WILEY ACING THE GATE ELECTRICAL ENGINEERING WILEY ACING THE GATE ELECTRICAL ENGINEERING Dr Debashis Chatterjee Professor Department of Electrical Engineering Jadavpur University Kolkata Dr J S Lather Professor Department of Electrical Engineering National Institute of Technology Kurukshetra Dr Lalita Gupta Assistant Professor Maulana Azad National Institute of Technology Bhopal WILEY ACING THE GATE ELECTRICAL ENGINEERING Copyright © 2016 by Wiley India Pvt Ltd., 4435-36/7, 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, Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 First Edition: 2016 ISBN: 978-81-265-5087-6 ISBN: 978-81-265-8235-8 (ebk) www.wileyindia.com PREFACE Wiley’s Acing the GATE Examination in Electrical Engineering is intended to be the complete book for those aspiring to compete in the Graduate Aptitude Test in Engineering (GATE) in Electrical Engineering discipline It comprehensively covers all the topics as prescribed in the GATE 2016 syllabus in terms of study material, quick reference material and an extensive question bank, complete with solutions The book offers a number of useful features and the approach is logical concept building rather than only formula based, as offered by the other books generally published in this domain The objective has been to structure this book as a complete reference covering fundamental aspects of theory before proceeding to relevant questions A three tier approach has been adopted to create this manuscript, that is, coverage of the basic building blocks for each of the subjects followed by the exhaustive solved examples, and then practice exercises with one and two marks questions Lastly the set of questions from GATE previous years’ papers starting 2003 to current have been provided along with their solutions and explanations as a supplement to the formal subject coverage The book presumes basic understanding of the fundamentals of Electrical Engineering and related basic electronics The book is divided into 10 chapters based on the units of the syllabus, wherein each chapter constitutes a subject The chapters are divided into various sections which are self-sufficient and easy to read and understand The sequence of the chapters has been arranged in such a way that almost no cross-referencing to subsequent chapter is needed The systematic coverage of the book by the aspirant enhances their knowledge with valuable insights into problem solving approach The authors are of the view that the systematic coverage of this text will not only enhance the GATE cracking skills but also impart a level of knowledge to the students which might have been missed by them during formal study of the subjects in the classroom In addition, the subject centric approach of the books trains the reader to crack other parallel examinations like UPSC Exams etc Though adequate precautions have been taken to ensure correctness of theoretical concepts, equations and related questions, we appreciate communication regarding any inadvertent mistakes that you might come across during the course of your study FM.indd 3/24/2016 12:36:48 PM vi Preface Last but not the least, the authors wish to convey thanks to the entire editorial and production team for being patient to wait for the completion of manuscript by the authors and understanding the technicalities involved in the writing such a comprehensive book and finally for all their efforts to make this dream a reality The continuous inputs from the editorial team of Wiley India and systematic coverage and compilation by the authors, have resulted into this cohesive and complete reference text for GATE examination in Electrical Engineering FM.indd D Chatterjee, J S Lather and Lalita Gupta 3/24/2016 12:36:48 PM ABOUT THE AUTHORS Dr Debashis Chatterjee is Professor at the Department of Electrical Engineering, Jadavpur University, Kolkata He received his B.E in Electrical Engineering from Jadavpur University, Kolkata, M Tech in Machine Drives and Power Electronics from IIT-Kharagpur and Ph.D from Jadavpur University, Kolkata Dr Chatterjee is the author and co-author of more than 60 journal articles and conference papers in reputed