feedback control system 5th chapter

5000Solutions Manual: Chapter 5 7th Edition Feedback Control of Dynamic Systems Gene F. Franklin J. David Powell Abbas Emami-Naeini . Assisted by: H.K. Aghajan H. Al-Rahmani P. Coulot P. Dankoski S. Everett R. Fuller T. Iwata V. Jones F. Safai L. Kobayashi H-T. Lee E. Thuriyasena M. Matsuoka J.K. Lee © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Chapter 5 The Root-Locus Design Method Problems and solutions for Section 5.1 1. Set up the listed characteristic equations in the form suited to Evans’s root-locus method. Give L(s); a(s); and b(s) and the parameter K in terms of the original parameters in each case. Be sure to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than that of a(s). (a) s + (1= ) = 0 versus parameter (b) s2 + cs + c + 1 = 0 versus parameter c (c) (s + c)3 + A(T s + 1) = 0 i. versus parameter A, ii. versus parameter T , iii. versus the parameter c, if possible. Say why you can or can not. Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all? kI kD s c(s) + G(s) = 0: Assume that G(s) = A , where s s+1 d(s) c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s). (Note: The …rst printing of the 7th edition kI had an error in the equation above where the term above was s incorrectly stated to be kI (s):) (d) 1 + kp + i. versus kp ii. versus kI iii. versus kD 5001 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 5002 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD iv. versus Solution: (a) K = 1= ; a = s; b=1 (b) K = c; a = s2 + 1; b=s+1 (c) (d) i. K = AT ; a = (s + c)3 ; b = s + 1=T ii. K = AT ; a = (s + c)3 + A; b=s iii. The parameter c enters the equation in a nonlinear way and a standard root locus does not apply. However, using a polynomial solver, the roots can be plotted versus c: i. K = kp A; a = s(s + 1= )d(s) + AkI (s + 1= )c(s) + kD b = s(s + 1= )c(s) ii. K = AkI ; a = s(s + 1= )d(s) + Akp s(s + 1= )c(s) + s2 Ac(s); kD s2 Ac(s); b = (s + 1= )c(s) AkD iii. K = ; a = s(s + 1= )d(s) + Akp s(s + 1= )c(s) + AkI (s + 1= )c(s); b = s2 c(s) iv. K = 1= ; a = s2 d(s) + Akp s2 c(s) + AkI sc(s); b = sd(s) + Akp sc(s) + AkI c(s) + AkD s2 c(s) © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. a rough evaluation of arrival and departure angles for complex poles and zeros. electronic. Upper Saddle River. Solution: We had to make up some numbers to do it on Matlab. write to: Rights and Permissions Department. but the idea here is just get the basic rules right. b(s) = (s + 1) Breakin(s): -2 (d) a(s) = s2 + s.366 Breakaway(s): -0.5 Angles of asymptotes: 45 .44 without the aid of a computer. b(s) = s + 1 Center of asymptotes: -0. or likewise. Each pole-zero map is from a characteristic equation of the form b(s) = 0. For information regarding permission(s).5003 Problems and solutions for Section 5. photocopying. storage in a retrieval system.. NJ. 180 Angle of departure: 56:3 (f) a(s) = s5 + 3s4 + s3 5s2 0:5. b(s) = s + 2 Breakin(s): -3. (a) a(s) = s2 + s. Inc. 5. mechanical. 1+K a(s) where the roots of the numerator b(s) are shown as small circles o and the roots of the denominator a(s) are shown as 0 s on the s-plane. Pearson Education.. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. 5. Roughly sketch the root loci for the pole-zero maps as shown in Fig.44(c) there are two poles at the origin. NJ 07458. . 135 Angle of departure: 115:8 at s = 2:01 + 1:01j 70:5 at s = 0:01 + 0:31j Breakin(s): 0.342 (c) a(s) = s2 . Inc.05 Breakaway(s): 0:652 © 2015 Pearson Education. b(s) = s + 1 Angle of departure: 137:9 Breakin(s): -2. Upper Saddle River.2 2.586 (b) a(s) = s2 + 0:2s + 1. or transmission in any form or by any means. and the loci for positive values of the parameter K. recording. b(s) = s2 + 5s + 6 Breakin(s): -2. Note that in Fig.414 Breakaway(s): -0. so the results partly depend on what was dreamed up. Show your estimates of the center and angles of the asymptotes.634 (e) a(s) = s3 + 3s2 + 4s 8 Center of asymptotes: -1 Angles of asymptotes: 60 . write to: Rights and Permissions Department. or likewise. For information regarding permission(s).5 -2 0 -1 -4 2 -2 Real Axis 2 4 Imaginary Axis Imaginary Axis 0 Real Axis 10 5 0 -5 -10 -10 2 1 Imaginary Axis Imaginary Axis 2 0 2 0 -2 -5 0 Real Axis 5 -4 -4 -2 0 Real Axis 2 4 © 2015 Pearson Education. recording.. THE ROOT-LOCUS DESIGN METHOD Root loci for Problem 5. NJ. Upper Saddle River. Upper Saddle River. or transmission in any form or by any means. storage in a retrieval system. electronic. All rights reserved. mechanical. Inc.5004 CHAPTER 5. Pearson Education.5 0 -0. Inc. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.2 2 Imaginary Axis Imaginary Axis 2 1 0 -1 -2 -10 1 0 -1 -5 0 -2 -6 5 -4 -2 Real Axis Real Axis 1 0 -1 -2 -4 0. .. photocopying. NJ 07458. Solution: (a) The real axis segment is (b) = 6=4 = 1:5. > 5.5005 3. Pearson Education. Inc. 135 (c) The plot is shown below. or likewise.. Root Locus 6 Imaginary Axis 4 2 0 -2 -4 -6 -8 -6 -4 -2 Real Axis 0 2 4 Solution for Problem 5.. (d) Verify your sketch with a Matlab plot. mechanical.3 © 2015 Pearson Education. (b) Sketch the asymptotes of the locus for K ! 1. Upper Saddle River. . or transmission in any form or by any means. (c) Sketch the locus. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. All rights reserved. electronic. Upper Saddle River. NJ 07458. i = 1> 45 . NJ. write to: Rights and Permissions Department. photocopying. For information regarding permission(s). recording. storage in a retrieval system. Inc. For the characteristic equation 1+ K = 0. s2 (s + 1)(s + 5) (a) Draw the real-axis segments of the corresponding root locus. NJ 07458. Inc. For information regarding permission(s). photocopying. NJ.. i = 45 . All rights reserved. . After completing each hand sketch. Real poles and zeros. or likewise. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. (b) = 4:33. Inc.. i = 90 = 135 60 . Sketch the root locus with respect to K for the equation 1+KL(s) = 0 and the listed choices for L(s). write to: Rights and Permissions Department. i (c) = 5. storage in a retrieval system. i = 90 (d) = 4. recording. THE ROOT-LOCUS DESIGN METHOD 4.5006 CHAPTER 5. (a) = 4. verify your results using Matlab. Pearson Education. electronic. 180 © 2015 Pearson Education. mechanical. or transmission in any form or by any means. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. Turn in your hand sketches and the Matlab results on the same scales. Upper Saddle River. Upper Saddle River. (a) L(s) = 2 s(s + 1)(s + 5)(s + 10) (b) L(s) = (s + 3) s(s + 1)(s + 5)(s + 10) (c) L(s) = (s + 2)(s + 4) s(s + 1)(s + 5)(s + 10) (d) L(s) = (s + 2)(s + 6) s(s + 1)(s + 5)(s + 10) Solution: All the root locus plots are displayed at the end of the solution set for this problem. photocopying. mechanical.5007 b 15 10 10 5 5 Imaginary Axis Imaginary Axis a 15 0 0 -5 -5 -10 -10 -15 -15 -10 -5 Real Axis 0 -15 -15 5 -10 15 10 10 5 5 0 -5 -10 -10 -10 -5 Real Axis 5 0 5 0 -5 -15 -15 0 d 15 Imaginary Axis Imaginary Axis c -5 Real Axis 0 5 -15 -15 -10 -5 Real Axis Root loci for Problem 5. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. For information regarding permission(s). Inc. NJ. recording. Inc. storage in a retrieval system. write to: Rights and Permissions Department. electronic. or transmission in any form or by any means. Upper Saddle River. NJ 07458. or likewise. Upper Saddle River.. .. All rights reserved.4 © 2015 Pearson Education. Pearson Education. d = 180 at s = 2j. THE ROOT-LOCUS DESIGN METHOD 5. After completing each hand sketch. verify your results using Matlab. (a) L(s) = 1 s2 + 3s + 10 (b) L(s) = 1 s(s2 + 3s + 10) (c) L(s) = (s2 + 2s + 8) s(s2 + 2s + 10) (d) L(s) = (s2 + 2s + 12) s(s2 + 2s + 10) (e) L(s) = (s2 + 1) s(s2 + 4) (f) L(s) = (s2 + 4) s(s2 + 1) Solution: All the root locus plots are displayed at the end of the solution set for this problem. i = 180 . Pearson Education. storage in a retrieval system. photocopying.5008 CHAPTER 5.. All rights reserved. Upper Saddle River.. Sketch the root locus with respect to K for the equation 1+KL(s) = 0 and the listed choices for L(s). = 0. (f) = 0. Inc. d = 0 at s = j. 1:5 + 2:78j a (d) a = 18:4 at s = 1 + 3:32j d (e) = 0. (b) = 1. (c) i i = = 90 . 180 . Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. . Turn in your hand sketches and the Matlab results on the same scales. = 16:8 at s = 1 + 3j. electronic. d = 1:5 + 2:78j 28:3 at s = = 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. (a) = 1:5. i = 180 . d = 161:6 at s = = 200:7 at s = 1 + 2:65j 1 + 3j. Complex poles and zeros. Inc. write to: Rights and Permissions Department. NJ. recording. or transmission in any form or by any means. NJ 07458. i = 180 . mechanical. i = 180 . Upper Saddle River. For information regarding permission(s). a a = 180 at s = j = 0 at s = 2j © 2015 Pearson Education. or likewise. d = 90 at s = 60 . Upper Saddle River. mechanical. write to: Rights and Permissions Department..5 4 Imaginary Axis Imaginary Axis -1 Real Axis f 4 2 0 -2 -4 -2 10 -1 0 1 Real Axis 2 0 -2 -4 -1. electronic.5 -1 -0. or transmission in any form or by any means.5 Real Axis Root loci for Problem 5. NJ. or likewise. . Inc.5009 a b 20 Imaginary Axis Imaginary Axis 20 10 0 -10 -20 -3 -2 -1 Real Axis 0 10 0 -10 -20 -20 1 -10 0 Real Axis c d 4 Imaginary Axis Imaginary Axis 4 2 0 -2 -4 -3 -2 -1 Real Axis 0 2 0 -2 -4 -3 1 -2 e 0 1 0 0. photocopying. Inc. Upper Saddle River. All rights reserved.. storage in a retrieval system. Pearson Education.5 © 2015 Pearson Education. recording. NJ 07458. For information regarding permission(s). This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. (a) = 2:67. NJ. mechanical. Upper Saddle River. i = 36 . (a) L(s) = 1 s2 (s + 8) (b) L(s) = 1 s3 (s + 8) (c) L(s) = 1 s4 (s + 8) (d) L(s) = (s + 3) s2 (s + 8) (e) L(s) = (s + 3) s3 (s + 4) (f) L(s) = (s + 1)2 s3 (s + 4) (g) L(s) = (s + 1)2 s3 (s + 10)2 Solution: All the root locus plots are displayed at the end of the solution set for this problem. Multiple poles at the origin. verify your results using Matlab. i = 90 (g) = 6. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. THE ROOT-LOCUS DESIGN METHOD 6. Pearson Education.5010 CHAPTER 5. 180 © 2015 Pearson Education. For information regarding permission(s). write to: Rights and Permissions Department. storage in a retrieval system. All rights reserved.. i = 90 (e) = 0:33. (c) = 1:6. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. = 135 108 . Inc. . 180 60 . photocopying.. (d) = 2:5. Inc. Turn in your hand sketches and the Matlab results on the same scales. i = 60 . electronic. (b) = 2. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). (f) = 3. 180 i i = = i 60 . or transmission in any form or by any means. NJ 07458. Upper Saddle River. After completing each hand sketch. 180 45 . recording. or likewise. 5 0 -0. or transmission in any form or by any means. For information regarding permission(s). This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5011 b Imaginary Axis Imaginary Axis a 20 10 0 -10 -20 -15 -10 -5 0 Real Axis 5 5 0 -5 -10 10 -10 -5 0 d Imaginary Axis Imaginary Axis 5 Real Axis c 10 5 0 50 0 -5 -15 -10 -5 Real Axis 0 5 -50 -10 10 -5 0 f Imaginary Axis e Imaginary Axis 5 Real Axis 0.. NJ. or likewise. Upper Saddle River. recording. Inc. All rights reserved. electronic..6 © 2015 Pearson Education. Pearson Education.5 10 5 0 -5 -10 -1 -5 -4 -3 -2 -1 Real Axis 0 -5 Real Axis 0 1 2 Imaginary Axis g -5 -4 -3 -2 Real Axis -1 0 5 0 -5 -10 -10 5 Solution for Problem 5. mechanical. storage in a retrieval system. write to: Rights and Permissions Department. Upper Saddle River. NJ 07458. . Inc. photocopying. THE ROOT-LOCUS DESIGN METHOD 7.. All rights reserved. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. i = 90 . (d) 60 . (a) = 3. d = 100:2 at s = = 99:8 at s = 2 + 8j 3 + 4j 3 + 4j 2 + 9j. After completing each hand sketch. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. i = 60 . Mixed real and complex poles. NJ 07458. i = (b) = 3:25. Inc. Inc. d = 135 . recording. photocopying. Upper Saddle River. 24:8 at s = d d = = 1+j 103:5 at s = 13:5 at s = = 3:5.. a (e) = 1:5. write to: Rights and Permissions Department. storage in a retrieval system. verify your results using Matlab. Pearson Education. or transmission in any form or by any means.5012 CHAPTER 5. For information regarding permission(s). . electronic. mechanical. Upper Saddle River. 180 . a = 71:6 at s = 1+j © 2015 Pearson Education. Turn in your hand sketches and the Matlab results on the same scales. (a) L(s) = (s + 3) s(s + 10)(s2 + 2s + 2) (b) L(s) = (s + 3) s2 (s + 10)(s2 + 6s + 25) (c) L(s) = s2 (s (s + 3)2 + 10)(s2 + 6s + 25) (d) L(s) = (s + 3)(s2 + 4s + 68) s2 (s + 10)(s2 + 4s + 85) (e) L(s) = [(s + 1)2 + 1] s2 (s + 2)(s + 3) Solution: All the plots are attached at the end of the solution set. i = 45 . i = 90 . 