Tóm Tắt Lý Thuyết Môn Anten Và Truyền Sóng

8/9/2019 Tm Tt L Thuyt Mn Anten V Truyn Sng

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Tm tt l thuyt anten v truyn sng

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Tm tt l thuyt mn anten v truyn sng


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Mc Lc

Tm tt l thuyt mn anten v truyn sng.......................................................................... 1

CHNG 1 : C SL THUYT ANTEN .......................................................................... 3CHNG 3: HTHNG BC X ....................................................................................... 6

CHNG 4: GII THIU CHUNG VTRUYN SNG...................................................... 9

CHNG 5: NH HNG CA MT T N TRUYN SNG V TUYN ................. 12

CHNG 6: NH HNG CA TN I LU VI TRUYN SNG............................. 14

CHNG 7:NH HNG CA TNG IN LY N TRUYN SNG V TUYN ........ 16

CHNG 8: TRUYN SNG TRONG THNG TIN VTINH ........................................... 19


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CHNG 1 : C SL THUYT ANTEN1.1.C sbc xin t C sbc xin tda vo phng trnh maxwel

Di V.= p Rot V. = -Ht Di V. = 0 Rot V. =+ Et

Sbin i ca E theo t to ra H xoy v ngc li ( l c sto sng in t)

1.2.Vai tr v nhim vca anten1.2.1. Nhim v Bc xin ttanten pht in tanten thu

Anten l thit bu ra ca my pht thanhhoc u vo ca my thu sng in t, lm nhim vchuyn i tn hiu. Anten

bao gm nhiu phn t. Tn hiu n cc phn tny c tnh ton v xl gipAnten xc nh c hng ca ngun tn hiu, tp trung bc xtheo hng mongmun v iu chnh theo sthay i ca mi trng tn hiu. Cng vic tnh tonny i hi thc hin theo thi gian thc ( realtime ) Anten c ththu hay phtngun tn hiu. Vi tnh cht nh vy, Anten phi m bo c khnng gim thiunh hng ca hin tng a ng v can nhiu.

1.2.2. Vai tr+ Anten pht : Bin i tn hiu in cao tn tmy pht thnh sng in ttdo

lan truyn trong khng gian.+ Anten thu : Tp trung nng lng sng in ttrong khng gian thnh tn hiu in

cao tn a n my thu

1.3.Kho st bc xca dng in v dng ttrong khng gian tdo Mc ch xc nh E v H Cch lm l gii hphng trnh Maxwell

Tnh cht tng qut trng khu xa, trong khng gian tdo ca mt hthngngun hn hp :+ Trng bc xc dng sng chy ( biu thbi hm ), lan truyn tngun ra xa v tn. Bin cng trng suy gim tlnghch vi khongcch.+ Vc t mt cng sut c hng ph hp vi hng bn knh ca htacu v phn btrong khng gian theo hm s

|, Gm,|+

|, m,|


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1.5.5.

Nguyn tbc xhn hp+ L phn tbc xbao gm mt ipol in v mt ipol tt vung gc vinhau

1.5.6.

Cp ipol vung gc

+ Nguyn tTuanike l mt thp ca hai ipol (in hoc t) t vung gcnhau trong khng gian, v c tip in sao cho dng in hay dng tchy trongcp ipol y c bin bng nhau, cn gc pha lch nhau 90. S


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CHNG 3: HTHNG BC X3.1.Gii thiu

- c c thphng hng, ngi ta chto anten gm nhiu anten bc xtonn 1 hthng bc x.

- Nu cc phn tbc xsp xp trn 1 ng thng ta c hthng bc xthng.- Xp cc phn ttrong 1 mt phng ta c hthng bc xmt phng.- Hay t cc phn ttrong 1 khi ta c hthng bc xkhi.

3.2.Hthng bc xthng, mt phng, khi3.2.1.

Hthng bc xthngCs: da trn l thuyt nhn thphng hng.L thuyt nhn thphng hng: Nu hthng bc xthng c N phn tging ht nhau t thng hng cch u nhau, dng in trn cc phn ttha mn.

1= Trong : an: l tsbin 2 dng in In v I1.

l gc sai pha 2 dng in In v I1.Trng bc xca hthng c xc nh:

EM = E1 + E2 + +En = - fN (,)

Vi fN (,)= f1(,)* fNk(,).Trong : f1 (, ) l hm phng hng bc xca phn tth1

fNk (,) l hm phng hng bc xca N phn t.3.2.2. Mt phng v khi

Coi N phn tca hng trn cng l 1 bc xmi khi hthng bc xphngc coi nh hthng bc xthng gm M phn t.Coi M mt phng trn cng l 1 hng bc xmi khi

3.3.Kho st trng bc xca 1 hthng bc xthng c N phn t- N phn tging nhau, t thng hng cch u nhau 1 khong l d sao cho tha

mn hthc:

1= - Githit sai pha trn dng in knhau v bng .- Dng in trn phn tbng nhau(do khng tnh tn hao) nn c an= 1.

