The state variable description of an LTI system is given by 3 1O. Statement for Linked Answer Questions 3 and 4 :

Transcription

1 CHAPTER 6 CONTROL SYSTEMS YEAR TO MARKS MCQ 6. The tate variable decription of an LTI ytem i given by Jxo N J a NJx N JN K O K OK O K O xo a x + u Kxo O K 3 a3 OKx O K 3 O L P L J PL P L P x N K O y _ i x Kx O 3 L P where y i the output and u i the input. The ytem i controllable for (A) a!, a, a! (B) a, a!, a! 3 3 (C) a, a!, a (D) a!, a!, a MCQ 6. The feedback ytem hown below ocillate at rad/ when (A) K and a.75 (B) K 3 and a.75 (C) K 4 and a.5 (D) K and a.5 MCQ 6.3 () Statement for Linked Anwer Quetion 3 and 4 : The tranfer function of a compenator i given a Gc () + a + b Gc i a lead compenator if (A) a, b (B) a 3, b (C) a 3, b (D) a 3, b MCQ 6.4 The phae of the above lead compenator i maximum at (A) rad/ (B) 3 rad/ (C) 6 rad/ (D) / 3 rad/ GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

2 PAGE 34 CONTROL SYSTEMS CHAP 6 YEAR ONE MARK MCQ 6.5 The frequency repone of a linear ytem Gjω ( ) i provided in the tubular form below Gjω ( ) Gj ( ω) 3c 4c 5c 6c 8c (A) 6 db and 3c (B) 6 db and 3c (C) c 6dB and 3c (D) 6dB and 3c MCQ 6.6 The teady tate error of a unity feedback linear ytem for a unit tep input i.. The teady tate error of the ame ytem, for a pule input rt () having a magnitude of and a duration of one econd, a hown in the figure i (A) (B). (C) (D) MCQ 6.7 An open loop ytem repreented by the tranfer function ( ) G () i ( + )( + 3) (A) Stable and of the minimum phae type (B) Stable and of the non minimum phae type (C) Untable and of the minimum phae type (D) Untable and of non minimum phae type YEAR TO MARKS MCQ 6.8 The open loop tranfer function G () of a unity feedback control ytem i given a K b + 3 l G () ( + ) From the root locu, at can be inferred that when K tend to poitive infinity, (A) Three root with nearly equal real part exit on the left half of the GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

3 CHAP 6 CONTROL SYSTEMS PAGE 35 -plane (B) One real root i found on the right half of the -plane (C) The root loci cro the jω axi for a finite value of KK! ; (D) Three real root are found on the right half of the -plane MCQ 6.9 A two loop poition control ytem i hown below The gain K of the Tacho-generator influence mainly the (A) Peak overhoot (B) Natural frequency of ocillation (C) Phae hift of the cloed loop tranfer function at very low frequencie ( ω " ) (D) Phae hift of the cloed loop tranfer function at very high frequencie ( ω " 3) YEAR MCQ 6. The frequency repone of G () ( + )( + ) Gjω ( ) plane (for < ω < 3) i TO MARKS plotted in the complex GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

4 PAGE 36 CONTROL SYSTEMS CHAP 6 MCQ 6. The ytem X AX+ Bu A, > B H > H i (A) Stable and controllable (B) Stable but uncontrollable (C) Untable but controllable (D) Untable and uncontrollable MCQ 6. The characteritic equation of a cloed-loop ytem i ( + )( + 3) k ( + ), k>.hich of the following tatement i true? (A) It root are alway real (B) It cannot have a breakaway point in the range < Re[] < (C) Two of it root tend to infinity along the aymptote Re[] (D) It may have complex root in the right half plane. YEAR 9 ONE MARK MCQ 6.3 The meaurement ytem hown in the figure ue three ub-ytem in cacade whoe gain are pecified a G, G,/ G3. The relative mall error aociated with each repective ubytem G, G and G 3 are ε, ε and ε 3. The error aociated with the output i : (A) ε + ε+ (B) ε 3 εε ε (C) ε + ε ε 3 (D) ε+ ε+ ε3 3 MCQ 6.4 The polar plot of an open loop table ytem i hown below. The cloed loop ytem i (A) alway table (B) marginally table (C) un-table with one pole on the RH -plane (D) un-table with two pole on the RH -plane GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

5 CHAP 6 CONTROL SYSTEMS PAGE 37 MCQ 6.5 MCQ 6.6 The firt two row of Routh tabulation of a third order equation are a follow Thi mean there are (A) Two root at! j and one root in right half -plane (B) Two root at! j and one root in left half -plane (C) Two root at! j and one root in right half -plane (D) Two root at! j and one root in left half -plane The aymptotic approximation of the log-magnitude v/ frequency plot of a ytem containing only real pole and zero i hown. It tranfer function i ( + 5) (A) ( + )( + 5 ) ( + 5) (C) ( + )( + 5 ) (B) (D) ( + 5) ( + )( + 5) 8( + 5) ( + )( + 5) YEAR 9 TO MARKS MCQ 6.7 The unit-tep repone of a unity feed back ytem with open loop tranfer function G ( ) K/(( + )( + )) i hown in the figure. The value of K i (A).5 (B) (C) 4 (D) 6 GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

6 PAGE 38 CONTROL SYSTEMS CHAP 6 MCQ 6.8 The open loop tranfer function of a unity feed back ytem i given by. G () ( e - )/. The gain margin of the i ytem i (A).95 db (B) 7.67 db (C).33 db (D) 3.9 db MCQ 6.9 MCQ 6. Common Data for Quetion 9 and : A ytem i decribed by the following tate and output equation dx() t 3x() t + x() t + u() t dt dx() t x( t) + u( t) dt yt () x() t when ut () i the input and yt () i the output The ytem tranfer function i (A) (B) (C) + 5 (D) The tate-tranition matrix of the above ytem i -3t e (A) -t -3t -tg (B) e -3t e -t e -3t -t G e + e e e -3t -t -3t 3t -t -3t (C) e e + e -t G (D) e e e -t G e e YEAR 8 ONE MARK MCQ 6. A function yt () atifie the following differential equation : dy() t + yt () δ() t dt where δ () t i the delta function. Auming zero initial condition, and denoting the unit tep function by ut (), yt () can be of the form (A) e t (B) e -t t (C) eu() t -t (D) e u() t YEAR 8 TO MARK MCQ 6. The tranfer function of a linear time invariant ytem i given a G () GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

