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Trang 20------51K t khi ba iu khon trong phng trnh (6,45) l tt c cc N matrn N, mt b nghch o ma trn [Bie771 c th c s dng lyc mt bn cp nht quy cho R ^ N vgi tr nghch o trc,RN^(n - 1).

An- Y)v,An)v^(n)R.L,(n~ n"(6.46)-1 . , 1-i , nR~NN(n-l)yN(n)yl{n)R^N(n - 1) RNN{n-\)--

X+ u(n)WhereM(") = y^(ra)B^Ar(n-l)jN(n) (6.47)Da trn cc phng trnh quy, hn ch tia RLS dn n fol lowing phng trnh vi phn cp nht trng lng:wN(n) = wN(n-\) +kN(n)e*{n,n-l) (6.48)whereX + u(n)gii thut RLS c th c tm tt nhsau:M") = wr. ..7-, (6-49)

1. khi to2. N*N: nhn dng ma trn, v 5 l tch cc ln hng s3. 2. quy tnh nh sau4. d(n)5. w (n-\)y(n)6. (6.50)7. e(n)

8. x(n)-d(n)9. (6.51)10. k(n)11. R"\n-\)y(n)12. (6.52)13.X + yT(n)R \n-\)y(n)14. R (n)15.i [R~] (n - 1) ~k(n)yT(n)R'] (n-\)]16. (6.53)17. w(n)

18. w(n-}) +k{n)e*(n)19. (6.54)

Trong phng trnh (6,53), X l h s trng lng c th thay ihiusut ca b cn bng. Nu mt knh l thi gian bt bin, X c th c thitlp 1. Thng thng 0,8

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nh [Lin84.Cc gii thut RLSc m t trn, c gi l giithut Kalman RLS, s dng ccphp tnh s hc 2.5N2 + 4.5N mi ln lp.

6.8.4 Tm tt v gii thutC rt nhiu cc bin th ca cc thut ton LMS, RLS tn ti thch nghi

vi mt b cn bng.Bng 6.1 cho thy cc yu cu tnh ton ca cc thutton khc nhau, v lit k mt s u im vnhc im ca mi thutton.Lu rng cc thut ton RLS chi t tng tv theo di cc mntrnh din tt hn nhiu so vithut ton LMS. Tuy nhin, cc thutton RLS thng c yu cutnh ton v cu trc chng trnh phc tp..Ngoi ra, mt s thut ton RLS c xu hng khng n nh. Nhanhngang b lc (FTF) thut ton i hi stnh ton t nht trong s cc thutton RLS, v n c th s dng mt bin gii cu trnh mt nnh. Tuy nhin, k thut cu h c s dng kh tinh t cho rngri cc knh v tuyn di ng khc nhau, v FTF l khng c s dng rngri.6,9 b cn bng fractionally khong cch unhauCc b cn bng c tho lun cho n nay khai thc spacings vi tl tng trng . N cng c bit l my thu tt nht cho tnhiu thng tin lin lc b hng bi ting n Gaussian bao gm mt blc ph hp ly mu nh k vi t l biu tng ca tin nhn. Trong s cmt ca knh bin dng, cc b lc ph hp trc khi cn bng phi c kthp vi cc knh v tn hiu b li.Bng 6.1 So snh cc thut ton khc nhau cho thchngEqualization [Pro9l]Thut ton S nhn

hot ngu im Nhc im

LMSGradientDFE

2N+ 1 Tnh ton phctp thp, chng trnhngin

Hit chm, theodi km

KalmanRLS

2.5N2 +4.5.V

Hit nhanh, khnng theodi tt

Tnh ton phc tp cao

FTF 7^+ 14 Hit nhanh, theodi tt, tnhton phc tp

Lp trnh phctp, khng n nh(nhng c th s dngphng php gii cu)

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GradientLattice

13N-8 n nh, tnhton phctp,cu trclinh hot

Hiu sut khng ttnhcc RLS, lp trnhphc tp

GradientLatticeDFE

+ 33iV2-36 tnh ton vi phc

tp thp

Lp trnh phctp

FastKalmanDFE

207V+ 5 C th c s dngcho DFE,hi t nhanhchng v theo di tt

chng trnh phctp , tnh ton khngthp, khng n nh

SquareRoot RLSDFE

1.57V2 +6.5N

Better numericalproperties

tnh ton vi phc

tp caoTrong thc t, p ng knh l cha bit, v do cc blc phhp ti u nht phi c c tnh thch nghi. Mt giiphp ti u, trong cc b lc ph hp l khp vi cc xung tnhiu truyn c th dn n mt s xung cp ng k v hiusut. Ngoi ra, mt b lc ti u l rt nhy cm vi bt kli thi gian ly mu kt qu ca n [Qur77]. Mt b cnbng fractionally khong cch u nhau(FSE) c da trn vic lymu tn hiu u vo t nht l tc cng nhanh cngttNyquist Pro91].FSE b p cho s bindng knh trc khi hiu ng rng caxy ra do t l lymu biu tng. Ngoi ra, b cn bng c th b p cho bt

k s chm tr thi gian cho bt k giai on thi gianty . Trongthc t, FSE kt hp cc chc nng ca mt b lc phhp v cnbng thnh mt cu trc b lc duy nht. Kt qu mphng th hin tnh hiu qu ca FSE trn mt b cn bng tcbiu tng c a ra trong cc giyt do Qureshi v ForneyQur77, v Gitlin v Weinstein [Git81.Cc b cn bng phi tuyn da vo cc k thut MLSE xut hinphbin trong cc h thng khng dy hin i (nhng m ttrongmc 6.7.2). Ngi c quan tm c th tm thy Chng 6 [Ste94huch cho cng vic tip theo trong ny.

6,10 a dng k thuta dng l mt thng tin lin lc k thut my thu mnh m cung cpcc ci tin lin kt khng dy vi chi ph tng ithp. Khngging nhequalization, a dng khng yu cu chi ph oto tmt trnh t o to l khng cn thit bi my pht. Hn na, cmtphm vi rng ca vic trin khai a dng, nhiu m rt thc t v cungcp lin kt ci thin ng k vi chi ph gia tng t.a dng khai thc tnh cht ngu nhin ca cc i pht thanh tuyntruyn bng cch tm ng dn tn hiu c lp (hoc t nht l cao chahiu chnh) cho truyn thng.Trong hu nhtt c cc ng

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dng, quyt nh s a dng ca ngi nhn, v khng bit n cc mypht.Khi nim a dng c th c gii thch mt cch n gin Nu mt i phtthanh ng dn tri qua mt phadinh vng su, mt con ng c lp cth c mt tn hiu mnh. Bi c nhiu con ng la chn, chai SNRs tc thi trung bnh ngi nhnc th c ci thin, thng

bng 20 dB n 30 dB.Nh th hin trong chng 3 v 4, c hai loi - fading quy m nhv quy mln. fades quy m nh c c trng bi s bin ng bin su v nhanhchng xy ra nhdi chuyn in thoidi ng trn mt khong cch ch l vibc sng.Nhng fadesc gy ra bi s phn x nhiu ln tmi trngxung quanhti vng ph cn ca in thoi di ng. Fading quy mnhthng kt qu trong mt phn b x Rayleigh fading ca mnh tnhiu trn mt khong cch nh. ngn chn fades su xy ra, k thut adng knh hin vi c th khai thc cc tn hiu thay i nhanh. Vd, fading quy m nh c hin th trong hnh 3.1 cho thy rng nu haing-ten c phn cch bng mt phn nh camt mt, ngi ta c thnhn c null trong khi nhn c mttn hiu mnhm.Bng cch la chn tn hiu tt nht ti mi thi im, mt my thu cth gim thiu hiu ng fading quy m nh(iu ny c gi l ng ten adng, a dng khng gian).Fading quy m ln c gy ra bi che chn do cc bin thtrong ahnh v bn cht ca mi trng xung quanh. Trong iu kin su sc m, mnh tn hiu nhn c mt in thoi di ng c th gim xung v diy l khng gian t do. Trong Chng 3, Fading fading quy m ln c thhin l ng nhp-phn phi chun vi lch chun khong 10 dB trongmi trng th.Bng cch chn mt trm c s m khng phi lm khinhng ngi khc l, in thoi di ng c th ci thinng k tl tn hiunting n trung bnh vo ng lin ktchuyntip.iu ny c gi l v m s a dng, tin thoi di ng ang tndng li th s phn ly ln gia cc trm c sphc v.V m s a dng cng rt hu ch my thu trm gc. Bng cch sdng ng-ten trm c s tch bit trong khng gian, cc trm c s c th ci thin cc ng link ngc bng cch chncc ng-ten vi cc tn hiumnh nht tin thoi di ng.

