[chuck-users] Converting dB to chuck .gain

[Ge Wang]

I think the relationship is:

   fun float dbtogain( float db )
   { return math.pow( 2, db/3 ); }

   fun float gaintodb( float gain )
   { return 3 * math.log(gain) / math.log(2); }

This is using 1.0 gain as reference (0 dB down).
* You may have to normalize your input data.
   (offset all dB values by constant amount)
* Both functions probably could be optimized.
* gaintodb prefers positive values.
* feel free to call me an idiot as always.

Best,
Ge!

On Wed, 8 Mar 2006, Perry R Cook wrote:

> I think gain is just linear amplitude multiplication,
> like 0.0 is nothing, and 1.0 is nominal amplitude,
> and 0.5 is halfway between in amplitude space, thus
> -3dB from 1.0.
>
> PRc
>
> On Wed, 8 Mar 2006, Graham Percival wrote:
>
>> What is chuck's .gain measured in?  Neither std.dbtorms() nor std.dbtopow() 
>> seem to work...
>> 
>> I have a bunch of data about a sound, (frequency, dB):
>> 86.132812, 9.996495,
>> 172.265625,  6.996109,
>> 258.398438,  -0.955794,
>> 344.531250,  -11.900859,
>> ...
>> 
>> and I'd like to try some additive synthesis in chuck.  The first value 
>> (frequency) is obvious, but I'm at a loss as to what to do with the second 
>> value (dB).  What's the conversion factor between dB and .gain ?
>> 
>> Cheers,
>> - Graham
>> 
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