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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
$ 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^{-2*z} = \sum_{k = 0}^{+ \infty} \frac{-(2z)^k}{k!} $
$ = \sum_{k = 0}^{+ \infty} \frac{ (-2 * z) ^k}{k!} $
Substitute -n for k,
$ = \sum_{ n = - \infty }^{+ \infty} \frac{ (- 2 * z) ^-n}{(-n)!} u[-n] $
Change the integration and reorder the expression
$ e^{-2z} = \sum_{ k = - \infty }^{+ \infty} \frac{ (-2) ^{-n}}{(-n)!} z^{-n} u[-n] $
Compare with the Z transform equation x[n]
$ x[n] = \frac{ (-2) ^{-n}}{(-n)!} u[-n] $
