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Redbook CD: What would specs be if it were invented today?
Old 28th January 2013 | Show parent
Lives for gear
🎧 10 years
Originally Posted by haberdasher ➡️
And there's the, er, bit I forgot, thanks! So in terms of minimum pit/land length EFM allows the amount of data stored to be 8/17*3 =~ 1.4 times as much, not 1.7.
I do not quite get your math. An actual 8-bit data byte (which already includes redundancy and interleaving) needs 17 bits of space on the disk surface. This means that the amount of data* on the disk surface is bloated 17/8 = 2.125 fold just because it is safest for the laser to read just the transitions from reflection/no reflection and vice versa and interpret those as 1, and clearly there can not be two ones/transitions too close to each other, or too far for each other (clock accuracy). Thus the minimum and maximum pit/land lengths specified. Simple and it works.

*) each 17 bits hold only 8 bits worth of data actually, so the amount of data is not increased at all, coding is "wasteful" but necessary.
Old 28th January 2013
Gear Addict
🎧 10 years
Because the minimum length between transitions (i.e. minimum wavelength of the digital signal) that can be reliably read by a given laser/system remains the same regardless of the coding scheme used.

So as we've said, if EFM were not implemented, the minimum wavelength stored is a transition every bit (i.e. 1111), whereas with EFM there are at least three bits between every transition (I think is actually there is one exception to that in non-audio data, but I can't remember what - possibly something to do with subcode!)

For a given acceptable minimum wavelenth then you can then pack bits three times as tightly using EFM as without. So yes you have 2.1x as many bits for a given amount of data, but by squeezing each bit into a third of the space you can have 1.4x as much data.

Yes the maximum distance between transitions is important too for tracking, but I am ignoring that for this point as it doesn't really affect things
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