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Flash Memories Part 5

Tham khảo tài liệu 'flash memories part 5', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Error Correction Codes and Signal Processing in Flash Memory 69 Here r is the received codeword and H is defined as the parity matrix Each element of GF 2m a can be represented by a m-tuples binary vector hence each element in the vector can be obtained using mod-2 addition operation and all the syndromes can be obtained with the XOR-tree circuit structure. Furthermore for binary BCH codes in flash memory even-indexed syndromes equal the squares of the other one i.e. S2 S therefore only odd-indexed syndromes Si S3 . .S2t-1 are needed to compute. Then we propose a fast and adaptive decoding algorithm for error location. A direct solving method based on the Peterson equation is designed to calculate the coefficients of the errorlocation polynomial. Peterson equation is show as follows Í3 11 s 2 . ÍS1 S2 . 52 . . 53 . . . . S . St 1 1 1 1 1 1 . b tT 1 . S.1 _ St . St 1 . . . S2t-1 _ . a 18 For DEC BCH code t 2 with the even-indexed syndrome S1 S3 the coefficient ơ1 ơ2 can be obtained by direct solving the above matrix as a S1 a2 S S3 S1 Hence the error-locator polynomial is given by 2 . 2 S 2 ơ x 1 Ơ1x a2x2 1 S1x S3 3 x2 S1 19 20 To eliminate the complicate division operation in above equation a division-free transform is performed by multiplying both sides by S1 and the new polynomial is rewritten as 21 . Since it always has S1 0 when any error exists in the codeword this transform has no influence of error location in Chien search where roots are found in ơ x 0 that is also ơ x 0. a x a0 ax a2x2 S1 S2x S3 S3 x2 21 The final effort to reduce complexity is to transform the multiplications in the coefficients of equation 21 to simple modulo-2 operations. As mentioned above over the field GF 2w each syndrome vector S 0 S 1 S m-1 has a corresponding polynomial S x S 0 S 1 x S m-1 xm 1. According to the closure axiom over GF 2m each component of the coefficient ơ1 and ơ2 is obtained as aj i 2 S1 j for 0 i j m -1 a2 i S3 i 2S1 j S1 k for 0 i j k m -1 22 It can be seen that