Exponential approach to zero:
1 / exp(t) == exp (-t)
we "stretch" it by a time-constant "c":
gain(t) = exp (-t * c)
To find the time t, at which the exponential approach reaches gain "g":
exp (-c * t) = g
take the log of both sides: log (exp (-c * t) = log (g)
since log (exp (x)) == x : -c t = log (g)
divide by -c : t = -log (g) / c
set g = 1e-5 and c = _a/sr and we get: t = -log (1e-5) / (_a/sr)
The iterative approach using g += c * (target_gain - g);
converges faster than the exact exp() calculation.
Except with 32-bit float, if target-gain is 1.0f and "c" is small.
With 32bit float (1.0 - 1e-5) = .9999900 is represented as
sign: +1 | mantissa: 0x7fff58 | exponent: 126
there are only 126 "steps" to 1.0. Rounding of the lowest
mantissa bit does matter. We have to assume worst-case,
and increase the required loop_fade_length buffersize.
vs. approaching 0, where there are over 2^110 steps between
zero and 1e-5.
Robin Gareus
6f755c3c02