publications His areas of interest are Parameter Estimation and Speed Control of Induction Machines, Control of Induction Generators, Development of Improved Harmonic Elimination Techniques of Inverters, Control of Permanent Magnet Machines Dr J S Lather is Professor at the Department of Electrical Engineering, National Institute of Technology, Kurukshetra He received his B.E in Electrical Engineering from SVNIT, Surat, M Tech in Control Systems from NIT, Kurukshetra and Ph.D in the area of Robust Control from NIT, Kurukshetra Dr Lather has more than 21 years of teaching, research and industrial experience to his credit He has published more than 40 research papers in national and international journals and in conferences proceedings His areas of specialisation are Control Systems, Power Systems and Semantic Computing Dr Lalita Gupta is Assistant Professor at the Department of Electronics and Communication Engineering, Maulana Azad National Institute of Technology, Bhopal She received her B.E in Electronics and Telecommunication from Pt Ravi Shankar Shukla University Raipur, M Tech in Digital Communication from Maulana Azad National Institute of Technology, Bhopal and Ph.D from the same institute Since July 2004 she has been associated as a faculty member with the Maulana Azad National Institute of Technology, Bhopal She is a member of IEEE, IETE, ICEIT, IE Dr Gupta has 35 research publications in national and international journals of repute Her area of specialisation is Signal Processing FM.indd 3/24/2016 12:36:48 PM FM.indd 3/24/2016 12:36:48 PM ACING THE GATE ABOUT GATE EXAMINATION The Graduate Admission Test in Engineering (GATE) is an All-India level competitive examination for engineering graduates interested in pursuing Masters or Ph.D programmes in India The examination tests the examinees in General Aptitude, Engineering Mathematics and the discipline (subject) of study in the undergraduate course The objective of GATE is to identify meritorious and motivated candidates for higher studies in Engineering and Sciences The examination serves as a benchmark for normalisation of the undergraduate engineering education in the country The level of competitiveness can be gauged from the fact that close to ten lakh students appear in this competitive examination every year Admission to Higher Learning Courses A valid GATE score is essential to become eligible for admission to the post-graduate courses in Engineering, that is, M.Tech, M.E or direct doctoral programme in the Indian higher education institutes Although qualifying the GATE examination entitles you to apply for the higher degrees; achieving qualifying score is definitely not enough if one is aspiring for admission to top institutes like the IITs, the NITs, the Indian Institute of Science (IISc) and some of the high ranked universities For this, a high GATE score is important Needless to say, a percentile of greater than 95 is perhaps the least one needs to secure admission to a top institute A total of 804463 candidates appeared for GATE 2015 out of which about 125851 students belonged to Electrical branch of engineering It is important to mention here that only 15.05% of those who appeared could qualify according to the qualifying marks set by the GATE examination committee Financial Assistance Selected GATE qualified candidates admitted to M.Tech programmes in Colleges/Universities all over India are eligible for obtaining financial assistance The financial assistance is awarded to Indian nationals doing the M.Tech programmes, subject to institute rules It is also available in the form of Half-Time Teaching Assistantship (HTTA) FM.indd 3/24/2016 12:36:48 PM 598 CHAPTER 6: CONTROL SYSTEM G(jω) (dB) 20 −20 The RH array is -40 dB/decade 0.1 −20 dB/decade ω (rad/s) (log scale) s3 30 s2 13 K s1 (30 × 13) − K 13 s0 K dB/decade This transfer function has (a) three poles and one zero (b) two poles and one zero (c) two poles and two zeros (d) one pole and two zeros (GATE 2008: Marks) Solution: The slope at 40 dB/decade shows presence of poles The slope at 20 dB/decade shows presence of one zero The slope at dB/decade shows presence of one zero Therefore, the