Given the continuous time state space model:

$\dot{x}(t)=Ax(t)+Bu(t)$, $\quad y(t)=Cx(t), \quad t\in R^{+}$ with: $\left[ \begin{array}{c|c} A & B \\ \hline C & \\ \end{array} \right]$ = $ \left[ \begin{array}{ccc|c} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \end{array} \right]$

Using: $C(sI-A)^{-1}B$ yields the following transfer function:


I'm used to seeing $s$-terms in the denominator, for instance: $\frac{1}{s+7}$. Which then provides the pole location(s) and thus the stability.

What does this zero say about stability?


Your equations give: $$y = x_2\\sx_2=x_3\\sx_3=0$$ Which implies that: $$s^2y=0$$ Thus the transfer function is indeed zero.

Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent: $$\exists n | A^n = \mathbb{0} $$ Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.

  • $\begingroup$ So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right? $\endgroup$ – user463102 Aug 3 '18 at 12:02
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    $\begingroup$ Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time. $\endgroup$ – Niki Di Giano Aug 3 '18 at 12:07
  • $\begingroup$ I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki. $\endgroup$ – user463102 Aug 3 '18 at 12:13
  • $\begingroup$ I'm really glad I could help. $\endgroup$ – Niki Di Giano Aug 3 '18 at 12:17
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    $\begingroup$ $s^2\,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable). $\endgroup$ – Kwin van der Veen Aug 4 '18 at 0:37

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