z-transforms and Laplace transforms

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 $f_{n} \equiv f (n T)$ , as

$F (z) = \sum_{n = 0}^{\infty} f (n T) z^{- n}$

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

$F (s) = \int_{0}^{\infty} f (t) e^{- s t} d t .$

Thus, for example, if

\begin{array}{rcl} f (t) & = & e^{- α t} (continuous time exponential) \\ 𝕃 {f (t)} = F (s) & = & \frac{1}{s + α} \end{array}

which has a (simple) pole at $s = - α = s_{1}$ say.

As we have seen, sampling $f (t)$ gives the sequence ${f (n T)} = {e^{- α n T}}$ with z-transform

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

The z-transform has a pole when $z = z_{1}$ where

$z_{1} = e^{- α T} = e^{s_{1} T}$

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

Task!

The continuous time function $f (t) = t e^{- α t}$ has Laplace transform

$F (s) = \frac{1}{{(s + α)}^{2}}$

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

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

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

Consider $f (n T) = n T e^{- α n T} = (n T) {(e^{- α T})}^{n}$

The ramp sequence ${n T}$ has z-transform $\frac{T z}{{(z - 1)}^{2}}$

$∴$ $f (n T)$ has z-transform

$F (z) = \frac{T z e^{α T}}{{(z e^{α T} - 1)}^{2}} = \frac{T z e^{- α T}}{{(z - e^{- α T})}^{2}}$ (see Key Point 8)

This has a (second order) pole when $z = z_{1} = e^{- α T} = e^{s_{1} T}$ .

We have seen in both the above examples a close link between the pole $s_{1}$ of the Laplace transform of $f (t)$ and the pole $z_{1}$ of the z-transform of the sampled version of $f (t)$ i.e.

$z_{1} = e^{s_{1} T} (1)$

where $T$ is the sample interval.

Multiple poles lead to similar results i.e. if $F (s)$ has poles $s_{1}, s_{2}, \dots$ then $F (z)$ 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 $δ (t)$ can be defined informally as follows:

Figure 21

$The graph shows a function, $P_{\epsilon}(t)$, that is equal to one over epsilon for the interval from zero to epsilon, and zero elsewhere.$

Long description

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

$P_{ε} (s) = \int_{0}^{ε} \frac{1}{ε} e^{- s t} = \frac{1}{ε s} (1 - e^{- ε s}) (2)$

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

$δ (t) = lim_{ε \to 0} P_{ε} (t)$

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

\begin{array}{rcl} Δ (s) & = & lim_{ε \to 0} \frac{1}{ε s} (1 - e^{- ε s}) \\ = & lim_{ε \to 0} \frac{1 - (1 - ε s + \frac{{(ε s)}^{2}}{2!} - \dots)}{ε s} (Using the Maclaurin seies expansion of e^{- ε s}) \\ = & lim_{ε \to 0} \frac{ε s - \frac{{(ε s)}^{2}}{2!} + \frac{{(ε s)}^{3}}{3!} + \dots}{ε s} \\ = & 1 \end{array}

i.e. $𝕃 δ (t) = 1$ (3)

Task!

A shifted unit impulse $δ (t - n T)$ is defined as $lim_{ε \to 0} P_{ε} (t - n T)$ as illustrated below.

$A graph showing a rectangular pulse function that starts at time $nT$ and ends at time $nT + \frac{1}{\text{constant}}$, with a height of $\frac{1}{\text{constant}}$ on the vertical axis.$

Long description

Obtain the Laplace transform of this rectangular pulse and, by letting $ε \to 0$ , obtain the Laplace transform of $δ (t - n T)$ .

\begin{array}{rcl} 𝕃 {P_{ε} (t - n T)} = \int_{n T}^{n T + ε} \frac{1}{ε} e^{- s t} d t & = & \frac{1}{ε s} {[- e^{- s t}]}_{n T}^{n T + ε} \\ = & \frac{1}{ε s} (e^{- s n T} - e^{- s (n T + ε)}) \\ = & \frac{1}{ε s} e^{- s n T} (1 - e^{- s ε}) \to e^{- s n T} as ε \to 0 \end{array}

Hence $𝕃 {δ (t - n T)} = e^{- s n T}$ (4)

which reduces to the result (3)

$𝕃 {δ (t)} = 1$ when $n = 0$

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

$ℤ {δ_{n}} = 1$

$ℤ {δ_{n - m}} = z^{- m}$

for discrete impulses of height 1.

