giao trinh tin hieu so

8/6/2019 Giao Trinh Tin Hieu So

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HC VIN CNG NGH BU CHNH VIN THNG

GHP KNH TN HIU S

(Dng cho sinh vin ho to i hc txa)

Lu hnh ni b

H NI - 2007


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HC VIN CNG NGH BU CHNH VIN THNG

GHP KNH TN HIU S

Bin son : TS. CAO PHN

THS. CAO HNG SN


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1

LI NI U

Ghp knh tn hiu s l mt lnh vc rt quan trng. Khi u ca ghp knh tn hiu sl iu xung m (PCM) v iu ch Delta (DM), trong PCM c s dng rng ri hn. TPCM, cc nh ch to thit b vin thng cho ra i thit b ghp knh cn ng b (PDH) v

sau l thit b ghp knh ng b (SDH). Mng thng tin quang SDH mra mt giai onmi ca cng ngh truyn thng nhm p ng nhu cu tng trng rt nhanh ca cc dch vvin thng, c bit l dch v Internet.

Vi tc bit hin ti ca SDH l 10 Gbit/s vn cha p ng mt cch y cho truynlu lng Internet , ang v s pht trin theo cp s nhn. V vy cng ngh ghp knh theo

bc sng (WDM) xut hin. c th tn dng bng tn truyn dn ti min ca s th haica si quang n mode, k thut ghp cht cc bc sng DWDM ang ng vai tr quan trngtrn mng thng tin quang ton cu.

Tuy nhin, thng tin quang SDH l cng ngh ghp knh cnh. V vy rng bng tnvn khng c tn dng trit . Theo c tnh th hiu sut s dng rng bng tn kh dngca h thng thng tin quang SDH mi t c 50%. Trc thc t mt mt rng bng tnng truyn cn b lng ph, mt khc cng ngh truyn gi IP v ATM i hi h thng thngtin quang SDH phi tho mn nhu cu trc mt v c cho tng lai, khi m cc dch v gia tng

pht trin trnh cao. Ch c th tho mn nhu cu v tc truyn dn v nng cao hiu suts dng bng tn ng truyn bng cch thay i cc phng thc truyn ti lu lng s liu.

Vn mu cht ng dng cc phng thc truyn ti tin tin l kt chui cc ccconten, s dng cc phng thc ng gi s liu thch hp, truyn ti gi linh hot theo cch tis dng khng gian v chuyn mch bo v thng minh nng cao tin cy ca mng v rtngn thi gian phc hi ca h thng khi c s c. Nhng vn ny sc phn tch k trong

cc ch sau y:1) Trnh by mt s khi nim cbn trong truyn dn tn hiu, c bit l tn hiu s v

cc phng php ghp knh s.2) Cc phng php duy tr mng. Ni dung ch yu ca chuyn ny l cc phng

php chuyn mch bo v mng ng thng v mng vng SDH.3) Cc chun Ethernet, mng vng th bi v FDDI.4) Cc phng thc truyn ti s liu bao gm cc phng thc ng khung s liu, kt

chui, iu chnh dung lng tuyn, cc giao thc ti s dng khng gian v.v.Sau mi chng c cc bi tp hoc cu hi sinh vin t kim tra v nh gi kin thc ca

mnh khi i chng vi p s v tr li trong phn ph lc.Ti liu ging dy ny c bin son theo cng mn hc "Ghp knh tn hiu s"ca chng trnh o to i hc chnh quy hin nay ca Hc vin Cng ngh Bu chnh Vinthng. Tuy nhin, y l ln bin son u tin nn khng trnh khi thiu st v ni dung v hnhthc. Rt mong cc c gi gp ti liu ngy cng hon thin hn.

kin ng gp ca cc c gi xin vui lng gi trc tip cho Phng o to i hc txa Hc vin Cng ngh Bu chnh Vin thng.

Xin chn thnh cm n!

Nhm tc gi


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CHNG I

MT S KHI NIM CBN

TRONG TRUYN DN TN HIU

1.1. GII THIU CHUNGTrong chng ny gii thiu cc ni dung chnh sau y:

- Mt s khi nim cbn trong truyn dn tn hiu s.

- Cc phng php s ho tn hiu analog nh: PCM, DPCM v DM. Trong phngphp PCM c s dng rng ri trong cc h thng ghp knh PDH.

- Cc phng php ghp knh: im qua cc phng php ghp knh theo tn s, theotn s trc giao, theo thi gian, theo m, ghp knh thng k v.v. trong ghp knh theo thigian c s dng trong ghp knh PDH, SDH.

- ng b trong vin thng:

tin hnh phn tch cc phng thc ng b nh: ng b sng mang, ng b khiu, ng b bit, ng b khung, ng b gi, ng b mng, ng ba phng tin v ng

bng h thi gian thc. Tu thuc vo tng trng hp c th m s dng mt trong ccphng thc ng b hoc s dng ng thi mt s phng thc ng b. Chng hn trongmng thng tin quang SDH s dng cng b mng, ng b sng mang, ng b khung, ng

b k hiu.

1.2. NHP MN GHP KNH S

1.2.1. Tn hiu v cc tham s

1.2.1.1. Cc loi tn hiu(1) Tn hiu analog: tn hiu analog (tng t) l loi tn hiu c cc gi tr bin lin

tc theo thi gian, th d tn hiu thoi analog.

Mt dng in hnh ca tn hiu analog l sng hnh sine, c th hin di dng:

S(t) = Asin (t + )

trong : A l bin tn hiu, l tn s gc ( = 2f, f l tn s), l pha ca tn hiu.

Nu tn hiu l tp hp ca nhiu tn s th ngoi cc tham s trn y cn c mt tham skhc, l di tn ca tn hiu.

(2) Tn hiu xung: tn hiu xung l loi tn hiu c cc gi tr bin l hm ri rc cathi gian. in hnh ca tn hiu xung l tn hiu xung ly mu tn hiu analog da vo nh l lymu.

(3) Tn hiu s: y cng l loi tn hiu c cc gi tr bin l hm ri rc ca thi giannh tn hiu xung. Tuy nhin, khc vi tn hiu xung ch bin ca cc xung bng 0 hoc 1,mt khc tp hp ca mt nhm xung i din cho mt ch s, hoc mt k t no . Mi mtxung c gi l mt bit. Mt vi loi tn hiu sin hnh nh: tn hiu 2 mc (0 v 1), cn ctn l tn hiu xung nh phn hay tn hiu xung n cc; v tn hiu ba mc (-1, 0 v +1), cnc gi l tn hiu xung tam phn hay tn hiu xung lng cc.

(4) Tn hiu iu bin xung, iu tn xung hoc iu pha xung: y l trng hp m sngmang xung ch nht c bin , hoc tn s, hoc pha bin i theo quy lut bin i ca bin tn hiu iu ch. Ba dng tn hiu ny thng c s dng trong mng thng tin analog.


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1.2.1.2. Cc tham sca tn hiu

(1) Mc in

Mc in tng i: ( )0

log10P

PdBL x=

trong : Px l cng sut tn hiu (mW) ti im cn xc nh mc in, P0 l cng sut tn hu tiim tham kho (mW).

Mc in tuyt i: ( )W1

log10m

PdBL xm =

L(dB)m= 0 dBm khi cng sut ti im x bng 1 mW, L(dBm) > 0 khi cng sut tn hiu ti im xln hn 1 mW, L(dBm) < 0 khi cng sut tn hiu ti im x b hn 1 mW.

(2) T s tn hiu trn nhiu

( )n

s

P

PdBSNR log10=

n

s

n

s

I

I

V

Vlog20log20 ==

trong : Ps, Vs, Is tng ng l cng sut, in p v dng in tn hiu; Pn, Vn, In tng ng lcng sut, in p v dng in nhiu.

1.2.2. ng truyn v rng bng tn truyn dn

1.2.2.1.ng truyn

L mi trng truyn dn c s dng truyn ti tn hiu, th dng truyn cpkim loi, ng truyn cp si quang, ng truyn Radio, v.v. ng truyn cn c phn chiathnh tuyn (Path), knh v.v.

1.2.2.2. rng bng tn truyn dn

Mun o rng bng tn truyn dn ca tn hiu no phi cn c vo cc quy nhsau y:

(1) rng bng tn in (BW)e

rng bng tn in l bng tn t tn s tn hiu bng zero n tn s tn hiu m ti p ng ca tn hiu (h s khuch i, in p, dng in) gim cn 0,707 so vi gi tr cc ica p ng tn hiu(hnh 1.1).

Hnh 1.1- rng bng tn in

(2) rng bng tn quang (BW)o

rng bng tn quang l bng tn t tn siu ch bng zero n tn siu ch mti mc cng sut quang gim 50% (3dBm) so vi cng sut quang cc i, nh minh hohnh 1.2.

1

0,707

f0

(BW)e

fmax

V/Vmax


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Hnh 1.2. rng bng tn quang

1.2.3. Truyn dn n knh v a knh

Truyn dn n knh v a knh c ng l h thng truyn dn quang c mt hay nhiubc sng. Th d: h thng thng tin quang thng thng ch c mt bc sng ti 1310 nmhoc 1550 nm; trong khi , h thng thng tin quang ghp bc sng (WDM) c th truyn ngthi hng chc bc sng khc nhau nm trong min ca s th hai (1300 nm) hoc ca s th ba

(1550 nm) ca si quang n mode.1.2.4. H thng truyn dn s v cc tham s

1.2.4.1. Hthng truyn dn s

H thng truyn dn s bao gm h thng truyn dn cp si quang v h thng truyndn vi ba s. H thng truyn dn vi ba s l h thng a im ng thng. H thng truyn dns cp si quang c th s dng cu trc ng thng, vng hoc hn hp. Cc cu hnh ny sc trnh by chi tit trong chng III. Di y ch gii thiu khi qut mt vi cu trc cbnca h thng.

(1) H thng truyn dn ng thngCc cu hnh ca h thng truyn dn ng thng nh hnh 1.3.

Ch thch: TRM- B ghp u cui, ADM- B ghp xen/ r, REG - B ti sinh (b lp).

Hnh 1.3. Cc cu hnh ng thng

Trong cu hnh im ni im ch c hai b ghp u cui kt ni trc tip vi nhau hocqua b lp bng ng truyn s, to thnh mt ng thng, v vy gi l h thng ng thng.Ngoi ra cn c tn gi khc l h thng h. Cu hnh a im, xen/ r ngoi hai b ghp u cuicn c thm mt hoc nhiu b ghp xen rc kt ni vi nhau bi ng truyn s thnh mtng thng. Cu hnh a im, r nhnh cng l h thng h. Ti a im xen/r, cc lung s

c tip tc truyn ti mt b ghp u cui khc to thnh mt nhnh ca h thng chnh.Cc cu hnh ng thng p dng cho vi ba s v thng tin cp si quang PDH hoc SDH.

TRM ADM

b) Cu hnh a im, xen/ r

ng truynTRM

ng truyn

TRM REG

a) Cu hnh im ni im

ng truynTRM

ng truyn

Pmax

f0

(BW)o

fmax

P(dBm)

3 dBm


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Cc cu hnh trn y khng c kh nng t duy tr khi ng truyn c s c, chng hn t cphoc hng nt.

(2) H thng truyn dn vng (ring)

Trong cu hnh ny ch c cc ADM v c th c cc REG. Cc nt c kt ni vi nhaubi hai hoc bn si quang to thnh mt vng kn, nh trn hnh 1.4.

Hnh 1.4. Cu hnh vng ca h thng truyn dn s

1.2.4.2.Cc tham s

(1) Tc bit: s bit pht i trong mt giy.

Cc n vo tc bit: bit/s, kbit/s (1kbit/s = 103 bit/s), Mbit/s (1Mbit/s = 103 kbit/s =106 bit/s), Gbit/s (1Gbit/s = 103 Mbit/s = 106 kbit/s = 109 bit/s), Tbit/s (1Tbit/s = 103 Gbit/s = 106Mbit/s = 109 kbit/s = 1012 bit/s). Tn hiu sc s dng trong cc mng thng tin s.

(2) T s li bit BER: s bit b li chia cho tng s bit truyn.

- PDH: BER 10-6 cht lng ng truyn bnh thng, 10-6 < BER < 10-3 cht lngng truyn gim st (cnh bo vng), BER 10-3 cht lng ng truyn rt xu (cnh bo ).

- SDH: BER 10-9 cht lng ng truyn bnh thng, BER = 10-6 cht lng ngtruyn gim st (cnh bo vng), BER = 10-3 cht lng ng truyn rt xu (cnh bo ).

(3) Rung pha (Jitter)

Rung pha l siu ch pha khng mong mun ca tn hiu xung xut hin trong truyndn s v l s bin i nh cc thi im c ngha ca tn hiu so vi cc thi im l tng.Khi rung pha xut hin th thi im chuyn mc ca tn hiu s s sm hn hoc mun hn sovi tn hiu chun, nh minh ho trn hnh 1.5.

Hnh 1.5. Tn hiu s b rung pha

ADM

ADM

ADM

ADM

Ring STM-N

Bin ng bao b rung pha

t

b) Tn hiu s b rung pha

ng bao chun

Bin

t

a) Xung nhp chun


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Rung pha xut hin l do c ly ng truyn khc nhau nn tr khc nhau, lch tn sng h ngun v ng h thit b thu trong cng mt mng, lch tn s gia ng h ca thit bSDH v tn s ca lung nhnh PDH.

1.3. S HO TN HIU ANALOG

S ho tn hiu analog l chuyn i tn hiu analog thnh tn hiu s. Mun vy c th s

dng mt trong cc phng php sau y:- iu xung m (PCM)

- iu xung m vi sai (DPCM)

- iu ch Delta (DM)

Sau y trnh by cc phng php s ho tn hiu analog.

1.3.1. iu xung m PCM

PCM c c trng bi ba qu trnh. l ly mu, lng t ho v m ho. Ba qutrnh ny gi l chuyn i A/D.

Mun khi phc li tn hiu analog t tn hiu s phi tri qua hai qu trnh: gii m vlc. Hai qu trnh ny gi l chuyn i D/A.

S khi ca cc qu trnh chuyn i A/D v D/A nh hnh 1.6.

Hnh 1.6- S khi qu trnh chuyn i A/D v D/A trong h thng PCM

1.3.1.1. Chuyn i A/D

(1) Ly mu

Hnh 1.7 th hin ly mu tn hiu analog. y l qu trnh chuyn i tn hiu analogthnh dy xung iu bin (VPAM). Chu k ca dy xung ly mu (Tm) c xc nh theo nh lly mu ca Nyquist:

max2

1

fTm (1.1)

trong f-max l tn s ln nht ca tn hiu analog.

Hnh 1.7- Ly mu tn hiu analog

B mho-nn s

B lymu

B lngt ho

B gii m- dn s

B lcthpng

truynTn hiuanalog

VPAM Tn hiuanalog

Chuyn i A/D Chuyn i D/A

Tn hiu analog

S(t)

t

Xung ly mu

Tm


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Tn hiu thoi c bng tn hu hiu t 0,3 n 3,4 kHz. T biu thc (1.1), c th ly gitr fmax = 4000 Hz. Do chu k ly mu tn hiu thoi l:

sHz

Tm 12540002

1=

= (1.2)

Hoc tn s ly mu tn hiu thoi:

kHzffm 82 max == (1.3)

(2) Lng t ho

Lng t ho l lm trn bin xung ly mu ti mc lng t gn nht. C ngha lgn cho mi xung ly mu mt s nguyn ph hp. Mc ch ca lng t ho m ho gi trmi xung ly mu thnh mt t m c s lng bt t nht.

C hai phng php lng t ho: u v khng u.

Lng t ho u

Hnh 1.8 minh ho lng t ho u. Lng t ho u l chia bin cc xung ly mu

thnh cc khong u nhau, mi khong l mt bc lng tu, k hiu l . Cc ng songsong vi trc thi gian l cc mc lng t. Sau lm trn bin xung ly mu ti mc lngt gn nht s nhn c xung lng t.

Nu bin ca tn hiu analog bin thin trong khong t -a n a th s lng mc

lng t Q v c mi quan h sau y:

=Q

a2(1.4)

Hnh 1.8- Lng t ho u

Lm trn bin xung ly mu gy ra mo lng t. Bin xung mo lng t nm

trong gii hn t - /2 n +/2. Cng sut mo lng t PMLTc xc nh theo biu thc sauy:

( )daaaPMLT +

=2/

2/LT

2W (1.5)

trong : a l bin ca tn hiu analog, WLT(a) l xc sut phn b gi tr tc thi ca bin

xung ly mu trong mt bc lng t. WLT

(a) = 1/. Thay biu thc (1.4) vo kt qu ly tchphn nhn c:

Tn hiu analog

S(t)

t

Xung lng t

Tm

- Bc lng tu

Mc lng t

01

23

4

5

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12

2=MLTP (1.6)

T biu thc (1.6) thy rng cng sut mo lng t ch ph thuc vo , khng phthuc vo bin tn hiu. Nh vy t s cng sut tn hiu c bin ln trn cng sut nhiulng t s ln hn t s cng sut tn hiu c bin yu trn cng sut mo lng t. Theo

phn tch ph th tn hiu thoi ch yu do cc thnh phn tn hiu c cng yu to thnh. Vth nu s dng lng t ho u s lm gim cht lng tn hiu thoi ti u thu. Mun khc

phc nhc im ny, trong thit b ghp knh PCM ch s dng lng t ho khng u.

Lng t ho khng u

Tri vi lng t ho u, lng t ho khng u chia bin xung ly mu thnh cckhong khng u theo nguyn tc khi bin xung ly mu cng ln th di bc lng tcng ln, nh trn hnh 1.9. Lng t ho khng u c thc hin bng cch s dng b nn.

Hnh 1.9- Lng t ho khng u

(3) M ho - nn s

c tnh bin b m ho - nn s

Chc nng ca m ho l chuyn i bin xung lng t thnh mt t m gm mt sbit nht nh. Theo kt qu nghin cu v tnh ton ca nhiu tc gi th trong trng hp lng

t hou, bin cc i ca xung ly mu tn hiu thoi bng 4096 . Do mi t m phicha 12 bit, dn ti hu qu l tc bit mi knh thoi ln gp 1,5 ln tc bit tiu chun 64kbit/s. Mun nhn c tc bit tiu chun, thng s dng b nn c c tnh bin dng

logarit, cn c gi l b nn analog. Biu thc ton hc ca b nn analog theo tiu chun chuu c dng:

0 v m ho V thnh +1. Ngc li, ti thi im ly mu m gi tr ca X(t)

b hn gi tr hm bc thang th V < 0 v c m ho thnh -1. Trong qung thi gian sn tnhiu tng hoc gim nhanh th hm bc thang tng hoc gim khng kp v gy ra qu ti sn(phn c cc ng t nt ti hnh 1.12).

Hnh 1.12- Chuyn i A/D trong DM

1.3.3.2. Chuyn i D/A

Ti pha thu ti lp li hm bc thang da vo kt qu gii m. Nhn c mt dy cc bit1, b tch phn ti my thu to ra dy bc thang i ln, nhn c dy cc bit 1 v -1 an xen nhauth b tch phn to ra dy bc thanh nm ngang v nhn c dy cc bit -1 th b tch phn tolp dy bc thang i xung. Tn hiu dng bc thang qua b lc tch ra gi tr trung bnh ca hm

bc thang v l ng tc khi phc li tn hiu analog. V tn hiu analog ti u ra b lc l gitr trung bnh ca hm bc thang nn trong qung thi gian qu ti sn th dng sng tn hiuanalog thu c b lch so vi dng sng analog ti pha pht. Do qu ti sn gy ra mo tnhiu. khc phc mo tn hiu do qu ti sn cn s dng k thut iu ch Delta thch ng(ADMo).

1.4. CC PHNG PHP GHP KNH1.4.1. Ghp knh phn chia theo tn s FDM

t

Bin Tn hiu analog

Hm bc thang

0Tn hi u DM

Qu ti sn


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Khi nim: ghp knh theo tn s l tn s (hoc bng tn) ca cc knh khc nhau, nhngc truyn ng thi qua mi trng truyn dn. Mun vy phi s dng biu ch, gii iuch v b lc bng.

1.4.1.1. S khi v nguyn l hotng b FDM

S khi h thng ghp knh v tch knh theo tn s nh hnh 1.13.

S c N nhnh, mi nhnh dnh cho mt knh. S ch c mt cp iu ch, nhng trongthc t c nhiu cp iu ch. Tu thuc mi trng truyn dn l v tuyn, dy trn, cp ixng hay cp ng trc m s dng mt s cp iu ch cho thch hp.

Pha pht: tn hiu ting ni qua b lc thp hn ch bng tn t 0,3 n 3,4 kHz. Bngtn ny c iu ch theo phng thc iu bin vi sng mang fNc hai bng bn. Trongghp knh theo tn s ch truyn mt bng bn, loi b bng bn th hai v sng mang nhb lc

bng, nh biu din trn hnh 1.14. Trong hnh 1.14 th d truyn bng di. Ti cp iu chknh, khong cch gia hai sng mang k nhau l 4 kHz.

Hnh 1.13- S khi h thng ghp knh theo tn s

Hnh 1.14- Tn hiu iu bin trong cp iu ch knh

Cp iu ch knh hnh thnh bng tn cs60 108 kHz. T bng tn csto ra bngtn nhm trung gian nhsng mang nhm trung gian. T bng tn nhm trung gian to ra bng

tn ng truyn nhmt sng mang thch hp. N b lc bng ti u ra nhnh pht ni songsong vi nhau.

Biuch

B lcthp

B lcbng

B lcthp

Biuch

B lcbng

f1

B lcbng

Biuch

B lcthp

fN

f2

fN

B lcbng

B giiiu ch

B lcthp

f1

B lcbng

B giiiu ch

B lcthp

f2

B lcbng

B giiiu ch

B lcthp

fN

c tnh suy hao - tn s ca b lc bng

0,3 3,4

Bng trnBng di

f (kHz)

Bng tn thoi


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Pha thu: cc b lc bng ti nhnh pht v nhnh thu ca mi knh c bng tn nh nhau.u vo nhnh thu c N b lc bng ni song song v ng vai tr tch knh. Biu ch tinhnh pht s dng sng mang no th b gii iu ch ca knh y cng s dng sng mang nhvy. Tn hiu knh c gii iu ch vi sng mang v u ra b gii iu ch ngoi bng m tncn c cc thnh phn tn s cao. B lc thp loi b cc thnh phn tn s cao, ch gi li bng

m tn.Ghp knh theo tn s c u im l cc biu ch v gii iu ch c cu to n gin

(s dng cc diode bn dn), bng tn mi knh ch bng 4 kHz nn c th ghp c nhiu knh.Chng hn, my ghp knh cp ng trc c th ghp ti 1920 knh. Tuy nhin do s dng iubin nn kh nng chng nhiu km.

1.4.1.2. Ghp phn chia theo tn strc giao OFDM

(1) Mu

Ghp phn chia theo tn s trc giao l mt cng ngh trong lnh vc truyn dn p dngcho mi trng khng dy, th d truyn thanh radio. Khi p dng vo mi trng c dy nh

ng dy thu bao s khng i xng (ADSL), thng s dng thut nga m ri rc (DMT).Tuy thut ng c khc nhau nhng bn cht ca hai k thut ny u pht sinh t cng mt tng. V vy trong phn ny xt trng hp s dng cho mi trng khng dy.

Nh trnh by trong phn FDM, bng tn tng ca ng truyn c chia thnh Nknh tn s khng chng ln nhau. Tn hiu mi knh c iu ch vi mt sng mang ph ringv N knh c ghp phn chia theo tn s. trnh giao thoa gia cc knh, mt bng tn bovc hnh thnh gia hai knh k nhau. iu ny gy lng ph bng tn tng. khc phcnhc im ny ca FDM, cn s dng N sng mang ph chng ln, nhng trc giao vi nhau.iu kin trc giao ca cc sng mng ph l tn s ca mi mt sng mang ph ny bng s

nguyn ln ca chu trnh (T) k hiu, nh biu th trn hnh 1.15. y l vn quan trng ca kthut OFDM.

Hnh 1.15. Ba sng mang ph trc giao trong mt k hiu OFDM

(2) M hnh h thngiu ch cc sng mang trc giao cn s dng phng php bin i Fourier ri rc

ngc (IDFT). Hnh 1.16 l s biu ch OFDM.

0.2

0.4

0.6

0.8

1

0-0.2

-0.4

-0.6

-0.8

-1

Bin chun ho

Thi gianchun ho

(t / T)0.4 10.80.6

0.2


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u vo biu ch c dy s liu d0, d1,...., dN-1 trong dn l k hiu phc (c th nhntu ra biu ch phc nh QAM, PSK, v.v.). Gi thit thc hin bin i Fourier ngc trndy 2dn s nhn c N s phc Sm (m = 0,1,...., N-1):

( )[ ]1,....1,02exp22exp21

0

1

0

==

=

=

=Nmtfjd

N

nmjdS

N

nnn

N

nnm (1.10)

trong

S

nNT

nf = v t = mTS

trong TS l chu k ca cc k hiu gc. Cho phn thc ca dy k hiu trong biu thc (1.10)i qua b lc ly thp i vi tng k hiu ring trong qung thi gian TS s nhn c phin bn

bng gc ca tn hiu ODFFM:

( )

=

=

1

0

2expRe2N

nn t

T

njdty khi 0 t T (1.11)

trong , T = NTS

1.4.2. Ghp phn chia theo thi gian TDM

Khi c nhiu tn hiu c tn s hoc bng tn nh nhau cng truyn ti mt thi im phis dng ghp knh theo thi gian. C th ghp knh theo thi gian cc tn hiu analog hoc cctn hiu s. Di y trnh by hai phng php ghp knh ny.

1.4.2.1. TDM tn hiu tng t

(1) S khi b ghp

S khi TDM 4 knh nh hnh 1.17.

Sm

Chuynni tip

thnh songsong

ej2f1tm

ej2fN-1tm

dn

Hnh 1.16. Biu ch OFDM

ngtruyn

B lcthp

B lcthp

B lcthp

B lcthp

Thu

xungB

B lcthp

B lcthp

B lcthp

B lcthp

Pht

xungB

1

2

3

4

Tn hiu

analogTn hiuanalog

1

2

3

4

B chuynmch

B phnphi

Hnh 1.17. S khi ghp 4 knh theo thi gian


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(2) Nguyn l hot ng

B lc thp hn ch bng tn tn hiu thoi analog ti 3,4 kHz. B chuyn mch ng vai

tr ly mu tn hiu cc knh, v vy chi ca b chuyn mch quay mt vng ht 125 s, bngmt chu k ly mu. Chi tip xc vi tip im tnh ca knh no th mt xung ca knh y ctruyn i. Trc ht mt xung ng bc truyn i v tip theo l xung ca cc knh 1, 2, 3

v 4. Kt thc mt chu k ghp li c mt xung ng b v ghp tip xung th hai ca cc knh.Qu trnh ny c tip din lin tc theo thi gian. pha thu hot ng ng b vi pha pht,yu cu chi ca b phn phi quay cng tc v ng pha vi chi ca b chuyn mch. Nghal hai chi phi tip xc vi tip im tnh ti v tr tng ng. Yu cu ng b gia my pht vmy thu sc p ng nhxung ng b.

Pha thu, sau khi tch dy xung ca cc knh cn khi phc li tn hiu analog nh sdng b lc thp ging nh b lc ny ti pha pht.

Hnh nh ghp knh theo thi gian tn hiu 3 knh c minh ho ti hnh 1.18.

XR(t) l dy xung ghp ti u ra b chuyn mch.

1.3.2.2. TDM tn hiu s

(1) S khi b ghp

S khi b ghp TDM tn hiu sc th hin ti hnh 1.19.

(2) Nguyn l hot ng

Qu trnh hot ng ca b chuyn mch v b phn phi c trnh by trong phn

TDM tn hiu tng t (analog). Sau y trnh by hot ng TDM tn hiu s.Pha pht: sau khi ly mu tn hiu thoi analog ca cc knh, xung ly mu c a vo

b m ho tin hnh lng t ho v m ho mi xung thnh mt t m nh phn gm 8 bit.

t

tXR(t)

XB XB XB3

33 22

2

1

1

1

125s

S1(t)

tS2(t)

t

S3(t)

Hnh 1.18- Dng sng ca TDM


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Cc bit tin ny c ghp xen byte to thnh mt khung nhkhi to khung. Trong khung cnc t m ng b khung t ti u khung v cc bit bo hiu c ghp vo v tr quy nhtrc. B to xung ngoi chc nng to ra t m ng b khung cn c chc nng iu khin cckhi trong nhnh pht hot ng.

Pha thu: dy tn hiu si vo my thu. Dy xung ng hc tch t tn hiu thu

ng b b to xung thu. B to xung pha pht v pha thu tuy thit k c tc bit nh nhau,nhng do t xa nhau nn chu s tc ng ca thi tit khc nhau, gy ra sai lch tc bit. Vvy di s khng ch ca dy xung ng h, b to xung thu hot ng n nh. Khi ti tokhung tch t m ng b khung lm gc thi gian bt u mt khung, tch cc bit bo hiu x l ring, cn cc byte tin c a vo b gii m chuyn mi t m 8 bit thnh mt xung.Do b phn phi hot ng ng b vi b chuyn mch nn xung ca cc knh ti u ra b giim c chuyn vo b lc thp ca knh tng ng. u ra b lc thp l tn hiu thoi analog.B to xung pha thu iu khin hot ng ca cc khi trong nhnh thu.

Hnh 1.19- S khi h thng TDM tn hiu s

1.4.2.3. Ghp knh thng k

(1) Mu

Trong ghp phn chia theo thi gian ng b trnh by trn y vic phn b khe thigian cho cc ngun l tnh, ngha l cnh; do khi cc ngun khng c s liu th cc khe b

b trng, gy lng ph. khc phc nhc im ny cn s dng phng php ghp thi gianthng k.

(2) c im ca TDM thng k- Phn b cc khe thi gian linh ng theo yu cu;

- B ghp knh thng k r sot cc ng dy u vo v tp trung s liu cho n khighp y khung mi gi i;

- Khng gi cc cc khe thi gian rng nu cn c s liu t ngun bt k;

- Tc s liu trn ng truyn thp hn tc s liu ca cc ng dy u vo;

- Nu c n cng I/O a vo b ghp thng k, ch c kkhe thi gian kh dng, trong k


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Hnh 1.20. S khi b ghp knh thng k

(4) Nguyn l hot ng

Th d s c ba ngun s liu. B ghp tin hnh ghp s liu ca cc ngun theonguyn tc trnh by trong phn c im trn y to thnh mt khung s liu nh hnh1.21. Cc gi s liu c gi qua ng truyn. B tch x l cc gi v da vo a ch phn

pht s liu n my thu tng ng.

Hnh 1.21. Khun dng khung TDM thng k

C hai la chn khun dng khung con TDM thng k:

- Trng hp th nht (hnh 1.21 b):

Trong khung con ch c mt ngun s liu, chiu di s liu thay i v hot ng khi titrng thp.

- Trng hp th hai (hnh 1.21c):

Trong khung con c nhiu ngun s liu, c nhiu mo u, hot ng khi ti trng cao.

c im th t nu r tc s liu ng truyn thp hn tc s liu tng cacc ngun u vo. Sdnh vy l v phi hn ch kch cca bm gim gi thnh, nhngquan trng hn l gim tr ca s liu. Vn ny c kim nghim qua o th v ktquc trnh by ti cc hnh 1.22 v 1.23.

Ngun 1

Ngun 2

My thu1

My thu2ng truyn

My thu3

B tchB ghpng dyu vo

Ngun 3

C a ch iu khin Khung con TDM thng k FCS C

a ch S liu

b) Khung con ch c mt ngun s liu

FCS- dy kim tra khunga) Khung tng qut

S liuChiu dia chS liuChiu dia ch

c) Khung con c nhiu ngun s liu


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T hnh 1.22 thy rng mun tng h s s dng ng phi tng kch cbm. Nhngt hnh 1.23 li cho bit khi tng h s s dng, tc l tng kch cbm th tr li tng rtnhanh.

1.4.3. Ghp knh phn chia theo m CDM

Ghp knh phn chia theo m chnh l a truy nhp phn chia theo m (CDMA). Nguyn

l chung ca CDMA c th hin nh hnh 1.24.

Hnh 1.24. Nguyn l a truy nhp phn chia theo m

t: thi gian

S: m & E

f: tn s1

N

tr (ms)

40

100

400

200 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

H s s dng ng truyn

M= 25 kbit/s

M= 50 kbit/sM= 100 kbit/s

M- Tc bit ca ng truyn

Hnh 1.23- tr ph thuc vo h s s dng ng truyn

4

6

10

Kch cbm(s khung c m)

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

H s s dng ng truyn

8

Hnh 1.22- Kch ctrung bnh ca bmph thuc vo h s s dng ca ng truyn


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Trong CDMA, nhiu ngi s dng c th dng chung tn s v trong cng thi gian. khng gy nhiu cho nhau, mi ngi s dng chc php pht i mt nng lng bit (Eb) nhtnh m bo t s Eb/ N0 quy nh, trong Eb l nng lng bit ca tn hiu cn thu v N0 lmt ph tp m tng ng gy ra do cc tn hiu ca ngi s dng khc. gim mt

ph tp m cn phi tri ph tn hiu ca ngi s dng trc khi pht. Ngoi ra, my thu c

th phn bit c tn hiu cn thu vi cc tn hiu khc, mi tn hiu pht i phi c ci khung ring theo mt m nht nh. C th so snh CDMA nh l nhiu ngi trong phng nichuyn vi nhau tng i mt theo cc ngn ng khc nhau (cc m khc nhau). Nu ni kh (N0nh) th h hon ton khng gy nhiu cho nhau. Hnh 1.24 biu th N ngi s dng, mi ngic m ho bng mt m ring, c k hiu t 1 n N. Mi khi con c trng cho s chimtim nng v tuyn ca ngi s dng: tn s, thi gian v E0.

Do c th ca di ng nn khi mt ngi s dng no n gn trm gc, N0 ca ngiy gy ra cho my thu ngi khc s ln hn (ting ca ngi y nghe to hn) v gy nhiu nhiuhn cho my thu ngi khc. Hin tng ny c gi l hin tng gn - xa. gim nh hng

ca hin tng gn - xa, cn iu chnh cng sut my di ng thp hn khi n tin n gn trmgc. Trong h thng CDMA, qu trnh iu khin cng sut c tin hnh tng. CDMA l

phng thc a truy nhp c nhiu u im so vi cc phng thc a truy nhp khc.

1.5. KHUNG V A KHUNG TN HIU

1.5.1. Khi nim v khung v a khung

Khung tn hiu l tp hp ca mt s bit hoc mt s byte c chiu di cnh hockhng cnh, bao gm cc bit ng b khung t ti u khung, trng tin ghp tn hiu cangi s dng v mt s bit phng vai tr chn, gim st, iu khin, v.v.

a khung l tp hp ca mt s khung. u a khung c t m ng ba khung lm gcthi gian ghp cc khung theo th t quy nh. Pha thu tch t m ng ba khung lm gcthi gian tch cc khung theo trnh t nh ghp pha pht. Ngoi t m ng ba khungv cc khung, trong a khung cn c cc bit ph nh bo hiu, cnh bo v.v.

a khung c to lp khi cn cc khe thi gian chuyn ti bo hiu cc knh hoc dngchung cc byte mo u cho cc khung trong a khung.

1.5.2. Cu trc cbn ca mt khung tn hiu

Cu trc cbn ca mt khung tn hiu nh hnh 1.25. Trong thi gian TKghp cc bitng b khung, cc bit ph v cc thng tin u vo b ghp.

1.6. NG B TRONG VIN THNG

1.6.1. Mu

Tin hnh ng b hot ng ca cc thit b khc nhau hoc s tin trin ca cc qutrnh khc nhau bng cch ng chnh thang thi gian ca chng gi l ng b.

TK- di khung (thi hn khung)

Hnh 1.25- Cu trc cbn ca khung tn hiu

Trng tinCcbit ph

Cc bitng bkhung


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Nhiu hot ng trong h thng s cn phi tun theo mi quan h tin nh. Nu hai hot ngtun theo mt s tin nh, th ng b nhm m bo cho cc hot ng din ra theo mt trnh tchnh xc. Ti mc phn cng, ng bc iu tit bng cch phn phi mt tn hiu thi gianchung ti tt c cc mun ca h thng. mc tru tng cao, cc qu trnh phn mm cng b nhtrao i thng bo.

Ph thuc vo phm vi ng dng, cc h thng tru tng khc nhau c chp nhn chiu qu v c cu trc theo kiu phn cp, trong mi mc tru tng lin h vi cc ctnh ca mc tru tng thp hn v che du cc chi tit khng cn thit i vi mc cao hn.Tru tng cho php cc nh thit k b qua cc chi tit khng cn thit v tp trung vo cc cim cn thit. V vy d dng thc hin mt bn thit k h thng phc tp hn.

Trong cc h thng phn cng s, gii php chung l cu trc h thng c din gii theocc mc tru tng nh mc vt l, mc mch, mc phn t v mc mun. Trong mc vt l,nh thit k quan tm n cc quy tc vt l chi phi cc c tnh ca bn dn. Mc mch linquan n transistor, resistor, v.v. Mc phn t tp trung vo cc cng, cc cng logic v.v. Trong

mc mun, cc phn tc phn chia thnh cc thc th phc tp hn nh cc b nh, cckhi logic, cc CPU v.v.

Cc giao thc thng tin c thc hin nh cc mun phn mm, c cu trc ph hpvi m hnh lp. Cc ngn xp giao thc c xy dng theo cch cc giao thc ti mc chotrc cung cp cc dch v cho cc giao thc mc trn v s dng cc dch v ca mt s mcthp hn. Trong m hnh giao thc tham kho kt ni h thng m(OSI) c by mc (lp) trutng. Cc tiu chun ca mc 1 (lp vt l) quy nh cc giao din vt l v khung bit cs, cngha l quy nh cc bit c truyn trn mi trng vt l nh th no nhm cung cp mt knhtruyn dn sim ni im y . Cc tiu chun mc 2 (lp kt ni d liu) quy nh cc giaothc nhm cung cp mt knh sim ni im khng c li bng cch pht li cc khung b lihoc nhk thut sa li. Cc giao thc ca cc lp trn cung cp cc dch vnh tuyn mng(lp mng), cc dch v truyn ti qua mng (lp truyn ti) v cung cp cho ngi s dng ucui cc dch vng dng trc tip.

Nhng ci g l tiu chun tru tng c s dng m t cc h thng phn cng vphn mm u lin quan vi nhau v ti mc bt k s hot ng chnh xc ca chng u phthuc vo thi gian. Cc thc th ca cc mc tru tng khc nhau trong h thng phn cng v

phn mm thng yu cu chc nng ng bc lp khc nhau. Th d, ng b cc qu trnhgiao thc ti mc cho trc v nguyn tc l c lp vi ng b hot ng cc qu trnh mcthp. Tuy nhin, t th d trn y thy rng vn ng b c th khc nhau hon ton v mc

tru tng v tnh cht ca cc phn t hoc qu trnh ng b.

Mi quan tm ny lm xut hin s nghi ngv mc thch hp ca s chp nhn thut ng"ng b" lin quan n mt tp hp y ca nhng vn c tnh cht khc nhau, trong thi gian l cn thit. Tuy nhin, s nghin cu y vng b nu ln mt sc imchung trong bi cnh khc nhau. V vy a ra l do ti sao thut ng c tnh lch s ny c chp nhn.

i vi nhiu k s thng tin s, vic cm nhn thut ngng b cn b hn ch. H chorng n ch lin quan n hot ng tch ng h ti my thu v cc thng tin cha trong tn hiuthu c. Thc ra vn ny ch lin quan n ng b sng mang hoc ng b k hiu. Tri

li, ng bng vai tr quan trng trong mt s lnh vc vin thng.


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Gii iu ch kt hp ca tn hiu iu bin da vo cu trc li sng mang, ngha l davo tch tn hiu kt hp vi sng mang c tn s v pha cho trc. l ng b sng mang.

Trong trng hp bt k, gii iu ch s yu cu nhn bit cc thi im ly mu v quyt nh tch thng tin logic t tn hiu analog thu c, do a ra quyt nh hnh thnh bit 0 hay

bit 1. y l ng b k hiu.

Sau khi tch c thng tin logic, bc tip theo, ti mc tru tng cao l sp xp li cckhung t cc bit thu c. y chnh l ng b khung. ng b khung cho php thit b thu hiuc vai tr cc byte ti cc v tr khc nhau trong khung (th d 30 knh dnh cho cc cuc giin thoi khc nhau trong b ghp PCM-30).

Khi thng tin ngun c phn chia thnh cc gi truyn hoc nh tuyn c lp tich (trong mng chuyn mch gi) th c th m phng knh nu thit b thu c kh nng cn

bng tr khc nhau ca cc gi thu c. Do ti to li c lung bit gc nu lung ny truyn qua mng chuyn mch knh. Vic cn bng tr gi l ng b gi v c thc hin

bng cch khi phc li nh thi gc t dy cc gi thu c thng qua k thut thch nghi hoc

bng cch x l thng tin nh thi ngun c ghi trong u gi.Nhng khi nim trn y lin quan n cc mc khc nhau ca ng b trong truyn dn

im ni im. Mt mc khc ca ng b l ng b mng: t p trung vo hot ng ca hthng cc nt mng. H thng ny c th phn phi ng h chung ti tt c cc nt mng truyn dn v chuyn mch trong khun dng s, sao cho mi phn t mng c th hot ngng b vi cc phn t mng khc v ng b cc lung bit n.

Ti mc tru tng cao nht, ng ba phng tin lin quan n vic sp xp cn thncc phn t hn tp (hnh nh, vn bn, audio, video, ...) thnh thng tin a phng tin ti ccmc tch hp khc nhau.

Mt loi khc ca ng b mng l ng bng h thi gian thc truyn qua mng vinthng, trong vic phn phi thi gian tuyt i (thi gian theo tiu chun quc gia) c linquan ti mc ch qun l mng.

1.6.2. ng b sng mang

Trong cc h thng iu bin (AM), khi nhn tn hiu iu ch s(t) vi sng mang

cos2f0t c tn hiu iu bin X(t) dng:

X(t) = s(t) . cos2f0t (1.12)

hoc [1+ m s(t)] . cos2f0t (1.13)

Trong trng hp sau, ng bao ca tn hiu iu bin X(t) t l vi s(t) nu ( ) 1tms .iu ny cho php thit k d dng b gii iu ch (gii iu chng bao).

Trong trng hp trc c kh nng gii iu ch bng cch nhn tn hiu iu ch visng hnh sine c tn s v pha ca sng mang v sau cho qua b lc loi tr cc thnh

phn tn s cao:

X(t). cos2f0t = s(t) . cos20t = [s(t)/2] (1+ cos20t) (1.14)

Loi iu ch ny yu cu tn hiu nhn cos0t c s dng trong my thu phi c cngtn s v pha ca sng mang iu ch thu c. S dch pha bt k ca s gy suy hao tn

hiu mt i lng [s(t)/2]cos ti u ra b lc thp (nu = /2 th tn hiu ra bng zero).T cc biu thc trn y thy rng iu bin trong min tn s tng ng vi s

chuyn dch ph tn hiu iu ch ti tn s sng mang f0. Tht vy, ph ca tn hiu iu bin l


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d tha, gm hai phn chnh nm v hai pha ca sng mang f0. iu ch mt bng bn (SSB) chtruyn mt trong hai phn chnh (mt trong hai bng bn). iu ch SSB phi kt hp, trong sng chnh pha thm ch cn cht ch hn, v mt lng dch pha bt k cng gy ra mo tn hiuiu bin.

Nh trnh by trn y, gii iu ch kt hp l da vo ti cu trc sng mang, ngha

l da vo vic khi phc tn hiu kt hp vi sng mang v tn s v pha. Hot ng ny chnhl ng b sng mang.

C th d dng ti cu trc sng mang, nu trong ph tn hiu thu c mt ng ph tisng mang f0, thng xy ra khi tn hiu iu ch c gi tr trung bnh bng zero. Trong trngh p ny, c th thc hin tch sng mang nh s dng b lc bng h p hoc vng kho pha(PLL). PLL c thit k c bng thng hp, do b to dao ng iu khin bi in p ngoi(VCO) c th kho v theo di thng dng tn s xung quanh tn s danh nh.

ng tic l trong nhiu trng hp khng c vch ph ti f0. Mt mt, theo quan imtruyn thng tin th iu ny l c hiu qu, bi v cng sut ca sng mang nu c truyn i s

gy lng ph. Mt khc, trng hp ny cn h thng ng b tinh vi hn c kh nng khi phcsng mang v tn s v pha.

Mt th dn gin ca ng b sng mang: xem xt trng hp truyn dn s kho dchpha nh phn (BPSK), trong k hiu 1 v 0 l c lp vi nhau, c cng xc sut xut hin v

c m ho thnh cc xung vung i cc nhau. V vy, sng iu bin c dng cost v phcng sut lin tc, khng c cc vch ri rc ti f0. Tt nhin, ch bin i phi tuyn mi c thto ra vch ph f0 mong mun t tn hiu thu c. Trong trng hp n gin ny, bnh phngv chia tn mi gii quyt c vn (xem hnh 1.26). Bnh phng sng iu ch xo b

iu ch v to ra thnh phn (1+ cos20t)/2 c vch ph ti tn s 2f0 xut hin v thu c

sng mang nhchia tn.

Hnh 1.26- ng b sng mang cho h thng BPSK

Trong iu ch pha cu phng (h thng QPSK truyn cc nhm k hiu 2 bit), thit bng b da vo tng tn s tn hiu gp 4 xo iu ch v sau to ra vch ph ti tn s 4f0.

1.6.3. ng b k hiu (symbol)Trong truyn dn s thng s dng dy xung i din cho cc k hiu cn truyn v pht

i vi tc khng i R= 1/ T, trong T khong cch gia hai k hiu k nhau (chu k).

Trong mi trng hp, pha thu c th gii iu ch kt hp hoc khng kt hp bitc nh thi dy, ngha l v tr thi gian ca cc k hiu v tch thng tin logic t tn hiuanalog thu c. Thng tin nh thi dy cho php c k hiu ti cc thi im ng.

Khi phc nh thi dy k hiu t tn hiu analog thu c gi l ng b k hiu. ikhi cn lin quan n khi phc ng h.

Hnh 1.27 minh ho nguyn tc thu bng gc nh phn. Tn hiu analog thu c r(t) c

ly mu to ra dy cc gi tr thc r(kT), t tch ra dy bit nhquyt nh logic. B ly muc iu khin bi h thng ng b thch hp. H thng ny nh gi cc thi im c t = kT

bng cch kim tra r(t).

B lcbng (...)

2 PLL vchia tn

s(t) f0


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Khi ng b k hiu c thc hin sau khi chuyn i tn hiu thnh bng gc, c th sdng mt s k thut khi phc nh thi k hiu ging nh k thut ng b sng mang.

Xem xt truyn bng gc nhiu mc: nu ph ca tn hiu truyn dn c dng:

( ) ( ) =k

k kTtgats (1.15)

c mt ng ph ti tn s k hiu 1/ T v nm ti trung tm b lc bng hp th c kh nngkhi phc sng hnh sine, t tch ra dy xung nh thi c tn s ca dy k hiu. Nu khngc ng ph ti tn s 1/ T, vn c kh nng to ra n bng cch chuyn i phi tuyn thch hp.

Th d nh chuyn i bnh phng u = s2 hoc chnh lu u = s.

Cng c thng b k hiu bng cch khi phc trc tip t tn hiu ly bng m khngcn khi phc sng mang v chuyn i thnh bng gc. Th d, tn hiu iu ch c dng:

( ) ( ) tkTtgats k k 0cos = (1.16)l c th c c ng bao hoc bnh phng tn hiu nhn c ng ph ti tn s k hiu1/ T. S dng ng ph ny c thi gian k hiu.

Tuy nhin, cn c cc k thut ng b sng mang v ng b k hiu khc da vo ccnguyn tc khc nhau to ra cc ng ph. Sau y tm tt ba lnh vc ng b k hiu:

(1) Da vo bm li;

(2) Da vo tm kim cc i v lc;

(3) Da vo chuyn i phi tuyn v lc.

Lnh vc th nht s dng cc h thng PLL. Lnh vc th hai so snh dy k hiu pht iban u vi cc k hiu lp lu tr nh gi dch pha. Lnh vc th ba c trnh bytrn y.

1.6.4. ng b khung

Sau khi hon thnh ng b sng mang v ng b k hiu v thng tin logic ctch ra t tn hiu n, bc tip theo l xc nh im u v im cui ca t m hoc canhm cc t m, nh vy gi l ng b t m. ng thi sp xp li cc t m thu c thnhkhung theo ng trnh t nh khung pha pht, nh vy gi l ng b khung.

Trong truyn dn s, cc bit thng c t chc thnh khung n nh ngha khc nhau cho

cc byte. Cc byte cc v tr khc nhau trong khung c th dnh cho cc knh ngi s dngkhc nhau c chung mi trng vt l trong ghp knh phn chia thi gian (TDM), chng hn nh

B lc cnbng knh

Quyt nhk hiu

1011r(kT)

B ly mu

Bng b k hiu(khi phc ng h)

t = kT

Hnh 1.27- ng b k hiu trong my thu bng gc nh phn

r(t)


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trong b ghp PCM-30 hoc phn nh cc chc nng mo u (kim tra li, truyn ti thng tinqun l v iu khin v.v.). V vy ng b khung l ht sc quan trng trong truyn dn s. Tchcc lung nhnh ng c bt u t m t chnh xc cc khung.

K hoch ng b khung bt k (cng c quan hn ng chnh khung) gm hai hotng cbn:

(1) Tm kim: xy ra khi thit b (bng chnh) chch khi ng b khung v ngchnh khung ang d tm lung bit thu c.

(2) Duy tr: mi khi thit b tha nhn ng b khung v kim tra lin tc ranh giikhung.

T m ng chnh khung t u khung trgip ng b khung v t m ny c ci tmt gi trc bit. Tm kim c thc hin bng cch d tm mu t m ng chnh ti v tr

bt k ca lung bit thu c v c duy tr nhkim tra t m ng chnh khung, ti bt umt khung. Trong khi tm kim mu t m ng chnh khung c th gp trng hp t m ngchnh khung b phng to t lung bit s liu. V vy cn tin hnh kim tra t m ng chnh

khung ti mt s v tr trc khi cng nhn c ng b. Mc tiu la chn k hoch ng chnhkhung c hiu qu l:

(1) Di iu kin ng chnh khung chnh xc, ti thiu ho xc sut mt ng chnhkhung do li ng truyn (mt ng chnh cng bc);

(2) Di iu kin chch ng chnh khung, ti thiu ho xc sut ng chnh khung gimo do phng to mu t m ng chnh khung trong lung bit ngu nhin thu c;

(3) Ti thiu ho thi gian khi phc ng chnh khung.

C th phn tch qu trnh phng on khi m t mt v khi phc ng chnh khung phhp vi k hoch ng chnh khung chn nhs dng m hnh chui Markov thch hp nhhnh 1.28, trong P l xc sut nhn bit ng t m ng b khung. Tt nhin, P c biu thkhc nhau di cc iu kin khc nhau v trong min khc nhau ca biu .

Hnh 1.28- M hnh chui Markov ca k hoch ng chnh khung

A0

A1 C-1

C1

A2

C0

B

A-1

PP

P

P

1-P

1-P

1-P

1-P1-P

1-P

1-PP

P

P

P1-P

1-P

P


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T trng thi ng chnh ng A0, trong qu trnh duy tr c thc hin, bng

chnh chuyn ti trng thi chch ng b B ch khi pht hin li trong t m ng chnh lintip. Ti trng thi B, bng chnh thc hin qu trnh tm kim v khi pht hin c mu bitging t m ng chnh th chuyn sang trng thi ng chnh ng tm thi C0. Ti y bngchnh thc hin qu trnh duy tr v s chuyn sang trng thi bnh thng A0 ch khi khng pht

hin li trong t m ng chnh lin tip. Ngc li, nu pht hin li trong t m ng chnhu tin th quay trv trng thi B bt u li qu trnh tm kim.

Khi trng thi ng chnh ng A0, nu pht hin c mt ng chnh th chuyn sangtrng thi B. Nguyn nhn gy ra s chuyn ny l:

(1) Li trong cc t m ng chnh (mt ng chnh cng bc);

(2) Mt nh thi bit hoc trt khung n (mt ng chnh thc); nu mt nh thi bit,bng chnh bt u qu trnh tm kim t trng thi B.

Cc tham sc trng cho cht lng k hoch ng b l:

(1) Tc bin c trung bnh R ca mt cng bc;

(2) Thi gian khi phc ng chnh trung bnh rt v phng sai2tr ca thi gian khi

phc ng chnh tr (thi gian ti lp khung) c xc nh nh l khong cch gia thi im btu qu trnh tm kim trong trng thi B v thi im ti chim ng chnh thc trong trng thiA0;

(3) Xc sut ng chnh gi mo pfa, c ngha l xc sut chuyn t trng thi B sang trngthi A0 do mu t m ng chnh b phng to, mc d vn cn trong iu kin chch ng chnh.

Cn nhn mnh rng trong trng hp mt ng chnh cng bc, thi im bt u caqu trnh tm kim xy ra ng thi vi thi im bt u khung; cn trong trng hp mt thc

th n xy ra ng thi vi thi im ti chim nh thi k hiu. V vy, thi gian khi phc camt cng bc, theo thng k, khng chm hn thi gian khi phc ca mt thc.

Mt ng chnh cng bc l lin quan, xc sut P ca cng nhn t m ng chnh ngvi gi thit l h thng trong iu kin ng chnh bnh thng c xc nh theo biu thc sauy:

( ) apP a == 111 (1.17)

trong , l tc khng i ca li bit ng truyn (gi thit khng tng quan) v a l slng bit ca t m ng chnh.

Mt khc, qu trnh tm kim l lin quan, xc sut P by gil xc sut phng to mu tm ng chnh. cho n gin, gi thit rng cc bit ny l c lp thng k v c cng xc sutxut hin th xc sut P c xc nh nh sau:

apP

2

12 == (1.18)

Theo phn tch m hnh ti hnh 1.20, nhn c cc biu thc xc nh cc tham sctrng cho cht lng ca k hoch ng b l:

( )( )

( )( )( )

a

aa

L

R

p

pp

L

RR

=1

1

11

1 0

1

110 (1.19)


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( )( )

+

++=

11

11

11 1

2

12

2

2

0

pp

p

pL

R

Ltr (1.20)

( )( )

( )

( )( ) ( )( )[ ]

( )( )( )[ ]

21111

2211121

21

21

212

222

2

221

212

12

122

2

12

20

22

+

+

++

+

=

++

+

pppp

p

pppp

p

pp

paL

R

Ltr

(1.21)

( ) ( ) aLaL

i

i

fa pppp+

==

= 12

022 1111

(1.22)

trong L l chiu di khung (s bit trong khung) v R0 l tc bit danh nh ca tn hiu ghp.

V L >>1 nn s bin i ngu nhin ca tr c xu hng phn b chun (phn b Gauss). V vy

thi gian khi phc ng chnh cc i c xc nh theo biu thc sau y:trrr tt 3max/ += (1.23)

Da vo s trn y c th lp k hoch ng chnh khung khc nhau i vi tn hiughp ti cc mc khc nhau ca phn cp s cn ng b (PDH). Th di vi PCM-30, gi tr

tiu chun ca = 3 v = 1. S thay i mt t khi thit kng b khung i vi thit bSDH. Trong t m ng chnh (gm 96 byte i vi STM-16) cn ch hai tp hp con ca cc

byte trong qu trnh tm kim v duy tr: chn t m ng chnh di gim xc sut phng to,trong khi chn t m ng chnh ngn gim bt xc sut mt ng chnh cng bc. Biu trng thi s dng cho k hoch ng b SDH c th hin ti hnh 1.29.

S c ba trng thi chnh cn xem xt:

(1) Trng thi trong khung (IF) l trng thi hot ng bnh thng di cc iu kinng chnh (tng ng trng thi ng chnh chnh xc A0);

(2) Trng thi mt khung (LOF) l trng thi cnh bo ca mt ng chnh (tng ngtrng thi B);

(3) Trng thi chch khung (OOF) l trng thi trc cnh bo (tng ng cc trng thi

Ai vi 0


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T trng thi IF, trong ang tin hnh qu trnh duy tr, bng chnh chuyn sangtrng thi OOF sau khi pht hin li trong t m ng chnh ca M khung lin tip. Sau K khunglin tip c t m ng chnh b li th bng chnh chuyn sang trng thi LOF. ang trongtrng thi OOF, nu pht hin J khung lin tip khng c li trong t m ng chnh khung th bng chnh quay tr li trng thi IF. Trong trng thi LOF ang tin hnh qu trnh tm kim,

nu khng pht hin li trong N t m ng chnh khung lin tip th quay trv trng thi ngchnh bnh thng IF.

Biu trng thi ti hnh 1.29 khng phi l m hnh Markov. trnh chuyn mchgin on gia hai trng thi OOF v IF, b ghi dch phi m cc khung c li trong t m ngchnh khung khi m h thng ang trong trng thi OOF (ngha l b ghi dch iu khin chuynt trng thi OOF sang trng thi LOF v m t 0 n K) khng ci t ti zero khi m h thngang trong trng thi IF i vi L khung lin tip.

Cc gi tr tiu chun ca cc tham s khi lp k hoch ng b khung SDH l: M 5, J2, K = 24, N = 24 v L = 24.

Sau khi ng b khung, cc khung ng bc sp xp c trt t thnh a khung nht m ng ba khung t u a khung. y chnh l ng ba khung.

1.6.5. ng b bit

Trong vin thng ng b bit c din t theo hai ngha chnh. Th nht, ng b bitc lc c hiu c lin quan n ng b k hiu c trnh by trc y. Th hai, tng quthn, ng b bit c s dng biu thng b lung bit cn ng b theo tn sng h cathit b ti ch. Vn ny c thc hin bng cch ghi cc bit ca lung bit cn ng b vo bnhn hi (bm) theo tn s ca lung vo v sau c ra theo tn s ca ng h thit bti ch. ng b bit c hiu ch yu theo cch gii thch th hai ny.

ng b bit c s dng sp xp cc bit v khi u khung ca tn hiu PCM ti uvo tng i in t s, cho php chuyn cc octet t mt khe thi gian ti mt khe thi giankhc.

Ngoi ra, ng b bit cn c thc hin trong b ghp tn hiu s, ti khi ng b ho. Ti ycc nhnh c ng b bit chuyn lung s cn ng b thnh lung ng b bng cch chn

bit.

1.6.6. ng b gi

Chuyn mch gi bao gm phn chia thng tin ngun thnh thng bo hoc cc gi truyn i, hoc nh tuyn ti ch. Cc gi cha mt son s liu ngun v b sung thm

mt vi thng tin mo u. Gi c chiu di cnh hoc thay i. Gi c chiu di cnh gi lt bo.

Chuyn mch gi l mt cng ngh c hiu qu lin kt s liu vi thoi hoc vi lulng thi gian thc khc trong mt mng duy nht. thc hin mng s lin kt a dch v

bng rng (B-ISDN), cc t chc tiu chun ho quc t chn k thut chuyn mch t bo:kiu chuyn ti khng ng b (ATM).

Phng php chuyn ti chuyn mch gi c cc c im sau y:

(1) Do tnh cht thng k ca chuyn mch gi, c bit l xp hng bn trong mng, ccgi m c tr nht nh khi chuyn ti qua mng v c cc thi gian n trung bnh thng k;

(2) Nu cc gi ca cuc gi n c nh tuyn c lp (mi gi i qua mt tuyn khcnhau xuyn qua mng) th chng n ch khng theo th t;


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(3) Ti my thu c kh nng khi phc tn sng h ca ngun thng tin khi da volung bit n.

Chuyn ti trong sut tn hiu thoi xuyn qua mng chuyn mch gi i hi ng uvi cc vn trn y ti to ting ni c cht lng chp nhn c t cc gi n c trkhc nhau. V vy yu cu b sung cc chc nng ti giao din thu. Cn bng tr ngu nhin

ca gi thng lin quan n ng b gi.Nhim v cn bng tr gi c thc hin theo mt s cch v c chia lm hai nhm:

da vo mi trng mng khng ng b, ti cc nt khc nhau c nh thi bi ng hc lp ti ch; hoc da vo mi trng mng ng b c h thng phn phi ng h chung ticc nt. Ph hp vi cch phn chia ny, c cc phng php ch yu sau y khi phc nhthi:

(1) Khi phc nh thi khngng b

- nh gi tr khng nhn thy

K hoch n gin nht nh gi thi gian to ra mt gi n c tin hnh trong

trng hp xu nht: my thu cho rng gi c nh gi da vo tr truyn dn cc tiu vda vo cc gi khc c th b tr khng vt qu mt lng thi gian cc i cho trc. Nh vygi l nh gi khng nhn thy. Sau khi nh gi thi gian kt thc ca gi th nht, my thu sdng s th t dy trong cc gi tip theo xc nh mt cch chnh xc thi gian kt thc cami gi. Cc gi n c tr ln s b loi.

- o hnh trnh

Mc d nh gi tr khng nhn thy l n gin nhng khng y trong mngng di. K thut nh gi tr thc t tt nht l o tr hnh trnh gia gi chuyn i v githu c v s dng gi tr ny nh gi tr mt hng ca cc gi khc vi gi thit trc

phn b nh nhau gia hai hng.

- Tr thay i do b sung

Trong trng hp ny, o tr thc t khi truyn cc gi qua mng. S thay i ca trc o nhdu hiu tr tch lu ca mi gi. Mi phn t mng b sung tr vo du hiutr khi o theo ng h ti ch v lng chnh lch gia thi im n v i. Bit tr gi, chophp xc nh thi gian kt thc l thi gian thc t cng vi lng chnh lch gia gi tr duhiu tr cc i v gi tr du hiu tr thc t.

- K hoch thch ng

Khng c phng php no trn y o tr hon ton chnh xc. V vy cc thut tonkhc nhau c s dng thay i thch ng tr khi thu lung gi da vo mc y bm thu hoc da vo lp li hnh trnh o tr.

- K hoch thch nghi da vo PLL

Tt c cc phng php trn y ch ph hp vi truyn dn thoi trn mng chuyn mchgi bng hp. Cc mng ATM B-ISDN yu cu nghim ngt hn do tc chuyn mch cao vdo c nhiu dch v. Trong cc mng ATM khng ng b, thng s dng k thut lc jitter t

bo nhkhi phc nh thi PLL. K thut n gin ny c thc thc hin nhlc trc cm s liu mc y hoc cc gi tr tc thi n ca t bo sc a vo b lc trc. PPL l

b lc thp lc Jitter t bo. Mc d k thut ny c cht lng tt hn v linh hot hn cc kthut m t trc y, nhng m phng knh ti u ra mng ATM vn gp kh khn trongvic tun th cc tiu chun hin hnh v jitter.


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(2) Khi phc nh thi ng b

Khi nim cbn ca k thut ny da vo tnh kh dng ca ng h tham kho chung(ng h mng ng b). y khng phi l vn trong mng SDH, m trong mng quang ngb (SONET) ca Bc M chn lp vt l truyn ATM. V l do ny m k thut ng bgim jitter t bo c thit ki vi mng ATM B -ISDN.

- K thut m ho tn sng b (SFET)ng h ngun khng ng bc so snh vi ng h tham kho mng. S khng

ng nht gia hai ng hc o v m ho trong mo u lp p ng ATM (AAL). Ti mythu, ng h mng chung v thng tin m ho c s dng cu trc li ng h ngun.

- Du hiu thi gian (TS)

Bm 16 bit do ng h tham kho mng iu khin. Hai byte ca mo u lp con hit (CS) mang gi tr tc thi ca bm trong mi nhm 16 t bo. Ti pha thu, ng h ngunc cu trc li t TS thu c v ng h mng.

- Du hiu thi gian dng b (SRTS)

Phng php ny l TS ci tin v da vo s quan st thy rng i vi ng h ngunchnh xc, cc bit c ngha thp ca TS 16 bit chuyn ti hu ht thng tin c ch. V vy SRTSch cn 4 bit. iu ny cho php lin kt SRTS vo trong mo u ca AAL hin c m vnkhng lm tng kch cca n. SRTS c ITU-T chp nhn nh l k thut tiu chun khi

phc nh thi i vi AAL-1 (m phng knh).

1.6.7. ng b mng

ng b mng lin quan n phn phi thi gian v tn s trong mng cung cp ng htri khp trn mt vng rng ln. Mc ch l ng chnh thi gian v tn s ca tt cng h

nhkh nng thng tin ca cc tuyn kt ni gia chng (chng hn cp ng, cp si quang, cctuyn radio). Sau y l mt sng dng c hiu qu:

(1) ng bng h ni b ca cc im ghp v chuyn mch khc nhau trong mngvin thng s.

(2) ng bng h trong mng vin thng yu cu mt vi dng a truy nhp phn chiathi gian, chng hn nh mng v tinh, u cui di ng ca thng tin di ng GSM v.v.

(3) Ngi s dng mng o khong cch gia hai nt trong mng, xc nh v tr v hotng ca chng.

ng b mng ng vai tr trung tm trong thng tin s, c nh hng nht nh n cht

lng hu ht cc dch v m nh iu hnh cung cp cho khch hng. ng b mng thng tin ssc trnh by trong cc chuyn ca chng sau.

1.6.8. ng ba phng tin

a phng tin lin quan n tch hp cc thnh phn khng ng nht nh vn bn, hnhnh, audio v vidio trong sa dng ca cc mi trng ng dng. S liu c th ph thuc rtnhiu vo thi gian nh audio v vidio hnh nh ng v i hi trnh by theo th t thi giankhi s dng. Nhim v ca tch hp nh vy gi l ng ba phng tin. ng b c th sdng chm dt tranh chp gia cc lung s liu v cc bin c bn ngoi do ngi s dngto ra. Ni mt cch khc, mun m ch mi tng quan thi gian gia mi trng, nh xem

video kt hp vi m thanh, hoc c th trnh by r rng nh trng hp ti liu a phng tinshu vn bn ch thch thoi hoc trong trng hp siu vn bn a phng tin.


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S ph thuc thi gian ca cc dy s liu c th l tuyn tnh, nh trong trng hp trnhdin file audio trn dy hnh nh. Nhng cng c kh nng trnh by s liu theo kiu khc nhtruy nhp thun nhanh, truy nhp ngc nhanh v truy nhp ngu nhin.

Vn ng ba phng tin c gii thiu rng ri trong cc ti liu. Tuy nhin,vic tho lun chi tit ch ny khng nm trong ni dung cun ti liu ny.

1.6.9. ng bng h thi gian thcMt kiu khc ca ng b mng l phn phi thi gian tham kho tuyt i (thi gian

theo tiu chun quc gia) ti cc ng h thi gian thc ca thit b trong mng vin thng (ngb ca cc ng h thi gian thc).

Phn phi ng h tiu chun quc gia nhm mc ch qun l v iu khin mng. Ccs kin bt ku c h thng gim st thit b , chng hn nh vt ngng t s li bit(BER), cc cnh bo ng, hng phn cng v.v. c lu tr bo co. Khi mng vin thngc qun l bi h thng qun l (mng qun l vin thng tiu chun TMN), cc s kin sc ch nhthit b truyn thng bo qun l ti h thng iu hnh (OS). Trong trng hp

khc, vic lu tr thng tin phi bao gm c ngy giv ng h thi gian thc ca thit b scly ra.

iu cn thit l cc ng h thi gian thc ton mng phi c ng b theo thi giantuyt i nh nhau, nu khng s khng lin quan vi cc thng bo khc nhau mt cch c ngha theo mt nhn chung. Ch khi ng h thi gian thc thit bc ng b vi thi giantiu chun th mi c kh nng sp xp mi tng quan thi gian v logic trong s cc s kinkhc nhau v v vy mi dn n suy on c t s liu cha x l s kin tp hp v lutr.

ng bng h thi gian thc khc vi ng b mng. ng bng h thi gian thc

phn phi thng tin thi gian tuyt i (th d 10.32.05 AM ngy 23 thng 6 nm 2006, hoc duhiu thi gian khc) v a ra cc yu cu khc nhau ca chnh xc. i vi qun l, rt cnvn nu trn y, chnh xc thi gian n vi mili giy l hon ton c kh nng. Phmvi ngy gitrong qun l cn lu tr khng g khc l xc nh r ngy, thng, nm v gi, pht,giy.

Mc ch ca ng b mng l ti thiu ho thng dng li thi gian trong s cc ng h,tr khi ng b pha. iu ny c ng l tn hiu nh thi vt l ng b (th d sng hnhsine) c phn phi ti cc ng h mng. ng b mng vin thng s thc hin lch thi giankhng ln hn 10 ns hoc 100 ns.

Mt khc, ng b thi gian thc thng c thc hin nhtrao i thng bo v thngtin thi gian (cc du hiu thi gian) theo giao thc ph hp c chuyn ti trn cc tuyn giacc nt mng.

Th d giao thc thi gian mng (NTP) c s dng trong cc dch v thi gian Internetv khch hng ng bng h thi gian thc cng nh t chc v duy tr tng mngcon ng b thi gian. NTP c pht trin t cc giao thc n gin hn, nhng c thit kc bit c chnh xc, n nh v tin cy cao, thm ch khi s dng trn cc tuynInternet in hnh lin quan n cc cng ghp v cc mng khng tin cy.

Giao thc da vo thng bo c chuyn ti trn giao thc Internet (IP) cc gi giao thc

datagram ngi s dng (UDP) cung cp dch v chuyn ti khng kt ni. Tuy nhin, n snsng p ng i vi cc b giao thc khc. Cc c trng khc khng bt buc gm xc nhn vmt m ho thng bo cng nh cung cp iu khin v gim st t xa.


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Trong NTP, mt hoc nhiu dch v s cp c ng b trc tip t cc ngun thamkho bn ngoi. Cc dch v thi gian th cp c ng b t cc dch v scp ph hp vi

phn cp. Cu hnh li cc ng ng b thay th l c kh nng khc phc s gin on v ccs c. Thut ton c kh nng nh gi v b tr truyn dn ngu nhin ca cc gi truyn quamng v v vy thc hin chnh xc thi gian tuyt i ti mc vi mili giy.

1.7. NGU NHIN HO TN HIU

1.7.1. Khi nim

Trong qu trnh truyn tn hiu s c th gp mt s trng hp khng mong mun, nhhng n cht lng ca tn hiu thu. Trng hp th nht, xut hin dy cc bit 0 hoc bit 1ko di, dn ti khng tch c ng h t dy xung thu trong qung thi gian xut hin cc dy

bit nh vy. Trng hp th hai, xut hin lp i lp li nhiu ln mt t m nn lm tng tch lurung pha ca tn hiu thu. C hai nguyn nhn ny sc khc phc nu s dng ngu nhin hotn hiu, tc l s dng b trn ti pha pht v b gii trn ti pha thu.

1.7.2. Cu to v hot ng ca b trn v b gii trn

Hnh 1.30 l cu trc ca b trn v b gii trn.

B trn v b gii trn u c b ghi dch, mi b ghi dch cha 5 t tr. Thi gian trca mi t tr bng 1/x v bng rng mt bit. Th d dy tn hiu u vo b trn l Di th khiqua hai t s l Di .(1/x). (1/x), ngha l tr hai bit, v (1/x). (1/x) = x

-2.

Tn hiu nh phn u vo b trn i ti b cng mun 2 th nht v c b sung thmtn hiu nh phn n t b cng mun 2 th hai. Do tn hiu nh phn u ra b trn Ds= Di

Ds(x -3 x -5) hoc Di = Ds (1 x -3 x -5) v v vy

Ds = Di / (1 x -3 x -5)

Dy tn hiu nh phn u ra b trn Ds qua ng truyn, a ti u vo b gii trn. Vvy dy nh phn u ra b gii trn c dng:

x-1x-1

x-1

x-1

x-1

x-1

x-1

S liu ra DsS liu vo Di

a) B trn

B cngmun 2

B ghi dchphn hi

S liu vo

DsS liu ra Di

'

b) B gii trn

Khi phc ng h

B cngmun 2

x-1

x-1x-1

B ghi dchphn hi

Hnh 1.30- B trn v gii trn


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( ) ( ) ( ) iisi DxxxxDxxDD === 55353 1/11

Nh vy l sau khi gii trn nhn c dy tn hiu nh phn nhu vo b trn.

Nh trnh by trn y, mc ch ca trn l xo trn cc bit trong mt t m theo mtquy lut nht nh. Da vo quy lut , b gii trn hot ng ngc li ti to tn hiu banu.

TM TT

Trong s ho tn hiu analog th phng php PCM l n gin nht, tuy nhin s dngphng php ny th tc bit mi knh thoi ln hn cc phng php khc. Phng php ghpknh theo thi gian l phng php ghp knh c s dng rng ri trong cc h thng thng tins. V vy phi s dng cc gii php ng b v ngu nhin ho tn hiu nhm m bo chtlng tn hiu thu, c th l t s li bit khng vt qu ngng ci t trc.

Cn phn bit cc loi ng b trong mng vin thng:

ng b sng mang l tch sng mang t tn hiu iu ch trong gii iu ch kt hp.

ng b k hiu l nhn bit cc thi im ly mu v quyt nh trong gii iu ch s tchthng tin logic t tn hiu analog thu c. ng b t m v ng b khung l nhn bit thiim bt u v kt thc t m hoc khung ti to khung t dy bit thu. ng b gi l cn

bng tr cc thi im n ca gi nhm ti cu trc bn tin trong mng chuyn mch gi.ng b mng l phn phi ng h chung n cc nt trong mt mng rng ln iu khinng h cc nt chy cng tc bit v pha vi ng h chung. ng ba phng tin l spxp cc phn t hn tp nh hnh nh, vn bn, audio, vodeo, v.v. trong truyn thng a phngtin ti cc mc tch hp khc nhau. ng bng h thi gian thc l phn phi thi gian tuyti (thi gian tiu chun quc gia) trong mng vin thng qun l mng.

BI TP(1) u vo b m ho - nn s c mt xung lng t VPAM = 875 , xc nh gi tr 8 bit

u ra ca b m ho - nn s.

(2) u vo b m ho - nn s c mt xung lng t VPAM = -1898 , xc nh 8 bit ura ca b m ho - nn s.

(3) u vo b m ho - nn s c mt xung lng t VPAM = 209, tm bin xung b nn tng ng vi 8 bit ti u ra b m ho - nn s.

(4) u vo b gii m - dn s c t m 0110 1101, tm bin xung u ra b gii m -dn s.

(5) V nhnh dng ca c tnh bin b m ho- nn s theo trc to y = f(x). Chox = 0,5; xc nh gi tr ca mi bit trong t m 8 bit ti u ra b m ho- nn s.

(6) Tnh tc bit ca mt knh thoi trong trng hp khng s dng b nn v c sdng b nn A = 87,6/ 13.

(Xem p s ti phn ph lc).


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CHNG II

GHP KNH PCM, PDH V SDH

2.1. GII THIU CHUNG

Ni dung chnh ca chng II gm c:

- Cu trc cc khung trong ghp knh PDH: cc loi khung 2/8, 8/34, 34/140 u c hailoi khung. l loi khung ch s dng chn dng v loi khung khc s dng c chn dngv chn m. Ring khung 140/565 ch s dng chn dng.

- Ghp knh SDH: s dng phng php xen byte sp xp hoc ghp cc byte tn hiuvo cc khung. Trong qu trnh ghp s dng con trng chnh tc v pha ca cckhung tn hiu n v khung ghp thng qua vic s dng chn dng v chn m.

2.2. GHP KNH PCM

2.2.1. S khi b ghp PCM-N

S khi b ghp PCM-N nh hnh 2.1.

Hnh 2.1- S khi b ghp knh PCM-N

2.2.2. Nguyn l hot ng

Theo tiu chun ca chu u th N = 30, ngha l ghp c 30 knh thoi. Theo tiuchun bc M N = 24. Pha m tn c N b sai ng (S) ng vai tr chuyn hai dy m tnthnh bn dy m tn v ngc li. C th l mt pha b sai ng kt ni vi my in thoi qua

LM MHNSGhpknh

Lp mng

S1

LMB TX

thu

u ra

u vo

X lbo hiu

B TX

pht CXK

1

1

N

Gii mng

Tchknh

GM-DS

SN

CXK

N 1

N

TX B


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hai si dy ng ca cp m tn, pha khc kt ni vi hai si thuc nhnh pht v hai si thucnhnh thu ca thit b PCM-N. u ra v u vo pha mng kt ni vi thit b ghp bc cao quacp ng trc.

Qu trnh chuyn i tn hiu ca PCM- 30 nh sau:

(1) Nhnh pht

Tn hiu thoi analog qua S, qua b lc thp hn ch bng tn ting ni n 3,4 kHz.Khi LM c chc nng ly mu tn hiu thoi vi tc 8 kHz. Khi m ho - nn s MH-NSthc hin lng t ho khng u v m ho mi xung lng t thnh 8 bit nhb m ho - nns A = 87,6/13. Tn hiu nh phn u ra khi MH-NS c a vo khi ghp knh. Ti y,ngoi tn hiu s ca 30 knh thoi cn c tn hiu s ca mt knh ng b v mt knh bo hiuc ghp xen bit, to thnh lung E1 c tc bit l 2048 kbit/s. Cui cng dy s liu nh phnc khi lp m ng chuyn thnh dy xung ba mc HDB-3.

Ngoi cc khi trn y, trong nhnh pht cn c b to xung pht hot ng ti tc bit2048 kbit/s v u ra ca n c khi chia tn to dy xung c tc bit theo yu cu iu

khin cc khi lin quan hot ng. Khi TXB to ra xung ng b khung v a khung. Khi xl bo hiu tip nhn tn hiu gi ca cc knh thoi chuyn thnh cc bit v c ghp vo vtr quy nh trong lung s E1.

(2) Nhnh thu

Dy tn hiu 2048 kbit/s HDB-3 t mng ti trc ht c khi gii m ng chuyni thnh dy xung hai mc. Trong tn hiu thu c cc t m ca 30 knh thoi, knh ng b vknh bo hiu. Cc loi tn hiu ny c tch ra nhkhi tch knh. Tn hiu ng b khung ivo khi to xung thu khi ng khi chia tn, nhm hnh thnh cc khe thi gian ng b vi

pha pht. Ngoi ra, khi tch knh cn c chc nng tch ng h t dy bit vo ng b tc

bit ca b to xung thu. Cc bit tn hiu gi c tch ra, i vo khi x l tn hiu gi chuyn thnh sng m tn rung chung my in thoi. B to xung thu cng c b phn chia tn hnh thnh dy xung iu khin hot ng ca cc khi nhnh thu.

Mi byte (8 bit) ca tn hiu thoi qua khi gii m - dn s GM-DS chuyn thnh mtxung c bin tng ng v a ti khi chn xung knh (CXK), u ra khi CXK l tp hpxung ca ring tng knh. Dy xung iu bin u ra khi CXK qua b lc thp khi phc tnhiu thoi analog, qua S ti my in thoi.

2.2.3. Cu trc khung v a khung

(1)i vi PCM-30Tn hiu su ra thit b PCM-30 c sp x p thnh khung v a khung trc khi

truyn. Cu trc ca khung v a khung nh hnh 2.2.

(a) Cu trc khung

Mi khung c thi hn l 125 s, c chia thnh 32 khe thi gian v nh s th t t

TS0n TS31. Mi TS c thi hn l 3,9 s v ghp 8 bit s liu. T m ng b khung c cutrc ring 0011011 v c ghp vo TS0 ca khung F0 v cc khung chn (F2, F4,..., F14). TrongTS0 ca cc khung l (F1, F3,..., F15) ghp cc bit nh sau: bit th nht s dng cho quc gia (Si),

bit th hai cnh bng 1 phn bit t m ng b khung vi t m ng b khung gi to khi7 bit cn li trong TS0 ca cc khung l trng vi 7 bit tng ng ca t m ng b khung, bitth ba cnh bo mt ng b khung (A). Tn hiu cc knh thoi th nht n th 15 ghp vo cc


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khe thi gian TS1n TS15; tn hiu cc knh thoi th 16 n th 30 ghp vo cc khe thi gianTS17n TS31. Tn hiu gi ca mi knh thoi c 4 bit (a, b, c, d) ghp vo mt na ca khe thi

gian TS16 ca cc khung F1 F15 trong a khung.

(b) Cu trc a khung

PCM-30 ghp c 30 knh thoi. V vy c tt c l 30 tn hiu gi. Mi khe TS16 ghpc tn hiu gi ca hai knh thoi. Do cn phi c tt c l 15 khe thi gian TS16 chuynti tn hiu gi ca tt c cc knh thoi. Ngoi ra cn thm mt TS16 na ghp xung ng b

a khung v cnh bo mt ng ba khung. Nh vy yu cu a khung phi cha 16 khung (mikhung c mt TS16).

Cc khe thi gian TS16 ca cc khung trong a khung c b tr chuyn ti s liu nhsau:

TS16 ca khung zero (F0) ghp cc bit ng ba khung 0000 v bit cnh bo mt ng bakhung Y.

Na bn tri ca TS16 khung th nht ghp 4 bit tn hiu gi ca knh thoi th nht, nabn phi ghp 4 bit tn hiu gi ca knh thoi th 16. Na bn tri ca TS16 khung th hai ghptn hiu gi ca knh thoi th hai v knh thoi th 17. C tip tc nh vy cho n TS16 cui

cng ca khung th 15 ghp tn hiu gi ca knh thoi 15 v knh thoi 30.(2)i vi PCM-24

Mi khung c mt bit c(F) t u khung v 24 khe thi gian, mi khe ghp 8 bit.

Tng s bit trong khung bng 8 bit 24 + 1 bit = 193 bit. Tc bit u ra PCM-24 ctnh nh sau:

RPCM-24 = 193 bit/ khung 8. 103 khung /s = 1544 kbit/s

a khung ca PCM-24 gm 24 khung, nh s th t t F1n F24, nh trn hnh 2.3.Mi bit ca t m ng b khung 001011 c ghp vo v tr bit th nht ca cc khung F4, 8, 12,

16, 20, 24. Cc bit th nht ca cc khung l truyn t m ng ba khung (cc bit m). Bit th nhtcc khung F2, 6, 10, 14, 18, 22, l cc bit kim tra s d chu trnh (cc bit e1 e6). Bit th tm ca cckhe thi gian trong khung F6, F12 , F18 v F24 truyn tn hiu gi (A, B, C, D).

TK= 125 s

Cc khung chn

Si

Si

SnSnSnSnSn1 A

0 0 1 1 0 1 1

Khung F0Khung F0

a

0

dcbadb c

0 0 0 x Y x x

Khung F1 F15Cc khung l

TS0 TS1 TS2 TS15 TS17TS16 TS31TS30TS29

Fo F1 F15F2 F3 F4 F5 F7F6 F8 F9 F10 F12F11 F13 F14

TK= 125 s 16 = 2 ms

a khung

Khung

Hnh 2.2- Cu trc khung v a khung PCM-30

A= 0 -c ng b khungA = 1- mtng b khungSi - sdng cho quc tSn, x- sdng cho quc gia

Y= 0- c ng ba khungY= 1- mtng ba khungabcd - 4 bit bo hiu


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2.3. GHP KNH PDH

2.3.1. Cc tiu chun tc bit

Hin nay trn th gii tn ti ba tiu chun tc bit. l cc tc bit theo tiu chun

Chu u, tiu chun Bc M v tiu chun Nht Bn. Cc tiu chun ny c trnh by didng phn cp s cn ng b nh hnh 2.4.

(1) Tiu chun chu u (CEPT)Tiu chun chu u bao gm 5 mc. Tc bit ca mc sau c to thnh bng cch

ghp bn lung s ca mc ng trc lin k. Mc th nht c tc bit 2048 Mbit/s c tothnh t thit b ghp knh PCM-30 hoc t tm mch trung k ca tng i in t s. Tc bitca mc th hai l 8448 kbit/s, gm c 120 knh. Mc th ba c 480 knh v tc bit bng34368 kbit/s. Mc th t c 1920 knh v tc bit l 139368 kbit/s. Bn mc ny c CCITT(hin nay i tn thnh ITU-T) chp nhn lm cc tc bit tiu chun quc t. Mc th nm ctc bit bng 564992 kbit/s v bao gm 7680 knh.

(2) Tiu chun Bc M

Tiu chun Bc M gm 5 mc. Tc bit ca mc th nht bng 1544 kbit/s, c hnhthnh t thit b ghp knh PCM-24 hoc t tng i in t s v c 24 knh. Ghp bn lung smc th nht c tc bit mc hai l 6312 kbit/s v gm c 96 knh. Mc th ba c tc bit

2048kbit/s

8448kbit/s

34368kbit/s

139264kbit/s

564992kbit/s

4 4 4 4

E1 E2 E3 E4 E5

CEPT

ITU-T

TK= 125s 24 = 3 ms

m

F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 F21 F22 F24F23

m m m m m m mm m me1 e2 e3 e4 e5 e6

0 10 0 1 1

A B C D

m

Hnh 2.3- Cu trc a khung ca PCM-24

ITU-T

1544kbit/s

6312kbit/s

32064kbit/s

97728kbit/s

400352kbit/s

44736kbit/s

274176kbit/s

560160kbit/s

4

5 3 4

7 6 2

T1 T2

T3 T4 T5

Bc M

Nht Bn

Hnh 2.4- Phn cp s cn ng b


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l 44736 kbit/s l kt qu ca ghp by lung s mc hai v bao gm 672 knh. Ba mc ny cITU-T chp nhn lm tiu chun quc t. Mc th t c c bng cch ghp su lung s mcba, tc bit bng 274176 kbit/s v bao gm 4032 knh. Mc th nm l kt qu ca ghp hailung s mc bn nhn c 8064 knh v tc bit l 560160 kbit/s.

(3) Tiu chun Nht Bn

Hai mc u tin hon ton ging tiu chun Bc M. Mc th ba c hnh thnh tghp nm lung s mc hai, c tc bit l 32064 kbit/s v 480 knh. Ba mc u tin ny c ITU-T chp nhn. Ghp ba lung s mc ba c lung s mc bn vi tc bit bng97728 kbit/s, 1440 knh. Mc cui cng ghp bn lung s mc bn nhn c 5760 knh vtc bit bng 400352 kbit/s.

2.3.2. K thut ghp knh PDH

2.3.2.1. S khi b ghp knh PDH

Nh trnh by trong mc 2.2.1, theo tiu chun chu u, c nm mc khi ghp bnlung vo sc mt lung ra. V vy s khi tng qut ca b ghp knh PDH nh hnh 2.5.

Mi lung s dng ring mt s khi nh: b nhn hi (M1), khi tch ng h (H),khi so pha v khi iu khin chn. Cc khi dng chung gm c: khi to xung ng b(TXB), khi to xung (TX) v khi ghp xen bit.

Lung nhnh c a ti b nhn hi v a vo khi tch ng h to ra tn siu khin ghi fG. C mi mt xung iu khin ghi tc ng vo M1 th mt bit ca lung nhnhc ghi vo mt nh. Cc bit ghi sc c ly ra theo ng hiu khin c f1 da

vo nguyn tc mt bit iu khin c tc ng vo M1 th mt bit c ly ra. Dy bit u ra bnhi vo khi ghp. Dy xung iu khin ghi v iu khin c i ti khi so pha. Cn c vo lch pha (lch thi gian) gia hai dy xung ny m u ra khi so pha xut hin xung dng

M1Tch H

Khi sopha

Khi iukhin chn

Khighpxenbit

1aLung nhnh 1

fG1 f1

2b3a3b

4b4a

Lung nhnh 2

Lung nhnh 3

Lung nhnh 4

TX BKhiTX

1 3 42

Hnh 2.5- S khi b ghp PDH

1b+_

2a


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hay m. Nhn c xung dng, khi iu khin chn pht lnh chn dng v nhn c xungm s pht lnh chn m. Khi ghp xen bit tin hnh chn xung theo lnh iu khin. Ngoi dy

bit ca bn lung vo cn c xung ng b t khi to xung ng b v cc bit bo hiu (khngth hin trong hnh v) u c a vo khi ghp ghp xen bit to thnh lung ra. Hot ngghp xen bit, so pha v hot ng chn c gii thiu trong cc phn sau.

Pha thu tin hnh tch knh theo trnh t ngc li vi qu trnh ghp. Trc tin tchxung ng b v tch ng h t dy bit thu c. Xung ng b lm gc thi gian tch cc bitca cc lung thnh phn, xung ng hc s dng iu khin b to xung thu. Dy xungknh ca mi lung c tch ring bit v cc t m tm bit ln lt c gii m v dn trthnh dy xung lng t nh pha pht. B lc thp khi phc tn hiu analog t dy xung lngt.

2.3.2.2. Phng php ghp xen bit

Qu trnh ghp xen bit c minh ho ti hnh 2.6.

Hnh 2.6- Ghp xen bit bn lung E1 thnh lung E2

Gi thit ghp bn lung mc 1 thnh lung mc 2. Trc khi ghp s liu cc lung,phi ghp mt xung hoc mt nhm xung ng b khung. Sau xung ng b khung l bit th nhtca lung E1#1, bit th nht ca lung E1#2, bit th nht ca lung E1#3, bit th nht ca lungE1# 4. Tip ghp cc bit th hai ca cc lung vo theo trnh t nh ghp cc bit th nht. Ctip tc ghp nh vy cho ht cc bit ca bn lung vo trong chu k ghp TGH. Ghp xung ng

b khung trc khi ghp tip cc bit s liu ca bn lung nhnh.

B ghp phi sp xp cc bit st li vi nhau v cn phi hnh thnh cc bit c rng bhn trong mt chu k ghp TGH ngoi xung ng b v cc bit ph khc phi cha ht cc bitca bn lung nhnh. V vy tc bit lung ra lun lun ln hn tc bit tng ca bn lungvo. Thi hn ca chu k ghp TGH ph thuc vo cp ghp.

t

E # 1

E # 2

E # 3

E # 4t

t

t

t

E2

t

XB

T = 125s

TGH


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Trong qu trnh ghp xen bit c th xy ra trng h p trt bit. Nguyn nhn ca hintng ny l do ng h tch t lung vo c tn s khc vi tn s ca ng h ni (hnh 2.7).

Nu tn sng h ni ti b hn tn s xung nh thi cha trong lung vo th mt bittrong b nhn hi c c hai ln, nhng ln sau l c khng nn gim tc bit u ra.

Ngc li, nu tn sng h ni ti ln hn tn s xung nh thi cha trong lung vo th mt

s bit c c thm nn lm tng tc bit ca lung ra. Tng thm hoc gim s bit u ra bnhm c quan hn trt. Trong thc t c hai dng trt, l trt iu khin c vtrt khng iu khin c. Trt iu khin c c ngha l iu khin c phm vi tnghoc gim s bit, chng hn trt mt octet hoc mt khung. Trt khng iu khin c l dolch nh thi v do khng iu khin c phm v tng hoc gim s bit. Nu phm vi lchtn s gia ng h ni ti v tn s lung bit vo duy tr phm vi 10-9 v tn s ly mu bng 8kHz th trt c th xy ra sau mi qung thi gian l 34 gi. Tng thm dung lng b nhnhi s hn ch trt khng iu khin c nhchuyn thi im trt n khong gia hai khis liu. Bin php quan trng hn ch trt l n nh tn s b to xung ca cc nt trongmng thng tin PDH.

2.3.2.3. Kthut chn trong PDH

(1) Khi nim

T hnh 2.7b bit c trong trng hp tn s (nghch o ca chu k) ng h ni cab ghp nh hn tn s ca lung nhnh th mt s bit tin bnh mt ti u ra (do gn trngthi im xut hin vi xung c trc). V vy bo ton thng tin ca lung nhnh, cn ti tocc bit b mt ny ca lung bit u ra b ghp v ghp chng vo mt v tr quy nh trongkhung. Hot ng nh vy gi l chn m.

Tri li, trong trng hp tn sng h ni ca b ghp ln hn tn s lung nhnh nhhnh 2.7a th mt s ln c khng lm gim tc bit lung ra. m bo tc bit nh mc,cn b sung mt s bit khng mang tin v ghp vo v tr quy nh trong khung. Nh vy gi lchn dng.

(2) Chn dng

B ghp knh PDH phi nhn bit c thi im c xung c nhng khng c xung u

ra b nhn hi, ng thi phi m c s bit khng mang tin cn b sung vo lung ra bnhny trong mt n v thi gian. Yu cu th nht c thc hin nhkhi so pha v yu cuth hai do bm m nhim.

ng h ni ti

Lung vob nh

t

t

c khng (gim)

t

a b

a) Tn sng h ni ti ln hn tn s lung vob) Tn sng h ni ti nh hn tn s lung vo

Hnh 2.7- Hin tng trt bit

Lung u rab nh

t

tThiu bit Tha bit

c thm (tng)


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u vo khi so pha c c dy bit iu khin ghi c tch ra t lung bit thu v dy bitiu khin c ly tng h ni (xem hnh 2.5). Khi so pha theo di mc lch pha (lch thigian) gia dy bit ghi v dy bit c v nhn bit quy lut bin thin ny ca lch pha xc nhnthi im thiu bit trong lung ra b nhn hi. T hnh 2.7b cho bit lch pha gia hai dy bitghi v c gim dn t gi tr cc i n gi tr cc tiu v sau thi im dch pha cc tiu ng

mt chu k ca dy bit c s xut hin thi im chn dng. Ti thi im u ra khi sopha c mt xung dng a ti khi iu khin chn, khi ny pht lnh iu khin chn dng.Nhn c lnh chn dng, khi ghp xen bit chn mt bit khng mang tin vo v tr quy nhca khung sau. Cn nu khng chn dng th v tr bit chn dng l bit tin.

Lnh iu khin chn dng hoc khng chn cng chnh l thng bo chuyn ti phathu. Nhn c thng bo ny, my thu xo bit chn dng trc khi gii m. Lnh iu khinchn dng trong khung ch s dng chn dng l 111 c ghp vo khung hin ti. i vikhung s dng chn dng v chn m th lnh iu khin chn dng l 111 111. Trong , ba

bit 111 trc c ghp vo khung hin ti v ba bit 111 sau ghp vo khung tip theo.

(3) Chn mCng nh trng hp chn dng, b ghp knh PDH phi nhn bit thi im m mt

bit c tc ng vo b nhn hi ly ra hai bit gn trng nhau. Nu khng c gii php g cbit th bit c thm trong cp bit ny s b mt v do mt thng tin. V vy mi ln c thml mt ln xy ra chn m.Khi so pha cn c vo lch pha gia dy bit ghi v dy bit c bitc thi im chn m. T hnh 2.7b bit c lch pha tng dn t gi tr cc tiu n gi trcc i. Ti thi im lch pha t gi tr cc i, mt xung m xut hin ti u ra khi so pha,i ti khi iu khin chn v khi ny pht lnh chn m. Nhn c lnh ny, khi ghp xen bitghp mt bit mang thng tin ca bit c ra sau (0 hoc 1) vo v tr quy nh trong khung tiptheo. My thu nhn c thng bo chn m, tin hnh tch bit chn m x l nh cc bitthng tin khc. Lnh iu khin chn m gm 000 000. Trong ba bit 000 trc c ghp vokhung hin ti v ba bit 000 sau ghp vo khung tip theo.

(4) Khng chn

i vi khung ch s dng chn dng, khi khng chn th cc bit iu khin chn l 000c ghp vo khung hin ti; trong trng hp ny cc bit chn l cc bit tin ly t cc lungnhnh. i vi khung s dng chn dng v chn m th th lnh iu khin khng chn l111000, trong ba bit 111 ghp vo khung hin ti v ba bit 000 ghp vo khung tip theo. Nhnc lnh khng chn, b ghp ci t bit chn dng l bit tin v bit chn m l bit khng mangtin.

2.3.3. Cu trc khung PDH in hnh

2.3.3.1. Cu trc khung b ghp 2/8

B ghp ny ghp bn lung nhnh 2048 kbit/s 50 ppm thnh lung s mc 2 c tc

bit 8448 kbit/s 30ppm. Cu trc khung trong trng hp ch s dng chn dng nh hnh 2.8.

Cu trc khung b ghp 2/8 khi s dng chn dng, chn m v khng chn nh hnh2.9.

Trong khung ghp cc bit knh nghip v 32 kbit/s kt ni gia hai b ghp u cui lkt qu ca iu ch delta thch ng (ADMo). Bit th t trong phn khung 3 ghp bit gi chungca knh dch v. Bn bit u tin ca cc PK2, 3, 4 ghp cc bit iu khin chn. Bit th 5 n


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bit th 8 ca phn khung 4 ghp bn bit chn m. Bit th 9 n bit th 12 ca phn khung 4 ghpbn bit chn dng.

Hnh 2.8- Cu trc khung b ghp 2/8 s dng chn dng v khng chn

T hnh 2.9 kim tra li nhng vn phn tch i vi khung ca b ghp 2/8 khichn m, chn dng v khng chn.

Tc bit nh mc tng ca bn lung nhnh l:

V= 2048. 103 4 = 8192. 10 3 bit/s (2.1)

Khi khng chn m v cng khng chn dng th cc bit 5 8 ca phn khung 4 l cc

bit khng mang tin, cc bit 9 12 ca phn khung 4 l cc bit tin. Vy tng s bit ca 4 lungnhnh ghp trong khung khi c bn lung khng chn l:

T = 256 bit 3 + 252 + 4 bit = 1024 bit / khung

Tc bit truyn cc bit tin ca khung khi c bn lung khng chn l:

Vtruyn = 1024 bit / khung 8. 103 khung / s = 8192. 103 bit/s (2.2)

Kt qu ca cc biu thc (2.1) v (2.2) nh nhau, vy vic quy nh v tr cc bit trn yhon ton hp l.

Khi c bn lung u chn dng th cc bit 5 12 ca phn khung 4 l cc bit khngmang tin.

Hnh 2.9- Cu trc khung b ghp 2/8 khi s dng chn dng, chn m v khng chn

TK= 1056 bit = 125 s

ng b khung

8 256 4

11100110

4 4 2522564 4

Cc bit iukhin chn Cc bit chn m

Cc bit chn dngKnh dch v 32 kbit/s

4

D tr

Cnh bo mt ng b khungKnh dch v chung

PK1 PK2 PK3 PK4

4256

TK= 848 bit = 100,4 s

44 4

PK1 PK2 PK3 PK4

12 200 4 204208

1111010000 b11b12

Cnh bo mt ng b khung

Bit d trCc bit chn dng

Cc bt iu khin chn

208

ng b khung


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Vy tng s bit tin ca bn lung nhnh ghp xen bit trong khung khi chn dng l:

T(+) = 256 bit 3 + 252 bit = 1020 bit / khung

Khi c bn lung u chn m th cc bit 5 12 ca phn khung 4 l cc bit tin.

Vy tng s bit tin ghp xen bit trong khung khi c bn lung u chn m l:

T(-)

= 256 bit 3 + 252 bit + 4 bit + 4 bit = 1028 bit / khung.

Cn ch l cc lung nhnh hot ng c lp vi nhau nn c th lung ny chnnhng lung khc khng chn.


B ghp 8/34 c hai kiu cu trc khung. Kiu cu trc khung th nht s dng chndng v khng chn. Kiu cu trc khung th hai s dng chn dng, chn m v khng chn.

B ghp ny ghp bn lung nhnh 8448 kbit/s 30 ppm thnh lung mc ba 34368 kbit/s 20ppm. Cu trc khung khi s dng chn dng v khng chn nh hnh 2.10.

Tng s bit trong khung bng 1536 bit v c chia lm 4 phn khung. Cc bit iu khinchn ghp vo u cc phn khung th hai, th ba v th t ca khung hin ti. Cc bit chndng ghp vo v tr bit th 5 v bit th 8 ca PK4. Lnh iu khin chn dng ca mi lungnhnh gm 3 bit 111 v khng chn l 000. Lung no c yu cu chn dng th chn mt bit

khng mang tin vo v tr bit dnh ring cho mnh ti v tr bit th 5 8 trong PK4. Khi khngchn th bit chn c thay bng bit thng tin ly t lung nhnh y.

Cu trc khung 8/34 khi s dng chn dng, chn m v khng chn nh hnh 2.11.

Cc bit chn dng

4 4

PK2 PK3 PK4

376380

Cc bit iu khin chn

380

TK= 1536 bit = 44,9 s

4

PK1

Bit d trCnh bo mtng b khung

1111010000

ng b khung

Hnh 2.10- Cu trc khung b ghp 8/34 s dng chn dng

372 410 1 1

Hnh 2.11- Cu trc khung b ghp 8/34 khi s dngchn dng, chn m v khng chn

TK= 2148bit = 62,5 s

ng b khung

12 704

111110100000

4 4 4 7004

Cc bit iu khin chn

Cc bit chn m

Cc bit chn dng

Cc bit dch v

4

D tr

PK1 PK2 PK3

70444


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Khung bao gm 2148 bit, c thi hn 62,5 s v c chia lm 3 phn khung. S bit trongmi phn khung l 716. H thng c thit k ghp xen bit 4 lung nhnh vo cc v tr bit 13n 716 trong phn khung 1 v 2 v v tr bit 17 n 716 trong phn khung 3. T m ng bkhung 111110100000 (12 bit) chim v tr bit 1 n 12 ca phn khung 1. Cc bit iu khin chn

v cc bit dch v chim v tr cc bit 1 n 4, 9 n 12 trong phn khung 2 v 1 n 4 trong phnkhung 3. Chn dng c ch th bi t m 111 trong hai khung lin tip. Trong khi chn mc ch th bi t m 000 trong hai khung lin tip. Ch th khng chn gm cc bit 111 trongkhung hin ti v cc bit 000 trong khung tip theo. Cc bit 9 n 12 trong phn khung 3 c sdng chuyn ti cc bit chn m. Cc bit chn dng chim v tr bit 13 n 16 ca phn khung3. Trong phn khung 2 c cc bit 5 v 6 l cc bit ca knh dch v s ADMo 32 kbit/s, bit 7 chth cnh bo ti b ghp u xa, bit 8 l tn hiu rung chung ca knh dch v.


C hai loi cu trc khung: loi th nht ch s dng chn dng, loi th hai c c chndng v chn m. Cu trc khung s dng chn dng nh hnh 2.12.

C hai phng php hnh thnh tc bit mc 4. Phng php th nht s dng 4 lung34368 kbit/s. Phng php th hai ghp trc tip 16 lung 8448 kbit/s nhn c lung mc bn.C hai phng php u s dng chn dng. Cu trc khung ca phng php ghp th nhtc th hin ti hnh 2.12. Khung bao gm 2928 bit, chia thnh 6 phn khung 488 bit v thi hn

bng 44,9 s. Trong phn khung1, bit 1 n 12 truyn t m ng b khung 111110100000 (12bit), bit 13 l bit cnh bo truyn ti u xa (bng1 khi c cnh bo, bng 0 khi khng c cnhbo). Bit 14 n 16 trong phn khung 1 s dng cho quc gia v ci t bng 1 khi truyn qua

bin gii quc gia. Trong cc phn khung 2, 3, 4, 5, 6 l cc bit iu khin chn. Khi c lnh iukhin chn 11111 th chn mt bit khng mang tin vo v tr cc bit chn dng trong khung sau.Khi khng chn th truyn 00000 v bit chn trong khung sau l bit tin. Cc bit cn li trong cc

phn khung l ca bn lung nhnh ghp xen bit.

Cu trc khung b ghp 34/140 khi s dng chn dng v chn m nh hnh 2.13.

Cu trc khung ny l ca b ghp khi ghp 4 lung mc 3 thnh lung mc 4 c tc

bit 139264 kbit/s 15ppm. Khung c 2176 bit, thi hn 15,625 s c chia lm 4 phn khung544 bit. Trong PK1, bit 1 n bit 10 dnh cho t m ng b khung 1111010000, bit 11 knh dchv 32 kbit/s iu ch Delta thch ng (ADMo), bit 12 s dng rung chung cho knh dch v.

PK5

Cc bit chn dng

PK2 PK3 PK4

484

Cc bit iu khin chn

4 4 480

TK= 2928 bit = 44,9 sPK1

S dng cho quc giaCnh bo mtng b khung

111110100000

ng b khung

Hnh 2.12- Cu trc khung b ghp 34/140 s dng chn dng

472 412 31 4844 4844 4844

PK6


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Trong PK2, 3, 4, bit 1 n bit 4 s dng cho iu khin chn. Khi chn dng cu trc iu khinchn 111 c truyn trong hai khung lin tip. Tri li, khi chn m th cc bit iu khin chn l000 truyn trong hai khung lin tip. Khi khng chn, cc bit iu khin chn 111 truyn trongkhung hin ti v 000 truyn trong khung sau. Trong PK4, cc bit 5 n 8 l cc bit chn m mangtin khi chn v khng mang tin khi khng chn; cc bit 9 n 12 mang tin khi khng chn dng

v khng mang tin khi c chn dng. Cc bit cn li trong khung l ca 4 lung nhnh ghp xenbit.


B ghp ny s dng bn lung 139264 kbit/s 15 ppm ghp xen bit nhn c lung

mc 5 c tc bit 564992 kbit/s 15 ppm. Cu trc khung ca b ghp nh hnh 2.14.

Vo nm 1986 s dng tc bit 564992 kbit/s c 7680 knh thoi trn trung k hoch thng dung lng cao. H thng ny bao gm thit b thng tin quang v b ghp bn lung139264 kbit/s m CMI v chn bit to ra lung mc nm 564992 kbit/s. Lung s ny cchuyn thnh m 5B6B v kt hp vi cc bit mo u to ra tc bit ng truyn xp x680 Mbit/s trn cp si quang a mode. Cc bit mo u bao gm knh dch v, iu khinchuyn mch bo v, gim st v.v. H s suy hao ca si quang a mode khong 0,6 dB/km ti

bc sng 1310 nm nn khong lp cht c 30 km. t lu sau si quang n mode xut hinv s dng tc bit mc nm ny t chc mng thng tin quang PDH c khong lp t

gn 100 km ti bc sng 1550 nm.Khung c 2688 bit v c chia lm 7 phn khung, mi phn khung 384 bit. Trong qu

trnh ghp ch s dng chn dng. Lnh iu khin chn dng ca mi lung nhnh l 11111

Cc bit chn dng

4

PK2 PK3 PK4

4 532540

Cc bit iu khin chn

540

TK= 2176 bit = 15,625 s

4

PK1

Cc bit dch v1111010000ng b khung

Hnh 2.13- Cu trc khung b ghp 34 /140 s dng chn dng, chn m v khng chn

532 410 2 4

Cc bit chn m

PK6PK5

Cc bit chn dng

PK2 PK3 PK4

Cc bit iu khin chn

4 4 376

TK= 2688 bitPK1

111110100000

Hnh 2.14- Cu trc khung b ghp 140/565 s dng chn dng v khng chn

380437212 3804 3804 3804

PK7

3804 4

Cc bit cnh bong b khung


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c truyn ti v tr bit th nht n th nm ca PK2 n PK6. Nhn c lnh ny, b ghpchn mt bit khng mang tin vo mt trong bn v tr ti bit th 5 n th 8 trong phn khung 7ca khung sau. Khi khng chn, lnh iu khin c cu trc 00000 nn bit chn c thay th bi

bit tin ca chnh lung nhnh . Bn bit u tin trong PK7 l cc bit cnh bo ca cc lungnhnh.

2.4. GHP KNH SDH2.4.1. Cc tiu chun ghp knh SDH

2.4.1.1. Cc khuyn nghca ITU-T vSDH

G.707 Cc tc bit SDH

G.708 Giao din nt mng SDH

G.709 Cu trc ghp ng b

G.773 Cc b giao thc ca giao din Q

G.774 M hnh thng tin qun l SDH

G.782 Cc kiu v cc c tnh chung ca thit b ghp SDH

G.783 Cc c tnh ca cc khi chc nng thit b ghp SDH

G.784 Qun l SDH

G.803 Cu trc mng truyn dn da vo SDH

G.957 Cc giao din quang ca thit b v h thng lin quan n SDH

G.958 Cc h thng s SDH s dng cho cp si quang

2.4.1.2. Tc bit ca SDH

Mng SDH l mng ng b, trong mi phn t mng s dng tn hiu ng bccung cp t mt ngun ng h chun quc gia. Theo khuyn ngh G.707/Y.1322 th tc bit

phn cp SDH c 6 mc. Mc 0 c tc bit l 51, 84 Mbit/s. Mc 1 c tc bit l 155,52Mbit/s. Tc bit cc mc cao l bi s nguyn ca tc bit mc 1. Su mc tc bit baogm:

STM-0 = 51,840 Mbit/s

STM-1 = 155,520 Mbit/s

STM- 4 = 622,08 Mbit/s

STM- 16 = 2048,32 Mbit/sSTM- 64 = 9953,28 Mbit/s

STM- 256 = 39813,120 Mbit/s

Cc lung nhnh PDH u vo thit b ghp SDH c ITU-T chp nhn gm c:

- Theo tiu chun chu u: 2,048 Mbit/s; 8,448 Mbit/s; 34,368 Mbit/s v 139,264 Mbit/s.

- Theo tiu chun Bc M: 1,544 Mbit/s; 6,312 Mbit/s v 44,376 Mbit/s.

2.4.1.3. Quy nh vconteno (VC)

Tn hiu lung nhnh PDH a n thit b ghp SDH trong khong thi gian 125 sc cha trong mt hp c dung lng nht nh v gn nhn ch r trong hp cha loi tn hiulung nhnh no, hp nh vy gi l conteno. C hai loi conteno: conteno mc thp


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VC-11, VC-12, VC-2 v conteno mc cao VC-3, VC-4. Ngoi cc conteno, khuyn nghG.707/Y.1322 cn quy nh cc loi conteno kt chui nh: VC-4-4c, VC-4-16c, VC-4-64c vVC-4-256c. Tc bit tng v tc bit ti trng ca cc conteno n v kt chui c litk nh bng 2.1.

Bng 2.1- Dung lng cc VC-nLoi VC-n Tc bit tng (kbit/s) Tc bit ca ti trng (kbit/s)

VC-11 1664 1600

VC-12 2240 2176

VC-2 6848 6784

VC-3 48960 48384

VC-4 150336 149760

VC-4-4c 601344 599040

VC-4-16c 2405376 2396160

VC-4-64c 9621504 9584640

VC-4-256c 38486016 38338560

2.4.1.4.Quy nh vng, tuyn v on

Khi tm hiu thit b cng nh cu trc v hot ng ca mng SDH c lin quan n khinim vng, tuyn v on nh biu th trn hnh 2.15, v vy trong mc ny trnh by cc quynh .

(1) on (section)

C hai loi on, l on ghp v on lp. on ghp l mi trng truyn dn giahai trm ghp knh k tip nhau, trong mt trm to ra tn hiu STM-N v trm kia kt cui tnhiu STM-N ny. on lp l b phn truyn dn gia hai trm lp k tip nhau, hoc gia trmlp v trm ghp knh k tip.

(2) Tuyn (Path)

Tuyn l b phn truyn dn c tnh tim nhp vo mt tn hiu c hnh thnh biconteno (VC) n im tch ra chnh tn hiu y. C hai loi tuyn, l tuyn mc thp linquan n tn hiu VC-11, VC-12, VC-2 v tuyn mc cao lin quan n tn hiu VC-3 v VC-4.

(3) ng (Line) l tp hp ca tt c cc tuyn ca h thng truyn dn thng sut tnhiu STM-N.

VC

VC

VC

VC

VCMUX REG REGMUXMUX Tuyn

Tuyn (VC)on lp on lp

ng (STM-N) on ghp

Tuyn

Hnh 2.15- M hnh xc nh ng, on v tuyn

VC


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2.4.2. S khi ghp cc lung PDH vo khung STM-N

2.4.2.1. S khi b ghp knh SDH

S khi b ghp knh SDH nh hnh 2.16.

2.4.2.2. Chc nng cc khi

u vo b ghp l cc lung nhnh PDH ca chu u v Bc M. Cc khi ca thit bghp c phn thnh cc nhm C-n, VC-n, TU-n, TUG-n, AU-n, AUG v STM-N. Chc nngca cc khi trong cc nhm ny l:

(1) C-n: contenmc n (n = 1, 2, 3, 4).

Mc 1 ca Bc M k hiu C-11 v ca chu u k hiu C-12. Cc mc cn li c mtch s. C-n c chc nng sp xp lung nhnh PDH tng ng, n thm cc byte khng mangtin cho s byte nh mc ca khung chun C-n.

(2) VC-n: conteno mc n.

VC-n c chc nng sp xp tn hiu C-n, chn thm bit chuyn lung vo cn ng bthnh lung ra ng b, b sung cc byte mo u tuyn (VC-n POH).

(3) TU-n: con tr khi nhnh mc n (n = 11, 12 v 3).

Con tr khi nhnh c chc nng ng chnh tc bit v tc khung tn hiu ghp VC-

n mc thp cho ph hp vi tc bit cng nh tc khung ca tn hiu VC-n mc cao hn.(4) TUG-n: nhm khi nhnh mc n (n = 2, 3)

Nhm khi nhnh ghp xen byte cc tn hiu TU-n mc thp thnh khung chun TUG-2hoc ghp cc tn hiu TUG-2 thnh khung chun TUG-3. Cng c th sp xp tn hiu TU-3thnh khung TUG-3.

(5) AU-n: con tr khi qun l mc n (n = 3, 4).

Con tr khi qun l ng chnh tc bit v tc khung ca tn hiu ghp VC-3 hocVC-4 cho ph hp vi tc bit v tc khung ca tn hiu AUG.

(6) STM-N: mun truyn dn ng b mc N (N = 1, 4, 16, 64 v 256).

STM-N ghp xen byte N tn hiu AUG, mo u on v con tr khi qun l AU-n thnhkhung STM-N.

Ch thch:

X l con trng ghp cc lung nhnh PDH chu uN = 1, 4, 16, 64 v 256

3 TUG-3

C-4STM-N AUG AU-4 VC-4

C-3

TU-3 VC-3

VC-3AU-3

C-11

TUG-2

C-2

C-12

TU-2 VC-2

VC-12TU-12

139,264Mbit/s

VC-11TU-11

44,736 Mbit/s34,368 Mbit/s

6,312 Mbit/s

2,048 Mbit/s

1,544 Mbit/s

77

3

3

4

1

N 1

1

Hnh 2.16- S khi thit b ghp knh SDH

STM-01


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2.4.3. Qu trnh ghp cc lung

giao trinh tin hieu so

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