tai lieu tham khao anten va truyen song

7/31/2019 Tai Lieu Tham Khao Anten Va Truyen Song

1/147

HC VIN CNG NGH BU CHNH VIN THNG

TRUYN SNG & ANTEN

Bin son : TS. NGUYN PHM ANH DNG

THS. PHM TH THY HIN


2/147

LI NI U

Cc h thng thng tin v tuyn c bit l cc h thng thng tin di ng v ang pht

trin rt mnh m. Qu trnh truyn sng v anten l nhng phn kin thc khng th thiu khinghin cu v cc h thng ny.

Mc ch ca ti liu ny l cung cp cho sinh vin cc kin thc cn bn nht v truynlan sng v tuyn in v anten.Ti liu bao gm cc bi ging v mn hc "Truyn sng vanten" c bin son theo chng trnh i hc cng ngh vin thng ca Hc vin Cng nghBu chnh Vin thng.

Ti liu ny c xy dng trn cssinh vin hc cc mn: L thuyt trng in t,K thut siu cao tn.

Do hn ch ca thi lng nn ti liu ny ch bao gm cc phn cn bn lin quan ncc kin thc cn bn v truyn sng v anten. Tuy nhin hc k ti liu ny sinh vin c th honchnh thm kin thc ca mn hc bng cch c cc ti liu tham kho dn ra cui ti liu ny.

Ti liu ny c chia lm su chng. c kt cu hp l sinh vin c th t hc.Mi chng u c phn gii thiu chung, ni dung, tng kt, cu hi vi bi tp. Cui ti liu lp n cho cc bi tp.

Ngi bin son: TS.Nguyn Phm Anh Dng

ThS. Phm Th Thy Hin

i


3/147

Chng 1: Cc vn chung v truyn sng

CHNG 1

CC VN CHUNG V TRUYN SNG

1.1 GII THIU CHUNG

1.1.1 Cc chc trnh by trong chng

- S phn cc ca sng v tuyn in

- Phn chia sng v tuyn in theo tn s v bc sng

- Cc phng php truyn lan sng trong mi trng thc

- Cng thc truyn sng trong khng gian t do

1.1.2 Hng dn

- Hoc k cc phn c trnh by trong chng

- Tham kho thm [1], [2], [3]

- Tr li cc cu hi v bi tp

1.1.3 Mc ch ca chng

- Nm c cc dng phn cc ca sng v tuyn in v cc bng sng v tuyn

- Hiu v cc phng php truyn lan sng trong mi trng thc

- Nm c cch tnh ton cc tham s khi truyn sng trong khng gian t do

1.2 NHC LI MT S TNH CHT CBN CA SNG IN T.

Sng in t bao gm hai thnh phn: in trng, k hiu E (V/m) v t trng, k hiu

H (A/m). Chng c quan h mt thit vi nhau trong qu trnh truyn lan v c m t bng hphng trnh Maxwell, vit cc dng khc nhau.

Gi s ta xt mt sng phng truyn lan trong mi trng in mi ng nht v ng

hng c cc tham s: h sin mi v h s t thm , khi khng c dng in v in tchngoi, th h phng trnh Maxwell biu th mi quan h gia in trng v t trng c vitdi dng vi phn nh sau:

yx

yx

HE

t z

HE

t t

=

=

(1.1)

3


4/147


Nghim ca h phng trnh ny cho ta dng ca cc thnh phn in trng v t trngl mt hm bt k.

xE Fz

t F tv v

= + +

1 2

z

(1.2a)

y Gz

H t G tv v

= + +

1 2

z

(1.2b)

Trong : F1, F2, G1, G2 l cc hm sng ty .

vz

t=

=

1(m/s) l vn tc pha ca sng.

T (1.2) ta c : G1 = F1/ Z v G2 = F2/ Z vi Z =

() l trkhng sng ca mi

trng.

Nu mi trng truyn sng l chn khng (cn c gi l khng gian t do) cc thams ca mi trng c gi tr:

0 = 109/36 (F/m) ; 0 = 4.10

-7 (H/m)

Do : v . (m / s) c= =

8

0 0

13 10 = (vn tc nh sng)

Z = =0

0

0

120 ()

Trong thc t sng in t thng bin i iu ha theo thi gian. i vi cc sng in

t phc tp ta c th coi n l tng v s cc dao ng iu ha, ngha l c th p dng phpphn tch Fourier biu th. Trong trng hp ny khi gi thit ch c sng thun, tc l sngtruyn t ngun theo phng trc z v mi trng m khng c sng nghch th cc thnh phnin trng v t trng c biu th nh sau:

( ) ( )m mE E zcos t E cos t kzv= =

( ) ( )m mH E Ezcos t cos t kzvZ Z= = (1.3)

Trong k = /v = 2/ gi l h s pha hay hng s sng.

Sng in t c mt cng sut ( hay cn gi l thng lng nng lng), c biu th bi vc

t nng lng . Nh vy sng in t c cc vc tk [E H]= G G G

EG

v HG

nm trong mt phng

vung gc vi phng truyn sng . Bi vy sng in t truyn i trong mi trng ng nhtng hng l sng in t ngang TEM.

kG

4


5/147


Hnh 1.1. Struyn lan sng in t

1.3 SPHN CC CA SNG V TUYN IN.

Trng in t ca sng v tuyn in khi i trong mt mi trng s dao ng theo mthng nht nh. Phn cc ca sng in t chnh l hng dao ng ca trng in t. Vic sdng cc phn cc khc nhau ca sng in t c mt ngha rt ln trong vic s dng hiu qutn s trong thng tin v tuyn.

Trng ti vng xa ca anten c dng sng phng TEM v c xc nh bng vect

Pointing: . iu ny c ngha l cc vectk [E H]= G G G

EG

v HG

nm trong mt phng vung gc

vi phng truyn sng k .G

Phng ca ng do u mt ca vc t trng in v ln s xc nh phn cc sng.Trng in v trng t l cc hm thay i theo thi gian. Trng t thay i ng pha vitrng in v bin ca n t l vi bin ca trng in, v th ta ch cn xt trng in.C ba loi phn cc sng v tuyn in: phn cc thng, phn cc trn v phn cc elip.

1.3.1 Phn cc thng.

Hu ht truyn dn v tuyn s dng phn cc tuyn tnh, trong phn cc ng c

gi l phn cc trong trng in vung gc vi mt t v phn cc ngang c gi l phncc trong trng in song song vi mt t. Gi thit rng phng ngang v ng c coi ltrc x v y (hnh1.2a). Ti mt im no trong khng gian, vect trng ca sng c biuth bi cc thnh phn thng ng v nm ngang nh sau:

= EyEG

yaG

ysint (1.4)

= ExEG

xaG

xsint (1.5)

trong , l cc vectn v trong phng ng v phng ngang; EyaG

xaG

y, Ex l gi trnh

(hay bin ) ca trng in trong phng ng v phng ngang.

5


6/147


Trng tng s l vectE hp vi trc ngang mt gc c xc nh nh sau:

y

x

Earctan g

E = (1.6)

Trong trng hp ny ta thy vect khng bin i. di ca vect thay i theo thi giannhng u mt ca vectlun nm trn ng thng cnh trng vi phng ca vectc gc

nghing (hnh 2c). l hin tng phn cc ng thng ca sng in t. Khi = 0

EG

0 ta c

sng phn cc ngang, lc ny vect lun song song vi mt t; cn khi = 90EG

0 ta c sng

phn cc ng, vect lun vung gc vi mt y.EG

yaG

xaG

y x| | E E= +2 2

G

E

Hnh 1.2. Cc thnh phn ngang v ng ca phn cc thng

1.3.2 Phn cc trn

Khi cc thnh phn thng ng v nm ngang c bin bng nhau ( k hiu l E0) nhngmt trng nhanh pha hn 900. Cc phng trnh th hin chng trong trng hp ny nh sau:

= EyEG

yaG

0 sint (1.7a)

= aE

G

x

G

E0 cost (1.7b)

p dng ptr. (1.6) cho trng hp ny ta c =t. Bin vect tng l E0. Trong

trng hp ny, vect c bin khng i nhng hng ca n thay i lin tc theo thi

gian vi quy lut t. Ni cch khc, vect

EG

EG

quay quanh gc ca n trong mt phng xy vi vn

tc . u mt ca vecttrng in v ln ng trn c bn knh bng di vect. l hintng phn cc trn.

6


7/147


t =0

90

t =0

270

t =0

180 t

im nhn theo

IEEE

z

RHC

im nhn theo

IEEE

z

LHC

Hnh 1.3. Phn cc trn

Hng ca phn cc trn c nh ngha bi phng quay ca vectin nhng iuny i hi ta phi quan st c chiu quay ca vect. Theo nh ngha ca IEEE th phn cc trntay phi (RHC) l phn cc quay theo chiu kim ng h khi nhn dc theo phng truyn sng(hnh 1.3), cn phn cc trn tay tri (LHC) l phn cc quay ngc chiu kim ng h khi nhn

dc theo phng truyn sng. Phng truyn sng dc theo trc z dng.

1.3.3 Phn cc elip

Trong trng hp tng qut hn sng in t c dng phn cc elip. iu ny xy ra khihai thnh phn tuyn tnh l:

= EyEG

yaG

ysint (1.8a)

E = EG

xaG

xcos(t+) (1.8b)

7


8/147


T s sng phn cc elip l t s gia trc chnh v trc ph ca elip. Phn cc elip trcgiao xy ra khi mt sng c cng t s phn cc nhng phng quay ngc chiu.

1.4 PHN CHIA SNG V TUYN IN THEO TN S V BC SNG

1.4.1 Nguyn tc phn chia sng v tuyn in

Sng in t ni chung c ng dng rt rng ri trong i sng nhiu lnh vckhc nhau nh y hc, quc phng, thm d ti nguyn khong sn, nghin cu v tr, thng tinlin lc...Da vo tnh cht vt l, c im truyn lan phn chia sng v tuyn in thnh cc

bng sng khc nhau.

Sng cc di: Nhng sng c buc sng ln hn 10.000 m (tn s thp hn 30 kHz).

Sng di: Nhng sng c buc sng t 10.000 n 1.000 m (Tn s t 30 n 300 kHz)

Sng trung: Nhng sng c buc sng t 1.000 n 100 m (Tn s t 300 kHz n 3MHz)

Sng ngn: Nhng sng c buc sng t 100 n 10 m (Tn s t 3 n 30 MHz). Sdng cho thng tin pht thanh iu tn, truyn hnh.

Sng cc ngn: Nhng sng c buc sng t 10 m n 1mm (Tn s t 30 n 300.000MHz).Sng cc ngn c chia nh hn thnh mt s bng tn s .

Tip n l cc bng sng gn nh sng, hng ngoi, nh sng trng, tia cc tm, tia X

Khong tn s t 30 Hz n 3000 GHz c chia thnh 11 bng tn nh trong bng 1.1

1.4.2. Cc bng sng v tuyn in v ng dng

Mi bng sng c ng dng cho cc h thng thng tin khc nhau do c im truynlan sng trong cc mi trng thc.

Bng sng cc di s dng lnh vc vt l, thng tin v tuyn o hng, thng tin trnbin.

Bng sng di v bng sng trung c s dng cho thng tin pht thanh ni a, iubin; thng tin hng hi; v tuyn o hng.

Bng sng ngn s dng cho pht thanh iu bin c ly xa v mt s dng thng tin cbit.

Bng sng mt c s dng cho pht thanh iu tn v truyn hnh.

Bng sng decimt c s dng cho truyn hnh, cc h thng thng tin vi ba s bnghp, thng tin di ng.

Bng sng centimt c s dng cho thng tin vi ba s bng rng, thng tin v tinh.

Bng sng milimt c s dng hn ch cho thng tin v tinh vi bng Ka, dng chothng tin v tr.

8


9/147


Bng 1.1

Tn bng tn (Bng sng) K hiu Phm vi tn s

Tn s v cng thp ULF 30 - 300 Hz

Tn s cc thp ELF 300 - 3000 Hz

Tn s rt thp VLF 3 - 30 kHz

Tn s thp (sng di) LF 30 - 300 kHz

Tn s trung bnh (sng trung) MF 300 - 3000 kHz

Tn s cao (sng ngn) HF 3 - 30 MHz

Tn s rt cao (sng mt) VHF 30 - 300 MHz

Tn s cc cao (sng decimet) UHF 300 - 3000 MHz

Tn s siu cao (sng centimet) SHF 3 - 30 GHz

Tn s v cng (sng milimet) EHF 30 - 300 GHz

Di milimet 300 - 3000 GHz

1.5 CC PHNG PHP TRUYN LAN SNG TRONG MI TRNG THC.

Slc v bu kh quyn.

Bu kh quyn ca tri t c chia lm 3 vng chnh: tng i lu, tng bnh lu v tngin ly. Bin gii gia cc tng ny khng r rng v thay i theo ma v theo vng a l. Tnhcht ca cc vng ny rt khc nhau.

Tng i lu l khong khng gian tnh t b mt tri t ln n cao 6 n 11 km.Nhit ca khng kh trong tng i lu thay i theo cao (nhit gim khi cao tng).V d nhit trn b mt tri t l 100C c th gim n -550C ti bin trn ca tng i lu.

Tng bnh lu bt u t bin trn ca tng i lu v c phm vi khong 50 km. cim ca tng ny l nhit hu nh khng thay i theo cao.

