### 1 Sampling theory

If a continuous-time signal $f\left(t\right)$ is sampled at terms $t=0,T,2T,\dots nT,\dots$ then a sequence of values

$\left\{f\left(0\right),\phantom{\rule{1em}{0ex}}f\left(T\right),\phantom{\rule{1em}{0ex}}f\left(2T\right),\dots f\left(nT\right),\dots \right\}$

is obtained. The quantity $T$ is called the sample interval or sample period .

Figure 18

In the previous Sections of this Workbook we have used the simpler notation $\left\{{f}_{n}\right\}$ to denote a sequence. If the sequence has actually arisen by sampling then ${f}_{n}$ is just a convenient notation for the sample value $f\left(nT\right)$ .

Most of our previous results for z-transforms of sequences hold with only minor changes for sampled signals.

So consider a continuous signal $f\left(t\right)$ ; its z-transform is the z-transform of the sequence of sample values i.e.

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

We shall briefly obtain z-transforms of common sampled signals utilizing results obtained earlier. You may assume that all signals are sampled at $0,T,2T,\dots nT,\dots$

Unit step function

$u\left(t\right)=\left\{\begin{array}{cc}\hfill 1\hfill & \hfill t\ge 0\hfill \\ \hfill 0\hfill & \hfill t<0\hfill \end{array}\right\$

Since the sampled values here are a sequence of 1’s,

$\begin{array}{rcll}ℤ\left\{u\left(t\right)\right\}=ℤ\left\{{u}_{n}\right\}& =& \frac{1}{1-{z}^{-1}}& \text{}\\ & =& \frac{z}{z-1}\phantom{\rule{2em}{0ex}}\left|z\right|>1\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

where $\left\{{u}_{n}\right\}=\left\{1,\phantom{\rule{1em}{0ex}}1,\phantom{\rule{1em}{0ex}}1,\phantom{\rule{1em}{0ex}}\dots \right\}$ is the unit step sequence.

$\phantom{\rule{0.3em}{0ex}}↑$

Ramp function

$r\left(t\right)=\left\{\begin{array}{ccc}t\hfill & \hfill \hfill & t\ge 0\hfill \\ 0\hfill & \hfill \hfill & t<0\hfill \end{array}\right\$

The sample values here are

$\left\{r\left(nT\right)\right\}=\left\{0,\phantom{\rule{1em}{0ex}}T,\phantom{\rule{1em}{0ex}}2T,\phantom{\rule{1em}{0ex}}\dots \right\}$

The ramp sequence   $\left\{{r}_{n}\right\}=\left\{0,\phantom{\rule{1em}{0ex}}1,\phantom{\rule{1em}{0ex}}2,\phantom{\rule{1em}{0ex}}\dots \right\}$  has z-transform $\frac{z}{{\left(z-1\right)}^{2}}$ .

Hence $\phantom{\rule{1em}{0ex}}ℤ\left\{r\left(nT\right)\right\}=\frac{Tz}{{\left(z-1\right)}^{2}}$  since $\left\{r\left(nT\right)\right\}=T\left\{{r}_{n}\right\}$ .

Obtain the z-transform of the exponential signal

$f\left(t\right)=\left\{\begin{array}{ccc}{e}^{-\alpha t}\hfill & \hfill \hfill & t\ge 0\hfill \\ 0\hfill & \hfill \hfill & t<0.\hfill \end{array}\right\$

[Hint: use the z-transform of the geometric sequence $\left\{{a}^{n}\right\}$ .]

The sample values of the exponential are

$\left\{1,\phantom{\rule{1em}{0ex}}{e}^{-\alpha T},\phantom{\rule{1em}{0ex}}{e}^{-\alpha 2T},\dots ,{e}^{-\alpha nT},\dots \right\}$

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

But $ℤ\left\{{a}^{n}\right\}=\frac{z}{z-a}$

$\therefore \phantom{\rule{2em}{0ex}}ℤ\left\{{\left({e}^{-\alpha T}\right)}^{n}\right\}=\frac{z}{z-{e}^{-\alpha T}}=\frac{1}{1-{e}^{-\alpha T}{z}^{-1}}$

#### 1.1 Sampled sinusoids

Earlier in this Workbook we obtained the z-transform of the sequence $\left\{cos\omega n\right\}$ i.e.

