On Tue, Apr 15, 2025 at 02:08:02PM +0100, Gordonjcp wrote: > On Tue, Apr 15, 2025 at 01:43:10PM +0200, Fons Adriaensen wrote: > > > > Now, to be 'constructive', the correct way to compute c > > would be > > > > f = cutoff_frequency / sample_rate; > > a = 2 * (1 - cos (2 * pi * f); > > c = (sqrt (a * (a + 4)) - a) / 2; > > > > > > A good approximation (less than 1% relative error) > > for 0 <= f <= 0.5 could be > > > > c = 2 * pi * f * (1 + f * (3.45f + 4.27f * f)); > > Can you explain how you get there, in terms that a 35-years-ago > Straight-Cs highschool maths student might understand?
It's really just basic digital filter theory and algebra. I found an online version: go to <https://www.moseleyinstruments.com/blog/first_order_lowpass> and scroll down to 'Determining the coefficient ‘c’ when the cutoff is given'. The approximation is found by playing with gnuplot for a few minutes. Ciao, -- FA _______________________________________________ Linux-audio-dev mailing list -- linux-audio-dev@lists.linuxaudio.org To unsubscribe send an email to linux-audio-dev-le...@lists.linuxaudio.org
