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.m...@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.m...@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.bran...@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 > >> > >> > > > -- Brandon Zeeb Columbus, Ohio
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