| Line 37: | Line 37: | ||
<math>x[n]=u[-n](-3^{n-1}-2^{n-1})</math> | <math>x[n]=u[-n](-3^{n-1}-2^{n-1})</math> | ||
=== Answer 2=== | === Answer 2=== | ||
| − | + | Kyungjun Kim | |
| + | Using a partial fraction expansion, we can change the original equation to | ||
| + | <math>X(z) = \frac{A}{3-z}+\frac{B}{2-z}</math> | ||
| + | Where A = 1, B = -1, so we get | ||
| + | <math>= -\frac{1}{3-z}-\frac{1}{2-z}</math> | ||
| + | By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of <math> \frac{1}{1-r} </math>, which is equal to <math> sum_{n=0}^{+\infty} (\frac{1}{r})^n </math> | ||
| + | <math>= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}})</math> | ||
| + | <math>= -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n</math> | ||
| + | <math>= \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n</math> | ||
| + | Then let k=-n | ||
| + | <math>= \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k}</math> | ||
| + | Comparing it with z-transform formula, we can get | ||
| + | |||
| + | <math>x[n]=u[-n](-3^{n-1}-2^{n-1})</math> | ||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
Revision as of 15:52, 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 |z|<2 $.
(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{A}{3-z}+\frac{B}{2-z} $
$ = -\frac{1}{3-z}-\frac{1}{2-z} $
$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $
$ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $
Let k=-n
$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $
by comparison with z-transform formula
$ x[n]=u[-n](-3^{n-1}-2^{n-1}) $
Answer 2
Kyungjun Kim Using a partial fraction expansion, we can change the original equation to $ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ Where A = 1, B = -1, so we get $ = -\frac{1}{3-z}-\frac{1}{2-z} $ By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of $ \frac{1}{1-r} $, which is equal to $ sum_{n=0}^{+\infty} (\frac{1}{r})^n $ $ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $ $ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $ $ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $ Then let k=-n $ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $ Comparing it with z-transform formula, we can get
$ x[n]=u[-n](-3^{n-1}-2^{n-1}) $
Answer 3
Write it here.
Answer 4
Write it here.
