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The Z Transform

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

$\displaystyle X(z) \isdefs \sum_{n=-\infty}^\infty x(n) z^{-n} \qquad\hbox{(bilateral {\it z} transform)} \protect$ (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,7.1 the lower summation limit given above may be written as 0 rather than $ -\infty$ to yield the unilateral z transform:

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

The unilateral z transform is most commonly used. For inverting z transforms, see §6.8.

Recall (§4.1) that the mathematical representation of a discrete-time signal $ x(n)$ maps each integer $ n\in\mathbb{Z}$ to a complex number ( $ x(n)\in\mathbb{C}$ ) or real number ( $ x(n)\in\mathbb{R}$ ). The z transform of $ x$ , on the other hand, $ X(z)$ , maps every complex number $ z\in\mathbb{C}$ to a new complex number $ X(z)\in\mathbb{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 may say $ 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 as

$\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\mathbb{Z}$ and $ z\in\mathbb{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 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, (September 2007 Edition)
Copyright © 2024-05-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University