# Poles and zeroes - $\mathcal Z$-transform

I just have a small question, something that I am unsure about. I have a difference equation for a filter:

$$y[n] = x[n] - x[n-1]$$

I have worked out the $\mathcal Z$-transform:

$$\mathcal Z\left\{y[n]\right\}= Y(z) = X(z)\left(1 - z^{-1}\right)$$ \begin{align} H(z) &= \frac{Y(z)}{X(z)}\\ &= 1 - z^{-1}\\ &= 1 - \frac 1z\\ &= z - \frac zz\\ &= z - 1 \end{align}

There is no denominator? Technically, there is: $\displaystyle H(z) = \frac{z - 1}{1}$. Does this simply mean that there are no poles?

## 1 Answer

Your term manipulation is incorrect:

\begin{align} H(z)&=1-z^{-1}\\ &= \left(1-z^{-1}\right)\frac{z}{z}\\ &= \frac{z-1}{z} \end{align}

So, there is a pole at $z=0$, which you can also see from the initial equation $H(z)=1-z^{-1}$ if you insert $z=0$.

• Silly mistake on my part! Makes sense now. Thanks for your time! – embedded.95 Feb 1 '17 at 18:15