. Topic: LTI Systems: Frequency Response (d)  The given input signal is x(t ) = 2 cos ( 2t ) whose Fourier transform is given by X ( jω ) = 2p [δ (ω - 2) + δ (ω + 2) ] It is given that the transfer function is 1 H ( jω ) = 2 + jω The Fourier transform of the output signal is given by Y ( jω ) = H ( jω ) X ( jω ) Therefore, Y ( jω ) = = = =

= = =

1 ⋅ 2p [δ (ω - 2) + δ (ω + 2)] 2 + jω 2p 2p δ (ω - 2) + δ (ω + 2) 2 + j2 2 - j2 2p [(2 - j 2)δ (ω - 2) + (2 + j 2)δ (ω + 2)] 8 p [δ (ω - 2) + δ (ω + 2)] 2 p - j [δ (ω - 2) - δ (ω + 2)] 2 cos 2t sin 2t + 2 2 2⎡ 1 1 ⎤ cos 2t + sin 2t ⎥ ⎢ 2 ⎣ 2 2 ⎦ 1 2

I' = 12 - I

cos ( 2t - 0.25p )

133. Topic: Time Domain Analysis of Simple Linear Circuits (c)  We know that P = v(t) · i(t) The energy is given by P · t = v(t) · i(t) · t Now, i(t) = I = 2 A (given) and v(t) · t = Area under v(t)-t curve. Therefore, ⎛1 ⎞ v(t ) ⋅ t = ⎜ × 2 × 600 ⎟ + (10 × 600) ⎝2 ⎠ = 600 + 6000 = 6600 Therefore, E = (6600) ×´ 2 J = 13200 J = 13.2 kJ 134. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (c)  The parallel combination of RL and C is replaced by impedance Z as shown in the following figure. The impedance is Z= R

Z

Vi −

Vo −

H ( s) =

(a)  Since the power is absorbed by 60 V source, I´> 0.

+ −

RL Z = Z + R ( R L + R) + sRR LC

If R = R L , then H ( s) =

20 V

+ −

+

Therefore, the transfer function is

132. Topic: Network Theorems: Maximum Power Transfer

60 V

1 + sR LC

+

= 2 -0.5 cos ( 2t - 0.25p )

I

RL

1 2 + sRC

Therefore, RL= R 135. Topic: Continuous-Time Signals: Fourier Series and Fourier Transform Representations (a)  The Fourier series of a real periodic function has only cosine terms if it is even and only sine terms if it is odd.

I´

12 A

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136. Topic: Continuous-Time Signals: Fourier Series and Fourier Transform Representations (b)  Given that the function f (t ) = sin 2 t + cos 2t Therefore, 1 f (t ) = (1 - cos 2t ) + cos 2t 2 Hence, the frequency components are 0, 1/p Hz. 137. Topic: Discrete-Time Signals: z-Transform (a)  It is given that n

n

⎛1⎞ ⎛1⎞ x( n) = ⎜ ⎟ u( n) - ⎜ ⎟ u( -n - 1) 3 ⎝ ⎠ ⎝2⎠

151

139. Topic: Steady State Sinusoidal Analysis Using Phasors (b) Let I be the current through the circuit. Therefore, 20 10 I= = = 2∠ - 36.87° 8+6 j 4+3j The reactive power is Q =| I 2| Real part of X L = 4 × 4 = 16 VAR 140. Topic: Solution of Network Equations Using Laplace Transform (b). At t = 0-, the switch is in position ‘a’ as shown in the following figure. 10 kΩ

n

Here, (1 / 3) u( n) is right-sided signal; so the ROC will be 1 / 3 < z . Also -(1 / 2) u( - n - 1) is left-sided signal 1 so ROC will be z < . Combining this information, 2 the ROC of the given function will be given as

+ 0.2 µF 100 V 0.5 µF

1 1 < z < 3 2 138. Topic: Network Theorems: Maximum Power Transfer (c)  The circuit given in the problem can be replaced by the figure shown below after applying 1 V source to the load terminals and short circuiting the voltage source of 100 V. Vx − +

4Ω

4Ω

4Ω

−

+ Vx

100 V = v(0−) = v(0+)

⇒ 0.8 µF

−

At t > 0, the switch is in position ‘b’, as shown in the following figure. I(s)

I1

1/sC I (I − I1)

R = 5 kΩ + −

v(0+) 100 = s s + 1V −

C = (0.5 + 0.3)  0.2 mF = 0.16 mF The current through the capacitor is

Req = 1 I

For the maximum power transfer, RL = Req Applying Kirchhoff’s voltage law to the outer loop, we get 1 = 4I1 + Vx Also, Vx = 4(I - I1) From the above two equations, we get

Therefore,

0.3 µF

+ −

= 0.16 µF

n

1 I= A 4 R eq =

1 =4Ω I

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 151

i (t ) =

v(0 + ) - t / RC e u (t ) R

Now, v(0+) = 100 V,

1 1 = RC 5 × 103 × 0.16 × 10 -6

and   R = 5 kΩ Therefore, i(t) = 20e-1250tu(t) mA 141. Topic: Solution of Network Equations Using Laplace Transform (a)  It is given that L

di + Ri = V0 (1 + Be - Rt / L sint )u(t ) dt

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Taking Laplace transform on both sides, we get LsI ( s) - Li(0 - ) + RI ( s) =

V0 V0 B + s [ s + ( R/L)]2 + 1

Substituting the value of i(0—) and solving for I(s), we get V0 B ⎤ ⎛ 1 ⎞ ⎡ LV0 V0 I ( s) = ⎜ + + ⎢ ⎥ ⎟ 2 s [ s + ( R/L)] + 1 ⎦ ⎝ Ls + R ⎠ ⎣ R By the final value theorem, lim i(t ) = lim sI ( s) = t →∞

s →0

V0 R

142. Topic: Solution of Network Equations Using Laplace Transform t (b)  The Laplace transform of ⎡ ∫ f (τ ) dτ ⎤ is ⎢⎣ 0 ⎥⎦

F ( s) f -1 (0 + ) F ( s) + = s s s with zero initial condition. 143. Topic: LTI Systems: Frequency Response (a)  Given that: Input x(t) = sin (ω t + 1) = sin (t + 1) Hence, ω = 1

1

1 ⇒ 2 3

—1 From the given figure, we can see that it is finite impulse response (FIR) filter and not a low-pass filter. 146. Topic: Steady State Sinusoidal Analysis Using Phasors (d)  The characteristic equation for a parallel RLC circuit is 1 1 s2 + s+ =0 RC LC where bandwidth is 1 RC It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased. At resonant frequency, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity. In parallel RLC circuit, the admittance is minimum at resonance. Hence, magnitude of input impedance attains its maximum value at resonance. 147. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (a)  Short-circuit admittance parameters for a two port π-network shown in the following figure are Y11 = Ya + Yb Y12 = Y21 = -Yb

The transfer function is given as s2 + 1 2 s + 2s + 1 -ω 2 + 1 T ( jω ) = -ω 2 + 1 + 2 jω

Y22 = Yb + Yc

T ( s) =

Yb 1

2

=0 ω =1

Ya

Yc

So, y(x) is zero for all sampling frequencies w s. 144. Topic: LTI Systems: Definition and Properties (b)  It is given that y(t ) =

∞

1´ For the given network,

∫ x(t - τ )h(2τ )dτ . We know

Ya = Yb = Yc =

-∞

that h(t) is not the impulse response of a low-pass filter. Therefore, the system is not a low-pass system. However, the system is both LTI and BIBO. 145. Topic: LTI Systems: Impulse Response (a)  It is given that h( 2) = 1, h (3) = -1 and h( k ) = 0 otherwise. h(t) is shown in the following figure.

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 152

2´ 1 = 2S 0.5

Therefore, Y11 = 2 + 2 = 4 S Y12 = Y21 = -2 S Y22 = 2 + 2 = 4 S Therefore, the short-circuit admittance matrix is ⎡ 4 -2 ⎤ ⎢ -2 4 ⎥ S ⎣ ⎦

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148. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (c)  For the reciprocal networks S12 = S21 For the symmetrical networks, S11 = S22 For anti-symmetrical networks, S11 = −S22 For lossless reciprocal networks, S11 = S22 2

Applying Kirchhoff’s voltage law in the outer loop, we get (3 + i)2 + (2 + i)2 = 10 Solving the above equation, we get i=0 The power supplied by the voltage source is P = Vi = 10 ´× 0 = 0 W 152. Topic: Steady State Sinusoidal Analysis Using Phasors (a)  The circuit shown in the given figure is redrawn as shown in the following figure (L = 20 mH, C = 50 mH):

2

S11 + S12 = 1

and

VA

Therefore, the network is not lossless but reciprocal.

jω L Vin 20∠0° V ω =103 rad/s

149. Topic: Discrete-Time Signals: z-Transform (a)  It is given that

1 jω C

1Ω

I

X ( z ) = 5 z 2 + 4 z -1 + 3; 0 < z < ∞ We know that

Applying Kirchhoff’s current law at node A, we get

z δ [ n + n0 ] ←⎯ → z n0

V A − V in V A VA + + =0 jωL 1 1/jωC

Therefore, x [ n] = 5δ [ n + 2] + 4δ [ n - 1] + 3δ [ n]

Therefore,

150. Topic: LTI Systems: Impulse Response (c)  It is given that h1 [ n] = δ [ n - 1] and h2 [ n] = δ [ n - 2]. Therefore,

⎡ ⎤ 1 + 1 + j103 × 50 × 10 -6 ⎥ VA ⎢ 3 -3 10 20 10 j × × ⎣ ⎦ 20 = j103 × 20 × 10 -3

z h1 [ n] = δ [ n - 1] ←⎯ → H1 ( z ) = z -1 z and    h2 [ n] = δ [ n - 2] ←⎯ → H 2 ( z ) = z -2

Solving the above equation, we get

The overall impulse response in z-domain is

VA = - j1 V

H ( z ) = H1 ( z ) H 2 ( z ) = z -1 z -2 = z -3

Also,

The overall impulse response in discrete-time domain is h [ n] = δ [ n - 3] 151. Topic: Network Solution Methods: Nodal and Mesh Analysis (a)  The given network can be redrawn as shown in the following figure. 1Ω

3+i

i

I=

VA 1

Therefore, I = - j1 A 153. Topic: Solution of Network Equations Using Laplace Transform (a)  At t = 0-, the circuit is shown below. 10 Ω

1Ω

1Ω 0.75 A 3A

2+i 1A 1Ω

0.75 A + −

10 V 1.5 A

2A

10 Ω

1Ω

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The current will divide equally in both 10 Ω resistors as shown in the circuit. Therefore, i(0-) = i(0+) = 0.75 A + At t ≥ 0 , the circuit in s-domain is as shown below.

Taking Laplace transform on both sides (assuming zero initial conditions), we get s 2Y ( s) + 4 sY ( s) + 3Y ( s) = 2 sX ( s) + 4 X ( s) Therefore, Y ( s) 2s + 4 2( s + 2) = 2 = X ( s) s + 4 s + 3 ( s + 1)( s + 3)

10 Ω I(s) Ls

It is also given that x(t ) = e -2t u(t )

10 Ω

10 Ω

15 s

Li(0+)

Therefore,

− +

X ( s) =

1 s+2

Hence, The above circuit can be rearranged as shown below. 10 Ω

V(s)

2( s + 2) ( s + 1)( s + 3)( s + 2) 2 1 1 = = ( s + 1)( s + 3) ( s + 1) ( s + 3)

Y ( s) =

10 Ω

I(s) Ls

Taking inverse Laplace transform on both sides, we get y(t ) = (e - t - e -3t )u(t )

+ 15 − s

10 Ω

Li(0+)

− +

Applying KCL, we get V ( s) - (15/s) V ( s) V ( s) + 15 × 10 -3 × 0.75 + + =0 10 10 10 + 15 × 10 -3 s Solving the above equation, we get 10000 + 7.5s 2 s( s + 1000) {10000 + 7.5( -1000)} 10000 = + 2 × 1000 s 2(( -1000)( s + 1000) 5 1.25 = s s + 1000

V ( s) =

Taking inverse Laplace transform, we get v(t) = 5 - 1.25 e-1000t V Now, i (t ) =

v (t ) = 0.5 - 0.125e -1000 t A 10

154. Topic: Solution of Network Equations Using Laplace Transform (b)  It is given that d 2 y(t ) 4 dy(t ) 2dx(t ) + + 3 y (t ) = + 4 x (t ) 2 dt dt dt

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 154

155. Topic: LTI Systems: Definition and Properties (c)  A discrete-time LTI system is causal if and only if the ROC of its system function is the exterior of a circle, including infinity. A discrete-time LTI system is stable if and only if the ROC of its system function includes the unit circle, z = 1. It is given that H (z) =

2 - (3 / 4) z -1 1 - (3 / 4) z -1 + (1 / 8) z -2

Therefore, (1 - (1 / 4) z -1 ) + (1 - (1 / 2) z -1 ) (1 - (1 / 4) z -1 )(1 - (1 / 2) z -1 ) 1 1 = + 1 - (1 / 4) z -1 1 - (1 / 2) z -1

H ( z) =

1 , the system is stable and causal. 2 1 1 1 For the ROC : z < and the ROC : < z < , the 4 4 2 ROC does not include the unit circle. So, the system is 1 not stable. Also the ROC is not the exterior of z = , 2 so it is not causal.

For the ROC : z >

156. Topic: Network Theorems: Norton (a)  The Norton equivalent current is obtained by short circuiting the terminals PQ as shown in the circuit depicted in the following figure.

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Chapter 2  •  Networks, Signals and Systems

Isc

Isc P

Given that ⎛ t ⎞ vi (t ) = Vp cos ⎜ ⎝RC ⎟⎠

j30 Ω Therefore,

25 Ω −j50 Ω

16∠0°A

ω=

15 Ω Q The short circuit current is 25 25 I sc = × 16 ∠0° = × 16 ∠0° 15 + j 30 + 25 40 + j 30 ( 25 × 16)∠0° = = 8∠ - 36.86° 50 ∠36.86°

Substituting ω =

1 and s = j w in the ­expression for RC

⎡ (1 + j ) R R ⎤ I ( s) Vi ( s) = ⎢ + 1 + j ⎥⎦ j ⎣ Therefore, Vi ( s) 3R = I ( s) (1 + j )

I N = I sc = 8∠ - 36.86° = (6.4 - j 4.8) A

I ( s) =

or

157. Topic: Network Theorems: Maximum Power Transfer (c)  For maximum power transfer, RL = RTH To calculate RTH, short circuit the voltage sources and open circuit the current sources as depicted in the following figure. 10 Ω 10 Ω oc sc

RTH

⎛ 1 ⎞ R ⋅ (1/ sC ) Vo ( s) = ⎜ R I ( s) ⎟ I ( s) = sC + (1/ sC ) R ⎝ ⎠ Therefore, V ( s) R ⋅ i (1 + j ) 1 + sCR 3R V ( s) R Vi ( s) = ⋅ (1 + j ) = i 1 + j 3R 3 In time domain, Vo ( s) =

vo (t ) =

Therefore, RTH = 10 + 10  10 = 15 Ω 158. Topic: Steady State Sinusoidal Analysis Using Phasors (a)  The following figure shows the circuit in the given figure which is drawn in s-domain. 1 R+ sC + + (R1/sC) −

I(s)

Vo(s) −

From this figure, we have 1 ⎞ R ⋅ (1/ sC ) ⎛ Vi ( s) = ⎜ R + I ( s) ⎟⎠ I ( s) + ⎝ sC R + (1/ sC ) 1 + sCR R = I ( s) + I ( s) sC 1 + sCR

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 155

Vi ( s) × (1 + j ) 3R

Now,

sc

Vi(s)

1 RC

Vi(s), we get

Hence, the Norton current is

10 Ω

155

Vp 1 ⎛ t ⎞ vi (t ) = cos ⎜  ⎝ RC ⎟⎠ 3 3

159. Topic: LTI Systems: Impulse Response (a)  The unit step response of the given network is s(t) = (1 − e−at) Therefore, the unit impulse response is h(t ) =

d d s(t ) = (1 - e -a t ) = a e -a t dt dt

160. Topic: Continuous-Time Signals: Fourier Series and Fourier Transform Representations (c)  Trigonometric Fourier series of an even function has DC and cosine terms only and does not have the sine terms. 161. Topic: LTI Systems: Stability (b)  Given that h(n) = 2nu(n − 2) For the causal system, h(n) = 0 for n < 0. The given sequence h(n) = 0 for n < 0. Therefore, the given system is causal.

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GATE ECE Chapter-wise Solved papers ∞

∑2

Therefore,

=∞

n

X ( s) =

n=2

therefore, for bounded inputs, the output is not bounded. Hence, the given system is not stable. 162. Topic: Network Solution Methods: Nodal and Mesh Analysis (b)  Converting delta into star, the circuit can be redrawn as shown below. j4 Ω

I

−j4 Ω

2Ω

14∠0°

2Ω

It is also given that H ( s) =

=

1⎛1 1 ⎞ 1 -6 s ⎜ ⎟+ e 2 ⎝ s s + 2⎠ s Taking inverse Laplace transform, we have =

165. Topic: Solution of Network Equations Using Laplace Transform 2( s + 1) (b)  F ( s) = L[ f (t )] = 2 s + 4s + 7 From the initial value theorem,

Z = ( 2 + j 4) ( 2 - j 4) + 2 Solving the above equation, we get Z =7Ω

lim f (t ) = lim sF ( s) = lim s ⋅

Therefore, the current I is obtained as

t →0

14∠0° = 2∠0°A 7

s →∞

Q 2.5 × 10 -3 == -50 V C 50 × 10 -6

Therefore,

Vc(0+) = -50 V In steady-state, the capacitor behaves as an open circuit; therefore, Vc(∞) = 100 V Now, Vc (t ) = Vc (∞) + ⎡⎣Vc (0 + ) - Vc (∞) ⎤⎦ e - t / RC 3

= 100 - 150e -2 ×10 t The current through the capacitor is dV (t ) ic (t ) = C c dt 3

-2 × 103 t

3

164. Topic: Solution of Network Equations Using Laplace Transform (d)  It is given that

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 156

2( s + 1) s + 4s + 7 2

2 s 2 [1 + (1 s)] 2

s [1 + ( 4 s) + (7 s )] 2

=

2(1 + 0) =2 (1 + 0 + 0)

From the final value theorem, lim f (t ) = lim sF ( s) = lim s ⋅ t →∞

s→0

s→0

2( s + 1) =0 s + 4s + 7 2

166. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (d)  From the circuit, we have I 2 = Y21V1 + Y22V2 Therefore, I 2 = 0.01V1 + 0.1V2 (1) From given figure, Therefore,

= 15e -2 ×10 t A

x (t ) = e u (t ) + δ (t - 6 )

s →∞

V2 = - I 2 R L = -100 I 2

= 50 × 10 × 150 × 2 × 10 × e

-2 t

s →∞

= lim

163. Topic: Time Domain Analysis of Simple Linear ­Circuits (a) At t = 0−, voltage across the capacitor is

-6

1 1 + e -6 s s ( s + 2) s

y(t ) = 0.5(1 - e -2t )u(t ) + u(t - 6)

The equivalent impedance of the circuit is

Vc (0 - ) = -

1 s

Therefore, Y ( s) = X ( s) ⋅ H ( s)

2Ω

I=

1 + e -6 s s+2

V2 100 Substituting the value of I2 in Eq. (1), we get I2 = -

-

V2 = 0.01V1 + 0.1V2 100

Therefore, V2 1 =V1 11

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Chapter 2  •  Networks, Signals and Systems

167. Topic: LTI Systems: Parallel and Cascade Structure (b)  It is given that y[ n] = x[ n - 1] . Therefore, Y ( z ) = z -1 X ( z ) or

Note: Power is consumed only by resistance, that is, by the real part of impedance. 170. Topic: Network Theorems: Thevenin (a)  To find the Thevenin’s equivalent impedance across nodes 1 and 2, connect a 1 V source and find the current through the voltage source as shown in the following figure.

Y ( z) = z -1 . X ( z)

For the cascaded system, H ( z ) = H1 ( z ) H 2 ( z ) Therefore, z -1 =

ib

(1 - 0.4 z -1 ) H 2 ( z) (1 - 0.6 z -1 )

99 ib

1 kΩ or,

157

z -1 (1 - 0.6 z -1 ) H 2 ( z) = (1 - 0.4 z -1 )

9 kΩ 1

168. Topic: Network Solution Methods: Nodal and Mesh Analysis (c)  The given network can be redrawn as shown in the following figure.

100 kΩ 2

+ A i

j1 Ω

1Ω

−

ib

ia

(9+1) kΩ

100 Ω

iTH

1A

1∠0 V B

1

(i+1)

99ib

+

D

−

+

C 2

1A

−j1 Ω

1Ω i

(i+1) E

According to Kirchhoff’s current law at node D, there will be no current in voltage sources. According to Kirchhoff’s current law at node A, current through inductor will be i1 = i + 1 (1) Applying Kirchhoff’s voltage law in loop ABDC, we have 1 ´× i + (i + 1)j1 - 1 ∠ 0 + 1 ∠ 0 = 0 Therefore, i=

-j (2) 1+ j

Therefore from Eqs. (1) and (2), we have i1 = i + 1 =

V =1 V

1∠0 V

-j 1 +1 = A j +1 j +1

169. Topic: Time Domain Analysis of Simple Linear Circuits (b)  The average power is the same as RMS power.

Thevenin’s impedance is 1 I By applying Kirchhoff’s current law at node 1, we get ib - ia + 99ib = iTH Therefore, 100ib - ia = iTH Z TH =

By applying Kirchhoff’s voltage law to the outer loop, we get 10 ´× 103 ib = 1 Therefore, ib = 10-4 A Applying Kirchhoff’s voltage law in the first loop, we get 10 ×´ 103ib = -100 ia Therefore, ia = -100 ib Substituting the value of ia in the expression, we get

2

25 ⎛ 5 ⎞ 2 P = I rms R=⎜ ⎟ × 4 = × 4 = 50 W 2 ⎝ 2⎠

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 1.indd 157

Therefore,

100ib - ia = iTH 100ib + 100ia = iTH iTH = 200ib = 200 ×´ 10-4 = 0.02 A

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GATE ECE Chapter-wise Solved papers

Hence, Z TH =

1 iTH

i=

1 = = 50 Ω 0.02

171. Topic: Time Domain Analysis of Simple Linear ­Circuits (d)  Since there is no resistance, hence the time constant will be zero. This means as soon as the switch will be closed, voltages across C1 and C2 will become equal. We know that capacitor allows sudden change of voltage only if impulse of current passes through it. Therefore, the current i(t) is an impulse current. 172. Topic: Discrete-Time Signals: z-Transform (c)  It is given that n

n

⎛ 1⎞ ⎛ 1⎞ x [ n] = ⎜ ⎟ - ⎜ ⎟ u [ n] ⎝ 3⎠ ⎝ 2⎠

n

-n

n

⎛ 1⎞ u [ - n - 1] - ⎜ ⎟ u [ n] ⎝ 2⎠

Applying Kirchhoff’s current law at node D, we have i5 = i2 + i3 + i4 = i2 + i1 Therefore, i5 = 3 A Applying Kirchhoff’s current law at node F, we have

n

1 ⎛ 1⎞ ⎜⎝ ⎟⎠ u [ n] ↔ 3 1 - (1 / 3) z -1

ROC z >

VC - VD = 1 ´× i6 = -5 V 1 3

The z-transform of -n

1 ⎛ 1⎞ ⎜⎝ ⎟⎠ u [ - n - 1] ↔ 3 1 - 3 z -1

ROC z < 3

The z-transform of n

1 ⎛ 1⎞ ⎜⎝ ⎟⎠ u [ n] ↔ 2 1 - (1 / 2) z -1

1 ROC z > 2

So the overall ROC will be intersection of three ROCs, that is, 1 < z 1 and g[k] is zero for k < 0 as it is causal sequence. 1 1 = × 1 + 1g[1] 2 2 Therefore, g[1] = 0.

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179. Topic: LTI Systems: Stability (d)  Given that y (t ) = Therefore,

Taking Laplace transform, we get

t

∫ x(τ ) cos (3τ )dτ −∞

y (t − t 0 ) =

t −t0

∫

−∞

x(τ ) cos (3τ ) dτ

For the input x (t − t0 ), y ′(t ) is t

∫ −∞

x(τ − t0 ) cos 3τ dτ =

( t − t0 )

∫ −∞

t

∫ −∞

cos 2 (3τ )dτ =

t

Y ( s) = H ( s) , therefore X ( s) Y(s) = H(s) · X(s)

−s −s Hence, Y ( s) = 1 . e = e 2 s s s3 Taking the inverse Laplace transform, we get

⎡1 + cos (6τ ) ⎤ ⎥ dτ 2 ⎦

y (t ) =

180. Topic: Network Theorems: Maximum Power Transfer (c)  For the pure resistive load (RL) to extract the maximum power, RL = Rs2 + X s2 = 4 2 + 32 = 5 Ω 181. Topic: Wye-Delta Transformation Rb Rc (b)  RA = Ra + Rb + Rc After scaling, Ra′ = kRa , Rb′ = kRb  and  Rc′ = kRc After scaling, the new value of RA is kRb × kRc kRb Rc = kRa + kRb + kRc ( Ra + Rb + Rc )

Therefore, RA′ = kRA Similarly, RB′ = kRB and RC′ = kRC 182. Topic: Solution of Network Equations Using Laplace Transform (c)  It is given that h(t) = tu(t) Taking Laplace transform, we get 1 H ( s) = 2 s It is also given that x(t) = u(t − 1)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 160

(t − 1) 2 u(t − 1) 2

183. Topic: Solution of Network Equations Using Laplace Transform (b)  It is given that x(t) = u(t) and

∫⎢ −∞ ⎣

Since for the bounded input, the output is not bounded, the system is not stable.

RA′ =

Since

x(τ ) cos 3 (τ + t0 )dτ

Since y ′(t ) ≠ y(t − t0 ) , the system is not time invariant. For the input x(τ ) = cos (3τ ) (bounded input), y (t ) =

e−s s

X ( s) =

H ( s) =

1 s

Taking Laplace transform, we get 1 s

X ( s) =

1 1 1 × = 2 s s s Taking inverse Laplace transform, we get

Y(s) = H(s) · X(s) =

y(t) = tu(t) 184. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Function (d)  V2 ( s) / × 10 −6 s) 10 × 103 + (1100 = V1 ( s) 10 × 103 + (1100 / × 10 −6 s) + (1100 / × 10 −6 s) = Therefore,

s × 10 4 + 10 4 10 4 (1 + s) = 4 4 4 4 s × 10 + 10 + 10 10 ( s + 2) V2 ( s) s + 1 = V1 ( s) s + 2

185. Topic: Continuous-Time Signals: Fourier Series and Fourier Transform Representations 2

2

(d)  It is given that g (t ) = e − π t . Therefore, g ( f ) = e − π f . As h(t) is a filter matched to g(t). Therefore, h( f ) = e − π f

2

When g(t) is applied as input to h(t), output in frequency domain is represented as y( f ) = g ( f )h( f ) 2

2

Therefore,   y( f ) = e − π f ⋅ e − π f = e −2π f

2

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Chapter 2  •  Networks, Signals and Systems

186. Topic: Continuous-Time Signals: Sampling Theorem and Applications (a)  The minimum sampling frequency is (fS)min = 2fM

Therefore,

( f S ) min = 2 × 5 × 103 Hz = 10 kHz So f S ≥ 10 kHz. Hence, option (a) is not valid. 187. Topic: LTI Systems: Impulse Response (c)  Convolution of h1(t) and h2(t). 188. Topic: LTI Systems: Causality and Stability (c)  All poles must lie within s = 1. 189. Topic: Network Solution Methods: Nodal and Mesh Analysis (c)    VYZ1 = 100 × 1.25 × 0.8 = 100 V In the second case when 100 V is applied at YZ terminals, this whole 100 V will appear across the secondary winding. Hence, VWX2 = 100 V Therefore, VYZ1 100 VWX2 100 , = = VWX1 100 VYZ2 100 190. Topic: Network Solution Methods: Nodal and Mesh Analysis (c)  The charge across a capacitor is Q = CV The breakdown charge of capacitor C1 is Q1 = C1V1 = 10 ´× 10 ´× 10 = 100 µC The breakdown charge of capacitor C2 is -6

Q2 = C2V2 = 5 ×´ 10-6 ´× 5 = 25 µC The breakdown charge of capacitor C3 is Q3 = C3V3 = 2 ´× 10-6 ´× 2 = 4 µC When the capacitors are in series, the charge is the same. So, the maximum charge on C2 and C3 will be minimum of (Q2, Q3) = min(25 µC, 4 µC) = 4 µC = Q23. In series, the equivalent capacitance of C2 and C3 is C23 =

C 2 C3 5 × 10 −6 × 2 × 10 −6 10 = = μF C2 + C3 5 × 10 −6 + 2 × 10 −6 7

So, the equivalent voltage is Q 4 × 10 −6 28 = = 2.8 V V23 = 23 = −6 C23 (10 / 7) × 10 10 In parallel, the voltage is same: V1 = V23 = 2.8 V

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 161

161

Therefore, the maximum safe voltage that can be applied across the combination is 2.8 V. The charge on capacitor C1 is Q1 = C1V1 = 10 ×´ 10-6 ×´ 2.8 = 28 µC In parallel, the total charge Q is obtained as Q = Q1 + Q23 = 4 ×´ 10-6 + 28 ×´ 10-6 = 32 µC Therefore, the effective capacitance across the terminals is 32 µC. 191. Topic: Network Solution Methods: Nodal and Mesh Analysis (d)  The network given in the problem can be redrawn as shown in the following figure. A I1

I2

5Ω Is + Vs

10 V 2A

+ 2 Ω V2 −

+ 1 Ω V1 −

− The voltage across 1 Ω resistance is V1 = 10 V The current through 1 Ω resistance is 10 = 10 A 1 The voltage across 2 Ω resistance. I1 =

V2 = 10 V The current through 2 Ω resistance is 10 = 5A 2 Applying Kirchhoff’s current law at node A, we get -2 + Is + I2 + I1 = 0 Is = 2 - I1 - I2 = 2 - 10 - 5 Therefore, Is = -13 A The voltage at node A is VA = 10 V Applying Kirchhoff’s voltage law on the left loop, we get Vs - 10 - 10 = 0 Hence, Vs = 20 V I2 =

192. Topic: Network Solution Methods: Nodal and Mesh Analysis (c)  The current in the 1 Ω resistor is I1 =

10 = 10 A 1

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193. Topic: Network Theorems: Thevenin (c)  To find VTH, open circuit the load resistance RL. Then, i2 = 0. VL1 =

Vs × ( j 4) 100 ∠53.13° × 4 ∠90° = 80 ∠90° = 3 + j4 5∠53.13° VTH = 10VL1 + i2 × j 6 + i2 × 3 = 10 × 80 ∠90° + 0 × j 6 + 0 × 3 = 800 ∠90°

194. Topic: Steady State Sinusoidal Analysis Using ­Phasors (c) ω L1 q1 = R1 q2 =

and

196. Topic: LTI Systems: Impulse Response (b)  It is given that the impulse response of the continuous time system is h(t ) = d ( t − 1) + d ( t − 3) We know that the step response of a system is the integration of its impulse response

∫ d (t − 1)dt = u(t − 1) and ∫ d (t − 3)dt = u(t − 3) Therefore, the output of the given system for step input is y(t ) = u(t − 1) + u(t − 3) The following figure shows the signal y(t). y(t)

ω L2 R2

2 1

Therefore,

ω L1 = q1 R1 and ω L2 = q2 R2 The coils are connected in series; therefore, ω ⋅ L = ω L1 + ω L2 = q1 R1 + q2 R2 and R = R1 + R2 Therefore, qR +q R q= 1 1 2 2 R Since R = R1 + R2 , we get q=

1 e −2 s 1 − e −2 s − = s s s

It is also given that d 2 y (t ) dy(t ) +5 + 6 y (t ) = x (t ) 2 dt dt Taking Laplace transform on both the sides and assuming dy(t ) y(0) = 0 and =0 dt t =0 we get s 2Y ( s) + 5sY ( s) + 6Y ( s) = X ( s) Therefore, Y ( s) =

1 − e −2 s 1 − e −2 s = s( s 2 + 5s + 6) s( s + 2)( s + 3)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 162

2

3

From the given figure, we see that at t = 2, y(t) = 1. 197. Topic: Network Solution Methods: Nodal and Mesh Analysis (a)  Given that i1 = 2 A, i4 = −1 A, and i5 = −4 A.

q1 R1 + q2 R2 R1 + R2

195. Topic: Solution of Network Equations Using Laplace Transform (b)  It is given that x(t ) = u(t ) − u(t − 2) Therefore, X ( s) =

t 1

i2

i5 B

1W

1W i3

i4

A i1 1 W

C

i6

Applying KCL at node A, we get i1 + i4 = i2 Therefore, i2 = 2 − 1 = 1 A Applying KCL at node B, we get i2 + i5 = i3 Therefore, i3 = 1 − 4 = −3 A Applying KCL at node C, we get i3 + i6 = i1 Therefore, i6 = 2 − (− 3) = 5 A 198. Topic: Network Solution Methods: Nodal and Mesh Analysis (2.8)  By source transformation, that is, replacing the current source and parallel resistance into equivalent voltage source and series resistance, we get

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Chapter 2  •  Networks, Signals and Systems

2kΩ

1kΩ

4kΩ I − + 8V

+ 20 V −

163

201. Topic: Network Theorems: Norton (d)  According to Norton’s theorem, a complex network connected to a load is replaced with an equivalent impedance in parallel with a current source as shown in the figure below.

3kΩ Applying KVL in the above circuit, we get

Load

Zeq

IN

20 − 2 × 103 I − 1 × 103 I − 4 × 103 I + 8 − 3 × 103 I = 0 Therefore,

I=

28 A = 2.8 mA 10 × 103

199. Topic: Network Theorems: Maximum Power Transfer (c)  Figure below shows two cascaded sections of one electrical network:

202. Topic: Steady State Sinusoidal Analysis Using Phasors (b)  The following figure shows the 230 V rms source supplying power to two loads connected in parallel. +

230 V Section 1

Z1ZL Z2

Load 1

Load 2

Section 2 −

where, Z1 = output impedance of section 1 Z2 = input impedance of section 2 For maximum power transfer between these two cascaded sections, the condition is

Therefore,  

Z 2 = Z1*

S1 = P − jQ

5W + −

5W 1A

Applying KCL at node A, we get VA − 5 V −1+ A = 0 5 15 where, VA is the voltage at node A.

Current I =

Q = P tan ϕ = 10 × 103 tan(cos −1 0.8) = 7.5 k VAR The complex power delivered by source to Load 1 is therefore given by

A

Therefore,VA =

The reactive component of power is

Z L = Z1* and ZL = Z2

200. Topic: Network Solution Methods: Nodal and Mesh Analysis (0.5)  Consider the circuit given below.

5V

Given that Load 1 draws power P = 10 kW and the leading power factor cos ϕ = 0.8

30 V 4

VA 30 = A = 0.50 A 15 4 × 15

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 163



= 10 − j7.5 kVA

Given that Load 2 draws a reactive power, so S = 10 kVA and the lagging power factor cos ϕ = 0.8. Now,

I 10 W

sin ϕ =

Q P and cos ϕ = S S

Therefore, P = 8 kW and Q = 6 kVAR The complex power delivered by the source to Load 2 is S2 = P + jQ = 8 + j  6 Therefore, complex power delivered by the source to both the loads is S1 + S2 = 18 − j1.5 kVA 203. Topic: Time Domain Analysis of Simple Linear ­Circuits (1.1 mA)  The equivalent circuit at t = 0+ is shown in the figure below

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5 kW

10 V

4 kW

1 kW

From the above circuit, b1 = s11a1 + s12a2 b2 = s21a1 + s22a2

+ −

Writing in the matrix form, we get

i(0+)

⎡b1 ⎤ ⎡ s11 ⎢b ⎥ = ⎢ s ⎣ 2 ⎦ ⎣ 21

For t = 0+, i(0 + ) =

(5 × 10

10 3

+ 4 × 103

)

= 1.11 mA

204. Topic: Time Domain Analysis of Simple Linear C ­ ircuits (a)  In a series RC circuit, initially at t = 0, capacitor charges with a current of Vs/R and in steady state at t = ∞, capacitor behaves like open circuit and no current flows through the circuit. Hence, the current i(t) can be represented as an exponentially decaying function as shown in the figure below.

Therefore, s11 =

b1 a1

a2 = 0

Applying this condition, we get s11 − s11 s22 − s12 s21 (1 + s22 )

s11 =

207. Topic: Linear 2-Port Network Parameters: State Equations for Networks (d)  Given state model, ⎡ x1 ⎤ ⎡0 1 ⎤ ⎡ x1 ⎤ ⎢ x ⎥ = ⎢0 0 ⎥ ⎢ x ⎥ ⎦⎣ 2⎦ ⎣ 2⎦ ⎣

i(t)

⎡0 1 ⎤ A= ⎢ ⎥ ⎣0 0⎦

Let

Let the state transition matrix be ϕ (t ) . Therefore,

t

0

205. Topic: Solution of Network Equations Using Laplace Transform . (a) Given y(t) + 5(t) = u(t) and y(0) = 1. Applying Laplace transform to the given differential equation, we get 1 sY ( s) − y(0) + 5Y ( s) = s Therefore, Y ( s)[ s + 5] = Hence,

1 + y ( 0) s

⎛1 ⎞ +1 1 4 ( s + 1) ⎜ ⎟ Y ( s) = ⎜ s ⎟ = = + ⎝ s + 5 ⎠ s( s + 5) 5s 5( s + 5) Applying inverse Laplace transform to the above equation, we get y (t ) =

1 4 −5t + e = 0.2 + 0.8e −5t 5 5

206. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (b)  Refer to the circuit shown below. a1 b1

s12 ⎤ ⎡ a1 ⎤ s22 ⎥⎦ ⎢⎣ a2 ⎥⎦

Two port network

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 164

a2 b2

ϕ (t ) = L−1 ⎡⎣( sI − A) −1 ⎤⎦

[ sI − A]

−1

−1

⎡ s −1⎤ 1 ⎡s 1⎤ = 2⎢ =⎢ ⎥ s ⎣0 s ⎥⎦ ⎣0 s ⎦

Therefore, ⎡1 / s 1 / s 2 ⎤ ⎡1 t ⎤ ϕ (t ) = L−1 ⎢ ⎥=⎢ ⎥ 1 / s ⎦ ⎣0 1⎦ ⎣0 208. Topic: Continuous-Time Signals: Sampling Theorem and Applications (3)  Given that, the signal x(t) is band limited to [−500 Hz, 500 Hz] and signal y(t) is band limited to [−1000 Hz, 1000 Hz]. Also z(t) = x(t)y(t) We know that multiplication in time domain results in convolution in frequency domain. The range of convolution in frequency domain is [−1500 Hz, 1500 Hz]

−500

=

Hz 500

−1500

∗ −1000

Hz 1000

Hz 1500

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Chapter 2  •  Networks, Signals and Systems

So, maximum frequency present in z(t) is 1500 Hz. Nyquist sampling frequency is twice the maximum frequency component of the sampled signal. Therefore, Nyquist rate = 3000 Hz or 3 kHz. 209. Topic: Continuous-Time Signals: Sampling Theorem and Applications (a)  Given that x(t ) = cos(10π t ) + cos(30π t ) and sampling frequency fs = 20 Hz The frequency spectrum of x(t) is shown in the below figure: x(t)

f(Hz)

−15 −5

5

15

The frequency spectrum of sampled version of x(t) is shown in the following figure. xs(t)

−35 −25 −15 −5

f(Hz) 5

15

25

35

Since the LPF has a cut-off frequency of 20 Hz, therefore, after the LPF the signal will contain only 5 Hz and 15 Hz components. 210. Topic: Continuous-Time Signals: Sampling Theorem and Applications (10)  m(t) is a baseband signal with maximum frequency 5 kHz. let us assume that the frequency spectrum M(f) of signal m(t) is as shown below. M(f)

f(Hz)

−5000

+5000

Since, y (t ) = m(t )cos(40000π t ) . therefore, frequency spectrum Y(f) of y(t) is Y( f ) = M( f )*

1 [d ( f − 20000) + d ( f + 20000)] 2

Therefore, Y( f ) =

1 [ M ( f − 20000) + M ( f + 20000)] 2

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 165

Thus the spectrum of the modulated signal is as follows: Y(f )

−25000

−20000

f(Hz)

−15000 15000

20000

25000

If y(t) is sampled with a sampling frequency fs then the resultant signal is a periodic extension of successive replica of y(t) with a period fs. It is observed that 10 kHz and 20 kHz are the two sampling frequencies which cause a replica of M(f ) which can be filtered out by a LPF. Hence, the minimum sampling frequency (fs) which extracts m(t) from y(t) is 10 kHz. 211. Topic: Discrete-Time Signals: Discrete-Time Fourier Transform (DTFT) ⎛ π n⎞ (10) Given: x[n] = sin ⎜ ⎟ ⎝ 5 ⎠ The given signal is a periodic signal with period N = 10. Therefore, the Fourier series coefficients of the signal are also periodic with period N = 10. Signal x[n] can also be re-written as 1 j 210π n 1 − j 210π n e − e 2j 2j 1 1 1 ⇒ a−1 = a−1+10 = a9 = − a1 = ; a−1 = − 2j 2j 2j a1 = a1+10 , a1 = a1+ 20 a−1 = a−1+10 , a−1 = a−1+ 20 x[n] =

Extending the concept, we get that for coefficient ak, k = 10m +1 or 10m − 1. Therefore, B = 10. 212. Topic: Discrete-Time Signals: z-Transform (b)  Given that x[n] = x[−n]. Therefore, by time reversal property of z-transform X(z) = X(z−1). If one zero is 0.5 + j 0.25, then other zero will be 1 . (0.5 + j 0.25) 213. Topic: Discrete-Time Signals: Discrete-Time Fourier Transform (DTFT) (3.375)  Given that n

⎛ 2⎞ x[n] = ⎜ ⎟ u[n + 3] ⎝ 3⎠ Therefore, Fourier transform of x[n] is −3 ⎛ 2⎞ ⋅ e − j 3ω n ⎜ ⎟ ∞ ⎛ 2 ⎞ − jω n ⎝ 3 ⎠ iω X (e ) = ∑ ⎜ ⎟ e = 2 n =−3 ⎝ 3 ⎠ 1 − e −i ω 3

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Comparing the expressions, we get

Therefore, the steady state output is

3

27 ⎛ 3⎞ A= ⎜ ⎟ = = 3.375 ⎝ 2⎠ 8

3

y(t ) = ∫ 3 × 5dτ = 45 0

214. Topic: Discrete-Time Signals: Discrete-Time Fourier Transform (DTFT) (4)  y[n] = x[n] * x[n] Let Y(e j w) be the Fourier transform of y[n], X(e j w) be the Fourier transform of x[n]. Therefore, Y(e j w) = X(e j w) · X(e j w) Hence, Y (e jω ) =

1 1 . 1 − 0.5e − j ω 1 − 0.5e − jω

Also  Y (e j ω ) =

∑ y[n] ⋅ e n =−∞

Therefore,

∞

∞

∑ y[n] = y[e n =−∞

j0

217. Topic: LTI Systems: Poles and Zeros (c)  Minimum phase system has all the zeros inside the unit circle. Maximum phase system has all the zeros outside the unit circle. Mixed phase systems have some zeros outside the unit circle and some zeros inside the unit circle. For the given system, H ( z) = 1 +

7 −1 3 −2 z + z , 2 2

− jω n

]=

one zero is inside the unit circle and one zero outside the unit circle. Hence, it is mixed phase system.

1 1 . =4 0.5 0.5

215. Topic: LTI Systems: Definition and Properties (d)  Let us assume that signal x[n] is periodic with period N. Therefore, x[n] = x[n + N] Hence, for the given signal to be periodic

218. Topic: LTI Systems: Phase Delay (c)  Phase response of passband waveform

ϕ ( f ) = −2πa ( f − f c ) − 2πb f c Group delay t g =

− dϕ ( f ) =a 2π df

Thus, a is actual signal propagation delay from transmitter to receiver.

sin (π 2 n) = sin[π 2 ( n + N )] We also know that every trigonometric function repeats itself after 2π interval. Hence, sin (π 2 n + 2π k ) = sin(π 2 n + π 2 N )

219. Topic: LTI Systems: Frequency Response (d) Let X( f ) be the Fourier transform of x(t). The LTI system is shown in the figure below. x(t ) h(t )

Therefore, 2π k = π 2 N or

2π k 2k = N or N = π π2

Hence, if k is an integer, N cannot be an integer. Hence, the given signal is non-periodic. 216. Topic: LTI Systems: Digital Filter Design Techniques (45)  The given filter can be represented as x(t)

h(t)

y ( t ) = x ( t ) ∗ h (t ) Given that x(t) 3 h(t) t

t 3

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 166

Output y(t) is given by the convolution of x(t) with h(t). Therefore, y(t) = x(t) * h(t) Therefore,

Y( f ) = X( f  ) · H( f  )

or, Y( f  ) = e−j4πf  · X( f  ) Transforming into time domain, we get y(t) = x(t − 2)

y(t)

5

y(t )

220. Topic: Network Solution Methods: Nodal and Mesh Analysis (2.49)  Given 56% of the total flux emanating from one coil links to other coil, that is, K = 56% = 0.56. We have, M K= L1 L2 as L1 = 4 H; L2 = 5 H, we have M = (0.56) 20 H = 2.49H

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Chapter 2  •  Networks, Signals and Systems

221. Topic: Network Solution Methods: Nodal and Mesh Analysis (d) 4W

Super node

10Ð0°V +−

V1

6W

jωC ⎤ 1 ∠ 0° ⎡1 V ⎢ + jωC + = R 2 ⎥⎦ R ⎣

Therefore,

V=

Hence,

y (t ) =

j6 W

Given that A(ω ) =

Appling KVL for V1 and V2, we get

Therefore,

+−

+

V2 Hence, ω =

+

−

10Ð0°

−

1 4 + 9R C ω

V1 = V2 + 10 ∠0°

Applying KCL at super node, we get −4 ∠0° +

V1 V2 V2 + + =0 − j3 6 j6

=

1 4

3RC

Therefore, −10 +

From the above two equations,

Therefore,V2 = ( 2 − j 22) V 222. Topic: Time Domain Analysis of Simple Linear ­Circuits (b)  Let the voltage at the top junction be V as shown in the figure below:

Therefore, vo (t ) =

R

R R

y(t) C C

Appling KCL at this node, we get V − 1 ∠ 0° V V + + =0 R (1 jωC ) ( 2 / jωC )

250 = 31.25 V 8

224. Topic: Network Solution Methods: Nodal and Mesh Analysis (2.618)  Let us consider the infinite ladder network shown below.

V C

3ix + ix = 0 5

The voltage vo(t) is given by vo (t ) = 5ix

⎛ 1 1 1⎞ 10 + + ⎟ = 4 ∠0° + V2 ⎜ ⎝ − j3 6 j6 ⎠ j3

R

(VA − 2ix ) + ix = 0 5

Hence, ix = 50 /8 A

V2 + 10 ∠0° V2 V2 + + = 4 ∠0° 6 j6 − j3

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 167

2

2

−10 +

V1 V2 V2 + + = 4 ∠0° − j3 6 j6

+ sin(ω t) −

2

223. Topic: Time Domain Analysis of Simple Linear ­Circuits (31.25)  Let the voltage at the top node be VA. For t → ∞, that is, at steady state, the inductor will behave as a short circuit. Therefore, VA = 5ix Applying KCL at top node, we get

V1 − V2 = 10 ∠0° or,

V 1 = 2 2 + jω 3 RC

1 4

2

V1

2 2 + 3 jω RC

Voltage y(t) is given by,

V2

−j3 W

4Ð0°A

167

R R

R R

R

Req For an infinite ladder network, if all resistances are having the same value of R, then equivalent resistance Req is given by ⎛1+ 5 ⎞ Req = ⎜ ⎟ × R = 1.618 R ⎝ 2 ⎠ The given network can be considered as series network of R and Req as shown in figure below:

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Differentiating the above equation w.r.t time, we get

R

Rdi(t ) d 2 i (t ) i (t ) +L + =0 dt C dt 2

Req

Dividing both LHS and RHS by L, we get d 2 i(t ) R di(t ) i(t ) + + =0 L dt LC dt 2

Re Therefore, R e = R + Req = R + 1.618 R Hence,

Re = 2.618 R

225. Topic: Network Theorems: Norton (2)  The figure below shows the Norton’s equivalent circuit.

2

The values can be rewritten as 2

D1, 2 =

−R ⎛ R⎞ ⎛ 1 ⎞ ± ⎜ ⎟ −⎜ ⎝ 2 L ⎠ ⎝ LC ⎟⎠ 2L

For critically damped response,

1F 0.5 H

2

b¢ Z n The Norton’s equivalent impedance is given by 1 2+ 1 Zn = 1 jω ⋅1 1 + jω . 2 1 jω = + 2 + jω jω 1 × jω ⋅

=

( 2 − ω 2 ) + jω [2 jω − ω 2 ]

Therefore, ⎡(ω 2 − 2) − jω ⎤⎦ ⎡⎣ω 2 + 2 jω ⎤⎦ Zn = ⎣ ⎡⎣ω 4 + 4ω 2 ⎤⎦ Equating the imaginary term to zero, we get, ω 3 − 4ω = 0

Therefore, ω (ω 2 − 4) = 0. Hence, ω = 2 rad /s 226. Topic: Wye-Delta Transformation (10) (7.5)(5) + (3)(5) + (7.5)(3) R1 = = 10 Ω 7.5 227. Topic: Time Domain Analysis of Simple Linear ­Circuits (10) Let R = 40 Ω and L = 4 H. Applying kVL to the loop, we get Ri(t ) +

D1, 2 =

4

( R / L )2 − LC

−R / L ±

b

1W



The roots of the above equation are

Ldi(t ) 1 + ∫ i(t )dt + Vo = 0 dt C

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 168

⎛ R⎞ ⎛ 1 ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ ⎟ 2L LC ⎠ Therefore,     C = 4 L F R2 Given L = 4 H; R = 40 Ω, substituting values, we get C=

4×4 F = 10 mF ( 40) 2

228. Topic: Frequency Domain Analysis of RLC Circuits (0.45)  Let the operating frequency (ω), at which current leads the voltage, be ωx. Therefore ωx < ωr. Again, magnitude of current is half the value at resonance. At ω = ωx. V Ix = Z At ω = ωr, I resonance = Given that, I x =

V R

I resonance 2

It implies,

V V = Z 2R

Therefore, Z = 2R Now, given R = 1 Ω, L = 1 H, C = 1 F 2

⎛ 1 ⎞ Z = R +⎜ − ω x L⎟ = 2 ⎝ω C ⎠ 2

x

2

⎞ ⎛ 1 − ω x L⎟ = 4 Therefore, R 2 + ⎜ ⎠ ⎝ ω xC

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V3 ( s)

R⋅ R

=

Chapter Signals 1 ⎡ 1 ⎤ 1 Systems V1 (2  s)•  Networks, ⎤ ⎡and

+ R⎢ + R⎥ R+ R+ sC ⎢⎣ sC ⎥⎦ ⎦ ⎣ sC

By substituting values for R, L and C, we get ωx = 0.45 rad/s or ωx = 2.18 rad/s. Since ωx < ωr, therefore, ωx = 0.45 rad/s.

=

229. Topic: Time Domain Analysis of Simple Linear ­Circuits (a)  Voltage across the capacitor at time t is given by

=

=

2 where, τ = Req Ceq = × 103 × 1 × 10 −6 3

)

Therefore,

2 ms 3

2 × 103

(1 × 10

3

+ 2 × 10

3

)

=

× 10 × 103 × 10 × 103

(

s 2 × 100 × 10 −6

10 V 3

3 2

⎡ −1 0 ⎤ ⎡ 1⎤ x ( 0) = ⎢ ⎥ , A = ⎢ ⎥ ⎣ −1⎦ ⎣ 0 −2⎦

10 10 − t / τ vC (t ) = − e 3 3

⎡s + 1 0 ⎤ s + 2⎥⎦ ⎣

Voltage across resistor R2 is the same as that across capacitor C, therefore,

[ sI − A]−1 = ⎢0

10 10 − t / τ − e 3 3 10 = ⎡⎣1 − e − t / τ ⎤⎦ V 3

VR 2 (t ) =

Therefore, I (t ) =

2

V3 ( s) s2 = V1 ( s) 1 + 3s + s 2

Therefore,



) × (10 × 10 )

231. Topic: Linear 2- Port Network Parameters: State Equations of Networks (c)  The solution of state equation of x(t) = L−1 [sI − A−1] × x(0)

Vinitial = 0 V Vfinal = 5 ×

s 2C 2 ⋅ R ⋅ R [1 + 2 R( sC )] + RsC + R2 s2C 2

+ 3s × 100 × 10 −6 × 10 × 103 + 1

2 ⎛ ⎞ 3 3 ⎜⎝ Req = 1 × 10  2 × 10 Ω = kΩ and Ceq = 1μF⎟⎠ 3 Therefore, τ =

R⋅ R 1 1 R . [ 2 R( sC ) + 1] + [1 + RsC ] sC sC sC

s 2 × 100 × 10 −6 × 100 × 10 −6

vC (t ) = Vfinal + (Vinitial − Vfinal )e −t /τ

(

169

−1

⎤ ⎡ 1 0 ⎢ ⎥ s 2 0 + ⎤ ⎡ 1 s +1 =⎢ = ⎥ s + 1⎥⎦ ⎢ 1 ⎥ ( s + 1)( s + 2) ⎢⎣ 0 0 ⎢⎣ s + 2 ⎥⎦

VR 2 (t ) 5 A = ⎡⎣1 − e − t / τ ⎤⎦ mA 3 3 2 × 10

⎤ ⎡ −1 ⎡ 1 ⎤ ⎢ L ⎢ s + 1⎥ 0 ⎥ ⎣ ⎦ ⎥ L−1 ⎡⎣( sI − A) −1 ⎤⎦ = ⎢ ⎢ ⎥ −1 ⎡ 1 ⎤ L ⎢ ⎢0 ⎥ ⎥ ⎣ s + 2 ⎦⎦ ⎣

where τ = 2/3 ms 230. Topic: Solution of Network Equations Using Laplace Transform (b)  Given that the two blocks are connected in cascade. The cascaded networks in s-domain are shown below

⎡e − t =⎢ ⎣0



0 ⎤ ⎥ e −2t ⎦

Now +

+ 1 sC

1 sC R

V1(s)

R

V3(s) −

−

V3 ( s) = 1 V1 ( s) sC =

R⋅ R 1 ⎤ ⎤ ⎡ 1 ⎡ ⎢⎣ R + R + sC ⎥⎦ + R ⎢⎣ sC + R ⎥⎦

R⋅ R 1 1 R . [ 2 R( sC ) + 1] + [1 + RsC ] sC sC sC

s 2C 2 ⋅ R ⋅ R Ch wise GATE_ECE_CH02_Networks, Signals[and 2.indd+169 R 2 s 2C 2 1 +Systems_Expln_Part 2 R( sC )] + RsC =

⎡ x1 (t ) ⎤ ⎡e − t ⎢ x (t ) ⎥ = ⎢ ⎣ 2 ⎦ ⎣0

0 ⎤ ⎡1 ⎤ ⎥⎢ ⎥ e −2t ⎦ ⎣ −1⎦

Therefore, x1 (t ) = e − t and x2 (t ) = −e −2t 232. Topic: Solution of Network Equations Using Laplace Transform LT (d) If f(t) ←⎯ → F(s), then t ⋅ f (t ) ↔

−d F ( s) ds

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=

=−

− ( 2 s + 1)

( s + s + 1) 2

2

=

2s + 1

−1 / 2⎤ ⎡ 1 y2 = ⎢ 1 ⎥⎦ ⎣ −1 / 2 ⎡ 5 / 3 −5 / 6 ⎤ yeq = ⎢ 5 / 3⎥⎦ ⎣ −5 / 6 Therefore the relation between h parameters and y parameters is given by − y12 / y11 ⎤ Δy / y11 ⎥⎦

Δy ⎡⎛ 5 ⎞ ⎛ 5 ⎞ ⎤ ⎡⎛ −5 ⎞ ⎛ −5 ⎞ ⎤ = ⎜ ⎟ ×⎜ ⎟ − ⎜ ⎟ ×⎜ ⎟ 5 3 y11 ⎢⎣⎝ 3 ⎠ ⎝ 3 ⎠ ⎥⎦ ⎢⎣⎝ 6 ⎠ ⎝ 6 ⎠ ⎥⎦

Therefore, h22 = 1.24 234. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (c)  For the p two-port network shown in the figure, the admittance matrix Y is given by 1 ⎤ 30 ⎥ ⎥ 1 1⎥ + 60 30 ⎥⎦

Now impedance matrix Z is given by inverse of admittance matrix Y, that is. Z = [Y ]−1



−1

⎡9 6 ⎤ Z=⎢ ⎥ ⎣6 24 ⎦ 235. Topic: Wye- Delta Transformation (29.09)  Figure below shows the given Y-network and its equivalent Delta network.

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 170

W

11 W

10 W

Z       Star network     Equivalent delta network

(10)(10) + (10)(11) + (10)(11) = 29.09 Ω 11

Value of resistance Y is given by (10)(10) + (10)(11) + (10)(11) = 32 Ω 10 Value of resistance Z is given by Y =

(10)(10) + (10)(11) + (10)(11) = 32 Ω 10 Lowest value among three resistances = 29.09 Ω Z=

236. Topic: Discrete-Time Signals: Interpolation of ­Discrete-Time Signals (0.408)  The root mean square (rms) value of x(t) is given by T

X rms =

1 x(t ) 2 dt T ∫0

From the given curve, x(t) is given by ⎧2 ⎪ t x (t ) = ⎨ T ⎪⎩0

0 ≤ t ≤ T /2 T /2 ≤ t ≤ T

Therefore,

−

⎡0.1333 −0.0333⎤ Z=⎢ ⎥ ⎣ −0.0333 0.05 ⎦

10

11 W

X =

y − parameter matrix of network 2 is

1 ⎡1 ⎢ 30 + 10 Y= ⎢ ⎢ −1 ⎢⎣ 30

Y

X

Value of resistance X is given by

⎡ 2 / 3 −1/ 3⎤ y1 = ⎢ ⎥ ⎣ −1/ 3 2 / 3⎦

h22 =

10 W

( s + s + 1)2 2

233. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (1.24)  If two π networks are connected in parallel, their y-parameters are added. y − parameter matrix of network 1 is

⎡ 1 / y11 h=⎢ ⎣ y21 / y11 where Δy = y11y22 - y12y21 The value of

10 W

−d ⎛ 1 ⎞ ⎜⎝ 2 ⎟ ds s + s + 1⎠

X rms =

1 T

T ⎡T / 2 ⎛ 2 ⎞ 2 ⎤ 2 ⎢ ∫ ⎜ t ⎟ dt + ∫ (0) dt ⎥ ⎢⎣ 0 ⎝ T ⎠ ⎥⎦ T /2 T /2

=

1 4 ⎛ t3 ⎞ . T T 2 ⎜⎝ 3 ⎟⎠ 0

=

4 ⎛T ⎞ ⎜ ⎟ 3T 3 ⎝ 2 ⎠

=

1 = 0.408 6

3

237. Topic: Discrete-Time Signals: DFT (b)  This can be solved directly using options and satisfying the condition given in equation X = DFT(x)

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Chapter 2  •  Networks, Signals and Systems

DFT ( DFT( x ) ) = DFT( X ) =

1

N −1

∑ x[n]e N

−j

2π nk N

n= 0

DFT of y[1 2 3 4] 1 1 1⎤ ⎡1 ⎤ ⎡1 ⎢1 − j −1 j ⎥⎥ ⎢⎢ 2 ⎥⎥ 1 ⎢ 1 = X = ⎢ ⎥ ⎢ ⎥ 1 1 1 1 3 − − 4 4 ⎢ ⎥⎢ ⎥ 1 1 4 + j − − j ⎣ ⎦⎣ ⎦ DFT of (x) will not result in [1 2 3 4] Let us try with DFT of y[1 2 3 2]

240. Topic: LTI Systems: Impulse Response (a)  Given that system equation is y[n] = a y[n − 1] + b x[n]

⎤ ⎡10 ⎢2 + 2 j ⎥ ⎢ ⎥ ⎢2 ⎥ ⎢ ⎥ ⎣ −2 − j 2⎦

Therefore, Y ( z) b = X ( z ) 1 − a z −1 Hence, the transfer function H(z) is given by H ( z) =

1 1 1⎤ ⎡1 ⎤ ⎡ 8⎤ ⎡ 4 ⎤ ⎡1 ⎢1 − j −1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ j ⎥ ⎢ 2⎥ 1 ⎢ 1 ⎢ −2⎥ ⎢ −1⎥ = = X = 4 ⎢1 −1 1 −1⎥ ⎢3 ⎥ 4 ⎢ 0⎥ ⎢ 0⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ j ⎦ ⎣ 2⎦ ⎣1 + j −1 ⎣ −2⎦ ⎣ −1⎦ 1 1 1⎤ ⎡ 4 ⎤ ⎡ 2 ⎤ ⎡1 ⎤ ⎡ 4⎤ ⎡1 ⎢ −1⎥ ⎢1 − j −1 ⎥ ⎢ ⎥ j ⎥ ⎢ −1⎥ 1 ⎢⎢ 4 ⎥⎥ ⎢⎢ 2⎥⎥ ⎢ ⎥= 1 ⎢ = = ⎢ 0⎥ 4 ⎢1 −1 1 −1⎥ ⎢ 0 ⎥ 2 ⎢6 ⎥ ⎢3 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −1⎦ ⎣1 + j −1 − j ⎦ ⎣ −1⎦ ⎣ 4 ⎦ ⎣ 2⎦ This is same as X. Therefore, the correct option is (b).

∞

∑ h[n] = 2 n= 0

Therefore, ⎡ 1 ⎤ b⎢ ⎥=2 ⎣1 − a ⎦ Rearranging the terms, we get

a = 1−

⎧1 ⎪ , SX ( f ) = ⎨ w ⎪0, ⎩

1 1 1 = −1 2 −1 (1 − 2 z ) (1 − 2 z ) (1 − 2 z −1 )

Therefore, x[n] = 2 u[n]* 2 u[n] = ∑ 2 ⋅ 2 n

n

k

w

Rx (τ ) =

(n−k )

k =0

= 2

k =0

= 20 ⋅ 22 + 21 ⋅ 21 + 22 ⋅ 20 = 4 + 4 + 4 = 12 239. Topic: LTI Systems: Convolution ⎧n for 0 ≤ n ≤ 10 (10) Given x[n] = ⎨ ⎩0 otherwise ] =x[xn[]n]* x[n] y[n] = xy[[nn]*



n

n

k =0

k =0

4

4

k =0

k =0

= ∑ x[k=]∑ ⋅ x[xn[−k ]k⋅]x[n − k ] y[4x][k=]∑ y[4] = ∑ ⋅ x[x4[−k ]k⋅ ]x[4 − k ] = x[0] ⋅ =x[x4[]0+] ⋅xx[1[]4⋅]x+[3x][1+] ⋅xx[2[3]]⋅ + x[x2[]2] ⋅ x[2] 3]x⋅ [x4[1]]⋅ + x[x0[]4] ⋅ x[0] + x[3] ⋅ x+[1x][+ = 0.4 + 1=.30.+4 2+.21.+3 3+.12.+2 4+.03.1 + 4.0 3 + 0 = 10 = 0 + 3 += 40 ++ 33++ 04 =+ 10

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 171

1

∫ w .e

j 2π ft

⎫ f ≤ w⎪ ⎬ f > w ⎪⎭

df

−w

Therefore, x[2] = ∑ 2k ⋅ 2( 2 − k )

b 2

241. Topic: LTI Systems: Frequency Response (4) Given

238. Topic: Discrete -Time Signals: z-Transform (12)  Given that

n

b 1 − a z −1

Also given that

DFT of

X ( z) =

171

1 e j 2π wt − e − j 2π wt 1 ⎛ sin( 2π wt ) ⎞ = ⎜ ⎟⎠ w j 2π t w⎝ πt

Now, ⎡ 1 ⎞⎤ ⎛ 1 ⎞ ⎛ E ⎢π x(t ) ⋅ x ⎜ t − ⎟ ⎟⎠ ⎥ = π Rx ⎜⎝ ⎝ 4 4w ⎠ w ⎣ ⎦ 1 ⎞ ⎛ sin ⎜ 2π w ⋅ ⎟ ⎝ 1 4w ⎠ =π⋅ =4 1 w π⋅ 4w 242. Topic: LTI Systems: Parallel and Cascade Structure (b)  Given that, H ( s) =

1 1 = s + s − 6 ( s + 3)( s − 2) 2

It is given that the system is stable. Therefore, its ROC includes jw axis. This implies it cannot be causal, because for causal system, ROC is on right side of the right most pole.

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GATE ECE Chapter-wise Solved papers

Therefore, pole at s = 2 must be removed so that the system becomes causal and stable simultaneously. Hence, a zero must be introduced at s = 2. So the transfer function of the total system is 1 1 ( s − 2) = ( s + 3)( s − 2) s+3

a R R R

d Now, g (t ) = a ∫ h(τ )dτ + h(t ) + a h(t ) dt 0 Taking Laplace transform of the above equation, we get G ( s) = a 2

H ( s) + sH ( s) + a H ( s) s

1 1 +s 2 s( s 2 + a s + a 2 ) (s + a s + a 2 ) a + 2 (s + a s + a 2 )

= a2

=

2R

|VC| = 25 V 245. Topic: Network Solution Methods: Nodal and Mesh Analysis (100)  The given circuit can be considered equivalent to the following circuit:

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 172

R

R

2R

b Rab= R/3 = 100 Ω 246. Topic: Network Solution Methods: Nodal and Mesh Analysis (a)  It is given that current through the 4 Ω resistance in the first branch is I. Therefore, the current through the 4 Ω resistor in second branch is also I. Applying KCL at the top node, we get -5 + I + I + 2I = 0. Therefore, 5 4 I = 5 or I = A 4 From the circuit, V1 = I × 4 = 4 × Similarly,

5 =5V 4

V2 = (4 I ) + V1 = 4 (5) + V1 = 20 + 5 = 25 V

247. Topic: Network Solution Methods: Nodal and Mesh Analysis (8) 0.5 Vx

ω ⋅ L 1 L 1 0.1 × 10 −3 10 = = = = 2.5 4 R R C 4 1 × 10 −6

Therefore, VC = 2.5 × 10 ∠ − 90° = 25∠ − 90°

R

Rab

Therefore, the number of poles = 1

Q=

R R/2

a

a 2 + a s + s2 1 = s( s 2 + a s + a 2 ) s

244. Topic: Network Solution Methods: Nodal and Mesh Analysis (25) Let VC be the voltage across the capacitor. We have VC = QV ∠ − 90° Also,

R R/2

Using bridge condition, the above circuit can be redrawn as follows:

s 2Y ( s) + a sY ( s) + a 2Y ( s) = X ( s)

t

R

b

243. Topic: LTI Systems: Poles and Zeros (1)  Taking Laplace transform of the given differential equation, we get

2

R/2

Rab

Therefore, H1(s) = (s − 2)

Y ( s) 1 Therefore, = = H ( s) X ( s) s 2 + a s + a 2

R/2

P

10 W

+ 5A

20 W

Vx

8W

+ −

0.25 Vx

− Applying KCL at point P in the circuit, we get Vx Vx − 0.25Vx + − 5 + 0.5Vx = 0 20 10

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Chapter 2  •  Networks, Signals and Systems

251. Topic: Steady State Sinusoidal Analysis Using Phasors (100)  Let the input voltage be represented as V.

Therefore, ⎛ 1 0.75 ⎞ + 0.5⎟ = 5 Vx ⎜ + ⎝ 20 10 ⎠ Therefore, Vx = 8 V

V = VR2 + (VC − VL ) 2

248. Topic: Frequency Domain Analysis of RLC Circuits (0.5)  Impedance of a capacitor C is given by 1 ZC = jω C

(100)2 = (80)2 + (VC - 40)2 (VC - 40)2 = (100)2 − (80)2 = 3600 (VC - 40) = ± 60 Therefore,

As ω →∞, Z C → 0 Replacing all capacitors by short circuit, the equivalent circuit of the given circuit is

VC = ± 60 + 40 Since VC cannot be negative, therefore, VC = 60 + 40 = 100 V.

+ 1 kW

1 kW

Vo −

Vi 1.0 sin w t 1 kW

1 kW

From the above circuit V Vo = i 2 1⋅sin ωt = = 0.5 sin ω t 2

252. Topic: Steady State Sinusoidal Analysis Using Phasors (5)  The capacitor 1mF acts as a short circuit for the AC voltage and an open circuit for DC voltage under steady state condition. Hence, voltage Vab under steady state condition = 5 V. 253. Topic: Time Domain Analysis of Simple Linear ­Circuits (c)  The capacitor is initially uncharged, therefore VC(0) = 0. The capacitor will be charged to supply voltage 3 V when the switch is in position B for infinite time. The capacitor voltage vC(t) is given by

Therefore, the peak value of output voltage is 0.5 V.

vC (t ) = VC (∞) + [VC (0) − VC (∞)]e − t / τ = 3 − 3e − t / τ

249. Topic: Network Theorems: Maximum Power Transfer (1.66) 4×2j VTh(rms) = = 2 2∠45° 2+2j

τ = RC = 120 × 0.1 × 10 −6 s = 12 μs iC (t ) = C

Z Th = 2 || 2 j = 1 + j

=

Maximum power transfer to RL is Pmax = I × RL =

2 2∠45° 2 +1+ j

2

× 2 = 1.66 W

250. Topic: Network Theorems: Thevenin (10)  Let the Thevenin’s equivalent voltage be Vth. Applying KCL at terminal a, we get Vth − 12 Vth + −1 = 0 3 6 ⎛ 1 1⎞ Therefore,Vth ⎜ + ⎟ = 5 ⎝ 3 6⎠ or,

dvC (t ) dt

= (0.1 × 10 −6 )

RL = Z Th = 2 Ω

2

173

Vth = 10 V

d [3 − 3e −t /τ ]A dt

0.3 × 10 −6 −t /τ 3 × 0.1 × 10 −6 × e − t /τ = e A τ 12 × 10 −6

Therefore, instantaneous power of the source p(t) = viC(t) ⎡ 0.3 − t / τ ⎤ 0.9 − t / τ p(t ) = 3 × ⎢ e ⎥= e ⎣ 12 ⎦ 12 Therefore, the energy taken from the 3 V source to charge the 0.1 mF capacitor is

∞

E = ∫ p (t )d (t ) 0

=∫

∞ 0.9 − t /τ

e dt 12 0.9 ⎡ −t /τ ⎤ ∞ = ( −τ ) e ⎦0 12 ⎣ 0

⎡ 0.9 ⎤ ⎡ 0.9 ⎤ = τ ⎢ ⎥ = 12 × 10 −6 ⎢ ⎥ 12 ⎣ 12 ⎦ ⎣ ⎦ = 0.9 μJ Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 173

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∞

E = ∫ p (t )d (t ) 0

174

=∫

∞ 0.9 − t /τ

e dt 12 GATE ECE Chapter-wise Solved papers 0.9 ⎡ −t /τ ⎤ ∞ = ( −τ ) e ⎦0 12 ⎣ ⎡ 0.9 ⎤ ⎡ 0.9 ⎤ = τ ⎢ ⎥ = 12 × 10 −6 ⎢ ⎥ ⎣ 12 ⎦ ⎣ 12 ⎦ = 0.9 μJ 0

254. Topic: Solution of Network Equations Using Laplace Transform (c)  For a finite duration time domain signal, RoC of the signal’s bilateral Laplace transform F(s) is the entire s-plane. 255. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (*) ⎡ 0.3 −0.2⎤ ⎢ −0.2 0.3⎥⎦ ⎣ The correct answer is not given in the options. 256. Topic: Continuous-Time Signals: Fourier Series and Fourier Transform Representations (2)  The average power of g(t) =

1 1 ( −3t ) 2 dt 3 ∫−1

=

9 ⎡t3 ⎤ ⎢ ⎥ 3 ⎣ 3 ⎦ −1

( z − 1) Y ( z) z ⎤ ( z − 1) ⎡ 2 = ⎢ 2⎥ z ⎣ z ( z − 1) ⎦

X ( z) =

Therefore, x[n] = 2u[n - 3]. Hence, x[2] = 0 259. Topic: Discrete-Time Signals: Interpolation of ­Discrete-Time Signals (a)  ∠ak = −π only changes the sign of the magnitude |ak|. Since the magnitude spectrum |ak| is even, the corresponding time-domain signal is real. 260. Topic: LTI Systems: Convolution (d)  x(−t)*d(−t − t0) = x(−t)*d(t + t0) = x[(−(t + t0)] = x(−t − t0) 261. Topic: LTI Systems: Impulse Response (b) Let h(t) be the impulse response of the system

t

y (t ) =

1 −2

5 −1 z z + 6 6

=

z2 5 1 z2 − z + 6 6

If we need to get h(t), then we have to differentiate y(t). Thus differentiating the unit-step response gives the impulse response for a LTI system. 262. Topic: LTI Systems: Digital Filter Design Techniques 3

(a) Given y[n] = ∑ a i x( n − i ) i=0

z2 = 1⎞ ⎛ 1⎞ ⎛ ⎜⎝ z − ⎟⎠ ⎜⎝ z − ⎟⎠ 2 3

y[n] =

n

∑ x[m]

m=0

According to accumulation property of z-transform Y ( z) =

y[n] = a 0 x[n] + a1 x[n − 1] + a 2 x[n − 2] + a 3 x[n − 3]

1 1 , z= 2 3

258. Topic: Discrete-Time Signals: z-Transform (0)  Given that

X ( z) zX ( z ) = (1 − z −1 ) ( z − 1)

Therefore, ( z − 1) Y ( z) z ⎤ ( z − 1) ⎡ 2 = z ⎢⎣ z ( z − 1) 2 ⎥⎦

X ( z) =

2 2 z −3 = 1) 174(1 − z −1 ) z 2 ( z −2.indd Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part

∫ h(τ )dτ

−∞

257. Topic: LTI Systems: Parallel and Cascade Structure (c)

So, poles are located at z =

y(t)

y(t) is unit step response of the system

9 2 = ⋅ = 2 Watts 3 3

1−

h(t)

u(t)

1

Y ( z) = H ( z) = X ( z)

2 2 z −3 = z 2 ( z − 1) (1 − z −1 )

=



Getting a null at zero frequency implies that given filter can be a high pass filter but it cannot be a low pass filter. High pass filter is possible if we have negative coefficients. Let us consider the first option (a), where, a1 = a2 = 0, a0 = −a3

y[n] = −a3x[n] + a3x[n - 3] H(z) = −a3 (1 − z−3) H(ej w) = −a3(1 - e−j3 w) = −a 3 e

−

j 3ω 2

3ω −j ⎡ j 32ω −e 2 ⎢e ⎢ 2j ⎢⎣

⎛ 3ω ⎞ − j = −a 3 2 j sin ⎜ e ⎝ 2 ⎟⎠

⎤ ⎥2j ⎥ ⎥⎦

3ω 2

=

12/4/2018 11:10:05 AM

Chapter 2  •  Networks, Signals and Systems

 3ω  − j = −α 3 2.sin  e  2 

265. Topic: Frequency Domain analysis of RLC Circuits (c)  In a series RLC circuit

3ω π j 2 .e 2

ξ=

H ( e jω ) ω = 0 = 0 In other cases it is not possible. Hence, option (a) is the correct option.

(1.333)  Connecting a source of 1A across the terminals a−b, we get the circuit as shown below. VTh

a +

+ −

4I

4W

2W

I

VTh

1A

=

1 L R C

R C 2 L

266. Topic: Time Domain Analysis of Simple Linear ­Circuits (2.528)  The circuit for time t→∞ can be redrawn shown as follows. As the capacitor behaves as an open circuit, VC(∞) = 4 V.

3W RTh =

VTh 1

Applying KCL at node a, we get VTh − 4 I VTh VTh + + −1 = 0 2 2 4 4 V 3 Norton’s equivalent resistance is equal to the Thevenin’s equivalent resistance. Also, VTh = 4I, therefore, VTh =

4 Ω = 1.333 Ω 3

264. Topic: Network Solution Methods (120.22) 2π El C= ln(b /a) C 2π E = l ln(b /a) Therefore, C1 ln(b2 /a2 ) = C2 ln(b1 /a1 ) =

1 2

− b

RTh =

1 2Q

=

263. Topic: Network Theorems: Thevenin and Norton’s

2W

175

ln(10 /1) ln(5)

+ + −

2W

10 V

Vc(¥) = 4 V −

At t = 0, the capacitor acts as a short circuit. Therefore, VC(0−) = 0  V 5 6 5 τ = RTh C = (3  2) × = × = 1s 6 5 6 vC (t ) = VC (∞) + [VC (0 − ) − VC (∞)]e − t / τ = 4 − 4e − t Therefore, voltage vC(t) at t = 1 s is vC = 4 − 4e −1 = 2.528 V 267. Topic: Frequency Domain Analysis of RLC Circuits (b)  The given LC tank circuit is shown in figure below. Y = YC + YLR

Y = jωC +

1 ( jω L + R)

( R − jω L) ( R 2 + ω 2 L2 ) Placing imaginary part to zero, we get the resonant frequency of the tank circuit as given in option (b). = jωC +

C

Where C1 and C2 are the capacitances per unit length of the two coaxial capacitors of the original and the ­modified capacitor respectively. ln(5) × 172 pF/m ln(10) = 120.22 pF/m

C2 =

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 175

L

R

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GATE ECE Chapter-wise Solved papers

268. Topic: Frequency Domain Analysis of RLC Circuits (20)  The equivalent circuit at the given input frequency is shown below. 50 W

j5

P

j5

I 5p ∠−90° w = 5000

−j 5000C

j5

Applying KCL at node P, we get ⎛ ⎞ ⎜ 1 ⎟ 1 Vi ⎜ + ⎟ −I =0 ⎜ j5 j ⎛ 5 − 1 ⎞ ⎟ ⎜⎝ ⎟ ⎜⎝ 5000C ⎠ ⎟⎠ Given that I = 0. Hence,

Therefore, 1 −1 = 1 ⎞ j5 ⎛ j ⎜5 − ⎟ ⎝ 5000C ⎠ 1 = 5

1 1 −5 5000C

Solving the above equation, we get C=

1 F = 20 μF 5 × 10 4

269. Topic: Solution of Network Equations Using Laplace Transform (−2)  Given that x(t) = as(t) + s(−t) and s(t) = be−4tu(t) x(t) = abe−4tu(t) + be4tu(−t)

ab s+4 b b e 4 t u( −t ) ⎯LT ⎯→ − s−4

ab e −4 t u(t ) ⎯LT ⎯→

Therefore,

X ( s) =

b = −2 270. Topic: Solution of Network Equations Using Laplace Transform (a) 

⎡ 2 −1⎤ ⎡ 3⎤ A= ⎢ B = ⎢ ⎥ C = [3  -2] ⎥ ⎣0 −4 ⎦ ⎣ −1⎦ −1 H(s) = C(sI - A) B −1 1 ⎤ ⎡3⎤ ⎡s − 2 = [3  -2] ⎢ s + 4 ⎥⎦ ⎢⎣ −1⎥⎦ ⎣ 0 = [3  -2]

1 ⎤⎡ 3 ⎤ ⎡s + 4 1 ⎢ s − 2⎥⎦ ⎢⎣ −1⎥⎦ ( s − 2)( s + 4) ⎣ 0

= [3  -2]

⎡3s + 12 + 1⎤ 1 ( s − 2)( s + 4) ⎢⎣ − s + 2 ⎥⎦

=

⎛ ⎞ ⎜ 1 ⎟ 1 Vi ⎜ + ⎟ =0 ⎜ j5 j ⎛ 5 − 1 ⎞ ⎟ ⎜⎝ ⎟ ⎜⎝ 5000C ⎠ ⎟⎠



On solving the numerator, we get

ab b ⎡ a ( s − 4) − ( s + 4) ⎤ − = b⎢ ⎥ s+4 s−4 s 2 − 16 ⎣ ⎦

Hence, 16 ⎡ a ( s − 4) − ( s + 4) ⎤ ; −4 < σ < +4 b⎢ = 2 2 ⎥ s − 16 ⎣ ⎦ s − 16

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 2.indd 176

9 s + 39 + 2 s − 4 11s + 35 = ( s − 2)( s + 4) ( s − 2)( s + 4)

271. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (b)  Let us obtain the impedance (Z) matrix of the given network. 2 ⎤ ⎡7 + j 4 [Z ] = ⎢ 7 − j 4 ⎥⎦ ⎣ 2 ΔZ = [(7 + j 4) (7 − j 4)] − 4 = 49 + 16 − 4 = 61 Z 7 + j4 A = 11 = = 3.5 + j 2 2 Z 21 B=

ΔZ 61 = = 30.5 2 Z 21

C=

1 1 = = 0.5 Z 21 2

D=

Z 22 7 − j 4 = = 3.5 − j 2 2 Z 21

30.5 ⎤ ⎡3.5 + j 2 T=⎢ 3.5 − j 2⎥⎦ ⎣ 0.5 272. Topic: Continuous-Time Signals: Sampling Theorem and Applications (0.4)  Given that ∞ ⎛ sin(π t / 2) ⎞ x (t ) = ⎜ ∗ ∑ d (t − 10 n) ⎟ ⎝ (π t / 2) ⎠ n = −∞

Convolution in time domain becomes multiplication in frequency domain. 1 ∞ ∑ d ( f − kf s ) 10 k = −∞

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Chapter 2  •  Networks, Signals and Systems

fs =

For n = 16

1 = 0.1 Ts

1 j 2π n 1 - j 2π n x[n] = 1 + e 16 + e 16 2 2 1 1 a-1 = , a1 = , a0 = 1 2 2

The relation sin(π t / 2) can be represented as (π t / 2)

a-1 = a-1+16 Therefore, a-1 = a15 = −1 4 Similarly

a0 = 1, a1 =

1 4

f

1 2

1 1 , a2 to a14 = 0, a15 = 2 2

DFS coefficients are also periodic with period 16.

∞

∑ δ (t - 10n) can be represented as

n = -∞

a31 = a15 + 16 a31 = a15 a31 =



−0.2 −0.1

177

0.1

0.2

0.3

f

1 2

275. Topic: Discrete-Time Signals: z-Transform (c) Since x[n] is absolutely summable, its ROC must include the unit circle.

Multiplication in frequency domain will result in maximum frequency of 0.2.

+2j Unit circle

0.1 0.2 −0.2 −0.1 Thus Nyquist rate = 0.4 samples/seconds 273. Topic: Discrete-Time Signals: DFT (11)  Given that x1[n] = {1, 2, 3, 0}, x2[n] = {1, 3, 2, 1}, X3(k) = X1(k)X2(k) Based on the properties of DFT, x1[n] ∗ x2[n] = X1(k)X2(k) = x3[n] Circular convolution between two 4-point signals is as follows: ⎡ 1 0 3 2⎤ ⎡ 1⎤ ⎡ 9⎤ ⎢ 2 1 0 3⎥ ⎢ 3⎥ ⎢ 8⎥ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢ 3 2 1 0 ⎥ ⎢ 2⎥ ⎢11⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 3 2 1⎦ ⎣ 1⎦ ⎣14 ⎦ Therefore, x3[2] = 11 274. Topic: Discrete-Time Signals: Discrete-Time Fourier Transform (DTFT) (0.5)  Given that ⎛π ⎞ x[n] = 1 + cos ⎜ n⎟ ⎝8 ⎠

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 177

−2j ROC must be inside the circle of radius 2. Therefore, x[n] must be a non-causal signal. 276. Topic: LTI Systems: Parallel and Cascade Structure (c) x[n]

v[n] + z−1 −5/3 +

+

y[n]

1 z−1 −2/9

5/3

From the graph 2 v[n] = x[n] + v[n - 1] - v[n - 2] 9

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278. Topic: LTI Systems: Impulse Response (b)  The given differential equation is,

5 5 y[n] = - v[n - 1] + v[n - 2] 3 3

y ′′(t ) - y ′(t ) - 6 y(t ) = x(t )

2 ⎤ ⎡ V [ z ] ⋅ ⎢1 - z -1 + z -2 ⎥ = X ( z ) 9 ⎣ ⎦ Therefore, V ( z) = X ( z)

1 1 - z1 +

(1)

2 -2 z 9 

Y ( z ) -5 -1 5 -2 = z + z (2) V ( z) 3 3

On applying Laplace transform on both sides, s2Y(s) – sy(0) – y ′(0) – [sY(s) – y(0)] – 6Y(s) = X(s) To calculate the transfer function all initial conditions are taken as ‘0’. Therefore (s2 - s - 6) Y(s) = X(s) 1 1 Y ( s) H ( s) = = 2 = X ( s) ( s - s - 6) ( s - 3)( s + 2)

Multiplying Eqs.(1) and (2), we get

5 - z -1 (1 - z -1 ) Y ( z) = 3 2 X ( z) 1 - z -1 + z -2 9 For unit step response,

1 ↔ -e - at u( -t ); σ < - a s+a

Therefore, h(t)

1 1 = - e 3t u( -t ) + e -2t u( -t ) 5 5 So option (b) is correct.

On solving, A = 5; B = −5 n

n

⎛ 1⎞ ⎛ 2⎞ y[n] = 5 ⎜ ⎟ u[n] - 5 ⎜ ⎟ u[n] ⎝ 3⎠ ⎝ 3⎠ 277. Topic: LTI Systems: Poles and Zeros (a) z4 [( z + 0.5) + (0.5) ][( z − 0.5) 2 + (0.5) 2 ] 2

z4 z4 = [( z 2 + 0.5 − z )( z 2 + 0.5 + z )] z 4 + 0.25 −4

H ( z ) = 1 − 0.25 z  h[n] = [1, 0, 0, 0, −0.25] Now h[0] = 1 so h[n] is real for all n Hence, the correct option is (a).

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 178

1 ⎡ -e 3t u( -t ) + e -2t u( -t ) ⎤⎦ 5⎣

=

A B = + 1 2 1 - z -1 1 - z -1 3 3

=

Using the following standard pair:

1 ↔ -e at u( -t ); σ < a s-a

5 - z -1 3 Y ( z) = 2 -1 1 - z + z -2 9

2

1⎡ 1 1 ⎤ ⎢ 5 ⎣ s - 3 s + 2 ⎥⎦

It is given that system is non-causal and unstable. To satisfy both the conditions, ROC should be to the left of the left most pole.



1 X ( z) = 1 - z -1

H ( z) =

=

279. Topic: LTI Systems: Impulse Response (a) Let N(t) be the noise at the output filter. Variance of N(t) = E[N 2(t)] - E[N(t)]2. Since the input noise has zero mean, output noise mean is also zero. ∞

E [ N (t )] = E [W (t )]. ∫ h(t )dt -∞

E [W (t )] = 0 where W(t) is white noise. Var[N(t)] = E[N2(t)] = RN(0) Since Rn(t) = h(t)*h(−t)*Rw(t) Since RN (τ ) =

N0 . δ (τ ) 2

RN (τ ) = [h(τ ) * h( -τ )]. RN (τ ) =

N0 2

N0 2

α

∫ h(k ). h(τ + k )dk

-α

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Chapter 2  •  Networks, Signals and Systems

RN (0) = =

N0 2

282. Topic: Network Solution Methods: Nodal and Mesh Analysis (1) 

α

∫

h2 ( k )dk

-α

N0 3 (3 A2 ) = A2 N 0 2 2

5Ω 8V

C + 8V − 5A

1A

x2[n] = { 1.5 , 1.5 , 0, .....}

B V1

A 0V

1Ω

Energy of x1 [n] = ∑ x [n] n= 0

5A

4A

280. Topic: Discrete- Time Signals (1.5)  x1[n] = a(0.5)nu[n]

∞

179

←

1Ω

→

4A + 1A 8V −

2 1

D 1Ω

0V i

1Ω

5A ∞

= ∑ α 2 (0.5) 2 n n= 0

0V

⎡ 1 1 ⎤ = α 2 ⎢1 + + + .......⎥ ⎣ 4 16 ⎦ ⎡ ⎤ ⎢ 1 ⎥ 2 4 =α ⎢ ⎥=α 1 3 ⎢1 - ⎥ ⎣ 4⎦ Energy of x2 [n] =

(

1.5

) +(

1.5

)

2

Therefore,

=3

4 = 3 or, 3 α = 1.5

Let the charge stored when the capacitor is electrically isolated be Q2, capaticance be C2 and energy stored be E2 and the plate separation is 2d Now, Q2 = Q1 As the plate separation in this case is twice that in first case, therefore, C2 = C1/2 Q22 Q22 Q2 = = 1 = 2E 2C2 2(C1 / 2) C1

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 179

4 A = 4A 1 4 Current in D branch = A = 4A 1 8−4 Current in B branch = A = 4A 1 Hence, Current in C branch = 4 + 1 = 5A Again applying nodal analysis, we get Current in A branch =

281. Topic: Network Solution Methods (a)  Let the capacitance when the plate separation is d be C1, energy is E and charge stored is Q1. Therefore, Q2 E= 1 2C1

Hence, E2 =

V1 - 0 V1 - 8 V1 - 0 V1 - 8 + + + =0 1 1 1 1 4V1 - 16 = 0 Hence, V1 = 4 V

If is given that energy of x1[n] is same as energy of x2[n]. Therefore, α 2 .

On applying nodal analysis, we get

That is,

2

2

0V i

⎛ 0 - V1 ⎞ i+⎜ +5= 0 ⎝ 1 ⎟⎠ Therefore, i = V1 - 5 = 4 - 5 = -1A 283. Topic: Time Domain Analysis of Simple Linear Circuits (b)  For the maximum power transfer, the resistance must have half the voltage drop with respect to the V source. Therefore, the drop across R is S . That is, 2 V V I L R = S Therefore, I L = S 2 2R 284. Topic: Time Domain Analysis of Simple Linear Circuits (d)  When the switch is in position 1 for long time, we have

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GATE ECE Chapter-wise Solved papers

⎛ 10 ⎞ VC = ⎜ 2V = 4 V ⎝ 3 + 2 ⎟⎠ Therefore, VC (0 - ) = VC (0 + ) = 4 V • At t = 0, the switch goes to position 2. • At t → ∞: We have VC(∞) = 5 × 2 = 10 V 4Ω

VC

+ −

2Ω 2Ω

Req = 4 Ω + 2 Ω  = 6 Ω   t = CReq = 0.1 × 6 = 0.6 s Thus, VC (t ) = VC (∞) + [VC (0) - VC (∞)]e - t / τ = 10 + [4 - 10]e - t / 0.6 = 10 - 6e - t / 0.6 285. Topic: Frequency Domain Analysis of RLC Circuits (0.316)  For the resonance in a parallel RLC circuit, we have IR = I I L = QI ∠- 90° I C = QI ∠+ 90° Therefore,

IR

The two-port network is reciprocal if AD − BC = 1. Therefore, the ­determinant of the transmission matix is (T) = 1 for reciprocal networks. 288. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (3)  We have Z11 = Z a + Z c = 2 jω Z 22 = Z b + Z c = 3 + 2 jω

Therefore,

IL

287. Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (d)  It is given that ⎛ A B⎞ T =⎜ ⎝ C D ⎟⎠

=Q=R

Z12 = Z 21 = Z c = jω

Therefore,

Z b + jω = 3 + 2 jω Z b = 3 + jω

Hence,

Therefore, Rb = 3 Ω. 289. Topic: Continuous-Time Signals: Sampling Theorem and Applications (b)  Any discrete time signal such as x(n) = cos(ω0n) is said to be periodic if ω0/2π is a rational number. 290. Topic: Continuous-Time Signals: Sampling Theorem and Applications (13) Let x(t) be the continous time sine wave of frequency 33 Hz. Let X(ω) be its frequency spectrum. Therefore, X(ω )

C 10 × 10 -6 = 10 = 0.316 L 10 × 10 -3

286. Topic: Time Domain Analysis of Simple Linear Circuits (b)  It is given that vi(t) = 2 cos(200t) + 4 sin(500t) •  Case 1: When input is 2 cos(200t): At ω = 200 rad/s, the input series combination of LC circuit acts like short circuit because L and C are at resonance. Therefore, vo = vi = 2 cos (200t) •  Case 2: When input is 4 sin (500t), at ω = 500 rad/s, then the output parallel combination of L and C acts like an open circuit because of resonance. Thus, there is no flow of current. Hence,   vo = vi = 4 sin(500t) Therefore, the overall output is vo = 2 cos(200t) + 4 sin(500t)

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 180

−33 Hz

33 Hz

When this sinusoid is sampled using a Delta impulse train frequency of 46 Hz, the sinusoid ­spectrum shifts to left and right by multiples of 46 Hz. Hence, the spectrum of sampled signal xp(t) is Xp(ω) as shown below: Xp(ω )

−59

−13

Hz 13

59

After passing Xp(ω) through a low pass filter (LPF) with cutoff frequency of 23 Hz, we have Xp(ω)(LPF), which is given by

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Chapter 2  •  Networks, Signals and Systems

ω

181

Therefore, ⎛ sin(t ) ⎞ ⎛ sin(t ) ⎞ sin(t ) ⎜⎝ ⎟ ∗⎜ ⎟= πt ⎠ ⎝ πt ⎠ πt

Hz

−13

293. Topic: Discrete-Time Signals: z-Transform (b)  It is given that x[n] = a n u[n] + b n u[n] z z + X ( z) = z −a z −b

13

Thus, the fundamental frequency of Xp(ω) (LPF) is 13 Hz. 291. Topic: Continuous-Time Signals: Sampling Theorem and Applications (c)  It is given that x(t) = cos(6πt) + sin(8πt) The bandwidth of the signal is 8π = 4 Hz 2π ⎡ ⎛ 5⎞ ⎤ y(t ) = x( 2t + 5) = x ⎢ 2 ⎜ t + ⎟ ⎥ ⎣ ⎝ 2⎠ ⎦ Taking Fourier transform, we get

For ROC, |z| > a and |z| > b. However, it is given that 0  0

(2)

At t = 0-, switch is in open state, so the loop is not closed. Hence iL(0-) = 0 A.



Putting Eq. (2) in Eq. (1), we get

At t = ∞, switch is closed and inductor is short-circuited. 2W

V1 = 2( - 1.4 I 2 ) - 2 I 2 = -4.8 I 2 ⇒

V1 = B = 4.8 -I2

334. Topic: Frequency Domain Analysis of RLC Circuits (c) Given w = 5,  C = 1 mF. XC =

1 = −200 j kΩ jωC

1V

+ -

iL(¥) = 12 A

Time constant, τ =

100 kΩ

1 1 L = = Req 1 + 1 2

So we have iL(0-) = 0 A ⇒ iL (∞)= + 1 µF VC −

5 sin(5t)

Now,

100 kΩ

=   =  

5∠0° × ( −200 j ) 200 − j 200

5∠ − 90° 2∠ − 45°

π⎞ ⎛ ∠ - 45° = 2.5 2 sin ⎜ 5t - ⎟ V ⎝ 4⎠ 2

5

VC = 2.5 2 sin(5t - 0.25π ) V

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 193

1 1 ⇒τ = 2 2

iL (t ) = iL (∞) + [iL (0 - ) - iL (∞)]e - t / τ ; t > 0 =

VC =

1H

(1)

⎛ -2 ⎞ - I1 + ⎜ I + ( - I2 ) = 0 ⎝ 5 2 ⎟⎠

2Ω

t=0

-I2

5W

(-2I2)

335. Topic: Time Domain Analysis of Simple Linear Circuits (0.316 A)  Assuming switching to be happening at t = 0, we can redraw the given circuit as follows:

2W

X

193

1 ⎛ 1⎞ 1 + ⎜ 0 - ⎟ e -2t = (1 - e -2t ) 2 ⎝ 2⎠ 2

So, iL (t = 0.5) = =

1 1 (1 - e -2 × 0.5 ) = (1 - e -1 ) 2 2 1 (1 - 0.368) = 0.5 × 0.632 = 0.316 A 2

336. Topic: Continuous-Time Signals: Fourier Series and Fourier Transform Representation (8)  Hilbert Transform affects the phase by 90° phase shift and has no effect on magnitude. Therefore,

12/4/2018 11:11:22 AM

R

I

1

A

B

R1

3

C 194

D

GATE ECE Chapter-wise Solved papers

R

R3

1

∞

∞

2

y(t ) dt =

∫ −∞

∞

2

x(t ) dt =

∫ −∞

∫

2

X ( f ) df

−∞

x(t) = 4 sinc(2t)



+ 11 V −

R1

R

R3

R1

R1

FT

sinc(t ) → rect ( f ) 4  f   f  4sinc( 2t ) FT → rect   = 2 rect   2 2 2 ∞

R1

R3

+ 11 V −

R3

2

X ( f ) df = Area under |X(f  )|2 curve

∫  

R1

R1

R1

−∞

=4×2=8 ∞

Therefore,

∫

Req = (R1 || R1) + (R1 || R1 || R3 || R3) + (R1 || R1)

2

y(t ) dt = 8

Req =

-∞

R1  R1 R3  R1 + || + 2  2 2  2

= R1 +

X(f)  

  = 1+

2  

  IS =

f (Hz)

−1

1

2

( R1 × R3 ) / 4 R1 R3 = R1 + R1 R3 2( R1 + R3 ) + 2 2

1× 3 3 11 = 1+ = 2(1 + 3) 8 8

VS 11 = =8A Req 11/8

338. Topic: Continuous-Time Signals: Sampling Theorem and Applications (13)  Given Bandwidth of x(t) = 5 kHz Sampling rate = fs fs min = ? |H(f)|

337. Topic: Network Solution Methods: Nodal and Mesh Analysis (8)  Given R1 = 1 Ω, R2 = 2 Ω, R3 = 3 Ω

K

Network is symmetric; so VA = VB and VC = VD. So the circuits reduces to −8 −6

R

R1

1

A R1

R1

R 3

C

0

6

8

f(kHz)

Assume an arbitrary spectrum for x(t ) as shown below.

B

R3

x(f) + 11 V −

D

R

R3

1

R1

I

-5

5

f(kHz)

R1

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 194

+

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Chapter 2  •  Networks, Signals and Systems

Spectrum of sampled signal:

195

x[2n] = {x(0), x( 2), x( 4), x(6)} For n = 0, 1 x(0) = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 4.5 8

K

For n = 2, −fs−5 −fs −fs+5 −8 −6 −5

0

5

6

8

fs−5

fs

fs+5

e

For proper reconstruction: fs - 5 ≥ 8 kHz ⇒



x[2] = −0.5 − 0.5 j

339. Topic: Discrete-Time Signals: DFT (3)  Given X(k) = k + 1 for {0 ≤ k ≤ 7} ⇒ X(k) = {1, 2, 3, 4, 5, 6, 7, 8} 1 x[n] = N

x[n] =

7

∑ X ( k )e

Similarly, we get

j 2π nk 8

1 ∑ X ( k )e 8 n= 0

π j nk 4

Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 195

x[4] = –0.5

(1)

x[6] = –0.5 + 0.5j 3

∑ x[2n] = x[0] + x[2] + x[4] + x[6]

n= 0

7

= ( j )k

1 x[2] = {1( j )0 + 2( j )1 + 3( j ) 2 + 4( j )3 + 5( j ) 4 8 + 6( j )5 + 7( j )6 + 8( j )7 }

 fs ≥ 13 kHz fsmin = 13 kHz



π j × 2k 4

(2)

n= 0

= 4.5 – 0.5 – 0.5j – 0.5 – 0.5 + 0.5j = 3

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Ch wise GATE_ECE_CH02_Networks, Signals and Systems_Expln_Part 3.indd 196

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Electronic Devices

CHAPTER3

Syllabus Energy bands in intrinsic and extrinsic silicon; Carrier transport: diffusion current, drift current, mobility and resistivity; Generation and recombination of carriers; Poisson and continuity equations; P-N junction, Zener diode, BJT, MOS capacitor, MOSFET, LED, photo diode and solar cell; Integrated circuit fabrication process: oxidation, diffusion, ion implantation, photolithography and twin-tub CMOS process.

Chapter Analysis Topic

GATE 2009

Energy bands in intrinsic and extrinsic silicon

1

Carrier transport

1

GATE 2010

Generation and recombination of carriers

GATE 2011

1

GATE 2012

GATE 2013

1

1

GATE 2014

GATE 2015

GATE 2016

GATE 2017

5

1

5

3

1

4

1

2

2

1

5

1

2

3

GATE 2018

1

Poisson and continuity equations P-N junction

2

1

1

Zener diode

1

1

1

BJT

1

1

1

3

1 8

MOS capacitor MOSFET

3

4

3

1

2

1

1

1

4

1

2

1

1

5 2

LED

1

Photo diode and solar cell Integrated circuit fabrication process

1 1

1

2

1

1

Important Formulas 1.

In an intrinsic semiconductor, n = p = ni

2.

⎛ -E ⎞ ni2 = AT 3 exp ⎜ G 0 ⎟ ⎝ kT ⎠

3.

Current density in any material is J = ( nm n + pm p ) qe Where e is eleectric field intensity.

4. Conductivity σ = ( nm n + pm p ) q 5.

For silicon, the energy band gap [Eg(T )] at temperature T (K) is Eg (T ) = 1.21 - 3.60 × 10 -4 T

6.

For germanium, the energy band gap [Eg(T )] at temperature T (K) is Eg (T ) = 0.785 - 2.23 × 10 -4 T

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 197

7.

The probability that an energy level in a semiconductor is occupied by an electron is f (E) =

8.

1 ⎛ E - EF ⎞ 1 + exp ⎜ ⎝ kT ⎟⎠

The Fermi level for N-type semiconductor is EF = EC - kT ln

⎛ 2π mn kTq ⎞ where N C = 2 ⎜ ⎟⎠ ⎝ h2

NC ND

3/ 2

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  9. According to law of mass action, product of concentration of holes and electrons in any semiconductor is constant and is given by np = ni2 10. For an N-type semiconductor, p≅

ni2 ND

J p = pm p qe - qDp

I D = I 0 (eVD

12. The Hall voltage (VH) is given by BIRH VH = d 13. For conductors, the value of Hall’s coefficient RH is given by 1 RH = nq 14. For semiconductors, the value of Hall’s coefficient RH is given by pm p2 q( nm n + pm p ) 2

15. Current density in an N-type semiconductor is J ≅ N D m n qe 16. Conductivity in an N-type semiconductor is

σ ≅ N D mn q

26. VT =

J ≅ NA mp q e 18. Conductivity for P-type semiconductor is

σ ≅ NA mp q 19. Drift current in the semiconductor is given by J = ( nm n + pm p )qe 20. The hole diffusion current density is given by dp dx

21. The electron diffusion current density is dn dx

22. The diffusion constant of a carrier is related to its mobility and is given by Einstein equation Dp

mp

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 198

=

Dn = VT mn

ηVT

- 1)

kT q

27. Variation of reverse saturation current with t­emperature for a diode is I 0 (T ) = I 0 (T1 ) × 2(T -T1 )/10 28. The variation of cut-in voltage and the forward voltage of a diode with temperature is dV = -2.5 mV/ °C dT 29. Static resistance of a diode is R=

VD1 I D1

30. Dynamic resistance of a diode is rd =

17. Current density in a P-type semiconductor is

J n = qDn

dn dx

25. V–I characteristic equation or the Shockley’s diode equation is

n2 n≅ i NA

J p = - qDp

dp dx

24. The total electron current density is given by J n = nm n qe + qDn

11. For a P-type semiconductor,

RH = nm n2 -

23. The total hole current density is given by

ΔVd ΔI d

31. Dynamic resistance (r) of a diode in the forward-biased region ≅ 26η I D . 32. Average AC resistance of a diode is (VD 2 - VD1 ) ( I D 2 - I D1 ) 33. The relation between the transition capacitance and the applied reverse bias of a varactor diode is CT =

K (Vγ + VR ) n

34. For an LED, the relation between the wavelength (in nm) of emitted light and band gap energy (in eV) of the ­semiconductor material is

λ=

1240 Δ Eg

35. For a BJT, emitter current I E = I C + I B , where IC is the collector current and IB is the base current.

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Chapter 3  • Electronic Devices

36. For the common-base configuration in the active region, collector current I C = α I E + I CO . 37. The generalized expression for the collector current in a transistor in common-base configuration is ⎡ ⎛ -V ⎞ ⎤ I C = α I E + I CO ⎢1 - exp ⎜ CB ⎟ ⎥ ⎝ VT ⎠ ⎦ ⎣

199

46. The relationship between the output current ID in the saturation region for a given value of input gate-source voltage (VGS) in a JFET is given by Shockley’s equation ⎛ V ⎞ I D = I DSS ⎜1 - GS ⎟ V ⎠ ⎝

2

P

(I - I ) 38. α = C CO ( I E - 0)

47. Shockley’s equation as defined for JFETs is applicable for the depletion MOSFET also in both the depletion and the enhancement regions.

39. AC alpha (aac) is defined as

48. For an enhancement MOSFET,

α ac =

ΔI C ΔI E

VDS( sat ) = VGS - VTh For an enhancement MOSFET, the drain current is zero for gate–source voltage less than the threshold voltage VTh. For voltages greater than the threshold voltage, the drain current is given by I D = k (VGS - VTh ) 2

VCB = const

40. For common-emitter configuration, collector current I C = β I B + (β + 1) I CO , where

β=

α (1 - α )

Static drain resistance (RD) is defined as

41. b is referred to as the DC forward current transfer ratio or the DC current gain of the transistor. 42. For AC applications AC beta (bac) is defined as βac = Δ I C Δ I B for constant common-emitter voltage. 43. I C = (β + 1) I CO

RD =

49. Dynamic drain resistance (rd) is defined as rd =

I = CO (1 - α ) I C = γ I B + γ I CO

γ = (β + 1) =

gm =

1 (1 - α )

VGS = const

ΔI D ΔVGS

VDS = const

⎛ V ⎞ = g m0 ⎜1 - GS ⎟ VP ⎠ ⎝

51. Amplification factor (m) is defined as

45. The drain resistance (rd) in the saturation region of a JFET is rd =

ΔVDS ΔI D

50. Transconductance (gm) is defined as

44. For common-collector configuration

where

VDS ID

m=

ro (1 - VGS VP ) 2

ΔVDS ΔVGS

= rd × g m I D = const

QUESTIONS 1.

For the circuit shown in the following figure, the voltage Vo is s

2Ω + 4V −

(a) (b) (c) (d)

2Ω − 2Ω

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 199

Vo +

(GATE 2000: 2 Marks)

− 2V +

2V 1V −1 V None of the above

2.

In the circuit shown in the following figure, assume that the transistor is in active region. It has a large β and its base-emitter voltage is 0.7 V. The value of IC is

12/4/2018 11:17:07 AM

200

GATE ECE Chapter-wise Solved papers

15 V

7.

A material has conductivity of 10−2 mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is (a) 45 MHz (b) 90 MHz (c) 450 MHz (d) 900 MHz (GATE 2001: 2 Marks)

8.

In the following figure, a silicon diode is carrying a constant current of 1 mA. When the temperature of the diode is 20°C, VD is found to be 700 mV. If the temperature rises to 40°C, VD becomes approximately equal to

RC 10 kΩ

Ic

5 kΩ 430 Ω

+ (a) (b) (c) (d) 3.

Indeterminate since RC is not given 1 mA 5 mA 10 mA (GATE 2000: 2 Marks)

1 mA

−

The current gain of a BJT is (a) gmr0

gm (b) r0

(c) gm rπ

(d)

gm rπ

VD

(a) 740 mV (c) 680 mV 9.

(b) 660 mV (d) 700 mV (GATE 2002: 1 Mark)

If the transistor in the following figure is in saturation, then

(GATE 2001: 1 Mark) 4.

5.

IC

MOSFET can be used as a (a) current controlled capacitor (b) voltage controlled capacitor (c) current controlled inductor (d) voltage controlled inductor (GATE 2001: 1 Mark) The effective channel length of a MOSFET in saturation decreases with increase in (a) gate voltage (b) drain voltage (c) source voltage (d) body voltage (GATE 2001: 1 Mark)

6. An npn BJT has gm = 38 mA/V, Cμ = 10−14 F, Cπ = 4 × 10−13 F, and DC current gain β0 = 90. For this transistor fT and fβ are (a) fT = 1.64 × 108 Hz and fβ = 1.47 × 1010 Hz (b) fT = 1.47 × 1010 Hz and fβ = 1.64 × 108 Hz (c) fT = 1.33 × 1012 Hz and fβ = 1.47 × 1010 Hz (d) fT = 1.47 × 1010 Hz and fβ = 1.33 × 1012 Hz (GATE 2001: 2 Marks)

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 200

C

IB βdc denotes the dc current gain

B

E (a) (b) (c) (d)

IC is always equal to βdcIB IC is always equal to –βdcIB IC is greater than or equal to βdcIB IC is less than or equal to βdcIB (GATE 2002: 1 Mark)

10. A Zener diode regulator in the following figure is to be designed to meet the specifications: IL = 10 mA, Vo = 10 V and Vin varies from 30 V to 50 V. The Zener diode has VZ = 10 V and IZK (knee current) = 1 mA. For satisfactory operation

12/4/2018 11:17:08 AM

Chapter 3  • Electronic Devices

R +

+ IL = 10 mA

I2 Vin

Vo

16. In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using stem or water vapour) produces (a) (b) (c) (d)

RL

Dz −

−

superior quality oxide with a higher growth rate inferior quality oxide with a higher growth rate inferior quality oxide with a lower growth rate superior quality oxide with a lower growth rate (GATE 2003: 1 Mark)

17. An N-type silicon bar 0.1 cm long and 100 mm2 in cross-sectional area has a majority carrier concentration of 5 × 1020/m3 and the carrier mobility is 0.13 m2/V⋅s at 300 K. If the charge of an electron is 1.6 × 10-19 Coulomb, then the resistance of the bar is

(a) R ≤ 1800 Ω (b) 2000 Ω ≤ R ≤ 2200 Ω (c) 3700 Ω ≤ R ≤ 4000 Ω (d) R > 4000 Ω

(a) 106 ohm (c) 10-1 ohm

(GATE 2002: 2 Marks) 11. N-type silicon is obtained by doping silicon with (a) Germanium (b) Aluminium (c) Boron (d) Phosphorus (GATE 2003: 1 Mark) 12. The band gap of silicon at 300 K is (a) 1.36 eV (b) 1.10 eV (c) 0.80 eV (d) 0.67 eV (GATE 2003: 1 Mark) 13. The intrinsic carrier concentration of silicon sample at 300 K is 1.5 × 1016/m3. If after doping, the number of majority carriers is 5 × 1020/m3, the minority carrier density is (a) 4.50 × 1011/m3 (b) 3.33 × 104/m3 20 3 (c) 5.00 × 10 /m (d) 3.00 × 10-5/m3 (GATE 2003: 1 Mark) 14. Choose proper substitutes for X and Y to make the following statement correct. Tunnel diode and avalanche photodiode are operated in X bias and Y bias, respectively. (a) X: reverse, Y: reverse (b) X: reverse, Y: forward (c) X: forward, Y: reverse (d) X: forward, Y: forward (GATE 2003: 1 Mark) 15. For an N-channel enhancement-type MOSFET, if the source is connected at a higher potential than that of the body (i.e. VSB > 0), the threshold voltage VTh of the MOSFET will (a) remain unchanged (b) decrease (c) change polarity (d) increase (GATE 2003: 1 Mark)

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 201

201

(b) 104 ohm (d) 10-4 ohm (GATE 2003: 2 Marks)

18. The electron concentration in a sample of uniformly doped N-type silicon at 300 K varies linearly from 1017/ cm3 at x = 0 to 6 × 1016/cm3 at x = 2 μm. Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 × 10-19 Coulomb and the diffusion constant Dn = 35 cm2/s, the current density in the silicon, if no electric field is present, is (a) zero (b) -506 A/cm2 (c) -560 A/cm2 (d) -1120 A/cm2 (GATE 2003: 2 Marks) 19. At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions stated above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (a) 1 (c) 4 × 103

(b) 5 (d) 8 × 103 (GATE 2003: 2 Marks)

20. A particular green LED emits light of wavelength ° . The energy band gap of the semicon 5490  A ductor material used there is (Planck’s constant = 6.626 × 10−34 J ⋅ s) (a) 2.26 eV (c) 1.17 eV 

(b) 1.98 eV (d) 0.74 eV (GATE 2003: 2 Marks)

21. When the gate-to-source voltage (VGS) of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current is observed to be 1 mA. Neglecting the channel width modulation effect and

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GATE ECE Chapter-wise Solved papers

assuming that the MOSFET is operating at saturation, the drain current for an applied VGS of 1400 mV is (a) 0.5 mA (c) 3.5 mA 

Vout

(b) 2.0 mA (d) 4.0 mA (GATE 2003: 2 Marks)

22. Match items in Group 1 with items in Group 2, most suitably. Group 1 Group 2 P. LED 1. Heavy doping Q.  Avalanche photodiode 2.  Coherent radiation R.  Tunnel diode 3.  Spontaneous emission S. LASER 4. Current gain (a) P-1; Q-2; R-4; S-3 (b) P-2; Q-3; R-1; S-4 (c) P-3; Q-4; R-1; S-2 (d) P-2; Q-1; R-4; S-3  (GATE 2003: 2 Marks) 23. If P is passivation, Q is N-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard N-well CMOS fabrication process is (a) P-Q-R-S (b) Q-S-R-P (c) R-P-S-Q (d) S-R-Q-P (GATE 2003: 2 Marks) 24. The impurity commonly used for realizing the base region of a silicon N-P-N transistor is (a) Gallium (b) Indium (c) Boron (d) Phosphorus (GATE 2004: 1 Mark) 25. If for a silicon NPN transistor, the base-to-emitter voltage (VBE) is 0.7 V and the collector-to-base voltage (VCB) is 0.2 V, then the transistor is operating in the (a) normal active mode (b) saturation mode (c) inverse active mode (d) cut-off mode (GATE 2004: 1 Mark) 26. Consider the following statements S1 and S2. S1: The b of a bipolar transistor reduces if the base width is increased. S2: The b of a bipolar transistor increases if the doping concentration in the base is increased. Which one of the following is correct? (a) S1 is FALSE and S2 is TRUE (b) Both S1 and S2 are TRUE (c) Both S1 and S2 are FALSE (d) S1 is TRUE and S2 is FALSE (GATE 2004: 1 Mark)

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27. Given figure is the voltage transfer characteristic of

Vin 0 (a) an NMOS inverter with enhancement mode transistor as load (b) an NMOS inverter with depletion mode transistor as load (c) a CMOS inverter (d) a BJT inverter (GATE 2004: 1 Mark) 28. The resistivity of a uniformly doped N-type silicon sample is 0.5 Ω⋅cm. If the electron mobility (mn) is 1250 cm2/V⋅s and the charge of an electron is 1.6 × 10−19 C, the donor impurity concentration (ND) in the sample is (a) 2 × 1016/cm3 (c) 2.5 × 10l5/cm3

(b) 1 × 1016/cm3 (d) 2 × 1015/cm3 (GATE 2004: 2 Marks)

29. The longest wavelength that can be absorbed by silicon, which has the band gap of 1.12 eV, is 1.1 mm. If the longest wavelength that can be absorbed by another material is 0.87 mm, then the band gap of this material is (a) 1.416 eV (c) 0.854 eV

(b) 0.886 eV (d) 0.706 eV (GATE 2004: 2 Marks)

30. In an abrupt P–N junction, the doping concentrations on the P-side and N-side are NA = 9 × 1016/cm3 and ND = 1 × 1016/cm3, respectively. The P–N junction is reverse biased and the total depletion width is 3 mm. The depletion width on the P side is (a) 2.7 mm (c) 2.25 mm 

(b) 0.3 mm (d) 0.75 mm (GATE 2004: 2 Marks)

31. Consider an abrupt P junction. Let Vbi be the built-in potential of this junction and VR be the applied reverse bias. If the junction capacitance (Cj) is 1 pF for Vbi + VR = 1 V, then for Vbi + VR = 4 V, Cj will be (a) 4 pF (c) 0.25 pF

(b) 2 pF (d) 0.5 pF (GATE 2004: 2 Marks)

32. Consider the following statements S1 and S2. S1: The threshold voltage (VTh) of a MOS capacitor decreases with increase in gate oxide thickness.

