Signals in Natural Domain
Chapter 7 :  The Z-transform
 
Properties of the z-transform
We use the notation
$\displaystyle \{x[n]\}\leftrightarrow X(z),\qquad ROC=R_{x}$

to denote z-transform of the sequence.

 
1. Linearity
The z-transform of a linear combination of two sequence is given by
$\displaystyle a\{x[n]\}+b\{y[n]\} \leftrightarrow aX(z)+bY(z), \,$ ROC contains$\displaystyle (R_{x}\cap Ry)$

The algebraic form follows directly from the definition, equation (7.2). The linear combination is such that some zero's can cancel the poles, then the region of convergence may be larger. For example if the linear combination $ \{a^{n}u[n]-a^{n}u[n-N]\}$ is a finite-length sequence, the ROC is entire z-plane except at $ a=0$, like individual ROCs are. If the intersection of $ R_{x}$ and $ R_{y}$ is null set, the z-transform of the linear combination will not exist.

 
2. Time shifting
If we shift the time sequence, we get
$ \{x[n-n_{0}]\}\leftrightarrow z^{-n_{0}}X(z)$, $ ROC =R_{x}$ except for possible addition or deletion of $ z=0$ and/or $ z=\infty$
We have

$\displaystyle Y(z)=\sum\limits^{\infty}_{n=-\infty}x[n-n_{0}]z^{-n}$

changing variable, $ m=n-n_{0}$

$\displaystyle Y(z)$ = $\displaystyle \sum\limits^{\infty}_{m=-0}x[m]z^{-(m+n_{0})}$  
  = $\displaystyle z^{-n_0}\sum\limits^{\infty}_{m=-\infty}x[m]z^{-m}$  
  = $\displaystyle z^{-n_0}X(z)$  


The factor $ z^{-n_{0}}$ can affect the poles and zeros at $ z=0$, $ z=\infty$

 

3.  Multiplication by a exponential sequence

$\displaystyle \{z^{n}_{0}x[n]\} \leftrightarrow X(z/z_{0}),\quad ROC=\{z:\, z/z_{0}\,\epsilon \, R_{x}\}$

This follows directly from defining equation (7.2).

 

4.   Differentiation of $ X(z)$:
If we differentiate $ X(z)$ term by term we get

$\displaystyle \frac{dX(z)}{dz}=\sum\limits^{\infty}_{n=-\infty}x[n](-n)z^{-n-1}$

Thus

$\displaystyle -z\frac{dX(z)}{dz}=\sum\limits^{\infty}_{n=-\infty}nx[n]z^{-n}$

$ \{nx[n]\}\leftrightarrow - z\frac{dX(z)}{dz},\,\,ROC=R_x$, except possibly $ z=0,\,z=\infty$
The ROC does not change (except $ z=0$, $ z=\infty$). This follows from the property that $ X(z)$ is an analytic function.

 

5. Conjugation of a complex sequence
$\displaystyle \{x^{*}[n]\}\leftrightarrow X^{*}(z^{*}),\,\, ROC=R_x$
$\displaystyle Y(z)$ = $\displaystyle \sum\limits^{\infty}_{n=-\infty} x^{*}[n]z^{-n}$  
  = $\displaystyle \left(\sum\limits^{\infty}_{n=-\infty}x[n](z^{*})^{-n}\right)^{*}$  
  = $\displaystyle X^{*}(z^{*})$  

Since ROC depends only an magnitude $ \vert z \vert$ it does not change.

 

6.  Time Reversal

$\displaystyle \{x[-n]\}\leftrightarrow X(1/z)$

$\displaystyle ROC=\{z:\,\frac{1}{z}\,\,\epsilon R_{X}\}$

We have                                                      $\displaystyle Y(z)=\sum\limits^{\infty}_{n=-\infty}x[-n]z^{-n}$

putting $ m=-n$
$\displaystyle y(z)$ = $\displaystyle \sum\limits^{\infty}_{m=-\infty}x[m]z^{m}$  
  = $\displaystyle X(1/z)$  

If we combine it with the previous property, we get

$\displaystyle \{x^{*}[-n]\} \leftrightarrow X^{*}(1/z^{*}),\,\, ROC=\{z:\frac{1}{z}\epsilon R_{x}\}$

 
7.  Convolution of sequence

$\displaystyle \{x[n]\}*\{y[n]\}\leftrightarrow X(z)Y(z),\,\,$   ROC contains$\displaystyle \,\, R_{x}\cap R_{y}$

The z-transform of the convolution is       $\displaystyle \sum\limits^{\infty}_{n=-\infty}(\sum\limits^{\infty}_{k=-\infty}x[k]y[n-k])z^{-n}$

Interchanging the order of summation     $\displaystyle =\sum\limits^{\infty}_{k=-\infty}x[k]\sum\limits^{\infty}_{n=-\infty}y[n-k]z^{-n}$

using time shifting property (or changing index of summation)
  = $\displaystyle \sum\limits^{\infty}_{k=-\infty} x[k]z^{-k}Y(z)$  
  = $\displaystyle X(z)Y(z)$  


If there is pole-zero cancelation, the ROC will be larger than the common ROC of two sequence.
Convolution property plays an important role in analysis of LTI system. An LTI system, which produces a delay of $ n_{0}$, has the transfer function $ z^{-n_{0}}$, therefore delay of $ n_o$ units is often depicted by $ z^{-n_{0}}$


Fig 7.3

 

8.  Complex convolution theorem
If we multiply two sequences then

$\displaystyle \{x[n]y[n]\}\leftrightarrow \frac{1}{2 \pi j} \oint X(v) Y(z/v)v^{-1}dv,\,\,$ROC contains$\displaystyle \,\, \{zw,z \epsilon R_{x},w\epsilon R_{y}\}$

This can be proved using inverse z-transform definition.

 

9.  Initial value Theorem
If $ x[n]$ is zero for $ n<0$, i.e. $ \{x[n]\}$ is causal, then

$\displaystyle x[0]=\lim\limits_{z \rightarrow \infty}X(z)$

Taking limit term by term in $ X(z)$, we get the above result.

10.  Parseval's relation

$\displaystyle \sum\limits^{\infty}_{n=-\infty} x[n]y^{*}[n]\leftrightarrow \frac{1}{2 \pi j}\oint X(v)Y^{*}(1/v^{*})v^{-1}dv$


These properties are summarized in table 7.1

$\displaystyle \begin{tabular}{l\vert l\vert l} \hline Sequence & Transform & ... ...j}\oint X(v)Y(z/v)u^{-1}_{dv}$&Contains $R_{x}R_{y}$\\ \hline \end{tabular}$

Table 7.1     z-transform properties
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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