http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering rd Edition By Norman S Nise John Wiley & Sons Copyright © 2000 by John Wiley & Sons, Inc All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any from or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 7508400, fax (978) 750-4470 Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, e-mail: PERMREQ@WILEY.COM To order books please call (800) 225-5945 ISBN 0-471-36601-3 John Wiley & Sons, Inc 605 Third Avenue New York, NY 10158 USA Solutions to Skill-Assessment Exercises Chapter 2.1 The Laplace transform of t is F(s) = using Table 2.1, Item Using Table 2.2, Item 4, s2 (s + 5)2 2.2 Expanding F(s) by partial fractions yields: F(s) = A B C D + + + s s + (s + 3) (s + 3) where, A= 10 ( s + 2)(s + 3)2 D = (s + 3)2 = S→0 10 B= s(s + 3)2 = −5 C = S→ −2 10 10 = , and s(s + 2) S→ −3 40 dF(s) = ds s→ −3 Taking the inverse Laplace transform yields, f (t) = 10 40 − 5e −2t + te −3t + e −3t 9 2.3 Taking the Laplace transform of the differential equation assuming zero initial conditions yields: s3C(s) + 3s2C(s) + 7sC(s) + 5C(s) = s2R(s) + 4sR(s) + 3R(s) Collecting terms, (s + 3s + 7s + 5)C(s) = (s + 4s + 3)R(s) Thus, Solutions to Skill-Assessment Exercises s + 4s + C(s) = R(s) s + 3s + 7s + 2.4 G(s) = 2s + C(s) = R(s) s + 6s + Cross multiplying yields, dc dr d 2c + + 2c = + r dt dt dt 2.5 C(s) = R(s)G(s) = s A B C * = = + + s (s + 4)(s + 8) s(s + 4)(s + 8) s (s + 4) (s + 8) where A= 1 1 1 = B= = − , and C = = (s + 4)(s + 8) S→0 32 s(s + 8) S→ −4 16 s(s + 4) S→ −8 32 Thus, c(t) = 1 − e −4t + e −8t 32 16 32 2.6 Mesh Analysis Transforming the network yields, Now, writing the mesh equations, Chapter (s + 1)I1 (s) − sI2 (s) − I3 (s) = V(s) −sI1 (s) + (2s + 1)I2 (s) − I3 (s) = −I1 (s) − I2 (s) + (s + 2)I3 (s) = Solving the mesh equations for I2(s), (s + 1) V(s) −s −1 I2 (s) = (s + 1) −s −1 0 −s −1 −1 (s + 2) −1 = (s + 2s + 1)V(s) s(s + 5s + 2) (2s + 1) −1 −1 (s + 2) But, V L (s) = sI2 (s) Hence, V L (s) = (s + 2s + 1)V(s) (s + 5s + 2) or V L (s) s + 2s + = V(s) s + 5s + Nodal Analysis Writing the nodal equations, ( + 2)V1 (s) − V L (s) = V(s) s −V1 (s) + ( + 1)V L (s) = V(s) s s Solving for V L (s) , ( + 2) V(s) s V(s) −1 (s + 2s + 1)V(s) s V L (s) = = (s + 5s + 2) ( + 2) −1 s −1 ( + 1) s or V L (s) s + 2s + = V(s) s + 5s + Solutions to Skill-Assessment Exercises 2.7 Inverting Z2 (s) −100000 = = −s Z1 (s) (10 / s) Noninverting G(s) = − 10 + 10 ) ( [Z1 (s) + Z2 (s)] G(s) = = s = s +1 10 Z1 (s) ( ) s 2.8 Writing the equations of motion, (s + 3s + 1)X1 (s) − (3s + 1)X2 (s) = F(s) −(3s + 1)X1 (s) + (s + 4s + 1)X2 (s) = Solving for X2 (s) , (s + 3s + 1) F(s) −(3s + 1) (3s + 1)F(s) X2 (s) = = s(s + 7s + 5s + 1) (s + 3s + 1) −(3s + 1) −(3s + 1) (s + 4s + 1) Hence, X2 (s) (3s + 1) = F(s) s(s + 7s + 5s + 1) 2.9 Writing the equations of motion, (s + s + 1)θ1 (s) − (s + 1)θ (s) = T(s) −(s + 1)θ1 (s) + (2s + 2)θ (s) = where θ1 (s) is the angular displacement of the inertia Solving for θ (s) , (s + s + 1) T(s) −(s + 1) (s + 1)F(s) θ (s) = = (s + s + 1) −(s + 1) 2s + 3s + 2s + −(s + 1) (2s + 2) From which, after simplification, Chapter θ (s) = 2s + s + 2.10 Transforming the network to one without