17 Jul
2012
17 Jul
'12
12:13 a.m.
Hi Simon,
On Mon, Jul 16, 2012 at 11:26 PM, Simon Iten itensimon@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.