bÀi tẬp dÀi truyen dan

Bi tp di k thut truyn dn (nhm 5)

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BI TP DIMN HC: K THUT TRUYN DNB MN: IN T VIN THNGSinh vin.............................................................................. Lp Ngnh: in t vin thng Gio vin hng dn: ThS. o Huy Du Ngy giao : ............................................ Ngy hon thnh: Tn ti: Thit k tuyn truyn dnT cc knh khc

(1)

Bin i A /D

X l t /h bng gc

Ghp knh

iu ch

(2)

(3)

(4)

(5)Knh truyn

(6)

Bin i D /A

(11) (10)

Khi phc t /h gc s

Gii ghp knh

Gii iu ch

Yu cu: 1. Tn hiu analog u vo c ph tn nm trong khong 100Hz n 8000Hz. Di bin tn hiu 0 mV n 40mV 2. Bin i A/D: s dng phng php bin i PCM. Vi 3 mu u vo lin tip 12mV ; 20mV; 37mV. 3. X l tn hiu bng gc: S dng phng php m ha HDB3 vi tn hiu u vo l kt qu ca m ha mu 1,2,3.

(9)

(8)

(7)

4. Ghp knh s dng phng php ghp knh ng b SDH. Trnh byphng php ghp lung 140M ln STM1. Hot ng con tr AU4-PTR. 5. iu ch s dng phng php iu ch QPSK. Vi lung bit u vo l kt qu ca m ha mu 1,2,3. GIO VIN HNG DN (K v ghi r h tn)

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Nhn xt ca gio vin

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

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Mc lc

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Phn I phn tch ti thit k tuyn truyn dnT cc knh khc

(1)

Bin i A /D

X l t /h bng gc

Ghp knh

iu ch

(2)

(3)

(4)

(5)Knh truyn

(6)

Bin i D /A

(11) (10)

Khi phc t /h gc s

Gii ghp knh

Gii iu ch

Trn s khi h thng thng tin s bao gm mt lot cc khi x l d liu c bn,cc khi ny thc hin cc chc nng sau y: Khi bin i A/D,D/A:Bin i tn hiu tng t thnh tn hiu s thuc bn pht. N chuyn i tn hiu tng t lin tc thnh chui cc t m biu din bng cc xung in p nh phn; v thnh phn chuyn t s sang tng t (DAC) trong b thu, n chuyn i cc xung in p tr thnh tn hiu tng t lin tc Vic m ha c th c thc hin bi mt trong cc phng php nh sau: iu xung m (PCM); iu xung m Logarit ( log(PCM)); iu xung m vi sai (DPCM); iu xung m vi sai t thch nghi (ADPCM); iu ch Delta (DM); iu ch delta t thch nghi (ADM) hoc m d on tuyn tnh (LPC). Thng thng phng php c s dng ph bin l m ho PCM. ADC gm c mt mch ly mu, lng t v m ho PCM. Mch ly mu to ra cc mu in p ri rc t cc qung thi gian u nhau ca tn hiu tng t. Mch lng t lm xp x cc in p ny nh mc gn nht ca tp hp cc mc in p ( l qu trnh lng t chuyn i tn hiu tng t thnh tn hiu s). Mch m ho PCM chuyn i tng mc lng t thnh m t m nh phn vi tng s 1 v 0 biu din bng mt trong hai mc in p. Mch lc chng tp gim thiu s mo c th xy ra do kt qu ca qu trnh ly mu.

(9)

(8)

(7)

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Bi tp di k thut truyn dn (nhm 5) Trong mch DAC ca khi thu, cc xung in p nh phn c chuyn thnh cc mc in p lng t nh mch gii m PCM v sau c lm bng phng nh mch lc thng thp ti to (t nht l xp x) li thnh tn hiu tng t gc. S s ho tn hiu thng lun tng rng bng tn truyn dn tn hiu nhng n cho php nhn c t l tn hiu/tp thp hn nhng trng hp khc. y l mt v d mt ngun ( rng bng) c th c i ly ti nguyn khc nh th no (cng sut pht). Cc thit b CODEC lm cho cc k thut x l tn hiu s phc tp c s dng rng ri m ho hiu qu tn hiu trc khi truyn i v cng gii m cc tn hiu nhn khi chng b hng do tp mo v nhiu. iu ny lm tng phc tp ca thit b thu pht nhng cho php truyn dn hu nh khng c li, c chnh xc v kh nng lp cao.

