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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.

Regards
--
Don Cox
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