Given the following system function, determine all of the possible inverse Laplace Transforms:

$$
\hat{H}(s) = \frac{ (s-1) }{ (s+1)(s-2) } 
$$

We can see that the system has two poles, one at -1 and the other at 2. The system has one zero at 1, and the other at $\infty$. We know that the system can be broken up into components:

\begin{align*}
\hat{H}(s) = \frac{A}{s+1} + \frac{B}{s-2} \\
\end{align*}

We can find the coefficients using the Heaviside method:

\begin{align*}
A &= \left. (s+1)\hat{H}(s) \right|_{s = -1} = \frac{2}{3} \\
B &= \left. (s-2)\hat{H}(s) \right|_{s = 2} = \frac{1}{3} \\
\hat{H}(s) &= \frac{2/3}{s+1} + \frac{1/3}{s-2} \\
\end{align*}


We now have three cases with non-trivial regions of convergence. The first is where we have two right-sided signals. In this case $RoC = \{ s \st \Real{s} \gt 2 \}$:

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This system is causal but not BIBO stable. Here the possible inverse transform is 

\begin{align*}
h(t) &= \frac{2}{3} e^{-t}u(t) + \frac{1}{3} e^{2t}u(t) \\
\end{align*}

Here we have $ \frac{1}{3} e^{2t}u(t)$ is the unstable term. The next possibility is where we have two left-sided signals. In this case $RoC =  \{ s \st \Real{s} \lt -1 \}$:

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This system is anti-causal and not BIBO stable. Here the inverse transform is

\begin{align*}
h(t) &= -\frac{2}{3} e^{-t}u(-t) - \frac{1}{3} e^{2t}u(-t) \\
\end{align*}

In this case we have $- \frac{1}{3} e^{2t}u(-t)$ is the stable term. The last possibility is where we have right and left sided signals whose regions of convergence have a non-trivial overlap. In this case $RoC = \{ s \st -1 \lt \Real{s} \lt 2 \}$:

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This system is two-sided and is BIBO stable. Here the inverse transform is 

\begin{align*}
h(t) &= \frac{2}{3} e^{-t}u(t) - \frac{1}{3} e^{2t}u(-t) \\
\end{align*}

Now both terms are stable, which can be verified by the fact that the $RoC$ includes the $i\omega$-axis implying BIBO stability.


Example