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S2: The threshold voltage (VTh) of a MOS capacitor decreases with increase in substrate doping concentration. Which one of the following is correct? (a) S1 is FALSE and S2 is TRUE (b) Both S1 and S2 are TRUE (c) Both S1 and S2 are FALSE (d) S1 is TRUE and S2 is FALSE (GATE 2004: 2 Marks) 33. The drain of an N-channel MOSFET is shorted to the gate so that VGS = VDS. The threshold voltage (VTh) of MOSFET is 1 V. If the drain current (ID) is 1 mA for VGS = 2 V, then for VGS = 3 V, ID is (a) 2 mA (c) 9 mA

(b) 3 mA (d) 4 mA (GATE 2004: 2 Marks)

e0 = 8.85 × 10−12 F/m. The depletion capacitance of the diode per square metre is (a) 100 mF (b) 10 mF (c) 1 mF (d) 20 mF (GATE 2005: 2 Marks) 39. A MOS capacitor made using P-type substrate is in the accumulation mode. The dominant charge in the channel is due to the presence of (a) holes (b) electrons (c) positively charged ions (d) negatively charged ions (GATE 2005: 2 Marks) 40. For an N-channel MOSFET and its transfer curve shown in the figure, the threshold voltage is

34. The band gap of silicon at room temperature is (a) 1.3 eV (c) 1.1 eV

VD = 5V D

(b) 0.7 eV (d) 1.4 eV (GATE 2005: 1 Mark)

ID

VG = 3V

35. The primary reason for the widespread use of silicon in semiconductor device technology is (a) abundance of silicon on the surface of the Earth (b) larger band gap of silicon in comparison to germanium (c) favourable properties of silicon dioxide (SiO2) (d) lower melting point (GATE 2005: 1 Mark) 36. A silicon P–N junction at a temperature of 20°C has a reverse saturation current of 10 pA. The reverse saturation current at 40°C for the same bias is approximately (a) 30 pA (b) 40 pA (c) 50 pA (d) 60 pA (GATE 2005: 1 Mark) 37. A silicon sample A is doped with 1018 atoms/cm3 of boron. Another sample B of identical dimensions is doped with 1018 atoms/cm3 of phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is (a) 3 (b) 1/3 (c) 2/3 (d) 3/2 (GATE 2005: 2 Marks) 38. A silicon P–N junction diode under reverse bias has depletion region of width 10 mm. The relative permittivity of silicon, er = 11.7 and the permittivity of free space

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Transfer characteristics

G

S 1V

(a) (b) (c) (d)

VGS

VS = 1V

1 V and the device is in active region. -1 V and the device is in saturation region. 1 V and the device is in saturation region. −1 V and the device is in active region. (GATE 2005: 2 Marks)

41. The concentration of minority carriers in an extrinsic semiconductor under equilibrium is (a)  directly proportional to the doping concentration (b) inversely proportional to the doping concentration directly proportional to the intrinsic concentration (c)  (d) inversely proportional to the intrinsic concentration (GATE 2006: 1 Mark) 42. Under low-level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the (a) diffusion current (c) recombination current

(b) drift current (d) induced current (GATE 2006: 1 Mark)

43. The values of voltage (VD) across a tunnel diode corresponding to peak and valley currents are Vp and Vv ,

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and a storage time ts for 0 < t ≤ ts. VR is given by (all in volts)

respectively. The range of tunnel diode voltage VD for which the slope of its I–VD characteristics is negative would be (a) VD < 0 (b) 0 ≤ VD < VP (c) VP ≤ VD < VV (d) VD ≥ VV (GATE 2006: 1 Mark)

1 + 2 5V

44. An N-channel depletion MOSFET has following two points on its ID - VGS curve (i)  VGS = 0 at ID = 12 mA and (ii)  VGS = −6 Volts at ID = 0

Which of the following Q-points will give the highest transconductance gain for small signals? (a) VGS = −6 V (b) VGS = −3 V (c) VGS = 0 V (d) VGS = 3 V (GATE 2006: 1 Mark)

1 kΩ

5V

VR

− (a) VR = −5 V (c) 0 ≤ VR < 5 V

(b) VR = +5 V (d) −5 V < VR < 0 V (GATE 2006: 2 Marks)

49. Find the correct match between Group 1 and Group 2.

45. The phenomenon known as ‘early effect’ in a bipolar transistor refers to a reduction of the effective base width caused by (a) electron-hole recombination at the base (b) the reverse biasing of the base-collector junction (c) the forward biasing of emitter-base junction (d) the early removal of stored base charge during saturation to cut-off switching (GATE 2006: 1 Mark)



46. The majority carriers in an N-type semiconductor have an average drift velocity v in a direction perpendicular to a uniform magnetic field B. The electric field E induced due to Hall effect acts in the direction

50. The electron and hole concentrations in an intrinsic semiconductor are ni per cm3 at 300 K. Now, if acceptor impurities are introduced with a concentration of NA per cm3 (where NA >> ni), the electron concentration per cm3 at 300 K will be (a) ni (b) ni + NA

(a) (b) (c) (d)

v×B B×v along v opposite to v

Group 1 Group 2 E.  Varactor diode 1.  Voltage reference F.  PIN diode 2.  High-frequency switch G.  Zener diode 3.  Tuned circuits H.  Schottky diode 4.  Current-controlled attenuator (a) E-4, F-2, G-1, H-3 (b) E-2, F-4, G-1, H-3 (c) E-3, F-4, G-1, H-2 (d) E-1, F-3, G-2, H-4 (GATE 2006: 2 Marks)

(c) NA – ni

(d)

ni2 NA

(GATE 2006: 2 Marks)

(GATE 2007: 1 Mark)

47. A heavily doped N-type semiconductor has the following data: Hole–electron mobility ratio = 0.4, doping concentration = 4.2 × 108 atoms/m3, intrinsic concentration = 1.5 × 104 atoms/m3. The ratio of conductance of the N-type semiconductor to that of the intrinsic semiconductor of same material and at the same temperature is given by

51. In a P+–N junction diode under reverse bias, the magnitude of electric field is maximum at (a) the edge of the depletion region on the P – side (b) the edge of the depletion region on the N – side (c) the P+–N junction (d) the centre of the depletion region on the N – side (GATE 2007: 1 Mark)

(a) 0.00005 (c) 10000

(b) 2000 (d) 20000 (GATE 2006: 2 Marks)

48. In the circuit shown below the switch was connected to position 1 at t < 0, and at t = 0 it is changed to position 2. Assume that the diode has zero voltage drop

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52. A P+–N junction has a built-in potential of 0.8 V. The depletion layer width at a reverse bias of 1.2 V is 2 mm. For a reverse bias of 7.2 V, the depletion layer width will be (a) 4 mm (b) 4.9 mm (c) 8 mm (d) 12 mm (GATE 2007: 2 Marks)

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53. The DC current gain (b ) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base transport factor is (a) 0.980 (b) 0.985 (c) 0.990 (d) 0.995 (GATE 2007: 2 Marks) 54. Group I lists four types of PN junction diodes. Match each device in Group I with one of the options in Group II to indicate the bias condition of that device in its normal mode of operation.

Group I P.  Zener diode Q.  Solar cell R.  LASER diode S.  Avalanche photodiode (a) P-1, Q-2, R-1, S-2 (c) P-2, Q-2, R-2, S-1



Group II 1.  Forward bias 2.  Reverse bias

(b) P-2, Q-1, R-1, S-2 (d) P-2, Q-1, R-2, S-2 (GATE 2007: 2 Marks)

55. Group I lists four different semiconductor devices. Match each device in Group I with its characteristic property in Group II.

Group I P. BJT Q.  MOS capacitor R.  LASER diode S. JFET (a) P-3, Q-1, R-4, S-2 (c) P-3, Q-4, R-1, S-2

Group II 1. Population inversion 2.  Pinch-off voltage 3.  Early effect 4. Flat-band voltage (b) P-1, Q-4, R-3, S-2 (d) P-3, G-2, R-1, S-4 (GATE 2007: 2 Marks)

56. Which of the following is true? (a)  A silicon wafer heavily doped with boron is a P+ substrate (b) A silicon wafer lightly doped with boron is a P+ substrate (c)  A silicon wafer heavily doped with arsenic is a P+ substrate (d) A silicon wafer lightly doped with arsenic is a P+ substrate (GATE 2008: 1 Mark) 57. Which of the following is NOT associated with a P–N junction? (a) Junction capacitance (b) Charge storage capacitance (c) Depletion capacitance (d) Channel length modulation (GATE 2008: 1 Mark)

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 205

205

58. The drain current of a MOSFET in saturation is given by ID = K(VGS − VTh)2 where K is a constant. The magnitude of the transconductance gm is (a)

K (VGS - VTh ) 2 VDS

(c)

K (VGS - VTh ) 2 ID (d) VGS VGS - VDS

(b) 2K(VGS − VTh)

(GATE 2008: 1 Mark) 59. A silicon wafer has 100 nm of oxide on it and is inserted in a furnace at a temperature above 1000°C for further oxidation in dry oxygen. The oxidation rate (a)  is independent of current oxide thickness and temperature (b)  is independent of current oxide thickness but depends on temperature (c) slows down as the oxide grows (d)  is zero as the existing oxide prevents further oxidation (GATE 2008: 1 Mark) 60. Silicon is doped with boron to a concentration of 4 × 1017 atoms/cm3. Assume the intrinsic carrier concentration of silicon to be 1.5 × 1010/cm3 and the value of kT to be 25 mV at 300 K. Compared to undoped silicon, the Fermi level of doped silicon (a) (b) (c) (d)

goes down by 0.13 eV goes up by 0.13 eV goes down by 0.427 eV goes up by 0.427 eV (GATE 2008: 2 Marks)

61. Consider the following assertions: S1: For Zener effect to occur, a very abrupt junction is required. S2:  For quantum tunnelling to occur, a very narrow energy barrier is required. Which of the following is correct? (a) Only S2 is true. (b) S1 and S2 are both true but S2 is not a reason for S1. (c) S1 and S2 are both true and S2 is a reason for S1. (d) Both S1 and S2 are false. (GATE 2008: 2 Marks) 62. Two identical NMOS transistors M1 and M2 are connected as shown below. Vbias is chosen so that both transistors are in saturation. The equivalent gm of the pair is ∂Iout defined to be at constant Vout. ∂Vin

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Iout Vout Vbias

M2 gm 2

Vin

gm1

64. The cross-section of a JFET is shown in the following figure. Let VG be −2 V and let VP be the initial pinch-off voltage. If the width W is doubled (with other geometrical parameters and doping levels remaining the same), then the ratio between the mutual transconductances of the initial and the modified JFET is VG Gate

M1

P+



The equivalent gm of the pair is (a) the sum of individual gm’s of the transistors (b) the product of individual gm’s of the transistors (c) nearly equal to the gm of M1 (d) nearly equal to the gm/g0 of M2 (GATE 2008: 2 Marks)

63. The measured transconductance gm of an NMOS transistor operating in the linear region is plotted against the gate voltage VG at a constant drain voltage VD. Which of the following figures represents the expected dependence of gm on VG? (a)

N

Source

W

Drain

P+ Gate (a) 4

(c)

VG 1 ⎛ 1 - 2 /VP ⎞ (b) ⎜ ⎟ 2 ⎝ 1 - 1/ ( 2VP ) ⎠

1 - ( 2/ V ) (d) P 1 - (1/ ( 2VP )) 1 - 1/ ( 2VP ) 1 - 2 /VP

(GATE 2008: 2 Marks) 65. In an N-type silicon crystal at room temperature, which of the following can have a concentration of 4 × 1019 cm−3? (a) Silicon atoms (b) Holes (c) Dopant atoms (d) Valence electrons (GATE 2009: 1 Mark)

gm

VG

(b)

66. The ratio of the mobility to the diffusion coefficient in a semiconductor has the unit (a) V–1 (b) cm⋅V–1 (c) V⋅cm–1 (d) V⋅s (GATE 2009: 1 Mark)

gm

VG

Common Data for Questions 67 and 68: Consider a silicon p–n junction at room temperature having the following parameters: Doping of the N-side = 1 × 1017 cm−3 Depletion width on the N-side = 0.1 µm Depletion width on the P-side = 1.0 µm Intrinsic carrier concentration = 1.4 × 1010/cm3

(c) g m

VG

Thermal voltage = 26 mV Permittivity of free space = 8.85 × 10−14 F/cm Dielectric constant of silicon = 12

(d) g m

VG

(GATE 2008: 2 Marks)

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67. The built-in potential of the junction (a) is 0.70 V (b) is 0.76 V

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(c) is 0.82 V (d) cannot be estimated from the data given 

(GATE 2009: 2 Marks)

68. The peak electric field in the device is (a) 0.15 mV/cm, directed from P-region to N-region (b) 0.15 mV/cm, directed from N-region to P-region (c) 1.8 mV/cm, directed from P-region to N-region (d) 1.80 mV/cm, directed from N-region to P-region (GATE 2009: 2 Marks) 69. Consider the following two statements about the internal conditions in an N-channel MOSFET operating in the active region. S1: The inversion charge decreases from source to drain. S2: The channel potential increases from source to drain. Which of the following is correct? (a) Only S2 is true (b) Both S1 and S2 are false (c) Both S1 and S2 are true, but S2 is not a reason for S1 (d) Both S1 and S2 are true, and S2 is a reason for S1 

(GATE 2009: 2 Marks)

Linked Answer for Questions 70 and 71: Consider the CMOS circuit shown in the figure below, where the gate voltage VG of the N-MOSFET is kept constant at 3 V. Assume that, for both transistors, the magnitude of the threshold voltage is 1  V and W  the product of the transconductance parameter and the   L W ratio, that is, the quantity mCox   is 1 mA V-2.  L



207

(c) N-MOSFET is in triode and P-MOSFET in saturation region (d) N-MOSFET is in saturation and P-MOSFET is in triode region (GATE 2009: 2 Marks)

71. Estimate the output voltage Vo for VG = 1.5 V. (Hint: Use the appropriate current-voltage equation for each MOSFET, based on the answer to Question 70.) 1 1 4+ V (a) 4 − V (b) 2 2 3 3 4+ V (c) − V (d) 2 2  (GATE 2009: 2 Marks) 72. At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon N-channel MOSFET is (a) 450 cm2/V⋅s (b) 1350 cm2/V⋅s 2 (c) 1800 cm /V⋅s (d) 3600 cm2/V⋅s (GATE 2010: 1 Mark) 73. Thin gate oxide in a CMOS process is preferably grown using (a) wet oxidation (b) dry oxidation (c) epitaxial deposition (d) ion implantation (GATE 2010: 1 Mark) Linked Answer Questions 74 and 75: The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following i­nformation is given: T = 300 K, electronic charge = 1.6 × 10-19 C, thermal voltage = 26mV and electron mobility = 1350 cm2/V⋅s. 1V

5V ND = 1016/cm3 3V

vo VG

70. For small increase in VG beyond 1 V, which of the following gives the correct description of the region of operation of each MOSFET? (a) Both the MOSFETs are in saturation region (b) Both the MOSFETs are in triode region

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x=0

x = 1 µm

74. The magnitude of the electric field at x = 0.5 mm is (a) 1 kV/cm (b) 5 kV/cm (c) 10 kV/cm (d) 26 kV/cm (GATE 2010: 2 Marks) 75. The magnitude of the electron drift current density at x = 0.5 mm is (b) 1.08 × 104 A/cm2 (a) 2.16 × 104 A/cm2 (c) 4.32 × 103 A/cm2 (d) 6.48 × 102 A/cm2 (GATE 2010: 2 Marks)

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76. Compared to a P–N junction with NA = ND = 1014/cm3, which one of the following statements is TRUE for a P–N junction with NA = ND = 1020/cm3?

(a) 1 V (c) 3 V

6V

(a)  Reverse breakdown voltage is lower and depletion capacitance is lower. (b) Reverse breakdown voltage is higher and depletion capacitance is lower. (c)  Reverse breakdown voltage is lower and depletion capacitance is higher. (d) Reverse breakdown voltage is higher and depletion capacitance is higher. (GATE 2010: 2 Marks) 77. In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector dopings in atoms/ cm3, respectively. If the emitter injection efficiency of the BJT is close to unity, which one of the following conditions is TRUE? (a) NE = NB = NC (b) NE >> NB and NB > NC (c) NE = NB and NB < NC (d) NE < NB < NC (GATE 2010: 2 Marks) 78. Drift current in semiconductors depends upon (a) only the electric field (b) only the carrier concentration gradient (c)  both the electric field and the carrier concentration (d) both the electric field and the carrier concentration gradient (GATE 2011: 1 Mark) 79. A silicon P–N junction is forward-biased with a constant current at room temperature. When the temperature is increased by 10°C, the forward-bias voltage across the P–N junction (a) increases by 60 mV (b) decreases by 60 mV (c) increases by 25 mV (d) decreases by 25 mV (GATE 2011: 1 Mark) 80. A Zener diode, when used in voltage stabilization circuits, is biased in (a)  reverse-bias region below the breakdown voltage (b) reverse breakdown region (c) forward-bias region (d) forward-bias constant current mode (GATE 2011: 1 Mark) 81. In the circuit shown below, for the MOS ­ transistor, mN Cox = 100 mA/V2 and the threshold voltage VTh = 1 V. The voltage Vx at the source of the upper transistor is

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 208

(b) 2 V (d) 3.67 V

W/L= 4

5V

Vx

W/L= 1



(GATE 2011: 2 Marks)

82. For a BJT, the common-base current gain a = 0.98 and the collector-base junction reverse-bias saturation current ICO = 0.6 m A. This BJT is connected in the common-emitter mode and operated in the active region with a base drive current IB = 20 m A. The collector current IC for this mode of operation is (a) 0.98 mA (b) 0.99 mA (c) 1.0 mA (d) 1.01 mA (GATE 2011: 2 Marks) Common Data for Questions 83 and 84: The channel resistance of an N-channel JFET shown in the figure below is 600 Ω, when the full channel thickness (tch) of 10 mm is available for conduction. The built-in voltage of the gate P+N junction (Vbj) is −1 V. When the gate-to-source voltage (VGS) is 0 V, the channel is depleted by 1 mm on each side due to the built-in voltage and hence the thickness available for conduction is only 8 mm. +

Gate

VGS -

P+ Source tch N

Drain

P+

83. The channel resistance when VGS = 0 V is (a) 480 Ω (b) 600 Ω (c) 750 Ω (d) 1000 Ω (GATE 2011: 2 Marks)

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209

84. The channel resistance when VGS = −3 V is (a) 360 Ω (b) 917 Ω (c) 1000 Ω (d) 3000 Ω (GATE 2011: 2 Marks)

88. The gate-source overlap capacitance is approximately (a) 0.7 fF (b) 0.7 pF (c) 0.35 fF (d) 0.24 pF  (GATE 2012: 2 Marks)

85. The I–V characteristics of the diode in the circuit given below are V - 0.7 i= A, V ≥ 0.7 V 500

89. In a forward-biased P–N junction, the sequence of events that best describes the mechanism of current flow is

1 kΩ i + V −

+ −



10 V

The current in the circuit is (a) 10 mA (c) 6.67 mA

(b) 9.3 mA (d) 6.2 mA (GATE 2012: 1 Mark)

86. The source of a silicon (ni = 1010 per cm3) N-channel MOS transistor has an area of 1 sq mm and a depth of 1 mm. If the dopant density in the source is 1019/cm3, the number of holes in the source region with the above volume is approximately (a) 107 (b) 100 (c) 10 (d) 0 (GATE 2012: 2 Marks) Common Data for Questions 87 and 88: In the threedimensional view of a silicon N-channel MOS transistor shown in the figure, d = 20 nm. The transistor is of width 1 mm. The depletion width formed at every P–N junction is 10 nm. The relative permittivities of Si and SiO2, respectively, are 11.7 and 3.9, and e0 = 8.9 × 10−12 F/m. 1 µm G D 0.2 µm

1 nm

1 µm

0.2 µm D d d S 0.2 µm P-substrate 0.2 µm B

87. The source-body junction capacitance is approximately (a) 2 f F (b) 7 f F (c) 2 pF (d) 7 pF (GATE 2012: 2 Marks)

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(a)  injection and subsequent diffusion and recombination of minority carriers (b) injection and subsequent drift and generation of minority carriers (c)  extraction and subsequent diffusion and generation of minority carriers (d) extraction and subsequent drift and recombination of minority carriers (GATE 2013: 1 Mark) 90. In a MOSFET operating in the saturation region, the channel length modulation effect causes (a) (b) (c) (d)

an increase in the gate-source capacitance a decrease in the transconductance a decrease in the unity-gain cut-off frequency a decrease in the output resistance (GATE 2013: 1 Mark)

91. A silicon bar is doped with donor impurities ND = 2.25 × 1015 atoms/cm3. Given the intrinsic carrier concentration of silicon at T = 300 K is ni = 1.5 × 1010 cm3. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are (a) n0 = 1.5 × 1016 cm−3, p0 = 1.5 × 105 cm−3 (b) n0 = 1.5 × 1010 cm−3, p0 = 1.5 × 1015 cm−3 (c) n0 = 2.25 × 1015 cm−3, p0 = 1.5 × 1010 cm−3 (d) n0 = 2.25 × 1015 cm−3, p0 = 1 × 105 cm−3 (GATE 2014: 1 Mark) 92. The donor and accepter impurities in an abrupt junction silicon diode are 1 × 1016 cm−3 and 5 × 1018 cm−3, respectively. Assume that the intrinsic carrier concentration in silicon ni = 1.5 × 1010 cm−3 at 300 K, kT = 26 mV and the permittivity of silicon eSi = 1.04 q × 10−12 F/cm The built-in potential and the depletion width of the diode under thermal equilibrium conditions, respectively, are (a) (b) (c) (d)

0.7 V and 1 × 10−4 cm 0.86 V and 1 × 10−4 cm 0.7 V and 3.3 × 10−5 cm 0.86 V and 3.3 × 10−5 cm (GATE 2014: 1 Mark)

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93. At T = 300 K, the band gap and the intrinsic carrier concentration of GaAs are 1.42 eV and 106 cm−3, respectively. In order to generate electron hole pairs in GaAs, which one of the wavelength (lc) ranges of incident radiation, is most suitable? (Given that: Planck’s constant is 6.62 × 10−34 J-s, velocity of light is 3 × 1010 cm/s and charge of electron is 1.6 × 10−19 C) (a) 0.42 mm < lc < 0.87 mm (b) 0.87 mm < lc < 1.42 mm (c) 1.42 mm < lc < 1.62 mm (d) 1.62 mm < lc < 6.62 mm (GATE 2014: 1 Mark) 94. In the figure, ln (ri) is plotted as a function of 1/T, where ri is the intrinsic resistivity of silicon, T is the temperature, and the plot is almost linear.

In(ρi)

(c) majority carrier concentration (d) excess minority carrier concentration (GATE 2014: 1 Mark) 98. If the emitter resistance in a common-emitter voltage amplifier is not bypassed, it will (a) reduce both the voltage gain and the input impedance (b)  reduce the voltage gain and increase the input impedance (c)  increase the voltage gain and reduce the input impedance (d) increase both the voltage gain and the input impedance (GATE 2014: 1 Mark) 99. Two silicon diodes, with a forward voltage drop of 0.7 V, are used in the circuit shown in the figure. The range of input voltage Vi for which the output voltage Vo = Vi is R +

+

D1

D2

Vi

1/T The slope of the line can be used to estimate (a) band gap energy of silicon (Eg) (b)  sum of electron and hole mobility in silicon (mn + mp) (c) reciprocal of the sum of electron and hole mobility in silicon (mn + mp)−1 (d) intrinsic carrier concentration of silicon (ni) (GATE 2014: 1 Mark) 95. At T = 300 K, the hole mobility of a semiconductor kT = 26 mV. The hole diffusion mp = 500 cm2/V-s and q constant Dp in cm2/s is . (GATE 2014: 1 Mark) 96. An increase in the base recombination of a BJT will increase the (a) common emitter dc current gain b (b) breakdown voltage BVCEO (c) unity-gain cut-off frequency fT (d) transconductance gm (GATE 2014: 1 Mark) 97. A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to the (a) minority carrier mobility (b) minority carrier recombination lifetime

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 210

Vo

+ −1 V − −

(a) −0.3 V < Vi < 1.3 V (c) −1.0 V < Vi < 2.0 V

+ − 2V −

(b) −0.3 V < Vi < 2 V (d) −1.7 V < Vi < 2.7 V (GATE 2014: 1 Mark)

100. In CMOS technology, shallow P-well or N-well regions can be formed using (a) low pressure chemical vapour deposition (b) low energy sputtering (c) low temperature dry oxidation (d) low energy ion-implantation (GATE 2014: 1 Mark) 101. In MOSFET fabrication, the channel length is defined during the process of (a) isolation oxide growth (b) channel stop implantation (c) poly-silicon gate patterning (d) lithography step leading to the contact pads (GATE 2014: 1 Mark) 102. The doping concentrations on the P-side and N-side of a silicon diode are 1 × 1016 cm−3 and 1 × 1017 cm−3, respectively. A forward bias of 0.3 V is applied to the diode. At T = 300 K, the intrinsic carrier concentration of silicon ni = 1.5 × 1010 cm−3 and kT q = 26 mV. The electron concentration at the edge of the depletion region on the P-side is

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Chapter 3  • Electronic Devices

(a) 2.3 × 109 cm−3 (c) 1 × 1017 cm−3

(b) 1 × 1016 cm−3 (d) 2.25 × 106 cm−3 (GATE 2014: 2 Marks)

103. Consider an abrupt PN junction (at T = 300 K) shown in the figure. The depletion region width Xn on the N-side of the junction is 0.2 mm and the permittivity of silicon (eSi) is 1.044 × 10−12 F/cm. At the junction, the approximate value of the peak electric field (in kV/cm) is .

P+-region NA>>ND

211

is the Fermi level, which one of the following represents the energy band diagram for the biased N-type semiconductor? (a)

EC (b) EF EV

(c)

EC (d) EF EV

EC EF EV

EC EF EV

(GATE 2014: 2 Marks)

Xn

N-region ND = 1016/cm3 (GATE 2014: 2 Marks)

104. When a silicon diode having a doping concentration of NA = 9 × 1016 cm−3 on p-side and ND = 1 × 1016 cm−3 on N-side is reverse biased, the total depletion width is found to be 3 mm. Given that the permittivity of silicon is 1.04 × 10−12 F/cm, the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are (a) 2.7 mm and 2.3 × 105 V/cm (b) 0.3 mm and 4.15 × 105 V/cm (c) 0.3  mm and 0.42 × 105 V/cm (d) 2.1  mm and 0.42 × 105 V/cm (GATE 2014: 2 Marks) 105. Consider a silicon sample doped with ND = 1 × 10 / cm3 donor atoms. Assume that the intrinsic carrier concentration ni = 1.5 × 1010/cm3. If the sample is additionally doped with NA = 1 × 1018/cm3 acceptor atoms, the approximate number of electrons/cm3 in the sample, at T = 300 K, will be . (GATE 2014: 2 Marks)

108. A BJT is biased in forward active mode. Assume VBE = 0.7 V, kT/q = 25 mV and reverse saturation current IS = 10−13 A. The transconductance of the BJT (in A/V) is . (GATE 2014: 2 Marks) 109. For the amplifier shown in the figure, the BJT parameters are VBE = 0.7 V, b = 200, and thermal voltage VT = 25 mV. The voltage gain (vo/vi) of the amplifier is . VCC = +12 V R1 33 kW

106. Consider two BJTs biased at the same collector current with area A1 = 0.2 mm × 0.2 mm and A2 = 300 mm × 300 mm. Assuming that all other device parameters are identical, kT/q = 26 mV, the ­intrinsic carrier concentrations 1 × 1010 cm−3, and q = 1.6 × 10−19 C, the difference between the base-emitter voltages (in mV) of the two BJTs (i.e., VBE1 - VBE2) is . (GATE 2014: 2 Marks) 107. An N-type semiconductor having uniform doping is biased as shown in the figure V

N-type semiconductor If EC is the lowest energy level of the conduction band, EV is the highest energy level of the valance band and EF

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 211

vo

1 mF

15

RC 5 kW 1 mF

vi R2 11 k W

Rs 10 W

RE 1 kW

CE 1 mF

(GATE 2014: 2 Marks) 110. An ideal MOS capacitor has boron doping-­ concentration of 1015 cm−3 in the substrate. When a gate voltage is applied, a depletion region of width 0.5 mm is formed with a surface (channel) potential of 0.2 V. Given that eo = 8.854 × 10-14F/cm and the relative permittivities of silicon and silicon dioxide are 12 and 4, respectively, the peak electric field (in V/mm) in the oxide region is . (GATE 2014: 2 Marks) 111. In the circuit shown, the silicon BJT has b = 50. Assume VBE = 0.7 V and VCE(sat) = 0.2 V. Which one of the following statements is correct?

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212

GATE ECE Chapter-wise Solved papers

10 V RC 50 kW 5V RB

(a) For RC = 1 kΩ, the BJT operates in the saturation region (b) For RC = 3 kΩ, the BJT operates in the saturation region (c) For RC = 20 kΩ, the BJT operates in the ­cut-off region (d) For RC = 20 kΩ, the BJT operates in the linear region (GATE 2014: 2 Marks) 112. For the MOSFET M1 shown in the figure, assume W/L = 2, VDD = 2.0 V, mnCox = 100 mA/V2 and VTh = 0.5 V. The transistor M1 switches from saturation region to linear region when Vin (in Volts) is . VDD

R = 10 kW Vout Vin

M1

(GATE 2014: 2 Marks) 113. Consider the common-collector amplifier in the figure (bias circuitry ensures that the transistor operates in forward active region, but has been omitted for simplicity). Let IC be the collector current, VBE be the base-emitter voltage and VT be the thermal voltage. Also, gm and ro are the small-signal transconductance and output resistance of the transistor, respectively. Which one of the following conditions ensures a nearly constant small signal voltage gain for a wide range of values of RE?

VCC

Vin Vout RE

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 212

(a) gmRE > 1

(b) ICRE >> VT (d) VBE >> VT (GATE 2014: 2 Marks)

114. A BJT in a common-base configuration is used to amplify a signal received by a 50 Ω antenna. Assume kT/q = 25 mV. The value of the collector bias current (in mA) required to match the input impedance of the amplifier to the impedance of the antenna is . (GATE 2014: 2 Marks) 115. A depletion type N-channel MOSFET is biased in its linear region for use as a voltage controlled resistor. Assume threshold voltage VTh = −0.5 V, VGS = 2.0 V, VDS = 5 V, W/L = 100, Cox = 10−8 F/cm2 and mn = 800 cm2/ V-s. The value of the resistance of the voltage c­ ontrolled resistor (in Ω) is . (GATE 2014: 2 Marks) 116. A region of negative differential resistance is observed in the current-voltage characteristics of a silicon PN junction if (a) both the P-region and the N-region are heavily doped (b) the N-region is heavily doped compared to the P-region (c)  the P-region is heavily doped compared to the N-region (d) an intrinsic silicon region is inserted between the P-region and the N-region (GATE 2015: 1 Mark) 117. A silicon sample is uniformly doped with donor type impurities with a concentration of 1016/cm3. The electron and hole mobilities in the sample are 1200 cm2/ V-s and 400 cm2/V-s respectively. Assume complete ionization of impurities. The charge of an electron is 1.6 × 10−19C. The resistivity of the sample (in Ω-cm) is . (GATE 2015: 1 Mark) 118. Assume electronic charge q = 1.6 × 10-19 C, kT/q = 25 mV and electron mobility mn = 1000 cm2/V-s. If the concentration gradient of electrons injected into a P-type silicon sample is 1 × 1021/cm4, the magnitude of electron diffusion current density (in A/cm2) is . (GATE 2015: 1 Mark) 119. A piece of silicon is doped uniformly with phosphorous with a doping concentration of 1016/cm3. The expected value of mobility versus doping concentration for silicon assuming full dopant ionization is shown below. The charge of an electron is 1.6 × 10−19C. The conductivity (in S cm−1) of the silicon sample at 300 K is .

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Chapter 3  • Electronic Devices

10 V

Hole and electron mobility in silicon at 300 K 1400 Electron

1200

Hole

1000

N − Si

800 600

1 cm (GATE 2015: 2 Marks)

400

+

20

19 1. E

+

18 1. E

+

17 1. E

+

16 1. E

+

15 1. E

+ 1. E

+ 1. E

+ 1. E

14

200 0 13

Mobility (cm2 V−1 s−1)

213

Doping concentration (cm−3)

(GATE 2015: 1 Mark) 120. An N-type silicon sample is uniformly illuminated with light which generates 1020 electron-hole pairs per cm2 per second. The minority carrier lifetime in the sample is 1 μs. In the steady state, the hole concentration in the sample is approximately 10x, where x is an integer. The value of x is . (GATE 2015: 1 Mark) 121. In the circuit shown in the figure, the BJT has a current gain (b ) of 50. For an emitter-base voltage VEB = 600 mV, the emitter collector voltage VEC (in Volts) is .

124. In a MOS capacitor with an oxide layer thickness of 10 mm, the maximum depletion layer thickness is 100 nm. The permittivities of the semiconductor and the oxide layer are es and eox respectively. Assuming es/eox = 3, the ratio of the maximum capacitance to the minimum capacitance of this MOS capacitor is . (GATE 2015: 2 Marks) 125. For a silicon diode with long P and N regions, the accepter and donor impurity concentrations are 1 × 1017 cm−3 and 1 × 1015 cm−3, respectively. The lifetime of electrons in P region and holes in N region are both 100 µs. The electron and hole diffusion coefficients are 49 cm2/s and 36 cm2/s, respectively. Assume kT/q = 26 mV, the ­intrinsic carrier concentration is 1 × 1010 cm−3 and q  = 1.6 × 10−19C. When a forward voltage of 208 mV is applied across the diode, the hole current density (in nA/cm2) injected from P region to N regions is . (GATE 2015: 2 Marks)

3V

60 kW

500 W (GATE 2015: 1 Mark)

126. The energy band diagram and electron density profile n(x) in a semiconductor are shown in the figure. Assume that n( x ) = 105 e ( q α x / kT ) cm -3 with a = 0.1 V/cm and x kT expressed in cm. Given = 0.026 V, Dn = 36 cm2 s−1, q and D /m = kT /q. The electron current density (in A/ cm2) at x = 0 is E(eV)

122. Which one of the following processes is preferred to form the gate dielectric (SiO2) of MOSFETs? (a) Sputtering (b) Molecular beam epitaxy (c) Wet oxidation (d) Dry oxidation (GATE 2015: 1 Mark) 123. A dc voltage of 10 V is applied across an n-type silicon bar having a rectangular cross-section and a length of 1 cm as shown in figure. The donor doping concentration ND and the mobility of electrons mn are 1016cm−3 and 1000 cm2V−1s−1, respectively. The average time (in μs) taken by the electrons to move from one end of the bar to other end is .

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 213

log(n(x))

Slope = −0.1eV/cm

EC EV x=0 (a) −4.4 × 10−2 (c) 0

x

x=0

x

(b) −2.2 × 10−2 (d) 2.2 × 10−2 (GATE 2015: 2 Marks)

127. A MOSFET in saturation has a drain current of 1 mA for VDS = 0.5 V. If the channel length modulation coefficient

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214

GATE ECE Chapter-wise Solved papers

is 0.05 V−1, the output resistance (in kΩ) of the MOSFET is . (GATE 2015: 2 Marks) 128. A small percentage of impurity is added to an intrinsic semiconductor at 300 K. Which one of the following statements is true for the energy band diagram shown in the following figure?



Ec

Ev

(a)  Intrinsic semiconductor doped with p­entavalent atoms to form n-type semiconductor. (b) Intrinsic semiconductor doped with trivalent atoms to form n-type semiconductor. (c)  Intrinsic semiconductor doped with p­entavalent atoms to form p-type semiconductor. (d) Intrinsic semiconductor doped with trivalent atoms to form p-type semiconductor. (GATE 2016: 1 Mark) 129. Consider a silicon sample at T = 300 K, with a uniform donor density Nd = 5 × 1016 cm−3, illuminated uniformly such that the optical generation rate is Gopt = 1.5 × 1020 cm−3 s−1 throughout the sample. The incident radiation is turned off at t = 0. Assume low-level injection to be valid and ignore surface effects. The carrier lifetimes are tp0 = 0.1 ms and tn = 0.5 ms. 0

n-type Si

The hole concentration at t = 0 and the hole concentration at t = 0.3 µs, respectively, are (a) 1.5 × 1013 cm−3 and 7.47 × 1011 cm−3 (b) 1.5 × 1013 cm−3 and 8.23 × 1011 cm−3 (c) 7.5 × 1013 cm−3 and 3.73 × 1011 cm−3 (d) 7.5 × 1013 cm−3 and 4.12 × 1011 cm−3 (GATE 2016: 1 Mark)

130. Light from the free space is incident at an angle qi to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are n1 = 1.5 and n2 = 1.4, respectively.

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 214

(Cladding)

n1

(Core)

Light

Conduction band

0.01 eV

n2

The maximum value of qi (in degrees) for which the incident light will be guided in the core of the fibre is ______. (GATE 2016: 1 Mark)

131. A long-channel NMOS transistor is biased in the linear region VDS = 50 mV and is used as a resistance. Which one of the following statements is NOT correct? (a) If the device width W is increased, the resistance decreases. (b) If the threshold voltage is reduced, the resistance decreases. (c) If the device length L is increased, the resistance increases. (d) If VGS is increased, the resistance increases. (GATE 2016: 1 Mark) 132. Assume that the diode in the figure has Von = 0.7 V, but is otherwise ideal. i2

R1

i1

i2

2 kΩ + 2V −

R2 6 kΩ

The magnitude of the current i2 (in mA) is equal to _______. (GATE 2016: 1 Mark)

133. Resistor R1 in the circuit below has been adjusted so that I1 = 1 mA. The bipolar transistor Q1 and Q2 are perfectly matched and have very high current gain, so their base currents are negligible. The supply voltage VCC is 6 V. The thermal voltage kT/q is 26 mV. VCC R1 I2

I1

R2

The value of R2 (in Ω) for which I2 = 100 mA is _____. (GATE 2016: 1 Mark)

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Chapter 3  • Electronic Devices

134. Transistor geometries in a CMOS inverter have been adjusted to meet the requirement for worst case charge and discharge times for driving a load capacitor C. This design is to be converted to that of a NOR circuit in the same technology, so that its worst case charge and discharge times while driving the same capacitor are similar. The channel length of all transistors are to be kept unchanged. Which one of the following statements is correct? VDD

215

136. The figure shows the I−V characteristics of a solar cell illuminated uniformly with solar light of power 100 mW/cm2. The solar cell has an area of 3 cm2 and a fill factor of 0.7. The maximum efficiency (in %) of the device is _____. I

VDD

In2 In1 In

V

(0, 0)

Out

VOC = 0.5 V Out C

C

(a) Widths of PMOS transistors should be doubled, while widths of NMOS transistors should be halved. (b) Widths of PMOS transistors should be doubled, while widths of NMOS transistors should not be changed. (c)  Widths of PMOS transistors should be halved, while widths of NMOS transistors should not be changed. (d) Widths of PMOS transistors should be unchanged, while widths of NMOS transistors should be halved. (GATE 2016: 1 Mark)

(GATE 2016: 1 Mark) 137. Consider a silicon pn-junction with a uniform acceptor doping concentration of 1017 cm−3 on the p-side and a uniform donor doping concentration of 1016 cm−3 on the n-side. No external voltage is applied to the diode. Given: kT/q = 26 mV, ni = 1.5 × 1010 cm−3, e Si = 12e 0 , e 0 = 8.85 × 10 -14 F/m , and q = 1.6 × 10−19 C. The charge per unit junction area (nC cm−2) in the depletion region on the p-side is  . (GATE 2016: 2 Marks) 138. Consider avalanche breakdown in a silicon p+n junction. The n-region is uniformly doped with a donor density ND. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field Ecrit. Assume Ecrit to be independent of ND. If the built-in voltage of the p+n junction is much smaller than the breakdown voltage, VBR, the relationship between VBR and ND is given by

135. The figure shows the band diagram of a Metal Oxide Semiconductor (MOS). The surface region of this MOS is in

(a) VBR ×  N D  = constant (b) ND ×  VBR  = constant (c) ND × VBR = constant

(d) ND/VBR = constant (GATE 2016: 2 Marks)

139. Consider a region of silicon devoid of electrons and holes, with an ionized donor density of N d+ = 1017 cm -3 . The electric field at x = 0 is 0 V/cm and the electric field at x = L is 50 kV/cm in the positive x-direction. Assume that the electric field is zero in the y and z directions at all points.

SiO2 EM

ΦB ΦB

EC EFS Ei

Nd+=1017 cm−3

EV (a) inversion (c) depletion

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 215

(b) accumulation (d) flat band (GATE 2016: 1 Mark)

x=0

x=L

Given q = 1.6 × 10–19 Coulomb, e0 = 8.85 × 10–14 F/cm, er = 11.7 for silicon, the value of L (in nm) is _____. (GATE 2016: 2 Marks)

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GATE ECE Chapter-wise Solved papers

140. The figure below shows the doping distribution in a p-type semiconductor in log scale. The magnitude of the electric field (in kV/cm) in the semiconductor due to non-uniform doping is .

then the collector current IC (in mA) at room temperature is _____. (Given: thermal voltage VT = 26 mV at room temperature, electronic charge q = 1.6 × 10−19 C.) IB p

n

1016

−3

n

NA (cm ) 1014

IC

IE 1

2 Position (µm) (GATE 2016: 2 Marks)

141. The switch S in the circuit shown has been closed for a long time. It is opened at time t = 0 and remains open after that. Assume that the diode has zero reverse current and zero forward voltage drop. 1Ω + 10 V −



t=0 S 1 mH

10 µF

− V + C

The steady-state magnitude of the capacitor voltage VC (in volts) is _____. (GATE 2016: 2 Marks)

142. A voltage VG is applied across a MOS capacitor with metal gate and p-type silicon substrate at T = 300 K. The inversion carrier density (in number of carriers per unit area) for VG = 0.8 V is 2 × 1011 cm–2. For VG = 1.3 V, the inversion carrier density is 4 × 1011 cm–2. What is the value of the inversion carrier density for VG = 1.8 V?. (a) 4.5  × 1011 cm–2 (b) 6.0  × 1011 cm–2 11 –2 (c) 7.2  × 10 cm (d) 8.4  × 1011 cm–2 (GATE 2016: 2 Marks) 143. Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of VGS and VDS. Given, gm = 0.5 µA/V for VDS= 50 mV and VGS = 2 V, gd = 8 µA/V for VGS = 2 V and VDS = 0 V, where g m =

0.5 µm (GATE 2016: 2 Marks) 145. Figures I and II show two MOS capacitors of unit area. The capacitor in Figure I has insulator materials X (of thickness t1 = 1 nm and dielectric constant ε1 = 4) and Y (of thickness t2 = 3 nm and dielectric constant ε2 = 20). The capacitor in Figure II has only insulator material X of thickness tEq. If the capacitors are of equal capacitance, then the value of tEq (in nm) is _____.

t2 t1

Metal e2 e1 Si

Metal

eε 1

tEq

Si

∂I D ∂I and gd = D .The thresh∂VGS ∂VDS

old voltage (in volts) of the transistor is ______. (GATE 2016: 2 Marks) 144. The injected excess electron concentration profile in the base region of an npn BJT, biased in the active region, is linear, as shown in the figure. If the area of the emitterbase junction is 0.001 cm2, μn = 800 cm2/(V-s) in the base region and depletion layer widths are negligible,

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 216

0

(GATE 2016: 2 Marks) 146. A bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statements is true? (a) Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites.

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Chapter 3  • Electronic Devices

(b) Silicon atoms acts as n-type dopants in Arsenic sites and p-type dopants in Gallium sites. (c) Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites. (d) Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites. (GATE 2017: 1 Mark) 147. An n+-n silicon device is fabricated with uniform and non-degenerate donor doping concentrations of N D1 = 1 × 1018 cm -3 and N D 2 = 1 × 1015 cm -3 corresponding to the n+ and n regions, respectively. At the operational temperature T, assume complete impurity ionization, kT/q = 25 mV, and intrinsic carrier concentration to be ni = 1 × 1010 cm-3. What is the magnitude of the built-in potential of this device? (a) 0.748 V (b) 0.460 V (c) 0.288 V (d) 0.173 V (GATE 2017: 1 Mark)

Normalized Excess Carrier Concentratoion

148. For a narrow base PNP BJT, the excess minority carrier concentrations (∆nE for emitter, ∆pB for base, ∆nC for collector) normalized to equilibrium minority carrier concentrations (nE0 for emitter, pB0 for base, nC0 for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in? 105

∆pB pB0 0

∆nE nE0

∆nC nC0

−1

Emitter (P)

Base (N)

Collector (P)

(a) Forward active (b) Saturation (c) Inverse active (d) Cut-off (GATE 2017: 1 Mark)

(c) a decrease in the low-frequency cut-off frequency (d) a decrease in the high-frequency cut-off frequency (GATE 2017: 1 Mark) 151. An N-channel enhancement mode MOSFET is biased at VGS > VTH and VDS > (VGS - VTH), where VGS is the gate-tosource voltage. VDS is the drain-to-source voltage and VTH is the threshold voltage. Considering channel length modulation effect to be significant, the MOSFET behaves as a (a) voltage source with zero output impedance (b) voltage source with non-zero output impedance (c) current source with finite output impedance (d) current source with infinite output impedance (GATE 2017: 1 Mark) 152. An npn bipolar junction transistor (BJT) is operating in the active region. If the reverse bias across the base-collector junction is increased, then (a)  the effective base width increases and common-­ emitter current gain increases (b)  the effective base width increases and common-­ emitter current gain decreases (c)  the effective base width decreases and common-­ emitter current gain increases (d)  the effective base width decreases and common-­ emitter current gain decreases (GATE 2017: 1 Mark) 153. Consider an N-channel MOSFET having width W, length L, electron mobility in the channel mn and oxide capacitance per unit area Cox. If gate-to-source voltage VGS = 0.7 V, drain-to-source voltage VDS = 0.1 V, mnCox = 100 mA/V2, threshold voltage VTH = 0.3 V and W/L = 50, then the transconductance gm (in mA/V) is ___________. (GATE 2017: 1 Mark) 154. The dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron concentration of n = 1 × 1016 cm−3 and electronic charge q = 1.6 × 10−19 C. If a bias of 5 V is applied across a 1 mm region of this semiconductor, the resulting current density in this region, in kA/cm2, is .

149. A good transconductance amplifier should have (a) high input resistance and low output resistance (b) low input resistance and high output resistance (c) high input and output resistances (d) low input and output resistances (GATE 2017: 1 Mark)

107

150. The Miller effect in the context of a common emitter amplifier explains (a) an increase in the low-frequency cut-off frequency (b) an increase in the high-frequency cut-off frequency

0

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 217

217

Drift velocity (cm/s)

Constant

linear

5 × 105 Electic field ( V/cm)

(GATE 2017: 2 Marks)

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GATE ECE Chapter-wise Solved papers

155. As shown, a uniformly doped Silicon (Si) bar of length L = 0.1 μm with a donor concentration ND = 1016 cm-3 is illuminated at x = 0 such that electron and hole pairs are generated at the rate of ⎛ GL = GL 0 ⎜ 1 ⎝

x⎞ ⎟ , 0 ≤ x ≤ L, L⎠

where GL0 = 1017 cm-3s-1. Hole lifetime is 10-4 s, electronic charge q = 1.6 × 10-19 C, hole diffusion coefficient DP = 100 cm2/s and low level injection condition prevails. Assuming a linearly decaying steady state excess hole concentration that goes to 0 at x = L, the magnitude of the diffusion current density at x = L/2, in A/cm2, is __________.

158. An abrupt pn junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which one of the following figures represents the electric field profile near the pn junction? (a) 

E(x)

n-side

p-side

(0, 0) x W

(b) 

E(x)

Light n-side

p-side

Si (ND = 1016 cm−3) (0, 0) X

=0

L=

0.1 µm

(GATE 2017: 2 Marks) 156. As shown, two Silicon (Si) abrupt p-n junction diodes are fabricated with uniform donor doping concentrations of ND1 = 1014 cm−3 and ND2 = 1016 cm−3 in the n-regions of the diodes, and uniform acceptor doping concentrations of NA1 = 1014 cm−3 and NA2 = 1016 cm−3 in the p-regions of the diodes, respectively. Assuming that the reverse bias voltage is >> built-in potentials of the diodes, the ratio C2/C1 of their reverse bias capacitances for the same applied reverse bias is . p

n

14

10 -3 cm

14

10 -3 cm

p

W

(c) 

E(x)

n-side

p-side W (0, 0)

(d) 

E(x)

n 16

10 -3 cm

16

10 -3 cm

n-side

p-side W

C1 Diode 1

C2 Diode 2

(0, 0)

(GATE 2017: 2 Marks) 157. A MOS capacitor is fabricated on p-type Si (Silicon) where the metal work function is 4.1 eV and electron affinity of Si is 4.0 eV. EC - EF = 0.9 eV, where EC and EF are the conduction band minimum and the Fermi energy levels of Si, respectively. Oxide er = 3.9, eo = 8.85 × 10−14 F/cm, oxide thickness tox = 0.1 μm and electronic charge q = 1.6 × 10−19 C. If the measured flat band voltage of this capacitor is -1 V, then the magnitude of the fixed charge at the oxide-semiconductor interface, in nC/cm2 is . (GATE 2017: 2 Marks)

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 218

(GATE 2017: 2 Marks) 159. An electron (q1) is moving in free space with velocity 105 m/s towards a stationary electron (q2) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is × 10−8 m. [Given, mass of electron m = 9.11 × 10−31 kg, charge of electron e = -1.6 × 10−19C, and permittivity e 0 = (1 / 36π ) × 10 -9 F/m] (GATE 2017: 2 Marks)

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160. For a particular intensity of incident light on a s­ilicon P-N junction solar cell, the photocurrent density (JL) is 2.5 mA/cm2 and the open-circuit voltage (VOC) is 0.451 V. Consider thermal voltage (VT) to be 25 mV. If the intensity of the incident light is increased by 20 times, assuming that the temperature remains unchanged, VOC (in volts) will be _________. (GATE 2017: 2 Marks) 161. In the circuit shown, transistors Q1 and Q2 are biased at a collector current of 2.6 mA. Assuming that transistor current gains are sufficiently large to assume collector current equal to emitter current and thermal voltage of 26 mV, the magnitude of voltage gain V0/Vs in the midband frequency range is ________ (up to second decimal place). 5V 1 kΩ V0 Q1 Vs

Q2 RB2 —5 V (GATE 2017: 2 Marks) 162. In a p-n junction diode at equilibrium, which one of the following statements is NOT TRUE? (a) The hole and electron diffusion current components are in the same direction. (b) The hole and electron drift current components are in the same direction. (c) On an average, holes and electrons drift in opposite direction. (d) On an average, electrons drift and diffuse in the same direction. (GATE 2018: 1 Mark) 163. A p-n step junction diode with a contact potential of 0.65 V has a depletion width of 1 μm at equilibrium. The forward voltage (in volts, correct to two decimal places) at which this width reduces to 0.6 μm is . (GATE 2018: 1 Mark) 164. The circuit shown in the figure is used to provide regulated voltage (5 V) across the 1 kΩ resistor. Assume that the Zener diode has a constant reverse breakdown voltage for a current range, starting from a minimum

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 219

required Zener current, IZ min = 2 mA to its maximum allowable current. The input voltage V1 may vary by 5% from its nominal value of 6 V. The resistance of the diode in the breakdown region is negligible. R

5V

VI

1 kΩ

The value of R and the minimum required power dissipation rating of the diode, respectively, are (a) 186 Ω and 10 mW (b) 100 Ω and 40 mW (c) 100 Ω and 10 mW (d) 186 Ω and 40 mW (GATE 2018: 2 Marks) 165. Red (R), Green (G) and Blue (B) Light Emitting Diodes (LEDs) were fabricated using p-n junctions of three different inorganic semiconductors having different bandgaps. The built-in voltages of red, green and blue diodes are VR, VG and VB respectively. Assume donor and acceptor doping to be the same (NA and ND, respectively) in the p and n sides of all the three diodes. Which one of the following relationships about the built-in voltages is TRUE? (a) VR > VG > VB (b) VR < VG < VB (c) VR = VG = VB (d) VR > VG < VB (GATE 2018: 2 Marks) 166. A dc current of 26 μA flows through the circuit shown. The diode in the circuit is forward biased and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of 0.5 nF. Its neutral region resistances can be neglected. Assume that the room temperature thermal equivalent voltage is 26 mV.

5 sin(wt)mV ∼

100 Ω V

For ω = 2 × 106 rad/s, the amplitude of the small-signal component of diode current (in μA, correct to one decimal place) is . (GATE 2018: 2 Marks)

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Given: Boltzmann constant k = 1.38 × 10–23 J ⋅ K–1, electronic charge q = 1.6 × 10–19 C. Assume 100% acceptor ionization.

167. A solar cell of area 1.0 cm2, operating at 1.0 sun intensity, has a short circuit current of 20 mA, and an open-circuit voltage of 0.65 V. Assuming room temperature operation and thermal equivalent voltage of 26 mV, the open circuit voltage (in volts, correct to two decimal places) at 0.2 sun intensity is . (GATE 2018: 2 Marks)

At room temperature (T = 300 K), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be ________. (GATE 2018: 2 Marks)

168. A junction is made between p– Si with doping density NA1 = 1015 cm–3 and p Si with doping density NA2 = 1017 cm–3.

Answer Key   1. (d)

 2. (d)

 3. (c)

 4. (b)

 5. (b)

 6. (b)

 7. (a)

 8. (b)

 9. (d)

 10. (a)

 11. (d)

 12. (b)

 13. (a)

 14. (c)

 15. (d)

 16. (d)

 17. (a)

 18. (d)

 19. (c)

 20. (a)

 21. (d)

 22. (c)

 23. (b)

 24. (c)

 25. (a)

 26. (d)

 27. (c)

 28. (b)

 29. (a)

 30. (b)

 31. (d)

 32. (d)

 33. (d)

 34. (c)

 35. (a)

 36. (b)

 37. (b)

 38. (b)

 39. (a)

 40. (c)

 41. (b)

 42. (a)

 43. (c)

 44. (c)

 45. (b)

 46. (b)

 47. (d)

 48. (a)

 49. (c)

 50. (d)

 51. (c)

 52. (a)

 53. (b)

 54. (b)

 55. (b)

 56. (a)

 57. (d)

 58. (b)

 59. (d)

 60. (c)

 61. (b)

 62. (c)

 63. (d)

 64. (a)

 65. (c)

 66. (a)

 67. (b)

 68. (b)

 69. (d)

 70. (d)

 71. (d)

 72. (b)

 73. (b)

 74. (c)

 75. (a)

 76. (c)

 77. (b)

 78. (c)

 79. (d)

 80. (b)

 81. (c)

 82. (d)

 83. (c)

 84. (c)

 85. (d)

 86. (d)

 87. (a)

 88. (a)

 89. (a)

 90. (d)

 91. (d)

 92. (d)

 93. (a)

 94. (a)

 95. (13)

 96. (b)

 99. (d)

100. (d)

101. (c)

102. (a)

103. (30.66) 104. (b)

111. (b)

112. (1.5) 113. (b)

114. (0.5)

121. (2)

122. (d)

124. (4.33) 125. (28.59) 126. (c)

131. (d)

132.  (0.25) 133. (598.67) 134. (b)

123. (100)

 97. (d)

 98. (b)

105.  (225.2) 106.  (381)

107. (d)

108. (5.785) 109. (–237.76) 110. (2.4)

115. (500)

117. (0.52) 118. (4000) 119. (1.92)

120. (14)

127. (20)

129. (a)

130.  (32.6)

135. (a)

141. (100) 142. (b)

143. (1.2)

144. (6.656) 145. (1.6)

151. (c)

153. (0.5)

154. (1.6)

152. (c)

161. (50) 162. (d)

116. (a) 136. (21)

137. (–49.6) 138. (c)

139. (32.36)

140. (0.0133)

146. (a)

147. (d)

148. (c)

149. (c)

150. (d)

157. (6.9)

158. (a)

159. (5.06)

160. (0.52)

155.  (16. 0) 156. (10.0)

163.  (0.416) 164. (b)

165. (b)

128. (a)

166. (6.40 mA) 167. (0.608) 168. (0.1192)

Answers with Explanation 1.

Topic: P-N Junction (d)  Diode is in the forward biased state.

Applying KCL at node A, we get:

A 2Ω + 4V −

2Ω − 2Ω

Vo +



− 2V +

2.

⇒ Vo = -

2 V 3

Topic: BJT (d)  We know that,

Ch wise GATE_ECE_CH03_Electronic Devices_Qns.indd 220

-Vo - 4 -Vo -Vo + 2 + + =0 2 2 2 ⇒ -3Vo = 2

β=

IC IB

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Ele es Chapter 3  • Electronic Devices

As b is large so, I B = 0 Now, the equivalent circuit can be drawn as following:

6.

15 V

RC IC R th I = 0 B + VBE − IC Vth

IC = 3.

Vth − 0.7 5 − 0.7 = = 10 mA RE 430

Therefore,

I V rπ = T and g m = C VT IB

b=

4.

5.

5 × 10 10 = kΩ 15 3

Topic: BJT (c)  In BJT, current gain is related with gm and rπ in following manner b = g m rπ where b, gm and rπ are the current gain, transconductance and input resistance respectively. Also,

Topic: MOSFET (b)  Effective channel length will decrease due to increase in drain voltage in MOSFET. This phenomenon is known as channel length modulation.

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 221

fb =

fT b

1.47 × 1010 Hz = 1.64 × 108 Hz 90

Topic: Carrier Transport (a)  Conduction current, J C = σE Displacement current is the rate of change of electric flux density, ∂D JD = = ω D = ωe E ( ∵ e = D /E ) ∂t where D is the electric flux density and e is the permittivity of the substance. When | J C | = | J D |

8.

I C VT I C × = ⇒ I C = bI B VT I B I B

Topic: MOSFET (b)  MOSFET can be used as a voltage controlled capacitor. It has three modes which are obtained due to variation in gate voltage, that is, (i) accumulation mode (ii) depletion mode (iii) inversion mode

38 × 10 −3 Hz = 1.47 × 1010 Hz 2 π (10 −14 + 4 × 10 −13 )

fb =

5 × 103 × 15 = 5 V (10 × 103 ) + (5 × 103 )

Rth = (10 × 103 ) || (5 × 103 ) =

fT =

Therefore,

7.

Vth =

Topic: BJT (b)  For the given BJT, gm = 38 mA/V, Cp = 4 × 10−13 F, Cμ = 10−14F and b0 = 90. gm fT = 2 π (C μ + C π ) Therefore, Now,

430 Ω

221

9.

σE = ωeE = 2 π f e0 er E where e0 is permittivity of free space and er is relative permittivity. Therefore, 10 −2 σ f = = = 45 MHz 2 πe0 er 2 π × 8.85 × 10 −12 × 4 Topic: P-N Junction (b)  As we know the reduction in forward bias voltage with rise in temperature is as following: dVD ≅ −2 mV/ °C dT So, on a rise of 20°C, the drop in voltage, = 2 × 20 = 40 mV Then, total diode voltage VD ≅ 700 − 40 = 660 mV Topic: BJT (d)  A transistor can be in operated in three regions – cut-off region, active region and saturation region. In cut-off region, the collector current IC is zero. In active region, the collector current varies linearly with the base current IB, that is, I c = bdc I B . IC increases with increase in base current IB. After sometime the transistor operates in saturation region and IC does not increase with the increase in IB. In this region, the collector current

12/4/2018 11:17:38 AM

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GATE ECE Chapter-wise Solved papers

being at higher potential than the substrate, implies that there is reverse bias potential between the source and the body (VSB > 0  V).

becomes constant ( I c sat = bdc I B min where I B min is the minimum base current IB which is required to keep the transistor in saturation region). Therefore, IC is less than or equal to bdc I B in saturation region.

Substrate 10. Topic: Zener Diode (a)  For Zener regulator,

D

Vin − Vo ≥ Iz + IL R Vin − 10 ≥ 1 × 10 −3 + 10 × 10 −3 R Vin − 10 ≥ 11 × 10 −3 R

G P S Induced N channel

Case (i): When Vin = 30 V

30 − 10 ≥ 11 × 10 −3 ⇒ R ≤ 1818 Ω (1) R



Case (ii): When Vin = 50 V 50 − 10 ≥ 11 × 10 −3 ⇒ R ≤ 3636 Ω (2) R

From Eqs. (1) and (2), we get

R ≤ 1818 Ω 11. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (d)  N-type silicon is obtained by doping silicon with Phosphorus. 12. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (b)  The band gap of silicon at 300 K is 1.10 eV. 13. Topic: Energy Bands in Intrinsic and Extrinsic Silicon

(d)  To achieve high-quality oxide growth, dry oxidation alone is employed. However, dry ­oxidation growth rate is slower than the wet oxidation. 17. Topic: Carrier Transport: Mobility and Resistivity (a)  We know that conductivity s = nemn + pemp

As, silicon-bar is N-type, conductivity is given by s ≈ nemn



Therefore, resistivity,



1 nemn and the resistance of the bar r =

R =

ni 2 1.5 × 1016 × 1.5 × 1016 = n 5 × 10 20 = 45 × 1010 = 4.5 × 1011/m3



14. Topic: P-N Junction (c)  Tunnel diode and avalanche photodiode are operated in forward bias and reverse bias respectively.



p=

15. Topic: MOSFET (d)  Consider the following figure. As the MOSFET is an N-channel enhancement type MOSFET, the source

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 222

Depletion region

The reverse-bias voltage will result in widening of the depletion region which in turn leads to reduction in channel depth as shown in the figure above. To return the channel to its former state, the gate-source voltage (VGS) has to be increased. Hence, increase in VSB, results in increase in threshold voltage Vth.

where ni is the intrinsic concentration Therefore,

N+

VSB

16. Topic: Integrated Circuit Fabrication Process: Oxidation

(a)  From the law of mass action, ni2 = np

N+

rl l = A nemnA

Substituting the given values, we get R =

10−3 5 × 1020 × 1. 6 × 10 −19 × 0. 13 × 100 × 10−12

= 0. 96 × 106 Ω ≅ 106 Ω

18. Topic: Carrier Transport: Diffusion Current (d)  The current density for N-type semiconductor is given by dn Jn = q mn n e + qD n (1) dx

12/4/2018 11:17:39 AM

Chapter 3  • Electronic Devices





Given that no electric field is present, so e = 0. Therefore, Eq. (1) reduces to dn dx

J n = qDn

(2)

From the problem, we have (0,10 ) (2 × 10−6, 6 × 1016)

(d)  For a MOSFET, the drain current (ID), gatesource voltage (VGS) and threshold voltage (VTh) are related as 16

16

dn 6 × 10 − 10 × 10 = dx 2 × 10 −4







Substituting in Eq. (2), we get

I D = K (VGS − VTh ) 2 = −2 × 1020

Hence,  1 × 10 −3 = K(0.9 − 0.4) 2 Therefore,

(

)

J = 1.6 × 10 −19 × 35 × 10−4 × −2 ×× 1020 A/m n = −1120 A/cm2

2

19. Topic: P-N Junction (c)  for silicon at low value of current h = 2. Therefore, for silicon diode, the diode current is given by V

I Si = I 0Si ( n DSi



ηVT

I Ge = I 0 Ge (e



(

VDSi 2VT

− 1) = I 0 Ge (eVDGe /VT − 1)

)

(

VDGe VT

− 1 = I 0 Ge e

−1

K=

1 × 10 −3 1 A/V 2 = mA/mV 2 0.25 25 × 10 4

When ID =

VGS = 1400 mV, then 1 × (1400 − 400) 2 = 4 mA 25 × 10 4

22. Topic: LED, Photo Diode, Tunnel Diode and LASER (c)  Option (c) gives the correct matching. 23. Topic: Integrated Circuit Fabrication Process

where, I0Ge is the reverse saturation current of the Ge diode. Given that the current through the two diodes are equal, therefore I 0Si e



− 1) = I 0Si (eVDSi / 2VT − 1)

where, I0Si is the reverse saturation current of the Si diode. h = 1 for germanium. Therefore, for germanium diode, the diode current is given by VDGe ηVT





21. Topic: MOSFET

(0,0)



1.24 l ° . Therefore, the Given that the LED emits at 5490 A −4 emission wavelength is 5490 × 10 mm. Therefore, band gap energy of the given semiconductor material (in eV) is 1.24 Eg = = 2.26 eV 5490 × 10 −4 Eg =



17

223

)

(b)  N-well implant, source/drain diffusion, metalization and passivation is the order of steps followed in N-well CMOS fabrication process. 24. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (c)  The impurity commonly used for realizing the base region of a silicon N-P-N transistor is Boron. 25. Topic: BJT (a)  Given that the BJT is a Si–NPN t­ransistor. as base-emitter voltage is 0.7V, the base-emitter junction is forward-biased. As the collector-base voltage is 0.2V, the base–collector junction is reverse-biased as shown in the following figure. Therefore, the give NPN transistor is o­ perating in normal active mode.

Substituting different values in the above equation, we get

( (

V

e DSi I 0Ge = V I 0Si e DGe

VT VT

)=e − 1) e

−1

0.718 2 × 26 ×10 −3 0.1435 26 ×10 −3

−1 −1

= 4 × 10

−

3

20. Topic: LED (a)  Band gap energy of a semiconductor material (in eV), which emits light at wavelength l (in mm) is given by

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 223

+

C

VCB = 0.2 V N P

B + N VBE = 0.7 V

−

E

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32. Topic: MOS Capacitor (d)  We have

26. Topic: BJT (d)  We have IC α = IB 1 − α When base width increases, recombination in base region increases and a decreases, hence b decreases. If doping in base region increases, then recombination in base increases and a decreases, thereby decreasing b. Hence, statement S1 is TRUE and statement S2 is FALSE.

b=



27. Topic: MOS Capacitor (c)  The given figure shows voltage transfer characteristics of a CMOS invertor. 28. Topic: Carrier Transport: Mobility (b)  We know that resistivity 1 ρ= nqμ n For the given sample, n = ND. Therefore, ND =

1 1 = = 1 × 1016/cm3 qμ n ρ 1.6 × 10 −19 × 1250 × 0.5

29. Topic: LED (a)  We have 1.24 1.24 eV = eV = 1.42 eV Eg = l ( μm ) 0.87 30. Topic: P-N Junction (b)  We know that WN N A = WP N D where, WN is the depletion width of n-side, WP is the depletion width of p-side, NA is the doping ­concentration of p-side and ND is the doping concentration of n-side. Therefore, WP =

WN × N D (3 × 10 −6 − WP ) × 1016 = NA 9 × 1016

Therefore, WP = 0.3 mm

31. Topic: P-N Junction (d)  We know that junction capacitance (Cj) is related to the applied reverse bias (VR) by

Therefore,

Hence,

VR 2 = = C j2 VR1 C j2 =

C j1 2

=

VTh = ϕGC − 2ϕ F −

QR Qox − Cox Cox

If Cox decreases, QB/Cox and Qox/Cox increases and VTh decreases. Also, Cox decreases when tox increases. Therefore, the threshold voltage (VTh of a MOS capacitor decreases with increase in the gate oxide thickness. Moreover, the threshold voltage (VTh) of a MOS capacitor decreases with increase in substrate doping concentration.

33. Topic: MOSFET

(d)  For a MOSFET, ID = K(VGS − VTh)2 Therefore, 1 × 10−3 = K(2 − 1)2. Hence, K = 1 mA/V2 For VGS = 3 V, drain current is given by I D = 1 × 10 −3 (3 − 1) 2 = 4 mA

34. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (c)  The band gap of silicon at room temperature is 1.1 eV. 35. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  The primary reason for the widespread use of silicon in semiconductor device technology is abundance of silicon on the surface of the Earth. 36. Topic: P-N Junction (b)  The reverse saturation current of a P−N junction diode, doubles itself for every 10 °C rise in ­temperature. Therefore, the reverse saturation current at 40°C is four times the reverse saturation current at 10°C. Therefore, reverse saturation current at 40°C is 40 pA. 37. Topic: Carrier Transport: Mobility (b)  Conductivity of an N-type semicon­ductor is

σ n = nqμ n 4 =2 1

1 × 10 −12 F = 0.5 pF 2

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 224

e ox tox

where tox is the gate oxide thickness, eox is the permittivity of the gate oxide and Cox is the value of MOS capacitor.

C j ∝ VR −1/ 2 C j1



Cox =



Conductivity of a P-type semiconductor is sp = pqmp



Therefore,

σp σn

=

μp μn

=

1 3

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Chapter 3  • Electronic Devices

38. Topic: P-N Junction (b)  Depletion capacitance of a diode ee A C= 0 r d where, d is the width of the depletion region and A is the area. The depletion capacitance of the diode per square metre is given by C/A. From the given data,

43. Topic: P-N Junction (c) The I−VD characteristics of a tunnel diode are shown in the following figure. I Peak point IP Negative resistance region

C e 0 e r 8.85 × 10 −12 × 11.7 = = F = 10.35 μ F A d 10 × 10 −6 ≅ 10 μ F

Valley point IV

39. Topic: MOS Capacitor (a)  In accumulation mode for N-channel MOS having P-substrate, VG is negative. When negative VG is applied to the gate electrode, the holes in the P-type substrate are attracted to the semiconductor oxide interface. This condition is called carrier accumulation on the surface. 40. Topic: MOSFET

VD VP



VGS − VTh = 2 − 1 = 1 V

As, VDS ≥ (VGS − VTh ), therefore N-channel MOSFET will operate in saturation region.

41. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (b)  We know that np = ni2, where ni is intrinsic carrier concentration. For N-type semiconductor, p is minority carrier concentration. Hence, for a N-type semiconductor. ni2 1 1 or p ∝ or p ∝ ND n n where, ND is the donor atom concentration. Similarly, for a P-type semiconductor, n is the minority carrier concentration and is given by p=



ni2 1 1 or n ∝ or n ∝ , p p NA where, NA is the acceptor atom concentration.

   n =

42. Topic: Carrier Transport: Diffusion Current (a)  Under low-level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the diffusion current.

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 225

From the above curve, we can see that the slope of I−VD characteristics is negative for Vp ≤ VD < Vv. This region is also referred to as the negative resistance region.

gm =

VGS = VG − VS = 3 − 1 = 2 V VDS = VD − VS = 5 − 1 = 4 V Also, from the figure, threshold voltage VTh = 1 V Then,

VV

44. Topic: MOSFET (c)  We have

(c)  From the given figure



225

∂I D ∂VGS

gm =

VDS = const

 V  and I D = I D SS 1 − GS  VP  

2I ∂I D = − DSS ∂VGS VP

2

 VGS   1 − V  P

So gm will be maximum when VGS = 0 and given by 2I gm 0 = − DSS VP

45. Topic: BJT (b)  The phenomenon known as ‘early effect’ in a bipolar transistor refers to a reduction of the effective base width caused by the reverse biasing of the base-collector junction. 46. Topic: Carrier Transport (b)  According to Hall effect, electric force + magnetic force = 0 Therefore, qE + qv × B = 0, or E = −v × B, or E = B × v 47. Topic: Carrier Transport: Mobility and Resistivity (d)  For N-type semiconductor, sn = nqmn For intrinsic semiconductor, si = niq (mn + mp) Therefore,

σn nμ n = σ i ni ( μ n + μ p ) =

4.2 × 108 × μ n = 20000 1.5 × 10 4 × μ n [1 + ( μ p μ n )]

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GATE ECE Chapter-wise Solved papers

48. Topic: P-N Junction (a)  Diode retains resistance of forward-bias condition in reverse bias (ideally zero resistance) for the time interval of storage time. Therefore, during storage time, VR = −5 V. 49. Topic: Zener Diode (c)  Option (c) is the correct match between Group 1 and Group 2. 50. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (d)  By the law of electrical neutrality, p + ND = n + NA Given that ND = 0, NA >> ni. Therefore, p = NA Using law of mass action, np = ni2

ni2 n2 = i or,      n = p NA

51. Topic: P-N Junction (c)  Electrical field is always maximum at the junction.

54. Topic: Photo- Diode and Solar Cell (b)  Option (b) is the correct match between Group 1 and Group 2. 55. Topic: MOSFET (c)  Option (c) is the correct match between Group 1 and Group 2. 56. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  Boron is an acceptor impurity, so silicon wafer doped with high concentration of boron is a P+ substrate. It may be mentioned here that N+ and P+ refers to heavily doped semiconductor such that its resistivity levels are of the order of few milli-ohm-cm. 57. Topic: P–N Junction (d)  Channel length modulation is not associated with a P–N junction 58. Topic: MOSFET (b)  We have

52. Topic: P-N Junction gm =

(a)  Junction potential = Built-in potential + Reversebias voltage. Therefore, Vj = Vo + VR, Now for abrupt P–N junction, depletion width



W = kVj1/2 Let the depletion width at reverse bias of 1.2 V be W1 and 7.2 V be W2. Therefore, W1 = k (1.2 + 0.8)1/2 and W2 = k (7.2 + 0.8)1/2



Therefore,



Hence,     W2 = 2W1 = 4 mm



(d)  The oxidation rate is zero as the existing oxide prevents further oxidation. 60. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (c)  As boron is P-type impurity, therefore Fermi level goes down. If Ei is the Fermi level of intrinsic silicon, then Ei − EF = kT ln



We know that, a = ( b *)g



where, b * = base transport factor and g = emitter injection efficiency Substituting values, we get b* =

a  50  1 =   = 0. 9853 ≈ 0. 985  51  0. 995 g

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 226

NA ni

17     = 25 × 10 −3 ln 4 × 10 = 0.427 eV 1.5 × 1010

b = 50 b 50 a = = b + 1 51

VDS = const .

∂K (VGS − VTh ) 2 ∂VGS

59. Topic: Integrated Circuit Fabrication Process: Oxidation

53. Topic: BJT (b)  Given that, Therefore,

=

= 2 K (VGS − VTh )

W2 k (8)1/ 2 = =2 W1 k ( 2)1/ 2



∂I D ∂VGS

61. Topic: Zener Diode (b)  S1 and S2 are both true but S2 is not a reason for S1. 62. Topic: MOSFET (c)  The equivalent gm is given by 1 1 1 = + gm gm1 gm2

From the given figure, we have that transistor M2 is always in saturation due to bias applied but transconductance of transistor M1 (gm1) changes in accordance with Vin.

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Chapter 3  • Electronic Devices



Now, gm 1 gm 2 gm = = gm 1 + gm 2



gm 1 gm 2  g  gm 2  1 + m 1  gm 2  

But gm2 >> gm1, therefore, gm = gm1

63. Topic: MOSFET (d)  Given that the NMOS transistor is operating in the linear region. Therefore, VDS < (VGS − VP). The transconductance gm is given by, gm =

ID VGS

or, g mVGS = I D which is an expression of hyperbola.

64. Topic: MOSFET (a)  The figure shown is of an N-channel JFET. Hence, the pinch-off voltage VP is positive. The transconductance gm is given as −2 I DSS ⎛ VG ⎞ gm = 1− VP ⎜⎝ VP ⎟⎠

Given that



Therefore,

VG = −2 V



67. Topic: P-N Junction (b)  Given that: ND = 1 × 1017 × 106/m3 = 1 × 1023/m3 Depletion width WN = 0.1 μm Depletion width WP = 1.0 μm ni = 1.4 × 1010 × 106/m3 = 1 × 1016/m3 We know that: WP N A = WN N D Therefore, N A = 1 × 10 22/m3 Built-in potential, kT ⎛ N A N D ⎞ Vo = ln q ⎜⎝ n2 ⎟⎠ kT = VT ( thermal voltage) = 26 mV q Therefore, Vo = 26 × 10 −3 ln

−2 I DSS VP

⎛ 2⎞ ⎜⎝1 + V ⎟⎠ P

From the above expression, it is clear that the gm is inversely proportional to pinch-off voltage VP. When the width W is increased then VP becomes more negative as more reverse bias is required at gate–drain junction to reach the pinch-off condition. Hence, as the width W is increased, magnitude of VP increases and the value of gm reduces. Therefore, the ratio of initial gm and that of modified gm is greater than 1.

Only option (a) is greater than 1 and values of other options is less than 1. Hence, (a) is the correct answer.



10 22 × 10 23 = 0.76 V (1.4 × 1016 ) 2

e = q × N D    (W = WN + WP) Therefore, e E = −∫ dx e 1.1×10 −6

=

1.6 ×10−19 ×1×1017 ×1.1×10−6 ×10 2 12×8.25×10 −14 = 0.146 mV/cm � ≅ 0.15 mV/cm



It is directed from N- to P-region, from positive charges of transition region dipole toward the negative charges.

69. Topic: MOSFET (d)  Consider the following figure of an N-channel MOSFET. D N+ P

G

D = VT μ

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 227

μ 1 = D VT

−e dx e

=

66. Topic: Carrier Transport: Mobility (a)  We know that

Therefore,

∫ 0

65. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (c)



i

where,

68. Topic: P-N Junction

g m (initial) >1 g m ( modified )

Therefore, the ratio of mobility to diffusion coefficient in a semiconductor has the units of (volt)-1.

(b)  We know that gm =





227

Substrate

N+ S

Inversion layer

As we can see from the figure, inversion layer width is more towards the source side and it is lesser towards

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228



GATE ECE Chapter-wise Solved papers

the drain side, so the inversion charge decreases from source to drain. Therefore, in order to attract electrons from source to drain, large positive voltage needs to be applied at drain terminal as compared to source terminal. Hence, VD > VS



I D1 = I D2 = 0. 125 mA 0.125 × 10 -3 A =

In other words, the channel potential increases from source to drain. This causes the inversion layer width to increase towards source side as compared to drain side. Hence, the inversion charge decrease from source to drain.

Common Solution for Questions 70 and 71: Topic: MOSFET 70. (d) and 71. (d) Assuming that both the transistors are in saturation, then the drain current of N-type MOSFET is

mP Co xW æ 1 ö ç VGS2 − VT hP − VDS2 ÷ VD S2 L 2 è ø

V −5ö æ 0. 125 ×10−3 = 1 ×10−3 ç 3 − 5 + 1 − D (V − 5) 2 ÷ø D è

or,        VDS > 0

As the two drain currents are equal, therefore

0. 25 ×10−3 = (− VD + 3)( VD − 5)

(

= −VD2 + 8VD − 15

)

61 =0 4 



or,



Solving the quadratic equation, we get

VD2 − 8VD +

61  8 ± 64 −  4 ×   4 3 VD = =4± V 2 2

For the value

5V M2

VD = 4 − N-MOSFET

3V

Vo M1

VGS

Using VDS = VD − 5 and VGS = 3 − 5 = 2, we have, VDS ≤ VGS − VThP



which implies that P-MOSFET will be in saturation. Hence, the valid solution for the quadratic equation is

P-MOSFET VD = 4 +

I D1 =

mN C oxW (VGS1 − VT hN )2 2L

1 × 10−3 (1. 5 − 1)2 2 = 0. 125 mA [usin g VGS1 = 1 . 5V ]



Also, drain current of P-type MOSFET is I D2 =

mP Co xW (VGS2 − VT hP )2 2L

1 × 10 -3 (3 − 5 + 1) 2 2 = 0. 50 mA



The two drain currents have to be equal to satisfy Kirchhoff’s current law, as the gate currents are zero. Further both the MOSFETs, cannot be in saturation simultaneously. N-MOSFET which has saturation current of 0.125 mA, cannot carry current of 0.50 mA. Hence P-MOSFET is not in saturation. Therefore, N-MOSFET is in saturation and P-MOSFET junction is in triode region.

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 228

This will result in

which implies that the P-MOSFET is in the triode region.

72. Topic: MOSFET (b)  At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon N-channel MOSFET is 1350 cm2/V s.

=



3 V 2

VDS ≥ VGS − VThP

=



3 V 2

73. Topic: Integrated Circuit Fabrication Process



(b)  To achieve high-quality oxide growth (i.e. a uniform film with good dielectric properties), dry oxidation alone is employed. Hence, when gate oxide is grown, dry oxidation process is used. Wet oxidation is used to grow field oxide because the quality of the dielectric properties of the field oxide is not as critical as they are for the gate oxide. Dry oxidation is slower than the wet oxidation.

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Chapter 3  • Electronic Devices

74. Topic: Generation and Recombination of Carrierss

IZ

(c)  Given that the sample is in thermal ­equilibrium, therefore

e=

VZ VR

v 1 = = 106 V/m = 10 kV/cm d 1 × 10 −6

75. Topic: Carrier Transport: Drift Current (a)  We have J = NDq mne Subsituting different given values in the above equation, we get J=  1016 × (1.6 × 10–19) × (1350) × (10 × 103) = 2.16 × 104 A/cm2 76. Topic: P-N Junction (c)  With increase in doping, the reverse breakdown voltage decreases and depletion width d also decreases. As depletion capacitance, C = εA/d, so depletion capacitance increases with increase in doping.

Breakdown region

Both zener and avalanche diodes are used in voltage stabilization circuits, to regulate the load voltage against variations in load current and input voltage. They are used in these applications, because in the breakdown region large change in diode current produces only a small change in the diode voltage.

81. Topic: MOSFET (c)  Given that for the upper transistor

77. Topic: BJT (b)  NE >> NB and NB > NC

W1 =4 L1

78. Topic: Carrier Transport: Drift Current (c)  Drift current density

So, VDS1 = 6 - Vx and VGS1 = 5 - Vx

J = neme Hence, drift current

VGS1 - VTh = 5 - Vx - 1 = 4 - Vx

I = nemeA

where, A is the cross-sectional area of the semi-­ conductor Therefore, I depends upon carrier concentration and electric field. 79. Topic: P-N Junction (d)  The rate of change of forward-bias v­ oltage of a P–N junction with temperature is given by dV = −2.5 mV/ °C dT

229

From the above equation, we see that for 10°C increase in temperature, forward voltage across P–N junction decreases by 25 mV.

80. Topic: Zener Diode (b)  The following figure shows the characteristics of a Zener diode.

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 229



Therefore, VDS1 > ( VGS1 − VTh ) Hence, the transistor is in saturation region. Also, given that for the transistor at the bottom W2 =1 L2 Here, the drain is connected to the gate, hence, the transistor is in saturation. The current flowing through both the transistors is the same. Hence, æ W öæ V − VTh ö mN C ox ç 1 ÷ ç GS1 ÷ 2 ø è L1 ø è

2

æW ö æV − VTh ö = mN C ox ç 2 ÷ . ç GS 2 ÷ L 2 ø è 2 ø è

2

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Substituting the different values in the equation above, we get



R2 = 600 ×

(5 − Vx −1) 2 (V −1) 2 4 (∵VGS2 = Vx − 0) =1 x 2 2



On simplifying the above equation, we get

(

)

4 Vx2 − 8Vx + 16 = Vx2 − 2Vx + 1 3Vx2

− 30Vx + 63 = 0

Therefore,





0.98 α = = 49 1 − α 1 − 0.98 IC = 49 × 20 × 10−6 + 50 × 0.6 × 10−6 = 1.01 mA

b=

83. Topic: MOSFET (c)  Given that the channel resistance Ro at tch = 100 µm is 600 Ω. Also, at VGS = 0 V, tch = 8 mm

R ∝

1 1 R ∝ A = tch W. Therefore, As R ∝ , and A tch 1 1 ,R∝ A tch (A is the cross-sectional area of the channel and W is the channel width.) Therefore, channel resistance R1 at VGS = 0 V R1 =

−6

(10 × 10 ) × 600 = 750 Ω (8 × 10 −6 )

84. Topic: MOSFET (c)  The depletion width of the channel on one side is directly proportional to the square-root of sum of the built-in voltage of the P+N junction and gate-source voltage. Therefore, W∝ V VGS1 + Vb j1 W2 = W1 VGS2 + Vbj2

= 1000 Ω

The diode in the given circuit is forward-biased and the current through the diode is V − 0.7 A 500 Substituting this value of i in the above equation we have

tch 2 = [10 − 2( 2)] μ m = 6 μ m 1 tch

R2 = 600 × Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 230

(10 ×10−6 ) (6 ×10−6 )



Therefore, current through the diode is i=

V − 0.7 3.8 − 0.7 = = 6.2 mA 500 500

86. Topic: MOSFET (d)  Given that A = 1 × 10−12 m2 and d = 10−6 m Therefore, V = Ad = 10−18 m3 = 10−12 cm3 Also, given that ND = n = 1019/cm3 and ni = 1010/cm3 Therefore, p=

ni 2 10 20 = = 10 /cm3 N D 1019

Therefore, holes in volume V is H = pV = 10−11



As the number of holes cannot be a decimal number, therefore, number of holes H = 0. 87. Topic: MOSFET (a)  Source-body junction capacitance Cj =

e A e0er A = d d

Here, er = 11.7 as the channel is of Si. A = 1 mm × 0.2 mm = 0.2 × 10−12 m2 d = Depletion width of P–N junction = 10 nm = 10−8 m Therefore, Cj =

W2 = 2 μ m R∝



(V − 0.7) − V  = 0 500 Solving the above equation, we get 11.4 V= = 3.8 V 3 10 − 1000



W2 = 4=2 W1

As,

(6 ×10−6 )

i=

IC = bIB + (1 + b )ICO



(10 ×10−6 )

85. Topic: P-N Junction (d)  Applying KVL to the given circuit, we get 10 − 1000i − V = 0

Vx = 3 V

82. Topic: BJT (d)  We have



Therefore, channel resistance R2 at VGS = −3 V is

8.9 × 10 −12 × 11.7 × 0.2 × 10 −12 = 2 fF 10 −8

88. Topic: MOSFET (a)  Gate-source capacitance Cg =

e1 A1 e 0 e r1 A1 = d1 d1

= 1000 Ω 12/4/2018 11:17:48 AM

Chapter 3  • Electronic Devices



er1 = 3.9 as between gate and source there is SiO2



A1 = 1 mm × d = 1 × 10  × 20 × 10  = 2 × 10 d1 = 1 nm = 10−9 m Therefore, −6

Cg =

−9

−14



m

2

8.9 × 10 −12 × 3.9 × 2 × 10 −14 = 0.7 pF 10 −9

231

Therefore, the wavelength of incident radiation should be less than 0.87 mm. Hence, (a) is the correct option.

94. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  We know that intrinsic concentration ni is given by ni ∝ T 3/ 2 e

− E g / kT

1 ni

89. Topic: P-N Junction (a)  In a forward-biased P−N junction diode, due to application of forward bias voltage, minority carriers are injected from either side of diode, followed by ­diffusion and finally recombination.



Also,



Therefore, from the graph, band gap energy graph of silicon can be estimated.

90. Topic: MOSFET (d)  In a MOSFET operating in the saturation region, the channel length modulation effect causes a decrease in the output resistance.

95. Topic: Carrier Transport: Mobility (13)  From Einstein relation,

91. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (d)  Donor impurity concentration, ND = 2.25 × 1015 atoms/cm3 Intrinsic concentration, ni = 1.5 × 1010/cm3 Since complete ionization has taken place, electron concentration n0 = ND = 2.25 × 1015/cm3 Hole concentration p0 is given by n 2 (1.5 × 1010 ) 2 = 1 × 105/cm3 p0 = i = n0 2.25 × 1015 92. Topic: P-N Junction (d)  Built in potential Vbi is given by N N Vbi = VT ln A 2 D ni



⎡ 5 × 1018 × 1 × 1016 ⎤ = 26 × 10 −3 ln ⎢ ⎥ 10 2 ⎣ (1.5 × 10 ) ⎦ = 0.859 V ≅ 0.86 V Depletion width W is given by W =

2e SiVbi ⎛ N A + N D ⎞ = 3.34 × 10 −5 cm q ⎜ N AND ⎟ ⎝ ⎠

93. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  Band gap energy is given by hc Eg = l Therefore,

lc =

6.62 × 10 −34 × 3 × 108 = 0.87 μm 1.42 × 1.6 × 10 −19

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 231

ρi ∝

Dp

μp

=

kT q

Therefore, Dp = 26 × 10–3 × 500 × 10–4 m2/s = 13 cm2/s 96. Topic: BJT (b)  An increase in the base recombination of a BJT will increase the breakdown voltage BVCEO. 97. Topic: Generation and Recombination of Carriers (d)  Recombination rate, R = B(nn0 + nn) (pn0 + pn) nn0 and pn0 = Electron and hole concentrations respectively under thermal equilibrium nn and pn = Excess electron and hole concentrations, respectively. 98. Topic: BJT (b)  When a CE amplifier’s emitter resistance is not bypassed, due to the negative feedback, the voltage gain decreases. As it has current-series feedback configuration, therefore, the input impedance increases. 99. Topic: P-N Junction (d) When Vi < −1.7 V, diode D1 is ON and diode D2 is OFF. Therefore, output voltage Vo = −1.7 V. When Vi > 2.7 V; diode D1 is OFF and diode D2 is ON. Therefore, output voltage Vo = 2.7 V. When −1.7 < Vi < 2.7 V, both the diodes D1 and D2 are OFF. Therefore, for this input voltage range, Vo = Vi. 100. Topic: Integrated Circuit Fabrication Process: Ion Implantation (d)  In CMOS technology, shallow P-well or N-well regions can be formed using low energy ion-implantation.

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GATE ECE Chapter-wise Solved papers

101. Topic: Integrated Circuit Fabrication Process (c)  In MOSFET fabrication, the channel length is defined during the process of poly-silicon gate patterning

106. Topic: BJT (381)  Given that the two collector currents are equal. Therefore, I C1 = I C 2 Hence,

102. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  Electron concentration, n≅

ni2 Vbi /VT e NA

(1.5 × 1010 ) 2 0.3/ 26 ×10−3 = (e ) (1 × 1016 )



VBE1

or,

1.6 × 10 −19 × 1016 × 0.2 × 10 −6 × 10 2 1.044 × 10 −12 = 30.66 kV/cm 104. Topic: P-N Junction (b) Given NA = 9 × 1016/cm3; ND = 1 × 1016/cm3 Total depletion width x = xn + xp = 3 mm e = 1.04 × 10−12 F/cm xn N A 9 × 1016 = = x p N D 1 × 1016



Therefore, xn = 9xp Total depletion width, xn + xp = 3 mm. Therefore, 9xp + xp = 3 mm. Hence, xp = 0.3 mm. Hence, Depletion width on the p-side = 0.3 mm. Maximum electric field qN A x p E= e 1.6 × 10 −19 × 9 × 1016 × 0.3 × 10 −4 V/cm = 1.04 × 10 −12

= 4.15 × 105 V/cm

105. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (225.2)  Number of holes/cm3 is given by p = NA − ND = 1 × 1018 − 1 × 1015 = 9.99 × 1017 Number of electrons/cm3 is given by n=

ni2 (1.5 × 1010 ) 2 = = 225.2 /cm3 p 9.99 × 1017

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 232

VT

=

IS2 I S1



IS2 I S1

As IS ∝ A, therefore, ⎡ 300 × 10 −6 × 300 × 10 −6 ⎤ VBE1 − VBE 2 = 26 × 10 −3 ln ⎢ −6 −6 ⎥ ⎣ 0.2 × 10 × 0.2 × 10 ⎦ VBE1 − VBE 2 = 381 mV or,

103. Topic: P-N Junction

=

(VBE1 −VBE 2 )

e

VBE1 −VBE 2 = VT ln

= 2.3 × 109 cm3

(30.66)  Given that Xn = 0.2 mm, eSi = 1.044 × 10−12 F/ cm and ND = 1016/cm3 Peak electric field, qN D X n E= e Si

VBE 2

I S1e VT = I S2 e VT

107. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (d)  Option (d) represents the energy band diagram for the biased N-type semiconductor. 108. Topic: BJT (5.785)  Given that: kT = 25 mV, IS = 10−13 A VBE = 0.7 V, q

⎛ IC ⎞ Transconductance, g m = ⎜ ⎟ ⎝ VT ⎠



I C = I S ⎡⎣eVBE /VT −1⎤⎦ −3 = 10−13 ⎡ e 0.7 / 25×10 −1 ⎤ = 144.625 mA ⎣ ⎦ Therefore, I 144.625 × 10 −3 gm = C = = 5.785 A/V VT 25 × 10 −3

109. Topic: BJT



(−237.76)  Given that: VBE = 0.7 V, b = 200, VT = 25 mV For the DC analysis of the circuit, the capacitors are considered as open circuit. 11 × 103 Base voltage, VB = 12 × = 3V 11 × 103 + 33 × 103



Emitter voltage, VE = 3 − 0.7 = 2.3 V



Emitter current, I E =



Base current, IB =



        = 11.334 μA

2.3 A = 2.271 mA (10 + 1 × 103 )

IE 2.271 × 10 −3 = (b + 1) 201

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Chapter 3  • Electronic Devices

Collector current IC = βIB = 200 × 11.334 × 10 −6 A = 2.26 mA Resistor, re =

113. Topic: BJT (b)  Small signal voltage gain

−3

25 × 10 = 10.98 Ω 2.2771 × 10 −3

AV =

Voltage gain vo −b RC = vi b re + (1 + b )( Rs )



Therefore, AV ≅

−200 × 5 × 10 ( 200 × 10.98) + ( 201)10 = −237.76

110. Topic: MOS Capacitor (2.4)  The peak electric field in the silicon region is



2 × 0.2 ESi = = 0.8 V/μ m 0.5 The peak electric field in the oxide region is Eox =

ESi e Si = 2.4 V/μm e ox

111. Topic: BJT (b)  Applying KVL in base-emitter loop, we get 5 − IB (50 × 103) − 0.7 = 0 5 − 0.7 A = 86 μ A 50 × 103



Therefore, I B =



Now, I C = b I B = 50 × 86 × 10 A = 4.3 mA



Assuming that the BJT is in saturation and applying KVL in the collector-emitter loop, we get 10 −VCE(sat) 10 − 0.2 RC = = = 2279 Ω IC 4.3×10−3 For RC ≥ 2279 Ω, the BJT is in saturation. Hence, option (b) is the correct option.



−6



(1.5)  Given that MOSFET M1 switch from saturation to linear region. Now, VDS = VGS - VTh; where VDS = Vout and VGS = Vin Therefore,VDS = Vout = Vin − VTh



1 W 2 Drain current I D = μ n Cox (VGS − VTh ) 2 L VDD − Vout ID = 10 × 103 1 2 = × 100 × 10 −6 × 2 × (VGS − 0.5) 2 Solving the above equations, we get Vin = 1.5 V

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 233

I C RE VT + I C R E

(∵ I C  I E )

For a nearly constant small signal voltage gain for a wide range of values of RE, ICRE >> VT

114. Topic: BJT (0.5)  Input impedance of CB amplifier, Z i = rc =

VT IE

As the signal is received from 50 Ω antenna, Zi = 50 Ω. Given that VT = kT/q = 25 mV. Therefore, 50 = Hence, I E =

25 × 10 −3 IE

25 × 10 −3 = 0.5 mA 50 Ω

115. Topic: MOSFET (500) Given VTh = −0.5 V, VGS = 2 V, VDS = 5 V, W/L = 100, Cox = 10 −8 F/cm 2 , mn = 800 cm2/V-s Drain current ID =

1 W μ n Cox [2(VGS −VTh )VDS −VDS 2 ] 2 L −1

⎡ ∂ ⎧1 ⎡ ∂I D ⎤ W ⎨ Cox ⎡⎣ 2 (VGS − VTh )VDS ⎥ = rds = ⎢ ⎢ L ⎣ ∂VDS ⎩ 2 ⎣ ∂VDS ⎦

}

⎤ − VDS 2 ⎤⎦ ⎥ ⎦

112. Topic: MOSFET



RE RE I E RE Vout = = = rc + RE VT VT + I E RE Vin + RE IE

3

=

233

−1

W W ⎡ ⎤ = ⎢ μ n Cox (VGS − VTh ) − μ n Cox VDS ⎥ L L ⎣ ⎦

−1

Therefore,

rds =

=

1 W μ n Cox (VGS −VTh −VDS ) L 1 = 500 Ω 800 ×10 ×100 ×( 2 + 0.5 − 5) −8

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116. Topic: P-N Junction (a)  A region of negative differential resistance is observed in the current-voltage characteristics of a silicon PN junction if both the P-region and the N-region are heavily doped.

121. Topic: BJT (2) Given that the VEB = 0.6 V, Therefore IB =

117. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (0.52)  The silicon sample doped with donor type impurity will act as N-type semiconductor. The resistivity is given by inverse of conductivity, that is

ρ=

1 1 = σ N N D qμ n

Substituting values for donor concentration, electron mobility and charge on electron, we have 1 1016 × 1.6 × 10 −19 × 1200 = 0.52 Ω- cm

ρ=

118. Topic: Carrier Transport: Diffusion Current kT = 25 mV and mn = (4000) Given q = 1.6 × 10 −19 C; q 2 1000 cm /V-s From Einstein relation, Dn kT = q μn Therefore, Dn = 25 × 10 −3 × 1000 cm 2 /V-s = 25 cm2/s Diffusion current density

μ n = 1200

(

122. Topic: Integrated Circuit Fabrication Process: Oxidation (d)  Dry oxidation is preferred to form the gate dielectric (SiO2) of MOSFETs. 123. Topic: Generation and Recombination of Carriers V 10 (100)  e= = = 10 V/cm d 1 Vd = mne = 1000 × 10 = 104 cm/s Vd =



T=

1 × 10 −2 L = 4 s = 100 μs Vd 10 × 10 −2

124. Topic: MOS Capacitor (4.33)

Cmax ⎡ X d max e ox ⎤ = ⎢1 + × ⎥ Cmin ⎣ tox es ⎦ ⎡ 100 1 ⎤ = ⎢1 + × ⎥ = 4.33 ⎣ 10 3 ⎦



125. Topic: Carrier Transport: Diffusion Current (28.59)  Given that Nd = 1 × 1015 cm−3; tp = tn = 100 µs; Va = 208 mV, Dp = 36 cm2/s, Dn = 49 cm2/s, ni= 1 × 1010 cm−3 Hole current density Jp = pn 0 =

q Dp × pn 0

τp

[e q Va

kT

− 1]

ni2 Nd

Therefore, Jp = =

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 234

L T

Therefore,

= 1016 × 1.6 × 10−19 × 1200 = 1.92 s cm−1

120. Topic: Generation and Recombination of Carriers (14)  The concentration of hole-electron pair at t = 1 μs = 1020 × 10−6 = 1014/cm3 Hence, x = 14

)

VEC = 3 − I C R C = 3 − 2 × 10 −3 × 500 = 2 V

σ N = N D qμ n

is

I C = b I B = 50 × 0.04 × 10 −3 A = 2 mA

dn dx = 1.6 × 10−19 × 25 × 1 × 1021 = 4000 A/cm2

cm 2 V-s The conductivity of silicon So,

voltage

3 − 0.6 = 0.04 mA 60 × 103

J = qDn

119. Topic: Carrier Transport: Mobility (1.92)  As per the graph, mobility of electrons at the concentration 1016/cm3 is 1200 cm/Vs .

emitter-base

qni2 Nd

Dp

τp

[e q Va

kT

−1]

1.6 ×10−19 ×10 20 × 1×1015 −3 ⎡ 208×10−3 ⎤ 36 26×10 e −1⎥ A/cm 2 ⎢ −6 100 ×10 ⎢⎣ ⎥⎦ 2

12/4/2018 11:17:55 AM

Jp =

Dp

qni2 Nd

τp

[e q Va

kT

Chapter 3  • Electronic Devices

−1]

1.6 ×10−19 ×10 20 × 1×1015 −3 ⎡ 208×10−3 ⎤ 36 26×10 −1⎥ A/cm 2 ⎢e −6 100 ×10 ⎢⎣ ⎥⎦ 2 = 28.59 nA/cm

=

⎛ qα x ⎞ ⎜ ⎟

Given that: n( x ) = 1015 e ⎝ kT ⎠ , so



x=0

= 1.5 × 1013 e − ( 0.3 ×10

= 1015 × 1.6 × 10 −19 × 1384.5 × E x

sin α max = n12 − n22 = (1.5) 2 − (1.4) 2 Therefore,

Substituting, we have J n (drift ) = −2.2 × 10

A/cm

amax = sin−1(0.5385) ≈ 32.6°

2

J = J n (diff) + J n (drift) = 0 A/cm 2

131. Topic: MOSFET (d)  We have

127. Topic: MOSFET (20)  Drain current is given by I D = I D ( sat ) (1 + lVDS ) where l is the channel length modulation coefficient. Output resistance, ro = Hence, ro =

dVDS 1 = l I D ( sat ) dI D

1 Ω = 20 kΩ 0.05 × 10 −3

128. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  As we see that the new Fermi level is above the mid band energy level. Therefore, the impurity added must be an n-type semiconductor. Hence, a pentavalent impurity is added and the resulting semiconductor is an n-type semiconductor. 129. Topic: Generation and Recombination of Carriers (a)  We have We know that

Gopt = 1.5 × 1020 cm−3 s−1

RDS =



Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 235

NA τP

1 μ n C (W /L)(VGS − VT )

Therefore, if VGS increases, then RDS should decrease. Hence, option (d) is incorrect.

132. Topic: P-N Junction (0.25)  We have 2 − VD 2 − 0.7 1.3 i2 = = = A 3 R2 6 × 10 6 × 103 i1 =

VD 0.7 = A R1 2 × 103

Therefore, iD = i2 − i1 =



1.3 0.7 − 3 6 × 10 2 × 103

which is negative and this implies that the diode is in OFF mode. Hence, i2 = i1 =

2 = 0.25 mA ( 2 × 103 ) + (6 × 103 )A

133. Topic: BJT (598.67)  R2 =

Gopt = R =

/ 0.1 × 10 −6 )

130. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (32.6)  We have

− kT 1 dn( x ) ⋅ ⋅ = −α = −0.1 V/cm q n( x ) dx −12

−6

= 7.46 × 1011 cm −3

2

J n (drift ) x = 0 = n(0)qμ n E x Ex =

Here pn = NA. At t = 0, hole concentration = pn = 1.5 × 1013 cm−3. At t = 0.3 ms, hole concentration is given by ∆p(t = 0.3 ms) Now,

= 3.846 × 1015 /cm 4

J n (diff) = 2.2 × 10 A /cm

ΔP (t ) = Pn e − t / τ P



−2

Hence,

NA 0.1 × 10 −6

N A = 1.5 × 1013 cm −3

126. Topic: Carrier Transport: Mobility dn( x ) J n (diff) = qDn (c)  dx

d ( n( x )) dx

Therefore, 1.5 × 10 20 =



235

VT ⎛ I1 ⎞ 26 × 10 −3 ⎛ 1 × 10 −3 ⎞ = ln ln I 2 ⎜⎝ I 2 ⎟⎠ 100 × 10 −6 ⎜⎝ 100 × 10 −6 ⎟⎠

= 598.67 Ω

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134. Topic: MOS Capacitor (c)  Since NMOS transistors are in parallel, the width of PMOS transistors should be halved while the width of NMOS transistors should not be changed. If any of the transistors is in the ON mode, the output becomes LOW. 135. Topic: MOS Capacitor (a)  When the semiconductor in a MOS is in inversion, the intrinsic energy level crosses over the Fermi level and goes as deep into the other side of the Fermi level at the interface as distant it is from the Fermi level in the bulk. Here, the intrinsic Fermi level is below EFS by Φ B in the bulk and above EFS by Φ B at the interface. Thus, the MOS is in inversion.



Now, the charge per unit volume in the depletion region is – qN a = −(1.6 × 10 −19 ) × 1017 C cm −3 = −1.6 × 10 −2 C cm −3



The charge per unit area Q in depletion region on p-side is –qN a X p . Q = −1.6 × 10−2 × 3.1 × 10−6   = −49.6 nC cm−2

138. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (c)  We have

136. Topic: Photo Diode and Solar Cell (21)  The total power incident on the solar cell is

VBR

Pin = 3 × 100 = 300 mW

VBR =

Fill factor is given by

VBR × ND = constant

I mVm = 0.7 (180 × 10 −3 )(0.5 V ) Therefore, I mVm � = 0.063 W � � = � 63 � mW F=

139. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (32.36)  We have ⎛ q⎞ E = ⎜ ⎟ N d+ X N ⎝ e⎠

Therefore, the maximum efficiency is I mVm 63 ×10 −3 × 100% = × 100% = 21% Pin 300 × 10 −3

Substituting values, we get ⎛ ⎞ 1.6 × 10 −19 50 × 103 = ⎜ × 1017 × X N −14 ⎝ 8.85 × 10 × 11.7 ⎟⎠ Therefore,

137. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (−49.6)  We have 1/ 2

XN  = 3.236 × 10−6 cm = 3.236 × 10−8 m = 32.36 nm 140. Topic: Generation and Recombination of Carriers (0.0133)  The current density equation is given by

⎡⎛ 2e Si ⋅Vbi ⎞ ⎛ 1 ⎞⎤ X p = ⎢⎜ ⎟⎥ ⎜ ⎟ ⎢⎣⎝ q ( N d /N a ) ⎠ ⎝ N a + N d ⎠ ⎥⎦ Now,

eE2 2qN D

Therefore,

I V F� = m m I SCVOC

η=

1 Doping concentration 1 ∝ ND

VBR ∝

J = Jdiffusion + Jdrift

⎛ kT ⎞ ⎛ N N ⎞ Vbi = ⎜ ⎟ ln ⎜ a 2 d ⎟ ⎝ q ⎠ ⎝ ni ⎠

NA(cm−3)

ln(NA) 1016

⎛ 1016 × 1017 ⎞ = 0.757 V = 26 × 10 ln ⎜ ⎝ (1.5 × 1010 ) 2 ⎟⎠

36.84

−3

1014



= 3.1 × 10−6 cm

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 236

2

1

1/ 2

⎤ ⎡ 2 × 12 × 8.85 × 10 −14 × 0.757 Now, X p = ⎢ −19 17 16 16 17 ⎥ ⎣ (1.6 × 10 )(10 / (10 ))(10 + 10 ) ⎦

32.23 0.001

0.01

µm(log scale)

When there is no net flow of current, we have J = 0; hence, for holes, we have

12/4/2018 11:17:58 AM

Chapter 3  • Electronic Devices

⎛ dp ⎞ − qD P ⎜ ⎟ + qm P ( pE ) = 0 ⎝ dx ⎠ ⎛ dp ⎞ − qD P ⎜ ⎟ = qm P ( pE ) ⎝ dx ⎠

Therefore,

E=

Therefore,

⎛ dN A ⎞ ⎜⎝ dx ⎟⎠

Therefore, VG1 − VT Q1 = VG 2 − VT Q2

= VT

d ln [ N A ( x )] dx

0.8 − VT 2 × 1011 = 1.3 − VT 4 × 1011

( p = NA )



Since it is given that the doping is measured in log scale, we get log10 ( x1 ) = 1 μ m 1 Therefore,   x1 = 10 μ m = 0.001 cm



Now,



142. Topic: MOS Capacitor (b)  For a MOS capacitor, we have (VG1 − VT ) ∝ Q

VT ⎛ dp ⎞ ⎜ ⎟ p ⎝ dx ⎠

V = T p

VT = 0.3 V and

Q3 = 6 × 1011 cm −2

⇒ x2 = 102 µm = 0.01 cm From the given data, we get

143. Topic: MOS Capacitor (1.2)  For the linear operation, we have

ln(1014) = 32.23

I D = k N′ (VGS − VT )VDS

ln(1016) = 36.84

dI D = k N′ VDS = g m dVGS

⎛ 36.84 − 32.23 ⎞ = 0.0133 kV/cm There, E = 0.026 ⎜ ⎝ 0.01 − 0.001 ⎟⎠

Substituting values, we get

141. Topic: P-N Junction (100) At t = 0+ and t > 0, the simplified circuit is as shown in the following figure.

10 A







0.5 × 10 −6 = k N′ (50 × 10 −3 ), therefore k N′ = 10 −5 dI D = k N′ (VGS − VT ) = gd dVDS 8 × 10 −6 = 10 −5 ( 2 − VT )

+ 10 µF −

VT = 1.2 V

VC

After the switch is opened at t = 0, the current flowing through the inductor at t = 0+ is equal to 10 A.

144. Topic: BJT (6.656)  We have diffusion current density in the base region as follows: ⎛d ⎞ J nB = qDn ⎜ nB ⎟ ⎝ dx ⎠

Energy stored in the inductor =

1 2 1 LI = × 10 −3 × 10 2 = 50 × 10 −3 J 2 2

This energy is transferred to the capacitor through the diode. Now, energy stored in the capacitor 1 = × 10 × 10 −6 × VC2 = 5 × 10 −6 × VC2 2 Therefore, 5 × 10 −6 × VC2 = 50 × 10 −3 VC =

50 × 10 −3 = 10 4 = 100 V 5 × 10 −6

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 237

VG 2 − VT Q2 = VG 3 − VT Q3 1.3 − 0.3 4 × 1011 = 1.8 − 0.3 Q3

log10(x2) = 2 µm

1 mH

237

Now,

dnB 0 −1014  =   = −2 × 1018/cm4 dx 0.5 ×10 −4 − 0

Also,

Dn = μnVT = 800 [cm2/s] × 26 × 10−3

Therefore, Dn = 20.8 cm2/s Thus, JnB = 1.6 × 10−19 × 20.8 × 2 × 1018

Therefore, JnB = 6.656 A/cm2 Since the area of emitter−base junction is 0.001 cm2, the area of base−collector junction is ABC = 0.001 cm2

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GATE ECE Chapter-wise Solved papers

Hence, the collector current is IC = ABC × JnB

148. Topic: BJT (c) Normalized Excess Carrier Concentratoion

= 0.001 × 6.656 = 6.656 mA 145. Topic: MOS Capacitor (1.6)  For the MOS capacitor shown in Figure I, since there are two capacitors of thickness t1 and t2 in series, we have CI = =

(e1 /t1 ) × � (e 2 /t 2 ) e1e 2 /t1t 2 � = (e1 /t1 )� + (e 2 /t 2 ) (e1t 2 + � e 2 t1 ) / (t1t 2 ) �



e1 t Eq

Since both the capacitors are of the same value, on comparing, we get

⎡⎛ 4 ⎞ ⎤ t Eq = ⎢⎜ ⎟ × 3 + 1⎥ nm ⎣⎝ 20 ⎠ ⎦ ⎡⎛ 3 ⎞ ⎤ 8 t Eq = ⎢⎜ ⎟ + 1⎥ nm = = 1.6 nm ⎝ ⎠ 5 ⎣ 5 ⎦ 146. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  Valency order Ga = 3

0

∆nE nE0

−1 Base (N)

149. Topic: BJT (c) Transconductance, gm =

I O Output current = Vi Input voltage

Therefore, it should have high input and output resistances.

150. Topic: BJT (d)  Miller effect increases input capacitance and thereby decreases the higher cut-off frequency. Cu

As = 5 Si = 4

C

GaAs crystal atom will form 5 bonds with Ga atom and every Ga atom will form 3 bonds with As atom. Note: Si acts as p-type dopant in As site. Si acts as an n-type dopant in Ga site.

147. Topic: P-N Junction (d) N D1 = 1018 /cm3 , N D 2 = 1015 /cm3 ,

RC

RL

RC

RL

E

Breaking the above circuit, we have

Cπ

Built in voltage is given by

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 238

gmVπ

Rπ

kT = 25 mV and n1 = 1 × 1010 /cm3 q

⎛N ⎞ ⎛ 1018 ⎞ Vbt = VT ln ⎜ D1 ⎟ ⇒ Vbt = 25 × 10 −3 ln ⎜ 15 ⎟ = 0.173 V ⎝ 10 ⎠ ⎝ N D2 ⎠

Collector (P)

The above figure shows the minority carrier distribution in the PNP transistor for the inverse active mode. In this case, the B-E junction is reverse-biased and the B-C junction is forward-biased. Electrons from the collector are now injected into the base. The gradient in the minority carrier electron concentration in the base is in the opposite direction compared with the forward active mode. So, the emitter and collector currents will change direction.

For the capacitor shown in Figure II, we have

t Eq = (e1 /e 2 )t 2 + t1

105

∆nC nC0

Emitter (P)

e1e 2 e1 = e1t 2 + e 2 t1 (e1 /e 2 )t 2 + t1

CII =

∆pB pB0

Cm

E

Now, Cm = Cu[1 + gm(RC || RL)]

12/4/2018 11:18:01 AM

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Chapter 3  • Electronic Devices



Total capacitance (equivalent),



Cf = C π + C m For 3-dB-frequency, 1 Cf Due to Miller effect, Cf increases. Hence, F3dB will decrease. F3dB ∝





Vd = μE





Hence, MOSFET is in saturation region and therefore it will act as current source because channel length modulation effect is considerable. Thus, output impedance will exist. So, MOSFET acts as a current source with finite output impedance.

152. Topic: BJT

Therefore, μn =



Current density is given by J = N0 μn qE ⇒ J =1017 × 20 × 1.6 × 10−19 ×



⎛ x⎞ GL = GLo ⎜1 − ⎟ , ⎝ LP ⎠ where GLo = 1017 /cm3 Also, DP = 100 cm 2/ Vs, τ p = 10 −4 sec, q = 1.6 × 10 −19 C . G(n) GLo

x LP

We know dp J p = + qDP (1) dx Now,

153. Topic: MOSFET

LP = DP τ P = 0.1 cm

(0.50) For N-channel MOSFET, VDS = 0.1 V, VGS = 0.7 V, VTH = 0.3 V, mnCox = 100 mA/V2 = 0.1 mA/V2, W/L = 50, VGS > VTH and VDS < VGS - VTH

N-Channel MOSFET is in linear region.    Therefore, transconductance for N-channel MOSFET is given by gm =



∂I 0 μC W = n ox VDS L ∂VGS

⇒ gm = 0.1 ´× 50 ´× 0.1 = 0.5 mA/V

154. Topic: Carrier Transport: Diffusion Current



(1.6) Given N0 = 1 ´× 1016 atoms/cm3, q = 1.6 ×´ 10−19 C, V = 5 V, d = 10−6 m = 10−4 cm Electric field is given by E=

5 V = −4 V/cm d 10

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 239

5 = 1.6 kA/cm 2 10−4

155. Topic: Generation and Recombination of Carriers (16.0) Given,



(c)  If the reverse bias across the base-­collector junction increases, then (i)  depletion width across base-collector junctions will increase (ii) effective base width will decrease (iii)  recombination of carrier in base region will decrease ⎛ I ⎞ (iv)  common-emitter current gain ⎜ b = C ⎟ will IB ⎠ ⎝ increase Therefore, option (c) is correct.

Vd 107 = = 20cm 2/Vs E 5 ×105



151. Topic: MOSFET (c)  Given that

VGS > VTH and VDS > VGS − VTH

Drift velocity is given by

and dp GLo = dx LP dp 1017 = = 1018 dx 0.1 Therefore, from Eq. (1), we have ⇒



J P = 1.6 × 10 −19 × 100 × 1010 = 16 A/cm 2 156. Topic: P-N Junction (10.0) p 14

p

n -3

10 cm

14

-3

10 cm

C1 Diode 1

16

n -3

10 cm

16

-3

10 cm

C2 Diode 2

12/4/2018 11:18:03 AM

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GATE ECE Chapter-wise Solved papers



Since



eA qNW 2 C= , Vbi = 2e W 1 1 N ∝ 2 and C ∝ W W 1 1 1 and C ∝ ( N ) 2 (N )2 ∝ W where N = NA || ND Then 1016 × 1016 = 5× 1015 /cm3 N 2 = N A2 || N D2 = 2 × 1016 1014 × 1014 N1 = N A1 || N D1 = = 5× 1013 /cm3 2 × 1014 Therefore,





1

then electric field triangle will be upwarded. Therefore, option (a) is the correct answer. 159. Topic: Energy Bands in Intrinsic and Extrinsic





1

C2 ⎛ N 2 ⎞ 2 ⎛ 5 × 1015 ⎞ 2 = =⎜ = 10 C1 ⎜⎝ N1 ⎟⎠ ⎝ 5 × 1013 ⎟⎠



157. Topic: Energy Bands in Intrinsic and Extrinsic Silicon (6.9)



efm

1 q1q2 4π e 0 r

1 2 mv v = 0 = 0 2 1 q1q2 Final potential energy ( PE) = 4π e 0 r Energy conservation theorem is (KE)initial + (PE)initial = (KE)final + (PE)final Final kinetic energy ( KE) =

⇒Þ r =

0.9 eV

Semiconductor 9 −19 2 2 −29 ⇒Þ r = 2 × 9 × 10 × ( −1.6 × 10 ) = 18 × 1.6 × 10 9.1 × 10 −31 × 1010 9.1 × 10 −21

eϕ m = 4.1 eV, e χ = 4 eV EC − EFP = 0.9 eV ⇒ eϕs = e χ + 0.9 = 4.9 eV

⇒Þ r =

ϕ m = 4.1 V, ϕs = 4.9 V ϕ ms = ϕ m − ϕs = 4.1 − 4.9 = −0.8 V Now,

and



VFB = ϕms

ϕ' − ox Cox

ϕ'ox = − 0.8 + 1 = 0.2 Cox

Hence, ϕ'ox = 0.2 × Cox = 0.2 × 34.515 × 10−9 = 6.9 nC/cm 2 158. Topic: P-N Junction (a)  If the left side is p-region and the right side is n-region, then electric field triangle will be downwarded. If the left side is n-region and the right side is p-region,

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 240

18 × 1.6 2 × 10 −8 = 5.063 × 10 −8 m 9.1

160. Topic: Photo Diode and Solar Cell (0.52)  Open circuit voltage of solar cell

e ox 3.9 × 8.85 × 10−14 = = 34.515 × 10−9 F/cm 2 tox 0.1 × 10−14

Therefore

2q1q2 4π e 0 mv 2

Here, q1 = q2 = e

EFP

Cox =

1 2 1 q1q2 mv = 2 4π e 0 r

⇒Þ

EC

Metal

=0 r →∞

ec

efs

EFm



( PE)initial =

Evac

Evac



Silicon (5.06)  Mass of electron, m = 9.11 × 10−31 kg; charge of electrons, e = -1.6 × 10−19 C; velocity of electron of q1(v) = 105 m/sec Initial kinetic energy, (KE)initial = 1/2 mv2 Initial potential energy,



⎛ J ⎞ VOC = VT ln ⎜1 + L ⎟ (1) JS ⎠ ⎝

where JL = photo current density, JS = reverse saturation current density and VT = thermal voltage. As intensity of the incident light is increased 20 times, J L1 = 20 J L As temperature is constant, J S′ = J S , VT′ = VT



Given that VOC = 0.451. From Eq. (1), ⎛J ⎞ ⎛ J ⎞ V = VT ln ⎜1 + L ⎟ ≈ VT ln ⎜ L ⎟ JS ⎠ ⎝ JS ⎠ ⎝

12/4/2018 11:18:05 AM

Chapter 3  • Electronic Devices



Now,

⇒

⎛ J′⎞ ⇒ VOC ′ ≈ VT ln ⎜ L ⎟ ⎝ JS ⎠

S

⇒ VOC ′ = 0.451 + (0.025 ln 20) = 0.525 V

Wdep2 Wdep1

=

Vbi − Vf 0.65 − Vf 0.6 ⇒ = Vbi 0.65 1

⇒ 0.65 − Vf = 0.6 0.65

⎛ 20 J L ⎞ ⇒ VOC ′ ≈ VT ln ⎜ ⎝ J S ⎟⎠ ⎛J ⎞ ⇒ VOC ′ ≈ VT ln ⎜ L ⎟ + VT ln 20 ⎝J ⎠

2e ⎛ 1 1 ⎞ + (Vbi − Vf ) ⎜ q ⎝ N A N D ⎟⎠

Wdep =

⎛ J′⎞ VOC ′ = VT′ ln ⎜1 + L ⎟ J S′ ⎠ ⎝

241

⇒ 0.65 – Vf = 0.36 × 0.65 ⇒ Vf = 0.416 V 164. Topic: Zener Diode (b) IZ min = 2 mA Vi = (6 ± 0.3) × 10 −3 = 5.7 V → 6.3 V I S(min) = I Z min + I L

161. Topic: BJT (50)  Consider the given figure, voltage gain of this combination of BJT is



V0 − RC = VS re1 + re 2 where, RC = 1 kΩ = 1000 Ω re1 =

VT V 26 × 103 = T = = 10 Ω I C1 I C 2 2.6 × 103



and



V V 26 × 103 = 10 Ω re 2 = T = T = I e1 I c 2 2.6 × 103 Therefore, V0 −1000 = = −50 Vs 20

162. Topic: P-N Junction (d)  Since the concentration of holes on the p-side is much greater than that in the n-side, a very large hole diffusion current tends to flow across the junction from the p to the n material. Hence an electric field must build up across the junction in such a direction that a hole drift current will tend to flow across the junction from n-side to p-side in order to counterbalance the diffusion current. After depletion region is formed and electric field is set-up; holes on p-side and electrons on n-side move away from junction. However, diffusion current components (before formation of depletion region) and drift current components (after formation of depletion region) are in the same direction. 163. Topic: P-N Junction (0.416) Vbi = 0.65 Wdep1 = 1 μ m at equilibrium Vf = ?

Ch wise GATE_ECE_CH03_Electronic Devices_Exp.indd 241

Wdep2 = 0.6 μ m

I S(min) = 2 × 10 −3 + Rmax =

5 = 7 mA 1 × 103

5.7 − 5 = 100 Ω 7 × 10 −3

6.3 − 5 = 13 mA 100 IZ max = IS max – IL = 8 mA Pdiss = VZ IZ max = 5 × 8 × 10-3 = 40 mW

For Rmax; I S max =

165. Topic: LED (b)  Using concept of VIBGYOR

lR > lG > lB 1 We know EG ∝ . Also Vbi ∝ EG . So l 1 Vbi ∝ ⇒ VR > 1. The open circuit small signal voltage gain is approximately

Vin Vout RE

(a) gmRE > VT

(c) gmro >> 1

(d) VBE >> VT (GATE 2014: 2 Marks)

137. For the common collector amplifier shown in the figure, the BJT has high b, negligible VCE(sat), and VBE = 0.7 V. The maximum undistorted peak-to-peak output voltage vo (in Volts) is .