transfer function has two poles and two zeros Ans (c) 51 The following figure shows a feedback system where K > + S K s(s+3)(s+10) For a stable system, first column of elements should be positive, that is 00 ⇒ s1 : 390 − Kss1>:: 0390 ⇒− 390 −KK K >390 ⇒K K< < 390 390 0 :K >0 s s :K>0 s :K>0 Therefore, Therefore Therefore,, K< 390 < K < 390 00 < 390 Solution: Given that transfer function is 100 s + 20s + 100 (GATE 2008: Marks) Solution: Applying Routh-Hurwitz criterion to obtain the range of K for a stable system, we get C(s) = R(s) K K = s(s + 3)(s + 10) s(s + 3)(s + 10) + K K 1+ s(s + 3)(s + 10) The characteristic equation is given by K + s[s2 + 13s + 30 ] = s3 + 13s2 + 30s + K = Chapter (578-614).indd 598 Type depends on the value of damping ratio (z) Standard transfer function of second-order system w n2 = s + 2zw n s + w n2 Therefore, w n2 = 100 ⇒ w n = 10 2zw n = 20 z = 20 =1 20 z = shows that the system is critically damped Ans (c) 3/23/2016 4:08:04 PM s 0] 0 = [1 −1 − 0 s + 2 1 −1 1 s s(s + 2) 0 = [1 ] 1 QUESTIONS SOLVED GATE PREVIOUS1YEARS’ 599 0 s + Statement for Linked Answer Questions 53 and 54: The state space equation of a system is described by s(s + 2) 0] s + = [1 x = Ax + Bu y = Cx = where x is state vector, u is input, y is output and 0 0 1 , B = , C = [1 ] A= −2 1 Ans (d) 53 The transfer function G(s) of this system will be s (a) (s + 2) s (c) (s − 2) 54 A unity feedback is provided to the above system G(s) to make it a closed-loop system as shown in the figure below s+1 (b) s(s − 2) s(s + 2) r(t) + Y(t) Σ G(s) (d) s(s + 2) (GATE 2008: Marks) For a unit step input r(t), the steady-state error in the output will be (a) 0 (b) 1 (c) 2 (d) ∞ Solution: Given that: x = Ax + Bu y = Cx 0 A= 0 C = [1 (i) (GATE 2008: Marks) (ii) 0 1 B= 1 − 2 1 ] I= 0 Solution: We have H(s) = R(s) = 1/s 0 1 Steady-state error is given by ess = lim s ⋅ E(s) Taking Laplace transform of Eq (i), we get s→ R(s) = lim s ⋅ + G(s)H(s) s→ s = lim s s→ ⋅ 1 1 + s(s + 2) sX(s) = AX(s) + BU (s) sX(s) − AX(s) = BU (s) X(s) [sI − A ] = BU (s) −1 X(s) = [sI − A ] BU (s) Taking Laplace transform on Eq (ii), we get s(s + 2) =0 = lim s→ s(s + 2) + Y(s) = C X(s) Y(s) = C [sI − A]-1 BU(s) Ans (a) Transform function is given by 55 The measurement system shown in the following figure uses three sub-systems in cascade whose gains are specified as G1, G2 and The relative G3 small errors associated with each respective subsystem G1, G2 and G3 are e1, e2 and e3 The error associated with the output is Y (s) = C(sI − A)−1 B U (s) = [1 s 0] 0 −1 − 0 s + 2 1 = [1 1 s 0] 0 s(s + 2) s + = [1 Chapter (578-614).indd 599 0 1 Input G1 G2 G3 Output s(s + 2) 0] s + = −1 3/23/2016 4:08:21 PM 600 CHAPTER 6: CONTROL SYSTEM (a) e + e + e3 (c) e + e − e e1 ⋅ e e (b) (d) e + e + e -1.42 (GATE 2009: Mark) Solution: Given that: e1 is the error associated with G1 e2 is the error associated with G2 e3 is the error associated with 1/G3 Number of closed-loop poles in right half of s place, Z = P −N Overall gain in series combination= G1G2 G3 Therefore, error as applicable for product of more than two quantities is given by e + e + e Ans (d) 56 The polar plot of an open-loop stable system is shown in the following figure The closed-loop system is Im w=∞ −1.42 Re w=0 (a) always stable (b) marginally stable (c) unstable with one pole on the RH s-plane (d) unstable with two poles on the RH s-plane (GATE 2009: Mark) where P is the number of open loop poles and N is the number of encirclements (-1, j 0) From above sketch N = −2 and P = 0, therefore Z = - (−2) ⇒Z = 2, So the system is unstable with poles on right half of the s-plane Ans (d) 57 The first two rows of Routh’s tabulation of a thirdorder equation are as follows: s3 2 s2 4 This means there are (a) two roots at s = ± j and one root in right s-plane (b) two roots at s = ± j 2 and one root in left s-plane (c) two roots at s = ± j 2 and one root in right s-plane (d) two roots at s = ± j and one root in left s-plane half half half half (GATE 2009: Mark) Solution: From the given Routh’s array, the auxiliary equation is, 2s2 + = Solution: Mapping the mirror image of the given plot, we have 2s2 = −2 s2 = −1 s = ±j