CHNG 1

TNG QUAN V H THNG THNG TIN S1.1 V tr ca m ha knh trong h thng thng tin sM ha knh l mt khu rt quan trng trong h thng thng tin s khng dy cng vi m ha ngun, ghp knh, iu ch, to ra mt tn hiu ph hp cho vic truyn dn v tuyn v tn hiu c kh nng iu khin c s sai bit v sa cc li xy ra nu c c th khi phc li gn nh nguyn dng tn hiu tin tc m mnh truyn i.

Hnh 1.1: V tr ca m ha knh truyn trong h thng thng tin s

M ho knh: mc ch l lm gim xc sut sai thng tin khi truyn qua knh truyn.Vic gim thiu xc sut sai da vic pht hin sai v sa sai c th dn n vic gim t s tn hiu trn nhiu (SNR) cn thit nh gim c cng sut, tit kim nng lng. Vic sa sai hu hiu cho tn hiu SNR nh s thun li cho vic bo mt, tri ph v tng chnh xc ca thng tin nhn- mc ch quan trng nht ca truyn thng.1.2 Khi nim m ha knh v phn loi1.2.1 Khi nimM ha knh l vic a thm cc bit d vo tn hiu s theo mt quy lut no y, nhm gip cho bn thu c th pht hin v thm ch sa c c li xy ra trnknh truyn.Thc hin b gii m Viterbi trn FPGATrang 22

Chng 1: Tng quan v h thng thng tin s

Mt s h thng c th khc phc li bng cch gi mt yu cu cho bn pht gi li tn hiu nu pht hin li, l ch ARQ. Nhng vic ny ch thch hp cho cc h thng truyn dn hu tuyn v mt s h thng v tuyn khng yu cu v thi gian tr. Thay vo , vi cc h thng thng tin khng dy ngy nay, ngi ta hay s dng mt loi m c th pht hin v khc phc li mt cch t ng. Vic ny gim thiu thi gian tr so vi cc h thng yu cu truyn li. B m ny thng c gi l m iu khin li (ECC), hay chnh xc hn l FEC.Mc ch ca l thuyt M ha trn knh truyn l tm nhng m c th truyn thng nhanh chng, cha ng nhiu t m t hp l v c th sa li hoc t nht pht hin cc li xy ra. Cc mc ch trn khng ph thuc vo nhau, v mi loi m c cng dng ti u cho mt ng dng ring bit. Nhng c tnh m mi loi m ny cn cn tu thuc nhiu vo xc sut li xy ra trong qu trnh truyn thng.i vi mt a CD thng thng, li trong m thanh xy ra ch yu l do bi v nhng vt xc trn mt a. V th, cc m c lng vo vi nhau. D liu c phn b trn ton b mt a. Tuy khng c tt cho lm, song mt m ti din n gin c th c dng lm mt v d d hiu. Chng hn, chng ta ly mt khi s liu bit (i din cho m thanh) v truyn gi chng ba ln lin. Bn my thu, chng ta kim tra c ba phn lp li trn, tng bit tng bit mt, ri ly ci no c s bu cao nht. im khc bit y l, chng ta khng ch truyn gi cc bit theo th t. Chng ta lng n vo vi nhau. Khi d liu ny, trc tin, c chia ra lm 4 khi nh. Sau chng ta gi mt bit khi u tin, tip theo mt bit khi th hai v.v tun t qua cc khi. Vic ny c lp i lp li ba ln phn b s liu ra trn b mt a. Trong ng cnh ca m ti din n gin trn, vic lm ny hnh nh khng c hiu qu cho lm. Song hin nay c nhng m c hiu ng cao, rt ph hp vi vic sa li xy ra t ngt do mt vt xc hay mt vt bi, khi dng k thut lng s liu ni trn.Mi m thng ch thch hp cho mt ng dng nht nh. Vin thng trong v tr b gii hn bi nhiu nhit trong thit b thu. Hin trng ny khng xy ra mt cch t pht bt thng, song xy ra theo mt chu trnh tip din. Tng t nh vy, modem vi di tn hp b hn ch v nhiu m tn ti trong mng li in thoi. Nhng nhiu m ny c th c biu hin r hn bng mt m hnh tp m tip din. in thoi di ng hay c vn do s suy sng nhanh chng xy ra. Tn s cao c dng c th gy ra s suy sng tn hiu mt cch nhanh chng, ngay c khi my nhn ch di ch vi phn Anh. Mt ln na, ngi ta hin c mt loi m ha trn knh truyn c thit k i u vi tnh trng suy sng.1.2.2 Phn loi m ha knhL thuyt m ha i s c chia ra lm 2 loi m chnh1. M khi.

2. M trellis.Chng phn tch ba c tnh sau ca m (ni chung) l: Chiu di ca m. Tng s cc t m hp l. Khong cch Hamming ti thiu gia hai t m hp l.

