tong quan ve thong tin viba 879

7/31/2019 Tong Quan Ve Thong Tin Viba 879

1/66

Tng quan v thng tin Viba s

1


2/66

CHNG 1TNG QUAN V VI BA S.

1.1. Gii thiu chung.H thng truyn dn l 1 h thng bao gm cc thit b, phng tn dng truyn tn hiu

t ni ny n ni khc.

Ngy nay, theo phng tin truyn dn, cc HTTT bao gm cc loi h thng ch yu sau:+ HTTT dng cp ng trc, trong mi trng truyn dn l cp ng trc (coaxial cable)Cc h thng s dng cp ng trc c dung lng khng cao, c ly khong lp ngn v kh

nng c ng km. Cc h thng loi ny ang dn c thay th v c s dng ch trongnhng tnh hung c th nht nh.

+ HTTT sng cc ngn (microwave) vi mi trng truyn dn v tuyn trn gii sng ccngn, bao gm cc loi h thng thng tin v tinh, thng tin v tuyn tip sc (radio-relay) vthng tin di ng;

Cc h thng thng tin v tinh c dung lng trung bnh song b li c c ly lin lc t lnn rt ln. Cc h thng ny c s dng lm trc xuyn lc a hoc phc v cho cc tuynkh trin khai cc loi hnh lin lc khc (nh tuyn lin lc t lin-hi o, t lin-cc gin

khoan du, t lin-cc tu vin dng...). Ngoi ra, cc h thng v tinh a tnh cn c sdng cho cc h thng pht qung b truyn hnh. Trong tng lai gn, khi h thng cc v tinhqu o thp v trung bnh c trin khai, cc h thng v tinh c th c s dng cho cthng tin di ng ph sng ton cu.

Cc h thng thng tin di ng phc v cc u cui di ng, ni chung c dung lng thp.Kh nng di ng l u th ln nht ca cc h thng ny.

Cc h thng v tuyn tip sc mt t (terrestrial radio-relay) c dung lng t thp ticao, c kh nng thay th tt cc tuyn cp ng trc trong cc mng ni ht ln ng trc. Vithi gian trin khai tng i thp, tnh c ng ca cc h thng v tuyn tip sc mt t hnhn mt s loi h thng khc. Mt u im na ca cc h thng ny l rt d trin khai, ngayc trong cc iu kin a hnh gy nhiu tr ngi cho vic trin khai cc loi h thng dunglng cao khc nh trong cc th, hoc qua cc vng c a hnh rng ni vi c ly chnglin lc ln n 70 km, trung bnh l t 40 dn 45 km.

+ HTTT quang si (fiber-optic) vi mi trng truyn dn l cp si quang hc (gi tt l cpquang).

H thng cp quang c dung lng ln nht, gi r (theo chi ph tnh trn knh thoi) do thng c s dng lm ng trc quc gia, xuyn quc gia, xuyn lc a. Nhc im c

bn ca HTTT cp quang l kh nng c ng h thng km, chi ph lp t ban u kh cao, vvy trong mt s trng hp c th th vic trin khai c xem l rt kh khn.

+ HTTT v tuyn lm vic trong gii tn s t 60MHz tr xung.

1.2. H thng thng tin vi baT ting Anh microwave c ngha l sng cc ngn hay vi ba theo cch dch qua ting TrungQuc. T vi ba c s dng chung cho cc h thng v tinh, di ng hay v tuyn tip sc mtt, song nc ta t vi ba c s dng t trc ch ch cc h thng v tuyn tip sc.Do , hin nay trong cc ti liu k thut ca ta, ni vi ba l ni ti h thng v tuyn tip scmt t.

Thng tin sng cc ngn gia hai im bt u xut hin vo nhng nm 30 ca th k trctuy nhin lc by gi do kh khn v mt k thut nn ch lm vic di sng mt do vy uim ca thng tin siu cao tn cha c pht huy. Nm 1935 ng thng tin VTTS u tinc thnh lp Newyooc v Philadenphi chuyn tip qua 6 a im v chuyn c 5 knhthoi. V TTVTTS bng n sau chin tranh th gii ln th hai. H thng vi ba s bt u hnh

thnh t u nhng nm 50 v pht trin mnh m cng vi s pht trin ca k thut vinthng.

2


3/66

Trm chuyn tip Trm chnh Trm u cui

chng (hop)

M hnh ca mt h thng thng tin vi ba

1.2.1. Gii tn s ca cc h thng vi ba:Tng quan v phn chia cc bng tn

Bng tn K hiu c tnh lan truyn Phm vi ng dng3-30KHz(Chc km)

Tn s rt thp (VLF) - Sng mt t- Lan truyn c ly xa- Mc tp nhiu khquyn ln

ng dng nhiu chothng tin di nc(solar)

30-300KHz

(Km)

Tn s thp (LF) - Tng t nh VLF

nhng b hp th vo banngy

V tuyn hng hi

300-3000KHz(Trm mt)

Tn s trung bnh(MF)

- Sng mt t v sngtri ban m- Suy hao thp ban mcao vo ban ngy- Tp kh quyn

V tuyn v nh vhng hi, cc tn s chocu h v v tuynqung b AM

3-30MHz(Chc mt)

Tn s cao(HF)

Phn x tng in ly thayi theo thi gian trongngy, ma v tn s

V tuyn nghip d;pht sng quc t; thngtin qun s, hng khng

ng di.30-300MHz(mt)

Tn s rt cao (VHF) Lan truyn theo tn nhnthng (LoS)

Truyn hnh VHF, phtthanh FM, thng tin ohng AM, thng tin vi

ba0,3-3GHz(dm)

Tn s cc cao(ultrahigh frequency-UHF)

Lan truyn theo tm nhnthng

Truyn hnh UHF,radar, thng tin vi ba

3-30GHz(cm)

Tn s siu cao(superhigh frequency-SHF)

Lan truyn Los, suy haonhanh theo lng ma,suy hao kh quyn do xi

v hi nc, hp th hinc cao 22GHz.

Thng tin v tinh, thngtin vi ba.

30-300GHz(mm)

Tn s siu siu cao(Extremely HighFrequency EHF)

LoS, hp th hi nc ti183GHz v hp th xiti 60 v 119GHz

Rada, v tinh thnghim.

103-107GHz Hng ngoi, nh sngnhn thy v tia cctm

LoS Thng tin quang

Thng tin siu cao tn lm vic di sng cc ngn dng truyn tn hiu c di tn rng.V l thuyt, gii sng dng cho cc h thng vi ba l t 60MHz cho ti 60/80GHz. Trong thc

t, i vi cc h thng vi ba dng thng phm thng lm vic trn gii sng t 60MHzn 20 GHz, cc h thng cng tc vi gii tn s cao hn (60 80 GHz) hin vn ang cntrong giai on th nghim.

3


4/66

H thng thng tin siu cao tn lm vic di tn s:

60 MHz 80 GHz do c di tn lm vic rt rng v cao so vi thng tin cao tn v vy c

s dng lm phng tin truyn dn chnh trong vin thng cng cng siu ngn

siu cao

Cf

= siu cao,siu ngn

Gi l Viba

Trong h thng thng tin siu cao tn bao gm:

+ H thng thng tin vi ba l h thng thng tin siu cao tn cc trm chuyn tip c

t trn mt t.

+ H thng thng tin v tinh l h thng thng tin siu cao tn cc trm chuyn tip c

t trn v tinh nm ngoi qu t.

+ Thng tin di ng l giao tip vin thng gia my di ng MS vi trm thu pht gc

BTS cng lm vic di sng siu cao tn.

1.2.2. c im+ Do lm vic di sng siu cao tn nn m bo c vic truyn nhng tn hiu di

rng.

+ rng di tn siu cao khong 30GHz do nhiu i c th lm vic ng thi.

+ Hu nh khng b can nhiu kh quyn v cng nghip

+ Trong di sng SCT d dng to ra cc h thng anten c tnh nh hng cao, bp

sng hp nh vy my pht c th gim cng sut v trn cng mt phm vi ta c th trin khai

nhiu h thng cng lm vic m khng gy nhiu ln nhau.

+ Trin khai nhanh v gi thnh r hn so vi cc h thng thng tin dng cp (cp

quang hoc cp ng trc) v vic trin khai h thng cp l rt tn km v trong khu vc ng

dn c c nhiu cng trnh xy dng th vic trin khai mt h thng cp l rt kh khn.

+ D dng qun l v h thng vi ba ch gii hn qun l trong phm vi ca trm v

tuyn dc theo trc (trong khi h thng cp phi qun l ton b tuyn cp v cbit phi

i u vi nguy c t cp).

+ Di sng SCT c nhc im l ch truyn c chc chn trong tm nhn thng c ly

khng qu 50 km. V vy khi mun thng tin i xa cn thc hin chuyn tip nhiu ln.

+ C tc nh hn nhiu so vi h thng cp quang v hin nay ng trc ch cn

s dng nhng khu vc cha ko c cp quang do a hnh phc tp.

+ Chu tc ng ca ng truyn: hp th do hi nc v xi, suy hao do ma v hin

tng pha inh c bit i vi cc h thng bng rng phi chu tc ng ca pha inh a

ng chn lc theo tn s.

1.3. Phn loi- C nhiu phng php, cn c phn loi:

+ Theo dung lng (tc bt tng cng B u vo) cc h thng vi ba s c phnthnh:

4


5/66

+ Cc h thng dung lng thp: B100 Mb/s.+ Theo c ly lin lc (haul)+ Tuyn di (c ly lin lc ln hon 400km): thng l nhng ng trc c dung lng

ln so snh c vi cp quang. Di tn c s dng rng ri t 4 n 6 GHz.+ Tuyn ngn (c ly lin lc di 400km): dung lng thp, thng thng 1DS1, 4DS1,

1E1, 4E1 dng ni cc trung tm chuyn mch di ng. Di tn thng s dng khong 15GHz v di tn ny cho php thu gn kch thc ca an ten v thit b. Do chng ngn nnkhng cn phn tp chng li hin tng pha ing. Nguyn nhn gy gin on lin lc chyu gy do ma nn cn c h s khuych i ln v chng ngn. Vi chng ln hn thng sdng di tn L6GHz, U6GHz hoc 11GHz dung lng thp. Di tn ny khng chu nh hng

pha ing do ma nn c th bo m c ly lin lc xa hn.- Cn c phn loi theo mc ch s dng:

+ H thng vin thng ring ( ni b ) l mng thng tin phc v cho thng tin ring, ni

b ca cc c quan, n v khng dng kinh doanh.

