No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, photocopying, or otherwise without the prior permission of the author.GATE SOLVED PAPERElectrical EngineeringCONTROL SYSTEMSCopyright By NODIA & COMPANYInformation contained in this book has been obtained by authors, from sources believes to be reliable.However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineeringor other professional services.NODI A AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, J aipur 302039Ph :+91 - 141 - 2101150www.nodia.co.inemail : enquiry@nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount*GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inYEAR 2013ONE MARKQ. 1The Bode plot of a transfer function G s ^ h is shown in the figure below.The gainlogG s 20 ^_h i is 32dB and8dB -at 1 / rad s and 10 / rad s respectively. The phase is negative for all w. Then G s ^ h is(A) .s398(B) .s3982(C) s32(D) s322Q. 2Assuming zero initial condition, the response y t ^ h of the system given below to a unit step input u t ^ h is(A) u t ^ h(B) tu t ^ h(C) t u t22^ h(D) eu tt -^ hYEAR 2013TWO MARKSQ. 3The signal flow graph for a system is given below. The transfer function U sYs^^hh for this system isGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A) s ss5 6 212+ ++(B) s ss6 212+ ++(C) s ss4 212+ ++(D) s s 5 6 212+ +Q. 4The open-loop transfer function of a dc motor is given as V sss 1 1010aw=+^^hh. When connectedinfeedbackasshownbelow,theapproximatevalueofKathatwill reducethetimeconstantoftheclosedloopsystembyonehundredtimesas compared to that of the open-loop system is(A) 1(B) 5(C) 10(D) 100Common Data For Q. 5 and 6The state variable formulation of a system is given asxxxxu2001111212=--+oo> > > > H H H H , x 0 01= ^ h , x 0 02= ^ hand yxx1 012=6>@HQ. 5The response y t ^ h to the unit step input is(A)e2121 t 2--(B)e e 12121 t t 2- -- -(C) e et t 2-- -(D)e 1t--Q. 6The system is(A) controllable but not observable(B) not controllable but observable(C) both controllable and observable(D) both not controllable and not observableYEAR 2012TWO MARKSQ. 7The state variable description of an LTI system is given by xxx123oooJLKKKNPOOO 0aaaxxxu00 00 0001312123= +JLKKKJLKKKJLKKKNPOOONPOOONPOOOy xxx1 0 0123=JLKKK_NPOOOiGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inwhere y is the output and u is the input. The system is controllable for(A)0, 0, 0 a a a1 2 3! ! = (B)0, 0, 0 a a a1 2 3! ! =(C)0, 0, 0 a a a1 3 3! = = (D)0, 0, 0 a a a1 2 3! ! =Q. 8The feedback system shown below oscillates at 2 / rad s when(A)2 0.75 and K a = = (B)3 0.75 and K a = =(C)4 0.5 and K a = = (D)2 0.5 and K a = =Statement for Linked Answer Questions 9 and 10 :The transfer function of a compensator is given as( ) Gsc s bs a=++Q. 9( ) Gsc is a lead compensator if(A)1, 2 a b = = (B)3, 2 a b = =(C)3, 1 a b =- =- (D)3, 1 a b = =Q. 10The phase of the above lead compensator is maximum at(A)/ rad s 2 (B)/ rad s 3(C)/ rad s 6 (D) 1/ / rad s 3YEAR 2011ONE MARKQ. 11The frequency response of a linear system( ) G jwis provided in the tubular form below( ) G jw 1.3 1.2 1.0 0.8 0.5 0.3( ) G j + w 130c - 140c - 150c - 160c - 180c - 200c -(A) 6 dB and 30c(B) 6 dB and30c -(C)6dB -and 30c(D)6dB -and30c -Q. 12The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input( ) r thaving a magnitude of 10 and a duration of one second, as shown in the figure is(A) 0(B) 0.1(C) 1(D) 10GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 13An open loop system represented by the transfer function( ) G s( )( )( )s ss2 31=+ +- is(A) Stable and of the minimum phase type(B) Stable and of the nonminimum phase type(C) Unstable and of the minimum phase type(D) Unstable and of nonminimum phase typeYEAR 2011TWO MARKSQ. 14The open loop transfer function( ) G sof a unity feedback control system is given as( ) G s( ) s sKs2322=++b lFrom the root locus, at can be inferred that when Ktends 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 the s-plane(C) The root loci cross thejw axis for a finite value of; 0 KK!(D) Three real roots are found on the right half of the s-planeQ. 15A two loop position control system is shown belowThe gain Kof the Tacho-generator influences mainly the(A) Peak overshoot(B) Natural frequency of oscillation(C) Phase shift of the closed loop transfer function at very low frequencies ( ) 0 " w(D) Phase shift of the closed loop transfer function at very high frequencies ( ) "3 wYEAR 2010TWO MARKSQ. 16The frequency response of ( )( 1)( 2)G ss s s1=+ + plotted in the complex( ) G jwplane0 ) (for < < 3 wisGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 17The systemA Bu X X = +o with, A B102201=-= > > H H is(A) Stable and controllable(B) Stable but uncontrollable(C) Unstable but controllable(D) Unstable and uncontrollableQ. 18Thecharacteristicequationofaclosed-loopsystemis ( )( ) ( ) , s s s k s k 1 3 2 0 0 > + + + = .Which of the following statements is true ?(A) Its root are always real(B) It cannot have a breakaway point in the range[ ] Re s 1 0 < < -(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.YEAR 2009ONE MARKQ. 