15 Apr
2025
15 Apr
'25
5:19 p.m.
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:
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
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?