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A question on frame bitrate, Frame bitrate calculation on lossless formats
MondayMassacre
post Oct 20 2013, 20:45
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Hi there,

I've been reading up on some of the principles of lossless compression, yet I fail to grasp the possibly obvious. As far as I understand, there are several possibilities and steps to compress audio data; blocking, prediction, residual coding and some other stuff I'm not even sure I wanna know of, otherwise it would make my head spin even further. But probably this might be the problem...

So there I have this piece of music, moderately complex (meaning neither total silence nor pure noise) and when I open this (e.g.) flac-file in VLC and open the media information dialog, I see the bitrate fluctuating heavily. Now my question is: Is this bitrate indeed reflecting a combination of the three "basic" elements of samplerate * depth * channels or is this simply the amount of encoded data VLC is currently reading from the flac-file?
If a lossless format uses a bitrate as high as necessary to restore the signal, which part of the signal can be left out? What's variable? The samplerate? The bit depth?

Confused greetings from Austria
M
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saratoga
post Oct 20 2013, 21:00
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QUOTE (MondayMassacre @ Oct 20 2013, 15:45) *
Now my question is: Is this bitrate indeed reflecting a combination of the three "basic" elements of samplerate * depth * channels or is this simply the amount of encoded data VLC is currently reading from the flac-file?


Its the average size of the compressed audio over some interval.

QUOTE (MondayMassacre @ Oct 20 2013, 15:45) *
If a lossless format uses a bitrate as high as necessary to restore the signal, which part of the signal can be left out? What's variable? The samplerate? The bit depth?


Nothing is left out. Lossless formats are lossless. Zero loss.

This may be helpful to you: http://en.wikipedia.org/wiki/Lossless_compression
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MondayMassacre
post Oct 20 2013, 21:45
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So EVERY uncompressed sample weighs in at 44.1kHz * 16 Bit depth * 2 channels = 1.411.200 bits, no matter if noise or silence, yet the size of the stored data of the sample is dependent on the compression, and hence the bitrate is, as mentioned above, the amount of encoded data for some time interval? So the bitrate reflects the amount of data (i.e. compression ratio) used by an algorithm to store 1.411.200 bits of information (per time interval)?

Therefore: 1.411 kbit/s mean, the signal in that second is too complex to be compressed even a bit. And any bitrate below equals a compression rate of ([bitrate] / (1.411.200 / 100))% of the signal in that second.

Thanks! I think I will go and let Huffman make my head spin a bit more.

This post has been edited by MondayMassacre: Oct 20 2013, 21:46
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saratoga
post Oct 20 2013, 22:01
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That's correct. An incompressible signal would have a bit rate of 1.411 kbit/s or perhaps even higher due to overhead. Most music signals can be compressed to perhaps two thirds of their original size for stero cd pcm.
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Nystagmus
post Feb 21 2014, 01:26
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QUOTE (saratoga @ Oct 20 2013, 15:01) *
That's correct. An incompressible signal would have a bit rate of 1.411 kbit/s or perhaps even higher due to overhead. Most music signals can be compressed to perhaps two thirds of their original size for stero cd pcm.


Unless the original audio is greater than 44.1 kHz 16-bit (1411 kbps) also! Each sample rate / bit resolution has it's own characteristics. Not everything in the audio world is RedBook audio. I realise it's obvious to many of us but it's worth saying. Other than that, I totally agree with Saratoga. Interesting thread too.


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