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.


1 Answer 1


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 $c_2=(1-a)/(1+a)$.

  • $\begingroup$ 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, 2015 at 14:11
  • $\begingroup$ great, thank you! would you be able to complete the math to obtain $c_1$, $c_2$ coefficients? $\endgroup$ Aug 27, 2015 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, 2015 at 15:12
  • 1
    $\begingroup$ @ChesnokovYuriy: I've added the necessary steps. $\endgroup$
    – Matt L.
    Aug 27, 2015 at 15:23
  • $\begingroup$ thank you very much for helping me getting into theoretical DSP part $\endgroup$ Aug 28, 2015 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.