BCH Code & Reed Solomon Code

8/22/2019 BCH Code & Reed Solomon Code

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Raj


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Large class of powerful random error correctingcyclic codes

Discovered by

(1) Hocquenghem in 1059(2) Bose in 1960

(3) Chaudhuri in 1960


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For any positive integer m (m3) and t (t


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Specified in terms of roots from GF(2

m

)g(X) is lowest degree polynomial over GF(2)which has , 2, 3, .. ,2t as roots

g(i)=0 for 1 i 2t

Let i(X) be the minimal polynomial ofi

g(X) = LCM{1(X),

2(X), .. ,

2t(X)}


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If is even theni =(

i)2l

iis odd andl 1iis the conjugate ofiEvery even powers ofis the conjugate of

some odd power of

Even powers ofandsome odd power ofhas the same (X)

i(X)=i(X)g(X)=LCM{1(X), 3(X), 5(X).,.2t - 1(X)}


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Maximum degree ofi(X) is m.There are t such polynomials

Degree of g(X) = tm


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Code defined for m

10


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n=2m-1, m

3, t=1, dmin

=2t+1=3

is the primitive element of GF(2m)is the only root (t=1)g(X)=

1(X)

1(X) is the primitive polynomial of degree 3

Single Error correcting BCH code with lengthn=2m-1 is the Hamming Code


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n=2m-1, m3, t=1, dmin=2t+1=3

is the primitive element of GF(2m)is the only root (t=1)g(X)=1(X)

1(X) is the primitive polynomial of degree 3Single Error correcting BCH code with lengthn=2m-1 is the Hamming Code


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m=4, n=2m-1=15 is the primitive element of GF(24)minimal polynomials are

: 1(X) = 1 + X + X4

3 : 3(X) = 1 + X + X2 + X3 + X45 : 5(X) = 1 + X + X2


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Double Error Correcting Code t=2

, 3 are the roots (2t-1=2x2-1=3)g(X)= LCM{1(X), 3(X)} = 1(X). 3(X)

g(X)= 1 + X4 + X6 + X7 + X8

n-k mt =(4 x 2) = 8k = n-(n-k)=15-8=7

(15, 7) Code

dmin=2t+1=2x2+1=5


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Triple Error Correcting Code t=3

, 3 , 5 are the roots (2t-1=2x3-1=5)g(X)= LCM{1(X), 3(X) , 5(X)} = 1(X) . 3(X) . 5(X)

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

n-k = 10k = n-(n-k)=15-10=5

(15, 5) Code

dmin=2t+1=2x3+1=7


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v(X)=v0 + v1X+ + vn-1Xn-1

if v(X) has , 2, 3, 2t as roots v(X) isdivisible by 1(X), 2(X), 3(X), .. ,2t(X)

v(X) divisible g(X) and it is a code polynomial

Let v(X) be a code polynomial

v(i)= v0 + v1(i)+ + vn-1(

i)n-1 for 1i2t


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v(i)= v0 + v1(i)+ v2(

i)2 + vn-1(i)n-1 for 1i2t

v(i)= v0 + v1i

+ v22i + vn-1

(n-1)i

2 3 n 1

2 3 n 12 2 2 2

2 3 n 13 3 3 3

2 3 n 12 t 2 t 2t 2t

1

1

H 1

1


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Conjugate roots can be omitted

2 3 n 1

2 3 n 13 3 3 3

2 3 n 12t 1 2t 1 2t 1 2t 1

1

1

H

1

n - columns

t-rows

Tv.H 0


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m=4, t=1, dmin=2t+1=5, n=15, k=7

H contain elements from GF(24)GF(24) tuples 4-tuples

, 3 are the roots

2 3 4 5 6 7 8 9 10 11 12 13 14

3 6 9 12 15 18 21 24 27 30 33 36 39 42

1H

1


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power tuple

00001 1000

2

01003 11004 00105 10106 01107 1110

power tuple

8 00019 100110

010111 110112 001113 101114 011115 1111


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Parameter 2t+1 is called designed distance

dmin of BCH code may or may not be equal todesigned distance


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Multiple error correction capability

