Solution Manual For Feedback Control Systems, 5th Edition

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CHAPTER 22.1(a) An ideal voltage source model implies constant voltage regardless of the external connections. Thiscould result in current approaching infinity in the case of a short circuit and therefore would require infinite powerto supplied by the source. Similarly, an ideal current source implies constant current regardless of externalconnections. This cannot possibly be true in the case of an open circuit.(b) An appropriately sized resistor in series with the ideal voltage source, or in parallel with the idealcurrent source, will provide a more accurate model of a physical power supply.2.2The operational amplifiers must be operating in their linear amplification range. This means that the outputvoltage can never exceed the power supply voltage. Usually, the ideal op-amp assumptions are used to develop alinear model.2.3(a)1221123322312( )( ),( )( )( )0RRRIsVsVsR IsRRRL sIs  1212212212112132322312231121121323( )( )0( )( )( )RRVsR VsRIsRRRRRL sR RR RR RRRRL sVsR RVsRRL sR RR RR R(b) ReplaceR3with123232223221121231212122123()( )( )()[()()]RRR L sRL sVsR R L sVsL LRRsRLLRRR R LsR R R(c)123232012112132312132310( ),10()10limsssVsssR RR RsvRRL sR RR RR RR RR RR R

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2.4(a)2222211( )( )( )( )( )( )11110,( )( )10105101010105( )3iiVsV sVsV sVsVssssVsVsVsss(b)22222211( )( )( )( )( )( )( )11110,( )( )10105101010105( )6322iiVsV sVsV sVsVsVssssVsVsVsssss(c)22220010(1)10(1)1010( )( ), lim( )( )( ),lim( )33(3)(63)ssssa VssVsb VssVss sss2.5( )( )( )( )( )1.( )( )1( )1( )( )( )1( )( )()( )( )ffoooioiiiiitiofoiRRVsVsa vtv tb vtv tVsRVsRdv tc vtR Cd vtvddtRC 2.6From the solution to Problem 2.5(a), The gain of the first op-amp stage is one. For the second stage,21010( 12)102.4150ossKKveeKK . For the third stage,22204.81oosKvveK.2.7(a)10,10, let10, then100.ffiifiZZZRKRKZ (b)110, let10, then10.fiZsCRKCFZsR(c)

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Let R3= R4= 100K. Gain= -1 for second stage. For first stage,211211121111112110( )10,101110Let10, then100, and10.RsRRsGsR CRRsRssCR CRKRKCF2.7(d)Let R3== R4=100K. Gain = -1 for second stage. For first stage,1212111211111121110( )10,1110Let10, then10.R CsCGsR CR CsRssCR CRKCCF(e)211221121111(1)(101)10,1,let100,100.fffiiZRRR C sRR C ssR CRRK CFZRRR  

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2.7(f)1112121212210.11010,10Let1, then10, and100.fiZCCR CRMCFCFZCR C ssC  2.81a22,818,22, and8181281121222281( )1281181828122121818183434( ),121217178181()[1, 2;8,1],[2;18]ABBAABABABAbBc ABdMV   () *ABinv MV2.9(a)(b)122121Figure P2.9(a):Figure P2.9(b):andaababCG G EG HFCG EG G FHGG GGG122Figure P2.9(c):andcdcdCG EG FGG GGG H2.10 (a)(b)121122Figure P2.10(a):,Figure P2.10(b):,andaababCG G EFG HECG EFG G EHGG GGG21Figure P2.10(c):,andcdccdCG G EFG EGGG HGH

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2.11 423,3 ,1ABCBACABa(4)(1)( 3)(1)(1)b1( 693)1,111(3)(1)(4)(1)( 3)(3)(4)(1)(1)(4)(1)(3)( 1)(1)(1)[1( 6)]3,111111ABC    (c)>> M=[1 2 3;3 -1 0;1 -1 -1], V=[4;0;-1]2.12(a)Δ= 1(-G1-G2-G4-G3G4-G1G2G3G4) + (G1G2+G1G4+G2G4)(-G1G2G4)Δ= 1+G1+G2+G4+G3G4+G1G2+G1G4+G2G4+G1G2G4+G1G2G3G4M1=G1G2G3G4,Δ1= 1.M2=G3G4,Δ2= 1+G1M3=G1,,Δ3= 1+G4M4=-1,Δ4= 1+G1+G4+G1G4M5=-G3G4G1,Δ5= 112343441G G G GG GGDR(b)2132344BRG BG ACG RG G BG CDBG C 2.13.(a)12323112323123123121312312312132(1)32( )( )1()1G G GG GGG G GG GC sR sGGGG G GG GG GGGGG G GG GG G (b)1213,(2),ARG ACBGRG ABCCG B1122231232311231231213122310112003210111101GG GGGGG G GG GCGGGGG G GG GG GG GGG2.14(a)111211211112211221[1283][26]41723,12813(12)8314( )4172sssssssMsMsssssT sss  (b)111( )( )2( )( );( );( )4( )3( )11R sC sA sA sB sC ssB sA sss

