ĐỒ ÁN TỐT NGHIỆP _Nghiên cứu mã Turbo trong hệ thống CDMA

7/30/2019 N TT NGHIP _Nghin cu m Turbo trong h thng CDMA

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MC LCMC LC

CHNG 1 : TNG QUAN V CDMA.................................................................11.1 GII THIU CHNG..................................................................................11.2 TNG QUAN .............................................................................................11.3 TH TC THU/PHT TN HIU.................................................................21.4 CC C TNH CA CDMA.......................................................................2

1.4.1 Tnh a dng ca phn tp......................................................................21.4.2 iu khin cng sut CDMA...................................................................31.4.3 Cng sut pht thp..................................................................................51.4.4 B m-gii m thoi v tc s liu bin i.........................................51.4.5 Bo mt cuc gi......................................................................................61.4.6 Chuyn giao mm(Soft Handoff)..............................................................71.4.7 Dung lng...............................................................................................71.4.8 Tch tn hiu thoi....................................................................................81.4.9 Ti s dng tn s v vng ph sng ..............................................81.4.10 Gi tr Eb/E0 thp v chng li ........................................................81.4.11 Dung lng mm .......................................................................9

1.5 KT LUN CHNG...................................................................................9CHNG 2 : KHI NIM M TURBO..............................................................10

2.1 GII THIU CHNG...............................................................................102.2 S KT NI M V RA I CA M TURBO(TURBO CODE)..........102.3 B M HO TCH CHP H THNG QUY (RSC) .........................11

2.3.1 M chp tuyn tnh.................................................................................12

2.3.2 M tch chp h thng quy.................................................................132.3.3 Cc b m ho tch chp quy v khng quy..................................142.3.4 Kt thc Trellis.......................................................................................15

2.4 KT LUN CHNG ............................................................................16CHNG 3 : M TURBO KT NI SONG SONG.............................................17

3.1 GII THIU CHNG................................................................................173.2 B M HO .......................................................................................... 173.3 K THUT XO (PUNCTURE).................................................................193.4 B CHN (INTERLEAVER).......................................................................20

3.4.1 B chn ma trn .....................................................................................21

3.4.2 B chn gi ngu nhin...........................................................................213.4.3 B chn dch vng..................................................................................223.4.4 B chn chn-l(Odd-Even) ...................................................................223.4.5 B chn Smile.........................................................................................233.4.6 B chn khung........................................................................................243.4.7 B chn ti u.........................................................................................243.4.8 B chn ng dng.................................................................................253.4.9 B chn S ...............................................................................................25

3.5 B GII M.................................................................................................263.5.1 Khi nim v cc thut ton gii m ......................................................26

3.5.2 Tng quan v cc thut ton gii m ......................................................273.5.3 Thut ton Log-MAP ............................................................................29


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3.5.4 Thut ton SOVA...................................................................................303.5.4.1 tin cy ca b gii m SOVA tng qut....................................303.5.4.2 B gii m thnh phn SOVA..........................................................343.5.4.3 S khi ca b gii m SOVA....................................................34

3.6 CI TIN CHT LNG PCCC QUA THIT K B CHN..................373.6.1 Thit k b chn mi..............................................................................393.6.2 Cc phng php ti u ho cu trc b chn........................................42

3.7 S KHC NHAU GIA M CHP V M PCCC..................................423.8 SO SNH CHT LNG CC H THNG M HO.............................................................................................................................423.9 KT LUN CHNG.................................................................................43

CHNG 4 :NG DNG CA M TURBO.......................................................444.1 GII THIU CHNG................................................................................444.2 CC NG DNG TRUYN THNG A PHNG TIN(MMC) .........44

4.2.1 Hn ch khi ng dng TC vo MCC.......................................................444.2.1.1 Bng thng gii hn.........................................................................444.2.1.2 Khi lng d liu ln..................................................................... 444.2.1.3 Tnh thi gian thc...........................................................................444.2.1.4 Cc c tnh ca knh truyn ..........................................................44

4.2.2 Cc xut khi ng dng TC vo MCC.................................................454.2.2.1 Kch thc khung ln......................................................................454.2.2.2 Ci tin qu trnh gii m ................................................................45

4.3 CC NG DNG TRUYN THNG KHNG DY ...............................464.3.1 Cc hn ch khi ng dng TC trong truyn thng khng dy.................46

4.3.1.1 Knh truyn .....................................................................................464.3.1.2 Hn ch v thi gian........................................................................474.3.1.3 Kch thc khung nh......................................................................474.3.1.4 Bng thng gii hn.........................................................................47

4.3.2 Ci tinvic thc hin gii m PCCC bng cch tng h s Scalling vkhong cch t do theo chun CDMA2000.....................................................47

4.3.2.1 B m ho PCCC theo chun CDMA2000......................................474.3.2.2 Phn b trng s 2,3 m PCCC trong CDMA2000.......................504.3.2.3 H s Scalling................................................................................51

4.4 KT LUN CHNG.................................................................................52CHNG 5 CHNG TRNH M PHNG V KT QU...............................54

5.1 GII THIU CHNG................................................................................545.2 CHNG TRNH M PHNG...................................................................54

5.2.1 Cu trc chng trnh.............................................................................545.2.2 Chng trnh chnh.................................................................................54

5.3 KT QU M PHNG................................................................................565.4 KT LUN CHNG.................................................................................62


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DANH MC CC HNH V

Hnh 1.1: Ph trong qu trnh thu v pht CDMA.....................................................1Hnh 1.2: Cc qu trnh phn tp trong CDMA.........................................................3Hnh 1.3: iu khin cng sut trong CDMA...........................................................5Hnh 1.4: Chuyn giao mm......................................................................................7Hnh 2.5: M kt ni ni tip................................................................................... 10Hnh 2.6: M kt ni song song............................................................................... 11Hnh 2.7: Thanh ghi dch cho s m ha .......................................................12Hnh 2.8: Cc v d v m chp...............................................................................13Hnh 2.9: B m ho tch chp c r=1/2 ; K=3........................................................13Hnh 2.10: B m ho RSC ca hnh 2.5.................................................................14

Hnh 2.11: B m ho tch chp khng quy r = 1/2 va K = 3vi chui ng vo v ng ra.................................................................................14Hnh 2.12: B m ho tch chp quy c r = 1/2 v K = 3 ca hnh 2.6 cng vichui ng vo v ra .................................................................................................15Hnh 2.13: Cch thc kt thc trellis b m RSC.................................................15Hnh 3.14: B m ho PCCC tng qut...................................................................18Hnh 3.15: M PCCC tc 1/3 gm 2 b m ho chp h thng quy...............18Hnh 3.16: S chi tit m ho PCCC tc 1/3..................................................19Hnh 3.17: B chn lm tng trng s m ca b m ho RSC2 khi so snh vi bm ho RSC1...........................................................................................................20

Hnh 3.18: B chn ma trn.....................................................................................21Hnh 3.19: B chn gi ngu nhin vi di chui ng vo L= 8.........................21Hnh 3.20: B chn dich vng vi L=8, a=3, s=0....................................................22Hnh 3.21: M t b chn Smile..............................................................................24Hnh 3.22: B chn bn ngu nhin vi L = 16 v S = 2.........................................26Hnh 3.23: Cc thut ton gii m da trn Trellis..................................................27Hnh 3.24: B gii m lp Log-MAP.......................................................................29Hnh 3.25: Cc ng survivor v ng cnh tranh c on tin cy.........30Hnh 3.26: V d trnh by vic gn tin cy bng cch s dng cc gi tr metrictrc tip....................................................................................................................32

Hnh 3.27: Tin trnh cp nht cho thi im t 2 (MEMlow = 2).........................33Hnh 3.28: B gii m thnh phn SOVA .....................................................34Hnh 3.29: S khi b gii m SOVA................................................................35Hnh 3.30: B gii m SOVA lp............................................................................36Hnh 3.31: Qu trnh to thng tin extrinsic............................................................40Hnh 3.32: Cu trc b gii m lp vi cc trng s................................................41Hnh 3.33: So snh h thng m ho.......................................................................43

Hnh 4.34: S gii m lp.............................................................................46Hnh 4.35: B gii m Pipeline..........................................................................46

Hnh 4.36: S b m ho PCCC theo chun CDMA2000..................................48

Hnh 4.37: Th tc tnh a chi ng ra ca b chn...........................49Hnh 4.38: Khong cch t do hiu dng................................................................52


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Hnh 5.39: Chng trnh m phng chnh...............................................................55Hnh 5.40: Kt qu ln 1..........................................................................................59Hnh 5.41: Kt qu ln 2..........................................................................................59Hnh 5.42: Kt qu ln 3..........................................................................................60Hnh 5.43: Kt qu ln 4..........................................................................................60Hnh 5.44: Kt qu ln 5..........................................................................................61Hnh 5.45: Kt qu ln 6..........................................................................................61Hnh 5.46: Kt qu ln 7..........................................................................................62


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Chng 1: Tng quan v CDMA

CHNG 1 : TNG QUAN V CDMA

1.1 GII THIU CHNG

L thuyt v CDMA c xy dng t nhng nm 1950 v c p dng

trong thng tin qun s t nhng nm 1960. Cng vi s pht trin ca cng ngh

bn dn v l thuyt thng tin trong nhng nm 1980, CDMA c thng mi

ho t phng php thu GPS v Ommi-TRACS, phng php ny cng c

xut trong h thng t ong ca Qualcomm - M vo nm 1990.

Trong chng ny ta cp n nhng vn sau:

Tng quan v CDMA. Th tc thu pht tn hiu trong CDMA.

Cc c im ca CDMA.

1.2 TNG QUAN

Hnh 1.1:Ph trong qu trnh thu v pht CDMA

T/L: l thi hn ct.

1

f0 1/T

T

Ph tin

T/L

Fc

Fc+T/L

Ph tn hiu

pht

T/L

fc

fc+T/L

Ph tn hiuthu c

f0 1/T

T

Ph tin tc

Tri ph

Nn ph

My pht dng m PN tri ph


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Trong thng tin CDMA th nhiu ngi s dng chung thi gian v tn s, m

PN (tp m gi ngu nhin) vi s tng quan cho thp c n nh cho mi

ngi s dng. Ngi s dng truyn tn hiu nh tri ph tn hiu truyn c s

dng m PN n nh. u thu to ra mt dy gi ngu nhin nh u pht vkhi phc li tn hiu d nh nh vic tri ph ngc cc tn hiu ng b thu

c.

1.3 TH TC THU/PHT TN HIU

Tn hiu s liu thoi (9,6 Kb/s) pha pht c m ho, lp, chn v

c nhn vi sng mang f0 v m PN tc 1,2288 Mb/s (9,6 Kb/s x

128).

Tn hiu c iu ch i qua mt b lc bng thng c rng bng

1,25 MHZ sau pht x qua anten.

u thu, sng mang v m PN ca tn hiu thu c t anten c a

n b tng quan qua b lc bng thng rng bng 1,25 MHz v s

liu thoi mong mun c tch ra ti to li s liu thoi nh s

dng b tch chn v gii m.

1.4 CC C TNH CA CDMA1.4.1 Tnh a dng ca phn tp

Trong h thng iu ch bng hp nh iu ch FM analog s dng trong

h thng in thoi t ong th h u tin th tnh a ng to nn nhiu fading

nghim trng. Tnh nghim trng ca vn fading a ng c gim i trong

iu ch CDMA bng rng v cc tn hiu qua cc ng khc nhau c thu nhn

mt cch c lp.

