Question

Given the data-bits m = 11010110, determine the number of k (parity-bits) by using Hamming Code requirements. Illustrate the error detection and correction scheme using Hamming code method, for both the sender and receiver to detect an error at the following positions:

a.6thbit position.

b.11thbit position.Assume an odd-parity scheme for this problem.

Answer 1

According to the question the data-bits m = 11010110 => n= 8, the number of parity bits used in the hamming code can be calculated as follows:-

n+p+1 <= 2^p

=> 8+p+1 <= 2^p

=> 9+p <= 2^p

=> for p=4, the above condition will satisfy therefore, we will take p=4, that is 4 parity bits will be used, hence the size of hamming code will be 4+8 = 12.

The hamming code will be as follows:-

Power of 2					23				22		21	20
position	12	11	10	9	8	7	6	5	4	3	2	1
message	1	1	0	1	p8	0	1	1	p4	0	p2	p1

Since it is given that odd-parity scheme is used, therefore the value parity bits will be calculated as follows:-

Calculation of p1:-

Positions used = 1, 3, 5, 7, 9, 11

values used = p1, 0, 1, 0, 1, 1

=> p1 = 0

Calculation of p2:-

Positions used = 2, 3, 6, 7, 10, 11

values used = p2, 0, 1, 0, 0, 1

=> p2 = 1

Calculation of p4:-

Positions used = 4, 5, 6, 7, 12

values used = p4, 1, 1, 0, 1

=> p4 = 0

Calculation of p8:-

Positions used = 8, 9, 10, 11, 12

values used = p8, 1, 0, 1, 1

=> p8 = 0

Therefore the hamming code that the sender will send be 110100110010

(a.) The there is an error at 6th bit position therefore 1 at 6th position will change to 0, and message that receiver will receive will be 110100010010.

Error detection :- we use the odd parity scheme to detect error

Calculation of c1:-

Positions used = 1, 3, 5, 7, 9, 11

values used = 0, 0, 1, 0, 1, 1

=> c1 = 0

Calculation of c2:-

Positions used = 2, 3, 6, 7, 10, 11

values used = 1, 0, 0, 0, 0, 1

=> c2 = 1

Calculation of c4:-

Positions used = 4, 5, 6, 7, 12

values used = 0, 1, 0, 0, 1

=> c4 = 1

Calculation of c8:-

Positions used = 8, 9, 10, 11, 12

values used = 0, 1, 0, 1, 1

=> c8 = 0

Since c2 & c4 has value 1 therefore there is an error in received message.

Error correction :- the bit which has the error is c8c4c2c1

=> position of error bit = 0110 =(6)₁₀

For error correction change the value of 6th position bit, therefore the message will become 110100110010.

(b.) The there is an error at 11th bit position therefore 1 at 11th position will change to 0, and message that receiver will receive will be 100100110010.

Error detection :- we use the odd parity scheme to detect error

Calculation of c1:-

Positions used = 1, 3, 5, 7, 9, 11

values used = 0, 0, 1, 0, 1, 0

=> c1 = 1

Calculation of c2:-

Positions used = 2, 3, 6, 7, 10, 11

values used = 1, 0, 1, 0, 0, 0

=> c2 = 1

Calculation of c4:-

Positions used = 4, 5, 6, 7, 12

values used = 0, 1, 1, 0, 1

=> c4 = 0

Calculation of c8:-

Positions used = 8, 9, 10, 11, 12

values used = 0, 1, 0, 0, 1

=> c8 = 1

Since c8, c2 & c1 has value 1 therefore there is an error in received message.

Error correction :- the bit which has the error is c8c4c2c1

=> position of error bit = 1011 =(11)₁₀

For error correction change the value of 6th position bit, therefore the message will become 110100110010.