+ H thng vin thng cng cng: l h thng ginh cho mi i tng s dng v sdng kinh doanh, yu cu cht lng cao, thun tin, n gin, d s dng.

- Cn c phn loi theo quy m ca mng:

+ Mng ni ht: phc v cho 1 khu vc a l, khu vc dn c ( tnh, thnh ph)

+ Mng quc gia ( mng lin tnh) phc v thng tin gia cc vng, cc khu vc, cc tnh

thnh.

+ Mng quc t: dng phc v thng tin gia cc nc.

- Cn c phn loi theo a l:

+ Mng vin thng nng thn: mt tha, khng tp trung.+ Mng vin thng thnh ph: mt dy c, tp trung nhiu.

- Cn c phn loi theo phng php x l truyn dn tn hiu:

+ H thng vin thng tng t

+ H thng vin thng s.

Ch yu i vo vin thng cng cng, vin thng s.

1.4. Cc ch tiu cht lng c bn ca h thng vi ba:

i vi cc h thng thng tin s hin ti, cc tn hiu s l cc tn hiu nhn gi tr trong tphu hn cc gi tr c th c v c thi gian tn ti hu hn. Khi tp cc gi tr c th c ca tnhiu gm hai phn t 0 v 1 th h thng c gi l nh phn v tn hiu khi c gi l bt.Gi gi tr ca bt th klDk v thi gian tn ti ca n l Tk (Tk =Tv l hng s vi mi k).

u thu tn hiu khi phc li l kD

v c rng l T

k, nu D

k Dk th tn hiu th k c gi

l b li, nu T

k Ttn hiu th k c gi l c jitter. Cng nh cc h thng truyn dn s khc,ch tiu cht lng c bn ca h thng vi ba s l xc sut bt li v jitter (rung pha hay cn c gil trt trong mt s ti liu). Xc sut li bt BER (Bit-Error Ratio) c nh ngha l:

BER=P{D

k Dk}, vi P{.} l xc sut (1.1)

Khi T

k= T+Tth T. c gi ljitter (1.2)Tu tng loi dch v m cc h thng c cc i hi khc nhau v BER vjitter. i vi cc

h thng truyn thoi, yu cu BER


6/66

kh cao. i vi tn hiu truyn hnh, nu s dng iu ch xung m (PCM) thng th BER ihi cng nh i vi thoi song cn lu l tc truyn vi truyn hnh l kh cao. Khi sdng ADPCM (Adaptive Differential Pulse Coded Modulation: iu ch xung m vi sai t thchnghi) truyn hnh th yu cu BER 10-6

Tnh khng kh dng ca h thng l khong thi gian khng th lm vic c, bt u khiBER >10-3 trong mi giy v ko di 10 s lin tip (10 s ny l khong thi gian khng lm vicc). Thi gian khng lm vic c kt thc khi BER


7/66

Chng 2Cu trc H thng viba s

I. S khi c bn ca h thng vi ba s:1. S khi ca mt trm u cui thc t:

S khi c bn tuyn pht:Nhim v:

- Bin i tn hiu bng gc thnh tn hiu dng sng- Chuyn i tn hiu ln bng tn cng tc- Khuch i tn hiu, hn ch ph tn hiu v bc x qua an ten.

Hnh 1. S khi c bn tuyn pht.

Chc nng:i vi h thng viba s tn hiu vo tuyn pht bao gm d liu di dng s c a n

t tng i hoc t trm viba s khc.

-M ho: M ho knh nhm sa li bng cch a vo mt lngthng tin d.

-Khi iu chnh x t tn hiu s bng gc thnh tn hiu dng sng: Cc dng iu ch

c bn FSK, PSK, QAM. Z

FSK : + Khng nhy cm vi mo bin

+ Thit b n gin.+ Hiu qu ph thp

PSK v QAM hiu qu ph ph thuc vo mc iu ch M

Hiu qu ph tng k=log2(M) ln

Khi M=4 th 4-PSK v 4-QAM nh nhau

Khi M=8 thng s dng 8-PSK

Khi M>8 thng s dng M-QAM

Cc b lc pht c lp ngay sau b iu ch quyt nh ph tn ca knh. Thng thng

l b lc cosine nng vi h s un lc t 0.2 n 0.7.- B trn tn nhm a tn hiu ln bng tn cng tc. Tu thuc tn s cng tc m hthng c th thc hin trn nhiu ln hoc thc hin iu ch ngay cao tn. Tuy s iu ch

trc tip t cao tn nhng ch c s dng nhng thit b c tn s cng tc thp khong 1

GHz viu ch FSK donhc im ca s ny l kh t c mt c tuyn iu ch

tuyn tnh v hn na tn s trung tm ca my pht khng n nh. Khi iu ch trung tn

yu cu i vi tn s trung tn l ftt > 3Rb (Rb l tc ca lung s liu)

- Khuch i cng sut: C th dng n sng chy (TWT) hoc bn dn Gallium-Arsenide

(GaAs FET) cho ra cng sut khong vi W. Thng thng sau b khuch i cng sut thng

lp mt mch lc ph nhm hn ch s m rng ph do tnh phi tuyn ca b khuch i.

7


8/66

- H thng an ten phid c s dng dn sng v bc x sng in t vo mi trng.

Dy phi thng gy ra mt lng tn hao nht nh t l vi di ca phi. An ten thng

c dng parabol c tnh nh hng cao v tng ch ln khong vi chc dBi.

Hnh 2. S khi c bn tuyn thu.

Tuyn thuNhim v:- Thu nhn v chuyn i tn hiu thu c v trung tn- Chuyn i thnh tn hiu bng gc- Khi phc xung clockChc nng: tuyn thu, tn hiu thu c a n my thu chuyn i tn

hiu thu c v tn s trung tn. My thu thc cht l mt thit b x l tnhiu cao tn bao gm mt khi khuch i cao tn v cc b trn tn.

B khuch i tn hiu cao tn c tc dng tng nhy my thu. (Giithch)

Cc b trn tn kt hp vi cc b lc thng gii bin i tn hiu siucao tn thu c v tn s trung tn. Tu theo yu cu cht lng v di tncng tc my thu c th thc hin i tn 1 ln hoc nhiu ln nhm loi

b cc tn s nhiu nh, nhiu trung gian v nhiu ln cn (Gii thch)Tn hiu trung tn s c a vo b gii iu ch chuyn t tnhiu dng sng v tn hiu s. Trn c s ca chui tn hiu s bng gcsau khi gii iu ch, tn hiu xung clock c khi phc v cung cp chocc khi ti to xung v khi gii m thu c d liu nh pht i upht.

S khi ca mt trm u cui thc t:

8


9/66

Hnh 3. S khi tuyn thu pht ca trm u cui.Mt trm u cui bao gm cc thnh phn: phn x l tn hiu bng gc,

phn v tuyn, phn nghip v v phn h thng phi-/ng-ten nh trn(hnh 2.15). Tn hiu nghip v c truyn i bng vic thm vo chc nngiu ch tn s vo b to dao ng ch sng phn pht. Pha thu s thchin gii iu ch tn s thu c tn hiu iu khin xa. Nh h thngiu khin ra lnh t xa cho php cc trm u cui c th iu khin c

cc trm trung gian m khng c ngi phc v. Khi chuyn mch dphng nhn tn hiu iu khin chuyn mch t khi iu khin cht lng.Khi cht lng h thng gim xung qu mt ngng cho php hoc gin onlin lc th h thng c iu khin chuyn sang knh d phng nhmtng kh dng ca h thng.

2. S khi trm trung gian:Nhim v ca trm trung gian- Khuch i tn hiu nhm b li nhng tiu hao trn ng truyn- Dch tn s nhm trnh hin tng t kch (tn hiu pht lt vo u thu)

- Ti to tn hiu s bng gc, loi b tp m tch lu. Chc nng ny khng nht thit phic tt c cc trm trung gian.C nhng s ca trm trung gian nh sau: My thu v my pht thc hin chuyn i tn s, tn hiu cao tn c chuyn i v tn

s trung tn my thu. Tn hiu c gii iu ch chuyn tn hiu thu c thnh tnhiu s bng gc sau c a n khi ti to xung nhm gt b tp m tch lu. Tnhiu u ra ca khi ti to xung c a c a vo khi iu ch chuyn tnhiu bng gc thnh tn hiu dng sng. Ti my pht s thc hin vic chuyn i tnhiu trung tn thnh tn hiu cao tn v bc x ra anten. y l s trm trung gian ph

bin s dng cho nhng h thng c dung lng trung bnh v cao.

Hnh 4. My thu pht i tn vi b ti to tn hiu trung tn

My thu thc hin chuyn i tn s cao tn thnh tn hiu trungtn v gii iu ch nhm thu c tn hiu tn hiu s bng gc. Timy pht thc hin iu ch trc tip ti cao tn v pht ra an ten.S trm trung gian c th hin trn hnh v 3.1. thng c pdng cho nhng h thng dung lng nh v lm vic tn s cao hn.

Hnh 5. My thu i tn vi b ti to tn hiu trung tn/bng gc vmy pht iu ch trc tip

9


10/66

Ti trm trung gian khng thc hin vic ti to xung, gii iu ch viu ch. My thu thc hin vic chuyn i tn hiu v trung tnv chuyn sang my pht. Ti my pht li chuyn i tn s tn hiutrung tn ln tn s tn s cao tn tng ng vi phn hoch tn s. vivic s dng trm trung gian theo s hnh 3.1. h thng thng tincho php s dng chung knh d phng vi h thng v tuyn tng t

v ct gim nhng thit b cn thit.

