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5

Ma hoa knh

Theo quan im cua nganh thng tin, tai nguyn thng tin chu yu la cng sut, thi gian va

bng thng cua tn hiu. i vi mt mi trng thng tin cho trc, ba tai nguyn nay co thmu thun ln nhau. Vic cn i cac mu thun nay tuy vao tng trng hp cu th. Tuynhin, nhn chung th ta co th at c tc truyn s liu cao nht trong mt bng thngnho nht trong khi gi cho cht lng truyn dn mc chp nhn c. Trong thng tin sth cht lng truyn dn co lin quan mt thit vi xac sut li bitPbtai u thu.

nh ly v thng lng knh cua Shannon- Hartley:

)N/S1(logBC2

+= (bit/s)

a ch ra gii han ly thuyt cua tc truyn s liu t b phat co cng sut cho trc, qua

mt knh vi bng thng cho trc, hoat ng trong mi trng co nhiu a bit. Tuy nhin, thc hin c gii han ly thuyt nay, ta phai tm c mt phng phap ma hoa phu hp(theo Shannon th phng phap nay co tn tai).

Trong thc t, yu cu cua vic thit k la phai thc hin c mt tc truyn s liu yucu (thng c xac nh bi dch vu cung cp) trong mt bng thng han ch cua mt knhtruyn sn co va mt cng sut han ch tuy ng dung cu th. Hn na, con phai at c tc nay vi mt ty sBER (Bit Error Rate) va thi gian tr chp nhn c. Nu mt tuyntruyn dn PCM khng at c ty s BER yu cu vi cac rang buc nay th cn phai sdung cac phng phap ma hoa iu khin li (error control coding).

Ma hoa iu khin li, con c goi lama hoa knh(channel encoding) c s dung phat hin va sa cac ky t hay cac bit thu b li. Ma hoa phat hin li (error detectioncoding) c s dung nh la bc u tin cua qua trnh sa li bng cach kch cho u cuithu phat ra tn hiu yu cu lp lai t ng ARQ (Automatic Repeat reQuest), truyn theohng ngc lai v cho u cui phat. Nu qua trnh truyn lai thanh cng th coi nh la asa c li. Nu ky thut ARQ khng thch hp, chng han nh khi tr truyn dn qua lnth se s dung ky thut ma hoa sa li khng phan hi FECC (Forward Error CorrectionCoding). Ca ma phat hin li va ma sa li u a thm d vao d liu phat, trong o d thm vao trong ma sa li nhiu hn trong ma phat hin li. Ly do la i vi ma sa li, d thm vao phai u cho bn thu khng ch phat hin c li ma con sa c li, khngcn phai truyn lai.

Phn u cua chng nay se trnh bay tng quan v iu khin li ap dung trong h thngthng tin s, bao gm gii thiu v cac phng phap iu khin li, phn loai cac ma iukhin li.

Phn sau cua chng tp trung vao cac loai ma iu khin li, bao gm hai loai chnh lamakhi (block code) vama chp (convolutional code).

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Phn ma khi se nu mt loai ma khi n gian nht lama kim tra chn le parity.

Phn ma khi tuyn tnh (linear block code) se tp trung trnh bay vma vong (cyclic code)va mt loai ma vong n gian lama Hamming.

Phn ma chp cui chng se trnh bay phng phap dungs cy (tree diagram),s li (trellis diagram) vas trang thai (state diagram) minh hoa cho qua trnh ma hoa

ma chp. Phn giai ma ma chp trnh bay thut toan Viterbi dung s li.Cac ni dung v ma hoa c trnh bay y theo quan im la danh cho ngi a nm cly thuyt ma hoa, ch nu thut toan ma hoa va giai ma, a ra v du minh hoa, ch khngnu c s toan hoc.

5.1 Tng quan v iu khin li

5.1.1 Cac phng phap iu khin li

ai lng o li thng thng laty l li bit BER (Bit Error Rate) hayxac sut li bit (Pb).Pbn gian la xac sut mt bit nh phn bt ky truyn i b li. BER laty s li trung bnh, ctnh la tch cuaPb vaRb, yRb la tc bit trong knh.Pb in hnh trong mt h thngPCM tuyn tnh la 10 -7, trong h thng PCM nen phi tuyn la 10 -5, trong h thng ADPCMla 10 -4.

iu khin li nhm muc ch la lam giam ty l li trong mt h thng khi ty l nay ln quamc cho phep. Nhn chung co nm phng phap iu khin li.

Giai phap u tin va d thy nht latng cng sut phat, nhng khng phai luc nao cung coth thc hin c. V du nh, i vi mt may in thoai bo tui th khng chp nhn khilng cua pin qua ln.

Giai phap th hai, rt hiu qua trong vic chng lai li chum gy bi fading, la s dung phntp (diversity). Co ba kiu phn tp chnh la phn tp khng gian, phn tp tn s va phn tpthi gian. Ca ba kiu phn tp nay u a thm d vao trong d liu phat bng cachtruyn gp i: qua hai ng, tai hai tn s hay vao hai thi im khac nhau. Trongphntp khng gian, s dung hai hay nhiu antenna t tai nhng v tr u xa co mt trong cacantenna o thu c tn hiu tt nht, t b fading nht.Phn tp tn s s dung hai hay nhiutn s khac nhau phat cung mt tin. Phn tp tn s co th la trong bng hay ngoai bngtuy vao khoang cach tn s gia cac song mang. Trong h thng phn tp thi gian, phatcung mt tin nhng vao hai hay nhiu thi im khac nhau.

Giai phap th ba la truyn song cng, hay con goi lakim tra echo (echo checking). y,khi b phat phat tin n b thu, tin c phat ngc v b phat trn mt knh hi tip ring.Nu tin phat ngc v khac vi tin phat i th bit la co li. Phng phap nay co khuyt imla yu cu bng thng gp i so vi truyn trn mt hng nn khng chp nhn khi cn tndung ph.

Phng phap th t i pho vi BER cao layu cu lp lai t ng ARQ (AutomaticRepeat reQuest). Trong h thng ARQ, ma phat hin li (error detecting code)c s dung bn thu kim tra li trong khi s liu thu va tra li cho bn phat trn mt knh hi tip.

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Tn hiu tra li lachp nhn ACK (ACKnowledgment) khi s liu thu ung vakhng chpnhn NAK (Non - AcKnowledgment)khi s liu thu sai. Nu bn phat nhn NAK, bn phatphai tin hanh truyn lai khi s liu b li. Co hai ky thut ARQ chnh laARQ dng va i(stop and wait ARQ) vaARQ lin tuc (continuous ARQ). Trong h thng ARQ dng va i,sau khi phat khi s liu i, bn phat dng lai va ch nhn tra li t bn thu, ri tuy theo trali o la ACK hay NAK ma bn phat phat khi s liu tip theo hay phat lai khi s liu vari. Nu thi gian ch qua thi gian quy nh (goi la time-out), bn phat coi nh la khi sliu va phat b li va vn tin hanh phat lai. Han ch cua phng phap nay la thi gian trtruyn dn ln. Trong h thng ARQ lin tuc, cac khi s liu u mang s th t - N - vaban tin tra li ACK/NAK cung mang s th t N tng ng. Bn phat lin tuc phat i cackhi s liu ma khng ch nhn tra li t bn thu. Bn thu kim tra li cac khi s liu thu vatra li v cho bn phat ban tin ACK/NAK kem theo s th t cua khi tin tng ng. Khi naobn phat nhn tra li NAK t bn thu, bn phat se phat lai tt ca cac khi s liu k t khi sliu b li i viARQ lui lai N (go-back-N ARQ), hoc bn phat se ch phat lai khi s liub li i vi ARQ chon loc (selective ARQ). Mc du ARQ chon loc rt hiu qua trong s

dung bng thng nhng yu cu dung lng b nh ln hn ARQ lui lai N, c bit trong cackt ni tc cao. ARQ phu hp vi cac h thng thng tin may tnh, v o co sn knhsong cng bn thu co th phat lai cho bn phat ban tin ACK/NAK. Tuy nhin, trong cacng truyn dai vi tc cao, in hnh nh thng tin v tinh th rt kho thc hin ARQ.

Phng phap th nm giam BER la thc hin ma hoa sa li khng phan hi FECC(Forward Error Correction Coding). Trong lch s, vic chp nhn s dung rng rai FECCco tr hn so vi cac phng phap khac, bi v phc tap va gia ca cua no cao hn. Ngaynay, phc tap a giam xung nh vao s gia tng cac chip ma hoa/ giai ma VLSI. FECCli dung s khac nhau gia tc truyn dn va thng lng knh giam xac sut liPb.

Vic giam xac sut li b tra gia bng vic tng thi gian tr truyn dn, do tng d cho u ma co th phat hin va sa c li va do mt thi gian kim tra khi s liu thu sali. Tuy nhin, li ch cua FECC co c thng nhiu hn khuyt im v tr ln.

Ma khi

Ma khng vong

Ma Golay RS

Hamming (e=1) e>1

BCH nh phn

Ma vong

Ma tuyn tnh

(Ma nhom)

Ma khng

tuyn tnh

Ma chp

Ma hoa iu khin li

Hnh 5.1Phn loai ma iu khin li

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5.1.2 Phn loai ma iu khin li

Nhn chung, co th phn loai ma phat hin va sa li (goi chung la ma hoa knh - ma hoaiu khin li) theo s trong hnh 5.1.

a) Ma khi

Ma khic c trng bi hai s nguyn n va k, va mt ma trn sinh hay a thc sinh.Hnh 5.2 minh hoa mt b ma hoa ma khi vi k bit tin vao va n bit ma hoa ra. T ma n bitc tao ra duy nht t k bit tin va (n-k) la s bit kim tra d. Ty l ma (coder rate) la R =k/n, la tiu chun anh gia d cua ma. Ty l ma thng t 1/2 n 1. Ma h thng(systematic code) la ma co mt cac bit tin cung vi cac bit d trong t ma. Trong cac tai liuv ma hoa th co hai nh ngha v ma h thng. nh ngha nghim ngt hn cho rng ma cotnh h thng khi k bit tin phai nm lin tuc thanh mt khi va cac bit d phai nm lin tuctrong mt khi khac. nh ngha t nghim ngt hn th ch yu cu trong t ma co mt cacbit tin ch khng cn phai nm lin tuc thanh khi.

B ma hoa khi n bitma hoa

kbit

tin

T ma n bit

(n-k) bitk bit

Phn tin Phn d

Hnh 5.2 Ma khi h thng (n, k)

Ma khi tuyn tnh (liear block code) - con goi lama nhom (group code) - co cac t ma cotng ng 1-1 vi cac phn t thuc nhom toan hoc. Ma tuyn tnh co cha t ma gm toans 0 va co tnh cht ong, chng han i vi ma tuyn tnh nh phn, vi hai t ma va

bt ky, ta lun co , cung la mt t ma. Vic co cha t ma gm toan s

0 va tnh cht ong lam cho vic tnh toan i vi ma tuyn tnh c bit d. Hnh 5.3 la mtv du n gian v ma tuyn tnh. No minh hoa cho tnh cht ong cua ma. Co 4 ky t ngunla a, b, c, va d, k = 2, n = 5. y la ma (5, 2)

iC

jC

kjiCCC =+

kC

a = 00 00000

b = 01 00111c = 10 11100

Ma hoa

d = 11 11011

cdb,bdc,dbc ===

Hnh 5.3 Minh hoa tnh cht ong cua ma khi tuyn tnh

Ma vong (cyclic code) la mt lp con cua ma khi tuyn tnh khng co t ma gm toan s 0.

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Mt ma khi tuyn tnh c goi la ma vong nu sau mt ln dch vong mt t ma th cungc mt t ma thuc cung b ma. V du cac t ma sau y c goi la ma vong:

1101000, 0110100, 0011010, 1000110, 0001101, 1010001, 0100011.

Ma Golay la mt loai ma vong sa c sai nhiu li. Ma Golay (23, 12) co kha nng sac 3 li cho t ma dai 23 bit. Ma nay c Golay phat minh nm 1949 va c nhiu

chuyn gia quan tm nghin cu ti cu truc va c ch giai ma. Thc t ang co hai phngphap giai ma laphng phap Kasami vagiai ma tm kim co h thng (systematic searchdecoding). Ma Golay (23, 12) c s dung kha ph bin trong mt s h thng thng tin.

Ma BCH nh phn (binary BCH code) la mt loai ma vong c Hocquenghem tm ra nm1959, sau o c Bose va Chaudhuri tm ra mt cach c lp vao nm 1960. Ma BCH co

th sa c t li trong t ma dai n bit, vi . V du ma

BCH (15, 7) co th sa sai ti a 2 li.

1t2d,mtkn,12nmin

m

+=

Ma RSc Reed va Solomon gii thiu ln u tin vao nm 1960. Theo ly thuyt ma, co

th xem ma RS la ma BCH khng nh phn. Ma RS c t chc theo ky t. Ma RS taothanh n ky t, mi ky t dai m bit, m tuy thuc vao ng dung cu th, v du m = 8 th mi kyt chnh la mt byte. Ma RS hoat ng trn ky t nhiu bit ch khng phai trn tng bit nhcac ma vong khac. Mt c im quan trong cua ma RS la kha nng sa li chum. Ma RS co

th sa sai t li, vi2

knt

= . y n va k la s ky t ma hoa va s ky t mang tin ch

khng phai s bit. V du ma RS (31, 15) co 15 ky t vao, mi ky t 5 bit, tc la 75 bit tin va31 ky t ma hoa, mi ky t 5 bit. Ma nay co th sa c 8 li bit c lp hoc 4 li chum daikhng qua 5 bit. Ma RS c dung rng rai trong cac u CD va trong b nh may tnh.

Ma Hamming la mt trng hp ring n gian nht cua ma BCH nh phn. Ma nay cR.W. Hamming a ra va c dung trong mt s h thng thng tin. Ma Hamming co kha

nng sa sai 1 li. Quan h gia n va k thoa man bt ng thc:1n

22

nk

+

b) Ma chp

Ma chp cung c c trng bi hai s nguyn la n va k nh ma khi, nhng n bit ra khoib ma hoa khng ch phu thuc vao k bit vao ma con phu thuc vao K-1 b k bit vao trco. K c goi la dai rang buc (constraint length). Ma chp (n, k, K) c xy dng tcac thanh ghi dch kK bit. Vy co th xem ma chp la ma co nh, o la im khac bit c bancua ma chp so vi ma khi.

Ma chp c Elias xut ln u tin vao nm 1955. Sau o, Wozencraft a ra mt thuttoan giai ma tng i hiu qua. Nm 1963, Massey a ra cach giai ma t hiu qua hnnhng d thc hin. Nm 1967, Viterbi a a ra thut toan giai ma ti u c goi lathuttoan Viterbi. T y, ma chp c ng dung rng rai trong nganh vin thng.

5.1.3 Kha nng phat hin va sa li cua ma khi

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a) Mi quan h gia khoang cach Hamming va kha nng phat hin va sa li

Ly thuyt ma a chng minh rng: khoang cach Hamming gia cac t ma trong mt b ma co

lin quan n kha nng phat hin sai va sa sai cua b ma o, cu th la:

1srd ++

trong o d la khoang cach Hamming, r la s li phat hin c, s la s li sa c, rs .Ta kim tra iu nay qua mt v du minh hoa sau y:

Gia s ta co b ma u M. M co 8 t ma nh sau:

Ky t A B C D E F G H

T ma 000 001 010 011 100 101 110 111

TM, ta lp b maM1

co khoang cach Hamming u la2. Nu chon t ma B (001) lam tma xut phat th b maM1bao gm 4 t ma sau:

Ky t B C E H

T ma 001 010 100 111

Goi 4 t ma trn la t ma dung va 4 t ma con lai la t ma cm.

Trong trng hp sai 1 li, ro rang cac t ma dung c truyn i se chuyn thanh cac t ma

cm bn thu. Cu th la B (001) chuyn thanh F (101), D (011), A (000); C chuyn thanhG(110), A (000), D (011); E chuyn thanh A (000), G (110), F (101); H chuyn thanh D(011), F (101), G (110). Luc nay co th d dang phat hin c li. Nu bn thu nhn c tma la A, co th kt lun la t ma truyn i b li nhng khng th kt lun c t ma nao (B,C hay E) a truyn i. Noi cach khac, khi s trng hp sai nhiu hn s t ma cm thkhng th phat hin c li. Trong trng hp sai 2 li, ta thy t ma dung nay se chuynthanh t ma dung khac nn khng th phat hin c li.

TM, ta lp b maM2 co khoang cach Hamming u la 3. Nu chon t ma B (001) lam tma xut phat th b maM2bao gm 2 t ma sau:

Ky t B G

T ma 001 110

Trong trng hp sai 1 li, ro rang cac t ma dung c truyn i se chuyn thanh cac t macm bn thu. Cu th la B (001) chuyn thanh F (101), D (011), A (000); G chuyn thanhC(010), E (100), H (111). Luc nay co th d dang phat hin c li va do s trng hp sai

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khng trung nhau va bng s t ma cm nn co th sa c li. Trong trng hp sai 2 li,ta thy t ma dung chuyn thanh t ma cm nhng trung vi t ma cm trong trng hp sai1 li ch phat hin c li ch khng sa c li.

Tom lai, t v du trn ta co th kt lun: nu khoang cach Hamming la 2 th co kha nng phathin c 1 li, nu khoang cach Hamming la 3 th co kha nng phat hin va sa c 1 li

va phat hin c 2 li. iu nay hoan toan ung chng minh trn v mi quan h giakhoang cach Hamming va kha nng phat hin va sa li cua ma.

Cung qua v du trn v b maM2 ta thy rng: vi s lng t ma trong b ma la 2 thc strong mi t ma ch co 1 bit tin. Nhng y chiu dai t ma la 3. Nh vy trong 3 bit o co2 bit d. "D" y hiu theo ngha la khng mang tin nhng c thm vao nhm muc chkim tra li. Phn sau ta se xet tng quat v mi quan h gia dai tng cng cua t ma vas bit tin.

b) Mi quan h gia dai tng cng cua t ma va s bit tin

tm ra mi quan h gia dai tng cng cua t ma va s bit tin, trc ht ta a ra khainim vector li e. Vector li la vector biu din v tr cac bit li xut hin trong t ma thu, quic bit khng li c biu din la0 va bit li c biu din la1.

V du t ma phat la1110010 va t ma thu la1100110. Luc nay vector li la e = 0010100

Goi dai tng cng cua t ma la: n; suy ra s t ma tng cng la: 2n

Goi s bit tin trong t ma la : k; suy ra s t ma dung la: 2k

Vy s t ma cm la: 2n - 2k

Goi E la s lng vector li, ta co:

n321E...EEEE ++++=

y Ei la vector li biu din trng hp sai i li.

)!in(!i

!nCE

i

ni ==

Vi mi t ma dung truyn i th ti a co th xay ra E trng hp li. Vy vi s t ma dungla2k th ti a co th xay ra Ex2k trng hp li. co th phat hin va sa ht tt ca cac linay th yu cu mi trng hp sai phai chuyn t ma dung sang mt t ma cm khac nhau,

noi cach khac, s trng hp sai khng c vt qua s lng t ma cm, ngha la:knk

222Ex .

Trong trng hp sai 1 li, ta co:

nEE1==

Vy quan h gia n vakphai thoa man bt ng thc sau:

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1n

22

nk

+

5.2 Ma khi

5.2.1 Ma kim tra chn le (parity)

y la loai ma khi n gian nht. Ma nay c dung ph bin trong truyn s liu dangASCII. Vi phng phap nay, mi ky t trc khi truyn i c thm vao mt bit chn le,goi labit parity (P). BitPc tnh toan da vao ky t phat sao cho tng s bit 1 trong ky t(k ca bitP) la s chn nu parity laloai chn ( even parity) va la s le nu parity laloai le(odd parity). Dung ma parity le se tranh c trng hp truyn t ma gm toan s 0, tuynhin, ma parity chn lai c dung ph bin hn.

Hnh 5.4 la mt v du minh hoa cho ma kim tra chn le. Bit parity chn la 1, bit parity le la0 vi ky t 1001001. Ty l ma la 7/8, mt mc d rt thp.

1 0 0 1 0 0 1 P

Hnh 5.4 V du ma parity

B0 B1 B2 B3 B4 B5 B6

P chn = 1P le = 0R = k/n = 7/8

7 bit tin 1bitkim tra

B6 B5 B4 B3 B2 B1 B0

P leP chn

Hnh 5.5 Mach tnh toan bit parity

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Khi nhn ky t, bn thu se thc hin tnh toan bit parity tng t nh bn phat va so sanh.Nu chung bng nhau th kt lun khng co li, nu khac nhau th kt lun co li.

Mach tnh toan bit parity cho ca bn phat va bn thu n gian la tp cac cng XOR nh trnhnh 5.5.

By gi ta xet n kha nng phat hin li cua ma parity. Gia s dung P chn, cac t ma mang

tin la 7 bit t 0000000 n 1111111, cac t ma lin tip trong b ma nay se la:0000000 0

0000001 1

0000010 1

..................

Ta thy khoang cach Hamming cua b ma nay la 2. Vy theo ly thuyt ma, ma nay ch phathin c 1 li. Tuy nhin, thc t ma nay phat hin c tt ca cac li n hay cac li xut

hin vi s li le, khng phat hin c cac li xut hin vi s li chn.5.2.2 Ma kim tra tng khi BCC(Block sum Check Character)

Khi truyn i mt khi ky t, mt ky t trong khi co th b li, va v vy coi nh khi o bli. Xac sut khi ky t b li goi laxac sut li khi (block error rate). Khi truyn i khi kyt, ta co th cai thin kha nng kha nng phat hin li cua ma parity bng cach khng chthm bit P cho ring tng ky t n le ma con thm tp cac bit P tnh trn ca mt khi hoanchnh.

Vi phng phap nay, mi ky t trong khi c thm vao mt bit P, goi la bit parity hang

(row parity), mi v tr cua bit trong khi c thm mt bit P goi labit parity ct (columnparity). Tp cac bit parity ct tao thanh ky t kim tra tng khi BCC.

Hnh 5.6 trnh bay mt v du dung parity chn cho hang va parity le cho ct.

B0 B1 B2 B3 B4 B5 B6 P

1 0 0 1 1 1 0 0

0 1 1 0 0 0 1 1

1 1 0 0 0 0 0 01 1 1 1 0 0 1 1

0 0 1 1 0 0 1BCC =

Cac bit parity ct

Cac

bit

Parityhang

Ni

dung

khung

1

Bit P cho BCC

Hnh 5.6 V du kim tra tng modulo-2

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Qua v du trn ta thy mc du hai bit li trong mt ky t khng c phat hin nh bit parityhang nhng se c phat hin nh bit parity ct. Ma nay khng phat hin c hai bit litrong cung ky t xay ra cung ct vao cung thi im (v du nh cac bit li xut hin cacv tr nh anh du trong hnh). Tuy nhin, kha nng nay rt t xay ra nn ma kim tra tng acai thin c kha nng phat hin li cua ma parity n. Nu xay ra li n th cn c vao bitP hang va P ct thu sai khac so vi P hang va P ct tnh, ta co th xac nh c v tr bit li,do o co th sa li.

Phng phap kim tra tng thng c dung trong trng hp s liu truyn i la mt mangky t ASCII.

Phng phap nay co mt bin th la s dung tng bu 1 tnh BCC thay cho tng modulo-2.Trong phng phap mi nay, xem cac ky t trong khi cn truyn nh la cac s nh phnkhng du. Trc ht, cng cac s nay lai dung thut toan bu 1, sau o ao ngc kt qua laitao thanh BCC. Bn thu tin hanh tnh tng bu 1 cua tt ca cac ky t trong khi (k ca BCC),nu khng co li xut hin th kt qua phai bng 0. Lu y rng, vi thut toan bu 1, bit nh

cui cung c quay vong ri cng vao tng ang co cho nn s 0 se c biu din hoc latoan s 0 hoc la toan s 1.

Phng phap mi nay c trnh bay qua v du trn hnh 5.7. T v du ta thy phng phapmi nay phat hin li tt hn phng phap tng modulo-2.

Bn phat Bn thu

0 1 1 1 0 0 1 0 1 1 1 0 0 1

1 0 0 0 1 1 0 1 0 0 0 1 1 0

0 0 0 0 0 1 1 0 0 0 0 0 1 1

1 0 0 1 1 1 1 1 0 0 1 1 1 1

1 0 1 0 0 0 1 = tng bu 1 0 1 0 1 1 0 1

0 1 1 1 1 1 1 0

1 0 1 0 0 1 0 1

1 1 1 1 1 1 1 = s 0 trong s bu 1

0 1 0 1 1 0 1 = BCC

ao bit

1

1

Hnh 5.7 V du kim tra tng bu 1Ma kim tra tng bu 1 thng c tnh toan bng phn mm, dung kim tra li cho cacban tin giao thc qua internet.

5.2.3 Ma khi tuyn tnh

a) V du v ma khi tuyn tnh

Hnh 5.8 minh hoa mt mach tao ma khi tuyn tnh (4, 7) gm 4 bit tin (I1 n I4) va 3

bit kim tra chn le (P1 n P3).

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Gia s dung parity chn, mi quan h gia cac bit tin va bit kim tra la:

3213

4212

4311

IIIP

IIIP

IIIP

=

=

=

Nu cac bit tin la I1 = 1, I2 = 0, I3 = 1, I4 = 1, cac bit P tnh c se la P1 = 1, P2 = 0 va P3 = 0.

I1 I2 I3 I4 P1 P2 P3

Hnh 5.8 Mach tao ma khi (4, 7)

Ta co th vit lai quan h gia cac bit tin va bit kim tra trong v du trn nh sau:

43211

43211

43211

xI0xI1xI1xI1P

xI1xI0xI1xI1P

xI1xI1xI0xI1P

=

=

=

T cac phng trnh quan h nay, ta rut ra ma trn kim tra H nh sau:

=

100:0111

010:1011

001:1101

H

Phn bn trai cua ng chm chm la cac h s cua cac bit tin I1 n I4, phn bn phaing chm chm la ma trn 3x3 co ng cheo la 1 (ma trn n v).

b) Ma trn sinh (generator matrix)

Ma trn sinh, ky hiu laG, la mt ma trn4 x 7, c tao ra bng cach kt hp mt ma trn

n v 4 x 4 vi ma trn hoan v cua ma trn bn trai ng chm chm cua H. Trong v dutrn, ma trn sinh la:

=

011:1000

101:0100

110:0010

111:0001

G

Nh ma trn sinh G, ta co th tnh toan c t ma bng cach nhn vector hang m biu din

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cho t ma mang tin vi G. Trong v du trn, day mang tin la 1011, t ma khi tuyn tnh (4,7)c tao ra la:

[ ] [1011100

0001110

0010101

0100011

1000111

1011 =

]

Ta nhn thy ma trn kt qua chnh la vector biu din cho t ma khi (4, 7) gm co haiphn: 4 bit bn trai la 4 bit tin I1 n I4, 3 bit bn phai chnh la 3 bit kim tra P1 n P3.

Theo cach lp ma nay, ta nhn thy khoang cach Hamming ti thiu la 3, do o ma nay sasai c 1 li.

c) Bang syndrome giai ma sa li

Syndrome la mt t ma c lp vi t ma phat va ch phu thuc vao day thu b li, ky hiu

vector syndrome la s. Bang syndrome la tp hp tt ca cac syndrome co th co. Goi c lavector biu din cho t ma khi (n, k) . Ta co quan h:

cGxm =

Goi e la vector li va r la t ma thu, ta co:

ecr =

Syndrome c tnh nh sau:TTTTTT

HxeHxe0HxeHxcHx)ec(Hxrs =====

Bang syndrome c tnh sn vi gia thit la truyn i t ma gm toan bit 0. V du bangsyndrome trong trng hp sai 1 li nh hnh 5.9:

Vector li Syndrome

0010000001

0100000010

10000001001100001000

1010010000

0110100000

1111000000

0000000000

Hnh 5.9 Bang syndrome hoan chnh cho tt ca cac li n

Nhn vao bang hnh 5.9 ta thy: khi khng co li syndrome la0, khi co li syndrome khac 0

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va cac syndrome khng ging nhau nn co th cn c vao syndrome bit v tr bit li, to sa c li. V du trn, gia s thu c t ma1011101, t ma ung sa c se lac nhsau:

[ ]

[ ] [ ] [1011100c001

001

010

100

110

101

011

111

1011101s

1011101r

==

=

=

]

Trng hp xut hin li chum, ngi ta s dung ky thut tao loan (interleaving): xao trn

th t cac bit tin trong ban tin ma hoa trc khi phat va sp xp lai sau khi thu, tach chumli thanh cac li n ri mi a n b giai ma.

5.3 Ma vong

5.3.1 c im cua ma vong

Nh a gii thiu trn (5.1.2), ma vong la mt lp con cua ma khi tuyn tnh. Ma vong cocac c im sau:

- Cu truc toan hoc cua ma vong cho phep kha nng sa li cao.

- Co th thc hin ma vong d dang bng phn cng, bng cac thanh ghi dch va cac cngXOR

- Dch vong mt t ma cung c mt t ma thuc cung b ma.

- Co th biu din ma vong bng a thc

- Co th tao ra t ma vong bng cach nhn modulo-2 vector mang tin vi a thc sinh. Lucnay ma vong c goi la ma vong khng h thng

5.3.2 Ma kim tra d vong CRC (Cyclic Redundancy Check)

Ma CRC la mt loai ma vong c s dung rng rai trn cac knh truyn ni tip bit phathin li (khng sa li). Trong CRC, mt tp bit kim tra c tnh toan cho mi khung tinda vao ni dung khung, sau o c gn thm vao ui khung truyn i. Bn thu thchin tnh toan tng t nh bn phat phat hin li. Cac bit kim tra goi laday kim trakhung FCS (Frame Check Sequence).Thut toan cu th nh sau:

a) Tnh toan tao ma CRC bn phat va kim tra li bn thu

Goi M(x) la a thc tin bc k-1, G(x) la a thc sinh bc r

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Thc hin phep chia cho G(x), se c:r

x)x(M

)x(G

)x(R)x(Q

)x(G

x)x(Mr

+= , vi Q(x) la thng s va R(x) la s d

T y suy ra:

)x(Q)x(G

)x(Rx)x(Mr

=+

t la a thc biu din cho t ma CRC phat. Ro rang la nu

khng co li xut hin th bn thu, sau khi chia t ma thu cho a thc sinh ta se c phn dla0.

)x(Rx)x(M)x(Tr

+=

b) V du

V du cn truyn i mt khung tin 8 bit 11100110 qua ng truyn s liu, s dung ma CRC phat hin li, a thc sinh s dung la11001.

T ma CRC c tao ra nh hnh 5.10 sau:

1 1 1 0 0 1 1 0 0 0 0 0 1 1 0 0 1

1 1 0 0 1 1 0 1 1 0 1 1 0

0 0 1 0 1 1 1

1 1 0 0 1

1 1 1 0 0

1 1 0 0 1

0 0 1 0 1 0 0

1 1 0 0 1

0 1 1 0 1 0

1 1 0 0 1

0 0 0 1 1 0D (FCS)

Hnh 5.10 V du tao ma CRC

Sau khi thc hin tnh toan nh trn, ta tm c t ma CRC la: 11100110 0110, trong o 8bit u la 8 bit tin va 4 bit sau la 4 bit kim tra.

Gia s tai bn thu, ta thu c t ma: 111001101111. Hnh 5.11 trnh bay vic thc hinphep chia a thc thu cho a thc sinh nh trn.

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1 1 1 0 0 1 1 0 1 1 1 1 1 1 0 0 1

1 1 0 0 1 1 0 1 1 0 1 1 0

0 0 1 0 1 1 1

1 1 0 0 1

1 1 1 0 01 1 0 0 1

0 0 1 0 1 1 1

1 1 0 0 1

0 1 1 1 0 1

1 1 0 0 1

0 0 1 0 0 1

D 0 nn phat hin c li

Hnh 5.11 V du giai ma CRC va phat hin li

Vic la chon a thc sinh rt quan trong v no xac nh cac kiu li co th phat hin. Mt athc sinh bc r co t nht 3 s 1 se phat hin c tt ca cac li n, tt ca cac li i, tt cacac li xay ra vi s le, tt ca cac li chum ngn hn r va hu ht cac li chum dai hn hocbng r. Sau y la mt vai a thc sinh thng dung trong thc t:

CRC - 16: G(x) = x16 + x15 + x2 + 1

CRC - CCITT: G(x) = x16

+ x12

+ x5

+ 1CRC - 32: G(x) = x32 + x26 + x23 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1

CRC - 16 va CRC - CCITT c dung rng rai trong WAN, CRC - 32 c dung trong huht cac LAN. Kha nng t sa li cua CRC thp nhng kha nng phat hin li rt cao nnthng c dung kt hp vi ARQ sa li.

c) Mach thc hin ma CRC

Hnh 5.12 trnh bay s thc hin CRC bn phat.

Ta xet cu th v du trn. S bit trong FCS la4 nn cn mt thanh ghi dch 4 bit (goi la thanh

ghi FCS) biu din x3, x2, x1va x0trong a thc sinh (goi la cac bit tch cc - active bit).Vi a thc sinh nay th cac sx3 va x0la 1 con x2va x1 la0. Trang thai mi cua x1 vax2 cthay bng trang thai trc o cua x0 vax1. Trang thai mi cua x0 vax3 c xac nh bi trangthai cua ng phan hi a XOR vi s trc o.

Mach hoat ng nh sau: trc tin xoa thanh ghi dch FCS va nap song song 8 bit u tintrong khung tin vao thanh ghi vao song song - ra ni tipPISO (Parallel Input - SerialOutput). Tn hiu iu khin phan hi la1.Theo tc cua ng h phat TxC, cac bit nayc dch ra ng truyn ln lt t msb n lsb. Cung luc nay, dong bit nay c XOR vi

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Tn hiu in khin phan hi(t1 thanh 0 sau 8 N TxC)

FCS

PISO

lsb msb

x0 x1 x2 x3

TxD

Nap song song tng byte trong N bytetrong khung tin

TxC

Thanh ghi PISO Thanh ghi FCSTxC

lsb msb X0 X1 X2 X3

0 0 1 1 0 0 1 1 1 0 0 0 0

1 0 0 1 1 0 0 1 1 1 0 0 1

2 0 0 0 1 1 0 0 1 0 1 0 0

3 0 0 0 0 1 1 0 0 1 0 1 1

4 0 0 0 0 0 1 1 0 1 1 0 0

5 0 0 0 0 0 0 1 1 0 1 1 0

6 0 0 0 0 0 0 0 1 1 0 1 0

7 0 0 0 0 0 0 0 0 1 1 0 0

8 0 0 0 0 0 0 0 0 0 1 1 09 0 0 0 0 0 0 0 0 0 0 1 1

10 0 0 0 0 0 0 0 0 0 0 0 1

11 0 0 0 0 0 0 0 0 0 0 0 0

Hnh 5.12 S mach tao CRC bn phat

x3 qua ng phan hi tr lai cac u vao chon loc cua FCS. Sau khi 8 bit u (byte u tin

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trong khung tin) i qua ht thanh ghi PISO, thu tuc nay c lp lai. Sau khi xut ra byte tincui cung trong khung tin, thanh ghi PISO c nap toan la s 0, tn hiu iu khin phan hit1 tr thanh 0, do o ni dung cua thanh ghi FCS - la cac bit kim tra - i theo sau khungtin phat ra ng truyn. Trong hnh 5.12, ta a gia s khung tin ch co1 byte (N = 1)

Thanh ghi SIPO Thanh ghi FCSRxC RxDlsb msb X0 X1 X2 X3

0 1 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 0 0 0

2 1 1 1 1 1 0 0

3 0 1 1 1 1 1 1 0

4 0 0 1 1 1 0 1 1 1

5 1 0 0 1 1 1 1 0 1 0

6 1 1 0 0 1 1 1 1 1 0 1

7 0 1 1 0 0 1 1 1 0 1 1 1

8 0 0 1 1 0 0 1 1 1 1 0 1 0

9 1 0 1 0 110 1 0 0 1 1

11 0 0 0 0 0

12 0 0 0 0

Hnh 5.13 S mach kim tra CRC bn thu

oc byte

FCS

SIPO

lsb msb

x0 x1 x2 x3

RxC

oc song song byte (xN)

RxD

D = 0

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Hnh 5.13 trnh bay s kim tra CRC bn thu. S liu thu RxD ln lt c dch vaothanh ghi vao ni tip - ra song song SIPO ( Serial Input - Parallel Output) gia mi bit.Cac bit RxD ln lt c XOR vi x3 va phan hi tr lai thanh ghi SIPO, mi khi nhn cbyte 8 bit, thit b iu khin se iu khin oc byte.

d) Giai ma ma CRC bng phng phap by li

Xet trng hp ma CRC c thanh lp vi s bit tin kva s bit tng cng n thoa man bt

ng thc:1n

22

nk

+ . V duk = 4, n = 7

Phn tch ra cac tha s nguyn t ri chon tha s bc r lam a thc sinh. V du:1xn

+

)1xx)(1xx)(1x(1x3237

+++++=+ .

Chon a thc sinh la hoc la1xx)x(G23

++= 1xx)x(G3

++=

Ma CRC trong trng hp nay co kha nng sa c li, cu th la1 li.

Y

N

M = 0

Dch vong T'(x) sang trai

Chia T'(x) cho G(x) - d laR'(x)

T'(x) = T'(x) + R'(x)

Dch vong T'(x) sang phai M ln

Tng Mln 1

W 1 ?

Tnh trong lng d W

End

Begin

Hnh 5.14 Thut toan sa li ma CRC bng phng phap by li

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Hnh 5.14 trnh bay thut toan sa li cho ma CRC bng phng phap by li. Goi T'(x) laa thc biu din cho t ma thu. Qua trnh sa li bt u bng vic dch vong T'(x) sangtrai, sau o chia cho G(x) tm s d R'(x). Vic dch vong se din ra nhiu ln cho n khitrong lng d nho hn hoc bng 1. Trong lu trn, ta dung bin M m s lndch vong trai. Sau o cng T'(x) vi R'(x), li c by va sa sau phep cng nay. Tuynhin, ta cn phai sa lai v tr cac bit trong T'(x) do nhng ln dch vong trai. Do o, phaidch vong T'(x) ngc lai sang phai vi s ln bng s ln trc o a dch vong sang trai (sln dch vong phai y chnh laM).

V du tin la 4 bit 1100, a thc sinh chon la 1xx23

++ . Thc hin ma hoa ta c t maphat la1100 101. Gia s bn thu thu c t ma1110 101. Kim tra li bng cach chia T'(x)cho G(x), ta thy phep chia con d nn kt lun co li. Vy tin hanh sa li nh sau:

Dch vong trai ln th nht c 1101011, chia cho G(x) d la011.

Dch vong trai ln th hai c 1010111, chia cho G(x) d la110.

Dch vong trai ln th ba c 0101111, chia cho G(x) d la001.Thc hin phep cng modulo-2: 0101111001 ta c 0101110, ri dch vong sang phai 3ln ta c lai t ma1100101 ging nh bn phat.

5.3.3 Ma Hamming

Ma Hamming la mt trng hp ring n gian nht cua ma vong. Ma Hamming cod = 3, cokha nng sa c 1 li.

Mt t ma Hamming c biu din di dang tng quat . y i la cac

bit tin vac la cac bit kim tra.

....iiciicicc8421

Cac bit c chnh la kt qua cua phep XOR gia tr ch v tr cua cac bit 1 vi nhau. Qua trnhkim tra li bn thu din ra tng t nh bn phat. Nu kt qua cua phep XOR la mt gia trkhac 0 th o chnh la v tr cua bit li.

V du xet kha nng sa li n cua ma Hamming (7, 11) trong trng hp t ma mang tin la1011101.

T ma Hamming co dang: .101c011c1cc8421

Cac bit 1 cac v tr: 3, 6, 9 va11. i cac s nay sang nh phn:

101111,10019,01117,01106,00113

Tnh XOR:

100110111001011101100011 = .

Vy t ma Hamming phat i la: 10100110101

Gia s bn thu, thu c t ma: 10000110101. i gia tr ch v tr cua cac bit 1 sang nhphn ri tnh XOR tng t nh bn phat:

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101111,10019,01117,01106,00011

3001110111001011101100001 =

T y xac nh c bit li la bit v tr th3. Vy t ma thu c sa lai la: 10100110101ging nh t ma phat.

5.4 Ma chp

5.4.1 Ma hoa ma chp

Nh trnh bay muc 5.1.2, ma chp c c trng bi ba s nguyn lan, kvaK. Ma chp(n, k, K) c xy dng t cac thanh ghi dch kK bit. y ta xet loai ma chp ph binnht la ma chp cok = 1. B ma hoa la thanh ghi dch K bit. u ra cua cac v tr trong thanhghi c la chon cng modulo-2 vi nhau. S lng b cng modulo-2 chnh lan. Mtb chuyn mach se ln lt ly mu mi u ra cua b cng modulo-2 theo nhp cua ng hthanh ghi dch.

Hnh 5.15 minh hoa mt b ma hoa ma chp vi k = 1, K = 3, n = 2.

a) Biu din ma chp bng a thc sinh

2

1

T0 T0 T0

Vao 1 1 0 1Ra 11 10 11 01

Hnh 5.15 V du b ma hoa ma chp ty l 1/2

Co th biu din b ma hoa ma chp bng cac a thc sinh. Mi a thc sinh biu din chomt b cng modulo-2. a thc sinh co bc 1K miu ta s kt ni gia u ra cua mtv tr trong thanh ghi dch vi b cng modulo-2. Theo v du trn, hai a thc sinh

la va2

1x1)x(G += x1)x(G

2+=

Gia s day tin vao b ma hoa la1100, day ma hoa se la11101101 . . ., ngha la ng vi mtbit tin vao co hai bit ma hoa ra. Do o, ty l ma la1/2.

nh ngha ap ng xung cua ma hoa la ap ng cua b ma hoa khi bit vao la1. Trong v dutrn, ap ng xung se la: 110110. Vi day vao la1101, ta thy day ra co th c tnh la chpday vao vi ap ng xung. Do o ma nay co tn la ma chp.

b) Biu din ma chp bng s cy

Hnh 5.16 trnh bay s cy biu din ma chp cho v du trn. Gia s ban u toan bthanh ghi c xoa v0. oc s cy theo phng ngang t trai qua phai, mi nhanh cy

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biu din mt t ma hai bit ra ng vi mt bit vao. Mi khi co bit vao la0, i sang nhanhphai tip theo pha trn, nu bit vao la1 th i sang nhanh phai tip theo pha di.

01

11

00

10

10

11

01

00

10

01

11

00

11

00

0

1

Hnh 5.16 S cy biu din b ma hoa ma chp hnh 5.15

Gia s day vao la110, i theo ng net m trn s cy, ta c day ra la111011.Nus bit vao la L th s nhanh trong s cy se la 2L. Nh vy, khi s bit vao tng th s cy rt cng knh.

c) Biu din ma chp bng s li

Nhn trong s cy ta thy thc t la b ma hoa ma chp ch co4 trang thai phn bit, kyhiu laa, b, c vad tng ng vi cac cp bit nh phn 00, 10, 01 va 11.

T s cy, ta thy: ln phn nhanh u tin, tao ra hai nut, ln phn nhanh th hai tao rabn nut va c sau mi ln phn nhanh s nut tng gp i. Sau ln phn nhanh th ba, ta thy

na trn va na di cua cy ging ht nhau. Nh vy, vao thi im ti nao o, hai nut bt kyco cung trang thai u co th kt hp vi nhau thanh mt nut. Ap dung iu nay cho s cy trn hnh 5.16, ta c s li trn hnh 5.17.

Cac nut trong li biu din trang thai cua b ma hoa. Cac nut cung hang biu din cungtrang thai. T mi nut li co hai nhanh ra: mt nhanh ng vi bit vao la0 (ng net lin),mt nhanh ng vi bit vao la1 (ng net t). Tng quat, sau ct nut thK, cu truc lic lp lai.

10

010101

101010

1111

11111111

0000

00000000

d

c

b

10

a

Hnh 5.17S li biu din b ma hoa ma chp hnh 5.15- 123 -

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5.4.2 Giai ma ma chp bng thut toan Viterbi

Khac vi ma khi co dai t ma c nh, ma chp khng co kch thc c thu. Tuy vy,ma chp cung b ep vao mt cu truc khi bng cach gn thm mt s bit 0 vao cui mt daytin am bao ui day tin c dch ht qua thanh ghi dch. Cac bit 0 nay khng mang tinnn ty l ma se nho hn k/n. gi cho ty l ma xp x vi k/n, chu ky gn thm bit 0

thng rt dai. Chng han trong v du trn y, sau 300 bit tin mi gn thm 2 bit 0. Vy ty lma la300/604 xp x 1/2.

Co ba kiu giai ma chp chnh la kiu tun t, ngng va Viterbi, trong o Viterbi la phbin nht.

Thut toan Viterbi da trn c s giai maln cn gn nht (nearest neighbour). Thut toantnh khoang cach Hamming (goi lametric) gia tn hiu thu vao thi im ti va tt ca cacng trong li dn n mi trang thai cung thi im ti. Khi hai ng cung dn n mttrang thai, chon ra ng co khoang cach Hamming ngn hn, goi lang sng (survivingpath). Vic chon ng sng c thc hin cho tt ca cac trang thai vao tt ca cac thi im.

Ta xet lai v du ma hoa ma chp hnh 5.15. Gia s day thu la1010001010, day vao b ma hoala5 bit, trong o co3 bit tin va2 bit 0 thm vao.

Trc ht ta xy dng li giai ma nh hnh 5.18.

1

1

0

0

2

1

1

1

212

010

12

1211

10

1011

0

a = 00

b = 10

c = 01

d = 11

Hnh 5.18 S li giai ma

Thc hin so sanh, chon ng co metric thp hn, cui cung ta con lai ng sng la ngin m (net t va net lin) trn hnh 5.19. T y suy ra day tin giai ma la: 11100

1

2

0

1

0

Hnh 5.19 ng sng va kt qua giai ma

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Trong thc t, b giai ma Viterbi gm co ba khi chnh. Th nht la khi tnh gia tr metricnhanh BMV (Branch Metric Value), th hai la khi tnh metric ng PMV (Path MetricVaue) - la tng cac metric nhanh doc theo mt ng trong li va th ba la khi xac nhu ra - chon ng co metric nho nht.

TOM TT CHNG1. ai lng o li thng thng laty l li bit BER hayxac sut li bit (Pb).

2. iu khin li nhm muc ch la lam giam ty l li trong mt h thng khi ty l nay lnqua mc cho phep. Nhn chung co nm phng phap iu khin li.

3. Giai phap u tin va d thy nht latng cng sut phat, nhng khng phai luc nao cungco th thc hin c.

4. Giai phap th hai, rt hiu qua trong vic chng lai li chum gy bi fading, la s dungphn tp. Co ba kiu phn tp chnh laphn tp khng gian, phn tp tn sva phn tpthi gian

5. Giai phap th ba la truyn song cng, hay con goi lakim tra echo

6. Phng phap th t i pho vi BER cao layu cu lp lai t ng ARQ Trong hthng ARQ, ma phat hin li c s dung bn thu kim tra li trong khi s liu thuva tra li cho bn phat trn mt knh hi tip. Tn hiu tra li lachp nhn ACKkhi sliu thu ung vakhng chp nhn NAKkhi s liu thu sai. Nu bn phat nhn NAK, bnphat phai tin hanh truyn lai khi s liu b li.

7. Phng phap th nm giam BER la thc hin ma hoa sa li khng phan hi FECC

(Forward Error Correction Coding).8. Ma hoa knh, con c goi la ma hoa iu khin li, c s dung phat hin va sa

cac ky t hay cac bit thu b li, bao gm ma hoa phat hin li vama hoa sa li khngphan hi FECC

9. Nhn chung, co th phn loai ma phat hin va sa li thanh ma khi vama chp

10. Ma khic c trng bi hai s nguyn n va k, va mt ma trn sinh hay a thc sinh.Ma khi tuyn tnh - con goi lama nhom - co cac t ma co tng ng 1-1 vi cac phn tthuc nhom toan hoc.

11. Ma kim tra chn le (parity)la loai ma khi n gian nht. Ma nay c dung ph bintrong truyn s liu dang ASCII

12. Ma vong (cyclic code) la mt lp con cua ma khi tuyn tnh khng co t ma gm toan s0. Mt ma khi tuyn tnh c goi la ma vong nu sau mt ln dch vong mt t ma thcung c mt t ma thuc cung b ma.

13. Ma CRCla mt loai ma vong c s dung rng rai trn cac knh truyn ni tip bit phat hin li. Trong CRC, mt tp bit kim tra c tnh toan cho mi khung tin da vao

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ni dung khung, sau o c gn thm vao ui khung truyn i. Bn thu thc hintnh toan tng t nh bn phat phat hin li.

14. Ma Hammingla mt trng hp ring n gian nht cua ma vong. Ma Hamming co khanng sa sai 1 li.

15. Ma chp cung c c trng bi hai s nguyn la n va k nh ma khi, nhng n bit ra cua

b ma hoa khng ch phu thuc vao k bit vao ma con phu thuc vao K-1 b k bit vaotrc o.

16. Ma chp (n, k, K) c xy dng t cac thanh ghi dch kK bit. Vy co th xem ma chpla ma co nh, o la im khac bit c ban cua ma chp so vi ma khi.

17. Co nhiu cach khac nhau biu din b ma hoa ma chp nh a thc sinh, s cy, s li.

18. Thut toan giai ma chp c dung rng rai nht lathut toan Viterbi. Cng vic c bannht trong thut toan Viterbi la la chon ng co metric nho nht va loai bo cac ng

khac.