Hnh 1.2: S phn chia m ha knh thnh hai nhnh ring bitTrong mi loi m li c phn tch thnh 2 nhnh na l m tuyn tnh v m khng tuyn tnh.Thng th cc m khng tuyn tnh khng c ng dng trong thc t v cc nhc im ca n, nn y chng ta ch cp n cc m tuyn tnh.Trong phn tip theo chng ta s khi qut s lc v m khi v m trellis.1.3 Khi qut v m khi v m trellis1.3.1 M khiM khi tuyn tnh mang tnh nng tuyn tnh, chng hn tng ca hai t m no y li chnh l mt t m; v chng c ng dng vo cc bit ca ngun trn tng khi mt; ci tn m khi tuyn tnh l v vy. C nhng khi m bt tuyn tnh, song kh m chng minh c rng mt m no l mt m tt nu m y khng c c tnh ny.Bt c m khi tuyn tnh no cng c i din l (n,m,dmin), trong 1. n, l chiu di ca t m, trong k hiu,2. m, l s k hiu ngun c dng m ha tc thi,

3. dmin, l khong cch hamming ti thiu ca m. C nhiu loi m khi tuyn tnh, nh1. M vng (M Hamming l mt b phn nh ca m tun hon).2. M chn l.3. M Reed-Solomon.4. M BCH.5. M Reed-Muller.6. M hon ho.M khi c gn lin vi bi ton ng gi ng xu l bi ton gy mt s ch trong nhiu nm qua. Trn b din hai chiu, chng ta c th hnh dung c vn mt cch d dng. Ly mt nm ng xu, nm trn mt bn, ri dn chng li gn vi nhau. Kt qu cho chng ta mt mu hnh lc gic tng t nh hnh t ong. Cc m khi cn da vo nhiu chiu khc na, khng d g m hnh dung c. M Golay c hiu ng cao, dng trong truyn thng qua khong khng v tr, s dng nhng 24 chiu. Nu c dng l m nh phn (thng thy), cc chiu m ch n chiu di ca t m nh nh ngha trn.1.3.2 M trellisM trellis hay cn gi l m chp (kt hp) c s dng trong cc modem di tn m (V.32, V.17, V.34) v trong cc in thoi di ng GSM, cng nh trong cc thit b truyn thng ca qun i v trang v trong cc thit b truyn thng vi v tinh.Mc ch ca vic to ra m chp l nhm lm cho tt c cc k hiu t m tr thnh tng trng s ca nhiu loi k hiu thng ip trong nhp liu. N tng t nh ton kt hp c dng trong cc h tuyn tnh bt bin dng tm xut liu ca mt h thng, khi chng ta bit nhp liu v cc p ng xung.Ni chung chng ta tm xut liu ca b m chp h thng, tc s kt hp ca nhp liu bit, i chiu vi trng thi ca b m ha kt hp, hoc trng thi ca cc thanh ghi.V c bn m ni, m chp khng gip thm g trong vic chng nhiu hn mt m khi tng ng. Trong nhiu trng hp, chng ni chung cho chng ta mt phng php thc thi n gin hn, hn hn mt m khi c hiu qu tng ng. B m ha thng l mt mch in n gin, c mt b nh, mt vi bin php truyn thng tin phn hi bo tnh hnh, thng l cc cng loi tr XOR. B m ha c th c thc thi trong phn mm hay phn sn.Thut ton Viterbi l mt thut ton ti u nht c dng gii m cc m chp. Hin c nhng phng php gim c gip vo vic gim khi lng tnh

ton phi lm. Nhng phng php ny phn ln da vo vic tm tuyn ng c kh nng xy ra cao nht. Tuy khng ngn gn, song trong mi trng nhiu thp hn, ngi ta thng thy chng cho nhng kt qu kh quan. Cc b iu hnh vi x l hin i c kh nng thc hin nhng thut ton tm gim c ni trn vi t l trn 4000 t m trong mt giy. ti ch yu nghin cu v thut ton gii m Viterbi thy c u im ca thut ton trong vic gim ti thiu sai s khi m ha v gii m tn hiu. Do , trong cc phn tip theo ca n, chng ta ch tm hiu vic m ha tin tc dng m chp v gii m da trn thut ton Viterbi cng nh nhng u khuyt im ca chng. ng thi ta tin hnh m phng thut ton trn Matlab v trn Kit FPGA kim chng thc t hn. Cn i vi cc m trellis cn li th ta s khng phn tch trong phm vi cun n ny.

