# A convoluted transfer function (from prof. Dullerud's robust control testbook)

The following proble is from the book:

A Course in Robust Control Theory (a convex approach), middle of p. 200

Consider the following general feedback loop:

, ie $\dot x(t) = Ax(t) + B_1w(t) + B_2u(t)$
$z(t) = C_1x(t) + D_{12}u(t)$
$y(t) = C_2x(t) + D_{21}2(t)$

Suppose:
1. $v(t) = u(t) - Fx(t)$

Question:
How to show the transfer function: $w$ to $v$ is $S(\hat H,K)$
where,
1. $S(\hat H,K)$ = $H_{11} + H_{12}K(I-H_{22}K)^{-1}H_{21}$
2. $\hat H$ has a state space representation:

$$H = \left[ \begin{array}{c|c c} A & B_1 & B_2 \\ \hline -F & 0 & I \\ C_2 & D_{21} & 0 \end{array} \right]$$

This is an interesting question. Here is how I showed it.

\begin{align} \dot{x} &= Ax + B_1 w + B_2 K (C_2 x + D_{21} w) \\ &= (A + B_2 K C_2) x + (B_1 + B_2 K D_{21}) w \\ v &= K(C_2 x + D_{21} w) - Fx \\ &= (K C_2 - F) x + K D_{21} w \end{align}

Also, writing down $\hat{H}$, helps

\begin{align} H_{11} &= -F(sI - A)^{-1}B_1 \\ H_{12} &= I - F(sI - A)^{-1}B_2 \\ H_{21} &= D_{21} + C_2(sI - A)^{-1}B_1 \\ H_{22} &= C_2(sI - A)^{-1}B_2 \end{align}

Now, we need to show that $S(\hat{H},K)$ equals to

$$T_{wv} = K D_{21} + (K C_2 - F)(sI - A - B_2 K C_2)^{-1} (B_1 + B_2 K D_{21})$$

The main idea in here is to use the Matrix Inversion Lemma to obtain

$$([sI - A] + [- B_2 K] I [C_2])^{-1} = (sI - A)^{-1} + (sI - A)^{-1} B_2 K (I - C_2 (sI - A)^{-1} B_2 K)^{-1} C_2 (sI - A)^{-1}$$

Now, we just need to manipulate $T_{wv}$ and replace $H_{ij}$ in proper places, which is easy but time consuming so I will skip this part. I couldn't find a clever way to overcome the calculations.

• All papers and books I have seen always say "it is routine to show...(w/o showing)". In fact these derivations are not quite trivial. – sleeve chen May 25 '15 at 22:24