6.10.1 ngun gc ca ci thin adng la chn

Trc khi tho lun v cc k thut a dng c s dng, n lgi tr xcnh s lng cc u im l c th t c bngcch s dng s adng Hy xem xt M c lp knh x RayleighFading c sn mt my

thu. Mi knh c gi l mt chi nhnh a dng. Hn na gisrng mi chi nhnh c SNR trung bnh c a ra biSNR = r = ~ a (6,55)Nu mi chi nhnh c mt SNR tc thi = y, sau t phngtrnh (5,154), pdf ca yyp (yt) =-e r v.> 0 (6,56trong T l SNR trung bnh ca tng chi nhnh. Xc l mt chi nhnh duynht c SNR t hn so vi s y ngngPr [YL

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By gi, xc sut m tt c cc chi nhnh M a dng c lp nhn c tnhiu ng thi t hn so vi mt s c th SNR ngng yMYI ...... YW RJ = 0-ey / T) M = PM (y) (6,58)m cng thc {6,58) l xc sut ca tt c cc chi nhnh khng tc SNR - y. Nu mt chi nhnh duy nht t c SNR> y sau xcsut m SNR> y cho mt hoc nhiu chi nhnh c cho bi

Pr [yt> y = 1-PM (y) = 1 - (1-e "" T V) M (6,59)Phng trnh (6,59) l mt biu hin cho kh nng ti a dng mtngng khi la chn s a dng c dng dng [Jak71], vc vtrong Hnh 6.11. xc nh t l tn hiu-ting n trung bnh ca tn hiu nhn c khi s adng a dng c s dng, n l thit yu u tin tm pdf ca tnhiu Fading. i vi a dng chn lc, SNR trung bnh c tm thy bi tnhton u tin phi sinh tPM (Y). Cn c theo nhng ng ny,P (T) = ^ v (T) = (l- "r" "V" r 16,60) Sau , c ngha l SNR,y, cdng c th hin nh -40-SO -20 n 0 1010 L06 (r / r). dB

Hnh 6.11 th phn b xc sut ca SNR - y ngng i vi s a dngchnlc chi nhnh M. T i din cho SNR trung bnh trn tng chi nhnh T[Jak71] IEEE.(6.61)

Y = J \ pm (Y) Y = r | AFX (l-e *) M 'e xdx0 0trong x - y / T. Phng trnh (6,61) c nh gi l mang linhnh s cithin SNR trung bnh c cung cp bi tnh a dngchn lc.M- V I r LTK(6.62)V d sau y minh ha u im l tnh a dng cung cp.V d 6,4Gi nhnh bn a dng chi nhnh s dng, ni m mi chi nhnhnhnc mt tn hiu Fading x Rayleigh c lp. Nu SNR trung bnh l 20 dB,xc nh xc sut m SNR s gim xung di 10 dB. So snh iu ny vitrng hp ca mt my thu duy nht m khng c tnh a dng.Gii php V d 6,4i vi v d ny, ngng quy nh y = 10 dB, r = 20 dB, v c bnchinhnh. Do y / T = 0,1 v s dng cng thc (6,58),P4 (10dB) = (l-tf01) 4 = 0,000082 Khi khng c s dng a dngc, cng thc (6,58) c th c nh gi bng cch sdng M = 1,

P (J0dB) = (1 -

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thin tnh a dng c th trc tip lin quan n t l li bit trungbnh cho iu ch khc nhau bng cch s dng cc nguynl c tho luntrong phn 5.11.1.La chn tnh a dng d dng thc hin bi v tt c nhngg cnthit l vic gim st trm v chuyn i mt ng-ten nhn. Tuynhin, n khng phi l mt k thut tnh a dng ti u bi v n khng s

dng tt c cc chi nhnh c th ng thi mt lc. Ti a t l kthp s nh mi chi nhnh M mt cch ng b theo tng giai on v trngs nh vy SNR c th t c ln nhtc sn b thu tt c cc thi im

6.10.2 ngun gc ca T l Kt hp Citin ti a

Trong t l ti a kt hp, cc tn hiu in p rt t mi chi nhnhadng M phi hp theo tng giai on cung cp b sung in p thngnht v ring bit trng lng cung cp SNR ti u.Nu tng chi nhnh t c th phong b tn hiu kt qu p dng i vi my d

MRM = bng G, r,. (6.63)Gi srng mi chi nhnh c nng lng ting n trung bnh N,nnglng ting n tng s NT c p dng pht hin ch n ginl tng trng s ca ting n trong tng chi nhnh. Nh vyNT = N (bng Anh (6,64)kt qu trong mt SNR p dng cho cc my d, YM, c a ra bi->

Ym = ^ (6,65)Sdng s bt bnh ng Chebychev Cou93, YM l ti a khi Gt =r / N, dnnM-i M

V vy, cc SNR ca b kt hp a dng (xem trong hnh 6,14) ch n ginl tng ca cc SNRs trong tng chi nhnh.Gi tr ca y (r ~ / 2N, trong r ang bng r {t) theo quy nh tiphngtrnh (4,67). Nh th hin trong chng 4, v bc tn hiu nhn c mt tnhiu v tuyn in thoi di ng fading c thc m hnh ha t hai clp Gaussian bin ngu nhin Tc vTQ, mi c ngha l khng c v phngsai bng mt " l,

Y> = = TRIT2C + T '(667)Do , YM l mt phn phi Chi-bnh phng ca 2Af Gaussiancc bin ngunhin vi phng sai c ~ / {2N) = V / 2, trong fc nh ngha trongphng trnh (6,55). Pdf kt qu cho YM cth c hin th cP (Ym) = "- ^ ryM> 0 (6,68)

F (M-1)!Xc sut YM l t hn mt s ngng SNR yy M K,Pr YM

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M MYm = = ^ V = MV (6,70)Cc thut ton iu khin thit lp c v giai on cho thu kt hp tl cc i l tng t nh nhng yu cu trong b cnbng v thu Rake. Hnh 6,14 v 6,16 Hnh minh ha cu trc kt hp t l cci. Ti a t l kt hp c tho lun trong mc6.10.3.3, v c th c p

dng chohu nh bt k ng dng a dng, mc d thng chi ph ln hnnhiu v phc tp hn so vi cc k thut a dng khc.

6.10.3 Khng gian thc tin a dng cnnhca dng v khng gian, cn c gi l ng ten a dng v, l mt trongnhng hnh thc ph bin nht ca s a dng c sdng trong cc hthng khng dy. Thng thng cc h thng v tuyn di ng bao gm mtng-ten trm c s v mt ng-ten ding gn mt t nng cao. S tn tica mt con ng trc tip gia my pht v nhn l khng m bo v kh

nng ca mt s scatterers trong vng ln cn ca in thoi ding cho thymt tn hiu fading Rayleigh. Tm hnh ny fJak70], Jakes suylun rng cc tn hiu nhn c tcc ng-ten khng gian tchbit trn in thoi di ng s c c bn cha c sa chaphongb cho cc phn tch ng-ten ca mt na bc sng hochn.Khi nim v a dng v khng gian ng-ten cng c s dngtrong thitk trm gc. Ti mi trang web di ng, nhiu c strm ng-ten nhn cs dng cung cp tip nhn a dng v. Tuy nhin, k tkhi scatterers quan trng ni chung l trn mt t trong vng ln cn cain thoi di ng, cc c s trm ng-ten phi c t cch nhau ngk xa nhau t cdecorrelation. Phn tch trn th t ca vi chc bcsng c yu cu ti cc trm c s. a dng v khng gian do c

thc s dng ti nh ga hoc in thoi di ng hoc c s, hoc chai. Hnh 6,12 cho thy mt s khi chung ca mtchng trnh a dngv khng gian [Cox83a].AntennamGlG2* GmVariable GainChuyn mch logic hoc Demodulatorsu raHnh 6,12

S khi tng qut v s a dng khng gian.Khng gian a dng v cc phng php tip nhn c th c phnloi thnh bn loi [Jak71]:1. la chn s a dng2. Thng tin phn hi v tnh a dng3. T l kt hp cc i4. t c a dng v Bnh ng

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6.10.3.1 la chn a dngLa chn v tnh a dng l k thut v tnh a dng n gin phntch trong phn 6.10,3. Mt s khi ca phng php ny ltng tnhth hin trong hnh 6,12, ni demodulators m c s dng cung cp cc chi nhnh v tnh a dng m c li nhunc iu chnh

cung cp SNR trung bnh i vi tng chi nhnh. Khi xut pht trongphn 6.10.1, chi nhnh thu c SNR tc thi cao nht l kt ni vi cc b giiiu ch. Cc tn hiu ng-ten c th c ly mu v tt nht gi nmt b gii iu chduy nht. Trong thc t, cc chi nhnh ln nht (5 + N) /N c s dng, v n rt kh o lng SNR. La chn v tnh a dngthct h thng khng th hot ng trn c s thc stc thi, nhng phi cthit k cc hng s thi gian ni b ca mch la chn ngn hn sovi i ng ca t l fading tn hiu.

