Question 3.P.41: Determine the inverse Z-transform of x(z) = 1/2-4z^-1 +2z^-2......
Determine the inverse Z-transform of
x(z)=\frac{1}{2-4 z^{-1}+2 z^{-2}} by long division method if
(a) ROC |z| > 1
(b) ROC |z| < 1/2
(a) For ROC |z| > 1, that is, ROC is exterior of the circle, it is causal sequence. So power series expansion should be in negative power of z.
\begin{matrix} 2-4z^{-1}+2z^{-2} \Biggr)& \begin{matrix}\quad\qquad\qquad1/2-z^{-1}+3/2z^{-2}\\\hline \not{1}& \\ \begin{matrix} \qquad\qquad\not{1}\pm 2z^{-1}\pm z^{-2}\\-\\\hline \qquad\cancel{2z^{-1}}-z^{-2} \end{matrix} \\\begin{matrix} \cancel{\begin{matrix} 2z^{-1} \\ – \end{matrix}} \begin{matrix} \mp 4z^{-2} \pm 2z^{-3}\\ & \end{matrix}\\\hline\cancel{3z^{-2}}-2z^{-3} \end{matrix} \end{matrix} \\ & \begin{matrix} \cancel{\begin{matrix} -3z^{-2} \\- \end{matrix}} \begin{matrix} \mp 6z^{-3}\pm 3z^{-2} \\ & \end{matrix} \\\hline 4z^{-3}-z^{-2} \end{matrix} \end{matrix}
Therefore,
X(z)=\frac{1}{2-4 z^{-1}+2 z^{-2}}=\frac{1}{2}+z^{-1}+\frac{3}{2} z^{-2}+\ldots
Then the inverse Z-transform of X(z)
\Rightarrow x(n)=\operatorname{IZT}[X(z)]=\underset{\uparrow}{\{1 / 2,1,3 / 2, \ldots\}}
(b) If ROC |z| < 1/2, that is., ROC is interior of the circle, then the integral x(n) is non-causal and requires a power series expansion in positive power of Z.
\begin{matrix} 2z^{-2}-4z^{-1}+2 )& \begin{matrix} 1/2z^2+2z^3+7/2z^4+…. \\\hline \not{1} \end{matrix} \end{matrix}
\begin{matrix} \cancel{\begin{matrix} 1 \\ – \end{matrix} }\begin{matrix} \mp 4z\pm z^2 \\ & \end{matrix} \\\hline\cancel{4z}+z^2 \\ \begin{matrix} \cancel{4z} \\ – \end{matrix} \begin{matrix} \mp 8z^2\pm 4z^3 \\ & \end{matrix}\\\hline \cancel{7z^2}-4z^3 \\\begin{matrix} \cancel{7z^2} \\ – \end{matrix} \begin{matrix} \mp 14z^3\pm 72^4 \\ & \end{matrix}\\\hline 10z^3-7z^4 \end{matrix}
X(z)=\frac{1}{2} z^2+2 z^3+\frac{7}{2} z^4+\ldots
Therefore, Z-transform form n ≤ 0 is
\begin{gathered} X(z)=x(0) z^0+x(-1) z^1+x(-2) z^2+\ldots \\ x(n)=\{\ldots 7 / 2,2,1 / 2,0,\underset {↑}{0}\} \end{gathered}