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
62 CHAPTER 5.STABILIZATION yielding T2=(1+q192)m-92n. So we should take x=、更 1+92 T2 T2 that is, x(A)=-30λ+19，y()=5λ+1. Next is a procedure for doing a coprime factorization of G.The main idea is to transform variables,s→λ，so that polynomials inλyield funct ions in S. Procedure B Input:G Step 1 If G is stable,set N=G,M=1,X=0,Y 1,and stop;else,continue. Step 2 Transform G(s)to G(A)under the mapping s =(1-A)/A.Write G as a ratio of coprime poly nomials: G(A)=zQ) m() Step 3 Using Euclid's algorithm,find polynomials r(A),y(A)such that nx my =1. Step 4 Transform n(),m(),x(),y()to N(s),M(s),X(s),Y(s)under the mapping λ=1/(s+1). The mapping used in this procedure is not unique;the only requirement is that polynomialsn,and so on,map to N,and so on,in S. Example 4 For 1 G(8)=8-08-2 the algorit hm gives A2 G()= 6λ2-5λ+1’ n(A)=2, m(A)=6λ2-5λ+1， x()=-30λ+19， y()=5+1 (from Example 3), 1 N(s)= (8+1)2
CHAPTER STABILIZATION yielding r qq m qn So we should take x q r y qq r that is x y Next is a procedure for doing a coprime factorization of G The main idea is to transform variables s so that polynomials in yield functions in S Procedure B Input G Step If G is stable set N G M X Y and stop else continue Step Transform Gs to G under the mapping s Write G as a ratio of coprime polynomials G n m Step Using Euclids algorithm nd polynomials x y such that nx my Step Transform n m x y to Ns Ms Xs Y s under the mapping s The mapping used in this procedure is not unique the only requirement is that polynomials n and so on map to N and so on in S Example For Gs s s the algorithm gives G n m x y from Example Ns s
,.X COPRIME FACTORIZATION BY STATE-SPACE METHODS (OPTIONAL) 63 M(s)= (8-1)(8-22 (8+1)2 X(s)= 19s-11 8+1 Y()= s+6 s+1 513 Coprime Factorization by State4space Methods (Optional) This optional section presents a state space procedure for computing a coprime factorizat ion over S of a proper G.This procedure is more efficient than the poly nomial method in the preceding sect ion. We start with a new data structure.Suppose that A,B,C,D are real matrices of dimensions n×n2n×121×n21×1. The transfer funct ion going along with this quartet is D+C(sI-A)B. Note that the const ant D equals the value of the transfer funct ion at s=oo;if the transfer funct ion is strictly proper,then D=0.It is convenient to write AB instead of D+C(sI-A)B. Beginning with a realizat ion of G, G(s)= A B LGD the goal is to get state space realizations for four functions N,M,X,Y,all in S,such that G-- N M2 NX+MY =1 First,we look at how to get N and M.If the input and output of G are denoted u and y, respectively,then the state model of G is 元=Ax+Bu2 (.2) y=Cx+Du. (.3) Choose a real matrix F,1x n,such that A+BF is stable (i.e.,all eigenvalues in Res <0).Now define the signal v:=u-Fr.Then from (..2)and (..3)we get 元=(A+BF)x+B2 u=Fx+v2 =(C+DF)x+Dv
COPRIME FACTORIZATION BY STATESPACE METHODS OPTIONAL Ms s s s Xs s s Y s s s Coprime Factorization by StateSpace Methods Optional This optional section presents a statespace procedure for computing a coprime factorization over S of a proper G This procedure is more ecient than the polynomial method in the preceding section We start with a new data structure Suppose that A B C D are real matrices of dimensions n n n n The transfer function going along with this quartet is D CsI A B Note that the constant D equals the value of the transfer function at s if the transfer function is strictly proper then D It is convenient to write A B C D instead of D CsI A B Beginning with a realization of G Gs A B C D the goal is to get statespace realizations for four functions N M X Y all in S such that G N M NX M Y First we look at how to get N and M If the input and output of G are denoted u and y respectively then the state model of G is x Ax Bu y Cx Du Choose a real matrix F n such that A BF is stable ie all eigenvalues in Res Now dene the signal v u F x Then from and we get x A BF x Bv u F x v y C DF x Dv
