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Writing W 1 (I − GK)−1 as W 1 I + GK(I − GK)−1 , gives W 1 (I − GK)−1 W 2 K(I − GK)−1 = W1 0 + W 1G W2 K(I − GK)−1 .

X z1 . A more pedestrian approach: s G−1 1 = Hence, by the series formula,  s G2 G−1 1 = A − BD1−1 C1 −D1−1 C1 A − BD1−1 C1  −BD1−1 C1 −D2 D1−1 C1 BD1−1 D1−1 0 A C2 .  BD1−1 BD1−1  . D2 D1−1 . 29 SIGNALS AND SYSTEMS Now apply the state transformation T = G2 G−1 1  I −I A − BD1−1 C1 s 0 =  C2 − D2 D1−1 C1 s = A − BD1−1 C1 C2 − D2 D1−1 C1 0 I to obtain 0 A C2  BD1−1  0 D2 D1−1 BD1−1 D2 D1−1 . The final step follows since the states associated with A are uncontrollable. 7. 1. Since any state-space system has finite L2 [0, T ] induced norm, we may set ǫ2 = G [0,T ] < ∞.

It follows that Z is strictly positive real. 43 SIGNALS AND SYSTEMS 3. The fact that W ∈ RH∞ and W −1 ∈ RH∞ follows trivially from the asymptotic stability of the matrices A and A − BR−1 (C − B ′ P ) = A − BW −1 L. Verify that the Riccati equation can be written as P A + A′ P + L′ L = 0. ) Now verify W ∼W = W ′ W + B ′ (−sI − A′ )−1 L′ W + W ′ L(sI − A)−1 B +B ′ (−sI − A′ )−1 L′ L(sI − A)−1 B = D + D′ + B ′ (−sI − A′ )−1 (C ′ − P B) + (C − B ′ P )(sI − A)−1 B +B ′ (−sI − A′ )−1 L′ L(sI − A)−1 B = D + D′ + B ′ (−sI − A′ )−1 C ′ + C(sI − A)−1 B +B ′ (−sI − A′ )−1 L′ L − P (sI − A) − (−sI − A′ )P (sI − A)−1 B = D + D′ + B ′ (−sI − A′ )−1 C ′ + C(sI − A)−1 B = Z + Z ∼.

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