By Prof. Bruce A. Francis (eds.)

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**Sample text**

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.

30 Ch. 4 Lemma 1. For each matrix F in RH~ there exist matrices G, H, and F 1 in RH~ satisfying the equation F=G 0 H and having the properties that G and H are invertible in RH~ and F t is nonsingular. ~'l coprime, then the matrix / N I can be filled out to yield a square matrix which is invertible in RH~. Lemma 2. Let M and N be RH=-matrices with equal number of columns. Then M and N are right-coprime iff there exist matrices U and V in RH~ such that Proof. (It') Define where a question mark denotes an irrelevant block.

Prove equivalence of (i) and (iii) in Theorem 1. 3 Stabilizability Let's say that G is stabilizable if there exists a (proper real-rational) K which stabilizes it. 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 .