[PD] vcf~ producing output without input for 0Hz cutoff?
Christof Ressi
info at christofressi.com
Fri Apr 12 09:35:24 CEST 2024
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
>> <antoine at metalu.net <mailto:antoine at 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. <peterparker at fastmail.com
>> <mailto:peterparker at fastmail.com>> a écrit :
>>
>> * Antoine Rousseau <antoine at metalu.net
>> <mailto: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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