Hnh 6. Trm trung gian dch chuyn tn s, khng c b ti to tn hiu

Trm trung gian khng thc hin ti to tn hiu v dch chuyn tns. V bn cht trm trung gian nh hnh v 3. l mt b khuch itn hiu cao tn c bng tn gii hn cho mi hng truyn dn.

Hnh 7. Trm trung gian khng c b ti to tn hiu v khng dchchuyn tn s

II. Cc phng n tn s:

1. K hoch b tr tn s cho cc trm a lung v tuyna. Phng n lun phin:Mi 1 knh cao tn s dng phng n 2 tn s. Ton b h thng bao gm cc tn s

thu v pht xen k nhau.

V d: H thng 3 knh cao tn phi s dng 6 tn s

Knh cao tn 1 s dng f1, f2.Knh cao tn 1 s dng f3, f4.

Knh cao tn 1 s dng f5, f6.

10


11/66

Cc tn s f1, f3, f5 l cc tn s thu hoc pht

Cc tn s f2, f4, f6 l cc tn s pht hoc thu.

+ K hoch lun phin (Interleaved Plan)

Hnh 8. K hoch tn s lun phin.

+ K hoch ti dng tn s (Co-channel Plan)

Hnh 9. K hoch c ti dng tn s.Trong k hoch ti s dng tn s mi knh c s dng hai ln nh s phn bit phn cc

cho gia hai phn cc. Trong k hoch lun phin tn s cc knh c thm vo gia cc knhcng phn cc. K hoch lun phin tn s thun li hn khi thit k cc b lc phn bit

phn cc cho.Khong cch gia cc knh XS l khong cch gia cc knh ln cn cng phn cc. S l tc

symbol v X l khong cch knh c chun ho theo tc symbol. Tng ng nh vy Yv Z l bng tn bo v trung tm v hai u bng tn c chun ho. Vic la chn k hochtn s no hon ton ph thuc vo nhy ca h thng vi nhiu. Nhn chung k hoch ti sdng tn s c hiu qu s dng ph cao hn nhng vic vic s dng k hoch ti s dng tn

s thng c s dng rng ri trong nhng h thng cc chng ngn (20-30km) vi nhngdng iu ch n gin nh QPSK (y l dng iu ch c tnh chng nhiu cao). Tuy nhinvic s dng k hoch ti s dng tn s nhng chng di hn v nhng dng iu ch phctp l khng kh thi.

Nhn xt: Cc tn s pht v cc tn s thu xen k nhau

Cc tn s pht c cng sutln, cc tn s thu c cng sut nh cho nn khng th s

dng 1 nten thu v pht tn hiu theo 1 hng v nh vy tn hiu pht s nh hng n tn

hiu thu. M mi 1 hng ti s dng 2 anten pht v thu ring bit.

Ti trm trung gian lc ny phi s dng 4 anten. 4 anten ny th phi dng b lc ghp cc my thu vo 1 anten v cc my pht vo 1 anten

11


12/66

- u im: yu cu vi b lc ghp n gin ( cc tn s xa nhau s d lc hn cc tn

s xa nhau)

-Nhc im: S anten s dng nhiu, 1 trm trung gian phi c 4 anten nn hiu qu

kinh t thp.

b. Phng n phn nhm:

Mi 1 knh cao tn s dng phng n 2 tn s. Ton b h thng c chia thnh 2nhm tn s thu v pht ring bit.

V d: H thng 3 knh tn s dng 6 tn s.

Knh cao tn 1 s dng f1, f4.Knh cao tn 2 s dng f2, f5.

Knh cao tn 3 s dng f3, f6.

Cc tn s f1, f2, f3 l nhm cc tn s thu hoc pht.

Cc tn s f4, f5, f6 l nhm cc tn s pht hoc thu.

Ton b h thng c chia thnh 2 nhm tn s s dng thu v pht ring bit.

Khong cch gia 2 nhm tn s chn cch xa nhau.

V vy ti trm trung gian theo 1 hng cho php s dng 1 anten thu v pht ( cho 2

hng) s anten gim i 1 na.u im: S anten s dng t, hiu qu kinh t cao

Nhc im: Yu cu vi b lc pht phc tp v khong cch gia cc tn s thu v tn

s pht gn nhau.

Phng n ny thng c s dng cc tuyn thng tin c c ly di , tit kim

c rt nhiu anten.

2. B tr tn s cho trm lp:

+ K hoch 2 tn s: Ti mt trm lp (A) s dng 2 tn s sng mang cho lin lc haihng. My thu trn c hai hng cng lm vic trn tn s f1, trong khi my pht trn c haihng cng tc trn cng tn s f2 (xem hnh 1.7).

12


13/66

Rx

RxTx

Tx Rx

RxTx

Tx

f1

f1

f1

f1f2

f2 f2

f2

Trm lp A Trm lp B

Hnh 10. Phng n b tr 2 tn s.

+ K hoch 4 tn s: Ti mt trm lp (A), theo mt hng thu trn tn s f1 pht trn tn sf2, theo hng ngc li thu trn tn s f3 pht trn tn s f4 (xem h1.8).

Rx

RxTx

Tx Rx

RxTx

Tx

f1

f4

f1

f4f3

f2 f2

f3

Trm lp A Trm lp B

Hnh 11. Phng n b tr 4 tn s.

i vi phng n b tr 4 tn s th thit b trm phc tp hn do phi lm vic trn 4 tn ssong b li xuyn nhiu gia cc hng thu-pht rt nh.

III. Hiu qu ph ca h thng viba s:Khi nim hiu qu ph ca h thng vi ba s c phn bit vi rng bng tn m trn

l sng mang c iu ch khng tnh n nh hng ca cc can nhiu ni b h thngv nhiu gia cc h thng. Thng thng c xc nh qua t s gia tc truyn dn v rng bng tn.

E=R/B [bit/s/Hz]Trong R l dung lng truyn dn tnh theo bit/sB l rng bng tn

1. Hiu qu ph vi cc trm a lung

Tuy nhin trong trng hp i vi cc trm a lung c k hoch tn s gm c N knhgo v N knh return, khong cch knh l XS, v rng bng tn tng cng l B th hiuqu s dng ph phi tnh n tn hao do cc bng tn bo v:

21kkEE n=

Trong BNXSkXSbkbREn 5.0/;/;/ 21 === A- tc s liu truyn trong mi knh

b - rng bng tn ca knh k1 cho ta thy lp y trn bng tn v k2 th hin s tn hao hiu qu ph do bng tn

bo v. Ta c th vit li nh sauBRNBNXSXSbbRE 5.0/)5.0/)(/)(/( ==

RN la tc truyn thng tin tng cng ca h thng trn mt hng v 0.5B l rngbng tn s dng tng ng.

13


14/66

2. Hiu qu ph ca knh v tuyn s:Tc d ca h thng viba s c chun ha theo nhng khuyn ngh G.702, G.703 v

G.704 cho h thng phn cp s cn ng b v khuyn ngh G.707, G.708 v G.709 cho hthng phn cp s ng b (SDH hoc SONET). Tc d bt fb bao gm DS1 (1,544 Mbps) DS3(44.736 Mbps), E1 (2,048 Mbps) v E3 (34,368 Mbps), STS1 hoc Sub-STM1 (51,84 Mbps) vSTM-1 (155,52Mbps). Tc bt c truyn thc t fbr qua h thng viba s thng cao hnkhong 6% (fbr= 1.06fb) so vi tc chun ha do s thm vo cc bt thng tin phc v cho

m sa sai v nhng thng tin ph tr khc phc v cho cng tc qun tr ni b hoc thchin ghp knh v tuyn theo chun ring. tng hiu qu s dng bng tn thng dng nhng dng iu ch nhiu mc. Dng iu

ch thng dng nht l iu ch M-QAM trong M l s mc iu ch hay s trng thi cadng iu ch trn mt phng Constellation ca tn hiu. Gi thit b lc cosin nng c h s unlc l (0


15/66

Nhiu knh ln cn l nhiu t cc knh cao tn c cng phn cc nhng tn s ln cn. Nhiuny c quan h cht ch n vic chn khong cch gia cc knh. Nhiu knh ln cn cng cth gy ra bi cc h thng s v tng t khc nh h thng vi ba tng t, thng tin v tinh:

15


16/66

CHNG 3:LAN TRUYN SNG V HIN TNG PHA INH

I. nh hng ca kh quyn ti vic truyn sng cc ngn.1. Truyn sng trong khng gian t do.ng truyn v tuyn tm nhn thng di sng SCT b nh hng bi kh quyn v

khong cch gia hai trm. Khng gian t do l mi trng truyn sng l tng. Trong thc t

khng tn ti mi trng truyn sng l khng gian t do. Ch c cc knh thng tin gia cc vtinh c th coi nh l gn vi khng gian t do. Tuy vy cc kt qu nghin cu truyn sngtrong khng gian t do l c s pht trin nghin cu truyn sng trong tng kh quyn gnmt t. Tn hao truyn dn c bn trong khng gian t do l tn hao truyn dn nu an tenc thay bng anten ng hng, t trong mt mi trng in mi hon ho ng nht nghng v v hn vi cc khong cch gia cc anten gi nguyn. Khi tn hao truyn dnsng v tuyn t im pht n im thu trong khng gian t do c tnh theo cng thc sau:

=

d

L4

lg20 (2.1)

y d v tng ng l di tia v di bc sng c tnh cng mt loi n v. L

thng thng c din t bng dB di dng:L = 32,4 + 20 lg (fMHz . dkm) (2.2)Khng gian gii hn nng lng truyn t ng-ten pht ti ng-ten thu l cc i Fresnel c

dng elipsoid nh trn hnh v 2.1.