Nhng hin tng fading xy ra mt cch lin tc trong h thng ny do

fading a ng khng th loi tr hon ton c v vi cc hin tng fading a

ng xy ra lin tc th b gii iu ch khng th x l tn hiu thu mt cch

c lp c.

Phn tp l mt hnh thc tt lm gim fading, c 3 loi phn tp l theo

thi gian, theo tn s v theo khong cch. Phn tp theo thi gian t c nh s

dng vic chn v m sa sai. H thng CDMA bng rng ng dng phn tp theo

tn s nh vic m rng kh nng bo hiu trong mt bng tn rng v fading lin

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hp vi tn s thng c nh hng n bng tn bo hiu (200 - 300) KHz. Phn

tp theo khong cch hay theo ng truyn c th t c theo 3 phng php

sau:

* Thit lp nhiu ng bo hiu (chuyn vng mm) kt ni my di ng ngthi vi 2 hoc nhiu BS.

* S dng mi trng a ng qua chc nng tri ph ging nh b thu qut thu

nhn v t hp cc tn hiu pht vi cc tn hiu pht khc tr thi gian.

* t nhiu anten ti BS.

Cc loi phn tp nng cao hot ng ca h thng CDMA c ch ra trn

hnh 1.2 v c tm tt nh sau:

+Phn tp theo thi gian : Chn m, tch li v m sa sai.

+Phn tp theo tn s:Tn hiu bng rng 1,25 MHz.

+Phn tp theo khong cch (theo ng truyn) :Hai cp anten thu

ca BS, b thu a ng v kt ni vi nhiu BS (chuyn vng mm).

Hnh 1.2:Cc qu trnh phn tp trong CDMA

1.4.2 iu khin cng sut CDMA

H thng CDMA cung cp chc nng iu khin cng sut 2 chiu (t BS

n my di ng v ngc li) cung cp mt h thng c dung lng lu lng

ln, cht lng dch v cuc gi cao v cc li ch khc. Mc ch ca iu khin

cng sut pht ca my di ng l iu khin cng sut pht ca my di ng sao

cho tn hiu pht ca tt c cc my di ng trong mt vng phc v c th c

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thu vi nhy trung bnh ti b thu ca BS. Khi cng sut pht ca tt c cc my

di ng trong vng phc v c iu khin nh vy th tng cng sut thu c ti

b thu ca BS tr thnh cng sut thu trung bnh ca nhiu my di ng.

B thu CDMA ca BS chuyn tn hiu CDMA thu c t my di ngtng ng thnh thng tin s bng hp. Trong trng hp ny th tn hiu ca cc

my di ng khc cn li ch nh l tn hiu tp m ca bng rng .Th tc thu hp

bng c gi l li s l nhm nng cao t s tn hiu/ giao thoa (db) t gi tr

m ln n mt mc ln cho php hot ng c vi li bit chp nhn c.

Mt mong mun l ti u cc li ch ca h thng CDMA bng cch tng s

lng cc cuc gi ng thi trong mt bng tn cho trc. Dung lng h thng l

ti a khi tn hiu truyn ca my di ng c thu bi BS c t s tn hiu/giao

thoa mc yu cu ti thiu qua vic iu khin cng sut ca my di ng.

Hot ng ca my di ng s b gim cht lng nu tn hiu ca cc my

di ng m BS thu c l qu yu. Nu cc tn hiu ca cc my di ng kho

th hot ng ca cc my ny s c ci thin nhng giao thoa i vi cc my di

ng khc cng s dng mt knh s tng ln lm cho cht lng cuc gi ca cc

thu bao khc s b gim nu nh dung lng ti a khng gim.

Vic ng, m mch iu khin cng sut t my di ng ti BS v iu

khin cng sut t BS ti my di ng s dng trong h thng CDMA c ch trn

hnh 1.3. Mch m ng iu khin cng sut t my di ng ti BS l chc nng

hot ng c bn ca my di ng. My di ng iu chnh ngay cng sut pht

theo s bin i cng sut thu c t BS. My di ng o mc cng sut thu c

t BS v iu khin cng sut pht t l nghch vi mc cng sut o c. Mch

m ng iu khin cng sut lm cho cc tn hiu pht ca tt c cc my di ngc thu vi cng mt mc ti BS. BS cung cp chc nng mch m ng iu

khin cng sut qua vic cung cp cho cc my di ng mt hng s nh c cho

n. Hng s nh c lin quan cht ch ti yu t ti v tp m ca BS, tng ch

anten v b khuych i cng sut. Hng s ny c truyn i t BS ti my di

ng nh l mt phn ca bn tin thng bo .

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Hnh 1.3: iu khin cng sut trong CDMA

1.4.3 Cng sut pht thp

Vic gim t s Eb/No (tng ng vi t s tn hiu trn nhiu) chp nhn

c khng ch lm tng dung lng h thng m cn lm gim cng sut pht yucu khc phc tp m v giao thoa. Vic gim ny ngha l gim cng sut pht

yu cu i vi my di ng. N lm gim gi thnh v cho php hot ng trong

cc vng rng ln hn vi cng sut thp khi so vi cc h thng analog hoc

TDMA c cng sut tng t. Hn na, vic gim cng sut pht yu cu s lm

tng vng phc v v lm gim s lng BS yu cu khi so vi cc h thng khc.

Mt tin b ln hn ca vic iu khin cng sut trong h thng CDMA l

lm gim cng sut pht trung bnh. Trong a s trng hp th mi trng truyn

dn l thun li i vi CDMA. Trong cc h thng bng hp th cng sut pht cao

lun lun c yu cu khc phc fading to ra theo thi gian. Trong h thng

CDMA th cng sut trung bnh c th gim bi v cng sut yu cu ch pht i khi

c iu khin cng sut v cng sut pht ch tng khi c fading.

1.4.4 B m-gii m thoi v tc s liu bin i

B m - gii m thoi ca h thng CDMA c thit k vi cc tc bin

i 8 Kb/s. Dch v thoi 2 chiu ca tc s liu bin i cung cp thng tin

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thoi c s dng thut ton m - gii m thoi tc s liu bin i ng gia BS

v my di ng. B m - gii m thoi pha pht ly mu tn hiu thoi to ra cc

gi tn hiu thoi c m ho dng truyn ti b m - gii m thoi pha thu. B

m - gii m thoi pha thu s gii m cc gi tn hiu thoi thu c thnh cc mutn hiu thoi.

Hai b m - gii m thoi thng tin vi nhau 4 nc tc truyn dn l

9600 b/s, 4800 b/s, 2400 b/s, 1200 b/s, cc tc ny c chn theo iu kin hot

ng v theo bn tin hoc s liu.

B m - gii m thoi bin i s dng ngng tng thnh chn tc

s liu. Ngng c iu khin theo cng ca tp m nn v tc s liu s

ch chuyn i thnh tc cao khi c tn hiu thoi vo. Do , tp m nn b trit

i to ra s truyn dn thoi cht lng cao trong mi trng tp m.

1.4.5 Bo mt cuc gi

H thng CDMA cung cp chc nng bo mt cuc gi mc cao v v c

bn l to ra xuyn m, vic s dng my thu tm kim v s dng bt hp php

knh RF l kh khn i vi h thng t ong s CDMA bi v tn hiu CDMA

c scrambling (trn). V c bn th cng ngh CDMA cung cp kh nng bo

mt cuc gi v cc kh nng bo v khc, tiu chun xut gm kh nng xc

nhn v bo mt cuc gi c nh r trong EIA/TIA/IS-54-B. C th m ho

knh thoi s mt cch d dng nh s dng DES hoc cc cng ngh m tiu

chun khc.

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1.4.6 Chuyn giao mm(Soft Handoff)

Hnh 1.4: Chuyn giao mm

Chuyn giao mm l chuyn giao trong trm di ng bt u thng tin vi

mt trm gc mi m vn cha ct thng tin vi trm gc c .Chuyn giao mm ch

c th c thc hin khi c trm gc c ln trm gc mi u lm vic cng mt

tn s .MS thng tin vi hai on ca hai nh khc nhau (chuyn giao hai ng)

hoc vi ba on ca ba nh khc nhau (chuyn giao ba ng ) .BS iu khintrc tip qu trnh x l cuc gi trong qu trnh chuyn giao c gi l BS s

cp .BS s cp c th khi u bn tin iu khin ng xung .Cc BS khc khng

iu khin x l cuc gi l cc BS th cp .Chuyn giao mm kt thc khi hoc

BS s cp hoc BS th cp b loi b .Nu BS s cp b loi b th BS th cp tr

thnh BS s cp cho cuc gi ny . Chuyn giao ba ng c th kt thc bng cch

loi b mt trong s cc BS v tr thnh chuyn giao hai ng.

1.4.7 Dung lng

Vic ti s dng tn s trong h thng t bo to ra mt mc giao thoa

nht nh m rng dung lng h thng mt cch c iu khin. Do CDMA c

c tnh gt giao thoa mt cch c bn nn c th thc hin iu khin giao thoa c

hiu qu hn trong h thng FDMA v TDMA .Trong h thng CDMA mt bng

tn rng c s dng chung bi tt c cc BS.

Cc thng s xc nh dung lng ca h thng CDMA gm li x l, chu

k cng sut thoi, hiu qu ti s dng tn s v lng bp sng ca anten BS .Khi

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c nhiu knh thoi c cung cp trong h thng CDMA trong cng mt t l cuc

goi b chn v hiu qu trung k cng tng ln th cng nhiu dch v thu bao c

cung cp trn mt knh.

1.4.8 Tch tn hiu thoiTrong thng tin 2 chiu song cng tng qut th t s chim dng ti ca tn

hiu thoi khng ln hn khong 35%. Trong trng hp khng c tn hiu thoi

trong h thng TDMA v FDMA th kh p dng yu t tch cc thoi v tr thi

gian nh v li knh tip theo l qu di. Nhng do tc truyn dn s liu gim

nu khng c tn hiu thoi trong h thng CDMA nn giao thoa ngi s dng

khc gim mt cch ng k. Dung lng h thng CDMA tng khong 2 ln v

suy gim truyn dn trung bnh ca my di ng gim khong 1/2 v dung lng

c xc nh theo mc giao thoa nhng ngi s dng khc.

1.4.9 Ti s dng tn s v vng ph sng

Tt c cc BS u ti s dng knh bng rng trong h thng CDMA.Giao

thoa tng tn hiu my di ng thu c t BS ,giao thoa to ra trong cc my di

ng ca cng mt BS v giao thoa to ra trong cc my di ng ca BS bn

cnh,giao thoa tng t tt c cc my di ng bn cnh bng 1/2 ca giao thoa tng

t cc my di ng khc trong cng BS.Hiu qu ti s dng tn s ca cc BS

khng nh hng khong 65%, chnh l giao thoa tng t cc my di ng khc

trong cng mt BS vi giao thoa t tt c cc BS.Trong trng hp anten ca BS l

khng nh hng th giao thoa trung bnh gim xung 1/3 v mi anten kim sot

nh hn 1/3 s lng my di ng trong BS.Do dung lng cung cp bi ton

b h thng tng xp x 3 ln.

1.4.10 Gi tr Eb/E0 thp v chng liEb/E0 l t s ca nng lng trn mi bit i vi mt ph cng sut tp

m , l gi tr tiu chun so snh hiu sut ca phng php iu ch v m

ho s.M sa sai c s dng trong h thng CDMA cng vi gii iu ch hiu

sut cao,c th tng dung lng v gim cng sut yu cu i vi my pht gim

Eb/E0 .

