2 z-transforms and Laplace transforms

In this Workbook we have developed the theory and some applications of the z-transform from first principles. We mentioned much earlier that the z-transform plays essentially the same role for discrete systems that the Laplace transform does for continuous systems. We now explore the precise link between these two transforms. A brief knowledge of Laplace transform will be assumed.

At first sight it is not obvious that there is a connection. The z-transform is a summation defined, for a sampled signal $f_n\equiv f\left(nT\right)$ , as

$F\left(z\right)={\sum }_{n=0}^{\infty }f\left(nT\right)z^{-n}$

while the Laplace transform written symbolically as $𝕃\left\{f\left(t\right)\right\}$ is an integral , defined for a continuous time function $f\left(t\right),t\ge 0$ as

$F\left(s\right)={\int \; <!--nolimits\; -->}_0^{\infty }f\left(t\right)e^{-st}dt.$

Thus, for example, if

which has a (simple) pole at $s=-\alpha =s_1$ say.

As we have seen, sampling $f\left(t\right)$ gives the sequence $\left\{f\left(nT\right)\right\}=\left\{e^{-\alpha nT}\right\}$ with z-transform

$F\left(z\right)=\frac{1}{1-e^{-\alpha T}{z}^{-1}}=\frac{z}{z-e^{-\alpha T}}.$

The z-transform has a pole when $z=z_1$ where

$z_1=e^{-\alpha T}=e^{s_{1}T}$

[Note the abuse of notations in writing both $F\left(s\right)$ and $F\left(z\right)$ here since in fact these are different functions.]

Task!

The continuous time function $f\left(t\right)=t{e}^{-\alpha t}$ has Laplace transform

$F\left(s\right)=\frac{1}{{\left(s+\alpha \right)}^2}$

Firstly write down the pole of this function and its order:

Answer

Now obtain the z-transform $F\left(z\right)$ of the sampled version of $f\left(t\right)$ , locate the pole(s) of $F\left(z\right)$ and state the order:

Answer

We have seen in both the above examples a close link between the pole $s_1$ of the Laplace transform of $f\left(t\right)$ and the pole $z_1$ of the z-transform of the sampled version of $f\left(t\right)$ i.e.

$z_1=e^{s_{1}T}\left(1\right)$

where $T$ is the sample interval.

Multiple poles lead to similar results i.e. if $F\left(s\right)$ has poles $s_1,s_2,\dots$ then $F\left(z\right)$ has poles $z_1,z_2,\dots$ where $z_i=e^{s_{i}T}$ .

The relation (1) between the poles is, in fact, an example of a more general relation between the values of $s$ and $z$ as we shall now investigate.

The rectangular pulse $P_{ϵ}\left(t\right)$ of width $\epsilon$ and height $\frac{1}{\epsilon }$ shown in Figure 21 encloses unit area and has Laplace transform

$P_{\epsilon }\left(s\right)={\int \; <!--nolimits\; -->}_0^{\epsilon }\frac{1}{\epsilon }{e}^{-st}=\frac{1}{\epsilon s}\left(1-e^{-\epsilon s}\right)\left(2\right)$

As $\epsilon$ becomes smaller $P_{\epsilon }\left(t\right)$ becomes taller and narrower but still encloses unit area. The unit impulse function $\delta \left(t\right)$ (sometimes called the Dirac delta function) can be defined as

$\delta \left(t\right)=\underset{\epsilon \to 0}{lim}{P}_{\epsilon }\left(t\right)$

The Laplace transform, say $\Delta \left(s\right)$ , of $\delta \left(t\right)$ can be obtained correspondingly by letting $ϵ\to 0$ in (2), i.e.

i.e. $𝕃\delta \left(t\right)=1$ (3)

Task!

A shifted unit impulse $\delta \left(t-nT\right)$ is defined as $\underset{\epsilon \to 0}{lim}{P}_{\epsilon }\left(t-nT\right)$ as illustrated below.

Obtain the Laplace transform of this rectangular pulse and, by letting $\epsilon \to 0$ , obtain the Laplace transform of $\delta \left(t-nT\right)$ .

Answer

These results (3) and (4) can be compared with the results

$\mathbb{Z}\left\{{\delta }_n\right\}=1$

$\mathbb{Z}\left\{{\delta }_{n-m}\right\}=z^{-m}$

for discrete impulses of height 1.

Now consider a continuous function $f\left(t\right)$ . Suppose, as usual, that this function is sampled at $t=nT$ for   $n=0,1,2,\dots$

Figure 22

This sampled equivalent of $f\left(t\right)$ , say $f_{\ast }\left(t\right)$ can be defined as a sequence of equidistant impulses, the ‘strength’ of each impulse being the sample value $f\left(nT\right)$ i.e.

$f_{\ast }\left(t\right)={\sum }_{n=0}^{\infty }f\left(nT\right)\delta \left(t-nT\right)$

This function is a continuous-time signal i.e. is defined for all $t$ . Using (4) it has a Laplace transform

$F_{\ast }\left(s\right)={\sum }_{n=0}^{\infty }f\left(nT\right)e^{-snT}\left(5\right)$

If, in this sum (5) we replace $e^{sT}$ by $z$ we obtain the z-transform of the sequence $\left\{f\left(nT\right)\right\}$ of samples:

${\sum }_{n=0}^{\infty }f\left(nT\right)z^{-n}$

2.1 Table 2: z-transforms of some sampled signals

This table can be compared with the table of the z-transforms of sequences on the following page.

Note: $R$ is such that the closed forms of $F\left(z\right)$ (those listed in the above table) are valid for $\left|z\right|>R$ .

2.2 Table of z-transforms