Cyclic Codes BCH Codes

Transcription

1 Cycc Codes BCH Codes Gaos Feds GF m A Gaos fed of m eements can be obtaned usng the symbos 0,, á, and the eements beng 0,, á, á, á 3 m,... so that fed F* s cosed under mutpcaton wth m eements. The operator + s defned by ddng by p where p s a prmte rreducbe poynoma n GF m. = q. p + a where a s a poynoma of degree m- or ess oer GF. The outcome s a set of m non zero poynomas of á oer GF m wth degree m- or ess. Exampe.. tartng wth m = 4, p = , whch s a prmte poynoma oer GF and a factor of 5 +, et pá = á 4 + á 3 + = 0. Hence á 4 = + á 3 and the GF 4 can be constructed and s gen by Tabe 4. Eements of GF 4 usng p = oer GF 4 Power Eements Poynoma 4-Tupe á á á á á 3 á á 4 + á á 5 + á + á 3 0 á 6 + á + á + á 3 á 7 + á + á 0 á 8 á + á + á 3 0 á 9 + á 0 0 á 0 á + á á + á + á 3 0 á + á 0 0 á 3 á + á 0 0 á 4 á + á Tabe s rreducbe oer GF and does not hae roots oer GF, but t has 4 roots oer GF 4. These are gen by á 7, á, á 3 and á 4. It can be shown, usng Tabe 4., that + á 7 + á + á 3 + á 4 =

2 Further Feds A new fed eement â s ntroduced n an extenson fed of GF wth â a root of a poynoma f so that fâ = 0. L L L For any 0, s aso a root of f so that f = 0. The eement s the conugate of â. Ths aso mpes that f â, an eement n GF m s a root of f oer GF, then a the dstnct conugates of â, aso eements of GF m are roots of f. Exampe Usng f = = , á 7 s a root. The conugates of á 7 are á 7, 7 3, 7 4. Note that 7 s á = á / á 05 = á 7 and hence coses the group. The conugates of á 7 are á 4, á 8 /á 5 = á 3, and á 56 /á 45 = á. The other prmte roots are á 5 and á 0. Further snce â s an eement of GF m m m, n genera,, and 0, and the m nonzero eements of GF m m form a the prmte roots of 0. Aso snce the zero eement 0 of GF m s the root of, then the m eements form a the roots of m Mnma poynomas. The fed eement â can aso be a root of a poynoma of degree ess than m. The poynoma of smaest degree oer GF for whch f = fâ = 0 s known as the mnma poynoma of â, and denoted by Ö. Ths poynoma s aso rreducbe. Further f f, a poynoma oer GF has â as a root, then f n genera s dsbe by Ö, the mnma poynoma. If f tsef s an rreducbe poynoma then f = Ö. It foows that the conugates of â, â, roots of Ö = Öâ. It can be shown that f L 0, 3,.. are aso Exampe: For GF 4 and the fed eements of Tabe 4., startng from â = á 3, we obtan â = á 6, â 4 = á, â 8 = á 4. These resut n the poynoma + á á + á + á á + + á resutng n ++ á+ á + + á + á+ á + + á + á + á 3 fnay resutng n the mnma poynoma Another way to obtan the mnma poynoma s the foowng. Let ã = á n GF 4 be used as the prmte eement. Hence ã = á, ã 4 = á 4, ã 8 = á 8, and ã 6 = á 6 = ã coses the group. Hence Ö of degree 4 must hae the foowng form. Ö = a 0 + a + a + a a 4 4. Usng the poynoma representaton and substtutng for ã = á, Ö = a 0 + a á+ a á + a 3 á 3 + a 4 á 4.

3 Ths resuts n a 0 + a á+ a á + a 3 á 3 + a 4 á 3 + = 0. Ths s rearranged to get a 0 + a 4 + a á+ a á + a 3 + a 4 á 3 = 0 Hence a 0 + a 4 = 0 a = 0 a = 0 a 3 + a 4 = 0 Ths resuts n a 3 = a 4 = a 0 = ; a = a = 0; and therefore the poynoma = Ö. The Tabe 4. shows the mnma poynomas wth the prmte eements as the prmte roots of the mnma poynomas usng p = Conugate roots Mnma poynoma 0 + Ö 0 á, á, á 4, á 8, Ö á 3, á 6, á 9, á, Ö 3 á 5, á 0, ++ Ö 5 á 7, á, á 3, á 4, Ö 7 Tabe 4. Tabe 4. shows that the degree of each mnma poynoma n GF 4 usng as the prmte poynoma g. Note that budng up other generator poynomas g from g, st uses g so that g w aways ncude the prmte root á. BCH Code It s charactersed by the foowng: Bock ength n = m ; Party checks n-k mt; Mnmum dstance d mn t+; The generator poynoma g s specfed n terms of ts roots n GF m. Eery prmte eement á s a root of a mnma poynoma Ö. It can be shown that a een powers of á, beong to a mnma poynoma wth a precedng odd power as one of ts roots. Ths s ustrated by Tabe 4. aboe. BCH Bound: The mnmum dstance of the code generated by g s greater than the argest number of consecute prmte roots of g. Usng a generator poynoma g = Ö 0. Ö. Ö 7 yeds the set of prmte roots whose ndex s 0,., 4,,, 7, 8,,,,, 3, 4. Note that there are 5 consecute prmte roots n the sequence so that g has a mnmum dstance of at east 6. Lookng at Tabe 4., t can be seen that eery odd root s n the same poynoma as. Hence t consecute odd roots guarantee t consecute roots. Aso t can be shown 3

