The Z Transform

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

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

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

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

or, using operator notation,

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

which can be abbreviated

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

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

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

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