Error Control Code

Widely used in many areas, like communications, DVD, data storage… In communications, because of noise, you can never be sure that a received bit is right In physical layer, what we do is, given k data bits, add n-k redundant bits and make it into a n -bit codeword. We send the codeword to the receiver. If some bits in the codeword is wrong, the receiver should be able to do some calculation and find out – There is something wrong – Or, these things are wrong (for binary codes, this is enough) – Or, these things should be corrected as this for non-binary codes – (this is called Block Code)

Error Control Codes You want a code to – Use as few redundant bits as possible – Can detect or correct as many error bits as possible

Error Control Code Repetition code is the simplest, but requires a lot of redundant bits, and the error correction power is questionable for the amount of extra bits used Checksum does not require a lot of redundant bits, but can only tell you “something is wrong” and cannot tell you what is wrong

(7,4) Hamming Code The best example for introductory purpose and is also used in many applications. (7,4) Hamming code. Given 4 information bits, (i0,i1,i2,i3), code it into 7 bits C=(c0,c1,c2,c3,c4,c5,c6). The first four bits are just copies of the information bits, e.g., c0=i0. Then produce three parity checking bits c4, c5, and c6 as (additions are in the binary field): – c4=i0+i1+i2 – c5=i1+i2+i3 – c6=i0+i1+i3 For example, (1,0,1,1) coded to (1,0,1,1,0,0,0). Is capable of correcting one bit error.

Generator matrix Matrix representation. C=IG where G is called the generator matrix.

Parity check matrix It can be verified that CH=(0,0,0) for all codeword C

Error Correction Given this, suppose you receive R=C+E. You multiply R with H:S=RH=(C+E)H=CH+EH=EH. S is called the syndrome. If there is only one `1’ in E, S will be one of the rows of H. Because each row is unique, you know which bit in E is `1’. The decoding scheme is: – Compute the syndrome – If S=(0,0,0), do nothing. If S!=(0,0,0), output one error bit.

How G is chosen How is G chosen such that it can correct one error? The key is – First: ANY linear combinations of the row vectors of G has weight at least 3 (having at least three `1’s) ANY codeword is a linear combination of the row vectors So a codeword has weight at least 3. Second: The sum of any two codeword is still a codeword So the distance (number of bits that differ) is also at least 3. So if one bit is wrong, won’t confuse it with other codeword. Because if there is only one bit wrong, the received vector will have one bit difference with the original codeword C0. If it is also only one bit away from another codeword C1, it means that C0 and C1 have at most two bit difference, a contradiction.

Error Detection What if there are 2 error bits? Can the code correct it? Can the code detect it? What if there are 3 error bits? Can the code correct it? Can the code detect it?

Exercise

Answer: (a). The error vector is a codeword.

The existence of H We didn’t compare a received vector with all codewords. We used H. The existence of H is no coincidence (need some basic linear algebra). Let Ω be the space of all 7-bit vectors. The codeword space is a subspace of Ω spanned by the row vectors of G. There must be a subspace orthogonal to the codeword space spanned by 3 vectors. So, we can take these 3 vectors and make them the column vectors of H. Given we have chosen a correct G (can correct 1 error), any vector with only one `1’ bit or two `1’ bits multiplied with H will result in a non-zero vector.

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