Signal Processing and Linear Systems-B.P.Lathi copy

88 6 7 3 root locus t he e xample in sec 67 2 gives a

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Unformatted text preview: ate Errors Steady-st~te specifications impose additional constraints on the closed-Ioo t~ansfer functIon T(s). T he s teady-state error is t he difference between t he d p :~~d o utput [reference J(t») a nd t he a ctual o utput yet). Thus e(t) = J(t) _ y(t)~ E (s) = F(s) - Y(s) = F (s) = [1 _Y(S)] F(s) F(s)[1 - T(s») (6.90) 442 6 Continuous-Time System Analysis Using t he Laplace Transform 443 6.7 Application t o Feedback a nd C ontrol S tate E quations y (1) .0,(9) y (t) (a) F ig. 6 .44 A u nity f eedback s ystem w ith v ariable g ain K . (b) K=48 K=O assuming T(O) = 1 a nd frO) = O. M any systems in practice have unity feedback, as depicted in Fig. 6.44. In such a case, t he s teady-state error analysis is g reatly simplified. Let us define p ositional e rror c onstant K p , v elocity e rror c onstant K v , a nd a cceleration e rror c onstant K a as K=O -2 -4 Ka = lim s2[KG(s)] Kp = lim[KG(s)], .-0 8 -0 Because T (s) = K G(s)/1 + KG(8), (6.96) from Eq. (6.90), we o btain 1 E (s) = 1 + K G(s)F(s) T he s teady-state errors are given b y F ig. 6 .43 A t hird-order f eedback s ystem a nd i ts r oot l ocus. T he s teady-state e rror ess is t he value of e(t) as t - + obtained from t he final-value theorem [Eq. (6.68)]: 0 0. T his value can b e readily e r = hm e 8 = lim [ 1- T(s)] = 1 - T(O) 8 --'--- 8 -0 (6.91) 8-0 P (6.92) s -o (6.97b) 1/ s3 1 + K G(s) (6.97c) = 8im [ -irs)] = - i(o) l _0 (6.94) 3. Using a similar argument, we c an show t hat for a unit parabolic i nput t ( t2/2)u(t), a nd F(s) = l /s 3 a nd ep, t he s teady s tate e rror is ep = f ro) - -2- (6.95) 1 1 G (s)=-s(s + 8) Hence, from Eq. (6.96) K S I f T(O) i= 1, e r = 0 0. Hence for a finite steady-state error to r amp i nput, a necessary condition is T(O) = 1, implying zero steady-state error to s tep i nput. Assuming T(O) = 1 a nd a pplying L'Hopital's rule t o Eq. (6.93), we have er 1 l ims_o s2[KG(8)] = Ka For t he s ystem in Fig. 6.36a, I f T(O) = 1, t he s teady-state error t o u nit-step input is zero. 2. For a u nit r amp i nput, F (s) = 1 / s2 a nd e r, t he s teady-state error, is given by (6.93) (6.97a) 1 1 1 + K G(s) - lim 8 _0 s[KG(s)] = Kv 8 -0 1 - T(s) . e r= I 1m - - - 1 + Kp _ 1 /8 2 8 - :---,:_...,.....,. . 1. F or t he u nit s tep i nput, t he s teady-state error e 8 is given by 8-0 1 1 + l ims_o [KG(s)] e = lim e BS = lim s E(s) = lim s F(s)[l- T(s)] 8 -0 l /s . e 8 = I1m 8 -:---,:_.,.....,s-O 1 + K G(s) Kp = 0 0, K v=S' Ka = 0 (6.98) S ubstitution of these values in Eq. (6.97) yields e s = 0, 8 er = K' A system where G (s) has one pole a t t he origin (as t he p resent case) is d esignated as t ype 1 s ystem. Such a s ystem can track position of a n o bject with zero error (e s ~ 0), a nd yields a constant error in tracking a n o bject moving with constant velOCIty (e r = a c onstant). B ut t he t ype 1 s ystem is n ot suitable for tracking a constant acceleration object. 444 6 Continuous-Time System Analysis Using t he Laplace Transform I f G (s) has n o poles a t t he origin, t hen K p is finite and K v = K a = o. 6.7 Application to Feedback a nd C ontrol S tate EquatIOns T hus, for Compensator f (l) G(s) = (...
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