How can I get the inverse of the transfer function $\Phi$? a question from Page 56 of Lemma 4.5 in the book "ESSENTIALS OF ROBUST CONTROL"

Question

Let $\Phi(s) = = \gamma^2 I + B^*\left( (sI-A)(C^*C)^{-1}(sI+A^*) \right)^{-1}B + B^*(sI+A^*)^{-1}C^*D - D^*C(sI-A)^{-1}B - D^*D.$ How to compute the inverse of $\Phi$, in Page 56 of Zhou's book "ESSENTIALS OF ROBUST CONTROL", the answer is directly given as follows $\begin{bmatrix} R^{-1}D^*C & R^{-1}B^* \end{bmatrix} \left[\begin{matrix} sI-A-B R^{-1} D^* C & -B R^{-1} B^* \\ C^*\left(I+D R^{-1} D^*\right) C & sI+\left(A+B R^{-1} D^* C\right)^* \end{matrix}\right]^{-1} \begin{bmatrix} BR^{-1} \\ -C^*DR^{-1} \end{bmatrix} + R^{-1}.$ with $R = \gamma^2 I - D^*D$. How to obtain this answer, thanks.

Answer 1

So, we have that

$$ \Phi(s)=\gamma^2I-H^{\sim}(s)H(s) $$ where $H(s)=C(sI-A)^{-1}B+D$ and $H^\sim(s)=-B^*(sI+A^*)^{-1}C^*+D^*$. The state-space representation for $\Phi(s)$ is given by $$ \left[\begin{array}{cc|c} A & 0 & B\\ -C^*C & -A^* & -C^*D\\ \hline -D^*C & -B^* & R \end{array}\right]. $$

Now, we use that the inverse of a system $(A,B,C,D)$ is given by $(A-BD^{-1}C,BD^{-1},-D^{-1}C, D^{-1})$

This yields the following state-space representation for $\Phi(s)^{-1}$: $$ \left[\begin{array}{cc|c} A+BR^{-1}D^*C & BR^{-1}B^* & BR^{-1}\\ -C^*(I+DR^{-1}D^*)C & -(A+BR^{-1}D^*C)^* & -C^*DR^{-1}\\ \hline R^{-1}D^*C & R^{-1}B^* & R^{-1} \end{array}\right], $$ from which the result follows.