CHNG 2THUT GII M VITERBI2.1 Khi nim m chpM chp l mt k thut m ha sa sai. M chp thuc h m li (m ha theo Trellis) v c xy dng da trn mt a thc sinh hoc mt s chuyn trng thi (trellis m) c trng. Qu trnh gii m ca m chp phi da vo trellis m thng qua cc gii thut khc nhau, trong ni ting nht l gii thut Viterbi.Ti sao gi l m chp v cu trc m ha c th biu din di dng php tnh chp gia a thc sinh m v chui tn hiu c m ha.M ha chp v thut ton gii m Viterbi c s dng trong khong hn mt t in thoi, c th l ln nht trong cc loi thut ton c ng dng. Tuy nhin, hin ti th thut ton x l viterbi c ng dng nhiu nht trong cc thit b m thanh v hnh nh k thut s. Ngy nay, chng cn c s dng trong cc thit b bluetooth.Mc ch ca m ha knh truyn l nhm tng dung lng knh truyn, bng cch cng thm vo tn hiu nhng d liu d tha c thit k mt cch cn thn trc khi truyn ln knh truyn. M ha chp v m ha khi l 2 dng chnh ca m ha knh truyn. M ha chp th da trn d liu ni tip, 2 hoc mt vi bit c truyn mt lc, cn m ha khi th da trn mt khi d liu ln tng quan (c trng l khong vi trm bytes). V d, m Redsolomon l mt m ha khi.S khc nhau c bn gia m ha khi v m ha chp l m ha khi l m ha khng nh. Cho mt chui d liu K bit, th ng ra ca b m ha khi l mt khi d liu n bit duy nht. M ha chp khng kt ni cc khi bit ring vo trong mt khi t m, thay vo n s chp nhn mt chui bit lin tc v ta thnh mt chui ng ra. Hiu qu hay tc d liu ca m ha chp c nh gi bng t l ca s bit ng vo k, v s bit ng ra n. Trong m ha chp l c mt vi b nh dng ghi nh dng bit vo. Thng tin ny c s dng m ha cc bit tip theo.2.2 Biu din m chpC ba phng php biu din m chp l: s li, s trng thi, v s hnh cy. lm r phng php ny ta tp trung phn tch da trn v d hnh 2.1 vi b m (2,1,3), a thc sinh (7,5).

* S hnh cy:Gi thit trng thi ban u ca cc thanh ghi dch trong b m u l trng thi ton 0. Nu bit vo u tin l bit 0 th u ra ta nhn c chui 00, cn nu bit vo u tin l bit 1 th u ra ta nhn c chui 11. Nu bit vo u tin l bit 1 v bit vo tip theo l bit 0 th chui th nht l 11 v chui th hai l chui 10. Vi cch m ho nh vy, ta c th biu din m chp theo s c dng hnh cy (xem hnh 2.1).Vi hng ln l hng ca bit 0 i vo b m, nhnh i xung biu hin cho bit 1 c dch vo. T s hnh cy ta c th thc hin m ho bng cch da vo cc bit u vo v thc hin ln theo cc nhnh ca cy, ta s nhn c tuyn m, t ta nhn c dy cc chui u ra.

*S hnh li:

Do c tnh ca b m chp, cu trc vng lp c thc hin nh sau: chui n bit u ra ph thuc vo chui k bit u vo hin hnh v (N-1) chui u vo trc hay (N-1) k bit u vo trc . T v d hnh 2.1 ta c chui 2 bit u ra ph thuc vo 1 bit u vo l 1 hoc 0 v 4 trng thi c th c ca hai thanh ghi dch, l 00; 01; 10; 11. T s hnh cy trn, ta thy rng ti tng th 3, c mi trng thi 00, 01, 10, 11 u c 2 nhnh n t cc trng thi trc ty thuc vo bit c dch vo b m l bit 0 hay bit 1. Vi tnh cht ta c th biu din m chp bng s c dng hnh li gn hn, trong cc ng lin nt c k hiu cho bit u vo l bit 1 v ng t nt c k hiu cho cc bit u vo l bit 0 (xem hnh 2.9). Ta thy rng t sau tng th hai hot ng ca li n nh, ti mi nt c hai ng vo nt v hai ng ra khi nt. Trong hai ng i ra th mt ng vi bit u vo l bit 0 v ng cn li ng vi bit u vo l bit 1.

