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Which of the systems are Lyapunov, asymptotically or BIBO stable: $1) \quad \left[ \begin{array}{c|c} A & B\\ \hline C & \end{array} \right]=\left[ \begin{array}{ccc|c} -2&1&0&2\\ 0&-2&0&1\\ 0&0&-5&-3\\ \hline 1&1&0 \end{array} \right]$ $2) \quad \left[ \begin{array}{c|c} A & B\\ \hline C & \end{array} \right]=\left[ \begin{array}{ccc|c} 0&-2&0&0\\ 2&0&0&1\\ 0&0&0&1\\ \hline 0&-2&0 \end{array} \right]$

$3) \quad \left[ \begin{array}{c|c} A & B\\ \hline C & \end{array} \right]=\left[ \begin{array}{ccc|c} 0&1&0&0\\ 0&0&1&-1\\ 0&0&0&2\\ \hline 1&0&0 \end{array} \right]$ $\qquad 4) \quad \left[ \begin{array}{c|c} A & B\\ \hline C & \end{array} \right]=\left[ \begin{array}{ccc|c} -2&0&0&1\\ 0&1&1&0\\ 0&-1&1&0\\ \hline 1&0&1 \end{array} \right]$

My approach:

First $\textbf{system}$ $1$: All the eigenvalues of $1$ are strictly smaller than $0$, which means this systems is asymptotically stable, which implies BIBO stability and Lyapunov stability.

$\textbf{system}$ $2$: The eigenvalues are $0$ and $\pm2i$ the Jordan block associated with $\pm2i$ is size $2$x$2$ and not $1$x$1$ so this system is not Lyapunov stable. The transfer function is determined using $C(SI-A)^{-1}B$ which yields: $\frac{-2s}{s^2+4}$. From the denominator we see that the poles are $\pm2i$. This means that not all $Re(\lambda)<0$ and therefore this system is not BIBO and not asymptotically stable. Also, the Hautus lemma shows that the $\pm2i$ eigenvalue is not controllable.

$\textbf{system}$ $3$: The eigenvalue $0$ appears three times, the Jordan block is size $3$x$3$ so this system is not Lyapunov stable. $C(SI-A)^{-1}B$ yields $-\frac{1}{s^2}+\frac{2}{s^3}$. The poles are on the Im-axis and therefore this system is not asymptotically and not BIBO stable.

$\textbf{system}$ $4$: The eigenvalues are $-2$ and $1\pm i$. So this system is not Lyapunov, not asymptotically and not BIBO stable. But we you look at the transfer function we get: $\frac{1}{s+2}$ which has an asymptotically stable pole and as such, also is BIBO and Lyapunov stable.

I'm confused regarding system $4$. The eigenvalues of the $A$ matrix imply instability and yet the input output transfer function is stable. So the unstable mode does not show in the output, so is this called stable or not?

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Your reasoning for system 1 and 3 seem correct. However your reasoning for system 2 is not entirely correct, namely the two by two matrix you are considering is not a Jordan block (but a real Jordan block). I am also not sure how you obtained that $\pm2i$ is not controllable, namely it is (only eigenvalue $0$ is not observable). Your reasoning for system 4 is not entirely correct. Namely not asymptotically stable does not mean not BIBO stable, since the unstable modes can be uncontrollable or unobservable. All modes are observable, but only the mode corresponding to the eigenvalue $-2$ is controllable and therefore the system is BIBO. You also concluded this from the transfer function, however a stable transfer function does not guarantee Lyapunov or asymptotic stability, since these stability definitions consider the entire state space.

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  • $\begingroup$ You're right, I made a mistake with $\pm 2i$ I did it by hand instead of checking with Matlab. So for BIBO, the controllable eigenvalues must already be stable? So i can use Hautus to test the stable eigenvalues and see if these are controllable? Thanks for your time and patience with my stupidity. $\endgroup$ – user463102 Aug 23 '18 at 9:55
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    $\begingroup$ @user463102 For BIBO stability all eigenvalues which are both controllable and observable need to be stable (in order words all unstable eigenvalues need to be either uncontrollable or unobservable). In this context a stable eigenvalue has a negative real part, so a real part of zero is considered unstable. And to test controllability or observability you can indeed use Hautus. $\endgroup$ – Kwin van der Veen Aug 23 '18 at 11:56

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