[PD] Non-Linear Quantization / Bitcrush

[Ludwig Maes]

And we want f' to be 1 (integer step) / (per) quantization size (for
that amplitude)


On 2 November 2010 19:41, Ludwig Maes <ludwig.maes at gmail.com> wrote:
> The reason you use the inverse is so that the amplitude remains the
> same albeit quantized. The reason we use another function before
> flooring is to distritube the floor levels.But afterwards we need to
> bring the values back to their "original" place
>
> On 2 November 2010 19:37, Ludwig Maes <ludwig.maes at 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 at 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 at 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
>>>>
>>>>
>>>
>>
>