tut5_sol - 0000001 ˆ = e and 1001001 ˆ ˆ = + = e r U...

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1 EE 4212 Information and Coding Solution to Tutorial 5 A (7,4) Systematic Linear Block Code = 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 0 1 1 1 0 G (a) r = 10010xx Case 1 : xx = 00, r = 1001000, rH T = 100 + 011 = 111 => 0000001 ˆ = e and 1001001 ˆ ˆ = + = e r U (0-bit outside erasures changed) Case 2 : xx = 11, r = 1001011, rH T = 100 + 011 + 110 + 111 = 110 => 0000010 ˆ = e and 1001001 ˆ ˆ = + = e r U (0-bit outside erasures changed) As both cases have 0 bit outside the erasures and give the same estimated codeword, the decoding output is 1001001 ˆ = U Erasure correction is guaranteed as d min = 3, α = 0, and γ = 2. The inequality d min 2 α + γ + 1 is satisfied. (b) r = 1001xxx Case 1 : xxx = 000, r = 1001000, rH T = 100 + 011 = 111 =>
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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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