*S trng thi:S trng thi c thc hin bng cch n gin s 4 trng thi c thc ca b m (00, 01, 10 v 11) v trng thi chuyn tip c th c to ra t trng thi ny chuyn sang trng thi khc, qu trnh chuyn tip c th l:

Next State/output symbol, if

Current StateInput = 0:Input = 1:

0000/0010/11

0100/1110/00

1001/1011/01

1101/0111/10

Kt qu ta thu c s trng thi trong hnh 2.10 nh sau:

Hnh 2.10: S trng thi ca b m chp (2,1,3)

T s trng thi hnh 2.10, cc ng lin nt c k hiu cho bit u vo l bit 0 v ng t nt c k hiu cho cc bit u vo l bit 1. So vi s hnh li v s hnh cy th s trng thi l s n gin nht.2.3 u nhc im ca m chp2.3.1 u imCng nh cc m sa sai khc, m chp cho php chng ta c th sa li d liu b sai lch khi truyn qua knh truyn khi phc chnh xc tn hiu gc.Vic thc hin m ha dng m chp tng i n gin hn cc loi m sa sai khc m cht lng m ha li tt.Vic thc hin m ha dng m chp c th c thc hin bng phn cng v phn mm.Da trn hnh thc m ha m chp cng thut gii Viterbi cho n, cc b m ha sau ny u k tha nhng c tnh u vit ca n.2.3.2 Nhc imVic m ha v gii m lin quan n m chp ch gii quyt c cc li mt bit cn i vi cc knh truyn xut hin nhiu bit lin tip th thut ton m ha v gii m ny s khng cn hon ho na.Knh truyn y phi l knh truyn t nhiu, v nu knh truyn nhiu qu ln, m ha chp s khng cn tt na. Khi ta phi cn ti tri ph tn hiu a tn hiu xung di mc nhiu gim thiu nh hng.

2.4 nh nghia thut ton ViterbiThut ton Viterbi l mt gii php c s dng ph bin gii m chui bit c m ha bi b m ha tch chp. Chi tit ca mt b gii m ring ph thuc vo mt b m ha tch chp tng ng. Thut ton Viterbi khng phi l mt thut ton n l c th dng gii m nhng chui bit m c m ha bi bt c mt b m ha chp no.Thut ton Viterbi c khi xng bi Andrew Viterbi nm 1967 nh l mt thut ton gii m cho m chp qua cc tuyn thng tin s c nhiu. N c s dng trong c hai h thng CDMA v GSM, cc modem s, v tinh, thng tin v tr, v cc h thng mng cc b khng dy. Hin nay cn c s dng ph bin trong k thut nhn dng ging ni, nhn dng t m, ngn ng hc my tnh.Thut ton gii m Viterbi l mt trong hai loi thut ton gii m c s dng vi b m ha m chp- mt loi khc l gii m tun t. u im ca gii m tun t so vi Viterbi l n c th hot ng tt vi cc m chp c chiu di rng buc ln, nhng n li c thi gian gii m bin i.

Cn u im ca thut ton gii m Viterbi l n c thi gian gii m n nh. iu rt tt cho vic thc thi b gii m bng phn cng. Nhng m yu cu v s tnh ton ca n tng theo hm m nh l mt hm ca chiu di rng buc, v vy, trong thc t, ngi ta thng gii hn chiu di rng buc ca n K = 9 hoc nh hn. Stanford Telecom to ra mt b gii m Viterbi K = 9 hot ng tc n 96 kbps, v mt b gii m vi K = 7 hot ng vi tc ln n 45 Mbps. Cc k thut khng dy nng cao c th to ra mt b gii m Viterbi vi K = 9 hot ng tc ln n 2 Mbps. NTT tuyn b rng h to c b gii m Viterbi hot ng tc 60 Mbps, nhng tnh kh thi ca n vn cha c kim chng.2.5 Phn tch thut gii ViterbiChng ta s ly v d v m chp c tc m l k/n =

Hnh 2.11: B m chp tc FF: thanh ghi dch. Ti mi xung clock, ni dung ca thanh ghi dch c dch qua phi 1 bit. Bit u tin s l ng vo, v bit cui cng s l ng ra. Mt thanh ghi dch c th s xem xt vic cng tr vo ng vo. Cc thanh ghi dch c th c hiu nh l b nh ca b m ha. N ghi nh nhng bit u ca chui.Thanh ghi dch c khi u vi tt c gi tr l 0. Thut ton XOR: 1 1= 0; 1 0=1; 0 1=1; 0 0=0Nu chng ta lm vic trn mt chui ng vo l 01011101, ng ra l 00 11 10 00 01 10 01 002. B m ha ny cng c th c m hnh bi mt bng trng thi hu hn. Mi mt trng thi c quy nh bi 2 bit nh phn- trng thi ca 2 thanh ghi dch. Mi mt s chuyn trng thi c quy nh bi w/v1v2 vi w i din cho bit ng vo, v v1v2 l i din cho 2 bit ng ra, trong trng hp ny chng ta lun lun c w = v1.

Bng 2.1: Trng thi ng vo v ng ra ca b m ha tc

Next State/output symbol, if

Current StateInput = 0:Input = 1:

0000/0010/11

0100/1110/00

1001/1011/01

1101/0111/10

Hnh 2.12: hnh trng thi ca m chp By gi chng ta c th m t thut ton gii m, phn chnh l thut ton Viterbi. C l, khi nim quan trng nht h tr cho vic hiu c thut ton Viterbi l s Trellis. Hnh bn di cho chng ta thy s trellis cho v d ca chng ta tc , m ha chp vi chiu di rng buc K = 3 vi bn tin 8 bit.

