09 Sistemas de Control Para Ingenieria 3ra Edicion Norman S Nise Sol - Redação (2024)

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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTASIGUENOS EN:VISITANOS PARA DESARGALOS GRATIS.http://librosysolucionarios.nethttps://twitter.com/Libros_y_Soluhttps://www.facebook.com/pages/Solucionarios-de-Libros/345772498866324https://plus.google.com/b/113394888343830071226/113394888343830071226Solutions toSkill-AssessmentExercisesTo AccompanyControl Systems Engineering3rd EditionByNorman S. NiseJohn Wiley & Sonshttp://librosysolucionarios.netCopyright © 2000 by John Wiley & Sons, Inc.All rights reserved.No part of this publication may be reproduced, stored in a retrieval system ortransmitted in any from or by any means, electronic, mechanical, photocopying,recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the1976 United States Copyright Act, without either the prior written permission of thePublisher, or authorization through payment of the appropriate per-copy fee to theCopyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470. Requests to the Publisher for permission should beaddressed to the Permissions Department, John Wiley & Sons, Inc., 605 ThirdAvenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, e-mail:PERMREQ@WILEY.COM. To order books please call 1 (800) 225-5945.ISBN 0-471-36601-3John Wiley & Sons, Inc.605 Third AvenueNew York, NY 10158USAhttp://librosysolucionarios.netSolutions to Skill-AssessmentExercisesChapter 22.1.The Laplace transform of t is 1s2 using Table 2.1, Item 3. Using Table 2.2, Item 4,F(s) = 1(s + 5)2 .2.2. Expanding F(s) by partial fractions yields:F(s) = As+ Bs + 2+ C(s + 3)2 + D(s + 3)where,As s S=+ +=→102 35920( )( )B = 10s(s + 3)2S→−2= −5 C = 10s(s + 2) S→−3= 103, andD = (s + 3)2 dF(s)ds s→−3= 409Taking the inverse Laplace transform yields,f (t) = 59− 5e−2t + 103te−3t + 409e−3t2.3. Taking the Laplace transform of the differential equation assuming zero initialconditions yields:s3C(s) + 3s2C(s) + 7sC(s) + 5C(s) = s2R(s) + 4sR(s) + 3R(s)Collecting terms,(s3 + 3s2 + 7s + 5)C(s) = (s2 + 4s + 3)R(s)Thus,http://librosysolucionarios.net2 Solutions to Skill-Assessment ExercisesC(s)R(s)= s2 + 4s + 3s3 + 3s2 + 7s + 52.4.G(s) = C(s)R(s)= 2s +1s2 + 6s + 2Cross multiplying yields,d 2cdt2 + 6dcdt+ 2c = 2drdt+ r2.5.C(s) = R(s)G(s) = 1s2 *s(s + 4)(s + 8)= 1s(s + 4)(s + 8)= As+ B(s + 4)+ C(s + 8)whereA = 1(s + 4)(s + 8) S→0= 132B = 1s(s + 8) S→−4= − 116, and C = 1s(s + 4) S→−8= 132Thus,c(t) = 132− 116e−4t + 132e−8t2.6.Mesh AnalysisTransforming the network yields,Now, writing the mesh equations,http://librosysolucionarios.netChapter 2 3(s +1)I1(s) − sI2 (s) − I3(s) = V(s)−sI1(s) + (2s +1)I2 (s) − I3(s) = 0−I1(s) − I2 (s) + (s + 2)I3(s) = 0Solving the mesh equations for I2(s),I2 (s) =(s +1) V(s) −1−s 0 −1−1 0 (s + 2)(s +1) −s −1−s (2s +1) −1−1 −1 (s + 2)= (s2 + 2s +1)V(s)s(s2 + 5s + 2)But, VL (s) = sI2 (s)Hence,VL (s) = (s2 + 2s +1)V(s)(s2 + 5s + 2)orVL (s)V(s)= s2 + 2s +1s2 + 5s + 2Nodal AnalysisWriting the nodal equations,(1s+ 2)V1(s) − VL (s) = V(s)−V1(s) + (2s+1)VL (s) = 1sV(s)Solving for VL (s),VL (s) =(1s+ 2) V(s)−11sV(s)(1s+ 2) −1−1 (2s+1)= (s2 + 2s +1)V(s)(s2 + 5s + 2)orVL (s)V(s)= s2 + 2s +1s2 + 5s + 2http://librosysolucionarios.net4 Solutions to Skill-Assessment Exercises2.7.InvertingG(s) = − Z2 (s)Z1(s)= −100000(105 / s)= −sNoninvertingG(s) = [Z1(s) + Z2 (s)]Z1(s)=(105s+105 )(105s)= s +12.8.Writing the equations of motion,(s2 + 3s +1)X1(s) − (3s +1)X2 (s) = F(s)−(3s +1)X1(s) + (s2 + 4s +1)X2 (s) = 0Solving for X2 (s),X2 (s) =(s2 + 3s +1) F(s)−(3s +1) 0(s2 + 3s +1) −(3s +1)−(3s +1) (s2 + 4s +1)= (3s +1)F(s)s(s3 + 7s2 + 5s +1)Hence,X2 (s)F(s)= (3s +1)s(s3 + 7s2 + 5s +1)2.9.Writing the equations of motion,(s2 + s +1)θ1(s) − (s +1)θ2 (s) = T(s)−(s +1)θ1(s) + (2s + 2)θ2 (s) = 0where θ1(s) is the angular displacement of the inertia.Solving for θ2 (s),θ2 (s) =(s2 + s +1) T(s)−(s +1) 0(s2 + s +1) −(s +1)−(s +1) (2s + 2)= (s +1)F(s)2s3 + 3s2 + 2s +1From which, after simplification,http://librosysolucionarios.netChapter 2 5θ2 (s) = 12s2 + s +12.10.Transforming the network to one without gears by reflecting the 4 N-m/rad springto the left and multiplying by (25/50)2, we obtain,1 kg1 N-m-s/rad1 N-m/radT(t)θa(t)θ1(t)Writing the equations of motion,(s2 + s)θ1(s) − sθa (s) = T(s)−sθ1(s) + (s +1)θa (s) = 0where θ1(s) is the angular displacement of the 1-kg inertia.Solving for θa (s) ,θa (s) =(s2 + s) T(s)−s 0(s2 + s) −s−s (s +1)= sT(s)s3 + s2 + sFrom which,θa (s)T(s)= 1s2 + s +1But, θ2 (s) = 12θa (s).Thus,θ2 (s)T(s)= 1 / 2s2 + s +12.11.First find the mechanical constants.Jm = Ja + JL (15*14)2 = 1 + 400(1400) = 2Dm = Da + DL (15*14)2 = 5 + 800(1400) = 7http://librosysolucionarios.net6 Solutions to Skill-Assessment ExercisesNow find the electrical constants. From the torque-speed equation, set ωm = 0 tofind stall torque and set Tm = 0 to find no-load speed. Hence,Tstall = 200ωno−load = 25 which,KtRa= TstallEa= 200100= 2Kb = Eaωno−load= 10025= 4Substituting all values into the motor transfer function,θm (s)Ea (s)=KTRa Jms(s + 1Jm(Dm + KT KbRa)= 1s(s + 152)where θm (s) is the angular displacement of the armature.Now θL (s) = 120θm (s) . Thus,θL (s)Ea (s)= 1 / 20s(s + 152))2.12.Lettingθ1(s) = ω1(s) / sθ2 (s) = ω2 (s) / sin Eqs. 2.127, we obtain(J1s + D1 + Ks)ω1(s) − Ksω2 (s) = T(s)− Ksω1(s) + (J2s + D2 + Ks)ω2 (s)From these equations we can draw both series and parallel analogs by consideringthese to be mesh or nodal equations, respectively.http://librosysolucionarios.netChapter 2 7 Series analog Parallel analog2.13.Writing the nodal equation,Cdvdt+ ir − 2 = i(t)But,C = 1v = vo + δvir = evr = ev = evo +δvSubstituting these relationships into the differential equation,d(vo + δv)dt+ evo +δv − 2 = i(t) (1)We now linearize ev .The general form isf (v) − f (vo ) ≈ dfdv voδvSubstituting the function, f (v) = ev , with v = vo + δv yields,evo +δv − evo ≈ devdv voδvSolving for evo +δv ,http://librosysolucionarios.net8 Solutions to Skill-Assessment Exercisesevo +δv = evo + devdv voδv = evo + evo δvSubstituting into Eq. (1)dδvdt+ evo + evo δv − 2 = i(t) (2)Setting i(t) = 0 and letting the circuit reach steady state, the capacitor acts like anopen circuit. Thus, vo = vr with ir = 2. But, ir = evr or vr = ln ir .Hence, vo = ln 2 = 0.693. Substituting this value of vo into Eq. (2) yieldsdδvdt+ 2δv = i(t)Taking the Laplace transform,(s + 2)δv(s) = I(s)Solving for the transfer function, we obtainδv(s)I(s)= 1s + 2orV(s)I(s)= 1s + 2 about equilibrium.http://librosysolucionarios.net9Chapter 33.1.Identifying appropriate variables on the circuit yieldsWriting the derivative relationsC1dvC1dt= iC1LdiLdt= vLC2dvC2dt= iC2 (1)Using