gate Control Systems

CHAPTER 6 CONTROL SYSTEMS

YEAR 2012 MCQ 6.1

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L Jx1NPL P L P K O y = _1 0 0iKx2O Kx 3O L P where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0

MCQ 6.2

TWO MARKS

(D) a1 ! 0, a2 ! 0, a 3 = 0

The feedback system shown below oscillates at 2 rad/s when

(A) K = 2 and a = 0.75

(B) K = 3 and a = 0.75

(C) K = 4 and a = 0.5

(D) K = 2 and a = 0.5

Statement for Linked Answer Questions 3 and 4 :

MCQ 6.3

The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a =− 3, b =− 1

MCQ 6.4

(D) a = 3, b = 1

The phase of the above lead compensator is maximum at (B) 3 rad/s (A) 2 rad/s (C)

6 rad/s

(D) 1/ 3 rad/s

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PAGE 314

CONTROL SYSTEMS

CHAP 6

YEAR 2011 MCQ 6.5

The frequency response of a linear system G (jω) is provided in the tubular form below G (jω)

1.3

+G (jω) − 130c

MCQ 6.6

MCQ 6.7

ONE MARK

1.2

1.0

0.8

0.5

0.3

− 140c

− 150c

− 160c

− 180c

− 200c

(A) 6 dB and 30c

(B) 6 dB and − 30c

(C) − 6 dB and 30c

(D) − 6 dB and − 30c

The 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 (t) having a magnitude of 10 and a duration of one second, as shown in the figure is

(A) 0

(B) 0.1

(C) 1

(D) 10

An open loop system represented by the transfer function (s − 1) is G (s) = (s + 2) (s + 3) (A) Stable and of the minimum phase type (B) Stable and of the non–minimum phase type (C) Unstable and of the minimum phase type (D) Unstable and of non–minimum phase type

YEAR 2011 MCQ 6.8

TWO MARKS

The open loop transfer function G (s) of a unity feedback control system is given as K bs + 2 l 3 G (s) = 2 s (s + 2) From the root locus, at can be inferred that when K tends to positive infinity, (A) Three roots with nearly equal real parts exist on the left half of the s GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 315

-plane (B) One real root is found on the right half of the s -plane (C) The root loci cross the jω axis for a finite value of K; K ! 0 (D) Three real roots are found on the right half of the s -plane MCQ 6.9

A two loop position control system is shown below

The gain K of 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 (ω " 0) (D) Phase shift of the closed loop transfer function at very high frequencies (ω " 3) YEAR 2010 MCQ 6.10

TWO MARKS

1 The frequency response of G (s) = plotted in the complex s (s + 1) (s + 2) G (jω) plane (for 0 < ω < 3) is



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PAGE 316

MCQ 6.11

MCQ 6.12

CONTROL SYSTEMS

CHAP 6

o = AX + Bu with A = >− 1 2H, B = >0H is The system X 0 2 1 (A) Stable and controllable

(B) Stable but uncontrollable

(C) Unstable but controllable

(D) Unstable and uncontrollable

The characteristic equation of a closed-loop system is s (s + 1) (s + 3) k (s + 2) = 0, k > 0 .Which of the following statements is true ? (A) Its root are always real (B) It cannot have a breakaway point in the range − 1 < Re [s] < 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 2009 MCQ 6.13

MCQ 6.14

ONE MARK

The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as G1, G2, 1/G3 . The relative small errors associated with each respective subsystem G1, G2 and G3 are ε1, ε2 and ε3 . The error associated with the output is :

(A) ε1 + ε2 + 1 ε3

(B) ε1 ε2 ε3

(C) ε1 + ε2 − ε3

(D) ε1 + ε2 + ε3

The polar plot of an open loop stable system is shown below. The closed loop system is

(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 -plane GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

MCQ 6.15

CONTROL SYSTEMS

PAGE 317

The first two rows of Routh’s tabulation of a third order equation are as follows. s3 2 2 s2 4 4 This means there are (A) Two roots at s = ! j and one root in right half s -plane (B) Two roots at s = ! j2 and one root in left half s -plane (C) Two roots at s = ! j2 and one root in right half s -plane (D) Two roots at s = ! j and one root in left half s -plane

MCQ 6.16

The asymptotic approximation of the log-magnitude v/s frequency plot of a system containing only real poles and zeros is shown. Its transfer function is

(A)

10 (s + 5) s (s + 2) (s + 25)

(C)

100 (s + 5) s (s + 2) (s + 25)

1000 (s + 5) s (s + 2) (s + 25) 80 (s + 5) (D) 2 s (s + 2) (s + 25) (B)

2

YEAR 2009 MCQ 6.17

TWO MARKS

The unit-step response of a unity feed back system with open loop transfer function G (s) = K/ ((s + 1) (s + 2)) is shown in the figure. The value of K is

(A) 0.5

(B) 2

(C) 4

(D) 6



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MCQ 6.18

CONTROL SYSTEMS

CHAP 6

The open loop transfer function of a unity feed back system is given by G (s) = (e - 0.1s) /s . The gain margin of the is system is (A) 11.95 dB (B) 17.67 dB (C) 21.33 dB

(D) 23.9 dB

Common Data for Question 19 and 20 : A system is described by the following state and output equations dx1 (t) =− 3x1 (t) + x2 (t) + 2u (t) dt dx2 (t) =− 2x2 (t) + u (t) dt y (t) = x1 (t) when u (t) is the input and y (t) is the output MCQ 6.19

MCQ 6.20

The system transfer function is (A) 2 s + 2 (B) 2 s + 3 s + 5s − 6 s + 5s + 6 (C) 2 2s + 5 (D) 2 2s − 5 s + 5s + 6 s + 5s − 6 The state-transition matrix of the above system is e - 3t 0 e - 3t e - 2t − e - 3t (B) = (A) = - 2t G G e + e - 3t e - 2t 0 e - 2t e - 3t e - 2t + e - 3t (C) = G 0 e - 2t

e 3t e - 2t − e - 3t (D) = G 0 e - 2t

YEAR 2008 MCQ 6.21

ONE MARK

A function y (t) satisfies the following differential equation : dy (t) + y (t) = δ (t) dt where δ (t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u (t), y (t) can be of the form (A) et (B) e - t (C) et u (t)

(D) e - t u (t)

YEAR 2008 MCQ 6.22

TWO MARK

The transfer function of a linear time invariant system is given as G (s) = 2 1 s + 3s + 2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 319

The steady state value of the output of the system for a unit impulse input applied at time instant t = 1 will be (A) 0 (B) 0.5 (C) 1 MCQ 6.23

(D) 2

The transfer functions of two compensators are given below : 10 (s + 1) C1 = , C2 = s + 10 (s + 10) 10 (s + 1) 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 compensator

MCQ 6.24

MCQ 6.25

The asymptotic Bode magnitude plot of a minimum phase transfer function is 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 zeros

Figure shows a feedback system where K > 0

The range of K for which the system is stable will be given by (A) 0 < K < 30 (B) 0 < K < 39 (C) 0 < K < 390 MCQ 6.26

(D) K > 390

The transfer function of a system is given as 100 2 s + 20s + 100 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 320

CONTROL SYSTEMS

CHAP 6

The system is (A) An over damped system

(B) An under damped system

(C) A critically damped system

(D) An unstable system

Statement for Linked Answer Question 27 and 28.

MCQ 6.27

The state space equation of a system is described by Xo = AX + Bu,Y = CX where X is state vector, u is input, Y is output and 0 1 0 A == , B = = G, C = [1 0] G 0 −2 1 The transfer function G(s) of this system will be s (A) (B) s + 1 (s + 2) s (s − 2) (C)

MCQ 6.28

s (s − 2)

(D)

1 s (s + 2)

A unity feedback is provided to the above system G (s) to 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) 3

YEAR 2007 MCQ 6.29

ONE MARK

The system shown in the figure is

(A) Stable (B) Unstable (C) Conditionally stable (D) Stable for input u1 , but unstable for input u2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 321

YEAR 2007 MCQ 6.30

If x = Re [G (jω)], and y = Im [G (jω)] then for ω " 0+ , the Nyquist plot for G (s) = 1/s (s + 1) (s + 2) is (B) x =− 3/4 (A) x = 0 (C) x = y − 1/6

MCQ 6.31

TWO MARKS

(D) x = y/ 3

The system 900/s (s + 1) (s + 9) is to be such that its gain-crossover frequency 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 of 3 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/s

MCQ 6.32

If the loop gain K of a negative feed back system having a loop transfer function K (s + 3) / (s + 8) 2 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 K does not exist

MCQ 6.33

The system shown in figure below

can be reduced to the form GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 322

CONTROL SYSTEMS

CHAP 6

with (A) X = c0 s + c1, Y = 1/ (s2 + a0 s + a1), Z = b0 s + b1 (B) X = 1, Y = (c0 s + c1) / (s2 + a0 s + a1), Z = b0 s + b1 (C) X = c1 s + c0, Y = (b1 s + b0) / (s2 + a1 s + a0), Z = 1 (D) X = c1 s + c0, Y = 1/ (s2 + a1 s + a), Z = b1 s + b0 MCQ 6.34

Consider the feedback system shown below which is subjected to a unit step input. The system is stable and has following parameters Kp = 4, Ki = 10, ω = 500 and ξ = 0.7 .The steady state value of Z is

(A) 1

(B) 0.25

(C) 0.1

(D) 0

Data for Q.35 and Q.36 are given below. Solve the problems and choose the correct answers. R-L-C circuit shown in figure

MCQ 6.35

For 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 %

MCQ 6.36

(D) 48 %

If 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 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 323

(B) Square wave of 50 Hz (C) Square wave of 300 Hz (D) Square wave of 2.0 KHz

YEAR 2006 MCQ 6.37

ONE MARK

For a system with the transfer function 3 (s − 2) , H (s) = 2 4s − 2s + 1 the matrix A in the state space form Xo = AX + Bu is equal to V V R R S1 0 0 W S0 1 0 W (A) S 0 1 0 W (B) S 0 0 1 W SS− 1 2 − 4 WW SS− 1 2 − 4 WW X X T T V V R R 0 1 0 1 0 0 W W S S (C) S3 − 2 1 W (D) S 0 0 1 W SS1 − 2 4 WW SS− 1 2 − 4 WW X X T T YEAR 2006

MCQ 6.38

TWO MARKS

Consider the following Nyquist plots of loop transfer functions over ω = 0 to ω = 3 . Which of these plots represent a stable closed loop system ?