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1.4.11Dung lng mm

H thng CDMA c mi lin quan linh hot gia s ngi s dng v loi

dch v.Trong h thng Analog v TDMA th s cuc gi c n nh i vi

ng truyn lun phin hoc s tt cuc gi xy ra trong trng hp tc nghnknh trong trng thi chuyn giao.Nhng trong h thng CDMA cuc gi c tho

mn nh tng t l li bit cho ti khi cuc gi khc hon thnh .H thng CDMA

cn s dng lp dch v cung cp dch v cht lng cao ph thuc vo gi

thnh dch v v n nh cng sut (dung lng) nhiu cho cc ngi s dng dch

v lp cao.C th cung cp th t u tin cao hn i vi dch v chuyn giao ca

ngi s dng lp dch v cao so vi ngi s dng thng thng.

1.5 KT LUN CHNG

Vi nhng u im vt tri ca CDMA nn ngy nay cng ngh ny

c s dng rng ri trn ton th gii ni chung v Vit Nam ta ni ring.

Nhng u im ca cng ngh ny c da trn nhng k thut v k thut

m ho v gii m ni ring l mt phn quan trng to nn u im ca cng ngh

ny.Ta s i su tm hiu vn m Turbo nhng chng tip theo.

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Chng 2: Khi nim m Turbo

CHNG 2 : KHI NIM M TURBO

2.1 GII THIU CHNGM Turbo l s kt ni gm hai hay nhiu b m ring bit to ra mt m

tt hn v cng ln hn.M hnh ghp ni m u tin c Forney nghin cu

to ra mt loi m c xc sut li gim theo hm m ti tc nh hn dung lng

knh trong khi phc tp gii m ch tng theo hm i s.M hnh ny bao gm

s kt ni ni tip mt b m trong v mt b m ngoi.

Chng ny trnh by:

S kt ni cc m v s ra i ca m Turbo( TC).

Gi thiu v m chp h thng quy (Recursive Systematic

Convelutional Code_RSC), l c s ca vic tao ra m TC.

2.2 S KT NI M V RA I CA M TURBO(TURBO CODE)

Forney s dng mt b m khi ngn hoc mt b m tch chp vi gii

thut gii m Viterbi xc sut ln nht lm b m trong v mt b m Reed-

Salomon di khng nh phn tc cao vi thut ton gii m sa li i s lm bm ngoi.

Mc ch lc u ch l nghin cu mt l thuyt mi nhng sau ny m hnh

ghp ni m tr thnh tiu chun cho cc ng dng cn li m ln .C hai

kiu kt ni c bn l kt ni ni tip (hnh 1.1) v kt ni song song ( hnh 1.2)

Hnh 2.5: M kt ni ni tip

B m ho 1 c gi l b m ngoi ,cn b m ho 2 l b m trong.i

vi m kt ni ni tip ,tc m ho: Rnt=k1k2/n1n2

i vi m song song , tc m ho tng : Rss= k/( n1+n2)

B m ho 1

r = k1/n1

B m ho 2

r = k2/n2Ng vo Ng ra

10


15/68


Hnh 2.6: M kt ni song song

Trn ch l cc m hnh kt ni l thuyt.Thc t cc m hnh ny cn phi s

dng thm cc b chn gia cc b m ho nhm ci tin kh nng sa sai.

Nm 1993, Claude Berrou, Alain Glavieux, Puja Thitimajshima cng vit

tc phm Near Shannon limit error correcting coding and decoding :TURBO

CODE nh du mt bc tin vt bc trong nghin cu m sa sai.Loi m m

h gii thiu thc hin trong khong 0.7dB so vi gii hn ca Shannon cho knh

AWGN.Loi m m h gii thiu c gi l m Turbo ,thc cht l s kt ni

song song cc b m tch chp c bit cng vi cc b chn .Cu hnh ny gi

l :Kt ni song song cc m tch chp ( Parallel Concatenated ConvolutionalCode_ PCCC)

Ngoi ra cng c Kt ni ni tip cc m tch chp(Serial Concatenated

Convolutional Code_SCCC) v dng Kt ni hn hp cc b m tch chp(

Hybrid Concatenated Convolutional Code_HCCC).Cc loi m ny c nhiu c

im tng t nhau v cng xut pht t m hnh ca Berrou nn gi chung l:

TURBO CODE ( TC).

2.3 B M HO TCH CHP H THNG QUY (RSC)

Trong b m TC s dng mt b m tch chp c bit : m tch chp h

thng quy ( Recursive Systematic Convolutional Code_RSC ).Tnh h thng c

ngha l u vo ca b m ho cng c ngha l mt phn ca ng ra .V th ,mt

bit trong nbit ng ra ca mt vng lp m ho n l 1 bit trong thng ip i vo

b m ho .

Ng vo Ng raB ghp

(Multiplexer)

B m ho1r = k/n

1

B m hor = k/n

2

11


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Tnh quy c ngha l c hi tip t ng ra b m ho v ng vo .Cc b

m ho tch chp truyn thng khng c hi tip nn c th c coi nh mt b lc

FIR cn cc b lc RSC nh hi tip nn c th coi nh l b lc IIR.

2.3.1 M chp tuyn tnhB m ha s dng cc thanh ghi dch a thm d vo lung d liu .

B phn c bn ca phn cng trong vic m ha ny l thanh ghi dch vi

(m+1) ngn ( stages), nh hnh sau

Hnh 2.7: Thanh ghi dch cho s m haMi mt k hiu gi trn hnh l mt s nh phn i din cho s ngn mch

hoc h mch( gi=1 l ngn mch, gi=0 l h mch). Cc bits thng tin trn thanh

ghi c kt hp bi b cng modulo 2 to nn cc bit u ra ta gi cc bit

u ra l cc bit m ha. Cng thc biu din cc bit u ra ng vi cc bit vo l:

yj =Sj-m gl . . . . Sj-1 g1 Sj g0 (mod 2)

m= Sj i gi

i=0 m t b m ha m chp ngi ta a ra cc thng s ca b m ha

nh sau : (n,k, K) , Trong

k : s u vo

n :s u ra

K:chiu di constraint lengths(s ngn ln nht trn thanh ghi)

Trong k < n ta c th thm d vo lung d liu thc hin pht hin

sai v sa sai.Hnh 2.4 bn di cho ta thy r hn v b m chp

gl

g1

g0

Sj-m Sj-1

..... Sj

Message bits xi

encoded bit yj

12


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+

+

1 0 0

(1+1+0+0)

(1+0+0)

1

0

1

1

1

1

(1+0+0)

(1+0+0+0)

000

+

+

Hnh 2.8: Cc v d v m chp

2.3.2 M tch chp h thng quy

M tch chp h thng quy c ly t b m ho tch chp thng thng

bng cch hi tip mt trong nhng ng ra m ho thnh ng vo ca n

Hnh 2.9: B m ho tch chp c r=1/2 ; K=3 Mt b m tch chp thng thng c biu din qua cc chui g 1= [1 1 1] v

g2 =[ 1 0 1] v c th c vit l G =[ g1,g2] .B m ho RSC tng ng b m ho

tch chp thng thng c biu din l G = [ 1, g2/g1 ] trong ng ra u tin

( biu din bi g1) c hi tipv ng vo , g1l ng ra h thng , g2lng ra

feedforward.Hnh 2.6 trnh by b m ho RSC

D D

+

+

x

c(2)

c(1)

13


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Hnh 2.10: B m ho RSC ca hnh 2.5

2.3.3 Cc b m ho tch chp quy v khng quyMt b m ho tch chp quy c khuynh hng cho ra cc t m c trng

s tng so vi b m ho khng quy ,ngha l b m tch chp quy cho ra t t

m c trng s thp v cng dn n vic thc hin sa sai tt hn

i vi m Turbo ,mc ch ca vic thc hin cc b m ho RSC l tn

dng bn cht quy ca cc b m ho v tn dng s kin b m ho l h thng

kim tra b m ho tch chp quy hay khng quy ,ta xt v d

sau ,b m tch chp khng quy n gin c ma trn sinh g1 = [1 1] v g2 =[1 0] ,

( hnh 2.7)

Hnh 2.11: B m ho tch chp khng quy r = 1/2 va K = 3vi chui ng vo v ng ra

Hnh 2.8 trnh by mt b m ho tch chp tng ng ca hnh 2.7 c:

G =[ 1, g2/ g1]

D

+

+

x

c2

c1

+ D

D

+

c2=[1 0 0 0]

c1=[1 1 0 0]

x=[1 0 0 0]

14


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Hnh 2.12: B m ho tch chp quy c r = 1/2 v K = 3 ca hnh 2.6 cngvi chui ng vo v ra

Hai m c cng khong cch t do ti thiu v c th m t bng cu trc

trellis. V vy cc m c cng xc sut li s kin u tin ,tuy nhin cc m ny c

cc mc li bit khc nhau do BERph thuc vo s tng ng ng vo - ra ca

b m ho. BER ca m tch chp h thng quy th thp hn BER ca m tch

chp khng quy tng ng vi cng t s tn hiu trn nhiu(SNR) thp .

2.3.4 Kt thc Trellis

i vi b m tch chp thng thng, Trellis c kt thc bng( m= k -1)

cc bit zero thm vo sau chui ng vo. Cc bit thm vo ny li b m tch chp

thng thng n trng thi tt c zero ( l trng thi kt thc trellis) .Nhng cch

ny khng th p dng cho b m ho RSC do c qu trnh hi tip .Cc bit thmvo kt thc cho b m ho RSC ph thuc vo trng thi ca b m ho v rt

kh d on .Ngay c khi tm c cc bit kt thc cho mt trong cc b m ho

thnh phn th cc b m ho thnh phn khc c th khng c li n trng thi

tt c zero vi cng cc bit kt thc do c s hin din ca b chn gia cc b m

ho thnh phn. Hnh 2.9 l kt thc trellis :

Hnh 2.13: Cch thc kt thc trellis b m RSC

m ho chui ng vo ,kho chuyn bt n v th A , kt thc trellis thkho chuyn bt n v tr B.

D+

c2=[1 1 1 1]

c1=[1 0 0 0]

x=[10 0 0]

15


20/68


2.4 KT LUN CHNG

Qua chng ny ta bit c cc khi nim v m tch chp ,m tch chp h

thng quy.T cc m ny kt ni vi nhau theo mt kiu kt ni nht nh ta

c m Turbo, kt ni y l kt ni song song cc b m RSC.Tuy nhin, hnh thnh nn m Turbo khng phi kt ni n gin nh vy m cn phi c b

chn cng nh l s dng k thut xo mang li hiu qu cao hn.Chng sau ta

s tm hiu v m ho ,gii m cng nh l cu trc ca m PCCC

16


21/68

Chng 3: M Turbo kt ni song song

CHNG 3 : M TURBO KT NI SONG SONG

(PARALLEL CONCATENATED CONVOLUTIONAL CODE)

3.1 GII THIU CHNG

M Turbo kt ni song song l s kt ni song song cc b m tch chp h

thng quy.

Chng ny i chi tit trnh by c th cc vn lin quan n m PCCC

nh:

Cu trc b m ha, tng qut v c th Cc phng php v thut ton gii m, ch yu l thut ton:Log-MAP

v SOVA.

Phn tch cht lng m PCCC, tnh u vit v cc yu t nh hng n

cht lng thc hin m PCCC.