64 CHAPTER 5.STABILIZATION Evidently from these equations,the transfer function from v to u is (5.4) and that from v to y is N(s):= A+BFB C+DF D (5.5) Therefore, u=Mv,y=Nv, so that y=NM-u,that is,G=N/M.Clearly,N and M are proper,and they are stable because A+BF is.Thus N,ME S.Suggestion:Test the formulas above for the simplest case,G(s)=1/s (A=0B=1,C=1,D=0). The theory behind the formulas for X and Y is beyond the scope of this book.The procedure is to choose a real matrix H,n x 1,so that A+HC is stable,and then set X(s):= 「A+HC|H] (5.6) Y(s):= A+HC-B-HD F 1 (5.7) In summary,the procedure to do a coprime factorization of G is this: Step 1 Get a realization (A,B,C,D)of G. Step 2 Compute matrices F and H so that A+BF and A+HC are stable. Step 3 Using formulas(5.4)to (5.7),compute the four functions N,M,X,Y. 5.4 Controller Parametrization:General Plant The transfer function P is no longer assumed to be stable.Let P=N/M be a coprime factorization over S and let X,Y be two functions in S sat isfying the equation NX+MY =1. (5.8) Theorem 2 The set of all Cs for which the feedback system is internally stable equals ∫X+MQ Y-NQ It is useful to note that Theorem 2 reduces to Theorem 1 when P ES.To see this,recall from Section 5.2(Step 1 of Procedure B)that we can take N=P,M=1,X=0,Y=1 when P∈S.Then X+MQ Q Y-NQ=1-PQ The proof of Theorem 2 requires a preliminary result
CHAPTER STABILIZATION Evidently from these equations the transfer function from v to u is Ms A BF B F and that from v to y is Ns A BF B C DF D Therefore u Mv y Nv so that y NMu that is G NM Clearly N and M are proper and they are stable because A BF is Thus NM S Suggestion Test the formulas above for the simplest case Gs s A B C D The theory behind the formulas for X and Y is beyond the scope of this book The procedure is to choose a real matrix H n so that A HC is stable and then set Xs A HC H F Y s A HC B HD F In summary the procedure to do a coprime factorization of G is this Step Get a realization A B C D of G Step Compute matrices F and H so that A BF and A HC are stable Step Using formulas to compute the four functions N M X Y Controller Parametrization General Plant The transfer function P is no longer assumed to be stable Let P NM be a coprime factorization over S and let X Y be two functions in S satisfying the equation NX M Y Theorem The set of al l Cs for which the feedback system is internal ly stable equals X MQ Y NQ Q S It is useful to note that Theorem reduces to Theorem when P S To see this recall from Section Step of Procedure B that we can take N P M X Y when P S Then X MQ Y NQ Q P Q The proof of Theorem requires a preliminary result
55)5 CONTROLLER PARAMETRIZATION2 GENERAL PLANT 6. Lemma Let C-NC/MC be a coprime factorization over S.Then the feedback system is internally stable iff (NNC+MMO1 S. The proof of this lemma is almost identical to the proof of Theorem 31,and so is omitted/ Proof of Theorem 2 ()Suppose that Q1 S and X+MQ C:-Y-NQ To show that the feedback system is internally stable,define NC:=X+MQ,MC:=Y-NQ. Then from the equation NX+MY=1 it follows that NNC+MMC=1. Therefore,C=NC/MC is a coprime factorization,and from Lemma 1 the feedback system is internally stable/ (2)Conversely,let C be any controller achieving internal stability/We mst find a Q in S such that C= X+MQ Y-NQ Let C=NC/MC be a coprime factorization over S and define V:=(NNC+MMO so that NNCV+MMCV =1. () By Lemma 1,V 1 S/Let Q be the solution of MCV=Y-NQ. (.10) Substitute (/10)into (9)to get NNCV+M(Y-NQ)=1. (.11) Also,add and subtract NMQ in (8)to give N(X+MQ)+M(Y-NQ)=1. (.12) Comparing (/11)and (/12),we see that NCV=X+MQ. (.13) Now (10)and (13)give NCV X+MQ C=MOY-NO
