Linear control system analysis and design with matlab

Free ebooks ==> www.Ebook777.com Linear Control System Analysis and Design ® with MATLAB Sixth Edition Constantine H Houpis Stuart N Sheldon www.Ebook777.com LINEAR CONTROL SYSTEM ANALYSIS AND DESIGN WITH MATLAB® Sixth Edition Free ebooks ==> www.Ebook777.com AUTOMATION AND CONTROL ENGINEERING A Series of Reference Books and Textbooks Series Editors FRANK L LEWIS, Ph.D., Fellow IEEE, Fellow IFAC Professor The Univeristy of Texas Research Institute The University of Texas at Arlington SHUZHI SAM GE, Ph.D., Fellow IEEE Professor Interactive Digital Media Institute The National University of Singapore PUbLIshED TITLEs Linear Control System Analysis and Design with MATLAB®, Sixth Edition, Constantine H Houpis; Stuart N Sheldon Real-Time Rendering: Computer Graphics with Control Engineering, Gabriyel Wong; Jianliang Wang Anti-Disturbance Control for Systems with Multiple Disturbances, Lei Guo; Songyin Cao Tensor Product Model Transformation in Polytopic Model-Based Control, Péter Baranyi; 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not warrant the accuracy of the text or exercises in this book This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S Government works Version Date: 20130916 International Standard Book Number-13: 978-1-4665-0427-1 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under U.S Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400 CCC is a not-for-profit organization that provides licenses and registration for a variety of users For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com www.Ebook777.com To the memory of John J D’azzo An educator, researcher, and leader who motivated, encouraged, and inspired his students and colleagues to achieve their greatest potential A perfect gentleman, he never raised his voice in anger It is these attributes that inspired others, who knew and worked with him, to continue to contribute to the education of students May his memory be eternal Free ebooks ==> www.Ebook777.com www.Ebook777.com Contents Preface xix Authors xxiii Part I  Introductory Material Chapter Introduction 1.1 Introduction 1.2 Introduction to Control Systems 1.2.1 Classical Examples 1.2.2 Modern Examples .6 1.3 Definitions 1.4 Historical Background 11 1.5 Control System: A Human Being 13 1.6 Digital Control Development 15 1.7 Mathematical Background 16 1.8 Engineering Control Problem 18 1.9 Computer Literacy 21 1.10 Outline of Text 21 References 22 Chapter Unmanned Aircraft Vehicles 25 2.1 Introduction 25 2.2 Twentieth-Century UAV R&D .25 2.3 Predator 25 2.3.1 Introduction .25 2.3.2 Mission 27 2.3.3 Features 27 2.3.4 Background .28 2.3.5 General Characteristics .28 2.4 Grim Reaper (U.S Air Force Fact Sheet MQ-9 Reaper, Posted on January 5, 2012) 29 2.4.1 Mission 29 2.4.2 Features 29 2.4.3 Background 30 2.5 RQ-4 Global Hawk (U.S Air Force Fact Sheet RQ-4 Global Hawk, Posted on January 19, 2012) 30 2.5.1 Mission 30 2.5.2 