Topic: Computing an inverse z-transform

## Question

Compute the inverse z-transform of

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

Ruofei

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

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

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

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

Since $|2|<Z<|3|$

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

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

Thus,

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

$X(Z) = -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}$

$X(Z) = -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} -\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1}$

In $-\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n}$, Let k=-n, then -k=n

In $\frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}$, Let i=n+1, then n=i-1

$-\sum_{n=-\infty}^{+\infty} u[n] (\frac{1}{3})^{n+1} Z^{n}-\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1}$

$-\sum_{n=-\infty}^{+\infty} u[-k] (\frac{1}{3})^{-k+1} Z^{-k}-\sum_{n=-\infty}^{+\infty} u[i-1] 2^{i-1} Z^{-i}$

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

Grader's comment: Made a mistake in the last step . It should be 2 instead of 3

Li-Pang Mo

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

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

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

$|2|<Z<|3|$, which makes $\frac{z}{3}<1, \frac{2}{z}<1$

Use geometric series:

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

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

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

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

$let p = -n , n = -p, q = n+1 , n = q-1$

$X(z) = -\sum_{n=-\infty}^{+\infty} u[-p] (z)^{-p} (\frac{1}{3})^{-p+1} - \sum_{n=-\infty}^{+\infty} u[q-1] (2)^{q-1} (z)^{q-1}$

By observation:

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

Write it here.