Topic: Computing an inverse z-transform

## Question

Compute the inverse z-transform of

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

$X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n}$

$= \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}$
NOTE: Let n=-k

$= \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}$ (compare with $\sum_{n=-\infty}^{+\infty} x[n] z^{-k}$)

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

Therefore, x[n] = 3 − 1 + nu[ − n]

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

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

Let n = -k

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

By comparison with the x-transform formula,

x[n] = 3n − 1u[ − n]

By Yeong Ho Lee

$X[z] = \sum_{n=0}^{\infty}z^{n}3^{-n-1}$

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

Now, let n = -k

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

Using the z-transform formula, x[n] = 3n − 1u[ − n]

Gena Xie

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

since |z|<3,

|z|/3 < 1

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

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

substitute n by -n,

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

based on the definition

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

By Yixiang Liu

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

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

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

by geometric series

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

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

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

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

By comparison with the x-transform formula

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

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

x[n] = 3n − 1u[ − n]

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

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

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

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

n=-k

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

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

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

by formula

X[n] = 3n − 1u[ − n]

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

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

$X(z) = \frac{1}{3} \frac{1-\left( \frac{z}{3} \right) ^n}{1- \frac{z}{3}}$ $, \quad \text{As n goes to} \quad \infty \quad \text{since} \quad |z|<3$

$X(z) = \frac{1}{3} \sum_{n=0}^{\infty} \left( \frac{z}{3} \right)^n$

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

$X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} \left(\frac{1}{3} \right)^{-n} u[-n] z^{-n}$ $, \quad \text{By comparing with DTFT equation, we get} \quad$

$x[n] = \frac{1}{3} * \frac{1}{3}^{-n} u[-n]$

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

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

$=\frac{1}{3}\frac{1}{1-\frac{z}{3}}$

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

let n=-k

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

by comparison to inverse z-transform formula,

x[n] = 3 − 1 + ku[ − k]

Grader's comment: Right hand side is in 'k' while the left hand side in 'n'

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

we have |z| < 3, so

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

sum will look like this:

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

with unit step:

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

substituting n with -k we get:

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

finally we get:

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

using the formula we get:

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

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

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

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

let -k = n,

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

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

so by comparison $, x[n] = (\frac{1}{3})^{-n+1} u[-n]$

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

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

Since z/3 < |1|, base on geometric series:

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

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

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

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

$X(z) =\sum_{m=-\infty}^{\infty} {z}^{-m} u[-m]{3}^{m-1}$

base on observation: x[n] = u[ − n]3n − 1

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

for $\quad \text{ROC} \quad |z|<3$

we can use geometric series

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

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

so, by comparing to z-transform formula,we have

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