Kirchhoff’s current and voltage laws,iC1= iL + iR = iL + 1R(vL − vC2)vL = −vC1+ viiC2= iR = 1R(vL − vC2)Substituting these relationships into Eqs. (1) and simplifying yields the stateequations asdvC1dt= − 1RC1vC1+ 1C1iL − 1RC1vC2+ 1RC1vidiLdt= − 1LvC1+ 1LvidvC2dt= − 1RC2vC1− 1RC2vC21RC2viwhere the output equation isvo = vC2Putting the equationsin vector-matrix form,http://librosysolucionarios.net10 Solutions to Skill-Assessment Exercisesx•=− 1RC11C1− 1RC1− 1L0 0− 1RC20 − 1RC2x +1RC11L1RC2vi (t)y = 0 0 1[ ]x3.2.Writing the equations of motion(s2 + s +1)X1(s) − sX2 (s) = F(s) − sX1(s) + (s2 + s +1)X2 (s) − X3(s) = 0 − X2 (s) + (s2 + s +1)X3(s) = 0Taking the inverse Laplace transform and simplifying,x1••= −x1•− x1 + x2•+ fx2••= x1•− x2•− x2 + x3x3••= −x3•− x3 + x2Defining state variables, zi,z1 = x1; z2 = x1•; z3 = x2 ; z4 = x2•; z5 = x3; z6 = x3•Writing the state equations using the definition of the state variables and theinverse transform of the differential equation,z1•= z2z2•= x1••= −x1•− x1 + x2•+ f = −z2 − z1 + z4 + fz3•= x2•= z4z4•= x2••= x1•− x2•− x2 + x3 = z2 − z4 − z3 + z5z5•= x3•= z6z6•= x3••= −x3•− x3 + x2 = −z6 − z5 + z3The output is z5. Hence, y = z5 . In vector-matrix form,http://librosysolucionarios.netChapter 3 11z•=0 1 0 0 0 0−1 −1 0 1 0 00 0 0 1 0 00 1 −1 −1 1 00 0 0 0 0 10 0 1 0 −1 −1z +010000f (t); y = 0 0 0 0 1 0[ ]z3.3.First derive the state equations for the transfer function without zeros.X(s)R(s)= 1s2 + 7s + 9Cross multiplying yields(s2 + 7s + 9)X(s) = R(s)Taking the inverse Laplace transform assuming zero initial conditions, we getx••+ 7 x•+ 9x = rDefining the state variables as,x1 = xx2 = x•Hence,x1•= x2x2•= x••= −7 x•− 9x + r = −9x1 − 7x2 + rUsing the zeros of the transfer function, we find the output equation to be,c = 2 x•+ x = x1 + 2x2Putting all equation in vector-matrix form yields,x•=0 1−9 −7x +01rc = 1 2[ ]x3.4.The state equation is converted to a transfer function usingG(s) = C(sI − A)−1 B (1)wherehttp://librosysolucionarios.net12 Solutions to Skill-Assessment ExercisesA =−4 −1.54 0, B =20, and C = 1.5 0.625[ ] .Evaluating (sI − A) yields(sI − A) =s + 4 1.5−4 sTaking the inverse we obtain(sI − A)−1 = 1s2 + 4s + 6s −1.54 s + 4Substituting all expressions into Eq. (1) yieldsG(s) = 3s + 5s2 + 4s + 63.5.Writing the differential equation we obtaind 2xdt2 + 2x2 = 10 + δf (t) (1)Letting x = xo + δx and substituting into Eq. (1) yieldsd 2 (xo + δx)dt2 + 2(xo + δx)2 = 10 + δf (t) (2)Now, linearize x2.(xo + δx)2 − xo2 = d(x2 )dx xoδx = 2xoδxfrom which(xo + δx)2 = xo2 + 2xoδx (3)Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation givesus the linearized intermediate differential equation,d 2δxdt2 + 4xoδx = −2xo2 +10 + δf (t) (4)The force of the spring at equilibrium is 10 N. Thus, since F = 2x2,10 = 2xo2from whichxo = 5http://librosysolucionarios.netChapter 3 13Substituting this value of xo into Eq. (4) gives us the final linearized differentialequation.d 2δxdt2 + 4 5δx = δf (t)Selecting the state variables,x1 = δxx2 = δx•Writing the state and output equationsx1•= x2x2•= δx••= −4 5x1 + δf (t)y = x1Converting to vector-matrix form yields the final result asx•=0 1−4 5 0x +01δf (t)y = 1 0[ ]xhttp://librosysolucionarios.net14Chapter 44.1.For a step inputC(s)  10(s ) 4)(s ) 6)s(s ) 1)(s ) 7)(s ) 8)(s ) 10)= As+ Bs +1+ Cs + 7+ Ds + 8+ Es +10Taking the inverse Laplace transform,c(t) = A + Be−t + Ce−7t + De−8t + Ee−10t4.2.Since a = 50 , Tc = 1a= 150= 0.02s; Ts = 4a= 450= 0.08 s; and Tr = 2.2a= 2.250= 0.044 s.4.3.a. Since poles are at –6 ± j19.08, c(t) = A + Be−6t cos(19.08t + φ).b. Since poles are at –78.54 and –11.46, c(t) = A + Be−78.54t + Ce−11.4t .c. Since poles are double on the real axis at –15 c(t) = A + Be−15t + Cte−15t .d. Since poles are at ±j25, c(t) = A + Bcos(25t + φ).4.4.a. ωn = 400 = 20 and 2ζωn = 12; ∴ ζ = 0.3 and system is underdamped. b. ωn = 900 = 30 and 2ζωn = 90; ∴ ζ = 1.5 and system is overdamped. c. ωn = 225 = 15 and 2ζωn = 30; ∴ ζ = 1 and system is critically damped. d. ωn = 625 = 25 and 2ζωn = 0; ∴ ζ = 0 and system is undamped. 4.5.ωn = 361 = 19 and 2ζωn = 16; ∴ ζ = 0.421.Now, Ts = 4ζωn= 0.5 s and Tp = πωn 1 − ζ 2= 0.182 s.From Figure 4.16, ωnTr = 1.4998. Therefore, Tr = 0.079 s.Finally, %os = e-ζπ1−ζ 2*100 = 23.3%http://librosysolucionarios.netChapter 4 154.6.a. The second-order approximation is valid, since the dominant poles have a real part of–2 and the higher-order pole is at –15, i.e. more than five-times further.b. The second-order approximation is not valid, since the dominant poles have a real partof –1 and the higher-order pole is at –4, i.e. not more than five-times further.4.7.a. Expanding G(s) by partial fractions yields G(s) = 1s+ 0.8942s + 20− 1.5918s +10− 0.3023s + 6.5.But –0.3023 is not an order of magnitude less than residues of second-order terms (term 2and 3). Therefore, a second-order approximation is not valid.b. Expanding G(s) by partial fractions yields G(s) = 1s+ 0.9782s + 20− 1.9078s +10− 0.0704s + 6.5.But 0.0704 is an order of magnitude less than residues of second-order terms (term 2 and3). Therefore, a second-order approximation is valid.4.8.See Figure 4.31 in the textbook for the Simulink block diagram and the output responses.4.9.a. Since sI − A =s −23 s + 5, (sI − A)−1 = 1s2 + 5s + 6s + 5 2−3 s. Also,BU(s) =01 / (s +1).The state vector is X(s) = (sI − A)−1[x(0) + BU(s)] = 1(s +1)(s + 2)(s + 3)2(s2 + 7s + 7)s2 − 4s − 6 .The output is Y(s) = 1 3[ ]X(s) = 5s2 + 2s − 4(s +1)(s + 2)(s + 3)= − 0.5s +1− 12s + 2+ 17.5s + 3.Taking the inverse Laplace transform yields y(t) = −0.5e−t −12e−2t +17.5e−3t .b. The eigenvalues are given by the roots of sI − A = s2 + 5s + 6, or –2 and –3.http://librosysolucionarios.net16 Solutions to Skill-Assessment Exercises4.10.a. Since (sI − A) =s −22 s + 5, (sI − A)−1 = 1s2 + 5s + 4s + 5 2−2 s. Taking the Laplacetransform of each term, the state transition matrix is given byΦ(t) =43e−t − 13e−4t 23e−t − 23e−4t− 23e−t + 23e−4t − 13e−t + 43e−4t.b. Since Φ(t − τ ) =43e−(t −τ ) − 13e−4(t −τ ) 23e−(t −τ ) − 23e−4(t −τ )− 23e−(t −τ ) + 23e−4(t −τ ) − 13e−(t −τ ) + 43e−4(t −τ ) and Bu(τ ) =0e−2τ,Φ(t − τ )Bu(τ ) =23e−τe−t − 23e2τe−4t− 13e−τe−t + 43e2τe−4t.Thus, x(t) = Φ(t)x(0) + Φ(t − τ )Bu(τ )0t∫ dτ =103e−t − e−2t − 43e−4t− 53e−t + e−2t + 83e−4t.c. y(t) = 2 1[ ]x = 5e−t − e−2thttp://librosysolucionarios.net17Chapter 55.1.Combine the parallel blocks in the forward path. Then, push 1s to the left past thepickoff point.1ssss 2 +1s+--R( s)C(s)Combine the parallel feedback paths and get 2s. Then, apply the feedbackformula, simplify, and get, T sss s s( ) = ++ +34 212 2.5.2.Find the closed-loop transfer function, T(s) = G(s)1 + G(s)H(s)= 16s2 + as +16,where G(s) = 16s(s + a) and H(s) = 