Syndrome may be many componentsith component of syndrome

s i=r(i)= r0 + r1(

i)+ r2(i)2 + rn-1(

i)n-1 1i2t

Syndrome components are elements in GF(2m)

There are 2t syndrome components


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Dividing r(X) by polynomial i(X)

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

i)=0

s i=r(i)= bi(

i)

Syndrome component s i is obtained byevaluating bi(X) with X=

i


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Double error correcting BCH Code, t=2

r = ( 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 )r(X) = 1 + X8

s=(s1, s2, s3, s4)

minimal polynomials are : 1(X) = 1 + X + X42 : 2(X) = 1 + X + X43

: 3(X

)= 1 + X + X

2

+ X

3

+ X

4

4 : 4(X) = 1 + X + X4


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b1(X)=b2(X)=b4(X) = rem(r(X) / 1(X) ) = X2

b3(X) = rem(r(X) / 3(X) ) = 1+X3

s1=b1()= (1)2 = 4

s2=b1()= (2)2 = 4

s3=b3()= 1+(3)3 =1+9=1+ +3 = 7s4=b1()= (

4)2 = 8

s=(2, 4, 7, 8)


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r(X)=v(X)+e(X)

si=r(i)=v(i)+e(i)=e(i) 1 i 2t

Let e(X)= Xj1 + Xj2 + . + Xjv 0 j1 j1 . j1 n

s1

= j1 + j2 + .. + jv

s2 = (j1)2 + (j2)2 + .. + (jv)2

s3 = (j1)3 + (j2)3 + .. + (jv)3

...

s2t = (j1)2t + (j2)2t + .. + (jv)2t

j1, j2, . , jv are unknown


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Let e(X)= Xj1 + Xj2 + . + Xjv 0 j1 j1 . j1 n

s1

= j1 + j2 + .. + jv

s2 = (j1)2 + (j2)2 + .. + (jv)2

.........................

s2t = (j1)2t + (j2)2t + .. + (jv)2t

j1, j2, . , jv are unknown

Any method of solving above equations is thedecoding algorithm.

Solutions yields the smallest number of errors is

the right solution.

This is the most probable error pattern e(X)caused by channel.


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Let l= ji 1 l v

s1 = 1 + 2 + .. + v

s2 = (1)2 + (2)

2 + .. + (v)2

...

s2t = (1)2t + (2)

2t + .. + (v

)2t

Above 2t equations are called power sumsymmetric functions

1,2, .. , v are error location numbers


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Define (X) = (1+ 1X) (1+ 2X).. (1+ vX)

(X) = 0 + 1X + 2X + + vXV

Roots are 1-1,2

-1, .. , v-1

(X) is the unknown polynomial whose

coefficients must be determined(X) is know as error location polynomial


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Define (X) = (1+ 1X) (1+ 2X).. (1+ vX)

0 = 11 = 1 + 2 + ..+v

2 = 12+ 23 ++V-1v

..v = 12. vis

are knows are elementary symmetric

functions ofi

s.


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Newtons Identities

s1 + 1 =0s2 + 1s1 + 22 =0

s3 + 1s2 +2s1 + 33 =0

..sv + 1sv-1 + +v-1s1 + vv =0

sv+1 + 1sv + +v-1s2 + vs1 =0

ii= i for a odd i ii= 0 for even iSolutions that yield (X) of minimal degree giveinformation about most probable error pattern

if v

t

(X)will give the actual error pattern e(X).


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Compute the syndrome s=(s1, s2,,s2t) from r(X)

Determine error location polynomial (X) fromthe syndrome components s1, s2,,s2tDetermine the error location numbers

1,2, .. , v by finding the roots of the (X)and correct the errors in r(X)


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Compute the syndrome s=(s1, s2,,s2t) from r(X)

Determine error location polynomial (X) fromthe syndrome components s1, s2,,s2tDetermine the error location numbers

1,2, .. , v by finding the roots of the (X)and correct the errors in r(X)


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Berlekamp iterative algorithm

Find minimum degree polynomial (1)(X) whosecoefficients satisfy the first Newtons identitys1 + 1 =0

Next step is to test whether the coefficients of(1)(X) satisfy the second Newtons identitys2 + 1s1 + 22 =0

If the coefficients of(1)(X) satisfy second

Newtons identity we set(2)(X)=

(1)(X)


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If the coefficients of(1)(X) does not satisfy

second Newtons identity a correction term isadded to (1)(X) to form (2)

(X) such that (2)

(X)has minimum degree and its coefficients satisfyfirst two Newton's identities

There are at the end of second step we obtain aminimum degree polynomial (2)(X) whosecoefficients satisfy first two Newtons identifies


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The third step is to find the minimal polynomial

(3)

(X) from (2)

(X) such that (3)

(X) coefficientssatisfy first three Newtons identities

Again we test (2)(X) satisfy third Newton'sidentity s3 + 1s2 +2s1 + 33 =0

If they do not, a correction term is added to(2)(X) to from

(3)(X)

Iteration continues until (2t)(X) is obtained

The (2t)(X) is taken to be error locationpolynomial (X)

(X) =(2t)

(X)


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Let ()(X) =1 +1()X + 2

()X2 + .+ l()Xl

be the minimum polynomial determined at theth step of iteration whose coefficients satisfyNewtons identities

To determine

(+1)

(X) we compute the

following quantity

d = s+1+ 1()s+1+ 2

()s-1+..+l()s+1l

The discalleddiscrepancy

If d=0 the coefficients of()

(X) satisfy(+1)th Newton's identity and (+1)(X) =

()(X)


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If d0 the coefficients of()

(X) do not satisfy

(+1)th

Newton's identity and a correction termmust be added to ()(X) to obtain (+1)

(X)

Go back to the steps prior to th step anddetermine a polynomial ()(X) such that theth discrepancy d0 and ( -l)[lis thedegree of()(X)] has the largest value.

(+1)(X)= ()

(X) + dd-1X (-)()(X)


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To carry out iteration of finding ()(X) we begin

with following table and proceed to fill out thetable wherel is the degree ()

(X)

The (+1)th is can be fill out as followsIf d=0 then

(+1)(X)=

()(X) andl+1=l

If d0, find another row prior to the throwsuchthat d0 and the number (-l) in the lastcolumn of the table has the largest value.

(+1)

(X)= ()

(X) + dd-1

X(-)

()

(X)l+1 = max(l,l+ - )


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()

(X) d l -l-1 1 1 0 -1

0 1 s1 0 0

1

2

2t


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Consider a (15,5) triple error correcting code.

v = ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 )r = ( 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 )

r(X)= X3 + X5 + X12

Minimal Polynomials : 1(X) = 1 + X + X

4

2 : 2(X) = 1 + X + X4

3 : 3

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

4 : 4(X) = 1 + X + X4

5 : 5(X) = 1 + X + X2

6 : 6(X) = 1 + X + X2 + X3 + X4


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Dividing r(X) by 1(X), 3(X) and 5(X)