Khi x l bng gc : X l tn hiu gc truyn trn knh thng tin,Xl tn hiu bng gc l yu cu ch yu i vi cc knh thng tin truyn trn cp ng trc hoc cp i xng, nhng n cng cn thit i vi cc phng php truyn dn iu ch cao tn, ni m tn hiu cng c a xung bng gc ti cc trm lp. Trong mt h thng thng tin s, thit b lp cn c b lc, cn bng v ti sinh. Tuy nhin truyn dn chng, cn phi bin i cc tn hiu nh phn t thit b ghp knh thnh cc m ng gim li cho knh truyn dn. Bao gm:M ho ngun, m ho bo mt v m iu khin li. Ngoi vic m ho v gii m PCM, CODEC c th c 3 chc nng b sung. u tin, chc nng m ho ngun (bn pht) lm gim s bit nh phn d tha nh vo qu trnh tnh ton cc xc sut xut hin ca tin tc. Tip n l m ho cc tn hiu bit bng cch s dng m bo mt nhm bo v thng tin c nhn ca khch hng. M mt ny c th thuc dng b mt hoc cng khai

Phn cui cng l m ho iu khin li (cn gi l m ho knh) gim st li (xc nh li v sa).C mt cch m ho iu khin li, m n b sung bit d tha vo cc lung bit, ngc vi qu trnh m ho ngun (loi b d tha). Tuy nhin c hai qu trnh ny c th c s dng trong cng mt h thng, bi v kiu d tha, m n xut hin t nhin trong thng tin truyn i khng nht thit ging vi kiu dng cho xc nh v hiu chnh li bn thu. B ghp knh:Tp hp cc tn hiu bng gc s v phn chia tn hiu s t tn hiu bng gc s c thc hin nh b ghp knh v b tch knh.

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Bi tp di k thut truyn dn (nhm 5) Hin nay mng vin thng tn ti nhiu h thng ghp tch theo tn s (FDM) hoc theo thi gian (TDM). i vi thit b FDM, cc tp hp nhm, siu nhm, ch nhm hoc 16 siu nhm FDM ca cc knh m tn thng cn thit phi giao tip vi cc h thng truyn dn s. iu ny c thc hin nh s dng cc Codec c bit. i vi h thng mi y , vic tp hp v phn chia tn hiu s thng c thc hin nh s dng cc s ghp-tch theo thi gian TDM . Trong h thng thng tin s, ghp knh cung cp cc truyn dn ng thi, thng l ghp knh theo thi gian (TDM). Cc b ghp knh TDM chn hoc mt s t m PCM hoc cc s nh phn PCM cho php cc tuyn thng tin c truyn i trn cng mt ng truyn vt l (nh cp, si quang, knh tn s v tuyn). Nu vic thng tin theo thi gian thc th iu ny c ngha l: Tc bit ca tn hiu c ghp knh gp t nht N ln so vi tng tn hiu PCM ring bit v nh vy yu cu di thng tng ln Tch knh l chia dng bit thu c tr li thnh cc tn hiu PCM thnh phn. iu ch v gii iu ch: Cc Modem (b iu ch v gii iu ch) tc ng ln cc dng xung nh phn thng tin n mang c th truyn qua mt thit b vt l no , mt tc no , vi mt mo c th chp nhn c, trong mt di tn xc nh hay c phn b. B iu ch bn pht c th thay i cc mc in p ring l, cc bit. c bit b iu ch cng sa dng xung tn hiu hay lc gii hn rng bng thng v cn thay i ph hp vi bng tn cho php. V vy, u vo ca b iu ch l tn hiu s di gc trong khi u ra thng l dng sng thng di. B gii iu ch bn thu chuyn i t dng sng thng di thu c thnh tn hiu di gc. B cn bng hiu chnh li mo tn hiu m n c th xut hin trong khi truyn. Phn tch sng chuyn i tn hiu di gc iu ch tr thnh dng k hiu nh phn. Trong b tch sng c b lc thch ng, y l mt kiu x l tn hiu trc khi tn hiu a ti x l mch quyt nh vi mc tiu l ci thin c tnh li. Mt s phng php iu ch v gii iu ch hin ang c s dng c phn loi theo kiu iu ch ch yu. Trong chng 6 s xem xt mt vi loi sau y: iu bin: iu bin xung (PAM) ; iu bin xung M mc (PAM M mc); kho ng m (OOD) tch kt hp; kho ng m tch ng bao; iu bin cu phng M trng thi (QAM M trng thi). iu tn: Kho dch pha tn s-tch khng kt hp (FSK tch kt hp); Pa lien tc-kho dch tn s-tch kt hp (CP-FSK-CD); pha lin tc-kho dch tn s-tch khng kt hp (CP-FSK-NCD); kho dch cc tiu (MSK).

