CHNG I. GII THIU 1.1 LCH S

* Thng tin in : - Telegraphy (1884) - Telephony (1878) * Nn tng l thuyt : LT trng in t Maxwell (1854) * H thng Telegraphy khng dy dng bc x in t (Marconi 1897) * n in t v pht dao ng (1904 1915)

Mng ni ht Anten pht anten thu

Thnh ph ng dn c

H thng yu cu ph TH hp (truyn thanh)

Dng bc x in t

Suy hao ph thuc khong cch theo quy lut ly tha

C ly thng tin ln

Dy i suy hao 23 dB(10 kHz/km) Truyn d liu tn s thp

Cp ng trc Tn hiu Video Tn hao 4 5 dB, quy lut hm m Khong cch thng tin hn ch

Thng tin di ng (tu b) Cc phng tin giao thng ng b, my bay.

CC H THNG THNG TIN

1

- Thng tin v tinh - Kinh t - Bo mt * Nhc im : Hiu sut thp

1.2 CC H THNG ANTEN

+ Anten thng dng : - Anten ru trn t - Anten tai th trn tivi - Anten vng cho UHF - Anten Log-chu k cho TV - Anten Parabol thu sng v tinh

+ Trm tip sng vi ba (Microwave Relay) - Anten mt - Anten Parabol bc nha

+ H thng thng tin v tinh : - H anten loa t trn v tinh - Anten cho thu sng v tinh - Mng cc loa hnh nn chiu x (20-30GHz)

+ Anten phc v nghin cu khoa hc QUY C V CC DI TN S Di tn s Tn, k hiu ng dng 3 30 kHz Very low freq. (VLF) o hng, nh v 30 300kHz Low freq. (LF) Pha v tuyn cho mc ch o

hng 300 3000kHz

Medium freq. (MF) Pht thanh AM, hng hi, trm thng tin duyn hi, ch dn tm kim

3 30 MHz High Freq. (HF) in thoi, in bo, pht thanh sng ngn, hng hi, hng khng

30 300MHz Very High Freq. (VHF) TV, pht thanh FM, iu khin giao thng, cnh st, taxi, o hng

0,3 3 GHz Ultrahigh (UHF) TV, thng tin v tinh, do thm, Radar gim st, o hng

3 30 GHz Superhigh freq. (SHF) Hng khng, Viba (microwave links), thng tin di ng, thng tin v tinh

30 300GHz Extremly high freq. (EHF)

Radar, nghin cu khoa hc

2

1.3 KHI QUT V TRUYN SNG IN T + Di pht thanh AM chun (0,55 1,6 MHz): Dng thp anten + Di sng di :

- Anten n gin vi li thp, t trn mt t. - Mode truyn: sng mt, suy hao ~ R-4. - Mc nhiu cao do nhiu cng nghip - Cn my pht cng sut ln (50-500kW) - Mc nhiu v suy hao cao - C ly thng tin c vi trm dm - Suy hao tng nhanh theo tn s (khng s dng cho TS>20MHz) - Chiu cao ca anten cn c la chn thch hp. - C th c hin tng Fading trong thi gian hng giy, pht, chu nh

hng ca nhit v m khng kh. khc phc Fading phn tp theo khng gian v tn s.

+ Di sng 30 40 MHz : - C th s dng s phn x t tng in ly - C ly thng tin hng ngn km cc dch v truyn thng quc t - S phn x ph thuc mt in t to bi bc x mt tri - Khng c s dng trn 40MHz (do xuyn qua v fading)

+Trn 40MHz - Truyn thng (TV, Viba) - Kch thc anten phi ln gp mt s ln bc sng - di sng Viba ( 3 30cm) c th dng nhng anten gng c li cao

(40-50dB), cng sut my pht gim, nhiu kh quyn gim, c th dng tn hiu bin nh

+ Di sng mm : - Suy gim sng do kh quyn hoc do ma tng - C ly thng tin b gii hn

3

CHNG 2 C S L THUYT AN TEN, CC THNG S C BN CA ANTEN

2.1 M U Mt s qui c v k hiu: ch nt mvector, ch nghingthng s + nh ngha anten: l mt cu trc c lm t nhng vt liu dn in tt, c thit k c hnh dng kch thc sao cho c th bc x sng in t theo mt kiu nht nh mt cch hiu qu. + Nguyn l hot ng: dng in thay i theo thi gian trn b mt anten bc x sng in t Anten l mt cu trc m dng thay i theo thi gian, c cp t mt ngun thch hp qua ng truyn hoc ng dn sng, c th b kch thch vi bin ln trn b mt anten. + Yu cu v cu trc anten: n gin, kinh t (v d : anten na sng) + Bi ton chnh ca l thuyt v k thut anten: xc nh phn b mt dng in J trn b mt anten sao cho trng bc x tha mn cc iu kin bin trn anten. Bi ton ny thng ch c th gii gn ng. + Phn b dng trn anten c th c xc nh chnh xc hn khi xc nh c c trng tr khng ca anten. + T c tnh tuyn tnh ca h phng trnh Maxwell, v nguyn tc c th xc nh c phn b trng tng khi bit phn b trng ca phn t dng. + Cc phng trnh Maxwell, th vector v th v hng l nhng cng c ton hc ch yu gii bi ton v anten. + Cc c trng c bn ca mt anten:

- Kiu bc x (hm phng hng). - rng tia, h s nh hng, in tr bc x.