180 . NJ. or likewise. (c) = 3:33. Inc. or transmission in any form or by any means. or likewise. mechanical.. storage in a retrieval system. Pearson Education. NJ. All rights reserved. NJ 07458. For information regarding permission(s).. write to: Rights and Permissions Department. Upper Saddle River. .7 © 2015 Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5013 a b c 10 10 3 8 8 6 6 1 0 4 Imaginary Axis Imaginary Axis Imaginary Axis 2 2 0 -2 -1 -8 -8 -10 -10 -10 -5 Real Axis 0 0 -2 -6 -6 -3 2 -4 -4 -2 4 -15 d -10 -5 Real Axis 0 5 -15 -10 -5 Real Axis 0 e 5 4 15 3 10 Imaginary Axis Imaginary Axis 2 5 0 1 0 -1 -5 -2 -10 -3 -4 -15 -5 -15 -10 -5 Real Axis 0 5 -3 -2 -1 Real Axis 0 Solution for Problem 5. electronic. Inc. photocopying. Upper Saddle River. recording. i (c) = 1. Upper Saddle River. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. i = (b) = 2:67. (a) L(s) = s+2 1 . NJ 07458. After completing each hand sketch. 180 = 180 = 12. Turn in your hand sketches and the Matlab results on the same scales. All rights reserved. i = 60 . From the locus. Upper Saddle River. For information regarding permission(s). (b) L(s) = (c) L(s) = s 1 s2 s2 + 2s + 1 : What is the largest value that can s(s + 20)2 (s2 2s + 2) be obtained for the damping ratio of the stable complex roots on this locus? (d) L(s) = (e) L(s) = (f) L(s) = (s + 2) . 1 s+2 .5014 CHAPTER 5. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. or likewise. mechanical. storage in a retrieval system. or transmission in any form or by any means. 180 .. s(s 1)(s + 6)2 1 1)[(s + 2)2 + 3] (s Solution: (a) = 4. Pearson Education. (d) i 90 = 60 . (e) = 3. NJ. i = 60 . write to: Rights and Permissions Department. . recording. Inc. photocopying. THE ROOT-LOCUS DESIGN METHOD 8. 180 (f) = 1. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). 180 . the maximum damping is 0:31 when K 2600. verify your results using Matlab.. RHP and zeros. Inc. d = 92:7 at s = 1 + j The maximum damping ratio is obtained at a point at the smallest angle o¤ the horizontal. i = 60 . the magnetic levitation system with inte2 s(s + 10) (s 1) gral control and lead compensation. the model for a case of magnetic levitation s + 10 s2 1 with lead compensation. d = 60:0 at s = 2 + 1:73j © 2015 Pearson Education. electronic. or likewise. Inc. storage in a retrieval system..5 -10 -1 -1 -5 0 Real Axis 10 10 plot f Imag Axis 0 -5 -10 2 Imag Axis 5 5 2 4 plot e 15 Imag Axis 0 1 Real Axis 0 -5 0 -2 -15 -20 -5 0 Real Axis 5 -10 -8 -6 -4 -2 0 Real Axis 2 -4 -2 0 Real Axis 2 Solution for Problem 5.5 Imag Axis 5 Imag Axis Imag Axis 5 -10 plot c plot b 0 -5 -10 -10 -5 0 Real Axis plot d 0 -0.5015 10 10 1 plot a 0 -5 0. Pearson Education. For information regarding permission(s). All rights reserved. NJ. recording. . photocopying.8 © 2015 Pearson Education. Upper Saddle River. write to: Rights and Permissions Department. NJ 07458.. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Inc. Upper Saddle River. electronic. mechanical. or transmission in any form or by any means. or likewise.4 -1. and b(s): Sketch the root locus with respect to the parameter . Upper Saddle River. estimate the closed-loop pole locations.5 -1 0.5 0. a(s). Inc. photocopying. mechanical.8 0. and sketch the corresponding step responses when = 0.45: Control system for Problem 5.6 -0..5016 CHAPTER 5. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.45 in root locus form with respect to the parameter . 5. and identify the corresponding L(s). 5 a l p h a = 2 0. Inc.5 Amplitude Imaginary Axis 1 0 0.4 2 a l p h a = 0 1. Upper Saddle River. Put the characteristic equation of the system shown in Fig.5 1. For information regarding permission(s). recording. NJ. NJ 07458.2 -2 -2.5 1 a l p h a = 0 . or transmission in any form or by any means. All rights reserved. THE ROOT-LOCUS DESIGN METHOD 9.9 Solution: The characteristic equation is s2 +2s+5+5 s = 0 and L(s) = the root locus and step responses are plotted below. and 2.5 -6 -5 -4 -3 -2 Real Axis -1 0 1 2 0 0 5 10 Time(sec) 15 © 2015 Pearson Education. write to: Rights and Permissions Department. storage in a retrieval system. Pearson Education. Figure 5.. electronic. Use Matlab to check the accuracy of your approximate step responses. 0:5. s2 s : + 2s + 5 root locus StepResponse 2. 20 .2 1. write to: Rights and Permissions Department. Verify that a multiple root occurs at a complex value of s for some value of a in this range. . 1 curve o¤ toward the asymptotes. or transmission in any form or by any means.. storage in a retrieval system. Inc. NJ 07458. mechanical. photocopying. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. paying particular attention to the region between 2:5 and 3:5.5017 10. the locus branch from 0. recording..5 Imaginary Axis a=3 10 Imaginary Axis Imaginary Axis a=0 10 -8 -1 2 -1 0 -8 R e a l A x is -6 -4 R e a l A x is -2 0 2 4 -1 0 -1 4 -1 2 -1 0 -8 -6 -4 R e a l A x is Solution for Problem 5. For information regarding permission(s). Use the Matlab function rltool to study the behavior of the root locus of 1 + KL(s) for (s + a) L(s) = s(s + 1)(s2 + 8s + 52) as the parameter a is varied from 0 to 10. Solution: For small values of . At about = 3:11 these loci touch corresponding to complex multiple roots. a = 2 . Inc. For large values of the branches from the complex roots break into the real axis and those from 0. Upper Saddle River. Upper Saddle River. or likewise.10 © 2015 Pearson Education. 1 makes a circular path around the zero and the branches from the complex roots curve o¤ toward the asymptotes.5 8 8 8 6 6 6 4 4 4 2 2 2 0 Imaginary Axis 10 0 0 -2 -2 -2 -4 -4 -4 -6 -6 -6 -8 -8 -1 0 -1 4 -1 2 -1 0 -8 -6 -4 -2 0 2 -1 0 -1 4 4 -8 -1 2 -1 0 -8 -6 R e a l A x is -4 -2 0 2 -1 0 -1 4 4 8 6 6 4 4 4 2 2 2 0 Imaginary Axis 8 6 Imaginary Axis 8 0 -2 -2 -4 -4 -4 -6 -6 -6 -8 -8 -6 -4 -2 0 2 4 -1 0 -1 4 -4 -2 0 2 4 -2 0 2 4 0 -2 -8 -6 a=10 10 -1 0 -8 a=5 10 -1 2 -1 0 R e a l A x is 10 -1 0 -1 4 -1 2 R e a l A x is a = 3 . electronic. All rights reserved. Pearson Education. NJ. .. or likewise. recording.11 Solution: (a) The system is stable for 0 K 478:234. photocopying. For information regarding permission(s).5018 CHAPTER 5. or transmission in any form or by any means. Pearson Education. write to: Rights and Permissions Department. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. The root locus of the system and the location of the roots at the crossover points are shown in the left plot. electronic. Inc. © 2015 Pearson Education. storage in a retrieval system. NJ 07458.46: Feedback systems for Problem 5. Upper Saddle River. Figure 5.46 are stable. mechanical. NJ. All rights reserved. Use Routh’s criterion to …nd the range of the gain K for which the systems in Fig. . THE ROOT-LOCUS DESIGN METHOD 11. and use the root locus to con…rm your calculations. Upper Saddle River. (b) There is a pole in the right hand plane thus the system is unstable for all values of K as shown in the right plot. 5. Inc. electronic..5 -2 -1. or likewise. photocopying. storage in a retrieval system.5 Real Axis Solution for Problem 5. NJ 07458.5 Real Axis 0. . NJ.5019 Root Locus plot a Imag Axis Imag Axis 5 Root Locus 5 0 plot b 0 -5 -5 -10 -5 Real Axis 0 -3 -2.5 -2 -1 -0. For information regarding permission(s).5 plot d 2 0 -2 -1 -1. Inc. or transmission in any form or by any means. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5 -6 -4 Real Axis -2 0 -4 -3 -2. All rights reserved. recording.11 © 2015 Pearson Education. write to: Rights and Permissions Department. Upper Saddle River.5 0 -0.5 1 -1.. Upper Saddle River. mechanical.5 0 Root Locus 4 plot c Imag Axis Imag Axis Root Locus 1. Pearson Education. Inc.5 0 -1 -0. THE ROOT-LOCUS DESIGN METHOD 12. electronic.0 8 5 -1 0 .2 6 0 . s (s + 5) and determine the value of the root-locus gain for which the complex conjugate poles have the maximum damping ratio.1 9 0 . All rights reserved. R oot Loc us 15 0 . and use [K]= rloc…nd(sys) to pick the gain for the maximum damping. .3 8 0 .5020 CHAPTER 5.275 when K 10:7. For information regarding permission(s). or transmission in any form or by any means.8 -5 6 8 -10 0 .1 3 -2 0 . Inc. What is the approximate value of the damping? Solution: Plot the system on Matlab using rlocus(sys). or likewise. recording.1 3 0 .5 2 10 12 0 . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.. Inc.275 damping marked © 2015 Pearson Education. storage in a retrieval system.0 4 14 0 1 R e a l Ax is Root locus with 0. We …nd that the maximum damping. NJ 07458.8 4 Imaginar y Axis 2 0 2 4 0 . Upper Saddle River. Sketch the root locus for the characteristic equation of the system for which (s + 2) L(s) = 2 .0 4 14 12 10 10 0 . NJ. Pearson Education.2 6 0 . photocopying. mechanical.3 8 -15 -6 -5 -4 0 . Upper Saddle River.0 8 5 0 . write to: Rights and Permissions Department.5 2 8 6 5 0 ..1 9 -3 0 . = 0. .4 -4 -6 -8 -20 0.5 5 2.84 -15 0. mechanical.74 0. For information regarding permission(s).74 0.99 Amp litu d e Imag Axis StepResponse 1.22 2 200 17. All rights reserved.965 4 1 0. Pearson Education.6 0. NJ 07458.8 0.92 6 RootLocus 0.5 15 12. photocopying.5 but none that will make the poles from the resonance have that much damping.4 0. or transmission in any form or by any means.5 10 7. 0. .6 1.5 -2 0.5021 13. NJ. Figure 5. the gain is about 35.22 -10 -5 RealAxis 0 0 0 1 2 3 Time(sec) 4 5 6 © 2015 Pearson Education. write to: Rights and Permissions Department.92 0.2 0. Inc.47: Feedback system for Problem 5.965 0.42 0. recording.6 0.13 (a) Find the locus of closed-loop roots with respect to K. storage in a retrieval system. (d) Use Matlab to plot the response of the resulting design to a reference step. (c) Using rloc…nd. 5. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Upper Saddle River. For the system in Fig.42 0. Solution: (a) The locus is plotted below (b) There is a K which will make the ’dominant’ poles have damping 0. (d) The step response shows the basic form of a well damped response with the vibration of the response element added. (b) Is there a value of K that will cause all roots to have a damping ratio greater than 0:5? (c) Find the values of K that yield closed-loop poles with the damping ratio = 0:707.84 8 0.47. Inc. Upper Saddle River.2 0.. or likewise.99 0. All rights reserved. Inc. Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.13 © 2015 Pearson Education. . write to: Rights and Permissions Department. NJ. Upper Saddle River.5022 CHAPTER 5. recording. Inc.. or likewise. For information regarding permission(s).. Upper Saddle River. or transmission in any form or by any means. electronic. THE ROOT-LOCUS DESIGN METHOD Root locus and step response for Problem 5. mechanical. NJ 07458. photocopying. storage in a retrieval system. For the feedback system shown in Fig.48. Upper Saddle River. photocopying. NJ..14 Solution: Use block diagram reduction to …nd the characteristic equation of the closed loop system. Pearson Education. electronic.5023 14. NJ 07458. recording. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Inc.. 5. or likewise. All rights reserved. …nd the value of the gain K that results in dominant closed-loop poles with a damping ratio = 0:5. Inc. write to: Rights and Permissions Department. Upper Saddle River. then divide that up into terms with and without K to 10s …nd the root locus form. For information regarding permission(s). Figure 5. .48: Feedback system for Problem 5. where L(s) = 2 : Plugging into Matlab s + s + 10 and using rloc…nd produces the required gain to be K = 0:22:The locus is © 2015 Pearson Education. storage in a retrieval system. or transmission in any form or by any means. mechanical. or likewise. Inc. NJ. All rights reserved. Inc.3 -1 0.84 -4 -6 -5 -4 0. Upper Saddle River.92 -3 0.84 0.3 0.5024 CHAPTER 5.5 damping marked © 2015 Pearson Education.58 -3 0. recording.14 0 1 R eal Axis Root locus with 0. THE ROOT-LOCUS DESIGN METHOD R oot Locus 4 0. For information regarding permission(s). This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. storage in a retrieval system. write to: Rights and Permissions Department. photocopying..92 2 0. Pearson Education. NJ 07458.14 3 0.58 0.44 0. Upper Saddle River.98 Imaginary Axis 1 6 0 5 4 3 2 1 -1 0. or transmission in any form or by any means.72 0. . electronic. mechanical..72 0.98 -2 0.44 -2 0. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Inc.15(b)(c) © 2015 Pearson Education.. storage in a retrieval system.3 15.. For information regarding permission(s). All rights reserved. Use axes -4 x 4. Upper Saddle River. Solution: (a) d a = = 180 25:26 90 + 266:5 + 92:6 = 63:83 . Upper Saddle River. Pearson Education. or transmission in any form or by any means. (b) Sketch the root locus for this system for parameter K = 9:8kp . 90 + 86:5 + 69:9 + 87:4 180 = 26:11 (b) (c) Root Locus 3 2 Imaginary Axis 1 0 -1 -2 -3 -4 -3 -2 -1 0 1 2 3 4 Real A xis Problem 5. . Inc. Use the command axis([-4 4 -3 3]) to get the right scales. photocopying.5025 Problems and solutions for Section 5. NJ. NJ 07458. (a) Compute the departure and arrival angles at the complex poles and zeros. Let Dc (s) = kp at …rst. or likewise. A simpli…ed model of the longitudinal motion of a certain helicopter near hover has the transfer function G(s) = 9:8(s2 0:5s + 6:3) : (s + 0:66)(s2 0:24s + 0:15) and the characteristic equation 1 + Dc (s)G(s) = 0. mechanical. (d) Suggest a practical (at least as many poles as zeros) alternative compensation Dc (s) which will at least result in a stable system. write to: Rights and Permissions Department. recording. (c) Verify your answer using Matlab. 3 y 3. .5026 CHAPTER 5. Let (s + :66)(s + :33) : Dc (s) = (s + 5)2 Root Loc us 3 2 Imaginary Axis 1 0 -1 -2 -3 -4 -3 -2 -1 0 1 2 3 4 Real A xis Problem 5. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. THE ROOT-LOCUS DESIGN METHOD (d) For this problem a double lead is needed to bring the roots into the left half-plane. Upper Saddle River. or likewise. recording. storage in a retrieval system. The plot shows the rootlocus for control for. NJ. photocopying. mechanical. or transmission in any form or by any means. Inc. Upper Saddle River. . All rights reserved. NJ 07458. electronic.15(d) © 2015 Pearson Education. Inc. For information regarding permission(s). write to: Rights and Permissions Department. Pearson Education.. NJ 07458. 5.49. Is there anything special about this (c) Repeat part (a) for …xed K1 = 2 with the parameter K = from 0 to 1. All rights reserved. Give the corresponding L(s). storage in a retrieval system. electronic.. photocopying.. NJ. mechanical. a(s). or likewise. For information regarding permission(s). and b(s): (b) Repeat part (a) with value? = 5. recording. Pearson Education. © 2015 Pearson Education. . Upper Saddle River. (a) Substituting = 2 and divide the equation above up into terms with and without K1 to …nd Evans form: 1 + K1 2(s + 5) =0 s(s + 2)(s + 11) ) L(s) = 2(s + 5) s(s + 2)(s + 11) (b) Substituting = 5 and rewrite the equation in Evans form with respect to K1 : 1 + K1 2(s + 5) =0 s(s + 5)(s + 11) ) We have a pole-zero cancellation at s = L(s) = 2 s(s + 11) 5. Inc.16 16. write to: Rights and Permissions Department.49: Control system for Problem 5. varying Solution: Use block diagram reduction to …nd the characteristic equation of the closed-loop system: 1+ 10K1 s(s + ) 0:1 + 0:2s + 1 s(s + 10)(s + ) K1 or s(s + )(s + 11) + 2K1 (s + 5) = 0 =0 The root locus for each part is attached at the end. Inc. or transmission in any form or by any means. Upper Saddle River.5027 Figure 5. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. (a) Plot the root locus of the characteristic equation as the parameter K1 is varied from 0 to 1 with = 2. For the system given in Fig. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction..5028 CHAPTER 5. electronic. or likewise.5 0 -0. write to: Rights and Permissions Department.5 -20 -1 -30 -1. Upper Saddle River.5 1 Imaginary Axis Imaginary Axis 20 10 0 -10 0. All rights reserved. Pearson Education.. Inc.5 0 -0. For information regarding permission(s).5 -1 -1. recording. Inc. NJ 07458. THE ROOT-LOCUS DESIGN METHOD (c) Substituting K1 = 2 and divide the characteristic equation up into terms with and without to …nd Evans form: 1+ s(s + 11) =0 s2 (s + 11) + 4(s + 5) ) L(s) = s(s + 11) s3 + 11s2 + 4s + 20 plot a plot b 40 2 30 1. NJ. .5 -40 -15 -10 -5 Real Axis 0 5 -2 -40 -30 -20 -10 Real Axis 0 10 20 plot c 2 1. Upper Saddle River. storage in a retrieval system.16 © 2015 Pearson Education.5 Imaginary Axis 1 0.5 -2 -30 -25 -20 -15 -10 Real Axis -5 0 5 10 Solution for Problem 5. photocopying. or transmission in any form or by any means. mechanical. Upper Saddle River. Inc. recording.17 © 2015 Pearson Education. NJ.. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. or likewise. For the system shown in Fig. 5. and b(s) and be sure to show with arrows the direction in which c increases on the locus.50: Control system for Problem 5. Inc.50. Upper Saddle River. write to: Rights and Permissions Department. Give L(s). All rights reserved. Solution: L(s) = a(s) s2 + 9 = + 144s b(s) s3 R oot Loc us 10 Imag Axis 5 0 -5 -10 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 R eal Ax is Solution for Problem 5. For information regarding permission(s).17 17. storage in a retrieval system. a(s). .. mechanical. NJ 07458.5029 Figure 5. or transmission in any form or by any means. photocopying. determine the characteristic equation and sketch the root locus of it with respect to positive values of the parameter c. Pearson Education. photocopying. storage in a retrieval system. (s + p)2 where z and p are real and z > p. NJ. THE ROOT-LOCUS DESIGN METHOD 18. mechanical. Then L(s) can be written as: L = = = (z p)ej z + p) j (z p)e (z p)2 (ej 1)2 1 ((z (z = p)ej p)( 4) 1 4(z ej =2 2 e 2j j =2 2 1 p) (sin( =2))2 Because z > p. Inc. electronic. Upper Saddle River. For information regarding permission(s). This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. write to: Rights and Permissions Department. All rights reserved. NJ 07458. and therefore © 2015 Pearson Education. . Show that the root locus for 1+KL(s) = 0 with respect to K is a circle centered at z with radius given by r = (z p): Hint: Assume s + z = rej and show that L(s) is real and negative for real under this assumption. or transmission in any form or by any means. Pearson Education. or likewise. this function is real and negative for real these points are on the locus. Upper Saddle River. recording.. Solution: Assume s + z = (z p)ej . Suppose you are given a system with the transfer function L(s) = (s + z) . Inc..5030 CHAPTER 5. Solution: Root Locus Root Locus 0. or likewise. NJ.19 © 2015 Pearson Education.5 0 Real Axis Solution for Problem 5.. The extreme cases occur when the pole is located at in…nity or when it is located at s = 2. or transmission in any form or by any means. electronic. Give values for p and sketch the three distinct types of loci. Pearson Education. write to: Rights and Permissions Department. Upper Saddle River. Inc. The loop transmission of a system has two poles at s = 1 and a zero at s = 2. mechanical.5 -100 -80 -60 -40 -20 0 -2 Real Axis -1.. photocopying.5 -2 -1. There is a third real-axis pole p located somewhere to the left of the zero. All rights reserved.5 20 pole is at negativ e inf inity pole is at s=-2 Imag Axis Imag Axis 10 0 0 -10 -20 -0. storage in a retrieval system. depending on the exact location of the third pole. . For information regarding permission(s). Inc.5 -1 -0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5 0 Real Axis Root Locus 5 Imag Axis pole is slightly to the lef t of zero 0 -5 -3 -2.5031 19. NJ 07458. Upper Saddle River. recording. Several di¤erent root loci are possible.5 -1 -0. 5032 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD 20. For the feedback con…guration of Fig. 5.51, use asymptotes, center of asymptotes, angles of departure and arrival, and the Routh array to sketch root loci for the characteristic equations of the listed feedback control systems versus the parameter K: Use Matlab to verify your results. K ; s(s + 1 + 3j)(s + 1 3j) K (b) G(s) = 2 ; s K(s + 5) (c) G(s) = ; (s + 1) K(s + 3 + 4j)(s + 3 4j) (d) G(s) = ; s(s + 1 + 2j)(s + 1 2j) (a) G(s) = s+2 s+8 s+1 H(s) = s+3 s+7 H(s) = s+3 H(s) = H(s) = 1 + 3s Figure 5.51: Feedback system for Problem 5.20 Solution: The root locus for each part is attached at the end. (a) L(s) - = (s + 2) s(s + 1 + 3j)(s + 1 Asymptotes: 4 1 = 3 Center of asymptotes: = 2:67 Angle of asymptotoes: = 60 ; 180 Angle of departure: d = 29:93 at s = Imaginary-axis crossings: (s) 3j)(s + 8) 1 + 3j = s4 + 10s3 + 26s2 + (80 + K) s + 2K s4 : 1 26 2K s3 : 10 80 + K K s2 : 18 10 2K 2 K 100K+14400 s: 180 K s0 : 2K Routh’s test gives 0 < K < 80 for stability. Solving K = 80, the crossings are s = 4j. (s) with © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 5033 (b) L(s) - Asymptotes: 3 (s + 1) s2 (s + 3) = 1=2 - Center of asymptotes: = 1 - Angle of asymptotoes: = 90 - Imaginary-axis crossings: (s) = s3 + 3s2 + Ks + K s3 : 1 K s2 : 3 K s : 2K 3 s0 : K Routh’s test gives K > 0 for stability. Solving the crossings are s = 0. (s) with K = 0, (c) L(s) - Asymptotes: 2 = (s + 5)(s + 7) (s + 1)(s + 3) 2=0 - Breakin/Breakaway: dL(s) = 0 =) 8s3 + 64s + 104 = 0 ds Therefore the breakin/breakaway points are at s = 2:27; 5:73. (d) L(s) = - Asymptotes: 3 (1 + 3s) (s + 3 + 4j) (s + 3 4j) s(s + 1 + 2j)(s + 1 2j) 3=0 - Angle of departure: - Angle of arrival: a d = = 108:4 at s = 23:4 at s = 1 + 2j 3 + 4j © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 5034 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Root Locus Root Locus 5 plot a plot b Imag Axis Imag Axis 5 0 0 -5 -5 -10 -5 Real Axis 0 -3 -2.5 -2 Root Locus -1.5 -1 Real Axis -0.5 0 -0.5 0 Root Locus 4 1.5 plot c plot d 2 Imag Axis Imag Axis 1 0.5 0 -0.5 0 -2 -1 -1.5 -4 -6 -4 Real Axis -2 0 -3 -2.5 -2 -1.5 -1 Real Axis Solution for Problem 5.20 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 5035 21. Consider the system in Fig. 5.52. Figure 5.52: Feedback system for Problem 5.21 (a) Using Routh’s stability criterion, determine all values of K for which the system is stable. (b) Use Matlab to …nd the root locus versus K. Find the values for K at imaginary-axis crossings. Solution: (a) (s) = s4 + 5s3 + 9s2 + (5 + K)s + 3K s4 : 1 9 3K s3 : 5 5+K s2 : 8 K 3K 5 2 K 40K+200 s: 40 K s0 : 3K For the system to be stable, 0 < K < 4:49. (b) The imaginary axis crossings are at s = 1:38j when K = 4:49. Root locus is shown below. Root Locus 6 4 Im ag Axis 2 0 -2 -4 -6 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 Real Axis Root locus for Problem 5.21 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 5036 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Problems and solutions for Section 5.4 22. Let G(s) = 1 (s + 2)(s + 3) and Dc (s) = K s+a : s+b Using root-locus techniques, …nd values for the parameters a; b, and K of the compensation Dc (s) that will produce closed-loop poles at s = 1 j for the system shown in Fig. 5.53. Figure 5.53: Unity feedback system for Problems 5.22, 5.28, and 5.33 Solution: Since the desired poles are slower than the plant, we will use PI control. The solution is to cancel the pole at -3 with the zero and set the gain to K = 2: Thus, a = 3; b = 0; and K = 2: © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. electronic. mechanical. Upper Saddle River.Angle of departure: d = 90 at s = 1 + 2j . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Therefore d (s) = 4s3 + 12s2 + 18s + 10 = 0 ds Thus the repeated roots are at s = 1.Angle of asymptotoes: = 45 .Center of asymptotes: = 1 .. . Find the value of K at that point. Solution: The root locus for the system is attached at the end. its derivative is equal to zero at the multiple roots.5037 23. 16:25.Imaginary-axis crossings: (s) = s4 + 4s3 + 9s2 + 10s + K s4 : 1 9 K s3 : 4 10 s2 : 6:5 K s : 10 8K 13 s0 : K Routh’s test gives 0 < K < 16:25 for stability. 135 . For information regarding permission(s). All rights reserved. respectively. (s) with . photocopying. Suppose that in Fig. (s) = . 1 1:225j. recording. write to: Rights and Permissions Department. or transmission in any form or by any means. and how many multiple roots there are. NJ. G(s) = 1 s(s2 + 2s + 5) and Dc (s) = K : s+2 Without using Matlab. Inc. the corresponding value of K is K = 4. © 2015 Pearson Education. storage in a retrieval system. Upper Saddle River. 5.53.. paying particular attention to points that generate multiple roots. Plugging the roots into the characteristic equation. Inc. or likewise. Pearson Education. sketch the root locus with respect to K of the characteristic equation for the closed-loop system. the crossings are s = 1:58j.Asymptotes: 4 1 s(s + 2)(s2 + 2s + 5) 0=4 .Location of multiple roots: If a polynomial has repeated roots. Solving K = 16:25. NJ 07458. state what the location of the multiple roots is. or transmission in any form or by any means.. electronic. Pearson Education. NJ 07458.23 © 2015 Pearson Education. write to: Rights and Permissions Department.. Inc. Upper Saddle River. photocopying.5038 CHAPTER 5. mechanical. Upper Saddle River. Inc. NJ. storage in a retrieval system. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. All rights reserved. or likewise. For information regarding permission(s). recording. . THE ROOT-LOCUS DESIGN METHOD Root Locus 4 3 I magi nary Axi s 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 Real Axis 0 1 2 Root locus for Problem 5. Upper Saddle River. electronic. 5. NJ.74 -4 -3 0.92 0. Design a lead compensation Dc (s) = K s+p added in series with the plant so that the dominant poles of the closed-loop system are located at s = 2 2j.84 0.74 0.5039 24..53 has an open-loop plant s+z to be given by G(s) = 1=s2 .22 2 0. mechanical. NJ 07458. the zero is at z = with a gain of K = 72: The locus is plotted below.99 0. Inc.965 -2 -3 -7 0. Solution: Setting the pole of the lead to be at p = 20. 1:78 Root Loc us 3 0. Pearson Education. or likewise. All rights reserved.84 0. Suppose the unity feedback system of Fig. storage in a retrieval system.965 Imag Axis 1 0.99 7 0 -1 6 5 -6 -5 4 3 2 1 0. .22 -1 0 1 Real Axis Root locus for Problem 5.92 0.42 0. For information regarding permission(s). Upper Saddle River.6 -2 0.24 © 2015 Pearson Education. or transmission in any form or by any means.. Inc. photocopying. recording.6 0. write to: Rights and Permissions Department.42 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. 6 0. NJ. Inc. Pearson Education.99 0. electronic. photocopying. THE ROOT-LOCUS DESIGN METHOD 25. or transmission in any form or by any means. it is necessary to select a smaller K and set p = 0:01: Other choices are of course possible.53 has the open-loop plant 1 G(s) = : s(s + 3)(s + 6) Design a lag compensation to meet the following speci…cations: The step response settling time is to be less than 5 sec.74 RootLocus 0. the Kv = 28=18 = 1:56: To get a Kv = 10.4 0. or likewise.6 0. The steady-state error to a unit ramp input must not exceed 10%.22 -5 RealAxis 0 00 5 10 15 20 Root locus and step response for Problem 5. Assume that the unity feedback system of Fig.92 -6 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.965 2 10 2 -2 1 0. these can be met at K = 28 where the ! n = 2: With this gain. storage in a retrieval system. Upper Saddle River.84 0. write to: Rights and Permissions Department.25 © 2015 Pearson Education. The step response of this design is plotted below. recording. we need a lag gain of about 6:5: Selecting the lag zero to be at 0:1 requires the pole to be at 0:1=6:5 = 0:015: To meet the overshoot speci…cations.84 -10 0. 25 . The step response overshoot is to be less than 17%..99 0. 6 0. For information regarding permission(s).42 0.965 -4 0. Inc.5040 CHAPTER 5.42 0. mechanical. All rights reserved.2 4 Imag Axis 0.4 0.2 0.74 0. 5.6 1.92 1. Upper Saddle River.. NJ 07458. Solution: The overshoot speci…cation requires that damping be 0:5 and the settling time requires that ! n > 1:8: From the root locus plotted below.8 10 8 6 4 2 0.22 0. electronic. Upper Saddle River. .5041 26.. or transmission in any form or by any means. p 5. we select the pole to be at p = zero and gain to be z = 3.53 are satis…ed if the closed-loop poles are located at s = 1 j 3. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. K = 12. recording. Inc.. Upper Saddle River. Pearson Education. NJ 07458. photocopying. 10 and …nd the © 2015 Pearson Education. s+z that will meet the speci(b) Design a lead compensator Dc (s) = K s+p …cation. write to: Rights and Permissions Department. the complex poles have real part at s = 0:5. Solution: (a) With proportional control. (a) Show that this speci…cation cannot be achieved by choosing proportional control alone. (b) To design a lead. Dc (s) = kp . NJ. or likewise. All rights reserved. storage in a retrieval system. Inc. For information regarding permission(s). mechanical. A numerically controlled machine tool positioning servomechanism has a normalized and scaled transfer function given by G(s) = 1 : s(s + 1) Performance speci…cations of the system in the unity feedback con…guration of Fig. NJ 07458.. Upper Saddle River. The steady-state error to a unit ramp at the reference input must be less than 0. or transmission in any form or by any means. NJ. Inc. recording. storage in a retrieval system. the overshoot is reduced to 3:64% and the rise time is 0:35 sec. Upper Saddle River. electronic. Inc. (e) Give the Matlab response of your …nal design to a reference step. we need a new Kv = 20. or likewise. write to: Rights and Permissions Department.. . which is an increase of a factor of 5. © 2015 Pearson Education. the pole needs to be at p = 0:08: (d) The root locus is plotted below. (b) Kv = lim sGDc = lim s s!0 s!0 245(s + 1) 10 = 4:083 s(s + 1)(s + 10) (s + 6) (c) To meet the steady-state requirement. All rights reserved. photocopying. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. mechanical. (a) Design a lead compensation that will cause the system to meet the dynamic response speci…cations. Solution: (a) Setting the lead pole at p = 60 and the zero at z = 1. THE ROOT-LOCUS DESIGN METHOD 27. (b) What is the velocity constant Kv for your design? Does it meet the error speci…cation? (c) Design a lag compensation to be used in series with the lead you have designed to cause the system to meet the steady-state error speci…cation.5042 CHAPTER 5. Pearson Education. (e) The step response is plotted below.4 sec. If we set the lag zero at z = 0:4. ignoring the error requirement. With the lead compensator. (d) Give the Matlab plot of the root locus of your …nal design. For information regarding permission(s).05. A servomechanism position control has the plant transfer function G(s) = 10 : s(s + 1)(s + 10) You are to design a series compensation transfer function Dc (s) in the unity feedback con…guration to meet the following closed-loop speci…cations: The response to a reference step input is to have no more than 16% overshoot. the dynamic speci…cations are met with a gain of 245. The response to a reference step input is to have a rise time of no more than 0. .27 © 2015 Pearson Education.74 0.42 0.92 0.22 0. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5 -2 -3 -4 0.965 0. photocopying. NJ. Pearson Education. or transmission in any form or by any means.22 0.5 0.74 0.5 Lead-lag step response 1.99 0.22 0 -8 -6 -4 Real Axis -2 0 2 0 1 2 3 4 Time (sec) 5 6 7 Solution to Problem 5. Upper Saddle River.92 -10 0.84 0.92 0.22 0 -8 -6 -4 Real Axis -2 0 2 0 0. Upper Saddle River.6 0. NJ 07458.42 0.99 0.. recording. Inc.965 0. For information regarding permission(s).5043 Lead root locus 4 3 Lead Step response 1.42 0.92 3 1 Time (sec) 0.84 0. .99 1 10 0 8 6 4 Amplitude Imaginary Axis 2 2 -1 0. All rights reserved. Inc.74 0.5 -2 -3 -4 -10 0.6 0.74 0.99 1 10 0 8 6 4 Amplitude Imaginary Axis 2 2 -1 0.84 0. storage in a retrieval system.42 0. or likewise.6 0.5 Lead-lag root locus 1.84 0. write to: Rights and Permissions Department.965 1 0. mechanical.5 4 0.965 1 0.6 0. THE ROOT-LOCUS DESIGN METHOD 28. 5. Solution: The poles can be put in the desired location with proportional control alone. Assume the closed-loop system of Fig. Upper Saddle River.5044 CHAPTER 5. Inc.2. All rights reserved. or transmission in any form or by any means. mechanical. recording. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Upper Saddle River. with a gain of kp = 2 resulting in a Kv = 1: To get a Kv = 5. NJ 07458. we add s + 0:1 : a compensation with zero at 0:1 and a pole at 0:02: Dc (s) = 2 s + 0:02 © 2015 Pearson Education. photocopying. storage in a retrieval system. NJ. Inc. Pearson Education. . write to: Rights and Permissions Department.53 has a feed forward transfer function G(s) given by 1 : G(s) = s(s + 2) Design a lag compensation so that the dominant poles of the closed-loop system are located at s = 1 j and the steady-state error to a unit ramp input is less than 0. electronic. For information regarding permission(s)... or likewise. An elementary magnetic suspension scheme is depicted in Fig. and the gravitational force is 9. For small motions near the reference position.8 N/kg. The power ampli…er is a voltage-to-current device with an output (in amperes) of i = u + V0 . write to: Rights and Permissions Department.. The mass of the ball is 20 g. U (e) Assume that a lead compensation is available in the form E = s+z Dc (s) = K s+p : Give values of K. . Upper Saddle River. (c) What is the transfer function from u to e? (d) Suppose the control input u is given by u = Ke. or likewise. Inc. Figure 5. mechanical. recording.5045 29. The upward force (in newtons) on the ball caused by the current i (in amperes) may be approximated by f = 0:5i + 20x. and p that yields improved performance over the one proposed in part (d). Solution: (a) The equations of motion can be written as X m• x = f orces = 0:5i + 20x mg = 0:5(u + Vo ) + 20x mg Substituting numbers. photocopying. For information regarding permission(s). All rights reserved.54. NJ. the voltage e on the photo detector is related to the ball displacement x(in meters) by e = 100x. z. electronic. Inc.. (b) Give the value of the bias V0 that results in the ball being in equilibrium at x = 0. NJ 07458. or transmission in any form or by any means. storage in a retrieval system. Sketch the root locus of the closed-loop system as a function of K. Upper Saddle River. Pearson Education. we have 0:02• x = 0:5(u + Vo ) + 20x 0:196: © 2015 Pearson Education. 5.54: Elementary magnetic suspension (a) Write the equations of motion for this setup. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. All rights reserved. photocopying. E 2500 = 2 U s 1000 (d) The locus starts at the two poles symmetric to the imaginary axis. we can pick z = 1000 to cancel one of the open-loop plant poles. and pick p = 150 to pull the locus into the left-hand plane. x • = 0 and u = 0. any p lead will improve its performance. © 2015 Pearson Education. electronic.. say 0:7. Root Locus 40 30 Imaginary Axis 20 10 0 -10 -20 -30 -40 -40 -30 -20 -10 0 Real Axis 10 20 30 40 Root loci for Problem 5. mechanical. For example. THE ROOT-LOCUS DESIGN METHOD (b) In equilibrium at x = 0. See the plot below. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.. write to: Rights and Permissions Department. For information regarding permission(s). Pearson Education. K can be selected to give a desired amount of damping. Inc. Upper Saddle River. storage in a retrieval system. or transmission in any form or by any means. meet at the origin and cover the imaginary axis. Inc. or likewise. The locus is plotted below. . NJ. NJ 07458. recording. Therefore to have the bias cancel gravity.5046 CHAPTER 5. 0:5Vo 0:196 = 0 or Vo = 0:392: (c) Taking Laplace transforms of the equation and substituting e = 100x. K = 4:75 gives a damping of 0:7.29d (e) Since the system with a proportional gain is on the stability boundary. Upper Saddle River. NJ 07458. Pearson Education.29e © 2015 Pearson Education. storage in a retrieval system. Inc. Upper Saddle River. recording.. . All rights reserved. NJ. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. write to: Rights and Permissions Department. mechanical. photocopying. For information regarding permission(s).5047 Root Locus 50 40 30 Imaginary Axis 20 10 0 -10 -20 -30 -40 -50 -160 -140 -120 -100 -80 -60 Real Axis -40 -20 0 20 40 Root loci for Problem 5.. or likewise. electronic. Upper Saddle River. Inc. or transmission in any form or by any means. . Upper Saddle River.74 -6 0.22 -2 0 2 -1 0 1 2 Time(sec) 3 Solutions for Problem 5.4 0. (b) Use Matlab to plot the system’s response to a reference step. All rights reserved. The value of gain for closed loop roots at damping of 0:7 is K = 1:04 (b) The …nal value of the step response plotted below is 0:887. To get a positive output we would use a positive gain in positive feedback. or likewise. Inc. photocopying. For information regarding permission(s). mechanical. NJ. the problem statement might be confusing.2 0. RootLocus 0. or transmission in any form or by any means.6 StepResponse 0.99 -0.4 -1 0. which is the regular positive locus for G: The locus is plotted below. With the G(s) as given. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. 4 . THE ROOT-LOCUS DESIGN METHOD 30. write to: Rights and Permissions Department.99 1 100 8 6 4 2 -0. recording.8 0.965 -0.6 -2 -3 -4 -10 0.84 4 0.5048 CHAPTER 5.92 3 0.6 -4 RealAxis 0. A certain plant with the non minimum phase transfer function G(s) = 4 2s .. Solution: (a) With all the negatives.42 0. Inc.74 0. NJ 07458. Matlab needs to plot the negative locus. storage in a retrieval system.84 -8 0. Pearson Education.22 0.30 © 2015 Pearson Education.92 0.42 0. s2 + s + 9 is in a unity positive feedback system with the controller transfer function Dc (s): (a) Use Matlab to determine a (negative) value for Dc (s) = K so that the closed-loop system with negative feedback has a damping ratio = 0:707.965 2 0 Amplitude Imag Axis 0. Upper Saddle River.2 -0. .65 8 0. Upper Saddle River.36 0.76 -3-3 2 0. …nd a value for the gain K that will provide the maximum damping ratio.48 2 8 5 0. Inc. the lead compensator s+2 H(s) = K s+4 stabilizes the system. Pearson Education.24 0. So the system is stable for all K.62 0.13 -2 RealAxis 1. electronic.5 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Using the root-locus procedure.13 0. 5.03 12 -1 0 0. NJ 07458.09 0.36 0.97 -1 1 0. Inc.12 1 Imag Axis Imag Axis 4 2 0. storage in a retrieval system.5 -2 10 -4 0.12 -1 RealAxis 30 1 Loci for Problem 5. All rights reserved.19 -3 0.55. . Solution: (a) The root locus is plotted below and lies entirely in the left-half plane.88 1.24 0.48 -2 0. Figure 5. photocopying.4 6 0 4 0.31 © 2015 Pearson Education.09 0. recording. (b) Assume that the sensor transfer function is modeled by a single pole with a 0:1 sec time constant and unity DC gain.03 12 10 0..5 2 1 0.06 0.88 0 6 -10 3 2. or transmission in any form or by any means.4 0. mechanical.26 0.5049 31. or likewise.65 0. NJ.97 2 -5 RootLocus 0. the gain is K = 6:25 but the damping of the complex poles is only 0:073: A practical design would require much more lead.26 10 3 0. write to: Rights and Permissions Department. Consider the rocket-positioning system shown in Fig.62 0. RootLocus 0.76 2.5 0. For information regarding permission(s).5 0. Upper Saddle River.5 0. (b) At maximum damping.55: Block diagram for rocket-positioning control system (a) Show that if the sensor that measures x has a unity transfer function.19 0.06 0. Upper Saddle River. L(s) = 100(s + 1) : The locus is plotted below. Inc. THE ROOT-LOCUS DESIGN METHOD 32.84 0.99 0.32 (a) Find the locus of closed-loop roots with respect to K. write to: Rights and Permissions Department. or transmission in any form or by any means. recording. (b) Find the maximum value of K for which the system is stable. NJ 07458. NJ.99 14 0 -2 -4 -6 -14 12 10 -12 -10 8 6 4 2 0. photocopying. s2 (s2 + 12s + 40) Root Locus 6 0.22 -2 0 2 Real Axis Locus for Problem 5..5050 CHAPTER 5..32 © 2015 Pearson Education. All rights reserved. For the system in Fig. . or likewise. 5. Assume K = 2 for the remaining parts of this problem.965 0. electronic.42 0. storage in a retrieval system.92 0. (c) What is the steady-state error (e = r y) for a step change in r? (d) What is the steady-state error in y for a constant disturbance w1 ? (e) What is the steady-state error in y for a constant disturbance w2 ? (f) If you wished to have more damping.6 0.84 0.56: Figure 5. Upper Saddle River.56: Control system for Problem 5.74 -8 -6 0.42 0.965 Im ag Axis 2 0.74 0. For information regarding permission(s). mechanical. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.22 4 0.92 0. what changes would you make to the system? Solution: (a) For the locus. Pearson Education. Inc.6 -4 0. Inc. photocopying. the controller needs to have a lead compensation. . or transmission in any form or by any means. or likewise.. Inc. Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. NJ. recording. mechanical. write to: Rights and Permissions Department. Upper Saddle River. storage in a retrieval system. © 2015 Pearson Education. All rights reserved.. Upper Saddle River. NJ 07458. electronic. For information regarding permission(s).5051 (b) The maximum value of K for stability is K = 3:36: (c) The transfer function from R to Y is Y 200 = 2 2 : R s (s + 12s + 40) + 200(s + 1) Therefore the steady-state error for a step change in r is estep (1) Y 1 R s 2 2 s (s + 12s + 40) + 200s =0 = lim 2 2 s!0 s (s + 12s + 40) + 200(s + 1) = lim s 1 s!0 (d) The transfer function from W1 to Y is: Y 100s2 = 2 2 W1 s (s + 12s + 40) + 200(s + 1) Therefore the steady-state error for a constant disturbance w1 is estep (1) = lim s s!0 Y W1 1 =0 s (e) The transfer function from W2 to Y is: Y 100 = 2 2 W2 s (s + 12s + 40) + 200(s + 1) Therefore the steady-state error for a constant disturbance w1 is estep (1) = lim s s!0 Y W2 1 = 0:5 s (f) To get more damping in the closed-loop response. In this problem. This is the transfer function relating the input force u(t) and the position y(t) of mass M in the non-collocated sensor and actuator problem. NJ.7. write to: Rights and Permissions Department.29 indicates that we’d better make the damping greater than 0.. or likewise. THE ROOT-LOCUS DESIGN METHOD 33. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. If not. and Dc (s) = K. we will use root-locus techniques to design a controller Dc (s) so that the closedloop step response has a rise time of less than 0. Pearson Education. The locus in this case is the imaginary axis and cannot meet the specs for any K: (b) The specs require that > 0:6. storage in a retrieval system.1 sec and an overshoot of less than 10%. Solution: (a) The approximate plant transfer function is G(s) = s12 . Then the resulting overshoot is 9:8% and the rise time is 0:05 sec. All rights reserved. the only chance we have is to introduce a notch as well as a lead. but is given by m = M=10. Upper Saddle River. For information regarding permission(s). and K = 30. The compensation resulting in the plots s + 4 s2 =9:25 + s=9:25 + 1 : The overshoot shown is Dc (s) = 12 (:01s + 1) s2 =3600 + s=30 + 1 is 7% and the rise time is 0:04 sec : © 2015 Pearson Education. and let M = 1. the overshoot will be increased and Figure 3. Consider the plant transfer function G(s) = bs + k s2 [mM s2 + (M + m)bs + (M + m)k] to be put in the unity feedback loop of Fig.5052 CHAPTER 5. b = 0:1. (d) Now suppose that the small mass m is not negligible. mechanical. and show that K and z can be chosen to meet the speci…cations. k = 1. You may use Matlab for any of the following questions: (a) Approximate G(s) by assuming that m = 0. The locus will be a circle with radius 15: Because of the zero. The plot shows the result when Dc = 25(s + 4): The resulting overshoot is 9:9% and the rise time is 0:06 sec. experimentation shows that we can lower the overshoot of less than 10% only by setting the zero at a low value and putting the poles on the real axis. photocopying.. adjust the controller parameters so that the speci…cations are met. As a matter of fact. (c) In this case. z = 4. NJ 07458. . or transmission in any form or by any means. 5. electronic. Dc (s) = K s+p and pick p so that the values for K and z computed in part (b) remain more or less valid. (c) Repeat part (b) but with a practical controller given by the transfer function p(s + z) . we pick p = 150. Can K be chosen to satisfy the performance speci…cations? Why or why not? (b) Repeat part (a) assuming Dc (s) = K(s + z).53. Inc. (d) With the resonance present. Upper Saddle River. recording. Check to see if the controller you designed in part (c) still meets the given speci…cations. Inc. ! n > 18: Select z = 15 for a start. . Inc. NJ 07458.6 0.5 1 1. For information regarding permission(s). NJ.2 0.6 0. storage in a retrieval system.5 Imaginary Axis 20 1 10 0 0.8 1 1.5053 Root Locus 1.4 0. All rights reserved.5 3 3. write to: Rights and Permissions Department.5 -10 -20 -60 -50 -40 -30 Real Axis -20 -10 0 0 Root Locus 0 0.5 -10 -20 -60 -50 -40 -30 Real Axis -20 -10 0 0 Root Locus 1.5 2 2. or likewise. Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5 Imaginary Axis 4 1 2 0 0.5 Imaginary Axis 20 1 10 0 0. or transmission in any form or by any means. Upper Saddle River.5 Root loci and step responses for Problem 5..8 1 0 0.. Inc. photocopying.33 © 2015 Pearson Education.4 0. mechanical. electronic. recording.2 0. Upper Saddle River.5 -2 -4 -10 -8 -6 -4 Real Axis -2 0 2 0 0 0. Inc. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. electronic. Upper Saddle River. (c) Find values of the gains kp and kD for Dc (s) = kp + kD s that meet the design speci…cations with at least a 10% margin. With this K. Consider the Type 1 system drawn in Fig. We would like to design the compensation Dc (s) to meet the following requirements: (1) The steady-state value of y due to a constant unit disturbance w should be less than 54 . mechanical. and (2) the damping ratio = 0:7. Using root-locus techniques.. © 2015 Pearson Education. photocopying. Solution: (a) To meet the error requirement. 5. storage in a retrieval system. or likewise. Upper Saddle River.57: Control system for Problem 5.34 (a) Show that proportional control alone is not adequate. NJ 07458. For information regarding permission(s). Figure 5. NJ. we need ystep (1) = lim s s!0 1 Y 1 = lim W s s!0 s2 + s + K 0:8: Thus K must be at least K 1:25.57. the damping ratio will be 1 = p ) 0:45 2 K So we can’t meet both requirements with proportional control. The 0:7 1..5054 CHAPTER 5. Inc. All rights reserved. ystep (1) = lim s s!0 Y 1 1 = lim W s s!0 s2 + (1 + kD )s + kp So the error requirement can be met by setting kp damping ratio requirement can be written as = By choosing kD p 1:4 kp 1 + kD p 2 kp 0:8: 1:25. Pearson Education. or transmission in any form or by any means. write to: Rights and Permissions Department. . recording. THE ROOT-LOCUS DESIGN METHOD 34. we can satisfy both speci…cations. (b) Show that proportional-derivative control will work. (b) With PD control. 3 5 0 .2 0 .3 5 0 -4 -3 -2 -1 0 1 R e a l Ax is 0 1 2 3 4 5 6 7 Time ( s e c ) Solution for Problem 5. Pearson Education. For information regarding permission(s).1 . Inc.8 0 .5055 (c) Setting kp = 1:4 and kD = 0:85.9 2 0 . Upper Saddle River. we get ystep (1) = 0:714 and = 0:782.34 © 2015 Pearson Education.9 8 4 0 .9 6 .9 2 0 .9 6 0 .4 0 . storage in a retrieval system. recording.5 0 . .9 8 4 -1 0 .9 9 6 . mechanical..6 0 . Upper Saddle River. or transmission in any form or by any means. photocopying.1 0 .5 8 0 . Inc.5 0 .7 6 0 .7 6 0 .3 0 .5 5 0 1 0 .. All rights reserved.7 1 0 .9 9 6 4 3 2 Amplitude Imaginary Axis 0 . write to: Rights and Permissions Department.5 -5 0 .5 0 . The root loci and disturbance step response are plotted below.5 8 0 . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.3 4 1 . NJ. NJ 07458.0 . or likewise. electronic. R o ot L oc u s Ste p r e s p o n s e fo r p r o b le m 5 .8 6 0 .8 6 0 . . Inc. It is interesting to note. The discrete equivalent 2 z 1 in the Dc (s): for Ts = 0:1 sec is given by substituting s = 0:1 z + 1 Dd (z) = 0:85 2 z 1 18:4z 15:6 + 1:4 = : 0:1 z + 1 z+1 To evaluate this discrete controller..8 s =0. Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. the damping ratio requirement of 0:7 is still met with the digital control. Problem 5. Upper Saddle River. however. All rights reserved. It said to use the Dc (s) from Problem 5..5056 CHAPTER 5. that there is an oscillation of the output at 5 Hz. …nd the Dc (z) that is the discrete equivalent to your Dc (s) from Problem 5. which has created an extraneous oscillatory root in the closed-loop system.FPE7e.T 0. 1 Continuousdesign Discreteequivalentdesign. mechanical. However. photocopying.5 at www. which suggests a decrease in the damping due to the digital implementation.7 rather than the correct one. © 2015 Pearson Education. we have Dc (s) = 0:85s + 1:4.This is also a result of having a pure di¤erentiation in the D(s). Evaluate the time reponse using Simulink.com or in Chapter 8. (Note: The material to do this problem is covered in the Appendix W4. This arises because the use of the trapezoidal equivalent places a root of the compensation at z = -1. electronic. NJ 07458.34.6 0.2 00 2 4 6 Time(sec) 8 10 12 14 Solution for Problem 5. or transmission in any form or by any means.) (Note: The …rst printing of the 7th edition had an error in the problem statement.34) Solution: From Problem 5. The results of the step responses are shown below.1sec 0. . recording.4 0. write to: Rights and Permissions Department. For information regarding permission(s).35 Note that there is slightly greater overshoot in the digital system. Inc. THE ROOT-LOCUS DESIGN METHOD 35. and determine whether the damping ratio requirement is met with the digital implementation. storage in a retrieval system. Upper Saddle River. Using a sample rate of 10 Hz. NJ.34 using the trapezoid rule. we use Simulink to compare the two implementations. or likewise. which is half the sample rate. Upper Saddle River. recording. write to: Rights and Permissions Department. Using Matlab’s C2D function for this discrete equivalent produces Dd2 (z) = 9:219z 7:819: which eliminates the oscillation at 5 hz. Pearson Education. mechanical.. Inc. Inc. NJ 07458. For information regarding permission(s). NJ. or likewise. or transmission in any form or by any means.. . photocopying. Upper Saddle River. Using the Matched Pole-Zero discrete equivalent will help this situation since it does not create the oscillatory pole in the compensator. © 2015 Pearson Education. All rights reserved.. storage in a retrieval system. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.e. electronic. a zero in the numerater with no pole.5057 i. Ke = back emf constant = 0:1V sec Ra = armature resistance = 10 . Inc. Figure 5. Upper Saddle River. write to: Rights and Permissions Department. NJ. photocopying. electronic. km = Kt = torque constant = 0:1 N m=A. Where are all three closed-loop root locations for this value of KA ? Solution: (a) Neglecting viscous friction and the e¤ect of inductance. mechanical. or transmission in any form or by any means. Gear ratio = 1 : 1. . Upper Saddle River. (b) Choose a gain KA that gives roots at = 0:7.58: Positioning servomechanism (a) What is the range of the ampli…er gain KA for which the system is stable? Estimate the upper limit graphically using a root-locus plot. JL + Jm = total inertia = 10 va = KA (ei ef ): 3 kg m2 . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Inc. the transfer function of the DC motor is o Va = (JL + Jm Kt Ra )s2 + Kt Ke Ra s = 10 s2 + s © 2015 Pearson Education. which is not correct) T = motor torque = Kt ia . recording.58. Ko = 10V=rad. NJ 07458.5 36. Pearson Education. storage in a retrieval system. All rights reserved.5058 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Problems and solutions for Section 5. Consider the positioning servomechanism system shown in Fig. 5... (note: 1st printing of book had eo = Kpot o . eo = Ko o . or likewise. where ei = Ko i . For information regarding permission(s). mechanical.6 -2 0. or transmission in any form or by any means. electronic. photocopying. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.36 (b) The damping is 0:7 when K = 0:046. 