Bin php tng cht lng bng thit k b chn

So snh gia PCCC vi cc loi m bit trc y.

3.2 B M HO

M PCCC l s kt ni song song ca 2 hay nhiu m RSC. Thng thng

ngi ta s dng ti thiu 2 b m ho tch chp .S khi m PCCC tng qut

c trnh nh hnh 3.1

Mi b m ho RSCi c gi l cc b m thnh phn (constituent

code).Cc b m thnh phn c th khc nhau ,tc m khc nhau nhng c cng

c khi bit ng vo l k ,cc chui m ho ng ra bao gm mt chui h thng(chui bit vo) . cc b m ho th hai tr i ,chui bit nhn vo m ho trc

ht phi qua mt b chn .Tt c cc chui m ho ng ra s c hp li thnh

mt chui bit duy nht n bit trc khi truyn .

17


22/68


Hnh 3.14: B m ho PCCC tng qut

Tc m ho (code rate) ca b m ho PCCC l: r = k/n

Mi bit thng tin ng vo s tr thnh mt phn ca t m ng ra (tnh h

thng ) v s c km theo bng ( 1/r - 1) bit ( gi l bit parity) sa li nu

c.Nu r cng nh tc s bit parity i km s ln v dn n kh nng sa li cao

hn rt nhiu nhng tc truyn gim i , s bit truyn nhiu hn c ngha l bngthng ln hn v tr tng ln .Theo khuyn co ca cc t chc nh chun th

gi tr r ch nn nh nht l 1/6 .

Trong qu trnh hp cc chui m ho thnh mt chui m ho duy nht ta c

th dng mt k thut kh mi m l k thut xo (puncture) .

Mt m Turbo tiu biu l loi c kt ni theo kiu PCCC. S khi c

biu dintrong hnh 3.2

Hnh 3.15: M PCCC tc 1/3 gm 2 b m ho chp h thng quy

...

...

.

.

.

c0

B m hoRSC1

B m hoRSC2

B chn 1x

c1,i

c2,i

B m hoRSCn

B chn n-1c

n+1,i

.

.

.

.

.

.

Chuyni song

song sangni tiphoc

puncturec

c3

B m hoRSC1

B m hoRSC2B chn

xc

2

c1

18


23/68


24/68


chui ny cc bit 0 ti nhng ch b xo bt .Nh vy c th lm sai lch bit

parity nn gim cht lng.

3.4 B CHN (INTERLEAVER)

i vi m Turbo , c mt hay nhiu b chn c s dng gia cc b mho thnh phn .B chn c s dng ti b m ho nhm mc ch hon v tt c

cc chui ng vo c trng s thp thnh chui ra c t m ng ra trng s cao hay

ngc li .Lun m bo vi mt chui ng vo th ng ra mt b m ho s cho t

m trng s cao cn b m ho kia s cho ra t m trng s thp lm tng

khong cch t do ti thiu.

B chn khng nhng c s dng ti b m ho m n cng vi b gii

chn (deinterleaver) c trong b gii m ng mt vai tr quan trng .Vai tr ca b

chn chnh ti b gii m mi bc l ht .Mt b chn tt s lm cho cc ng vo

ca b gii m SISO t tng quan vi nhau tc l mc hi t ca thut ton gii

m s tng ln ,ng ngha vi vic gii m chnh xc hn.

V d b chn c s dng tng trng s ca cc t m nh trong hnh 3.2

Hnh 3.17: B chn lm tng trng s m ca b m ho RSC2 khi so snh vi

b m ho RSC1

T hnh 3.4 ,i vi b m ho RSC1 th chui ng vo x cho ra chui m

tch chp quy c trng s thp c2 . trnh b m ho RSC2 cho ra chui ng ra

quy khc cng c trng s thp ,b chn hon v chui ng vo x thnh 1 chui

mi vi hi vng cho ra chui m tch chp quy c trng s cao c 3 .V vy , trng

s m ca m PCCC l va phi , n c kt hp t m trng s thp ca b m

ho 1 v trng s cao ca b m ho 2.

c2M trng s thp

M trng s cao

M h thng

B m ho RSC 1

B m ho RSC 2B chn

xc

1

c3

20


25/68


3.4.1 B chn ma trn

B chn ma trn khi c s dng thng nht trong cc h thng lin lc .

Hnh 3.18: B chn ma trn

Vi b chn nh trn , nu ta cho chui vo l ( x 1 x2. . . . . .x15), nh bng 3.1

Bng 3.1 Bng ma trn vo

Th chui ra l:x1 x6 x11 x2 x7 x12 . . . x5 x10 x15

3.4.2 B chn gi ngu nhin

Hnh 3.19: B chn gi ngu nhin vi di chui ng vo L= 8

B chn gi ngu nhin s dng tnh ngu nhin c nh v sp xp chui ng

vo theo th t hon v.Nh hnh trn b chn vit vo [ 01101001] v c ra

[ 01011001]

c ra0 1 1 00 0 1 0

1 1 0 00 0 1 1

Vit vo

x1 x6 x11x2 x7 x12x3 x8 x13x4 x9 x14x5 x10 x15

0 1 1 0 1 0 0 1

1 3 6 8 2 7 4 5

0 1 0 1 1 0 0 1

Vit vo

Hon v ngu nhin c nh

c ra

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3.4.3 B chn dch vng

Hnh 3.20: B chn dich vng vi L=8, a=3, s=0

Php hon v ca b chn dch vng l: p(i)= ( a*i+ s)mod L

Vi a


27/68


Mt b chn khi 3x3 c dng hon v chui tin tc x cho b m ho

RSC2 nh sau:

x1 x4 x7x2 x5 x8x3 x6 x9

Bng 3.3 B chn khi

Chui tin tc x c vit theo ct c ra theo hng .Chui tin c hon v

cho ra chui m ho c3 .T chui c3 ch c cc bit m ho v tr chn c lu tr

nh trong bng 3.4

x1 x4 x7 x2 x5 x8 x3 x6 x9- c34 - c32 - c38 - c26 -

Bng 3.4 Cc bit m ho l ca chui c3

i vi m ho PCCC r= 1/2, cc chui bit m ho sau phi c ghp

vi nhau nh trong bng3.5 sau y:

x1 x4 x7 x2 x5 x8 x3 x6 x9c21 c34 c27 c32 c25 c38 c23 c36 c29

Bng 3.5 Chui tin v chui m ho c ghp

Mi bit tin c bit m ho ring ca n.

3.4.5 B chn Smile

B chn chn l nh trn cho duy nht mt bit kim tra i km theo mt bit

m ho .Hn ch ca b chn ny l: sau khi m ho c hai chui bit thng tin

( chui tin tc gc v chui sau khi qua b chn ) trng thi ca c hai b m phi

ging nhau .Ta vn c th thm vo sau chui thng tin mt s bit tail bits hoc

kt thc Treliss lm cho hai b m ho u kt thc cng mt trng thi zerobng cch dng mt b chn c bit l Simile.

tng ca b chn ny xut pht t tng mt khi thng tin K bit c th

c chia thnh m+1 chui vi m l tham s nh ca b m ho .V d m=2 ta c

chui:

Chui 0 = { dk |k mod ( m+1) =0}

Chui 1 = { dk |k mod ( m+1) =1}

23


28/68


Chui 2 = { dk |k mod ( m+1) =2}

Hnh 3.21: M t b chn Smile

V d nh i vi b RSC nh trn vi mt K cho trc , trng thi cui

cng ca b m ho m t bng trng ca hai D flip-flop s l s kt hp ca cc

chui va nu trn th hin trong bng 3.6 sau :

K mod(m+1) S0K S1K0 Chui 1+chui 2 Chui 0+chui 11 Chui 2+chui 0 Chui 1+chui 22 Chui 0+chui 1 Chui 2+chui 0

Bng 3.6 Trng thi cui ca b m ho

Th t ca cc bit n l trong mi chui khng cn quan trng ,ch cn cc bit

trong cng mt chui .Mt b chn Simile phi thc hin vic hon v cc bittrong mi chui a c b m ho v cng trng thi nh khi khng s dng

b chn .

3.4.6 B chn khung

Nu b chn Simile cn s dng thm tail bit li c hai b m ho n

cng mt trng thi th b chn khung li khng cn tail bit.Mi mt b RSC do

tnh hi quy ca n c th c trng bng mt a thc sinh chu k L .Trong trng

hp ny N bit thng tin sau khi c chn s c lu hai ln trong b nh kch

thc 2N ti nhng a ch m vic c chng ra sau ny b ngn cch bng mt

khong thi gian bng vi s nguyn ln ca L .Bng cch ny, nu b m ho bt

u bng mt trng thi zero th s kt thc ti mt trng thi zero m khng cn

thm bt k mt tail bit no.

3.4.7 B chn ti u

B chn ti u l b chn cho ra cc chui m ho ng ra c trng s thp tnht .B chn ny thit k di dng v phc tp , thut ton m t nh sau:

24


29/68


Bc 1 : pht ra chn ngu nhin .

Bc 2 : pht tt c cc chui tin ng vo c th.

Bc 3: i vi tt c cc chui tin ng vo c th m ho thnh t m v xc

nh kt qu trng s ca t m tm c phn b trng s ca t m .Bc 4: xc nh trng s t m nh nht v s cc t m vi trng s .

3.4.8 B chn ng dng

Mt b chn ng dng theo nh ngha l mt b chn trung bnh ca tt c

cc b chn c th s dng. Ta xt mt chui kbit gm w bit 1 v kw bit cn li

ng nhin l bit 0. Mt b chn ng dng chiu di k l mt thit b xc sut s

nh x chui ny thnh tt c cc hon v

wk ring bit vi xc sut nh nhau l

w

k

1

.

K thut ny cho php kho st mt b m PCCC bt k nh mt b m

PCCC ch gm 2 b m thnh phn da trn phn b ng dng to bi b chn.

3.4.9 B chn S

B chn loi S (S = 1,2,3,...) l b chn bn ngu nhin c xy dng nh

sau : Chn s nguyn ngu nhin em so snh vi cc s nguyn ngu nhin chn

trc S. Nu vic chn hin ti v chn trc S khc nhau nh hn S th s

nguyn ngu nhin b loi b. Qu trnh ny c lp li cho n khi N s nguyn

phn bit c chn. Thit k b chn ny m bo trnh c cc s kin chu k

ngn. Mt s kin chu k ngn xy ra khi 2 bit gn nhau trong c trc v sau khichn.

Thut ton hon v cho b chn bn ngu nhin c m t nh sau:

Bc 1: Chn ch s ngu nhin i[0,L1].

Bc 2: Chn s nguyn dng2

LS< .

25


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31/68


3.5.2 Tng quan v cc thut ton gii m

Hnh 3.10, Trnh by cc h thut ton gii m da trn s Trellis

Hnh 3.23: Cc thut ton gii m da trn Trellis

H th nht l h cc thut ton MAP cn gi l thut ton BCJR (Bahl-

Cocke-Jelinek-Raviv, tn bn ngi tm ra thut ton ny). Thut ton ny lin

quan n cc thut ton gii m kh nng xy ra ln nht (ML) nhm lm gim tia xc sut li bit.H ny bao gm cc thut ton symbol-by-symbol MAP, l

phng php ti u tnh cc thng tin APP(a posteriori), y l thut ton dng

tch, phc tp rt cao. Trong h ny cn c hai loi thut ton lm gn ng

thut ton MAP tr thnh thut ton dng tng phc tp t hn m cht lng

gii m gn nh tng ng l Log-MAP v phin bn gn ng ca Log-MAP l

Max-log-MAP.H thut ton gii m khc l mt h thut ton da trn vic sa

i thut ton Viterbi (VA) c s dng thm metric b sung v VA truyn thngkhng tnh cc thng tin APP, metric b sung lm iu . H thut ton gii m

ny bao gm thut ton ni ting l thut ton Viterbi ng ra mm (SOVA) v thut

ton t c bit n hn l thut ton Viterbi ng ra lit k ni tip .

