Yes, eight data bits are converted to 14 channel bits that are recorded onto
the disc. Channel bit ones are represented by pit/land or mark/land
transitions. Channel bit zeros are not recorded. Eight-to-fourteen
modulation is accomplished using a lookup table based on a 2-of-10 RLL code.
This means channel bit ones are always separated by at least two and no more
than ten channel bit zeros. This controls the minimum and maximum pit/mark
or land lengths that vary from 3T to 11T.
Pits or marks return a low amount of light while lands return a high amount
of light. This analog signal is converted to a digital waveform, and the
transitions are recognized as channel bit ones. A 1T read data clock is
phase locked to this digital waveform, and inserts channel bit zeros where a
one is absent. This recreates the original 14 channel bits that are then
converted back to 8 data bits (one byte.)
DVD is similar, except for an 8-to-16 code.
Media Sciences, Inc.
From: Association for Recorded Sound Discussion List
[mailto:[log in to unmask]] On Behalf Of Don Cox
Sent: Thursday, April 22, 2004 12:06 PM
To: [log in to unmask]
Subject: Re: [ARSCLIST] What type of file are music CD's saved as
On 22/04/04, Steven C. Barr wrote:
> Then is it the bits (pits and lands) on the CD surface that represent
> the ones and zeros after being processed by an algorithm so that
> sequences of either can be compressed?
There is no compression on a standard audio CD.
The ones are represented by a bump beginning or ending, and the zeros by
no change. For technical reasons such as avoiding a long sequence of
ones or zeros, each 8 bit byte is converted to a 14-bit code for actual
storage on the disc. Then there is a great deal of rearrangement of data
to make error correction robust.
(Similar things happen on hard drives.)
> Thus, reprocessing the ones and
> zeroes with the same algorithm creates the .wav file? (decimal:) (0)
> (63) (255) (252) Thus, if the waveform in digital is 00000000 00111111
> 11111111 11111100... the actual bits on the CD surface represent the
> lengths of those strings of ones and strings of zeroes rather than the
> bits that compose the digitized
> waveform themselves...right?
Not exactly the lengths, but a long string of zeros (in the 14 bit code)
would give a long bump or a long gap, yes. Codes are chosen to avoid
sequences of ones.
The reference for all this is "Digital Audio and Compact Disc Technology"
edited by Baert et al, sponsored by Sony.
The CD was way ahead of its time when it first came out.
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