0:475 0:482j: © 2015 Pearson Education.965 -2 -3 -7 0.965 Imaginary A xis 1 0..42 0. Root Locus 3 0.84 0.. write to: Rights and Permissions Department. . storage in a retrieval system.92 0. NJ. Upper Saddle River. poles are at s = 10:05. For the value of KA . Inc. NJ 07458.74 -4 -3 Real Axis 0. Pearson Education.74 0. All rights reserved. Upper Saddle River.6 0. the upper limit of KA for stability is 0:11.99 0.22 -1 0 1 Root locus for Problem 5.5059 From the root locus plotted below.92 0. Inc. recording.22 2 0.84 0.99 7 0 -1 6 5 -6 -5 4 3 2 1 0. For information regarding permission(s). or likewise.42 0. indicate the closed-loop poles with a dot ( ) and include the boundary of the region of acceptable root locations. the zero was put at s = 1:7. n X 1 = Kv i=1 m X 1 1 + pi i=1 zi ) Kv = 0:7344 © 2015 Pearson Education. On your plot. . and kp was set to 0:036. 0:45 0:773j. The transfer function from current I(s) to tape velocity (s) (in millimeters per millisecond per ampere) is 15(s2 + 0:9s + 0:8) (s) = : I(s) (s + 1)(s2 + 1:1s + 1) We wish to design a Type 1 feedback system so that the response to a reference step satis…es tr 4msec. Inc. NJ 07458. photocopying. (b) Assume a proportional-integral compensator of the form kp (s + )=s. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. THE ROOT-LOCUS DESIGN METHOD 37.5060 CHAPTER 5. (b) Using rltool. giving values for kp and . 0:5451 0:5589j. Inc. Upper Saddle River. Here. The root locus and the step response are plotted in the second row below. NJ. Mp 0:05 (a) Use the integral compensator kI =s to achieve Type 1 behavior. and select the best possible values of kp and you can …nd. or likewise. Show on the same plot the region of acceptable pole locations corresponding to the speci…cations. storage in a retrieval system. Sketch the root-locus plot of your design. write to: Rights and Permissions Department. With the PI compensator. This will improve the transient response.. ts 15msec. the closed-loop poles are at s = 0:7749 0:7774j.. recording. Thus the velocity constant can be calculated from Truxal’s formula. mechanical. All rights reserved. and the velocity constant Kv your design achieves. Upper Saddle River. we can choose the location of zeroto pull the locus to the left-hand plane. and sketch the root-locus with respect to kI . Pearson Education. For information regarding permission(s). and the closed-loop zeros are at s = 1:7. Solution: (a) The root locus with respect to kI and the step response with kI = 0:036 are plotted in the …rst row below. We wish to design a velocity control for a tape-drive servomechanism. or transmission in any form or by any means. . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5 0. photocopying.0.2 -1 .2 0. For information regarding permission(s).5 -1 .5 0 0.2 0 0. or transmission in any form or by any means.4 0.4 .0.6 1 0..6 .4 . or likewise. Pearson Education. storage in a retrieval system.1. Inc.0.8 0. NJ.8 Amplitude Imaginary Axis 0..0.8 Amplitude Imaginary Axis 0.6 0. write to: Rights and Permissions Department.036 R oot Locus p 0. Upper Saddle River.5 0 R eal Axis 0 2 4 6 T ime ( sec ) Solution for Problem 5.036 R oot loc us I 1 1. All rights reserved. NJ 07458. recording.1.5 0 0.0.2 .5 0 2 R eal Axis 4 6 8 10 8 10 T ime ( sec ) PI c ontr ol with k = 0. Upper Saddle River. mechanical.0.5061 Integ r al c ontr ol with k= 0.8 0 -2 .4 .5 0 -1 . electronic. Inc.6 0.37 © 2015 Pearson Education.0.2 1 0. Eq. with this loop closed. .89) and then. recording. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. s+z for the loop to cancel (b) Design a lead compensation Dc (s) = K s+p the pole at s = 1 and place the two remaining poles at 4 j4: The new control is U (s). scaled equations of a cart as drawn in Fig. (c) Compute the transfer function of the new plant from U to Y with Dc (s) in place.90) In this problem you are to design a control for the system by …rst closing a loop around the pendulum. Time is 3g(m +m ) measured in terms of = ! o t where ! 2o = `(4mcc+mpp) : The cart motion. mechanical. Upper Saddle River. storage in a retrieval system. Upper Saddle River.(5. NJ 07458. photocopying. For this problem. is measured in units of pendulum length as y = 3x 4` and the input is force normalized by the system weight. y. electronic. NJ. © 2015 Pearson Education. Inc. Inc. For information regarding permission(s).59 of mass mc holding an inverted uniform pendulum of mass mp and length ` with no friction are • = v (5. THE ROOT-LOCUS DESIGN METHOD 38.59: Figure of cart-pendulum for Problem 5.90). The normalized.. where the force is V (s) = U (s) + Dc (s) (s): Draw the root locus of the angle loop.5062 CHAPTER 5. or transmission in any form or by any means. closing a second loop around the cart plus pendulum. All rights reserved. v = g(mcu+mp ) : These equations can be used to compute the transfer functions V = 1 s2 1 s2 1 + Y = 2 2 V s (s 1) (5.38 (a) Draw a block diagram for the system with V input and both Y and as outputs. Pearson Education. write to: Rights and Permissions Department. Eq.89) (5.. or likewise. let the mass ratio be mc = 5mp : Figure 5.88) y• + =v where = 3mp 4(mc +mp ) is a mass ratio bounded by 0 < < 0:75.(5. 5. cart position. = 0:125.5063 (d) Design a controller Dc (s) for the cart position with the pendulum loop closed. electronic. NJ 07458. s2 + 2s + 1 (e) The step responses are shown below.. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. All rights reserved. the remaining poles can be placed at s+1 The root locus is shown s = 4 4j. Upper Saddle River. Therefore Dc (s) = 41 s+9 below. NJ. Therefore the transfer function from U to Y with Dc (s) is Y 41 s2 0:875 = 2 U s + 8s + 32 s2 (d) Dc = kc s2 + 0:2s + 0:01 : The root locus is shown below. Draw the root locus with respect to the gain of Dc (s) (e) Use Matlab to plot the control. we set z = 1. A more reasonable alternative choice would be to place the pendulum roots at s = 0:5 j0:5: © 2015 Pearson Education. and pendulum position for a unit step change in cart position. Inc. (c) Since mc = 5mp . Inc. Upper Saddle River. storage in a retrieval system.. . or transmission in any form or by any means. Pearson Education. or likewise. photocopying. recording. Solution: (a) (b) To cancel the pole at s = 1. The pendulum position control is rather fast for this problem. mechanical. write to: Rights and Permissions Department. Then the closed loop transfer function for the loop becomes V = (s K = 2 1)(s + p) + K s + (p K 1)s + K p : Setting p = 9 and K = 41. For information regarding permission(s). Inc. storage in a retrieval system.2 -1 0 Real Axis 1 2 0 20 40 60 80 T ime (sec) Root loci and step responses for Problem 5. THE ROOT-LOCUS DESIGN METHOD Inner pendulum loop Outer cart loop Final design 2 2 0 -2 1 Amplitude 4 Imag Axis Imag Axis 1. write to: Rights and Permissions Department.6 0.2 -1 -1. Upper Saddle River.2 1 Amplitude Imag Axis 1 0. electronic.5 -10 -5 0 -8 -6 -4 -2 0 2 Real Axis Real Axis Alternative design Alternative design 4 0 50 100 150 T ime (sec) 1. mechanical. For information regarding permission(s).5 0 -0. Pearson Education.5 0 -4 -0. or transmission in any form or by any means. or likewise. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. .5 4 0 -2 -4 0.5 -2 0 -0.5064 CHAPTER 5.8 0.5 0. All rights reserved.4 0. Inc. recording.38 © 2015 Pearson Education.. NJ 07458.. Upper Saddle River. photocopying. NJ.5 1. . .S. For information regarding permission(s). rad. Inc. Parameter identi…cation based on sea-trials data (Trankle. Upper Saddle River. electronic. or transmission in any form or by any means. NJ.60. All rights reserved. 1987) was used to estimate the hydrodynamic coe¢ cients in the equations of motion. (s) s(s + 0:2647)(s + 0:0063) (s) 0:0000064 Gw (s) = = .5065 39. The result is that the response of the heading angle of the ship to rudder angle and wind changes w can be described by the second-order transfer functions 0:0184(s + 0:0068) (s) = . rad=sec.. rad: r = yaw rate. rad r = reference heading angle. recording. Upper Saddle River. NJ 07458. Inc. w(s) s(s + 0:2647)(s + 0:0063) G (s) = where = heading angle. storage in a retrieval system. Consider the 270-ft U.60: USCG cutter Tampa (902) © 2015 Pearson Education. mechanical. m=sec: Figure 5. w = wind speed. 5. photocopying. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Pearson Education. = rudder angle. write to: Rights and Permissions Department. Coast Guard cutter Tampa (902) shown in Fig. or likewise. or transmission in any form or by any means. mechanical. (b) The rate feedback from a yaw-rate gyroscope is giving us a derivative control for free. Using root-locus technique. by _ = r). i. and. NJ.) If the steady-state value of the heading due to this wind gust is more than 0:5 .5066 CHAPTER 5. photocopying. the controller gain will also be negative.39b Let’s set KD = 1 and see what we can do with a simple proportional controller. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Checking the step response with this proportional control using Matlab. Dc (s) = Kp . we can use the stepinfo command in Matlab. which meets the settling time requirement. Thus the block diagram of the system will look like shown below. All rights reserved. And for a settling time less than 50 sec. (Since the plant has a negative sign. Therefore Dc (s) = 1:5 is adequate for the problem. Inc. The value from Matlab is ts = 314:64 sec. Block diagram for Problem 5. Solution: (a) To determine the open-loop settling time to 1% of the …nal value. Upper Saddle River. (b) In order to regulate the heading angle . And the rudder angle de‡ection for a 5 input is less than 7:5 . Upper Saddle River.. write to: Rights and Permissions Department. Pearson Education. for a 5 change in heading the maximum allowable rudder angle de‡ection is speci…ed to be less than 10 . NJ 07458. Inc. (c) Check the response of the closed-loop system you designed in part (b) to a wind gust disturbance of 10 m= sec (Model the disturbance as a step input. design a compensator that uses and the measurement provided by a yaw-rate gyroscope (that is. modify your design so that it meets this speci…cation as well. we need > 4:6=50 = 0:092 from the design relations for the standard second order system. storage in a retrieval system. recording. The settling time of to a step change in r is speci…ed to be less than 50 sec. For information regarding permission(s). we …nd ts = 28:2 sec. electronic.e. or likewise.) With the proportional control. . © 2015 Pearson Education. Kp = 1:5 was picked.. the maximum de‡ection of the rudder is almost surely at the initial instant. when it is (0) = Kp r (0): Thus to keep below 10 for a step of 5 . THE ROOT-LOCUS DESIGN METHOD (a) Determine the open-loop settling time of r for a step change in . we need jKp j < 2. electronic. Upper Saddle River.2 0 δ ψ 1 0. Inc. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. write to: Rights and Permissions Department. the steady-state value of the heading angle due to a disturbance of 10m= sec is (1) = lim s s!0 (s) 10 = 0:3386 rad = 19:4 > 0:5 w(s) s So we need to modify our design.4 0. Pearson Education.5 0 -0. mechanical.. All rights reserved.5067 Root Locus 1 Imaginary Axis 0. recording. or likewise. all speci…cations are met as shown below.6 -5 0.5 -1 -4 -3 -2 -1 Real Axis 0 1 2 Heading angle to a step input Rudder angle to 5 1. © 2015 Pearson Education. storage in a retrieval system. photocopying. or transmission in any form or by any means. . let’s add an integral term to the controller.2 0 0 10 20 Time (sec) 30 40 -10 0 10 20 Time (sec) 30 40 Root locus and Step response for Problem 5. Inc.8 0.39b (c) With the compensator from part (b).4 o input 5 1. NJ. the closed-loop transfer function from w to is (s) Gw = w(s) 1 + Kp G (1 + KD s) = 0:0000064 s(s + 0:2647)(s + 0:0063) + 0:0278(s + 0:0068)(1 + s) Using the Final Value Theorem. Upper Saddle River. Dc (s) = Kp + KI =s. NJ 07458.. Using rltool. To reject the disturbance completely. For information regarding permission(s). we …nd that when Dc (s) = 1:87 0:11=s. .6 -5 0.5 0 0 0 Heading angle to a step input 200 400 600 800 Time (sec) Rudder angle to 5 1.05 -0.2 0 0 20 40 60 Time (sec) 80 100 -10 0 20 40 Time (sec) 60 80 Root locus and Step response for Problem 5. Inc. For information regarding permission(s). Upper Saddle River.4 o 1000 1200 1400 input 5 1. NJ 07458.6 -0. write to: Rights and Permissions Department.5068 CHAPTER 5.39c © 2015 Pearson Education.8 0. NJ. electronic.2 P control ψ Imaginary Axis 0.6 0 0.5 -1 Real Axis -0.4 0.15 0. storage in a retrieval system. Upper Saddle River.8 -2 -1.2 0.8 0.. recording.2 0 δ ψ 1 0. or likewise. THE ROOT-LOCUS DESIGN METHOD Root Locus Heading angle to 10 m/s disturbance 0..4 0. All rights reserved. photocopying. Pearson Education.4 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. or transmission in any form or by any means.2 -0. Inc. mechanical.1 PI control -0. (c) Based on your root locus. it provides a perfect measure of _ . Show that this value yields an unstable system with roots at s= 2:9.6. Inc.5069 40.. 