# [PD] signal math explanation

Pall Thayer pallthay at gmail.com
Sat Jan 18 18:59:04 CET 2014

```Ah... I've seen this in some of the help patches. Why would someone do it
with multiple [+~] instead of a single [*~]? There's no difference?

On Sat, Jan 18, 2014 at 12:49 PM, IOhannes m zmölnig <zmoelnig at iem.at>wrote:

> 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.
>
> IOhannes
>
>
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--
*****************************
Pall Thayer
artist
http://pallthayer.dyndns.org
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