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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
alec green
an exponential can be expanded into the series:
$ e^{x} = \sum_{n=0}^{+\infty}\frac{x^{n}}{n!} $
$ X(z) = e^{-2z} = \sum_{n=0}^{+\infty}(\frac{(-2z)^{n}}{n!} = \frac{(-2)^{n}}{n!}z^{n}) $
$ = \sum_{n=-\infty}^{+\infty}u[n]\frac{(-2)^{n}}{n!}z^{n} $
letting k = -n:
$ = \sum_{k=-\infty}^{+\infty}u[-k]\frac{(-2)^{-k}}{(-k)!}z^{-k} $
and by comparison with:
$ X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n} $
$ x[n] = u[-n]\frac{(-2)^{-n}}{(-n)!} $
due to the step function in the x[n], the factiorial in x[n] is never evaluated on a negative argument (which would be undefined).
Answer 3
Write it here.
Answer 4
Xiang Zhang
From the formula of exponential function of Taylor series we can find that
$ e^x = \sum_{ n = 0 }^{+ \infty} \frac{x^n}{n!} $
Hence we can find in our expression that
$ x = 2z $
