Chapter 6 BCH code -...

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Chapter Chapter 66

BCH codeBCH code

Ammar Abu-Hudrouss Islamic University Gaza

Spring 2009

Slide ٢Channel Coding Theory

Zeros of cyclic code

Many of the most important block codes for random-error correction fall into the family of BCH codes, named after their discoverers Bose, Chaudhuri and Hocquenghem.

BCH codes include Hamming codes as a special case.

The construction of a t-error correcting binary BCH code starts with an appropriate choice of length:

n = 2m - 1 (m is integer 3)

It is given that n - k mt and dmin 2t + 1

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Slide ٣Channel Coding Theory

Zeros of cyclic code

A t-error correcting q-ary BCH code of length qm - 1 is a cyclic code whose roots include 2t consecutive powers of , the primitive element of GF(qm).

Two main cases of interests

Binary BCH codes will consist of binary sequences of length 2m —1 with roots including 2t consecutive powers of the primitive element of GF(2m).

Reed Solomon codes are the special case where m = 1. They therefore consist of sequences of length q — 1 whose roots include 2t consecutive powers of the primitive element of GF(q).

Slide ٤Channel Coding Theory

Construction of Binary BCH code

We construct GF(16) so that we can define codes of length 15. The primitive polynomial to be used is p(X) = X4 + X + 1. The polynomial basis representation of the elements is

ValueElementValueElement101170000

0101800010

1010900101

01111001002

11101110003

11111200114

11011301105

10011411006

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Slide ٥Channel Coding Theory

Single error correcting code

we choose 1 and 2 as the consecutive roots for this code.

We know that there will be conjugate roots, the full set of conjugates being 12 48

The generator polynomial is

842)( XXXXXg

0013117121069532

84234)(

XXXXXg

1)( 4 XXXg

Slide ٦Channel Coding Theory

Single error correcting code

Alternatively we find that the generator polynomial is the LCM of the minimum polynomials of 1 and 2.

Both these elements have X4 + X + 1 as their minimum polynomial, therefore this is the generator of the code. The code has 4 parity checks (from the degree of the generator) and so is a (15, 11) code.

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Slide ٧Channel Coding Theory

Double error correcting code

Suppose we choose 1 2 3 and 4 as the consecutive roots for this code.

Starting of and including the conjugates we have roots 1 2

4 8

Which include 2 and 4 but not 3 . The conjugates of 3 are 3 6 12 9

The degree of the generator is 8, so the code is (15, 7).

91263842)( XXXXXXXXXg

11)( 2344 XXXXXXXg

1)( 4678 XXXXXg

Slide ٨Channel Coding Theory

Triple error correcting code

For triple correcting code we want to have 1 2 3 4 5 and 6 all the roots are included in the double code but 5 with conjugates 5 10

The degree of the generator is 10, so the code is (15, 5).

1054678 1)( XXXXXXXg

11)( 24678 XXXXXXXg1)( 245810 XXXXXXXg

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Slide ٩Channel Coding Theory

Roots and Parity Check Matrix

Consider the code generator polynomial

And if is a root of v (X ) then

Or in vector form

012

21

1 vXvXvXvXv nn

012

21

10 vvvvv nn

0

1

2

1

n

v

Slide ١٠Channel Coding Theory

Roots and Parity Check Matrix

Similarly if there are j roots from 1 to j

But vHT = 0

0

1111121

221

22

21

111

12

11

jj

jj

nj

nj

nn

v

TTj

Tj

Tnj

TTj

Tj

Tnj

TTTTn

TTTTn

H

1

1

11

211

21

11

222

12

121

11

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Slide ١١Channel Coding Theory

Parity Check Matrix for Single Error

If we take element as a root, then

The powers of a are just all the nonzero elements of the field

For example taking the code based on GF(8), for which 3 + + 1 = 0, gives

TTTTnH 121

100101101011100010111

H

TTTTTTT

H 1123456

Slide ١٢Channel Coding Theory

Parity Check Matrix for Double Errors

Suppose we choose 1 2 3 and 4 as the consecutive roots for this code, and 3 are independent, then