7 CHAP 6 CONTROL SYSTEMS PAGE 39 The teady tate value of the output of the ytem for a unit impule input applied at time intant t will be (A) (B).5 (C) (D) MCQ 6.3 The tranfer function of two compenator are given below : ( + ) C, C + ( + ) ( + ) hich one of the following tatement i correct? (A) C i lead compenator and C i a lag compenator (B) C i a lag compenator and C i a lead compenator (C) Both C and C are lead compenator (D) Both C and C are lag compenator MCQ 6.4 The aymptotic Bode magnitude plot of a minimum phae tranfer function i hown in the figure : Thi tranfer function ha (A) Three pole and one zero (C) Two pole and two zero (B) Two pole and one zero (D) One pole and two zero MCQ 6.5 Figure how a feedback ytem where K > MCQ 6.6 The range of K for which the ytem i table will be given by (A) < K < 3 (B) < K < 39 (C) < K < 39 (D) K > 39 The tranfer function of a ytem i given a + + GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

8 PAGE 3 CONTROL SYSTEMS CHAP 6 The ytem i (A) An over damped ytem (C) A critically damped ytem (B) An under damped ytem (D) An untable ytem MCQ 6.7 Statement for Linked Anwer Quetion 7 and 8. The tate pace equation of a ytem i decribed by Xo AX+ Bu, Y CX where X i tate vector, u i input, Y i output and A, B, C [ ] G G The tranfer function G() of thi ytem will be (A) (B) + ( + ) ( ) (C) (D) ( ) ( + ) MCQ 6.8 A unity feedback i provided to the above ytem G () to make it a cloed loop ytem a hown in figure. For a unit tep input rt, () the teady tate error in the input will be (A) (B) (C) (D) 3 YEAR 7 ONE MARK MCQ 6.9 The ytem hown in the figure i (A) Stable (B) Untable (C) Conditionally table (D) Stable for input u, but untable for input u GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

9 CHAP 6 CONTROL SYSTEMS PAGE 3 YEAR 7 TO MARKS MCQ 6.3 If x Re[ G( jω)], and y Im[ G( jω)] then for ω " +, the Nyquit plot for G () /( + )( + ) i (A) x (B) x 34 / (C) x y 6 / (D) x y/ 3 MCQ 6.3 The ytem 9/ ( + )( + 9) i to be uch that it gain-croover frequency become ame a it uncompenated phae croover frequency and provide a 45c phae margin. To achieve thi, one may ue (A) a lag compenator that provide an attenuation of db and a phae lag of 45c at the frequency of 3 3 rad/ (B) a lead compenator that provide an amplification of db and a phae lead of 45c at the frequency of 3 rad/ (C) a lag-lead compenator that provide an amplification of db and a phae lag of 45c at the frequency of 3 rad/ (D) a lag-lead compenator that provide an attenuation of db and phae lead of 45c at the frequency of 3 rad/ MCQ 6.3 If the loop gain K of a negative feed back ytem having a loop tranfer function K ( + 3)/( + 8) i to be adjuted to induce a utained ocillation then (A) The frequency of thi ocillation mut be 4 3 rad/ (B) The frequency of thi ocillation mut be 4 rad/ (C) The frequency of thi ocillation mut be 4 or 4 3 rad/ (D) Such a K doe not exit MCQ 6.33 The ytem hown in figure below can be reduced to the form GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

10 PAGE 3 CONTROL SYSTEMS CHAP 6 with (A) X c + c, Y /( + a + a), Z b + b (B) X, Y ( c+ c)/( + a+ a), Z b+ b (C) X c + c, Y ( b + b )/( + a + a ), Z (D) X c + c, Y /( + a + a), Z b + b MCQ 6.34 Conider the feedback ytem hown below which i ubjected to a unit tep input. The ytem i table and ha following parameter K 4, K, 5 and ξ 7..The teady tate value of Z i p i ω (A) (B).5 (C). (D) Data for Q.35 and Q.36 are given below. Solve the problem and chooe the correct anwer. R-L-C circuit hown in figure MCQ 6.35 MCQ 6.36 For a tep-input e i, the overhoot in the output e will be (A), ince the ytem i not under damped (B) 5 % (C) 6 % (D) 48 % If the above tep repone i to be oberved on a non-torage CRO, then it would be bet have the e i a a (A) Step function GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

11 CHAP 6 CONTROL SYSTEMS PAGE 33 (B) Square wave of 5 Hz (C) Square wave of 3 Hz (D) Square wave of. KHz YEAR 6 ONE MARK MCQ 6.37 For a ytem with the tranfer function 3( ) H (), 4 + the matrix A in the tate pace form Xo AX+ Bu i equal to R V R V S S (A) S (B) S S 4 S 4 T X T X R V R V S S (C) S3 (D) S S 4 S 4 T X T X YEAR 6 TO MARKS MCQ 6.38 Conider the following Nyquit plot of loop tranfer function over ω to ω 3. hich of thee plot repreent a table cloed loop ytem? GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

12 PAGE 34 CONTROL SYSTEMS CHAP 6 (A) () only (B) all, except () (C) all, except (3) (D) () and () only ( + jω) MCQ 6.39 The Bode magnitude plot Hj ( ω) ( + jω)( + jω) 4 i MCQ 6.4 A cloed-loop ytem ha the characteritic function ( 4)( + ) + K( ). It root locu plot againt K i GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

13 CHAP 6 CONTROL SYSTEMS PAGE 35 YEAR 5 ONE MARK MCQ 6.4 MCQ 6.4 A ytem with zero initial condition ha the cloed loop tranfer function. T () + 4 ( + )( + 4) The ytem output i zero at the frequency (A).5 rad/ec (B) rad/ec (C) rad/ec (D) 4 rad/ec Figure how the root locu plot (location of pole not given) of a third order ytem whoe open loop tranfer function i MCQ 6.43 (A) K (B) K 3 ( + ) (C) K (D) K ( + ) ( ) The gain margin of a unity feed back control ytem with the open loop ( ) tranfer function G () + i (A) (B) (C) (D) 3 GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

14 PAGE 36 CONTROL SYSTEMS CHAP 6 YEAR 5 TO MARKS MCQ 6.44 A unity feedback ytem, having an open loop gain K( ) GH () (), ( + ) become table when (A) K > (B) K > (C) K < (D) K < MCQ 6.45 hen ubject to a unit tep input, the cloed loop control ytem hown in the figure will have a teady tate error of (A). (B). 5 (C) (D).5 MCQ 6.46 In the GH-plane, () () the Nyquit plot of the loop tranfer function e GH () () π -5. pae through the negative real axi at the point (A) (.5, j) (B) (.5, j) (C) (D).5 MCQ 6.47 If the compenated ytem hown in the figure ha a phae margin of 6c at the croover frequency of rad/ec, then value of the gain K i (A).366 (B).73 (C).366 (D).738 Data for Q.48 and Q.49 are given below. Solve the problem and chooe the correct anwer. A tate variable ytem Xo () t X() t + u() t 3 G G with the initial condition X() [, 3] T and the unit tep input ut () ha GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

