# Zeros and poles from transfer function

I have a transfer function $$H(z) = \frac{Y(z)}{X(z)} = 1 - 0.5z^{-1} \text{.}$$ I'm interested in zeros and poles. I know I need to adjust the function to $$H(z) = \frac{\prod_i(z-n_i)}{\prod_i(z-p_i)} \text{.}$$ My attempt is as follows: $$H(z) = z^{-1} (z - 0.5)$$ So I guess a zero is at $$z = 0.5$$. Is it correct? Moreover, what should I do with $$z^{-1}$$? Ignore it?

Thank you.

• why would you ignore it? You've found your zeros, great. What about your poles? – Marcus Müller Jun 7 at 13:17
• @MarcusMüller Oh, I didn't think of that. So there is a pole at $z=0$, right? – DaBler Jun 7 at 13:59
• Well, let's try that out! What happens to the original $H(z)$ when you set $z=0$? – Marcus Müller Jun 7 at 13:59
• another way to write $z^{-1}$ is $$\frac{1}{z-0}$$ – robert bristow-johnson Jun 7 at 14:46

If you put the transfer function into the $$z$$ form, you get $$H(z) = \frac{Y(z)}{X(z)} = \frac{z - 0.5}{z}$$
Then you can immediately see that $$Y(z) = z - 0.5$$, and $$X(z) = z$$. Thus, by inspection, the transfer function has a pole at $$z = 0$$, and a zero at $$z = 0.5$$.