The characteristic equation is -1.42 2s3 + 4s2 + 2s + = Dividing both sides by 2, we get s3 + 2s2 + s + = Mapping the circle of R → to an infinite radius R → ∞ gives Chapter (578-614).indd 600 To find the roots 3/23/2016 4:08:37 PM SOLVED GATE PREVIOUS YEARS’ QUESTIONS 601 Therefore, transfer function is s2 (s + 2) + 1(s + 2) = 1000(s + 5) (s + 2) (s2 + 1) = s = −2 s = ± j s (s + 2)(s + 25) Ans (b) Ans (d) 58 The asymptotic approximation of the log-magnitude vs frequency plot of a system containing only real poles and zeros is shown in the following figure Its transfer function is −40dB/decade 80dB −60dB/decade 59 The unit-step response of a unity feedback system with open-loop transfer function G(s) = K (s + 1)(s + 2) is shown in the following figure The value of K is Response w rad/s 0.1 0.75 0.5 0.25 0 25 10(s + 5) (a) s(s + 2)(s + 25) 1000(s + 5) (b) s (s + 2)(s + 25) Time(s) (a) 0.5 (b) 2 (c) 4 (d) 6 (GATE 2009: Marks) 100(s + 5) (c) s(s + 2)(s + 25) 80(s + 5) (d) s (s + 2)(s + 25) Solution: Given that K (s + 1)(s + 2) H(s) = G(s) = (GATE 2009: Mark) Solution: From the given data: (i) Bode plot starts with slope −40 dB/dec which implies the presence of two poles at origin (ii) Slope change from −40 dB/dec to −60 dB/ dec (−20 dB/dec added) implies that there is a pole in the transfer function wc1 = rad/s (iii) Slope change from −60 dB/dec to −40 dB/dec (+20 dB/dec added) implies that there exists a zero in the transfer function wc2 = rad/s (iv) Slope change from −40 dB/dec to −60 dB/ dec (−20 dB/dec added implies that there is a pole in the transfer function wc3 = 25 rad/s Therefore, K(s + 5) T (s) = s (s + 2)(s + 25) To find K, substitute s = jw We get T ( jw ) = T (jw)|at w =0 K(s + 5) s→ R(s) 0.25 = lim s ⋅ s→ + G ( s ) H ( s ) 1/s 0.25 = lim s ⋅ K s→ 1+ (s + 1)(s + 2) (s + 1)(s + 2) 0.25 = lim s→ K + (s + 1)(s + 2) 0.25 = K +2 = 0.25K + 0.5 K= =6 6.25 ( jw ) ( jw + 2)( jw + 25) K(5) 0.1 × × 25 Chapter (578-614).indd 601 ess = lim s ⋅ E(s) ⇒ [∵ R(s) = 1/s] Ans (d) 60 The open-loop transfer function of a unity feedback system is given by = 80 Therefore, 80 = 20 log From the figure, Steady-state error (ess) = − 0.75 = 0.25 We know that K = 1000 G(s) = (e−0.1s ) s 3/23/2016 4:08:56 PM 602 CHAPTER 6: CONTROL SYSTEM The gain margin of this system is (a) 11.95 dB (b) 17.67 dB (c) 21.33 dB (d) 23.9 dB 2s − 2s + (c) (d) s + 5s − 2 s + 5s + (GATE 2009: Marks) (GATE 2009: Marks) Solution: Given that: Solution: Given that: G(s) = (e−0.1s ) s (i) (i) dx1(t) = −3x1(t) + x2 (t) + 2u(t) dt (ii) dx2 (t) = −2x2 (t) + u(t) dt y(t) = x1(t) (iii) Gain margin = 20 log (ii) G( jw pc ) where wpc is the phase crossover frequency ∠G( jw pc ) = −180° (iii) Taking Laplace transform on Eqs (i), (ii) and (iii), we get Substituting j = jw in the transfer function given in Eq (i), we get e−0.1jw jw 180° − 90° ∠G( jw ) = −0.1w × p G( jw ) = sX1(s) = −3X1(s) + X2 (s) + U(s) X1(s) (s + 3) = X2 (s) + U(s) sX2 (s) = −2 × (s) + U (s) X2 (s) (s + 2) = U (s) Substituting w = wpc, we have 180° − 90° ∠G( jw pc ) = −0.1w pc × p (iv) X2 (s) = U (s) s+2 (v) (iv) Substituting value from Eq (iv) in Eq (iii), we get 180° −0.1w pc × − 90° = −180° p Substituting Eq (v) in Eq (iv), we get U (s) + 2U (s) s+2 [1 + 2(s + 2)] X1(s)(s + 3) = U (s) s+2 U (s)(2s + 5) X1(s) = (s + 2)(s + 3) X1(s)(s + 3) = ⇒ w pc = 15.7 rad/s Substituting value in Eq (ii), we get gain margin as 15.7 = 20 log 15.7 = 23.9 dB GM = 20 log From Eq (iii), Y (s) = X1(s) Ans (d) Common Data for Questions 61 and 62: A system is described by the following state and output equations dx1(t) = −3x1(t) + x2 (t) + 2u(t) dt dx2 (t) = −2x2 (t) + u(t) dt y(t) = x1(t) where u(t) is input and y(t) is the output 61 The system transfer function is s+2 (a) s + 5s − Chapter (578-614).indd 602 s+3 (b) s + 5s + Y (s) = U (s)(2s + 5) (s + 2)(s + 3) Transfer function is Y (s) (2s + 5) = U (s) (s + 2)(s + 3) Ans (c) 62 The state-transition matrix of the above system is e−3t 0 (a) −2t −3 t −2 t +e e e e−3t e−2t + e3t (c) −2 t e e−3t e−2t − e−3t (b) e2t e3t e−2t − e−3t (d) −2 t e (GATE 2009: Marks) 3/23/2016 4:09:28 PM SOLVED GATE PREVIOUS YEARS’ QUESTIONS 603 Solution: From Eqs (i) and (ii) given in