15 CHAP 6 CONTROL SYSTEMS PAGE 37 MCQ 6.48 MCQ 6.49 The tate tranition matrix t 3 ( e 3 t 3t ) 3 ( e e ) (A) 3t G (B) > t H e e 3 t 3t t 3 ( e e ) ( e ) (C) > 3t H (D) > t H e e The tate tranition equation t t e (A) X() t - t e -t G (B) X() t - -3t G e 3e 3t t e (C) X() t t t e -3t G (D) X() t -3 -t G 3e e YEAR 4 ONE MARK MCQ 6.5 The Nyquit plot of loop tranfer function GH () () of a cloed loop control ytem pae through the point (, j ) in the GHplane. () () The phae margin of the ytem i (A) c (B) 45c (C) 9c (D) 8c MCQ 6.5 Conider the function, F () 5 ( + 3+ ) where F () i the Laplace tranform of the of the function ft. () The initial value of ft () i equal to 5 (A) 5 (B) (C) 3 5 (D) MCQ 6.5 For a tachometer, if θ () t i the rotor diplacement in radian, et () i the output voltage and K t i the tachometer contant in V/rad/ec, then the E () tranfer function, will be Q () (A) K t (B) Kt (C) K t (D) K t YEAR 4 TO MARKS 3 MCQ 6.53 For the equation, the number of root in the left half of -plane will be (A) Zero (B) One (C) Two (D) Three GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

16 PAGE 38 CONTROL SYSTEMS CHAP 6 MCQ 6.54 For the block diagram hown, the tranfer function C () R () i equal to (A) (C) (B) + + (D) + + MCQ 6.55 The tate variable decription of a linear autonomou ytem i, Xo AX where X i the two dimenional tate vector and A i the ytem matrix given by A G. The root of the characteritic equation are (A) and + (B) j and +j (C) and (D) + and + MCQ 6.56 The block diagram of a cloed loop control ytem i given by figure. The value of K and P uch that the ytem ha a damping ratio of.7 and an undamped natural frequency ω n of 5 rad/ec, are repectively equal to (A) and.3 (B) and. (C) 5 and.3 (D) 5 and. MCQ 6.57 The unit impule repone of a econd order under-damped ytem tarting -6t from ret i given by ct (). 5e in 8t, t$. The teady-tate value of the unit tep repone of the ytem i equal to (A) (B).5 (C).5 (D). MCQ 6.58 In the ytem hown in figure, the input xt () int. In the teady-tate, the repone yt () will be (A) in( t 45 c) (B) in( t + 45 c) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

17 CHAP 6 CONTROL SYSTEMS PAGE 39 (C) in( t 45c) (D) in( t + 45 c) MCQ 6.59 The open loop tranfer function of a unity feedback control ytem i given a G () a +. The value of a to give a phae margin of 45c i equal to (A).4 (B).44 (C).84 (D).4 YEAR 3 ONE MARK MCQ 6.6 MCQ 6.6 A control ytem i defined by the following mathematical relationhip dx dx t x ( e - ) dt dt The repone of the ytem a t " 3 i (A) x 6 (B) x (C) x 4. (D) x A lead compenator ued for a cloed loop controller ha the following tranfer function K( + a ) ( + ) b For uch a lead compenator (A) a < b (B) b < a (C) a > Kb (D) a < Kb MCQ 6.6 A econd order ytem tart with an initial condition of 3 G without any external input. The tate tranition matrix for the ytem i given by e -t -tg e. The tate of the ytem at the end of econd i given by (A). G (B). 368 G (C). 736 G (D). G YEAR 3 TO MARKS MCQ 6.63 A control ytem with certain excitation i governed by the following mathematical equation dx dx 4t 5t + + x + 5e + e dt dt 8 GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

18 PAGE 33 CONTROL SYSTEMS CHAP 6 The natural time contant of the repone of the ytem are (A) ec and 5 ec (B) 3 ec and 6 ec (C) 4 ec and 5 ec (D) /3 ec and /6 ec MCQ 6.64 The block diagram hown in figure give a unity feedback cloed loop control ytem. The teady tate error in the repone of the above ytem to unit tep input i (A) 5% (B).75 % (C) 6% (D) 33% MCQ 6.65 The root of the cloed loop characteritic equation of the ytem hown above (Q-5.55) MCQ 6.66 MCQ 6.67 (A) and 5 (B) 6 and (C) 4 and 5 (D) 6 and The following equation define a eparately excited dc motor in the form of a differential equation d ω + B dω + K ω K Va dt J dt LJ LJ The above equation may be organized in the tate-pace form a follow R d V S ω dω S dt d P dt S ω > H + QV a ω S dt here Tthe P X matrix i given by B J K K B LJ LJ J (A) G (B) G (C) K B G (D) B G K LJ J The loop gain GH of a cloed loop ytem i given by the following expreion K ( + )( + 4) The value of K for which the ytem jut become untable i GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at: J LJ

19 CHAP 6 CONTROL SYSTEMS PAGE 33 (A) K 6 (B) K 8 (C) K 48 (D) K 96 MCQ 6.68 The aymptotic Bode plot of the tranfer function K/[ + ( / a)] i given in figure. The error in phae angle and db gain at a frequency of ω.5a are repectively (A) 4.9c,.97 db (C) 4.9c, 3 db (B) 5.7c, 3 db (D) 5.7c,.97 db MCQ 6.69 The block diagram of a control ytem i hown in figure. The tranfer function G () Y ()/ U () of the ytem i (A) (C) 8 ` + j` + 3j 7 ` + j` + 9 j (B) (D) 7 ` + 6 j` + 9 j 7 ` + 9 j` + 3j YEAR ONE MARK MCQ 6.7 The tate tranition matrix for the ytem X o AX with initial tate X () i (A) ( I A) - GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

20 PAGE 33 CONTROL SYSTEMS CHAP 6 At (B) e X() (C) Laplace invere of [( I A) - ] (D) Laplace invere of [( I A) - X()] YEAR TO MARKS 3 MCQ 6.7 For the ytem Xo X+ u 5 G G, which of the following tatement i true? (A) The ytem i controllable but untable (B) The ytem i uncontrollable and untable (C) The ytem i controllable and table (D) The ytem i uncontrollable and table MCQ 6.7 A unity feedback ytem ha an open loop tranfer function, G () K. The root locu plot i MCQ 6.73 MCQ 6.74 The tranfer function of the ytem decribed by dy dy + du + u dt dt dt with u a input and y a output i ( + ) ( + ) (A) (B) ( + ) ( + ) (C) (D) ( + ) ( + ) For the ytem Xo X+ u 4 G G ; Y 84 B X, with u a unit impule and with zero initial tate, the output y, become (A) e t (B) 4e t GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