CONTROLLER PARAMETRIZATION GENERAL PLANT Lemma Let C NC MC be a coprime factorization over S Then the feedback system is internal ly stable i NNC MMC S The proof of this lemma is almost identical to the proof of Theorem and so is omitted Proof of Theorem Suppose that Q S and C X MQ Y NQ To show that the feedback system is internally stable dene NC X MQ MC Y NQ Then from the equation NX M Y it follows that NNC MMC Therefore C NC MC is a coprime factorization and from Lemma the feedback system is internally stable Conversely let C be any controller achieving internal stability We must nd a Q in S such that C X MQ Y NQ Let C NC MC be a coprime factorization over S and dene V NNC MMC so that NNC V MMC V By Lemma V S Let Q be the solution of MC V Y NQ Substitute into to get NNC V MY NQ Also add and subtract NMQ in to give NX MQ MY NQ Comparing and we see that NC V X MQ Now and give C NC V MC V X MQ Y NQ
66 CHAPTER 2 STABILIZATION It remains to show that Q.S/Multiply (/10)by X and (/13)by Y,then subtract and switch sides (NX MY)Q=YNCV-XMC5 But the left,hand side equals Q by (8),while the right,hand side belongs to S/So we are done/ Theorem 2 gives an automatic way to stabilize a plant/ Example Let 1 P6)=s-1)s-2 5 Apply Procedure B to get N(s)= (s+1)5 M(s)= s-0s-25 (s+1)e X(s)= 19s-11 s+1 Y(s)= 8+6 8+1 According to the theorem,the controller X(s) 19s-11 C(s)= Y(s) s+6 achieves internal stability/ As before,when P was stable,all closed,loop transfer functions are affine functions of Q if C is parametrized as in the theorem statement/For example,the sensitivity and complementary sensitivity functions are S-M(Y-NQ)5 T =N(X+MQ)5 Finally,it is sometimes useful to note that Lemma 1 suggests anot her way to solve the equation NX +MY 1 given coprime N and M/First,find a controller C achieving internal stability for P =N.M-this might be easier than solving for X and Y/Next,write a coprime factorization of C:C NC MC/Then Lemma 1 sayys that V:=NNC+MMC is invert ible in S/Finally,set X =NCV--and Y MCV- 5.5 Asymptotic Properties How to find a C to achieve internal stability and asymptotic properties simultaneously is perhaps best shown by an example/ Let 1 P()=8-s-习5 The problem is to find a proper C so that
CHAPTER STABILIZATION It remains to show that Q S Multiply by X and by Y then subtract and switch sides NX M Y Q Y NC V XMC V But the lefthand side equals Q by while the righthand side belongs to S So we are done Theorem gives an automatic way to stabilize a plant Example Let P s s s Apply Procedure B to get Ns s Ms s s s Xs s s Y s s s According to the theorem the controller Cs Xs Y s s s achieves internal stability As before when P was stable all closedloop transfer functions are ane functions of Q if C is parametrized as in the theorem statement For example the sensitivity and complementary sensitivity functions are S MY NQ T NX MQ Finally it is sometimes useful to note that Lemma suggests another way to solve the equation NX M Y given coprime N and M First nd a controller C achieving internal stability for P NMthis might be easier than solving for X and Y Next write a coprime factorization of C C NC MC Then Lemma says that V NNC MMC is invertible in S Finally set X NC V and Y MC V Asymptotic Properties How to nd a C to achieve internal stability and asymptotic properties simultaneously is perhaps best shown by an example Let P s s s The problem is to nd a proper C so that