gears by reflecting the N-m/rad spring to the left and multiplying by (25/50)2, we obtain, T(t) θ1(t) N-m-s/rad θa(t) kg N-m/rad Writing the equations of motion, (s + s)θ1 (s) − sθ a (s) = T(s) −sθ1 (s) + (s + 1)θ a (s) = where θ1 (s) is the angular displacement of the 1-kg inertia Solving for θ a (s) , (s + s) T(s) −s sT(s) θ a (s) = = s +s +s (s + s) −s −s (s + 1) From which, θ a (s) = T(s) s + s + But, θ (s) = θ a (s) Thus, θ (s) 1/ = T(s) s + s + 2.11 First find the mechanical constants 1 Jm = Ja + J L ( * )2 = + 400( )=2 400 1 )=7 Dm = Da + DL ( * )2 = + 800( 400 Solutions to Skill-Assessment Exercises Now find the electrical constants From the torque-speed equation, set ωm = to find stall torque and set T m = to find no-load speed Hence, T stall = 200 ω no−load = 25 which, Kt T stall 200 = = =2 100 Ra Ea Kb = Ea ω no−load = 100 =4 25 Substituting all values into the motor transfer function, KT Ra Jm θ m (s) = = K K 15 Ea (s) s(s + (Dm + T b ) s(s + ) Jm Ra where θ m (s) is the angular displacement of the armature Now θ L (s) = θ m (s) Thus, 20 θ L (s) / 20 = ) Ea (s) s(s + 15 ) 2.12 Letting θ1 (s) = ω1 (s) / s θ (s) = ω (s) / s in Eqs 2.127, we obtain K K )ω (s) − ω (s) = T(s) s s K K − ω (s) + (J2 s + D2 + )ω (s) s s (J1s + D1 + From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively Chapter Series analog Parallel analog 2.13 Writing the nodal equation, C dv + ir − = i(t) dt But, C =1 v = vo + δ v ir = e vr = e v = e vo + δv Substituting these relationships into the differential equation, d(vo + δ v) + e vo + δv − = i(t) dt (1) We now linearize e v The general form is f (v) − f (vo ) ≈ df δv dv vo Substituting the function, f (v) = e v , with v = vo + δ v yields, e vo + δv − e vo ≈ de v dv Solving for e vo + δv , δv vo 56 Solutions to Skill-Assessment Exercises The uncompensated system’s phase margin measurement is taken where the magnitude plot crosses dB We find that when the magnitude plot crosses dB, the phase angle is -144.80 Therefore, the uncompensated system’s phase margin is -1800 + 144.80 = 35.20 The required phase margin based on the required damping ratio is Φ M = tan −1 2ζ −2ζ + + 4ζ = 48.1o Adding a 100 correction factor, the required phase margin is 58.10 Hence, the compensator must contribute φ max = 58.10 - 35.20 = 22.90 Using φ max = sin −1 − sin φ max 1− β = 0.44 The ,β= + sin φ max 1+ β compensator’s peak magnitude is calculated as Mmax = = 1.51 Now find the β frequency at which the uncompensated system has a magnitude 1/ Mmax , or –3.58 dB From the Bode plot, this magnitude occurs at ω max = 50 rad/s The compensator’s zero is at zc = 1 But, ω max = Therefore, zc = 33.2 The T T β compensator’s pole is at pc = z = c = 75.4 The compensator gain is chosen to βT β yield unity gain at dc Hence, Kc = 75.4 / 33.2 = 2.27 Summarizing, Gc (s) = 2.27 300,000 (s + 33.2) , and G(s) = s(s + 50)(s + 120) (s + 75.4) 11.4 A 10% overshoot requires ζ = % − log 100 = 0.591 The required bandwidth 2 % π + log 100 is then calculated as ω BW = π 1− ζ2 Tp (1 − 2ζ ) + 4ζ − 4ζ + = 7.53 rad/s In order to meet the steady-state error requirement of Kv = 10 = K , we (8)(30) calculate K = 2400 The uncompensated Bode plot