21 CHAP 6 CONTROL SYSTEMS PAGE 333 (C) e 4 t (D) e 4 4t MCQ 6.75 MCQ 6.76 MCQ 6.77 The eigen value of the ytem repreented by R V S Xo S S X are (A),,, S (B),,, (C),,, T X (D),,, *A ingle input ingle output ytem with y a output and u a input, i decribed by dy dy + + y 5 du 3u dt dt dt for an input ut () with zero initial condition the above ytem produce the ame output a with no input and with initial condition dy( ) 4, y( ) dt input ut () i (A) δ() t e u() t (/)t 35 (B) () t e u() t δ 3t 7 (/)t (C) e u() t 5 35 (D) None of thee *A ytem i decribed by the following differential equation dy dy + - y ute () dt dt t dy the tate variable are given a x y and x b y e dt l t, the tate varibale repreentation of the ytem i xo t e x (A) > t ut () xo H > H> + e x H > H xo x (B) > ut () xo H > + H> x H > H xo t e x (C) > ut () xo H > + H> x H > H (D) none of thee Common Data Quetion Q.78-8*. The open loop tranfer function of a unity feedback ytem i given by ( + α) G () ( + )( + ) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

22 PAGE 334 CONTROL SYSTEMS CHAP 6 MCQ 6.78 MCQ 6.79 Angle of aymptote are (A) 6 c, c,3c (B) 6c, 8c, 3c (C) 9c, 7c, 36c (D) 9c, 8c, 7c Intercept of aymptote at the real axi i (A) 6 (B) 3 (C) 4 (D) 8 MCQ 6.8 Break away point are (A). 56, (B)., (C). 56, (D). 56, YEAR ONE MARK MCQ 6.8 The polar plot of a type-, 3-pole, open-loop ytem i hown in Figure The cloed-loop ytem i (A) alway table (B) marginally table (C) untable with one pole on the right half -plane (D) untable with two pole on the right half -plane. 3 MCQ 6.8 Given the homogeneou tate-pace equation xo x G the teady tate value of x lim x() t, given the initial tate value of x() T t " 3 8 B i 3 (A) x G (B) x G (C) x 3 G (D) x G 3 YEAR TO MARKS MCQ 6.83 The aymptotic approximation of the log-magnitude veru frequency plot GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

23 CHAP 6 CONTROL SYSTEMS PAGE 335 of a minimum phae ytem with real pole and one zero i hown in Figure. It tranfer function i ( + 5) (A) ( + )( + 5 ) ( + 5) (C) ( + )( + 5) ( + 5) (B) ( ) + ( + 5 ) 5( + 5) (D) ( + )( + 5) Common Data Quetion Q.84-87*. A unity feedback ytem ha an open-loop tranfer function of G () ( + ) MCQ 6.84 Determine the magnitude of Gjω ( ) in db at an angular frequency of ω rad/ec. (A) db (B) db (C) db (D) db MCQ 6.85 MCQ 6.86 The phae margin in degree i (A) 9c (B) c (C) c (D) 9c The gain margin in db i (A) 3.97 db (B) 6. db (C) db (D) None of thee MCQ 6.87 The ytem i (A) Stable (C) Marginally table (B) Un-table (D) can not determined MCQ 6.88 *For the given characteritic equation K+ K The root locu of the ytem a K varie from zero to infinity i GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

24 PAGE 336 CONTROL SYSTEMS CHAP 6 ************ GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

25 CHAP 6 CONTROL SYSTEMS PAGE 337 SOLUTION SOL 6. Option (D) i correct. General form of tate equation are given a xo Ax+ Bu yo Cx+ Du For the given problem R a V R S V S A S a, B S Sa 3 S T X T X R a VR V R V S S S AB S as Sa Sa 3 S S TR XT aa X VT R XV R aa V S S S AB Saa 3 S S S aa 3 S S T XT X T X For controllability it i neceary that following matrix ha a tank of n 3. R aa V S U 6 B : AB : A B@ S a S T X So, a! aa! & a! a 3 may be zero or not. SOL 6. Option (A) i correct. K ( + ) Y () 3 [ R ( ) Y ( )] + a + + K ( + ) K ( + ) Y (); a + + E 3 R () + a Y ()[ + a+ ( + k) + ( + k)] K ( + ) R ( ) Y () K ( + ) Tranfer Function, H () 3 R () + a + ( + k) + ( + k) Routh Table : GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

26 PAGE 338 CONTROL SYSTEMS CHAP 6 For ocillation, a( + K) ( + K) a a K K + + Auxiliary equation A () a + ( k + ) k + a k + ( k ) ( k + ) ( k + ) + j k+ jω j k+ ω k + (Ocillation frequency) k and a SOL 6.3 Option (A) i correct. GC () + a jω + a + b jω + b Phae lead angle, φ tan ω tan ω a a k a b k J ω ωn tan K a b O tan K + ω O L ab P For phae-lead compenation φ > b a > ω( b a) c ab + ω m b > a Note: For phae lead compenator zero i nearer to the origin a compared to pole, o option (C) can not be true. SOL 6.4 Option (A) i correct. φ tan ω a tan a k ω a b k GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

27 CHAP 6 CONTROL SYSTEMS PAGE 339 dφ / / dω a b ω ω + a + a k a b k + ω a + ω ab b b a a b ω ω ab b a b l ab # rad/ec SOL 6.5 Option (A) i correct. Gain margin i imply equal to the gain at phae cro over frequency ( ω p ). Phae cro over frequency i the frequency at which phae angle i equal to 8c. From the table we can ee that + Gj ( ω p) 8c, at which gain i.5. GM log e Gj ( ωp) o logb.5 l 6 db Phae Margin i equal to 8c plu the phae angle φ g at the gain cro over frequency ( ω g ). Gain cro over frequency i the frequency at which gain i unity. From the table it i clear that Gj ( ω ), at which phae angle i 5c g φ PM 8 c + + Gj ( ω g ) 8 5 3c SOL 6.6 Option (A) i correct. e know that teady tate error i given by R() e lim " + G () where R ()" input G ()" open loop tranfer function For unit tep input R () So e + G( ) G() 9 b l lim. " + G () Given input rt () [ μ( t) μ( t )] or R () : e D So teady tate error e : D GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

28 PAGE 34 CONTROL SYSTEMS CHAP 6 el ( e ) # lim " + G () ( e ) + 9 SOL 6.7 Option (B) i correct. Tranfer function having at leat one zero or pole in RHS of -plane i called non-minimum phae tranfer function. G () ( + )( + 3) In the given tranfer function one zero i located at (RHS), o thi i a non-minimum phae ytem. Pole, 3, are in left ide of the complex plane, So the ytem i table SOL 6.8 Option (A) i correct. K b + 3 l G () ( + ) Step for plotting the root-locu () Root loci tart at, and () n > m, therefore, number of branche of root locu b 3 (3) Angle of aymptote i given by ( q + ) 8c, q, n m ( (I) # + ) 8c 9c ( 3 ) ( (II) # + ) 8c 7c ( 3 ) (4) The two aymptote interect on real axi at centroid / Pole / Zeroe b 3 l x n m 3 3 (5) Between two open-loop pole and there exit a break away point. ( + ) K + b 3 l dk d Root locu i hown in the figure GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