6.10.3.2 Thng tin phn hi hoc qut adngQut a dng rt ging vi a dng la chn ngoi tr vic thay vlun lun

s dng tt nht cc tn hiu M, cc tn hiu M c quttrong mtchui c nh cho n khi mt ngi c tm thy trnmt ngng nhtrc. Tn hiu ny sau nhn c cho n khi n ri xungdi ngng v mt ln na li bt u qu trnh qut. Cc thngk fading kt qu c phn km hn so vi nhngngi thu c bngcch cc phng php khc, nhng li thvi phng php ny l rng n rtn gin thc hin - ch l mt trong nhng ngi nhn yu cu. Mt s khi ca phng php ny c th hin trong hnh 6,13.Hnh 6,13C bn di hnh thc a dng qut.

6.10.3.3 ti a Ratio Kt hp

Trong phng php ny ln u tin c xut bi Kahn[Kah54], cc tnhiu ttt c cc ngnh M trng theo in p tn hiu c nhn ca h tl nng lng ting n v sau tng hp. Hnh 6,14 cho thy mt s khi ca k thut ny. y,cc tn hiu c nhn phi c phi hp theotng giai on trc khi c tng hp2Ch. 6 Equalization. Tnh a dng, v knh m(khng ging nhla chn a dng) m thng yu cu mt my thu cnhn v mch gim dn cho mi phn t ng-ten. Ti a t l kthp sn xut mt SNR u ra bng tng hp ca SNRs cnhn, nh c giithch trong phn 6.10.2. V vy, n c li thsn xut mt u ravi mt SNR chp nhn c ngay c khikhng c tn hiu c nhn l tchp

nhn c. K thut ny chophp gim thng k ca fading ca bt k b kthp a dngtuyn tnh c bit n. DSP k thut hin i,thu k thut s hin nay hnh thc ti u ca s a dng thc t ny.* u rathch ng kim sotHnh 6,14B kt hp t l ti a.6.10.3.4 Gain bnh ng Kt hp

Trong mt s trng hp, n khng phi l thun tin cung cpkh

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nng trng thay i cn thit cho t l ti a thc s kt hp.Trong trnghp ny, trng lng chi nhnh c tt c cc thit lp s thngnht nhng cc tn hiu t mi chi nhnh c phi hp theo tng giaion cung cp c bng cch kt hp a dng. iu ny cho php nginhn khai thc cc tn hiung thi nhn c trn tng chi nhnh. Khnng sn xut mt tn hiu chp nhn c t mt s cc yu t u

vo khng th chp nhn c l vn cn gi li, v hiu sut l chnhnh km hn so vi t l ti a kt hp v tt hn s a dng la chn.6.10.4 phn cc a dng

Ti trm c s, khng gian a dng thc t l ng k t hn sovi in thoidi ng v thu hp ca cc lnh vc s c i hi phic ng-ten ln SPAC-ings [Vau90. Chi phtng i cao ca vic s dng a dng khnggian ti cc trm c s nhc nh xem xtsdng phn cc trc giao khaithc s a dng phn cc.Trong khi iu ny ch cung cp hai chi nhnh adng, n chophp cc yu t ng-ten c phi hp v tr.

Trong nhng ngy u ca i pht thanh di ng, tt c cc n v thubao c gn trong xe v s dng ng-ten roi dc. Ngy nay, tuy nhin,tn

ti hn mt na cc n v thu bao di ng.iu ny c ngha rng hu htcc thu bao khng cn sdng phn cc thng ng do nghing tay khi inthoi cm tay di ng c s dng.Hin tng ny gn y gyra quan tmn tnh a dng phn cc ti cc trm c s.o ng dn phn cc ngang v dc gia mt in thoi ding v trm cs c bo co l khng tng quan ca o din Lee v

Yeh [Lee72]. Deeorrelation cho cc tn hiu trong miphn cc c gyra bi s phn x nhiu trong cc knh giacc ng-ten trm di ng v cs. Chng 3 cho thy h s phn x cho mi phn cc khc nhau, m ktqu trong bin v giai on khc nhau i vi tng, hoc t nht l mt sngi, s phn x. Sau khi phn nh y ngu nhin, trng thi phnccca tn hiu s c c lp ca s phn cc truyn qua ng.Trong

thc t, tuy nhin, c mt s ph thuc ca s phn ccnhn c s phncc truyn.Thng t ny v anten phn cc tuyn tnh c s dng m t ngi nhiu bn trong ta nh [Haw91], [Rap92a], [Ho94.Khi ng i b tcnghn, s a dng phn cc c tm thy lm gim ng k s lylan chm tr nhiu ng i mkhng lm gim ng k nng lng nhnc.

Trong phn cc a dng c nghin cu trong qu kh, ch yu c sdng cho cc lin kt v c nh m thay i t ttrong thi gian.Line-of-linkt l vi sng, v d, thng s dngphn cc a dng h tr ngi dngng thi trn cng knh v tuyn.K t khi cc knh khng thay inhiu trong lin kt nh,c rt t kh nng can thip phn cc cho. Khi

ngi dng di ngtng ln nhanh chng, s phn cc a dng c khnng tr nn quan trng nng cao li nhun v kh nng lin kt.Mt phctho ca mt m hnh trn l thuyt cho vic tip nhn trm c sphncc a dng theo ngh ca Kozono [Koz85] c a ra di y.M hnh l a dng phn ccN c gi nh rng tn hiu c truyn t mt in thoi di ng vi sphn cc thng ng (hoc ngang). Nhn c ti cc trm c s bi mt ngten a dng phn vi 2 chi nhnh. Hnh6,15 cho thy m hnh l thuyt v hthng ta . Nh thy trong hnh, mt ng-ten phn a dng bao

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gm hai yu t ng-ten V", vV2, m lm cho mt ct gc (gc phncc) vi trc Y trm di ngnm trong s ch o ca p gc b p t hngchm tia chnhca ng ten a dng nhtrong hnh 6,15 (b).Mt s ca cc tn hiu theo chiu dc phn cc truyn c chuyn ithnh tn hiu phn cc ngang v tuyn truyn a ng.Cc tn hiu n cctrm gc c c din t nh

x = rjeosCcor + j) (6.7La)y = r2cos (n * + 2) (6.71.b)trong x v y l mc tn hiu nhn c khi p =0. N l gi rngr] v r2 c Rayleigh phn phi clp, v , v 2 phn c lp thng nht.a ^ / ^ -in thoi di ngXchm tia chnh (b) x-z my bayHnh 6,15M hnh l a dng trm c s phn cc da trn [Koz85J.Cc gi tr tn hiu nhn c ti cc yu t V, v V2 c th c vit l: t / j= (ar, costj), + r26cos 2) costor-(ar, ti li, ti lir2b (J> 2) Sinco *v2 = (-ar, cos , + r, 6cos 2) coscor - (-ar, ti li t + r26sin2) sinwf

Trng mt ti li = a p cos b = cos a.P h s tng quan c th c vit l(6.72) (6.73)- Rtan2 (q) cos2 (p) ~ V \ 2 tan2 (a) cos2 (P) + Tnir ={R \) (R2>(6.74)(6,75)ni/? = Jr2xat + r2b ~ + 2R, r2a6cos {, + 4> 2) i22 = ^ / r a2 + r22-2R, r2a & cos ( [+ 2)(6.76)(6.77) y, f l phn bit i xphn cc cho ca ng truyn giamt in v trm c s.H s tng c xc nh bi ba yu t: gc phn cc, b trgc t hngchm tia chnh ca ng ten a dng v s phn ccphn bit cho. Hs tng ni tr nn cao hn nhgc b tr[3tr nn ln hn. Ngoi ra, p nichung tr nn thp hn l gc phn cc mt tng.iu l bi v cc thnhphn phn cc ngangtr nn ln hn nh l mt gia tng.Bi cc yu t V ng-ten, v V2 l phn cc vo thng ng,mc tnhiu nhn c l thp hn nhn c mt ng-tenphn cc theo chiudc. Gi tr trung ca L mt tn hiu, lin quan n nhn bng phncc theo chiu dc c cho biL = a2 / r + b2 (6,78) Kt qu th nghim thc thc hin bng phn cc adng [Koz85] cho thy rng s phn cc a dng l mt s a dng k tipnhn kh thi.

