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“CHAPTER 3 BS. RO ” iza Ra Gt Ga ” | Gs -G [e Cs) - Gt Ge RO (GA GG- Ga) B32. & . o - co) z Ha Ha [e Lc? NM poe Ga cê lá 1% GaltHe) CO - (tG)Gr RO) St GalHy-tha) B33. g Gs ce). Ro, G x x Ga Ex Gs , | He a -0- e. Rs), & LA Gr ? EXA us Ga % He RS) GF (Gard) co CRM) Gt) Gts [66] “ G GaGs + G GM o RO Gama TG GH Gr MHa E GGG + G,G3H) B-3-4. Im the following diagrams, (a) denotes the unit-step response and (b) “Corresponds to the unit-ramp response. 7) ue) é E) dd od e p ty É) to z Po qor ss 4% EEE K p= fed deb E = Kad uia [o] tm b) é é O) o + O) + a at CITE SA e CM as 4 Be mirdDe ndo) Ame geno = s(re0) - 12 — When only the disturbance input D(s) is present, the output Cp(s) is given by Se) GO D6) 1% Gl) Gute) Since the desired output to the disturbance input D(s) is zero, the error En(s) - can be given by E6)=0- Cab) =— Co 6) Gale) =— = DE BO o gu PO For the stable system, the steady-state error esen(t) is given by -3 Gr) DO) =Li Este) = bi = boto Esep tt) £: Dlt) Lim 566) ft TE GET) The steady-state error when both the reference input R(s) and disturbance input. D(s) are present is the sm of essp(t) and esen(t) and is given by Est) = Essa lt) + Essp tt) [ sRG) .S&E DO ] 1H GH) GU 1 Ge Gas) s> = La s “É | og [88-66 00]] B-3-7.. when D(s) = 0, the block diagram of the system can be simplified as follows: Re) 6. & & Gr RO [GG H, H fe The closed-loop transfer function Cp(s)/R(s) can be given by GG Ga Gs Cats) = (HG Ga dt — Ge Gr Ga Gs R62 + Le GG Os Ma 14 GG Hit Gol Ga Gs 14 G,G Hg - 14 — When R(s) = 0, the biock E dlegram of the system shown in Figure 3-76 can be modified as follows: | De). e * 4. BS) 6, *| & [| & a Hy a En po Ha ja DE) Co) GG - Gy Pe) demo Hence CE) = Ga Ga o Ga Gs DO) tags G (am + E 1H Ga Ga Ge Mk FG Gay B-3-8. There are infinitely many state-space representations for this system. We shall give tvo of the possible state-space representations. State-spaçe representation 1: From Figure 3-77, we obtain / Yo. = =. sez E E) vt) DE sSppsps+E which is equivalent to Erpyritig=mlrau Comparing this equation with the standard equation Grajtayrag=hiirhi+rhurka ve obtain , Up, e2=/, as=E, b=0, b=0, hm=l, b=E -15- ujm 4 ep] % vs 7 s+p . 5 From this block disgram ve get the following equations: A VaT-xt»M X EP =x, s+p Xz, = V-x +Xs s AL Xa s from which ve obtain %+ px =(2-p)u-(2-p)x, Mn -M+rã du d= Rewriting, ve have X =X 2 h=-Mrmara Z=-(2-p)x —pxs +(2-p)u = or FA 9 (o Ala o gj=|-/ 0 eles %| [4/7 o —plim] |2-p -17- “ % Pal, o ap % Bão. vd “ E A tIjrg=u Define . X =s %=4 Xmy aten H+30 + 2% =é n=% FARTA or &l [o 4 oh [9] dtl=|0 o «llul+role % lo -2 -3ls) do no) g=[' e elx % B30. The transfer function G(s) of the system is given by - co gex-arg=u aU VT] - 18 — The transfer matrix of the system can be given by - tl -f s -f o - 10 o D=CÓsI-A E=[ ]2 Ss - Go=g6st-arB=[s 7 ole s Isfeseg sté (00 Lo 2 ses o vol] s+esteas+2] 235 dis —a . / fo s+6 Sreste este |S ss h se Sposa pst2 Si pos es+2 Ss , SS º Srostrgs+a strssaps+z A MATLAB solution, to this problem is given below. - A=[0 1 00.0 1-2 4 6]; B=[0 00 11 0]; C-[1 0 0601 0]; D=[0 00 0]; [nom den] = ss2H7A,B,C,D,1) mm = O 0.0000 0.0000 1.0000 O 0.0000 1.0000 0.0000 dem = 10000 60000 4.0000 2.0000 [nen den] = ss21(A/B,CD,2) mam = “O 0.0000 1.0000 6.0000 O 1.0000 6.0000 0.0000 den= 1.0000 60000 4.000 2.000 - 20 - [A [4 / o [4 0 t Ijoe s ol séljso, B-3-13. Since the same force transnits the shaft, we have f=hlê-i)=b(5-2)+ ba (9-8) (1) where displacement z is defined in the figure belov. a x + z In terms of the equivalent viscous friction coefficient, the force f is