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HC VIN CNG NGHBU 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 NGHBU 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 sl mt lnh vc rt quan trng. Khi u ca ghp knh tn hiu sl iu xung m (PCM) v iu chDelta (DM), trong PCM c sdng rng ri hn. TPCM, cc nh chto thit bvin thng cho ra i thit bghp knh cn ng b(PDH) v

sau l thit bghp 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 vInternet.

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

bc sng (WDM) xut hin. c thtn dng bng tn truyn dn ti min ca sthhaica si quang n mode, kthut 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 nghghp knh cnh. V vy rng bng tnvn khng c tn dng trit . Theo c tnh th hiu sut sdng rng bng tn khdngca hthng thng tin quang SDH mi t c 50%. Trc thc tmt mt rng bng tnng truyn cn blng ph, mt khc cng nghtruyn gi IP v ATM i hi hthng thngtin quang SDH phi thomn nhu cu trc mt v ccho tng lai, khi m cc dch vgia tng

pht trin trnh cao. Chc ththomn nhu cu vtc truyn dn v nng cao hiu sutsdng bng tn ng truyn bng cch thay i cc phng thc truyn ti lu lng sliu.

Vn mu cht ng dng cc phng thc truyn ti tin tin l kt chui cc ccconten, sdng cc phng thc ng gi sliu thch hp, truyn ti gi linh hot theo cch tisdng khng gian v chuyn mch bo vthng minh nng cao tin cy ca mng v rtngn thi gian phc hi ca hthng khi c sc. Nhng vn ny sc phn tch ktrong

cc chsau y:1) Trnh by mt skhi nim cbn trong truyn dn tn hiu, c bit l tn hiu sv

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

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

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

mnh khi i chng vi p sv trli trong phn phlc.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 nghBu chnh Vinthng. Tuy nhin, y l ln bin son u tin nn khng trnh khi thiu st vni dung v hnhthc. Rt mong cc c gigp ti liu ngy cng hon thin hn.

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

Xin chn thnh cm n!

Nhm tc gi


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3

CHNG I

MT SKHI NIM CBN

TRONG TRUYN DN TN HIU

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

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

- Cc phng php sho tn hiu analog nh: PCM, DPCM v DM. Trong phngphp PCM c sdng rng ri trong cc hthng 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 sdng trong ghp knh PDH, SDH.

- ng btrong vin thng:

tin hnh phn tch cc phng thc ng bnh: ng bsng mang, ng bkhiu, ng bbit, ng bkhung, ng bgi, ng bmng, 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 sdng cng bmng, ng bsng mang, ng bkhung, ng

bk 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 trbin lin

tc theo thi gian, th dtn hiu thoi analog.

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

S(t) = Asin (t + )

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

Nu tn hiu l tp hp ca nhiu tn sth ngoi cc tham strn 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 trbin 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 trbin l hm ri rc ca thi giannhtn hiu xung. Tuy nhin, khc vi tn hiu xung chbin ca cc xung bng 0 hoc 1,mt khc tp hp ca mt nhm xung i din cho mt chs, hoc mt k tno . 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 nhphn 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 chnht 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 sdng 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 : Pxl cng sut tn hiu (mW) ti im cn xc nh mc in, P0l cng sut tn hu tiim tham kho (mW).

Mc in tuyt i: ( )W1

log10m

PdBL xm =

L(dB)m= 0 dBmkhi 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) Tstn hiu trn nhiu

( )n

s

P

PdBSNR log10=

n

s

n

s

I

I

V

Vlog20log20 ==

trong : Ps, Vs, Istng ng l cng sut, in p v dng in tn hiu; Pn, Vn, Intng 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 sdng 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 cvo cc quy nhsau y:

(1) rng bng tn in (BW)e

rng bng tn in l bng tn ttn stn hiu bng zero n tn stn hiu m ti p ng ca tn hiu (hskhuch i, in p, dng in) gim cn 0,707 so vi gi trcc 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 ttn siu chbng zero n tn siu chmti mc cng sut quang gim 50% (3dBm) so vi cng sut quang cc i, nhminh 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 hthng 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 , hthng thng tin quang ghp bc sng (WDM) c thtruyn ngthi hng chc bc sng khc nhau nm trong min ca sthhai (1300 nm) hoc ca sthba

(1550 nm) ca si quang n mode.

1.2.4. Hthng truyn dn sv cc tham s

1.2.4.1. Hthng truyn dn s

H thng truyn dn sbao gm h thng truyn dn cp si quang v h thng truyndn vi ba s. Hthng truyn dn vi ba sl hthng a im ng thng. Hthng truyn dnscp si quang c thsdng cu trc ng thng, vng hoc hn hp. Cc cu hnh ny sc trnh by chi tit trong chng III. Di y chgii thiu khi qut mt vi cu trc cbnca hthng.

(1) Hthng truyn dn ng thngCc cu hnh ca hthng truyn dn ng thng nhhnh 1.3.

Ch thch: TRM- Bghp u cui, ADM- Bghp xen/ r, REG - Bti sinh (blp).

Hnh 1.3. Cc cu hnh ng thng

Trong cu hnh im ni im chc hai bghp u cui kt ni trc tip vi nhau hocqua blp bng ng truyn s, to thnh mt ng thng, v vy gi l hthng ng thng.Ngoi ra cn c tn gi khc l hthng h. Cu hnh a im, xen/ rngoi hai bghp u cuicn c thm mt hoc nhiu bghp xen rc kt ni vi nhau bi ng truyn sthnh mtng thng. Cu hnh a im, rnhnh cng l hthng h. Ti a im xen/r, cc lung s

c tip tc truyn ti mt bghp u cui khc to thnh mt nhnh ca hthng chnh.Cc cu hnh ng thng p dng cho vi ba sv 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 khnng tduy tr khi ng truyn c sc, chng hn t cphoc hng nt.

(2) Hthng truyn dn vng (ring)

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

Hnh 1.4. Cu hnh vng ca hthng truyn dn s

1.2.4.2.Cc tham s

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

Cc n vo tc bit: bit/s, kbit/s (1kbit/s = 103bit/s), Mbit/s (1Mbit/s = 103kbit/s =106bit/s), Gbit/s (1Gbit/s = 103Mbit/s = 106kbit/s = 109bit/s), Tbit/s (1Tbit/s = 103Gbit/s = 106

Mbit/s = 109kbit/s = 1012bit/s). Tn hiu sc sdng trong cc mng thng tin s.

(2) Tsli bit BER: sbit bli chia cho tng sbit truyn.

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

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

(3) Rung pha (Jitter)

Rung pha l siu chpha khng mong mun ca tn hiu xung xut hin trong truyndn sv l sbin i nhcc 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 sssm hn hoc mun hn sovi tn hiu chun, nhminh hotrn hnh 1.5.

Hnh 1.5. Tn hiu sbrung pha

ADM

ADM

ADM

ADM

Ring STM-N

Bin ng bao brung pha

t

b) Tn hiu sbrung 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 hngun v ng hthit bthu trong cng mt mng, lch tn sgia ng hca thit bSDH v tn sca lung nhnh PDH.

1.3. SHO TN HIU ANALOG

Sho tn hiu analog l chuyn i tn hiu analog thnh tn hiu s. Mun vy c ths

dng mt trong cc phng php sau y:

- iu xung m (PCM)

- iu xung m vi sai (DPCM)

- iu chDelta (DM)

Sau y trnh by cc phng php sho tn hiu analog.

1.3.1. iu xung m PCM

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

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

Skhi ca cc qu trnh chuyn i A/D v D/A nhhnh 1.6.

Hnh 1.6- Skhi qu trnh chuyn i A/D v D/A trong hthng PCM

1.3.1.1. Chuyn i A/D

(1) Ly mu

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

max2

1

fTm (1.1)

trong f-maxl tn sln nht ca tn hiu analog.

Hnh 1.7- Ly mu tn hiu analog

Bmho-nn s

Blymu

Blngtho

Bgii m- dn s

Blcthpng

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 t0,3 n 3,4 kHz. Tbiu thc (1.1), c thly gitrfmax = 4000 Hz. Do chu kly mu tn hiu thoi l:

sHz

Tm 12540002

1=

= (1.2)

Hoc tn sly mu tn hiu thoi:

kHzffm 82 max == (1.3)

(2) Lng tho

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

C hai phng php lng tho: u v khng u.

Lng tho u

Hnh 1.8 minh holng tho u. Lng tho 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 lngtgn nht snhn c xung lng t.

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

lng tQ v c mi quan hsau y:

=Q

a2 (1.4)

Hnh 1.8- Lng tho u

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

trong gii hn t- /2 n +/2. Cng sut mo lng tPMLTc 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 bgi trtc thi ca bin

xung ly mu trong mt bc lng t. WLT

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

Tn hiu analog

S(t)

t

Xung lng t

Tm

- Bc lng tu

Mc lng t

01

23

4

5

67


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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. Nhvy tscng sut tn hiu c bin ln trn cng sut nhiulng ts ln hn tscng sut tn hiu c bin yu trn cng sut mo lng t. Theo

phn tch phth tn hiu thoi chyu do cc thnh phn tn hiu c cng yu to thnh. Vthnu sdng lng tho u slm gim cht lng tn hiu thoi ti u thu. Mun khc

phc nhc im ny, trong thit bghp knh PCM chsdng lng tho khng u.