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Bi tp di k thut truyn dn (nhm 5) iu pha: Kho dch pha nh phn (BPSK)-tch kt hp; kho dch pha nh phan-m ho vi sai (DE-BPSK); kho dch pha vi sai (DPSK); kho dch pha cu phng (QPSK); kho dch pha M trng thi (M-PSK).

PHN II THIT K H THNG TRUYN DN2.Tn hiu analog u vo c ph tn nm trong khong 100Hz n 8000Hz. Di bin tn hiu 0 mV n 40Mv. 2.1:bin i A/D nguyn l bin i A/D : s dng phng php bin i PCM. Vi 3 mu u vo lin tip 12mV ; 20mV; 37mV

Cu hnh c bn ca h thng truyn tin PCM

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Bi tp di k thut truyn dn (nhm 5) nh ngha PCM: PCM l qu trnh chuyn i c bn mt tn hiu tng t sang tn hiu s m thng tin cha ng trong cc mu tn hiu tng t lin tc c thay th bng cc bit s ni tip. Tn hiu tng t u tin c ly mu tn s ln hn tn s Nyquyst sau c lng t ho. - PCM chnh quy: Cc bc lng t cn bng. - PCM khng chnh quy: Cc bc lng t khng bng nhau. Nguyn tc m ha PCM nh sau : PCM c thc hin theo 1 qui trnh 4 bc c nguyn tc nh sau: Bc 1: Lc nhm hn ch ph tn ca tn hiu lin tc cn truyn. Bc 2: Ly mu tn hiu thoi l khai trin c chu k tn hiu thu c bin c gi tr tc thi (tc l ly gi tr ca bin ti thi im nht nh). Bc 3: Lng t ho bin bng cch chia bin ca xung ly mu thnh cc mc v ly trn bin xung n mc gn nht. Bc 4: M ho l thay th 1 xung c lng t thnh mt dy bit nh phn gi l t m.

Tn hiu nguyn thu7 /2 /2Ga tr mu Xp x

5

3 /2

/2 - /2-3 /2

-5 /2- 7 /2

V d m ha 3bit/mu

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Bi tp di k thut truyn dn (nhm 5) Thc hin ly mu: Tn hiu lin tc sau khi lc, c ri rc ho nh ly mu bng chui xung nhpc tn s f (theo nh l ly mu) c c cc tn hiu PAM. Minh ho theo hnh:

Qu trnh ly mu cc tn hiu tng t nh l ly mu Nyquist: tn hiu c bng tn hn ch c th c c trng chnh xc bi cc tr s ly mu, khong cch gia cc tr s ny khng c vt qu 1 na chu k ca tn s cao nht ca tn hiu: : flm 2fmax (Ts Tmax/2). fmax : tn s ln nht ca tn hiu. . Ts: gi l khong cch Nyquist v l khong cch thi gian di nht c dng ly mu tn hiu c bng tn thp b hn ch v vn cho php khi phc li tn hiu m khng b mo. Vi 3 tn hiu bi ra c fmax=8000hz suy ra tn s ly mu s l 16000hz Qu trnh thc hin ly mu c thc hin theo gin thi gian bn:

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u(t)a: t/h tng t

G(t ) Ux(t ) Um(t ) Gm(t ) f1 f2 fm-f2 fm-f1 fm fm+f1 fm+f2 2fmQu trnh ly mu tn hiu tng t

tb: ph tn t/h c: dy xung iu khin ly

t

d: t/h u ra b ly mu

mu t

t e: ph tn ca t/h lylsaaavamamac cmuxahahxkhxk ahxmymu t

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T tch in

+Ngun tng t

Tn hiu PAM

CKhuych i tr khng cao To xung Mch ly mu v duy tr

Lng t ha: Ni dung ca lng t ho l: - Bin ca tn hiu c chia thnh cc khong u hoc khng u. - Mi khong l 1 bc lng t bin tn hiu ng vi u hoc cui mi bc lng t gi l mt mc lng t. - Sau khi c cc mc lng t th bin cc xung ly mu c lm trn n mc lng t gn nht. - C hai loi lng t ho bin ; lng t ho u v lng t ho khng u.T 5 / 2 /2 /2 5 / 2 T

