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I have a issue to find out, is there any way that one can determine a group delay from frequency response of the FIR filter. Let's say I have I type FIR filter with order of 51. Frequency response obtained from freqz MATLAB's function is presented below. Frequency response of I type FIR filter with order of 51 I know group delay is negative derivative of linear (in this example) phase response, but how can one determine that from the picture above.

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  • $\begingroup$ You should review the form of the frequency response of linear phase filters. Given the filter order, it's straightforward to determine the (constant) group delay. No need to look at the plots. $\endgroup$
    – Matt L.
    Mar 3 at 12:59
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The group delay is the negative derivative of the phase response as the OP has stated, and specifically for the delay of one clock sample, the phase will go linearly negative to $2\pi$ radians as the frequency goes from 0 to $f_s$

From the picture we see the phase is going approximately 800 degrees at a frequency of $.17\pi$ rad/sample (where $2\pi$ rad/sample is the sampling rate). So this would be equivalent to

$800/360*(2\pi)$ rad /$(.17\pi)$ rad/sample = $26$ sample delay

For linear phase filters, the group delay is half the order which would be 25.5 in this case (we can't resolve that from the graphic alone). The OP wrote that the order of the filter was 51 which implies 52 taps.

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    $\begingroup$ isn't it $\frac{N-1}{2}$ so a filter with 2 taps will have a group delay of 0.5 $\endgroup$
    – Ben
    Mar 3 at 13:21
  • $\begingroup$ yes N = 52 right? $\endgroup$ Mar 3 at 13:22
  • $\begingroup$ $N = 51$ (and cut off frequency $0.1$) in this specific example. So from the Ben's formula it's should be, I guess, $25$ as you Dan wrote before. But I must admit I don't get why the formula $800(2\pi/360).17\pi$ equals to $26$, when I try calculate it straight forward all I get is $7.457$ I'm sure I'm making some rookie mistake $\endgroup$
    – Gal_Anonim
    Mar 3 at 13:29
  • $\begingroup$ The order of the filter is 1 less than the number of taps, so is it a 51 order filter or 51 tap filter? (I wrote the formula incorrectly, sorry!--- it is simply comparing 2pi radians over the sampling rate of 2pi (in radians/sample) which ends us being a normalized slope of 1, so the slope is the delay in samples if you divide the phase in radians by the frequency in radians/sample. $\endgroup$ Mar 3 at 13:31
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    $\begingroup$ I have filter with order of 51, so 52 taps. Based on formula $\frac{N-1}{2}$ group delay should be $25.5$ samples. $\endgroup$
    – Gal_Anonim
    Mar 3 at 14:46

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