19The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as, ,1/ GG G1 2 3. The relative small errors associated with eachrespectivesubsystem, GG1 2and G3are,1 2e e and 3e .Theerrorassociated with the output is :(A) 11 23e ee+ + (B) 31 2ee e(C) 1 2 3e e e + - (D) 1 2 3e e e + +Q. 20Thepolarplotofanopenloopstablesystemisshownbelow.Theclosedloop system isGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A) always stable(B) marginally stable(C) un-stable with one pole on the RH s-plane(D) un-stable with two poles on the RH s-planeQ. 21The first two rows of Rouths tabulation of a third order equation are as follows. ss242432This means there are(A) Two roots at s j ! =and one root in right half s-plane(B) Two roots at2 s j ! =and one root in left half s-plane(C) Two roots at2 s j ! =and one root in right half s-plane(D) Two roots at s j ! =and one root in left half s-planeQ. 22Theasymptoticapproximationofthelog-magnitudev/ sfrequencyplotofa system containing only real poles and zeros is shown. Its transfer function is(A) ( )( )( )s s ss2 2510 5+ ++(B) ( )( )( )s s ss2 251000 52+ ++(C) ( )( )( )s s ss2 25100 5+ ++(D) ( )( )( )s s ss2 2580 52+ ++YEAR 2009TWO MARKSQ. 23Theunit-stepresponseofaunityfeedbacksystemwithopenlooptransfer function( ) / (( 1)( 2)) G s K s s = + +is shown in the figure. The value of KisGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in(A) 0.5(B) 2(C) 4(D) 6Q. 24Theopenlooptransferfunctionofaunityfeedbacksystemisgivenby ( ) ( )/ G s e s. s 01=-. The gain margin of the is system is(A) 11.95 dB(B) 17.67 dB(C) 21.33 dB(D) 23.9 dBCommon Data for Question 25 and 26 :A system is described by the following state and output equations ( )( ) ( ) ( )( )2 ( ) ( )( ) ( )dtdxtxt xt u tdtdxtxt u ty t xt3 211 2221=- + +=- +=when( ) u tis the input and( ) y tis the outputQ. 25The system transfer function is(A) s ss5 622+ -+(B) s ss5 632+ ++(C) s ss5 62 52+ ++(D) s ss5 62 52+ --Q. 26The state-transition matrix of the above system is(A) ee e e0tt t t32 3 2+-- - -= G(B) e e ee 0t t tt3 2 32-- - --= G(C) e e ee 0t t tt3 2 32+- - --= G(D) e e ee 0t t tt3 2 32-- --= GYEAR 2008ONE MARKQ. 27A function( ) y tsatisfies the following differential equation : ( )( )dtdy ty t + ( ) t d =where( ) t dis the delta function. Assuming zero initial condition, and denoting the unit step function by( ), ( ) u t y tcan be of the form(A) et(B) et -(C)( ) eu tt(D)( ) e u tt -YEAR 2008TWO MARKQ. 28The transfer function of a linear time invariant system is given as( ) G ss s 3 212=+ +Thesteadystatevalueoftheoutputofthesystemforaunitimpulseinput applied at time instant t 1 =will be(A) 0(B) 0.5(C) 1(D) 2GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 29The transfer functions of two compensators are given below :( )( ),( )Cs sCss1010 110 1101 2=++=++Which one of the following statements is correct ?(A) C1 is lead compensator and C2 is a lag compensator(B) C1 is a lag compensator and C2 is a lead compensator(C) Both C1 and C2 are lead compensator(D) Both C1 and C2 are lag compensatorQ. 30TheasymptoticBodemagnitudeplotofaminimumphasetransferfunctionis shown in the figure :This transfer function has(A) Three poles and one zero(B) Two poles and one zero(C) Two poles and two zero(D) One pole and two zerosQ. 31Figure shows a feedback system where K 0 >The range of Kfor which the system is stable will be given by(A) 0 30 K < < (B) 0 39 K < Q. 32The transfer function of a system is given as s s 20 1001002+ +The system is(A) An over damped system(B) An under damped system(C) A critically damped system(D) An unstable systemStatement for Linked Answer Question 33 and 34.Thestatespaceequationofasystemisdescribedby, A B C X X u Y X = + =o where Xis state vector, u is input, Yis output and, , [1 0] A B C001201=-= = = = G GQ. 33The transfer function G(s) of this system will be(A) ( ) s s2 +(B) ( ) s ss21-+GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in(C) ( ) s s2 -(D) ( ) s s 21+Q. 34A unity feedback is provided to the above system( ) G sto make it a closed loop system as shown in figure.For a unit step input( ) r t , the steady state error in the input will be(A) 0(B) 1(C) 2(D) 3YEAR 2007ONE MARKQ. 35The system shown in the figure is(A) Stable(B) Unstable(C) Conditionally stable(D) Stable for input u1, but unstable for input u2YEAR 2007TWO MARKSQ. 36If[ ( )] Re x G jw = ,and[ ( )] Im y G jw = thenfor0 " w+,theNyquistplotfor ( ) / ( )( ) G s s s s 1 1 2 = + +is(A) x 0 = (B)/ x 3 4 =-(C)/ x y 1 6 = - (D)/ x y 3 =Q. 37Thesystem900/ ( 1)( 9) s s s + + istobesuchthatitsgain-crossoverfrequency becomes same as its uncompensated phase crossover frequency and provides a 45c phase margin. To achieve this, one may use(A) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45c at the frequency of 3 3 rad/ s(B) a lead compensator that provides an amplification of 20 dB and a phase lead of 45c at the frequency of 3 rad/ s(C) a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45c at the frequency of3 rad/ s(D) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45c at the frequency of 3 rad/ sGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 38If the loop gain Kof a negative feed back system having a loop transfer function ( 3)/ ( 8) Ks s2+ +is to be adjusted to induce a sustained oscillation then(A) The frequency of this oscillation must be 4 3 rad/ s(B) The frequency of this oscillation must be 4 rad/ s(C) The frequency of this oscillation must be 4 or 4 3 rad/ s(D) Such a Kdoes not existQ. 