Lng tho khng u

Tri vi lng tho u, lng tho 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, nhtrn hnh 1.9. Lng tho khng u c thc hin bng cch sdng bnn.

Hnh 1.9- Lng tho khng u

(3) M ho - nn s

c tnh bin bm ho - nn s

Chc nng ca m ho l chuyn i bin xung lng tthnh mt tm gm mt sbit nht nh. Theo kt qunghin cu v tnh ton ca nhiu tc gith trong trng hp lng

thou, bin cc i ca xung ly mu tn hiu thoi bng 4096 . Do mi tm phicha 12 bit, dn ti hu qul tc bit mi knh thoi ln gp 1,5 ln tc bit tiu chun 64kbit/s. Mun nhn c tc bit tiu chun, thng sdng bnn c c tnh bin dng

logarit, cn c gi l bnn analog. Biuthc ton hc ca bnn analog theo tiu chun chuu c dng:

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

b hn gi trhm bc thang th V


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Khi nim: ghp knh theo tn sl tn s(hoc bng tn) ca cc knh khc nhau, nhngc truyn ng thi qua mi trng truyn dn. Mun vy phi sdng biu ch, gii iuchv blc bng.

1.4.1.1. Skhi v nguyn l hot ng bFDM

Skhi hthng ghp knh v tch knh theo tn snhhnh 1.13.

Sc N nhnh, mi nhnh dnh cho mt knh. Schc mt cp iu ch, nhng trongthc tc nhiu cp iu ch. Tu thuc mi trng truyn dn l v tuyn, dy trn, cp ixng hay cp ng trc m sdng mt scp iu chcho thch hp.

Pha pht: tn hiu ting ni qua blc thp hn chbng tn t0,3 n 3,4 kHz. Bngtn ny c iu chtheo phng thc iu bin vi sng mang fNc hai bng bn. Trongghp knh theo tn schtruyn mt bng bn, loi bbng bn thhai v sng mang nhblc

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 knhau l 4 kHz.

Hnh 1.13- Skhi hthng ghp knh theo tn s

Hnh 1.14- Tn hiu iu bin trong cp iu chknh

Cp iu chknh hnh thnh bng tn cs60 108 kHz. Tbng tn csto ra bngtn nhm trung gian nhsng mang nhm trung gian. Tbng 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

Blcthp

Blcbng

Blcthp

Biuch

Blcbng

f1

Blcbng

Biuch

Blcthp

fN

f2

fN

Blcbng

Bgiiiu ch

Blcthp

f1

Blcbng

Bgiiiu ch

Blcthp

f2

Blcbng

Bgiiiu ch

Blcthp

fN

c tnh suy hao - tn sca blc bng

0,3 3,4

Bng trnBng di

f (kHz)

Bng tn thoi


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Pha thu: cc blc bng ti nhnh pht v nhnh thu ca mi knh c bng tnnhnhau.u vo nhnh thu c N b lc bng ni song song v ng vai tr tch knh. Biu ch tinhnh pht sdng sng mang no th bgii iu chca knh y cng sdng sng mang nhvy. Tn hiu knh c gii iu chvi sng mang v u ra bgii iu chngoi bng m tncn c cc thnh phn tn scao. Blc thp loi bcc thnh phn tn scao, chgili bng

m tn.Ghp knh theo tn sc u im l cc biu chv gii iu chc cu to n gin

(sdng cc diode bn dn), bng tn mi knh chbng 4 kHz nn c thghp c nhiu knh.Chng hn, my ghp knh cp ng trc c thghp ti 1920 knh. Tuy nhin do sdng iubin nn khnng chng nhiu km.

1.4.1.2. Ghp phn chia theo tn strc giao OFDM

(1) Mu

Ghp phn chia theo tn strc giao l mt cng nghtrong 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 skhng i xng (ADSL), thng sdng thut nga m ri rc (DMT).Tuy thut ng c khc nhau nhng bn cht ca hai k thut ny u pht sinh tcng mt tng. V vy trong phn ny xt trng hp sdng cho mi trng khng dy.

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

nguyn ln ca chu trnh (T) k hiu, nhbiu thtrn hnh 1.15. y l vn quan trng ca kthut OFDM.

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

(2) M hnh hthngiu chcc sng mang trc giao cn sdng phng php bin i Fourier ri rc

ngc (IDFT). Hnh 1.16 l sbiu chOFDM.

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 chc dy sliu d0, d1,...., dN-1trong dnl k hiu phc (c thnhntu ra biu chphc nhQAM, PSK, v.v.). Githit thc hin bin i Fourier ngc trndy 2dnsnhn c N sphc 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 TSl chu kca cc k hiu gc. Cho phn thc ca dy k hiu trong biu thc (1.10)i qua blc ly thp i vi tng k hiu ring trong qung thi gian TSsnhn 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 shoc bng tn nhnhau cng truyn ti mt thi im phisdng ghp knh theo thi gian. C thghp 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) Skhi bghp

Skhi TDM 4 knh nhhnh 1.17.

Sm

Chuynni tip

thnh songsong

ej2f1tm

ej2fN-1tm

dn

Hnh 1.16. Biu chOFDM

ngtruyn

Blcthp

Blcthp

Blcthp

Blcthp

Thu

xungB

Blcthp

Blcthp

Blcthp

Blcthp

Pht

xungB

1

2

3

4

Tn hiu

analogTn hiuanalog

1

2

3

4

Bchuynmch

Bphnphi

Hnh 1.17. Skhi ghp 4 knh theo thi gian


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(2) Nguyn l hot ng

Blc thp hn chbng tn tn hiu thoi analog ti 3,4 kHz. Bchuyn mch ng vai

tr ly mu tn hiu cc knh, v vy chi ca bchuyn mch quay mt vng ht 125 s, bngmt chu kly 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 kghp li c mt xung ng bv ghp tip xung thhai ca cc knh.Qu trnh ny ctip din lin tc theo thi gian. pha thu hot ng ng bvi pha pht,yu cu chi ca bphn phi quay cng tc v ng pha vi chi ca bchuyn mch. Nghal hai chi phi tip xc vi tip im tnh ti vtr tng ng. Yu cu ng bgia 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 blc thp ging nhblc ny ti pha pht.

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

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

1.3.2.2. TDM tn hiu s

(1) Skhi bghp

Skhi bghp TDM tn hiu sc thhin ti hnh 1.19.

(2) Nguyn l hot ng

Qu trnh hot ng ca bchuyn mch v bphn 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

bm ho tin hnh lng tho v m ho mi xung thnh mt tm nhphn 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 tm ng bkhung t ti u khung v cc bit bo hiu c ghp vo vtr quy nhtrc. Bto xung ngoi chc nng to ra tm ng bkhung 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 ttn hiu thu

ng bbto xung thu. Bto xung pha pht v pha thu tuy thit kc tc bit nhnhau,nhng do t xa nhau nn chu stc ng ca thi tit khc nhau, gy ra sai lch tc bit. Vvy di skhng chca dy xung ng h, b to xung thu hot ng n nh. Khi ti tokhung tch tm ng bkhung lm gc thi gian bt u mt khung, tch cc bit bo hiu xl ring, cn cc byte tin c a vo bgii m chuyn mi tm 8 bit thnh mt xung.Do bphn phi hot ng ng bvi bchuyn mch nn xung ca cc knh ti u ra bgiim c chuyn vo blc thp ca knh tng ng. u ra blc thp l tn hiu thoi analog.Bto xung pha thu iu khin hot ng ca cc khi trong nhnh thu.

Hnh 1.19- Skhi hthng 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 bkhe thigian cho cc ngun l tnh, ngha l cnh; do khi cc ngun khng c sliu th cc khe b

btrng, gy lng ph. khc phc nhc im ny cn sdng phng php ghp thi gianthng k.

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

- Bghp knh thng k r sot cc ng dy u vo v tp trung sliu cho n khighp y khung mi gi i;

- Khng gi cc cc khe thi gian rng nu cn c sliu tngun bt k;

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

- Nu c n cng I/O a vo bghp thng k, chc kkhe thi gian khdng, trong k


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Hnh 1.20. Skhi bghp 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 nhhnh1.21. Cc gi sliu c gi qua ng truyn. Btch xl cc gi v da vo a chphn

pht sliu 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 thnht (hnh 1.21 b):

Trong khung con chc mt ngun sliu, chiu di sliu thay i v hot ng khi titrng thp.

- Trng hp thhai (hnh 1.21c):

Trong khung con c nhiu ngun sliu, 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. Sdnhvy l v phi hn chkch cca bm gim gi thnh, nhngquan trng hn l gim trca sliu. Vn ny c kim nghim qua o thv ktquc trnh by ti cc hnh 1.22 v 1.23.

Ngun 1

Ngun 2

My thu1

My thu2ng truyn

My thu3

BtchBghpng dyu vo

Ngun 3

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

a ch S liu

b) Khung con chc mt ngun sliu

FCS- dy kim tra khunga) Khung tng qut

SliuChiu dia ch Sliu Chiu dia ch

c) Khung con c nhiu ngun sliu


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Thnh 1.22 thy rng mun tng hssdng ng phi tng kch cbm. Nhngthnh 1.23 li cho bit khi tng hssdng, tc l tng kch cbm th trli 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 thhin nhhnh 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

Hssdng ng truyn

M= 25 kbit/s

M= 50 kbit/sM= 100 kbit/s

M- Tc bit ca ng truyn

Hnh 1.23- trphthuc vo hssdng ng truyn

4

6

10

Kch cbm(skhung c m)

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Hssdng ng truyn

8

Hnh 1.22- Kch ctrung bnh ca bmphthuc vo hssdng ca ng truyn


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Trong CDMA, nhiu ngi sdng c thdng chung tn sv trong cng thi gian. khng gy nhiu cho nhau, mi ngi sdng chc php pht i mt nng lng bit (Eb) nhtnh m bo tsEb/ N0quy nh, trong Ebl nng lng bit ca tn hiu cn thu v N0lmt phtp m tng ng gy ra do cc tn hiu ca ngi sdng khc. gim mt

phtp m cn phi tri phtn hiu ca ngi sdng trc khi pht. Ngoi ra, my thu c

thphn 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 ngkhc nhau (cc m khc nhau). Nu ni kh(N0nh) th hhon ton khng gy nhiu cho nhau. Hnh 1.24 biu thN ngi sdng, mi ngic m ho bng mt m ring, c k hiu t1 n N. Mi khi con c trng cho schimtim nng v tuyn ca ngi sdng: tn s, thi gian v E0.

Do c th ca di ng nn khi mt ngi sdng no n gn trm gc, N0ca ngiy gy ra cho my thu ngi khc sln 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 vkhung 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 bkhung t ti u khung, trng tin ghp tn hiu cangi sdng v mt s bit phng vai tr chn, gim st, iu khin, v.v.

a khung l tp hp ca mt skhung. u a khung c tm ng ba khung lm gcthi gian ghp cc khung theo tht quy nh. Pha thu tch tm ng ba khung lm gcthi gian tch cc khung theo trnh tnh ghp pha pht. Ngoi tm ng ba khungv cc khung, trong a khung cn c cc bit phnhbo 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 nhhnh 1.25. Trong thi gian TKghp cc bitng bkhung, cc bit phv cc thng tin u vo bghp.

1.6. NG BTRONG VIN THNG

1.6.1. Mu

Tin hnh ng bhot ng ca cc thit bkhc 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 hthng scn phi tun theo mi quan htin nh. Nu hai hot ngtun theo mt stin nh, th ng bnhm 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 ccc mun ca hthng. mc tru tng cao, cc qu trnh phn mm cng bnhtrao i thng bo.

Phthuc vo phm vi ng dng, cc hthng tru tng khc nhau c chp nhn chiu quv c cu trc theo kiu phn cp, trong mi mc tru tng lin hvi 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 kbqua cc chi tit khng cn thit v tp trung vo cc cim cn thit. V vy ddng thc hin mt bn thit khthng phc tp hn.

Trong cc hthng phn cng s, gii php chung l cu trc hthng c din gii theocc mc tru tng nhmc vt l, mc mch, mc phn tv mc mun. Trong mc vt l,nh thit kquan 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 ttp trung vo cc cng, cc cng logic v.v. Trong

mc mun, cc phn tc phn chia thnh cc thc thphc tp hn nhcc bnh, cckhi logic, cc CPU v.v.

Cc giao thc thng tin c thc hin nhcc 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 vcho cc giao thc mc trn v sdng cc dch vca mt smcthp hn. Trong m hnh giao thc tham kho kt ni hthng 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 nhthno nhm cung cp mt knhtruyn dn sim ni im y . Cc tiu chun mc 2 (lp kt ni dliu) quy nh cc giaothc nhm cung cp mt knh sim ni im khng c li bng cch pht li cc khung blihoc nhkthut sa li. Cc giao thc ca cc lp trn cung cp cc dch vnh tuyn mng(lp mng), cc dch vtruyn ti qua mng (lp truyn ti) v cung cp cho ngi sdng ucui cc dch vng dng trc tip.

Nhng ci g l tiu chun tru tng c sdng m tcc hthng phn cng vphn mm u lin quan vi nhau v ti mc bt kshot ng chnh xc ca chng u phthuc vo thi gian. Cc thc thca cc mc tru tng khc nhau trong hthng phn cng v

phn mm thng yu cu chc nng ng bc lp khc nhau. Th d, ng bcc qu trnhgiao thc ti mc cho trc vnguyn tc l c lp vi ng bhot ng cc qu trnh mcthp. Tuy nhin, tth dtrn y thy rng vn ng bc thkhc nhau hon ton vmc

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

Mi quan tm ny lm xut hin snghi ngvmc thch hp ca schp 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, snghin cu y vng b nu ln mt sc imchung trong bi cnh khc nhau. V vy a ra l do ti sao thut ngc tnh lch sny c chp nhn.

i vi nhiu ksthng tin s, vic cm nhn thut ngng bcn bhn ch. Hchorng n chlin quan n hot ng tch ng hti my thu v cc thng tin cha trong tn hiuthu c. Thc ra vn ny chlin quan n ng bsng mang hoc ng bk hiu. Tri

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


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Gii iu chkt 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 sv pha cho trc. l ng bsng mang.

Trong trng hp bt k, gii iu chsyu cu nhn bit cc thi im ly mu v quyt nhtch thng tin logic ttn hiu analog thu c, do a ra quyt nh hnh thnh bit 0 hay

bit 1. y l ng bk hiu.

Sau khi tch c thng tin logic, bc tip theo, ti mc tru tng cao l sp xp li cckhung tcc bit thu c. y chnh l ng bkhung. ng bkhung cho php thit bthu hiuc vai tr cc byte ti cc vtr khc nhau trong khung (th d30 knh dnh cho cc cuc giin thoi khc nhau trong bghp 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 thm phng knh nu thit b thu c khnng cn

bng trkhc 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 trgi l ng bgiv c thc hin

bng cch khi phc li nh thi gc tdy cc gi thu c thng qua kthut thch nghi hoc

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

im ni im. Mt mc khc ca ng b l ng b mng: tp trung vo hot ng ca hthng cc nt mng. H thng ny c thphn phi ng hchung ti tt ccc nt mng truyn dn v chuyn mch trong khun dng s, sao cho mi phn tmng c th hot ngng bvi cc phn tmng khc v ng bcc lung bit n.

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

Mt loi khc ca ng bmng l ng bng hthi gian thctruyn 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 bsng mang

Trong cc hthng iu bin (AM), khi nhn tn hiu iu chs(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) tlvi s(t) nu ( ) 1tms .iu ny cho php thit kddng bgii iu ch(gii iu chng bao).

Trong trng hp trc c khnng gii iu chbng cch nhn tn hiu iu chvisng hnh sine c tn sv pha ca sng mang v sau cho qua b lc loi trcc thnh

phn tn scao:

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

Loi iu chny yu cu tn hiu nhn cos0t c sdng trong my thu phi c cngtn sv pha ca sng mang iu chthu c. Sdch pha bt kca sgy suy hao tn

hiu mt i lng [s(t)/2]costi u ra blc 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 phtn hiu iu chti tn ssng mang f0. Tht vy, phca tn hiu iu bin l


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dtha, gm hai phn chnh nm vhai pha ca sng mang f0. iu chmt bng bn (SSB) chtruyn mt trong hai phn chnh (mt trong hai bng bn). iu chSSB phi kt hp, trong sng chnh pha thm ch cn cht chhn, v mt lng dch pha bt kcng gy ra mo tn hiuiu bin.

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

l da vo vic khi phc tn hiu kt hp vi sng mang vtn sv pha. Hot ng ny chnhl ng bsng mang.

C thddng ti cu trc sng mang, nu trong phtn hiu thu c mt ng phtisng mang f0, thng xy ra khi tn hiu iu chc gi trtrung bnh bng zero. Trong trnghp ny, c th thc hin tch sng mang nh s dng b lc bng hp hoc vng kho pha(PLL). PLL c thit kc bng thng hp, do bto dao ng iu khin bi in p ngoi(VCO) c thkho v theo di thng dng tn sxung quanh tn sdanh nh.

ng tic l trong nhiu trng hp khng c vch phti 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 hthng ng btinh vi hn c khnng khi phcsng mang vtn sv pha.

Mt th dn gin ca ng bsng mang: xem xt trng hp truyn dn skho dchpha nhphn (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, chbin i phi tuyn mi c thto ra vch phf0mong mun ttn 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 chxo b

iu chv to ra thnh phn (1+ cos20t)/2 c vch ph ti tn s2f0 xut hin v thu c

sng mang nhchia tn.

Hnh 1.26- ng bsng mang cho hthng BPSK

Trong iu chpha cu phng (hthng QPSK truyn cc nhm k hiu 2 bit), thit bng bda vo tng tn stn hiu gp 4 xo iu chv sau to ra vch phti tn s4f0.

1.6.3. ng bk hiu (symbol)Trong truyn dn sthng sdng 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 knhau (chu k).

Trong mi trng hp, pha thu c thgii iu chkt 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 ttn hiu analog thu c gi l ng bk hiu. ikhi cn lin quan n khi phc ng h.

Hnh 1.27 minh honguyn tc thu bng gc nhphn. Tn hiu analog thu c r(t) c

ly mu to ra dy cc gi trthc r(kT), t tch ra dy bit nhquyt nh logic. Bly muc iu khin bi hthng ng bthch hp. Hthng ny nh gi cc thi im c t = kT

bng cch kim tra r(t).

Blcbng (...)

2 PLL vchia tn

s(t) f0


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Khi ng bk hiu c thc hin sau khi chuyn i tn hiu thnh bng gc, c thsdng mt skthut khi phc nh thi k hiu ging nhkthut ng bsng mang.

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

( ) ( ) =k

k kTtgats (1.15)

c mt ng phti tn sk hiu 1/ T v nm ti trung tm b lc bng hp th c khnngkhi phc sng hnh sine, t tch ra dy xung nh thi c tn sca dy k hiu. Nu khngc ng phti tn s1/ T, vn c khnng to ra n bng cch chuyn i phi tuyn thch hp.

Th dnhchuyn i bnh phng u = s2hoc chnh lu u = s.

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

( ) ( ) tkTtgats k k 0cos = (1.16)l c thc c ng bao hoc bnh phng tn hiu nhn c ng phti tn sk hiu1/ T. Sdng ng phny c thi gian k hiu.

Tuy nhin, cn c cc kthut ng bsng mang v ng bk hiu khc da vo ccnguyn tc khc nhau to ra cc ng ph. Sau y tm tt ba lnh vc ng bk 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 thnht sdng cc hthng PLL. Lnh vc thhai so snh dy k hiu pht iban u vi cc k hiu lp lu tr nh gi dch pha. Lnh vc thba c trnh bytrn y.

1.6.4. ng bkhung

Sau khi hon thnh ng bsng mang v ng bk hiu v thng tin logic ctch ra t tn hiu n, bc tip theo l xc nh im u v im cui ca tm hoc canhm cc tm, nhvy gi l ng btm. ng thi sp xp li cc tm thu c thnhkhung theo ng trnh tnhkhung pha pht, nhvy gi l ng bkhung.

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

cc byte. Cc byte cc vtr khc nhau trong khung c thdnh cho cc knh ngi sdngkhc nhau c chung mi trng vt l trong ghp knh phn chia thi gian (TDM), chng hn nh

Blc cnbng knh

Quyt nhk hiu

1011r(kT)

Bly mu

Bng bk hiu(khi phc ng h)

t = kT

Hnh 1.27- ng bk hiu trong my thu bng gc nhphn

r(t)


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trong bghp 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 bkhung l ht sc quan trng trong truyn dn s. Tchcc lung nhnh ng c bt u tm tchnh xc cc khung.

Khoch ng bkhung 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 bkhung v kim tra lin tc ranh giikhung.

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

bt kca lung bit thu c v c duy tr nhkim tra tm ng chnh khung, ti bt umt khung. Trong khi tm kim mu tm ng chnh khung c thgp trng hp tm ngchnh khung bphng to t lung bit s liu. V vy cn tin hnh kim tra tm ng chnh

khung ti mt svtr trc khi cng nhn c ng b. Mc tiu la chn khoch ng chnhkhung c hiu qul:

(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 tm ng chnh khung trong lung bit ngu nhin thu c;

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

C thphn tch qu trnh phng on khi m tmt v khi phc ng chnh khung phhp vi khoch ng chnh khung chn nhsdng m hnh chui Markov thch hp nhhnh 1.28, trong P l xc sut nhn bit ng tm ng bkhung. 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 khoch 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 bB chkhi pht hin li trong tm ng chnh lintip. Ti trng thi B, bng chnh thc hin qu trnh tm kim v khi pht hin c mu bitging tm ng chnh th chuyn sang trng thi ng chnh ng tm thi C0. Ti y bngchnh thc hin qu trnh duy tr v schuyn sang trng thi bnh thng A0chkhi khng pht

hin li trong tm ng chnh lin tip. Ngc li, nu pht hin li trong tm ng chnhu tin th quay trvtrng 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 schuyn ny l:

(1) Li trong cc tm 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 ttrng thi B.

Cc tham sc trng cho cht lng khoch ng bl:

(1) Tc bin ctrung 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 nhl 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 gimo pfa, c ngha l xc sut chuyn ttrng thi B sang trngthi A0do mu tm ng chnh bphng 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 tm ng chnh ngvi githit l hthng 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 (githit khng tng quan) v a l slng bit ca tm 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, githit rng cc bit ny l c lp thng k v c cng xc sutxut hin th xc sut P c xc nh nhsau:

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 khoch ng bl:

( )( )

( )( )( )

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

22

2

2

221

21

2

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 (sbit trong khung) v R0l tc bit danh nh ca tn hiu ghp.

V L >>1 nn sbin i ngu nhin ca trc xu hng phn bchun (phn bGauss). 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 strn y c thlp khoch ng chnh khung khc nhau i vi tn hiughp ti cc mc khc nhau ca phn cp scn ng b(PDH). Th di vi PCM-30, gi tr

tiu chun ca = 3 v = 1. Sthay i mt t khi thit kng bkhung i vi thit bSDH. Trong tm 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 tm ng chnh di gim xc sut phng to,trong khi chn tm ng chnh ngn gim bt xc sut mt ng chnh cng bc. Biutrng thi sdng cho khoch ng bSDH c thhin ti hnh 1.29.

Sc 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

Aivi 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 tm ng chnh ca M khung lin tip. Sau K khunglin tip c tm 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 tm 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 tm ng chnh khung lin tip th quay trvtrng 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, bghi dch phi m cc khung c li trong tm ngchnh khung khi m hthng ang trong trng thi OOF (ngha l bghi dch iu khin chuynttrng thi OOF sang trng thi LOF v m t0 n K) khng ci t ti zero khi m hthngang trong trng thi IF i vi L khung lin tip.

Cc gi trtiu chun ca cc tham skhi lp khoch ng bkhung SDH l: M 5, J2, K = 24, N = 24 v L = 24.

Sau khi ng bkhung, cc khung ng bc sp xp c trt tthnh a khung nhtm ng ba khung t u a khung. y chnh l ng ba khung.

1.6.5. ng bbit

Trong vin thng ng bbit c din t theo hai ngha chnh. Thnht, ng bbitc lc c hiu c lin quan n ng bk hiu c trnh by trc y. Thhai, tng quthn, ng bbit c sdng biu thng blung bit cn ng btheo tn sng hcathit bti ch. Vn ny c thc hin bng cch ghi cc bit ca lung bit cn ng bvo bnhn hi (bm) theo tn sca lung vo v sau c ra theo tn sca ng hthit bti ch. ng bbit c hiu chyu theo cch gii thch thhai ny.

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

Ngoi ra, ng bbit cn c thc hin trong bghp tn hiu s, ti khi ng bho. Ti ycc nhnh c ng bbit chuyn lung scn ng bthnh lung ng bbng cch chn

bit.

1.6.6. ng bgi

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 sliu ngun v bsung thmmt vi thng tin mo u. Gi c chiu di cnh hoc thay i. Gi c chiu di cnh gi ltbo.

Chuyn mch gi l mt cng nghc hiu qulin kt sliu 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 tchc tiu chun ho quc t chn k thut chuyn mch tbo: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 trnht 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 tht;


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(3) Ti my thu c khnng khi phc tn sng hca 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 tcc gi n c trkhc nhau. V vy yu cu bsung cc chc nng ti giao din thu. Cn bng trngu nhin

ca gi thng lin quan n ng bgi.Nhim vcn bng trgi c thc hin theo mt scch 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 bc hthng phn phi ng hchung ticc nt. Ph hp vi cch phn chia ny, c cc phng php chyu sau y khi phc nhthi:

(1) Khi phc nh thi khng ng b

- nh gi trkhng 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 trtruyn dn cc tiu vda vo cc gi khc c thbtrkhng vt qu mt lng thi gian cc i cho trc. Nhvygi l nh gi khng nhn thy. Sau khi nh gi thi gian kt thc ca gi thnht, my thu sdng sthtdy trong cc gi tip theo xc nh mt cch chnh xc thi gian kt thc cami gi. Cc gi n c trln sbloi.

- o hnh trnh

Mc d nh gi tr khng nhn thy l n gin nhng khng y trong mngng di. Kthut nh gi trthc ttt nht l o trhnh trnh gia gi chuyn i v githu c v sdng gi trny nh gi trmt hng ca cc gi khc vi githit trc

phn bnhnhau gia hai hng.

- Trthay i do bsung

Trong trng hp ny, o trthc tkhi truyn cc gi qua mng. Sthay i ca trc o nhdu hiu trtch luca mi gi. Mi phn tmng bsung trvo du hiutrkhi o theo ng hti chv lng chnh lch gia thi im n v i. Bit trgi, chophp xc nh thi gian kt thc l thi gian thc tcng vi lng chnh lch gia gi trduhiu trcc i v gi trdu hiu trthc t.

- Khoch thch ng

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

- Khoch thch nghi da vo PLL

Tt ccc phng php trn y chph 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 sdng kthut lc jitter t

bo nhkhi phc nh thi PLL. Kthut n gin ny c thc thc hin nhlc trc cm sliu mc y hoc cc gi trtc thi n ca tbo sc a vo blc trc. PPL l

blc thp lc Jitter tbo. Mc d kthut 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 thcc tiu chun hin hnh vjitter.


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(2) Khi phc nh thi ng b

Khi nim cbn ca kthut ny da vo tnh khdng ca ng htham kho chung(ng hmng 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 kthut ng bgim jitter tbo c thit ki vi mng ATM B -ISDN.

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

ng nht gia hai ng hc o v m ho trong mo u lp p ng ATM (AAL). Ti mythu, ng hmng chung v thng tin m ho c sdng cu trc li ng hngun.

- Du hiu thi gian (TS)

Bm 16 bit do ng htham kho mng iu khin. Hai byte ca mo u lp con hit(CS) mang gi trtc thi ca bm trong mi nhm 16 tbo. Ti pha thu, ng hngunc cu trc li tTS thu c v ng hmng.

- Du hiu thi gian dng b(SRTS)

Phng php ny l TS ci tin v da vo squan st thy rng i vi ng hngunchnh 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 nhl kthut tiu chun khi

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

1.6.7. ng bmng

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

nhkhnng 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 hni bca cc im ghp v chuyn mch khc nhau trong mngvin thng s.

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

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

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

lng hu ht cc dch vm nh iu hnh cung cp cho khch hng.ng bmng 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 nhvn bn, hnhnh, audio v vidio trong sa dng ca cc mi trng ng dng. Sliu c thphthuc rtnhiu vo thi gian nhaudio v vidio hnh nh ng v i hi trnh by theo thtthi giankhi sdng. Nhim vca tch hp nhvy gi l ng ba phng tin. ng bc thsdng chm dt tranh chp gia cc lung sliu v cc bin cbn ngoi do ngi sdngto ra. Ni mt cch khc, mun m ch mi tng quan thi gian gia mi trng, nhxem

video kt hp vi m thanh, hoc c thtrnh by r rng nhtrng 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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Sphthuc thi gian ca cc dy sliu c thl tuyn tnh, nhtrong trng hp trnhdin file audio trn dy hnh nh. Nhng cng c khnng trnh by sliu 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 chny khng nm trong ni dung cun ti liu ny.

1.6.9. ng bng hthi gian thcMt kiu khc ca ng bmng l phn phi thi gian tham kho tuyt i (thi gian

theo tiu chun quc gia) ti cc ng hthi gian thc ca thit btrong mng vin thng (ngbca cc ng hthi gian thc).

Phn phi ng htiu chun quc gia nhm mc ch qun l v iu khin mng. Ccskin bt ku c hthng gim st thit b, chng hn nhvt ngng tsli bit(BER), cc cnh bo ng, hng phn cng v.v. c lu trbo co. Khi mng vin thngc qun l bi h thng qun l (mng qun l vin thng tiu chun TMN), cc skin sc ch nhthit btruyn thng bo qun l ti hthng iu hnh (OS). Trong trng hp

khc, vic lu trthng tin phi bao gm cngy giv ng hthi gian thc ca thit bscly ra.

iu cn thit l cc ng hthi gian thc ton mng phi c ng btheo thi giantuyt i nhnhau, nu khng skhng lin quan vi cc thng bo khc nhau mt cch c ngha theo mt nhn chung. Chkhi ng h thi gian thc thit bc ng bvi thi giantiu chun th mi c khnng sp xp mi tng quan thi gian v logic trong scc skinkhc nhau v v vy mi dn n suy on c tsliu cha xl skin tp hp v lutr.

ng bng hthi gian thc khc vi ng bmng. ng bng hthi gian thc

phn phi thng tin thi gian tuyt i (th d10.32.05 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 khnng. Phmvi ngy gitrong qun l cn lu trkhng g khc l xc nh r ngy, thng, nm v gi, pht,giy.

Mc ch ca ng bmng l ti thiu ho thng dng li thi gian trong scc ng h,trkhi ng b pha. iu ny c ng l tn hiu nh thi vt l ng b (th dsng hnhsine) c phn phi ti cc ng hmng. ng bmng vin thng sthc hin lch thi giankhng ln hn 10 ns hoc 100 ns.

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

Th dgiao thc thi gian mng (NTP) c sdng trong cc dch vthi gian Internetv khch hng ng bng hthi gian thc cng nhtchc v duy tr tng mngcon ng bthi gian. NTP c pht trin tcc 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 bgiao thc khc. Cc c trng khc khng bt buc gm xc nhn vmt m ho thng bo cng nhcung cp iu khin v gim st txa.


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Trong NTP, mt hoc nhiu dch v s cp c ng b trc tip tcc ngun thamkho bn ngoi. Cc dch vthi gian thcp c ng btcc dch vscp ph hp vi

phn cp. Cu hnh li cc ng ng bthay thl c khnng khc phc sgin on v ccsc. Thut ton c khnng 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 sc thgp mt strng hp khng mong mun, nhhng n cht lng ca tn hiu thu. Trng hp thnht, xut hin dy cc bit 0 hoc bit 1ko di, dn ti khng tch c ng htdy xung thu trong qung thi gian xut hin cc dy

bit nhvy. Trng hp thhai, xut hin lp i lp li nhiu ln mt tm nn lm tng tch lurung pha ca tn hiu thu. Chai nguyn nhn ny sc khc phc nu sdng ngu nhin hotn hiu, tc l sdng btrn ti pha pht v bgii trn ti pha thu.

1.7.2. Cu to v hot ng ca btrn v bgii trn

Hnh 1.30 l cu trc ca btrn v bgii trn.

Btrn v bgii trn u c bghi dch, mi bghi dch cha 5 t tr. Thi gian trca mi t trbng 1/x v bng rng mt bit. Th ddy tn hiu u vo btrn l Dith khiqua hai t sl Di.(1/x). (1/x), ngha l trhai bit, v (1/x). (1/x) = x

-2.

Tn hiu nhphn u vo btrn i ti bcng mun 2 thnht v c bsung thmtn hiu nhphn n tbcng mun 2 thhai. Do tn hiu nhphn u ra btrn Ds= Di

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

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

Dy tn hiu nhphn u ra btrn Dsqua ng truyn, a ti u vo bgii trn. Vvy dy nhphn u ra bgii trn c dng:

x-1x-1

x-1

x-1

x-1

x-1

x-1

Sliu ra DsSliu vo Di

a) Btrn

Bcngmun 2

Bghi dchphn hi

Sliu vo

DsSliu ra Di

'

b) Bgii trn

Khi phc ng h

Bcngmun 2

x-1

x-1x-1

Bghi dchphn hi

Hnh 1.30- Btrn v gii trn


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

Nhvy l sau khi gii trn nhn c dy tn hiu nhphn nhu vo btrn.

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

TM TT

Trong sho tn hiu analog th phng php PCM l n gin nht, tuy nhin sdngphng 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 sdng rng ri trong cc hthng thng tins. V vy phi sdng cc gii php ng bv ngu nhin ho tn hiu nhm m bo chtlng tn hiu thu, cthl tsli bit khng vt qu ngng ci t trc.

Cn phn bit cc loi ng btrong mng vin thng:

ng bsng mang l tch sng mang ttn hiu iu chtrong gii iu chkt hp.

ng bk hiu l nhn bit cc thi im ly mu v quyt nh trong gii iu chstchthng tin logic t tn hiu analog thu c. ng b tm v ng bkhung l nhn bit thiim bt u v kt thc tm hoc khung ti to khung tdy bit thu. ng bgi l cn

bng trcc thi im n ca gi nhm ti cu trc bn tin trong mng chuyn mch gi.ng bmng l phn phi ng hchung n cc nt trong mt mng rng ln iu khinng hcc nt chy cng tc bit v pha vi ng hchung. ng ba phng tin l spxp cc phn thn tp nhhnh nh, vn bn, audio, vodeo, v.v. trong truyn thng a phngtin ti cc mc tch hp khc nhau. ng bng hthi 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 bm ho - nn sc mt xung lng tVPAM = 875 , xc nh gi tr8 bit

u ra ca bm ho - nn s.

(2) u vo bm ho - nn sc mt xung lng tVPAM = -1898 , xc nh 8 bit ura ca bm ho - nn s.

(3) u vo bm ho - nn sc mt xung lng tVPAM= 209, tm bin xung bnn tng ng vi 8 bit ti u ra bm ho - nn s.

(4) u vo bgii m - dn sc tm 0110 1101, tm bin xung u ra bgii m -dn s.

(5) Vnhnh dng ca c tnh bin bm ho- nn stheo trc toy = f(x). Chox = 0,5; xc nh gi trca mi bit trong tm 8 bit ti u ra bm ho- nn s.

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

(Xem p sti phn phlc).


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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 chsdng chn dng v loi khung khc sdng cchn dngv chn m. Ring khung 140/565 chsdng chn dng.

- Ghp knh SDH: sdng 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 sdng chn dng v chn m.

2.2. GHP KNH PCM

2.2.1. Skhi bghp PCM-N

Skhi bghp PCM-N nhhnh 2.1.

Hnh 2.1- Skhi bghp 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 MN = 24. Pha m tn c N bsai ng (S) ng vai tr chuyn hai dy m tnthnh bn dy m tn v ngc li. Cthl mt pha bsai ng kt ni vi my in thoi qua

LM MHNSGhpknh

Lp mng

S1

LMBTX

thu

u ra

u vo

Xlbo hiu

BTX

pht CXK

1

1

N

Gii mng

Tchknh

GM-DS

SN

CXK

N 1

N

TXB


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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 bPCM-N. u ra v u vo pha mng kt ni vi thit bghp bc cao quacp ng trc.

Qu trnh chuyn i tn hiu ca PCM- 30 nhsau:

(1) Nhnh pht

Tn hiu thoi analog qua S, qua blc thp hn chbng 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 sMH-NSthc hin lng tho khng u v m ho mi xung lng tthnh 8 bit nhbm ho - nnsA = 87,6/13. Tn hiu nhphn u ra khi MH-NS c a vo khi ghp knh. Ti y,ngoi tn hiu sca 30 knh thoi cn c tn hiu sca mt knh ng bv mt knh bo hiuc ghp xen bit, to thnh lung E1 c tc bit l 2048 kbit/s. Cui cng dy sliu nhphnc khi lp m ng chuyn thnh dy xung ba mc HDB-3.

Ngoi cc khi trn y, trong nhnh pht cn c bto 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 bkhung 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 sE1.

(2) Nhnh thu

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

pha pht. Ngoi ra, khi tch knh cn c chc nng tch ng htdy bit vo ng btc

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. Bto xung thu cng c bphn chia tnhnh 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 sGM-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 blc thp khi phc tnhiu thoi analog, qua Sti my in thoi.

2.2.3. Cu trc khung v a khung

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

truyn. Cu trc ca khung v a khung nhhnh 2.2.

(a) Cu trc khung

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

TS0n TS31. Mi TS c thi hn l 3,9 s v ghp 8 bit sliu. Tm ng bkhung c cu

trc ring 0011011 v c ghp vo TS0 ca khung F0v cc khung chn (F2, F4,..., F14). TrongTS0ca cc khung l(F1, F3,..., F15) ghp cc bit nhsau: bit thnht sdng cho quc gia (Si),

bit thhai cnh bng 1 phn bit tm ng bkhung vi tm ng bkhung gito khi7 bit cn li trong TS0ca cc khung ltrng vi 7 bit tng ng ca tm ng bkhung, bitthba cnh bo mt ng bkhung (A). Tn hiu cc knh thoi thnht n th15 ghp vo cc


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khe thi gian TS1n TS15; tn hiu cc knh thoi th16 n th30 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 F1F15 trong a khung.

(b) Cu trc a khung

PCM-30 ghp c 30 knh thoi. V vy c tt cl 30 tn hiu gi. Mi khe TS16ghpc tn hiu gi ca hai knh thoi. Do cn phi c tt cl 15 khe thi gian TS16chuynti tn hiu gi ca tt ccc knh thoi. Ngoi ra cn thm mt TS16na ghp xung ng b

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

Cc khe thi gian TS16ca cc khung trong a khung c btr chuyn ti sliu nhsau:

TS16ca khung zero (F0) ghp cc bit ng ba khung 0000 v bit cnh bo mt ng bakhung Y.

Na bn tri ca TS16khung thnht ghp 4 bit tn hiu gi ca knh thoi thnht, nabn phi ghp 4 bit tn hiu gi ca knh thoi th16. Na bn tri ca TS16khung thhai ghptn hiu gi ca knh thoi thhai v knh thoi th17. Ctip tc nhvy cho n TS16cui

cng ca khung th15 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 sbit trong khung bng 8 bit 24 + 1 bit = 193 bit. Tc bit u ra PCM-24 c

tnh nhsau:

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

a khung ca PCM-24 gm 24 khung, nh s th t tF1n F24, nh trn hnh 2.3.Mi bit ca tm ng b khung 001011 c ghp vo vtr bit thnht ca cc khung F4, 8, 12,

16, 20, 24. Cc bit thnht ca cc khung ltruyn tm ng ba khung (cc bit m). Bit thnhtcc khung F2, 6, 10, 14, 18, 22,l cc bit kim tra sdchu trnh (cc bit e1 e6). Bit thtm ca cc

khe thi gian trong khung F6, F12 , F18 v F24truyn 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 F1F15Cc 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 bkhungA = 1- mt ng bkhungSi- sdng cho quc tSn, x- sdng cho quc gia

Y= 0- c ng ba khungY= 1- mt ng 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 thgii tn ti ba tiu chun tc bit. l cc tc bit theo tiu chun

Chu u, tiu chun Bc Mv tiu chun Nht Bn. Cc tiu chun ny c trnh by didng phn cp scn ng bnhhnh 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 sca mc ng trc lin k. Mc thnht c tc bit 2048 Mbit/s c tothnh tthit bghp knh PCM-30 hoc ttm mch trung kca tng i in ts. Tc bitca mc thhai l 8448 kbit/s, gm c 120 knh. Mc thba c 480 knh v tc bit bng34368 kbit/s. Mc thtc 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 thnm ctc bit bng 564992 kbit/s v bao gm 7680 knh.

(2) Tiu chun Bc M

Tiu chun Bc Mgm 5 mc. Tc bit ca mc thnht bng 1544 kbit/s, c hnhthnh tthit bghp knh PCM-24 hoc ttng i in tsv c 24 knh. Ghp bn lung smc thnht c tc bit mc hai l 6312 kbit/s v gm c 96 knh. Mc thba 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

53 4

76 2

T1 T2

T3 T4 T5

Bc M

Nht Bn

Hnh 2.4- Phn cp scn ng b


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l 44736 kbit/s l kt quca ghp by lung smc hai v bao gm 672 knh. Ba mc ny cITU-T chp nhn lm tiu chun quc t. Mc thtc c bng cch ghp su lung smcba, tc bit bng 274176 kbit/s v bao gm 4032 knh. Mc thnm l kt quca ghp hailung smc 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 thba c hnh thnh tghp nm lung smc hai, c tc bit l 32064 kbit/s v 480 knh. Ba mc u tin ny c ITU-T chp nhn. Ghp ba lung smc ba c lung smc bn vi tc bit bng97728 kbit/s, 1440 knh. Mc cui cng ghp bn lung smc bn nhn c 5760 knh vtc bit bng 400352 kbit/s.

2.3.2. Kthut ghp knh PDH

2.3.2.1. Skhi bghp knh PDH

Nh trnh by trong mc 2.2.1, theo tiu chun chu u, cnm mc khi ghp bnlung vo sc mt lung ra. V vy skhi tng qut ca bghp knh PDH nhhnh 2.5.

Mi lung sdng ring mt skhi nh: bnhn 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 bnhn hi v a vo khi tch ng hto ra tn siu khin ghi fG. Cmi mt xung iu khin ghi tc ng vo M1th mt bit ca lung nhnhc ghi vo mt nh. Cc bit ghi sc c ly ra theo ng hiu khin c f1da

vo nguyn tc mt bit iu khin c tc ng vo M1th 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 cvolch 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- Skhi bghp 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 spht 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 btkhi to xung ng bv cc bit bo hiu (khngthhin 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 tngc li vi qu trnh ghp. Trc tin tchxung ng bv tch ng htdy bit thu c. Xung ng blm gc thi gian tch cc bitca cc lung thnh phn, xung ng hc sdng iu khin bto xung thu. Dy xungknh ca mi lung c tch ring bit v cc tm tm bit ln lt c gii m v dn trthnh dy xung lng tnhpha pht. Blc thp khi phc tn hiu analog tdy xung lngt.

2.3.2.2. Phng php ghp xen bit

Qu trnh ghp xen bit c minh hoti hnh 2.6.

Hnh 2.6- Ghp xen bit bn lung E1thnh 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 bkhung. Sau xung ng bkhung l bit thnhtca lung E1#1, bit thnht ca lung E1#2, bit thnht ca lung E1#3, bit thnht ca lungE1# 4. Tip ghp cc bit thhai ca cc lung vo theo trnh tnhghp cc bit thnht. Ctip tc ghp nhvy cho ht cc bit ca bn lung vo trong chu kghp TGH. Ghp xung ng

bkhung trc khi ghp tip cc bit sliu ca bn lung nhnh.

Bghp phi sp xp cc bit st li vi nhau v cn phi hnh thnh cc bit c rng bhn trong mt chu kghp TGHngoi xung ng bv cc bit phkhc 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 kghp TGHphthuc 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 thxy ra trng hp trt bit. Nguyn nhn ca hintng ny l do ng htch tlung vo c tn skhc vi tn sca ng hni (hnh 2.7).

Nu tn sng hni ti b hn tn sxung nh thi cha trong lung vo th mt bittrong bnhn hi c c hai ln, nhng ln sau l c khng nn gim tc bit u ra.

Ngc li, nu tn sng hni ti ln hn tn sxung nh thi cha trong lung vo th mt

sbit c c thm nn lm tng tc bit ca lung ra. Tng thm hoc gim sbit u ra bnhm c quan hn trt. Trong thc tc 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 sbit, chng hn trt mt octet hoc mt khung. Trt khng iu khin c l dolch nh thi v do khng iu khin c phm vtng hoc gim sbit. Nu phm vi lchtn sgia ng hni ti v tn slung bit vo duy tr phm vi 10-9v tn sly mu bng 8kHz th trt c thxy ra sau mi qung thi gian l 34 gi. Tng thm dung lng bnhnhi shn chtrt 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 sb to xung ca cc nt trongmng thng tin PDH.

2.3.2.3. Kthut chn trong PDH

(1) Khi nim

Thnh 2.7b bit c trong trng hp tn s(nghch o ca chu k) ng hni cabghp nhhn tn sca lung nhnh th mt sbit 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 bmt ny ca lung bit u ra bghp v ghp chng vo mt v tr quy nh trongkhung. Hot ng nhvy gi l chn m.

Tri li, trong trng hp tn sng hni ca bghp ln hn tn slung nhnh nhhnh 2.7a th mt sln c khng lm gim tc bit lung ra. m bo tc bit nh mc,cn bsung mt sbit khng mang tin v ghp vo vtr quy nh trong khung. Nhvy gi lchn dng.

(2) Chn dng

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

ra bnhn hi, ng thi phi m c sbit khng mang tin cn bsung vo lung ra bnhny trong mt n vthi gian. Yu cu thnht c thc hin nhkhi so pha v yu cuthhai do bm m nhim.

ng hni ti

Lung vobnh

t

t

c khng (gim)

t

a b

a) Tn sng hni ti ln hn tn slung vob) Tn sng hni ti nhhn tn slung vo

Hnh 2.7- Hin tng trt bit

Lung u rabnh

t

tThiu bit Tha bit

c thm (tng)


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bit th8 ca phn khung 4 ghp bn bit chn m. Bit th9 n bit th12 ca phn khung 4 ghpbn bit chn dng.

Hnh 2.8- Cu trc khung bghp 2/8 sdng chn dng v khng chn

Thnh 2.9 kim tra li nhng vn phn tch i vi khung ca bghp 2/8 khichn m, chn dng v khng chn.

Tc bit nh mc tng ca bn lung nhnh l:

V= 2048. 1034 = 8192. 10 3bit/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 sbit ca 4 lung

nhnh ghp trong khung khi cbn lung khng chn l:

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

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

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

Kt quca cc biu thc (2.1) v (2.2) nhnhau, vy vic quy nh vtr cc bit trn yhon ton hp l.

Khi cbn lung u chn dng th cc bit 5 12 ca phn khung 4 l cc bit khng

mang tin.

Hnh 2.9- Cu trc khung bghp 2/8 khi sdng chn dng, chn m v khng chn

TK= 1056 bit = 125 s

ng bkhung

8 256 4

11100110

4 4 2522564 4

Cc bit iukhin chn Cc bit chn m

Cc bit chn dngKnh dch v32 kbit/s

4

Dtr

Cnh bo mt ng bkhungKnh dch vchung

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 bkhung

Bit dtrCc bit chn dng

Cc bt iu khin chn

208

ng bkhung


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

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

Khi cbn lung u chn m th cc bit 512 ca phn khung 4 l cc bit tin.

Vy tng sbit tin ghp xen bit trong khung khi cbn 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.

2.3.3.2. Cu trc khung bghp 8/34

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 thhai sdng chn dng, chn m v khng chn.

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

Tng sbit 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 vtr bit th5 v bit th8 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 vtr bit dnh ring cho mnh ti v tr bit th5 8 trong PK4. Khi khng

chn th bit chn c thay bng bit thng tin ly tlung nhnh y.

Cu trc khung 8/34 khi sdng chn dng, chn m v khng chn nhhnh 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 dtrCnh bo mtng bkhung

1111010000

ng bkhung

Hnh 2.10- Cu trc khung bghp 8/34 sdng chn dng

372 410 1 1

Hnh 2.11- Cu trc khung bghp 8/34 khi sdngchn dng, chn m v khng chn

TK= 2148bit = 62,5 s

ng bkhung

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

Dtr

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. Sbit trong

mi phn khung l 716. Hthng c thit kghp xen bit 4 lung nhnh vo cc vtr bit 13n 716 trong phn khung 1 v 2 v v tr bit 17 n 716 trong phn khung 3. Tm ng bkhung 111110100000 (12 bit) chim vtr bit 1 n 12 ca phn khung 1. Cc bit iu khin chn

v cc bit dch vchim vtr cc bit 1 n 4, 9 n 12 trong phn khung 2 v 1 n 4 trong phnkhung 3. Chn dng c chthbi tm 111 trong hai khung lin tip. Trong khi chn mc chthbi tm 000 trong hai khung lin tip. Chthkhng 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 vtr bit 13 n 16 ca phn khung3. Trong phn khung 2 c cc bit 5 v 6 l cc bit ca knh dch vsADMo 32 kbit/s, bit 7 chthcnh bo ti bghp u xa, bit 8 l tn hiu rung chung ca knh dch v.


C hai loi cu trc khung: loi thnht chsdng chn dng, loi thhai c cchndng v chn m. Cu trc khung sdng chn dng nhhnh 2.12.

C hai phng php hnh thnh tc bit mc 4. Phng php thnht sdng 4 lung34368 kbit/s. Phng php thhai ghp trc tip 16 lung 8448 kbit/s nhn c lung mc bn.Chai phng php u sdng chn dng. Cu trc khung ca phng php ghp thnhtc thhin 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 tm ng bkhung 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 sdng 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 vtr 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 bghp 34/140 khi sdng chn dng v chn m nhhnh 2.13.

Cu trc khung ny l ca bghp 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 tm ng bkhung 1111010000, bit 11 knh dchv32 kbit/s iu chDelta 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

Sdng cho quc giaCnh bo mtng bkhung

111110100000

ng bkhung

Hnh 2.12- Cu trc khung bghp 34/140 sdng chn dng

472 412 31 4844 4844 4844

PK6


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Trong PK2, 3, 4, bit 1 n bit 4 sdng 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.


Bghp ny sdng 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 bghp nhhnh 2.14.

Vo nm 1986 sdng tc bit 564992 kbit/s c 7680 knh thoi trn trung khochthng dung lng cao. Hthng ny bao gm thit bthng tin quang v bghp bn lung139264 kbit/s m CMI v chn bit to ra lung mc nm 564992 kbit/s. Lung sny 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. Hssuy 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 sdng tc bit mc nm ny tchc 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 chsdng 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 bkhung

Hnh 2.13- Cu trc khung bghp 34 /140 sdng 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 bghp 140/565 sdng chn dng v khng chn

380437212 3804 3804 3804

PK7

3804 4

Cc bit cnh bong bkhung


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c truyn ti vtr bit thnht n thnm ca PK2 n PK6. Nhn c lnh ny, bghpchn mt bit khng mang tin vo mt trong bn vtr ti bit th5 n th8 trong phn khung 7ca khung sau. Khi khng chn, lnh iu khin c cu trc 00000 nn bit chn c thay thbi

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 bgiao 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 bghp SDH

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

G.784 Qun l SDH

G.803 Cu trc mng truyn dn da vo SDH

G.957 Cc giao din quang ca thit bv hthng lin quan n SDH

G.958 Cc hthng sSDH sdng cho cp si quang

2.4.1.2. Tc bit ca SDH

Mng SDH l mng ng b, trong mi phn tmng sdng tn hiu ng bccung cp tmt ngun ng hchun quc gia. Theo khuyn nghG.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 snguyn 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 bghp 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 chr trong hp cha loi tn hiulung nhnh no, hp nhvy 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 nhbng 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 bcng nhcu trc v hot ng ca mng SDH c lin quan n khinim vng, tuyn v on nhbiu thtrn 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 ktip nhau, trong mt trm to ra tn hiu STM-N v trm kia kt cui tnhiu STM-N ny. on lp l bphn truyn dn gia hai trm lp ktip nhau, hoc gia trmlp v trm ghp knh ktip.

(2) Tuyn (Path)

Tuyn l bphn 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 ccc tuyn ca hthng 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. Skhi ghp cc lung PDH vo khung STM-N

2.4.2.1. Skhi bghp knh SDH

S khi bghp knh SDH nhhnh 2.16.

2.4.2.2. Chc nng cc khi

u vo bghp 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 Mk hiu C-11 v ca chu u k hiu C-12. Cc mc cn li c mtchs. C-n c chc nng sp xp lung nhnh PDH tng ng, n thm cc byte khng mangtin cho sbyte 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, bsung cc byte mo u tuyn (VC-n POH).

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

Con trkhi 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 nhtc 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 trkhi qun l mc n (n = 3, 4).

Con trkhi 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 bmc N (N = 1, 4, 16, 64 v 256).

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

Ch thch:

Xl 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/s

34,368 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- Skhi thit bghp knh SDH

STM-01


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2.4.3. Qu trnh ghp cc lung nhnh PDH vo khung STM-1

Trong SDH, khung ca tt ccc mc ghp u c 9 hng v thi hn l 125 s, sct

nhiu hay t l phthuc vo dung lng byte trong khung nhiu hay t.

Khung STM-1 c cu trc nhhnh 2.17.

Khung STM-1 c 9 hng 270 ct. Khi nim ct ng nht vi khi nim byte (8 bit).

Khung bao gm 4 phn: mo u on lp (RSOH) chim 3 hng v 9 ct, mo u on ghp(MSOH) chim 5 hng v 9 ct. Con trAU-4, k hiu l AU-4 PTR ghp vo hng 4, ct 1 nct 9. Phn ti trng dnh ghp tn hiu VC-4 c 261 ct v 9 hng.

Tng sbyte trong khung STM-1 = 270 byte 9 = 2430 byte.

Tc bit truyn ca khung STM-1: STM-1 = 8 bit / byte 2430 byte /khung 8.103

khung/s = 155,52 Mbit/s. y chnh l tc bit mc 1 ca SDH.

Trong mc ny chtrnh by ghp cc lung nhnh PDH ca chu u vo khung STM-1.Mun to thnh khung STM-1 c thp dng mt sphng php sau y:

(1) Ghp mt lung nhnh 139,264 Mbit/s;

(2) Ghp 3 lung nhnh 34,368 Mbit/s;

(3) Ghp 63 lung nhnh 2,048 Mbit/s;

(4) Ghp 1 lung nhnh 34,368 Mbit/s v 42 lung nhnh 2,048 Mbit/s;

(5) Ghp 2 lung nhnh 34,368 Mbit/s v 21 lung nhnh 2,048 Mbit/s.

2.4.3.1. Sp xp cc lung nhnh 139,264 Mbit/s vo khung STM-1(1) Sp xp lung nhnh 139,264 Mbit/s vo khung VC-4

Trc ht khi C-4 chuyn i m ba mc ca lung nhnh thnh m hai mc v chuyngiao cho khi VC-4. c thbin lung nhnh cn ng bthnh lung ng btheo ng hca thit bSDH, khi VC-4 cn tin hnh chn m. Ct thnht ca khung VC-4 ghp 9 byteVC-POH, cn li 260 ct c chia thnh 20 khi, mi khi 13 byte nh biu th trn hnh 2.18.

Mi hng ca khung VC-4 c 260 byte nhhnh 2.18b, ghp cc loi bit nhsau:

130 bit n khng mang tin (R); 10 bit mo u (O); 5 bit iu khin chn (C); 1 bit chn

S; v (241 byte + 6 bit) tin I.Slng byte I m lung nhnh 139,264 Mbit/s cung cp cho khung VC-4 trong thi hn

125 s c xc nh nhsau:

MSOH

Ti trng

Hnh 2.17- Cu trc khung STM-1

AU-4 PTR

RSOH

9 ct 261 ct270 ct

9 hng


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BE4= 139264.103bit : 8 bit : 8.103 = 2176 byte (2.3)

Sbyte tin I sp xp cnh trong khung VC-4 l:

B0= (241 byte + 6 bit) 9 = 2169 byte + 6 byte + 6 bit = 2175 byte + 6 bit. (2.4)

So snh cc biu thc (2.3) v (2.4) thy rng lung nhnh E4 cung cp cho khung VC-4nhiu hn 2 bit I so vi slng bit I ghp cnh trong khung ny.V vy phi tin hnh chn

bit v y l chn m. 2 bit chn m ny c chn vo 2 hng ca khung VC-4. Hng no cnchn th lnh iu khin chn m c ci t trong 5 bit iu khin CCCCC = 11111 v bit chnS ca ca dng trong khung sau l bit tin I. Dng no khng chn th c lnh iu khinCCCCC = 00000 v bit S l bit n khng mang tin.

(2) Ghp khung VC-4 vo khung STM-1

Sau khi sp xp khung VC-4 th khung ny c ghp vo phn ti trng ca khungSTM-1 nhhnh 2.18a. Khi STM-1 ghp thm cc byte mo u SOH v cc byte con trAU-4hnh thnh khung STM-1 hon chnh.

2.4.3.2. Sp xp lung nhnh 34,368 Mbit/s vo khung STM-1

(1) Sp xp lung nhnh 34,368 Mbit/s vo khung VC-3Qu trnh sp xp c thhin ti hnh 2.19.

Hnh 2.18- Sp xp lung nhnh 139,264 Mbit/s vo VC-4

VC-4 POH

9 hngC2G1F2H4F3

N1K3

20 khi 13 byte1 byte

J1J1

13 byte

RSOH

MSOH

PTR

STM-1

POH W 96 I X 96 I Y 96 I Y 96 I Y 96 I

X 96 I Y 96 I Y 96 I Y 96 I X 96 I

Y 96 I Y 96 I Y 96 I X 96 I Y 96 I

Y 96 I Y 96 I X 96 I Y 96 I Z 96 I

1 12 byte1

Ch thch: I bit tin W = I I I I I I I IO- mo u X = C R R R R R O OC- iu khin chn Y = R R R R R R R RS- bit chn Z = I I I I I I S RR- bit n

a)

b)

B3


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Trc ht mi khi C-3 chuyn i tn hiu xung ba mc 34,368 Mbit/s thnh hai mc vchuyn giao cho khi VC-3. Khi VC-3 chuyn lung nhnh cn ng b34,368 Mbit/s thnh

lung ng bbng cch sdng chn dng v chn m. Khung VC-3 c 85 ct 9 hng, trong ct u tin l VC-3 POH c cu trc nhVC-4 POH. Chia khung VC-3 thnh 3 phn khungT1, T2 v T3, mi phn khung chim 3 hng nhhnh 2.19a v c cthho nhhnh 2.19b.

Trong mi phn khung c:(71 byte + 5 bit) n cnh R; 5 bit C1 v 5 bit C2; mt bit S1 v mt bit S2; (178 byte

+ 7 bit) tin I.

Sbyte khi tc lung nhnh E3 t nh mc 34,368 Mbit/s cung cp cho mt phnkhung l:

BE3 = 34368.103bit : 8 : 8.103/3 = 179 byte. (2.5)

Trong thi gian tc lung nhnh E3 t nh mc th S1 l bit n, S2 l bit I vC1C1C1C1C1= 00000, C2C2C2C2C2 = 00000.

- Chn m

Khi tc lung nhnh tng 30 ppm th sbit tng thm trong mi phn khung l:

bTng= 34368. 103bit 30. 10-6/ 8. 103/3 = 0,4 bit

C ngha l csau 5 phn khung tng thm 2 bit I. Khi c lnh C1C1C1C1C1 = 11111chuyn bit chn m S1 ca hai phn khung sau tbit n thnh bit I v C2C2C2C2C2 = 00000cc bit S2 ginguyn trng thi bit I.

- Chn dng

Chn dng xy ra trong trng hp tc bit ca lung nhnh gim 30 ppm, ngha ltrong mi phn khung gim 0,4 bit v sau 5 phn khung gim 2 bit. Khi c ch th chn

VC-3 POH

3 hngJ1J1C2G1F2H4F3

N1K3

B3J1

T1

T2

T3

3 hng

3 hng

a)

38I 38I 38I38I 38I 38I 38I

38I 38I 38I 38I 38I

C 38I 38I

38I 38I 38I 38I 38I C 38I

Ch thch

= RRRRRRRRI bit tin, R bit n,S1bit chn thnht, S2 bit chn thhaiC1 iu khin bit S1, C2 iu khin bit S2

R R R R R R C1C2

R R R R R R R S1 S2 I I I I I I I

Hnh 2.19- Sp xp lung nhnh 34,368 Mbit/s vo khung VC-3

38I 38I 38I38I 38I 38I 38I 38I 38I 38I 38I 38I C 38I 38I 38I 38I 38I 38I 38I C 38I

38I 38I 38I38I 38I 38I 38I 38I 38I 38I 38I 38I C 38I 38I 38I 38I 38I 38I 38I

b)

I I I I I I I I


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C1C1C1C1C1= 00000 cc bit S1 trong hai phn khung sau ginguyn trng thi l cc bitn, cn C2C2C2C2C2 = 11111 chuyn cc bit I ti bit chn dng S2 ca 2 phn khung sauthnh cc bit n.

(2) Ghp 3 khung TU-3 vo VC-4

Trnh tghp 3 khung TUG-3 vo khung VC-4 nhhnh 2.20.

Khung TUG-3 c 86 ct v 3 TUG-3 c 258 ct. Trong khi , khung VC-4 c 261 ct.V vy phi n 18 byte cnh vo ct thhai v th3 ca khung VC-4. Tct th4 n ctth261 dnh ghp 3 khung TUG-3. Qu trnh ghp xen byte t cc byte trong cng mtkhung TUG-3 vo mt c

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