5 / 2

/2 /2 5 /2

Lng t ho

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Bi tp di k thut truyn dn (nhm 5) Tn hiu ri rc theo thi gian tn hiu s PCM Chn m ha 3 tn hiu trn theo phng php lng t ha u vi s bt l 8 nn S cQ= 2^8=256 mc lng t : Bin theo bi ra l:0 mV n 40mV40 =0.15625 mV 256 12 M ha tn hiu th nht:Q1= =77 mc 0.15625

Bc lng t: =

Ta c chui bit nh phn ca tn hiu th nht l:01001101 M ha tn hiu th 2:Q2= M ha tn hiu th 3:Q3=20 =128 mc 0.15625 37 =237 mc 0.15625

Ta c chui bit ca tn hiu th 2 l:10000000 Ta c chui bit ca tn hiu th 3 l:11101101 Vy ta c chui bit ca 3 tn hiu trn lin tip sau khi m ha PCM l :010011011000000011101101 2.2 x l tn hiu bng gc :S dng phng php m ha HDB3 vi tn hiu u vo l kt qu ca m ha mu 1,2,3.X l tn hiu gc truyn trn knh thng tin,X l tn hiu bng gc l yu cu ch yu i vi cc knh thng tin truyn trn cp ng trc hoc cp i xng, nhng n cng cn thit i vi cc phng php truyn dn iu ch cao tn, ni m tn hiu cng c a xung bng gc ti cc trm lp. Trong mt h thng thng tin s, thit b lp cn c b lc, cn bng v ti sinh. Tuy nhin truyn dn chng, cn phi bin i cc tn hiu nh phn t thit b ghp knh thnh cc m ng gim li cho knh truyn dn.

Phng php m ha HDB3 u im ca m HDB3 : HDB3 s dng rng ri trong thc t cho cc giao tip ghp knh bc cao. Loi m ho ny quan trng khi truyn s liu c tc cao trn i dy kim loi c di tn hn ch. Vic gim rng bng tn cn thit ca knh

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Bi tp di k thut truyn dn (nhm 5) hoc tng tc bit vi mt rng bng tn cho s lm tng t s tn hiu trn tp m t c xc sut li cho trc . Bo mt tin tc nhng ni yu cu tnh an ton cao c bit trn cc tuyn truyn s liu ca cc h thng my tnh x l s liu. To ph tn hiu nhm ng dng cho nhng mc ch c bit nh ng b gim thnh phn bin tn s khng n 0, hoc gim cc thnh phn cao v thp ca tn s trc lc lc.C th a nhng s " 0" c bit vo cc lung s c m lng cc bc cao b chn v cc lung s b chn. Trong qu trnh m ho PCM nh trnh by trn, tt c cc bit thng tin c ngm gi thit l nh phn n cc. Gi thit ny l hp l min l cc bit thng tin c xc nh trong mt cng on nht nh no ca thit b x l v dy ni khng c daig qu vi mt. Vi nhng ng dy ni tng i di,ng cp i xon bc kim, hoc cp ng trc th khng nn s dng lung bit nh phn,l do l : 1. N c cha thnh phn mt chiu. 2. N c cc thnh phn tn s thp mc cao. 3. Khi truyn mt dy "0" lin tip th khng c im chuyn tn hiu. Nguyn tc m ha : - M HDB3 l m nh phn mt cao c nhiu nht 3 s " 0" lin tip. m mt tn hiu nh phn thnh mt tn hiu HDB3 theo cc quy tc sau y: Quy c : - S 0 nh phn c m bng mt trng thi trng vi s bt 0 lin tip khng vt qu 3 bt. - Mt bt 1 nh phn c m bng k hiu B theo lut lng cc. - Cc dy 4 s 0 lin tip trong tn hiu nh phn c m nh sau : a) Bt 0 u tin ca dy c m bng nu du B v du V trc ca dy m tam phn c cc tnh ngc vi nhau b) Bt 0 u tin ca dy c m bng B nu du B v du V trc ca dy m tam phn cng cc tnh vi nhau.

c) Cc bt 0 th 2 v th 3 c m thnh . d) Bt 0 th 4 bin i thnh Ve) Da trn bipolar-AMI Chui 4 s 0 lin tip c thay th theo quy lut nh sau *Nu trc l l bit 1 + 0 0 0 0

