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I'm learning a few things about codec, and now I'm confused by G.722. There's a way to calculate codec bit rate that I read here on wikipedia.

It said: bit rate = sampling rate x bit depth

G.722 has 16 kHz sampling rate and 14 bit depth. If I calculate them accurately, G.722 has 224 kbps bit rate.

16 x 14 = 224

But another wikipedia article said that G.722 has 64 kbps bit rate.

How is that possible? I've googled through this matter and got nothing.

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Compression? Your calculation shows what data rate is require to capture the audio. The 64kbps is what it transmits at post-compression. –  Paul May 17 '12 at 14:28

2 Answers 2

up vote 1 down vote accepted

In G.722, the 64 kbps (7 kHz) audio encoder includes a transmit audio part which converts an audio signal to a uniform digital signal which is coded using 14 bits with 16 kHz sampling and a SB-ADPCM encoder which reduces the bit rate to 64 kbps.

Source

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oh, I see. Just when I thought that digital signal which is coded using 14 bits with 16 kHz sampling was already a compressed form of analog signal. Thanks –  pepito May 22 '12 at 14:41
    
That's an uncompressed digital representation of the analog signal. –  pjc50 May 22 '12 at 16:25

Your math would be correct if there was no compression/encoding. Since G.722 uses encoding (SB-ADPCM, see spec for details: http://www.itu.int/rec/dologin_pub.asp?lang=e&id=T-REC-G.722-198811-I!!PDF-E&type=items), the data rate is lower than the one of the raw pcm stream.

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