+ Cc phn t bc x c bn: Phn t dng in nguyn t, vng in nguyn t, dng t nguyn t, vng t nguyn t.

4

2.2 PHNG TRNH MAXWELL V CC IU KIN BIN 2.2.1 H PHNG TRNH MAXWELL + i tng ch yu ca thuyt v k thut anten l kho st s bc x v thu trng iu ha ~ejwt. + Dng in v trng s c biu din di dng cc vector m cc thnh phn ca chng l cc s phc. Khi , trng thc c dng:

tjt )e(Re),( rEr = (2.1) + Cc phng trnh Maxwell: (2.2.a e)

+ Trong chn khng :

j

Dj

j

===

+==

J

B

D

JH

BE

0

(2.2a)

(2.2b)

(2.2e)

(2.2d)

(2.2c)

(2.3a);B (2.3a); , 00 ,HE ==D + );/(36

10 90 metFara

= )/(10.4 70 metHenry= + Trong mi trng c hng s in mi v dn in : dng dn EJc = (2.2b) => ( ) JJH +

+=++= E

jjEj

2.2.2 CC IU KIN BIN BIN CA MT VT DN L TNG ( = ): (2.5) Bn trong vt dn: E , H = 0 Trn b mt: n x E = 0, n . H = 0 Mt dng in mt: sJ = n x H Mt in tch mt: Dns .=BIN CA MT VT DN KHNG L TNG: Trng in t xuyn qua b mt vi bin gim theo hm m: e-z/ ( = (2/o)1/2 vi ng , = 6.6x10

mS /108.5 7=-3cm tn s 1MHz, v 2.1x10-4cm 1GHz (2.7)

V d: vi ng, = 5.8x107 S/m, = 6.6x10-3 cm tn s 1MHz, v 2.1x10-4 cm tn s 1GHz.

Trong a s cc trng hp thc t c th coi trng in t khng xuyn qua cc vt dn tt nh kim loi. Tuy nhin, khi tnh n in tr ca cc vt dn kim loi

5

th cn tnh ti tn hao Joule theo nh lut Ohm (tn hao ca ng truyn, ng dn sng) TNH TN HAO: T trng H to ra dng mt HnJ s = ( nh lut Ampere) Thnh phn tip tuyn ca in trng lin quan vi mt dng in mt:

ss JnZEn = (L Ohm) (2.8) Trong Zs l tr khng b mt ca vt dn: ( )

ss

jZ += 1 (Ohm/dt) (2.9) Bao gm thnh phn thun tr 1/s (in tr ca lp da c chiu su s) v thnh phn cm ng do s xuyn qua ca t trng. Tn hao trn n v din tch c cho bi phn thc ca vector Poynting hng vo vt dn ti b mt vt dn:

s

sJP

2

21

= (2.10) - Nu = v cng, th chiu su lp da, v do tr khng b mt v tn hao = 0 - Thng ngi ta so snh tr khng b mt vi tr khng ca khng gian t do:

OhmZ 3772

1

0

00 =

= (2.11)

- Vi Cu, ti 1MHz, Zs = 2.6x10-4(1+j) Ohm - Kt qu trn c th p dng cho cc vt dn tt khc v cho cc b mt c bn knh cong ln hn nhiu so vi su lp da. BIN GIA HAI IN MI: 21 EnEn = , 21 HnHn = , 21 DnDn = 2.2.3 TH VECTOR V TH V HNG T (2.2a), (2.2b) v (2.3) => ,020 JjEkE = (2.12) Vi l s sng ca khng gian t do ( ) 2/1000 =k- Theo phng trnh ny in trng c th c tm trc tip khi bit phn b dng. Trong thc t c th n gin ha bi ton nh th vect A v th v hng : Mt khc bt c vect no vi zero curl u c th biu din di dng gradient ca mt hm v hng. Do c th t : AB = (2.13) - V 0= A nn A c gi l th vector. - S dng cng thc ca gii tch vector => ( )++=+ 000202 . jAJAkA (2.14) - n gin ta chn : = 00jA (iu kin Lorentz) (2.15) - Khi (2.14) tr thnh : JAkA 0202 =+ (2.16) - Thay cc phng trnh (2.14) v (2.15) vo (2.2c) => (2.17) 0202 / =+ k 6

- S dng iu kin Lorentz v (2.14) => 00/. jAAjE += (2.18)

- Trng hp ngun dng : zz aJJ .= th zz aJJ .= v ( ) zz JAk 0202 =+ (2.19)

2.3 BC X CA PHN T DNG IN - nh ngha phn t dng in: dlI thng, rt mng, rt ngn. Gi thit d liu //

(z). - Th vector ch c mt thnh phn theo phng (z) tun theo PT (2.19). trong

Jz=I/dS, vi dS l tit din ca phn t dng. Th tch dV

- Cc mt sng ng pha c dng hnh cu bn knh r, tm = gc to .

