# 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)?

• of course the inverse of a stable system can be unstable. it matters not where the zeros are, but poles must be in the stable region ($\mathrm{Re}\{s\}<0$ or $|z|<1$). when a system is inverted, poles become zeros and zeros become poles. any zeros that were not in the stable region for poles will become unstable poles when the system is inverted. classic example is an all-pass filter or a maximum-phase filter. invert that and it blows up. – robert bristow-johnson Jan 21 '15 at 1:30

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.
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.