y = x * (1 + (1 - x))0 <= x <= 10 <= y <= 1
My lack of skill with mathematics probably means I’m confusing things with a lot of vague descriptions and inconsistent/misnamed terminology, but I did manage to mess around enough to obtain something that resembles the curve I’m talking about. It looks like this:Code: [Select]y = x * (1 + (1 - x))0 <= x <= 10 <= y <= 1
y = x * (1 + (1 - x))y = x * (2 - x)y = 2x - x^2
y = 2x - x^2
dy = (2 - 2x)dx
u = 20*ln(y)/ln(10)u = 20*ln(2x - x^2)/ln(10)
du = 20/ln(10) * (2-2x)dx/(2x-x^2)
I would love to try to help, but I'm having a very hard time understanding your problem. You are talking about this https://en.wikipedia.org/wiki/Synthesizer#ADSR_envelope, correct?Your question seems to concern the attack part, but then you mention decay. In what way are those two related? Do you set the decay on the synth and it automatically sets the attack? Are they connected by some ratio? I'm aware of the first statement in bold, but I can't make sense of it, mainly due to relationship to the decay.
The rest sort of sounds like you are describing a linear differential equation... the solution to which involves exponential functions.
Is any of this helping you at all?P.S.: note that I have no idea of the actual inner workings of classic analog synthesizers.
Just take the derivative[…]Hope that helps!
is K( 1 - e ^ (-at)) the function you were looking for?
Quote from: Woodinville on 01 May, 2013, 04:40:38 PMis K( 1 - e ^ (-at)) the function you were looking for?I have no idea! What does that do? I can speculate about the curve slightly, but not much without knowing the identities of the variables. Thanks anyway, I guess. Thanks to everyone who replied. I guess I might need to do more boring recording and measuring, heh.