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Solution Manual For Modern Control Engineering, 5th Edition

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    B-2-1. . : . RG) — [Zo ’ 0 <) : - B . - B . pa a8 GF G2 . cs) G +6 : RE) /t (G+ G2)(Gs~ Ge) : CS — B-2-2. : e6) re 62) 6) [¢.] . a 7 — RE [- a6) - +6, ——— en : 14 Go (Hy —He) C6) _ _(/+G06G- . RS) 1+ GulH =e) - I SE—————————— . B-2-3. . } ’ Co 2%) < =! [622 = & = x) : [] EE : Ld 1 StudyXY ©2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding peslission(s), write to: Rights and Permissions Department, - Pearson Education, Inc., Upper Saddle River, NJ 07458. -

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    “.g © : - £Hat— = —) ns rrr me Ki mg 4 N.C . 6:93 (Ga tH) w«@) . ’ I 4G) (Gt) Bs 3 Ci GGG + G Gz Hy BN BC eB — 2 £y : RO 1+ Gell ¥ G61 + Gy HH: + G6, G, + G,G3H, B-2-4. In the following diagrams, (a) denotes the unit-step response and (b) ~~ corresponds to the unit-ramp response. R . : - . al) “e) ‘ © : P3 ® ’ : . . Rd . ) - H 2 ) . | 6/2 8 + % oe 7 2 3 # ¢ © ’ Yl KL =k-r I baz . gt) ue. @ . : é .# [44 ® ] . R * '® + * - - . B so Py . 2 - - . 6 7 2 3 F£F . o a 3 § ¢ . Loening)=trnd) fhet(rnd =tlread ’ + StudyXxy ©2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, niechanical, photocopying, recording, or likewise. For information regarding permi@sion(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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    wo . : “ ) | ® oo ’ oo . ® f R £ ) : 0 rz 2 gp & = [3 . » =k (ts +s) = #( rth + ars) B-2-5. When D(s) is zero, the closed-loop transfer function Cr(s)/R(s) is i Gl) _ _Ge€) GE) Rs) - At Gels) Gp é) : When R(s) = 0, the closed-loop transfer function Cp(s)/D(s) is - : C6) _ / - . pe) 1+ Gel® Gb) When both the reference input and disturbance input are present, the output C(s) a is the sum of Cg(s) and Cp(s). Hence - Lo / - : CE)=Ce fo) + C, p= —L—| 9 Gol RE) + PO) | CL % p65) PEP) 59 Gl? B-2-6. When only the reference input R(s) is present, the output Cg(s) is given Rs) 1 G05) 626) For the reference input R(s), the desired output is R(s) for the unity-feedback . system such as the present system. Thus, the error Ep(s) is the difference ‘between R(s) and the actual output Cp(s). The error Ep(s) is given by . : ) Cels) BE =RO-G@) = RO [ /- =| — re G,6) G26) ] — / yy ’ . #&) [ & 146,6)6209 1G ,6) 5:09 4 . Assuming the system to be stable, the steady-state error eggp(t) can be given by 2 AR) Css (8) = Li, Cp) = Li SEH =. ITY) © 2010 Pearson Education, Inc., Upper Sadd FRAN. Al rights reserved Ths publication is protectZd By opyright a id Feission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permjssion(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. .

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    When only the disturbance input D(s) is present, the output Cp(s) is given by ’ . ’ Spb) _ G5 . 26) 1+ G(s) G5) since the desired output to the disturbance input D(s) is zero, the error Ep(s) - can be given by . E6)=0— Cpls) =— C6) . Hence - Gels? R - . Ep) =—Cp 6) = — ———————— Df) ° id 146,09 6.5) For the stable system, the steady-state error eggp(t) is given by ’ &; &) = Lom 5) = Da. = = -5 G2(5) DE) . wr) = Lz Cott) = Ly 560) = Looe 5500 The steady-state error when both the reference input R(s) and disturbance input D(s) are present is the sum Of eggp(t) and eggp(t) and is given by : Cssft) = Crp (@) + Essp (BD : = /; [ sRE) _ SG (5) PE) ] BE Edd 136,67 6,9 /1G,6)G2(3) ~ = La. = [ #8) . = —————r — 6,6) PB) : . sso f It 6,6)6:(9 % I . B-2-7 .. when D(s) = 0, the block diagram of the system can be simplified as - . follows: : ER 2 W vs | sa G® > e /+G,6.H, 3 The closed-loop transfer function Cg(s)/R(s) can be given by . - Ge Gr Gz Gs So Cris) = I+G) Ga #7 _ Ge G1 G2 Gy : RE) yp SG Gs ll 16,6. H+ Go, BY Studyxy | 1+ 6,6, Hy ©2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding persassion(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. . ~

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    When R(s) = 0, the block diagram of the system shown in Figure 2-34 can be ified as follows: . modi. DE) . Cpl : $C Lara se f— 4, ~B—% 1% . Dé&) G6) | & | 7 Hence %6) _ % 6, oo G2 8s —_—— = 26) (+G:G3G, (6th +4) IEG. GG Mt G Gl, EO, B-2-8. There are infinitely many state-space representations for this system. We shall give two of the possible state-space representations. State-space representation 1: From Figure 2-35, we obtain + . Yo Ss sez = BE — . TE) 43 LL Sppstys+E : St P 52 . which is equivalent to ~ ase - . . RN Frpytyr2y=d+zu : Comparing this equation with the standard equation . fod o oop oe . A dray+ay rary = 4+ hi +hutbha . we obtain © g | ~v; . : UY=p, a:=/, az=2, by=0, b=0, be=1, 1 StudyXY © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding péignission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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    X= Y- Pe «@€ - Tamj-pid —fu=*"fi% . %=d fw where oo by =be=0 . | p= bape =o . b=be=a fa: fo =/ ] : Also, define Aa=by—a; fs — 4a fr— p= Z—p Then, state-space equations can be given by . - : p4 eo 1 oflxl |4 : H=1° e / +p lu : [= 1 —rIps] Lad . % Yalr 0 x|+ fk : : Xs or - % eo + oll | 2 ’ . %|=lo ¢o / +} 7 ju : 5) |2 -1 rls] [=r : y=[/ ¢ lx © |x : Siz _ st+p+2-p 1+ EL . ) _ . S+p s+p P + Studyxy we can redrav the block diagram as shown below. : ©2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is-protected by Copyright and written permission should be ~~ obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by. any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. -6-

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    v ay v x Xo R - % 2 7 2 ht) . . ® Q +4 +h oo From this block disgram we get the following equations: . V= v-X + Xz R N : T=X, s+p ’ —_— - T-xi + Xs s : : XL Xs from which we obtain g./ + py =(2~ -— p a + PR = (2p) a= (ep) x, Hw —x+A tu : : ° - . N= : Rewriting, we have . = [J A AE TT . . . . A =u-— (z-p) x, —p Xs +(2-plu . Lo 7=% Co or ’ . - A £1 4 7 dfx ° Zl=1-/ 0 /|le|+] 1 |e . | Vv EA pz eo —=plixs 2- 1 StudyXY § © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permissions). write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. ee B

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    EF i 7= C / eo. 2] EA) : . x ee B-2-9 . . Ld . . ’ Lt3y+2y=« Define . m= # Ld Gg ‘Then . o [+30 + 2X =u oo . ; EA =; R =A, . or : . CL x e ( oflx [J Ll=|e oo lx t+|ole AE) le —2 =3)m] (4 - Z ¥=[7 o o]|% : . Xz - B-2-10 . R -¢ -/ an = Rh B . 4 [5 ~| Br &=U °1 The transfer fimction G(s) of the system is given by Co : s+¢ 77 1 StudyXY . -/ 1 udyXY . )= C{ sT—- A r=[7 o : : G6) gC I 4) E L 1 -3 S+¢ / ©2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. -8- - R

