CHNG 3:M LDPCCui th k 20 v u th k 21, hng lot cc loi m c pht minh. Trong phi k ti mt s loi m in hnh nh : Reed-Solomon, Turbo, m xon, m TCMChnh v vy c ngi ni rng y l th k ca l thuyt m ha. Cng trong thi gian , mt loi m c tn l m kim tra chnlmt thpLDPC(LowDensityParityCodes) rai. Khi Mackay chng minh n c kh nng tim cn ti hn Shannon ngay lp tc no gy c s ch ti cc nh khoa hc.3.1 Khi qut v m LDPCNu nh trong khong 20 nm trc, ngi ta thng ni nhiu ti m Turbo nh l mt loi m tt nht ci tin hiu sut knh th gi y iu khng cn ng na. Mt hng pht trin mi m ra c tn l LDPC_ m kim tra chn l mt thp, ang tr thnh mt la chn tt nht thay th cho m Turbo trong tng lai khng xa. MLDPCchophpccnhthit kcthtimcncti hn Shannon. V l thuyt ngi ta chng mnh c rng : m LDPC khng c di m ln c th t c iu . Tuy nhin, t ngi bit c rng m LDPC cGallager xut t nhng nm 60 ca th k trc. Nhng vo thi , khoa hc my tnh cha c pht trin, kh nng tnh ton ca cc chic computer thi vn cn hn ch. Chnh iu ny khng th nhn thy c nhng u im vt tri ca m LDPC, v lm n ri vo qun lng. Mi ti nhng nm 90 th Mackay bng thut gii tng tch(Sum-productAgorithm)michngminhcrngccmLDPC khng u trn knh Gauss ch cch ti hn Shannon 0.054 dB.V c bn chng ta co th nh ngha m LDPC nh sau. LDPC (Low Density Parity Check Code), hay gil m kim tra chn l mt thp , k thut m ha m sa li trc FEC (Forward-Error Correction) thuc h m khi (Block Codes). LDPC c trng bi ma trn sa sai kch thc ln gm cc gi tr 0 v 1 vi mt gi tr 1 thp (low density). Theo nh inh ngha ca Gallager, th m LDPC (n,,) nh mt m khi tuyn tnh nh phn c di n c c trng bi ma trn kim tra H vi mi ct cha phn t 1 v mi hng cha phn t 1. S lng ct ca H bng di khi l n v s lng hng bng s du kim tra chn l (j=n-kn, trong k l di ca bn tin). ng thi, , phi rt nh so vi n. V t l m haR1.DiylvdcamGallagerLDPC[1]ccutrcthhin trong hnh3.1 th hin mt m LDPC (20,3,4 ) : Hnh 3.1.Ma trn kim tra chn l cho m LDPC ( 20,3,4)Do m LDPC thuc lp m khi truyn tnh do n mang y c tnh cht ca mt m khi tuyn tnh. Chui bt tin chiu di l k sau khi m ha s thu c mt t m c di tng ng l n. T l gia R=k/ns c coi l t l m. Trong mt t m LDPC bt k u c n-k bt m kim tra. Kch thc ma trn kim tra Hcng c kch thc khng ngoi l chnh l (n-k) x n. iu kin mt m LDPC c coi la tha mn cng ging nh m khi tuyn tnh C.HT=0. Trong , HT l ma trn chuyn v ca ma trn H.3.2 hnh TannerLc u, m LDPC c biu din qua ma trn chn l H. Tuy nhin, hin nay, mt trong nhng cch c coi l hiu qu nht biu in m LDPC chnh l thng qua hnh Tanner. Trc khi, tm hiu cch biu din ca m LDPC th chng ta s tm hiu qua v hnh Tanner. y l mt th hai pha, bn tri gi l nut bit cn bn phi gi l nut kim tra. i vi m khi tuyn tnh th hnh tanner t ra rt hiu qu.Tht vy, ta i xt v d 3.1. V d 3.1: Cho ma trn chn l H nh sau: 1111]1

