Topic: Computing an inverse z-transform

## Question

Compute the inverse z-transform of

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

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

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.)

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

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).

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

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:

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

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]

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

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...

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