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PAGE 324

MCQ 6.39

MCQ 6.40

CONTROL SYSTEMS

CHAP 6

(A) (1) only

(B) all, except (1)

(C) all, except (3)

(D) (1) and (2) only

10 4 (1 + jω) The Bode magnitude plot H (jω) = is (10 + jω) (100 + jω) 2

A closed-loop system has the characteristic 2 (s − 4) (s + 1) + K (s − 1) = 0 . Its root locus plot against K is



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function

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CHAP 6

CONTROL SYSTEMS

PAGE 325

YEAR 2005 MCQ 6.41

A system with zero initial conditions has the closed loop transfer function. s2 + 4 T (s) = (s + 1) (s + 4) The system output is zero at the frequency (A) 0.5 rad/sec (B) 1 rad/sec (C) 2 rad/sec

MCQ 6.42

(D) 4 rad/sec

Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is

(A) K3 s

MCQ 6.43

ONE MARK

K s2 (s + 1) K K (C) (D) s (s2 + 1) s (s2 − 1) The gain margin of a unity feed back control system with the open loop (s + 1) is transfer function G (s) = s2 (A) 0 (B) 1 2 (C) 2 (D) 3 (B)



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PAGE 326

CONTROL SYSTEMS

CHAP 6

YEAR 2005 MCQ 6.44

TWO MARKS

A unity feedback system, having an open loop gain K (1 − s) , G (s) H (s) = (1 + s) becomes stable when

MCQ 6.45

MCQ 6.46

(A) K > 1

(B) K > 1

(C) K < 1

(D) K < − 1

When subject to a unit step input, the closed loop control system shown in the figure will have a steady state error of

(A) − 1.0

(B) − 0.5

(C) 0

(D) 0.5

In the G (s) H (s)-plane, the Nyquist plot of the loop transfer function G (s) H (s) = πes passes through the negative real axis at the point (B) (− 0.5, j0) (A) (− 0.25, j0) -0.25s

(C) 0 MCQ 6.47

(D) 0.5

If the compensated system shown in the figure has a phase margin of 60c at the crossover frequency of 1 rad/sec, then value of the gain K is

(A) 0.366

(B) 0.732

(C) 1.366

(D) 2.738

Data for Q.48 and Q.49 are given below. Solve the problem and choose the correct answer. 0 1 1 X (t) + = Gu (t) with the initial G 0 −3 0 T condition X (0) = [− 1, 3] and the unit step input u (t) has A state variable system Xo (t) = =



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CHAP 6

MCQ 6.48

CONTROL SYSTEMS

The state transition matrix − 3t 1 1 ) 3 (1 − e (A) = G − 3t 0 e 1 (C) > 0

MCQ 6.49

1 3

(e3 − t − e− 3t) H e− 3t

PAGE 327

1 (B) > 0 1 (D) > 0

1 3

(e− t − e− 3t) H e− t

(1 − e− t) H e− t

The state transition equation t − e-t (A) X (t) = = - t G e

1 − e-t (B) X (t) = = - 3t G 3e

t − e 3t (C) X (t) = = - 3t G 3e

t − e - 3t (D) X (t) = = - t G e

YEAR 2004 MCQ 6.50

The Nyquist plot of loop transfer function G (s) H (s) of a closed loop control system passes through the point (− 1, j 0) in the G (s) H (s)plane. The phase margin of the system is (A) 0c (B) 45c (C) 90c

MCQ 6.51

(D) 180c

Consider the function, 5 F (s) = 2 s (s + 3s + 2) where F (s) is the Laplace transform of the of the function f (t). The initial value of f (t) is equal to (A) 5 (B) 25 (C)

MCQ 6.52

ONE MARK

5 3

(D) 0

For a tachometer, if θ (t) is the rotor displacement in radians, e (t) is the output voltage and Kt is the tachometer constant in V/rad/sec, then the E (s) will be transfer function, Q (s) (A) Kt s2 (B) Kt s (C) Kt s

(D) Kt

YEAR 2004 MCQ 6.53

TWO MARKS

For the equation, s3 − 4s2 + s + 6 = 0 the number of roots in the left half of s -plane will be (A) Zero (B) One (C) Two

(D) Three



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PAGE 328

MCQ 6.54

MCQ 6.55

MCQ 6.56

MCQ 6.57

CONTROL SYSTEMS

For the block diagram shown, the transfer function

C (s) is equal to R (s)

2 (A) s +2 1 s

2 (B) s + s2 + 1 s

2 (C) s + s + 1 s

(D)

(C) − 2 and − 2

(D) + 2 and + 2

1 s +s+1 o = AX The state variable description of a linear autonomous system is, X where X is the two dimensional state vector and A is the system matrix 0 2 given by A = = . The roots of the characteristic equation are 2 0G (A) − 2 and + 2 (B) − j2 and + j2 2

The block diagram of a closed loop control system is given by figure. The values of K and P such that the system has a damping ratio of 0.7 and an undamped natural frequency ωn of 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 0.2

The unit impulse response of a second order under-damped system starting from rest is given by c (t) = 12.5e - 6t sin 8t, t $ 0 . The steady-state value of the unit step response of the system is equal to (A) 0 (B) 0.25 (C) 0.5

MCQ 6.58

CHAP 6

(D) 1.0

In the system shown in figure, the input x (t) = sin t . In the steady-state, the response y (t) will be

(A)

1 sin (t − 45c) 2

(B)

1 sin (t + 45c) 2



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CHAP 6

CONTROL SYSTEMS

(C) sin (t − 45c) MCQ 6.59

PAGE 329

(D) sin (t + 45c)

The open loop transfer function of a unity feedback control system is given as 1. G (s) = as + 2 s The value of ‘a ’ to give a phase margin of 45c is equal to (A) 0.141 (B) 0.441 (C) 0.841

(D) 1.141

YEAR 2003 MCQ 6.60

A control system is defined by the following mathematical relationship d2 x + 6 dx + 5x = 12 (1 − e - 2t) dt dt2 The response of the system as t " 3 is (A) x = 6 (B) x = 2 (C) x = 2.4

MCQ 6.61

(D) x =− 2

A lead compensator used for a closed loop controller has the following transfer function K (1 + as ) (1 + bs ) For such a lead compensator (B) b < a (A) a < b (C) a > Kb

MCQ 6.62

ONE MARK

(D) a < Kb

2 A second order system starts with an initial condition of = G without any 3 e - 2t 0 external input. The state transition matrix for the system is given by = G 0 e-t . The state of the system at the end of 1 second is given by 0.271 0.135 (A) = (B) = G 1.100 0.368G 0.271 (C) = 0.736G

0.135 (D) = 1.100 G

YEAR 2003 MCQ 6.63

TWO MARKS

A control system with certain excitation is governed by the following mathematical equation d2 x + 1 dx + 1 x = 10 + 5e− 4t + 2e− 5t 2 dt 18 dt2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 330

CONTROL SYSTEMS

CHAP 6

The natural time constant of the response of the system are (A) 2 sec and 5 sec (B) 3 sec and 6 sec (C) 4 sec and 5 sec MCQ 6.64

MCQ 6.65

MCQ 6.66

The block diagram shown in figure gives a unity feedback closed loop control system. The steady state error in the response of the above system to unit step input is

(A) 25%

(B) 0.75 %

(C) 6%

(D) 33%

The roots of the closed loop characteristic equation of the system shown above (Q-5.55)

(A) − 1 and − 15

(B) 6 and 10

(C) − 4 and − 15

(D)− 6 and − 10

The following equation defines a separately excited dc motor in the form of a differential equation d2 ω + B dω + K2 ω = K V dt J dt LJ LJ a The above equation may be organized in the state-space form as follows R 2 V Sd ω W dω S dt2 W = P dt + QV > H a S dω W ω S dt W Where Tthe PX matrix is given by K K − BJ − LJ − LJ − BJ (A) = (B) = G G 1 0 0 1 2

0 1 (C) =− K − B G LJ J 2

MCQ 6.67

(D) 1/3 sec and 1/6 sec

2

1 0 (D) =− B − K G J LJ 2

The loop gain GH of a closed loop system is given by the following expression K s (s + 2) (s + 4) The value of K for which the system just becomes unstable is GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

MCQ 6.68

MCQ 6.69

CONTROL SYSTEMS

PAGE 331

(A) K = 6

(B) K = 8

(C) K = 48

(D) K = 96

The asymptotic Bode plot of the transfer function K/ [1 + (s/a)] is given in figure. The error in phase angle and dB gain at a frequency of ω = 0.5a are respectively

(A) 4.9c, 0.97 dB

(B) 5.7c, 3 dB

(C) 4.9c, 3 dB

(D) 5.7c, 0.97 dB

The block diagram of a control system is shown in figure. The transfer function G (s) = Y (s) /U (s) of the system is

1 18`1 + s j`1 + s j 12 3 1 (C) s 27`1 + j`1 + s j 12 9 (A)

1 27`1 + s j`1 + s j 6 9 1 (D) s 27`1 + j`1 + s j 9 3 (B)

YEAR 2002 MCQ 6.70

ONE MARK

The state transition matrix for the system Xo = AX with initial state X (0) is (A) (sI − A) - 1 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 332

CONTROL SYSTEMS

CHAP 6

(B) eAt X (0) (C) Laplace inverse of [(sI − A) - 1] (D) Laplace inverse of [(sI − A) - 1 X (0)] YEAR 2002 MCQ 6.71

TWO MARKS

2 3 1 For the system Xo = = X + = Gu , which of the following statements is true G 0 5 0 ? (A) The system is controllable but unstable (B) The system is uncontrollable and unstable (C) The system is controllable and stable (D) The system is uncontrollable and stable

