Help finding the transfer matrix for this system

Question

Find the transfer function of the following system: \begin{eqnarray} \dot{x}_1&=&ax_1+bx_2 + u\\ \dot{x}_2 &=&-bx_1 +ax_2\\ \dot{x}_3&=&cx_3\\ y &=& x_1+x_3 \end{eqnarray} Now I am given the formula for the transfer matrix as $T(s)=C(sI-A)^{-1}B+D$. Here the 4 matrices are $$A = \left(\begin{matrix} a & b&0\\-b&a&0\\ 0&0 & c \end{matrix} \right)$$ $$B = \left(\begin{matrix} 1\\0\\0 \end{matrix} \right)$$ $$C = \left(\begin{matrix}1&0&1 \end{matrix} \right)$$ $$D = \left(\begin{matrix} 0 \end{matrix} \right)$$. Now the transfer matrix I am looking for is thus given as: $$T(s) = \left(\begin{matrix}1&0&1 \end{matrix} \right) \left(\begin{matrix} s- a & -b&0\\b&s-a&0\\ 0&0 & s-c \end{matrix} \right)^{-1} \left(\begin{matrix} 1\\0\\1 \end{matrix} \right)$$ Now I am stuck on finding $\left(\begin{matrix} s- a & -b&0\\b&s-a&0\\ 0&0 & s-c \end{matrix} \right)^{-1}$. Can anyone help with this? Or did I make a mistake earlier on? Thanks for any help in advance!

Answer 1

Here is another way:

First note that $x_3$ is uncontrollable, and so will not contribute to the transfer function. (Another way to see this is to note that the transfer function gives the zero-state response to an input $u$, and we see that $x_3(t) = 0$ whenever the initial state is zero.) Hence $\hat{x_3} = 0$.

You have $s \hat{x_1}(s) = a \hat{x_1}(s) + b \hat{x_2}(s) + \hat{u}(s)$, $s \hat{x_2}(s) = a \hat{x_2}(s) - b \hat{x_1}(s)$. Solving for $\hat{x_1}$ (that is, eliminating $\hat{x_2}$) gives $\hat{x_1}(s) = \frac{s-a}{(s-a)^2+b^2} \hat{u}(s)$. Since $\hat{y} = \hat{x_1}+ \hat{x_3}$, we have $$\hat{h}(s) = \frac{s-a}{(s-a)^2+b^2}$$