Features 31 2.5.3 Background 31 2.6 Summary 31 Reference 32 ix 673 Problems 19.6 Repeat Problem 19.5 by the PCT DIG technique (a) By use of Equation 19.8, obtain [Dc(z)]TU and the control ratio [C(z)/R(z)]TU (b) Obtain the values of Mp, tp, and ts and compare these values with those of Problem 19.5 19.7 Repeat Problem 19.4 for T = 0.025 19.8 Repeat Problem 19.5 for T = 0.025 19.9 Repeat Problem 19.6 for T = 0.025 and compare the results with those of Problem 19.8 19.10 (CAD Problem) For the basic system of Figure 19.1, where Gx(s) = K x/s(s + 4): (a) Determine Mp, tp, ts, and Km for ζ = 0.6 and T = 0.01 s by the DIR method (b) Obtain the PCT control system of part (a) and the corresponding figures of merit (c) Generate a controller Dc(z) (with no ZOH) that reduces the value of ts of part (a) by approximately one-half for ζ = 0.6 and T = 0.01 s Do first by the DIG method (s plane) and then by the DIR method Obtain Mp, tp, ts, and Km for c(t) of the PCT and c*(t) (for both the DIG and the DIR designs) Plot c*(t) for both designs (d) Find a controller Dc(z) (with no ZOH) that increases the ramperror coefficient of part (a), with ζ = 0.6 and T = 0.01 s with a minimal degradation of the transient-response characteristics of part (a) Do first by the DIG method (s plane) and then by the DIR method Obtain Mp, tp, ts, and Km for c(t) of the PCT and c*(t) (for both the DIG and the DIR designs) Plot c*(t) for both designs (e) Summarize the results of this problem in tabular form and analyze 19.11 Using the PCT DIG technique, design a controller [Dc(z)]TU for Problem 18.13 that will achieve the desired performance specifications 19.12 For the digital control system of Figure 19.1, the plant transfer function is Gx (s ) = K x (s + 3) s(s + 1)(s + 4) The desired figures of merits (FOM) are Mp = 1.043 and ts = s (a) By use of the PCT DIG technique, where T = 0.02 s, determine Dc(s) that will yield the desired dominant poles, based upon satisfying the specified FOM, for the system’s control ratio (b) Determine the resulting FOM for the compensated system of part (a) (c) By use of Equation 19.8, obtain [Dc(z)]TU and the control ratio [C(z)/R(z)]TU (d) Obtain the values of Mp, tp, and ts for part (c) and compare these values with those of part (b) Free ebooks ==> www.Ebook777.com www.Ebook777.com Answers to Selected Problems CHAPTER     5.1 (a) Loop e =  R1 + + LD  i1 −  + LD  i2 CD    CD      =  R2 + LD +  i2 −  LD + CD  i1 CD     1  (b) Node va :  + CD +  va − CDvb = e R R R 2    vb :  + CD  vb − CDva =  LD  (c) State   x1    =    x2   −  L   y1    =    y2   −1       x1   C e  + R2  R1R2   x2   −  L ( R1 + R2 )   L ( R1 + R2 )  R1R2   R2  ( R1 + R2 )   x1   ( R1 + R2 )   + e R1R2   x2   R2  −  (R + R )  ( R1 + R2 )    − 5.3 Circuit (b)   x1    =    x2   −  L  1  0 C1   x1     + e   R  x −     L1  L1  5.6 (d) G( D) = y K = f (t ) D + BD + KD + KBD + 3K 675 Free ebooks ==> www.Ebook777.com 676 Answers to Selected Problems (e)   −2 K A=    