Tuy cng l cc thut ton ng ra mm da trn s trellis nhng khc vi

VA l mt thut ton gii m trellis ML v gim thiu xc sut li t m, thut ton

MAP li nhm ti gim ti a xc sut li bit. MAP l mt phng php ti u

c on cc trng thi v ng ra ca cc qu trnh Markov trong iu kin nhiu

Cc thut ton giim da trnTrellis

Viterbi

Max-Log-MAP

SOVA ci tin

SOVA

Log-MAP

MAP

27


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trng. Tuy nhin MAP t kh nng c ng dng thc t bi cc kh khn v s

hc lin quan n vic biu din xc sut, cc hm phi tuyn cng mt s cc php

nhn v cng khi tnh ton cc gi tr ny.Log-MAP l mt bin th ca MAP, cht

lng gn nh tng ng m khng gp tr ngi trong vic ng dng trong thct. Log-MAP c thc hin hon ton trong min logarit, nh php nhn

chuyn thnh php cng v ta c c mt hm tng i d thc hin hn.

Max-Log-MAP v SOVA l thut ton gn ti u dng gim bt phc tp

tnh ton nhng trong knh nhiu Gauss th cht lng hai loi ny cng khng cao,

c bic trong vng SNR thp. Max -Log-MAP hu nh ging vi Log-MAP ch c

duy nht mt im khc l s dng mt hm n gin hn rt nhiu. Cc nghin

cu cho thy Max-Log-MAP lm gim cht lng khong 0.5 dB so vi MAP/Log-

MAP trong knh nhiu Gauss.

Mc d thut ton MAP tt hn thut ton SOVA nhng n c cu trc phn

cng v qu trnh tnh ton gii m li phc tp hn nhiu.Do gii hn ca n v

phc v cho chng trnh m phng nn ta ch tp trung tm hiu v :Log-MAP

v SOVA.

28


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3.5.3 Thut ton Log-MAP

Hnh 3.24: B gii m lp Log-MAP

Gii thut gii m c thc hin nh sau:

1. Tch tn hiu nhn ra thnh 2 chui tng ng cho b gii m 1 v b gi

m 2 .

2. vng lp u tin ,thng tin a priori ca b gii m 1 c a v 0.Sau

khi b gii m 1 a ra c thng tin extrinsic th s c chn v a ti

b gii m 2 ng vai tr l thng tin a priori ca b gii m ny.B gii

m 2 sau khi a ra thng tin extrinsic th vng lp kt thc.Thng tin

extrinsic ca b gii m th 2 s c gii chn v a v b gii m 1 nh

l thng tin a priori .

3. Qu trnh gii m gii m c lp li nh vy cho n khi thc hin s lnlp qui nh .

Harddecision

Deinter.

Deinter.

Inter.

Inter.DEC1 DEC2( )I;c )(1

( )I;c )(2

( )I;c )(1

( )I;c )(2

)O;u(Ak1 )u( ke1)u( ke2

)O;u(Ak2

)(1 ka u )u( ke2

29


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4. Sau vng lp cui cng ,gi tr c on c c tnh bng cch gii chn

thng tin b gii m th 2 v a ra quyt nh cng.

3.5.4 Thut ton SOVA3.5.4.1 tin cy ca b gii m SOVA tng qut

tin cy trong gii m SOVA c tnh ton t biu trellis nh hnh :

Hnh 3.25: Cc ng survivor v ng cnh tranh c on tin cy

Trong Hnh 3.12 trnh by biu trellis 4 trng thi. ng lin nt ch ra

ng survivor (gi thit y l mt phn ca ng ML) v ng t nt ch ra

ng cnh tranh (xy ra ng thi) ti thi im t i vi trng thi 1. n gin

th cc ng survivor v cnh tranh cho cc nt khc khng c v ra. Nhn S 1,t

biu din trng thi 1 ti thi im t. Cng vy, cc {0,1} c vit trn mi ng

ch ra quyt nh nh phn c c on cho cc ng. Mt metric tch ly

Vs(S1,t)gn cho ng survivor i vi mi nt v metric tch ly Vc(S1,t) gn cho

ng cnh tranh i vi mi nt. Thng tin c bn cho vic gn gi tr tin cy L(t)

n ng survivor ca nt S1,t l gi tr tuyt i ca 2 metric tch ly.

L(t) = |Vs(S1,t) Vc(S1,t)| (3.1)

Gi tr ny cng ln th ng survivor cng ng tin cy. tnh ton tin

cy ny, gi thit metric tch ly ca survivor th lun lun ln hn metric tch ly

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ca cnh tranh. gim phc tp, cc gi tr tin cy ch cn c tnh cho ng

survivor ML v khng cn thit tnh cho cc ng survivor khc bi v chng s

c b qua sau ny.

minh ha khi nim tin cy, 2 v d c a ra bn di. Trong cc v dny, thut ton Viterbi chn ng survivor nh l ng c metric tch ly nh

nht. Trong v d u tin, gi thit ti nt S1,t c metric survivor tch ly l Vs(S1,t)

= 50 v metric cnh tranh tch ly l Vc(S1,t) = 100. Gi tr tin cy lin kt n vic

chn ng survivor ny l L(t) = |50 100| = 50. Trong v d th 2, gi thit

metric survivor tch ly khng i Vs(S1,t) = 50 v metric cnh tranh tch ly l

Vc(S1,t) = 75. Kt qu gi tr tin cy l L(t) = |5075| = 25. Hnh 3.13 minh ha

vn s dng tr tuyt i gia cc metric survivor v cnh tranh tch ly nh l

php o tin cy ca quyt nh.

Trong Hnh 3.13, cc ng survivor v cc ng cnh tranh ti S1,t tch ra ti

thi im t 5. Cc ng survivor v cc ng cnh tranh cho ra cc quyt nh

nh phn c on i lp ti cc thi im t, t2 v t4 trong Hnh3.13.

minh ha, chng ta gi thit cc metric tch ly ca survivor v cnh tranh ti S1,t l

bng nhau, Vs(S1,t) = Vc(S1,t) = 100. iu ny c ngha l c hai ng survivor v

cnh tranh c cng xc sut l ng ML. Hn na, chng ta gi thit l metric tch

ly survivor th tt hn metric tch ly cnh tranh ti thi im t2 v t4 nh

c trnh by trong Hnh 3.13. gim bt phc tp ca hnh v, cc ng

cnh tranh ny ti cc thi im t2 v t4 khng a ra. T gi thit ny, chng

ta thy rng gi tr tin cy gn cho ng survivor ti thi im t l L(t) = 0, iu

ny c ngha l khng c tin cy lin kt vi vic chn ng survivor. Ti ccthi im t2 v t4, cc gi tr tin cy gn cho ng survivor th ln hn 0 (L(t

2)= 25 vL(t4)=10) ngha l kt qu cc metric tch ly tt hn cho ng

survivor. Tuy nhin, ti thi im t, ng cnh tranh cng c th l ng survivor

bi v chng c cng metric. V vy, c th c cc quyt nh nh phn c c

on tri ngc nhau ti cc thi im t, t2, v t4 m khng c lm gim cc

gi tr tin cy lin kt sut dc theo ng survivor.

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Hnh 3.26: V d trnh by vic gn tin cy bng cch s dng cc gi trmetric trc tip

ci tin cc gi tr tin cy ca ng survivor, mt php tnh truy ngc

cp nht cc gi tr tin cy c gi thit. Th tc cp nht ny c tch hp vo

trong thut ton Viterbi nh sau:

i vi nt Sk,t trong biu trellis (p ng n trng thi k ti thi im t),

lu L(t) = |Vs(Sk,t)Vc(Sk,t)|.

Nu c nhiu hn mt ng cnh tranh, th sau nhiu gi tr tin cy phi

c tnh v gi tr tin cy nh nht c ly l L(t).

Khi to gi tr tin cy Sk,tbng + (tin cy nht).

So snh cc ng survivor v cnh tranh ti Sk,t v lu li cc cp nh

(MEM) trong cc quyt nh nh phn c c on ca 2 ng l khc

nhau.Cp nht cc gi tr tin cy ti cc MEM ny vi th tc nh sau:o Tm MEM dng thp nht, coi nh l MEMlow, m gi tr tin cy ca n

khng c cp nht.

o Cp nht gi tr tin cy ca MEMlow L(tMEMlow) bng cch gn gi tr

tin cy thp nht gia MEM = 0 v MEM = MEMlow.

Tip theo t v d, cc c on bit i lp gia cc ng ca bit survivor

v cnh tranh ca S1,t c nh v v lu tr nh l MEM={0, 2, 4}. Vi thng tin

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38/68


3.5.4.2 B gii m thnh phn SOVA

B gii m thnh phn SOVA c on chui tin tc qua vic s dng mt

trong 2 lung bit m ha c sinh ra bi b m ha TC. Hnh 3.15 trnh by ng

vo v ng ra ca b gii m thnh phn SOVA.

Hnh 3.28: B gii m thnh phn SOVA

B gii m thnh phn SOVA x l cc ng vo (t l log kh nng xy ra)

L(u) v Lcy trong :

+ chui a priori information ca chui tin u c k hiu li l L(u) cho ph

hp vi cc k hiu s dng trong gii m SOVA tng qut

+ Lcy l chui nhn c qua cn bng cng nh gii m Log-MAP

Chuiy c nhn qua knh truyn. Tuy nhin, chui L(u) c sinh ra v c

ly t b gii m thnh phn SOVA c trc . Nu khng c b gii m thnh

phn SOVA trc th sau khng c cc gi tr a priori. V vy, chui L(u)

c khi ng n chui tt c zero.B gii m thnh phn SOVA cho ra cc ng ra u v L(u) trong :

+u l chui tin c on.

+L(u) l chui thng tinposteriori.

3.5.4.3 S khi ca b gii m SOVA

B gii m SOVA c th c thc hin theo nhiu cch khc nhau. Nhng

c l theo hng tnh ton th d dng thc hin b gii m SOVA cho cc m c

chiu di bt buc K ln v kch c khung di bi v s cn thit cp nht tt c ccng survivor. Do th tc cp nht ch c ngha cho ng ML, nn vic thc

hin ca b gii m SOVA ch thc hin th tc cp nht i vi ng ML c

trnh by trong Hnh 3.16

Lcy

L(u)

SOVA

L(u)

u

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Hnh 3.29: S khi b gii m SOVA

B gii m SOVA ly ng vo l L(u) v Lcy, l gi tr tin cy v gi tr nhnc qua cn bng tng ng, v cho ra u v L(u), tng ng l cc quyt nh

bit c on v cc thng tin a posteriori L(u). Vic thc hin b gii m SOVA

ny bao gm 2 b gii m SOVA ring bit. B gii m SOVA u tin ch tnh cc

metric ca ng ML v khng tnh (gi li) cc gi tr tin cy. Cc thanh ghi dch

c s dng m cho cc ng vo trong khi b gii m SOVA u tin ang x

l ng ML. B gii m SOVA th hai (c thng tin ca ng ML) tnh li

ng ML v cng tnh v cp nht cc gi tr tin cy. Ta thy rng phng php

thc hin ny lm gim phc tp trong tin trnh cp nht. Thay v truy ngc v

cp nht 2m ng survivor, th ch c ng ML cn c x l.

