Well, let's simplify a bit, forget all the filter complexity (Q, slope, definition of the cutoff frequency...).

Let's just say that the output of a lowpass filter cannot move faster than the cutoff frequency: a 1Hz filter output cannot move faster than 1Hz (so it can't go back and forth in less than a second or so), a 1kHz can't go back and forth in less than about 1ms, etc. The output of a 0Hz filter can't move... at all. When you set the cutoff to 0Hz, the output freezes to its current value. It won't magically decay to 0.

Hey, if you set the framerate of a movie to 0 frame/second, it will just stop, and will show the same image forever; it won't fade to black!

Antoine

Le jeu. 11 avr. 2024 à 14:08, Peter P. <peterparker@fastmail.com> a écrit :

* Antoine Rousseau <antoine@metalu.net> [2024-04-11 13:40]:
> That doesn't seem incorrect to me; after all, a lowpass filter at 0Hz
> implies that its output is constant (any change would involve frequencies >
> 0Hz).

Thanks Antoine,

Why does a lowpass filter, that has a cutoff frequency of 0Hz imply that
it's output is constant?

I will describe the problem again hoping that I will understand it
better myseld:
I have an oscillating input signal that has some DC offset (unipolar
sawtooth from phasor~). I fade this signal's amplitude to -inf dB using
[line~].

I also fade down the filter cutoff (defined as the -3dB point of the
filter curve) from 400Hz to 0Hz. The filter will then continue to produce an
non-decaying output.

If I fade down the filter cutoff down to only 1Hz, it's output will decay (somehow
counterintuitively to me). This is the part I don't get.

I understand that vcf~ is a resonant filter, and it can have a gain
greater 1 around the cutoff frequency, especially for high Q values. The
above behavior can also be observed for Q=1.

Thanks for all hints!
Peter

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