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1、Chapter 10Design of Control Systems in State-Space (Chapter 12)10.1IntroductionThis chapter has been organized into four sections. In section 10.2 we give a complete solution to the state feedback pole placement. In section 10.3 we solve the pole placement problem using observer-based state feedback
2、. In section 10.5, a more practical design problem, called the regulator problem, will be solved by augmented state feedback.10.2Pole Placement10.2.1State FeedbackState feedback is accomplished by letting the control input in the state equation x = Ax + Bube u = Rv Fx.Control system with state feedb
3、ack is depicted in Fig. 10.1. In general, the external control input v has the same dimension as u, The matrix R is therefore square. F is an m n matrix called the state feedback gain matrix. Without loss of generality, R is chosen nonsingular.v-R- h -B- h - xR-C- h-yxA6F6u-D?Figure 10.1: System wit
4、h state-feedbackWith the state feedback the state equation of the closed-loop system is x = (A BF)x + BRvIn the state feedback systems, performance specifications are achieved by properly chosen matrices R and F. State feedback has the following properties.1. State feedback doesnt change the control
5、lability, that is (A, B) and (A BF , BR) have the same controllability. This is becauseI A + BFBR = I AB?I0 FR?23210.2. POLE PLACEMENT2332. The uncontrollable poles cannot be altered, that is, if pr+1, pr+2, ., pnare uncontrollabel eigenvalues of A, then they are also eigenvalues of ABF and is uncon
6、trollable, which will be shown immediately.10.2.2Necessary and Sufficient Condition for Pole PlacementThe pole placement problem is to find a state feedback gain matrix F so that the closed-loop poles are placed at the n desired locations 1, 2, ., n. We discuss here only SISO systems. In this case F
7、 is a row vector and denoted by fT, R is a scalar.We show first that the controllability of (A, b) is a necessary condition for arbitrary pole placement. Indeed, suppose that (A, b) is uncontrollable. Then, there exists a nonsingular matrix T such thatT1AT =?A11A12 0A22? ,T1b =?b1 0? .Since u = v fT
8、x = v fTT x, the A-matrix of the closed-loop system with the state vector x isT1AT T1bfTT|z=fT=T1(A bfT)T=?A11A12 0A22? ?b1 0?h fT 1fT 2i =“ A11 b1fT 1A12 b1fT 2 0A22#.The above equation shows that the uncontrollable part A22is independent of the state feedback gain fT.We show in the following that
9、the controllability of (A, b) is also sufficient for the pole placement problem.It suffices to show that for any given closed-loop poles there exists a state feedback gain vector fTsolving the pole placement problem. In the previous chapter we have shown that the transform x = Tc,2xc,2in the state s
10、pace with Tc,2= QcS?f(n)?will yield a state equation in the controllable canonical form II:Ac,2=010.0001. .0 0.001 a0a1an2an1,bc,2=0 0 . 0 1The closed-loop with the state feedback has the A-matrixA bfT=Tc,2h T1 c,2ATc,2 T1 c,2bfTTc,2iT1 c,2=Tc,2h Ac,2 bc,2fTc,2i T1 c,2where fTc,2= fTTc,2=hf0f1.fn1iI
11、t is readily seen thatAc,2 bc,2fTc,2=010.0001. .0 0.001 (a0+f0)(a1+f1)(an2+fn2)(an1+fn1)is of row companion form, hence the coefficient vector of the characteristic polynomial fcl(s) of the closed-loop A matrix is a + fc,2, that isfcl(s) = sn+ (an1+fn1)sn1+ + (a1+f1)s + (a0+f0)which should produce t
12、he desired eigenvalues i, i = 1,2,.,n. The characteristic polynomial of the closed- loop system is then(s) =nYi=1(s i) = sn+ n1sn+ + 1s + 0234CHAPTER 10. DESIGN OF CONTROL SYSTEMS IN STATE-SPACE (CHAPTER 12)whose coefficient vector = 01.n2n1Tcan be fully determined by the n desired poles i. Equating
13、 fcl(s) with (s) we get = a + fc,2fc,2= aHence, necessary and sufficient condition for pole placement is that (A, b) is completely controllable.The proof of the sufficiency also gives an algorithm for constructing a state feedback gain vector fc,2to accomplish a pole placement requirement. However,
14、fc,2is the gain vector for the controllable canonical form II (Ac,2, bc,2). The gain vector for the original form (A, b) is fT= fc,2T1 c,2. Using the formulaT1 c,2=lT lTA . lTAn1we getfT=?T aT?T1 c,2 =lT? 0I + 1A + + n1An1? lT?a 0I + a1A + + an1An1?|z =An, from CayleyHamilton Theorem=lT(A) = 0 0.0 1
15、Q1c(A)The last equation is known as Ackermanns formula for the state feedback gain matrix. Remember that lTis the last row vector of the inverse of the controllability matrixQc=?b Ab A2b.An1b?.Example 10.1. For the system x =?23 49? x +?3 1? ufind state feedback gain vector fTso that the closed-loop poles are placed to s1,2= 1 j2. step 1. Calculate the desired characteristic polynomial (s).(s) = (s + 1 + j2)(s + 1 j2) = s2+ 2s + 5step 2. Calculate the controllability matrix QcQc=?39 13?Qcis of f
