DSP - Z-Transform Existence

A system, which has system function, can only be stable if all the poles lie inside the unit circle. First, we check whether the system is causal or not. If the system is Causal, then we go for its BIBO stability determination; where BIBO stability refers to the bounded input for bounded output condition.

This can be written as;

$Mod(X(Z))< \infty$

$= Mod(\sum x(n)Z^{-n})< \infty$

$= \sum Mod(x(n)Z^{-n})< \infty$

$= \sum Mod[x(n)(re^{jw})^{-n}]< 0$

$= \sum Mod[x(n)r^{-n}]Mod[e^{-jwn}]< \infty$

$= \sum_{n = -\infty}^\infty Mod[x(n)r^{-n}]< \infty$

The above equation shows the condition for existence of Z-transform.

However, the condition for existence of DTFT signal is

Example 1

Let us try to find out the Z-transform of the signal, which is given as

$x(n) = -(-0.5)^{-n}u(-n)+3^nu(n)$

$= -(-2)^nu(n)+3^nu(n)$

Solution − Here, for $-(-2)^nu(n)$ the ROC is Left sided and Z<2

For $3^nu(n)$ ROC is right sided and Z>3

Hence, here Z-transform of the signal will not exist because there is no common region.

Example 2

Let us try to find out the Z-transform of the signal given by

$x(n) = -2^nu(-n-1)+(0.5)^nu(n)$

Solution − Here, for $-2^nu(-n-1)$ ROC of the signal is Left sided and Z<2

For signal $(0.5)^nu(n)$ ROC is right sided and Z>0.5

So, the common ROC being formed as 0.5<Z<2

Therefore, Z-transform can be written as;

$X(Z) = \lbrace\frac{1}{1-2Z^{-1}}\rbrace+\lbrace\frac{1}{(1-0.5Z)^{-1}}\rbrace$

Example 3

Let us try to find out the Z-transform of the signal, which is given as $x(n) = 2^{r(n)}$

Solution − r(n) is the ramp signal. So the signal can be written as;

$x(n) = 2^{nu(n)}\lbrace 1, n<0 (u(n)=0)\quad and\quad2^n, n\geq 0(u(n) = 1)\rbrace$

$= u(-n-1)+2^nu(n)$

Here, for the signal $u(-n-1)$ and ROC Z<1 and for $2^nu(n)$ with ROC is Z>2.

So, Z-transformation of the signal will not exist.

Z -Transform for Causal System

Causal system can be defined as $h(n) = 0,n<0$. For causal system, ROC will be outside the circle in Z-plane.

$H(Z) = \displaystyle\sum\limits_{n = 0}^{\infty}h(n)Z^{-n}$

Expanding the above equation,

$H(Z) = h(0)+h(1)Z^{-1}+h(2)Z^{-2}+...\quad...\quad...$

$= N(Z)/D(Z)$

For causal systems, expansion of Transfer Function does not include positive powers of Z. For causal system, order of numerator cannot exceed order of denominator. This can be written as-

$\lim_{z \rightarrow \infty}H(Z) = h(0) = 0\quad or\quad Finite$

For stability of causal system, poles of Transfer function should be inside the unit circle in Z-plane.

Z-transform for Anti-causal System

Anti-causal system can be defined as $h(n) = 0, n\geq 0$ . For Anti causal system, poles of transfer function should lie outside unit circle in Z-plane. For anti-causal system, ROC will be inside the circle in Z-plane.