| Line 23: | Line 23: | ||
<math> =\frac{-1}{z} \frac{1}{1-\frac{3}{z}} </math> | <math> =\frac{-1}{z} \frac{1}{1-\frac{3}{z}} </math> | ||
<math> =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n </math> | <math> =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n </math> | ||
| − | <math> =\sum_{n=0}^{+\infty} (-z^- | + | <math> =\sum_{n=0}^{+\infty} (-z^-1)) (3z^-1)^n </math> |
| − | NOTE: <math> (3z^-1)^n = (3^n) (z^- | + | NOTE: <math> (3z^-1)^n = (3^n) (z^-n) (-z^-1) </math> |
Revision as of 13:56, 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) (-z^-1) $
Answer 2
Write it here.
Answer 3
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Answer 4
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