I have the z-transform of a sequence, and I want to find the sequence.
The transform is:
\begin{equation} V(z)=\frac{z}{2z^2-7z-4} \end{equation}
Instead of using tables of inverse z-transforms, I thought of using Cauchy theorem, since the formula for the inverse z-transform very familiar to it:
\begin{equation} v(n)=\frac{1}{2i\pi}\oint V(z)z^{n-1}dz \end{equation}
So we start, listing up the form of the inverse transform:
\begin{equation} v(n)=\frac{1}{2\pi i}\oint \frac{z}{2(z-4)(z+\frac{1}{2})}\cdot z^{n-1}=\frac{1}{4\pi i}\oint \frac{z^n}{(z-4)(z+\frac{1}{2})} \end{equation}
and use the residue theorem, with all poles included:
\begin{equation} Res(f; z_0)=\lim_{z\rightarrow z_0}(z-z_0)f(z) \end{equation}
\begin{equation} Res(f;-1/2)=\lim_{z\rightarrow -\frac{1}{2}}(z+\frac{1}{2})\frac{z^n}{(z-4)(z+\frac{1}{2}))}=\lim_{z\rightarrow -\frac{1}{2}}\frac{z^n}{(z-4)}=-\bigg(-\frac{1}{2}\bigg)^n\bigg(\frac{2}{9}\bigg)\frac{1}{4i\pi} \end{equation}
\begin{equation} Res(f;4)=\lim_{z\rightarrow 4}(z-4)\frac{z^n}{(z-4)(z+\frac{1}{2}))}=\lim_{z\rightarrow 4}\frac{z^n}{(z+\frac{1}{2})}=4^n\frac{2}{9}\frac{1}{4i\pi} \end{equation}
So from here, where do the residues fit in the inverse z-transform integral, and how do we complete this procedure?
UPDATE:
Recalling Cauchy's theorem:
\begin{equation} \oint \frac{f(z)}{z-a}=Res(z_0)2\pi i, \end{equation}
where $a$ is simply the pole, so for the two residues, we have:
Series at the residue at pole $z=4$ \begin{equation} v(n)_{Res(f;4)}=-\bigg(-\frac{1}{2}\bigg)^n\bigg(\frac{2}{9}\bigg)\frac{1}{4i\pi}\cdot 2\pi i=-\bigg(-\frac{1}{2}\bigg)^n\bigg(\frac{2}{18}\bigg) \end{equation}
Residue at pole $z=-\frac{1}{2}$ \begin{equation} v(n)_{Res(f;-\frac{1}{2})}=4^n\frac{2}{9}\frac{1}{4i\pi}\cdot 2\pi i=4^n\frac{2}{18} \end{equation}
summing these together, we obtain the final answer:
\begin{equation} v(n)=4^n\frac{2}{18}-\bigg(-\frac{1}{2}\bigg)^n\bigg(\frac{2}{18}\bigg)=\frac{4^n-2^{-n}}{9} \end{equation}
The inverse z-transform (the n-series of the function $V(z)=\frac{z}{2z^2-7z-4}$) is therefore:
\begin{equation} v(n)=\frac{4^n-2^{-n}}{9} \end{equation}
Here is a plot of the original function, V(z):
and this is the plot of the sequence, V(n):