Ta c:fkN(, ) = 1. + = , vi = kdcos() +

- Hm phng hng bin chun ha.


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- Tcng thc trn ta c thsuy ra: hm t cc i khi = 0. Khi th

cos(max) = --

- Tcng thc ta xt cc trng hp sau:

+ ng bin ng pha.+ Lun phin o pha

+ Gc pha ca cc phn tbin i theo quy lut sng chy.3.4.Hthng hai chn t

Trng hp: ||= 1, = 0+ Hng bc xcc i c xc nh tiu kin:

cos= 2Hoc

cos= y, = 0,1,2 vi +

Trng hp: ||= 1, = 180(hai chn tc kch thch bi cc dngin ng bin, ngc pha nhau)

+ Hng bc xcc i c xc nh tiu kin:

cos= 2 1Hoc

cos= 2 12

y, = 0,1,2 vi + +Hng bc xbng khng c xc nh tiu kin

cos= 2


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y = 0,1,2 vi < Trng hp: ||= 1, = 90

Ta c:

= 1 +Bin i vdng

= 2 cos 2 cos 4

+

Ta c hm bin phng hng:

||= 2 cos (2 cos

4)

Khi d= khi y

= sc:

+ Cc i bng 2 khi = 180+ Cc tiu bng 0 khi = 0

3.5.nh hng tng hgia cc phn tanten trong hthng bc x

Ta c:

C thni dng in trn 2 phn tnh hng tng hln nhau.


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CHNG 4: GII THIU CHUNG VTRUYN SNG4.1.Nhim vtruyn sng v tuyn

+ Xc nh trng ti im thu v tm phng php trng nhn c ti im thul ln nht.

+ Xc nh nh hng ca mi trng ti truyn sng, t a ra cc bin php hnchti a nhng nh hng xu khng mong mun.

4.2.Phn chia di tn v tuynLoi sng Bc sng () Tn s(f)

Cc di >10000m < 30KHzDi 10000 -> 1000 m 30-> 300KHzTrung 1000 -> 100m 300 -> 30000KHz

Ngn 100 -> 10m 3 -> 30MHzCc ngn 10m -> 1 mm 30 MHz -> 300GHz

4.3.Cc phng php truyn4.3.1.

Truyn sng bmt Nguyn l

+ Bmt tri t l mi trng dn khp kn ng sc in trng.

+ Ngun bc xnm thng ng trn mt t, sng in t truyn lan dc theo

mt t n im thu.

c im

+ Nng lng sng b hp th t i vi tn s thp, c bit vi mt tm, mt bin (dn ln)

+ Khnng nhiu xmnh, chophp truyn lan qua cc vt chn

+ Sdng cho bng sng di v trung vi phn cc ng

Ph hp vi sng cc di, di, v mt phn song trung.

4.3.2.

Truyn sng nhphn xtng ion Nguyn l

+ Li dng c tnh phn xsng ca tng in ly vi cc bng sng ngn


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+ Sng in tphn xsquay trvtri t

c im

+ Khng n nh do sthay i iu kin phn xca tng in ly.

Ph hp vi sng ngn

4.3.3.

Truyn sng trc tip trong tm nhn thng- Hai anten thu v pht phi c t cao trn mt t trnh bche chn bi

cc vt cn trn ng truyn hay cong ca tri t.- Sng truyn tpht n thu trong min khng gian nhn thy trc tipgia hai anten.

- c im: t phthuc vo iu kin thin nhin, sdng phbin.

Ph hp vi sng cc ngn


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CHNG 5: NH HNG CA MT T N TRUYN SNGV TUYN

5.1.Cc c im gy nh hng- Mt t khng bng phng (chcoi l phng vi cli nhv ln).- Thng sin ca mt t thay i theo su v khong cch truyn lan (trong

nghin cu ly thng sin tng ng)

5.2.Cc c im ca mt t n truyn sngPhng php truyn sng bmt : ph hp vi sng di,sng cc di v 1 phn sngtrung.

5.2.1. nh hng ca mt t n anten t cao (h>>)

Trng ti im thu:

1 2Ebthu E E

Ebthu max khi E1 v E2 cng pha

Cng thc vedenski:2

2,18 [ ]1[ ] 2[ ]

[ ] [ ]

P km DEh h m h m

r km m (mV/m)

5.2.2. nh hng ca mt t n anten t thpAnten t vung gc mt t

[ ].245

[km]

P km DEh

r (mV/m)

Anten t song song mt t

E//=0

Dng in trong anten nh cng bin nhngngc chiu anten tht

5.2.3. nh hng ca mt t n cli thng tincc i trong tm nhn thng

max 3,57( 1[ ] 2[ ])AB h m h m


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5.2.4.

nh hng ca mt t n hp thsng truyn

[ ]245

[ ]

P km DEh F

r km (mV/m)

F: nh hng ca mt t lm suy hao sng lan truyn ( 0 1F )

5.2.5.

nh hng ca mt t lm lch ng i ca sngSng truyn theo phng php bmt khi i qua nhng min t c thng sinkhc nhau th bi hng

5.2.6.