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6.10.5 a dng tn sTn s a dng truyn thng trn nhiu hn mt tn s sng mang.L do cbn pha sau k thut ny l tn s c phn cch binhiu hn bngthng gn kt ca knh s khng nhng mt dncng [Lem91. V mt lthuyt, nu cc knh khng tng quan,xc sut ca fading ngthi s c sn phm ca xc fadingc nhn (xem phng trnh (6,58)).

a dng tn s thng c s dng trong vi ba ng ca tmnhn lin thc mt s knh trong ch multiplex phn chia tns(FDM). Do tuyn truyn tng i lu v kt qu khc x, vngsu,fading i khi xy ra. Trong thc t, 1: N chuyn mch bo vc cungcp bi ngi c cp php pht , trong mt tn s danh ngha l nhnri nhng l c sn trn mt c stand-by cung cp a dng tn chuyn icho bt k mt trong cc tu sn bay khc N (tn s) c s dngtrn cng lin kt, mi lu lng truy cp thc hin c lp. Khi s a dng lcn thit, m bo lu lng thch hp l ch chuyn sang tn s sao . Kthut ny c nhng bt li m n khng ch i hi bng thng phtngnhng cng i hi c c nhn nhiu knh c cho s a dng tns. Tuy nhin, i vi lu thng quan trng, chi ph c l hp l.6.10.6 Thi gian a dng

Thi gian a dng lin tc truyn thng tin ti spacings thi gianvt qu thigian gn kt ca cc knh, do , lp i lp li nhiutn hiu s c thc nhn c vi iu kin c lp fading, do cung cp cho s adng. Mt hin i thc hin a dng thi gian lin quan n vic sdng my thu Rake tri ph CDMA, ni m cc knh a cung cp kh nngd phng trong cc thng ip truyn.6,11 rake nhn

Trong sysiems CDMA tri ph (xem Chng 5), tc chipthng ln hnnhiu so vi bng thng ca knh fading phng.Trong khi , k thut iuch thng thng i hi mt b cn bng c th xa b s can thip lin ktgia cc k hiu lin k,CDMA m lan rng c thit k cung cp stng quan rt thp gia cc chip lin tip. Nh vy, tuyntruyn, schm tr ly lan trong knh v tuyn ch n thunl cung cp nhiu phin bnca tn hiu truyn nhn. Nu cc thnh phnny nhiu a ngc tr hon trong thi gian nhiu hn lmt thi gian con chip,chng xut hin ging nhting n cha c sacha mt my thu CDMA, v cn bng l khng cn thit.

Tuy nhin, k t khi c thng tin hu ch trong cc thnh phn nhiu ang, thu CDMA c th kt hp cc phin thi gian tr honca vic truynti tn hiu ban u ci thin tn hiu ting n t l ngi nhn. Mtmy thu Rake khng ch ny - n c gng thu thp cc phin thigian chuyn ca tn hiu ban u bng cch cung cp mt mythu tng quan ring bit cho mi tn hiu nhiu ang. Nhn Rake, th hin trong hnh 6,16, v bn cht l mt my thu adng c thit k c bit cho CDMA, a dng c cung cp bi mt thct rng cc thnh phn nhiu a ng lthc t khng tng quan vinhau tng i chm tr tuyn truyn ca h vt qu mt khong thigian chip.RIT)IF hoc tn hiu baseband CDMA vi nhiu a ng* Tng quan 1

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tng quan 2tng quan Mm '(t)Hnh 6,16

Thc hin thu Rake mt M-chi nhnh (Mt finger /-). Tngquantng pht hin thi gian chuyn phin bn ca vic truyn ti ban

u ca CDMA, v mi finger ca Rake lin quan n mt phnca tn hiu btr hon t nht mt con chip trong thi gian tcc finger khc.Mt my thu Rake s dng correlators nhiu cch ring bit phthin M nhiu a ng thnh phn mnh nht. Cc kt qu u raca mi tng quan trng s cung cp mt c lng ca tn hiutruyn hn l c cung cp bi duy nhtmt thnh phn. gii iu ch v quyt nh bit sau da trn kt qu ura trng s ca correlators M.

tng c bn ca mt my thu Rake ln u tin c xutbi Gic v Mu xanh l cy [Pri58]. Trong mi trng ngoi tri, s chm tr giacc thnh phn nhiu a ng thng ln, v nu tc chip c lachn, cc thuc tnh t tng quanca mt chui ly lan CDMA thp c thm bo rng cc thnh phn nhiu a ng s xut hin gn nhkhngtng quan vi nhau.Gi scorrelators M c s dng trong mt my thu CDMA chp M nhiua ng thnh phn mnh nht. Mt mng litrng s c s dng cung cp mt s kt hp tuyn tnh cau ra tng quan phthin bit.b tng quan c ng b cho m nhiu a ng mnhnht,. a m2 n r, sau so vi thnh phn m ,. Cc tng quan th hail ng b ha vi m2. Ntng quan mnh m vi m2 nhng c mi tngquan thp vim,.Lu rng nu ch c mt tng quan duy nht c sdngtrong u thu (xem Hnh 5,52), mt khi u ra ca tngquan duynht b hng bng s m dn, ngi nhn khng c thsa gitr. Bit quyt nh da trn s tng quan ch duy c th to ra mttl bit li ln.Trong mt my thu Rake, nu u ra t mt trong nhng tngquan b hng bng s m dn, nhng ngi khckhng th c, v cc tnhiu b li c th c gim thng quacc quy trnh trng s. Quyt nhda trn s kt hp ca s liuthng k M quyt nhring c cung cp bi Rake cung cp mt hnh thc a dng m cth khc phc fading v do nng cao nhn CDMA.Cc s liu thng k M quyt nh trng s hnh thnh mt s liuthng k quyt nh tng th nhtrong hnh 6,16. Cc kt qu ura ca correlators M c k hiu l ZL, Z2l ... ZM. Chng ltrngs,,,, ...v AMln lt. Cc h s trng s l da vo scmnh hoc cc SNR tmi u ra b tng quan. Nu nng lng

hoc SNR nh trong mt b tng quan c th, n s c ch nhmt yu t trng s nh.Cng nhtrong trng hp ca mt t l tia kt hp chng trnh a dng, tng th tn hiu Zc cho biMz = Z -z-

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ng h hin cc d liu hng thng minh, mt t (hng) ti mtthi im.S AN XEN xon c th c s dng v trca S ANXEN khi trong thi trang tng t. S AN XEN xon ph hp cho sdng vi m xon.

C mt s chm tr vn c hin lin quan vi mt S AN XENk tkhi khi tin nhn nhn khng th c gii m hontoncho n khi tt c cc bit nm n ngi nhn v de-Interleaved.Trong thc t, bi pht biu ca con ngi l chp nhnc lng nghec bit hng mt b iu bin ti mt thi imc trong bit c m ha tb m ha

I A

1 m+1

2 m+2

m 2m

nm

wm hngn ctHnh 6,17Chn S AN XEN bit ngun c c vo ct v c nh chng n-bit.

cho n khi xy c sc c ln hn 40 ms c. l v l dony mtt c cc S AN XEN d liu khng dy c s chm tr khng vtqu 40ms. TS AN XEN kch thc v chiu c lin quan chtch loi coder bi pht biu c sdng, ngun codlingtc v schm tr ti a chp nhn c.

6,13 Nguyn tc c bn ca knh m hoKnh m ha bo v d liu k ting n ting n t cc li bngcch ting n chn lc gii thiu d tha trong vic truyn d liu.ms knh c s dng pht hin li c gi l m s phthin li, trong khi ting n m s c th pht hin v sa li c gi

l m s ting n sa li.Nm 1948, Shannon ting n ting n rng bng vic m tingn thch hp ca thng tin, li gy ra bi mt knh nhiu ting ncth c gim xung bt k cp mong mun m khng phi hysinh tc chuyn giao thng tin [Sha48]. Knh ca cngthcShannon nng lc p dng vi knh AWGN v c cho bi1 +N0B

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) = Slog2 (* 4(6.81)ting n C l dung lng knh (bit mi giy), B l bngthngtruyn (Hz), P l nng lng ting n tn hiu nhnc (watt) vN0 l mt mt ting n mt cng sut (watt /

Hz). nng lngnhn ti mt b tip nhn(6.82)ni Eb l nng lng bit trung bnh, v Rb l tc truyn bit.Phngtrnh (6,81) c th c chun ha bng bng thngtruyn ting n vc cho biC / B biu th hiu qu sdng bng thng.