39The system shown in figure belowcan be reduced to the formwith(A), 1/ ( ), X c s cY s a s a Z b s b0 120 1 0 1= + = + + = +(B)1, ( )/ ( ), X Y c s c s a s a Z b s b0 120 1 0 1= = + + + = +(C), ( )/ ( ), 1 X c s cY bs b s a s a Z1 0 1 021 0= + = + + + =(D), 1/ ( ), X c s cY s a s aZ bs b1 021 1 0= + = + + = +Q. 40Considerthefeedbacksystemshownbelowwhichissubjectedtoaunitstep input. The system is stable and has following parameters4, 10, 500 K Kp iw = = =and. 07 x = .The steady state value of Z is(A) 1(B) 0.25(C) 0.1(D) 0GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inCommon Data For Q. 41 and 42R-L-C circuit shown in figureQ. 41For a step-input ei, the overshoot in the output e0 will be(A) 0, since the system is not under damped(B) 5 %(C) 16 %(D) 48 %Q. 42If the above step response is to be observed on a non-storage CRO, then it would be best have the ei as a(A) Step function(B) Square wave of 50 Hz(C) Square wave of 300 Hz(D) Square wave of 2.0 KHzYEAR 2006ONE MARKQ. 43For a system with the transfer function ( )( )H ss ss4 2 13 22=- +-, the matrix A in the state space formA B X X u = +o is equal to(A) 101012004 - -RTSSSSVXWWWW(B) 001102014 - -RTSSSSVXWWWW(C) 031122014--RTSSSSVXWWWW(D) 101002014 - -RTSSSSVXWWWWYEAR 2006TWO MARKSQ. 44ConsiderthefollowingNyquistplotsoflooptransferfunctionsover0 w = to 3 w = . Which of these plots represent a stable closed loop system ?GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A) (1) only(B) all, except (1)(C) all, except (3)(D) (1) and (2) onlyQ. 45The Bode magnitude plot( )( )( )( )H jj jj10 10010 124ww ww=+ ++ isQ. 46A closed-loop system has the characteristic function ( )( ) ( ) s s Ks 4 1 1 02- + + - =. Its root locus plot against KisGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inYEAR 2005ONE MARKQ. 47A system with zero initial conditions has the closed loop transfer function.( ) Ts( )( ) s ss1 442=+ ++The system output is zero at the frequency(A) 0.5 rad/ sec(B) 1 rad/ sec(C) 2 rad/ sec(D) 4 rad/ secQ. 48Figureshowstherootlocusplot(locationofpolesnotgiven)ofathirdorder system whose open loop transfer function is(A) sK3(B) ( ) s sK12+(C) ( ) s sK12+(D) ( ) s sK12-GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 49The gain margin of a unity feed back control system with the open loop transfer function( ) G s( 1)ss2=+ is(A) 0(B) 21(C)2(D) 3YEAR 2005TWO MARKSQ. 50A unity feedback system, having an open loop gain ( ) ( ) G s H s( )( )sK s11=+-,becomes stable when(A)K 1 > (B) K 1 >(C)K 1 < (D) K 1 H(C) ( ) e ee10t tt31 3 33-- --> H(D) ( ) ee101tt---> HQ. 55The state transition equation(A)( ) tt eeXtt=---= G(B)( ) teeX13tt 3=---= G(C)( ) tt eeX3tt33=--= G(D)( ) tt eeXtt3=---= GYEAR 2004ONE MARKQ. 56TheNyquistplotoflooptransferfunction( ) ( ) G s H s ofaclosedloopcontrol systempassesthroughthepoint( , ) j 1 0 - inthe( ) ( ) G s H s plane.Thephase margin of the system is(A) 0c(B) 45c(C) 90c(D) 180cQ. 57Consider the function,( )( )Fss s s 3 252=+ + where( ) Fsis the Laplace transform of the of the function( ) f t . The initial value of( ) f tis equal to(A) 5(B) 25(C) 35(D) 0Q. 58For a tachometer, if( ) t qis the rotor displacement in radians,( ) e tis the output voltage and Kt is the tachometer constant in V/ rad/ sec, then the transfer function, ( )( )Q sEs will be(A) K st2(B) K st(C) K st(D) KtYEAR 2004TWO MARKSQ. 59Fortheequation,s s s 4 6 03 2- + + = thenumberofrootsinthelefthalfofs-plane will be(A) Zero(B) One(C) Two(D) ThreeQ. 60For the block diagram shown, the transfer function ( )( )RsC s is equal to(A) ss 122+(B) ss s 122+ +(C) ss s 12+ +(D) s s 112+ +GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 61The state variable description of a linear autonomous system is,A X X =o where X isthetwodimensionalstatevectorandAisthesystemmatrixgivenby A0220== G. The roots of the characteristic equation are(A)2 -and2 + (B)2 j -and2 j +(C)2 -and2 - (D)2 +and2 +Q. 62The block diagram of a closed loop control system is given by figure. The values of Kand Psuch that the system has a damping ratio of 0.7 and an undamped natural frequency nwof 5 rad/ sec, are respectively equal to(A) 20 and 0.3(B) 20 and 0.2(C) 25 and 0.3(D) 25 and 0.2Q. 63The unit impulse response of a second order under-damped system starting from restisgivenby( ) . , sin c t e t t 125 8 0t 6$ =-.Thesteady-statevalueoftheunit step response of the system is equal to(A) 0(B) 0.25(C) 0.5(D) 1.0Q. 64Inthesystemshowninfigure,theinput( ) sin x t t = .Inthesteady-state,the response( ) y twill be(A)( 45 ) sin t21c - (B)( 45 ) sin t21c +(C)( ) sin t 45c - (D)( 45 ) sin t c +Q. 65The open loop transfer function of a unity feedback control system is given as( ) G ssas 12=+.The value of a to give a phase margin of 45c is equal to(A) 0.141(B) 0.441(C) 0.841(D) 1.141YEAR 2003ONE MARKQ. 