1. Thus,ωn = 4 and 2ζωn = a , from whichζ = a8. But, for 5% overshoot, ζ =− ln(%100)π2 + ln2 (%100) = 0.69. Since, ζ = a8,a = 5.52.5.3.Label nodes.http://librosysolucionarios.net18 Solutionsto Skill-Assessment ExercisesN1 (s) N2 ( s) N3(s ) N4 ( s)N5 (s) N6 (s)N7 (s)Draw nodes.R( s ) N1 (s) N2 (s) N3( s ) N4 ( s) C (s)N5 ( s) N6 ( s)N7 ( s)Connect nodes and label subsystems.R(s ) N2 ( s) N3( s) N4 ( s) C ( s)N5 (s) N6 ( s)N7 ( s)11ss−1ss1 1−111sN1 ( s)Eliminate unnecessary nodes.R(s) C(s)1 s s 1s1s-s-15.4.Forward-path gains are G1G2G3 and G1G3.http://librosysolucionarios.netChapter 5 19Loop gains are −G1G2H1, −G2H2 , and −G3H3.Nontouching loops are [−G1G2H1][−G3H3] = G1G2G3H1H3and [−G2H2 ][−G3H3] = G2G3H2H3.Also, ∆ = 1 + G1G2H1 + G2H2 + G3H3 + G1G2G3H1H3 + G2G3H2H3.Finally, ∆1 = 1 and ∆2 = 1.Substituting these values into T(s) = C(s)R(s)=Tk∆kk∑∆ yieldsT(s) = G1(s)G3(s)[1 + G2 (s)][1 + G2 (s)H2 (s) + G1(s)G2 (s)H1(s)][1 + G3(s)H3(s)]5.5.The state equations are,x1•= −2x1 + x2x2•= −3x2 + x3x3•= −3x1 − 4x2 − 5x3 + ry = x2Drawing the signal-flow diagram from the state equations yields1s1s1s1 1 11-5-4-3 -2-3r x1x2x3 y5.6.From G(s) = 100(s + 5)s2 + 5s + 6 we draw the signal-flow graph in controller canonicalform and add the feedback.http://librosysolucionarios.net20 Solutions to Skill-Assessment Exercises1-5-6100500-1yrWriting the state equations from the signal-flow diagram, we obtainx xx.=− −+ = [ ]105 5061 010100 500ry5.7.From the transformation equations,P−1 =3 −21 −4Taking the inverse,P =0.4 −0.20.1 −0.3Now,P−1AP =3 −21 −41 3−4 −60.4 −0.20.1 −0.3=6.5 −8.59.5 −11.5P−1B =3 −21 −413=−3−11CP = 1 4[ ] 0.4 −0.20.1 −0.3= 0.8 −1.4[ ]Therefore,z•=6.5 −8.59.5 −11.5z +−3−11uy = 0.8 −1.4[ ]zhttp://librosysolucionarios.netChapter 5 215.8.First find the eigenvalues.λI − A =λ 00 λ −1 3−4 −6=λ −1 −34 λ + 6= λ2 + 5λ + 6From which the eigenvalues are –2 and –3.Now use Axi = λxi for each eigenvalue, λ . Thus,1 3−4 −6x1x2= λx1x2For λ = −2,3x1 + 3x2 = 0−4x1 − 4x2 = 0Thus x1 = −x2For λ = −34x1 + 3x2 = 0−4x1 − 3x2 = 0Thus x1 = −x2 and x1 = −0.75x2 ; from which we letP =0.707 −0.6−0.707 0.8Taking the inverse yieldsP−1 =5.6577 4.24335 5Hence,D = P−1AP =5.6577 4.24335 51 3−4 −60.707 −0.6−0.707 0.8=−2 00 −3P−1B =5.6577 4.24335 513=18.3920CP = 1 4[ ] 0.707 −0.6−0.707 0.8= −2.121 2.6[ ]http://librosysolucionarios.net22 Solutions to Skill-Assessment ExercisesFinally,z•=−2 00 −3z +18.3920uy = −2.121 2.6[ ]zhttp://librosysolucionarios.net23Chapter 66.1.Make a Routh table.s7 3 6 7 2s6 9 4 8 6s5 4.666666667 4.333333333 0 0s4 -4.35714286 8 6 0s3 12.90163934 6.426229508 0 0s2 10.17026684 6 0 0s1 -1.18515742 0 0 0s0 6 0 0 0Since there are four sign changes and no complete row of zeros, there are fourright half-plane poles and three left half-plane poles.6.2.Make a Routh table. We encounter a row of zeros on the s3 row. The evenpolynomial is contained in the previous row as −6s4 + 0s2 + 6 . Taking thederivative yields −24s3 + 0s. Replacing the row of zeros with the coefficients ofthe derivative yields the s3 row. We also encounter a zero in the first column atthe s2 row. We replace the zero with ε and continue the table. The final result isshown now ass6 1 -6 -1 6s5 1 0 -1 0s 4 -6 0 6 0s 3 -24 0 0 0 ROZs 2 ε 6 0 0s 1 144/ε 0 0 0s 0 6 0 0 0There is one sign change below the even polynomial. Thus the even polynomial(4th order) has one right half-plane pole, one left half-plane pole, and 2 imaginaryaxis poles. From the top of the table down to the even polynomial yields one signchange. Thus, the rest of the polynomial has one right half-plane root, and one lefthttp://librosysolucionarios.net24 Solutions to Skill-Assessment Exerciseshalf-plane root. The total for the system is two right half-plane poles, two lefthalf-plane poles, and 2 imaginary poles.6.3.Since G(s) = K(s + 20)s(s + 2)(s + 3), T(s) = G(s)1 + G(s)= K(s + 20)s3 + 5s2 + (6 + K)s + 20KForm the Routh table.s3 1 (6 + K)s2 5 20Ks130 −15K5s0 20KFrom the s1 row, K < 2. From the s0 row, K > 0. Thus, for stability, 0 < K < 2 .6.4.First findsI − A =s 0 00 s 00 0 s−2 1 11 7 1−3 4 −5=(s − 2) −1 −1−1 (s − 7) −13 −4 (s + 5)= s3 − 4s2 − 33s + 51Now form the Routh table.s3 1 -33s2 -4 51S1 -20.25S0 51There are two sign changes. Thus, there are two rhp poles and one lhp pole.http://librosysolucionarios.net25Chapter 77.1.a. First check stability.T(s) = G(s)1 + G(s)= 10s2 + 500s + 6000s3 + 70s2 +1375s + 6000= 10(s + 30)(s + 20)(s + 26.03)(s + 37.89)(s + 6.085)Poles are in the lhp. Therefore, the system is stable. Stability also could bechecked via Routh-Hurwitz using the denominator of T(s) . Thus,15u(t): estep(∞) = 151 + lims→0G(s)= 151 + ∞= 015tu(t): eramp(∞) = 15lims→0sG(s)= 1510 * 20 * 3025* 35= 2.187515t2u(t): eparabola (∞) = 15lims→0s2G(s)= 300= ∞, since LL [15t2 ] = 30s3b. First check stability.T(s) = G(s)1 + G(s)= 10s2 + 500s + 6000s5 +110s4 + 3875s3 + 4.37e04s2 + 500s + 6000 = 10(s + 30)(s + 20)(s + 50.01)(s + 35)(s + 25)(s2 − 7.189e − 04s + 0.1372)From the second-order term in the denominator, we see that the system isunstable. Instability could also be determined using the Routh-Hurwitz criteria onthe denominator of T(s) . Since the system is unstable, calculations about steady-state error cannot be made.7.2.a. The system is stable, sinceT(s) = G(s)1 + G(s)= 1000(s + 8)(s + 9)(s + 7) +1000(s + 8)= 1000(s + 8)s2 +1016s + 8063 and is ofType 0. Therefore,Kp = lims→0G(s) = 1000 * 87 * 9= 127; Kv = lims→0sG(s) = 0; and Ka = lims→0s2G(s) = 0b. estep(∞) = 11 + lims→0G(s)= 11 +127= 7.8e − 03http://librosysolucionarios.net26 Solutions to Skill-Assessment Exerciseseramp(∞) = 1lims→0sG(s)= 10= ∞eparabola (∞) = 1lims→0s2G(s)= 10= ∞7.3.System is stable for positive K. System is Type 0. Therefore, for a step inputestep(∞) = 11 + Kp= 0.1. Solving for Kp yields Kp = 9 = lims→0G(s) = 12K14 *18; fromwhich we obtain K = 189 .7.4.System is stable. Since G1(s) = 1000, and G2 (s) = (s + 2)(s + 4),eD(∞) = − 1lims→01G2 (s)+ limG1(s)s→0= − 12 +1000= −9.98e − 047.5.System is stable. Create a unity-feedback system, where He (s) = 1s +1−1 = −ss +1.The system is as follows:+-R(s) Ea(s) C(s)100s + 4-−ss +1Thus,Ge (s) = G(s)1 + G(S)He (s)=100(s + 4)1 − 100s(s +1)(s + 4)= 100(s +1)S2 − 95s + 4Hence, the system is Type 0. Evaluating Kp yieldshttp://librosysolucionarios.netChapter 7 27Kp = 1004= 25The steady-state error is given byestep(∞) = 11 + KP= 11 + 25= 3.846e − 027.6.Since G(s) = K(s + 7)s2 + 2s +10, e(∞) = 11 + Kp= 11 + 7K10= 1010 + 7K.Calculating the sensitivity, we getSe:K = Ke∂e∂K= K1010 + 7K(−10)7(10 + 7K)2 = − 7K10 + 7K7.7.GivenA =0 1−3 −6; B =01; C = 1 1[ ]; R(s) =1s.Using the final value theorem,estep(∞) = lims→0sR(s)[1 − C(sI − A)−1B] = lims→0[1 − 1 1[ ] s −13 s + 6−1 