Chapter 5 Stabilization In this chapter we study the unity-feedback system with block diagram shown in Figure 5.1.Here Figure 5.1:Unity-feedback system. P is strictly proper and C is proper. Most sy nthesis problems can be formulated in this way:Given P,design C so that the feedback system (1)is internally stable,and(2)acquires some additional desired property;for example, the output y asy mptotically tracks a step input r.The method of solution is to parametrize all Cs for which (1)is true,and then to see if there exists a parameter for which(2)holds.In this chapter such a parametrization is derived and then applied to two problems:achieving asy mptot ic performance specs and internal stabilization by a stable controller. 5.1 Controller Parametrization:Stable Plant In this section we assume that P is already stable,and we parametrize all Cs for which the feedback system is internally stable.Introduce the symbol S for the family of all stable,proper, real-rational functions.Notice that S is closed under addition and multiplication:If FGE S,then F+G,FGES.Also,1E S.(Thus S is a commutat ive ring with identity.) Theorem 1 Assume that P E S.The set of all Cs for which the feedback system is internally stable equals {-9noQes} 57
Chapter Stabilization In this chapter we study the unityfeedback system with block diagram shown in Figure Here C P r u y d e Figure Unityfeedback system P is strictly proper and C is proper Most synthesis problems can be formulated in this way Given P design C so that the feedback system is internally stable and acquires some additional desired property for example the output y asymptotically tracks a step input r The method of solution is to parametrize all Cs for which is true and then to see if there exists a parameter for which holds In this chapter such a parametrization is derived and then applied to two problems achieving asymptotic performance specs and internal stabilization by a stable controller Controller Parametrization Stable Plant In this section we assume that P is already stable and we parametrize all Cs for which the feedback system is internally stable Introduce the symbol S for the family of all stable proper realrational functions Notice that S is closed under addition and multiplication If F G S then F G F G S Also S Thus S is a commutative ring with identity Theorem Assume that P S The set of al l Cs for which the feedback system is internal ly stable equals Q P Q Q S
58 CHAPTER 5 STABILIZATION &tt品a.风知.Q, C Q:=1+P元5 Then,s aa (cCrse,supp to,s aade Q C:= 1-PQ5 (5.1) AceGail tOte den liticintimn2.te reedka systm i te rin rIster fuld 1 1+PC tftan Jainti 「1-PQ-P(1-PQ)-(1-PQ)1 Q 1-PQ -Q 5 PQ P(1-PQ) 1-PQ clay,tese rinertis batos. Nct ttalritater untaao a aingnaotrpaaetrt ecdhek The fen T5+T.Q fs Ghe T5 T.I.IIatcua the se IWI alac p emeIay se IWI fuldIsae S=1-PQ- T-PQ5 toaiu!apiaiotonac cttda ystm Iastt as arupkey taks apr (when=0).Pachecam the tecem.Thgn aympka tas atp It the belter fultGhr (le.,s)ha a3rCts=0,ti丛， t6a益Linzwr从aadii P(0)Q(0)=15 c=90Q,sQ0= P而5 oRrve tcte iatp atac.ALOy G caned ttacrcer tetr rGn naa c ts =0,aItmusthThe crem 3 (Cho.ter 3. Example Fte p P(S)=8+1)s+2
CHAPTER STABILIZATION Proof Suppose that C achieves internal stability Let Q denote the transfer function from r to u that is Q C P C Then Q S and C Q P Q Conversely suppose that Q S and dene C Q P Q According to the denition in Section the feedback system is internally stable i the nine transfer functions P C P C C P C P all are stable and proper After substitution from and clearing of fractions this matrix becomes P Q P P Q P Q Q P Q Q P Q P P Q P Q Clearly these nine entries belong to S Note that all nine transfer functions above are ane functions of the free parameter Q that is each is of the form T TQ for some T T in S In particular the sensitivity and complementary sensitivity functions are S P Q T P Q Let us look at a simple application Suppose that we want to nd a C so that the feedback system is internally stable and y asymptotically tracks a step r when d Parametrize C as in the theorem Then y asymptotically tracks a step i the transfer function from r to e ie S has a zero at s that is P Q This equation has a solution Q in S i P Conclusion The problem has a solution i P when this holds the set of all solutions is C Q P Q Q S Q P Observe that Q inverts P at dc Also you can check that a controller of the latter form has a pole at s as it must by Theorem of Chapter Example For the plant P s s s
5.2.COPRIME FACTORIZATION 59 suppose that it is desired to find an internally stabilizing controller so that y asymptotically tracks a ramp r.Parametrize C as in the theorem.The transfer function S from r to e must have (at least)two zeros at s =0,where r has two poles.Let us take Q(s)=9s+b 8+1 This belongs to S and has two variables,a and b,for the assignment of the two zeros of S.We have as+b S(8)=1- (s+1)2(s+2) s3+4s2+(5-a)s+(2-b) (8+1)2(s+2) so we should take a=5,6 =2.This gives Q(s)= 5s+2 8+1? C(s)= (5s+2)(s+1)(s+2) s2(s+4) The controller is internally stabilizing and has two poles at s =0. 5.2 Coprime Factorization Now suppose that P is not stable and we want to find an internally stabilizing C.We might try as follows.Write P as the ratio of coprime poly nomials, N P二M By Euclid's algorithm (reviewed below)we can get two other poly nomials X,Y satisfy ing the equation NX+MY =1. Remembering Theorem 3.1 (the feedback system is internally stable iff the characteristic poly nomial has no zeros in Re s 0),we might try to make the left-hand side equal to the characteristic polynomial by setting c= The trouble is that Y may be 0;even if not,this C may not be proper. Example 1 For P(s)=1/s,we can take N(s)=1,M(s)=s.One solution to the equation NX +MY =1 is X(s)=1,Y(s)=0,for which X/Y is undefined.Another solution is X (s)= -s+1,Y(s)=1,for which X/Y is not proper. The remedy is to arrange that N,M,X,Y are all elements of S instead of polynomials.Two functions N and M in S are coprime if there exist two other functions X and Y also in S and satisfying the equation NX+MY=1