1 0 1 1 0 01 1 0 0 0 10 1 0 1 1 00 0 1 0 1 1H (3.1)Khi , hnh s biu din nh sau:Hnh 3.2. hnh Tanner vi ma trn HTrong hnh Tanner trn chng ta c th thy rng : Cc nt bt v cc nt kim tra c b tr 2 bn i xng nhau. Nt kim tra zi ni vi nt bt cj khi v ch khi H(i,j)=1Thc cht ca vic biu din ny l biu din h phng trnh c suy ra t ma trn kim tra H. Trong , cc nt kim tra zi tng ng vi cc hng cn cc nt bit cj tng ng vi cc ct ca ma trn kim tra H. Ch rng, zi y l tng nhng m l tng modul2 ca tt c cc phn t cj khi H(I,j)=1.Xt v d 3.1 ta s c h phng trnh sau:4 2 1 1c c c z 5 3 2 2c c c z (3.2)6 5 1 3c c c z 6 4 3 4c c c z Ti y, chngtacngaramt khi nimlchuktanner nhm phc v nh gi cht lng xy dng m LDPC ti cc phn tip theo.z1z2z3z4c1c2c3c4c5c6Bit nodesCheck nodes Chu k Tanner: l mt ng khp kn lin kt t mt nt bt k i mt vng ri li quay li chnh n.Gi s nh trong v d 3.1 ta c mt trong nhng chu k Tanner chnh l c6 z4 c4 z1 c1 z3 c6. Chiu di ca chu k ny l 6 tc l s bc khp kn mt vng ca mt chu k Tanner.Ngi ta chng minh c rng nu nh mt m LDPC c chiu di chu k Tanner nh hn 4 th hiu qu ca m ny s rt thp. Do , trong khi xy dng mt m LDPC chng ta phi t bit quan tm ti iu ny. C th coi y l mt trong nhng tiu chun quan trng xy dng mt m LDPC tt.3.3 Xy dng m LDPC Khi ni v cc cch xy dng m LDPC th hin nay tn ti rt nhiu phng php. Nhng trong khun kh n ny, em xin ch trnh by mt s cch in hnh l phng php c din ca Gallager, phng php ca Mackay-Neal, v phng php ca Eleftheriou v Olcer.3.3.1 Phng php ca GallagerM LDPC c Gallager a ra trong nhng nm 60 vi cu trc cng rt n gin. l mt loi m c c trng bi 1 ma trn rt nhiu s 0 v rt t s 1. xy dng loi m ny th ng a ra cu trc ca ma trn kim tra H nh sau: 111111]1

cwHHHHH..321 (3.3)Trong , cc ma trn con Hd vi 1dwc tha mn iu kin sau: Cc s nguyn bt k v wr ln hn 1 hay ni cch khc, ma trn con Hd s c kch thcrw trong wrl s hng v trng s ct bng 1. MatrnconH1cdngvicchngth i=1,2,3, ,th hng th i bao gm tt c wrgi tr 1 trong cc ct t (i-1)wr+1 ti iwr. Cc ma trn con khc l hon v ca ma trn H1.iu ny s c r rng hn khi xt v d sau: V d 3.2: Cho t m di 12vi ma trn H c wc=3, wr=41 1 1 1 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 0 0 0 00 0 0 0 0 0 0 0 1 1 1 11 0 1 0 0 1 0 0 0 1 0 00 1 0 0 0 0 1 1 0 0 0 10 0 0 1 1 0 0 0 1 0 1 01 0 0 1 0 0 1 0 0 1 0 00 1 0 0 0 1 0 1 0 0 1 00 0 1 0 1 0 0 H 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ] (3.4)Tuy nhin, cu trc ny tn ti mt nhc im nh hng rt ln ti cht lng ca m LDPC. l n c chiu di chu k Tanner nh hn 4.3.3.2. Phng php ca MackayDa trn phng php ca Gallager, Mackay a ra mt s cc phng php nhm ci tin hn na hiu qu khi xy dng m LDPC. Cng chnh ng l ngi u tin chng minh li ch khi s dng ma trn nh phn tha. Cc phng php ca ng a ra rt nhiu, c thim qua tn mt s phng php nh sau:1. Khi to ma trn H ton 0 sau cho gi tr 1 trt ngu nhin trn cc ct2. Khi to ngu nhin ma trn H c trng lng ct bng wc.3. Khi to ma trn H ngu nhin c trng s ct l wc v trng lng hng xp x hc bng wc.4. Khi to ma trn H nh trn nhng trong H khng c 2 ct no trng nhau cc v tr ca s 1.5. Khi to ma trn H nh phng php 4 nhng chiu di chu k Tannner phi ln hn 4.6. Khi to ma trn H nh phng php 5 nhng H c dng H=[H1, H2] trong H2phi l ma trn kh nghch tc l tn ti ma trn nghch o.Phng php caMackayara cthtrnh c chukTannerngn. Nhng xt v tnh cu trc th phng php ny cha c hon ho. V mc ch mun t c l c th to ra mt b m ha t ng tc l c th lp trnh c, ng thi phc tp thp. Do phng php ny thiu tnh cu trc do kh m c th m bo phc tp ca thut ton nh mong mun. 3.3.3 Phng php ca Eleftheriou v OlcerThc cht ca cc phng php ny u tp trung i vo xy dng ma trn H. Tuy nhin, cc phng php trn vn tn ti cc nhc im hoc l phc tp thut ton cao. Tht vy, n c nh phng php ca Gallager v Mackay th n khng c tnh cu trc do d rt kh khn trong vic xy dng b m ha vi phc tp thp. Cn phng php thit k khi khng cn bng BIBD th qu phc tp a ra mt ma trn H mong mun. Do , chng ta s i xt mt thut ton xy dng ma trn H kh n gin m hiu qu. l thut ton ca Eleftheriou v Olcer [2].Thut ton tm chia lm 2 bcchnh sau: Bc 1 : Chng ta c ma trn H dng nh sau :

