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 > > > > > > > > > > > > _______________________________________________ > > > Pd-list@lists.iem.at <mailto:Pd-list@lists.iem.at> mailing list > > > UNSUBSCRIBE and account-management -> > > > https://lists.puredata.info/listinfo/pd-list > > > <https://lists.puredata.info/listinfo/pd-list> > > > > > > _______________________________________________ > > > Pd-list@lists.iem.at <mailto:Pd-list@lists.iem.at> mailing list > > > UNSUBSCRIBE and account-management -> > > > https://lists.puredata.info/listinfo/pd-list > > > <https://lists.puredata.info/listinfo/pd-list> > > > > > > > > > _______________________________________________ > > > Pd-list@lists.iem.at mailing list > > > UNSUBSCRIBE and account-management -> > > > https://lists.puredata.info/listinfo/pd-list > > > > > > > > _______________________________________________ > > Pd-list@lists.iem.at mailing list > > UNSUBSCRIBE and account-management -> > > https://lists.puredata.info/listinfo/pd-list > > > > _______________________________________________ > Pd-list@lists.iem.at mailing list > UNSUBSCRIBE and account-management -> > https://lists.puredata.info/listinfo/pd-list _______________________________________________ Pd-list@lists.iem.at mailing list UNSUBSCRIBE and account-management -> https://lists.puredata.info/listinfo/pd-list
