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Existence of the Z Transform

The z transform of a signal $ x$ will always exist provided (1) the signal starts at a finite time and (2) it is asymptotically exponentially bounded, i.e., there exists a finite integer $ n_f$, and finite real numbers $ A\geq 0$ and $ \sigma$, such that $ \left\vert x(n)\right\vert<A\exp(\sigma n)$ for all $ n\geq n_f$. The bounding exponential may be growing with $ n$ ($ \sigma>0$). These are not the most general conditions for existence of the z transform, but they suffice for most practical purposes.

One would naturally expect the z transform $ X(z)$ to be defined only in the region $ \left\vert z\right\vert>\exp{\sigma}$ of the complex plane, where $ A\exp(\sigma n)$ is the asymptotically bounding exponential envelope for $ x(n)$ discussed in the previous paragraph. This expectation is reasonable because the infinite series

$\displaystyle \sum_{n=0}^\infty e^{\sigma n} z^{-n}
= \sum_{n=0}^\infty \left(\frac{e^{\sigma}}{z}\right)^n

requires $ \left\vert z\right\vert>\exp{\sigma}$ to ensure convergence. Since $ \sigma<0$ for a decaying exponential, we see that the region of convergence of the $ z$ transform always includes the unit circle of the $ z$ plane.

More generally, it turns out that, in all practical cases, the domain of $ X(z)$ can be extended to include the entire complex plane except for isolated ``singular'' points called poles at which $ H(z)$ approaches infinity. The mathematical technique for doing this is known as analytic continuation, and it is discussed in §B.1 as applied to Laplace transforms (the continuous-time counterpart of z transforms).

The z transform is discussed more fully elsewhere [52,60], and we will derive below only what we will need.

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (August 2006 Edition).
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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