# A cource in H Control Theory by Bruce A. Francis

By Bruce A. Francis

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Example text

2b) Then we have from (1) and (2) that [I -RX] x] [[0 MI]N 0 Y I (3) v The two matrices on the left in (3) have inverses in RH()(), hence so does the matrix on the right in (3). The next step is to show that [ [0 MI]N [I [~]u1 V = all the matrices noting that ~ ~ G 21 G 22 in (4) at S We have [~]u [MO] V 0 [0 I] G [ Evaluate V is nonsingular. I ~][~ ] ~ (4) . V =00; then take determinants of both sides G 22 is strictly proper and the matrix on the left-hand side of (4) is invertible in RH()().

QM)] [M Y -MQ] __ [ Equating -N M N X -NQ -- 1 the (1,2)-blocks on each side in (4) gives (X -QN)( Y -MQ ) = Or -QM )(X -NQ ) , which is equivalent to (3). 4 39 Next, we show that if K is given by (2), it stabilizes U := Y-MQ, V:= - ~ Define X --NQ .... U := Y -QM, G. 2). Also from (5) [~~r E RHoo So from Lemma 1 K stabilizes G. Finally, suppose [( stabilizes Let K in RHoo' G. vVe must show K satisfies (2) for some Q l = UV- be a right-coprime factorization. ] [M u] [I ~¥U J'V] = -N M The two matrices N V 0 D .

Matrices X and Y such that YN = I. Hence ~= XM E + YN ~. (2) 28 Ch. = Substitution [ y -v z 2 ] . into (2) gives Hence the three transfer A similar argument matrices from w , v l' V 2 to f. belong to RHoo. works for the remaining three transfer matrices to 'rJ. [] Proof of Theorem 1. We shall prove the equivalence of (i) and see that the matrix displayed in a rational matrix. (ii) is indeed nonsingular, (ii). e. its inverse exists as We have (3) Now is nonsingular because both M and V are. Also, since G 22 is strictly proper, we have that 1o 0 0] [ 1 K G 21 is nonsingular G 22 1 when evaluated at Thus both matrices on the right-hand The equations corresponding S =00: its determinant equals side of (3) are nonsingular.