Question 61, we have x 1(t) = −3x1(t) + x2 (t) + 2u(t) x 2 (t) = −2x2 (t) + u(t) Gain without error = 10% Gain with 10% error = 10 + 0.1= 10.1 Gain with −10% error = 10 − 0.1 = 9.9 Therefore, 10 ± 1% Ans (a) 64 For the system , the approximate time taken (s + 1) for a step response to reach 98% of its final value is Therefore, state matrix is x 1 −3 1 x1 2 = + u(t) x − x 1 2 ⇒ x (t) = Ax + Bu (a) 1 s (b) 2 s (c) 4 s (d) 8 s State transition matrix f(t) = L−1[f (s)] (GATE 2010: Mark) f(s) = (sI − A)−1 Solution: Given that s sI − A = 0 [sI − A ]−1 0 −3 1 s + − − = s − 2 s + 2 s + = s + 3 (s + 3)(s + 2) Therefore, s+2 (s + 3)(s + 2) (s + 3)(s + 2) f(s) = s + 0 (s + 3)(s + 2) s + (s + 2)(s + 3) = 0 s + e−3t f = L−1 f (s) = −2 t e Y (s) = Convert in s domain to time domain, we have, step response, y(t) = 2(1 − et )u(t) = 0.98 × ⇒ t = s ≈ s Ans (c) 65 If the electrical circuit of Fig (b) is an equivalent of the coupled tank system of Fig (a), then h1 −3 t −e −2 t e s+1 h2 (a) Coupled tank Ans (b) 63 As shown in the following figure, a negative feedback system has an amplifier of gain 100 with ± 10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path The overall system gain is approximately + B A (b) 100 ± 10% − D C Electrical equivalent (a) A, B are resistances and C, D capacitances (b) A, C are resistances and B, D capacitances 9/100 (a) 10 ± 1% (c) 10 ± 5% t (b) 10 ± 2% (d) 10 ± 10% (GATE 2010: Mark) Solution: Given that Gain = 100 with ± 10% tolerance and 9/100 attenuator Chapter (578-614).indd 603 (c) A, B are capacitances and C, D resistances (d) A, C are capacitances and B, D resistances (GATE 2010: Mark) Solution: From the given figure, h1 and h2 are compared as voltage across the capacitors 3/23/2016 4:09:39 PM 604 CHAPTER 6: CONTROL SYSTEM −270° B and D are compared as the resistance across the two points Therefore, A and C are capacitances and B and D are resistances Ans (d) 66 The frequency response of G(s) = [s(s + 1)(s + 2)] Im (b) ω=0 Re G( jw ) = ω= (c) Im −90° To find the point of intersection at the real axis, we divide and multiply Eq (1) with complex conjugate of the expression Im Re −3/4 (d) ( jw )(1 + jw )(2 + jw ) (−jw )(1 − jw )(2 − jw ) × (−jw )(1 − jw )(2 − jw ) = −jw (2 − jw − jw − w ) = −2 jw − 3w + jw w (1 + w )(4 + w ) Im ω =0 Re G( j w) = Re ω= −1/6 −1/6 w (1 + w )(4 + w ) −3w w (1 + w )(4 + w ) − jw (2 − w ) w (1 + w )(4 + w ) At real axis, imaginary part |G (jw)| = Therefore, w (2 − w2) = (GATE 2010: Marks) w2 = 2; w = rad/s At w = rad/s Solution: Given that G(s) = 0° w=0 plotted in the complex G(jw) plane (for < w < ∞ is (a) −3/4 w=∞ −180° s(s + 1)(s + 2) G( jw ) = ( + 1) + = < 3/4 Substituting s = jw in the transfer function, we get −3/4 G( jw ) = jw ( jw + 1)( jw + 2) Im (1) Re G( jw ) = and w + w2 + w2 ∠G( jw ) = −90° − tan−1 w − tan−1 w 67 The (i) At w = 0, G( jw ) = ∞ and ∠G( jw ) = −90° −1 2 ,B = A= (ii) At w − ∞, G( jw ) = and ∠G( jw ) = −270° 2 Chapter (578-614).indd 604 Ans (a) system 0 is 1 x = Ax + Bu with 0 −1 2 , B = A= 1 2 3/23/2016 4:10:01 PM SOLVED GATE PREVIOUS YEARS’ QUESTIONS 605 (a) stable and controllable (b) stable but uncontrollable (c) unstable but controllable (d) unstable and uncontrollable By Routh-Hurwitz criterion s(s2 + 4s + 3) + Ks + 2K = s3 + 4s2 + 3s + Ks + 2K = (GATE 2010: Marks) s3 + 4s2 + s(3 + K ) + 2K = To form Routh array, Solution: Given: x = Ax + Bu −1 2 0 B = A= 0 2 1 s3 s s0 2K Thus, there is no sign change in the first column of Routh array which indicates the absence of righthalf poles To sketch the root locus: Equation (i) can be written as, A − lI = −1 2 − l 0 2 l 2 = −1 − l − l 1+ + G(s)H(s) = = −2 + l − l + l = l − l − For unity feedback system H(s) = Therefore, A − lI = 0; ⇒ l = 2, −1 G(s) = Eigen values are of opposite sign, so the system is unstable Ans (c) 68 The characteristic equation of a closed-loop system is s(s + l)(s + 3) + k(s + 2) = 0, k > Which of the following statements is true? (a) Its roots are always real (b) It cannot have a breakaway point in the range -l < Re[s] < (c) Two of its roots tend to infinity along the asymptotes Re[s] = −1 (d) It may have complex roots in the right half plane (GATE 2010: Marks) Solution: Given that s(s + 1)(s + 3) + K(s + 2) = 0; Chapter (578-614).indd 605 K(s + 2) =0 s(s + 1)(s + 3) Standard form = (−1 − l )(2 − l ) − 2K 12 + 2k > A − lI = 12 + 2K It is given that K > 0, therefore, from s1 Therefore, system is controllable Next check for stability l2 − l − = 3+K s1 For × matrix, first check for [B : AB] 2 AB = −1 2 = 2 1 2 [B : AB ] = 2 ≠ 2 K > (i) K(s + 2) s(s + 1)(s + 3) Poles: Zeros N = m=1 s = s = −2 s = −1 s = −3 jw −3 −2 −1 s Number of root locus branches (n) = Number of root locus branches ending at zero at infinity = n − m = Number of root locus branches ending at first zero = Breakaway point occurs between two open-loops poles say between s = and s = −1 Therefore, 3/23/2016 4:10:14 PM 606 CHAPTER 6: CONTROL SYSTEM SReal part of poles − SReal part of zeros n−m −2) − − − (− = = −1 Centroid = Solution: For step input ess = 0.1 We know that 1+ K 1 = 1+ K ⇒K =9 ess = As N = 3, the root loci branches terminate at infinity along asymptotes, so +1 < Re(s) < Ans (c) 69 The frequency response of a linear system G(jw) is provided in the tabular form below Also, K = s+1 s+1 G(s) = |G(jw)| 1.3 1.2 1.0 0.8 0.5 0.3 ÐG(jw) −130° −140° −150° −160° −180° −200° The gain margin and phase margin of the system are (a) 6 dB and 30° (b) 6 dB and −30° (c) −6 dB and 30° (d) −6 dB and −30° (GATE 2011: Mark) Solution: Gain margin = −20 log G( jw ) w =w r (t) = 10 [ u(t) − u(t − 1)] Taking Laplace transform e−s = 10 R(s) = 10 − s s − e−s s Therefore, steady-state error for pulse input is, pc ess (p) = lim s ⋅ E(s) = 20 log From the figure, input G( jw pc ) Phase margin = 180° + fgc where fgc is the gain cross-over frequency or angle at wgc From the given data, the frequency at which the response crosses −180° is 0.5 G( jw ) = 0.5, at ∠G( jw ) = −180° s→ R(s) s→ + G(s)H(s) R(s) = lim s ⋅ s + 100 s→ s+1 = lim s ⋅ = lim s ⋅ s→ = dB 0.5 Phase margin = 180 + fgc 10(1 − e−s )/s s + 10 s+1 ( ∵ R(s) = 10(1 − e−s ) and H(s) = 1) Therefore GM = 20 log = = 180° + (−150°) = 30° 10 (1 − e0 ) =0 10/1 Ans (a) Ans (a) 70 The steady state error of a unity feedback linear system for a unit step input is 0.1 The steadystate error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the following figure is 71 A point z has been plotted in the complex plane, as shown in the following figure Im r(t) z · Unit circle Re 10 t 1s (a) 0 (b) 0.1 (c) 1 (d) 10 (GATE 2011: Mark) Chapter (578-614).indd 606 The plot of the complex number y = is z 3/23/2016 4:10:28 PM SOLVED GATE PREVIOUS YEARS’ QUESTIONS 607 Im (a) y = 1/z y > Unit circle [∵ z < 1] z − is positive real and imaginary part Therefore, `y’ should have positive real and negative imaginary part Ans (d) Re y· (b) 72 An open-loop system represented by the transfer (s − 1) function G(s) = is (s + 2)(s + 3) Im (a) stable and of the minimum phase type Unit circle (b) stable and of the non-minimum phase type (c) unstable and of the minimum phase type Re (d) unstable and of the non-minimum phase type (GATE 2011: Marks) y· Solution: Given that (c) Im G(s) = · Unit circle y Re (d) Im (s − 1) (s + 2)(s + 3) Zeros (m) =1; s = Poles (n) = 2; s = −2; s = −3 (i) All the poles lie on the left half of the s-plane, therefore open-loop system is stable (ii) One zero lies on the right half of the s-plane, therefore non-minimum phase type Ans (b) 73 The open-loop transfer function G(s) of a unity feedback control system is given as Unit circle 2 K s + 3 Re G(s) = s2 (s + 2) ·y (GATE 2011: Mark) Solution: Given • z Unit circle y = 1/z From the root locus, it can be inferred that when K tends to positive infinity, (a) three roots with nearly equal real parts exist on the left half of the s-plane (b) one real root is found on the right half of s-plane (c) the root loci cross the jw axis for a finite value of K; K ≠ (d) three real roots are found on the right half of the s-plane (GATE 2011: Marks) Solution: K(s + 2/3) Point z lies inside the unit circle hence the value of z < Chapter (578-614).indd 607 G(s) = s2 (s + 2) 3/23/2016 4:10:35 PM 608 CHAPTER 6: CONTROL SYSTEM Sum of poles − Sum of zeros No of poles − No of zeros −2 − (−2/3) = = −2/3 −1 Overall transfer function Centroid(s A ) = Y (s) s(s + 