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===Answer 1=== | ===Answer 1=== | ||
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| + | Gena Xie | ||
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| + | <math>X(z) = e^{-2z}. \</math> | ||
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| + | By Taylor Series, | ||
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| + | <math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty}{\frac{{-2z}^n}{n!}}<\math> | ||
| + | |||
| + | substitute n by -n | ||
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| + | <math>X(z) = \sum_{-\infty}^{n=0){\frac{{-2}^-n}{{-n!}}}{z^{-n}} = \sum_{-\infty}^{+\infty){\frac{{-2}^-n}{{-n!}}}u[-n]{z^{-n}}<\math> | ||
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| + | based on the definition, | ||
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| + | <math>X(z) = \frac{{-2}^-n}{{-n!}}}u[-n] | ||
=== Answer 2=== | === Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 16:09, 19 September 2013
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) = e^{-2z}. \ $
(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
Gena Xie
$ X(z) = e^{-2z}. \ $
By Taylor Series,
$ X(z) = e^{-2z} = \sum_{n=0}^{+\infty}{\frac{{-2z}^n}{n!}}<\math> substitute n by -n <math>X(z) = \sum_{-\infty}^{n=0){\frac{{-2}^-n}{{-n!}}}{z^{-n}} = \sum_{-\infty}^{+\infty){\frac{{-2}^-n}{{-n!}}}u[-n]{z^{-n}}<\math> based on the definition, <math>X(z) = \frac{{-2}^-n}{{-n!}}}u[-n] === Answer 2=== Write it here. ===Answer 3=== Write it here. ===Answer 4=== Write it here. ---- ---- [[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]] $
