Solutions of Exercise 6.2#
Given the state-space model:
(28)#\[\begin{split}
\begin{dcases}
\dot{x} = \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} x + \begin{bmatrix} 1 \\ 0 \end{bmatrix} u \\
y = \begin{bmatrix} 1 & 1\end{bmatrix} x
\end{dcases}
\end{split}\]
Compute the poles of the system
Compute its transfer function \(G(s)\)
Compute the poles and zeros of \(G(s)\)
Solution#
Question 1#
The poles of (28) are given by the roots of the characteristic polynomial of \(A\):
\[\begin{split}
\varphi(\lambda) = \det(\lambda I - A) = \det \begin{bmatrix} \lambda - 1 & 1 \\ -2 & \lambda - 1 \end{bmatrix} = (\lambda-1)^2 + 2 = \lambda^2 - 2 \lambda + 3
\end{split}\]
The roots of \(\varphi(\lambda)\) are \(\lambda = 1 \pm \sqrt{2}\). Note that one of them is unstable (\(\lambda = 1 + \sqrt{2}\)), which means (28) is unstable.
Question 2#
The transfer function can be computes as
\[\begin{split}
\begin{aligned}
G(s) &= C (sI - A)^{-1} B + D = \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} s - 1 & 1 \\ -2 & s - 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\
&= \frac{1}{s^2 - 2s + 3} \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} s - 1 & -1 \\ 2 & s - 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\
&= \frac{1}{s^2 - 2s + 3} \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} s - 1 \\ 2 \end{bmatrix} \\
&= \frac{s + 1}{s^2 - 2s + 3}
\end{aligned}
\end{split}\]
Question 3#
The poles of \(G(s)\) are still \(s = 1 \pm \sqrt{2}\). The system is therefore not input-output stable.
There is one zero in \(s = -1\).