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Unstable Poles--Unit Circle Viewpoint

We saw in §8.4 that an LTI filter is stable if and only if all of its poles are strictly inside the unit circle ($ \vert z\vert=1$ ) in the complex $ z$ plane. In particular, a pole $ p$ outside the unit circle ($ \vert p\vert>1$ ) gives rise to an impulse-response component proportional to $ p^n$ which grows exponentially over time $ n$ . We also saw in §6.2 that the z transform of a growing exponential does not not converge on the unit circle in the $ z$ plane. However, this was the case for a causal exponential $ u(n)p^n$ , where $ u(n)$ is the unit-step function (which switches from 0 to 1 at time 0). If the same exponential is instead anticausal, i.e., of the form $ u(-n)p^n$ , then, as we'll see in this section, its z transform does exist on the unit circle, and the pole is in exactly the same place as in the causal case. Therefore,to unambiguously invert a z transform, we must know its region of convergence. The critical question is whether the region of convergence includes the unit circle: If it does, then each pole outside the unit circle corresponds to an anticausal, finite energy, exponential, while each pole inside corresponds to the usual causal decaying exponential.



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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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