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We also know that the Z transform of an impulse <math>\delta (n - n0)</math> is:
 
We also know that the Z transform of an impulse <math>\delta (n - n0)</math> is:
  
<math></math>
+
<math> X(z) = \sum_{{n=-\infty}{+\infty}} delta (n - n0) z^{-n}</math>
  
 
<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>
 
<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>

Revision as of 18:44, 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

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

By Taylor Series,

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

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} $

$ 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 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] $




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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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