Re: [pyo-discuss] Digest for pyo-discuss@googlegroups.com - 1 update in 1 topic

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kjel sidloski

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19 Sept 2022, 19:20:1619/09/2022
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Thank you Aaron! This is great to know and would have certainly taken me a long time to figure out without your help.

Kjel

On Thu, Sep 15, 2022 at 11:53 PM <pyo-d...@googlegroups.com> wrote:
Aaron Krister Johnson <akjm...@gmail.com>: Sep 15 01:22PM -0700

N.B.: if you apply `Pow()` to `MidiAdsr()`, I think the latter is 0-1
scaled, and depending on your base that you're raising from, you'll want to
re-normalize, likely by subtraction and multiplication
 
Consider that `Pow()` is like python's `math.pow()`, but applied to a
vector signal. So, you can see that:
 
math.pow(2, 0) = 1
math.pow(2, 1) = 2
 
and
 
math.pow(10, 0) = 1
math.pow(10, 1) = 10
 
so in general, after any signal ranging from 0-1 is applied as the
exponent, you'll want to subtract 1 and divide by the (base - 1)
 
so if the general form is
 
math.pow(b, x)
 
and x is your control or audio signal, you'll do:
 
base = Sig(<your_real_base>)
scaled_control = (Pow(base, MidiAdsr(...)) - 1) / (base - 1)
 
and your scaled control will now be between 0-1 in range, but have
analogous curvature that you want per a given power base.
 
This works similarly with `Log()`
 
Cheers,
 
AKJ
 
 
 
 
Aaron Krister Johnson
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