Ecc_ Bch Codes

8/10/2019 Ecc_ Bch Codes

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Subject Seminaron

BCH Codes

(Bose , Chaudhuri, Hocquenghem Codes)

Presented byUmamaheswar.V

4JN12LDS17MTech DECS, III Sem

JAWAHARLAL NEHRU NATIONAL COLLEGE OF ENGINEERING,SHIMOGA.
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CONTENTS first slide

CONTENTS

Introduction

Binary primitive BCH codes

Generator polynomial of Binary BCH CodesProperties of Binary BCH CodesParity Check Matrix of the BCH code

Decoding procedures

Syndrome computation.

Determination of the error pattern.Error correction.

Umamaheswar.V (III Sem M.tech) BCH Codes November 11, 2013 2 / 35
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Introduction

Introduction

The Bose, Chaudhuri, and Hocquenghem (BCH) codes form a large

class of powerful random error-correcting cyclic codes.

This class of codes is a remarkable generalization of the Hammingcodes for multiple-error correction.

Binary BCH codes were first discovered by Hocquenghem in 1959 and

independently by Bose and Chaudhuri in 1960.The original applications of BCH codes were restricted to binarycodes of length 2m 1 for some integer m. These were extended laterby Gorenstein and Zieler (1961) to the nonbinary codes with symbolsfrom Galois field GF(q).

Among the nonbinary BCH codes, the most important subclass is theclass of Reed-Solomon (RS) codes.

Among all the decoding algorithms for BCH codes, Berlekampsiterative algorithm, and Chiens search algorithm are the most efficient

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For any integer m 3 and t


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Binary primitive BCH codes Generator polynomial of Binary BCH Codes:

Generator polynomial of Binary BCH Codes

Let be a primitive element in GF (2m) .For 1 i 2t, let i(x) be the minimum polynomial of the fieldelement i.

The degree ofi(x) is m or a factor of m.The generator polynomial g(x) of a t-error-correcting primitive BCHcodes of length 2m -1 is given by

g(x) =LCM{1(x), 3(x)..., 2t1(x)}

Since deg[i(x)]m,the deg[g(x)]mt. Hence the number of parity-check digits n-k of the code is atmost mt.


Bi i iti BCH d G t l i l f Bi BCH C d


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Binary primitive BCH codes Generator polynomial of Binary BCH Codes:

Note that the generator polynomial of the binary BCH code is originally

found to be the least common multiple of the minimum polynomials1(x), 2(x), 3(x)..., 2t(x)

g(x) =LCM{1(x), 2(x), 3(x)..., 2t(x)}

However, generally, every even power of in GF( 2m

) has the sameminimal polynomial as some preceding odd power of in GF( 22m ).As aconsequence, the generator polynomial of the t-error-correcting binaryBCH code can be reduced to

g(x) =LCM{1(x), 3(x)..., 2t1(x)}


Binary primitive BCH codes Example 1


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Example:Let be a primitive element of the Galois field GF(24)given by table belowsuch that 1++4=0 Find the generator polynomial of Single

error,double error and tripple error correcting BCH code.

The minimal polynomials of,3 and 5 are1(x) = 1 +X+X

4

3(x) = 1 +X+X2 +X3 +X4

5(x) = 1 +X+X2

The single-error-correcting BCH codes of length n=24-1=15 is generatedby g(x)=1(x) since t =1.

The double-error-correcting BCH code of length 15 is generated byg(x) =LCM{1(x), 3(x)}Since 1(x) and 3(x) are two distinct irreducible polynomials,g(x)=(1 +X+X4)(1 +X+X2 +X3 +X4)=1 +X4 +X6 +X7 +X8


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Thus the code is a (15,7)cyclic code with dmin 5.Because the weight of

the generator plolynomial is 5, the minimum distance of this code isexactly 5.

The triple-error-correcting BCH code of length 15 is generated by

g(x) =LCM{1(x), 3(x), 5(x)}

=(1 +X+X4)(1 +X+X2 +X3 +X4)(1 +X+X2)=1 +X+X2 +X4 +X5 +X8 +X10

Its a (15,5) cyclic code with dmin 7.Because the weight of the generatorplolynomial is 7, the minimum distance of this code is exactly 7.




