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I'm not sure how to check bibo stability. On a test there were two state space representations:

A1 = $\begin{bmatrix} 1 & 0 \\ 0 & -2 \\ \end{bmatrix}$, B1 = $\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$ and C1 = $\begin{bmatrix} 1 & 1 \\ \end{bmatrix}$

A2 = $\begin{bmatrix} -3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -5 \\ \end{bmatrix}$, B2 = $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ and C2 = $\begin{bmatrix} 1 & 1 & 1 \\ \end{bmatrix}$

Question: For $1$ and $2$ determine the stability of the systems, i.e. BIBO stable or not, Lyapunov stable or not and asymptotically stabel or not.

Here's what I thought: Eigenvalues of $1$ are: $1$ and $-2$. Eigenvalues of $2$ are: $-3$, $0$ and $-5$.

Eigenvalues $< 0$ means asymptotic stability. So $1$ and $2$ are not asymptotically stable.

Eigenvalues $\leq 0$ means Lyapunov stability. So $1$ is not Lyapunov stable and $2$ is Lyapunov stable.

And for some reason 1 is BIBO and 2 is not BIBO.

The transfer function of $1$ is: $\frac{1}{s+2}$

The transfer function of $2$ is: $\frac{1}{s+3} + \frac{1}{s} + \frac{1}{s+5}$

The time domain impulse response for 1 = $e^{-2t}$ and for 2 = $e^{-3t}+1+e^{-5t}$ Is it possible that the $1$ in the time domain response of system 2 makes it not BIBO?

Thanks in advance.

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For bounded input bounded output (BIBO) stability zero initial conditions are assumed. So if all the unstable modes/eigenvalues of a system are not controllable then those states can not blow up to infinity by themself, since they start at zero and will remain their. This is the case for the unstable mode of system 1.

Also in order for a system to be BIBO stable the controllable (and observable) modes need to have eigenvalues with negative real parts. Since system 2 has one controllable eigenvalue on the imaginary axis (zero real part), it is not BIBO stable.

Also a note on Lyapunov stability. A linear time invariant (LTI) system is Lyapunov stable if it does not have any eigenvalues with positive real part and if it does not have Jordan blocks associated with eigenvalues with zero real part greater than size one. For example this is system is not Lyapunov stable (also not BIBO stable)

$$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 \end{bmatrix}. $$

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I just saw that if all the poles of the transfer function have a negative real part, then the system is BIBO. Which would mean that $1$ is indeed BIBO and $2$ is not. Becuase of the $\frac{1}{s}$ in the transfer function of $2$. Is this correct?

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