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Bit representation on a CD

C

c.j[dot]w

Jan 1, 1970
0
Hello,

This question has some connections with the other one that I recently
posted.

When data is stored on a CD, a straight-forward approach would be to
store a 0 as a land and a 1 as a pit, or vice-versa. But the
representation used is that a 1 is representated by a change from pit to
land or from land to pit, and a 0 is represented by a pit or a land.

My questions is: Why is it better to use the transition/not-transition
approach? I have read an explanation, but don't understand it. It is
quoted below this post. The whole text can be found at
http://pauillac.inria.fr/~lang/hotlist/cdrom/Documents/tech-summary.html.

Thanks in advance!
/Carl




To ensure accurate recovery, the disc data must be encoded to optimize
the analog-to-digital conversion process that the radio frequency signal
must undergo. Goals of the low level data encoding include:

1. High information density. This requires encoding that makes the best
possible use of the high, but limited, resolution of the laser beam and
read head optics.
2. Minimum intersymbol interference. This requires making the minimum
run length, i.e. the minimum number of consecutive zero bits or one
bits, as large as possible.
3. Self-clocking. To avoid a separate timing track, the data should be
encoded so as to allow the clock signal to be regenerated from the data
signal. This requires limiting the maximum run length of the data so
that data transitions will regenerate the clock.
4. Low digital sum value (the number of one bits minus the number of
zero bits). This minimizes the low frequency and DC content of the data
signal which permits optimal servo system operation.

A straightforward encoding would be to simply to encode zero bits as
land and one bits as pits. However, this does not meet goal (1) as well
as the encoding scheme actually used. The current CD scheme encodes one
bits as transitions from pit to land or land to pit and zero bits as
constant pit or constant land.

To meet goals (2) to (4), it is not possible to encode arbitrary binary
data. For example, the integer 0 expressed as thirty-two bits of zero
would have too long a run length to satisfy goal (3). To accommodate
these goals, each eight-bit byte of actual data is encoded as fourteen
bits of channel data. There are many more combinations of fourteen bits
(16,384) than there are of eight bits (256). To encode the eight-bit
combinations, 256 combinations of fourteen bits are chosen that meet the
goals. This encoding is referred to as Eight-to-Fourteen Modulation
(EFM) coding.

If fourteen channel bits were concatenated with another set of fourteen
channel bits, once again the above goals may not be met. To avoid this
possibility, three merging bits are included between each set of
fourteen channel bits. These merging bits carry no information but are
chosen to limit run length, keep data signal DC content low, etc. Thus,
an eight bit byte of actual data is encoded into a total of seventeen
channel bits: fourteen EFM bits and three merging bits.

To achieve a reliable self-clocking system, periodic synchronization is
necessary. Thus, data is broken up into individual frames each beginning
with a synchronization pattern. Each frame also contains twenty-four
data bytes, eight error correction bytes, a control and display byte
(carrying the subcoding channels), and merging bits separating them all.
Each frame is arranged as follows:

Sync Pattern24 + 3channel bits
Control and Display byte14 + 3
Data bytes12 * (14 + 3)
Error Correction bytes 4 * (14 + 3)
Data bytes12 * (14 + 3)
Error Correction bytes 4 * (14 + 3)

TOTAL588channel bits
 
R

Rich Grise

Jan 1, 1970
0
Hello,

This question has some connections with the other one that I recently
posted.

When data is stored on a CD, a straight-forward approach would be to
store a 0 as a land and a 1 as a pit, or vice-versa. But the
representation used is that a 1 is representated by a change from pit to
land or from land to pit, and a 0 is represented by a pit or a land.

My questions is: Why is it better to use the transition/not-transition
approach? I have read an explanation, but don't understand it. It is
quoted below this post. The whole text can be found at
http://pauillac.inria.fr/~lang/hotlist/cdrom/Documents/tech-summary.html.

It's called NRZ (Non-Return to Zero) recording; maybe drawing a little
timing diagram would help you see the point - it ensures that there are
enough edges to recover the clock, and stuff like that. I put "nrz data"
in google and got this results page:
http://www.google.com/search?q=nrz+data&btnG=Google+Search

Hope This Helps!
Rich
 
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