01] = lims→0[1 − 1 1[ ]s + 6 1−3 ss2 + 6s + 301] = lims→0s2 + 5s + 2s2 + 6s + 3= 23Using input substitution,step(∞) = 1 + CA−1B = 1 − 1 1[ ] 0 1−3 −6−1 01 = 1 + 1 1[ ]−6 −13 0301= 1 + 1 1[ ] -130= 23http://librosysolucionarios.net28Chapter 88.1.a.F(−7 + j9) = (−7 + j9 + 2)(−7 + j9 + 4)0.0339(−7 + j9)(−7 + j9 + 3)(−7 + j9 + 6)= (−5 + j9)(−3 + j9)(−7 + j9)(−4 + j9)(−1 + j9)= (−66 − j72)(944 − j378)= −0.0339 − j0.0899 = 0.096 < −110.7ob. The arrangement of vectors is shown as follows:jωσs-planeX X-2-4-6 -3 -1-5-7XM1 M2 M3 M4M5(-7+j9)0From the diagram,F(−7 + j9) = M2M4M1M3M5= (−3 + j9)(−5 + j9)(−1 + j9)(−4 + j9)(−7 + j9) = (−66 − j72)(944 − j378)= −0.0339 − j0.0899 = 0.096 < −110.7o8.2.a. First draw the vectors.http://librosysolucionarios.netChapter 8 29jωσs-planeXX-2-3 -1 0j1j2j3-j1-j2-j3From the diagram,angles = 180o − tan−1 −3−1 −∑ tan−1 −31 = 180o −108.43o +108.43o = 180o .b. Since the angle is 1800, the point is on the root locus.c. K = Π pole lengthsΠ zero lengths=12 + 32( ) 12 + 32( )1= 108.3.First, find the asymptotes.σa =poles - zeros∑∑# poles-# zeros= (−2 − 4 − 6) − (0)3 − 0= −4θa = (2k +1)π3= π3, π, 5π3Next draw root locus following the rules for sketching.http://librosysolucionarios.net30 Solutions to Skill-Assessment Exercises-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3-5-4-3-2-1012345Real AxisImag Axis8.4.a.j3σjωs-planeXXO-2 2-j30b. Using the Routh-Hurwitz criteria, we first find the closed-loop transferfunction. T(s) = G(s)1 + G(s)= K(s + 2)s2 + (K − 4)s + (2K +13) Using the denominator of T(s), make a Routh table.http://librosysolucionarios.netChapter 8 31s2 1 2K+13s1 K-4 0s0 2K+13 0We get a row of zeros for K = 4. From the s2 row with K = 4, s2 + 21 = 0. Fromwhich we evaluate the imaginary axis crossing at 21.c. From part (b), K = 4.d. Searching for the minimum gain to the left of –2 on the real axis yields –7 at again of 18. Thus the break-in point is at –7.e. First, draw vectors to a point ε close to the complex pole.At the point ε close to the complex pole, the angles must add up to zero. Hence,angle from zero – angle from pole in 4th quadrant – angle from pole in 1st quadrant= 1800, or tan−1 34 − 90o − θ = 180o . Solving for the angle of departure, θ = -233.1.http://librosysolucionarios.net32 Solutions to Skill-Assessment Exercises8.5.a.jω4-3XXs-planeo0ζ = 0.5j4-j4σ2ob. Search along the imaginary axis and find the 1800 point at s = ± j4.06.c. For the result in part (b), K = 1.d. Searching between 2 and 4 on the real axis for the minimum gain yields thebreak-in at s = 2.89 .e. Searching along ζ = 0.5 for the 1800 point we find s = −2.42 + j4.18.f. For the result in part (e), K = 0.108.g. Using the result from part (c) and the root locus, K < 1.8.6.a. σjωζ = 0.591-2-4-6XXX0s-planehttp://librosysolucionarios.netChapter 8 33b. Searching along the ζ = 0.591 (10% overshoot) line for the 1800 point yields - 2.028+j2.768 with K = 45.55.c. Ts = 4Re= 42.028= 1.97 s; Tp = πIm= π2.768= 1.13 s; ωnTr = 1.8346 from the rise-time chart and graph in Chapter 4. Since ωn is theradial distance to the pole, ωn = 2.0282 + 2.7682 = 3.431. Thus, Tr = 0.53 s;since the system is Type 0, Kp = K2 * 4 * 6= 45.5548= 0.949. Thus,estep(∞) = 11 + Kp= 0.51.d. Searching the real axis to the left of –6 for the point whose gain is 45.55, wefind –7.94. Comparing this value to the real part of the dominant pole, -2.028, wefind that it is not five times further. The second-order approximation is not valid.8.7.Find the closed-loop transfer function and put it the form that yields pi as the rootlocus variable. Thus,T(s) = G(s)1 + G(s)= 100s2 + pis +100= 100(s2 +100) + pis=100s2 +1001 + piss2 +100Hence, KG(s)H(s) = piss2 +100. The following shows the root locus.http://librosysolucionarios.net34 Solutions to Skill-Assessment Exercisesσjωj10-j10XXOs-plane08.8.Following the rules for plotting the root locus of positive-feedback systems, weobtain the following root locus:σjω-2-3-4Xs-plane-1XXo0http://librosysolucionarios.netChapter 8 358.9.The closed-loop transfer function is T(s) = K(s +1)s2 + (K + 2)s + K. Differentiating thedenominator with respect to K yields2s∂s∂K+ (K + 2)∂s∂K+ (s +1) = (2s + K + 2)∂s∂K+ (s +1) = 0Solving for ∂s∂K, we get ∂s∂K= −(s +1)(2s + K + 2). Thus, Ss:K = Ks∂s∂K= −K(s +1)s(2s + K + 2).Substituting K = 20 yields Ss:K = −10(s +1)s(s +11).Now find the closed-loop poles when K = 20 . From the denominator of T(s) , s1,2= -21.05, - 0.95, -when K = 20 .For the pole at –21.05,∆s = s(Ss:K )∆KK= −21.05−10(−21.05 +1)−21.05(−21.05 +11)0.05 = −0.9975.For the pole at –0.95,∆s = s(Ss:K )∆KK= −0.95−10(−0.95 +1)−0.95(−0.95 +11)0.05 = −0.0025.http://librosysolucionarios.net36Chapter 99.1.a. Searching along the 15% overshoot line, we find the point on the root locus at –3.5+ j5.8 at a gain of K = 45.84. Thus, for the uncompensatedsystem, Kv = lims→0sG(s) = K / 7 = 45.84 / 7 = 6.55.Hence, eramp_uncompensated (∞) = 1 / Kv = 0.1527.b. Compensator zero should be 20x further to the left than the compensator pole.Arbitrarily select Gc (s) = (s + 0.2)(s + 0.01).c. Insert compensator and search along the 15% overshoot line and find the root locusat–3.4 + j5.63 with a gain, K = 44.64. Thus, for the compensatedsystem, Kv = 44.64(0.2)(7)(0.01)= 127.5 and eramp_compensated (∞) = 1Kv= 0.0078.d. eramp_uncompensatederamp_compensated= 0.15270.0078= 19.589.2.a. Searching along the 15% overshoot line, we find the point on the root locus at–3.5 + j5.8 at a gain of K = 45.84. Thus, for the uncompensated system,Ts = 4Re= 43.5= 1.143 s.b. The real part of the design point must be three times larger than theuncompensated pole’s real part. Thus the design point is 3(-3.5) + j 3(5.8) = -10.5+ j17.4. The angular contribution of the plant’s poles and compensator zero at thedesign point is 130.80. Thus, the compensator pole must contribute 1800 – 130.80= 49.20. Using the following diagram,http://librosysolucionarios.netChapter 9 37-pcσjωs-plane-10.5j17.449.20we find 17.4pc −10.5= tan 49.2o , from which, pc = 25.52. Adding this pole, we findthe gain at the design point to be K = 476.3. A higher-order closed-loop pole isfound to be at –11.54. This pole may not be close enough to the closed-loop zeroat –10. Thus, we should simulate the system to be sure the design requirementshave been met.9.3.a. Searching along the 20% overshoot line, we find the point on the root locus at–3.5 + 6.83 at a gain of K = 58.9. Thus, for the uncompensated system,Ts = 4Re= 43.5= 1.143 s.b. For the uncompensated system, Kv = lims→0sG(s) = K / 7 = 58.9 / 7 = 8.41. Hence,eramp_uncompensated (∞) = 1 / Kv = 0.1189 .c. In order to decrease the settling time by a factor of 2, the design point is twicethe uncompensated value, or –7 + j13.66. Adding