laplace transform - Can the inverse system of a stable system be unstable?

Can the inverse system of a stable system be unstable?

For the class of LTI systems, the criteria for stability of a system with impulse response $h(t)$ and systems function $H(s)$ are:

• $h(t)$ be absolutely integrable.

or in s-domain:

• $H(s)$ have the $j\omega$ axis (or $s=0$) in its convergence region, and do not have any derivatives of impulse in it (which means for rational system functions like $P(s)/Q(s)$ the order of the nominator be less than or equal to the denominator).

So I thought that if we consider the following system $$H(s)=\frac{s-2}{s^2-1}$$ with $ROC:s>1$, it is stable, but its inverse $H_i(s)=1/H(s)=\frac{s^2-1}{s-2}$ with $ROC:s>2$ has the first derivative of delta in it and thus is not stable.

Is my reasoning correct? and is there any way to answer the question for all of the systems (not only LTI)?

Answer

The ROC is determined by a condition on the real part of $s$, so for the given transfer function $H(s)$ you could define the ROC as $\Re\{s\}>1$, but then the system wouldn't be stable because the ROC does not include the $j\omega$ axis. You could as well have the ROC $-1<\Re\{s\}<1$. In this case the system would be stable but not causal.

And now for your main question: yes, the inverse of a stable system can be unstable. If we consider systems described by rational transfer functions (i.e. lumped elements) then the location of the system's poles determine its stability. The poles of the inverse system are the zeros of the current system, so there's no relation between the stability of a system and the stability of its inverse, because the zeros do not at all influence the system's stability, but they do determine the stability of the inverse system.

For your example, the inverse system has a pole at $s=2$, and there are two possible ROCs: $\Re\{s\}>2$, which corresponds to a causal but unstable system, and $\Re\{s\}<2$, corresponding to a stable but anti-causal system.