Control Systems DRILL (200 plus New Questions with Meticulous Solutions) for GATE/ESE-2018 9788193007884, 8193007883

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Control systems problems and solutions

conceptually empowered learning A K Tripathi MD, Mechasoft Publishers and Educators and Asstt. Prof. Electronics Engineering Department IERT, Degree Division, Allahabad (INDIA) Akhilesh Kumar Pandey Asstt. Prof. Electronics Engineering IERT, Degree Division, Allahabad (INDIA)

Mechasoft Publishers, Allahabad (INDIA)

Second Edition : 2021 MRP : 275.00 Printed at Gurukripa Printers ,Allahabad ISBN : 978-81-930078-8-4 Copyright © : 2015,2021 Mechasoft Publishers No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording or otherwise or stored in a database or retrieval system without the prior written permission of the publisher. Mechasoft Publishers, Allahabad (INDIA)

E-mail : [email protected]

Dedicated to Promotion and Elevation of Engineering Education

and Way of Life as Gifted by

ALMIGHTY "of which We are an Integral Part"

M

e

h c

s a

t f o

Preface MECHASOFT Publishers and Educators, are vehemently and resolutely dedicated to promotion and elevation of Engineering education in India and abroad. In sync with this endeavor and owing to ever growing demand , we would like to facilitate the students and professionals of Electronics, Electrical and Instrumentation Engineering with conceptually empowered and accelerated pace of learning by means of a set of 265 new GATE pattern problems together with meticulous solutions. The key idea behind this book is to portray a 'dialogue between a teacher and taught' in the form of questions and answers to unfold the intricacies of Control Systems so as to bring cryptic concepts within the grasp of students and put them at natural ease to take competitive/university exams with ample enthusiasm and condence. The second edition of the Control Systems has been enriched with two Drill Tests; Test I consisting of 25 problems of 1mark each and Test II consisting of another 25 problems of 2 marks each. We are potentially condent that the students shall be able to grade their preparedness through these tests and the grade achieved therefrom , will enable them to further strengthen the areas of good understanding while decimating the weak areas therein. This might fascinate readers to know that our previously published titles: GATE Solutions to Enhance Solving Skills for EC, EE, Digital Electronics text book ,Analog Circuits Drill, Digital Electronics Drill , Control Systems Drill , Communication Systems Drill and Electrical Machines Drill are in wide circulation. Thousands of units of these titles, have been shipped through www.amazon.in/kindle and www.ipkart.com till now and over 500 plus reviews have been sought from top GATE rankers and renowned teachers. The reviews so far, have been unarguably inspiring and do render potential impetus to further develop allied texts. We look forward with ample eagerness, to more reviews for this text as well. We would not forget to thank Mr. Trilok Nath Mishra, a sincere personality for his consistent persuasion to complete the task.We express thanks from our deep within, to Shreyansh Mishra for his intelligent and quality typing. The nal script has been prepared with utmost care. However, thinking on line that, there is always a room for improvement in anything done, we would welcome and greatly appreciate suggestions for further improvements. A K Tripathi Akhilesh Kumar Pandey

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Contents

Preface

1.

Introductory Concepts (Feedback Properties, Block Diagram and Signal Flow Graph) 2. Time Response 3. Routh Stability 4. Root Locus 5. Frequency Response (Bode and Nyquist Plots) 6. State Space Model 7. Controllers 8. Answer Key (Drill problems) 9. Test Drill I and Answer Key 10. Test Drill I1 and Answer Key

IV Questions

Solutions

1-8

49-56

8-13 14-17 17-23 24-37

57-65 65-69 69-78 78-94

38-44 44-47 48 110-114 115-121

94-103 103-107

Questions 1. Introductory Concepts (Feedback Properties, Block Diagram and Signal Flow Graph) Q1. In the feedback control system shown below, T(s) = Y(s)/R(s). The nominal values of b and c are 2 and 3 respectively. SaT represents sensitivity of T(s) to changes in a about its nominal value. If, in steady state, |SaT| = 0.4, the nominal value of a is _____________.

+ –

R(s)

a s+b

G(s) =

Y(s)

H (s) = c

Q2. The systems S1 and S2 shown in Fig. I and Fig. II have time constants τ1 and τ2 respectively.

10 s+3

+

The ratio τ1/τ2 is (a) 2 / 53

4s - 1 3s + 2

–

Fig.I

Fig.II

(b) 159 / 2

Q3. In the block diagram of control system shown below, Y(s) = YR(s) + YD(s) where YR(s) is the output component due to reference input R(s) and Y D (s) is output component due to disturbance input D(s). Which one of the following will ensure that YD(s) = 0?

(c) 2 / 159

(d) 53 / 2 D(s) Gff(s)

+ R(s) +

+

A(s)

+

–

1 (a) G (s) = ff A(s)

(c) G ff (s) = -

5

A(s) B(s)

Common data for Q4 and Q5 K A system with open loop transfer function G(s) = s+5 below.

Y(s)

C(s) (b) G ff (s) = -

1

B(s)

(d) G ff (s) = -

1 B(s) 1 C(s)

is placed in unity feedback loop as shown + –

K s+5

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Q4. Let τOL be the time constant of open loop system and τCL be that of closed loop system. Then, in order that τOL = 20 τCL, the value of K is _____________. Q5. For the value of K found in Q4, the open loop dc gain gOL and closed loop dc gain gCL are (a) gOL = 19 , gCL = 0.95 (b) gOL = 100 , gCL = 5 (c) gOL = 5 , gCL = 100 (d) gOL = 0.95 , gCL = 19 Q6. Consider the following statements in respect of introducing feedback in a control system. I. The feedback in a control system is employed to reduce sensitivity, improve transient response and minimize the effect of disturbance signals. II. The signal to noise ratio of feedback sensor must be high. III. The feedback increases the range of frequencies over which the system can be made to respond desirably. Of these , (a) only I and II are true. (b) only I is true. (c) only I and III are true. (d) I, II and III all are true. Q7. The signal flow graph and equivalent block diagram shown below in Fig.1 and Fig. II respectively represent the same system. –s X1(s)

–1

0.5

–s 4

2

–1

X2(s)

–p

(b) s

X2(s)

Fig.II -block diagram

Fig.I -signal flow graph The value of p should be (a) 1

4s3 s3 + 3s 2 + 3s + 1

X1(s)

1

−1

(c) s

(d) s

−2

Q8. The block diagrams shown below in Fig.I and Fig.II represent the same system. + –

X1(s)

P(s)

X2(s)

X1(s)

+ –

A(s)

Q(s) Fig. I

Fig. II

Choose the correct option. (a) A(s) =

P(s) 1 + P(s)[Q(s) - 1]

(b) A(s) =

P(s) Q(s) 1 + Q(s)[P(s) - 1]

(c) A(s) =

P(s) 1 + P(s) Q(s)

(d) A(s) =

Q(s) 1 + P(s) Q(s)

X2(s)

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Q9. The transfer function of a continuous time system shown below , has the poles located at s= –2 and s = –3 . The values of λ1 and λ2 are respectively (a) –5 and –6

X(s)

+

Y(s)

+ +

+

(b) 5 and 6

λ1

λ2

( c)1 and 5 (d) –1 and –5

s

–1

s–1

Q10. A system has block diagram as shown in Fig. I. A partially reduced block diagram of same system is shown in Fig. - II. s s+3 R(s) + 1 3s + + –

–

s+2

s+4

C(s)

4

X(s)

Y(s) C(s)

R(s) 5

Fig.I If Y(s) =

4s 2 + 13s

(s + 3)(s + 4) s+4 (a) X(s) = 2 s + 10s + 39

( c) X(s) =

1 s

Fig.II

, then 4s + 1 (b) X(s) = 2 s + 24s + 15

4s + 1 (d) X(s) = 2 s + 10s + 39

s+4 s + 24s + 15 2

Q11. In the block diagram shown below, the overall transfer function is desired to be Y(s) -s T (s ) = = . K U (s ) s + 20 The value of K is ________.

+ U(s)

–

10 s + 10

– +

Y(s)

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Q12. In the system shown below, r1(t) = 3e−t and r2(t) = 4u(t); u(t) is unit step function. R2(s) Assume zero initial conditions. R1(s)

The output Y(s) is s 2 (s + 1)(s + 4) (a) 2 4s + 25s + 12

+

3 s+3

+ +

(b)

3s 2 + 25s + 4

+

1 s +1

4s 2 + 25s + 12 (c) 2 s (s + 1)(s + 4)

s(s + 1)(s + 4)

(d)

Y(s)

s(s + 1)(s + 4) 3s 2 + 25s + 4

Q13. Consider the following statements regarding SIGNAL FLOW GRAPH. 1.The signal at any node is the sum of the signals coming into the node through branches. 2.The cofactor of a forward path is determinant of signal flow graph formed by deleting all the loops NOT touching the forward path. 3.The signal flow graph that does not contain any loop, has unity determinant. Of these, (a) 1,2 and 3 all are correct. (b) only 1 and 2 are correct. ( c) only 2 and 3 are correct. (d) only 1 and 3 are correct. b

Q14. For the signal flow graph shown below, y2 / y1 is (a) a ( c)

(b) a(1 + d)

1 + d + eg + cbg + edg

(d)

a

a y1

1 + d + eg + cbg

e y2 –g

a(1 + d)

y4 y3

c –d

1 + d + eg + bce + edg

Q15. The signal flow graph of a LTI system is shown in Fig.I and input x(t) to the system is shown in –1 –1 1 1 s s Fig.II. The response y(t) is Y(s) X(s) –4 1 1 - t -2 1 (a) u(t - 2) - e ( ) u(t - 2) + e -3(t -2) u(t - 2) –3 3 2 6 Fig.I: Signal Flow Graph (b) (c) (d)

1 1 - t 1 -3t - e + e 3 2 6

x(t)

1 1 1 u(t - 2) - e - t + e -3t 3 2 6

1

1 1 - (t -2) 1 - e u(t - 2) + e -3(t -2) u(t - 2) 3 2 6

0

2 Fig. II : Input

t(sec)

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Q16. The unit impulse response of unity feedback control system is y(t) = - t e - 2t +2e - 2t for t ³ 0. Then, the open loop transfer function G(s) is (a)

2s + 3 (s + 2) 2

(b)

2s + 3 (s + 1) 2

(c)

1 (s + 1) 2

(d)

2s + 3 s2 + 1

t -5(t -t ) u(t)dt ; Q17. The output y(t) of a system is related to its input u(t) as y(t) = (3 + t - t) e The transfer function Y(s)/U(s) is 0

ò

(a)

3e -5s (s + 5) 2

(b)

3s + 16 s + 7s + 9 2

(c)

3s + 16 (s + 5)

(d)

3s + 16 (s + 5) 2

Q18. Two identical first order low pass filters are cascaded non interactively. The unit step response of such a composite filter will be (a) oscillatory (b) critically damped (c) underdamped (d) overdamped Q19. The unit step response of a system is a unit impulse function. The transfer function of such a system will be (a) 1 (b) 1/s2 (c) s (d) 1/s Q20. A LTI system when excited by unit ramp function, generates the response y(t) = 0.8t −u(t) + 0.1 e−5t . Which one of the following is the unit impulse response of this system ? (a) 2.5e−5t

(b) 0.1 e−5t

(c) 0.8 − 0.5 e−5t

(d) 0.8 + 0.1 e−5t

Q21. The block diagrams shown in Fig. I and Fig. II represent the same system. Y(s) + 1 1 Y(s) 1 + s s +1 X(s) R(s) – s(s + 1)(s + 2) R(s) – 1 s+2

Fig.-I

Fig.-II

Then, X(s) is (a) s + 2 (b) s + 1 (c)1/(s+1) (d) 1/(s + 2) Q22. The block diagram of a control system is shown below. The transmittance C(s) / R(s) is + P (a) P / 1 – PQ + Q + (b) 2P / 1 – PQ R(s) + (c) PQ / 1 – Q Q + (d) PQ / 1 – PQ P +

C(s)

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Q23. Match List I with List II and select the correct answer using the codes given below the lists. List I List II (signal flow graph) (simplified signal flow graph) A A/(1 – B) x1 x2 1. A. x1 x2 B A B.

x1

x1

3.

x1

A+B x2

x2 B

B A

C. x1

2.

A/(1 + AB) x2

x2 x1 B

D. x1

1

A

x2 or

4. x2

AB

AB

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

A 2 4 2 2

B 4 2 4 4

C 1 3 3 1

D 1 3 3 3

Q24. Consider the following two statements in respect of a system whose signal flow graph is shown in fig.1 .

d x1 a

x2

b

x3

c

x4

Fig.1 d/k

Statement I: If node variable x3 becomes kx3 (k is a constant) while x1 , x2 and x4 remain same, then, the signal flow graph so that system dynamics remains unchanged can be redrawn as shown in fig.2. Statement II: If node variables x2 and x3 become m2x2 and m3x3 respectively while x1 and x4 remain same then , the signal flow graph so that system dynamics remains same, can be redrawn as in fig.3. (m2 and m3 are constants)

x1

a

c/k x2

kb

kx3

x4

Fig.2 m2 d m3

x1

m2a

Fig.3

c/m3 m2x2

m3 b m2

m3x3

x4

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Of these, (a) both statements I and II are correct. (b) both statements I and II are NOT correct. (c) Statement I is correct while II is not correct. (d) Statement II is correct while I is not correct. Q25. In the signal flow graph shown below, the gain x2 / x1 will be (a) a1a2 / (1+ a1a2b1)

1/a1

(b) a1a2 / (1+ a1b1)

a1

1

x1

1/a2 a2

1

1

x2

(c) 1 / (1+ a1b1) - b1 (d) 1 / b1 Q26. The design of a unity feedback control system shown below, is so attempted that R(s) + G(s) (I) the steady state error exhibited by system is 1.5 when C(s) – excited by unit ramp input. (II) the dominant roots of characteristic equation of third order system, are −1± j. In order that the above two conditions are satisfied, the third order open loop transfer function G(s) is (a)

4 s(s + 4s + 6)

(b)

2

2 s(s + 2s + 2)

(c)

2

4 s(s + 2s + 2)

(d)

2

2 s(s + 2s + 2) 2

Q27. The signal flow graph and the equivalent block diagram of a system, are shown below. –a 1

1

X(s)

Let P(s) =

s

–1

2

1

s

–1

1 Y(s)

–1

+ X(s) –

2s + 1 .The value of a is ___________. 5s 2 –1

+

P(s) –

Q28. The system described by the block diagram as shown below, has transfer function

Y(s) 10 = 2 X(s) s + 21s + 10

10

+ X(s)

–

s -1

s +1 l

The value of λ is __________.

+ +

Y(s)

Y(s)

Mechasoft

Control Systems

08

Q29. A control system has two equivalent block diagrams as shown below. For H(s) = 1, 1 G(s) = 2 . The value of |b1 – b2| is _________________ . s + b1s + b 2

+

–

1 s+5

R(s)

+ –

1 s + 10

Y(s) – –

+ R(s) –

G(s)

Y(s)

H(s)

Q30. Consider the unity feedback system shown below. It is desired that the maximum value of response y(t) to a unit step D(s) disturbance, D(s) = 1/s be less than 1 + + 0.01.In order to meet this requirement, R(s) + K s(s + 4.4) (a) K > 10 (b) K > 100 – ( c) K < 80 (d) K < 100

Y(s)

Time Response

2

Q31. Match List-I with List-II and select the correct answer using the codes given below the lists. List-I List-II (system function) (nature of damping)

64 3s + 4s + 5

A. T(s) =

1. underdamped

2

s 2 - 2s B. T(s) = 2 s + 6s + 9

2. overdamped

C. T(s) =

9s 2 + 3s + 10 s 2 + 5s + 2

3. critically damped

D. T(s) =

19s - 20 s + s + 100

4. oscillatory

A B C (a) 1 3 2 (c) 2 1 3

D 1 2

2

A B C (b) 1 2 3 (d) 2 3 2

20s Q32. The system with transfer function T(s) = 2 ξ = 0.1. The constant K is ______________. 3s + 2s + K + 4

D 1 1

is to have damping ratio

Q33.In the control system shown below, the steady state error e(¥) = l t e(t) is required to be t ®¥ less than or equal to 0.01 for unit ramp input r(t) = t u(t). What is restriction on K ?

Control Systems

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(a) K ≥ 1

R(s) (c) K ≥ 100

(d) K ≥ 10

E(s)

+

(b) K ≤ 10

–

09

+

K/s

10/s

–

Y(s)

0.1 Q34. The roots of a second order system lie in shaded region in s plane as shown below. If ξ represents damping ratio and ts represents settling time (in sec) for 2% tolerance, then (a)

1 3 6 (c) Ki > 20

1 (s + 1)(s + 2)

Y(s)

(d) no such Ki exists

Q53. Consider the unity feedback system shown below. The input x(t) = u(t); u(t) is unit step function. The value of λ for which system will exhibit overshoot, is (a) 0.25 2(1 + l s) + (b) 0.5 X(s) (s + 1)(s + 2) – Y(s) ( c) 0.75 (d) 1.0 s 2 + as + b Q54. The unit step response y(t) of a system with transfer function G(s) = 2 is

8 3 1 y(t) = - e - t - e -3t ; t ³ 0 . The value of |a – b| is ____________. 3 2 6

s + 4s + 3

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55. The transfer function of a thermocouple, relating its output voltage to input temperature,

3 ´10-5 is given as G(s) = V/0C 10s + 1 The thermocouple is suddenly immersed in a water bath at 100˚C at t = 0. The voltage output (in millivolt) at t = 1 sec, is ________________. Q56. A system with transfer function G(s) = k / (ps +1) has steady state gain equal to 3 and time constant equal to 2sec. Let td be the delay time, that is, the time taken by unit step response to reach 50% of its final steady state value. The values of k, p and td(in sec) are (a) k = 3, p = 2 and td = 1.37 (b) k = 2, p = 3 and td = 0.69 (c) k = 3, p = 2 and td = 0.69 (d) k = 2, p = 3 and td = 1.37 Q57. Consider a multi-loop system as shown below. The value of K such that the steady state tracking error to a unit step input R(s) = 1 /s is zero, is ___________ . s–1

+ – R(s) +

–

– K / s2

– s

–1

Q58. A multi loop feedback system is shown below. The gains K1 and K2 are so selected that the closed loop response to a unit step input is critically damped with two equal roots s = – 10. The values of K1 and K2 are respectively (a) K1 = 100, K2 = 0.19 (c) K1 = 25, K2 = 0.96

Y(s)

+ R(s)

–

Y(s)

K2 + +

(b) K1 = 10, K2 = 1.9 (d) K1 =2.5, K2 = 0.6

Q59. A first order system has transfer function G(s) =

1 s

K1 s +1

-1 and is excited by r(t) = 2δ(t) to 2 + 0.1s

y(t) G(s) r(t) generate response y(t); t ≥ 0; δ(t) is unit impulse function. Let the steady state gain be S, the time constant be τ and the time at which the impulse response is – 7.36 be t0 . Then, (a) |S| = 0.5, τ = 0.5, t0 = 0.01 (b) |S| = 0.5, τ = 0.05, t0 = 0.05 (c) |S| = 0.5, τ = 10, t0 = 0.05 (d) |S| = 0.5, τ = 0.1, t0 = 0.01 Q60. A system and its response to a unit step for K = 1, are shown below. Input R(s)

K

+

G(s) –

output Y(s)

y(t) 1.0 0.8

The value of K so that the steady state error is equal to zero, is ___________. t

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Routh Stability

3 Q61. A system with zero input as shown below, is used to generate a sinusoidal input=0 + output signal. The value of adjustable K + in terms of constant a for which the system has sinusoidal output and hertz frequency f of oscillation in terms of constant a, are respectively (a) K =

a a and f = Hz 2 2p

(c) K = 2a and f =

s s+a

a 2a and f = Hz 2 p

(d) K = 2 a and f = 4

output

K

(b) K =

2a Hz p

1 s+a

3

a Hz 2p

2

Q62. For the adjustable K, the polynomial P(s) = s + 2s + 4s + Ks + 6 (a) has all LHP roots for K < 8. (b) has all LHP roots for K > 3/2. (c) always has RHP root/roots for any real K. (d) always has LHP roots for K > 0. 5

3

Q63. For the polynomial Q(s) = 3s + 2s + s, how many roots are in the LHP, how many are in the RHP and how many are on imaginary axis (IA)? (a) 2 LHP, 2 RHP and 1 IA (b) 2 LHP, 1 RHP and 2 IA (c) 4 LHP, 0 RHP and 1 IA (d) 3 LHP, 1 RHP and 1 IA Q64.The stability boundary diagram on k1− k2 plane of a system with characteristic equation 2 s + (2 + 0.1k1) s + k1+ k2 = 0 , is k2 k2 (a) (b) +20 20 k1 k1 –20 –20 k2 +20

k2

(c)

+20

(d) –20

k1

k1

–20 k1 + k2 = 0

Q65. A system with transmittance G(s) =

as + 1 is placed in unity feedback configuration. s (s + b) 2

Identify the correct stability region on a−b plane.

Control Systems

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(a)

b

(b)

15

b

a

a b

b ( c)

(d)

a

1 Q66. The system with transfer function model H(s) = 3 is stable. 2 s + as + ks + 3 The constraints on α and k are (a) α > 0 and α k < 3 (c) α < 0 and α k > 3

a

(b) α > 0 and α k > 3 (d) α > 0 and α k < 3

Q67. The polynomial Q(s) =Ts2 + s + K is to have all the roots with real part more negative than −2. The region on T−K plane, that satisfies this condition, is K K (a) (b) 2 2 1

1 0.5 K

(c )

T 0.5

T

K

(d)

2

2

1

1

T 0.25 T 0.25 Q68. The impulse responses hi(t) ; i = 1, 2, 3, 4, 5 given below belong to five systems −t −t −3t Sj ; j = 1, 2, 3, 4, 5 respectively. h1(t) = e ; h2(t) = t e ; h3(t) = 1; h4(t) = e sin 3t ; h5(t) = cos4t. Of these, (a) only S1, S2 and S4 are stable. (c) only S1 and S3 are stable.

(b) only S3 and S5 are stable. (d) only S1, S2 and S3 are stable.

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Q69. For the system shown below, k can be appropriately chosen so as to force the system to oscillate sinusoidally at frequency (a) (c)

1 Hz p

(b)

3 Hz p

(d)

2 Hz p

+ R(s)

–

k s(s + 3) 2

Y(s)

3 Hz 2p

Q70. Which of the following statement is true for the system shown below? (a) Both open loop and closed loop systems are stable. (b) Both open loop and closed loop 4 + systems are unstable. 2 – s + 2s + 2 (c) Open loop system is unstable but closed loop system is stable. s -1 (d) Open loop system is stable but closed s +1 loop system is unstable. Q71. Consider the following statements. I. A continuous system excited by a step signal, generates an output of form y = t. y& = u This is an unstable system. II. The dynamics of a initially relaxed system with input u and output y is described by This system is stable. III. A system has poles located as s = −1 and s = −5 together with zeros located at s = 1 and s = −2. This is an unstable system. IV. A system with characteristic roots −3, −2 and 0, is stable. Of these statements, (a) only II and III are true . (b) only III and IV are true. (c) only IV is true. (d) only I is true. Q72. Using approximation of form e –x = 1 – x, the range of k that preserves the stability of a system with block diagram as shown below. is k e -2s + (a) k > 3 (b) 0 < k < 30/11 s – (s + 2)(s + 3) (c) 0 < k < 11 (d) no such range exists. Q73. The algebraic signs of first element of each of the rows of a Routh array are as given below. Rows 1 2 3 4 5 6 7 Signs + − + + + + − The number of LHP roots of the system, is ____________________. Common data for Q74 and Q75 The dynamics of a system with input x(t) and output y(t), is given as

&& & + (2G + 5) y(t) = x(t) y(t) - (G + 2) y(t)

Q74. The constraints on variable parameter G for the system to remain stable, are (a) −2.5 < G < −2 (b) G < −2 (c) G > −2.5 (d) G > −2

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Q75. The value of G that will force critical damping in the system, is _____________. Q76. Consider feedback control system shown below. 5 The closed loop stability region R on λ - μ plane, is + R(s) s(s + m ) – m m l 2 + (a) (b) mR s

Y(s)

R

l

l m

m (c)

(d)

R

R l

l

Q77. A feedback control system has characteristic equation s3 + (1+k)s2 +10s + (5+15k) = 0. Consider the following statements. I. The system is stable for |k| < 1. II. For k = 1, the system oscillates at frequency of approximately 0.5 Hz. III. for k = 2, the system is unstable due to two rhp roots. Of these, CORRECT statements are (a) I, II and III (b) II and III (c) I and III (d) I and II

4.