2 k-12 4 2(k-1)j-1 2( j-1) ( j-1)(k-1)I I I L II A A L AH I A A L AM M M O MI A A L A 1 1 1 1 1 1 1 ](3.5)Trong , I l ma trn n v kch thc pp.A l ma trn thu c khi dch vng tri hoc phi 1 bt cc hng I.Chng ta thc hin dch vng ma trnnh sau :- Hng th nht gi nguyn- Hng th 2 dch 1 ln v pha phi - Hng th 3 dch 2 ln v pha phiC lm nh vy s thu c ma trn Hs nh sau (3.6) Bc 2: Tam gic ho ma trn HsTa chia ma trn ra lm 2 ma trnHs=[H1s|H2s], trong H1s l ma trn jpjpv H1s l ma trn kch thc jp(k-j)p. Sau , chng ta thay b na di hnh tam gic ca ma trn H1s bng 0.Do , ta thu c ma trn Hsu kh n gin v c th lp trnh bng my tnh a ra mt cch d dng.

j 2 j 1 k 2s 2( j 3) 2( j 2) 2( k 3)uj 1 ( j 1)( k 1)I I I I I I0 I A A A AH 0 0 I A A A0 0 0 I A A 1 1 1 1 1 1 1 ]L LL LL LM MM O M M O ML L(3.7)Mt c im ng ch ca thut ton ny l thu c c ma trn kim tra trnh c lp vng 4 (free 4-cycles). Mt iu rt nguy him trong thnh lp m LDPC v n s l gim hiu qu ca m. Bc 3:a ra t m Khi c ma trn H hon ton c th tnh c t m C nh sau: . 0 s TuH c (3.8) [ ] 1| 2 * 0TH H C [ ] H1| H2 * 0PU 1 1 ] 1 2**0T TH P H S 1 2**T TH P H S (3.9)Nn1 2 1T TP H S H hoc ( )12 1PTTH S H,Cui cng ta thu c t m :C=[p|u] (3.10)Vi ( )12 1TTP H S H.3.4 M ha3.4.1 M ha dng ma trn sinh GCc m LDPC c nh ngha trn c s l ma trn kim tra chn l H, t ma trn H ta xy dng ma trn sinh G theo phng php kh Gaus Jordan. Phng php ny a ma trn H v dng: H=[In-k|P] (3.11)Vi P l ma trn kch thc (n-k)k v In-k l ma trn n v kch thc (n-k)(n-k). Ma trn sinh G c xc nh theo cng thc: G=[PT|Ik] (3.12) Qu trnh m ha n y ch n gin l thc hin php nhn gia ma trn hng n biu th chui tin u vo vi ma trn sinh tm c.c= uG= [u1 u2.uk][PT|Ik] (3.13)c= [c1 c2 ..cn] V d 3.3:Thc hin m LDPC vi chiu di t m bng 10, t l m bng 1/2, ma trn kim tra H c dng:

1 1 0 1 1 0 0 1 0 00 1 1 0 1 1 1 0 0 00 0 0 1 0 0 0 1 1 11 1 0 0 0 1 1 0 1 00 0 1 0 0 1 0 1 0 1H 1 1 1 1 1 1 1 ] (3.14)Trcht taamatrnH vdngmatrnbcthanghng(row-echelon)bng cc php ton trn cc phn t hng trong trng nh phn GF(2). Theo i s tuyn tnh, php ton trn cc phn t hng khng lm thay i cu trc ca m. Vi ma trn H trn ta thay th hng th 4 bng tng modul2 ca hng 1 v hng 4, hon v hng 3 v hng 5. Cui cng thay th hng 5 bng tng modul2 ca hng 5 v hng 4, kt qu ta c ma trn dng row-echelon.1 1 0 1 1 0 0 1 0 00 1 1 0 1 1 1 0 0 00 0 1 0 0 1 0 1 0 10 0 0 1 1 1 1 1 1 00 0 0 0 1 1 1 0 0 1rH 1 1 1 1 1 1 1 ] (3.15) Bc tip theo ta a ma trn v dng bc thang thu gn (reduced row echelon) bng cch thc hin phng php kh kh cc phn t pha trn ng cho. Th t thc hin nh sau: kh pha trn ng cho ct 2 bng cch thay th hng 1 bng tng modul2 ca hng 1 v hng 2, tng t cho ng cho ct 3, thay th hng 2 bng tng modul2 ca hng 2 v hng 3, vi ct 4 thay th hng 1 bng tng modul2 ca hng 1 v hng 4. Cui cng vi ct 5 ta thay th hng 1 bng tngmodul2 ca hng 1 hng 5 v hng 3, hng 2 bng tng modul2 ca hng 2 v hng 5, hng 4 bng tng modul2 ca hng 4 v hng 5. Kt qu ta c ma trn dng thu gon :

1 0 0 0 0 0 1 1 1 00 1 0 0 0 1 0 1 0 00 0 1 0 0 1 0 1 0 10 0 0 1 0 0 0 1 1 10 0 0 0 1 1 1 0 0 1rrH 1 1 1 1 1 1 1 ] (3.16)T y suy ra :G = 0 1 1 0 1 1 0 0 0 01 0 0 0 1 0 1 0 0 01 1 1 1 0 0 0 1 0 01 0 0 1 0 0 0 0 1 00 0 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 ](3.17) Ch Trongtrnghpsaukhi thchinccbccaphpkhGauss-Jordan m cho ta mt ma trn bc thang hng c mt hng ton 0 th ta c php b hng . iu ny s lm gim s hng ca ma trn sinh G t lm gim t l m.Tuy nhin vn kh khn phng php ny l ma trn G khng bo m c tnh tha nh ma trn H. Phng trnh m ha c = uGc thc hin b m ha c phc tpgn chnh xc bng n2 php tnh. i vi cc m c di t m ln, hng ngn n hng trm ngn bt th b m ha s tr nn cc k phc tp. gim bt tnh phc tp trong m ha ta c th s dng cc ma trn c dng cu trc. Tuy nhin vi nhng ma trn c tnh ngu nhin ta c th s dng phng php m ha trc tip trn ma trn H thng qua bin i H v dng ma trn tam gic di. Phng php ny c trnh by phn sau y.3.4.2 M ha LDPC dng ma trn kim tra chn l HKhc vi phng php trn l tm ma trn G t ma trn H cho trc sau thc hin m ha vi G. Mt m LDPC cng c th c m ha bng vic s dng trc tip ma trn H nh bin i v dng gn tam gic di. tng ca phng php ny l s dng ch yu cc hon v hng v ct sao cho vn gi c c im tha ca ma trn H. Trc ht ch hon v hng v ct a ma trn v dng gn nh tam gic di.Ht=1]1

E D CT B A (3.18)Vi T l ma trn tam gic di,ngha l T c cc gi tr 1 trn ng cho t tri qua phi, cc phn t trn ng cho bng 0, kch thc (m-g)(m-g). B l ma trn kch thc (m-g) g v A l ma trn kch thc (m-g) k, C c kch thc l gk v D c kch thc l gg, E c kch thc l g(m-g). Trong k l chiu di bn tin, n l di khi ca m, m l s bt kim tra m=n-k v g gi l gap, ni mt cch gn ng th g cng nh phc tp ca m ha cng thp. minh ha phng php ny ta xt v d sau:Thc hin m ha bn tin u=[1 1 0 0 1] vi chiu di t m bng 10, t l 1/2,vi H cho trc nh v d trn:1 1 0 1 1 0 0 1 0 00 1 1 0 1 1 1 0 0 00 0 0 1 0 0 0 1 1 11 1 0 0 0 1 1 0 1 00 0 1 0 0 1 0 1 0 1H 1 1 1 1 1 1 1 ](3.19) a v dng gn tam gic di ta hon v cc hng 2 v hng 3, ct 6 v ct 10, v chn gap bng 2.1 1 0 1 1 0 0 1 0 00 0 0 1 0 1 0 1 1 00 1 1 0 1 0 1 0 0 11 1 0 0 0 0 1 0 1 10 0 1 0 0 1 0 1 0 1tH 1 1 1 1 1 1 1 1 ] (3.20)Qu trnh nh dng tam gic trn, php kh Gauss Jordan c ng dng mt ln tng ng vi vic nhn ma trn 1]1

