chuong3_vibaso

7/25/2019 chuong3_ViBaSo

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Vi ba s


2/12

N

i dung

Khi nim thng tin vi ba

Cu trc ca h thng

nh hng ca mi trng truyn sng

Cc bin php nng cao cht lng ng truyn


3/12

Khi nim thng tin vi ba

nh ngha: Thng tin vi ba s l thng tin trong tm nhn thng, s dng sng siu cao

tn, tn hiu truyn l tn hiuiu ch s. uim

Di tn 300 MHz - 30 GHz =>truync dng s tc cao Nng lng sngi tp trung, nn cng sut yu cu nh (800mW n 5

W), thit b gn nh. Hu ht cc thit b vi ba s nhp v Vit nam c tn s lm vic t 1GHZ

n 10GHZ. y l phm vi ca s v tuyn, tp m thp C th p dng cc phng thciu chphc tp, truyn sng song cng,

thch hp vi mng thng tin cng cng. Nhcim

L thng tin trong tm nhn thng anten pht " nhn thy " anten thu, nn cly thng tin cci b gii hn bi cong ca mtt.

Do truyn sng trong kh quyn nn chunh hng nhiu ca mi tr ng:s thayi ca chit sut kh quyn theo cao, nh hng ca ma, nhhng ca hin tng fadinh, suy hao do hp thbi kh quyn.


4/12

Cu trc ca h thng

Thng tin vi ba s im -im

Khuych i vpht sng

Thu sng vkhuych i

Khi GiIIu ch

x l tnhiu thu

Khi Iuch

X l tnhiu pht

Truyn sng khng gian

ng truyn nh hng


5/12

nh hng ca mi trng truyn sng

C hai phng thc truyn tn hiu t ni pht n ni

thu: Truyn sng qua khng gian S dng cc ng truyn nh hng

Thng tin vi ba s dng truyn sng qua khng gian =>

chu nh h

ng ca mi tr

ng truyn sng Cc nh hng chnh:

thay i chit sut kh quyn theo cao=>tia sng b un cong nh hng ca cong mt t lm gii hn c ly truyn sng

nh hng hp th sng ca kh quyn nh hng ca ma nh hng ca fadinh


6/12

S thay i chit sut kh quyn theo cao lm tia sng b un cong Mt khng kh gim theo cao => thay i chit

sut kh quyn t t f>30 MHz: nc trong khng kh ng vai tr ch yu cong ca tia sng ph thuc s thay i v nhit

, p sut v m. Trong iu kin truyn lan bnh

thng, ng un cong xa dn b mt tri t. C: N = (n -1).106 vi: n-chit sut kh quyn, N-ch s

chit sut

Vi f


7/12

S thay i chit sut kh quyn theo cao lm tia sng b un cong dn/dh: bin thin ca chit sut kh quyn theo cao. Khi dn/dh > 0 (chit sut kh quyn tng theo cao)=>khc x

m=>tia sng b un cong ln bu tri- quay b lm ln trn Khi dn/dh < 0 =>khc x dng=>quay b lm xung di iu kin kh quyn thng: dn/dh = - 4.108 1/m (khc x

dng)=> di ng truyn tng 15% so vi ng truyn

thng khi tia sng khng b un cong. Tng ng vi trng hpny bn knh cong ca tia sng l R= 25000 km

Khc x m Khc x dng


8/12

nh hng ca cong mt t lm giihn c ly truyn sng Mt t: mt cu bn knh a = 6378 km; chiu cao anten pht, thu: h1,

h2 => c ly thng tin cc ai trong tm nhn thng AB:

m

p dng bng s, cng thc gn ng: [km]

Xt n nh hng ca s thay i chit sut kh quyn theo cao:

)(2 21 hhaAB +=

][][75,321max mhmhAB +=

][,))()(.(15,421max

kmmhmhAB +=

h

1

h

2

A B

a


9/12

nh hng hp th sng ca kh quyn,ca ma Hp th sng ca kh quyn: sng truyn trong kh quyn b

suy hao. Trong cc iu kin khng gian t do, mc suyhao ca sng (dB):

(dB)

tn s cng cao suy hao cng ln

Ma: sng in t, c bit l i vi bc sng nh (


10/12

nh hng ca fading

Hin tng Fainh:

Gi tr trng nhn c a im thu thay i theo thigian.

Do bin ng ng truyn, giao thoa ca cc tia sng(fadinh nhiu tia), nh hng ca cc i ln cn (fadinh la

chn)


11/12

Cc bin php nng cao cht lngng truyn

Khc phc hin tng fadinh trong thng tin vi ba Phn tp theo khng gian: s dng 2 hay nhiu anten

pht hoc 2 hay nhiu anten thu thu pht cng mttn hiu trn cng mt tn s

Dng b cn bng t thch nghi Phn tp theo tn s: truyn v thu ng thi cng mt

tn hiu trn 2 hoc hn 2 knh tn s v tuyn trong

cng mt di tn. Khc phc da vo tnh ton min phn x.


12/12

Cc bin php nng cao cht lngng truyn

S dng nng lng sng cc min Fresnel bc

cao nng cao cht lng truyn sng

T

R

l

O

MinFresnenth nht

h

2

h

1

Ei

H

d1 d2

a

F1

hih1

h2 h2

h1

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