Z Transforms

Difference Equations

A difference equation is a discrete equivalent of a differential equation, used in situations where only discrete values can be measured:

$\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 2 x = 3 t$

becomes

$x [n + 2] + 5 x [n + 1] + 2 x [n] = 3 n n = 0, 1, 2, ...$

These can be solved numerically by just evaluating the output for each value of n. For example:

$y [n] - 0.5 y [n - 1] = x [n]$

$x [n] = {01 n < 0 n \geq 0$

This evaluates to:

$y [0] = 0.5 y [- 1] + x [0] = 0.5 (0) + 1 = 1 y [1] = 0.5 y [0] + x [1] = 0.5 (1) + 1 = 1.5 y [2] = 0.5 y [1] + x [2] = 0.5 (1.5) + 1 = 1.75 \dots$

Alternatively, there is an analytical solution...

The z Transform

Consider a discrete sequence $f [k], n \in N$ . The z transform of this sequence is defined as:

$F (z) = Z {f [k]} = k = 0 \sum \infty f [k] z^{- k} = F (z) = f [0] + \frac{f [ 1 ]}{z} + \frac{f [ 2 ]}{z ^{2}} + \frac{f [ 3 ]}{z ^{3}} ...$

A closed-form expression can generally be found by the sum of the infinite series. For example, the z transform of the unit step $f [k] = 1$ :

$k = 0 \sum \infty f [k] z^{- k} = 1 + \frac{1}{z} + \frac{1}{z ^{2}} + ...$

This is a geometric series with $a = 1$ , $r = z^{- 1}$ , hence the sum is $\frac{a}{1 - r}$

$F (z) = \frac{z}{z - 1} for ∣ z ∣ > 1$

Taking a z transform of a difference equation converts it to a continuous function. The z domain is similar to the laplace domain, but for discrete time signals instead.

z Transform Properties

z transforms have linearity, the same as laplace and fourier transforms.

First Shift Theorem

If $f [k]$ is a sequence and $F (z)$ it's transform, then

$Z {f [k + i]} = z^{i} F (z) - (z^{i} f [0] + z^{i - 1} f [1] + ... + z f [i - 1])$

For example, if $i = 1$ :

$Z {f [k + 1]} = z F (z) - z f [0]$

For $i = 2$ :

$Z {f [k + 2]} = z^{2} F (z) - z^{2} f [0] - z f [1]$

Second Shift Theorem

The function $f (t) u (t)$ is defined:

$f (t) u (t) = {f (t) 0 t \geq 0 t < 0$

Where $u (t)$ is the unit step function. The function $f (t - i T) u (t - i T)$ , where $i$ is a positive integer, represents a shift to the right of this function by $i$ sample intervals. If this shifted function is sampled, we have $f [k - i] u [k - i]$ . The second shift theorem states:

$Z {f [k - i] u [k - i]} = z^{- i} F (z)$

Inverse z Transforms

z transforms are inverted using lookup tables, but to get them into a recognisable form, some manipulation is often needed, including partial fractions. For example, finding the inverse transform of $F (z) = \frac{z + 3}{z - 2}$ :

$F (z) = \frac{z + 3}{z - 2} = \frac{z}{z - 2} + \frac{3}{z - 2}$

The first term can be seen immediately from the table:

$Z {\frac{z}{z - 2}} = 2^{k}$

The second term rearranges to give:

$\frac{3}{z - 2} = \frac{3}{z} \frac{z}{z - 2} = 3 z^{- 1} \frac{z}{z - 2}$

This is in the form of the second shift theorem, so this can be applied to give:

$Z {3 (2)^{k - 1} u [k - 1]} = 3 z^{- 1} \frac{z}{z - 2}$

Thus,

$Z^{- 1} {\frac{z + 3}{z - 2}} = 2^{k} + 3 (2)^{k - 1} u [k - 1]$

Example

Solve $y [k + 2] - 5 y [k + 1] + 6 y [k] = 0$ , where $y [0] = 0$ , $y [1] = 2$ .

$Z {y [k + 2]} - 5 Z {y [k + 1]} + 6 Z {y [k]} = 0$

Taking z transforms:

$z^{2} Y (z) - zy [0] - zy [1] - 5 z Y (z) + 5 zy [0] + 6 Y (z) = 0$

Rearranging and using initial conditions:

$(z^{2} - 5 z + 6) Y (z) = 2 z$

$Y (z) = \frac{2 z}{z ^{2} - 5 z + 6}$

Using partial fractions:

$Y (z) = \frac{2}{z - 3} + \frac{2}{z - 2}$

Using inverse transforms straight from the table to get the solution:

$y [k] = 2 \times 3^{k} - 2 \times 2^{k}$

Computer Systems Engineering Notes