13:5. Figure 5. . Pearson Education. Inc. electronic.02 rad(= 1 )? (Assume the system is stable. storage in a retrieval system. or likewise. photocopying. (Include transfer functions in boxes. Upper Saddle River. (g) What is the maximum damping factor of the complex roots obtainable with the con…guration in part (e)? (h) What is the value of KT for part (g)? © 2015 Pearson Education. Golden Nugget Airlines has opened a free bar in the tail of their airplanes in an attempt to lure customers. write to: Rights and Permissions Department.) to (b) Draw a root locus with respect to K. recording. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. We will model the passenger moment as a step disturbance Mp (s) = M0 =s. All rights reserved. +1:2 6:6j: (e) You are given a black box with rate gyro written on the side and told that when installed. what is the value of K when the system becomes unstable? (d) Suppose the value of K required for acceptable steady-state behavior is 600.61 shows the block diagram of the proposed arrangement. In order to automatically adjust for the sudden weight shift due to passengers rushing to the bar when it …rst opens. Figure 5.) (f) For the rate gyro in part (e). mechanical. sketch a root locus with respect to KT . with a maximum expected value for M0 of 0. Upper Saddle River. Assume K = 600 as in part (d) and draw a block diagram indicating how you would incorporate the rate gyro into the auto pilot. the airline is mechanizing a pitch-attitude auto pilot. NJ. NJ 07458.61: Golden Nugget Airlines Autopilot (a) What value of K is required to keep the steady-state error in less than 0. For information regarding permission(s).. with output KT _ . or transmission in any form or by any means. we de…ne the transfer function from Mp to : (s + 3)(s + 10) (s) = Mp (s) s(s + 10)(s2 + 4s + 5) + K(s + 3) Using the Final Value Theorem. Root locus for problem 5. Upper Saddle River. recording. . NJ. For information regarding permission(s).5070 CHAPTER 5. Upper Saddle River. mechanical. Solution: (a) Since any error is due to the disturbance Mp . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. write to: Rights and Permissions Department. (b) The characteristic equation of the system in Evans form is 1+K (s + 3) =0 s(s + 10)(s2 + 4s + 5) The root locus is plotted below.41 4 3 2 Imag Axis 1 0 -1 -2 -3 -4 -10 -8 -6 -4 -2 0 2 Real Axis © 2015 Pearson Education. electronic. NJ 07458. THE ROOT-LOCUS DESIGN METHOD (i) Suppose you are not satis…ed with the steady-state errors and damping ratio of the system with a rate gyro in parts (e) through (h). All rights reserved.. Inc. photocopying. the steady-state error is e(1) = lim s s!0 10Mo (s) Mo = Mp (s) s K (10)(0:6) < 0:02 K Therefore we need K > 300. or transmission in any form or by any means. Discuss the advantages and disadvantages of adding an integral term and extra lead networks in the control law. Inc. storage in a retrieval system. Pearson Education. Support your comments using Matlab or with rough root-locus sketches.. or likewise. All rights reserved. For information regarding permission(s). Inc. 1:22 6:63: (e) The output of the rate gyro box would be added at the same spot as the attitude sensor output.965 -10 0. mechanical.22 0.84 -15 -30 -25 0.6 -10 0. the characteristic equation is s4 + 14s3 + 45s2 + 650s + 1800 = 0 The roots of the equation are s = 13:5. or likewise. 2:94.42 0.84 0.99 30 0 25 20 15 10 5 0.40f © 2015 Pearson Education.6 0. Upper Saddle River. we can do the Routh test or solve the characteristic equation for the j! crossings. (f) With the rate feedback. .92 0. photocopying. storage in a retrieval system. Upper Saddle River. The characteristic equation of the system is (s) = 1+K (s + 3) s(s + 10)(s2 + 4s + 5) = s4 + 14s3 + 45s2 + (50 + K)s + 3K = 0 Plugging s = j!. Here. recording. the characteristic equation in Evans form is 1 + KT s (s + 600s (s + 3) =0 + 4s + 5) + 600 (s + 3) 10) (s2 The root locus is shown below. Pearson Education. we can write (j!) = ! 4 45! 2 + 3K + j 14! 3 + (50 + K)! = 0 q 50+K From the imaginary part. we have ! = 0. write to: Rights and Permissions Department. So the system goes unstable if K > 143:7. 14 .22 -5 0 5 10 Real Axis Root locus for Problem 5. NJ 07458. Root Locus 15 0.74 -20 -15 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. This agrees with the root locus.965 5 Imag Axis 0. the latter method is used. (d) When K = 600. we see that the roots are on the imaginary axis when K = 143:7..92 10 0. electronic. Substituting these into the real part.5071 (c) To …nd the stability boundary. or transmission in any form or by any means. Inc. NJ..42 0.74 0.99 -5 0. 41 4 3 2 Imag Axis 1 0 -1 -2 -3 -4 -10 -8 -6 -4 -2 0 2 Real Axis © 2015 Pearson Education. (h) From the root locus in Matlab. For information regarding permission(s). Upper Saddle River. Inc.. write to: Rights and Permissions Department. mechanical. photocopying. Upper Saddle River. it is KT = 0:288. we can draw a point around and …nd that the maximum damping ratio occurs at s = 3:54 12j with a max damping of = 0:282. or likewise.5072 CHAPTER 5. recording. NJ 07458. All rights reserved. NJ. Pearson Education. or transmission in any form or by any means. The root locus and step responses for the cases with an integrator and with an integrator+extra lead network are shown below. (i) Integral (PI) control would reduce the steady-state error to the moment to zero but would make the damping less and the settling time longer. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. A lead network could improve the damping of the response. Root locus for problem 5.. THE ROOT-LOCUS DESIGN METHOD (g) From the root locus in Matlab. we can immediately get the gain value at the maximum damping. . electronic. storage in a retrieval system. Inc. or likewise.. or transmission in any form or by any means. Kv 2 16 sec 3 1 : (b) Output-velocity (tachometer) feedback : Let H(s) = 1 + KT s and Dc (s) = K: Select KT and K so that the dominant roots are in the same location as those of part (a). This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. photocopying. All rights reserved. s+p p = 6: z Select z and K so that the roots nearest the origin (the dominant roots) yield 0:4. is it possible to obtain a Kv = 12 at = 0:4? Select K and p so that the dominant roots correspond to the proportional-control case but with Kv = 100 rather than Kv = 12. Compute Kv . write to: Rights and Permissions Department. If you can. mechanical. recording. (c) Lag network : Let H(s) = 1 and D(s) = K s+1 : s+p Using proportional control. Figure 5. Inc. NJ 07458.62. Pearson Education. storage in a retrieval system. © 2015 Pearson Education. . electronic. Upper Saddle River. 7.5073 41. Dc (s) = K s+z . give a physical reason explaining the reduction in Kv when output derivative feedback is used. For information regarding permission(s).. and indicate the location of the roots corresponding to your …nal design. 5. For each of the following cases. Upper Saddle River. Inc.41 (a) Lead network : Let H(s) = 1. draw a root locus with respect to the parameter K. Consider the instrument servomechanism with the parameters given in Fig. NJ.62: Control system for Problem 5. 1+z 6s(s2 + 51s + 550) + 40000 =0 s2 (s2 + 51s + 550) + 40000s Root locus with res pect to z 30 20 Imaginary Ax is 10 0 -10 -20 -30 -45 -40 -35 -30 -25 -20 Real Axis -15 -10 -5 0 5 Root locus for Problem 5. electronic. recording. All rights reserved. THE ROOT-LOCUS DESIGN METHOD Solution: (a) Setting p = 6z. s + 100:8 (b) With H(s) = 1 + KT s and Dc (s) = K. NJ. With these value. photocopying. a root locus can be drawn with respect to z. write to: Rights and Permissions Department.5074 CHAPTER 5..41(a) At the point of maximum damping. . NJ 07458. For information regarding permission(s). Pearson Education. mechanical. K = 8410. and KT = 0:045. or likewise. Inc. the values are z = 16:8 and the dominant roots are at s = 11 13j: So the compensator is s + 16:8 Dc (s) = 40000 . the velocity constant is 1 = lim s 1 s!0 Kv Y R 1 550 + KKT = s2 K ) Kv = 9:058 © 2015 Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Upper Saddle River. storage in a retrieval system. the characteristic polynomial should be in the form of (s + p)(s2 + 22s + 290) = s3 + (p + 22)s2 + (22p + 290)s + 290p Equating the coe¢ cients leads to p = 29. the closed-loop transfer function is Y K = 3 2 R s + 51s + (550 + KKT )s + K For this system to have poles at s = 11 13j:.. or transmission in any form or by any means. Upper Saddle River. the velocity constant is Kv = lim s s!0 K(s + z) 1 K = 2 s + 6z s(s + 51s + 550) 3300 Thus the Kv requirement leads to K 35200: With K = 40000. Inc. All rights reserved. and = 0:37. . Upper Saddle River. the dominant roots are at s = 4:7 11:69j. the velocity constant is Kv = lim s s!0 K(s + 1) 1 K = s + p s(s2 + 51s + 550) 550p So Kv = 100 can be obtained by setting K p = 55000. Therefore. NJ 07458. storage in a retrieval system. Inc. For information regarding permission(s). electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.. a root locus can be drawn with the parameter p 1+p s(s2 + 51s + 550) + 55000(s + 1) =0 s2 (s2 + 51s + 550) Root loc us with res pec t to p 20 15 10 Imaginary Axis 5 0 -5 -10 -15 -20 -40 -35 -30 -25 -20 -15 Real Ax is -10 -5 0 5 10 Root locus for Problem 5. s+1 With Dc (s) = K s+p . write to: Rights and Permissions Department. Setting K = 55000p. Thus we can choose p = 0:11 to place the poles near the desired s+1 . Pearson Education. the derivative action will minimize the deviation from the reference because the input signal is continuously increasing. (c) Using proportional control (Dc (s) = K). mechanical.. With this value. Inc. recording.5075 The output derivative feedback is acting only when there is a change in the output. the desired pole locations are marked with a dot ( ). locations. Upper Saddle River. photocopying.41(c) In the plot. or likewise. or transmission in any form or by any means. NJ. the velocity constant is Kv = lim sK s!0 K 1 = s(s2 + 51s + 550) 550 Therefore Kv = 12 can be obtained by setting K = 6600. Thus the compensator is Dc (s) = 6050 s + 0:11 © 2015 Pearson Education. for a ramp input. NJ 07458. THE ROOT-LOCUS DESIGN METHOD Problems and solutions for Section 5. Plot the loci for the 0 locus or negative K for each of the following: (a) The examples given in Problem 5. or likewise. Inc.5 (d) The examples given in Problem 5. All rights reserved. photocopying.5076 CHAPTER 5.42(a) © 2015 Pearson Education.7 (f) The examples given in Problem 5. mechanical. For information regarding permission(s).. or transmission in any form or by any means. storage in a retrieval system.6 42.. NJ. electronic. Inc.3 (b) The examples given in Problem 5.6 (e) The examples given in Problem 5.8 Solution: plot a 15 10 Imaginary A xis 5 0 -5 -10 -15 -15 -10 -5 0 Real Axis 5 10 15 (a) Problem 5. . Pearson Education. Upper Saddle River.4 (c) The examples given in Problem 5. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Upper Saddle River. recording. write to: Rights and Permissions Department. Inc.5 0 -0..6 0. write to: Rights and Permissions Department. Pearson Education. recording.2 0. Upper Saddle River.5 0 -0.5 -1 0.5 Imaginary Axis Imaginary Axis 1 0.. NJ. storage in a retrieval system. Inc.2 0 0. All rights reserved.8 1 -3 -0.4 Problem 5.5 plot f 3 Imaginary Axis 3 Imaginary Axis 0 2 1 0 -1 -2 2 1 0 -1 -2 -3 -0.5077 plot a plot b 10 20 8 15 Imaginary Axis Imaginary Axis 6 4 2 0 -2 -4 10 5 0 -5 -10 -6 -15 -8 -15 -10 -5 Real Axis 0 5 10 -15 -10 -5 Real Axis plot c 0 5 10 -2 0 2 6 8 plot d 2 1 1. mechanical.2 0 Real Axis 0.6 -0.5 Real Axis plot e 0. or transmission in any form or by any means.4 Real Axis 0. For information regarding permission(s).5 -20 -15 -10 -5 Real Axis 0 -1 5 -10 -8 -6 -4 Real Axis Problem 5.5 Real Axis 0 0. NJ 07458. photocopying.4 -0.5 -1 -0. Upper Saddle River.42(b) plot a plot b 10 Imaginary Axis Imaginary Axis 5 0 5 0 -5 -5 -4 -3 -2 -1 Real Axis 0 1 2 -6 -4 -2 0 2 Real Axis plot c 4 plot d 3 Imaginary Axis Imaginary Axis 4 2 1 0 -1 2 0 -2 -2 -3 -4 -1 -0.42(c) © 2015 Pearson Education. or likewise. .2 0. electronic. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.5 -1.4 -0. Inc.42(e) © 2015 Pearson Education. photocopying.4 -6 -0.5 Real Axis plot e 10 0 0. NJ.. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Upper Saddle River. Inc. write to: Rights and Permissions Department. Pearson Education.2 0 -0.6 -8 -10 -20 5 plot e 10 Imaginary Ax is Imaginary Ax is plot c 10 Imaginary Ax is Imaginary Ax is plot a 10 -0.2 -4 -0. or likewise.42(d) plot b 8 8 8 6 6 6 4 4 4 2 0 -2 Imaginary Ax is 10 2 0 -2 2 0 -2 -4 -4 -4 -6 -6 -6 -8 -8 -10 -5 0 Real Ax is 5 10 -8 -10 -5 0 Real Ax is plot d 1 8 0. Upper Saddle River.6 4 0. recording.5 -1 10 -10 Real Axis plot d 5 Real Axis 20 10 0 -10 -20 -20 -10 0 Real Axis 10 Problem 5. THE ROOT-LOCUS DESIGN METHOD 0 20 10 0 -10 -20 -50 -20 -10 0 10 Real Axis plot c 20 30 40 -20 Imaginary Axis Imaginary Axis plot b Imaginary Axis Imaginary Axis plot a 50 20 10 0 -10 -20 -20 -10 0 Imaginary Axis Imaginary Axis 5 0 -5 -10 -10 -5 0 0 -15 -10 0 5 10 0 -5 0 Real Axis g Imaginary Axis -5 Real Axis plot f 0.4 2 0 -2 10 -10 -5 0 Real Ax is 5 0. .8 6 0. mechanical. For information regarding permission(s). or transmission in any form or by any means.5 -0.. storage in a retrieval system. electronic.5078 CHAPTER 5.5 -10 5 10 -0.8 -10 0 Real Ax is 10 20 -3 -2 -1 Real Ax is 0 1 Problem 5. All rights reserved. NJ 07458. photocopying.6 3 0.5 -0. .2 0 -0.4 0. Upper Saddle River.5 -1 -1. NJ 07458. or transmission in any form or by any means.. Inc.8 4 1. mechanical.8 -4 -15 -10 -5 Real Axis 0 5 -10 plot d -5 Real Axis -2 -2 0 plot e 8 4 8 3 6 2 0 -2 4 2 Imaginary Axis Imaginary Axis 4 0 2 Real Axis plot f 10 6 Imaginary Axis 0. Inc. write to: Rights and Permissions Department. Upper Saddle River. Pearson Education. For information regarding permission(s). All rights reserved. storage in a retrieval system. electronic.2 1 Imaginary Axis Imaginary Axis Imaginary Axis plot a 1 2 1 0 -1 -0. or likewise.5 4 2 0 -2 -4 1 0 -1 -2 -4 -6 -6 -8 -30 -3 -8 -20 -10 Real Axis 0 -10 -5 0 Real Axis -4 -5 0 Real Axis 5 Problem 5. NJ..5 0.5079 plot b plot c 5 2 0.42(f) © 2015 Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.6 -3 0 -0.4 -2 -0. recording. Suppose you are given the plant L(s) = 1 . Pearson Education. storage in a retrieval system. recording. or likewise. Upper Saddle River. THE ROOT-LOCUS DESIGN METHOD 43. electronic. mechanical. or transmission in any form or by any means.5080 CHAPTER 5.2 0 -0.8 1 0.4 Imaginary Axis Imaginary Axis 0. For information regarding permission(s). Positive root locus Negative root locus 1.6 -1 -0.5 -4 -3 -2 -1 Real Axis 0 1 2 -1 -2 -1.5 Real Axis 0 0.5 -0.2 -0. NJ 07458. Inc. .5 0 0.5 -1 -0. write to: Rights and Permissions Department. All rights reserved.6 0.5 1 0..4 -0.5 1 Positive(left) and Negative(right) root locus for Problem 5. s2 + (1 + )s + (1 + ) where is a system parameter that is subject to variations. we see that the system is stable for all > 1.. © 2015 Pearson Education.8 -1. Solution: The characteristic polynomial in Evans form with respect to 1+ is s+1 =0 s2 + s + 1 The positive and negative root locus are shown below.43 From the root locus. Inc. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. photocopying. Use both positive and negative root-locus methods to determine what variations in can be tolerated before instability occurs. Upper Saddle River. NJ. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. photocopying. All rights reserved. or likewise. mechanical. Consider the system in Fig.. The region of stability is the area under the parabola and above the kp axis. storage in a retrieval system. . Pearson Education. and 4kp2 3kp + 9kI < 0 The third of these represents a parabola in the [kp. kI ] plane plotted below. Inc. or transmission in any form or by any means. electronic. © 2015 Pearson Education. Solution: (a) De…ne kp = K1 and kI = K1 K2 and the characteristic polynomial is a(s) = s4 + 1:5s3 + 0:5s2 + kp s + kI The Routh array for this polynomial is (s) = s4 + 1:5s3 + 0:5s2 + kp s + kI s4 : s3 : 1 1:5 3 4kp s2 : 6 9kI s : kp 3 4kp s0 : kI 0:5 kI kp kI For the system to be stable. K2 ) plane for which the system is stable.44 (a) Use Routh’s criterion to determine the regions in the (K1 . it is necessary that kI > 0. kp < 0:75. NJ. recording.5081 44. Upper Saddle River. (b) Use rltool to verify your answer to part (a). write to: Rights and Permissions Department. Figure 5. NJ 07458. Inc. Upper Saddle River.. 5. For information regarding permission(s).63: Feedback system for Problem 5.63. . Inc. .03 0. recording.kI plane for problem 5.43 0. storage in a retrieval system. or transmission in any form or by any means. consider kp = 3=8 and kI = 1=16: The roots of the characteristic equation are 1:309.6 0. 0:191. For points on the parabola. Inc.01 0 0 0.02 0.44(a) (b) When kI = 0. write to: Rights and Permissions Department.7 0.06 0. For information regarding permission(s). or likewise. Upper Saddle River. THE ROOT-LOCUS DESIGN METHOD Stability region in the kp.1 0. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. photocopying. mechanical. Upper Saddle River. Pearson Education.4 kp 0.8 Stability region for Problem 5. NJ. All rights reserved..5 0.07 0.05 kI 0.2 0.5082 CHAPTER 5. there is obviously a pole at the origin. and 0:5j: © 2015 Pearson Education.04 0. electronic. NJ 07458.3 0. For these locations. or transmission in any form or by any means. (b) Indicate the root locations corresponding to K = 16 on the locus of part (a). (b) With K = 16. The values from the step command in Matlab are Mp = 44:2%. recording. photocopying.64: Control system for Problem 5. . With K = 16. and a settling time of 4:6 sec. (c) For K = 16.5083 Figure 5. Upper Saddle River. For information regarding permission(s).64. the characteristic equation of the system is 1+K 1 =0 s(s + 2) The root locus is plotted at the end. it can be written in Evans form as s 1 + KT 2 =0 s + s + 16 The root locus is shown below. (e) For the values of K and KT in part (d). estimate tr and ts . Upper Saddle River. (d) For K = 16 and with KT set so that Mp = 0:05( = 0:707). Using the design relations. and ts . and ts = 4:32 sec : The pole locations are indicated with ( ) on the plot for (a). or likewise. mechanical. tr = 0:32 sec. storage in a retrieval system. electronic.. NJ. estimate the transient-response parameters tr .45 45. Pearson Education. The block diagram of a positioning servomechanism is shown in Fig. ! n = 4 and = 0:25. Compare your estimates to measurements obtained using the step command in Matlab. (c) The characteristic equation of the system is s2 + (2 + KT )s + K = 0.. 5. what is the velocity constant Kv of this system? Solution: (a) When KT = 0. write to: Rights and Permissions Department. All rights reserved. draw the root locus with respect to KT . © 2015 Pearson Education. Mp . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Inc. Inc. Compare your estimates to the actual values of tr and ts obtained using Matlab. NJ 07458. (a) Sketch the root locus with respect to K when no tachometer feedback is present (KT = 0). we’d estimate the overshoot to be Mp = 45% and a rise time of 0:45 sec. Upper Saddle River.45 (e) The velocity constant is Kv = lim s s!0 K K = = 2:83 s(s + 2 + KT ) 2 + KT © 2015 Pearson Education. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. and ts = 1:62 sec : The values from the step command in Matlab are Mp = 4:3%. storage in a retrieval system.5 Time (sec) 2 2. tr = 0:45 sec. NJ. Inc.16 0 -5 -4 -3 -2 Real Axis -1 0 1 2 0 0.45(a) Step response for Problem 5. NJ 07458.5 -3 0.5 1 1. For information regarding permission(s).34 0. or transmission in any form or by any means.985 6 0 5 -1 0.76 0. or likewise. write to: Rights and Permissions Department.76 0. All rights reserved..5 0. recording.5 Plots for Problem 5. Problem 5. Root locus for Problem 5.5 0.34 0.64 0.5 2 1 Amplitude Imaginary Axis 4 0 -2 0.5084 CHAPTER 5. Upper Saddle River. and ts = 1:65 sec :.86 -4 -6 0.94 4 3 2 Amplitude Imaginary Axis 2 0.45(c) 3 Time (sec) 4 5 6 step response for Problem 5.86 0. .45(d) 4 1. mechanical..94 1 0. Inc.7 damping.64 0. photocopying. tr = 0:54 sec.45(b) 6 1.5 -4 -6 -6 0 -4 -2 0 2 4 0 1 2 Real Axis Root locus vs KT . electronic.16 3 1 1 0. Pearson Education.5 0. the locations are marked with ( ). This shows that KT = 3:66: Using the formulas inside the back cover yields Mp = 5%. THE ROOT-LOCUS DESIGN METHOD (d) Use rloc…nd on the locus vs KT to …nd the KT value that yields 0.985 -2 0. storage in a retrieval system. Solution: (a) The roots are complex for a0 > 0:25: We select a0 = 1 and the roots are at s = 0:5 0:866j.46 (a) With g = 0 and complex. NJ. electronic. (b) With respect to g. where g and a0 are gains. write to: Rights and Permissions Department. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. …nd a value for a0 such that the poles are (b) Fix a0 at this value.65: Control system for Problem 5. photocopying. or transmission in any form or by any means.5085 46. = 1. recording. Figure 5. adjusting g corresponds to varying the location of a zero in the s-plane. Upper Saddle River. or likewise. All rights reserved. For information regarding permission(s). .. Pearson Education. the root-locus form of the characteristic equation s is 1 + g 2 = 0: The locus is plotted below. For a …xed value of a0 . and construct a root locus that demonstrates the e¤ect of varying g. Consider the mechanical system shown in Fig. NJ 07458. Upper Saddle River.65. Inc. The feedback path containing gs controls the amount of rate feedback. mechanical. 5. s +s+1 © 2015 Pearson Education. Inc.. 99 -0. From the Appendix W5.FPE7e.47 Solution: Matlab cannot directly plot a root locus for a transcendental function. Pearson Education.66: Control system for Problem 5. photocopying.22 0 Real Axis 47.22 0.5 0.8 -2 0. All rights reserved. write to: Rights and Permissions Department. storage in a retrieval system. Inc.8 0.25 -0.42 0.42 -0. recording.5 1..5086 CHAPTER 5.6.4 0.5 0. . we see that the Padé(1. electronic.6 0.84 0.75 1. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction.92 0.1) approximation for © 2015 Pearson Education.99 Imag Axis 0.25 1 0.6 -1 0.74 0.92 0.66 using the Padé(1.4 0. NJ 07458. or transmission in any form or by any means.3.2 2 0 1.6.) Figure 5. Inc. mechanical.6 -0.965 -0.2 0. Sketch the root locus with respect to K for the system in Fig.6 0. what is the range of values of K for which the system is unstable?(Note: The material to answer this question is contained in Appendix W5.. NJ. Upper Saddle River.46 0. Upper Saddle River. For information regarding permission(s).com. or likewise.84 -1.5 0.74 0.1) approximation and the …rst-order lag approximation.75 0.3 discussed in www.965 0. For both approximations. 5. THE ROOT-LOCUS DESIGN METHOD Rootlocus for problem 5. Inc. For information regarding permission(s).1) Pade aproxim ates 3 2 Imaginary Axis 1 0 -1 -2 -3 -6 -5 -4 -3 -2 Real Axis -1 0 1 2 Solutions for Problem 5.1)) is e s = 1 : s+1 With the Padé(1. All rights reserved. photocopying.. however. The locus for the …rst-order lag is also shown in green. However. NJ. This demonstrates a limitation for this approximation of the delay.1). a locus valid for small values of s can be plotted. for higher order systems. Inc.46 with the (1. recording. write to: Rights and Permissions Department. . storage in a retrieval system. Pearson Education. s+2 and the …rst-order lag approximation (Padé(0. This yields Kmax = 2 for the Padé(1.1) approximation. as shown below by the red curve.5087 e s is e s = 1 (s=2) = 1 + (s=2) s 2 . This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. NJ 07458.47 © 2015 Pearson Education. Upper Saddle River. The rloc…nd routine is used by placing the cursor on the j! axis to …nd the maximum value of K at the instability boundary. electronic. mechanical. Upper Saddle River. or transmission in any form or by any means. or likewise.. the …rst order lag can be useful.1)and (0. Root loci f or problem 5. it produces a locus with two branches going to in…nity at s = 1=2: thus there is no value of K that produces instability for this approximation for a …rst order system. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. Such a system must be conditionally stable for it will be unstable if the gain is small enough. photocopying. or transmission in any form or by any means. Solution: The angles of departure from a triple pole are 180 and 60 for the negative locus and 0 and 120 for the positive locus. or likewise. Upper Saddle River. electronic..5088 CHAPTER 5. at least one pole starts out into the right-half plane. Upper Saddle River. For information regarding permission(s). recording. NJ. Inc. Pearson Education. Inc. NJ 07458. mechanical. write to: Rights and Permissions Department.. . storage in a retrieval system. All rights reserved. THE ROOT-LOCUS DESIGN METHOD 48. In either case. © 2015 Pearson Education. Prove that the plant G(s) = 1=s3 cannot be made unconditionally stable if pole cancellation is forbidden. 49 Problem 5. Pearson Education. G(s) = 1 . For information regarding permission(s). Solution: The root loci for four values are given in the …gure. electronic. write to: Rights and Permissions Department.49 Problem 5. making sure to include the point p = 2. Inc. NJ 07458.49 10 p=5 2 4 p = 10 20 5 10 Imag Axis Imag Axis -2 Real Axis 0 0 -10 -5 -20 -10 -10 -5 0 5 Real Axis -20 -10 0 10 20 Real Axis Solutions for Problem 5. Inc. or transmission in any form or by any means. Upper Saddle River. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction. ... Upper Saddle River. or likewise.49 6 4 4 2 2 Imag Axis Imag Axis p=2 p=1 6 0 -2 -4 0 -2 -4 -6 -6 -5 0 5 -6 -4 0 Real Axis Problem 5.5089 49. storage in a retrieval system. NJ. photocopying. mechanical. For the equation 1 + KG(s) where. s(s + p)[(s + 1)2 + 4] use Matlab to examine the root locus as a function of K for p in the range from p = 1 to p = 10. recording. The point is that the locus for p = 2 has multiple roots at a complex value of s: Problem 5.49 © 2015 Pearson Education. All rights reserved.