F1

F2

d+2/2

d+/2d

ng ngbin cai bin ca i

Fresnel th haiFresnel th nht

Hnh1: Cc i Fresnel

16


17/66

i fresnel th nht cha tt c cc im nh ngha bi hai on c di hp li ln hn

di ca tia thng mt khong nh hn /2, gii hn ca i fresnel th nht l elipsoid. Cci cn li c nh ngha theo kiu tng t. i fresnel th 2 cha tt c cc im nh

ngha bi hai on c di hp li ln hn ng thng mt khong ln hn /2 nhng nhhn 2/2. K t y cc fresnel c dng t chim gi l v elipsoid, vic xc nh i fresnel gipcho vic tnh ton khong h khi thit k tuyn vi ba.

2. S truyn dn sng cc ngn qua kh quyn trn u (well-mixed)a. Hin tng khc xi vi vi ba LOS, tia sng truyn dn xuyn qua cc thnh phn dy c ca khng kh v

chu nh hng bi s thay i ch s chit sut khng kh dc theo tia. Ch s chit sut l t sca vn tc nh sng trong khng gian t do vi vn tc ca nh sng trong mi trng. S thayi ch s chit sut trong kh quyn ph thuc vo cc hin tng kh tng, khi m rngkhong khng so snh c vi di ca chng v tuyn, c bit phi tnh n s thay ichit sut theo chiu thng ng. y l mt yu t nh hng rt ln n c tnh lan truynca tia sng. S thay ch s chit sut gy ra hin tng khc x sng nh trn hnh 2, lm tiasng b cong i.

S thay i ch s khc x n ca khng kh rt gn 1 v ch cn thay i vi phn triu cng cth gy nh hng ti s lan truyn sng v tuyn. c trng cho s thay i chit sut theo cao ngi ta a ra tnh khc x (refractivity) N thay th cho ch s khc x n (index ofrefraction) [7]:

N(h) = [n (h) - 1] x 106

y n(h) l chit sut ti cao h (h tnh bng km). S khc x ph thuc vo nhit , psut, m ca khng kh, in hnh c gi tr b mt t 200 n 500 (n vN). Theo quy nhca CCIR kh quyn tham chiu l mt mi trng c tnh khc x cho bi:

N(h) = 315 exp (- 0,136h)

y h l cao (tnh bng km) so vi mc nc bin.Biu thc trn din t tnh khc x gim theo hm m trong mt ct nghing. y l gi trm chng ta mong mun tnh c trong ngy v trong iu kin nhit kh hu khi m khquyn l gn mt t. Tnh khc x ca kh quyn cng b thay i theo thi gian trong ngy vthay i t v tr ny sang v tr khc v cng c th thay i theo ma trong nm [7].

Mt phn cch

n2

n1

(n1>

n2)

n1

Hnh 2: S khc x sng

17


18/66

Vic tnh ton s thay i ca tnh khc x l mt bc quan trng ca qu trnh thit ktuyn. S nh hng ca Gradient tnh khc x trong mt ct nghing l c s bn v hintng pha inh a ng.

Tia sng truyn ngang qua kh quyn khng ng nht c cong cho bi:

1/ = - dn/dh=-dN/dh.106

y l bn knh ca tia b un cong. Gi tr tham chiu ca dN/dh l - 40 n v N/kmtng ng vi bn knh ca ng cong bng 4a; (a = 6,37. 106m l bn knh tri t). Hnh 3 a)biu din mt tia nh vy (vi mt cung 1/) trn mt t c cong l 1/a. y trong iukin kh quyn tiu chun tia trc tip t ng-ten pht n ng-ten thu b un cong.

Nu ta chuyn trc to sao cho loi b cong ca tia sng th cong biu kin ca tri

t s l (dh

dn

a

1+ ). Hnh 3 b) biu th cong biu kin ca mt t khi tia truyn l thng, Nh

vy gi thit tia sng thng th tri t nh c nh ln vi bn knh ka:

dh

dn

a

ka

+

=1

1

v 610.1

1

1

1

+

=

+

=

dh

dNa

dh

dna

k

trong k l h s bn knh ca tri t tng ng thng gi l h s k.Thay a = 6,37.103km ta c:

( )dhdNk /157/157 +=i vi kh quyn tiu chun th dn/dh=-1/(4a) v gi tr tiu chun ca k = 4/3. iu ny c

ngha l vic lan truyn trn mt t qua kh quyn tiu chun tng ng vi vic lan truyn

trn mt mt t nh ln qua mi trng khng c kh quyn.Ch li ca qu t ti mt im trn ng v tuyn c cho bi:

kxdxxB ).().51/4()( = (m)

hdN/dh= -

40/km

300Nh

dN/dh= -

40/km

300Nh dM/dh=

117/km

300M

a.

b.

c.Hnh 3: Truyn dn qua kh quyn tiu chun.

a. Biu tri t tiu chun.b. Biu tia truyn thng.c. Biu tri t phng.

18


19/66


20/66

Hnh 4. Gi tr ke l mt hm ca di chng.Chng ta c th chuyn biu hnh 3.a thnh dng biu tng ng vi mt t phng

hnh 3c). Vi biu mt t phng th coi nh lc ny tia sng l mt ng lng cho. iukin lan truyn cho mt t phng c th c m t ngn gn bng vic a ra mt ch s khcx sa i. i vi mt cao cho trc, ch s khc x sa i c tnh bng tng ca ch skhc x khng kh vi t s gia cao v bn knh tri t. Ch s khc x sa i M cn cl modul khc x (refractive modulus)

M = N + (h/a) 106

cong ca tia truyn l:dM/dh = dN/dh + (106/a)

Kh quyn tiu chun c dN/dh = - 40 n v N/km. Gi tr ca dM/dh l khong 117 n vM/km bi v cong ca tri t c ly l 157 n v N hoc M/km.

H s kvi khc x chun l:

610)/(

1

=dhdMa

k

Bt k biu no trong hnh 3 u c th c s dng nh gi nh hng ca s thayi tnh khc x ln khong h ca tia sng v tuyn (Khi phn tch thng gi thit rnggradient ca tnh khc xl khng i). Biu hnh 3.a khng thun tin nht cho vic tnhton do h to l khng trc tuyn. Biu 3.b c nhc im l khi gradient ca tnh khcx thay i th thuc tnh ca ta sng cng thay i theo di ca ng truyn. Tuy nhin

biu ny cng thng c s dng m t hin tng lan truyn sng. Biu mt tphng hnh 3.c c s dng rng ri nht cho vic nghin cu tia sng trn c s ca tham skhc x sa i M.

Trong mt chng mc no s khc x cng l mt k qu c li, v d s un cong tiasng c th lm tng c ly lin lc.

Hin tng thiu khc x v siu khc x:

Gi tr tiu chun ca K = 4/3, nhng tu thuc vo tng vng khc nhau ga tr ca K cng

khc nhau. Theo s liu thng k cho thy: K < 4/3 nhng khu vc lnh kh, khi ny xut hinhin tng thiu khc x, hin tng pha-ing c th xy ra v khi K < 4/3 tri t coi nh cnh cao gy cn tr tn hiu n im thu. K > 4/3 nhng vng kh hu nng m, trng hpny gi l siu khc x, c bit khi dM/dh tin gn n 0, dN/dh gn n - 157 n v N/km th

20


21/66

cc tia hu nh song song b mt tri t, nng lng c th truyn i xa. Hin tng ny c thdn n pha-ing do nhiu x.

Trong thit k tuyn v tuyn tip sc thng chn K trong khong [1,1 - 1,6] tu thuc voiu kin a l.

Trong nhng iu kin c bit gi tr k c th nh hn 1 hoc k, tuy nhin iu kin nyt khi xy ra, nu c th thi gian xut hin khng ng k. Dng hnh hc ca tia sng vi ccgi tr khc nhau ca k cho trong hnh v 5.

k=1

k=4/3

k=

k=2/3

k=1

k=

k=2/3

k=4/3

k=1k=4/3

k=

k=2/3

Hin tng ng sngy l mt trng hp c bit ca khc x sng. Trong mt iu kin no , gia hai lp

kh c mt khng kh nh nhau, hnh thnh gia hai lp mt lp kh c mt khngkh khc hn. iu ny tng t nh hnh thnh mt ng dn sng m tia sng khi pht ra vimt gc ti no ch truyn dn trong "ng sng", lm cho tia sng b i lc theo hng khcm khng n c im thu.

ng sng thng xy ra ti cc vng c v thp, mt khng kh cao, cc v tr gn mtnc v c nhit thay i thng xuyn vi tc thay i kh nhanh. Hin tng ny chra trn hnh 6.

21


22/66

b. Hin tng nhiu x (difraction)Hin tng nhiu x l hin tng sng v tuyn b un cong quanh cc vt cn nh l cc

vt cn c nh nhn hay cc b mt c dng hnh cu. Tnh cht nhiu x s ph thuc vo kchthc ca chng ngi vt v bc sng v tuyn v chng gy ra mt tiu hao gi l tiu hao

nhiu xNh bit min Fresnel th nht cha hu ht cng sut tn hiu n my thu. Nu tn ti

mt vt th ra min Fresnel th nht th tia phn x s ngc pha vi tia trc tip ti im thuv gy ra mt s suy gim tn hiu ti im thu. Trong thc t nu khng c vt cn no trongmin Fresnel th nht th s khng c bt c mt tiu hao tp m x no. Cc min Fresnel th2, 3, t nh hng n vic to ra tiu hao nhiu x v cng sut cha trong min l khngng k.

Bn knh ca h elip xung quanh ng trc tip thay i dc theo ng truyn c biudin theo cng thc:

fd

dndFn

2.13,17= [m] n = 1,2,3 ...