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Power Polynomial 4 Tuplerepresentation representation representation

0 0 (0000)

1 1 (1000) (0100)

2 2 (0010)3 3 (0001)4 1 + (1100)

5 + 2 (0110)6 2 + 3 (0011)7 1 + + 3 (1101)8 1 + 2 (1010)9 + 3 (0101)

10 1 + 2 + 3 (1110)11 + 2 + 3 (0111)12 1 + + 2 + 3 (1111)13 1 + 2 + 3 (1011)

14

1 + 3

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Binary primitive BCH codes Properties of Binary BCH Codes
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Properties of Binary BCH Codes

Consider a t-error-correcting BCH code of length n =2m-1 withgenerator polynomial g(x).

g(x) has as , 2, 3,...,2t roots, i.e.g(i)=0 for 1 i 2t

The code polynomial v(x) is a multiple of g(x), v(x) also has, 2, 3,...,2t as roots, i.e. v(i)=0 for 1 i 2t

We have a new definition for t-error-correcting BCH code:A binary n-tuple v= (v0, v1, v2, ..., vn1) is a code word if and only if

the poly. v(x) =v0+v1x+ ... +vn1xn1

has , 2, 3,...,2t as rootsi.e. v(i) =v0+v1

i +v22i + ... +vn1

(n1)i=0




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y p p y

v= (v0, v1, v2, ..., vn1).

1i

2i

.

.

.

(n1)i

= 0 for 1 i 2t


Binary primitive BCH codes Parity Check Matrix of the BCH code
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y p y

H matrix (Parity Check Matrix) of the BCH code

H=

1 ... n1

1 (2)...(2)n1

.

.

.

1 (2t)...(2t)n1

If v is a code word in the t-error-correcting BCH code, then v HT = 0j is a conjugate ofi, then v(j)=0 iff v(i)=0(Thm. 2.11) j-th row ofH can be omitted. As a result H can be reduced to the following form :

H=

1 2... n1

1 3 (3

)2

... (3

)n

1.

.

.

1 (2t1) (2t1)2... (2t1)n1


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y y

Note that the entries of H are elements in GF(2m). Each element inGF(2m) can be represented by a m-tuple over GF(2). If each entry of H isreplaced by its corresponding m-tuple over GF(2) arranged in columnform, we obtain a binary parity-check matrix for the code.

EX 6.2For Double-error-correcting (15,7) BCH code,find the parity check

matrix. Let be a primitive element in GF(24)

The parity-check matrix is

H= 1 234 5 67 8 910 1112 13 14

13691215182124273033363942

Using 15=1 , and representing each entry of H by its 4-tuple,


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H=

100010011010111

010011010111100

001001101011110

000100110101111100011000110001000110001100011001010010100101

011110111101111

The t-error-correcting BCH code defined above indeed has minimumdistance at least 2t + 1.The parameter 2t + 1 is usually called the designed distance of the

t-error-correcting BCH code.The true minimum distance of a BCH code may or may not be equalto its designed distance. There are many cases where the trueminimum distance of a BCH code is equal to its designed distance.However, there are also cases where the true minimum distance is

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Binary BCH code with length n = 2m 1 can be constructed in thesame manner as for the case n= 2m 1.

Let be an element of order n in GF(2m) ,n | 2m 1 and g(x) be thebinary polynomial of minimum degree that has , 2,...,2t as roots.

Let 1(x), 2(x),...,2t(x) be the minimal poly. of, 2,...,2t

respectively then

g(x) =LCM{1(x), 2(x),...,2t(x)}

n = 1, , 2,...,2t are roots ofXn + 1

g(x)|(xn + 1)


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We see that g(x) is a factor ofXn + 1.

The cyclic code generated by g(x) is a t-error-correcting BCH code oflength n.

The number of parity-check digits mt

dmin 2t+ 1

Ifnot a primitive element ofGF(2m), the code is called a nonprimitiveBCH code.