Tng in ly tn ti cao khong t 60 km n 600 km. Lp kh quyn tng ny rtmng v b ion ha rt mnh ch yu l do bc x ca mt tri, ngoi ra cn c bc x ca cc vsao, cc tia v tr, chuyn ng ca cc thin thch to thnh mt min bao gm ch yu l ccin t t do v cc ion.

Bn cnh , do tnh cht vt l ca mi bng sng m mi bng sng c phng thc truynlan thch hp t c hiu qu nht.

Do , ty theo mi trng truyn sng c bn phng thc truyn lan sau: truyn lan sngb mt, truyn lan sng khng gian, truyn lan sng tri (sng in ly), v truyn lan sng t do.Sng b mt v sng khng gian u c gi l sng t (cng truyn lan trong tng i lu) tuynhin chng c s khc nhau r rt.

1.5.1 Truyn lan sng b mt

Sng b mt truyn lan tip xc trc tip vi b mt tri t. B mt qut l mt mitrng bn dn in, khi mt sng in t bc x t mt anten t thng ng trn mt t, cc

9


10/147


ng sc in trng c khp kn nhdng dn trn b mt qut nh ch ra hnh 1.4.Nu gp vt chn trn ng truyn lan, sng s nhiu x qua vt chn v truyn lan ra pha sauvt chn.

ThuPhtA B

Hnh 1.4: Qu trnh truyn lan sng b mt

Nh vy s truyn lan sng b mt c th dng truyn tt c cc bng sng. Tuy nhin,sng b mt b suy gim nhiu do s hp th ca tri t. S suy gim ph thuc vo tn s, khitn s tng th s suy gim cng ln. Hn na kh nng nhiu x qua vt chn trn ng truyn

ph thuc vo cao tng i ca vt chn so vi bc sng.Vi cc loi t c dn in ln nh mt bin, t m th sng t b suy hao trong t,

lm cho cng trng ti im thu tng ln. Cc sng v tuyn in c bc sng ln khnng nhiu x mnh v b mt t hp th nh. Bi vy sng b mt c s dng truyn lancc bng sng di v sng trung nh trong h thng pht thanh iu bin, hay s dng cho thngtin trn bin

1.5.2 Truyn lan sng khng gian

Lp kh quyn bao quanh qut c cao t 0 n 11km (vi tng i lu tiu chun),gi l tng i lu. Cc hin tng thi tit nh sng m ma, bo, tuyt... u xy ra trong tngi lu v nh hng rt ln n qu trnh truyn lan sng v tuyn in.

Nu hai anten thu v pht t cao (nhiu ln so vi bc sng cng tc) trn mt t thsng c th truyn trc tip t anten pht n anten thu, hoc phn x t mt t nh ch ra hnh1.5a, hoc li dng s khng ng nht ca mt vng no trong tng i lu tn x sng vtuyn dng cho thng tin gi l thng tin tn x tng i lu nh ch ra hnh 1.5b. Cc phngthc thng tin nh trn gi l truyn lan sng khng gian hay sng tng i lu.

Phng thc truyn lan sng khng gian thng c s dng cho thng tin bng sngcc ngn (VHF, UHF, SHF), nh truyn hnh, cc h thng vi ba nh h thng chuyn tip trnmt t, h thng thng tin di ng, thng tin v tinh... Phng thc truyn lan sng khng giansc nghin cu k trong chng II

Hnh 1.5: Truyn lan sng khng gianPht Thu

a)

ThuPht

Vngkhng

ng nhtb)

10


11/147


1.5.3 Truyn lan sng tri

Lp kh quyn cao khong 60 km n 600 km b ion ho rt mnh ch yu do nnglng bc x ca mt tri, to thnh mt lp kh bao gm ch yu l in t t do v cc ion. Lpkh quyn c gi l tng in ly. Tnh cht c bit ca tng in ly l trong nhng iu kin

nht nh c th phn x sng v tuyn in. Li dng s phn x s dng cho thng tin vtuyn bng cch phn x mt hoc nhiu ln t tng in ly, nh ch ra hnh 1.6. Phng thc gi l phng thc truyn lan sng tri hay tng in ly.

Pht Thu

Tng in li

Phn x nhiu ln t tng in li

ThuPht

Khuch tn t tng in ly

Tng in ly

Hnh 1.6: Truyn lan sng tri

1.5.4 Truyn lan sng tdo

Trong mt mi trng ng nht, ng hng v khng hp th v d nh mi trng chn khng,sng v tuyn in khi truyn lan tim pht n im thu si theo ng thng, nh ch ra

trn hnh 1.7, khng nh hng n qu trnh truyn sng.

Trong thc t mt mi trng l tng nh vy ch tn ti ngoi khong khng v tr. Vi lpkh quyn qut ch trong nhng iu kin nht nh, khi tnh ton cng c th coi nh mitrng khng gian t do.

Mc tiu tron

Trm trn mt

Hnh 1.7 Struyn lan sng tdo

11


12/147


Hnh 1.8 cho ta thy cc phng php truyn lan sng trong mi trng thc ca kh quyn qut

Sng t do

Sn tr i

Sng t Mt t

Khn ian t do

Tng in ly

Tng bnh lu

Hnh 1.8 Cc phng thc truyn sng v tuyn in

1.6 CNG THC TRUYN SNG TRONG KHNG GIAN TDO

1.6.1 Mt thng lng cng sut, cng in trng

Gi thit c mt ngun bc x v hng (ng hng) c cng sut pht PT(W) t ti

im A trong mt mi trng khng gian t do l mi trng ng nht ng hng v khnghp th, c h sin mi tng i ' = 1. Xt trng ti mt im M cch A mt khong r (m).

Hnh 1.9: Bc x ca ngun bc x v hng trong khng gian tdo

V ngun bc x l v hng, mi trng ng nht v ng hng nn nng lng sngin t do ngun bc x s ta u ra khng gian thnh hnh cu. Nh vy mt cng sut (mt thng lng nng lng) im M cch ngun mt khong r sc xc nh bng cng thcsau:

TiSP

r= 24 (W/m2) (1.9)

Theo l thuyt trng in t ta c:

Tng i luSng khng gian

1 m2r

A M(PW

10


13/147


i hS E (W/mhH= 2) (1.10)

hh

EH = (A/m) (1.11)

120

Trong : Eh (V/m), Hh (A/m) l gi tr hiu dng ca cng in trng v t

trng ; 120 l trkhng sng ca khng gian t do ()

Thay cng thc (1.11) vo (1.10) c

hi

ES = (W/m

2

120

2) (1.12)

So snh cng thc (1.12) v (1.9) ta c

ThE (V/m) (1.13)Pr

= 230

Nhn xt: cng in trng ca sng v tuyn in truyn lan trong mi trng ngnht ng hng v khng hp th t l thun vi cn hai cng sut bc x, t l nghch vikhong cch. Khong cch tng th cng trng gim v nng lng sng to rng ra khnggian, cn gi l s khuych tn tt yu ca sng. hn ch s khuch tn ny ngi ta s dngcc b bc x c nng lng tp trung v hng cn thng tin lm tng cng trng ln. chnh l cc anten c hng, vi h s hng tnh D hoc h s khuch i G.

Nu ngun bc x c hng, lc nng lng ca sng v tuyn in sc tp trung

v hng im M c biu th bng h s hng tnh hay h s khuch i nh ch ra trn hnh1.10.

bc x vh

ng M

A

Hnh1.10: Ngun bc x c hng

Trong trng hp ny mt cng sut c xc nh theo cng thc

T TSP G

= (W/mr 24

2) (1.14)

khi cng in trng sc tnh theo cng thc:

T ThE = (V/m) (1.15)P G

r30

11


14/147


Nu sng in t do ngun bc x bin i iu ho theo thi gian, ngha l theo quy lut

sint, cost, hoc vit di dng phc s eit th gi tr tc thi ca cng in trng scbiu th bi cng thc

( ) (T TEP G

t cosr

= 60

)t kr (V/m) (1.16)

Trong : tn s gc ca sng

k = /c =2/ h s sng (h s pha)

Nu vit dng phc cng thc (1.16) c dng:

( )(j t krT TE

P Gt e

r

=

60 )(V/m) (1.17)

Nu biu th c ly r (km), cng sut pht PT(kW), ta s c gi tr hiu dng ca cng

trng:

( )

( )

TT kW

h

km

EP G

r=

173

(mV/m) (1.18a)

Bin ca trng l

( )

( )

TT kW

m

km

EP G

r=

245

(mV/m) (1.18b)

v gi tr tc thi ca cng trng

( )( )

( )

(TT kW j t kz

km

EP G

t er

=

245 )(mV/m) (1.19)

1.6.2 Cng sut anten thu nhn c

Trong khi tnh ton tuyn ta cn phi xc nh cng sut anten thu nhn c PRa vo

u vo ca my thu sao cho my thu c th lm vic c. Cng sut anten thu nhn c bngmt thng lng cng sut ti ni t anten thu nhn vi din tch hiu dng ca anten thu:

PR= S.Ah (W) (1.20)

Din tch hiu dng c anten thu bng din tch thc t nhn vi hiu sut lm vic Ah = A. a.Trong cc h thng thng tin v tuyn s dng anten gng parabol trn xoay quan h gia tnhhng v din tch hiu dng ca anten c cho bi cng thc

Rh

GA

=

2

4 (m2

) (1.21)

Thay cng thc (1.14) v (1.21)vo cng thc (1.20) ta c

12


15/147


( )T T R

RP G G

Pr

=

2

2

4

(W) (1.22)

Nu biu th c ly r (km), cng sut bc x PT(kW), ta c cng thc

( ) ( )

( )

T RT kW mR

km

P G GP , .r

=2

3

26 33 10 (mW) (1.23)

1.6.3 Tn hao truyn sng

Khi sng v tuyn in truyn trong mt mi trng, ngoi tn hao do mi trng gy ranh b hp th trong cc phn t kh, trong hi nc..., tn hao do tn x do my ma, tn hao dovt chn v.v... th s suy hao ln nht chnh l do s khuych tn tt yu ca sng ra mi phng

v c gi l tn hao khng gian t do.Nu ta bc x ra mi trng mt cng sut PT, anten thu ch nhn c mt cng sut PR,

th h s tn hao truyn sng c nh ngha bng t s ca cng sut bc x trn cng sut antenthu nhn c, c biu th bng biu thc:

( )TR T R

rPL

P G G

= =

2

2

4(s ln) (1.24)

Trong trng hp khng c tc ng tnh hng ca ngun, ngha l GT=1, GR=1, tn haoc gi l tn hao truyn sng cbn trong khng gian t do, v bng:

( )rL

=

2

0 2

4(s ln) (1.25)

Tnh theo n v dB ta c:

10log10L = 20log10(4r) - 20log10 - 10log10GT - 10log10GR

L(dB)= 20log10(4r) - 20log10 - GT(dBi) - GR(dBi) (1.26)

v:

L0 = 20log10(4r) - 20log10 (dB) (1.27)

L0 = 20log10r(km) + 20log10f(GHz) + 92,45 (dB) (1.28)

L0 = 20log10r(km) + 20log10f(MHz) + 32, 5 (dB) (1.29)

1.7 NGUYN L HUYGHEN V MIN FRESNEL

1.7.1 Nguyn l Huyghen

13


16/147


hiu r mt sc im truyn lan ca sng trn mt t cn bit nhng khi nim vmin Fresnel. Vic biu th min c da trn nguyn l Huyghen.

Nguyn l Huyghen cho bit rng mi im ca mt sng gy ra bi mt ngun bc x scp c thc coi nh mt ngun sng cu th cp mi. V vy nguyn l ny cho php ta cth tnh trng in tmt im bt k trong khng gian khi bit c trng mt b mt

no . Gi s ngun ca sng scp t ti im A (hnh 1.11). K hiu S l mt mt kn bt kbao quanh ngun sng. By gi ta xc nh trng ca sng ti im bt k nm ngoi mt kn,theo cc tr s ca trng trn mt S.

S r

A

M

Hnh 1.11: Xc nh trng theo nguyn l Huyghen

K hiu l thnh phn ca trng cn tm ti im M v S l tr s ca thnh phn ytrn mt S. Khong cch t mi im trn mt S n M c k hiu l r.

Theo nguyn l Huyghen, trng th cp to bi mt nguyn t b mt dS ti im M c

xc nh theo cng thcikr

S

ed A d

r

= S (1.30)

Trong A l mt h s t l.

Trng tng ti im M s l trng to bi ton mt Sikr

SS

eA

r

= dS (1.31)

Nu S l mt phng th

( )i

A cos n,r =

n l php tuyn ngoi ca mt phng;

l bc sng cng tc

Thay vo (1.31) ta c cng thc Huyghen i vi mt phng

( )

ikr

SS

i ec ros n,r

= dS (1.32)

14


17/147


Trong trng hp mt S c dng bt k, cng thc Huyghen c dng tng qut

ikr ikr S

S

1 e edS

4 n r r n

=

(1.33)

Nguyn l Huyghen cng nu ln rng nng lng t mi im truyn theo tt c cc hngv to thnh nhiu mt sng cu scp c gi l cc sng con. ng bao ca cc sng conny s to ra mt mt sng mi. Vi chnh xc cao, mi mt sng c thc biu din bimt mt phng c php tuyn chnh l vc tmt thng lng nng lng k (hnh 1.12, ngAAc coi l v tr bt u ca sng). Cc sng scp bt ngun t mi im trn AA to ramt mt sng mi BB. Mt BB c v tip tuyn vi tt c cc sng scp c cng bn knh.