For length 15 we have

TTTTn

TTTTn

H11

32313

21

11

36912036912036912

12356789101112131415

TTTTTTTTTTTTTT

TTTTTTTTTTTTTT

H

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Slide ١٣Channel Coding Theory


Which gives

100011000110001000110001100011001010010100101011110111101111100010011010111010011010111100001001101011110000100110101111

H

Slide ١٤Channel Coding Theory

Parity Check Matrix for Triple Errors

Suppose we choose 1 2 3 4 5and 6 as the consecutive roots for this code, , 3 and 5 are independent, then

For length 15 we have

TTTTn

TTTTn

TTTTn

H

1

11

52515

32313

21

TTTT

TTTT

TTTT

H

1

11

51010

32312

214

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Slide ١٥Channel Coding Theory


After deleting all zeros row and duplicated row

101101101101101011011011011011100011000110001000110001100011001010010100101011110111101111100010011010111010011010111100001001101011110000100110101111

H

Slide ١٦Channel Coding Theory

Algebraic Decoding

BCH codes allow an algebraic method of decoding. Consider the case of a double error correcting code where there are errors at positions i and j .The syndrome

The syndrome has two components

Substituting the first in the second

Roots can be found by trying all possible values of i.

TeHs

jj

ji

s

s33

3

1

0331

21

21 ssss ii

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Slide ١٧Channel Coding Theory

Algebraic Decoding

Example: For the double-error correcting BCH code of length 15, if the received sequence is 101010110010101

Subsisting in equation in the previous slide

The equality holds for i = 2 and i = 13.

The errors are therefore at positions 2 and 13 giving a transmitted sequence m = 111010110010001

51410010110 TRHs

051221413 ii

00000000 TmHs

Slide ١٨Channel Coding Theory

BCH Decoding and the BCH bound

We want to prove here that if we want to 2t consecutive zeros in the frequency domain are sufficient to guarantee t error corrections.

Assume d = 2t + 1

Suppose we have a code vector c (X ) with fewer than d nonzero components (i.e dmin = weight(c) < d)

As c (X ) is a valid codevector, it has 2t consecutive roots. Which means that the spectrum C (z ) will have 2t = d -1.

Assume (X ) which is the inverse of c. i.e it equals zero when c =1 and vice versa.

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Slide ١٩Channel Coding Theory


The zero components in (X ) means that each ij will be root of the transform (z ) of (X ) . ij is the position of zeros in (X )

This mean

Now in the time domain, i ci = 0, therefore in the frequency domain , replacing the multiplication with a convolution,

v

j

i jzz1

1

1

0

n

jjkjC

Slide ٢٠Channel Coding Theory


But the degree v of (z ) is at most d -1 and = 0, which leaves us with

Which means that the components Ck can be generated from the previous d – 1 components as in the graph

1

1

d

jjkjk CC

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Slide ٢١Channel Coding Theory


But as C (z ) has d – 1 consecutive zero components. If these are happened to initialize the register all the next components will equals zero.

This mean that the minimum weight should equal at least 2t + 1 = d to avoid the generation of zero codeword.

Slide ٢٢Channel Coding Theory

Decoding in the Frequency Domain

Assume that we transmit a code sequence c (X) and that it is received with the addition of an error sequence e(X).

We take the transform of the received sequence, giving

We know that C(z) is zero in 2t consecutive spectral locations, so we can use these locations to give us a window on E(z),

2t components of E(z) can easily be obtained and can be considered to form a syndrome S(z).

zEzCzR

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Slide ٢٣Channel Coding Theory


We assume that there are v < t errors and define an error locator polynomial (X) such that it is zero in the positions where e (X ) is nonzero.

The product of the received sequence and the error locator sequence in the time domain will therefore be zero, which means that in the frequency domain the convolution between (z) and E (z) will be zero

00

t

jjkjE

Slide ٢٤Channel Coding Theory


If we know the error spectrum from positions m to m + 2t - 1, we can form t equations by letting k take values from m + t to m + 3t — 1.

