Hamming code
The hamming
codes are used extra parity bits to identify a single bit error. To get from
one-bit pattern to the other, few bits are to be changed in the data. Such a number of bits can be termed as Hamming
distance. It can only detect a single bit error correction. If
the parity has a distance of 2, a one-bit flip can be detected. But this can't be
corrected. Also, any two-bit flips cannot be detected.
7-bit Hamming code is commonly used which
contains 4 data bits and 3 Parity bits. The position of the parity bit is determined by the formula – 2^n (where
n=0,1,2…).
For n=0, the position of
first parity, P1 = 2^0=1
For n=1, the position of
first parity, P2 = 2^1=2
For n=2, the position of
first parity, P4 = 2^2=4
The value of P1
depends on the data bits D3, D5, and D7.
The value of P2
depends on the data bits D3, D6, and D7.
The value of P3 depends
on the data bits D5, D6, and D7.
Example: If data is 1011 then find the value of P1,
P2 & P4 ( Even parity)?
Solution: The data bit with the corresponding symbol is shown in the table:
7
|
6
|
5
|
4
|
3
|
2
|
1
|
D7
|
D6
|
D5
|
P4
|
D3
|
P2
|
P1
|
1
|
0
|
1
|
?
|
1
|
?
|
?
|
Now, P1
depends on the data bits D3, D5 and D7 i.e
(111) which contains the odd number of 1’s.
So, for even parity P1
= 1.
P2 depends on
the data bits D3, D6 and D7 i.e (101) which
contains the even number of 1’s.
So, for even parity P2=
0.
P4 depends on
the data bits D5, D6 and D7 i.e (101) which
contains the even number of 1’s.
So, for even parity P3
= 0.
It's very helpful. Thanks for sharing.
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