g1g mI ET0 I vi ma trn Ht ~H1]1

g1g mI ET0 IHt =1]1

0 D CT B A~ ~ (3.21) y C A ET C1~+ (3.22)

~1D ET B D +(3.23)

Trong v d ta xt 111]1

1 0 00 1 10 0 1T1 v1]1

g1g mI ET0 I=111111]1

1 0 1 0 10 1 1 1 10 0 1 0 00 0 0 1 00 0 0 0 1(3.24)Suy ra1 1 0 1 1 0 0 1 0 00 0 0 1 0 1 0 1 1 00 1 1 0 1 0 1 0 0 10 1 1 0 0 1 0 0 0 01 0 0 1 0 1 1 0 0 0H 1 1 1 1 1 1 1 1 ]%(3.25)Khi thc hin php kh Gauss Jordan xa E th ch c ~C,~Db nh hng cn cc phn khc ca ma trn kim tra chn l vn gi nguyn c tnh tha ca n.Cui cng m ha bn tin s dng ma trn ~H, t m c=[c1,c2,,cn] c chia thnh cc phn nh c=[u, p1, p2] vi u=[u1,u2,uk] l k bt thng tin,1 21 1 1 1[ , ... ]gp p p p lgbt kimtrauv1 22 2 2 2[ , ... ]m gp p p plccbt kim tra cn li. T m c=[u, p1, p2] phi tha mn phng trnh kim tra chn l ~0TH c . Do :AuT+ Bp1T+Tp2T=0 (3.26)

0 p 0 p D u CT2T1~T~ + + (3.27)V E c kh thnh ton 0, cc bt kim tra p1ch cn ph thuc vo cc bt thng tin c th tnh c lp. NuD%l ma trn kh o ta tnh c p1 theo cng thc:T~ ~1 T1u C D p(3.28)Nu ~D khng kh ngch th ta hon v cc hng ca ~H n khi c th. Khi tm c p1 ta tnh p2 theo phng trnh:12 1( )T T Tp T Au Bp +(3.29)Cc ma trn A, B v T rt tha do phc tp ca phng trnh ny rt thp, khiTl ma trn dng tam gic trn nn p2c th c tnh bng php thay th ngc li.Tr li v d trn ta c t m c=[c1,c2,c10]=[u,p1,p2] y p1=[c6,c7] vp2=[c8,c9,c10].Ccbt kimtrap1=[c6,c7], p2=[c8,c9,c10]ctnhnh sau:T~ ~1 T1u C D p= 1]1

