d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

Transcription

1 and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or muliplied by a posiive number in order o simplify he subsequen numerical calculaion wihou alering he sabiliy conclusion. Rouh s sabiliy crierion saes ha he number of roos of Equaion ( 6) wih posiive real pars is equal o he number of changes in sign of he coefficiens of he firs column of he array. I should be noed ha he exac values of he erms in he firs column need no be known; insead, only he signs are needed. The necessary and sufficien condiion ha all roos of Equaion ( 6) lie in he lef-half s plane is ha all he coefficiens of Equaion ( 6) be posiive and all erms in he firs column of he array have posiive signs. EXAMPLE Le us apply Rouh s sabiliy crierion o he following hird-order polynomial: a + a + a s + a = where all he coefficiens are posiive numbers. The array of coefficiens becomes a a a a - a a a a a a The condiion ha all roos have negaive real pars is given by a a 7 a a EXAMPLE Consider he following polynomial: s + = Le us follow he procedure jus presened and consruc he array of coefficiens. (The firs wo rows can be obained direcly from he given polynomial. The remaining erms are 4 Chaper / Transien and Seady-Sae Response Analyses

2 obained from hese. If any coefficiens are missing, hey may be replaced by zeros in he array.) The second row is divided by. In his example, he number of changes in sign of he coefficiens in he firs column i. This means ha here are wo roos wih posiive real pars. Noe ha he resul is unchanged when he coefficiens of any row are muliplied or divided by a posiive number in order o simplify he compuaion. Special Cases. If a firs-column erm in any row is zero, bu he remaining erms are no zero or here is no remaining erm, hen he zero erm is replaced by a very small posiive number and he res of he array is evaluaed. For example, consider he following equaion: + + s + = ( 6) The array of coefficiens is L If he sign of he coefficien above he zero ( ) is he same as ha below i, i indicaes ha here are a pair of imaginary roos. Acually, Equaion ( 6) has wo roos a s=; j. If, however, he sign of he coefficien above he zero ( ) is opposie ha below i, i indicaes ha here is one sign change. For example, for he equaion he array of coefficiens is - s + = (s - ) (s + ) = One sign change: L - One sign change: - - There are wo sign changes of he coefficiens in he firs column. So here are wo roos in he righ-half s plane. This agrees wih he correc resul indicaed by he facored form of he polynomial equaion. Secion 6 / Rouh s Sabiliy Crierion

3 If all he coefficiens in any derived row are zero, i indicaes ha here are roos of equal magniude lying radially opposie in he s plane ha is, wo real roos wih equal magniudes and opposie signs and/or wo conjugae imaginary roos. In such a case, he evaluaion of he res of he array can be coninued by forming an auxiliary polynomial wih he coefficiens of he las row and by using he coefficiens of he derivaive of his polynomial in he nex row. Such roos wih equal magniudes and lying radially opposie in he s plane can be found by solving he auxiliary polynomial, which is always even. For a n-degree auxiliary polynomial, here are n pairs of equal and opposie roos. For example, consider he following equaion: The array of coefficiens is The erms in he row are all zero. (Noe ha such a case occurs only in an oddnumbered row.) The auxiliary polynomial is hen formed from he coefficiens of he row. The auxiliary polynomial P(s) is which indicaes ha here are wo pairs of roos of equal magniude and opposie sign (ha is, wo real roos wih he same magniude bu opposie signs or wo complexconjugae roos on he imaginary axis). These pairs are obained by solving he auxiliary polynomial equaion P(s)=. The derivaive of P(s) wih respec o s is The erms in he row are replaced by he coefficiens of he las equaion ha is, 8 and 96. The array of coefficiens hen becomes s s s s - = d Auxiliary polynomial P(s) P(s) = dp (s) ds - - = s d Coefficiens of dp (s) ds We see ha here is one change in sign in he firs column of he new array.thus, he original equaion has one roo wih a posiive real par. By solving for roos of he auxiliary polynomial equaion, we obain or = =, =- s =;, s =;j 6 Chaper / Transien and Seady-Sae Response Analyses