- Vn tc pha = (cng thc) - Bc sng

fC

wC

koo===

2

2 (2.25)

Tm biu thc ca ca trng: - S dng (2.13) v (2.18) v h to cu. - Biu din A

r theo cc thnh phn trong h to cu v lu rng:

Ta c: ( )Aaer

IdlA rjkt sina-Acos

400

= (2.26) Dng (2.13):

aerrjkIdAH rjk02

0

0

14sin.1

+== l (2.27)

T (2.18) => aEaEjAAjE rr +=+=

00

. (2.28)

- Nu r rt ln so vi bc sng th : (vng xa) b qua cc s 21r

, 31r

arekIdjZE

rjk

4sin

0

00

= l (2.29a)

arekjIdH

rjk

4sin

0

0

= l (2.29b)

* Nhn xt: - Vy khu xa, trng bc x ch c thnh phn ngang, in trng v t trng

vung gc vi nhauv vung gc vi phng truyn sng. t s bin ca chng

chnh bng tr khng sng ca khng gian t do Z0; 21

0

00

= Z

- Dng vector:

HaZE r = 0 (2.30a) EaYH r = 0 (2.30b) Trong : 100 Z=Y- Trng khng c tnh i xng cu. ( E Z v H ph thuc sin ) * Vector Poynting phc:

( ) 2222020** 32sin.21

rakdZIIHE rl= (2.31b)

8

C dng thun thc, (trng bc x) c hng trng vi hng lan truuyn, v cng sut bc x gim t l nghch vi r2

* Cc s hng cn li ca (2.27) v (2.28): chim u th khi r < o v to ra trng phn ng khu gn v tnh thun o ca vector Poynting.

- Nu kor rt nh sao cho c th thay th: (khu gn) 10 rjke

a

rkIdH4

sin0l= (2.32a)

++

+= arjkrarjkr

IdZE r0

20

20 11sin11cos24

l (2.32b)

Cho k0rphng trnh (2.32b) tr thnh

+= ararQdE r 3

sincos24 3

l (2.32c)

Lu : - Tng t nh phn b trng tnh ca mt dipole in. - Mc d trng khu gn khng ng gp vo cng sut bc x, ch

lin quan n s tch t nng lng khu vc bao quanh ngay gn anten, nhng cn c tnh n khi tnh tr khng anten.

- Biu thc ca vector Poynting phc, c tnh bi vic s dng cc biu thc tng qut ca trng s c phn thc (phn lin quan trc tip n bc

x) ch bao gm trng bc x cho bi biu thc (2.31) __________________________________________________

2.4 MT S CC THNG S C BN CA ANTEN Bc x ca mt phn t dng in cn c gi l bc x lng cc. c dng

nh ngha cc thng s c bn ca anten ni chung. Kiu bc x: Phn b tng i ca cng sut bc x nnh l hm ca hng bc x trong

khng gian - Cng sut bc x ca dipole nguyn t t l vi sin2 (2.31). Kiu bc x c

dng hnh s 8 nh hnh sau: (hnh v) -a) Mt 3 chiu -b) Mt E -c) Mt H * Tia na cng sut: Gia cc im m cng sut bc x = cng sut cc i

9

H s nh hng v li: - Cc anten thng khng bc x dng u theo mi hng. - S thay i ca cng bc x theo hng khng gian c m t bi hm h

s nh hng D(,) ca anten. - Cng bc x l cng sut bc x gc t (hay gc khi). Chnh bng tch ca

vector Poynting vi r2. - i vi dipole nguyn t: (lu (31))

( ) 22

20

20

*

32sin.

kdZIIddPr l= (2.33)

nh ngha h s nh hng:

( )r

r

Pd

dPD = 4, (2.34)

Vi Pr l cng sut bc x ton phn. - Vi dipole nguyn t: t (2.33)=>

( )12

. 200* kdZIIPr

l= (2.35) V d =sin d d. T (2.33) v (2.34) => (2.36) ( ) 2sin5,1, =DCc i t gi tr 1.5 khi =/2. H s nh hng cc i (thng vit tt l h s nh hng) c trng cho

kh nng ca anten tp trung nng lng bc x theo mt hng cho trc. Anten v hng: Bc x ng u theo mi hng. li G(,)ca 1 anten c nh ngha tng t nh h s nh hng, nhng

cng sut bc x c thay bng cng sut ton phn t vo anten Pin. Hiu sut ca anten: inr PP = (2.37) ( ) ( ) ,, GG = Vy : (2.38) * Effectve isotropic radiated power: (EIRP)=(input power)x(maximum gain). chng hn 1 anten c li =10, cng sut ngun = 1W ch t hiu qu nh 1

anten c li 2 v cng sut 5W. C hai anten c sng 1 ch s EIRP.vy c th gim cng sut my pht nu s dng anten c li cao.