MCQ 6.72

MCQ 6.73

MCQ 6.74

A unity feedback system has an open loop transfer function, G (s) = K2 . The s root locus plot is

The transfer function of the system described by d2 y dy + = du + 2u dt dt dt2 with u as input and y as output is (s + 2) (s + 1) (A) 2 (B) 2 (s + s) (s + s) (C) 2 2 (D) 22s (s + s) (s + s) For the system 2 0 1 Xo = = X + = Gu ; Y = 84 0B X, G 0 4 1 with u as unit impulse and with zero initial state, the output y , becomes (A) 2e2t (B) 4e2t GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

(C) 2e 4t

PAGE 333

(D) 4e 4t

MCQ 6.75

The eigen values of the system represented by R0 1 0 0 V S W 0 0 1 0W S Xo = S X are 0 0 0 1W S W (A) 0, 0, 0, 0S (B) 1, 1, 1, 1 0 0 0 1W T 1 X (D) 1, 0, 0, 0 (C) 0, 0, 0, −

MCQ 6.76

*A single input single output system with y as output and u as input, is described by d2 y dy du 2 + 2 dt + 10y = 5 dt − 3u dt for an input u (t) with zero initial conditions the above system produces the same output as with no input and with initial conditions dy (0−) =− 4 , y (0−) = 1 dt input u (t) is (B) 1 δ (t) − 7 e− 3t u (t) (A) 1 δ (t) − 7 e(3/5)t u (t) 5 25 5 25 (C) − 7 e− (3/5)t u (t) 25

MCQ 6.77

(D) None of these

*A system is described by the following differential equation d2 y dy + - 2y = u (t) e− t dt dt2 dy the state variables are given as x1 = y and x2 = b − y l et , the state dt varibale representation of the system is 1 xo1 1 e− t x 1 (A) > o H = > − tH> H + > H u (t) 0 x2 0 e x2 1 1 x1 1 xo1 (B) > o H = > H> H + > H u (t) 0 1 x2 0 x2 1 xo1 1 e− t x 1 (C) > o H = > >x H + >0H u (t) H x2 0 −1 2 (D) none of these

Common Data Question Q.78-80*. The open loop transfer function of a unity feedback system is given by 2 (s + α) G (s) = s (s + 2) (s + 10) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 334

MCQ 6.78

MCQ 6.79

MCQ 6.80

CONTROL SYSTEMS

CHAP 6

Angles of asymptotes are (A) 60c, 120c, 300c

(B) 60c, 180c, 300c

(C) 90c, 270c, 360c

(D) 90c, 180c, 270c

Intercepts of asymptotes at the real axis is (A) − 6

(B) − 10 3

(C) − 4

(D) − 8

Break away points are (A) − 1.056 , − 3.471

(B) − 2.112, − 6.9433

(C) − 1.056, − 6.9433

(D) 1.056, − 6.9433

YEAR 2001 MCQ 6.81

ONE MARK

The polar plot of a type-1, 3-pole, open-loop system is shown in Figure The closed-loop system is

(A) always stable (B) marginally stable (C) unstable with one pole on the right half s -plane (D) unstable with two poles on the right half s -plane. MCQ 6.82

−3 1 Given the homogeneous state-space equation xo = = x the steady state 0 − 2G

value of xss = lim x (t), given the initial state value of x (0) = 810 − 10B

T

t"3

0 (A) xss = = G 0

−3 (B) xss = = G −2

− 10 (C) xss = = 10 G

3 (D) xss = = G 3

YEAR 2001 MCQ 6.83

is

TWO MARKS

The asymptotic approximation of the log-magnitude versus frequency plot GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 335

of a minimum phase system with real poles and one zero is shown in Figure. Its transfer functions is

(A)

20 (s + 5) s (s + 2) (s + 25)

(C)

20 (s + 5) s (s + 2) (s + 25) 2

10 (s + 5) (s + 2) 2 (s + 25) 50 (s + 5) (D) 2 s (s + 2) (s + 25) (B)

Common Data Question Q.84-87*.

MCQ 6.84

MCQ 6.85

MCQ 6.86

MCQ 6.87

MCQ 6.88

A unity feedback system has an open-loop transfer function of G (s) = 10000 2 s (s + 10) Determine the magnitude of G (jω) in dB at an angular frequency of ω = 20 rad/sec. (A) 1 dB (B) 0 dB (C) − 2 dB

(D) 10 dB

The phase margin in degrees is (A) 90c

(B) 36.86c

(C) − 36.86c

(D) − 90c

The gain margin in dB is (A) 13.97 dB

(B) 6.02 dB

(C) − 13.97 dB

(D) None of these

The system is (A) Stable

(B) Un-stable

(C) Marginally stable

(D) can not determined

*For the given characteristic equation s3 + s2 + Ks + K = 0 The root locus of the system as K varies from zero to infinity is GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 336

CONTROL SYSTEMS

CHAP 6

************



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CHAP 6

CONTROL SYSTEMS

PAGE 337

SOLUTION SOL 6.1

Option (D) is correct. General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R 0 a 0V R0V 1 W S S W A = S 0 0 a2W, B = S0W SSa SS1WW 0 0WW 3 X T T X R 0 a 0VR0V R 0V 1 S WS W S W AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 RT 0 0XT aX1 a2VWTRS0XVW RSa1 a2VW S A2 B = Sa2 a 3 0 0WS0W = S 0W SS 0 a a 0WWSS1WW SS 0WW 3 1 X XT X T T For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S U = 6B : AB : A2 B@ = S0 a2 0W SS1 0 0WW X T So, a2 ! 0 a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not.

SOL 6.2

Option (A) is correct. K (s + 1) Y (s) = 3 [R (s) − Y (s)] s + as2 + 2s + 1 K (s + 1) K (s + 1) R (s) Y (s) ;1 + 3 = 3 E 2 s + as2 + 2s + 1 s + as + 2s + 1 Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s) K (s + 1) Y (s) Transfer Function, H (s) = = R (s) s3 + as2 + s (2 + k) + (1 + k) Routh Table :



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PAGE 338

CONTROL SYSTEMS

CHAP 6

For oscillation, a (2 + K) − (1 + K) =0 a a = K+1 K+2 Auxiliary equation A (s) = as2 + (k + 1) = 0 s2 =− k + 1 a − k + 1 (k + 2) =− (k + 2) s2 = (k + 1) s = j k+2 jω = j k + 2 ω = k+2 = 2 (Oscillation frequency) k =2 and a = 2 + 1 = 3 = 0.75 2+2 4 SOL 6.3

Option (A) is correct. jω + a GC (s) = s + a = s+b jω + b Phase lead angle, φ = tan−1 a ω k − tan−1 a ω k a b Jω − ωN −1 K a b O = tan−1 ω (b − a) = tan c ab + ω 2 m 2 KK O 1+ ω O ab P L For phase-lead compensation φ > 0 b−a > 0 b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true.

SOL 6.4

Option (A) is correct. φ = tan−1 a ω k − tan−1 a ω k a b GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 339

1/a 1/b dφ = − =0 2 2 ω dω 1 +a k 1 +aωk a b 2 2 1 + ω = 1+1ω a ab2 b b a2 1 − 1 = ω2 1 − 1 a b ab b a b l ω = ab = 1 # 2 = SOL 6.5

2 rad/ sec

Option (A) is correct. Gain margin is simply equal to the gain at phase cross over frequency ( ωp ). Phase cross over frequency is the frequency at which phase angle is equal to − 180c. From the table we can see that +G (jωp) =− 180c, at which gain is 0.5. 1 GM = 20 log 10 e = 20 log b 1 l = 6 dB 0.5 G (jωp) o Phase Margin is equal to 180c plus the phase angle φg at the gain cross over frequency ( ωg ). Gain cross over frequency is the frequency at which gain is unity. From the table it is clear that G (jωg) = 1, at which phase angle is − 150c φPM = 180c + +G (jωg) = 180 − 150 = 30c

SOL 6.6

Option (A) is correct. We know that steady state error is given by sR (s) ess = lim s " 0 1 + G (s) where R (s) " input G (s) " open loop transfer function For unit step input R (s) = 1 s sb 1 l s So ess = lim = 0.1 s " 0 1 + G (s) 1 + G (0) = 10 G (0) = 9 Given input r (t) = 10 [μ (t) − μ (t − 1)] −s or R (s) = 10 :1 − 1 e−sD = 10 :1 − e D s s s So steady state error GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CONTROL SYSTEMS

CHAP 6

(1 − e−s) s # 10 10 (1 − e0) s = el = =0 lim ss 1+9 s"0 1 + G (s) SOL 6.7

Option (B) is correct. Transfer function having at least one zero or pole in RHS of s -plane is called non-minimum phase transfer function. s−1 G (s) = (s + 2) (s + 3) • 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 stable

SOL 6.8

Option (A) is correct. K bs + 2 l 3 G (s) = 2 s (s + 2) Steps for plotting the root-locus (1) Root loci starts at s = 0, s = 0 and s =− 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n−m (2 # 0 + 1) 180c (I) = 90c (3 − 1) (2 # 1 + 1) 180c (II) = 270c (3 − 1) (4) The two asymptotes intersect on real axis at centroid − 2 − b− 2 l 3 Poles − / Zeroes / x = = =− 2 n−m 3−1 3 (5) Between two open-loop poles s = 0 and s =− 2 there exist a break away point. s2 (s + 2) K =− 2 bs + 3 l dK = 0 ds s =0 Root locus is shown in the figure



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CHAP 6

CONTROL SYSTEMS

PAGE 341

Three roots with nearly equal parts exist on the left half of s -plane. SOL 6.9

Option (A) is correct. The system may be reduced as shown below

1 s (s + 1 + K ) Y (s) 1 = 2 = 1 R (s) 1 + s + s (1 + K ) + 1 s (s + 1 + K ) This is a second order system transfer function, characteristic equation is s2 + s (1 + K) + 1 = 0 Comparing with standard form s2 + 2ξωn s + ωn2 = 0 We get ξ = 1+K 2 Peak overshoot M p = e− πξ/