Mc ch c bn ca k thut pht hin li v sa li l giithiu trng thi d tha trong cc d liu ci thin hiu sut kt nikhng dy. Vic gii thiu ca cc bit d tha lm tng tc dliu c s dng trong lin kt, do ha lm tng yu ha bng

thng cho mt t l c nh d liu ngun. iu ny lmgim hiuha bng ha ca lin kt trong ha iukin SNR cao, nhngcung cp hiu sut BER tuyt ha ti ha gitr SNR thp.N cng c bit rng vic s dng tn hiu trc giao cho haxc sutca li tr ha ty nh bng cch m rng cc thit lp tnhiu, tc ha, bng cch lm cho s lng dng sng M ->GO, cungcp SNR cho mi bit vt qu gii hn Shannon caSNRb > -1,6 dB Vit79]. Trong mc gii hn, kt quca Shannonch ha rng tn ha bng thng rng c th c s dng t c ha thng tin lin lc li min ph, min l SNR tn

ti. Likim sot m ha dng sng, mt khc, c nhng yut m habng thng s ln dn tuyn tnh vi chiu di khi m. Sali m ha do cung ha li ha trong cc ng dng bng thng hnch, v cng c th cung cp s bo v lin kt trong ng dng nnglng hn ch.Mt coder knh hot ng trn tin nhn k thut s (source) d liubng cch m ha cc thng tin ngun vo mt chui m truynthng ha knh. C hai loi c bn ca m s sa li v phthin:m s khi v m s xon.6,14 Khi Mm s khi ra m s sa li (FEC) cho php mt s lng hn ch ccsai st c pht hin v sa cha m khng truyn li.m s khi cth c s dng ci thin hiu sut ca mt h thng thng tinlin lc khi cc phng tin khc ci tin (chng hn nhtng nng lng truyn hoc s dng mt b gii iu ch phc tphn) l khng thc t.Trong m s khi, bit chn l c thm vo cc lng ca ccbit thng lng lm cho codewords hoc cc khi m. Trongmt b

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m ha khi, k bit thng tin c m ha thnh lng bit m n. Tngcng c n - k bit d tha c thm vo cc bit thnglng k cho mcch pht hin v sa li [Lin83]. Cc m khic gi l mtm (n, k), v t l ca m ny c nh ngha lRe = k / n v bng tl ca thng tin c chia t l knh nguyn.

nng lc ca mt m khi sa li l mt chc nng ca khongcch m. Nhiu gia nh m s tn ti cung cp cc mc khcnhau ca bo v li [Cou93], [Hay94J, [Lin83], [Skl93], v [Vit79].

V d 6,5Interleavers v m khi thng c kt hp truyn ting nikhngdy. Hy xem xt mt interleaver vi hng m v n-bit t. Gismitca interleaver thc s l bit ngun k v (nk) bit tmtm khi. Cc interleaver / coder kt qu kt hp s ph v s bngn ca cc li knh chiu di / = mb vo m bursts ca b. chiu diDo , mt m (n, k) c th x l cc li burst chiu di b

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mt lnh vc ca cc yu t pm c gi l mttrng m rng GF (p) v c k hiu l GF (pm), trong mc mts nguyn dng.M s vi cc biu tng ttrng nhphn GF (2) hoc m rng lnh vc GF {2m) l ph bin nht c sdng trong truyn ti dliu k thut s v h thng lu

tr, k tkhi thng tin trong cc h thng ny lunlun c mha di dng nh phn trong thc t.Trong s hc nh phn, modulo-2 Ngoi ra v nhn rng c sdng. S hc ny thc s l tng ng vi s hc bnhthngngoi tr2 m c coi l bng nhau (1 + 1 =2 = 0). Lu rng k t khi 1 +1 = 0, 1 = -1, v do shc c sdng to ra cc m kim sot li, b sung tng ngvi php tr.M Reed-Solomon s dng nonbinary trng GF {2m), Nhng trngny c nhiu hn 2 yu t v phn m rng ca trng nhphn GF {2) = {0, 1}. Cc yu t b sung trong lnh vc m

rng GF{2m) khng th l 0 hoc 1 k t khi tt c cc yu t l duynht,do , mt biu tng mi c sdng i din cho cc yut khc trong lnh vc ny. Mi phn tnonzero c th c i dinbi cng sut a.Cc hot ng nhn "" cho cc lnh vc m rng phi c xcnh cc thnh phn cn li ca lnh vc ny c th c biu dinnh l chui cc ngun ca a. Cc hot ng nhn c thc s dng sn xut cc thit lp gii hn ca cc yu t Fhin th di yF = {0, l, a, a2,..... V,.....} = J0, mt , a ', a2 ......, cr \ .....} (6.87) c c cc tp hp hu hn ca cc yu t ca GF (2) tF,mtiu kin phi c i vi F n c th cha ch 2m yu t vl mt

tp ng di php nhn (tc l, php nhn ca hai yu tlnhvc c thc hin m khng li thit lp). iu kin ngtp hpcc yu t lnh vc do nhn c gi l cc a thc ti gin, v nthng c dng nh sau [Rhe89]:aU "+ 1 = 0 hoc tng ng (2 = 1 = a (6,88) S dng cc athc ti gin, bt k yu t trong c mt sc mnh ln hn 2m-2 cth c gim n mt phn tvi mt t nng lng hn2m ~ 2 nhsau:a = ct a = a (6,89)Trnh tca cc yu t F nhvy, tr thnh trnh t sau y F *, vicciu khon nonzero c ng di php nhn:

F * = {0, l, mt, a2,...... a2m ~ 2, a2m ~ ', a2m,.....} (6,90)_ M 0 1 2 2m-2 0 2,= {0, a, a, a,,...... a, a, a,.....Hy 2 k u tin ca F * v bn c cc yu t ca trng hu hnGF(2m) trong i din sc mnhGF (2m) = (0,1, a, a2,....., a2 ~ 2} = {0V,, a \ ...... a2 "" "2} (6,91)N c th c hin th mi yu t 2 ca trng huhn hn cth c biu din nh l mt a thc khc bit ca mc

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trong1 hoc t hn (s 0 yu t c i din bi cc a thc bngkhng, mt a thc khng c iu khon nonzero) [Rhe89J. Mica ccyu t nonzero ca GF (2'n), c th c k hiu l mt athc a, r, ni t nht mt trong cc m. h s l khc khng.a '= a, (x) = a [0 + al] x + al2x' +.....+ a (m (.6.92)

Ngoi ra hai yu t ca trng hu hn c nh ngha nhphp cngmodulo-2 ca mi ca cc h s a thc ca cc ngun nh,nh hnhdi y.'+ AJ = (alAi + cjAI) + (a,] + AJA) x +.....+ (aim_, + a; m _" "(6,93)Nhvy GF {2m) c th c xy dng v sdng phng trnh(6,92)v (6,93) i din cc a thc cho cc yu t 2m ca lnh vcny c th c c.6.14.1 V d v cc M s khi Hamming M sM Hamming l nhng ngi u tin v m sa li khng tmthng [Ham50]. Cc m s ny v cc bin th ca h c sdng kim sot li trong cc h thng truyn thng k thut

s. C c hai nh phn v m Hamming nonbinary. Mt m nhphn Hamming c thuc tnh m(n, k) = {2m-\, 2m-\-m) (6,94) k l s bit thng tin c sdng tothnh mt t m bit, v m l bt k s nguyn dng. S lngcc biu tng tnh chn l l n - k = m. Hadamard M sM Hadamard thu c bng cch chn nhcodewords cc hngca mt ma trn Hadamard. Mt ma trn Hadamard l mt matrnN x N l, O mi hng khc nhau t bt k hng khc trong chnhxcN / 2 a im. Mt hng cha tt c cc s vi phn cnli ccha N. / 2 s khng v N / 2. Khong cch ti thiu i vi ccm s ny c N / 2.

i vi N = 2, ma trn Hadamard lA =0 0 (6,95;Ngoi trng hp c bit xem xt trn khi N = 2m (m l mt snguyn dng), Hadamard m c di khi khc u cth,nhng cc m ny khng tuyn tnh.

