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

```Thank you Christof,

your equation helped me a lot, and now I see why some DC component will
be circulating the filter forever.```
```
Thank you Edwin,

explaining that "0 Hz is an infinite amount of octaves below 20 Hz" made
me understand that a cutoff at 0Hz will not reduce DC by -3dB.

Looking at [vcf~]'s help patch closer, I read that the filter's real and
complex outputs may be combined to allow for other filtering possibilies
than band-pass and low-pass. Does anyone have a quick example of such a
combination? I reckon that I make a highpass filter by subtracting the
lowpass-filtered signal from the same unfiltered signal?

Thanks again to everyone,
Peter

* Christof Ressi <i...@christofressi.com> [2024-04-12 09:36]:
> Just expand on Antoine's post, let's look at the formula of Pd's 1-pole
> lowpass filter:
>
> k = freq * 2pi / sr
>
> y[i] = x[i] * k + y[i-1] * (1 - k)
>
> For freq=0 this becomes:
>
> y[i] = y[i-1]
>
> As you can see, this would just repeat the previous output infinitely,
> ignoring the input altogether. There is no decay to zero!
>
> The same reasoning applies to bandpass filters such as [vcf~].
>
> Christof
>
> On 12.04.2024 09:10, cyrille henry wrote:
> > I don't think it's weird for a lowpass filter to go under 20Hz. They are
> > not restricted to audio signals.
> > I use them a lot to smooth control signals, or to replace line~.
> > (I really hate line~ to control sound amplitude or preset transition,
> > it's way too robotic)
> >
> > cheers
> > c
> >
> > Le 12/04/2024 à 08:01, Alexandre Torres Porres a écrit :
> > > and you got a strong DC component over there :)
> > >
> > > anyway, it also seems weird to have a lowpass or a bandpass going as
> > > low as in the 20hz range. If you wanna do it just so it fades out to
> > > silence, you need a DC filter, something like a [hip~ 5] object, so
> > > when the lowpass, bandpass gets there, then you have nothing.
> > >
> > > cheers
> > >
> > > Em qui., 11 de abr. de 2024 às 15:40, Antoine Rousseau
> > > <anto...@metalu.net <mailto:anto...@metalu.net>> escreveu:
> > >
> > >     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. <peterpar...@fastmail.com
> > > <mailto:peterpar...@fastmail.com>> a écrit :
> > >
> > >         * Antoine Rousseau <anto...@metalu.net
> > > <mailto:anto...@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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