The Z Transform Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search


The Z Transform

The bilateral z transform of the discrete-time signal $ x(n)$ is defined to be

$\displaystyle X(z) \isdef \sum_{n=-\infty}^\infty x(n) z^{-n} \qquad\hbox{(bilateral {\it z} transform)} \protect$ (7.1)

where $ z$ is a complex variable. Since signals are typically defined to begin (become nonzero) at time $ n = 0$, and since filters are often assumed to be causal, the lower summation limit given above may be written as 0 rather than $ -\infty$ to yield the unilateral z transform:

$\displaystyle X(z) \isdef \sum_{n=0}^\infty x(n) z^{-n} \qquad\hbox{(unilateral {\it z} transform)}$ (7.2)

For inverting z transforms, see §6.8.

Note that the mathematical representation of a discrete-time signal $ x(n)$ maps each integer $ n\in{\bf Z}$ to a complex number ( $ x(n)\in{\bf C}$) or real number ( $ x(n)\in{\bf R}$). The z transform of $ x$, on the other hand, $ X(z)$, maps every complex number $ z\in{\bf C}$ to a new complex number $ X(z)\in{\bf C}$. On a higher level, the z transform, viewed as a linear operator, maps an entire signal $ x$ to its z transform $ X$. We think of this as a ``function to function'' mapping. We express the fact that $ X$ is the z transform of $ x$ by writing

$\displaystyle \zbox {X \leftrightarrow x}
$

or, using operator notation,

$\displaystyle X(z) = {\cal Z}_z\{x(\cdot)\}
$

which can be abbreviated

$\displaystyle X = {\cal Z}\{x\}.
$

One also sees the convenient but possibly misleading notation $ X(z)
\leftrightarrow x(n)$, in which $ n$ and $ z$ must be understood as standing for the entire domains $ n\in{\bf Z}$ and $ z\in{\bf C}$, as opposed to denoting particular fixed values.

The z transform of a signal $ x$ can be regarded as a polynomial in $ z^{-1}$, with coefficients given by the signal samples. For example, the finite-duration signal

$\displaystyle x(n) = \left\{\begin{array}{ll}
n+1, & 0\leq n \leq 2 \\ [5pt]
0, & \mbox{otherwise} \\
\end{array}\right.
$

has the z transform $ X(z) = 1 + 2z^{-1}+ 3z^{-2} = 1 + 2z^{-1}+ 3(z^{-1})^2$.


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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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