Root Locus

Common data for Q78 and Q79 The signal flow graph of a control system is shown below. R(s)

Q78. With switch ‘S’ open

–Ks switch

S

1

Im

1

K=¥

K=0 K=¥ K=0

Re

s

Im K=¥

(b) K=¥

1

Y(s) Y

1

(a) system oscillates at 1 rad/sec. –1 (b) system is stable. (c) system is unstable with one root in the RHP. (d) system is unstable with two roots in the RHP. Q79. With switch ‘S’ closed and K as variable parameter, the root locus is

(a)

1 / s2

K=0 K=¥

Re

Control Systems

18

( c)

Mechasoft

(d)

Im K=¥ K=¥

K=0

Im K=0 K=¥

Re

K=¥

Re

K=0

K=¥

Common data for Q80 and Q81 A unity feedback control system has forward path transfer function G(s) =

K . (s + 2)3

Q80. which one of the following is correct root locus? Im

(a)

Im

(b)

I

k

K=0

I

K=¥

j2

K=0

´´ –2´

K=¥

Re

K=0

K=¥ ´ ´

–2 ´

k

R

Re

– j2 K=¥

K=0

k

Im K=¥

( c)

Im

(d)

K=¥ k

k

j2 3 k

K=¥

K=0 ´ ´ –2´

Re

K=¥ k

K=0

´ ´ ´ –2

Re

–j2 3 k

K=¥

K=¥ k

Q81. Assuming second order approximation to be valid , the value of K for which system exhibits damping ratio of 0.5 , is (a) 4

(b) 8

(c) 16

(d) 2

Control Systems

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19

k

Q82. The G(s) H(s) product of a feedback control system is G(s) H(s) = s(s + 1) (s + P) For given k and variable P (a) only one branch of root locus will terminate at ∞. (b) none of the root locus branch will terminate at ∞. (c) only two branches of root locus will terminate at ∞. (d) the number of branches of root locus terminating at ∞, is not determinable. k(s + 4 / 3) Q83. A unity feedback control system has open loop transfer function G(s) = .

s 2 (s + 12)

The designer seeks value of k > 0 for which all the roots are equal. This value of k and corresponding roots are (a) k = 12 at s = − 2, − 2, − 2 (b) k = 48 at s = − 2, − 2, − 2 (c) k = 12 at s = 0, 0, 0 (d) k = 48 at s = − 4, − 4, − 4 Q84.Consider the following statements in regard to a unity feedback system with loop function

G(s) =

k(s 2 + 16) . s2 + 4

k =¥

I. The correct root locus is as shown below. II. System is always oscillatory. III. System is unstable. Of these, correct statements are (a) I and III (b) II and III (c) I, II and III (d) I and II

Im +j4

k =0 ´ +j2 Re k =0 ´ –j2 k =¥ –j4

Q85. The root locus in the range 0 ≤ k ≤ ∞ for the unity feedback control system shown below, is

k(s + 2) (s + 1)(s + 3)

+

R(s)

–

(a)

´

(c)

´

Im

´

Re

(d)

Im

´

Y(s)

(b)

Im

´

–1

Re

´

Re

Im

´

´

Re

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Common statement for Q86 and Q87 The root locus for a feedback control system in the range 0 ≤ k ≤ ∞ is shown below. Im k=∞ k=0´

+j2

k=0

k=∞

´

-4

Re

-1 k=0

´

– j2 k=∞

Q86. The value of k for which one of the closed loop pole is located at s = −5, is ___________ . Q87. For the value of k found in Q86 and assuming second order approximation, the nature of system dynamics is (a) oscillatory (b) unstable (c) overdamped (d) underdamped Q88. The dynamics of a closed loop system is described by a characteristic equation (s+1) (s+2) (s+3) + k(s+4) = 0 ; 0 ≤ k ≤ ∞ . The root locus of this system has centroid located at (a) s = −1 + j0 (b) s = −2.5 + j0 (c) s = −5 + j0 (d) −1 + j1 1 Q89. The signal flow graph of a control system is shown below. s(s + k) 1 1 U(s) The root locus for 0 ≤ k ≤ ∞, is –1 Im ´+j1

Im

(a)

´ –1

´

Re

(b)

–1

Re

´ –j1 Im Im

+j1´ 1

( c) –j1´

Re

(d)

´

–1

Re

Y(s)

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21

Q90. A control system with unity feedback has the root locus for 0 ≤ k ≤ ∞ as shown below. Which one of the following will ensure Im k=∞ under-damped dynamics in the system? (a) k > a2/4 (c) k > a/2

k=0 ´ –a

(b) k < a2/4 (d) k < a/2

´ k=0

Re

k=∞ Common statement for Q91 and Q92 k(s 2 + 6s + 8) A feedback control system has loop transmittance G(s) H(s) = s 2 + 6s + 18 Q91. Identify the correct root locus for 0 ≤ k ≤ ∞ . Im +j3 × × (a) –4

–2

(b)

Re

–j3

×

+j1 –j1

–3

×

–j3

×

Im +j3

× (c)

Re

–3

Im +j3

Im +j3

× (d)

Re

+j1 –j1

–3

×

–j3

Q92. At break in point, if any, k has the value (a) 9 (b) 3

(c)

Re

–j3

(d) no suck k exists

3/2

Q93. The signal flow graph for a control system is shown below. U(s)

P

s

–1

1 The correct root locus for 0 ≤ P ≤ ∞, is

s

–1

–4 –P

Im

Im

×+j2

×+j2

(a)

Re

×–j2

1

(b)

Re

×–j2

Y(s)

22

Mechasoft

Control Systems

Im

Im (c)

×+j2

×+j2

(d)

Re

Re

×–j2

×–j2

Q94. The block diagram of a control system as shown in Fig.- I, is given to a student for the purpose of analysis. The student draws root locus as shown in Fig.- II for 0 ≤ k ≤ ∞. + –

k

Im j2 k = ¥ j1´

s2 + 4 s(s 2 + 1)

k =¥

k= 0

´

k =0

- j1´ k = 0

Fig.I ;Block diagram

Then, (a) the root locus is correct and system is always stable for k > 0. (b) the root locus is correct and system is always unnstable for k >0. (c) the root locus is not correct and the system is always stable. (d) the root locus is not correct and the system is always oscillatory. Q95. The loop function of a unity feedback system is G(s) =

- j2 k = ¥ Fig. II; root locus

ke -s

s(s + 1)(s + 2) The root locus is constructed using approximation e = 1 − s. Consider the following statements in respect of root locus. I. The break away / break in points are located at s = −0.35 and s = +1.88. II. The root locus intersects imaginary axis at s = ± j 0.7. III. The root locus includes real axis segment between s = −2 and s = −1 and between s = 0 and s = 1 IV. Three root locus branches terminate at ∞. Of these statements, (a) only II and III are correct (b) only III and IV are correct. ( c) only I and II are correct. (d) only IV is correct. k(s + 1) ; 0 £ k £ ¥, has all Q96. The root locus of a unity feedback system with G(s) = 2 s (s + 9) −s

three roots real and equal at s = –b with corresponding k = k1. The value of k1 / b is _________.

Re

Control Systems

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23 2

4(s + 1)

;0£a £¥ Q97. The root locus of a unity feedback system with loop function G(s) = s(s + a) has both the roots equal and real for some value of a. This value of a is _______________. & = Ax(t) + Bu(t) and y(t) = C x(t) Q98. Consider the system described by x(t)

é 0 where A = ê ë3 - k

1 ù é0 ù , B = ê ú , C = [1 - 1] ú - 2 - kû ë1 û

The root locus as 0 ≤ k ≤ ∞ (a) always lies on real axis and system is stable for k > 3. (b) always lies on real axis and system is stable for k < 3. (c) crosses jw axis at s = ± j3 with corresponding k = 3. (d) has one break away point at s = –3. Q99.Consider the following statements about root locus of a system is described by 3 2 characteristic equation s + ks +3ks +2k = 0 ; 0 ≤ k ≤ ∞. I. For k > 10.4, all the three roots will be real. II. system is stable for K > 2/3. 1 III. system may exhibit oscillations with corresponding frequency equal to Hz 2p Of these, INCORRECT statement (s) is/are (a) I (b) II (c)III (d) none Q100. A feedback system is shown below. which one of the following is the correct root locus for 0≤K≤∞? 10 + R(s) Y(s) (s + 2)(s + 5) –

K s+K

jω K=∞ (a)

jω K=0 (b)

K=0 K=∞

K=0

K=∞

K=0 K=∞

K=0 σ

K=∞

K=∞ σ

K=0 jω

K=∞

jω

( c)

(d) K=∞ K=0

K=0

σ

K=0 K=∞

K=∞

K=0

σ

Control Systems

24

Mechasoft

5

Frequency Response

Q101. The gain margin and phase margin of a feedback system with loop transmittance

s

G(s) H(s) = (a) 0 dB and 0˚

s ö æ ç1 + ÷ è 10 ø

3

are approximately

(b) ∞ and ∞

(c) −11.5 dB and 127˚

(d) ∞ and 53˚

Q102. Match List I (complex function) with List II (polar plot) and select the correct answer using the codes given below the lists. List I List II Im I w=¥

2 o A. F1 ( jw) = w Ð30

w=0 0 .5

1. 2 0

2.

w 2

w = w0 45˚

w=0

w=¥

2

3.

Re

w02 2

Im

1 C. F3 ( jw) = 4 ( jw) ( jw + 0.2)

Re

w= ¥

Im B. F2 ( jw) = w2 (cos 45o + jsin 45o )

◦

45

0.5w0

w = w0 30˚

w=0

3w02 2

Re

Im D. F4 ( jw) = 0.6w2 ( j + 1) + 0.5

4.

w=¥ Re w=0

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

A 3 2 4 3

B 2 3 1 2

C 1 4 2 4

D 4 1 3 1

Control Systems

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25 Im

Q103. The open loop transfer function G(jω) of a unity feedback system has the polar plot for 0 ≤ ω ≤ ∞ as shown below.The phase margin and gain margin of the system are respectively (a) π rad and 4 (b) π rad and 3/4 ( c) π /2 rad and 4 (d) π /2 rad and 3/4

–1

G(jw) plane w=0

w=∞ 1

Re

– 0.25

Q104. A unity feeddback system with loop transmittance G(s) =

1 has phase s(1 + sp1 )(1 + sp 2 )

cross over frequency (in rad/sec) and gain margin respectively

p +p 1 and 1 2 p1p 2 p1p 2 p1p 2 1 ( c) and p1 + p 2 p1p 2

(a)

(b)

p1p 2 and

p1p 2 p1 + p 2

(d)

p1p 2 and

p1 + p 2 p1p 2

Q105. The unit step response data of a typical second order system is given below. time (sec) unit step response

0 0

0.05 0.25

0.1 0.8

0.15 1.08

0.20 1.12

0.25 1.02

0.30 0.98

0.35 0.98

0.4 1.0

0.45 1.0

The corresponding frequency response indices peak resonance Mr and resonant frequency ωr respectively are (a) not determinable with given data (b) 1.08 and 11.58 rad/sec (c) 1.7 and 18.96 rad/sec (d) Mr and ωr do not exist. Q106. The transfer function of a system is H(s) =

Y(s) s+2 = 2 . In the steady state ,the X(s) s + s + 4

system generates response y(t) = cos (2t + ϕ) for x(t) = cos2t. The angle ϕ (in rad) is (a) 0.25π (b) 0 (c) − 0.25π (d) − 0. 5π Q107. The dB magnitude and phase angle plot of a continuous type minimum phase system is shown below. The system is phase angle 6 0

Inc

g sin a e r

ω

–6 –12 –18

–240º–210º –180º–150º –120º –90º

phase angle

dB magnitude

dB magnitude

12

(a) stable with gain margin equal to o 12 dB and phase margin equal to 30 . (b) unstable with gain margin equal o to 12 dB and phase margin equal to −30 . ( c) unstable with gain margin equal to −12 dB and o phase margin equal to 30 . (d) unstable with gain margin equal to −12 dB and o phase margin equal to −30 .

Control Systems

26

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Q108. Consider the following statements. I. If G(s) is a stable transfer function ,then F(s) = G –1 (s) is always a stable transfer function. II. The open loop frequency responses at two particular frequencies are observed to be 1.2Ð - 180o and 1.0 Ð - 190o . The corresponding closed loop unity feedback control system is, stable. Of these statements, (a) I and II both are true. (b) I is true but II is false. (c) I is false but II is true . (d) I and II both are false. Q109. The dB magnitude plot of a minimum phase system is shown below. The system transfer function is

10(s + 0.1) (a) 2 s (s + 10) (b)

10(s + 0.5) s2 (s + 2)

(s + 0.1) (c) 2 s (s + 10)

dB –12dB/oct +20

–6dB/oct ω2

0

ω1

1

ω (log scale)

–20 –12dB oct

4(s + 0.5) (d) 2 s (s + 2) Q110. The Nyquist plot of an all pole second order open loop system is shown below. The transfer function is

(a)

8 s + 2s + 4 2

Im

4 s + 2s + 2

w= ¥

(c)

4 (s + 1)(s + 2)

w=2

(d)

8 (s + 1)(s + 4)

(b)

2

–2

2 Re w=0

Mechasoft

Control Systems

27

Q111. The phase-frequency responses of three transfer functions G1(jω), G2(jω), and G3(jω) are shown below. It is also known that dB magnitude-frequency responses of G1(jω) and G2(jω) are same but ÐG 3 (jw) = ÐG1 (jw) + ÐG 2 (jw) w օ 0 Then, G1(jω), G2(jω) and G3(jω) represent respectively ÐG1 (j w) Phase օ (a) minimum phase, all pass and non minimum –90 phase systems. ÐG 2 (j w) (b) minimum phase, non minimum phase and all pass systems –180օ ( c) all pass, minimum phase and non minimum ÐG 3 (j w) phase systems. օ –270 (d) all pass, non minimum phase and minimum w phase systems. Q112. The amplitude ratio and phase shift function for a transmittance are A(ω) = 4 and ϕ(ω) = −3ω (rad). The forced sinusoidal response in the steady state to the o input r (t) = 10 cos( 5t − 30 ) is o (a) 40 cos (5t − 170 ) (b) 30 cos(5t − 170o) o o (c) 40 cos (5t − 45 ) (d) 30 cos(5t − 45 ) Q113. The amplitude ratio A(ω) and phase shift function ϕ(ω) for irrational transfer function

e - 4s G(s) = , are respectively s (a) A(ω) = − 4ω,ϕ(ω) = −π/2 ( c)A(ω) = 1/ω, ϕ(ω) = − 4ω

(b) A(ω) = 1/ω, ϕ(ω) = − 4ω − π/2 (d) A(ω) = −4ω,ϕ (ω) = − 4ω

Q114. For G(s) = s p 2 p (c) G( jw) = w and ÐG( jw) = 4

(a) G( jw) = w and ÐG( jw) = -

p (b) G( jw) = w and ÐG( jw) = 2 p (d) G( jw) = w and ÐG( jw) = 4

Q115. A system has six LHP poles, two LHP zeros together with one pole at origin. The slope of its highest frequency asymptote in its magnitude db Bode plot, is (a) − 30 dB /octave (b) − 42 dB /octave (c) − 36 dB /octave Q116. The polar plot of a conditionally stable system for open loop gain k = 1, is shown below. The open loop system is known to be stable. The closed loop system is stable for (a) k 1 (b) k < 4 and 1 < k < 5 (c) k >1/5 and k < 4 (d) k < 1/5 and 1< k < 4

(d) + 30 dB /octave

–5 c

–0.25 . b –1

w=0

Im w=¥ a o

Re

28

Control Systems

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Q117. The dB magnitude plot of a stable system is shown below. The transfer function of the system, is

(b) G(s) = (c)G(s) = (d) G(s) =

200 (s + 1) (s + 2) 10-1

40

(s + 10) (s + 100)

20

–6dB/oct

50

–12dB/oct

(s + 1) (s + 2) 105

(s + 10) (s + 100) Q118. The frequency response of an electrical machine is shown below. What are gain margin and phase margin?

(b) − 60 dB and 60

o

o

(c) 60 dB and 120o (d) 20 dB and 60o Q119. The transfer function for which the asymptotic dB magnitude plot is shown below, consists of

Phase

30 20 Gain(dB)

(a) − 20 dB and 120

log10ω

1

gain

180o 120o o

10

60

0

0o

–10

–60o

–20

–120o

–30

–180o

Phase(degrees)

(a) G(s) =

|G(jω)|dB

40

(a) 3 poles and 2 zeros

30

(b) 2 poles and 2 zeros

20

(c) 2 poles and 1 zero

Gain 10 (dB) 0 100Hz

100kHz

1kHz

Frequency

10kHz 1MHz –10 –20 Q120. Consider a control system shown below. The gain margin and phase margin are (d) 3 poles and 1 zero

o

(a) 200 and 90 (c) 200 and 10o

o

(b) 1/200 and 90 (d)1/200 and 10o

+ –

10 (s + 10) 2 1/ s

Control Systems

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29

Q121. The straight line approximated phase plot for a system with transfer function s + 1000 G(s) H(s) = , is (s + 10) 2 (a)

10

1 0

10

2

10

3

10

(b)

4

1

o

0

10

102

103

10

2

3

104

o

Phase o –90

Phase –90o

–180o

–180o ω

( c)

2

10

1

10

10

3

ω

(d)

4

10

1

0o

0o

Phase o –90

Phase o –90

o

–180

–180

10

10

104

o

ω

ω

Q122. The frequency response of a linear control system G(jω) is tabulated below.

|G(jω)|

1.3

1.2

1.0

0.8

0.6

0.25

0.15

ÐG(j w) –120o –130o –138o –150o –170o –180o –200o 120

130

The gain margin and phase margin of this system are respectively 7p 7p rad rad (a) 12.04 dB and (b) 6.02 dB and 15 30 7p

(d) 12.04 dB and

7p

rad 15 30 Q123. The open loop transfer function G(s) has three lhp poles. The gain margin of KG(s) is found to be 28 dB for K = 1. If gain margin is found to be 34 dB for K = K1 and 16 dB for K = K2 , then ratio K2 / K1 is_______________

( c) 6.02 dB and

rad

Q124. The phase crossover frequency of a unity feedback system with transmittance

G(s) = (a) 2 Hz

K will be s (1 + s / 4) (1 + s) (b) 4 Hz

(c) 2πHz

–1 (d) p Hz

Control Systems

30

Mechasoft

Q125. The dB magnitude plot of a system with no RHP pole , is shown below. The transfer function G(s) will be dB (a)

106 s 2 (s + 10)3

100

(b)

103 s 2 (s + 20)3

60

(c)

+12dB/oct. –18dB/oct.

104 s (s + 20)3

in regard to the dB magnitude plot of a unity feedback system with forward path transfer function G(s), as shown below. I. This system exhibits steady state error e(∞) =0 for input r(t) = t u(t) . II. This system has damping ratio ξ = 0.3 III. This is type 2 system. Of these correct statements are (a) I and II (b) I and III

ω(log scale)

1

108 s 2 (d) (s + 10)5 Q126. Consider the following statements

dB –6dB/oct

10

0

3.6

–12dB/oct (c) II and III

(d) I, II and III

Common statements for Q127 and Q128 The open loop transfer function of a control system is given as G(s) H(s) = Q127. Which one of the following is the correct Nyquist plot for Im (a) –2

w=¥ Re

(b)

K(s + 8) (s + 2)(s - 2)

1 [G( jw) H( jw)]? K Im w= ¥

w=0 –2

Re

w=0 Im

Im –2 ( c) w=0

w=¥ Re

ω(log scale)

(d)

–2 w=¥

w=0

Re

Control Systems

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31

Q128.Interpreting the correct Nyquist plot of Q127, it is observed that the closed loop system is stable for 1 1 (a) K > (b) K < (c) K > 2 (d) K < 2 2 2 Common statements for Q129 and Q130 A unity feedback control system has loop transmittance G(s) =

k(2 + s) ;k > 0. (2 – s)

Q129. Which one of the following is the correct Nyquist plot for G(s)/k? Im 1 (a) (b) Re

Re w=∞

w=0

w=0

w=∞

1

–1

1

–1

Im 1

–1 –1 Im

Im 1

( c) w=∞

1

(d)

Re

w=∞

w=0

Re

w=0

Q130. The closed loop system is (a) stable for k > −1 ( c) stable for −1 < k < 0

(b) stable for k > 1 (d) always unstable +

Q131.If x = Re[G(jω)] and y = Im[G(jω)], then for ω = 0 , the Nyquist plot for G(s) =

1 is s(s + 1)(s + 2)

(a) x=0

(b) x = –3/4

( c) x=y–1/3

(d)

x= y/ 3 Q132. In the block diagram of control system shown below, G(s) has three poles located at s = −1 and G (0) = 1. The gain margin and phase margin of this system, are respectively (a) 1/4 and π/2 radians ( c) 8 and π radians

(b) 4 and π/4 (d) 1/8 and π/2 radians

+ –

G(s)

Control Systems

32

Common data for Q.133 and Q.134 Consider the system whose signal flow graph is shown below.

1

U(s)

Mechasoft s

1

s

1

–1

1 1

Y(s)

–1

–1

–1

Q133. The average value of delay τd(ω) over the range 0 ≤ ω ≤ 5 rad/sec, is (a) 2.75 sec. (b) 1.37 sec (c) 0.137 sec (d) 0.275 sec Q134. The system under consideration is modified as shown below. Y (s ) then, the bandwidth of system with transfer function m , is U (s )

(a)

2 Hz p

(b)

1 2p

(c)

Hz

Common data for Q135 and Q136 Consider the system shown below.

R(s)

+ –

1 Hz p

(d)

A

d

y(t)

ym(t)

dt

1 Hz 2p

1 s(s + k)

Y(s)

Q135. This system is to be designed to exhibit phase margin. ϕm = 40o and resonant frequency wr = 68 rad/sec. The values of k and A to meet the design goal, are respectively (a) 12 and 50

(b) 4 and 50

(c) 8 and 100

(d) 40 and 50

Q136. With values of K and A found in Q135 and with input r(t) = 2 u(t), the peak value of the output ymax , is____________. Q137. Which one of the following, is the correct Nyquist plot for a system with

G(s) H(s) =

(a)

s-a ; a > 0, b > 0? s(s + b) Im

Im

(b)

Re

Re

Control Systems

Mechasoft

Im

( c)

33

Im

(d) Re

Re

Q138. Consider the following statements in relation with segments of dB and phase Bode plots of minimum phase function of a control system shown below. I. The gain margin is −x dB and phase o o margin is 180 − y . II. ωgc < ωpc ; ωgc is gain cross over frequency and ωpc is phase cross over frequency. III. The system is stable. Of these , the correct statements are (a) I and III

gain (dB) 0

w

x

phase f o

–180

(b) I and II

y

(c) II and III

o

w (d) I, II and III

Q139. The dB magnitude plot of a minimum phase system is shown below. The transfer function G(s) of this system, will be (a)

5.97(1 + 0.5s) 2 s(1 + 0.1s)(1 + 0.05s)

(b)

2.99 (1 + 0.5s) s(1 + 0.1s)(1 + 0.05s)

Gain (dB) +6dB /oct –6dB /oct 9 .5

(c)

2.99 s (1 + 0.1s) (1 + 0.5s) (1 + 0.05s)

(d)

5.97(1 + 0.1s) s(1 + 0.5s) 2 (1 + 0.05s)

–6dB /oct

1

2

10

20

frequency (rad / sec)

Q140. The dB magnitude-phase angle plot for a minimum phase system with open loop transfer function G(jω) H(jω) is shown below. If ωπ is phase cross over frequency, ωg is gain cross over frequency, Gm is gain margin and ϕm is phase margin, then

Control Systems

Increasing w

0.1

+20 +15 +10 +5 օ 0/0 –5 –10 –15 –20

0.2

(b) ωπ = 1 rad/sec, ωg = 1.8 rad/sec, Gm = 30 dB and ϕm = 15o

0.3 0.6 օ օ –60օ –180 1 –120 օ –150օ –90 –30օ 1.2 phase angle 1.5 1.8 2.0

(c) ωπ = 1.8 rad/sec, ωg = 1.8 rad/sec, Gm = 20 dB and ϕm = 15o (d) ωπ = ωg = 1 rad/sec, o Gm = 30dB and ϕm = 30

Common data for Q141 and Q142 The dB magnitude-phase angle plot of the open loop continuous time system with frequency response function increasing é æ w ö2 ù k ê1 - ç ÷ + jw / 2 ú ω

G( jw) H( jw) =

êë è 2 ø úû 2 jw ö æ jw ç 1 + ÷ (1 + jw / 4) è 0.5 ø

is shown below for k = 1.

0.25

10

0.6

1 1.5 –135o –200o –180o

20

0.2

0.5

o

–120 –150o Phase angle

0 –3 –10 –20 –30 –40 –90o

Q141. Value of k for which phase margin is 45o, will be (a) 0.1 (b) 10 (c) 3.16

`(d) 0.316

Q142. Value of k for which gain margin is 10 dB, will be (a) 3.16 (b) 0.316 (c) 2.24

(d) 0.447

Q143. The open loop continuous time frequency response function

G(j w) H(j w) =

10(1 + jw / 5) . (j w)(1 + j w)[(1 - w2 /121) + j0.04w]

The phase contribution of this function at f = 10 / p Hz, will be (a) − 82o

(b) − 182o

(c) − 98o

dB magnitude

(a) ωπ = 1.8 rad/sec, ωg = 1 rad/sec, o Gm = 15dB and ϕm = 30

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(d) − 262o

dB magnitude

34

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Q144. Assuming minimum phase characteristics, the transfer function model G(s) of the system with asymptotic dB plot shown below, is dB 20 dB / dec (s + 1) (s + 100) (a) G(s) = 15 (s + 5.62) (s + 17.78) (b) G(s) =

(s + 1) (s + 100) (s + 31.62) (s + 35.56)

(c) G(s) =

(s + 5.62) (s + 17.78) (s + 1) (s + 100)

–20 dB / dec ω (log Scale) 0

a

1

b

100

(s + 31.62) (s + 35.56) (s + 1) (s + 100) s 2 + 2s + 100 Q145. For the open loop function G (s ) = 2 , the dB plot is shown in s + 10s + 100 Fig. I and phase plot in Fig. II. (d) G(s) =

dB 0

20օ Phase

օ

10

1

10 Fig. . I

100

ω (log scale)

0 –20

օ

10

1

100

ω (log scale)

Fig. II

Of these , (a) dB plot is correct but phase plot is not correct. (b) dB plot is not correct but phase plot is correct. (c) dB plot and phase plot both are not correct. (d) dB plot and phase plot both are correct. Q146. The value ofµ such that a unity feedback control system with loop transmittance G(s) = (a) 9.16

(1 + s / a) 2 exhibits phase margin of 30o , will be s3 (b) 0.916

(c) 0.438

(d) 4.38

2

Q147. For loop transmittance G(s) = (a) 1 / 4ξ

wn , the value of |G(jωn) |, is s + 2xwn s + wn 2 2

(b) 1 / 2ξ

(c) ξ / 2

(d) 2 / ξ

Control Systems

36

Mechasoft - s/2

Q148. The Nyquist plot for loop transmittance G(s) =

pe s

is sketched on G(s) plane.

This plot intersects negative real axis at the point

æ

(a) ç -

è

1 ö , j0 ÷ 2 ø

æ 1 è 2

ö ø

(c) ç - , j0 ÷

(b) (-1, j0)

(d) (- 2, j0)

Q149. The root locus of a unity feedback system is shown below. For design value of gain K = 8, the root locations are shown by small square. The gain margin of the system, is jw

(a) 36.12 dB

K=∞ K=64

(b) 34.96 dB

K=8

(c) 18.06 dB

K=8 × × ×

σ

K=8

(d) 9.03 dB K=∞

Q150. Consider the following statements in regard to the dB magnitude plot of a minimum phase transmittance G(s)as shown below. 1 dB 25 Hz. I. G(s) has corner frequencies at f = and 10p 12 p –20 dB/dec II. G(s) has one zero and two poles on real axis.

