[PD] Non-Linear Quantization / Bitcrush

[Ludwig Maes]

Watch out in my numeric example, I was a bit careless and the q I
chose continues to increase for more and more negative amplitudes!

On 2 November 2010 19:44, Ludwig Maes <ludwig.maes at gmail.com> wrote:
> 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
>>>>>
>>>>>
>>>>
>>>
>>
>