the /~ operator equivalent
Mar 03 2014 | 7:12 pm
Hi there people!
I want to build an hardware version of a patch I wrote, it's a relatively simple one, but while experimenting with various sounds I decided to try the /~ operator, using it to divide a signal by another. It made to the final version of the patch.
NOW. I'm guessing what does really happen inside that operator, because it's not multiplicating the left signal by the right one's inverse, in fact if I try that it just sounds completely different...
While the *~ operator hardware is simply made with a ring modulator, what really is the /~ operator???
I hope to have been clear...uhm, if I wasn't, just ask and I'll try to explain myself better!
cheers!
- Mar 03 2014 | 8:36 pmHaha, I think it's called clipping. Multiplying two signals that are both in the -1 to 1 range always results in a signal that's within that range, but think about what happens when you divide 0.9 by 0.1.
- Mar 04 2014 | 11:56 amawwww you're right! It is actually multiplicating the left signal by 1/right signal! And 1/right signal is not the same as the right signal's inverse...I tried to replicate the clipping with some real world simulations, I generated a square wave and multiplicated it by 15 to get to clip it, but still they don't look the same... (in the image the upper graph shows a 1/sinewave, the lower graph shows the squarewave*15)Someone has any clue on how to break this in basic blocks, or "what is the real-world implementation of 1/signal"?? I know the question might not belong here, but who knows...thanks!
