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12 April, 07:42

Assuming we are using the Hamming algorithm presented in your text and even parity to design an error-correcting code, find the code word to represent the 8-bit information word 10011011

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  1. 12 April, 10:20
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    Before redundancy : 100101011100

    After checking parity redundancy: 100101010011

    Explanation:

    First calculate redundant bits,

    We know that the number of redundant bits can be calculated using the following formula:

    2^r ≥ m + r + 1

    r = redundant bit, m = data bit

    total data bits = 8

    so,

    2^4 ≥ 8 + 4 + 1

    16 ≥ 13

    so, redundant bits is 4

    Now let they denoted by r1, r2, r4, and r8, as the redundant bits are placed at positions corresponding to power of 2: 1, 2, 4, and 8.

    All the redundant bits are initialized by zero.

    For the data word 10011011, we can let the bits of the data word as w8, w7, w6, w5, w4, w3, w2, w1

    Now, put data bits and redundancy bits as follows.

    d12 d11 d10 d9 d8 d7 d6 d5 d4 d3 d2 d1

    w8 w7 w6 w5 r8 w4 w3 w2 r4 w1 r2 r1

    1 0 0 1 0 1 0 1 0 0 0 0

    r1 = d1 xor d3 xor d5 xor d7 xor d9 xor d11

    r1 = 0 xor 0 xor 1 xor 1 xor 1 xor 0

    r1 = 1

    r2 = d2 xor d3 xor d6 xor d7 xor d10 xor d11

    r2 = 0 xor 0 xor 0 xor 1 xor 0 xor 0

    r2 = 1

    r4 = d4 xor d5 xor d6 xor d7

    r4 = 0 xor 1 xor 0 xor 1

    r4 = 0

    r8 = d8 xor d9 xor d10 xor d11 xor d12

    r8 = 0 xor 1 xor 0 xor 0 xor 1

    r8 = 0

    So, the data transferred is

    d12 d11 d10 d9 d8 d7 d6 d5 d4 d3 d2 d1

    w8 w7 w6 w5 r8 w4 w3 w2 r4 w1 r2 r1

    1 0 0 1 0 1 0 1 0 0 1 1

    The bits give the binary number as 0011 whose decimal representation is 3. Thus, the bit 3 contains an error. To correct the error the 3th bit is changed from 1 to 0.
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