111111]1

1]1

1]1

01100110 1 0 0 10 0 1 1 01 10 1 (3.30)p1=[1 0], theo cng thc (3.11) ta c p2=[1 0 0] nnc=[1 1 0 0 1 1 0 1 0 0].3.5 Gii m 3.5.1 Phng php tng tch (SPA) :Sau khi Mackay v Neal chng minh tnh vt tri ca m LDPC th cnhiuphngphpgiimrainhThuttontruynBeliefBPA (Belief PropagationAlgorithm),MPAThut tonchuyntin(Message Passing Algorithm) v mt s thut ton khc. Tuy nhin, mc ny chng tach ni nthut tontng-tch(SumProduct Algorithm) [3] ca Mackay v Neal.. Trc ht, chng ta a ra mt s nh ngha : Vi ={tp hp cc nt v ni tijc}jV \i= {tp hp cc nt c ni ti nt f j}\ {v nt ci}iC ={tp hp cc c ni tiiv }iC \j={tp hp cc nt c ni ti nt ci}\ {v nt cj}Mv(~i)={thng tin t mi nt v tr nt ci}Mc(~j)={ thng tin t mi nt c tr n fj}Pi=Pr(ci=1|yi)Si= s kin phng trnh bao gm ci tha mn.qij(b)=Pr(ci=b| Si,yi,Mc(~j)), vi b{0,1}.rij(b)=Pr (phng trnh kim trafj tha mn ci=b,Mv(~i)),vi b{0,1}. Bc 1: Tnh xc sut qij (xc sut bit code th i vi iu kin kim tra chn l ca n phi tha mn k vng ca nt kim tra th j):qij(0) =( ) ( )0 | , , ~i i i cPr c y S M j (3.31)= ( ) ( ) ( ) ( ) 1 | 0, , ~ /i i i i c iP Pr S c y M j Pr S = ( )' \1 ' (0)ij ij Ci jK P rj i(3.32)( )' \1 ' (1)ij ij ij Ci jq K P rj iTrong , K c chn sao cho ( ) ( ) 0 1 1. ij ijq q + Bc 2:Tnh xc sut rij T kt qu ca Gallager[1], ta c chui M nh phn c lp ai tha mn Pr(ai=1)= pi. Do , s thu c xc sut ca {ai}Mibao gm cc s 1: +Mi l2121(1-2pi). Trong kt qu trn th pi tng ng vi qij(1) (hnh 1.3), suy ra :( ) ( ) ( )' \1 10 1 2 12 2Mji i ji Vj ir q + (3.33)V( ) ( )-110ji jir r (3.34)Hnh 3.3: Minh ha bn tin i qua vng lp tnh qij(b)Suy ra: ta c xc sut hu nghim l:( ) ( ) ( ) 0 1 0 .i i i jij CiQ K p r (3.35)

( ) ( ) 1 1 .i i i jij CiQ K p r(3.36)TiKi phi chn tha mn rng Qi(0) + Qi(1) = 1.(3.37)Cui cng thu c t m nh sau :Ci= 1 nu Qi(1)>Qi(0) (3.38)Ngc li thCi=0.Tuy nhin, phng php gii m ny cho thi gian x l rt lu. Nh vy, ko theo vn i hi cu hnh thit b cao hn m bo thi gian thc. Do , trong mc th III chng ta s tip cn mt phng php gii m mi nhm gim thi gian x l ca h thng. iu ny cng ng ngha gim phc tp ca thut ton.S a dng ca cc k thut gii m LDPC cng chnh l u im ca loi m ny. Trong mt s phng php cho kt qu c xem l tt nht hin nay v kh nng sa li. Chnh v iu ny LDPC ang ni ln v c kh nng s thay th m turbo trong tng lai. Tuy nhin, khuyt im ny dn dn c khc phc bi cch thut ton gii m phc tp thp hn m trong c th ni MSA l mt v du in hnh.3.5.2 Thut ton MSAHin nay rt nhiu hng c t ra nhm gim s phc tp thut ton v Min-Sum (MSA) cng l mt trong s , MSA c hnh thnh trn c spht trinthut tonSPA(SumProduct Algorithm)- thut tonm Hnh 3.4: minh ha bn tin i qua na vng lp tnh rij(b)Mackay-Neal s dng chng minh LDPC code c th tim cn gii hn Shannon. Do , tin cho vic tmhiu MSA chng ta khi qut li thut ton SPA [1, 2]. SPA c th m phng 5 bc nh sau: Bc 1: For0 i to1n do set Pr( 1| )i i iP c y (3.39)viiy l k t th i m knh nhn c. Sau set (0) 1ij iq P (3.40)(1)ij iq P (3.41)vi , i j tha.1ijh Bc 2: Cp nht ( )ijr b:( ) ( ) ( )' \1 10 1 2 12 2Mji i ji Vj ir q + (3.42)( ) ( )'110ji jir r (3.43) Bc 3: Cp nht ( )ijq b:( )ij' \(0) 1 ' (0)ij ij Ci jq K P rj i (3.44)ij ij '' \(1) (1)ii j ij C jq K P r (3.45)ViijKl hng s. Bc 4: For i=0 to n-1 (0) (1 ) (0)ii i i jij CQ K P r (3.46)(1) (1)ii i i jij CQ K P r(3.47) Bc 5: For i=0 to n n-1 If (1) (0)i iQ Q then 1ic else0icIf ( * 0Tc H )or (s vng lp = max) thenStopElse go to step 2. 3.5.3 Thut ton gii m MSA Nh trn nu MSA hnh thnh t thut ton SPA, c th hnh thnh lp ln MSA ta thc hin mt s ci tin cho cc bc nh sau [2]: Bc 1: Thay v tnhijq ta i tnh ijij(0)( ) log .(1)ijqL qq _ ,(3.48) Bc 2: V ( ) ( ) 11 0ij ijr r ( ) ( ) 01 1ij ijr r Thay vo phng trnh (4) ta c : ( ) ( )' \1 2 (1) 1 2 1.Mij i ji Vj ir q (3.49)Mt khc trong thc t:00 1 111tanh log 1 2 .2pp p pp 1 _+ 1 , ](3.50)Thay (12) vo (11) ta c:( ) ( )ij '' '\1 1tanh tanh ,2 2i ji Vj iL r L q _ _+ , , (3.51)suy ra:( )' ' '' '\ ' '\1 1tanh tanh ,2 2i j i j i ji Vj i i Vj iL q _ _ , , (3.52)vi:( )ij ij 'L qi j ; ( ) ( ) ij ijsign L q ; ( )'.i j ijL q Cui cng: ( )' '' '1. tanh2ji i j i ji iL r _ , = 1 1' '' '1.2tanh log log tanh2i j i ji i _ _ , , = 1 1' '' '1.2tanh log log tanh2i j i ji i _ _ , , =( )' '' \ '. .i j i ji Vj i i _ , Vi exp( ) 1( ) log tanh log2 exp( ) 1x xxx _ _ + _ , , , V d: phng trnh cho knh BI-AWGN: ( ) ( )2 2 /ij i iL q L c y Mt khc, chng ta tip n phng trnh (16) tnh ( )ijL r. Ta c:( )( ) ( ) ' ''mini j i jii _ ,( ),min'' \i ji Vj i (3.53)V( ) x tng ng vi gi tr nh nht ca' i jtrong ( )',i ji do :( )ij i'j '' \'r . minji ji V iiL (3.54) Bc 3 v 4: logarit hai v ta thu c:( ) ( ) ( )' \

