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Rectangular Codes

Rectangular Codes. Triplication codes:. m 1 m 2 m 3 • • • •. Repeated 3 times. m 1 m 1 m 1 m 2 m 2 m 2 m 3 m 3 m 3 • • • •. At receiving end, a majority vote is taken. Error detection and correction. Slides based on unknown ous contributor on the web….

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Rectangular Codes

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  1. Rectangular Codes • Triplication codes: m1 m2m3 • • • • Repeated 3 times m1m1m1 m2m2m2 m3m3m3 • • • • At receiving end, a majority vote is taken.

  2. Error detection and correction Slides based on unknown ous contributor on the web…

  3. Rectangular codes: • Redundancy: m -1 o o o • • • o xo o o• • • o xo o o • • • o x• • • • • • • • • • • • o o o • • • o xx x x x x o = message positionx = check position n -1 sum mod 2 It’d better use even-parity checking to avoid contradiction

  4. For a given size mn, the redundancy will be smaller the more the rectangle approaches a square. • For square codes of size n,we have (n -1)2bits of information. And 2n-1 bits of checking along the sides. • Note that: Rectangular codes also can correct bursty error. • (k2+1)x(k1+1) array codeIf k2  2(k1-1) we can correct k1 size of bursty errors

  5. 3.4 Hamming Error-correcting codes • Find the best encoding scheme for single-error correction for white noise. • Suppose there are m independent parity checks. →It means no sum of any combination of the checks is any other check. • Example: check 1 : 1 2 5 7 --- (1) check 2 : 5 7 8 9 --- (2) check 3 : 1 2 8 9 --- (3) It is not independent, because (1)+(2)=(3) • So third parity check provides no new information over that of the first two, and is simply wasted effort.

  6. The syndrome which results from writing a 0 for each of the m parity checks that is correct and 1 for each failure can be viewed as an m-bit number and can represent at most 2m things. • For n bits of the message, 2m n + 1 It is optimal when meets the equality condition. ( Hamming Codes ) • Using Syndrome to find out the position of errors. The ideal situation is to use the value of Syndrome to point out the position of errors.

  7. Example: 0 0 0  no error 0 0 1  error happened in the first position

  8. Locate errorcheck1 m1+ m3+ m5+ m7=0check2 m2+ m3+ m6+ m7=0check3 m4+ m5+ m6+ m7=0 Viewing m1 , m2 , m4 as check bit • Note that the check positions are equally corrected with the message positions. The code is uniform in its protection. Once encoded there is no different between the message and the check digits.

  9. Hamming code when m = 10, then n = 1023original message length : 1023 – 10 = 1013Redundancy:

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