Signals in Natural Domain
Chapter 7 :  The Z-transform
 
Methods of inverse z-transform
We can use the contour integration and the equation (7.6) to calculate inverse z-transform. This equation has to be evaluated for all values of $ n$, which can be quite complicated in many cases. Here we give two simple methods for the inverse transform computation.
 
1. Inverse transform by partial fraction expansion

This is method is useful when z-transform is ratio of polynomials. A rational $ X(z)$ can be expressed as

$\displaystyle X(z)=\frac{N(z)}{D(z)}$

where $ N(z)$ and $ D(z)$ are polynomials in. If degree $ M$ of the numerator polynomial $ N(z)$ is greater than or equal to the degree N of the denominator polynomial $ D(z)$, we can divide $ N(z)$ by $ D(z)$ and re-express $ X(z)$ as

$\displaystyle X(z)=\sum\limits^{M-N}_{k=0}a[k]z^{-k}+\frac{N_{1}(z)}{D(z)}$

where the degree of polynomial $ N_{1}(z)$ is strictly less than that of. For simplicity let us assume that all poles are simple. Then                        $\displaystyle X(z)=\sum\limits^{M-N}_{k=0}B_{k}z^{-k}+\sum\limits^{N}_{k=1}\frac{A_{k}}{1-d_{k}Z^{-l}}$

where $ A_{k}=(1-d_{k}Z^{-1})\frac{N_{1}(z)}{D(z)}\vert _{Z=d_k}$

Example: Let

$\displaystyle X(z)=\frac{1+2z^{-1}}{(1-\cdot 2\,z^{-1})(1+\cdot \,6z^{-1})}$

The partial fraction expression is

$\displaystyle X(z)$ = $\displaystyle \frac{A_{1}}{1-\cdot2\,z^{-1}}+\frac{A_{2}}{(1+\cdot 6\,z^{-1})}$  
$\displaystyle A_{1}$ = $\displaystyle (1-\cdot 2\,z^{-1})X(z)\vert _{Z=\cdot 2}=\frac{1+2z^{-1}}{1+\cdot 6\,z^{-1}}\vert _{z=\cdot2}=2.75$  
$\displaystyle A_{2}$ = $\displaystyle (1-\cdot 6\,z^{-1})X(z)\vert _{Z=-\cdot 6}=\frac{1+2z^{-1}}{1-\cdot 2z^{-1}}\vert _{z=\cdot-6}=-1.75$  
$\displaystyle X(z)$ = $\displaystyle \frac{2.75}{1-\cdot 2z^{-1}}-\frac{1.75}{1+\cdot 6z^{-1}}$  

The inverse z-transform depends on the ROC. If ROC is $ \vert z\vert>\cdot 6$, then ROCs associated with each term is outside a circle(so that common ROC is outside a circle), sequences are causal. Using linearity property and z-transform of $ a^{n}u[n]$ we get

$\displaystyle x[n]=2.75(0.2)^{n} u[n]-1.75(-.6)^{n}u[n]$

If the ROC is $ .2 < \vert z\vert< .6$, the ROC of the term    $ \frac{1}{1-.2z^{-1}}$   should be outside the circle $ \vert z\vert=.2$, and ROC for $ \frac{1}{1+.6 z^{-1}}$ should be. Hence we get the sequence as

        $\displaystyle x[n]=2.75(.2)^{n}u[n]+1.75(-.6)^{n}u[-n-1]$

Similarly if ROC is $ \vert Z\vert<.2$ we get a noncausal sequence

                    $\displaystyle x[n]=-2.75(.2)^{n}u[-n-1]+1.75(-.6)^{n}u[-n-1]$

If $ X(z)$ has multiple poles, the partial fraction has slightly different form. If $ X(z)$ has a pole of order s at $ z=d_i$, and all other poles are simple Then