Trong :f: L tn s cng tc, tnh bng (GHz)d1: L khong cch t im phn x ti im u (cui), tnh bng (km).d2: L khong cch t im phn x ti im cui (u) tnh bng (km).d: L khong cch t ng-ten pht ti ng-ten thu tnh bng (km).i vi mi loi a hnh khc nhau c th chn min Fresnel th nht s c cc tn hao

nhiu x khc nhau. Cc loi a hnh ny c th l mt t phng, hnh nm, hnh trn

Hnh 7: im c bit ca a hnh nm trong khng gian truyn sng

i Fresnel 2 i Fresnel1

d F1tia LOS

d1

H0

d2

22

n0

n

v

Tia LOSn

0

Hnh 6. Hin tng ng dn sng


23/66

Hnh 8. S ph thuc ca tn hao nhiu x theo khong h ca ng truyn.Hnh v ch ra s ph thuc ca tn hao nhiu x theo khong h ta thy rng nu khong h

tho mn iu kin ti thiu 0,6F1 th c th coi ng truyn l khng gian t do. C ngha ltn hao nhiu x bng khng.

Tn hao nhiu x do lt trn mt t phng:Cc tham s c tnh in ca mt mi trng bt k bao gm:

t thm (Henry/mt - H/m) Hng s in mi (Farad/mt - F/m) in dn (Siemen/ mt - S/m)

t thm ca t v bin thng thng c th coi bng t thm ca khng gian t do vvy cc bi ton thc t ch xt n hng s in mi v dn in. Thng thng s dng

hng s in mi tng i rso vi khng gian t do.S ph thuc ca hng s in mi tng i v dn in c ch ra trn hnh 9. Trongthc t i khi c th s dng cc biu thc kinh nghim tnh ton hng s in mi tngi rda trn in dn

5/1.50 =r

Hnh 9. Hng s in mi tng i v in dn l hm ca tn s.

xc nh tn hao nhiu x ng v tuyn b chn bi mt t phng cn xc nh dnnp b mt K theo sng phn cc ngang v ng.

( ) ( )[ ] 4/1223/13 /181).(10.942,1 += ffkK rh

23


24/66

( )[ ] 2/122 /18 fKK rhv +=Trong k l h s bn knh hiu dng thng c xc nh theo h s k cc tiu; f l tns tnh theo n v GHz.

Nu K < 10-3 th hon ton c th b qua cc c tnh in ca mt t.Nu K> 10-3 khi tiu hao nhiu x do mt t phng c th c tnh gn ng nh sau:

)()()( 21 YGYGXFL ++=

Trong F v G l hm ca X v YXXXF 6,17lg1011)( +=X l di ng truyn c chun ho ctnh theo cng thc

( ) 3/12/...064,0 kfdX =

42

42

35,15,41

75,06,11

KK

KK

++++

=

d l di tia sng tnh theo km, f tnh theo GHzVi phn cc ngang ti tt c cc tn s v i vi phn cc ng f> 20MHz trn t linhocf> 300 MHz trn bin c th ly =1 cao ca cc anten chun ho c th c xc nh theo cng thc sau:

( ) 3/12 /...05178,0 kfhY =h l cao anten tnh theo mtG l mt hm ca Yc xc nh theo cng thc sau:

[ ]

[ ]

0

>0

1>0 2>0

d1d2

h


26/66

Tn hao nhiu x gy ra bi vt chn trn duy nht

da db

>0

1>0 2>0

R

Hnh 11. Vt chn hnh trn duy nht trn ng truyn.)()()()( XQTLrL ++=

L() c xc nh ging nh vt chn hnh nm vi h s[ ][ ] 2/1)./()2/.).(2/.(2)2/sin(2 dRdRd ba ++=

Cc tham s c xc nh nh hnh v 11T() l tn hao trn mt cong ca vt chn

432 8,06,322,7)( +=T2/1

3/1)/(

+

=

R

R

dd

dd

ba

ba

Q(X) l tn hao c tnh theo cng thc

=4lg29617

405,12

0/).(

)(

XXX

XX

XXT

XQ

[ ] ( ) ..2/./ 2/13/1 = RX nu 1


27/66

Hnh 12 S hp th ca xy v ma vi sng v tuyn

Vic xc nh cc tiu hao c trng c th xc nh c da trn biu hnh 12. Tiu haogy ra bi y v hi nc c xc nh theo cng thc sau:

ddA woaa ).(. +==dl di ng truyn (km).o , w l tiu hao c trng ca xy v hi nc (dB/km).Bi tp: Tnh tiu hao gy ra bi cc cht kh v hi nc ca mt tuyn vi ba c di 50

km lm vic ti tn s 15 GHz.(A = 0,03.50=1,5 dB)

4. S phn x sng in tCc ng-ten lm vic trong h thng vi ba u c kh nng nh hng cao. Tuy nhin chmtia bc x t ng-ten cng ch dng hnh nn trn xoay vi mt gc m trn gin hngno . Gi s vi cc gc m hp t 10 - 20, vi c ly lin lc mt chng t 40 - 50 km, th tiim thu nng lng chm tia sng in t cng tri ra kh rng (c t 0,7 - 1,4 km) v caoca thp l rt nh so vi phm vi tri rng ca nng lng in t.

V vy trong qu trnh truyn dn sng t im pht ti im thu, mt phn nng lng sngv tuyn s gp mt tri t v do mt t c tnh cht phn x sng, nn trong cc tia phn x sc tia n c im thu. Trong trng hp tia phn x ny c cng pha vi tia trc tip, th tnhiu thu c tng cng. Tuy nhin khi tia phn x c pha sai khc pha tn hiu ca tia trc tip,th tn hiu thu s b suy yu. Trong trng hp tia phn x l ngc pha vi tia trc tip, th tn hiu

thu thm tr c th b trit tiu.S phn x cng c th xy ra cc vng kh quyn bt ng nht (vng i lu), tuy nhintrong h thng vi ba LOS, do truyn sng thng nn chng ta khng cn phi xt n trng hp

27


28/66

ny. S phn x sng t mt t xy ra khong gia tuyn hoc nhng ch a hnh c cutrc c bit ch ra trn hnh 13.

Tn hiu n im thu l tng tn hiu tia trc tip v tia phn x.

drdEEE +=

Tn hiu ca tia phn x s b chm pha so vi tia trc tip mt lng /.2 r=

trong r l s chnh lch qung ng gia tia trc tip v tia phn x

tn hiu phn x ti im thu c th c tnh theo cng thc sau:)exp(. = jERE dRTrong R l h s phn x ca b mt

)exp(. RjRR =Nh vy tn hiu ti im thu c th c vit li nh sau:

( )( )[ ])exp.1 Rd jREE ++= Thng thng ngi ta biu din lch pha gia tia trc tip v tia phn x ging nh s

gi chm v thi gian . Nu nh tn s gc ca sng mang l th c th biu din gc lchpha nh sau:

=Hm truyn t ca knh v tuyn c th biu din s b nh sau:)(exp1/)( +== RjREdEH

Cc gi tr ca ,, RR thay i thng ging theo thi gian do s thay i b mt hoc cc

iu kin kh hu. th ca H() s c mt ch trng 1800. su ca ch trng ph thucvo gi tr R . Nu R =1 th ti 1800 tia trc tip s b kh hon ton v ta c pha ding su.

Trong trng hp ,, RR gi nguyn ta thy rng H() ph thuc tn s v do knh vtuyn c c tnh ph thuc tn s.

i vi mt mt t g gh th tn hiu ti im thu s l tng ca tt c cc thnh phn phn

chiu. Hin tng phn x cng l mt trong cc nguyn nhn gy ra hin tng thng ging tnhiu ti im thu hay cn gi l hin tng pha ing.

II. Hin tng Pha inhPha ing l hin tng thay i tn hiu v tuyn mt cch bt thng ti im thu do s tc

ng ca mi trng truyn dn. Cc yu t gy ra hin tng pha ing bao gm:- S hp th ca cc cht kh hi nc, ma . y l nhng yu t ch yu i vi

nhng tn s ln hn 10 GHz.- Pha ing do hin tng lan truyn a ng:

S thay i gradient ch s khc x theo thi gian cng gy ra hin tngthng ging tn hiu ti im thu. Trng hp cc oan c th gy ra hin tng ng

sng lm tia sng khng n c im thu. Hin tng ny thng xy ra ti nhngni c v thp v gn b mt nc v c nhit thay i nhanh. Trong trnghp gradient thay i ln th y l nguyn nhn ch yu gy ra hin tng truyn

Hnh 13. Hin tng phn x sng t mt t

28


29/66

lan a ng. Hin tng ny thng xy ra nhng gi gia tra khi khng khb xo trn hon ton do cc dng i lu v gi. Khi gn ti v c bit trong nhngthng ma h, lng gi gim, nhit v m phn b khng u, to nn ccgradient ch s khc x bin i S phn x sng t b mt tri t, c bit l t b mt nc v s phn xt nhng bt ng trong kh quyn s nhiu x bi nhng vt cn trn ng truyncng l nhng nguyn nhn gy ra hin tng lan truyn a ng.

Suy hao ng truyn trong h thng vi ba mt t bao gm hai thnh phn: Thnh phn Lc trng cho suy hao truyn sng trong khng gian t do v thnh phn A(t.f) c trng choyu t nh hng ca mi trng. Nh vy suy hao ca h thng vi ba mt t trong trnghp ny c th vit:

a(t,f) = L . S(t,f)L l tn hao gy ra khi lan truyn trong khng gian t do v S(t,f) c gi l h s suy hao do pha-

ing. Ni chung S(t, f) l khng th ly gi tr trung bnh c, m n c xem nh mt qu trnh ngunhin vi hai bin c lp. Xt mt cch cht ch qu trnh ny khng dng. Nhng thun tin cho vickho st trong thc t ta gi thit S(t,f) l qu trnh dng.

H s suy hao do pha-ing S(t,f) l hm ca bin thi gian t v tn s f. Trong mt s trng hp s phthuc ca pha-ing vo tn s l khng ng k. Trong trng hp ny suy hao do pha-ing hu nh

khng i trong c bng tn . i vi cc h thng c bng tn hp (h thng dung lng nh), khi pha-ing c gi l pha-ing phng. i vi cc h thng bng rng (dung lng cao), gi tr S(t,f) thay i rtkhc nhau cc tn s khc nhau trong bng tn, trng hp ny pha-ing c gi l chn lc theotn s.