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2.151113221233211122321[234]2616942,1;,135( )694sssssssssMsMsssT ssss   2.16(a)21122113121212212131231211221[];,1;,1( )1G HG G HMG GG HMG GG GG GG G G HT sG HG G H   (b)2326(2)248(1)248472( )920246(2)218(1)sss ssssT sssssss s(c)22436( )812sT sss2.17(a)112212();0()MxB xxB xxKx  (b)21122112()0;()0MxK xB xxB xxK x(c)111112211111()0;()( );( )M xB xK xxK xxf tM xB xf t(d)111121121222121221()();()()M xb xxk xkxxM xb xxkxxF2.18(a)2121221122132()( )( )( );( )()( )0( )( )( )0;( )( )0( )( )MsBs XsBsXsF sBsXsBsK XsF sBsXsF sBsKMsBsBsXsXsBsBsKMsBsBsBsBsKXsBsKF sMBsMKsBKs (b)

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221211211113221121221()( )( )( );( )()( )0( )0( )( );( )()MsBsKXsBsXsF sBsXsBsKXsF sBsBsKXsBsKXsF sMBsMK sB KKsK KMsBsKBsBsBsK(c)21111211( )12( )( )( );( )(4)XsM s XscXsF sF sM sB ss s(d)21112112221212122122212121112122122121124312112122()( )()( )0()( )()( )( );0()( )()( );()()()()( )( )()[m sb skkXsb skXsb skXsm sb skXsF sb skF sm sb skXsm sb skkb skb skm sb skXsb skF sm m sbmmsk mk2212112()]mmsb k sk2.19(a)12()MyBByKyf(b)212212( )1[()] ( )( );( )()Y sMsBBsK Y sF sF sMsBBsK2.20(a)2112151122121234211( )( )( )1( )( )( )( )( )M sBsKXsXsG XsBsKM sBsKKXsF sXsG F sG XsBsKBsK(b)See Example 2.11.2.21(a)11112222214252()0;6372()xxxxxxxxxxf(b)22122121143222427( )2( )0;639( )2( )( )02( )639( )2( );( )242484395942722639ssXsXsssXsXsF sF sssXsXsF sssssssss

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(c)22224322242702( )( )427( );( )242484395942722639ssF sXsssXsF sssssssss2.22(a)112111222121( )();0()tJBBJB(b)2111212212121121212322121121221211212121( )( )()( )0()( )0( )( );( )()()ssJ sBBsB ssB sJ sB sJ sBBssB ssBssJ J sJ BJ BJ BsB B sJ sBBsB sB sJ sB s 2.23(a)12()JBB(b)212( )1( )()ssJsBBs2.24(a) From (2-51): Note that( )( ).sss 147422( )0.005714285714.2857( )710+ 1.289(s+4.9910 ) (s+369)3.5191010asGEsss(b)406714285714.2857( )( );lim( )232.7535(rad/s)369(4.9910 )ssssGssss2.25(a)2222;;;();0()mammam ammammLLmmmLLLmddieKeR iLeK idtdtddddJBKJBKdtdtdtdt(b)2222( )( );( )( )( )( )( )( )( );( )( )( )( )( )( )( )(( )( ))( )( )0( )( )(( )( ))( )mmmamamamamammaLmmmmmLmmmmLLLLmLLEsK ssEsEsEsR IsL sIsEsIsL sRsK IssKssJ ssB ssKsssJ sB sKKsJ ssBssKsssJ sBs    K

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(c)121223( )( )1masG GEsG G HKG G(d)1231223( )( )1LaG G GsEsG G HKG G(e) From (c) and (d), or by inspection,13mLLmaaGEE2.26From Example 2.11 write analogous differential equation models.1111211211212122221121221212121()()0()011()()()diM xB xxKxxLR iii dti dtdtCdiM xB xxKxxfLR iii dti dti dtvdtCC2.27(a)12121122(). Letand.1(), let,,, and=.MyBByKyfiyvfdiMBBiKidtvLMRBRBKCdt(b)1122112()0;()0MxB xxB xxK x11221Let,,,, and.ixixKLMRBC(c)

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1112211212211221211()0()01()0()0diMxB xxK xLR iii dtdtCB xxK xR iii dtC2.28(a) From Problem 2.17(c) solution,111112211111()0;()( );( )M xB xK xxK xxf tM xB xf t.11111,,,( ),andLLMRBixvf tvM x(b) From Problem 2.17(d) solution, let1122,,.ixixvF11111212112221112112122221212212121221()()0()0()()()diM xbxxkkxb xk xMb iiki dtkiidtdtdiM xbxxkxxFMb iiki dti dtvdtLet112211122,,,1,1LMLMbRCkCk2.29(a)223323312212121221121131131221131332();0(),0()KKBJrrrrrrrrKKrrrrrKBJr   