T=0

T=1

T=2

T=3

T=4

T=5T=6T=7T=8T=9T=10

State 00

State 01

State 10

State 11

Hnh 2.13: Cc nhnh trong b m ha

Bn trng thi c th ca b m ha c m t nh 4 hng ca nhng du chm theo chiu ngang. C mt ct ca 4 chm cho trng thi khi u ca b m ha v mt mi thi im ca bn tin. Cc ng in m kt ni cc im trong s biu din cho s chuyn trng thi khi ng vo l mt bit 1. ng chm chm l biu din cho s chuyn trng thi khi ng vo l bit 0. Ta c th thy r s ph hp gia s trellis v hnh trng thi ni trn.Hnh v bn di cho ta thy trng thi trellis cho ton b 8 bit ng vo. Cc bit ng vo b m ha v k hiu ng ra c th hin bn di ca hnh.

State 00

State 01

State 10

State 11

T=0

T=1

T=2

T=3

T=4

T=5T=6T=7T=8T=9T=10

ENC IN =0101110100

ENC OUT =00111000011001001011

Hnh 2.14: ng i hon chnh khi phc chnh xc tn hiu ti ng ra.Cc bit ng vo v cc k hiu ng ra ca b m th c th xem di cng ca hnh trn. Ch s ph hp gia cc k hiu ng ra v bng ng ra chng ta cp trn. Hy xem xt mt cch chi tit hn, s dng phin bn m rng ca s chuyn i t mt trng thi tc thi n mt trng thi k tip nh hnh bn di:

State 000011110010010110

State 01

State 10

State 11

Gi chng ta s xem xt cch thc gii m ca thut ton Viterbi. By gi chng ta gi s l chng ta c mt mu tin m ha (c th c vi li) v chng ta mun khi phc li tn hiu gc.Gi s chng ta nhn c mu tin m ha trn vi 1 bit li.

State 00

State 01

State 10

State 11

T=0

T=1

T=2

T=3

T=4

T=5T=6T=7T=8T=9T=10

ENC IN =0101110100

ENC OUT =00111000011001001011

RECEIVED =00111100011001001011

ERRORS =X

Hnh 2.15: Tn hiu nhn c 1 bit sai ti t =2 mi thi im chng ta nhn c 2 bit trong k hiu. Chng ta s tnh ton thng s metric o khong cch gia nhng g m chng ta nhn c vi tt c cc cp bit k hiu knh truyn c th m chng ta c th nhn c. i t thi im t=0 n t=1, ch c 2 trng thi m chng ta c th nhn c l 00 v 11. l bi v chng ta bit c b m ha tch chp bt u vi trng thi tt c u l 0 v cho 1 bit vo l 0 hay 1 th ch c 2 trng thi m chng ta c th i n v 2 ng ra ca b m ha. Nhng ng ra ny c trng thi l 00 v 11.Thng s metric m chng ta s s dng l khong cch Hamming gia cp bit ca k hiu nhn c v cp bit c th ca knh truyn. Khong cch Hamming c tnh mt cch n gin bng cch m c bao nhiu bit khc gia cp bit nhn c t knh truyn v cp bit so snh. Kt qu ch c th l 0, 1, 2. Gi tr ca khong cch Hamming (hay thng s metric) m chng ta tnh ton mi khong thi gian cho ng dn ca trng thi ti thi im trc v trng thi hin ti c gi l metric nhnh (branch metric). thi im u tin, chng ta s lu nhng kt qu ny nh l thng s metric tch ly, c lin kt n cc trng thi. thi im th 2, thng s metric tch ly s c tnh ton bng cch cng thm thng s metric tch ly trc vo metric nhnh hin ti. thi im t=1, ta nhn c 2 bit 00. Ch c mt cp k hiu knh truyn m chng ta c kh nng nhn c l 00 v 11. Khong cch Hamming gia 00 v 00 l bng 0. Khong cch Hamming gia 00 v 11 l 2. Do , gi tr thng s metric nhnh cho nhnh ng vi s chuyn trng thi t 00 n 00 l 0 v cho nhnh t 00 n 10 l 2. Khi m thng s metric tch ly trc l 0 th thng s metric tng s chnh bng thng s metric ca nhnh va xt. Tng t ta tnh c thng s metric cho 2 trng thi kia. Hnh bn di minh ha kt qu ti thi im t= 1

