Inverse Z-Transform -Methods

Long Division  (or) Power Series ExpansionMethod:-

X(Z) = \sum_{n=-\infty }^{\infty }x[n]\ 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)} =\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 | Z \right |> \left | r \right |  then X(Z) can be expressed as

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

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


Partial fraction Method:-

Let X(Z)=\frac{N(Z)}{D(Z)} =\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<N  ,  X(Z)   is a proper function .

if   M\geq 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)}+ \frac{N_{1}(Z)}{D(Z)}.

X(Z) = polynomial + rational \ proper \ function .

express    X(Z )  into powers of Z  as follows

X(Z ) = \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} = \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 ) = Z.(partial fraction \ expansion ) .

Convolution Method:-

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

then find the inverse Z- transforms of individual functions

x_{1}[n]\leftrightarrow X_{1}(Z)

x_{2}[n]\leftrightarrow X_{2}(Z)

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

i.e, x[n] = x_{1}[n] * 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 j}\oint_{c} \frac{f(Z)dZ}{Z-Z_{o}}=\left\{\begin{matrix} f(Z_{o}) \ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \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 j}\oint_{c} \frac{f(Z)}{(Z-Z_{o})^k}dZ=\left\{\begin{matrix}\frac{1}{(k-1)!} \frac{d^{k-1}f(Z)} {dZ^{k-1}}\ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \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 j}\oint_{c} \frac{f(Z)dZ}{(Z-Z_{1})(Z-Z_{2})(Z-Z_{3})..(Z-Z_{n})}=\sum_{i=1}^{n}\lim_{Z\rightarrow Z_{i}} \left  .



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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.