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

[Alexandre Torres Porres]

Em sáb., 13 de abr. de 2024 às 06:30, Peter P. 
escreveu:

> 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 don't, and this information was there before I revised all help files. I
suspect Miller wrote this, but I can't see how this is helpful or
insightful information. This would not be something particular to [vcf~],
any filter can be cascaded or combined in parallel, and I guess this is
beter suited for the raw filters, that can be used and combined to more
properly design filters. I would vote for just suppressing this
information, unless someone convinces me there's a good point and example.



> I reckon that I make a highpass filter by subtracting the
> lowpass-filtered signal from the same unfiltered signal?
>

Well, this is not MAX, we do have [hip~] :)


>
> Thanks again to everyone,
> Peter
>
> * Christof Ressi  [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
> > > > 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.  > > > > a écrit :
> > > >
> > > > * Antoine Rousseau  > > > > [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!
> > > >   

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

[Thomas Mayer]

Hi,

On 13.04.24 10:57, Peter P. wrote:

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?


Yes, but not as simple as that, because there is a phase shift at the 
cutoff frequency, so that you will get artifacts.


https://dsp.stackexchange.com/questions/75064/what-is-phase-shift



Thanks again to everyone,
Peter

* Christof Ressi  [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
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. mailto:peterpar...@fastmail.com>> a écrit :

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

[Peter P.]

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  [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
> > > 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.  > > > a écrit :
> > > 
> > >     * Antoine Rousseau  > > > [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  mailing list
> > >     UNSUBSCRIBE and account-management ->
> > > https://lists.puredata.info/listinfo/pd-list
> > > 
> > > 
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> > > 

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

[Edwin van der Heide]

0 Hz is an infinite amount of octaves below 20 Hz (10 Hz, 5 Hz, 2.5 Hz, 1.25 Hz, …).A low pass filter “normally” has a slope of a certain amount dB per octave. However, a filter with a frequency of 0 Hz doesn’t  have a slope with a certain amount of dB/oct because an octave up or down from 0 Hz (by doubling or halving the frequency) is also 0 Hz. The filter simply can’t move ;-)Best!EdwinOn 12 Apr 2024, at 09:49, Christof Ressi  wrote:

  

  
  
Alex made a great point there!
It's that very reason why DAW plugins usually don't let you go
  down all the way to zero. (Typically, they stop at 20 Hz or
  something.) If your signal has components below that, you'd need a
  high pass filter.

On 12.04.2024 09:35, Alexandre Torres
  Porres wrote:


  
  oh sure, but that's more like a slew or glide and
you don't go down to zero anyway :)


and if you're filtering audio, you don't want to keep
  inaudible stuff. All I'm saying is that if you are soing this
  to fade to slience you need a DC filter
  
  
  
Em sex., 12 de abr. de 2024 às
  04:12, cyrille henry 
  escreveu:

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
  >
  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. >
  a écrit :
  > 
  >         * Antoine Rousseau >
  [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
 

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

[Christof Ressi]

Alex made a great point there!

It's that very reason why DAW plugins usually don't let you go down all 
the way to zero. (Typically, they stop at 20 Hz or something.) If your 
signal has components below that, you'd need a high pass filter.


On 12.04.2024 09:35, Alexandre Torres Porres wrote:
oh sure, but that's more like a slew or glide and you don't go down to 
zero anyway :)


and if you're filtering audio, you don't want to keep inaudible stuff. 
All I'm saying is that if you are soing this to fade to slience you 
need a DC filter


Em sex., 12 de abr. de 2024 às 04:12, cyrille henry  
escreveu:


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
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.
mailto:peterpar...@fastmail.com>> a écrit :
>
>         * Antoine Rousseau 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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Re: [PD] vcf~ producing output without input for 0Hz cutoff?

[Christof Ressi]

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 
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. > a écrit :


    * Antoine Rousseau > [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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Re: [PD] vcf~ producing output without input for 0Hz cutoff?

