hello,
in the patch F01.pulse (and so for F02, F03)
why do we calibrate the resulting signal from 0 to 1 ? and not from -1 to 1 (full amplitude) like an audio signal ?

thx.

@mx It is because a pulse (useful for testing polarity without a time component) is not a symmetrical wave....... the author has created a positive pulse (rising and falling to zero) and there is no negative compliment to the positive pulse. Leaving the "base" of the waveform at -1 would give the output a negative DC offset and any DC offset is a bad thing for your speakers.......
https://en.wikipedia.org/wiki/Pulse_wave
David.

hummmm,
so for every non symetrical waveform we have to adapt the output so that there is no DC offset ...
thx.

@mx It is simpler just to put a [hip~] but it depends on the low end of the dynamic range that you are looking for and the hip could drastically distort your pulse......... so......... yes.
The [hip~] is useful for normal symmetrical (sine based) audio (such as a microphone input), as it will catch any small dc offsets that might have crept into a patch......
http://en.flossmanuals.net/pure-data/ch025_dc-offset/
David.

@whale-av I think I understood everything about that floss tutorial EXCEPT the bit about the [hip~ ]. How does [hip~ } effectively translate a signal in the same way {-~ 1] does?

@rjp9 Hello there..... it doesn't. But it filters low frequencies and it sees dc as a frequency of 0Hz. It might not completely remove the dc component because it has a roll off below the cut-off frequency that it is given. But it is close enough. If you set the cut-off frequency too high it will adversely affect the pulse.........
F03.pulse.spectrum_modded.pd
David.

Hello. I see your point @rjp9,
the think difficul to get is why with this signal

[phasor 60] + [phasor 50]

a [hip 1] will works well to remove the DC

and why with this little pulse train

[phasor 60]
[-~ 0.5]
[*~ 2]
[clip~ -0.5 0.5]
[cos~ ]

a [hip~ 1] will not works well to remove this DC (the min value will be -0.5) ?

by playing with the pulse wave patch, I see this behavior : less a signal is complexe (not many partials) less a [hip 1] will be efficient to remove the DC.

@mx Hello @mx
It is still working well....... it is just that the "balance of power" of the pulse train has shifted...... the pulses are wider, and so the flat base line is narrower, so the dc component is smaller, so to balance the unsymmetrical signal the shift towards "0" is less.
If you reduce your [x~ 2] to [x~ 1.1] you nearly have a sine wave and the [hip~ 1] therefore sees no low frequency to cut......... there is no dc component and so no shift / correction......
Maybe the dc component could be described as a "power offset"? and so better understood?........
David.

okay, its clear,
so we don't have to think of frequency domain.

thx for your time.
cheers