State 00

State 01

State 10

State 11

ENC IN = ENC OUT =

T=00011

0

00

T=1

Accumulated Error Metric =0

2

RECEIVED =00

Hnh 2.16: Ti thi im t = 1iu g s xy ra thi im t=2, chng ta nhn c mt cp k hiu knh truyn l 11, trong khi cp k hiu knh truyn m chng ta c th nhn c l 00 nu chuyn t trng thi 00 sang trng thi 00 v 11 khi chuyn t trng thi 00 n trng thi 10, 10 khi chuyn t trng thi 10 n trng thi 01, 01 khi chuyn t trng thi 10 n trng thi 11. Khong cch Hamming gia 00 v 11 l 2, gia 11 v 11 l 0, gia 01 hoc 10 vi 11 l 1. Chng ta cng cc thng s metric mi nhnh li vi nhau. thi im t=1 th trng thi ch c th l 00 hoc 10, thng s metric tch ly s c cng vo l 0 hoc l 2 mt cch tng ng. Hnh bn di th hin s tnh ton thng s metric tch ly thi im t=2.

State 00

State 01

State 10

State 11

ENC IN = ENC OUT =

T=0000011111001

0

00

T=1

T=2

1

11

Accumulated Error Metric =0 + 2 = 2

2 + 1 = 3

0 + 0 = 0

2 + 1 = 3

RECEIVED =0011

Hnh 2.17: Ti thi im t = 2 l tt c s tnh ton cho thi im t=2. ng nt m l metric nhnh c la chn v theo cc nhnh , thng s metric l nh nht. Gi chng ta s tnh thng s metric tch ly cho mi trng thi ti thi im t=3.Gi chng ta hy nhn vo hnh minh ha cho thi im t=3. Chng ta s gp phi mt t phc tp hn y, ti mi trng thi trong 4 trng thi ti t=3 s c 2

ng n t 4 trng thi ca thi im t=2. Chng ta s xoay s th no? Cu tr li l, chng ta s tnh ton thng s metric tch ly lin quan ca mi nhnh, v chng ta s b i gi tr metric ln hn, tc l s loi b nhnh i. Nu cp thng s metric mi trng thi l bng nhau th chng ta s gi li c 2 trng thi. Chng ta s k tha nhng tnh ton thc hin thi im t=2. Thut ton cng thng s metric tch ly trc vo nhnh mi, so snh kt qu v chn thng s metric nh hn (nh nht) tip tc dng cho thi im k tip, c gi l thut ton cng-so snh-chn. Hnh bn di cho thy kt qu ca vic x l ti thi im t=3.

State 00

State 01

State 10

State 11

ENC IN = ENC OUT =

T=00000001111111100101001010110

0

00

T=1

T=2

10

1110

T=3

Accumulated Error Metric =2+2 , 3+0 : 3

0+1 , 3+1 : 1

2+0 , 3+2 : 2

0+1 , 3+1 : 1

RECEIVED =00

1111

Hnh 2.18: Ti thi im t = 3Ch l cp k hiu knh truyn th 3 m chng ta nhn c s c mt li.Thng s metric nh nht hin ti l 1.Chng ta hy xem iu g xy ra thi im t=4. Tin trnh x l cng ging nh thi im t=3. Kt qu xem hnh bn di

State 00

State 01

State 10

State 11

ENC IN = ENC OUT =

T=000000000111111111111000010101001010101011010

0

00

T=1

T=2

10

1110

T=3

1

00

T=4

Accumulated Error Metric =3+0 , 1+2 : 3

2+1 , 1+1 : 2

3+2 , 1+0 : 1

2+1 , 1+1 : 2

RECEIVED =00

111100

Hnh 2.19: Ti thi im t = 4

Ch l thi im t=4, ng trellis ca tin tc thc s truyn i c xc nh bng ng in m, vi thng s metric tch ly l nh nht. Hy xem xt thi im t=5:

State 00

State 01

State 10

State 11

T=0000000000011111111111111110000001010101001010101010101101010

T=1

T=2

T=3

T=4

T=5

Accumulated Error Metric =3+1 , 2+1 : 3

1+2 , 2+0 : 2

3+1 , 2+1 : 3

1+0 , 2+2 : 1

ENC IN =01011

ENC OUT =0011100001

RECEIVED =0011110001

Hnh 2.20: Ti thi im t = 5 thi im t=10, s trellis s nh th ny, cc nhnh ko c chn c b i:

State 00

State 01

State 10

State 11

T=0

T=1

T=2

T=3

T=4

T=5T=6T=7T=8T=9T=10

ENC IN =0101110100

ENC OUT =00111000011001001011

RECEIVED =00111100011001001011

ERRORS =X

Hnh 2.21: Tt c d liu c gii m v sa sai chnh xcKt qu y cho thy chng ta gii m ng chui d liu gc. Nu chng ta nhn li con ng chng ta tm ra d liu gc l bng cch so snh d liu nhn c vi d liu so snh ca b gii m c c t bng trng thi. iu ny cho thy chng ta ang s dng thut ton gii m da trn s ging nhau ln nht.

