# A Course in H_infinity Control Theory by Bruce A. Francis

By Bruce A. Francis

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**Example text**

By controllability and observability Wc is surjective and Wo is injective. Hence tPotPĀ¢ has rank n, so FF does too. [] We need another definition. Let (I) be an operator from X to X, a Hilbert space. A complex number X is an eigenvalue of q) if there is a nonzero x in X satisfying 56 Ch. 5 CI~x= ~ . Then x is an eigenvector corresponding to %. In general an operator may not have any eigenvalues! For the remainder of this section let F~ RL~. The self-adjoint operator I'~FF maps Hz to itself and its rank is finite by Theorem 1, This property guarantees that it does in fact have eigenvalues.

To get a right-coprime factorization K = YX-1 we first choose /3 so that ,4F :=,4+/~F is stable. It is convenient to take/3 :=CF, so that AF=AF. 6) we get K = YX-1 , where ^ ^ ^ X(s):=[AF, B,F,I] =[AF,-1-1, CF, I] 01 = [AF, -H, F, 0]. A similar derivation leads to a left-coprime factorization K = ~ - 1 ~ , where X(s) := [All, -BH, F, I] IZ(s ) := [AH, -H, F, 0]. 7). By Lemma 1 we know that and Hence the product must be invertible in RH~. The only surprise is that the product equals the identity matrix, as is verified by algebraic manipulation.

Not every G is stabilizable; an obvious non-stabilizable G is G12=0, G2t=0, G22=0, G 11 unstable. In this example, the unstable part of G is disconnected from u and y. In terms of a state-space model G is stabilizable iff its unstable modes are controllable from u (stabilizability) and observable from y (detectability). The next result is a stabilizability test in terms of left- and right-coprime factorizations G = N M -1 = ~ I - 1 N . Theorem 1. The following conditions are equivalent: (i) G is stabilizable, (it) M, [0 I ] N axe right-coprime and M , I ~ ] are left-coprime, (iii) [0] M, N I are left-coprime and M, [0 I] are right-coprime.