Hi list, If I take a circle, cut it in half horizontally, then move the bottom half r*2 units to the right, what is the waveform that results?
-Jonathan
Hi Jonathan,
The half cycle would be written: sqrt(1-x^2) since in a circle x²+y²=1
Then depending on how you want to use it, you'd have to figure out a way to have the lower portion. See attached patch with a [phasor~]
Hope this helps.
Cheers Pierre-Olivier
On 28/06/2014 21:40, Jonathan Wilkes via Pd-list wrote:
Hi list, If I take a circle, cut it in half horizontally, then move the bottom half r*2 units to the right, what is the waveform that results?
-Jonathan
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Hi Jonathan,
On Sat, Jun 28, 2014 at 9:40 PM, Jonathan Wilkes via Pd-list < pd-list@lists.iem.at> wrote:
Hi list, If I take a circle, cut it in half horizontally, then move the bottom half r*2 units to the right, what is the waveform that results?
A phase corrected triangle wave at least looks like half a circle (the green wave: http://funsseelen.eu/images/gnuplot_100_1024.png).
Partials for a normal triangle wave: 1/(1^2) * sin(x) + 0 + -1/(3^2) * sin(3x) + 0 + 1/(5^2) * sin(5x) + 0 + -1/(7^2) * sin(7x) et cetera.. --> infinity
Partials for a phase corrected triangle wave: 1/(1^2) * sin(x) + 0 + 1/(3^2) * sin(3x) + 0 + 1/(5^2) * sin(5x) + 0 + 1/(7^2) * sin(7x) + et cetera.. --> infinity
(http://funsseelen.eu/C/fourier_series/triangle.c / http://funsseelen.eu/C/fourier_series/triangle2.c C code to calculate samples that create both wave forms)
Regards,
Funs
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Funs Seelen via Pd-list -
Jonathan Wilkes via Pd-list
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Lorenzo Sutton via Pd-list
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Pierre-Olivier Boulant via Pd-list