Practice Question on "Digital Signal Processing"
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)
Answer 1
$ 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.
Answer 2
$ \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.
Answer 3
$ 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 $.
- Another Instructor's comment: Can you state your answer in simplified form?
Answer 4
$ \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 $
- Instructor's comment: Looks good!