K 0 0 K −2 K 0 0     0 b =   c = 0   0    −B 0   K1 x1 x2 5.11 (a) B1 M2 f(t) K2 M1 B2 (b) ( M1D + B1D + K1 ) x1 − ( B1D + K1 ) x2 = −( B1D + K1 ) x1 + [ M D + ( B1 + B2 ) D + ( K1 + K )]x2 = f (c) Let x1 = x1, x2 = x2, x3 = x1, x4 = x2, u = f(t)     K x = −  M1  K1   M2 0 K1 M1 K + K2 − M2 B − M1 B1 M2           B1 x+  f M1      B1 + B2   M  −  M2  y = [1 5.18   C1ad C1b  (a + b)(c + d ) + D  y1 ≈ a + b x1   x1 a a+b x΄2 + – x2 C1 y1 D x2˝ ad (a + b) (c + d) 5.21 e= La J R J + La B R B + K b KT D θm + a D θm + a Dθm KT KT KT www.Ebook777.com 0 0]x 677 Answers to Selected Problems CHAPTER 6.3 y(t) = −0.005305e−100t + 1.1483e−3.125t cos(6.3431t + 150.026°) + 6.4 From problem 5.18,   C1ad C1b y1  (a + b)(c + d ) + D  y1 ≈ a + b x1 ⇒ [0.1 + D]y1 = x1 ⇒ x = D + 0.1   y1 y 0.4 = ⇒ = 0.1u−1 (t ) D + 0.1 u−1 (t ) D + 0.1 y(t ) = − 4e −0.1t 6.7 (a) Physical variables: Let x1 = vC = vo, x2 = iL and u = E = 50; thus  x1   −10  =  x2   0.1 −106   x1  10     +   E y = x1 −465  x2    (b) vC (t ) = 48.95 + A1e m1t + A2e m2 t where DvC (t ) = m1 A1e m1t + m2 A2e m2 t Inserting initial conditions yields vC (t ) = 48.95 + 1.16e −475.5t − 50.1e −9989.5t 6.13 Q(λ) = |λI − A| = λ2 + 6λ + = (λ + 2)(λ + 4) e −2 t = α − 2α1, e −4 t = α − 4α1 α = 2e −2 t − e −4 t, α1 = 0.5e −2 t − 0.5e −4 t  −e −2 t + 2e −4 t Φ(t ) = e At = α I + α1A =  −2 t −4 t  −e + e 2e −2( t − τ) Φ(t − τ)Bu(τ) =  −2( t − τ)  2e t 2e −2 t − 2e −4 t   2e −2 t − e −4 t  −2e −4( t − τ)   −e −4( t − τ)   0.5 − 3e −2 t + 4.5e −4 t  x(t ) = Φ(t )x(0) + Φ(t − τ)Bu(τ)dτ =  −2 t −4 t  0.75 − 3e + 2.25e  ∫ y(t ) = x1 Free ebooks ==> www.Ebook777.com 678 Answers to Selected Problems CHAPTER 7.1 (a) f (t ) = e −2 t [5t + 1.25 sin(4t + 180°)] (c) f(t) = 16te−t + 4e−t + 8.1976[e−t sin(2.2361t + 209.2°)] 7.3 ω(t) = 200 rad/s + 300e−0.75t rad/s 7.4 (b) Problem 6.3 (D + 100)[(0.1)D2 + 0.625D + (5)]xb = 50e X b (s) = 500 5000 = s(.1s + 10.625s + 67.5s + 500) s(s + 100)(s + 6.25s + 50) y(t ) = −0.005305e −100 t + 1.1483e −3.125t cos(6.3431t + 150.026°) + 7.5 (d) F (s) = − + − 6s 4(s + 1) 12(s + 3) s (h) F (s) = 1 1 + − − s ( s + 4) ( s + + j ) ( s + − j ) 7.6 (b) x (t ) = 10 −3t + e − 5e −2 t 3 7.7 (c) X= −3s + s (s + 2s + 10) x(t ) = −0.02 + 0.1t + 1.0269e − t cos(3t + 88.8884 ) 7.8 (a) Problem 7.1 (a) 0  (b) 1.3846  (c) 7.12 System (a) s + Φ(s) = [ sI − A]−1 =   −2 −1 −1  s +  =∆ s + 3    s + 2 ∆ = s + 5s + = (s + 1)(s + 4) X(s) = [ sI − A]−1 x(0) + [ sI − A]−1 bU(s) s +1    s +1      s(s + 1)((s + 4)   s(s + 4)     1  s =  + =  X( s ) =  = ∆  s +  ∆  s +  s ∆  (s + 1)(s + 2)   (s + 1)(s + 2)   (s + 2)     s(s + 1)(s + 4)   s(s + 4)  s www.Ebook777.com 679 Answers to Selected Problems G(s) = c[ sI − A]−1 b (b) X( s ) = s + 0]   [1 ∆  0    = s +  1  (s + 1)(s + 4)   1 −1  y(t ) = x1 (t ) = ᏸ −1  = ᏸ −1  +    s( s + 4)   4s 4(s + 4)  (c) y(t ) = 0.25 − 0.25e −4 t λ1 = −1, λ2,3 = −1 ± j2 7.15 (a) CHAPTER θ −60(s + 0.06333) = δ E (s + 10)(s + 0.15s + 8.005) 8.5 8.10 (e)  ( BRm + KT K m ) − JRm  