What is the name/family behind the following high-pass IIR filter:

$ T = 1/f_s \\ f_c = 5 \\ c_1 = \displaystyle \frac{1}{1 + \tan(\pi f_cT)}\\ c_2 = \displaystyle \frac{1 - \tan(\pi f_cT)}{1 + \tan(\pi f_cT)} \\ b = \left[c_1, -c_1\right] \\ a = \left[1, -c_2\right] $

Where $f_s$ is the sampling rate and $f_c$ the cut-off frequency.


This is a first-order discrete-time Butterworth filter, which can be obtained via the bilinear transform from the following analog prototype filter (with normalized cut-off frequency):


The bilinear transform replaces $s$ by $k\frac{z-1}{z+1}$, where $k$ is chosen such that the discrete-time filter has the specified cut-off frequency $f_c$, resulting in

$$k=\frac{1}{\tan(\pi f_c T)}\tag{2}$$

where $T=1/f_s$ is the sampling interval.

So if you replace $s$ in $(1)$ by

$$s=\frac{1}{\tan(\pi f_c T)}\frac{z-1}{z+1}\tag{3}$$

you obtain the corresponding discrete-time filter with the coefficients given in your question.


OK, here is how you obtain your final coefficients. Let $a=\tan(\pi f_c T)$. Plugging $(3)$ into $(1)$ then gives for the transfer function of the discrete-time filter


which is exactly in the given form with $c_1=1/(1+a)$ and $c_2=(1-a)/(1+a)$.

| improve this answer | |
  • $\begingroup$ Although I do not disagree with the Butterworth statement, one can note that every first-order filter would look like that, independent of approximation method. Still, with the lack of any design parameters, Butterworth is most likely the best bet, if any approximation is used. $\endgroup$ – Oscar Aug 27 '15 at 14:11
  • $\begingroup$ great, thank you! would you be able to complete the math to obtain $c_1$, $c_2$ coefficients? $\endgroup$ – Chesnokov Yuriy Aug 27 '15 at 14:30
  • $\begingroup$ @Oscar: You are absolutely right and I also thought about it, but for the very reason that you gave I decided to stick with "Butterworth". $\endgroup$ – Matt L. Aug 27 '15 at 15:12
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    $\begingroup$ @ChesnokovYuriy: I've added the necessary steps. $\endgroup$ – Matt L. Aug 27 '15 at 15:23
  • $\begingroup$ thank you very much for helping me getting into theoretical DSP part $\endgroup$ – Chesnokov Yuriy Aug 28 '15 at 6:54

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