Golay M sGolay m tuyn tnh nh phn (23,12) m vi mt khong cch tithiu 7 v mt kh nng sa li v 3 bit [Gol49]. y c mtcbit, mt trong mt m loi ny l khng tm thng v d ca mt

m hon ho. (Hamming m s m lp i lp li mt s cnghonho.) T m Mi nm trong khong cch 3 ca t m bt k,do lmcho ti a kh nng c th gii m. theo chu k M sM theo chu k l mt tp hp con ca lp m s tuyn tnh c ctnh theo chu k nh c tho lun trc y. Nhl mt ktqu cac tnh ny, nhng m ny c mt s lng ng k cccu trc cth c khai thc.Mt m tun hon c th c to ra bng cch s dng mt my pht

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in a thc g {p) bng (n - k). a thc pht in v mt mvng(n, k) l mt yu t v p "+ I v c hnh thc chungg (p) = p ~ k + gn_k_ 1 + ..... + S, P + 1 (6,96)Mt thng ip a thc x (p) cng c th c nh ngha lp (x) = xk, p * ~ 1 + ..... + Xxp + xn (6,97)

{xk_ ],..... * 0) i din cho cc bit thng tin k. Kt qu t m c {p)cth c vit nhc (p) = x (p) g (p) (6,98) c {p) l mt a thc ca mc t hn n.M ha cho mt m s theo chu k thng c thc hin bimt sthay i ng k thng tin phn hi tuyn tnh da trn my phtin hoc a thc chn l.BCH M sBCH chu k m l mt trong cc m khi quan trng nht k tkhichng tn ti cho mt lot cc t l, t c nhng li ch quantrng m ha, v c th c thc hin ngay c tc cao[Bos60]. Chiu di khi ca m c n = 2m - 1 cho m> 3, v s

lng li m h c th sa c bao bc bi t

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kch c lnh vc, do , p (X) cn c xem xt mt slng c bit n.a thc s khai ca CDPDPCDPD (X) = \ + X + xB (6,100)Trong bn cc k hiu yu t lnh vc, thit lp cc a thcnguynthy p (a) = 0. iu ny mang li kt qu sau, ng li cc thit

lp ca cc yu t lnh vc:a6 a 1 = 0 (6,101) Bng 6.2 cho thy cc bn thch hp ca 6biutng cht cc yu t lnh vc. Nhng yu t ny c tora bngcch bt u vi cc yu t 1 (a) v nhn c c nhng yu t lnhvc tip theo. Bt k yu t, trong c mt thut ngs mangli mt thut nga6 phn t tip theo, nhng a6 lkhng ctrong GF (64). Nguyn tc a thc nguyn thy c s dng chuyn i a6 mt 1. Cng lu rng a62 a = a63 = a o= 1. Ktqu ny n gin l rt quan trng khi thc hin php nhnlnhvc hu hn trong phn mm. Php nhn c th thc hinnhanhchng v hiu qu bng cch s dng modulo 2m - 1 b sung quyn

hn ca cc ton hng nguyn t. i vi mReed-Solomon 63,47 sdng trong cc h thng CDPD, php nhn ca hai yu t lnhvc tng ng thm cc quyn hn cahai ton hng modulo-63 in sc mnh ca sn phm.Ngoi ra trong GF (2 ) tng ng vi modulo-2 b sung thmcc hs a thc ca cc i din ca cc yu t. K t khi cc h s l mttrong hai l * s hoc 0 ca (v lnh vcny l phn mrng ca cc GF lnh vc nh phn {2)), iu ny chn gin l ththc hin vi cc bit-khn ngoan c quyn-OR ca cc i dinbiu tng 6 bit ca cc ton hng nguyn t . Mts v d b sung lnh vc hu hn trong GF (64) c hin th di y.

A27 ot5 = (001.110) 2XOR (100000) 2 = (101.110) 2 = a55(6.102.a) A19 cc62 = (011.110) ^ Oi ^ 100001) 2 = (111111), =A58 (6.102.b)6.14.2.1 Reed-Solomon M hoTrong cc cuc tho lun ca mt b m ha Reed Solomon, ccathc sau y c s dng thng xuyn:d (x): nguyn liu thng tin a thc p (x): a thc chn l c (x): tm a thc g (x): my pht in a thc q (x): thng athc r {x):cn li a thcHyd (x) = cn, xn] + cn ^ 2xn 2 +.....+ c2t + ix2 '*] + c2tx2t (6,103)l a thc thng tin vip (x) = c0 + c, x +.....+ c2t-\ x (6,104)l a thc chn l {ci l tt c cc yu t ca GF (64)). RS mhapolyno mial do c th c din t nhn - Ic (x) = d (x) + p (x) = X0 '*' (6,105)t = o

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Avectorofn yu t lnh vc (c0, c,,..... l mt tm nu v ch nul mt bi s ca my pht in a thc g (x). a thc phtinnh li, sa cha, m Reed-Solomon c dngg {x) = (x + a) (x + a2 ).....( x + a2t) = J ^ g ^ x (6,1063I = o

Mt phng php ph bin m ha mt m tun hon l lyc p (x) bng cch chia d (x) g (x). iu ny mang li mtthng athc q khng lin quan {x) v mt a thc quan trng r(x) nh sau:d (x) = g {x) q {x) + r (x) (6,107) V vy, cc a thc t m c thcth hin nhc (x) = p (x) + g (x) q (x) + r (x) (6,108) Nu a thc tnh chn lc denned l tng ng vi tiu cc ca cc h s r (x), sau n sau mc (x) = g {x) q (x) (6,109) Nh vy m bo rng cc a thc tml mt bi s ca cc a thc my pht, mt b m ha Reed-Solomon c th c xy dng bng cch thc hin qu trnh phn

chia trn c c p (X) . Phng php n gin c c phncn li tqu trnh phn chia cc a thc monic g (x) l kt ni mt sthay i ng k theo g (x) nhtrong hnh 6,18.Mi "+ M i dinchoBng 6.2 i din ca cc yu t ca GF (64)Ngun a

thc6-Bit K hiu

i din idin

idin

0a" 0 0 0 0 0 0 0I-a" I 0 0 0 0 0 1a X 0 0 0 0 1 0a1 x' 0 0 0 1 0 0a' X 0 0 1 0 0 0a* X 0 1 0 0 0 0a' x' 1 0 0 0 0 0a" 0 0 0 0 1 I

a x' x' 0 0 0 1 1 0X* x' X* 0 0 1 1 0 0a* X x' r

>1 1 0 0 0

alu x' X 1 I 0 0 0 0a" X x1 I 1 0 0 0 1 1aIJ X* I 0 0 a i 1a" xJ x' 0 0 ! 0 1 0

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a." X x' 0 1 0 1 0 0a'' X" X 1 0 1 0 0 0a11 X X* 1 0 1 0 0 1 Ia" x' x' X* I 0 0 t 1 0a11 X x' x' I 0 0 1 1 1 ]a" X x' x* x' 0 1 1 1 1 0a" X' X xJ xJ 1 1 1 1 0 0a*' X' X xJ x1 1 1 I 1 0 I 1a" X' X x' . 1 1 ] 0 1 0 ]a" X X* I 1 0 1 0 0 1a" x' 1 0 1 0 0 0 1a*J x' x' t 0 0 0 1 0a" x' x' 1 0 0 0 I 1 \aJ' x1 x' x' 0 0 1 1 1 0

a

1

' X x' x' 0 1 1 1 0 0aiP X" K xJ 1 1 1 0 0 0a" X X x' 1 I 1 0 0 1 1a" x' x' 1 1 0 0 t 0 1a,J X' 1 0 0 1 0 0 ]a" X x' 0 1 0 0 1 0a" x' xl 1 0 0 1 0 0a' xJ x' 1 0 0 1 0 ] 1a" X X x' 0 1 0 1 1 0a" Jt' x' X* 1 0 1 1 0 0

o" X xJ

X 1 0 1 1 0 1 1a" x' X x1 x' 1 1 0 t 1 0a* X" K J xl x' 1 I 0 I 1 1 \a11 X x' x' I 0 1 1 1 0 1a" X X x' x' 1 1 1 0 1 0a,J *' X xl x' 1 1 1 0 1 ] 1aM X x' x* 1 1 0 1 1 0 1a41 X* x' 1 0 1 ] 0 0 1a" x' X x' 1 1 0 0 1 0a x5 x' x' 1 1 0 0 1 1 1

a" x' xJ 1 0 0 1 1 0 1a X x' x' 0 1 1 0 1 0or X X x' 1 1 0 1 0 0a'1 xJ x' X 1 1 0 1 0 ] 1a' X x' I 0 1 0 1 0 1a" X x1 x' 1 0 1 0 1 0a" X X* x' 1 0 1 0 I \ 1

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a" xJ xJ x' X 1 0 1 1 1 0a* X xJ xJ x' 1 0 1 1 1 1 1a X X x' x' x' 1 1 1 ] 1 0a" X X X* x' x' 1 1 1 1 ) 1 1a X' X X xl 1 1 1 1 I 0 ]a" x' X X" I 1 1 1 0 0 1a11 X' X 1 1 1 0 0 0 1a" K' 1 1 0 0 0 0

mt c quyn - HAY ca hai m-bit s, mi X tng trng cho mtphpnhn ca hai m-bit di GF (2m), v mi ng k m-bit c cha mt slng m-bit biu th bi bt.