for this gain is shown below Chapter 11 57 Bode Diagrams 40 20 Phase (deg); Magnitude (dB) -20 -40 -60 -80 -100 -100 -150 -200 -250 10 -1 10 10 10 10 Frequency (rad/sec) Let us select a new phase-margin frequency at 0.8ω BW = 6.02 rad/s The required phase margin based on the required damping ratio is Φ M = tan −1 2ζ −2ζ + + 4ζ = 58.6o Adding a 50 correction factor, the required phase margin is 63.60 At 6.02 rad/s, the new phase-margin frequency, the phase angle is – which represents a phase margin of 1800 – 138.30 = 41.70 Thus, the lead compensator must contribute φ max = 63.60 – 41.70 = 21.90 Using φ max = sin −1 − sin φ max 1− β = 0.456 ,β = + sin φ max 1+ β We now design the lag compensator by first choosing its higher break frequency one decade below the new phase-margin frequency, that is, zlag = 0.602 rad/s The lag compensator’s pole is plag = β zlag = 0.275 Finally, the lag compensator’s gain is Klag = β = 0.456 58 Solutions to Skill-Assessment Exercises Now we design the lead compensator The lead zero is the product of the new phase margin frequency and plead = β , or zlead = 0.8ω BW β = 4.07 Also, zlead = 8.93 Finally, Klead = = 2.19 Summarizing, β β Glag (s) = 0.456 (s + 0.602) (s + 4.07) ; Glead (s) = 2.19 ; and K = 2400 (s + 0.275) (s + 8.93) 59 Chapter 12 12.1 We first find the desired characteristic equation A 5% overshoot requires ζ = % − log 100 π = 0.69 Also, ω n = = 14.47 rad/s Thus, the Tp − ζ 2 2 % π + log 100 characteristic equation is s + 2ζω n s + ω n = s + 19.97s + 209.4 Adding a pole at –10 to cancel the zero at –10 yields the desired characteristic equation, (s + 19.97s + 209.4)(s + 10) = s + 29.97s + 409.1s + 2094 The compensated system The matrix in phase-variable form is A − BK = −(k1 ) −(36 + k2 ) −(15 + k3 ) characteristic equation for this system is sI − (A − BK)) = s + (15 + k3 )s + (36 + k2 )s + (k1 ) Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as K = [ k1 k2 k3 ] = [2094 373.1 14.97] 12.2 2 1 The controllability matrix is CM = B AB A B = 1 −9 Since CM = 80 , 1 −1 16 [ ] CM is full rank, that is, rank We conclude that the system is controllable 12.3 First check controllability The controllability matrix is CMz 0 = B AB A B = −17 Since CMz = −1 , CMz is full rank, that is, rank 1 −9 81 [ ] We conclude that the system is controllable We now find the desired characteristic equation A 20% overshoot 60 Solutions to Skill-Assessment Exercises requires ζ = % − log 100 = 4.386 rad/s Thus, the = 0.456 Also, ω n = ζ Ts 2 % π + log 100 characteristic equation is s + 2ζω n s + ω n = s + 4s + 19.24 Adding a pole at –6 to cancel the zero at –6 yields the resulting desired characteristic equation, (s + 4s + 19.24)(s + 6) = s + 10s + 43.24s + 115.45 Since G(s) = s+6 (s + 6) = , we can write the phase2 (s + 7)(s + 8)(s + 9) s + 24s + 191s + 504 0 variable representation as A p = 0 ; Bp = ; C p = [ ] −504 −191 −24 1 The compensated system matrix in phase-variable form is The characteristic equation for A p − Bp K p = 0 −(504 + k1 ) −(191 + k2 ) −(24 + k3 ) this system is sI − (A p − BpK p )) = s + (24 + k3 )s + (191 + k2 )s + (504 + k1 ) Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as K p = [ k1 k2 k3 ] = [ −388.55 −147.76 −14] We now develop the transformation matrix to transform back to the z-system CMz = [Bz [ CMp = Bp A z Bz 0 A z Bz ] = 0 −17 and 1 −9 81 A p Bp 0 A p Bp = −24 1 −24 385 Therefore, P = CMz CMx Hence, −1 ] 191 24 0 0 = −17 24 = 1 −9 81 0 56 15 Chapter 12 61 0 1 K z = K p P = [−388.55 −147.76 −14] −7 = [ −40.23 62.24 −14] 49 −15 −1 12.4 −(24 + l1 ) For the given system e x = (A − LC)e x = −(191 + l2 ) e x The characteristic −(504 + l3 ) 0 • polynomial is given by [sI − (A − LC) = s + (24 + l1 )s + (191 + l2 )s + (504 + l3 ) Now we find the desired characteristic equation The dominant poles from Skill-Assessment Exercise 12.3 come from (s + 4s + 19.24) Factoring yields (-2 + j3.9) and (-2 - j3.9) Increasing these poles by a factor of 10 and adding a third pole 10 times the real part of the dominant second-order poles yields the desired characteristic polynomial, (s + 20 + j39)(s + 20 − j39)(s + 200) = s + 240s + 9921s + 384200 Equating coefficients of the desired characteristic equation to the system’s characteristic 216 equation yields L = 9730 383696 12.5 C The observability matrix is OM = CA = −64 −80 −78 , where CA 674 848 814 25 28 32 A = −7 −4 −11 The matrix is of full rank, that is, rank 3, since OM = −1576 77 95 94 Therefore the system is observable 12.6 The system is represented in cascade form by the following state and output equations: −7 0 z = −8 z + 0 u 0 −9 1 y = [1 0]z • 62 Solutions to Skill-Assessment Exercises The observability matrix is OMz 0 Cz = Cz A z = −7 , where Cz A z 49 −15 49 −15 1 = , we A = 64 −17 Since G(s) = (s + 7)(s + 8)(s + 9) s + 24s + 191s + 504 0 81 z can write the observable canonical form as −24 x = −191 −504 y = [1 • 0 0 x + 0 u 1 0 0]x The observability matrix for this form is OMx 0 Cx = Cx A x = −24 , where Cx A x 385 −24 −24 385 A = 4080 −191 12096 −504 x We next find the desired characteristic equation A 10% overshoot requires ζ = % − log 100 = 67.66 rad/s Thus, the = 0.591 Also, ω n = ζ Ts 2 % π + log 100 characteristic equation is s + 2ζω n s + ω n = s + 80s + 4578.42 Adding a pole at –400, or 10 times the real part of the dominant second-order poles, yields the resulting desired characteristic equation, (s + 80s + 4578.42)(s + 400) = s + 480s + 36580s + 1.831x10 For the system represented in observable canonical form −(24 + l1 ) e x = (A x − L x Cx )e x = −(191 + l2 ) e x The characteristic polynomial is given −(504 + l3 ) 0 • by [sI − (A x − L x Cx ) = s + (24 + l1 )s + (191 + l2 )s + (504 + l3 ) Equating coefficients of the desired characteristic equation to the system’s characteristic equation yields Chapter 12 456 L x = 36,389 1,830, 496 Now, develop the transformation matrix between the observer canonical and cascade forms −1 P = OMz −1OMx 0 0 0 0 1 = −7 −24 = −24 49 −15 385 −24 56 15 385 −24 0 = −17 81 −9 0 456 456 456 Finally, L z = PL x = −17 36,389 = 28,637 ≈ 28,640 81 −9 1,830, 496 1,539,931 1,540,000 12.7 We first find the desired characteristic equation A 10% overshoot requires ζ= % − log 100 = 0.591 2 % π + log 100 Also, ω n = Tp π = 1.948 rad/s Thus, the characteristic equation is 1− ζ2 s + 2ζω n s + ω n = s + 2.3s + 3.79 Adding a pole at –4, which corresponds to the original system’s zero location, yields the resulting desired characteristic equation, (s + 2.3s + 3.79)(s + 4) = s + 6.3s + 13s + 15.16 x• (A − BK) BKe x 0 x Now, • = + r; and y = C [ ] x , x N 1 N x N −C where 0 A − BK = − [ k1 −7 −9 1 C = [ 1] 0 k2 ] = − −7 −9 k1 0 = k2 −(7 + k1 ) −(9 + k2 ) 63 64 Solutions to Skill-Assessment Exercises 0 0 Bke = ke = 1 ke x• x1 x1 •1 0 Thus, x2 = −(7 + k1 ) −(9 + k2 ) ke x2 + r ; y = [ 0] x2 • 1 x N −1 x N x N −4 Finding the characteristic equation of this system yields 0 s 0 (A − BK) BKe sI − = 0 s − −(7 + k1 ) −(9 + k2 ) ke −C 0 s −4 −1 −1 s = (7 + k1 ) s + (9 + k2 ) −ke = s + (9 + k2 )s + (7 + k1 + ke )s + 4ke s Equating this polynomial to the desired characteristic equation, s + 6.3s + 13s + 15.16 = s + (9 + k2 )s + (7 + k1 + ke )s + 4ke Solving for the k’s, K = [2.21 −2.7] and ke = 3.79 65 Chapter 13 13.1 ∞ f (t) = sin(ω kT); f * (t) = ∑ sin(ω kT)δ (t − kT); k =0 (e jωkT − e − jωkT )e − kTs ∞ T ( s − jω ) − k = (e ) − (e T ( s + jω )− k ∑ j k =0 2j k =0 ∞ ∞ F * (s) = ∑ sin(ω kT)e − kTs = ∑ k =0 ∞ But, ∑x −k k =0 = 1 − x −1 Thus, F * (s) = 1 e −Ts e jωT − e −Ts e − jωT = − −T (s − j ω ) −T (s + j ω ) −Ts j ω T −Ts − j ω T −2Ts j 1 − (e e − e e j 1 − e 1− e ) + e =e −Ts sin(ω T ) z −1 sin(ω T ) 1 − e −Ts cos(ω T ) + e −2Ts = − 2z −1 cos(ω T ) + z −2 13.2 F(z) = z(z + 1)(z + 2) (z − 0.5)(z − 0.7)(z − 0.9) (z + 1)(z + 2) F(z) = (z − 0.5)(z − 0.7)(z − 0.9) z = 46.875 F(z) = 46.875 1 − 114.75 + 68.875 z − 0.5 z − 0.7 z − 0.9 z z z , − 114.75 + 68.875 z − 0.5 z − 0.7 z − 0.9 f (kT) = 46.875(0.5)k − 114.75(0.7)k + 68.875(0.9)k 13.3 Since G(s) = (1 − e −Ts ) , s(s + 4) B z − 2 z −1 A G(z) = (1 − z −1 )z z + z + = = z s s + 4 z s s + 4 s(s + 4) Let G2 (s) = 2 Therefore, g2 (t) = − 2e −4t , or g2 (kT) = − 2e −4kT + s s+4 Hence, G2 (z) = 2z 2z 2z(1 − e −4T ) − = z − z − e −4T (z − 1)(z − e −4T ) 66 Solutions to Skill-Assessment Exercises Therefore, G(z) = For T = z −1 2(1 − e −4T ) G2 (z) = z (z − e −4T ) 1.264 s, G(z) = z − 0.3679 13.4 Add phantom samplers to the input, feedback after H(s) , and to the output Push G1 (s)G2 (s) , along with its input sampler, to the right past the pickoff point and obtain the block diagram shown below Hence, T(z) = G1G2 (z) + HG1G2 (z) 13.5 Let G(s) = G(s) 20 4 20 = = − Let G2 (s) = Taking the inverse s s(s + 5) s s + s+5 Laplace transform and letting t = kT , g2 (kT) = − 4e −5kT Taking the z-transform yields G2 (z) = Now, G(z) = 4z 4z 4z(1 − e −5T ) − = z − z − e −5T (z − 1)(z − e −5T ) z −1 4(1 − e −5T ) G(z) 4(1 − e −5T ) G2 (z) = T(z) = = Finally, z (z − e −5T ) + G(z) z − 5e −5T + The pole of the closed-loop system is at 5e −5T − Substituting values of T , we find that the pole is greater than if T > 0.1022 s Hence, the system is stable for < T < 0.1022 s 13.6 Substituting z = s +1 into D(z) = z − z − 0.5z + 0.3 , we obtain s −1 D(s) = s − 8s − 27s − The Routh table for this polynomial is shown below Chapter 13 s3 -27 s2 -8 -6 s1 -27.75 s0 -6 67 Since there is one sign change, we conclude that the system has one pole outside the unit circle and two poles inside the unit circle The table did not produce a row of zeros and thus, there