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    [ro] / s+/ Aik : Gro+)+2| 3 self s | Logs, ) stp &s+7 s+7 : — s o . S24 Ss+7 3 ee B-2-11, : S— 5 -/ 2 , = - = = z © oa=[T 0] [Fer a | The transfer function G(s) of the system is given by : oo ~ s+ 7172 ) Ge)=CC:I~27B = 2] |; 5] |s . / s+ -/ |l2}| . = [7s — del [r=] +s) (s+) +3 | 3 sts 1s . / I 2] 25-3 = /25+ 5 . Fst gsts Fo Td | ssazs s*+gs+ EE ’ A MATIAB solution to this problem is given below. - : EEE : A=[5 13 I; . : B=[25]; . c=0 2% - D=0; : . * | frum de} = ss 2ABCD) | : i . 012 5 - . dean 16 8 . ) EE B-2-12 ) . - o / o V7 c= / 0 A={o 2 7s | + B= 1001s S%1s 1 5 : — -2 -¢ —§ | / 0 - ) © 2010 Pearson Education, Inc., Upper $addle River, NJ. All rights reserved. This PublicatioTs protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permissions), write to: Rights and Permissions Department, ~~ _ Pearson Education, Inc., Upper Saddle River, NJ 07458. I= -

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    The transfer matrix of the system can be given by : co -1 s =/ [4 a o . - Lo Clip s — o )=0(sI-A B= ] k ao=gr-oe=[s 0 ole = lol Tstesee ste [eo ' -[1° d / -z sess {lo 7] a S345 rgs +2 ’ —25 —es-2 SEO, ) / / St€ = —— 2, : SS rssi+astz |S SHES : / S+& Sp F554 4s5tT2 . Streep gst = = . S%gs ) Sites gs+2 s¥+fs+gst+2 | - . A MATLAB solution to this problem is given below. - . A=[0 1 00.0 1;2 4 6} . Co : B=[0 00 L1 of i C=[1 0 00 1 0}; : : D=[0 0;0 0 [oum, den] = ss2{(A,B,C.D,1) ’ om = : : - 0 0.0000 0.0000 1.0000 . - 0 00000 10000 0.0000 : ) 1.0000 6.0000 4.0000 2.0000 i [num,den] = ss2{A.B,C.D,2) i co ) - num = ) } ’ -0 0.0000 1.0000 6.0000 . 0 1.0000 6.0000 0.0000 : } ) : ) + StudyXxy den= ©2010 Pearson Education, Inc., Upper Saddle RiverJNJ. NOs 6.0000 714.9000. 0090. ted by Copyright and written permission should be obtained from the publisher prior to any prohibited refkoductioavsiorage-a-a-retreve ~ortramehission in any form or by any means, electronic, © mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. ~10-

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    B-2-13, Define ZF=z*t Pry +35 = Fy) CL 25xX<4, rosy </z Let us choose X = 3 and ¥ = 11. Then F=T™+ 535 +39 = pr ade +365 =65¢ - We shall obtain a linearized equation for the nonlinear equation near the point X=3, ¥= 11. Expanding the nonlinear equation into a Taylor series about point x = X, y = ¥ and neglecting the higher-order terms, we obtain E-F =k (x=Z)+ t= (=F) - where 5 = ee =2% re =6+ 88 =2 £; [£23 l= y=F Tey I # = 2£ = PTET = 24+ 48 = 90 ke= #... gaz TET BE. g=u “ Hence the linearized equation is F—g36 = FF (x=3)+ 70 (y—1/) or R=F7¢x+70y — 634 « B-2-14. Define Co Y=o2=fx), E=2 — 2 = ¥=do = f+ xx). Since the higher-order terms in this equation are small, neglecting those terms, we obtain. 4-f&) =o. 2 (x-2) or . Y-0.2x2% = 0.4 X2° (X-2) Thus, linear approximation of the given nonlinear equation near the operating point is + StudyXY : 4 =2.4X~- 3.2 i © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be . obtained fremmte-pubtSHerprortoamy Prottotetreprottetom-Storage fra rete vat-SySteT Tor Tasso ITE ny Tor Or BY 3ny means, slectonic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, - Pearson Education, Inc., Upper Saddle River, NJ 07458. ~11-

