Block codes - Codes and Turbo Codes - page 157

From this there results the recurrence relation:
S k = α 2 S k− 1 + α 9 S k− 2 ,

k =3 , 4 , ··· , 15

We thus obtain the coecients of the extended syndrome:

S 5 = α 4 ,S 6 = α 13 ,S 7 = α 6 ,S 8 = α 11 ,S 9 = α 6 ,S 10 = α 4
S 11 = α 13 ,S 12 = α 6 ,S 13 = α 11 ,S 14 = α 6 ,S 15 = α 4
Another way to obtain the extended polynomial involves dividing Γ( x )(1 + x n )
by Λ( x ) by increasing power orders.
The errors being with monomials x 3 and x 6 , let us calculate coecients e 3 and
e 6 .

e 3 = S ∗ ( α 12 )= α 2 + α 4 + α 10 + α 7 = α 7
e 6 = S ∗ ( α 9 )= α 4 + α 7 = α 3
The values found for coecients e 3 and e 6 are obviously identical to those ob-
tained in example 4.16. We can verify that the other e j coecients are all null.

Berlekamp-Massey algorithm for binary cyclic codes

For binary BCH codes the Berlekamp-Massey algorithm can be simplified since
it is no longer necessary to determine the error evaluator polynomial, and since
it is possible to show that the Δ j terms are null for j even. This implies:

δ 2 p =0
L 2 p = L 2 p− 1
Λ (2 p ) ( x )=Λ (2 p− 1) ( x )
Θ (2 p ) ( x )= x Θ (2 p− 1) ( x )

Hence the algorithm in t iterations:
Initial conditions :

L − 1 = 0
Λ ( − 1) ( x )=1Θ ( − 1) ( x )= x − 1

Recursion : 0 ≤ p ≤ t − 1

= j Λ (2 p− 1)

j

Δ 2 p +1 S 2 p +1 −j

δ 2 p +1 =1 if Δ 2 p +1 =0 and L 2 p− 1 ≤ p

=0 if not

L 2 p +1 = δ 2 p +1 (2 p +1 − L 2 p− 1 )+(1 − δ 2 p +1 ) L 2 p− 1

Λ (2 p +1)

Θ (2 p +1)

δ 2 p +1 ) x 2 Λ (2 p− 1)

Θ (2 p− 1)

Δ 2 p +1 x 2

1

=

Δ − 1

2 p +1 δ 2 p +1

(1 −