# How to take the inverse of this signal?

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?

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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.
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