the angles from the plant’spoles and the compensator’s zero at –3 to the design point, we obtain –100.80.Thus, the compensator pole must contribute 1800 – 100.80 = 79.20. Using thefollowing diagram,http://librosysolucionarios.net38 Solutions to Skill-Assessment Exercises-pcσjωs-plane79.20-7j13.66we find13.66pc − 7= tan 79.2o , from which, pc = 9.61. Adding this pole, we find thegain at the design point to be K = 204.9.Evaluating Kv for the lead-compensated system:Kv = lims→0sG(s)Glead = K(3) / [(7)(9.61)] = (204.9)(3) / [(7)(9.61)] = 9.138.Kv for the uncompensated system was 8.41. For a 10x improvement in steady-state error, Kv must be (8.41)(10) = 84.1. Since lead compensation gave us Kv =9.138, we need an improvement of 84.1/9.138 = 9.2.Thus, the lag compensator zero should be 9.2x further to the left than thecompensator pole. Arbitrarily select Gc (s) = (s + 0.092)(s + 0.01).Using all plant and compensator poles, we find the gain at the design point to beK = 205.4. Summarizing the forward path with plant, compensator, and gainyieldsGe (s) = 205.4(s + 3)(s + 0.092)s(s + 7)(9.61)(s + 0.01).Higher-order poles are found at –0.928 and –2.6. It would be advisable tosimulate the system to see if there is indeed pole-zero cancellation.9.4.The configuration for the system is shown in the figure below.http://librosysolucionarios.netChapter 9 391s(s + 7)(s +10)R(s) C(s)+-K+-Kf sMinor-Loop Design:For the minor loop, G(s)H(s) =K f(s + 7)(s +10). Using the following diagram, wefind that the minor-loop root locus intersects the 0.7 damping ratio line at –8.5 +j8.67. The imaginary part was found as follows: θ = cos-1 ζ = 45.570. Hence,Im8.5= tan 45.570 , from which Im = 8.67.σjωs-plane-7ζ = 0.7X X−10 −8.5(-8.5 + j8.67)θImThe gain, K f , is found from the vector lengths asK f = 1.52 + 8.672 1.52 + 8.672 = 77.42Major-Loop Design:Using the closed-loop poles of the minor loop, we have an equivalent forward-path transfer function ofGe (s) = Ks(s + 8.5 + j8.67)(s + 8.5 − j8.67)= Ks(s2 +17s +147.4).http://librosysolucionarios.net40 Solutions to Skill-Assessment ExercisesUsing the three poles of Ge (s) as open-loop poles to plot a root locus, we searchalong ζ = 0.5 and find that the root locus intersects this damping ratio line at–4.34 + j7.51 at a gain, K = 626.3.9.5.a. An active PID controller must be used. We use the circuit shown in thefollowing figure:where the impedances are shown below as follows:Matching the given transfer function with the transfer function of the PIDcontroller yieldsGc (s) = (s + 0.1)(s + 5)s= s2 + 5.1s + 0.5s= s + 5.1 + 0.5s= − R2R1+ C1C2+ R2C1s +1R1C2sEquating coefficients1R1C2= 0.5 (1)R2C1 = 1 (2)R2R1+ C1C2= 5.1 (3)In Eq. (2) we arbitrarily let C1 = 10−5. Thus, R2 = 105. Using these values alongwith Eqs. (1) and (3) we find C2 = 100 µF and R 1 = 20 kΩ .http://librosysolucionarios.netChapter 9 41b. The lag-lead compensator can be implemented with the following passivenetwork, since the ratio of the lead pole-to-zero is the inverse of the ratio of thelag pole-to-zero:Matching the given transfer function with the transfer function of the passive lag-lead compensator yieldsGc (s) = (s + 0.1)(s + 2)(s + 0.01)(s + 20)= (s + 0.1)(s + 2)s2 + 20.01s + 0.2=s + 1R1C1s + 1R2C2s2 + 1R1C1+ 1R2C2+ 1R2C1s + 1R1R2C1C2Equating coefficients1R1C1= 0.1 (1)1R2C2= 2 (2)1R1C1+ 1R2C2+ 1R2C1= 20.01 (3)Substituting Eqs. (1) and (2) in Eq. (3) yields1R2C1= 17.91 (4)Arbitrarily letting C1 = 100 µF in Eq. (1) yields R1 = 100 kΩ .Substituting C1 = 100 µF into Eq. (4) yields R2 = 558 kΩ .Substituting R2 = 558 kΩ into Eq. (2) yields C2 = 900 µF.http://librosysolucionarios.net42Chapter 1010.1.a.G(s) = 1(s + 2)(s + 4); G(jω) =1(8 - ω2 ) + j6ωM(ω) = (8 - ω2 )2 + (6ω)2For ω < 8, φ(ω) = -tan-1 6ω8 - ω2 .For ω > 8, φ(ω) = - π+ tan-1 6ω8 - ω2.b.Frequency (rad/sec)Phase (deg); Magnitude (dB)Bode Diagrams10-1 100 101 102-200-150-100-500 -100-80-60-40-200 http://librosysolucionarios.netChapter 10 43c.Real AxisImaginary AxisNyquist Diagrams-0.05 0 0.05 0.1 0.15 0.2-0.08-0.06-0.04-0.0200.020.040.060.0810.2.20 log M-120-100-80-60-400.1 1 10 100 1000-40 dB/dec-20 dB/dec-20 dB/dec-40 dB/decFrequency (rad/s)ActualAsymptoticFrequency (rad/s)Phase (degrees)-150-100-50-200-45o/dec-90o/dec-45o/dec-90o/dec-45o/dec-45o/dec0.1 1 10 100 1000ActualAsymptotichttp://librosysolucionarios.net44 Solutions to Skill-Assessment Exercises10.3.The frequency response is 1/8 at an angle of zero degrees at ω = 0 . Each polerotates 900 in going from ω = 0 to ω = ∞. Thus, the resultant rotates –1800 whileits magnitude goes to zero. The result is shown below.ReIm0 18ω = ∞ ω = 010.4.a. The frequency response is 1/48 at an angle of zero degrees at ω = 0 . Each polerotates 900 in going from ω = 0 to ω = ∞. Thus, the resultant rotates –2700 whileits magnitude goes to zero. The result is shown below.ImRe148ω = 0ω = ∞0ω = 6.631480-b. Substituting jω into G(s) = 1(s + 2)(s + 4)(s + 6)= 1s3 +12s2 + 44s + 48 andsimplifying, we obtain G( jω) = (48 −12ω2 ) − j(44ω − ω3 )ω6 + 56ω4 + 784ω2 + 2304. The Nyquisthttp://librosysolucionarios.netChapter 10 45diagram crosses the real axis when the imaginary part of G( jω) is zero. Thus, theNyquist diagram crosses the real axis at ω2 = 44, or ω = 44 = 6.63 rad/s. Atthis frequency G( jω) = − 1480. Thus, the system is stable for K < 480 .10.5.If K = 100, the Nyquist diagram will intersect the real axis at –100/480. Thus,GM = 20 log480100= 13.62 dB. From Skill-Assessment Exercise Solution 10.4, the1800 frequency is 6.63 rad/s.10.6.a.Frequency (rad/s)Phase (degrees)1 10 100 1000-300-250-200-150-100-501 10 100 1000-180-160-140-120-100-80Frequency (rad/s)20 log M-600b. The phase angle is 1800 at a frequency of 36.74 rad/s. At this frequency thegain is –99.67 dB. Therefore, 20 log K = 99.67, or K = 96,270 . We conclude thatthe system is stable for K < 96,270.c. For K = 10,000 , the magnitude plot is moved up 20 log10,000 = 80 dB.Therefore, the gain margin is 99.67- 80 = 19.67 dB. The 1800 frequency is 36.7http://librosysolucionarios.net46 Solutions to Skill-Assessment Exercisesrad/s. The gain curve crosses 0 dB at ω = 7.74 rad/s, where the phase is 87.10.We calculate the phase margin to be 1800 – 87.10 = 92.90.10.7.Using ζ =-ln(% / 100)π2 + ln2 (% / 100), we find ζ = 0.456, which corresponds to 20%overshoot. Using Ts = 2, ωBW = 4Tsζ(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 = 5.79rad/s.10.8.For both parts find thatG( jω) = 16027*(6750000 −101250ω2 ) + j1350(ω2 −1350)ωω6 + 2925ω4 +1072500ω2 + 25000000. For a range ofvalues for ω , superimpose G( jω) on the a. M and N circles, and on the b.Nichols chart.a.M = 1.31.41.51.61.82.0M = 0.70.60.50.4M = 1.0-2-1123ImRe1 2-1-2-3-3F = 20oo2530o-20 