Ch wise GATE_ECE_CH04_Analog Circuits.indd 273

vout

vin

(GATE 2014: 2 Marks)

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GATE ECE Chapter-wise Solved papers

140. In the circuit shown, the op-amp has finite input impedance, infinite voltage gain and zero input offset voltage. The output voltage Vout is

143. For the circuit with ideal diodes shown in the figure, the shape of output (vout) for the given sine wave input (vin) will be

R2 + +

R1

I1

−

vin 0

0.5T

Vout I2 (a) (a)

(a)  -I2 (R1 + R2)

(b)  I2R2

(c)  I1R2

(d)  -I1 (R1 + R2)

(GATE 2014: 2 Marks) (a) 141. Assuming that the op-amp in the circuit shown is ideal, 0.5T 0 Vo is given by

−

0.5T

T

(a) (b) (b) (c) 0.5T 0.5T 0.5T

T T T

(c) (c) (b) (d) 0 0

0.5T 0.5T

T T

Vo V2

0

0.5T

T

0 0

0.5T 0.5T

T T

0

0.5T

T

(b) (d) 0 00

T

(d) 0 0

0.5T 0.5T 0.5T

T T

0

0.5T

T

(d) (d)

+ (c) 2R

(b) 0

3R (a) (c)

R

−

−

+

V1

vout

T

R 0

0.5T

T

(GATE 2015: 1 Mark) (a)

5 5 V1 − 3V2 (b) 2V1 − V2 2 2

3 7 11 −3V1 + V2 (c) − V1 + V2 (d) 2 2 2

144. In the circuit shown the Zener diode is ideal and the Zener voltage is 6 V. The output voltage Vo (in Volts) is .

1 kΩ

(GATE 2014: 2 Marks) 142. In the voltage regulator circuit shown in the figure, the op-amp is ideal. The BJT has VBE = 0.7 V and b = 100, and the Zener voltage is 4.7 V. For a regulated output of 9 V, the value of R (in Ω) is . Vo = 9 V

Vi = 12 V + 1 kΩ

−

1 kΩ

10 V

1 kΩ

+ Vo −

(GATE 2015: 1 Mark) 145. An N-type silicon sample is uniformly illuminated with light which generates 10 20 electron-hole pairs per cm2 per second. The minority carrier life time in the sample is 1 μs. In the steady state, the hole concentration in the sample is approximately 10 x, where x is an integer. The value of x is . (GATE 2015: 1 Mark)

VZ = 4.7 V

R

(GATE 2014: 2 Marks)

Ch wise GATE_ECE_CH04_Analog Circuits.indd 274

146. In the circuit shown in the figure, the BJT has a current gain (b) of 50. For an emitter-base voltage VEB = 600 mV, the emitter collector voltage VEC (in Volts) is .

12/4/2018 11:23:43 AM

Chapter 4  • Analog Circuits

3V

275

R2 = 20 kW R1

− vout +

+ vin −

500 Ω

60 kΩ

(GATE 2015: 1 Mark) (GATE 2015: 1 Mark) 147. If the circuit shown has to function as a clamping circuit, which one of the following conditions should be satisfied for sinusoidal signal of period T ? C +

−

150. In the circuit shown using an ideal op-amp, the 3-dB cut-off frequency (in Hz) is . 10 kW 10 kW + vi vo − 0.1 mF 10 kW

10 kW

(GATE 2015: 1 Mark) R

V

151. The complex envelope of the band pass signal

(a) RC > T (GATE 2015: 1 Mark) 148. In the circuit shown, VO = VOA for switch SW in position A and VO = VOB for SW in position B. Assume that the V op-amp is ideal. The value of OB is . VOA

⎛ sin(p t / 5) ⎞ p⎞ ⎛ x (t ) = − 2 ⎜ sin ⎜ p t − ⎟ , centered about ⎟ ⎝ 4⎠ ⎝ p t /5 ⎠ 1 f = Hz is 2 ⎛ sin(pt / 5) ⎞ j p4 e (a) ⎜ ⎝ pt / 5 ⎟⎠ ⎛ sin(pt / 5) ⎞ − j p4 (b) ⎜ e ⎝ pt / 5 ⎟⎠ (c)

⎛ sin(pt / 5) ⎞ j p4 e 2⎜ ⎝ pt / 5 ⎟⎠

(d)

⎛ sin(pt / 5) ⎞ − j p4 e 2⎜ ⎝ pt / 5 ⎟⎠

1 kW

1 kW

5V

B

A

− VO

1 kW

152. A MOSFET in saturation has a drain current of 1 mA for VDS = 0.5 V. If the channel length modulation coefficient is 0.05 V−1, the output resistance (in kΩ) of the MOSFET is . (GATE 2015: 2 Marks)

+ SW

1 kW

1V

1 kW

(GATE 2015: 1 Mark) 149. In the bistable circuit shown, the ideal op-amp has saturation level of ± 5 V. The value of R1 (in kΩ) that gives a hysteresis width of 500 mV is .

Ch wise GATE_ECE_CH04_Analog Circuits.indd 275

(GATE 2015: 1 Mark)

153. For the NMOSFET in the circuit shown, the threshold voltage is Vth, where Vth > 0. The source voltage VSS is varied from 0 to VDD. Neglecting the channel length modulation, the drain current ID as a function VSS is represented by

12/4/2018 11:23:44 AM

276

GATE ECE Chapter-wise Solved papers

VDD

Vs I1

I2 −

+

V1

V2 V12

T2

VSS

T1

(GATE 2015: 2 Marks) (a) (a)

ID

(b) ID

155. The electric field profile in the depletion region of a p-n junction in equilibrium is shown in the figure, which one of the following statements is NOT TRUE?

VSS

VSS

VDD −Vth

Vth

(a) ID (b) (b) I (c) IDD

ID

VSS VDD −Vth

ID ID

E (V/cm) 104

(b) ID (d) ID

VDD −Vth Vth VDD −Vth (c) (b) ID (c) I (d) IDD

VSS VSS VSS

VSS VSS Vth VDD −Vth (d)

−0.1 0

0.5

1.0

X(mm)

(a) The left side of the junction is N-type and the right side is P-type (b) Both the N-type and P-type depletion regions are uniformly doped

ID

(c) The potential difference across the depletion region is 700 mV

VDD −Vth VDD −Vth

VSS VSS

VSS V VSS SS (d) (d)

ID

ID

156. An NPN BJT having reverse saturation current IS = 10−15 A is biased in the forward active region with VBE = 700 mV. The thermal voltage (VT) is 25 mV and the current gain ( b ) may vary from 50 to 150 due to manufacturing variations. The maximum emitter current (in mA) is .

VSS

VSS

VDD −Vth

VDD −Vth

Vth VDD −V Vth DD −Vth

VDD −Vth (GATE 2015: 2 Marks) 154. In the circuit shown I1 = 80 mA and I2 = 4 mA. Transistors T1 and T2 are identical. Assume that the thermal voltage VT is 26 mV at 27°C. At 50°C, the value of the voltage V12 = V1 − V2 (in mV) is .

Ch wise GATE_ECE_CH04_Analog Circuits.indd 276

VSS (d) If the P-type region has a doping concentration of 1015 cm−3, then the doping concentration in the N-type region will be 1016 cm−3 (GATE 2015: 2 Marks)

(GATE 2015: 2 Marks) 157. The diode in the circuit given below has VON = 0.7 V but is ideal otherwise. The current (in mA) in the 4 kΩ resistor is .

12/4/2018 11:23:45 AM

Chapter 4  • Analog Circuits

2 kW

277

161. The circuit shown in the figure has an ideal op-amp. The oscillation frequency and the condition to sustain the oscillations, respectively, are

3 kW D 1 kW

R1

1 mA 4 kW

R2

6 kW

− Vout +

(GATE 2015: 2 Marks)

2R 158. In the circuit shown, assume that diodes D1 and D2 are ideal. In the steady-state condition the average voltage Vab (in Volts) across the 0.5 mF capacitor is .

2C

1 µF

D1

D2

50 sin(wt) 0.5 µF b

C

R

(a)

1 1 and R1 = R2 (b) and R1 = 4 R2 CR CR

(c)

1 1 and R1 = R2 (d) and R1 = 4 R2 2CR 2CR (GATE 2015: 2 Marks)

a −

+ Vab (GATE 2015: 2 Marks)

162. In the circuit shown, assume that the op-amp is ideal. If the gain (vo/vin) is −12, the value of R (in kΩ) is .

159. In the circuit shown, assume that the diodes D1 and D2 are ideal. The average value of voltage Vab (in volts), across terminals ‘a’ and ‘b’ is .

R vin

D1 10 k W a +

6π sin(wt)

10 k W

Vab

−

+ b 20 k W

(GATE 2015: 2 Marks) 163. Assuming that the op-amp in the circuit shown below is ideal, the output voltage Vo (in Volts) is . 2 kW

+

12 V 1V

250 (1−δ) W −

− vo

160. In the circuit shown, assume that the op-amp is ideal. The bridge output voltage Vo(in mV) for d = 0.05 is .

250 (1+δ) W

10 kW

D2

(GATE 2015: 2 Marks)

100 W

10 kW

10 kW

−

1 kW

−

+

Vo

Vo 250 (1−δ) W

+ 250 (1+δ) W

−12 V

100 W

1V 50 W

(GATE 2015: 2 Marks)

Ch wise GATE_ECE_CH04_Analog Circuits.indd 277

(GATE 2015: 2 Marks) 164. For the voltage regulator circuit shown, the input voltage (Vin) is 20 V ± 20% and the regulated output (Vout)

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GATE ECE Chapter-wise Solved papers

is 10 V. Assume the op-amp to be ideal. For a load RL drawing 200 mA, the maximum power dissipation in Q1 (in watts) is . Q1 Vin

Vout Vref

4V +

R1

167. The I−V characteristics of three types of diodes at the room temperature, made of semiconductors X, Y and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. If Egx, Egy and Egz are the bandgaps of X, Y and Z, respectively, then I

RL

− X

Y

Z

R2 = 10 kW (GATE 2015: 2 Marks) V

165. Resistor R1 in the circuit below has been adjusted so that I1 = 1 mA. The bipolar transistor Q1 and Q2 are perfectly matched and have very high current gain, so their base currents are negligible. The supply voltage VCC is 6 V. The thermal voltage kT/q is 26 mV.

(a) (b) (c) (d)

Egx > Egy > Egz Egx = Egy = Egz Egx < Egy < Egz No relationship among these bandgaps exists. (GATE 2016: 1 Mark)

VCC 168. The figure shows the band diagram of a Metal Oxide Semiconductor (MOS). The surface region of this MOS is in

R1 I2

I1

R2 SiO2

The value of R2 (in Ω) for which I2 = 100 µA is

.

(GATE 2016: 1 Mark) 166. Which one of the following statements is correct about an AC-coupled common-emitter amplifier operating in the mid-band region? (a) The device parasitic capacitances behave like open circuits, whereas coupling and bypass capacitances behave like short circuits. (b) The device parasitic capacitances, coupling capacitances and bypass capacitances behave like open circuits. (c) The device parasitic capacitances, coupling capacitances and bypass capacitances behave like short circuits. (d) The device parasitic capacitances behave like short circuits, whereas coupling and bypass capacitances behave like open circuits. (GATE 2016: 1 Mark)

Ch wise GATE_ECE_CH04_Analog Circuits.indd 278

EM

ΦB ΦB

EC EFS Ei EV

(a) inversion (b) accumulation (c) depletion (d) flat band (GATE 2016: 1 Mark) 169. The diodes D1 and D2 in the below figure are ideal and the capacitors are identical. The product RC is very large compared to the time period of the AC voltage. Assuming that the diodes do not breakdown in the reverse bias, the output voltage Vo (in volt) at the steady state is .

12/4/2018 11:23:48 AM

Chapter 4  • Analog Circuits

thickness t1 = 1 nm and dielectric constant ε1 = 4) and Y (of thickness t2 = 3 nm and dielectric constant ε2 = 20). The capacitor in Figure II has only insulator material X of thickness tEq. If the capacitors are of equal capacitance, then the value of tEq (in nm) is .

D1

+ C

10 sin w t

279

R VO Metal

−

AC C

t2

2

t1

1

D2 (GATE 2016: 1 Mark)

Si

170. Consider the circuit shown in the following figure. Assuming VBE1 = VEB2 = 0.7 V, the value of the DC voltage VC2 (in volt) is . Metal

VCC = 2.5 V

ε1

β 1=100

tEq

Q2 Q1

β2 = 50 Si

10 kΩ VC2 1V 1 kΩ

(GATE 2016: 1 Mark)

(GATE 2016: 1 Mark) 171. The injected excess electron concentration profile in the base region of an npn BJT, biased in the active region, is linear, as shown in the figure. If the area of the emitter-base junction is 0.001 cm2, μn = 800 cm2/ (V-s) in the base region and depletion layer widths are negligible, then the collector current IC (in mA) at room temperature is . (Given: thermal voltage VT = 26 mV at room temperature, electronic charge q = 1.6 × 10−19 C.)

I −80 V −70 V

D1

IB p

n

173. The I−V characteristics of the zener diodes D1 and D2 are shown in Figure I. These diodes are used in the circuit given in Figure II. If the supply voltage is varied from 0 to 100 V, then breakdown occurs in

0—100 V

D2

n Figure I (a) (b) (c) (d)

0 0.5 µm

Figure II

D1 only D2 only Both D1 and D2 None of D1 and D2 (GATE 2016: 1 Mark)

(GATE 2016: 1 Mark) 172. Figures I and II show two MOS capacitors of unit area. The capacitor in Figure I has insulator materials X (of

Ch wise GATE_ECE_CH04_Analog Circuits.indd 279

D1

D2

IC

IE

V

174. Consider the constant current source shown in the figure below. Let β represent the current gain of the transistor. The load current I0 through RL is

12/4/2018 11:23:49 AM

280

GATE ECE Chapter-wise Solved papers

Vcc

1 nF 158 kΩ

R 158 kΩ

+VCC

1 nF Vref

− +

+ −

R2

−VCC 22.1 kΩ

RL

R1

100 kΩ ⎛ b ⎞ Vref ⎛ b + 1⎞ Vref (a) I 0 = ⎜ (b) I0 = ⎜ ⎟ ⎝ b + 1⎟⎠ R ⎝ b ⎠ R 10 kΩ ⎛ b + 1⎞ Vref ⎛ b ⎞ Vref (c) I 0 = ⎜ (d) I0 = ⎜ ⎝ b ⎟⎠ 2 R ⎝ b + 1⎟⎠ 2 R (GATE 2016: 1 Mark) 175. The following signal Vi of peak voltage 8 V is applied to the non-inverting terminal of an ideal op-amp. The transistor has VBE = 0.7 V, β = 100; VLED = 1.5 V, VCC = 10 V and −VCC = −10 V. The number of times the LED glows is . 10 V

10 V

100 Ω 8 kΩ

+Vcc − Vi

177. Consider an n-channel metal oxide semiconductor field effect transistor (MOSFET) with a gate-to-source voltage W of 1.8 V. Assume that = 4, μ N COX = 70 × 10 −6 AV −2 L the threshold voltage is 0.3 V, and the channel length modulation parameter is 0.09  V−1. In the saturation region, the drain conductance (in micro seimens) is . (GATE 2016: 2 Marks)

LED

+

2 kΩ

15 kΩ −Vcc

Vi 6V 4V 2V t −2 V −4 V −6 V (GATE 2016: 1 Mark) 176. Consider the oscillator circuit shown in the below figure. The function of the network (shown in dotted lines) consisting of the 100 kΩ resistor in series with the two diodes connected back-to-back is to:

Ch wise GATE_ECE_CH04_Analog Circuits.indd 280

(a) introduce amplitude stabilization by preventing the op-amp from saturating and thus producing sinusoidal oscillations of fixed amplitude. (b) introduce amplitude stabilization by f­orcing the op-amp to swing between positive and negative saturation and thus producing square wave oscillations of fixed amplitude. (c) introduce frequency stabilization by forcing the circuit to oscillate at a single frequency. (d) enable the loop gain to take on a value that produces square wave oscillations. (GATE 2016: 1 Mark)

178. The Ebers–Moll model of a BJT is valid (a) only in active mode. (b) only in active and saturation modes. (c) only in active and cut-off modes. (d) in active, saturation and cut-off modes. (GATE 2016: 2 Marks) 179. A long-channel NMOS transistor is biased in the linear region VDS = 50 mV and is used as a resistance. Which one of the following statements is NOT correct?

12/4/2018 11:23:50 AM

Chapter 4  • Analog Circuits

(a) If the device width W is increased, the resistance decreases. (b) If the threshold voltage is reduced, the resistance decreases. (c) If the device length L is increased, the resistance increases. (d) If VGS is increased, the resistance increases. (GATE 2016: 2 Marks) 180. Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of VGS and VDS. Given, gm = 0.5 µA/V for VDS = 50 mV and VGS = 2 V, gd = 8 µA/V for VGS ∂I ∂I = 2 V and VDS = 0 V, where g m = D and gd = D . ∂VGS ∂VDS The threshold voltage (in volts) of the transistor is . (GATE 2016: 2 Marks) 181. The figure shows a half-wave rectifier with a 475 µF ­filter capacitor. The load draws a constant current Io = 1 A from the rectifier. The figure also shows the input voltage Vi, the output voltage VC and the peak-to-peak voltage ripple u on VC. The input voltage Vi is a ­triangle-wave with an amplitude of 10 V and a period of 1 ms.

Vi

+ −

475 µF

100 µF

S t=0

+10 V

− +

10 kΩ

−

Vo −10 V

Z1 4 kΩ

Z2

1 kΩ 0V

0V

The time t = t1 (in seconds) at which Vo changes state is ______. (GATE 2016: 2 Marks) 183. An op-amp has a finite open loop voltage gain of 100. Its input offset voltage Vios ( = +5 mV) is modeled as shown in the circuit below. The amplifier is ideal in all other respects. Vinput is 25 mV. 15 kΩ

1 kΩ

+ − + −

− A0=100 + Vios = 5 mV Vinput

The output voltage (in millivolts) is ______. (GATE 2016: 2 Marks) 184. An ideal op-amp has voltage sources V1, V3, V5, …, VN−1 connected to the non-inverting input and V2, V4, V6, …, VN connected to the inverting input as shown in the figure below (+VCC = 15 V, −VCC = −15 V). The voltages V1, V2, V3, V4, V5, V6, … are 1, −1/2, 1/3, −1/4, 1/5, −1/6, …, V, respectively. As N approaches infinity, the output voltage (in volt) is .

t

−10 V VC

u

V2 10 kΩ 10 kΩ V4

10 kΩ VCC

VN

0

470 Ω

−10 V

Io =1 A

+10 V Vi 0

+10 V

+ VC

281

10 kΩ

t

1 kΩ

The value of the ripple m (in volts) is . (GATE 2016: 2 Marks)

V1

182. In the op-amp circuit shown, the Zener diodes Z1 and Z2 clamp the output voltage Vo to +5 V or –5 V. The switch S is initially closed and is opened at time t = 0.

VN−1

Vo VCC

1 kΩ V3 1 kΩ

Ch wise GATE_ECE_CH04_Analog Circuits.indd 281

1 kΩ

(GATE 2016: 2 Marks)

12/4/2018 11:23:51 AM

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GATE ECE Chapter-wise Solved papers

185. A p-i-n photodiode of responsivity 0.8 A/W is connected to the inverting input of an ideal op-amp as shown in the figure, +VCC = 15 V, −VCC = −15 V, load resistor RL = 10 kΩ. If 10 µW of power is incident on the photodiode, then the value of the photocurrent (in µA) through the load is ________. 1 MΩ VCC Vo 1 MΩ VCC

10 kΩ

VCC (GATE 2016: 2 Marks) 186. In the astable multivibrator circuit shown in the following figure, the frequency of oscillation (in kHz) at the output pin 3 is . VCC

RA = 2.2 kΩ 7

8

4

VCC

Res

(a) Forward active (c) Inverse active

(b)  Saturation (d)  Cutoff (GATE 2017: 1 Mark)

188. A good transconductance amplifier should have (a) high input resistance and low output resistance (b) low input resistance and high output resistance (c) high input and output resistances (d) low input and output resistances (GATE 2017: 1 Mark) 189. The Miller effect in the context of a common emitter amplifier explains (a) an increase in the low-frequency cutoff frequency (b) an increase in the high-frequency cutoff frequency (c) a decrease in the low-frequency cutoff frequency (d) a decrease in the high-frequency cutoff frequency (GATE 2017: 1 Mark) 190. For the DC analysis of the Common-Emitter amplifier shown, neglect the base current and assume that the emitter and collector currents are equal. Given that VT = 25 mV, VBE = 0.7 V, and the BJT output resistance ro is practically infinite. Under these conditions, the midband voltage gain magnitude, AV = |Vo/Vi| V/V, is . VCC = 12 V 73 kΩ

R1

RC

2 kΩ

3

6

+

10 µF

Out

C2

2 C1

Gnd 2 kΩ

C = 0.022 µF

Vi

1

47 kΩ

(GATE 2016: 2 Marks)

Normalized Excess Carrier Conecntration

187. For a narrow base PNP BJT, the excess minority carrier concentrations (∆nE for emitter, ∆pB for base, ∆nC for collector) normalized to equilibrium minority carrier concentrations (nE0 for emitter pB0 for base, nC0 for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in?

105 pB pB0 0

nC nC0

nE −1 nE0 Emitter (P) Base (N) Collecter (P) X and Y axes are not to scale

Ch wise GATE_ECE_CH04_Analog Circuits.indd 282

R2

RE

CE 100 µF

RL 8 kΩ

V0

-

(GATE 2017: 1 Mark) 191. An n-channel enhancement mode MOSFET is biased at VGS > VTH and VDS > (VGS - VTH), where VGS is the gateto-source voltage. VDS is the drain-to-source voltage and VTH is the threshold voltage. Considering channel length modulation effect to be significant, the MOSFET behaves as a (a) voltage source with zero output impedance (b) voltage source with non-zero output impedance (c) current source with finite output impedance (d) current source with infinite output impedance (GATE 2017: 1 Mark) 192. An npn bipolar junction transistor (BJT) is operating in the active region. If the reverse bias across the basecollector junction is increased, then

12/4/2018 11:23:52 AM

Chapter 4  • Analog Circuits

(a) the effective base width increases and commonemitter current gain increases (b) the effective base width increases and commonemitter current gain decreases (c) the effective base width decreases and commonemitter current gain increases (d) the effective base width decreases and commonemitter current gain decreases (GATE 2017: 1 Mark)

283

(c) 7.5 V and −21.2 V (d) 6.1 V and −22.6 V (GATE 2017: 1 Mark) 196. Consider the circuit shown in the figure. Assume that base-to-emitter voltage, VBE, is 0.8 V and common-base current gain (α) of the transistor is unity. +18 V

193. Consider an n-channel MOSFET having width W, length L, electron mobility in the channel µn and oxide capacitance per unit area Cox. If gate-to-source voltage VGS = 0.7 V, drain-to-source voltage VDS = 0.1 V, µnCox = 100 µA/V 2, threshold voltage VTH= 0.3 V and W/L = 50, then the transconductance gm (in mA/V) is .

44 kΩ

44 kΩ

16 kΩ

2 kΩ

(GATE 2017: 1 Mark) 194. The output Vo of the diode circuit shown in the figure is connected to an averaging DC voltmeter. The reading on the DC voltmeter in Volts, neglecting the voltage drop across the diode, is .

The value of the collector-to-emitter voltage VCE (in volt) is . (GATE 2017: 1 Mark)

+

1Ω

10 sinwt

197. For the operational amplifier circuit shown, the output saturation voltages are ±15 V. The upper and lower threshold voltages for the circuit are, respectively,

Vo

t = 50 Hz

-

(GATE 2017: 1 Mark) 195. In the figure, D1 is a real silicon pn-junction diode with a drop of 0.7 V under forward bias condition and D2 is a Zener diode with breakdown voltage of −6.8 V. The input Vin(t) is a periodic square wave of period T, whose one period is shown in the figure.

T

Vin(t)

t (sec) −14 V

10 Ω

D1 Vout(f) D2

Vout

−

+

10 kΩ

+ −

3V

(a) +5 V and -5 V (b) +7 V and -3 V (c) +3 V and -7 V (d) +3 V and -3 V (GATE 2017: 1 Mark)

+14 V 0

−

5 kΩ

10 µF

Vin(t)

Vin+

198. In the figure shown, the npn transistor acts as a switch. +5 V

4.8 kΩ

Vin(t)

Assuming 10τ 1.0 V.

The voltage (in volts, accurate to two decimal places) at Vx is (GATE 2018: 2 Marks)

Answer Key 1. (a)

2. (d)

3. (c)

4. (a)

5. (b)

6. (c)

7. (d)

8. (a)

9. (d)

10. (c)

11. (a)

12. (c)

13. (c)

14. (b)

15. (b)

16. (a)

17. (c)

18. (b)

19. (a)

20. (c)

21. (c)

22. (a)

23. (d)

24. (d)

25. (d)

26. (a)

27. (b)

28. (c)

29. (d)

30. (a)

31. (b)

32. (c)

33. (b)

34. (a)

35. (d)

36. (c)

37. (a)

38. (b)

39. (c)

40. (a)

41. (d)

42. (b)

43. (c)

44. (d)

45. (a)

46. (a)

47. (d)

48. (b)

49. (b)

50. (c)

51. (b)

52. (b)

53. (a)

54. (b)

55. (c)

56. (c)

57. (d)

58. (b)

59. (a)

60. (a)

61. (d)

62. (d)

63. (b)

64. (c)

65. (b)

66. (a)

67. (d)

68. (c)

69. (b)

70. (d)

71. (a)

72. (b)

73. (c)

74. (a)

75. (c)

76. (d)

77. (d)

78. (a)

79. (c)

80. (c)

81. (c)

82. (a)

83. (d)

84. (b)

85. (c)

86. (b)

87. (b)

88. (b)

89. (c)

90. (a)

91. (d)

92. (d)

93. (a)

94. (b)

95. (a)

96. (b)

97. (b)

98. (b)

99. (b)

100. (a)

101. (d)

102. (b)

103. (a)

104. (b)

105. (c)

106. (a)

107. (c)

108. (d)

109. (b)

110. (a)

111. (b)

112. (b)

113. (d)

114. (c)

115. (a)

116. (c)

117. (8)

118. (a)

119. (d)

120. (1.075)

121. (1.125) 122. (b)

123. (d)

124. (d)

125. (b)

126. (b)

127. (b)

128. (34.722)

129. (d)

130. (3.18)

131. (500)

132. (0.07)

133. (2.4)

134. (b)

135. (1.5)

136. (b)

137. (9.4)

138. (-237.76) 139. (-1)

140. (c)

141. (d)

142. (1093)

143. (c)

144. (5)

145. (14)

146. (2)

147. (d)

148. (1.5)

149. (1)

150. (159.15)

151. (c)

152. (20)

153. (a)

154. (83.88)

155. (c)

156. (1475)

157. (0.6)

158. (100)

159. (5)

160. (250)

161. (d)

162. (1)

163. (12)

164. (2.8056) 165. (598.67) 166. (a)

167. (c)

168. (a)

169. (0)

170. (0.5)

171. (6.656) 172. (1.6)

173. (a)

174. (b)

175. (3)

176. (a)

177. (28.35) 178. (d)

179. (d)

180. (1.2)

181. (2.105) 182. (0.798)

183. (413.79) 184. (15)

185. (-800)

186. (5.65)

187. (c)

188. (c)

189. (d)

190. (128)

191. (c)

192. (c)

193. (0.5)

194. (3.18)

195. (a)

196. (6)

197. (b)

198. (0.9)

199. (2)

200. (0.52)

201. (b)

202. (c)

203. (50)

204. (44.37)

205. (1.145)

206. (c)

207. (c)

208. (d)

209. (0.5)

210. (0.4226)

Ch wise GATE_ECE_CH04_Analog Circuits.indd 286

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Chapter 4  • Analog Circuits

Answers with Explanation 1.

Topic: BJT and MOSFET Amplifiers (a)  Common mode gain, AC =

ate a large range of frequencies. This oscillator is based on a bridge circuit originally developed by Max Wein in 1891 for the measurement of impedances.

α RC REE

8.

If REE = ∞, then, AC = 0 2.

3.

Topic: Simple Op-Amp Circuits (d)  As the feedback is applied on non-inverting terminal of the op-amp, so this circuit works as a Schmitt trigger. Schmitt trigger gives output voltage equal to the saturation voltage. Hence, Vo = +Vsat = +15 V Topic: BJT and MOSFET Amplifiers (c)  A resistor is introduced in the emitter of common amplifier to stabilize its dc operating point against variations in both temperatues and b of the transistor.

4.

Topic: Single-Stage BJT and MOSFET Amplifiers (a)  Capacitance effect is introduced in a bipolar transistor at high frequencies. Due to this, the current gain of bipolar transistor reduces at higher frequencies.

5.

Topic: BJT and MOSFET amplifiers (b)  For a stable amplifier, the poles in its open-loop transfer function should be on left half of s-plane.

6.

Topic: Simple Op-Amp Circuits (c)  Due to virtual ground, the voltage at node A will be zero.

Topic: Function Generators, Wave-Shaping Circuits and 555 Timers (a) R C

−

Vi A

Vo

+

This circuit acts as a differentiator and the output voltage of the differentiator is given as following: dVi dt So, the output voltage is the differentiation of applied input voltage. The following figure shows the output of different waves applied to a differentiator. Vo = − RC

Input signal Input signal

Output signal Output signal Spikes Spikes

Square wave Square wave

C V1 sinw t

C

V2 sinw t

C

Rectangular Rectangular A − +

Vo

Triangular wave Triangular wave

Now, applying KCL at node A, we get: 0 − V1 sin w t 0 − V2 sin w t 0 − Vo + + =0 1/sC 1/sC 1/sC ⇒ − sC (V1 sin w t ) − sC (V2 sin w t ) − sCVo = 0 ⇒ Vo = −(V1 + V2 ) sin w t 7.

Topic: Sinusoidal Oscillators (d)  The configuration shown in the figure is Weinbridge oscillator. It generates sine waves and can gener-

Ch wise GATE_ECE_CH04_Analog Circuits.indd 287

Sine wave Sine wave

Cosine wave Cosine wave

Therefore, differentiation of a triangular wave is a square wave.

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GATE ECE Chapter-wise Solved papers

  (i)  infinite input resistance Rin = ∞   (ii)  zero output resistance Ro = 0 (iii)  infinite open loop gain A = ∞   (iv)  infinite common mode rejection ratio; CMRR = ∞

Topic: Active Filters (d) At w → 0 Rs

+ RL Vo −

Vs

12. Topic: Function Generators, Wave-Shaping Circuits and 555 Timers (c)  In case of an astable multivibrator, neither of the two states is stable. Both output states are quasi stable. The output switches from one state to the other and the circuit functions like a free running square wave oscillator. In a bistable multivibrator both LOW and HIGH output states are stable. It can be used for storing and counting binary information.

At w = 0, VO = AVs (Pass band) At w → ∞ +

Rs Vs

13. Topic: Simple Diode Circuits (c)  The given circuit for transistor shunt regulator can be represented as

RL Vo −

At w = ∞, VO = AVs (Pass band) At w →

I1 20 Ω

1 √LC +

Vs

Iz

Vin = 20 − 30 V

+

Vo

1 LC

Vo = 10 V −

RB

− At w =

Ic

IE

Vin max − Vo 20 30 − 10 = = 1A 20 Given that Vo = 10 V, VZ= 9.5 V, VBE = 0.3 V, b = 99. From the circuit, I1 max =

, VO = 0 (Stop band)

Gain

Stop band

I E = I B + I C = ( b + 1) I B IB = IZ Pass band w =0

I E = 100 I Z

Pass band w =∞

w

Hence, the given circuit is band reject filter. 10. Topic: Simple Op-Amp Circuits (c)  We have Vo= ±Vio × A Here, Vio = 5 mV and A = 10,000; therefore, Vo = ±5 × 10-3 × 10000 V = ± 50 V However, as the supply of the opamp is ± 15 V, the output voltage is limited to ± 15 V. 11. Topic: Simple Op-Amp Circuits (a)  An ideal op-amp has following characteristics:

Ch wise GATE_ECE_CH04_Analog Circuits.indd 288

As I1 = I E . Therefore I1 = 100 I Z = 1 (∵ I1 = 1 A) ⇒ I Z = 0.01 A I C = b × I Z = 99 I Z = 0.99 A PZ = VZ I Z = 9.5 × 0.01 = 95 mW PT = VC I C = 10 × 0.99 = 9.9 W 14. Topic: Sinusoidal Oscillators (b)  Colpitts oscillator uses a pair of capacitors and an inductor in the tank circuit to produce the regenerative feedback signal. The frequency of oscillation is given by f osc =

1 2p LCeq

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Chapter 4  • Analog Circuits

where Ceq =

Applying KCL at node (3),

C1C2 C1 + C2

V3 − V2 dV =C o 100 dt

Substituting values: f osc =

1 2 × 10 −12 × 2 × 10 −12 × 10 × 10 −6 2p 2 × 10 −12 + 2 × 10 −12

Vo = −

17. Topic: BJT and MOSFET Amplifiers: Feedback (c)  In voltage series amplifier de-sensitivity factor (D) is given by D = 1 + Ab Where, b is feedback factor and A is open loop gain of amplifier. Therefore, Rif = Ri (1 + Ab ) ⇒ Increases

− Vo

+

Rof =

But Vo = AOL (Vi ) ⎛ V ⎞ ⎛ V ⎞ Vs − ⎜ o ⎟ ⎜ o ⎟ − Vo ⎝ AOL ⎠ ⎝ AOL ⎠ Hence, = R1 R2 The actual voltage gain for an op-amp based inverter amplifier is ACL =

Vo AOL ⋅ R2 ≅− =− Vs ( R1 + R2 ) + R1 AOL

R2 ⎛ ( R1 + R2 ) ⎞ R1 + ⎜ ⎟⎠ ⎝ A

10 × BW = 1 × 106 BW = 100 kHz

19. Topic: BJT and MOSFET Amplifiers

16. Topic: SimpleOp-Amp Circuits

(a)     f L = 20 Hz, f H = 1 kHz

(a) 10 µF

10 mH 100 Ω V2

18. Topic: Simple Op-Amp Circuits (b)  -3 dB bandwidth is the range of frequency at which the gain of amplifier drops by 3dB. Gain in dB = 20 log (gain) ⇒ 20 dB = 20 log (gain) ⇒ Gain = 10 ⇒ ⇒

100 × 10 × 103 ≈ −9 11 × 103 + 1 × 103 × 100

10 Ω Vs Vs = 10 cos(100t)

V3

Vo

For cascaded connection, the lower and higher frequencies are given by: fL f L* = , f H* = f H 21/n − 1 1 /n 2 −1 Where n is the number of amplifiers in cascade. Therefore, f L* =

Applying KCL at node (1), Vs 1 = − ∫ V2 dt 10 L



Ch wise GATE_ECE_CH04_Analog Circuits.indd 289

Vs = −

Ro ⇒ Decreases 1 + Ab

Given, Gain × BW = 1 × 106

OL

ACL = −

1 V2 dt 10 −3 ∫

Vo = 10 cos 100t

R2 = 10 kΩ

Vi

dVo dt

From Eq. (1),

Vs − Vi Vi − Vo = R1 R2

R1 = 1 kΩ

−V2 = 100C

= 79.6 MHz

15. Topic: Simple Op-Amp Circuits (b)  Apply KCL at the virtual ground (short)

Vs

289

10 V2 dt (1) L∫

20 1/ 3

2

−1

= 39.2 Hz ≈ 40 Hz

f H* = 103 21/ 3 − 1 = 0.5 KHz 20. Topic: Active Filters (c)  Ts = 100 μ sec fs =

1 1 = = 10 kHz Ts 100 × 10 −6

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When n = 0, y(t ) =

5 × 10 −6 [10 cos(8p × 103 )t ] Ts

Transconductance, g m =

y(t ) = 5 × 10 −1 cos(8p × 103 t ) 21. Topic: Simple Op-Amp Circuits



(c)  Slew Rate (SR ) = 1V/μ sec = 10 V/ sec 6

2 × 10 × 10 −3 5

⎡ | VGS | ⎤ ⎢1 − ⎥ ⎢⎣ | V p | ⎥⎦

⎡ | VGS | ⎤ ⎢1 − 5 ⎥ (2) ⎣ ⎦

VGS = RS I D = 2.5 × 103 Ω × 1 × 10 −3 A = 2.5 V

f in = 0 − 20 kHz, Gain = 40 dB

Putting the value of VGS in Eq. (2), we get

Vi = Vm sin w t

⎡ 2.5 ⎤ ⎢1 − 5 ⎥ = 2 mA/V ⎣ ⎦ Putting the value of gm in Eq. (1), we get Av = −2 × 10 −3 × 3 × 103 = −6

Gain in dB = 20 log (gain) 40 = 20 log (gain) gain = 10 2 = 100

Output of amplifier with gain A is as following Vo = AVi = AVm sin w t

Generalized output of LPF

3

⇒ Vm = 79.5 mV

(a)     V f ( f ) = bVO , b =

1 6

Due to virtual ground, the potential at inverting terminal will be Vf ( f  ). Appling KCL at the inverting terminal, we get

bV0 − 0 bV0 − V0 + =0 R1 R2 ⇒

bV0 bV0 V0 + = R1 R2 R2

⇒

⎡ R + R2 ⎤ 1 b⎢ 1 ⎥= ⎣ R1 R2 ⎦ R2 R1 1 = ⇒ R2 = 5 R1 R1 + R2 6

23. Topic: BJT and MOSFET Amplifiers (d)  I D = 1 mA, I D SS = 10 mA, VP = −5 V, R D = 3 kΩ Voltage gain AV =

Ch wise GATE_ECE_CH04_Analog Circuits.indd 290

= nf s ± f m = 12 kHz, 8 kHz, 32 kHz,... Since Cut-off frequency is 15 kHz so the frequencies present at the filter output are 12 kHz and 8 kHz. 25. Topic: Simple Diode Circuits (d)  The circuit shown in the figure is voltage doubler.

22. Topic: Sinusoidal Oscillators

⇒

1 1 = = 20 kHz t s 50 × 10 −6

Cut-off frequency, f c = 15 kHz

max

⇒ 10 = 100Vm × 2p × 20 × 10

fs =

2p f m = 24p × 103 ⇒ f m = 12 kHz

= AVm × (1) × ( 2p × 20 × 103 )

6

24. Topic: Active Filters (d)  Sampling period, t s = 50 μsec

Modulating frequency,

dV0 = AVm cos w t × w dt dV0 = AVm cos w t × ( 2p f ) dt dV SR = 0 dt

2 × 10 × 10 −3 5

gm =

Gain = 40 dB We know that ⇒ ⇒

gm =

2 I DSS Vp

Vo = − g m RD = − g m (3 × 103 kΩ) (1) Vi

26. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (a)  Gain of the transistor amplifier falls at high frequency due to internal capacitance of the device. 27. Topic: Simple Op-Amp Circuits (b)  Given that the input is a sine wave with peak-topeak voltage of 8 V, that is, Vi = 4 sin w t From the figure, we can see that the output is HIGH when the input voltage is less than 2 V. Crossover point when the input voltage is increasing and Vi = 2 V occurs at point sin w t = 1/2. Therefore, w t = p /6. Crossover point when the input voltage is decreasing and Vi = 2 occurs at point w t = p −(p /6) = 5p /6. Let Ton be the ON time of the output waveform, that is, when the output is HIGH and T be the total time period of the output waveform. The duty cycle is TON [(5p / 6) − (p / 6)] 1 = = T 2p 3

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Chapter 4  • Analog Circuits

28. Topic: BJT and MOSFET Amplifiers: Differential (c)  We know that CMRR = Ad − Ac where Ad and Ac are the differential and common mode gains in dBs, respectively. Therefore, CMRR = (48 − 2) dB = 46 dB 29. Topic: Small Signal Equivalent Circuits of Diode BJTs and MOSFETs (d)  The action of a JFET in its equivalent circuit can be best represented as a voltage-controlled current source. 30. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability

⎛ V − VBE ⎞ (a)    I CQ = b ⎜ CC , VCEQ = VCC − I CQ RC RB ⎟⎠ ⎝ ICQ increases directly with b, with all other parameters remaining the same. Therefore, the new value of ICQ is obtained as follows: ⎛ 200 ⎞ I CQ = 1.5 × ⎜ mA = 2 mA ⎝ 150 ⎟⎠ For the old value of b = 150, VCEQ = 3 V. Therefore, ICQ × R2 = 6 V − 3 V = 3 V Therefore, ⎛ ⎞ 3 R2 = ⎜ Ω = 2 kΩ ⎝ 1.5 × 10 −3 ⎟⎠ Therefore, the new value of VCEQ is obtained as follows: VCEQ = 6 − (2 × 10−3 × 2 × 103) = 2 V

31. Topic: BJT and MOSFET Amplifiers: Feedback (b)  In the current-shunt amplifier, Ri Rif = 1 + b Ai where ⎞ ⎛R 1 × 103 Ai = ⎜ i Av ⎟ = × 50 = 20 ⎝ Ro ⎠ 2.5 × 103 Therefore, 1 × 103 ⎛ 1⎞ Rif = = ⎜ ⎟ kΩ 1 + (0.2 × 20) ⎝ 5 ⎠ 32. Topic: BJT and MOSFET Amplifiers: Multi-Stage (c)  The voltage gain of the first stage is 50 ×

Ch wise GATE_ECE_CH04_Analog Circuits.indd 291

(1 × 103 ) = 40 250 + (1 × 103 )

291

The voltage gain of second stage is 50 ×

(1 × 103 ) = 40 250 + (1 × 103 )

The voltage gain of third (output) stage is 50. The total gain of the amplifier is 40 × 40 × 50 = 80,000 The total gain of the amplifier (in dB) is 20 log 80000 ≅ 98 dB 33. Topic: Simple Op-Amp Circuits (b) ⎤ ⎡ 5 × 103 ⎤ ⎡ 8 × 103 Vout = ⎢ × ⎢1 + ×3 3 3 ⎥ 3 ⎥ ⎣ (8 × 10 ) + (1 × 10 ) ⎦ ⎣ 1 × 10 ⎦ ⎛ 5 × 10 3 ⎞ −2×⎜ ⎝ 1 × 103 ⎟⎠ = 16 − 10 = 6 V 34. Topic: Sinusoidal Oscillators: Criterion for Oscillation (a)  The frequency of oscillation of the given circuit is 1 . 2p 6 RC 35. Topic: Function Generators, Wave Shaping Circuits and 555 Timers (d)  A sawtooth waveform is generated across the capacitor when it is charged by a constant current source. The time period of the sawtooth waveform is T=

1 1 = s = 2 ms f 500

The voltage across the capacitor when it is charged by a constant current source (I ) is given by I =C

dV dt

Substituting the different values in the equation above, we get I = 2 × 10 −6 ×

3 = 3 mA 2 × 10 −3

Hence, option (d) is correct. 36. Topic: Power Supplies: Regulation (c)  The voltage at the non-inverting terminal of the op-amp is 3 V due to the Zener diode. The voltage at the inverting terminal of the op-amp is the same as that at the non-inverting terminal due to virtual earth. The current flowing through the 20 kΩ resistor is 3 3 A= mA 20 20 × 103

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The regulated DC output voltage is 3 3 × 10 −3 × 20 × 103 + × 10 −3 × 40 × 103 = 9 V. 20 20

43. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (c)  The Thevenin’s equivalent circuit is shown in the following figure.

37. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (a)  VCE = VCC − I C RC 0.2 = 3 − I C × 1 × 10

VCC = 5 V 2.2 kΩ

3

IB R TH

Therefore, IC = 2.8 mA. Also I b= C IB

VTH + −

Therefore, ⎛ 2.8 × 10 −3 ⎞ IB = ⎜ A = 56 μ A. 50 ⎟⎠ ⎝ 38. Topic: BJT and MOSFET Amplifiers: Operational (b)  The ideal op-amp is a voltage-controlled voltage source. 39. Topic: BJT and MOSFET Amplifiers: Feedback (c)  In a voltage-series feedback amplifier, the input impedance increases by a factor (1 + Ab ) and the output impedance decreases by the factor (1 + Ab ). Hence, option (c)  is the correct answer. 40. Topic: Active Filters (a)  The filter configuration is a second-order low pass filter. 41. Topic: Simple Diode Circuits (d)  Let Vin be the input voltage, Vo be the output voltage, IZ the diode current and IL the load current. Then ⎛ Vin − Vo ⎞ ⎜⎝ R ⎟⎠ ≥ ( I Z + I L ) Here, Vin = 12 V; Vo = 5 V; IZ = 0. For IL = 100 mA, we have ⎛ 12 − 5 ⎞ R≤⎜ ≤ 70 Ω ⎝ 100 × 10 −3 ⎟⎠ For IL = 500 mA, ⎛ 12 − 5 ⎞ R≤⎜ ≤ 14 Ω ⎝ 500 × 10 −3 ⎟⎠ Therefore, the maximum value of R is 14 Ω. 42. Topic: Simple Diode Circuits (b)  VDC =

2Vm and PIV = 2Vm p

Ch wise GATE_ECE_CH04_Analog Circuits.indd 292

IC

+ VBE − 300 Ω

IE

The Thevenin’s voltage is ⎡ ⎤ 1 × 103 VTH = ⎢ × 5 = 1V 3 3 ⎥ ⎣ (1 × 10 ) + ( 4 × 10 ) ⎦ As the value of b is very large, IB can be ignored. Therefore,

IE =

VTH − VBE ⎛ 1 − 0.7 ⎞ =⎜ A = 1 mA  I c ⎝ 300 ⎟⎠ RE

Applying Kirchhoff’s voltage law in the collector–emitter loop, we get 5 − 2.2 × 103IC − VCE − 300IC = 0 Therefore, VCE = 5 − 2.2 × 103IC − 300IC = 2.5 V 44. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (d) I gm = C VT Therefore, gm =

1 × 10 −3 A/V = 40 mA/V 25 × 10 −3

Also, b = g m rp . Therefore, rp =

100 Ω = 2.5 kΩ 40 × 10 −3

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Chapter 4  • Analog Circuits

293

45. Topic: Simple Op-Amp Circuits (a)  Let the output voltage of the op-amp be Vo and the voltage at the non-inverting terminal be V. Applying KCL at the non-inverting terminal of the op-amp, we get

47. Topic: BJT and MOSFET Amplifiers: Feedback (d)  In a current-shunt feedback amplifier, the input impedance decreases by the factor (1 + Ab ) and the output impedance increases by the factor (1 + Ab ). Hence, option (d) is the correct answer.

V V V − Vo + + =0 R2 RL R2

48. Topic: BJT and MOSFET Amplifiers: Multi-Stage (b)  The cascade amplifier is a multistage configuration of CE–CB.

Solving the above equation, we get V 2V V + = o R2 RL R2 Applying KCL at the inverting terminal of the op-amp, we get

49. Topic: BJT and MOSFET Amplifiers: Operational (b)  Since the inverting terminal is at virtual ground, the current flowing through the voltage source is Is =

V − Vs V − Vo + =0 R1 R1

Input resistance Ri of the amplifier is given by

Solving the above equation, we get

Ri =

Vs − 2V = −Vo Substituting the value of Vo in the above equation, we get 2V − Vs 2V V + = R2 R2 RL

I E = I o (eVBE / hVT −1) Given that, VBE = 0.7 V, Io = 10−13 A, VT at 300 K = 26 mV and h = 1.

V iL = − s R2

Vi − Vo Vi Vi + + =0 R 1/jw C R + (1/jw C ) On solving the above equation, we get Vo jw C ⎛ 1 2⎞ = ⎜⎝ − 2 2 + R ⎟⎠ + 3 Vi R w C For oscillations to occur, imaginary part should be zero. Therefore, jw C ⎛ 1 ⎞ + R2 ⎟ = 0 ⎜− ⎠ R ⎝ w 2C 2 Hence, for oscillations to occur, 1 1 1 C= = = mF wR 2 × p × 103 × 103 2p

Ch wise GATE_ECE_CH04_Analog Circuits.indd 293

Vs = 10 kΩ Is

50. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (c)  The transistor acts as a diode when its two terminals are shorted. IC = IB ≈ IE. Therefore,

The current iL is given by V/RL. Therefore, from the above equation, we get

46. Topic: Sinusoidal Oscillators: Single- Transistor and Op-Amp Configuration (a)  Let the voltage at the non-inverting input of the op-amp be Vi. Let the capacitor be denoted as C and both the 1 kΩ resistors as R. Applying KCL at the non-inverting input of the op-amp, we get

Vs 10 × 103

Therefore, I E = 10 −13 [e 0.7 /1× 26 ×10

−3

− 1]

= 49 mA 51. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability

⎛ VCC − VBE ⎞ (b)  I B = ⎜ ⎝ RB + (b + 1) R E ⎟⎠ ⎡ ⎤ 20 − 0.7 =⎢ A = 40 μ A 3 3 ⎥ ⎣ ( 430 × 10 ) + (51 × 1 × 10 ) ⎦ VC = VCC − b I B RC = 20 − 50 × 40 × 10 −6 × 2 × 103 = 16 V

52. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability ⎛ 20 ⎞ (b)  Z i = 2 MΩ and Z o = ( 20 × 103 2 × 103 ) = ⎜ ⎟ kΩ ⎝ 11 ⎠ 53. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (a)  Under DC conditions, VDS= −2 V

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⎛ V ⎞ I D = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝

Differential mode gain

2

ADM = − g m RC 2

VDS

⎡ ( −2) ⎤ = 10 × 10 −3 ⎢1 − ⎥ A = 5.625 mA ⎣ ( −8) ⎦ = VDD − I D RD = ( 20 − 5.625 × 10 −3 × 2 × 103 ) V = 8.75 V

54. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability

(b)  g m =

2 VP

I D × I DSS

⎛2 ⎞ = ⎜ × (5.625 × 10 −3 × 10 × 10 −3 ) ⎟ S ⎝8 ⎠ = 1.875 mS Also,

Av = − g m ( rd RD )

Therefore, ⎛ 20 ⎞ Av = −(1.875 × 10 −3 ) × ⎜ × 103 ⎟ = −3.41 ⎝ 11 ⎠ 55. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (c)  From the graph, it is clear that the threshold voltage is VTH = 1 V From the given figure, VGS = 3 − 1 = 2 V

Thus only common-mode gain depends on RE. From the expression for common mode gain, it is clear that it is inversely proportional to RE. Therefore, for large values of RE, common mode gain decreases. 58. Topic: Simple Op-Amp Circuits (b)  Let us assume that the output voltage is positive, that is, +10 V. The lower diode conducts in this situation and the voltage at the non-inverting terminal of the op-amp is 2 × 103 × 10 = 8 V 2 × 103 + 0.5 × 103 When the input voltage Vi at the inverting terminal exceeds +8 V, the output goes from positive saturation voltage, +10 V, to the negative saturation voltage, −10 V. The upper diode conducts in this situation and the voltage at the non-inverting terminal of the op-amp is 2 × 103 × ( −10) = −5 V 2 × 103 + 2 × 103 When the input voltage Vi at the inverting terminal goes below −5 V, the output goes from negative saturation voltage, −10 V, to positive saturation voltage, +10 V. Therefore, the transfer characteristics shown in option (b) are correct. 59. Topic: Active Filters (a)  From the diagram it is clear that the circuit is a high-pass filter. The cut-off frequency in rad/s is given by 1 rad/s = 1000 rad/s 1 × 103 × 1 × 10 −6

VDS = 5 − 1 = 4 V Since VDS > (VGS − VTH), the MOSFET is in saturation region. 56. Topic: BJT and MOSFET Amplifiers: Operational (c)  Let IB1 and IB2 be the currents through the noninverting and inverting terminals, respectively. Let V1 and V2 be the voltages at the non-inverting and inverting terminals. V1 = −IB1 × 1 × 10

6

Due to virtual earth, V2 = V1. Applying Kirchhoff ’s voltage law around the inverting terminal loop, eo = V2 + IB2 × 1 × 106 = (IB2 − IB1) × 1 × 106 Therefore, output voltage is directly proportional to the difference of the two currents and hence is a measure of the offset current. 57. Topic: BJT and MOSFET Amplifiers: Differential (d)  Common mode gain ACM = −

Ch wise GATE_ECE_CH04_Analog Circuits.indd 294

RC 2 RE

60. Topic: Power Supplies: Regulation (a)  The maximum load current will be drawn when the input voltage is maximum, that is, Vi = 30 V Therefore, 30 − 5.8 = (IL + 0.5 × 10−3) × 1000 Hence, IL = 23.7 mA 61. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (d)  For an N-channel depletion MOSFET, 2

⎛ V ⎞ I D = I DSS ⎜1 − GS ⎟ . V ⎠ ⎝ P

The depletion MOSFET works for both positive as well as negative values of VGS. Its transconductance gain increases as VGS increases, therefore the maximum gain is at VGS = 3 V for the values of VGS given.

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62. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (d)  For a true transconductance amplifier, the input and the output resistances of the amplifier are infinite. 63. Topic: Simple Diode Circuits (b)  During the positive half cycle, the Zener diode is reverse biased. It acts as an open circuit till the input voltage is less than 6  V and output ­voltage is 0  V. When the input voltage is greater than 6  V, the diode starts conducting and the voltage drop across the diode is 6  V. When the input voltage is negative, the diode is forward biased and the whole of the input appears across the resistance R. 64. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (c)  Applying Kirchhoff’s voltage law to the base emitter loop, we get ⎡ I ⎤ 12 − 0.7 = 1 × 103 × I E + 53 × 103 × ⎢ E ⎥ ⎣ 60 + 1⎦ Therefore, IE = 6 mA. Applying Kirchhoff’s voltage law to the collector-emitter loop, we get

If we make hie and hfe equal to ∞, then circuit will be same as the ideal op-amp with Ri = ∞ and AV = ∞. Assuming of the vitual ground condition and applying KCL at node B, we get 0 − Vs 0 − Vc + =0 5.3 × 103 53 × 103 Therefore,

Vc = −10 Vs

67. Topic: Simple Op-Amp Circuits (d)  Applying KCL at the inverting node of the op-amp, we get dVC VC ⎛ 10 ⎞ = 10 −6 ⎜⎝ ⎟ =C dt 1 × 10 −3 1 × 103 ⎠ Vc at t = 1 ms is −3 ⎛ 10 ⎞ 1 × 10 VC = ⎜ = 10 V −6 3⎟ ⎝ 1 × 10 ⎠ 1 × 10

68. Topic: Power Supplies: Regulation (c)  The voltage at the non-inverting input terminal of the op-amp is 6 V. Therefore, the voltage at the inverting terminal of the op-amp is 6 V. Therefore,

VCE = 12 − 6 ´× 10−3 ´× 1 ×´ 103 = 6 V

Vout =

65. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (b)  New bDC = 66. from the calculation in solution of question 64, we get new values of IE = 6.31 mA and VCE = 5.7 V. The percentage change in VCE is ⎛ 6 − 5.7 ⎞ ⎜⎝ ⎟ × 100 = 5% 6 ⎠

295

6 × (12 × 103 + 24 × 103 ) =9V ( 24 × 103 )

Now, VCE = 15 − Vout = 15 − 9 = 6 V Therefore,   IC =

VE V 9 9 + E = + = 0.9 A 3 3 12 × 10 + 24 × 10 10 36 × 10 10 3

The power dissipated in the transistor is

66. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (a)  For small-signal gain of the amplifier, capacitance CC is considered as short circuited. Given that hie → ∞ and hfe → ∞, therefore amplifier circuit can be drawn as shown below

VCEIC = 6 × 0.9 = 5.4 W 69. Topic: Power Supplies: Regulation (b)  New unregulated input voltage is 18 V. Therefore, VCE = 18 − 9 = 9 V Also, IC = 0.9 A. The power dissipated in the transistor is VCEIC = 9 × 0.9 = 8.1 W The percentage increase in the power dissipated is therefore

53 kW 5.3 kW

C +

B ib

Vs

Vc

hre hfe ib E

Ch wise GATE_ECE_CH04_Analog Circuits.indd 295

−

1 kW

8.1 − 5.4 × 100 = 50% 5.4 70. Topic: Simple Diode Circuits (d)  The circuit shown in figure (d) is the correct fullwave rectifier.

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71. Topic: BJT and MOSFET Amplifiers: Feedback (a)  For a true transconductance amplifier, the input and the output resistances of the amplifier are infinite.

Therefore,

72. Topic: Single-Stage and MOSFET Amplifiers: Biasing (b)  Since b is large, IB ≈ 0 and IC ≈ IE. Applying Kirchhoff’s voltage law in the base-emitter loop, we get

where d is the gate oxide thickness (in cm). Therefore, d = 50 nm.

2 − 0.7 = 1 × 103 × IE Therefore, IE ≈ IC = 1.3 mA. The value of s­aturation collector-emitter voltage of a transistor VCE(sat) is approximately 0.2 V. Therefore, 10 − 0.2 ⎛ ⎞ I C(sat) = ⎜ A = 0.9 mA ⎝ 1 × 103 + 10 × 103 ⎟⎠

7 × 10 −12 =

75. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (c)  The capacitance is given by C=

⎛ 10 − 7 ⎞ ⎛ 1⎞ I =⎜ A=⎜ ⎟A ⎝ 210 ⎟⎠ ⎝ 70 ⎠ The same current flows through the Zener resistance. Therefore, the voltage across Zener resistance is ⎛ 10 ⎞ ⎜⎝ ⎟⎠ V = 0.14 V 70 Therefore, the output voltage is Vo = 7 V + 0.14 V = 7.14 V When Vi = 16 V, the current passing through 200 Ω is given by ⎛ 16 − 7 ⎞ ⎛ 3⎞ I =⎜ A=⎜ ⎟A ⎝ 210 ⎟⎠ ⎝ 70 ⎠ The same current flows through the Zener resistance. Therefore, the voltage across Zener resistance is ⎛ 3⎞ ⎜⎝ ⎟⎠ V = 0.43 V 70

1 × 10 −12 =

76. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (d)  Both S1 and S2 are false. 77. Topic: Simple Op-Amp Circuits (d)  As the input impedance of the op-amp is very high, the current through the resistor 2 kW (IR) is equal to the current through the diode (ID). Applying KCL at the inverting node of the op-amp, we get ⎛ VD ⎞ ⎛ 0 − Vo ⎞ ⎛ 0 − Vi ⎞ ⎜⎝ 2 × 103 ⎟⎠ = I s exp ⎜ V ⎟ = I s exp ⎜ V ⎟ ⎝ T⎠ ⎝ T ⎠ Solving the above equation and rearranging the terms, we get ⎞ ⎛ Vi Vo = −VT ln ⎜ 3 ⎝ 2 × 10 × I s ⎟⎠ It is given that when Vi = 2 V, we get Vo = Vo1 and when Vi = 4 V, we get Vo = Vo2. Therefore, ⎞ ⎛ 2 Vo1 = −VT ln ⎜ and 3 ⎝ 2 × 10 × I s ⎟⎠

Vo = 7 V + 0.43 V = 7.43 V

C=

Ch wise GATE_ECE_CH04_Analog Circuits.indd 296

eA d

1 × 10 −12 × 10 −4 d

where d is the depletion width (in cm). Therefore, d = 10−4 cm = 1 µm.

Therefore, the output voltage is

74. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (a)  The capacitance is given by

eA d

Therefore,

Since IC > IC(sat), the transistor is in saturation. 73. Topic: Simple Diode Circuits (c)  When Vi = 10 V, the current through 200 Ω is given by

3.5 × 10 −13 × 10 −4 d

⎞ ⎛ 4 Vo2 = −VT ln ⎜ 3 ⎝ 2 × 10 × I s ⎟⎠ Therefore, ⎛ ⎞ ⎛ ⎞ 2 4 + VT ln ⎜ Vo1 − Vo2 = −VT ln ⎜ 3 3 ⎟ ⎝ 2 × 10 × I s ⎟⎠ ⎝ 2 × 10 × I s ⎠ = VT ln 2

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78. Topic: Simple Op-Amp Circuits (a)  The output voltage Vo is given by

297

VCC = 9 V

⎛ R⎞ 1/sC ⎛ R1 ⎞ 1+ Vo ( s) = ⎜ − 1 ⎟ Vi ( s) + Vi ( s) R + 1/sC ⎜⎝ R1 ⎟⎠ ⎝ R1 ⎠

IB

VTH

IC

+ VBE −

IE

R TH

Solving the above equation, we get the transfer function as Vo ( s) 1 − sRC = Vi ( s) 1 + sRC

3 kΩ

+ −

2.3 kΩ

79. Topic: Simple Op-Amp Circuits (c)  w = ∠

Vo (s) = ∠(1 − sRC ) − ∠(1 + sRC ) Vi (s)

= − tan −1 w RC − tan −1 w RC = −2 tan −1 w RC The minimum value of f occurs when w → ∞. Therefore, fmin = −p. The maximum value of f occurs when w = 0. Therefore, fmax = 0. 80. Topic: Simple Op-Amp Circuits (c)  The voltage at the non-inverting input of the op-amp is ⎛ ⎞ 1 × 103 1× ⎜ = 0.5 V 3 3⎟ ⎝ 1 × 10 + 1 × 10 ⎠ The output voltage is 3 Vo = − ⎛ 2 × 10 ⎞ × 1 + ⎜⎝ 1 × 103 ⎟⎠

⎛ 2 × 103 ⎞ × 0.5 ⎜⎝1 + 1 × 103 ⎟⎠

= −2 + 1.5 = −0.5 V. 81. Topic: Simple Diode Circuits (c)  During the positive half cycle, Diode D1 is in the forward-biased region, Zener diode Z and diode D2 are operating in the reverse-biased region. Therefore, the maximum output voltage is 6.8 V + 0.7 V = 7.5 V During the negative half cycle, diode D1 is in the reverse-biased region, Zener diode Z and diode D2 are operating in the forward-biased region. Therefore, the minimum output voltage is -0.7 V. 82. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (a)  The Thevenin’s equivalent circuit is shown in the following figure.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 297

The Thevenin’s resistance is RTH =

( 20 × 103 ) × (10 × 103 ) = 6.67 kΩ ( 20 × 103 ) + (10 × 103 )

The Thevenin’s voltage is VTH =

10 × 103 × 9 = 3V ( 20 × 103 ) + (10 × 103 )

Since the value of b is very large, IB can be ignored. Therefore, IE =

VTH − VBE 3 − 0.7 = = 1 mA RE 2.3 × 103

83. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (d)  Midband voltage gain Av = − =−

RC 2re 3 × 103 = − 60 ( 2 × 25 × 10 −3 ) / (1 × 10 −3 )

84. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (b)  It is a current mirror circuit. Therefore, I X = IS For NMOS transistors, gate current = 0 Therefore, I G1 = I G 2 = 0 From the circuit, we see that I bias = I S + I G1 + I G2 Substituting the values of IG1 and IG2 in the above equation, we get I bias = I S + 0 + 0 Therefore, I X = I bias

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85. Topic: Simple Op-Amp Circuits (c)  Let the voltage at the non-inverting input be Vin. Applying KCL at the non-inverting node of the amplifier, we get ⎛ Vin − 15 ⎞ ⎛ Vin − ( −15) ⎞ ⎛ Vin − Vo ⎞ ⎜⎝ 10 × 103 ⎟⎠ + ⎜⎝ 10 × 103 ⎟⎠ + ⎜⎝ 10 × 103 ⎟⎠ = 0 Therefore, Vo 3 Since the output voltage swings from −15 V to +15 V, the input voltage at the non-inverting input of the op-amp swings from − 5 V to +5 V. Vin =

86. Topic: Simple Op-Amp Circuits (b)  As the input impedance of the op-amp is very high, the current flowing through the diode is the same as the current flowing through the 100 kW resistor. Therefore, 0 − Vi = I o (eV /VT − 1) 100 × 103 Substituting the value of Vi = −1 V, VT = 25 mV and Io = 1 mA and solving the above equation, we get V = 59.9 mV Applying KCL at the inverting terminal of the op-amp, we get −3 ⎡ 0 − ( −1) ⎤ ⎡ 0 + 59.9 × 10 − Vo ⎤ + ⎢ ⎥=0 ⎢100 × 103 ⎥ 4 × 103 ⎣ ⎦ ⎣ ⎦

Solving the above equation, we get Vo = 0.1 V 87. Topic: Active Filters (b)  Applying KCL at the inverting terminal of the op-amp, we get 0 − Vo (s) 0 − Vi (s) + =0 R1 + Ls ⎡⎣ R2 ( R2Cs + 1)⎤⎦ Therefore, the transfer function of the circuit is Vo (s) R2 =− Vi (s) ( R1 + Ls) ( R2Cs + 1) =−

R2 R2 LCs 2 + ( R1 R2C + L) s + R1

This is equivalent to the transfer function of a secondorder low pass filter, which is given by

Ch wise GATE_ECE_CH04_Analog Circuits.indd 298

A Bs 2 + Cs + D Therefore, the circuit given is that of a low-pass filter. 88. Topic: Function Generators, Wave Shaping Circuits and 555 Timers: 555 Timers (b)  In the astable multivibrator circuit based on 555 timer IC, the capacitor voltage VC varies from Vcc/3 to 2Vcc/3. Here Vcc = 9 V, therefore voltage Vc varies between 3 V to 6 V 89. Topic: Function Generators, Wave Shaping Circuits and 555 Timers: Function Generators (c)  Let the voltage at the non-inverting input be V1. Applying KCL at non-inverting input end, we get V −V 15 − V1 V − ( −15) + o 13 = 1 3 10 × 10 10 × 10 10 × 103 Therefore, 15 − V1 + Vo − V1 = V1 + 15 or V1 =

Vo 3

Since Vo swings from −15 V to +15 V, V1 switches between −5 V and +5 V. 90. Topic: Simple Diode Circuits (a)  Due to the current source of 1 A, the diode is reverse biased by 1 V. When the input voltage Vi is less than -1 V, the diode gets forward biased and the voltage V is equal to Vi. When the input voltage Vi is greater than -1 V, the diode is reverse biased and the voltage V is equal to 1 V. Therefore, the voltage V is min (Vi, 1). 91. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid- Frequency Small Signal Analysis and Frequency Response (d)  The best approximate answer for the output voltage vo(t) is vo (t ) = Av vi (t ) = −

hfe RC vi (t ) hie

−150 × 3 × 103 ( A cos 20t + B sin 106 t ) 3 × 103 = −150( A cos 20t + B sin 106 t ) =

Since, the output coupling capacitor is large, we get vo (t ) = −150 ( B sin 106 t ) 92. Topic: Simple Op-Amp Circuits (d)  From the circuit, we see that the feedback is negative feedback. The given circuit can be redrawn as shown in the following figure.

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10V I

IR =

5 kΩ

9. 3 V 9. 3 × 103 Ω

= 1 mA = I C1

Given that the emitter area of transistor Q1 is equal to half the emitter area of transistor Q2, that is,

5V V

+

+ VBE

−

AQ2

AQ1 =

2

+ Therefore, we have

5V 1.4 kΩ

V´ −

(b2 )effective = 2 × b2 = 1430

I

Since the effective value of b for transistor Q 2 is double that for transistor Q 1, so collector current of transistor Q 2 will also be nearly the double of that of transistor Q 1 that is

The current I is I=

10 − 5 = 1 mA 5 × 103

I o = I C2 = 2 × I C1 = 2 mA

The voltage V' is V ′ = 1.4 × 10 × 1 × 10 3

−3

= 1.4 V

The voltage V is V = V ′ + VBE = 1.4 + 0.6 = 2 V 93. Topic: Function Generators, Wave Shaping Circuits and 555 Timers: Function Generators (a)  For the given astable multivibrator circuit, only the frequency of v0(t) depends on R2.

95. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (a)  The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited). Vo + Vs

rπ

RB

gm vπ

vπ

RC

−

94. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (b)  The given circuit can be represented as

Input impedance is  Ri = RB || rp Voltage gain is    AV = gmRC Now, if CE is disconnected, resistance RE appears in the circuit

IR Io R = 9.3 kΩ

Vo Q2 (b = 715) 2

Q1 (b 1 = 700)

+ Vs

VBE = 0.7 V

rπ

RB

gm vπ

vπ

RC

−

−10V RE

Assuming that both the transistors are in active region, the voltage at Q1 base

(VBase )Q1 = (0.7 − 10) V Current through registor R

Ch wise GATE_ECE_CH04_Analog Circuits.indd 299

= −9.3 V

Input impedance Rin = RB || [rp + ( b + 1)RE] 

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From the above expression it is clear that when capacitance CE is decreased, input impedance increases. Voltage gain AV =

g m RC 1 + g m RE

Hence, voltage gain decreases when the capacitance CE is disconnected. 96. Topic: Simple Op-Amp Circuits (b)  The given circuit is that of an inverting amplifier with a load resistor (R3) connected at the ouput. ThereR fore, the gain of the amplifier is − 3 . R1 97. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (b)  The equivalent model of the BJT-based circuit is given in the following figure. Rs Vo Rin

Is RB rπ

Vs

gm Vπ

+ Vπ −

Co RC

RL

99. Topic: Simple Op-Amp Circuits (b) For vi < −5 V, diode D2 is conducting and diode D1 is not conducting. At vi = −10 V, applying KCL at the inverting terminal of the op-amp, we get 0 − 20 0 − ( −10) 0 − vo =0 + + 4R R R Therefore, −vo = +5 V. At −vi = −5 V, applying KCL at the inverting terminal of the op-amp, we get 0 − 20 0 − ( −5) 0 − vo =0 + + 4R R R Therefore, vo = 0 V. For vi > −5 V, both diodes are conducting. So, vo = 0 V. 100. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (a)  Whenever we use a bypass capacitor in parallel with emitter resistor RE in a common ­emitter configuration with emitter resistor it increases the voltage gain with increase in frequency as compared to common emitter configuration without emitter resistor. The resonant frequency for the circuit is 1

w=

LC

=

1 10 × 10 −6 × 10 −9

= 10 M rad/s

The gain is maximum at the resonant frequency; hence, option (a) is correct. We know that hie =

b gm

101. Topic: Active Filters (d) At w = 0, XL1 = jwLL1 = 0. Hence, the circuit can be redrawn as shown below

Therefore,

R1

⎛ 100 ⎞ hie = ⎜ Ω = 259 Ω ⎝ 0.3861⎟⎠

ii −

The resistance seen by the source Vs is

+

+ vo −

Rs + (RB || hie) = [1000 + (93000 || 259)] Ω = 1258 Ω 98. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (b)  The lower cut-off frequency due to C2 f LC2 =

1 2p ( RC  RL )C2

Substituting RC = 250 Ω, RL = 1000 Ω and C2 = 4.7 µF in the above equation, we get

Therefore, the output voltage vo = 0 at w = 0. At w = ∞, XL1 = ∞. Hence, the circuit can be redrawn as shown in the following figure: ii

R1 − +

+ vo −

f LC2 = 27.1 Hz

Ch wise GATE_ECE_CH04_Analog Circuits.indd 300

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Chapter 4  • Analog Circuits

Therefore, the output voltage is vo = R1iL. Hence, the given circuit is a high-pass filter. 102. Topic: Voltage Reference Circuits (b)  A Zener diode, when used in voltage stabilization circuits, is biased in reverse breakdown region. 103. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (a)  For DC biasing, the AC source is ­considered as a short circuit. Therefore, the DC voltage across the diodes is 4 × 0.7 V = 2.8 V The DC current is ⎛ 12.7 − 2.8 ⎞ ⎜⎝ ⎟ A = 1 mA 9900 ⎠ 104. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (b)  The diode is replaced by its dynamic resistance for the AC analysis. rd =

hVT I

where h = 1, VT = 25 mV and I = 1 mA. Therefore, rd = 25 Ω. The AC voltage across the diodes is vi v × 100 = i 9900 + 100 100 Vp cos w t = 100 100 cos w t = mV 100 = 1 cos (w t ) mV 105. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (c)  Applying Kirchhoff’s voltage law at the base– emitter junction, we get −5 − 0.7 − IE ´× 4.3 ×´ 103 + 10 = 0 Solving the above equation, we get, IE = 1 mA. The value of emitter current remains the same irrespective of whether the switch is closed or open. Applying Kirchhoff’s current law at the collector junction, Icap + 0.5 mA = 1 mA Therefore, Icap = 0.5 mA. The value of capacitor current remains the same irrespective of whether the switch is closed or open. The collector voltage is given by VC = VCE + VE Given that, when the transistor is in saturation, VCE(sat) = 0.7 V. Therefore, the collector voltage when the transistor is in saturation is

Ch wise GATE_ECE_CH04_Analog Circuits.indd 301

301

VC = 0.7 + IERE = 0.7 + 1 ´× 10−3 ´× 4.3 ×´ 103 = 5 V The collector voltage VC is equal to the voltage across the capacitor (Vcap). The voltage across the capacitor is related to the capacitor current by Vcap =

I cap × t C

Substituting the values of Icap, C and Vcap in the above equation, we get t=

5 × 5 × 10 −6 s = 50 ms 0.5 × 10 −3 s

Therefore, the time required by the transistor to leave the active region and reach the saturation region is the time required by the collector voltage or the capacitor voltage (as both are equal) to reach 5 V which is equal to 50 ms. 106. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (a)  The capacitor C1 will trap the negative peak voltage of the input signal (which is −1 V in this case). Therefore, the voltage across diode D1 will be cos (w t ) −1. 107. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (c)  rp = (b + 1)re = (b + 1)

VT VT = Ie Ib

where Ib is the DC current through the base t­erminal. Given that, Ib = 1 mA. Also, VT = 25 mV at room ­temperature. Therefore, rp =

25 × 10 −3 = 25 Ω 1 × 10 −3

108. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (d)  The equivalent circuit is shown below. 10 kW

io

iL

+ + Vi −

Vi¢

100 kW 1−Av¢

hie hfe ib

100 kW 1−1 Av¢

12 kW Vo

−

The resistor 100 kΩ between the collector and the base terminals is equivalent to the connection of ⎡ 100 ⎤ ⎢ ⎥ kΩ ⎣1 – Av ′ ⎦

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resistance at the input terminal and the connection of ⎡ 100 ⎤ ⎢ ⎥ kΩ ≈ 100 kΩ ⎣1 − (1/Av ′ ) ⎦ resistance at the output terminal. Now, load resistance RL ′ = (100 × 103 )  (12 × 103 ) = 10.7 kΩ

110. Topic: BJT and MOSFET Amplifiers: Feedback (a)  The given configuration is a voltage–series feedback configuration. The input impedance is given by Rif = Ri(1 + A0k) Therefore, the input impedance increases when k is increased. The output impedance is given by

Current gain i Ai = o = − hfe = −b = 100 ib Input resistance Ri ′ =

Vi ′ hie = 1.1 kΩ ib

Voltage gain Av ′ =

Vo io RL Ai RL = = Vi ib Ri ′ Ri ′

−100 × 10.7 × 103 1.1 × 103 = −972 =

Therefore, the equivalent resistance of 100 kΩ resistor on input side is −100 × 103 = 102.77 Ω 1 − ( −972)

Ro 1 + Ao k

Therefore, the output impedance decreases when k is increased. 111. Topic: Simple Op-Amp Circuits (b)  The output of the op-amp is negative as the input appearing at the inverting input of the op-amp is more than the input at the non-inverting terminal. the transistor Q is conducting as Vout, which is also the emitter voltage is negative and the base terminal of the transistor Q is at zero potential. Since the transistor Q is conducting, VBE = 0.7 V. Since VB = 0 V, VE = Vout = −0.7 V. 112. Topic: Simple Diode Circuits (b)  It is given that the voltage across the load resistance is RL = 5 V. Therefore, the voltage across 100 Ω resistance is V100 = 10 × 5 = 5 V The current through 100 Ω resistance is

Therefore, Vi ′=

Rof =

Vi × 102.77 = Vi × 10.172 × 10 −3 10 × 103 + 102.77

Voltage gain Av =

Vo Vo Vi ′ = × Vi V ′ Vi i

I100 =

The voltage across load RL is VR L = 5 V = I L (max) × R(min) Therefore, R (min) =

= Av ′ × 10.172 × 10 −3 = −972 × 10.172 × 10 −3 = −9.88

109. Topic: Active Filters (b)  The transfer function of the given circuit is − R2 ( O /P ) T ( s) = = ( I /P ) R1 + (1/Cs) − R2 sC −( R2 /R1 ) s = R1 sC + 1 s + (1/R1C )

It is the transfer function of the high-pass filter with cutoff frequency

w=

Ch wise GATE_ECE_CH04_Analog Circuits.indd 302

Now,

5 I L(max)

I100 = I Z + I L (max) = 50 mA Therefore,

Hence, Av ≅ 10.

=

V100 5 = = 50 mA 100 100

1 rad/s R1C

I L (max) = 50 − I Z = 50 × 10 −3 − 10 × 10 −3 = 40 mA Therefore, 5 = 125 Ω 40 × 10 −3 The minimum power rating of Zener diode is PZ = VZ I Z (max) Rmin =

The maximum current through the Zener diode is I Z (max) = 50 mA Therefore, PZ = 5 × 50 × 10 −3 = 250 mW

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Chapter 4  • Analog Circuits

113. Topic: Simple Diode Circuits (d)  When VYZ is positive, then all the four diodes are reverse biased. Therefore, VW = VX = 0 Hence, VWX = 0 V When VYZ is negative, then all the four diodes are forward biased. Since all the diodes are given to be ideal, they will act as short circuit. Therefore, VW = VX Therefore,

The output voltage of the second op-amp is ⎛ 1 × 103 ⎞ =8V 4 ⎜1 + ⎝ 1 × 103 ⎟⎠ 117. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (8)   Responsivity (R) =

114. Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability (c)  It is given that b  is very large. Therefore, IC = IE = 1 mA VE = IERE = 1 × 10−3 × 500 = 0.5 V and VBE = 0.7 V Therefore, VR2 = VBE + VE = 0.7 + 0.5 = 1.2 V It is a self-bias circuit and hence R2 VR2 = VCC × R1 + R2 Therefore, R2 1.2 = 3 × 60 × 103 + R2 Solving the above equation, we get R2 = 40 kΩ 115. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (a)  The lower cut-off frequency fL is given by fL =

1 2p ( RL + RD )C

1 = = 8 Hz 3 2 × 3.14 × (10 × 10 + 10 × 103 ) × 1 × 10 −6 116. Topic: Simple Op-Amp Circuits (c)  The output voltage of the first op-amp is ⎛ 1 × 103 ⎞ ⎛ −1 × 103 ⎞ 1⎜1 + − 2⎜ =4V 3⎟ ⎝ 1 × 10 ⎠ ⎝ 1 × 103 ⎟⎠

Ch wise GATE_ECE_CH04_Analog Circuits.indd 303

IP Po

where, IP = photocurrent generated and Po = incident optical power. Therefore,

VWX = VW − VX = 0 Hence, for all conditions, VWX = 0

303

0.8 = Hence,

IP 10 × 10 −6

I P = 8μA 118. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (a)  It will lead to a decrease in the threshold voltage 119. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (d)  Recombination rate, R = B(nn0 + nn) (pn0 + pn) nn0 and pn0 = Electron and hole concentrations respectively under thermal equilibrium nn and pn = Excess electron and hole concentrations, respectively 120. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (1.075)  Applying KVL in base-emitter loop, we get IBRB + 0.7 − 10 = 0 Therefore, IB =

10 − 0.7 10 − 0.7 = = 93 μA RB 100 × 103

Now, I C = b I B = 50 × 93 × 10 −6 A = 4.65 mA From the figure, Vo = I C RC . Therefore, RC =

Vo 5 = Ω = 1.075 kΩ I C 4.65 × 10 − 3

121. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (1.125)  The expression for band gap energy is given by hc Eg = λ

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Therefore,

λc =

6.6 × 10 −34 × 3 × 108 = 1.125 μm 1.1 × 1.6 × 10 −19

122. Topic: Small Signal Equivalent Circuits Of Diodes, BJTs and MOSFETs: BJT (b)  When a CE amplifier’s emitter resistance is not bypassed, due to the negative feedback, the voltage gain decreases. As it has current-series feedback configuration, therefore, the input impedance increases. 123. Topic: Simple Diode Circuits (d)  Assume that both the diodes are ON. Then, the equivalent circuit will be as shown in the figure below.

20 Ω

1 kΩ ID A ID1

ID2

0.7 V

0.3 V

10 V

From the above circuit diagram, ID =

(10 − 0.7) A = 9.3 mA 1 × 103

I D2 =

(0.7 − 0.3) A = 20 mA 20

Applying KCL at node A, we get I D1 = I D − I D2 = −10.7 mA which is not possible. This implies that diode D1 is OFF. Hence, ID1 ≅ 0. Therefore, all the current is flowing through D2. Therefore, diode D2 is ON. 124. Topic: Simple Diode Circuits (d)  When Vi < −1.7 V, diode D1 is ON and diode D2 is OFF. Therefore, output voltage Vo = −1.7 V. When Vi > 2.7 V; diode D1 is OFF and diode D2 is ON. Therefore, output voltage Vo = 2.7 V. When −1.7 < Vi < 2.7 V, both the diodes D1 and D2 are OFF. Therefore, for this input voltage range, Vo = Vi. 125. Topic: BJT and MOSFET Amplifiers: Feedback (b)  Output voltage is sampled and added at the input as current. Therefore, it is voltage-shunt negative feedback, that is, voltage-current negative feedback.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 304

126. Topic: BJT and MOSFET Amplifiers: Feedback (b)  By opening the output feedback resistor, feedback signal becomes zero. Hence, it is current sampling. As the feedback signal is subtracted from the input signal (Vs) it is series mixing. Hence, the feedback topology is an example of current series feedback. 127. Topic: BJT and MOSFET Amplifiers: Differential (b)  Differential mode gain Ad does not depend on the value of RE. Common mode gain Acm decreases as the value of RE is increased. Therefore, the common-mode rejection ratio (CMMR) = ( Ad /Acm ) increases as the value of resistance RE is increased. 128. Topic: BJT and MOSFET Amplifiers: Multi-Stage (34.722)  Overall voltage gain, Av =

⎡ Zi 2 ⎤ ⎡ RL ⎤ Vout = Av1 Av2 ⎢ ⎥ ⎥⎢ Vin ⎣ Zi 2 + Z o1 ⎦ ⎣ RL + Z o2 ⎦

⎤ ⎡ 1 × 103 ⎤ ⎡ 5 × 103 = 10 × 5 ⎢ ⎥ = 34.722 3 3⎥⎢ 3 ⎣ 5 × 100 + 1 × 10 ⎦ ⎣1 × 10 + 200 ⎦ 129. Topic: Simple Op-Amp Circuits (d)  The circuit shown is a monostable multivibrator. 130. Topic: Active Filters (3.18)  Given that the cut-off frequency of the filter is 5 kHz. 1 Cut-off frequency ( LPF) = = 5 × 103 2p R2C Therefore, R2 =

1 = 3.18 kΩ 2p × 5 × 103 × 10 × 10 −9

131. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (500) Given VTh = −0.5 V, VGS = 2 V, VDS = 5 V, W/L = 100, Cox = 10 −8 F/cm 2 , mn = 800 cm2/V-s Drain current 1 W I D = μ n Cox ⎡⎣ 2(VGS − VTh )VDS − VDS 2 ⎤⎦ 2 L −1

⎡ ∂ ⎧1 ⎡ ∂I D ⎤ W ⎨ Cox ⎡⎣ 2 (VGS − VTh )VDS ⎥ = rds ⎢ ⎢ L ⎣ ∂VDS ⎩ 2 ⎣ ∂VDS ⎦ − VDS

2

}

⎤ ⎤⎦ ⎥ ⎦

−1

W W ⎡ ⎤ = ⎢ μ n Cox (VGS − VTh ) − μ n Cox VDS ⎥ L L ⎣ ⎦

−1

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Chapter 4  • Analog Circuits

Therefore, rds =

=

RC =

10 − VCE(sat) IC

1

μ n Cox

1 = 500 Ω 800 × 10 × 100 × ( 2 + 0.5 − 5)

132. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (0.07)  In linear region ⎡ V 2⎤ I D = k ⎢(VGS − VT )VDS − DS ⎥ 2 ⎦ ⎣

135. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (1.5)  Given that MOSFET M1 switches from saturation to linear region. Now, VDS = VGS - VTh; where VDS = Vout and VGS = Vin Therefore, VDS = Vout = Vin − VTh Drain current I D =

∂I D = kVDS ∂VGS

VDD − Vout 10 × 103 1 2 = × 100 × 10 −6 × 2 × (VGS − 0.5) 2

Solving the above equations, we get Vin = 1.5 V.

10 −3 = 0.01 0.1

In saturation region, I D =

1 k (VGS − VT ) 2 2

ID =

k (VGS − VT ) 2

∂ ID ∂VGS

=

k = 2

0.01 = 0.07 2

133. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (2.4)  The peak electric field in the silicon region is ESi =

2 × 0.2 = 0.8 V/μm 0.5

The peak electric field in the oxide region is Eox =

ESi e Si = 2.4 V/μm e ox

134. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (b)  Applying KVL in base-emitter loop, we get 5 − IB (50 × 103) − 0.7 = 0 Therefore, I B =

5 − 0.7 A = 86 μA 50 × 103

Now, I C = b I B = 50 × 86 × 10 − 6 A = 4.3 mA Assuming that the BJT is in saturation and applying KVL in the collector-emitter loop, we get

Ch wise GATE_ECE_CH04_Analog Circuits.indd 305

1 W 2 μ nCox (VGS − VTh ) 2 L

ID =

∂I D = 10 −3 ∂VGS

Therefore, k =

10 − 0.2 = 2279 Ω 4.3 × 10 −3

For RC ≥ 2279 Ω, the BJT is in saturation. Hence, option (b) is the correct option.

W (V − VTh − VDS ) L GS −8

Given that

=

305

136. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (b)  Small signal voltage gain AV =

RE RE I E RE Vout = = = rc + R E VT VT + I E RE Vin + RE IE

Therefore, AV ≅

I C RE  VT + I C R E

(∵ I C  I E )

For a nearly constant small signal voltage gain for a wide range of values of RE, ICRE >> VT 137. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (9.4)  Given that, b = high. Therefore, base current IB can be neglected. Base voltage VB = 12 ×

10 × 103 = 8V 10 × 103 + 5 × 103

VE = VB − 0.7 = 7.3 V Therefore, VCE = 12 − 7.3 = 4.7 V Hence, maximum undistorted peak-to-peak output voltage vo(p−p) = 2 × 4.7 V = 9.4 V 138. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (−237.76)  Given that: VBE = 0.7   V, b = 200, VT = 25 mV

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For the DC analysis of the circuit, the capacitors are considered as open circuit. Base volatage, VB = 12 ×

11 × 103 = 3V 11 × 103 + 33 × 103

Emitter voltage,  VE = 3 - 0.7 = 2.3 V Emitter current,    I E =

2.3 A = 2.271 mA (10 + 1 × 103 )

Base current,    IB =

IE 2.271 × 10 −3 = (b + 1) 201

= 11.334 μA Collector current   IC = b I B = 200 × 11.334 × 10 −6 A = 2.26 mA re = Resistor,        

25 × 10 −3 = 10.98 Ω 2.2771 × 10 −3

Applying KCL at inverting input of the op-amp, we get V2 V2 − Vout + =0 R1 R2 Therefore,

⎡1 Vout 1⎤ = V2 ⎢ + ⎥ R2 ⎣ R1 R2 ⎦

Vout ⎛ R1 R2 ⎞ ⎛ R1 + R2 ⎞ I1 = R2 ⎜⎝ R1 + R2 ⎟⎠ ⎜⎝ R1 R2 ⎟⎠ Therefore, Vout = I1R2 Hence,

141. Topic: Simple Op-Amp Circuits (d)  Due to virtual earth concept, voltage at the inverting input of the op-amp is equal to V2. Applying KCL at inverting terminal of the op-amp, we get V2 − V1 V2 V2 − Vo + + =0 R 2R 3R Rearranging the terms, we get Vo V2 V2 V2 V1 = + + − 3R R 3R 2 R R

Voltage gain vo −b RC = vi b re + (1 + b )( Rs ) =

Therefore,Vo = −3V1 +

−200 × 5 × 103 ( 200 × 10.98) + ( 201)10

= −237.76 139. Topic: Single-Stage BJT and MOSFET Amplifiers: Mid-Frequency Small Signal Analysis and Frequency Response (−1)  When base and collector terminals of a transistor are shorted, it act as a diode. So vout = −0.7 V. v −0.7 V = −1 Gain = out = vin 0.7 V 73 140. Topic: Simple Op-Amp RTh =Circuits = 28.59 kΩ 47 (c)  Given that Z i = ∞, A0L 47= ∞, Vio = 0. VTh = × 12 = 4.7 V 120 Refer to the figure given below.

11 V2 2

142. Topic: Power Supplies: Regulation (1093)  Given that VBE = 0.7 V; b = 100; VZ = 4.7 V and Vo = 9 V Voltage across resistor R (VR) is given by R ⎛ ⎞ VR = 9 × ⎜ 3⎟ ⎝ R + 1 × 10 ⎠ Due to virtual ground at the inputs of the op-amp, the voltages at the positive input and negative input of op-amp are equal. Therefore VR = VZ. So R ⎛ ⎞ 4.7 = 9 × ⎜ ⎝ R + 1 × 103 ⎟⎠ Therefore, R = 1093 W 143. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (c)  The given circuit can be redrawn as a

d+

+

R2

D1

V2 R1

I1

− Vout

I2

+ V1

⎛ RR ⎞ V2 = ( R1  R2 ) I1 = ⎜ 1 2 ⎟ I1 ⎝ R1 + R2 ⎠

Ch wise GATE_ECE_CH04_Analog Circuits.indd 306

− b

− vout +

R

c

−

D2

During positive half cycle, both diodes are forward biased. So output half wave of positive polarity is produced at points d–c. As polarity at points d and c is opposite to that of vout, hence vout is of negative polarity. During negative half cycle, both diodes are reverse biased. So, vout = 0 V

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Chapter 4  • Analog Circuits

144. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (5)  The Zener diode is not in the breakdown region. Hence it acts as an open circuit. Therefore, 1 × 103 Vo = 10 × =5V (1 × 103 + 1 × 103 ) 145. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (14)  The concentration of hole-electron pair at t = 1 μs is = 1020 × 10−6 = 1014/cm3 Hence, x = 14 146. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJTs (2)  Given that the emitter base voltage is VEB = 0.6 V, Therefore 3 − 0.6 IB = = 0.04 mA 60 × 103 I C = b I B = 50 × 0.04 × 10 −3 A = 2 mA

(

Hysteresis = VTH − VTL ⎛R⎞ ⎛R⎞ = −( −5) ⎜ 1 ⎟ + 5 ⎜ 1 ⎟ ⎝ R2 ⎠ ⎝ R2 ⎠ ⎛ R1 ⎞ ⎛ R1 ⎞ + 5⎜ 500 × 10 −3 = −( −5) ⎜ ⎝ 20 × 103 ⎟⎠ ⎝ 20 × 103 ⎟⎠ =

R1 2 × 103

Therefore, R1 = 500 × 10−3 × 2 × 103 = 1000 Ω = 1 kΩ 150. Topic: Simple Op-Amp Circuits f 3dB =

(159.15) 

1 2p × 10 × 103 × 0.1 × 10 −6 = 159.15 Hz

)

148. Topic: Simple Op-Amp Circuits

= −4 V Therefore,

=

sin(pt / 5) sin(pt / 5) cos pt − sin(pt ) pt / 5 (pt / 5)

Also x(t ) = xc (t ) cos 2p f c t − xs (t ) sin( 2p f c t ) (Low pass representation of band-pass signals)

VOB −6 = = 1.5 VOA −4

xc (t ) =

149. Topic: Function Generators, Wave- Shaping Circuits and 555- Timers (1) vout

vin VTH

sin(pt / 5) sin(pt / 5) , x s (t ) = (pt / 5) (pt / 5)

xce(t) is the complex envelope of x(t) xce(t) = xc(t) + jxs(t) =

L+

sin(pt / 5) [1 + j ] = pt / 5

2 sin(pt / 5) jp / 4 e (pt / 5)

152. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (20)  Drain current is given by I D = I D(sat) (1 + λVDS )

L−

Ch wise GATE_ECE_CH04_Analog Circuits.indd 307

⎛ sin(pt / 5) ⎞ p⎞ ⎛ (c)  x(t ) = − 2 ⎜ sin ⎜ pt − ⎟ ⎝ 4⎠ ⎝ pt / 5 ⎟⎠ ⎛ sin(pt / 5) ⎞ ⎛ p p ⎞ x (t ) = − 2 ⎜ ⎜ cos sin pt − sin cos pt ⎟⎠ ⎝ pt / 5 ⎟⎠ ⎝ 4 4

3

⎞ ⎛ 1 × 103 ⎞ ⎛ 1 × 103 ⎞ ⎛ 1 × 103 1+ +⎜ VOA = −5 ⎜ 3⎟ 3 3⎟ ⎜ ⎝ 1 × 10 ⎠ ⎝ 1 × 10 + 1 × 10 ⎠ ⎝ 1 × 103 ⎟⎠

VTL

151. Topic: Active Filters

We can write the above expression as

⎛ 1 × 10 ⎞ ⎛ 1 × 10 ⎞ = −6 V −1 (1.5)  VOB = −5 ⎜ ⎝ 1 × 103 ⎟⎠ ⎜⎝ 1 × 103 ⎟⎠ 3

1 2p RC

=

VEC = 3 − I C RC = 3 − 2 × 10 −3 × 500 = 2 V 147. Topic: Simple Diode Circuits (d)  For the circuit to be functioned as clamping circuit, RC >> T.

307

where l is the channel length modulation coefficient. Output resistance,

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ro =

dVDS 1 = dI D λ I D(sat)

156. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (1475)    I B =

Hence, ro =

1 Ω = 20 kΩ 0.05 × 10 −3

I E = (b + 1) I B

153. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (a)  The gate-source voltage (VGS) is equal to drainsource voltage (VDS) in the given circuit. Hence, the NMOSFET transistor is in saturation. In saturation, ID = K(VGS − Vth)2 = K(VDD − VSS − Vth)2 As VSS increases, ID decreases as a square factor. Hence option (a) is correct.

I C I S VBE /VT = e b b

=

b +1 I .eVBE /VT b S 700 ×10 −3

= (1.02)(10

−15

)e 25×10

−3

= 1475 mA 157. Topic: Simple Diode Circuits (0.6)  The given circuit can be represented as following bridge structure.

154. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT 2 kW

(83.88)  I 2 = I E 2 = I s eVBE2 / hVT

3 kW D

From the given circuit

I

1 mA

VBE2 = V2

1 kW

Similarly,

4 kW

6 kW

I1 = I E1 = I s eVBE1 / hVT From the given circuit, VBE1 = V1 Ratio of currents I1 and I2 is given by

The bridge is balanced, so I = 0. Therefore the circuit can be simplified as follows.

⎛ V1 −V2 ⎞ hVT ⎟⎠

⎜ I1 = e⎝ I2

Now, VT = kT/Q. Given that VT at 27°C = 26 mV. Therefore at 50°C, VT = 27.8 mV. Therefore, 80 × 10 − 3 =e 4 × 10 − 3

1 mA

3 kW

Ix 4 kW

6 kW

⎛ V1 −V2 ⎞ ⎝⎜ 27.8 ⎠⎟

Therefore, V1 − V2 = 83.88 mV 155. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (c)  The maximum value of electric field = 104 V/cm = 106 V/m. The width of the depletion region = (0.1 + 1)mm = 1.1 × 10−6 m Built in potential 1 ψ 0 = × (106 ) × (1.1 × 10 −6 ) 2 = 0.55 V Hence, option (c) is not correct.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 308

2 kW

the current Ix through 4 kΩ resistor can be calculated as: Ix =

1 × 10 −3 × 9 × 103 = 0.6 mA 9 × 103 + 6 × 103

158. Topic: Simple Diode Circuits (100)  The average voltage Vab across the 0.5 mF capacitor is 100 V. 159. Topic: Simple Diode Circuits (5)  During the positive half-cycle of the input, diode D1 is ON and diode D2 is OFF. Therefore, Vab =

6p sin w t = 2p sin w t 3

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Chapter 4  • Analog Circuits

During the negative half-cycle of the input, diode D1 is OFF and diode D2 is ON. Therefore Vab =

1 2 jw oC Z2 = 1 R+ 2 jw oC R×

6p sin w t = 3p sin w t 2

=

2p 3p Average of Vab = + = 5V p p

1 A 50

R 2 /j R + jR

For a Wein oscillator

160. Topic: Simple Op-Amp Circuits (250)  Due to the concept of virtual earth, the voltages at the inverting and non-inverting inputs of the op-amp are the same. Therefore, voltage at inverting terminal of the op-amp is 1 V. Hence, current through the 50 Ω resistor is I 50 Ω =

309

b= Therefore, 1 +

Z2 1 = Z1 + Z 2 5

R1 = 5 or R1 = 4 R2 R2

162. Topic: Sinusoidal Oscillators (1)  The given circuit can be redrawn as given below. 10 kW

As the input impedance of the ideal op-amp is infinite, no current flows through its input terminals. Hence, the current through the 50  Ω and 100  Ω resistors connected at the inverting input of the op-amp is the same. Hence, I50 Ω = I100 Ω.

vx 10 kW X R

10 kW vin

− vo

1 [250(1 + δ ) − 250(1 − δ )] Vo = 100 1 = × 250 × 2δ 100 1 = × 250 × 2 × 0.05 = 0.25 V 100 = 250 mV 161. Topic: Sinusoidal Oscillators (d)  The given oscillator circuit is that of a Wein bridge oscillator. Frequency of a Wein bridge oscillator is w o = 1 RC , where RC is the time constant. In the question, time constant = 2RC. Therefore, oscillation frequency in the given case is w o = 1 2RC Impedance of the series connected 2R and C is Z1 = 2 R +

1 jw C

At the oscillation frequency, Z1 = 2 R +

1 = 2( R − jR) jw oC

Impedance of the parallel connected 2C and R is 1 R× jw 2C Z2 = 1 R+ jw 2C At the oscillation frequency,

Ch wise GATE_ECE_CH04_Analog Circuits.indd 309

+ Applying KCL at the inverting terminal of the op-amp, we get ν in − 0 0 − νx = . Therefore, ν in = −ν x 10 × 103 10 × 103 Applying KCL at node X, we get 0 − νx ν − 0 νx − νo = x + 3 R 10 × 10 10 × 103 Putting the value of vx = −vin and vo= −12vin in the above equation, we get R = 1 kΩ 163. Topic: Simple Op-Amp Circuits (12)  The inverting terminal of the op-amp is grounded and the positive terminal of the op-amp is at 1 V. Therefore, V+ > V− The op-amp acts as a comparator, so Vo = Vsat = 12 V 164. Topic: Power Supplies: Regulation (2.8056) PQ1(max) = VCE (max) × I C(max)

VCE (max) =

(24 − 10) V = 14 V

(1)

I C((max) = ( 200 + 0.4 ) mA = 200.4 mA

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Here, the intrinsic Fermi level is below EFS by Φ B in the bulk and above EFS by Φ B at the interface. Thus, the MOS is in inversion.

4−0 ⎛ ⎞ A⎟ ⎜⎝∵ I R 2 = I R1 = 10 × 103 ⎠ Substituting values in Eq. (1), we get PQ1 (max) = 14 × 200.4 × 10−3 = 2.8056 w 165. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (598.67) R2 =

VT ⎛ I1 ⎞ 26 × 10 −3 ⎛ 1 × 10 −3 ⎞ = ln ln I 2 ⎜⎝ I 2 ⎟⎠ 100 × 10 −6 ⎜⎝ 100 × 10 −6 ⎟⎠

= 598.67 Ω 166. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (a)  While the parasitic capacitances are in pF (picofarad), the coupling and bypass capacitors are in pF (microfarad). Therefore, for the mid-frequency region, the parasitic capacitances behave like open circuits, whereas coupling and bypass capacitances behave like short circuits.

169. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (0)  Let the voltage across the top capacitor be V1 and that of the bottom capacitor be V2 as shown in the figure. During the positive half cycle, diodes both D1 and D2 are ON. The capacitors are charged to open circuit voltage, that is,

−10 + V1 = 0 That is

V1 = 10 V

Also, 10  + V2 = 0 Therefore,

V2 = −10 V

Once the capacitors charge to V1 = 10 V and V2 = −10 V, then the diodes become open-circuited (i.e. OFF condition). Also, during the negative half-cycle, both diodes D1 and D2 are OFF. D1

167. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (c)  The current in a diode is given by ⎡ ⎛ eV ⎞ ⎤ I D = I S ⎢exp ⎜ D ⎟ −1⎥ ⎝ kT ⎠ ⎦ ⎣ ⎡⎛ eDp pno ⎞ ⎛ eDn npo ⎞ ⎤ and      I S = A ⎢⎜ ⎟ +⎜ ⎟⎥ ⎢⎣⎝ Lp ⎠ ⎝ Ln ⎠ ⎥⎦ n2 n2 pno = i , npo = i NA ND ⎛ − Eg ⎞ ni2 = N C N D exp ⎜ ⎝ kT ⎟⎠ ⎛ − Eg ⎞ ⇒ I S ∝ exp ⎜ ⎝ kT ⎟⎠ Thus, for greater Eg (bandgap), Is is less at a given bias. Thus, Egx < Egy < Egz 168. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (a)  When the semiconductor in a MOS is in inversion, the intrinsic energy level crosses over the Fermi level and goes as deep into the other side of the Fermi level at the interface as distant it is from the Fermi level in the bulk.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 310

−

+

10 sin ω t

+ V1 −

C

+ R VO −

+ V2

−

C

D2 On applying Kirchhoff’s voltage law at the output side of the circuit, we get Therefore,

−V2 − V1 + Vo = 0 Vo = V1 + V2 = 10 + (−10) = 0

170. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (0.5)  On applying Kirchhoff’s voltage law to the circuit, we get −VCC + VBE1 + VEB2 + IB2(10 × 103) + 1 = 0 −2.5 + 0.7 + 0.7 + IB2(10 × 103) + 1 = 0 Solving the above equation, we get IB2 = 10 μA Therefore, IC2 = b2⋅IB2 = 50(10 × 10−6) = 500 μA VC2 = IC2 × 1000 = 500 × 10−6 × 1000 = 0.5 V

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171. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (6.656)  We have diffusion current density in the base region as follows: ⎛d ⎞ J nB = qDn ⎜ nB ⎟ ⎝ dx ⎠ Now,

dnB 0 −1014  =   = −2 × 1018/cm4 dx 0.5 ×10 −4 − 0

Also,

Dn = μnVT = 800 [cm2/s] × 26 × 10−3 ⇒Dn = 20.8 cm2/s

173. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (a)  Breakdown voltage of Zener diode D1 = 80 V Breakdown voltage of Zener diode D2 = 70 V When Vin exceeds 80 V, then Zener diode D1 is in breakdown region, then the given circuit becomes as follows: + − 80 V (0 − 100 V) D2

JnB = 1.6 × 10−19 × 20.8 × 2 × 1018

Thus,

Therefore,  JnB = 6.656 A/cm2 Since the area of emitter−base junction is 0.001 cm2, the area of base−collector junction is ABC = 0.001 cm2 Hence, the collector current is IC = ABC × JnB = 0.001 × 6.656 = 6.656 mA 172. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (1.6)  For the MOS capacitor shown in Figure I, since there are two capacitors of thickness t1 and t2 in series, we have

e1e 2 /t1t 2 ( e /t ) × ( e 2 /t 2 ) CI = 1 1 = (e1 /t1 ) + (e 2 /t 2 ) (e1t 2 + e 2 t1 ) / (t1t 2 ) =

e1e 2 e1 = e1t 2 + e 2 t1 (e1 /e 2 )t 2 + t1

Zener diode D2 never get a chance to enter into the breakdown region. At the maximum input voltage Vin max = 100 V, the voltage across Zener diode D2 is given by VD2 = (100 − 80)V = 20 V < Vbreakdown Therefore, only Zener diode D1 enters into breakdown region. 174. Topic: Simple Op-Amp Circuits (b)  Voltage at non-inverting terminal = Voltage at inverting terminal = (VCC − Vref). Therefore, IE =

Therefore, the collector current is obtained as follows: IE =

e1 t Eq

Since both the capacitors are of the same value, on comparing, we get t Eq = (e1 /e 2 )t 2 + t1 ⎡⎛ 4 ⎞ ⎤ t Eq = ⎢⎜ ⎟ × 3 + 1⎥ nm ⎝ ⎠ 20 ⎣ ⎦ ⎡⎛ 3 ⎞ ⎤ 8 t Eq = ⎢⎜ ⎟ + 1⎥ nm = = 1.6 nm ⎝ ⎠ 5 5 ⎦ ⎣

Ch wise GATE_ECE_CH04_Analog Circuits.indd 311

(1 + b ) I C b

IC = I0 =

For the capacitor shown in Figure II, we have CII =

VCC − (VCC − Vref ) Vref = R R

b ⎛ Vref ⎞ 1 + b ⎜⎝ R ⎟⎠

where, IE is the emittor current of the transistor and IC is the collector current of the transistor. 175. Topic: Simple Op-Amp Circuits (3)  The voltage at inverting terminal is V− =

10( 2 × 33 ) =2V ( 2 × 103 + 8 × 103 )

Whenever V+ > V−, the BJT is ON which turns ON the LED. Therefore, when Vi > 2 V, then LED turns ON and is turned off when Vi < 2 V.

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ON

ON

180. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (1.2)  For the linear operation, we have

ON

6V 4V 2V

I D = k N′ (VGS − VT )VDS dI D = k N′ VDS = g m dVGS Substituting values, we get OFF

OFF

0.5 × 10 −6 = k N′ (50 × 10 −3 ) ⇒ k N′ = 10 −5

OFF

dI D = k N′ (VGS − VT ) = gd dVDS

Thus, from the given wave form, we can say that the LED turns ON (or glows) for 3 times. 176. Topic: Sinusoidal Oscillators: Op-Amp Configuration (a)  In the combination of two-diode circuit, at a given time, only one diode turns ON. • When Vo = + Vsat, then D1 is ON and D2 is OFF. • When Vo = −Vsat, then D2 is ON and D1 is OFF.

8 × 10 −6 = 10 −5 ( 2 − VT ) VT = 1.2 V 181. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (2.105)  The ripple voltage is

Thus, the given circuit introduces amplitude stabilization by preventing the op-amp from saturating and thus it produces sinusoidal oscillations of fixed amplitude. 177. Topic: Small Signal Equivalent Circuits of Diodes, BJTS and MOSFETs: MOSFETs (28.35)  The equation of current in saturation is given by ⎞ 1⎞ ID = ⎟⎠ 2 ⎟⎠

N

C ox

⎞W ⎞ (V − VT ) 2 1 + VDS ⎟⎠ L ⎟⎠ GS

−3 ⎛ T ⎞ 1 × 1 × 10 μ = I DC ⎜ ⎟ = = 2.105 V ⎝ C ⎠ 475 × 10 −6

182. Topic: Simple Op-Amp Circuits (0.798)  Initially, when the switch is closed, VB = 10 V. V01 = −10 V V0 = −V2 = −5 V +10 V

The drain conductance is ∂I D ⎞W ⎞ ⎞1⎞ = ⎟ ⎟ N C ox ⎟ ⎟ (VGS − VT ) 2 2 ∂VDS ⎠L ⎠ ⎠ ⎠ 1 = 7 0 10 −6 4 (1.8 − 0.3) 2 2 = 28.35 10 −6 S = 28.35 µS

100 µF 0. 09

178. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (d)  The Ebers–Moll model of a BJT is valid for all regions (active, saturation and cut-off) of operation. 179. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (d)  We have 1 RDS = μ n C (W /L)(VGS − VT ) Therefore, if VGS increases, then RDS should decrease. Hence, option (d) is incorrect.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 312

−

VOL

470 Ω

VB 10 kΩ

Vo

+

4 kΩ 1 kΩ

• At t = 0, the switch opens. • At t → ∞, VB → −10 V. VB (∞) + [VB (0) − VB (∞)]e − t / τ = −10 + (10 + 10)e − t / RC • At t = t1, let VB = −1 V therefore, −1 = −10 + 20e -t1 / RC ⎛ 20 ⎞ t1 = RC ⎜ ln ⎟ = 0.798 s ⎝ 9⎠

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Therefore, Vo = -8 V Hence, current through load resistor 10 kΩ

183. Topic: Simple Op-Amp Circuits (413.79)  The total input is Vios + Vinput = 5 mV + 25 mV = 30 mV

10 kΩ = −

Therefore, 1 + ( RF /R1 ) × 30 mV 1 + (1/AOL )[1 + ( RF /R1 )]

1 + (15 × 103 /1 × 103 ) 30 × 10 −3 V 1 + (1/100)[1 + (15 × 103 /1 × 103 )] = 413.79 mV =

1 1 1  V o = +  + + +  + ∞ 2 4 6 

Overall, we get T = TON + TOFF = 0.693(RA + 2RB)C = 0.693(2.2 ´× 103 + 9.4 ×´ 103) ×´ 0.022 ×´ 10-6 = 1.769 ×´ 10-4 s The frequency of oscillations is f =

Hence, Vo → ∞ . Therefore, the output voltage will be positive supply voltage Vo = +15 V 185. Topic: Simple Op-Amp Circuits (−800)  By virtual ground concept, the voltage at the inverting input of the op-amp (VI) = non-inverting input of the op-amp (VNI). Responsivity =

IP P0

Hence, for this circuit VNI = VI = 0 ⇒ Photodiode current IP = 0.8 × 10 × 10 −6 = 8 µA Applying KCL at the inventing terminal of the op-amp, we get IP =

Ch wise GATE_ECE_CH04_Analog Circuits.indd 313

0 − Vo = 8 × 10 −6 1 × 106

1 1 = Hz = 5.65 kHz T 1.769 × 10 −4

187. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (c)

Now, 1 1 1 1 + + + + = ∞ 2 3 4 5

TOFF = 0.693RBC

and

1 1 1 1  = +  1 + + + +   2 2 3 4

1+

A = −800 µA

TON = 0.693(RA + RB)C

184. Topic: Simple Op-Amp Circuits (15)  If only V1 is present and the remaining voltages are OFF, then the voltage at non-inverting terminal = 0 V (because parallel connection of infinite resistances leads to be short circuit). Thus, even if all V1, V3, … , VN−1 are present, the voltage at non-inverting terminal = 0 V By using virtual ground concept, the voltage at inverting terminal = 0 V. On applying KCL at inverting terminal, we get 0 − V2 0 − V4 0 − V6 0 − Vo + + ++ =0 10 × 103 10 × 103 10 × 103 10 × 103 V o = −(V2 + V 4 + V6 + )

10 × 103

186. Topic: Function Generators, Wave Shaping Circuits and 555 Timers: 555 Timers (5.65)  The given circuit acts like an astable multivibrator with

0

Normalized Excess Carrier Conecntration

V0 =

8

105 pB pB0 nC nC0 −1 Base (N)

Collecter (P)

nE nE0 Emitter (P) X and Y axes are not to scale

The above figure shows the minority carrier distribution in the PNP transistor for the inverse active mode. In this case, the B-E junction is reverse biased and the B-C junction is forward biased. Electrons from the collector are now injected into the base. The gradient in the minority carrier electron concentration in the base is in the opposite direction compared with the forward active mode. So, the emitter and collector currents will change direction. 188. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (c)  Transconductance, Gm =

I o Output current = Input voltage Vi

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RS

IL

IO

12 V

2 kΩ

+ VS

Vi

-

Ri

RO

GmVi

IL

73 kΩ

189. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (d)  Miller effect increases input capacitance and thereby decreases the higher cut-off frequency. Cu C rp

G mV p

RC

47 kΩ 2 kΩ

Replacing the above circuit by its Thevenin’s equivalent circuit,

RL

12 V 2 kΩ

E

Breakingthe above circuit, we have 28.59 kΩ C cp

CM RC

OA

RL

4.7 V

2 kΩ

E

Now,

CM = Cm[1 + gm(RC || RL)]

Total capacitance (equivalent), Cf = Cp + CM For 3-dB-frequency, F3dB ∝

1 Cf

Using Thevenin’s theorem, we have RTh = 73 × 103 || 47 × 103 = 28.59 kΩ VTh =

47 × 103 × 12 V = 4.7 V ( 47 × 103 + 73 × 103 )

Due to Miller effect, Cf increases. Hence, F3dB will decrease.

Applying KVL in the loop shown in the second circuit above, we have

190. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (128.0) The midband gain of common emitter configuration is V AV = o = − g m RL Vi

     −4.7 + 0.7 + 2 × 103 I E = 0 (Given: IB is negligible) 

Magnitude of voltage gain, Vo = gm RL Vi Using DC analysis (for calculation of gm), (i)  All capacitors are open circuit. (ii)  AC voltage is replaced by short circuit.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 314

⇒ IE =

4 4.7 − 0.7 = = 2 mA 3 2 × 10 2 × 103

Hence, IE = IC = 2 mA Now, gm =

IC 2 mA = = 0.08 A/V VT 25 mV

Using AC analysis, (i)  All capacitors are short circulated. (ii) All DC voltage source are placed by short circuit.

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⎛ I ⎞ (iv)  common emitter current gain ⎜ b = C ⎟ IB ⎠ ⎝ increase Therefore, option (c) is correct.

2 kΩ

73 kΩ

V0

193. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (0.50) For n-channel MOSFET,

8 kΩ

V in 47 kΩ

will

VDS = 0.1 V, VGS = 0.7 V, VTH = 0.3 V,

2 kΩ

μnCox = 100 μA/V2 = 0.1 mA/V2, W/L = 50, VGS > VTH and VDS < VGS - VTH Common emitter without RE The modified figure is

n-Channel MOSFET is in linear region. Therefore, transconductance for n-channel MOSFET is given by

V0

RL = 2 kΩ  8 kΩ = 1.6 kΩ 47 kΩ  73 kΩ

So, AV =

gm = ⇒

gm = 0.1 × 50 × 0.1 = 0.5 mA/V

194. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (3.18)  The circuit is a half-wave rectifier and the DC voltmeter will read the average value of Vo. +

Vo = − g m RL = g m RL′ Vin

10 sin ωt

⇒ AV = 0.08 × 1.6 × 103 = 128

Ch wise GATE_ECE_CH04_Analog Circuits.indd 315

Vo -

Hence, the average value of Vo, i.e. output of half-wave rectifier, is given by

VGS > VTH and VDS > VGS − VTH

192. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (c)  If the reverse bias across the base-collector junction increases, then (i) depletion width across base-collector junction will increase (ii) effective base width will decrease (iii)  recombination of carrier in base region will decrease

1Ω

t = 50 Hz

191. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (c)  Given that

Hence, MOSFET is in saturation region and therefore it will act as current source because channel length modulation effect is considerable. Thus, output impedance will exist. So, MOSFET acts as a current source with finite output impedance.

∂I o μC W = n ox VDS L ∂VGS

Vavg =

Vm 10 = = 3.18 V p p

195. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (a)  Case I: 0 < t < T/2, Vin(t) = 14 V D1 → Forward bias (0.7 V) D2 → Breakdown (6.8 V) The equivalent circuit is as shown below. + + +14 V

V

-

10 µF

+ 0.7 V 10 Ω

+ - 6.8 V

Vout(t)

-

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Maximum Vout = 7.5 V On applying Kirchhoff’s voltage law (KVL) in the loop, we have −14 + VC + 7.5 = 0 VC = 6.5 V

VTH =

VCC 18 × 16 × 103 × R2 = = 4.8 V R1 + R2 ( 44 × 103 + 16 × 103 )

a = 1, β =

α = ∞ (IB = 0 and IC = IE) 1− α

Applying KVL in loop 1, we get

T , capacitor will charge up to 6.5 V due to small 2 time constant. At t =

IE =

VTH − VBE 4.8 − 0.8 = = 2 mA ≅ I C RE 2 × 103

Applying KVL in collector- emitter loop

T Case II: < t < T , Vin(t) = −14 V 2

  VCE = 18 - IC (4 × 103 + 2 × 103) = 18 - 2 × 10-3 × 6 × 103 = 6 V

D1 → Reverse bias (open circuit) D2 → Forward bias The equivalent circuit is as shown below.

197. Topic: Simple Op-Amp Circuits (b) Vout − V + in

+

-

+

10 kΩ

Vsat = +15 V - Vsat = -15 V

V = 6.5 V 10 µF 10 Ω

-14 V

5 kΩ + −

3V

Vout(t) Short circut

Given that Vsat = +15 V, - Vsat = -15 V, Vout = + Vsat = 15 V, VF = VUT For calculating upper threshold voltage, we have

On applying KVL in the loop, we have 14 + Vout + 6.5 = 0 ⇒ Vout = -20.5 V (minimum) 196. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (6.0)  Given that VBE = 0.8 V and a = 1 From Thevenin’s equivalent circuit, we have 18 V 4 kΩ IC RTH

Vref × 10 × 103 Vout × 5 × 103 3 × 2 15 × 1 + = + 3 3 3 3 1+ 2 1× 2 5 × 10 + 10 × 10 5 × 10 + 10 × 10 6 15 = + = 7V 3 3 For calculating lower threshold voltage, we have

VUT =

Vout = −Vsat = −15, VF = VLT Vref × 10 × 103 Vout × 5 × 103 3 × 2 15 + = − 5 × 103 + 10 × 103 5 × 103 + 10 × 103 1 + 2 3 6 15 = − = −3 V 3 3

VLT =

198. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (0.90)  5V IC

IE VTH

4.8 kΩ

2 kΩ

12 kΩ

C B

Vin

RTH =

R1 R2 44 × 103 × 16 × 103 = = 11.73 kΩ R1 + R2 ( 44 × 103 + 16 × 103 )

Ch wise GATE_ECE_CH04_Analog Circuits.indd 316

IB

E

(i)

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Chapter 4  • Analog Circuits

Given,

IC =

−5 + VGS + I D = 0

VBE ON = 0.7 V

Therefore,

VCE sat = 0.2 V

Given MOSFET is in saturation region, we have

VCC − VCEsat 4.8 × 103

5 − 0.2 = 1 mA 4.8 × 103

=

VGS = 5 − I D

Kn 2 VGS − VTN ) ( 2 1 2 Therefore, I D = (5 − I D − 1) 2 ID =

Applying KVL in input loop, we have Vin − I B × 12 × 103 − 0.7 = 0 ⇒ IB =

or, 2 I D = ( 4 − I D )

Vin − 0.7 = 0.108 mA = 108.33 μA 12 × 103

I D2 − 10 I D + 16 = 0

−3

IC 1 × 10 = = 9.25 I B 0.108 × 10 −3

9.25 b α= = = 0.90 1 + b 10.25

( I D − 8)( I D − 2) = 0 Hence, I D = 8 mA, 2 mA If I D = 8 mA ,

199. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs

VGS = (5 − 8)V = −3 V < VTN

VDD = 8 V

VGS = (5 − 2)V = +3 V > VTN So, I D = 2 mA is the correct answer.

