# The Z-Transform

## Introduction

A linear system can be represented in the complex frequency domain (s-domain where s = s + jw) using the LaPlace Transform.

Where the direct transform is:

And x(t) is assumed zero for t ≤ 0

The Inversion integral is a contour integral in the complex plane (seldom used, tables are used instead)

Where s is chosen such that the contour integral converges.

If we now assume that x(t) is ideally sampled as in:

Where:

and

Analyzing this equivalent system using standard analog tools will establish the z-Transform.

## Sampling

Substituting the Sampled version of x(t) into the definition of the LaPlace Transform we get

But

(For x(t) = 0 when t < 0 )

Therefore

Now interchanging the order of integration and summation and using the sifting property of d-functions

(We are assuming that the first sample occurs at t = 0+)

if we now adjust our nomenclature by letting:

z = esT , x(n*Ts) = xn , and

Which is the direct z-transform (one-sided; it assumes xn = 0 for n < 0).

The inversion integral is:

(This is a contour integral in the complex z-plane)

(The use of this integral can be avoided as tables can be used to invert the transform.)

To prove that these form a transform pair we can substitute one into the other.

Now interchanging the order of summation and integration (valid if the contour followed stays in the region of convergence):

If “C” encloses the origin (that’s where the pole is), the Cauchy Integral theorem says:

And we get xk = xk                   Q.E.D

## An Example

Find the z-transform of

This is the “Unit Pulse” at n = k (assume k > 0)

(Note: dividing by z is equivalent to a delay of one sample time)

## A Short Table of z-Transforms

 f(t) (sampled) F(z) Region of Convergence U(t) |z| > 1 dn-k |z| > 1 t |z| > 1 t2 |z| > 1 eat |z| > eat sin(bt) |z| > 1 cos(bt) |z| > 1

## Properties of the z-Transform

The z-transform has properties that are analogous to those of the LaPlace Transform.  The following table has some of the more useful ones listed.

 Û where C is a closed contour that includes z=0 Signal z-Transform Û Superposition Û Time Shifting Û Û Û Û Time inversion Û Time Convolution (convolution) Frequency Differentiation Û Summation Û

You should familiarize yourself with these as they will be used, along with the table of transforms to move between time series and the z-domain.

## Finding the Inverse z-Transform

There are three common ways to find the time series, xn when X(z) is given:

1. Infinite Series – done by dividing out the rational polynomial in z
2. Partial Fraction Expansion – Same as in LaPlace
3. The Inversion Integral – a contour integral in the complex z-plane

### Example:  , determine fn

#### A.                 By Infinite Series

Now divide (long division) with the polynomials written in descending powers of z

2z-2+8z-3+22z-4+52z-5+114z-6+…

Z3-4z2+5z-2|2z

2z-8+10z-1-4z-2

8-10z-1+04z-2

8-32z-1+40z-2-16z-3

22z-1-36z-2+016z-3

22z-1-88z-2+110z-3-44z-4

52z-2-094z-3+044z-4

52z-2-208z-3+260z-4-104z-5

114z-3-216z-4+104z-5

\

And the time sequence for fn is

 n 0 1 2 3 4 5 6 … fn 0 0 2 8 22 52 114 …

Note that this method does NOT give a closed form for the answer, but it is a good method for finding the first few sample values or to check out that the closed form given by another method at least starts out correctly.

#### B.                 By Partial Fraction Expansion

To find k1 multiply both sides of the equation by (z-2), divide by z, and let z®2

or

k1 = 2

Similarly to find k3 multiply both sides by (z-1)2, divide by z, and let z®1

Equation A

And

k3 = -2

Finding k2 requires going back to Equation A above and taking the derivative of both sides

Now again let z®1

k2 = -2

\

#### C.                 Using the Inversion Integral

TBD

H.W. Find the inverse z-Transform of

We can check the answer by putting the three terms over the common denominator

It checks out!