99.5(s + 0.2) s(s + 50) 200 Hz, is −12 dB. IV. The dB magnitude at f = p

6

III. G(s) =

Of these, the correct statements are (a) only II and III (b) Only I and III

0 0.1 (c) Only I and IV

ω 100 (log scale) (d) I, II, III and IV

Common data for Q151 and Q152 The design of a phase locked loop system which has loop transmittance of form

G(s) =

k o , is attempted so that system has phase margin of 60 s æ ö sç + 1÷ è 1000 3 ø

at cross over frequency of 3 k rad/sec.

Q151. The value of k should be (a) 1000 2

(b) 2000 3

(c) 6000

(d) 3000

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Q152. The phase detector of phase locked loop often contributes a time delay where by phase margin decreases . The maximum permissible time delay so that phase margin dose not go below 45◦, is (a) 349µs (b) 1.5 ms ( c) 87.3µs (d) 0.5ms Q153. The corner frequencies (in rad/sec) in Bode plot of system with loop transmittance

G(s) = (a) 0.8 and 2

10 , are 0.625s + 1.75s + 1 2

(b) 0.625 and 1.75

(c) 1.6 and 1.75

Common statement for Q154 and Q155. Consider the Nyquist plot of a minimum phase function G(s) as shown below. c(t) r(t) G(s) input output

(d) 0.3125 and 1

Im w=¥

0.5

1

Re

w=0 w = 10

Q154. For input r(t) = 2 cos (10t + π /18) , the output c(t) = A cos(10t + δ ). Then, (a) A=2 and δ=π/3 (b) A=10 and δ=5π/9 ( c) A=0.1 and δ=–5π/9 (d) A=1 and δ=–4π/9 K Q155. Let G(s)= . Then, the values of K and a are respectively (s + a)2 (a) 100 and 10

(b) 10 and 10

( c) 2 and 0.1

(d) 1 and 0.1

Q156. The impulse response g(t) of a linear time invariant continuous time system is −3t h(t) = e u(t) ; u(t) is unit step function. The response of this system to sinusoidal input -1 -2 (b) 2 cos ( 3 t - 2 p)

(a) 2-2 cos ( 3 t - 2-1 p)

-1

-2

( c) 3-1 cos ( 3 t - 2-1 p) (d) 2 3 cos ( 3 t - 2 p) –bs Q157. A system has transfer function G(s) = ae / s. Let Re[G(jω)] = x and Im[G(jω)] = y. As ω →0 , its polar plot will be asymptotic to (a) x = –ab (b) x = ab (c) y = b / a (d) y = – b/a Q158. An input sin(t + θ) is applied to a series combination of R = 1MΩ and C = 1μF. The output is taken from across capacitor. The value of θ (in rad)so that there is no transient in the output, is π/m. The value of m is________________. Q159. A system has transfer function G(s) = 1 / (1 + bs)n . For an input r(t) = sinωt, the system generates the response y(t) = a sin(ωt – ϕ). If ϕ = π rad for n = 6, the value of a (rounded off to two decimal places) is _________________. Q160. Consider the system described by x(t) & = Ax(t) + B u(t) and y(t) = Cx(t) where u is input, y is output.

é 0 A=ê ë -5 - k

1ù é0 ù , B = ê ú , C = [6 ú - 2û ë1 û

3].

For k = 1, the bandwidth (in rad / sec) of the system is _________________.

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State Space Model

Q161. A linear time invariant system is described by state space model

dx = Fx + Gu ; y = Hx dt

é0 1ù F=ê ú ë -4 -2 û

é0 ù , G = ê ú and H = [1 0] ë1 û

x is state vector, u is unit step input and y is output. How long (in seconds) the system response takes to reach and stay within 2% of final steady state value? (a) 2 (b) 4 (c) 1 (d) 0.5 Q162. The dynamics of a system with zero external input, is described by state equation

é -1 1 ù dx = Ax where A ê and 0 -3úû dt ë é2ù the initial state vector x(0) = ê ú .The steady state value of x(¥) = lt x(t),will be t ®¥ ë -2 û é -1ù é2ù é0 ù é¥ ù (a) ê ú (b) ê ú (c) ê ú (d) ê ú ë -3û ë -2 û ë0 û ë¥ û -t ée 0 ù Q163. A linear time invariant system has state transition matrix (STM), ê -2t ú and ë0 e û é x1 (0) ù é 1 ù é x (t) ù é0.905ù = ê ú . After time t, the state vector is ê 1 ú = ê initial condition ê ú ú . x (t) 1.637 ë x 2 (0) û ë -2 û ë û ë 2 û The value of ‘t’ is nearly equal to (a)100ms (b) 200ms

(c)150ms

(d) 300ms

Q164. The signal flow graph together with assignment of state variables x1 and x2, is shown below. The condition for complete state controllability and complete state observability of the state model, is (a) d ≠ 0 and a, b, c can be any thing c (b) a ≠ 0 and b, c, d can be any thing (c) b ≠ 0 and a, c, d can be any thing 1 d 1/s 1/s (d) c ≠ 0 and a, b, d can be any thing r 1 y x2 x1 –a –b Common data for Q165 and Q166 Consider the model of system S1 as

é2 0 ù é1 ù x& = ê x + ê ú u and y = [1 0] x ú ë 0 -1û ë0 û é2 0ù é1 ù z + ê ú u and w = [1 0]z and model of system S2 as z& = ê ú ë0 1û ë0 û Q165. What can be said about stabilizability of these systems? (a) S1 and S2 both are stabilizable. (b) only S1 is stabilizable. (c) only S2 is stabilizable . (d) neither S1 nor S2 is stabilizable.

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Q166. What can be said about detectability of these systems? (a) S1 and S2 both are detectable. (b) only S1 is detectable . (c) only S2 is detectable. (d) neither S1 nor S2 is detectable. Q167. The state variable description of a linear autonomous system is x& = Fx where x is a state

é 0 gù ú ë -g 0 û

vector and F = ê

10

The value of g for which the system exhibits sustained oscillation with frequency Hz, p is _________________. Q168. A system has state transition matrix ϕ(t) whose Laplace transform is

é s+6 ê s 2 + 6s + 5 f (s) = ê -5 ê 2 ëê s + 6s + 5

1 ù s + 6s + 5 ú ú s ú s 2 + 6s + 5 ûú 2

The eigen values of the system are (a) 0 and − 6 (b) 1 and − 5

(c) 0 and + 6

(d) − 1 and − 5

Common data for Q169 and Q170 The state space representation of a system is given by x& = Ax + Bu and y = Cx where x is state vector, u is input and y is output.

é0 k ù é0 ù A=ê , B = ê ú and C = [1 0] ú ë0 -2 û ë1 û Q169. The transfer function G(s) of this system, is (a)

k s (s + 2)

(b)

ks (s - 2)

(c)

k s (s - 2)

(d)

ks (s + 1) (s + 2)

Q170. On placing G(s) in unity feedback configuration and applying unit step input as shown below, the settling time of system remains unchanged for k ≥ k´. The value of k´ is (a) 1 (b) 2 + r(t)=u(t) G(s) y(t) (c) 1.5 (d) 3 – Common data for Q171, Q172 and Q173 The signal flow graph together with assignment of state variables x1 and x2, is shown below. u

1 1

s–1 x2 –2

s

1

–1

–1 1

x1

1

1

y

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40

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Q171. The state space representation of this system, is

é0 -1ù é1ù x+ê úu ú ë1 -2 û ë1û y = [0 1]x

é0 1ù é0 ù x+ê úu ú ë -1 -2 û ë1 û y = [1 0] x

(b) x& = ê

é -1 1 ù é1ù x+ê úu ú ë 0 -2 û ë1û y = [1 1] x

(d) x& = ê

(a) x& = ê

é -1 0 ù é0 ù x+ê úu ú ë 0 -2 û ë1 û

( c) x& = ê

y = [1 0] x

Q172. The state transition matrix (STM) of the system is -2t (a) é e ê -t -2t ëe - 2e

ée- t

(c) ê

ë0

é 2e - t - e - t (b) ê 0 ë

0ù ú e- t û

e - t - 2e -2t ù ú e -2t û

e- t ù -2t ú ë1 e û é0

(d) ê

Q173. With zero initial condition, the response y(t) (a) 0.632

0ù ú e- t û

(b) 1.528

t =1

= y(l) for unit step input, will be

(c) 1.264

(d) 1.632

& = Fx(t) + Gu(t) and y(t) = Hx(t) of a dynamical system Q174. In the state space model x(t)

é -2 0 ù é2ù F=ê , G = ê ú , H = [1 1] ú ë 0 -3û ë0û x(t) is state vector, u(t) is input and y(t) is output. The transfer function G(s) =

Y(s) is U(s)

2 2 (s + 2) (d) (s + 2)(s + 3) (s + 3) & = Ax + Bu and y = Cx Q175. In the state space representation of a dynamical system x(t) é0 1ù é0 ù A=ê , B = ê ú and C = [1 1] ú ë -2 -3û ë1 û (a)

2 s+2

(b)

2 s+3

(c)

This system is placed in unity feedback configuration as shown below. In the steady state, this system tracks unit step input r1(t) = u(t) with error E1 and unit velocity input r2(t) = t u(t) with error E2. Then, E1 and E2 are respectively

(a) ( c)

2 and ¥ 3 3 and ¥ 2

(b) 0 and ¥ (d) ¥ and 0

ri(t) i = 1,2

+

G(s) = –

Y(s) U(s)

y(t)

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Q176. In the electrical circuit shown below, R1= 10 kΩ, C1 = C2 = 100μF and L =10 mH. + The voltages across C1 and C2 are assigned as state variables x1 and x2 R1 C2 respectively while current through L VL(t) ≡ y(t) u(t) ≡ Vi(t) + C1 L is assigned as state variable x3 . The state space model is –

104 ù é0 ù 4ú -10 ú x + êê0 úú u êë1 úû 0 úû y = [1 1 0] x é 0 0 104 ù é0 ù ê 4ú 0 10 ú x + êê1 úú u ( c) x& = ê -1 ê -102 102 0 ú êë0 úû ë û y = [0 1 1] x é1 0 ê 1 (a) x& = ê0 ê1 10-2 ë

-104 ù é1 ù 4 ú 0 10 ú x + êê0 úú u êë0 úû -102 0 úû y = [1 -1 0] x é102 -102 0 ù é1 ù ê 4ú 0 10 ú x + êê0 úú u (d) x& = ê 0 ê 1 êë1 úû 0 104 úû ë é -1 ê (b) x& = ê 0 ê102 ë

0

y = [-1 0 1] x

Common data for Q177 and Q178

dy d2 y = -1. = u has initial conditions y(0) = 1 and 2 dt t =0 dt dy(t) , the state variable Q177. With state variable assignment x1(t) = y(t) and x 2 (t) = dt & = A x(t) + Bu(t) and y(t) = Cx(t) is representation in the form x(t) The differential equation

é0 ë0 é1 & =ê (b) x(t) ë0 (a) x& (t ) = ê

1ù é0 ù x(t) + ê ú u(t) and y(t) = [1 0] x(t) ú 0û ë1 û 0ù é0 ù x(t) + ê ú u(t) and y(t) = [0 1] x(t) ú 1û ë1 û

é0 -1ù é1 ù x(t) + ê ú u(t) and y(t) = [0 1] x(t) ú ë0 -1û ë0 û é -1 0 ù é1 ù & =ê (d) x(t) x(t) + ê ú u(t) and y(t) = [0 1] x(t) ú ë 0 -1û ë0 û & =ê ( c) x(t)

Q178. For u(t) = 1; t ≥ 0, y(t)|t=1 = y(1) is (a) 1/2 (b) 1

(c) 3/2

(d) 2

Q179. Consider the state model of a system with input r(t) = δ(t) ;δ(t) is unit impulse and output y(t) .

é -4 1 ù é1 ù é0 ù & =ê x(t) x + ê ú r ; and y = [1 0]x and x(0) = ê ú ú ë -3 0 û ë0 û ë0 û

Control Systems

42

The system response y(t) at t = 0.25 sec is (a) 0.319 (b) – 0.109 Common data for Q180 and Q181 Consider an electrical network shown below. + vs(t) –

~

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(c) − 0.477

(d) 0.239

R1 = R2 = R3 = 1MΩ C1 = C2 =4.7μF

R2

R1

C1

R3

C2

v0(t), output

Q180. The voltage vc1 and vc2 across capacitors C1 and C2 respectively are assigned as state variables x1 and x2 respectively. vs(t) ≡ u and v0(t) ≡ y. The state space representation of the network will be

é10 / 47 20 / 47 ù é10 ù x+ê úu ú ë 20 / 47 10 / 47 û ë10 û

é -20 / 47 10 / 47 ù é10 / 47 ù x+ê ú úu ë 10 / 47 -20 / 47 û ë10 / 47 û

(b) x& (t ) = ê

(a) x& (t ) = ê

y = [1 0] x

y = [0 1] x

é 4 / 47 -2 / 47 ù é 2 / 47 ù x+ê ú úu ë -2 / 47 4 / 47 û ë 2 / 47 û

é -4 / 47 2 / 47 ù é 2 / 47 ù x+ê ú úu ë 2 / 47 -4 / 47 û ë 2 / 47 û

(d) x& (t ) = ê

( c) x& (t ) = ê

y = [0 1] x

y = [1 1] x

Q181. What can be said about state controllability and observability? (a) system is completely state controllable and observable. (b) system is neither state controllable nor observable. (c) system is state controllable but not observable because the bridge is balanced. (d) system is not state controllable but observable because the bridge is balanced.

Q182. A system with single input u and two outputs y1 and y2, is described by equations é x& 1 ù é - 3 4 ù é x 1 ù é 2 ù é y1 ù é - 4 ê x& ú = ê - 2 0 ú ê x ú + ê 1 ú u ; ê y ú = ê 5 û ë 2û ë û ë 2û ë ë 2û ë Now, consider the following statements.

I. T2 (s) =

y 2 (s) 9s + 21 = 2 U(s) s + 3s + 8

II. T1 (s) =

6ù - 1úû

é x1 ù êx ú ë 2û

y1 (s) -2(s + 11) = U(s) s 2 + 3s + 8

III. The eigen values are −1.5 ± j 2.4 Of these, correct statements are (a) only II and III

(b) only I and III

(c) only I and II

(d) all I, II and III

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ée

Q183. A system is described by state transition matrix f(t) = ê

-t

0 ù ú and has initial e -2t û

ë0 é1 ù condition x(0) = ê ú .The state of system after 0.5 seconds will be ë2û é 0.6 ù é0.74 ù é0.6 ù é 1.2 ù (a) ê (b) ê (c) ê ú (d) ê ú ú ú ë0.74 û ë 0.6 û ë1.2 û ë0.37 û

Q184. A single input multiple output system is shown below. u is input and yi ; i = 1, 2, 3 are outputs. The state space model of this system in the form x& = Ax + Bu and y = Cx, will be (a)x& = [b]x + [1]u and y = [a1

a2

a 3 ]x

é a1 ù (b) x& = (- b)x + [1]u and y = êa 2 ú x ê ú ëê a 3 ûú éb o o ù é1ù é a1 ù ê ú ê ú ê ú (c) x& = ê o b o ú x + ê1ú u and y = êa 2 ú x ëê o o b ûú ëê1ûú ëê a 3 ûú

u a1 s+b

a2 s+b

a3 s+b

y1

y2

y3

où é -b o é1ù ê ú (d) x& = o - b o x + ê1ú u and y = [a1 a 2 ê ú êú êë o êë1úû o -b úû

a 3 ]x

Common data for Q185 and 186 A system has state space model

é x& 1 ù ê x& ú = ë 2û

é1 ê4 ë

–2 ù é x 1 ù é 0 ù é x1 ù + ê ú u and y = [1 l ] ê ú ê ú ú –5 û ë x 2 û ë 1 û ëx2 û

Q185. Assume λ = 1 and an initial state vector [2 response y(t) |t=1= y(1) is ____________________.

T

1] ; T stands for transpose. The zero input

Q186. The values of λ for which the system model is unobservable, are (a) ±1 (b) ± 1 / 2 (c) – 1 / 2 and – 1 (d) –1 and –2 Q187. Consider a system in state variable from x& = Ax + Bu ; y = Cx +Du

1 0ù é 0 é -1 ù ê ú where A = ê 2 0 1ú , B = êê 0 úú , C = [1 2 0] and D = [0] êë -k - 3 - 2 úû êë 1 úû

Control Systems

44

The system is stable for (a) k < 6 (b) k > 4 (c) 4< k < 6 Q188. Consider a system whose signal flow graph with marked state variables x1 and x2, is shown below. This system is 1 (a) unstable but stabilizable . (b) completely controllable and observable. U(s) (c) not controllable because the state –1 variables do not depend on input u. (d) neither controllable nor observable but stable.

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(d) k > 6 2 –1 s

x1 1

–1

Y(s) s

1

–1

x2

1

7

Controllers

Common data for Q189, Q190 and Q191 The dB magnitude vs ω plot of a controller is shown below. dB Q189. The dB plot corresponds to 12.04 (a) lag compensator with pole corner frequency 40 rad/sec. (b) lead compensator with pole corner frequency 40 rad/sec. (c) lag compensator with pole corner frequency 20 rad/sec. 0 (d) lead compensator with pole corner frequency 20 rad/sec. Q190. The maximum phase contributed by compensator, is (a) 36.87o lag (b) 36.87o lead (c) 30o lag

+6dB/oct 10

ω (rad/sec)

(d) 30o lead

Q191. If this compensator is inserted in forward path of control system, then (a) both rise time and peak overshoot will decrease. (b) both rise time and peak overshoot will increase. (c) rise time will decrease while peak overshoot will increase. (d) rise time will increase while peak overshoot will decrease. Q192. A system has transfer function G(s) =

50 . A suitable compensation is desired to s (s + 10) 2

meet the following specifications. I. The system should follow constant velocity input with zero steady state error. II. The system should exhibit maximum of 25% peak overshoot in the transient state . Which one of the following will be better option? (a) two lead compensators (b) two lag compensators (c) PI compensator (d) single lead compensator

Q193. The maximum phase lead that can be contributed by the network shown below, is C æ R ö -1 æ R ö sin (a) sin -1 ç (b) ç ÷ è 2R 'C ø è R + 3R ' ÷ø æ R ö è 2R ' ÷ø

( c) sin -1 ç

(d) sin

R ö çè ÷ R + 2R ' ø

-1 æ

R

R'

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Q194. Consider the following statements pertaining to a compensator with transfer function

G C (s) =

s+a ; a < b and a, b > 0 s+b

I. GC(s) is a lead compensator that contributes maximum phase lead at frequency wm =

ab.

o -1 a / b. II. Maximum phase lead contributed by lead compensator fmax = 90 - 2 tan

III. The attenuation contributed by compensator at frequency wm = ab ,is equal to a / b. Of these the correct statements are (a) only I and II (b) only I and II ( c)all I, II, and III (d) only II and III Q195. A phase lag compensator has transfer function G C ( jw) =

1 + jw / b where a, b are 1 + jw / a

positive constants. If ϕlag is maximum phase lag contributed at ω =ω*, then

b - 90o a = 90o - 2 tan -1 b / a

-1 (a) a < b , w* = ab and flag = 2 tan

(b) b > a , w* = a / b and flag

o -1 b/a (c) b > a , w* = b / a and flag = 90 - tan -1

o

b / a - 90 (d) a < b , w* = a / b and flag = tan Q196. A PD compensated system is shown below. The design effort is made so that system exhibits phase margin of π/3 radian at cross over frequency of 1/2π Hz. The tuned parameter Kp is____________________. + X1(s)

–

Common data for Q197 and Q198 A unity feedback system shown below, is designed to exhibit phase margin ϕm = 45o at cross over frequency ωgc = 3 rad/sec.

1 s(s + 1)

KP + 0.5s

+

G(s) = –

k s(s + p)

Q197. The design values of k and p are respectively (a) 10 3 and 3

(b) 9 2 and 3

Q198. If a filter with transmittance H(s) = margin with values found in Q197, will be (a) 15o (b) − 15o

( c) 9 and 3 3

(d) 18 and 3 3

s-3 3 is inserted into the loop, the new phase s+3 3 (c) 30o

(d) − 30o

X2(s)

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Q199. In the control system shown below, the choice of controller is such that the response y(t) + follows the input r(t) = t u(t) with 10 controller zero steady state error. s (s + 2) r(t) y(t) – The type of controller should be (a) proportional plus integral (b) proportional (c) derivative (d)proportional plus derivative. Q200. In the control system shown below, G(s) has four poles and a zero in the left half of s plane. The controller GC(s) is a proportional plus integral plus derivative (PID). The plot G c ( jw) G( jw) dB vs w has slope of M dB/octave in the highest frequency (ω) range. The value of M, will be + GC(s) G(s) (a) − 6 (b) − 12 – (c) − 18 (d) − 24 Q201. A system is being controlled using proportional controller. If a time delay is introduced in the loop, then (a) phase margin will increase. (b) gain margin will increase. (c) phase margin will decrease. (d) phase margin will remain unchanged.

Q202. A compensator is to be designed for system with transfer function G(s) =

900 s(s + 1)(s + 9)

to satisfy the following. I. Compensated gain cross over frequency is equal to uncompensated phase cross over frequency. II. Compensated phase margin ≥ 45o. To achieve this goal, the designer may employ (a) a lag compensator that contributes attenuation of 20 dB and phase lag of 45o at w = 3 3 rad / sec. o (b) a lead compensator that contributes amplification of 20 dB and phase lead of 45 at

w = 3 3 rad / sec.

o

(c) a lag-lead compensator that contributes amplification of 20 dB and a phase lag of 45 at

w = 3 rad / sec.

(d) a lag-lead compensator that contributes an attenuation of 20 dB and phase lead of 45o at

w = 3 rad / sec. Q203. Consider a PD compensated marginally stable system shown below. The design goal is that the damping of closed loop system, x £

1 and steady state error for unit step input, 2

e(∞) ≤ 0.1 . The values of λ and μ should be respectively (a) λ = 5 and μ = 10 (b) λ = 5 and μ =

5

(c) λ = 9 and μ = 10 (d) λ = 9 and μ =

5

R(s)

+ –

controller λ + μs

4 s +4 2

Y(s)

Control Systems

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Q204. Consider a negative feedback control system as shown below , where

G(s) =

1 a and G c (s) = 1 + s+2 s +

X(s)

Gc(s)

k

–

Y(s) G(s)

The value k and a are tuned to force the system to exhibit overshoot of 5% and settling time of 1 sec to a unit step input. The value of k – a is ____________. o

Q205. An attempt is made to control a system as shown below, knowing that phase margin of 30 is acceptable. system controller + 10 R(s)

–

G c (s) = k

G(s) =

s(s+ 10)

Y(s)

The value of |Gc(s)| is __________.

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Answer Key Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 Q41 Q42 Q43

1 b a 95 a d b a a a 1 c d c a b d b c a a b a a d a 2 2 47 b a 29.33 d a c d d c a a d b c

Q44 Q45 Q46 Q47 Q48 Q49 Q50 Q51 Q52 Q53 Q54 Q55 Q56 Q57 Q58 Q.59 Q60 Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q70 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q86

a a b b a c c a d a 2 0.285 a 1 a b 1.25 d c a d a b a a d d d b 3 a –2.47 a a a a d 8 a d d c 20

Q87 Q88 Q89 Q90 Q91 Q92 Q93 Q94 Q95 Q96 Q97 Q98 Q99 Q100 Q101 Q102 Q103 Q104 Q105 Q106 Q107 Q108 Q109 Q110 Q111 Q112 Q113 Q114 Q115 Q116 Q117 Q118 Q119 Q120 Q121 Q122 Q123 Q124 Q125 Q126 Q127 Q128 Q129 Q130 Q131

d a b a a a d b c 9 8.95 a d a d d c a b c a d a a a a b c a d d d a a a a 8 d d a c a a b b

Q132 Q133 Q134 Q135 Q136 Q137 Q138 Q139 Q140 Q141 Q142 Q143 Q144 Q145 Q146 Q147 Q148 Q149 Q150 Q151 Q152 Q153 Q154 Q155 Q156 Q157 Q158 Q159 Q160 Q161 Q162 Q163 Q164 Q165 Q166 Q167 Q168 Q169 Q170 Q171 Q172 Q173 Q174 Q175 Q176

c d d c 2.5 a c a a d d d a d b b b c d c c a d a d a 4 0.42 5.34 b c a a b b 20 d a a c c c a a b

Q177 Q178 Q179 Q180 Q181 Q182 Q183 Q184 Q185 Q186 Q187 Q188 Q189 Q190 Q191 Q192 Q193 Q194 Q195 Q196 Q197 Q198 Q199 Q200 Q201 Q202 Q203 Q204 Q205

a a a a d d a b 2.058 c c c b b a a d c a 1.866 b b a b c d d 0.4 34.64

Solutions

Q1.

