The following question is from a System Theory test without answers or solutions. It concerns the BIBO stability of the following 4 systems:
$1) \left[ \begin{array}{l|l} A&B\\ \hline C & \end{array} \right] = \left[ \begin{array}{lll|l} -2&1&0&2\\ 0&-2&0&1\\ 0&0&-5&-3\\ \hline 1 & 1 & 0 \end{array} \right] \qquad 2) \left[ \begin{array}{l|l} A&B\\ \hline C & \end{array} \right] = \left[ \begin{array}{lll|l} 0&-2&0&0\\ 2&0&0&1\\ 0&0&0&1\\ \hline 0 & -2 & 0 \end{array} \right]$
$ 3) \left[ \begin{array}{l|l} A&B\\ \hline C & \end{array} \right] = \left[ \begin{array}{lll|l} 0&1&0&0\\ 0&0&1&-1\\ 0&0&0&-2\\ \hline 1 & 0 & 0 \end{array} \right] \qquad 4) \left[ \begin{array}{l|l} A&B\\ \hline C & \end{array} \right] = \left[ \begin{array}{lll|l} -2&0&0&1\\ 0&1&1&0\\ 0&-1&1&0\\ \hline 1 & 0 & 1 \end{array} \right]$
The following list of possible answers is provided:
$ A) \ \text{Only 1}\\ B) \ \text{1 and 2}\\ C) \ \text{1 and 4}\\ D) \ \text{1, 2 and 3}\\ E) \ \text{All of them}\\ $
My approach:
Taking $\mathcal{L}^{-1}\{C(sI-A)^{-1}B\}$ yields the following time domain responses. All of which have been verified with Matlab.
$1) \quad e^{-2t}(t+3)\\2) \quad -2 \cos(2t)\\3) \quad t(t-1)\\4) \quad e^{-2t}$
This shows that all repsonses except for 3 must be bounded for a bounded input (the matlab function "impulse" verifies this). But that option is not among the possible choices.
What's going on here? Am I doing something wrong?
The definition you used for BIBO stability is slightly off. Namely it can be defined as the integral of the absolute value of the impulse response has to be bounded. Using this then it can be shown that system 2 is not BIBO.
Another way of showing that a LTI state space model is BIBO is by looking at whether all modes that are both controllable and observable have an eigenvalue with a negative real part. Seeing whether a mode is both controllable and observable can be done with the Hautus lemma. You only need to check all eigenvalues with a non-negative real part (so do check eigenvalues with a zero real part) and if you know that a mode is uncontrollable you don't have to check observability anymore. The same holds the other way around. After some practice one can spot this quite quickly, since all $A$ matrices are in their real Jordan form.