We obtain the following remaindersb1(X) = 1

b3(X) = 1+X2+X3

b5(X) = X2Syndrome Components are

s1=s2=s4=1

s3=1 + 6 + 9 = 10 and s6=1 + 12 + 18 = 5s5=

10


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0

1 1 + X 0 1 0 take = -1


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0

1 1 + X 0 1 0 take = -1

2 1 + X 5 1 1


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0

1 1 + X 0 1 0 take = -1

2 1 + X 5 1 1

3 1 + X + 5X2 0 2 1 take = 0


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0

1 1 + X 0 1 0 take = -1

2 1 + X 5 1 1

3 1 + X + 5X2 0 2 1 take = 0

4 1 + X + 5X2 10 2 2


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0

1 1 + X 0 1 0 take = -1

2 1 + X 5 1 1

3 1 + X + 5X2 0 2 1 take = 0

4 1 + X + 5X2 10 2 2

5 1 + X + 5X3 0 3 2 take = 2


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s1 = 1

s2 = 1

s3 = 10

s4 = 1

s5 = 10

s6 = 5

()(X) d l -l Remarks-1 1 1 0 -1

0 1 1 0 0

1 1 + X 0 1 0 take = -1

2 1 + X 5 1 1

3 1 + X + 5X2 0 2 1 take = 0

4 1 + X + 5X2 10 2 2

5 1 + X + 5X3 0 3 2 take = 2

6 1 + X + 5X3 - - 2


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(X)=(6)(X) = 1 + X + 5X3

roots of(X) are 3, 10, 12There inverses are 12, 5, 3

3= 3,5=

5,12=

12

j3=3, j5=4, j12=12e(X)=Xj1 + Xj2 + . + Xjv

Therefore error pattern is

e(X)=X3 + X5 + X12Adding e(X) to v(X) we obtain all zero codevector


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If p is prime number and q is a any power of p

there are codes with symbols from Galois fieldGF(q)

Codes are called q - ary codes

Concepts and properties developed for binarycodes apply q - ary codes with littlemodifications


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An (n, k) linear code with symbols from GF(q) is

a k dimensional subspace of vector space of alln-tuples over GF(q)

A q-ary (n, k) cyclic code generated by apolynomial of degree (n k) with coefficients

from GF(q) which is a factor Xn + 1

Encoding and decoding are similar to that of abinary cods


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In binary BCH codes defined can be generalized

to non-binary codes in a straight forwardmanner.

For any choice of positive integer s and t, thereexists a q-ary BCH code of length n=qs-1 which is

capable of correcting any combination of t orfewer errors and requires no more than 2stparity check digits


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Let be a t-error correcting q-ary BCH code is

the polynomial of lower degree with coefficientform GF(q) for which , 2, .. ,2t are the roots

Let i(X) be the minimal polynomial ofi then

g=(X) = LCM{1

(X), 2

(X), .. ,2t

(X)}

Degree of each minimal polynomial is s or less

Therefore degree of g(X) is at most 2st.

Hence number of parity check digits of codegenerated by g(X) is no more than 2st

For q=2 we obtain BCH codes


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Special subclass of q-ary BCH code for which s=1

A t-error correcting Reed-Solomon code withsymbols from GF(q) has the followingparameters

Block length: n = q 1

No. parity check digits n-k = 2t

Minimum distance dmin = 2t + 1


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Length of the code is one less than the size of

code symbols (q-1) and minimum distance isone greater than the no: of parity check digits(2t + 1)

Codes with code symbols from Galois field

GF(2m), q=2m

A (n, k) RS-code have dmin > (n-k+1) and itbelongs to maximum distance separable code

category of LBCRS code are particularly useful in situationswhere errors tends to happen in bursts ratherthan randomly


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Let be the primitive element in GF(2m)

The generator polynomial g(X)g(X)= (X+) (X+ 2) .. (X+ 2t)

g(X)=g0 + g1X +.. + g2t-1X2t-1 + X2t

g(X) has , 2, .. ,2t as all roots and hascoefficients form GF(2m)

k=n-(n-k)=n-2t

Code generated by g(X) is an (n, n t)


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Cyclic code which consists of those polynomials

of degree (n 1) or less with coefficients fromGF(2m) that are multiples of g(X)


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Similar to binary case

Encoder operates on multiple bits rather thanindividual bits

Let message, a(X) = a0 + a1X +.. + ak-1Xk-1

Where k=n 2tIn systematic form the 2t parity check digits arethe coefficients of remainder b(X) resulting fromdividing message polynomial X2ta(X) by

generator polynomial g(X)

b(X) = rem(X2ta(X)/g(X))= b0 + b1X +.. + b2t-1X2t-1


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Incoming data stream bits are grouped into

blocks of m symbols GF(2m

) of block lengthn=2m-1

Each block is treated as k symbols with eachsymbol having m bits

Encoding algorithm expands a block of ksymbols to a n symbols by adding n k paritycheck digits


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Similar to binary code

BCH Code & Reed Solomon Code

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