T (s ) =

STa =

G (s ) a / (s+ b ) a = = ac 1 + G(s ) H( s) s + b + ac 1+ s+b

¶T a s+b ´ = ¶a T s + b + ac

Substituting nominal values of b and c

STa =

s+ 2 and S Ta = s + 2 + 3a

w2 + 4 w 2 + (2 + 3a )

2

In steady state

S Ta Q2. (b)

w= 0

=

2 = 0. 4 gives a = 1 2 + 3a

3 4s - 1 4s - 1 = gives time constant . t 1 = sec. 2 3s + 2 æ 3s ö 2 ç1 + ÷ 2ø è

The system S2 (Fig.II) has overall transfer function

10 / (s + 3) 10 ´ 5 1+ (s + 3)

=

10 s + 53

=

10 / 53 s 1+ 53

1 sec . 53 τ 159 The ratio 1 = (3 / 2 ) (1 / 53) = τ2 2 and time constant t 2 =

/

Q3. (a) To ease out finding

YD (s) the signal flow graph is drawn below. , D(s)

Forward paths P1 =B(s) and P2 = Gff(s) A(s) B(s) Cofactors: Δ1 = Δ 2 = 1 Determinant: Δ = 1−(− A(s) B(s) C(s) ) = 1+A(s) B(s) C(s)

YD (s) P1D1 + P2 D 2 = D(s) D

=

B(s) + G f f (s) A(s) B(s) 1 + A(s) B(s) C(s)

1 R(s)=0

For YD(s) = 0 , B(s) + Gff (s) A(s) B(s) = 0 or Gff (s) = −1/A(s)

Gff(s)

D(s) 1 YD(s)

A(s)

B(s) – C(s)

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K K 1 = and t OL = s+5 5 æ sö 5 ç1 + ÷ è 5ø G(s) K K 1 Overall transfer function T(s) = = = and t CL = . s ù 1 + G(s) s + 5 + K 5+ K é (5 + K )ê1 + 1 20 ë 5 + K úû t OL = 20 t CL gives = or k = 95 5 5+ K k Q5. (a) g OL = k =95 = 19 s + 5 s=0 Q4. G (s ) =

g CL =

k s+5+ k

k =95 s=0

= 0.95

Q6. (d) The feedback in a control system reduces sensitivity, increases speed of response by decreasing time constant, reduces the effects of disturbances and increases bandwidth. Since, the changes in feedback path affects the control system performance more adversely than changes in forward path , the design demands that the feedback sensor has high signal to noise ratio. Q7. (b) Use Mason's gain Rule to get forward path P1 = 0.5 × 2 × 4 = 4 ; cofactor Δ1 = 1 and determinant

D = 1 - éë -s -1 - s -1 - p ùû + éë ps -1 + ps -1 + s -2 ùû - éë(-p) (-s -1 ) (-s -1 ) ùû = 1 + p + 2 (1 + p) s -1 + (1 + p) s -2

X 2 (s) 4 4s 2 = = X1 (s) (1 + p) + 2(1 + p)s -1 + (1 + p)s -2 (1 + p)[s 2 + 2s + 1] But

4s 2 4s3 = (1 + p)(s 2 + 2s + 1) s3 + 3s 2 + 3s + 1

or 1 + p =

4s 2 (s3 + 3s 2 + 3s + 1) s3 + 3s 2 + 3s + 1 = 4s3 (s 2 + 2s + 1) s3 + 2s 2 + s

s3 + 3s 2 + 3s + 1 s 2 + 2s + 1 1 or p = -1 = 2 = 3 2 s + 2s + s s(s + 2s + 1) s or p = s –1

Q8. (a) Fig.I :

X 2 (s) P(s) X 2 (s) A(s) = and Fig.II : = X1 (s) 1 + P(s) Q(s) X1 (s) 1 + A(s)

Equating the two, P (s) A(s) = 1 + P (s) Q(s) 1 + A(s) P(s) + P(s) A (s) = A (s) + P(s) Q(s) A (s) P(s) P (s) A(s) = = 1 + P (s) Q (s) - P(s) 1 + P(s)[Q(s) - 1]

or Q9. (a)

51

Control Systems

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Y(s) 1 = –1 X(s) 1 – l1s – l 2s –2

1

2

1

1

Y(s)

X(s)

s N(s) = 2 º s – l1s – l 2 D(s)

l2

l1

But D(s) = (s + 2)(s + 3) = s 2 + 5s + 6

s

s–1

–1

Compare to get λ1 = –5 and λ2 = – 6 Q10. (a) The blocks are reduced as follows.

R(s)

+ –

1/ (s+ 2) 1 + 4 / (s+ 2)

s s+3 3s s+4

+

C(s)

+

5 s

R(s)

+ –

1 s+6 5 3s ´ s s+4

X(s) R(s)

1 s+4 s+6 = 2 1 15 æ öæ ö s + 10s + 39 1+ ç è s + 6 ÷ø çè s + 4 ÷ø

s s+3 3s s+4

+ +

C(s)

Y(s) s 3s 4s 2 + 13s + = s + 3 s + 4 (s+ 3)(s+ 4)

C(s)

52

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Control Systems

Q11. (c)

K +

U(s)

10 s + 10

Y(s)

+

K(s + 10) 10 –

U(s)

+

+

10 s + 10

Y(s)

é ù é K(s + 10) ù ê10 / (s + 10) ú æ 10 - Ks - 10K ö æ 10 ö = ÷ø çè ÷ êë1 - 10 úû ê 10 ú çè 10 s + 20 ø ê 1+ ú s + 10 û ë

U(s)

Now,

Q12. (c)

–

Y(s)

10 - Ks - 10K -s = gives K = 1 s + 20 s + 20

Y(s) R1 (s)

R 2 (s) = 0

3 / (s + 1)(s + 3) 3 3 = = 3 (s + 1) (s + 3) - 3 s (s + 4) 1(s + 1) (s + 3) 1/ (s + 1) s+3 s+3 = = = 1 3 (s + 1) (s + 3) - 3 s (s + 4) 1´ (s + 1) (s + 3)

Y(s) R 2 (s)

R1 (s) = 0

Y(s) =

3R1 (s ) (s + 3)R 2 (s ) 3 4 + where R1 (s) = and R 2 (s ) = s (s + 4) s (s + 4) s +1 s

Y (s ) =

9 4(s + 3) 4s 2 + 25s + 12 + 2 = s (s + 1)(s + 4) s (s + 4) s 2 (s + 1)(s + 4)

=

Q13.(d) Statements 1 and 3 are correct. Statement 2 is not correct. In fact, cofactor of a forward path is determinant of signal flow graph formed by deleting all the loops that are touching the forward path. Q14.( c) Use Mason's gain rule,T =

åPD i

D

i

to get forward path P1= a , cofactor ∆1 = 1+ d

and determinant ∆ = 1–[ –eg – d – bcg] + (– eg) (–d) = 1+ d + eg + bcg + edg

y2 a (1 + d) = y1 1 + d + eg + cbg + edg

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Q15. (a) Use Mason's gain rule to get Y(s) s -2 1 1 = = 2 = -1 -2 X(s) 1 + 4s + 3s s + 4s + 3 (s + 1) (s + 3) and

Y(s) =

é1/ 3 1/ 2 1/ 6 ù X(s) e -2s = = e -2s ê + (s + 1) (s + 3) s(s + 1) (s + 3) (s + 1) s + 3 úû ë s

1 1 1 y(t) = u(t - 2) - e -(t -2) u(t - 2) + e -3(t -2) u(t - 2) 3 2 6

Q16. (b) The closed loop transfer function T(s) is Laplace transform of impulse response

T(s) =

-1 2 2s + 3 N(s) + = º 2 2 (s + 2) (s + 2) (s + 2) D(s)

Then, open loop transfer function G(s) = Q17. (d)

Y (s ) U (s )

N(s) 2s + 3 = D(s) - N(s) (s + 1) 2

= G (s ) gives Y(s) = G(s) U(s) t

or y(t) = g(t) * u(t) = ò g(t - t) u(t) dt where * denotes convolution 0

comparing it with given y (t ) =

t

ò (3 + t - t )e

-5(t -t )

u (t )dt

0

g(t) = (3 + t) e -5t = 3e -5t + te -5t and G(s) =

3 1 3s + 15 + 1 3s + 16 + = = 2 s + 5 (s + 5) (s + 5) 2 (s + 5) 2

Q18.(b) Let the two identical low pass filters have transfer functions G1 (s) = G 2 (s) = G(s)= Then, the transfer function of composite filter G * (s) =

k2 has two poles, both located at s =−a. (s+a) 2

The unit step response will be critically damped. Q19.( c)Y(s) R (s) = 1 = 1 ; Laplace transform of unit impulse function δ(t) is 1. s

1 Y(s) = G(s) ´ R(s) = G(s) ´ = 1 Þ G(s) = s s Q20.(a) Unit impulse response =

k . s+a

d (unit step response) dt

d éd ù d éd ù (unit ramp response) ú = ê (0.8t - u (t) + 0.1e -5t ) ú ê dt ë dt û dt ë dt û d [0.8 - 0.5e -5t ] = 2.5e -5t dt

Control Systems

54

Q21.(a)

Y(s) R(s)

Y(s) R(s)

Fig.II

Mechasoft

1 s+2 s (s + 1) = = 1 s(s + 1)(s + 2) + 1 1+ s(s + 1)(s + 2)

Fig.I

1 é ù ê s (s + 1) (s + 2) ú é ù 1 = X(s) ê ú = X(s) ê ú 1 ê1 + ú ë s(s + 1)(s + 2) + 1 û ëê s(s + 1)(s + 2) ûú

X(s) = s + 2

P

Q22.(b) For ease in determination of the transmittance C(s)/R(s), the block diagram is transformed into signal flow graph as shown below. Now, use Mason's gain formula to get forward paths P1= P, P2 = P, P3 = P2Q, P4 = P2Q together with corresponding cofactors Δ1 = Δ2 = Δ3 = Δ4 =1 and determinant Δ = 1 − (P2Q2).

1

1 R(s)

P

Q Q

1

1 C(s) P

1

1

4

C(s) = R (s)

åPD i

i

i =1

=

D

P + P + P 2Q + P 2Q 2P (1 + PQ) 2P = = 2 2 1- P Q (1 + PQ)(1 - PQ) 1 - PQ

A

Q23.(a)

x 2 = (A + B) x1 x1

B

x2

x2

B A

x1

A+B

x1

x 2 = Bx1 = BAx 2

x1 or

or x1 = Ax 2 = ABx1

x2

x2

AB

AB

B A

x1

x 2 = Ax1 + Bx 2 Þ x 2 =

x2

A x1 1- B

B x1

A

1

x2

x2 A A = Þ x2 = x1 x1 1 - B 1- B A/(1 – B) x1

x2

Control Systems

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Q24.(a) For the system dynamics to remain same when a node variable is multiplied by a constant, all signals entering the node must be multiplied by the same constant and all signals leaving the node be divided by the same constant. Q25.(d) Use Mason's gain rule to get forward path P1 = a1a2 , cofactor Δ1 = 1 and determinant Δ = 1 − [−a1a2b1+1+1]+1 = a1a2b1.

Now,

x 2 P1D1 aa = = 1 2 = 1/ b1 x1 D a1a 2 b1

Q26.(a) Inserting the third insignificant root at s = – a, the characteristic equation is

1 + (s + a) (s + 1 + j1) (s + 1 - j1) = 0 or (s + a) (s 2 + 2s + 2) = 0 or s3 + (2 + a)s 2 + (2 + 2a)s + 2a = 0 or 1 +

2a =0 s[s + (2 + a)s + 2(a + 1)] 2

2a s[s + (2 + a)s + 2(a + 1)] a The ramp error constant K V = lt s G (s) = s® 0 a +1 Compare with 1 + G(s) = 0 to get , G (s) =

Steady state error e( ¥ ) =

and G(s) =

Q27.

2

1 a +1 = = 1.5 gives a = 2 Kv a

4 s (s + 4s + 6) 2

Y(s) P1 D 1 + P2 D2 = ; P1 = s–2 , P = 2s–1 and D1 = D2 = 1 X(s) D D = 1 – éë –2a – a s–1 – s–2 – 2 s–1 ùû = s–2 + (a + 2)s–1 + 1 + 2a Y(s) s–2 + 2s–1 2s + 1 N(s) = –2 = º –1 2 X(s) s + (a + 2)s + (1 + 2a) (1 + 2a)s + (a + 2)s + 1 D(s) P(s) N(s) 2s + 1 = = 1 + P(s) D(s) – N(s) (1 + 2a)s2 + as 2s + 1 2s + 1 P(s) = = 2 2 (1 + 2a)s + (a – 2)s –1 éë(1 + 2a)s + as ùû – (2s + 1) compare with P(s) =

2s + 1 to get a = 2 5s2 –1

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Q28. For ease in calculation, use the equivalent signal flow graph to get

10 , D1 = 1 s(s + 1) 10/(s+1) 1 X(s) 10 10l D = 1+ + –1 s(s + 1) (s + 1) Y(s) 10 / s(s + 1) 10 = = 2 10 10 l X(s) 1 + s + (10l + 1)s + 10 + s(s + 1) (s + 1) 10l + 1 = 21 Þ λ = 2 –1

Y(s) P1D1 = X(s) D

P1 =

1/s

Y(s)

λ

1

1

1(s+10)

Q29.Construct equivalent signal flow graph. R(s)

1

1/(s+5)

Y(s) P1D1 1 = , P1 = , D1 = 1 –1 R(s) D (s + 5)(s + 10) 1 1 (s + 5)(s + 10) + (s + 10) + (s + 5) D = 1+ + = s + 5 (s + 10) (s + 5)(s + 10) Y(s) 1 N(s) 1 = 2 and for H(s) = 1, G(s) = = 2 R(s) s + 17s + 65 D(s) - N(s) s + 17s + 64 b1 = 17, b 2 = 64 and | b1 - b 2 | = 47 Q30.(b)

1

Y(s)

Y(s) 1 1 1 = and for D(s) = , Y(s) = 2 D(s) s(s + 4.4) + K s s(s + 4.4 s + K)

For the response due to unit disturbance to be less than 1, in fact, less than 0.01, the system must be overdamped such that y(∞) < 0.01.

s 1 = s®0 s(s + 4.4s + K) K

y(¥) = l t y(t) = l t sY(s) = l t t ®¥

s® 0

1 < 0.01 Þ K > 100 K

2

d(t) = u(t)

1 y(t) 0.01 t

T(s) =

s 2 - 2s s 2 + 6s + 9

T(s) =

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Control Systems

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Q31.(a) System function 64 T(s) = 2 3 s + 4s + 5

9s 2 + 3s + 10 s 2 + 5s + 2

Characteristic equation

Characteristic roots

Nature of damping

3s 2 + 4s + 5 = 0

s 1 , s 2 = − 0.67 ± j 1 .1

underdamped

s 2 + 6s + 9 = 0

s 1 , s 2 = -3, - 3

critically damped

s 2 + 5s + 2 = 0

s 1 , s 2 = -0 .4, - 4 .6

over damped

19s - 20 underdamped s 1 , s 2 = -0 .5 ± j 9.99 s 2 + s + 100 = 0 s + s + 100 2 æK+4ö Q32. The characteristic equation is 3s2 + 2s + K + 4 = 0 or s 2 + s + ç ÷=0 T(s) =

2

3

wn =

x=

è

3

ø

K+4 2 and 2 xw n = 3 3

1 1 100 = 0.1 or = 0.1 Þ K = - 4 = 29.33 3w n 3 3 (K + 4) / 3

Q33(d). Eliminating minor loop, the control system can be drawn as shown below. The GH product =

10K and ramp error coefficient s (s + 1)

R(s) é 10K ù K v = lt s ´ ê = 10K. ú s® 0 ë s (s + 1) û 1 1 e(¥) = = £ 0.01 Þ K ³ 10 K v 10K

+

-

E(s)

æ ö K ç 10 / s ÷ ´ç ÷ s ç 1 + 0.1´ 10 ÷ s ø è

Q34.(a) The complex conjugate roots lie between

1 3 x = and x = locus 2 2 1 3 and therefore, < x < . 2 2 4 . The settling time,t s = Re[roots] Re[roots] = 4 gives ts = 1 sec. Re[roots] = 5 gives ts = 4/5 = 0.8 sec. 0.8 < ts < 1 sec.

x = cos 30 o =

Y(s)

o 1 x = cos 60 = locus 2

Im

3 locus 2 60 –5 – 4 ts=0.8

ts=1

◦

30◦

Re

Control Systems

58

6

Q35.( c) Damping ratio, x = cos q =

2

2

Mechasoft

Im

= 0.6

6 +8 Undamped natural frequency, wn = 62 + 82 = 10 rad / sec p-j Rise time, t r = wn 1 - x 2

+j8

× ωn

θ

Re

–6

2

1- x = 0.927 x p - 0.927 p - 0.927 So, t r = = = 0.277 sec @ 275 msec. 8 10 1 - 0.62 where j = tan -1

×

–j8

Q36.(d) Damping ratio ξ = 1.25 and undamped natural frequency ωn =10 rad/sec The transfer function T(s) = The unit step response Y(s) where a1 =

100 (s + 5)(s + 20)

wn 2 Y(s) 100 = 2 = R (s) s + 2 xw n s + w n 2 s 2 + 25s + 100 1 R (s)= s

=

100 a a2 a3 = 1+ + s (s + 5) (s + 20) s (s + 5) (s + 20)

= 1 , a2 = s =0

100 -4 100 1 = and a 3 = = s (s + 20) s =-5 3 s (s + 5) s =-20 3

4 1 y(t) = 1 + a 2 e -5t + a 3e -20t = 1 - e -5t + e -20t 3 3 Q37.(d) The steady state error e (¥) =

A A = = 0.4 Þ A = 2.Since, the system exhibits 1+ Kp 1+ 4

finite Kp, this is obviously type 0 system. The type ‘0' system has velocity error constant KV = 0 and steady state error due to ramp input, is E = ∞. Q38.( c)

C (s) 1 = éë1 - e -5s ùû ´ ; r (t) = d(t) and R(s) = 1 R(s) s C(s)

R (s) =1

Q39.(a) G(s) =

=

1 e -5s and c (t) = u(t) - u (t - 5) ; u(t) is unit step function. s s

1 K , position error constant K p = lt KG(s) = s ® 0 p1 p 2 (s + p1 ) (s + p 2 )

Control Systems

Mechasoft

e(¥) =

SeK( ¥ ) =

59

p1p 2 1 1 = = 1+ Kp 1+ K p1p 2 + K p1p 2

- p1 p 2 ¶ [e ( ¥ )] K K -K ´ = ´ ´ (p1p 2 + k) = 2 ¶K e ( ¥ ) (p1p 2 + K) p1p 2 p1 p 2 + K

Q40.(a) Closed loop transfer function T(s) =

as + b G(s) º s + as + b 1 + G(s) 2

as + b 2 and acceleration error constant K a = lt s G(s) = b s® 0 s2 2 2 2t 2 2 u ( t ), Steady state error, e (¥ ) = = For r(t) = t u(t) = Ka b 2

G(s) =

Q41.(d) The characteristic equation s2 +8 = 0 has characteristic roots located at s = ± j 8= ± j2.83 as shown below. System is oscillatory (time period T = 2π/ω) with ωn = ω = 2.83 rad/sec and damping ratio ξ = 0. Response T=

2p 2.83

Im

ξ=0

× +j2.83 Re

t

0

× –j2.83

Q42.(b) The system has over all transfer function

K2 ù é 2 ù é K + 1 ê 2 (K1s + K 2 ) Y(s) s úû êë s + 2 úû T(s) = = ë = 2 K ùé 2 ù U(s) s + 2(1 + K1 ) s + 2K 2 é 1 + ê K1 + 2 ú ê ú s û ës + 2û ë 2

and characteristic equation s + 2(1 + K1)s + 2K2 = 0 w n = 2K 2 = 10 gives K 2 = 50 2xw n = 2(1 + K1) gives 1 + K1 = 0.7 ´ 10 or K1 = 6

Control Systems

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–1

Q43.(c) Impulse response = L [G(s)] = g(t) 8 4´2 b G(s) = 2 = ; e– at sin bt ƒ 2 2 s + 4s + 8 (s + 2) + 2 (s + a)2 + b2 Impulse response, g(t) = 4e–2t sin 2t

Q44.(a)

Y(s) 4 = Þ sY(s) + 3 Y(s) = 4R(s) R(s) s + 3 & + 3y (t ) = 4r (t ) or y(t) Laplace transform gives

sY(s) - y(0- ) + 3Y(s) = 4 R(s) where R(s) = 1/ s Y(s) = =

4 y(0- ) + where y(0- ) = -2 s (s + 3) s + 3 4 / 3 4 / 3 -2 + s s+3 s+3

and y(t) =

4 4 -3t 4 -10 -3t - e + (-2) e -3t = + e 3 3 3 3

zero state zero input response response

natural forced response response

Q45.(a) r (t) = d(t) Þ R (s) = 1

Y(s) =

10 and y (t) = 10 e - 4 t s +4

natural response Forced component of response = 0 Since y(0– ) = 0, zero input response = 0 –4t Zero state response = 10e Q46.(b)

Y(s) 3s 6 = = 3R(s) s + 2 s+2

complete response

Control Systems

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For r(t) =δ (t) ; R(s) = 1, Y(s) = 3 -

and y zs (t)

t³0

= 3d (t) - 6e -2 t

y zs (t)

y zs (t)

t =0

61

6 s+2

t³0

Q47.(b) E(s) = R(s) – C(s) = R(s) – R(s) T(s) = R(s) [1 – T(s) ] 1 e(¥) = s lt ur0 s E(s) = s lt ur0 éës R(s) [1 – T(s) ]ùû ; R(s) = 2 s é1 – T(s) ù e(¥) = s lt ur0 ê ë s úû It is obvious that e(¥) = 0 when T(s) s=0 = T(0) = 1 and T(0) = 1 Þ

kz =1 8

For T(0) = 1 , use of L ' Hospital's rule gives é dT(s) ù e(¥) = s lt =0 ur0 ê – ë ds úû s

é k(s2 + 4s + 8) – k(s + z)(2s + 4) ù 8 lt = 0 Þ z = 2 and k = = 4 ur0 ê – ú 2 2 (s + 4s + 8) z ë û

Q48.(a) The overall transfer function

K1 Y(s) = 2 R (s) s + K 2 s + K 3 + 0.5 K 1

é

ù sK1R(s) ú ë s + K 2s + K 3 + 0.5K1 û

In the steady state, y(¥) = lt y(t) = lt sY(s) = lt ê t ®¥ s ®0 s ®0

or y(¥) =

2

K1 K1 é lt sR(s) ù = r(¥) û K 3 + 0.5K1 K 3 + 0.5K1 ë s®0

For y(¥) = r(¥) , K1 = K 3 + 0.5K1 Þ K 3 = 0.5K1

The peak overshoot, e -px / The 2% settling time,

1-x 2

= 0.05 gives x = 0.69

4 4 = 1 gives w n = = 5.8 xw n x

Now, from characteristic equation s 2 + K 2s + (K 3 + 0.5K1 ) = 0 w n = K 3 + 0.5K1 = K1 = 5.8 Þ K1 = 33.6 and K 3 = 0.5 K1 = 16.8 dampling ratio, x =

K2 K2 = = 0.69 gives K 2 @ 8 2w n 2 ´ 5.8

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62

Q49.(c) The impulse response curve and tangent to it at t = 0 is drawn below. Let the tangent intersect time axis at t = t'.

Mechasoft

g(t) (0, k)

d k d k [g(t)] = - e - t/ t and [g(t)] = dt t dt t t =0

–t/τ

ke

k -0 k = - Þ t'= t 0-t' t t d 0 (t , 0) (unit step response) Q50.( c) Unit impulse response = dt d éd ù d é1 1 ù = ê (unit ramp response) ú = ê - e -4t ú = e -4t dt ë dt û dt ë 4 4 û k Q51.(a) The root locus for G(s) = s (s + 2) (s + 4) k=¥ Slope of tangent =

Im

with relevant information found as follows, is sketched below. (i) Number of separate RL branches = 3 (ii) Real axis segments: between s = 0 and s = −2, and between s = − ∞ and s = −4. o o (iii) Asymptotic angles, θA = 60 , 180 k=¥ k=0 -2 - 4 ´ (iv) Centroid, s A = = -2 -4 3 (v) Break away points : The characteristic equation 1 + 3

k =0 s (s + 2)(s + 4)

k = 3 .1 k=0 ´ -2 - 0 .85

´

k=0

Re

k = 48

2

gives − k = s(s+2)(s+4) = s +6s +8s and

-

+j2 2 k = 48

dk = 3s 2 + 12s + 8 = 0 ds

gives possible break away points s = − 3.15, − 0.85

-j2 2 k=¥

But break away point is only s = − 0.85. The point s = − 3.15 does not satisfy real axis segment rule. (vi) Imaginary axis crossing point is determined from characteristic polynomial P(s) = s3+6s2+8s+k P(jω) = (k − 6ω2) + j(8ω−ω3) 3 2 2 Im[P(jω)] = 8ω−ω = 0 gives ω =2 2and Re[P(jω)] = k − 6ω = 0 gives k = 6ω = 48 So, root locus crosses imaginary axis at s = ± j 2 2 for k = 48. The system has damped oscillatory response in the range 3.1 < k < 48. For k > 48 system is unstable. For k = 48 the system exhibits sustained oscillations with corresponding frequency of oscillations equal to 2 2 rad/sec. For 3.1 < k < 48 the system has complex conjugate dominant roots. For 0 < k ≤ 3.1 the system has all real roots exhibiting exponentially damped response. For k = 3.1 system exhibits critically damped response in second order approximation.

Control Systems

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K Q52.(d) K v = lt s G(s)H(s) = lt s ´ i ´ s®0

s®0

s

63

K

1 (s + 1)(s + 2)

=

i 2

K

i > 20 or K > 40 i 2 K i The characteristic equation 1 + G(s) H(s) = 0 gives 1 + =0 s(s + 1)(s + 2) Routh array s3 1 2 or s3 + 3s 2 + 2s + K = 0 2 i s 3 K Velocity error constant, K v > 20 demands

i

s

1

s

0

6 - Ki 3

Ki

The closed loop stability demands 6 − Ki > 0 or Ki < 6 while , as discussed above , Ki should be larger than 40 for Kv to be large than 20.So, the two design goals can not be met simultaneously.

Q53.(a) Compare the characteristic equation : 1 +

2(1 + l s) = 0 or s 2 + (2l + 3)s + 4 = 0 s 2 + 3s + 2

with s2 + 2ξ ωn s + ωn2 = 0 to get ωn = 2 and ξ = (2λ + 3) / 4. For overshoot to occur, damping ratio ξ < 1, that is, (2λ + 3) / 4 < 1 or λ < 1/2. System will exhibit overshoot for λ = 0.25.

Q54.

Y(s) s 2 + as + b s 2 + as + b = = ; R(s) = 1/ s R(s) s 2 + 4s + 3 (s+ 1)(s+ 3)

Y(s) =

s 2 + as + b C C C s 2 + as + b b = 1 + 2 + 3 where C1 = = s(s+ 1)(s+ 3) s (s+ 1) s + 3 (s+ 1)(s+ 3) s=0 3

s 2 + as + b a - b -1 s 2 + as + b 9 - 3a + b C2 = = and C3 = = s(s+ 3) s=-1 2 s(s+ 1) s=-3 6 b (a - b- 1) / 2 (9 - 3a + b) / 6 b æ a - b - 1ö - t 9 - 3a + b -3t + + and y(t) = + ç e ÷e + 3 s +1 s+3 3 è 2 ø 6 8 3 1 Compare with y(t) = - e- t - e-3t to get b = 8 , a - b - 1 = -3 or a = 6 and | a – b |= 2 3 2 6 Y(s) =

Q55.