o-40o-50o-30o-70o40o50 o70o-25oG-plane-4http://librosysolucionarios.netChapter 10 47b.Open-Loop Phase (deg)Open-Loop Gain (dB)Nichols Charts-350 -300 -250 -200 -150 -100 -50 0-200-150-100-500 6 dB3 dB1 dB0.5 dB0.25 dB0 dB-1 dB-3 dB-6 dB-12 dB-20 dB-40 dB-60 dB-80 dB-100 dB-120 dB-140 dB-160 dB-180 dB-200 dB-220 dB-240 dBPlotting the closed-loop frequency response from a. or b. yields the followingplot:http://librosysolucionarios.net48 Solutions to Skill-Assessment ExercisesFrequency (rad/s)1 10 100 1000-300-250-200-150-100-500Phase (degrees)-120-100-80-60-40-200Frequency (rad/s)20 log M1 10 100 100010.9.The open-loop frequency response is shown in the following figure:http://librosysolucionarios.netChapter 10 49Frequency(rad/sec)Phase (deg); Magnitude (dB)Bode Diagrams-40-2002040 10-1 100 101 102-160-140-120-100 The open-loop frequency response is –7 at ω = 14.5 rad/s. Thus, the estimatedbandwidth is ωWB = 14.5 rad/s. The open-loop frequency response plot goesthrough zero dB at a frequency of 9.4 rad/s, where the phase is 151.980. Hence,the phase margin is 1800 – 151.980 = 28.020. This phase margin corresponds toζ = 0.25. Therefore, %OS = e− ζπ / 1−ζ 2( )x100 = 44.4%,Ts = 4ωBWζ(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 = 1.64 s andTp = πωBW 1 − ζ 2(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 = 0.33 s10.10.The initial slope is 40 dB/dec. Therefore, the system is Type 2. The initial slopeintersects 0 dB at ω = 9.5 rad/s. Thus, Ka = 9.52 = 90.25 and Kp = Kv = ∞ .http://librosysolucionarios.net50 Solutions to Skill-Assessment Exercises10.11.a. Without delay, G( jω) = 10jω( jω +1)= 10ω(−ω + j), from which the zero dBfrequency is found as follows: M = 10ω ω2 +1= 1. Solving for ω ,ω ω2 +1 = 10, or after squaring both sides and rearranging, ω4 + ω2 −100 = 0.Solving for the roots, ω2 = −10.51, 9.51. Taking the square root of the positiveroot, we find the 0 dB frequency to be 3.08 rad/s. At this frequency, the phaseangle, φ = -∠ (−ω + j) = -∠ (−3.08 + j) = −162o . Therefore the phase margin is1800 – 1620 = 180.b. With a delay of 3 s,φ = -∠ (−ω + j) − ωT = -∠ (−3.08 + j) − (3.08)(3) = −162o − 9.24o = −171.24o.Therefore the phase margin is 1800 – 171.240 = 8.760.c. With a delay of 7 s,φ = -∠ (−ω + j) − ωT = -∠ (−3.08 + j) − (3.08)(7) = −162o − 21.56o = −183.56o.Therefore the phase margin is 1800 – 183.560 = -3.560. Thus, the system isunstable.10.12.Drawing judicially selected slopes on the magnitude and phase plot as shownbelow yields a first estimate.http://librosysolucionarios.netChapter 10 51ExperimentalFrequency (rad/sec)-40-30-20-10010201 2 3 4 5 6 7 8 10 20 30 40 50 70 100 200 300 500 1000-95-90-85-80-75-70-65-60-55-50-45Phase(deg)Gain(dB)We see an initial slope on the magnitude plot of –20 dB/dec. We also see a final–20 dB/dec slope with a break frequency around 21 rad/s. Thus, an initial estimateis G1(s) = 1s(s + 21).Subtracting G1(s)from the original frequency response yields the frequencyresponse shown below.http://librosysolucionarios.net52 Solutions to Skill-Assessment Exercises4050607080901 2 3 4 5 6 78 10 20 30 40 50 70 100 200 300 500 1000020406080100Experimental Minus 1/s(s+21)Frequency (rad/sec)Gain(dB)Phase(deg)Drawing judicially selected slopes on the magnitude and phase plot as shownyields a final estimate. We see first-order zero behavior on the magnitude andphase plots with a break frequency of about 5.7 rad/s and a dc gain of about 44 dB= 20 log(5.7K) , or K = 27.8. Thus, we estimate G2 (s) = 27.8(s + 7). Thus,G(s) = G1(s)G2 (s) = 27.8(s + 5.7)s(s + 21). It is interesting to note that the originalproblem was developed from G(s) = 30(s + 5)s(s + 20).http://librosysolucionarios.net53Chapter 1111.1.The Bode plot for K = 1 is shown below.Frequency (rad/sec)Phase (deg); Magnitude (dB)Bode Diagrams-180-160-140-120-100-80-60 10-1 100 101 102 103-250-200-150-100 A 20% overshoot requires ζ =− log%100π2 + log2 %100= 0.456. This damping ratioimplies a phase margin of 48.10, which is obtained when the _ = -1800 + 48.10 =131.90. This phase angle occurs at ω = 27.6rad/s. The magnitude at thisfrequency is 5.15 x 10-6. Since the magnitude must beunity K = 15.15x10−6 = 194,200.http://librosysolucionarios.net54 Solutions to Skill-Assessment Exercises11.2.To meet the steady-state error requirement, K = 1,942,000. The Bode plot for thisgain is shown below.Frequency (rad/sec)Phase (deg); Magnitude (dB)Bode Diagrams-40-200204060 10-1 100 101 102 103-250-200-150-100 A 20% overshoot requires ζ =− log%100π2 + log2 %100= 0.456. This damping ratioimplies a phase margin of 48.10. Adding 100 to compensate for the phase anglecontribution of the lag, we use 58.10. Thus, we look for a phase angle of –1800 +58.10 = -129.90. The frequency at which this phase occurs is 20.4 rad/s. At thisfrequency the magnitude plot must go through zero dB. Presently, the magnitudeplot is 23.2 dB. Therefore draw the high frequency asymptote of the lagcompensator at –23.2 dB. Insert a break at 0.1(20.4) = 2.04 rad/s. At thisfrequency, draw –20 dB/dec slope until it intersects 0 dB. The frequency ofintersection will be the low frequency break or 0.141 rad/s. Hence thehttp://librosysolucionarios.netChapter 11 55compensator is Gc (s) = Kc(s + 2.04)(s + 0.141), where the gain is chosen to yield 0 dB atlow frequencies, or Kc = 0.141 / 2.04 = 0.0691. In summary,Gc (s) = 0.0691(s + 2.04)(s + 0.141) and G(s) = 1,942,000s(s + 50)(s +120).11.3.A 20% overshoot requires ζ =− log%100π2 + log2 %100= 0.456. The requiredbandwidth is then calculated as ωBW = 4Tsζ(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 = 57.9rad/s. In order to meet the steady-state error requirement of Kv = 50 = K(50)(120),we calculate K = 300,000 . The uncompensated Bode plot for this gain is shownbelow.Frequency (rad/sec)Phase (deg); Magnitude (dB)Bode Plot for K = 300000-60-40-2002040 10-1 100 101 102 103-250-200-150-100 http://librosysolucionarios.net56 Solutions to Skill-Assessment ExercisesThe uncompensated system’s phase margin measurement is taken where themagnitude plot crosses 0 dB. We find that when the magnitude plot crosses 0 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 dampingratio is ΦM = tan−1 2ζ−2ζ 2 + 1 + 4ζ 4= 48.1o. Adding a 100 correction factor, therequired phase margin is 58.10. Hence, the compensator must contribute φmax =58.10 - 35.20 = 22.90. Using φmax = sin−1 1 − β1 + β, β = 1 − sinφmax1 + sinφmax= 0.44 . Thecompensator’s peak magnitude is calculated as Mmax = 1β= 1.51. Now find thefrequency at which the uncompensated system has a magnitude 1/ Mmax , or –3.58dB. From the Bode plot, this magnitude occurs atωmax = 50 rad/s. Thecompensator’s zero is at zc = 1T. But, ωmax = 1T β. Therefore, zc = 33.2 . Thecompensator’s pole is at pc = 1βT= zcβ= 75.4. The compensator gain is chosen toyield unity gain at dc. Hence, Kc = 75.4 / 33.2 = 2.27. Summarizing,Gc (s) = 2.27(s + 33.2)(s + 75.4), and G(s) = 300,000s(s + 50)(s +120).11.4.A 10% overshoot requiresζ =− log%100π2 + log2 %100= 0.591. The required bandwidthis then calculated as ωBW = πTp 1 − ζ 2(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 = 7.53 rad/s.In order to meet the steady-state error requirement of Kv = 10 = K(8)(30), wecalculate K = 2400 . The uncompensated Bode plot for this gain is shown below.http://librosysolucionarios.netChapter 11 57Frequency (rad/sec)Phase (deg); Magnitude (dB)Bode Diagrams-100-80-60-40-2002040 10-1 100 101 102 103-250-200-150-100 Let us select a new phase-margin frequency at 0.8ωBW = 6.02 rad/s. The requiredphase margin based on the required damping ratiois ΦM = tan−1 2ζ−2ζ 2 + 1 + 4ζ 4= 58.6o . Adding a 50 correction factor, therequired 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 φmax = 63.60 – 41.70 = 21.90. Usingφmax = sin−1 1 − β1 + β,β = 1 − sinφmax1 + sinφmax= 0.456.We now design the lag compensator by first choosing its higher break frequencyone decade below the new phase-margin frequency, that is, zlag = 0.602 rad/s. Thelag compensator’s pole is plag = βzlag = 0.275. Finally, the lag compensator’s gainis Klag = β = 0.456.http://librosysolucionarios.net58 Solutions to Skill-Assessment ExercisesNow we design the lead compensator. The lead zero is the product of the newphase margin frequency and β , or zlead = 0.8ωBW β = 4.07. Also,plead = zleadβ= 8.93. Finally, Klead = 1β= 2.19. Summarizing,Glag (s) = 0.456(s + 0.602)(s + 0.275); Glead (s) = 2.19(s + 4.07)(s + 8.93); and K = 2400 .http://librosysolucionarios.net59Chapter 1212.1.We first find the desired characteristic equation. A 5% overshootrequiresζ =− log%100π2 + log2 %100= 0.69. Also, ωn = πTp 1 − ζ 2= 14.47 rad/s. Thus, thecharacteristic equation is s2 + 2ζωns + ωn2 = s2 +19.97s + 209.4 . Adding a pole at –10to cancel the zero at –10 yields the desired characteristic equation,(s2 +19.97s + 209.4)(s +10) = s3 + 29.97s2 + 409.1s + 2094. The compensated systemmatrix in phase-variable form is A − BK =0 1 00 0 1−(k1) −(36 + k2 ) −(15 + k3 ). Thecharacteristic equation for this system issI − (A − BK)) = s3 + (15 + k3 )s2 + (36 + k2 )s + (k1). Equating coefficients of thisequation with the coefficients of the desired characteristic equation yields the gains asK = k1 k2 k3[ ] = 2094 373.1 14.97[ ] .12.2.The controllability matrix is CM = B AB A2B[ ] =2 1 11 4 −91 −1 16. Since CM = 80 ,CM is full rank, that is, rank 3. We conclude that the system is controllable.12.3.First check controllability. The controllability matrix isCMz = B AB A2B[ ] =0 0 10 1 −171 −9 81. Since CMz = −1, CMz is full rank, that is, rank3. We conclude that the system is controllable.We now find the desired characteristic equation. A 20% overshoothttp://librosysolucionarios.net60 Solutions to Skill-Assessment Exercisesrequiresζ =− log%100π2 + log2 %100= 0.456. Also, ωn = 4ζTs= 4.386 rad/s. Thus, thecharacteristic equation is s2 + 2ζωns + ωn2 = s2 + 4s +19.24. Adding a pole at –6 tocancel the zero at –6 yields the resulting desired characteristic equation,(s2 + 4s +19.24)(s + 6) = s3 +10s2 + 43.24s +115.45.Since G(s) = (s + 6)(s + 7)(s + 8)(s + 9)= s + 6s3 + 24s2 +191s + 504, we can write the phase-variable representation as Ap =0 1 00 0 1−504 −191 −24; Bp =001; Cp = 6 1 0[ ] .The compensated system matrix in phase-variable form isAp − BpKp =0 1 00 0 1−(504 + k1) −(191 + k2 ) −(24 + k3 ). The characteristic equation forthis system is sI − (Ap − BpKp )) = s3 + (24 + k3 )s2 + (191 + k2 )s + (504 + k1) . Equatingcoefficients of this equation with the coefficients of the desired characteristic equationyields the gains as Kp = k1 k2 k3[ ] = −388.55 −147.76 −14[ ] .We now develop the transformation matrix to transform back to the z-system.CMz = Bz AzBz Az2Bz[ ] =0 0 10 1 −171 −9 81 andCMp = Bp ApBp Ap2Bp[ ] =0 0 10 1 −241 −24 385.Therefore, P = CMzCMx−1 =0 0 10 1 −171 −9 81191 24 124 1 01 0 0=1 0 07 1 056 15 1Hence,http://librosysolucionarios.netChapter 12 61Kz = KpP−1 = [−388.55 −147.76 −14]1 0 0−7 1 049 −15 1= −40.23 62.24 −14[ ] .12.4.For the given system ex•= (A − LC)ex =−(24 + l1) 1 0−(191 + l2 ) 0 1−(504 + l3 ) 0 0ex . The characteristicpolynomial is given by [sI − (A − LC) = s3 + (24 + l1)s2 + (191 + l2 )s + (504 + l3 ) . Nowwe find the desired characteristic equation. The dominant poles from Skill-AssessmentExercise 12.3 come from (s2 + 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 partof the dominant second-order poles yields the desired characteristic polynomial,(s + 20 + j39)(s + 20 − j39)(s + 200) = s3 + 240s2 + 9921s + 384200. Equatingcoefficients of the desired characteristic equation to the system’s characteristicequation yields L =2169730383696.12.5.The observability matrix is OM =CCACA2=4 6 8−64 −80 −78674 848 814, whereA2 =25 28 32−7 −4 −1177 95 94. The matrix is of full rank, that is, rank 3, since OM = −1576.Therefore the system is observable.12.6.The system is represented in cascade form by the following state and output equations:z•=−7 1 00 −8 10 0 −9z +001uy = 1 0 0[ ]zhttp://librosysolucionarios.net62 Solutions to Skill-Assessment ExercisesThe observability matrix is OMz =CzCzAzCzAz2=1 0 0−7 1 049 −15 1, whereAz2 =49 −15 10 64 −170 0 81. Since G(s) = 1(s + 7)(s + 8)(s + 9)= 1s3 + 24s2 +191s + 504, wecan write the observable canonical form asx•=−24 1 0−191 0 1−504 0 0x +001uy = 1 0 0[ ]xThe observability matrix for this form is OMx =CxCxAxCxAx2=1 0 0−24 1 0385 −24 1, whereAx2 =385 −24 14080 −191 012096 −504 0.We next find the desired characteristic equation. A 10% overshootrequiresζ =− log%100π2 + log2 %100= 0.591. Also, ωn = 4ζTs= 67.66 rad/s. Thus, thecharacteristic equation is s2 + 2ζωns + ωn2 = s2 + 80s + 4578.42. Adding a pole at–400, or 10 times the real part of the dominant second-order poles, yields the resultingdesired characteristic equation,(s2 + 80s + 4578.42)(s + 400) = s3 + 480s2 + 36580s +1.831x106 .For the system represented in observable canonical formex•= (Ax − LxCx )ex =−(24 + l1) 1 0−(191 + l2 ) 0 1−(504 + l3 ) 0 0ex . The characteristic polynomial is givenby [sI − (Ax − LxCx ) = s3 + (24 + l1)s2 + (191 + l2 )s + (504 + l3 ) . Equating coefficientsof the desired characteristic equation to the system’s characteristic equation yieldshttp://librosysolucionarios.netChapter 12 63Lx =45636,3891,830,496.Now, develop the transformation matrix between the observer canonical and cascadeforms.P = OMz−1OMx =1 0 0−7 1 049 −15 1−11 0 0−24 1 0385 −24 1=1 0 07 1 056 15 11 0 0−24 1 0385 −24 1 =1 0 0−17 1 081 −9 