[PD] signal math explanation
IOhannes m zmölnig
zmoelnig at iem.at
Sat Jan 18 18:49:12 CET 2014
On 01/18/2014 06:24 PM, Pall Thayer wrote:
> Can anyone tell me what one is accomplishing when doing something like this:
>
> [osc~ 440]
> 
> [+~]
> \ x1
> [+~]
> \ x2
> [+~]
> \ x3
> [+~]
> x4
>
> In other words, the chain of [+~] that feed the previous object's output
> into both inlets of the next... what does this do exactly?
it adds a signal with itself: y=x+x=2*x
so the output of the 1st [+~] is
x1=x0 (as the 2nd inlet~ is not connected)
and the following [+~] will output:
x2=x1+x1=2*x1=2*x0
x3=x2+x2=2*x2=2*2*x0=4*x0
x4=x3+x3=2*x3=2*4*x0=8*x0
so you could write the patch as:
[osc~ 440]

[*~ 8]
more often you see [*~] instead of [+~], which is a simple way to square
the input.
fgmadsr
IOhannes
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