Let

V0 (s) =

V0 (s) 3 ´10-5 3 ´10-6 100 = = ; Ti (s) = Ti (s) 10s + 1 s + 0.1 s

3 ´10-4 1 ù é1 = 3 ´10-3 ê and v 0 (t) = 3 ´10-3 [1 - e -0.1t ] ú s(s + 0.1) ë s s + 0.1 û

v 0 (t) t =1 = 3 ´ 10 - 3 [1 - e - 0.1 ] = 0.285 ´ 10 - 3 V =

0.285mV

Mechasoft

Control Systems

64

Q56.(a) G(s) =

k ps + 1

has steady state gain = G(0) = k = 3

and time constant = p = 2 ; Let

Y(s) R(s)

=

k/p s + (1 / p)

=

1.5 s + 0.5

1 1.5 1 ù é1 For R(s) = , Y(s) = =3 ê s s(s + 0.5) ë s s + 0.5 úû y(t) = 3[1 - e

-0.5t

y(t) t = t = 3[1 - e d

] and y(¥ ) = 3

-0.5t

d ] = 3 Þ t = ln(0.5) = 1.37 d 2 -0.5

Q57. Use equivalent signal flow graph as depicted below, to get

Y(s) ( -1)(- K/ s 2 ) = R(s) 1 + 1 + 1 + 1 + 1 s s s s2 1 R(s) K = 2 s + 3s + 1 1 K For R(s) = , Y(s) = 2 s s(s + 3s + 1)

–1/s 1

–1

–K/s2

1

Y(s)

–1

1

1/s

Since, sY(s) has all poles in the left half of s plane , FVT is applicable. K y( ¥ ) = l t sY(s) = l t 2 =K s®0 s ® 0 s + 3s + 1 The system will track unit step input with zero error in steady state for K = 1. Q58.(a) Use equivalent signal flow graph as depicted below to get K1/s+1 K1 R(s)

Y(s) Kl s(s + 1) = = 2 R(s) 1 + K1K 2 + K1 s + (1 + K1K 2 )s + K1 s + 1 s(s + 1)

1

1/s K2

1

Y(s)

1

–1 For the two equal roots to be located at s = –10, characteristic equation should be (s+10)2 = 0 or 2 s +20s+100 = 0. 2 Compare it with s + (1+ K1K2)s + K1 = 0 to get K1 = 100 and K2 = (20 – 1) / K1 = 0.19. Q59.(b) G(s) =

-1

-0.5

has steady state gain, | S |=| G(0) |= 0.5 2 + 0.1s 1 + 0.05s and time constant τ = 0.05. The pole is located at s = -20. =

2d (t) ƒ 2 ; G(s) =

jω

-10 s + 20

–20

×

σ

Control Systems

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65

2 ´ -10 Y(s) = for R(s) = 2 (s + 20) y(t) = - 20e -20t ; y(t 0 ) = -7.36 Þ -7.36 = -20e -20t 0 Þ t0 = 0.05 Q60. For steady state error, e(∞) = 0 or y(∞) = 1 to unit step input, the value of K = 1 ×( 1 / 0.8) = 1.25. Q61(d). The characteristic equation is 1 -

Ks = 0 Þ s 2 + (2a - K) s + a 2 = 0 (s + a) 2

For the characteristic roots to lie on imaginary axis (a requirement for sinusoidal oscillations) K = 2a and for K = 2a, the characteristic roots will be located at s = ±ja. So, frequency of oscillations ω = a rad/sec and Hertz frequency f = a/2π Hz. Q62.(c) The Routh array is constructed as follows: s4

1

4

s3

2

K

s2

8-K 2

6

s

1

s0

6

K 2 - 8K + 24 K -8

6

8-K > 0 or K < 8 and K 2 - 8K + 24 > 0. 2 8 ± 64 - 4 ´ 24 2 = 4 ± j2 2 The roots of quadratic function K – 8K + 24 are K = 2 For polynominal P(s) to have all LHP roots

that are complex. The polynomial, thus, has r h p root/roots for all K. Q63.(a) Q(s) = 3s5 +2s3 + s = s (3s4 +2s2 +1). Routh array for polynomial 3s4 + 2s2 + 1 is constructed as s

4

3

2

s

3

0 12

04

s

2

1

1

s

1

−8

s0

1

1 premature d (3s 4 + 2s 2 + 1) = 12s 3 + 4s termination ds

Control Systems

66

Mechasoft

There are two changes in algebraic sign in the left column.RHP roots = 2 , LHP roots = 2 (by symmetry) , I A roots = 1 (at origin) k2 Q64.(d) Use coefficient test which is necessary and sufficient both for second order system stability 2 investigation. The stability boundary diagram for s +(2+0.1k1) s +k1+k2=0 , is determined as follows. 2 + 0.1 k1 > 0 or k1 > − 20 and k1 + k2 > 0. Q65.(a) The characteristic equation is 1+G(s) = 0 Þ 1 + Construct Routh array as follows: 3

s 2 s s

1

s

0

1 b

+20 k1

–20

k1 + k2 = 0

as + 1 = 0 or s3 + bs 2 + as + 1 = 0 s (s + b) 2

b

a 1

ab - 1 b

a

1

Now, stability demands b > 0 and ab−1 > 0 or ab > 1. The stability region is outward the hyperbola in the first quadrant. Q66.(b) The Routh array for characteristic polynomial (the denominator polynomial of H(s)) D(s) = s3 + αs2 + ks + 3 is constructed below: 3 s 1 k 2 s α 3 s

1

ak - 3 a

0

s 3 The constraints for system stability are α > 0 and α k >3. Note that stability demands no change in algebraic sign in the first column of Routh array. Q67.(a) Replace s by p – 2 to get a new polynomial in p as Q (p) = T (p - 2) 2 + (p - 2) + K = T p 2 + (1 - 4T ) p + (4T + K - 2)

The coefficient test for all the roots to have real part more negative than −2 , gives T > 0 and 1 - 4T > 0 or - 4T > - 1 or T < 0.25 Þ 4T + K − 2 > 0 or K > − 4T + 2

0 < T < 0.25

Control Systems

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67

Using these inequalities, the region is sketched below. K T = 0.25 2 1 K = – 4T + 2 T

0.25

Q68.(a) Stable systems have impulse response h(t) that decays to 0 as t → ∞ . Only systems S1 S2 and S4 possess this property. Q69.(d)

Y(s) k / s (s + 3) 2 k = = 3 2 2 R(s) 1 + k / s(s + 3) s + 6s + 9s + k 3

2

The Routh array for characteristic polynomial s + 6s + 9s + k is constructed as below. s3 1 9 2 s 6 k s

1

54 - k 6

s0 k 54 - k 1 For the system to oscillate, s row must have zero entries , that is , =0

6

2

or k = 54 and for k = 54, the auxiliary equation 6s + k = 0 gives s = ±j3. Thus, the system oscillates at 3 rad/sec =

3 Hz for k = 54. 2p

Q70.(d) The open loop transfer function

4(s - 1) has all poles with negative real part (s + 1) (s 2 + 2s + 2)

and therefore , open loop system is stable. The closed loop transfer function T(s) is

T(s) =

4 / (s 2 + 2s + 2) 4(s + 1) N(s) = 3 º 2 4(s - 1) s + 3s + 8s - 2 D(s) 1+ 2 (s + 1) (s + 2s + 2)

Since, the denominator polynomial D(s) does not have all coefficients of same sign, it has at least one root in right half and closed loop system is unstable. Q71.(d) I. The output of form y = t is unbounded while step input is bounded. The system is unstable. Statement is true. II. y& = u Þ s Y(s) = U(s) Þ

Y(s) 1 = . The characteristic equation s = 0 does not have the root U(s) s

Control Systems

68

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with negative real part and also does not have root with positive real part. The system is marginally stable. The statement is not true. III. The system is stable because the poles which are also characteristic roots have negative real part. The system does have a zero with positive real part but this does not affect the stability. The statement is not true. IV. The system is marginally stable. All the roots do not have negative real parts and also no root has positive real part. The statement is not true.

k e -2s =0 s (s + 2)(s + 3) -2 s Use the approximation e = 1 - 2s to get s 3 + 5s 2 + 6s + k (1 - 2s ) = 0 or s3 + 5s 2 + (6 - 2k)s + k = 0 Q72.(b) The characteristic equation is 1 +

Routh array 3 s 2 s s1

1 5

6 − 2k k

30 - 10k - k 5

s0 k Stability demands k > 0 and 30 – 11k > 0 or k < 30/11 Thus, 0 < k < 30/11 preserves system stability. Q73. Routh array has 7 rows and system has 6 roots. Note that there are three changes in algebraic sign . So, the system has 3 RHP and 3 LHP roots. Q74.(a)

Y(s) 1 2 and characteristic equation is s − (G + 2)s+(2G + 5) = 0 = 2 X(s) s - (G + 2) s + (2G + 5)

For the system to remain stable − (G+2) > 0 or G < −2 and 2G + 5 > 0 or G > − 2.5. So, the constraints on G for system stability is −2.5 < G < −2. Q75. For critical damping, the characteristic roots must be negative real and equal. This condition is satisfied if (G + 2) 2 - 4(2G + 5) = 0 or G 2 - 4G - 16 = 0 Þ G = 6.47, - 2.47 But for G = 6.47, system is unstable. So, G = − 2.47 will force the critical damping in the system. Q76.(a) The Characteristic equation is 1 + Routh array 3 s 1 2 s μ 1 s 10μ – 5λ 0 s 5λ

10 5λ

5 (2s + l) ´ =0 s(s + m) s

; s3 + ms 2 + 10s + 5l = 0

Closed loop stability demands that all the elements of left column of array be of same algebraic sign, that is, μ > 0, λ > 0 and 10μ – 5λ > 0 or μ > λ / 2 μ

R μ > λ/2 λ

Control Systems

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69

Q77.(a) Routh array for characteristic polynomial P(s) = s3 + (1+k)s2 + 10s + (5 + 15k) 3 s 1 10 For stable system, there must be no change in algebraic 2 s 1+k (5+15k) sign in the left column of array, that is,

5 - 5k 1+ k

s1 s

0

1 + k > 0 or k > -1ü ý Þ | k | 0 or k < 1 þ

10 1

For k = 1, s row has zero elements (a case of premature termination). Auxiliary polynomial,

A(s) = éë(1 + k)s 2 +(5 + 15k) ùû

= 2s 2 + 20 = 0 Þ s = ± j 10 10 The system oscillates with frequency f = Hz = 0.5Hz 2p k =1

1

For k = 2, s row has element equal to –5 / 3 resulting in two changes in algebraic sign in the left column and system will be unstable due to two rhp roots. Q78.(a) With switch (S) open,

Y(s) 1/ s 2 1 = = 2 . The characteristic equation s2+1 = 0 has 2 R(s) 1 + 1/ s s +1

roots s = ± j1 located on imaginary axis and the system oscillates at 1 rad/sec.

Y(s) 1 / s2 1 = = 2 . 1 Ks R (s) 1 + + s + 1 + Ks s2 s2 1 Ks , The G(s) H(s) product for which the root locus is to be sketched with

Q79.(a) With switch (S) closed,T(s) =

Since , T(s) = 1+

s2 + 1

Ks k as variable parameter, is G(s) H(s) = 2 . s +1

s2 + 1 dK To evaluate break in point, K = and = 0 gives s ds 2 2s ´ s - (s + 1) = 0 or s 2 - 1 = 0 or s = ± 1 but s = + 1 2 s

Root locus

Im ´ +j1 90

–1 K=∞

does not satisfy real axis segment rule of root locus. The angle of departure at pole o o o o s = + j1, is ϕd = 180 − (90 ) +90 = 180 .

Re K=∞ 90 ´ –j1

Q80.(d) The correct root locus is identified using the information as follows: Real axis segment between s = −2 and s = −∞ lies on root locus. Centroid s A =

o

-2 - 2 - 2 = -2 3

and asymptotic angle, θA = ±60o, 180o . Imaginary axis crossing point is obtained from

o

Control Systems

70

characteristic equation 1 + polynomial P(s)

Mechasoft

K = 0 or s3 + 6s2 +12s + 8 + K = 0. The characteristic 3 (s + 2)

s = jw

= ( jw )3 + 6( jw ) 2 + 12( jw ) + 8 + K = (8 + K - 6w 2 ) + jw (12 - w 2 )

Im[P ( jw)] = w(12 - w2 ) gives w = 2 3 and Re[P ( jw)] = 8 + K - 6w2 = 0 gives K = 6w2 - 8 = 6(2 3) 2 - 8 = 64. Thus, the root locus crosses imaginary axis at s = ±j2 3 with corresponding K = 64. The break away point is determined as 1 +

K =0 (s + 2)3

Þ K = -(s + 2)3 = - (s3 + 6s 2 + 12s + 8) Im

dK = 0 gives 3s 2 + 12s + 12 = 0 or s = -2, - 2. ds Q81.(b) The value of K is determined at point of intersection of root locus and a line making an −1 o angle of cos (0.5) = 60 from negative real axis as shown below. ∆ OAB is an equilateral triangle. OA = OB = BA = 2 and K= product of phasor lengths from poles = (AB)3 = 23 = 8.

+ j2 3 A 60o B ´ ´ –2 ´

0

Re

- j2 3 Q82.(a) The characteristic equation is 1 + G(s) H(s) = 0

k = 0 Þ s 3 + s 2 + k + ps(s + 1) = 0 s(s + 1)(s + P) ps (s + 1) or 1 + 3 2 =0 s +s +k So, the equivalent GH product for sketching root locus with variable P and given k, is p s (s + 1) GH = 3 2 which has three poles and two zeros suggesting that only one branch of s +s +k root locus will terminate at ∞ along real axis. k(s + 4 / 3) Q83.(d) The root locus for G(s) = 2 is sketched below using the relevant information s (s + 12) as follows. or

(I) (ii) (iii) (iv)

1+

Number of separate branches = 3 Real axis segment : between s = −12 and s = − 4/3 Starting points are s = 0, 0, −12. Terminating points are s = −4/3, ∞ , ∞ .

Control Systems

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(v) Centroid, s A =

71

0 + 0 - 12 - (-4 / 3) 16 =- . 2 3 o

(vi) Asymptotic angles, θA = ± 90 . (vii) Break away /Break in points: The characteristic equation

æ 4ö k çs + ÷ s 2 (s + 12) s3 + 12s 2 3ø è Þ k = = 1+ 2 =0 (s + 4 / 3) s + 4 / 3 s (s + 12 )

and -

dk = ds

æ 4ö (3s 2 + 24s) ç s + ÷ - (s3 + 12s 2 ) è 3ø æ 4ö çè s + ÷ø 3

2

Im

k =¥

k = 48 - 16 / 3 k=0 × - 12

=0

k =¥ k=0 × Re - 4/3 × -4

k =¥

gives s(s 2 + 8s + 16) = 0 or s (s + 4) 2 = 0 Þ s = 0, - 4, - 4

(viii) k

s =-4

=

Product of phasor lengths from poles to the point s = -4 product of phasor lengths fom zeros to point s - 4

=

8´ 4´ 4 = 48 (8 / 3)

So the roots are located as s = − 4,– 4,– 4 for k = 48. Q84.(d) The following information in respect of root locus of the transfer function

G(s) =

k(s 2 + 16) suggest that the locus drawn is correct. s2 + 4

(i) Starting points ; s = ± j 2 (ii) Terminating points : s = ± j 4 (iii) Number of root locus branches = 2 (iv) Angle of departure from pole at s = j 2, o ϕd = 180 − ∑ angles of other poles of G(s) + ∑ angles of zeros of G(s) o o o o o o = 180 − (90 ) + (90 + 270 ) = +450 = + 90 and angle of arrival at zero at s = j4, o ϕa = 180 +∑ angles of poles of G(s) − ∑ angles of other zeros of G(s) o = 180 + (90o + 90o) − (90o) = + 270o (v) The characteristic equation 1 + 1/2

æ 1 + 4k ö ÷ . è 1+ k ø

k(s 2 + 16) = 0 or (1 + k) s 2 + 4(1 + 4k) = 0 2 (s + 4)

gives s = ± j2 ç

The roots are always complex conjugate with zero real part.

72

Control Systems

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So, the system is always marginally stable or marginally unstable. This is always oscillatory, varying k only varies frequency of oscillation from 2 rad/sec to 4rad/sec . System is not unstable because it neither has any root with positive real part nor any repeated IA roots . Thus, statement II is correct while statement III is not correct.

Im ϕa = + 270º

k=∞ k=0

×

ϕd = +90º

Re k=0

×

k=∞

Q85.( c)

Y(s) k(s + 2) / (s + 1)(s + 3) -k(s + 2) = (-1) = R(s) 1 + k(s + 2) / (s + 1)(s + 3) (s + 1) (s + 3) + k (s + 2)

Now, the characteristic equation whose roots are to be sketched in the range 0 ≤ k ≤ ∞ , is P(s) = (s + 1) (s + 3) + k (s + 2) = 0 Im or 1 +

k (s + 2) = 0 and the equivalent GH product is (s + 1)(s + 3)

GH =

k (s + 2)

k=0 –2 k=0 × × –3 k=∞ –1

k=∞

. The corresponding root locus

(s + 1)(s + 3) that includes only real axis segment, is sketched.

Q86. k

s =-5

= P (phasor lengths from poles )

Im

= 1´ 20 ´ 20 = 20

k = 58 ´

+ j2

20

k = 20 -5

+ j 13

´ -4

Re

-1 20 ´

- j2 - j 13

k = 58

Re

Q87.(d) The GH product =

73

Control Systems

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k (s + 4) (s + 1 + j2) (s + 1 - j2) 3

=

k (s + 4) (s 2 + 2s + 5)

2

and the characteristic equation P(s) = 0 is s + 6s + 13s + 20 + k = 0 p(jω) = (20 + k − 6ω2) + j(13ω − ω3) 3 Im[P(jω)] = 0 gives 13ω − ω = 0 or ω = 0 , 13 2 Re[P(jω)] = 0 gives 20 + k − 6ω = 0 or k = 58 The imaginary axis crossing point at s = ± j 13 for k = 58 is depicted. Thus, for k = 20, the system will have complex conjugate (dominant) roots and exhibit underdamped dynamics. Increasing k, system will exhibit sustained oscillation for k = 58 and will be unstable for k > 58. Q88.(a) Write the characteristic equation (s+1)(s+2)(s+3) + k(s+4) = 0 in the form as follows:

k(s + 4) = 0 º 1 + GH = 0 (s + 1)(s + 2)(s + 3) k(s + 4) The equivalent GH product is G H = (s + 1)(s + 2)(s + 3) å poles - å zeros whose root locus has centroid s A = P-Z 1+

=

-1 - 2 - 3 - (-4) = -1 3 -1

Note that centroid is always real.

Y(s) [s(s + k)]-1 1 1 = = = 2 -1 U(s) 1 + [s(s + k)] s (s + k) + 1 s + 1 + ks ks 2 where characteristic equation s +1+ks = 0 can be written as 1 + 2 = 0. s +1 ks The equivalent GH product isGH = whose root locus is sketched as (s + j1) (s - j1) demonstrated below. Q89.(b)

(i) Root locus is symmetrical about real axis. (ii) Number of root locus branches = 2 (iii) Starting points : s = ± j 1 (iv) Terminating points: s = 0 and − ∞. (v) Real axis segments: between s = 0 and s = − ∞. (vi) Break away/Break in points : The characteristic equation

s2 + 1 ks 1+ 2 = 0 gives - k = s s +1 dk 2s ´ s - (s 2 + 1) and = =0 ds s2 2

gives s −1= 0 or s = ±1. But s = +1 does not satisfy real axis segment rule. (vii) The angle of departure at s + j1,

Im

´ j1

fd =180º

90º

k=¥ -1

Re Break in point

´ - j1

ϕd = 180º − ∑ angles of poles of GH+∑ angles of zeros of GH = 180º − 90º+90º=180º.

90º

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Q90.(a) The root locus of control system corresponds to a second order system with two poles. For the system to exhibit underdamped dynamics, the roots must be complex conjugate with negative real part. The value of k at break away point is, let us say, k1: k1 = ∏ ( phasor lengths

a a a2 ´ = . The break away point is determinable from chracterteristic equation from poles) = 2 2 4 k dk 1+ = 0 Þ -k = s 2 + as and = 0 gives 2s + a = 0 or s = - a / 2. s (s + a) ds For k


Q91.(a) G(s) H(s) =

k(s 2 + 6s + 8) k(s + 2)(s + 4) = 2 s + 6s + 18 (s + 3 + j3) (s + 3 - j3)

locus is demonstrated below. (I) Root locus is symmetrical about real axis. (ii) Starting points : s = – 3 ± j 3 (iii) Terminating points: s = −2 and s = −4. (vi) Break away/in points : The characteristic equation

and identification of correct root 270º Im j3 ´

108 .4º

71 .6º k(s 2 + 6s + 8) s 2 + 6s + 18 = 0 gives k = 2 2 s + 6s + 18 s + 6s + 8 –4 –3 2 dk (2s + 6)(s + 6s + 8) - (2s + 6)(s 2 + 6s + 18) and = 0 gives =0 ds (s 2 + 6s + 8) 2 or s = -3

1+

´

–2

90º

(v) Break in angle = ±π /2 ≡ ± 90º (vi) Angle of departure at pole (s = − 3 + j3) : ϕd = 180º − ∑ angles of other poles + ∑angle of zeros = 180º − (90º) + (71.6º +108.4º) = 270º

Re

– j3

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Q92.(a) k (at break in point) =

75

P (phasor lengths from poles) 3´3 = =9 P (phasor lengths from zeros) 1´1

Y(s) P1D 1 Ps -2 P = = = 2 Q93.(d) Use Mason's gain rule to get -2 -2 U(s) D 1 - ( - 4s - Ps ) s + 4 + P The characteristic equation s2 + 4 + P = 0 can be written as1 + product for sketching root locus as GH =

P = 0 to get equivalent GH (s + 4) 2

P P = s + 4 (s + j2)(s - j2) 2

The root locus sketching is demonstrated as follows: (i) Root locus is symmetrical about real axis. (ii) Number of root locus branches = 2 (iii) Starting points : s = ±j2 (iv) Terminating points: both branches terminate at ∞. (v) Asymptotic angles, θA = ±90º (vi) Centroid, s A =

åP -åZ P-Z

P = ¥ Im j2 ´

Re –j2 ´ P = 0

(vii) Angle of departure at pole s = j2 : ϕd = 180º − ∑angle of other poles =180º − 90º = 90º Q94.(b) The product G(s) H(s) =

P=0 90º

0 =0 2

=

fd = 90o

P= ¥

k(s 2 + 4) and correctness of root locus is investigated as follow. s (s 2 + 1) Im

(i) Root locus is symmetrical about real axis k = ¥, j2 (ii) Number of root locus branches = 3 (iii) Starting points : s = 0 and s = ±j1 k = 0 , j1 ´ (iv) Terminating points : Two branches terminate at finite zeros s = ± j2 and one branch terminates at ∞ along real axis. k=¥ ´ (v) Angle of departure from pole at s = +j1 : k= 0 ϕd = 180º − ∑angles of other poles + ∑angles of zeros = 180º − (90º+90º) +(90º+270º) =360o = 0º k = 0 , - j1 ´ (vi) Angle of arrival at zero at s = +j2: ϕa = 180º + ∑angles of poles − ∑angles of other zeros k = ¥ , - j2 = 180º + (90º + 90º + 90º) −90º = 360o = 0º

fa = 0o

f = 0o d

90º Re 90º

90º

The root locus shown in Fig. II is correct and system is always unstable for k > 0 due to 2 R H P roots. Q95.( c) G (s ) =

ke -s s (s + 1) (s + 2)

@

k(1 - s) s (s + 1) (s + 2)

=

- k(s - 1) s (s + 1) (s + 2)

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The root locus construction is demonstrated below using the rules pertaining to negative parameter variation. (i) The root locus is symmetrical about real axis. (ii) Number of separate root locus branches = 3 (iii) Starting points : s = 0, −1 and −2 (iv) End points : One branch terminates at zero located at s = 1 and remaining two branches terminate at ∞. (v) Root locus segments on real axis: between s = − ∞ and s = −2, s = −1 and s = 0, s = 1 and s = ∞ ±i 360o (vi) Asymptotic angles : qA = = ± j180o

3 -1

(vii) Break away/Break in points: The characteristic equation 1 +

gives k =

- k (s - 1) =0 s (s + 1) (s + 2)

s (s + 1) (s + 2) dk and = 0 gives s3 - 3s + 1 = 0 (s - 1) ds

or s = −0.35, +1.88 and −1.53. The break away and break in points are located at s =−0.35 and s = 1.88. The real axis segments of root locus does not include s = – 1.53 (viii) Imaginary axis intersection points: The characteristic polynomial is Im k=1.5 P(s) = s (s + 1) (s + 2)−k( s−1) = s3 + 3s2 + (2−k)s + k 2 3 j0.7 1.88 P(jω) = (k – 3ω ) + j[ω(2 – k) – ω ] 2 k=0 k=0 Re[P(jω)] = 0 gives k – 3ω = 0 k=0 k=∞ ´ ´ k=∞ k=∞Re ´ –1 and Im[P(jω)] = 0 gives ω (2−k) −ω3 = 0 or 2−k = ω2 –2 Solve these two equations to get k = 1.5 and ω = 1/ 2 @ 0.7 –j0.7 –0.35 So, the root locus intersects the imaginary axis at s ± j 0.7. Using the informations listed above, the complete root locus is sketched . Q96. The root locus for G(s) =

-9 - ( -1) = -4 3 -1 Asymptotic angles, θA = ±90º

k(s + 1) is s 2 (s + 9)

sketched below.

controid, s A =

Break away / in points : 1 +

Im

k =¥

k(s + 1) =0 s 2 (s + 9) B

2

k=

-1 A

C –4 × –9,k=0

s (s + 9) dk and = 0 Þ s = 0, -3, -3 s+ 1 ds

–3,k=27

All three real and equal roots are located 2

at s = - 3 Þ b = 3 and k1 =

k1 27 = =9 b 3

OB ´ OB ´ BC 3 ´ 6 = = 27 AB 2

k =¥

× ×

k=0 0

Re

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2

Q97. Characteristic equation, 1 + G(s) = 0 or 1 +

4(s + 1) = 0 or s 2 + as + 4s 2 + 4 = 0 s(s + a)

as a *s = 1+ 2 ; a* = a / 5 The characteristic polynomial, 1 + 2 5s + 4 s + 0.8 a *s The root locus for loop function, GH = 2 is sketched below. 2 s + 0.8 a *s s + 0.8 * 1+ 2 = 0 Þ a* = a =∞ A s + 0.8 s d * 0.8 (a ) = 0 Þ s ± 0.8 ds

but s = +0.89 does not lie on real axis root locus segment. The break in point is s = – 0.89 at which both the roots are real and equal. Corresponding a * =

Im B´ j 0.8, a * = 0 a*=∞ Re ´ - j 0.8, a * = 0 B

AB2 1.6 = = 1.79 and a = 5 ´1.79 = 8.95 OA 0.8 jω

Q98.(a) The characteristic equation : |sI – A| = 0

és 0 ù é 0 sI - A = ê ú-ê ë 0 s û ë3 - k

1 ù -1 ù é s = ê ú ú - 2 - kû ëk - 3 s + 2 + k û

and |sI-A| = s2 + 2s + ks + k – 3 = s2 + 2s – 3 + k(s + 1)

k=∞ k=0 k=∞ ´ –3 –1 k(s + 1)

k(s + 1) | sI- A |= 0 Þ 1 + = 0 Þ GH product = (s - 1)(s + 3) (s - 1)(s + 3) The root locus for GH =

k=0 ´ 1

σ

k=3

k(s + 1) is sketched below. (s - 1)(s + 3)

jω

The locus always lies on real axis and system is stable for k > 3.

k(s 2 + 3s + 2) =0 Q99.(d) s + ks + 3ks + 2k = 0 Þ 1 + s3 k(s + 1)(s + 2) k=∞ k=∞ k=∞ GH product = s3 –2 –1 s3 Break away / in point : - k = 2 –4.73, s + 3s + 2 k=10.4 dk 2 2 3 2 2 = 0 Þ 3s (s + 3s + 2) - s (2s + 3) = 0 or s (s + 6s + 6) = 0 ds 3

2

s = 0,0, - 3 ± 3 or 0, 0, - 1.27, 4.73 but s = – 1.27 does not lie on real axis segment of root locus.

