The final code after encoding data bits 1101 into 7-bit even parity Hamming Code is

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UGC NET Paper 2: Electronic Science Nov 2020 Official Paper
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  1. 1110101
  2. 1011101
  3. 1010101
  4. 0110101

Answer (Detailed Solution Below)

Option 3 : 1010101
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UGC NET Paper 1: Held on 21st August 2024 Shift 1
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Detailed Solution

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Concept:

Hamming (7, 4) code: It is a linear error-correcting code that encodes four bits of data into seven bits, by adding three parity bits.

Example: It is used in the Bell-Telephone laboratory, error-prone punch caret reader to detect the error and correct them.

Hamming code:

Bits #

1

2

3

4

5

6

7

Transmitted bits

P1

P2

d1

P3

d2

d3

d4

 

P1 = d1 ⊕ d2 ⊕ d4

P2 = d1 ⊕ d4 ⊕ d3

P3 = d2 ⊕ d4 ⊕ d3

Solution:

Given data 1101 i.e.

d1 = 1, d2 = 1, d3 = 0, d4 = 1

We can write:

P1 = d1 ⊕ d2 ⊕ d4 = 1 ⊕ 1 ⊕ 1 = 1

P2 = d1 ⊕ d4 ⊕ d3 = 1 ⊕ 1 ⊕ 0 = 0

P3 = d2 ⊕ d4 ⊕ d3 = 1 ⊕ 1 ⊕ 0 = 0

Then transmitted final code is

P1

P2

d1

P3

d2

d3

d4

1

0

1

0

1

0

1


i.e. 1010101

Option 3 correct

Extra points:

Hamming Distance: The number of bits in which two codewords vary is called hamming distance.

\(\frac{{\begin{array}{*{20}{c}} {1011001}\\ {1101101} \end{array}}}{{3\;bits\;vary}}\)

i.e. 3 Hamming distance.

For Hamming code (n, k):

\(rate\left( R \right) = \frac{K}{n}\)

Where K = 2r – r – 1 (message length)

n = 2r – 1 (Block-length) (r ≫ 2)
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