0
$\begingroup$

I recently had this question in a quiz and was quite confused as I don't think I can assume there are more zeros from just one, so how should I interpret it?

Assuming a linear phase FIR filter with real value coefficients and a zero at $2e^{j0.5\pi}$, which of these is true?

a)There are another 3 zeros at $2e^{-j0.5\pi}$ , $0.5e^{j0.5\pi}$ and $0.5e^{-j0.5\pi}$


b)There are more than 3 other zeros.


c)Theres just another zero at $2e^{-j0.5\pi}$


d)All the zeros in the filter are within the unit circle

Thank you

$\endgroup$
4
  • $\begingroup$ why do you assume you can? There's more to this problem than you're telling us! $\endgroup$ – Marcus Müller Feb 24 at 23:16
  • $\begingroup$ @MarcusMüller Hi Marcus. It was a multiple answer in my last test an the options where that there were either multiple more zeros like 3 or more or just one. As it was a very weirdly posed question I don't really know if I can assume the filter to have more zeros or not $\endgroup$ – HelpMeBro Feb 24 at 23:19
  • $\begingroup$ Can you reproduce the actual wording of the question. It seems to me there was a hint in how the question was worded, but you don't "transport" that hint. Generally, I don't think any statement but symmetrical spectrum from the real-valuedness can be inferred. $\endgroup$ – Marcus Müller Feb 24 at 23:27
  • $\begingroup$ @MarcusMüller I'll try to change it up to be more readable thank you $\endgroup$ – HelpMeBro Feb 24 at 23:29
3
$\begingroup$
  1. For a real value FIR filter, all its zeros should be conjugate pairs. So $2e^{-j0.5π}$ is also a zero.

  2. For a linear phase FIR filter, it should satisfy $h(n) = \pm h(N - 1 - n)$, and its transfer function equals to $$ H(z) = \sum_{n=0}^{N-1} h(n)z^{-n} = \sum_{n=0}^{N-1} \pm h(N - 1 - n)z^{-n} $$ Let $m = N-1-n$, we can derive $$ H(z) = \sum_{m=0}^{N-1} \pm h(m)z^{-(N-1-m)} = \pm z^{-(N-1)} \sum_{m=0}^{N-1} h(m)z^{-m} = \pm z^{-(N-1)} H(z^{-1}) $$ Therefore, if $z=2e^{j0.5π}$ is a zero, $z^{-1} = 0.5e^{-j0.5π}$ must be a zero. And according to the first point, $0.5e^{j0.5\pi}$ is also a zero.

$\endgroup$
2
  • $\begingroup$ Thank you. So the 0.5 comes from -(0.5-1) right? $\endgroup$ – HelpMeBro Feb 25 at 16:50
  • $\begingroup$ @HelpMeBro which 0.5? The reciprocal of a complex number $z = re^{j\theta}$ equals to $z^{-1} = \frac{1}{r} e^{-j\theta}$. $\endgroup$ – ZR Han Feb 26 at 1:01

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.