$\displaystyle X(z)=\sum\limits^{M-N}_{k=0}B_{k}z^{-k}+\sum\limits^{N}_{k=1,k\ne... ...frac{A_{k}}{1-d_{k}Z^{-1}}+\sum\limits^{s}_{m=1}\frac{C_m}{(1-d_{i}z^{-1})^{m}}$

where $ A_{k}$ and $ B_{k}$ are obtained as before, the coefficients $ C_{m}$ are given by

$\displaystyle C_{m}=\frac{1}{(s-m)!(-d_{i})^{s-m}}\left\{\frac{d^{s-m}}{dw^{s-m}}(1-d_{i}w)^{s}X(w^{-1})\right\}_{w=d^{-1}_{i}}$

If there are more multiple poles, there will be more terms like the third term.
Using linearity and differentiation properties we get some useful z-transform pairs given in Table 7.2

Sequence
Transform
ROC
1. $ \{\delta [n]\}$ $ 1$ All $ \,{z}$
2. $ \{\delta [n-m]\}$ $ z^{-m}$ All$ \,{z}$, except 0(if $ m > 0$) or $ \infty ($if$ \, m < 0)$
3. $ \{a^{n}u[n]\}$ $ \frac{1}{1-az-1}$ $ \vert z\vert > \vert a\vert$
4. $ \{-a^{n}u[-n-1]\}$ $ \frac{1}{1-az^{-1}}$ $ \vert z\vert <\vert a\vert$
5. $ \{na^{n}u[n]\}$ $ \frac{az^{-1}}{(1-az^{-1})^{2}}$ $ \vert z\vert > \vert a\vert$
6. $ \{-na^{n}u[-n-1]\}$ $ \frac{az^{-1}}{(1-az^{-1})^{2}}$ $ \vert z\vert <\vert a\vert$
7. $ \{r^{n} \cos w_{0}n\,u[n]\}$ $ \frac{1-r \cos w_{0}z^{-1}}{1-2r\, \cos w_{0}z^{-1}+r^{2}\,z^{-2}}$ $ \vert z\vert > r$
8. $ \{r^{n} \sin w_{0}n\,u[n]\}$ $ \frac{\sin w_{0}\, z^{-1}}{1-2r \cos w_{0}z^{-1}+r^{2}\,z^{-2}}$ $ \vert z\vert > r$
9. $ \{a^{n},\,0 \leq n\leq N-1\}$ $ \frac{1-a^{N}\,z^{-N}}{1-a\,z^{-1}}$ $ \vert z\vert > 0$
Table 7.2  Some useful z-transform pairs
 
2.  Inverse Transform via long division
For causal sequence the z-transform $ X(z)$ can be exported into a pure series in. In the series expansion, the coefficient multiplying the term $ z^{-n}$ is. If $ X(z)$ is anticausal then we expand in terms of poles of.
Example 1: Let

$\displaystyle X(z)=\frac{1+2z^{-1}}{(1-.2z^{-1})(1+.6z^{-1})},\,\,\, ROC \,\,\vert z\vert >\, .6$

This is a causal sequence, long division gives


This gives $ x[0]=1,\,x[1]=1.6,\,x[2]=-.52,\,x[3]=.4$,.....
We can see that it is not easy to write the $ n^{th}$ term.

Example 2:

$\displaystyle X(z)=ln(1+az^{-1}),\,\,\vert z\vert \, > \, \vert a\vert$

Using the pure series expansion for $ ln(1+x)$ with $ \vert x\vert\, < \,\vert$, we obtain
$\displaystyle X(z)$ = $\displaystyle \sum\limits^{\infty}_{n=1}\frac{(-1)^{n+1}a^{n}z^{-n}}{n}$  
$\displaystyle x[n]$ = $\displaystyle \left \{ \begin{aligned}(-1)^{n+1}\frac{a^{n}}{n},\,\,n\geq 1 \\ 0,\quad\qquad \mbox{otherwise} \end{aligned} \right.$  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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