Practice Question on "Digital Signal Processing"

Topic: Computing an inverse z-transform


Question

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

$ 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] $

Grader's comment: Correct Answer

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).

Grader's comment: Correct Answer

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] $

Grader's comment: Correct Answer

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] $

Grader's comment: Correct Answer

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] $

Grader's comment: Correct Answer







Back to ECE438 Fall 2013 Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang