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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.)
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 + nu[ − 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] = 3n − 1u[ − 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] = 3n − 1u[ − 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] = 3n − 1u[ − 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] $