Vic x l gii m bt u vi xy dng mt thng s metric tch ly cho mt s cp k hiu knh truyn nhn c, v lu gi trng thi mi thi im t m thng s metric l nh nht. Mt khi thng tin ny c dng ln th b gii m viterbi sn sng khi phc li chui bit a vo b m ha chp, khi mu tin c m ha truyn i. iu ny t c bng nhng bc sau: u tin, chn mt trng thi c thng s metric nh nht v lu li s trng thi ca trng thi . S dng lp li cho nhng bc k tip mi cho n khi bt u ca trellis t c: da vo bng ghi nh trng thi cho trng thi c chn, chn mt trng thi mi c lit k trong bng ghi nh trng thi khi chuyn t trng thi trc n trng thi . Lu s trng thi ca mi trng thi c chn. Bc ny c gi l truy hi (traceback). Chng ta lm vic tip vi danh sch nhng trng thi c chn c lu trong bc x l trc . Chng ta tra xem bit ng vo no ph hp vi s truyn dn t mi trng thi trc n trng thi k tip. y l bit m phi c m ha bng m tch chp.Bng sau cho chng ta thy ma trn tch ly ca y 8 bit (cng vi 2 bit ph thm) ca bn tin mi thi im t:Bng 4.2: Bng ma trn tch ly ca c 8 bit ca bn tin

Ch rng v d v b gii m Viterbi ng vo quyt nh cng ny, thng s metric tch ly nh nht trng thi cui ch ra c c bao nhiu li k hiu knh truyn xy ra.Bng lch s trng thi bn di cho thy trng thi tn ti trc cho mi trng thi ti thi im t:

Bng 2.3: Bng lch s trng thi

Tng ng 0,1,2,3 l cc v tr 00,01,10,112.Bng sau cho thy trng thi c la chn khi truy hi ng dn t bng trng thi tn ti trn:

Bng 2.4: Bng cc trng thi c la chn khi truy hi

S dng bng ta thy c s chuyn i trng thi n cc ng vo gy ra chng, gi chng ta c th to li bn tin gc. Bng ny rt ging vi v d ca chng ta b m ha chp tc v K= 3.Bng 2.5: Bng trng thi k tip

Ghi ch: trong bng trn, x ch ra l mt s chuyn trng thi khng th xy ra t mt trng thi n mt trng thi khc.

By gi chng ta c tt c cc cng c cn thit ti to li bn tin gc t bn tin m chng ta nhn c.Bng 2.6: Bng cha cc d liu ca bn tin gc c khi phc

Hai bit ph c b qua.Lm th no m thut ton truy hi cui cng cng tm ra con ng i ng nht ca n thm ch nu n chn trng thi ban u l sai. iu ny c th xy ra nu c hn mt trng thi c thng s metric tch ly l nh nht. Chng ta s dng li hnh 2.18 lm sng t iu ny:

State 00

State 01

State 10

State 11

ENC IN = ENC OUT =

T=00000001111111100101001010110

0

00

T=1

T=2

10

1110

T=3

Accumulated Error Metric =2+2 , 3+0 : 3

0+1 , 3+1 : 1

2+0 , 3+2 : 2

0+1 , 3+1 : 1

RECEIVED =00

1111

thi im t=3, c 2 trng thi 01 v 11 u c thng s metric l 1. ng i ng i n trng thi 01, ch l ng in m l ng i thc s ca bn tin n trng thi ny. Nhng gi s chng ta chn trng thi 11 bt u qu trnh truy hi ca chng ta. Trng thi trc ca trng thi 11 l trng thi 10, cng l trng thi trc ca trng thi 01. iu ny l bi v thi im t=2, trng thi 10 c thng s metric tch ly l nh nht. V vy, sau trng thi bt u sai, chng ta c th ngay lp tc tr li vi tuyn ng ng.Vi v d v bn tin 8 bit, chng ta tin hnh xy dng mt s trellis cho ton b bn tin trc khi bt u qu trnh truy hi. Vi cc bn tin di hn hoc cc chui d liu lin tc, iu ny l khng thc t, bi v b nh chiu di rng buc v s tr hon trong gii m. Nghin cu cho thy l su truy hi ca Kx5 ch cho vic gii m viterbi vi loi b m m chng ta ang tho lun. Bt c mt s truy hi su hn s lm tng thi gian delay gii m v b nh yu cu cho