29 CHAP 6 CONTROL SYSTEMS PAGE 34 Three root with nearly equal part exit on the left half of -plane. SOL 6.9 Option (A) i correct. The ytem may be reduced a hown below Y () R () ( + + K) + + ( + K) + ( + + K) Thi i a econd order ytem tranfer function, characteritic equation i + ( + K) + Comparing with tandard form + ξωn+ ωn e get ξ + K Peak overhoot M p e πξ/ ξ So the Peak overhoot i effected by K. SOL 6. Option (A) i correct. Given G () ( + )( + ) Gjω ( ) jω( jω+ )( jω+ ) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

30 PAGE 34 CONTROL SYSTEMS CHAP 6 Gjω ( ) ω ω + ω Gj ( ω ) 9 tan c ( ω) tan ( ω/ ) In nyquit plot For ω, Gj ( ω) 3 + Gj ( ω ) 9c For ω 3, Gj ( ω) + Gj ( ω ) 9c 9c 9c 7c Interection at real axi Gjω ( ) jω( jω+ )( jω+ ) jω( ω + j3ω+ ) At real axi So, Im[ Gjω ( )] ω( ω) 4 9ω + ω ( ω) 3ω jω( ω) # 3ω + jω( ω) 3ω jω( ω) 3ω jω( ω) 4 9ω + ω ( ω) 3ω jω( ω) 4 9ω + ω ( ω) 9ω + ω ( ω) ω & ω rad/ec At ω rad/ec, magnitude repone i Gj ( ω) < at ω SOL 6. Option (C) i correct. Stability : Eigen value of the ytem are calculated a A λi λ A λi > H > λ H λ > λ H A λi ( λ)( λ) # & λ, λ, Since eigen value of the ytem are of oppoite ign, o it i untable Controllability : A > H, B >H AB >H GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

31 CHAP 6 CONTROL SYSTEMS PAGE 343 [ BAB : ] > H 6 BAB Y So it i controllable. SOL 6. Option (C) i correct. Given characteritic equation ( + )( + 3) + K ( + ) ; K > ( ) + K( + ) (3 + K) + K From Routh tabulation method K 4 K 43 ( + K) K() K > K There i no ign change in the firt column of routh table, o no root i lying in right half of -plane. For plotting root locu, the equation can be written a K ( + ) + ( + )( + 3) Open loop tranfer function G () K ( + ) ( + )( + 3) Root locu i obtained in following tep:. No. of pole n 3, at, and 3. No. of Zeroe m, at 3. The root locu on real axi lie between and, between 3 and. 4. Breakaway point lie between open loop pole of the ytem. Here breakaway point lie in the range < Re[] <. 5. Aymptote meet on real axi at a point C, given by / real part of pole / real part of zeroe C n m ( 3) ( ) 3 GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

32 PAGE 344 CONTROL SYSTEMS CHAP 6 A no. of pole i 3, o two root loci branche terminate at infinity along aymptote Re() SOL 6.3 SOL 6.4 Option (D) i correct. Overall gain of the ytem i written a G GG G 3 e know that for a quantity that i product of two or more quantitie total percentage error i ome of the percentage error in each quantity. o error in overall gain G i 3 G ε + ε+ ε 3 Option (D) i correct. From Nyquit tability criteria, no. of cloed loop pole in right half of -plane i given a Z P N P " No. of open loop pole in right half -plane N " No. of encirclement of (, j) Here N (` encirclement i in clockwie direction) P (` ytem i table) So, Z ( ) Z, Sytem i untable with -pole on RH of -plane. SOL 6.5 Option (D) i correct. Given Routh tabulation So the auxiliary equation i given by, GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

33 CHAP 6 CONTROL SYSTEMS PAGE 345! j From table we have characteritic equation a ( + ) + ( + ) ( + )( + ),! j SOL 6.6 Option (B) i correct. Since initial lope of the bode plot i 4 db/decade, o no. of pole at origin i. Tranfer function can be written in following tep:. Slope change from 4 db/dec. to 6 db/dec. at ω rad/ec., o at ω there i a pole in the tranfer function.. Slope change from 6 db/dec to 4 db/dec at ω 5 rad/ec., o at thi frequency there i a zero lying in the ytem function. 3. The lope change from 4 db/dec to 6 db/dec at ω 3 5 rad/ec, o there i a pole in the ytem at thi frequency. Tranfer function K ( + 5) T () ( + )( + 5) Contant term can be obtained a. Tj ( ω) 8 at ω. K() 5 So, 8 log (.) 5 K therefore, the tranfer function i T () # ( + 5) ( + )( + 5) SOL 6.7 Option (D) i correct. From the figure we can ee that teady tate error for given ytem i e Steady tate error for unity feed back ytem i given by R() e lim " + G ( ) G ^ h lim " > + K H ; R () (unit tep input) ( + )( + ) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

34 PAGE 346 CONTROL SYSTEMS CHAP K So, e.5 + K K.5 +.5K K SOL 6.8 Option (D) i correct. Open loop tranfer function of the figure i given by, G () Gjω ( ) e.. e jω j ω Phae cro over frequency can be calculated a, + Gj ( ω p ) 8c. 8 b ω p # 9c π l 8c. ω 8 p # π c 9c. ω p 9 c # π 8 c ω p 5. 7 rad/ec So the gain margin (db) log e Gj ( ω p ) o log b5. 7 l log db > H SOL 6.9 Option (C) i correct. Given ytem equation dx() t dt dx() t dt 3 x ( t) + x ( t) + u( t) x () t + u() t yt () x() t Taking Laplace tranform on both ide of equation. X () 3X() + X() + U() ( + 3) X( ) X() + U()...() Similarly X () X () + U() ( + ) X ( ) U ()...() GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

35 CHAP 6 CONTROL SYSTEMS PAGE 347 From equation () & () U () ( + 3) X( ) + U () + U () + ( + ) ( + 5) X () + 3; + E U () ( + )( + 3 ) From output equation, Y () X() ( + 5) So, Y () U () ( + )( + 3 ) Sytem tranfer function Y () ( + 5) ( + 5) T.F U () ( + )( + 3) SOL 6. Option (B) i correct. Given tate equation in matrix form can be written a, xo 3 x > xo H > u() t H> + x H > H dx() t AX() t + Bu() t dt State tranition matrix i given by φ () t L 6 Φ () ( I A) ( I A) 3 > H > H ( I A) 3 > + + H ( I A) + ( 3 )( ) > H R V S ( 3) ( 3)( ) So Φ () ( I A) S S S ( + ) T X 3t t 3t e e e φ() t L [ Φ()] > t H e SOL 6. Option (D) i correct. Given differential equation for the function dy() t + yt () δ() t dt GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

