Let's define the relaxation time of a filter as the delay needed for the impulse response of a filter to reach and remain under an amplitude threshold.

Is there a closed-form solution to the computation of an estimate or boundary of the relaxation time of a discrete recursive (IIR) filter from its poles/zeros or from the coefficients of its difference equation, except by computing a reasonable length of its response and checking the absolute value of the response?

  • $\begingroup$ If you have the transfer function then you can derive its time response (similar to this, for example), and the result will be a product or sum of exponentials and sines/cosines. The exponentials will give you what you're looking for. $\endgroup$ Jun 3 '20 at 13:54
  • $\begingroup$ I'm not sure I'm following your thought. I don't want to expand the impulse response. If I derive for example the z-domain expression for a 2nd order IIR filter, I obtain an expression with a 4th order denominator, i.e. I've doubled the poles. $\endgroup$
    – moala
    Jun 3 '20 at 16:01
  • $\begingroup$ If you know the 3 dB bandwidth you can use the rule of thumb that the 90%/10% fall time is approximately 0.35/BW in Hz. This is true for a first order system but reasonably accurate for any IIR system due to its dominant pole dominating the long term response. $\endgroup$ Jun 3 '20 at 17:00
  • $\begingroup$ So that said, you can identify the dominant pole and then use a simpler equation based on that. (Or if there isn't one dominant pole you can identify the dominant pole/zeros and use a simpler equation based on those) $\endgroup$ Jun 3 '20 at 17:01
  • $\begingroup$ @moala Well, that was a bit of a brainfart. I was thinking in terms of the inverse Z-transform, but what you get is not like what you get in s domain (what I expected, an $e^{-t}\cos t$-like response). You get a sum of coefficients with powers of the indices, something like $\sum_k A_k\,p_k^k\,u_k$, where $A$, $p$, and $u$ are defined, but $k$ is increasing. Maybe it can be deduced, mathematically, but I don't know how to (though, if it can be done, I wouldn't mind seeing it). $\endgroup$ Jun 3 '20 at 18:41

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