1 − ξ2

So the Peak overshoot is effected by K . SOL 6.10

Option (A) is correct. Given

1 s (s + 1) (s + 2) 1 G (jω) = jω (jω + 1) (jω + 2) G (s) =



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PAGE 342

CONTROL SYSTEMS

CHAP 6

1 ω ω2 + 1 ω 2 + 4 +G (jω) =− 90c − tan− 1 (ω) − tan− 1 (ω/2) In nyquist plot For ω = 0, G (jω) = 3 G (jω) =

+G (jω) =− 90c For ω = 3, G (jω) = 0 +G (jω) =− 90c − 90c − 90c =− 270c Intersection at real axis 1 1 G (jω) = = 2 jω (jω + 1) (jω + 2) jω (− ω + j3ω + 2) − 3ω2 − jω (2 − ω2) 1 2 # − 3ω + jω (2 − ω ) − 3ω2 − jω (2 − ω2) − 3ω2 − jω (2 − ω2) = 9ω4 + ω2 (2 − ω2) 2 2 jω (2 − ω2) = 4 −23ω − 9ω + ω (2 − ω2) 2 9ω4 + ω2 (2 − ω2) 2 =

2

At real axis Im [G (jω)] = 0 ω (2 − ω2) So, =0 9ω4 + ω2 (2 − ω2) 2 − ω2 = 0 & ω = 2 rad/sec At ω = 2 rad/sec, magnitude response is 1 G (jω) at ω = 2 = =1<3 6 4 2 2+1 2+4 SOL 6.11

Option (C) is correct. Stability : Eigen value of the system are calculated as A − λI = 0 −1 2 λ 0 −1 − λ 2 −> => A − λI = > H H 0 2 0 λ 0 2 − λH A − λI = (− 1 − λ) (2 − λ) − 2 # 0 = 0 & λ1, λ2 =− 1, 2 Since eigen values of the system are of opposite signs, so it is unstable Controllability : 0 −1 2 , B=> H A => H 1 0 2 2 AB = > H 2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 343

0 2 [B: AB] = > H 1 2 Y 0 6B: AB@ = So it is controllable. SOL 6.12

Option (C) is correct. Given characteristic equation s (s + 1) (s + 3) + K (s + 2) = 0 ; s (s2 + 4s + 3) + K (s + 2) = 0 s3 + 4s2 + (3 + K) s + 2K = 0 From Routh’s tabulation method

K>0

s3

1

3+K

s2

4

2K

s1

4 (3 + K) − 2K (1) 12 + 2K = >0 4 4

s0

2K

There 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 K (s + 2) =0 1+ s (s + 1) (s + 3) Open loop transfer function G (s) =

K (s + 2) s (s + 1) (s + 3)

Root locus is obtained in following steps: 1. No. of poles n = 3 , at s = 0, s =− 1 and s =− 3 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 − 1 < Re [s] < 0 .

5.

Asymptotes meet on real axis at a point C , given by C =

/ real part of poles − / real parts of zeroes

n−m (0 − 1 − 3) − (− 2) = 3−1



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PAGE 344

CONTROL SYSTEMS

CHAP 6

=− 1 As no. of poles is 3, so two root loci branches terminates at infinity along asymptotes Re (s) =− 1 SOL 6.13

Option (D) is correct. Overall gain of the system is written as G = G1 G 2 1 G3 We know that for a quantity that is product of two or more quantities total percentage error is some of the percentage error in each quantity. so error in overall gain G is 3 G = ε1 + ε2 + 1 ε3

SOL 6.14

Option (D) is correct. From Nyquist stability criteria, no. of closed loop poles in right half of s -plane is given as Z = P−N P " No. of open loop poles in right half s -plane N " No. of encirclement of (− 1, j0)

Here N =− 2 (` encirclement is in clockwise direction) P = 0 (` system is stable) So, Z = 0 − (− 2) Z = 2 , System is unstable with 2-poles on RH of s -plane. SOL 6.15

Option (D) is correct. Given Routh’s tabulation. s3

2

2

s2

4

4

s1

0

0

So the auxiliary equation is given by, 4s 2 + 4 = 0 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 345

s2 =− 1 s =! j From table we have characteristic equation as 2s3 + 2s + 4s2 + 4 = 0 s3 + s + 2s2 + 2 = 0 s (s2 + 1) + 2 (s2 + 1) = 0 (s + 2) (s2 + 1) = 0 s =− 2 , s = ! j SOL 6.16

Option (B) is correct. Since initial slope of the bode plot is − 40 dB/decade, so no. of poles at origin is 2. Transfer function can be written in following steps: 1. Slope changes from − 40 dB/dec. to − 60 dB/dec. at ω1 = 2 rad/sec., so at ω1 there is a pole in the transfer function. 2.

Slope changes from − 60 dB/dec to − 40 dB/dec at ω2 = 5 rad/sec., so at this frequency there is a zero lying in the system function.

3.

The slope changes from − 40 dB/dec to − 60 dB/dec at ω3 = 25 rad/sec, so there is a pole in the system at this frequency.

Transfer function K (s + 5) s (s + 2) (s + 25) Constant term can be obtained as. T (s) =

2

T (jω) at ω = 0.1 = 80 So,

80 = 20 log

K (5) (0.1) 2 # 50

K = 1000 therefore, the transfer function is 1000 (s + 5) T (s) = 2 s (s + 2) (s + 25) SOL 6.17

Option (D) is correct. From the figure we can see that steady state error for given system is ess = 1 − 0.75 = 0.25 Steady state error for unity feed back system is given by sR (s) ess = lim = G s " 0 1 + G (s) s ^ 1s h ; R (s) = 1 (unit step input) = lim s s"0> K H 1+ (s + 1) (s + 2) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 346

CONTROL SYSTEMS

CHAP 6

1 = 2 2+K 1 + K2 ess = 2 = 0.25 2+K =

So,

2 = 0.5 + 0.25K K = 1.5 = 6 0.25 SOL 6.18

Option (D) is correct. Open loop transfer function of the figure is given by, − 0.1s G (s) = e s G (jω) = e

− j0.1ω

jω

Phase cross over frequency can be calculated as, +G (jωp) =− 180c 180 b− 0.1ωp # π l − 90c =− 180c 0.1ωp # 180c = 90c π 0.1ωp = 90c # π 180c ωp = 15.7 rad/sec 1 So the gain margin (dB) = 20 log e = 20 log G (jωp) o >

1 1 b 15.7 l H

= 20 log 15.7 = 23.9 dB SOL 6.19

Option (C) is correct. Given system equations dx1 (t) =− 3x1 (t) + x2 (t) + 2u (t) dt dx2 (t) =− 2x2 (t) + u (t) dt y (t) = x1 (t) Taking Laplace transform on both sides of equations.

Similarly

sX1 (s) (s + 3) X1 (s) sX2 (s) (s + 2) X2 (s)

=− 3X1 (s) + X2 (s) + 2U (s) = X2 (s) + 2U (s) =− 2X2 (s) + U (s) = U (s)



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CHAP 6

CONTROL SYSTEMS

PAGE 347

From equation (1) & (2) U (s) + 2U (s) s+2 U (s) 1 + 2 (s + 2) (2s + 5) X1 (s) = = U (s) ; E s+3 s+2 (s + 2) (s + 3) From output equation, (s + 3) X1 (s) =

Y (s) = X1 (s) Y (s) = U (s)

So,

(2s + 5) (s + 2) (s + 3)

System transfer function Y (s) (2s + 5) (2s + 5) T.F = = = 2 U (s) (s + 2) (s + 3) s + 5s + 6 SOL 6.20

Option (B) is correct. Given state equations in matrix form can be written as, − 3 1 x1 2 xo1 > o H = > 0 − 2H>x H + >1H u (t) x2 2 dX (t) = AX (t) + Bu (t) dt State transition matrix is given by φ (t) = L− 1 6Φ (s)@ Φ (s) = (sI − A) − 1 s 0 −3 1 (sI − A) = > H − > 0 s 0 − 2H s + 3 −1 (sI − A) = > 0 s + 2H s+2 1 1 > 0 s + 3H (s + 3) (s + 2) R V 1 S 1 W (s + 3) (s + 3) (s + 2)W S −1 So Φ (s) = (sI − A) = S W 1 S 0 (s + 2) W T − 3t − 2t X e e − e− 3t −1 φ (t) = L [Φ (s)] = > H 0 e− 2t (sI − A) − 1 =

SOL 6.21

Option (D) is correct. Given differential equation for the function dy (t) + y (t) = δ (t) dt GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 348

CONTROL SYSTEMS

CHAP 6

Taking Laplace on both the sides we have, sY (s) + Y (s) = 1 (s + 1) Y (s) = 1 Y (s) =

1 s+1

Taking inverse Laplace of Y (s) y (t) = e− t u (t), t > 0 SOL 6.22

Option (A) is correct. Given transfer function G (s) =

1 s + 3s + 2 2

r (t) = δ (t − 1) R (s) = L [δ (t − 1)] = e− s Output is given by −s Y (s) = R (s) G (s) = 2 e s + 3s + 2 Steady state value of output −s lim y (t) = lim sY (s) = lim 2 se =0 t"3 s"0 s " 0 s + 3s + 2 Input

SOL 6.23

Option (A) is correct. For C1 Phase is given by θC = tan− 1 (ω) − tan− 1 a ω k 10 Jω − ω N 9ω −1 K 10 O = tan− 1 = tan 2 c 10 + ω2 m > 0 (Phase lead) KK OO ω 1+ 10 P Similarly for C2 , phase is L θC2 = tan− 1 a ω k − tan− 1 (ω) 10 J ω − ωN − 1 K 10 O = tan− 1 − 9ω < 0 (Phase lag) = tan 2 c 10 + ω2 m KK O 1+ ω O 10 P L Option (C) is correct. From the given bode plot we can analyze that: 1. Slope − 40 dB/decade"2 poles 1

SOL 6.24

2.

Slope − 20 dB/decade (Slope changes by + 20 dB/decade)"1 Zero

3.

