[PD] vcf~ producing output without input for 0Hz cutoff?

[Antoine Rousseau]

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 at fastmail.com> a écrit :

> * Antoine Rousseau <antoine at 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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