# Why doesn't the delay of an FIR filter equal the number of taps but half of them? [duplicate]

I have seen the mathematical proofs that show that the delay of an FIR filter is half the filter order, i.e N/2. Yet I am still confused:

When I look in the time domain, the output of the filter is: \begin{align} y[n] &= b_0 x[n] + b_1 x[n-1] + \cdots + b_N x[n-N] \\ &= \sum_{i=0}^{N} b_i\cdot x[n-i], \end{align} where N is the filter order. So clearly the output at a given index n depends on the input of N preceding input values, so why is the delay N/2 ?

• Peter was right to mark this as duplicate. You're not paying attention to the wording: Half the filter order is the group delay, and not that of just any FIR filter, but that of linear phase filters only. The exact things that you make statements about make a difference, so be a bit more careful when talking about specific things. In a mathematical context like this, qualifiers such as "group" or "linear phase" are almost never used in vain; they are essential to the meaning of the words. – Marcus Müller Feb 7 '18 at 14:29
• Let me be more specific: – D.Cohen Feb 7 '18 at 17:40