Spin Semiconductor

I like to take a sine wave, and SOF it into the 0 to 1.0 range to use as a tremolo LFO (or any other LFO application). I also like to square or cube or beyond this sine wave so that it is "warped" but still smooth. What I'd like to do is have a knob that controls the level of warp from, say, 1 to 5. 1 would be the original signal, and 5 would be the signal to the 5th power.

Now of course I could do pot skipping and have each step be a different power step (via cascaded MULX instructions), but I want it to be smooth across pot rotation rather than jump between steps.

In regular math, I can do powers by taking the log, multiplying, and then taking the exponent of the result.

Super trivial example:

starting value is 100. I want to square this.

log10(100) = 2.
2 x 2 = 4.
10^4 = 10,000.

Suppose I want to do the "1.5" power:

log10(100) = 2.
2 x 1.5 = 3
10^3 = 1,000.

Anyone who wants to chime in with the answer is welcome to do so.

The goal:

Take an input control signal between 0.0 and 1.0.
Take a pot control signal and use this to continuously control "variable power" of the control signal from 1 to 5.

TIA,

DL

The quick and maybe-not-so-dirty solution would be to crossmix between the "pure sine wave" and the "sine wave to the highest power you want available". I did a simulation in GeoGebra. You wont get the exact curve of the powers inbetween, but I doubt you'll notice much difference unless you set a very excessive "largest power".

Interesting strategy! Those sine waves might have to be scaled down prior to running through the crossfade to avoid clipping but I may just give this a try.

(though since they are always in phase, perhaps not)!

Thanks,

DL

Raise to power between 1 and 5

DrAlx wrote: ↑Tue Apr 02, 2019 11:38 pm Raise to power between 1 and 5

Code: Select all

;ACC contains something between 0 and 1
LOG 1,0
WRAX temp,1
SOF -2,0
SOF -2,0
MULX POT0
RDAX temp,1
EXP 1,0

y = exp((1 + (pot0 * 4)) * log(x)) seems straightforward.

If pot0 = 0 then y = exp(log(x)) = x
If pot0 =1 then y = exp(log(5 * x)) = x^5

My mind always broke on the log instruction, not sure why. If I sat down and really penciled in the result at different input values I could get it. You are the log-meister! Thanks.