1) + Ks = = R(s) + s(s + 1) + Ks + s(s + 1) + Ks Y (s) 1 = = R(s) s + s + Ks + s + s(K + 1) + (2k ± 1) × 180° ; n−m k = 0, 1 n − m Angle of asymptote q A = 180° = 90° × 180° k = 1; q = = 270° For k = 0; q1 = Standard second-order characteristic equation is given by s2 + 2zw n s + w n2 = Therefore, w n2 = 2zw n = + K wn = −2 z = 1+ K Therefore, peak overshoot = e−z p/1 −z −2/3 Ans (a) 75 A system with transfer function So, three roots with nearly equal real parts exist on the left half of the s-plane Ans (a) G(s) = is excited by sin (wt) The steady-state output of the system is zero at (a) w = rad/s (b) w = rad/s 74 A two-loop position control system is shown in the following figure R(s) + − + − Motor s(s + 1) (c) w = rad/s (d) w = rad/s (GATE 2012: Mark) Y (s) Ks Tacho-generator The gain K of the Tacho-generator influences mainly (a) peak overshoot (b) natural frequency of oscillation (c) phase shift of the closed-loop transfer function at very low frequencies (w → 0) (d) phase shift of the closed-loop transfer function at very high frequencies (w → ∞) (s2 + 9)(s + 2) (s + 1)(s + 3)(s + 4) Solution: The steady state error shall be zero when the overall function is stable and the factor w due to sinusoidal input, that is, gets can2 s + w2 celled by zero at (s2 + 9) It implies that s2 + w = s2 + ⇒ w = ⇒ w = rad/s Ans (c) 76 The state variable description of an LTI system is given by (GATE 2011: Marks) Solution: Solving for inner loop, transfer function is s(s + 1) = + + K(s) s ( s ) 1+ ⋅ K(s) s(s + 1) Chapter (578-614).indd 608 x 1 a1 0 x1 0 x 2 = 0 a1 x2 + 0 u x 3 a3 0 x3 1 x1 y = (1 0)x2 x3 3/23/2016 4:10:52 PM SOLVED GATE PREVIOUS YEARS’ QUESTIONS 609 where y is the output and u is the input, The system is controllable for (a) a1 ≠ 0, a2 = 0, a3 ≠ R(s) + s3 + as2 + 2s + − (b) a1 = 0, a2 ≠ 0, a3 ≠ Y (s) K(s + 1) (c) a1 = 0, a2 ≠ 0, a3 = (a) K = and a = 0.75 (b) K = and a = 0.75 (c) K = and a = 0.5 (d) K = and a = 0.5 (d) a1 ≠ 0, a2 ≠ 0, a3 = (GATE 2012: Marks) Solution: Given that (GATE 2012: Marks) x 1 a1 0 x1 0 x = 0 a2 x + 0 u − (1); x a3 0 x 1 3 Solution: Overall transfer function of the system is, Y (s) G(s) = R(s) + G(s)H(s) K(s + 1) = (∵ H(s) = 1) s + as2 + 2s + K(s + 1) 1+ s + as2 + 2s + K(s + 1) = s + as + 2s + + K(s + 1) K(s + 1) = s + as + s(2 + K ) + (K + 1) x1 y = (1 0) x2 x For a × matrix to check the countability, the condition is U = [B : AB : A2 B ] (i) We know that x = Ax + Bu (ii) By Routh array, the characteristic equation is On comparing Eqs (i) and (ii), s3 + as2 + s (2 + k) + (k + 1) = a1 0 0 a A = 0 B = 0 a3 0 1 0 a1a2 AB = a2 A = a2a3 0 0 a3a1 0 s3 : s : a2 a1a2 This implies that a1 ≠ 0, a2 ≠ and a3 may or may not be zero Ans (d) s1 : a K +1 a(2 + K ) − K + a s0 : K +1 For an oscillatory system, we have s1: which implies a(2 + K ) − K + =0 (i) a (K + 1) a= (K + 2) 0 a1a2 0 a1a2 0 = A2B = a2a3 0 a3a1 1 0 U = 0 1 Auxiliary equation is, as2 + K + = ⇒ s2 = − (K + 1) a Substituting value of a from Eq (i), we have s2 = −(K + 1) ⋅ 77 The feedback system shown in the following figure oscillates at rad/s when Chapter (578-614).indd 609 2+K s= j K +2 (K + 2) = −(K + 2) (K + 1) (ii) 3/23/2016 4:11:05 PM 610 CHAPTER 6: CONTROL SYSTEM df = dw Substituting s = jw in Eq (ii), we get jw = j K + w = K +2 − w a 1/b 1/b + (w /b)2 =0 1/a = + (w /a)2 + (w /b)2 2 w 2 = + w + a b b a 1 w 1 w + = + a a b2 b b a2 = K +2 ⇒ K = Therefore, a = 1/a K +1 +1 = = = 0.75 K +2 2+2 Ans (a) Statement for Linked Answer Questions 78 and 79: The transfer function of a compensator is given as s+a Gc (s) = s+b 1 w 1 − = − a b ab a b w = ab ⇒ w = ab = × = rad/s Ans (a) 78 Gc(s) is a lead compensator if (a) a = 1, b = (c) a = −3, b = −1 80 The transfer function V2 (s) of the circuit shown in V1(s) the following figure is (b) a = 3, b = (d) a = 3, = (GATE 2012: Marks) 100 µF Solution: Substituting s = jw in the given transfer function, we get Gc ( jw ) = + jw + a jw + b V1(s) − For a lead compensator f > f = tan−1 w w − tan−1 a b + 10 kΩ V2(s) 100 µF − (i) 0.5s + (a) s+1 w w − −1 b = tan−1 w (b − a) = tan a 2 w + ab w 1 + ab s+2 (c) s+1 3s + (b) s+2 s+1 (d) s+2 (GATE 2013: Mark) f > implies that Solution: The given circuit can