Op 18 jan 2009, om 02:02 heeft Franco Amato het volgende geschreven:
> 2009/1/17, Arjan <[email protected]>:
>>
>>
>> Op 18 jan 2009, om 00:54 heeft Franco Amato het volgende geschreven:
>>
>>> Hi to all,
>>> I have a problem that I can not solve.
>>> In my encoder I'm encoding video and audio, the problem comes with
>>> the audio
>>> part.
>>> My original audio raw samples are pcm float -1/+1 stereo. So I have
>>> 4 bytes
>>> (2 for the L channel and 2 for the R channel TOTAL = 4 bytes).
>>
>> floats are in fact 4 bytes -> 32bit
>> for a second of audio stored using float with a sample rate of 48Khz,
>> you would need 48000 * 2 chan * 4 bytes = 384000 bytes.
>
>
> I don't understand. Which is this size?
let's break this down
float sourceSamples[64];
int16_t destSamples[64];
int i;
for(i=0; i<64; i++) {
destSamples[i] = sourceSamples[i] * 0x7FFF;
}
it's really very simple.. just let the compiler care about the sizes
of the types
In regard to your question about the low audio level, try the following.
(please try to condense your messages, as not to litter the mailing
list)
you first need to find the maximum amplitude
float max = 0.0;
int i;
for(i=0; i<64; i++) {
if( abs( sourceSamples[i] ) > max ) max = abs( sourceSamples[i] );
}
now max holds the largest amplitude in your collection of samples.
Your theoretical max is 1.0, so by calculating the difference, you can
check how much you can amplify your samples while remaining in the
16bit range.
(0x7FFF is the max of a 16 bit signed int)
float multiply = 0x7FFF;
if( max ) multiply *= 1.0 / max;
so instead of multiplying with 0x7FFF, you now multiply with 'multiply'
Please note that you can't keep calculating this max value, as every
collection of samples will be differently amplified. You have to
calculate this max value once for all samples belonging to the same
piece of audio.
- Arjan
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