* in tr bc x Ra : - nh ngha: l in tr tng ng tiu th cng 1 lng cng sut nh anten

bc x khi dng cung cp nh nhau. - i vi anten dipode :

( ) 20

22

00a 806

R

==

ll dkdZ=> (2.39)

10

Trong : 1200 =Z , 0

02 =k

V d: dl = 1m, )1(3000 MHzfm == , Ra = 0,0084 . Nhn xt: - Ra thng rt nh - T l thun vi din tch ca anten Cc anten dipode thng c in khong ln v hiu sut thp, do li thp.

Mt anten c hiu sut cao phi c kch thc so snh c vi bc sng. Trong di sng pht thanh (500-1500kHz, tng ng 600-200m )cn anten vi cu

trc n gin nh cc thp cao. ______________________________________________

2.5 Bc x ca vng in nguyn t : + Phn t dng bn knh r0, cng I , trc ca phn t //z. 20.. rdt + Nu r0

*Cng sut bc x ton phn :

12Pr =400

2 kZM (2.45)

* in tr bc x tng ng:

2

0 a

02320 = rR (2.46) .10-3 (rt nh).

ng c hiu sut cao nhng c ph tn ng

_______________________________________________

2.6 BC X T CC PHN B DNG BT K

hn b dng

V d : ro = 10cm , ti 1MHz , Ra = 3,8* Nu dng N vng y Ra N2 ln Dng cho anten thu (radio).Anten vng kh hiu r . li > Ra.

Xt th tch V vi p )( 'rJ . Phn t dng ')( ' dVJ r ng gp vo th vector 1 lng : (2.24)

Rjke

RdVrJ

0

4)( ''0

(2.47) 'rrR = Vi

* Vng xa:

'.rarR r (2.47 )

=>

=Vr4

rajkr

rjk

dVeJerA r '.)(0'

0'0

)( (2.48) T (2.13) v (2.18) khi ch tnh n cc s hng cha 1/r =>

[ ] = V

rrr4rajk

rr

rjk

dVeJaJaeZjkE r '.)()(00'

0''0

. (2.49)

Khi dng in I phn b trn ng cong C, th PT(2.49) =>

( ) [ ] = C

rajkrr

rjk

r deIaaaareZjkE r '.00

'0

0

)'().(4

ll (2.50) n v dc theo C theo hng ca dng in

* Tng qut :

Vi a : vector

( ) ),(4

000

= freZjkE

rjk

r (2.51)

12

),( f :hm phng ng ca trng bc x. ________________________________________________

2.7 NGHIN CU THC NGHIM TR KHNG ANTEN * Mc ch : - Phi hp tr khng vi ng truyn tn hiu . hiu sut cao * Trng hp l tng : tr khng vo Ra ni trc tip anten vi ng

truyn c tr khng c trng Zc Zc = Ra* Xt : Anten c tr khng Za ni ngun qua ng truyn c Zc+ H s phn x sng ti u vo :

ca

ca

ZZZZ

+= (2.59)

VSWR ( Voltage Standing Wave Ratio )

+=

11

VSWR (2.60)

* iu kin phi hp tr khng : VSWR 1,5 gi tr VSWR = 1,5 tng ng vi || = 0.2 hoc h s phn x cng sut = 0.04 ( 4%)

* Tr khng anten :

*00

m

21

)W(2II

WjPPZ edra++= (2.61)

Vi : Pr : Cng sut bc x Pd : Tn hao Ohmic Wm : T nng trung bnh We : in nng trung bnh c tch tr vng cm ng (vng gn) I0 : Dng cp vo u vo anten => Khi Wm = We -> Phn cm ng ca Za = 0 (k cng hng) + Vi anten dipole : iu kin cng hng xy ra khi chiu di anten = n ( bc

sng) + Tnh in tr thun ca dipole na sng :

- Vt liu : Cu - Bn knh ng ng : ro- Dng trn anten : => mt dng in mt : zkI 00 cos

0

00

2cos

rzkI

- Tn hao Ohmic:

13

SS

d rRI

rIrP

0

020

2

0

000 82

1128

2 ==

= (2.62)