4 These wo pairs of roos of P(s) are a par of he roos of he original equaion. As a maer of fac, he original equaion can be wrien in facored form as follows: Clearly, he original equaion has one roo wih a posiive real par. Relaive Sabiliy Analysis. Rouh s sabiliy crierion provides he answer o he quesion of absolue sabiliy. This, in many pracical cases, is no sufficien. We usually require informaion abou he relaive sabiliy of he sysem. A useful approach for examining relaive sabiliy is o shif he s-plane axis and apply Rouh s sabiliy crierion. Tha is, we subsiue ino he characerisic equaion of he sysem, wrie he polynomial in erms of ŝ; and apply Rouh s sabiliy crierion o he new polynomial in ŝ. The number of changes of sign in he firs column of he array developed for he polynomial in ŝ is equal o he number of roos ha are locaed o he righ of he verical line s= s.thus, his es reveals he number of roos ha lie o he righ of he verical line s= s. Applicaion of Rouh s Sabiliy Crierion o Conrol-Sysem Analysis. Rouh s sabiliy crierion is of limied usefulness in linear conrol-sysem analysis, mainly because i does no sugges how o improve relaive sabiliy or how o sabilize an unsable sysem. I is possible, however, o deermine he effecs of changing one or wo parameers of a sysem by examining he values ha cause insabiliy. In he following, we shall consider he problem of deermining he sabiliy range of a parameer value. Consider he sysem shown in Figure. Le us deermine he range of for sabiliy. The closed-loop ransfer funcion is The characerisic equaion is The array of coefficiens becomes (s + )(s - )(s + j)(s - j)(s + ) = s = ŝ - s (s = consan) C(s) = s + = sa + s + B(s + ) + + s( + s + ) (s + ) C(s) Figure Conrol sysem. Secion 6 / Rouh s Sabiliy Crierion 7

5 For sabiliy, mus be posiive, and all coefficiens in he firs column mus be posiive. Therefore, When = 4 9, he sysem becomes oscillaory and, mahemaically, he oscillaion is susained a consan ampliude. Noe ha he ranges of design parameers ha lead o sabiliy may be deermined by use of Rouh s sabiliy crierion. 7 EFFECTS OF INTEGRAL AND DERIVATIVE CONTROL ACTIONS ON SYSTEM PERFORMANCE In his secion, we shall invesigae he effecs of inegral and derivaive conrol acions on he sysem performance. Here we shall consider only simple sysems, so ha he effecs of inegral and derivaive conrol acions on sysem performance can be clearly seen. Inegral Conrol Acion. In he proporional conrol of a plan whose ransfer funcion does no possess an inegraor s, here is a seady-sae error, or offse, in he response o a sep inpu. Such an offse can be eliminaed if he inegral conrol acion is included in he conroller. In he inegral conrol of a plan, he conrol signal he oupu signal from he conroller a any insan is he area under he acuaing-error-signal curve up o ha insan. The conrol signal u() can have a nonzero value when he acuaing error signal e() is zero, as shown in Figure 6(a).This is impossible in he case of he proporional conroller, since a nonzero conrol signal requires a nonzero acuaing error signal. (A nonzero acuaing error signal a seady sae means ha here is an offse.) Figure 6(b) shows he curve e() versus and he corresponding curve u() versus when he conroller is of he proporional ype. Noe ha inegral conrol acion, while removing offse or seady-sae error, may lead o oscillaory response of slowly decreasing ampliude or even increasing ampliude, boh of which are usually undesirable. Figure 6 (a) Plos of e() and u() curves showing nonzero conrol signal when he acuaing error signal is zero (inegral conrol); (b) plos of e() and u() curves showing zero conrol signal when he acuaing error signal is zero (proporional conrol). e() u() (a) e() u() (b) 8 Chaper / Transien and Seady-Sae Response Analyses

6 Figure 7 Proporional conrol sysem. E(s) C(s) + Ts + Proporional Plan conroller Proporional Conrol of Sysems. We shall show ha he proporional conrol of a sysem wihou an inegraor will resul in a seady-sae error wih a sep inpu.we shall hen show ha such an error can be eliminaed if inegral conrol acion is included in he conroller. Consider he sysem shown in Figure 7. Le us obain he seady-sae error in he uni-sep response of he sysem. Define Since E(s) he error E(s) is given by For he uni-sep inpu =/s, we have The seady-sae error is = - C(s) E(s) = + G(s) = + Ts + E(s) = G(s) = Ts + = - C(s) = + G(s) Ts + Ts + + s Ts + e ss = lim e() = lim se(s) = lim Sq s S s S Ts + + = + Such a sysem wihou an inegraor in he feedforward pah always has a seady-sae error in he sep response. Such a seady-sae error is called an offse. Figure 8 shows he uni-sep response and he offse. c() Offse Figure 8 Uni-sep response and offse. Secion 7 / Effecs of Inegral and Derivaive Conrol Acions on Sysem Performance 9