Topic: Computing a z-transform

## Question

Compute the z-transform of the following signal.

$x[n]=u[n]$

Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --Cmcmican 22:05, 16 April 2011 (UTC)

$X(z)=\sum_{n=-\infty}^\infty u[n]z^{-n}=\sum_{n=0}^\infty z^{-n}$

$X(z)=\frac{z}{z-1} \mbox{, ROC: }\Big|z\Big|>1$

--Cmcmican 22:05, 16 April 2011 (UTC)

TA's comment: Correct!
Instructor's comment: Exactly where do you get that the norm of z must be greater than one for convergence? It is important to clearly state it.

\begin{align} X(z) &= \sum_{n=-\infty}^{\infty}u[n]z^{-n} \\ &= \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n \\ &= \begin{cases} \frac{1}{1-\frac{1}{z}}, & |z| > 1 \\ diverges, & else \end{cases} \end{align}

If $z \leq 1$ then $\frac{1}{z} \geq 1$, then the sum would diverge.

Instructor's comment: You forgot the norm around z in the last line. The statement z<1 does not make any sense in the complex plane, so you would lose points for that.

$X(z) = \sum_{n=-\infty}^{\infty}u[n]z^{-n} = \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n$

So if $|\frac{1}{z}| \leq 1$, then by the geometric series formula we have $X(z)=\frac{1}{1-\frac{1}{z}}$.

On the other hand, if $|\frac{1}{z}| > 1$, then X(z) diverges.

Instructor's comment: The series does not converge when $|\frac{1}{z}| =1$.

\begin{align} X(z) &= \sum_{n=-\infty}^{\infty}u[n]z^{-n} \\ &= \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n \\ &= \begin{cases} \frac{1}{1-\frac{1}{z}}, & |z| > 1 \\ diverges, & else \end{cases} \end{align}
So $X(z)= \frac{1}{1-\frac{1}{z}}$ with ROC $|z|>1$