R0 = 1 kΩ

RD

ID

200. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: Diodes (0.52)  Open circuit voltage of a solar cell

R2 = 5 MΩ RC = 1 kΩ

RS

Taking the Thevenin's equivalent voltage VTH and resistance RTH, the equivalent circuit is shown below. 5 × 106 VTH = × 8 = 5 V, 3 × 106 + 5 × 106 3 × 106 × 5 × 106 RTH = = 1.875 MΩ 3 × 106 + 5 × 106

ID

J L1 = 20 J L As temperature is constant, J S′ = J S , VT′ = VT Given that VOC = 0.451V From Eq. (1), ⎛J ⎞ ⎛ J ⎞ Voc = VT ln ⎜1 + L ⎟ ≈ VT ln ⎜ L ⎟ JS ⎠ ⎝ JS ⎠ ⎝

1 kΩ

Now, ⎛ J′⎞ VOC ′ = VT′ ln ⎜1 + L ⎟ J S′ ⎠ ⎝

OA

5V + -

Ch wise GATE_ECE_CH04_Analog Circuits.indd 317

⎛ J ⎞ VOC = VT ln ⎜1 + L ⎟ (1) JS ⎠ ⎝

where JL = photo current density, JS = reverse saturation current density and VT = thermal voltage As intensity of the incident light increased 20 times,

8V

1.875 MΩ

(not possible)

If I D = 2 mA ,

(2.0) 

R1 = 3 MΩ

2

I D2 − 8 I D + 16 = 2 I D

Therefore,

b=

317

+ VGS -

⇒

⎛ J′⎞ VOC ′ ≈ VT ln ⎜ L ⎟ ⎝ JS ⎠

⇒

⎛ 20 J L ⎞ VOC ′ ≈ VT ln ⎜ ⎝ J S ⎟⎠

1 kΩ

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⇒

⎛J ⎞ VOC ′ ≈ VT ln ⎜ L ⎟ + VT ln 20 ⎝ JS ⎠

⇒

VOC ′ = 0.451 + (0.025 ln 20) = 0.525 V

201. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs

VGS = 2.5 − Vx VON = VGS − VTh = 1.5 − Vx VGS > VTh and VDS > VON Therefore, M2 is in saturation and M1 is also conducting. So,

(b)  Given that W2 = 2W1, V0V 2 = 2V0V 1 We know I D = kn (VGS − VTH ) 2 , g m = 2k n (VGS − VTH ) where and

μ n Cox w 2L

I D1

μC w 2 = n ox (VGS1 − VTH1 )(1) 2L

and ⎛μ C w ⎞ I D2 = 8 ⎜ n ox 1 ⎟ (VGS1 − VTH1 ) 2 (2) ⎝ 2L ⎠



From Eqs. (1) and (2), we have

or     1.5 - Vx = ± 1

VTh > VGS

I D2 = 8 I D1

So, M2 is in cut off region. Vx = 2.5 V (Not possible) For Vx = 0.5 V For transistor M2, VGS = 2 V VTh > VGS and VDS > VON So, M2 is in saturation region. For transistor M1, VGS = 2 V VDS = Vx = 0.5 V

Also g m1 = ⇒

= I D1 saturation

Therefore, kn (2.5 - Vx - 1)2 = kn(2 - 1)2

and

Therefore,

saturation

Hence, Vx = 0.5 V or 2.5 V For Vx = 2.5 V, for transistor M2 VGS = 0 V

VON = VGS − VTH kn =

I D2

μ n Cox w1 (VGS1 − VTH1 ) L

g m2 = 4 g m1

202. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (c)

VTh < VGS and VDS < VON So, M1 is in linear region. 203. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: BJT (50) 5V

3V

1Ω 2.5 V

M2

Vo Q1

Vx

2V

VS

M1 Q2 RB2

For transistor M2, VDS = 3 − Vx

Ch wise GATE_ECE_CH04_Analog Circuits.indd 318

-5 V

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Chapter 4  • Analog Circuits

Voltage gain of this combination of BJT is

Avf × BW ′ = Avfo × BW

Vo − Re = Vs re1 + re2

Therefore, 80 × BW ′ = 105 × 8 or BW ′ = 10 kHz

Now, re1 =

Therefore,

VT VT 26 = = = 10 Ω I e1 I c1 2.6

Avf =

and re2 =

VT VT 26 = = = 10 Ω I e2 I c2 2.6

Therefore,

Avfo ⎛ f ⎞ 1+ ⎜ ⎝ BW ′ ⎟⎠

204. Topic: Simple Op-Amp Circuits (44.37) For practical non-inventing op-amp, gain is given by

—

(1)

20 kΩ Applying KCL at a,

1 + R2 1 + 79 × 10 Vo = AF = = = 80 Vi R1 1 × 103 3

Vout − Vx Vx − 0.7 = 20 × 103 5 × 103 Therefore, Vo − Vx = 4Vx − 2.8    

+

or Vo = 5Vx − 2.8 Vο

−

(1)

Now, I1 = 31I Vx /VT

Ise

79 Ω 1 kΩ

V 1 × 103 1 b= f = = 3 3 80 Vo 1 × 10 + 79 × 10 A0 = 1 + A0 b

105 1 + 105 ×

1 80

= 80

or (High gain value) R2 = 1 + 79 = 80 R1

Gain bandwidth product of close loop sys­tem = Gain bandwidth product of open loop system

Ch wise GATE_ECE_CH04_Analog Circuits.indd 319

Vout +

I

The circuit is a non-inventing op-amp

Avfo = 1 +

= 44.37

a Vx

1 A0 = 10 , AF ≈  For   b

Avfo = 105

2

31 I

5

Ao =

⎛ 15 × 10 ⎞ 1+ ⎜ ⎝ 10 × 103 ⎟⎠ 3

5 kΩ Vp

A0 Vo = AF = Vi 1 + A0 b

+ Vf −

80

=

2

205. Topic: Simple Op-Amp Circuits (1.145) 20 kΩ

V0 −1000 = = −50 Vs 20

Vi

319

= 31I s eVP /VT

Therefore,

Vx V = ln 31 + P VT VT

or

Vx − VP = ln 311 VT

Hence,

Vx = 0.789 V

From Eq. (1), we have Vo = 5 × 0.789 − 2.8 = 1.145 V 206. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (c)  Given that, the two transistors are in saturation. Therefore iD1 = iD2 = iD . (1) Transconductance gm =

ΔiD i i = D = D1 ΔVgs Vgs Vgs

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When vout = -10 V, using KVL we have

⇒ g m = g m1

−VC1 − 2.5 − vout = 0

D iD

⇒ −VC1 − 2.5 − ( −10) = 0 ⇒ VC1 = 7.5

M2 Now

iD

t

G +

1 (1 × 10 −3 )dt = 7.5 V 1 × 10 −6 ∫0

M1 Vgs

−

⇒

S (2) Output resistance 1. Replace Vin with short circuit.

1 × 10 −3 [t ] = 7.5 1 × 10 −6

⇒      t = 7.5 × 10-3 s = 7.5 ms

2. Disconnect RL′ (here RL′ = ∞) 3. Apply Vx at output port after drawing medium frequency AC equivalent circuit. Shorting the input voltages gives Vp 1 = 0 V and Vp 2 = − I x ro1

208. Topic: Single-Stage BJT and MOSFET Amplifiers (d)  Transimpedance amplifier contains input of Norton’s circuit and output of Thevenin circuit. A good transimpedance amplifier has low input impedance and low output impedance. Ro

Applying KCL at A

I1

(V − I x ro1 ) I x = g m 2Vp 2 + x ro2 = − g m 2 ro1 I x +

⎛r ⎞ ⎡ Vx r ⎤ − I x ⎜ o1 ⎟ Vx = ro2 ⎢1 + ro1 g m 2 + o1 ⎥ I x ro2 ro2 ⎦ ⎝ ro2 ⎠ ⎣

Vx = ro2 + ro1ro2 g m 2 + ro1 ≅ ro1ro2 g m 2 Ix

1 kΩ − 0V

v in(t)

+

I

RL

209. Topic: Simple Op-Amp Circuits (0.5) V0 =

207. Topic: Simple Op-Amp Circuits (c) 1 µF − + I VC1 + 2.5 V −

+ R I − m 1

Ri

Rs

Is

−31 × 103 × 1 V = –31 V 1 × 103

As V0 < −Vsat, we have Vo = −15 V. +

31 kW

VC 2 1 µF

+15 V

−

1 kW vout(t)

-

+

+ 1V Given vin(t) = u(t). At t = 0, voltage across each capacitor is zero. Therefore at t > 0 I=

Vin − 0 1 = = 1 mA 3 1 × 10 1 × 103

VC 1 and VC 2 increase till t = 2.5 ms After t = 2.5 m sec, we have V2 = 2.5 V. VC 1 → increasing with time.

Ch wise GATE_ECE_CH04_Analog Circuits.indd 320

Va

-

+

V0 -15 V

-

Now using superposition principle, Va =

−15 × 1 × 103 + 1 × 31 × 103 32 × 103 =

31 − 15 16 = = 0.5 V 32 32

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Chapter 4  • Analog Circuits

210. Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs: MOSFETs (0.4226) Given

⎡ 2 × 1 × Vx − Vx 2 ⎤ 2 2 ⇒⎢ ⎥ = ( 2 − Vx − 1) 2 ⎣ ⎦ 2 ⇒ 2Vx − Vx2 = 2(Vx − 1) 2

⎛W ⎞ ⎛W ⎞ ⎜⎝ ⎟⎠ = 2 ⎜⎝ ⎟⎠ L 2 L 1

⇒ 2Vx − Vx2 = 2(Vx2 + 1 − 2Vx ) ⇒ 2Vx − Vx2 = 2Vx2 + 2 − 4Vx

VT2 = VT1 = 1 V

⇒ 3Vx2 − 6Vx + 2 = 0

VGS2 > 1.0 V For M2,

⇒ Vx =

VGS 2 − Vr2 = 2 − Vx − 1 = 1 − Vx < 1 V

M2 → saturation region Also, ID1 = ID2 V2 ⎤ ⎛W ⎞ ⎡ μ nCox ⎜ ⎟ ⎢(VGS1 − VT )VDS1 − DS1 ⎥ ⎝ L ⎠1 ⎣ 2 ⎦ =

6 ± 36 − 4 × 3 × 2 2×3

⇒ Vx = 1 ±

VDS2 = (3.3 − Vx ) > (VGS2 − VT ) ⇒   M1 → linear region

321

1 3

Now So,

VGS 2 = ( 2 − Vx ) > 1 ⇒ 1 - Vx > 0 Vx = 1 −

1 3

= 0.4226

μ nCox ⎛ W ⎞ 2 ⎜ ⎟ (VGS 2 − VT ) 2 ⎝ L ⎠2

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Digital Circuits

CHAPTER5

Syllabus Number systems; Combinatorial circuits: Boolean algebra, minimization of functions using Boolean identities and Karnaugh map, logic gates and their static CMOS implementations, arithmetic circuits, code converters, multiplexers, decoders and PLAs; Sequential circuits: latches and flip-flops, counters, shift-registers and finite state machines; Data converters: sample and hold circuits, ADCs and DACs; Semiconductor memories: ROM, SRAM, DRAM; 8-bit microprocessor (8085): architecture, programming, memory and I/O interfacing.

Chapter Analysis Topic

GATE 2009

GATE 2010

GATE 2011

GATE 2012

GATE 2013

GATE 2014

GATE 2015

GATE 2016

GATE 2017

GATE 2018

Combinatorial circuits

5

3

2

2

1

13

8

9

6

4

Sequential circuits

2

1

3

2

4

5

3

4

2

1

1

1

1

1

1

1

1

1

1

2

3

2

2

Number systems

1

Data converters Semiconductor memories 8-bit microprocessor (8085)

1 1

1

1

Important Formulas   1. X + Y = Y + X

14. X + YZ = ( X + Y ) ⋅ ( X + Z )

  2. XY = YX

15. XY + X Y = X

  3. X ⋅ X = 0

16. ( X + Y )( X + Y ) = X

  4. X + X = 1

17. ( X + Y ) ⋅ Y = XY

  5. X ⋅ X ⋅ X … X = X

18. X Y + Y = X + Y

  6. X + X + X +  + X = X

19. X + XY = X

  7. 1⋅ X = X

20. X ( X + Y ) = X

  8. 0 + X = X

21. ZX + Z XY = ZX + ZY

  9. 0 ⋅ X = 0

22. ( Z + X ) ⋅ ( Z + X + Y ) = ( Z + X ) ⋅ ( Z + Y )

10. 1 + X = 1

23. XY + X Z + YZ = XY + X Z

11. X + (Y + Z ) = Y + ( Z + X ) = Z + ( X + Y )

24. ( X + Y ) ⋅ ( X + Z ) ⋅ (Y + Z ) = ( X + Y ) ⋅ ( X + Z )

12. X (YZ ) = Y ( ZX ) = Z ( XY )

25. [ X 1 + X 2 + X 3 +  + X n ] = X 1 ⋅ X 2 ⋅ X 3 … X n

13. X ⋅ (Y + Z ) = X ⋅ Y + X ⋅ Z

26. [ X 1 ⋅ X 2 ⋅ X 3 … X n ] = [ X 1 + X 2 + X 3 +  + X n ]

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27. XY + X Z = ( X + Z )( X + Y )

Z = A3 ⋅ B3 + x3 ⋅ A2 ⋅ B2 + x3 ⋅ x2 ⋅ A1 ⋅ B1 + x3 ⋅ x2 ⋅ x1 ⋅ A0 ⋅ B0

28. ( X + Y )( X + Z ) = XZ + XY

where xi = Ai Bi + Ai Bi .

29. X ⋅ f ( X , X , Y , Z , …) = X ⋅ f (1, 0, Y , Z , …)



30. X + f ( X , X , Y , Z , …) = X + f (0, 1, Y , Z , …)

48. Multiplexer with n data inputs, one output and m control inputs: n = 2m.

31.

f ( X , X , Y , Z , …) = X ⋅ f (1, 0, Y , Z , …)

49. Demultiplexer with one input, n outputs and m control inputs: n = 2m.

+ X ⋅ f (0, 1, Y , Z , …) 32.

f ( X , X , Y , Z , …)

50. Retriggerable monoshot: Output pulse width is (n − 1)Tt + T where n is the number of trigger pulses, Tt is the time period of trigger pulses, T is the output pulse width for a single trigger pulse.

= [ X + f (0, 1, Y , Z , …)][ X + f (1, 0, Y , Z , …)] 33. X = X 34. Fan-out of a logic family = IOH/IIH or IOL/IIL, whichever is lower. 35. Figure-of-merit of a logic family = tP × PD, where tP = propagation delay and PD = power dissipation per gate. 36. AND gate: Y = A ⋅ B ⋅ C ⋅…⋅ D

51. IC timer 555 based astable multivibrator: Time period is T = 0.69(R1 + 2R2)C

52. IC timer 555 based monostable multivibrator: Time period is T = 1.1RC, where R is the charging path resistance, C is the capacitance.

37. OR gate: Y = A + B + C +  + D 38. NOT gate: Y = A

53. Characteristic equation (R-S latch/flip–flop): Qn +1 = S + RQn and S + R = 1 (active LOW inputs) and

39. NAND gate: Y = A ⋅ B ⋅ C ⋅ D …

Qn +1 = S + R Qn and S · R = 0 (active HIGH inputs).

40. NOR gate: Y = A + B + C + D … 41. EX-OR gate: Y = A ⊕ B = AB + AB

54. Characteristic

44. Full-adder: Sum, S = ABCin + ABC in + ABC in + ABCin and Carry, Cout = BCin + AB + ACin

ABBin + ABBin and Borrow, Bo = AB + ABin + BBin 47. Four-bit magnitude comparator: If the two four-bit numbers are represented by A3A2A1A0 and B3B2B1B0, then the conditions X(A = B), Y(A > B) and Z(A < B) are given by the following Boolean functions.

Y = A3 ⋅ B3 + x3 ⋅ A2 ⋅ B2 + x3 ⋅ x2 ⋅ A1 ⋅ B1 + x3 ⋅ x2 ⋅ x1 ⋅ A0 ⋅ B0

GATE_ECE_CH05_Digital_Circuits.indd 324

latch/flip–flop):

55. C  haracteristic equation (T latch/flip–flop): Qn +1 = T ⋅ Qn + T ⋅ Qn (active HIGH T-input) and Qn +1 = T ⋅ Qn + T ⋅ Qn (active LOW T-input).

and

46. Full-subtractor: Difference, D = ABBin + ABBin +

X = x3 · x2 · x1 · x0

(J-K

Qn +1 = J ⋅ Qn + K ⋅ Qn (active LOW J and K inputs)

43. Half-adder: Sum, S = AB + AB and Carry, C = AB

D = AB + AB

equation

Qn +1 = J ⋅ Qn + K ⋅ Qn (active HIGH J and K inputs) and

42. EX-NOR gate: Y = A ⊕ B = A ⋅ B + A ⋅ B

45. Half-subtractor: Difference, Borrow, Bo = AB

where R1 is the charging path resistance, R2 is the discharge path resistance and C is the capacitance.

56. Characteristic equation (D latch/flip–flop): Qn+1 = D. 57. Modulus of a shift counter: 2n where `n’ is number of flip–flops. 58. Resistive divider type n-bit D/A converter: The analog output voltage is VA = 59.

V1 × 20 + V2 × 21 + V3 × 22 +  + Vn × 2n −1 2n − 1

Binary ladder type n-bit D/A converter: VA =

V1 × 20 + V2 × 21 + V3 × 22 +  + Vn × 2n −1 2n

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Chapter 5  •  Digital Circuits

2n − 1. (2n including one for polarity.)

60. Percentage resolution in an n-bit D/A converter: 1 × 100 % Percentage resolution = n 2 −1 61. Percentage resolution in an n-bit BCD-input D/A converter: 1 ⎞ ⎛ Percentage resolution = ⎜ n/ 4 ⎟ × 100 % ⎝ 10 − 1⎠ 62. The analog voltage at op-amp output for current steering mode of operation for RF = R is − DVref,

325

where D is the fractional binary value of input digital word and Vref is the reference voltage.

63. Flash-type A/D converters: The number of comparators required in an n-bit A/D converter is

64. Half-flash A/D converter: The number of comparators required in n-bit A/D converter is 2 × (2n/2) 65. Counter-type A/D converter: The average conversion time in an n-bit counter-type A/D converter is 2n−1 clock cycles. 66. Successive approximation A/D converter: The conversion time in an n-bit successive approximation type A/D converter equates n clock cycles. 67. S/N ratio of a n-bit Nyquist frequency A/D converter is (6.02n + 1.76)dB.

QUESTIONS 1.

An 8-bit successive approximation analog to digital converter has full scale reading of 2.55 V and its conversion time for an analog input of 1 V is 20 µs. The conversion for a 2 V input will be (a) 10 µs (b) 20 µs (c) 40 µs (d) 50 µs

A B

X C

(GATE 2000: 1 Mark) 2.

The number of hardware interrupts (which require an external signal to interrupt) present in an 8085 microprocessor are (a) 1 (b) 4 (c) 5 (d) 13 (GATE 2000: 1 Mark)

3.

The most commonly used amplifier in sample and hold circuits is (a) a unity gain inverting amplifier (b) a unity gain non-inverting amplifier (c) an inverting amplifier with a gain of 10 (d) an inverting amplifier with a gain of 100 (GATE 2000: 1 Mark)

4.

The number of comparators in a 4-bit flash ADC is (a) 4 (b) 5 (c) 15 (d) 16

(a) 1, 0, 1 (c) 1, 1, 1 6.

(b) 0, 0, 1 (d) 0, 1, 1 (GATE 2000: 1 Mark)

In the 8085 microprocessor, the RST6 instruction transfers the program execution to the following location (a) 30 H (c) 48 H

(b) 24 H (d) 60 H (GATE 2000: 1 Mark)

7.

For the logic circuit shown in the following figure, the simplified Boolean expression of the output Y is

A B

(GATE 2000: 1 Mark) 5.

For the logic circuit shown in the following figure, the required input condition (A, B, C) to make the output (X) = 1 is

GATE_ECE_CH05_Digital_Circuits.indd 325

Y

C

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GATE ECE Chapter-wise Solved papers

(a) A + B + C (c) B 8.

(b) A (d) C (GATE 2000: 2 Marks)

For the 4-bit DAC shown in the following figure, the output voltage Vo is 1K

11. In the following figure, the J and K inputs of all the four Flip-Flops are made high. The frequency of the signal at output Y is

F = 10 kHz

7K +15 V

R

2R

2R

R

2R

2R

1V (a) 10 V (c) 4 V 9.

+ −

R

Vo

(a) 0.833 KHz (c) 0.91 KHz

(b) 1.0 KHz (d) 0.77 KHz

2R −15 V

(GATE 2000: 2 Marks)

1V (b) 5 V (d) 8 V (GATE 2000: 2 Marks)

A sequential circuit using D flip-flop and logic gates is shown in the following figure, where X and Y are the inputs and Z is the output. The circuit is

12. The 2’s complement representation of −17 is (a) 01110 (b) 01111 (c) 11110 (d) 10001 (GATE 2001: 1 Mark) 13. An 8085-microprocessor based system uses a 4K × 8-bit RAM whose starting address is AA00. The address of the last byte in this RAM is (a) 0FFFH (b) 1000H (c) B9FFH (d) BA00H (GATE 2001: 1 Mark)

X

Z

14. In the following figure, the LED

CLK Z

Vcc = 5 V

Y

1K

1K

1K

LED 1K

(a) (b) (c) (d)

S – R Flip-Flop with inputs X = R and Y = S S – R Flip-Flop with inputs X = S and Y = R J – K Flip-Flop with inputs X = J and Y = K J – K Flip-Flop with inputs X = K and Y = J (GATE 2000: 2 Marks)

10. The contents of Register (B) and Accumulator (A) of 8085 microprocessor are 49H and 3AH respectively. The contents of A and the status of carry flag (CY) and sign flag (S) after executing SUB B instruction are (a) A = F1, CY = 1, S = 1 (b) A = 0F, CY = 1, S = 1 (c) A = F0, CY = 0, S = 0 (d) A = 1F, CY = 1, S = 1 (GATE 2000: 2 Marks)

GATE_ECE_CH05_Digital_Circuits.indd 326

S1

(a) (b) (c) (d)

S1

emits light when both S1 and S2 are closed. emits light when both S1 and S2 are open. emits light when only S1 or S2 is closed. does not emit light, irrespective of the switch positions. (GATE 2001: 2 Marks)

15. In the TTL circuit shown in the following figure, S2 to S0 are select lines and X7 and X0 are input lines. S0 and X0 are LSBs. The output Y is

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Chapter 5  •  Digital Circuits

1

327

Word Line (WL) 0 X0 X 1 X2 X3 X 4 X5 X6 X7 Bit Line (BL) E C

C

S2 S1 S0

B A

(a) 5 V; 3 V; 7 V (c) 5 V; 5 V; 5 V

Y

(b) 4 V; 3 V; 4 V (d) 4 V; 4 V; 4 V (GATE 2001: 2 Marks)

(a) indeterminate (b) A ⊕ B (c)

A⊕ B

(d) C .( A ⊕ B) + C .( A ⊕ B) (GATE 2001: 2 Marks) 16. The digital block shown in the following figure is realized using two positive edge triggered D-flip-flops. Assume that of t < t0, Q1 = Q2 = 0. The circuit in the digital block is given by:

18. 4-bit 2’s complement representation of a decimal number is 1000. The number is (a) +8 (b) 0 (c) −7 (d) −8 (GATE 2002: 1 Mark) 19. If the input to the digital circuit shown in the following figure consisting of a cascade of 20 XOR-gates is X, then the output Y is equal to 1

X

Y DIGITAL BLOCK

t0

t1

t2

t3

Y t0 t1

(a) Figure (a) (c) Figure (c)

t2 t3 t4

X (a) 0

(b) Figure (b) (d) Figure (d)

(c)

(b) 1

X (d) X (GATE 2002: 1 Mark)





(a)

(b)

1 X

D1

1 X

D1

1

D1



(c)

X



(d)

1 X

Q1

1

D2

Q1 Q1

1

D2

Q1 Q1

Q2

Y

Q2 D2

Q1 D1

Y

Q2

Q1 Q1

Q2

Q2

Y

Q2 D2

Q2

20. The number of comparators required in a 3-bit comparator type ADC is (a) 2 (b) 3 (c) 7 (d) 8 (GATE 2002: 1 Mark) 21. The gates G1 and G2 in the following figure have propagation delays of 10 n sec and 20 n sec respectively. If the input Vi makes an abrupt change from logic 0 to 1 at time t = t0, then the output waveform Vo is

Y G1

Q2

G2

0

Vo

(GATE 2001: 2 Marks) Vi

1 17. In the DRAM cell shown in the following figure, the Vt of the NMOSFET is 1 V for the following three combinations of WL, C and BL voltages.

GATE_ECE_CH05_Digital_Circuits.indd 327

V1 1

0 t0

Vi

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GATE ECE Chapter-wise Solved papers

Which of the following is correct? (a) Only Statement 1 is TRUE (b) Only Statement 2 is TRUE (c) Both the statements are TRUE (d) Both the statements are FALSE (GATE 2002: 2 Marks)

1

(a) 0 t0

t1

t2

t3

t0

t1

t2

t3

t0

t1

t2

t3

t1

t2

t3

1

(b) 0

24. If the input X3, X2, X1, X0 to the ROM in the following figure are 8-4-2-1 BCD numbers, then the outputs Y3Y2Y1Y0 are

1

(c) 0

1 X3

1

0 X2

1 X1

0 X0

(d) 0 t0

(GATE 2002: 2 Marks)

ROM

22. The circuit shown in the following figure has two CMOS NOR gates. This circuit functions as a/an:

BCD-to-Decimal DECODER D0 D1

D8 D 9 0 1

R

Y3

Vss

Vo(output)

Y2 Y1 Y0 Vss

Vi

0

C

(a) Flip-flop (b) Schmitt trigger (c) Monostable multivibrator (d) Astable multivibrator

(a) (b) (c) (d)

gray code numbers 2-4-2-1 BCD numbers excess 3 code numbers none of the above (GATE 2002: 2 Marks)

(GATE 2002: 2 Marks) 23. Consider the following statements in connection with the CMOS inverter shown in the following figure where both the MOSFETs are of enhancement type and both have a threshold voltage of 2 V. Statement 1: T1 conducts when Vi ≥ 2 V. Statement 2: T1 is always in saturation when Vo = 0 V.

25. Consider the following assembly language program. MVI, B, 87H MOV A, B START: JMP NEXT MVI, B, 00H XRA B OUT PORT1 HLT NEXT: XRA B JP START OUT PORT2

+5 V

T2 V1



Vo T1

GATE_ECE_CH05_Digital_Circuits.indd 328

HLT

The execution of the above program in an 8085 microprocessor will result in (a) an output of 87H at PORT1 (b) an output of 87H at PORT2

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Chapter 5  •  Digital Circuits

with Y = P ⊕ Q ⊕ R and Z = RQ + PR + QP . The circuit acts as a

(c) infinite looping of the program execution with accumulator data remaining at 00H (d) infinite looping of the program execution with accumulator data alternating between 00H and 87H (GATE 2002: 2 Marks) 26. The number of distinct Boolean expressions of four variables is (a) 16 (b) 256 (c) 1024 (d) 65536 (GATE 2003: 1 Mark) 27. Without any additional circuitry, an 8:1 MUX can be used to obtain (a)  some but not all Boolean functions of three ­variables (b) all functions of three variables, but none of four variables (c) all functions of three variables and some but not all of four variables (d) all functions of 4 variables (GATE 2003: 1 Mark) 28. A 0 to 6 counter consists of three flip-flops and a combination circuit of two-input gate(s). The combination circuit consists of (a) one AND gate (b) one OR gate (c)   one AND gate and one OR gate (d) two AND gates (GATE 2003: 1 Mark) 29. The minimum number of comparators required to build an eight-bit-flash ADC is (b) 63 (c) 255 (d) 256 (a) 8 (GATE 2003: 1 Mark) 30. If the functions W, X, Y and Z are as follows: W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + PQ Z = R + S + PQ + PQR + PQS Then (a) W = Z , X = Z

(b) W = Z, X = Y

(c) W = Y (d) W = Y = Z (GATE 2003: 2 Marks) 31. The circuit shown in the following figure has four boxes each described by inputs P, Q, R and outputs Y, Z

GATE_ECE_CH05_Digital_Circuits.indd 329

Q

P P Q Z Y

R

P Q Z Y

R

P Q Z Y

P Q R

Z

R Y

Output

(a) (b) (c) (d)

four-bit adder giving P + Q four-bit subtractor giving P – Q four-bit subtractor giving Q – P four-bit adder giving P + Q + R (GATE 2003: 2 Marks)

32. The circuit shown in the following figure converts (a) BCD to binary code (b) Binary to excess–3 code (c) Excess–3 to Gray code (d) Gray to Binary code

MSB

OUTPUTS MSB (GATE 2003: 2 Marks) 33. A four-bit ripple counter and a four-bit synchronous counter are made using flip–flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S, respectively, then (a)  R = 10 ns, S = 40 ns   (b) 

R = 40 ns, S = 10 ns

(c)  R = 10 ns, S = 30 ns   (d) 

R = 30 ns, S = 40 ns

(GATE 2003: 2 Marks) 34. The circuit shown in the following figure is a four-bit DAC. The input bits 0 and 1 are represented by 0 and 5 V, respectively. The op-amp is ideal, but all the resistances and the 5 V inputs have a tolerance of ±10% . The specification

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Clk

A

1 ROM E

330

GATE ECE Chapter-wise Solved papers

(rounded to the nearest multiple of 5%) for the tolerance of the DAC is R

R 2R

Clk

−

4R

t1

(a) 1111 (c) 1000

+ R

8R

± 20% (a) ± 35% (b) (c) ±10% (d) ± 5% (GATE 2003: 2 Marks) 35. In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator is less than that of register B. As a result (a) Carry flag will be set but Zero flag will be reset (b) Carry flag will be reset but Zero flag will be set (c) both Carry flag and Zero flag will be reset (d) both Carry flag and Zero flag will be set (GATE 2003: 2 Marks)

36. In the circuit shown in the following figure, A is parallel-in, parallel-out four-bit register, which loads at the rising edge of the clock C. The input lines are connected to a four-bit bus, W. Its output acts as the input to a 16 × 4 ROM whose output is floating when the enable input E is 0. A partial table of the contents of the ROM is as follows: Address Data

0

2

4

0011

1111

0100

6

8

10

12

14

1010 1011

1000

0010

1000

The clock to the register is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t 2 is W MSB

Clk

t2

A

1 ROM E

time

(b) 1011 (d) 0010 (GATE 2003: 2 Marks)

37. The range of signed decimal numbers that can be represented by six-bit 1’s complement number is (a) −31 to +31 (b) −63 to +63 (c) −64 to +63 (d) −32 to +31 (GATE 2004: 1 Mark) 38. The following figure shows the internal schematic of a TTL AND-OR-Invert (AOI) gate. For the inputs shown in the flowing figure, the output Y is (b) 1 (a) 0 AB (c) AB (d) (GATE 2004: 1 Mark)

A B Y

Inputs are floating

{

39. A master–slave flip–flop has the characteristic that (a) change in the input is immediately reflected in the output (b) change in the output occurs when the state of the master is affected (c)   change in the output occurs when the state of the slave is affected (d) both the master and the slave states are affected at the same time (GATE 2004: 1 Mark) 40. Choose the correct one from among the alternatives A, B, C and D after matching an item from Group 1 with the most appropriate item in Group 2.

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Chapter 5  •  Digital Circuits

Group 1

Group 2

P.  Shift register

1.  Frequency division

Q. Counter

2. Addressing in memory chips

R. Decoder

3. Serial-to-parallel data conversion

45. The minimum number of 2-to-1 multiplexers required to realize a 4-to-1 multiplexer is (a) 1 (c) 3

(d) P − 1, Q − 2, R − 2

(b) 2 (d) 4 (GATE 2004: 2 Marks)

46. In the modulo-6 ripple counter shown in the following figure, the output of the two-input gate is used to clear the J-K flip—flops. The two-input gate is a/an

P − 3, Q − 1, R − 2 (a) P − 3, Q − 2, R − 1 (b) (c) P − 2, Q − 1, R − 3

331

1 C

J

B

J

A

J

C

K

B

K

A

K

Clock input

(GATE 2004: 1 Mark) 41. A digital system is required to amplify a binary-encoded audio signal. The user should be able to control the gain of the amplifier from a minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is (b) 6 (a) 8 (d) 7 (c) 5 (GATE 2004: 1 Mark) 42. 11001, 1001 and 111001 correspond to the 2’s complement representation of which one of the following sets of number? (a) 25, 9 and 57, respectively (b) −7, −7, and −7, respectively (c) −6, −6, and −6, respectively (d) −25, −9 and −57, respectively (GATE 2004: 2 Marks) 43. The Boolean expression AC + BC is equivalent to (a) AC + BC + AC (b) BC + AC + BC + AC B (c) AC + BC + BC + ABC (d) ABC + ABC + ABC + ABC (GATE 2004: 2 Marks) 44. A Boolean function f of two variables x and y is defined as follows: f(0, 0) = f(0, 1) = f(1, 1) = 1; f(1, 0) = 0

Assuming that the complements of x and y are not available, a minimum cost solution for realizing f using only two-input NOR gates and two-input OR gates (each having unit cost) would have a total cost of (a) 1 unit (b) 4 unit (c) 3 unit (d) 2 unit (GATE 2004: 2 Marks)

GATE_ECE_CH05_Digital_Circuits.indd 331

2-input gate

(a) NAND gate (c) OR gate

(b) NOR gate (d) AND gate (GATE 2004: 2 Marks)

47. The number of memory cycles required to execute the following 8085 instructions I.  LDA 3000 H II.  LXI D, FOF1 H would be (a) 2 for (I) and 2 for (II) (b) 4 for (I) and 3 for (II) (c) 3 for (I) and 3 for (II) (d) 3 for (I) and 4 for (II) (GATE 2004: 2 Marks) 48. Consider the sequence of 8085 instructions given below. LXI H, 9258, MOV A, M, CMA, MOV M, A Which one of the following is performed by this sequence? (a) Contents of location 9258 are moved to the accumulator (b) Contents of location 9258 are compared with the contents of the accumulator (c) Contents of location 9258 are complemented and stored in location 9258 (d) Contents of location 5892 are complemented and stored in location 5892 (GATE 2004: 2 Marks) 49. It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of the 8085 program for this purpose is given below:

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MVI A, 00H Loop; _____.    _____.    _____.    HLT END

53. The Boolean expression for the truth table shown below is

The sequence of instructions to the complete the ­program would be (a) (b) (c) (d)

JNZ LOOP, ADD B, DCR C, ADD B,

ADD B, DCR C JNZ LOOP, DCR C JNZ LOOP, ADD B DCR C, JNZ LOOP (GATE 2004: 2 Marks)

50. The 8255 programmable peripheral interface is used as described below. I. An A/D converter is interfaced to a microprocessor through an 8255. The conversion is initiated by a signal from the 8255 on Port C. A signal on Port C causes data to be stored into Port A. II. Two computers exchange data using a pair of 8255s. Port A works as a bidirectional data port supported by appropriate handshaking signals. The appropriate modes of operation of the 8255 for (I) and (II) would be (a) (b) (c) (d)

Mode 0 for (I) and Mode 1 for (II) Mode 1 for (I) and Mode 0 for (II) Mode 2 for (I) and Mode 0 for (II) Mode 2 for (I) and Mode 1 for (II) (GATE 2004: 2 Marks)

A

B

C

f

0

0

0

0

0

0

1

0

0

1

0

0

0

1

1

1

1

0

0

0

1

0

1

0

1

1

0

1

1

1

1

0

(a) B( A + C )( A + C )

 (b) B( A + C )( A + C )

(c) B ( A + C )( A + C )

 (d) B ( A + C )( A + C ) (GATE 2005: 2 Marks)

54. The transistor used in a portion of the TTL gate shown in the figure has a b = 100. The base-emitter voltage of is 0.7 V for a transistor in active region and 0.75 V for a transistor in saturation. If the sink current I = 1 mA and the output is at logic 0, then current IR will be equal to 5V 1.4 kΩ

I Vo

51. Decimal 43 in hexadecimal and BCD number system is, respectively, (a) B2, 01000011 (c) 2B, 00110100

(b) 2B, 01000011 (d) B2, 01000100 IR

(GATE 2005: 1 Mark)

1 kΩ

52. The Boolean function f implemented in the following figure using two input multiplexers is (a)

ABC + ABC (b) ABC + ABC

(c)

ABC + ABC

(a) 0.65 mA (c) 0.75 mA

(d) ABC + ABC (GATE 2005: 1 Mark)

C

0 0

0 f A

C 1

1

B

GATE_ECE_CH05_Digital_Circuits.indd 332

E

(b) 0.70 mA (d) 1.00 mA (GATE 2005: 2 Marks)

55. The present output Qn of an edge-triggered J-K flip– flop is at logic 0. If J = 1, then Qn+1 (a) cannot be determined (b) will be logic 0 (c)   will be logic 1 (d) will race around (GATE 2005: 2 Marks) 56. The following figure shows a ripple counter using positive edge-triggered flip–flops. If the present state of the counter is Q2 Q1 Q0 = 011 , then its next state (Q2 Q1 Q0 ) will be

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Chapter 5  •  Digital Circuits

1

(a) 0100-02FF (c) F900-FAFF

1

1 T0

Q0

T1

Q1

T2

Q0 (a) 010 (c) 111

Q1

(b) 1500-16FF (d) F800-F9FF (GATE 2005: 2 Marks)

Q2

Clk Q2

60. The number of product terms in the minimized sumof-product expression obtained through the following K-map is (where d denotes ‘don’t care’ states)

(b) 100 (d) 101 (GATE 2005: 2 Marks)

Statement for Linked Answer Questions 57 and 58: Consider an 8085 microprocessor system. 57. The following program starts at location 0100 H. LXI SP, 00 FF LXI H, 0107 MVI A, 20 H SUB M The content of accumulator when the program counter reaches 0109 H is (a) 20 H (b) 02 H (c) 00 H (d) FFH (GATE 2005: 2 Marks) 58. If in addition following code exists from 0109 onwards, ORI 40 H ADD M What will be the result in the accumulator after the last instruction is executed? (a) 40 H (b) 20 H (c) 60 H (d) 42 H (GATE 2005: 2 Marks) 59. What memory address range is NOT represented by chip#1 and chip#2 in the following figure. A0 to A15 in this figure are the address lines and CS means chip select.

333

1

0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

(a) 2 (c) 4

(b) 3 (d) 5 (GATE 2006: 1 Mark)

61. A new binary coded pentary (BCP) number system is proposed in which every digit of a base-5 number is represented by its corresponding three-bit binary code. For example, the base-5 number 24 will be represented by its BCP code 010100. In this numbering system, the BCP code 100010011001 corresponds of the following number in base-5 system (b) 1324 (a) 423 (d) 4231 (c) 2201 (GATE 2006: 2 Marks) 62. The point P in the following figure is stuck at 1. The output f will be A f

B P

A0—A7

256 bytes Chip #1 C

A8 A9 A9 A8 A0—A7

A10—A15

GATE_ECE_CH05_Digital_Circuits.indd 333

256 bytes Chip #2

not used

(a)

ABC (b) A

(c)

ABC (d) A (GATE 2006: 2 Marks)

63. For the circuit shown in the following figure, two fourbit parallel-in serial-out shift registers loaded with the data shown are used to feed the data to a full adder. Initially, all the flip-flops are in CLEAR state. After applying two clock pulses, the outputs of the full adder should be

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1 0 1 1 MSB LSB Shift registers

D

0 0 1 1 MSB LSB

D

Q

A

S Fulladder

Clk Q

Clk

(a)

(b)

(c)

(d)

B Co

Ci Q

D

(GATE 2006: 2 Marks)

Clk

66. Following is the segment of a 8085 assembly language program LXI SP, EFFFH

Clock (a) S = 0

Co = 0 (b) S = 0 Co = 1

(c) S = 1

Co = 0 (d) S = 1 Co = 1 (GATE 2006: 2 Marks)



CALL 3000H : :



3000 H : LXI H, 3CF4 H



PUSH PSW



SPHL



POP PSW



Q1

RET On completion of RET execution, the contents of SP is (b) 3CF8 H (a) 3CF0 H (c) EFFD H (d) EFFF H (GATE 2006: 2 Marks)

(a) Q1 and Q0 (b) Q0 and Q1

67. An I/O peripheral device shown in figure (b) below is to be interfaced to an 8085 microprocessor. To select the I/O device in the I/O address range D4 H–D7 H, its chip–select (CS ) should be connected to the output of the decoder shown in figure (a) below.

64. Two D-flip–flops, as shown in the following figure, are to be connected as a synchronous counter that goes through the following Q1Q0 sequence 00 → 01 → 11 → 10 → 00 → … . The inputs D0 and D1 , respectively, should be connected as

D0

Q0

D1

Q1

LSB Clk

Q0

MSB Clk

Clock

(c) Q1Q0 and Q1Q0 (d) Q1Q0 and Q1Q0 (GATE 2006: 2 Marks)

A2

65. A four-bit D/A converter is connected to a free-running three-bit UP counter, as shown in the following figure.

A3

1 kΩ Q2

D3 D2

A4 − Vo

Q1

D1

+

Q0

D0

1 kΩ

0 1 2 3−8 3 decoder 4 5 6 MSB 7

LSB

4-Bit DAC

Which of the waveforms shown in the following four options will be observed at Vo ?

GATE_ECE_CH05_Digital_Circuits.indd 334

IORD IOWR

I/O peripheral

A1 A0

A7 A6 A5 (a)

Clock 3-Bit counter

Data

(a) output 7 (c) output 2

CS (b) (b) output 5 (d) output 0 (GATE 2006: 2 Marks)

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68. X = 01110 and Y = 11001 are two five-bit binary numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is (a) 100111 (b) 001000 (c) 000111 (d) 101001 (GATE 2007: 1 Mark)

72. In the circuit shown in the following figure, X is given by

0

I0

0

I0

1

I1 4-to-1 I2 MUX I3 S1 S0

1

I1 4-to-1 Y I2 MUX I3 S1 S0

1

69. The Boolean function Y = AB + CD is to be realized using only two-input NAND gates. The minimum number of gates required is (a) 2 (b) 3 (c)   4 (d) 5 (GATE 2007: 1 Mark)

0

A (a)

70. The Boolean expression Y = ABCD + ABCD + ABCD + ABCD can be minimized to

1 0

B

X

C

X = ABC + ABC + ABC + ABC

(b) X = ABC + ABC + ABC + ABC (c)

(a) Y = ABCD + ABC + ACD

X = AB + BC + AC

(d) X = AB + BC + AC

(b) Y = ABCD + BCD + ABCD

(GATE 2007: 2 Marks)

(c) Y = ABCD + BCD + ABCD (d) Y = ABCD + BCD + ABCD (GATE 2007: 2 Marks) 71. Circuit diagram of a standard TTL NOT gate is shown in the figure. When Vi = 2.5 V, the modes of operation of the transistor will be

73. The following binary values were applied to the X and Y inputs of the NAND latch as shown in the following figure in the sequence indicated below: X = 0, Y = 0; X = 1, Y = 1.

The corresponding stable P, Q outputs will be

VCC = 5 V X P 100 W

1.4 kW

4 kW

Q4 Q2

+

D

Q1 Q

+ Y Q3 Vi

Vo 1 kW

−

−

(a) P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1 (b)

(a) Q1 : reverse active; Q2 : normal active; Q3 : saturation; Q4 : cut-off (b) Q1 : reverse active; Q2 : saturation; Q3 : saturation; Q4 : cut-off (c) Q1 : normal active; Q2 : cut-off; Q3 : cut-off; Q4 : saturation (d) Q1 : saturation; Q2 : saturation; Q3 : saturation; Q4 : normal active (GATE 2007: 2 Marks)

GATE_ECE_CH05_Digital_Circuits.indd 335

P = 1, Q = 0; P = 0, Q = 1 or P = 0, Q = 1; P = 0, Q = 1

(c) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1 (d) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 1 (GATE 2007: 2 Marks) 74. For the circuit shown in the following figure, the counter state Q1 Q0 follows the sequence

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as the IO/M signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is

D0

Q0

D1

A7 A6 A5 A4 A3 IO/M

Q1

(a) F8H–FBH (c) F8H–FFH

Clk

(a) (b) (c) (d)

00, 01, 10, 11, 00, ... 00, 01, 10, 00, 01, ... 00, 01, 11, 00, 01, ... 00, 10, 11, 00, 10, ...

Line 1: MVI A, B5H

(GATE 2007: 2 Marks)



Line 2: MVI B, 0EH

Statement for Linked Answer Questions 75 and 76: In the D/A converter circuit shown in the following figure, VR = 10 V and R = 10 kΩ.



Line 3: XRI 69H



Line 4: ADD B



Line 5: ANI 9BH



Line 6: CPI 9FH



Line 7: STA 30101H



Line 8: HLT

R

R

i

2R

VR 2R

2R

2R

2R

A1

A1

A0

A0

(b) F8H–FCH (d) F0H–F7H (GATE 2007: 2 Marks)

R 78. The contents of the accumulator just after execution of the ADD instruction in line 4 will be

− +

Vo

(a) C3H (c) DCH

(b) EAH (d) 69H (GATE 2007: 2 Marks)

79. After execution of line 7 of the program, the status of the CY and Z flags will be

75. The current i is (a) 31.25 µA (b) 62.5 µA (c) 125 µA (d) 250 µA

(a) CY = 0, Z = 0 (c) CY = 1, Z = 0

(GATE 2007: 2 Marks) 76. The voltage Vo is (a) −0.781 V (b) −1.562 V (c) −3.125 V (d) −6.250 V (GATE 2007: 2 Marks) 77. An 8255 chip is interfaced to an 8085 microprocessor system as an I/O mapped I/O as shown in the figure. The address lines A0 and A1 of the 8085 are used by the 8255 chip to decode internally its three ports and the control register. The address lines A3 to A7 as well

GATE_ECE_CH05_Digital_Circuits.indd 336

CS

Statement for Linked Answer Questions 78 and 79: An 8085 assembly language program is given below

R

8255

(b) CY = 0, Z = 1 (d) CY = 1, Z = 1 (GATE 2007: 2 Marks)

80. The two numbers represented in signed 2’s complement form are P = 11101101 and Q = 11100110. If Q is subtracted from P, the value obtained in signed 2’s complement form is (a) 100000111 (c) 11111001

(b) 00000111 (d) 111111001 (GATE 2008: 2 Marks)

81. Refer to the following figure. Which of the following Boolean Expressions correctly represents the relation between P, Q, R and M1?

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Chapter 5  •  Digital Circuits

337

P X Q Z Clk 1

M1 0

Y T t

R t1

(a) (b) (c) (d)

M1 = (P OR Q) XOR R M1 = (P AND Q) XOR R M1 = (P NOR Q) XOR R M1 = (P XOR Q) XOR R

Which of the following waveforms correctly represents the output at Q1? (GATE 2008: 2 Marks)

1 (a)

82. For the circuit shown in the following figure, I0 to I3 are the inputs to the 4:1 multiplexer R (MSB) and S are control bits. The output Z can be represented by

0 2T t1 + ∆T

I3

1 (b) 0

I2 4T 4-to-1 MUX I1

t1 + 2∆T

Z

1 (c)

I0

0 2T t1 + 2∆T R

S

(a) PQ + PQS + QRS

1

(b) PQ + PQR + PQS

(d) 0

(c) PQR + PQR + PQRS + QRS

4T

(d) PQR + PQRS + PQRS + QRS

t1 + ∆T

(GATE 2008: 2 Marks)

(GATE 2008: 2 Marks)

 83. For each of the positive edge-triggered J-K flip–flop used in the following figure, the propagation delay is ΔT .

84. For the circuit shown in the following figure, D has a transition from 0 to 1 after Clk changes from 1 to 0. Assume gate delays to be negligible. Which of the following statements is true? (a)  Q goes to 1 at the Clk transition and stays at 1.

Q0 1

J0

Q1 1

J1

(b)  Q goes to 0 at the Clk transition and stays at 0. (c)   Q goes to 1 at the Clk transition and goes to 0 when D goes to 1.

Clk

1

K0

1

K1

(d) Q  goes to 0 at the Clk transition and goes to 1 when D goes to 1. (GATE 2008: 2 Marks)

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1 0

Clk

1 0

D

Which of the following statement is correct?

Q



(a) PC = 2715 H HL = 30A0 H



(b) PC = 30A0 H HL = 2715 H



(c) PC = 6140 H HL = 6140 H



(d) PC = 6140 H HL = 2715 H

Q

Statement for Linked Answer Questions 85 and 86: In the circuit shown in the following figure, the comparator output is logic ‘1’ if V1 > V2 and is logic ‘0’ otherwise. The D/A conversion is done as per the relation: ⎛ 3 ⎞ VDAC = ⎜ ∑ 2n −1 bn ⎟ V ⎝ n= 0 ⎠

4-Bit D/A converter Binary to BCD

+5 V Clr − +

Clk

88. The full forms of the abbreviations TTL and CMOS in reference to logic families are (a) triple transistor logic and chip metal oxide semiconductor (b) tristate transistor logic and chip metal oxide semiconductor

where b3 (MSB), b2 ,  b1 and b0 (LSB) are the counter outputs. The counter starts from the clear state. VDAC

(GATE 2008: 2 Marks)

2 Digit LED display

4-Bit up counter

Vin = 6.2 V Clock 85. The stable reading of the LED display is (a) 06 (b) 07 (c) 12 (d) 13 (GATE 2008: 2 Marks)

(c) transistor transistor logic and complementary metal oxide semiconductor (d) tristate transistor logic and complementary metal oxide semiconductor (GATE 2009: 1 Mark) 89. In a microprocessor, the service routine for a certain interrupt starts from a fixed location of memory which cannot be externally set, but the interrupt can be delayed or rejected. Such an interrupt is (a) non-maskable and non-vectored (b) maskable and non-vectored (c) non-maskable and vectored (d) maskable and vectored (GATE 2009: 1 Mark)

90. If X = 1 in the logic equation ⎡⎣ X + Z {Y + ( Z + XY )}⎤⎦ { X + Z ( X + Y )} = 1

X +and Z {YV+in ( Zat+ XY )}⎤⎦ { X + Z ( X + Y )} = 1, then 86. The magnitude of the error between V⎡⎣DAC steady state in volts is Y =Z (a) Y = Z (b) (a) 0.2 (b) 0.3 (c) Z = 1 (d) Z = 0 (c) 0.5 (d) 1.0 (GATE 2009: 2 Marks) (GATE 2008: 2 Marks) 87. An 8085 executes the following instructions 2710 LXI H, 30A0 H 2713 DAD H 2714 PCHL All addresses and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL.

GATE_ECE_CH05_Digital_Circuits.indd 338

Statement for Linked Answer Questions 91 and 92: Two products are sold from a vending machine, which has two push buttons P1 and P2. When a button is pressed, the price of the corresponding product is ­displayed in a seven-segment display. If no buttons are pressed, ‘0’ is displayed, signifying ‘ ` 0’. If only P1 is pressed, ‘2’ is displayed, signifying ‘ ` 2’. If only P2 is pressed, ‘5’ is displayed, signifying ‘ ` 5’. If both P1 and P2 are pressed, ‘E ’ is displayed, signifying ‘Error’. The

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Chapter 5  •  Digital Circuits

names of the segments in the seven-segment display, and the glow of the display for ‘0’, ‘2’, ‘5’ and ‘E ’ are shown in the following figure.

P1

339

Q1

a f

b Q2

g P2

e

c P1

Q1

d 0

2

5

E

Q2 P2

Consider

(i) Push button pressed/not pressed is equivalent to logic 1/0, respectively.



(ii) A segment glowing/not glowing in the display is equivalent to logic 1/0, respectively.

91. If segments a to g are considered as functions of P1 and P2, then which of the following is correct? (a) g = P1 + P2 , d = c + e (b) g = P1 + P2 , d = c + e (c) g = P1 + P2 , e = b + c (d) g = P1 + P2 , e = b + c (GATE 2009: 2 Marks) 92. What are the minimum numbers of NOT gates and two-input OR gates required for designing the logic of the driver for this seven-segment display? (a) (b) (c) (d)

3 NOT and 4 OR 2 NOT and 4 OR 1 NOT and 3 OR 2 NOT and 3 OR



The inputs ( P1 , P2 ) for both the latches are first made (0, 1) and then, after a few seconds, made (1, 1). The corresponding stable outputs (Q1 , Q2 ) are (a) NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0) (b) NAND: first (1, 0) then (1, 0) NOR: first (1, 0) then (1, 0) (c) NAND: first (1, 0) then (1, 0) NOR: first (1, 0) then (0, 0) (d) NAND: first (1, 0) then (1, 1) NOR: first (0, 1) then (0, 1) (GATE 2009: 2 Marks)

95. What are the counting states (Q1 , Q2 ) for the counter shown in the following figure? Q1

Clock

J1

Q2 J2

Q1

J-K Flip-flop K1

Q1

Q2

J-K Flip-flop 1

K2

Q2

(GATE 2009: 2 Marks)  93. What are the minimum number of 2-to-1 multiplexers required to generate a two-input AND gate and a twoinput EX-OR gate? (a) 1 and 2 (c) 1 and 1

(b) 1 and 3 (d) 2 and 2 (GATE 2009: 2 Marks)

94. Refer to the NAND and NOR latches shown in the ­following figure.

GATE_ECE_CH05_Digital_Circuits.indd 339

(a) (b) (c) (d)

11, 10, 00, 11, 10 ... 01, 10, 11, 00, 01 ... 00, 11, 01, 10, 00 ... 01, 10, 00, 01, 10 ... (GATE 2009: 2 Marks)

96. Match the logic gates in Column I with their e­ quivalents in Column II shown in the following table.

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Column I

Column II

P.

1.

Q.

2.

R.

(a) 2000 − 20FF (c) 2E00 − 2EFF

(b) 2D00 − 2DFF (d) FD00 − FDFF (GATE 2010: 1 Mark)

99. The Boolean function realized by the logic circuit shown in the following figure is C

I0

D

I1 4-to-1 MUX

3. I2

I3

4.

S.

(a) P-2, Q-4, R-1, S-3 (c) P-2, Q-4, R-3, S-1

(b) P-4, Q-2, R-1, S-3 (d) P-4, Q-2, R-3, S-1 (GATE 2010: 1 Mark)

F (A,B,C,D)

S1

S0

A

B

(a) F = Σ m (0, 1, 3, 5, 9, 10, 14) (b) F = Σ m (2, 3, 5, 7, 8, 12, 13)

97. For the output F to be 1 in the logic circuit shown in the following figure, the input combination should be

(c) F = Σ m (1, 2, 4, 5, 11, 14, 15) (d) F = Σ m (2, 3, 5, 7, 8, 9, 12) (GATE 2010: 2 Marks)

A

100. Assuming that all flip–flops are in reset conditions initially, the count sequence observed at QA in the circuit shown in the following figure is

B

Output

F

DA

C (a) A = 1, B = 1, C = 0 (c) A = 0, B = 1, C = 0

(b) A = 1, B = 0, C = 0 (d) A = 0, B = 0, C = 1 (GATE 2010: 1 Mark)

98. In the circuit shown, the device connected to Y5 can have address in the range A8 A9 A10

A B C

748138 3—8 decoder A11 A12 A13 A14 A15

G2A G2B G1 IO/M

GATE_ECE_CH05_Digital_Circuits.indd 340

Y5 To device chip select

QA

DB

QA

QB QB

DC

QC QC

Clock

(a) 0010111 ... (c) 0101111 ...

(b) 0001011 ... (d) 0110100 ... (GATE 2010: 2 Marks)

101. For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is 3000 MVI A, 45H 3002 MOV B, A 3003 STC 3004 CMC 3005 RAR 3006 XRA B (a) 00 H

(b) 45 H

(c) 67 H (d) E7 H (GATE 2010: 2 Marks)

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Chapter 5  •  Digital Circuits

102. The output Y in the circuit shown in the following figure is always “1” when

341

105. The output of a three-stage Johnson (twisted-ring) counter is fed to a digital-to-analog (D/A) converter as shown in the following figure:

P Vref D/A Converter

Q Y

D2

D1

D0

Q2

Q1

Q0

Vo

R (a) (b) (c) (d)

two or more of the inputs P, Q, R are “0” two or more of the inputs P, Q, R are “1” any odd number of the inputs P, Q, R is “0” any odd number of the inputs P, Q, R is “1” (GATE 2011: 1 Mark)

103. The logic function implemented by the circuit shown in the following figure is (ground implies a logic ‘0’)

Clock



Johnson counter

Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output Vo is (a) (a) Vo Vo

4-to-1 MUX I0 I1 Y

F

I2

I3

(b) V (b) o Vo

S1 S0 P

Q

F = OR (P, Q) (a) F = AND (P, Q) (b) (c) F = XNOR (P, Q) (d) F = XOR (P, Q) (GATE 2011: 1 Mark)

(c) V (c) o Vo

104. When the output Y in the circuit shown in the ­following figure is ‘1’, it implies that data has

Y Data Clock

(a) (b) (c) (d)

D

Q Q

D

Q Q

Changed from ‘0’ to ‘1’ Changed from ‘1’ to ‘0’ Changed in either direction Not changed (GATE 2011: 1 Mark)

GATE_ECE_CH05_Digital_Circuits.indd 341

(d) V (d) o Vo

(GATE 2011: 2 Marks) 106. Two D-flip–flops are connected as a synchronous counter that goes through the following QB QA sequence

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00 → 11 → 01 → 10 → 00 → …. The connections to the inputs DA and DB are (a) DA = QB , DB = QA (b) DA = Q A , DB = Q B (c)    DA = (QB Q B + Q A QB ), DB = QA

(c) occurs when Clk = 1 and A = B = 1 (d) occurs when Clk = 1 and A = B = 0 (GATE 2012: 1 Mark) 111. The state transition diagram for the logic circuit shown in the following figure is

(d) DA = (QA QB + Q A Q B ), DB = Q B 2-1 MUX

(GATE 2011: 2 Marks) 107. An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is MVI A, 07 H RLC MOV B, A RLC RLC ADD B RRC (a) 8 CH (c) 23 H

(b) 64 H (d) 15 H (GATE 2011: 2 Marks)

108. In the sum of products function f(X, Y, Z) = ∑(2, 3, 4, 5), the prime implicates are (a) XY , XY (b) XY , XYZ , XYZ (c)   XYZ , XYZ , XY (d) XYZ , XYZ , XYZ , XYZ (GATE 2012: 1 Mark) 109. The output Y of a two-bit comparator is logic 1 whenever the two-bit input A is greater than the two-bit input B. The number of combinations for which the output is logic 1, is (a) 4 (b) 6 (c) 8 (d) 10 (GATE 2012: 1 Mark)

D

Q

X1 Y X0

Clk

Q

Select

A A=0

A=1 A=1 (a) Q=0

A=0

Q=1

A=0

A=0 A=1 (b) Q=0

A=1

Q=1

A=1

A=0 A=0 (c) Q=0

110. Consider the circuit shown in the following figure.

A=1

Q=1

A=1

A=1 A=0

A (d) Q=0

A=0

Q=1

Clk (GATE 2012: 2 Marks) B

In this circuit, the race around (a) does not occur (b) occurs when Clk = 0

GATE_ECE_CH05_Digital_Circuits.indd 342

112. A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles

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(a) an AND gate (c) an XOR gate

(b) an OR gate (d) a NAND gate (GATE 2013: 1 Mark)

115. For an n-variable Boolean function, the maximum number of prime implicants is (a) 2(n - 1) (b) n/2 (d) 2(n-1) (c) 2n (GATE 2014: 1 Mark)

113. For 8085 microporcessor, the following program is executed. 116. The number of bytes required to represent the decimal MVI A, 05 H; number 1856357 in packed BCD (binary coded deciMVI B, 05 H; mal) form is PTR: ADD B; (GATE 2014: 1 Mark) DCR B; JNZ PTR; 117. Consider the Boolean function, F ( w , x, y, z ) = wy + xy + wxyz + w xy + x F ( w , x , y , z ) = wy + xy + wxyz + w xy + xz + x y z . Which one of the followADI 03 H; ing is the complete set of essential prime implicants? HLT; At the end of program, accumulator contains (a) w , y, xz , x z (b)  w, y, xz (a) 17 H (b) 20 H y, x y z y, xz , x z (c)  (d)  (c) 23 H (d) 05 H (GATE 2014: 1 Mark) (GATE 2013: 1 Mark) 118. The Boolean expression ( X +Y )( X +Y ) + ( XY ) + X 114. There are four chips each of 1024 bytes connected to simplifies to a 16 bit address bus as shown in the following figure. (a)  X (b)  Y (c)  XY (d) X + Y RAMs 1, 2, 3 and 4, respectively, are mapped to address (GATE 2014: 1 Mark) 119. In the following circuit employing pass transistor logic, all NMOS transistors are identical with a threshold voltage of 1 V. Ignoring the body-effect, the output voltages at P, Q and R are

RAM #4 1024B E

E RAM #2 1024B

A0—A9

5V 8 Bit data bus

RAM #3 1024B

A10 A11 A12 A13 A14 A15

11 Input 10 01 S1 S0 00

E

5V

5V

P (a) 4 V, 3 V, 2 V (c) 4 V, 4 V, 4 V

E RAM #1 1024B

5V

Q

R

(b) 5 V, 5 V, 5 V (d) 5 V, 4 V, 3 V (GATE 2014: 1 Mark)

120. In the circuit shown in the figure, if C = 0, the expression for Y is C

(a)  0C00H-0FFFH, 3C00H-3FFFH (b)  1800H-1FFFH, 4800H-4FFFH (c)   0500H-08FFH, 5500H-58FFH (d)  0800H-0BFFH, 3800H-3BFFH

1C00H-1FFFH, 2C00H-2FFFH,

Y

2800H-2FFFH, 3800H-3FFFH, 1500H-18FFH, 3500H-38FFH, 1800H-1BFFH, 2800H-2BFFH, (GATE 2013: 2 Marks)

GATE_ECE_CH05_Digital_Circuits.indd 343

A B

A B

(a) Y = AB + AB (b) Y=A+B Y = A + B (c) (d) Y = AB (GATE 2014: 1 Mark)

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121. The output (Y ) of the circuit shown in the figure is VDD

B

A

applied as shown. The frequency (in kHz) of the waveform at Q3 is . (GATE 2014: 1 Mark) 125. The circuit shown in the figure is a D Q Q D Latch D Latch En En Q Q

C

Output (Y )

Clk

A (a) (b) (c) (d)

B

Toggle flip-flop JK flip-flop SR latch Master-slave D flip-flop (GATE 2014: 1 Mark)

C

(a)

A + B + C (b) A + B ⋅C + A⋅C

(c)

A + B + C (d) A⋅ B ⋅C (GATE 2014: 1 Mark)

122. In a half-subtractor circuit with X and Y as inputs, the Borrow (M) and Difference (N = X - Y) are given by (a) M = X ⊕Y, N = XY (b) M = XY, N =X ⊕Y (c) M = XY , N =X ⊕Y (d) M = X Y , N = X ⊕ Y

126. For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then (a) droop rate decreases and acquisition time decreases (b) droop rate decreases and acquisition time increases (c) droop rate increases and acquisition time decreases (d) droop rate increases and acquisition time increases (GATE 2014: 1 Mark) 127. The output F in the digital logic circuit shown in the figure is XOR X AND Y

(GATE 2014: 1 Mark)

F

123. Consider the multiplexer based logic circuit shown in the figure. Z

0 MUX

W

1

0 MUX 1

F

S1 S2

XNOR (a)  F = XYZ + XYZ

(b)  F = XYZ + XYZ

(c)  F = XYZ + XYZ

(d)  F = XYZ + XYZ (GATE 2014: 2 Marks)

128. In the circuit shown, W and Y are MSBs of the control inputs. The output F is given by

Which one of the following Boolean functions is realized by the circuit? (a) F = WS1 S2 (b) F = WS1 + WS2 + S1S2

4:1 MUX

I0

I0

I1

I1 F

VCC

(c) F = W + S1 + S2 (d) F = W ⊕ S1 ⊕ S2 (GATE 2014: 1 Mark) 124. Five J-K flip-flops are cascaded to form the circuit shown in the figure. Clock pulses at a frequency of 1 MHz are

GATE_ECE_CH05_Digital_Circuits.indd 344

4:1 MUX

Q

Q

I2

I2

I3

I3

W

X

Y

Z

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131. A 16-bit ripple carry adder is realized using 16 identical full adders (FA) as shown in the figure. The carrypropagation delay of each FA is 12 ns and the sum-propagation delay of each FA is 15 ns. The worst . case delay (in ns) of this 16-bit adder will be

(a) F = WX + W X + Y Z (b) F = WX + W X + Y Z (c) F = WX Y + W X Y (d) F = (W + X )Y Z

A0 B 0

(GATE 2014: 2 Marks) 129. If X and Y are inputs and the Difference (D = X  –  Y) and the Borrow (b) are the outputs, which one of the following diagrams implements a half-subtractor?

FA0

C0

FA1

S0

I0

X

2:1 MUX

S1

2:1 MUX

D

I1

S14

FA15

C15

S15

132. In the circuit shown, choose the correct timing diagram of the output (Y ) from the given waveforms W1, W2, W3 and W4.

Y S

S I0

2:1 MUX

X

I0

2:1 MUX

B

I1

X1

D

X1

D FF1 Q

I1

(d) X

I0 2:1 MUX

Q

B FF1

Clk

Q Output (Y )

Clk

I0 2:1 MUX

B

I1

Q Output (Y )

B

I1 S

X

S

X2

D

S

X2

D FF2 Q

Q

Y S

Y

C14

S

S X

(c) Y

FA14

A15 B15

(GATE 2014: 2 Marks) I0

D

I1

Y

C1

(b)

(a) Y

A14 B14

A1 B 1

I0

2:1 MUX

I1

X

I0

2:1 MUX

D

FF2

D

Q

I1 Q (GATE 2014: 2 Marks)

Clk

130. An 8-to-1 multiplexer is used to implement a logical function Y as shown in the figure. The output Y is given by

Clk X1

0 D 0 D 0 0 1 0

I0 I1 I2 I3 I4 I5 I6 I7

X X12 X2 W1

Y W1 W2 W3 W2

S2 S1 S0

W3 W4

A B C

W4

(a) Y = ABC + ACD (b) Y = ABC + ABD (c) Y = ABC + ACD (d) Y = ABD + ABC (GATE 2014: 2 Marks)

GATE_ECE_CH05_Digital_Circuits.indd 345

(a) W1 (c) W3

(b) W2 (d) W4 (GATE 2014: 2 Marks)

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GATE ECE Chapter-wise Solved papers

133. The outputs of the two flip-flops Q1, Q2 in the figure shown are initialized to 0, 0. The sequence generated at Q1 upon application of clock signal is

(a)

WL VDD BL

BL

Q1 J1 Q1

J2 Q2

K1 Q1

K2 Q2

Clk

(b) (a) 01110…. (c) 00110…

WL VDD

(b) 01010…. (d) 01100…

BL

BL

(GATE 2014: 2 Marks) 134. For the 8085 microprocessor, the interfacing circuit to input 8-bit digital data (DI0 - DI7) from an external device is shown in the figure. The instruction for correct data transfer is

A2 A1 A0

3−to−8 7 Decoder 6 5 C 4 3 B 2 1 A 0 G2AG2B G1

RD

I/O Device

(c) Digital inputs

VDD

DI0−DI7 DO0−DO7 Data bus (D0−D7) DS1

WL BL

BL

DS2

A8 A9 A10 A11 A12 A13 A14 A15

A3 A4 A5 A6 A7

(a) MVI A, F8H (c) OUT F8H

(b) IN F8H (d) LDA F8F8H (GATE 2014: 2 Marks)

135. An 8085 microprocessor executes “STA 1234H” with starting address location 1FFEH (STA copies the contents of the accumulator to the 16-bit address location). While the instruction is fetched and executed, the sequence of values written at the address pins A15 − A8 is (a) 1FH, 1FH, 20H, 12H (b) 1FH, FEH, 1FH, FFH, 12H (c) 1FH, 1FH, 12H, 12H (d) 1FH, 1FH, 12H, 20H, 12H (GATE 2014: 2 Marks) 136. If WL is the Word Line and BL the Bit Line, an SRAM cell is shown in

GATE_ECE_CH05_Digital_Circuits.indd 346

(d)

WL VDD BL

BL

(GATE 2014: 2 Marks) 137. In the figure shown, the output Y is required to be Y = AB + CD. The gates G1 and G2 must be, respectively,

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Chapter 5  •  Digital Circuits

(a) (b) (c) (d)

A G1 B G2

Y

C D (a) NOR, OR (c) NAND, OR

(b) OR, NAND (d) AND, NAND (GATE 2015: 1 Mark)

138. In the circuit shown, diodes D1, D2 and D3 are ideal and the inputs E1, E2 and E3 are ‘0 V’ for logic ‘0’ and ‘10 V’ for logic ‘1’. What logic gate does the circuit ­represent? D1

E1

D2 E2

Vo 1 kW 10 V

(a) 3-input OR gate (c) 3-input AND gate

(b) 3-input NOR gate (d) 3-input XOR gate (GATE 2015: 1 Mark)

 139. A mod-n counter using a synchronous binary up-counter with synchronous clear input is shown in the figure. The value of n is . 4-Bit binary counter CLOCK

CLK

QA QB QC QD

a modulo - 5 binary up counter a modulo - 6 binary down counter a modulo - 5 binary down counter a modulo - 6 binary up counter (GATE 2015: 1 Mark)

141. Consider a four bit D to A converter. The analog value corresponding to digital signals of values 0000 and 0001 are 0 V and 0.0625 V respectively. The analog value (in Volts) corresponding to the digital signal 1111 is . (GATE 2015: 1 Mark) 142. In an 8085 microprocessor, the shift registers which store the result of an addition and the overflow bit are, respectively (a) B and F (b) A and F (c) H and F (d) A and C (GATE 2015: 1 Mark) 143. In an 8085 microprocessor, which one of the following instructions changes the content of the accumulator? (a) MOV B, M (b) PCHL (c) RNZ (d) SBI BEH (GATE 2015: 1 Mark)

D3

E3

347

QA QB QC QD

CLEAR

144. Which one of the following 8085 microprocessor programs correctly calculates the product of two 8-bit numbers stored in registers B and C? (a) MVI A, 00H (b) MVI A, 00H JNZ LOOP CMP C CMP C LOOP : DCR B LOOP: DCR B JNZ LOOP HLT HLT (c) MVI A, 00H LOOP: ADD C DCR B JNZ LOOP HLT

(d) MVI A, 00H ADD C JNZ LOOP LOOP : INR B HLT (GATE 2015: 1 Mark)

(GATE 2015: 1 Mark)

145. The Boolean expression F ( X , Y , Z ) = XYZ + XY Z + XYZ + XYZ 140. The circuit shown consists of J-K flip-flops, each F ( Xwith , Y , Zan ) = XYZ + XY Z + XYZ + XYZ converted into the canonical prodactive low, asynchronous reset ( Rd input). The counter uct of sum (POS) form is corresponding to this circuit is (a) ( X + Y + Z )( X + Y + Z )( X + Y + Z )( X + Y + Z ) Q Q Q Q 01 J Q 11 J Q 2 1 J (b) ( X + Y + Z )( X + Y + Z )( X + Y + Z )( X + Y + Z ) Clock 1 K Rd 1 K Rd 1 K Rd (c) ( X + Y + Z )( X + Y + Z )( X + Y + Z )( X + Y + Z ) (d) ( X + Y + Z )( X + Y + Z )( X + Y + Z )( X + Y + Z ) (GATE 2015: 2 Marks)

GATE_ECE_CH05_Digital_Circuits.indd 347

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146. A function of Boolean variables X, Y and Z is expressed in terms of the min-terms as F(X, Y, Z) = ∑(1, 2, 5, 6, 7) Which one of the product of sums given below is equal to the function F(X, Y, Z)?

Which one of the following statements is TRUE? (a) Gate 1 is a universal gate (b) Gate 2 is a universal gate (c) Gate 3 is a universal gate (d) None of the gates shown is a universal gate (GATE 2015: 2 Marks)

(a) ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Y + Z ) 150. A 1-to-8 demultiplexer with data input Din, address (b) ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Y + Z ) + Z ) S0, S1, S2 (with S0 as the LSB) and Y0 to Y7 as the (c) ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Yinput eight demultiplexed outputs, is to be designed using two ( X + Y + Z)⋅( X + Y + Z)⋅( X + Y + Z)⋅( X +Y + Z)⋅( X +Y + Z) 2-to-4 decoders (with enable input E and address input ) A1) as shown in the figure. Din, S0, S1 and S2 are to (d) ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + Y + Z ) ⋅ ( X + YA+0 Zand be connected to P, Q, R and S but not necessarily in this ( X +Y + Z)⋅( X +Y + Z)⋅( X + Y + Z)⋅( X +Y + Z)⋅( X + Y + Z) order. The respective input connections to P, Q, R and S terminals should be (GATE 2015: 2 Marks) 147. All the logic gates shown in the figure have a propagation delay of 20 ns. Let A = C = 0 and B = 1 until time t = 0. At t = 0, all the inputs flip (i.e. A = C = 1 and B = 0) and remain in that state. For t > 0, output Z = 1 for a duration (in ns) of

P Q R S

1E

1Y0 2-to-4 Decoder 1Y1 1A0 1Y2 1Y3 1A1

Y0 Y1 Y2 Y3

2Y0 2-to-4 Decoder 2Y1 2A0 2Y2 2Y3 2A1

Y4 Y5 Y6 Y7

A Z B 2E

C (GATE 2015: 2 Marks) 148. A 3-input majority gate is defined by the logic function M(a, b, c) = ab + bc + ca. Which one of the following gates is represented by the function M ( M ( a, b, c), M ( a, b, c ), c) ? (a) (b) (c) (d)

3-input NAND gate 3-input XOR gate 3-input NOR gate 3-input XNOR gate

(b) S1, Din, S0, S2

(c) Din, S0, S1, S2

(d) Din, S2, S0, S1 (GATE 2015: 2 Marks)

(GATE 2015: 2 Marks) 149. A universal logic gate can implement any Boolean function by connecting sufficient number of them appropriately. Three gates are shown. X Y

(a) S2, Din, S0, S1

F 1 = X +Y

151. The figure shows a binary counter with synchronous clear input. With the decoding logic shown, the counter works as a Binary counter Q3 Q2 CLK Q1 Q0

Gate 1 CLR F 2 = X ×Y

X Y Gate 2 X Y

F 3 = X +Y Gate 3

GATE_ECE_CH05_Digital_Circuits.indd 348

(a) mod – 2 counter (c) mod – 5 counter

(b) mod – 4 counter (d) mod – 6 counter (GATE 2015: 2 Marks)

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Chapter 5  •  Digital Circuits

152. A three bit pseudo random number generator is shown. Initially the value of output Y = Y2Y1Y0 is set to 111. The value of output Y after three clock cycles is Y1

Y2

D2

Q2

D1

Q1

349

156. What is the voltage Vout in the following circuit? VDD

Y0

D0

Q0 10 kΩ Vout

CLK (a) 000 (b) 001 (c) 010 (d) 100 (GATE 2015: 2 Marks) 153. An SR Latch is implemented using TTL gates as shown in the figure. The set and reset pulse inputs are provided using the push-button switches. It is observed that the circuit fails to work as desired. The SR latch can be made functional by changing

Q

(a) 0 V (b) (|VT of PMOS| + VT of NMOS)/2 (c) Switching threshold of inverter (d) VDD (GATE 2016: 1 Mark) 157. The logic functionality realized by the circuit shown below is B

Set 5V A

M1

Y

Q M2

Reset (a) NOR gates to NAND gates (b) inverters to buffers (c)  NOR gates to NAND gates and inventers to buffers (d) 5 V to ground (GATE 2015: 2 Marks) 154. A 16 kB (=16,384 bit) memory array is designed as a square with an aspect of one (number of rows is equal to the number of columns). The minimum number of address lines needed for the row decoder is (GATE 2015: 2 Marks) 155. The output of the combinational circuit given below is

(a) OR (c) NAND

(b) XOR (d) AND (GATE 2016: 1 Mark)

158. The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is (a) 4 (b) 5 (c) 6 (d) 7 (GATE 2016: 1 Mark) 159. A 4 : 1 multiplexer is to be used for generating the output carry of a full adder. A and B are the bits to be added while Cin is the input carry and Cout is the output carry. A and B are to be used as the select bits with A being the more significant select bit.

A I0 I1 Y C

(a) A + B + C (b) A(B + C) (c) B(C + A) (d) C(A + B) (GATE 2016: 1 Mark)

GATE_ECE_CH05_Digital_Circuits.indd 349

I2 I3

4:1 Cout MUX S1 S0

A B Which one of the following statements correctly describes the choice of signals to be connected to the inputs I0, I1, I2 and I3 so that the output is Cout?

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(a) (b) (c) (d)

I0 = 0, I1 = Cin, I2 = Cin and I3 = 1 I0 = 1, I1 = Cin, I2 = Cin and I3 = 1 I0 = Cin, I1 = 0, I2 = 1 and I3 = Cin I0 = 0, I1 = Cin, I2 = 2 and I3 = Cin

(c) QSX + QS X (d) QS + QS (GATE 2016: 2 Marks)

(GATE 2016: 1 Mark) 160. Assume that all the digital gates in the circuit shown in the figure are ideal, the resistor R = 10 kΩ and the supply voltage is 5 V. The D flip-flops D1, D2 D3, D4 and D5 are initialized with logic values, 0, 1, 0, 1 and 0, respectively. The clock has a 30% duty cycle. The average power dissipated (in mW) in the resistor R is ______.

D

Clock

(GATE 2016: 1 Mark) 161. In an 8085 microprocessor, the contents of the accumulator and the carry flag are A7 (in hex) and 0, respectively. If the instruction RLC is executed, then the contents of the accumulator (in hex) and the carry flag, respectively, will be (a) 4E and 0 (b) 4E and 1 (c) 4F and 0 (d) 4F and 1 (GATE 2016: 1 Mark) 162. Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sum-of-product (SOP) expression for the function is

01

11

10

0

0

0

PQ RS 00 00 0

01

11

10

1

1

0

01

1

0

0

1

01

0

0

0

0

11

1

0

0

1

11

0

0

0

0

10

0

0

0

0

10

0

1

1

0

X=0 (a) PQS X + PQS X + QRS X + QRS X (b) QS X + QS X

GATE_ECE_CH05_Digital_Circuits.indd 350

OP0 OP1 OP2 3:8 OP3 Decoder OP4 OP5 OP6 OP7

IP0 IP1 IP2 IP3 8:3 IP4 Encoder IP5 IP6 IP7

Y2 Y1 Y0

R = 10 kΩ

Q D Q D Q D Q D Q D4 D5 D1 D2 D3

PQ RS 00 00 0

163. Identify the circuit below.

X=1

(a) (b) (c) (d)

Binary to Gray code converter Binary to XS3 converter Gray to Binary converter XS3 to Binary converter (GATE 2016: 2 Marks)

164. The functionality implemented by the circuit below is P Q R S

Y O0

C1

O1

C0

O2 O3 Enable = 1 is a tristate buffer

(a) 2-to-1 multiplexer (c) 7-to-1 multiplexer

(b) 4-to-1 multiplexer (d) 6-to-1 multiplexer (GATE 2016: 2 Marks)

165. For the circuit shown in the figure, the delays of NOR gates, multiplexers and inverters are 2 ns, 1.5 ns and 1 ns, respectively. If all the inputs P, Q, R, S and T are applied at the same time instant, the maximum propagation delay (in ns) of the circuit is _____.

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(a) mod-5 counter (c) mod-7 counter

P Q

0

R

MUX

1

S

(GATE 2016: 2 Marks)

0 MUX 1 So

So

T (GATE 2016: 2 Marks) 166. The state transition diagram for a finite state machine with states A, B and C, and binary input X, Y and Z is shown in the figure.

Y=1

168. In an N bit flash ADC, the analog voltage is fed simultaneously to 2N – 1 comparators. The output of the comparators is then encoded to a binary format using digital circuit. Assume that the analog voltage source Vin (whose output is being converted to digital format) has a source resistance of 75 Ω as shown in the circuit diagram below and the input capacitance of each comparator is 8 pF. The input must settle to an accuracy of 1/2 LSB even for a full scale input change for proper conversion. Assume that the time taken by the thermometer to binary encoder is negligible. − +

X=0, Y=0, Z=0 B Z=

1

Y=

Z=

X=0, Z=1

1,

1,

X=

Y=0, Z=0

Y=1

1

A

(b) mod-6 counter (d) mod-8 counter

Y=0, Z=1

X=1, Y=0

− +

Digital output

C Z=0 Which one of the following statements is correct? (a)  Transitions from State A are ambiguously defined (b) Transition from State B are ambiguously defined (c)  Transitions from State C are ambiguously defined (d)  All of the state transitions are defined unambiguously. (GATE 2016: 2 Marks) 167. For the circuit shown in the figure, the delay of the bubbled NAND gate is 2 ns and that of the counter is assumed to be zero. Q0 (LSB) 3-bit Synchronous counter

Q1 Q2 (MSB)

Clk RESET



If the clock (Clk) frequency is 1 GHz, then the counter behaves as a

GATE_ECE_CH05_Digital_Circuits.indd 351

+ 75 Ω −

− +

If the flash ADC has 8 bit resolution, which one of the following alternatives is closest to the maximum sampling rate? (a) 1 megasamples per second (b) 6 megasamples per second (c) 64 megasamples per second (d) 256 megasamples per second (GATE 2016: 2 Marks)

169. In an 8085 system, a PUSH operation requires more clock cycles than a POP operation. Which one of the following options is the correct reason for this? (a) For POP, the data transceivers remain in the same direction as for instruction fetch (memory to processor), whereas for PUSH their direction has to be reversed. (b) Memory write operations are slower than memory read operations in an 8085 based system. (c)  The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP operation uses the address already in the stack pointer.

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(d)  Order of registers has to be interchanged for a PUSH operation, whereas POP uses their natural order.



The logic implemented by the circuit is (a) XNOR (b) XOR (c) NOR

(d) OR

(GATE 2017: 1 Mark)

(GATE 2016: 2 Marks) 170. An 8 Kbyte ROM with an active low Chip Select input (CS) is to be used in an 8085 microprocessor based system. The ROM should occupy the address range 1000H to 2FFFH. The address lines are designated as A15 to A0, where A15 is the most significant address bit. Which one of the following logic expression will generate the correct CS signal for this ROM? (a) A15 + A14 + (A13 ⋅ A12 +  A13 ⋅ A12 )

TCK

Q D D-Latch Output CK

CLK1

CLK1 CLK2

(b) A15 ⋅ A14 ⋅ (A13 + A12) (c)

173. Consider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% cycle and CLK2 is a one-fifth period delayed version of CLK1. The duty cycle at the output of the latch in percentage is ________.

CLK2

A15 ⋅ A14 ⋅ ( A13 ⋅ A12 + A13 ⋅ A12 )

TCK/5

(d) A15 ⋅ A14 + A13 ⋅ A12

(GATE 2017: 1 Mark) (GATE 2016: 2 Marks)

171. In the latch circuit shown, the NAND gates have nonzero, but unequal propagation delays. The present input condition is: P = Q = ‘0’. If the input condition is changed simultaneously to P = Q = ‘1’, the outputs X and Y are

174. Which one of the following gives the simplified sum of products expression for the Boolean function F = m0 + m2 + m3 + m5, where m0, m2, m3 and m5 are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB? (a) AB + ABC + ABC (b) AC + AB + ABC

Gate 1

(c)

AC + AB + ABC

(d) ABC + AC + ABC

P X

(GATE 2017: 2 Marks) 175. Consider the circuit shown in the figure. Y Q 0

Y Gate 2 (a) X = ‘1’, Y = ‘1’ (b) either X = ‘1’, Y = ‘0’ or X = ‘0’, Y = ‘1’ (c) either X = ‘1’, Y = ‘1’ or X = ‘0’, Y = ‘0’ (d) X = ‘0’, Y = ‘0’ (GATE 2017: 1 Mark)

0

Q

GATE_ECE_CH05_Digital_Circuits.indd 352

NMOS

MUX

F

Z The Boolean expression F implemented by the ­circuit is (a) XY Z + XY + YZ (b) XYZ + XZ + YZ (c)

P

1

X



Y

0

1

172. For the circuit shown in the figure, P and Q are the inputs and Y is the output.

PMOS

MUX

XYZ + XY + YZ (d) XY Z + XZ + YZ (GATE 2017: 2 Marks)

176. Figure I shows a 4-bit ripple carry adder realized using full adders and Figure II shows the circuit of a full-adder (FA). The propagation delay of the XOR, AND and OR gates in Figure II are 20 ns, 15 ns and

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Chapter 5  •  Digital Circuits

10 ns, respectively. Assume all the inputs to the 4-bit adder are initially reset to 0. Y3 X 3

Z4

FA

Y2 X 2 Z3

FA

S3

Y1 X1 Z2

FA

S2

(c) PQR + PQR + PQR (d) ( P + Q + R) + ( P + Q + R) + ( P + Q + R)

Y0 X0 Z1

S1

(GATE 2017: 2 Marks) Z0

FA

353

S0

178. A 4-bit shift register circuit configured for right-shift operation, i.e. Din → A, A → B, B → C, C → D, is shown. If the present state of the shift register is ABCD = 1101, the number of clock cycles required to reach the state ABCD = 1111 is ____________.

Figure I Xn Yn

Sn A

Din

B

C

D

Clock (GATE 2017: 2 Marks)

Zn + 1 Zn Figure II

At t = 0, the inputs to the 4-bit adder are changed to X3X2X1X0 = 1100, Y3Y2Y1Y0 = 0100 and Z0 = 1. The output of the ripple carry adder will be stable at t (in ns) = ____________. (GATE 2017: 2 Marks)

179. A finite state machine (FSM) is implemented using the D flip-flops A and B, and logic gates, as shown in the figure below. The four possible states of the FSM are QAQB = 00, 01, 10, and 11.

177. A programmable logic array (PLA) is shown in the figure.

Q

D

QA

D

QB A

P

Q B

P Q Q R R CK x

x

Q

XIN

CK

Q

x CLK

x

x

x

x

x

x

F



Assume that XIN is held at a constant logic level throughout the operation of the FSM. When the FSM is initialized to the state QAQB = 00 and clocked, after a few clock cycles, it starts cycling through (a) (b) (c) (d)

P

Q

all of the four possible states if XIN = 1 three of the four possible states if XIN = 0 only two of the four possible states if XIN = 1 only two of the four possible states if XIN = 0 (GATE 2017: 2 Marks)

R

The Boolean function F implemented is (a) PQR + PQR + PQR (b) ( P + Q + R) + ( P + Q + R) + ( P + Q + R )

GATE_ECE_CH05_Digital_Circuits.indd 353

180. The state diagram of a finite state machine (FSM) designed to detect an overlapping sequence of three bits is shown in the figure. The FSM has an input “In” and an output “Out.” The initial state of the FSM is S0.

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00 S0 In = 0 Out = 0

VDD

In = 0 Out = 0 In = 1 Out = 0

In = 1 Out = 0 01 S1

x

y

y

In = 0 Out = 0

S2

f(x, y) In = 1 Out = 0

10 In = 0 Out = 0

x

In = 1 Out = 1 11

S3

181. The following FIVE instructions were executed on an 8085 microprocessor.



(a) NOR (b) AND (c) NAND (d) XOR

If the input sequence is 10101101001101, starting with the left-most bit, then the number of times “Out” will be 1 is ____________. (GATE 2017: 2 Marks)

MVI A, 33H MVI B, 78H ADD B CMA ANI 32H The accumulator value immediately after the execution of the fifth instruction is (a) 00H (b) 10H (c) 11H (d) 32H (GATE 2017: 2 Marks)

182. The clock frequency of an 8085 microprocessor is 5 MHz. If the time required to execute an instruction is 1.4 μs, then the number of T-states needed for executing the instruction is (a) 1 (c) 7

(b) 6 (d) 8 (GATE 2017: 2 Marks)

183. In a DRAM, (a) (b) (c) (d)

periodic refreshing is not required information is stored in a capacitor information is stored in a latch both read and write operations can be performed simultaneously (GATE 2017: 2 Marks)

184. The logic function f(x, y) realized by the given circuit is

GATE_ECE_CH05_Digital_Circuits.indd 354

(GATE 2018: 1 Mark) 185. A function F(A, B, C) defined by three Boolean variables A, B and C when expressed as sum of products is given by F = A ⋅ B ⋅C + A ⋅ B ⋅C + A⋅ B ⋅C

where, A, B , and C are the complements of the respective variables. The product of sums (POS) form of the function F is (a) F = ( A + B + C ) ⋅ ( A + B + C ) ⋅ ( A + B + C ) (b) F = ( A + B + C ) ⋅ ( A + B + C ) ⋅ ( A + B + C ) (c) F = ( A + B + C ) ⋅ ( A + B + C ) ⋅ ( A + B + C ) ⋅

( A + B + C )⋅( A + B + C )

(d) F = ( A + B + C ) ⋅ ( A + B + C ) ⋅ ( A + B + C ) ⋅

( A + B + C )⋅( A + B + C )

(GATE 2018: 1 Mark) 186. A traffic signal cycles from GREEN to YELLOW, YELLOW to RED and RED to GREEN. In each cycle, GREEN is turned on for 70 seconds, YELLOW is turned on for 5 seconds and RED is turned on for 75 seconds. This traffic light has to be implemented using a finite state machine (FSM). The only input to this FSM is a clock of 5 second period. The minimum number of flip-flops required to implement this FSM is . (GATE 2018: 1 Mark) 187. A four-variable Boolean function is realized using 4 × 1 multiplexers as shown in the figure.

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Chapter 5  •  Digital Circuits

VCC

I0

I1 4 × 1 MUX I2

I1 4 × 1 MUX I2

I3

I3

S1 S0

(GATE 2018: 2 Marks) 189. In the circuit shown below, a positive edge-triggered D Flip-Flop is used for sampling input data Din using clock CK. The XOR gate outputs 3.3 volts for logic HIGH and 0 volts for logic LOW levels. The data bit and clock periods are equal and the value of ΔT /TCK = 0.15, where the parameters ΔT and TCK are shown in the figure. Assume that the Flip-Flop and the XOR gate are ideal.

S1 S0

U V

⎡1 1 ⎤ ⎡1 0 ⎤ ⎢0 0⎥ (c) ⎢1 0 ⎥ (d) ⎣ ⎦ ⎣ ⎦

F (U, V, W, X)

I0

W X

The minimized expression for F(U, V, W, X) is (a) (UV + UV )W (b) (c) (d)

355

X

(UV + UV ) (WX + WX ) (UV + UV )W (UV + UV ) (WX + WX )

D

Q

Din D Flip-Flop CLK (GATE 2018: 2 Marks)

188. A 2 × 2 ROM array is built with the help of diodes as shown in the circuit below. Here W0 and W1 are signals that select the word lines and B0 and B1 are signals that are output of the sense amps based on the stored data corresponding to the bit lines during the read operation. B0

CK TCK CK Din

B1 DT

Sense amps W0

W1 VDD

B0 W0 ⎡ D00 W1 ⎢⎣ D10

B1

If the probability of input data bit(Din) transition in each clock period is 0.3, the average value (in volts, accurate to two decimal places) of the voltage at node X is . (GATE 2018: 2 Marks)

X0

Bits stored in the ROM Array During the read operation, the selected word line goes high and the other word line is in a high impedance state. As per the implementation shown in the circuit diagram above, what are the bits corresponding to Dij (where i = 0 or j = 0 or 1) stored in the ROM?

GATE_ECE_CH05_Digital_Circuits.indd 355

DT

190. The logic gates shown in the digital circuit below use strong pull-down nMOS transistors for LOW logic level at the outputs. When the pull-downs are off, high-value resistors set the output logic levels to HIGH (i.e. the pull-ups are weak). Note that some nodes are intentionally shorted to implement “wired logic”. Such shorted nodes will be HIGH only if the outputs of all the gates whose outputs are shorted are HIGH.

D01 ⎤ D11 ⎥⎦

⎡1 0 ⎤ ⎡0 1⎤ ⎢1 0 ⎥ (a) ⎢0 1⎥ (b) ⎣ ⎦ ⎣ ⎦

DT

X1 X2 Y X3

The number of distinct values of X3  X2  X1  X0 (out of the 16 possible values) that give Y = 1 is . (GATE 2018: 2 Marks)

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ANSWER KEY  1.  (b)

 2.  (c)

 3.  (b)

 4.  (c)

 5.  (d)

 6.  (a)

 7.  (*)  8.  (b)

 9.  (d)

 10. (a)

 11. (b)

 12. (b)

 13. (c)

 14. (d)

 15. (d)

 16. (c)

 17. (b)  18. (d)

 19. (b)

 20. (c)

 21. (b)

 22. (c)

 23. (a)

 24. (b)

 25. (b)

 26. (d)

 27. (c)  28. (d)

 29. (c)

 30. (a)

 31. (b)

 32. (d)

 33. (b)

 34. (a)

 35. (a)

 36. (c)

 37. (a)  38. (a)

 39. (c)

 40. (b)

 41. (d)

 42. (b)

 43. (d)

 44. (d)

 45. (c)

 46. (c)

 47. (b)  48. (c)

 49. (d)

 50. (d)

 51. (b)

 52. (a)

 53. (a)

 54. (c)

 55. (c)

 56. (b)

 57. (c)  58. (c)

 59. (d)

 60. (a)

 61. (d)

 62. (d)

 63. (d)

 64. (a)

 65. (b)

 66. (b)

 67. (b)  68. (c)

 69. (b)

 70. (d)

 71. (b)

 72. (a)

 73. (*)

 74. (b)

 75. (b)

 76. (c)

 77. (c)  78. (b)

 79. (c)

 80. (b)

 81. (d)

 82. (a)

 83. (b)

 84. (c)

 85. (d)

 86. (b)

 87. (c)  88. (c)

 89. (d)

 90. (d)

 91. (b)

 92. (d)

 93. (c)

 94. (b)

 95. (a)

 96. (d)

 97. (d)  98. (b)

 99. (d)

100.  (d)

101.  (c)

102.  (b)

103.  (d)

104.  (a)

105.  (a)

106.  (d)

107.  (c) 108.  (a)

109.  (b)

110.  (a)

111.  (d)

112.  (c)

113.  (a)

114.  (d)

115.  (d)

116.  (4)

117.  (d) 118.  (a)

119.  (c)

120.  (a)

121.  (a)

122.  (c)

123.  (d)

124.  (62.5) 125.  (d)

126.  (b)

127.  (a) 128.  (c)

129.  (a)

130.  (c)

131.  (195)

132.  (c)

133.  (d)

134.  (d)

135.  (a)

136.  (b)

137.  (a) 138.  (c)

139.  (6)

140.  (a)

141.  (0.9375) 142.  (b)

143.  (d)

144.  (c)

145.  (a)

146.  (b)

147.  (40) 148.  (b)

149.  (d)

150.  (d)

151.  (b)

152.  (d)

153.  (d)

154.  (7)

155.  (c)

156.  (c)

157.  (d) 158.  (a)

159.  (a)

160.  (1.5)

161.  (d)

162.  (b)

163.  (c)

164.  (b)

165.  (7)

166.  (c)

167.  (d) 168.  (a)

169.  (c)

170.  (a)

171.  (b)

172.  (b)

173.  (30)

174.  (b)

175.  (b)

176.  (120) 177.  (c) 178.  (10)

179.  (a), (d)

180.  (4)

181.  (b)

182.  (c)

183.  (b)

184.  (d)

185.  (c)

186.  (5)

189.  (0.8415) 190.  (8)

187.  (c) 188.  (a)

*  None of the options is correct.

ANSWERS WITH EXPLANATION 1. Topic: Data Converters: ADCs and DACs (b)  Conversion time of successive approximation analog to digital converter depends on the number of bits but does not depend on the input voltage. So, the conversion time for a 2 V input will be same as that for a 1 V, that is, 20 µs. 2.

Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (c)  There are five hardware interrupts which are present in a 8085 microprocessor. These are TRAP, ­ RST 7.5, RST 6.5, RST 5.5 and INTR.

3. Topic: Data Converters: Sample and Hold Circuits (b)  The polarity and amplitude of the samples should remain same. So, the unity gain non-inverting amplifier is the most commonly used amplifier in sample and hold circuits. 4. Topic: Data Converters: ADCs and DACs (c)  Number of comparators in M-bit flash ADC

= 2 M − 1 = 24 − 1 = 15

GATE_ECE_CH05_Digital_Circuits.indd 356

(Since M = 4)

5. Topic: Combinatorial Circuits (d) 1 A B 2

3 X

C

The output (X) will be 1 only if all the inputs to the AND gate 3 are 1. So, C must be 1. For the output of X-NOR gate 2 to be 1, B must be 1. Now as B is 1, A must be 0 to get output 1 from XOR gate 1. Hence, the required input conditions must be (0, 1, 1). 6. Topic: 8-bit Microprocessor (8085) (a)  The vector address of the interrupts can be ­calculated using the following formula:

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9. Topic: Sequential Circuits: Latches and Flip-Flops (d)  The next state of D flip-flop is the present value itself, that is,

Vector address = Interrupt number × 8 = 6 × 8 = ( 48)10 = (30) H 7. Topic: Combinatorial Circuits (*)  Following figure explains the output of the given logic circuit: AB

A B

AB ⋅ B = B

Qn+1 = D Therefore, from the figure given in the question, we have Qn+1 = D = X Z + Y Z (1)

B⋅B = 1

The equation of a JK flip-flop is given by the expression: Y=0

B

Qn+1 = KQ + JQ (2)



Comparing Eq. (1) and (2), we get:

B+C B⋅C

C

X = K, Y = J

8. Topic: Code Converters: ADCs and DACs (b)  The given figure is 4 bits digital to analog converter, that is, it accepts 4-bit digital signal and converts into analog signal. R=1

RF = 7

−

2R

R

2R

b0 = 0 (LSB)

2R

R Vx

Vo

Instruction SUB B subtracts the content of B from A and stores the result in A. That is, A ← A− B A ← A + ( − B) 0011 1010 1011 0111

1V b3 = 1 (MSB)

From the above figure, 4-bit input, b3b2b1b0 = 1010 Input voltage Vx, at the non-inverting terminal of op-amp is as following:

(

)

(

)

1 n −1 2 bn −1 + 2n − 2 bn − 2 + ... + 20 b0 2n 10 1 3 Vx = V 2 × 1 + 22 × 0 + 21 × 1 + 20 × 0 = 16 16 Vx =

A = 3 AH = 00111010 B = 49 H = 01001001

2R −15 V

2R

1 V = VR b 1 = 1 b2 = 0

+

10. Topic: 8-Bit Microprocessor (8085): Memory and I/O Interfacing (a) 

+15 V

R

Hence, the given sequential circuit resembles a JK flipflop with inputs X = K and Y = J.

1111 0001 F 1 H Therefore, the contents of A, CY, S are as following: A = F1H, CY = 1, S = 1 11. Topic: Sequential Circuits: Latches and Flip-Flops (b) J

Now, output voltage of non-inverting op-amp can be c­ alculated as following: ⎛ Rf ⎞ Vo = Vx ⎜1 + ⎟ R⎠ ⎝ =

GATE_ECE_CH05_Digital_Circuits.indd 357

Q0

J

Q1

J

Q2

J

Q3

Y

F = 10 kHz K CLR

K CLR

K CLR

K CLR

10 ⎛ 7 × 103 ⎞ 1+ =5V 16 ⎜⎝ 1 × 103 ⎟⎠

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The given circuit is a ripple or mod-N counter. The ­output frequency of mod-N counter is given as : f clk N This circuit will up count till N before resetting of flipflops. Resetting of flip-flops takes place only if the clocks given to each flip-flop or the output of NAND gate is zero. A NAND gate gives zero output only if all its inputs are high. Therefore, resetting of flip-flops takes place when both outputs Q1 and Q3 are high. At this moment, the counter value (Q3Q2Q1Q0) is (1010), that is, 10. Hence, the given circuit is a mod-10 counter.

15. Topic: Combinatorial Circuits (d)  The given TTL circuit is 8:1 multiplexer. 1

fy =

0 X0 X1 X2 X3 X4 X5 X6 X7 OR C C

B A

E MSB 8:1 MUX S2 S1 S0 LSB Y = f(A, B, C)

fy =

f clk 10 kHz = = 1 kHz N 10

12. Topic: Number Systems (b)  The binary equivalent of 17 = 10001. Therefore, binary equivalent of −17 = 10001 ↓ 1’s complement 01110 +1 01111 2’s complement Hence, 2’s complement representation of –17 is 01111.

Y = f ( A, B, C ) = f (C , B, A) = Σmin(0,3,5,6) f = C B A + CBA + CBA + CBA =C ( B A + BA) + C ( BA + BA) = C ( B  A) + C ( A ⊕ B) = C ( A ⊕ B) + C ( A ⊕ B) 16. Topic: Sequential Circuits: Latches and Flip-Flops (c) 17. Topic: Semiconductor Memories: DRAM (b)  For the given DRAM cell: Word line (WL) = VG

13. Topic: 8-Bit Microprocessor (8085) (c)  For the given 8085 microprocessor 4K × 8-bit RAM = 4 kB RAM = 4096 bytes = 212 bytes This is equivalent to 1000 bytes that corresponds to space from 0000 to 0FFF. So, the spacing is 0FFFH and the number of address lines required for addressing the memory is 12 (A0 - A11). FFF

D

AA00 Starting address 000 End address = AA00H + 0FFFH = B9FF 14. Topic: Logic Gates and their Static CMOS Implementations (d)  Because inputs to Ex-OR gate are always high, output of Ex-OR gate will be zero. Therefore, output of NAND gate will always be 1. Thus, LED does not ever emit light, irrespective of the position of the switches.

VGS ≥ VT S

VD ON C = VS

Condition for ON state of N-MOSFET VGS ≥ Vth ⇒ VG − VS ≥ Vth

Last byte address

FFF − − − − − − − − − − 001

GATE_ECE_CH05_Digital_Circuits.indd 358

G



VG − Vth ≥ VS ⇒ VS(max) = VG − Vth Two conditions: (i) If VD < VG − Vth , then VS = VD (ii) If VD ≥ VG − Vth , then VS = VG − Vth = VG − 1 Looking at the options, (b) satisfies the conditions.

18. Topic: Number Systems (d)  2’s complement of a number represents negative of that number. Therefore, the original number can be obtained by again taking 2’s complement of the 2’s complement. 2’s complement of (1000)2 = 1’s complement of (1000)2 +1 = (0111)2 + 1 = (1000)2

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Chapter 5  •  Digital Circuits

Since, the number is in the 2’s complement form and decimal equivalent of (1000)2 is ‘8’. Therefore, the given number is –8. 19. Topic: Combinatorial Circuits (b)  Truth table of an XOR-gate can be written as ­following: Inputs

Output

A

B

Y

0

0

0

0

1

1

1

0

1

22. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  As the circuit requires an external triggering and it has one stable and one quasistable state, it is monostable multivibrator. 23. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a)  Both the MOSFETS have a threshold voltage of 2 volts. Threshold voltage for NMOS is positive while it is negative for PMOS. So, Vtn =| Vt p | = 2 V For turning on NMOS, the gate to source voltage should be greater than its threshold voltage, that is, VGSn ≥ 2 V

1 1 0 From the above truth table, we can observe the following three points: (i) If A = 0, then Y = B

VGn − VSn ≥ 2 V Vi − 0 ≥ 2 V ⇒ Vi ≥ 2 V So, Statement 1 is TRUE. Now, consider Statement 2 to be TRUE. That is, Vo = 0 V. For Vo = 0 V, the output terminal will be connected to ground and hence NMOS will be in ON condition and PMOS will be in off condition. For NMOS to act in saturation region,

(ii) If A = 1, then Y = B (iii) If A = B, then Y = 1 1

359

X 1

VDSn ≥ VGSn − Vt n VDn − VSn ≥ VGn − VSn − Vt n

X We see that a pair of XOR-gate gives output ‘1’. Therefore, a cascade of 20 XOR-gate will give output ‘1’. 20. Topic: Data Converters: ADCs and DACs (c)  For n-bit flash ADC, number of comparator required = 2n − 1 Therefore, for 3-bit ADC, number of comparator required = 23 − 1 = 7 21. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (b)  Propagation delay of G1= 10 n sec Propagation delay of G2= 20 n sec Input Vi makes an abrupt change from logic 0 to 1 at time t = t0. The following figure shows the input waveform, outputs of G1 and G2: V0 G1

VDn ≥ VGn − Vt n (1) As VDn = 0 and VGn > Vt n so Eq. (1) is not true. Hence, Statement 2 is not TRUE. 24. Topic: Code Converters (b)  Truth table of the given circuit can be tabulated as following: Input

Output

8

4

2

1

2

4

2

1

X3

X2

X1

X0

Y3

Y2

Y1

Y0

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

0

1

1

0

0

1

0

0

0

1

0

2

0

0

1

1

0

0

1

1

3

0

1

0

0

0

1

0

0

4

0

1

0

1

0

1

0

1

5

0

1

1

0

1

1

0

0

6

0

1

1

1

1

1

0

1

7

Vi t0

t1

t2

t3

Therefore, option (b) is the correct waveform.

GATE_ECE_CH05_Digital_Circuits.indd 359

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excludes one comparator required for polarity. So, 28 - 1 = 63 comparators are required.

1

0

0

0

1

1

1

0

8

1

0

0

1

1

1

1

1

9

1

0

1

0

x

x

x

x

1

0

1

1

x

x

x

x

1

1

0

0

x

x

x

x

(a)  The K-maps for all four Boolean functions, namely, W, X, Y and Z are drawn as follows:

1

1

0

1

x

x

x

x

• The following figure shows the K-map for

1

1

1

0

x

x

x

x

1

1

1

1

x

x

x

x

30. Topic: Combinatorial Circuits: Minimization of Functions using Boolean Identities and Karnaugh Map

W = R + PQ + RS :

01

11

10

1

1

1

1

1

1

11

1

1

1

10

1

1

1

00

RS



Hence, the given figure gives the output as 2-4-2-1 BCD numbers.

PQ 00

25. Topic: 8-bit Microprocessor (8085) (b)  MVI, B, 87H (Moves 87H to B) MOV A, B  (Moves the content of A to B) START: JMP NEXT (Jumps to NEXT unconditionally) NEXT: XRA B (A ← = 00H, s = 0) JP START (will jump to start as s = 0) JMP NEXT  (jump unconditionally to label next) NEXT: XRA B (A = ← 87H) JP START (Will not jump as D7 = 1 ) OUT PORT2 (Display 87H at PORT2)

01

1

• The following figure shows the K-map for X = PQRS + PQRS + PQRS :

RS

00

01

11

10

PQ 00

1

26. Topic: Combinatorial Circuits: Boolean Algebra 01

(d)  Number of distinct boolean expressions for n varin ables = 22 . Here, n = 4. Therefore, number of distinct n 4 boolean expressions = 22 = 22 = 216 = 65536 27. Topic: Combinatorial Circuits: Multiplexers (c)

11

1

10

1

• The following figure shows the K-map for Y:

28. Topic: Sequential Circuits: Counters (d)  It is a modulo-7 counter and the count sequence is 000, 001, 010, 011, 100, 101, 110 and 000 and so on. The counter is reset the moment it goes to 111. Since there is no three-input gate here; two two-input AND gates will be required to simulate a three-­input AND gate that in turn will generate the LOW-to-HIGH CLEAR signal. 29. Topic: Data Converters: ADCs and DACs (c)  The number of comparators needed for an n-bit flash A/D converter is given by 2n−1. This, of course,

GATE_ECE_CH05_Digital_Circuits.indd 360

RS

00

01

11

10

PQ 00

1

01

1

1

1

11

1

1

1

10

1

1

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Chapter 5  •  Digital Circuits

Y = RS + PR + PQ + PQ

0

1 a

= RS + ( P + R )( P + Q )( P + Q ) = RS + ( P + PQ + PR + QR )( P + Q )

0

1

b

c 1

d 1

1

= RS + PQ + PQ + PQR + PQR + QR

0 0

= RS + PQ + QR ( P + P ) + QR

1 1 1

1

1

0

1

= RS + PQ + QR

1 0

• The following figure shows the K-map for Z: 0 w RS

00

01

11

1 1

1

1

1

1

1

11

1

1

1

10

1

1

1

1

01

1

0

z 0

Now, w=a x = a⊕b y = c ⊕ x ( a + b)

Z = R + S + PQ + PQR + PQS = R + S + PQ.PQR.PQS = R + S + ( P + Q )( P + Q + R)( P + Q + S ) = R + S + ( PQ + PR + PQ + QR)( P + Q + S ) = R + S + PQ + PQ + PQS + PR + PQR + PRS + PQS + PQR + QRS = R + S + PQ + PR + PQS + PQR + PRSS + PQS + PQR + QRS = R + S + PQ(1 + S ) + PR(1 + Q ) + PRS + PQS + PQR + QRS = R + S + PQ + PR + PRS + PQS + PQR + QRS = R + S + PQ + PR(1 + S + Q ) + PQS + QRS = R + S + PQ + PR + PQS + QRS From an examination of K-maps, it can be concluded that W = Z and X = Z 31. Topic: Combinatorial Circuits: Arithmetic Circuits (b)  Y = D = P ⊕ Q ⊕ R and Z = Borrow = RQ + PR + QP 32. Topic: Combinatorial Circuits: Code Converters (d)  The logic circuit is redrawn as shown in the following figure:

GATE_ECE_CH05_Digital_Circuits.indd 361

y

10

PQ 00

x

z = d ⊕ y( a + b + c) These are Boolean functions for converting Gray code number to Binary code number. 33. Topic: Sequential Circuits: Counters (b)  Each flip–flop has a propagation delay of 10 ns. Total propagation delay in a ripple counter is sum of propagation delays of all flip–flops. In the case of synchronous counters, it is equal to propagation delay of each flip–flop. Therefore, R = 40 ns and S = 10 ns. 34. Topic: Data Converters: ADCs and DACs (a) 

R R R⎤ ⎡ R + d1 + d0 Vo = −VR ⎢ d3 + d2 2 4 8 R R R R ⎥⎦ ⎣ R Therefore, Vo = −VR [constant ] R

The worst-case tolerance in Vo is ⎛ 1.1 × 1.1⎞ ± ⎜1 − ⎟ × 100% = ± 35% ⎝ 0.9 ⎠ 35. Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (a)  CMP B subtracts the contents of register B from the accumulator A. Since accumulator has content less than that of register B, therefore CY = 1 as it shows negative result. Also, the result being non-zero, the zero flag will be reset. 36. Topic: Semiconductor Memories: ROM (c) When W has the data 0110 (i.e., 6 in decimal), its data value at that address is 1010. Now, 1010, that is, 10 is acting as address at time t 2 and data at that moment is 1000.

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37. Topic: Number Systems (a)  The required range is given by  − (2n−1 − 1) to + (2n−1 − 1) = − (26−1 − 1) to + (26−1 − 1) = −31 to +31 38. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a)  For TTL logic, the floating input = 1. Therefore, the output is given by (1 + A · B)′ = 1′ = 0 39. Topic: Sequential Circuits: Latches and Flip–Flops (c)  The output of the master-slave flip–flop is basically the output of the slave flip–flop. When the clock goes HIGH, master flip–flop is enabled and slave flip–flop is disabled. Data input affects operation of master flip–flop only. When the clock goes LOW, master gets disabled and slave gets enabled and the slave output gets affected according to data input.

The following figure shows that the logic implementation in the absence of availability of complements. Therefore, the number of units is 2. y f x x 45. Topic: Combinatorial Circuits: Multiplexers (c)  Two 2-to-1 multiplexers are required at the input and one 2-to-1 multiplexer is required at the output. 46. Topic: Sequential Circuits: Counters (c)  This counter counts as 000, 001, 010, 011, 100, 101, 000, .... Thus, the logic gate should be such that it produces a ‘0’ output for B = 1, C = 1 (B = 0, C = 0) and a ‘1’ output for every other input. Note that inputs to the logic gate are from B and C. Therefore, the logic gate is an OR gate.

40. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (b)  Counter is used for frequency division, decoder is used for addressing in memory chips and shift register can be used for serial-to-parallel data conversion.

47. Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (b)  4 memory cycles are required to execute(I) and 3 memory cycles are required to execute(II).

41. Topic: Data Converters: ADCs and DACs (d)  From the given data, we have 2n ≥ 100, which gives n = 7.

48. Topic: 8-bit Microprocessor (8085): Programming (c)  The given instructions perform the following ­functions:

42. Topic: Number Systems (b) • 11001: The sign is negative. 2’s complement of magnitude bits is 0111. The decimal equivalent is -7. • 1001: The sign is negative. 2’s complement of magnitude bits is 111. The decimal equivalent is -7. • 111001: The sign is negative. 2’s complement of magnitude bits is 00111. The decimal equivalent is -7. Also note that 11001 and 111001 are extensions of 1001 where additional bit/s equal to MSB have been added to the left. 43. Topic: Combinatorial Circuits: Minimization of Functions Using Boolean Identities (d)  From options, ABC + ABC + ABC + ABC

LXI H, 9258

→

Memory location load data into the registor HL 9258

MOV A, M

→

Data moved from memory to accumulator

CMA

→

Data complemented in accumulator

MOV M, A

→

Complemented data moved to memory from accumulator

Hence, as a result of the given sequence of instructions, contents of location 9258 are complemented and stored in location 9258. 49. Topic: 8-bit Microprocessor (8085): Programming (d) Comments

= AC ( B + B ) + BC ( A + A) = AC + BC 44. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  From the truth table, one can draw the Karnaugh map. The simplified Boolean function is x+y

GATE_ECE_CH05_Digital_Circuits.indd 362

MVI A, 00H

→

A → 00 H

Loop: ADD B

→ →

A → 0A H Decreases the contents of register C 0BH times, ­therefore adding

DCR C

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JNZ Loop

→

0AH and 0BH eleven times This instruction causes the functioning of loop 11 times. Thus multiplication accomplished.

HLT END 50. Topic: 8-bit Microprocessor (8085): Programming (d)  The modes of operations of the 8255 would be Mode 2 for (I) and Mode 1 for (II). 51. Topic: Number Systems (b)  Hexadecimal equivalent can be determined by ­successively dividing the decimal number by 16 and recording the remainders. Remainders written in reverse order give the equivalent number. BCD equivalent is found by replacing each decimal digit by its four-bit binary equivalent. 52. Topic: Combinatorial Circuits: Multiplexers (a)  We know that f = E · A where E = BC + BC . Therefore, output f is given by ABC + ABC 53. Topic: Combinatorial Circuits: Boolean Algebra (a)  We have, f = ABC + ABC = B( AC + AC ) = B( A + C )( A + C ) 54. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  From the given circuit: +5 V

sistor Q3 is in saturation, then VBE3 = 0.75 V. Applying KVL in base-emitter loop of transistor Q3, we get I R × 1 × 103 − VBE3 = 0 I R = 0.75 × 10 −3 A = 0.75 mA 55. Topic: Sequential Circuits: Latches and Flip–Flops (c)  It is given that Qn = 0 and J = 1. K is either ‘0’ or ‘1’. If K = 0, the output will go to logic ‘1’ once triggered. If K = 1, the flip–flop will toggle and again go to logic ‘1’ state. 56. Topic: Sequential Circuits: Counters (b)  Initially, Q2Q1Q0 = 011. Therefore, Q2 = 0, Q1 = 0 and Q0 = 0. With the first clock pulse, all flip–flops change state. Therefore, the counter goes to 100. 57. Topic: 8-bit Microprocessor (8085): Programming (c) Location Mnemonics Operation 0100H LXI SP, 00FF SP → 00FF H LXI H, 0107 H-L → 0107 H 0103H MVI A, 20H A → 20H 0106H SUB M A → 00 H 0108H 0109H Address of M specified by contents of H-L memory and H-L → 0107 H, and 0107H → 20H . When PC reaches 0109 H, contents of accumulator reduces to zero after having operation of instruction SUB M. 58. Topic: 8-bit Microprocessor (8085): Programming (c) ORI 40 H A → 40 H 0109 H ADD M A → 40 H + 20 H = 60 H 010BH A → 0000 0000 H = 00 H M → 0107 H 0100 0000 H = 40 H ORI → 0100 0000 H = 40 H 0107 H → 20 H → 0010 0000 H = 20 H ADD M → 0110 0000 H = 60 H 59. Topic: Semiconductor Memories: SRAM (d)

1.4 kΩ

Chip #1

I = 1 mA Vo

Q2 Q1

A15 … … A12

A11

A10 A9 A8 A7 … … A0

×

×

×

×

×

×

0

1

0

0

0

0

×

×

×

×

×

×

0

1

1

1

1

1

Q3 + VBE3

Chip #2

−

1 kΩ IR When the output is at logic is zero, Vo = 0 V. Also, Vo = 0 V, when transistor Q3 is in saturation. When tran-

GATE_ECE_CH05_Digital_Circuits.indd 363

A15 … … A12

A11

A10 A9 A8 A7 … … A0

×

×

×

×

×

×

1

0

0

0

0

0

×

×

×

×

×

×

1

0

1

1

1

1

Therefore, F800 - F9FF cannot be the memory range for chip#1 and chip#2.

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60. Topic: Combinatorial Circuits: Karnaugh Map (a)  Four extreme entries of 1’s form one group and the other two 1’s in the rightmost column form the second term. Therefore, there are two number of product terms in minimized sum-of-product expression.

SP → EFFF H LXI SP, EFFFH CALL 3000H program transfers to memory location 3000H 3000H : LXI H, 3CF4H HL → 3CF4 PUSH PSW SP → EFFF → ////////

61. Topic: Number Systems (d)  Given BCP number = 100010011001. This n­ umber can be rewritten as 100 010 011 001 by splitting it in groups of three bits starting from extreme right. Replacing each three-bit group by its corresponding pentary equivalent, we get the answer as 4231.

    SP-1 → EFFE → X     SP-2 EFFD → Y     (X andY are accumulator’s contents and flag register contents loaded into memory location EFFE and EFFD, respectively.) SPHL     SP → 3CF4 contents of H-L register loaded into SP. POP PSW     SP → 3CF4 Flag register → Z     SP → 3CF5 Accumulator → W     SP → 3CF6      (Z and W are contents of memory locations 3CF4 and 3CF5 respectively.) RET

62. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  With point P stuck in logic ‘1’ state, we have to look at only the last three NAND gates appearing on the right side. The last but one NAND gate produces an output equal to (1·A)′ = A′. The last NAND gate inverts this to produce A. 63. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (d)  After the first clock pulse, A = 1, B = 1, Ci = 0, S = 0, Co = 1. After the second clock pulse, A = 1, B = 1, Ci = 1, S = 1, Co = 1.