Slope 0 dB/decade (Slope changes by + 20 dB/decade)"1 Zero

So there are 2 poles and 2 zeroes in the transfer function. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

SOL 6.25

CONTROL SYSTEMS

PAGE 349

Option (C) is correct. Characteristic equation for the system K =0 1+ s (s + 3) (s + 10) s (s + 3) (s + 10) + K = 0 s3 + 13s2 + 30s + K = 0 Applying Routh’s stability criteria s3

1

30

s2

13

K

s1

(13 # 30) − K 13 K

s0 For stability there should be no sign change in first column So, 390 − K > 0 & K < 390 K >0 0 < K < 90 SOL 6.26

Option (C) is correct. Given transfer function is 100 s + 20s + 100 Characteristic equation of the system is given by H (s)) =

or

2

s2 + 20s + 100 = 0 ωn2 = 100 & ωn = 10 rad/sec. 2ξωn = 20 ξ = 20 = 1 2 # 10

(ξ = 1) so system is critically damped. SOL 6.27

Option (D) is correct. State space equation of the system is given by, o = AX + Bu X Y = CX Taking Laplace transform on both sides of the equations. sX (s) = AX (s) + BU (s) (sI − A) X (s) = BU (s) X (s) = (sI − A) − 1 BU (s) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CONTROL SYSTEMS

So

CHAP 6

` Y (s) = CX (s) Y (s) = C (sI − A) − 1 BU (s) Y (s) = C (sI − A) − 1 B T.F = U (s)

s 0 0 1 s −1 => (sI − A) = > H − > H 0 s 0 −2 0 s + 2H R V 1 W S1 1 >s + 2 1H = Ss s (s + 2)W (sI − A) − 1 = S0 1 W s (s + 2) 0 s S (s + 2) W T X Transfer function V V R R 1 W S 1 W S1 s s (s + 2)W 0 Ss (s + 2)W G (s) = C [sI − A] − 1 B = 81 0BSS >1H = 81 0BS 1 W W 1 S (s + 2) W S0 (s + 2) W X X T T 1 = s (s + 2) SOL 6.28

Option (A) is correct. Steady state error is given by,

Here

SOL 6.29

sR (s) ess = lim = G s " 0 1 + G (s) H (s) R (s) = L [r (t)] = 1 (Unit step input) s 1 G (s) = s (s + 2)

H (s) = 1 (Unity feed back) V R sb 1 l W S s W = lim = s (s + 2) G = 0 S So, ess = lim 1 s"0S W s " 0 s (s + 2) + 1 S1 + s (s + 2) W X T Option (D) is correct. For input u1 , the system is (u2 = 0)

System response is



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CHAP 6

CONTROL SYSTEMS

PAGE 351

(s − 1) (s + 2) (s − 1) H1 (s) = = (s − 1) 1 (s + 3) 1+ (s + 2) (s − 1) Poles of the system is lying at s =− 3 (negative s -plane) so this is stable. For input u2 the system is (u1 = 0)

System response is 1 (s − 1) (s + 2) H2 (s) = = − 1 s ( ) ( s − 1) (s + 3) 1+ 1 (s − 1) (s + 2) One pole of the system is lying in right half of s -plane, so the system is unstable. SOL 6.30

Option (B) is correct. Given function is. 1 s (s + 1) (s + 2) 1 G (jω) = jω (1 + jω) (2 + jω) G (s) =

By simplifying 1 − jω 2 − jω − jω 1 1 G (jω) = c 1 # jω − jω mc 1 + jω # 1 − jω mc 2 + jω # 2 − jω m − jω (2 − ω2 − j3ω) jω 1 − j ω 2 − j ω = ω2 mc 1 + ω2 mc 4 + ω2 m ω2 (1 + ω2) (4 + ω2) jω (ω2 − 2) − 3ω2 = 2 + ω (1 + ω2) (4 + ω2) ω2 (1 + ω2) (4 + ω2) G (jω) = x + iy x = Re [G (jω)] ω " 0 = − 3 =− 3 1#4 4 = c−

+

SOL 6.31

Option (D) is correct. Let response of the un-compensated system is 900 H UC (s) = s (s + 1) (s + 9) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 352

CONTROL SYSTEMS

CHAP 6

Response of compensated system. 900 HC (s) = G (s) s (s + 1) (s + 9) C Where GC (s) " Response of compensator Given that gain-crossover frequency of compensated system is same as phase crossover frequency of un-compensated system So, (ωg) compensated = (ωp) uncompensated − 180c = +H UC (jωp) − 180c =− 90c − tan− 1 (ωp) − tan− 1 a

ωp 9k

J ω + ωp N p 9 O 90c = tan KK 2 O ω K1− p O 9 P L ω2p 1− =0 9 −1

ωp = 3 rad/sec. So, (ωg) compensated = 3 rad/sec. At this frequency phase margin of compensated system is φPM = 180c + +HC (jωg) 45c = 180c − 90c − tan− 1 (ωg) − tan− 1 (ωg /9) + +GC (jωg) 45c = 180c − 90c − tan− 1 (3) − tan− 1 (1/3) + +GC (jωg) R 1 V 3 + S 3 WW + +GC (jωg) 45c = 90c − tan− 1 S SS1 − 3 b 1 lWW 3 45c = 90c − 90c +T+GC (jωgX) +GC (jωg) = 45c The gain cross over frequency of compensated system is lower than uncompensated system, so we may use lag-lead compensator. At gain cross over frequency gain of compensated system is unity so.

ωg

HC (jωg) = 1 900 GC (jωg) =1 ωg2 + 1 ωg2 + 81 GC (jωg) = 3 9 + 1 9 + 81 = 3 # 30 = 1 900 10 900 in dB GC (ωg) = 20 log b 1 l 10



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CHAP 6

CONTROL SYSTEMS

PAGE 353

=− 20 dB (attenuation) SOL 6.32

Option (B) is correct. Characteristic equation for the given system, K (s + 3) =0 1+ (s + 8) 2 (s + 8) 2 + K (s + 3) = 0 s2 + (16 + K) s + (64 + 3K) = 0 By applying Routh’s criteria. s2

1

64 + 3K

s1

16 + K

0

s0

64 + 3K

For system to be oscillatory 16 + K = 0 & K =− 16 Auxiliary equation A (s) = s2 + (64 + 3K) = 0 &

SOL 6.33

s2 + 64 + 3 # (− 16) = 0 s2 + 64 − 48 = 0 s2 =− 16 & jω = 4j ω = 4 rad/sec

Option (D) is correct. From the given block diagram we can obtain signal flow graph of the system. Transfer function from the signal flow graph is written as c 0 P + c1 P (c 0 + c1 s) P s s2 T.F = = 2 a Pb a Pb (s + a1 s + a 0) − P (b 0 + sb1) 1 + 1 + 20 − 2 0 − 1 s s s s (c 0 + c1 s) P 2 ^s + a1 s + a 0h = P (b + sb1) 1− 2 0 s + a1 s + a 0 from the given reduced form transfer function is given by T.F = XYP 1 − YPZ by comparing above two we have X = (c 0 + c1 s) 1 Y = 2 s + a1 s + a 0 Z = (b 0 + sb1) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 354

SOL 6.34

CONTROL SYSTEMS

CHAP 6

Option (A) is correct. For the given system Z is given by Z = E (s) Ki s Where E (s) " steady state error of the system Here sR (s) E (s) = lim s " 0 1 + G (s) H (s) Input

R (s) = 1 (Unit step) s ω2 G (s) = b Ki + K p le 2 s s + 2ξωs + ω2 o H (s) = 1 (Unity feed back)

So,

SOL 6.35

R V sb 1 l S W s Wb Ki l Z = lim S 2 s"0S K ω W s i S1 + b s + K p l (s2 + 2ξωs + ω2) W T X Ki = lim = Ki = 1 2 s"0 ω >s + (Ki + K p s) 2 H Ki (s + 2ξωs + ω2)

Option (C) is correct. System response of the given circuit can be obtained as. 1 bCs l e 0 (s) H (s) = = 1 ei (s) bR + Ls + Cs l H (s) =

1 b LC l

1 = LCs2 + RCs + 1 s2 + R s + 1 L LC

Characteristic equation is given by, s2 + R s + 1 = 0 L LC Here natural frequency ωn = 1 LC 2ξωn = R L Damping ratio ξ = R LC = R 2 2L Here

ξ = 10 2

C L

1 # 10− 3 = 0.5 (under damped) 10 # 10− 6



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CHAP 6

CONTROL SYSTEMS

PAGE 355

So peak overshoot is given by % peak overshoot = e

− πξ 1 − ξ2

# 100 = e

− π # 0.5 1 − (0.5) 2

# 100 = 16%

SOL 6.36

Option ( ) is correct.

SOL 6.37

Option (B) is correct. In standard form for a characteristic equation give as sn + an − 1 sn − 1 + ... + a1 s + a 0 = 0 in its state variable representation matrix A is given as R V 1 0 g 0 W S 0 S 0 0 1 g 0 W A =S W Sh h h h h W S− a 0 − a1 − a2 g − an − 1W T X Characteristic equation of the system is 4 s 2 − 2s + 1 = 0 So, a2 = 4, a1 =− 2, a 0 = 1 R 0 1 0 VW RS 0 1 0 VW S A =S 0 0 1 W=S 0 0 1 W SS− a − a − a WW SS− 1 2 − 4WW 0 1 2 X X T T

SOL 6.38

Option (A) is correct. In the given options only in option (A) the nyquist plot does not enclose the unit circle (− 1, j0), So this is stable.

