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<math> = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math> | <math> = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math> | ||
| − | Let n=-k | + | NOTE: Let n=-k |
<math> = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math> | <math> = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math> | ||
Revision as of 14:11, 18 September 2013
Contents
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
Write it here.
Answer 3
Write it here.
Answer 4
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