Nh ch ra trong hnh cc sng th cp bt ngun t cc im dc AA c bin khng cng t

l theo tt c cc hng. Nu gi l gc gia hng ca im C bt k trn mt cu scp v

vc tphp tuyn ca mt sng th bin sng scp theo hng s t l vi (1+ cos). Nhvy bin sng theo hng k s t l vi (1+ cos0) = 2, cn trong hng khc bin s nh

hn 2. Sng ngc tr li c bin bng 0 v (1+ cos) = 0. Do khng c sng truyn theohng ngc tr li. Cc sng truyn v pha trc theo hng php tuyn vi mt sng. S saikhc pha gia cc dao ng ti cc im ln cn ca cc ng AA v BBph thuc vo khong

cch r gia chng theo t l k.r = 2r/. Nu r = th tt c cc im ca AA v BB s dao ng

cng pha, cn nu r = /2 th cc im s dao ng ngc pha.

C

A

A

B

B

Hnh 1.12. Biu din nguyn l Huyghen trong khng gian tdo

1.7.2Min Fresnel

15


18/147


Nguyn l Huyghen cho php xc nh phn khng gian thc s tham gia vo qu trnhtruyn lan sng. Gi s c mt ngun bc xc t ti im A v my thu c t tiim B. Ly A lm tm, ta v mt hnh cu bn knh r1. Hnh cu ny l mt trong s cc mtsng. Trn hnh 1.13 k hiu r2 l khong cch t B n mt cu bn knh r1. T B v mt h

cc ng thng ct mt cu cc im cch B mt khong bng r2 +/2. H cc ng

thng ny s to thnh mt hnh chp nn ct mt cu ti N1 v N1

. Bng cch tng t ta lpcc mt nn bc cao c

BN1 = + /22r

BN2 = + 2/2, 2r

BNn = + n/22r

Giao ca cc mt nn vi mt cu l cc ng trn ng tm. Min gii hn bi ccng trn gi l min Fresnel. Mim gii hn bi ng trn N1 l min; min gii hn bi cc

ng trn N1 v N2 l min Fresnel th hai(Min Fresnel bc cao)p dng nguyn l Huyghen, ta coi mt cu l tp hp nhng ngun im th cp v ta

tnh trng to bi nhng ngun y ti im B. Cc ngun im th cp trong min Fresnel th

nht s to ra trng ti B c pha khc pha vi trng do im N0 to ra B mt gc < 1800.Pha ca trng to bi ngun im th cp trong min Fresne bc hai khc pha vi trng do

im N0 to ra B mt gc 1800 < < 3600. Mt cch tng qut c th thy rng Pha ca

trng to bi min Fresne bc hai khc pha vi trng to bi min Fresnel th nht 1800. Phaca trng to bi min Fresne bc ba khc pha vi trng to bi min Fresnel th hai 1800 ...s khc nhau y c biu th bi cc du cng, tr trn hnh v.

Nn

1r

A B

Nn

N0

2r

N0

N1N2

N4

+++ + +

++

+

+

+

+

+

+

+

+ ++

+

+

-

-- -

-

--

--

--

-

- -

-

-

--

N3

N1

N1

Hnh 1.13: Nguyn l cu to min Fresnel trn mt sng cu

16


19/147


Ngi ta chng minh c rng tc dng ca cc min Fresnel bc cao nm k nhau s btr cho nhau do pha ca chng ngc nhau nn cui cng tc dng tng hp ca tt c cc minFresnel bc cao gn nh ch tng ng tc dng ca khong na min Fresnel th nht. Nhvy, khong khng gian c tham gia vo qu trnh truyn sng c th xem nhc gii hn bimt na min Fresnel th nht.

Cc vng Fresnel c thc xy dng trn cc b mt c hnh dng bt k. thun tinta chn b mt lp min Fresnel l mt phng S0. Mt phng ny vung gc vi phng truynlan AB (hnh 1.14)

1r

n

A

bn

2r

S0

0

B

Hnh 1.14: Xc nh bn knh min Fresnel

Theo nh ngha ta c:

n n 1 2AN + BN = r + r + n

2

Mt khc ANn v BNn c thc xc nh theo hnh hc

22 nn 1

22 nn 2

bb r

br b r

2

n 1

1

2

n 2

2

AN = r2r

BN =2r

+ +

+ +

Ta c bn knh min Fresnel tnh gn ng bng

1 2

1 2

r rb n

r rn=

+(1.34)

i vi vng Fresnel th nht, n = 1, nn

1 2

1 2

r rb

r r1=

+(1.35)

Khi ta dch chuyn mt phng S0 dc theo ng truyn lan t A n B, gii hn ca minFresnel s vch ra mt mt elipsoit. y, ta ch xt min th nht. Ta c

c1 1 1 2AN + BN = r + r + osnt2

=

17


20/147


y chnh l phng trnh ca hnh elipsoit vi cc tiu im A v B (hnh 1.15). Khong khnggian tham gia vo qu trnh truyn lan sng c gii hn bi min Fresnel th nht. Tronghnh v, khong khng gian ny c nh du bi cc ng k song song.

b1max

BA

Hnh 1.15: Vng tham gia vo qu trnh truyn lan sng

1.8 TNG KT

Chng ny xt cc vn chnh lin quan n qu trnh truyn lan sng v tuyn. Thnht l v phn cc ca sng v tuyn in, vic s dng cc phn cc khc nhau ca sng in tc ngha rt ln trong vic s dng mt cch hiu qa tn s trong thng tin v tuyn. Th hai lv cch phn chia cc bng sng v tuyn v ng dng. Th ba cp n cc phng phptruyn lan sng, c bn phng php truyn lan sng trong mi trng thc l: truyn lan sng

b mt, truyn lan sng khng gian, truyn lan sng tri v truyn lan sng t do. Mi phng

thc truyn sng sc s dng truyn lan cho bng sng nht nh t c hiu qu lnnht. Trong chng cng a ra cc cng thc tnh ton cc thng s cbn ca qu trnh truynsng l mt cng sut, cng in trng, cng sut nhn c ti im thu, v tn haotruyn sng. Cui cng cp n khi nim min Fresnel v t xc nh khong khng giantrc tip tham gia vo qu trnh truyn lan sng tim pht n im thu c gii hn bi mtna min Fresnel th nht.

1.9 CU HI V BI TP

1.Trnh by cc dng phn cc ca sng v tuyn in.

2. Trnh by cc phng php truyn lan sng trong mi trng thc.

3. Xy dng cng thc tnh mt cng sut bc x v cng in trng khi truyn sngtrong mi rng khng gian t do.

4. Pht biu nh ngha v vit biu thc tnh t hao truyn sng.

5. Trnh by v min Fresnel.

6. Mt tri c cng sut bc x theo mi hng khong 3,85.1020 W, khong cch nh nht t qu

t n mt tri l 147.098.090 km (vo thng ging) v ln nht l 152.097.650 km. Tnh:- Mt cng sut bc x cc tiu v cc i ca mt tri ln b mt qut?

18


21/147


- Mt cng sut bc x mt tri khong cch trung bnh v t l phn trm sai s cabc x cc i v cc tiu so vi gi tr trung bnh?

7. Mt my pht c cng sut 3 W, anten pht c h s khuch i l 30 dBi. c ly 40 km tmt anten thu c din tch hiu dng l 3,5 m2, hiu sut lm vic 100%. Tnh cng sut sngmang nhn c anten thu.

(a) 0,164.10-5 W; (b) 0,164.10-4 W; (c) 0,154.10-5 W ; (d) 0,154.10-4 W8. Xc nh cng sut my pht cn thit thc hin tuyn thng tin c cc iu kin: c lythng tin 50 km, tn s cng tc 2GHz, h s khuych i ca anten thu v anten pht l 30 dBi,cng sut anten thu nhn c l 10-6W.

(a) 1,63W; (b) 2,63W; (c) 3,63W; (d) 4,63W

9. Mt my pht c cng sut 50 W. Biu din cng sut my pht sang n v dBm v dBW?

(a) 15 dBW v 45 dBm; (b) 16 dBW v 46 dBm; (c) 17 dBW v 47 dBm; (d) 18dBW v 48 dBm

10. Cng sut bi 9 c cp cho anten v hng lm vic vi sng mang c tn s 900 MHz,tm cng sut thu (tnh theo dBm) ti im cch anten pht mt khong 10 km. Gi s anten thuc h s khuch i l 2 v sng truyn trong khng gian t do.

(a) - 45,5 dBm; (b) - 51,5 dBm; (c) - 55,5 dBm; (d) - 61,5 dBm

11. S liu nh bi 9 v 10, tnh bin cng in trng hiu dng ti im t anten thu.

(a) 2,9 mV/m; (b) 3,9 mV/m; (c) 4,9 mV/m; (d) 5,9 mV/m

12. Tnh tn hao khi truyn sng trong khng gian t do (theo dn v dB) bit c ly truyn sng50 km, tn s cng tc 2 GHz, vi anten v hng.

(a) 132,5 dB; (b) 135,5 dB; (c) 142,5 dB; (d) 145,5 dB

13. S liu nh bi 12, nu c hai anten c h s khuych i l 30 dBi th tn hao l bao nhiu?

(a) 72,5 dB; (b) 75,5 dB; (c) 82,5 dB; (d) 85,5 dB

14. Mt ngun v hng c cng sut bc x 100W. Mi trng truyn sng l khng gian t do.Hy xc nh:

a, Mt cng sut ti im cch xa ngun 1000 m.

(a) 6,96 W; (b) 6,96 mW; (c) 7,96 W; (d) 7,96 mW

b, Mt cng sut ti im cch xa ngun 20 km.

(a) 19,9 pW; (b) 19,9 W ; (c) 20,9 pW; (d) 20,9 W

15. Xc nh cng in trng hiu dng ti im thu vi cc gi thit cho trong bi 13.

(a) 44,7 mV/m v 1,74 mV/m; (b) 44,7 mV/m v 2,74 mV/m; (c) 54,7 mV/m v 1,74 mV/m; (d)

54,7 mV/m v 2,74 mV/m

16. Xc nh mt cng sut ti im cch anten 30 km ca mt anten c cng sut bc x 5 Wv h s khuch i ca anten l 40 dBi.

(a) 4,42 pW; (b) 4,42 W; (c) 5,42 pW; (d) 5,42 W

17. Mt anten pht c h s khuych i 30 dBi, hiu sut lm vic 60%. c cng intrng hiu dng ti im thu cch anten pht 100 km bng 3,46 mV/m th cn phi a vo antencng sut l bao nhiu? Vi iu kin sng truyn trong khng gian t do.

(a) 3 W; (b) 3,5W; (c) 4 W; (d) 5 W

19


22/147

Chng2: Truyn lan sng cc ngn

CHNG 2

TRUYN LAN SNG CC NGN

2.1 GII THIU CHUNG


- Cc phng php truyn lan sng cc ngn

- Truyn sng trong gii hn nhn thy trc tip vi cc iu kin l tng

- Truyn sng trong gii hn nhn thy trc tip khi kn nh hng ca a hnh

- nh hng ca tng i lu khng ng nht

2.1.2 Hng dn


- Tham kho thm [1], [2], [3]


2.1.3 Mc ch ca chng

- Nm c cc phng php truyn lan sng cc ngn

- Nm c cc cng thc tnh ton trng khi truyn sng trong gii hn nhn thy trctip vi iu kin l tng v trong cc iu thc t (c xt n nh hng ca a hnh vca tng i lu)

- Hiu v hin tng pha inh v bin php chng

2.2 CC PHNG PHP TRUYN LAN SNG CC NGN

Nh gii thiu chng 1, sng cc ngn l nhng sng c tn s t 3 MHz n 300GHz (ng vi bc sng nh hn 10 m) v c chia thnh 4 bng:

Sng mt: bc sng t 10 m n 1m (30 - 300 MHz)

Sng decimt: bc sng t 1m n 10 cm (300 - 3000 MHz)

Sng centimt: bc sng t 10 cm n 1cm (3000- 30.000 MHz)

Sng milimtt: bc sng ngn hn 1cm (tn s cao hn 30.000 MHz)

2.2.1 Truyn sng do khuch tn trong tng i lu

20


23/147


Tng i lu l lp kh quyn tri t b mt tri t ln n cao khong 8 - 10 km vtuyn cc, khong 10 - 12 km cc vtuyn trung bnh v 16 - 18 km vng nhit i. Tng ilu l mt mi trng c cc tham s thay i theo thi gian v khng gian. Cc hin tng khtng nh ma, bo, tuyt... u xy ra trong tng i lu. Bi vy tng i lu l mt mitrng khng ng nht. Nu mt vng no trong tng i lu khng ng nht vi mi

trng xung quanh, theo nguyn l quang, mt tia sng i vo vng khng ng nht s k khuchtn ra mi pha. S tuyn thng tin theo phng thc tn x tng i lu c vhnh 2.1

A B

C

CD V

Hnh 2.1. S khuch tn sn tron tn i lu

Gi s anten pht t ti A, gin tnh hng ca n c gii hn bi hai ng AC vAC1 v chim mt th tch nht nh ca tng i lu. An ten thu t ti B, gin tnh hng can c gii hn bi hai ng BC v BD. Hai gin ny giao nhau ti th tch V, th tch nys tham gia vo qu trnh truyn sng tn x v c gi l th tch tn x. Nu trong th tch Vcu to ca kh quyn khng ng nht, ngha l trong c nhng min m h sin mi cc

b khc vi h sin mi ca mi trng xung quanh th sng i vo vng ny s b khuch tnra mi pha v mt phn sc truyn ti anten thu.