$ℤ\left\{cos\omega n\right\}=\frac{{z}^{2}-zcos\omega }{{z}^{2}-2zcos\omega +1}$

Hence, since sampling the continuous sinusoid

$f\left(t\right)=cos\omega t$

yields the sequence $\left\{cosn\omega T\right\}$ we have, simply replacing $\omega$ by $\omega T$ in the z-transform:

$\begin{array}{rcll}ℤ\left\{cos\omega t\right\}& =& ℤ\left\{cosn\omega T\right\}& \text{}\\ & =& \frac{{z}^{2}-zcos\omega T}{{z}^{2}-2zcos\omega T+1}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Obtain the z-transform of the sampled version of the sine wave $f\left(t\right)=sin\omega t$ .

$ℤ\left\{sin\omega n\right\}=\frac{zsin\omega }{{z}^{2}-2zcos\omega +1}$

$\begin{array}{rcll}\therefore \phantom{\rule{2em}{0ex}}ℤ\left\{sin\omega t\right\}& =& ℤ\left\{sinn\omega T\right\}& \text{}\\ & =& \frac{zsin\omega T}{{z}^{2}-2zcos\omega T+1}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

#### 1.2 Shift theorems

These are similar to those discussed earlier in this Workbook but for sampled signals the shifts are by integer multiples of the sample period $T$ . For example a simple right shift, or delay, of a sampled signal by one sample period is shown in the following figure:

Figure 19

The right shift properties of z-transforms can be written down immediately. (Look back at the shift properties in Section 21.2 subsection 5, if necessary:)

If $y\left(t\right)$ has z-transform $Y\left(z\right)$ which, as we have seen, really means that its sample values $\left\{y\left(nT\right)\right\}$ give $Y\left(z\right)$ , then for $y\left(t\right)$ shifted to the right by one sample interval the $z$ -transform becomes

$ℤ\left\{y\left(t-T\right)\right\}=y\left(-T\right)+{z}^{-1}Y\left(z\right)$

The proof is very similar to that used for sequences earlier which gave the result:

$\phantom{\rule{2em}{0ex}}ℤ\left\{{y}_{n-1}\right\}={y}_{-1}+{z}^{-1}Y\left(z\right)$

Using the result

$ℤ\left\{{y}_{n-2}\right\}={y}_{-2}+{y}_{-1}{z}^{-1}+{z}^{-2}Y\left(z\right)$

write down the result for $ℤ\left\{y\left(t-2T\right)\right\}$

$ℤ\left\{y\left(t-2T\right)\right\}=y\left(-2T\right)+y\left(-T\right){z}^{-1}+{z}^{-2}Y\left(z\right)$

These results can of course be generalised to obtain $ℤ\left\{y\left(t-mT\right)\right\}$ where $m$ is any positive integer.

In particular, for causal or one-sided signals $y\left(t\right)$ (i.e. signals which are zero for $t<0$ ):

$ℤ\left\{y\left(t-mT\right)\right\}={z}^{-m}Y\left(z\right)$

Note carefully here that the power of $z$ is still ${z}^{-m}$ not ${z}^{-mT}$ . Examples:

For the unit step function we saw that:

$ℤ\left\{u\left(t\right)\right\}=\frac{z}{z-1}=\frac{1}{1-{z}^{-1}}$

Hence from the shift properties above we have immediately, since $u\left(t\right)$ is certainly causal,

$\begin{array}{rcll}ℤ\left\{u\left(t-T\right)\right\}& =& \frac{z{z}^{-1}}{z-1}=\frac{{z}^{-1}}{1-{z}^{-1}}& \text{}\\ & & & \text{}\\ ℤ\left\{u\left(t-3T\right)\right\}& =& \frac{z{z}^{-3}}{z-1}=\frac{{z}^{-3}}{1-{z}^{-1}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

and so on.

Figure 20