# 3.1 The z transform: definition

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A brief definition of the z-transform, explaining its relationship with the Fourier transform and its region of convergence, ROC.

## Basic definition of the z-transform

The z-transform of a sequence is defined as

$X(z)=\sum_{n=()}$ x n z n
Sometimes this equation is referred to as the bilateral z-transform . At times the z-transform is defined as
$X(z)=\sum_{n=0}$ x n z n
which is known as the unilateral z-transform .

There is a close relationship between the z-transform and the Fourier transform of a discrete time signal, which is defined as

$X(e^{i\omega })=\sum_{n=()}$ x n ω n
Notice that that when the $z^{-n}$ is replaced with $e^{-(i\omega n)}$ the z-transform reduces to the Fourier Transform. When the Fourier Transform exists, $z()=e^{i\omega }$ , which is to have the magnitude of $z$ equal to unity.

## The complex plane

In order to get further insight into the relationship between the Fourier Transform and the Z-Transform it is useful to lookat the complex plane or z-plane . Take a look at the complex plane:

The Z-plane is a complex plane with an imaginary and real axis referring to the complex-valued variable $z$ . The position on the complex plane is given by $re^{(i\omega )}$ , and the angle from the positive, real axis around the plane is denoted by $\omega$ . $X(z)$ is defined everywhere on this plane. $X(e^{i\omega })$ on the other hand is defined only where $\left|z\right|=1$ , which is referred to as the unit circle. So for example, $\omega =1$ at $z=1$ and $\omega =\pi ()$ at $z=-1$ . This is useful because, by representing the Fourier transformas the z-transform on the unit circle, the periodicity of Fourier transform is easily seen.

## Region of convergence

The region of convergence, known as the ROC , is important to understand because it defines the region wherethe z-transform exists. The ROC for a given $x(n)$ , is defined as the range of $z$ for which the z-transform converges. Since the z-transform is a power series , it converges when $x(n)z^{-n}$ is absolutely summable. Stated differently,

$\sum_{n=()}$ x n z n
must be satisfied for convergence. This is best illustratedby looking at the different ROC's of the z-transforms of $\alpha ^{n}u(n)$ and $\alpha ^{n}u(n-1)$ .

For

$x(n)=\alpha ^{n}u(n)$

$X(z)=\sum_{n=()}$ x n z n n α n u n z n n 0 α n z n n 0 α z 1 n
This sequence is an example of a right-sided exponential sequence because it is nonzero for $n\ge 0$ . It only converges when $\left|\alpha z^{(-1)}\right|< 1$ . When it converges,
$X(z)=\frac{1}{1-\alpha z^{(-1)}}=\frac{z}{z-\alpha }$
If $\left|\alpha z^{(-1)}\right|\ge 1$ , then the series, $\sum_{n=0}$ α z n does not converge. Thus the ROC is the range of values where
$\left|\alpha z^{(-1)}\right|< 1$
or, equivalently,
$\left|z\right|> \left|\alpha \right|$

For

$x(n)=-\alpha ^{n}u(-n-1)$

$X(z)=\sum_{n=()}$ x n z n n α n u -n 1 z n n -1 α n z n n -1 α -1 z n n 1 α -1 z n 1 n 0 α -1 z n
The ROC in this case is the range of values where
$\left|\alpha ^{-1}z\right|< 1$
or, equivalently,
$\left|z\right|< \left|\alpha \right|$
If the ROC is satisfied, then
$X(z)=1-\frac{1}{1-\alpha ^{-1}z}=\frac{z}{z-\alpha }$

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