+ j 2 ,k=2/3 × × ×

k=0 σ

-j 2

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2

3

2

jω axis crossing point : P(s) = s + ks +3ks +2k and P(jω) = – jω – kω + j3kω + 2k 2

2

Re[P(jω)] = (2k – kω ) = 0 Þ ω = 2 or ω =

2 and f =

2 1 = Hz 2p 2p

Im[P(jω)] =(3kω – ω ) = 0 Þ3k – ω = 0 Þ k= ω / 3 = 2 / 3. For k > 2 / 3, the roots lies in left half of s plane and system is stable. 3

k s= – 4.73 =

2

3

4.73 ´ 4.73 ´ 4.73 = 10.4 and for k > 10.4, all the three roots will be real. 3.73 ´ 2.73

Q100.(a) The characteristic equation : 1 +

10K =0 (s + K)(s + 2)(s + 5)

K(s + 7 s + 20) =0 s(s + 2)(s + 5) 2 K(s + 7 s + 20) K=∞ GH product = ;0£K £¥ s(s + 2)(s + 5) for which typical root locus is sketched.

jω

K=∞

2

s3 + 7s2 + 10s + K[s2 + 7s +20] = 0 or 1 +

K=0 × –5

K=0 × –2

K=0 ×

σ

K=∞

jw

101.(d) The Bode plots(both magnitude and phase) for G(jω) H(jω) =

jw ö æ ç1 + ÷ è 10 ø

3

are sketched and

gain margin, phase margin determination is demonstrated below. To find gain crossover frequency ωgc, let us find equation of line segment AB. (x1,y1)≡(log10, 20) dB y - y1 A y - y1 = 2 (x - x 1 ) ; y º dB and x º log10 w +20 x 2 - x1 +20dB/dec wgc=31.62rad/sec -20 - 20

dB - 20 =

(log 100 - log10)

(log w - log10)

100 0

1

Noting that dB = 0 at ω = ωgc 0 - 20 =

- 40 (2 - 1)

(log w gc - 1) Þ w gc = 31.62 rad / sec

ÐG( jw ) H( jw ) w=w = +90o - 3 tan -1 gc

w gc

10 = 90o - 3 tan -1 3.162 = -127.3o @ - 127o Phase cross over frequency ωpc = ∞ and gain margin GM = ∞.

PM = 180o + ÐG( jw ) H( jw ) = 180o - 127 o = 53o

w=w gc

10

–20 (x2,y2)≡(log100, –20)

w (log scale) B – 40dB/dec

|G(jω) H(jω)| 90º 0º

wpc= ∞ GM = ∞

–90º –127º –180º

PM=53º

w(log scale)

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3 2 w + j 0.5 w 2 2 3 2 x º Re éë F1 ( jw)ùû = w and y º Im éë F1 ( jw)ùû = 0.5w2 2 y 0.5w2 x represents a straight line passing through origin and making an angle = Þy= 2 x 0.5 3 w 3 of tan –1(1/ 3) = 30º with real axis. Polar plot is shown in (3). 1 2 1 2 F2 (jω) = w 2 (cos 45o + jsin 45o ) = w +j w 2 2 2 w w2 Þ y = x where x º Re[F2 ( jw)] = and y º Im[F2 ( jw)] = 2 2 Q102.(d) F1(jω) = w 2Ð30o = w 2 (cos30o + jsin 30o ) =

represents a straight line passing through origin and making an angle of 45º with real axis. Polar plot is shown in (2).

1 1 gives lt F3 ( jw) @ lt = ¥Ð0o w® 0 w® 0 ( jw) 4 ( jw ) ( jw + 0.2) 1 lt F3 ( jw) @ lt = 0Ð - 450o = 0Ð - 90o. 4 w®¥ w®¥ ( jw ) ´ jw

F3 (jω) = and

4

Also, phase angle remains negative and monotonically increases negatively form 0º at ω = 0 to −90º at ω = ∞. The polar plot is shown in (4). 2 2 2 F4(jω) = 0.6 ω (j + 1) + 0.5 = 0.5 + 0.6 ω + j 0.6 ω 2 where Re[F4(jω)] = x = 0.5 + 0.6 ω and Im[F4(jω)] = y = 0.6 ω2 gives x = 0.5 + y or y = x−0.5. The polar plot is a straight line passing through (x = 0.5, y = 0) and making an angle of 45º with real axis (x axis) as shown in (1). I

1 1 Q103.(c) GM = = =4 G( jw ) w=w 0.25

1/4

pc

o

PM = 180 + ÐG( jw ) = 90o º

p rad 2

o

w=w gc

= 180 - 90

o

–1 ω = ωpc ω = ωgc

1 1 + 3 s (1 + sp1 )(1 + sp 2 ) p1p 2s + (p1 + p 2 )s 2 + s 1 G( jw) = 3 jw - jp1p 2 w - w2 (p1 + p 2 ) 3 Im[G(jω)] = 0) for ω −p1p2ω = 0 or ω = 0 , 1/ p1p 2 that is, 1 phase cross over frequency wpc = rad / sec. p1p 2 Q104.(a) G(s) =

Im

w= ¥ +1 o 90 90

w=0

Re

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GM =

1 Re éëG ( jw)ùû

w=wpc

Mechasoft

p +p = w2pc (p1 + p 2 ) = 1 2 p1p 2

Q105.(b) From given data, peak time tp = 0.2 sec and peak over shoot Mp = 1.12−1 = 0.12. 2 p Now, M p = e -px / 1-x = 0.12 gives damping ratio x = 0.56 and t p = = 0.2 wn 1 - x 2 p gives wn = = 18.96 rad / sec. 0.2 1 - 0.562

Since, ξ = 0.56 < 1/ 2 , resonance will occur in frequency response. 1 1 Peak resonance M r = = = 1.08 2x 1 - x 2 2 ´ 0.56 1 - 0.56 2 and resonant frequency w r = w n 1 - 2x 2 = 18.96 1 - 2 ´ 0.562 = 11.58 rad / sec. Q106.(c) ÐH( jw )

w= 2

=Ð

jw + 2 æ 2 ö = tan -1 (2 / 2) - tan -1 ç = 45o - 90o 2 2÷ è ø 4 - w + jw w=2 4-2

= - 45o º - 0.25p rad. 12

= 20log

1 G( jw )

= 20log (1) - G( jw pc ) w=w pc

= 0 - (-12) = 12dB and phase margin, PM = 180o + ÐG (jw ) w=w

6 0 –6

gc

ωgc

dB

= 180o + ( -150o ) –12

GM=12dB

ωpc PM =30º

dB Magnitude

Q107.(a) Gain margin GM(in dB)

PM

= 30o where ωpc and ωgc are phase crossover and gain crossover frequencies respectively as demonstrated below. Both GM and PM are positive and therefore system is stable.

Q108.(d) Statement I is false. For example, G(s) =

–18

30

–240º –210º –180º –150º –120º –90º Phase Angle

s -1 is stable, pole is located in the LHP of s+2

1 s+2 is unstable due to pole being located at s = 1 in the RHP of s -plane. = G(s) s - 1 Statement II is also false. Gain margin GM = 20 log 1/1.2 = –1.58dB Phase margin PM = 180º + phase corresponding to unity magnitude = 180º − 190º = − 10º. The GM and PM both are negative and therefore , unity feedback control system is unstable. s-plane. But F(s) =

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Q109.(a) The initial segment of dB plot has slope of −12dB/oct, indicating presence of two poles at origin. The corner frequency ω1 which corresponds to a zero, is one decade below ω = 1 rad/sec. So, ω1 = 0.1 rad/sec.The corner frequency ω2 which corresponds to a pole, is one decade above ω = 1 rad/sec . So, ω2 = 10 rad/sec. Noting the point (dB = 20, ω = 0.1) on initial segment, the value of constant k is evaluated as

20 log

k w2

= 20 = 20 log10 Þ k = 10 ´ 0.1´ 0.1 = 0.1 s ö æ 0.1ç1 + ÷ 10 (s + 0.1) è 0.1 ø Now, the system transfer function G(s) = = s ö s 2 (s + 10) æ s 2 ç1 + ÷ è 10 ø w= 0.1

Q110.(a) The Nyquist plot corresponds to a system of type zero and order two.

let G(s) =

k k k = 2 and G ( jw) = 2 (s + a) (s + b) s + (a + b) s + ab (ab - w ) + j(a + b) w

G( jw) w=0 =

k = 2 Þ k = 2 ab ab

Re[G(jw )] = 0 for w2 = ab Þ ab = 4 ; w = 2 is read from plot. ab = 4 gives k = 2ab = 8

-k = -2 Þ a + b = 2 2(a + b) Solve ab = 4 and a + b = 2 to get a = 1 ± j 3 and b = 1 m j 3 Im[G( jw)]

w= 2

So, G(s) =

= -2

Þ

8 8 8 = = 2 2 (s + 1 + j 3)(s + 1 - j 3) (s + 1) + 3 s + 2s + 4

Q111.(a) G2(jω) represents an all pass system having transfer function of from G 2 ( jw) = Note that G 2 ( jw ) =

1 - jw . 1 + jw

1 + w2 −1 −1 −1 = 1 ,that is , 0 dB, but G2(jω)=tan (−ω) −tan ω= −2 tan ω 1 + w2

The phase varies from 0º at ω = 0 to − 180º at ω = ∞.

Since G 3 ( jw) dB = G1 ( jw) dB + G 2 ( jw) dB and ÐG 3 ( jw) = ÐG1 ( jw) + ÐG 2 ( jw) G1(jω) is minimum phase, G2(jω) is an all pass and G3(jω) which is obtained by cascading G1(jω) and G2(jω), is non minimum phase by virtue of zero of all pass system in RHP. Note that an all pass system is also non minimum phase. On cascading a system with an all pass system, the dB magnitude response of resulting system remains unchanged while phase response of resulting system is sum of phase responses of both. Q112.(a) A( w)

w= 5

= 4 and j ( w)

w= 5

= - 3 ´ 5 radian =

- 15 ´ 180 p

Forced response = 10 × 4 cos(5t−30º−140º) = 40 cos (5t−170º).

@ - 860 o º - 140 o

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cos 4w - jsin 4w jw jw 1 p A(w) = G( jw) = and j(w) = ÐG( jw) = - - 4w w 2 1 p p 1/2 1/2 ; G( jw) = w and ÐG( jw) = ´ = Q114.( c) G( jw) = ( jw) 2 2 4 Q113.(b) G( jw) =

e

Mechasoft

- j4 w

=

Q115.(a) Each pole whether at origin or in the LHP, contributes slope of −6dB/octave and each zero whether at origin or in the LHP, contributes slope of +6dB/octave. So, the slope of highest frequency asymptote = −6 × 6 + 2 × 6 − 6 × 1 = −30dB/octave. Q116.(d) For the closed loop system to be stable −1+j0 point must lie to the left of traversal of polar plot from ω = 0 to ω = ∞ , that is , it should lie between points a and b or to the left of point c. −k < −1 < − 0.25 k or 0.25k < 1 < k or 1 < k < 4 and −5k > −1 or k < 1/5 Q117.(d) From initial flat segment , 20 log k = 40 gives k = 100. The pole corner frequencies are log10ω = 1 or ω = 10 and log10ω =2 or ω = 100.

|G(jω)|dB –6dB/oct ≡ –20dB/dec 40

100 105 G(s) = = s öæ s ö (s + 10)(s + 100) æ ç1 + ÷ ç1 + ÷ è 10 ø è 100 ø

20

–40 dB/dec One decade log10ω

log10ω=2 log10ω=1 ω=100 ω=10 Q118.(d) Gain(dB) where phase equals –180º is −20 dB. This gives gain margin = 20dB. The phase(degrees) where dB gain equals zero, is −120º. This gives phase margin = 180º −120º = 60º. Q119.(a) Initial low frequency asymptote has slope of −20dB/dec indicating presence of one pole at origin. Mid frequency asymptote has slope of +20 dB/dec indicating presence of two zeros(slope changes from −20dB/dec to +20dB/dec). At frequency 10kHz slope again changes from +20 dB/dec to −20dB/dec indicating presence of two more poles. So, the transfer function consists of all together three poles and two zeros. Q120.(a) The loop transmittance G(s) H(s) =

0.1

and G( jw) H( jw) =

jw ö æ jw ç 1 + ÷ è 10 ø Note that gain margin GM =

2

10 0.1 = 2 2 s(s + 10) s ö æ s ç1 + ÷ è 10 ø

1 ; w pc is phase cross over frequency. G( jw) H( jw) w=w pc

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= -180o

ÐG( jw ) H( jw ) w=w pc w pc - 90o - 2 ´ tan -1 = -180o Þ w pc = 10 ´ tan 45o = 10. 10 0.1 1 G( jw ) H( jw ) w = w =10 = = and GM = 200 pc é 100 ù 200 10 ´ ê1 + ë 100 úû

|G(jω)H(jω)|dB , phase ω=0.1 ω=1

Note that G( jw) H( jw) =1 w=w gc

where ωgc is gain cross over frequency. It appears easy to determine ωgc from dB magnitude plot as demonstrated below. The initial segment of dB magnitude plot has slope of −20 dB/dec and passes through the point 20 log 0.1= −20 dB at ω = 1. At ω = 10, slope changes from −20 dB/dec to − 60dB/dec due to double pole corner frequency.

0o –20o –40o

ω=10 –20dB/dec

ωgc=0.1

ω=100 corner frequency for double poles

gc

–60o –80o

ω (log scale) – 60dB/dec

–100o

0.1 @ - 90o 10 and phase margin, PM = 180o - ÐG( jw ) H( jw ) w=w =0.1 = 180o - 90o = 90 o gc ÐG( jw) H( jw)

w=wgc = 0.1

= -90o - 2 ´ tan -1

Q121.(a) For a LHP pole, the approximate phase plot is 0º up to one tenth of corner frequency, falls 45º per decade through − 45º at corner frequency and continues at −45º per decade slope up to ten times the corner frequency. Beyond ten times the corner frequency, the approximate angle is −90º. Note two poles at corner frequency ω = 10. For a LHP zero, approximate phase plot is 0º up to one tenth of corner frequency, rises 45º /dec through +45º at corner frequency and continues at +45º /dec slope up to ten times the corner frequency. Beyond ten times the corner frequency, the approximate angle is +90º . Note a zero at corner frequency ω = 1000 rad/sec.

é

1 ù ú ëê G( j w) ûú

Q122.(a) Gain margin, GM = 20 log ê

; ωpc is phase cross-over frequency and w = w pc

æ 1 ö = -180o. So, GM = 20log ç = 12.04 dB è 0.25 ÷ø Phase margin, PM = 180o + ÐG( jw) w=wgc ; wgc is gain cross over frequency and ÐG( jw )

w=w pc

G( jw) w=w = 1.So, PM = 180º − 138º = 42º =( π/180) × 42 rad = 7π/30 rad. gc

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Q123.(a) For gain margin to be 34 dB, K is to be decreased from initial K = 1 by 6 dB (=34-28) or divided by antilog(6/20) @ 2. So, K1= 1/2. Similarly, for gain margin to be 16 dB, K is to be increased from initial K = 1 by (28−16) =12dB or multiplied by antilog (12/20)@ 4.So, K2 = 1×4 = 4

and

K2 4 = = 8. K1 (1 / 2)

o Q124.(d) Note that ÐG( jw ) w=w = -180 , w pc is phase crossover frequency. pc w pc w pc - 90 o - tan -1 - tan -1 w pc = - 180 o gives tan -1 + tan -1 w pc = tan 90 o 4 4 (w pc / 4) ´ w pc or = tan 90 o = ¥ Þ w 2pc = 4 Þ w pc = 2 rad / sec [1 - (w pc / 4) ´ w pc ]

and phase crossover frequency (in Hz) =

2 1 = Hz. 2p p

Q125.(d) The initial segment of dB plot has slope of +12dB/oct ( @ +40dB/dec ) indicating presence of two zeros at origin. The corner frequency corresponding to dB =100 will be one decade above frequency ω = 1 as demonstrated below. Value of k is determinable from initial segment as 20 log kw2 w=1 = 60 Þ k = 103. dB 100 Since, the slope at corner frequency ω = 10 w=10 rad/sec, changes by −30dB/oct (−18−12 =−30), +40dB/dec presence of five poles is indicated. dB=60 103 s 2 108 s 2 G(s) = = Initial 5 w=1 (s + 10)5 s ö æ Segment

ç1 + ÷ è 10 ø

Q126.(a) Initial segment has slope of −6dB/oct ( º − 20dB/dec) indicating presence of one pole at origin. Then value of k is determinable as 20 log

k w

w=10

= 0 = 20 log1 Þ k = 10

Since, ω = 3.6 corresponds to a pole corner frequency,

G(s) =

10

, velocity error constant s ö æ s ç1 + ÷ è 3.6 ø 1 K v = lt s G(s) = 10 and e (¥) = = 0.1; This is type 1 system. s® 0 Kv

Characteristic equation: 1 + G(s) = 0 gives

s2 + s + 10 = 0 or s 2 + 3.6s + 36 = 0 3.6

undamped natural frequency, ωn = 36 =6 rad sec and damping ratio x =

3.6 = 0.3 2´ 6

Control Systems

Mechasoft

Q127.(c)

lt

w® 0

lt

w®¥

é ù ( jw + 8) é G( jw) H( jw) ù = lt = -2 = 2Ð - 180o ê ú êë ú w® 0 k û ë ( jw + 2)( jw - 2) û é ù ( jw + 8) é G( jw ) H( jw ) ù = lt = 0 Ð - 90o ê ú êë ú w®¥ k û ë ( jw + 2)( jw - 2) û

Q128.(a) For 2K > 1, number of encirclements of (−1+j0) point, N = −1 and number of open loop poles in the RHP, P = 1. Applying principle of argument, N = Z − P where Z is number of closed loop poles (characteristic roots), Z = N +P = − 1+1 = 0 and therefore , closed loop system is stable for K > 1/2. Q129.(a)

G( jw) 2 + jw G(j w) = ; = k 2 - jw k

G( jw) = 1Ð 0 o and w® 0 k lt

Also

85

N = –1

ω=0 –2K

Im G(jω)H(jω) plane Re ω=∞

–1

4 + w2 = 1 independent of w. 4 + w2

G( jw) = -1 w®¥ k lt

G( jw) (2 + jw) 2 4 - w2 4w = = +j 2 k (2 - jw)(2 + jw) 4 + w 4 + w2

é G( jw) ù Re ê = 0 for 4 - w2 = 0 or w = 2 (consider only positive frequency) ú ë k û 4´ 2 é G( jw) ù and Im ê for w = 2 = =1 ú 4+4 ë k û Q130.(b) For k > 1, numbers of encirclements of point

- k(s + 2) – 1 + j0 is N = –1 and G(s) = has one pole s-2

N= –1 –k

in the RHP, that is, P = 1. The number of closed loop poles, Z= N + P = – 1 + 1 = 0 and closed loop system is stable. Q131.(b) G(j w ) =

1 (1 - j w )(2 - j w ) = j w (j w + 1)(j w + 2) j w (1 + w 2 )(4 + w 2 )

-3 w2 - 2 + j º x + jy (1 + w 2 )(4 + w 2 ) w (1 + w 2 )(4 + w 2 ) 3 3 G(j w ) w=0 = - + j¥ Þ x = – 4 4 =

Im

–1

+k

Re

Control Systems

86

Q132.( c) G(s) =

Mechasoft

1

has three poles and G(0) = 1 (s + 1)3 1 1 (1 - 3w 2 ) - jw (3 - w 2 ) G( jw ) = = = ( jw + 1)3 (1 - 3w 2 ) + jw (3 - w 2 ) (1 - 3w 2 ) 2 + w 2 (3 - w 2 ) 2 1 - 3w 2 -w (3 - w 2 ) = +j (1 - 3w 2 ) 2 + w 2 (3 - w 2 ) (1 - 3w 2 ) 2 + w 2 (3 - w 2 ) 2

Re éëG ( jw)ùû = 0 for1 - 3w2 = 0 or w = 1/ 3 and Im[G( jw )]

w=1/ 3

=

- (1 / 3)(3 - 1 / 3) = -3 3 / 8 2 2 1ö 1æ 1ö æ çè1 - 3 ´ ÷ø + çè 3 - ÷ø 3 3 3

1- 3 ´ 3 1 Im[G( jw )] = 0 for w = 3 and Re[G( jw )] = =w= 3 (1 - 3 ´ 3) 2 + 3(3 - 3) 1 8 Also , lt G( jw ) = lt = 1Ð0o 3 Im w®0 w®0 (1 + jw )

and

1 lt G( jw ) = lt = 0Ð - 270o w®¥ w®¥ (1 + jw )3

ωpc= 3

w=¥

pc

Using the information found above, the polar plot of G(jω) is sketched . Now, it is easy to read from polar plot that1 phase cross GM = = 8. 3 G( jw ) w =w over frequency ωpc= and pc

wgc=0 gc

Re w=0

– 1/ 8

unit circle ω=1 / 3

– 3 3 /8

The gain crossover frequency ωgc = 0 and

PM = 180o + ÐG( jw)

w=wgc

= 180o = π radians

Alternate way to find GM and PM, ÐG( jw )

w=w pc

or w pc = tan 60o = 3 and G( jw ) w=w = pc

So, GM =

= -180o gives - 3tan -1 w pc = -180o 1

é1 + ( 3) 2 ù ë û

3/2

=

1 8

1 = 8. G( jw ) w=w pc

G( jw ) w=w = 1 gives gc

PM = 180o + ÐG( jw )

1 = 1 or w gc = 0 and 2 3/2 (1 + w gc ) w=w gc

= 180o + tan -1 (0) = 180o = p radians .

Control Systems

Mechasoft

Q133.(d) Use Mason's gain rule to get Y(s) U(s)

= T(s) =

T(j w ) = t d (w ) =

s

-1

(1 + s

1 j w (j w + 1) d dw

-1

)-s -1 (1 + s )

-1

=

s

87

-2

1 -1 = s(s + 1) (1 + s )

and ÐT(j w ) = f(w ) = -

p 2

- tan

-1

w

1

f (w ) =

1+ w

2

5 15 1 1 -1 and average t d (w ) = ò ´ dw = ´ tan w @ 0.275 sec. 5 0 (1 + w2 ) 5 0

Ym (s) 1 1 Y 1 = ´s = and m ( jw ) = U(s) s(s + 1) s +1 U jw + 1

Q134.(d)

Ym 1 = U w2 + 1 1 2 b

w +1

Q135.(c)

and =

Ym U

= w=w b

Ym U

w= 0

2

gives

1 1 or w b = 1 or 2pf b = 1 or f b = Hz 2π 2

C(s) A / s (s + k) A = = 2 R(s) 1 + A / s(s + k) s + ks + A

The characteristic equation s2 + ks + A = 0, on comparison with general characteristic equation 2 2 s + 2ξωns + ωn = 0 gives undamped natural frequency ωn = A and damping ratio ξ=

k k = . 2ωn 2 A

The phase margin ϕm @ 100ξ = 40 gives ξ = 0.4 and resonant frequency wr = wn 1 - 2x 2 ο

gives wn =

wr 1 - 2x 2

=

k 68 = 0.4 gives k = 8. = 10 Now, A = ωn2 = 100 and 2 A 1 - 2 ´ 0.4 ´ 0.4

-px/ 1-x2 = e -p´0.4/ 1-0.4´0.4 = 0.254 Q136 (c) Peak overshoot = e ymax= 2 × (1 + 0.254) @ 2.5

Q137(a) Since, G(s) H(s) has a pole at origin, the Nyquist contour is chosen as shown below. Map of path ab : put s = jω so that G( jw ) H( jw ) =

jw - a jw ( jw + b)

Control Systems

88

Mechasoft

jw - a -a lt G( jw) H( jw) = lt @ lt w® 0 w® 0 jw ( jw + b) w® 0 jw´ b

b

= ¥Ð - 270o (map of point a) jw - a jw lt G( jw ) H( jw ) = lt @ lt w®¥ w®¥ jw ( jw + b) w®¥ jw ´ jw

a f ´ e

Im

= 0Ð - 90o (map of point b) ( jw - a) ( jbw + w 2 ) Also,G( jw ) H( jw ) = ( jbw - w 2 ) ( jbw + w 2 ) -w 2 (a + b) w (w 2 - ab) = + j - b 2w 2 - w 4 - b 2w 2 - w 4

d Nyquist Contour Im a f

Im[G( jw) H( jw)] = 0 for w2 - ab = 0 or w = ab

b,c,d

s plane

c

Re Semicircle of radius R®¥ Semicircle of radius ρ®0 G(s)H(s) plane w = ab

Re 1/b

e ab(a + b) a+b 1 Nyquist Plot = = 2 2 2 2 w= ab b ´ ab + a b b + ab b R e jq - a jq put s = lt R e so that G( j w ) H( j w ) = lt =0 Map of path b c d : R ®¥ R ®¥ Re jq (R e jq + b)

and Re[G( jw) H( jw)]

=

Map of path d e : mirror image of map of path ab

lt r e jq so that G( jw ) H( jw ) = lt Map of path e f a : put s = r® 0 r® 0

(r e jq - a) r e jq (r e jq + b) = -¥Ð - q

and map is semicircle of ∞ radius in the left half of complex plane.