Trong thc t phng thc thng tin ny t c s dng do tin cy km, pha inh su,yu cu cng sut my pht ln v tnh hng anten cao.

2.2.2 Truyn sng trong iu kin siu khc x tng i lu.

mt khong chiu cao no ca tng i lu nu chit sut bin thin theo quy lutdN

,dh < 0 157 (1/m) th tia sng i vo tng i lu s b un cong vi cong ln hn cong

qut, minh ha trong hnh 2.2. Hin tng gi l hin tng siu khc x tng i lu.

Gi thit min siu khc x tri t mt t ln n cao h0, ng thi cao ln hnh0 chit sut bin thin theo quy lut ging nhi vi tng i lu thng. t ti A mt ngun

bc x, nhng tia sng c gc xut pht ln hn so vi mt phng nm ngang (tia 1 v tia 2) s b

khc x t v n xuyn qua min siu khc x m khng b gi li. Ta k hiu th l gc gii hnm khi sng xut pht theo gc s b un cong theo ng gii hn cao h0 (bn knh cong

ca tia sng bng bn knh tri t, tia 3). Tt c cc tia c gc xut pht < th u b un congtrv mt t v phn x nhiu ln truyn i xa. Hnh nh sng truyn i xa khi c hin tngsiu khc x ging vi qu trnh truyn sng trong mt ng dn sng m thnh trn ca ng dn lgii hn trn ca min siu khc x v thnh di l mt t.

21


24/147


Li dng tnh cht trn ca min siu khc x truyn lan sng cc ngn i xa. Tuynhin min siu khc x xy ra bt thng, cao v chiu di ca min siu khc x cng lunlun thay i nn s dng phng php truyn lan bng siu khc x tng i lu thng tin b thtthng v khng lin tc. Chnh v th phng php ny cng khng s dng cho thng tin vi ba.

1 2 3

45 45

4gh

h0

A

Hnh 2.2 Hin tng siu khc x tng i lu

2.2.3 Truyn lan sng trong gii hn nhn thy trc tip

Hai phng php thng tin trn khng c s dng rng ri v cc nhc im ca n,ph thuc nhiu vo iu kin thin nhin. Bi vy, thng tin vi ba thng s dng phng phptruyn lan trong phm vi nhn thy trc tip. Ngha l hai anten thu v pht phi t cao trn mtt khng b che chn bi cc chng ngi vt c trn mt t, nh ch ra trong hnh v.

Hnh 2.3. Truyn sng trong gii hn nhn thy trc tip

Phn dui y ta s xem xt k phng php truyn lan ny.

2.3 TRUYN LAN SNG TRONG GII HN NHN THY TRC TIPVI CC IU KIN L TNG

2.3.1 Tnh cng trng trong trng hp tng qut - cng thc giao thoa

n gin trc ht ta nghin cu qu trnh truyn sng trong gii hn nhn thy trctip vi gi thit mi trng cc iu kin l tng. l: mt t phng, b qua cong v

22


25/147


gh gh ca mt t, kh quyn ng nht, khng hp th v anten t cao so vi mt t t nhtvi bc sng cng tc. Lc ny qu trnh truyn sng c m t nh hnh 2.4.

Tia 2

Tia 1B

A

ht

hr

r

Hnh 2.4. M hnh truyn sng trong iu kin l tng

Nh vy ti im thu B c mt tia i thng trc tip trong tng i lu (c gi thit lkhng gian t do) (tia 1), gi l tia ti trc tip, v mt tia phn x t mt t ti im C (tia 2) in. Ch c mt tia phn xn im B v vi gi thit mt t phng, ch c tia 2 l tho mniu kin gc ti bng gc phn xi vi im B.

Cng trng ti im B s l s tng cng trng ca tia ti 1 v tia phn x 2gy ra. Hin tng gi l hin tng giao thoa.

Gi thit di ng truyn l r, chiu cao anten pht v thu l ht, hr. Bng php tnh

hnh hc c th tm c im phn x C t mt t, gc nghing v hiu sng i gia tia

phn x t mt t v tia ti trc tip r.Cng in trng ti im thu do tia ti trc tip truyn trong khng gian t do s l:

( )

( )

TT kW j t

km

EP G

er

=1

1

1

245

(mV/m) (2.1)

Chn h to sao cho gc pha u ca tia ti trc tip bng 0.

Cng in trng ti im thu ca tia phn x s l :

( )

( )

( )TT kW j t k r

km

EP G

Rr

e

=2

2

2

245

(mV/m) (2.2)

Trong :

- r1 : on ng i ca tia ti trc tip, bng AB trn hnh

- r2 : on ng i ca cc tia phn x, bng AC + BC hnh

- r: l hiu sng i ca tia phn x v tia trc tip r = r1-r2

- k : h s sng bng 2/

23


26/147


- R: H s phn x phc t mt t: jR Re = , R l m un, gc pha ph thuc vo loit ti im phn x v phn cc ca sng. Cc gi tr ny thng c tnh sn bng

bng hoc th.

- GT1 v GT2 l h s khuch i ca anten pht theo hng tia trc tip v tia phn xTrong cng thc h s G hng tia ti v tia phn x coi nh bng nhau v bng GT, v

trong thc t mt tuyn vi ba bao gicng tho mn iu kin cao anten ht, hr> ht, hr nn c th coi r1 r2 r, phn bin . Nhng v bc sng cng tcgii sng vi ba rt b, gc sai pha do ng i khc nhau gia tia trc tip v tia phn x li

khng th b qua c v r, thay cc iu kin trn vo cc cng thc (2.1) v (2.2) ta snhn c:

( )

( )

TT kW j t

km

EP G

er

=1245

(mV/m) (2.3)

Cng in trng ca tia phn x s l :

( )

( )

j t rTT kW

km

EP G

R er

=2

2

245 (mV/m) (2.4)

Cng in trng tng hp ti B bng:

( )

( )

j rTT kW j t

km

E E EP G

R.e e

r

+ = + = +

2

1 2

245

1 (mV/m) (2.5)

hay

( ) ( )

( )

(TT kW j t

km

EP G R cos r / R

er

) + + +=

2

245 1 2 2(mV/m) (2.6)

Trong : +r2/ gc sai pha ton phn.

( )( )

R sin r /tg

R cos r /

+ =

+ +

2

1 2

Gi tr hiu dng ca cng in trng ti B c xc nh theo cng thc:

( ) ( )

( )

TT kW

h

km

EP G R cos r / R

r

+ + +=

2

173 1 2 2(mV/m) (2.7)

Cng thc trn c gi l cng thc giao thoa, xc nh cng trng ti im thukhi sng truyn lan trn mt t phng v anten t cao so vi mt t.

So snh (2.7) vi cng thc (1.18) ca truyn lan sng trong khng gian t do, trongtrng hp sng truyn trn mt t phng c h s suy gim F bng:

24


27/147


( )F R cos r /= + + +2

1 2 2 R (2.8)

F biu hin cho nh hng ca mt t phng ln qu trnh truyn lan sng khng gian c ly nhn thy trc tip trc tip, khi anten t cao trn mt t. Ch rng thut ng h s suygim y ch c ngha tng i v c iu kin, bi v gi tr cc i ca F c th ln hn 1.

Trong cng thc R l modun h s phn x v l gc sai pha khi phn x, chng ph thuc vogc ti, tnh cht ca t v s phn cc ca sng. Cc gi tr ny thng c tnh sn theo bnghay th.

Hiu ng i ca tia phn x t mt t v tia ti trc tip c xc nh theo phngphp hnh hc.

B

B

hr-ht

Tia 2

Tia 1

C

ht

hr

Hnh 2.5. Xc nh hiu sng i r

hr+ht

r

hr

( )

( )

/

r t r t r tr t

/

' t r t t r t r

t r

h h h h h hr AB h h r r r

r r

h h h h h hr AB h h r r r

r r

h hr r rr

= = + = +

r

+

+ + + = = + + = + +

=

1 22 2 2

2 2

1

1 22 2 2

2 2

2

2 1

21

2

21

2

2

(2.9)

Cc cng thc trn cha tnh n yu t phn cc, hoc ni chnh xc hn chng visng c phn cc ngang, lc vectcng trng ca tia ti v vectcng trng catia phn x l cng phng.

Nu sng c phn cc thng ng (hnh 2.6) th lc vect 1E vung gc vi tia AB cn

vect 2E vung gc vi tia CB, nh vy chng s c phng kkhc nhau. Tnh ton chnh xc

trong trng hp ny theo tng hp vect 1E v 2E vi gc lch tng ng gia chng c gi tr

gn ng l t rh hr . Trng tng hp c th xem cng phng vi trc ng.

25


28/147


Hnh 2.6. VectE1 v E2 trong trng hp sng phn cc thng ng

E2 E1

B

Tia 2

Tia 1

C

A

ht

hr

Nh vy, nu sng phn cc ngang th trng tng hp s l phn cc ngang v nu sng phncc ng th trng tng hp c th xem l phn cc ng.

Thay gi tr ca rcng thc (2.9) vo cng thc (2.8) ta c

t rh hF R cosr

= + + +

241 2 R (2.10)

V d 2.1. Xc nh h s suy gim v cng in trng hiu dng ti im thu khi ngtruyn c cc thng s sau: cng sut pht 15W, bc sng truyn lan l 35cm, h s khuch ica anten pht l 200, chiu cao anten pht l 80m, chiu cao anten thu l 20m, c ly ng

truyn l 8km. Bit khi sng phn cc ngang R = 0,91 v = 1800 v khi sng phn cc ng R =

0,68 v = 1800

Gii

Theo cng thc (2.9) hiu s qung ng i l

t rh h . .r ,r

= = =2 2 80 20

0 48000

(m)

Thay cc gi tr vo cng thc (2.10) tm c

Khi sng phn cc ngang:

F = 0,83 hay F = 20 lg 0,83 = -1,6 (dB)

Khi sng phn cc ng:

F = 0,783 hay F = 20 lg 0,712 = -2,1 (dB)Theo cng thc (2.7) gi tr cng trng ti im thu s l:

Khi sng phn cc ngang:

hE. .

. , ,

= =3

173 15 10 2000 83 31 1

8(mV/m)

Khi sng phn cc ng:

hE. .

. , ,

= =3

173 15 10 2000 783 29 3

8(mV/m)

2.3.2 Cc dng n gin ca cng thc giao thoa

26


29/147


Trong thc t cao ca anten pht v thu nh hn rt nhiu so vi khong cch gia

chng nn gc nghing ca tia phn x t mt t s nhn mc c th xem R = 1 v = 1800.Thay vo cc cng thc (2.6) v (2.8) v bin i ta s nhn c

( )

t rr

F cos r / sin sin r

= + + + = =

21 2 2 1 2 2

h h

(2.11)

V

( )

( )

( ) ( )

( ) ( )

TT kW t m r mh

km m m

EP G h h

sinr r

=

346 2 (mV/m) (2.12)

Cng thc (2.12) c thn gin hn c na nu t c iu kin

t rh h

r

2

9

Hay t r rh h 18

Khi y c th thay th gn ng

t r t r h h h hsinr r

2 2

Lc ny cng thc giao thoa c dng n gin nht

( ) ( ) ( )

( ) ( )

TT kW t m r m

h

km m

E, P G h h

r=

2

2 17

(mV/m) (2.13)

Cng thc (2.13) do Vdenski a ra nm 1922 nn c gi l cng thc Vdenski.

V d 2.2. Xc nh h s suy gim v cng trng ti im t anten thu vi cc thng ssau: cng sut pht 50 W, bc sng 10 cm, h s khuch i ca anten pht l 60; chiu cao ca

anten pht v anten thu ln lt l 25m v 10m; khong cch gia hai anten l 10km, R = 1 v =1800.

Gii: H s suy gim c xc nh theo cng thc (2.11)

t rh h . .F sin sin sinr , .

= = =

0

4

2 2 25 102 2 2

0 1 10

=90 2

Nh vy, h s suy gim y khng ng nh tn gi, m trng tng hp ti im thu s tngln gp hai ln.

Cng in trng ti im thu c cc nh theo (2.12)

hE. .

sin

= =3

0346 50 10 6090 60

10(mV/m)

V d 2.3 Nh v d 2.2 nhng bc sng bng 1m.

Gii: Ta c (mt rh h .= =25 10 2502) v r / =18 555 m2 ngha l tha mn iu kin t r

rh h

18

nn

cng trng c tnh theo cng thc (2.13)

27


30/147


hE, . . . .

,

= =3

2

2 17 50 10 60 25 109 4

10

(mV/m)

2.3.3 iu kin truyn sng tt nht

Qua vic kho st cng thc giao thoa trn ta thy tia phn x t mt t thng l gytc dng xu, lm gim cng trng ti im thu. Nu chn quan h gia cc thng s cang thng tin mt cch thch ng, c th lm cho tia phn x hoc s khng gy tc dng xulm yu trng hoc s tng thn cng trng ti im thu.

Gi tr hiu dng cng trng ca tia ti trc tip c xc nh bi biu thc

( )

( )

TT kW

h

km

EP G

r

=1173

(mV/m) (2.14)

Trong khi , gi tr hiu dng cng trng tng im thu c xc nh bi

( )

( )

( ) ( )

( ) ( )

TT kW t m r mh

km m m

EP G h h

sinr r

=

346 2 (mV/m) (2.15)

Trng tng s bng trng ca tia ti trc tip khi thc hin ng thc:

t rh hsin ,r =2

0 5

T nhn ct r

rh h

12

= (2.16)

Biu thc ny chng t rng, vi khong cch gia cc trm thng tin v bc sng chotrc, nu chn cao anten thch hp sao cho ng thc (2.16) c tha mn th tia phn x tmt t s khng gy tc dng lm yu trng ca tia ti trc tip. V mt ngha vt l, iu nyc th gii thch l trong trng hp trn gc lch pha do hiu sng i gia hai tia bng 600,thm vo l gc chm pha 1800 khi sng phn x t mt t nn gia cc vec tE1 v E2 s cgc lch pha chung 2400. Do vy ln ca vc t tng bng ln ca cc vc t thnh phn(hnh 2.7).