Hnh 6,18Reed-Solomon m ha mch.d liuBan u, tt c cc thanh ghi c thit lp l 0, v chuyn ic thit lp v tr d liu. M cn_ biu tng, thng qua k cntuntchuyn mch v ng thi chuyn n dng u ra. Ngaynhm biu tng CB "4 i vo cc mch, chuyn i l ln n v trchnl, v m ca mng li thng tin phn hi rng khngcthng tin phn hi c cung cp thm. Lc ngay lp tc, ngk BQ thng qua b2t_l c cc biu tng chn l p0 qua p2t-\tngng trc tip n cc h s ca a thc chn l. H cthc chuyn tun tvi sn lng hon thnh qu trnh gii mReed-Solomon.6.14.2.2 Gii m Reed SolomonGi srng mt t mc (x) = UQ ^ v] x +.....+ vn_txc truyn i v kt qu knh li trong t m nhn c(6,110)n-1r (x) = r0 + RLX +.....+ rB_, x * "(6,111)E mu li (x) l s khc bit gia c (x) v r (x). Sdng phngtrnh (6,110) v (6,111),e (x) = r (x)-c (x) = en + e, x +.....+ in t,(6,112)Hy cc 2t mt phn hi chng 1

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XJ \ x \ xk, ....... trong 0

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l phn cn li thu c khi chia nhn c t m r (x) x + a.S: ~ remr (x)I 1,2 ....... (6,115)x + a ".

Vic phn chia hai a thc kt qu trong mt a thc q thng {x)v athc cn li mt remix). Mc ca rem cn li (x) phi nhhn mc chia a thc p (x). V vy, nu p {x) c bng 1 (tcl,p (x) = x + a1), rem (x) phi c bng 0. Ni cch khc, rem (x) ch n ginl mt yu t trng v c th c biu th nhrem. Do , xc nh cc hichng ny mt phn bt u vi tnh tonH * L = (6.H6)x + a x + aSp xp li nng sut phng trnh trnr (x) = q {x) - (x + a) + rem (6,117)Cho x = a1,r (a) = q (a) - (a + a) + rem (6,118) = rem = S (

V vy, vic tnh ton ca St 2t hi chng mt phn c th c n ginha tmt phn chia a thc ton din (c tnh toncng cao) ch nthun l nh gi a thc nhn c r (x)ti x - a [Rhe89]:St - r (a '), i = 1,2 ......, 2 * (6,119)Ngoi rar (x) = r0 + r, x +.....+ rn_] xn'i (6,120)V vy r (a) c dngr (a) = r0 + ria '+ r2a2' +.....+ rll_1atn "I) (6,121)Vic nh gi r (a1) c th c thc hin rt hiu qu trong phnmm bng cch b trchc nng n c dng nhsaur (a-) = {.....^ r "V + r" _2ja '+ r "_3jai +..... Ja''+ r0 (6,122)tnh ton a thc Li Locator

Cc qu trnh gii m Reed-Solomon ch n gin l btk thchin gii quyt cc phng trnh(6.114.a) (6.114.c). Nhngphng trnh 2t l i xngchc nng trong p, p2,....., (3 ^, c gi l chc nng i xngscmnh tng hp By gi chng taxc nh cc a thcc (x) = (1 + P, x) (1 + p2x ).....( 1 + Pftx) = o0 + a, x +.....+ akxk(6,123)Ngun gc ca (x) l p ~ ", p2,....., p ^, m ngc ca s li vtr p;Nh vy, (x) c gi l cc a thc locator li bi v n gintipcha cc a im chnh xc ca mi sai st trong r (x) Lu

rng(x) l mt a thc khng r c h s cng phi c xcnh trong qu trnh gii m Reed-Solomon.H s ca (x) v p s li v tr c lin quansau phng trnha0 = 1 (6.124.a)a = P + p2 +.....+ p * (6.124.b)a2 = piP2 + p2P3 +.....+ PFT, PFC (6,124-c)o * = P, P2P3 ..... P * (6.124.d)

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S lng khng r ai v pj c th c lin quan n mt phncbit n hi chng Sj thit lp cc phng trnh sau yc gil Identities Newton.Sx + b = 0 (6.125.a)S2 + o-JS, + 2a2 = 0 (6.125.b)

S3 + a2S2 + Phn mm OJS, + 3c3 = 0 (6,125-c)Sk + o, SA_, +.....+ Ok-l \ + kck = 0 (6.125.d)Phng php ph bin nht xc nh (x) l thut tonBerlekamp-Massey [Lin83].6,15 chp MM chp c bn khc nhau t cc m khi trong cc chuithng tin khngc phn nhm thnh cc khi ring bit v m ha [Vit793. Thay vo , mtchui lin tc ca cc bit thng tinc nh x vo mt chui lin tc ca ccbit u ra b m ha.Lp bn ny l cu trc cao cho php mt phngphp giim khc nhau ng k t cc m khi c s dng.N c th c lp lun rng m ha xon c th t c mtc m

ha ln hn c th t c bng cch s dng mt khi m ha cng vi sphc tp.M xon c to ra bng cch i qua cc trnh t thng tin thng qua mt ngk thay i hu hn trng thi. Ni chung, ng kthay i c cha N (k bit) giaion v chc nng my pht in mtuyn tnh i s da trn gener athc ator nh th hin trong hnh 6,19. Cc d liu u vo c chuyn vo vdc theo ccthay i ng k bit k ti mt thi im. S lng cc bit u rachomi chui k bit d liu u vo l n bit. T l m RC ~ k / n.Tham s V c gil di rng buc v cho bit s bit d liuu vo sn lng hin ti phthuc vo. N xc nh cch mnhm v phc tp m. Tip theo, l mt phctho nhng cch khc nhau ca i din cho m xon,

N giai onk bit d lium ha chuiHnh 6,19S khi ca b m ha xon.My pht in Matrix - Ma trn my pht cho mt m xon bn v hn k tkhi u vo l bn v hn chiu di. Do , iu ny c th l mt cch thuntin i din cho mt m xon.

a thc pht in - y, chng ti ch nh mt tp hp ca nvect, mt chomi ca cc b cng modulo-2 n c s dng.Mi vector kchthc 2k cho thy s kt ni ca cc b m havi b cng modulo-2. A 1 v

tr th i ca vector ch ra rng s thay i tng ng vi giai on ng k ckt ni v mt 0cho thy khng c kt ni.Bng logic - Mt bng logic c th c xy dng hin th cc kt qu ura ca b m ha xon v trng thi ca b m ha chochui u vo c mttrong s thay i ng k.S trng thi - K t khi u ra ca b m ha c xc nh bi cc uvo v trng thi hin hnh ca b m ha, mt s trng thi c th c sdng i din cho qu trnh m ha.Cc s trng thi ch n gin l mt

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th ca cc trng thic th c ca b m ha v qu trnh chuyni c th t mt trng thi khc.Cy S - Cc s cy th hin cu trc ca cc b m hatrong cc hnhthc ca mt cy vi cc ngnh, i din cc tiu bang khc nhau v cc ktqu u ra ca coder.

S li mt co quan st cht ch ca cy cho thy cu trc lp i lpli chnh n mt khi s lng cc giai on ln hn chiu di hnch. c quan st thy rng tt c cc ngnh pht ra thaint c cng mt trng thi ging ht nhau trong ngha mchng to ra ccchui u ra ging nhau. iu ny c ngha l hai nt c cng mt nhn hiu cth c sp nhp. Bng cch lmvic ny trong s cy, chng ta c th cc mt s khcc gi l mt s li mt co m l mt i din nhgn hn.6.15.1 Gii m m xonChc nng ca b gii m l c tnh cc thng tin u vo c mha bng cch s dng mt quy tc hay phng php m kt qu

trong s c th c ti thiu cc sai st. C mt s tng ngmt-mt gia cctrnh t thng tin v trnh t m. Hn na, bt kthng tin v mtcp chui m duy nht kt hp vi mt con ng thng qua cc li mtco. V vy, cng vic ca cc b gii mxon l c tnh ng dn thngqua cc li mt co ctheo sau bi b m ha.C mt s k thut gii m m xon. iu quan trng nht canhng phngphp ny l thut ton Viterbi, thc hin ti a kh nng gii m m xon. Thutton ln u tin c m t bi A.J.Viterbi [Vit67], [For78], quyt nh giim C hai cng v mm c th c thc hin cho m xon. Quytnh mm gii m cao cp bng cch khong 2-3 dB.6.15.1.1 Cc thut ton Viterbi

Cc thut ton Viterbi c th c m t nh sau:Hy nt li mt co tng ng ca trng thi S ti thi im c k hiul S - . Mi nt trong li mt co l c giao mt gi tr V (bng ) datrn mt thc o. Cc gi tr ntc tnh ton theo cch sau y.1 Set V (S0Q) - 0 v i - 1.2 Vo thi im i, tnh ton cc s liu ng dn mt phn cho tt ccc con ng vo mi nt.3 Thit lp V (SJJ) bng phn ng dn nh nht s liu vo ccnt tng ngvi trng thi Sj ti thi im i. Mi quan h c thb ph v bi cc nt trc la chn mt ng dn ngunhin. Cc ngnh nonsurviving cxa t li mt co. Theo cch ny, mt nhm cc ng dn ti thiu cto ra t SQ 0.4 Nu ti

