# Finding remaining zeros of linear phase FIR filter

Question given is: if $$h(n)$$ is a linear phase causal FIR of order $$10$$ with real coefficients, find the remaining zeros of this filter if the zeros given are

$$q_{1,2} = -2 \pm 2j$$

$$q_{3} = -\frac12 + \frac{j}{2}$$

$$q_{4} = -1$$

I know for sure that $$-\frac {1} 2 - \frac j 2$$ has to be another zero, but there are 5 more zeros unaccounted for. How am I supposed to find them?

• I don't want to give you the complete answer since this looks like an obvious homework or test question, but certainly want to help. I believe the question is checking if you understand a well documented characteristic of the placement of zeros for a LINEAR PHASE filter. It may help if you understand the zero placement for MINIMUM PHASE and MAXIMUM PHASE filters as well, and the difference between all three filters. (I hope that was enough hints on what to dig into) Oct 25, 2019 at 2:59
• Are you sure it's order $10$, not length $10$ (i.e., $10$ filter coefficients)? Oct 25, 2019 at 6:48
• @MattL. it is order 10. Just so I understand what you're talking about for exam prep, if it has 10 zeros, shouldn't it have 11 coefficients and be of order 10? Would that make the length 11? Oct 25, 2019 at 15:14
• @S'Danc: Yes, filter length = no. of coefficients = order + 1. But with the given information, do you manage to come up with 10 zeros? Oct 25, 2019 at 16:03
• @MattL., yeah, I didn't know linear phase FIR filters had the reciprocal rule for zeros until someone commented it. It was pretty straightforward once I learned that. Oct 25, 2019 at 16:23