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This question is from a system theory exam without answers. So I was wondering if my resoning is correct.

Consider the discrete-time state-space realization

$x(t+1)=Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t \in N$

with $\frac{1}{2} \begin{bmatrix}1&1&0\\1&1&0\\0&1&1\end{bmatrix}, \quad B=\begin{bmatrix}1\\-1\\2\end{bmatrix}, \quad$ and $C=\begin{bmatrix} -1&0&1\end{bmatrix}$

Note that the eigenvalues of the matrix $A$ are given by $\lambda_1=0, \lambda_2=\frac{1}{2},$ and $\lambda_3=1$.

Which of the following statements is true?

A) $\quad$ The system is Lyapunov stable and asymptotically stable, but not BIBO stable.

B) $\quad$ The system is Lyapunov stable, but not asymptotically stable and not BIBO stable.

C) $\quad$ The system is Lyapunov stable and BIBO stable, but not asymptotically stable.

D) $\quad$ The system is not Lyapunov stable and not asymptotically stable, but is BIBO stable.

E) $\quad$ The system is not Lyapunov stable, not asymptotically and not BIBO stable.

My reasoning: two eigenvalues are inside the unit disk and one eigenvalue is on the unit disk. $\lambda_3 = 1$ has geometric multiplicity $1$ so the system is BIBO stable and Lyapunov stable, but not asymptotically stable, which is answer $C$.

Is this correct?

Thanks in advance.

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You are correct about Lyapunov and asymptotic stability, but incorrect about BIBO stability. Namely Lyapunov stability requires all eigenvalues to lie inside the unit circle or on the unit circle with a geometric multiplicity of one. Asymptotic stability requires all eigenvalues to lie inside the unit circle, which for LTI systems also means exponential stability. However BIBO stability means that a dynamical system that starts of with all its states at zero and is subjected to any possible bounded input will always also have a bounded output. This means that it requires all the eigenvalues of the controllable and observable modes to lie inside the unit circle. So on the unit circle is not BIBO, for example consider the system \begin{align} x(t+1) &= x(t) + u(t) \\ y(t) &= x(t) \end{align} with a bounded input $u(t)=1$ for $t>0$. In order to check whether the eigenmode corresponding to the eigenvalue 1 is controllable and observable one can use the Hautus test.

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