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My question is indicated in the title: What is the relation between delay of a digital FIR filter and step response of it?

Is the delay equal to the stablization time of the step response of the FIR?

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2 Answers 2

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You can easily see the relationship between delay and step response of an FIR filter if you recognise that the step response is the integral of the impulse response. All linear phase FIR filters are symmetric, so the peak of the impulse response will be the centre tap (for an even order), so this will be the middle of the step response, which is the delay of the filter. To see that this only depends on the filter order and the condition of symmetry see below.

Different responses. Same delay. The delay is the sample delay to the middle tap (NT/2). (or 1/2 way between the two for an odd order).

4.8kHz passband 10th order FIR impulse response: enter image description here

9.6kHz passband 10th order FIR impulse response: enter image description here

4.8kHz passband 10th order FIR Step response: enter image description here

9.6kHz passband 10th order FIR Step response: enter image description here

For completeness here are the Mag/phase plots:

4.8kHz passband 10th order FIR Mag/Phase: enter image description here

9.6kHz passband 10th order FIR Mag/Phase: enter image description here

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  • \$\begingroup\$ That's a neat answer +1 \$\endgroup\$
    – Andy aka
    Mar 25, 2015 at 16:04
  • \$\begingroup\$ Do you mean step response = 2*delay ? \$\endgroup\$ Mar 26, 2015 at 4:03
  • \$\begingroup\$ I mean the group delay = the delay to the middle tap of the filter. So if the filter is 10th order (as above) and the sample period was 10us then the delay = 50us \$\endgroup\$
    – akellyirl
    Mar 26, 2015 at 9:16
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No. For a symmetric FIR filter, the delay is related ONLY to the width of the filter, and has zip to do with the transfer function. A high pass filter of width 5 and cutoff pi/4 with have the same delay as a width 5 low pass filter with cutoff pi/42.

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