Mt s chi tit ca mt b gii m SOVA lp c trnh by Hnh 3.17

B GII M SOVA

Chui trng thi MLSOVAkhng cth tccp nht

Thanh ghi dch

Thanh ghi dch

SOVA L(u)

u

L(u)

Lcy

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Hnh 3.30: B gii m SOVA lp

B gii m x l cc bit knh nhn c trn mt khung c bn. Nh c

trnh by trong Hnh 3.17, cc bit knh nhn c tch thnh dng bit h thng y1

v 2 dng bit parity y2 v y3 t cc b m ha 1v 2 tng ng. Cc bit ny c

cn bng bi gi tr tin cy knh v c ly ra qua cc thanh ghi CS. Cc thanh ghi

trnh by trong hnh c s dng nh cc b m lu tr cc chui cho n khi

chng ta cn. Cc kha chuyn c t v tr m nhm ngn nga cc bit t cc

khung k tip i x l cho n khi khung hin hnh c x l xong.

B gii m thnh phn SOVA cho ra thng tin a posteriori L(ut) v bit c

c on ut ( thi im t). Thng tin a posteriori L(ut) c phn tch thnh 3 s

hng

L(ut)=L(ut) + Lcyt,1 + Le(ut) (3.2)

L(ut) l gi tr a priori v c sinh ra bi b gii m thnh phn SOVAtrc .

Lcyt,1 l gi tr knh h thng nhn c qua cn bng.

Le(ut) l gi tr extrinsic c sinh ra bi b gii m thnh phn SOVA hin

ti. Tin tc i xuyn qua gia cc b gii m thnh phn SOVA l gi tr extrinsic.

Le(ut)=L(ut) Lcyt,1 L(ut) (3.3)

Gi tr a priori L(ut) c tr i t s b tr l thng tin a posteriori L(ut)

ngn nga tin tc i ngc li b gii m m t sinh ra n. Cng vy, gi tr

Le2

(u)

-

--

I{L2(u)}

y2

y1

y3

tincy

knh4E

b/N

0

SOVA1

SOVA2

thanh ghi CS

thanh ghi CS

thanhghi CS

thanhghi CS

2 thanh ghi dchsong song

2 thanh ghi dchsong song

II-1

I-1+

+

I

u

Le1

(u)L1(u)

-

CS : dch vngI : b chnI-1: b gii chn

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knh h thng nhn c qua cn bng L cyt,1 c tr i nhm xa tin tc

thng thng trong cc b gii m thnh phn SOVA. Hnh 3.17 trnh by b

gii m m PCCC l s kt ni theo th t vng kn ca cc b gii m thnh phn

SOVA. Trong s gii m vng kn ny, mi mt b gii m thnh phn SOVAc on chui tin bng cch s dng dng bit parity qua cn bng. Hn na, b

gii m PCCC thc hin gii m lp nhm cho ra cc c on a priori / tin cy

ng tin tng hn t 2 dng bit parity qua cn bng khc nhau, vi hy vng

thc hin gii m tt hn. Thut ton m Turbo lp vi ln lp th n nh sau:

1. B gii m SOVA1 c ng vo l chui Lcy1(h thng), Lcy2 (parity), v cho

ra chui Le1(u). i vi ln lp u tin, chui Le2(u)=0 bi v khng c gi tr a

priori (khng c gi tr extrinsic t SOVA2). Thng tin extrinsic t SOVA1 c

tnh bng

( ) ( ) ( )1211

''' yLuLuLuL cee = (3.4)

trong rateN

EL bc

=0

4

2. Cc chui Lcy1 v Le1(u) c chn v l I(Lcy1)v I{Le1(u)}.

3. B gii m SOVA2 c ng vo l cc chui L cy1 (h thng), v I(Lcy3)

(parity c chn b gii m) v I{L e1(u)} (thng tin a priori) v cho ra cc

chui I{L2(u)} v I{u} .

4.Thng tin extrinsic t SOVA2 c ly l:

I{Le2(u)} = I{L2(u)} - I{Le1(u)} - I(Lcy1)

Cc chui I{Le2(u)} vI{u} c gii chn v l Le2(u) v u.

5. Le2(u) c hi tip v SOVA1 nh l thng tin a priori cho ln lp k tip

v ul ng ra ca cc bit c c on cho ln lp th n.

3.6 CI TIN CHT LNG PCCC QUA THIT K B CHN

Cht lng m PCCC ph thuc vo 2 c tnh: ph khong cch ca n v

s thch hp ca n c gii m lp. C 2 c tnh ny b nh hng bi vic

chn b chn trong m PCCC.

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M PCCC c c im quan trng l vic thc hin sa sai li ca n gn t

n gii hn Shannon. Cht lng ca cc m PCCC tt hn cc m tch chp ch

khi chiu di ca b chn l rt ln, c th t n vi ngn bit. i vi cc b chn

c chiu di khi ln, th hu nh cc b chn ngu nhin u thc hin tt. Cn ivi b chn ngu nhin c chiu di khi ngn, th vic thc hin ca m PCCC b

gim st mt cch ng k n mc m tc li bit (BER) ca n ln hn so

vi m tch chp vi cng phc tp tnh ton nh nhau. i vi cc b chn c

chiu di khi ngn, th vic chn b chn c nh hng ng k n vic thc hin

ca m PCCC. Trong nhiu ng dng nh thoi, s tr hon l yu t quan trng

chn chiu di khi ca b chn. i vi cc ng dng ny th cn phi thit k cc

b chn kch c khi ngn c c BER c th chp nhn c.

C 2 tiu chun ln nht trong vic thit k mt b chn:

Cc c tnh ph khong cch (phn b trng s) ca m

S tng quan gia ng ra mm ca mi b gii m tng ng vi

cc bit kim tra chn l ca n v chui d liu ng vo thng tin.

Tiu chun th 2 i khi c xem nh l c tnh ca s thch hp gii m

lp (iterative decoding suitability IDS). y l thc o nh gi nh hng cathut ton gii m lp v nu 2 chui d liu t tng quan, th ta s c th s dng

thut ton gii m lp.

Vic thc hin ca m PCCC c BER thp b nh hng ln bi khong cch

t do ti thiu (dmin). Vic thc hin ca m PCCC t n ng tim cn dmin

c chng minh. Sn nhiu xy ra ti cc t s tn hiu/nhiu t va phi n cao

l kt qu ca dminnh. Sn nhiu c th c lm thp xung bng cch tng kch

c b chn hay dmin.nh gi vic thc hin ca m PCCC thng c da vo gi thit rng b

thu l b gii m kh nng xy ra ln nht (ML). Tuy nhin, cc m PCCC s dng

thc s l thut ton lp gn ti u. Vic thc hin gii m lp c li dng nu

thng tin c gi n mi b gii m t cc b gii m khc t tng quan vi

chui d liu thng tin ng vo.

Kt thc trellis ca m PCCC c cp n, c bit khi b chn c

thit k t cc i dmin. Nu vn ny khng c cp trong khi thit k b

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chn, th c th dn n gi tr dmin rt nh do s tn ti ca cc chui d liu khng

c kt thc trellis v trng s ng ra thp, kt qu a n l s suy gim cht

lng m PCCC.

Khng c mt cng thc chung cho b chn. B chn tt nht l b chn cthit k cho cc yu cu ring bit ca h thng. V d nh nu kch thc khi d

liu ngn v SNR thp th b chn chn l tt hn b chn ngu nhin v ngc li

khi SNR cao. Vi kch thc khi d liu ln th b chn S s tt hn b chn ngu

nhin.

3.6.1 Thit k b chn mi

Thit k b chn mi vi mc ch l tng khong cch t do hiu qu ti

thiudmin ca m PCCC trong khi gim i hay ti thiu l khng tng c tnh

tng quan gia chui d liu ng vo thng tin v thng tin a priori (ng ra mm

ca b gii m). Tuy nhin, vic thit k b chn mi ny rt phc tp v i hi

phi bit c thng tin ca ton b cc khi c th xy ra tnh kh nng tng

quan ca chng vi ng ra mm. Thit k b chn phi p ng c 2 ch tiu :

ph khong cch ca m v s tng quan gia thng tin d liu ngvo v ng ra

mm ca mi b m ha tng ng vi cc bit kim tra chn l ca n.

Mt khi cc iu kin trn c tha th vn cn mt yu t phi xem xt

tng cht lng ca PCCC hn na l ngu nhin m b chn v b gii

chn tch cc li chm t ng ra b gii m ny n ng vo ca b gii m tip

theo. Theo quan im ny th b chn l tng l b chn ngu nhin thc s

(khng phi gi ngu nhin). Ta s thy mt s tng vt v cht lng khi s dng

mt b chn tt.

B chn l b hon v i (i) nhm thay i th t ca chui d liu Nk t d1, d2,...,dN. Nu chui d liu ng vo l d = [d1, d2,...,dN] th chui hon v l

dP, trong P l ma trn chn . Mi b chn c b gii chn tng ng ( -1) tc

ng ln cc chui d liu c chn v khi phc th t ban u ca chui d

liu thng tin.

B chn ngu nhin n gin l b hon v ngu nhin . i vi N c gi tr

ln th hu ht cc b chn ngu nhin dng trong m PCCC u thc hin tt.

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Hnh 3.31: Qu trnh to thng tin extrinsic

Trong hnh trn, Ek l thng tin extrinsic tng ng vi bit thng tin ukti thi

im k to ra bi thut ton MAP s dng cc tnh ton thun v nghch trn ton

b chui nhn c.

T kt qu ca rt nhiu cuc m phng cho thy 95% gi tr ca thng tin

extrinsic c xc nh t trc Nm v sau Nm so vi k hiu nhn trong chui

nhn, vi Nm bng 10 ln chiu di bt buc K ca m. B chn kch thc cng ln

th cng c nhiu k hiu khng tng quan vi 2Nm k hiu trong chui d liu.

Qu trnh ny bao gm 2 tiu chun c lp c lng gi tr mm ca cngmt bit v sau cc c lng ny c a n b gii m khc vi lng thng

tin ln hn nhiu. iu ny dn n cht lng ca m PCCC tt hn rt nhiu. Tuy

nhin, khi kch c khi ca b chn gim th cht lng ca m PCCC b gim st,

n mc m BER ca n ln hn BER ca m tch chp c cng phc tp tnh

ton nh nhau. V vy, thit k b chn kch c ngn l cng vic quan trng.

Thit k b chn mi da trn vic thc hin gii m lp ca cc m PCCC.

Ti mi bc lp, mt s thng tin lin quan n cc bit kim tra chn l ca mt b

gii m c a n b gii m khc cng vi chui d liu h thng v cc bit

kim tra chn l tng ng n b m ha . Hnh 3.19 trnh by s gii m

lp.