Vi r0 = 0,5cm, =>>>==>== 13,73062,010.6,6),100(3 60 aS RRRmMHzm ____________________________________

2.8. TR KHNG TNG H

+ Khi 2 anten dipole t gn nhau phn b dng trn mi anten chu nh hng

bi trng bc x ca anten cn li. z1, z2 : to dc theo b mt z1, z2 : to dc theo trc Gi : - A11(z1) : th vector ti z1 gy bi dng I2(z2) - A12(z1) (cng thc)

- Th vector tng cng ti z1: ( ) ( )1121111 zAzAAZ += (2.63) - Cng trng :

11 )(1

21

220

00)(1 ZZ Az

kj

E +=

iu kin bin : Ez = -Eg khi arbzb =>> ,22 (2.64) Ez = 0 khi arzb =>> ,2 l Vi b : rng khe gia hai chn t Eg : in trng gia hai mp khe gia hai chn t.

agg Z

IbV

IV ==

)0()0(

. : tr khng vo ca dipole ( khi b>> c th biu din ggb VbE =0lim )

V )(zEE gg = vi )(z : hm delta Dirac )(z = 0 khi 0z (2.65)

1')'( =

dzz

zz

C th vit li (2.63) cho c 2 b mt dipole 1 v 2 : [ ] )()()()( 11001121112

1

220 zVjzAzAz

k =++ (2.66a)

[ ] )()()()( 220022222122

220 zVjzAzAz

k =++ (2.66b)

14

H (2.66) c nghim dng :

101101000

112111 cossin2)()( zkCzkVYjkzAzA +=+ (2.67a)

202202000

222221 cossin2)()( zkCzkVYjkzAzA +=+ (2.67b)

Cc hng s C1, C2 phi tho mn iu kin bin :

0)(2)(1 21 == ll II Khi (2.67) tr thnh :

2,1,''0 )(

4)(

0

=

= jijjjij

Rjk

iij dzzIRezA

j

j

ijl

l

(2.68)

Vi : [ ] 21221111 )'( azzR += [ ] 21222112 )'( dzzR += (2.69) [ ] 21221221 )'( dzzR += [ ] 21222222 )'( azzR += T (2.69) v (2.67) c th vit

22)0(221)0(12

12)0(211)0(11

ZIZIVZIZIV

+=+=

(2.70)

T nguyn l thun nghch => Z21 = Z12 Tr khng tng h Z11 v Z22 Tr khng ring, khc mt mc no vi tr khng vo ca mi anten c lp.

- Nu chiu di cc dipole 2

0 , v cch nhau 5

0 th tr khng ring tr khng vo ca mi anten c lp.

+ nh gi tr khng tng h : T nguyn l thun nghch => Tng tc ca trng gy bi dng I1(z1)vi dng I2(z2) v ngc li =>

(2.71) 111112222221 )()()()(1

1

2

2

dzzIzEdzzIzE zz

=l

l

l

l

Hoc di dng vector :

+=+

1

1

2

2

2 111221

220)1(122212

2

220)(2 )()()()(

l

l

l

ldzzA

zkIdzzA

zkI zz (2.72)

Nhn (2.66a) vi I1(z1) ri ly tch phn theo z1 (2.66b) vi I2(z2) ri ly tch phn theo z2

15

=> 1

'2

1221

220

)0(2)0(1

)(2)1(1)0(2

1'

1121

220

)0(1)0(1

)(1)1(1)0(11)1(

)0(1

)1(1100

4)(

4)(

1202

2

2

2

'2

1

1

1

1101

1

1

1

'1

dzdzR

ez

kII

III

dzdzR

ez

kII

IIIdz

II

VYjk

Rjkzz

Rjkzz

zz

++

+=

l

l

l

l

l

l

l

l

l

l

(2.73)

-Gi thit phn b dng chun ho I1(z1)/I1(0) v I2(z2)/I2(0) khng thay i do tng tc gia cc dipole =>cc tch phn. Trong (2.73) khng ph thuc vo dng vo anten (v c chun ho) . I1(0) v I2(0) c th xem nh cc bin c lp.

So snh (2.73) vi (2.70) =>

1212

21

220

)0(2)0(1

)(2)1(1

0012

1201

1

2

2

2 )(4

dzdzR

ez

kIIII

YkjZ

Rjkzz

+=l

l

l

l (2.74) * Nu 222

021

= ll : thc nghim v l thuyt chng minh

20

220

2

22

10

110

1

11

sin)(sin

)0()(;

sin)(sin

)0()(

ll

ll

kzk

IzI

kzk

IzI ==

(2.75) => (2.74) tr thnh :

22200

10212010

012

2

2

002010

)(sin)cos(2)sin()sin(4

dzzkR

ekR

eR

ekk

jZZRjkRjkRjk

+=

l

lllll (2.76)

