Practice Question, ECE438 Fall 2013, Prof. Boutin

On computing the inverse z-transform of a discrete-time signal.

Compute the inverse z-transform of

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

(Write enough intermediate steps to fully justify your answer.)

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

Answer 1

$ 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 + n}u[ − n]

Answer 2

$ 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] = 3^{n − 1}u[ − n]

Answer 3

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] = 3^{n − 1}u[ − n]

Answer 4

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] $

Answer 5

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] = 3^{n − 1}u[ − n]

Answer 6 - Ryan Atwell

$ 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] = 3^{n − 1}u[ − n]

Answer 7

$ 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] $

Answer 8

$ 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 + k}u[ − k]

Back to ECE438 Fall 2013 Prof. Boutin </math>

Answer 8

$ 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] $

Answer 9

$ 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] $

Answer 10

$ 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]3^{n − 1}

Answer 11

$ 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] $