[Alexandre Torres Porres]

oh sure, but that's more like a slew or glide and you don't go down to zero
anyway :)

and if you're filtering audio, you don't want to keep inaudible stuff. All
I'm saying is that if you are soing this to fade to slience you need a DC
filter

Em sex., 12 de abr. de 2024 às 04:12, cyrille henry  escreveu:

> 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 > 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.  > a écrit :
> >
> > * Antoine Rousseau  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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Re: [PD] vcf~ producing output without input for 0Hz cutoff?

[cyrille henry]

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 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. mailto:peterpar...@fastmail.com>> a écrit :

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

[Alexandre Torres Porres]

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 
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.  a
> écrit :
>
>> * Antoine Rousseau  [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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Re: [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.  a écrit :

> * Antoine Rousseau  [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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Re: [PD] vcf~ producing output without input for 0Hz cutoff?

[Peter P.]

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

[Antoine Rousseau]

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).

Antoine


Le jeu. 11 avr. 2024 à 13:08, Peter P.  a écrit :

> * Alexandre Torres Porres  [2024-04-10 22:19]:
> > the problem is your input, with a DC component (0 hz). This doesn't seem
> to
> > happen with [osc~] and seems like normal behaviour and not something to
> be
> > documented
>
> I see, phasor~'s waveform is not symmetric around 0, good point. But why
> will vcf~'s output evnetually decay to -inf dB for a cutoff of 1Hz but
> not for a cutoff of 0Hz?
>
>
>
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Re: [PD] vcf~ producing output without input for 0Hz cutoff?

[Peter P.]

* Alexandre Torres Porres  [2024-04-10 22:19]:
> the problem is your input, with a DC component (0 hz). This doesn't seem to
> happen with [osc~] and seems like normal behaviour and not something to be
> documented

I see, phasor~'s waveform is not symmetric around 0, good point. But why
will vcf~'s output evnetually decay to -inf dB for a cutoff of 1Hz but
not for a cutoff of 0Hz?



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

[Alexandre Torres Porres]

the problem is your input, with a DC component (0 hz). This doesn't seem to
happen with [osc~] and seems like normal behaviour and not something to be
documented

cheers

Em qua., 10 de abr. de 2024 às 11:56, Peter P. 
escreveu:

> Howdy list,
>
> just discovered, that [vcf~] produces output after its input has decayed
> for a cutoff frequency of 0Hz, but not for values higher 0.
>
> A minimal working example is attached. Is this behavior known? Could it
> be mentioned in the help patch maybe?
>
> cheersz, P
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[PD] vcf~ producing output without input for 0Hz cutoff?

[Peter P.]

Howdy list,

just discovered, that [vcf~] produces output after its input has decayed
for a cutoff frequency of 0Hz, but not for values higher 0.

A minimal working example is attached. Is this behavior known? Could it
be mentioned in the help patch maybe? 

cheersz, P
#N canvas 416 334 716 365 12;
#X obj 78 213 vcf~ 12;
#X obj 78 131 *~;
#X obj 123 249 env~;
#X floatatom 123 277 5 0 0 3 - - - 0;
#X obj 78 57 phasor~ 300;
#X obj 130 111 line~;
#X msg 188 81 1 \, 0 1000;
#X obj 100 182 line~;
#X obj 317 42 bng 20 250 50 0 empty empty 1.start\ note 0 -10 0 12 #fcfcfc 
#00 #00;
#X obj 526 22 loadbang;
#X msg 526 52 \; pd dsp 1;
#X obj 70 286 env~;
#X floatatom 70 314 5 0 0 3 2.\ observe\ output\ after\ amplitude\ envelope\ 
has\ finished - - 0;
#X msg 133 151 400 \, 0 1000;
#X text 225 153 setting vcf~ cutoff to 0Hz will make it produce a signal.;
#X connect 0 0 11 0;
#X connect 0 1 2 0;
#X connect 1 0 0 0;
#X connect 2 0 3 0;
#X connect 4 0 1 0;
#X connect 5 0 1 1;
#X connect 6 0 5 0;
#X connect 7 0 0 1;
#X connect 8 0 6 0;
#X connect 8 0 13 0;
#X connect 9 0 10 0;
#X connect 11 0 12 0;
#X connect 13 0 7 0;
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