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Unformatted text preview: 0000001 ˆ = e and 1001001 ˆ ˆ = + = e r U (0-bit outside erasures changed) Case 2 : xxx = 111, r = 1001111, rH T = 100+011+101+110+111 = 011 => 0001000 ˆ = e and 1000111 ˆ ˆ = + = e r U (1-bit outside erasures changed) As the two cases give different estimated codeword, the decoding output is NOT guaranteed. We take the output in Case 1 as the decoding output as it has fewer bit outside the erasures changed. U ˆ is probably 1001001, but not guaranteed. Erasure correction is NOT guaranteed as d min = 3, = 0, and = 3. The inequality d min ≥ 2 + + 1 is NOT satisfied....
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- Spring '11