Suppose that there's a linear time-invariant system with the following transfer function: $$H(s) = \frac{1}{3(s-2)}-\frac{1}{3(s+1)}$$
If the system is causal and stable, I can determine \$h(t)\$ by simply evaluating the Laplace transform of \$H(s)\$.
Now, if the system is neither stable or causal, how would I determine \$h(t)\$? I can't figure how the response in time domain is affected by causality and stability, in general.
Technically, the Laplace transform already implies a causal system by its definition:
$$F(s)=\int_0^\infty f(t)e^{-st}dt$$
As you can see, the integral starts from 0, so this implies that \$f(t < 0) = 0\$. In other words, a non-causal system would react before it received the Dirac pulse or \$f(t < 0) \neq 0\$. A regular Laplace transform would not work in this case.
If the system is causal, the method of finding \$h(t)\$ does not change whether or not the system is stable or unstable. The example you gave has a pole in both left- and right-half plane and is therefore unstable, but the inverse Laplace transform can still be calculated (IIRC) to be
$$\frac{e^{2t}}{3} - \frac{e^{-t}}{3}$$
As you can see, the first term explodes as \$t\$ increases, hence the system is unstable, but I didn't have to use any weird techniques or something.
If you are given the information that the transfer function is a two-sided Laplace transform, then you cannot assume anymore that the system is causal while you might still reconstruct the impulse response from it, but I have never encountered such a situation.
I am writing this answer because in accepted answer it was assumed that Laplace transform is one sided which is not always the case ,so for more general case we can approach like this-
Assuming Bilateral Laplace transform and for Rational function (as given in question )-
Three region of convergence is possible (for rational function)
1.$$\left(-\infty , -1 \right) $$
In this case system is neither causal nor stable
2.$$\left(-1 , 2 \right)$$
In this case system is stable because it contains imaginary axis but not causal .
3.$$\left(2 , \infty\right)$$
In this case system is causal because ROC is right of right most pole but unstable
From above all three conditions , you can conclude how changing stability or causality conditions implies different ROC and corresponding to each ROC we get different inverse Laplace transform (h(t)) .
For calculation of h(t) for these different ROC ,there are many straight forward methods and you can found them in any mathematics book