K  − g  Lc   X=  Kc   Ms    − Rc Lc 0 0 0 0 K − s Ms Kx Lf Bs Ms KT K g  Rm   0     1   Lc  X +  r  0  0     0 Rf  − L f  8.13 System (1) (d)  −2  z=  0 −3 0  2     z +  −2  u, y = [1   −4  CHAPTER 9.2 (a) < K < 1280 /9 s1,2 = ± j 2.5827 for K = 142.22 + (c) < K < 30 s1,2 = ± j2.2365 for K = 30 9.3 (b) (s + ± j1.414)(s − ± j1.414) (f) (s − 1)(s + 2)(s − 3)(s + 5) 1]z Free ebooks ==> www.Ebook777.com 680 Answers to Selected Problems 9.7 s4 s3 s2 s1 s0 6−K 12 + 4K − K K 2+K 4K K Stable for < K < Kp Kv Ka 20 ∞ 36 0 9.12 (a) (d) 9.13 r(t) (a) (c) 2t t2 5/2 ∞ ∞ ∞ 9.20 C (s) K ( s + 4) = R(s ) s + 5s + 5s + (4 + K )s + K CHAPTER 10 10.1  (g) R  eal axis loci: −2 to −∞ for K > and −2 to ∞ for K < Branches: γ = 180° for K > 0, and γ = 0° for K < Imaginary crossing: at the origin for K = −0.1176 Angle of departure: 116.5° for K > and −63.5° for K < at pole −1 + j1 Angle of approach: 49.1° for K > and −130.9° for K < at pole −3 + j5 10.4 K = 0.05427 10.6 Km = 29.6684 10.7 (b) −175 < K < 1105 10.12 A = CHAPTER 11 11.1 (1) ωc = 0.867, ωϕ = 2.24 (3) ωc = 6.22 11.5 (b) Total correction −2.01 www.Ebook777.com 681 Answers to Selected Problems 11.9 Case II G(s) = 20(1 + 0.5s) s(1 + 5s)(1 + 0.125s) 11.14 (b) For K > PR = 1: N = thus Z R = 1; therefore, the system is unstable CHAPTER 12 12.1 Plant Mm ωm ωc ωϕ K K1 γ (2) (4) 1.32 1.32 8.02 0.658 15.8 1.51 6.94 0.606 3.92 0177 7.84 0.708 45° 45° 12.5 (a) System (1) at K = Mm = 1.44 at ωm = 0.828 System (4) at K = Mm = 0.102 at ωm = 1.32 12.9 System (b) For K = 0.082, Mm = 1.12, ωm = 0.761, γ = 53.2° 12.11 (a) Mm = dB = (b) For K = 10.7, γ = 45° (c) Gain margin = 13 dB = 4.49 CHAPTER 13 13.3 System (1) Basic Lag compensated: α = 10, T = Lead compensated: α = 0.1, T = 0.25 Lag–lead: α = 10, T = 2, 0.25 Dominant Roots Other Roots K1(s−1) Tp (s) ts (s) Mo −1.25 ± j2.166 −1.03 ± 1.787 −6.501 −5.267 −0.726 −3.123 1.72 16.16 1.67 1.8 3.36 5.37 0.14 0.41 3.13 0.916 1.78 0.10 −0.598 −2.897 −40.40 30.18 0.967 5.15 0.244 −2.23 ± j8.86 −2.08 ± j3.6 13.6 (a) ϕb = 78.7°, b = 2.4 (b) α = 1.25 (c) A lag compensator 13.18 (a) For ζ = 0.5, the uncompensated system yields the following data: K = 25.01, K1 = s−1, Mp = 1.163, = 0.725 s, and ts = 1.615 s (b) Based upon the desired values of ζ and σ = −7, the desired dominant poles are p1,2D = −7 ± j12.124 One possible solution is as follows: let a = 40, b = 0.01, c = 0.1 By applying the angle condition and the use of a CAD package, the following results are obtained: A = 6.0464, Kh = 11.18, K1 = 126.1 s−1, Mp = 1.61, = 0.27 s, and ts = 0.426 s 13.21 A = 274.396, Kh = 50, K x = 0.020993 Free ebooks ==> www.Ebook777.com 682 Answers to Selected Problems CHAPTER 14 14.1 1/T a b 0.02 c d 0.02, Kx ωϕ γ(°) Mm Mp (s) ts (s) K1 1270 1250 1270 27,000 23,700 26,900 23,700 2.1 51.5 51.5 51 51.5 57.6 51.5 57.4 1.15 1.14 1.16 1.43 1.16 1.42 1.16 1.16 1.16 1.17 1.25 1.16 1.25 1.17 1.27 1.3 1.27 0.5 0.52 0.5 0.53 2.69 2.02 2.0 1.62 1.32 1.58 1.34 2.335 22.98 23.35 4.96 4.36 