# Zeros in FIR Filter

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

• why do you assume you can? There's more to this problem than you're telling us! Feb 24, 2021 at 23:16
• @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 Feb 24, 2021 at 23:19
• 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. Feb 24, 2021 at 23:27
• @MarcusMüller I'll try to change it up to be more readable thank you Feb 24, 2021 at 23:29

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.
• @HelpMeBro which 0.5? The reciprocal of a complex number $z = re^{j\theta}$ equals to $z^{-1} = \frac{1}{r} e^{-j\theta}$. Feb 26, 2021 at 1:01