Vi

[ ] 2122111 )( dzR += l [ ] 2122212 )( dzR += l [ ] 212220 dzR +=

___________________________________________

16

CHNG III : CC LOI ANTEN DIPOLE 3.1 ANTEN DIPOLE NA SNG * Nui= dy song hnh * Gm 2 nhnh 4

0 * Th nghim +LT phn b dng c dng sng ng hnh sin :

zkII 00 cos=

44 00 z (2.52)

S dng (2.50) vi zz azraa ', == v cos=zr aa =>

=

4

4

cos'0

000

0

0

00

')'cos()cos(4

dzezkaar

eZIjkE zjkzrrjk

areZjI rjk

sin

)cos2

cos(

2

000

= (2.53)

=>

aerZjIaHH rjk

sin

)cos2

cos(

2. 000 == (2.54)

* Mt dng cng sut :

2

220

20

sin

)cos2

cos(

8E

21

=

r

ZIHRe (2.55)

* Cng sut bc x ton phn : tch phn (2.55) trn mt cu r

ddZI

Pr ..sinsin

)cos2

cos(

8

2

0 0

2

02

0

= (2.56)

Tch phn (2.56) c tnh theo tch phn cosine =>

20565,36 IPr = (2.57)

* in tr bc x ca anten dipole na sng 73,13 => Dy song hnh nui anten cn c tr khng 73,14 * H s nh hng :t (2.55)v(2.57) =>

17

( )2

sin

)cos2

cos(64,1,

=

D (2.58)

Dmax = 1,64 Phn t dng Gc na cng sut = 780 * Ra = 73,13 l rt ln tr khng vo (b qua cm khng vo)

3.2. ANTEN HNH NN + Gm 2 hnh nn i nh, gc m 0 , c kch thch ti tm gia 2 m tip xc

hnh cu, bi ngun in p hnh sin. (hnh v) + Nghin cu l thuyt bi tc gi Schelkunoff chng minh : cu trc hnh nn

s cho sng in t ngang hnh cu TEM vi cc thnh phn E , H , ch ph thuc vo r v . Khi cc phng trnh Maxwell s tr thnh :

=

aHjrErra

or )( (3.1a)

=

aEjrHrr

aHrr

ao

r )().(sinsin (3.1b)

V gi thit Er = 0 nn s hng u tin trong (3.1b) phi =0 => c th t :

constC

rfCH=

= sin)(.

(3.2)

=> (3.1a,b) tr thnh :

sin)(..)( rfrCjrE

r o=

(3.3a)

rEjrfr

rC o= )

sin)(..( (3.3b)

* Vi phn (3.3a) theo r v thay vo (3.3b) =>

rEkrEr202

2

)( =

(3.4)

=> sinsin00

)(2)(1 reC

reCrE

rjkrjk

+=

Ch v phi ca (3.3a) t l vi sin1 =>

18

sinsin00

reC

reCE

rjkrjk

+ += (3.5)

Cc sng cu lan trun ra xa v vo trong ngun vi bin C+ v C- , tng ng .

S dng (3.1a) => sinsin00

00 reYC

reYCH

rjkrjk

+ += (3.6)

2

1

0

00

= Y : dn np sng ca khng gian t do

* in p gia hai hnh nn = tch phn ng ca E t o n - o : (3.7)

rjkrjk eVeVV 00 + += Vi )

2(cotln2 0gVV = , V c dng sng in p.

* Mt dng mt trn hai hnh nn trn v di l:

sinsin00

00 reYC

reYCJ

rjkrjk

s

+ = Hng theo trc z

dng ton phn trn mi hnh nn l I = 2rsinoJs )( 0000 rjkrjkc

rjkrjk eVeVYeIeII ++ == (3.8) I c dng sng dng:

(3.9) )

2(cotln 0

0

g

YYc = : Dn np c trng ca ng truyn hnh nn

Tr khng c trng:

)2(cotln120)

2(cotln 0001 gg

ZYZ cc === (3.10) * Nu ti , cc mt nn h mch l tng th I = 0 v 0l=r (3.11) 0000

ll jkjk eVeV + =

00

00

ll

tgkjZZtgkjZZZZ

tc

ctca +

+= (3.12) Zt : Tr khng u cui hiu dng, do dng cm ng (cng thc) * Khi o

azZ zc

2ln120)( = (3.13)

=>

= 1ln120)( aZ zcl

(3.14)

in tr bc x ca anten tr (xilanh) :

2220

=

oaR

l (3.15)

* Thc t t s dng anten hnh nn c gc m nh thay cho anten xilanh v kh ch to v ph hp

* Anten nn vi gc m rng thng c ng dng nhiu hn v ph rng

*V d : o = 30o , 23200 >> l => in khng hoc 50

Tr khng vo 130 20 Nu ni vi ng truyn c tr khng c trng 158 th s phi hp tr

khng rt tt trong di tn 3 -1:

min

maxf

f

* 1 dng gn ging trong thc t l anten tam gic, s dng trong di UHF t knh 14 n 83 (450900 MHz) (anten c o)