SOL 6.39

Option (A) is correct. Given function is, 10 4 (1 + jω) H (jω) = (10 + jω) (100 + jω) 2 Function can be rewritten as, 0.1 (1 + jω) 10 4 (1 + jω) H (jω) = = 2 ω ω 2 ω ω 4 10 91 + j C 10 91 + j a1 + j 10 ka1 + j 100 k 10 100 C The system is type 0, So, initial slope of the bode plot is 0 dB/decade. Corner frequencies are ω1 = 1 rad/sec ω 2 = 10 rad/sec ω 3 = 100 rad/sec As the initial slope of bode plot is 0 dB/decade and corner frequency ω1 = 1 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 356

CONTROL SYSTEMS

CHAP 6

rad/sec, the Slope after ω = 1 rad/sec or log ω = 0 is(0 + 20) =+ 20 dB/dec. After corner frequency ω2 = 10 rad/sec or log ω2 = 1, the Slope is (+ 20 − 20) = 0 dB/dec. Similarly after ω3 = 100 rad/sec or log ω = 2 , the slope of plot is (0 − 20 # 2) =− 40 dB/dec. Hence (A) is correct option. SOL 6.40

Option (B) is correct. Given characteristic equation. (s2 − 4) (s + 1) + K (s − 1) = 0 K (s − 1) or =0 1+ 2 (s − 4) (s + 1) So, the open loop transfer function for the system. K (s − 1) , no. of poles n = 3 G (s) = (s − 2) (s + 2) (s + 1) no of zeroes m = 1 Steps for plotting the root-locus (1) Root loci starts at s = 2, s =− 1, s =− 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n−m (I)

(2 # 0 + 1) 180c = 90c (3 − 1)

(II)

(2 # 1 + 1) 180c = 270c (3 − 1)

(4) The two asymptotes intersect on real axis at / Poles − / Zeroes = (− 1 − 2 + 2) − (1) =− 1 x = 3−1 n−m (5) Between two open-loop poles s =− 1 and s =− 2 there exist a break away point. (s2 − 4) (s + 1) K =− (s − 1) dK = 0 ds s =− 1.5 SOL 6.41

Option (C) is correct. Closed loop transfer function of the given system is, s2 + 4 T (s) = (s + 1) (s + 4) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

T (jω) =

PAGE 357

(jω) 2 + 4 (jω + 1) (jω + 4)

If system output is zero 4 − ω2 T (jω) = =0 ^ jω + 1h (jω + 4) 4 − ω2 = 0 ω2 = 4 & ω = 2 rad/sec SOL 6.42

Option (A) is correct. From the given plot we can see that centroid C (point of intersection) where asymptotes intersect on real axis) is 0 So for option (a) G (s) = K3 s Centroid =

SOL 6.43

/ Poles − / Zeros = 0 − 0 = 0 n−m

3−0

Option (A) is correct. Open loop transfer function is. (s + 1) G (s) = s2 jω + 1 − ω2 Phase crossover frequency can be calculated as. G (jω) =

+G (jωp) =− 180c tan− 1 (ωp) =− 180c ωp = 0 Gain margin of the system is. G.M =

SOL 6.44

1 G (jωp)

=

1 = ω2p + 1 ω2p

ω2p =0 ω2p + 1

Option (C) is correct. Characteristic equation for the given system 1 + G (s) H (s) = 0 (1 − s) 1+K =0 (1 + s) (1 + s) + K (1 − s) = 0 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 358

CONTROL SYSTEMS

CHAP 6

s (1 − K) + (1 + K) = 0 For the system to be stable, coefficient of characteristic equation should be of same sign. 1 − K > 0, K + 1 > 0 K < 1, K > − 1 −1 < K < 1 K <1 SOL 6.45

Option (C) is correct. In the given block diagram

Steady state error is given as ess = lim sE (s) s"0

E (s) = R (s) − Y (s) Y (s) can be written as Y (s) = :"R (s) − Y (s), 3 − R (s)D 2 s s+2 = R (s) ; 6 − 2 E − Y (s) ; 6 E s (s + 2) s + 2 s (s + 2) Y (s) ;1 +

6 = R (s) ; 6 − 2s E s (s + 2)E s (s + 2)

(6 − 2s) (s + 2s + 6) (6 − 2s) So, E (s) = R (s) − 2 R (s) (s + 2s + 6) 2 = R (s) ; 2 s + 4s E s + 2s + 6 For unit step input R (s) = 1 s Y (s) = R (s)

2

Steady state error ess = lim sE (s) s"0

(s2 + 4s) ess = lim =s 1 2 =0 s (s + 2s + 6)G s"0



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CHAP 6

SOL 6.46

CONTROL SYSTEMS

PAGE 359

Option (B) is correct. When it passes through negative real axis at that point phase angle is − 180c So +G (jω) H (jω) =− 180c − 0.25jω − π =− π 2 − 0.25jω =− π 2 j0.25ω = π 2 jω=

π 2 # 0.25

s = jω = 2π Put s = 2π in given open loop transfer function we get − 0.25 # 2π =− 0.5 G (s) H (s) s = 2π = πe 2π So it passes through (− 0.5, j0) SOL 6.47

Option (C) is correct. Open loop transfer function of the system is given by. G (s) H (s) = (K + 0.366s) ; 1 E s (s + 1) G (jω) H (jω) =

K + j0.366ω jω (jω + 1)

Phase margin of the system is given as φPM = 60c = 180c + +G (jωg) H (jωg) Where ωg " gain cross over frequency = 1 rad/sec 0.366ωg So, 60c = 180c + tan− 1 b − 90c − tan− 1 (ωg) K l = 90c + tan− 1 b 0.366 l − tan− 1 (1) K = 90c − 45c + tan− 1 b 0.366 l K 15c = tan− 1 b 0.366 l K 0.366 = tan 15c K K = 0.366 = 1.366 0.267



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PAGE 360

SOL 6.48

CONTROL SYSTEMS

CHAP 6

Option (A) is correct. Given state equation. o (t) = >0 1 H X (t) + >1H u (t) X 0 −3 0 Here 0 1 1 ,B = > H A => H 0 −3 0 State transition matrix is given by, φ (t) = L− 1 [(sI − A) − 1] s 0 0 1 s −1 => [sI − A] = > H − > H 0 s 0 −3 0 s + 3H R V 1 W S1 1 >s + 3 1H = Ss s (s + 3)W [sI − A] − 1 = S0 1 W s (s + 3) 0 s S (s + 3) W T X φ (t) = L− 1 [(sI − A) − 1] 1 => 0

SOL 6.49

1 3

(1 − e−3t) H e−3t

Option (C) is correct. State transition equation is given by X (s) = Φ (s) X (0) + Φ (s) BU (s) Here Φ (s) " state transition matrix R V 1 W S1 s s (s + 3)W Φ (s) = SS 1 W S0 (s + 3) W T X X (0) " initial condition −1 X (0) = > H 3

So

1 B => H 0 V R R1 1 W 1 VW S1 S s s (s + 3)W − 1 Ss (s + 3) s W 1 1 X (s) = SS 1 W> 3 H + S 1 W>0H s 0 0 S (s + 3) W S s+3 W X V T X V TR R 1 3 1 S− + W 1 s s (s + 3)W 1 1 SS− s + 3WW s =S + = + s > 2H S 0+ 3 W >0H s S 3 W 0 S s+3 W S s+3 W T X T X



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CHAP 6

CONTROL SYSTEMS

PAGE 361

R V S 12 − 1 W s + 3W X (s) = Ss S W 3 S s+3 W T X Taking inverse Laplace transform, we get state transition equation as, t − e− 3t X (t) = > − 3t H 3e SOL 6.50

Option () is correct Phase margin of a system is the amount of additional phase lag required to bring the system to the point of instability or (− 1, j0) So here phase margin = 0c

SOL 6.51

Option (D) is correct. Given transfer function is 5 F (s) = 2 s (s + 3s + 2) F (s) =

5 s (s + 1) (s + 2)

By partial fraction, we get 5 F (s) = 5 − 5 + 2s s + 1 2 (s + 2) Taking inverse Laplace of F (s) we have f (t) = 5 u (t) − 5e− t + 5 e− 2t 2 2 So, the initial value of f (t) is given by lim f (t) = 5 − 5 + 5 (1) = 0 2 2 t"0 SOL 6.52

Option (C) is correct. In A.C techo-meter output voltage is directly proportional to differentiation of rotor displacement. e (t) \ d [θ (t)] dt e (t) = Kt

dθ (t) dt

Taking Laplace tranformation on both sides of above equation E (s) = Kt sθ (s) So transfer function E (s) T.F = = ^Kt h s θ (s) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 362

SOL 6.53

CONTROL SYSTEMS

CHAP 6

Option (B) is correct. Given characteristic equation, s3 − 4s2 + s + 6 = 0 Applying Routh’s method, s3

1

1

s2

−4

6

s1

− 4 − 6 = 2.5 −4

0

6 s0 There are two sign changes in the first column, so no. of right half poles is 2. No. of roots in left half of s -plane = (3 − 2) = 1 SOL 6.54

Option (B) is correct. Block diagram of the system is given as.