3CF7

64. Topic: Sequential Circuits: Counters (a)

3CF8

Q1

Q0

D1( Q0 )

D0 ( Q1)

0

0

0

1

0

1

1

1

1

1

1

0

1

0

0

0

→

1 2 3 4 5 6 7 8 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ Counter → 001 010 011 100 101 110 111 000 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ Input → 0001 0010 0011 1000 1001 1010 1011 0000 of D/A converter 1 2 3 8 9 10 11 0 Therefore, the waveform given in (b) is the correct answer. 66. Topic: 8-bit Microprocessor (8085): Programming (b) Location Instruction Operation

GATE_ECE_CH05_Digital_Circuits.indd 364

SP initial address

SP address after RET execute

65. Topic: Data Converters: ADCs and DACs (b)  C/P

3CF6

Hence SP → 3CF8 H. 67. Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (b) A7

A6

A5

A4

A3

A2

A1

A0

1

1

0

1

0

1

0

0

D4

1

1

0

1

0

1

0

1

D5

1

1

0

1

0

1

1

0

D6

1

1

0

1

0

1

1

1

D7

I/P lines to decoder From the table we can see that the line A5 is low for all the address values. Therefore, the output is taken from the fifth line. 68. Topic: Number Systems (c)  In binary addition using 2’s complement format, CARRY is disregarded in the SUM. 01110 + 11001 = 1000111. By disregarding the CARRY (the leftmost bit), we get 000111.

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Chapter 5  •  Digital Circuits

69. Topic: Combinatorial Circuits: Boolean Algebra (b)  A ⋅ B + C ⋅ D = [( A ⋅ B ) ⋅ (C ⋅ D )] . The following figure shows the NAND implementation and it requires three NAND gates.

I1 =

VB2 = (5 − 0.7 − I1 × 4 × 103 )V = (5 − 0.7 − 0.72 × 10 −3 × 4 × 103 )V = 1.42 V

Y



70. Topic: Combinatorial Circuits: Minimization of Functions using Boolean Identities and Karnaugh Map (d)  The K-map corresponding to given Boolean expression is shown in the following figure:



Since VB2 > 0.7 V, it implies that the transistor Q2 operates in saturation region. Since transistor Q2 is in saturation, I2 =

CD 00

01

AB 00

11

10

1



01 11



10



1

The simplified expression from the K-map is given by Y = ABCD + ABCD + BCD 71. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (b)  From the given circuit, when Vi is at logic HIGH (2 V - 5 V), base-emitter junction of Q1 becomes reverse biased and current flows through first resistor and base-collector junction of Q1 into the base of Q2. Hence, Q1 operates in the reverse active region. Considering the following figure, we have + VCC = 5 V I1 4 kW 0.7 V Q2 Q1

0.7 V

VB2

1 kW

GATE_ECE_CH05_Digital_Circuits.indd 365

VCC R2 + R3

=

5 1.4 × 10 + 1 × 103

=

5 A = 2.03 mA 2.4 × 103

3

Also, VB3 = I 3 R3 = ( 2.03 × 10 −3 ) × 1 × 103 V = 2.03 V

1 1

5 − 0.7 − 0.7 A = 0.72 mA 4 × 103 + 1 × 103

Also,

A B

C D

365

Since VB3 > 0.7 V, it implies that Q3 also operates in ­saturation region. Q3 and Q4 together form a totem pole output. Only one of the transistors will be ON. As Q3 is in saturation, ­therefore Q4 will be in cut off.

72. Topic: Combinatorial Circuits: Multiplexers (a)  Let the output of the first MUX be Y. Therefore, Y = AB + AB = A ⊕ B X = YC + YC = Y ⊕ C Thus, X = A ⊕ B ⊕ C = ABC + ABC + ABC + ABC 73. Topic: Sequential Circuits: Latches and Flip–Flops (*)  The circuit shown is that of an R-S latch (X = S, Y = R) with active LOW inputs. The input entry X = 0, Y = 0 is a forbidden combination. Therefore, none of the answers is correct. 74. Topic: Sequential Circuits: Counters (b) Initially Q1Q0 are 00. NOR gate output is 1 and therefore D0 is 1 and D1 is 0. With first clock edge, Q0 becomes 1 and Q1 remains 0. This drives NOR gate output to 0 making D0 as 0 and D1 as 1. With second clock edge, Q0 becomes 0 and Q1 becomes 1. This does not change status of NOR gate output. Therefore with third clock edge, both Q0 remains 0 and Q1 becomes 0. The process repeats afterwards

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75. Topic: Data Converters: ADCs and DACs

79. Topic: 8-bit Microprocessor (8085): Programming

(b)  The negative terminal of the op-amp shown in the circuit can be considered at virtual earth as the non-inverting input is ­connected to ground. The resistive network can be simplified to prove VR is connected across a resistance equal to R. Therefore, the current drawn from VR = 10/(10 × 103) = 1 mA. This c­ urrent is successively divided between two equal resistance paths of 2R each. It can be proved that current ⎛ 1⎞ i = ⎜ ⎟ mA = 62.5 µA ⎝ 16 ⎠

(c)  Line 5: ANI 9BH. This line ANDs the contents of accumulator A with 9B. accumulator with 9BH A 11 10 10 10 9BH 10 01 10 11 10 00 10 10 Accumulator stores 8AH in line 6 and compares its contents with 9FH. Since 9FH is greater than 8AH, so carry flag will be generated while zero flag remains unaffected. 80. Topic: Number Systems

76. Topic: Data Converters: ADCs and DACs (c)  The net current in the inverting terminal of the op-amp is equal to ⎛1 1 ⎞ ⎛ 5⎞ ⎜⎝ + ⎟⎠ mA = ⎜⎝ ⎟⎠ mA 4 16 16

(b)  The two numbers P and Q are represented in 2’s complement form. (P-Q) can be found out by adding 2’s complement of Q to P and disregarding the CARRY. The answer will also be in 2’s complement notation. P = 11101101 and Q = 11100110

Therefore,

The 2’s complement of Q = 00011010

⎛ 5⎞ Vo = − ⎜ ⎟ × 10 −3 × 10 × 103 = −3.125 V ⎝ 16 ⎠

P − Q = P + (2’s complement of Q). Addition of the two numbers gives 00000111.

77. Topic: 8-bit Microprocessor (8085): Architecture (c)  8255 chip will select in I/O mapped when A7

A6

A5

A5

A3

1

1

1

1

1

A0, A1 and A2 are in don’t care conditions.

81. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  (PQ)′ = (P ′ + Q ′) Therefore, M1 = [ PQ( P + Q )] ⊕ R = [( P + Q )( P + Q )] ⊕ R

A7

A6

A5

A4

A3

A2

A1

A0

1

1

1

1

1

0

0

0 = F8H

1

1

1

1

1

1

1

1 = FFH

= ( P ⊕ Q) ⊕ R 82. Topic: Combinatorial Circuits: Multiplexers (a)  The output can be written as

So, range of chip selection is F8H – FFH 78. Topic: 8-bit Microprocessor (8085): Programming (b) After line (1) A contains B5 H After line (2) B contains 0EH After line (3) A is XOR with 69 H A 10 11 01 01 69 01 10 10 01 DC 11 01 11 00 After this contents of registor B are added to accumulator A. 11 01 11 00 00 00 11 10 11 10 10 10 So after line 4, A = EAH

GATE_ECE_CH05_Digital_Circuits.indd 366

Z = PRS + PQRS + PRS + ( P + Q ) RS The following figure shows the respective Karnaugh map. RS PQ 00

00

01

11

10

1

1

01

11

1

1

1

10

1

1

1

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Therefore, the output is Z = PQ + PQS + QRS 83. Topic: Sequential Circuits: Latches and Flip–Flops (b)  The arrangement is a divide-by-4 circuit. Therefore, the output waveform will have a time period of 4T. Also, the total propagation delay is 2∆T. 84. Topic: Sequential Circuits: Latches and Flip–Flops (c)  At high to low clock transition, the input to the top cross-coupled NOR gate is 1, while the input to the bottom cross-coupled NOR gate is 0. Therefore output Q = 1. When D = 1, the input to the top cross-coupled NOR gate is 0 and to the bottom cross-coupled NOR gate is 1. Therefore Q = 0. 85. Topic: Data Converters: ADCs and DACs (d)  VDAC = 2 −1 b0 + 20 b1 + 21 b2 + +22 b3 = 0.5b0 + b1 + 2b2 + 4b3 The counter output will start from 0000 and will increase by 1 LSB at every clock pulse as it moves from 0000 to 1111. The corresponding D/A converter output voltage levels would be 0 V, 0.5 V, 1.0 V, 1.5 V, 2.0 V, 2.5 V, 3.0 V, 3.5 V, 4.0 V, 4.5 V, 5.0 V, 5.5 V, 6.0 V, 6.5 V, 7.0 V and 7.5 V. The moment D/A converter output goes from 6.0 V to 6.5 V, the comparator output goes to LOW state thereby disabling the clock signal. The corresponding counter output is 1101, which gives the reading of ‘13’ in LED display. 86. Topic: Data Converters: ADCs and DACs (b)  The magnitude of the error between D/A converter output and Vin is 6.5 − 6.2 = 0.3 V 87. Topic: 8-bit Microprocessor (8085): Programming

(c)  PC = 6140 H HL = 6140 H

88. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  TTL stands for transistor transistor logic and CMOS stands for complementary metal oxide semiconductor. 89. Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (d)  If an interrupt has an address, then it is vectored interrupt, and if a interrupt can be rejected, then it is maskable interrupt. 90. Topic: Combinatorial Circuits: Boolean Algebra (d)  Substituting for X = 1 and X = 0 , we get the answer. We know that

GATE_ECE_CH05_Digital_Circuits.indd 367

1 + Boolean expression = 1 Therefore, expression inside first bracket reduces to 1 and expression in the second bracket reduces to X + Z . This reduces the given Boolean equation to Z = 1, which gives Z = 0. 91. Topic: Combinatorial Circuits: Decoders and PLAs (b)  The truth table can be drawn as follows: P1

P2

a

b

c

d

e

f

g

0

0

1

1

1

1

1

1

0

0

1

1

0

1

1

0

1

1

1

0

1

1

0

1

1

0

1

1

1

1

0

0

1

1

1

1

From the truth table, we can write the simplified Boolean functions for a, b, c, d, e, f and g as follows: a =1 b = P2 c = P1 d =1 e = P1 + P2 f = P1 + P2 g = P1 + P2 From the above expressions, we get c+e=1=d and hence the answer. 92. Topic: Combinatorial Circuits: Decoders and PLAs (d)  The answer is obvious from the Boolean functions of a to g. Implementation of b and c require NOT gates. Implementation of e, f and g require OR gates. 93. Topic: Combinatorial Circuits: Multiplexers (c)  One 2-to-1 multiplexer is required to implement AND gate. In the case of AND gate implementation, select line is tied to A input, I0 input line is tied to logic 0 and I1 input line is tied to logic 1. Similarly, EX-OR gate also requires one 2-to-1 multiplexer. Select line is connected to A. Input lines I0 and I1, respectively, are connected to B and B. 94. Topic: Sequential Circuits: Latches and Flip–Flops (b)  Both circuits are R-S latches with active LOW inputs. Therefore, in both cases, for (P1, P2) = (0, 1), the output (Q1, Q2) will be 10. For (P1, P2) = 11; both inputs are inactive. Therefore, the previous output will be retained. That is, (Q1, Q2) = (1, 0).

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95. Topic: Sequential Circuits: Counters (a)  Initially the count is 00. J and K inputs of both flip–flops are in logic ‘1’ state. With first clock pulse, count sequence becomes 11. Now, J and K inputs of the first flip–flop are both in logic ‘0’ state. J and K inputs of second flip–flop, respectively, are in logic ‘0’ and logic ‘1’ states. Therefore, with the second clock pulse, the count sequence becomes 10. With the third clock pulse, the first–flop toggles and the count sequence is 00. So, the counting sequence is 11, 10, 00, 11, 10, 00 ... 96. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  NOR gate is equivalent to a bubbled AND gate. NAND gate is equivalent to bubbled OR gate. EX-OR gate is equivalent to a EX-NOR gate with one input bubbled. EX-NOR gate is equivalent to EX-OR gate with one input bubbled. 97. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  For F = 1, even number of inputs to the EX-NOR gate at the output should be in logic ‘1’ state. In the given logic circuit, other than C input, the other two inputs cannot be simultaneously in logic ‘1’ state. Only one of them can be ‘1’ at a time. Therefore, C must be in logic ‘1’ state and hence the answer. 98. Topic: Semiconductor Memories: SRAM (b)  From the given circuit: A15 A14 A13 A12 A11 A10 A9 A8 . 0  0 1  0  1  1 0 1   for y 5

To enable chip



A7

A6

A5

A4

A3

A2

A1

A0

0   1

0   1

0   1

0   1

0   1

0   1

0   1

0   1

Hence, the range of the address is 2D00 – 2DFF

99. Topic: Combinatorial Circuits: Multiplexers (d) F ( A, B, C , D ) = ABC + ABD + ABC + AB(CD ) = ABC ( D + D ) + AB(C + C ) D + ABC ( D + D ) + ABCD which gives F = Σm( 2, 3, 5, 7, 8, 9,12)

GATE_ECE_CH05_Digital_Circuits.indd 368

100. Topic: Sequential Circuits: Latches and Flip–Flops (d)  Initially, QA, QB, QC are ‘0’. This makes DA = ‘1’. With the first clock pulse, QA becomes ‘1’. QB and QC remain in logic ‘0’ state. DA is still ‘1’.With the second clock pulse, QA remains in logic ‘1’ state. QB becomes ‘1’. QC remains ‘0’. DA becomes ‘0’. With the third clock pulse, QA becomes ‘0’. The first four entries of QA therefore are 0110. This matches with only the answer given in option (d). 101. Topic: 8-bit Microprocessor (8085): Programming (c) MVI A, 45 H A = 45 H MOV B, A B = 45 H STC set carry ⇒ carry flag = 1 CMC compliment carry ⇒ carry Flag = 0 RAR ……….. Rotate right with carry 0

1

0

0

0

1

0

1

0

Therefore, carry flag = 1

A=

0010 0010 = 22H 2 2

XRA B → XOR with 45 H 22 H = 00100010 45 H = 01000101 Output = 01100111 Therefore, A = 67 H 102. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (b)  Look at the three NAND gates appearing on the extreme left side. First NAND has P and Q as inputs. Second NAND has Q and R as inputs and third NAND has R and P as inputs. Output of any of these NAND gates produces logic ‘1’ at the output provided that either P = Q = 1 or Q = R = 1 or R = P = 1 or P = Q = R = 1 and hence the answer is option (b). 103. Topic: Combinatorial Circuits: Multiplexers (d) I1 and I2 are tied to logic ‘1’ state. I0 and I3 are tied to logic ‘0’ state. Therefore, the output is logic ‘1’ for P = 0, Q = 1 and P = 1, Q = 0. Therefore, the output is F = P Q + P Q = XOR (P, Q) 104. Topic: Sequential Circuits: Latches and Flip–Flops (a)  Presently, Y = 1. This implies that prior to clock pulse; both the D-inputs were in logic ‘1’ state. This

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Chapter 5  •  Digital Circuits

further implies that Q-output of the first flip–flop was in logic ‘0’ state. This can result from its D-input being previously in logic ‘0’ state. Hence, the data input has changed from 0 to 1. 105. Topic: Sequential Circuits: Counters (a)  Sequence of Johnson counter in one complete cycle is 000, 100, 110, 111, 011, 001 and 000. The corresponding analog outputs will be 0 V, 4 V, 6 V, 7 V, 3 V, 1 V and 0 V if LSB = 1 V. 106. Topic: Sequential Circuits: Latches and Flip–Flops (d)  Present State

Next State

QB

QA

QB

QA

0

0

1

1

1

1

0

1

0

1

1

0

1

0

0

0

0

0

1

1

Now, using the excitation table of D-flip-flop DA = QA QB + Q A Q B DB = Q B



107. Topic: 8-bit Microprocessor (8085): Programming (c)  MVI A, 07 H 0000 0111 ⇒ content of A RLC 0000 1110 ⇒ content of A MOV B, A 0000 1110 ⇒ content of B RLC 0001 1100 ⇒ content of A RLC 0011 1000 ⇒ content of A ADD B B 0000 1110 + A 0011 1000 0100 0110 RRC → 0010 0011 ⇒ Contents of A Therefore, content of A = 23 H 108. Topic: Combinatorial Circuits: Minimization of Functions using Boolean Identities and Karnaugh Map (a) Substituting f ( X , Y , Z ) = Σ ( 2, 3, 4, 5) Figure below shows the K-map for this Boolean ­function.

GATE_ECE_CH05_Digital_Circuits.indd 369

YZ

YZ

X X

1

YZ

YZ

1

1

369

1

Therefore, the minimized Boolean function is given by f ( X , Y , Z ) = XY + XY So, the prime implicants are XY and X Y . 109. Topic: Combinatorial Circuits: Arithmetic Circuits (b)  Output will be 1 if A > B. Now, A can be greater than B for the following conditions:  (   i) If B = 00, there will be three combinations for which O/P will be 1, that is, when A = 01, 10, or 11.    (ii) If B = 01, there will be two conditions, that is, A = 10 and 11. (iii) If B = 10, there will be one condition, that is, A = 11. Therefore, there are a total of six combinations for which output is in logic ‘1’ state. 110. Topic: Sequential Circuits: Latches and Flip–Flops (a)  The circuit shown is that of an active-HIGH clocked R-S flip–flop with S = A and R = B. The race condition does not occur in these flip–flops. 111. Topic: Sequential Circuits: Latches and Flip–Flops (d)  When A = 1, X1 = Q will be selected as feedback to D-input. Therefore, Q-state (0 or 1) will be retained. When A = 0, X0 = Q will be selected as feedback to D-input. Therefore, Q-output will toggle. 112. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  It is clear from the truth table of an EX-OR gate that both switches can be used to either turn ON or turn OFF the switch. Let us assume that one is at the ground floor. If the two switches are in different states, operating the ground floor switch will put them in the same state thereby changing the status of lamp. If the switches are in same state initially, then operating the switch puts them in different states again changing the status of lamp. Similar explanation holds for first floor switch. 113. Topic: 8-bit Microprocessor (8085): Programming (a)  Accumulator changes as follows (05 + 05 + 04 + 03 + 02 + 01)H At the end of Loop accumulator contains = 14 H ADI 03 H → A = (14 + 03) = 17 H

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114. Topic: Semiconductor Memories: SRAM (d)  Since the range of RAM-1 is different in all four options, we will first check for RAM-1 only and then the same procedure can be followed for RAM-2, RAM-3 and RAM-4 if needed. RAM-1 will be selected when S0 = 0 and S1 = 0 , ( S0 = A12 = 0, S1 = A13 = 0 ). Now, RAM-1 will be enabled when the input of MUX is 1, or the output of AND gate is 1. So, A10 = 0 , A11 = 1 , A14 = 0 and A15 = 0 In view of the above description, START ADDRESS A15-A0 for RAM-1 will be 0000100000000000. The LAST ADDRESS for the same can be determined by adding 0000001111111111 to the above number as size of RAM is 1024 bytes = 210 bytes. The LAST ADDRESS is therefore 0000101111111111. Hence, the range of RAM1-1 is 0800H to 0BFFH.



As per transposition theorem, ( A + BC ) = ( A + B)( A + C )



So, ( X + Y )( X + Y ) = X + YY = X + 0 = X



Hence, the given expression reduces to X + XY + X

= X + XY ⋅ X = X (1+ XY ) Using absorption theorem, (1 + A) = 1, we get

X (1 + X Y ) = X ⋅1 = X

119. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  Refer to the figure shown below.

5V

115. Topic: Combinational Circuits: Boolean Algebra (d)  For an n-variable Boolean Function, the maximum number of prime implicants = 2(n - 1)

5V

116. Topic: Number Systems (4)  In packed BCD (binary coded decimal) two decimal digits are encoded within a single type by ­ ­taking advantage of the fact that four bits are enough to represent the range 0 to 9. Therefore, decimal number 1856357 requires 4 bytes in packed BCD form.

5V

M1 P M2 Q

5V

M3 R

117. Topic: Combinatorial Circuits: Karnaugh Map (d)  The given Boolean function is

F ( w , x, y, z ) = ( wy + xy + wxyz + wxy + xz + xy z ) By using K-map, we have that the essential prime ­implicants (EPS) are y, xz , xz

wx

00

01

1

11

10

1

1



1

1

1

11

1

1

1

1

1

10

1

Hence, the transistor is in saturation. Therefore, ID1 = K(VGS - VT)2 = K(4 - VP)2 For NMOS transistor M2 : ID1 = K(5 - VQ - 1)2 = K(4 - VQ)2

01

Assume all NMOS transistors are in saturation. Therefore, VDS ≥ (VGS − VT ) For NMOS transistor M1 : (5 - VP) ≥ (5 - VP - 1) or (5 - VP) > (4 - VP)



—x —z yz 00



As ID1 = ID2, therefore, (4 - VP)2 = (4 - VQ)2

y



Hence, VP = VQ and VP + VQ = 8 Therefore, VP = VQ = 4 V For NMOS transistor M3 : ID3 = K(5 - VR - 1)2

xz



118. Topic: Combinatorial Circuits: Boolean Algebra (a)  The given Boolean expression is ( X + Y )( X + Y ) + XY + X

GATE_ECE_CH05_Digital_Circuits.indd 370

As ID3 = ID2, therefore, (4 - VQ)2 = (4 - VR)2



Hence, VR = VQ = 4 V Therefore, VP = VQ = VR = 4 V

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Chapter 5  •  Digital Circuits

120. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a)  The following figure shows the equivalent circuit, for C = 0 C=0 1 A B

122. Topic: Combinatorial Circuits: Arithmetic Circuits (c)  Truth table for half-subtractor is

A.B A.B Y

A B

A.B + A.B =A . B

A.B

371

Difference (N )

Borrow (M)

X

Y

0

0

0

0

0

1

1

1

1

0

1

0

1

1

0

0

Hence, N = X ⊕ Y and M = XY

From the figure above, Y = 1⋅ A  B

123. Topic: Combinatorial Circuits: Multiplexers (d)  Output of first MUX = WS1 + W S1 = W ⊕ S1

= A B = A ⊕ B = AB + AB



Let Y = W ⊕ S1

121. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a)  If the NMOS transistors are connected in series and PMOS transistors are connected in parallel, then the output expression is ‘or’ of the complement of all inputs. Therefore,



Output of second MUX = Y S 2 + YS2



= Y ⊕ S2



= W ⊕ S1 ⊕ S2

Y = A⋅ B ⋅C = A + B + C 124. Topic: Sequential Circuits: Latches and Flip–Flops (62.5)  The given circuit is a ripple (asynchronous) counter. In ripple counter, output frequency of each flip-flop is half of the input frequency applied to it, if all the states of the counter are used. Otherwise output ­frequency of the counter is: ⎤ ⎡ Input frequency ⎢ Modulus of the counter ⎥ ⎣ ⎦

So, the frequency at

1

1

Q3 =

Input clock frequency 16

=

1 × 106 Hz = 62.5 kHz 16

1 J4 Q4 Clk K4

1

1 J3 Q3 Clk K3

1

J2 Q2 Clk 1 K2

1

1 J1 Q1 Clk K1

J0 Q0 Clk K0 1

Clock

GATE_ECE_CH05_Digital_Circuits.indd 371

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125. Topic: Sequential Circuits: Latches and Flip–Flops (d)  Latches are level triggered devices. Hence, if we use two latches in cascade with inverted clock, then one latch will behave as master and another latch which is having inverted clock will be used as a slave. The combined behaviour is that of a flip-flop. Hence, the given circuit is implementing a master-slave D flip-flop. 126. Topic: Data Converters: Sample and Hold Circuits (b)  Capacitor droop rate =

dv dt

For a capacitor, dv ∝ 1 dt C Therefore, droop rate decreases as capacitor value is increased. For a capacitor, Q = CV = i × t. Therefore, acquisition time t ∝ C Hence, the acquisition time increases as the capacitor value is increased.



Let Q = W ⊕ X . The output of the second MUX



F = QY Z + QY Z



= QY ( Z + Z )



= QY = (W X + WX )Y



= W X Y + WX Y

129. Topic: Combinatorial Circuits: Arithmetic Circuits (a)  Truth table of a half-subtractor is

127. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a)  Let the output of XOR gate be denoted as A, as shown in the figure below.

Y

X

Y

D

B

0

0

0

0

0

1

1

1

1

0

1

0

1

1

0

0

I0 2:1 MUX

D D = X×Y + XY =X Å Y

I1 S

XOR

X S

X Y

Y

A

I0 2:1 MUX

F

I1

B

B = X×Y + X×0 = X×Y + 0 = X×Y

Z So, D = X ⊕ Y = XY + XY and B = XY

XNOR

Therefore, A = X ⊕Y = XY + YX F = A( A  Z ) = A( AZ + AZ ) = 0 + AZ = AZ



Y = ABCD + ABCD + ABC

Therefore, F = ( XY + XY ) Z F = ( XYZ + XYZ )

128. Topic: Combinatorial Circuits: Multiplexers (c)  The output of the first MUX = W XVCC + WXVCC = W X + WX (Since VCC = logic 1) =W ⊕ X

GATE_ECE_CH05_Digital_Circuits.indd 372

130. Topic: Combinatorial Circuits: Multiplexers (c) Output Y of the multiplexer is given by (Remaining combinations of the select lines produce output 0.) Therefore, Y = ACD( B + B) + ABC = ABC + ACD 131. Topic: Combinatorial Circuits: Arithmetic Circuits (195)  The given figure shows a 16-bit ripple carry adder circuit. In this circuit carry signal propagates from first stage FA0 to the last stage FA15. Hence, their propagation delays get added together but the sum result is not propagating. Hence, the next stage sum result depends upon the previous stage carry.

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Chapter 5  •  Digital Circuits



So, last stage carry (C15) will be produced after 16 × 12 ns = 192 ns



Second last stage carry (C14) will be produced after 15 × 12 ns = 180 ns



For last stage sum result (S15) total delay



= 180 ns + 15 ns = 195 ns

132. Topic: Sequential Circuits: Latches and Flip–Flops (c)  This circuit is negative edge trigged, so output of the D-flip-flop will change only when Clk signal is going from HIGH to LOW (1 to 0). This is synchronous circuit, so both the flip-flops will trigger at the same time and will respond on falling edge of the clock. So, the correct output (Y) waveform is associated to W3 waveform.

1

1

0

0

G2 B is provided active low signal through ( RD ). This means to read the status of the I/O device, LDA instruction is appropriate with device address. Again, to enable the decoder, O/P of AND gate must be 1. DS2 signal should be 1 (which is the O/P of multi-input AND gate) to enable the I/O device. So,

A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0

So, worst case delay = 195 ns



373

1 0

1



F 8 F 8 Hence, the device address = F8F8H So, the correct instruction used is LDA F8F8H.

135. Topic: 8-bit Microprocessor (8085): Programming (a)  Let the opcode of STA be XXH and the content of accumulation be YYH. For the STA 1234 H, given that the starting address = 1FFEH So, the sequence of data and addresses is given below: Address (in hex) : Data (in hex)

0

Clk A15 — A8

A7 — A0

1F FEH 1F FFH

X1

20 00 H 12 34 H

XXH 34 H 12 H YYH

X2

Y(W3)

133. Topic: Sequential Circuits: Latches and Flip–Flops (d) Clock

J1(Q2) K1(Q2)

J2(Q1) K2(Q1)

Q1

Q2

Initial

-

-

-

-

0

0

1 CP

1

0

0

1

1

0

2nd CP 1

0

1

0

1

1

3rd CP

0

1

1

0

0

1

4th CP

0

1

0

1

0

0

st



So, the output sequence generated at Q1 is 01100…….

134. Topic: 8-bit Microprocessor (8085): Architecture (d)  This circuit diagram indicates that the I/O device is memory mapped. In order to enable the 3- to -8 decoder, G2 A is provided active low signal through (IO/M ) and

GATE_ECE_CH05_Digital_Circuits.indd 373

136. Topic: Semiconductor Memories: SRAM (b)  For an SRAM construction, four MOSFETS are required (2 - PMOS and 2 - NMOS) with interchanged outputs connected to each CMOS inverter. So option (b) is correct. 137. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a)  Given expression is Y = AB + CD

The first term AB can be obtained by considering G1 as NOR gate, and second term (CD ) is obtained from another lower NOR gate. So, final expression can be implemented by considering G2 as OR gate.

138. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  If any input is logic ‘0' (i.e., 0 V) then the corresponding diode is ON and if any input is logic ‘1' (i.e. 10 V) then the corresponding diode is OFF. Case (i): If any of the inputs is low (i.e. 0 V), the corresponding diode is ON and therefore the output voltage Vo = 0 V.

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Case (ii): If all the inputs are high (i.e. 10 V) then all the diodes are reverse biased (OFF). Therefore, the output voltage Vo = 10 V. Hence, the circuit represents a positive logic 3 - input AND gate. 139. Topic: Sequential Circuits: Counters (6)  To find the modulus of the counter, consider the status of the inputs (QB, QC) as 1. So, QA QB QC QD = 0110 Hence, it is a MOD-6 counter 140. Topic: Sequential Circuits: Latches and Flip–Flops (a)  On analysis, we have   (i) Clock is taken from normal output of the flip-flop. Hence, it is a, up-counter.   (ii) Input of the NAND-gate is taken from Q2 and Q0. So Q2 = 1 and Q0 = 1 (iii) To find the modulus (Q2, Q1, Q0) = (101) So, it is modulo-5 binary up-counter. 141. Topic: Data Converters: ADCs and DACs (0.9375)  Analog output = [Resolution] × [Decimal equivalent of binary] = (0.0625) (15) = 0.9375 V 142. Topic: 8-bit Microprocessor (8085): Architecture (b)  In an 8085 microprocessor, after performing the addition, result is stored in accumulator and if any carry (overflow bit) is generated, it updates the flag register. 143. Topic: 8-bit Microprocessor (8085): Programming (d)  Generally arithmetic or logical instructions update the data of accumulator and flags. In the given options only SBI BEH is an arithmetic instruction. SBI BEH instructs to add the content of accumulator with immediate data BE H and store the result in accumulator. 144. Topic: 8-bit Microprocessor (8085): Programming (c)  MVI A, 00H ← Load accumulator by 00H LOOP: ADD C ← Adds the content of a­ ccumulator with contents of C register and stores the result in a­ ccumulator. DCR B JNZ LOOP HLT This will continue till B register reaches to 00H. So, repetitive addition of a number as many times as the other number will give the product of these two ­numbers. 145. Topic: Combinatorial Circuits: Boolean Algebra (a)  The given Boolean expression in terms of minterms is F ( X , Y , Z ) = XYZ + XY Z + XYZ + XYZ

GATE_ECE_CH05_Digital_Circuits.indd 374

Therefore, in terms of minterms, F(X, Y, Z) is given by F ( X , Y , Z ) = Σ ( 2, 4, 6, 7) In terms of maxterms, F(X, Y, Z) is F ( X , Y , Z ) = Π(0,1, 3, 5) Hence, F(X, Y, Z) in POS form is ( X + Y + Z )( X + Y + Z )( X + Y + Z )( X + Y + Z ) 146. Topic: Combinatorial Circuits: Boolean Algebra (b)  Given that F(X, Y, Z) = ∑(1, 2, 5, 6, 7) Therefore F(X,Y, Z) in terms of maxterms = ∏(0, 3, 4) Product of sums = ( X + Y + Z )( X + Y + Z )( X + Y + Z ) 147. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (40)  As per the information given in the question, the waveforms A, B, C are as follows:

A B

C t= 0 For the given logic gates, the output waveforms X and Y are first obtained and then used to obtain Z. A Y Z B X C B X=B (with 20 ns delay) t=0 B with zero delay t=0 X = B with 20 ns delay t = 20 ns Y= XA A t=0 t = 20

AB without delay t = 20 ns

Y = AB with delay t = 40 ns

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Chapter 5  •  Digital Circuits

R → S0 S → S1 Input to both the decoders should be same so

Z= Y Å C Y t = 40 C

t = 20

P → Din

t = 40

t=0 Y ÅC (without delay) t = 0

20

NOT gate along with OR gate is used to select one decoder at a time so Q → S2.

40

P → Din Q → S2 R → S0 S → S1

Z 0

20

40

60

The output Z is HIGH during a period of 20 ns to 60 ns for a duration of 40 ns. 148. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (b) We have, M(a, b, c) = ab + bc + ac = ∑m(3, 5, 6, 7) M ( a, b, c) = Σm(0, 1, 2, 4) = x M ( a, b, c ) = ab + bc + ac = Σm( 2, 4, 6, 7) = y c = ∑m(1, 3, 5, 7) = z f ⎡⎣ M ( a, b, c), M ( a, b, c ), c ⎤⎦ = f (x, y, z) = xy + yz + zx = [(∑m(0, 1, 2, 4)(∑m(2, 4, 6, 7)] + [(∑m(2, 4, 6, 7) ∑m(1, 3, 5, 7)] + [∑m(1, 3, 5, 7) ∑m(0, 1, 2, 4)] = ∑m(2,4) + ∑m(7) + ∑m(1) = ∑m(1, 2, 4, 7) = a⊕b⊕c Hence, the given function represents a 3-input Xor gate. 149. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  Only NAND and NOR are universal gates. Hence, option (d) is correct. 150. Topic: Combinatorial Circuits: Decoders (d) We need to implement 1:8 demultiplexer. Select lines of DEMUX should be mapped to address lines of decoder. So, LSB of DEMUX should be connected to LSB of address lines of decoder. Y0 Y1 Din

S2 S1

GATE_ECE_CH05_Digital_Circuits.indd 375

Y7 S0

375

151. Topic: Sequential Circuits: Counters (b) Q3 Q2 Q1 Q0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 Once the output of Ex-NOR gate is 0, then counter will be RESET. So, Ex-NOR-gate will produce logic 0 for Q3 = 0, Q2 = 1. So, the counter will show the sequence like 0

1

2

3

Hence, it is MOD-4 counter. 152. Topic: Sequential Circuits: Latches and Flip–Flops (d) D2(Q1 Å Q0) D1(Q2) D0(Q1) Y2(Q2) Y1(Q1) Y0(Q0) − − − 1 1 1 st 0 1 1 1 1 1 clock 0 2nd clock 3rd clock

0 1

0 0

1 0

0 1

0 0

1 0

After three clock pulses output Y2Y1Y0 = 100 153. Topic: Sequential Circuits: Latches and Flip–Flops (d)  The S- R Latch can be made functional by changing 5V to ground. 154. Topic: Semiconductor Memories: ROM (7)  Generally the structure of a memory chip = ­Number of rows × Number of columns = M × N The number of address line required for row decoder is n where M = 2n or n = log2M As M = N, so, M × N = M × M = M 2 = 16 kB = 24 × 103 ≅ 24 × 210

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Therefore, M2 = 214 or M = 128 Therefore, n = log2128 = 7

between the input and output. Therefore, the output is the mid-point of the characteristic curve, that is,

155. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c) We have Y = ABC ⊕ (AB ⊕ BC) Here, ( AB ⊕ BC ) = AB ⋅ BC + AB ⋅ BC = ( A + B ) BC + AB( B + C ) = ABC + ABC

VIL + VIH 2 where Va is the switching threshold of the inverter. Va =

157. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d) • When B = 1, then M1 is ON and M2 is OFF, therefore, output Y = A. • When B = 0, then M1 is OFF and M2 is ON; therefore, output Y = 0. Therefore, the given circuit implements the following:

Now, Y = ABC ⊕ (AB ⊕ BC) = = = = = = = = = = = = = = = = = =

ABC ⊕ ( ABC + ABC ) ABC ⊕ ( ABC + ABC ) ABC ( ABC + ABC ) + ABC ( ABC + ABC ) ABC ( ABC + ABC ) + ABC ( ABC + ABC ) ABC ( ABC ⋅ ABC ) + ( A + B + C ) ⋅ ABC ( ABC ⋅ ABC ) + ( A + B + C ) ⋅ ( ABC + ABC ) ( ABC + ABC ) ABC ( A + B + C )( A + B + C ) ABC ( A + B + C )( A + B + C ) + ( A + B + C )( ABC + ABC ) + ( A + B + C )( ABC + ABC ) ABC + ABC + ABC ABC + ABC + ABC AB(C + C ) + ABC AB(C + C ) + ABC AB + ABC AB + ABC B( A + AC ) B( A + AC ) B( A + C ) B( A + C )

156. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (c)  The given CMOS inverter’s transfer characteristics are as shown in the figure below. Vout

Y = AB  Therefore, the given circuit functions as an AND gate 158. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (a) •  The number of 2-input NAND gates required to implement 2-input XOR gate is 4. Figure below shows the implementation circuit. A AB AB + AB =

AB

⊕

B AB • Similarly, the number of 2-input NOR gates required to implement 2-input XNOR gate is 4. 159. Topic: Combinatorial Circuits: Multiplexers (a)  For a full adder, we have Cout = Σ(3, 5, 6, 7) = ABCin + ABCin + ABCin + ABCin Now, 4 × 1 MUX output is S1S0 I 0 + S1S0 I1 + S1S0 I 2 + S1S0 I 3

Va

where S1 = A and S0 = B. Therefore, I0 = 0; I1 = Cin; I2 = Cin; I3 = 1 Vin VIL

VR

VIH

As we see from the given figure, the inverter is connected in feedback loop formed by connecting 10 kΩ resistor

GATE_ECE_CH05_Digital_Circuits.indd 376

160. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (1.5)  The table below shows the Q1, Q2, Q3, Q4, Q5 and Y outputs for different clock cycle inputs.

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Chapter 5  •  Digital Circuits

Clk

Q1

Q2

Q3

Q4

Q5

Y = Q3 + Q5

0

0

1

0

1

0

0

1

0

0

1

0

1

1

2

1

0

0

1

0

0

3

0

1

0

0

1

1

4

1

0

1

0

0

1

5

0

1

0

1

0

0

The Clk and Y waveforms are as shown below.

0

T

2T

Average power ( Pavg ) =

3T

1 Ttotal

∫

Ttotal

0

4T

5T

V 2 (t ) dt R

2T 4T 5T 1 ⎡ T 25 25 ⎤ + ∫ 0 dt + ∫ + ∫ 0 dt ⎥ ∫ 3 3 ⎢ T 2T 10 × 10 4T 5T ⎣ 0 10 × 10 ⎦ 25 75 1 ⎡ 25 ⎤ + × 2⎥ = = 1.5 mw = ⎢ 3 5 ⎣10 × 103 10 × 103 ⎦ 5 × 10

Pavg =

377

Then, the output of decoder, OP7 is HIGH. This output is connected to input IP5 of encoder [(5)10 = (101)2]. Therefore, the output of the encoder Y2 Y1 Y0 = 1 0 1. Therefore, the input to the circuit = 1 1 1 and its output = 1 0 1. Consider a Gray to Binary converter. If the input = 111, then, the output is 101. Therefore, the given circuit acts like Gray to Binary ­converter. 164. Topic: Combinatorial Circuits: Decoders and PLAs (b)  When C1 = 0 and C0 = 0, then O0 is high → Y = P. When C1 = 0 and C0 = 1, then O1 is high → Y = Q. When C1 = 1 and C0 = 0, then O2 is high → Y = R. When C1 = 0 and C0 = 1, then O3 is high → Y = S. Therefore, the given circuit acts like 4:1 multiplexer with inputs P, Q, R and S and selection inputs C1 and C0. 165. Topic: Combinatorial Circuits: Multiplexers (7)  The maximum propagation delay can be calculated by using critical path (which is shown in the figure here as dotted line). P

161. Topic: 8-bit Microprocessor (8085): Programming (d)  It is given that A = (A7)H and CF = 0, where A is the accumulator and CF is the carry flag: A = A7H = 1 0 1 0 0 111 0

Q

0

R

0 MUX

MUX

1

S

1 So

After executing RLC (Rotate left accumulator content with carry), we get

Y

So

A = 0 1 0 0 1111 1 Therefore, content in accumulator is (4F)H and CF = 1 162. Topic: Combinatorial Circuits: Minimization of Functions Using Boolean Identities and Karnaugh Map (b)  The minimum SOP expression for the function is X ( SQ ) + X ( SQ ) = QS X + QS X PQ 00 RS 00

01

11

10

PQ 00 RS 00

01 1

1

01

11 1

1

11

10

10 X=0

01 1

11 1

10

tP = tNOR + tMUX + tNOR + tMUX = (2 + 1.5 + 2 + 1.5) ns = 7 ns 166. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (c) Present State

X

Y

Z

Next State

A

0

0

0

B

0

0

1

A

0

1

0

A

0

1

1

A

1

1

0

0

C

X=1

1

0

1

C

1

1

0

A

1

1

1

A

1

163. Topic: Combinatorial Circuits: Decoders and PLAs (c)  Assume that X2 X1 X0 = (1 1 1)2 = (7)10.

GATE_ECE_CH05_Digital_Circuits.indd 377

The maximum propagation delay of the circuit is

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B

C

0

0

0

A

0

0

1

C

0

1

0

B

0

1

1

B

1

0

0

A

1

0

1

C

1

1

0

B

1 0 0 0 0 1 1 1 1

1 0 0 1 1 0 0 1 1

1 0 1 0 1 0 1 0 1

B C C C B C A C A, B

168. Topic: Data Converters: ADCs and DACs (a)  The total capacitance is (2n – 1) × C = (28 − 1) × 8 pF = 2.04 nF Now, t = RC = 75 × 2.04 × 10−9 s = 153 ns The settling time is 5τ = 765 ns 1 Sampling rate= settling time ≈ 1 MHz or 1 Megasamples/ss 169. Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (c)  PUSH is a pre-decremented operation. Since it is 16-bit instruction, it takes 6 clock cycles for fetch and decode operation. PUSH Instruction:

The last row (shown in bold border lines) in the above table shows the ambiguous condition. • If X = 1, Z = 1, next state is A • If Y = 1, Z = 1, next state is B 167. Topic: Sequential Circuits: Counters (d)

T2

T3

6 clock cycles

3 clock cycles

3 clock cycles

Write operation

Write operation

Fetch and decode

Total number of clock cycles = 6 + 3 + 3 = 12 clock cycles. POP is a post-incremented instruction and it takes only 4 clock cycles fetch and decode operation.

Clk

Q2

Q1

Q0

Output of NAND

−

0

0

0

1

0

0

1

1

0

0

1

1

0

1

0

2

0

1

0

1

0

1

1

3

0

1

1

1

1

0

0

4

1

0

0

1

1

0

1

5

1

0

1

1

1

1

0

6

1

1

0

0

0

0

0

POP Instruction: Q2′

Q1′

Q0′

If the delay of the NAND gate is 0, then the given circuit acts like mod-6 counter. However, here the delay of NAND gate is 2 ns; thus therefore the counter will count two more clocks before the RESET pin goes LOW to reset the counter (the clock period = 1 ns and the gate delay = 2 ns). Therefore, the given counter acts like mod-8 counter.

GATE_ECE_CH05_Digital_Circuits.indd 378

T1

T1

T2

T3

4 clock cycles

3 clock cycles

3 clock cycles

Write operation

Write operation

Fetch and decode

Total number of clock cycles = 4 + 3 + 3 = 10 clock cycles. 170. Topic: Semiconductor Memories: ROM (a)  8 KB Þ 213 bytes. Therefore, 2n = 213 or n = 13. Hence, 13 address lines are required.. The given range is 1000H to 2FFFH. Therefore, A14 , A15 = 0 ⎫ ⎬ For CS = 0 A13 = 0 /1, A12 = 1/ 0⎭ Hence, CS = A15 + A14 + ( A13 ⋅ A12 + A13 ⋅ A12 )

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171. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (b) Gate 1 P X

1

0

OFF

ON

1 (NMOS ON)

1

0

OFF

ON

0 (NMOS ON)

173. Topic: Sequential Circuits: Latches and Flip–Flops (30) CLK1 CLK1

Y

CLK2

Q

Q D D-Latch CK

Output

CLK2 TCLK/5

Gate 2 When the initial input is

Truth Table (D-FF)

P=Q=0 The corresponding outputs are X=Y=1 When the first input condition is changed to P=Q=1 There are two possibilities: • Possibility 1: When Gate 1 is faster than Gate 2, the possible outputs are

CK

D

Q

0

0

0

0

1

0

1

0

0

1

1

1

TCLK

X = 0, Y = 1

CLK1

 ossibility 2: When Gate 2 is faster than Gate 1, the • P possible outputs are

CLK2 Output

X = 1, Y = 0

TCLK/5

172. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (b)  P is connected to gate terminal of PMOS as well as NMOS. If P is at logic 0, then PMOS turns ON and NMOS turns OFF. If P is at logic 1, then PMOS turns OFF and NMOS turns ON. If PMOS is turned ON, then outputY is connected to Q. If NMOS is turned ON, then outputY is connected to Q . Output =

Q( PMOS ON )

P

Q

PMOS

NMOS

0

0

ON

OFF

0 (PMOS ON)

0

1

ON

OFF

1 (PMOS ON)

GATE_ECE_CH05_Digital_Circuits.indd 379

Q( NMOS ON )

For Q, TCLK TCLK 5T − 2T 3T − = = 2 5 10 10 TCLK TCLK 7T + = TOFF = 2 5 10 T TON ×100% Duty cycle, Q(%) = ON = T TON + TOFF TON =

=

3T 10 3T 7T + 10 10

× 100% = 30%

174. Topic: Combinatorial Circuits: Boolean Algebra (b)  Given A is MSB and C is LSB. So function is F0(A, B, C), i.e. F0 ( A, B, C ) = A B C + A B C + A B C + A B C

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Therefore, F0 ( A, B, C ) = A B C + A B C + A B C + A B C + A B C

(Since X + X = X ⇒ A B C + A B C = A B C ) ⇒ F0 ( A, B, C ) = AC ( B + B) + A B (C + C ) + AB C

176. Topic: Combinatorial Circuits: Arithmetic Circuits (120) Propagation delay of the XOR gate = 20 ns Propagation delay of the AND gate = 15 ns Propagation delay of the OR gate = 10 ns Internal diagram of full adder is Xn Yn

⇒ F0 ( A, B, C ) = AC + A B + AB C

B Sn

175. Topic: Combinatorial Circuits: Multiplexers (b)  The output equation for a multiplexer is

Zn + 1 Zn

0

I0

Internal diagram of 4-bit ripple carry adder is Y X0 Y0

I1

40

1

S0

C0

S

35

Y = SI 0 + SI1

Y

A0 20

Z0 X1

0

Y1

15

10

Z1

10

Z2

A1 45 20

S1

P

0

MUX

0

1

MUX

15

F

15

B1 1 X2 Y2

A2 20 20

S2

Z The output of first MUX is P = XY + X ⋅ 0 ⇒ P = XY The output of second MUX is

F = Z ⋅P + Z⋅P



⇒ F = Z ⋅ XY + Z ⋅ XY



⇒ F = ZXY + Z ( X + Y )

15

15 10

X3 Y3

A3 20 25

GATE_ECE_CH05_Digital_Circuits.indd 380

⇒ F = XYZ + XZ + YZ

S3

C3 15



Z3

B2

15 10

B3

Z4

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Chapter 5  •  Digital Circuits

Correct Output

Time at Which Correct O/P Obtained

Correct Output

Time at Which Correct O/P Obtained

A0

20

C3

Z3 + 15 = 110

B0

15

Z4

C3 + 10 = 120

S0

40

C0

A0 + 15 = 35

Z1

C0 + 10 = 45

A1

20

B1

15

S1

Z1 + 20 = 65

C1

Z1 + 15 = 60

Z2

C1 + 10 = 70

A2

20

B2

15

S2

Z2 + 20 = 90

C2

Z2 + 15 = 85

Z3

C2 + 10 = 95

A3

20

B3

15

S3

Z3 + 20 = 115

381

The final correct output of 4-bit ripple carry adder is obtained at 120 ns. 177. Topic: Combinatorial Circuits: Decoders and PLAs (c) Function F from the circuit F = PQR + PQR + PQR 178. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (10.0) For right shift SIPO shift register, Bn+1 = An Cn+1 = Bn Dn+1 = Cn As per the given circuit,

A

Din

B

C

D

Clock An +1 = An ⊕ Dn The state table for given right shift SIPO shift register is

Present state Clock pulse

An

Bn

Cn

Dn

An+1 = An ⊕ Bn

Bn+1 = An

Cn+1 = Bn

Dn+1 = Cn

Clock pulse

0

1

1

0

1

0

1

1

0

1

1

0

1

1

0

0

0

1

1

2

2

0

0

1

1

1

0

0

1

3

3

1

0

0

1

0

1

0

0

4

4

0

1

0

0

0

0

1

0

5

5

0

0

1

0

0

0

0

1

6

6

0

0

0

1

1

0

0

0

7

7

1

0

0

0

1

1

0

0

8

8

1

1

0

0

1

1

1

0

9

9

1

1

1

0

1

1

1

1

10

Therefore, the number of clock cycles required to reach the state 1111 is 10.

GATE_ECE_CH05_Digital_Circuits.indd 381

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179. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (a, d) Given possible states of the FSM are

S2

1

S3

1

S3

1

S1

0

S1

0

S2

0

S2

1

S3

1

QDA = QA ⊕ QB

S3

0

S2

0

DB = QA ⋅ X in

S2

0

S0

0

S0

1

S1

0

QAQB = 00, 01, 10, 11 At constant logic level throughout the operation, i.e. XIN = 0 or 1 constantly

(i) Take XIN = 0 QA

QB

DA  =  QA   ⊕  QB

DB  =  QA  ˙XIN

QA+1

QB+1

S1

1

S1

0

0

0

0

1

0

1

S1

0

S2

0

0

1

1

1

1

1

S2

1

S3

1

1

1

0

1

0

1

1

0

1

1

1

1

From the above truth table, we can see that there are only 2 distinct states out of 4, if XIN = 0. (ii) Take XIN = 1 QA

QB

DA  =  QA   ⊕  QB

0

0

0

1

0

1

0

1

1

1

1

1

1

1

0

0

0

0

1

0

1

0

1

0

DB  =  QA  ˙XIN

QA+1

QB+1

Therefore, the number of times “Out” will be 1 is 4. 181. Topic: 8-bit Microprocessor (8085): Programming (b) Instruction

Operation

MVIA, 33H

Register A is loaded with 33H

MVIB, 78H

Register B is loaded with 78H (A + B) → A

ADD B

Initial State

In

Next Value

Out

S0

1

S1

0

S1

0

S2

0

S2

1

S3

1

S3

0

S2

0

GATE_ECE_CH05_Digital_Circuits.indd 382

78H

0111 1000

+A

+33H

+0011 0011

1010  1011  → A

From the above truth table, we can see that there are 4 distinct states out of 4, if XIN = 1. Both options (a) and (d) are correct. 180. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (4)  Input sequence is 10101101001101.

B

All bits of Register A complemented individually CMA

1

0

1

0

1

0

1

1

0

1

0

1

0

1

0

0

A =

(Register A AND 33H) → A ANI 33H

A= AND

33H

0101

0100

0011

0011

0001  0000  → A

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Chapter 5  •  Digital Circuits

182. Topic: 8-bit Microprocessor (8085): Memory and I/O Interfacing (c) Clock frequency of microprocessor 8085 is 5 MHz. Time required to execute an instruction is 1.4 μs. So, time period of each clock cycle or duration of one 1 T-state = = 0.2 µs 5 MHz

Green

So, number of possible T-state or number of clock cycle is Tootal time required for instruction 1.4 µs = =7 Duration of one T state 0.2 µs

Yellow

Red

183. Topic: Semiconductor Memories: DRAM (b)  Inside a dynamic RAM chip, each memory cell holds one bit of information and is made up of two parts, transistor and capacitor. In DRAM, capacitor holds bit of information, 0 or 1.

Green is on for 70 seconds Yellow is on for 5 seconds Red is on for 75 seconds TCLK = 5 seconds

184. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (d)  f ( x, y ) = x y + xy = x  y

70 + 5 + 75 = 30 5 No. of flip-flops required, n = log2(30) = 4.91 Therefore, no. of flip-flops = 5 Total no. of unique states =

⇒ f ( x, y ) = x  y = x ⊕ y = XOR 185. Topic: Combinatorial Circuits: Boolean Algebra (c)  F = ABC + AB.C + A. BC

SOP = ∑ m(0, 2, 4)



POS = Π (1, 3, 5, 6, 7) = ( A + B + C )( A + B + C )

187. Topic: Combinatorial Circuits: Multiplexers (c) Y

0

Y

I0 1 I1

( A + B + C )( A + B + C )( A + B + C )



BC BC A 1 A A

383

VCC

I0 4 ×1 MUX

Y I1

4×1 MUX

F (U, V,W,X)

0 I 2

1

I2

0

I3 S S 1 2

0 BC

BC

BC

I3

1

1 U V

W X

SOP form BC B+C B+CB+C B+C A 0 A 0 A

0

0

Y = UV ⋅ 0 + UV ⋅1 + UV ⋅1 + UV ⋅ 0 = UV + UV

= WY [ X + X ] = WY = YW

POS form

GATE_ECE_CH05_Digital_Circuits.indd 383

Output of the second MUX is F = W XY + W XY + W X ⋅ 0 + WX ⋅ 0

0

186. Topic: Sequential Circuits: Shift–Registers and Finite State Machines (5)

Output of the first MUX is

= [UV + UV ]W

Both options (c) and (d) represent the same function, but the minimized expression is option (c).

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188. Topic: Semiconductor Memories: ROM ⎧V (a)  B0 = ⎨ DD ⎩ 0

when W0 = VDD otherwise

⎧V B1 = ⎨ DD ⎩ 0

when W1 = VDD otherwise

Therefore, B0 = W0 ⇒ B0/W0 = 1 and B1 = W1 ⇒ B1/W1 = 1 It can be represented as B0 B1 W0 ⎡1 0 ⎤ W1 ⎢⎣0 1 ⎥⎦

Average value of voltage at node X is ⎛ ΔT ⎞ VX ( avg ) = 0.3 × 3.3 ⎜1 − + 0.7 × 0 ⎝ TCK ⎟⎠ = 0.3 × 3.3 (1 - 0.15) + 0 VX(avg) = 0.8415 V 190. Topic: Combinatorial Circuits: Logic Gates and their Static CMOS Implementations (8) X0 P X1 Q X2

189. Topic: Sequential Circuits: Latches and Flip–Flops (0.8415) For Logic 1; XOR|o/p = 3.3 V Logic 0; XOR|o/p = 0 V ΔT = 0.15 TCK Probability of Din transition in each clock period = 0.3 Probability of not transition = 1 − 0.3 = 0.7

Y X3 P = [( X 1 ⊕ X 2 ) X 3 X 0 ] ⋅ X 0 = 0 Q = [( X 1 ⊕ X 2 ) X 3 ] Y = P + X3 = X3 Since X3 is high for 8 combinations, Y will also be high for 8 combinations.

3.3 V 0V DT

GATE_ECE_CH05_Digital_Circuits.indd 384

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Control Systems

CHAPTER6

Syllabus Basic control system components; Feedback principle; Transfer function; Block diagram representation; Signal flow graph; Transient and steady-state analysis of LTI systems; Frequency response; Routh-Hurwitz and Nyquist stability criteria; Bode and root-locus plots; Lag, lead and lag-lead compensation; State variable model and solution of state equation of LTI systems.

Chapter Analysis Topic

GATE 2009

GATE 2010

Feedback principle

GATE 2011

GATE 2012

GATE 2013

GATE 2014

GATE 2015

GATE 2016

1

Transfer function

GATE 2017

GATE 2018

1

1

2

Block diagram representation

1

1

Signal flow graph

1

Transient and steady-state analysis of LTI systems

1

Routh-Hurwitz and Nyquist stability criteria

2

Bode and root-locus plots

1

1

Lag, lead and lag-lead compensation

1

1

State variable model and solution of state equation of LTI systems

1

2

1 1

2

1 1

1

1 2

1

1

2

1

1

5

4

3

1

4

3

6

3

1

5

5

2

1

1

2 5

2

1

1

2

1

Important Formulas 1. In the case of a linear control system,

4. Mason’s formula:

y(t) = y1(t) + y2(t) + y3(t) + ... + yn(t) where y1(t) is the system response due to x1(t) alone, y2(t) is the system response due to x2(t) alone, y3(t) is the system response due to x3(t) alone and yn(t) is the system response due to xn(t) alone. 2. In the case of a linear control system, y(ax) = a⋅y(x) 3. The transfer function is C ( s) G ( s) = R( s) 1 + G ( s) ⋅ H ( s)

Ch wise GATE_ECE_CH06_Control Systems.indd 385



T=

∑P Δ i

i

i

Δ

where Pi = the ith forward path gain; (i) is the number of forward paths; Δ = 1 - (Sum of loop gains) + (Sum of all gain products of two non-touching loops) − (Sum of gain products of three non-touching loops) + (Sum of gain products of four non-touching loops) − … Δ i = ( Δ ) (evaluated with all those loops touching Pi eliminated).

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5. The transfer function of canonical form of feedback control system: (for negative feedback)



G  1+ GH



G  1- GH

(for positive feedback)

15. The steady-state error for a unit ramp input is 1 Kv 16. The acceleration error constant Ka is given by K a = lim s 2 G ( s) s→0

17. The steady-state error for a unit parabolic input is

6. Mass element: F (t ) = M ⋅

1 Ka

du(t ) d 2 x (t ) = M⋅ dt dt 2

7. Spring element: F (t ) = K ⋅ ( x1 - x2 ) = K ⋅ x(t )

18. The sensitivity of magnitude of T(K) with respect to K is written as

8. Damping element: ⎛ dx (t ) dx (t ) ⎞ F (t ) = f ⋅ [u1 (t ) - u2 (t )] = f ⋅ ⎜ 1 - 2 ⎟ ⎝ dt dt ⎠

SK T =

where f is the damping constant. 9. Inertia element: d θ (t ) d w (t ) = J⋅ dt dt 2 2

where J is the inertia constant.

Kr =

s→0

Ch wise GATE_ECE_CH06_Control Systems.indd 386

C⎞ ⎛ lim ⎜ Td - ⎟ s→0 ⎝ R⎠

1 1⎛ C⎞ lim ⎜ Td - ⎟ s→0 s ⎝ R⎠

1 Kr

13. The steady-state error for a unit step input is

14. The velocity error constant Kv is given by

1

The steady-state error for a ramp input is

s→0

1 1+ K p

K T (K )

22. For a generalized feedback control system, the ramp error constant is

12. The position error constant Kp is given by K p = lim G ( s)

.

1 Ks

11. Torsional damping element:

where f is the torsional damping constant.

dk

The steady-state error for step input is

where K is the torsional spring constant.

⎛ dθ (t ) dθ (t ) ⎞ T (t ) = f ⋅ (w1 - w 2 ) = f ⋅ ⎜ 1 - 2 ⎟ ⎝ dt dt ⎠

dϕ T

21. For a generalized feedback control system, the step error constant is

T (t ) = K ⋅ (θ1 - θ 2 ) = K ⋅ θ (t ) = K ⋅ ∫ (w1 - w 2 )dt = K ⋅ ∫ w (t )dt

K T (K )

20. SKT ( K ) = SKT ( K ) + jϕ T Skϕ T

Ks =

10. Torsional spring element:

K v = lim sG ( s)

dk

.

19. The sensitivity of phase ϕ T with respect to K is written as ϕ

I (t ) = J ⋅

d T (K )

SKT ( K ) ≡

where K is the spring constant.

23. For a generalized feedback control system, the parabolic error constant is K pa =

1 1⎛ C⎞ lim 2 ⎜ Td - ⎟ s→0 s ⎝ R⎠

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Chapter 6  •  Control Systems

The steady-state error for a unit parabolic input is 1 K pa

A + kA2 T= 1 A3 + kA4 The sensitivity of the system with respect to parameter k is given by k ( A2 A3 - A1 A4 ) dT k Sk = ⋅ = dk T ( A1 + kA2 ) / ( A3 + kA4 ) T

25. A linear constant coefficient second-order control system is represented by d2 y dy + 2 ζw n + w n2 y = w n2 x dt dt 2 where z is the damping ratio and wn is the undamped natural frequency. 26. The damped frequency is

wd = wn 1 - ζ

2

27. The damping factor is a = zwn 28. The time constant is

τ=

33. Phase margin is

ϕ

24. For a generalized feedback control system with the transfer function of the form

1 ζw n

PM

30. In the case of a linear second-order control system, the time for first peak is given by tp =

π wn 1 - ζ 2

31. In the case of a linear second-order control system, overshoot is given by ⎡ ⎛ ζπ ⎞ ⎤ exp ⎢ - ⎜ ⎟⎥ ⎢ ⎝ 1-ζ 2 ⎠ ⎥ ⎣ ⎦ 32. Gain margin is

34. Nyquist stability plot: Number of infinite semicircles in ‘type l’ system is l. 35. Nyquist stability criterion: A closed-loop control system is stable only if N = - P0 ≤ 0. P0 = number of poles of GH in the RHP. N= total number of clockwise encirclements of the point ( -1, 0) in the GH(s) plane. 36. Magnitude criterion is K =

where w π is phase cross-over frequency.

D( s1 ) N ( s1 )

37. Angle criterion is ⎡ N ( s1 ) ⎤ Arg ⎢ ⎥ = ( 2l + 1)π rad, (for K > 0) ⎣ D( s1 ) ⎦ = 2lπ rad, (for K < 0) 38. Number of loci is equal to number of poles of open-loop transfer function. 39. Center of asymptotes is n

σc =

m

∑ p i - ∑ zi i =1

i =1

n-m

where n and m, respectively, are the number of poles and zeros of the open-loop transfer function. - pi and -zi , respectively, are the poles and zeros of the open-loop transfer function. Angles between the asymptotes and the real axis are given by ⎡ ( 2l + 1)π ⎢ n - m for K > 0 β=⎢ ⎢ ( 2l )π for K < 0 ⎢⎣ n - m where l = 0, 1, 2, …, (n - m - 1). 40. Breakaway points can be computed from n

1 GH ( jw π )

Ch wise GATE_ECE_CH06_Control Systems.indd 387

= [180° + Arg GH ( jw 1)]

w 1 is gain crossover frequency.

29. A linear second-order control system has its poles at -α ± jw d

387

∑σ i =1

m 1 1 =∑ + p σ i =1 b i b + zi

- pi and -zi are poles and zeros of the open-loop transfer function.

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GATE ECE Chapter-wise Solved papers

41. Departure and arrival angles can be computed from the following: Departure angle, qD = 180° + Arg(GH ′)

where x, r and c, respectively, represent state, input and output variables. 51. Homogeneous state equation:

Arg(GH′) is the phase angle of the open-loop t­ransfer function GH computed at the complex pole in question.

dx(t ) = Ax(t ) dt

qA = 180° − Arg(GH ′′) Arg(GH ′′) is the phase angle of the open-loop transfer function GH at the complex zero, but ignoring the contribution of that particular zero. 42. Gain and phase margin can be computed from the following: Gain margin =

Value of K at imgainary cross-over Design value of K

Phase margin = 180° + Arg[ GH ( jw1 ) ] where w1 is the gain cross-over frequency and is given by GH ( jw1 ) = 1

f(t) = L-1[(sI - A)-1] … t ≥ 0 and ⎛ 1⎞ ⎛ 1⎞ ϕ (t ) = e At = 1 + At + ⎜ ⎟ ! A2 t 2 + ⎜ ⎟ ! A3t 3 +  ⎝ 2⎠ ⎝ 3⎠ 53. Properties of state transition matrix: (a)  f(0) = I, where I is identity matrix. (b)  f-1(t) = f(-t). (c)  f(t2 - t1)f(t1 - t0) = f(t2 - t0) for any t0, t1 and t2.

43. Damping ratio (ζ) is related to angle q as follows:

θ = cos -1 ζ

(d) [f(t)]k = f(kt), where k is an integer. 54. Linear time-invariant system:

where q is the angle made by the line drawn from origin to the point on root locus corresponding to gain factor K of interest with the negative real axis. ⎛ Ki ⎞ 44. Integral controller: I out ( s) = ⎜ ⎟ E ( s) ⎝ s⎠ 45. Derivative controller: Dout ( s) = K d sE ( s) 46. PID controller: U ( s) = K p E ( s) + K i

52. State transition matrix:

E ( s) + K d s E ( s) s

(a) x(t) = L-1[(sI - A)-1] x(0) + L-1 [(sI - A)-1 BR(s)] t

(b)  x(t ) = ϕ (t ) x(0) + ∫ ϕ (t - τ ) Br (τ ) dτ t

(c)  x(t ) = ϕ (t - t0 ) x(t0 ) + ∫ ϕ (t - τ ) Br (τ )dτ t0

(d)  c(t ) = Dϕ (t - t0 ) x(t0 ) t

+ ∫ ϕ (t - τ ) Br (τ ) dτ + Er (t )

⎛ s + a⎞ 47. Lead compensator: PLead = ⎜ ,a 0, αk < 3 (b) α > 0, αk > 3 (c) α > 0, αk > 0 (d) α > 0, αk < 0 (GATE 2000: 2 Marks)

(a) over damped (c) underdamped

5. The root-locus diagram for a closed loop feedback system is shown in the following figure. The system is overdamped

2. The transfer function of a system is given by 1 H ( s) = 2 . The impulse response of the system s ( s - 2) is: (* denotes convolution, and u(t) is unit step function) (a) (t2 * e−2t) u(t) (b) (t * e2t) u(t) −2 (c) (te t) u(t) (d) (te−2t) u(t) (GATE 2001: 1 Mark)

jw

K=5

3. The equivalent of the block diagram in the following figure is given in E

F

(a) E

G1

K=0 K=0

−3

−2

−1 K=1

(b) E

G1G2

F

HG2

(a) (b) (c) (d)

only if 0 ≤ K ≤ 1 only if 1 < K < 5 only if K > 5 if 0 ≤ K < 1 or K > 5

C 6. The Nyquist plot for the open-loop transfer function G(s) of a unity negative feedback system is shown in the following figure. If G(s) has no pole in the right half of s-plane, the number of roots of the system characteristic equation in the right half of s-plane is Im(s)

C

G(s)-Plane

(c) E

G1

C Re(s)

−1 F

HG2

(d) E

G1G2

F

H/G2





s

(GATE 2001: 1 Mark) G1

H/G2



∞←K

K=∞

C

H

F



G2

(b) critically damped (d) undamped (GATE 2001: 1 Mark)

C (a) 0 (c) 2 (GATE 2001: 1 Mark)

Ch wise GATE_ECE_CH06_Control Systems.indd 389

(b) 1 (d) 3 (GATE 2001: 1 Mark)

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390

GATE ECE Chapter-wise Solved papers

7. An electrical system and its signal-flow graph representations are shown in the figures 1 and 2 respectively. The values of G2 and H, respectively are Z1(s)

Z2(s)

I1(s)

Vo(s)

10. Consider

a system with the transfer function s+6 G ( s) = 2 . Its damping ratio will be 0.5, when ks + s + 6 the value of k is (a)

2 6

(c)

1 6

I2(s) Z3(s)

Vi

Z4(s)

I2(s) G3

G2

(a) s = - j 3 (b) s = −1.5

Vo(s)

V1(s)

(d) 6 (GATE 2002: 1 Mark)

11. Which of the following points is NOT on the root locus of a system with the open loop transfer function k G ( s) H ( s) = ? s( s + 1)( s + 3)

Figure 1 G1 I1(s)

(b) 3

(c) s = −3

(d) s = -∞ (GATE 2002: 1 Mark)

H Figure 2 (a)

Z3 ( s) - Z 3 ( s) , Z 2 ( s) + Z3 ( s) + Z 4 ( s) Z1 ( s) + Z3 ( s)

(b)

- Z3 ( s) - Z 3 ( s) , Z 2 ( s) - Z3 ( s) + Z 4 ( s) Z1 ( s) + Z3 ( s)

(c)

Z3 ( s) Z 3 ( s) , Z 2 ( s) + Z3 ( s) + Z 4 ( s) Z1 ( s) + Z3 ( s)

(d)

12. The phase margin of a system with the open-loop trans(1 - s) fer function G ( s) H ( s) = is (1 + s)( 2 + s) (a) 0° (c) 90°

(GATE 2002: 1 Mark) 13. The transfer function Y ( s) /U ( s) of a system described by the state y(t) = 0.5x(t) is

- Z3 ( s) Z 3 ( s) , Z 2 ( s) - Z3 ( s) + Z 4 ( s) Z1 ( s) + Z3 ( s) (GATE 2001: 2 Marks)

8. The open-loop DC gain of a unity negative feedback syss+4 tem with closed-loop transfer function 2 is s + 7 s + 13 (a)

4 4 (b) 13 9

(c) 4

(d) 13 (GATE 2001: 2 Marks)

9. The feedback control system shown in the following figure is stable R(s) + −

K≥0

s−2 (s + 2)2

Ch wise GATE_ECE_CH06_Control Systems.indd 390

(b) only if K ≥ 1 (d) only if 0 ≤ K ≤ 1 (GATE 2001: 2 Marks)

equations

•

x (t ) = -2 x (t ) + 2u (t ) and

(a)

0.5 1 (b) ( s - 2) ( s - 2)

(c)

0.5 1 (d) ( s + 2) ( s + 2) (GATE 2002: 1 Mark)

14. If the impulse response of a discrete-time system is h [ n] = -5n u[ - n - 1] , then the system function H(z) is equal to (a)

-z and the system is stable z -5

(b)

z and the system is stable z -5

(c)

-z and the system is unstable z -5

C(s)

s−2 (a) for all K ≥ 0 (c) only if 0 ≤ K < 1

(b) 63.4° (d) ∞

(d)

z and the system is unstable z -5 (GATE 2002: 2 Marks)

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Chapter 6  •  Control Systems

15. The system shown in the following figure remains stable when 1

s−1

K

(a)

sin 2π t t

(b)

sin 2π t sin π t cos 3π t + t t

(c)

sin 2π t sin 0.5π t cos 1.5π t + t t

(d)

sin 2π t sin π t cos 0.75π t + t t

1

R(s)

Y(s)

391

3 1 (a) K < −1 (c) 1 < K < 3

(b) −1 < K < 1 (d) K > 3 (GATE 2002: 2 Marks)

16. The

transfer function of a system is 100 G ( s) = . For a unit step input to the sys( s + 1)( s + 100) tem the approximate settling time for 2% criterion is (a) 100 sec (b) 4 sec (c) 1 sec (d) 0.01 sec (GATE 2002: 2 Marks)

17. The characteristic polynomial of a system is q(s) = 2s5 + s4 + 4s3 + 2s2 + 2s + 1. The system is (a) stable (b) marginally stable (c) unstable (d) oscillatory (GATE 2002: 2 Marks)

(GATE 2002: 2 Marks) 20. In the following figure, the Nyquist plot of the openloop transfer function G(s).H(s) of a system is shown. If  G(s).H(s) has one right-hand pole, the closed-loop system is Im(GH)

Re(GH) (−1,0)

18. The system with the open loop transfer function 1 G ( s) H ( s) = has a gain margin of 2 s( s + s + 1) (a) −6 dB (c) 3.5 dB

(b) 0 dB (d) 6 dB (GATE 2002: 2 Marks)

2 sin 2π t 19. In the following figure m(t ) = , s(t) = cos t sin 199π t 200π t and n(t ) = . The output y(t) will be t

m(t)

×

+

×

Lowpass filter

y(t)

Cutoff frequency = 1 Hz passband gain = 1 s(t)

n(t)

Ch wise GATE_ECE_CH06_Control Systems.indd 391

(a) (a) Always stable

Im(GH) (b) Unstable with one closed-loop right hand pole (c) Unstable with two closed-loop right hand poles (d) Unstable with three closed-loop right hand poles (GATE 2003: 1 Mark) Re(GH) 21. A PD controller used to compensate a system. Com−3,0is−1,0 pared to the uncompensated system, the compensated system has (a) a higher type number (b) reduced damping (b) (c) higher noise amplification (d) larger transient overshoot (GATE 2003: 1 Mark)

s(t)

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GATE ECE Chapter-wise Solved papers

22. The signal flow graph of a system is shown in the following figure. The transfer function C ( s) /R( s) of the system is 1 s

1

R(s)

⎡ xl ⎤ ⎡1 0 ⎤ ⎡ xl ⎤ ⎢ x ⎥ = ⎢1 1 ⎥ ⎢ x ⎥ and ⎦⎣ 2⎦ ⎣ 2⎦ ⎣

1 s

6

−2

25. The zero-input response of a system given by the statespace equation

−3

−4

1 C(s) (a)

6 6s (b) 2 s + 29 s + 6 s + 29 s + 6

(c)

s( s + 2) (d) s( s + 27) s + 29 s + 6 s 2 + 29 s + 6

⎡te t ⎤ (a) ⎢ ⎥ ⎣ t ⎦

⎡et ⎤ (b) ⎢ ⎥ ⎣t ⎦

⎡ et ⎤ (c) ⎢ t ⎥ ⎣te ⎦

⎡t ⎤ (d) ⎢ t ⎥ ⎣te ⎦ (GATE 2003: 2 Marks)

2

2

(GATE 2003: 2 Marks) 23. The approximate Bode magnitude plot of a minimum-phase system is shown in the following figure. The transfer function of the system is dB

160 140

⎡ x1 (0) ⎤ ⎡1 ⎤ ⎥=⎢ ⎥ ⎢ ⎣ x2 ( 0 ) ⎦ ⎣ 0 ⎦

26. The gain margin for the system with open-loop transfer 2(1 + s) function G(s ) H (s ) = , is s2 (b) 0 (a) ∞ (d) −  ∞ (c) 1 (GATE 2004: 1 Mark) 27. Given G ( s) H ( s) = K /[ s( s + 1)( s + 3)] , the point of intersection of the asymptotes of the root loci with the real axis is (a) - 4 (b) 1.33 (c) -1.33 (d) 4 (GATE 2004: 1 Mark) 28. Consider the signal flow graph shown in the following figure. The gain x5/x1 is

20 w 0.1 (a) 108

( s + 0.1)3 ( s + 10) 2 ( s + 100)

(b) 10 7

( s + 0.1)3 ( s + 10)( s + 100)

10

x1

a

x2

b

x3

c

x4

x5

d

100

e

f

(a)

1- (be + cf + dg ) abc

( s + 0.1) 2 (c) 10 ( s + 10) 2 ( s + 100)

(b)

bedg 1- (be + cf + dg )

( s + 0.1)3 (d) 10 ( s + 10)( s + 100) 2

(c)

abcd 1- (be + cf + dg ) + bedg

8

9

g

1- (be + cf + dg ) + bedg (d) abcd 24. The root locus of the system G ( s) H ( s) = K /[ s( s + 2)( s + 3)] (GATE 2004: 2 Marks) G ( s) H ( s) = K /[ s( s + 2)( s + 3)] has the break- away point located at (a) (-0.5, 0) (b) (-2.548, 0) 29. A system described by the following differential equa(c) (-4, 0) (d) (-0.784, 0) tion is initially at rest. For input x(t) = 2u(t), the output y(t) is (GATE 2003: 2 Marks) (GATE 2003: 2 Marks)

Ch wise GATE_ECE_CH06_Control Systems.indd 392

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Chapter 6  •  Control Systems

The system is (a) controllable, but not observable (b) observable, but not controllable (c)  neither controllable nor observable (d) controllable and observable

d 2 y (t ) dy(t ) +3 + 2 y (t ) = x (t ) dt dt 2 (a) (1 - 2e-t + e-2t)u(t) (b) (1 + 2e-t - 2e-2t)u(t) (c) (0.5 + e-t + 1.5e-2t)u(t) -t

393

(GATE 2004: 2 Marks)

-2t

(d) (0.5 + 2e +2e )u(t) (GATE 2004: 2 Marks) 30. Consider the Bode magnitude plot shown in the following figure. The transfer function H(s) is

⎡1 0 ⎤ the state transition matrix eAt is given 34. Given A = ⎢ 0 1 ⎥⎦ ⎣ by ⎡0 (a) ⎢ - t ⎣e

e -t ⎤ ⎥ 0⎦

⎡et (b) ⎢ ⎣0

0⎤ ⎥ et ⎦

⎡e - t (c) ⎢ ⎣0

0⎤ ⎥ e -t ⎦

⎡0 (d) ⎢ t ⎣e

et ⎤ ⎥ 0⎦

20 log H(jw) 0

−20dB/decade

−20

(GATE 2004: 2 Marks)

1

10

100

w (rad/s)

(a)

( s + 10) 10( s + 1) (b) ( s + 1)( s + 100) ( s + 10)( s + 100)

(c)

10 2 ( s + 1) 103 ( s + 100) (d) ( s + 10)( s + 100) ( s + 1)( s + 10) (GATE 2004: 2 marks)

31. A system has poles at 0.01 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is (a) -90° (b) 0° (c) 90° (d) -180° (GATE 2004: 2 Marks) 32. The open-loop transfer function of a unity feedback K system is G(s ) = . The range of K 2 s(s + s + 2) (s + 3) for which the system is stable is 21 (a) > K > 0 (b) 13 > K > 0 4 21 (c) < K < ∞ (d) −6 < K < ∞ 4 (GATE 2004: 2 Marks) 33. The state variable equations of a system are given as follows: (i) x1 = -3 x1 - x2 + u (ii) x2 = 2 x1

Ch wise GATE_ECE_CH06_Control Systems.indd 393

y = x1 + u

35. Despite the presence of negative feedback, control systems still have problems of instability because the (a) components used have non-linearities (b) dynamic equations of the systems are not known exactly (c) mathematical analysis involves approximations (d) system has large negative phase angle at high frequencies (GATE 2005: 1 Mark) 36. A linear system is equivalently represented by two sets of state equations: X = AX + BU and W = CW + DU The Eigen values of the representations are also computed as [l] and [m]. Which one of the following statements is true? (a) [l] = [m] and X = W (b) [l] = [m] and X ≠ W (c) [l] ≠ [m] and X = W

(d) [l] ≠ [m] and X ≠ W (GATE 2005: 1 Mark)

37. A ramp input applied to a unity feedback system results in 5% steady-state error. The type number and zero frequency gain of the system are, respectively, (a) 1 and 20 (b) 0 and 20 (c) 0 and 1/20 (d) 1 and 1/20 (GATE 2005: 2 Marks) 38. The polar diagram of a conditionally stable system for open-loop gain K = 1 is shown in the f­ollowing figure. The open-loop transfer function of the system is known to be stable. The closed-loop system is stable for

12/4/2018 11:33:43 AM

394

GATE ECE Chapter-wise Solved papers

Im jw

jw

−0.2

−8

s

s

Re −2

(b)

(a) GH plane jw

jw

(c)

1 1 (b) K < and < K < 5 8 2 (c) K
R1C1. The transfer function Vo /Vi represents a kind of controller. Match the impedances in Group I with the types of controllers in Group II.

(b) Q → 1; R → 3 (d) Q → 3, R → 2 (GATE 2008: 2 Marks)

71. A signal flow graph of a system is given in the ­following figure.

(GATE 2008: 2 Marks)

(a) 0 (c) 2

R

⎛ x1 ⎞ ⎡ -α d ⎜ ⎟ ⎢ x2 = - β dt ⎜ ⎟ ⎢ ⎝ x3 ⎠ ⎢⎣ α

β -γ γ

γ ⎤ ⎛ x1 ⎞ ⎡0 0 ⎤ ⎛u ⎞ -γ ⎥⎥ ⎜ x2 ⎟ + ⎢⎢0 1 ⎥⎥ ⎜ 1 ⎟ ⎜ ⎟ ⎝u ⎠ -β ⎦⎥ ⎝ x3 ⎠ ⎣⎢1 0 ⎥⎦ 2 0 ⎤ ⎛ x1 ⎞ ⎡1 0 ⎤ ⎛u ⎞ 0 ⎥⎥ ⎜ x2 ⎟ + ⎢⎢0 1 ⎥⎥ ⎜ 1 ⎟ ⎜ ⎟ ⎝u ⎠ 0 ⎥⎦ ⎝ x3 ⎠ ⎢⎣0 0 ⎥⎦ 2

⎛ x1 ⎞ ⎡ -γ 0 β ⎤ ⎛ x1 ⎞ ⎡0 1 ⎤ d ⎜ ⎟ ⎢ ⎥ ⎛ u1 ⎞ ⎥⎜ ⎟ ⎢ (d) dt ⎜ x2 ⎟ = ⎢ γ 0 α ⎥ ⎜ x2 ⎟ + ⎢0 0 ⎥ ⎜ u ⎟ ⎝ 2⎠ ⎝ x3 ⎠ ⎢⎣ - β 0 -α ⎥⎦ ⎝ x3 ⎠ ⎢⎣1 0 ⎥⎦ (GATE 2008: 2 Marks)

Ch wise GATE_ECE_CH06_Control Systems.indd 398

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Chapter 6  •  Control Systems

72. The magnitude plot of a rational transfer function G(s) with real coefficients is shown in the following figure. Which of the following compensators has such a magnitude plot? G(jw) 20 dB log w −40 dB

75. The gain and phase margins of G(s) for closed-loop stability are (a) 6 dB and 180° (b) 3 dB and 180° (c) 6 dB and 90° (d) 3 dB and 90° (GATE 2009: 2 marks) 76. The feedback configuration and the pole-zero locations 2 2 of G ( s) H ( s) = ( s - 2 s + 2)/( s + 2 s + 2) are shown in the following figures (a) and (b), respectively. The root locus for negative values of K, that is, for −∞ < K < 0, has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to + + + +− −

(a) Lead compensator (b) Lag compensator (c) PID compensator (d) Lead-lag compensator

(a) Im(s) (a) Im(s)

⎡1 0 ⎤ 73. Consider the system dx /dt = Ax + Bu with A = ⎢ ⎥ ⎣0 1⎦ ⎡ p⎤ and B = ⎢ ⎥ , where p and q are arbitrary real numbers. ⎣q⎦

Common Data for Questions 74 and 75: The Nyquist plot of a stable transfer function G(s) is shown in the following figure. We are interested in the stability of the closed-loop system in the feedback configuration shown.

0 0 Re(s) Re(s) P P (b) (b) (a) ± 2 and 0° (b) ± 2 and 45° (c) ± 3 and 0° (d) ± 3 and 45° (GATE 2009: 2 Marks) 77. The transfer function Y(s)/R(s) of the system shown in the following figure is R(s)

Im

+ −

Σ

+ −1

−0.5

Re

1 (s + 1)

Y(s)

+

1 (s + 1)

G(s)

+

G(s) G(s)

K K

(GATE 2009: 1 Mark)

Which of the following statements about the controllability of the system is true? (a) The system is completely state controllable for any non-zero values of p and q. (b) Only p = 0 and q = 0 result in controllability. (c) The system is uncontrollable for all values of p and q. (d) We cannot conclude about controllability from the given data. (GATE 2009: 1 Mark)

399

−

−

Σ

−j 74. Which of the following statements is true? (a) G(s) is an all-pass filter (b) G(s) has a zero in the right-half plane (c) G(s) is the impedance of a passive network (d) G(s) is marginally stable (GATE 2009: 2 marks)

Ch wise GATE_ECE_CH06_Control Systems.indd 399

(a) 0 (c)

(b)

1 s +1

2 2 (d) s +1 s+3 (GATE 2010: 1 Mark)

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400

GATE ECE Chapter-wise Solved papers

1 78. A system with the transfer function

π⎞ ⎛ an output y(t ) = cos ⎜ 2t - ⎟ ⎝ 3⎠

Y ( s) s has = X ( s) s + p

2 3 (b) 3

(c) 1

(d)

3 2

(GATE 2010: 1 Mark)

Magnitude (dB)

79. For the asymptotic Bode magnitude plot shown in the following figure, the system transfer function can be

0.5

1/s

Y(s) 1

−1

for the input signal

π⎞ ⎛ x(t ) = p cos ⎜ 2t - ⎟ . The, the system parameter `p’ is ⎝ 2⎠ (a)

1/s

2 U(s)

−1 81. The state variable representation of the system can be ⎡0⎤ ⎡ 1 1⎤ x + ⎢ ⎥ u;    (a) x = ⎢ y = [0 0.5] x ⎥ ⎣ -1 0 ⎦ ⎣ 2⎦ ⎡0 ⎤ ⎡ -1 1 ⎤ (b) x = ⎢ ⎥ x + ⎢ 2⎥ u;      y = [0 0.5] x 1 0 ⎣ ⎦ ⎣ ⎦ (c)

⎡0 ⎤ ⎡ 1 1⎤ x = ⎢ x + ⎢ ⎥ u; y = [0.5 0.5] x ⎥ ⎣ -1 0 ⎦ ⎣ 2⎦   

⎡0 ⎤ ⎡ -1 1 ⎤ x + ⎢ ⎥ u;   (d) x = ⎢ ⎥ ⎣ -1 0 ⎦ ⎣ 2⎦

y = [0.5 0.5] x

40 (GATE 2010: 2 Marks) 0 0.001

(a)

(c)

0.1

10

w 1000

10 s + 1 100 s + 1 (b) 0.1s + 1 0.1s + 1 100 s 10 s + 1

(d)

0.1s + 1 10 s + 1

(GATE 2010: 1 Mark) 80. A unity negative feedback closed loop system has a plant with the transfer function G ( s) = 1/ ( s 2 + 2 s + 2) and a controller Gc(s) in the feed forward path. For a unit step input, the transfer function of the controller that gives minimum steady-state error is s+2 (a) Gc ( s) = s + 1 (b) Gc ( s) = s+2 s +1

82. The transfer function of the system is (a)

s + 1 (b) s -1 2 s +1 s2 + 1

(c)

s -1 s + 1 (d) 2 s + s +1 s + s +1 2

(GATE 2010: 2 Marks) Common Data for Questions 83 and 84: The input−output 100 transfer function of a plant H ( s) = . The plant is s( s + 10) 2 placed in a unity negative feedback configuration as shown in the following figure. R(s) +

(c) Gc ( s) =

2 ( s + 1)( s + 4) (d) Gc ( s) = 1 + + 3s s ( s + 2)( s + 3)

Σ

U(s) H(s) =

100 s(s + 10)2

Y (s)

− Plant

(GATE 2010: 2 Marks) Common Data for Questions 81 and 82: The signal flow graph of a system is shown in the following figure.

Ch wise GATE_ECE_CH06_Control Systems.indd 400

83. The signal flow graph that does not model the plant transfer function H(s) is

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Chapter 6  •  Control Systems

86. The root locus plot for a system is given in the following figure. The open-loop transfer function corresponding to this plot is given by

(a) 1s

1s

1s

100

U(s)

Y (s) −10

−10 jw

−100

(b) 1s

1s

401

1s

100

U(s)

Y (s) −20

−3

s

−2 −1

0

−100 (c) 1s

1s

1s

100

U(s)

Y (s) −20 −100

(d) 1s

1s

1s

(a) G ( s) H ( s) = K

s( s + 1) ( s + 2)( s + 3)

(b) G ( s) H ( s) = K

( s + 1) s( s + 2)( s + 3) 2

(c) G ( s) H ( s) = K

1 s( s - 1)( s + 2)( s + 3)

(d) G ( s) H ( s) = K

( s + 1) s( s + 2)( s + 3)

100

U(s)

Y (s) (GATE 2011: 2 Marks)

84. The gain margin of the system under closed loop unity negative feedback is (a) 0 dB (b) 20 dB (c) 26 dB (d) 46 dB (GATE 2011: 2 Marks) 85. For the transfer function G(jw) = 5 + jw, the ­corresponding Nyquist plot for positive frequency has the form jw

(GATE 2011: 1 Mark) 87. The block diagram of a system with one input u and two outputs y1 and y2 is given in the following figure.

jw u j5

1 s+2

y1

2 s+2

y2

s

s 5

A state space model of the above system in terms of the state vector x and the output vector y = [y1  y2]T is

(b)

(a)

(a) x = [2]x + [1]u; y = [1 2] x jw

jw

⎡1⎤ (b) x = [ -2]x + [1]u; y = ⎢ ⎥ x ⎣ 2⎦ 1/5

s

s 1/5

(c)

(d) (GATE 2011: 1 Mark)

Ch wise GATE_ECE_CH06_Control Systems.indd 401

⎡1⎤ ⎡ -2 0 ⎤ (c) x = ⎢ x + ⎢ ⎥ u; y = [1 2] x ⎥ ⎣ 0 -2⎦ ⎣1⎦ ⎡1 ⎤ ⎡2 0⎤ (d) x = ⎢ ⎥ x + ⎢ 2⎥ u; y = 0 2 ⎣ ⎦ ⎣ ⎦

⎡1 ⎤ ⎢ 2⎥ x ⎣ ⎦ (GATE 2011: 2 Marks)

12/4/2018 11:33:56 AM

402

GATE ECE Chapter-wise Solved papers

88. A

system

with

transfer

function.

G(s)

=

( s + 9)( s + 2) is excited by sinwt. The steady- state ( s + 1)( s + 3)( s + 4) output of the system is zero at (a) w = 1 rad/s (b) w = 2 rad/s (c) w = 3 rad/s (d) w = 4 rad/s (GATE 2012: 1 Mark) 2

93. The Bode plot of transfer function G(s) is shown in the following figure. Gain (dB)

40 32

Statement for Linked Answer Questions 89 and 90: The transfer function of compensator is given as Gc ( s) =

20

( s + a) ( s + b) 0

89. Gc(s) is a lead compensator if (a) a = 1, b = 2 (b) a = 3, b = 2 (c) a = −3, b = −1 (d) a = 3, b = 1 (GATE 2012: 2 Marks) 90. The phase of the above lead compensator is the maximum at (a) 2 rad /s (b) 3 rad/s (c)

6 rad/s (d) 1/ 3 rad/s

⎛ x1 ⎞ ⎛ 0 ⎜ x ⎟ = ⎜ 0 ⎜ 2⎟ ⎜ ⎝ x3 ⎠ ⎝ a3

a1 0 0

0⎞ a2 ⎟ ⎟ 0⎠

⎛ x1 ⎞ ⎛ 0⎞ ⎜ x ⎟ + ⎜ 0⎟ u ⎜ 2⎟ ⎜ ⎟ ⎝ x3 ⎠ ⎝ 1⎠

⎛ x1 ⎞ y = (1 0 0 ) ⎜ x2 ⎟ ⎜ ⎟ ⎝ x3 ⎠

100 w (rad/s)

1

−8

The gain 20 log G ( s) | is 32 dB and -8 dB at 1 rad/s and 10 rad/s, respectively. The phase is negative for all w. Then G(s) is (a)

39.8 39.8 (b) s s2

(c)

32 32 (d) s s2

(GATE 2012: 2 Marks) 91. The state variable description of an LTI system is given by

10

(GATE 2013: 1 Mark) 94. The signal flow graph for a system is shown in the following figure. The transfer function Y ( s) /U ( s) for this system is 1

1

where y is the output and u is the input. The system is controllable for (a) a1 ≠ 0, a2 = 0, a3 ≠ 0

(b) a1 = 0, a2 ≠ 0, a3 ≠ 0

(c) a1 = 0, a2 ≠ 0, a3 = 0

(d) a1 ≠ 0, a2 ≠ 0, a3 = 0

s−1

s −1

1

U(s)

Y(s) −4 −2

(GATE 2012: 2 Marks) 92. Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is U(s) 1 s

Y(s)

(a)

s +1 s +1 (b) 2 5s + 6 s + 2 s + 6s + 2

(c)

s + 1 (d) 1 2 s + 4s + 2 5s + 6 s + 2

2

2

(GATE 2013: 2 Marks) (a) u(t) (b) t u(t) (c)

t2 u(t ) (d) e-t u(t) 2 (GATE 2013: 1 Mark)

Ch wise GATE_ECE_CH06_Control Systems.indd 402

95. A system is described by the differential equation d  2y/dt2 + 5(dy/dt) + 6y(t) = x(t). Let x(t) be a rectangular pulse given by x(t) = 1 for 0 < t < 2 and x(t) = 0 otherwise. Assuming that y(t) = 0 and dy/dt = 0 at t = 0, the Laplace transform of y(t) is

12/4/2018 11:33:58 AM

403

Chapter 6  •  Control Systems

(a)

e -2 s 1 - e -2 s (b) s( s + 2)( s + 3) s( s + 2)( s + 3)

(c)

e -2 s ( s + 2)( s + 3)

(d)

98. The state transition matrix eAt of the system shown in the given figure is

1 - e -2 s ( s + 2)( s + 3)

(GATE 2013: 2 Marks) 96. The open-loop transfer function of a DC motor is given as w (s)/Va(s) = 10/(1 + 10s). When connected in feedback as shown in the f­ ollowing figure, the approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is Va(s) + −

Σ

⎡ e -t 0⎤ 0⎤ (b) ⎢ -t ⎥ -t ⎥ e ⎦ e -t ⎦ ⎣ -te

⎡e - t (c) ⎢ - t ⎣e

⎡e - t -te - t ⎤ 0⎤ (d) ⎥ ⎢ -t ⎥ e -t ⎦ e -t ⎦ ⎣e (GATE 2013: 2 Marks)

99. Consider the following block diagram in the figure. R(s)

+

(b) 5 (d) 100 (GATE 2013: 2 Marks)

Statement for Linked Answer Questions 97 and 98: The state diagram of a system shown in the following figure is described by the state-variable equations:

−1

1

−1

1

u

y 1 s

1 s

97. The state-variable equations of the system shown in the figure are ⎡ -1⎤ ⎡ -1 0 ⎤ X + ⎢ ⎥ u; y = [1 - 1] X + u (a) X = ⎢ ⎥ ⎣1⎦ ⎣ 1 -1⎦ ⎡ -1⎤ ⎡ -1 0 ⎤ X + ⎢ ⎥ u; y = [ -1 - 1] X + u (b) X = ⎢ ⎥ ⎣ -1 -1⎦ ⎣ 1⎦ (c)

(a)

C ( s) is R( s)

G1G2 (b) G1G2 + G1 + 1 1+ G1G2

(c) G1G2 + G2 + 1

(d)

G1 1+ G1G2

(GATE 2014: 1 Mark)

101. The input -3e2tu(t), where u(t) is the unit step function, s-2 is applied to a system with transfer function . If s+3 the initial value of the output is -2, then the value of the output at steady state is . (GATE 2014: 1 Mark) 102. For the second order closed-loop system shown in the figure, the natural frequency (in rad/s) is U(s) + −

⎡ -1⎤ ⎡ -1 0 ⎤ X = ⎢ X + ⎢ ⎥ u; y = [ -1 - 1] X - u ⎥ ⎣ -1 -1⎦ ⎣ 1⎦

⎡ -1⎤ ⎡ -1 -1⎤ X + ⎢ ⎥ u; y = [1 - 1] X - u (d) X = ⎢ ⎥ ⎣ 0 -1⎦ ⎣ 1⎦ (GATE 2013: 2 Marks)

Ch wise GATE_ECE_CH06_Control Systems.indd 403

+

100. The natural frequency of an undamped second-order system is 40 rad/s. If the system is damped with a damping ratio 0.3, the damped natural frequency in rad/s is . (GATE 2014: 1 Mark)

X = AX + Bu y = CX + Du 1

G2

+

The transfer function (a) 1 (c) 10

C(s)

+

G1

w

10 (1 + 10s)

Ka

⎡ e -t (a) ⎢ - t ⎣te

(a) 16 (c) 2

4 s(s + 4)

Y(s)

(b) 4 (d) 1 (GATE 2014: 1 Mark)

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404

GATE ECE Chapter-wise Solved papers

103. The forward path transfer function of a unity negative feedback system is given by G ( s) =

K ( s + 2)( s - 1)

The value of K which will place both the poles of the closed-loop system at the same location is .

Which one of the following compensators C(s) achieves this? ⎛ 1 ⎞ (a)  3⎜ ⎝ s + 5 ⎟⎠

⎛ 0.03 ⎞ 5⎜ + 1⎟ (b)  ⎝ s ⎠

(c) 2(s + 4)

⎛ s + 8⎞ (d) 4 ⎜ ⎝ s + 3 ⎟⎠

(GATE 2014: 1 Mark)

(GATE 2014: 2 Marks)

104. Consider the feedback system shown in the figure. The Nyquist plot of G(s) is also shown. Which one of the following conclusions is correct?

107. The state equation of a second-order linear system is given by

+ G(s)

k −

+1

⎡e - t ⎤ 1 For x0 = ⎡⎢ ⎤⎥ , x(t ) = ⎢ and -t ⎥ ⎣ -1⎦ ⎣ -e ⎦ ⎡e - t - e -2t ⎤ 0 for x0 = ⎡⎢ ⎤⎥ , x(t ) = ⎢ -t -2 t ⎥ ⎣1 ⎦ ⎣ - e + 2e ⎦

Im G(jw)

−1

x (t ) = Ax(t ), x(0) = x0

Re G(jw)

⎡ 3⎤ When x0 = ⎢ ⎥ , x(t ) is ⎣ 5⎦ ⎡ -8e - t + 11e -2t ⎤ ⎡11e - t - 8e -2t ⎤ (a) ⎢ (b) ⎢ -t -2 t ⎥ -t -2 t ⎥ ⎣8e - 22e ⎦ ⎣ -11e + 16e ⎦

(a)  G(s) is an all-pass filter (b)  G(s) is a strictly proper transfer function (c)  G(s) is a stable and minimum-phase transfer function (d) The closed-loop system is unstable for sufficiently large and positive k (GATE 2014: 1 Mark) 105. In a Bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4th order all-pole system? (a) −80 dB/decade (b) −40 dB/decade (c) +40 dB/decade (d) +80 dB/decade (GATE 2014: 1 Mark) 1 106. For the following feedback system G ( s) = . ( s + 1)( s + 2)

⎡3e - t - 5e -2t ⎤ ⎡5e - t - 3e -2t ⎤ (c) ⎢ (d) ⎢ -t -2 t ⎥ -t -2 t ⎥ ⎣ -3e + 10e ⎦ ⎣ -5e + 6e ⎦ (GATE 2014: 2 Marks) 108. The steady-state error of the system shown in the figure for a unit step input is . R(s) +

C(s)

E(s) 1 s+2

K=4 r(t)

−

e(t)

c(t) 2 s+4

(GATE 2014: 2 Marks) 109. The

phase

margin

in

degrees

of

G ( s) =

10 The 2%-settling time of the step response is required to G ( s) = calculated using the asymptotic ( s + 0.1)( s + 1)( s + 10) be less than 2 seconds. Bode plot is . r + y (GATE 2014: 2 Marks) C(s) G(s) 110. The Bode asymptotic magnitude plot of a minimum − phase system is shown in the figure.

Ch wise GATE_ECE_CH06_Control Systems.indd 404

10 ( s + 0.1)( s + 1)( s + 1

12/4/2018 11:34:02 AM

405

Chapter 6  •  Control Systems

(a)

s +1 ( s + 2)( s + 4)( s + 7)

(b)

s+4 ( s + 1)( s + 2)( s + 7)

(c)

s+7 ( s + 1)( s + 2)( s + 4)

(d)

( s + 1)( s + 2) ( s + 7)( s + 4)

26.02 |G(jω)| 6.02 (dB) 0 0.1

1

2

10

20

−6.02 ω(rad/s) in log scale If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is . (GATE 2014: 2 Marks)

(GATE 2014: 2 Marks) 114. The characteristic equation of a unity negative feedback system is 1 + KG(s) = 0. The open loop transfer function G(s) has one pole at 0 and two poles at −1. The root locus of the system for varying K is shown in the figure.

111. Let h(t) denote the impulse response of a causal sys1 tem with transfer function . Consider the following s +1 three statements. S1: The system is stable. S2:

jω

h(t +1) is independent of t for t > 0. h(t )

ξ = 0.5 A

S3: A non-causal system with the same transfer function is stable.

(0,0) −1 O 3

−1

For the above system, (a) only S1 and S2 are true (b) only S2 and S3 are true

σ

(c) only S1 and S3 are true (d) S1, S2 and S3 are true (GATE 2014: 2 Marks) ps 2 + 3 ps - 2 s + (3 +The p) s constant + ( 2 - p)damping ratio line, for x = 0.5, intersects ps 2 + 3 ps - 2 the root locus at point A. The distance from the origin with p a positive real para­meter. G p ( s) = 2 to point A is given as 0.5. The value of K at point A is s + (3 + p ) s + ( 2 - p ) . The maximum value of p until which Gp remains stable is . (GATE 2014: 2 Marks) (GATE 2014: 2 Marks) 115. Consider the state space system expressed by the signal flow diagram shown in the figure. 113. In the root locus plot shown in the figure, the pole/zero marks and the arrows have been removed. Which one of c3 c2 the following transfer functions has this root locus? 112. Consider

a

transfer

function

G p ( s) =

2

i

jw

s−1

s−1

s−1 x3

u

x2

x1

c1

y

a3 a2

a1

2 1 s

Ch wise GATE_ECE_CH06_Control Systems.indd 405

The corresponding system is (a) always controllable (b) always observable (c) always stable (d) always unstable (GATE 2014: 2 Marks)

12/4/2018 11:34:03 AM

406

GATE ECE Chapter-wise Solved papers

116. Consider the state space model of a system, as given below 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡ x1 ⎤ ⎡ -1 ⎡ x1 ⎤ ⎢ x ⎥ = ⎢ 0 -1 0 ⎥ ⎢ x ⎥ + ⎢ 4 ⎥ u; y = 1 1 1 ⎢ x ⎥ [ ]⎢ 2 ⎥ ⎢ 2⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ x3 ⎥⎦ ⎢⎣ 0 0 -2 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ x3 ⎥⎦ The system is (a) controllable and observable (b) uncontrollable and observable (c) uncontrollable and unobservable (d) controllable and unobservable (GATE 2014: 2 Marks) 117. An unforced linear time invariant (LTI) system is represented by ⎡ x1 ⎤ ⎡ -1 0 ⎤ ⎡ x1 ⎤ ⎢ x ⎥ = ⎢ 0 -2⎥ ⎢ x ⎥ ⎦⎣ 2⎦ ⎣ 2⎦ ⎣ If the initial conditions are x1(0) = 1 and x2(0) = −1, the solution of the state equation is (a) (b) (c) (d)

x1(t) = −1, x2(t) = 2 x1(t) = −e−t, x2(t) = 2e−t x1(t) = e−t, x2(t) = −e−2t x1(t) = −e−t, x2(t) = −2e−t

(GATE 2014: 2 Marks)

118. The state transition matrix ⎡ x1 ⎤ ⎡0 1 ⎤ ⎡ x1 ⎤ ⎢ x ⎥ = ⎢0 0 ⎥ ⎢ x ⎥ is ⎦⎣ 2⎦ ⎣ 2⎦ ⎣

ϕ (t )

120. For the signal flow graph shown in the figure, the value C ( s) is of R( s)

−H3

1 X1

⎡1 t ⎤ ⎡0 1⎤ (c) ⎢ (d) ⎥ ⎢ ⎥ ⎣1 t ⎦ ⎣0 1⎦ (GATE 2014: 2 Marks) 119. By performing cascading and/or summing/differencing operations using transfer function block G1(s) and G2(s), one CANNOT realize a transfer function of the form

X4

G 1 X2 G 2

R(s)

C(s) G 3 X5 G 4

−H1

−H2

(a)

G1G2G3G4 1- G1G2 H1 - G3G4 H 2 - G2G3 H 3 + G1G2G3G4 H1 H 2

(b)

G1G2G3G4 1+ G1G2 H1 + G3G4 H 2 + G2G3 H 3 + G1G2G3G4 H1 H 2

(c)

1 1 + G1G2 H1 + G3G4 H 2 + G2G3 H 3 + G1G2G3G4 H1 H 2

(d)

1 1 - G1G2 H1 - G3G4 H 2 - G2G3 H 3 + G1G2G3G4 H1 H 2 (GATE 2015: 1 Mark)

121. A unity negative feedback system has an open-loop K transfer function G ( s) = . The gain K for the s( s +10) . system to have a damping ratio of 0.25 is (GATE 2015: 1 Mark) 122. Consider the Bode plot shown in figure. Assume that all the poles and zeros are real-valued.

40 dB

−40 dB/dec

(a) G1(s)G2(s)

40 dB/dec 0 dB

G1 ( s) G2 ( s)

fL

300

900

fH Freq. (Hz)

The value of fH - fL (in Hz) is

⎤ ⎡ 1 + G2 ( s) ⎥ (c) G1 ( s) ⎢ ( ) G s ⎣ 1 ⎦

(GATE 2015: 1 Mark) 123. The phase margin (in degrees) of the system 10 is . G ( s) = s( s + 10)

⎤ ⎡ 1 - G2 ( s) ⎥ (d) G1 ( s) ⎢ ( ) G s ⎣ 1 ⎦ (GATE 2015: 1 Mark)

Ch wise GATE_ECE_CH06_Control Systems.indd 406

1

of a system

⎡t 1 ⎤ ⎡1 0 ⎤ (a) ⎢ (b) ⎥ ⎢ ⎥ ⎣1 0 ⎦ ⎣t 1 ⎦

(b)

X3

(GATE 2015: 1 Mark)

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Chapter 6  •  Control Systems

124. A unity negative feedback system has the open-loop

the gain K (> 0) at which the root locus crosses the imaginary axis is . (GATE 2015: 1 Mark)

−11.4 −21.5 −32.8 −45.3

2 3 5 10

K transfer function G ( s) = . The value of s( s + 1)( s + 3)

407

−89.4 −151 −167 −174.5

The unit step response of the system approaches a steady state value of . (GATE 2015: 2 Marks)

10( s + 1) 125. The polar plot of the transfer function G ( s) = s + 10 for 0 ≤ w ≤ ∞ will be in the (a) first quadrant (b) second quadrant (c) third quadrant (d) fourth quadrant (GATE 2015: 1 Mark)

129. The position control of a DC servo-motor is given in the figure. The values of the parameters are KT = 1 N-m/A, Ra = 1 Ω, La = 0.1 H, J = 5 kg-m2, B = 1 N-m (rad/s) and Kb = 1 V/(rad/s). The steady-state position response (in radians) due to unit impulse disturbance torque Td is .

126. The transfer function of a first order controller is given as

Td(s)

GC ( s) =

K ( s + a) s+b

+

where, K, a and b are positive real numbers. The condition for this controller to act as a phase lead compensator is (a) a < b (b) a > b (c) K < ab (d) K > ab (GATE 2015: 1 Mark)

Va(s) −

− KT 1 Ra + Las + Js + B

1 s

θ(s)

Kb

(GATE 2015: 2 Marks) 130. A plant transfer function is given as

K ⎞ 1 ⎛ . G ( s) = ⎜ K P + I ⎟ 127. The output of a standard second-order system for ⎝ s ⎠ s ( s + 2) 2 -t p⎞ ⎛ e cos ⎜ 3t - ⎟ . a unit step input is given as y(t) = y(t ) = 1 When ⎝ 6 ⎠ the plant operates in a unity feedback configu3 p⎞ 2 -t ⎛ ration, the condition for the stability of the closed loop y (t ) = 1 e cos ⎜ 3t - ⎟ . The transfer function of the system is ⎝ 6⎠ 3 system is (a) (c)

2 ( s + 2)( s + 3 )

1 (b) 2 s + 2s + 1

3 4 (d) 2 2 s + 2s + 3 s + 2s + 4 (GATE 2015: 2 Marks)

128. The transfer function of a mass-spring damper system is given by 1 G ( s) = Ms 2 + Bs + K The frequency response data for the system are given in the following table. w in rad/s 0.01 0.1 0.2 1

Ch wise GATE_ECE_CH06_Control Systems.indd 407

|G(jw)| in dB −18.5 −18.5 −18.4 −16

Arg (G(jw)) in deg −0.2 −1.3 −2.6 −16.9

(a) K P >

KI >0 2

(b) 2KI > KP > 0 (c) 2KI < KP (d) 2KI > KP (GATE 2015: 2 Marks) 131. The characteristic equation of an LTI system is given by F(s) = s5 + 2s4 + 3s3 + 6s2 - 4s -8 = 0. The number of roots that lie strictly in the left half s-plane is . (GATE 2015: 2 Marks) 132. The open-loop transfer function of a plant in a unity feedback configuration is given as G ( s) =

K ( s + 4) ( s + 8)( s 2 - 9)

The value of the given K (> 0) for which −1 + j 2 lies on . the root locus is (GATE 2015: 2 Marks)

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GATE ECE Chapter-wise Solved papers

133. For the system shown in the figure, s = −2.75 lies on the root locus if K is . + −

e(t) y(t)

−

Y(s)

s+3 s+2

K

+ x(t)

(a) 0 (c) 1.0 10 (GATE 2015: 2 Marks)

134. A lead compensator network includes a parallel combination of R and C in the feed-forward path. If the s+2 transfer function of the compensator is GC ( s) = , +4 s the value of RC is . (GATE 2015: 2 Marks) 135. The state variable representation of a system is given as 1⎤ ⎡ 1⎤ ⎡0 x = ⎢ x; x ( 0 ) = ⎢ ⎥ ⎥ ⎣0⎦ ⎣0 -1⎦ y = [0 1]x The response y(t) is (a) sin (t) (c) 1 - cos(t)

138. A closed-loop control system is stable if the Nyquist plot of the corresponding open-loop transfer function (a) encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number of right-half s-plane poles. (b) encircles the s-plane point (0 − j1) in the ­clockwise direction as many times as the number of right-half s-plane poles. (c) encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number of left-half s-plane poles. (d) encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number of right-half s-plane zeros. (GATE 2016: 1 Mark) 139. The number and direction of encirclements around the

(b) 1 − e (d) 0 (GATE 2015: 2 Marks) t

136. A network is described by the state model as x1 = 2 x1 - x2 + 3u x2 = -4 x2 - u y = 3 x1 - 2 x2

point −1 + j0 in the complex plane by the Nyquist plot of 1- s G ( s) = is 4 + 2s (a) zero (b) one, anti-clockwise (c) one, clockwise (d) two, clockwise (GATE 2016: 1 Mark)

⎛ Y ( s) ⎞ The transfer function H ( s) = ⎜ is ⎝ U ( s) ⎟⎠ (a)

11s + 35 11s - 35 (b) ( s - 2)( s + 4) ( s - 2)( s + 4)

(c)

11s + 38 11s - 38 (d) ( s - 2)( s + 4) ( s - 2)( s + 4) (GATE 2015: 2 Marks)

137. For the unity feedback control system shown in the figure, the open-loop transfer function G(s) is given as 2 . The steady state error ess due to a unit G ( s) = s( s + 1) step input is

Ch wise GATE_ECE_CH06_Control Systems.indd 408

(b) 0.5 (d) ∞ (GATE 2016: 1 Mark)

140. Match the inferences X, Y and Z, about a system, to the corresponding properties of the elements of first column in Routh’s Table of the system characteristic equation. X: The system is stable … P: … when all elements are positive Y: The system is unstable … Q: … when any one element is zero Z: The test breaks down … R: … when there is a change in sign of coefficients (a) X  → P, Y → Q, Z → R (b) X  → Q, Y → P, Z → R (c) X  → R, Y → Q, Z → P (d) X  → P, Y → R, Z → Q (GATE 2016: 1 Mark)

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Chapter 6  •  Control Systems

409

k In the feedback system shown below G ( s) = 141. The block diagram of a feedback control system is 145. ( s + 1)( s + 2)( s + 3) shown in the following figure. The overall closed-loop k G ( s ) = . gain G of the system is ( s + 1)( s + 2)( s + 3) + + G2 G1 Y + y G(s) k − − −

H1

(a) G =

The positive value of k for which the gain margin of the loop is exactly 0 dB and the phase margin of the loop is exactly zero degree is . (GATE 2016: 2 Marks)

G1G2 1 + G1 H1

146. The

G1G2 (b) G = 1 + G1G2 + G1 H1 (c) G =

G ( s) =

asymptotic

phase

plot

positive, is shown below.

G1G2 1 + G1G2 H1

0.1

1

10

100

0° −45°

G1G2 1 + G1G2 + G1G2 H1 (GATE 2016: 1 Mark)

−135°

142. Which one of the following is an eigen function of the class of all continuous-time, linear, time-invariant systems [u(t) denotes the unit-step function]?

−225°

(a) e jw0 t u(t )

(b) cos(ω0t)

(c) e jw0 t

(d) sin(ω0t)

−270°

(GATE 2016: 2 Marks)

143. The open-loop transfer function of a unity-feedback K control system is given by G ( s) = . For the peak s ( s + 2) overshoot of the closed-loop system to a unit step input to be 10%, the value of K is . (GATE 2016: 2 Marks)

147. The transfer function of a linear time invariant system is given by H(s) = 2s4 − 5s3 + 5s − 2. The number of zeros in the right half of the s-plane is . (GATE 2016: 2 Marks) 148. The first two rows in the Routh table for the characteristic equation of a certain closed-loop control system are given as

1 . ( s 2 + 2 s)

s3

1

(2K + 3)

The step response of the closed-loop system should have minimum setling time and have no overshoot.

s2

2K

4

144. In the feedback system shown below G ( s) =

The range of K for which the system is stable is (a) −2.0  −4 ⇒ K > -1 Therefore, 0 ≤ K 0 ⇒ K > 3 16. Topic: Transient and Steady State Analysis of LTI Systems 100 1 = (b)  G ( s) = ( s + 1)( s + 100) ⎛ s ⎞ ( s + 1) ⎜ +1 ⎝ 100 ⎟⎠

1

G ( jw ) H ( jw ) =

jw (( jw ) + jw + 1) π w ∠G ( jw ) H ( jw ) = - - tan -1 2 1- w2 2

For phase crossover frequency -180° = -90° - tan -1 tan -1

wϕ 1 - wϕ 2

wϕ 1 - wϕ 2

= 90°

1 - w ϕ 2 = 0 ⇒ w ϕ = 1 rad/sec Ch wise GATE_ECE_CH06_Control Systems.indd 415

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416

w ϕ Solved papers -GATE 180° =ECE -90Chapter-wise ° - tan -1 1 - wϕ 2

tan -1

wϕ 1 - wϕ

2

= 90°

Gain margin= - 20 log10 | G ( jw ϕ ) H ( jw ϕ ) |

(

)

= -20 log 1 = 0 19. Topic: Feedback Principle (*)  Input applied to low pass filter = [m(t ) s(t ) + n(t )] s(t ) sin 199π t ⎤ ⎡ 2 sin 2π t =⎢ ⋅ cos 200π t + ⎥ × cos 200π t t t ⎣ ⎦ 1 = [(sin 202π t - sin 198π t ) + sin 199π t ] × cos( 200π t ) t Cut-off frequency of LPF, f c = 1 Hz ⇒ w c = 2π rad/sec Output of LPF, y (t ) = y (t ) =

sin 2π t sin 2π t - sin π t + + 2t 2t 2t sin 2π t sin 0.5π t ⋅ cos 1.5π t + 2t t

20. Topic: Routh-Hurwitz and Nyquist Stability Criteria (a)  We know that N = 1 (from Nyquist plot) and P =1 (from given data). Therefore, Z = P-N = 0 Note that the encirclement of point (−1, 0) is ­counter clockwise. Due to this reason, N has been taken to be negative. There is no zero in right half plane of s-plane. Therefore, the system is stable. 21. Topic: Lag, Lead and Lag-Lead Compensation (c)  The derivative element reduces signal-to-noise ratio. 22. Topic: Transfer Function (d)  The transfer function (from Mason’s gain formula) is P1 Δ1 Δ There is only one forward path with gain equal to 1. Therefore, P1 = 1 ⎛ -3 -24 -2 ⎞ ⎛ -2 -3 ⎞ Δ = 1- ⎜ - ⎟ +⎜ × ⎟ ⎝ s s s ⎠ ⎝ s s⎠ s 2 + 29 s + 6 s2 3 24 s + 27 Δ1 = 1 + + = s s s

⎛ -3 -24 -2 ⎞ ⎛ -2 -3 ⎞ - ⎟ +⎜ × ⎟ Δ = 1- ⎜ ⎝ s s s ⎠ ⎝ s s⎠ s 2 + 29 s + 6 s2 3 24 s + 27 Δ1 = 1 + + = s s s =

1 - w ϕ 2 = 0 ⇒ w ϕ = 1 rad/sec

= -20 log10 w ϕ w ϕ 2 + (1 - w ϕ 2 )

P1 = 1

Therefore, the transfer function is ( s + 27) /s s( s + 27) = ( s 2 + 29 s + 6) /s 2 s 2 + 29 s + 6 23. Topic: Bode and Root-Locus Plots (a) 

•  For w = 0.1 to 10: The magnitude of slope is 60 dB per decade. Therefore, there are three zeros at w = 0.1. •  For w = 10 to 100: The magnitude slope decreases from 60 dB per decade to 20 dB per decade, which is equivalent to −40 dB per decade. Therefore, there are two poles at w = 10. •  For w =100: The magnitude slope decreases from 20 dB per decade to 0 dB per decade, which is equivalent to −20 dB per decade. Therefore, there is one pole at w = 100. Therefore, the transfer function is given by K ( s + 0.1)3 ( s + 10) 2 ( s + 100) For w = 0, we get 20 log10 10−7K = 20 ⇒ log10 10−7K = 1 ⇒ 10−7K = 10 which gives K = 108 Therefore, the transfer function is 108 × ( s + 0.1)3 ( s + 10) 2 ( s + 100) 24. Topic: Bode and Root-Locus Plots (d)  If (x, 0) is the breakaway point, then 1 1 1 + + =0 x x+2 x+3 ( x + 2)( x + 3) + x( x + 3) + x( x + 2) = 0 3 x 2 + 10 x + 6 = 0 Solving the quadratic equation, we get x = − 2.55 and − 0.784 Since (−2.55, 0) cannot lie on the root locus, the breakaway point is (−0.784, 0).

=

Ch wise GATE_ECE_CH06_Control Systems.indd 416

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Chapter 6  •  Control Systems

25. Topic: State Variable Model and Solution of State Equation of LTI Systems (c)  We have ⎡ s 0 ⎤ ⎡1 0 ⎤ ⎡s - 1 0 ⎤ -⎢ ( sI - A) = ⎢ = ⎢ ⎥ ⎥ ⎥ ⎣0 s ⎦ ⎣1 1 ⎦ ⎣ -1 s - 1⎦

Therefore, the transfer function is abcd 1- (be + cf + dg ) + bedg 29. Topic: Transient and Steady-State Analysis of LTI Systems (a) 

Therefore, ⎡s - 1 0 ⎤ ⎡ 1 ⎢ +1 s - 1⎥ ⎢ ⎦ = ⎢ s -1 ( sI - A) -1 = ⎣ ( s - 1) 2 ⎢ +1 ⎢ ( s - 1) 2 ⎣

⎤ 0 ⎥ ⎥ 1 ⎥ s - 1⎦⎥

Now, ⎡ et L-1 [ sI - A ]-1 = e At = ⎢ t ⎣te

0⎤ ⎥ et ⎦

d 2 y(t ) 3dy(t ) + + 2 y (t ) = x (t ) dt dt 2

Taking Laplace transform on both sides, we get s2Y(s) + 3sY(s) + 2Y(s) = X(s) For x(t) = 2u(t), X(s) = 2 s This gives the output as 2 s( s + 2)( s + 1) 2 A B C = + + s( s + 2)( s + 1) s s + 2 s + 1 Y ( s) =

Now, x(t) = eAt[x(t0)] ⎡ et or,      ⎢ t ⎣te

0 ⎤ ⎡1 ⎤ ⎡ et ⎤ ⎥ ⎢ ⎥=⎢ ⎥ e t ⎦ ⎣0 ⎦ ⎣te t ⎦

26. Topic: Transfer Function (b)  For the given transfer function, ArgG(s)H(s)s=jw = -180° + tan-1 w At the phase cross-over frequency w = wp ArgG (s)H (s ) s = jw = − 180° + tan −1(wp ) p

= -180° Therefore, phase cross-over frequency, wp = 0 rad/s Now, G(s) H(s) at w = wp = ∞

Therefore, 2 = A(s + 2)(s + 1) + Bs(s + 1) + C(s + 2)s By substituting s = 0, -1 and -2, one at a time, we find A = 1, B = 1 and C = -2. This gives Y ( s) =

⎛ 0 +1+ 3 ⎞ , 0⎟ = ( -1.33, 0) ⎜⎝ ⎠ 3-0 28. Topic: Signal Flow Graph (c)  There is only one forward path with gain equal to abcd. Therefore, P1 = abcd Δ = 1 − (be + cf + dg) + bedg Δ1 = 1

Ch wise GATE_ECE_CH06_Control Systems.indd 417

1 1 2 + s s + 2 s +1

y(t ) = (1 + e -2t - 2e - t )u(t ) 30. Topic: Bode and Root-Locus Plots (c)  As is evident from the Bode magnitude plot, there is one zero at w = 1 and one pole each at w = 10 and 100. Therefore, the transfer function is given by

Therefore, gain margin = 1/∞ = 0 27. Topic: Bode and Root-Locus Plots (c)  The point of intersection of the asymptotes of the root loci with the real axis is centroid. The center of asymptotes is given by

417

K ( s + 1) ( s + 10)( s + 100) For w = 0, we have 20 log10 10−3K = −20 dB or       K = 100 Therefore, the transfer function is 100( s + 1) ( s + 10)( s + 100) 31. Topic: Routh-Hurwitz and Nyquist Stability Criteria (a)  Poles at 0.01 and 1 Hz give -180° phase. Zero at 5 Hz gives +90° phase. Therefore, at 20 Hz, the phase shift is approximately -90°.

