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

  • $\begingroup$ why do you assume you can? There's more to this problem than you're telling us! $\endgroup$ Feb 24, 2021 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, 2021 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. $\endgroup$ Feb 24, 2021 at 23:27
  • $\begingroup$ @MarcusMüller I'll try to change it up to be more readable thank you $\endgroup$
    – HelpMeBro
    Feb 24, 2021 at 23:29

1 Answer 1

  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.

  • $\begingroup$ Thank you. So the 0.5 comes from -(0.5-1) right? $\endgroup$
    – HelpMeBro
    Feb 25, 2021 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, 2021 at 1:01

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