ij i j ij Cj jL q L c L r + (3.55)( ) ( ) ( )

i i jij CiL Q L c L r +(3.56)Thut ton MSA c th tm tt bng cc bc sau:Bc 1: For 0i to 1n do Khi to( )ij iL q y Bc2: Cp nht( )ijL r theo (19).Bc 3: cp nht( )ijL qtheo (20).Bc 4: Cp nht ( )iL Q theo (21).Bc 5: For 0i to 1n do:If ( ) 0iL Q < then 0ic Else1icIf ( * 0Tc H )or (s vng lp = max) thenstopElse go to step 2 . Nhn xt c im thut ton MSA:1 . T (19) ta thy, s phc tp ca qu trnh tnh ton vi MSA c kh nng gim s ph thuc vo knh truyn i, iu ny t hin ch khng cn k ti thng tin v 2, y chnh l u im ni tri ca thut ton ny so viSPA. VdiviknhBI-AWGN( )2 2 /ij ijL q y cththaybng ( ) ij iL q y . 2. Mt khc, thay v vi SPA, ta phi tnh xc sut xut hin t m (Pxi) i vi xi vi vic s dng cc php nhn, trong thut ton nhaanta chi phi tnh gi tr min (x) gi tr t m nh nht bng cc php cng v so snh nn khi lng tnh ton gim i rt nhiu.T cc nhn xt trn cho thy: Khi s dng thut ton MPA c th gim khi lng tnh ton rt nhiu so vi thut ton SPA.3.6 M phng h thng thng tin s s dng m LDPC3.6.1. M hinh knhTrong muc nay chung ta sechitim hiu anh hng cua tin hiu qua knh phadinh Rayleigh. y laknh c coi lati nht trong thng tin di ng, do onocn c tim hiu mt cach su sc. Ham pdf cua noc thhin trong cng thc22 2( ) *exp( ),2p ( 0, 0) trong l tham s ca phn b Rayleigh. Gi tr trung bnh v phng sai ca bin ngu nhin c phn b Rayleigh s l:;2xm 2 222x _ ,Thng tin qua knhRayleigh fading c th hin trong qua hinh 3. Thng tin thu c se co dang sau :k k k ky x a n +Trong o yk la tin hiu thu xk la tin hiu u vao akl bin ngu nhin theo phn b Rayleigh biu din tc ng ca knhtruyn fading ln tn hiu. nk la tap m gaussaxxknxykHnh3.5: M hnh knhFadinh