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Example 6.3:Consider the Galois field GF(26).The element =3 has ordern=21.Let t=2.Let g(x) be the binary polynomial of minimum degree thathas , 2, 3, 4 as roots.

The elements , 2 and 4 have the same minimal polynomial,which is

1(x) = 1 +X+X2 +X4 +X6

The minimal polynomial of3

is

3(x) = 1 +X2 +X3

Therefore,

g(x) =1(x)3(x)= 1 +X+X4 +X5 +X7 +X8 +X9

The (21,12) code generated by g(x) is a double-error- correctingnonprimitive BCH code.


Decoding of BCH Codes
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Decoding of BCH Codes


Decoding of BCH Codes Decoding
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Suppose that a code word v(x) =v0+ v1x+ ... + vn1xn1 is transmittedand the transmission errors result :

r(x) =r0+r1x+r2x2 + ... +rn1x

n1

Let e(x) be the error pattern. Then

r(x) =v(x) +e(x)

Decoding procedure:(1) Syndrome computation.(2) Determination of the error pattern.(3) Error correction.

(1)Syndrome computation:The syndrome is 2t-tuple,

S= (S1, S2, ..., S2t) =r.HT

Si =

r(

i

) =r

0+r

1

i

+ ... +rn

1(

i

)

n1

for1 i 2

t


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Computation:Let i(x) be the minimum polynomial of

i.

Dividing r(x) by i(x),we obtain

r(x) =ai(x)i(x) +bi(x)

Where bi(x) is the remainder and i(i)=0.

Then Si=r(i) =bi(

i)

Thus, the syndrome component Sican be obtained by computing bi(x)with x=i.


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EX6.4Let be a primitive element of the Galois field GF(24) such that1 + + 4 = 0.Consider double-error-correcting (15, 7) BCH code. If r =(100000001000000)is the received vector, find the syndrome components.

The corresponding received polynomial is

r(x) = 1 +x8

The syndrome consists of four components:

S= (S1, S2, S3, S4)

The minimal polynomials for , 2 and 4 are identical and

1(x) =2(x) =4(x) =1 +x+x4

The minimal polynomials of3 is

3(x) =1 +x+x2 +x3 +x4


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Dividing r(x) = 1 +x8 by1(x) =1 +x+x4,we obtain the remainder

b1(x) =x2

Dividing r(x) = 1 +x8 by3(x) =1 +x+x2 +x3 +x4,we obtain the

remainder

b3(x) =1+x3

Substituting , 2 and 4 into b1(x),we obtainS1 =b1() =

2

S2 =b1(2) =4

S4 =b1(4) =8

Substituting 3 into b3(x),we obtainS3 =b3(

3) = 1 + 9 = 1 + + 3 =7

S= (2, 4, 7, 8)


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(2)Determination of error pattern:

Since r(x)=v(x)+e(x) then Si=r(i) =v(i) +e(i) =e(i) for 1

i 2tThis gives a relationship between the syndrome and the error pattern.Suppose

e(x) =xj1+xj2+....+xjv 0 j1


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Once j1 , j2 ,....,jvhave been found, the powers j1,j2, ...,jv tell usthe error locations in e(x).

If the number of errors in e(x) is t or less, the solution that yields anerror pattern with the smallest number of errors is the right solution.

For convenience,let l=jl , 1 l vbe the error location numbers.

S1=1+2+...+vS2=

21 +

22 +...+

2v

...

S2t=

2t

1 +

2t

2 +...+

2t

v

power sum symmetricfunction


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Define(x) = (1 + 1x)(1 + 2x)...(1 + vx)

=0+ 1x+ 2x2

+ ... + vxv

Where (x)is called the error-location polynomial.

The roots of(x) are 11 , 12 ,...,

1v which are the inverse of the

error location numbers.

Coefficients of(x):

0 = 11 =1+ 2+ ... + v2 =12+ 23+ ... + v1v

...v =12...v




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These coefficients are known as elementary symmetric functions. is arerelated to sjsby Newtons identities.