49.4 43.57 14.5 (a) G(s) = 2.1 4.62 4.62 B  + Ts  with T = and α = 0.5 K 8K  + αTs  14.10 For the basic system—with Mm = 1.16: K1 = 0.557, ωm = 0.51, ωϕ = 0.49, γ = 51.1°, Mp = 1.17, = 5.51 s, ts = 11.9 s 14.17 (a) K2x = 2.02, ωϕ = 2.15, Mm = 1.4, ωm = 1.33, Mp = 1.29, tp, = 1.39 s, and ts = 3.03s  (c) For γ = 50°, u = ωϕ T = 8.6T = 1.19 yields T = 0.1384 From the graphical construction, with ω1 ≤ ωϕ / 20 and ω2 ≥ 20ωϕ = 172, obtain Kt = 24.28 and A = 180, thus C (s) 2907(s + 1)(s + 7.225) = E (s ) s (s + 1.108)(s + 406.3) The dB crossing slope is −12 dB/oct at ω = 6, but the zero at −7.225 will have a moderating effect on the degree of stability (d) Mm = 1.47, Mp = 1.30, = 0.34 s, ts = 1.03 s, K2 = 46.65; since the values of Mp are essentially the same, a good improvement in system performance has been achieved, γ = 51° and ωϕ = 9.12 CHAPTER 15 15.2 (a) 2.26(s + 20) (s + 21) Gc (s) = (b) Figures of merit for a unit step input Rise time Peak value Peak time Settling time Final value MT(s) Y(s)/R(s) 0.68964 1.1181 1.5016 2.9509 1.0011 0.67142 1.1312 1.4773 3.7099 15.4 (a) Gc (s) = 2.465(s + 2.1) (s + 2.58) www.Ebook777.com 683 Answers to Selected Problems (b) Figures of merit for a unit step input Rise time Peak value Peak time Settling time Final value MT(s) Y(s)/R(s) 0.95415 1.1001 2.0945 3.0553 0.95638 1.0969 2.0941 3.0418 15.8 For y(t)ss = 0, H(s) must have a pole at the origin Make the pole c nondominant so that it has no effect on Lm (1/H) in the region of ≤ ω ≤ 16 → c = 200 With KH = 3350, αp = 0.00377 @ = 0.168 CHAPTER 16 16.3 System (1) (a) Not completely observable (b) Completely controllable (c) G(s) = s−2 (d) One observable state, two controllable states (e) Unstable System (4) (a) Not completely observable (b) Not completely controllable (c) G(s) = s +1 (d) One observable states, two controllable states (e) Stable 16.5 (a) Insert a cascade compensator: Gc (s) = s +1 k3 s + (k3 + k2 )s + k1 (d) A desired control ratio is (b) H eq (s) = desired Y (s ) 3A 625 = = R(s ) (s + ± j 4)(s + 25) s + 31s + 175s + 625 (e) With A = 208.33, k = [1 0.1456 0.1296] 625 K1 = 3.57 (g) Geq = s + 31s + 175s (h) Mp = 1.09, = 0.822 s, and ts = 1.23 s Free ebooks ==> www.Ebook777.com 684 Answers to Selected Problems MT ( s ) = 16.8 (a) 11, 700 11, 700 = (s + 12 ± j 21)(s + 20) s + 44s + 1, 065s + 11, 700 which yields the FOM: Mp = 1.06, = 0.215 s, and ts = 0.273 s, K1 = 10.99 16.15 Qd (λ) = λ + 9λ + 18, kcT = [ −5 x1 (t ) = −1] −3t −6 t e − e ⋅ x2 (t ) = −5e −3t + 4e −6 t 3 CHAPTER 17 17.1 17.8 (a.1) SKM (s) = = K =15 s + 7s + 25ss + 39s s + 7s + 25s + 54s + 75 (a) KG = 30.8 (b) No, use a compensator of the form Gc (s) = 17.9 s + 7s + 25s + 39s s + 7s + 25s + (39 + K )s + 5K A(s + a) (s + b)(s + c) (b) Setting dx1/dt = dx2/dt = yields equilibrium points: x1a = x2a = x1b = x2b = x1c = −2, x2c =  − x2 Jx =   −2 x1 −2 x1    x = x0 CHAPTER 18 18.3 (a) z−1 + z−2 + 0.1875 z−3 − 0.4063 z−4 + … 18.8 (a) C(z) = C(z)[1.5z−1 −0.8125z−2 + 