Nu chn quan h gia cc thng s ca mt Eng thng tin nh th no thc hin c ng

thc

E2t rh hsinr

=

2

1 E1

Hay t rr

h h

=4

(2.17) 060

1800

Hnh 2.7

28


31/147


th trng tng ti im thu s ln gp hai ln trng ca tia ti trc tip to ra. V ngha vt l,iu ny c gii thch l trong trng hp ny, gc lch pha do hiu sng i gia hai tia

bng 1800, thm vo l gc lch pha 1800 khi sng phn x t mt t, trng ca tia phn xti im thu sng pha vi trng ca tia ti trc tip.

2.4 TRUYN SNG TRONG GII HN NHN THY TRC TIP KHIKN NH HNG CA I HNH

2.4.1 Truyn sng trong gii hn nhn thy trc tip khi tnh n cong ca mt t

Khi khong cch gia anten pht v anten thu kh ln, ta khng th coi mt t l phng mphi coi n l mt cu, do trong cc tnh ton cn phi tnh n cong ca mt t.

Mt thng s quan trng ca ng thng tin trong trng hp ny l khong cch tm nhn

thng. Khong cch ny c xc nh bi di ca on ng thng ni gia anten pht,anten thu v tip tuyn vi mt t, k hiu l r0 (hnh 2.8 ).

0r AC CB= +

O

hrhtA B

a

ro

Hnh 2.8 Cly nhn thy trc tip

C

y ( )

2 2

t tAC a h a 2ah= + =

( )2 2

r rCB a h a 2ah= + =

Do ( )0 tr 2a h h= + r (m) (2.18)

Trong a l bn knh tri t, ht v hr l cao ca anten pht v anten thu so vi mt t.

Thay a = 6370 km v biu th r0 bng km, ht, hrbng mt ta c :

( )0 tr 3,57 h h= + r (km) (2.19)

By gita kho st bi ton truyn sng trn mt t cu (hnh 2.9).

29


32/147


A

B

A1 B1C

O

ahrht

hthr

Hnh 2.9 M hnh truyn sng trn mt t cu

Qu trnh truyn lan sng trn mt t cu tng t nh mt t phng. Trng ti im thul kt qu giao thoa ca hai tia: tia trc tip v tia phn x t mt t. Nu qua im phn x casng trn mt t ta v mt mt phng tip tuyn vi mt t v tnh chiu cao anten k t mt t

phng y (gi l chiu cao ginh) th cng trng ti im thu s tnh theo cng thc giaothoa nh mt t phng nhng cn thay chiu cao thc ht v hrbng chiu cao gi h

't v hr

'.

Cng thc truyn sng trong gii hn nhn thy trc tip khi anten t cao trn mt t cu

s l:

( ) ( ) ( )

( ) ( )

' 'TT kW t m r m

h

km m

E, P G h h

r=

22 17

(mV/m) (2.20)

tnh cng trng cn tm cc gi tr chiu cao gi ca anten theo chiu cao thc vkhong cch r.

T hnh v ta tm c

( )' 't t t t(A C)

A C a h h h h

a

= = 2

1

1 22

(2.21a)

( )' 'r r r r (CB )

CB a h h h ha

= = 2

1

1 22

(2.21b)

Vic tnh ton chnh xc im phn x tng i phc tp, ta ch xt trng hp c ly thngtin ln gn bng khong cch tm nhn thng. Ta c th coi

tA C ah1 2 v rC B a h1 2

Cho rng ( )0 tr r 2a h h = + r

Ta tm c tt r

r hA Ch h

=+

1 (2.22a)

30


33/147


r

t r

r hCB

h h=

+1

(2.22b)

Tuy nhin trong cc cng thc tnh ton cng trng ta thy ch c tch s cao thtca anten l ht v hr. Do vy tnh ton khi kn cong qut ta a vo h s b m (mthng c xc nh theo th), lc

' 't r t r h h mh h=

Nh vy cng thc truyn sng trong gii hn nhn thy trc tip khi anten t cao trn mt tcu s l:

( ) ( ) ( )

( ) ( )

TT kW t m r m

h

km m

E, P G m.h h

r=

22 1 7

(mV/m) (2.23)

2.4.2 Truyn sng trong gii hn nhn thy trc tip khi tnh n mp m ca mt

t

Trong cc trng hp nu trn chng ta u coi mt t l bng phng nhng thc t mtt l khng bng phng. Khi sng ti gp mt b mt mp m th sng phn x s c tnh chttn x v ch c mt phn nng lng sng phn x ti c im thu lm nh hng n cng trng ti im thu. Nh vy vic cn thit l phi xc nh s mp m ca b mt. R rngrng b mt c coi l mp m mt vi tn s v gc ti no , nhng khi cc tham s nythay i th b mt ny li c th coi l bng phng. nh gi mp m ca mt t ta sdng tiu chun Rayleigh. Hnh 2.10a minh ha b mt thc v b mt ny c l tng trong

hnh 2.10b.

BB

C

CA

A

h

Tia BTia A

Hnh 2.10b.M hnh l tng haca b mt mp m

Mt sng 1

Mt sng 2

Hnh 2.10a.Mt ct a hnh thc

Tiu chun Rayleigh c xy dng trn csb mt c l tng ha vi tia A cphn x t phn trn ca b mt mp m cn tia B c phn x t phn di. Cc mt sngtng ng AA' v CC' c biu din trong hnh 2.10b. Ty ta c s sai khc v qung ngca hai tia ny khi t ti cc im C v C' ti mt sng CC' sau khi phn x ti B v B' l:

( ) ( )' ' ' 'r AB BC A B B C = + + ( )h

c h ssin

os2 in= =

1 2

Do s sai lch v pha l:

31


34/147


2 4 sinhr

= =

Ta thy rng nu cao h l nh so vi bc sng th s sai lch v pha cng nh v do

b mt c coi l bng phng. Thc t s sai lch v pha chy t 0 n . Khi = cc tia

phn x s trit tiu nhau, trng tng bng 0. Khi gc sai pha > /2 th s phn x sng c

tnh cht tn x. Nh vy tiu chun Rayleigh nhn c t

hsin =

4

2

hay h

sin

8(2.24)

Nu mp m ca mt t tha mn tiu chun Rayleigh th c th coi mt t l phng.

Vi tia ti trc tip ta phi xt n vng tham gia vo qu trnh truyn lan sng.

2.5 NH HNG CA TNG I LU KHNG NG NHT

2.5.1 H sin mi v chit sut ca tng i lu

Tng i lu l mt mi trng khng ng nht theo mi phng, th hin cc tham sca mi trng: nhit , m v p sut lun thay i theo khng gian v thi gian.

Tnh cht quan trng ca tng i lu l nhit gim theo cao, khong 60/km. Nhit trung bnh gii hn trn ca tng i lu trong cc min cc khong - 550C v min nhiti khong - 800C.

p sut trung bnh ca kh quyn mt t l 1041 mbar (1 mbar = 1/1000 bar; 1bar c p

lc bng 105 N/m2 ), cao 5 km tr s gim i gn mt na cn 538 mbar. Ti cao 11km, p sut trung bnh l 225 mbar, ln n cao 17 km l gii hn trn ca tng i lu vngnhit i tr s ca n ch cn khong 90 mbar.

Hi nc trong tng i lu l do s bc hi nc ti dng, bin hay sng h,... ditc dng bc x ca mt tri. V vy tng kh quyn i dng m hn tng kh quyn trn tlin, lng hi nc gim nhanh theo cao.

Trong phn kho st sau ta dng khi nim tng i lu tiu chun hay tng i luthng, c tnh cht sau: mt t c p sut P = 1013 mbar, nhit T = 150C, m tng i60 %. Mi khi chiu cao tng 100 m th p sut gim i 12 mbar, nhit gim i 0,550C, m

tng i c bo ton sut cao. Gii hn trn ca tng i lu thng l 11 km.H sin mi ca khng kh vn c coi gn ng bng 0 nhng thc ra n ln hn 0

mt cht v ph thuc vo p sut, nhit v m ca khng kh

' hPPT T

= + +

6156 48101

10 (2.25)

Chit sut ca mi trng c xc nh bi'

'n

= = +1

12

Hay ' hPn PT T

= = + +

678 48101

10 (2.26)

32


35/147


Thc t gi tr n ch ln hn 1 rt t nn s dng thun tin ngi tan thng dng khinim ch s chit sut biu th chit sut. Ch s chit sut c nh ngha bng

( )N n= 610 1

T cng thc (2.26) ta c

hP

N PT T= +

78 4810

(2.27)

Ly vi phn biu thc ny theo h ta c

hdN dP P P dT dP

dh T dh dh dhT T T

= + +

2 3 2

1 9600 480078

h (2.28)

Tng i lu thng (hay cn gi l tng i lu tiu chun) s c 2dN

4,3.10dh

= (1/m).

Trong cc tnh ton thng dng gi tr 2dN

4.10dh

2.5.2 Hin tng khc x kh quyn

Tng i lu khng ng nht cho nn nu c mt tia sng truyn i khng song song vimt t th n s b khc x lin tip. Kt qu l tia sng b un cong, hin tng ny gi l hintng khc x kh quyn. Ta s xc nh bn knh cong ca quo sng khi c khc x khquyn. Kho st hai lp kh quyn k nhau c chit sut khc nhau mt lng dn, v dh l b dy

ca lp kh quyn c chit sut n + dn (hnh 2.11)

Quo sng

d

+ d

R

O

ca

b

dh

n = const

n + dn = const

Hnh 2.11. M t cc thng s tnh bn knh cong ca tia

Gi thit ta bc x mt tia sng c gc ti i xuyn qua khong dh ti lp c chit sut

n + dn vi gc ti + d. Bn knh cong ca tia sng s bng:

ab

R d=

Xt tam gic abc ta c

33


36/147


( )dh dh

abc c

= os +d os

Do dh

Rc .dos

=

(2.29)

p dng nh lut khc x ti im a ta c

( ) ( )nsin n dn sin d = + +

Khai trin v phi v b qua cc i lng nh bc hai ta c

n sin n sin ncos d +sin dn = +

Nnsin dn

cn

os d

=

Thay gi tr ny vo (2.29) ta c

nR

dnsindh

=

(m) (2.30)

Trong thc t chit sut ca lp kh quyn n 1, thng tin vi ba c c ly truyn sng lnhn nhiu ln chiu cao ca anten nn tia sng truyn t anten pht n anten thu gn nh nm

ngang, bi vy sin 1, khi bn knh cong ca quo sng c tnh theo cng thc ngin

Rdn dN

dh dh

= =

6

1 10 (m) (2.31)

Nhn xt: Bn knh cong ca tia sng khi i qua tng i lu ph thuc vo tc binthin ca chit sut theo cao m khng ph thuc vo gi tr tuyt i ca n. Nu chit sut

tng theo cao (dn

0dh

> ) th bn knh cong c gi tr m, quo sng s c b lm hng ln

trn (tia sng b un cong ln) v c gi l khc x m. Nu chit sut gim theo cao

(dn

0dh

< ), bn knh cong c gi tr dng, quo sng s c b lm quay xung di v c

gi l khc x dng. Nu chit sut khng thay i theo cao, tia sng si thng.

2.5.3 nh hng ca khc x kh quyn khi truyn sng trong tm nhn thng.

Nh cp phn trc, khi p dng cng thc giao thoa, trng im thu ph thucvo hiu s hnh hc ca ng i gia tia ti trc tip v tia phn x t mt t. Hin tng khcx kh quyn lm cho tia sng i cong, do hiu s hnh hc ca tia ti trc tip v tia phn xs khc trong trng hp sng truyn trong kh quyn ng nht.

34


37/147


AB

C

Hnh 2.12 Quo ca tia sng trc tip v tia phn x tmtt trong tng kh quyn thc

xt n nh hng ca khc x kh quyn, phng php n gin nht l coi c hai tiati trc tip v tia phn x trn mt t u c truyn theo quo thng nhng khng phitruyn lan trn mt t cu bn knh a m trn mt cu tng tng c bn knh atd.

Hnh 2.13 Cc quo ca sng v tuyna) Quo thc vi tri t bn knh thc

b) Quo ng thng vi tri t c bn knh tng ng

Bn knhtri t

Bn knhcon u

a)

aR

b)

at

Vic thay th tia sng thc v mt t thc bng tia sng i thng v mt t tng ngphi tha mn iu kin: cong tng i gia mt t thc v tia sng thc phi bng congtng i gia mt t tng ng v tia sng i thng. Ngha l phi tha mn phng trnh:

tda R a =

1 1 1 1

T rt ratd

aa

aR

=1

(2.32)

Thay bn knh cong R bi cng thc (2.31) vo (2.32) ta c:

tda a

adn dN

a adh dh

= =

+ + 61 1 10(2.33)

Khi nim bn knh tng ng ca tri t cho php p dng cc cng thc phn trcnhng cn thay th bn knh a ca mt t trong cc cng thc bng bn knh at.