1. Pha inh phng:a. Pha inh phng do truyn dn a ng:Trong thng tin v tuyn vn truyn dn nhiu tia c cp nghin cu t nhiu thp

k trc v nhng m hnh ton hc v knh nhiu tia c xut. M hnh truyn dnnhiu tia vi s lng tia ln th p dng nh l gii hn trung tm kho st, theo nh l nyvi s lng tia ln th cng trng thu c phn b Gauss, bin trng thu s c phn

b Rayleigh. Trong m hnh kho st nu mt tia ln hn cc tia cn li th phn b ny chuynv phn b Rice hoc Nakagami. Trong h thng vi ba tn s lm vic cao, truyn sng thngnn s lng tia trong knh l khng nhiu, do tnh ng ca knh thng tin nh hn so vicc knh ngvi phn b Rayleigh, Rice v Nakagami.

xc nh d tr pha ing c hai phng php:Phng php MajoliH s xut hin pha inh phng nhiu tia cho thng xu nht P0 ph thuc vo cc tham s

ca tuyn nh tn s cng tc, di ca chng, kh hu, g gh ca a hnh. Theo phngphp Majoli P0 c xc nh theo cng thc sau:

( )53

0 10.4/

= dfCP 30 )50/)(4/.(..3,0 dfCaP =

trong :dl di khong tuyn (km)

fl tn s cao tn (GHz)c l hng s a hnh. c = 1; 4; 0,25 tng ng vi iu kin a hnh trung bnh kh hu ni kh hu m t hoc mt nc, kh hu min ni kh. thch ng vi s g gh ca a hnh, Majoli a vo cng thc mt h s c trng chos g gh ca a hnh

30 )50/)(4/.(..3,0 dfCaP =

a l h s c trng cho gh gh ca a hnh c th xc nh nh sau:( ) 3,115/ = ua

29


30/66

Trong cc h thng bng hp th s thay i suy hao pha inh trong bng tn hiu l kh nhv c th xem nh l pha inh phng. khc phc phaing phng c th thc hin c nh sdng mt lng d tr cng sut thch hp gi l d tr pha inh phng (flat fading margin). Cth hiu d tr pha inh phng l s chnh lch gia mc tn hiu thu c khi khng c phainhPR v ngng lm vic ca my thu RX tng ng vi BER cho trc.

Thng thng su pha ing phng nhiu tia phn b theo hm logarit do vy xc sut phaing nhiu tia t ngng d tr pha ing phng FM c xc nh nh sau:10FM/10=P

Khi cng sut thu c ti may thu nh hn mc ngng RX th h thng gin on lin lc.Xc sut gin on lin lc c th c xc nh l tch ca xc sut xut hin pha inh phng vxc sut pha inh vt d tr pha inh phng FM nh sau:

10/0 10.)(

FMm PFMFADP

=>Mt s tiu chun p dng h thng thng tin nh sau:

Vi L10-6 trong ti a 0,045% ca thng bt k (thi gian hpthnh l 1 pht) v BER>10-3 trong ti a 0,006% ca thng bt k (thi gian hp

thnh l1 giy). Vi 28010-3 trong ti a 0,054.(L/2500)% ca thng bt k(khong thi gian hp thnh l 1 giy)

T l thi gian gin on thng tin (%) trong thng xu nht do pha ing nhiu tia l:10/

010.100FMP =

tho mn cc tiu chun cho mt h thng thng tin nh nu trn cn xc nh c d tr pha ing ti thiu:Vi L


31/66

gim v khng ng vai tr quan trng. tn s t 7 n 15 GHz cn xc nh c c tnhtiu hao ca c pha inh do ma v pha inh phng nhiu tia. Tm li vi tn s thp c lychng lp di th pha inh nhiu tia l ch yu v trong trng hp ngc li khi tn s cao, cly chng lp ngn th tiu hao do ma l nh hng ch yu.

Phng php xc nh trc tip tn hao do ma nh sau:Mc tiu xc nh lng tiu hao gy ra do ma vt 0.01% thi gianBc 1: Cn bit c cng ma vt 0.01 % theo thi gian (thi gian hp thnh 1 pht)

R (mm/h). Tham s ny c th nhn c t nhng s liu thng k ti a phng hoc c thtra gn ng theo nhng bo co ca CCIRBc 2: trn c s xc nh tiu hao c trng

RkR .=Trong

2/2cos.cos)( 2 vhvh kkkkk ++=

( ) kkkkk vvhhvvhh 2/2cos.cos...2 ++=

: gc ngng ca anten.: l gc nghing phn cc i vi phn cc ngang=450i vi phn cc vng.

=00i vi phn cc ngang.=900i vi phn cc ng.Bng 2. cc h s qui i tnh ton tiu hao c trng

Tn s (GHz) k H kV H V1 0,0000387 0,0000352 0,912 0,8802 0,000154 0,000138 0,963 0,9234 0,000650 0,000591 1,121 1,0756 0,00175 0,00155 1,308 1,2657 0,00301 0,00265 1,332 1,3128 0,00454 0,00395 1,327 1,310

10 0,0101 0,00887 1,276 1,26412 0,0188 0,0168 1,217 1,20015 0,0367 0,0335 1,154 1,12820 0,0751 0,0691 1,099 1,06525 0,124 0,113 1,061 1,03030 0,187 0,167 1,021 1,00035 0,263 0,233 0,979 0,96340 0,350 0,310 0,939 0,929

Bc 3:Tnh di ng truyn hiu dng

0/1 dd

dde +=

( 01,00 .015,0exp.35 Rd = vi hmmR /10001,0

kmd 81,70 = vi hmmR /10001,0 Bc 4:nh gi tiu hao ng truyn vt trong 0,01% thi gian

eR dA .01,0 =

Bc 5:

Xc inh tiu hao vt trong p % thi gian)lg.043,0546,0(

01,0 ).(12.0p

p pAA+=

31


32/66

V r rng xc sut gin on lin lc s l p% khi Ap=FM

Bi tp: Xc nh tiu hao do ma ch vt trong khong 0,0001 % thi gian i vi chng

tip pht di 53 km hot ng tn s 12GHz vi phn cc ngang. Gi s rng lng ma vt

trong 0,01 % thi gian l 30 mm/h.

2. Pha-ing nhiu tia chn lc tn s trong h thng vi ba s bng rng

a. Cc m hnh pha-ing nhiu tia chn lc tn s c xut nghin cu vn ny trc ht chng ta gi thit rng h thng c thit k theo ngtiu chun thit k tuyn, tc l khong h ng truyn ln hn cho php. H thng c githit nh vy s khng b nh hng pha-ing do nhiu x. Pha-ing xy ra c nguyn nhn ltruyn dn nhiu tia, gm tia ti trc tip, tia khc x t kh quyn v cc tia phn x t mt ttruyn n ng-ten thu. Kt qu sng v tuyn ti ng-ten thu theo nhiu ng khc nhau, hintng ny c gi l truyn dn nhiu tia, ch ra trn hnh v 14

Khi nghin cu hin tng knh nhiu tia v vn pha-ing chn lc theo tn s c khnhiu m hnh knh nhiu tia c a ra nh: M hnh ba tia, m hnh ba tia n gin ho, mhnh hai tia, m hnh a thc. Trong m hnh ba tia n gin ho ca Rummler c coi lhay nht v n tr thnh m hnh l thuyt cho vic nghin cu hin tng ny. M hnh ny c kim nghim bng o lng thc t v c thng bo rng ri l chnh xc.

Khi knh c xc nh l truyn dn nhiu tia th vic m t cc m hnh hm truyn dnpha-ing ca knh l cng vic ht sc quan trng t ta mi c th xut cc bin phpkhc phc.

Xut pht im kho st l cc phn ng xung v phn ng tn s ca knh, cc phn ngny khng k n suy lc do pha-ing. Vi vic gi thit y l qu trnh dng v khng kn suy lc do pha-ing th phn ng xung ca knh c m t nh l tng trng s cc hmDelta.

=

=N

1nnn )t()t(c

Tng ng phn ng tn s c dng:

=

=N

1n

f2jn

ne)f(C

Trong n v n tng ng l h s suy hao v tr ca tia th n; N l s tia trong knhnhiu tia. Bng cc kt qu nghin cu qua o lng thc t ngi ta a ra mt s m hnhknh nhiu tia nh sau:

M hnh ba tia tng qut: M hnh ny mc nhn rng, trong mt chng vi ba s lun tn ti ba tia truyn lan t u

pht n u thu.

- Tia trc tip (tia LOS)- Tia c bin nh i gn v bn di tia LOS (tc chnh l tia phn x b mt) cbin nh v b mt t, cc ht kh v hi nc hp th mt phn nng lng.

- Mt tia c bin ln hn cch xa tia LOS hn l tia khc x.

32

Hnh 14. Dng truyn dn nhiu tia


33/66

Hm truyn ca knh trong trng hp ny c dng:21 2

22

11)( fjfj eefC ++=

Trong s hng th nht m t tia trc tip v ly lm tham chiu, s hng th hai v th ba

m t hai tia cn li ca knh nhiu tia. 1 , 2 l cc gi tr bin tham chiu c trng do suyhao bin , 1 v 2 l cc gi chm ca hai tia tng ng.

Thc t 1 c gi tr rt nh v vec t [1 e-j2f

] c lch pha so vi vec t tn hiu tia trctip l rt nh. Mt khc thc t cho thy gi tr bin ca vec t ny cng khng ln nn nhhng ca thnh phn ny n trng thu l c th b qua c.

M hnh 3 tia tng qut ny c bit ngha khi truyn sng qua vng c b mt nc. Tuyvy m hnh ba tia tng qut li khng a ra c dng ton hc y ca m hnh knh. Mhnh 3 tia tng qut cha qu nhiu tham s nn vic kho st theo m hnh ny l rt kh khn.M hnh c xem l i din cho knh nhiu tia li l m hnh ba tia n gin ho caRummler a ra, m hnh ny ph hp hn vi cc s hiu quan trc v o lng.

