| Line 11: | Line 11: | ||
Compute the inverse z-transform of | Compute the inverse z-transform of | ||
| − | <math>X(z) = e^{-2z}. | + | <math>X(z) = e^{-2z}. </math> |
(Write enough intermediate steps to fully justify your answer.) | (Write enough intermediate steps to fully justify your answer.) | ||
| Line 22: | Line 22: | ||
Gena Xie | Gena Xie | ||
| − | <math>X(z) = e^{-2z}. | + | <math>X(z) = e^{-2z}. </math> |
By Taylor Series, | By Taylor Series, | ||
| − | <math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty} | + | <math>X(z) = e^{-2z} = \sum_{n=0}^{+\infty}\frac{{(-2 z)} ^ {n}}{n!}</math> |
substitute n by -n | substitute n by -n | ||
| − | <math>X(z) = \sum_{-\infty}^{ | + | <math>X(z) = \sum_{n=-\infty}^{0}\frac { { (-2) } ^ {-n} } { (-n)! } {z^{-n}} = \sum_{n=-\infty}^{+\infty} \frac { { (-2) } ^ {-n} } { (-n)! } u[-n]{z^{-n}}</math> |
based on the definition, | based on the definition, | ||
| − | <math>X(z) = \frac{{-2}^-n}{ | + | <math>X(z) = \frac{ {(-2)} ^ {-n} } { (-n)! } u[-n] </math> |
| + | |||
=== Answer 2=== | === Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 15:50, 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) = 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{{(-2 z)} ^ {n}}{n!} $
substitute n by -n
$ X(z) = \sum_{n=-\infty}^{0}\frac { { (-2) } ^ {-n} } { (-n)! } {z^{-n}} = \sum_{n=-\infty}^{+\infty} \frac { { (-2) } ^ {-n} } { (-n)! } u[-n]{z^{-n}} $
based on the definition,
$ X(z) = \frac{ {(-2)} ^ {-n} } { (-n)! } u[-n] $
Answer 2
Write it here.
Answer 3
Write it here.
Answer 4
Write it here.
