Hi Simon, On Mon, Jul 16, 2012 at 11:26 PM, Simon Iten <[email protected]> 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. _______________________________________________ [email protected] mailing list UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list