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m. JVhen cng quyt nhgii m c thc hin, s liu c s dng l khongcchHamming, trong khi khong cch Euclide c s dng gii mquyt nh mm.6,15.1.2 khc Gii m thut ton cho Gii m tun t xon MFanoFano thut ton tm kim cho cc con ng c th xy ra nhtthng quacc li mt co bng cch kim tra mt con ng timt thigian [Fan63]. Tng thm s liu theo tng chi nhnh l t l thun vi xc sutca tn hiu nhn c cho chi nhnh , nh trong Viterbi giim, vi ngoi l m mt hng s b sung tiu cc l thm vo mi chi nhnh sliu. Gi tr ca hng s nyc la chn nh vy cc s liu cho cc ngdn chnh xcs tng trung bnh trong khi cc s liu cho bt k con ngkhng chnh xc s gim trung bnh. Bng cch so snh cc s liu ca mtng dn ng c vin vi mt ngng ngy cng tng,thut ton pht hin vloi b cc ng dn khng chnh xc.Vic thc hin t l li ca thutton Fano l so snh vi Viterbigii m. Tuy nhin, so vi Viterbi gii m, trnht gii m c mt s chm tr ln hn ng k. l li th hn Viterbi giim l n i hi phi lu tr t hn, v do m s vi di rng bucln hn c th c s dng. Thut ton StackNgc li vi cc thut ton Viterbi theo di 2 {N * ng dn v cc sliu tng ng ca h, cc giao dch thut ton chng vicc ng dn thn v cc s liu tng ng ca h. Trong mt thut ton chng, cc ngdn c th xy ra nhiu hn c sp xp theo th t theo s liu cah, vi con ng pha trn cng ca ngn xp c s liu ln nht. Timi bc ca thut ton, ch ng dn trn cng ca ngn xp c ko dithmmt chi nhnh. iu ny mang li 2 k tha v cc s liu ca h.2k cngvi cc ng dn khc sau c sp xp li theocc gi tr ca cc s liu,v tt c cc ng dn vi cc s liugim xung di chn trc mt slng t s liu ca con ng trn cng c th c loi b.Sau , qutrnh m rngng dn vi cc s liu ln nht c lp i lp li. So vi giim Viterbi, cc thut ton ngn xp yu cu tnh ton s liu thn,nhng ny tit kim tnh ton c b p n mt mc lnbi cc tnhton lin quan n vic sp xp li ngn xp sau khilp i lp li mi. Trong sosnh vi cc thut ton Fano, cc thut ton ngn xp tnh ton n gin vkhng c retracing trn cng mt con ng, nhng mt khc, cc thutton ngn xp yu culu tr nhiu hn so vi thut ton Fano.Phn hi gii m , b gii m lm cho mt quyt nh kh khn trn cc bit thng jon datrn cc s liu tnh ton t on j on / + m, m lmt s nguyndng chn trc. Do , quyt nh xem liu bitthng l 1 hoc 0 ph thucvo cho d con ti thiu cc khong cch Hamming bt u bng on j v kt thc giai on; + m cha mt 0 hoc 1 trong ngnh pht rat on j. Mt khi quyt nh c thc hin, ch l mt phn ca cy m btngun t cc bit chn on j, c lu gi v phn cn li l b i. y l tnhnng phn hi ca b gii m. Bc tip theo l m rng mt phn cacy c sng st n on / + 1 + m v xem xtcc ng t on j + ti j + 1 +m trong quyt nh cc bit on / + 1. Thtc ny c lp mi giai on. Cc m tham ch n gin l s lng cc trongcy b gii m xem xt trin vng trc khi a ra mt quyt nh kh

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khn. Thay vo tnh ton cc s liu, cc b gii m thng tin phn hi cth c thc hin bng tnh ton cc hi chng ttrnh t nhn v s dng mtbng tm kim phng php sa li. i vi mt s m xon, cc b giim thng tin phn hi c th c n gin ha mt b gii m logic a shoc mt b gii m ngng.

6,16 m tngu im ca cc m kim sot li, cho d h c m chn hocm xon, l hcung cp c m ha cho lin kt thng tin lin lc. c m ha m baonhiu tin nhn gii m thc hin tt hnso vi nguyn hiu sut li bit ca victruyn ti c m ha. M ha c nhng g cho php mt t lli knh 10 h tr tc d liu gii m c I0 hoc tt hn .Mi m kim sot li c m ha c bit. Ty thucvomnh m ticular, thc cc b gii m, v xc sut BER knh, Pc.N cth c hin th rng mt xp x tt cho xc sut li gii mtin nhn,PB l ni t biu s li c th c sa cha trong mtm khi (n, k). Nh ,vi mt knh BER c bit n, l c xc nh t thng bo li gii

m mt cch d dng . c m hacc binphp lng thm SNR s c yu cu cung cp cnghiu sut BER cho tnhiu thng uncoded.6,17 li mt co Coded ModulationTrellis m ha iu ch (TCM) l mt k thut m kt hp c hai m ha v iuch t c nhng li ch quan trng m ha m khng nh hngn hiu qu bng thng [Ung87]. TCM n s dng d phng iuch nonbinary kt hp vi mt b m ha hu hn trng thi quyt nh vic lachn tn hiu iu ch to ra cc trnh tn hiu c m ha. TCM sdng m rng b tn hiu d tr, m ha v thit k m ha v chc lpbn tn hiuphi hp ti a ha trc khong cch min ph (khong cch

ti thiu Euclide) gia cc tn hiu c m ha.Trong nhn, cc tn hiu cgii m bi mt b gii m trnh tkh nng phn mm quyt nh tia. Tng m ha ln l 6 dB c th thu c m khng c bt k m bng ,gim t l thng tin hiu qu.V d 6,6s ngang nhauIS-54 USDC tiu chun xc nh vic s dng cc b cn bngphn hi quytnh(DFE).

a dng1. AMPS h thng lm cho vic s dng a dng la khng gian.

2. Cc tiu chun PACS quy nh c th vic s dng a dngng-ten cho cctrm c s v n v di ng.knh m ho1. Cc tiu chun IS-95 nh xut lm cho vic s dng mt t l1 / 3, hnch chiu di L = 9 xon m vi khi an xen. Interleavers dng l 32* 18 khi interleaver.2. H thng AMPS s dng mt m BCH (40,28) cho knh iu khin pha trcv mt m s (48,30) BCH cho knh iu khinngc .

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Cn bng, a dng, v m ha knh, nh c tho lun, c cng mt mctiu nng tin cy v cht lng ca dch v thng tin lin lc. Mi kthut c li th v bt li ring. nh i c xem xt l nhng phc tp /cng sut / chi ph so hiu nng h thng. Mi k thut c kh nng cithin hiu nng h thng mt cch ng k .

6,18 vn 6,1 S dng k hiu ging nhau c m t trong Phn 6.3, ngoi tr by gihy dk = ^ wnkynk, v xc minh rng MSE l ging ht nhau cho u vo nhiun = (]b lc tuyn tnh th hin trong hnh P6.1 (cu trc ny c s dng t l tia kt a dng, thu Rake, v anten thch nghi).Hnh P6.1: Nhiu u vo thch nghi tuyn tnh b kt hp.6,2 Xem xt cc b cn bng hai vi thch nghi th hin trong hnhP6.2.(a) Tm mt biu cho MSE trong iu khon ca u; 0, ii, i> v N.(b) Nu N> 2, tm MSE ti thiu(c) Nu WQ, = 0, u'i =- 2 v A = 4 sanipk-s/cycle, nhng g l MSE td> i cc

thng s trong

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6,6 S dng my m phng thc giai on 2 (j tap 'cn bngLMS datrn mch hin th trong hnh 6.2. Gi s mi yu t chm10 micro giy ca schm tr, v truyn tn hiu baseband xftf l mt nh phn o to xung hnh xenkI v s 0, ni m mi xungc thi 10 micro giy . Gis rng x> tt i thng mt knh phn tn thi gian trc c p

dng equalizerNu knh l mt 2-rayvn phng phm knh,. vi xung bin bng nhau ti t0 v r = 15micro giy, s dng phng trnh (6,35) i6.37 'xc minh rngXIT ban u) c c ti to bng cch cn bng. S dng ccxung vi c sin nng ln rolloff vi mt - 1.(a) Cung cp d liu ha minh ha hi t cn bng.

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na, cho Piy) biu th mt chc nng m t BER cho mt iu ch c bitkhi SNR = y. Sau:y = Pr \ P (y)> x \ = Pr \ y

Khi 4 a dng chi nhnh c s dng, xc nh cc EblN0 trung bnh cnthit cho BPSK, duy tr mt 10 "BER trong mt knh fading Rayleigh.