Revision as of 18:29, 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.)

Share your answers below

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 $

Back to ECE438 Fall 2013 Prof. Boutin