Revision as of 22:47, 19 September 2013

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.)

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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] $