M hnh hai tianh hng ca tia phn x do i gn tia trc tip, nng lng thp, nn c lch pha nh v

trong trng hp c bit th c th b qua c. M hnh hai tia a ra gi thit knh thng tinch truyn di dng hai tia, khng k n tia phn x b mt. N gn ging m hnh 3 tia ngin ho nhng li khng k n bt k nh hng no ca tia gn tia LOS v hm truyn caknh c dng:

fjefC += 21)(

Vi m hnh ny cha 3 tham s f, , . Ngi ta tin hnh p dng m hnh ny kho st knh thng tin thc t, cc php o lng trong bng tn 55MHz cho thy 3 tham

s ca m hnh l c lp thng k v c phn b u. Khi n gin ho, m hnh nych cn cha 2 tham s (f,) nn vic tnh ton kho st n gin hn nhiu. y cxem l m hnh n gin nht kho st h thng pha-ing nhiu tia chn lc tn s. Tuynhin m hnh ny li khng khi qut y dng truyn lan nhiu tia thc t ca knh.

M hnh a thc:Ngi ta biu din bin v thi gian tr nhm l mt hm ca tn s di dng mt a

thc tng qut nh sau:n

nfCfCfCfCCf +++++= ...)(3

32

210iu ny tng ng vi vic m t hm truyn ca knh di dng mt a thc phc nh

sau:

n

nn

N

n fjJBAAfC )2)(()( 10 ++= =f l tn s cao tn, c o lng t tn s trung tm bng tn ca knh, A l h s c trngcho tn hao trung bnh ca tia th n, Bn c trng cho khe pha-ing th n. Vi N = 1 th a thctrin khai kh ph hp vi s liu o lng t knh c rng bng 26,4 MHz, RF = 6GHz, d =42,5 km. M hnh ny cng a ra qu nhiu tham s vi N = 1 m c ti 5 tham s. Do vic kho st theo m hnh ny cng rt kh khn.

M hnh ba tia n gin ho ca RummlerM hnh c xy dng t rt nhiu s liu quan trc trn mt chng vi ba LOS tiu chun,

vi RF = 6GHz, trn chng 42,5 km t Atlanta ti Palmetto bang Georgia (Hoa k). Cc s liu

o lng c Rummler thc hin cho trn hnh 15

33


34/66

Cc s liu ny c thu c t cc o lng vi cc thit b o lng c bit c chnhxc rt cao . Hnh 15 m t c tnh bin v pha o lng c ca knh nhiu tia, cng tcti gin tn 6GHz tc 140Mb/s, iu ch 64 QAM do Rummler thc hin.

C s kho st pht trin dng m hnh ba tia n gin ho l phng trnh hm truyn

vi m hnh ba tia tng qut, vi gi thit l B.1


35/66

])(2cos21[)( 0222 ffbbafC +=

)(2cos21

])(2[cos.)(

2

1)(

02

0

ffbb

bffb

df

fdfD

+

=

=

Mt th d tiu biu ca A(f) v D(f) ca knh c cho trn hnh. Hm truyn knh theo m

hnh Rummler c dng tun hon, chu k 1/ v c cc khe pha-ing nhiu tia ti tn s khe f0.

Trong thc t hm truyn ca knh thng tin khng hon ton tun hon, nh nu trnhnh 16. Phn ng ca knh do s khng m t hon ton chnh xc theo (*) trn ton b trctn s m ch ng trong gii tn cng tc ca h thng, tc l ch ng trong di tn Nyquisthoc trong di tn ln hn i cht. Ngoi ra phn ng ca knh cng khng th biu din hon

ton chnh xc theo (*) nu rng ca knh qu ln (> 1/), do tnh chu k nu. Tuy nhiny l mt m hnh kh ni ting c s dng rng ri v chnh xc trong cc iu kin thc t.M hnh ny cng c s dng lm m hnh chun cho cc phn mm m phng h thngthng tin.

Trong trng hp b


36/66

Vi 0 = 6,3 ns. iu ny c ngha l xc sut0

e0 4

1ff

ln gp 5 ln xc sut

0

e0 2

1ff

4

1

0 v pha-ing khng cc tiu khi < 0 vi b


37/66

Trn hnh 18 do chng ta ch quan tm n tn s ln cn sng mang fc nn bin tn s f0c thay bng (f0 - fc).

T ng cong signature cho thy: ti mt tn s khe pha-ing f0 cho trc, nu B nhn mtgi tr ng vi ng cong signature, th li bt ca h thng l BER = 10-3, nu B ln hngi tr gii hn th h thng b gin on. So snh ng cong signature ca h thng 16QAM v 64 QAM ta thy khi rng bng tn cng ln th gi tr B thay i cng nhiu vr rng l min gin on thng tin do pha-ing cng ln. iu ny chng t pha-ing ivi h thng vi ba bng rng l pha-ing chn lc tn s v mc nguy hi ca n ln hnrt nhiu so vi pha-ing phngKt lun:- ng cong signature ca h thng c hnh M hay W tu thuc vo v th hng lntrn hay quay xung ding cong signature cng thp v cng rng th kh nng chu ng pha ing ca hthng cng ti.

- ng cong signature ca h thng o c cng mt i xng th c tnh lc ca hthng cng mt i xngc. Phng php xc nh xc sut gin on lin lc do pha inh nhiu tia chn lc theo

tn s:ng cong signature l qu tch cc im trn mt phng B-f0 ng vi trng thi gin on

lin lc ca h thng, chia mt phng B-f0 thnh hai phn ng vi trng thi gin on lin lcdo pha inh na trn v khng gin on lin lc na di.

Nh vy ta hon ton c th xc nh c xc sut gin on ca h thng nh s dngng cong Signature. Xc sut gin on lin lc ca h thng l tch ca xc sut xy ra phainh reyleigh v xc sut gin on lin lc do pha inh nhiu tia Pos.

Xc sut xy ra pha ing a ng lin quan n lng tn hao gy ra bi pha inh phng

)19( 0 dBAP >=Xc sut gin on lin lc l xc sut s kin thng s B vt qu gi tr ca ng cong

signature v c tnh theo cng thc sau

37

(f0-

fc)*T

s

b)

(f0 -fe)

0 21-12

a)

0,5

0,5

10-1

Hnh 18 - ng cong signature theo m phng ASTRAS:

a. Cho h thng 16 QAM

b. Cho h thng 64 QAM

*Ts


38/66

00

2/1

2/1

0 )(.)(

exp dffpM

fP

c

c

f

f Bos

+

=

Trong :

)( 0f l hm s ng cong signature ca h thng.MB l gi tr k vng ca su khe pha ing theo m hnh Rumler th M B=3.8 dBp(f0) l hm mt xc sut ca tn s khe pha ing

fc l tn s sng mangNh vy xc sut gin on lin lc do pha ing nhiu tia gy ra l

stotal PP 0=Vic xc nh xc sut gin on linlc theo phng php trn hon ton ph thuc vo

ng cong signature o c nh cc thit b chuyn dng.Phng php 2: Phng php tnh th.Gi thit phn b thi gian tr tng i ca tia gin tip l hm m (gi tr trung bnh l 0 )

v phn b i vi c bin ng i ca tia gin tip v dch tn ca khe l phn b uth xc sut gin on lin lc do pha inh nhiu tia chn lc theo tn s s c xc nh nhsau:

[ ]2

0 /).2( TKPos =Trong

=

PSKfor

PSKfor

QAMfor

QAMfor

K

4

8

16

64

0,1

0,7

5,5

4,15

( ) 5,10 50/.7,0 d= [ns]d l di ca tuyn tnh theo km

Nu tc ca ng truyn l rb (Mbit/s), M l s mc iu ch th xc sut gin on

thng tin gy ra bi pha inh nhiu tia chn lc theo tn s s l( )[ ]2625,1 10log/.2)2( = MrdKP bos

Nh bit xc xut sut xut hin pha inh nhiu tia s l P0=PR , trong PR l xc sutpha inh Reyleigh v c hiu nh h s hot ng ca pha inh nhiu tia Reyleigh

10s,P(10)

( ) ( )( )aTerfcP /10ln548.0.5.010 =

40. Xc sut khong pha ing fdg>60s,P(60)

( ) ( )( )bTerfcP /10ln548.0.5.060 =

41. Xc sut BER vt 10-3 10/0 10. aFMP

42. Xc sut mch tr ln khngdng c Pu

P0 Pa P(10)

43. s dng ca ng truyn % 100(1-Pu) %

44. Xc sut BER>10-6 10/0 10. bFMP

45. Xc sut BER>10-6

trong khong >60s

P0 Pa P(60)

46. H s ci thin phn tp khng gianIos

( ) ( ) 10/22

30 1010.2.1

FMrad

fSI =

Bng tra cu cc tham s dng cho tnh ton ng truynMt s nc 2 2 C2Php 0.5 -1 -an mch 0.33 - -M 0.5 -0.5 56.6.sqrt(d)Thy s 0.59 -1.38 -

II. Cc tnh ton ch tiu cht lng v kh nng s dng1. Tnh ton ch tiu cht lng

Cc tnh ton ch tiu cht lng

1. di on (km) Tng di ca chng tip pht

2. S chng tip pht trong mt on N3. S b lp S chng tip pht c ti sinh tn

hiu

4. Xc sut BER >10-3 =

N

i

FMi

aiP1

)10/(0 10

5. Xc sut BER >10-3 (Pha ing la chn)( )[ ]

2

1

62

5.1 10log.24=

N

ibiiii MrdK

6. Tng gin on thng tin (BER>10-3) (4)+(5)7. Ch tiu cc giy li trm trng 0.006% L < 280km

57


58/66

0.054.L/2500 % 280km < L 10-3) 10lg(%ch tiu / %tnh ton)

