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Consider the following three discrete-time state-space realizations $(A_1,B_1,C_1), (A_2,B_2,C_2) \ \text{and} \ (A_3,B_3,C_3)$ with

$A_1=\begin{bmatrix}0&1\\0&1 \end{bmatrix}, \ \ \ \quad B_1=\begin{bmatrix}0\\2 \end{bmatrix}, \quad C_1=\begin{bmatrix}0&1 \end{bmatrix}$

$A_2=\begin{bmatrix}0&1\\-1&1 \end{bmatrix}, \quad B_2=\begin{bmatrix}0\\2 \end{bmatrix}, \quad C_2=\begin{bmatrix}0&1 \end{bmatrix}$

$A_3=\begin{bmatrix}0&1\\0&1 \end{bmatrix}, \ \ \ \quad B_3=\begin{bmatrix}1\\1 \end{bmatrix}, \quad C_3=\begin{bmatrix}1&1 \end{bmatrix}$

The question is: Which discrete-time state space realizations are $0$-controllable? The answers state that all the systems are $0$-controllable.

The ranks of the controllability matrices are: $\text{rank (}\mathcal{C}_1) = 2$, $\text{rank (}\mathcal{C}_2) = 2$ and $\text{rank (}\mathcal{C}_3) = 1$.

I know that for discrete-time systems controllability and $0$-controllability are equivalent if the $A$-matrix is invertible. $A_2$ is the only invertible matrix so $A_2$ is $0$-controllable.

But I don't know how to test the other matrices. I've read several definitions but these don't give the correct answer.

Thanks in advance.

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When a system is controllable then it is indeed also 0-controllable, but as you stated the reverse does not have to hold. The term 0-controllable is related to stabilizability, namely the uncontrollable modes have to be stable. But for 0-controllable the state has to be able to reach zero in finite time, instead of eventually converge to zero for stabilizability. For continuous LTI systems every stable uncontrollable mode would die out to zero exponentially, so would take an infinite amount of time to actually reach zero. Therefore a continuous LTI system is only 0-controllable if it is controllable as well. For discrete LTI systems this is not completely the case, since when an uncomtroppable mode has an eigenvalue of zero, it will die out in finite amount of time.

In order to check whether a discrete LTI system is 0-controllable you either want a mode to be controllable or have an eigenvalue of zero, such any uncontrollable modes would die out in finite time. The easiest tool to check this in my opinion would be the Hautus lemma, namely for all $\lambda\neq0$ the following matrix should have full rank $[A-\lambda\,I \quad B]$.

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  • $\begingroup$ Did you recieve the bounty? I've never awarded bounty before so I wanted to know if it went right. $\endgroup$ – user463102 Oct 10 '18 at 15:24
  • $\begingroup$ Yes. And if you have any further question about 0-controllability feel free to ask. $\endgroup$ – Kwin van der Veen Oct 10 '18 at 21:21

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