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BCH codes are not limited to binary codes, but may be used with multilevel phase-shift keying whenever the number of levels is a prime number or a power of a prime number. A BCH code in 11 levels has been used to represent the 10 decimal digits plus a sign digit.
BCH codes make use of field theory and polynomials over that field. The way the check polynomial is constructed provides the key to indicating that an error has occurred.
If we wish to construct a BCH code to detect and correct 2 errors we use the finite field GF(16) or Z2[x]/<x4+x+1>
Now if we have α a root of x4+x+1, m1(x)=x4+x+1. Now m1 is minimal for α since
If we wish to construct a code to correct 1 error we use m1(x). Our codewords will be
- C(x)≡0 (mod m1(x))
which has roots α, α2, α4, α8.
This does not allow us to choose many codewords - so we look for the minimal polynomial for the missing power of α from above - α3, and then the minimal polynomial for this is
which has roots α3, α6, α12, α24=α9.
We take codewords having all of these as roots, so we form the polynomial
and take codewords corresponding to polynomials of degree 14 such that
- C(x)≡0 (mod m1,3(x))
So now C(α)=C(α2)=...=C(α8)=0. (*)
Now in GF(16) we have 15 nonzero elements, and thus our polynomial will be of degree 14 with 8 check and 7 information bits - we have 8 check bits since we have (*).
Construct our information codeword as
- (c14, c13, ..., c8)
so our polynomial will be
Call this CI.
We then want to find a CR such that CR=CI (mod m1,3(x))=c7+c6+...+c0
So we have the following codeword to send C(x) = CI+CR (mod m1,3(x)) = 0
For example, if we are to encode (1,1,0,0,1,1,0)
and using polynomial long division of m1,3(x) and CI to get CR(x), in Z2 we obtain CR to be
So then the codeword to send is
- (1,1,1,0,0,1,1,0, 0,0,0,0,1,0,0,1)
Suppose we receive a codeword vector r (the polynomial R(x)).
If there is no error R(α)=R(α3)=0
If there is one error, ie r=c+ei where ei represents the ith basis vector for R14
so we can recognize one error. A change in the bit position shown by α's power will aid us correct that error.
If there are two errors
- = (α3)i+(α3)j
which is not the same as S13 so we can recognize two errors. Further algebra can aid us in correcting these two errors.
Original source (first two paragraphs) from Federal Standard 1037C
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