Topic: Computing an inverse z-transform

## Question

Compute the inverse z-transform of

$X(z) =\frac{1}{(3-z)(2-z)}, \quad \text{ROC} \quad |z|>3$.

$X(z) =\frac{1}{(\frac{3z}{z}-z)(\frac{2z}{z}-z)} \quad$

$=-\frac{1}{z}\frac{1}{1-\frac{3}{z}}(-\frac{1}{z}\frac{1}{1-\frac{2}{z}}) \quad$

$=(\sum_{n=0}^{+\infty}-\frac{1}{z}(\frac{3}{z})^n)(\sum_{n=0}^{+\infty}-\frac{1}{z}(\frac{2}{z})^n)$

$=(-\sum_{n=0}^{+\infty}3^nz^{-n-1})(-\sum_{n=0}^{+\infty}2^nz^{-n-1})$

$=(-\sum_{n=-\infty}^{+\infty}3^nu[n]z^{-n-1})(-\sum_{n=-\infty}^{+\infty}2^nu[n]z^{-n-1})$

Let $n=k-1$

$=(-\sum_{k=-\infty}^{+\infty}3^nu[k-1]z^{-k})(-\sum_{k=-\infty}^{+\infty}2^nu[k-1]z^{-k})$

By observing that $X(z) =\sum_{n=-\infty}^{+\infty}x[n]z^{-n}$

$x[n] =(-3^{n-1}u[n-1])(-2^{n-1}u[n-1])$

$=6^{n-1}u[n-1]$

Grader's comment: You should use partial fractions to split up into two parts

alec green

$X(z) = \frac{1}{(3-z)(2-z)} = \frac{A}{(3-z)} + \frac{B}{(2-z)} = -\frac{1}{(3-z)} + \frac{1}{(2-z)}$

given the ROC, rewrite as:

$= -(\frac{-1}{z})(\frac{1}{1-\frac{3}{z}}) + (\frac{-1}{z})(\frac{1}{1-\frac{2}{z}}) = (\frac{1}{z})(\frac{1}{1-\frac{3}{z}}) - (\frac{1}{z})(\frac{1}{1-\frac{2}{z}})$

$= \sum_{n=0}^{+\infty}\frac{1}{z}(\frac{3}{z})^{n} - \sum_{n=0}^{+\infty}\frac{1}{z}(\frac{2}{z})^{n}$

$= \sum_{n=-\infty}^{+\infty}u[n]3^{n}z^{-n-1} - \sum_{n=-\infty}^{+\infty}u[n]2^{n}z^{-n-1}$

letting -k = -n-1, and therefore n = k-1:

$= \sum_{k=-\infty}^{+\infty}u[k-1]3^{k-1}z^{-k} - \sum_{k=-\infty}^{+\infty}u[k-1]2^{k-1}z^{-k}$

$= \sum_{k=-\infty}^{+\infty}u[k-1](3^{k-1} - 2^{k-1})z^{-k}$

finally, by comparison with:

$X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n}$

$x[n] = u[n-1](3^{n-1} - 2^{n-1})$

Write it here.