0.3125z−3] + E(z)[z−1 + 0.5z−2] c(kT) = 1.5c[(k − 1)T] − 0.8125c[(k − 2)T] + 0.3125c[(k − 3)T] + e[(k − 1)T] + 0.5e[(k − 2)T] 18.12 (a) Gz ( z ) = K (3T − + e −3T ) Te −3T ) K G ( z − z1 ) −(1 − e −3T − 3T ⋅ KG = ⋅ z1 = −3T ( z − 1)( z − e ) (3T − + e −3T ) K G ( z − z1 ) K G ( z − z1 ) C ( z) = = R( z) K G ( z − z1 ) + ( z − 1)( z − e −3T ) z + ( K G − − e −3T )z + (e −3T − K G z1 ) www.Ebook777.com 685 Answers to Selected Problems CHAPTER 19 19.1 (a) z Domain Value s Plane Value −1.5 −2 −20 −4 ± j3 −6 ± j8 Equation 19.8 esT 0.941748 0.923077 0.428571 0.84615 ± j 0.10256 0.75 ± j 0.25 0.941765 0.923116 0.449329 0.84602 ± j 0.10201 0.74670 ± j 0.24745 19.3 System  C (s)  K ( s + 6) Ks + K (a)   = (s + ± j 6)(s + 8) = s + 16s + 116s + 416 → K = 69.333 R ( s )  M (b) For T = 0.1 → ± j(0.6114/T) = ± j6.114 and (−0.1)/T = −1; thus, the zero and poles of the desired control ratio not lie in the good Tustin region of Figure 19.16 Thus, 0.6114 0.1 ≥ → T ≤ 0.1019, ≥ → T ≤ 0.0125, thus T ≤ 0.0125 T T 19.10 (a) G( z) = 4.934 E − 5K ( z + 0.9868) ⋅ K = 10.8 for ζ = 0.6 ( z − 1)( z − 0.9608) C ( z) 0.00053287( z + 0.9868) = ( z − 1.96 z + 0.9613) R( z ) Mp = 1.0946, = 1.2 s, ts = 1.81 s, and K1 = 2.7 (b) The PCT system is G pc (s) = G A (s )Gx (s) = 200 K and K = 10.8 s(s + 4)(s + 200) C (s) 2160 = R(s ) (s + 3.945s + 10.8)(s + 200.1) Mp = 1.0946, = 1.2 s, ts = 1.81 s, and K1 = 2.7 (c) DIG—To reduce ts by about 1/2 select Dc (s) = K c ( s + 4) (s + 10) G pc (s) = Dc (s)G A (s)Gx (s) = For K = 64.9071 200 K s(s + 10)(s + 2000) C (s) 12981 = R(s ) (s + 9.66s + 64.8)(s + 200.3) Mp = 1.0947, = 0.49 s, ts = 0.8 s, and K1 = 6.49 Free ebooks ==> www.Ebook777.com 686 Answers to Selected Problems (d) DIG—To increase the ramp error coefficient Dc (s) = K c (s + 0.01) (s + 0.001) G pc (s) = Dc (s)G A (s)Gx (s) = For K = 10.7969 200 K s(s + 10))(s + 200) C (s) 2159.38(s + 0.01) = R(s ) (s + 3.936s + 10.76)(s + 0.001003)(s + 200.3) Mp = 1.0984, = 1.2023 s, ts = 1.8469 s, and K1 = 26.99 (e) The results are shown in the following table System Domain K Mp (s) ts (s) K1 (a) Basic (b) PCT (c) DIG (c) DIG (c) DIR (d) DIG (d) DIG s s s z z s z 11.11 10.8 64.9 64.9 64.9 10.7969 10.7969 1.0948 1.0946 1.0947 1.0949 1.0946 1.0984 1.0984 1.18 1.2 0.49 0.49 0.49 1.2023 1.2 1.78 1.81 0.8 0.739 0.738 1.8469 1.84 2.78 2.7 6.49 www.Ebook777.com 26.99 SYSTEMS & CONTROLS Linear Control System Analysis and Design with MATLAB ® Sixth Edition Thoroughly classroom-tested and proven to be a valuable self-study companion, Linear Control System Analysis and Design: Sixth Edition provides an intensive overview of modern control theory and conventional control system design using in-depth explanations, diagrams, calculations, and tables Keeping mathematics to a minimum, the book is designed with the undergraduate in mind, first building a foundation, then bridging the gap between control theory and its real-world application 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