Chui nhnR

k

Chui truyn

Tk

Nm

Nm

Thng tin extrinsic ng ra

Ek

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Hnh 3.32: Cu trc b gii m lp vi cc trng s

Cc ng vo ca mi b gii m l chui d liu ng vo dk, cc bit kim tra

chn l 1ky v2ky , v thut ton t s kh nng xy ra (LLR) c lin kt vi cc

bit kim tra chn l t b gii m khc ( 1kW hay2

kW ) c dng nh thng tin

a priori. B gii m dng tt c cc ng vo ny to ra 3 ng ra tng ng vi

trng s ca cc ng vo ny. Trong Hnh 3.19, kdbiu th trng s ca chui d

liu ng vo dk. dn minh ha chui d liu ng vo c a vo b gii m th hai

sau khi chn. Ng vo ca mi b gii m ly t b gii m khc c dng nh

thng tin a priori trong bc gii m tip theo v ph hp vi trng s ca cc bit

kim tra chn l. Thng tin ny s c nhiu hiu qu trong vic thc hin gii m

lp nu n t tng quan vi chui d liu ng vo (hay chui d liu ng vo cchn). V vy, y l tiu chun thit k b chn. i vi cc b chn c kch c

khi ln, th hu ht cc b chn ngu nhin a n s tng quan thp gia ikW

v chui d liu ng vo dk. H s tng quan1

,2

11 kk

dWr c nh ngha nh l s

tng quan gia 11k

W v 2kd . H s tng quan c tnh gn bng nh sau:

=

21

21211

,0

,)exp(1

2,1 kk

kkkkcar

kdkW(3.5)

trong a v c l cc hng s ph thuc vo cc a thc hi tip v n

(feedforward) ca b m ha. H s tng quan ca ng ra b gii m th hai

22 d,W

r tnh gn bng:

)(2

1 1

,

1

,

2

, 112 dWdWdWrIPrr +=

(3.6)

Gii chn

B gii

m 1

B gii

m 2

Chn

1

nW

2

nW

1

kW2

kW

2k

W 1n

W

r

ndr

kd1r

ky

kd

2r

ny

nd

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1

'k

V c nh ngha tng t s dng2

,2' dWr . o s ph hp gii m lp

(IDS) c nh ngha nh sau:

( )=

+=N

k

kk VV

N

IDS

11

11'

2

1

( 3.7)

Gi tr IDS thp l biu th cc c tnh tng quan gia W1 v d tri dc sut

chui d liu chiu di N. Thit k b chn l da trn iu kin IDS.

3.6.2 Cc phng php ti u ho cu trc b chn

C mt phng php c gii thiu ti u ha cu trc b chn mang

tn phng php Hungari (vn phn tng tuyn tnh). Mc ch ca phng

php ny l ph v tt c cc chui trng s 2 ngha l c t nht mt ng ra trnh

c chui kt thc zero.

Mt phng php khc ngh kiu b chn kinh in vi tr nh nht,

gi l b hon v trng thi hu hn (FSP). Hai thut ton c pht trin cho mt

cu trc lp v mang tnh h thng ca b chn vi phc tp tng tuyn tnh vi

kch thc b chn. Mi mt vect chuyn i c mt hm tr gi km theo, v cc

thut ton nhm ti mc tiu l lm gim ti a hm tr gi ny. phc tp tnhton ph thuc vo chiu di ca cc mu li c xem xt.

Cc nghin cu gn y cho thy cc tail bit c th b m khng lm nh

hng lm n cht lng khi kch thc b chn ln.

3.7 S KHC NHAU GIA M CHP V M PCCC

Ta c s khc nhau nh bng 3.7 di:

Tiu chun M tch chp M PCCC

Chiu di bt buc ln hn Tt XuKhong cch t do ln hn Tt Khng khcTc m ho thp hn Khng khc TtCc b m ho quy Khng khc TtCc b gii m ng ra mm Khng khc Tt

Bng 3.7 So snh m chp v m PCCC

3.8 SO SNH CHT LNG CC H THNG M HO

T l li bit (BER) v t s tn hiu/nhiu (SNR) ca qu trnh truyn dn xc

nh cht lng ca knh truyn. Tuy nhin, s khc nhau l bao nhiu cng sut

(SNR) th cn thit thc hin BER thp. T l li bit l xc sut ca bt c bit no

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b li trong qu trnh truyn. T s tn hiu/nhiu ( Eb/No) l t s cng sut knh/

cng sut nhiu. BER cng thp th cng tt bi v c ngha l c mt vi li trong

d liu cui cng SNR cng thp th cng tt bi v c ngha l t cng sut cn thit

truyn tn hiu.Nh trn gii thiu m PCCC RSC l m c thc t tt nht bi v n c

th thc hin SNR thp ti BER thp v gn vi gii hn Shannon l thuyt ti a

ca thc hin knh. li m xc nh thc hin m ha nh th no. li m

l s khc nhau trong SNR gia 1 knh c m ha v 1 knh khng c m ha.

li m c th c xc nh bng cch o khong cch gia gi tr SNR ca bt

k mt trong nhng knh c m ha no v knh khng c m ha ti BER

cho trc. V d li m i vi m PCCC RSC c tc ti BER 10-5 l

khong 8,5 dB. iu ny c ngha l tn hiu c m ha PCCC c th hoc l

c nhn 2,65 ln xa hn tn hiu khng c m ha ( cng mt cng sut

pht), hoc l n ch cn 1/7 cng sut pht (cho cng mt khong cch).

Hnh 3.33: So snh h thng m ho

3.9 KT LUN CHNG

Qua chng ny ta c c ci nhn c th hn v m Turbo , bit c cch

m ho v gi m ca m TC , cc phng php nhm lm tng cht lng ca m.

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Vic so snh gia m Turbo vi cc m khc nhm lm ni bt c u im ca

m ny.Tuy nhin m TC c ng dng nh th no?

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Chng 4: ng dng ca m Turbo

CHNG 4 :NG DNG CA M TURBO

4.1 GII THIU CHNG

M Turbo ra i gii quyt c nhng hn ch trong lnh vc vin thng

m tiu biu l hai h thng :Truyn thng khng dy(wireless communication) v

truyn thng a phng tin(multimedia communication).

4.2 CC NG DNG TRUYN THNG A PHNG TIN(MMC)

4.2.1 Hn ch khi ng dng TC vo MCC

4.2.1.1 Bng thng gii hn

Bng thng l vn quan tm hng u v lng thng tin truyn ti ngy

cng ln i hi phi lun thay i nhng bng thng l ti nguyn quc gia kh

thay i. Bng thng s dng trong cc MCC thng rt ln nn dng cc b m

tc thp s khng cho hiu qu cao.

4.2.1.2 Khi lng d liu ln

Cc ng dng ca MCC thng l cc khi d liu ln , nhng yu cu

phi c x l trong khong thi gian rt ngn .V vy yu cu cc b m ho v

gii m TC phi cao.4.2.1.3 Tnh thi gian thc

D liu multimedia c bn cht thi gian thc , s chm tr ca d liu s

lm gim cht lng thng tin .Trong khi TC cn phi c cu trc gii m lp d

tng cht lng th tnh thi gian thc l mt thch thc khi gii quyt mu thun

gia thi gian p ng v t l li bit( BER) . c tnh thi gian thc buc cc m

TC ng dng trong MCC khng th c s vng lp ln.

4.2.1.4 Cc c tnh ca knh truyn

T cc c tnh ca knh truyn ta c th a ra cc gii php thch hp cho

tng h thng .Chnh v vy vic tm hiu c tnh ca knh truyn l rt cn thit ,

mt trong s cc cch tm hiu knh truyn l dng mng Bayes.

Phng php thc hin nh sau:

D liu c truyn qua knh truyn v s dng nhiu lai mhnh m TC vi cc thng s khc nhau .

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Ghi li cc gi tr BER.

Thnh lp mt mng Bayes vi cc chnh xc ca m kt

qu v cc yu t nh hng l cc nt mng .Mt ng ni

trc tip s i t cc yu t n cc chnh xc ca m ktqu.

S dng cc gi tr BER ghi nhn th cho mng ny .

Tm ra mt tp hp cc thng s ti u ho cc s iu

chnh.

4.2.2 Cc xut khi ng dng TC vo MCC

4.2.2.1 Kch thc khung ln

Do cc ng dng MCC l khi d liu ln nn s dng kch thc khung

ln cho m TC .Kch thc khung ln ng ngha vi kch thc b chn ln v

lm cho cht lng m TC tng ln .

4.2.2.2 Ci tin qu trnh gii m

Gii m ng: c c im nh sau:

t mt ngng vng lp l s ln lp ti a cho mt

khung.

S vng lp thc s gii m mt khung s nh hn hay

bng gi tr ngng v ph thuc vo kt qu gii m .

iu kin ngng qu trnh gii m l khung ht li

.Trong qu trnh gii m , kt qu gi tr c lng ca

vng lp gii m trc s c lu li v so snh vi kt

qu ca vng lp gii m k tip .Nu hai kt qu ging

nhau th ht li v tip tc gii m cho khung tip theo.

Gii m u tin: Cc ng dng MCC s thm cc thng tin v tin vo

trong khung tu theo tm quan trng ca khung .Sau khi nhn c chui tin t

knh truyn ,b gii m s gii m tm ra cc t m. Sau vng lp u tin b gii

m c th nhn c thng tin v mc u tin ca khung v s quyt nh s

vng lp ph hp vi khung .Theo phng php ny th thi gian dng cho cc

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khung c u tin thp s thp , ng thi lm gim BER ca cc khung c u

tin cao v tng tc p ng ca h thng .

Gii m Pipeline :mc ch lm gim ti a tr ca h thng do cc vng lp

gii m gy ra .S n gin ca mt b gii m lp nh sau

Hnh 4.34: S gii m lp

B gii m thng thng s lp li qu trnh gii m n ln cho mi t m y tm

c c on gn ng vi t m x nht .

Cu trc n gin ca b gii m Pipeline l:

BGM: l b gii m

Hnh 4.35: B gii m Pipeline

B chui cc b gii m c s dng cho mi vng lp.B nh dng lu tr

thng tin h thng cho cc vng lp tng ng, tc tng n ln.4.3 CC NG DNG TRUYN THNG KHNG DY

4.3.1 Cc hn ch khi ng dng TC trong truyn thng khng dy

4.3.1.1 Knh truyn

Knh truyn trong truyn thng khng dy c mc nhiu cao hn mi

trng truyn dy. V vy cc m knh phi c kh nng ng u vi mc

nhiu ln .Trong mi trng khng dy c fading ca cc tn hiu truyn ,fading l

hu qu bn cht vt l ca knh truyn vi tng bin .Trit fading bng cch

46

B gii m 1 B gii m 2xy

Vng lp1BGM 1y

Vng lp 1BGM 2

Vng lp 2BGM 1

Vng lp 2BGM 2

B nh 1 B nh 2


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Hnh 4.36: S b m ho PCCC theo chun CDMA2000

Hm truyn ca mi b m ho thnh phn l:

( )

( )

( )

( )

=

Dd

Dn

Dd

DnDG

o 11)(

Trong d(D)=1+D2 + D3, n0(D) = 1+D + D3 , n1(D) = 1+ D2 +D3 .

Cc k t ng ra d liu m ho c chn lc vi cc mu lc tng ng vi

tc m ho ,s 0 c ngha l k t s b xo b v k t 1 c cho qua.

48

+

D D D+

+

+

+

D D D+

+

+

Lp liv

chnlc k

t

B chn

X

Yo

Y1

X

Yo

Y1

NgvoN

turbo

bit

Ng ra

K t m

no

n1

d

no

n1

d

d


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Ng ra

Tc m ho1/2 1/3 1/4

X 11 11 11Y0 10 11 11Y1 00 00 00X 00 00 00Y0 01 11 01Y1 00 00 11

Bng 4.1 : Cc mu lc cho cc chu k bit d liu

B chn trong CDMA2000 l b chn m trong ton b chui bit ng vo cab chn c vit ln lt vo mt dy theo chui a ch ,sau c ra theo chui

a ch c nh ngha.Hnh 4.4 m t qu trnh sp xp a ch ng vo /ng ra

ca b chn TC:

Hnh 4.37: Th tc tnh a chi ng ra ca b chn

Th tc tnh a ch ng ra ca b chn trong CDMA2000:

Xc nh thng s ca b chn n , vi n l s nguyn ( cho km theo bng

di).

Reset b m (n+5) v 0.

Trch ra n bit cao t b m v cng thm 1 to ra gi tr mi .Sau lai

b tt c ngoi tr n bit thp ca gi tr ny .

49

n bit

n bit

5 bit(i

4

i0)

n bit(t

n-1t

0)

5 bit thp(i

4i

0)

n bit cao(i

n+4i

5)

Cng 1 vchn n bit

thp

Bng tra

o bit

Nhn vchn n

bit thp

Loi ranu ng

vo N

turbo

B

m(n+5)bit

a ch

ng rabchn(n+5)

bit ktip

Bitcao

Bitthp


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Ly 5 bit thp t b m lm a ch c qua tra bng c ng ra n bit.

Nhn gi tr ly c bc 3 v 4 v loi b tt c ngoi tr n bit thp.

o ngc 5 bit thp ca b m .

To mt a ch ng ra th c cc bit cao ca n bng gi tr ly trong bc 6

v cc bit thp trong bc 5.

Chp nhn a ch ng ra th nh a ch ng ra nu n nh hn Nturbo cn

nu khng th loi.

Tng b m v lp li t bc 3 n bc 6 cho n khi tt c cc a ch

ng ra ca b chn Nturbo c ly .

Bng thng s ca b chn:

Kch c b chn Nturbo Thng s b chn n378 4570 5762 51145 61530 62298 73066 74602 86138 89210 912282 920730 10

Bng 4.2 Bng thng s ca b chn.

4.3.2.2 Phn b trng s 2,3 m PCCC trong CDMA2000

Hm xc sut li bit ca m PCCCl:

exp2

00

==

n

dd

b

d

n

freedd

b

db

freeN

REdD

N

REdQDP

Vi:

j

dj

dA

ND

,

=+

=

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Trong Dd l ph khong cch ca m PCCC, N:kch thc b chn , dfree :

l khong cch t do ti thiu ca m PCCC, R: tc m ho , A,j :l h s ca

hm m trng s c iu kin.

Mt khi t m xc nh c lin kt vi mt ng vo trng s 2 phi c mtkhi thng tin trng s 2 v mt khi chn l c trng s jzcycle + t , vi zcyclel trng

s chn l lin kt vi chu k trellis, j: l s nguyn v t:l trng s chn l kt hp

vi con ng chuyn tip dn n s bt u ca chu k u tin v t chu k cui

cng n trng thi ton zero .

Mt khc zmin= zcycle + t , vi zminl trng s ca chui chn l c trng s nh

nht c tng hp bi b m ho PCCC vi mt ng vo trng s 2. Khong cch

t do dfree l thng s quan trng ca m sa sai v dfree,eff = 2 + 2zmin nn zmin l

mt thng s quan trng c tc ng n c tnh ca b gii m .

Nu s dng b chn xu vi dfree = 9 ,ng vo c dng Dj (1+D2+ D3) tc

m 1/3. Ta c dfree,eff = 14 vi ng vo c dng Dj (1+D7) .V vy cn s dng

b chn c dfree = 24 v dfree = dfree,eff ,ng vo c dng Dj (1+D7) lm ng ra c

dng Dj-k (1+D14) , knguyn .

Nu tc m ho l 1/4 th dfree,eff = 38 khi ng vo c dng Dj

(1+D7

) v ngra s c dng Dj-k (1+D14) .Minh ho cho nhng iu ny bng nhng hnh v sau:

4.3.2.3 H s Scalling

Thng tin extrinsic Le c nhn vi h s Scalling Sk .Nhn h s ny

vo b mo do thut ton gii m chu mo ln gy ra bi cc ng ra mm.H s

Sk c nhn vo cc gi tr symbol nhn c trong mi ln lp gii m .Ngoi ra

gi tr tin cy ca knh Lc mt s symbol c thay th bng SkLc .

Nu t s tin cy LLR ca mt s symbol ln th cc symbol ny xem nh

cc symbol tin cy.

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.

Dj(1+ D2+D3)

.

Dj-k(1 +D2+D3)(a) khong cch t do

. Dj(1+D7)

. Dj-k(1+ D7)

(b) khong cch t do hiu dng

. Dj(1+D7)

. Dj-k(1+ D14)

(c) khong cch t do hiu dng


Hnh4.38: Khong cch t do hiu dng

4.4 KT LUN CHNG

Qua chng ny ta bit c nhng ng dng ca m TC vo trong truyn

thng a phng tin v truyn thng khng dy .Nhng kh khn khi thc hin

tm ra nhng cch khc phuc nhng nhc im nhm mang li nhng tin ch

hn na ca cng ngh CDMA i vi cuc sng con ngui .

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Chng 5: Chng trnh m phng v kt qu

CHNG 5 CHNG TRNH M PHNG V KT

QU

5.1 GII THIU CHNG

Chng ny trnh by m phng m Turbo PCCC vi tc 1/2 v 1/3 tu

theo s la chn k thut xo hay l khng.Chy chng trnh vi cc thng s u

vo khc nhau c nhng kt qu cui cng v t l li bit(BER) v t l li

khung(FER),cc kt qu ny c a ra hnh v c nhng nhn xt v mPCCC .

5.2 CHNG TRNH M PHNG

5.2.1 Cu trc chng trnh

Chng trnh m phng c vit bng Matlab:gm cc file:

1.bo_ma_hoa.m

2.chuong_trinh_chinh.m

3.logmapo.m4.ma_hoa.m

5.nhi_phan.m

6.rsc_ma_hoa.m

7.so_do_trellis.m

8.so_nguyen.m

9.sova.m

10.tach_kenh.m

11.Turbo.txt(lu kt qu chy chng trnh)

12.Hinh_kq.m( kt qu ra hnh)

5.2.2 Chng trnh chnh

chy chng trnh ta vo file :chuong_trinh_chinh.m nhp cc thng

s u vo,nh:

+Chn thut ton:hm Ttgiaima = 0 l ta chn thut ton Log-Map.hm Ttgiaima = 1 l ta chn thut ton SOVA.

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+Chn kch thc khung bng hm L_f=? bao gm thng tin v bit ui

(info+tail).

+Chn ma trn sinh :ta mc nh ma trn sinh g = [1 1 1; 1 0 1 ] .

+Chn dng k thut xo hay l khng:hm kt_xoa = 0 l dng k thut xohmkt_xoa=1lkhng dng k thut

xo .

+Chn s ln lp ca mi khung bng hm: lan_lap =?

+Nhp s khung b li kt thc chng trnh bng hm: ket_thuc=?

+Nhp gi tr Eb/N0 bng hm:EbN0db =?

Sau khi nhp cc thng s xong ta cho chay chng trnh nh sau:

Hnh 5.39: Chng trnh m phng chnh

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5.3 KT QU M PHNG

Thc hin chy chng trnh nhiu ln vi cc thng s u vo khc nhau,ta

c kt qu nh bng.

Ln 1 Ln 2 Ln 3 Ln 4 Ln 5 Ln 6 Ln 7Thut

ton

Log-Map Log-Map SOVA SOVA Log-Map SOVA Log-Map

Kch

thc

khung

300 300 300 300 500 1000 1000

K

thutxo

C C Khng Khng C Khng C

S ln

lp

mi

khung

5 10 5 10 5 5 5

S

khungb li

kt

thc

10 10 10 10 5 3 3

Eb/No

(dB)

2 2 3 2 3 3 3

Bng 5.1 Thng s cc ln chy chng trnh

V kt qu cui cng ca cc ln chy chng trnh nh sau:

Ln1 ***************************** Eb/N0 = 2.00 db ***********************************Kich thuoc khung = 300, rate 1/2.

132 Khung da trao doi , 10 Khung bi loi.

Ti le loi bit (Tu lan lap 1 cho den 5):

2.3337e-002 5.1098e-003 2.1100e-003 1.3219e-003 1.6270e-003

Ti le loi khung (Tu lan lap 1 cho den 5):

9.1667e-001 3.0303e-001 1.2879e-001 8.3333e-002 7.5758e-002

*******************************************************************************

56


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2.1310e-002 3.8149e-003 1.2414e-003 7.2160e-004 5.8536e-004 4.3902e-0044.4911e-004 3.9360e-004 4.4406e-004 3.4314e-004


8.7669e-001 2.6316e-001 7.6692e-002 3.4586e-002 2.8571e-002 1.9549e-002

1.6541e-002 1.5038e-002 1.3534e-002 1.5038e-002

*******************************************************************************




3.5705e-003 1.8030e-004 7.8819e-005 6.9952e-005 4.4335e-005


2.9008e-001 1.9671e-002 5.8720e-003 4.1104e-003 2.9360e-003

*******************************************************************************




2.4345e-002 4.9595e-003 2.1467e-003 1.5380e-003 1.0034e-003 8.0603e-004

8.1425e-004 7.5668e-004 7.1555e-004 6.2508e-004


8.0147e-001 2.3775e-001 9.8039e-002 6.1275e-002 4.4118e-002 3.6765e-002

3.4314e-002 2.9412e-002 2.4510e-002 2.4510e-002

*******************************************************************************




3.5505e-003 7.3512e-005 1.9345e-005 2.5794e-005 2.0635e-005


4.6692e-001 1.2845e-002 3.2113e-003 3.2113e-003 3.2113e-003

*******************************************************************************

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3.0484e-003 4.2338e-005 1.4113e-005 1.8347e-005 6.3507e-006Ti le loi khung (Tu lan lap 1 cho den 5):

6.1972e-001 1.9014e-002 7.0423e-003 4.2254e-003 2.1127e-003

*******************************************************************************




3.5560e-003 2.1283e-005 6.8106e-006 5.1079e-006 5.1079e-006

Ti le loi khung (Tu lan lap 1 cho den 5):7.3322e-001 1.0195e-002 4.2481e-003 2.5489e-003 2.5489e-003

*******************************************************************************

Bng 5.2 Kt qu cui cng

Hnh v c c sau khi chy file:Hinh_kq.m t cc kt qu cui cng

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Hnh 5.40: Kt qu ln 1


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Qua nhng hnh trn v mt s kt qu chy chng trnh khc,ta c nhnxt:

s ln lp cng ln th t l li bit cng nh t l li khung u

gim.

nu s lng khung a vo cng ln th BER v FER cng

thp(th hin r trong Hnh 5.7 v Hnh 5.8).

m s hot ng tt khi ta la chn kch thc khung ln.

t l li khung(FER) thng ln hn t l li bit(BER) nhng

ln lp cng ln th BER~FER(trong hnh 5.7 v 5.8 khi ln lp

th 5 th BER~ FER ~0).

5.4 KT LUN CHNG

Sau khi chy chng trnh vi cc thng s u vo cho nh bng 5.1 v

chy nhiu ln chng trnh vi cc thng s khc nhau v cc kt qu ny uc lu li trong file Turbo.txt.

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Documents

ĐỒ ÁN TỐT NGHIỆP _Nghiên cứu mã Turbo trong hệ thống CDMA