Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a little problem computing the inverse of this signal:

$$X(z) = \frac{(z-1)(z+\frac{3}{2})}{(z+\frac{j}{2})(z-\frac{j}{2})(z-\frac{1}{2})}$$ $$X(z^{-1}) = ?$$

I know how to take the inverse, $z=z^{-1}$ and then multiply the brackets and so on... But my problem is the numbers in the brackets, $1, -\frac{3}{2}, \frac{j}{2}$, are poles and zeros of a Pole-zero plot of the sequence $X(z)$.
And I think I loose that information, don't I?

share|cite|improve this question
up vote 1 down vote accepted

if $X(z)$ has a zero/pole at $z=w$ then $X(z^{-1}$ will have a zero at $z=w^{-1}$, so you don't lose any information about the poles and zeros. You end up with:

$$\begin{align} X(z^{-1}) & = \frac{(z^{-1}-1)(z^{-1}+\tfrac{3}{2})}{(z^{-1}+\tfrac{j}{2})(z^{-1}-\tfrac{j}{2})(z^{-1}-\tfrac{1}{2})} \\ & = \frac{z(1-z)(1+\tfrac{3}{2}z)}{(1+\tfrac{j}{2}z)(1-\tfrac{j}{2}z)(1-\tfrac{1}{2}z)} \\ & = \frac{-\tfrac{3}{2}z(z-1)(z+\tfrac{2}{3})}{\tfrac{j^2}{8}(z + \tfrac{2}{j})(z - \tfrac{2}{j})(z-2)} \\ & = -\frac{12}{j^2} \frac{z(z-1)(z+\tfrac{2}{3})}{(z + \tfrac{2}{j})(z - \tfrac{2}{j})(z-2)} \\ \end{align}$$

Note that $X(z)$ had a zero at infinity, and hence $X(z^{-1})$ has a zero at zero.

share|cite|improve this answer
You were faster, I erased my answer. ;-) – Luboš Motl Jun 26 '11 at 11:22
how about to replace $-1/j^2$ by $1$? – Ilya Jun 26 '11 at 12:26
Ok, so $-\frac{12}{j^2}=12$. When I need to plot the poles and zeros, what effect does the $12$ have on the plot or can I ignore it? And where is $\frac{2}{j}$ on the imaginary axis? – madmax Jun 26 '11 at 15:09
@Gortaur I wasn't sure that $j^2=-1$ (not being an engineer ;)) so I left it in this form. – Chris Taylor Jun 26 '11 at 18:38
@madmax To find $\frac {2} {j}$ multiply both numerator and denominator by $j$, giving $-2j$ which lies on $-2$ on the imaginary axis. – mbaitoff Jul 11 '11 at 8:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.