given by j= bus (9-*) (2) From Equation (1) we have bithêsbi=hkthg+bs E = a lbt+Chtb)g] 63) By substituting Equation (2) into Equation (1), we have O rnteturg]-2] f=hté-t) = A Rr Lz+lht 24] bot b> .s . =b, b, ESTES (4-*) (4) Hence, by comparing Equations (2) and (4), we obtain bath [A hey = by = bits th, l L ' ? Betha + % B-3-14. (a) NX +ÊX =e MO = 77) = qo (b) Detine the dispiacenent o€ a point betueen springs Xy and Xp as y. Then the equations of motion for thie system became mx +kz (x-4) = e ky = A (x-s) - 21 — Má tm de (X+L 10) et Samio = (um0)6'+ (ces) 5 -23- Pe, r x º A p) 9 % o o Z À a 4 cm Ledi al jm o fu %| o A o ,, % o ol x b ke , | ? ma Ba Tm | * *, FARRA ó oa Xs tj jo o to x . za B3-6. o Vo =T where T=-aÉga! -malano: J=mt* For small 0, . mlrô = — 268*0 —mato or . Zé 1)p= de +5)5 =o B3-17. Note that For x direction: - (Memx- mé uno) 5! rm (om 6) = te For small 6 and smail 62, we have (Mem úrml é me. (1) For rotational motion: JO =uplaio -mileno J=T+we* Lam” ae (Temo = mpluno - meleng For small 6, ve have 5 (Trnl)6 = mgdo-mtx (2) From Equation (2), .. Trml? “ x = 9º NA 8 Subetituting this iast equation into Equation (1), we obtain (Mem )(q8 - — Fm 7; Jemtô =“ =. Dim Ee um? $ (rm) pó me = Thus, ê= ml(Meme md “ (Mem)zemnlt. (Mim) Mme? Also, from Equation (1) we have “o mpto- mi (Mem)x rm Tome 63) [MItm(TAMtD]X + MltIO =e(Trnt) "2 from which ve get - 24 - fr=8=z, fi=x=As Then, from the definition of state variables and Equations (3) and (4), the state equation and output equation can be written as 1 Tr % é t o ellx 0 ; ml(Mpmga Ed t - PR ETC 72 RARA | ed DM 7 2007755777 u & o oo “a o . 4 Az X - q 5) x Itmdt | * MIMT+MES º ia MIrm(TAHA9 o x, % 1 o eo x . & o o / A % EA B-3-18. The equations for the system are Mm, =, Z,-hã, — ks (4X) + u Mr % = —ha To-brk, —les (xa x, ) Rewriting, we have mkt, th Ati, = ks + ma + nd, the do tha mm ks 26 Assuming the zero initial condition and taking the Laplace transforms of these two equations, we obtsin : (mstebs ht )IO = ks XD + VE) 1) Gm srha sthat EDITA = és Xi 6) (2) By elimínating X2(s) from Equations (1) and (2), we get . ha (o st hstk 1 As) XE) = 2 —WLKxwO+tr mast+b.s+ br thks ' o Hence ht) MeS+bastha + hs TO) (ms? k hist krtks)Cma st thastkaths) “RE From Equation (2), we obtain Xl) — ks Hence MO mst bisthatks Halo) SD MO ks VO MO VE” Ems"rhys the hs) Que Str das rh) RA B-3-19. The equations for the given circuit are as followw: E Rrl (GL =sS , , diz Ê Aitbfudt+l(-)=o É Jide = € Taking the Laplace transforms of these three equations, assuming zero initial conditions; gives AntO+LÍSI6) —sIls)) =E:6) (1) RLOD+ Elo +l[shy-sLtw]|=o (2) let) = E 6) 43 From Equation (2) we obtaín (Rr +Ls)T.6) = Ls 16) L - ACs* . 2€) LlCs*t RCSst/ Dé) Ro subetituting Equation (4) into Equation (1), ve get (hris-is 5] ) ht) = Es6) LCs"tRacs+] LC (RR)! + CRC IL)SE LCstrRes+] From Equations (3) and (4), ve have Ls = Lose RLOSFT 46) = Ests) (6) F, “LO=E6) 6) -27 - Et) — L)-Tats) (a) €s [—— R Le) 18-L6 [5 E Ab) Z6) &6) (b) as [| masai — (c) & > T6O-fts E) T E Ke TIS) É Ri o sal” Rest] Pitas > & E:6) 7 Gr E) te) ? Gs aco Res” ” Qs + PI- a Cas he (e) Et6) ER Cs I Elo) “is Masi Leesa Mas - Gs | BR ju Ri je Gs pe EM) 7 7 EMs) tg) é Reser | Lkaser - RiC23 [e (m) Et 7 Et) RG REG HMA) - - 29 - B-321. Impedance Z, is ! ZeRtgs -Impedance Z2 is Z=htar Hence . 1 EM Za t+ &s O Br Z Z ' Htgs +Ra Cas = Ras +/ BCG +RO)S+ + EL (& E 4 5) S 1£ ve change Rj to by, R2 to bz, Cy to 1/Xj, C2 to 1/k2, then ve obtain Relas +) bakes+/ BROS ++ o Cuth)tes ++ de XD bas tha = bastrks KO (brth)sthrk, (sr k)+(hstks) The analogous mechanical system is stiown belov. 