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(b)''''11131313131222'1112111332221();0();;;KJBKrrrrKKJJBBrrrr2.30221210.0110010zn zzK 2.31(a)1100 (rpm)100gmrr. The encoder produces 100 pulses per revolution of the gearboxoutput shaft. Therefore, the resolution is3603.6100per pulse.(b)22620.01 (Kg)0.0110(Kg)LmLJmJJm2.32(a)23822121( )11380.47519( )202044440(1)(10)122121LasssEssssss ssss(b)320.47520()( )9.5()(1)(10)20()0.475( )11(109.5)9.51(1)(10)PDLPDPDCPDK sKsK sKs ssK sKsssKsKs ss(c)32( )20()20() (1)(10)20()0.475( )11(109.5)9.51(1)(10)aPDPDPDCPDEsK sKK sKs ssK sKsssKsKs ss(d)( )( )( )0.475( )( )( )(1)(10)CLLaCasssEssEss ss

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8../.lc)F~)::~:li-z.5")e-stdk:J:'-$~=e-si/O::l_e-2.~5-Z.S-s2. ~--58.2.(b)Fb)=-5:slC)FlS):6,5~,+ )t~2.+'1)l.)iFt.s )=7l ~+t:J.S-)=11:5+3.~(:Sf~.5)~f-l3)~cjf-+ ~-f.9.zsIllii)::5 ~~UIt+3/JD)=9..1.33cs«J.f.:l-2.54iM.,.Ij;tFL~)s:~.33.52.5'lJJ)~~.33~-J~;52.+lip.:52+11,:5'+11.if)jU)::: (,e-lZ ~t.t-Jf51J)=~lJle-~ht.t-,.lfe-~1.4.rL~)=.LJ.1.1ILJ)_Jj.z'11.~fj2)=-"1.21/:5-~.2,LJ{~tl.l)l.~1(~.,..~)LflJ~l.+J.j5+5

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(bFe.•..)-JJ-I..L-v-stlS+l)=:52.+T+s.rl~ l;6}==;t -/+e-:Cvtc:J:I.-j)::Z~+J:.Z~+,=-Z{~rJ)4--Jjl.fls-l-IOl~+Il2.+3~(.'S+IJ4-31.(~+j)~+j'~~l.i)=.;,e-Ce.&.d3;/; - ~e-~~3,t8.3.(A)Ft..~)=:5:.2,5' .•.. -5.f-z.s"j(~H)(~+,)~~s+a~(1)::z.S- -5e-zTl,Se-2~I8.LJF~)=s+5=k.,+A~.-,.c )I..l:st2)1.+£3)2-:s.J..z.-j5~+l.TJ'.3kJ=-2+i3r5=~_,_I::(J,7tJ7/9Sb-2t'3+Z+j31-J1-:.Pl.:tJ:;.7tJ7e-j~e.l-2.:lj3)i+tJ.7tJ7eilfe..t-Z:;;T3)£2:1:[&.i-lf).l3i-1r')]() ~!:t)=CJ,7~7e-e.Jfe.-J7i"=/.LJJ~e.-2.Z elM(3~-.lf5°)I)F~l)h1.la.).Jk1=o.TIJ11_~5°.; +'it) ;:2.t.~.71)7) ~f3i_~5D)= /,.l.JJ~M:J.(3-i-4!t'j-1-

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8. ,.to:tLi)=.Ife-u;i;'3);dfje:Ji=.Jfe-lli--8)t.-l.J=_2e-lLi-3):.getf.e-2~:-yt.lJ1J3.J.j)e-2t ::-3.)2.27/-/e-2t.:J'[dr/dt].:-32.21~+2..(bd[~~/di]= ~r{:5)-~~t1+)=5J'[~e(,e-2t]-.lJe~=1.;4.)3-0"13:-~l2'7j+2~+2.(C+'it-)::-<Ie -2li -~)uti-3)alo/di=_ge-2.'i-.3)U.~-3)+.1Je-2ti:.-3)}~Li-3)1~:'3-1:[d~/d:t]~_9e:3~r#e-3$::216e-3sj+~~+2(ell;[[df/dt]::::sFl:5)-t~)=~rl:5)::J.j~e.-3~;:5+2.

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B.7,B,g " -rUL ~H~~tb)jHr.t-Juu",.-'/::,'~fk.f.J~~~&~Ar+it..tt.tJ~lhL1~).LtM-t'U}=~sF'ts)= ~.t,-';let>~-YOB,tt.to:S2.A'lS) T55XlS)T l.JXl~)=J~/SXl6)=10==2'.5+-1b/3+-)M2~(5+J)&t!l)Ss+1:5+4. 'X-l:t)::2.5 -~e-tT..2e-lltJt ~~.~~lb)l.A'l~) - S X~)-"itA)r5[~X~)-~ltJ)]+J..j-=)&/5Xis)=10+-Sr-L~)'*""Xu>}+5XlO)szsTI}l.S+lf)cs+1)(~+Li)_s2-J-1,6+/0_Z.J5-513y~- ~ (~+n(s+1.J)-s-+~t-It-~""LJ"'XU)=2.5--!e-.i:+1e-~.t1t{k)}=2.5"-5"13+-Y"=Lt=a1lLt:1}t:tf[~e-t: -~e-JJ*J-t:.;1-
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