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32. Topic: Routh-Hurwitz and Nyquist Stability Criteria (a)  The characteristic equation for the system can be written as 1 + G(s ) = 0 4

3

Therefore, ⎡ -3⎤ AT C T = ⎢ ⎥ ⎣ -1⎦ Hence,

2

⎡1 -3⎤ QO = ⎢ ⎥≠0 ⎣0 -1⎦

s + 4 s + 5 s + 6s + K = 0

or,

The Routh–Hurwitz array for the system is s4

1

5

K

s3

4

6

0

s

s1

7/2 21−4K

2/7

s

K

2

0

Since [QO ] is non-singular, the system is observable. 34. Topic: State Variable Model and Solution of State Equation of LTI Systems (b)  We have ⎡ s 0 ⎤ ⎡1 0 ⎤ [ sI - A] = ⎢ ⎥-⎢ ⎥ ⎣0 s ⎦ ⎣0 1⎦

For the system to be stable,

⎡s - 1 0 ⎤ [ sI - A] = ⎢ s - 1⎥⎦ ⎣ 0

21 − 4 K > 0 and        K > 0 Combining both the conditions, we get, 21 > K >0 4 33. Topic: State Variable Model and Solution of State Equation of LTI Systems (d)  We have ⎡ x1 ⎤ ⎡ -3 -1⎤ ⎡ x1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = ⎢ 2 0 ⎥ ⎢ x ⎥ + ⎢0 ⎥ u ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎡ x1 ⎤ ⎡1 ⎤ y = [1 0] ⎢ ⎥ + ⎢ ⎥ u ⎣ x2 ⎦ ⎣ 0 ⎦ QC = [ B

AB ]

⎡ -3⎤ AB = ⎢ ⎥ ⎣ 2⎦ Therefore, ⎡1 -3⎤ QC = ⎢ ⎥ ⎣0 2 ⎦ Since [QC ] is non-singular and has a rank of 2, the system is controllable. QO = [C TATC T ] where ⎡1⎤ C =⎢ ⎥ ⎣0⎦ ⎡ -3 2⎤ AT = ⎢ ⎥ ⎣ -1 0 ⎦ T

Ch wise GATE_ECE_CH06_Control Systems.indd 418

e At = L-1 [ sI - A]-1 Thus,

e At

⎡ 1 ⎢ s -1 =⎢ ⎢ 0 ⎢⎣

⎤ 0 ⎥ ⎡et ⎥=⎢ 1 ⎥ ⎣0 s - 1⎥⎦

0⎤ ⎥ et ⎦

35. Topic: Feedback Principle (a)  Despite the presence of negative feedback, control systems still have problems of instability because the components used have non-linearities. 36. Topic: State Variable Model and Solution of State Equation of LTI Systems (a), (b)  The Eigen values of A = [ λ ] The Eigen values of W = [ μ ] The Eigen values of a system are always unique. So, [l] = [ μ ] However, a system can be represented by different state models having different set of state variables X=W X≠W Both conditions are possible. 37. Topic: Transient and Steady-State Analysis of LTI Systems (a)  The steady-state error is 0.05 =

1 20

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Chapter 6  •  Control Systems



Also the steady-state error is 1/Kv where Kv is the velocity error constant. Now, Kv is finite for a ramp input for a Type ‘1’ system. The steady-state error of 0.05 implies a gain of 1 = 20 0.05

For phase-cross over frequency, w = wp, we have Arg G( jw ) w = w = −180° p

p

38. Topic: Bode and Root-Locus Plots (b)  The system is stable in region -0.2 to -2 and on the left side of -8 as number of encirclements there is zero (see the following figure).

(−2wp ) − 2 − ta n −1 + 2wp + ta n −1

wp

w=0

= −180° = −p

2 wp

=

2

æw (w / 2) 3 2wp + ç p − p + ç 2 3 è

Im

419

p 2

ö p ÷= ÷ 2 ø

Using, for x < 1, tan−1 x = x −

x3 x 5 x7 + − + 3 5 7

We neglect the higher order terms to get −8

5wp

Re

2

−2 −0.2 æ w2 wp ç 5 − p ç 12 è w=0

0.2 K < 1 ⇒ K < 5 and 2K > 1 ⇒ K > 0.5

G( j w p ) = a =

Therefore, 0.5 < K < 5

=

1 8K < 1 or K < 8 39. Topic: Routh-Hurwitz and Nyquist Stability Criteria (d)  For the given transfer function, we have

If

3e−2 jw jw (2 + jw)

−2 j w 1

3e

=

3 w1 w12 + 4

w14 + 4w12 − 9 = 0 w14 = 1. 6 Þ w1 = 1. 26 rad /s

Ch wise GATE_ECE_CH06_Control Systems.indd 419

p 2

3

wp2

(w

2 p

+ 4

)

3 2

( 0. 632) + ( 0. 632) 2 + 4

Gain margin = 20 log10

= 2. 263

1 a 1 = −7.09 dB 2.263

At gain cross-over frequency, w = w1

then, w1 is gain-cross over frequency. Therefore,

jw1(2 + jw1

24

=

ö ÷ = p Þ w p ≈ 0.632 rad /s ÷ ø

= 20 log10

G( jw1) = 1,

G( jw1 ) =

wp3

40. Topic: Routh-Hurwitz and Nyquist Stability Criteria (d)  At phase cross-over frequency, w = wp

Now,

G( jw ) =

-

=1

ÐGH ( jw1) = -2w1 −

p æw ö − tan −1 ç 1 ÷ 2 è 2 ø

Substituting values, we have p − t an −1( 0. 63) 2 = −2.52 - 1.57 − 0.562

Arg GH ( jw1) = -2 × 1. 26 −

= −4.662 rad = − 267. 5

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GATE ECE Chapter-wise Solved papers

Then phase margin is given by

Now, x (t) = sintu(t)

f PM = 180 + Arg GH ( j w1)

which gives

= 180 + (−267. 5 ) = −87. 5 41. Topic: Bode and Root-Locus Plots (c)  We have 1 + G(s)H(s) = 0 Now, K=

s 2 + 3s s -1

For breakaway and break-in points, we have dK =0 ds ⇒ - s 2 + 2s + 3 = 0 ⇒ s 2 - 2s - 3 = 0 ⇒ ( s - 3)( s + 1) = 0

y(t) =

44. Topic: Routh-Hurwitz and Nyquist Stability Criteria

(d)  Given that G ( s) =

Since it is a unity feedback control system, H(s) = 1. K Therefore, G ( s) ⋅ H ( s) = ( s + 1)( s + 2)

Phase cross-over frequency is given by - tan -1

wπ w - tan -1 π = -180° 1 2



This gives wp = ∞



Now open loop gain G ( s) ⋅ H ( s) at w = wp is given by K

s = 3, -1

42. Topic: Lag, Lead and Lag-Lead Compensation

wπ + 1 ⋅ wπ 2 + 4 2



This gives s = s + jw = −5 cos 60° + j5 sin 60° = −2.5 + j4.33 For compensating the lag network (K/s2), a lead network is required. Options given in options (b) and (d) are lag networks and the one given in option (c) is invalid. With option (a), the transfer function becomes K ( s + 3) s ( s + 9.9) 2

Substituting s = −2.5 + 4.33j in the transfer ­function, we get the phase angle as 53.3°. 43. Topic: Transient and Steady-State Analysis of LTI Systems (a)  y(t) = x(t) * h(t) which gives Y(s) = X(s)⋅H(s) H ( s) =

1 1 π 1 = ∠ - 45° = ∠s +1 4 2 2

Ch wise GATE_ECE_CH06_Control Systems.indd 420

=0

Therefore, gain margin 1 1 = =∞ G (s) ⋅ H (s) 0

(a)  x = 0.5; cos−10.5 = 60°. Therefore, q = 60°. Now, wn = 5 and q = 60°

K ( s + 1)( s + 2)



Therefore, where -1 is the breakaway point and 3 is the break-in point.

⎛ π⎞ sin ⎜ t - ⎟ ⎝ 4⎠ 2

1

45. Topic: Transient and Steady-State Analysis of LTI Systems (c)  1 G1 ( s) = 2 s + as + b G2 ( s) =

s s 2 + as + b

The 3-dB bandwidth is given by the determinant of the characteristic equation. The determinants in both cases are equal to

a 2 - 4b .

46. Topic: Transfer Function 1 1 2 = (b)  C ( s) = s s + 2 s( s + 2) R( s) =

1 s

Therefore, the transfer function is C ( s) 2 = R( s) s + 2

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Chapter 6  •  Control Systems

47. Topic: Transient and Steady-State Analysis of LTI Systems (c)  h(t) = e-t Therefore, 1 H ( s) = s +1 1 R( s) = s The output is 1 1 ⋅ s +1 s 1 A B = + s ( s + 1) s s + 1

By definition, G(j w) w = w = 1 , therefore, 1

(αw1 )

+1

w

Therefore, w1 = 21/ 4 or, α =

=1

1 = 0.84 21/ 4

50. Topic: Routh-Hurwitz and Nyquist Stability Criteria (c)  Using the value of a from the above solution, the transfer function cane be written as G(s ) =

0. 84s + 1 s2

Unit-impulse response is given by: c ( t ) = L−1[C (s) ] = 0. 84 u (t) + t

which gives Substituting s = 0 and -1 one at a time, we get

51. Topic: Routh-Hurwitz and Nyquist Stability Criteria

A = 1 and B = -1

(b)  The characteristics equation for the system can be written as

This gives 1 1 s s +1

1 + G(s ) = 0 or,

which gives y(t) = (1 - e-t)u(t)



For t = ∞,



p = t an −1 w1a 4

w1a = 1

Ch wise GATE_ECE_CH06_Control Systems.indd 421

K+2

s

a

K+1

s1

a(K + 2 ) − K − 1 a

0

s0

K+1

-w2a + K + 1 = 0 Substituting w = 2 rad/s, we get K + 1 = 4a

(1)

Also, a(K + 2) − K − 1 =0 a

= 180° + ta n −1 w1a − 180° Therefore ,

1

or, from the array above

1

s3

as 2 + K + 1 = 0

49. Topic: Routh-Hurwitz and Nyquist Stability Criteria (c)  The phase margin is given by

s + a s + s ( K + 2) + K + 1 = 0

For oscillations,

Here, the gain magnitude at phase crossover ­frequency is a = 1. Therefore, the gain margin is 20 log10 1 = 0 dB

2

2

48. Topic: Routh-Hurwitz and Nyquist Stability Criteria (d)  The gain margin ⎛ 1⎞ 20 log10 ⎜ ⎟ dB ⎝ a⎠

3

The Routh–Hurwitz array for the system is

y(t) = 1

f PM = 180° + Arg GH ( jw ) w = w

[∵ L [d ( t )] = 1]

At t = 1, c ( t ) = 1. 84

1 = A(s + 1) + Bs

or,

2

2 1

Y(s) =

Y(s) =

421

or, a(K + 2) = K + 1 (2)

Solving Eqs. (1) and (2), we get K = 2, a = 0.75

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H( f ) =

52. Topic: Lag, Lead and Lag-Lead Compensation (d)  The phase shift is

Therefore, H ( s) =

f = tan 3wT − tan wT −1

−1

For maximum phase shift,

1 1 A B × = + s s + (1/ 5) s s + (1/ 5)

T 3T = Therefore, 2 1 + (3T w ) 1 + (T w ) 2

1⎞ ⎛ Therefore, A ⎜ s + ⎟ + Bs = 1 ⎝ 5⎠

2

3 + 3T 2w 2 = 1 + 9T 2w 2 , w T =

1

Substituting s = 0 and −1/5, one at a time, we get A = 5, B = −5. This gives

3

Substituting the value of wT in the expression for f, we get ⎛ 1 ⎞ 1 ϕ max = tan -1 ⎜ 3 × ⎟ - tan -1 ⎝ 3⎠ 3

Y ( s) =

56. Topic: Bode and Root-Locus Plots (d) 

53. Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  We have

G ( s) =

ϕ (t ) = L-1 [ sI - A]-1 -1

⎡ -1⎤ 0⎤ ⎡ 0 1⎤ ⎤ -1 ⎡ s -1 ⎡ s -⎢ = L ⎢⎢ ⎥ =L ⎢ ⎥ ⎥ ⎥ ⎣1 s ⎦ ⎣ ⎣0 s ⎦ ⎣ -1 0 ⎦ ⎦

K × 20 K = s(1 + s)[1 + ( s / 20)] s( s + 1)( s + 20)

At w = 0.1, from Bode plot in (1 + sT) form, we can write 20 log10(K/w) = 60

-1

This gives K = 5. Therefore,

That is,

G(s) =

⎡ s ⎢ 2 L [SI - A] = ⎢ s + 1 ⎢ -1 ⎢⎣ s 2 + 1 -1

⎡ cos t = ⎢ ⎣ - sin t

1 ⎤ s + 1⎥ ⎥ [ D = s 2 + 1] s ⎥ s 2 + 1⎥⎦ 2

55. Topic: Transient and Steady- State Analysis of LTI Systems (b)  5 H( f ) = 1 + j10π f Therefore, 5 5 1 = = H ( s) = 1 + 5s 5[ s + (1/ 5)] s + (1/ 5) (by suubstituting j 2π f = s)

100 s( s + 1)(1 + 0.05s)

57. Topic: Bode and Root-Locus Plots (d)  The transfer function is G ( s) 1+ G ( s)

sin t ⎤ cos t ⎥⎦

54. Topic: Routh-Hurwitz and Nyquist Stability Criteria (d)  The given transfer function T(s) has one zero at s = 5. Thus, the system will have a zero on the on right half s-plane. So it is non-minimum phase system because a minimum phase system will not have any poles or zeros in right of s- plane

Ch wise GATE_ECE_CH06_Control Systems.indd 422

5 5 s s + (1/ 5)

⇒ y(t ) = 5[1 - e - t / 5 ]u(t )

π π π = - = 3 6 6

-1

5 5 1 = = 1 + 5s 5[ s + (1/ 5)] s + (1/ 5) (by suubstituting j 2π f = s)

The step response is

dϕ =0 dw

3 ⎡⎣1 + (T w ) 2 ⎤⎦ = 1 + (3T w )

5 1 + j10π f



as H ( s) = 1 . For the point s = - 1 + j1 to lie on root locus, we have 1 + G(s) = 0 ⇒ 1+

K =0 s( s + 7 s + 12) 2

⇒ s(s2 + 7s + 12) + K = 0 Substituting s = - 1 + j, we get (-1 +j)(1 -2j - 1 - 7 + 7j + 12) + K = 0 ⇒ K = +10

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sI a ( s) = -w ( s) - 10 I a ( s) + 10U ( s)

w ( s) = ( -10 - s) I a ( s) + 10U ( s) = ( -10 - s)6  ( s• +Control ( s) 1)w ( s) + 10U Chapter Systems = -( s 2 + 11s + 10)w ( s) + 10U ( s) 58. Topic: Lag, Lead and Lag-Lead Compensation (b)  K v = lim sG ( s) H ( s) s→0

1000 = lim s ×

( K p + K d s)100 s( s + 10)

s→0

Therefore, Kp = 100 Now characteristic equation 1 + G(s)H(s) = 0, which gives ( K p + K d s)100 =0 1+ s( s + 10) Substituting Kp = 100, we get s2 + 10s + 104+ 100Kds = 0 s2 + (10 + 100Kd)s + 104 = 0 Comparing with standard second-order equation, s2 + 2xwns + wn2 = 0 we get wn = 100; 2xwn = 10 + 100Kd Given x = 0.5; therefore, 2 × 0.5 × 100 = 10 + 100Kd That is, Kd = 0.9 59. Topic: Lag, Lead and Lag-Lead Compensation (a)  The lead compensator C(s) should first stabilize the plant. That is, it should remove the term 1/( s - 1) . The only option that satisfies this condition is the one given at (a).

w ( s)( s 2 + 11s + 11)w ( s) = 10U ( s) 10 w ( s) = 2 U ( s) ( s + 11s + 11) 61. Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  The sum of the Eigen values is the trace of the principle diagonal matrix. Therefore, the sum is -3. Only option (a) satisfies both conditions. 62. Topic: State Variable Model and Solution of State Equation of LTI Systems (d)  The multiplication of the Eigen values is the determinant of the matrix. Therefore, from given options, it seems determinant should be ±2. Only option (d) satisfies since the determinant is 2. 63. Topic: Transient and Steady-State Analysis of LTI Systems (c)  Overshoot is given by ⎡ ⎛ -πξ ⎞ ⎤ exp ⎢ - ⎜ ⎟⎥ ⎢ ⎝ 1 - ξ2 ⎠ ⎥ ⎣ ⎦ where z is the damping constant given by z = cos q and q is the angle made with respect to the real axis. For overshoot to be the same, damping constant (z ) should be same for all poles. This implies that (cos q) should be the same which is possible only in the figure given in option (c).

60. Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  We have

.

⎡ dw ⎤ ⎢ dt ⎥ ⎥ ⎢ ⎢ dia ⎥ ⎢⎣ dt ⎥⎦

This gives di dw = -w + ia and a = -w - 10ia + 10u dt dt Taking Laplace transforms of above equations, we get sw(s) = -w(s) + Ia(s) or, (s + 1) w(s) = Ia(s) Also, sI a ( s) = -w ( s) - 10 I a ( s) + 10U ( s)

w ( s) = ( -10 - s) I a ( s) + 10U ( s) = ( -10 - s)( s + 1)w ( s) + 10U ( s) = -( s + 11s + 10)w ( s) + 10U ( s) 2

423

jw q

s

64. Topic: Bode and Root-Locus Plots (d)  Generalized transfer function of the control system corresponding to the given pole-zero plot is given by G ( s) =

s 2 + as + b s 2 + ps + q

This is the transfer function of a notch filter. 65. Topic: Transient and Steady-State Analysis of LTI Systems (c)  The transfer function is H ( s) =

K ( s + 1) ( s + 2)( s + 4)

w ( s)( s + 11s + 11)w ( s) = 10U ( s) 10 w ( s) = 2 U ( s) ( s + 11s + 11) 2

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Output C ( s) = R( s) H ( s) Given R( s) =

1 s

Therefore, K ( s + 1) s( s + 2)( s + 4)

C ( s) =

Also, steady state value =1. Therefore, lim s→0



s ⋅ K ( s + 1) =1 s( s + 2)( s + 4)

8( s + 1) A B = + ( s + 2)( s + 4) s + 2 s + 4

magnitude of frequency response = 10 / 3 is satisfied for only option (c). Hence transfer function in option (c) satisfies the given magnitude response frequency for both the values of w. 68. Topic: Routh-Hurwitz and Nyquist Stability Criteria (c)  Given the transfer function

A and B can be solved to be equal to -4 and 12, respectively. Therefore, -4 12 + H ( s) = s+2 s+4 h(t ) = ( -4e -2t + 12e -4 t )u(t )



or,



This is also the impulse response of the system.

66. Topic: Transient and Steady-State Analysis of LTI Systems (d)  For P=

25 , ζ = 0, w n = 5 s + 25 2

Here the damping ratio = 0, so the transfer function corresponds to a sinusoidal function, so P corresponds to the graph in option (3). 36 20 > 1, w n = 6 ,ζ = 2×6 s + 20 s + 36 2

Here the damping ratio greater than 1, so transfer function corresponds to over-damped case represented by hyperbolic function, so Q corresponds to graph in option (4). 36 , ζ = 1, w n = 6 s 2 + 12 s + 36 Here the damping ratio is equal to 1, so transfer function corresponds to a critically damped case, thus R corresponds to graph represented in option (1).



s + as − 4 G(s ) 1 + G(s )

Then closed loop transfer function =



The characteristic equation can be written as s2 + a s − 4 + s + 8 = 0 s 2 + (a + 1) s + 4 = 0



This will be stable if (a + 1) > 0 and real.

69. Topic: Routh-Hurwitz and Nyquist Stability Criteria (c)  The characteristic equation for the given transfer function, 1 + G(s) = 0 5

s + 2s + 3s 3 + 6s 2 + 5s + 13 = 0



4

The Routh–Hurwitz table for the system is s5

1

3

5

s

2

6

13

4

s3

0 →K

−3 2

0

s2

6K + 3 K

13

0

s

−3 2

0

For, R=

2



or, Q=

s+8

G(s ) =

For,



Here, the damping ratio is less than 1, so the t­ransfer function corresponds to an under-damped case, thus S corresponds to graph represented in option (2).

67. Topic: Transient and Steady-State Analysis of LTI Systems (c)  Substituting w = 0, satisfies the given magnitude condition of frequency response = 5, for all the given transfer functions. However, for w = 3 2 rad/s, the

This gives K = 8. So, H ( s) =





s0

13

For, S=

49 1 , ζ =  , w n = 7 2 s 2 + 7 s + 49

Ch wise GATE_ECE_CH06_Control Systems.indd 424

 6K + 3  As K → 0+, lim +   = + ve K →0  K 

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Chapter 6  •  Control Systems



Since, there are two changes in the sign, it implies that there are two positive poles in the open right hand side plane.

70. Topic: Lag, Lead and Lag-Lead Compensation

73. Topic: State Variable Model and Solution of State Equation of LTI Systems (c)  We have

(b)  As the first step, we shall find the transfer function of the given circuit. V Vi ( R1C1 s + 1)1 = - o R1 Z Therefore,

⎡1 0 ⎤ ⎡ p⎤ A= ⎢ and B = ⎢ ⎥ ⎥ ⎣0 1⎦ ⎣q⎦ For controllability, the following condition should be met: QC = ⎣⎡ B

Vo Z ( R1C1 s + 1) =Vi R1

Case Q: Z =

R2C2 s + 1 C2 s

R2 Case R: Z = R2C2 s + 1 The transfer functions in these two cases are therefore given as under: Case Q: Z=

( R C s + 1) ( R2C2 s + 1) R2C2 s + 1 Vo = - 1 1 ⋅ , Vi R1 C2 s C2 s

Z=

R2 V R2 ( R C s + 1) , o = - 1 1 ⋅ R2C2 s + 1 Vi R1 ( R2C2 s + 1)

Case R:

It is given that R2C2 > R1C1. By examining the transfer functions in the case of Q and R, it is obvious that R corresponds to a lag controller (note the pole being closer to origin than the zero) and Q corresponds to a PID controller (note the transfer function being sum of a proportional term, a derivative term and an integral term). 71. Topic: State Variable Model and Solution of State Equation of LTI Systems (d)  It is self-explanatory. 72. Topic: Lag, Lead and Lag-Lead Compensation (d)  A lag compensator produces a low-pass filter like gain versus frequency response with a flat response at lower frequencies and the gain beginning to fall from pole frequency and eventually becoming zero at zero frequency. A lead compensator produces a high-pass filter like gain versus frequency response with gain rising from zero frequency and eventually becoming flat from pole frequency. If p2 > z2 > z1 > p1 (p1 and z1 are lag compensator pole and zero; p2 and z2 are lead compensator pole and zero), the answer is obvious.

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425

AB  An -1

B ⎤⎦ ≠ 0

⎡1 0 ⎤ ⎡ p ⎤ ⎡ p ⎤ AB = ⎢ ⎥⎢ ⎥ =⎢ ⎥ ⎣0 1⎦ ⎣ q ⎦ ⎣ q ⎦ This gives ⎡p QC = ⎢ ⎣q

p⎤ ⇒⎮QC⎮= 0 q ⎥⎦

So, the system is uncontrollable for all values of p and q. 74. Topic: Routh-Hurwitz and Nyquist Stability Criteria (b)  The Nyquist plot has one encirclement of origin in the clockwise direction. Therefore, G(s) has one zero in the right half plane. 75. Topic: Routh-Hurwitz and Nyquist Stability Criteria (c)  The gain margin is ⎛ 1⎞ ⎛ 1 ⎞ 20 log10 ⎜ ⎟ = 20 log10 ⎜ = 6 dB ⎝X⎠ ⎝ 0.5 ⎟⎠ The phase margin is 90°. 76. Topic: Bode and Root-Locus Plots (b)  We have 1 + G(s)H(s) = 0 Now, K ( s 2 - 2 s + 2) =0 s 2 + 2s + 2 s 2 + 2s + 2 K=- 2 s - 2s + 2

1+ or,

Substituting ∂K / ∂s = 0, we have 2( s 2 - 2 s + 2)( s + 1) - 2( s 2 + 2 s + 2)( s - 1) = 0 That is, 2s 2 - 4 s 2 + 4 = 0 Therefore, s=± 2

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The angle of departure is given by fD = 180° + f where

For a unit step input, we have R( s) =

f = ΣfZ - ΣfP Now, f = 135° and therefore fD = 180° + 135° = 315°. Departure angle with positive real axis = 45°. 77. Topic: Transfer Function (b)  The feedback in this block schematic is zero as the output of the output summing point is zero. A zero feedback implies no feedback. Therefore, closed loop transfer function is same as the open loop gain, which is 1 s +1 78. Topic: Transient and Steady-State Analysis of LTI Systems (b)  The phase difference between the input and the output is

ϕ=-

π ⎛ π⎞ π - ⎜ - ⎟ = = 30° 3 ⎝ 2⎠ 6

and w = 2 rad/s From the transfer function,

ϕ = 90° - tan -1

w p

2 = 30° p

or, 90° - tan -1

2 3

K [1 + ( s / 0.1)] 1 + ( s / 10)

Here, 20logK = 0 or, Therefore

K=1 G ( s) H ( s) =

10 s + 1 0.1s + 1

80. Topic: Lag, Lead and Lag-Lead Compensation (d)  The steady-state error is ess = lim s→0

Ch wise GATE_ECE_CH06_Control Systems.indd 426

s × (1/s) 1 + G ( s)Gc ( s)

ess = lim

1 1 + G ( s)Gc ( s)

s→0

For the given value of G(s) and the different values of Gc(s), one at a time, we find that for ⎛ 2⎞ Gc ( s) = 1 + ⎜ ⎟ + 3s ⎝ s⎠ we get steady-state error of zero. For all other cases, we get finite non-zero values of steady-state error. 81. Topic: State Variable Model and Solution of State Equation of LTI Systems (b)  It is self-explanatory. 82. Topic: State Variable Model and Solution of State Equation of LTI Systems (c)  The forward path gain is 1 ⎛ 1⎞ ⎛ 1⎞ p1 = 2 ⎜ ⎟ ⎜ ⎟ (0.5) = 2 ⎝ s⎠ ⎝ s⎠ s 1 ⎛ 1⎞ p2 = 2 ⎜ ⎟ (1)(0.5) = ⎝ s⎠ s Δ1 = 1 , Δ 2 = 1,

79. Topic: Bode and Root-Locus Plots (a)  The system transfer function is G ( s) H ( s) =

ess = lim s→0

After simplification, p=

1 s

sR( s) 1 + G ( s)Gc ( s)

⎧ 1 1⎫ Δ = 1 - ⎨- - 2 ⎬ = 1 + 1 + 1 ⎩ s s ⎭ s s2 The transfer function of the system is Y ( s) P1 Δ1 + P2 Δ 2 = Δ U ( s) =

(1/ s 2 ) + (1/ s) s +1 = 1 + (1/ s 2 ) + (1/ s) s 2 + s + 1

83. Topic: Signal Flow Graph (d)  For option (d), we have Y ( s) 100 /s3 = U ( s) 1 + (100 /s 2 ) This is not the transfer function of H(s). 84. Topic: Feedback Principle 100 (c)  G ( s) H ( s) = s( s + 10) 2

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Chapter 6  •  Control Systems

Y1 ( s) X 1 ( s) 1 = X 1 ( s)U ( s) s + 2

⎛w⎞ ϕ = -90° - 2 tan -1 ⎜ ⎟ ⎝ 10 ⎠

Y2 ( s) X 2 ( s) 2 = X 2 ( s)U ( s) s + 2

For the phase cross-over frequency, f = −180°. Therefore, ⎛w⎞ -180° = -90° - 2 tan ⎜ ⎟ ⎝ 10 ⎠

Also,

-1

X 1 ( s) Y ( s) 1 = and 1 =1 U ( s) s + 2 X 1 ( s)

⇒ w = 10 rad/s

X 2 ( s) Y ( s) 1 = and 2 =2 U ( s) s + 2 U ( s)

Substituting, s = jw, we get |G (jw)⋅H (jw)|w = 10 =

100 100 1 = = 2 w (w + 100) 10( 200) 20

The gain margin is given by 1 = 20 | G ( jw ) H ( jw ) |

Therefore, sX1(s) + 2X1(s) = U(s) and Y1(s) = X1(s) sX2(s) + 2X2(s) = U(s) and Y2(s) = 2X2(s) Taking inverse Laplace transform, we get x2 (t ) + 2 x1 (t ) = u(t ) and y1(t) = x1(t)

Therefore, the required gain margin (in dB) is 20 log 20 = 26 dB 85. Topic: Routh-Hurwitz and Nyquist Stability Criteria (a)  G( j w ) = 5+j w

x2 (t ) + 2 x2 (t ) = u(t ) and y2(t) = 2x2(t) From the given data, we have y = [ y1

x2 (t ) = -2 x2 (t ) + u(t )

For w = 0, we get

⎡ x1 ⎤ ⎡ -2 0 ⎤ ⎡ x1 ⎤ ⎡1⎤ ⎢ x ⎥ = ⎢ -2 0 ⎥ ⎢ x ⎥ + ⎢1⎥ u(t ) ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2⎦ ⎣

G(0) = 25 + 0 = 5

G(∞) = ∞ 86. Topic: Bode and Root-Locus Plots (b)  From the plot, we can observe that one pole terminates at one zero at position (-1, 0). A root locus path between (−2, 0) and (−3, 0) is also justified as it is to the right of odd number of poles and zeros (two poles and one zero). The root locus path extending from (−3, 0) to ∞ implies that there are two poles at (-3, 0). 87. Topic: State Variable Model and Solution of State Equation of LTI Systems (b)  We have Y1 ( s) 1 = U ( s) s + 2 Y2 ( s) 2 = U ( s) s + 2

Ch wise GATE_ECE_CH06_Control Systems.indd 427

T ⎡1 ⎤ y2 ] = ⎢ ⎥ x ⎣ 2⎦

x1 (t ) = -2 x1 (t ) + u(t )

G ( jw ) = 25 + w 2

For w = ∞, we get

427

or,

x = [ -2] x + [1] u

From the above computation, we conclude that option (b) is the correct answer. 88. Topic: Transient and Steady-State Analysis of LTI Systems (c)  The transfer function is G ( s) =

( s 2 + 9)( s + 2) ( s + 1)( s + 3)( s + 4)

We have x(t) = sin wt, which gives X ( s) =

w s2 + w 2

Therefore, the output Y(s) is given by Y ( s) =

w ( s 2 + 9)( s + 2) ⋅ 2 ( s + 1)( s + 3)( s + 4) s + w 2

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The steady-state value can be calculated by ­determining the limiting value of y(t) at t = ∞. This is given by

⎛ x1 ⎞ y = (1 0 0 ) ⎜ x2 ⎟ ⎜ ⎟ ⎝ x3 ⎠

lim y(t ) = lim sY ( s) t →∞

s→0

= lim s s→0

( s 2 + 9)( s + 2) w ⋅ ( s + 1)( s + 3)( s + 4) ( s 2 + w 2 )

In the above equation, we have applied the final value theorem. This will be zero if (s2 + w2) cancels (s2 + 9) term as only then final value t­ heorem will be applicable. Therefore, w = 3 rad/s

Now, QC = [B AB A2B] Therefore, ⎡0 AB = ⎢⎢ 0 ⎢⎣ a3

89. Topic: Lag, Lead and Lag-Lead Compensation (a)  ( s + a) Gc ( s) = ( s + b)

There is only one option that satisfies this condition, that is, option (a). 90. Topic: Lag, Lead and Lag-Lead Compensation (a)  For the phase to be maximum, we should have ∂P =0 ∂w

which gives

1/a 1/ b =0 2 2 1 + (w /a ) 1 + (w 2 / b 2 ) That is, 1 1/ 2 =0 2 1 + w 1 + (w 2 / 4) 1 2 Therefore, - 2 =0 2 1+ w w +4 or w 2 + 4 - 2 - 2w 2 = 0 Therefore, w 2 = 2 or w = 2 rad/s. 91. Topic: State Variable Model and Solution of State Equation of LTI Systems (d)  We have ⎛ x1 ⎞ ⎛ 0 ⎜ x ⎟ = ⎜ 0 ⎜ 2⎟ ⎜ ⎝ x3 ⎠ ⎝ a3

Ch wise GATE_ECE_CH06_Control Systems.indd 428

a1 0 0

0⎞ a2 ⎟ ⎟ 0⎠

⎛ x1 ⎞ ⎛ 0⎞ ⎜ x ⎟ + ⎜ 0⎟ u ⎜ 2⎟ ⎜ ⎟ ⎝ x3 ⎠ ⎝ 1⎠

a1 0 0

0 ⎤⎡0 ⎤ a2 ⎥⎥ ⎢⎢ a2 ⎥⎥ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦

⎡ a1a2 ⎤ A2 B = ⎢⎢ 0 ⎥⎥ ⎢⎣ 0 ⎥⎦

⎛w⎞ ⎛w⎞ P = tan ⎜ ⎟ - tan -1 ⎜ ⎟ ⎝ a⎠ ⎝ b⎠ -1

w w > Therefore, a < b a b

0 0

0 ⎤ ⎡0 ⎤ ⎡ 0 ⎤ a2 ⎥⎥ ⎢⎢0 ⎥⎥ = ⎢⎢ a2 ⎥⎥ 0 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ 0 ⎥⎦

⎡0 A2 B = AAB = ⎢⎢ 0 ⎢⎣ a3

The phase is given by

For a lead compensator, the phase is positive. For this,

a1

Thus, ⎡0 0 ⎢ Qc = ⎢0 a2 ⎢⎣1 0

a1a2 ⎤ 0 ⎥⎥ 0 ⎥⎦

For a system to be controllable, |QC | ≠ 0 |QC | = (0 - a1a22) ≠ 0 ⇒ a1 ≠ 0 and a2 ≠ 0 92. Topic: Transient and Steady-State Analysis of LTI Systems (b)  We have 1 s Y(s) is therefore given by 1/s2. Therefore, x(t) = u(t) and X(s) =

y(t) = tu(t) In general, Laplace transform of tnu(t) is n! s n +1 93. Topic: Bode and Root-Locus Plots (b)  The gain decreases at 40 dB per decade (from 32 dB at w = 1 rad/s to − 8 dB at ­w = 10 rad/s). This implies that there are two poles at origin. It means either option (b) or option (d) is c­ orrect. Substituting w = 1 rad/s in both the options,we get

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Chapter 6  •  Control Systems

The time constant is

⎡ 39.8 ⎤ 20 log ⎢ 2 ⎥ = 32 dB ⎣ 1 ⎦

topen loop = 10 The desired closed loop time constant is

⎡ 32 ⎤ 20 log ⎢ 2 ⎥ = 30.1 dB ⎣1 ⎦

10 = 0.1 100 The closed loop transfer function is t=

So option (b) is the correct option. 94. Topic: Signal Flow Graph (a)  Using Mason’s gain formula, we get

10 K a / (1 + 10 s) 10 K a = [1 + (10 K a ) / (1 + 10 s)] 1 + 10 s + 10 K a

Δ = 1 - [ -2 s -1 - 2 s -2 - 4 - 4 s -1 ] Δ=

429

=

5s 2 + 6 s + 2 s2

Ka s + K a + 0.1

The time response of closed loop system is Va e−(Ka + 0.1)t

1 s2 1 P2 = s -1 = s P1 = s -2 =

The closed loop time constant is 1 K a + 0.1

Δ1 = 1

The closed loop time constant is required to be equal to 0.1. Therefore,

Δ2 = 1 Therefore,

1 = 0.1 K a + 0.1

Y ( s) P1 Δ1 + P2 Δ 2 = Δ U ( s) (1/s 2 ) + (1/s) s +1 = = 2 2 2 5s + 6 s + 2 (5s + 6 s + 2) /s 95. Topic: Transient and Steady-State Analysis of LTI Systems (b)  x(t) = u(t) - u(t - 2) Therefore, 1 e -2 s 1 - e -2 s X ( s) = = s s s Taking Laplace transform of system differential equation, we get Y(s)(s2 + 5s + 6) = X(s) That is, (1 - e -2 s ) / s 1 - e -2 s 1 - e -2 s = = Y ( s) = 2 s + 5s + 6 s( s 2 + 5s + 6) s( s + 2)( s + 3) 96. Topic: Lag, Lead and Lag-Lead Compensation (c)  The open loop transfer function is 10 1 = 1 + 10 s s + 0.1 The time response is w(t) = Va(t)e−0.1t

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0.1 K a = 1 - 0.01 = 0.99 K = 9.9 ≅ 10 97. Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  Refer to the following figure; .

1

−1 x1

x1

1 x2

.

−1 x2

1

u

y

1 s

1 s

Now, x1 = - x1 - u  and x2 = -( x2 + x1 ) = -( x2 - x1 - u ) x2 = x1 - x2 + u and y = x2 This gives y = x1 - x2 + u ⎡ x1 ⎤ ⎡ -1 0 ⎤ ⎡ x1 ⎤ ⎡ -1⎤ ⎢ x ⎥ = ⎢ 1 -1⎥ ⎢ x ⎥ + ⎢ 1⎥ u ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎡ -1 0 ⎤ ⎡ -1⎤ x = ⎢ x + ⎢ ⎥u ⎥ ⎣ 1 -1⎦ ⎣ 1⎦

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98. Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  We have ⎡ -1 0 ⎤ A= ⎢ ⎥ ⎣ 1 -1⎦

Substituting the value of initial conditions, we get sY(s) - y(0-) + 3Y(s) = sX(s) - x(0-) - 2X(s) Given that, y(0-) = -2, x(0-) = 0 as x(t) = -3e2t u(t) ⎛ -3 ⎞ Therefore, sY ( s) + 2 + 3Y ( s) = ( s - 2) ⎜ ⎝ s - 2 ⎟⎠

⎡s + 1 0 ⎤ sI - A = ⎢ ⎥ ⎣ -1 s + 1⎦ ⎡s + 1 0 ⎤ s + 1⎥⎦ ⎣ 1

1

[ sI - A]-1 = ( s + 1) × ( s + 1) ⎢

Hence, Y ( s) = -

102. Topic: Transient and Steady-State Analysis of LTI Systems (c)  Transfer function

The state transition matrix f(t) is given by eAt = L-1[sI - A]-1 0⎤ ⎥ e -t ⎦

Y ( s) 4 = 2 U ( s) s + 4 s + 4

99. Topic: Block Diagram Representation (c)  Figure below shows the signal flow graph for the given block diagram. G1

G2

5 s+3

Therefore, y(t)   =   -5e-3tu(t). Hence, the steady-state value is y(∞) = 0

0 ⎤ ⎡ 1 / ( s + 1) [ sI - A] = ⎢ 2 1 s + 1 1 ( s + 1) ⎥⎦ / ( ) / ⎣ -1

⎡ e -t e At = ⎢ - t ⎣te

Therefore, sY(s) + 3Y(s) = sX(s) - 2X(s)

If we compare the above equation with standard second order system transfer function, that is,

w n2 s 2 + 2ξw n s + w n2

1 C(s)

R(s) 1

We get,

1

w n2 = 4 ⇒ w n = 2 rad/s

1 From the above figure, we see that there are three parallel paths. The gains of these paths are given by P1 = G1G2, P2 = G2, P3 = 1 By Mason’s gain formula, C ( s) = P1 + P2 + P3 = G1G2 + G2 + 1 R( s) 100. Topic: Transient and Steady-State Analysis of LTI Systems (38.15)  Given that the natural frequency wn = 40 rad/s and x = 0.3. Therefore, the damped natural frequency

wd = wn 1 - ξ

103. Topic: Routh-Hurwitz and Nyquist Stability Criteria (2.25)  Given that: G ( s) =

K ( s + 2)( s - 1)

Given that the system is a unity feedback system. Therefore, H(s) = 1. Characteristics equation [1 + G(s)H(s)] = 0 Therefore, 1 +

K =0 ( s + 2)( s - 1)

Poles of the given system are

2

S1, 2 = -1 ±

= 40 1 - (0.3) 2

9 - 4K 4

= 38.15 rad/s 101. Topic: Transient and Steady-State Analysis of LTI Systems (0)  Y ( s) s - 2 = X ( s) s + 3

Ch wise GATE_ECE_CH06_Control Systems.indd 430

9 - 4 K = 0, then only both poles of the closed loop 4 system are at the same location. If

Therefore, K =

9 = 2.25 4

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Chapter 6  •  Control Systems

104. Topic: Routh-Hurwitz and Nyquist Stability Criteria (d)  For larger values of k, it will encircle the critical point (−1 + j 0) which makes closed-loop system unstable. 105. Topic: Bode and Root-Locus Plots (a)  In a Bode diagram, in plotting the magnitude with respect to frequency, a pole introduces a line with slope of −20 dB/decade.

108. Topic: Transient and Steady-State Analysis of LTI Systems 4 2 (0.5)  Given G ( s) = ; H ( s) = s+2 s+4 For unit step input K p = lim G ( s) H ( s) s→0

⎛ 4 ⎞⎛ 2 ⎞ = lim ⎜ ⎟⎜ ⎟ =1 s→0 ⎝ s + 2 ⎠ ⎝ s + 4 ⎠

Therefore, 4th order all-pole system gives a slope of 4 × (−20 dB/decade), that is, −80 dB/decade. 106. Topic: Transient and Steady-State Analysis of LTI Systems (c)  By observing the options, if we put any option other than (c), the characteristic equation will have third order terms, where we cannot describe the setting time. If G(s) = 2(s + 4) is considered, then the characteristic equation is s2 + 3s + 2 + 2s + 8 = 0 s 2 + 5s + 10 = 0

or,

Steady-state error ess =

=

10 (1 + 10 s)(1 + s)(1 + 0.1s)

By approximation;

we get

⎛ 10 ⎞ G ( s) = ⎜ ⎝ 10 s + 1⎟⎠

w n = 10; ξw n = 2.5 2

Phase margin = θ = 180° + ∠GH at w = wgc At w gc , G ( s) = 1 Therefore, 10 =1 100w gc2 + 1

4 2% setting time = ξw n Given that 2% setting time 2 ξw n

Hence, w gc = 0.9949 rad/s w gc = 0.9949 rad/s ⎛ 10 × 0.99 ⎞ = 180margin ° - tan -1 ⎜ = 180° -⎟ tan -1 ⎛ 10 × 0.99 ⎞ Therefore, phase ⎜⎝ ⎟⎠ ⎝ ⎠ 1 1     = 98.73°

107. Topic: State Variable Model and Solution of Stable Equation of LTI Systems (b)  Applying the linearity principle, we get

⎡ e -t ⎤ ⎡e - t - e -2t ⎤ x (t ) = 3 ⎢ - t ⎥ + 8 ⎢ - t -2 t ⎥ ⎣ -e ⎦ ⎣ - e + 2e ⎦ ⎡11e - t - 8e -2t ⎤ x (t ) = ⎢ -t -2 t ⎥ ⎣ -11e + 16e ⎦

Ch wise GATE_ECE_CH06_Control Systems.indd 431

1 1 = = 0.5 1+1 2

109. Topic: Bode and Root-Locus Plots (98.73)  Given that 10 G ( s) = ( s + 0.1)( s + 1)( s + 10) 10 = s ⎞ s⎞ ⎛ ⎛ 0.1 ⎜1 + (1 + s) ⎜1 + ⎟ .10 ⎝ 10 ⎠ ⎝ 0.1⎟⎠

s2 + 2xwns + wn2 = 0

Therefore, a = 3; b = 8 Hence,

1 1+ K p

ess =

Comparing with the standard characteristic equation:

⎡0 ⎤ ⎡3⎤ ⎡ 1⎤ ⎢5⎥ = a ⎢ -1⎥ + b ⎢ 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

431

110. Topic: Bode and Root-Locus Plots (0.50) 

G(jw )

(−20 dB/decade)

26.02

dB 6.02 −6.02

20 0.1

1

2

10

w (rad/s)

(−20 dB(decade))

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From the figure it is clear that the initial slope is -20 dB/ decade. Due to initial slope, it is a type-1 system, and it has non-zero velocity error coefficient (kv) From the magnitude plot, the magnitude is 0 dB at w = 2 rad/s. Therefore, velocity error coefficient, kv = 2

113. Topic: Bode and Root-Locus Plots (b) The root locus plot of transfer s+4 is shown below ( s + 1)( s + 2)( s + 7)

function

jw A . The steady state error ess = kv Given that the input is a unit ramp signal, therefore, A = 1 and ess =

−7 −4 −2

1 = 0.5 2

111. Topic: Routh-Hurwitz and Nyquist Stability Criteria 1 (a)  Given that H ( s) = . Therefore, h(t) = e-t u(t). s +1 Hence, the system is stable. Therefore, statement S1 is true. h(t +1) is As h(t) is absolutely integrable, therefore, h(t )

114. Topic: Bode and Root-Locus Plots (0.375)  We know that the co-ordinate of point A of the given root locus, that is, magnitude condition |G(s)H(s)| = 1 Here, the damping factor x = 0.5 and the length of OA = 0.5.

independent of time for t > 0. Therefore, statement S2 is also true.

A

1 . For a non-causal s +1 system, h(t ) = -e - t u( -t ) . This is not absolutely integrable, hence the system is unstable. Therefore, statement S3 is not true. The transform function is H(s) =

112. Topic: Routh-Hurwitz and Nyquist Stability Criteria (2)  Given that G p ( s) =

ps + 3 ps - 2 s 2 + (3 + p ) s + ( 2 - p ) 2

The characteristic equation is

θ

−1

By forming Routh−Hurwitz array, ( 2 - p) s2 1 1 s (3 + p ) 0 s 0 ( 2 - p)

Therefore, the maximum value of p unit which Gp remains stable is 2.

O

OX 0.5

Hence, OX =

1 4

Now, sin θ =

AX OA

Therefore,

and (2) (2 − p) > 0 ⇒ p < 2 that is, −3 < p < 2

−1/3 X

cos 60° =

For stability, first column elements must be positive and non-zero, that is (1) (3 + p) > 0 ⇒ p > −3

−2/3

Refer to the figure above. Then in the right angle triangle OAX OX cos θ = . OA Therefore,

s2 + (3 + p)s + (2 − p) = 0

Ch wise GATE_ECE_CH06_Control Systems.indd 432

s

−1

sin 60° = Hence,

AX 0.5 AX

sin 60° = 3 0.5 AX = 43 AX = 4

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Chapter 6  •  Control Systems

Hence, the co-ordinate of point A is ( -1/ 4 + j 3 / 4). Substituting the above value of A in the transfer function and equating to 1, we get K s( s + 1) 2

1 3 + 16 16

=1 s = -1/ 4 + j 3 / 4

2

⎛ 9 3⎞ .⎜ + ⎟ = 0.375 ⎝ 16 16 ⎠

115. Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  From the given signal flow graph, the state model is . ⎡ x1 ⎤ ⎡0 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ x. ⎥ = ⎢0 0 1 ⎥ ⎢ x ⎥ + ⎢0 ⎥ u ⎢ .2 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ x3 ⎥⎦ ⎢⎣ a3 a2 a1 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣1 ⎥⎦ Y = [c1

⎡ x1 ⎤ c3 ] ⎢⎢ x2 ⎥⎥ + Bu ⎢⎣ x3 ⎥⎦

c2

Comparing the above model with the standard state model, we get ⎡0 1 0 ⎤ ⎡0 ⎤ ⎢ ⎥ ⎢ A = ⎢0 0 1 ⎥ , B = ⎢0 ⎥⎥ ⎢⎣ a3 a2 a1 ⎥⎦ ⎢⎣1 ⎥⎦ Now, QC = ⎣⎡ B

AB

Matrix QC is given by QC = C = [ B AB A2 B] If QC ≠ 0, then the system is controllable.

Therefore, K=

A B ⎤⎦ 2

Therefore,

4 -8 ⎤ ⎡0 ⎢ QC = ⎢ 4 -4 4 ⎥⎥ ⎢⎣0 0 0 ⎥⎦ QC = 0 Hence, the system is uncontrollable. ⎡C ⎤ ⎢ QO = ⎢CA ⎥⎥ ⎢⎣CA2 ⎥⎦ If QO ≠ 0 , then the system is observable. 1⎤ ⎡ 1 1 ⎢ QO = ⎢ -1 0 -2 ⎥⎥ ⎢⎣ 1 1 4 ⎥⎦ Therefore, QO = 1 . Hence, the system is observable. Therefore, the given system is uncontrollable and observable. 117. Topic: State Variable Model and Solution of State Equation of LTI Systems (c)  The solution of state equation of x(t) = L−1 [sI − A] − 1 × x(0) ⎡ -1 0 ⎤ ⎡ 1⎤ x ( 0) = ⎢ ⎥ , A = ⎢ ⎥ ⎣ -1⎦ ⎣ 0 -2⎦

[ sI - A]

-1

⎡0 0 QC = ⎢⎢0 1 ⎢⎣1 a1

1 ⎤ a1 ⎥⎥ a2 + a12 ⎥⎦

| QC | = 1 ≠ 0

⎡s + 1 0 ⎤ =⎢ s + 2⎥⎦ ⎣0

⎡s + 2 0 ⎤ 1 = = s + 1⎥⎦ ( s + 1)( s + 2) ⎢⎣ 0

Hence, the system is always controllable. 116. Topic: State Variable Model and Solution of State Equation of LTI Systems (b)  From the given state space model; 0⎤ ⎡ -1 1 ⎡0 ⎤ A = ⎢⎢ 0 -1 0 ⎥⎥ , B = ⎢⎢ 4 ⎥⎥ , C = [1 1 1] ⎢⎣ 0 ⎢⎣0 ⎥⎦ 0 -2 ⎥⎦

Ch wise GATE_ECE_CH06_Control Systems.indd 433

433

-1

⎤ ⎡ 1 ⎢ s +1 0 ⎥ ⎢ ⎥ 1 ⎥ ⎢0 ⎢⎣ s + 2 ⎥⎦

⎤ ⎡ -1 ⎡ 1 ⎤ ⎢ L ⎢ s + 1⎥ 0 ⎥ ⎣ ⎦ ⎥ L-1 ⎡⎣( sI - A) -1 ⎤⎦ = ⎢ ⎢ ⎥ -1 ⎡ 1 ⎤ L ⎢ ⎢0 ⎥ ⎥ ⎣ s + 2 ⎦⎦ ⎣ ⎡e - t =⎢ ⎣0

0 ⎤ ⎥ e -2t ⎦

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Now,

we get ⎡ x1 (t ) ⎤ ⎡e - t ⎢ x (t ) ⎥ = ⎢ ⎣ 2 ⎦ ⎣0

K = w n2

0 ⎤ ⎡1 ⎤ ⎥⎢ ⎥ e -2t ⎦ ⎣ -1⎦

Therefore, x1 (t ) = e - t and x2 (t ) = -e -2t 118. Topic: State Variable Model and Solution of State Equation of LTI Systems (d)  Given state model,

or, w n = K 2ξw n = 10



10 2 K

Substituting x = 0.25 , we have

⎡ x1 ⎤ ⎡0 1 ⎤ ⎡ x1 ⎤ ⎢ x ⎥ = ⎢0 0 ⎥ ⎢ x ⎥ ⎦⎣ 2⎦ ⎣ 2⎦ ⎣

K =

10 0.5

Therefore, K = 400

⎡0 1 ⎤ Let A = ⎢ ⎥ ⎣0 0⎦

122. Topic: Bode and Root-Locus Plots

Let the state transition matrix be f(t). Therefore,

ϕ (t ) = L ⎡⎣( sI - A) ⎤⎦ -1

[ sI - A]

-1

Therefore,

ξ=

Therefore,

-1

-1

⎡ s -1⎤ 1 ⎡s 1⎤ = 2⎢ =⎢ ⎥ s ⎣0 s ⎥⎦ ⎣0 s ⎦

⎡1/s 1/s 2 ⎤ ⎡1 t ⎤ ϕ (t ) = L-1 ⎢ ⎥=⎢ ⎥ ⎣0 1/s ⎦ ⎣0 1⎦

(8970)  40 =

⎛ 300 ⎞ =1 Hence,     log10 ⎜ ⎝ f L ⎟⎠ Therefore, fL = 30 Hz -40 =

119. Topic: Transfer Function G1 ( s) (b)  G2 ( s)

40 - 0 log10 (300) - log10 ( f L )

log10

0 - 40 f H - log10 (900)

Therefore, fH = 900 × 10 = 9000 Hz

120. Topic: Signal Flow Graph (b)  The transfer function can be found using Mason's gain formula. 121. Topic: Transient and Steady-State Analysis of LTI Systems (400)  Given that K G ( s) = 2 s + 10 s and damping ratio x = 0.25. The closed loop transfer function of the system is given by K 2 s + 10 s + K Characteristic equation of the system is s 2 + 10 s + K = 0 Comparing with the standard characteristic equation, s 2 + 2ξw n s + w n2 = 0

Ch wise GATE_ECE_CH06_Control Systems.indd 434

Hence, fH −fL = (9000 −30) Hz = 8970 Hz 123. Topic: Routh-Hurwitz and Nyquist Stability Criteria (84.32)  The phase margin of the given system is 84.32. 124. Topic: Bode and Root-Locus Plots (12)  The characteristic equation is given by 1+

K =0 s( s + 1)( s + 3)

This simplifies to s3 + 4s2 + 3s + K = 0 Using Routh Table, s1 row should be zero for poles to be an imaginary axis. Hence, 12 - K =0 4 Therefore, K = 12.

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Chapter 6  •  Control Systems

125. Topic: Bode and Root-Locus Plots (a)  Given that the transfer function is

435

Y ( s) = G ( s) ⋅ U ( s) Therefore, 1 1 Y ( s) = ⋅ Ms 2 + Bs + K s

10( s + 1) s + 10 Using s = jw in the transfer function, we get G ( s) =

y(∞) = lim sY ( s) = lim

10( jw + 1) G ( jw ) = ( jw + 10)

s→0

y(∞) =

So,

At w = 0, G ( jw ) = 1

s→0

1 Ms 2 + Bs + K

1 K

From the given table, for w = 0.01 rad/s, |G(jw)| dB = -18.5

At w = ∞, G ( jw ) = 10 jω

20 log G ( jw ) = -18.5 1 = -18.5 K 1 -18.5 log = K 20 -18.5 1 y(∞) = = 10 20 = 0.118 K

20 log ω= ∞ 10

ω= 0

σ

Since, zero is nearer to imaginary axis, therefore the plot will move in clockwise direction. Hence, the plot is in the first quadrant. 126. Topic: Lag, Lead and Lag-Lead Compensation (a)  The pole zero plot of the given controler is

Therefore, y(∞) ≈ 0.12 129. Topic: Transient and Steady-State Analysis of LTI Systems (−0.5)  Given that Td(s) = 1

θ ( s) = s = −b

s

s = −a

For phase lead compensator a < b. Hence, option (a) is the correct option. 127. Topic: Transient and Steady-State Analysis of LTI Systems (d) Here ξw n = 1 1 - ξ2 =



3 2

1 2 wn = 2 Hence, the transfer function of the system is represented by option (d). 128. Topic: Transient and Steady-State Analysis of LTI Systems (0.12) G(s)

Ch wise GATE_ECE_CH06_Control Systems.indd 435

Steady state value is lim sq (s) = −0.5 s→0            130. Topic: Routh-Hurwitz and Nyquist Stability Criteria (a)  Given that the plant transfer function K ⎞ 1 ⎛ G ( s) = ⎜ K P + I ⎟ ⎝ s ⎠ s ( s + 2) The characteristic equation is

ξ=

U(s)

-1 ⎡ K bK T ⎤ s ⎢( Js + B) + ⎥ Ra + La s ⎦ ⎣

Y(s)

s3 + 2 s 2 + sK P + K I = 0 The Routh-Hurwitz table is 1 s3 2 s2 s1 ( 2 K P - K I ) s

0

2 KI

KP KI

0 0

0

0

For a stable system, the first column elements must be positive. Therefore,

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⎛ 2K - K I ⎞ K I > 0 and ⎜ P ⎟⎠ > 0 ⎝ 2 Therefore, K P >

KI >0 2

Therefore,

131. Topic: Routh-Hurwitz and Nyquist Stability Criteria (2)  Given that s5 + 2s4 + 3s3 + 6s2 - 4s - 8 = 0 The Routh-Hurwitz array is 5

s

1

3

−4

s

4

2

6

−8

s

3

8

12

0

3

−8

0

0

0

s2 s1

−9.33

s0

−8

10K = 3 or K = 0.3 134. Topic: Lag, Lead and Lag-Lead Compensation (0.5) Given GC ( s) =

Zero = 2 =

X ( s) = ( sI - A) -1 x(0)

Hence, s2 = 1, −4 = s = ±1, ± j2

-1

-1⎤ ⎡ 1⎤ ⎡s X ( s) = ⎢ ⎥ ⎢ ⎥ ⎣0 s + 1⎦ ⎣0 ⎦

Hence, there are only two poles on the left half s-plane. 132. Topic: Bode and Root-Locus Plots (25.5)  By magnitude condition G ( s) H ( S ) s = -1+ 2 j = 1 K 2j+3 7 + 2 j 2 + 2 j -4 + 2 j 20 8 53 13

Solving the above, we get ⎡1/s ⎤ X ( s) = ⎢ ⎥ ⎣0⎦

=1

⎡1⎤ x (t ) = ⎢ ⎥ ⎣0⎦

= 25.5

⎡1 ⎤ y(t ) = [0 1] ⎢ ⎥ = 0 ⎣0 ⎦

133. Topic: Bode and Root-Locus Plots (0.3)  G ( s) H ( s) =

10 K ( s + 3) ( s + 2) =

1 1 = τ RC

Hence, RC = 0.5

2x2 + 6x - 8 = 0. Therefore, x = 1, −4

Therefore, K =

s+2 s+4

135 Topic: State Variable Model and Solution of State Equation of LTI Systems (d)  x = Ax

2s4 + 6s2 - 8 = 0 Let x = s2 then

So,

Poles length 0.75 = =3 Zero length 0.25

K′ =

136. Topic: Transfer Function

K ′( s + 3) ( s + 2)

(a) 

⎡ 2 -1⎤ ⎡ 3⎤ A= ⎢ ⎥ , B = ⎢ -1⎥ , C = [3  -2] 0 4 ⎣ ⎦ ⎣ ⎦

H(s) = C(sI - A)−1 B

jω

-1

1 ⎤ ⎡3⎤ ⎡s - 2 = [3  -2] ⎢ 0 s + 4 ⎥⎦ ⎢⎣ -1⎥⎦ ⎣

ξ = 0.5

= [3  -2]

1 ⎤⎡ 3 ⎤ ⎡s + 4 1 ⎢ s 0 2⎥⎦ ⎢⎣ -1⎥⎦ ( s - 2)( s + 4) ⎣

= [3  -2]

⎡3s + 12 + 1⎤ 1 ( s - 2)( s + 4) ⎢⎣ - s + 2 ⎥⎦

A

−1

(0, 0) −1 O 3

σ

=

Ch wise GATE_ECE_CH06_Control Systems.indd 436

9 s + 39 + 2 s - 4 11s + 35 = ( s - 2)( s + 4) ( s - 2)( s + 4)

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Chapter 6  •  Control Systems

137. Topic: Transient and Steady-State Analysis of LTI Systems (a)  It is given that G ( s) =

2 s( s + 1)

The type of system is 1 and order is 2. The given input is unit step (i.e. type 0). That is, Type of system > Type of input

on the right half of s-plane. It is given that closed-loop system is stable, that is, Z = 0 Therefore, N = P. Hence the Nyquist plot encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number of right half s-plane poles. 139. Topic: Routh-Hurwitz and Nyquist Stability Criteria (a)  We have G ( s) =

Therefore, ess = 0 Shortcut Method • Type of system = Type of input ⇒ ess = constant. • Type of system > Type of input ⇒ ess = 0. • Type of system  0 or K > 0 and 4K 2 + 6K - 4 > 0 ⇒ 2 K 2 + 3K - 2 > 0 2K ⎡ ⎛ 1⎞ ⎤ ( K + 2) ⎢ K - ⎜ ⎟ ⎥ > 0 ⎝ 2⎠ ⎦ ⎣ 1 1 Therefore, K > -2 andK > K->2 .and K > . 2 2 1 1 , K> Hence, K Hence > 2 2 1 Hence, K < -2 2 Therefore, the range of K is 0.5 0

Ch wise GATE_ECE_CH06_Control Systems.indd 440

1 =∞ s ( s + 3)( s + 2) p

1 =∞ 6 × sp

152. Topic: Bode and Root- Lotus Plots (a) Given G ( s) =

s p + b1 s q -1 +  + bp s q + a1 s q -1 +  + bq

Slope at w → ∞ is -60 dB Slope at ω → ∞ is given by -20 (q - p) where q = number of poles and p = number of zeros. Now, considering the different options, we have (a)  p = 0, q = 3 (Slope) ω → ∞ = -20(3 - 0) = - 60 dB (correctly match) (b)  p = 1, q  = 7 (Slope) ω → ∞ = -20(7 - 1) = -120 dB (Does not match) (c)  p = 2, q = 3 (Slope) ω → ∞ = -20(3 - 2) = -20 dB (Does not match) (d)  p = 3, q = 5 (Slope) w → ∞ = -20(5 - 3) = -40 dB (Does not match) Hence the correct answer is option (a). 153. Topic: Lag, Lead and Lag-Lead Compensation (a)  Phase lag compensator: For this compensator, pole is more close to origin as compared to zero. And it contains only one pole and one zero. So, the correct answer is option (a).

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Chapter 6  •  Control Systems

154. Topic: Lag, Lead and Lag-Lead Compensation (d)  Lag Compensator

Lead Compensator

1.

Low-pass filter

High-pass filter

2.

ξ decreases, wn decreases

ξ increases, wn increases

3.

Gain crossover frequency wgc decreases

Gain crossover frequency wgc increases

4.

Bandwidth decreases

Bandwidth increases

5.

Settling time Ts = 4/ξwn increases

Settling time Ts = 4/ξwn decreases

6.

Lag compensator improves the steady state performance of the system

Lead compensator improves the transient performance of the system

155. Topic: State Variable Model and Solution of State Equation of LTI Systems (5.0)  Given that ⎡ x1 (t ) ⎤ ⎡0 0 ⎤ ⎡ x1 (t ) ⎤ ⎡ 0 ⎤ ⎢ x (t ) ⎥ = ⎢0 -9⎥ ⎢ x (t ) ⎥ + ⎢ 45⎥ u(t ) (1) ⎦⎣ 2 ⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣

and

⎡ x1 (0) ⎤ ⎡0 ⎤ ⎢ x ( 0) ⎥ = ⎢ 0 ⎥ ⎣ 2 ⎦ ⎣ ⎦ Now, x˙ = Ax + Bu(2) From Eqs. (1) and (2), we have ⎡0⎤ ⎡0 0 ⎤ A= ⎢ , B=⎢ ⎥ ⎥ ⎣0 -9⎦ ⎣ 45⎦ x(t) = ZIR + ZSR where ZIR = f  [x(0)]|Input = 0 and ZSR = f (I/P)|x (0) = 0

t

⇒ x(t ) = exp[ A(t )]x(0) + exp[ A(t - τ )]Bu(τ )dτ ∫ 0

x(t) = ZSR

Ch wise GATE_ECE_CH06_Control Systems.indd 441

441

X ( s) = ϕ ( s) BU ( s) (3)



0 ⎤ ⎡s [ sI - A] = ⎢ ⎥ ⎣0 s + 9⎦ Laplace transform of state transition matrix is ⎡1 ⎢s L [ϕ (t )] = ϕ ( s) = [ sI - A]-1 = ⎢ ⎢0 ⎢⎣

⎤ 0 ⎥ ⎥ 1 ⎥ s + 9 ⎥⎦

⎡0⎤ BU ( s) = ⎢ 45 ⎥ ⎢ ⎥ ⎣ s⎦ From Eq. (3),

X ( s) = ϕ ( s) BU ( s)



⎡1/s X ( s) = ⎢ ⎢0 ⎣

0 ⎤⎡ 0 ⎤ ⎡ 0 ⎤ ⎢ ⎥ 1 ⎥⎥ ⎢⎢ 45 ⎥⎥ = ⎢ 45 ⎥ s + 9 ⎦ ⎣ s ⎦ ⎢⎣ s( s + 9) ⎥⎦

0 ⎤ ⎡ ⎢ ⎥ X ( s ) = ⇒ ⎢5 - 5 ⎥ ⎣ s s + 9⎦ Taking inverse Laplace transform, ⎡ 0 ⎤ x (t ) = ⎢ -9 t ⎥ ⎣5 - 5e ⎦ ⎡ x (t ) ⎤ ⎡ 0 ⎤ ⇒⎢ 1 ⎥ = ⎢ -9 t ⎥ ⎣ x2 (t ) ⎦ ⎣5 - 5e ⎦ ⇒ x1(t) = 0, x2(t) = 5 − 5e−9t ⇒ x1 (∞) = 0, x2 (∞) = 5 - 5(e -9 ×∞ ) = 5 Therefore, the value of lim t →∞



=

x12 (t ) + x22 (t )

x12 (∞) + x22 (∞) = 0 + 52 = 5

156. Topic: Routh-Hurwitz and Nyquist Stability Criteria (b)  (i)

G ( s) H ( s) =

K ( s 2 + 2 s + 2)( s + 2)

(ii) If K is 10, Nyquist plot does not enclose -1 + 0⋅j. (iii) If K is 100, Nyquist plot encloses -1 + 0⋅j.

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GATE ECE Chapter-wise Solved papers

From the second figure, we have

Im For K = 10

N = -2 From Eq. (1), we get -2 = 0 - Z Therefore, Z = 2 (System is unstable) Re

w=∞

157. Topic: Routh-Hurwitz and Nyquist Stability Criteria

(−1, 0)

(c)  G ( s) =

10 K ( s + 2) , H(s) = 1 s 2 + 3s 2 + 10 Im

Fig: (i) Im

+j 5.43 K

—jw (−1, 0) w =∞

Re w=0

−1 + j0

w=∞

—K 0 1

-1.5 ⎤ s + 4s + 6 ⎥ ⎥ s+4 ⎥ s 2 + 4 s + 6 ⎥⎦

So the closed loop transfer function is 3s + 5 T ( s) = 2 s + 4s + 6

2 = 2 ⇒ ξ 1- ξ =

Squaring both the sides, we get 1 16 1 ⇒ ξ2 - ξ4 = 16

ξ 2 (1 - ξ 2 ) =

    ⇒16ξ 4 - 16ξ 2 + 1 = 0 Let ξ 2 = x. Then we have 16x2 - 16x + 1 = 0

Ch wise GATE_ECE_CH06_Control Systems.indd 444

ξ2 =

1 3 = 0.0669 2 4

2

⇒ ξ2
0. So −20 − 20 log k > 0

⇒ 20 + 20 log k < 0  ⇒ 20 log k < −20  ⇒ k < 0.1 Therefore, k0 = 0.1

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Communications

CHAPTER7

Syllabus Random processes: autocorrelation and power spectral density, properties of white noise, filtering of random signals through LTI systems; Analog communications: amplitude modulation and demodulation, angle modulation and demodulation, spectra of AM and FM, superheterodyne receivers, circuits for analog communications; Information theory: entropy, mutual information and channel capacity theorem; Digital communications: PCM, DPCM, digital modulation schemes, amplitude, phase and frequency shift keying (ASK, PSK, FSK), QAM, MAP and ML decoding, matched filter receiver, calculation of bandwidth, SNR and BER for digital modulation; Fundamentals of error correction, Hamming codes; Timing and frequency synchronization, inter-symbol interference and its mitigation; Basics of TDMA, FDMA and CDMA.

Chapter Analysis Topic

GATE 2009

GATE 2010

GATE 2011

GATE 2012

GATE 2013

GATE 2014

GATE 2015

GATE 2016

GATE 2017

GATE 2018

Random processes

4

3

1

2

2

10

4

7

3

2

Analog communications

2

2

2

1

3

2

5

3

2

Information theory

1

Digital communications

2

2

3

1

1

4

3

7

4

3

3

4

5

2

3

Fundamentals of error correction

1

Basics of TDMA, FDMA and CDMA

1

1 1

1

Important Formulas 1. A random variable is discrete if

∑ P (x ) = 1 x

i

i

2. If random variable x can take values x1, x2, x3 … xn and the random variable y can take values y1, y2, y3 … ym, then

∑∑P

xy

i

4. Marginal probabilities

∑P

xy

∑P

xy

5. Cumulative distribution function Fx ( x ) = P ( x ≤ x ) 6. Probability density function (PDF)

3. Relation between the conditional probabilities is

∑P

xy

i

( xi y j ) = ∑ Py x ( y j xi ) = 1 j

dFx ( x ) = px ( x ) dx 7. Properties of PDF +∞

Pxy ( xi , y j ) = Px y ( xi y j ) Py ( y j ) = Py x ( y j xi ) Px ( xi )

Ch wise GATE_ECE_CH07_Communications.indd 445

( xi , y j ) = Px ( xi )

j

( xi , y j ) = 1

j

( xi , y j ) = Py ( y j )

i

∫

px ( x )dx = 1

-∞

px ( x ) ≥ 0

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GATE ECE Chapter-wise Solved papers

8. The CDF Fx(x) of the Gaussian random variable with zero mean and unit variance is x

1

Fx ( x ) =

∫e

2π

2

-x /2

16. For two independent random variables x and y, pxy ( x, y ) = px ( x ) p y ( y )

dx 17. For two independent random variables x and y,

-∞

9. For a Gaussian random variable with mean m and variance s, px ( x ) = and   Fx ( x ) =

1

e - ( x - m)

2

σ 2π x

1

σ 2π

∫e

18. The mean value of a random variable x, also referred to as the average value or the expected value, is defined as

/ 2σ 2

- ( x - m ) / 2σ 2 2

Fxy ( x, y ) = Fx ( x ) Fy ( y )

dx

-∞

x = E [ x ] = ∑ xi Px ( xi ) i

19. If the random variable x is continuous, then

10. For two random variables x and y, the CDF Fxy(x, y) is defined as

x = E[ x ] =

Fxy ( x, y ) =Δ P ( x ≤ x and y ≤ y ) 11. Joint PDF pxy(x, y) is defined as pxy ( x, y ) =

∂2 Fxy ( x, y ) ∂x ∂y

12. For two random variables x and y, given that the joint density is pxy(x, y), the individual probability densities or the marginal densities px(x) and py(y) can be obtained from pxy(x, y) as follows: px ( x ) =

Δ

xn =

n

px ( x )dx

21. The nth central moment of a random variable x is defined as Δ

( x - x)n =

+∞

∫ ( x - x)

n

px ( x )dx

+∞

∫

pxy ( x, y )dx

22. σ x is known as the standard deviation of a random variable x and is given by σ x2 = ( x - x ) 2 = x 2 - x

px y ( x y ) p y ( y ) = pxy ( x, y ) p y x ( y x ) px ( x ) = pxy ( x, y )

px y ( x y ) =

ψ g (t ) =

px y ( x y )dx = 1

-∞

15. Two continuous random variables x and y are said to be independent when p x y ( x /y ) = p x ( x ) and    p y x ( y /x ) = p y ( y )

+∞

∫ g (t ) g (t + t )dt

-∞

p y x ( y x ) px ( x ) py ( y)

2

23. The autocorrelation function yg(t) of a real energy signal g(t) is

14. Bayes’ rule for continuous random variables

Ch wise GATE_ECE_CH07_Communications.indd 446

∫x

-∞

13. Conditional PDF

∫

+∞

-∞

pxy ( x, y )dy

-∞

+∞

x

20. The nth moment of a random variable x is defined as the mean value of xn, that is,

-∞

and    py ( y) =

∫ xp ( x)dx

-∞

+∞

∫

+∞

24. For complex energy signals, the autocorrelation function is

ψ g (t ) =

+∞

+∞

-∞

-∞

∫ g (t ) g*(t - t )dt = ∫ g*(t ) g (t + t )dt

25. The autocorrelation function of an energy signal and its ESD form a Fourier transform pair, that is, FT ψ g (t ) ←⎯ →ψ g ( f )

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Chapter 7  •  Communications

26. The autocorrelation function of a real power signal g(t) is 1 T →∞ T

+T / 2

1 T →∞ T

+T / 2

Rg (t ) = lim = lim

∫

33. The PSD and corresponding autocorrelation function of band-limited white noise is given by

g (t ) g (t - t )dt

S X (ω ) =

-T / 2

∫

g (t ) g (t + t )dt

1 T →∞ T

+T / 2

1 T

+T / 2

= lim

T →∞

∫

g (t ) g*(t - t )dt

and RX (t ) =

No B sin Bt 2π Bt

34. The PSD and corresponding auto correlation function for coloured noise sequence are respectively given by

-T / 2

∫

g*(t ) g (t + t )dt

S X (ω ) =

-T / 2

28. For a power signal, the autocorrelation function and the PSD form a Fourier transform pair, that is, FT Rg (t ) ←⎯ → Sg ( f )

29. The PSD Sx(f) of a random process x(t) is defined as the ensemble average of the PSDs of all sample functions, that is, ⎡ X (f)2⎤ ⎥ W/Hz S x ( f ) = lim ⎢ T T →∞ ⎢⎣ T ⎥⎦ 30. The average power Px of a wide-sense random process x(t) is its mean square value x 2 , that is, Px = x = Rx (0) =

+∞

∫ S ( f )df

2

x

-∞

31. A white noise process W(t) is defined by N SW (ω ) = o 2

-∞ < ω < ∞

where SW (w) is the power spectral density, No is a real constant and called the intensity of the white noise. The corresponding autocorrelation function is given by RW (t ) =

No d (t ) 2

where d (t ) is the Dirac delta. 32. The average power of white noise Pavg = EW 2 (t ) =

Ch wise GATE_ECE_CH07_Communications.indd 447

No , -B < ω < B 2

-T / 2

27. The autocorrelation function of a complex power signal g(t) is Rg (t ) = lim

447

1 2π

2a 2 b 2 -b t and RX (t ) = a e b2 + ω 2

35. The PSD and corresponding auto correlation function for white noise are respectively given by SW (ω ) =

No dω → ∞ 2 -∞

No , -π ≤ ω ≤ π 2

and RW ( m) =

No d [ m] 2

36. Thermal noise is given by Pn ∝ TB or Pn = KTB where Pn is the maximum noise power output of a resistor (W), K is Boltzmann’s constant (J/K), T is absolute temperature (K) and B is the bandwidth (Hz) over which noise is measured. 37. The amplitude modulated signal is expressed mathematically by v(t) = Vc(1 + mcoswmt)coswct, where m is the modulation index = Vm/Vc. 38. When more than one sinusoidal or cosinusoidal signals with different amplitudes amplitude modulate a carrier, the overall modulation index for AM is m = ( m12 + m22 + m32 + ) 39. When the message signal has non-zero offset, that is, its maximum Vmax and its minimum Vmin are not symmetric (Vmax ≠ Vmin), then modulation index m for AM is given by

∞

∫

b>0

m=

Vmax - Vmin 2Vc + Vmax + Vmin

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GATE ECE Chapter-wise Solved papers

40. For AM, the modulated signal can also be expressed as mVc cos(ω c - ω m )t 2 mVc cos(ω c + ω m )t + 2

v(t ) = Vc cos ω c t +

41. The total power (Pt) in an AM signal is related to the unmodulated carrier power (Pc) by the following expression:

50. The resulting PM signal is vPM(t) = Acos[wct + kpm(t)] 51. The instantaneous frequency w (t) in this case is given PM by the following expression:

ω

PM

(t ) =

52. The information content Ii of message mi is ⎛ 1⎞ I i = log 2 ⎜ ⎟ ⎝ Pi ⎠

⎛ m2 ⎞ Pt = Pc ⎜1 + 2 ⎟⎠ ⎝ 42. For proper detection of the signal by envelope detector 1 1 1 < < RC < , or 2π B < 1 (for sinusoidal modulation with modulating frequency wm) or D >> 1 (for arbitrary modulation signal band ­limited to wM), the FM signal is termed as the wide-band FM and the bandwidth in this case is given by the following expression: Bandwidth = 2mf ω m or 2 D ω M 49. Instantaneous phase q(t) of the PM signal is mathematically expressed as follows: q(t) = wct + q0 + kpm(t)

Ch wise GATE_ECE_CH07_Communications.indd 448

i

i

+ J1 ( m f )[sin(ω c + ω m )t - sin(ω c - ω m )t ] + J 3 ( m f )[sin(ω c + 3ω m )t - sin(ω c - 3ω m )t ] +}

dθ (t ) . = ω c + k p m( t ) dt

i =1

56. The code efficiency h is

η=

H ( m) L

57. The channel capacity for a binary symmetric c­ hannel is ⎡ 1 1 ⎤ Cs = 1 - ⎢ Pe log + (1 - Pe ) log ⎥ 1 - Pe ⎦ Pe ⎣ 58. The average information received by the receiver in bits per symbol is I ( x ; y ) = H ( x ) - H ( x /y ) = ∑ ∑ P ( xi , y j ) log i

j

= ∑ ∑ P ( xi , y j ) log i

j

= ∑ ∑ P ( xi , y j ) log i

j

P ( xi /y j ) P ( xi ) P ( y j /xi ) P( y j ) P ( xi , y j ) P ( xi ) P ( y j )

59. I(x ; y) = I(y ; x) = H(y) − H(yú/x)

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Chapter 7  •  Communications

60. The channel capacity for a discrete memoryless channel is

69. 2 f PCM = nf s

Cs = max I ( x; y ) P ( xi )

H ( x) =

+∞

∫

p( x ) log

-∞

1

σ 2π

70. The mean square value or the power of the quantization noise for uniform quantization

1 dx p( x )

63. For a given mean square value x 2 , the entropy is maximum for a Gaussian distribution of x given by p( x ) =

⎛ n⎞ f PCM = ⎜ ⎟ f s ⎝ 2⎠

or,

61. The channel capacity in bits per second is C = kCs 62. The entropy H(x) for a continuous random ­variable x is

e- x

2

Nq =

2 So m = 3L2 2 No mp

and the corresponding entropy is

64. For a band-limited white Gaussian noise n(t) with PSD ζ /2, the entropy in bits per sample of each Nyquist symbol n(t) is H ( n) =

1 log( 2π eζ B) 2

65. The mutual information I(x ; y) of continuous random variables x and y is ⎡ 1 ⎤ ⎡ ⎤ 1 I ( x; y ) = log ⎢ ⎥ - log ⎢ p( x /y ) Δx ⎥ Δ p ( x ) x ⎣ ⎦ ⎣ ⎦ p( x / y ) = log p( x ) = H ( x ) - H ( x /y ) 66. The Shannon−Hartley theorem states

72. The m-law for positive amplitude is y=

⎛ µm ⎞ 1 m ln ⎜1 + ≤1 ⎟ ,0 ≤ ln(1 + µ ) ⎝ mp ⎠ mp

73. The A-law for positive amplitudes is ⎧ 1 ⎛ m⎞ m 1 ≤ ⎪ ⎜ ⎟, 0≤ mp A ⎪ 1 + ln A ⎝ mp ⎠ y=⎨ Am ⎞ 1 m ⎪ 1 ⎛ ⎪1 + ln A ⎜1 + ln m ⎟ , A ≤ m ≤ 1 ⎝ p ⎠ p ⎩ 74. The output SNR using m-law compander is So 3L2 ≅ N o [ln(1 + µ )]2 75. The SNR improvement in DPCM over PCM is at least

S⎞ ⎛ C = B log 2 ⎜1 + ⎟ bits/s ⎝ N⎠ 67. For a system with infinite bandwidth, the channel capacity C is lim C = 1.44

B →∞

68. In binary PCM, the number of bits to be transmitted per second = nfs where n = log2L and L is the number of standard levels.

Ch wise GATE_ECE_CH07_Communications.indd 449

Gp =

Pm Pd

76. For delta modulation, to avoid slope overload Δ dm(t ) > Ts dt max

S bits/s ζ

where N = ζ B

2 ( Δv ) 2 mp = 2 12 3L

71. The SNR (So/No) for uniform quantization is

/ 2σ 2

1 log( 2π eσ 2 ) 2

449

77. ASK signal is ⎧ A sin ω c t xc (t ) = ⎨ ⎩ 0

for bit ‘1’ for bit ‘0’

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GATE ECE Chapter-wise Solved papers

78. FSK signal is ⎧ A sin ω c1t xc (t ) = ⎨ ⎩ A sin ω c 2 t

82. The minimum distance between t error-correcting code words without overlapping is

for bit ‘1’ for bit ‘0’

dmin = 2t + 1 83. ML estimator has to satisfy

79. BPSK signal is xc 0 (t ) = A cos(ω c t + θ 0 ) for bit ‘0’

∞

- ∫ r (t )V (t ) sin( 2π fC t + f )dt = 0

xc1 (t ) = A cos(ω c t + θ1 ) for bit ‘1’

-∞

84. If T is the sampling time interval of the time-­multiplexed signal of n different signals each having a sampling interval of Ts, then

80. QPSK signal

π⎞ ⎛ 1 ⎞ ⎛ x (t ) = ⎜ ⋅ di (t ) ⋅ cos ⎜ ω c t + ⎟ ⎟ ⎝ ⎝ 2⎠ 4⎠ π⎞ ⎛ 1 ⎞ ⎛ +⎜ ⋅ d (t ) ⋅ sin ⎜ ω c t + ⎟ ⎝ ⎝ 2 ⎟⎠ q 4⎠

T=

81. For a noisy channel with capacity C, there exist codes of rate R < C such that maximum-likelihood (ML) decoding can lead to error probability

Ts n

85. If time-multiplexed signal is considered as a low-pass signal having a bandwidth of f and fm is the bandTDM width of individual signals, then f TDM = nf m

Pe ≤ 2 - nEb ( R )

QUESTIONS 1. The amplitude modulated waveform s(t ) = Ac [1 + K a m(t )] cos ω c t is fed to an ideal envelope detector. The maximum magnitude of K a m(t ) is greater than 1. Which of the following could be the detector output?

4. A message m(t) band-limited to the frequency fm has a power of Pm. The power of the output signal in the following figure is

(a) Ac m(t) (b) Ac2 [1+ K a m(t )] 2

(c) Ac [1+Kam(t)] (d) Ac [1+Kam(t)]2 (GATE 2000: 1 Mark) 2. The frequency range for satellite communication is (a) 1 kHz to 100 kHz (b) 100 kHz to 10 kHz (c) 10 MHz to 30 MHz (d) 1 GHz to 30 GHz (GATE 2000: 1 Mark) 3. In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 kHz and 25 kHz respectively. These waveforms will be orthogonal for a bit interval of (a) 45 µ sec (b) 200 µ sec (c) 50 µ sec (d) 250 µ sec (GATE 2000: 2 Marks)

Ch wise GATE_ECE_CH07_Communications.indd 450

(a) (c)

Pm cos θ Pm (b) 2 4 Pm sin 2 θ Pm cos 2 θ (d) 4 4 (GATE 2000: 2 Marks)

5. In an FM system, a carrier of 100 MHz is modulated by a sinusoidal signal of 5 kHz. The bandwidth by Carson’s approximation is 1 MHz. If y(t) = (modulated waveform)3, then by using Carson’s approximation, the bandwidth of y(t) around 300 MHz and the spacing of spectral components are, respectively. (a) 3 MHz, 5 kHz (b) 1 MHz, 15 kHz (c) 3 MHz, 15 kHz (d) 1 MHz, 5 kHz (GATE 2000: 2 Marks)

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Chapter 7  •  Communications

6. A band limited signal is sampled at the Nyquist rate. The signal can be recovered by passing the samples through (a) an RC filter (b) an envelope detector (c) a PLL (d) an ideal low-pass filter with appropriate bandwidth (GATE 2001: 1 Mark) 7. A video transmission system transmits 625 picture frames per second. Each frame consists of a 400 × 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is (a) 16 Mbps (b) 100 Mbps (c) 600 Mbps (d) 6.4 Gbps (GATE 2001: 2 Marks) 8. During transmission over a communication channel, bit errors occur independently with probability p. If a block of n bits is transmitted, the probability of at most one bit error is equal to (a) 1 – (1 – p)n (b) p + (n – 1) (1 – p) (c) np(1 – p)n – 1 (d) (1 – p)n + np(1 – p)n – 1 (GATE 2001: 2 Marks) 9. A 1 MHz sinusoidal carrier is amplitude modulated by a symmetrical square wave of period 100 μ sec. Which of the following frequencies will NOT be present in the modulated signal? (a) 990 kHz (b) 1010 kHz (c) 1020 kHz (d) 1030 kHz (GATE 2002: 1 Mark) 10. For a bit-rate of 8 kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are (a) 16 kHz and 20 kHz (b) 20 kHz and 32 kHz (c) 20 kHz and 40 kHz (d) 32 kHz and 40 kHz (GATE 2002: 1 Mark) 11. An angle modulated signal is given by s(t ) = cos 2π ( 2 × 106 t + 30 sin 150t + 40 cos 150t ). The maximum frequency and phase deviations of s(t) are (a) 10.5 kHz, 140π rad (b) 6 kHz, 80π rad (c) 10.5 kHz, 100π rad (d) 7.5 kHz, 100π rad (GATE 2002: 2 Marks)

Ch wise GATE_ECE_CH07_Communications.indd 451

451

12. The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is (a) raised cosine (c) parabolic

(b) flat (d) Gaussian (GATE 2003: 1 Mark)

13. The input to a coherent detector is DSBSC signal plus noise. The noise at the detector output is (a) in-phase component (b) quadrature component (c) zero (d) envelope (GATE 2003: 1 Mark) 14. At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by (b) 3 dB (a) 6 dB (c) 2 dB (d) 0 dB (GATE 2003: 1 Mark) 15. Let X and Y be two statistically independent random variables uniformly distributed in the ranges (-1, 1) and (-2, 1), respectively. Let Z = X + Y. Then the probability that (Z ≤ -2) is 1 (a) Zero (b) 6 (c)

1 1 (d) 12 3 (GATE 2003: 2 Marks)

Common Data for Questions 16 and 17: Let X be the Gaussian random variable obtained by sampling the process x(t) at t = ti and let ∞ 1 - y2 / 2 Q(α ) = ∫ e dy 2π α Autocorrelation function Rxx(t) = 4(e-0.21 |t| + 1) and mean = 0. 16. The probability that [X ≤ 1] is (a) 1 - Q(0.5) (b) Q(0.5) ⎛ 1 ⎞ ⎛ 1 ⎞ (c) Q ⎜ 1- Q ⎜ (d) ⎝ 2 2 ⎟⎠ ⎝ 2 2 ⎟⎠ (GATE 2003: 2 Marks] 17. Let Y and Z be the random variables obtained by sampling X(t) at t = 2 and t = 4, respectively. Let W = Y - Z. The variance of W is (a) 13.36 (b) 9.36 (c) 2.64 (d) 8.00 (GATE 2003: 2 Marks)

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GATE ECE Chapter-wise Solved papers

18. A DSBSC signal is to be generated with a carrier frequency fc = 1 MHz using a non-linear device with the input−output characteristic vo = a0vi + a1vi3 where a0 and a1 are constants. The output of the non-linear device can be filtered by an appropriate band-pass filter. Let vi = Ac′ cos( 2π f c′t ) + m(t ) be the message signal. Then the value of f c′ (in MHz) is (a) 1.0 (b) 0.333 (c) 0.5 (d) 3.0 (GATE 2003: 2 Marks) Common Data for Questions 19 and 20: m(t ) = cos[( 4π × 103 )t ] be the message signal c (t ) = 5 cos[( 2π × 106 )t ] be the carrier.

Let and

19. c(t) and m(t) are used to generate an AM signal. The modulation index of the generated AM signal is 0.5. Then the quantity (total sideband power)/(carrier power) is

6. Generation of DSBSC (a) (b) (c) (d)

P − 1; Q − 3; R − 2; S − 4 P − 6; Q − 5; R − 2; S − 3 P − 6; Q − 1; R − 3; S − 2 P − 5; Q − 6; R − 1; S − 3 (GATE 2003: 2 Marks)

22. A superheterodyne receiver is to operate in the frequency range 550 kHz to 1650 kHz, with the intermediate frequency of 450 kHz. Let R = Cmax/Cmin denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then (a) R = 4.41, I = 1600 (c) R = 3.0, I = 1600

(b) R = 2.10, I = 1150 (d) R = 9.0, I = 1150 (GATE 2003: 2 Marks)

23. Let x(t) = 2cos(800pt) + cos(1400pt). x(t) is sampled with the rectangular pulse train shown in the figure. The 1 1 (b) (a) only spectral components (in kHz) present in the sam2 4 pled signal in the frequency range 2.5 kHz to 3.5 kHz 1 are 1 (c) (d) 3 8 p(t) (GATE 2003: 2 Marks) 3 20. c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos[ 2π (1008 × 103 t )] in the FM signal (in terms of the Bessel coefficients) is t −T0/6 0 T0/6 −T0 T0 5 (a) 5 J 4 (3) (b) J 8 (3) 2 T0 = 10−3 s 5 (c) J 8 ( 4) (d) 5J4(6) (a) 2.7, 3.4 (b) 3.3, 3.6 2 (GATE 2003: 2 Marks) (c) 2.6, 2.7, 3.3, 3.4, 3.6 (d) 2.7, 3.3 21. Choose the correct one from among the alternatives a, b, c, d after matching an item in Group 1 with the most appropriate item in Group 2. Group 1 P. Ring modulator Q. VCO R. Foster−Seeley discriminator S. Mixer Group 2 1. Clock recovery 2. Demodulation of FM 3. Frequency conversion 4. Summing the two inputs 5. Generation of FM

Ch wise GATE_ECE_CH07_Communications.indd 452

(GATE 2003: 2 Marks) 24. A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantized into 128 levels using a mid-rise uniform quantizer. The quantization noise power is (a) 0.768 V (b) 48 × 10−6 V2 (c) 12 × 10−6 V2 (d) 3.072 V (GATE 2003: 2 Marks) 25. If Eb, the energy per bit of a binary digital signal, is 10−5 Ws and the one-sided power spectral density of the white noise, No = 10−6 W/Hz, then the output SNR of the matched filter is (a) 26 dB (b) 10 dB (c) 20 dB (d) 13 dB (GATE 2003: 2 Marks)

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Chapter 7  •  Communications

26. The input to a linear delta modulator having a step size ∆ = 0.628 is a sine wave with frequency fm and peak amplitude Em. If the sampling frequency fs = 40 kHz, the combination of the sine wave frequency and the peak amplitude, where slope overload will take place is Em   fm (a) (b) (c) (d)

0.3 V 1.5 V 1.5 V 3.0 V

8 kHz 4 kHz 2 kHz 1 kHz (GATE 2003: 2 Marks)

27. If S represents the carrier synchronization at the receiver and r represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is (a) r = 0.5, S is required (b) r = 1.0, S is required (c) r = 0.5, S is not required (d) r = 1.0, S is not required (GATE 2003: 2 Marks) 28. A signal is sampled at 8 kHz and is quantized using an 8-bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is (a) R = 32 kbps, SNRq = 25.8 dB (b) R = 64 kbps, SNRq = 49.8 dB (c)   R = 64 kbps, SNRq = 55.8 dB (d) R = 32 kbps, SNRq = 49.8 dB (GATE 2003: 2 Marks) 29. The distribution function Fx (x) of a random variable x is shown in the following figure. The probability that x = 1 is Fx(x)

1.0 0.55 0.25 −2

(a) zero (c) 0.55

x 0

1

3 (b) 0.25 (d) 0.30 (GATE 2004: 1 Mark)

30. An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz, respectively. An appropriate value for the time constant of the envelope detector is

Ch wise GATE_ECE_CH07_Communications.indd 453

(a) 500 ms (c) 0.2 ms

453

(b) 20 ms (d) 1 ms (GATE 2004: 1 Mark)

31. An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0.1 are added together. The resultant signal can be closely approximated by (a) broadband FM (b) SSB with carrier (c) DSBSC (d) SSB without carrier (GATE 2004: 1 Mark) 32. In a PCM system, if the code word length is increased from 6 to 8 bits, the ­signal-to-quantization-noise ratio improves by the factor (a) 8/6 (b) 12 (c) 16 (d) 8 (GATE 2004: 1 Mark) 33. In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t1 ≤ t ≤ t2. This indicates that during this interval (a) the input to the modulator is essentially constant. (b) the modulator is going through slope overload. (c) the accumulator is in saturation. (d) the speech signal is being sampled at the Nyquist rate. (GATE 2004: 1 Mark) 34. A 1mW video signal having a bandwidth of 100 MHz is transmitted to a receiver through a cable that has 40 dB loss. If the effective one-sided noise spectral density at the receiver is 10-20 W/Hz, then the signal-to-noise ratio (SNR) at the receiver is (a) 50 dB (b) 30 dB (c) 40 dB (d) 60 dB (GATE 2004: 2 Marks) 35. A random variable X with uniform density in the interval 0 to 1 is quantized as follows: If 0 ≤ X ≤ 0.3, xq = 0 If 0.3 W W

−fc−B −fc −fc+ B

1 1 (c) RC < (d) RC > ω ω (GATE 2006: 2 Marks) 60. A message signal with bandwidth 10 kHz is lower sideband SSB modulated with carrier frequency fc1 = 106 Hz. The resulting signal is then passed through a narrow-band frequency modulator with carrier frequency fc2 = 109 Hz. The bandwidth of the output would be (a) 4 × 104 Hz (b) 2 × 106 Hz (c) 2 × 109 Hz (d) 2 × 1010 Hz (GATE 2006: 2 Marks)

Ch wise GATE_ECE_CH07_Communications.indd 457



f

fc−B fc fc+B

(a)

25 25 (b) 8N o B 4N o B

(c)

25 25 (d) No B 2N o B (GATE 2006: 2 Marks)

65. A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols/s. Assuming independent generation of symbols, the most efficient source encoder would have an average bit rate of

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GATE ECE Chapter-wise Solved papers

(a) 6000 bit/s (c) 3000 bit/s

(b) 4500 bit/s (d) 1500 bit/s (GATE 2006: 2 Marks)

66. The minimum sampling frequency (in samples/s) required to reconstruct the following signal from its samples without distortion 3

⎛ sin 2π1000t ⎞ ⎛ sin 2π1000t ⎞ + 7⎜ x (t ) = 5 ⎜ ⎟ ⎟⎠ ⎝ ⎠ ⎝ πt πt would be (a) 2 × 103

(b) 4 × 103

(c) 6 × 103

(d) 8 × 103

2

(GATE 2006: 2 Marks) 68. In the following figure the minimum value of the constant C, which is to be added to y1(t) such that y1(t) and y2(t) are different, is

x(t) with −V V , range 2 2

(b)

(c) R(t ) = - R( -t )

71. If S(f) is the power spectral density of a real, wide-sense stationary random process, then which of the following is ALWAYS true? (a) S(0) ≥ S(f ) (b) S(f ) ≥ 0 ∞

(c) S(-f) = -S(f ) (d) S ( f ) df = 0

∫

-∞

(GATE 2007: 1 Mark) 72. A Hilbert transformer is a (a) non-linear system (b) non-causal system (c)  time-varying system (d) low-pass system (GATE 2007: 2 Marks) 73. In the following scheme, if the spectrum M(f) of m(t) is as shown in the following figure, then the spectrum Y(f) of y(t) will be

y2(t) M(f)

C

Δ (a) ∆ (b) 2 (c)

R(t ) ≤ R(0)

(GATE 2007: 1 Mark)

67. The minimum step size required for a delta modulator operating at 32 k samples/s to track the signal (here u(t) is the unit-step function) x(t) = 125t[u(t) − u(t − 1)] + (250 − 125t) [u(t − 1) − u(t − 2)] so that slope overload is avoided would be (a) 2−10 (b) 2−8 (c) 2−6 (d) 2−4

Same quantizer Q

(a) R(t ) = R( -t )

(d) The mean square value of the process is R(0)

(GATE 2006: 2 Marks)

Quantizer Q with L y1(t) levels, step size ∆ + allowable signal + dynamic range(−V,V )

70. If R(t) is the autocorrelation function of a real, widesense stationary random process, then which of the following is NOT true?

Δ2 Δ (d) 12 L

−B

f 0

+B cos(2πBt)

m(t)

(GATE 2006: 2 Marks)

+

69. If E denotes expectation, the variance of a random variable X is given by

+

(a) E[X 2] - E 2[X] (b) E[X 2] + E 2[X]

∑

y(t)

Hilbert transform

(c) E[X 2] (d) E 2[X] (GATE 2007: 1 Mark)

Ch wise GATE_ECE_CH07_Communications.indd 458

sin(2πBt)

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Chapter 7  •  Communications

Y(f)

Y(f)

(a)

√2a

(b)

f2

459

f2 a

0

+B

Y(f)

(c)

−2B

f

−B

f1

f

−B

0 +B

+2B

-2√2a

-√2a

0

-a

0

f1

-a

Y(f)

(d)

a

-√2a Constellation 1 −B

f 0

+B

−2B

−B

0 +B

+2B

(GATE 2007: 2 Mark) 74. In delta modulation, the slope overload distortion can be reduced by (a) decreasing the step size (b) decreasing the granular noise (c) decreasing the sampling rate (d) increasing the step size (GATE 2007: 2 Marks) 75. The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t) with unity rolloff factor is given by p(t ) =

sin 4π Wt 4π Wt (1 - 16W 2 t 2 )

The value of p(t) at t = 1 4W is (a) −0.5 (c) 0.5

Constellation 2

f

(b) 0 (d) ∞ (GATE 2007: 2 Marks)

77. The ratio of the average energy of constellation 1 to the average energy of constellation 2 is (a) 4a2 (b) 4 (c) 2 (d) 8 (GATE 2007: 2 Marks) 78. If these constellations are used for digital communications over an AWGN channel, then which of the following statements is true? (a) Probability of symbol error for constellation 1 is lower. (b) Probability of symbol error for constellation 1 is higher. (c) Probability of symbol error is equal for both the constellations. (d) The value of No will determine which of the two constellations has a lower probability of symbol error. (GATE 2007: 2 Marks) Statement for Linked Answer Questions 79 and 80: An input to a 6-level quantizer has the probability density function f(x) as shown in the given figure. Decision boundaries of the quantizer are chosen so as to maximize the entropy of the quantizer output. It is given that three consecutive decision boundaries are ‘−1’, ‘0’ and ‘1’.

76. During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by (a) pn

(b) 1 − pn

(c) np(1 − p)n − 1 + (1 − p)n

(d) 1 − (1 − p)n

f(x) a b −5

−1 0 1

x 5

(GATE 2007: 2 Marks) Common Data for Questions 77 and 78: Two 4-ary signal constellations are shown. It is given that f1, and f2 constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let N0/2 denote the power spectral density of white Gaussian noise.

Ch wise GATE_ECE_CH07_Communications.indd 459

79. The values of a and b are (a) a = 1/6 and b = 1/12 (b) a = 1/5 and b = 3/40 (c) a = 1/4 and b = 1/16 (d) a = 1/3 and b = 1/24 (GATE 2007: 2 Marks)

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GATE ECE Chapter-wise Solved papers

(a)

80. Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is (a)

(c)

152 64 (b) 9 3 76 3

(d) 28

CDF

CDF

−1

10

(b)

CDF

X 1

CDF −1

10

(GATE 2007: 2 Marks)

CDF −1

81. In a direct-sequence CDMA system, the chip rate is 1.2288 × 106 chips/s. If the processing gain is desired to X −1 0 CDF 1 be at least 100, the data rate 1

X 1

(GATE 2007: 2 Marks)

1

CDF 0 1

−1

1 CDF 1 X 10

−1

CDF −1

(b) must be greater than 12.288 × 103 bits/s.

X

0CDF

(c)

(a) must be less than or equal to 12.288 × 103 bits/s.

X 1 X 1X

0 0

1

−1

CDF 1

X 1

X 0

1

CDF −1 1

1 CDF 10

−1 −1

−1

X

10

−1

82. In a GSM system, eight channels can co-exist in (d) CDF 1 X 200  kHz bandwidth using TDMA. A GSM-based −1 0cellu1 a lar operator is allocated 5-MHz bandwidth. Assuming frequency reuse factor of 1/5, that is, a five-cell repeat pattern, the maximum number of simultaneous channels that can exist in one cell is X (a) 200 (b) 40 −1 0 1 (c) 25 (d) 5

−1

X 0

1

−1

X

−1

0

1

−1

(GATE 2007: 2 Marks)

(GATE 2008: 2 Marks)

83. Consider the amplitude modulated (AM) signal Accoswct + 2coswmtcoswct. For demodulating the signal using envelope detector, the minimum value of Ac should be

85. Px ( x ) = M exp( -2 x ) + N exp( -3 x ) is the probability density function for the real random variable X, over the entire x-axis. M and N are both positive real numbers. The equation relating M and N is

(a) 2 (c) 0.5

(b) 1 (d) 0

(a) M +

(GATE 2008: 1 Mark) 84. The probability density function (PDF) of a random variable X is as shown below. PDF 1

−1

X 0

2 1 N = 1 (b) 2M + N = 1 3 3

(c) M + N = 1

(d) M + N = 3 (GATE 2008: 2 Marks)

86. Noise with double-sided power spectral density of K over all frequencies is passed through an RC low-pass filter with 3 dB cut-off frequency of fc. The noise power at the filter output is (a) K (b) Kfc (c) Kpfc (d) ∞ (GATE 2008: 2 Marks)

1

The corresponding cumulative distribution f­ unction (CDF) has the form

Ch wise GATE_ECE_CH07_Communications.indd 460

X 1

1

1

(c) must be exactly equal to 12.288 × 103 bits/s. CDF −1 1 0 (d) can take any value less than 122.88 × 103 bits/s.

CDF 1

1

87. Consider the frequency modulated signal 10 cos[2p × 105t + 5 sin(2p × 1500t)+ 7.5 sin(2p × 1000t)] with carrier frequency of 105 Hz. The modulation index is

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Chapter 7  •  Communications

(a) 12.5 (c) 7.5

(b) 10 (d) 5 (GATE 2008: 2 Marks)

88. The signal cos(wct) + 0.5cos(wmt)sin(wct) is (a) FM only (b) AM only (c) Both AM and FM (d) Neither Am nor FM (GATE 2008: 2 Marks) 89. A memoryless source emits n symbols each with a probability p. The entropy of the source as a function of n (a) increases as log n (b) decreases as log(l/n) (c) increases as n (d) increases as nlogn (GATE 2008: 2 Marks) 90. Consider a binary symmetric channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is (a) p3 + 3p2(1−p) (b) p3 3 (c) (1−p) (d) p3 + p2(1−p) (GATE 2008: 2 Marks) Common Data for Questions 91, 92 and 93: A speed signal, band limited to 4 kHz and peak voltage varying between +5 V and −5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.

94. Four messages band-limited to W, W, 2W and 3W, respectively, are to be multiplexed using time-division multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is (a) W (b) 3W (c) 6W (d) 7W (GATE 2008: 2 Marks) 95. A white noise process x(t) with two-sided PSD 1 × 10-10 W/Hz is input to a filter whose magnitude squared response is shown in the following figure. H(f) 2 1 x(t)

y(t)

92. Assuming the signal to be uniformly distributed between its peak to peak value, the signal-to-noise ratio at the quantizer output is (a) 16 dB (b) 32 dB (c) 48 dB (d) 64 dB (GATE 2008: 2 Marks) 93. The number of quantization levels required to reduce the quantization noise by a factor of 4 would be (a) 1024 (b) 512 (c) 256 (d) 64 (GATE 2008: 2 Marks)

Ch wise GATE_ECE_CH07_Communications.indd 461

f

−10 kHz

10 kHz

The power of the output process y(t) is given by (a) 5 × 10-7 W (b) 1 × 10-6 W (c) 2 × 10-6 W (d) 1 × 10-5 W (GATE 2009: 1 Mark) 96. If the power spectral density of stationary random process is a sine-squared function of frequency, the shape of its autocorrelation is (a)

(b) R(τ)

R(τ) τ

τ (d)

(c) R(τ)

91. If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (a) 64 kHz (b) 32 kHz (c) 8 kHz (d) 4 kHz (GATE 2008: 2 Marks)

461

R(τ) τ

τ (GATE 2009: 1 Mark)

97. For a message signal m(t) = cos(2pfmt) and carrier of frequency fc, which of the following represents a single side-band (SSB) signal? (a) cos(2pfmt)cos(2pfct) (b) cos(2pfct) (c) cos[2p(fc + fm)t] (d) [1 + cos(2pfmt)]cos(2pfct) (GATE 2009: 1 Mark) 98. Consider two independent random variables X and Y with identical distributions. The variables X and Y take values 0, 1 and 2 with probabilities 1/2, 1/4 and 1/4, respectively. What is the conditional probability P ( X + Y = 2 X - Y = 0) ?

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GATE ECE Chapter-wise Solved papers

1 16 (d) 1

(a) 0 (c)

(b)

1 6

(GATE 2009: 2 Marks)

99. A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true? k

1

2

3

4

5

P(X = k)

0.1

0.2

0.4

0.2

0.1

(a) Both the student and the teacher are right. (b) Both the student and the teacher are wrong. (c)  The student is wrong but the teacher is right. (d) The student is right but the teacher is wrong. (GATE 2009: 2 Marks)

are uniformly quantized with a step size of 0.1 V, the resulting signal-to-quantization-noise ratio is approximately (a) 46 dB (b) 43.8 dB (c) 42 dB (d) 40 dB (GATE 2009: 2 Marks) 104. Suppose that the modulating signal is m(t)  =  2cos(2pfmt) and the carrier signal is xc(t) = Accos(2pfct). Which one of the following is a c­ onventional AM signal without over-modulation? (a) x(t ) = Ac m(t ) cos( 2π f c t ) (b) x(t ) = Ac [1 + m(t )]cos( 2π f c t ) Ac m(t ) cos( 2π f c t ) 4

(c) x(t ) = Ac cos( 2π f c t ) +

(d) x(t ) = Ac cos( 2π f m t ) cos( 2π f c t ) + Ac sin( 2π f m t ) sin( 2π f c t )

(GATE 2010: 1 Mark) ⎛ 1⎞ ⎛ 1⎞ sin ω 2 t 100. A message signal given by m(t ) = ⎜ ⎟ cos ω1t - ⎜ 105. Consider an angle-modulated signal x(t) = 6cos[2p × ⎝ 2⎠ ⎝ 2 ⎟⎠ 6 10 t + 2sin(8000pt) + 4cos(8000pt)]V. The average ⎛ 1⎞ ⎛ 1⎞ m(t ) = ⎜ ⎟ cos ω1t - ⎜ ⎟ sin ω 2 t is amplitude modulated with a carrier of power of x(t) is ⎝ 2⎠ ⎝ 2⎠ (a) 10 W (b) 18 W frequency wc to generate s(t ) = [1 + m(t )]cos ω c t . (c) 20 W (d) 28 W What is the power efficiency achieved by this modula(GATE 2010: 1 Mark) tion scheme? (a) 8.33% (b) 11.11% 106. Consider the pulse shape s(t) as shown. The impulse response h(t) of the filter matched to this pulse is (c) 20% (d) 25% (GATE 2009: 2 Marks) s(t) 101. A communication channel with AWGN operating at a 1 signal-to-noise ratio (SNR) >> 1 and bandwidth B has capacity Cl. If the SNR is doubled keeping B constant, the resulting capacity C2 is given by t 0 T C2 ≈ C1 + B (a) C2 ≈ 2C1 (b) h(t) (b) h(t) (a) C2 ≈ C1 + 0.3 B (c) C2 ≈ C1 + 2 B (d) 1 1 (GATE 2009: 2 Marks) Common Data for Questions 102 and 103: The amplitude of a random signal is uniformly distributed between −5 V and 5 V. 102. If the signal-to-quantization-noise ratio required in uniformly quantizing the signal is 43.5 dB, the step size of the quantization is approximately (a) 0.0333 V (b) 0.05 V (c) 0.0667 V (d) 0.10 V (GATE 2009: 2 Marks) 103. If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values

Ch wise GATE_ECE_CH07_Communications.indd 462

t

−T (c)

t

0

0 (d)

h(t) 1

T h(t)

1

t 0

T

t 0

T

2T

(GATE 2010: 1 Mark)

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Chapter 7  •  Communications

107. x(t) is a stationary process with the power spectral density Sx(f) > 0 for all f. The process is passed through a system shown as follows.

463

109. The probability of bit error is .

(a) 0.5 × e-3 5

(b) 0.5 × e-5

(c) 0.5 × e-7

(d) 0.5 × e-10 (GATE 2010: 2 Marks)

110. The Nyquist sampling rate for the signal

+ + +

x(t)

d dt

y(t) s( t ) =

D Delay = 0.5 ms Let SY(f ) be the power spectral density of y(t). Which one of the following statements is correct? (a) SY(f ) > 0 for all f (b) SY(f ) = 0 for f > 1 kHz (c) SY(f ) = 0 for f = nfo, fo = 2 kHz, where n is any integer (d) SY(f ) = 0 for f = (2n + 1)fo, fo = 1 kHz, where n is any integer (GATE 2010: 2 Marks) Statement for Linked Answer Questions 108 and 109: Consider a baseband binary PAM receiver shown in the following figure. The additive channel noise n(t) is white with power spectral density SN(f) = No/2 = 10-20 W/Hz. The low-pass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk). Yk = Nk if transmitted bit bk = 0, and Yk =a + Nk if transmitted bit bk = 1, where Nk represents the noise sample value. The noise sample has a probability density function PN ( n) = 0.5α e - α n . (This has k mean zero and variance 2/a2.) Assume transmitted bits to be equiprobable and threshold z is set to a/2 =10-6 V. n(t) r(t) x(t)

y(tk)

y(t)

+

LPF

bk

S/H

sin(500π t ) sin(700π t ) × πt πt

is given by (a) 400 Hz (c) 1200 Hz

(b) 600 Hz (d) 1400 Hz (GATE 2010: 2 Marks)

111. The List I (lists the attributes) and the List II (lists of the modulation systems). Match the attribute to the modulation system that best meets it. List I A. Power efficient transmission of signals B. Most bandwidth efficient transmission of voice signals C. Simplest receiver structure D. Bandwidth efficient transmission of signals with significant dc component List II 1. Conventional AM 2. FM 3. VSB 4. SSB-SC A    B    C     D (a) 4  2  1  3 (b) 2  4  1  3 (c) 3  2  1  4 (d) 2  4  3  1 (GATE 2011: 1 Mark)

Sampling time

Threshold z

Receiver where 1 if y(tk) ≥ z bk =

0 if y(tk) ≤ z

108. The value of the parameter a (in V-1) is (a) 1010 (b) 107 -10 (c) 1.414 × 10 (d) 2 × 10-20 (GATE 2010: 2 Marks)

Ch wise GATE_ECE_CH07_Communications.indd 463

112. An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate, and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we ­transmit two quantized samples/s, the information rate is (b) 2 bps (a) 1 bps (c) 3 bps (d) 4 bps (GATE 2011: 1 Mark) 113. x(t) is a stationary random process with autocorrelation function Rx(t ) = exp(-p t2). This process is passed through the following system. The power spectral density of the output process y(t) is

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GATE ECE Chapter-wise Solved papers

117. Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max [X, Y] is less than 1/2 is

H(f) = j2πf + x(t) −

Σ

y(t)

(a) ( 4π 2 f 2 + 1) exp( -π f 2 ) (b) ( 4π 2 f 2 - 1) exp( -π f 2 )

(a)

3 9 (b) 4 16

(c)

1 2 (d) 4 3

(c) ( 4π 2 f 2 + 1) exp( -π f )

(GATE 2012: 1 Mark)

(d) ( 4π 2 f 2 - 1) exp( -π f ) (GATE 2011: 2 Marks) 114. A message signal m(t) = cos2000pt + 4cos4000pt modulates the carrier c(t) = cos2pfct where fc = 1 MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy (a) 0.5 ms < RC < 1 ms (b) 1 ms 0.5 ms

118. Power spectral density of a real process X(t) for positive frequencies is shown in the following figure. The values of E[X2(t)] and E ⎡⎣ X (t )⎤⎦ , respectively, are Sx (ω) 400d(w -104) 6

(GATE 2011: 2 Marks) Statement for Linked Data Questions 115 and 116: A fourphase and an eight-phase signal constellation are shown in the following figure.

Q d

r1

0

r

r

10

11

w (103 rad/s)

(a)

6000 6400 , 0 (b) ,0 π π

(c)

6400 20 6000 20 , (d) , π π 2 π π 2

d r2

9

(GATE 2012: 1 Mark)

115. For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r1 and r2 of the circles are (a) r1 = 0.707d, r2 = 2.782d (b) r1 = 0.707d, r2 = 1.932d (c)   r1 = 0.707d, r2 = 1.545d (d) r1 = 0.707d, r2 = 1.307d (GATE 2011: 2 Marks) 116. Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is (b) 8.73 dB (a) 11.90 dB (c) 6.79 dB (d) 5.33 dB (GATE 2011: 2 Marks)

Ch wise GATE_ECE_CH07_Communications.indd 464

119. A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount e and decreases that of the second by e. After encoding, the entropy of the source (a) increases (b) remains the same (c) increases only if N = 2 (d) decreases (GATE 2012: 1 Mark) 120. In a baseband communications link, frequencies up to 3500 Hz are used for signalling. Using a raised cosine pulse with 75% excess bandwidth and for no intersymbol interference, the maximum possible signalling rate in symbols/s is (b) 2625 (a) 1750 (d) 5250 (c) 4000 (GATE 2012: 1 Mark) 121. The signal m(t) as shown is applied both to a phase modulator (with kp as the phase constant) and a frequency modulator (with kf as the frequency constant) having the same carrier frequency

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Chapter 7  •  Communications

m(t)

(a)

2

−2

0

2

4

6

8

t(seconds)

−2

1 4 2 5   (b)     (c)     (d)   2 9 3 9 (GATE 2013: 2 Marks)

126. Consider two identically zero-mean random variables U and V. Let the cumulative distribution functions of U and 2 V be F(x) and G(x), respectively. Then, for all values of x (a) F(x) - G(x) ≤ 0 (b) F(x) - G(x) ≥ 0 (c) [F(x) - G(x)] . x ≤ 0 (d) [F(x) - G(x)] . x ≥ 0 (GATE 2013: 2 Marks)

The ratio kp/kf (in radian/Hz) for the same maximum 127. Let U and V be two independent and identically dis1 tributed random variables such that P (U = +1) = P (U = -1) = . phase deviation is 2 1 (a) 8p   (b)  4p   (c)  2p  (d)   p P (U = +1) = P (U = -1) = . The entropy H(U + V) in bits is 2 (GATE 2012: 2 Marks) 122. A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) = 9/10, then the probability of error for an optimum receiver will be (a) 7/80 (b) 63/80 (c) 9/10 (d) 1/10 (GATE 2012: 2 Marks)

(a) 3/4 (c) 3/2

(GATE 2013: 2 Marks) Common Data Questions for 128 and 129: Bits 1 and 0 are transmitted with equal probability. At the receiver, the PDF of the respective received signals for both bits are as shown in the following figure:

123. A BPSK scheme operating over an AWGN channel with noise power spectral density of N0/2 uses equiprobable signals s1 (t ) =

1 0.5

2E 2E sin(ω c t ) and s2 (t ) = sin(ω c t ) T T

over the symbol internal (0, T ). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is ⎛ E ⎞ ⎛ 2E ⎞ Q⎜ (a) Q ⎜ (b) ⎟ ⎟ ⎝ No ⎠ ⎝ No ⎠ ⎛ E ⎞ ⎛ E ⎞ Q⎜ (c) Q ⎜ (d) ⎟ ⎟ ⎝ 2 No ⎠ ⎝ 4 No ⎠ (GATE 2012: 2 Marks) 124. The bit rate of digital communication system is R kbps. The modulation used is 32 QAM. The minimum bandwidth required for ISI-free transmission is (a) R/10 Hz (b) R/10 kHz (c) R/5 Hz (d) R/5 kHz (GATE 2013: 1 Mark) 125. Let U and V be two independent zero mean Gaussian random variables of variances 1/4 and 1/9, respectively. The probability P(3V ≥ 2U) is

Ch wise GATE_ECE_CH07_Communications.indd 465

(b) 1 (d) log23

−1

1

4

128. If the detection threshold is 1, the bit error rate (BER) will be (a)

1 1 1 1   (b)     (c)     (d)   4 2 8 16 (GATE 2013: 2 Marks)

129. The optimum threshold to achieve minimum bit error rate is (a)

1 3 4   (b)     (c)  1  (d)   2 2 5 (GATE 2013: 2 Marks)

130. If calls arrive at a telephone exchange such that the time of arrival of any call is independent of the time of arrival of earlier or future calls, the probability distribution function of the total number of calls in a fixed time interval will be (a) Poisson (b) Gaussian (c) Exponential (d) Gamma (GATE 2014: 1 Mark)

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131. Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E(X), is . (GATE 2014: 1 Mark) 132. Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index (m) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V, is . (GATE 2014: 1 Mark) 133. Consider an FM signal f(t)   =   cos[2pfct + b1sin 2pf1t + b2sin2pf2t]. The maximum deviation of the instantaneous frequency from the carrier frequency fc is (a) b1f1 + b2f2 (b) b1f2 + b2f1 (c) b1 + b2 (d) f 1 + f2 (GATE 2014: 1 Mark) 134. In a double side-band (DSB) full carrier AM transmission system, if the modulation index is doubled, then the ratio of total sideband power to the carrier power . increases by a factor of (GATE 2014: 1 Mark) 135. The capacity of a binary symmetric channel (BSC) with cross-over probability 0.5 is . (GATE 2014: 1 Mark) 136. The capacity of a band-limited additive white Gaussian noise (AWGN) channel is given by P ⎞ ⎛ C = W log 2 ⎜1 + 2 ⎟ bits per second (bps), where ⎝ σ W⎠ W is the channel bandwidth, P is the average power received and s 2 is the one-sided power spectral density of the AWGN. P = 1000, the channel capacity (in kbps) σ2 with infinite bandwidth (W → ∞) is approximately For a fixed

(a) 1.44 (c) 0.72 

(b) 1.08 (d) 0.36 (GATE 2014: 1 Mark)

137. In a code-division multiple access (CDMA) system with N = 8 chips, the maximum number of users who can be assigned mutually orthogonal signature sequences is . (GATE 2014: 1 Mark) 138. Let x be a real-valued random variable with E[X] and E[X2] denoting the mean values of X and X2, respectively. The relation which always holds true is

Ch wise GATE_ECE_CH07_Communications.indd 466

(a) (E[X])2 > E[X2] (c)  E[X2] = (E[X])2

(b)  E[X2] ≥ (E[X])2 (d)  E[X2] > (E[X])2 (GATE 2014: 2 Marks)

139. Consider a random process X (t ) = 2 sin( 2π t + φ ), where the random phase j is uniformly distributed in the interval [0, 2p]. The auto-correlation E[X(t1)X(t2)] is (a) cos[2p(t1 + t2)] (b) sin[2p(t1 - t2)] (c) sin[2p(t1 + t2)] (d) cos[2p(t1 - t2)] (GATE 2014: 2 Marks) 140. The input to a 1-bit quantizer is a random variable X with pdf fx(x) = 2e−2x for x ≥ 0 and fx(x) = 0 for x < 0. For outputs to be of equal probability, the quantizer threshold should be . (GATE 2014: 2 Marks) 141. Let X1, X2 and X3 be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X1 + X2 ≤ X3} is . (GATE 2014: 2 Marks) 142. Let X(t) be a wise sense stationary (WSS) random process with power spectral density SX(f). If Y(t) is the process defined as Y(t) = X(2t - 1), the power spectral density SY(f) is (a) SY ( f ) =

1 ⎛ f ⎞ - jπ f SX ⎜ ⎟ e 2 ⎝ 2⎠

(b) SY ( f ) =

1 ⎛ f ⎞ - jπ f / 2 SX ⎜ ⎟ e 2 ⎝ 2⎠

(c) SY ( f ) =

1 ⎛ f⎞ SX ⎜ ⎟ 2 ⎝ 2⎠

(d) SY ( f ) =

1 ⎛ f ⎞ - j 2π f SX ⎜ ⎟ e 2 ⎝ 2⎠ (GATE 2014: 2 Marks)

143. A real band-limited random process X(t) has two-sided power spectral density -6 ⎪⎧10 (3000 - f Watts/Hz for f ≤ 3 kHz SX ( f ) = ⎨ otherwise ⎪⎩0

where f is the frequency expressed in Hz. The signal X(t) modulates a carrier cos16000pt and the resultant signal is passed through an ideal band-pass filter of unity gain with centre frequency of 8 kHz and band-width of 2 kHz. The output power (in Watts) is . (GATE 2014: 2 Marks)

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Chapter 7  •  Communications

144. A binary random variable X takes the value of 1 with probability 1/3. X is input to a cascade of two independent identical binary symmetric channels (BSCs) each with crossover probability 1/2. The output of BSCs are the random variables Y1 and Y2 as shown in the figure. Y1

X

BSC

The value of H(Y1) + H(Y2) in bits is . (GATE 2014: 2 Marks) 145. Consider a communication scheme where the binary valued signal X satisfies P{X = +1} = 0.75 and P{X = −1} = 0.25. The received signal Y = X + Z, where Z is a Gaussian random variable with zero mean and variance s 2. The received signal Y is fed to the threshold detector. The output  is: of the threshold detector X  = ⎧+1, Y > t X ⎨ ⎩ -1, Y ≤ t

0/1

BPSK Modulator

+

0/1 BPSK Demodulator

AWGN Channel 2

(a) strictly positive (b) zero (c) strictly negative (d) strictly positive, zero, or strictly negative depending on the non-zero value of s 2 (GATE 2014: 2 Marks) 146. Consider the Z-channel given in the figure. The input is 0 or 1 with equal probability. 1.0 0

0 Output

0.25

If the BER of this system is Q(b γ ), then the value of b is . (GATE 2014: 2 Marks) 148. Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s1(t) = acos2pf1t and s2(t) = acos2pf2t, where a = 4 mV. Assume an AWGN channel with two-sided noise power spectral density N 0 / 2 = 0.5 × 10 -12 W/Hz. Using an optimal receiver and the relation Q( v ) =

1

∫ 2π

∞ v

2

e - u / 2 du, the

bit error probability for a data rate of 500 kbps is

  ≠ X}, To achieve a minimum probability of error P { X the threshold t should be

1

AWGN Channel 1

Y2

BSC

Input

467

(a) Q(2) (b) Q( 2 2) Q( 4 2 ) (c) Q(4) (d) (GATE 2014: 2 Marks) 149. In a PCM system, the signal m(t) = {sin (100pt) + cos(100pt)} V is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is . (GATE 2014: 2 Marks) 150. An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with bandwidth 100 kHz. The bit rate is 200 kbps and the system characteristic is a raised cosine spectrum with 100% excess bandwidth. The minimum value of M is . (GATE 2014: 2 Marks)

1 0.75

If the output is 0, the probability that the input is also 0 equals (GATE 2014: 2 Marks) 147. Let Q( γ ) be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter g is a function of bit energy and noise power spectral density. A system with two independent and identical AWGN channels with noise power spectral density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.