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    B-3-1. since the same force transmits the shaft, we have feb (3-2) =b, (§-8) + bs (7-2) Co (&3) where displacement z is defined in the figure below. 5 R = 1 by 0 | G By B 2 | . x ¥ z In terms of the equivalent viscous friction coefficient, the force f is given by $=. » ( P _ *) (2) From Equation (1) we have . . . . . by Zrby E+ bz =p Ztbag+ bs 7 ” Lh hth] s z= by +(batbs)Yy (3 © b, tba tbs ' i By substituting Equation (3) into Equation (1), we have ’ . . : 4 —Z . s Fy bd Th x (batb3)T] f f=4(2 x) Th Lh - by + bs 2 =p— (g-X (4) bgapes CF) Hence, by comparing Equations (2) and (4), we obtain ber . bat bs / = Bb) ———— = —— bit b.tb ts ! > ¢ by + bs &; ) eee ————————— rrr TTT B32, ’ (2) i . wx +X = : | ~, XO _ J ’ TE) ns*+% © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. -12-

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    (b) Define the displacement of a point between ings kj and ky as y. Then the equations of motion for this aystem becom ha 1 z 7 . . WX Fy (X-y) =u % Y= Xa (x2) From the second equation, we have . . #, =k . or yy +k H 2 y= _%_ i 8 Foyt ka : mE + _biks dH =u or i} ths : X& / Te mse Puke . . 7 +#sy - 8-8-3. . ) hd Ld Te mg, th Gof) hy =a, te . * mg th (Gg ~3 th, =u, x = F) ZL = : Ce [ ’ Ee 7 j pA Ww = #2 3 * =% Then . . . MX, +b (a=) +%& Xx =, . Wy Xp + (t= )r by = 4, . . . i XG =X : . B - == [b-z) tb x] +55, 4 =x Som i Ln Crome yrba] + © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permigs®a(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. E

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    . or atl CM wl as ) - [C/A 21 £L 4 wm Tw © wm 1%] | 9 [4] ’, = + % o ¢ [Hl le Oflu : 3 be _ ke 4 A Xs 4 me wz wm || %# e my x & _ / eo oo oo xX, Co : Ya 0 ¢ ( ollx; Xf ———— B34. . . Je =T . : . where T=—24a*—~mglawrt J =ml* : For small 6, : wd26 =~ 2ka*6 —my le or . 4 28a | 2 \ Fee 2) m0 mis vq)" B-3-5. Note that : Gxt dato, Yo=Aens ’ For x direction: ’ ed Ld - ’ MAX +2 Zs =u So ar oz eo z = (x+ =u MX +m pr (X+ fan 7) i. | | ©2010 P Education, Inc., Upper S: £2 bets. Bi @sresoffe &: cB Db d writt ion should by earson Education, Inc., Upper : ain) orf and written permission should be obtained from the publisher prior to any po 2 reproduction, storage ih a retrieval system, or transmission any form or by any ans, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. -14- h .

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    © we have . * ’ g2 p a (Mim) x — md (anf) E tml (en8)d =cc For small 6 and small 62, we have se [Ad (Mim) X+md 6 = (1) For rotational motion: R To =mpd aint —mx Leng where ’ . tween, =mg ~ Thus, o . (Tm?) = mpl ais — mil cn 8 For small 6, we have te .- (T+mL?) 6 =mgdo- mlx = @ From Equation (2), . . I+. YL } . X=9Q8—- ——___§¢ 7 ml co. Substituting this last equation into Equation (1), we obtain: Tr+ml® oo 15 . m)(38~ 25) +mlb =u _ (mm )(38 =~ ——— E IT (Mm) 36 — (pm) T+ pmd® me Thus, so snd (Mim) md PR {Ci Bp m=" (Mm )z + pm dL (Msi) IT +m d Also, from Equation (1) ve have - oe mgd ls — mlx rm) % rm EEE mex _ (+m) ’ I4+mLe . or . B N a” B [MTom zee] +7428 = 4( T+ rion . from which we get ’ © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permissions), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. = =

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