If the region of convergence (ROC) for system function $H(z)$ is $R_h$, what is the ROC of the inverse function $G(z)=\frac{1}{H(z)}$?



I'm going to go out on a limb and answer this based on my gut instinct.

Let's suppose $H(z)$ is casual and has a single zero (a) and a single pole (b):

$$ H(z) = \frac{(1-az^{-1})}{(1-bz^{-1})} $$

if the system is casual, then we know that the sequence is right-sided with ROC that is exterior of the circle. Thus, the ROC of H(z) is:

$$b < |z|$$

Now for the part I was uncertain about: the ROC of the inverse system G(z).

$$ G(z) = \frac{1}{H(z)} = \frac{(1-bz^{-1})}{(1-az^{-1})} $$

When we take the inverse of H(z), it is necessary for a portion of the ROC of G(z) to overlap with a portion of the ROC of H(z) in order for an inverse function G(z) to exist. The overlapping doesn't necessary need to include the entire ROC just a portion of it. This is the constraint that determines if G(z) has ROC that is casual / exterior-of-circle, or anti-casual / interior-of-circle. Its a matter of noting the poles of G(z) and checking the following cases to determine direction of the inequality for ROC:

(1) does the exterior of the largest pole of G(z) overlap with a portion of ROC of H(z)? if so then this represents a valid causal ROC (that is an exterior circle). ROC: a < |z|

(2) does the interior of the smallest pole of G(z) overlap with a portion of the ROC of H(z)? if so then this represents a valid anti-casual ROC (that is an interior circle). ROC: |z| < a

it is entirely possible that both a causal inverse and an anti-casual inverse both exist and are valid.

Another consideration for choosing a valid inverse function of H(z) is to make sure that the inverse function is stable, ie. the ROC includes the unit circle, which enables the system to be realizable in real hardware, and is more than a mere mathematical abstraction.


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