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
>>
>>
>



--
Brandon Zeeb
Columbus, Ohio