ENC3250 Solutions pg 38

ENC3250 Solutions pg 38 - 8.9 a. 00000 10101 01010 00000 0...

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Unformatted text preview: 8.9 a. 00000 10101 01010 00000 0 3 2 10101 2 0 5 01010 2 5 0 000000 010101 101010 110110 000000 0 3 3 4 b. 010101 3 0 6 6 101010 3 6 0 3 110110 4 6 3 0 8.10 a. p(v|w) = βd(w,v)(1 – β) (n – d(w,v) n− d 1− β d p (v|w1 ) β d (1 − β ) b. If we write di = d(w i ,v), then = n −d = p( v| w2 ) β d (1 − β ) β 1 1 2 2 −d1 2 c. If 0 < β < 0.5, then (1 – β )/β > 1. Therefore, by the equation of part b, p(v|w1 )/p(v|w2 ) > 1 if an only if d1 < d2 . 8.11 Suppose that the minimum distance between codewords is at least 2t + 1. For a codeword w to be decoded as another codeword w', the received sequence must be at least as close to w' as to w. For this to happen, at least t + 1 bits of w must be in error. Therefore all errors involving t or fewer digits are correctable. 8.12 C1 C2 C4 C8 = D1 ⊕ D2 ⊕ D4 ⊕ D5 ⊕ D7 = D1 ⊕ D3 ⊕ D4 ⊕ D6 ⊕ D7 = D2 ⊕ D3 ⊕ D4 ⊕ D8 = D5 ⊕ D6 ⊕ D7 ⊕ D8 8.13 The transmitted block and check bit calculation are shown in Table 8.2a and b. Now suppose that the only error is in C8. Then the received block results in the following table: Position Bits Block Codes 12 D8 0 11 D7 0 10 9 D6 D5 1 1 1010 1001 8 C8 1 7 D4 1 0111 6 D3 0 5 D2 0 4 C4 1 3 D1 1 0011 2 C2 1 1 C1 1 The check bit calculation after reception: Position Hamming 10 9 7 3 XOR = syndrome Code 1111 1010 1001 0111 0011 1000 The nonzero result detects and error and indicates that the error is in bit position 8, which is check bit C8. 8.14 Data bits with value 1 are in bit positions 12, 11, 5, 4, 2, and 1: -38- ...
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