1 The z-transform
If you have studied the Laplace transform either in a Mathematics course for Engineers and Scientists or have applied it in, for example, an analog control course you may recall that
- the Laplace transform definition involves an integral
- applying the Laplace transform to certain ordinary differential equations turns them into simpler (algebraic) equations
- use of the Laplace transform gives rise to the basic concept of the transfer function of a continuous (or analog) system.
The z-transform plays a similar role for discrete systems, i.e. ones where sequences are involved, to that played by the Laplace transform for systems where the basic variable $t$ is continuous. Specifically:
- the z-transform definition involves a summation
- the z-transform converts certain difference equations to algebraic equations
- use of the z-transform gives rise to the concept of the transfer function of discrete (or digital) systems.
Key Point 1
Definition:
For a sequence $\left\{{y}_{n}\right\}$ the z-transform denoted by $Y\left(z\right)$ is given by the infinite series
$Y\left(z\right)={y}_{0}+{y}_{1}{z}^{-1}+{y}_{2}{z}^{-2}+\dots ={\sum}_{n=0}^{\infty}{y}_{n}{z}^{-n}$ (1)
Notes:
- The z-transform only involves the terms ${y}_{n}$ , $n=0,1,2,\dots $ of the sequence. Terms ${y}_{-1},{y}_{-2},\dots $ whether zero or non-zero, are not involved.
- The infinite series in (1) must converge for $Y\left(z\right)$ to be defined as a precise function of $z$ . We shall discuss this point further with specific examples shortly.
- The precise significance of the quantity (strictly the ‘variable’) $z$ need not concern us except to note that it is complex and, unlike $n$ , is continuous.
Key Point 2
We use the notation $\phantom{\rule{1em}{0ex}}\mathbb{Z}\left\{{y}_{n}\right\}=Y\left(z\right)$ to mean that the z-transform of the sequence $\left\{{y}_{n}\right\}$ is $Y\left(z\right)$ .
Less strictly one might write $\phantom{\rule{1em}{0ex}}\mathbb{Z}{y}_{n}=Y\left(z\right)$ . Some texts use the notation $\phantom{\rule{1em}{0ex}}{y}_{n}\leftrightarrow Y\left(z\right)$ to denote that (the sequence) ${y}_{n}$ and (the function) $Y\left(z\right)$ form a z-transform pair.
We shall also call $\left\{{y}_{n}\right\}$ the inverse z-transform of $Y\left(z\right)$ and write symbolically
$\phantom{\rule{2em}{0ex}}\left\{{y}_{n}\right\}={\mathbb{Z}}^{-1}Y\left(z\right)$ .