Ch wise GATE_ECE_CH07_Communications.indd 467

151. Consider a discrete-time channel Y = X + Z, where the additive noise Z is signal-dependent. In particular, given the transmitted symbol X ∈ {-a, +a} at any instant, the noise sample Z is chosen independently from a Gaussian distribution with mean bX and unit variance. Assume a threshold detector with zero threshold at the receiver. When b = 0, the BER was found to be Q(a) = 1 × 10−8. (Q( v ) =

1

∞

∫e 2π

- u2 / 2

du, and for v > 1, use Q( v ) ≈ e

- v2 / 2

)

v

When b = −0.3, the BER is closest to

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GATE ECE Chapter-wise Solved papers

(a) 10−7

(b) 10−6

(c) 10−4

(d) 10−2 (GATE 2014: 2 Marks)

(a)

15 π⎞ ⎛ cos ⎜ 40π t - ⎟ ⎝ 2 4⎠

152. A fair coin is tossed repeatedly until a ‘Head’ appears for the first time. Let L be the number of tosses to get this first ‘Head’. The entropy H(L) in bits is . (GATE 2014: 2 Marks)

(b)

15 ⎛ sin(π t ) ⎞ π⎞ ⎛ cos ⎜10π t + ⎟ ⎝ 2 ⎜⎝ π t ⎟⎠ 4⎠

(c)

15 π⎞ ⎛ cos ⎜10π t - ⎟ ⎝ 2 4⎠

⎛ ⎞  (t ) sin((d) 153. Consider the signal s(t ) = m(t ) cos( 2π f c t ) + m 2π f c t15 ) sin(π t ) cos ⎛⎜ 40π t - π ⎞⎟ ⎜⎝ π t ⎟⎠ ⎝ 2 2⎠  (t ) denotes the Hilbert trans (t ) sin( 2π f c t ) where m s(t ) = m(t ) cos( 2π f c t ) + m form of m(t) and the bandwidth of m(t) is very small (GATE 2015: 1 Mark) compared to fc. The signal s(t) is a 158. The modulation scheme commonly used for transmis(a) high-pass signal sion for GSM mobile terminals is (b) low-pass signal (a) 4 - QAM (c) band-pass signal (b) 16 - PSK (d) double sideband suppressed carrier signal (c) Walsh - Hadamard orthogonal codes (GATE 2015: 1 Mark) (d) Gaussian Minimum Shift Keying (GMSK) (GATE 2015: 1 Mark) 154. A message signal m(t ) = A sin( 2 π f t ) is used to modum

m

late the phase of a carrier Accos(2p fct) to get the modulated signal y(t) = Accos(2p fct + m(t)). The bandwidth of y(t) (a) depends on Am but not on fm (b) depends on fm but not on Am (c) depends on both Am and fm

159. Let the random variable X represent the number of times a fair coin needs to be tossed till two ­consecutive heads appear for the first time. The expectation of X is . (GATE 2015: 2 Marks) 160. { X n }n = -∞ is an independent and identically distributed n= ∞

(i, i, d) random process with Xn equally likely to be +1

(d) does not depend on Am and fm (GATE 2015: 1 Mark) 155. A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size ∆ of the delta modulator are 20,000 samples per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is (a)

1 1 (b) 2π π

(c)

2 (d) p π (GATE 2015: 1 Mark)

156. A sinusoidal signal of amplitude A is quantized by a uniform quantizer. Assume that the signal utilizes all the representation levels of the quantizer. If the signal to quantization noise ratio is 31.8 dB, the number of levels in the quantizer is . (GATE 2015: 1 Mark) π⎞ ⎛ 157. The signal cos ⎜10π t + ⎟ is ideally sampled at ⎝ 4⎠ a sampling frequency of 15 Hz. The sampled signal is passed through a filter with impulse response ⎛ sin(π t ) ⎞ π⎞ ⎛ ⎜⎝ π t ⎟⎠ cos ⎜⎝ 40π t - 2 ⎟⎠ . The filter output is

Ch wise GATE_ECE_CH07_Communications.indd 468

or −1. {Yn }n = -∞ is another random process obtained as n= ∞

Yn =  Xn + 0.5 Xn - 1. The autocorrelation function of

{Yn }nn== ∞-∞

denoted by RY[k] is

(a)

(b) RY [k]

1.25

0.5

0.5

RY [k] 1

−3 −2 −1 0 1

2 3 k

(c)

−3 −2 −1 0 1

2 3 k

(d) RY [k] 0.5 0.25

1.25

RY [k]

0.5 0.25

−2 −1 0 1

2

0.25 k

1.25 0.25

−3 −2 −1 0 1

2 3 k

(GATE 2015: 2 Marks) 161. The variance of the random variable X with probability 1 density function f ( x ) = x e - x is . 2 (GATE 2015: 2 Marks)

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Chapter 7  •  Communications

162. A random binary wave y(t) is given by y (t ) =

∞

∑X

n = -∞

n

469

165. Let X ∈{0,1} and Y ∈{0,1} be two independent binary random variables. If P(X = 0) = p and P(Y = 0) = q, then P(X + Y ≥ 1) is equal to (a) pq + (1 - p)(1 - q) (b) pq

p(t - nT - f )

where p(t) = u(t) - u(t - T), u(t) is the unit step function and f is an independent random variable with uni(c) p(1 - q) (d) 1 - pq form distribution in (0, T). The sequence {Xn} consists (GATE 2015: 2 Marks) of independent and identically distributed binary valued random variables with P{Xn = +1} = P{Xn = −1} = 0.5 166. Consider a binary, digital communication system which for each n. uses ⎡ 3T ⎞ ⎤ g(t) and -g(t) for transmitting bits over an ⎛ pulses ⎛ 3T ⎞ The value of the autocorrelation Ryy ⎜ ⎟  E ⎢ y(t ) yAWGN t - ⎟channel. If the receiver uses a matched filter, ⎜ ⎝ ⎝ 4⎠ 4 ⎠ ⎥⎦ ⎣ which one of the following pulses will give the mini⎡ ⎛ 3T ⎞ ⎤ ⎛ 3T ⎞ Ryy ⎜ ⎟  E ⎢ y(t ) y ⎜ t - ⎟ ⎥ equals . mum probability of bit error? ⎝ ⎝ 4⎠ 4 ⎠⎦ ⎣ (GATE 2015: 2 Marks)

(a)

163. The input X to the Binary Symmetric Channel (BSC) shown in the figure is ‘1’ with probability 0.8. The crossover probability is 1/7. If the received bit Y = 0, the con. ditional probability that ‘1’ was transmitted is

(b) g(t)

g(t)

1

1

0

1

t 0

X 0

1

t

Y 0

6/7

P [X = 0] = 0.2 1/7

1/7

(d)

(c) g(t)

g(t)

1

1

P [X = 1] = 0.8 1

1 6/7

(GATE 2015: 2 Marks) 0 1 164. A source emits bit 0 with probability and bit 1 with 3 2 probability . The emitted bits are communicated to the 3 receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R are as

and

⎧1 ⎪ , -3 ≤ x ≤ 1 f R|0 ( r ) = ⎨ 4 ⎪⎩0, otherwise ⎧1 ⎪ , -1 ≤ x ≤ 5 f R|1 ( r ) = ⎨ 6 ⎪⎩0, otherwise

The minimum decision error probability is

1

t

0

1

t

(GATE 2015: 2 Marks) 167. The transmitted signal in a GSM system is of 200 kHz bandwidth and 8 users share a common bandwidth using TDMA. If at a given time 12 users are talking in a cell, the total bandwidth of the signal received by the base station of the cell will be at least (in kHz) . (GATE 2015: 2 Marks) 168. Consider binary data transmission at a rate of 56 kbps using baseband binary pulse amplitude modulation (PAM) that is designed to have a raised-cosine spectrum. The transmission bandwidth (in kHz) required for a roll-off factor of 0.25 is . (GATE 2016: 1 Mark)

1 12

169. A superheterodyne receiver operates in the frequency range of 58 MHz−68 MHz. The intermediate frequency fIF and local oscillator frequency fLO are chosen such 1 1 that fIF ≤ fLO. It is required that the image frequencies fall (c) (d) 9 6 fIF outside the 58 MHz−68 MHz band. The minimum required fIF (in MHz) is . (GATE 2015: 2 Marks)

(a) 0

Ch wise GATE_ECE_CH07_Communications.indd 469

(b)

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GATE ECE Chapter-wise Solved papers

(GATE 2016: 1 Mark)

(b)

170. The amplitude of a sinusoidal carrier is modulated by a single sinusoid to obtain the amplitude modulated signal s(t) = 5 cos 1600πt + 20 cos 1800πt + 5 cos 2000πt. The value of the modulation index is .

y(t) 0.5

t 0

(GATE 2016: 1 Mark) 171. A speech signal is sampled at 8 kHz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in kHz) required for transmission is . (GATE 2016: 1 Mark) 172. For a superheterodyne receiver, the intermediate frequency is 15 MHz and the local oscillator frequency is 3.5 GHz. If the frequency of the received signal is greater than the local oscillator frequency, then the image fre. quency (in MHz) is

(c)

2TS

TS/2

2TS

t 0 (d)

y(t) 1

t 0

(GATE 2016: 1 Mark) 176. Consider the plot of f(x) versus x as shown below. f(x) +2 −5 Suppose

0 −2

+5

x

. Which one of the ­following

is a graph of F(x)? (a)

F(x)

−5

175. A binary baseband digital communication system employs the signal ⎧ 1 , 0 ≤ t ≤ TS ⎪ p (t ) = ⎨ TS ⎪ 0, otherwise ⎩

TS

1

174. A discrete memoryless source has an alphabet (a1, a2, ⎧ 1 1 1 1⎫ a3, a4) with corresponding probabilities ⎨ , , , ⎬ . ⎩ 2 4 8 8⎭ The minimum required average codeword length in bits to represent this source for error-free reconstruction is _____. (GATE 2016: 1 Mark)

2TS

y(t)

(GATE 2016: 1 Mark) 173. An analog baseband signal, band limited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities p1 = p4 = 0.125 and p2 = p3. The information rate (bits/s) of the message source is . (GATE 2016: 1 Mark)

TS

(b)

0

+5

x

0

+5

x

F(x) −5

for transmission of bits. The graphical representation of the matched filter output y(t) for this signal will be (a)

(c)

y(t)

F(x) −5

1/TS

0

+5 x

t 0

Ch wise GATE_ECE_CH07_Communications.indd 470

TS

2TS

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471

Chapter 7  •  Communications

(d)

F(x) +

−5

+5 X(t) 0

(GATE 2016: 2 Marks) 177. An analog pulse s(t) is transmitted over an additive white Gaussian noise (AWGN) channel. The received signal is r(t) = s(t) + n(t), where n(t) is additive white Gaussian noise with power spectral density N 0 / 2. The received signal is passed through a filter with impulse response h(t). Let Es and Eℎ denote the energies of the pulse s(t) and the filter h(t), respectively. When the signalto-noise ratio (SNR) is maximized at the output of the filter (SNRmax), which of the following holds? (a) Es = Eh ; SNR max =

2 Es N0

(b) Es = Eh ; SNR max =

Es 2N 0

(c) Es > Eh ; SNR max >

2 Es N0

(d) Es < Eh ; SNR max >

2 Eh N0

n = -∞

⎧1 0 ≤ t ≤ T = ⎨ ⎩0 otherwise

If there is a null at f = 1/3T in the power spectral density . of X(t), then k is (GATE 2016: 2 Marks) 179. Consider random process X(t) = 3V(t) − 8, where V(t) is a zero mean stationary random process with autocorrelation Rv(t ) = 4e−5|t|. The power in X(t) is . (GATE 2016: 2 Marks) 180. A wide sense stationary random process X(t) passes through the LTI system shown in the following figure. If the autocorrelation function of X(t) is RX(t ), then the autocorrelation function RY(t ) of the output Y(t) is equal to

Ch wise GATE_ECE_CH07_Communications.indd 471

(a) 2RX(t ) + RX(t  − T0) + RX(t  + T0) (b) 2RX(t ) − RX(t  − T0) − RX(t  + T0) (c) 2RX(t ) + 2RX(t  − 2T0) (d) 2RX(t ) − 2RX(t  − 2T0) (GATE 2016: 2 Marks) 181. Two random variables x and y are distributed according to

. (GATE 2016: 2 Marks)

∑ β g (t - nT ), where g (t ) n

Delay = T0

The probability P(X + Y ≤ 1) is

178. An information source generates a binary sequence {an} ⋅ an can take one of the two possible values −1 and +1 with equal probability and are statistically independent and identically distributed. This sequence is precoded to obtain another sequence {bn}, as bn = an + kan−3. The sequence {bn} is used to modulate a pulse g(t) to generate the baseband signal ∞

−

⎧( x + y ) 0 ≤ x ≤ 1 0 ≤ y ≤ 1 f X ,Y ( x, y ) = ⎨ otherwise ⎩ 0,

(GATE 2016: 2 Marks)

X (t ) =

Y(t)

x

182. An ideal band-pass channel 500 Hz–2000 Hz is deployed for communication. A modem is designed to transmit bits at the rate of 4800 bits/s using 16-QAM. The rolloff factor of a pulse with a raised cosine spectrum that utilizes the entire frequency band is . (GATE 2016: 2 Marks) 183. The digital communication system uses a repetition code for channel encoding/decoding. During transmission, each bit is repeated three times (0 is transmitted as 000, and 1 is transmitted as 111). It is assumed that the source puts out symbols independently and with equal probability. The decoder operates as follows: In a block of three received bits, if the number of zeros exceeds the number of ones, the decoder decides in favor of a 0, and if the number of ones exceeds the number of zeros, the decoder decides in favor of a 1. Assuming a binary symmetric channel with crossover probability p = 0.1, the average probability of error is . (GATE 2016: 2 Marks) 184. Consider a discrete memoryless source with alphabet S = {s0, s1, s2, s3, s4, …} and respective probabilities of ⎧1 1 1 1 1 ⎫ occurrence P = ⎨ , , , , , …⎬ . The entropy of ⎩ 2 4 8 16 32 ⎭ the source (in bits) is . (GATE 2016: 2 Marks)

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(a)

186. The bit error probability of a memory less binary symmetric channel is 10−5. If 105 bits are sent over this channel, then the probability that not more than one bit will be in error is . (GATE 2016: 2 Marks)

(b)

0

1

P

0

1

P

0

1

P

Capacity

1

(c) 1 Capacity

187. A binary communication system makes use of the symbols ‘zero’ and ‘one’. There are channel errors. Consider the following events: x0 : a ‘zero’ is transmitted x1 : a ‘one’ is transmitted y0 : a ‘zero’ is received y1: a ‘one’ is received

1

Capacity

185. A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidth of 4.0 kHz and twosided noise power spectral density η/2 = 2.5 × 10-5 W/Hz. If information at the rate of 52 kbps is to be transmitted over this channel with arbitrarily small bit error rate, then the minimum bit-energy Eb (in mJ/bit) necessary is . (GATE 2016: 2 Marks)

The following probabilities are given:

The information in bits that you obtain when you learn which symbol has been received (while you know that a ‘zero’ has been transmitted) is ______. (GATE 2016: 2 Marks) 188. Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the ­distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link (a) increases by a factor of 2 (b) decreases by a factor of 2 (c) remains unchanged (d) decreases by a factor of 2 (GATE 2017: 1 Mark) 189. Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memoryless binary symmetric channel with crossover probability p?

Ch wise GATE_ECE_CH07_Communications.indd 472

(d) 1 Capacity

1 3 1 P ( x0 ) = , P ( y0 x0 ) = , and P ( y0 x1 ) = . 2 4 2

0

1

P

(GATE 2017: 1 Mark) 190. Which one of the following statements about differential pulse code modulation (DPCM) is true? (a) The sum of message signal sample with its prediction is quantized. (b) The message signal sample is directly quantized, and its prediction is not used. (c) The difference of message signal sample and a radom signal is quantized. (d) The difference of message signal sample with its prediction is quantized. (GATE 2017: 1 Mark) 191. A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal to quantization noise ratio (SQNR) at the output of the quan-

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Chapter 7  •  Communications

473

tizer is 40 dB. The minimum number of bits per sample needed to achieve the desired SQNR is ____________. (GATE 2017: 1 Mark)

2. Assume that U and V are statistically independent. The mean value of the random process at t = 2 is .

192. Let X(t) be a wide sense stationary random process with the power spectral density SX(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal lowpass filter with the frequency response

194. Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is .

⎧ ⎪⎪1, H( f ) = ⎨ ⎪0, ⎪⎩

1 Hz 2 1 f > Hz 2 f ≤

as shown in Figure (b). The output of the lowpass filter is Y(t). SX(f ) exp(|f |)

0 (a)

X(t)

(GATE 2017: 2 Marks)

Ideal lowpass filter h(t) cutoff = 1/2 Hz

Y(t)

(b)

Let E be the expectation operator and consider the following statements. I. E(X(t)) = E(Y(t)) II. E(X2(t)) = E(Y2(t))

(GATE 2017: 2 Marks) 195. The signal x(t) = sin (14000 π t), where t is in seconds, is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low-pass filter with frequency response H(f) as follows: ⎧1, H( f ) = ⎨ ⎩0,

f ≤ 12 kHz f > 12 kHz

What is the number of sinusoids in the output and their frequencies in kHz? (a) Number = 1, frequency = 7 (b) Number = 3, frequencies = 2, 7, 11 (c) Number = 2, frequencies = 2, 7 (d) Number = 2, frequencies = 7, 11 (GATE 2017: 2 Marks) 196. The unmodulated carrier power in an AM transmitter is 5 kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum unmodulated carrier power (in kW) that can be used without overloading the transmitter is . (GATE 2017: 2 Marks) 197. A modulating signal given by x(t) = 5 sin (4π 103t - 10π cos 2π 103t) V is fed to a phase modulator with phase deviation constant kp = 5 rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is . (GATE 2017: 2 Marks)

III. E(Y2(t)) = 2 Select the correct option: (a) Only I is true (b) Only II and III are true (c) Only I and II are true (d) Only I and III are true

X(t) = U + V(t),

198. Let (X1, X2) be independent random variables. X1 has mean 0 and variance 1, while X2 has mean 1 and variance 4. The mutual information I(X1; X2) between X1 and X2 in bits is __________. (GATE 2017: 2 Marks)

where U is a zero-mean Gaussian random variable and V is a random variable uniformly distributed between 0 and

199. Consider a binary memoryless channel characterized by the transition probability diagram shown in the figure.

(GATE 2017: 2 Marks) 193. Consider the random process

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GATE ECE Chapter-wise Solved papers

If the detection threshold is zero, then the probability of error (correct to two decimal places) is .

0.25 0

0 0.25

(GATE 2018: 1 Mark)

0.75 1

1 0.75

The channel is (a) lossless (c) useless

(b) noiseless (d) deterministic (GATE 2017: 2 Marks)

200. In binary frequency shift keying (FSK), the given signal waveforms are u0(t) = 5 cos (20000 p t);   0 ≤ t ≤ T u1(t) = 5 cos (22000 p t);   0 ≤ t ≤ T, where T is the bit-duration interval and t is in seconds. Both u0(t) and u1(t) are zero outside the interval 0 ≤ t ≤ T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milli­seconds) required to have u0(t) and u1(t) uncorrelated is (a) 0.25 ms (b) 0.5 ms (c) 0.75 ms (d) 1.0 ms (GATE 2017: 2 Marks) 201. Consider the following amplitude modulated signal: s(t) = cos(2000pt) + 4 cos(2400pt) + cos(2800pt) The ratio (accurate to three decimal places) of the power of the message signal to the power of the carrier signal is . (GATE 2018: 1 Mark) 202. Consider a binary channel code in which each codeword has a fixed length of 5 bits. The Hamming distance between any pair of distinct codewords in this code is at least 2. The maximum number of codewords such a code can contain is . (GATE 2018: 1 Mark)

204. Let c(t) = Ac cos(2pfct) and m(t) = cos(2pfmt). It is given that fc >> 5fm. The signal c(t) + m(t) is applied to the input of a non-linear device, whose output vo(t) is related to the input vi(t) as vo(t) = avi(t) + bvi2(t),where a and b are positive constants. The output of the non-linear device is passed through an ideal-pass filter with center frequency fc and bandwidth 3fm, to produce an amplitude modulated (AM) wave. If it is desired to have the sideband power of the AM wave to be half of the carrier power, then a/b is (a) 0.25 (b) 0.5 (c) 1 (d) 2 (GATE 2018: 2 Marks) 205. Consider a white Gaussian noise process N(t) with twosided power spectral density SN(f ) = 0.5 W/Hz as input to 2

a filter with impulse response 0.5e - t 2 (where t is in seconds) resulting in output Y(t). The power inY(t) in watts is (a) 0.11 (b) 0.22 (c) 0.33 (d) 0.44 (GATE 2018: 2 Marks) 206. A random variable X takes values -0.5 and 0.5 with probabilities 1/4 and 3/4, respectively. The noisy observation of X is Y = X + Z, where Z has uniform probability density over the interval (-1, 1). X and Z are independent. If the MAP rule based detector outputs X as  ⎧ -0.5, X =⎨ ⎩0.5,

Y ε 0 . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be five times the minimum field. The intrinsic impedance of the medium should be (a) 120p Ω (b) 60p Ω (c) 600p Ω (d) 24p Ω (GATE 2004: 2 Marks) 38. A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage of the power that is reflected back is (a) 57.73 (b) 33.33 (c) 0.11 (d) 11.11 (GATE 2004: 2 Marks)

(a) (b) (c) (d)

unit circles constant resistance circles constant reactance circles constant reflection coefficient circles (GATE 2005: 1 Mark)

42. Characteristic impedance of a transmission line is 50 Ω. Input impedance of the open circuited line is ZOC = 100 + j150 Ω. When the transmission line is short circuited, then the value of the input impedance will be (a) 50 Ω (b) 100 + j150 Ω (c) 7.69 + j11.54 Ω

(d) 7.69 - j11.54 Ω (GATE 2005: 2 Marks)

Common Data for Questions 43 and 44: Voltage standingwave pattern in a lossless transmission line with characteristic impedance 50 Ω and a resistive load is shown in the following figure.

39. In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz? (a) to increase the sensitivity of measurement (b) to transmit the signal to a far-off place (c) to study amplitude modulation (d) because crystal detector fails at microwave frequencies (GATE 2004: 2 Marks) 40. The magnetic field intensity vector of a plane wave is given by  H ( x, t ) = 10 sin(50000t + 0.004 x + 30)a y where a y denotes the unit vector in y direction. The wave is propagating with a phase velocity (a) 5 × 104 m/s (b) 3 × 108 m/s 7 (c) 1.25 × 10 m/s (d) 3 × 106 m/s (GATE 2005: 1 Mark) 41. Many circles are drawn in a Smith chart used for transmission line calculations. The circles shown in the following figure represent

V(Z)

4

1 2

l

l/2 Load

43. The value of the load resistance is (a) 50 Ω (b) 200 Ω (c) 12.5 Ω (d) 0 Ω (GATE 2005: 2 Marks) 44. The reflection coefficient is given by (a) -0.6 (b) -1 (c) 0.6 (d) 0 (GATE 2005: 2 Marks) 45. Which one of the following does represent the electric field lines for the TE02 mode in the cross-­section of a hollow rectangular metallic waveguide? (a) y

x

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Chapter 8  • Electromagnetics

(a) 10 W (c) 0.1 W

(b) y

x (c) y

x (d) y

x (GATE 2005: 2 Marks) 46. Two identical and parallel dipole antennas are kept apart   by a distance of l/4 in the H plane. They are fed with equal currents but the rightmost antenna has a phase shift of +90°. The radiation pattern is given as (a) (b)

(b) 1 W (d) 0.01 W (GATE 2006: 1 Mark)

49. A medium of relative permittivity er2 = 2 forms an interface with free space. A point source of electromagnetic energy is located in the medium at a depth of 1 m from the interface. Due to total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by (a) 2p m2 (b) p 2m2 π (c) (d) p m2 2m 2 (GATE 2006: 2 Marks) 50. A medium is divided into regions I and II about x = 0 plane, as shown in the following figure. An electro magnetic wave with electric field E 1 = 4 a x + 3a y + 5a z is incident normally on the interface from region I. The  electric field E2 in region II at the interface is Region I

Region II

σ1 = 0, µ1 = µ 0 εr1 = 3

(c)



(d)

E1

47. The electric field of an electromagnetic wave propagating in the positive z-direction is given by

π⎞ ⎛ E = a x sin(ω t - β z ) + a y sin ⎜ ω t - β z + ⎟ ⎝ 2⎠

The wave is (a) linearly polarized in the z-direction (b) elliptically polarized (c) left-hand circularly polarized (d) right-hand circularly polarized (GATE 2006: 1 Mark)

48. A transmission line is feeding 1 W of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated is

Ch wise GATE_ECE_CH08_Electromagnetics.indd 523

σ2 = 0, µ2 = µ0 εr2 = 4

E2

x0

  (a) E2 = E1 (b) 4 a x + 0.75a y - 1.25a z (c) 3a x + 3a y + 5a z (d) -3a x + 3a y + 5a z (GATE 2006: 2 Marks) 51. When a plane wave travelling in free space is incident normally on a medium having er = 4.0, the fraction of power transmitted into the medium is given by (a)

8 1 (b) 9 2

(c)

1 5 (d) 3 6 (GATE 2006: 2 Marks)

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GATE ECE Chapter-wise Solved papers

Common Data for Questions 52 and 53: A 30-V battery with zero source resistance is c­ onnected to a coaxial line of characteristic impedance of 50 Ω at t = 0 s and terminated in an unknown resistive load. The line length is such that it takes 400 ms for an electromagnetic wave to travel from source end to load end and vice versa. At t = 400 ms, the voltage at the load end is found to be 40 V. 52. The load resistance is (a) 25 Ω (c) 75 Ω

(b) 50 Ω (d) 100 Ω (GATE 2006: 2 Marks)

53. The steady-state current through the load resistance is (a) 1.2 A (b) 0.3 A (c) 0.6 A (d) 0.4 A (GATE 2006: 2 Marks) 54. A rectangular waveguide having TE10 mode as the dominant mode is having a cut-off frequency of 18 GHz for the TE30 mode. The inner broad-wall dimension of the rectangular waveguide is (a) 5/3 cm (b) 5 cm (c) 5/2 cm (d) 10 cm (GATE 2006: 2 Marks) 55. A mast antenna consisting of a 50 m long ­vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna in ohms is (a)

2π π (b) 5 5

(c)

4π 2 5

2

(GATE 2007: 1 Mark) 58. A right circularly polarized (RCP) plane wave is incident at an angle of 60° to the normal on an air-dielectric interface as shown in the following figure. If the reflected wave is linearly polarized, the relative dielectric constant ε r 2 is Linearly polarized RCP 60˚

60˚

Air

e r1 = 1

Dielectric

e r2 qt

2

(d) 20p 2 (GATE 2006: 2 Marks)

56. If C is a closed curve enclosing a surface S, then  the   H magnetic field intensity  , the current density J and  the electric flux density D are related by      ⎛  ∂ D ⎞  (a) ∫∫ H ⋅ dS =  ∫C ⎜⎝ J + ∂t ⎟⎠ ⋅ dl S     ⎛  ∂D ⎞  (b) ∫ H ⋅ dl =  ∫∫S ⎜⎝ J + ∂t ⎟⎠ ⋅ dS C      ⎛  ∂D ⎞  (c)  ∫∫S H ⋅ dS = ∫C ⎜⎝ J + ∂t ⎟⎠ ⋅ dl      ⎛  ∂ D ⎞  (d)  ∫C H ⋅ dl = ∫∫S ⎜⎝ J + ∂t ⎟⎠ ⋅ dS (GATE 2007: 1 Mark)

Ch wise GATE_ECE_CH08_Electromagnetics.indd 524

57. A plane wave of wavelength (l) is travelling in a direction making an angle 30° with  the positive x-axis and 90° with positive y-axis. The E field of the plane wave can be represented as (E0 is a constant)  (a) E = yE0 e j [ω t - ( 3 π l ) x - (π l ) z ]  (b) E = yE0 e j [ω t - (π l ) x - ( 3π l ) z ]  (c) E = yE0 e j [ω t + ( 3π l ) x + (π l ) z ]  (d) E = yE0 e j [ω t -π l ) x - ( 3π l ) z ]

(a) 2 (b) 3 (c) 2 (d) 3 (GATE 2007: 2 Marks) 59. A load of 50 Ω is connected in shunt in a two-wire transmission line of Z0 = 50 Ω as shown in the following figure. The two-port ­scattering parameter matrix (S-matrix) of the shunt element is

Z0 = 50 Ω

⎡ 1 ⎢(a) ⎢ 2 ⎢ 1 ⎢⎣ 2

50 Ω

Z0 = 50 Ω

1 ⎤ 2 ⎥ ⎥ 1 - ⎥ 2 ⎥⎦

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Chapter 8  • Electromagnetics



⎡0 1 ⎤ (b) ⎢ ⎥ ⎣1 0⎦ ⎡ 1 2 ⎤ ⎢⎥ (c) ⎢ 3 3 ⎥ ⎢ 2 - 1⎥ ⎢⎣ 3 3 ⎥⎦ ⎡ 1 ⎢ 4 (d) ⎢ ⎢- 3 ⎢⎣ 4

3⎤ - ⎥ 4 ⎥ 1 ⎥ 4 ⎥⎦

(GATE 2007: 2 Marks)

60. The parallel branches of a two-wire transmission line are terminated in 100 Ω and 200 Ω resistors as shown in the following figure. The characteristic impedance of the line is Z0 = 50 Ω and each section has a length of l/4. The voltage reflection coefficient Γ at the input is

525

H0 is a constant, a and b are the dimensions along the x-axis and y-axis, respectively. The mode of propagation in the waveguide is (a) TE20 (b) TM11 (c) TM20 (d) TE10 (GATE 2007: 2 Marks)

63. A l/2 dipole is kept horizontally at a height of l0/2 above a perfectly conducting infinite ground plane. The  radiation pattern in the plane of the dipole ( E plane) looks approximately as y y (a) (b)

z

z (c)

y



z

(d)

y

z (GATE 2007: 2 Marks)

l/4 l/4

50

Ω

100 Ω

= Z0

Z0 = 50 Ω

Z 0

Γ

= 50

l/4

Ω

200 Ω

-5 7 (a) - j (b) 7 5 (c)

j

5 5 (d) 7 7 (GATE 2007: 2 Marks)

61. An air-filled rectangular waveguide has inner dimensions of 3 cm × 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance h0 = 377 W) (a) 308 W (b) 355 W (c) 400 W (d) 461 W (GATE 2007: 2 Marks)  62. The E field in a rectangular waveguide of inner dimensions a × b is given by  ωm ⎛ π ⎞ ⎛ 2π x ⎞ sin(ω t - β z ) y E = 2 ⎜ ⎟ H 0 sin ⎜ ⎝ a ⎟⎠ η ⎝ a⎠

Ch wise GATE_ECE_CH08_Electromagnetics.indd 525

64. For static electric and magnetic fields in an inhomogeneous source-free medium, which of the following represents the correct form of two of Maxwell’s equations?   (a) ∇ ⋅ E = 0; ∇ × B = 0   (b) ∇ ⋅ E = 0; ∇ ⋅ B = 0   (c) ∇ × E = 0; ∇ × B = 0   (d) ∇ × E = 0; ∇⋅ B = 0 (G ATE 2008: 1 Mark) 65. For a Hertz dipole  antenna, the half power beam width (HPBW) in the E plane is (a) 360° (b) 180° (c) 90° (d) 45° (GATE 2008: 1 Mark) 66. A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant er = 9). The magnitude of the reflection coefficient is (a) 0 (b) 0.3 (c) 0.5 (d) 0.8 (GATE 2008: 2 Marks) 67. One end of a lossless transmission line having the characteristic impedance of 75 Ω and length of 1 cm is short circuited. At 3 GHz, the input impedance at the other end of the transmission line is (a) 0 (b) resistive (c) capacitive (d) inductive (GATE 2008: 2 Marks)

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GATE ECE Chapter-wise Solved papers

68. In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is NOT preserved if (a) radius as well as operating wavelength is halved. (b) radius as well as operating wavelength is doubled. (c) radius is halved and operating wavelength is doubled. (d)  radius is doubled and operating wavelength is halved. (GATE 2008: 2 Marks) 69. A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is (a) 6.25 GHz (b) 6.0 GHz (c) 5.0 GHz (d) 3.75 GHz (GATE 2008: 2 Marks) 70. At 20 GHz, the gain of a parabolic dish antenna of 1 m diameter and 70% efficiency is (a) 15 dB (b) 25 dB (c) 35 dB (d) 45 dB (GATE 2008: 2 Marks) 71. Two infinitely long wires carrying current are as shown in the following figure. One wire is in the yz plane and parallel to the y-axis. The other wire is in the xy plane and parallel to the x-axis. Which components of the resulting magnetic field are non-zero at the origin?

Q; Cylindrical

P; Coaxial

R; Rectangular (a) Only P has no cut-off frequency (b) Only Q has no cut-off frequency (c) Only R has no cut-off frequency (d) All three have cut-off frequencies (GATE 2009: 1 Mark)   73. If a vector field V is related to another vector field A through V = ∇ × A, which of the following is true? Note: C and SC refer to any closed contour and any surface whose boundary is C.     (a)  ∫ V ⋅ dl = ∫ ∫ A ⋅ dS C

(b) (c) (d)

∫

C

∫

C

∫

C

  A ⋅ dl =

SC





∫ ∫ V ⋅ dS

SC

  ∇ × V ⋅ dl =   ∇ × A ⋅ dl =





∫ ∫ ∇ × A ⋅ dS

SC





∫ ∫ V ⋅ dS

SC

(GATE 2009: 2 Marks)

74. A magnetic field in air is measured to be  ⎛ x ⎞ y B = B0 ⎜ 2 a y - 2 a x . What current dis⎝ x + y2 x + y 2 ⎟⎠ tribution leads to this field?

z 1A

 B a ⎛ 1 ⎞ (a) J = - 0 z ⎜ 2 ,r ≠ 0 m0 ⎝ x + y 2 ⎟⎠ y

1A x (a) x, y, z components (c) y, z components

(b) x, y components (d) x, z components (GATE 2009: 1 Mark)

72. Which of the following statements is true regarding the fundamental modes of the metallic waveguides shown in the following figures?

Ch wise GATE_ECE_CH08_Electromagnetics.indd 526

 B a ⎛ 2 ⎞ ,r ≠ 0 (b) J = - 0 z ⎜ 2 m0 ⎝ x + y 2 ⎟⎠  (c) J = 0, r ≠ 0  B a ⎛ 1 ⎞ (d) J = 0 z ⎜ 2 ,r ≠ 0 m0 ⎝ x + y 2 ⎟⎠ (GATE 2009: 2 Marks) 75. A transmission line terminates in two branches, each of length l/4, as shown in the following figure. The branches are terminated by 50 Ω loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zin as seen by the source.

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Chapter 8  • Electromagnetics

l/4

4 l/

0Ω

ZL1 = 50 Ω

Short Z0 = 30 Ω l/8

0

=1 Z0

Z0 = 50 Ω

Z0 = 60 Ω

Z 0

Zin

= 0Ω

ZL2 = 50 Ω

(b) 100 Ω (d) 25 Ω (GATE 2009: 2 Marks)

76. The electric field component of a time harmonic plane EM wave travelling in a non-magnetic lossless dielectric medium has amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time average power density vector (in W/m2) is 1 1 (b) 30π 60π 1 1 (c) (d) 120π 240π (GATE 2010: 1 Mark) (a)

77. A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m. If the line is distortion less, the attenuation constant (in Np/m) is (a) 500 (b) 5 (c) 0.014 (d) 0.002  (GATE 2010: 1 Mark) 78. A plane wave having the electric field component  Ei = 24 cos(3 × 108 t - β y )a z V/m and travelling in free space is incident normally on a lossless medium with m = m0 and e = 9e0, which occupies the region y ≥ 0. The reflected magnetic field component is given by 1 cos(3 × 108 t + β y )a x A/m 10π 1 (b) cos(3 × 108 t + β y )a x A/m 20π 1 (c)   cos(3 × 108 t + β y )a x A/m 20π 1 (d) cos(3 × 108 t + β y )a x A/m 10π (GATE 2010: 2 Marks) (a)

79. In the circuit shown in the following figure, all the transmission line sections are lossless. The VSWR on the 60 ohm line is

Ch wise GATE_ECE_CH08_Electromagnetics.indd 527

ZL = 30 Ω

10

l/

4

(a) 200 Ω (c) 50 Ω

Z0 = 30√2 Ω l/4

(a) 1.00 (c) 2.50

(b) 1.64 (d) 3.00 (GATE 2010: 2 Marks)

80. Consider a closed surface S surrounding a volume V. If  r is the position vector of a point inside S, with nˆ the  unit normal on S, the value of the integral  ∫∫ 5r ⋅ n dS is S

(a) 3 V (c ) 10 V

(b) 5 V (d) 15 V (GATE 2011: 1 Mark)

81. A transmission line of characteristic impedance 50 Ω is terminated by a 50 Ω load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be p/4 radians. The phase velocity of the wave along the line is (b) 1.2 × 108 m/s (a) 0.8 × 108 m/s 8 (c) 1.6 × 10 m/s (d) 3 × 108 m/s (GATE 2011: 1 Mark) 82. The modes in a rectangular waveguide are denoted by TEmn/TMmn, where m and n are the Eigen numbers along the larger and smaller dimensions of the waveguide, respectively. Which one of the following statements is TRUE? (a) The TM10 mode of the waveguide does not exist. (b) The TE10 mode of the waveguide does not exist. (c) The TM10 and the TE10 modes both exist and have the same cut-off frequencies. (d) The TM10 and the TM01 modes both exist and have the same cut-off frequencies. (GATE 2011: 1 Mark) 83. The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity er and relative permeability mr = 1 are given by  E = E p e j (ωt - 280π y ) u z V/m   H = 3e j (ωt - 280π y ) u x A /m

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GATE ECE Chapter-wise Solved papers

(a) 330 Ω (c) 143.3 Ω



Assuming the speed of light in free space to be 3 × 108 m/s, intrinsic impedance of free space to be 120p, the relative permittivity er of the medium and the electric field amplitude Ep are (a) er = 3, Ep =120p (b) er = 3, Ep = 360p (c) er = 9, Ep = 360p (d) er = 9, Ep = 120p (GATE 2011: 2 Marks)  84. A current sheet J = 10u y A/m lies on the dielectric interface x = 0 between two dielectric media with ε r1 = 5, m r1 = 1 in region 1 (x < 0) and er2 = 2, mr2 = 2 in region 2 (x > 0), as shown in the following figure. If the  magnetic field in region 1 at x = 0 is H1 = 3u x + 30u y A/m, the magnetic field in region 2 at x = 0+ is x > 0 (Region II) : ε r2 = 2, µ r2 = 2

87. The electric field of a uniform plane electromagnetic wave in free space, along the positive x-direction, is  given by E = 10( a y + ja z )e - j 25 x . The frequency and polarization of the wave, respectively, are (a) 1.2 GHz and left circular (b) 4 Hz and left circular (c) 1.2 GHz and right circular (d) 4 GHz and right circular (GATE 2012: 1 Mark) 88. A plane wave propagating in air with  E = (8a x + 6 a y + 5a z )e j (ω t + 3x - 4 y ) V/m is incident on a  perfectly conducting slab positioned at x ≤ 0, the E field of the reflected waves is (a) ( -8a x - 6 a y - 5a z )e j (ω t + 3 x + 4 y ) V/m

x

(b) ( -8a x + 6 a y - 5a z )e j (ω t + 3 x + 4 y ) V/m

J x=0

(c) ( -8a x - 6 a y - 5a z )e j (ω t - 3 x - 4 y ) V/m

y

(d) ( -8a x + 6 a y - 5a z )e j (ω t - 3 x - 4 y ) V/m (GATE 2012: 1 Mark)

x < 0 (Region I) : ε r1 = 5, µ r1 = 1  (a) H 2  (b) H 2  (c) H 2  (d) H 2

= 1.5u x + 30u y - 10u z A/m = 3u x + 30u y - 10u z A/m = 1.5u x + 40u y A/m = 3u x + 30u y + 10u z A/m (GATE 2011: 2 Marks)

85. A transmission line of characteristic impedance 50 Ω is terminated in a load impedance ZL. The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of l/4 from the load. The value of ZL is (a) 10 Ω (b) 250 Ω (c) (19.23 + j46.15) Ω (d) (19.23 - j46.15) Ω (GATE 2011: 2 Marks) 86. A coaxial cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. Given 10 -9 m0 = 4π × 10 -7 H/m, ε 0 = F/m, the characteristic 36π impedance of the cable is

Ch wise GATE_ECE_CH08_Electromagnetics.indd 528

(b) 100 Ω (d) 43.4 Ω (GATE 2012: 1 Mark)

89. The radiation pattern of an antenna in spherical coordinates is given by F(q) = cos4 q, 0 ≤ q ≤ p/2.

The directivity of the antenna is (a) 10 dB (b) 12.6 dB (c) 11.5 dB (d) 18 dB (GATE 2012: 1 Mark)  90. The direction of vector A is radially outward from the origin, with A= kr n where r2 = x2 + y2 + z2 and k is a  constant. The value of n for which ∇ ⋅ A = 0 is (a) -2 (b) 2 (c) 1 (d) 0 (GATE 2012: 2 Marks) Statement for Linked Answer Questions 91 and 92: An infinetely long uniform solid wire of radius ‘a’ carriers a uniform DC current of density J . 91. The magnetic field at a distance r from the centre of the wire is proportional to (a) r for r < a and 1/r2 for r > a (b) 0 for r < a and 1/r for r > a (c) r for r < a and 1/r for r > a (d) 0 for r < a and 1/r2 for r > a (GATE 2012: 2 Marks)

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Chapter 8  • Electromagnetics

92. A hole of radius b (where b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown in the following figure. The magnetic field inside the hole is

529

  (a)  ∫∫ (∇ × A) ⋅ d s over the closed surface bounded by  the loop

 (b)  ∫∫∫ (∇ ⋅ A) dV over the closed volume bounded by  the loop

(c)  ∫∫∫ (∇ ⋅ A) dV over the open volume bounded by the loop   (d)  ∫∫ (∇ × A) ⋅ d s over the closed surface bounded by

b d

the loop (GATE 2013: 1 Mark)  96. The divergence of a vector field A = x a x + y a y + z a z is

a

(a) (b) (c) (d)

uniform and depends only on d uniform and depends only on b uniform and depends on both b and d Non uniform (GATE 2012: 2 Marks)

93. A transmission line with a characteristic impedance of 100 Ω is used to match a 50 Ω section to a 200 Ω section. If the matching is to be done both at 429 MHz and 1 GHz, the length of the transmission line can be approximately (b) 1.05 m (a) 82.5 cm (c) 1.58 m (d) 1.75 m (GATE 2012: 2 Marks) 94. The magnetic field along the propagation direction inside a rectangular waveguide, with the cross-­section shown in the following figure, is Hz = 3 cos (2.094 × 102x) cos (2.618 × 102y) cos (6.283 × 1010 t − bz). The phase velocity vp of the wave inside the waveguide ­satisfies

(a) 0 (c) 1

(b) 1/3 (d) 3 (GATE 2013: 1 Mark)

97. The return loss of a device is found to be 20 dB. The VSWR and magnitude of reflection coefficient are, respectively, (a) 1.22 and 0.1 (b) 0.81 and 0.1 (c) -1.22 and 0.1 (d) 2.44 and 0.2 (GATE 2013: 1 Mark) Statement for Linked Answer Questions 98 and 99: A monochromatic plane wave of wavelength l = 600 mm is propagating in the direction as shown in the following figure.    Ei , Er and Et denote incident, reflected and transmitted electric field vectors associated with the wave. Ei

Er kr

Hi

Hr qi

ki

qr e r = 1.0

y

e r = 4.5

0 19.2°

Ht

x

Et

1.2 cm x

kt

3 cm z

(a) vp > c

(b) vp = c

(c) 0 < vp < c

(d) vp = 0

 98. The angle of incidence qi and the expression for Ei are

(GATE 2012: 2 Marks)

(a) 60° and

 95. Consider a vector field A( r ) . The closed loop line inte  gral  ∫ A ⋅ dl can be expressed as:

(b) 45° and

Ch wise GATE_ECE_CH08_Electromagnetics.indd 529

E0 2 E0 2

( a x - a z )e - j [π ×10

4

(x+ z) 3 2]

4

z 3)

( a x + a z )e - j (π ×10

V/m

V/m

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GATE ECE Chapter-wise Solved papers

(c) 45° and (d) 60° and

E0 2 E0 2

( a x - a z )e - j (π ×10

E0

(b) -

2

E0 2

(d)

E0 2

( a x + a z )e - j [π ×10

2

4

z 3)

4

V/m

(x-z) 3 2]

q2 towards the plate (d)  4π ε d 2

V/m

(GATE 2014: 1 Mark)

103. Consider the periodic square wave in the figure shown. x 1 0

V/m

1

2

3

4

t

−1

( a x + a z )e j (π ×10 E0

(c) 0.44

(x+ z) 3 2)

(GATE 2013: 2 Marks)

 99. The expression for Er is (a) 0.23

4

( a x - a z )e - j (π ×10

4

z 3)

( a x + a z )e - j [π ×10

( a x + a z )e - j [π ×10

4



V/m 4

( x - z ) 3]

( x + z ) 3]

V/m

V/m

(GATE 2013: 2 Marks) 100. The electric field (assumed to be one-dimensional) between two points A and B is shown. Let yA and yB be the electrostatic potentials at A and B, respectively. The value of yB − yA in Volts is .

The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is closest in value to . (GATE 2014: 1 Mark)

104. To maximize power transfer, a lossless transmission line is to be matched to a resistive load impedance via a l/4 transformer as shown. Lossless transmission line l/4 Transformer

Zin = 50 W



ZL = 100 W

The characteristic impedance (in Ω) of the l/4 transformer is . (GATE 2014: 1 Mark)

40 kV/cm 20 kV/cm 0 kV/cm A

5 mm

B

(GATE 2014: 1 Mark)  101. If E = -( 2 y 3 - 3 yz 2 ) x - (6 xy 2 - 3 xz 2 ) y + (6 xyz ) z is the electric field in a source free region, a valid expression for the electrostatic potential is (a) xy3 − yz2 (b) 2xy3 − xyz2 3 2 (c) y + xyz (d) 2xy3 − 3xyz2 (GATE 2014: 1 Mark) 102. The force on a point charge +q kept at a distance d from the surface of an infinite grounded metal plate in a medium of permittivity ε is (a) 0 q2 away from the plate (b)  16π ε d 2 2

q (c)  towards the plate 16π ε d 2

Ch wise GATE_ECE_CH08_Electromagnetics.indd 530

105. In the following figure, the transmitter Tx sends a wideband modulated RF signal via a coaxial cable to the receiver Rx. The output impedance ZT of Tx, the characteristic impedance Z0 of the cable and the input impedance ZR of Rx are all real. Transmitter ZT Tx

Characteristic impedance = Z0

Receiver ZR Rx

Which one of the following statements is TRUE about the distortion of the received signal due to impedance mismatch? (a) The signal gets distorted if ZR ≠ Z0, irrespective of the value of ZT (b) The signal gets distorted if ZT ≠ Z0, irrespective of the value of ZR (c) Signal distortion implies impedance mismatch at both ends: ZT ≠ Z0 and ZR ≠ Z0 (d) Impedance mismatches do NOT result in signal distortion but reduce power transfer efficiency (GATE 2014: 1 Mark)

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Chapter 8  • Electromagnetics

106. Which one of the following field patterns represents a TEM wave travelling in the positive x direction? (a) E = +8 y, H = -4 z (b) E = -2 y, H = -3 z (c) E = +2 z , H = +2 y (d) E = -3 y, H = +4 z (GATE 2014: 1 Mark) 107. Consider an air filled rectangular waveguide with a cross-section of 5 cm × 3 cm. For this waveguide, the cut-off frequency (in MHz) of TE21 mode is . (GATE 2014: 1 Mark) 108. For an antenna radiating in free space, the electric field at a distance of 1 km is found to be 12 mV/m. Given that intrinsic impedance of the free space is 120p W, the magnitude of average power density due to this antenna at a distance of 2 km from the antenna (in nW/m2) is . (GATE 2014: 1 Mark) 109. Match Column A with Column B.

112. If the electric field of a plane wave is  E ( z , t ) = x 3 cos(ω t - kz + 30°) - y 4 sin(ω t - kz + 45°)( mV/m) the polarization state of the plane wave is (a) left elliptical (c) right elliptical

(b) left circular (d) right circular (GATE 2014: 2 Marks)

113. A region shown below contains a perfect  conducting half-space and air. The surface current K s on the sur  face of the perfect conductor is K s = x 2 amperes per  meter. The tangential H field in the air just above the perfect conductor is

y

Ks

Air x Perfect conductor

Column A

Column B

1. Point electromagnetic source

P. Highly directional

2. Dish antenna

Q. End free

3. Yagi–Uda antenna

R. Isotropic

(a)    (b)    (c)   (d) 1 → P 1 → R 1 → Q 1 → R 2 → Q 2 → P 2 → P 2 → Q 3 → R 3 → Q 3 → R 3 → P (GATE 2014: 1 Mark) 110. Given the vector A = (cos x) (sin y) a x + (sin x) (cos y) a y , where a x , a y denote unit vectors along x, y directions,  respectively. The magnitude of curl of A is  . (GATE 2014: 2 Marks) 111. In spherical coordinates, let a q , a ϕ denote unit vectors along the q , ϕ directions. 100 sin q cos(ω t - β r )a q V/ m E= r and 0.265 sin q cos(ω t - β r )a ϕ A/m H= r

represent the electric and magnetic field components of the EM wave at large distances r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell located at r = 1 km, 0 ≤ q ≤ π /2 is (GATE 2014: 2 Marks)

Ch wise GATE_ECE_CH08_Electromagnetics.indd 531

(a) ( x + z )2 amperes per meter (b) x 2 amperes per meter (c) -z 2 amperes per meter (d) z 2 amperes per meter (GATE 2014: 2 Marks) 114. Assume that a plane wave in air with an electric field  E = 10 cos(ω t - 3 x - 3 z )a y V/m is incident on a non-magnetic dielectric slab of relative permittivity 3 which covers the region z > 0. The angle of transmission in the dielectric slab is degrees. (GATE 2014: 2 Marks)  115. Given F = za x + xa y + ya z . If S represents the portion   of the sphere x2 + y2 + z2 = 1 for z ≥ 0, the ∫ ∇ × F . ds is s . (GATE 2014: 2 Marks) ⋅

116. For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of the spacing between the plates by a factor of two results in (a) halving of v and no change in Z (b) no changes in v and halving of Z (c) no change in both v and Z (d) halving of both v and Z (GATE 2014: 2 Marks)

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GATE ECE Chapter-wise Solved papers

117. The input impedance of a l/8 section of a lossless transmission line of characteristic impedance 50 Ω is found to be real when the other end is terminated by a load ZL (= R + jX)Ω. If X is 30 Ω, the value of R (in Ω) is (GATE 2014: 2 Marks) 118. In the transmission line shown, the impedance Zin (in ohms) between node A and the ground is .

(b) Hf

r

(c) Hf

A Z0 = 50 W, L = 0.5 l 100 W

50 W r

Zin = ?

(d) Hf (GATE 2014: 2 Marks)

119. For a rectangular waveguide of internal dimensions a × b (a > b), the cut-off frequency for the TE11 mode is the arithmetic mean of the cut-off frequencies for TE10 mode and TE20 mode. If a = 5 cm, the value of b (in cm) is . (GATE 2014: 2 Marks)   120. A vector P is given by P = x 3 ya x - x 2 y 2 a y - x 2 yza z . Which one of the following statements is TRUE?  (a) P is solenoidal, but not irrotational  (b) P is irrotational, but not solenoidal  (c) P is neither solenoidal nor irrotational  (d) P is both solenoidal and irrotational (GATE 2015: 1 Mark) 121. In a source free region in vacuum, if the electrostatic potential j = 2x2 + y2 + cz2, the value of constant c must be . (GATE 2015: 1 Mark) 122. Consider a straight, infinitely long, current carrying conductor lying on the z-axis. Which one of the following plots (in linear scale) qualitatively represents the dependence of Hf on r, where Hf is the magnitude of the azimuthal component of magnetic field outside the conductor and r is the radial distance from the conductor?

r

(GATE 2015: 1 Mark) 123. The electric field component of a plane wave travelling in a lossless dielectric medium is given by  z ⎞ ⎛ E ( z , t ) = a y 2 cos ⎜108 t ⎟ V/m. The wavelength ⎝ 2⎠ (in m) for the wave is

. (GATE 2015: 1 Mark)

124. A coaxial cable is made of two brass conductors. The spacing between the conductors is filled with Teflon (er = 2.1, tan d = 0). Which one of the following circuits can represent the lumped element model of a small piece of this cable having length ∆z? (a)

RDz/2 LDz/2

RDz/2 LDz/2

GDz

CDz

Dz (b)

RDz/2 LDz/2

RDz/2 LDz/2

(a) Hf CDz

r

Ch wise GATE_ECE_CH08_Electromagnetics.indd 532

Dz

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Chapter 8  • Electromagnetics

(c)

LDz/2



LDz/2

GDz

533

The polarization of the wave is (a) right handed circular (b) right handed elliptical (c) left handed circular (d) left handed elliptical

CDz

(GATE 2015: 2 Marks) 129. Consider the 3 m long lossless air-filled transmission line shown in the figure. It has a characteristic impedance of 120 p Ω, is terminated by a short circuit, and is excited with a frequency of 37.5 MHz. What is the nature of the input impedance (Zin)?

Dz (d)

RDz

LDz

GDz

CDz ZL = 0

Dz

Zin

(GATE 2015: 1 Mark) 125. The electric field intensity of a plane wave travelling in free space is given by the following expression  E ( x, t ) = a y 24π cos(ω t - k0 x ) ( V/m )

In this field, consider a square area 10 cm × 10 cm on a plane x + y = 1. The total time averaged power (in mW) passing through the square area is . (GATE 2015: 2 Marks)

126. The electric field of a plane wave propagating in a lossless non-magnetic medium is given by the following expression  E ( z , t ) = a x 5 cos ( 2π × 10 9 t + β z )

(a) open (c) inductive

(b) short (d) capacitive (GATE 2015: 2 Marks)

130. A 200 m long transmission line having parameters shown in the figure is terminated into a load RL. The line is connected to a 400 V source having source resistance RS through a switch which is closed at t = 0. The transient response of the circuit at the input of the line (z = 0) is also drawn in the figure. The value of RL (in Ω) is . V (0,t)

100 V Rs = 150 W

+ a y 3 cos ( 2π × 10 9 t + β z - π 2)

3m

R0 = 50 W εr,eff = 2.25

RL

Vs = 400 V

The type of the polarization is (a) right hand circular (b) left hand Elliptical (c) right hand Elliptical (d) linear

62.5 V 200 m z=0

2.0

z=L

V (0,t) (GATE 2015: 2 Marks)

  127. A vector field D = 2r 2 a r + za z exists inside a cylindrical region enclosed by the surfaces r = 1, z = 0 and z = 5. s = 150 W this cylindricalRregion. 0 = 50 W Let S be the surfaceRbounding ε = 2.25 r,eff   ⎛ ⎞ D ⋅ ds The surface integralVsof= this field on S ⎜  400 V ∫∫ ∫ ⎟ is ⎝ S ⎠ . 200 m (GATE 2015: 2 Marks) z=0 128. The electric field of a uniform plane electromagnetic  7 wave is E = ( a x + j 4 a y ) exp[ j ( 2π × 10 t - 0.2 z )]

Ch wise GATE_ECE_CH08_Electromagnetics.indd 533

100 V RL 62.5 V

z=L

2.0

t (ms) (GATE 2015: 2 Marks)

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GATE ECE Chapter-wise Solved papers

131. The longitudinal component of the magnetic field inside an air-filled rectangular waveguide made of a perfect electric conductor is given by the following expression



The magnitude of the transmitted electric field component in (V/m) after it has travelled a distance of 10 cm inside the dielectric region is  . (GATE 2015: 2 Marks)

Hz(x, y, z, t) = 0.1 cos(25p x) cos(30.3p y) cos(12p × 109t −b z) (A/m) 135. Consider the time-varying vector I = x 15cos (ω t ) + y 5sin (ω t ) in Cartesian coordinates, where w  > 0 is a The cross-sectional dimensions of the waveguide are constant. When the vector magnitude |I| is at its mingiven as a = 0.8 m and b = 0.033 m. The mode of propimum value, the angle q that I makes with the x-axis agation inside the waveguide is (in degrees, such that 0 ≤ q ≤ 180) is ___________. (a) TM12 (b) TM21 (GATE 2016: 1 Mark) (c) TE21 (d) TE12  exists in (GATE 2015: 2 Marks) 136. A uniform and constant magnetic field B = zB the z direction in vacuum. A particle of mass m with 132. An air-filled rectangular waveguide of internal dimena small charge q is introduced into this region with an sions a cm × b cm (a > b) has a cutoff frequency of 6 GHz  z . Given that B, m, q, v and initial velocity v = x v x + zv x for the dominant TE10 mode. For the same waveguide, if vz are all non-zero, which one of the following describes the cutoff frequency of the TM11 modes is 15 GHz, the the eventual trajectory of the particle? cutoff frequency of the TE01 mode in GHz is . (a) Helical motion in the z direction (GATE 2015: 2 Marks) (b) Circular motion in the xy-plane 133. Two half-wave dipole antennas placed as shown in the (c) Linear motion in the z direction figure are excited with sinusoidally varying currents of (d) Linear motion in the z direction frequency 3 MHz and phase shift of π /2 between them (GATE 2016: 1 Mark) (the element at the origin leads in phase). If the maximum radiated E-field at the point P in the x-y plane 137. Concentric spherical shells of radii 2 m, 4 m, and 8 m occurs at an azimuthal angle of 60°, the distance d carry uniform surface charge densities of 20 nC/m2, (in meters) between the antennas is . −4 nC/m2 and ρs, respectively. The value of ρs (nC/m2)   z required to ensure that the electric flux density D = 0 at radius 10 m is . (GATE 2016: 1 Mark)

O y

d 60°

OP>>d P (GATE 2015: 2 Marks)

x

134. Consider a uniform plane wave with amplitude (E0) of 10 V/m and 1.1 GHz frequency travelling in air, and incident normally on a dielectric medium with complex relative permittivity (er) and permeability (mr) as shown in the figure. Air

Dielectric mr =1−j2 er =1−j2 •

E=? 10 cm

E0 = 10 V/m Freq = 1.1 GHz

Ch wise GATE_ECE_CH08_Electromagnetics.indd 534

138. Let the electric field vector of a plane electromagnetic wave propagating in a homogenous medium be  x e - j (ω t - β z ) , where the propagation expressed as E = xE constant b is a function of the angular frequency w. Assume that b (w) and Ex are known and are real. From the information available, which one of the following CANNOT be determined? (a) The type of polarization of the wave (b) The group velocity of the wave (c) The phase velocity of the wave (d) The power flux through the z = 0 plane (GATE 2016: 1 Mark) 139. Faraday’s law of electromagnetic induction is mathematically described by which one of the following equations?   (a) ∇ ⋅ B = 0 (b) ∇ ⋅ D = ρ V  →   ∂D → ∂B (c) ∇ × E = (d) ∇ × H = σ E + ∂t ∂t (GATE 2016: 1 Mark)

12/4/2018 11:47:30 AM

141. The propagation constant of a lossy transmission line is (2 + j5) m−1 and its characteristic impedance is (50 + j0) Ω at ω = 106 rad s−1. The values of the line constants L, C, R, G are, respectively, (a) L = 200 µH/m, C = 0.1 µF/m, R = 50 Ω/m, G = 0.02 S/m (b) L = 250 µH/m, C = 0.1 µF/m, R = 100 Ω/m, G = 0.04  S/m (c) L = 200 µH/m, C = 0.2 µF/m, R = 100 Ω/m, G = 0.02  S/m (d) L = 250 µH/m, C = 0.2 µF/m, R = 50 Ω/m, G = 0.04 S/m (GATE 2016: 1 Mark) 142. The current density in a medium is given by  400 sin q a r Am -2 J= 2π ( r 2 + 4) The total current and the average current density flowing through the portion of a spherical surface π π r = 0.8 m, ≤ q ≤ , 0 ≤ ϕ ≤ 2π are given, respec12 4 tively, by (a) 15.09 A, 12.86 Am−2 (b) 18.73 A, 13.65 Am−2 (c) 12.86 A, 9.23 Am−2 (d) 10.28 A, 7.56 Am−2 (GATE 2016: 2 Marks) 143. An antenna pointing in a certain direction has a noise temperature of 50 K. The ambient temperature is 290 K. The antenna is connected to a pre-amplifier that has a noise figure of 2 dB and an available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noise temperature Te for the amplifier and the noise power Pao at the output of the preamplifier, respectively, are (a) Te = 169.36 K and Pao = 3.73 × 10−10 W (b) Te = 170.8 K and Pao = 4.56 × 10−10 W (c) Te = 182.5 K and Pao = 3.85 × 10−10 W (d) Te = 160.62 K and Pao = 4.6 × 10−10 W (GATE 2016: 2 Marks) 144. A positive charge q is placed at x = 0 between two infinite metal plates placed at x = −d and at x = + d, respectively. The metal plates lie in the yz-plane.

Ch wise GATE_ECE_CH08_Electromagnetics.indd 535

at x=−d

140. If a right-handed circularly polarized wave is incident normally on a plane perfect conductor, then the reflected wave will be (a) right-handed circularly polarized. (b) left-handed circularly polarized. (c) elliptically polarized with a tilt angle of 45°. (d) horizontally polarized. (GATE 2016: 1 Mark)

535

at x=+d

Chapter 8  • Electromagnetics

The charge is at rest at t = 0, when a voltage +V is applied to the plate at −d and voltage −V is applied to the plate at x = +d. Assume that the quantity of the charge q is small enough that it does not ­perturb the field set up by the metal plates. The time that the charge q takes to reach the right plate is proportional to (a) d/V (b) d /V (c) d / V (d) d /V (GATE 2016: 2 Marks) 145. Consider the charge profile shown in the figure. The resultant potential distribution is best described by

ρ (x) ρ1

b

0 x

ρ2 →

∇V = (a)

− ρ( x)

ε

V(x)

b

0 a

x

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GATE ECE Chapter-wise Solved papers

(b)

T1

V(x)

T2 Z0 = 50 Ω

Z0 = 50 Ω Open

50 Ω

b

0 a

x 1 0.8 …

0.6

(c)

V(x)

0.4 0.2 0

0

b

a

(d)



x

0 a

x

(GATE 2016: 2 Marks) 146. A lossless microstrip transmission line consists of a trace of width w. It is drawn over a practically infinite ground plane and is separated by a dielectric slab of thickness t and relative permitivity er > 1. The inductance per unit length and the characteristic impedance of this line are L and Z0, respectively. w

e = e0 e = e0 er; er > 1



t

Which one of the following inequalities is always satisfied? (a) Z 0 >

Lt Lt (b) Z0 < ε0εr w ε0εr w

(c) Z 0 >

Lw Ltw (d) Z0 < ε0εrt ε0εrt (GATE 2016: 2 Marks)

147. A microwave circuit consisting of lossless transmission lines T1 and T2 is shown in the following figure. The plot shows the magnitude of the input reflection coefficient Γ as a function of frequency f. The phase velocity of the signal in transmission lines is 2 × 108 m/s.

Ch wise GATE_ECE_CH08_Electromagnetics.indd 536

1

1.5

2

2.5f (in GHz)

The length L (in meters) of T2 is . (GATE 2016: 2 Marks)

148. Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. At 10 GHz operating frequency, the value of the propagation constant (per meter) of the corresponding propagating mode is __________ (GATE 2016: 2 Marks)

V(x)

b

0.5

149. Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. The increasing order of the cut-off frequencies for different modes is (a) TE01  a, I = J × pa2

This gives B =



m0 J π a 2 2π r

That is, B ∝

1 for r > a r

92. Topic: Electrostatics (c)  It is uniform and depends on both b and d. 93. Topic: Transmission Lines: Impedance Matching (c)  In the given case, the transmission line used to match two transmission lines of characteristic impedance 50 Ω and 200 Ω has a characteristic impedance of 100 Ω, as Z 0 = ( Z1 × Z 2 ), that is, 100 = (50 × 200) Ω

Ch wise GATE_ECE_CH08_Electromagnetics.indd 551

3 × 108 = 0.075 m 4 × 1 × 10 9

Now, 1.58 =9 0.174

This gives B =



=

m0 I 2π r

For r < a

3 × 108 = 0.174 m 4 × 429 × 106

At an operating frequency of 1 GHz, quarter-wave length of line

n + 2 = 0. Therefore, n = -2 91. Topic: Electrostatics (c)  The magnetic flux density at a distance r from the wire is given by the following expression:

551

1.58 = 21 and    0.075 Thus, if the length of the line was chosen to be 1.58 m, it would be an odd multiple of quarter-wave length for both frequencies, thereby providing impedance matching. For none of the other given answer choices, it turns out to be an odd multiple for both frequencies. Hence, option (c) is the correct answer.

94. Topic: Waveguides (a)  The phase velocity inside the waveguide is always greater than the free-space velocity. 95. Topic: Electrostatics (a)  It is from the definition of Stoke’s theorem 96. Topic: Electrostatics  (d)  Divergence of vector ( A ) is given by  ∂A ∂Ay ∂Az + ∇⋅ A = x + ∂x ∂y ∂z =1+1+1= 3 97. Topic: Transmission Lines: Impedance Transformation (a)  We have Return loss = 20 log r, which gives r = 0.1. VSWR =

1+ r 1- r

Substituting | r| = 0.1, we get VSWR = 1.22

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GATE ECE Chapter-wise Solved papers

98. Topic: Plane Waves and Properties: Reflection and Refraction (c)  Electric field lies in the x-z plane, the plane of incidence in the case of parallel polarization.

40 - 20 ( x - 0) 5 × 10 -4 Therefore, E = 4 × 10 4 x + 20 E - 20 =

ε r1 sin qi = ε r 2 sin qt

sin qi = 2.12 × 0.3289 = 0.697 or    qi = sin -1 (0.697) ≈ 45°



Incidence electric field  Ei = E0 (cos qi a x - sin q j a z )e - jβ ( x sin qi + z cos qi ) 2π ⎛ ⎞ -j⎜ ( x sin 45 + z cos 45 ) ⎝ 600 ×10 -6 ⎟⎠

= E0 (cos 45 a x - sin 45a z)e = E0 =

E0 2

1 2

( a x - a z )e ( - jπ ×10

4

2 )( x + z )

3)(1

4

(x+ z) 3 2]

0

x + 20) dx

5 ×10 -4

0

−8

= −150 × 10−4 kV VAB = −15 V 101. Topic: Electrostatics (d)  Given that  E = -( 2 y 3 - 3 yz 2 )a x - (6 xy 2 - 3 xz 2 )a y + 6 xyz ⋅ a z By verification, option (d) satisfies E = -∇V 102. Topic: Electrostatics (c)  For two point charges, q1 and q 2 placed distance R apart, the force F acting between them is given by



E0r η2 cos qt - η1 cos qi = E0i η1 cos qi + η2 cos qt

F=

1 q1q2 4πε R 2

F=

1 q2 4πε ( 2d ) 2

In the given case,

Therefore,      F=

Now

η1 εr2 = = 2.12 η2 ε r1

4

= − (2 × 10 × 25 × 10 + 20 × 5 × 10−4)

V/m

99. Topic: Plane Waves and Properties: Reflection and Refraction (a)  The electric field of the reflected wave propagates in a direction opposite to that of the incident wave. Hence, the term z gets replaced by -z. Also, the magnitude of electric field in the reflected wave will be less than that of the incident wave. In the case of parallel polarization,

∫ (4 × 10

4



( a x - a z )e[ jπ ×10

A

⎛ ⎞ x2 = - ⎜ 4 × 10 4 + 20 x ⎟ 2 ⎝ ⎠

which gives



5 ×10 -4

VAB = - ∫ E ⋅ dl = -

1 sin qi = 4.5 sin(19.2°)

B

q2 16πεd 2

Since the charges are of opposite polarity, the force between them is attractive. +q d

Solving the above equation,

Metal plate E0r = 0.23 E0i

This explains the answer.

100.   Topic: Electrostatics (15) 

For A: (0 kV/cm, 20 kV/cm)



For B: (5 × 10−4 kV/cm, 40 kV/cm)

Ch wise GATE_ECE_CH08_Electromagnetics.indd 552

d −q 103. Topic: Maxwell’s Equations: Poynting Vector (0.5)  For a periodic sequence wave, the amplitude of nth harmonic component is ∝ 1/n

Therefore, for a periodic sequence wave, power in nth harmonic component is ∝ 1 / n2

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Chapter 8  • Electromagnetics



Ratio of the power is 7th harmonic to power in 5th harmonic for the given waveform is

Therefore, E2 = 6 mV/m

Power density is

2

1/ 7 25 = ≅ 0.5 1 / 52 49

P=

104. Topic: Transmission Lines: Impedance Transformation (70.7)  Here impedance is matched by using l/4 transformer. Therefore, Z o = Z L Z in = 100 × 50 = 50 2 = 70.7 Ω 105. Topic: Transmission Lines: Characteristic Impedance (c)  Signal distortion implies impedance mismatch at both ends, that is, ZT ≠ Z0 and ZR ≠ Z0 106. Topic: Waveguides: Modes (b)  For TEM wave, electric field (E), magnetic field (H) and direction of propagation (P) are orthogonal to each other. Given that the direction of propagation P is in the positive x- direction. Hence, the correct option is (b) 107. Topic: Waveguides: Cut- off Frequencies (7810)  2

f c (TE 21 ) = =

c ⎛ 2⎞ ⎛ 1⎞ ⎜ ⎟ +⎜ ⎟ 2 ⎝ a⎠ ⎝ b⎠

1 E 2 1 36 × 10 -6 = × = 47.7 nW/m 2 2 η 2 120π

109. Topic: Antennas: Radiation Pattern (b)  1. Point electromagnetic source, radiates field in all directions equally. Hence, it is isotropic. 2.  Dish antenna is highly directional. 3. Yagi−Uda antenna is end free. 110. Topic: Electrostatics (0)  a x

⎛ 2⎞ ⎛ 1⎞ ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ 5 3

∂ ∂ ∂x ∂y cos x sin y sin x cos y  =0

100 sin q e - jβ r r 0.265 Hq = sin q e - jβ r r r 1 Pavg = ∫ Eq H q* .ds 2s Eq =

= 7.81 × 10 Hz = 7810 MHz Pavg

108. Topic: Antennas: Radiation Pattern (47.7)  Electric field of an antenna is

Therefore, E ∝ 1 r Hence,

E1 r2 = E2 r1

Ch wise GATE_ECE_CH08_Electromagnetics.indd 553

1 100(0.265) 2 2 sin q r sin q dqd ϕ 2 ∫s r2 1 = ∫ ( 26.5) sin 3q dq dϕ 2 s

=

9

1 Jβ where, is radiation field , 2 is induced field and r r J is electrostatic field βr 3

∂ ∂z 0

111. Topic: Maxwell’s Equations: Wave Equation (55.5) 

2

⎡ Jβ 1 η Iq dI J ⎤ sin q ⎢ + 2 - 3⎥ 4π βr ⎦ r ⎣ r

a z

 Therefore, Curl A = 0

= 1.5 × 1010 0.16 + 0.111

Eq =

a y

 Curl A =

2

2

3 × 1010 2

553

π /2

2π

0

0

= 13.25 ∫ sin 3q dq ∫ dϕ = 13.25( 2 / 3)( 2π ) Therefore, Pavg = 55.5 W. 112. Topic: Plane Waves and Properties: Polarization (a)  Given that  E ( z , t ) = x 3 cos(ω t - kz + 30°) - y 4 sin(ω t - kz + 45°) Therefore, E x = 3 cos(ω t - kz + 30°) E y = -4 cos(ω t - kz + 45°)

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GATE ECE Chapter-wise Solved papers

At z = 0,

Ex = 3 cos(wt + 30°)



Ey = −4 sin (w t + 45°)

sin q i = sin q t Therefore

Ex ≠ E y

Hence, the wave is having elliptical polarization.



q = 30° − 135° = −105°. Therefore, the wave is left hand polarized.



Hence, the beam is left elliptical polarized (LEP)

113. Topic: Plane Waves and Properties: Propagation Through Various Media (d)  From the given figure, we have that medium (1) is a perfect conductor and medium (2) is air. Therefore, H1 = 0 From boundary conditions, we have (H1 − H2) × an = Ks

sin 60 3 = sin q t 1 Hence,  sin q t = 0.5 or q t = 30 115. Topic: Electrostatics (3.14)  Using Stoke’s theorem and the fact that C is a closed curve, that is, x2 + y2 = 1, z = 0, we have     ∫ ∇ × F ⋅ ds = ∫ F .dr s

Therefore, 

s

So, β x = β cos q x = 3

β y = β cos q y = 0 β z = β cos q z = 3

Squaring and adding, we have

βx 2 + β y 2 + βz 2 = β 2 9 + 3 = β 2 ⇒ β = 12

or,

Since β cos q z = 3 , therefore 3 12 Hence, q z = 60 = q i cos q z =

Ch wise GATE_ECE_CH08_Electromagnetics.indd 554

zdx + xdy + ydz

C

2π

∫ cos q (cos q dq ) 0

2π

=

- H x a x + H z a z = 2a x

E = E0 e - jβ ( x cos q x + y cos q y + z cos q z )

C

= ∫ xdy =

-( H x a x + H y a y + H z a z ) × a y = 2a x

114. Topic: Plane Waves and Properties: Reflection and Refraction (30)  Given that:  E = 10 cos(ω t - 3 x - 3 z )a y



∫ ∇ × F ⋅ ds = ∫

- H 2 × a y = 2a x

H z = 2 ⇒ H = 2a z

C

(Let, x = cos q, y = sin q (where q is 0 to 2p))

Since, H1 = 0,aˆn =aˆy and K s = 2a x . Therefore,

Therefore,

E2 E1

1⎛ sin 2q ⎞ ⎜q + ⎟ = π  3.14 2⎝ 2 ⎠0

116. Topic: Transmission Lines: Characteristic Impedance (b)  Characteristic impedance Z is given by Z=

276 ∈r

log ( d /r )

where d = distance between the two plates. From the above expression, Z changes, if the spacing between the plates changes.

speed of propagation v is given by v=

1 LC

Therefore, v is independent of spacing between the plates. 117. Topic: Transmission Lines: Characteristic Impedance (40)  Given that: l = l/8 and characteristic impedance Zo = 50 Ω. Then ⎡ Z + jZ o ⎤ Z in (l = l/8) = Z o ⎢ L ⎥ ⎣ Z o + jZ L ⎦

12/4/2018 11:48:02 AM

Chapter 8  • Electromagnetics



Substituting values, we have ⎡ Z + j 50 ⎤ ⎡ Z L + j 50 50 - jZ L ⎤ Z in = 50 ⎢ L × ⎥ = 50 ⎢ ⎥ ⎣ 50 + jZ L ⎦ ⎣ 50 + jZ L 50 - jZ L ⎦ ⎡ 50 Z L + 50 Z L + j (50 2 - Z L 2 ) ⎤ = 50 ⎢ ⎥ 50 2 + Z L 2 ⎣ ⎦



Given that Zin is real, therefore imaginary part of Zin = 0

Therefore,

1 1 9 + = 5 b 2 4(5) 1 9 1 - + = 5 20 b 2

Solving the above equation, we get b = 2 cm.

120. Topic: Electrostatics (a)  Given that  P = x 3 ya x - x 2 y 2 a y - x 2 yza z

50 2 - Z L 2 = 0 Z L2 = 50 2 or R 2 + X 2 = 50 2

∇ ⋅ P = 3x 2 - 2 x 2 y - x 2 y = 0  Therefore, P is solenoidal.

R 2 = 50 2 - X 2 = 50 2 - 30 2

a x

Therefore, R = 40 Ω.

∇× P =

118. Topic: Transmission Lines: Characteristic Impedance (33.33) Here L =

l 2

Therefore, Z in = (100 || 50)Ω =

100 Ω = 33.33 Ω 3

119. Topic: Waveguides (2)  f c10

2 C ⎛ 1⎞ = ⎜ ⎟ 2 ⎝ a⎠

⎛ 1⎞ ⎛ 2⎞ f c10 = k ⎜ ⎟ ; f c 20 = k ⎜ ⎟ ⎝ a⎠ ⎝ a⎠ f c11 = k

1 1 + a2 b2

Given that

∂ ∂ ∂x ∂y x3 y - x 2 y 2

f c10 + f c 20 2

1 1 k ⎡1 2⎤ k 2+ 2 = ⎢ + ⎥ 2 ⎣a a⎦ a b 1 1 3 + = a 2 b 2 2a Substituting the value of a in the above equation and taking square on both sides, we get

Ch wise GATE_ECE_CH08_Electromagnetics.indd 555

a z ∂ ∂z - x 2 yz

 Hence, P is not irrotational.  So, P is solenoidal but not irrotational. 121. Topic: Electrostatics (−3)  Given that φ = 2 x 2 + y 2 + cz 2 E = -∇φ = -4 xa x - 2 ya y - 2cza z

∇⋅ E = 0

Therefore, −4 − 2 − 2c = 0 Hence, c = −3 122. Topic: Electrostatics (c)  The azimuthal component of magnetic field outside the conductor is Hϕ =

f c11 =



a y

= a x ( - x 2 y ) - a y ( -2 xyz ) + a z ( -2 xy 2 - x 3 ) ≠ 0

Zin (L = l/2) = Z1 = 50 Ω



555

1 2π r

where r is the distance from current element. Therefore, Hϕ ∝

1 r

123. Topic: Plane Waves and Properties: Propagation Through Various Media (8.885)  Given that  z ⎞ ⎛ E ( z , t ) = 2 cos ⎜108 t ⎟ a y ⎝ 2⎠

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GATE ECE Chapter-wise Solved papers

Therefore, β =

127. Topic: Maxwell’s Equations: Differential Integral Forms and Their Interpretation (78.53)  Given that   D = 2r 2 a r + za z

1 2

2π Now,2πl = = 2π 2 m l= = 2βπ 2 m β Therefore, l = 8.885 m Therefore, l = 8.885 m

∫ D ⋅ ds = ∫ (∇ ⋅ D)dv S

124. Topic: Transmission Lines: Impedance Transformation (b)  Loss tangent tan δ =

∇⋅ D =

σ ω∈

=

Given that tan d = 0, therefore, s = 0.

V

1 ∂ 1 ∂Dϕ ∂Dz . ( r Dr ) + + r ∂r r ∂ϕ ∂z 1 ∂ . ( r 2r 2 ) + 0 + 1 r ∂r

1 2(3)r 2 + 1 r = 6r + 1 =

Let G be the conductivity of the dielectric material. Therefore, s = G = 0

Hence, option (b) is the correct answer. 1

1 E02 a x 2 η

=

( 24π ) 2 a x = 7.53a x 2η



Surface over the plane x + y = 1(x + y + 0z = 1)



Normal to this plane is n = (1,1, 0)  Power(P ) = ∫ P avg .ds

1

⎛ 6r3 r 2 ⎞ =⎜ + ⎟ ( 2π )(5) 2⎠0 ⎝ 3 1⎞ ⎛ = ⎜ 2 + ⎟ 10π = 78.53 ⎝ 2⎠  ⎛ ⎞ Therefore, S D ⋅ ds⎟ = 78.53 ∫∫ ∫ ⎜ ⎝ S ⎠ 128. Topic: Plane Waves and Properties: Polarization  7 (d)  Given that E = ( a x + 4 ja y )e j ( 2π ×10 t - 0.2 z )

From the expression:

S

= ∫ 7.53a x . S

=

7.53

2

w = 2p × 107 dydz

× 10 × 10 × 10 -2 × 10 -2

2 = 53.3 × 10 -3W = 53.3 mW

126. Topic: Plane Waves and Properties: Polarization (b)  Ex = 5cos(wt + bz)

b = 0.2

Ex = cos wt and Ey = 4 cos(wt + p/2) = −4 sinwt



So, it is left hand elliptical polarization

129. Topic: Transmission Lines: Characteristic Impedance (d)  Z in = jZ 0 tan β l

l=

π⎞ ⎛ E y = 3 cos ⎜ ω t + β z - ⎟ ⎝ 2⎠ π 2 So, the electric field is left hand elliptical. ϕ=-

Ch wise GATE_ECE_CH08_Electromagnetics.indd 556

5

r=0 ϕ=0 z=0

V

 P avg = average poynting vector =

2π

∫ (∇ . D)dv = ∫ ∫ ∫ (6r + 1)rd rdϕ dz

125. Topic: Maxwell’s Equations: Poynting Vector (53.3)  Given that  E ( x, t ) = a y 24π cos(ω t - k0 x )

a x

and

3 × 108 =8m 37.5 × 106

βl =

2π 2π 3π l= (3) = l 8 4

If for a short-circuit line, 0 < β l < π / 2 , then the nature of the input impedance is inductive.

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Chapter 8  • Electromagnetics



If for a short circuit line, π /2 < β l < π , then the nature of input impedance is capacitive.

130. Topic: Transmission Lines: Impedance Transformation (30)  S R

Therefore, 1 = 91.65 b

For TE10 fc =

100 0 0

3 × 108 1 × = 13.7 GHz 2 b

1

R (100)

133. Topic: Antennas: Antenna Arrays (50)  ψ = δ + β d cos q

100 2

R S (100) 62.5

3



For maximum field, y = 0 3 × 108 3 × 106 = 100 m

t (µs) z=0

l=

z=l

Given V(t = 2 ms, z = 0) = 62.5 62.5 = V(t = 0, z = 0) + V(t = 1, z = 0) + V(t = 2, z = 0)

δ + β d cos q = 0

62.5 = 100 + R(100) + R | S (100) R=

RL - 50 1 , S= 2 RL + 50

-

π 2π d cos 60° = 0 + 2 l π 2π 1 = (d ) 2 100 2

-1 So, R = 4 RL = 30 Ω

Therefore, d = 50 m

131. Topic: Waveguides: Modes (c)  From the given expression, we have

134. Topic: Electrostatics (90)  We have I = x 15 cos (ω t ) + y 5 sin (ω t )

mπ x = 25π x. Therefore, m = 25a = 2 a nπ y Also, = 30.3π y. Therefore, n = 30.3b = 1 b

The given magnetic field Hz means that the mode of propagation is TE mode. Therefore, mode = TE21

132. Topic: Waveguides: Cut- off Frequencies (13.7)  Given that for TE10, fc = 6 GHz fc =

a=

2

1 2 mε

⎛ m⎞ ⎛ n⎞ ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ a b

2

1 40

15 × 10 9 =

Therefore, I = (15 cos ω t ) 2 + (5 sin ω t ) 2 = 25 + 200 cos 2 ω t Here, | I | is minimum when cos(wt) = 0. Hence, q = wt = 90°. 135. Topic: Electrostatics (a)  The magnetic field is Baz. The velocity is vxax + vzaz. Therefore, F = Q(v × B) = Q(vxax + vzaz) × Baz= Qvx ⋅ B(−ay)

That is a helical motion in the direction of z-axis. 136. Topic: Electrostatics (−0.25)  For the flux density to be zero at radius r = 10 m, the total charge enclosed must be zero. The total charge enclosed on the sphere of radius 2 m is

For TM11

Ch wise GATE_ECE_CH08_Electromagnetics.indd 557

557

3 × 108 2

1 1 + a2 b2

4π(r2)20 nC/m2 = 320π nC The charge on the sphere with 4 m radius is

12/4/2018 11:48:08 AM

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GATE ECE Chapter-wise Solved papers

4π(r2) ⋅ (−4) nC/m2 = −256π nC The charge on the sphere with 8 m radius is

Now, R = 2Z0 = 2 × 50 = 100 Ω/m. Therefore, R = 100 Ω/m

4π(r2) ⋅ ρs ⋅ nC/m2 = 256ρsπ nC

G=

The total charge enclosed should be zero. Therefore, 320π − 256π + 256ρsπ = 0 Therefore, ⇒ρs = −0.25 nC/m2

We see only option (b) satisfies these two requirements. Confirming the value of β from option (b), we get

137. Topic: Maxwell’s Equations: Wave Equation (d)  We see that µr and er cannot be found, even though vp = w/b can be calculated. Therefore, the power flux (P) cannot be calculated. 1 E P= 2 η

2

where

β = ω LC = 106 250 × 10 -6 × 0.1 × 10 -6 = 106 × 10 -6 × 5 = 5 Therefore, β = 5 Thus, option (b) is true. 141. Topic: Maxwell’s Equations: Poynting Vector (d)  The current density at r = 0.8 m is  ⎡ 400 sin q ⎤ 2 J =⎢ ⎥ a r A/m π + 2 ( 0 . 64 4 ) ⎣ ⎦  ⇒ J = [ 400 sin q / 2π ( 4.64)] a r A/m 2

m η = 120π r εr 138. Topic: Maxwell’s Equations: Differential and Integral Forms and Their Interpretation (c)  From Faraday’s law of electromagnetism, we see that a time-varying magnetic field induces a non-conservative electric field. Mathematically,   ∂B ∇× E = ∂t 139. Topic: Plane Waves and Properties: Polarization (b)  When an EM wave undergoes reflection, the electric field orientation does not change. Whether a circular polarized wave is left circular or right circular is decided by observing the vector rotation with increasing time. Thus, when we observe the incident wave, we observe it by standing at the point of incidence and it is found to be right circular. When we observe reflected wave, we observe it by standing away from the point of incidence and reflected wave coming to us. Therefore, a wave that is right circular going into incidence is left circular after reflection. 140. Topic: Transmission Lines: Characteristic Impedance (b)  Since the characteristic impedance is purely real and propagation constant is complex, the transmission line is distortion-less. Thus,

100 R = = 0.04 S/m 2 2500 Z0



Thus, the total current I through the portion r = 0.8 m, where π/12 ≤ θ ≤ π/4 and 0 ≤ f ≤ 2π is  I = ∫∫ J ⋅ da I =∫

2π

∫

π /4

ϕ = 0 q = π /12

I=

⎛ 400 sin q ⎞ 2 ⎜⎝ ⎟ ar r sin q dq dϕ aˆr 2π × 4.64 ⎠

2π π /4 400 r 2 sin q dq dϕ ∫ ∫ 0 ϕ = q = π /12 ( 2π × 4.64) 2π

0.8

π /2

⎛ 400 × 0.64 ⎞ ⎡ q sin 2q ⎤ =⎜ (ϕ ) ⎢ ⎟ ⎝ 2π × 4.64 ⎠ 2 4 ⎥⎦ π /12 0 ⎣ ⎛ 400 × 0.64 ⎞ ⇒I =⎜ ⎝ 2π × 4.64 ⎟⎠ × 2π [(π /8 - π / 624) - 1/ 4(sin π / 2 - sin π / 6)] ⎡ π 1 ⎛ 1⎞ ⎤ ⎛ 400 × 0.64 ⎞ ⇒I =⎜ × 2π × ⎢ - ⎜ 1 - ⎟ ⎥ ⎟ ⎝ 2π × 4.64 ⎠ ⎣12 4 ⎝ 2 ⎠ ⎦ 400 × 0.64 ⎛ π 1 ⎞ ×⎜ - ⎟ ⎝ 12 8 ⎠ 4.64 ⇒ I = 7.55 A =

Now, the area of the portion of the spherical surface is S=∫

2π

∫

π /4

ϕ = 0 q = π /12

r 2 sin q dq dϕ

0.8

2π

π /4

⇒ S = 0.64 × (ϕ ) 0 × ( - cos q ) π /12

α = RG ; β = ω LC ; Z 0 = R /G Therefore,  α Z0 = R

Ch wise GATE_ECE_CH08_Electromagnetics.indd 558

⇒ S = 0.64 × 2π × (0.26) ⇒ S = 1.04 m 2

12/4/2018 11:48:09 AM

Chapter 8  • Electromagnetics

Now, the average current density is J avg =

Therefore,

α = 2ω m0 ε 0

7.55 = 7.25 A/m 2 1.04

Note: No option is matching. Option (d) can be treated as the final answer since it is the closest. 142. Topic: Antennas (a)  • Antenna’s noise temperature is TAN = 50 K.

• Ambient temperature is TA = 290 K.



• Preamplifier noise figure is FP = 2 dB.



• Amplifier gain is GP = 40 dB. TA = 50 K

2 × 2 p × 1.1 × 10 9 = 46.07 3 × 108 z = 10 cm =

Therefore, E3 = 10e -10 ×10

Te = (F − 1)TA = (102/10 − 1) × 290 = 169.36 K Now, the noise power at the input is kTeB = 1.38 × 10−23 × (169.36 + 50) × 12 × 106 = 3.62 × 10−14 The noise power at output of the preamplifier is



For air: h1 = 120pΩ and E1 = 10 V/m



For dielectric: mr = 1−j2; er = 1−j2 and h2 = 120pΩ

Since, h1 = h2, so E2 = E1 = 10 V/m Let E3 be the electric filed in the dielectric after travelling 10 cm. Then E3 = E2 e - α z

α + jβ =

jωm (σ + jωε )

= jω m0 ε 0 m r ε r = jω m0 ε 0 (1 - j 2) = jω m0 ε 0 + 2ω m0 ε 0

Ch wise GATE_ECE_CH08_Electromagnetics.indd 559

× 46.07

= 10e -4.6 = 0.1

(c)   We have 1 2 mv = qV 2 2qV v2 = m Now v =

x , where x = 2d. Therefore t 2qV m

x = t So

m m 1 = 2d ⋅ 2qV 2q V

t=x

Hence, t ∝

d V

145. Topic: Electrostatics (d)  From Poisson’s equation, we get For one-dimensional charge density, we get

Pao = (3.632 × 10−14) × 104 = 3.73 × 10−10 W 143. Topic: Plane Waves and Properties: Propagation Through Various Media (0.1)  Given that

-2

144. Topic: Electrostatics

Pre amp F = 2 dB G = 40 dB Now, the effective noise temperature of the amplifier input is

559

d 2V - r( x ) = ε dx 2 for x < 0, ρ(x) = −ρ2. Therefore ⇒

d 2V r2 = ε dx 2

Solving this we get ⎛r ⎞ V- ( x ) = ⎜ 2 ⎟ x 2 + c1 x + c2 < 0 ⎝ ε⎠ where c1 and c2 are arbitrary constants. Thus, V−(x) is an upward parabola. Similarly, V+(x) = −(ρ1/ε)x2 + c3x + c4 which is a downward parabola for x > 0. At x = 0, V(x) = V+(x) and V(x) will be constant for x < b and x > a and it will have no discontinuity. Thus, we see option (d) is correct.

12/4/2018 11:48:10 AM

560

GATE ECE Chapter-wise Solved papers

γ = ( mπ a) 2 + ( nπ b) 2 - k 2 Alternatively, observe that the charge density is the same as in a pn-junction depletion region. Hence, the potential difference V(x) will be the same as in pn-junction. The only difference being that zero potential is assumed to be at x = 0 instead of at x = b as is usually assumed in pn-junction.

146. Topic: Transmission Lines: Characteristic Impedance (a)  We have L . Therefore, Z 0 = C

Z0 =

Lt ε0εr w

Z0 >

(π

2

× 2.286 × 10 -2

)

2

⎛ 200π ⎞ -⎜ ⎝ 3 ⎟⎠

2

( using m = 1 andd n = 0) = j158.05 Therefore the value of propagation constant is 158.05. 149. Topic: Waveguides: Modes (c)  We have

Lt εw

where e  fc + 25% of fc f > 5/4 fc(1) Cut-off frequency is 2

⎛ m⎞ ⎛ n⎞ f c = c /2 ⎜ ⎟ + ⎜ ⎟ ⎝ a ⎠ ⎝ b⎠

Therefore, the component in a  ˆz has a magnitude of 2 and the component in xy plane has a magnitude of 2. These two components are out of phase by 90° and have unequal amplitudes. Hence, it is an elliptically polarized wave. 158. Topic: Electrostatics (0.78) Let x 2 + y 2 = l 2 . Volume revolution is about z-axis. Volume of region,





0



2

For TE10 mode: m = 1, n = 0 fc =

1

R = ∫ π l 2 dz

Ch wise GATE_ECE_CH08_Electromagnetics.indd 562

x2 + y2

3 × 108 c = = 6.55 GHz 2a 2 × 2.29 × 10 -2

From Eq. (1), f > 5/4 × 6.55 × 109 Hz > 8.19 GHz Case 2: f < 95% of fc f < 0.95 fc(2) For TE20 mode: m = 2, n = 0

12/4/2018 11:48:16 AM

Chapter 8  • Electromagnetics

fc =

3 × 108 c = = 13.1 GHz a 2.29 × 10 -2

⇒Z=

l0 1.5 = 4( nx - ny ) 4(1.5 ∼ 1.5001)

⇒ Z = 0.375 cm

From Eq. (2), f < 0.95 × 13.1 × 109 Hz f < 12.44 GHz



563

163. Topic: Antennas: Radiation Pattern (a)  ⎛π ⎞ jI 0 ⋅ ηe - jβ r cos ⎜ cos q ⎟ ⎝2 ⎠ along z -plane Ees = 2π r sin q ⎛π ⎞ cos ⎜ cos(90 - q )⎟ ⎝2 ⎠ - jβ r Ees = jI 0ηe in xy plane 2π r sin q (90 - q )

Range of operation frequency is 8.19 GHz < f < 12.44 GHz

161. Topic: Light Propagation in Optical Fibers (4.33) 

O

Ees = jI 0 ηe - jβ r

qC d

A

s

The smallest angle of incidence at which a light ray passing from one medium to another less refractive medium can be totally reflected from the boundary between the two critical angles is

qC = sin -1 ( n2 /n1 ) where n2 = refractive index of air = c m0 ε 0 and n1 = refractive index of water = c m0 ε 0 1.75 . Now, ⎛ c m0 ε 0 ⎞ qC = sin -1 ⎜ ⎟ = 49.1° ⎝ c m0 ε 0 1.75 ⎠ In ΔAOB, tan qC = 5/d ⇒ d=

5 = 4.33 m tan 49.1°

162. Topic: Light Propagation in Optical Fibers (0.37) Initially, the wave is circularly polarized. So, the initial phase difference between field components is p/2 to become linearly polarized. The wave must travel a distance such that the phase difference between the field components is p. Hence, path difference is p/2. So, π ZK x - ZK y = 2 F ⎛F ⎞ π ⇒ 2π Z ⎜ ε rx ε ry ⎟ = ⎝C ⎠ 2 C 4Z ( n - ny ) = 1 ⇒ l0 x

Ch wise GATE_ECE_CH08_Electromagnetics.indd 563

⎛π ⎞ cos ⎜ sin q ⎟ ⎝2 ⎠ cos q z

y Half wave dipole x

45°

For q = 90°, ⎛π ⎞ cos ⎜ sin 90°⎟ ⎝2 ⎠ 0 Ees = = cos 90° 0 d ⎛π ⎞ cos ⎜ sin q ⎟ ⎝2 ⎠ dq Ees = lim q → 90° d cos q dq ⎛π ⎞ sin ⎜ sin q ⎟ cos q ⎝2 ⎠ = lim q → 90° - sin q =0 Therefore, for q = 90° and ϕ = 45° , the direction of null will occur. 164. Topic: Transmission Lines: Smith Chart (c)  Constant resistance circle is given by 2

r ⎞ ⎛ ⎛ 1 ⎞ 2 ⎜⎝ x ⎟ + y = ⎜⎝ ⎟ 1+ r ⎠ 1+ r ⎠

2

1 ⎛ r ⎞ Here, center = ⎜ , r is normal, 0 , radius = ⎝ 1 + r ⎟⎠ 1+ r ized resistance.

12/4/2018 11:48:18 AM

564

GATE ECE Chapter-wise Solved papers

Constant reactance circle is given by

⎛E ⎞ ⇒ 10 x = ln ⎜ 0 ⎟ ⎝ Ex ⎠

2

1⎞ 1 ⎛ ( x - 1) 2 + ⎜ y - ⎟ = 2 ⎝ ⎠ x x

⎛E ⎞ Given 20 log10 ⎜ 0 ⎟ = 20 dB. So, ⎝ Ex ⎠

Here x is normalized reactance. For short circuit, r = x = 0

E0 = 10 ⇒ 10x = ln (10) = 2.3 Ex

For open circuit, r = x = ∞ For matched load, r = 1, x = 0

⇒ x = 0.23 m

Hence, the answer is option (c). 165. Topic: Light Propagation in Optical Fibers (0.48)  Lmin = minimum length

168. Topic: Waveguides (0.75 cm)  Cutoff frequency for TE01 = c/2b Cutoff frequency for TE10 = c/2a

Length ∝ wavelength Lmin1 l1 156 nm = = = 0.48 Lmin 2 l 2 325 nm 166. Topic: Transmission Lines: Characteristic Impedance

Given that c c a =2 ⇒ =2 2b 2a b The waveguide frequency is

(0.001)  Attenuation constant α = RG

f =

For distortionless transmission line L C L R = ⇒ = R G C G

This is 25% higher than the cutoff frequency of the dominant mode. So

Also, characteristic impedance Z 0 =

L R = C G

Given: Z0 = 50 Ω, R = 0.05 Ω/m. So, we have 50 =

0.05 ⇒ G

G=

5⎛ c ⎞ ⎜ ⎟ 4 ⎝ 2a ⎠

0.05 50

f = Now,

167. Topic: Plane Waves and Properties: Skin Depth (b) E after attenuation

l

l guide = =

0.05 0.05 α = R G = 0.05 = 50 50 a = 0.001 Np/m

1 - ( f c /f )

l=

l 1 - ( 4 / 5) 2

Therefore, c = 2.4 f = 2.4 × 1.25 fc = 3 fc ⇒ fc = Now, f c (for TE10 mode) = ⇒

where d is skin depth.

Ch wise GATE_ECE_CH08_Electromagnetics.indd 564

=

3× 4 c = 2.4 cm = 5 f

Also attenuation is

E Ex = e -10 x ⇒ e10 x = 0 E0 Ex

2

5l = 4 cm 3

E x = E0 e - α x 1 1 α= = = 10 Np/m δ 0.1

5 fc 4

Therefore, b =

c 3 c 2a

3 c c = ⇒ a = = 1.5 3 2a 2

a 1.5 = = 0.75 2 2

12/4/2018 11:48:20 AM

Chapter 8  • Electromagnetics

169. Topic: Plane Waves and Properties: Propagation Through Various Media (2)  E=

(

)

(

)

2 a x - a z cos ⎡6 3π × 108 t - 2π x + 2 z ⎤ V/m ⎣ ⎦

⇒ k = 2π ( x + 2 z )

ω = 6 3π × 108 = 3 × 2 3π × 108

ω = 2 3π × 3 × 108 rad/sec k can also be written as

Ch wise GATE_ECE_CH08_Electromagnetics.indd 565

565

⎛ 1 2 ⎞ k = 2 3π ⎜ x + z 3 ⎟⎠ ⎝ 3 ⇒ cos qix =

1 3

and tan qix = 2

Since there is no reflection, we have incident angle = Brewster’s angle = qB. tan q B =

ε0εr = ε r = 2 , i.e., tan qix ε0

So,     εr = 2

12/4/2018 11:48:21 AM

Ch wise GATE_ECE_CH08_Electromagnetics.indd 566

12/4/2018 11:48:21 AM

aPPENDIX

Solved GATE (ECE) 2019 QUESTIONS

Chapter 1: Engineering Mathematics

Again for n = 1, Eq. (1) becomes 1

1.

(c)

1 1− z

(d) cos (z) (1 Mark)

Topic: Complex Analysis: Analytic Functions (d)  cos (z) is analytic over the entire complex plane. 2.

⎛x⎞ dy = −⎜ ⎟ dx ⎝ y⎠ ⇒         y dy = −x dx ⇒  x dx + y dy = 0 Integrating both sides, we get x2 + y2 = c which is the equation of a circle.

Which one of the following functions is analytic over the entire complex plane? (a) ln (z) (b) e1/ z

The families of curves represented by the solution of n ⎛x⎞ dy the equation = − ⎜ ⎟ for n = −1 and n  = +1, dx ⎝ y⎠ ­respectively, are (a) Parabolas and Circles (b) Circles and Hyperbolas (c) Hyperbolas and Circles (d) Hyperbolas and Parabolas (1 Mark)

3.

The value of the contour integral 2

1 1⎞ ⎛ z + ⎟ dz 2π j ∫ ⎜⎝ z⎠ evaluated over the unit circle |z| = 1 is

. (1 Mark)

Topic: Complex Analysis 2 1 1⎞ ⎛ (0)  f = z + dz z ⎟⎠ 2π j ∫ ⎜⎝ 2

⎛ z2 +1 ⎞ 1 = ⎜ ⎟ dz  ∫ ( 2π j ) ⎝ z ⎠

Topic: Differential Equations: First Order Equations (c)  Given

j

n

⎛x⎞ dy = −⎜ ⎟  dx ⎝ y⎠

(1)

−1

1

−1

⎛x⎞ dy = − ⎜ ⎟ becomes Eq. (1). dx ⎝ y⎠ dy ⎛ y⎞ ⇒  = −⎜ ⎟ dx ⎝x⎠ dy dx ⇒  =− y x Integrating both sides, we get ln y = −ln x + ln c ⇒  ln (xy) = ln c

−j

For n = −1,

⇒  xy = c c    ⇒    y = x which is the equation of a hyperbola.

GATE_ECE_2019_new.indd 1

There are two poles at z = 0 (lying inside the contour). So by using f (z)

∫ ( z − z ) 0

n

dz =

d n −1 f (z) ( n − 1)! dz n −1 1

we get f =

1 ( z 2 + 1) 2 dz 2π j ∫ z2

=

1 1 d 2 ×2 ( z + 1) 2 2π j 1! dz

=

2 ( z 2 + 1)( 2 z ) =0 2π j z =0

z =0

6/22/2019 11:22:22 AM

A2 4.

GATE ECE Chapter-wise Solved papers

The number of distinct eigenvalues of the matrix ⎡2 2 3 3⎤ ⎢0 1 1 1 ⎥ ⎥ is equal to A=⎢ . ⎢0 0 3 3⎥ ⎢ ⎥ ⎣0 0 0 2 ⎦ (1 Mark)

Topic: Linear Algebra: Eigen Values and Eigen Vectors

(3) 

⎡2 ⎢0 A=⎢ ⎢0 ⎢ ⎣0

2 1 0 0

3 1 3 0

3⎤ 1 ⎥⎥ is an upper triangular matrix. 3⎥ ⎥ 2⎦

Eigenvalues will be the principle diagonal elements, which are 2, 1, 3, 2. Distinct eigenvalues are 2, 1, 3. 5.

⎧1 − e − x if x ≥ 0 FZ ( x ) = ⎨ if x < 0 ⎩ 0 The Pr(Z > 2 | Z > 1), rounded off to two decimal places, is equal to . (1 Mark) Topic: Probability and Statistics ⎧1 − e − x ; x ≥ 0 (0.367)  FZ ( x ) = ⎨ x 2 | Z > 1) = =

P  (Z > 2) = 1 - P(Z ≤ 2) = 1 - 1 + e−2 ⇒  P ( Z > 2) = e −2

If X and Y are random variables such that E[2X + Y] = 0 and E[X + 2Y] = 33, the E[X] + E[Y] = . (1 Mark)

⇒ 

P (Z ≤ 1) = 1 - e−2 P ( Z > 1) = e −1

So P ( Z > 2) e −2 = = e −1 = 0.367 P ( Z > 1) e −1

⇒  E[2X] + E[Y] = 0 ⇒  2E[X] + E[Y] = 0

(1)

E[X + 2Y] = 33 ⇒  E[X] + E[2Y] = 33 ⇒  E[X] + 2E[Y] = 33 Adding both equations, we get

(2)

8.

Consider a differentiable function f(x) on the set of real numbers such that f(−1) = 0 and f ′( x ) ≤ 2. Given these conditions, which one of the following inequalities is necessarily true for all x ∈ [−2, 2] ? (a)

3E[X] + 3E[Y  ] = 33 (c)

⇒         E[X] + E[Y  ] = 11 The value of the integral to

.

π

π

0

y

∫ ∫

sin x dx dy   is equal x (1 Mark)

Topic: Calculus: Multiple Integrals (+2)  The limits of the inner integral can be changed from (y to p) to (0 to x). Therefore the integral changes to π x

π

sin x sin x x ∫0 ∫0 x dx dy = ∫0 x [ y ]0 dx π

π

sin x x dx = ∫ sin x dx = [ cos x ]π0 x 0 0

=∫

= −cos p + cos 0 = +1 + 1 = +2 7.

Let Z be an exponential random variable with mean 1. That is, the cumulative distribution function of Z is given by

GATE_ECE_2019_new.indd 2

P ( Z > 2) P ( Z > 1)

P  (Z ≤ 2) = 1 - e−2

Topic: Probability and Statistics (11)  E[2X + Y] = 0

6.

P ( Z > 2) ∩ P ( Z > 1) P ( Z > 1)

1 x +1 2 1 f ( x) ≤ x 2 f ( x) ≤

Topic: Calculus (b) 

(b) f(x) ≤ 2|x + 1| (d) f(x) ≤ 2|x| (2 Marks)

f(−1) = 0, f ′( x ) ≤ 2 x ∈ [−2, 2]

⇒  −2 ≤ x ≤ 2 ⇒ 

− 2 ≤ f ′( x ) ≤ 2

⇒  −2 ≤

f ( 2) − f ( −1) ≤2 2 − ( −1)

⇒  −2 ≤

f ( 2) − f ( −1) ≤2 3

⇒  −6 ≤ f ( 2) − 0 ≤ 6 ⇒  −6 ≤ f ( 2) ≤ 6 Going through options for x = 2 , we have

6/22/2019 11:22:29 AM

Appendix  •  Solved GATE (ECE) 2019

with y(x) as a general solution. Given that y(1) = 1 and y(2) = 14

1 ( 2) ≤ 1 2 1 3 (b) f ( 2) ≤ ( 2 + 1) ≤ ≤ 1.5 2 2 (c) f(2) ≤ 2(2) ≤ 4 (a)

f ( 2) ≤

the value of y(1.5), rounded off to two decimal places, is . (2 Marks)

(d) f(2) ≤ 2(2 + 1) ≤ 6 All the options are satisfied. So, option (b) is necessarily true. 9.

Topic: Differential Equations d2 y dy − 3 x + 3 y = 0; x > 0 2 dx dx y(x) → General solution

2 (5.25)  x

Consider the line integral ∫ ( xdy − ydx)

y(1) = 1

C

the integral being taken in a counterclockwise direction over the closed curve C that forms the boundary of the region R shown in the figure below. The region R is the area enclosed by the union of a 2 × 3 rectangle and a semi-circle of radius 1. The line integral evaluates to y

⇒         [q 2 - q - 3q + 3]y = 0

⇒          q(q - 3) - 1(q - 3) = 0

R

⇒            (q - 3) (q - 1) = 0 So q = 3, 1.

1 0 1

2

3

4

5

CF = C1e3z + C2ez

x

(a) 6 + π/2

(b) 8 + π

(c) 12 + π

(d) 16 + 2π (2 Marks)

Topic: Calculus (c)  Line integral = ∫ ( xdy − ydx ) c

M = −y   ⇒ N = x    ⇒ So

∂N =1 ∂x

⎛ ∂N ∂M ⎞ Line integral = ∫ ⎜ − ⎟ dx dy = ∫ 2 dx dy ⎝ ∂x ∂y ⎠

π⎞ ⎛ = 2 (Area) = 2 ⎜ 6 + ⎟ = (12 + π ) 2⎠ ⎝ 10. Consider the homogeneous ordinary differential e­ quation x2

⇒         y = C1 x 3 + C2 x (1) For x = 1; y = 1 1 = C1 + C2(2) For x = 2; y = 14 14 = 8C1 + 2C2 ⇒ 7 = 4C1 + C2(3) Subtracting Eq. (2) from Eq. (3), we get

∂m = −1 ∂y

∂N ∂M − = 1 − ( −1) = 2 ∂x ∂y

GATE_ECE_2019_new.indd 3

[q(q - 1) - 3q + 3]y = 0

⇒            q 2 - 3q - q + 3 = 0

2



y(2) = 14 Let x = ez. Therefore ln x = z. The given equation can be written as

⇒               q 2 - 4q + 3 = 0

C 3



A3

d2 y dy − 3 x + 3 y = 0, x > 0 2 dx dx

6 = 3C1 ⇒ C1 = 2 7 = 8 + C2 ⇒ C2 = -1 Therefore,   y = 2x3 - x y(1.5) = 2(1.5)3 - 1.5 = 5.25

Chapter 2: Networks, Signals and Systems 11. Let H(z) be the z-transform of a real-valued discrete ⎛1⎞ time signal h[n]. If P ( z ) = H ( z ) H ⎜ ⎟ has a zero at ⎝z⎠ 1 1 z = + j , and P(z) has a total of four zeros, which 2 2 one of the following plots represents all the zeros ­correctly?

6/22/2019 11:22:33 AM

A4

GATE ECE Chapter-wise Solved papers

Topic: Discrete-time Signals: Z-Transform

(a) z-plane Imaginary

⎛1⎞ (d)  P ( z ) = H ( z ) H ⎜ ⎟ ⎝z⎠ Zeros come in ‘set of four’ for discrete time filters.

axis 2 |z| = 1 0.5 −2

z1 = z = 2 Real axis

0.5

z3 =

−2

(b) z-plane Imaginary z4 =

axis 2 |z| = 1

−2

−0.5

0.5

1

2 Real axis

−2

1 1 − j 2 2

z2 = z* =

−0.5

0.5

1 1 + j 2 2

1 1 = = 1− j z 1 1 + j 2 2 1 1 = = 1+ j z* 1 1 − j 2 2

12. Consider the two-port resistive network shown in the figure. When an excitation of 5 V is applied across Port 1, and Port 2 is shorted, the current through the short circuit at Port 2 is measured to be 1 A (see (a) in the figure).   Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see (b) in the figure), what is the current through the short circuit at Port 1?

(c) z-plane Imaginary

1Ω

axis

2Ω

2 |z| = 1

RΩ

Port 1

Port 2

0.5 −2

−0.5

1Ω

2 Real axis

0.5

2Ω

−0.5

RΩ

5V + −

−2

1A

(d) z-plane Imaginary (a)

axis 2 |z| = 1

1Ω

1

2Ω

0.5

RΩ

? −2

0.5

1

2 Real axis

−0.5

(b)

−1

(a) 0.5 A (c) 2 A

−2

(1 Mark)

GATE_ECE_2019_new.indd 4

+ 5V —

(b) 1 A (d) 2.5 A (1 Mark)

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Appendix  •  Solved GATE (ECE) 2019

Topic: Network Solution Methods: Nodal and Mesh Analysis (b)  Let the current be I. Using reciprocity theorem

14. It is desired to find three-tap causal filter which gives zero signal as an output to an input of the form

⎛5⎞ ⎛5⎞ ⇒ ⎜ ⎟=⎜ ⎟ ⎝1⎠ ⎝ I ⎠

⇒

y(t ) = u(t ) − 2e − t u(t ) + e −3t u(t ) Forced response = u(t).

⎛ Response ⎞ ⎛ Response ⎞ =⎜ ⎜ Excitation ⎟ ⎟ ⎠ Network 1 ⎝ Excitation ⎠ Network 2 ⎝

or

⎛ jπ n ⎞ ⎛ jπ n ⎞ x[n] = c1 exp ⎜ − ⎟ + c2 exp ⎜ 2 ⎟ ⎝ 2 ⎠ ⎝ ⎠ where c1 and c2 are arbitrary real numbers. The desired three-tap filter is given by

I =1A

13. Let Y(s) be the unit-step response of a causal system having a transfer function 3− s G ( s) = ( s + 1)( s + 3)

h[0] = 1, h[1] = a, h[2] = b and h[n] = 0 for n < 0 or n > 2 What are the values of the filter taps a and b if the output is y[n] = 0 for all n, when x[n] is as given above?

