3 Inversion of z-transforms using residues

This method has its basis in a branch of mathematics called complex integration. You may recall that the ‘ $z$ ’ quantity of z-transforms is a complex quantity, more specifically a complex variable. However, it is not necessary to delve deeply into the theory of complex variables in order to obtain simple inverse z-transforms using what are called residues . In many cases inversion using residues is easier than using partial fractions. Hence reading on is strongly advised.

3.1 Pole of a function of a complex variable

If $G\left(z\right)$ is a function of the complex variable $z$ and if

$G\left(z\right)=\frac{G_1\left(z\right)}{{\left(z-z_0\right)}^k}$

where $G_1\left(z_0\right)$ is non-zero and finite then $G\left(z\right)$ is said to have a pole of order $k$ at $z=z_0$ .

For example if

$G\left(z\right)=\frac{6\left(z-2\right)}{z\left(z-3\right){\left(z-4\right)}^2}$

then $G\left(z\right)$ has the following 3 poles.

1. pole of order 1 at $z=0$
2. pole of order 1 at $z=3$
3. pole of order 2 at $z=4$ .

(Poles of order 1 are sometimes known as simple poles.)

Note that when $z=2$ , $G\left(z\right)=0$ . Hence $z=2$ is said to be a zero of $G\left(z\right)$ . (It is the only zero in this case).

Task!

Write down the poles and zeros of

$G\left(z\right)=\frac{3\left(z+4\right)}{z^2\left(2z+1\right)\left(3z-9\right)}$ (18)

State the order of each pole.

Answer

The residue of a complex function $G\left(z\right)$ at a first order pole $z_0$ is

Res $\left(G\left(z\right),z_0\right)={\left[G\left(z\right)\left(z-z_0\right)\right]}_{z_0}$ (19)

The residue at a second order pole $z_0$ is

Res $\left(G\left(z\right),z_0\right)={\left[\frac{d}{dz}\left(G\left(z\right){\left(z-z_0\right)}^2\right)\right]}_{z_0}$ (20)

You need not worry about how these results are obtained or their full mathematical significance. (Any textbook on Complex Variable Theory could be consulted by interested readers.)

Example

Consider again the function (18) in the previous guided exercise.

The second form is the more convenient for the residue formulae to be used.

Using (19) at the two first order poles:

Using (20) at the second order pole

Res $\left(G\left(z\right),0\right)={\left[\frac{d}{dz}\left(G\left(z\right){\left(z-0\right)}^2\right)\right]}_0$

The differentiation has to be carried out before the substitution of $z=0$ of course.

Task!

Carry out the differentiation shown on the last line of the previous page, then substitute $z=0$ and hence obtain the required residue.

Answer

3.3 Inverse z-transform formula

Recall that, by definition, the z-transform of a sequence $\left\{f_n\right\}$ is

$F\left(z\right)=f_0+f_1{z}^{-1}+f_2{z}^{-2}+\dots f_n{z}^{-n}+\dots$

If we multiply both sides by $z^{n-1}$ where $n$ is a positive integer we obtain

$F\left(z\right)z^{n-1}=f_0{z}^{n-1}+f_1{z}^{n-2}+f_2{z}^{n-3}+\dots f_n{z}^{-1}+f_{n+1}{z}^{-2}+\dots$

Using again a result from complex integration it can be shown from this expression that the general term $f_n$ is given by

$f_n=$ sum of residues of $F\left(z\right)z^{n-1}$ at its poles (22)

The poles of $F\left(z\right)z^{n-1}$ will be those of $F\left(z\right)$ with possibly additional poles at the origin.

To illustrate the residue method of inversion we shall re-do some of the earlier examples that were done using partial fractions.

Example:

$Y\left(z\right)=\frac{z^2}{\left(z-a\right)\left(z-b\right)}a\ne b$

so

$Y\left(z\right)z^{n-1}=\frac{z^{n+1}}{\left(z-a\right)\left(z-b\right)}=G\left(z\right),$ say.

$G\left(z\right)$ has first order poles at $z=a$ , $z=b$ so using (19).

We need simply add these residues to obtain the required inverse z-transform

$\therefore f_n=\frac{1}{\left(a-b\right)}\left(a^{n+1}-b^{n+1}\right)$

as before.

Task!

Obtain, using (22), the inverse z-transform of

$Y\left(z\right)=\frac{6z^2-9z}{{\left(z-3\right)}^2}$

Firstly, obtain the pole(s) of $G\left(z\right)=Y\left(z\right)z^{n-1}$ and deduce the order:

Answer

Answer

In the above examples all the poles of the various functions $G\left(z\right)$ were real. This is the easiest situation but the residue method will cope with complex poles.

Example

We showed earlier that

$\frac{z^2}{z^2+1}$ and $cos\left(\frac{n\pi }{2}\right)$

formed a z-transform pair.

We will now obtain $y_n$ if $Y\left(z\right)=\frac{z^2}{z^2+1}$ using residues.

Using residues with, from (22),

$G\left(z\right)=\frac{z^{n+1}}{z^2+1}=\frac{z^{n+1}}{\left(z-i\right)\left(z+i\right)}$ where $i^2=-1.$

we see that $G\left(z\right)$ has first order poles at the complex conjugate points $\pm i$ .

Using (19)

$\text{ Res}\left(G\left(z\right),i\right)={\left[\frac{z^{n+1}}{z+i}\right]}_i=\frac{i^{n+1}}{2i}\text{ Res}\left(G\left(z\right),-i\right)=\frac{{\left(-i\right)}^{n+1}}{\left(-2i\right)}$

(Note the complex conjugate residues at the complex conjugate poles.)

Hence ${\mathbb{Z}}^{-1}\left\{\frac{z^2}{z^2+1}\right\}=\frac{1}{2i}\left(i^{n+1}-{\left(-i\right)}^{n+1}\right)$

But $i=e^{i\pi ∕2}$ and $-i=e^{-i\pi ∕2}$ , so the inverse $z$ -transform is

$\frac{1}{2i}\left(e^{i\left(n+1\right)\pi ∕2}-e^{-i\left(n+1\right)\pi ∕2}\right)=sin\left(n+1\right)\frac{\pi }{2}=cos\left(\frac{n\pi }{2}\right)\text{ as expected.}$

Task!

Show, using residues, that

${\mathbb{Z}}^{-1}\left\{\frac{z}{z^2+1}\right\}=sin\left(\frac{n\pi }{2}\right)$

Answer