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

Figure 22

A graph showing a periodic function that transitions into an exponentially decaying curve over time.

Long description

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

$f_{*} (t) = \sum_{n = 0}^{\infty} f (n T) δ (t - n T)$

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

$F_{*} (s) = \sum_{n = 0}^{\infty} f (n T) e^{- s n T} (5)$

If, in this sum (5) we replace $e^{s T}$ by $z$ we obtain the z-transform of the sequence ${f (n T)}$ of samples:

$\sum_{n = 0}^{\infty} f (n T) z^{- n}$

Key Point 20

The Laplace transform

$F (s) = \sum_{n = 0}^{\infty} f (n T) e^{- s n T}$

of a sampled function is equivalent to the z-transform $F (z)$ of the sequence ${f (n T)}$ of sample values with $z = e^{s T}$ .

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.

$f (t)$	$f (n T)$	$F (z)$	Radius of convergence
$t \geq 0$	$n = 0, 1, 2, \dots$		$R$


$1$	$1$	$\frac{z}{z - 1}$	$1$

$t$	$n T$	$\frac{z}{{(z - 1)}^{2}}$	$1$

$t^{2}$	${(n T)}^{2}$	$\frac{T^{2} z (z + 1)}{{(z - 1)}^{3}}$	$1$

$e^{- α t}$	$e^{- α n T}$	$\frac{z}{z - e^{- α T}}$	$\|e^{- α T}\|$

$sin ω t$	$sin n ω T$	$\frac{z sin ω T}{z^{2} - 2 z cos ω T + 1}$	$1$

$cos ω t$	$cos n ω T$	$\frac{z (z - cos ω T)}{z^{2} - 2 z cos ω T + 1}$	$1$

$t e^{- α t}$	$n T e^{- α n T}$	$\frac{T z e^{- α T}}{{(z - e^{- α T})}^{2}}$	$\|e^{- α T}\|$

$e^{- α t} sin ω t$	$e^{- α n T} sin ω n T$	$\frac{e^{- α T} z^{- 1} sin ω T}{1 - 2 e^{- α T} z^{- 1} cos ω T + e^{- 2 a T} z^{- 2}}$	$\|e^{- α T}\|$

$e^{- α T} cos ω t$	$e^{- α n T} cos ω n T$	$\frac{1 - e^{- α T} z^{- 1} cos ω T}{1 - 2 e^{- α T} z^{- 1} cos ω T + e^{- 2 a T} z^{- 2}}$	$\|e^{- α T}\|$

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

2.2 Table of z-transforms


$f_{n}$	$F (z)$	Name

$δ_{n}$	1	unit impulse

$δ_{n - m}$	$z^{- m}$

$u_{n}$	$\frac{z}{z - 1}$	unit step sequence

$a^{n}$	$\frac{z}{z - a}$	geometric sequence

$e^{α n}$	$\frac{z}{z - e^{α}}$

$sinh α n$	$\frac{z sinh α}{z^{2} - 2 z cosh α + 1}$

$cosh α n$	$\frac{z^{2} - z cosh α}{z^{2} - 2 z cosh α + 1}$

$sin ω n$	$\frac{z sin ω}{z^{2} - 2 z cos ω + 1}$

$cos ω n$	$\frac{z^{2} - z cos ω}{z^{2} - 2 z cos ω + 1}$

$e^{- α n} sin ω n$	$\frac{z e^{- α} sin ω}{z^{2} - 2 z e^{- α} cos ω + e^{- 2 α}}$

$e^{- α n} cos ω n$	$\frac{z^{2} - z e^{- α} cos ω}{z^{2} - 2 z e^{- α} cos ω + e^{- 2 α}}$

$n$	$\frac{z}{{(z - 1)}^{2}}$	ramp sequence

$n^{2}$	$\frac{z (z + 1)}{{(z - 1)}^{3}}$

$n^{3}$	$\frac{z (z^{2} + 4 z + 1)}{{(z - 1)}^{4}}$

$a^{n} f_{n}$	$F (\frac{z}{a})$

$n f_{n}$	$- z \frac{d F}{d z}$

2 z-transforms and Laplace transforms

Task!

Answer

Answer

Key Point 19

Task!

Answer

Key Point 20

2.1 Table 2: z-transforms of some sampled signals

2.2 Table of z-transforms