# Check bibo, lyapunov and asymptotic stability for 2 given systems.

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?

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