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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+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
Write it here.
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] $
