Obtain Even Parity hamming code for 1010. Prove that hamming code is an error detecting and correcting code.

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written 8.6 years ago by

teamques10 ★ 69k

For a 4-bit code there are 3 parity bits p1, p2 and p3 at location 1, 2 and 4 resp.
So, the code will be: “p1 p2 n1 p3 n2 n3 n4” where, n1, n2, n3, n4 are bits of the code and p1,p2 and p3 are parity bits to be calculated
Therefore, the code for even parity is calculated as below:

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Therefore the even parity hamming code is: 1011010.
Consider that the same calculated code is sent , but it is received with an error in one particular bit, say bit7. S the code word received is: 1011011.
Hence, to identify the error and correct it the receiver follows the procedure as below:
Parity check for bits 4,5,6,7 : 1011 = 1 (odd parity) = c1
Parity check for bits 2,3,6,7: 0111 = 1 (odd parity) = c2
Parity check for bits 1,3,5,7: 1101 = 1(odd parity) = c3
Therefore the receiver now arranges the error correcting bits as c1 c2 c3 = 111 and gets the error bit as bit7 and complements it.
Therefore the correct code is: 1011010

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