1.Finally, Lz = PLx =1 0 0−17 1 081 −9 145636,3891,830,496=45628,6371,539,931≈45628,6401,540,000.12.7.We first find the desired characteristic equation. A 10% overshoot requiresζ =− log%100π2 + log2 %100= 0.591.Also, ωn = πTp 1 − ζ 2= 1.948 rad/s. Thus, the characteristic equation iss2 + 2ζωns + ωn2 = s2 + 2.3s + 3.79 . Adding a pole at –4, which corresponds to theoriginal system’s zero location, yields the resulting desired characteristic equation,(s2 + 2.3s + 3.79)(s + 4) = s3 + 6.3s2 +13s +15.16.Now, x•xN•=(A − BK) BKe−C 0xxN+01r; and y = C 0[ ]xxN,whereA − BK=0 1−7 −9−01k1 k2[ ] =0 1−7 −9−0 0k1 k2=0 1−(7 + k1) −(9 + k2 )C = 4 1[ ]http://librosysolucionarios.net64 Solutions to Skill-Assessment ExercisesBke =01ke =0keThus, x1•x2•xN•=0 1 0−(7 + k1) −(9 + k2 ) ke−4 −1 0x1x2xN+01r ; y = 4 1 0[ ]x1x2xN.Finding the characteristic equation of this system yieldssI −(A − BK) BKe−C 0=s 0 00 s 00 0 s−0 1 0−(7 + k1) −(9 + k2 ) ke−4 −1 0=s −1 0(7 + k1) s + (9 + k2 ) −ke4 1 s= s3 + (9 + k2 )s2 + (7 + k1 + ke )s + 4keEquating this polynomial to the desired characteristic equation,s3 + 6.3s2 +13s +15.16 = s3 + (9 + k2 )s2 + (7 + k1 + ke )s + 4keSolving for the k’s,K = 2.21 −2.7[ ] and ke = 3.79.http://librosysolucionarios.net65Chapter 1313.1.f (t) = sin(ωkT); f *(t) = sin(ωkT)k =0∞∑ δ(t − kT);F*(s) = sin(ωkT)k =0∞∑ e−kTs = (e jωkT − e− jωkT )e−kTs2 jk =0∞∑ = 12 j(eT (s− jω) )−k − (eT (s+ jω )−kk =0∞∑But, x−kk =0∞∑ = 11 − x−1Thus,F*(s) = 12 j11 − e−T (s− jω) − 11 − e−T (s+ jω)= 12 je−Tse jωT − e−Tse− jωT1 − (e−Tse jωT − e−Tse− jωT ) + e−2Ts = e−Ts sin(ωT )1 − e−Ts 2cos(ωT ) + e−2Ts= z−1 sin(ωT )1 − 2z−1 cos(ωT ) + z−213.2.F(z) = z(z +1)(z + 2)(z − 0.5)(z − 0.7)(z − 0.9)F(z)z= (z +1)(z + 2)(z − 0.5)(z − 0.7)(z − 0.9) = 46.8751z − 0.5−114.751z − 0.7+ 68.8751z − 0.9F(z) = 46.875zz − 0.5−114.75zz − 0.7+ 68.875zz − 0.9,f (kT) = 46.875(0.5)k −114.75(0.7)k + 68.875(0.9)k13.3.Since G(s) = (1 − e−Ts )8s(s + 4),G(z) = (1 − z−1)z8s(s + 4)= z −1zzAs+ Bs + 4= z −1zz2s+ 2s + 4.Let G2 (s) = 2s+ 2s + 4. Therefore, g2 (t) = 2 − 2e−4t , or g2 (kT) = 2 − 2e−4kT .Hence, G2 (z) = 2zz −1− 2zz − e−4T = 2z(1 − e−4T )(z −1)(z − e−4T ).http://librosysolucionarios.net66 Solutions to Skill-Assessment ExercisesTherefore, G(z) = z −1zG2 (z) = 2(1 − e−4T )(z − e−4T ).For T = 14 s, G(z) = 1.264z − 0.3679.13.4.Add phantom samplers to the input, feedback after H(s), and to the output. PushG1(s)G2 (s), along with its input sampler, to the right past the pickoff point andobtain the block diagram shown below.Hence, T(z) = G1G2 (z)1 + HG1G2 (z).13.5.Let G(s) = 20s + 5. Let G2 (s) = G(s)s= 20s(s + 5)= 4s− 4s + 5. Taking the inverseLaplace transform and letting t = kT , g2 (kT) = 4 − 4e−5kT . Taking the z-transformyields G2 (z) = 4zz −1− 4zz − e−5T = 4z(1 − e−5T )(z −1)(z − e−5T ).Now, G(z) = z −1zG2 (z) = 4(1 − e−5T )(z − e−5T ). Finally, T(z) = G(z)1 + G(z)= 4(1 − e−5T )z − 5e−5T + 4.The pole of the closed-loop system is at 5e−5T − 4. Substituting values of T , wefind that the pole is greater than 1 if T > 0.1022 s. Hence, the system is stable for0 < T < 0.1022 s.13.6.Substituting z = s +1s −1 into D(z) = z3 − z2 − 0.5z + 0.3, we obtainD(s) = s3 − 8s2 − 27s − 6 . The Routh table for this polynomial is shown below.http://librosysolucionarios.netChapter 13 67s3 1 -27s2 -8 -6s1 -27.75 0s0 -6 0Since there is one sign change, we conclude that the system has one pole outsidethe unit circle and two poles inside the unit circle. The table did not produce a rowof zeros and thus, there are no jω poles. The system is unstable because of thepole outside the unit circle.13.7.Defining G(s) as G1(s) in cascade with a zero-order-hold,G(s) = 20 1 − e−Ts( ) (s + 3)s(s + 4)(s + 5)= 20 1 − e−Ts( ) 3 / 20s+ 1 / 4(s + 4)− 2 / 5(s + 5).Taking the z-transform yieldsG(z) = 20 1 − z−1( ) (3 / 20)zz -1+ (1 / 4)zz - e−4T − (2 / 5)zz - e−5T= 3 + 5(z -1)z - e−4T − 8(z -1)z - e−5T .Hence for T = 0.1 second, Kp = limz→1G(z) = 3, Kv = 1Tlimz→1(z -1)G(z) = 0, andKa = 1T 2 limz→1(z -1)2 G(z) = 0. Checking for stability, we find that the system isstable for T = 0.1 second, since T(z) = G(z)1 + G(z)= 1.5z −1.109z2 + 0.222z − 0.703 has polesinside the unit circle at –0.957 and +0.735.Again, checking for stability, we find that the system is unstable for T = 0.5second, since T(z) = G(z)1 + G(z)= 3.02z − 0.6383z2 + 2.802z − 0.6272 has poles inside and outsidethe unit circle at +0.208 and –3.01, respectively.13.8.Draw the root locus superimposed over the ζ = 0.5 curve shown below. Searchingalong a 54.30 line, which intersects the root locus and the ζ = 0.5 curve, we findthe point 0.587∠ 54.3o = (0.348+j0.468) and K = 0.31.http://librosysolucionarios.net68 Solutions to Skill-Assessment Exercises-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1.5-1-0.500.511.5Real AxisImag Axisz-Plane Root Locus54.30(0.348+j0.468)K=0.3113.9.LetGe (s) = G(s)Gc (s) = 100Ks(s + 36)(s +100)2.38(s + 25.3)(s + 60.2)= 342720(s + 25.3)s(s + 36)(s +100)(s + 60.2).The following shows the frequency response of Ge ( jω) .http://librosysolucionarios.netChapter 13 69Frequency (rad/sec)Phase (deg); Magnitude (dB)Bode Diagrams-60-40-2002040 10-1 100 101 102 103-250-200-150-100 We find that the zero dB frequency, ωΦM, for Ge ( jω)is 39 rad/s. Using Astrom’sguideline the value of T should be in the range, 0.15 / ωΦM= 0.0038 second to0.5 / ωΦM= 0.0128 second. Let us use T = 0.001 second.Now find the Tustin transformation for the compensator. Substituting s = 2(z −1)T(z +1)into Gc (s) = 2.38(s + 25.3)(s + 60.2)with T = 0.001 second yieldsGc (z) = 2.34(z − 0.975)(z − 0.9416).13.10.Gc (z) = X(z)E(z)= 1899z2 − 3761z +1861z2 −1.908z + 0.9075. Cross-multiply and obtain(z2 −1.908z + 0.9075)X(z) = (1899z2 − 3761z +1861)E(z). Solve for the highestpower of z operating on the output, X(z), and obtainz2 X(z) = (1899z2 − 3761z +1861)E(z) − (−1.908z + 0.9075)X(z) . Solving forhttp://librosysolucionarios.net70 Solutions to Skill-Assessment ExercisesX(z) on the left-hand side yieldsX(z) = (1899 − 3761z-1 +1861z−2 )E(z) − (−1.908z-1 + 0.9075z−2 )X(z). Finally, weimplement this last equation with the following flow chart:http://librosysolucionarios.netCover: Solutions to Skill-Assessment ExercisesCopyrightChapter 22.1, 2.2, 2.32.4, 2.5, 2.62.7, 2.8, 2.92.10, 2.112.122.13Chapter 33.13.23.3, 3.43.5Chapter 44.1, 4.2, 4.3, 4.4, 4.54.6, 4.7, 4.8, 4.94.10Chapter 55.1, 5.2, 5.35.45.5, 5.65.75.8Chapter 66.1, 6.26.3, 6.4Chapter 77.1, 7.27.3, 7.4, 7.57.6, 7.7Chapter 88.1, 8.28.38.48.5, 8.68.78.88.9Chapter 99.1, 9.29.39.49.5Chapter 1010.110.210.3, 10.410.5, 10.610.7, 10.810.910.1010.11, 10.12Chapter 1111.111.211.311.4Chapter 1212.1, 12.2, 12.312.4, 12.5, 12.612.7Chapter 1313.1, 13.2, 13.313.4, 13.5, 13.613.7, 13.813.913.10