COPRIME FACTORIZATION suppose that it is desired to nd an internally stabilizing controller so that y asymptotically tracks a ramp r Parametrize C as in the theorem The transfer function S from r to e must have at least two zeros at s where r has two poles Let us take Qs as b s This belongs to S and has two variables a and b for the assignment of the two zeros of S We have Ss as b s s s s a s b s s so we should take a b This gives Qs s s Cs s s s s s The controller is internally stabilizing and has two poles at s Coprime Factorization Now suppose that P is not stable and we want to nd an internally stabilizing C We might try as follows Write P as the ratio of coprime polynomials P N M By Euclids algorithm reviewed below we can get two other polynomials X Y satisfying the equation NX M Y Remembering Theorem the feedback system is internally stable i the characteristic polynomial has no zeros in Re s we might try to make the lefthand side equal to the characteristic polynomial by setting C X Y The trouble is that Y may be even if not this C may not be proper Example For P s s we can take Ns Ms s One solution to the equation NX M Y is Xs Y s for which XY is undened Another solution is Xs s Y s for which XY is not proper The remedy is to arrange that N M X Y are all elements of S instead of polynomials Two functions N and M in S are coprime if there exist two other functions X and Y also in S and satisfying the equation NX M Y
60 CHAPTER-5 STABILIZATION Noticetkatforths euation id,N,ad M ca laeno common zs in Re>o noratte oints =oo-if trelewelesucnaloints urelewould follow 0=N(sX(s+M(s)Y(s≠15 rtca pepvea tts condition is ap sufficiatforoime LEtG bealcaetiona tasferfunction.A IEpeetetion of trero I G-N.M,5, M weN ad M ameis acoprime factorization of Govs.TpIpseof hs seetion is peetametad fortreonstliction of fourrunc tions in s sctisfy ing tretwo uctions G=N NX+MY=15 Treconstluction of N ad M is (y. Example 1 Tec(s)=1.(s.1).To wilec =N.M with ad M in s,simpy dividetre numecorad d omincorpoly nomias,1 ad s.1,by acommon poly nomiawillo zs in Re≥0，s8(s+1): 女=器N=M式 s.1-M(s)' I teinkeerk is gleckrtka 1,ta N ad M aenotooplme-tre raeacommon ze at S=00.So N(问)=3十石 1 M=8 suffice 6 Euclid's agolithn compte treleetcommon divisorof two give poly nomias sa n) ad m(1).wre rad m aecoplimetreagorthn ca beused computeroly nomias x(1). y(1)sclisfying +my =15 Procedure A:Euclid's Algor Ipt poly nomids I mitaizer itis nottetrtderen dere(m),intercragend m. Step.Dividem int n getquotatq-ad IGhanderr_ Imq-+r deke_:dekem5
CHAPTER STABILIZATION Notice that for this equation to hold N and M can have no common zeros in Res nor at the point s if there were such a point s there would follow Ns Xs Ms Y s It can be proved that this condition is also sucient for coprimeness Let G be a realrational transfer function A representation of the form G N M NM S where N and M are coprime is called a coprime factorization of G over S The purpose of this section is to present a method for the construction of four functions in S satisfying the two equations G N M NX M Y The construction of N and M is easy Example Take Gs s To write G NM with N and M in S simply divide the numerator and denominator polynomials and s by a common polynomial with no zeros in Res say s k s Ns Ms Ns s k Ms s s k If the integer k is greater than then N and M are not coprimethey have a common zero at s So Ns s Ms s s suce More generally to get N and M we could divide the numerator and denominator polynomials of G by s k where k equals the maximum of their degrees What is not so easy is to get the other two functions X and Y and this is why we need Euclids algorithm Euclids algorithm computes the greatest common divisor of two given polynomials say n and m When n and m are coprime the algorithm can be used to compute polynomials x y satisfying nx my Procedure A Euclids Algorithm Input polynomials n m Initialize If it is not true that degree n degree m interchange n and m Step Divide m into n to get quotient q and remainder r n mq r degree r degree m
,COPRIME FACTORIZATION 61 steD.Divider into m to get quotientd and remainder m=rg+卫， degree:degree r5 stepi Divide p into r L=卫q十r÷ dlegreer degree 5 Cntinue Stop at Step k when rk is a nonzero constant. Then x,y are obtained as illustrated by the following example for k =3.The equations are nmc以+ry m=r以g+卫， =卫q+r that is, 100 15c4 410≥1y≥= 01之m] 511rL00 Solve for r by,say,Gaussian elimination: r1+4qn5q541+4gm5 Set x +49以 0541495 Ex an pe1 The algorit hm for )=1,m(1)=61 5 51 +1 goes like this: (1) = 2 6 1)= 56 4(1)= 15 5 6 y(1)= 分 Since p is a nonzero constant,we stop after Step 2.Then the equations are n=mgk+r人 m rd+
COPRIME FACTORIZATION Step Divide r into m to get quotient q and remainder r m rq r degree r degree r Step Divide r into r r rq r degree r degree r Continue Stop at Step k when rk is a nonzero constant Then x y are obtained as illustrated by the following example for k The equations are n mq r m rq r r rq r that is q q r r r q n m Solve for r by say Gaussian elimination r qq n q q qq m Set x r qq y r q q qq Example The algorithm for n m goes like this q r q r Since r is a nonzero constant we stop after Step Then the equations are n mq r m rq r