| Line 29: | Line 29: | ||
<math>X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}}</math> | <math>X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}}</math> | ||
| − | Since | + | Since <math>|2|<Z<|3|</math> |
<math>\frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> | <math>\frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> | ||
| Line 37: | Line 37: | ||
Thus, | Thus, | ||
| − | <math>X(Z) = \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> | + | <math>X(Z) = -\sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n}</math> |
| − | <math>X(Z) = \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}</math> | + | <math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}</math> |
| − | + | <math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + -\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1}</math> | |
| + | |||
| + | In <math>-\sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n}</math>, Let k=-n, then -k=n | ||
| + | |||
| + | In <math>\frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n}</math>, Let k=-n, then -k=n | ||
<math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[k-1] 3^{k-1}Z^{-k}</math> | <math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[k-1] 3^{k-1}Z^{-k}</math> | ||
Revision as of 14:36, 19 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)(2-z)}, \quad \text{ROC} \quad 2<|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
Ruofei
$ X(Z) = \frac{1}{(3-Z) (2-Z)} $
$ X(Z) = -\frac{1}{3-Z} + \frac{1}{2-Z} $
$ X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} + \frac{1}{Z} \frac{1}{\frac{2}{Z}-1} $
$ X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}} $
Since $ |2|<Z<|3| $
$ \frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n} $
$ \frac{1}{1-\frac{Z}{3}} = \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} $
Thus,
$ X(Z) = -\sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n} $
$ X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n} $
$ X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} + -\sum_{n=-\infty}^{+\infty} u[n] 2^{n} Z^{-n-1} $
In $ -\sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} $, Let k=-n, then -k=n
In $ \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n} $, Let k=-n, then -k=n
$ X(Z) = -\sum_{n=-\infty}^{+\infty} u[k-1] 3^{k-1}Z^{-k} $
Therefore, $ x(n) = -u[n-1] 3^{n-1} $
Answer 2
Write it here.
Answer 3
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Answer 4
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