Assuming the spectral zeros of the code are from positions 1 to 2t, the equations are as follows:

This set of t equations in t + 1 unknowns is called the key equation, this can be solved if we assume 0 = 1

00

0

12120

121120

1110

tttt

tttt

ttt

EEEEEE

EEE

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Slide ٢٥Channel Coding Theory


Example 1Take the example of the (15, 7) binary BCH code from Section 6.6 where we received a sequence 1010101 100 10101 which was found to have two errors, in positions 2 and 13.

We calculate the Fourier transform of the received sequence in the positions of the zeros, i.e. positions 1, 2, 3 and 4.

The key equation becomes

zzzzzS 1421335411

0214

1135

0213

1511

Slide ٢٦Channel Coding Theory


To solve, we can multiply each term in the second equation by and add to the first equation to eliminate 2. This gives

Hence,

Decoding can now be carried out by finding the roots of z 2 + l4z + 1,

The roots are 2 and 13. Hence the errors are at positions -2 and -13, or the error at the locations are 13,2.

141

12

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Slide ٢٧Channel Coding Theory


Example 2 Suppose now that only one error occurred, say the error in

position 13. The syndromes are found to be s1 = 13, s3 = 9, giving a full syndrome polynomial of

Hence,

Which leads

zzzzzS 132113947

0213

1119

0211

197

131 02

Slide ٢٨Channel Coding Theory


The connection polynomial is therefore 13 z + 1, which has root at -13 , indicating an error at position 13.

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Slide ٢٩Channel Coding Theory


Example 3 Suppose now that we have three errors at 2,10,13. The

syndromes are found to be s1 = a11, s3 = a10, giving a full syndrome polynomial of

Hence,

Which leads

zzzzzS 1127310414

0211

1710

027

11014

111 1

2

Slide ٣٠Channel Coding Theory


The Connection polynomial is therefor z 2 + l1 z + 1,

No roots can be found for this equation.

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Slide ٣١Channel Coding Theory

Polynomial Form of The Key Equation

We wish to find a solution to the key equation, which can be expressed as

This convolution can be expressed as multiplication of the sequence polynomial in the previous equation it represents the terms from t to 2t-1

The term of power more that 2t is represented by f(z)z2t

The term of power less than t is represented by (z)

12000

tkEt

jjkj

1202 tkzzzfzEzA t

Slide ٣٢Channel Coding Theory

Euclid's method

enables us to find minimum degree solutions for polynomials f(z) and g(z) such that

where r(z) is known to have degree less than some fixed value. In our case, r(z) will have degree < t, a(z) = z2t and b(z) is the

syndrome polynomial S(z).

The polynomial g(z) will give us (z), which is what we need to know. The method involves repeated division of polynomials until a remainder of degree < t is found.

zrzgzbzfza

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Slide ٣٣Channel Coding Theory

Euclid's method

The first step is to divide a(z) by b(z) to find the quotient q1(z) and remainder r1(z):

The degree of r1(z) is less than t then we have reached our solution with f(z) = 1, g (z) = q1 (z) and r (z) = r1 (z).

Otherwise set g1(z) = q1(z) and proceed to the next stage. The second step is to divide b(z) by r1(z) giving

Leading to

zrzrzqza 211

zrzrzqzb 212

)(1 2122 zrzbzgzqzazq

Slide ٣٤Channel Coding Theory

Euclid's method

Set g2 (z) = q2(z)g1(z) + 1. If the degree of r2(z) is less than t then g(z) = g2(z);otherwise, continue to the next step.