G ( s) that is, Y ( s) = . The forced response of the system s is (a) u(t) - 2e−tu(t) + e−3tu(t) (b) 2u(t) - 2e−tu(t) + e−3tu(t) (c) 2u(t) (d) u(t)

x[n]

n=0 ↓ h[n] = {1, a, b}

(a) a = 1, b = 1 (c) a = −1, b = 1

(b) a = 0, b = −1 (d) a = 0, b = 1 (2 Marks)

(1 Mark)

Topic: Fourier Series and Fourier Transform Representations (d)  Given that

Topic: LTI Systems: Causality 3− s (d)   G ( s) = ( s + 1)( s + 3)

⎛ πn⎞ ⎟ ⎜− j 2 ⎠

3− s G ( s) Y ( s) = = s s( s + 1)( s + 3)   A B C Y ( s) = + + s ( s + 1) ( s + 3)   3− s A( s + 1)( s + 3) + B( s + 3) s + C ( s + 1)( s) = s( s + 1)( s + 3) s( s + 1)( s + 3)

x[n] = C1e ⎝

Subtracting Eq. (1) from Eq. (2), we get 3A + 2B = −1 ⇒ 2B = −1 - 3 ⇒ 2B = -4 B = −2 ⇒ Putting values of A and B in Eq. (1), we get 1 - 2 + C = 0 ⇒ C =1 Now

GATE_ECE_2019_new.indd 5

1 2 1 Y ( s) = − + s ( s + 1) ( s + 3)

⎛ πn⎞ ⎟ ⎜j 2 ⎠

+ C2 e ⎝

Let the Fourier transform of h[n] be H(ω). Therefore H (ω ) = 1 + ae jω + be j 2ω For the given input −π ⎞ − jn2 π π ⎞ + jnπ ⎛ ⎛ y( x ) = C1 H ⎜ ω = e + C2 H ⎜ ω = ⎟ e 2 ⎟ 2 ⎠ 2⎠ ⎝ ⎝ For y[n] = 0

(3 - s) = A  (s2 + 4s + 3) + B(s2 + 3s) + C(s2 + s) Comparing LHS with RHS, we get A + B + C = 0 ⇒ 4A + 3B + C = -1(2) ⇒ 3A = 3 A =1 ⇒

y[n] = 0

(1)



−π ⎞ π⎞ ⎛ ⎛ H ⎜ω = = 0 and H ⎜ ω = ⎟ = 0 ⎟ 2 ⎠ 2⎠ ⎝ ⎝ From the first condition 1 + ae-jπ/2 + be-jπ = 0 ⇒ 1 - aj - b = 0 ⇒ (1 - b) - aj = 0 From the second condition 1 + ae jπ/2 + be jπ = 0 ⇒ (1 - b)aj = 0 Solving Eqs. (1) and (2) we get 1-b=0⇒b=1 and      a = 0

(1)

(2)

15. In the circuit shown, if v(t) = 2 sin (1000t) volts, R = 1 kΩ, and C = 1 μF, then the steady-state current i(t), in milliamperes (mA), is

6/22/2019 11:22:40 AM

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GATE ECE Chapter-wise Solved papers

i(t)

=

3 × 10 −3 + 2 1⎞ ⎛3 Yeq = ⎜ + j ⎟ 2⎠ ⎝2 =

C v(t)

1 3 + −2 j × 103 2 × 103

R

R C

C

× 10 −3 =

1 Z eq

1⎞ ⎛3 v ( t ) × Yeq = 2 sin(1000t ) × ⎜ + j ⎟ × 10 −3 A 2⎠ ⎝2 = [3sin(1000t) + jsin(1000t)] mA

R

= [3sin(1000t) + cos(1000t)] mA

(a) sin(1000t) + cos(1000t) (b) 2 sin(1000t) + 2 cos(1000t) (c) 3 sin(1000t) + cos(1000t) (d) sin(1000t) + 3cos(1000t)

16. The RC circuit shown below has a variable resistance R(t) given by the following expression: t ⎞ ⎛ R(t ) = R0 ⎜1 − ⎟ for 0 ≤ t < T ⎝ T⎠ where R0 = 1 Ω, C = 1 F. We are also given that T = 3R0C and the source voltage is Vs = 1 V. If the current at time t = 0 is 1 A, then the current I(t), in amperes, at time t = T/2 is (rounded off to 2 decimal places).

(2 Marks) Topic: Network Solution Methods (c)  Given v(t) = 2sin(1000t) V R = 1 kΩ i(t) = ? (mA) C = 1 μF The given circuit has to be changed into

R(t)

I(t) R

1 j × 10 −3 2

R ⇒ As the network forms Balanced bridge ⇓

−jXc −jXc

−jXc

R

C Vs + − t=0

R (2 Marks)

R

i(t)

v(t) −jXc 2R

−jXc

⇐ i(t)

Topic: Network Solution Methods t ⎞ ⎛ (0.25 A)  R(t ) = R0 ⎜1 − ⎟ ,  0 ≤ t < T ⎝ T⎠ R0 = 1Ω and C = 1 F At t = 0,   i(0) = 1 A

R

T = 3R0C = 3 s Vs = 1 V

−2jXc v(t) i(t)

R



v(t) 2 2R  R = R 3 1 1 XC = = = 103 ω C 103 × 1 × 10 −6 The equivalent conductance is given by Yeq = Y1 + Y2 =

GATE_ECE_2019_new.indd 6

1 1 + −2 jX C 2 / 3R

⎛ t⎞ R(t ) = ⎜1 − ⎟ ⎝ 3⎠ By KVL we have 1 idt = Vs C∫ di(t ) dR(t ) 1 ⇒    R(t) + i (t ) + i=0 dt dt C     i(t)R(t) +

⎛ t ⎞ di(t ) ⎛ −1 ⎞ + i (t ) ⎜ ⎟ + i = 0 ⇒    ⎜1 − ⎟ ⎝ 3 ⎠ dt ⎝ 3 ⎠ di ⇒                    (3 - t) = −2i dt

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Appendix  •  Solved GATE (ECE) 2019

⇒       

Since g(n) → is a length ‘3’ finite impulse response and values of g(−1) and g(1) must be known and for minimum value,

di −2 = dt i 3−t

⇒          ln i = 2 ln(t - 3)+ln c

g[0] = 4

⇒          ln i = ln[(t - 3)2 ⋅ C]

g[1] = 3 g[−1] = −3

⇒               i(t) = (t - 3)2C

g[n] = {−3, 4, 3}

⇒             i(0) = (0 - 3)2C ⇒     C =

Now, 10g[−1] + g[1] = 10(−3) + 3 = −27.

1 (∵i(0) = 1, given ) 9 1 i(t) = (t - 3)2 9 1 i(1.5) = ( −1.5) 2 = 0.25 A 9

Chapter 3: Electronic Devices 18. Which one of the following options describes correctly the equilibrium band diagram at T = 300 K of a Silicon pnn+ p++ configuration shown in the figure?

17. Let h[n] be a length-7 discrete-time finite impulse response filter, given by h[0] = 4, h[1] = 3, h[2] = 2, h[3] = 1

p

n

n+

p++

(a)

h[−1] = −3, h[−2] = −2, h[−3] = −1

EC

and h[n] is zero for |n| ≥ 4. A length-3 finite impulse response approximation g[n] of h[n] has to be obtained such that E ( h, g ) = ∫

π

−π

2

EF

is minimized, where H(ejω ) and G(ejω ) are the ­discrete-time Fourier transforms of h[n] and g[n], respectively. For the filter that minimizes E(h, g), the value of 10g[−1] + g[1], rounded off to 2 decimal places, is . (2 Marks)

E ( h, g ) =

π

∫

|n| ≥ 4;

jω

2

H ( e ) − G ( e ) dω (c)

  Filter, minimizes E(h, g) •  Value of 10g[−1] + g[1]

−π

EC EF

EV jω

EC

•

∫

(b)

h[n] = 0

−π

π

EV

H (e jω ) − G (e jω ) dω

Topic: Discrete-time Signals (-27)  h[0] = 4 h[−1] = −3 h[1] = 3 h[−2] = −2 h[2] = 2 h[−3] = −1 h[3] = 1

A7

3

H (e jω ) − G (e jω ) dω = 2π ∑ h( n) − g ( n) 2

2

EF

n =−3

For minimum value of E(h, g) 3

∑

h( n) − g ( n)

2

must be also minimum

EV

n =−3

⇒ |4 - g(0)|2 + |3 - g(1)|2 + |−3 - g(−1)| + 4 + 4 + 2

GATE_ECE_2019_new.indd 7

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A8

GATE ECE Chapter-wise Solved papers

(d)

(c)

C

EC B

EF

E (d)

C

EV B (1 Mark) Topic: Energy Bands in Intrinsic and Extrinsic Silicon (a)  EF will always be constant only, which eliminates options (b) and (c). Since p++ implies very heavy doping, so option (a) is ­correct here.

E (1 Mark) Topic: BJT (a)  It is an NPN transistor. C

EC B EV EF

19. The correct circuit representation of the structure shown in the figure is B

E

E 20. The figure shows the high-frequency C-V curve of a MOS capacitor (at T = 300 K) with Φms = 0 V and no oxide charges. The flat-band, inversion, and accumulation conditions are represented, respectively, by the points

C

C Q

P n++ p+

R

n++ n

0 n+

(a)

C

(a) P, Q, R (c) R, P, Q

VG (b) Q, R, P (d) Q, P, R (1 Mark)

B

E (b)

C B

E

GATE_ECE_2019_new.indd 8

Topic: MOS Capacitor (b)  Flat band is represented by Q Inversion is represented by R Accumulation is represented by P 21. In the circuit shown, VS is a square wave of period T with maximum and minimum values of 8 V and −10 V, respectively. Assume that the diode is ideal and R1 = R2 = 50 Ω. The average value of VL is volts (rounded off to 1 decimal place).

6/22/2019 11:22:51 AM

Appendix  •  Solved GATE (ECE) 2019

R1

+ +8 0 −10

VS + − T 2

R2

Topic: Photo Diode and Solar Cell (a)  Quantum efficiency = η Responsivity = R  Wavelength = λ

VL

η=

I out hv × Pin q

R=

I out Pin

T −

and

(1 Mark) Topic: P-N Junction (−3) When VS = 8 V, the diode is reverse biased and acts as open circuit as shown in the figure below.

A9

where    Iout = Output photodetector    Pin = Incident current light power

R1

   q = Change on one electron    h = Planck’s constant    v = c/λ

+ 8V + −

R 2 VL

Therefore R =η ×

− VL = 8 ×

q = 1.62 × 10−19 C  (charge on one electron)

R2 R1 + R2

h = 6.63 × 10−34    (Planck’s constant)

Since R1 = R2 = 50 Ω, we have VL = 8 × 1

c = 3 × 108 m/s    (speed of light) Solving, we get

R2 = 4V 2 R2

R=

When VS = −10 V, the diode is forward biased and acts as a short circuit as shown in the figure below. + − 10 V +

q q ⎛ q ⎞ =η × = ηλ ⎜ ⎟ c hv ⎝ hc ⎠ h× λ

R2

VL

ηλ 1.24

23. In the circuit shown, the breakdown voltage and the maximum current of the Zener diode are 20 V and 60 mA, respectively. The values of R1 and RL are 200 Ω and 1 kΩ, respectively. What is the range of Vi that will maintain the Zener diode in the ‘on’ state?

− R1 VL = −10 V 2

Average value of VL

Vi + −

⎛ 4 − 10 ⎞ (VL )avg = ⎜ ⎟ V = −3 V ⎝ 2 ⎠ 22. The quantum efficiency (η) and responsivity (R) at a wavelength λ (in μm) in a p-i-n photodetector are related by (a) R =

η ×λ 1.24

(b) R =

λ η × 1.24

(c) R =

1.24 × λ η

(d) R =

1.24 η ×λ

GATE_ECE_2019_new.indd 9

(2 Marks)

(a) 22 V to 34 V (c) 18 V to 24 V Topic: Zener Diode (b)  R1 = 200 Ω RL = 1000 Ω Range of Vi ?

RL

(b) 24 V to 36 V (d) 20 V to 28 V (2 Marks)   Vz = 20 V Iz max = 60 mA

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GATE ECE Chapter-wise Solved papers

200 Ω

Vi + −

VZ

1000 Ω

To maintain Zener diode in the ON state VZ = 20 V When Zener diode is in the OFF state, the voltage across the 1000 Ω (1K) resistor is V1k =

Vi min × 1000 1200

⇒ 20 =

Vi min × 1000 1200



P⎞ P 1⎛ SiO2 − Ge interface = ⎜ Pi − i ⎟ = i 3⎝ 4⎠ 4 Power obsorbed in Ge P 1⎛ = ⎜ Pi − i 3⎝ 4

⇒ 240 = Vi min × 10



Topic: P-N Junction (0.231)  Sample dimension = 1 cm × 1 cm Power incident Pi = 20 mW Power reflected from P SiO2 = i 4 Power reflected from

Power transmitted through sample

Minimum value of Vi is Vi min = 24 V

=

Zener diode is in break down state when ⎞ ⎛ Vi max − 20 − 20 ⎟ mA ≤ 60 mA ⎜ 0.2 ⎝ ⎠

Pt = Po e −α t ⇒

⇒                   Vi max ≤ 36 V Therefore 24 V ≤ Vi ≤ 36 V.

Pi 4

Pi 2 Pi −α t = e ⇒ α t = ln( z ) 4 4

⇒                   t = 0.231 μm

24. A Germanium sample of dimensions 1 cm × 1 cm is illuminated with a 20 mW, 600 nm laser light source as shown in the figure. The illustrated sample surface has a 100 nm of loss-less Silicon dioxide layer that reflects one-fourth of the incident light. From the remaining light, one-third of the power is reflected from the Silicon dioxide-Germanium interface, one-third is absorbed in the Germanium layer, and one-third is transmitted through the other side of the sample. If the absorption coefficient of Germanium at 600 nm is 3 × 104 cm−1 and the bandgap is 0.66 eV, the thickness of the Germanium layer, rounded off to 3 decimal places, is μm. 20 mW, 600 nm 1 cm

25. In an ideal p-n junction with an ideality factor of 1 at T = 300 K, the magnitude of the reverse-bias voltage required to reach 75% of its reverse saturation current, rounded off to 2 decimal places, is mV. −23 −1 [k = 1.38 × 10 JK , h = 6.625 × 10−34 J-s, −19 q = 1.602 × 10 C] (2 Marks) Topic: P-N Junction (35.83)  Given η = 1, T = 300 K. I = I 0 (eVBE /ηVT − 1)  −3 I0  4 From Eqs. (1) and (2) ⇒  I =

(1) (2)

3 − I 0 = I 0 ( eVBE /VT − 1) 4 ⇒ eVBE /VT = 0.25

1 cm 100 mm

⇒  VBE = -0.2384 × 1.386 V

T

⇒  VBE = -35.83 mV

(2 Marks)

GATE_ECE_2019_new.indd 10

Pi Pi 2 Pi + = 4 4 44

Power transmitted through thickness (t) =

⇒                Vi max ≤ 12 + 24

Germanium

P⎞ P 1⎛ Pi − i ⎟ = i 3 ⎜⎝ 4⎠ 4

Power entered in Ge sample =

⇒       Vi max - 20 - 4 ≤ 12

Silicon dioxide

⎞ Pi ⎟= ⎠ 4

Magnitude Ve = 35.83 mV

6/22/2019 11:22:59 AM

A11

Appendix  •  Solved GATE (ECE) 2019

26. Consider a long-channel MOSFET with a channel length 1 μm and width 10 μm. The device parameters are acceptor concentration NA = 5 × 1016 cm−3, ­electron mobility μn = 800 cm2/V-s, oxide capacitance/area Cox = 3.45 × 10−7 F/cm2, threshold voltage VT = 0.7 V. The drain saturation current (ID sat) for a gate voltage of 5 V is mA (rounded off to two decimal places). [ε0 = 8.854 × 10−14 F/cm, εSi = 11.9] (2 Marks) Topic: MOSFET (25.52)  L = 1 μm

(a) 1.8 and 1.2 (c) 1.8 and 2.4 Topic: BJTs and MOSFETs (c)  Vth = 0.6 V 3V

+ − 0.6 V

+ − 0.6 V Vout1

3V + −

W = 10 μm NA = 5 × 1016 cm−3        μn = 800 cm2 V-s Cox = 3.45 × 10−7 F/cm2 VT = 0.7 V ID sat = ? VG = 5 V ε0 = 8.854 × 10−14 F/cm εSi = 11.9 For MOSFET in saturation region, 1 2 ⎛W ⎞ I D sat = μ nCox ⎜ ⎟ (VGS − VT ) 2 ⎝L⎠

3V + − Vout1 = ( 3 − 0.6 − 0.6 ) V = 1.8 V 3V

3V + 0.6 V −

3V + 0.6 V −

+ 0.6 V − Vout2

3V + −

1 ⎛ 10 ⎞ × 800 × 3.45 × 10 −7 × ⎜ ⎟ (5 − 0.7) 2 2 ⎝ 1⎠ = 25.5162 mA ≈ 25.52 mA =

Chapter 4: Analog Circuits 27. In the circuits shown, the threshold voltage of each NMOS transistor is 0.6 V. Ignoring the effect of channel length modulation and body bias, the values of Vout1 and Vout2, respectively, in volts, are

Vout2 = ( 3 − 0.6 ) V = 2.4 V 28. In the circuit shown, V1 = 0 and V2 = Vdd. The other relevant parameters are mentioned in the figure. Ignoring the effect of channel length modulation and the body effect, the value of Iout is mA (rounded off to one decimal place). Vdd W = 10 L W = 10 L

3V

3V + −

(b) 2.4 and 2.4 (d) 2.4 and 1.2 (2 Marks)

Vdd

Vout1

V1 3V + −

W =5 L W =5 L

W = 40 L Iout V2

1 mA

3V

3V

3V Vout2

W =2 L

W =3 L

3V + − (2 Marks)

GATE_ECE_2019_new.indd 11

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GATE ECE Chapter-wise Solved papers

NML = VIL - VOLU

Topic: BJT and MOSFET Amplifiers (6) Vdd

W = 10 L W = 10 M4 L



M6 W = 40 L M7

W =5 L W =5 L

Vdd V 1 1 mA

MHH = VOHU - VIH Case(i):

M3

Iout V2

βn = β  p, let take VDD = 5, VTN = -VTP = 0.8, VI = 2.5

VIL = VTN +

1 [2VIH - VDD - VTN - VTP] = 0.45 2 (NM)L = VIL - VOLU = 1.625 VOLU =

VOUH =

M5

1 [2VIL + VDD - VTN - VTP] = 4.575 2

I2 W =2 L M1

VIH = VTN +

W =3 L M2

5 [V + VTP - VTN] = 2.95 8 DD

(NM)H = 1.625 Case (ii):

M1 and M2 have the same Vgs

(βn ≠ βp) Because width of PMOS increased

So, current flows in the ratio of W/L ⎛3⎞ So,          I2 = ⎜ ⎟ ⎝2⎠

3 [V + VTP - VTN] = 2.075 8 DD

⎛W ⎞ ⎛W ⎞ ⎜ L ⎟ >⎜ L ⎟ ⎝ ⎠ P ⎝ ⎠n

(1 mA )

= 1.5 mA

 V1 = 0; therefore M3 is in cut off and entire I2 current flows through M5 branch.

βp Pn

> 1,

So,       I5 = 1.5 mA [ I5 ]

⎛ 40 ⎞ Iout = ⎜ ⎟ ⎝ 10 ⎠ Therefore,

VIL = VTN +

[Ratio of W/L]

Iout = 4 × 1.5 mA = 6 mA

VDD + VTP − VTN ⎛ βn ⎞ − 1⎟ ⎜⎜ ⎟ ⎝ βp ⎠

⎡ ⎤ βn ⎢ ⎥ βp ⎢2 ⎥ 1 − ⎢ βn ⎥ −3 ⎥ ⎢ ⎥⎦ ⎢⎣ β p

= 2.216

Chapter 5: Digital Circuits 29. A standard CMOS inverter is designed with equal rise and fall times (βn = βp). If the width of the PMOS transistor in the inverter is increased, what would be the effect on the LOW noise margin (NML) and the HIGH noise margin (NMH)? (a) NML increases and NMH decreases. (b) NML decreases and NMH increases. (c) Both NML and NMH increase. (d) No change in the noise margins.

⎛ β VIH ⎜1 + n ⎜ β p ⎝ VOLU =

=

⎞ ⎛β ⎟⎟ − VDD − ⎜⎜ n ⎠ ⎝ βp ⎛β ⎞ 2⎜ n ⎟ ⎜β ⎟ ⎝ p⎠

⎞ ⎟⎟ (VTN − VTP ) ⎠

3.048 × 1.8 − 5 − 0.64 + 0.8 2 × 0.8

= 0.404 (NM )L = 2.216 - 0.404 = 1.812 (Increased) (1 Mark)

Topic: Combinatorial Circuits (a) Given, bn = bp for standard CMOS inverter. We know that standard definition of noise margin

GATE_ECE_2019_new.indd 12

βn < 1, let assume 0.8 βp

VOUH =

=

1 ⎡ ⎛ βn ⎢1+ ⎜ 2 ⎢⎣ ⎜⎝ β p

⎞ ⎟⎟VIL + VDD ⎠

⎛β −⎜ n ⎜β ⎝ p

⎤ ⎞ ⎟⎟VTN − VTP ⎥ ⎥⎦ ⎠

1 [( 2.216)1.8 + 5 − 0.64 + 0.8] = 4.574 2

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Appendix  •  Solved GATE (ECE) 2019

VIH = VTN +

= 0.8 +

(VDD

⎤ ⎡ β ⎢ 2 n ⎥ βp + VTP − VTN ) ⎢ ⎥ − 1⎥ ⎢ βn β −1 ⎢ 3 n +1 ⎥ βp ⎢⎣ β p ⎥⎦

A13

(2)  Let us consider a case when EN = 1. Output of NAND gate = ( D ⋅1) = D Output of NOR gate = ( D + 0) = D Therefore, output F = D = D . 31. In the circuit shown, A and B are the inputs and F is the output. What is the functionality of the circuit?

[5 − 1.6] × −0.132 −0.2

Vdd

= 2.248 + 0.8 = 3.048 (NM)H = 4.574 − 3.048 = 1.526 (decreased) 30. In the circuit shown, what are the values of F for EN = 0 and EN = 1, respectively?

F

Vdd

EN A

B

F (a) Latch (c) SRAM cell

(b) XNOR (d) XOR (1 Mark)

D (a) 0 and D

(b) Hi-Z and D

(c) 0 and 1

(d) Hi-Z and D (1 Mark)

Topic: Combinatorial Circuits (b)  (1) Let us consider a case EN = 0. When any input of a NAND gate is 0, its output = 1. Therefore, the upper PMOS FET is OFF. Output of NOR gate = (1 + D ) = 0 Therefore, the NMOS FET is also OFF. Therefore, the output F is in a high impedance (Hi-Z) state.

Topic: Combinatorial Circuits (b) A

B

Q1

Q2

Q3

Q4

F

0

0

ON

ON

OFF

OFF

High

0

1

OFF

ON

ON

OFF

Low

1

0

ON

OFF

OFF

ON

Low

1

1

OFF

OFF

ON

ON

High

From truth table Vdd

Q1

Vdd Q2 EN

F

F Q3

Q4

D A

GATE_ECE_2019_new.indd 13

B

6/22/2019 11:23:08 AM

A14

GATE ECE Chapter-wise Solved papers

32. In the circuit shown, the clock frequency, i.e., the frequency of the Clk signal, is 12 Hz. The frequency of the signal at Q2 is kHz.

(c)

A=0 A=0 A=1 Q=0

Q=1 A=1

D1

Q1

D2

Q2

(d) A = 1

A=0 A=0

Clk Q1

Clk Q2

12 kHz

Q=0

Q=1

(1 Mark)

(2 Marks)

Topic: Combinatorial Circuits (4)  D1 = Q1 Q2 , D2 = Q1 Present state

Topic: Sequential Circuits (c) 

Input

Q2

Q1

A=1

D1

D2

Next state Q1+

Q2+

0

0

1

0

1

0

1

0

0

1

0

1

0

1

0

0

0

0

D

Q

1

Q

0

f

There are only three states. Therefore, fQ = 2

CLK

12 kHz = 4 kHz 3

A Output of MUX f = A ⋅ Q + AQ

33. The state transition diagram for the circuit shown is

D = Q⋅ f Q+ = D = Q ⋅ f = Q⋅ A Q D

Q

1

Q

0

Q+ = Q + A ⊕ Q For Q = 0;

CLK

A

(a) A = 0

A=0 A=1

Q=0

Q=1 A=1

(b)

34. A CMOS inverter, designed to have a mid-point voltage V1 equal to half of Vdd, as shown in the figure, has the following parameters: Vdd = 3 V

A=0

μnCox = 100 μA/V2; Vtn = 0.7 V for nMOS

A=1

μpCox = 40 μA/V2; |Vtp| = 0.9 V for pMOS

A=0

Q=0

Q=1 A=1

GATE_ECE_2019_new.indd 14

Q+ = 1+ A ⊕1 = 1 If Q = 0, whatever the value of A, either 0 (or) next stage will be Q = 1. Therefore, option (c) satisfies only.

⎛W ⎞ ⎛W ⎞ The ratio of ⎜ ⎟ to ⎜ ⎟ is equal to L ⎝ ⎠n ⎝ L ⎠ p off to three decimal places).

(rounded

6/22/2019 11:23:12 AM

Appendix  •  Solved GATE (ECE) 2019

Vout

A15

Vdd = 4 V

Vdd

Vdd 2

Vout Vin

V1 =

Vdd 2

Vdd

Vin (2 Marks)

Topic: Combinatorial Circuits (0.225)  At saturation, mid-point voltage, 1 ⎛W ⎞ μnCox ⎜ ⎟ (VnS − Vtn ) 2 2 ⎝ L ⎠N 1 ⎛W ⎞ = μ pCox ⎜ ⎟ (VSG − Vtp ) 2 2 ⎝ L ⎠P ⎛W ⎞ ⎛W ⎞ 100 ⎜ ⎟ (1.5 − 0.7) 2 = 40 ⎜ ⎟ (1.5 − 0.9) 2 ⎝ L ⎠n ⎝ L ⎠P ⎛W ⎞ 2 ⎜L⎟ 40(0.6) 2 4 ⎛6⎞ ⎝ ⎠N ⇒          = = (W /L) P 100(0.8) 2 10 ⎜⎝ 8 ⎟⎠ ⎛W ⎜L ⎝ ⇒            ⎛W ⎜ ⎝L

⎞ ⎟ ⎠N = 0.225 ⎞ ⎟ ⎠P

35. In the circuit shown, the threshold voltages of the PMOS (|Vtp|) and NMOS (Vtn) transistors are both equal to 1 V. All the transistors have the same output resistance rds of 6 MΩ. The other parameters are listed below: ⎛W ⎞ μnCox = 60 μA/V 2 ; ⎜ ⎟ =5 ⎝ L ⎠ nMOS ⎛W ⎞ μ pCox = 30 μA/V 2 ; ⎜ ⎟ = 10 ⎝ L ⎠ nMOS μn and μp are the carrier mobilities, and Cox is the oxide capacitance per unit area. Ignoring the effect of channel length modulation and body bias, the gain of the circuit is (rounded off to 1 decimal place).

GATE_ECE_2019_new.indd 15

(2 Marks) Topic: Combinatorial Circuits (-900)  μnCox = 60 μA/V2; (W/L)NMOS = 5 μpCox = 30 μA/V2; (W/L)PMOS = 10 I SD1 = I SD2

μ pCox ⎛ W ⎞ μ C ⎛W ⎞ ( 4 − Vx − 1) 2 = n ox ⎜ ⎟ (Vx − 0 − 1) 2 2 ⎜⎝ L ⎟⎠ p 2 ⎝ L ⎠N 2

⇒         30 × 10 (3 − Vx ) 2 = 60 (Vx − 1) 2 × 5 ⇒          3 - Vx = Vx - 1 ⇒                   4 = 2Vx ⇒                   Vx = 2 V Vdd = 4 V

M1

M3

Vx M2

M1 and M2 will have ⎛W they are identical ⎜ ⎝L

M4

equal current flowing also since ⎞ ⎛W ⎞ ⎟ =⎜ L ⎟ ⎠1 ⎝ ⎠ 2

6/22/2019 11:23:16 AM

A16

GATE ECE Chapter-wise Solved papers

Np and the number of system zeros Nz in the frequency range 1 Hz ≤ f ≤ 107 Hz is ______.

So, VSG = VSG also, by KVL in loop 1 2 4 = VSG1 = VSG 2 So,

0 slope 1p 100

VSG1 = VSG 2 = 2V

Therefore, current through M1 and M2 ⎛1⎞ ⎛W I = ⎜ ⎟ ( μ pCox ) ⎜ 2 ⎝ ⎠ ⎝L

⎞ 2 ⎟ [VSG − VT ] ⎠

2p

104

105

106

2z

0 101

102 103 1z

2p 0 slope

⎛1⎞ = ⎜ ⎟ (30)(10)( 2 − 1) 2 = 150 μA ⎝2⎠

1p

(a) Np = 5, Nz = 2 (b) Np = 6, Nz = 3

Vdd = 4 V +V SG1 −

M1

107 f (in Hz)

(c) Np = 7, Nz = 4

M3

(d) Np = 4, Nz = 2 + VSG2 −

M2

(1 Mark)

Vout M4

Vin

Topic: Bode and Root-locus Plots (b)  Let p → pole, z → zeros. 1 pole gives a slope of −20 dB/dec. 1 zero gives a slope of +20 dB/dec. At low frequency region below 10 Hz, slope = 0. Therefore there is No pole or zero at origin.

M1 and M2 are matched with same ⎛W ⎞ ⎜ L ⎟ and same VSG . Hence, ⎝ ⎠ I M3 = I M 4 = 150 μA

0 slope 1p 100

For MOSFET M4, gm = =

2 μ nCox

W ID L

104

105

106

2z

0 101

102 103 1z

2p

107 f (in Hz)

0 slope

2 × 60 μ × 5 × 150 μ = 300 μ 1p

μA⎞ ⎛ Av = -gm(rd || rd) = ⎜ −300 (6 MΩ  16 MΩ) V ⎟⎠ ⎝ μA⎞ ⎛ = ⎜ −300 (3 MΩ) = −900 V/V V ⎟⎠ ⎝

Chapter 6: Control Systems 36. For an LTI system, the Bode plot for its gain is as illustrated in the figure shown. The number of system poles

GATE_ECE_2019_new.indd 16

37. Consider a six-point decimation-in-time Fast Fourier Transform (FFT) algorithm, for which the signal-flow graph corresponding to X[1] is shown in the figure. Let ⎛ j 2π ⎞ W6 = exp ⎜ − ⎟ . In the figure, what should be the ⎝ 6 ⎠ values of the coefficients a1, a2, a3 in terms of W6 so that X[1] is obtained correctly?

6/22/2019 11:23:19 AM

Appendix  •  Solved GATE (ECE) 2019

x[0]

X[0] a1

x[3]

X[1]

−1

x[1]

X[2] a3

−1

X[3]

x[2]

X[4]

x[5]

38. Consider a causal second-order system with the transfer 1 function G ( s) = 1 + 2s + s 2 1 as an input. Let C(s) be the s corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value lim c(t ), t →∞ rounded off to two decimal places, is (a) 5.25 (b) 4.50 (c) 3.89 (d) 2.81 with a unit-step R( s) =

a2

x[4]

(2 Marks)

X[5]

−1

Topic: Transient and Steady-state Analysis of LTI Systems 1 (b)   G ( s) = 2 s + 2s + 1 R(s) = 1/s

(a) a1 = −1, a2 = W6 , a3 = W

2 6

(b) a1 = 1, a2 = W62 , a3 = W6 (c) a1 = 1, a2 = W6 , a3 = W62

C(s) is corresponding output

(d) a1 = −1, a2 = W , a3 = W6 2 6

C(s) = G(s) R(s) (2 Marks)

Topic: Signal Flow Graph (c)  From the signal flow graph X[1] = a1x[0] - a1x[3] + a2x[1] - a2x[4] + a3x[2] - a3x[5](1)

C ( s) =

1 1 = s( s + 2 s + 1) s( s + 1) 2

C ( s) =

A B C + + s ( s + 1) ( s + 1) 2

2

When s = 0, we have A=

= a 1[x(0) - x(3)] + a2[x(1) - x(4)] + a3[x(2) - x(5)]

X [k ] = ∑ x[n]e

⎛ 2π ⎞ − j⎜ ⎟ nk ⎝ N ⎠

or

n=0

N −1

C=

∑ x (n)W n=0

kn N

X [1] = ∑ x[n]WNn

B=

n=0

For N = 6

=

5

X [1] = ∑ x[n]WNkn

d 1 ( s + 1) 2 ds s( s + 1) 2 1 s2

= ( s =−1)

−1 = −1 1

⇒ B = −1

k =0

Therefore,

Therefore,

X (1) = x ( 0 ) ⋅1 + [ x (1) − x ( 4 )]W + [ x ( 2 ) − x ( 5 )]W 1 6

2 6

1 1 1 c( s ) = − − s ( s + 1) ( s + 1) 2

+ x ( 3 )W

3 6

⎡⎣∵W = W63 ⋅W61 = −W61 and W65 = W62 ⋅W63 = −W62 ⎤⎦ 4 6

Comparing with Eq. (1), we get a1 = 1 a2 = W6 a3 = W62

GATE_ECE_2019_new.indd 17

1 = −1 ( −1)

Now

N −1

For k = 1;

1 =1 (0 + 1) 2

when s = −1, we have

For FFT N −1

A17

Hence c(t) = (1 − e−t − te−t) u(t) lim c(t ) = 1 t →∞

94% of steady state = 0.94 For 0.94 = 1 - e−t - te−t Solving for given options, option (b) is satisfying.

6/22/2019 11:23:25 AM

A18

GATE ECE Chapter-wise Solved papers

39. The block diagram of a system is illustrated in the figure shown, where X(s) is the input and Y(s) is the output. Y ( s) The transfer function H ( s) = is X ( s)

ensure that the transfer function of this LTI system is 1 H ( s) = 3 ? s + 3s 2 + 2 s + 1 (a)

+ X(s)

−

s

∑

+ ∑

−

1 s

1 s

Y(s)

+

⎡0 1 0⎤ ⎢ A = ⎢ 0 0 1 ⎥⎥ and C = [1 0 0] ⎢⎣ −1 −2 −3⎥⎦

⎡0 1 0⎤ ⎢ (b) A = ⎢ 0 0 1 ⎥⎥ and C = [1 0 0] ⎢⎣ −3 −2 −1⎥⎦ ⎡0 1 0⎤ ⎢ A = ⎢ 0 0 1 ⎥⎥ and C = [0 0 1] ⎢⎣ −1 −2 −3⎥⎦

s2 + 1 s3 + s 2 + s + 1

(c)

(a) H ( s) = (b) H ( s) =

s2 + 1 s + 2s 2 + s + 1

(c) H ( s) =

s +1 s + s +1

⎡0 1 0⎤ ⎢ (d) A = ⎢ 0 0 1 ⎥⎥ and C = [0 0 1] ⎢⎣ −3 −2 −1⎥⎦

(d) H ( s) =

s2 + 1 2s 2 + 1

3

2

(2 Marks)

(2 Marks)

Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  x (t ) = Ax(t ) + Bu(t ) y(t) = Cx(t) + du(t)

Topic: Block Diagram Representation (b)  The given block diagram can be simplified as

B = [0 0 1]T d=0

+ X(s)

− ∑

s + 1s

−

1 s

Y ( s) 1 Y ( s) . X 1 ( s) = H ( s) = 3 = U ( s) s + 3s 2 + 2 s + 1 X 1 ( s ) U ( s )

Y(s)

Y ( s) = 1 ⋅ ( Numerator ) X 1 ( s) + X(s)

∑ −

s2 + 1 = 1s s + s2 + 1

1 s

Y(s)

s2 + 1 X(s)

Y(s)

⇒

X 1 ( s) 1 = 3 (Denominator) 2 U ( s) ( s + 3s + 2 s + 1)

⇒

X 1 ( s) 1 = U ( s ) s 3 + 3s 2 + 2 s + 1

⇒ (s3 + 3s2 + 2s + 1) X1(s) = U(s) ⇒

H ( s) =

Y ( s) s2 + 1 = 3 X ( s) s + 2 s 2 + s + 1

40. Let the state-space representation of an LTI system be x (t ) = Ax(t ) + Bu(t ), y(t) = Cx(t) + du(t) where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let B = [0 0 1]T and d = 0. Which one of the following options for A and C will

GATE_ECE_2019_new.indd 18

d 3 x1 d 2 x1 dx + 3 + 2 1 + x1 (t ) = u(t ) dt dt 3 dt 2

Hence,

dx1 (t ) = x1 = x2 (1) dt d 2 x1 (t ) x1 = x 2 = x3 (2) =  dt 2 x3 + 3 x3 + 2 x2 + x1 = u(t )



x3 = −3 x3 − 2 x2 − x1 + u(t ) (3)

6/22/2019 11:23:32 AM

Appendix  •  Solved GATE (ECE) 2019

A19

For two poles to lie on jω axis

Now, Y ( s) =1 X 1 ( s)

6−K =0 ⇒6-K=0⇒K=6 3

⇒  Y(s) = X1(s) ⇒  y(t) = x1(t)(4) Representing in matrix form ⎡ x1 ⎤ ⎡ 0 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ x ⎥ = ⎢ 0 0 1 ⎥ ⎢ x ⎥ + ⎢0 ⎥ u(t ) ⎢ 2⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ x3 ⎥⎦ ⎢⎣ −1 −2 −3⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣1 ⎥⎦ ⎡ x1 ⎤ y(t ) = [1 0 0 ] ⎢⎢ x2 ⎥⎥ + 0 u(t ) ⎢⎣ x3 ⎥⎦ Therefore, ⎡0 1 0⎤ ⎡0 ⎤ A = ⎢⎢ 0 0 1 ⎥⎥ , B = ⎢⎢0 ⎥⎥ , C = [1 0 0] ⎢⎣ −1 −2 −3⎥⎦ ⎢⎣1 ⎥⎦ 41. Consider a unity feedback system, as in the figure shown, with an integral compensator K/s and open-loop transfer function 1 s 2 + 3s + 2 where K > 0. The positive value of K for which there are exactly two poles of the unity feedback system on the jω axis is equal to (rounded off to two decimal places). G ( s) =

X(s) +

Σ

Chapter 7: Communications 42. A linear Hamming code is used to map 4-bit messages to 7-bit codewords. The encoder mapping is linear. If the message 0001 is mapped to the codeword 0000111, and the message 0011 is mapped to the codeword 1100110, then the message 0010 is mapped to (a) 0010011 (b) 1100001 (c) 1111000 (d) 1111111 (1 Mark) Topic: Hamming Codes (b)   Messages = 4 bit Codeword = 7 bit

→ EXO Ring 

4 bit        7 bit 0001  →  0000111 0011  →  1100110  0010   →  1100001

43. Consider the signal

π⎞ ⎛ 2π ⎞ ⎛π f (t ) = 1 + 2 cos(π t ) + 3 sin ⎜ t ⎟ + 4 cos ⎜ t + ⎟ , 4⎠ ⎝ 3 ⎠ ⎝2 where t is in seconds. Its fundamental time period, in seconds, is . (1 Mark)

Y(s) K s

−

Topic: Analog Communications (12 s)

G(s)

π⎞ ⎛ 2π ⎞ ⎛π f (t ) = 1 + 2 cos π t + 3 sin ⎜ t ⎟ + 4 cos ⎜ t + ⎟ 4⎠ ⎝ 3 ⎠ ⎝2 (2 Marks) Topic: Block Diagram Representation 1 (6)  G ( s) = 2 s + 3s + 2 Y ( s) K = 3 ,K >0 X ( s) s + 3s + 2 + K s3

1

s

3 6−K 3 K

2

s s0

GATE_ECE_2019_new.indd 19

2 K

⎛ 2π π ⎞ π ω0 = HCF ⎜ π , , ⎟ = 3 2⎠ 6 ⎝ Fundamental period is 2π 2π = = 12 s ω0 π / 6 44. The baseband signal m(t) shown in the figure is phase-modulated to generate the PM signal ϕ(t) = cos (2πfct + km(t)). The time t on the x-axis in the figure is in milliseconds. If the carrier frequency is fc = 50 kHz and k = 10π, the ratio of the minimum instantaneous frequency (in kHz) to the maximum instantaneous frequency (in kHz) is (rounded off to 2 decimal places).

6/22/2019 11:23:36 AM

A20

GATE ECE Chapter-wise Solved papers

m(t) 1 0

1

2

3

4

5

6

7

8

9 t(in ms)

−1 (1 Mark) Topic: Digital Communications (0.75)  Given

ϕ(t) = cos(2πfct + km(t)) or

θ (t) = 2πfct + km(t)

Instantaneous angular frequency is dθ (t ) ωi = = 2π f c + km(t ) dt Therefore, fi = fc + = 50 +

If we keep p(t) = ϕ1(t), but take q(t ) = Eϕ2 (t ), for what value of E would we obtain the same bit-error probability Pb? (a) 0 (b) 1 (c) 2 (d) 3 (2 Marks) Topic: Digital Communications (d)  Bits 0 and 1 are equally likely 1 p(0) = p(1) = 2 When p(t) = ϕ1(t) and q(t) = -ϕ1(t), probability of error

10π d m(t ) 2π dt

From the given figure d m(t ) dt

max

⎞ ⎟⎟ ⎠

1

p(t)

f1(t)

q(t)

When p(t ) = ϕ1 (t ) and q(t ) = E ϕ2 (t ) dmin = E + 1 = 1 + E

=2

⎛ d2 Pe2 = Q ⎜ min ⎜ 2N 0 ⎝

fmax = 50 + 5 × 2 = 60 kHz

⎞ ⎟ ⎟ ⎠

⎡ E +1 ⎤ =Q⎢ ⎥ ⎢⎣ 2 N 0 ⎥⎦

Also min

⎛ 4 Pe1 = Q ⎜ ⎜ 2N 0 ⎝ dmin

−1

Therefore

d m(t ) dt

⎞ ⎟ ⎟ ⎠

Now dmin = 2. So

k d m(t ) 2π dt

d = 50 + 5 m(t ) dt

⎛ d2 Pe1 = Q ⎜ min ⎜ 2N 0 ⎝

= −1

f 2(t)

Therefore fmin = (50 - 5)kHz = 45 kHz Hence

√E

f min 45 = = 0.75 f max 60

dmin p(t)

f1(t)

1 45. A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density N0/2. Binary signaling, with 0  p(t) and 1  q(t), is used for the transmission, along with an optimal receiver that minimizes the bit-­ error probability.  Let ϕ1(t), ϕ2(t) form an orthonormal signal set. If we choose p(t) = ϕ1(t) and q(t) = −ϕ1(t), we would obtain a certain bit-error probability Pb.

GATE_ECE_2019_new.indd 20

Given Pe1 = Pe2  Therefore, E +1 = 2 ⇒  E + 1 = 4 ⇒  E = 4 - 1 = 3

6/22/2019 11:23:41 AM

Appendix  •  Solved GATE (ECE) 2019

46. Let a random process Y(t) be described as Y(t) = h(t) × X(t) + Z(t), where X(t) is a white noise process with power spectral density SX( f ) = 5 W/Hz. The filter h(t) has a magnitude response given by |H( f )| = 0.5 for −5 ≤ f ≤ 5, and zero elsewhere. Z(t) is a stationary random process, uncorrelated with X(t), with power spectral density as shown in the figure. The power in Y(t), in watts, is equal to W (rounded off to two decimal places). 1 SZ(f) (W/Hz)

−5

5

f(Hz) (2 Marks)

Topic: Random Processes (17.5 W)  Given Y(t) = h(t) × X(t) + Z(t) SX( f ) = 5 W/Hz

A21

H(f)12 (0.5)2 × 5

−5

5

f

47. A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude modulated to generate an AM signal f(t) = A(1 + m(t)) cos 2πfct, where fc = 600 kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is Mbps. (2 Marks) Topic: Analog Communications (0.5904)  m(t) is band limited between 5 kHz and 15 kHz. The case here is of band pass sampling.

|H( f )| = 0.5 for −5 ≤ f ≤ 5 Z(t) and X(t) are uncorrelated

Nyquist frequency = 1 SZ(f ) (W/Hz)

2 fH K

f ⎡ f ⎤ where K = ⎢ H ⎥ , that is integral part of H . Bw ⎣ Bw ⎦ Therefore

−5 Let ⇒

5

f(Hz)

SY1( f ) = |H( f )|2 × SX( f )

Y(t) = Y1(t) + Z(t)

PY (t ) =

∞

∫ [S

Y1

∞

( f ) + SZ ( f )] df

∫

SY 1 ( f )df +

−∞

∞

∫S

Z

2 × 615 = 61.5 kHz 20

Sampling frequency fs = 1.2 s

−∞

=

So, Nyquist frequency =

SY( f ) = SY1( f ) + SZ( f )

( f )df

−∞

1 = (0.5) × 5 × 10 + × 10 × 1 2 2

Power in y(t) = Py(t) = 12.5 + 5 = 17.5 W

GATE_ECE_2019_new.indd 21

⎡ 615 ⎤ K =⎢ ⎥ = 20 ⎣ 30 ⎦

Y1(t) = h(t) × x(t)

Nyquist frequency = 1.2 × 61.5 kHz = 73.8 kHz Number of quantization levels = 256 ⇒ n = 8 [as 2n = 256] where n = number of bits Rate of coded bit stream = nfs = 8 × 73.8 kbps = 0.5904 mbps

6/22/2019 11:23:44 AM

fN(x)

1/4 A22

GATE ECE Chapter-wise Solved papers

−2 48. A random variable X takes values −1 and +1 with probabilities 0.2 and 0.8, respectively. It is transmitted across a channel which adds noise N, so that the random variable at the channel output is Y = X + N. The noise N is independent of X, and is uniformly distributed over the interval [−2, 2]. The receiver makes a decision

fx −1

−Tx

1/4

−3

x

1Rx P

⎧ −1, if Y ≤ θ  X =⎨ ⎩+1, if Y > θ

1

−1Rx P

where the threshold θ ∈ [−1, 1] is chosen so as to minimize the probability of error Pr[  X ≠ X ]. The minimum probability of error, rounded off to 1 decimal place, is . (2 Marks)

Tx

2 1 = 10 5

8 4 P ( X = 1) = 0.8 = = 10 5

1/4

−

−1

3

Chapter 8: Electromagnetics

Topic: Fundamentals of Error Correction (0.1)  Given that P ( X = −1) = 0.2 =

2

49. What is the electric flux



( ∫ E ⋅ da )

through a

quarter-cylinder of height H  (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q?

Y=X+N

ε0

⎧−1 if Y ≤ θ  X =⎨ ⎩+1 if Y > θ

θ ∈ [−1,1] ⎛ 1R Pe = P ( X = −1) P ⎜ X ⎝ −TX

H ⎞ ⎛ −1 ⎞ ⎟ + P ( X = 1) P ⎜ ⎟ ⎠ ⎝ 1TX ⎠

1 1 Pe = 0.2 × (1 − θ ) + 0.8 × + (1 + θ ) 4 4 dPe = 0. dθ

For minimum Pe, Therefore Pe = −

Q

0.2 0.8 0.6 + = = 0.15 ≈ 0.1 4 4 4

(a)

HQ HQ (b) ε0 4ε 0

(c)

Hε0 4Q

(d)

Q  ar 2πε 0 r →

S

−2

2 fx −1

GATE_ECE_2019_new.indd 22

x

=

S

Q  a r r dϕ dz a r 2π ε 0 r

Q QH [ϕ ]π0 / 2 [ z ]0H = 2πε 0 4ε 0

1Rx P

1/4

→

Electric flux = ∫ E ⋅ ds = ∫

1/4

(1 Mark)

Topic: Electrostatics (b)  For infinite long line, E=

fN(x)

4H Qε 0

−Tx 6/22/2019 11:23:50 AM

Appendix  •  Solved GATE (ECE) 2019

50. In the table shown, List I and List II, respectively, contain terms appearing on the left-hand side and the righthand side of Maxwell’s equations (in their standard form). Match the left-hand side with the corresponding right-hand side. List I

List II

1.  ∇ ⋅ D

(P) 0

2.  ∇ × E

(Q)  ρ

3.  ∇ ⋅ B

(R)  − ∂B ∂t

4.  ∇ × H

(S)  J +

(a) 1-P, 2-R, 3-Q, 4-S (c) 1-Q, 2-S, 3-P, 4-R

∂D ∂t

% change in Rrad =

4.  ∇ × H = J +

{non-existence of monopole}

∂D ∂t

51. Radiation resistance of a small dipole current element of length l at a frequency of 3 GHz is 3 ohms. If the length is changed by 1%, then the percentage change in the radiation resistance, rounded off to two decimal places, is %. (1 Mark)



= (1.0201 - 1) × 100%



= 2.01%

52. Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the  magnetic field B between the wires at a distance r from W1 is W1 r



ΔRrad = ?%

⇒

6 μ0 I 5π r

μ02 I 2 (d) 2π r 2

5μ 0 I (c) 6π r

(2 Marks)

Topic: Electrostatics (c)  Since both lines carry current in opposite directions, hence →

→

→

B = B1 + B2 Magnetic field at distance r from W1 is

→

2

2

μ0 ( 2 I ) 2π (3r )

→ 2 μ0 I ⎞ ⎛μ I B=⎜ 0 + ⎟ r π 2 3 ( 2π r ) ⎠ ⎝

=

Rrad 2 ⎛ l2 ⎞ ⎛ 1.01l ⎞ =⎜ ⎟ = = 1.0201 Rrad1 ⎝ l1 ⎠ ⎜⎝ l ⎟⎠

GATE_ECE_2019_new.indd 23

(b)

B2 =

Now 2

μ0 I 6π r

μ0 I 2π r Magnetic field at distance 3r from W2 is

Length of antenna is related with radiation resistance by ⎛ dl ⎞ Rrad = 80π 2 ⎜ ⎟ ⎝λ⎠ 2 Rrad ∝ l

(a)

B1 =

Rrad = 3 Ω Δl = 1%

W2

→

Topic: Antennas (2.01%)  Frequency f = 3 GHz

× 100



(b) 1-Q, 2-R, 3-P, 4-S (d) 1-R, 2-Q, 3-S, 4-P (1 Mark)

{using Faraday’s Law}

Rrad1

⎞ ⎛ Rrad 2 =⎜ − 1⎟ × 100 ⎜R ⎟ ⎝ rad1 ⎠

Topic: Maxwell’s Equations (b) 1.  ∇ ⋅ D = ρ (or) ∇ ⋅ D = 0 {for charge free surface} −∂B 2.  ∇ × E = ∂t 3.  ∇ ⋅ B = 0

Rrad 2 − Rrad1

A23

μ 0 I ⎧ 2 ⎫ 5 ⎛ μ0 I ⎞ 5μ 0 I = ⎨1 + ⎬ = 2π r ⎩ 3 ⎭ 3 ⎜⎝ 2π r ⎟⎠ 6π r

53. The dispersion equation of a waveguide, which relates the wavenumber k to the frequency ω, is k (ω ) = (1/c) ω 2 − ω02

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GATE ECE Chapter-wise Solved papers

where the speed of light c = 3 × 108 m/s, and ω0 is a constant. If the group velocity is 2 × 108 m/s, then the phase velocity is (a) 1.5 × 108 m/s (b) 2 × 108 m/s 8 (c) 3 × 10 m/s (d) 4.5 × 108 m/s (2 Marks)

⇒        

1

⇒

Therefore

2

2

=

1 2

⎛w⎞ ⎛w⎞ ⎜ w ⎟ +⎜ h ⎟ = 2 ⎝ ⎠ ⎝ ⎠ 2

⎛w⎞ ⇒                 1 + ⎜ ⎟ = 4 ⎝h⎠ 2

⎛w⎞ ⇒              ⎜ ⎟ =3 ⎝h⎠ ⇒                 

k (ω ) = 1/c ω 2 − ω02 dk (ω ) 1 1 = × ω 2 − ω02 ( 2 ω ) dω c 2 ω dk (ω ) = dω c ω 2 − ω02

2

⎛ 1 ⎞ ⎛1⎞ ω ⎜ ⎟ +⎜ ⎟ ⎝w⎠ ⎝h⎠

⇒     

Topic: P  lane Waves and Properties: Phase and Group Velocity (d)  Phase velocity is given by ω vp = k (ω ) Group velocity is given by dω vg = dk (ω )

2

w = 3 = 1.732 h

55. In the circuit shown, VS is a 10 V square wave of period T = 4 ms with R = 500 Ω and C = 10 μF. The capacitor is initially uncharged at t = 0, and the diode is assumed to be ideal.   The voltage across the capacitor (VC) at 3 ms is equal to volts (rounded off to one decimal place).

v p ⋅ vg = c 2 vp =

R

(3 × 10 ) c = = 4.5 × 108 m/sec vg ( 2 × 108 ) 2

8 2

54. A rectangular waveguide of width w and height h has cut-off frequencies for TE10 and TE11 modes in the ratio 1 : 2. The aspect ratio w/h, rounded off to two decimal places, is . (2 Marks)

+10

+

0 T −10 2 t=0

+ −

T

C

VC −

(2 Marks) Topic: Electrostatics (3.3)  Given that

Topic: Waveguides (1.73)  Given that

T = 4 ms, R = 500 Ω, C = 10 μF Time constant τ = RC = 500 × 10 × 10-6 s f CTE10 1 = f CTE11 2

C 2w

= 10 + [0 − 10]e − t / 5



= 10(1 - e−t/5) At t = 2 ms,

Cut off frequency for TE11( fCTE11) 2

vc = v(∞) + [v(0 + ) − v(∞)]e − t /τ



Cut off frequency for TE10 ( fCTE10) =

C ⎛ 1 ⎞ ⎛1⎞ + = 2 ⎜⎝ w ⎟⎠ ⎜⎝ h ⎟⎠

= 5000 μs = 5 ms

2

Vcap = 10(1 - e−2/5) = 10 × 0.329 = 3.3 V

Therefore c 2w 2

c ⎛ 1 ⎞ ⎛1⎞ + 2 ⎜⎝ w ⎟⎠ ⎜⎝ h ⎟⎠

GATE_ECE_2019_new.indd 24

2

=

1 2

6/22/2019 11:24:01 AM

Solved GATE (ECE) 2020

aPPENDIX

QUESTIONS Chapter 1: Engineering Mathematics 1.

If v1, v2, … , v6 are six vectors in  4 , which one of the following statements is FALSE? (a) It is not necessary that these vectors span  4 . (b) These vectors are not linearly independent. (c) Any four of these vectors form a basis for  4 . (d) If {v1, v3, v5, v6} spans  4 , then it forms a basis for  4 .

Topic: Calculus (b)  The given function, f ( x, y, z ) = e1− x cos y + xze −1/(1+ y

= − cos ye1− x cos y + ze −1/(1+ y At point (1, 0, e), 4.

f ( x, y, z ) = e

+ xze

(d)

1 e (1 Mark)

Solved Chapter Wise GATE ECE 2020 Paper.indd 1

(1, 0 , e )

The general solution of

d2 y dy − 6 + 9 y = 0 is 2 dx dx

(a)

y = C1e 3 x + C2 e −3 x (b) y = (C1 + C2 x )e −3 x

(c)

y = (C1 + C2 x )e 3 x (d) y = C1e 3 x

D2 − 6D + 9 = 0 ⇒ ( D − 3) 2 = 0 ⇒ D = 3, 3 Hence, solution of the given differential equation is y = (C1 + C2 x )e 3 x 5.

Consider the following system of linear equations. Which one of the following conditions ensures that a solution exists for the above system? (a) b2 = 2b1 and 6b1 − 3b3 + b4 = 0 (b) b3 = 2b1 and 6b1 − 3b3 + b4 = 0 (c) b2 = 2b1 and 3b1 − 6b3 + b4 = 0 (d) b3 = 2b1 and 3b1 − 6b3 + b4 = 0

−1/(1+ y 2 )

with respect to x at the point (1, 0, e) is (a) −1 (b) 0 (c) 1

= −1 ⋅ e 0 + e ⋅ e −1 = 0

x1 + 2 x2 = b1 ; 2 x1 + 4 x2 = b2 ; 3 x1 + 7 x2 = b3 ; 3 x1 + 9 x2 = b4

The partial derivative of the function 1− x cos y

)

Topic: Differential Equations (c)  The given differential equation is d2 y dy − 6 + 9y = 0 2 dx dx The characteristic equation of differential equation is

Topic: Vector Analysis

3.

∂f ∂x

2

(1 Mark)

(1 Mark)    (c)  For a vector field A, A is irrotational if ∇ × A = 0.  So, ∇ 2 A = 0.

)

On partial differentiation 2 ∂f ∂ ∂ = [e1− x cos y ] + [ xze −1/(1+ y ) ] ∂x ∂x ∂x 2 ∂ = e1− x cos y ⋅ (1 − x cos y ) + ze −1/(1+ y ) ∂x

(1 Mark) Topic: Linear Algebra (c) Given, v1, v2, … , v6 are six vectors in  4 . For a four-dimensional vector space, 1. Any four linearly independent vectors form a basis (or). 2. Any set of four vector in  4 spans  4 , then it form a basis. Hence, options (a), (b) and (d) are true and option (c) is false.  2. For a vector field A, which one of the following is FALSE?   (a) A is solenoidal if ∇ ⋅ A = 0.  (b) ∇ × A is another vector field.   (c) A is irrotational if ∇ 2 A = 0.    (d) ∇ × (∇ × A) = ∇(∇ ⋅ A) − ∇ 2 A

2

(2 Marks)

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GATE ECE Chapter-wise Solved papers

Topic: Linear Algebra (a)  The system of linear equations is

dy =0 dx

At y = 1, we have



x1 + 2 x2 = b1 (1)



2 x1 + 4 x2 = b2 (2)



3 x1 + 7 x2 = b3 (3)



3 x1 + 9 x2 = b4 (4)

⇒ y = C (constant) ⇒ y =1 x2 + C and y = 1 are the s­ olutions of 2 given differential equation.

Hence, ln y − 1 =

From Eqs. (1) and (2), we get b2 = 2[ x1 + 2 x2 ] = 2b1

7.

For the solid S shown below, the value of

S

From Eqs. (1), (3) and (4), we get

(rounded off to two decimal places) is

⎡1 2 b1 ⎤ ⎢3 7 b ⎥ 3⎥ ⎢ ⎢⎣3 9 b4 ⎥⎦

.

z 1

b1 ⎤ ⎡1 2 ⎢ = ⎢0 1 b3 − 3b1 ⎥⎥ ( R2 → R2 − 3R1 , R3 → R3 − 3R1 ) ⎢⎣0 3 b4 − 3b1 ⎥⎦  

3

(2 Marks)

Therefore answer (A) is correct as it mentions both the conditions

Topic: Calculus (2.25) Here, x : 0 to 3, y : 0 to 1 and z : 0 to 1 − y. Therefore, I =∫

Which one of the following options contains two soludy tions of the differential equation = ( y − 1) x ? dx

=∫

(a) ln y − 1 = 0.5 x 2 + C and y = 1 (b) ln y − 1 = 2 x + C and y = 1 2

(d) ln y − 1 = 2 x 2 + C and y = −1 (2 Marks)

⇒

dy = ( y − 1) x dx

3 x =0

x dxdydz

3

⎡ x2 ⎤ [ z ]10− y dy y =0 ⎢ 2 ⎥ ⎣ ⎦0 1

=

9 1 (1 − y )dy 2 ∫0

=

9⎡ 9 1 y2 ⎤ y − ⎢ ⎥ = × = 2.25 2⎣ 2 ⎦0 2 2

8.

X is a random variable with uniform probability density function in the interval [−2, 10]. For Y = 2X − 6, the conditional probability P (Y ≤ 7 | X ≥ 5) (rounded off to three decimal places) is .

Topic: Probability and Statistics (0.3)  X follows uniform probability density distribution over [−2, 10]. Therefore

Integrating both sides, we get 1

∫ y − 1 dy = ∫ xdx

Solved Chapter Wise GATE ECE 2020 Paper.indd 2

1− y

z =0

∫ ∫

(2 Marks)

dy = xdx y −1

⇒ ln y − 1 =

1 y =0

1

(c) ln y − 1 = 0.5 x 2 + C and y = −1

(a)  Given that,

y

x

For solution to exist, b4 − 3b3 + 6b1 = 0

Topic: D  ifferential Equations

1

0

b1 ⎡1 2 ⎤ ⎢ ⎥ = ⎢0 1 b3 − 3b1 ⎥   ( R3 → R3 − 3R2 ) ⎢⎣0 0 b4 − 3b3 + 6b1 ⎥⎦

6.

∫∫∫ xdxdydz

x2 +C 2

≤ f x ( x) =

1 1 1 = = b − a 10 − ( −2) 12

When −2 ≤ x ≤ 10 and 0 otherwise.

8/10/2020 10:06:24 AM

B3

appendix  •  Solved GATE (ECE) 2020

Given: Y = 2X − 6 ⇒ X = For Y = 7, we have 

10. In the circuit shown below, the Thevenin voltage VTh is

Y +6 2

2V −+

7+6 X = = 6.5 2

1Ω

1A

⎡Y < 7 ⎤ ⎡ X < 6.5 ⎤ P⎢ ⎥ = P⎢ X >5 ⎥ ⎣ X > 5⎦ ⎣ ⎦ P[ X > 5 and X < 6.5] = P[ X > 5] P[5 < X < 6.5] = = P[ X > 5]

So,

∫ ∫

6.5

5 10 5

2Ω

4Ω

2Ω

2A

+ VTh −

(a) 2.4 V (c) 3.6 V

(b) 2.8 V (d) 4.5 V (1 Mark)

f ( x )dx

Topic: Network Theorems: Superposition, Thevenin and Norton’s, Maximum Power Transfer (c)  The given circuit is

f ( x )dx

1 ∫5 12 dx [ x]65.5 = = 10 10 1 [ x ]5 ∫5 12 dx (6.5 − 5) 1.5 = = = 0.3 5 (10 − 5) 6.5

2V −+

2Ω

1Ω

1A

4Ω

2Ω

2A

+ VTh −

Chapter 2: Networks, Signals and Systems 9.

By applying source transformation, we get 2V −+

The output y[n] of a discrete-time system for an input x[n] is

1Ω

y[n] = max x[k ] .

1V +−

−∞≤ k ≤ n

The unit impulse response of the system is (a) 0 for all n. (b) 1 for all n. (c) unit step signal u[n]. (d) unit impulse signal δ [n]. Topic: D  iscrete-Time Signals: Discrete-Time Fourier Transform (dtft), Dft, Fft, Z-transform, Interpolation of Discrete-time Signals max x[k ] and x[k ] = δ [k ] (c) Given, y[n] = −∞≤ k ≤n y[n] = max δ [k ] −∞≤ k ≤ n

For n < 0,

y[n] = max δ [k ] = 0 −∞≤ k ≤ n

For n = 0,

y[n] = δ [0] = 1

For n > 0,

y[n] = max δ [k ] = 1 −∞≤ k ≤ n

⎡0, −∞ < n < 0 ⎤ Therefore, y[n] = ⎢ ⎥ ⎣1, 0 ≤ n ≤ ∞ ⎦

4Ω

+

2Ω VTh

I + 4V −

−

Applying KVL in the loop, we get

(1 Mark)

So,

2Ω

Now, VTH

1 − 1× I + 2 − 2 × I − 2 × I − 4 = 0 −1 ⇒ 5 I = −1 ⇒ I = A 5 ⎛ 1⎞ = 2 I + 4 = 2 ⎜ − ⎟ + 4 = 3.6 V ⎝ 5⎠

11. Which one of the following pole-zero plots corresponds to the transfer function of an LTI system characterized by the input-output difference equation given below? 3

y[n] = ∑ ( −1) k x[n − k ] k =0

(a)

Im 1 3rd order pole −1

1 Re

⇒ y[n] = u[n]

Solved Chapter Wise GATE ECE 2020 Paper.indd 3

−1

8/10/2020 10:06:26 AM

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GATE ECE Chapter-wise Solved papers

(b)

Y [ z ] = X [ z ] − z −1 X [ z ] + z −2 X [ z ] − z −3 X [ z ] Y [ z] ⇒ = 1 − z −1 + z −2 − z −3 X [ z]

Im 1

3rd order pole

⇒ H [ z] = −1

1

z 3 − z 2 + z − 1 ( z − 1)( z 2 + 1) = z3 z3

Hence, pole-zero plot is given by

Re

Im 1 −1 (c)

3rd order pole Im

−1

1 Re

1 3rd order pole −1

−1

1 Re

12. In the given circuit, the two-port network has the imped⎡ 40 60 ⎤ ance matrix [ Z ] = ⎢ ⎥ . The value of ZL for which ⎣ 60 120 ⎦ maximum power is transferred to the load is Ω.

−1

(d)

Im 10 Ω

I1

I2

1 4th order pole −1

+ 120 V −

1

+ V1 −

+ [Z]

V2 −

ZL

Re (1 Mark) Topic: Linear 2-port Network Parameters: Driving Point and Transfer Functions; State Equations for Networks

−1 (1 Mark) Topic: L  TI Systems: Definition and Properties, Causality, Stability, Impulse Response, Convolution, Poles and Zeros, Parallel and Cascade Structure, Frequency Response, Group Delay, Phase Delay, Digital Filter Design Techniques 3

y[n] = ∑ ( −1) k x[n − k ] (a) Given, k =0

= x[n] − x[n − 1] + x[n − 2] − x[n − 3] Taking z-transform of the equation, we get

Solved Chapter Wise GATE ECE 2020 Paper.indd 4

⎡ 40 60 ⎤ ⎡ Z11 (48)  Given that, [ Z ] = ⎢ ⎥=⎢ ⎣ 60 120 ⎦ ⎣ Z 21 We know that,

⎡ V1 ⎤ ⎡ Z11 ⎢V ⎥ = ⎢ Z ⎣ 2 ⎦ ⎣ 21

Z12 ⎤ Z 22 ⎥⎦

Z12 ⎤ ⎡ I1 ⎤ Z 22 ⎥⎦ ⎢⎣ I 2 ⎥⎦

So,

V1 = 40 I1 + 60 I 2 (1)

And

V2 = 60 I1 + 120 I 2 (2)

For maximum power, transfer, ZL = ZTh. The given circuit can be redrawn as shown below to calculate RTh.

8/10/2020 10:06:28 AM

B5

appendix  •  Solved GATE (ECE) 2020

10 Ω

I1

So, impedance of the circuit is

I2

+ V1

+ V2

[Z]

−

v (t ) = 20∠45° i (t ) = 20 cos 45° + j 20 sin 45° 20 20 = +j 2 2

Z= ZTh

−

From the above circuit, we have V1 = −10I1 and Z Th

⇒ R + jX L =

V = 2 I2

2

+j

R = XL = So,      

From Eq. (1), we get V1 = 40 I1 + 60 I 2

⇒ ωL =

⇒ −10 I1 = 40 I1 + 60 I 2 ⇒ −50 I1 = 60 I 2

⇒L=

6 ⇒ I1 = − I 2 5

20 2

20 2

20 2 20 5 2

= 2.828 H

(Since ω = 5 rad/s)

14. The current I in the given network is

Substituting the value of I1 in Eq. (2), we get ⎛ 6⎞ V2 = 60 ⎜ − ⎟ I 2 + 120 I 2 ⎝ 5⎠

120∠

⇒ V2 = −72 I 2 + 120 I 2

−

90° V + −

V2 = 48 I2

Z I Z = (80

⇒ Z Th = 48 Ω 120∠ 13. The current in the RL circuit shown below is i(t ) = 10 cos(5t − π / 4) A. The value of the inductor (rounded off to two decimal places) is H. R

−

−

⇒

20

30° V +

−

j35)Ω

Z

(b) 2.38∠ − 96.37° A.

(a) 0 A.

(c) 2.38∠143.63° A. (d) 2.38∠ − 23.63° A.

i(t)

(2 Marks) +

200 cos(5t) V

L −

Topic: Steady State Sinusoidal Analysis Using Phasors (c)  Given that, I

(1 Mark)

120∠

−

Topic: S  teady State Sinusoidal Analysis Using Phasors (2.83)  Given that,

90° V + −

Z

A  

v(t ) = 200 cos 5t V =

200 2

Solved Chapter Wise GATE ECE 2020 Paper.indd 5

−

j35)Ω

I

∠0° V

10 π⎞ ⎛ ∠ − 45° A. and i(t ) = 10 cos ⎜ 5t − ⎟ A = 4⎠ 2 ⎝

Z = (80

120∠

−

30° V + −

Z I

8/10/2020 10:06:29 AM

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GATE ECE Chapter-wise Solved papers

From the circuit, we have

1 1 1⎤ ⎡ 1⎤ ⎡1 ⎢1 − j −1 j ⎥⎥ ⎢⎢ −1⎥⎥ = [0, 0, 4, 0] y[k ] = ⎢ ⎢1 −1 1 −1⎥ ⎢ 1⎥ ⎢ ⎥⎢ ⎥ j −1 − j ⎦ ⎣ −1⎦ ⎣1

120∠ − 90° 120∠ − 90° = Z 80 − j 35 = 1.347∠ − 66.37°A

I1 =

I2 =

And

120∠ − 30° 120∠ − 30° = Z 80 − j 35

DFT

  Then, x[n] = {a, 0, b, 0, c, 0, d , 0}   { A, B, C , D, A, B, C , D}

= 1.374∠ − 6.37°A



DFT

  x[n] = {a, 0, b, 0, c, 0, d , 0}   { A, B, C , D, A, B, C , D} 

I = −( I1 + I 2 )

Now,

DFT

  y[n] = {a, b, c, d}   { A, B, C , D}

Now, if

= −(1.374∠ − 66.37° + 1.374°∠ − 6.37°)

DFT

  Similarly, y[n] = {1, − 1, 1, − 1}    Y [ K ] = {0, 0, 4, 0}

= 2.38∠143.63° A

DFT

  x[n] = {1, 0, − 1, 0, 1, 0, − 1, 0}    X [ K ] = {0, 0, 4 , 0, 0, 0, 4 , 0} ↑ ↑ 15. A finite duration discrete-time signal x[n] is obtained by x[2] x[6] DFT   x[nπ] =   X [ K ] = {0, 0, 4 , 0, 0, 0, 4 , 0} t ){1, 0, − 1, 0, 1, 0, − 1, 0}  sampling the continuous-time signal x(t ) = cos( 200 ↑ ↑ at sampling instants t = n/400, n = 0, 1, … , 7. The 8-point x[2] x[6]  discrete Fourier transform (DFT) of x[n] is defined as π kn 7 16. For a 2-port network consisting of an ideal lossless trans−j X [k ] = ∑ x[n]e 4 , k = 0, 1, … , 7. former, the parameter S21 (rounded off to two decimal n=0 places) for a reference impedance of 10 Ω is . Which one of the following statements is TRUE? 2 Port Network (a) All X[k] are non-zero. (b) Only X[4] is non-zero. (c) Only X[2] and X[6] are non-zero. (d) Only X[3] and X[5] are non-zero.

(2 Marks) Topic: Discrete-Time Signals: Discrete-Time Fourier Transform (dtft), Dft, Fft, Z-transform, Interpolation of Discrete-time Signals (c)  Given that, 7

X [k ] = ∑ x[n]e

−j

π kn 4

, k = 0, 1, … , 7

n=0

x(t ) = cos( 200π t )

And For t =

2:1 Port 1

Port 2 (2 Marks)

Topic: Linear 2-Port Network Parameters: Driving Point and Transfer Functions (0.8)  For an ideal transformer of n : 1, the scattering matrix is given by

n , we have 100

⎡ S11 ⎢S ⎣ 21

⎛ 200π n ⎞ ⎛πn⎞ x[n] = cos ⎜ ⎟ = cos ⎜ 2 ⎟ , n = 0, 1, .... , 7 ⎝ 400 ⎠ ⎝ ⎠

{

π 3π 5π = cos 0, cos , cos π , cos , cos 2π , cos , 2 2 2 7π cos 3π , cos 2

}

= {1, 0, − 1, 0, 1, , 0 − 1, 0} Let y[n] = {1, − 1, 1, − 1}. So, the four point DFT will be

Solved Chapter Wise GATE ECE 2020 Paper.indd 6

⎡ n2 − 1 2n ⎤ S12 ⎤ ⎢ n2 + 1 n2 + 1 ⎥ ⎥ =⎢ S22 ⎥⎦ ⎢ 2n 1 − n2 ⎥ ⎢ 2 ⎥ ⎣ n + 1 1 + n2 ⎦

Now,   S21 =

2n n +1

       =

2× 2 4 = = 0.8 (Since n = 2) 22 + 1 5

2

17. X(ω) is the Fourier transform of x(t) shown below. The value of places) is

∫

∞

−∞

2

X (ω ) dω (rounded off to two decimal .

8/10/2020 10:06:31 AM

appendix  •  Solved GATE (ECE) 2020

18. The transfer function of a stable discrete-time LTI sysK(z −α ) tem is H ( z ) = , where K and α are real numz + 0.5 bers. The value of α (rounded off to one decimal place) with α > 1, for which the magnitude response of the system is constant over all frequencies, is .

x(t) 3 2 1

(2 Marks)

t 0 1

−3 −2 −1

2 3

4

(2 Marks) Topic: F  ourier Series and Fourier Transform Representations, Sampling Theorem and Applications (58.64)  We know that x(t ) ←⎯→ X (ω )

Topic: Digital Filter Design Techniques (−2)  The transfer function of the given stable discrete time LTI system is H ( z) =

K(z −α ) ( z + 0.5)

We know that for an all pass filter,

FT

∫

∞

−∞

FT [ x(t )]2 dt ←⎯ →

1 2π

∫

∞

−∞

zero = 2

X (ω ) dω

∞

2

X (ω ) dω = 2π × Energy [ x(t )]

−∞

zero =

∞

As, Energy [ x(t )] = ∫ [ x(t )]2 dt −∞

Here, −1 < t < 0;

x(t) = t + 1

0 < t < 1;

x(t) − 1 = 2(t − 0) ⇒ x(t) = 2t + 1

1 < t < 2; 2 < t < 3;

x(t) = −2t + 5 x(t) = −t + 3 0

∫

2

1

−1

0

3

( −2t + 5) dt + ∫ ( −t + 3) 2 dt 2

2

0

1

−1

0

= ∫ (t 2 + 1 + 2t )dt + ∫ ( 4t 2 + 1 + 4t )dt +

∫

2

1

3

( 4t 2 + 25 − 20t )dt + ∫ (t 2 + 9 − 6t )dt 2

0

2

3

⎡ 4t 3 ⎤ ⎡t3 ⎤ +⎢ + 25t − 10t 2 ⎥ + ⎢ + 9t − 3t 2 ⎥ 3 3 ⎣ ⎦1 ⎣ ⎦2 1 ⎡4 ⎤ ⎡4 ⎤ = + ⎢ (1) + 1 + 2 ⎥ + ⎢ (8 −1) + 25 −10( 4 −1) ⎥ 3 ⎣3 ⎦ ⎣3 ⎦ ⎡1 ⎤ + ⎢ ( 27 − 8) + 9 − 3(99 − 4) ⎥ 3 ⎣ ⎦

Therefore,

∫

1 13 13 1 28 + + + = = 9.33 3 3 3 3 3

∞

−∞

2

X (ω ) dω = 2π × 9.33 = 58.64

Solved Chapter Wise GATE ECE 2020 Paper.indd 7

Hence, from the given transfer function, α = −2.

19. A single crystal intrinsic semiconductor is at a temperature of 300 K with effective density of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic Fermi level is shifted from mid-bandgap energy level by (a) 18.02 meV. (b) 9.01 meV. (c) 13.45 meV. (d) 26.90 meV. (1 Mark)

1

⎡t3 ⎤ ⎡ 4t 3 ⎤ = ⎢ + t + t2 ⎥ + ⎢ + t + 2t 2 ⎥ ⎣3 ⎦ −1 ⎣ 3 ⎦0

=

1 = −2 ⎛ 1⎞ − ⎜ 2⎟ ⎝ ⎠

Chapter 3: Electronic Devices

So, E[ x(t )] = ∫ (t + 1) 2 dt + ∫ ( 2t + 1) 2 dt + −1

1 pole *

1 Pole of the system is at z = − . So, zero of the system 2 is

From this property, we get

∫

B7

Topic: Energy Bands in Intrinsic and Extrinsic Silicon (b)  Given that, VT = kT = 26 mV and NV = 2NC. The shift of intrinsic Fermi level from mid-bandgap energy level is given by EFi − Emidgap =

kT ⎛ NV ⎞ ln ⎜ ⎟ 2 ⎝ NC ⎠

=

26 × 10 −3 ⎛ 2 N C ⎞ ln ⎜ ⎟ 2 ⎝ NC ⎠

=

26 × 10 −3 ln 2 = 9.01 meV. 2

8/10/2020 10:06:33 AM

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GATE ECE Chapter-wise Solved papers

20. Consider the recombination process via bulk traps in a forward biased pn homojunction diode. The maximum recombination rate is Umax. If the electron and the hole capture cross-sections are equal, which one of the following is FALSE? (a)  With all other parameters unchanged, Umax decreases if the intrinsic carrier density is reduced. (b)  Umax occurs at the edges of the depletion region in the device. (c)  Umax depends exponentially on the applied bias. (d) With all other parameters unchanged, Umax increases if the thermal velocity of the carriers increases. (1 Mark) Topic: G  eneration and Recombination of Carriers (b)  The maximum recombination rate is given by qV 1 U max = σ VTh N t ni e 2 kT 2

Depletion or transition capacitance is given by

εA w For one-sided abrupt pn junction, width is given by CD =

w=

2ε (Vbi − V ) eN D

2εVB = eN D

where V is anode to cathode applied potential. Therefore, CD =

⇒

εA 2ε (Vbi − V ) eN D

1 2 = (Vbi − V ) 2 2 C D ε A eN D

Now, at Vbi = V , 1/C D2 becomes zero. So, let x1 = Vbi =

So, U max ∝ VTh , U max ∝ ni and U max ∝ e qV / 2 kT . Hence, statements (a), (c) and (d) are true. The Umax occurs in the depletion region, where sum of electrons and holes is minimum, not at the edges of depletion region. Hence statement (b) is false. 21. A one-sided abrupt pn junction diode has a depletion capacitance CD of 50 pF at a reverse bias of 0.2 V. The plot of 1/C D2 versus the applied voltage V for this diode is a straight line as shown in the figure below. The slope of the plot is ________ × 1020 F−2V−1. 2

1/CD

V

Not given in question and y1 = 0. Now, at x2 = −0.2 V, y2 =

1 1 = = 4 × 10 20. 2 C D (50 × 10 −12 ) 2

Therefore, slope of the plot is m=

y2 − y1 x2 − x1

Note: Vbi is not given in question. So, the answer cannot be calculated. 22. The band diagram of a p-type semiconductor with a band-gap of 1 eV is shown. Using this semiconductor, a ′ of 100 nF/cm2 MOS capacitor having VTh of −0.16 V, Cox and a metal work function of 3.87 eV is fabricated. There is no charge within the oxide. If the voltage across the capacitor is VTH, the magnitude of depletion charge per unit area (in C/cm2) is

0 (a) −5.7 (b) −3.8 (c) −1.2 (d) −0.4 (2 Marks)

Vacuum level

4 eV

Topic: P-N Junction, Zener Diode, Bjt, Mos Capacitor, Mosfet, Led, Photo Diode and Solar Cell (*)  Given that,

EC 0.5 eV

2

1/CD y2 = 4 × 1020 y1 = 0 V x1 = Vbi

x2 = −0.2 0 V

Solved Chapter Wise GATE ECE 2020 Paper.indd 8

Ei E Fs 0.2 eV E V (a) 1.70 × 10−8 (c) 1.41 × 10−8

(b) 0.52 × 10−8 (d) 0.93 × 10−8 (2 Marks)

8/10/2020 10:06:34 AM

appendix  •  Solved GATE (ECE) 2020

B9

Topic: P  -N Junction, Zener Diode, BJT, MOS Capacitor, MOSFET, LED, Photo Diode and Solar Cell

(a) approximately 2.0 times that of T1. (b) approximately 0.3 times that of T1. 2 approximately 2.5 times that of T1. ′ =100(c) (a)  Given that, EG =1 eV, VTh = − 0.16 V, ϕm = 3.87 eV, Cox nF/m . 2 (d) approximately 0.7 times that of T1. ′ =100 nF/m . = − 0.16 V, ϕm = 3.87 eV, Cox (2 Marks)

Vacuum level

Topic: P-N Junction, Zener Diode, BJT, MOS Capacitor, MOSFET, LED, Photo Diode and Solar Cell (*)  Ratio of current gain is given by

4 eV

β1 = β2

EC 0.5 eV Ei 0.3 eV E F s 0.2 eV E V

Therefore,   Ei − EFS = 0.5 eV − 0.2 eV = 0.3 eV ′ Qox Q′ − d + 2ϕFP (1) Cox Cox

ϕms = ϕm − ϕs

Now,

  And ϕFP = Ei − EFS

0 d 0

N B2 ( x )dx N B1 ( x )dx

∫

d

0

1017 dx

1 17 (10 − 1014 ) × d 2 2 × 1017 × d ≅2 = (1017 − 1014 ) × d ⇒ β 2 = 0.5β1

⇒ Ei − EV = 0.5 V

Now,   VTh = ϕms −

d

=

EG = 0.5 eV 2

Now,

∫ ∫

= ϕm − ( EV ,ac − EFS ) = 3.87 − 4.8 = −0.93 eV = 0.3 eV

Substituting all the values in Eq. (1), we get Q′ − 0.16 = −0.93 − 0 − d + 2 × 0.3 Cox Q′ ⇒ d = 0.6 + 0.15 − 0.93 = −0.17 Cox

24. A pn junction solar cell of area 1.0 cm2, illuminated uniformly with 100 mW/cm2, has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill factor = 0.8, and thickness = 200 μm. The charge of an electron is 1.6 × 10−19 C. The average optical generation rate (in cm−3s−1) is (a) 0.84 × 1019. (b) 5.57 × 1019. (c) 1.04 × 1019. (d) 83.60 × 1019. (2 Marks) Topic: P-N Junction, Zener Diode, BJT, MOS Capacitor, MOSFET, LED, Photo Diode and Solar Cell (a)  Given that, FF = 0.8, A = 1.0 cm2, S = 100 mW/cm2, η = 15%, VOC = 0.7 V, t = 200 μm and e = 1.6 × 10 −19 C. The efficiency of solar cell is given by

⇒ Qd′ = −0.17 × Cox = −0.17 × 100 × 10 −9 −8

= −1.70 × 10 C/cm

η=

2

23. The base of an npn BJT T1 has a linear doping profile NB(x) as shown below. The base of another npn BJT T2 has a uniform doping NB of 1017/cm3. All other parameters are identical for both the devices. Assuming that the hole density profile is the same as that of doping, the common-emitter current gain of T2 is

0.8 × 0.7 × I SC 100 mW 15 ⇒ I SC = mA 0.56 Now, the average optical generation rate is ⇒ 0.15 =

G=

1017 (/cm3) NB(x)

( FF) × VOC × I SC Pin

I SC q × A× t

15 × 10 −3 0.56 × 1.6 × 10 −19 × 1× 200 × 10 −4 15 × 1019 = = 0.837 × 1019 cm −3s −1 0.56 × 32 =

1014 (/cm3) Emitter 0

Solved Chapter Wise GATE ECE 2020 Paper.indd 9

Base

Collector d

8/10/2020 10:06:35 AM

B10

GATE ECE Chapter-wise Solved papers

Chapter 4: Analog Circuits 25. The components in the circuit shown below are ideal. If the op-amp is in positive feedback and the input voltage Vi is a sine wave of amplitude 1V, the output voltage Vo is

= 2 × 230 × 2 = 650.53 V

+5V + −

1kΩ Vi

Vo

Vmax is the voltage across capacitor C1. Vo = VC2 = 2Vmax

−5V (a) (b) (c) (d)

Vo = VC2 = 2Vmax = 2 × Vrms × 2

1kΩ 1V 0 −1V

Topic: Simple Diode Circuits: Clipping, Clamping and Rectifiers (650.53)  The given circuit is a voltage doubler circuit. Hence, the output voltage is

= 2 × Vrms × 2

a non-inverted sine wave of 2V amplitude. an inverted sine wave of 1V amplitude. a square wave of 5V amplitude. a constant of either +5V or −5V.

= 2 × 230 × 2 = 650.53 V 27. In the circuit shown below, all the components are ideal. If Vi is +2 V, the current Io sourced by the op-amp is mA.

(1 Mark) Topic: Simple Op-Amp Circuits (d)  The given circuit is a Schmitt trigger of non-inverting type. So, Vo = ± 5 V Now, V + =

Vo × 1 + Vi × 1 Vo + Vi = 1+1 2

Let Vo = −5V, then V + =

1 kΩ

1 kΩ Vi

5V + −5

Topic: Simple Op-Amp Circuits (6)  For the given circuit 1 kΩ

Similarly, Vo can change from +5V to −5V if Vi < −5V. Since, given input wave has peak value 1V, therefore output can’t change from +5V to − 5V or −5V to + 5V. Therefore, we get a constant output of +5V or −5V.

−

1 kΩ 2 V

26. In the circuit shown below, all the components are ideal and the input voltage is sinusoidal. The magnitude of the steady-state output Vo (rounded off to two decimal places) is V.

Io 1 kΩ

Output voltage of op-amp is ⎛ 1× 103 ⎞ Vo = ⎜1 + × 2 = 4V 3 ⎟ ⎝ 1× 10 ⎠

D2

+ C2 = 0.1 μ F Vo

− (1 Mark)

Solved Chapter Wise GATE ECE 2020 Paper.indd 10

Vo

+

Vi = 2V

D1

1 kΩ

(1 Mark)

−5 + Vi >0 2 ⇒ Vi > 5V

230 V(rms)

V

−5 + Vi 2

Now, Vo can change from −5V to + 5V if V  + > 0 or

C1 = 0.1μF

Io

−

Now, applying KCL at output node, we get 2−4 0−4 + Io + =0 3 1× 10 1× 103 ⇒ I o = ( 4 + 2) × 10 −3 = 6 mA

8/10/2020 10:06:37 AM

appendix  •  Solved GATE (ECE) 2020

28. For the BJT in the amplifier shown below. VBE = 0.7 V, kT/q = 26 mV. Assume that BJT output resistance (ro) is very high and the base current is negligible. The capacitors are also assumed to be short circuited at signal frequencies. The input vi is direct coupled. The low frequency voltage gain vo/vi of the amplifier is

Now, g m =

B11

I C 0.465 mA = = 0.0178 A/V VT 26 mV

Substituting these values in Eq. (1), we get ⎛ 10 × 10 ⎞ 3 AV = −(0.0178) × ⎜ ⎟ × 10 20 ⎝ ⎠

VCC = 10 V

= −89.42

RC 10 kΩ vo C1 1 μF

29. Using the incremental low frequency small-signal model of the MOS device, the Norton equivalent resistance of the following circuit is

RL 10 kΩ

VDD

vi R

CE 100μF

RE 20 kΩ

gm,rds

VEE = −10 V (a) −89.42 (b) −128.21 (c) −178.85 (d) −256.42 (2 Marks) Topic: Single-Stage BJT and MOSFET Amplifiers: Biasing, Bias Stability, Mid-Frequency Small Signal Analysis and Frequency Response (a)  The low frequency voltage gain AV of given amplifier is v Av = o = − g m ( RC || RL ) (1) vi For DC analysis, all capacitors are open circuited. So,

rds + R (a) rds + R + g m rds R (b) 1 + g m rds (c) rds +

1 rds + R + R (d) gm (2 Marks)

Topic: Small Signal Equivalent Circuits of Diodes, BJTs and MOSFETs (b)  The small signal AC equivalent model of the given circuit is G

VCC = 10 V

D

+ Vπ −

IC RC = 10 kΩ

gmVπ S

IE

9.3 IC = = 0.465 mA 20 × 103

Solved Chapter Wise GATE ECE 2020 Paper.indd 11

+ − Vx

R

Vx = ( I x − g mVx )rds + I x R ⇒ Vx (1 + g m rds ) = ( rds + R) I x

VEE = −10 V 0 – 0.7 – I C ( 20 × 103 ) + 10 = 0 (Assume I C ≅ I E )

Ix

Vπ = −Vx

Here,

RE = 20 kΩ

Applying KVL in base-emitter loop, we get

rds



⇒

Vx R + rds = I x 1 + g m rds

⇒ RNorton =



R + rds 1 + g m rds

8/10/2020 10:06:38 AM

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GATE ECE Chapter-wise Solved papers

30. The components in the circuit given below are ideal. If R = 2 kΩ and C = l μF, the −3 dB cut-off frequency of the circuit in Hz is

Vi = 15 V

Vo = 9 V

R3 = 1 kΩ

R

R1 = 1 kΩ C Vi( jω) R 2C

2R

− +

+ − VBE

Vo ( jω)

R2

Vz = 3.3 V (a) 14.92 (c) 59.68

(b) 34.46 (d) 79.58

Applying KCL in loop 1, we get 3.3 + 0.7 − VR2 = 0

(2 Marks)

⇒ VR2 = 4 V

Topic: Active Filters (d)  Given that, R = 2 kΩ and C = l μF . The given circuit is op-amp active filter of inverting type. Therefore, the 3 dB cut off frequency of the circuit is f 3dB

Now, the resistance R1 and R2 are in series as the base current of the transistor is approximately 0. Therefore, VR2 =

1 = 2π RC 1 = = 79.58 Hz 2π × 2 × 103 × 1× 10 −6

31. In the voltage regulator shown below, Vi ulated input at 15V. Assume VBE = 0.7V current is negligible for both the BJTs. If output Vo is 9V, the value of R2 is

is the unregand the base the regulated Ω.

Vo × R2 R1 + R2

⇒ VR2 =

9 R2 1 + R2

⇒4=

9 R2 1 + R2

⇒ 4 + 4 R2 = 9 R2 4 kΩ 5 4 ⇒ R2 = × 1000 = 800 Ω 5 ⇒ R2 =

Vo = 9 V

Vi = 15 V

R3 = 1 kΩ

1

Chapter 5: Digital Circuits

R1 = 1 kΩ

32. The figure below shows a multiplexer where S1 and S0 are the select lines, I0 to I3 are the input data lines, EN is the enable line, and F(P, Q, R) is the output, F is 0

EN

R

I0

0

I1

(2 Marks)

R

I2

Topic: P  ower Supplies: Ripple Removal and Regulation (800)  From the given circuit, we have VBE = 0.7V and I B  0 A.

1

I3 S1

S0

P

Q

VZ = 3.3 V

Solved Chapter Wise GATE ECE 2020 Paper.indd 12

R2

MUX

F

8/10/2020 10:06:39 AM

appendix  •  Solved GATE (ECE) 2020

(a) PQ + QR. (b) P + QR. Q + PR. (c) PQR + PQ. (d) (1 Mark) Topic: C  ombinatorial Circuits: Arithmetic Circuits, Code Converters, Multiplexers, Decoders and PLAs (a)  The output of the given multiplexer is

B13

35. An enhancement MOSFET of threshold voltage 3 V is being used in the sample and hold circuit given below. Assume that the substrate of the MOS device is connected to −10 V. If the input voltage V1 lies between ±10 V, the minimum and the maximum values of VG required for proper sampling and holding respectively, are Vi

Vo

F = PQR + PQR + PQ = RQ( P + P ) + PQ (Since, P + P = 1)

= PQ + QR

33. In an 8085 microprocessor, the number of address lines required to access a 16 K byte memory bank is . (1 Mark) Topic: 8 -bit Microprocessor (8085): Architecture, Programming, Memory and I/O Interfacing (14)  Let the number of address lines required to access 16 kB memory be n. So, 2n = 16 kB

VG

(a) 3 V and −3 V. (c) 13 V and −7 V.

Topic: Data Converters: Sample and Hold Circuits, ADCs and DACs (c)  For proper sampling and holding, we have VDS ≥ VGS − VT

⇒ 2n = 16 × 210 bits

⇒ V1 − Vo ≥ (VG − Vo ) − VT

⇒ 2n = 214

⇒ V1 ≥ VG − VT

⇒ n = 14

⇒ VG ≤ V1 + VT

34. A 10-bit D/A converter is calibrated over the full range from 0 to 10 V. If the input to the D/A c­ onverter is 13A (in hex), the output (rounded off to three decimal places) is V. (1 Mark) Topic: D  ata Converters: Sample and Hold Circuits, ADCs and DACs (3.069)  The resolution of n-bit D/A converter is Resolution =

VFS ( 2n − 1)

where VFS is full scale voltage. 10 10 For 10-bit D/A, Resolution = 10 = ( 2 − 1) 1023 Input voltage is given by (13A )16 = (1× 16 2 + 3 × 16 + 10) = (314)10 Hence, output voltage of D/A is ⎛ 10 ⎞ Vo = ⎜ ⎟ × 314 = 3.069 V ⎝ 1023 ⎠

Solved Chapter Wise GATE ECE 2020 Paper.indd 13

(b) 10 V and −10 V. (d) 10 V and −13 V. (2 Marks)

At V1 = 10 V,

VGmin = 10 + 3 = 13 V

At V1 = −10 V,

VGmax = −10 + 3 = −7 V

36. P, Q and R are the decimal integers corresponding to the 4-bit binary number 1100 considered in signed magnitude, 1’s complement, and 2’s complement ­representations, respectively. The 6-bit 2’s complement ­representation of (P + Q + R) is (a) 110101 (b) 110010 (c) 111101 (d) 111001 (2 Marks) Topic: Number Systems (a)  Given that, 4-bit binary number 1100 is represented in sign magnitude, 1’s complement and 2’s complement. So, P = sign magnitude of 1100 = −4 Q = 1’s complement of 1100 = −3 R = 2’s complement of 1100 = −4 Hence, P + Q + R = −4 − 3 − 4 = −11 Therefore, 6 bit 2’s complement of (P + Q + R) = 1st complement of 001011 + 1 = 110101

8/10/2020 10:06:40 AM

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GATE ECE Chapter-wise Solved papers

37. The state diagram of a sequence detector is shown below. State S0 is the initial state of the sequence detector. If the output is 1, then

Flip Flop 1

0/0

S0

S1

tpd = 2 ns

tpd = 3 ns tsetup = 5 ns thold = 1 ns

0/0 1/0

tpd = 2 ns

Clk IN

0/0

S3

S2 1/0

0/0 Flip Flop 2

1/0 S4

1/0

tpd = 8 ns tsetup = 4 ns thold = 3 ns

0/1

1/0 (a) (b) (c) (d)

the sequence 01010 is detected. the sequence 01011 is detected. the sequence 01110 is detected. the sequence 01001 is detected.

(2 Marks)

(2 Marks) Topic: Sequential Circuits: Latches and Flip-Flops, Counters, Shift-Registers and Finite State Machines (a)  The dotted lines in the state diagram shows the desired sequence to get output 1. So, the transition sequence is

Topic: Sequential Circuits: Latches and Flip-Flops, Counters, Shift-Registers and Finite State Machines (76.92)  From the circuit, for flip-flop 1 tpd = 3 ns, tsetup = 5 ns and thold = 1 ns, for flip-flop 2 tpd = 8 ns, tsetup = 4 ns and thold = 3 ns, for NOR gate tpd = 2 ns and for NAND gate tpd = 2 ns. Total propagation delay is given by = (t pd + tsetup ) max = 8 + 5 = 13 ns Hence the frequency of operations is

/0 /0 /0 /0 S0 ⎯0⎯ → S1 ⎯1⎯ → S2 ⎯0⎯ → S3 ⎯1⎯ → /1 S4 ⎯0⎯ → S3

f clock = S0 Initial state

S3

S2

S1 0/0

1/0

1 1000 = MHz 13 ns 13

= 76.92 MHz

0/0 1/0 0/1 S4 Final target

Now, the corresponding input sequence is 0, 1, 0, 1, 0. The corresponding output sequence is 0, 0, 0, 0, 1. Hence, if the output is 1, the sequence 01010 is detected. 38. For the components in the sequential circuit shown below, tpd is the propagation delay, tsetup is the setup time, and thold is the hold time. The maximum clock frequency (rounded off to the nearest integer), at which the given circuit can operate ­reliably, is MHz.

Solved Chapter Wise GATE ECE 2020 Paper.indd 14

Chapter 6: Control Systems 39. The pole-zero map of a rational function G(s) is shown below. When the closed contour Γ is mapped into the G(s)-plane, then the mapping encircles Im s-plane

Re

8/10/2020 10:06:41 AM

appendix  •  Solved GATE (ECE) 2020

(a)  the origin of the G(s)-plane once in the ­counter-clockwise direction. (b) the origin of the G(s)-plane once in the ­clockwise direction. (c) the point −1 + j0 of the G(s)-plane once in the counter-clockwise direction. (d) the point −1 + j0 of the G(s)-plane once in the clockwise direction.

41. For the given circuit, which one of the following is the correct state equation? 0.5 H

40. The loop transfer function of a negative feedback system is G ( s) H ( s) =

K ( s + 11) s( s + 2)( s + 8)

i1

2Ω

(1 Mark)

(a)

d dt

4 ⎤ ⎡ v ⎤ ⎡ 0 4 ⎤ ⎡ i1 ⎤ ⎡ v ⎤ ⎡ −4 ⎢i ⎥ = ⎢ −2 −4 ⎥ ⎢ i ⎥ + ⎢ 4 0 ⎥ ⎢i ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ 2⎦

(b)

d dt

⎡ v ⎤ ⎡ −4 −4 ⎤ ⎡ v ⎤ ⎡ 4 4 ⎤ ⎡ i1 ⎤ ⎢i ⎥ = ⎢ −2 4 ⎥ ⎢ i ⎥ + ⎢ 4 0 ⎥ ⎢i ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ 2⎦

(c)

d ⎡ v ⎤ ⎡ 4 −4 ⎤ ⎡ v ⎤ ⎡ 0 4 ⎤ ⎡ i1 ⎤ = + ⎢ ⎥ dt ⎢⎣i ⎥⎦ ⎢⎣ −2 −4 ⎥⎦ ⎢⎣ i ⎥⎦ ⎢⎣ 4 4 ⎥⎦ ⎣i2 ⎦

(d)

d ⎡ v ⎤ ⎡ −4 −4 ⎤ ⎡ v ⎤ ⎡ 4 0 ⎤ ⎡ i1 ⎤ = + ⎢ ⎥ dt ⎢⎣i ⎥⎦ ⎢⎣ −2 −4 ⎥⎦ ⎢⎣ i ⎥⎦ ⎢⎣ 0 4 ⎥⎦ ⎣i2 ⎦

0.5 H

i1

2Ω

i 2Ω 2i1 + −

Solved Chapter Wise GATE ECE 2020 Paper.indd 15

i2

+ v −

0.25 F

+ i − 2

Applying KVL in first loop, we get 2i1 − 2i − 0.5

For system to be marginally stable, we have 10(16 + K ) − 11K =0 10 ⇒ 160 + 10 K − 11K = 0 ⇒ K = 160

1Ω

1Ω

0.5 H a

16 + K 11K 0

0.25 F

Using source transformation of both the sources, we get

Using Routh − Hurwitz criteria, we get s s2 s1 s0

+ v −

i

⇒ s3 + 10 s 2 + 16 s + Ks + 11K = 0

1 10 10(16 + K ) − 11K 10 11K

i2

Topic: State Variable Model and Solution of State Equation of LTI Systems (a)  From the given circuit

Topic: Routh-Hurwitz and Nyquist Stability Criteria (160)  The characteristics equation of system is given by

3

1Ω

(2 Marks)

The value of K, for which the system is marginally stable, is .

1 + G ( s) H ( s) = 0 K ( s + 11) ⇒ 1+ =0 s( s + 2)( s + 8) ⇒ s( s + 2)( s + 8) + K ( s + 11) = 0

+ v − 0.25 F

i

(1 Mark) Topic: Routh-Hurwitz and Nyquist Stability Criteria (b) The s-plane contour is encircling two poles and three zeros in clockwise direction. So, N=Z−P=3−2=1 Hence, it encircles the origin once in clockwise direction.

B15



⇒

di −v = 0 dt

di = −2v − 4i + 4i1 (1) dt

Applying KCL at node a, we get i − 0.25

dv v − i2 =0 − 1 dt

8/10/2020 10:06:43 AM

B16

GATE ECE Chapter-wise Solved papers

1 dv = −4 v + 4i + 4i2 (2) 43. A system with transfer function G ( s) = ,a>0 ( s + 1)( s + a) dt 1 From Eqs.(1) and (2), we get G ( s) = , a > 0 is subjected to an input 5 cos 3t . The steady state ( s + 1)( s + a) 1 4 ⎤ ⎡ v ⎤ ⎡ 0 4 ⎤ ⎡ i1 ⎤ d ⎡ v ⎤ ⎡ −4 cos(3t − 1.892). The value output of the system is +⎢ =⎢ ⎢i ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ dt ⎣i ⎦ ⎣ −2 −4 ⎦ ⎣i ⎦ ⎣ 4 0 ⎦ ⎣ 2 ⎦ 10 of a is . 42. The characteristic equation of a system is (2 Marks) s3 + 3s 2 + ( K + 2) s + 3K = 0. Topic: Transient and Steady-State Analysis of Lti Systems In the root locus plot for the given system, as K var1 ies from 0 to ∞, the break-away or break-in point(s) lie (4) Given that, G ( s) = , a > 0 and ( s + 1)( s + a) within (a) (−1, 0). (b) (−2, −1). 1 c(t ) = cos(3t − 1.892). (c) (−3, −2). (d) ( −∞, − 3). 10

⇒

(2 Marks) Topic: Bode and Root-Locus Plots (a)  The characteristic equation of system is s3 + 3s 2 + ( K + 2) s + 3K = 0 ⇒ s3 + 3s 2 + 2 s + K ( s + 3) = 0 ⇒K =−

( s 3 + 3s 2 + 2 s ) ( s + 3)

dK =0 ds

For break-away points, Now,

dK ( 2 s3 + 12 s 2 + 18s + 6) =− ds ( s + 3) 2

1

G ( jω ) =

ω + 1 ω 2 + a2 Therefore, Amplitude of c(t) = Amplitude of r (t ) G ( jω ) ω =3 1

=5

10

⎧ (3s 2 + 6 s + 2)( s + 3) − ( s3 + 3s 2 + 2 s) ⋅1 ⎫ dK = −⎨ ⎬ ds ( s + 3) 2 ⎩ ⎭ 3 2 2 3 2 ⎧ 3s + 6 s + 2 s + 9 s + 18s + 6 − s − 3s − 2 s ⎫ dK ⇒ = −⎨ ⎬ ds ( s + 3) 2 ⎩ ⎭ ⇒

Now, r (t ) = 5 cos 3t and

1

⇒

10

=5

2

1

ω + 1 ω 2 + a2 2

ω =3

1



10 9 + a 2

⇒ 9 + a 2 = 25 ⇒ a 2 = 25 − 9 = 16 ⇒ a = 4 44. Consider the following closed loop control system + R(s)

C(s)

G(s)

Y(s)

−

Therefore, 2 s3 + 12 s 2 + 18s + 6 = 0 ⇒ s3 + 6 s 2 + 9 s + 3 = 0 ⇒ s = −0.46, − 3.87, − 1.65

where G ( s) =

1 s +1 and C ( s) = K . If the steady s( s + 1) s+3

state error for a unit ramp input is 0.1, then the value of K is .

Im

(2 Marks) Topic: Transient and Steady-State Analysis of LTI Systems (30)  Given that, × × −2

×

Re

G ( s) =

1 ( s + 1) and C ( s) = K s( s + 1) ( s + 3)

Now, open loop transfer function of the system = G(s)C(s) Hence, break-away point lies within (−1, 0).

Solved Chapter Wise GATE ECE 2020 Paper.indd 16

=

K s( s + 3)

8/10/2020 10:06:44 AM

appendix  •  Solved GATE (ECE) 2020

The system is type-1 system. So, for unit ramp input, the steady state error is given by 1 ess = KV where

KV = lim sG ( s)C ( s) = lim s ×

Now,

ess =

s →0

s →0

K K = s( s + 3) 3

∞

45. A digital communication system transmits a block of N bits. The probability of error in decoding a bit is α. The error event of each bit is independent of the error events of the other bits. The received block is declared erroneous if at least one of its bits is decoded wrongly. The probability that the received block is erroneous is (a) N(1 − α) (b) α N (d) 1 − (1 − α) (1 Mark) N

Topic: F  undamentals of Error Correction, Hamming Codes (d)  Given that, probability of error in decoding single bit = α. So, the probability of no error = 1 − α. The probability of no error in received block of N bits = (1 − α)(1 − α) … N times = (1 − α)N Hence, the probability that the received block is erroneous = 1 − (1 − α)N 46. A binary random variable X takes the value +2 or −2. The probability P(X = +2) = α. The value of α (rounded off to one decimal place), for which the entropy of X is maximum, is . (1 Mark) Topic: Information Theory: Entropy, Mutual Information and Channel Capacity Theorem (0.5)  Given that, the probability P ( X = +2) = α . The entropy of X will be maximum, when the probabilities of X = +2 and X = −2 are equal. So, P ( X = +2) = P ( X = −2) = α Therefore,

α +α =1 1 ⇒ α = = 0.5 2

Solved Chapter Wise GATE ECE 2020 Paper.indd 17

∞ ⎧1; 5 ≤ t ≤ 7 Y = ∫ W (t )ϕ (t )dt , where ϕ (t ) = ⎨ −∞ ⎩0; otherwise and W(t) is a real white Gaussian noise process with two-sided power spectral density Sw(f ) = 3 W/Hz, for all f. The variance of Y is .

Topic: Random Processes: Autocorrelation and Power Spectral Density, Properties of White Noise, Filtering of Random Signals Through LTI Systems

Chapter 7: Communications

(c) 1 − α 

47. The random variable

(1 Mark)

3 ⇒ 0.1K = 3 ⇒ K = 30 K

N

B17

(6)  Given that, Y = ∫ W (t )ϕ (t )dt −∞

where So,

⎧1, 5 ≤ t ≤ 7 ϕ (t ) = ⎨ ⎩0, otherwise 7

Y = ∫ W (t )dt (1) 5

Now, variance of Y is 7 7 E [Y 2 ] = E ⎡ ∫ W (t1 )dt1 ∫ W (t 2 )dt 2 ⎤ 5 ⎣⎢ 5 ⎦⎥

=∫

7

E[Y 2 ] = ∫

7

5

5

∫

7

∫

7

5

5

E (W (t1 )W (t 2 ))dt1dt 2 RW (t 2 − t1 )dt1dt 2 (2)

Given, spectral density Sw ( f ) = 3W/Hz. So, RW (τ ) = RW (t 2 − t1 ) = 3δ (t 2 − t1 ) Substituting the value in Eq. (2), we get E [Y 2 ] = ∫

7

5

∫

7

5

3δ (t 2 − t1 )dt1dt 2

7

= 3∫ [u(7 − t1 ) − u(5 − t1 )]dt1 5

= 3× 2 = 6 48. The two sides of a fair coin are labelled as 0 and 1. The coin is tossed two times independently. Let M and N denote the labels corresponding to the outcomes of those tosses. For a random variable X, defined as X = min(M, N), the expected value E(X) (rounded off to two decimal places) is . (1 Mark) Topic: Random Processes: Autocorrelation and Power Spectral Density, Properties of White Noise, Filtering of Random Signals Through Lti Systems (0.25)  When coin is tossed two times, then the possible value of X are X 1 = 0, 0 ⇒ min(0, 0) = 0 X 2 = 0, 1 ⇒ min(0, 1) = 0

8/10/2020 10:06:46 AM

B18

GATE ECE Chapter-wise Solved papers

X 3 = 1, 0 ⇒ min(1, 0) = 0

Therefore, z (t ) = x(t ) × cos( 2π ( f c + 40)t )

X 4 = 1, 1 ⇒ min(1, 1) = 1

= 2[cos(ωm + ωc )t ⋅ cos(ωc t + 2π 40t )] +

Therefore, P (0) =

2[cos(ωc − ωm )t ⋅ cos(ωc t + 2π 40t))]

3 1 and P (1) = 4 4

= cos[(ωm + 2ωc + 2π 40)t ] + cos[ωm t − 2π 40t ] + cos[( 2ωc t − ωm t + 2π 40t )] + cos[ωm t + 2π 40t ]

The expected value E(X) is given by E ( X ) = ∑ X i P( X i )

After passing through low pass filter, higher f­requency components will be attenuated. So,

i

3 1 = 0 × + 1× = 0.25 4 4

y(t ) = cos(ωm t − 2π 40t ) = cos[(1000π − 80π )t ] = cos 920π t

49. For the modulated signal x(t ) = m(t ) cos( 2π f c t ), the message signal m(t ) = 4 cos(1000π t ) and the carrier frequency fc is 1 MHz. The signal x(t) is passed through a demodulator, as shown in the figure below. The output y(t) of the demodulator is Ideal LPF with cut-off 510 Hz x(t)

y(t)

SPM (t ) = cos(1000π t + K p m(t ))

(

t

and SFM (t ) = cos 1000π t + K f ∫ m(τ )dτ −∞

)

where Kp is the phase deviation constant in ­radians/volt and Kf is the frequency deviation constant in radians/ second/volt. If the highest instantaneous frequencies of SPM(t) and SFM(t) are same, then the value of the ratio

1

−510

50. SPM(t) and SFM(t) as defined below, are the phase modulated and the frequency modulated waveforms, respectively, corresponding to the message signal m(t) shown in the figure.

f(Hz) 510

K p /K f is

seconds.

cos(2 π (fc + 40)t)

V

(a) cos(460π t ). (b) cos(920π t ).

m(t) 10

(c) cos(1000π t ). (d) cos(540π t ). (2 Marks) Topic: A  nalog Communications: Amplitude Modulation and Demodulation, Angle Modulation and Demodulation, Spectra of AM and FM, Superheterodyne Receivers, Circuits for Analog Communications (b)  Given that, LPF with cut-off x(t)

z(t)

y(t) 510 Hz

1

3

6 7 t(s) (2 Marks)

Topic: Analog Communications: Amplitude Modulation and Demodulation, Angle Modulation and Demodulation, Spectra of AM and FM, Superheterodyne Receivers, Circuits for Analog Communications (2)  Given that, SPM (t ) = cos(1000π t + K p m(t ))

1

−510

0

510

(

t

SFM (t ) = cos 1000π t + K f ∫ m(t )dt −∞

)

Highest instantaneous frequencies of SPM (t ) is cos(2 π (fc + 40)t) Now,     x(t ) = m(t ) cos( 2π f c t ) = 4 cos(ωm t ) cos(ωc t )   = 2[cos(ωm + ωc )t + cos(ωc − ωm )t ]

Solved Chapter Wise GATE ECE 2020 Paper.indd 18

f PMi =

1 dm(t ) 1000π KP + 2π 2π dt

Highest instantaneous frequencies of SFM (t ) is f FMi =

1 1000π × K f m(t ) + 2π 2π

8/10/2020 10:06:48 AM

P3 = 1 − (shaded area) −3

−2

−1

0

+1 +1.5

appendix  •  Solved GATE (ECE) 2020

( f PMi ) max = ( f FMi ) max

Given that,

dm(t ) 1 1 Kp = K f m(t ) max dt max 2π 2π

Therefore,

B19

d m(t ) max = 5 and m (t ) max = 10 dt

We have,

5 K p = 10 K f ⇒

Therefore,

Kp Kf

=

10 =2 5

P4 = 1 − (shaded area) −3

−2

−1

+1 +1.5 +2

By comparing the above graphs, we can conclude that P3 is larger among the four.

Chapter 8: Electromagnetics

51. In a digital communication system, a symbol S ­randomly chosen from the set {s1, s2, s3, s4} is ­transmitted. It is given that s1 = −3, s2 = −1, s3 = +1 and s4 = +2. The received symbol is Y = S + W . W is a zero-mean unit-variance Gaussian random variable and is independent of S. Pi is the conditional probability of symbol error for the maximum likelihood (ML) decoding when the transmitted symbol S = si. The index i for which the conditional symbol error probability Pi is the highest is . (2 Marks) Topic: D  igital Communications: PCM, DPCM, Digital Modulation Schemes, Amplitude, Phase and Frequency Shift Keying (ASK, PSK, FSK), QAM, MAP and ML Decoding, Matched Filter Receiver, Calculation of Bandwidth, SNR and BER for Digital Modulation (3)  Since the noise variable is Gaussian with zero mean and ML decoding is used, the decision boundary between two adjacent signal points will be their arithmetic mean. In the following graph, the shaded area indicates the conditional probability of decoding a symbol correctly when it is transmitted.

52. The impedances Z = jX, for all X in the range ( −∞, ∞), map to the Smith chart as (a) a circle of radius 1 with centre at (0, 0). (b) a point at the centre of the chart. (c) a line passing through the centre of the chart. (d) a circle of radius 0.5 with centre at (0.5, 0). (1 Mark) Topic: Transmission Lines: Equations, Characteristic Impedance, Impedance Matching, Impedance Transformation, Sparameters, Smith Chart (a)  Smith chart is constructed within a circle of unit radius Γ ≤ 1. 2

2

1⎞ ⎛1⎞ ⎛ (Γ − 1) 2 + ⎜ Γ − ⎟ = ⎜ ⎟ x⎠ ⎝x⎠ ⎝ For Z = jx, R = 0, that means, a circle of radius 1 with centre at (0, 0). y

(0, 0)

(1, 0) x

P1 = 1 − (shaded area) −3 −2 −1 +1 + 2

P1 = 1 − (shaded area)

−3 −2 −1 +1 + 2

P2 = 1 − (shaded area) −3

−2

−1

0

+1

+2

−3

−2

−1

0

+1

+2

P2 = 1 − (shaded area)

53. A transmission line of length 3λ/4 and having a characteristic impedance of 50 Ω is terminated with a load of 400 Ω. The impedance (rounded off to two decimal places) seen at the input end of the transmission line is Ω. (1 Mark) Topic: Transmission Lines: Equations, Characteristic Impedance, Impedance Matching, Impedance Transformation, Sparameters, Smith Chart (6.25)  Given that, l =

−3

−2

−1

0

+1 +1.5

−3

−2

−1

0

+1 +1.5

Solved Chapter Wise GATE ECE 2020 Paper.indd 19

P3 =3λ1 − (shaded λ 3π 2π 3area) l= , βl = × = , Z 0 = 50 Ω and Z L = 400 Ω. λ area) 4 2 P3 = 41 − (shaded

3λ 2π 3λ 3π , βl = × = , Z 0 = 50 Ω and Z L = 400 4 λ 4 2

8/10/2020 10:06:49 AM

B20

GATE ECE Chapter-wise Solved papers

The impedance seen at the input end of transmission line is given by ⎡ Z + jZ 0 tan β l ⎤ Z in = Z 0 ⎢ L ⎥ ⎣ Z 0 + jZ L tan β l ⎦ where l is the length of transmission line, Z0 is the characteristic impedance and ZL is the impedance of load. Therefore,

Z in =

Z 02 (50) 2 = = 6.25 Ω ZL 400

3π ⎞ ⎛ ⎜ Since tan 2 = ∞ ⎟ ⎝ ⎠

54. For an infinitesimally small dipole in free space, the electric field Eθ in the far field is proportional to (e − jkr /r ) sin θ , where k = 2π /λ . A vertical infinitesimally small electric dipole (δ l  λ ) is placed at a distance h(h > 0) above an infinite ideal conducting plane, as shown in the figure. The minimum value of h for which one of the maxima in the far field radiation pattern occurs at θ = 60°, is z

⎛ ψ⎞ ⎛ ψ⎞ sin ⎜ N ⎟ sin ⎜ 2 ⎟ ⎝ 2⎠= ⎝ 2⎠ ⇒ A.F. = ⎛ψ ⎞ ⎛ψ ⎞ sin ⎜ ⎟ sin ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎛ψ ⎞ ⎛ψ ⎞ 2 sin ⎜ ⎟ cos ⎜ ⎟ 2⎠ ⎝ ⎝ 2 ⎠ = 2 cos ⎛ ψ ⎞ = ⎜2⎟ ψ ⎛ ⎞ ⎝ ⎠ sin ⎜ ⎟ ⎝2⎠ Therefore, A.F.N

⎛ψ ⎞ 2 cos ⎜ ⎟ ⎝ 2 ⎠ = cos ⎛ ψ ⎞ ⎜2⎟ 2 ⎝ ⎠

⎛ 2π ⎞ ψ = β d cos θ = ⎜ ⎟ ( 2h) cos θ ⎝ λ ⎠

where

⎛ 2π ⎞ ⎛πh⎞ = cos ⎜ h cos 60° ⎟ = cos ⎜ ⎟ ⎝ λ ⎠ ⎝ λ ⎠ We know that cos θ is maximum, when θ = nπ ; for n = 0, 1, 2,… So, A.F.N

θ = 60°

πh = nπ ⇒ h = nλ λ For n = 1,  hmin = λ

55. The magnetic field of a uniform plane wave in vacuum is given by  H ( x, y, z , t ) = ( aˆ x + 2aˆ y + baˆ z ) cos(ωt + 3 x − y − z ).

θ y

iˆ

0 Infinite conducting plane (a) λ (c) 0.25λ

The value of b is

Topic: Electrostatics (a)  As plane is conducting, then from image theory, the image of small electric diode will be formed at the same distance under the plane. Iδ l

h

Topic: Maxwell’s Equations: Differential and Integral Forms and Their Interpretation, Boundary Conditions, Wave Equation, Poynting Vector (1)  Given that,  H = ( aˆ x + 2aˆ y + baˆ z ) cos(ωt + 3 x − y − z ) iˆ

For uniform plane wave, aˆ H ⋅ aˆ P = 0 (1) where aˆ H is unit vector in magnetic field direction and aˆ P is unit vector in power flow direction. aˆ H =

Now,

aˆ P =

h Iδ l

aˆ x + 2aˆ y + baˆ z 12 + 22 + b 2 −3aˆ x + aˆ y + aˆ z

= =

aˆ x + 2aˆ y + baˆ z 5 + b2 −3aˆ x + aˆ y + aˆ z

11 3 +1 +1 Substituting the value in Eq. (1), we get iˆ

Now, |Total E| = |(Esingle element)| |A.F.|

. (2 Marks)

(b) 0.5λ (d) 0.75λ (2 Marks)

Solved Chapter Wise GATE ECE 2020 Paper.indd 20

A.F. = = A.F.max

That is,  

Iδ l

h

where A.F. is antenna factor and given by

2

2

2

( aˆ x + 2aˆ y + baˆ z ) ⋅ ( −3aˆ x + aˆ y + aˆ z ) = 0

⇒ −3 + 2 + b = 0 ⇒ b = 1

8/10/2020 10:06:51 AM

The book comprises previous years’ (2000 to 2020) GATE questions in Electronics and Communication Engineering covered chapter-wise with detailed solutions. Each chapter begins with a detailed year-wise analysis of topics on which questions are based and is followed by listing of important formulas and concepts for that chapter. The book is designed to make the students well-versed with the pattern of examination, level of questions asked and the concept distribution of questions and thus bring greater focus to their preparation. It aims to be a must-have resource for an essential step in their preparation, that is, solving and practicing previous years’ papers.

Salient Features l

Chapter-wise coverage of GATE Electronics and Communication Engineering previous years’ questions (2000 to 2018). Chapter-wise 2019 and 2020 GATE papers available as appendix

l

Chapter on Engineering Mathematics included for complete coverage of the technical section of the GATE paper

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Detailed topic-wise analysis of questions for each chapter

l

Important Formulas and Concepts summarized for easy recall in each chapter

l

All questions marked for level of difficulty (i.e. 1 Mark or 2 Marks)

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Detailed solutions for all questions, tagged topic-wise

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Packaged with unique scratch code for free 7-day subscription for online GATE tests

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GATE

ELECTRONICS AND COMMUNICATION ENGINEERING

CHAPTER-WISE SOLVED PAPERS

(2000-2020)

Inside

The Graduate Aptitude Test in Engineering (GATE) is an All-India level examination for engineering graduates aspiring to pursue Master’s or Ph.D. programmes in India. The Public Sector Undertakings (PSU’s) also use GATE used as a recruiting examination. The examination is highly competitive, and the GATE score plays an important role in fulfilling the academic and professional aspirations of the students. This book is aimed at supporting the efforts of GATE aspirants in achieving a high GATE score.

GATE ELECTRONICS AND COMMUNICATION ENGINEERING CHAPTER-WISE SOLVED PAPERS (2000-2020)

About the Book

Dr Anil Kumar Maini Varsha Agrawal Nakul Maini

l

Detailed Exam Analysis Chapter-wise and Topic-wise.

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Questions from previous years’ (2000 – 2020) papers.

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Unique scratch code that provides access to w

Free online test with analytics.

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7-Day free subscription for topic-wise GATE tests.

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Instant correction report with remedial action.