ENC3250 Solutions pg 39

ENC3250 Solutions pg 39 - Position 12 Bit D8 Block 1 Codes...

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Unformatted text preview: Position 12 Bit D8 Block 1 Codes 1100 11 D7 1 1011 10 D6 0 9 D5 0 8 C8 7 D4 0 6 D3 0 5 D2 1 0101 4 C4 3 D1 0 2 C2 1 C1 Check bit calculation: Position 12 Code 1100 1011 0101 0010 11 5 XOR = C8 C4 C2 C1 8.15 The Hamming Word initially calculated was: bit number: 12 0 11 0 10 1 9 1 8 0 7 1 6 0 5 0 4 1 3 1 2 1 1 1 Doing an exclusive-OR of 0111 and 1101 yields 1010 indicating an error in bit 10 of the Hamming Word. Thus, the data word read from memory was 00011001. 8.16 Need n – k check bits such that 2(n – k) – 1 ≥ 1024 + (n – k). The minimum value of n – k that satisfies this condition is 11. 8.17 The calculation shows that g(X) divides f(X) with no remainder. This result is verified by multiplying the quotient by g(X) to get back f(X) exactly: -39- ...
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.

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