- C ly tm nhn thng:

35


38/147


( )' td t r r a h h= +0 2

Gi k l t s ca bn knh tng ng v bn knh thc k = atd//a ta c

( )' tr ka h h= +0 2 r (m)

Vi tng i lu thng k = 4/3 ta c

( )' t rr , h (m) h (m)= +0 4 15 (km)

- cao gi ca anten

( )2

1' A Ch h2ak

=

2.5.4 Cc dng khc x kh quyn

Cn c vo s bin i ca chit sut theo cao ta tin hnh phn loi cc dng khc xkh quyn. Khc x kh quyn c chia lm hai loi chnh

- Khc x m

ng vidN

0dh

> , trong trng hp ny chit sut tng theo cao v quo tia sng c

b lm hng ln trn, R< 0. Bn knh tng ng ca tri t s nh hn bn knh thc v iu s dn n gim cng in trng im thu.

- Khc x dng

ng vi dN 0dh

< , trong trng hp ny chit sut s gim theo cao v quo tia sng

c b lm hng xung di, R> 0.

Khc x dng c phn thnh mt s trng hp sau:

1) Khc x kh quyn thng: ng vi s khc x xy ra trong tng i lu thng.Cngin trng ti im thu trong trng hp ny ln hn so vi trng hp khng c khc x.

2) Khc x ti hn: xy ra khi6dN 10

0,157

dh a

= = (m-1)

Trng hp ny bn knh cong ca tia sng bng bn knh ca tri t, bn knh tng

ng ca tri t c gi tr. Sng truyn song song vi mt t cu.

3) Siu khc x: xy ra khi6dN 10

,dh a

< haydN

0,157dh

< (m-1)

Trng hp ny bn knh cong ca tia sng nh hn bn knh ca tri t, bn knh tngng ca tri t nh hn 0. Sng b un cong trv mt t v b phn x trn mt t.

36


39/147


Bng 2.1: Phn loi cc dng khc x kh quyn

Loikhc

x

dN

dh

(1/m)

R (m) at(m) Quo sng thct

Qu o sng tngng

Khc

x m>0 < 0 < 6,37.106

Khng

khc

x

0

6,37. 106

Khc

xthng

- 0,04 2,5. 107

8,5. 106

Khc

x tihn

- 0,157 6,37. 106

Siu

khc

x

< - 0,157 < 6,37.

106

< 0

at =

aat < 0

a

aat= 4a /3

at= aa

a at< a

37


40/147


2.5.5 Hp th sng trong tng i lu

Sng v tuyn in truyn lan trong tng i lu ngoi cc hin tng phn x, khc x, tnx cn b suy hao do hp th trong cc phn t, hp th do ma, sng m,... Cc hp th ny phthuc nhiu vo tn s, iu kin kh tng ca tng vng v phng ca tia sng.

Hp th phn t

Hp th phn t trong tng i lu ch yu do phn t hi nc (H2O) v phn t xy (O2).Hp th phn t ph thuc vo tn s. Khi tn s cng tc di 10 GHz hp th phn t c th bqua, cn t 10 GHz hp th phn t tng nhanh theo tn s. C cc gi tr cng hng hp th,tng ng vi cc bc sng 1,35cm; 1,5 cm v 0,75 mm i vi phn t hi nc, bc sng0,5 cm v 0,25 cm i vi phn t xy.

Hnh 2.15 ch ra s ph thuc ca hp th sng do phn t hi nc ti tn s 22 GHz vcc phn t oxy khong tn s 60 GHz

3 6 15 30 60 150 300

10+2

10+1

100

10

-1

10-2

10-3

Tn s GHz

Su

yhaodB/km

f = 22 GHz

O2

H2 O

Hnh 2.14. Sph thuc ca h s hp th sng ca O2 v H2O vo tn s

Hp th trong ma v sng m

Hp th sng trong ma ph thuc vo cng ma tnh theo n v mm/gi v tn s. nhng tn s nh hn 6 GHz th hp th trong ma khng ng k. Khi cng ma tng thhp th tng. Hnh 2.15 ch ra s ph thuc ca h s hp th trong ma ph thuc vo tn s casng v tuyn in v cng ma 100 mm/gi

38


41/147


42/147


Mt trong s nhng hin tng thng gp trong thng tin v tuyn l cng trng tnhiu thu ln khi t my thu v tr ny nhng li c th rt b hoc thm ch bng khng nu tachuyn i my thu sang v tr khc. Trong nhiu trng hp, khi my thu t mt v tr cnhth s dao ng cng trng ti im thu vn xy ra. Hin tng c gi l pha inh.

Ngi ta phn bit hin tng phainh thnh phainh phng v pha inh la chn tn s. Pha inhphng nh hng ch yu ln cc h thng dung lng nh bng tn hp. Pha inh la chn tn snh hng ch yu n cc h thng truyn dn dung lng cao, bng tn rng. Hai loi pha inhny c th xut hin c lp vi nhau hoc xut hin ng thi vi nhau. Khng mt loi pha inhno c th tin on c mt cch chnh xc bi s bin ng ca chng. Kinh nghim cho thyiu kin kh hu v a hnh l nguyn nhn chnh gy ra pha inh. Cc c trng quan trng ca

pha inh l tn s trung bnh ca pha inh (tr s trung bnh ca pha inh trong mt pht hoc mtgiy) v su pha inh.

Theo l thuyt, ta bit rng sng v tuyn truyn lan trong tng i lu c thn im thu

theo cc ng khc nhau, ngoi sng i thng cn nhiu sng phn x t cc im khc nhautrn ng truyn dn, ngi ta gi l hin tng truyn dn a ng. Do ng i khc nhaucho nn trng gia cc sng c mt tr (s lch pha) vi nhau. Trng ti im thu l tngvectca tt c cc trng sng cho nn c th tng, gim hoc bng khng. Hin tng phainh ny c gi l pha inh nhiu tia hay pha inh a ng. y l loi pha inh rt nguyhim. Pha inh nhiu tia vi hiu ng la chn lm mo dng bin v mo thi gian tr sut rng bng tn ca knh. iu ny lm tng t s li bit (BER) m khng c bt k biu hin g tn hiu thu.

khc phc hin tng pha inh c th bng cc phng php chng hoc thng.

Phng php thng c thc hin thng qua vic thit k cc mch trong my thu loi tr tc ng ca pha inh. V d nng cao cht lng ca tn hiu thu, my thu s dng ccb cn bng t thch nghi.

Cc phng php chng bao gm:

- S dng cc anten c hng tnh cao loi tr cc tia phn x gy giao thoa sng ti phapht cng nh pha thu.

- To ra s chnh lch v cao gia hai anten loi tr bt cc tia phn xn cim thu.

- S dng cc bin php phn tp anten: phn tp tn s, phn tp khng gian, phn tp phn

cc, phn tp gc v phn tp thi gian hoc kt hp cc bin php phn tp ny vi nhau.a, Phn tp khng gian

Phn tp khng gian l phng php s dng hai hay nhiu hn hai anten cho cc my thuhoc pht truyn dn ng thi cng mt tn hiu trn cng mt knh v tuyn.

Thng thng ngi ta s dng hai anten b tr cch nhau mt khong no ln hn nmln bc sng cng tc pht hoc thu cng mt thng tin t ni pht n ni nhn tin. Khongcch ny m bo sao cho cc tn hiu ring bit thu khng tng quan nhau. Nh vy ta lun thuc tn hiu tt v bng cch kt hp (hoc chn) tn hiu gia hai ng truyn ny ta scmt tn hiu tt.

40


43/147


f

f

Tx

Rx

Rx

KthpS liu

vo

S liura

Hnh 2.17 Phn tp khng gian

b, Phn tp tn s

Phn tp tn s l phng php truyn ng thi cng mt tn hiu trn hai tn s khc nhautrong cng mt di tn. Bi v xc sut xy ra ng thi pha inh hai tn s khng tng quanvi nhau l rt nh. V th ta lun thu c tn hiu tt v bng cch kt hp (hoc chn) tn hiugia hai ng truyn ny ta sc mt tn hiu tt.

SW

f

fS liuvo

Tx1 Rx1

Rx2

S liura

Tx2

Hnh 2.18 Phn tp tn s

c, Phn tp phn cc

L phng php s dng h thng anten ca c hai trm thu pht c cc dng phn cc khc

nhau nh: phn cc ng v phn cc ngang; phn cc trn quay tri v phn cc trn quay phi.

2.7 TNG KT

Sng cc ngn c truyn lan trong tng i lu bng ba phng php: tn x tng i lu,siu khc x tng i lu v truyn lan trong gii hn nhn thy trc tip. Tuy nhin, phng phptruyn lan cbn nht l truyn lan. Trong chng ny xt cc cng thc tnh ton cc thams ca ng truyn sng trong gii hn nhn thy trc tip vi iu kin l tng. T lm c

s xt cc nh hng thc t ca mi trng tng i lu ln qu trnh truyn lan sng nhnhhng ca a hnh, nh hng ca tng kh quyn khng ng nht. Cc nh hng ny u tcng trc tip n cc tham s ca tuyn truyn dn gia im pht v im thu nh c ly tmnhn thng, bn knh cong ca tia sng Chng ny cng cp n hin tng in hnhthng gp phi trong truyn lan sng v tuyn, lm gim cht lng truyn dn trong thng tinv tuyn l hin tng pha inh cng nh cc bin php c p dng trnh hin tng ny.

2.8 CU HI V BI TP.

1. Trnh by cc phng php truyn lan sng cc ngn.

41


44/147


2. Xy dng cng thc tnh cng in trng ti im thu khi truyn sng trong gii hn nhnthy trc tip trong iu kin l tng.

3. Xc nh iu kin truyn sng tt nht?

4. Xc nh cng thc truyn sng khi tnh n nh hng ca a hnh (mt t cu) ln qu trnhtruyn lan sng?

5. Tnh bn knh cong ca tia sng khi truyn sng trong tng i lu khng ng nht?6. Hin tng khc x kh quyn nh hng nh th no khi truyn sng trong tm nhn thng?

7. Trnh by cc dng khc x kh quyn?

8. Trnh by v cc dng pha inh v bin php chng.

9. Cho ng truyn c cc thng s sau: Cng sut bc x 15 W, bc sng cng tc 35 cm, hs khuch i ca anten pht l 100, cao ca anten pht v anten thu ln lt l 80 m v 20 m,

c ly ng truyn l 10 km. Vi R = 0,91 v = 1800 khi sng phn cc ngang v R = 0,68; =1800 khi sng phn cc ng. Xc nh h s suy gim?

(a) 0,42 v 0,44; (b) 0,52 v 0,54; (c) 0,62 v 0,64; (d) 0,72 v 0,7410. S liu nh bi 9, xc nh cng in trng hiu dng ti im thu?

(a) 10 mV/m v 11,5 mV/m; (b) 11 mV/m v 10,5 mV/m; (c) 11 mV/m v 11,5 mV/m; (d) 10,5

mV/m v 11,5 mV/m

11. S liu nh bi 9, xc nh tn hao truyn sng bit h s khuch i ca anten thu l 100.

(a) 76,78 dB v 76,45 dB; (b) dB; (c) 76,78 dB v 80,45 dB; (d) 80,78 dB v 80,45 dB

12. Xc nh h s suy gim khi ng truyn c cc tham s: cng sut pht 50 W, bc sngcng tc 10 cm, h s khuch i anten pht 60, cao anten pht 25 m, anten thu 10 m, c ly

truyn sng 10 km, h s phn x R = 1 v = 1800.

(a) 1,62 ; (b) 1,72; (c) 1,82; (d) 1,92

13. S liu nh bi 12, xc nh cng in trng ti im thu?

(a) 115,34 V/m; (b) 125,34 V/m; (c) 115,34 mV/m; (d) 125,34 mV/m

14. S liu nh bi 12 nhng bc sng cng tc l 1m. Hy xc nh cng in trng tiim thu?

(a) 15 mV/m; (b) 16 mV/m; (c) 17 mV/m; (d) 18 mV/m

15. Mt anten pht c t cao 49m v anten thu c t cao 25m. Khong cch tmnhn thng ca hai anten ny l gi tr no di y?

(a) 35,8 km; (b) 42,8 km; (c) 45,8 km; (d) 50,8 km

16. Mt anten pht c t cao 30m v anten thu c t cao 15m. Khong cch tmnhn thng ca hai anten ny l gi tr no di y?

(a) 27,4 km; (b) 30,4 km; (c) 33,4 km; (d) 35,4 km

17. Anten pht v tuyn truyn hnh t cao 64m. Tnh cao ca anten thu ti mt im tcch xa i pht mt khong 50 km c th thu c tn hiu.

(a) 2 m; (b) 2,5 m; (c) 3 m; (d) 3,5 m

18. Xc nh bn knh cong ca tia sng khi i trong tng i lu i lu tiu chun?

(a) 2.106 m; (b) 2,5.106 m; (c) 2.107 m; (d) 2,5.107 m

42


45/147

Chng 3: Knh truyn sng v tuyn trong thng tin ding

CHNG 3KNH TRUYN SNG V TUYN

TRONG THNG TIN DI NG

3.1 GII THIU CHUNG


- Truyn lan sng phng trong mi trng v tuyn di ng, cc c tnh ca knh- Knh truyn sng trong min thi gian- Knh truyn sng trong min tn s- Knh truyn sng trong min khng gian- Quan h gia cc thng s trong cc min khc nhau

- Cc loi phainh phm vi hp- Cc phn b Rayleigh v Rice- Cc m hnh knh trong min thi gian v min tn s- nh gi c tnh knh trong cc min khc nhau

3.1.2 Hng dn

- Hoc k cc phn c trnh by trong chng- Tham kho thm [3]


3.1.3 Mc ch ca chng

- Nm c c im truyn sng trong mi trng v tuyn di ng, cc c tnh caknh

- Nm c c tnh knh truyn sng trong cc min khng gian, thi gian v tn scng nh quan h gia cc thng s trong cc min ny- Hiu v cc loi phainh phm vi hp, cc phn b Rayleigh v Rice- Nm c cc m hnh knh trong min thi gian v min tn s

3.2 MU

3.2.1 Truyn lan sng phng trong mi trng v tuyn phainh di ng

Trong thng tin v tuyn, sng v tuyn c truyn qua mi trng vt l c nhiu cutrc v vt th nh ta nh, i ni, cy ci xe c chuyn ng. Ni chung qu trnh truynsng trong thng tin v tuyn rt phc tp. Qu trnh ny c th ch c mt ng truyn thng

(LOS: line of sight), hay nhiu ng m khng c LOS hoc c hai. Truyn sng nhiu ngxy ra khi c phn x, nhiu x v tn x. Hnh 3.1 m t mi trng truyn sng ny.