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-

0

0

0

0c m ha thnh

c m ha thnh + 0 0 0 + 0 0 0 * Nu trc l chn bit 1 + 0 0 0 0 0 0 0 0 c m ha thnh + + 0 0 0 0 +

c m ha thnh

Thc hin m ha HDB3:chui bit ca 3 tn hiu m

ha :010011011000000011101101

M NH PHN 010011011000000011101101 HDB3

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2.3 GHP KNH :s dng phng php ghp knh ng b SDH. Trnh by phng php ghp lung 140M ln STM1. Hot ng con tr AU4-PTR.2.3.1 Khi nim v SDH :phn cp s ng b l giai on tip theo ca giai on phn cp truyn dn ng b. SDH thc hin mt h cc tiu chun (hay giao thc). N c th v s giao tip vi cc mng hin c c xy dng trn cc tiu chun vng. SDH c th v s tr thnh mt tiu chun cng nghip cho thng tin si quang bng rng ton cu. Mng thng tin ny c th c b sung bi cc h thng cp ng v/hoc v tuyn cho cc ng dng ng truyn c ly ngn v/hoc tc thp. Khi mng s ca quc gia c trin khai mt phn thng qua cc chun SDH th cc vng SDH ny c th c kt ni ti nhng vng c trin khai vi cc tiu chun phi SDH (thng c ni ti l cc chun cn ng b PDH). Mt s nt c bit ca SDH gm: Cc tiu chun SDH STM-1 :155.52 Mb/s STM-4 :622.080 Mb/s STM-8 :1214,16Mb/s STM-12 :1866,24 Mb/s STM-16 :2488,32Mb/s STM-64 :9953.28Mb/s

Cung cp cc phn t mng duy nht cc mng PDH c th ckt ni ti hay giao tip vi cc mng SDH Cung cp cc kh nang tch hp OAM&P trong mi phn t mng NE Cc mng SDH SDH cung cp cc knh thng tin s liu truyn cc bn tin OAM&P gia cc Phn t mng NE hay gia cc H thng iu hnh (OS) vi NE trong cng mt khung tn hiu mang ti tin. iu ny tit kim c thit b v to ra s linh hot cho qu trnh vn hnh 2.3.2 Ghp t lung 140 ln STM-1

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1STM-1 AUG

1

AU-4

VC4

C4=139.264Mb/s

Trong : o C4 :khi container vai tr tip nhn cc tn hiu PDH,bao gm :cc lung d liu PDH,cc bit /byte chn c nh,cc bit/byte chn ng,khung tn hiu C4 gm 9 hng 260 ct o VC-4 :con ten n o c vai tr nhn khung tn hiu t C4 ln ,b xung thm cc byte POH qun l thng tin t ni gi n ni nhn VC ,ta c VC4=C4+9 bytePOH L loi VC bc cao Cc byte poh ca C4 k hiu l :VC4-POH

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o

AU-4 :n v qun l : AU4=VC4+pointer(9byte) N bao gm mt conteno o v mt con tr n v qun l

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Bi tp di k thut truyn dn (nhm 5) o AUG :nhm n v qun l :l mt hoc nhiu n v qun l ghp vi nhau theo phng thc xen byte to thnh mt nhm n v qun l AUG. o STM-N :m un chuyn ti ng b 2.3.3 xp xp cc lung nhnh 140Mb/s vo khung STM-1

1BYTE

1B

12BYTE

POH

W

96 I

Y

96 I

Y

96 I

Y

96 I

X

96 I

X Y Y

96I 96I 96I

Y Y Y

96I 96I 96I

Y Y X

96I 96I 96I

Y X Y

96I 96I 96I

X Y Z

96I 96I 96I

sp xp 140Mb/s vo VC-4 : - mi hng VC-4 c 260 byte 130 bit n (R)+10 bit mo u (o)+5 bit iu khin chn +1 bit chn S +(241byte+6bit )tin - byte I (lung 14o Mb/s)cung cp cho khung VC4 trong thi hn 125 s :2176 B. - byte I xp xp c nh trong khung VC-A 2175 byte+6bit - ghp khung VC-4 vo STM-1

RSOH 1- 19 -

Bi tp di k thut truyn dn (nhm 5) VC4 cng thm byte con tr to thnh AU-4 STM1=AU4+SOH

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