Number Systems and Codes Questions and Answers

Answer: A

The code word is 100111110.

Five data bits need four Hamming check bits (\(2^{4} \ge 5 + 4 + 1\)), so the code has 9 bits. Number the positions 9 down to 1 from left to right; check bits sit at positions 1, 2, 4 and 8 and the data 1 0 1 1 1 fills positions 9, 7, 6, 5 and 3.

For odd parity, each check bit makes the count of 1s in its group odd:

\[ \begin{aligned} P_{1} &: \text{bits } 3, 5, 7, 9 = 1, 1, 0, 1 \Rightarrow P_{1} = 0 \\ P_{2} &: \text{bits } 3, 6, 7 = 1, 1, 0 \Rightarrow P_{2} = 1 \\ P_{4} &: \text{bits } 5, 6, 7 = 1, 1, 0 \Rightarrow P_{4} = 1 \\ P_{8} &: \text{bit } 9 = 1 \Rightarrow P_{8} = 0 \end{aligned} \]

Reading positions 9 to 1 gives 1 0 0 1 1 1 1 1 0. Options (C) and (D) have only 8 bits, too few for single-error correction of 5 data bits, and 111000001 does not contain the data bits at the data positions.

Answer: C

The Hamming distance is 10.

The Hamming distance between two code words of equal length is the number of positions in which they differ, that is, the number of 1s in their bitwise XOR.

The XOR has 4 + 2 + 4 = 10 ones, so the words differ in 10 of their 12 positions. They agree only in positions 5 and 8. Counting only part of the word, or counting the agreeing positions, produces the smaller distractors.

Answer: B

The decimal equivalent is -13.

8D in hexadecimal is the 8-bit pattern 1000 1101. In sign-magnitude form the most significant bit is the sign and the remaining seven bits are the magnitude:

• Sign bit 1: the number is negative.
• Magnitude 000 1101 = 8 + 4 + 1 = 13.

So the value is -13. Reading the pattern as unsigned gives 141, and dropping the sign gives 13, the first distractor. Reading it in two's complement would give -115, so the -11 and 11 options correspond to no standard reading of 8D.

Answer: C

The error is in the 5th location.

In a 7-bit Hamming code the bit positions are numbered 7 down to 1 from left to right: D7 D6 D5 P4 D3 P2 P1, with parity bits at positions 1, 2 and 4. For the received word 0100011, b7 = 0, b6 = 1, b5 = 0, b4 = 0, b3 = 0, b2 = 1, b1 = 1. With even parity, each check over its group should be 0:

• C1 = b1 ⊕ b3 ⊕ b5 ⊕ b7 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
• C2 = b2 ⊕ b3 ⊕ b6 ⊕ b7 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
• C4 = b4 ⊕ b5 ⊕ b6 ⊕ b7 = 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1

The syndrome C4 C2 C1 = 101 = 5, so bit 5 is wrong. It is the third bit counted from the left, which is the trap in the '3rd location' option.

Answer: A

Both (A) and (R) are true and (R) is the correct explanation of (A).

Computers use 2's complement because the same adder then handles both addition and subtraction: A - B is computed as A + (2's complement of B). There is only one representation of zero, and no end-around carry is needed as in 1's complement.

The 2's complement itself is obtained with very simple hardware: invert every bit (an XOR or NOT gate per bit) and add 1 through the carry input of the adder. That simplicity is the reason the scheme is preferred, so the reason explains the assertion.