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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.)

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Answer 1

$ 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, (Instructor's comment: Definition of what? Be clear.)

$ 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

$ X(z) = e^{-2z}. $

By Taylor Series,

$ X(z) = e^{-2z} = \sum_{n=0}^{+\infty}\frac{{(-2 z)} ^ {n}}{n!} = 1 - 2 z + \frac{4 z^2}{2!} - \frac{8 z^3}{3!} ... $

We also know that the Z transform of an impulse $ \delta (n - n0) $ is:

$ X(z) = \sum_{n=-\infty}^{+\infty}\delta (n - n0) z^{-n} = z^{-n0} $

Therefore the inverse Z Transform of the signal will be given by:

$ n[n] = \delta[n] - 2 \delta[n+1] + \frac{4 \delta[n+2]}{2!} - \frac{8 \delta[n+3]}{3!} ... = \sum_{k=0}^{+\infty}\frac{(-2)^k}{k!} \delta[n+k] $

Since we know that a discrete signal can be written as a weighted impulse train, we can alternatively rewrite x[n] as:

$ x[n] = \frac{(-2)^n}{(-n)!} u[-n] $

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 $

Let's expand the original signal to the expression below

$ e^{-2z} = \sum_{ n = 0 }^{+ \infty} \frac{ (- 2 z) ^n}{n!} $

Replace $ n = 0 $ to $ n = - \infty $ by introducing u[n].

We can get that,

$ e^{-2z} = \sum_{ n = - \infty }^{+ \infty} \frac{ (- 2 z) ^n}{n!} u[n] $

Substitute in n = -k (k = -n)

$ e^{-2z} = \sum_{ k = + \infty }^{- \infty} \frac{ (- 2 z) ^{-k}}{(-k)!} u[-k] $

Change the integration and reorder the expression

$ e^{-2z} = \sum_{ k = - \infty }^{+ \infty} \frac{ u[-k] (-2) ^{-k}}{(-k)!} z^{-k} $

By comparison with original expression

$ X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n} $

Substitute k back to n (k = n)

$ e^{-2z} = \sum_{ n = - \infty }^{+ \infty} \frac{ u[-n] (-2) ^{-n}}{(-n)!} z^{-n} $

Then we can recover back x[n]

$ x[n] = \frac{ (-2) ^{-n}}{(-n)!} u[-n] $

Answer 5

By taylor series

$ e^{-2z} = \sum_{k = 0}^{+ \infty} \frac{(-2z)^k}{k!} $

$ = \sum_{k = 0}^{+ \infty} \frac{ (-2z) ^k}{k!} $

Substitute -n for k,

$ = \sum_{ n = - \infty }^{+ \infty} \frac{ (-2z)^{-n}}{(-n)!} u[-n] $

Pull z^-n out of expression,

$ = \sum_{ k = - \infty }^{+ \infty} \frac{ (-2) ^{-n}}{(-n)!} z^{-n} u[-n] $

Compare with the Z transform equation from RHEA table to get...

$ x[n] = \frac{ (-2) ^{-n}}{(-n)!} u[-n] $

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