36 PAGE 348 CONTROL SYSTEMS CHAP 6 Taking Laplace on both the ide we have, Y() + Y() ( + ) Y( ) Y () + Taking invere Laplace of Y () yt () e t u() t, t > SOL 6. Option (A) i correct. Given tranfer function G () Input rt () δ( t ) R () L[ δ( t )] e Output i given by Y () RG () () e Steady tate value of output lim yt () t " 3 lim Y() lim e " " SOL 6.3 SOL 6.4 Option (A) i correct. For C Phae i given by θ C tan ( ω) tan ω ak Jω ω N tan K O tan 9ω K + ω c > O + ω m (Phae lead) Similarly for C, phae i L P θ C tan ω a tan ( ω) k J ω ωn tan K O tan 9ω < K + ω c O + ω m (Phae lag) L P Option (C) i correct. From the given bode plot we can analyze that:. Slope 4 db/decade" pole. Slope db/decade (Slope change by + db/decade)" Zero 3. Slope db/decade (Slope change by + db/decade)" Zero So there are pole and zeroe in the tranfer function. GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

37 CHAP 6 CONTROL SYSTEMS PAGE 349 SOL 6.5 Option (C) i correct. Characteritic equation for the ytem + K ( + 3)( + ) ( + 3)( + ) + K K Applying Routh tability criteria K ( 3 # 3) K 3 K For tability there hould be no ign change in firt column So, 39 K > & K < 39 K > < K < 9 SOL 6.6 Option (C) i correct. Given tranfer function i H) () + + Characteritic equation of the ytem i given by or + + ω & ω rad/ec. n n n ξω ξ # ( ξ ) o ytem i critically damped. SOL 6.7 Option (D) i correct. State pace equation of the ytem i given by, X o AX+ Bu Y CX Taking Laplace tranform on both ide of the equation. X () AX() + BU() ( I A) X( ) BU() X () ( I A) BU( ) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

38 PAGE 35 CONTROL SYSTEMS CHAP 6 ` Y() CX() So Y () CI ( A) BU( ) Y() T.F CI ( A) B U() ( I A) > H > H > + H R V S ( I A) + ( ) ( + ) > H S + S S ( + ) T X Tranfer function R V R V S S ( + ) G () CI [ A] B 8 S S > ( ) B H S + 8 BS S ( + ) S ( + ) T X T X ( + ) SOL 6.8 SOL 6.9 Option (A) i correct. Steady tate error i given by, R() e lim " + GH ( ) ( ) G Here R () L[ rt ( )] (Unit tep input) G () ( + ) H () (Unity feed back) R V S b l So, e lims ( ) lim " S + + ( + ) + G " S ( + ) Option (D) i correct. T X For input u, the ytem i ( u ) Sytem repone i GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

39 CHAP 6 CONTROL SYSTEMS PAGE 35 ( ) ( + ) ( ) H () ( ) + ( + 3) ( + ) ( ) Pole of the ytem i lying at 3 (negative -plane) o thi i table. For input u the ytem i ( u ) Sytem repone i ( ) H () ( ) + ( )( + ) ( + ) ( )( + 3) One pole of the ytem i lying in right half of -plane, o the ytem i untable. SOL 6.3 Option (B) i correct. Given function i. G () Gjω ( ) ( + )( + ) jω( + jω)( + jω) By implifying Gjω ( ) jω j j j j ω j j ω c ω # ωmc + ω # ωmc + jω # jωm jω jω jω c ω mc + ω mc 4 + ω m jω( ω j3ω) ω ( + ω)( 4+ ω) 3ω jωω ( ) + ω ( + ω)( 4+ ω) ω ( + ω)( 4+ ω) Gjω ( ) x+ iy x Re[ Gj ( ω)] ω " 4 4 # SOL 6.3 Option (D) i correct. Let repone of the un-compenated ytem i HUC () 9 ( + )( + 9) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

40 PAGE 35 CONTROL SYSTEMS CHAP 6 Repone of compenated ytem. HC () 9 GC() ( + )( + 9) here GC () " Repone of compenator Given that gain-croover frequency of compenated ytem i ame a phae croover frequency of un-compenated ytem So, ( ω g ) compenated ( ω p ) uncompenated 8c + H UC ( jω p ) ωp 8c 9 c tan ( ωp) tan a 9 k J ωp ωp + N 9c tan K 9 O ω p K O L 9 P ω 9 p ω p 3 rad/ec. So, ( ω g ) compenated 3 rad/ec. At thi frequency phae margin of compenated ytem i φ 8 c + + H ( ω ) PM C j g 45c 8c 9 c tan ( ω ) tan ( ω / 9) + + G ( jω ) g g C g 45c 8c 9 c tan ( 3) tan ( / 3) + + GC( jωg) R 3 V S + 45c 9 tan 3 c S + + GC( jωg) 3 S b 3 l 45c 9c 9 c + T + G ( ω X ) + G C ( jω g ) 45c The gain cro over frequency of compenated ytem i lower than uncompenated ytem, o we may ue lag-lead compenator. At gain cro over frequency gain of compenated ytem i unity o. H C ( jω g ) 9 GC( jωg) ωg ωg+ ωg+ 8 G C ( jω g ) # in db G C ( ω g ) logb l C j g GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

41 CHAP 6 CONTROL SYSTEMS PAGE 353 db (attenuation) SOL 6.3 Option (B) i correct. Characteritic equation for the given ytem, K ( + 3) + ( + 8) ( + 8) + K( + 3) + (6 + K) + ( K) By applying Routh criteria K 6 + K K For ytem to be ocillatory 6 + K & K 6 Auxiliary equation A () + (64+ 3 K) & # ( 6) & jω4j ω 4 rad/ec SOL 6.33 Option (D) i correct. From the given block diagram we can obtain ignal flow graph of the ytem. Tranfer function from the ignal flow graph i written a cp cp + T.F ( c+ c) P a a Pb Pb + + ( + a+ a) P( b+ b) ( c+ c) P ^ + a+ ah Pb ( + b) + a + a from the given reduced form tranfer function i given by T.F XYP YPZ by comparing above two we have X ( c+ c) Y + a + a Z ( b + b ) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

42 PAGE 354 CONTROL SYSTEMS CHAP 6 SOL 6.34 SOL 6.35 Option (A) i correct. For the given ytem Z i given by Z E () K i here E ()" teady tate error of the ytem Here E () Input R () So, G () R() lim " + GH () () (Unit tep) Ki K ω b + p le o + ξω+ ω H () (Unity feed back) Z R V S b l lim K i S Ki K ω b l " S + + S b p l ( + ξω + ω ) T X lim Ki K " > + ( Ki+ Kp) ω H i Ki ( + ξω+ ω ) Option (C) i correct. Sytem repone of the given circuit can be obtained a. e () b C l H () ei() R+ L+ b C l H () b LC l LC + RC + + R + L LC Characteritic equation i given by, + R + L LC Here natural frequency ξω n Damping ratio R L ξ ω n R L LC GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at: LC R 3 Here ξ #.5 6 (under damped) # L C