S1+ 1 = 0S2+ 1S1+ 22 = 0S3+ 1S2+ 2S1+ 33 = 0...Sv+ 1Sv1+ ... + v1S1+vv = 0Sv+1+ 1Sv+ ... + v1S2+ vS1 = 0

Note that, for binary case,1+1=2=0,we have

ii=

i for odd i0 for even i


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Berlekamps iterative algorithm: To carry out the iteration of finding(X) , we begin with the following table:


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l is the degree of()(x).

Ifd= 0, then +1(x) =(x)and l+1 =l.

Ifd= 0, find another row prior to the th row such that d= 0and the number l in the last column of the table has the largestvalue.Then +1(x) is given by

(+1)(x) =()(x) +dd1

x()()(x)

and l+1 =max(l, l+ ).

In either case,d+1 =S+2+ (+1)1 S+1+ ... +

(+1)l+1

S+2l+1

The polynomial 2t(x) in the last row should be the required (x).



Ex 6 5: Let be a primitive element of the Galois field GF (24) such that


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Ex 6.5: Let be a primitive element of the Galois field GF(24) such that1 + + 4 = 0.Consider (15,5) triple-error-correcting BCH codes withp(x) = 1 +x+x4.

Assume that the codeword of all zeros,

v= (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

is transmitted and the received vector is

r= (0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0)

Then r(x) =x3 +x5 +x12

Obtain the error polynomial and corrected vector.

Solution:The minimal polynomials for , 2 and 4 are identical and

1(x) =2(x) =4(x) =1 +x+x4


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The minimal polynomials for 3

and 6

are identical and

3(x) =6(x) =1 +x+x2 +x3 +x4

The minimal polynomials of5 is

3(x) =1 +x+x2

1(x) = 1 +X+X4

3(x) = 1 +X+X2 +X3 +X4

5(x) = 1 +X+X2



Dividing r(x) by 1(x) 3(x)and 5(x) respectively we obtain the following


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Dividing r(x) by 1(x), 3(x)and5(x),respectively,we obtain the followingremainders:b1(x) = 1b3(x) = 1 +X

2 +X3

b5(x) =X2

Substituting , 2 and 4 into b1(x),we obtain the following syndromecomponents:

S1 =S2 =S4 = 1

Substituting 3 and 6 into b3(x),we obtain

S3 = 1 + 6 + 9 =10

S6 = 1 + 12 + 18 =5

Substituting 5 into b5(x),we obtain

S5 =10

S= (1, 1, 10, 1, 10, 5)


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d+1=S+2+ (+1)1 S+1+ ... +

(+1)l+1

S+2l+1

(+1)(x) =()(x) +dd1

x()()(x)

l+1 = max(l, l+ )

S1 = 1= 0; d0 = 1 = 0

= 1(1)(x) =(0)(x) +d0d11 x

(0+1)(1)(x) = 1 + 1.1.x.1 = 1 +xl1 = max(l0, l1+ ) = max(0, 0 + 0 + 1) = 1 l = 1 l1 = 1 1 = 0

d1 =S2+ (1)1 S1 = 1 + 1.1 = 0

(2)(x) =(1)(x) = 1 +xl2 =l1 = 1 l = 2 l2 = 2 1 = 1

d2 =S3+ (2)1 S2+

(2)2 S1 =

10 + 1.1 + 0.1 = (1 + + 2) + 1 =5


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d2 =5 = 0

= 0

(3)(x) =(2)(x) +d2d10 x

(20)(0)(x) = 1 +x+ 5.1.x2.1= 1 +x+ 5x2

l3 = max(l2, l0+ ) = max(1, 0 + 2 0) = 2 l= 3 l3 = 3 2 = 1

d3 =S4+ (3)1 S3+

(3)2 S2+

(3)3 S1 = 1 + 1.

10 + 5.1 + 0.1= 1 + (1 + + 2) + ( + 2) = 0


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(x) =(6)(x) = 1 +x+ 5x3

x=3

, 10

, 12

Error location numbers

1 =3 =12, 2 =

10 =5, 3 =12 =3

e(x) =x3 +x5 +x12

r(x) =r(x) +e(x) = 0Umamaheswar.V (III Sem M.tech) BCH Codes November 11, 2013 34 / 35

End
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