* Mt kim loi, Cu, Al hoc cu trc dy __________________________________

3.3 ANTEN GP * Cu to:

- Gm 2 vt vi nhau u cui - Mt trong hai c h ti tm v ni vi ng truyn. - Ra = 292 -> ni vi 300cZ (ph bin cho anten thu) - Do c im cu trc c th b c mt s thay i tr khng vo anten theo

tn s ph rng . - Khi l o/2: dng trn mi vt dn l nh nhau nu c cng ng knh (do

tr khng tng h);I1=Io cos koz. - Nu hai vt dn dt rt gn nhau c th b qua s khc pha ca trng bc x trng tng =2 ln trng ring , ( ) )(4 rPtP rr = (ring) =>

2

056,364 IPr = Trong : l dng cung cp bi ng truyn 0I

== 5,29213,734aR (3.19) 20

__________________________________________ 3.4 ANTEN DIPOLE NGN + Tn s thp bc sng di hn ch kh nng s dng dipole na sng

gim chiu i anten gim Ra phi p dng 1 s bin php b dung khng mc ni tip anten vi cun cm gim hiu sut v li.

(hnh v) Tng s phn b ng u ca dng trn anten tng Ra. + C th cc t ghp vo u cui ca mi nhnh anten . + C th ghp thm 4 hay nhiu hn cc thanh vt dn kiu hnh qut u cui

mi nhnh dng s khng =0 u cui mi nhnh, m =0 cui cc nhnh ca hnh qut in tr bc x s tng

2

1

1

24

++llll ln(l1=1/2 mi thanh hng qut)

_____________________________________________ 3.5 ANTEN N CC

+ Cu trc t mt na ca anten dipole c t trn mt t thng c chiu di

041 = c s dng ch yu cho pht thanh AM (500-1500kHz). L do : l loi

anten ngn hiu qu nht cho cc bc sng t 200 600m - S phn cc theo phng thng ng c tn hao t hn so vi phn cc theo

phng ngang (so vi t ), vng tng s AM. - Nui= cp ng trc c v ngoi ni t.

+ Cu trc khc : - Mt anten n cc t trn nh 1 ct . - 4 ng nm ngang c chiu di 0,3, to ra mt mt t o, sao cho kiu bc

x v li ca anten anten na sng (nh hiu ng th nh), Ra 36,56. + S dng ch yu lm cc trm c s thong tin di ng . + Mn chn o (mt t o ) phi c dn tt. Thng s dng 120 dy ng tm

v c chiu di tng /3 t di anten 1 khong vi inch ng gp 1 lng gia tng tng ng 2 vo tr khng vo ca anten hiu sut anten 95% .

+ Vi cc tng s thp hn thng dng cc phn ghp to cng hng _____________________________________________

21

3.6 BALUN B PHI HP TR KHNG + Kt ni 1 h cng bng vi 1 h khng cn bng . + Anten dipole nui bi ng dy song hnh c cn bng so vi t khi 2 na

ca dipole c cng nh hng v v tr so vi t . - Khi 2 na ca dipole c in th V v -V so vi t . - Khi anten dipole c nui bi cp ng trc th h khng cn bngdng

c kch thch trn mt ngoi ca v cp ng trc dng 2 trn na ca dipole hin tng giao thoa cc trng bc x thay i kiu bc x ca dipole cn PALUN.

- BALUN c cu trc theo rt nhiu kiu ph thuc vo di tng cng tc . - lm nght 1/4 bc sng : s dngtn s cao . + BALUN dng cho anten thu TV. ________________________________________________

22

CHNG4 ANTEN MNG - S dng trong cc h thng thng tin point_to_point i hi tnh nh hng rt

cao ca anten chm bc x t hp cc anten n gin theo 1 trt t nht nh : anten mng c li cao cng sut pht gim.

- Xt mng gm N anten ging nhau, c cng tnh nh hng, c kch thch vi bin

- Xt 1 anten chun t ti gc to c cng in trng bc x dng :

( ) refE

rjk

r 40

,

= (4.1)

( ),f : hm phng hng ca cc anten phn t ca mng.

- vng xa : irr -> coi cc tia t cc anten phn t n im kho st // ->

ir rar=iR

- Trng to bi phn t th i s chm pha 1 lng so vi anten chun gc to .

ii rak

0

- Trng tng c dng:

( ) =

+

=n

i

rajkiji

jkr

rireC

refE

1,

0

4 (4.2) + Nguyn tc nhn ging phng hng : (4.2) c th c vit di dng :

( ) ( ) refFE

jkr

r 4,,

=

(4.3) ( ) =

+ =n

i

rajkiji

ireCF1

,0

H s nh hng

2E

( ) ( ) ( )2

,

2

,, FfD= (4.4)

Nguyn l: hm phng hng ca 1mng = hm phng hng ca 1 anten phn t x hm phng hng c trng cho mng .