66A control system is defined by the following mathematical relationship( )dtd xdtdxx e 6 5 12 1t222+ + = --The response of the system as t " 3 is(A) x 6 = (B) x 2 =(C). x 24 = (D) x 2 =-GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 67A lead compensator used for a closed loop controller has the following transfer function (1 )(1 ) Kbsas++ For such a lead compensator(A) a b < (B) b a (D) a Kb VXWWWWWHWhere the P matrix is given by(A) 1 0JBLJK2- -= G(B) 0 1LJKJB2- -= G(C) 0 1LJKJB2- -= G(D) 1 0JBLJK2- -= GQ. 73The loop gain GH of a closed loop system is given by the following expression ( )( ) s s sK2 4 + + The value of K for which the system just becomes unstable is(A) K 6 = (B) K 8 =(C) K 48 = (D) K 96 =Q. 74The asymptotic Bode plot of the transfer function/ [1 ( / )] K s a +is given in figure. The error in phase angle and dB gain at a frequency of0.5a w =are respectively(A) 4.9c, 0.97 dB(B) 5.7c, 3 dB(C) 4.9c, 3 dB(D) 5.7c, 0.97 dBQ. 75The block diagram of a control system is shown in figure. The transfer function ( ) ( )/ ( ) G s Ys U s =of the system is(A) 18 1 11s s12 3+ +^ ^ h h(B) 27 1 11s s6 9+ +^ ^ h h(C) 1 27 11s s12 9+ +^ ^ h h(D) 1 1 271s s9 3+ +^ ^ h h***********GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inSOLUTI ONSol . 1Option (B) is correct.From the given plot, we obtain the slope asSlope log loglog logw wG G 20 202 12 1=--From the figurelogG 2028dB =-logG 20132dB =and 1w 1 / rad s = 2w 10 / rad s =So, the slope isSlope log log8 3210 1=-- -40 / dB decade =-Therefore, the transfer function can be given asG s ^ h Sk2=at1 w =G jw ^ hwkk2= =In decibel,logG j 20 w ^ h logk 20 32 = =or,k. 10 3983220= =Hence, the Transfer function isG s ^ h .sks3982 2= =Sol . 2Option (B) is correct.The Laplace transform of unit step funn is U s ^ h s1=So, the O/ P of the system is given asYs ^ h s s1 1=b b l l s12=For zero initial condition, we checku t ^ h dtdy t=^ h&U s ^ hSYs y 0 = - ^ ^ h h&U s ^ hs sy102= - c ^ m hor,U s ^ h s1= y 0 0 = ^^hhHence, the O/ P is correct which isYs^ h s12=its inverse Laplace transform is given byy t ^ htu t = ^ hGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSol . 3Option (A) is correct.For the given SFG, we have two forward pathsPk1s s s 1 11 1 2= =- - -^ ^ ^ ^ h h h hPk2s s 1 1 11 1= =- -^ ^ ^ ^ h h h hsince, all the loops are touching to both the paths Pk1and Pk2 so,k1D 1 k2D = =Now, we haveD1 = - (sum of individual loops) + (sum of product of nontouching loops)Here, the loops areL14 1 4 =- =- ^ ^ h hL2s s 4 41 1=- =- -^ ^ h hL3s s s 2 21 1 2=- =-- - -^ ^ ^ h h hL4s s 2 1 21 1=- =-- -^ ^ ^ h h hAs all the loop, , L L L1 2 3 and L4 are touching to each other so,DL L L L 11 2 3 4= - + + + ^ hs s s 1 4 4 2 21 2 1= -- - - -- - -^ h5 6 2 s s1 2= + +- -From Masons gain formulae U sYs^^hh Pk kDS D=s ss s5 6 21 22 1=+ ++- -- - s ss5 6 212=+ ++Sol . 4Option (C) is correct.Given, open loop transfer functionG s ^ h sKsK1 1010a a101=+=+By taking inverse Laplace transform, we haveg t ^ het 101=-Comparing with standard form of transfer function, Ae/ tt -, we get the open loop time constant, olt 10 =Now, we obtain the closed loop transfer function for the given system asH s ^ h 1 10 10G sG ss KK110aa=+=+ +^^hh s KKaa101=+ + ^ hBy taking inverse laplace transform, we geth t ^ h. keak t a101=- +^ hSo, the time constant of closed loop system is obtained as cltk1a101=+or, cltk1a= (approximately)Now, given that ka reduces open loop time constant by a factor of 100. i.e., clt100olt=or, k1a 10010=or,ka10 =GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inSol . 5Option (A) is correct.Given, the state variable formulation, xx12oo> H xxu20011112=--+ > > > H H H ....(1)andy xx1 012=6 > @ H....(2)From Eq. (1) we getx1ox u 21= +Taking Laplace transformsX x 01 1- ^ hXs211=- +(Here, X1 denotes Laplace transform of x1)So,s X 21+ ^ hs1=x 0 01= ^ ^ h hor,X1 s s 21=+ ^ h....(3)Now, from Eq. (2) we haveyx1=Taking Laplace transform both the sides,Y XL=or,Ys s 21=+ ^ h(from eq. (3))or,Ys s 21 121= -+; ETaking inverse Laplace transformyu t e u t21 t 2= --^ ^ h h 8 Be2121 t 2= --for t 0 > ^ hSol . 6Option (A) is correct.From the given state variable system, we haveA 2001=-> HandB 11=> H;C 1 0 =6 @Now, we obtain the controllability matrixCM: B AB =6 @ 1221=-> Hand the observability matrix is obtained asOM CCA=> H 1200=-> HSo, we getRank of the controllability matrix " RankC 2M= ^ hRank of the observability matrix " RankO 1M= ^ hSince, the order of state variable isand x x 21 2 ^ h. Therefore, we haveRankCM ^ h = order of state variablesbut,Rank ( ) OM < order of state variablesThus, system is controllable but not observableGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSol . 7Option (D) is correct.General form of state equations are given asxox u A B = +yox u C D = +For the given problemA0,aaa00 00 0312=RTSSSSVXWWWWB001=RTSSSSVXWWWWAB0aaa a00 00 000100312 2= =RTSSSSRTSSSSRTSSSSVXWWWWVXWWWWVXWWWWA B2a aa aa a a a 0000 00001002 33 11 2 1 2= =RTSSSSRTSSSSRTSSSSVXWWWWVXWWWWVXWWWWFor controllability it is necessary that following matrix has a tank of n 3 = .U : : B AB A B2=6 @ 0aa a 001 00021 2=RTSSSSVXWWWWSo,a20 !a a1 20 ! a 01& !a3 may be zero or not.Sol . 8Option (A) is correct.( ) Ys( )[ ( ) ( )]s as sKsRs Ys2 113 2=+ + ++-( )( )Yss as sKs12 113 2++ + ++; E ( )( )s as sKsRs2 113 2=+ + ++( )[ ( ) ( )] Ys s as s k k 2 13 2+ + + + + ( 1) ( ) Ks Rs = +Transfer Function,( )( )( )H sRsYs=( ) ( )( )s as s k kKs2 113 2=+ + + + ++Routh Table :For oscillation, ( ) ( )aa K K 2 1 + - +0 =a KK21=++Auxiliary equation( ) As ( ) as k 1 02= + + =s2 ak 1=-+s2 ( )( )kkk112 =+- ++ ( ) k 2 =- +GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.insj k 2 = +jwj k 2 = +wk 2 2 = + =(Oscillation frequency)k2 =anda.2 22 143075 =++= =Sol . 9Option (A) is correct.( ) G sC s bs aj bj aww=++=++Phase lead angle,ftan tana b1 1 w w= -- -a a k ktanaba b112ww w=+--JLKKKNPOOO

( )tanabb a12ww=+--c mFor phase-lead compensation0 > fb a - 0 >ba >Note:Forphaseleadcompensatorzeroisnearertotheoriginascomparedto pole, so option (C) can not be true.Sol . 10Option (A) is correct.ftan tana b1 1 w w= -- -a a k k ddwf / /aabb111102 2w w=+-+=a a k k aab122w+b ba1 122w= + a b1 1-ab a b1 12w= -b lwab = / sec rad 1 2 2#= =Sol . 11Option (A) is correct.Gain margin is simply equal to the gain at phase cross over frequency (pw ). Phase cross over frequency is the frequency at which phase angle is equal to180c - .From the table we can see that( ) G j 180pc + w =- , at which gain is 0.5.GM20( )logG j1p10w=e o200.516 log dB = =b lPhaseMarginisequalto180cplusthephaseangle gf atthegaincrossover frequency (gw ). Gain cross over frequency is the frequency at which gain is unity.From the table it is clear that( ) G j 1gw = , at which phase angle is150c - PMf 180 ( ) G jgc + w = + 180 150 30c = - =Sol . 12Option (A) is correct.We know that steady state error is given byess ( )( )limG ssRs1 s 0=+ "where( ) Rs input "( ) G s openlooptransfer function "GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inFor unit step input( ) Rss1=Soess ( )10.1 limG ss s1 s 0=+="b l( ) G 1 0 + 10 =( ) G 0 9 =Given input( ) r t 10[ ( ) ( 1)] t t m m = - -or( ) Rss se 101 1 s= --: D se101s=--: DSo steady state error ssel( )( )limG ssse1101ss0#=+-"- ( ) e1 910 10=+-0 =Sol . 13Option (B) is correct.Transfer function having at least one zero or pole in RHS of s-plane is called non-minimum phase transfer function.( ) G s( )( ) s ss2 31=+ +-In the given transfer function one zero is located at s 1 =(RHS), so this is a non-minimum phase system.Poles, 2 3 - - , are in left side of the complex plane, So the system is stableSol . 14Option (A) is correct.( ) G s( ) s sKs2322=++b lSteps for plotting the root-locus(1) Root loci starts at0, 0 2 s s s and = = =-(2) n m > , therefore, number of branches of root locus b 3 =(3) Angle of asymptotes is given by ( )n mq 2 1 180c-+,, q 0 1 =(I) ( )( )3 12 0 1 180#c-+90c =(I I) ( )( )3 12 1 1 180#c-+270c =(4) The two asymptotes intersect on real axis at centroidx n mPoles Zeroes=--/ / 2323 1 32=-- --=-b l(5) Between two open-loop poles s 0 =and s 2 =-there exist a break away point.K32( )ss s 22=-++b l dsdK0 =s0 =Root locus is shown in the figureGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inThree roots with nearly equal parts exist on the left half of s-plane.Sol . 15Option (A) is correct.The system may be reduced as shown below ( )( )RsYs ( )( )( )s s Ks s Ks s K111111 112=++ ++ +=+ + +This is a second order system transfer function, characteristic equation is(1 ) 1 s s K2+ + + 0 =Comparing with standard forms s 2n n2 2xw w + + 0 =We getx K21=+Peak overshootMpe/ 12=px x - -So the Peak overshoot is effected by K .Sol . 16Option (A) is correct.Given( ) G s( )( ) s s s 1 21=+ +( ) G jw( )( ) j j j 1 21w w w=+ +( ) G jw1 412 2w w w=+ +( ) G j + w ( ) ( / ) tan tan 90 21 1c w w =- - -- -In nyquist plotFor, ( ) G j 0 w w = 3 =( ) G j + w 90c =-For, ( ) G j 3 w w = 0 =( ) G j + w 90 90 90 c c c =- - - 270c =-Intersection at real axis( ) G jw( )( ) j j j 1 21w w w=+ + ( ) j j3 212w w w=- + +GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in ( ) ( )( )j jj3 213 23 22 2 2 22 2#w w w w w ww w w=- + - - - -- - - ( )( ) j9 23 24 2 2 22 2w w ww w w=+ -- - - ( ) ( )( ) j9 239 224 2 2 224 2 2 22w w www w ww w=+ ---+ --At real axis[ ( )] ImG jw 0 =So, ( )( )9 224 2 22w w ww w+ --0 =2 02& w - = 2 w =rad/ secAt2 w =rad/ sec, magnitude response is( ) G jat 2ww= 2 2 1 2 41=+ + 6143< =Sol . 