Chuong1_Tong Quan He Thong So Viba

8/9/2019 Chuong1_Tong Quan He Thong So Viba

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1

Chng 1

Tng quan v h thng vi ba sGii thiu chng

Chng ny trnh by tng quan v cc vn sau:

+ Khi nim v c im chung ca cc h thng vi ba s

+ Phn loi cc h thng Vi ba s

+ Cc u, nhc im ca h thng Vi ba s

+ Cc mng Vi ba s im-im v im-nhiu im+ iu ch v gii iu ch

+ Phng php gim rng bng tn truyn trong h thng Vi ba s

+ Cc m truyn dn ph bin trong h thng

1.1 c im

Thng tin vi ba s l mt trong 3 phng tin thng tin ph bin hin nay (bn

cnh thng tin v tin v thng tin quang). H thng vi ba s s dng sng v tuyn

v bin i cc c tnh ca sng mang v tuyn bng nhng bin i gin on v

truyn trong khng trung. Sng mang v tuyn c truyn i c tnh nh hng rt

cao nh cc anten nh hng.

H thng Vi ba s l h thng thng tin v tuyn s c s dng trong cc

ng truyn dn s gia cc phn t khc nhau ca mng v tuyn. H thng Vi ba

s c th c s dng lm:

+ Cc ng trung k s ni gia cc tng i s.

+ Cc ng truyn dn ni tng i chnh n cc tng i v tinh.

+ Cc ng truyn dn ni cc thu bao vi cc tng i chnh hoc cc tng i

v tinh.

+ Cc b tp trung thu bao v tuyn.

+ Cc ng truyn dn trong cc h thng thng tin di ng kt ni cc my di

ng vi mng vin thng.

Cc h thng truyn dn Vi ba s l cc phn t quan trng ca mng vin

thng, tm quan trng ny ngy cng c khng nh khi cc cng ngh thng tin


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v tuyn mi nh thng tin di ng c a vo s dng rng ri trong mng vin

thng.

1.1 M hnh h thng vi ba s

Mt h thng vi ba s bao gm mt lot cc khi x l tn hiu. Cc khi ny

c th c phn loi theo cc mc sau y:

+ Bin i tn hiu tng t thnh tn hiu s

+ Tp hp cc tn hiu s t cc ngun khc nhau thnh tn hiu bng tn gc

+ X l tn hiu bng gc truyn trn knh thng tin

+ Truyn tn hiu bng gc trn knh thng tin

+ Thu tn hiu bng gc t knh thng tin

+ X l tn hiu bng gc thu

c phn thnh cc ngun khc nhau t

ng ng+ Bin i tn hiu s thnh cc tn hiu tng t tng ng

- Bin i ADC v DAC c th c thc hin bng mt trong cc phng php sau

y: iu v gii iu xung m (PCM); xung m Logarit (Log(PCM)); xung m vi

sai (DPCM); xung m vi sai t thch nghi (ADPCM); iu v gii iu delta (DM);

Delta t thch nghi (ADM).

- Tp hp cc tn hiu s t cc ngun khc nhau thnh tn hiu bng gc v phn

chia tn hiu s t tn hiu bng gc c thc hin nh qu trnh ghp-tch. C haih thng ghp-tch ch yu: theo thi gian TDM v theo tn s FDM. Trong FDM

Codec

ThoiTng t

Ngun s

ADCB

Ghp sMy Pht

Codec

ThoiTng t

Ngun s

BTch s

MyThu

FDM

FDM

ng truyn

Hnh 1.1 M hnh ca h thng vi ba s tiu biu

DAC


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c cc tp hp nhm, siu nhm, ch nhm hoc 16 siu nhm. FDM ca cc knh

m tn thng cn thit giao tip vi h thng truyn dn s (nh cc b Codec)

- Vic x l tn hiu bng gc thnh dng sng v tuyn thch hp truyn trnknh thng tin ph thuc vo mi trng truyn dn v mi mi trng truyn dn

c c tnh v hn ch ring. Vic xc nh s iu ch v gii iu ch thch hp

yu cu nhy ca thit b tng ng vi t l li bit BER cho trc tc

truyn dn nht nh, ph thuc vo phc tp cng nh gi thnh ca thit b.

1.2 Phn loi

Ph thuc vo tc bt ca tn hiu PCM cn truyn, cc thit b v tuyn phi

c thit k, cu to ph hp c kh nng truyn dn cc tn hiu . C th

phn loi nh sau:

+ Vi ba s bng hp (tc thp): c dng truyn cc tn hiu c tc

2Mbit/s, 4 Mbit/s v 8 Mbit/s, tng ng vi dung lng knh thoi l 30 knh, 60

knh v 120 knh. Tn s sng v tuyn (0,4 - 1,5)GHz.

+ Vi ba s bng trung bnh (tc trung bnh): c dng truyn cc tn hiu c

tc t (8-34) Mbit/s, tng ng vi dung lng knh thoi l 120 n 480 knh.

Tn s sng v tuyn (2 - 6)GHz.

+ Vi ba s bng rng (tc cao): c dng truyn cc tn hiu c tc t

(34-140) Mbit/s, tng ng vi dung lng knh thoi l 480 n 1920 knh. Tn

s sng v tuyn 4, 6, 8, 12GHz.

Hnh 1.2. S khi thit b thu pht vi ba s.

Giao tipnhnh iu ch Chuyn itn s Khuch icng sutX lbng tn

gc

Tch vghp knh Gii

iu chChuyn i

tn s

Khuch im thpKnh

nghip v

B lcnhnh

X l s X l tng t

Dao ngni

Dao ngni

LO

LO


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1.4 Mt su im ca h thng vi ba s

1. Nh cc phng thc m ho v ghp knh theo thi gian dng cc vi mch tch

hp c ln nn thng tin xut pht t cc ngun khc nhau nh in thoi, my tnh,

facsimile, telex,video... c tng hp thnh lung bit s liu tc cao truyn

trn cng mt sng mang v tuyn.

2. Nh s dng cc b lp ti sinh lung s liu nn trnh c nhiu tch lu trong

h thng s. Vic ti sinh ny c th c tin hnh tc bit cao nht ca bng

tn gc m khng cn a xung tc bit ban u.

3. Nh c tnh chng nhiu tt, cc h thng vi ba s c th hot ng tt vi t s

sng mang / nhiu (C/N)>15dB. Trong khi h thng vi ba tng t yu cu (C/N)

ln hn nhiu (>30dB, theo khuyn ngh ca CCIR). iu ny cho php s dng li

tn s bng phng php phn cc trc giao, tng ph hiu dng v dung lng

knh.

4. Cng mt dung lng truyn dn, cng sut pht cn thit nh hn so vi h

thng tng t lm gim chi ph thit b, tng tin cy, tit kim ngun. Ngoi ra,

cng sut pht nh t gy nhiu cho cc h thng khc.

1.5 Mt s khuyt im ca h thng vi ba s

1. Khi p dng h thng truyn dn s, ph tn tn hiu thoi rng hn so vi h

thng tng t.

2. Khi cc thng s ng truyn dn nh tr s BER, S/N thay i khng t gi tr

cho php th thng tin s gi n on, khc vi h thng tng t thng tin vn tn ti

tuy cht lng km

3. H thng ny d b nh hng ca mo phi tuyn do cc c tnh bo ho, do cclinh kin bn dn gy nn, c tnh ny khng xy ra cho h thng tng t FM

Cc vn trn c khc phc nh p dng cc tin b k thut mi nh

iu ch s nhiu mc, dng thit b d phng (1+n) v s dng cc mch bo v.

1.6 Cc mng vi ba s

Thng cc mng vi ba s c ni cng vi cc trm chuyn mch nh l mt

b phn ca mng trung k quc gia hoc trung k ring, hoc l ni cc tuyn

nhnh xut pht t trung tm thu thp thng tin khc nhau n trm chnh. (ng

dng trong cc trung tm chuyn mch hoc t chc cc mng Internet)


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1.6.1 Vi ba s im ni im

Mng vi ba s im ni im hin nay c s dng ph bin. Trong cc mng

ng di thng dng cp si quang cn cc mng quy m nh hn nh t tnhn cc huyn hoc cc ngnh kinh t khc ngi ta thng s dng cu hnh vi ba

s im-im dung lng trung bnh hoc cao nhm tho mn nhu cu ca cc

thng tin v c bit l dch v truyn s liu. Ngoi ra, trong mt s trng hp vi

ba dung lng thp l gii php hp dn cung cp trung k cho cc mng ni ht,

mng thng tin di ng.

1.6.2 Vi ba s im ni n nhiu im

Mng vi ba s ny tr thnh ph bin trong mt s vng ngoi v nng thn.

Mng bao gm mt trm trung tm pht thng tin trn mt an ten ng h ng phc

v cho mt s trm ngoi vi bao quanh. Nu cc trm ngoi vi ny nm trong phm

vi (bn knh) truyn dn cho php th khng cn dng cc trm lp, nu khong

cch xa hn th s s dng cc trm lp a tn hiu n cc trm ngoi vi. Ty, thng tin s c truyn n cc thu bao. Thit b vi ba trm ngoi vi c th

t ngoi tri, trn ct.v.v... mi trm ngoi vi c th c lp t thit b cho nhiu

trung k. Khi mt cao c th b sung thm thit b; c thit k hot ng

trong cc bng tn 1,5GHz -1,8GHz v 2,4GHz s dng mt sng mang cho h

thng hon chnh.

Hin nay cc h thng im ni n a im 19GHz c ch to v lp

t Chu u cung cp cc dch v s liu (Kbit/s) Internet trong mng ni ht

RX/TX

RX/TX

MUX/ DEMUX

MUX/ DEMUX

Hnh 1.3 M hnh ca h thng vi ba s im ni im tiu biu.


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khong cch 10Km. Trm trung tm pht tc bit khong 8,2Mb/s v a ch mi

trm li s dng k thut TDMA.

1.7. iu ch v giI iu ch s

1.7.1. iu ch s

iu ch s l phng thc iu ch i vi tn hiu s m trong 1 hay

nhiu thng s ca sng mang c thay i theo sng iu ch. Hay ni cch khc,

l qu trnh gn tin tc (sng iu ch) vo mt dao ng cao tn (sng mang)

nh bin i 1 hay nhiu hn 1 thng s no ca dao ng cao tn theo tin tc.

Thng qua qu trnh iu ch s, tin tc vng tn s thp s c chuyn ln vng

tn s cao c th truyn i xa.

Hnh 1.4. S m t qu trnh iu ch v gii iu ch s.

Gi s c 1 sng mang hnh sin nh sau:

RX/TX

MUX/

DEMUX

MW

MW

MW

MW

Trung kNi ht

TX/RX

Trm trung tm

Trm ngoi vi 3

Trm ngoi vi 1

Trm ngoi vi 2

RX/TX

MUX/

DEMUX

RX/TX

MUX/

DEMUX

Hnh 1.3 M hnh ca h thng vi ba s im ni im tiu biu.

Tn hiu bng tnv tuyn

My thuMy pht

Tn hiu bngtn gc

Tn hiu bngtn gc

B iu ch

Sng mang

B gii iu ch


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)cos(.)( 00 += tAtf (1.1)

Trong :

+ A : bin ca sng mang

+ o = 2..fo : tn s gc ca sng mang

+ fo : tn s ca sng mang

+ (t) : pha ca sng mang

Tu theo tham s c s dng mang tin: c th l bin A, tn s fo, pha

(t) hay t hp gia chng m ta c cc kiu iu ch khc nhau:

+iu ch kha dch bin ASK (Amplitude Shift Keying): Sng iu

bin c to ra bng cch thay i bin ca sng mang tu thuc bng gc.

Sng iu bin c to ra bng cch nhn sng cao tn hnh sin vi bng gc.

+iu ch kha dch tn s FSK (Frequency Shift Keying): Sng iu bin

c to ra bng cch thay i tn s sng mang theo bin tn hiu bng gc.

+iu ch kha dch pha PSK (Phase Shift Keying): : Sng iu bin c

to ra bng cch thay i pha sng mang theo bin tn hiu bng gc.

+iu ch bin v pha kt hp hay iu ch cu phng QAM

(Quadrature Amplitude Modulation).

1.7.2. Gii iu ch s

Gii iu ch l qu trnh ngc li vi qu trnh iu ch, trong qu trnh

thu c c mt trong nhng tham s: bin , tn s, pha ca tn hiu sng mang

c bin i theo tn hiu iu ch v tu theo phng thc iu ch m ta c cc

phng thc gii iu ch thch hp ly li thng tin cn thit.

1.7.3. Cc phng thc iu ch v gii iu ch sHin nay hu ht cc thit b vi ba s u s dng phng php iu ch pha

(PSK) v iu ch cu phng (QAM), do vy chng ny trnh by v hai loi iu

ch ny.

1.7.3.1. Phng thc iu ch PSK

C s ton hc

PSK l phng thc iu ch m pha ca tn hiu sng mang cao tn bin i

theo tn hiu bng tn gc.Biu thc tn hiu sng mang: )cos()( 00 += ttf


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Biu thc tn hiu bng gc: s(t) l tn hiu dng nh phn (0,1) hay l mt

dy NRZ (Non-Return Zero).

Khi , tn hiu iu pha PSK c dng: }2/]).([cos{)( 0 ++= tsttP (1.2)

Trong : = 2/n l s sai pha gia cc pha ln cn ca tn hiu.

Biu din tn hiu theo kiu cu phng:

}2/]).([cos{)( 0 ++= tsttP

)sin(}.2/]).(sin{[)cos(}.2/]).(cos{[ 00 ++= ttstts

t

=

=

}2/]).(sin{[)(

}2/]).(cos{[)(

tstb

tsta

)sin().()cos().()( 00 +++= ttbttatP (1.3)

Vy, tn hiu iu pha l tng ca hai tn hiu iu bin vung gc nhau.

1.7.3.1.1. iu ch pha 2 mc 2-PSK

T biu thc (4.2), vi n = 2, = th ta c kiu iu ch 2-PSK hay cn gi

l PSK nh phn BPSK. Tn hiu 2-PSK c dng:

}

2

).(cos{)( 0

tsttP ++= (1.4)

iu ch

Tn hiu bng gc s(t) l xung NRZ lng cc v s iu ch ny s dng

mt trong hai pha lch nhau 180o v c gi l PSK nh phn (BPSK).

Sng mang

Xung vo

Dng sng iu ch

1

0

-1

1

0

-1

1

0

-1

t

t

t

Hnh 1.5. Tn hiu 2PSK


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+Vi cc bit 1: }2

cos{)( 01

++= ttP

+Vi cc bit -1: }2cos{)( 01 += ttP

Nh vy, bin ca ca tn hiu BPSK khng i trong qu trnh truyn dn,

nhng b chuyn i trng thi.

Hnh 1.6. Biu vector BPSK,2

=

Gii iu ch

Tn hiu 2-PSK c tng hp vi sng mang chun thng qua b lc thng

thp loi b thnh phn hi bc cao cho ta thu c tn hiu ban u.

Hnh 1.7. S nguyn l gii iu ch tn hiu 2-PSK.

Pha ca tn hiu sng mang chun bng vi pha ca tn hiu thu nhn c,

nn nu tn hiu thu l:

ttsttP 00 sin).(.2)2

cos(.2)(

== vi s(t) = 1 (1.5)

th tn hiu chun l: t0sin.2 v tn hiu gii iu ch l: s(t).

1.7.3.1.2. iu ch pha 4 trng thi 4-PSK

T biu thc (4.2), vi n = 4, = /2 th ta c kiu iu ch 4-PSK hay PSK

cu phng (QPSK). Tn hiu 4-PSK c dng:

}4

).(cos{)( 0

tsttP ++= (1.6)

Tn hiu bng gc s(t) l xung NRZ lng cc nhn 4 gi tr.

coso.t

sino.t

-1 1

LPF s(t)

Sng mang chun

BPSK


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iu ch

S nguyn l b iu ch 4-PSK s dng mt trong 4 pha lch nhau 90o,

c gi l 4-PSK hay PSK cu phng (QPSK).

Hnh 1.8. S nguyn l iu ch tn hiu QPSK.

Tn hiu bng gc c a vo b bin i ni tip thnh song song, u ra

c hai lung s liu c tc bit gim i mt na, ng thi bin i tn hiu n

cc thnh tn hiu 1. Hai sng mang a ti hai b trn lm lch pha nhau 90o.

Tng hp tn hiu u ra 2 b trn ta c tn hiu 4-PSK.

Tn hiu ra 2 b trn:

ttatM 01 cos).()( = ttbtM 02 sin).()( = vi a(t) = 1, b(t) = 1.

Tn hiu ra 4-PSK l: ttbttatP 00 sin).(.cos).()( += (1.7)

s(t)B quaypha 90o P(t)

Sng mang chun f0(t) = cos0t

b(t) = 1

a(t) = 1

SPC

a(t)

b(t)

P(t)

t

t

t

1

0

-1

1

0

-1

1

0

-1

Hnh 1.9. Tn hiu 4PSK

-1-1 1-1

11-11

Hnh 1.10 Biu vectorca iu ch QPSK


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Gii iu ch

Hnh 1.11. S nguyn l gii iu ch pha 4-PSK.

Gi s tn hiu thu c l: )](4

cos[.2)( 0 tttP

++=

ttbtta 00 sin).(cos).( +=

Vi (t) = n/2; n = 0,1,2,3. V a(t) = 1, b(t) = 1.

Hai tn hiu chun vo b trn:

)2

cos(.2)( 01

n

ttPref +=

)2

sin(.2)( 02

n

ttPref +=

Tn hiu sau khi qua cc b lc:

2

1

2

)(]

24)(cos[)(1 ==+=

tanttPLPF

(1.8.a)

2

1

2

)(]

24)(sin[)(2 ==+=

tbnttPLPF

(1.8.b)

1.7.3.1.3. iu ch pha 8 trng thi 8-PSKT biu thc (3.4), vi n = 8, = /4 th ta c sng iu ch 8-PSK. Tn hiu

8-PSK c dng: ]8

).(cos[)( 0

tsttP ++= (1.9)

Tn hiu bng gc s(t) nhn 8 gi tr.

iu ch

B iu ch 8-PSK l s kt hp tn hiu ca 2 b iu ch 4-PSK. Sng mang

ca 2 b iu ch cos mt s sai pha 45o

. Mt b m ho bin i tn hiu c tora t tn hiu bng gc s(t) sau khi i qua b SPC thnh cc tn hiu iu ch.

P(t)B quaypha 90o

Sng mang chun f0(t) = cos0t


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Vi (t) = n/2; n = 0,1,2,...,7. V a(t) = 1, b(t) = 1.

Tn hiu chun vo b trn:

)cos(.2)( 01 Rref ttP += , vi:8

2)( ntR =

Tn hiu c gii iu ch sau khi qua cc b lc thng thp :

]4

3

8)(cos[)(1

+=

RLPF ttP (1.20.a)

]4

2

8)(cos[)(2

+= RLPF ttP (1.20.b)

]48)(cos[)(3

+= RLPF ttP (1.20.c)

]8

)(cos[)(4 RLPF ttP

+= (1.20.d)

Sau b lc thng thp l cc b so snh nhm xc nh 4 tn hiu nh phn. Cc

mch logic to ra 3 tn hiu nh phn t 4 ng vo bng cc x l logic thch hp.

Nhn xt:

+Khi s pha tng ln th tc bit gim, iu ny s lm gim bng thng,

tit kim c ng truyn dn, cho php truyn c nhiu knh thng tin.

+Tuy nhin, khi s pha tng ln cc t hp bit s cng gn nhau hn, ngha

l tng kh nng mc li ca h thng.

Do vy, trong thng tin s tc cao s trng thi pha nhiu, gim kh

nng mc li c th s dng phng php iu ch bin cu phng QAM.

1.7.3.2. iu ch bin cu phng QAM

iu ch bin cu phng QAM l phng php iu ch kt hp gia iu

ch bin ASK v iu ch pha PSK. Trong phng thc iu ch ny, ta thc

hin iu ch bin nhiu mc 2 sng mang m 2 sng mang ny c dch pha 1

gc 90o. Tn hiu tng ca 2 sng mang ny c dng va iu bin va iu pha:

)](.cos[).()( 11 tttatQ o += v )](.sin[).()( 22 tttbtQ o +=

Tn hiu s(t) l tng ca 2 thnh phn ss(t) v sc(t) v c biu din nh sau:

)()()(21tQtQtQ += )](.sin[).()](.cos[).( 21 tttbttta oo +++= (1.21)

Nh c bin thay i m cc trng thi pha ca sng mang cch xanhau, do vy kh nng mc li s gim, y cng chnh l u im ca QAM.


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iu ch

Hnh 1.15. S nguyn l iu ch tn hiu M-QAM.

B chuyn i SPC chuyn i tn hiu iu ch vo thnh 2 chui tn hiu

NRZ song song. B bin i 2/L c chc nng chuyn i chui NRZ thnh chui

tn hiu c ML = mc. Vi L = 4 th M = 16, ta c iu ch 16-QAM, v vi L =

8 th M = 64, ta c iu ch 64-QAM.

Hnh 1.17. Biu khng gian tn hiu QAM nhiu trng thi.

2/L LPF

2/L LPF

SPC B quaypha 90o

s(t)

Sng mang

Tn hiu

M-QAM

L2 - QAM

Ccmc

Cc mc

Hnh 1.16. Biu khng gian tn hiu 16QAM


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Gii iu ch

Tn hiu M-QAM vo: ttbttatQ 00 sin).(cos).()( +=

Tn hiu chun: ttQref 01 cos.2)( = v ttQref 02 sin.2)( =

Sau khi loi b thnh phn hi bc cao cc b lc thng thp ta s c:

)()(1 tatQLPF = v )()(2 tbtQLPF =

Hnh 1.18. S nguyn l gii iu ch M-QAM.

Bin ca tn hiu gii iu ch c L = M mc, trong M l s trng thi

tn hiu. Tn hiu L mc c bin i bi b bin i ADC thnh n/2 tn hiu 2

mc, trong L = 2n/2

v M = 2n

. Vi 16-QAM th n = 4, L = 4 v vi 64-QAM thn = 6, L = 8. T n tn hiu ny, b bin i PSC s to nn tn hiu gii iu ch.

1.8 Gim rng bng tn truyn bng phng php iu ch

nhiu mc.

Theo nh l Nyquist: rng bng tn ca knh truyn(B ) (knh thng

thp) phi ln hn hoc bng tc k hiu chia 2 )2

( Sr

khng c hin tng

giao thoa gia cc k hiu.2

SrB (1.22)

Trong h thng PCM, bfr SS = (1.23)

bfS , : ln lt l tn s ly mu, s bit trong t m.

Thay (1.23) vo (1.22) ta c biu thc v rng bng tn cn thit ca knh

truyn trnh hin tng giao thoa gia cc k hiu nh sau:

22

bfr

BSS

= (1.24)

Q(t)

Sng mang chun

LPF

LPF

B quay

pha 90o

ADC

ADC

s(t)

PSC


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Gi s ta s dng phng php iu ch pha M trng thi. Lc tc k hiu

gim M2log ln. Do , rng bng tn cn thit ca knh truyn cng gim

M2log ln so vi iu ch nh phn hai mcnhbiu thc:

M

bfB S

2log2 (1.25)

V d: M ho PCM mt knh thoi KHzfS 8= vi s bit trong t m: b = 8bit th

bng tn ti thiu l:2

8.8

22min ===

bfrB SS = 32KHz. Trong khi , phng php

truyn dn tn hiu tng t yu cu bng tn thoi 3,1KHz (0,3-3,4) KHz. Suy ra,

phng php truyn dn tn hiu s c bng tn xp x 10 ln so vi phng php

tng t. Nu s dng phng php iu ch 16-PSK c M=16 mc th bng thng

yu cu gim 416loglog 22 ==M ln v tng ng 8 KHz.

1.9 Cc m truyn dn

Nu cng cc s liu c truyn i lin tc, li c th pht sinh khi nhn

chng. V th vic phc hi s liu cc k kh khn . Do , cc tn hiu nh phn t

thit b ghp knh c bin i thnh cc m truyn dn gim li tn hiu trong

qu trnh truyn.

t c iu , cc m truyn dn phi tho mn cc yu cu sau y:

+ Phi phi hp c tnh ph ca tn hiu vi c tnh ca knh truyn.

+ m bo cc dy bit phi c lp thng k vi nhau gim lng trt, gim s

ph thuc mu do cc mu lp gy ra.

+ D dng tch c xung ng h v ti sinh tn hiu

+ m bo d

cn thit gim st li truyn dn v pht hin

c s c cathit b.

+ Phi duy tr d tha thng tin mc thp c th c gim tc bt v

gim rng bng tn tn hiu

+ Gim thnh phn mt chiu ca tn hiu n mc bng 0.

+ Gim cc thnh phn tn s thp gim xuyn m v kch thc ca b phn v

cc linh kin trong mch. Tn hiu nh phn n cc c thnh phn mt chiu, c

cha nng lng ln trong trong ph tn thp v vy khng thch hp cho vic truyn


16/17

17

dn. Trong thc t ngi ta thng s dng cc m lng cc chng hn nh m

truyn dn HDB3 (m nh phn mt cao c cc i 3 s 0 lin tip), CMI...

1.9.1 Cc m ng truynTrong h thng truyn dn thng tin Vi ba thng s dng cc loi m HDB3,

CMI, v do vy ta ch xem xt 2 loi m ny.

M HDBN (High Density Binary with maximum of 3 consecutive

Zeros)

M HDBN l m lng cc mt cao c cc i N s 0, y l loi m ci

tin ca m AMI thc hin vic thay th N+1 s 0 lin tip bng N+1 xung nhp

cha xung phm lut V v xung phm lut ny s ti bit th N+1 ca cc m s 0

lin tc.

Vi loi m HDBN ny th dng HDB3 thng c s dng trong h thng

truyn n thng tin vi ba s.

M HDB3

M HDB3 l m lng cc mt cao c cc i l 3 s 0 lin tip.

Quy tc m ho:

+Mc logic 1 c m ho theo mc lng cc.

+Mc logic 0 c m ho theo trng thi 0 thng thng.

+i vi dy 4 s 0 lin tip th c m ho theo mt trong 2 trng hp

sau: OOOV hoc BOOV sao cho s bit B gia 2 bit Vl l.

Hnh 1.19. Dng sng HDB3.

M ny kh thng dng v ITU-T khuyn ngh s dng tc bit

2,048Mbps; 8,448Mbps; 34,368Mbps theo tiu chun chu u (khuyn ngh G-

703).

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

B 0 0 0 V B B B 0 0 V 0

t

t

t

Gi trnh phn

Quy lutm ho

Tn hiuHDB3

+V

-V

0


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M CMI (Code Mark Inversion)

M CMI l m o du m, y chnh l loi NRZ 2 mc.

Quy tc m ho:+Mc logic 0 c m ho thnh cc sng vung dng - m hoc m -

dng nhng mi mc ch chim 1 khong thi gian T/2.

+Mc logic 1 c m ho thnh cc sng vung dng - dng hoc m -

m nhng mi mc ch chim 1 khong thi gian T theo lut lun phin.

M CMI c ITU-T khuyn ngh s dng tc bit 140Mbps theo tiu

chun chu u (khuyn ngh G-703).

Ngoi ra, cn nhiu m khc nh: m Wal1, m Wal2, m Manchester, mchui, m 5B6B,... tuy nhin chng khng c s dng thng dng.

Theo khuyn ngh G703 v cc giao tip ca CCITT cho chi tit tr khng,

loi i dy dn mc tn hiu dng khung, ti khung phn b cng nh m truyn

dn nhng tc bit khc nhau dng cho h Chu u.

Bng 1.2 M truyn dn dng trong vi ba s

Tc bit (Mb/s) 2.048 8.448 34.368 139.246

Loi cp S/C C C C

Tr khng() 120/75 75 75 75

M ng HDB3 HDB3 HDB3 CMI

Dng xung chun Vung Vung Vung Vung

S: cp i xng. C: Cp ng trc.

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