# Long Division (or) Power Series Expansion Method:-

Power Series ExpansionMethod:-

$X(Z)&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}x[n]\&space;Z^{-n}$

$X(Z)$can be expressed either in positive powers of Z or negative powers of Z.

if the sequence is causal $X(Z)$ has negative powers of Z similarly the Non-causal sequence   $X(Z)$  negative powers of Z.

Let $X(Z)=\frac{N(Z)}{D(Z)}&space;=\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}}$ .

$X(Z)$ is causal it has ROC  $\left&space;|&space;Z&space;\right&space;|>&space;\left&space;|&space;r&space;\right&space;|$  then $X(Z)$ can be expressed as

$X(Z)&space;=&space;x(0)+x(1)Z^{-1}+x(2)Z^{-2}+x(3)Z^{-3}+........$

$X(Z)$ is non-causal it has ROC $\left&space;|&space;Z&space;\right&space;|<\left&space;|&space;r&space;\right&space;|$ then $X(Z)$ can be expressed as

$X(Z)=x(0)+x(-1)Z^{1}+x(-2)Z^{2}+x(-3)Z^{3}+...........$

Partial fraction Method:-

Let $X(Z)=\frac{N(Z)}{D(Z)}&space;=\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}}$     and    $a_{o}=1$

if     $M  ,  $X(Z)$   is a proper function .

if   $M\geq&space;N$   ,  $X(Z)$   is improper function  so convert the improper function to proper function as

$X(Z)=c_{o}+c_{1}Z^{-1}+c_{2}Z^{-2}+.........+c_{M-N}Z^{-(M-N)}+&space;\frac{N_{1}(Z)}{D(Z)}$.

$X(Z)&space;=&space;polynomial&space;+&space;rational&space;\&space;proper&space;\&space;function$ .

express    $X(Z&space;)$  into powers of Z  as follows

$X(Z&space;)&space;=&space;\frac{b_{o}Z^{N}+b_{1}Z^{N-1}+b_{2}Z^{N-2}+.......+b_{M}Z^{N-M}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}$

then divide   $X(Z)$  by   Z

$\frac{X(Z)}{Z}&space;=&space;\frac{b_{o}Z^{N-1}+b_{1}Z^{N-2}+b_{2}Z^{N-3}+.......+b_{M}Z^{N-M-1}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}$  .

Now  express   $\frac{X(Z)}{Z}$  into partial fractions using different cases and find out the inverse Z-transform  for the function

$X(Z&space;)&space;=&space;Z.(partial&space;fraction&space;\&space;expansion&space;)$ .

Convolution Method:-

express $X(Z)$   as a product of two functions $X_{1}(Z)$   and $X_{2}(Z)$  as follows $X(Z)&space;=X_{1}(Z)&space;.&space;X_{2}(Z)$

then find the inverse Z- transforms of individual functions

$x_{1}[n]\leftrightarrow&space;X_{1}(Z)$

$x_{2}[n]\leftrightarrow&space;X_{2}(Z)$

by using convolution method find convolution of $x_{1}[n]$  and $x_{2}[n]$

i.e, $x[n]&space;=&space;x_{1}[n]&space;*&space;x_{2}[n]$

now $x[n]$   is the inverse Z-transform of $X(Z)$ .

Cauchy Residue Theorem:-

$f(Z)$ a function in Z if the derivative $\frac{df(Z)}{dZ}$   exists on and inside contour C and $f(Z)$ has no poles at $Z=Z_{o}$   then.

$\frac{1}{2\pi&space;j}\oint_{c}&space;\frac{f(Z)dZ}{Z-Z_{o}}=\left\{\begin{matrix}&space;f(Z_{o})&space;\&space;if&space;\&space;Z_{o}&space;\&space;is\&space;inside&space;\&space;C\\&space;0&space;\&space;if&space;\&space;Z_{o}\&space;is\&space;outside&space;\&space;C&space;\end{matrix}\right.$ .

if   $(k+1)^{th}$ the derivative  of   $f(Z)$  exists on and has no poles at $Z=Z_{o}$   then.

$\frac{1}{2\pi&space;j}\oint_{c}&space;\frac{f(Z)}{(Z-Z_{o})^k}dZ=\left\{\begin{matrix}\frac{1}{(k-1)!}&space;\frac{d^{k-1}f(Z)}&space;{dZ^{k-1}}\&space;if&space;\&space;Z_{o}&space;\&space;is\&space;inside&space;\&space;C\\&space;0&space;\&space;if&space;\&space;Z_{o}\&space;is\&space;outside&space;\&space;C&space;\end{matrix}\right.$ .

the values on the right hand side are called Residue’s of the pole $Z=Z_{o}$ .

if there are n no of poles inside C$\frac{1}{2\pi&space;j}\oint_{c}&space;\frac{f(Z)dZ}{(Z-Z_{1})(Z-Z_{2})(Z-Z_{3})..(Z-Z_{n})}=\sum_{i=1}^{n}\lim_{Z\rightarrow&space;Z_{i}}&space;\left&space;$ .

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## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems.

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