43


46/147


Hnh 3.1: Truyn sng v tuyn

Phn x xy ra khi sng v tuyn p vo cc vt cn c kch thc ln hn nhiu so vibc sng. Ni chung phn x gy ra do b mt ca qut, ni v tng ca ta nh.

Nhiu x xy ra do sng in t gp phi cc b mt sc cnh v cc thnh gca cc cu

trc.Tn x xy ra khi kch thc ca cc vt th trong mi trng truyn sng nh hn bc

sng. Tn x thng xy ra khi sng v tuyn gp phi cc k hiu giao thng, ct n.Ngoi phn x, nhiu x v tn x, sng v tuyn cn b suy hao ng truyn. Cng

tn hiu cng b thay i theo thi gian do s chuyn ng ca my thu hoc my pht. phntch ta c thc trng nh hng truyn sng v tuyn thnh hai loi: suy hao tn hiu phm virng v mo tn hiu phm vi hp. Suy hao tn hiu phm vi rng gy ra do suy hao ng truynv s che ti my pht v my thu cn mo tn hiu phm vi hp xy ra do truyn sng nhiung. Di y ta s xt hai nh hng ny.

Ngoi ra, hiu ng Doppler cng nh hng xu ln cc c tnh truyn dn ca knh vtuyn di ng. Do chuyn ng ca my di ng, hiu ng Doppler gy ra dich tn si vi

tng sng mang thnh phn. Nu ta nh ngha gc ti i l gc hp bi phng ti ca sng tith i v phng chuyn ng ca my di ng nh thy hnh 3.2, th gc ny s xc nh tn sDoppler (dch Doppler) ca sng ti th i theo biu thc sau:

i df : f cos i= . (3.1)

Trong trng hp ny, fd l tn s Doppler cc i quan h vi tc my di ng v, tc nh sng c0 v tn s sng mang f0 theo cng thc sau

d

vf

c= 0

0

f (3.2)

44


47/147


i+ 1

Ad

i

Hnh 3.2. Gc ti i ca sng ti i minh ha hiu ng Doppler

Tn s Doppler cc i (cc tiu), fi = fd (fi = -fd) t c khi i=0 (i=). fi=0 khii=/2 v i=3/2. Do hiu ng Doppler ph ca tn hiu c pht trong qua trnh truyn dn s

b mrng. Hiu ng ny gi l tn tn. Gi tr ca tn tn ch yu ph thuc vo tn s Dopplercc i v cc bin ca cc sng mang thnh phn thu c. Trong min thi gian, hiu ngDoppler dn n p ng xung kim ca knh trnn thay i theo thi gian. C th ch ra rngcc knh v tuyn di ng tha mn nguyn l xp chng v v th cc h thng tuyn tnh. Dotnh cht thay i theo thi gian ca p ng xung kim, ni chung cc knh v tuyn di ngthuc loi cc h thng tuyn tnh thay i theo thi gian.

3.2.2. nh hng phm vi rng

Hnh 3.3 cho thy nh hng suy hao tn hiu phm vi rng trong mi trng truyn sngdi ng. T hnh v ny ta thy suy hao tn hiu phm vi rng bao gm suy hao hay tn thtng truyn v che ti (cn gi l phainh chm). Suy hao ng truyn xy ra do khong cchn my pht. Che ti l s thay i cng sut thu v suy hao tn hiu gy ra do cc vt cn giamy pht v my thu.

Suy hao ng truyn lin quan n t s gia cng sut pht v cng sut thu:

txPL

rx

PL

P

(3.3)

trong Ptx l cng sut pht v Prx l cng sut thu. Trong thc t LPL l mt hm ph thuc vokhong cch gia my pht v my thu.

n

PLL

r

1

(3.4)

trong r l khong cch gia my pht v my thu, n l s m suy hao ng truyn. Ty theomi trng truyn sng n nm trong di t 2,5 n 4. Chng hn trong vng thnh ph n = 3,8 -4,5 cn trong vng nng thn n = 2,5 - 3. Cc chng ngi gia my pht v my thu dn n sthay i cng sut thu xung quanh gi tr cng sut thu trung bnh Prx, hin tng ny c gi l

45


48/147


che ti. Che ti c coi l phainh chm v c c trng bi phn b chun log (phn bchun trong thang dB).

txP rx

P

Hnh 3.3. Suy hao ng truyn v che ti.

3.2.3. nh hng phm vi hp

Nh ni trn truyn sng a ng gy ra do phn x, nhiu x v tn x dn nnhiu ng truyn khng trc tip (khng phi LOS). Cc ng truyn khng trc tip ny nmy thu lch nhau theo thi gian v khng gian, iu ny gy ra cc hiu ng phm vi hp trongthng tin v tuyn di ng nh: tri tr, tri gc v tri Doppler. Hnh 3.4 cho thy ba nh hng

phm vi hp gy ra do truyn sng a ng khng trc tip trong knh v tuyn di ng.

0

1

2

Hnh 3.4 Cc nh hng phm vi hp trong knh v tuyn

Tri trl so tr truyn sng tng i gia cc ng truyn sng khng trc tip gyra do cc vt phn x nhi ni v cc ta nh.

Tri gc l so v dch gc cu cc ng truyn khng trc tip so vi ng truyntrc tip (xem hnh 3.4).

Tri Dopplerla so v tc thay i knh gy ra do s chuyn ng cu my pht v(hoc) my thu so vi cc vt th tn x trong mi trng truyn sng a ng.

46


49/147


Ngoi ra tng ca rt nhiu ng truyn khng trc tip trong truyn sng a ng dnn thng ging bin tn hiu thu v th gy ra phainh v mo tn hiu. Trong khi lp m hnhknh, ta tp trung ln cc nh hng truyn sng a ng (cc nh thng phm vi hp) i vicc my pht v (hoc) my thu s dng nhiu anten.

3.2.4 Cc c tnh ca knh

Trong thng tin v tuyn di ng, cc c tnh knh v tuyn di ng c tm quan trngrt ln, v chng nh hng trc tip ln cht lng truyn dn v dung lng. Trong cc hthng v tuyn thng thng (khng phi cc h thng v tuyn thch ng), cc tnh cht thng kdi hn ca knh c o v nh gi trc khi thit k h thng. Nhng trong cc h thng iuch thch ng, vn ny phc tp hn. m bo hot ng thch ng ng, cn phi lin tcnhn c thng tin v cc tnh cht thng k ngn hn thm ch tc thi ca knh.

C th phn cc knh v tuyn thnh hai loi: "phainh phm vi rng" v "phainh phm vihp". Cc m hnh truyn sng truyn thng nh gi cng sut trung bnh thu c ti cckhong cch cho trc so vi my pht. i vi cc khong cch ln (vi km), cc m hnhtruyn sng phm vi rng c s dng. Phainh phm vi hp m t s thng ging nhanh sngv tuyn theo bin , pha v tra ng trong khong thi gian ngn hay trn c ly di chuynngn. Phainh trong trng hp ny gy ra do truyn sng a ng.

Cc knh v tuyn l cc knh mang tnh ngu nhin, n c th thay i t cc ngtruyn thng n cc ng b che chn nghim trng i vi cc v tr khc nhau. Hnh 3.5 chothy rng trong min khng gian, mt knh c cc c trng khc nhau (bin chng hn) ticc v tr khc nhau. Ta gi c tnh ny l tnh chn lc khng gian (hay phn tp khng gian) v

phainh tng ng vi n l phainh chn lc khng gian. Hnh 3.6 cho thy trong min tn s,knh c cc c tnh khc nhau ti cc tn s khc nhau. Ta gi c tnh ny l tnh chn lc tns (hay phn tp tn s) v pha inh tng ng vi n l phainh chn lc tn s. Hnh 3.7 chothy rng trong min thi gian, knh c cc c tnh khc nhau ti cc thi im khc nhau. Tagi c tnh ny l tnh chn lc thi gian (hay phn tp thi gian) v phainh do n gy ra l

phainh phn tp thi gian. Da trn cc c tnh trn, ta c th phn chia phainh knh thnh:phainh chn lc khng gian (phadinh phn tp khng gian), phainh chn lc tn s (phainhphn tp tn s), phainh chn lc thi gian (phn tp thi gian ).

Tnh chn lc khng gian ca knh

Bin

Min khng gian

Hnh 3.5. Tnh cht knh trong min khng gian.

47


50/147


Tnh chn lc tn s ca knh

Bin

Min tn s

Hnh 3.6. Tnh cht knh trong min tn s.

Tnh chn lc thi gian ca knh

Bin

Min thi gian

Hnh 3.7. Tnh cht knh trong min thi gian.

3.3. KNH TRUYN SNG TRONG MIN KHNG GIAN

Cc thuc tnh trong min khng gian bao gm: tn hao ng truyn v chn lc khnggian. Tn hao ng truyn thuc loi phainh phm vi rng cn chn lc khng gian thuc loi

phainh phm vi hp. Cc m hnh truyn sng truyn thng nh gi cng sut thu trung bnh timt khong cch cho trc so vi my pht, nh gi ny c gi l nh gi tn hao ngtruyn. Khi khong cch thay i trong phm vi mt bc sng, knh th hin cc c tnh ngunhin rt r rt. iu ny c gi l tnh chn lc khng gian (hay phn tp khng gian).

Tn hao ng truyn.

M hnh tn hao ng truyn m t suy hao tn hiu gia anten pht v anten thu nh lmt hm ph thuc vo khong cch v cc thng s khc. Mt s m hnh bao gm c rt nhiuchi tit va hnh nh gi suy hao tn hiu, trong khi mt s m hnh ch xt n tn s vkhong cch. Chiu cao an ten l mt thng s quan trng. Tn hao ng truyn c xc nhtheo cng thc (3.4).

T l thuyt v cc kt qa o lng ta bit rng cng sut thu trung bnh gim so vikhong cch theo hm log cho mi trng ngoi tri v trong nh. Ngoi ra ti mi khong cchr, tn hao ng truyn L(r) ti mt v tr nht nh l qu trnh ngu nhin v c phn b logchun xung quanh mt gi tr trung bnh (ph thuc vo khong cch). Nu xt c s thay i theov tr, ta c th biu din tn hao ng truyn L(r) ti khong cch r nh sau:

48


51/147


( ) ( ) lg + = + +

[dB] = L(r)r

L r X L r n Xr

0

0

10 (3.5)

Trong ( )L r l tn hao ng truyn trung bnh phm v rng i vi khong cch pht

thu r; X l bin ngu nhin phn b Gauss trung bnh khng (o bng dB) vi lch chun (cngo bng dB), r0 l khong cch tham chun gia my pht v my thu, n l m tn hao ngtruyn.

Khi cc i tng trong knh v tuyn khng chuyn ng trong mt khong thi giancho trc v knh c c trng bi phainh phng i vi mt rng bng tn cho trc, ccthuc tnh knh ch khc nhau ti cc v tr khc nhau. Ni mt cch khc, phainh chn thunl mt hin tng trong min thi gian (mang tnh chn lc thi gian).

T cng thc 3.5 ta thy rng tn hao ng truyn ca knh c nh gi thng kphm vi rng cng vi hiu ng ngu nhin. Hiu ng ngu nhin xy ra do phainh phm vi hptrong min thi gian v n gii thch cho tnh chn lc thi gian (phn tp thi gian). nh hngca chn lc khng gian c thc loi b bng cch s dng nhiu anten. MIMO (MultipleInput Multiple Output: Nhiu u vo nhiu u ra) l mt k thut cho php li dng tnh cht

phn tp khng gian ny ci thin hiu nng v dung lng h thng.

3.4. KNH TRUYN SNG TRONG MIN TN S

Trong min tn s, knh bnh hng ca hai yu t: iu ch tn s v chn lc tn s.

3.4.1. iu ch tn s

iu ch tn s gy ra do hiu ng Doppler, MS chuyn ng tng i so vi BTS dnn thay i tn s mt cch ngu nhin. Do chuyn ng tng i gia BTS v MS, tng snga ng b dch tn s. Dch tn s trong tn s thu do chuyn ng tng i ny c gi ldch tn s Doppler, n t l vi tc chuyn ng, phng chuyn ng ca MS so vi phngsng ti ca thnh phn sng a ng. Dch Dopplerc xc nh theo cng thc (3.1). Tcng thc ny ta c th thy rng nu MS di chuyn v pha sng ti dch Doppler l dng v tns thu s tng, ngc li nu MS di chuyn ri xa sng ti th dch Doppler l m v tn s thuc s gim. V th cc tn hiu a ng n MS t cc phng khc nhau s lm tng rng

bng tn tn hiu. Khi v (hoc) i thay i dch Doppler thay i dn n tri Doppler.

