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The Math Behind the Compact Disc

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The Math Behind the Compact Disc. Linear Algebra and Error-Correcting Codes. william j. martin. mathematical sciences. wpi wednesday december 3. 2008 fairfield university. How the device works.

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### The Math Behind the Compact Disc

Linear Algebra and Error-Correcting Codes

william j. martin. mathematical sciences. wpi

wednesday december 3. 2008

fairfield university

How the device works

The compact disc is a complex system incorporating interesting ideas from engineering, physics, CS and math. We will focus only on the mathematics of the error- correction strategy.

For more info on the CD, see Kelin

Kuhn’s book “Laser Engineering”:

W J Martin Mathematical Sciences WPI

Borrowed from K J Kuhn’s book “Laser Engineering”

W J Martin Mathematical Sciences WPI

The Pits

- Each pit is 0.5 microns wide…
- and 0.83 to 3.56 microns long.
- Tracks are separated by 1.6 microns of “land”
- Wavelength of green light is about 0.5 micron
- 40 tracks under one strand of human hair

W J Martin Mathematical Sciences WPI

Modelling a CommunicationsChannel

Linear algebra model:

r = m+e (vector add.)

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Channel with Error Correction

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A number system that the computer can understand:

F = { 0, 1 }

Ordinary multiplication

Addition: 1+1=0

Now music is turned into binary vectors!

Turn it into an algebra problem!W J Martin Mathematical Sciences WPI

A bit (or a nibble?) of graph theory

- The n-cube is a type of Hamming graph
- Vertices are all binary n-tuples
- n-tuples are adjacent if they differ in only one coordinate
- Nice ‘eigenvalues’!

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Binary Vector Spaces

- The vectors are all possible binary n-tuples

0 0 1 0 1 1 1 0 1 0 1 1 0 0 0

+

0 0 1 1 1 1 0 0 0 0 0 0 0 0 1

=

0 0 0 1 0 0 1 0 1 0 1 1 0 0 1

W J Martin Mathematical Sciences WPI

This is a metric:

dist( x, y ) 0 with dist( x, y ) = 0 iff x=y

dist( x, y ) = dist( y, x )

Triangle inequality

dist( x, z ) dist( x, y ) + dist( y, z )

Hamming DistanceThe distance between two binary n-tuples x and y is the number of coordinates in which they differ

dist( 001100, 001011 ) = 3

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Theorem

n

Let C (the “code”) be a subset of F with minimum distance between any two codewords equal to d.

Then there exists an algorithm which corrects up to t errors per transmitted codeword if and only if d 2t + 1.

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Proof

If x and y are distinct codewords, then the balls of radius t around them are disjoint. So if the received vector is within distance t of x, it must be at distance > t from any other codeword. So decoding is unique.

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A Useful Extension of the Theorem

The above (computationally infeasible) decoding algorithm also correctly recovers from any t symbol errors and any s symbol erasures provided d > 2t+s.

transmit: 0 1 1 2 2 3 0

receive: 0 1 3 3 ? ? ?

(here, t=2 errors and s=3 erasures)

W J Martin Mathematical Sciences WPI

Small Example

- Let C denote the “rowspace” of the matrix
Then C = { 000000, 110100, 011010, 101110,

001101, 111001, 010111, 100011 }

and C has minimum distance 3 so C allows correction of any single-bit error in any transmitted codeword.

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The binary Hamming code

Codewords: 0 0 0 0 0 0 0 1 1 1 1 1 1 1

1 1 0 1 0 0 0 0 0 1 0 1 1 1

0 1 1 0 1 0 0 1 0 0 1 0 1 1

0 0 1 1 0 1 0 1 1 0 0 1 0 1

0 0 0 1 1 0 1 1 1 1 0 0 1 0

1 0 0 0 1 1 0 0 1 1 1 0 0 1

0 1 0 0 0 1 1 1 0 1 1 1 0 0

1 0 1 0 0 0 1 0 1 0 1 1 1 0

- Quadratic Residues!
- In we have
- = 1 6 = 1
- = 4 5 = 4
- 3 = 2 4 = 2

Z

Z

7

2

2

2

2

2

2

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The Fano projective plane

3

Vector Space: F

“Poynts”: 1-dim. subspaces

“Lynes”: 2-dim. subspaces

2

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All codewords:

0 0 0 0 0 0 0 1 1 1 1 1 1 1

0 0 0 1 1 1 1 1 1 1 0 0 0 0

0 1 1 0 0 1 1 1 0 0 1 1 0 0

0 1 1 1 1 0 0 1 0 0 0 0 1 1

1 0 1 0 1 0 1 0 1 0 1 0 1 0

1 0 1 1 0 1 0 0 1 0 0 1 0 1

1 1 0 0 1 1 0 0 0 1 1 0 0 1

1 1 0 1 0 0 1 0 0 1 0 1 1 0

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Codes from polynomials

Let’s replace F={0,1} with F={0,1,…,6} (with modular arithmetic). Now consider the vector space F[z] of all polynomials in z with coefficients in F. For any subset N of F, we have a linear transformation

L: F[z] F

via f(z) [ f(0), f(1), f(2), f(3), f(4), f(5) ] (Here, we use, N={0,1,2,3,4,5}.)

This is a Reed-Solomon code.

N

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Polynomials to Codewords

Example:

- Let the message be [1, 2, 2] (working mod 7)
- Polynomial is f(z) = z + 2 z + 2
- Codeword is
[f(0), f(1), f(2), f(3), f(4), f(5)] = [ 2, 5, 3, 3, 5, 2]

2

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Reed-Solomon Codes

FACT: Two polynomials of degree less than k having k points of intersection must be equal.

SO: Reed-Solomon code of length n<q and dim k has min. dist. n-k+1

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Compact Disc Parameters

SONY/Philips design (1980)

- Music is sampled 44,100 times per second
- Each sample consists of 32 bits, representing
left and right channel signal magnitude 0—65535 (Pulse Code Modulation – PCM)

- So chip must process 1,411,200 raw data bits per second
- But it gets much worse!

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Cross-Interleaved RS Codes

- Inner code is a 28-dimensional subspace of a
32-dimensional vector space over a finite field of size 256.

- Outer code is a 24-dimensional subspace of a 28-dimensional vector space.
- Six 32-bit samples make up a 192-bit frame which is encoded as a 224-bit codeword. (Eventually, codewords have length 588 bits!)

W J Martin Mathematical Sciences WPI

Encoding – The numbers

- The codewords from the first code are interleaved into a virtually infinite array of 28 rows of symbols over GF(256).
- We pull out 8 binary columns (one symbol) to obtain a 28x8=224-bit frame which is then encoded using another Reed-Solomon code to obtain a codeword of length 256 bits.

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Interleaving to disperse errors

- Codewords of first code are stacked like bricks
- 28 rows of vectors over GF(256)
- Extract columns and re-encode using second Reed-Solomon code

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Splitting Odd and Even Bits

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Back to the Pits

- Each pit is 0.5 microns wide…
- and 0.83 to 3.56 microns long.
- Tracks are separated by 1.6 microns of “land”
- Not all 01-sequences can be recorded

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EFM: Eight-to-Fourteen Modulation

- This encoding scheme can only store sequences where each consecutive pair of ones is separated by at least 2 and at most 10 zeros
- This is achieved by a mapping F F
which is given by a lookup table.

8

14

2

2

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Further Processing

- Three more ‘merge bits’ are added to each of these 14
- So 256+8=264=33x8 bits, carrying six samples, or 192 information bits, gets encoded as 588 channel bits on the disk
- This represents 0.000136 seconds of music

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What actually goes on the disc?

- We must do this 7,350 times per second
- So CD player reads 4,321,800 bits per second of music produced
- To get 74 minutes of music, we must store
74x60x4321800 = 19,188,792,000

bits of data on the compact disc!

W J Martin Mathematical Sciences WPI

When in doubt, erase

- Inner code has minimum distance 5 (over GF(256))
- Rather than correct two-symbol errors, the CD just erases the entire received vector.

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So…how good is it?

- The two Reed-Solomon codes team up to correct ‘burst’ errors of up to 4000 consecutive data bits (2.5 mm scratch on disc)
- If signal at time t cannot be recovered, interpolate
- With smart data distribution, this allows for recovery from burst errors of up to 12,000 data bits (7.5 mm track length on disc)
- If all else fails, mute, giving 0.00028 sec of silence.

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Space communications (Mariner,Voyager,etc.)

DVD, CD-R, CD-ROM

Cell phones, internet packets

Memory: chips, hard drives, USB sticks

RAID disk arrays

Quantum computing

Other ApplicationsW J Martin Mathematical Sciences WPI

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