How to interpret WAV data?
Peter Reichard
I am writting a program to draw a sample curve. I have read some online
documents about the WAV formate, however, my curve doesn't look like in
other sample editors.
If I have a 8bit mono sample, one article described that the data is saved
like:
channel 0 | channel 0 | channel 0
[BYTE] [BYTE] [BYTE]
The range for each value is therefore 0-255 (128 means zero). However, an
other article says that each channel ist stored as an integer. For me, that
means 2 bytes, like:
channel 0 | channel 0 | channel 0
[BYTE][BYTE] [BYTE][BYTE] [BYTE][BYTE]
Because it has the same range (0-255), it would be a waste of memory and
only applicable for 16bit mono samples. What is the correct storage scheme?
Thanks
Iain
You can find the format for wavs in the MSDN documentation.
Basically there is a preamble then a format descriptor (which tells you how
many bits per sample among other things).
then there is a sized chunk of data.
If you know that your data is 8 bit mono, it will be stored as a sequence
of signed bytes.
If it were 16 bit stereo it would be stored as pairs of signed shorts.
Iain
--
Iain Downs (DirectShow MVP)
Software Product Consultant
www.idcl.co.uk
Alessandro Angeli [MVP::DigitalMedia]
> If you know that your data is 8 bit mono, it will be
> stored as a sequence of signed bytes.
UNSIGNED bytes, not signed :-)
> If it were 16 bit stereo it would be stored as pairs of
> signed shorts.
--
// Alessandro Angeli
// MVP :: Digital Media
// a dot angeli at psynet dot net
Iain
wrote:
> Iain wrote:
>
>> If you know that your data is 8 bit mono, it will be
>> stored as a sequence of signed bytes.
>
> UNSIGNED bytes, not signed :-)
>
>> If it were 16 bit stereo it would be stored as pairs of
>> signed shorts.
Surely the values go +/- zero? So FF == -1 and so on. And if cast to a
larger integer value the sign bit should be extended?
Phil Frisbie, Jr.
> On Wed, 18 May 2005 18:42:59 +0200, Alessandro Angeli [MVP::DigitalMedia]
> wrote:
>
>>Iain wrote:
>>
>>>If you know that your data is 8 bit mono, it will be
>>>stored as a sequence of signed bytes.
>>
>>UNSIGNED bytes, not signed :-)
>>
>>>If it were 16 bit stereo it would be stored as pairs of
>>>signed shorts.
>
>
> Surely the values go +/- zero?
No. 8 bit PCM is unsigned, with 127 the 'zero' level.
> So FF == -1 and so on. And if cast to a
> larger integer value the sign bit should be extended?
>
> Iain
--
Phil Frisbie, Jr.
Hawk Software
http://www.hawksoft.com
Peter Reichard
>
> No. 8 bit PCM is unsigned, with 127 the 'zero' level.
That's right, 16bit has signed values and 0 is the zero level.
Alessandro Angeli [MVP::DigitalMedia]
> No. 8 bit PCM is unsigned, with 127 the 'zero' level.
128 is the zero level.
Iain
wrote:
> Phil Frisbie, Jr. wrote:
>
>> No. 8 bit PCM is unsigned, with 127 the 'zero' level.
>
> 128 is the zero level.
I sit corrected!
Alessandro Angeli [MVP::DigitalMedia]
> I think I missed the answer ;-)
> Say you are using DirextX to strip out the header details
> i.e. Formats; is there a simple algorithm that can be
> used on all wave formats, something that say can be
> iterated through, dependant on the number of channels and
> format of the buffer?
> Or is there more than 2 storage possibilities which stops
> this from being so simple?
If your data is PCM (wFormatTag == WAVE_FORMAT_PCM), the
samples are wBitsPerSample/8 bytes wide, in little endian
order, and there are nChannels samples per frame, in channel
order. The samples are signed and silence is 0 but for 8-bit
samples which are unsigned and silence is 128.