Great! Unfortunately, I'm not entirely sure how to realize this in Pd. Can you help me out with a little example?
Thanks
On Tue, Nov 2, 2010 at 2:37 PM, Ludwig Maes ludwig.maes@gmail.com wrote:
So you want amplitude 'a' dependant quantization size 'q' ? take your chosen q(a); in your example it seems you want a simple line: q=q(0)-k*a; define f(a) as integral of 1/q from a=0 to a; also calculate the inverse of f(a) i.e. a(f);
now for each sample do: out=a(round(f(in))) where round is any floor or the like...
have fun!
ps:
in your example: q=q0-k*a with for example q(0)=0.001 and q(0.8)=0.0001: q:=0.001-0.0009/0.8*a then f=2558.427881-1111.111111*ln(10.-9.*a) and inverse=easy
On 2 November 2010 19:20, Ludwig Maes ludwig.maes@gmail.com wrote:
This is pretty easy actually, I use such things mostly to guide my rhythmical quantization...
On 2 November 2010 19:19, brandon zeeb zeeb.brandon@gmail.com wrote:
This is even better. If I could minimize the jumps around Y = 0.5 to
-0.5
It'll be exactly what I'm looking for... or a start at least.
Do you see what I mean now? See how the amount of quantization changes
with
Y and a minimum quantization value?
I think I'm getting towards the answer now...
-- Brandon Zeeb Columbus, Ohio