17Option (C) is correct.Stability :Eigen value of the system are calculated asA I l - 0 =A I l -1022 00 ll=-- > > H H 1022ll=- --> HA I l - ( 1 )(2 ) 2 0 0#l l =- - - - =&,1 2l l, 1 2 =-Since eigen values of the system are of opposite signs, so it is unstableControllability :A 1022=-> H, B01=> HAB 22=> H[ : ] B AB0122=> H: B AB6 @0 =YSo it is controllable.Sol . 18Option (C) is correct.Given characteristic equation( 1)( 3) ( 2) s s s Ks + + + + 0 = ;K 0 >( 4 3) s s s K (s 2)2+ + + + 0 =4 (3 ) 2 s s Ks K3 2+ + + + 0 =From Rouths tabulation methods31K 3+s24K 2s10 >4(3 ) 212 2K KK4 4(1)=+ -+s0K 2GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inThere is no sign change in the first column of routh table, so no root is lying in right half of s-plane.For plotting root locus, the equation can be written as ( )( )( )s s sKs11 32++ ++0 =Open loop transfer function( ) G s( )( )( )s s sKs1 32=+ ++Root locus is obtained in following steps:1.No. of poles n 3 = , at0, 1 s s = =-and3 s =-2.No. of Zeroes m 1 = , at s 2 =-3.The root locus on real axis lies between s 0 =and s 1 =- , between s 3 =-and s 2 =- .4.Breakaway point lies between open loop poles of the system. Here breakaway point lies in the range[ ] Re s 1 0 < < - .5.Asymptotes meet on real axis at a point C , given byCn mreal part of poles real partsof zeroes=--/ / ( ) ( )3 10 1 3 2=-- - --1 =-Asno.ofpolesis3,sotworootlocibranchesterminatesatinfinityalong asymptotes( ) Re s 1 =-Sol . 19Option (D) is correct.Overall gain of the system is written asGG G G11 23=Weknowthatforaquantitythatisproductoftwoormorequantitiestotal percentageerrorissomeofthepercentageerrorineachquantity.soerrorin overall gain G isG 311 23e ee= + +Sol . 20Option (D) is correct.From Nyquist stability criteria, no. of closed loop poles in right half of s-plane is given asZ P N = -P " No. of open loop poles in right half s-planeN " No. of encirclement of ( 1, 0) j -HereN 2 =-(` encirclement is in clockwise direction)GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inP 0 =(` system is stable)So,( ) Z 0 2 = --Z 2 = , System is unstable with 2-poles on RH of s-plane.Sol . 21Option (D) is correct.Given Rouths tabulation.s32 2s24 4s10 0So the auxiliary equation is given by,s 4 42+ 0 =s21 =-sj ! =From table we have characteristic equation ass s s 2 2 4 43 2+ + + 0 =s s s 2 23 2+ + + 0 =( ) ( ) s s s 1 2 12 2+ + + 0 =( )( ) s s 2 12+ + 0 =s 2 =- , s j ! =Sol . 22Option (B) is correct.Since initial slope of the bode plot is40 -dB/ decade, so no. of poles at origin is 2.Transfer function can be written in following steps: 1.Slope changes from40 -dB/ dec. to60 -dB/ dec. at21w=rad/ sec., so at 1wthere is a pole in the transfer function.2.Slope changes from60 -dB/ dec to40 -dB/ dec at52w=rad/ sec., so at this frequency there is a zero lying in the system function.3.The slope changes from40 -dB/ dec to60 -dB/ dec at253w=rad/ sec, so there is a pole in the system at this frequency.Transfer function( ) Ts( )( )( )s s sKs2 2552=+ ++Constant term can be obtained as.( ) Tj. at 01ww=80 =So,80 ( . )( )logK2001 5052#=K 1000 =therefore, the transfer function is( ) Ts( )( )( )s s ss2 251000 52=+ ++Sol . 23Option (D) is correct.From the figure we can see that steady state error for given system isess. . 1 075 025 = - =Steady state error for unity feed back system is given byess 1 ( )( )limG ssRss 0=+ " = GGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inlims1( 1)( 2)ss sKs01=+"+ +^ h> H;( ) Rss1= (unit step input) 11K2=+ K 22=+So,ess0.25K 22=+=20.5 0.25K = +K..025156 = =Sol . 24Option (D) is correct.Open loop transfer function of the figure is given by,( ) G sse. s 01=-( ) G jwje. j01w=w -Phase cross over frequency can be calculated as,( ) G jp+ w 180c =-0.118090p #c wp- -b l180c =-. 01180p #cwp90c =. 01pw18090#cc p= pw . 157 =rad/ secSo the gain margin (dB) ( )logG j201pw=e o20log115.71=^ h= G. log 20 157 = . 239 =dBSol . 25Option (C) is correct.Given system equations ( )dtdxt13 ( ) ( ) 2 ( ) xt xt u t1 2=- + + ( )dtdxt2( ) ( ) xt u t 22=- +( ) y t ( ) xt1=Taking Laplace transform on both sides of equations.( ) sX s1( ) ( ) ( ) X s X s U s 3 21 2=- + +( ) ( ) s X s 31+ ( ) ( ) X s U s 22= + ...(1)Similarly( ) sX s2( ) ( ) X s U s 22=- +( ) ( ) s X s 22+ ( ) U s = ...(2)From equation (1) & (2)( ) ( ) s X s 31+( )( )sU sU s22 =++( ) X s1 ( ) ( )sU sss3 21 2 2=+ ++ +; E( )( )( )( )U ss ss2 32 5=+ ++From output equation,( ) Ys ( ) X s1=GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSo,( ) Ys ( )( )( )( )U ss ss2 32 5=+ ++System transfer function ( )( )U sYsT.F =( )( )( )s ss2 32 5=+ ++ ( )s ss5 62 52=+ ++Sol . 26Option (B) is correct.Given state equations in matrix form can be written as, xx12oo> H( )xxt u30122112=--+ > > > H H H ( )dtd t X( ) ( ) A t B t X u = +State transition matrix is given by( ) t f ( ) s L1F =-6 @( ) s F ( ) sI A1= --( ) sI A -ss 00 3012= ---> > H H( ) sI A -ss3012=+ -+> H( ) sI A1-- ( )( ) s sss 3 212013=+ +++> HSo( ) ( ) s sI A1F = -- ( ) ( )( )( )s s ss313 2121=+ + ++0RTSSSSSVXWWWWW( ) [ ( )] t s L1f F =- e e ee 0t t tt3 2 32=-- - --> HSol . 