43 CHAP 6 CONTROL SYSTEMS PAGE 355 So peak overhoot i given by πξ % peak overhoot ξ e # π# 5. e (.) 5 # 6% SOL 6.36 Option ( ) i correct. SOL 6.37 SOL 6.38 Option (B) i correct. In tandard form for a characteritic equation give a n n + a a + a n in it tate variable repreentation matrix A i given a R V S g S g A S h h h h h S S a a a g an T X Characteritic equation of the ytem i 4 + So, a 4, a, a R V R S S A S S S a a a S T X T V 4 X Option (A) i correct. In the given option only in option (A) the nyquit plot doe not encloe the unit circle (, j), So thi i table. SOL 6.39 Option (A) i correct. Given function i, Hjω ( ) 4 ( + jω) ( + jω)( + jω) Function can be rewritten a, 4 ( + jω). ( + jω) Hjω ( ) j ω j 4 ω 9 + C 9 + C j ω j ω a + ka + k The ytem i type, So, initial lope of the bode plot i db/decade. Corner frequencie are ω rad/ec ω rad/ec ω 3 rad/ec A the initial lope of bode plot i db/decade and corner frequency ω GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

44 PAGE 356 CONTROL SYSTEMS CHAP 6 rad/ec, the Slope after ω rad/ec or logω i( + ) + db/dec. After corner frequency ω rad/ec or log ω, the Slope i ( + ) db/dec. Similarly after ω 3 rad/ec or log ω, the lope of plot i ( # ) 4 db/dec. Hence (A) i correct option. SOL 6.4 Option (B) i correct. Given characteritic equation. ( 4)( + ) + K( ) K ( ) or + ( 4)( + ) So, the open loop tranfer function for the ytem. K ( ) G () ( )( + )( + ), no. of pole n 3 no of zeroe m Step for plotting the root-locu () Root loci tart at,, () n > m, therefore, number of branche of root locu b 3 (3) Angle of aymptote i given by ( q + ) 8c, q, n m ( (I) # + ) 8c 9c ( 3 ) ( (II) # + ) 8c 7c ( 3 ) (4) The two aymptote interect on real axi at / Pole / Zeroe ( + ) ( ) x n m 3 (5) Between two open-loop pole and there exit a break away point. ( 4)( + ) K ( ) dk d 5. SOL 6.4 Option (C) i correct. Cloed loop tranfer function of the given ytem i, T () + 4 ( + )( + 4) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

45 CHAP 6 CONTROL SYSTEMS PAGE 357 Tjω ( ) ( jω) + 4 ( jω+ )( jω+ 4) If ytem output i zero 4 ω Tjω ( ) ^jω+ ( h jω+ 4) 4 ω ω 4 & ω rad/ec SOL 6.4 Option (A) i correct. From the given plot we can ee that centroid C (point of interection) where aymptote interect on real axi) i So for option (a) G () K 3 / Pole Zero Centroid n m 3 / SOL 6.43 SOL 6.44 Option (A) i correct. Open loop tranfer function i. ( ) G () + jω + Gjω ( ) ω Phae croover frequency can be calculated a. + Gj ( ω p ) 8c tan ( ωp) 8c ω p Gain margin of the ytem i. G.M ωp Gj ( ωp) ωp + ωp + ω Option (C) i correct. Characteritic equation for the given ytem + GH () () ( ) + K ( + ) ( + ) + K( ) GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at: p

46 PAGE 358 CONTROL SYSTEMS CHAP 6 ( K) + ( + K) For the ytem to be table, coefficient of characteritic equation hould be of ame ign. K >, K + > K <, K > < K < K < SOL 6.45 Option (C) i correct. In the given block diagram Steady tate error i given a e lim E() " E () R () Y () Y () can be written a Y () R () Y () 3 R () :", D + Y () 6 ; + ( + ) E R () 6 ; ( + ) E R () 6 Y () 6 ; ( + ) + E ; ( + ) E ( 6 ) Y () R () ( ) ( 6 ) So, E () R () R () ( + + 6) R () + 4 ; E For unit tep input R () Steady tate error e lim E() e " ( 4 ) lim + " ( + + 6) G GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

47 CHAP 6 CONTROL SYSTEMS PAGE 359 SOL 6.46 Option (B) i correct. hen it pae through negative real axi at that point phae angle i 8c So + Gj ( ω) Hj ( ω ) 8c.5jω π π.5jω π j.5ω j ω π π #. 5 jω π Put π in given open loop tranfer function we get GH () () πe 5 π 5. # π π. So it pae through (.5, j) SOL 6.47 Option (C) i correct. Open loop tranfer function of the ytem i given by. GH () () ( K+.366 ) ; ( + ) E K+ j. 366ω Gj ( ω) Hj ( ω ) jω( jω+ ) Phae margin of the ytem i given a φ 6c 8 c + + Gj ( ω ) Hj ( ω ) PM g g here So, ω g " gain cro over frequency rad/ec. 366ωg 6c 8c+ tan b 9c tan K l 9c + tan. 366 b tan K l 9c 45c+ tan 5c tan. 366 K tan 5c. 366 b K l. 366 b K l () ( ω ) g K GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

48 PAGE 36 CONTROL SYSTEMS CHAP 6 SOL 6.48 Option (A) i correct. Given tate equation. X o () t > X() t + u() t 3 H > H Here A >,B 3 H > H State tranition matrix i given by, φ () t L [( I A) ] [ I A] > H > H > + 3 H R V S [ I A] + 3 ( 3) ( + 3 ) > H S + S S ( + 3) φ () t L [( I A) ] 3t 3 ( e ) > 3t H e T X SOL 6.49 Option (C) i correct. State tranition equation i given by X () Φ() X() + Φ() BU() Here Φ () " tate tranition matrix R V S ( 3) Φ () S + S S ( + 3) X () T " initial condition X () > 3 H B >H R V R V S ( + 3) S ( 3) So X () S + S + S > 3 H S > H S ( + 3) S + 3 TR X T X 3 V R S + V ( 3) S + S S > H S > S + 3 S + 3 T X T X X H GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

49 CHAP 6 CONTROL SYSTEMS PAGE 36 R V S X () + 3 S S 3 S + 3 T X Taking invere Laplace tranform, we get tate tranition equation a, 3t t e X () t > 3t H 3e SOL 6.5 Option () i correct Phae margin of a ytem i the amount of additional phae lag required to bring the ytem to the point of intability or (, j) So here phae margin c SOL 6.5 SOL 6.5 Option (D) i correct. Given tranfer function i F () 5 ( + 3+ ) F () 5 ( + )( + ) By partial fraction, we get F () ( + ) Taking invere Laplace of F () we have ft () 5 t ut () 5e 5 t + e So, the initial value of ft () i given by lim ft () () t " Option (C) i correct. In A.C techo-meter output voltage i directly proportional to differentiation of rotor diplacement. et ()\ d [ θ( t)] dt et () dθ() t Kt dt Taking Laplace tranformation on both ide of above equation E () K t θ() So tranfer function E () T.F Kt θ() ^ h GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

