UGC-NET | UGC NET CS 2015 Jun – III | Question 1

For the 8 – bit word 00111001, the check bits stored with it would be 0111. Suppose when the word is read from memory, the check bits are calculated to be 1101. What is the data word that was read from memory?
(A) 10011001
(B) 00011001
(C) 00111000
(D) 11000110


Answer: (B)

Explanation: 8 – bit word = 00111001, Check bits = 0111.
There are four bits and it’s position will be
2⁰ = 1; i.e. p₁ = 1
2¹ = 2; i.e. p₂ = 1
2² = 4; i.e. p₄ = 1
2³ = 8; i.e. p₈ = 0.
Encoded string will be:
d₁₂ = 0; d₁₁ = 0; d₁₀ = 1; d₉ = 1; d₈ = p₈ = 1; d₇ = 1; d₆ = 0; d₅ = 0; d₄ = p₄ = 1; d₃ = 1; d₂ = p₂ = 1; d₁ = p₁ = 1.
i.e.
d₁₂ = 0; d₁₁ = 0; d₁₀ = 1; d₉ = 1; d₈ = 1; d₇ = 1; d₆ = 0; d₅ = 0; d₄ = 1; d₃ = 1; d₂ = 1; d₁ = 1.
Check bits at other end = 1101.
XOR(0111, 1101) = 1010. 10_th bit be change.
So, new encoded string will be:
d₁₂ = 0; d₁₁ = 0; d₁₀ = 0; d₉ = 1; d₈ = 1; d₇ = 1; d₆ = 0
d₅ = 0; d₄ = 1; d₃ = 1; d₂ = 1; d₁ = 1.
And the data word that was read from memory will be – 00011001.
So, option (B) is correct.

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