How Causality and Stability Affect Linear Time-Invariant Systems


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.

Best Answer

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.