[PD] +=~ object for pd

[Funs Seelen]

Hi Simon,

On Mon, Jul 16, 2012 at 11:26 PM, Simon Iten <itensimon at gmail.com> wrote:
>
> is there a way to achieve this without the signal accumulator object? i feel
> like there should be an easy solution but i can't seem to find it. any
> hints?
>

If you mean the following, where x is the input and y the output..

y += x;

and that for each sample..

then [biquad~] might be a solution:

[sig~ 1]
|
|  [clear(
| /
[biquad~ 1 0 1 0 0]
|

This adds the last output to the current input. The [clear( message
resets biquad~ to 0. Now you just have to find a method to translate
your pulse to a bang.