Nhiu xsng quanh mt t cu- Hin tng nhiu x:quo sng bun cong i quanh vt trn ng truyn ( thyr khi kch thc vt ).

- Mt t hnh cu cng l chng ngi vi struyn sng.

2

[ ]1'[ ] 2 '[ ]

[ ] [ ]

P km DEh h m h m

r km m

(mV/m)

Trong : 1' 1 1h h h 2 ' 2 2h h h

2( 1 )1

2

A Ch

a

2( 1 )2

2

B Ch

a

11

1 2

hA C r

h h

2

11 2

hB C r

h h


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CHNG 6: NH HNG CA TN I LU VI TRUYN SNG6.1.c im tng i lu vi truyn sng

Do chit sut kh quyn thay i theo cao nn sng truyn trong tng i lu bkhc xlin tc(un cong).

R=

dh

dn

1= 25000 km(kh quyn thng)

Thay a= atd (atd = 8500km) ng vi cc cng thc truyn sng thng

6.2.Hin tng phadinh- Khi nim: l hin tng trng in tim thu thay i theo thi gian

-Nguyn nhn xy ra: do thay i chit sut dn n thay i khcx=> thay ing i => gy ra phadinh

-Khc phc:Phn tp khng gian: sdng 2 hay nhiu anten pht hoc thu thu pht ng thi

1 tn hiu trn cng 1 tn s. do ng i tia sng khc nhau nn t bphadinh 1 cchng thi

Phn tp tn s: 1anten pht1 anten thu thu pht 1 tn hiu trn nhiu tn skhcnhau nhvy lm gim nh hng ca i pht ln cn

a nh hng ca phadinh vo trong tnh ton ng truyn

Pthu[dB]=Pphat+Gphat+Gthu-

G : tng ch

:tng suy hao

6.3.nh hng ca ma- Ma nh hng n truyn sng, c bit


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d: khong cch ng truyn

: bc sng truyn

6.5.nh hng suy hao do kh quyn,do nhit

Hp thu kh N2 v O2> 10GHz


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CHNG 7: NH HNG CA TNG IN LY N TRUYNSNG V TUYN

7.1. c im ca tng in li

- Tng in li l tng ngoi cng ca tri t (60-600km)- Chu tc ng trc tip ca bc xmt tri v cc dng ht v trdn n s

phn ly cc phn tthnh nguyn t:N2 N + N

O2 O + O

- Mt in tch N=(102106) t/1cm3, N thay i theo cao, ngy m.

Tng in li c chia thnh 4lp: D ,E , F1, F2. Lp D v F1 chtn ti vo nhnggiban ngy, cn lp E v F2tn ti cban m v ban ngy nhng mt in tca

ban ngy ln hn ban m.

7.2.nh hng ca tng in li n truyn sng v tuyn7.2.1.

Phng thc phn xtng in li khng n nh theo ngy m- Truyn sng trong mt khong thi gian ngn, cnh trong ngy, c thxem

trong thi gian , mt in tch N thay i cha ng k.- Thay i bc sng lm vic.

7.2.2.

nh hng ca hin tng pha inh i vi sng ngnPha inh l hin tng trng nhn c im thu thay i theo thi gian, lcmnh, lc yu


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Hnh xuyn c bn knh trong AB, bn knh ngoi AC c gi l min im lng. chng ta khng thu c tn hiu.


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CHNG 8: TRUYN SNG TRONG THNG TIN VTINH8.1.Gii thiu vthng tin vtinh

- Phng thc truyn sng l truyn thng trc tip- Vtinh c vai tr chuyn tip v pht thu

- Quo vtinh: Quo trn v quo elip8.2.c im knh truyn sng trong thng tin vtinh

- Cli thng tin l rt ln vi thng tin vtin sdng quo a tnh,khong cc mt t n vtinh 36000 km nn tn hao nng lng tnhiu trong khng gian tdo l rt ln. trln 0,2s.

- Di tn slm vic 152 GHz khong m cc suy gim do tng khquyn gy ra vn c thchp nhn c. Mt sbng tn phbin:bng C, Ku.

- Sng truyn qua 2 tng kh quyn: tng i lu v tng in li nn chu

nh hng ca cc tng ny (suy hao do ma, hp thkh quyn vquay phn cc).

Tóm Tắt Lý Thuyết Môn Anten Và Truyền Sóng

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