9. ci thin ca b cn bng (BER>10-3) Ie

10. ci thin ca b cn bng v phn tp BER>10-3

Ios

11. Xc sut BER>10-6 =

N

i

FMi

biP1

)10/(0 10

12. Xc sut BER>10-6 (Pha ing la chn) Da vo ng cong signaturetng ng vi BER >10-6

13. Tng BER>10-6 (11)+(12)

14. Ch tiu cc pht suy gim cht lng 0.045% L < 280km

0.04.L/2500 % 280km < L 10-6) 10lg(%ch tiu / %tnh ton)

16. ci thin ca b cn bng BER>10-6 Ie

17. ci thin ca b cn bng v phn tp BER>10-6 Ios

2. Tnh ton kh nng s dng

18. khng s dng c ca thit b, % 0.15 10-3

19. khng s dng c do ma, p % FMa = 0.12(A0.01)p-(0.546+0.043lgp)

20. khng s dng c do pha ing nhiu tia phng,% iai

N

i

i PPP )10(1001

0=

21. khng s dng c do pha ing la chn theo tns, %

100.P(10).(Xc sut BER>10-3 dopha ing la chn theo tn s)

22. Tng khng s dng c, % (18)+(19)+(20)+(21)23. Ch tiu khng kh dng ca h thng, % 0.06.L/600 % vi L


59/66

THIT K TUYN THNG TIN (tham kho)

1: NHIM V1. C s ban u:

Khi thit k 1 tuyn thng tin cho trc im u cui

- Cc trm trung gian chnh

- Cc dch v

- Dung lng, tc

- Thit b ( c th)

V d: Thit k tuyn thng tin H Ni Lo Cai

Cc trm trung gian chnh: Sn La, Yn Bi

Cc dch v thoi, s liu, hnh

Dung lng: thoi v knh, s liu tc ..hnh x kinh

2. Nhim v

Chn tuyn:Chn v tr t trm trung gian, phi xy dng 2 hoc 3 phng n phn tch chn

phng n ti u, phi da vo bn a hnh t l 1/50.000; 1/100.000

Ch : Tuyn phi trnh i qua cc vng c a hnh phc tp, khu cng nghip, trung

tm thu pht i trnh nhiu v c gng cho tuyn i l ngn nht gim s trm trung gian.

- V tr t trm phi c gng tn dng cao ca a hnh, thun tin cho vic sinh hot,

trin khai, s dng c cc ngun in mng, phi nh du v tr to ca im t trm.

Chn thit b s dng:

Cn c vo dung lng nu khng dung lng th phi ghp song song nhiu thit b

gi l knh cao tn

Chn thit b cu hnh ( n, song song) phi m bo dung lng v tc thng tin v

phi c tnh kh thi.

* Tnh v kim tra cc ch tiu cht lng

- d tr phainh

- n nh thng tin ( thng cho trc *tt % )*T % khng n nh thng tin

Nu khng tho mn yu cu th phi chn li tuyn v tr t trm thay i ha - chiu

cao anten, tnh ton, kim tra li ti khi t yu cu.

c: Chn tuyn

Chn v tr t trm

59


60/66

2: C S TNH KIM TRA CHT LNG

1.2. Phng trnh c bn

a.P

p TT

PW ; p p a (db)

P = =

PP: Cng sut tn hiu u ra my pht

PT: Cng sut tn hiu u vo my thu

W: suy hao tn hiu

W: tng tiu hao tn hiu

Phng trnh ni ln quan h gia tn hiu u ra my pht v u vo my thu.

b.

th

Phng vtrat ra

Phng vtvoth

t vo

PNT

L i ch iu chNP

T

= = =

cho trong cc ch tiu k thut ca my

Pt ny cho bit mi quan h gia u vo v u ra my thu

2 phng trnh cho bit quan h gia u vo my pht v ?? u ra my thu.

c. Xc nh phng trnh th nng ca khong tuyn.

T b ta c:

ra vo

pth th

t t vtvo

PP PP P W P

= =

Vit di dng db

p P T Tp tvo SCT SCT A e K P P a a a a 2G P W

+ + =

V tri gm cc i lng bit k hiu l PE gi l th nng lng

V phi l i lng WKcha bit phi tnh

60


61/66

Phng trnh trn gi l phng trnh th nng lng in ca khong tuyn thng tin v

khi thit k tuyn phi tnh ton m bo c phng trnh th nng: < khng thng tin

c = khng c d tr phainh

Tip theo phi i tnh aK tho mn bt phng trnh th nng

aK l tng tiu hao khong cch gia 2 anten

aK=ak2+a a hnh + a phainh

ak2 l tiu hao trong khng gian t do = 32,4+20lgR(LM)

a a hnh: l tiu hao do nh hng a hnh mt t ca khong tuyn. Vic ny phi tnh

ton c th cho tng khong tuyn.

aphainh: tiu hao do nh hng ca phainh, l hin tng ngu nhin khng th tnh ton

c chnh xc m phi da vo l thuyt xc sut thng k v thc nghim xc dng quan

h gia tiu hao pha inh v thi gian mt lin lc do pha inh

Wpha inh = f(T%)

V th

Khi tnh ton thng phi gi thit chia u nh hng ca pha inh n v i vi cc

khong tuyn v s tnh c Wpha inh cho tng khong tuyn.

Sau khi tnh c aKK1, ah1, ap1 cho 1 khong tuyn c th thay vo phng trnh th

nng phi tho mn bt phng trnh th nng. Nu khng tho mn th phi thay i v tr t

trm thay i tuyn, ah tnh li. Nu tho mn 21 1 11 k pK(P a a a )> + +l ( c d phng)

v do pha inh chim

Tnh c d tr phainh

1 k

1

p h

p p

P P a a

P a

=

>l

T quan h Wp f(T%) s tnh c Tr%(max) thi gian gin on lin lc do nh hng

ca pha inh.

*1 1

1*

1 1

Yu cu: T % T %

Tongt tnh c H %Yu cu: H % H %

Khong tuyn t

3:TNH AH

tnh c phi bit c cu trc thc a hnh ca khong tuyn nhng bn a

hnh v trn mt phng khng cho bit c cu trc thc ca khoangr tuyn. V vy phi tin

hnh v biu mt ct.

Biu mt ct l phn giao gia b mt qu t vi mt phng qua 2 im t anten

v tm qu t v biu mt ct phi c cc dng c:

- Bn a

61


62/66

6262626262626262626262626262626262

JJ 62 62 62 62j 62

JJ JJ 62

JJ 62j 62 JJ J

J 2 62 62

62626262626262626262626262626262626262j 62 JJ JJ

62 JJ 262 62

62 JJ 262 62

62 JJ 62j 62 JJ

62 62

626262626262626262626262626262626262626262626262626262

6262626262626262626262 6262626262626262626262 6262626262626262626262

6262626262626262626262 6262626262626262626262 6262626262626262626262

6262626262626262626262 6262626262626262626262

6262626262626262626262626262 62 2 62 2

2 62 2 62 2

62 62 2

62 62 62j 62 JJ JJ

62 62 j 62 JJ JJ 62j JJ

626262626262626262626262 JJ

62 62 2 J

J 62 62 62 62j 62 62

62 62 62j 62 JJ

JJ

^6262626262 626262626262626262626262626262626262626262626

2626262626262626262626262626262626262626262626262626262626262626262626

2626262626262626262626262626262626262626262626262626262626262626262

62626262626262 6262626262626262626262 62626262626262626262626262 62

2 62 2 62 262 262

22

j

62 JJ JJ 62

6262 JJ 62 62 62

JJ62j 62 62 62

62


63/66

63 63 2 JJ

6363

j 63 63 63 63 63 63

63 63 63 JJ

26363 63j 63 JJ JJ

J63636363636363636363636363636363636363 2

J63

63j 63

JJ 63 63 63 JJ

63j 63 63 63 63 63 63

63 2 JJ v tm thi phn b c n nh u

cho cc khong tuyn.

V d: Tuyn c M khong tuyn

**1

T %T %

M=

Sau t th Wphainh = f(T%) th hnh 52 ( trang 55)

c T1% v f s tm c Wphainh

Tnh d tr th nng pha inh

21

2e h1 p1K

P W W W + +

Tho mn bt phng trnh c d tr, d tha ( nu khng tho mn th phi

tnh li)

Tnh c d tr pha inh

K

2p e h pmaxP P W W W = =

T th Wphainh = f(T%) hnh 52

Tnh ra c T1% ( cho php gin on T1% m vn b c)Tnh 1 1H % 100% T %=

*1H % H %> tho mn

c d tr th nng cho phainh th thit k tuyn l m bo

Ton tuyn c

M

11

M *1

1

T% T %

H% 100% T % H %

=

= >

63


64/66

CC BC THIT K

Bi tp:

- Trnh by cc bc v biu mt ct

- V biu mt ct ca khong tuyn theo cc s liu cho trc.

- Xc nh cc tham s ca khong tuyn

H, H0, P(g), K, Rchn

- Phng php tnh Wh cho khong tuyn v

- Phng php tnh Wpcho khong tuyn v chia u bnh qun khng n nh T*%

cho cc khong tuyn ( cho M khong tuyn)

- Phng php tnh d tr pha inh Pp

- Phng php tnh n inh thng tin H%

H% > H*%

T th 5.2

W*1p( tiu chun)

Cho 21

2e h1 p1K

P W W W + +

Tnh d trK

2p e h pmaxP P W W W = thc t

1 1N % 100% T %=

Ton tuyn

N

iH% Hi%=

Yu cu: H% > H*%

1. Cho R c ly khong tuyn

ri

Bnh

A R ri Bnh B350 59 km 320m

r1 4 Km 300r2 8 Km 200r3 12 Km 200

r4 18 Km 200r5 22 Km 200r6 28 Km 100r7 32 Km 100r8 36 Km 200r9 38 Km 300

64


65/66


66/66

tong quan ve thong tin viba 879

Documents