0
$\begingroup$

When finding the poles of something like the following transfer function, would I be able to write $z=\sqrt[L]{\mu}$ since square roots aren't technically defined on the complex plane?

$$Y(z) = \frac{a_L z^L + a_{L-1}z^{L-1} + ... +a_0}{z^L - \mu}$$

where $a_j,\mu \in\mathbb{C}$, $|\mu|<1$ and L is positive.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ You're mistaken if you think that roots of complex numbers aren't defined: link $\endgroup$ – Matt L. Mar 4 '20 at 14:09
  • $\begingroup$ I tried finding the roots of x^5 - 1 = 0 on Wolfram Alpha There is the obvious solution x = 1 and 4 non-obvious solutions x~~-0.30902 ±0.95106 i x~~0.80902 ± 0.58779 i $\endgroup$ – Ben Mar 4 '20 at 14:09
  • 1
    $\begingroup$ So to answer your question, some poles will have imaginary values. $\endgroup$ – Ben Mar 4 '20 at 14:10
2
$\begingroup$

For $\mu > 0$, $z^L \in \left ( \sqrt[L]{\mu} \right) e^{j 2\pi n/L} \,\forall\, n \in 1 \cdots L $.

I.e., the roots are evenly spaced on a circle $\sqrt[L]{\mu}$ in radius, and there's an $L$ of a lot of them.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ You may have made my day with that maths joke haha $\endgroup$ – luke glover Mar 5 '20 at 10:15
  • $\begingroup$ Would we have to consider the case for $\mu \leq 0$? ... well I guess $\mu = 0$ just gives $z=0$ but not sure for $\mu <0$. Thanks for your help :) $\endgroup$ – luke glover Mar 5 '20 at 10:17
  • 1
    $\begingroup$ I'm already concerned that I'm answering a homework problem. For $\mu$ equals any number, $z^L - \mu = 0$ is just an algebra problem -- can you work out what the answer is for $\mu < 0$? $\endgroup$ – TimWescott Mar 5 '20 at 14:43
  • $\begingroup$ Not a homework problem, I am researching the design of digital filters but I am new to the subject and its been a while since I formally studied any complex analysis. If I was to make a guess, I would assume that calculating the poles for $\mu < 0$ would just give the same roots of unity again but in reverse direction?(each subsequent root would work clockwise around the unit disc) $\endgroup$ – luke glover Mar 5 '20 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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