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[[Category:inverse z-transform]] | [[Category:inverse z-transform]] | ||
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| − | + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | |
| + | </font size> | ||
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| + | Topic: Computing an inverse z-transform | ||
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| + | </center> | ||
---- | ---- | ||
| + | ==Question== | ||
Compute the inverse z-transform of | Compute the inverse z-transform of | ||
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<math> =6^{n-1}u[n-1] </math> | <math> =6^{n-1}u[n-1] </math> | ||
| + | |||
| + | :<span style="color:blue"> Grader's comment: You should use partial fractions to split up into two parts </span> | ||
| + | |||
=== Answer 2=== | === Answer 2=== | ||
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<math>= \sum_{n=-\infty}^{+\infty}u[n]3^{n}z^{-n-1} - \sum_{n=-\infty}^{+\infty}u[n]2^{n}z^{-n-1}</math> | <math>= \sum_{n=-\infty}^{+\infty}u[n]3^{n}z^{-n-1} - \sum_{n=-\infty}^{+\infty}u[n]2^{n}z^{-n-1}</math> | ||
| + | |||
| + | letting -k = -n-1, and therefore n = k-1: | ||
| + | |||
| + | <math>= \sum_{k=-\infty}^{+\infty}u[k-1]3^{k-1}z^{-k} - \sum_{k=-\infty}^{+\infty}u[k-1]2^{k-1}z^{-k}</math> | ||
| + | |||
| + | <math>= \sum_{k=-\infty}^{+\infty}u[k-1](3^{k-1} - 2^{k-1})z^{-k}</math> | ||
| + | |||
| + | finally, by comparison with: | ||
| + | |||
| + | <math>X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n}</math> | ||
| + | |||
| + | <math>x[n] = u[n-1](3^{n-1} - 2^{n-1})</math> | ||
| + | |||
| + | :<span style="color:blue"> Grader's comment: Correct Answer </span> | ||
===Answer 3=== | ===Answer 3=== | ||
Latest revision as of 11:54, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Computing an inverse z-transform
Question
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] $
- Grader's comment: You should use partial fractions to split up into two parts
Answer 2
alec green
$ X(z) = \frac{1}{(3-z)(2-z)} = \frac{A}{(3-z)} + \frac{B}{(2-z)} = -\frac{1}{(3-z)} + \frac{1}{(2-z)} $
given the ROC, rewrite as:
$ = -(\frac{-1}{z})(\frac{1}{1-\frac{3}{z}}) + (\frac{-1}{z})(\frac{1}{1-\frac{2}{z}}) = (\frac{1}{z})(\frac{1}{1-\frac{3}{z}}) - (\frac{1}{z})(\frac{1}{1-\frac{2}{z}}) $
$ = \sum_{n=0}^{+\infty}\frac{1}{z}(\frac{3}{z})^{n} - \sum_{n=0}^{+\infty}\frac{1}{z}(\frac{2}{z})^{n} $
$ = \sum_{n=-\infty}^{+\infty}u[n]3^{n}z^{-n-1} - \sum_{n=-\infty}^{+\infty}u[n]2^{n}z^{-n-1} $
letting -k = -n-1, and therefore n = k-1:
$ = \sum_{k=-\infty}^{+\infty}u[k-1]3^{k-1}z^{-k} - \sum_{k=-\infty}^{+\infty}u[k-1]2^{k-1}z^{-k} $
$ = \sum_{k=-\infty}^{+\infty}u[k-1](3^{k-1} - 2^{k-1})z^{-k} $
finally, by comparison with:
$ X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n} $
$ x[n] = u[n-1](3^{n-1} - 2^{n-1}) $
- Grader's comment: Correct Answer
Answer 3
Write it here.
Answer 4
Write it here.