Q138.( c) GM = x dB and PM =y°. Both GM and PM are positive. So, the system is stable. Note ωgc < ωpc.

Q139.(a) Initial segment has slope of −6dB/oct indicating presence of a pole at origin. The open loop gain k is determinable from initial segment as 20log

k æ 9.5 ö = 9.5 or k = 2 ´ anti log ç ÷ = 5.97 è 20 ø w w=2

There are two zeros with corner frequency 2 rad/sec as slope changes from − 6dB/oct to + 6dB/oct. There is a pole with corner frequency 10 rad/sec and yet another pole with corner frequency 20 rad/sec.

G(s) =

5.97(1 + s / 2) 2 5.97 (1 + 0.5s) 2 = s öæ s ö s (1 + 0.1s)(1 + 0.05s) æ s ç1 + ÷ ç1 + ÷ è 10 ø è 20 ø

Control Systems

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89

Q140.(a) The curve crosses the 0 dB line at a phase angle of −150° with corresponding frequency 1 rad/sec. Therefore, ωg = 1 rad/sec and ϕm = 180°−150° = 30°. The curve crosses the −180° line at dB magnitude of −15dB with corresponding frequency 1.8 rad/sec. Therefore, ωπ = 1.8 rad/sec and Gm = 0 − (−15) = 15dB. Q141.(d) Phase margin f m = 180 o + Ð G ( j w ) H ( j w ) w = w gc ; w gc is gain cross over frequency. For f m = 45 o , Ð G ( j w ) H ( w )

w = w gc

= 45 o - 180 o = - 135 o.

The frequency at which ÐG( jw) H(w) = -135o , is 0.25 rad/sec and corresponding gain is 10 dB as shown in figure. So, k should be decreased by 10 dB where after new gain cross over frequency will be 0.25 rad/sec (the gain cross over frequency for k = 1, is 0.5 rad/sec). Since, antilog(10/20) = 3.162, the value of k for ϕm = 45°, will be equal to 1/3.162 = 0.316. 142.(d) The current gain margin GM = 0−(−3) = 3 dB and phase cross over frequency ωpc = 0.6 rad/sec.To raise GM from 3 dB to 10 dB, the plot should be shifted downward by 7 dB that can be done by decreasing k by 7 dB. Note that decreasing k will not change ωpc = 0.6 rad/sec. Variation in k, does not change phase. Since, antilog (7/20) = 2.239, the value of k for GM = 10dB, will be 1/2.239 = 0.447.

é ù ê ú 0.04 w ú o -1 -1 w - tan -1 ê Q143.(d) ÐG( jw) H( jw) = -90 - tan w + tan êæ 5 w2 ö ú ê ç1 ÷ú ë è 121 ø û but −180° has to be added for phase calculation for ω >11 due to quadratic factor. This is because inverse tangent function behaviour for complex quantities with real part negative or imaginary part negative, can not be identified on calculator. For ω > 11, the phase of quadratic factor becomes positive on calculation whereas the true angle must lie between −90° and −180°. So, the true angle is obtained by applying correction of −180°. é ù ê 0.04 ´ 20 ú ú - 180o @ - 262o ÐG( jw) H( jw) w= 20 = -90o - tan -1 20 + tan -1 (20 / 5) - tan -1 ê ê æ1 - 20 ´ 20 ö ú ÷ êë çè 121 ø úû Q144.(a) The initial segment is flat coincident with 0 dB line indicating no pole or zero at origin and 20 log k = 0 gives k = 1. The other factors are identified as follows: corner frequency change in slope type of factor ωc1=1 ωc2= a ωc3= b ωc4= 100

+20dB/dec –20dB/dec –20dB/dec +20dB/dec

1+jω ; (a real axis zero) –1 (1+jω / a) ; (a real axis pole) (1+jω / b) –1 ; (a real axis pole) (1+jω / 100) ; (a real axis zero)

Control Systems

90

Mechasoft

The corner frequency w C2 = a, lies on straight line having slope m = +20 dB/dec and passing through a point (x1 = log ω = 1, y1 = dB = 0). Use straight line equation y − y1 = m (x−x1) where x = logω and y = dB to get dB = 20 (logω − log1) and y = dB w=a = 15 gives log(a) – log(1) = 15/20 or a = antilog(15/20) = 5.62 Similarly, the corner frequency w C3 = b lies on straight line that has slope m = −20dB/dec and passes through the point ( ω =100, dB = 0). Use the straight line equation y − y1 = m(x − x1) to get dB = −20(logω− log100) and dB w= b = 15 gives 15= − 20 (logb −log100) and b = 100 antilog(−15/20) = 17.78 Collect the factors and use values of a and b to get transfer function model in time constant form as G( jw) =

(1 + jw) (1 + jw /100) (1 + jw / 5.62)(1 + jw /17.78)

(s + 1) (s + 100) (s + 5.62) (s + 17.78) é (100 - w2 ) 2 + 4w2 ù = 10 log ê 2 2 2ú ë (100 - w ) + 100w û

and in pole- zero format as G(s) = Q145.(d) G( jw)dB

1/2

é (100 - w2 ) + 4w2 ù G( jw) = ê 2 2 2ú ë (100 - w ) + 100w û

Observe the following table to justify that dB plot is correct.

ω

1

8

4

9

10

11

12

16

|G(jω)|dB – 0.04 –0.85 – 6.95 – 10.9 –13.98 – 11.4 – 8.13 – 2.94

50 – 0.18

é 2w ù é 10 w ù ÐG( jw) = tan -1 ê - tan -1 ê 2ú 2 ë100 - w û ë100 - w úû Observe the following table to justify that phase plot is also correct.

ω

1

ÐG(j w) – 4.6º

4 –20º

6

8

–32.5º –41.8º

Q146.(b) Phase margin, PM = 180o + ÐG( jw)

or tan -1

wgc a

= 60o

or wgc = a 3 but

10

14

16

50

100

–0º

39.3º

34.1

9.4º

4.6º

w=wgc

G( jw) w=w

gc

1/3

or

1 + (3a 2 / a 2 ) é 4 ù = 1 or α = ê ú 3 a ´3 3 ë3 3 û

gives 180o + 2 tan -1

= 0.916

wgc

- 270o = 30o a 1 + w2 gc / a 2 = 1 gives =1 w3gc

Control Systems

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91

2

Q147.(b) G( jw) =

wn [(wn - w ) + 4 x 2 w2 wn 2 ] 1/2 2

2 2

2

and

G( jw) w=w = n

wn 1 = 2 2xwn 2x

p p é w wù p w p w ´ e - jw / 2 = cos - jsin ú = - sin - j cos ê jw jw ë 2 2û w 2 w 2 w p Im[G( jw )] = 0 for = or w = p 2 2 p p and Re[G( jw )] w=p = - sin = -1 p 2

Q148.(b) G( jw ) =

gain K at stability boundary 64 = =8 design value of K 8

GM dB = 20 log 8 = 18.06 dB Q150.(d) Initial segment has slope of −20dB/dec indicating presence of a pole at origin. Slope of −20dB/dec is also equivalent to slope of −6dB/oct. The corner frequency w c1 where slope changes by +20dB/dec, is one octave above the frequency ω = 0.1 rad/sec, that is, w c1= 0.1 × 2 = 0.2 rad/sec. The corner frequency w c1 is associated with real axis zero by virtue of slope change by +20dB/dec. Another corner frequency w c2 where slope changes by −20dB/dec, is associated with real axis pole and is one octave below ω = 100 rad/sec, that is, w c2 = (1/2) × 100 = 50 rad/sec . The gain k is evaluated from initial segment as follows

one octave dB=12 ω=0.1 dB=6

ωc1 =ω=0.2

one octave dB=6

ωc2 =ω=50 dB=0 æ 12 ö w= 0.1 = 12 Þ k = 0.1´ anti log ç ÷ = 0.398 è 20 ø ω=100 s ö æ 0.398 ç1 + ÷ 0.398 ´ 50 (s + 0.2) 99.5 (s + 0.2) è 0.2 ø = Now, G (s ) = ´ = s ö 0.2 s (s + 50) s (s + 50) æ s ç1 + ÷ è 50 ø k 20 log w

Thus, G(s) has one zero and two poles on real axis. The corner frequencies are 0.2 rad/sec

and 50 rad/sec or

0.2 1 50 25 = Hz and = Hz. 2π 10π 2π π

–6dB/oct ≡ –20dB/dec

Q149.(c) GM =

–6dB/oct ≡ –20dB/dec

The Nyquist plot will interseet negative real axis at point(– 1,j0).

Control Systems

π

dB=0

and frequency 400 rad/sec is 2 octaves above frequency 100 rad/sec. So, dB magnitude at 400 rad/sec, will be 0 −2 × 6 = −12dB

gc

one octave

ω=100 dB= –6 ω=200 dB= –12 ω=400

k

Q151( c) G( jw ) w=w = 1 Þ

Mechasoft

one octave

–20dB/dec ≡ –6dB/oct

92

Note that f = 200 Hz = 400 rad/sec

= 1 ; w gc is gain cross over frequency

1

æ w2 ö2 wç + 1 è 3 ´ 106 ÷ø

w=3000

1/2

é 3000 ´ 3000 ù k = 3000 ´ ê + 1ú 6 ë 3 ´10 û

= 6000

Q152.(c)The time delay of τ seconds has the transmittance of from e−τs ≡ e−jωτ and e−jωτ=cosωτ – jsinωτ 1/2

e - jwt = éëcos 2 wt + sin 2 wt ùû Ðe - jwt

w=wgc

= 1 º 0 dB

and

æ - sin wt ö Ðe - jwt = tan -1 ç ÷ = - wt è cos wt ø

= -3000t radians

Since, phase margin reduces by 15º

180o 15p 3000t´ = 15o or t = ´10-3 = 0.0873×10 -3 seconds p 180 ´ 3 10 16 16 = 2 = 0.625s + 1.75s + 1 s + 2.8s + 1.6 (s + 0.8) (s + 2)

Q153.(a) G (s ) =

2

So, the corner frequencies (in rad/sec) are 0.8 and 2. Q154.(d) G( j w )

w=10

= G( j10) = 0.5 and ÐG( jw ) w=10 = ÐG( j10) = – p / 2

A = 2 ´ 0.5 = 1 and d =

Q155.(a) G( jw ) =

p p – 4p – = 18 2 9

k k k (a2 – w2 ) 2awk = = –j 2 2 2 2 2 2 2 2 2 ( jw + a) (a – w ) + j2aw (a – w ) + 4a w (a – w2 )2 + 4a2w2

Re [G( jw ) ] w=10 = 0 Þ a2 –102 = 0 or a = 10 Im [G( jw ) ] w=10 = – 0.5 Þ

2awk = 0.5 Þ k = aw = 10 ´ 10 = 100 (a – w2 )2 + 4a2w2 2

Control Systems

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93

1 1 and G( jw) = s+3 jw + 3 1 1 G( jw) w= 3 = = w= 3 2 2 3 w +9

Q156.(d) G(s) =

ÐG( jw)

w= 3

æ 3ö -1 æ 1 ö = - tan -1 çç = -30o = -p / 6 ÷÷ = - tan ç ÷ è 3ø è 3 ø

The response y(t) =

3 cos( 3 t - p /12 - p / 6) 2 3

= 2-1 cos( 3 t - 2-2 p ) a - jbw a -a e = [cosbω - jsinbw ] = [sinbω + jcos bw ] Q157.(a) G(j w ) = jw jw w a a Re[G(j w )] = x = - sin bω and Im[G(j w )] = y = - cos bw w w é a ù é -ab cos bω ù As w ® 0, y ® ¥ and lim ê - sin bw ú = lim ê úû = - ab ; L 'Hospital's rule w®0 1 ë w û w®0 ë As w ® 0, polar plot will be asymptotic to x = – ab Q158. H(s) =

H(j w) =

1/ Cs 1 1 = = R + 1/ Cs 1 + sRC s + 1

1 1 + jw

For w = 1, | H(j w) | =

1 2

1+ w 1 pö æ y(t) = sin ç t + q - ÷ 4ø 2 è 1 For q = p / 4, y(t) = sin t, 2

=

R=1MΩ 1μF sin(t + θ)

C

1 and ÐH(j w) = - tan -1w = - p / 4 2

r(t) = sin(t + θ)

1 s +1

and output starts from zero with no transient. Therefore, m = 4.

1 1 , G(j w) = and ÐG(j w) = - n tan -1 bw n 2 2 n/2 (1 + jb w) (1 + b w ) 1 For n = 6, G(j w) = and ÐG(j w) = -6 tan -1 (bw) (1 + b 2 w2 )3 1 For r(t) = sin wt, y(t) = sin[wt - 6 tan -1 (bw)] 2 2 3 (1 + b w )

Q159. G(j w) =

Output

y(t)

Control Systems

94

Mechasoft

-1

f = 6 tan (bw) = p Þ bw = tan(p / 6) = 1/ 3 1 1 27 a= = = = 0.42 2 2 3 3 (1 + b w ) (1 + 1/ 3) 64 Y(s) Q160. The transfer function, = C(sI- A) -1 B = T(s) U(s) és 0 ù é 0 1 ù é s - 1 ù és - 1 ù (sI - A) = ê -ê =ê ú ú ú=ê ú for k = 1 ë0 s û ë-5 - k - 2û ë5 + k s + 2û ë6 s + 2û For k = 1, (sI - A)

-1

1 és + 2 = 2 ê s + 2s + 6 ë - 6

1ù s

ú û

1 1 és + 2 1ù é0ù é1ù T(s) = 2 [6 3] ê = 2 [6 3] ê ú ú ê ú s + 2s + 6 ë - 6 s û ë1 û s + 2s + 6 ës û 6 + 3s 3(j w + 2) T(s) = 2 ; T(j w ) = and | T(j w ) |= 2 s + 2s + 6 (6 - w ) + j 2w

2 3 w +4 2 2 2 (6 - w ) + 4w

T(j w ) w=0 = T(j0) = 1;

|T(jω)|

Let w b be the frequency :| T(j w b ) |=

T(j w ) w=w = b

1

1

| T(j0) |

2

1 2

2 2 wb + 4 1 1 = T(j0) = 2 2 2 2 2 (6 - w b ) + 4w b

0

2 2 3 (w b + 4) 1 2 = 2 2 2 2 Þ w b = 28.5; negative frequency discarded (6 - w b ) + 4w b

5.34 BW

Bandwidth = w b = 28.5 = 5.34 rad / sec

Q161.(b) The characteristic equation of given second order system is

s I -F 2

= 0

Þ

s -1 4 s+2

= 0 Þ s 2 + 2s + 4 = 0. Compare with general equation

2

s + 2ξωns + ωn = 0, to get ωn = 2 and ξ = 1/2. The system exhibits underdamped dynamics with settling time, ts = 4 /ξωn = 4sec. Q162.(c) The characteristic equation is lI - A = 0 Þ

l +1

-1

0

l+3

= 0 Þ (l + 1) (l + 3) = 0

and eigen values(characteristic roots) are −1 and −3. The system is stable forcing the dynamics to

é0 ù ë0 û

decay to zero, that is x(¥) = ê ú in absence of an external input.

ω

Control Systems

Mechasoft -t

é x1 (t) ù é e ú=ê ë x 2 (t) û ë 0

0 ù ú e -2t û

Q163.(a) ê

-t

é x1 (0) ù ée ê x (0) ú = ê ë 2 û ë0

0 ù ú e -2t û

95

é1ù ê -2 ú = ë û

-t

é e ù ê -2t ú ë -2e û

e– t= 0.905 gives t = 0.0998 sec = 99.8 m s @100ms . −2t Note that −2e = −1.637 ,also, gives t = 0.1 sec = 100 ms. Q164.(a) Observe the signal flow graph to write state equation and output equation as

x& 1 = -bx1 + dx 2 ;

x& 2 = cx1 - ax 2 + r ;

y = x1

In matrix form

é x& 1 ù é - b d ù é x1 ù é0ù ê& ú = ê ú ê x ú + ê1 ú r x c a ë û ë 2û ë û ë 2û éx ù y = [1 0] ê 1 ú ëx 2 û é- b

where system matrix, F = ê

ëc

dù é0 ù , control matrix G = ê ú and ú -a û ë1 û

output matrix, H = [1 0]

é0 d ù The contrllability matrix Qc = [G FG ] = ê ú ë1 - a û éHù é 1 0ù and observability matrix Q0 = ê ú = ê ú ë HFû ë- b d û Since, Q c = - d and Q 0 = d, system is completely state controllable and state observable if d ≠ 0 while a, b, c can be anything. Q165.(b) Both system S1 and S2 are described by diagonal models. Mere inspection reveals that neither S1 nor S2 are completely state controllable. Both have one zero entry in control matrix. S1 has uncontrollable but stable mode at −1 while unstable mode at 2, is controllable. So S1 is stabilizable. Note that S2 has unstable and uncontrollable mode at 1 while mode at 2, is unstable but controllable. So, S2 is not stabilizable. Q166.(b) S1 and S2 both have diagonal models. Mere inspection reveals that S1 and S2 both are unobservable. The output matrix has one zero entry in both models. S1 has two modes: one at 2 which is unstable but observable and other at −1 which is stable but unobservable. So, S1 is detectable. S2 has two unstable modes out which one at 1, is unobservable also. So, S2 is not detectable. Q167. The characteristic equation of given second order system is

Control Systems

96

sI - F = 0 Þ

s

-g

g

s

Mechasoft

= 0 Þ s 2 + g 2 = 0 and roots (eigen values) are located at s = ±jg.

10 10 Hz or 2p´ = 20 rad / sec, roots must be located at p p s + 6 1ù é s=±j20. So, g = 20. ê Adj [sI - A] ë -5 s úû -1 = 2 gives Q168.(d) f (s) = [sI - A] = sI - A s + 6s + 5 For the system to oscillate at f =

2 characteristic equation sI - A = s + 6s + 5 = 0 or(s + 1) (s + 5) = 0 and characteristic roots (eigen values) = −1 and −5. é1

-1

é s -k ù 1 és + 2 = ú s (s + 2) êë 0 ë0 s + 2 û

Q169.(a) [sI - A]-1 = ê

k ù ê kù s s(s + 2) ú ê ú = s úû 1 ú ê êë 0 s + 2 úû é0 ù k ê1 ú = s (s + 2) ë û

é1/ s k / s(s + 2) ù 0] ê 1/ (s + 2) úû ë 0 k 2 Q170.(a) The characteristic equation is 1 + = 0or s + 2s + k = 0 and characteristic roots s (s + 2) are -1 ± 1 - k. For k = 1, both roots are located at s =−1 and for k >1, the roots are complex -1

G(s) = C (sI - A ) B = [1

conjugate but real part of the roots remain constant. Since, the settling time depends on real part of roots, it remains unchanged for k ≥ 1. Thus , k´ = 1. Q171.(c) For the state variable assignment as shown below

x& 1 = - x1 + x 2 + u x& 2 = -2x 2 + u y = x1 + x 2

1 u

1 . x2

s

x

s

–1

x

x

1

;

s

s

–1

x1 x

é é 1 0 ù é -1 1 ù ù é s + 1 -1 ù where f(s) = ês ê -ê ú =ê ú ú s + 2 úû ë 0 ë ë 0 1 û ë 0 -2 û û

-1

1 y

1

y = [1 1]x

Q172.(c) S T M f(t) = [L-1f(s)] -1

. x1

–1

–2

é -1 1 ù é1ù x+ê úu ú ë 0 -2 û ë1û

and state space model in matrix form is x& = ê

x2

1

Control Systems

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é 1 ês +1 =ê ê ê 0 ë ée- t f (t ) = ê ë0

97

1

ù (s + 1)(s + 2 )úú ú 1 ú s+2 û -t -2t e -e ù ú e -2t û t

ò

é0 ù

é1ù

ë û

ëû

Q173(c) x(t) = f(t) x(0) + f(t - t ) Bu(t ) dt where x(0) = ê ú , B = ê ú and f(t) from Q172. 0 1 0

t

- (t -t )

ée x(t) = ò ê 0 0ë

e

- (t -t )

-e

-2(t -t )

e -2(t -t )

ù ú û

é1ù ê1ú u(t) dt ëû t

é -(t -t) 1 -2(t -t) ù - ´e 1 -2t ù é3 -t ê 2e ú 2e + e ú 2 -(t -t ) -2(t -t ) t é ù ê2 2e -e ê ú 2 = òê ú dt = ê ú ú =ê 1 1 -2(t -t ) 2t e ú ê ú 0 ê ë û ê ú - e 1 -2(t -t) ê úû ´e ë 2 2 ê ú ë 2 û0

1 -2t ù é3 -t 2e + e ú ê2 2 -t y(t) = [1 1] ê ú = 2 - 2e and y(1) = y(t) ê 1 - 1 e -2t ú êë 2 2 úû -1

t =1

= 2[1 - e -1 ] = 1.264

0 ù 0 ù és + 2 é1/ s + 2 Q174.(a) G(s) = H(sI - F) G where (sI - F) = ê =ê ú s + 3û 1/ s + 3úû ë 0 ë 0 0 ù é2ù é 1 é1/ s + 2 1 ù é2ù 2 G(s) = [1 1] ê = = ú ê ú ê ú ê ú 1/ s + 3û ë 0 û ë s + 2 s + 3 û ë0û s + 2 ë 0 -1

-1

é s -1 ù Y(s) Q175.(a) G(s) = = C(sI - A) -1 B = [1 1] ê ú R(s) ë 2 s + 3û

-1

é0 ù ê1 ú ë û

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1 é s+3 ù é0 ù ê s 2 + 3s + 2 s 2 + 3s + 2 ú ê ú ê úê ú (s + 1) 1 G(s) = [1 1] ê = úê ú = (s + 1) (s + 2) (s + 2) ê úê ú -2 s ê 2 ú 1 ë s + 3s + 2 s 2 + 3s + 2 û ë û 1 1 1 1 2 K p = lt G(s) = lt = and E1 = = = s® s ®0 s + 2 1 2 1+ Kp 1+ 3 2 1 K v = lt s G(s) = 0 and E 2 = =¥ s ®0 Kv x2 Q176.(b) VC1 = x1 , VC2 = x 2 and I L = x 3 + Vi - VC1 R1

= C1

dVC1 dt

+ C2

dVC2 dt

R1 u=Vi(t)

dVc2 C2 = x3 dt x1 - x 2 - VL = 0 or x1 - x 2 - Lx& 3 = 0

+ x1 -

+ -

+

C2 C1

L x3

-

In terms of state variable x1, x2, x3, input u and output y

x& 1 = -

x x1 u - 3+ R1C1 C1 R1C1

, x& 2 =

x3 C2

, x& 3 =

x1 x 2 and y = x1 - x 2 L L

Put the component values to get

x& 1 = - x1 - 104 x 3 + u , x& 2 = 104 x 3 , x& 3 = 102 x1 - 102 x 2 and y = x1 - x 2 é -1 é x& 1 ù ê ê ú and in matrix form ê x& 2 ú = ê 0 ê102 êë x& 3 úû ë é x1 ù y = [1 - 1 0] êê x 2 úú êë x 3 úû

0 0 -102

-104 ù ú 104 ú 0 úû

VL(t)≡y(t)

é x1 ù é 1 ù ê x ú + ê0 ú u ê 2ú ê ú êë x 3 úû êë0 úû

Q177.(a) Choose state var iables x1 = y and x 2 = y& Then , x& 1 = x 2 ; x& 2 = u and y = x1 é x& 1 ù é0 1 ù é x1 ù é0 ù = x + u and y = [1 0] ê x& ú ê0 0 ú êx ú ê1 ú û ë û ë 2û ë ë 2û

Control Systems

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é é1 Q178.(a) f (s) = ê s ê êë ë 0

0 ù é0 1 ù 1 úû êë 0 0 úû

-1

.

