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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.

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    $\begingroup$ why would you ignore it? You've found your zeros, great. What about your poles? $\endgroup$ Jun 7, 2021 at 13:17
  • $\begingroup$ @MarcusMüller Oh, I didn't think of that. So there is a pole at $z=0$, right? $\endgroup$
    – DaBler
    Jun 7, 2021 at 13:59
  • $\begingroup$ Well, let's try that out! What happens to the original $H(z)$ when you set $z=0$? $\endgroup$ Jun 7, 2021 at 13:59
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    $\begingroup$ another way to write $z^{-1}$ is $$ \frac{1}{z-0} $$ $\endgroup$ Jun 7, 2021 at 14:46

1 Answer 1

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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 $z = 0.5$.

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