Question
Asked Sep 29, 2019
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12-bit (8,4) Hamming code whose hexadecimal value is 0xE4F arrives at a receiver.

What was the original value in hexadecimal?

Assume that not more than 1 bit is in error.

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Expert Answer

Step 1

Answer:

0xE4F = 111001001111

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1 1 1 0 0 1 0 1 1 1 1 bl b2 ь3 b4 b5 ь6 b7 ь8 ь9 b10 bl1 b12 Here, "b represents bit

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Step 2

Split the bits into parity bits and data bits as follows:

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1 1 1 1 0 0 0 1 1 1 1 pl р2 р3 р4 d1 d2 d3 d4 d5 d6 d7 d8 Here, "p" represents parity bit and "d"represents data bit

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Step 3

p1 = b3 + b5 + b7 + b9 + b11

     = 1 + 0 + 0 + 1 + 1

     = 1

p1 value is same bit value as in the table

p2 = b3 + b6 + b7 + b10 + b11

     = 1 + 1 + 0 + 1 + 1

     = 0

p2 value is not same bit value as in the table

p3 = b5 + b6 + b7 + b12

  &n...

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1 0 1 0 0 1 0 0 1 1 1 1 bl1 b10 b12 b1 b2 ьз b4 b5 ь6 b7 ь8 ь9 Therefore, original hexadecimal value is OXA4F

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