[PD] +=~ object for pd

[Simon Iten]

this seems like a great approach. thanks!
my only concern is, that there will be nothing fast enough to detect my 30-1200 pulses per second... and i also don't think bangs can be sent that precisely, or can they?
On Jul 17, 2012, at 12:13 AM, Funs Seelen wrote:

> 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.