50 PAGE 36 CONTROL SYSTEMS CHAP 6 SOL 6.53 Option (B) i correct. Given characteritic equation, Applying Routh method, There are two ign change in the firt column, o no. of right half pole i. No. of root in left half of -plane ( 3 ) SOL 6.54 Option (B) i correct. Block diagram of the ytem i given a. From the figure we can ee that C () R () R () : + + R () D C () R () : + + D C () R () + + SOL 6.55 Option (A) i correct. Characteritic equation i given by, I A ( I A) > H > H,! > H 4 SOL 6.56 Option (D) i correct. For the given ytem, characteritic equation can be written a, GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

51 CHAP 6 CONTROL SYSTEMS PAGE K ( ) ( + P + ) ( + ) + K( + P) + ( + KP) + K From the equation. ω n K 5 rad/ec (given) So, K 5 and ξω n + KP #. 7# 5 + 5P or P. o K 5, P. SOL 6.57 SOL 6.58 Option (D) i correct. Unit - impule repone of the ytem i given a, 6t ct (). 5e in8t, t $ So tranfer function of the ytem. H () L[ ct ()]. 5 # 8 ( + 6) + ( 8) H () + + Steady tate value of output for unit tep input, lim yt () lim Y() lim H() R() t " 3 " " Option (A) i correct. Sytem repone i. H () Hjω ( ) lim " ; + + E. + jω jω + Amplitude repone Hjω ( ) ω ω + Given input frequency ω rad/ec. So Hj ( ω) ω rad/ec + Phae repone θh( ω ) 9c tan ( ω) θh( ω) ω 9c tan () 45c GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

52 PAGE 364 CONTROL SYSTEMS CHAP 6 So the output of the ytem i yt () Hj ( ω) xt ( θh) in( t 45c) SOL 6.59 Option (C) i correct. Given open loop tranfer function jaω + Gjω ( ) ( jω) Gain croover frequency ( ω g ) for the ytem. Gj ( ω g ) a ωg + ω g ω+ g ωg 4 a 4 ωg a ωg...() Phae margin of the ytem i φ PM 45c 8c + + Gj ( ωg) tan 45c 8c+ tan ( ω a) 8c ( a) 45c ω g ω ga () From equation () and () 4 a a 4 & a. 84 g SOL 6.6 Option (C) i correct. Given ytem equation i. d x 6 dx + + 5x ( e ) dt dt t Taking Laplace tranform on both ide. X () + 6X () + 5X () : + D ( ) X( ) ; ( + ) E Sytem tranfer function i X () 4 ( + )( + 5)( + ) Repone of the ytem a t " 3 i given by lim ft () lim F( ) (final value theorem) t " 3 " GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

53 CHAP 6 CONTROL SYSTEMS PAGE 365 lim ; 4 " ( + )( + 5)( + ) E 4 4. # 5 SOL 6.6 Option (A) i correct. Tranfer function of lead compenator i given by. Ka + a k H () + a b k R j ω V S + a a k Hjω ( ) KS + j ω S a b k T X So, phae repone of the compenator i. θh( ω ) tan ω tan ω aa k a b k J ω ωn tan K a b O ω( b a) tan K + ω ; O ab + ω E L ab P θ h hould be poitive for phae lead compenation ω( b a ) So, θh ( ω ) tan ; > ab + ω E b > a SOL 6.6 Option (A) i correct. Since there i no external input, o tate i given by X () t φ() t X() φ() t " tate tranition matrix X [] "initial condition So xt () xt () t e > th> e 3 H t e > t H 3e At t, tate of the ytem xt () t e > H e. 7 >. H SOL 6.63 Option (B) i correct. GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

54 PAGE 366 CONTROL SYSTEMS CHAP 6 Given equation dx dx x dt dt e t + e 4 5t Taking Laplace on both ide we have X () + X () + X () ( + + ) X( ) 8 ( + 4)( + 5) + 5( + 5) + ( + 4) ( + 4)( + 5) Sytem repone i, X () ( + 4)( + 5) + 5( + 5) + ( + 4) ( + 4)( + 5) + b + 8 l ( + 4)( + 5) + 5( + 5) + ( + 4) ( + 4)( + 5) + + b 3 lb 6 l e know that for a ytem having many pole, nearne of the pole toward imaginary axi in -plane dominate the nature of time repone. So here time contant given by two pole which are nearet to imaginary axi. Pole nearet to imaginary axi, 3 6 τ 3 ec So, time contant ) τ 6 ec SOL 6.64 Option (A) i correct. Steady tate error for a ytem i given by R() e lim " + GH () () here input R () (unit tep) G () 3 5 b + 5 lb + l So e H () (unity feedback) b l lim 5 5 " ( + 5)( + ) %e 5 6 # 5% SOL 6.65 Option (C) i correct. Characteritic equation i given by + GH () () GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

55 CHAP 6 CONTROL SYSTEMS PAGE 367 Here H () (unity feedback) G () So, b + 5 lb + l ( + 5)( + ) ( + 6)( + ) 3 5 b + 5 lb + l 6, SOL 6.66 Option (A) i correct. Given equation can be written a, d ω β dω K ω + dt J dt LJ Here tate variable are defined a, dω x dt ω x So tate equation i xo xo B x K x J + LJ dω x dt LJ K V a LJ K V a In matrix form xo BJ / K/ LJ x KLJ / > xo H > H> + V a x H > H R d V S ω S dt P d ω S dω > dt H+ QV a S dt So matrix P i T X BJ / K/ LJ > H SOL 6.67 Option (C) i correct. Characteritic equation of the ytem i given by + GH + K ( + )( + 4) ( + )( + 4) + K K GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at:

56 PAGE 368 CONTROL SYSTEMS CHAP 6 Applying routh criteria for tability K K 48 6 K Sytem become untable if K 6 48 & K 48 SOL 6.68 Option (A) i correct. The maximum error between the exact and aymptotic plot occur at corner frequency. Here exact gain(db) at ω.5a i given by gain( db) ω5. a log K log + ω a / (. 5a) log K log; + E log K. 96 a Gain(dB) calculated from aymptotic plot at ω 5. a i log K Error in gain (db) log K ( log K.96) db 96. db Similarly exact phae angle at ω.5a i. θh ( ω) ω tan ω. 5a. aa k tan b 5 a a l.56 6 c Phae angle calculated from aymptotic plot at ( ω.5 a) i.5c Error in phae angle. 5 ( 6. 56c) 49c. SOL 6.69 Option (B) i correct. Given block diagram Given block diagram can be reduced a GATE Previou Year Solved Paper By RK Kanodia & Ahih Murolia Publihed by: NODIA and COMPANY ISBN: Viit u at: