Ebook "Truyền số liệu" - ĐH Phương Đông

8/14/2019 Ebook "Truyn s liu" - H Phng ng

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Hnh 3.1 M hnh OSI

Hnh 3.2 Cc lp OSI

X l ngang hng:

Vi mt my n, mi tng ch phc v tng trn v di ca n. V d tng 3 ch

c dng phc v bi tng 2 v yu cu phc v i vi tng 4. Gia cc my, tng x

trn my ny kt ni vi tng x trn my khc l vic kt ni mang ngha v mt logic.

Vic kt ni ny ch thc hin vi s tho thun ngm gi l giao thc. Trn mi my

vic x l gia cc lp gi l x l ngang hng (peer to peer processes).

Ti tn vt l th vic kt ni mi l trc tip nh hnh v 3.2. Khi chuyn mt dngbt t my A ti my B. Ti cc tng cao, vic kt ni ch di chuyn xung thng qua cc

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lp my A, v ngc li thng qua cc lp my B. Ti mi tng th dng b ny phi

thm mt s thng tin ca tng v chuyn ti tng thp v ngc li my B th qua mi

tng th thng tin ny c tch ra v cui cng c ti my B l dong bt ban u.

Giao din gia cc tng

D liu v thng tin mng qua cc tng ca my gi v tr li qua cc tng ca my

nhn c thc hin nh v giao din gia cc cp tng lin k. Mi giao din ny c

nh ngha bi cc thng tin ca tng . Vi giao din nh ngha ng v chc nng

lm ph hp ti mt mng. Nh vy mi tng ch phc v yu cu ti tng n mong

i.

T chc ca cc tng

By tng c th thuc v 3 nhm con. Tng 1, 2, v 3- tng ng vi tng vy l,

lin kt d liu v mng- gi l tng h tr mng; S tho thun vi din mo vt l cho

vic chuyn d liu t mt thit b ny ti mt thit b khc (nh phn in, kt ni vt

l, a ch vt l v thi gian giao vn v tin cy). Tng 5, 6 v 7 tng ng vi tng

phin, trnh din v ng dng- c th thng qua mt nhm l h tr s dng tng n lm

vic thng qua phn mm h thng, Tng 4, tng giao vn thc hin vic truyn d liu

gia hai u mt (end to end) thc hin vic kim sot li v kim sot lung d liu

gia 2 u mt. Di m hnh OSI phn lp l c xy dng bi phn mm ngoi tr

lp vt l l c xy dng ch bi phn cng.Hnh 3.3 Trao i thng tin trn m hnh OSI

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Hnh 3.3 Cho ta mt cch nhn ton b ca cc tng trong m hnh OSI. L7 l n

v d liu chnh ti tng 7, L6 l n v d liu chnh ti tng 6 bng cch thm vo L7

mt phn H6 v tng t nh vy cho ti tng vt l l mt chui bt truyn trn ng

truyn sang ti thit b khc v ngc li vi tin trnh trc l thm vo cc H th by

gi lc b c H tng ng khi i qua mi tng v n tng cui cng th c mt nv d liu nh ban u nh truyn sang.

3.2 Chc nng ca cc tng

Tng vt l

Lin quan n nhim v truyn dng bt khng c cu trc qua ng truyn vt l.

Vic truy cp ny nh cc phng tin c in, hm v th tc. Vi hnh 3.4 th hin v

tr tng vt l vi kha cnh ng truyn v li lin kt d liu.

Tnh cht vt l ca giao din v phng tin truyn thng

Tng vt l c nh ngha cc tnh cht ca giao din gia thit b v ng

truyn. Cn kiu truyn th ta c th xem xt chng 7

Biu din ca bt

Tng vt l d liu gm mt dng bt (c th l 0 hoc 1) khng theo bt c mt

cu trc no. V vic truyn ny thng qua tn hiu in hoc tn hiu quang, Tng vt l

nh ngha cc kiu m ho.Hnh 3.4 Tng vt l

Tc truyn d liu

Tc truyn (s bt truyn trn mt giy ) cng c xc nh bi tng vt l. Hay

ni cch khc tng vt l s xc nh thi gian truyn ca mt bt.

S ng b ho cc bt

Ngi gi v ngi nhn cn phi ng b ho ti cc mc.

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Cu hnh ng truyn

Tng vt l lin quan ti s kt ni ca cc thit b vi ng truyn. Trn cu hnh

im ti im, hai thit b c ni vi nhau thng qua mt ng dn. Trong cu hnh

nhiu im th mt ng ni c chia s gia mt s thit b

Cu trc lin kt vt l

Xc nh cch thc cc thit b lin kt nhau to thnh mt mng. Cc thit b c

th lin kt bng cch s dng mt cu trc mng li (tt c cc thit b c lin kt

vi nhau), mt cu trc hnh sao (cc thit b c lin kt thng qua mt thit b trung

tm), Mt cu trc hnh vng (thit b ny c ni tip thit b kia to thnh mt vng),

mt cu trc bus (mi thit b trn ng truyn thng thng)

Phng thc truyn

Tng vt l cng xc nh hng truyn gia hai thit b, simplex, half duplex ,

hoc full duplex. Theo phng thc simplex mt thit b ch gi v mt thit b ch nhn.

Phng thc simplex l cch kt ni mt chiu. Trong phng thc half- duplex c hai

thit b u c th gi v nhn nhng khng ng thi. Trong phng thc Full duplex

(hoc simply duplex), hai thit b c th gi v nhn ng thi ti mt thi im.

Tng lin kt d liu

Cung cp phng tin truyn thng tin qua lin kt vy l (tng vt l) m botin cy; gi cc khi d liu (frame) vi cc c ch ng b ho, kim sot li v kim

sot lung d liu cn thit. Hnh 3.5 th hin mi quan h gia tng lin kt d liu vi

mng v tng vt l

Cc c trng c bn ca tng lin kt d liu gm:

Framing. Tng lin kt d liu phn chia d liu nhn t tng mng cho vo mtn v d liu c th qun l gi l Frames.

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Hnh 3.5 Tng lin kt d liu

a ch vt l. Nu frames c phn chia vo cc h thng khc nhau trnmng, thm vo tng lin kt d liu mt header ca frames c nh ngha l a

ch vt l v gi a (a ch ngun) ch gc sang ni nhn (a ch ch) ca frame.

Nu frame c dng gi cho ngi nhn ngoi mng th a ch nhn l a ch

ca thit b kt ni ti mng tip.

iu khin lung. Nu tc d liu l tuyt i vi ngi nhn th lun lunnh hn tc ca th tc ngi gi, tng lin kt d liu li dng k thut iu

khin lung trnh vn trn d liu i vi ngi nhn.

iu khin li. Tng lin kt d liu tng thm tnh tin cy cho tng vt l bithm my d v pht li d liu li hoc frames mt. Nu s dng k thut ngn

chn khng lp li ca frames

iu khin truy nhp. Khi c hai hay nhiu thit b kt ni ging nhau, giaothc tng lin kt d liu l cn thit d thy thit b va b iu khin lin kt

bt k thi gian no.

V d 3.1 Trong hnh 3.6, mt nt vi a ch vt l 10 gi 1 frame n mt nt vi

a ch 87. Hai nt c kt ni bi mt lin kt. Ti tng lin kt d liu, frame ny

cha ng a ch vt l trong phn header. y ch l nhng a ch cn. Phn sau ca

header cha cc thng tin khc. Phn mc ni thng cha cc bit thm pht hin li.

Hnh 3.6 tng lin kt d liu (v d 3.1)

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Tng mng

L tng p ng s pht tn cc gi tin dc trn cc lin kt mng t ngun ti ch.

Bt c ni no tng lin kt d liu quan st thy s phn tn ca gi tin gia hai h

thng trong cng mt mng, tng mng m bo rng mi gi tin i t im xut pht ti

im kt thc.Nu hai h thng c ni vi cng mt lin kt, th khng cn thit phi c tng

mng. Tuy nhin nu hai h thng c kt ni vi hai lin kt mng khc nhau trong

mt mng ln thng qua cc thit b kt ni th cn phi c tng mng t c s

phn tn gi tin t ngun ti ch. Hnh minh ho l hnh 3.7

Hnh 3.7 Tng mng

Tnh cht ca tng mng bao gm hai thnh phn nh sau:

a ch logic. a ch logic c to bi tng lin kt d liu iu khin cc vn lin quan ti a ch cc b. Nu mt gi tin vt qua c hng ro mng

chng ta cn c mt h thng a ch khc gip phn bit h thng ngun v h

thng ch. Tng mng thm vo mt a ch cho gi tin i t tng trn xung n,

a ch logic th hin a ch ca ngi gi v ca ngi nhn.

Chn ng. Khi cc mng hoc lin kt c lp c ni vi nhau to nnmt mng ln hn, cc thit b kt ni (nh tuyn, hoc l cng) nh tuyn ng

truyn cc gi tin ny ti im kt thc. Mt trong cc chc nng ca tng mng l

cung cp k thut nh tuyn ng truyn.

V d 3.2. hy tng tng rng hnh 3.8, chng ta mun gi d liu t mt nt vi

a ch mng l A v a ch vt l l 10, trn mt mng cc b (LAN) n mt nt c a

ch mng l E v a ch vt l l 87 trn mt mng cc b khc. V hai thit b t trn

hai mng khc nhau, nn chng ta khng th ch dng a ch vt l c; a ch vt lch c tnh cc b. Ci chng ta cn y l cc a ch ph bin m c th vt qua ranh

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gii ca vng mng cc b. Cc a ch mng logic c c trng ny. Gi tin ti tng

mng cha ng cc a ch logic, phn cn li ging vi bn gc truyn n ch (A

v E, minh ho trong hnh v). Chng s khng i khi ta i t mng ny n mng kia.

Tuy nhin, a ch vt l s thay i khi gi tin di chuyn t mng ny n mng khc.

Hp R l b nh tuyn (router).

Hnh 3.8 Tng mng (v d 3.2)

Tng Chuyn ti:

Tng chuyn ti chu trch nhim cho ngun ti ch ca vic gi v nhn tin nhn.

Khi m tng mng nhn thy ti a ch cui ca cc gi tin, n khng c nhn dng

mt s mi quan h gia cc gi tin . cc gi tin thc s c lp gia khi gi i. Mt

khc, tng chuyn ti chc nhn thy c hai li iu khin v lun iu khin t ngun ti

ch. Hnh 3.9 th hin mi quan h gia tng chuyn ti, tng mng v tng phin.

V kha cnh bo mt, tng chuyn ti phi khi to mt kt ni gia hai cng. Mtkt ni l mt ng dn logic gia ngun ti ch kt hp cc gi tin c gi. Khi

to mt kt ni bao gm ba bc: Thit lp kt ni, truyn d liu v gii phng kt ni.

Hnh 3.9 Tng chuyn ti

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Cc kh nng c t ca tng chuyn ti gm: Dch v - im a ch: My tnh thng chy mt vi chng trnh trong

cng mt thi gian, v th vic pht t ngun ti ch khng phi t mt my tnh ti

my tnh khc m t mt chng trnh ring mt my tnh ti mt chng trnh ang

x l my khc.Dng u ca tng d liu phi bao gm mt kiu d liu c gi l

dch v im a ch hay cng a ch. Tng mng ly mi gi d liu, tng chuyn ti ly

cc gi tin vo x l.

Phn on v lp rp: Mt tin nhn c chia thnh nhiu on, mi on baogm mt s th t, s ny qua tng chuyn ti s c lp rp ng a ch n v nh

danh thay th cc gi b mt ca tin nhn.

iu khin kt ni: Tng chuyn ti c th khng kt ni hay kt ni. Mt kt

ni b t chi tng chuyn ti c xem nh mi on nh l c lp ca cc gi v

c gi n ti tng chuyn ti my nhn. Mt kt ni nh hng tng chuyn ti l

to mt kt ni vi tng chuyn ti my nhn d liu sau mi gi d liu, sau khi d

liu gi ht kt ni c ngt ra.

iu khin lung: Nh tng lin kt, tng chuyn ti chu trch nhim choiu khin lung, tuy nhin iu khin lung tng chuyn ti thc hin cui.

iu khin li: Nh tng d liu, tng chuyn ti chu trch nhim kim sot

li. Tuy nhin kim sot li tng chuyn ti thc hin cui nhanh hn nh mt lin kt

n. Khi gi d liu qua tng chuyn ti phi chc chn rng d liu n v i t tng

chuyn ti khng c li.

V d 3.3 Hnh 3.10 minh ho mt tng chuyn ti. D liu n t tng di c dch

v im a ch j v k (j l a ch ca ng dng gi v k l a ch ca ng dng nhn).

Khi kch thc d liu qu ln hn iu khin ca tng mng, d liu s c chia thnh

hai gi, mi gi u c a ch j v k. Sau tng mng, a ch mng c thm vo

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mi gi. Cc gi c th c chuyn qua cc ng khc nhau v n ch theo th t

khc nhau. Hai gi s c chuyn ti tng mng ch ni s chu trch nhim cho vic

chuyn cc dng tng mng. Hai gi ny c a qua tng chuyn ti ni chng c

ghp li.

Hnh 3.10 Tng chuyn ti (v d 3.3)

Tng phin

Dch v c cung cp bi ba tng vt l, lin kt d liu v mng khung cho

mt vi x l. Tng phin l mt ngi iu khin hp thoi mng, n thit lp, duy tr

v ng b li gia h thng kt ni.

Cc kh nng c t ca tng phin gm:

iu khin thoi: Tng phin cho php hai h thng vo cung mt hp thoi,

n cng cho php kt ni gia hai h x l cng mt thi im(bn song song) hay hai

thi im khc nhau (song song). V d hp thoi gia thit b kt ni u cui ti my

ch ln c th l bn song song.

ng b: Tng phin cho php mt x l thm mt kim tra mt im dngd liu. V d nu h thng gi mt file gm 2000 trang, n nn c kim tra sau khi c

100 trang c nhn mt cch c lp. Trong trng hp nu c li xy ra trong khi

truyn n trang 523 th qu trnh truyn li s bt u t trang 501: trang 1 n trang 500

khng cn thit phi truyn li.

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Hnh 3.11 minh ho quan h gia tng phin v tng chuyn ti, tng hin th.

Tng trnh din

Tng trnh din cp n vn c php v ngha ca thng tin c chuyn gia

hai h thng. Hnh 3.12 th hin mi quan h gia tng trnh din, tng ng dng v tng

phin.

Hnh 3.12 Tng trnh din

Cc kh nng c t ca tng th hin gm: Dch: Cc b x l hai h thng thng chuyn thng tin vo mt mu gm

cc xu k t, s. thng tin s b chuyn thnh dng bit trc khi b chuyn. Bi v hai h

thng m ho ca hai my tnh khc nhau nn tng trnh din s chu trch nhim thao tc

gia hai phng thc m ho khc nhau. Tng trnh din gi cc thng tin ph thuc vo

cc nh dng chung. Tng trnh din my nhn chuyn thnh thng tin ca my .

Bo mt: Khi mang cc thng tin nhy cm, h thng phi thit lp chc chn

mt vng ring. Bo mt c ngha l ngi gi thay i cc nh dng ca thng tinthnh nhng mu khc v gi kt qu ln mng. Vic gii m cc mu ny thnh thnh

nhng mu gi thng qua x l ca tng trnh din.

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Nn: D liu c nn gim s lng bit truyn, d liu nn tr thnh

mt phn quan trng trong khi truyn vn bn, m thanh, hnh nh.

Tng ng dng

Tng ng dng thit lp giao din gia ngi dng truy cp vo mng. N cung cp

giao din ngi dng v h tr cc dch v th th in t, chuyn file truy cp v

chuyn giao, chia s d liu qun l cc kiu d liu phn tn. Hnh 3.13 th hin mi

quan h ca tng ng dng ti ngi dng v tng hin th. Rt nhiu dch v ng dng

c sn; hnh th hin ba: X.400 (dch v iu khin tin); X.500(dch v th mc); v

chuyn, nhn v qun l file (FTAM). Ngi dng v d ny dng X.400 gi mt tin

nhn in t. Ch : u v ui ca gi tin khng c thm vo tng ny.

Hnh 3.13 Tng ng dng

Thit b cui mng o: mt thit b cui mng o l mt phin bn phn mm

ca mt thit b cui v cho php ngi dng truy cp t xa vo my ch. lm th,

ng dng to mt phn mm m phng u cui ca my ch t xa. My tnh ca ngi

dng dng phn mm thit b cui m c th truy cp ni chuyn vi my ch t xa. My

ch t xa tin tng rng n ang kt ni vi mt trong thit b cui ca n v cho phpngi dng truy cp vo.

Chuyn, nhn v qun l file: cho php ngi ngi dng truy cp, chuyn

v qun l file t my iu khin t xa.

Dch v mail: cung cp c bn cho gi v lu tr th in t.

Dch v th mc: cung cp h thng phn tn d liu ngun v truy cp cc

thng tin chung.

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Tm tt chc nng ca cc tng

Hnh 3.14 Tm tt chc nng ca cc tng

3.3 Giao thc TCP/IP

Hnh 3.15 TCP/IP v m hnh OSI

Giao thc TCP/IP c dng trn internet, n c pht trin thnh chun OSI.

Giao thc TCP/IP to ra nm tng: vt l, mng, chuyn ti v ng dng. T tng th

nht n tng bn tin cung cp cc chun vt l, giao din mng, mng trong v chc

nng chuyn ti.

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ApplicationApplication

PresentationPresentation

SessionSession

ApplicationApplicationSMTP

FTP

TELNET

DNS

SNMP

NFS

RPC

TFTP

TransportTransport

NetworkNetwork

Data linkData link

PhysicalPhysical

TCPTCP UDPUDP

IPIP

ICMPICMP IGMPIGMP

ARPARP RARPRARP

Protocols defined bythe underlying networks

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TCP/IP l mt trong cc giao thc to nn s tng tc gia cc modul, mi mt

modul c mt chc nng ring nhng chung khng nht thit hot ng c lp. chun

OSI cc chc nng thuc v cc tng, cc tng ca giao thc TCP/IP bao gm cc giao

thc tng i c lp v c trn v ni trong h thng khi cn thit. T kho Phn

h - hierarchical ngha l mi mc cao hn ca giao thc c h tr bng mt haynhiu mc thp hn.

tng chuyn ti, TCP/IP c nh ngha hai giao thc: TCP-Transmission

Control Protocol v UDP. tng mng, giao thc chnh c nh ngha bng TCP/IP l

IP, tuy nhin c mt vi giao thc h tr chuyn d liu tng ny. Xem chng 24 v

25 bit thm v giao thc TCP/IP.

3.4 T kho

Application layer error bit framedata link layer header destination address interfacelogical layer pressentation layer network layer session layer node to node delivery source addressopen system source destination deliveryOpen systems Interconnection trailer

peer to peer process transmission control protocol/

Internet-working Protocol(TCP/IP)Physical address transmission rate

physical layer transport layer port address

3.5 Tm tt:

1. T chc chun quc t to ra mt chun gi l h thng kt ni m cho php cc

h thng khc nhau kt ni.

2. M hnh by tng OSI cung cp cc nguyn tc cho s pht trin ca cc tnh

nng chung ca kin trc phn mm v phn cng.

3. Tng vt l, lin kt d liu, v tng mng l cc tng h tr mng.

4. Cc tng Phin, hin th v ng dng l cc tng h tr ngi dng.

5. Tng chuyn ti lin kt gia cc tng mng v tng ngi dng.

6. Tng lin kt d liu chu trch nhim gi cc n v d liu n cc vng khc

m khng c li.

7. Tng mng chu trch nhim gi cc gi d liu t ngun ti ch thng qua cc

lin kt mng.8. Tng chuyn ti chu trch nhim chuyn ton b thng tin t ngun ti ch.

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9. Tng phin chu trch nhim thit lp, duy tr v ng b cc tng tc gia hai

thit b kt ni.

10.Tng hin th thao tc gia cc phn gia hai thit b thng qua truyn d liu

vo mt nh dang ln nhau trn.

11.tng ng dng thit lp ngi dng truy cp vo mng12.TCP/IP giao thc kin trc 5 tng c pht trin trc chun OSI, n l giao

thc dng cho Internet.

CHNG 4

TN HIU

Trng tm ca lp vt l l chuyn thng tin dng cc tn hiu in thng quathit b truyn tin. Ta c th chn nhiu tiu chun truyn tin khc nhau gia hai my

tnh, gi nh, chung cnh bo, lm vic vi cc thng tin chuyn qua kt ni mng.

Thng tin c th l ting, nh, s, ch hoc dng m.

Thng tin c phn theo tng loi: d liu, ting, nh,...

gi thng tin, v d dng hnh nh, ta phi m ho nh thnh dng cc s 0 v 1

v chuyn n thit b thu, thit b thu khi phc li nh, lm c vic nh vy, thit

b gi phi thng bo cho thit b thu cch thc khi phc.Nhng cc tn hiu 0 v 1 khng th gi qua lin kt mng, chng phi c

chuyn thnh cc dng m thit b truyn tin chp nhn c nn cc s 0 v 1 phi c

chuyn thnh dng tn hiu in t.

truyn c, thng ti phi c chuyn thnh tn hiu in t.

4.1. Tng t v s

D liu v tn hiu phi c biu din di dng tng t hoc s. Tng t l ni

n i tng (thng tin) dng lin tc. S l ni n i tng ri rc.

D liu tng t v s.

D liu c th l tng t hoc s. V d ting ni ca con ngi l dng tng t.

D liu lu tr trong my tnh l dng s (lu tr di dng s 0 v 1).

Tn hiu tng t v s.

Tn hiu c th dng tng t hoc s. Tn hiu tng t l dng sng lin tc, thay

i lin tc theo thi gian. Tn hiu s l dng ri rc, c th tn hiu ri rc ch l cc s

0 v 1.

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Ta c th minh ho tn hiu bng vic chm im, trc ng xc nh ln ca tn

hiu, trc ngang xc nh thi gian tn ti ca tn hiu. th dng tng t l ng

cong lin tc, th dng s l ng gp khc.

Tn hiu c th dng tng t hoc s. Tn hiu tng t lun tn ti khng gii

hn tn hiu nm gia hai tn hiu; tn hiu dng s lun tn ti xc nh s tn hiu nhtnh nm gia hai tn hiu.

Hnh 4.1 So snh tn hiu tng t v tn hiu s

4.2. Tn hiu tun hon v khng tun hon.

Tn hiu tng t v s gm 2 dng: Tun hon v khng tun hon.

Tn hiu tun hon:

Tn hiu tun hon l tn hiu m n kt thc mt mu trong mt khong thi gian

xc nh, khong thi gian gi l chu k v tin shi lp li chu k khc. Chu k gi

li T (tnh theo giy), chu k cc tin shi c th khc nhau theo tng tn hiu. Hnh di

minh ho tn hiu tun hon.Tn hiu tun hon l tn hiu tn ti mu lp li lin tc. Thi gian lp li ngn

nht gi l chu k (T) tnh theo giy.

Hnh 4.2 V d v tn hiu tun hon

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Tn hiu khng tun hon:

L tn hiu thay i, khng lp li theo thi gian, khng tn ti chu k.

Tn hiu khng tun hon l tn hiu khng lp li mu.

Hnh 4.3 V d v tn hiu khng tun hon

Qua nghin cu, ng dng, bng k thut bin i Fourier, mt tn hiu c th c

chuyn thnh s xc nh cc tn hiu tun hon. Bng vic xem xt tn hiu tun hon

cho ta cch xem xt tn hiu khng tun hon d dng v tt hn.

Tn hiu khng tun hon c th c chuyn thnh tp cc tn hiu tun hon.

Sng hnh sin l tn hiu tun hon n gin nht.

4.3. Tn hiu tng t

Tn hiu tng t phn thnh hai loi: n gin hoc ghp hp. Mt tn hiu n

gin (tn hiu hnh sin) khng th chuyn thnh tn hiu n gin hn. Tn hiu ghp hp

l tn hiu c th chuyn thnh nhiu tn hiu hnh sin.

Tn hiu tng t n gin:

Sng hnh sin l dang c s ca tn hiu tng t tun hon. V mt trc quan tn

hiu hnh sin l ng cong, thay i theo chu k, lin tc.. Mi chu k l mt ng

cong n theo trc thi gian, tn hiu hnh sin c trng bi 3 tham s: bin , chu k

hoc tn s, pha.

Hnh 4.4 Sng hnh sin

Bin :

Nh trn hnh, bin ca tn hiu l gi tr ca tn hiu ti mi im trn sng.Bin c gi tr bng khong cch t im trn ng cong n trc ngang ca h to

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(l trc thi gian). Gi tr ln nht ca bin sng hnh sin bng gi tr cao nht m

theo trc ng trn hnh.

Bin c th c o theo n v: vn, wat ph thuc kiu tn hiu. Vn dng cho

ni n dng in, wat ni n cng sut.

Bin l cao ca tn hiu, n v o ph thuc vo kiu tn hiu (V, W).

Chu k v tn s:

Chu k (T) l khong thi gian theo giy m tn hiu hon thnh mt vng. Tn s

(f) l s chu k trong 1 giy. Hnh di minh ho chu k v tn s.

Hnh 4.5 Bin

Hnh 4.6 Chu k v tn s

n v o chu k: Theo giy. Trong k thut truyn tin s dng 5 n v o: s, ms,

s, ns, ps

n v Gi tr n v Gi trS 1s Hz 1Hz

ms 10-3s KHz 10-3Hz

s 10-6s MHz 10-6Hz

Ns 10-9s GHz 10-9Hz

Ps 10-12s THz 10-12Hz

n v o tn s: L Hz, trong k thut truyn tin s dng 5 loi n v o: Hz,

KHz, MHz, GHz, THz.

Chuyn tn s thnh chu k v ngc li:

Chu k = 1/(Tn s)

T=1/f hoc f=1/T

Chu k l khong thi gian tn hiu hon tt mt vng, tn s l s chu k trong

mt giy. Chu k v tn s c th chuyn i vi nhau f=1/T hoc T=1/f.

Nh chng ta bit tn s l tham s thit lp quan h gia tn hiu v thi gian, tn

s l s chu k trong mt giy. Theo mt cch nhn khc, tn s l tham s th hin tc

thay i, cc tn hiu in c th dao ng theo dng sng m mc nang lng bin

i quanh mc trung bnh. Tc sng sin thay i sng t mc thp nht ln mc cao

nht l tn s ca sng.

Tn s l tc thay i mc tn hiu theo thi gian. S thay i khong nng

lng trong thi gian cng ngn th tn s cng cao ngc li th tn s cng thp.

18


19/87

Nu gi tr tn hiu thay i trong thi gian cc ngn gi l tn s cao, ngc li gi

l tn s thp.

Hai gi tr cc: Nu tn hiu khng thay i hay mc in th khng thay i th

c tn s bng 0. Nu tn hiu thay i tc th tn s l khng xc nh.

Nu tn hiu khng thay i th tn s bng 0. Nu tn hiu thay i tc th tn skhng xc nh.

Pha: L v tr ca dng sng ti thi im 0. Pha xc nh trng thi ca chu k u

tin.

Pha xc nh v tr dng sng ti thi im 0.

Pha c o theo hoc radian. S dch chuyn ca pha 360 0 dn n dch chuyn

mt chu k. S dch chuyn ca pha 1800 s dch chuyn chu k..

Hnh 4.7 Mi quan h gia cc pha khc nhau

Quan st trn hnh so snh theo bin , tn s, pha v chc nng ca tng tham

s trn. S thay i ca 3 tham s lin quan n tn hiu v iu khin in t.

Hnh 4.8 Thay i bin

Hnh 4.9 Tn s thay i

4.4. Min thi gian v tn s

Sng hnh sin c th xc nh bi cc tham s: bin , tn s, pha. Ta c th v

sng hnh sin bng cch chm im trong mt thi gian no gi l min thi gian.

Min thi gian cho bit s thay i bin trong mt khong thi gian. Pha v tn s

khng hon ton c th o c theo min thi gian.

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20/87

Hnh 4.10 Thay i pha

a ra mi quan h gia bin v tn s ta c th khi nim min tn s. Hnh

sau so snh min thi gian (bin tn hiu theo thi gian) v min tn s (bin ln

nht vi mt tn s xc nh).

Hnh 4.11 Vng thi gian v tn s

Hnh sau m ta v d min thi gian v min tn s vi 3 tn hiu thay i v bin

v tn s.

Tn hiu c tn s thp trong min tn s tng ng vi tn hiu c chu k ln trong

min thi gian v ngc li. Mt tn hiu thay i nhanh trong min thi gian tng ng

c tn s cao trong min tn s.

Hnh 4.12 Vng thi gian v tn s cho cc tn hiu khc nhau

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21/87

4.5. Tn hiu hn hp

Ta xem xt tn hiu n gin hnh sin. Nhng nhiu dng sng khng thay i

theo ng cong trn t bin thp nht n bin cao nht. Nu tn hiu l tun hon

th n c th c chuyn mt cch logic v dng hnh sin. Trong thc t vi mi tn hiu

tun hon ta c th chuyn thnh ghp hp ca nhiu tn hiu hnh sin. Mi tn hiu snghnh sin c cc tham s bin , tn s, pha c th khc nhau.

xc nh mt tn hiu hn hp thnh tp cc tn hiu n gin, ta s dng php

phn tch Fourier. Ta xem xt khi nim hn hp trn v d trn hnh. Trn hnh cho thy

mt tn hiu tun hon c phn tch thnh 2 sng hnh sin. Sng th nht c tn s l 6

trong khi sng hnh sin th 2 c tn s l 0. Vic thm hai im ny bi kt qu trong

nh ca th. Ch rng tn hiu gc trng nh mt sng hnh sin m c trc thi gian

c dch chuyn xung. Bin trung bnh ca tn hiu ny khc 0. Yu t ny ch cho

bit s hin din ca thnh phn tn s bng 0, tc l dng din mt chiu. Thnh phn

DC ny l nguyn nhn ca vic dch chuyn v pha trn 10 n v ca sng hnh sin.

Hnh 4.13 Tn hiu vi thnh phn DC

Tri ngc vi th theo min thi gian m t tng tn hiu n l, th theo

min tn s a ra theo tn hiu hn hp l mt lot cc tn s, a ra c tp cc tn s

c lp nhau.

th min thi gian tin li trong vic xem xt chuyn t hai tn hiu thnh mt

tn hiu hn hp, th theo min tn s c cc thanh ng xc nh tn s vi ln

ca bin trong tn hiu hn hp.

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22/87

Hnh 4.14 Dng sng hn hp

Trn hnh m t mt tn hiu hn hp c phn tch thnh 4 tn hiu. Tn hiu ny

l gn ging vi tn hiu s. Cho mt tn hiu s chnh xc, chng ta cn mt lng v

hn cc tn hiu iu ho l (f, 3f, 5f, 7f, 9f,...), mi tn hiu iu ho vi mt bin khc nhau. th tn s c minh ho nh trn.

Ph tn s v bng thng:

Hai khi nim y l ph tn s v bng thng. Ph tn s ca mt tn hiu l tp

tt c cc tn s thnh phn, c th hin theo th min tn s. Bng thng ca mt

tn hiu l rng ph tn s, ni theo cch khc bng thng l phm vi cc tn s thnh

phn, ph tn s ni n cc thnh phn trong phm vi tn s.

Bng thng = (Tn s thnh phn ln nht) (Tn s thnh phn nh nht)

Ph tn s ca mt tn hiu t hp cc tn hiu sng hnh sin cu thnh

Hnh 4.15 Bng thng

4.6. Tn hiu s

Tn hiu c th c biu din di dng tn hiu s. V d s 1 l m ca in th

dng, 0 m ca in th 0.

22


23/87

Hnh 4.18 Mt tn hiu s

Bit Interval v Bit Rate:

Hu ht cc tn hiu s l khng tun hon, chu k v tn s khng xc nh. Hai

khi nim t ra y l Bit Interval v Bit Rate. Bit Interval l khong thi gian cn

thit gi mt bit. Bit Rate l s lng Bit Interval trong 1 giy, theo cch ni khc, Bit

Rate l s bit c gi i trong 1 giy (bps).

Hnh 4.19 Bit Interval v Bit Rate

Phn tch tn hiu s:

Mt tn hiu s c th phn tch thnh s xc nh cc sng sin c gi l tn hiu

iu ho. Mi tn hiu iu ho xc nh bi 3 tham s: bin , tn s, pha. iu ny c

ngha l khi ta gi mt tn hiu s ln mt phng tin truyn tin l ta ang gi mt s

cc tn hiu n gin. nhn c cc tn hiu truyn, tt c cc tn s thnh phn

phi c xc nh v truyn trn thit b truyn tin. Nu mt thnh phn tn s trong tn

hiu khng c truyn th tn hiu nhn c b sai lch.

Trong thc t mi thit b truyn tin ch thc hin truyn tin vi mt phm vi tn s

nht nh, nn ta lun nhn c tn hiu sai lch.

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24/87

Hnh 4.20 Tn hiu s iu ho

Mc d ph tn s ca tn hiu s cha mt s lng xc nh cc tn s vi cc

bin khc nhau, nu ta ch gi cc thnh phn tn s c ngha (theo ngng xc

nh) th ta s nhn c tn hiu vi chnh xc hp l (vi mo nh nht). Ta gi

phn ph xc nh l ph ngha, v bng thng ca vng ph tn s c ngha l bngthng c ngha.

Hnh 4.21 Cc ph tn quan trng v chnh xc

4.7. Cc thut ng v khi nim

Amplitude Code frequency periodAnalog composite signal gigahertz (GHz) periodic signalanalog data Cycle harmonics phaseanalog signal Digital hertz (Hz) picosecondaperiodic signal digital data kilohertz (KHz) signalBandwidth digital signal megahertz (MHz) sine wave

bit interval direct current DC microsecond spectrumbit rate Fourier analysis milisecond terahertz (THz)bits per second Fourier transform nanosecond time-domain plot

24


25/87

4.8 Tm tt

25


26/87

CHNG 5

M HO V IU CH

Nh chng ta tho lun chng 4; thng tin cn phi c bin i thnh tn

hiu trc khi n c th c truyn sang h thng thng tin.

Ta phi bin i d liu sang tn hiu truyn i t ni ny sang ni khc.

Tuy nhin thng tin c bin i ph thuc vo dng thc ban u ca n v trn

dng thc s dng bi phn cng truyn thng. Nu bn mun gi mt bc th tnh yu

bng tn hiu khi bn cn phi bit mu ca khi no ph hp vi nhng t no trong

thng ip ca bn trc khi bn thc s xy dng ng la ca bn.T ng l thng tin

v nhng lung khi pht ra l s miu t ca thng tin .

Mt tn hiu n gin bn thn n khng mang nhiu thng tin truyn t. Tn hiu

cn phi c vn dng n mc n cha ng nhng thay i c th nhn bit c v

s thay i ny phi c nhn ra bi ngi gi v ngi nhn ging nh vic miu t

thng tin d nh. u tin, thng tin phi c dch sang cc mu nh trc ca cc

bt 0 v 1; v d nh m trao i thng tin chun ca M (ASCII) c trnh by trong

ph lc A.

D liu lu tr trong mt my tnh l dng cc bt 0 v 1. c th mang c t

ni ny sang ni khc ( trong hoc ngoi my tnh), th d liu thng c chuyni sang dng tn hiu s. iu ny c gi l s chuyn i tn hiu s sang tn hiu s

(digital to digital D/D) hoc m ho d liu s sang tn hiu s.

i lc chng ta cn chuyn i mt tn hiu tng t (v d on ni chuyn trong

in thoi) sang tn hiu s v mt vi l do no nh gim bt hiu ng ca ting n.

iu ny c gi l s chuyn tn hiu tng t sang tn hiu s (A/D) hoc s ho mt

tn hiu tng t.

Vo lc khc chng ta li cn chuyn mt tn hiu s t u ra ca mt my tnhqua mt phng tin truyn thng c thit k cho dng tn hiu tng t. V d nh

vic gi tn hiu t ni ny n ni khc qua mng in thoi cng cng, tn hiu s cung

cp bi my tnh s c chuyn sang tn hiu tng t. iu ny c gi l bin i

tng t sang s hay iu ch tn hiu s.

Thng th mt tn hiu tng t c truyn qua mt khong cch di s dng

phng tin truyn thng tng t. V d m thanh, m nhc t mt trm radio, bn thn

n l mt tn hiu tng t c pht qua khng kh. Tuy nhin, tn s ca m thanhhoc m nhc khng thch hp cho vic truyn pht. Tn hiu pht i phi c mang bi

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27/87

mt tn hiu c tn s cao hn. iu ny c gi l bin i tng t sang tng t

(A/A) hay iu ch tn hiu tng t.

Hnh 5.1. Cc phng thc chuyn i tn hiu khc nhau

5.1. Bin i D/D.

Chuyn i hoc m ho s / s l s miu t thng tin dng s sang mt tn hiu

s. V d khi chng ta truyn tn hiu t my tnh n my in, c hai d liu gc v d

liu c truyn u dng s. Trong kiu m ho ny cc s nh phn 0 v 1 pht ra bi

my tnh c chuyn thnh cc xung in th, cc xung ny c th truyn c qua dy

dn in. Hnh 5.2 Ch ra mi quan h gia thng tin dng s, phn cng m ho s / s,

v kt qu tn hiu s:

Hnh 5.2: M ho s / s

Trong rt nhiu k thut m ho s / s, chng ta ch bn n nhng k thut hu

dng nht cho vic truyn thng d liu. C 3 loi ph bin: n cc, cc v lng cc

c ch ra nh hnh 5.3:

Hnh 5.3: Cc kiu m ho s / s

M ho n cc: l dng n gin vi ch mt k thut c s dng.

27

Cc phng thcchuyn i

S/s(D/D)

Tng t/s(A/D)

S/tng t(D/A)

Tng t/tngt (A/A)

01011101 M ho s /sM ho s /s

M ho s / s

n cc Cc Lng cc


28/87

M ho cc c 3 kiu con: NRZ, RZ, v biphase. Hai trong s chng c nhng s

bit i phc tp.

Dng th 3 l dng m ho lng cc c ba s bin i: AMI, B8ZS, v HDB3.

M ho n cc:

M ho n cc rt n gin v th s. Mc d ngy nay n gn nh li thi, tuy

nhin tnh n gin ca n cung cp s ch dn d dng lm c s pht trin cho cc h

thng m ho phc tp hn v cho php chng ta nghin cu cc loi bi ton m bt k

h thng truyn s no cng phi thc hin.

H thng truyn s lm vic da trn xung in cng vi mt kt ni trung gian,

thng l dy dn hoc cp. Trong hu ht cc kiu m ho, mc in p cao thp ng

vi gi tr nh phn 1 hoc 0. Tnh c cc ca mt xung m ch vic la chn l cc

dng hay cc m. M ho n cc c tn nh vy l bi v n ch s dng mt cc.Tnh c cc ny ch nh 1 trong 2 trng thi 0 hoc 1 (thng l 1). Trng thi cn li

(thng l 0) c i din bi in p 0.

M ho n cc ch s dng mt mc in p.

Hnh 5.4 ch ra tng vic m ho n cc. Trong v d ny, m nh phn 1 c

m ho ng vi gi tr dng v m nh phn 0 c m ho ng vi gi tr 0. Hn na

vic m ho n cc khng phc tp v d thc hin.

Hnh 5.4: M ho n cc

Tuy nhin, m ho n cc c t nht 2 vn lm cho n t mong mun: thnh

phn mt chiu v s ng b ho.

Thnh phn mt chiu (DC):

Bin trung bnh ca mt tn hiu m ho n cc l khc 0. iu ny to ra

thnh phn dng mt chiu (DC) mt thnh phn c tn s bng 0. Khi mt tn hiu

cha thnh phn DC, n khng th truyn i m khng x l.

28

thi gian

Bin

0 1 0 0 1 1 1 0


29/87

ng b ho:

Khi mt tn hiu khng n nh, bn nhn khng th xc nh im u v im

cui ca mi bit. V th vn ng b ho trong vic m ho n cc c th xy ra bt

c khi no dng d liu gm mt lot cc ch s 0 hoc 1. Qu trnh s ho dng s thay

i mc in p ch ra s thay i gi tr bit. S thay i tn hiu cng ch ra rng mtbit va kt thc v mt bit mi bt u. Tuy nhin trong m ho n cc mt lot cc

bt cng gi tr, nh 7 s 1, tc l khng c s thay i in p, mc in p dng

khng b ph v sau 7 ln min l nhn gi tr bit 1. Bt c khi no khng c tn hiu thay

i ch ra im bt u ca bit tip theo trong chui, bn nhn phi da trn mt mc

thi gian. Chng hn vi tc bit 1000 bps, nu bn nhn xc nh mt in p dng

tr 0.005s, m tc c 1 bt l 0.001s, hay 5 bit.

S thiu ng b gia ng h ca bn nhn v bn gi lm sai lch thi gian ca

tn hiu, v d 5 bt 1 b ko di thnh 0.006s, v do bn nhn s hiu thnh 6 bt 1.

Mt bit ph trong dng d liu gy ra mi th sau khi n c gii m nhm. Mt gii

php c pht trin iu khin vic ng b ho trong truyn pht mt cc l s

dng mt du tch, mc song song mt ng mang mt xung ng h v cho php bn

nhn phn chia ng b ho li thi gian ca n. Nhng vic nhn i s ng s

dng cho truyn pht ng ngha vi vic lm tng chi ph v v vy s khng kinh t.

M ho cc:

M ho cc s dng 2 mc in th, mt in p dng v mt in p m. Bng vic

s dng c 2 mc, trong phng php m ho cc, mc in th trung bnh trn ng

truyn c gim xung v vn v thnh phn DC ca m ho n cc v th c

gim nh. Trong m ho Manchester v Manchester vi sai (xem trang sau), mi bit gm

c c hai in th dng v in th m, v vy thnh phn DC hon ton c th loi ra.

M ho cc s dng 2 mc bin (mc dng v mc m)

Trong s rt nhiu kiu m ho cc a dng, chng ta s ch kim tra 3 kiu thng

dng nht: nonreturn to zero (NRZ), return to zero (RZ), v biphase. M ho NRZbao gm 2 cch: nonreturn to zero, level (NRZ-L), v nonreturn to zero, invest (NRZ-I).

Biphase cng c 2 phng php. u tin, Manchester l phng php c s dng bi

mng LAN. K n, Manchester vi sai, l phng thc c s dng bi mng Token

Ring LAN (xem hnh 5.5).

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30/87

Hnh 5.5: Kiu m ho cc

M ho Nonreturn to Zero (NRZ):

Trong m ho NRZ, mc ca tn hiu lun l dng hoc m. Hai phng thc

thng dng nht ca vic truyn pht NRZ c trnh by nh sau:

M ho NRZ-L: Trong m ho NRZ-L, mc ca tn hiu ph thuc vo kiu ca bit

m n trnh by. in th dng quy c l bit 0, tn hiu in th m quy c l bit 1;

theo cch mc ca tn hiu ph thuc vo trng thi ca cc bit.

Trong NRZ-L mc ca tn hiu ph thuc vo trng thi ca bit

Mt vn c th ny sinh khi c mt dy di cc bit 0 v 1 trong d liu. Bn

nhn nhn mt dng in th lin tc v c th xc nh c bao nhiu bit c gi da

vo ng h ca chng, iu ny c th c ng b hoc khng c ng b vi

ng h ngi gi.

M ho NRZ-I: Trong NRZ-I . Mt s o ngc ca in th miu t mt bit 1. S

chuyn i trng thi gia in th dng v in th m a ra mt bit 1. Mt bit 0 c

miu t nh mt s khng thay i. NRZ-I tt hn NRZ-L v s ng b ho cung cp

bi s thay i tn hiu trong mi thi im mt bit 1 gp phi. Hin trng ca chui bit 1trong lung d liu cho php bn nhn ng b ho thi gian ca n n ni nhn thc

s ca vic truyn. Mt chui bit 0 c th gy ra vn , tuy nhin do cc bt 0 khng hn

nh vy, chng gim thiu vn xy ra.

Trong NRZ-I cc tn hiu c o ngc nu mt bit 1 c gp.

Hnh 5.6 Miu t mu cc bit ca NRZ-L v NRZ-I. Trong chui NRZ-L, in th

dng v m c ngha r rng; dng i vi 0 v m i vi 1. Trong chui NRZ-I , bn

nhn tm kim s thay i t mt mc ny n mc khc nh l c s nhn ra bt 1.

30

Polar

NRZ RZ Biphase

NRZ-L NRZ-I Manchester Manchester

vi sai


31/87

Hnh 5.6: M ho NRZ-L v NRZ-I

M ho Return to Zero (RZ)

Nh chng ta c th thy, bt c thi im no th d liu gc cng cha ng

cc s 1 v khng lin tip nhau. Bn nhn c th mt v tr ca n. V nh chng ta

cp n trong phn tho lun v m ho n cc, mt cch m bo ng b ho l

gi cc tn hiu thi gian phn tch trn mt knh phn tch. Tuy nhin gii php ny s

lm tng chi ph ng thi d xy ra li ca bn thn chng. Mt gii php tt hn l

bng cch no cha ng vic ng b ho trong tn hiu m ho. Mt vi th gingnh gii php c cung cp bi NRZ-I, nhng kh nng trnh by trnh by chui 0 v 1

l nh nhau.

m bo vic ng b ho, cn phi c mt tn hiu thay i cho mi bit. Bn

nhn c th s dng nhng thay i ny xy dng, cp nht v ng b ho ng h

ca n. Nh chng ta bit trn, NRZ-I thc hin iu ny cho mt chui tun t cc

bt 1. Nhng thay i vi mi bit, chng ta cn phi c nhiu hn 2 mc. Mt gii

php l m ho theo kiu Return to Zero (RZ), bng vic s dng 3 gi tr: dng, m

v khng. Trong RZ, nhng thay i tn hiu khng phi gia cc bt, nhng trong mi

bit. Ging nh NRZ-L, in th dng c ngha l 0, v in th m c ngha l 1. Nhng

khng hn nh vy, trong khong thi gian ca mt na, mt na tn hiu cn li tr v

0. Mt bit 1 thc t c miu t l dng-0 v mt bit o c miu t l m-0 s tt hn

ch c mt mnh dng v m. Hnh 5.7 minh ho khi nim ny.

31

thi gian

thi gian

Bin

NRZ-L

NRZ-I

0 1 0 0 1 1 1 0

Thi gian

Gi tr

0 1 0 0 1 1 1 0


32/87

Hnh 5.7: M ho RZ

S bt li chnh ca m ho RZ l i hi 2 thay i tn hiu m ho 1 bit, v v

vy n chim gi gii rng hn. Tuy nhin c ba kh nng chng ta kim tra tt hn,

l hiu qu tt nht.

Biphase: C l gii php tt nht cho n nay gii quyt vn ng b ho lm ho Biphase. Trong phng php ny s thay i tn hiu trong khong thi gian ca

mi bit nhng khng tr v 0. Thay vo n chuyn sang cc i din. Ging nh RZ,

gia khong thi gian truyn cho php ng b ho.

M ho Biphase c b sung l: Manchester v Manchester vi sai.

Manchester: M ho Manchester s dng cch o ngc mi bt trong khong

thi gian ca n ng b v miu t bit. Vic chuyn trng thi m-dng tng ng

vi bt 1 v dng-m tng ng vi bt 0. y ta s dng vic mt chuyn trng thin cho hai mc ch. M ho Manchester t c theo mc ca ng b ho nh RZ,

nhng ch c 2 gi tr bin .

Manchester vi sai: Trong Manchester vi sai vic o ngc trong khong thi gian

ca mi bt c s dng cho vn ng b ho, nhng s c mt hoc thiu vng ca

vic bin i c thm vo u trong khong thi gian tm ngng c s dng

xc nh cho bit. Mt s bin i c ngha l bt 0 v s khng bin i c ngha l bt 1.

Manchester vi sai yu cu 2 s thay i tn hiu trnh by bt 0 nhng ch c 1 trnh

by bit 1. Hnh 5.8 Ch ra cc tn hiu Manchester v Manchester vi sai i vi cc mu

bt ging nhau:

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33/87

Hnh 5.8: M ho Manchester v Manchester vi sai

M ho lng cc:

M ho lng cc, ging nh RZ, s dng 3 mc in th: dng, m v 0. Tuy

nhin khng ging nh RZ, mc 0 trong m ho lng cc c s dng miu t bit

0, cn bt 1 ng vi in th m hoc dng. Nu u tin mt bit c miu t bi bin

dng, th bt 1 th hai s c miu t bin m, cn bt 1 th ba li c miu

t bng bin dng ... Vic lun phin ny xut hin c khi cc bit 1 ri rc nhau.

Ba kiu ca m ho lng cc thng thng truyn thng d liu l: AMI,B8ZS, v HDB3 (xem hnh 5.9).

Hnh 5.9: Cc kiu m ho lng cc

M ho AMI (o du xen k lng cc): y l kiu m ho lng cc n gin

nht; trong tn gi ca n; t du xut pht t in tn v c ngha l 1. V vy AMI c

ngha l o 1 xen k nhau. Mt v tr trung lp, in th 0 s trnh by bt 0. Nhng bt 1c miu t bi cc in p dng m an xen nhau. Hnh 5.10 a ra v d ny.

33

Lng ccLng cc

AMIAMI B8ZSB8ZS HDB3HDB3

Thi gian

Thi gian

Manchester

Manchestervi sai

Bin


34/87

Hnh 5.10: M ho AMI

Mt s bin i ca AMI c gi l gi ba bc vi bt 0 nm xen k gia in th

dng v in th m.Vi vic o ln trong mi ln xut hin ca 1, AMI t c 2 mc ch: th nht

cc thnh phn DC l 0, v th 2 mt dy di cc s 1 c ng b ho. Khng c k

thut no chc chn vic ng b ho cho mt chui di cc s 0.

Hai bin i ca AMI va c pht trin gii quyt vn ng b ho chui

0, c bit cho vic truyn pht khong cch ln. u tin, c s dng Bc M,

c gi l B8ZS (lng cc thay th 8-zero). Th hai, c s dng Nht v Chu

u, c gi l HDB3 (lng cc mt cao 3). C hai kiu ny u l s thch nghi viAMI m ch thay i mu gc trong trng hp c nhiu chui 0 lin tip.

B8ZS: l mt s tho thun c chp nhn Bc M cung cp vic ng b

ho cho chui 0. Trong tt c cc tnh hung, cc chc nng B8ZS tng t nh ca

AMI. AMI thay i cc vi mi 1 gp phi. Nhng thay i ny cung cp s ng b

ho cn thit bi bn nhn. Nhng tn hiu khng thay i trong sut chui 0, v vy vic

ng b ho thng b mt.

S khc nhau gia B8ZS v AMI xut hin bt c khi no c 8 hoc nhiu hn cc

bt 0 lin tip gp phi trong dng d liu. Gii php cung cp bi B8ZS l p t s thay

i tn hiu gi bn trong chui 0 (c gi l violation). mi thi im c 8 bit 0 xut

hin lin tip, B8ZS a vo nhng thay i trong mu da trn s khc bit ca bit 1

trc (1 ch xut hin pha trc ca chui 0). Xem hnh 5.11 hiu r hn.

34

Thi gian

Bin

0 1 0 0 1 1 1 0


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Hnh 5.11: M ho B8ZS

Nu bt 1 trc l dng, 8 bt 0 s c m ho l 0, 0, 0, +, -, 0, -, +. Hy nh

rng bn nhn ang tm kim s thay i xc nh 1. Khi n thy c 2 in tch dng

lin tip bao quanh 3 bt 0, n nhn ra mu, tnh ton v a vo violation khng gy

ra li. Sau n tm kim phn cn li ca nhng violation trng i. Khi tm thy

chng, bn nhn chuyn 8 bt thnh 0 v quay tr li ch AMI thng thng.

Nu cc ca bt 1 trc l m, mu ca cc violation s l tng t nhng o

ngc li cc. C mu dng v m c ch ra trong hnh 5.11.

HDB3: Vn ng b ho chui lin tip cc s 0 c gii quyt Nht v Chu

u khc vi M. Quy c ny, gi l HDB3, a s thay i vo mu AMI mi thi

im 4 bt 0 lin tip thay cho 8 bit nh ca Bc M. Mc d tn gi ca n l HDB3, cc

mu thay i bt c khi no c 4 bt 0 lin tip. Xem hnh 5.12.

Hnh 5.12: M ho HDB3

Nh trong B8ZS, mu ca cc violation trong HDB3 c da trn s phn cc ca

bt 1 trc . Tuy nhin khc vi B8ZS, HDB3 cng nhn vo s cc bt 1 va xut hin

trong dng bt k t ln thay th cui cng. Bt c khi no s ca bt 1 k t ln thay th

cui cng l l, B8ZS y mt violation vo v tr th 4 ca cc bt 0 lin tip. Nu s

phn cc ca bt trc l dng, violation l dng. Nu s phn cc ca bit trc

l m, violation l m.

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Bt c khi no s ca bt 1 k t ln thay th cui cng l chn, B8ZS y violation

vo v tr u tin v v tr th 4 ca cc bt 0 lin tip. Nu cc ca bt trc l dng, c

hai violation l m. Nu cc ca bit trc l m, c hai violation l dng. C 4 mu

ny c ch ra hnh 5.12.

V d 5.1: S dng B8ZS, m ho dng bt 10000000000100; p dng vi cc cabt 1 u tin l 1.

Gii php: Xem hnh 5.13

Hnh 5.13: Gii php cho v d 5.1

V d 5.2: S dng HDB3, m ho dng bt 10000000000100; p dng vi s cc

bt 1 trc n l chn v bt 1 u tin l dng.

Gii php: Xem hnh 5.14

Hnh 5.14: Gii php cho v d 5.2

5.2 Chuyn i tn hiu tng t(analog) thnh tn hiu s(digital)

i khi chng ta cn s ho mt tn hiu tng t. V d gi ging ni ca con

ngi trong mt khong cch xa. Chng ta cn s ho n gim nhiu v n. Vn

ny c gi l chuyn i tn hiu tng t v tn hiu s hoc l s ho tn hiu

tng t. iu ny cn gim bt s s lng bc x ln ca cc gi tr trong thng ipca tn hiu tng t. Nh vy cn a ra lung tn hiu s vi s mt mt thng tin l

36

Thi gian

Bin

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

Thi gian

Bin

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


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nh nht. Mt vi phng thc chuyn t tn hiu tng t v tn hiu s s c tho

lun phn sau.

Hnh 5.15 Bin i tng t sang s

Trong s chuyn i t tn hiu tng t v tn hiu s, chng ta a ra cc thng

tin di dng sng lin tc nh l mt dy nhp ca cc tn hiu s (1s hoc 0s).

Chuyn i tn hiu tng t v tn hiu s cn s dng mt vi tn hiu s c

trnh by ti phn 5.1. Cu trc ca s chuyn i l khng c vn g. Vn l lmnh th no chuyn cc thng tin t s lng ln cc gi tr ti s ri rc ca cc gi

tr m khng nh hng ti gic quan con ngi v cht lng ca tn hiu.

iu ch bin xung (Pulse Amplitude Modulation - PAM):

Bc u tin trong qu trnh chuyn i tn hiu tng t v tn hiu s c gi

l iu ch bin xung. K thut ny thao tc vi tn hiu thng t, n gin n v to

ra nhng di ca xung c bn c kt qu mu tn hiu tng t. Thut ng mu c

ngha l n v bin ca tn hiu trong 1 khong thi gian l bng nhau.

L thuyt mu c dng trong iu ch bin xung l hiu qu hn trong cc

vng khc ca nng lng, n l d liu truyn thng. Tuy nhin PAM l nn tng quan

trng trong chuyn i tn hiu tng t v tn hiu s l thuyt ny c gi l iu ch

xung theo m (pulse code modulation - PCM)

Trong PAM, tn hiu ban du l mu ti thi gian ngh bng nhau (hnh v 5.16

di y m t iu ny).

Hnh 5.16:PAM

PAM gi l k thut mu v gi. hin ti, cp tn hiu l c, tip theo l tr

gip tng hp. Nhng mu gi tr xy ra ch trong s xut hin ca sng, nhng nhn

chung kt qu trong PAM vn cn ngn.

37

Analog/Digitalconversion(codec)

Amplitud

e

Amplitud

e

Time Time

a. Anaglog signal b. PAM signal


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L do PAM khng hu dng trong d liu truyn thng l, mc d n truyn i

sng gc t mt di ca xung, nhng xung ny vn cn c bin (vn l tn hiu tng

t, khng phi l tn hiu s). to ra tn hiu s chng ta phi sa i chng bng cch

dng PCM.

Lu : PAM c mt vi ng dng, nhng bn thn n khng dng trong d liu truynthng. Tuy nhin y l bc quan trng ph bin u tin trong l thuyt chuyn i v c

gi l PCM.

iu ch xung theo m (PCM):

PCM sa i vn to ra xung bng PAM hon thnh mt tn hiu s. Lm nh

vy, PCM u tin lng t ho nhng xung ca PAM. S lng t l l thuyt tha

hng nhng gi tr trong mt dy bit ti mu c bit. Kt qu ca lng t ho

c a ra trong hnh 5.17

Hnh 5.17

Hnh 5.18 trnh by 1 mu phng thc ca tha k tn hiu v gi tr ln lng

t ho mu. Mi gi tr chuyn i vo tng ng 7 bt nh phn. Bt th 8 l bt du

(+ l 0 v l 1).

38

Amplitude

Time

+24+38

+48 +39+26

-15

-80

-50

+52

+127+125

+110+90

+88+77


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S nh phn sau truyn vo tn hiu s s dng mt k thut m ho tn hiu s

thnh tn hiu s. Hnh 5.19 a ra kt qu ca iu ch xung theo m ca tn hiu gc

khi m ho ln cui vo mt cc tn hiu vi gi tr ca 3 mu u.

Hnh 5.19:PCM

PCM trn thc t c to ra t 4 b x l song song: PAM, lng t ho, m nh

phn v m chuyn tn hiu s thnh tn hiu s. Hnh 5.20 th hin cc phn t x l.

PCM l mu phng thc dng s ho ging ni trong ng kiu T (T-line) trong hthng ca trm truyn thng bc M (xem chng 8).

Hnh 5.20 T tn hiu tng t sang m s PCM

+024 00011000 -015 10001111 +125 01111101

+038 00100110 -080 11010000 +110 01101110

+048 00110000 -050 10110010 +090 01011010

+039 00100111 +052 00110110 +088 01011000

+026 00011010 +127 01111111 +077 01001101

39

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

Hng dch chuyn

+024 +038 +048

PAM

Lng tha

M nhphn

0001100000100110

M has/s

+02400011000

+0380010011

0

Hng dch chuyn


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Tc ly mu(Sampling Rate).

Nh chng ta thy t cc hnh v trc, s chnh xc ca mi tn hiu s c ti

to li t cc tn hiu tng t ph thuc vo s mu em theo. S dng PAM v

PCM chng ta ti to li sng chnh xc bng cch em theo s lng mu khng xc

nh, hoc chng ta c th ti to s pht ra trng khng ca chnh n. R rng chng tamun tm mt s u trn trc s. Nh vy, cu hi l C bao nhiu mu th ?

Trn thc t, rt t thng tin ng ch cho vic gi v nhn to li cu trc ca

tn hiu tng t. Theo nh l Nyquist m bo tnh chnh xc trong s ti to tn

hiu tng t nguyn bn ta s dng PAM, s tc ly mu phi tin hnh hai ln tn

s cao nht ca tn hiu gc. Nh vy, nu chng ta mun mu ging ni in thoi vi

tn s ln nht 4000 Hz, chng ta cn tc ly mu ca 8000 mu trn 1 giy.

Lu : Theo nh l Nyquist, s tc ly mu t nht phi tin hnh 2 ln tn scao nht.

Mt s tc ly mu hai ln tn s x Hz c ngha tn hiu phi c ly mu

mi x giy. Dng ging ni qua ng dy in thoi trn l mt v d, iu ny c

ngha l mt mu mi 1/8000 giy. Hnh 5.21 minh ho cho khi nim ny.

40

Amplitude

Time

Tn s cao nht = x HzTc mu = 2x mu/giy

Thi gian gia 2 mu = x

Hnh 5.21 nh l Nyquist


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1. Hnh 5.22 ch ra mi quan h gia 3 thnh phn: u vo l thng tin s, b bin i

D/A, v kt qu ra l tn hiu tng t.

Hnh 5.22 iu ch s sang tng t

C nhiu thit b bin i D/A, nhng chng ta s ch bn n nhng thit b c li

nht cho vic truyn d liu.

Trong chng 4, mt tn hiu hnh sin c 3 c trng: bin , tn s v pha. Khi

thay i mt trong 3 c trng ny, ta s to c phin bn mi ca tn hiu hnh sin .

Chng hn tn hiu gc l mc 1, c th bin i thnh mc 0 hoc ngc li. V th,bng s thay i lin tc hnh dng ca mt tn hiu in n gin, ta c th dng n

m t d liu s. Bt c ba c trng trn c th c thay i theo cch ny, a cho ta

t nht 3 thit b bin i d liu s thnh tn hiu tng t: ASK (Amplitude Shift

Keying - di bin ), FSK (Frequency Shift Keying - di tn s) v PSK (Phase Shift

Keying - di pha). Hn na, thit b th t tt hn c l thay i hn hp ca c bin ,

tn s v pha c gi l b iu ch bin cu phng QAM (Quadrature Amplitude

Modulation). QAM l hiu qu nht so 3 thit b trc, v l thit b c dng trong tt

c cc MODEM hin i (hnh 5.23).

Cc thnh phn ca b bin i D/A:

Trc khi tho lun cc phng php c th bin i D/A, c hai vn c bn

phi c nh ngha: tc truyn bit/baud v tn hiu mang.

Bit Rate and Baud Rate:

Hai thut ng c dng thng xuyn trong vic truyn d liu l bit rate v baudrate. Bit rate l s bit c truyn trong mt giy. Baud rate ch ra s n v tn hiu trn

mt giy c yu cu m t nhng bit . Khi tho lun v hiu qu my tnh, th bit

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rate l quan trng hn, v ta mun bit thi gian x l tng mu tin.Tuy nhin, trong vic

truyn d liu, th ch trng n tnh hiu qu ca vic chuyn d liu t ni ny n ni

khc trong cc mu tin hay cc khi tin. Cc n v tn hiu t hn c yu cu l hiu

qu hn cho h thng v bng thng hp hn c yu cu truyn cc bit; v vy,

chng ta ch trng hn vo baud rate. Baud rate xc nh bng thng yu cu gi tnhiu.

Baud rate = Bit rate / (s bt trn mt n v tn hiu)

Sng mang:

iu ch bin (ASK):

Cng tn hiu mang c bin i sang dng s nh phn 0 hoc 1. Trng hp

ny tn s v pha l khng i trong khi bin thay i. in p biu din mc 1 v

m t mc 0 l sang tri vi h thng. Khong thi gian 1 bit l l giai on c nhngha l 1 bit. nh bin ca tn hiu trong sut khong thi gian mi bt l khng i

v s thay i ca n ph thuc vo bit (0 hoc 1). Tc truyn dng ASK b gii hn

bi cc tnh cht vt l ca mi trng truyn.

Khng may, vic truyn ASK l nhy cm cao vi nhiu. Thut ng ting n

(nhiu) m ch n in p khng nh trc can thip vo ng truyn bi cc hin

tng khc nh l s nng ln hay in t c sinh ra bi cc ngun khc. Cc in p

khng nh trc ny trn ln vi tn hiu lm thay i bin . Mc 0 c th b i

thnh mc 1 hoc ngc li. R rng nhiu l vn kh gii quyt i vi ASK, do ch

c da vo bin nhn bit. Nhiu thng tc ng ln bin ; v th ASK l

phng php iu ch b nh hng nht bi nhiu.

K thut ASK ph bin c gi l OOK (on-off-keying). Vi OOK, cc gi tr bit

c m t khng theo in p. S tin b l gim bt trong nng lc yu cu truyn

thng tin.

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Hnh 5.24 ASK

Bng thng ASK:

Nh trnh by trong chng 4, bng thng ca mt tn hiu l tng di tn ang

s dng bi tn hiu . Khi ta phn tch mt tn hiu iu ch ASK, ta s nhn c ph

ca nhiu tn s n. Tuy nhin, cc tn s quan trng nht nm trong khong fc Nbaud/2

v fc + Nbaud/2 vi tn s mang fc gia.

Bng thng yu cu cho ASK c tnh ton theo cng thc.

BW = (1+d)NbaudBW: bng thng

Nbaud: baud rate

d: h s truyn (d 0)

Hnh 5.25 Mi quan h gia tn s v bng thng trong ASK

Mc d ch c tn s mang, qu trnh iu ch mt tn hiu phc tp thc ra l s

ho trn bi nhiu tn hiu n vi tn s khc nhau.

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iu ch tn s (FSK)

Tn s tn hiu mang c bin i m t cc ch s nh phn 0 hoc 1. Tn s

tn hiu mang trong sut thi gian mi bit l khng i, gi tr ca ph thuc vo bit 0

hoc 1: cn li bin v pha l khng i.

FSK trnh c hu ht cc vn nhiu ca ASK. Bi v thit b nhn coi tn stiu biu thay i qua s chu k cho, n c th b qua cc nh in p. Cc yu t

gii hn ca FSK l mi trng vt l ca sng mang.

Bng thng FSK

Mc d FSK dch gia 2 tn s mang, n d phn tch thnh 2 tn s trong cng thi

im. Ta c th ni ph FSK l s hn hp ca 2 ph ASK tp trung quanh tn s fc0 v

fc1. bng thng yu cu cho vic truyn FSK bng tc baud rate ca tn hiu cng vi

tn s dch (khc nhau gia hai tn s mang): BW = (fc1-fc0) + Nbaud.

Hnh 5.27: FSK

Hnh 5.28 Quan h gia baud rate v bng thng FSK

iu ch pha (PSK)

Pha ca sng mang c bin i biu din sang s nh phn 0 hoc 1. C bin

v tn s l khng i, cn pha th thay i. V d, nu ta bt u vi mt pha 0o

m t bt 0, th ta c th thay i pha sang 180 o gi s nh phn 1. Pha ca tn hiu

trong sut thi gian mi bt l khng i v gi tr ca n ph thuc vo bit 0 hoc 1.

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Phng php trn thng c gi l 2-PSK hoc PSK nh phn, bi v 2 pha

khc nhau (0o v 180o) c dng. Hnh 5.30 ch r iu ny bng mi quan h gia pha

sang gi tr bit. S th hai, c gi l chm sao (constellation) hoc s trng thi

pha, ch ra cng mi quan h c minh ho ch bng cc pha.

Bng thng PSK

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CHNG 8: A KNH

Bt c khi no dung lng ng truyn ni hai thit b ln hn nhu cu

truyn ca cc thit b, th ng truyn c th c dng chung. Cng ging

nh mt ng ng nc ln c th dn nc ti nhiu nh trong cng mt lc. aknh l tp hp cc k thut cho php vic truyn ng thi nhiu tn hiu qua mt

ng truyn d liu n.

Do vic s dng, trao i d liu v thng tin lin lc ngy cng tng nn

ng truyn cng cn nh vy. Chng ta c th iu tit s gia tng ny bng

cch ni thm cc ng truyn n ring r mi khi c mt knh mi cn c

lin kt hoc chng ta c th thit lp cc ng truyn dung lng cao hn v

dng mi ng ny truyn nhiu tn hiu khc nhau. Cng ngh ngy nay sdng cc thit b bng thng rng nh nh cp ng trc, cp quang, sng mt t,

sng v tinh v.v.. Cc cng ngh ny c kh nng truyn vi dung lng cao hn

rt nhiu so vi nhu cu truyn trung bnh ca tn hiu. Nu dung lng truyn ca

mt lin kt ln hn nhu cu truyn ca cc thit b trong lin kt , th dung

lng cao l lng ph. Mt h thng hiu qu l s dng ti a tt c cc

phng tin ca n. Hn na, mt cng ngh t tin ch pht huy hiu qu khi cc

ng lin kt c dng chung.8.1 Nhiu mt/ mt nhiu

Trong mt h thng ghp knh, n thit b dng chung mt ng truyn.

Hnh 8.1b th hin m hnh c bn ca mt h thng ghp knh. Bn thit b bn

tri iu khin dng d liu ca chng ti mt thit b ghp knh (Multiplexer

MUX), thit b ghp knh hp cc dng d liu ny thnh mt dng d liu n

(nhiu ti mt). u nhn, dng d liu c a vo mt thit b phn knh

(demultiplexer DEMUX), thit b phn knh tch dng d liu thnh cc dngd liu thnh phn (mt ti nhiu) v iu khin chng ti thit b nhn.

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Hnh 8.1 H thng ghp knh v h thng khng ghp knh

Trong hnh 8.1b, t ng truyn lin quan n ng kt ni vt l. T

knh lin quan n mt phn ca ng truyn m n mang d liu gia hai cpthit b cho trc. Mt ng truyn c th c nhiu (n) knh.

Cc tn hiu c ghp knh s dng ba k thut c bn: Ghp tn (FDM),

Ghp thi gian (TDM), v ghp sng (WDM). TDM c phn chia thnh TDM

ng b (thng c gi l TDM) v TDM khng ng b, cn c gi l TDM

tnh.

Hnh 8.2 Cc dng ghp knh

8.2 Ghp tn

Ghp tn (FDM) l mt k thut c p dng khi bng thng ca ng

truyn ln hn bng thng kt hp ca cc tn hiu c truyn. Trong FDM, cc

tn hiu c sinh ra bi cc thit b gi iu ch tn hiu trn cc tn s sng

mang khc nhau. Nhng tn hiu c iu ch ny c kt hp thnh mt tn

hiu phc hp n truyn trn ng truyn. Cc tn s mang c tch bit

bi bng thng rng cha cc tn hiu iu ch. Nhng di bng thng ny l

nhng knh cho php cc tn hiu khc nhau c th truyn qua. Cc knh phi

c tch bit bi cc di bng thng khng s dng (bng thng ng gc - guard

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bands) ngn nga cc tn hiu khng b chng lp bng thng. Hn na cc tn

s sng mang phi khng b nhiu bi cc tn s d liu gc. Nu khng tun th

theo mt trong hai iu kin trn th c th khng khi phc c cc tn hiu gc.

Hnh 8.3 Ghp tn (FDM)

Hnh 8.3 cho ci nhn khi qut v FDM. Trong minh ho ny, ng truyn

c chia thnh ba phn, mi phn i din cho mt knh. Cng ging nh mt

nt ni m c ba ng nh hp vo hnh thnh nn mt con ng c ba

ln xe. Mi ng tng ng vi mt ln ng ca xa l. Mi chic xe khi i vo

xa l t mt ng no th n vn c ln ng dnh ring cho n v c th i

qua m khng b chen ln bi cc chic xe cc ln ng khc.

Mc d hnh 8.3 minh ho ng truyn nh l c chia thnh cc knh tch

bit, nhng s phn chia knh thc t c phn theo tn s ch khng phi theo

khng gian.

Qu trnh ghp tn FDM

Hnh 8.4 Qu trnh ghp tn theo min thi gian (FDM)

Hnh 8.4 l mt minh ho theo min thi gian ca qu trnh ghp knh. FDM

l mt qu trnh tng t v chng ta coi thit b vo/ra l nhng chic in thoi.

Mi chic in thoi sinh ra mt tn hiu c phm vi tn s ging nhau. Trong thit

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Hnh 8.6 Qu trnh tch knh FDM, min thi gian

Hnh 8.7 Qu trnh tch knh FDM, min tn s

8.3 Ghp sng (WDM)

Ghp sng cng ging nh ghp tn, ngoi tr qu trnh ghp v tch knh lin

quan n tn hiu nh sng c truyn qua cc knh quang. tng ca chng

ging nhau: l kt hp cc tn hiu c cc tn s khc nhau. Tuy nhin s khc

nhau l tn s ca cc tn hiu rt cao.

Hnh 8.8 cho ci nhn khi qut v mt qu trnh tch knh v ghp knh

WDM. Cc di tn rt hp ca nh sng t cc ngun khc nhau c kt hp

to ra mt di tn nh sng rng hn. u nhn, tn hiu c tch ra bi thit b

tch knh.

Ai c th vn cn phn vn v k thut ca WDM. Mc d cng ngh l rt

phc tp, nhng tng li rt n gin. Chng ta mun kt hp nhiu ngun nh

sng thnh mt nh sng n thit b ghp knh v lm ngc li thit b tch

knh. Hp v tch ngun nh sng c d dng x l bi mt lng knh. T kin

thc vt l c bn l lng knh khc x nh sng da trn gc ti v tn s nh

sng. S dng k thut ny, thit b ghp knh s kt hp nhiu tia nh sng vo

(mi tia cha mt di cc tn s hp) thnh mt tia ra c di tn rng hn. Thit btch knh s lm ngc li qu trnh trn.

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8.4 Ghp thi gian (TDM)

Ghp thi gian l mt tin trnh k thut s c p dng khi dung lng tc

truyn d liu ca phng tin truyn ln hn tc truyn c yu cu ca

cc thit b gi v nhn. Trong trng hp , nhiu nhim v truyn c th dng

mt ng kt ni n bng cch chia nh d liu cc thit b gi v an xen ccphn d liu vo vi nhau.

Hnh 8.10 Qu trnh ghp thi gian

Hnh 8.10 cho ci nhn khi qut v TDM. Ch rng ng truyn c s

dng ging nh FDM, tuy nhin y ng truyn c th hin bng cc

phn theo thi gian ch khng phi theo tn s.Trong hnh v, cc phn ca cc tn hiu 1, 2, 3 v 4 chim ng truyn mt

cch tun t. N ging nh mt ci thang trt, phc v nhiu dng khc nhau.

Mi dng c ng ring ca n, v nhng ngi trt trn mi ng ch n

lt ri ln thang. Khi ln n nh, ngi trt trt xung theo dng m trc

y anh ta i ln thang.

TDM c th c thc hin theo hai cch: TDM ng b v TDM khng ng

b.

TDM ng b

Trong ghp thi gian ng b, thut ng ng b c ngha hi khc so vi

ngha khi n c s dng trong cc lnh vc khc ca truyn thng. y ng

b c ngha l thit b ghp knh phn b chnh xc cc khe thi gian ging nhau

cho mi thit b trong mi thi im m khng quan tm thit b c d liu

truyn hay khng. V d khe thi gian A c phn b ring cho thit b A th cc

thit b khc khng th s dng khe A ny c. mi thi im, khe thi gian

c phn b ca n xut hin, mt thit b c c hi gi mt phn d liu ca n.

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Nu mt thit b khng th truyn hoc khng c d liu truyn th khe thi

gian ca n vn trng.

Khung frame

Cc khe thi gian c nhm li thnh nhng frame. Mi frame bao gm mt

vng trn vn cc khe thi gian, mt vng bao gm mt hoc nhiu khe thi giandnh cho mi thit b gi. Trong mt h thng vi n ng vo th mi frame s c

t nht n khe, mi khe c phn b mang d liu t mt ng vo xc nh.

Nu tt c cc thit b vo dng chung mt ng truyn v cng truyn vi mt

tc truyn ging nhau th mi thit b s c mt khe thi gian trn mi frame.

Khi tc truyn mi thit b l khc nhau, ta hon ton c th vn iu tit

c ng truyn. Mt t truyn vi hai khe thi gian trn mt frame s n

nhanh hn hai ln so vi t truyn vi mt khe thi gian trn mt frame. Nhngkhe thi gian c phn b cho mt thit b chim nhng v tr ging nhau trong

mi frame v cu thnh knh ca thit b . Trong hnh 8.11 th hin thit b ghp

knh vi 5 ng vo ghp thnh mt ng n dng TDM ng b. Trong v d

ny, tt c cc ng vo c cng tc truyn, do vy s khe thi gian trong mi

frame l bng s ng vo.

Hnh 8.11 TDM ng b

Qu trnh an xen - Interleaving

TDM ng b c th c so snh vi b chuyn mch c tc quay vng

rt nhanh. Khi b chuyn mch m trc mt thit b, thit b c c hi gi

mt s lng d liu xc nh (x bt) ln ng truyn. B chuyn dch dch

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chuyn t thit b ny n thit b khc vi mt tc khng i theo mt trt t

c nh. Qu trnh ny c gi l interleaving.

Interleaving c th c thc hin theo bt, byte hoc cc n v d liu khc.

Ni cch khc, thit b ghp knh c th ly ln lt t cc thit b 1 hoc 8 bt.

Trong mt h thng cho, cc n v d liu lun lun c cng kch thc.Hnh 8.12 minh ho qu trnh interleaving v xy dng frame. Trong v d,

chng ta an xen cc t truyn bng k t, nhng khi nim l ging nhau cho

nhng n v d liu c chiu di tu . Nh chng ta thy, mi thit b gi mt

thng ip khc nhau. Thit b ghp knh an xen cc thng ip khc nhau v

hnh thnh nhng frame trc khi a ln ng truyn.

u nhn, thit b tch knh phn tch mi mt frame bng cch ly ra ln

lt tng k t. Khi mt k t c loi b khi frame, n c chuyn ti thit bnhn thch hp.

Hnh 8.12 v 8.13 cng ch ra nhc im chnh ca TDM ng b. Bng vic

phn b mi mt khe thi gian cho mt ng vo xc nh, chng ta s c nhng

khe rng khi cc ng vo khng cng hot ng. Trong hnh 8.12, ch c 3

frame u l y. Ba frame cui c tng cng 6 khe rng. C 6 khe rng trn tng

s 24 khe c ngha l dung lng ng truyn b lng ph.

Hnh 8.12 TDM ng b, Qu trnh ghp knh

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Hnh 8.13 TDM ng b, Qu trnh tch knh

Cc bit khung

Bi v trt t khe thi gian trong h thng TDM ng b khng thay i trn

cc frame, rt t thng tin cn phi gp vo trong mi frame. Trt t ca cc thnhphn ch cho thit b tch knh bit ni iu khin mi khe thi gian, v vy

vic nh a ch l khng cn thit. Tuy nhin do nhiu nhn t khc nhau c th

gy ra s thiu nht qun v thi gian. V nhng l do ny, mt hoc nhiu bt

ng b thng c thm vo phn u ca mi frame. Nhng bt ny c gi

l bt frame, n cho php thit b tch knh ng b dng d liu n n c th

tch cc khe thi gian mt cch chnh xc. Trong phn ln cc trng hp, nhng

thng tin ng b ny thng bao gm mt bt trn mi frame, lun phin gia 0v 1.

Hnh 8.14 Cc bt khungV d v TDM ng b: Gi s rng chng ta c 4 ngun vo trn mt ng

truyn TDM ng b, cc t truyn c an xen bi k t. Nu mi mt ngun

to ra 250 k t/giy, v mi frame mang mt k t t mi ngun, ng truyn s

phi mang 250 frame/giy.

Nu chng ta gi s mi k t bao gm 8 bt, th mi frame l 33 bt, 32 bt cho

4 k t v mt bit ng b. Nh vy mi thit b to ra 2000 bps, nhng ng

truyn phi vn chuyn 8250 bps.

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Hnh 8.15 Tnh ton tc truyn i vi cc khung

Nhi bt

Nh lu trn, hon ton c th kt ni cc thit b c tc truyn khc

nhau vo mt h thng TDM ng b. V d, thit b A dng mt khe thi gian,

trong khi thit b B nhanh hn dng 2 khe. S khe mi frame v s khe c

phn b cho mi ng vo vn c nh trong h thng, nhng cc thit b c tc

khc nhau c th kim sot mt s lng khe khc nhau. Nh rng chiu di

khe thi gian l c nh. V vy, k thut ny hot ng c th nhng tc

truyn phi l bi s ca nhau.

V d chng ta c th iu tit mt thit b nhanh gp 5 ln thit b khc bng

vic phn b cho n 5 khe thi gian v mt khe thi gian cho cc thit b khc. Tuy

nhin chng ta khng th iu tit mt thit b nhanh hn 5 ln thit b khc theo

cch ny, bi v chng ta khng th a mt na khe thi gian vo mt frame.

Khi cc tc khng phi l bi s ca nhau, chng ta c th lm cho n hot

ng bng mt k thut c gi l nhi bt. Trong k thut nhi bt, b ghp knh

thm cc bt ph vo dng ngun ca thit b tc ca cc thit b tho mniu kin tc l bi s ca nhau. V d, nu chng ta c mt thit b c tc

gp 2.75 ln tc ca cc thit b khc, chng ta c th thm cc bt nng tc

ca n ln gp 3 ln tc ca cc thit b khc. Nhng bt thm s c thit

b tch knh loi b.

TDM khng ng b

Nh chng ta thy phn trc, TDM ng b khng m bo ton b dung

lng ca ng truyn c s dng. Thc t, mt thi im no ch c mts cc khe thi gian trong frame l c s dng. Bi v cc khe thi gian c

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phn b trc v c nh, bt c khi no thit b c kt ni m khng hot ng,

th khe thi gian tng ng l rng, v nh vy knh truyn b lng ph. V d

chng ta c mt thit b ghp knh u ra ca 20 my tnh ging nhau ln mt

ng truyn n. S dng TDM ng b, th tc ca ng truyn phi

gp t nht 20 ln tc ca mi ng vo. Nhng nu ch c 10 my tnh hotng th mt na dung lng ng truyn b lng ph.

TDM khng ng b c thit k trnh kiu lng ph ny. Cng ging nh

thut ng ng b, thut ng khng ng b c ngha hi khc so vi ngha

c dng trong cc lnh vc khc trong truyn thng. y n c ngha l

mm do hoc khng c nh.

Hnh 8.16 Ghp thi gian khng ng b

Cng ging nh TDM ng b, TDM khng ng b cho php mt s ng

vo tc thp hn c ghp vo mt ng n tc cao hn. Tuy nhin,

khng ging nh TDM ng b, trong TDM khng ng b tng tc ca cc

ng vo c th ln hn dung lng ng truyn. Trong h thng ng b, nu

chng ta c n ng vo, th frame cha mt s c nh cc khe thi gian t nht l

n. Trong h thng khng ng b, nu chng ta c n ng vo th frame cha

khng nhiu hn m khe, vi m < n. Theo cch ny, TDM khng ng b h tr

mt s ng vo nh TDM ng b nhng vi dung lng ng truyn thp

hn. Ni cch khc, cho trc mt ng truyn, TDM khng ng b c th h

tr nhiu thit b hn TDM ng b.

S khe thi gian m trong mt frame ca TDM khng ng b c da trn

mt s phn tch thng k theo s lng ng vo c th ng thi cng truyn

mi thi im. Cc khe thi gian khng c phn b trc m n sn sng phc

v cho bt k cc ng vo no m c d liu gi. Thit b ghp knh qut qua

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cc ng vo, chp nhn cc phn ca d liu cho n khi mt frame c

y, sau gi frame ln ng truyn. Nu d liu khng y tt

c cc khe trong frame, khi frame ch y mt phn s c truyn i; nh vy

dung lng ng truyn khng c s dng 100%. Nhng cc khe thi gian

c phn b ng, tc thp hn cho cc ng vo, gim bt ng k mc lng ph ng truyn.

Hnh 8.17 minh ho mt h thng c 5 my tnh dng chung mt ng truyn

d liu s dng TDM khng ng b. Trong v d ny, kch thc frame l ba khe

thi gian. Hnh v minh ho thit b ghp knh x l ba mc vo khc nhau. Trong

trng hp th nht, ch c 3 trong 5 my tnh c d liu gi. Trong trng hp

th hai, c 4 thit b ang cn gi d liu i, nhiu hn mt so vi s lng khe

trong mi frame. Trong trng hp th 3, tt c cc thit b u c d liu gi i(trng hp ny theo thng k l him khi xy ra). Trong mi trng hp thit b

ghp knh qut qua cc thit b theo th t t 1 n 5, vo cc khe thi gian khi

n gp d liu cn gi.

Hnh 8.17 Cc v d v cc khung TDM khng ng b

Trong trng hp th nht, c ba ng vo hot ng tng ng vi ba khe

trong mi frame. i vi 4 frame u tin, d liu vo c phn phi mt cch

cn i gia tt c cc thit b lin lc. Tuy nhin, trong frame th 5, thit b 3 v 5

hon thnh vic truyn d liu ca mnh, nhng thit b 1 vn cn hai k t cn

gi. Thit b ghp knh nht k t A t thit b 1, n qut sang cc thit b khc

nhng cc thit b ny u khng c d liu cn gi, n tr li thit b 1 nht k t

A cui cng. V lc ny khng c d liu y cc khe thi gian, thit b ghp

knh truyn frame th 5 khi ch c hai khe c d liu. Nu trong h thng

TDM ng b khi n s cn ti 6 frame, mi frame c 5 khe truyn ton b

d liu - tng cng c 30 khe. Nhng ch c 14 khe c d liu, khi ng

truyn ri hn na thi gian truyn. Vi h thng khng ng b, ch c mt frame

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l khng y. Trong sut thi gian truyn, ton b dung lng ca ng truyn

c s dng ti a.

Trong trng hp th hai, c mt ng vo hot ng nhiu hn cc ng

khc. Lc ny, v thit b ghp knh qut t 1 n 5, n y frame trc khi tt

c cc ng vo c kim tra. V vy frame th nht mang d liu ca thit b 1,

3 v 4 khng c d liu t thit b 5. Thit b ghp knh tip tc qut t ni n tm

dng, y phn u tin ca thit b 5 vo khe u tin ca frame tip theo, sau

thit b ghp knh tr li thit b 1 ly phn d liu th hai ca thit b 1 vo

khe th hai, qu trnh c tip tc nh vy. Nh chng ta thy, khi s ngi gi

khng bng s khe trong mi frame, th cc khe khng c mt cch cn i.

Trong v d ny, thit b 1 chim khe u tin ca frame th nht, nhng li chim

khe th hai trong frame th hai, v.v.

Trong trng hp th ba, cc khung c nh trn, nhng y 5 ng

vo u hot ng. Trong v d ny, thit b 1 chim khe u tin trong frame th

nht, khe th 3 trong frame th 2 v khng c khe no trong frame th 3.

Trong trng hp 2 v 3, nu tc ca ng truyn bng tng tc ca 3

ng vo th d liu cn truyn s n thit b ghp knh nhanh hn d liu m

thit b ghp knh a ln ng truyn. Trong trng hp cn phi c mt b

m lu d liu cho n khi thit b ghp knh sn sng a n ln ngtruyn.

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nh a ch v chi ph ph tri

Trng hp 2 v 3 trong v d trn cho thy yu im chnh ca TDM

khng ng b: Lm th no thit b tch knh bit c khe no thuc vo ng

ra no. Trong TDM ng b d liu ca mt thit b thuc vo khe thi gian c

ch nh bi v tr ca khe trong frame. Nhng trong TDM khng ng b, d liut mt thit b cho c th khe u ca mt frame v khe th ba ca frame k

tip. Do thiu s rng buc v tr c nh, mi khe phi mang a ch ch dn

cho thit b tch knh iu khin d liu. a ch ny ch c dng cc b, n

c gn vo bi thit b ghp knh v c loi b bi thit b tch knh. Trong

hnh 8.17, a ch c xc nh bi 1 s thp phn.

Vic thm nhng bt a ch vo mi khe lm tng tng chi ph ca h thng

khng ng b v hn ch phn no hiu qu tim nng ca h thng. hn chtc ng ca n, cc a ch thng ch bao gm mt s lng bt nh v c th

lm cho s lng bt a ch t hn bng vic m rng mt a ch y cho phn

u ca t truyn v c rt gn trong cc t truyn sau .

Nhu cu ca vic nh a ch lm cho TDM khng ng b thiu hiu qu

i vi vic an xen theo bt hoc byte. Nu an xen theo bt th mi bt d liu

cn mang theo mt a ch, gi s l 3 bt a ch. Nh vy phi mt 4 bt truyn

mt bt d liu. Cho d ng truyn lun y th ch dung lng l c sdng truyn d liu, phn cn li l chi ph ph tri. V nhng l do ny, TDM

khng ng b ch hiu qu khi kch thc ca cc khe thi gian l kh rng.

Khe thi gian c di thay i

TDM khng ng b c th iu tit qu trnh truyn vi nhng tc khc

nhau bng vic thay i di cc khe thi gian. Cc trm truyn c tc nhanh

hn c th c phn cho khe thi gian di hn. qun l cc trng di thay

i ny cn phi thm nhng bt iu khin vo phn u ca mi khe thi gian thng bo chiu di ca phn d liu ang ti. Nhng bt thm ny cng lm tng

chi ph ca h thng v h thng ch c hiu qu vi nhng khe thi gian rng hn.

a knh o

a knh o ngc li vi qu trnh ghp knh. H thng phn knh ly

dng d liu t mt ng truyn tc cao v chia n thnh cc phn c th

truyn ng thi qua cc ng truyn tc thp hn m khng lm gim tc

truyn d liu chung.

To sao chng ta cn a knh o? Gi s mt t chc no mun gi d

liu, m thanh, hnh nh, mi dng d liu i hi tc truyn d liu khc nhau.

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gi m thanh n cn mt ng truyn 64 Kbps, gi d liu n cn mt

ng truyn 128 Kbps, v gi hnh nh n c th cn mt ng truyn 1.544

Mbps. iu tit tt c cc nhu cu ny, t chc c hai s la chn. Th nht,

t chc c th thu mt knh 1.544 Mbps t mt cng ty vin thng, nhng rt

t khi s dng n dung lng ti a, r rng iu ny l s dng khng hiu quthit b. Th hai, t chc c th thu mt vi knh ring c tc thp hn.

Bng vic s dng mt tho thun c gi l bng thng khi yu cu, t chc ny

c th dng nhng knh ny bt c khi no n cn. Truyn d liu m thanh c th

c truyn trn bt k knh no. Tn hiu d liu v hnh nh c th c chia ra

v c gi i trn nhiu hn hai ng truyn. Ni cch khc tn hiu d liu v

hnh nh c th chia knh trn nhiu ng truyn.

Hnh 8.18 a knh thun v ngc

8.5 ng dng ghp knh: H thng in thoi:

Ghp knh l cng c thit yu trong ngnh cng nghip in thoi. Bng cchxem xt mt vi h thng in thoi ta s hiu ck cc ng dng ca FDM v

TDM. Tt nhin cc h thng in thoi khc nhau trn th gii s s dng cc

cng ngh khc nhau, nhng ta ch xem xt cc h thng Bc M.

Mt s nh cung cp dch v in thoi Bc M cung cp cc dch v cc b

v dch v ng di. Mt s nh cung cp nh: AT&T, MCI v Sprint.

V mc ch ny ca phn ny, ta coi cc nh cung cp dch v khc nhau nh

mt mng in thoi. ng dy ni t mt thu bao n mng in thoi gi l

ng dch v.

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Hnh 8.19: Mng in thoi.

H thng dch v chung

Khi u cc cng ty in thoi cung cp dch v tng t. Sau cng ngh

cho php h a ra dch v k thut s v v mng. Cho n ny, cc nh cungcp vn ang chuyn t dch v tng t sang dch v k thut s. V tng lai s

ton b mng s cung cp dch v k thut s. Cc cng ngh FDM v TDM vn

ang c s dng.

Hnh 8.20 Cc loi dch v in thoi.

Dch v tng t

C rt nhiu dch v tng t, trong c hai loi c cp y: Dch v

chuyn mch v dch v thu bao.

Hnh 8.21 Cc loi dch v dch v tng t

Dch v chuyn mch tng t

y chnh l dch v quay s quen thuc, thng gp khi s dng in thoi nh. N s dng mt i dy thng hoc cp i dy xon ni in thoi khch

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hng vi mng in thoi thng qua tng i cc b. Kt ni ny gi l kt ni cc

b (Local Loop). Mng in thoi khch hng kt ni n gi l mng in thoi

chuyn mch cng cng (PSTN).

Tn hiu kt ni cc b l tng t v c tn s t 0 n 4000 Hz. Khi m

ngi gi quay s n s in thoi khc, cuc gi c chuyn n cc b chuynmch cc tng i in thoi. B chuyn mch tng ng to ra mt knh truyn

dn ni ngi gi n ngi nhn. B chuyn mch kt ni hai ng dy trong

khong thi gian thc hin cuc gi.

Hnh 8.22 Dch v chuyn mch tng tDch v thu bao tng t:

Mt dch v to cho khch hng c hi thu ng truyn, cn gi l ng

chuyn dng, c kt ni lin tc. Mc d s kt ni vn i qua cc chuyn mch

trn mng in thoi nhng khch hng s coi nh mt ng n v b chuyn

mch lun ng v khng cn phi quay s.

Hnh 8.23 Dch v thu bao tng t

ng iu kin:

Cng ty vin thng cn a ra dch v gi l dch v iu kin (Conditioning).N s tng cht lng ca ng truyn bng cch gim s suy gim tn hiu,

gim mo, gim thay i tr. ng ny l ng tng t nhng v cht lng

ng truyn nn c th truyn d liu s nu n c ni vi modem.

H thng tng t phn cp:

ti a hiu nng ca c s h tng mng, cc hng in thoi ghp cc tn

hiu t cc ng c bng thng thp thnh ng c bng thng cao hn. Theo

cch ny th cc ng chuyn mch tng t, cc ng thu bao c th c kt

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hp li to ra s ng truyn t hn v to hn. i vi cc ng tn hiu

tng t, k thut FDM c s dng.

Trong h thng phn cp ca AT&T c to bi cc nhm, siu nhm, nhm

ch, hay nhm cc ln.

Hnh 8.24 Analog hierarchy

Trong h thng phn cp tng t ny, 12 ng m thanh c ghp li

thnh mt ng bng thng rng hn to thnh mt nhm (). Mt group c bng

thng 48 KHz v h tr 12 ng m thanh. mc tip theo, ti a 5 nhm c th kt hp vi nhau to nn tn hiu

ghp gi l siu nhm (Supergroup). Mt ng cp siu nhm c bng thng 240

KHz v h tr ti 60 ng m thanh. Mt ng siu nhm c th c kt hp

bi 5 nhm hoc 60 ng m thanh ring bit.

Mc tip theo, 10 siu nhm kt hp vi nhau to thnh mt nhm ch

(Mastergroup). Mt ng nhm ch phi c bng thng l 2,4 MHz. Nhng v

cn phi c di bng thng canh gc nn tng bng thng cn thit cho ng dynhm ch cn ti 2,52 MHz. Mt ng dy nhm ch cn ti 600 knh m thanh.

Mc cui cng, 6 nhm ch c th kt hp to ra mt nhm cc ln

(Jumbogroup). Mt ng dy nhm cc ln cn c bng thng 15,12 MHz nhng

c tng ln 16,984 MHz chn cc bng thng canh gc vo gia cc nhm

ch.

C nhiu s khc nhau trong kin trc ny gia cc hng in thoi khc nhau.

Trong tng lai, h thng tng t s c thay bng h thng s nn ta s hn chvic xem xt h thng tng t y.

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Dch v k thut s:

Gn y cc cng ty in thoi bt u cung cp cc dch v k thut s cho

khch hng. u im ca dch v k thut s l t nhy cm hn dch v tng t

i vi nhiu v cc dng khc nhau ca xuyn nhiu. Mt ng dy in thoi

hot ng nh ngten, n s b nhiu trong khi truyn tn hiu tng t ln tn hius. Trong vic truyn tn hiu tng t, c tn hiu v nhiu u dng tng t nn

khng th tch ra c. Trong truyn tn hiu k thut s, tn hiu dng s cn

nhiu vn dng tng t nn tn hiu c th d dng c phn bit v tch ra d

dng. u im khc ca truyn s l gi thnh r. Do h thng s ch cn phn bit

gia 2 hay 3 mc in th ca tn hiu thay cho mt di tn hiu lin tc nh trong

truyn tng t nn truyn s s tit kin in nng hn so vi truyn tng t.

Chng ta s xem xt cc dch v k thut s khc nhau: switch/56, DDS vDS.

Dch v chuyn mch 56 (switch/56):

Thiu trang 249.

Dch v tn hiu s (DS):

Sau khi cung cp cc dch v switch/56 v DDS, cc cng ty vin thng nhn

thy cn phi pht trin h thng dch v k thut s phn cp nh h thng tngt. V vy bc pht trin tip theo l dch v DS. y chnh l s phn cp ca h

thng tn hiu s. Hnh 8.28 ch ra tc d liu c h tr trong cc mc khc

nhau.

Hnh 8.28 S phn cp tn hiu s- Mt ng DS-0 ging nh DDS. N l mt knh n 64 Kbps.

- DS-1 l dch v 1,544 Mbps, gp 24 ln 64 Kbps cng thm 8 Kbps cho chi

ph ph tri. N c th c dng nh mt dch v n truyn 1,544

Mbps hoc c th dng kt hp 24 ng DS-0 khc nhau.

- DS-2 l dch v 6,312 Mbps, gp 96 ln 64 Kbps cng vi 168 Kps cho chi

ph ph tri. N c th c dng nh ng dch v n hoc c th dng

kt hp 4 ng DS-1 hoc 96 ng DS-0 hoc kt hp cc loi dchv trn.

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- DS-3 l dch v 44,376 Mbps, gp 672 ln 64 Kps cng thm 1,368 Mbps

cho chi ph ph tri. N c th c dng nh ng dch v n hoc kt

hp cc dch v trn.

- DS-4 l dch v 274,176 Mbps gp 4032 ln 64 Kbps cng thm 16,128

Mbps cho chi ph ph tri. N thng c dng kt hp cc ngdch v trn.

Cc ng T:

DS-0, DS-1 l tn dch ca dch v s. trin khai cc dch v ny, cng

ty vin thng phi s dng cc ng T (T-1 n T-4). Dung lng ca cc ng

ny chnh l tc ca cc dch v DS-1 n DS-4.

Service Line Rate(Mbps) Voice ChannelsDS-1 T-1 1,544 24DS-2 T-2 6,312 96DS-3 T-3 44,736 672DS-4 T-4 274,176 6032

Bng 8.1 Tc d liu ca DS v ng T.

T-1 trin khai dch v DS-1 v c nh vy. DS-0 khng c cung cp nh

mt dch v nhng c nh ngha l c s cho cc mc ch ph. Cc hng vinthng tin rng DS-0 s thay th DDS.

Cc ng T truyn tng t:

Cc ng T truyn tn hiu s nh: d liu s, m thanh, ting ni... N

cn c s dng truyn cc tn hiu tng t.

Kh nng s dng ng T truyn tn hiu m ra th h mi ca dch v

in thoi. Trc kia, mt t chc cn 24 ng dy in thoi cn ti 24 cp dy

cp ko t tng i n ni cng ty. Ngy nay c th kt hp 24 ng dy thnhmt ng T-1 v ch phi ko mt ng n tng i. Hnh 8.29 ch ra cch thc

25 ng m thanh c th kt hp thnh mt ng T-1.

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Hnh 8.29 ng T-1 ghp nhiu ng in thoi.

Khung T-1:

Nh cp trn, dch v DS-1 cn 8 Kbps cho chi ph ph tri. hiu

c chi ph ph tri ny c tnh th no, chng ta hy xem xt nh dng ca

mt khung d liu 24 knh m thanh.

Mt khung d liu ca ng T-1 c 193 bits bao gm: 24 khe thi gian - mikhe thi gian c 8 bits v 1 bit thm vo ng b (24x8+1=193).

Hnh 8.30 Cu trc khung d liu ca ng T-1Nu mt khc mi khe thi gian chuyn mt on d liu cho mt knh m

thanh. 24 knh m thanh s c xen k ri trn mt khung d liu. Nu ng T-

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1 c kh nng truyn 8000 khung/s th tng tc bits ca ng truyn l:

8000x193=1,544 Mbps.

Cc ng T nh:

C rt nhiu khch hng thu bao khng s dng ht kh nng truyn ca mt

ng T. m cung cp mt dung lng thch hp cho cc khch hng ny cccng ty vin thng cung cp dch v ng T nh. Dch v ny cho php cc

khch hng thu bao chia s mt ng dy T bng cch ghp cc knh truyn ca

khch hng vi nhau.

V d: Mt cng ty nh cn ng c dung lng bng ng T. Nu c 4

cng ty nh vy trong mt ta nh th h c th chia s nhau mt ng T. lm

nh vy, cc ng truyn ca cc cng ty c a n thit b gi l DSU/CSU

(Digital service unit/Channel service unit). Thit b ny cho php chia ng Tthnh 4 knh ring r.

Hnh 8.31 ng T nh

Cc ng E:

Chu u s dng h thng ng T gi l ng E (E-lines). Hai h thng ny

tng ng nhng khc nhau v dung lng truyn.

LineRate

(Mbps)Voice

ChannelsE-1 2,048 30E-2 8,448 120E-3 34,368 480E-4 139,264 1920

Bng 8.2 Tc ng ECc dch v ghp knh khc:

Chng ta cp n ghp knh trn ng cp vt l. Tuy vy vic ghp

knh cng rt cn thit cho hiu nng ca vic truyn sng v tinh v mt t.

Ngy nay, cc cng ty vin thng cn gii thiu v cung cp cc dch v rt mnh

khc da trn cng ngh ghp knh nh: ISDN, SONET, ATM. Cc dch v ny s

c cp chng 16 n chng 20.

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8.6 Digital subscriber line (DSL)

Mt v d v ghp knh, tch knh v iu ch l mt cng ngh c gi l

h DSL. DSL l mt cng ngh mi n s dng cc mng vin thng hin nay nh

ng in thoi a phng thc hin qu trnh phn phi d liu, m thanh,

hnh nh vi tc cao.DSL l mt h cc cng ngh; y chng ta s ch cp n 5 dng trong

nhng h ny: ADSL, RADSL, HDSL, VDSL v SDSL.

ADSL

Cc cng ty in thoi va lp t mng s din rng tc cao gii quyt

vn thng tin lin lc gia cc u mi trung tm ca n. Tuy nhin ng

truyn gia ngi dng (subscriber) v mng vn l mt ng tng t. Vn

t ra l to ra nhng ng truyn k thut s t ngi thu bao n mng mkhng phi thay i cc ng cc b (Local Loop). ng cc b l cc si cp

xon i vi bng thng tim nng l 1 MHz hoc cao hn.

ADSL l khng i xng, c ngha l n cung cp tc truyn cao hn theo

hng t u mi trung tm ti khu ca ngi dng so vi hng t ngi dng

n u mi trung tm. iu ny chnh l ci m ngi dng mong mun. H

mun nhn cc file c kch thc ln mt cch nhanh chng t Internet , nhng h

thng c nhng file nh, v d nh thng ip e-mail ngn gi.ADSL chia bng thng ca cp xon i thnh ba di. Di th nht thng t

0 25 KHz, c dng cho dch v in thoi thng thng (plain old telephone

service - POTS). Dch v ny ch s dng 4 KHz ca di bng, di bng thng cn

li lm di canh gc tch knh m thanh v knh d liu. Di th hai thng t

25 200 KHz, c dng cho lin lc ngc (t ngi s dng n u mi trung

tm). Di th ba thng t 250 KHz n 1 MHz c dng cho lin lc xui (t

u mi trung tm n ngi s dng). C th thc hin s chng lp gia dixui v di ngc to bng thng rng hn cho chiu xui. Hnh 8.32 ch ra cc

di bng thng:

Hnh 8.32 Di bng thng cho ADSL

Cc k thut iu ch

Phn ln vic thc hin ADSL u s dng mt k thut iu ch c gi l

iu ch bin (hoc pha) khng sng mang (carrierless amplitude/phase

CAP). Sau mt k thut khc c bit n l a sc ri rc (discrete multitone

- DMT) c chun ha bi ANSI.

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CAP l mt k thut iu ch ging nh QUAM nhng vi mt s khc bit

quan trng: tn hiu mang b loi b. K thut ny phc tp hn QUAM v cha

c chun ho.

DMT kt hp QUAM v FDM. Bng thng c sn cho mi hng c chia

thnh cc knh 4 KHz, mi knh c mt tn s sng mang ca n.Hnh 8.33 minh ho khi nim DMT vi Nknh. Cc bt c to ra bi cc

ngun c chuyn qua mt b chuyn i tun t sang song song (serial to

parallel converter), mt khiNbt c chia thnh Nng song song, mi

mt ng gm 1 bt. Cc tn hiu QUAM c to thnh t mi ng c

ghp tn thnh tn hiu kt qu v c a ln ng truyn.

Hnh 8.32 DMTChun ANSI quy nh tc 60 Kbps cho mi knh 4 KHz, iu c ngha

l mt iu ch QUAM c tc 15 bt trn mt baud.

Knh ngc thng chim 25 knh, tc bt s l 25 60 Kbps = 1.5

Mbps. Tuy nhin tc bt theo hng ny thng t 64 KHz n 1 Mbps

( trnh nh hng ca ting n).

Knh xui thng chim 200 knh, tc bt s l 200 60 Kbps = 12

Mbps. Tuy nhin tc bt theo hng ny thng t 500 KHz n 8 Mbps( trnh nh hng ca ting n).

RADSL

RADSL l mt cng ngh da trn ADSL. N cho php cc tc truyn d

liu khc nhau ph thuc vo tng dng lin lc: m thanh, d liu, hnh nh v.v.

Tc khc nhau cng c th c phn b cho ngi dng da trn nhu cu bng

thng ca h. RADSL c li cho ngi dng v gi thnh da trn tc truyn

cn thit.HDSL

HDSL c thit k bi cng ty Bellcore (hin nay l Tellcordia) nh l mt

s thay th cho ng T-1 (1.544 Mbps). ng T-1 dng cch m AMI rt d b

suy gim nhng tn s cao. iu ny gii hn chiu di ca ng T-1 l 1 Km.

i vi nhng khong cch xa hn th cn mt b khuch i, v vy lm tng gi

thnh ng truyn.

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HDSL s dng m 2B1Q t b suy gim hn. Mt tc gn 2Mbps c th t

c m khng cn mt b khuych i vi khong cch 3.6 Km. HDSL dng hai

dy xon i thc hin truyn song cng.

SDSL

SDSL cng ging nh HDSL nhng s dng mt cp xon i n tc cng tc nh HDSL. Mt k thut c gi l echo cancelation c s

dng to ra t truyn song cng.

VDSL

VDSL mt cch tip cn tng t nh ADSL, n s dng cp ng trc, cp

xon i, cp quang cho khong cch ngn (300 1800 m). K thut iu ch l

DMT vi tc bt t 50 55 Mbps cho chiu xui v t 1.5 n 2.5 Mbps cho

chiu ngc.8.7. FTTC

u im ca cng ngh cp quang l chng nhiu v dung lng bng thng

cao. Nh so vi cc cng ngh khc th gi thnh ca n kh cao. Cc cng ty in

thoi v truyn hnh cp a ra cng ngh mi gi l FTTC, tc l: thu ng cp

quang trong khi vn gi gi thnh h. ng cp quang trin khai t bu in

trung tm hay t cng ty cp quang n tuyn hnh lang. T tuyn hnh lang n

ni hp ng thu bao s dng cp xon hoc cp ng trc.FTTC trong mng in thoi:

ng cp quang kt ni v ghp nhiu knh m thanh khc nhau. Cp

xon ng t ni hp ng thu bao c ghp li vi nhau v chuyn sang dng

tn hiu quang. Ti trm chuyn mch, tn hiu quang c ghp knh sd dng

WDM to tn hiu quang bng thng rng.

Hnh 8.35 FFTC trong mng in thoiFTTC trong mng truyn hnh cp:

V cu trc cng ging nh h thng FTTC trong mng in thoi. Nhng

dy t ni hp ng thu bao n tuyn hnh lang s dng cp ng trc.

Hnh 8.35 FFTC trong mng in thoi

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CHNG 9 PHT HIN V SA SAI

Cc mng phi c kh nng truyn d liu mt cch chnh xc t

thit b ny ti thit b khc. Mt h thng khng bo m c s ton

vn d liu trong qu trnh truyn s t c s dng. Tuy nhin, bt klc no d liu c truyn t ni ny n ni khc, chng u c th b

thay i. Trn thc t, hu nh thng ip b sai lch mt s phn ca

n trong qu trnh truyn. Nhiu yu t, bao gm c nhiu ng truyn

thay i hoc xo 1 hay nhiu bt ca d liu. Cc h thng tin cy phi

c mt cch pht hin v sa nhng li nh vy.

D liu c th b thay i trong qu trnh truyn. H thng truyn

thng tin cy phi pht hin v sa c li.

Pht hin v sa sai c tin hnh c tng lin kt d liu v

tng giao vn.

9.1 Cc kiu li.

Mi khi tn hiu in t truyn tim ny ti im khc, n b tc

ng bi nhiu t cc ngun nhit, t v in. Nhiu ny c th thay i

hnh dng v thi gian ca tn hiu. Nu tn hiu ang mang d liu nh

phn c m ho, nhng thay i nh vy c th bin i nghaca d liu. Trong li bt n, mt bt 0 b thay i thnh mt bt 1 hoc

mt bt 1 b thay i thnh mt bt 0. Trong li chm, nhiu bt b thay

i. V d mt nhiu xung c di 0,01 giy tc ng vo tn hiu

truyn vi tc 1200 bps s lm thay i tt c hoc mt s trong

12 bt thng tin.

Li bt n

Thut ng li bt n c ngha l ch mt bt ca n vd liu cho (chng hn nh byte, k

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Ebook "Truyền số liệu" - ĐH Phương Đông