From the figure we can see that C (s) = :R (s) 1 + R (s)D 1 + R (s) s s C (s) = R (s) : 12 + 1 + 1D s s 2 C (s) = 1 + s2+ s R (s) s SOL 6.55

Option (A) is correct. Characteristic equation is given by, sI − A = 0 s 0 0 2 s −2 (sI − A) = > H − > H = > = s2 − 4 = 0 0 s 2 0 −2 s H s1, s2 = ! 2

SOL 6.56

Option (D) is correct. For the given system, characteristic equation can be written as, GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

1+

PAGE 363

K (1 + sP) = 0 s (s + 2)

s (s + 2) + K (1 + sP) = 0 s2 + s (2 + KP) + K = 0 From the equation. ωn = K = 5 rad/sec (given) So, K = 25 and 2ξωn = 2 + KP 2 # 0.7 # 5 = 2 + 25P or P = 0.2 so K = 25 , P = 0.2 SOL 6.57

Option (D) is correct. Unit - impulse response of the system is given as, c (t) = 12.5e− 6t sin 8t , t $ 0 So transfer function of the system. H (s) = L [c (t)] = 12.52# 8 2 (s + 6) + (8) 100 H (s) = 2 s + 12s + 100 Steady state value of output for unit step input, lim y (t) = lim sY (s) = lim sH (s) R (s)

t"3

s"0

s"0

100 1 = 1.0 = lim s ; 2 E s s"0 s + 12s + 100 SOL 6.58

Option (A) is correct. System response is. s s+1 jω H (jω) = jω + 1 H (s) =

Amplitude response ω ω+1 Given input frequency ω = 1 rad/sec. 1 So H (jω) ω = 1 rad/sec = = 1 1+1 2 Phase response H (jω) =

θh (ω) = 90c − tan− 1 (ω) θh (ω) ω = 1 = 90c − tan− 1 (1) = 45c GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 364

CONTROL SYSTEMS

CHAP 6

So the output of the system is y (t) = H (jω) x (t − θh) = 1 sin (t − 45c) 2 SOL 6.59

Option (C) is correct. Given open loop transfer function jaω + 1 G (jω) = (jω) 2 Gain crossover frequency (ωg) for the system. G (jωg) = 1 a2 ωg2 + 1 =1 − ωg2 a2 ωg2 + 1 = ωg4 ωg4 − a2 ωg2 − 1 = 0 Phase margin of the system is

...(1)

φPM = 45c = 180c + +G (jωg) 45c = 180c + tan− 1 (ωg a) − 180c tan− 1 (ωg a) = 45c ωg a = 1 From equation (1) and (2) 1 −1−1 = 0 a4 a 4 = 1 & a = 0.841 2 SOL 6.60

Option (C) is correct. Given system equation is. d 2 x + 6 dx + 5x = 12 (1 − e− 2t) dt dt 2 Taking Laplace transform on both side. s2 X (s) + 6sX (s) + 5X (s) = 12 :1 − 1 D s s+2 (s2 + 6s + 5) X (s) = 12 ; 2 E s (s + 2) System transfer function is X (s) =

24 s (s + 2) (s + 5) (s + 1)

Response of the system as t " 3 is given by lim f (t) = lim sF (s) (final value theorem) t"3

s"0



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CHAP 6

CONTROL SYSTEMS

PAGE 365

24 = lim s ; s"0 s (s + 2) (s + 5) (s + 1)E = SOL 6.61

24 = 2.4 2#5

Option (A) is correct. Transfer function of lead compensator is given by. K a1 + s k a H (s) = s a1 + b k R ω V S1 + j a a kW H (jω) = K S W SS1 + j a ω kWW b T X So, phase response of the compensator is. θh (ω) = tan− 1 a ω k − tan− 1 a ω k a b Jω − ωN ω (b − a) = tan K a b2 O = tan− 1 ; KK O ab + ω2 E 1+ ω O ab P L θh should be positive for phase lead compensation ω (b − a) So, θh (ω) = tan− 1 ; >0 ab + ω2 E −1

b >a SOL 6.62

Option (A) is correct. Since there is no external input, so state is given by X (t) = φ (t) X (0) φ (t) "state transition matrix X [0] "initial condition e− 2t 0 2 So x (t) = > H> H 0 e− t 3 2e− 2t x (t) = > − t H 3e At t = 1, state of the system 0.271 2e− 2 x (t) t = 1 = > − 1H = > 1.100H 2e

SOL 6.63

Option (B) is correct. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 366

CONTROL SYSTEMS

CHAP 6

Given equation d2 x + 1 dx + 1 x = 10 + 5e− 4t + 2e− 5t dt2 2 dt 18 Taking Laplace on both sides we have s2 X (s) + 1 sX (s) + 1 X (s) = 10 + 5 + 2 s 2 18 s+4 s+5 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) (s2 + 1 s + 1 ) X (s) = 2 18 s (s + 4) (s + 5) System response is,

X (s) =

10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) s (s + 4) (s + 5) bs2 + 1 s + 1 l 2 18

=

10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) s (s + 4) (s + 5) bs + 1 lbs + 1 l 3 6

We know that for a system having many poles, nearness of the poles towards imaginary axis in s -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 axis s1 =− 1 , s2 =− 1 3 6 So, time constants )

SOL 6.64

τ1 = 3 sec τ2 = 6 sec

Option (A) is correct. Steady state error for a system is given by sR (s) ess = lim s " 0 1 + G (s) H (s) 1 Where input R (s) = (unit step) s G (s) = b 3 lb 15 l s + 15 s + 1

So

H (s) = 1 (unity feedback) sb 1 l s ess = lim = 15 = 15 60 15 + 45 45 s"0 1+ (s + 15) (s + 1) %ess = 15 # 100 = 25% 60

SOL 6.65

Option (C) is correct. Characteristic equation is given by 1 + G (s) H (s) = 0 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

Here

So,

PAGE 367

H (s) = 1

(unity feedback) G (s) = b 3 lb 15 l s + 15 s + 1 1 + b 3 lb 15 l = 0 s + 15 s + 1 (s + 15) (s + 1) + 45 = 0 s2 + 16s + 60 = 0 (s + 6) (s + 10) = 0 s =− 6, − 10

SOL 6.66

Option (A) is correct. Given equation can be written as, d 2 ω =− β dω − K 2 ω + K V J dt LJ LJ a dt 2 Here state variables are defined as, dω = x 1 dt ω = x2 So state equation is 2 xo1 =− B x1 − K x2 + K Va J LJ LJ xo2 = dω = x1 dt In matrix form K/LJ xo1 − B/J − K 2 /LJ x1 >o H = > >x H + > 0 H Va H x2 1 0 2 R 2 V Sd ω W S dt2 W = P >dωH + QVa dt S dω W S dt W X So matrix P is T − B/J − K 2 /LJ > 1 H 0

SOL 6.67

Option (C) is correct. Characteristic equation of the system is given by 1 + GH = 0 K =0 1+ s (s + 2) (s + 4) s (s + 2) (s + 4) + K = 0 s3 + 6s2 + 8s + K = 0 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 368

CONTROL SYSTEMS

CHAP 6

Applying routh’s criteria for stability s3

1

8

2

6

K

s

s1 s0

K − 48 6 K

System becomes unstable if K − 48 = 0 & K = 48 6 SOL 6.68

Option (A) is correct. The maximum error between the exact and asymptotic plot occurs at corner frequency. Here exact gain(dB) at ω = 0.5a is given by 1 + ω2 a (0.5a) 2 1/2 = 20 log K − 20 log ;1 + E = 20 log K − 0.96 a2 Gain(dB) calculated from asymptotic plot at ω = 0.5a is = 20 log K Error in gain (dB) = 20 log K − (20 log K − 0.96) dB = 0.96 dB Similarly exact phase angle at ω = 0.5a is. θh (ω) ω = 0.5a =− tan− 1 a ω k =− tan− 1 b 0.5a l =− 26.56c a a Phase angle calculated from asymptotic plot at (ω = 0.5a) is − 22.5c Error in phase angle =− 22.5 − (− 26.56c) = 4.9c gain(dB) ω = 0.5a = 20 log K − 20 log

SOL 6.69

2

Option (B) is correct. Given block diagram

Given block diagram can be reduced as



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CHAP 6

CONTROL SYSTEMS

Where

G1 =

G2 =

1 bs l 3 1 + b1l s

=

1 bs l 1 + b 1 l 12 s

PAGE 369

1 s+3

=

1 s + 12

Further reducing the block diagram.

2G1 G2 1 + (2G1 G2) 9 (2) b 1 lb 1 l s + 3 s + 12 = 1 + (2) b 1 lb 1 l (9) s + 3 s + 12 2 2 = 2 = (s + 3) (s + 12) + 18 s + 15s + 54 1 2 = = s (s + 9) (s + 6) 27 a1 + ka1 + s k 9 6

Y (s) =

SOL 6.70

Option (C) is correct. Given state equation is, o = AX X Taking Laplace transform on both sides of the equation, sX (s) − X (0) = AX (s) (sI − A) X (s) = X (0) X (s) = (sI − A) − 1 X (0) = Φ (s) X (0) Where φ (t) = L− 1 [Φ (s)] = L− 1 [(sI − A) − 1] is defined as state transition matrix

SOL 6.71

Option (B) is correct. State equation of the system is given as, o = >2 3H X + >1H u X 0 5 0 Here

1 2 3 A = > H, B = > H 0 0 5

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PAGE 370

CONTROL SYSTEMS

CHAP 6

2 3 1 2 AB = > H> H = > H 0 5 0 0 1 2 U = [B : AB] = > H 0 0 U = (1 # 0 − 2 # 0) = 0 Matrix U is singular, so the system is uncontrollable. Check for Stability: Characteristic equation of the system is obtained as, sI − A = 0 s 0 2 3 (sI − A) = > H − > H 0 s 0 5 s − 2 −3 => 0 s − 5H sI − A = (s − 2) (s − 5) = 0 s = 2, s = 5 There are two R.H.S Poles in the system so it is unstable. SOL 6.72

Option (B) is correct. Given open loop transfer function, no of poles = 2 G (s) = K2 , s no of zeroes = 0 For plotting root locus: (1) Poles lie at s1, s2 = 0 (2) So the root loci starts (K = 0) from s = 0 and s = 0 (3) As there is no open-loop zero, root loci terminates (K = 3) at infinity. (4) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n−m So the two asymptotes are at an angle of (i)

(2 # 0 + 1) 180c = 90c 2

(ii)

(2 # 1 + 1) 180c = 270c 2

(5) The asymptotes intersect on real axis at a point given by / Poles − / zeros = 0 − 0 = 0 x= n−m 2 (6) Break away points 1 + K2 = 0 s GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 371

K =− s2 dK =− 2s = 0 & s = 0 ds So the root locus plot is.