4 m that the degree of eery dsor of, cannot exceed m. nce at most t mnmum poynomas are requred to guarantee that g has t consecute odd roots, the order of g s m.t and, at most, m.t party checks are requred. Encodng a BCH codeword. The encodng process s dentca to the standard cycc code. For a k-bt data d the resutant party bts are found from rem { n-k..d}/ g whch are appended to the front of the d to obtan the codeword. Eery codeword n a BCH code s a codeword f t s dsbe by the GF m roots, á, á,...á t. Decodng a BCH codeword. Assume a codeword sent, and r s receed. Then the error pattern can be dered from r = + e. The syndrome of a t-correctng BCH code s gen by =,,.. t, and = rá Dde r, n turn, by each of the mnma poynomas comprsng g. In each case a remander term b s obtaned. Ths remander s n GF. Ths s substtuted by the correspondng prmte root beongng to the mnma poynoma. Exampe: Usng g = n GF 4 the 5,7 code uses as prmte poynomas, Ö = , and Ö 3 = Ths ges g = for a 5,7 code. Usng a data pattern [0000] that ges d = , a code word s but gen by = Let r be Ths resuts n an e = To determne the syndrome =,, 3, 4 the r s dded by each of the mnma poynomas. Usng Ö = , the remander s b = Usng the roots of the mnma poynoma, á, á, á 4, Hence = + á + á 3 = á = + á 4 + á 6 = + á + á = á 7 4 = + á 8 + á = á + á 3 = á 4 3 s obtaned from Ö 3 = The remander s b 3 = + +. Usng the frst root of ths mnma poynoma, á 3, 3 = + á 3 + á 6 = á + á = á 3. Hence = á, á 7, á 4, á 3 4

5 5 The second step, after determnng the syndrome n terms of the prmte eements s to determne the error ocaton poynoma ó from the syndrome components. There are arous methods aaabe. They are based on a genera souton nong the foowng. Gen the errors, t, the error postons are denoted by..,, nce the syndromes = eá, eery syndrome s reated drecty to the error parameters. Ths ges rse to a set of equatons t t t t Defne the error ocator poynoma as The prmte eement roots of ths poynoma are the nerse error ocaton postons. It s easy to show from the aboe the set of Newton Identtes gen by + ó = 0 + ó + ó = ó + ó + 3ó 3 = ó ó - + ó = 0 Note that snce n GF + = = 0, ó = ó for odd, and 0 for een. The Berekamp-Massey Agorthm w be used for the souton of the Newton Identtes. The goa of the agorthm s to fnd at teraton + connecton poynoma ó + n terms of the error poynoma prmte eements, and gen by usng as the error dscrepancy that becomes a correcton factor the aue, d, usng d where the upper ndces assocated wth ó ndcate the coeffcent aue assocated wth an approprate n the equaton óat the th teraton. If d = 0, then there s no dscrepancy at that stage, and the present aue of ó, ó, s carred to the next teraton ó +. If d 0, fnd a preous teraton row, ñ, for whch d 0, and the aue of ñ- ñ where ñ denotes the order of ó ñ. Then work out the aue of the next teraton ó + usng, max d d 4.3 The teratons are contnued unt the quantty, +t becomes ad

6 Exampe: The BCH 5,5 code, whch has t=3, s generated usng Ö = ; Ö 3 = ; Ö 5 = ++. Ths resuts n a g = A code poynoma s but usng the data pattern [00] whch s d = The codeword s but by usng 0 d/ g to obtan the remander. In ths case the remander s gen by so that the codeword = The receed word s r = Ths mpes an error poynoma e = Of course the decoder does not know ths. The procedure for decodng starts wth the syndrome cacuaton, obtaned by ddng the receed word r by each mnma poynoma n turn to work out the correspondng prmte eement assocated wth the syndrome eement. In ths case = [,, 3, 4, 5, 6 ]. nce á, á, á 4, are obtaned from the same poynoma Ö = , r s dded by Ö, to obtan b = Hence = + á + á 3, and usng the GF 4 arthmetc, based on + 3 = 4, and Tabe 4. = á. Usng á, = + á 4 + á 6 = + á + á = á 7. Usng á 4, 4 = + á 8 + á = á + á 3 = á 4. á 3, á 6, are obtaned from Ö 3 = , to obtan b 3 = ++, and usng the prmte eements, á 3 3 = + á 3 + á 6 = á + á = á 3 and usng á 6 6 = + á 6 + á = + á + á 3 = á. Fnay á 5 s obtaned from Ö 5 = ++, to obtan b 5 =. Therefore 5 =. The Berkeamp-Massey Agorthm s now used. Intasaton Iteraton 0: ó - = ; d - =0; - =0; - =0-0=0; snce d - =0; ó 0 = ; Iteraton: =0; d 0 = = á and usng 4.3 ó = ó 0 + á. Therefore at end of teraton the entry s + á. 0 0 Check on d : d = + ó = á 7 + á.á = á 7 + á = á 7 + á 7 = 0 Iteraton : =; d = 0; Hence ó = ó ; =; - = ; d = 3 + ó = á 3 + á 9.á = á 3 + á 8 = á + á + á 3 = á 8. Entry + á. á 8 Iteraton 3: =; d = á 8 ; Hence update ó, usng row teraton 0, to obtan 6

