### 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 $f\left(t\right),\phantom{\rule{1em}{0ex}}t\ge 0$ as

$F\left(s\right)={\int }_{0}^{\infty }f\left(t\right){e}^{-st}dt.$

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.]

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:

$F\left(s\right)=\frac{1}{{\left(s+\alpha \right)}^{2}}$ has its pole at $s={s}_{1}=-\alpha$ . The pole is second order.

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:

Consider $f\left(nT\right)=nT{e}^{-\alpha nT}=\left(nT\right){\left({e}^{-\alpha T}\right)}^{n}$

The ramp sequence $\left\{nT\right\}$ has z-transform $\frac{Tz}{{\left(z-1\right)}^{2}}$

$\therefore$ $f\left(nT\right)$ has z-transform

$\phantom{\rule{2em}{0ex}}F\left(z\right)=\frac{Tz{e}^{\alpha T}}{{\left(z{e}^{\alpha T}-1\right)}^{2}}=\frac{Tz{e}^{-\alpha T}}{{\left(z-{e}^{-\alpha T}\right)}^{2}}$ (see Key Point 8)

This has a (second order) pole when $z={z}_{1}={e}^{-\alpha T}={e}^{{s}_{1}T}$ .

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.

##### Key Point 19

The unit impulse function $\delta \left(t\right)$ can be defined informally as follows:

Figure 21

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

${P}_{\epsilon }\left(s\right)={\int }_{0}^{\epsilon }\frac{1}{\epsilon }{e}^{-st}=\frac{1}{\epsilon s}\left(1-{e}^{-\epsilon s}\right)\left(2\right)$

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)

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 $\delta \left(t-nT\right)$ .

Hence $𝕃\left\{\delta \left(t-nT\right)\right\}={e}^{-snT}$ (4)

which reduces to the result (3)

$𝕃\left\{\delta \left(t\right)\right\}=1$ when $n=0$

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

$ℤ\left\{{\delta }_{n}\right\}=1$

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

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:

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

##### Key Point 20

The Laplace transform

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

of a sampled function is equivalent to the z-transform $F\left(z\right)$ of the sequence $\left\{f\left(nT\right)\right\}$ of sample values with $z={e}^{sT}$ .

#### 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.

 $f\left(t\right)$ $f\left(nT\right)$ $F\left(z\right)$ Radius of convergence $t\ge 0$ $n=0,1,2,\dots$ $R$ $1$ $1$ $\frac{z}{z-1}$ $1$ $t$ $nT$ $\frac{z}{{\left(z-1\right)}^{2}}$ $1$ ${t}^{2}$ ${\left(nT\right)}^{2}$ $\frac{{T}^{2}z\left(z+1\right)}{{\left(z-1\right)}^{3}}$ $1$ ${e}^{-\alpha t}$ ${e}^{-\alpha nT}$ $\frac{z}{z-{e}^{-\alpha T}}$ $\left|{e}^{-\alpha T}\right|$ $sin\omega t$ $sinn\omega T$ $\frac{zsin\omega T}{{z}^{2}-2zcos\omega T+1}$ $1$ $cos\omega t$ $cosn\omega T$ $\frac{z\left(z-cos\omega T\right)}{{z}^{2}-2zcos\omega T+1}$ $1$ $t{e}^{-\alpha t}$ $nT{e}^{-\alpha nT}$ $\frac{Tz{e}^{-\alpha T}}{{\left(z-{e}^{-\alpha T}\right)}^{2}}$ $\left|{e}^{-\alpha T}\right|$ ${e}^{-\alpha t}sin\omega t$ ${e}^{-\alpha nT}sin\omega nT$ $\frac{{e}^{-\alpha T}{z}^{-1}sin\omega T}{1-2{e}^{-\alpha T}{z}^{-1}cos\omega T+{e}^{-2aT}{z}^{-2}}$ $\left|{e}^{-\alpha T}\right|$ ${e}^{-\alpha T}cos\omega t$ ${e}^{-\alpha nT}cos\omega nT$ $\frac{1-{e}^{-\alpha T}{z}^{-1}cos\omega T}{1-2{e}^{-\alpha T}{z}^{-1}cos\omega T+{e}^{-2aT}{z}^{-2}}$ $\left|{e}^{-\alpha T}\right|$

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

 ${f}_{n}$ $F\left(z\right)$ Name ${\delta }_{n}$ 1 unit impulse ${\delta }_{n-m}$ ${z}^{-m}$ ${u}_{n}$ $\frac{z}{z-1}$ unit step sequence ${a}^{n}$ $\frac{z}{z-a}$ geometric sequence ${e}^{\alpha n}$ $\frac{z}{z-{e}^{\alpha }}$ $sinh\alpha n$ $\frac{zsinh\alpha }{{z}^{2}-2zcosh\alpha +1}$ $cosh\alpha n$ $\frac{{z}^{2}-zcosh\alpha }{{z}^{2}-2zcosh\alpha +1}$ $sin\omega n$ $\frac{zsin\omega }{{z}^{2}-2zcos\omega +1}$ $cos\omega n$ $\frac{{z}^{2}-zcos\omega }{{z}^{2}-2zcos\omega +1}$ ${e}^{-\alpha n}sin\omega n$ $\frac{z{e}^{-\alpha }sin\omega }{{z}^{2}-2z{e}^{-\alpha }cos\omega +{e}^{-2\alpha }}$ ${e}^{-\alpha n}cos\omega n$ $\frac{{z}^{2}-z{e}^{-\alpha }cos\omega }{{z}^{2}-2z{e}^{-\alpha }cos\omega +{e}^{-2\alpha }}$ $n$ $\frac{z}{{\left(z-1\right)}^{2}}$ ramp sequence ${n}^{2}$ $\frac{z\left(z+1\right)}{{\left(z-1\right)}^{3}}$ ${n}^{3}$ $\frac{z\left({z}^{2}+4z+1\right)}{{\left(z-1\right)}^{4}}$ ${a}^{n}{f}_{n}$ $F\left(\frac{z}{a}\right)$ $n\phantom{\rule{1em}{0ex}}{f}_{n}$ $-z\frac{dF}{dz}$