Signals in Natural Domain
Chapter 7 :  The Z-transform
 
Definition of the Z-transform
We saw earlier that complex exponential of the from $ \{e^{jwn}\}$ is an eigen function of for a LTI System. We can generalize this for signals of the form $ \{z^{n}\}$ where, $ z$ is a complex number.
$\displaystyle y[n]$ = $\displaystyle \sum\limits^{\infty}_{k=-\infty} h[k]x[n-k]$  
  = $\displaystyle \sum\limits^{\infty}_{k=-0}h[k]z^{n-k}$  
  = $\displaystyle \left(\sum\limits^{\infty}_{k=-\infty}h[k]z^{-k}\right)z^{n}$  
  = $\displaystyle H(z)z^{n}$  
where
$\displaystyle H(z)=\sum\limits^{\infty}_{k=-\infty}h[k]z^{-h}\,\,\,\,\,\qquad$ (7.1)
Thus if the input signal is $ \{z^{n}\}$ then output signal is. For $ z=e^{jw}$ $ w$ real (i.e for $ \vert z\vert = 1 $), equation (7.1) is same as the discrete-time fourier transform. The $ H(z)$ in equation (7.1) is known as the bilateral z-transform of the sequence. We define for any sequence of a sequence $ \{x[n]\}$ as
$\displaystyle X(z)=\sum\limits^{\infty}_{n=-\infty}x[n]z^{-n}$ (7.2)
where $ z$ is a complex variable. Writing $ z$ in polar form we get $ z=re ^{jw}$, where $ r$ is magnitude and $ \omega$ is angle of.
$\displaystyle X(re^{jw})$ = $\displaystyle \sum\limits^{\infty}_{n=-\infty}x[n](re^{jw})^{-n}$  
  = $\displaystyle \sum\limits^{\infty}_{n=-\infty}(x[n]r^{-n})e^{-jwn}\qquad\qquad\qquad (7.3)$ (7.3)

This shows that $ X(re^{jw})$ is Fourier transform of the sequence. When $ r=1$ the z-transform reduces to the Fourier transform of. From equation (7.3) we see that for convergence of z-transform that Fourier transform of the sequence $ \{r^{-n}x[n]\}$ converges. This will happen for some r and not for others. The values of z - for which $ \sum\limits^{\infty}_{n=-\infty} r^{-n} \vert x[n]\vert < \infty$ is called the region of convergence(ROC). If the ROC contains unit circle (i.e. $ r=1$ or equivalently $ \vert z\vert = 1 $ then the Fourier transform also converges. Following examples show that we must specify ROC to completely specify the z-transform.

 
Example 1: Let $ \{x[n]\}=\{a^{n}u[n]\}$, then
$\displaystyle X(z)$ = $\displaystyle \sum\limits^{\infty}_{n=-\infty}x[n]z^{-n}$  
  = $\displaystyle \sum\limits^{\infty}_{n=0}(a z^{-1})^{n}$  
This is a geometric series and converges if $ \vert a z^{-1}\vert < 1\,$ or. Then
$\displaystyle X[z]=\frac{1}{1-az^{-1}}=\frac{z}{z-a},\vert z\vert >\vert a\vert$ (7.4)

We see that $ X(z)=0$ at $ z=0$, and $ 1/X((z)=0$ at. Values of $ z$ where $ X(z)$ is zero is called zero of $ X(z)$ and value of $ z$ where $ 1/X(z)$ is zero is called a pole of. Here we see that ROC consists of a region in Z-plane which lies outside the circle centered at origin and passing through the pole.

Fig 7.1

Example 2: Let, $ \{y[n]\}=\{-a^{n}u[-n-1]\}$, then
$\displaystyle Y(z)$ = $\displaystyle \sum\limits^{-1}_{n=-\infty}-a^{n}z^{-n}$  
  = $\displaystyle \sum\limits^{\infty}_{m=1}-a^{-m}z^{m}$  
This is a geometric series which converges when $ \vert a^{-1}z\vert < 1$, that is $ \vert z\vert <\vert a\vert$ Then
$\displaystyle Y(z)=\frac{-a^{-1}z}{1-a^{-1}z}=-\frac{z}{a-z}$
$\displaystyle \qquad\,\,=\frac{z}{z-a},\qquad \vert z\vert<\vert a\vert$ (7.5)


Fig 7.2

 
Here the ROC is inside the circle of radius. Comparing equation (7.4) and (7.5) we see that algebraic form of $ X(z)$ and $ Y(z)$ are same, but ROC are different and they correspond to two different sequences. Thus in specifying z-transform, we have to give functional form $ X(z)$ and the region of convergence.
Now we state some properties of the region of convergence
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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