Re: [RFC] TLV entry for linear volume

[James Courtier-Dutton <James@superbug.co.uk>]

Takashi Iwai wrote:
> At Wed, 23 Aug 2006 19:22:37 +0100,
> James Courtier-Dutton wrote:
>> Takashi Iwai wrote:
>>> At Wed, 23 Aug 2006 17:31:10 +0200,
>>> I wrote:
>>>   
>>>>> So, 0 dB is when (volume - rec->min) / (max - min) == 1
>>>>> So, for +6 dB gain, the volume will need to be higher than max?
>>>>> Does this sound right? In that case, maybe "max" is not a good name for 
>>>>> it because volume can be greater than max.
>>>>>       
>>>> Hm, right, I didn't think of overload case with a linear volume
>>>> codec.  By min and max, I thought of a "segment" between mute and
>>>> 0dB (although not implemented rightly in alsa-lib).
>>>>
>>>> Maybe min and max should be in (0.01) dB expressions since the min and
>>>> max "values" are known from snd_ctl_elem_info.  The only problem is
>>>> that we have no standard definition of "mute" in dB expression.
>>>> In alsa-lib, -9999999 indicates mute.  But it should be defined as a
>>>> constant in a public header.
>>>>     
>>> It turned out that minimal dB makes the computation too complicated,
>>> and chips are very likely from mute to a certain dB.  So, I decided to
>>> drop the min dB there.
>>>
>>> The below are the revised patches.
>>>
>>>
>>> Takashi
>>>   
>> I think you will find that your second approach is still wrong.
>> You need to first identify the 0dB point, and then the scale factor and 
>> offset value for the linear component before the log10 conversion.
>> So, linear to log should be something like:
>> 20log10 ((linear / scale) - offset)
> 
> What is linear and scale here?
I explain that in the rest of my email.
but I will repeat differently:
"linear" is your X value that the hardware requires, that you say is a
linear scale and not a log10 scale.
"scale" is some multiplier to convert the integer X value into some
floating point value. Together with an offset value to reach the correct
0db, +6dB and -6dB gain points.

> 
> In my case, the calculation is so simple as below.
> 
> The value y at x [0,max] is
> 
>    y = (x / max) * maxVal
> 
> where maxVal = 10^(maxdB/20)
> 
> so, in dB expression,
> 
>   Y = 20 * log10(y) = 20 * log10((x / max) * 10^(maxdB/20))
>     = 20 * log10(x/max) + maxdB
> 
> 
> Takashi

It is unclear from your equation where the 0dB point is.
>From your equation, when x = max( i.e. what I would expect to be the 0dB
point as a result of comments by you in previous emails), one reaches
the maxdB point, and not the 0dB point. Getting the 0dB point on the
log10 curve is very important, otherwise the conversion from linear to
dB occurs at the wrong point on the log10 curve, resulting in the dB
values all being wrong.
Remember, these dB values are dB GAIN, not absolute dB levels.

James



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