The third step continues in similar vein, dividing r1 (z) by r2(z):

Again the degree of the remainder is decreasing. Using Equations (6.14) and (6.15)b to eliminate r1(z) and r2(z) gives

If the degree of r3 (z) is less than t then g3(z) = g3(z) = q3(z)g2(z) + g1(z).

zrzrzqzr 3221

zrzbzgzqzgzazqzq 3231321

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Slide ٣٥Channel Coding Theory

Euclid's method

The method continues in this way until a remainder of degree less than t is found, at each stage setting

0,1 1021 zgzgzgzgzqzg nnnn

Slide ٣٦Channel Coding Theory

Euclid's method

Example We shall now solve the key equation for the double-error

example. The syndrome was given as

Divide z 4 by S (z) to give Divide z 4 by S (z ) to give

Which leads

Divide S (z) by

141325311 zzzzS

125261413253111344 zzzzzzz

1341 zzg

12526 zz

81252695141325311 zzzzzz

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Slide ٣٧Channel Coding Theory

Euclid's method

Set

As the remainder is < 2, this is the end of Euclid's algorithm. The result for g2(z) is the connection polynomial (z)

9829134952 1 zzzzzg

Slide ٣٨Channel Coding Theory

Berlerkamp-Massey Algorithm

The algorithm is easy to implement but difficult to understand how it works.

We will focus on the implementation.

Consider l represents the degree of the error locator polynomial. n represents the degree of the syndrome polynomial

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Slide ٣٩Channel Coding Theory


Berlekamp-Massey Algorithm for binary BCH example Begin l : = 0; n : = 0; k = -1; (z) := 1; D (z ) := z (z); While (n < 2t) do

If ( 0) then If (l < n – k) l* := n – k k: = n – l D (z ) := (z)/ ; l := l* end

l

iiniS

0

:

Slide ٤٠Channel Coding Theory


(z) := *(z); end; D (z ) := z D (z ); n := n + 1; end end

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Slide ٤١Channel Coding Theory




Initializations l : = 0; n : = 0; k = -1; (z) := 1; and D (z ) := z (z); First iteration

l < n – k l* := 1; k :=0; D (z ) =, l : =1

141325311 zzzzS

014 1* 14 zz

1* 14 zzz

Slide ٤٢Channel Coding Theory


D (z) = z D (z) = z; n =1;

2nd iteration (1 < 4)

D (z) = z D (z) = z2; n =2;

3rd iteration (2 < 4)

l < n – k l* := 2; k :=1; D (z ) =z + , l : =2

0

014 1* 142 zzz

1* 142 zzzz

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Slide ٤٣Channel Coding Theory


D (z) = z D (z) = z2 + z; n =3;

4th iteration (3 < 4)

D (z) = z D (z) = z3 + z2; n =4; Stop

0

Slide ٤٤Channel Coding Theory


Inversionless Berlekamp-Massey Algorithm Begin l : = 0; n : = 0; k = -1; (z) := 1; D (z ) := z (z); * = 1; While (n < 2t) do

If ( 0) then *(z) = *x (z) + x D (z); If (l < n – k) l* := n – k; * = ;

l

iiniS

0

:

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Slide ٤٥Channel Coding Theory


k: = n – l; D (z ) := (z); l := l*; End; (z) := *(z); end; D (z ) := z D (z ); n := n + 1; End; End;

Slide ٤٦Channel Coding Theory




Initializations l : = 0; n : = 0; k = -1; (z) := 1; and D (z ) := z (z); * = 1; First iteration

l < n – k l* := 1; k :=0; D (z ) =0, l : =1

141325311 zzzzS

014 1* 14 zz

1* 14 zzz

14*

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Slide ٤٧Channel Coding Theory


D (z) = z D (z) = 0z; n =1;

2nd iteration (1 < 4)

D (z) = z D (z) = 0z2; n =2;

3rd iteration (2 < 4)

l < n – k l* := 2; k :=1; D (z ) = 14z +1, l : =2

0

014 1413214* zzz

14*

1413214* zzzz

Slide ٤٨Channel Coding Theory


D (z) = z D (z) = 14z2 +z; n =3;

4th iteration (3 < 4)

D (z) = z D (z) = 14z3 +z2; n =4; Stop

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