ù ú úû

-1

és =ê ë0

-1ù s úû

-1

99 2

é1 / s 1 / s ù é1 t ù =ê ú and f (t) = ê ú 1/ s û ë 0 1û ë 0

Note that x1(0) = y(0) = 1 and x2(0) = y(0)= –1 t

é0 ù x(t) = f(t) x(0) + ò f(t - t) Bu(t) dt ; B = ê ú ë1 û 0 é1 t ù é 1 ù t é1 (t - t) ù é0 ù é1 - t ù dt = ê ê0 1ú ê -1ú + ò ê0 ú ê ú ú+ 1 û ë1 û ë û ë û 0ë ë -1 û

and u (t) = 1 t

é(t - t) ù dt 1 úû 0

ò êë

t

é é t2 ù t2 ù é1 - t ù ê tt - ú 1 t + 2 ú = êê 2 úú ê -1 ú + ê ë û ê t ú ë û0 ëê -1 + t ûú é t2 ù 2 1 t + ú = 1 - t + t and y(1) = 1/ 2 y(t) = [1 0] ê 2 ê ú 2 êë -1 + t úû

s é ê (s + 1) (s + 3) -1 -1 ê é é1 0 ù é -4 1 ù ù és + 4 -1ù f(s) = ês ê =ê ú-ê úú = ê 3 s úû Q179.(a) ë ê ë ë0 1 û ë -3 0 û û -3 ê ëê (s + 1) (s + 3) é -0.5e - t + 1.5e -3t STM, f(t) = ê -t -3t ë -1.5e + 1.5e

1 ù (s + 1) (s + 3) ú ú ú ú s+4 ú (s + 1) (s + 3) úû

0.5e - t - 0.5e -3t ù ú 1.5e - t - 0.5e -3t û t

é1 ù The state response, x(t) = f(t) x(0) + ò f(t - t) Br(t)dt; B = ê ú ë0 û 0 t é -0.5e -(t -t) + 1.5e -3(t -t) 0.5e -(t -t) - 0.5e -3(t -t) ù é1 ù é0 ù = ê ú+òê ú ê ú d (t)dt - t -t -3 t -t - t -t -3 t -t ë0 û 0 ëê -1.5e ( ) + 1.5e ( ) 1.5e ( ) - 0.5e ( ) ûú ë0 û

é -0.5e - t + 1.5e -3t ù =ê -t -3t ú ë -1.5e + 1.5e û

; d (t) = 0 for t ¹ 0

é -0.5e - t + 1.5e -3t ù The system response y(t) = [1 0]x(t) = [1 0] ê -t -3t ú ë -1.5e + 1.5e û = -0.5e - t + 1.5e -3t and y(t) t =0.25 = -0.5e -0.25 + 1.5e -3´0.25 = 0.319

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Q180.(a) Write nodal equations to get

C1x& 1 +

x1 - x 2 x1 - u x - x1 x 2 - u + = 0 ; C2 x& 2 + 2 + =0 R3 R1 R3 R2

R + R3 1 1 and y = x 2 or x& 1 = - 1 x1 + x2 + u R1R 3C1 R 3C1 R1C1

R2

R1 u

+

~

R3

–

x1

R + R3 1 1 x& 2 = x1 - 2 x2 + u R 3C1 R 2 R 3C 2 R 2C2

C1

y x2 C2

Fit the component values and put equations in matrix form to get

é x& 1 ù é -20 / 47 10 / 47 ù ê x& ú = ê 10 / 47 -20 / 47 ú û ë 2û ë éx ù y = [0 1] ê 1 ú ëx2 û

é x1 ù êx ú ë 2û

é10 / 47 ù + ê úu ë10 / 47 û

é -20 / 47 10 / 47 ù é10 / 47 ù é -100 / 2209 ù AB = ê = ê ú ê ú ú ë 10 / 47 -20 / 47 û ë10 / 47 û ë -100 / 2209 û é -20 / 47 10 / 47 ù CA = [0 1] ê ú = [10 / 47 -20 / 47] ë 10 / 47 -20 / 47 û

Q181.(d)

é10 / 47 -100 / 2209 ù The controllability matrix , Q c = [B : AB] = ê ú ë10 / 47 -100 / 2209 û 1 ù é C ù é 0 and observability matrix, Q 0 = ê =ê ú ú ë CA û ë10 / 47 -20 / 47 û Since |Qc| =0 but |Q0| ≠ 0, the system is not completely state controllable but completely state observable. The bridge is balanced with component values. The voltage across R3 can not be influenced by input u º vs(t).

é -3 4 ù ú ë -2 0 û é -3 4 ù és + 3 -4 ù -ê ú =ê s úû ë -2 0 û ë 2 s é 2 ê s 4 é ù s + 3s + 8 ê -2 s + 3ú = ê -2 ë û ê 2 ëê s + 3s + 8

Q182.(d) System matrix, A = ê

é1 0 ù [sI - A] = s ê ú ë0 1 û [sI - A]-1 =

1 sI - A

4 ù s + 3s + 8 ú ú s+3 ú s 2 + 3s + 8 ûú 2

Control Systems

Mechasoft

101

-1

T(s) = C(sI - A) B s 4 é ù 2 2 é -4 6 ù ê s + 3s + 8 s + 3s + 8 ú é 2 ù é -2(s + 11) ù 1 =ê ê ú ê ú= 2 ú -2 s + 3 ú ë1 û s + 3s + 8 êë 9s + 21 úû ë 5 -1û ê êë s 2 + 3s + 8 s 2 + 3s + 8 úû Y (s) -2 (s + 11) Y (s) 9s + 21 T1 (s) = 1 = 2 and T2 (s) = 2 = 2 U(s) s + 3s + 8 U(s) s + 3s + 8 Also, lI - A =

l + 3 -4 = l 2 + 3l + 8 2 l

eigen values = -1.5 ± j2.398 @ - 1.5 ± j2.4 ée- t

é e- t ù 0 ù é1 ù = ê -2t ú -2t ú ê ú ë 0 e û ë2û ë 2e û é e -0.5 ù é 0.6 ù = ê -1 ú = ê ú ë0.74 û ë 2e û

Q183.(a) x(t) = f(t) x(0) = ê

x(t)

t = 0.5

Q184.(b) The simulation diagram (signal flow graph) together with state variable assignment at the output of integrator, is shown below .

x& = - bx + u y1 = a1x

; y 2 = a 2 x ; y3 = a 3 x

x

and in matrix form

x& = [- b][x] + [1]u é y1 ù é a1 ù ê y ú = ê a ú [x] ê 2ú ê 2ú êë y3 úû êë a 3 úû

u

1/ s

1

a1 a2

. x –b

a3

y1 y2 y3

é1 - 2 ù és 0 ù é1 - 2 ù és - 1 2 ù and sI - A = ê ú ú-ê ú=ê ú ë 4 - 5û ë 0 s û ë 4 - 5 û ë -4 s + 5 û -2 é s+5 ù ê (s + 1)(s + 3) (s + 1)(s + 3) ú és + 5 - 2 ù 1 -1 ê ú f(s) = [sI - A] = = s - 1úû ê 4 s -1 ú (s - 1)(s + 5) + 8 êë 4 ê (s + 1)(s + 3) (s + 1)(s + 3) ú ë û

Q185. The system matrix, A = ê

Control Systems

102

1 é 2 ês +1 - s + 3 f(s) = ê ê 2 - 2 ëê s + 1 s + 3

Mechasoft

-1 1 ù + é 2e - t - e -3t - e - t + e -3t ù s +1 s + 3ú ú and STM, f(t) = ê - t ú -1 2 ú 2e - 2e -3t - e - t + 2e -3t û ë + s + 1 s + 3 ûú

é 2e - t - e -3t - e - t + e -3t ù é 2 ù The unforced state response (no input), x(t) = f(t)x(0) = ê - t -3t -t -3t ú ê ú ë 2e - 2e - e + 2e û ë1 û é 4e - t - 2e -3t - e - t + e -3t ù é3e - t - e -3t ù x(t) = ê - t = ê -t -3t -t -3t ú -3t ú 4e 4e e + 2e ë û ë3e - 2e û y(t) -t -3t 3 é3e - e ù -t -3t For l = 1, y(t) = [1 l] x(t) = [1 1] ê - t = 6e 3e ú -3t ë3e - 2e û 2.058 -1 -3 and y(t) t =1 = 6e - 3e = 2.058 éC ù 0 1 Q186 ( c) The observability matrix, Q 0 = ê ; C = [1 l] and ú ëCA û é1 - 2 ù CA = [1 l] ê ú = [1 + 4l - 2 - 5l] ë 4 - 5û l ù é 1 Q0 = ê ú ; For the system model to be unobservable ë1 + 4l - 2 - 5l û

t(sec)

|Q0 |= 0, that is, (–2 – 5λ) – λ(1 + 4λ) = 0 or 2λ2 + 3λ +1 = 0 or λ = –1, –1 / 2. Q 187(c) The characteristic equation is |sI – A| = 0

1 0ù é s -1 és 0 0 ù é 0 ê ú ê (sI- A) = ê0 s 0 ú - ê 2 0 1úú = êê -2 s êë0 0 s úû êë -k - 3 - 2 úû êë k 3

0 ù - 1 úú s + 2 úû

2

|sI – A| = s(s + 2s + 3) + 1( –2s –4 + k ) = 0 3 2 s + 2s + s + k – 4 = 0 The Routh array is constructed below. 3 s 1 1 s s

s

0

2

1

2 (2 – k+4)/2

k–4

k–4

For the system to be stable, all elements of left column of array must have same algebraic sign , that is, k > 4 and –k + 6 > 0 or –k > –6 or k < 6 All together, 4 < k < 6.

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Control Systems

103

Q188.( c) x& 1 = 2x1 + u ; x& 2 = - x1 + x 2 - u ; y = x1 + x 2

é x& 1 ù é 2 0 ù é x1 ù é 1 ù é x1 ù ê x& ú = ê -1 1ú ê x ú + ê -1ú u and y = [1 1] ê x ú ûë 2û ë û ë 2û ë ë 2û 0ù é 2 0ù é 1ù és - 2 A=ê , B = ê ú , C = [1 1] and sI- A = ê ú ú ë -1 1û ë -1û ë 1 s - 1û | sI- A |= 0 Þ (s - 2)(s - 1) = 0 Þ s = 1, 2 The characteristic roots (eigen values) are positive real (located in rhp) and the system is unstable. . 2ù é 1 2 x1 x1 The controllability matrix, Qc =[B : AB] = ê ú

ë -1 - 2 û

é C ù é1 1ù The obserrvability matrix, Q o = ê ú=ê ú ëCA û ë1 1û

s

1

Since, |Qc| = 0, system is not completely Controllable .

–1

1

–1

U(s) –1 . x2

Y(s) s–1

1

x2 Since, |Qo| = 0, system is not completely observable. 1 The system being uncontrollable, is not slabilizabele. Since, x&1 + x& 2 = 2x1 + u – x1 + x2 – u = x1 +x2 , that is, dynamics of state variables is independent of u, the system is not controllable. Q189.(b) The dB plot corresponds to lead compensator with zero corner frequency ωz = 10 rad/sec. The gain increases at the rate of +20dB/dec (or +6dB/oct.) from 0 dB at zero corner frequency ωz = 10 rad/sec and attains peak value of 20 log(1/α) dB at pole corner frequency ωp where after gain remains constant 20 log

1 = 12.04 gives a = 0.25 a

In general, lead compensator has transfer function

1 + st 1 + jwt or G lead ( jw) = ; a 0

Q5. The input-output relationship of a system is given by

Q6. Consider a unity feedback system shown below. The stability range of K is (a) – 1 < K < 0 2 (c) 0 < K < 2

(b) –2 < K < 0 1 (d) 0 < K < 2

+ –

K(s –1) s(s + 2)

Control Systems

Mechasoft

3

111 2

Q7. Consider a system with characteristic equation s +14s +(45+k) s + k = 0. The centroid of root locus is located at ( – λ, 0).The value of λ is _________. Q8. Consider the following statements in relation with a cascade control system consisting of inner and outer loop. Assume that both the loops have single time constants. 1. The dynamics of inner loop must be significantly faster than that of outer loop. 2. The closed loop transfer function of inner loop must have time constant much larger than that of outer loop . 3. There arises a risk of interaction between the two loops leading to instability if the inner loop has time constant larger than that of outer loop. Of these , the correct statement(s) is (are) (a) only 1

(b) only 1 and 3

(c) only 2

Q9. Consider a unity feedback system with loop function G(s) =

(d) only 2 and 3 k ; (s + a) (s + b) (s + c) (s + d)

k > 0 and a, b, c, d, are positive real constants. On sketching the Bode plots , it is observed that the gain cross-over frequency ωgc is larger than the phase cross-over frequency ωpc . Which one of the following statement is correct? (a) The stability of the closed loop system can not be concluded from observation . (b) The closed loop system is stable. (c) The closed loop system is unstable . (d) the closed loop system is marginally unstable. Q10. Consider a third order system with transfer function G(s) =

75 . (s + 15)(s + 4s + 100) 2

Let its second order approximation be Ga (s) =

K . s + 4s + 100 2

The value of K is ________________ .

Q11. Consider the following statements regarding Routh stability investigation. 1. The presence of entire row of zeros always implies that system has poles on jω axis. 3 2. A seventh order system has a row of zeros at the s row and two sign changes below the 4 s row. This implies that the system has at least 2 poles one jω axis. 3. The criterion never gives the actual location of system’s closed loop poles. Of these ,INCORRECT statement (s) is / are

(a) only 1

(b) only 1 and 2

(c) all 1,2 and 3

(d) only 1 and 3

Q12. A unity feedback system has forward path transfer function G(s) = 16/s(s + a).

The value of a , rounded off to one decimal place, for which the closed loop response to unit step input , exhibits 5% overshoot, is __________.

Control Systems

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Q13. The control system with input r(t) = 5t u(t) as shown below, exhibits 10% error in the steady state. The value of λ is _________. 672(s + 5)

+ input r(t) = 5t u(t) –

output

s(s + 6)(s + 7)(s + l )

Q14. The system shown below, is stable for k > q and k > 0. Then, (a) p < 0 and q > 0

k(s - p) s(s - q)

+

(b) p < 0 and q < 0

–

(c) p > 0 and q < 0 (d) p> 0 and q > 0

Q15. Consider the root locus of a system shown below. For some gain k, the closed loop transfer function will have a real pole at s= –5. jω The value of k is __________________.

+ –

×

4k (s + 4)(s 2 + 2 s + 2)

× –4

The gain margin (in dB) is (a) ∞

(b) 0

(c) 1.58

(d) 10.46

In

σ

–1

× Q16. The unity feedback system shown below, has G(s) =

j1

–j1

6 (s + 2 s + 2)(s + 2) 2

+ –

G(s)

out

Q17. An input r(t) = sin (t + ϕ) is applied to a series RC circuit with R = 1MΩ and C = 1μF. The output is taken across capacitor. In order that there is no transient in the output, the value of ϕ (in degrees) is __________.

1 ù é 0 é0 ù x + ê úu ú ë -100 -20 û ë1 û

Q18. A system has state vector differential equation x& = ê The nature of system dynamics, is (a) overdamped (c) underdamped

(b) critically damped (d) oscillatory

Control Systems

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113

Q19. A system with transfer function G(s) = [50/(s+50)] has time constant tc and settling time ts such that |tc – ts| = λ. The value of λ (in ms) is ________________________. Q20. A system with input r(t)=5u(t) and output y(t), is shown below. G(s) Input r(t) = 5 u(t)

s+2 s+5

+ –

Output y(t)

Let λ be the forced component of response and μ be the natural response at t = 0.2 sec . The ratio λ / μ (correct up to two decimal places) is ___________________. Q21. The poles of a system are located as shown below. The time (in sec) at which the first undershoot occurs in its unit step response, is (a) 0.449 (b) 0.898 (c) 1.347

jω j7

× –3

(d) 1.223

σ

×

–j7

Q22. A system with transfer function G(s) = [(10 – s) / (10+s)] is excited at t = 0 by r(t) = 2u(t) ; u(t) is step function. The response of system at t = 100ms, is_____. Q23. A system is represented by block diagram as shown below. The transfer function [C(s) / R(s)] is (a)

s3 + 1 s(2s3 + s + 2)

s2 + 1 (b) 2 s3 + s+ 2 ( c)

s2 + 1 2 s 4 + s3 + s 2 + 2

(d)

s3 + 1 s(s 3 + 2 s + 1)

+ R(s)

–

s

s

–

+

(1/s)

C(s)

+ (1/s) ss

–5t

Q24. A control system generates response y(t) =e ( sin bt + cos bt ); b > 0 for input r(t) = δ(t) ; δ(t) is a unit impulse . The dc gain of control system is 0.2 . The value of b is __________.

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æ z – sö ; z > 0, p > 0 ,is placed in unity Q25. A system with transfer function L(s) = k ç è s – p ÷ø

feedback configuration . The closed loop system will be stable for (a) k

p regardless of whether p > z or p < z z

*

End of Test Drill I

(b) k > 1 regardless of value of p and z (d)

p p < k < 1 or 1 < k < z z

*

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Answer Key Q1 Q8 Q15 Q22

(3) Q2 (10.25) t (b) Q9 so (c)f (4.25) Q16 ha (d) ecQ23 (a) (0.53)

M

Q3 Q10 Q17 Q24

(1.43) (5) (45) (5)

Q4 (d) Q5 t Q11 (c) Q12 of s a Q19 Q18 ch (b) e Q25 (d)

M

Q6 (a) (5.5) Q13 (60) Q20

Q7 (6.5) (b) ft(a) Q14 (8) o as (b) (1.8) chQ21

e

M

have trust

you can do

dispel the fear

be at natural ease

Test Drill-II Time:1hour 20minutes

Maximum Marks :50

Instructions (i) Revise the subject thoroughly before embarking upon test drill. (ii) Do not open the pages of the test drill until you are fully prepared. (iii)Each question carries one mark. (iv) Beware of Negative Marking * Wrong answers for multiple choice type questions (MCQ) will result in Negative marks. For all MCQs, a wrong answer will result in deduction of 2/3 marks. * There is No Negative marking for questions of Numerical Answer Type. (v) Set the timer. (vi) Conclude the test within allotted time while being in examination like environment. (vii) Match your answers with key answers given on the last page of the test and grade your preparation. Marks Obtained > 40 – Excellent 35 to 40 – Very Good 30 to 35 – Good 25 to 30 – Fair

GOOD LUCK

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Q1. Consider the position control system shown below .The parameter λ can vary from 0 to 0.8. 16 + X(s) Y(s) s(s + 0.8) – Which one of the following is INCORRECT?

ls + 1

(a) For λ∈[0.45,0.8] , the system doss not exhibit sinusoidal dynamics. (b) For λ= 0.45 ,the response of the system settles within 2% of final value in 1 second. (c) For λ= 0.075 , the peak value of unit step response is 1.44. (d) For λ= 0, the peak value of unit step response will be less than 1.44. Q2. The two equivalent block diagrams as shown below, represent a closed loop system. + –

R (s)

G1 (s)

+

+

G2 (s)

C (s)

– H2 (s)

H1 (s)

R (s)

X(s)

1+G1(s)

C (s)

Choose the correct option. (a) X(s) =

G2 (s) 1 + G1 (s) H1 (s) + G2 (s) H2 (s)

(b) X(s) =

G2 (s) 1 + G1 (s) G2 (s) H1 (s) + G2 (s) H2 (s)

(c) X(s) =

G1 (s) G2 (s) 1 + G1 (s) H1 (s) + G2 (s) H2 (s)

(d) X(s) =

G1 (s) G2 (s) 1 + G1 (s) G2 (s) H1 (s) + G2 (s) H2 (s)

100 Q3. Consider the feedback control system shown below. r (t) = u(t) ; u(t) is unit step function. s(s + 2) 1 R(s) y(t) = 1– a e–bt sin(ct +d) where a,b,c and d 1+0.1s have usual dimensions The values of a , b, c, and d are respectively (a) 1.25 , 6 ,8 and 53.13◦ (b) 1.66 ,3 ,4 and 36.87◦ ◦ ◦ (c) 1.66 , 6 ,10 and 73.78 (d) 1.25 ,10, 8 and 26.57

Q4. The system with block diagram shown below, is designed to exhibit underdamped unit step response with peak overshoot less than 10% . + In order to meet the design demand X(s) – (a) l Î[1.75, 4.3] (b) l Î[–2,0] ( c) l £ 1.74

(d) l ³ 4.3

10

+ –

1

1 s(s + 2) ls

Y(s)

Y(s)

Control Systems

Mechasoft

117

Q5. Refer to the signal flow graph and the typical response y (t) of a system shown below. r(t) = 12u(t) ; u(t)is unit step function. The values of t1 , t2 and ∆are respectively y(t) 10 (a) 0.32 , 0.52 and 2.1 ∆ (b) 0.63 , 1.05 and 4.2 s(s + 2) 12 y(t) r(t) ( c) 0.23, 0.33 and 3.2 1 1 (d) 0.95 , 1.57 and 6.3 0 t1 t2 –1 t(sec) Q6. A unity feedback system with closed loop transfer function Y(s) k s+b = 2 is excited by input x(t) = t ; t ³ 0. Let S e(a ¥ ) and S be( ¥ ) be the sensitivities X(s) s + as + b

of e (∞)to variation in a and b respectively ; e (∞) is the error in steady state . The values of S e(a ¥ ) and S be( ¥ ) are respectively –1

–2

(a) b and –1

(b) b and a

a (d) a–1 and b–1 and –1 a–k Q7. Consider a system with characteristic equation s5 +s4 +6s3 +6s2 +25s+25=0. This system has L roots in the LHP , M roots in the RHP and N roots on the imaginary axis of s-plane. Let P = | L–M–N |. Then , value of P is___________.

(c)

Q8. The first two rows of Routh array for a control system are given below.

1 1

3 3

3 2

1

The system has p LHP roots, q RHP roots and r roots on imaginary axis of s-plane. Then, (a) p =q = r =2 (b) p =3 ,q=1 , r=2 (c) p=3 ,q= 0, r=2 (d) p=3, q=2, r = 0 Q9. Consider the system with constant transportation lag T as depicted below. The parameters a and b are positive real constants. + 1 k R(s) e–sT C(s) s(1+bs) (1+as) – The correct region ‘R’ of stability on T-k plane is (a)

k

(b)

a+b ab

k ab a+b R

R

T

T

Control Systems

118 k

( c) ab a+b

(d) R

Mechasoft k

a+b ab

R

T T Q10. A unity feedback control system is described by characteristic equation 2 s +(k–1) s + 3k–2=0 Which one of the following statements is NOT correct ? (a) The system is stable for k >1. (b) The system oscillates at 1/2π Hz for k=1. ( c) The unit step response will be critically damped for k=0.675 and k = 13.325. (d) The unit step response will be overdamped for k >13.325. 2ö æ Kçs + ÷ è 3ø ; K > 0. Q11. A unity feedback system has open loop transfer function G(s) = 2 s (s + 6) The value of K for which all the characteristic roots are equal, is_______________. Q12. The dynamics of a system with input r(t) is described by state model é x&1 ù é 0 1 ù é x1 ù é x1 (0) ù é0 ù éa ù ê x& ú = ê –2 –3ú ê x ú + ê b ú r ; ê x (0) ú = ê0 ú and r(t) = u(t);u(t)is unit step function. û ë 2û ë û ë 2û ë ë 2 û ë û é x1 (¥) ù é1ù The system is designed to exhibit ê ú = ê ú . The values of a and b are respectively ë x2 (¥) û ë1û (a) –1 and 1 (b) 0.75 and –2.5 (c) –2 and3 (d) –1 and 5 5

Q13. Consider the Routh array constructed below. Notice that the s row was originally all zeros. The original polynomial has p roots in the LHP , q roots in the RHP and r roots on jω axis. Then, (a) p = 1, q = 4, r = 2 (b) p = 1, q = 2, r = 4 (c) p = 2, q = 1, r = 4 (d) p = 3, q = 2, r = 2

7

s s6 5 s 4 s s3 2 s 1 s 0 s

1 1 3 1 7 –15 –4 –21

2 2 4 –1 8 –21

–1 –1 –1 –3

–2 –2

Control Systems

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119

Q14. A system and the pattern of its closed loop poles are given below. The value of k is ______________. jω

Input

k s2

+ –

s2

× ×

+ +

1 s

σ

×

Output

1 s+3

Q15. The transfer function T(s) =

k1s + k 2 is to have only two poles on jω axis. s + k1 s 3 + s 2 + k 2 s + 1 4

The constraints on k1 and k2 is (a) k1 = k2 (b) k12 + k22 = k1k2

(d) k 1 + k 2 =

(c) k1k2 = 1

k 1k 2

Q16. Consider the system with input r(t) =u(t); u(t) is unit step function as shown below, where Ske(∞) e(∞) represents sensitivity of steady state error, e(∞) to changes in k. For nominal value of k = 10, |Sk | is (a) 9/17 (b) 7/8 ( c)10/17

Input r(t)

+ –

k(s + 7) s + 2s + 10

Output y(t)

2

(d) 1/8 Q17. What is range of k for which the system shown below, is stable? (a) k > –1 + In k(s2– 2s + 4) (b) k < 1 – (c) k > – (1 / 2) (d) – (1 / 2) < k < 1

Out

1 s + 2s + 2 2

k(s + 2)

. Q18. A unity feedback system has forward path transfer function G(s) = 2 s - 4s + 13 The value of k, in the nearest integer , at break in point of root locus, is ______________. Q19. Consider the system shown below. The break in point of root locus for 0 ≤ p ≤ ∞, lies at s = –λ. In The value of λ ,rounded off to two decimal places, is _____________.

+ –

10 (s + 2)(s + p)

Out

120

Control Systems

Q20. A system with transfer function G(s) =

10 has phase margin ϕm1 and e–0.1s G(s) has phase s(s + 1)

Mechasoft

margin ϕm2. The value of ϕm1 – ϕm2 , rounded off to two decimal places ,in degrees, is _____________. Q21. A system is represented by state - space model

é -2 x& = ê ë - K1

1

ù é0 ù x + ê ú r and y = [1 0]x ú -(K 2 + 1) û ë1 û

where r is the input and y is the output. The system is to exhibit settling time ts = 0.5 sec and damping ratio ξ = 0.5. The value of K1 + K2 is ______. Q22. The signal flow graph of a system together with chosen state variables x1 and x2, is given below. Which one of the following is true if the state model is UNCONTROLLABLE? b1

(a) b1 = – b2 (b) b1 = b2

u

(input) (c) b1 = –2b2

s–1

s–1

1 x2

b2

x1

1

y(output)

–5

–3 –1

(d) b1 = 2b2

é0 1 ù ú and x1 (0) = x 2 (0) = 1 ë0 0 û The value of x12 (t) + x 22 (t) at t = 1,is ________ .

Q23. A system is represented by x& = Ax where A = ê

Q24. A multi loop feedback control system with input r and output y, and labelled state variables x1 and x2 is shown below. In the state model 1 x1 é l11 l12 ù s x& = Ax + Br, A = ê + ú y r – – ël 21 l 22 û + the value of λ11 + λ12 +λ21 +λ22 is_____________.

1 2

x2

1 s

Control Systems

Mechasoft

121 Q25. Consider the block diagram of feedback control system shown below, where the transfer functions are represented by frequency response curves. Assume that the systems have minimum phase transfer functions. Im

input R(s)

Polar plot G1(jω)

+ –

G1(s)

+ –

Re

10

G3(s)

dB

ω increasing

Bode plot G2(jω)

dB –6dB/oct

output Y(s)

G2(s)

log magnitude Vs phase plot G3(jω)

9.54

ω=1 ο

–360 –270ο –180ο 0

1

0.6

ϕ

–90ο

ω(log scale)

–12dB/oct The damping ratio ξ , rounded off to one decimal place , of the system is ___________.

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End of Test Drill II

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Answer Key Q1 Q8 Q15 Q22

M

(b) (d) Q2 t (a) (a) Q9 f o s (b) (b) Q16 ha (5) (b) Q23 ec

Q3 Q10 Q17 Q24

(a) (c) (d) (–1)

Q4 Q11 Q18 Q25

M

(a) Q6 Q5 (b) ft (d) Q13 (12) Q12 o s (18)haQ19 (5.16) Q20 c e(0.5)

(c) Q7 (1) t (c) Q14 f(3) o s (17.65) Q21 (241) a

h

ec M