Nathen Peterson

2022-03-25

Determine Z-transform for the sequence
$x\left[n\right]={\left(\frac{1}{4}\right)}^{\nu }\left[3-n\right]$

### Answer & Explanation

Aaliyah Phillips

$x\left[n\right]={\left(\frac{1}{4}\right)}^{\nu }\left[3-n\right]$
Apply Z transform of $x\left[n\right]$
$Z\left(z\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]{z}^{-n}$
$=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^{\nu }\left[3-n\right]{x}^{-n}$
$=\sum _{n=-\mathrm{\infty }}^{3}{\left(\frac{1}{4}\right)}^{n}{z}^{-n}$
$=\sum _{n=-\mathrm{\infty }}^{3}{\left(\frac{1}{4}{z}^{-1}\right)}^{n}$
$={\left(\frac{1}{4}{z}^{-1}\right)}^{3}+{\left(\frac{1}{4}{z}^{-1}\right)}^{2}+{\left(\frac{1}{4}{z}^{-1}\right)}^{1}+1+{\left(\frac{1}{4}{z}^{-1}\right)}^{-1}+\dots$
$=\frac{{\left(\frac{1}{4}{z}^{-1}\right)}^{3}}{1-{\left(\frac{1}{4}{z}^{-1}\right)}^{-1}}$
$=\frac{\frac{1}{64}{z}^{-3}}{1-4z}$
Thus, $X\left(z\right)=\frac{1}{64}\frac{{z}^{-3}}{1-4z}$
Region of convergence $|4z|<1⇒|z|<\frac{1}{4}$
$⇒1-4z=0⇒z=\frac{1}{4}$ is pole

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