vic gii m v cng ko lm tng hiu qu vic gii m, ngoi tr b m ha thng (punctured code) m chng ta s ni sau. thc thi mt b gii m Viterbi bng phn mm, bc u tin l phi xy dng mt vi cu trc d liu xoay quanh thut gii m s c thc thi. Nhng cu trc d liu ny c thc thi tt nht khi l cc mng. Su mng chnh m chng ta cn cho b gii m viterbi l: Mt bn sao ca Bng tri thi k tip ca b m ha m chp, bng chuyn trng thi ca b m ha. Kch c ca bng ny (hng x ct) l 2(K-1) x 2K. Mng ny phi c khi u trc khi bt u tin trnh gii m. Mt bn sao ca bng ng ra ca b m ha m chp. Kch c ca bng ny l 2(K-1) x 2K. Mng ny cng cn phi c khi u trc khi bt u tin trnh gii m. Mt mng lu tr trng thi hin ti v trng thi k ca b m ha m chp, vi gi tr ng vo (0 hoc 1) s cho ra trng thi k tip v trng thi hin ti. Chng ta s gi bng ny l bng ng vo. Kch thc ca bng l 2(K-1) x 2(K-1). Mng ny cng cn phi c khi u trc khi bt u tin trnh gii m. Mt mng lu tr lch s cc trng thi trc cho mi trng thi ca b m ha cho Kx5 + 1 cp k hiu knh truyn nhn c. Chng ta s gi bng ny l bng lch s trng thi. Kch thc ca mng ny l 2(K-1) x (Kx5 +1). Mng ny khng cn khi ng trc khi bt u tin trnh gii m. Mt mng lu tr thng s metric tch ly cho mi trng thi c tnh ton s dng nguyn tc cng- so snh- la chn. Mng ny s c gi l mng thng s metric tch ly. Kch thc ca mng ny l 2(K-1) x 2. Mng ny khng cn khi ng trc khi bt u tin trnh gii m. Mt mng dng lu tr danh sch cc trng thi c quyt nh trong sut qu trnh truy hi. N c gi l mng chui trng thi. Kch thc ca mng ny l (Kx5 + 1). Mng ny khng cn khi ng trc khi bt u tin trnh gii m.Gi chng ta hy ni v tc ca nhng b m ha chp m c th c gii m bi cc b gii m Viterbi. trn chng ta cp n b m ha thng, l mt hng chung ca b m ha tc cao, tc ln hn t k n n. Punctured code c to ra bi d liu m ha u tin s dng mt b m ha tc 1/n nh l b m ha th d c m t trc y v sau xa b mt vi k hiu knh truyn ng ra ca b m ha. Qu trnh ny c gi l puncturing. V d, to ra m tc t m tc , th n gin l s xa k hiu knh truyn theo mu punctured sau y:

Bng 2.7: V d v punctured code

Trong , bit 1 ch ra rng mt k hiu knh truyn s c truyn, v bit 0 ch ra rng mt k hiu knh truyn s c xa. xem lm th no m vic ny c th to ra b m tc . Hy ngh l mi ct ca bng trn tng ng vi mt bit ng vo n b m ha v mi mt bit 1 trong bng tng ng vi mt k hiu knh ng ra. C 3 ct trong bng v 4 bit 1. Thm ch bn c th to ra b m tc 2/3 s dng mt b m ha vi mu puncturing sau:

vi 2 ct v 3 bit 1. gii m mt punctured code, bit 1 phi thay th nhng k hiu rng cho nhng k hiu b xa ng vo ca b gii m Viterbi. K hiu rng c th l k hiu c lng t n mc 1 yu v mc 0 yu hoc hn na c th l mt k hiu c c bit, m khi c x l bng mch ACS trong b gii m, kt qu l khng thay i thng s metric tch ly t trng thi trc.D nhin, n khng phi bng 2. V d, mt m tc 1/3 v K=3 (7,7,5) c th c m ha s dng b m ha nh bn di:

Hnh 2.22: B m tc 1/3 v K= (7,7,5)

B m ha ny c 3 b cng modulo, v vy vi mi mt bit ng vo, c th to ra 3 ng ra k hiu knh truyn. D nhin, vi mu puncturing ph hp, bn c th to ra nhng m tc cao hn s dng b m ha ny.2.6 Gii m quyt nh cng v gii m quyt nh mmGii m quyt nh mm v quyt nh cng da vo loi lng t ha c s dng cc bit nhn c. Gii m quyt nh cng s dng loi lng t ha 1 bit trn cc gi tr knh nhn c. Gii m quyt nh mm s dng loi lng t ha nhiu bit trn cc gi tr knh nhn c. i vi gii m quyt nh mm l tng (lng t ha khng xc nh), cc gi tr knh nhn c c s dng trc tip trong b gii m ha knh. Hnh 2.23 biu din gii m quyt nh cng v quyt nh mm.

Hnh 2.23: Gii m quyt nh cng v mm