SOL 6.73

Option (A) is correct. System is described as. d2 y dy = du + 2u + dt dt2 dt Taking Laplace transform on both sides. s2 Y (s) + sY (s) = sU (s) + 2U (s) (s2 + s) Y (s) = (s + 2) U (s) So, the transfer function is (s + 2) Y (s) = 2 T.F = U (s) (s + s)

SOL 6.74

Option (A) is correct. Here, we have 1 2 0 A = > H, B = > H, C = [4, 0] 1 0 4 We know that transfer function of the system is given by. Y (s) G (s) = = C (sI − A) − 1 B U (s) s 0 2 0 s−2 0 [sI − A ] = > H − > H = > 0 s 0 4 0 s − 4H

R V S 1 W 0 (s − 4) 0 (s − 2) S W 1 −1 = (sI − A) = > 1 W (s − 2)H S (s − 2) (s − 4) 0 S 0 (s − 4)W TR X R V V S 1 W S 1 0 W1 (s − 2) Y (s) S(s − 2)W W So, = [4 0] SS >1H = [4 0] S W 1 U (s) 1 W S(s − 4)W S 0 (s − 4)W T T X X GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CONTROL SYSTEMS

CHAP 6

Y (s) = 4 U (s) (s − 2) Here input is unit impulse so U (s) = 1 and output Y (s) = 4 (s − 2) Taking inverse Laplace transfer we get output y (t) = 4e2t SOL 6.75

Option (D) is correct. Given state equation R V S0 1 0 0W S W o = S0 0 1 0W X X S0 0 0 1W S0 0 0 1W T X R V 0 0 1 0 S W S0 0 1 0W Here A =S W S0 0 0 1W S0 0 0 1W T obtained X as Eigen value can be

or SOL 6.76

A − λI = 0 R V R S0 1 0 0W Sλ S0 0 1 0W S0 (A − λI) = S W−S S0 0 0 1W S0 S0 0 0 1W S0 T X T 3 A − λI = λ (1 − λ) = 0 λ1, λ2, λ3 = 0 , λ4 = 1

0 λ 0 0

0 0 λ 0

V V R 0W S− λ 1 0 0 W 0W S 0 − λ 1 0 W =S W 0W S 0 0 − λ 1 WW λW S 0 0 0 1 − λW X X T

Option (A) is correct. Input-output relationship is given as d 2y dy du 2 + 2 dt + 10y = 5 dt − 3u dt Taking Laplace transform on both sides with zero initial condition. s 2 Y (s) + 2sY (s) + 10Y (s) = 5sU (s) − 3U (s) (s2 + 2s + 10) Y (s) = (5s − 3) U (s) (5s − 3) Output Y (s) = 2 U (s) (s + 2s + 10) With no input and with given initial conditions, output is obtained as d 2y dy 2 + 2 dt + 10y = 0 dt GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 373

Taking Laplace transform (with initial conditions) [s2 Y (s) − sy (0) − y' (0)] + 2 [sY (s) − y (0)] + 10Y (s) = 0 Given that y' (0) =− 4 , y (0) = 1 [s2 Y (s) − s − (− 4)] + 2 (s − 1) + 10Y (s) = 0 Y (s) [s2 + 2s + 10] = (s − 2) (s − 2) Y (s) = 2 (s + 2s + 10) Output in both cases are same so (5s − 3) (s − 2) U (s) = 2 2 (s + 2s + 10) (s + 2s + 10) U (s) =

(s − 2) (5s − 10) =1 5 (5s − 3) (5s − 3)

(5s − 3) 7 = 1= − 5 5s − 3 (5s − 3)G 7 U (s) = 1 ;1 − 5 (5s − 3)E Taking inverse Laplace transform, input is u (t) = 1 :δ (t) − 5 e3/5t u (t)D = 1 δ (t) − 7 e3/5t u (t) 5 5 5 25 SOL 6.77

Option (C) is correct. d 2 y dy + − 2y = u (t) e− t dt 2 dt State variable representation is given as o = AX + Bu X Or Here

...(1)

x1 xo1 > o H = A >x H + Bu x2 2 x1 = y , x 2 = b

dy − y l et dt

dx1 = dy = x e− t + y = x e− t + x 2 2 1 dt dt dx1 = x + x e− t + (0) u (t) or 1 2 dt Similarly 2 dx2 = d y et + dy et − et dy − yet dt dt dt dt 2 2 d y Put from equation (1) dt 2

...(2)



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CONTROL SYSTEMS

So,

CHAP 6

dx2 = u (t) e− t − dy + 2y et − yet : D dt dt = u (t) −

dy t e + 2yet − yet = u (t) − [x2 e− t + y] et + yet dt

= u (t) − x2 dx2 = 0 − x + u (t) 2 dt

...(3)

From equation (2) and (3) state variable representation is 0 xo1 1 e− t x 1 >o H = > >x H + >1H u (t) H x2 0 −1 2 SOL 6.78

Option (B) is correct. Characteristic equation of the system 1 + G (s) = 0 2 (s + α) =0 1+ s (s + 2) (s + 10) s (s + 2) (s + 10) + 2 (s + α) = 0 s3 + 12s2 + 20s + 2s + 2α = 0 s3 + 12s2 + 22s + 2α = 0 2α =0 1+ 3 s + 12s2 + 22s No of poles n = 3 No. of zeros m = 0 Angle of asymptotes (2q + 1) 180c φA = , q = 0, 1, 2 n−m φA =

(2q + 1) 180c = (2q + 1) 60c 3

φA = 60c, 180c, 300c SOL 6.79

Option (A) is correct. Asymptotes intercepts at real axis at the point / real Parts of Poles − / real Parts of zeros C = n−m Poles at

s1 = 0 s2 =− 2 s 3 =− 10



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CHAP 6

CONTROL SYSTEMS

So SOL 6.80

PAGE 375

C = 0 − 2 − 10 − 0 =− 4 3−0

Option (C) is correct. Break away points dα = 0 ds α =− 1 [s3 + 12s2 + 22s] 2 dα =− 1 [3s2 + 24s + 22] = 0 2 ds s1, s2 =− 1.056, − 6.9433

SOL 6.81

Option ( ) is correct.

SOL 6.82

Option (A) is correct. Given state equation o = >− 3 1 H X X 0 −2 Or

o = AX , where A = >− 3 1 H X 0 −2

Taking Laplace transform on both sides. sX (s) − X (0) = AX (s) X (s) (sI − A) = X (0) X (s) = (sI − A) − 1 X (0) Steady state value of X is given by xss = lim sX (s) = lim s (sI − A) − 1 X (0) s"0

s"0

s 0 −3 1 s + 3 −1 => (sI − A) = > H − > H 0 s 0 −2 0 s + 2H s+2 1 1 > 0 s + 3H (s + 3) (s + 2) R V 1 S 1 W (s + 3) (s + 2) (s + 3)W S =S W 1 S 0 (s + 2) W T X So the steady state value R V 1 S 1 W (s + 3) (s + 2) (s + 3)W 10 S xss = lim s S W>− 10H 1 s"0 0 S (s + 2) W T X GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia (sI − A− 1) =

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PAGE 376

CONTROL SYSTEMS

CHAP 6

R V 10 S 10 − W 0 (s + 3) (s + 2) (s + 3)W S => H = lim s S W 0 s"0 − 10 S W (s + 2) T X SOL 6.83

Option (D) is correct. Initial slope of the bode plot is − 40 dB/dec. So no. of poles at origin is 2. Then slope increased by − 20 dB/dec. at ω = 2 rad/sec, so one poles lies at this frequency. At ω = 5 rad/sec slope changes by + 20 dB/dec, so there is one zero lying at this frequency. Further slope decrease by − 20 dB/dec at ω = 25 so one pole of the system is lying at this frequency. Transfer function K (s + 5) H (s) = 2 s (s + 2) (s + 25) At ω = 0.1, gain is 54 dB, so 5K 54 = 20 log (0.1) 2 (2) (25) K = 50 H (s) =

SOL 6.84

50 (s + 5) s (s + 2) (s + 25) 2

Option (B) is correct. Open loop transfer function of the system is 10 4 G (s) = s (s + 10) 2 10 4 10 4 = jω (jω + 10) 2 jω (100 − ω2 + j20ω) 10 4 G (jω) = ω (100 − ω2) 2 + 400ω2

G (jω) = Magnitude

At ω = 20 rad/sec 10 4 10 4 =1 = 20 # 5 # 102 20 9 # 10 4 + 16 # 10 4 Magnitude in dB = 20 log 10 G (j20) = 20 log 10 1 = 0 dB G (j20) =

SOL 6.85

Option (C) is correct. Since G (j ω) = 1 at ω = 20 rad/sec, So this is the gain cross-over frequency ωg = 20 rad/sec Phase margin φPM = 180c + +G (jωg) GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 6

CONTROL SYSTEMS

PAGE 377

20 ωg +G (jωg) =− 90c − tan− 1 = 100 − ωg2 G φPM = 180 − 90c − tan− 1 ; 20 # 20 2 E =− 36.86c 100 − (20) SOL 6.86

Option (C) is correct. To calculate the gain margin, first we have to obtain phase cross over frequency (ωp). At phase cross over frequency +G (jωp) 20ωp − 90c − tan− 1 = 100 − ω2p G 20ωp tan− 1 = 100 − ω2p G 100 − ω2p

=− 180c =− 180c = 90c

= 0 & ωp = 10 rad/sec. 1 Gain margin in dB = 20 log 10 e G (jωp) o G (jωp) = G (j10) =

10 4 10 (100 − 100) 2 + 400 (10) 2

10 4 =5 10 # 2 # 102 G.M. = 20 log 10 b 1 l =− 13.97 dB 5 =

SOL 6.87

Option (B) is correct. Since gain margin and phase margin are negative, so the system is unstable.

SOL 6.88

Option (C) is correct. Given characteristic equation s3 + s2 + Ks + K = 0 K (s + 1) =0 1+ 3 s + s2 K (s + 1) =0 1+ 2 s (s + 2) so open loop transfer function is K (s + 1) G (s) = 2 s (s + 1) root-locus is obtained in following steps: 1. Root-loci starts(K = 0 ) at s = 0 , s = 0 and s =− 2 2.

There is one zero at s =− 1, so one of root-loci terminates at s =− 1 and other two terminates at infinity



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PAGE 378

CONTROL SYSTEMS

3.

No. of poles n = 3 , no of zeros ,m = 1

4.

Break - Away points

CHAP 6

dK = 0 ds Asymptotes meets on real axis at a point C / poles − / zeros C = n−m (0 + 0 − 2) − (− 1) =− 0.5 = 3−1

***********



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