27Option (D) is correct.Given differential equation for the function ( )( )dtdy ty t + ( ) t d =Taking Laplace on both the sides we have,( ) ( ) sYs Ys + 1 =( 1) ( ) s Ys + 1 =( ) Yss 11=+Taking inverse Laplace of( ) Ys( ) y t ( ) eu tt=-, t 0 >Sol . 28Option (A) is correct.Given transfer function( ) G ss s 3 212=+ +Input( ) r t ( ) t 1 d = -( ) Rs [ ( )] t e 1 Lsd = - =-Output is given by( ) Ys ( ) ( ) Rs G s =s se3 2s2=+ +-GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inSteady state value of output( ) limy tt"3( ) limsYss 0="lims sse3 2ss02=+ +"-0 =Sol . 29Option (A) is correct.For C1 Phase is given by C1q ( ) tan tan101 1ww= -- -a ktan11010 12www=+--JLKKKNPOOOtan1090 >12ww=+-c m (Phase lead)Similarly for C2, phase is C2q ( ) tan tan101 1 ww = -- -a ktan11010 12www=+--JLKKKNPOOO tan1090 H1KKii= =Sol . 41Option (C) is correct.System response of the given circuit can be obtained as.( )( )( )H sesesi0=R LsCsCs11=+ +bbll( ) H s1 LCs RCs12=+ + sLRsLCLC112=+ +b lCharacteristic equation is given by,sLRsLC1 2+ + 0 =Here natural frequency LC1nw=2nxwLR=Damping ratiox LRLC2=RLC2=Herex0.521010 101 1063##= =-- (under damped)So peak overshoot is given by% peak overshoot e100 12#= xpx--100 e ( . ).1 05052#=# p--% 16 =Sol . 42Option ( ) is correct.Sol . 43Option (B) is correct.In standard form for a characteristic equation give as... s a s a s annn111 0+ + + +--0 =in its state variable representation matrix A is given asA a a a a00100100n 0 1 2 1h h hgghgh=- - - --RTSSSSSSVXWWWWWWCharacteristic equation of the system is s s 4 2 12- + 0 =So,, , a a a 4 2 12 1 0= =- =A a a a0010010011020140 1 2=- - -=- -RTSSSSRTSSSSVXWWWWVXWWWWSol . 44Option (A) is correct.In the given options only in option (A) the nyquist plot does not enclose the unit circle ( , ) j 1 0 - , So this is stable.GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inSol . 45Option (A) is correct.Given function is,( ) H jw( )( )( )j jj10 10010 124w ww=+ ++Function can be rewritten as,( ) H jw( )j jj1011010 110010 1424w ww=+ ++9 9 C C . ( )j jj110110001 12w ww=+ ++a a k kThe system is type 0, So, initial slope of the bode plot is 0 dB/ decade.Corner frequencies are 1w 1 =rad/ sec 2w 10 =rad/ sec 3w 100 =rad/ secAs the initial slope of bode plot is 0 dB/ decade and corner frequency11w=rad/sec, the Slope after1 w =rad/ sec or log 0 w =is(0 20) 20 + =+dB/ dec.After corner frequency102w=rad/ sec or log 12w= , the Slope is ( ) 20 20 0 + - =dB/ dec.Similarly after1003w=rad/ sec or log 2 w = , the slope of plot is ( ) 0 20 2 40#- =-dB/ dec.Hence (A) is correct option.Sol . 46Option (B) is correct.Given characteristic equation.( )( ) ( ) s s Ks 4 1 12- + + - 0 =or ( )( )( )s sKs14 112+- +-0 =So, the open loop transfer function for the system.( ) G s( )( )( )( )s s sKs2 2 11=- + +-,no. of poles n 3 =no of zeroes m 1 =Steps for plotting the root-locus(1) Root loci starts at, , s s s 2 1 2 = =- =-(2) n m > , therefore, number of branches of root locus b 3 =(3) Angle of asymptotes is given by ( )n mq 2 1 180c-+,, q 0 1 =(I) ( )( )3 12 0 1 180#c-+90c =(I I) ( )( )3 12 1 1 180#c-+270c =(4) The two asymptotes intersect on real axis atx n mPoles Zeroes=--/ / ( ) ( )3 11 2 2 1=-- - + -1 =-(5)Betweentwoopen-looppoless 1 =- ands 2 =- thereexistabreakaway point.GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inK( )( )( )ss s14 12=--- + dsdK0 =s. 15 =-Sol . 47Option (C) is correct.Closed loop transfer function of the given system is, ( ) Ts( )( ) s ss1 442=+ ++( ) Tjw( )( )( )j jj1 442w ww=+ ++If system output is zero( ) Tjw1 ( ) j j 442w ww=+ +-^ h0 =42w - 0 = 2w 4 =&w2 =rad/ secSol . 48Option (A) is correct.FromthegivenplotwecanseethatcentroidC (pointofintersection)where asymptotes intersect on real axis) is 0So for option (a)( ) G ssK3=Centroid0n m 3 00 0Poles Zeros=--=--=/ /Sol . 49Option (A) is correct.Open loop transfer function is.( ) G s( )ss 12=+( ) G jwj 12ww=-+Phase crossover frequency can be calculated as.( ) G jp+ w 180c =-( ) tanp1w-180c =- pw 0 =Gain margin of the system is.G.M ( ) G j111ppp22www= =+ 10pp22ww=+=Sol . 50Option (C) is correct.Characteristic equation for the given system( ) ( ) G s H s 1+ 0 = ( )( )Kss111++-0 =(1 ) (1 ) s K s + + - 0 =GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in( ) ( ) s K K 1 1 - + + 0 =Forthesystemtobestable,coefficientofcharacteristicequationshouldbeof same sign.K 1 0 > - , K 1 0 > +K 1 < , K 1 >-1 - K 1 < > H H ss 22=--> Hs 42= - 0 =, ss1 22 ! =GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inSol . 62Option (D) is correct.For the given system, characteristic equation can be written as,1( )(1 )s sKsP2+++ 0 =( 2) (1 ) s s K sP + + + 0 =(2 ) s s KP K2+ + + 0 =From the equation. nw 5 K = =rad/ sec (given)So,25 K=and2nxw 2 KP = +. 2 07 5# #2 25P = +orP . 02 =so K 25 = , P . 02 =Sol . 63Option (D) is correct.Unit - impulse response of the system is given as,( ) c t . sin e t 125 8t 6=-, t 0 $So transfer function of the system.( ) [ ( )] H s c t L =( ) ( ).s 6 8125 82 2#=+ +( ) H ss s 12 1001002=+ +Steady state value of output for unit step input,( ) limy tt"3( ) ( ) ( ) lim lim sYs sH s Rss s 0 0= =" "lims s ss12 100100 1s 02=+ +"; E. 10 =Sol . 