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===Answer 1=== | ===Answer 1=== | ||
| − | + | <math>X(z) =\frac{1}{(\frac{3z}{z}-z)(\frac{2z}{z}-z)} \quad </math> | |
| + | |||
| + | <math> =-\frac{1}{z}\frac{1}{1-\frac{3}{z}}(-\frac{1}{z}\frac{1}{1-\frac{2}{z}}) \quad </math> | ||
| + | |||
| + | <math> =(\sum_{n=0}^{+\infty}-\frac{1}{z}(\frac{3}{z})^n)(\sum_{n=0}^{+\infty}-\frac{1}{z}(\frac{2}{z})^n) </math> | ||
| + | |||
| + | <math> =(-\sum_{n=0}^{+\infty}3^nz^{-n-1})(-\sum_{n=0}^{+\infty}2^nz^{-n-1}) </math> | ||
| + | |||
| + | <math> =(-\sum_{n=-\infty}^{+\infty}3^nu[n]z^{-n-1})(-\sum_{n=-\infty}^{+\infty}2^nu[n]z^{-n-1}) </math> | ||
| + | |||
| + | Let <math> n=k-1 </math> | ||
| + | |||
| + | <math> =(-\sum_{k=-\infty}^{+\infty}3^nu[k-1]z^{-k})(-\sum_{k=-\infty}^{+\infty}2^nu[k-1]z^{-k}) </math> | ||
| + | |||
| + | By observing that <math> X(z) =\sum_{n=-\infty}^{+\infty}x[n]z^{-n} </math> | ||
| + | |||
| + | <math>x[n] =(-3^{n-1}u[n-1])(-2^{n-1}u[n-1]) </math> | ||
| + | |||
| + | <math> =6^{n-1}u[n-1] </math> | ||
=== Answer 2=== | === Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 13:31, 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|>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{2z}{z}-z)} \quad $
$ =-\frac{1}{z}\frac{1}{1-\frac{3}{z}}(-\frac{1}{z}\frac{1}{1-\frac{2}{z}}) \quad $
$ =(\sum_{n=0}^{+\infty}-\frac{1}{z}(\frac{3}{z})^n)(\sum_{n=0}^{+\infty}-\frac{1}{z}(\frac{2}{z})^n) $
$ =(-\sum_{n=0}^{+\infty}3^nz^{-n-1})(-\sum_{n=0}^{+\infty}2^nz^{-n-1}) $
$ =(-\sum_{n=-\infty}^{+\infty}3^nu[n]z^{-n-1})(-\sum_{n=-\infty}^{+\infty}2^nu[n]z^{-n-1}) $
Let $ n=k-1 $
$ =(-\sum_{k=-\infty}^{+\infty}3^nu[k-1]z^{-k})(-\sum_{k=-\infty}^{+\infty}2^nu[k-1]z^{-k}) $
By observing that $ X(z) =\sum_{n=-\infty}^{+\infty}x[n]z^{-n} $
$ x[n] =(-3^{n-1}u[n-1])(-2^{n-1}u[n-1]) $
$ =6^{n-1}u[n-1] $
Answer 2
Write it here.
Answer 3
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Answer 4
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