3.4.2. Chn lc tn s (phn tp tn s).

Trong phn ny ta s phn tch chn lc tn s cng vi mt thng s khc trong min tns: bng thng nht qun. Bng thng nht qun l mt so thng k ca di tn s trn mtknh phainh c coi l knh phainh "phng" (l knh trong tt c cc thnh phn phctruyn qua vi khuych i nh nhau v pha tuyn tnh). Bng thng nht qun cho ta di tn

trong cc thnh phn tn s c bin tng quan. Bng thng nht qun xc nh kiuphainh xy ra trong knh v v th n ng vai tr cstrong vic thch ng cc thng siuch. Bng thng nht qun t l nghch vi tri tr (xem phn 3.6). Phainh chn lc tn s rt

49


52/147


khc vi phainh phng. Trong knh phainh phng, tt c cc thnh phn tn s truyn qua bngthng knh u chu nh hng phainh nh nhau. Tri li trong phainh chn lc tn s (cn gil phainh vi sai), mt son ph ca tn hiu qua knh phainh chn lc tn s bnh hngnhiu hn cc phn khc. Nu bng thng nht qun nh hn rng bng tn ca tn hiu c

pht, th tn hiu ny chu nh hng ca phainh chn lc (phn tp tn s). Phainh ny s lm

mo tn hiu.

3.5. KNH TRUYN SNG TRONG MIN THI GIAN

Mt trong s cc khc bit quan trng gia cc knh hu tuyn v cc knh v tuyn l ccknh v tuyn thay i theo thi gian, ngha l chng chu nh hng ca phainh chn lc thigian. Ta c th m hnh ha knh v tuyn di ng nh l mt b lc tuyn tnh c p ng xungkim thay i theo thi gian. M hnh knh truyn thng s dng m hnh p ng xung kim, yl mt m hnh trong min thi gian. Ta c th lin h qu trnh thay i tn hiu v tuyn phm

vi hp trc tip vi p ng xung kim ca knh v tuyn di ng. Nu x(t) biu din tn hiu pht,

y(t) biu din tn hiu thu v h(t,) biu din p ng xung kim ca knh v tuyn a ng thayi theo thi gian, th ta c th biu din tn hiu thu nh l tch chp ca tn hiu pht vi png xung kim ca knh nh sau:

( ) ( ) ( , ) ( ) ( , )

= = y t x h t d x t h t (3.6)

trong t l bin thi gian, l tr thi gian ca mt ng truyn a ng (cn gi l tra ng) ca knh i vi mt gi tr t cnh.

nh hng a ng ca knh v tuyn thng c bit n dng phn tn thi gianhay tri tr. Phn tn thi gian (gi tt l tn thi) hay tri tr xy ra khi mt tn hiu c truynt anten pht n anten thu qua hai hay nhiu ng c cc di khc nhau. Mt mt tn hiuny c truyn trc tip, mt khc n c truyn t cc ng phn x khc nhau c dikhc nhau vi cc thi gian n my thu khc nhau. Tn hiu ti anten thu chu nh hng ca tnthi ny s b mo dng. Trong khi tit k v ti u ha cc h thng v tuyn s truyn s liutc cao ta cn xt cc phn x ny.

Tn thi c thc c trng bng tr tri, tr tri trung bnh hay tr tri trung bnh qunphng.

3.5.1. Tr tri trung bnh qun phng.

Mt thng s thi gian quan trng ca tn thi l tri tr trung bnh qun phng (RDS:Root Mean Square Delay Spread): cn bc hai mment trung tm ca l lch tr cng sut. RDS lmt so thch hp cho tri a ng ca knh. Ta c th s dng n nh gi nh hng canhiu giao thoa gia cc k hiu (ISI).

= 2

2 , (3.7)

( )

( )

=

k k

k

k

k

P

P , (3.8)

50


53/147


( )

( )

=

k

k

k

k

kP

P

2

2

, (3.9)

trong P(k) l cng sut trung bnh a ng ti thi im k.

3.5.2. Tr tri cc i

Tr tri cc i (X dB) ca l lch tr cng sut c nh ngha l tr thi gian m nng lng a ng gim X dB so vi nng lng cc i.

3.5.3. Thi gian nht qun

Mt thng s khc trong min thi gian l thi gian nht qun. Thi gian nht qun xcnh tnh "tnh" ca knh. Thi gian nht qun l thi gian m knh tng quan rt mnh vi

bin ca tn hiu thu. Ta k hiu thi gian nht qun l Tc. Cc k hiu khc nhau truyn quaknh trong khong thi gian nht qun chu nh hng phainh nh nhau. V th ta nhn cmt knh phainh kh chm. Cc k hiu khc nhau truyn qua knh bn ngoi thi gian nhtqun s bnh hng phainh khc nhau. Khi ny ta c mt knh phainh kh nhanh. Nh vydo nh hng ca phainh nhanh, mt s phn ca k hiu s chu tc ng phainh ln hn cc

phn khc

3.6.QUAN H GIA CC THNG S TRONG CC MIN KHCNHAU

Ta nghin cu cc c tnh knh v cc thng s ca n trong cc min khng gian, tns v thi gian. Cc c tnh ny khng tn ti ring bit, hay ni mt cc khc chng lin quanvi nhau. Mt s thng s trong min ny nh hng ln cc c tnh ca min khc.

3.6.1. Bng thng nht qun v tri tr trung bnh qun phng.

Bng thng nht qun Bc l so thng k di tn s m trong di ny knh cho qua tt ccc thnh phn ph vi suy gim gn nh bng nhau v pha tuyn tnh. Bng thng nht qun thhin di tn m trong di tn ny hoc cc bin hoc cc pha ca hai tn hiu thu c tng quanrt cao. Cc thnh phn ph ca mt tn hiu trong di ny chu tc ng ca knh theo mt cchging nhau (knh pha inh hay khng pha inh)

Ta bit rng l lch tr cng sut v p ng tn s bin ca knh v tuyn di ngquan h vi nhau qua bin i Fourrier. V th ta c th trnh by knh trong min tn s bngcch s dng cc c tnh p ng tn s ca n. Tng t nh cc thng s tri tr trong min

thi gian, ta c th s dng bng thng nht qun c trng knh trong min tn s. Tri trtrung bnh qun phng t l nghch vi bng thng nht qun v ngc li, mc d quan hchnh xc ca chng l mt hm ph thuc vo cu trc a ng. Ta k hiu bng thng nht

51


54/147


qun l BC v tri tr trung bnh qun phng l . Khi hm tng quan ng bao ln hn 90%,bng thng nht qun c quan h sau y vi tri tr trung bnh qun phng:

Thi gian nht qun chu nh hng trc tip ca dch Doppler, n l thng s knh trongmin thi gian i ngu vi tri Doppler. Tri Doppler v thi gian nht qun l hai thng s t lnghch vi nhau. Ngha l

CD

TB

1(3.11)

Khi thit k h thng ta ch cn xt mt trong hai thng s ni trn.

3.7. CC LOI PHA INH PHM VI HP

Ph thuc vo quan h gia cc thng s tn hiu ( rng bng tn, chu k k hiu,) vcc thng s knh (tri tr trung bnh qun phng, tri Doppler, ), ta c th phn loi phainh

phm vi hp da trn hai c tnh: tri tra ng v phainh chn lc tn s. Tri tra ngl mt thng s trong min thi gian, trong khi vic knh l phainh phng hay chn lc tn sli tng ng vi min tn s. V th thng s min thi gian, tri tra ng, nh hng lnc tnh knh trong min tn s. Tri Doppler dn n tn tn v phainh chn lc thi gian, vth lin qun n tri Doppler ta c th phn loi phainh phm vi hp thnh phainh nhanh v

phainh chm. Tri Doppler l mt thng s trong min tn s trong khi hin tng knh thayi nhanh hay chm li thuc min thi gian. Vy trong trng hp ny, tri Doppler, thng strong min tn s, nh hng ln c tnh knh trong min thi gian. Hiu bit c cc quan hny s h trta trong qu trnh thit k h thng. Bng 3.1 lit k cc loi phainh phm vi hp.

Bng 3.1. Cc loi phainh phm vi hp

Csphn loi Loi Phainh iu kin

Phainh phng BBS


55/147


dng b lc tuyn tnh. V th ta cn c gng chuyn vo knh phainh phng cho tn hiu truyndn. Tuy nhin do khng th thay i tri tra ng v bng thng nht qun, nn ta ch c ththit k chu k k hiu v rng bng tn tn hiu t c knh phainh phng.

Da trn tri Doppler, ta c th phn loi knh thnh phainh nhanh v phainh chm.Nu p ng xung kim knh (trong min thi gian) thay i nhanh trong cho k k hiu, ngha l

nu thi gian nht qun knh nh hn chu k k hiu ca tn hiu pht, knh s gy ra phainhnhanh i vi tn hiu thu. iu ny s dn n mo dng tn hiu. Nu p ng xung kim knhthay i vi tc chm hn nhiu so vi k hiu bng gc pht, knh s gy ra phainh chm ivi tn hiu thu. Trong trng hp ny knh t ra tnh i vi mt s chu k k hiu. Tt nhin tamun c phainh chm v n h trcht lng truyn dn n nh hn. Ta khng th xc nhDoppler khi thit k h thng. V th, khi cho trc tri Doppler, ta cn chn rng bng tn tnhiu (bng thng sng mang con) trong gii thut iu ch thch ng nhn c knh phainhchm thay v knh phainh nhanh. Nh vy ta st c cht lng truyn dn tt hn.

3.8. CC PHN B RAYLEIGH V RICE

Khi nghin cu cc knh v tuyn di ng, thng cc phn b Rayleigh v Rice c sdng m t tnh cht thng k thay i theo thi gian ca tn hiu phainh phng.

3.8.1. Phn b phainh Rayleigh

Ta c th coi phn b phainh Rayleigh l phn bng bao ca tng hai tn hiu phn

b Gauss vung gc. Hm mt xc sut (PDF) ca phn b phainh Rayleigh c biu dinnh sau:

,( )

,

=

>BD)

Ch thch d: khong cch thu pht; MIMO: Multile Input Multiple Output; BD: triDoppler; BC: rng bng nht qun ca knh xt cho trng hp tng

quan ln hn 90%; T: chu k k hiu; : tri tr trung bnh qun phng;TC: thi gian nht qun ca knh; BS: rng bng tn hiu pht

Cc thng s knh trong min tn s l tri Doppler v rng bng nht qun (xem bng3.2). Cc thng s knh min thi gian l thi gian nht qun v tri tr trung bnh qun phng.Tri Doppler gy ra do chuyn ng tng i gia MS v BTS. Cc thng s ny c th dn n

phainh chn lc thi gian (hay phn tp thi gian) trong min thi gian v tri Doppler t lnghch vi thi gian nht qun ca ca knh.Tri tr xy ra do tra ng. rng bng nhtqun ca knh t l nghch vi tri tr trung bnh qun phng. V th tri tr trung bnh qun

phng c th dn n phainh chn lc tn s (hay phn tp tn s) trong min tn s.

Chng ny trnh by ngn gn phn b Raylegh v Rice. Cc m hnh knh trongmin tn s v thi gian c tng kt t cc ti liu tham kho. Ngoi ra ta cng ch trngthuc tnh ca knh trong min tn s. Chng ny phn tch nh hng ca mt s thng s(tha s K, tri tr trung bnh qun phng) ln hm truyn t tn s ca knh. Tri tr cng ln

60


63/147


th tc bin thin bin trong hm truyn t knh min tn s cng ln. Tha s K xc nh ln bin thin trong hm truyn t knh min tn s. K cng ln th bin thin cng nh.

3.12 CU HI

1. Trnh by cc nh hng truyn sng trong mi trng v tuyn di ng.

2. Trnh by thuc tnh ca knh trong min thi gian.

3. Trnh by thuc tnh ca knh trong min khng gian.

4. Trnh by thuc tnh ca knh trong min tn s.

5. Trnh by mi quan h gia cc thng s trong cc min.

6. Phn loi pha inh phm vi hp

7. Trnh by m hnh knh trong min tn s.

8. Trnh by m hnh knh trong min tn s.

61


64/147

Chng 4: L thuyt chung v anten

CHNG 4

L THUYT CHUNG V ANTEN

4.1 GII THIU CHUNG


- V tr ca anten trong thng tin v tuyn

- Cc tham s cbn ca anten

- Cc ngun bc x nguyn t

4.1.2 Hng dn


- Tham kho thm [1], [2]


4.1.3 Mc ch ca chng

- Hiu c v tr ca anten trong thng tin v tuyn- Hiu v cc tham s ca anten

- Hiu v cc ngun bc x nguyn t (cc anten n gin nht)

4.2. MU

Sng in t c th truyn dn bng hai phng php:

- Truyn dn trong cc thit bnh hng nhng dy song hnh, cp ng trc, ng

dn sang, cp si quang... Khi truyn lan trong cc h thng ny sng in t b gii hn trongkhong khng gian ca thit b v c gi l sng in t rng buc.

- Bc x sng ra khng gian sng truyn i trong cc mi trng thc v c gi lsng in t t do.

Thit b dn

tai lieu tham khao anten va truyen song

Documents