I'm having a little trouble understanding reed solomon encoding. I'm currently creating a datamatrix encoder which uses reed solomon to generate error correction code words. As an example I'm basically trying to generate 5 check characters for the numbers: 142 164 186. The polynomial divisor for generating 5 check characters is: g(x) = x^5 + 62x^4 + 111x^3 +144x^2 + 68x + 23 however I'm not quite following how I use this on my 3 numbers to generate those check characters. The 5 check characters that should be generated are 114 25 5 88 102. Could anyone possibly explain to me how this works? Kind regards, Deon

# reed solomon encoding.

Started by ●November 16, 2005

Reply by ●November 16, 20052005-11-16

"decipher" <dxb04u@cs.nott.ac.uk> wrote in message news:k9OdnY0LHo86o-benZ2dnUVZ_sednZ2d@giganews.com...> I'm having a little trouble understanding reed solomon encoding....> As an example I'm basically trying > to generate 5 check characters for the numbers: 142 164 186. > > The polynomial divisor for generating 5 check characters is: > > g(x) = x^5 + 62x^4 + 111x^3 +144x^2 + 68x + 23 > > however I'm not quite following how I use this on my 3 numbers to generate > those check characters. The 5 check characters that should be generated > are > 114 25 5 88 102. Could anyone possibly explain to me how this works?Assuming that your information polynomial is u(x) = 142x^2 + 164x + 186, multiply u(x) by x^5 and then divide the result by g(x) to get a quotient q(x) and a remainder r(x) = r4 x^4 + r3 x^3 + r2 x^2 + r1 x + r0o f degree 4 or less. Your codeword then is x^5 u(x) - r(x) = 142x^7 + 164x^6 + 186x^5 - r4 x^4 - r3x^3 - r2x^2 - r1x - r0. Since x^5 u(x) = q(x)g(x) + r(x), we see that x^5 u(x) - r(x) = q(x)g(x) is a multiple of g(x). Note: if you don't get the same answer 114 25 5 88 102, it may be that u(x) should have been 186x^2 + 164x + 142

Reply by ●November 16, 20052005-11-16

thanks very much for the quick reply. Ok I think i kind of follow you. Is it possible to work these 5 values out by hand as thats what I'm about to try?? ... :)

Reply by ●November 17, 20052005-11-17

decipher wrote:> thanks very much for the quick reply. Ok I think i kind of follow you. Is > it possible to work these 5 values out by hand as thats what I'm about to > try?? ... :)The worked examples in this might help: http://www.bbc.co.uk/rd/pubs/whp/whp031.shtml