Assume that a 12-bit Hamming codeword consisting of 8-bit data and 4 check…
2021
Assume that a 12-bit Hamming codeword consisting of 8-bit data and 4 check bits is \(d_8 d_7 d_6 d_5 c_8 d_4 d_4 d_3 d_2 c_4 d_1 c_2 c_1\),where the data bits and the check bits are given in the following tables:
\(\overset{\textbf{Data bits}}{\begin{array}{|c|c|c|c|} \hline d_{8} & d_{7} & d_{6} & d_{5} & d_{4} & d_{3} & d_{2} & d_{1} \\\hline 1 & 1 & 0 & x & 0 & 1 & 0 & 1 \\\hline \end{array}}\qquad \overset{\textbf{Check bits}}{\begin{array}{|c|c|c|c|} \hline c_{8} & c_{4} & c_{2} & c_{1} \\\hline y & 0 & 1 & 0 \\\hline \end{array}}\)
Which one of the following choices gives the correct values of \(x\) and \(Y\) ?
- A.
\(x\)is 0 and\(y\)is 0 - B.
\(x\)is 0 and\(y\)is 1 - C.
\(x\)is 1 and\(y\)is 0 - D.
\(x\)is 1 and\(y\)is 1
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Correct answer: A
Solution: compute the unknown bits using even parity checks.
Map the 12 positions (position 1 is the least significant / rightmost):
pos1 = c1 = 0, pos2 = c2 = 1, pos3 = d1 = 1, pos4 = c4 = 0,
pos5 = d2 = 0, pos6 = d3 = 1, pos7 = d4 = 0, pos8 = c8 = y,
pos9 = d5 = x, pos10 = d6 = 0, pos11 = d7 = 1, pos12 = d8 = 1.
Use the parity checks:
c1 covers positions 1,3,5,7,9,11. Sum = 0 + 1 + 0 + 0 + x + 1 must be even, so x = 0.
c8 covers positions 8,9,10,11,12. Sum = y + x + 0 + 1 + 1 must be even. With x = 0 this gives y = 0.
Therefore the correct values are x = 0 and y = 0.
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