In case of FIR filter design, is the following relation always applicable or there are other relations/formulas also?
If window length is denoted by $M$ ,then filter order will be $M-1$
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$\begingroup$ this answer has more relations and formulas but here $N=M-1$ is the filter order, one less than the number of nonzero taps. $\endgroup$– robert bristow-johnsonCommented Jul 19, 2020 at 0:05
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3$\begingroup$ Does this answer your question? Filter order vs number of taps vs number of coefficients $\endgroup$– Matt L.Commented Jul 19, 2020 at 8:01
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$\begingroup$ "one less than the number of nonzero taps." I don't think that's correct. The $y[n] = ax[n] + bx[n-50]$ is a 50-order filter, while having just 2 nonzero taps. $\endgroup$– Michael GrunerCommented Jul 23, 2020 at 21:46
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1 Answer
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Yes. The number of taps or coefficients in a FIR filter (or window length as you call it) tell the order of the filter. A more general way to see it, to avoid confusions, is to think of the order as the largest delay in the filter. For example, consider the following naive reverb:
$$y[n] = a x[n] + (1 - a) x[n-D]$$
Even though there are only two (visible) coefficients, the order of the filter is $D$ because it’s the largest delay. The window length is $D+1$ but all the middle coefficients are $0$, hence not visible.
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1$\begingroup$ calling a single delay tap (plus an undelayed tap) a "reverb" is a bit of a stretch. $\endgroup$ Commented Jul 18, 2020 at 23:54
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$\begingroup$ Yeah, you’re right. I hope the “naive” makes it a bit more appropriate. $\endgroup$ Commented Jul 19, 2020 at 0:56
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$\begingroup$ Is a zero coefficient a "tap"? $\endgroup$ Commented Jul 19, 2020 at 23:03