are no jω poles The system is unstable because of the pole outside the unit circle 13.7 Defining G(s) as G1 (s) in cascade with a zero-order-hold, 1/ 2/5 (s + 3) / 20 G(s) = 20(1 − e −Ts ) + − = 20(1 − e −Ts ) (s + 4) (s + 5) s(s + 4)(s + 5) s Taking the z-transform yields 5(z -1) 8(z -1) (3 / 20)z (1 / 4)z (2 / 5)z = 3+ G(z) = 20(1 − z −1 ) + − − −4T −5T z - e −4T z - e −5T z-e z-e z -1 Hence for T = 0.1 second, K p = lim G(z) = , K v = z→1 Ka = lim(z -1)G(z) = , and T z→1 lim(z -1)2 G(z) = Checking for stability, we find that the system is z→1 T stable for T = 0.1 second, since T(z) = 1.5z − 1.109 G(z) = has poles + G(z) z + 0.222z − 0.703 inside the unit circle at –0.957 and +0.735 Again, checking for stability, we find that the system is unstable for T = 0.5 second, since T(z) = 3.02z − 0.6383 G(z) = has poles inside and outside + G(z) z + 2.802z − 0.6272 the unit circle at +0.208 and –3.01, respectively 13.8 Draw the root locus superimposed over the ζ = 0.5 curve shown below Searching along a 54.30 line, which intersects the root locus and the ζ = 0.5 curve, we find the point 0.587∠54.3o = (0.348+j0.468) and K = 0.31 Solutions to Skill-Assessment Exercises z-Plane Root Locus 1.5 (0.348+j0.468) K=0.31 0.5 Imag Axis 68 54.30 -0.5 -1 -1.5 -3 -2.5 -2 -1.5 -1 -0.5 Real Axis 0.5 1.5 13.9 Let Ge (s) = G(s)Gc (s) = 100K 342720(s + 25.3) 2.38(s + 25.3) = s(s + 36)(s + 100) (s + 60.2) s(s + 36)(s + 100)(s + 60.2) The following shows the frequency response of Ge ( jω ) Chapter 13 69 Bode Diagrams 40 20 Phase (deg); Magnitude (dB) -20 -40 -60 -100 -150 -200 -250 10 -1 10 10 10 10 Frequency (rad/sec) We find that the zero dB frequency, ω Φ M , for Ge ( jω ) is 39 rad/s Using Astrom’s guideline the value of T should be in the range, 0.15 / ω Φ M = 0.0038 second to 0.5 / ω Φ M = 0.0128 second Let us use T = 0.001 second Now find the Tustin transformation for the compensator Substituting s = into Gc (s) = 2.38(s + 25.3) with T = 0.001 second yields (s + 60.2) Gc (z) = 2.34 (z − 0.975) (z − 0.9416) 2(z − 1) T(z + 1) 13.10 G c (z) = X(z) 1899z − 3761z + 1861 = Cross-multiply and obtain E(z) z − 1.908z + 0.9075 (z − 1.908z + 0.9075)X(z) = (1899z − 3761z + 1861)E(z) Solve for the highest power of z operating on the output, X(z) , and obtain z X(z) = (1899z − 3761z + 1861)E(z) − (−1.908z + 0.9075)X(z) Solving for 70 Solutions to Skill-Assessment Exercises X(z) on the left-hand side yields X(z) = (1899 − 3761z -1 + 1861z −2 )E(z) − (−1.908z -1 + 0.9075z −2 )X(z) Finally, we implement this last equation with the following flow chart: ... dB dB -1 dB -3 dB -6 dB -1 2 dB -2 0 dB -4 0 dB Open-Loop Gain (dB) -5 0 -6 0 dB -8 0 dB -1 00 -1 00 dB -1 20 dB -1 40 dB -1 50 -1 60 dB -1 80 dB -2 00 -2 00 dB -2 20 dB -3 50 -3 00 -2 50 -2 00 -1 50 -1 00 -5 0 -2 40... Asymptotic -2 0 dB/dec -4 0 dB/dec -2 0 dB/dec 20 log M -4 0 Actual -6 0 -8 0 -4 0 dB/dec -1 00 -1 20 0.1 10 Frequency (rad/s) 100 1000 Phase (degrees) -4 5o/dec -9 0o/dec -5 0 -4 5o/dec -9 0o/dec -1 00 -4 5o/dec Actual... Exercises Imag Axis 30 -1 -2 -3 -4 -5 -8 -7 -6 -5 -4 -3 -2 Real Axis -1 8.4 a jω X j3 s-plane σ O -2 -j3 X b Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function T(s)