The Art of Error Correcting Coding phần 10 potx

... 1990. E. L. Blokh and V. V. Zyablov, Coding of Generalized Concatenated Codes,” Probl. Pered. Inform., vol. 10, no. 1, pp. 45–50, 1974. The Art of Error Correcting Coding, Second Edition Robert H. ... ed., Key Papers in the Development of Coding Theory, pp. 109 –120, IEEE Press, 1974. W. W. Peterson and E. J. Weldon, Jr., Error Correcting Codes, 2nd ed., MIT Press, 1...

Ngày tải lên: 14/08/2014, 12:20

... 11 00 01 10 11 00 01 10 000 100 011 111 000 100 011 111 010 110 001 101 001 101 010 110 00 01 10 000 100 011 111 000 100 011 111 010 110 001 101 001 101 010 110 00 01 10 000 100 011 111 000 100 011 111 010 110 001 101 001 ... 2. Before the third-stage decoding, each two-dimensional coordinate (r xj ,r yj ) of the received COMBINING CODES AND...

Ngày tải lên: 14/08/2014, 12:20

... standard array of the binary linear (4, 2, 2) code is the following: ¯s 00 01 10 11 00 0000 0 110 1011 1101 11 100 0 1 110 0011 0101 10 0100 0 010 1111 100 1 01 0001 0111 101 0 1100 Decoding with the standard ... respect to the decoding algorithms. 3 Matlab is a registered trademark of The Mathworks, Inc. 1 Introduction The history of error correcting coding...

Ngày tải lên: 14/08/2014, 12:20

... representation of integers, then the value of the syndrome directly gives the position of the error. This is the idea in the algorithms for encoding and decoding presented below. The columns of the parity-check ... single error as ¯r. Show that the syndrome of the vector π c (¯r) is the binary representation of the position of the error, where π c i...

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... (3.17) • Find the coefficients of the error- locator polynomial σ(x). • Find the inverses of the roots of σ(x), that is, the locations of the errors, α j 1 , ,α j ν . • Find the values of the errors e j 1 , ... +1. The roots of each of the factors of the polynomial x 7 + 1 are shown in the following table. The reader is invited to verify that in fact...

Ngày tải lên: 14/08/2014, 12:20

... P s denotes the probability of a symbol error at the input of the RS decoder, P S = 1 − (1 − p) m , and p denotes the probability of a bit error at the input of the RS decoder. The probability of a ... of elements of GF(2 4 ), p(x) = x 4 + x 3 + 1. Power Vector 0 0000 1 0001 α 0 010 α 2 0100 α 3 100 0 α 4 100 1 α 5 101 1 α 6 1111 α 7 0111 α 8 1 110 α 9...

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... = 10 ¯u = 01 ¯u = 11 000 00000 101 10 0101 1 1 1101 110 10000 00 110 1101 1 0 1101 011 0100 0 11 110 00011 101 01 100 0 0100 100 10 01111 1100 1 010 00 010 1 0100 0100 1 11111 001 00001 101 11 0101 0 1 1100 101 ... (11,01,01,00 ,10) y = (10, 01,11,00,11) 00 01 10 11 00 11 00 10 01 10 00 01 10 11 10 00 01 10 11 01 00 01 10 11 01 00 01 10 11 00 00 01 10 11...

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... to 7.8. The evolution of the metric values with respect to the 00 01 10 11 11 00 10 01 00 11 10 01 00 01 10 11 00 01 10 11 00 01 10 11 00 01 10 11 00 01 10 11 00 01 10 11 11 01 00 10 1101 Figure ... (11,01,01,00 ,10, 11) y = (11,11,01,01,00 ,10) y = (11,01,01,00 ,10, 00) y = (11,11,01 ,10, 10 ,10) 00 01 10 11 00 01 10 11 00 01 10 11 00 01 10 11 00 01 10...

Ngày tải lên: 14/08/2014, 12:20

...  1 (G) =     101 0100 101 1 010 01 1100 0 100 1001     −→ G  =     100 0111 0100 101 0 0100 11 0001 110     . Therefore, G 1 =  2 (G  ) = G  , with  2 = I =  1234567  (the identity permutation). ... needed in the last part of the algorithm, to compute the soft outputs. In the second part of SOVA decoding, the VA transverses the trellis backward...

Ngày tải lên: 14/08/2014, 12:20

... 1111 100 0 100 0 100 0 100 0 0000 0100 0100 0100 0000 100 0 0 010 0 010 0000 0100 0100 0001 0000 0 010 0 010 0 010 0000 0001 0001 0001 0001 100 0 0100 0001 0000 0100 0100 0 010 0 010 0001 0000 0 010 0001 0000 100 0 ... observed clearly as the length of the code increases. The longer the code, the higher the code rate and the steeper the BER curve. Example...

Ngày tải lên: 14/08/2014, 12:20