be represented as w(b − a) > b−a > ⇒ b > a C1 100 µF + This condition is satisfied by options (a) and (c) However, in case of lead compensator, zero is nearer to the origin as compared to pole, therefore option (c) is not valid and the correct option is (a) Ans (a) V1(s) I(s) V2(s) − For loop 1: Let C1 = C2 =100 mF = C V1(s) = (GATE 2012: Marks) Chapter (578-614).indd 610 C2 100 µF − 79 The phase of the above lead compensator is maximum at (a) rad/s (b) rad/s (c) rad/s (d) rad/s Solution: From Eq (i) of Question 78, we have + R 10 kΩ 1 I (s) I (s) + R.I (s) + C1s C2s 1 V1(s) = I (s) +R+ Cs Cs (i) 3/23/2016 4:11:29 PM SOLVED GATE PREVIOUS YEARS’ QUESTIONS 611 w = K 1/N For loop 2: I (s) Cs ⇒ V2 (s) = I (s) R + Cs 6.309 = K 1/2 (∵ N = 2) K = 39.8 39.8 G(s) = s V2 (s) = R I (s) + I (s) = V2 (s) (ii) Cs) (R + 1/C Ans (b) Substituting Eq (ii) in Eq (i), we get 82 The signal flow graph for a system is given below V (s) ⋅ Cs RCs + V1(s) = Rcs + Cs V2 (s) 10 × 103 × 100 × 10−6 s + s+1 = = − V1(s) 10 × 10 × 100 × 10 s + s + s−1 U(s) s−1 −2 Y(s) −4 Ans (d) 81 The Bode plot of a transfer function G(s) is shown in the following figure The transfer function s+1 (a) 5s + 6s + Gain (dB) 40 s+1 (c) s + 4s + 32 Y (s) for this system is U (s) s+1 (b) s + 6s + (d) 5s + 6s + (GATE 2013: Marks) 20 Solution: Using Mason's gain formula, −8 10 100 ω (rad/s) Y (s) s−2 + s−1 = U (s) − [−2s−2 − 4s−1 − 2s−1 − 4] + s−2 (1 + s) The gain (20 log|G(s)|) is 32 dB and -8 dB at rad/s and 10 rad/s, respectively The phase is negative for all w, then G(s) is 39.8 39.8 32 32 (a) (b) (c) (d) s s s s = + 2s−2 + 6s−1 + s−2 (s + 1) = s−2 (5s2 + 6s + 2) s+1 = 5s2 + 6s + (GATE 2013: Mark) Ans (a) Solution: Any two points on same line segment of Bode plot satisfies the equation of a straight line The slope of straight line is G2 − G1 = −40 dB/decade log w − log w1 − 32 = −40 w log w1 − 32 = −40 w log 1 Chapter (578-614).indd 611 x 1 −2 0 x1 1 = + u, x −1 x 1 2 1 x and y = [1 ] x x2 (0) = 2 x1(0) = 83 The system is (Set w = w ) −32 = −40 log(w ) ⇒ w = 6.309 We know that Common Data for Questions 83 and 84: The state variable formulation of a system is given as (a) controllable but not observable (b) not controllable but observable (c) both controllable and observable (d) both not controllable and not observable (GATE 2013: Marks) 3/23/2016 4:11:51 PM 612 CHAPTER 6: CONTROL SYSTEM Solution: On inverse Laplace transform, we get −2 0 A= − 1 1 B= 1 C = [1 0] To check for controllability: 0 1 = −2 AB = −2 −1 1 −1 [BAB ] = 1 −2 ≠ 1 −1 Therefore, system is controllable To check for observability: 0 = [−2 CA = [1 ] −2 −1 e−2t L−1[f(s)] = e−t e2x e−Ax = ex e2 x e−Ax ⋅ B = x e 1 = 1 t t ∫ [using x x = −t ] e−Ax ⋅ B(x) = 0 e2x x ⋅ dx e e2 t − = et − 0] C = =0 CA −2 0 Therefore, the system is not observable ∫ e2x x e Therefore, Ans (a) 84 The response y(t) to a unit step input is 1 (a) − e−2t 2 1 (b) 1 − e−2t − e−t 2 (c) e−2t − e−t (d) 1 − e−t x(t) = eAt e2 t − et − e−2t = (GATE 2013: Marks) Solution: Solution for the given state equation in time domain is, e−t e2 t − et − − e−2t = − e−t t x(t) = eAt x(0) + eAt ∫ e−Ax BU (x)dx Given x(0) = We know that eAt = L−1 [sI − A ]−1 eAt = L−1 [sI − A ]−1 s 0 −2 0 = sI − A = s 0 − −2 sI − A = s − − 1 = s − 1 s + 1 −1 s + [sI − A ] = [sI − A ]−1 = (s + 1)(s + 2) (s + 1)(s + 2) = s + = s + 1/(s + 2) 1/(s + 2) Chapter (578-614).indd 612 Therefore, y(t) = [1 s + s + 0 s + 1 0 s + 1 0 s + 2 s + 2 0]x(t) − e−2t = [1 0] 1 − e−t 1 = − e−2t 2 Ans (a) 3/23/2016 4:12:06 PM ... 2 .18 .1 Reluctance 89 95 11 2 12 4 14 2 16 9 16 9 16 9 17 0 17 0 17 0 17 0 17 0 17 1 17 1 17 1 17 1 17 2 17 2 17 2 17 3 17 3 17 3 17 3 17 3 17 4 17 4 17 4 17 4 17 4 17 4 17 5 17 5 17 5 17 5 17 5 17 5 17 5 17 6 17 6 17 6 17 7 17 7 17 8... 10 03 10 04 10 05 10 07 10 08 10 09 10 10 10 11 1 012 10 12 10 15 10 15 10 15 10 16 10 17 10 17 10 18 3/24/2 016 12 :36:49 PM xxx CONTENTS 10 .12 Inverters 10 .12 .1 Voltage Source Inverters 10 .12 .2 Current... Solved GATE 2 014 10 77 Solved GATE 2 015 11 29 Solved GATE 2 016 11 61 Index FM.indd 30 10 19 10 19 10 21 1022 10 22 10 23 10 23 10 24 10 25 10 26 10 26 10 26 10 27 10 27 10 32 10 38 10 44 10 48 I1 3/24/2 016 12 :36:49