Pravin Paikrao
asked
Jul 14, 2017
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HAMMING CODE (ERROR DETECTION AND CORRECTION)MORRIS MANO 5ED ,PAGE NO 313
P1 = XOR of bits (3, 5, 7, 9, 11) = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0 P2 = XOR of bits (3, 5, 7, 10, 11) = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 P4 = XOR of bits (5, 6, 7, 12) = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1 P8 = XOR of bits (9, 10, 11, 12) = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1 my doubt is how to take min-terms of xor bits shown in parenthesis ??
P1 = XOR of bits (3, 5, 7, 9, 11) = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0P2 = XOR of bits (3, 5, 7, 10, 11) = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0P4 = XOR of bits (5, 6, 7, 12) = 1 ...