(18 pts, 2x9) A 12-bit Hamming code word containing 8 bits of data and 4 parity

Question

(18 pts, 2x9) A 12-bit Hamming code word containing 8 bits of data and 4 parity bits is read from memory. What was the original 8-bit data word that was written into memory if the 12-bit word read out is

Note: the data may be correct or it may have an error. I’ve broken up the numbers and the answers into groups of 4 bits to make it easier to answer and to grade.

(a) 0110 1011 1010                Bit # in error (1-12)? _______

Corrected version of 12 bits transmitted: _______ _______ _______

Original 8 bits of data (corrected as needed):_______ _______

[side note: I know the answer to this problem but a ton of people in my class are struggling with this concept so I thought I would post it so people can get help on it]

Explanation / Answer

Original data: 0110 1011 1010

Applying XOR operation on the carry or parity bits and these bits are:

C1(1, 3, 5, 7, 9, 11) = 0, 1, 1, 1, 1, 1 = 1

C2(2, 3, 6, 7, 10, 11) = 1, 1, 0, 1, 0, 1 = 0

C4(4, 5, 6, 7, 12) = 0, 1, 0, 1, 0 = 0

C8(8, 9, 10, 11, 12) = 1, 1, 0, 1, 0 = 1

Therefore, C = 1001, which means error bit is 5.

Data bits are: 3, 5, 6, 7, 9, 10, 11, 12 i.e, 1 0 1 1 1 0 1 0

Corrected 8-bit data- 1 0 0 1 1 0 1 0.

Corrected 12 bits - 0 1 1 0 0 1 1 1 1 0 1 0

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