64Option (A) is correct.System response is.( ) H ss s1=+( ) H jwjj1 ww=+Amplitude response( ) H jw1 ww=+Given input frequency1 w =rad/ sec.So( ) H j1rad/ secww= 1 11=+ 21=Phase response( )hq w ( ) tan 901c w = --( )h1q ww=( ) tan 90 1 451c c = - =-So the output of the system is( ) y t ( ) ( ) H j x thw q = - ( ) sin t2145c = -Sol . 65Option (C) is correct.Given open loop transfer function( ) G jw( ) jja 12ww=+GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inGain crossover frequency ( )gwfor the system.( ) G jgw 1 = a 1gg22 2ww-+1 =a 1g2 2w+g4w =a 1g g4 2 2w w - - 0 = ...(1)Phase margin of the system is45PMc f = ( ) G j 180gc + w = +45c( ) tan a 180 180g1c c w = + --( ) tan ag1w-45c =agw 1 = (2)From equation (1) and (2) a11 14- - 0 =a4 21= a & . 0841 =Sol . 66Option (C) is correct.Given system equation is.6 5dtd xdtdxx22+ + ( ) e 12 1t 2= --Taking Laplace transform on both side.( ) ( ) ( ) s Xs sXs Xs 6 52+ + 12 s s121= -+: D( ) ( ) s s Xs 6 52+ + 12( 2)2s s=+; ESystem transfer function is( ) Xs( )( )( ) s s s s 2 5 124=+ + +Response of the system as t "3 is given by( ) limf tt"3lim ( ) sFss 0=" (final value theorem) ( 2)( 5)( 1)24lims s s s s s 0=+ + + "; E 2 524#= . 24 =Sol . 67Option (A) is correct.Transfer function of lead compensator is given by.( ) H sbsKas11=++aakk( ) H jw Kjbja11ww=++aakkRTSSSSVXWWWWSo, phase response of the compensator is.( )hq w tan tana b1 1 w w= -- -a a k kGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in 1tanaba b21ww w=+--JLKKKNPOOO ( )tanabb a12ww=+--; Ehqshould be positive for phase lead compensationSo,( )hq w( )0 tanabb a>12ww=+--; Eba >Sol . 68Option (A) is correct.Since there is no external input, so state is given by( ) t X ( ) (0) t X f =( ) t " f state transition matrix[0] X "initial conditionSo( ) x tee 00 23tt2=--> > HH( ) x tee23tt2=--> HAt t 1 = , state of the system( ) x tt 1 = ee2221=--> H ..02711100=> HSol . 69Option (B) is correct.Given equation dtd xdtdxx2118122+ + e e 10 5 2t t 4 5= + +- -Taking Laplace on both sides we have( ) ( ) ( ) s Xs sXs Xs21181 2+ +s s s104552= ++++( ) ( ) s s Xs21181 2+ +( )( )( )( ) ( ) ( )s s ss s s s s s4 510 4 5 5 5 2 4=+ ++ + + + + +System response is,( ) Xs( )( )( )( ) ( ) ( )s s s s ss s s s s s4 52118110 4 5 5 5 2 42=+ + + ++ + + + + +b l ( )( )( )( ) ( ) ( )s s s s ss s s s s s4 5316110 4 5 5 5 2 4=+ + + ++ + + + + +b b l lWeknowthatforasystemhavingmanypoles,nearnessofthepolestowards imaginary axis ins-plane dominates the nature of time response. So here time constant given by two poles which are nearest to imaginary axis.Poles nearest to imaginary axiss1 31=- , s612=-So, time constants secsec3612tt==)GATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSol . 70Option (A) is correct.Steady state error for a system is given byess ( ) ( )( )limG s H ssRs1 s 0=+ "Where input( ) Rss1=(unit step)( ) G ss s 153115=+ +b b l l( ) H s 1 = (unity feedback)Soess ( )( )1lims ss s115 145 s 0=++ +"b l 15 4515=+ 6015=%ess 6015100#= 25% =Sol . 71Option (C) is correct.Characteristic equation is given by( ) ( ) G s H s 1+ 0 =Here( ) H s 1 = (unity feedback)( ) G ss s 153115=+ +b b l lSo,1s s 153115++ +b b l l0 =( )( ) s s 15 1 45 + + + 0 =16 60 s s2+ + 0 =( 6)( ) s s 10 + + 0 =s6, =- 10 -Sol . 72Option (A) is correct.Given equation can be written as, dtd22w J dtdLJKLJKVa2b ww =- - +Here state variables are defined as, dtdwx1=wx2=So state equation isx1o JBxLJKxLJKVa 122=- - +x2o dtdw= x1=In matrix form xx12oo> H / / / BJ K LJ xxKLJV1 0 0a212=- -+ > > > H H H dtddtd22wwRTSSSSVXWWWWP ddtQVaw= + > HSo matrix P isGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in / / BJ K LJ1 02- -> HSol . 73Option (C) is correct.Characteristic equation of the system is given byGH 1+ 0 = ( )( ) s s sK12 4++ +0 =( )( ) s s s K 2 4 + + + 0 =s s s K 6 83 2+ + + 0 =Applying rouths criteria for stabilitys31 8s26 Ks1K648 -s0KSystem becomes unstable if K6480-= 48 K & =Sol . 74Option (A) is correct.Themaximumerrorbetweentheexactandasymptoticplotoccursatcorner frequency.Here exact gain(dB) at0.5a w =is given by( ) gain dB. a 05 w=log log Ka20 20 122w= - + ( . )log log Kaa20 20 105/22 1 2= - +; E . logK 20 096 = -Gain(dB) calculated from asymptotic plot at. a 05 w =islogK 20 =Error in gain (dB)20 (20 0.96) log log K K dB = - - . 096 =dBSimilarly exact phase angle at0.5a w =is.( ).ha 05q ww=tana1 w=--a k .tanaa 05 1=--b l2 .56 6 c =-Phase angle calculated from asymptotic plot at ( 0.5 ) a w =is22.5c -Error in phase angle. ( . ) 225 2656c =- -- . 49c =Sol . 75Option (B) is correct.Given block diagramGiven block diagram can be reduced asGATE SOLVED PAPER - EECONTROL SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inWhereG1 1ss1 31=+^^hh s 31=+G2 ss11121=+bbll s 121=+Further reducing the block diagram.( ) Ys( ) G GG G1 2 921 21 2=+ ( ) ( )( )s ss s1 2311219231121=++ ++ +b bb bl ll l ( )( ) s s 3 12 182=+ + + s s 15 5422=+ + ( 9)( 6) s s2=+ + s s27 19161=+ +a a k k***********