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===Answer 4=== | ===Answer 4=== | ||
Write it here. | Write it here. | ||
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| + | === Answer 5 === | ||
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| + | <math>X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math> | ||
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| + | <math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n </math> | ||
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| + | Let n = -k | ||
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| + | <math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} </math> | ||
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| + | By comparison with the x-transform formula, | ||
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| + | <math>x[n] = 3^{n-1} u[-n] </math> | ||
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---- | ---- | ||
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[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]] | [[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]] | ||
Revision as of 15:18, 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] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n} $
$ = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n} $
NOTE: Let n=-k
$ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $ (compare with $ \sum_{n=-\infty}^{+\infty} x[n] z^{-k} $)
$ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $
Therefore, $ x[n]= 3^{-1+n} u[-n] $
Answer 2
Write it here.
Answer 3
Write it here.
Answer 4
Write it here.
Answer 5
$ X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} $
$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n $
Let n = -k
$ X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} $
By comparison with the x-transform formula,
$ x[n] = 3^{n-1} u[-n] $
