| Line 25: | Line 25: | ||
<math> = \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1} </math> | <math> = \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1} </math> | ||
NOTE: Let k=n+1 | NOTE: Let k=n+1 | ||
| − | <math> = -3^{k-1} u[k-1] z^{-k} | + | |
| + | <math> = -3^{k-1} u[k-1] z^{-k}</math> | ||
Therefore, <math> x[n]= -3^{n-1} u[n-1] </math> | Therefore, <math> x[n]= -3^{n-1} u[n-1] </math> | ||
Revision as of 14:07, 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) =\frac{1}{\frac{3z}{z} - z} = \frac{-1}{z} \frac{1}{1-\frac{3}{z}} $
$ =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n = \sum_{n=0}^{+\infty} (-z^{-1}) (3z^{-1})^n $ NOTE: $ (3z^{-1})^n = (3^n) (z^{-n}) $ $ = \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1} $ NOTE: Let k=n+1
$ = -3^{k-1} u[k-1] z^{-k} $
Therefore, $ x[n]= -3^{n-1} u[n-1] $
Answer 2
Write it here.
Answer 3
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Answer 4
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