7 ó 3 = ó + d.d ó 0 = ó + á 8. / á.. = ó + á.. Hence ó 3 = + á + á.. Entry on Iteraton 3 s 3 + á + á.. 0 Check on d 3 : d 3 = ó 3 + ó 3 = á 4 + á 3.á + á 7.á = á 4 + á 9 + á 4 = 0 Iteraton 4: : =3; d 3 = 0; Hence ó 4 = ó 3 ; 4 = 3 =; - = ; Check : = 4. therefore 3 s not ad. Contnue. Current entry 4 + á + á.. d 4 d 4 = ó ó 4 = + á 4.á + á 3.á = + á 5 + á 5 = Iteraton 5: =4; d 4 = ; Hence update ó 4 to ó 5, usng row teraton, to obtan ó 5 = ó 4 + d 4.d ó =. / á 8. + á = ó 4 + á 7 + á 8 3 = + á + á + á 7 + á 8 3 = + á + á + á 3 3. Entry for teraton á + á + á Check on d 5 : d 5 = ó ó ó 3 5 = á +.á + á 4.á + á 3.á 3 = 0 Iteraton 6: =5; d 5 = 0; Hence ó 6 = ó 5 ; 6 = 5 =3; - = 3; Check : = 5. therefore 5 s true. Iteraton stopped. The outcome of the agorthm s ó 6 = + á + á + á 3 3. The roots of ths cubc poynoma are found to be n ths case by a process of tra and error on the ffteen prmte eements = ; = á 3 ; = á 9 ; eg for =; + á + á + á 3 = + á + á 3 + á + á 3 = 0 The error ocatons n e are the nerse of these roots. o error ocatons are at poston, 6,. Ths s the expected resut. The oera teratons are gen n the Tabe 4.3 beow 7

8 I ó d á á á á á + á á + á á + á + á á + á + á Other BCH Codes Tabe 4.3 Bnary BCH codes wth ength n m can be constructed as for those wth n = m. Let â be an eement of order n n GF m. Consder a poynoma that has as roots â, â, â 3,..., â t. n tsef s a factor of some m. A the eements are roots of n +. Therefore ths s a cycc code. In partcuar, for a sequence of t roots, the g that s the LCM of the mnma poynomas of a the roots, generates a t-error correctng BCH code. nce â s not a prmte eement of GF m and n m, the BCH code generated s caed a nonprmte BCH code. Non-bnary BCH codes Reed oomon Codes Bnary BCH codes can be generazed to any GFq where p s a prme number and q any power of p to obtan a q-ary code. An n,k q-ary cycc code s generated by a poynoma of degree n-k wth coeffcents from GFq whch s a factor of n +. Let á be a prmte eement, n GFq s, where n = q s. For a t error correctng code the generator poynoma g has t roots from GFq gen by á, á,.., á t. The degree of each mnma poynoma s s or ess, and hence the number of party check dgts generated by by g s no more than st. The speca subcass for whch s= s the most mportant subcass of q-ary BCH codes. These codes are usuay caed Reed-oomon codes. A t-error correctng R code wth symbos from GFq has the foowng parameters Bock ength n = q Number of party-check dgts n-k = t Mnmum dstance d mn = t+. Usng GFq = GF m, and usng á as a prmte eement n GF m, a Reed- oomon code, t-error correctng, can be generated usng a g = +á+á +á 3. +á t so that g = g 0 + g + g + + g t- t- + t, so that the g s are now not from GF but from GF m. Generatng a codeword s st the process of ddng t d by g and usng the remander to bud up the systematc codeword. 8

9 Decodng foows the nes of a BCH code nong:. yndrome cacuaton. Error ocaton usng an error ocaton poynoma, and an agorthm for the souton such as the Berekamp-Massey agorthm for the souton of ó 3. Obtan from the error ocaton poynoma, the error aues, Z, n terms of á s usng Newton s denttes 4. Fnay obtan the error aues at the ocatons obtaned from the error ocaton poynoma usng an equaton reatng the error ocatons and Z. 9