+ Ngm nh : B qua tc ng tng h. ______________________________________________

23

4.1 MNG NG NHT 1 CHIU

Xt mng N +1 phn t cc dipole na sng cch nhau cckhong =d, c kch

thch bi cc dng coa cng bin C = Io lch pha lin tip .dn=n..d. => Kiu ca trng bc x Fc dng (cn gi l h s mng hay nhn t mng)

( ) ( )( )

+

+

+

=2

cossin

cos21sin

0

0

0 dk

dkN

IF

- Khong cch gia cc tia chnh v tia ph u tin :

13+

=N

U - Khi N>>: bin tia ph u tin = 3

2 (hay 0,21) bin tia chnh.

- C N-1 tia ph gia 2 tia chnh - Kiu mng Ftun hon vi chu k 2 theo bin u - V : dkdkudk 000 cos = nn ch c mt khong ca u c

ngha vt l gi l khng gian kh bin : 00

22

dud

- Thc t thng yu cu ch c 1 tia chnh trong vng kh kin chn dipole nh 2 trng hp :

1) Mng ng pha:

- Khi = 00 0 =u => tia chnh xy ra khi u = 0 hay 20cos ==

- Gc ga hai im khng ca tia chnh xc nh t iu kin:

=+ cos2

10dk

N (khi gc ca hm sin t s ca =F )

- Vi 2 >>N , t = 2 => 0

-> ==

sin2

cos => rng tia chnh BW :

LdN

BW 00 2)1(

22 =+== , vi L = (N+1)d : chiu di mng Nhn xt * c trng ca mng ng pha l rng tia t l nghch vi chiu di

ca mng * BW=6 hay 0.1 rad khi L=20 o, kh thi tng s cao.

24

+ vic tnh chnh xc h s nh hng ca mng l rt kh . Trong trng hp dn gin ca mng ng pha th cn phi tnh tch phn sau:

( )

dddk

dkN

..sincossin2sinsin

cossin21sin)cos

2cos(2

0 0 0

0

+

- Cng c th nh gi gn ng h s nh hng =(4/ gc c chim bi chm tia chnh ) tch ca gc gii hn bi cc tia na cng sut trong mt phng E v mt phng (Hnh v) .

- Gc na cng st trong mt phng E=78o(1,36rad). - Gc na cng sut trong mp (Hnh v) c xc nh t iu kin

( ) 2220 )1(22

12

1sin21 +

=

+=>+= NuuNINF

(v mu s ca F thay i chm hn nhiu so vi t s) Dng php khai trin chui

165,2

21 += Nu

=> rng tia na cng sut :

dNdkNBW

)1(

65,2)1(

265,22 002

12

1 +=+==

=> H s nh hng :

02

1

)1(48,536,12

4

dNBW

D += (Tha s 2 mu s tnh cho 2 tia ) - Nu cc phn t ca mng l cc anten v hng th kiu bc x s c tnh i

xng trc quanh trc ca mng. Khi gc na cng sut = .

002

1

)1(37,2)1(65,2

22

4

dNdN

BWD +=+=

2/ Mng c pha dng in bin i theo quy lut sng chy: n ging xt trng hp u0 = -k0d bp chnh cc i khi :

0cos00 ==>=== dkdkuu hng bc x cc i trc ca mng (trc x)

H s mng F c dng:

25

( ) ( )( )

+

=1cos2sin

1cos21sin

0

0

0

dk

dkN

IF

* Trung bc x, kiu mng c tnh i xng trc quanh trc ca mng - Tia chnh hay bp sng chnh = 0 khi: ( ) ( ) = + 1cos21 0dkN - Khi N>> ln cn im khng c th vit ( ) 21cos

2 =

( ) 2102 22

)1(2

2

==>+=

LBW

dN

L = (N+1)d : Chiu di mng => BW t l nghch vi chiu di mng (o theo bc sng) * nh gi h s nh hng: theo gc na cng sut:

( ) 2

1cos,2

12

21

210

=+=

NIF

( )2

1

0

21 1

63,1

+= Nd

=> - Gc c gii hn bi chm tia na cng sut:

2

21

21

2

0 0

cos12..sin2

1

==

dd

=> 0

)1(73,44 dND +=

+ iu kin Hansen Woodyard:

NdkuddkNdN ===>= 000 .....

=> Cc i chum tia chnh (hay bp sng chnh ) xy ra khi

Ndkuu +== 00

- th F(u) c dng tng t mng End fire nhng b dch tra mt on N

=> phn kh kin ca bp sng chnh b thu hp h s nh hng tng v cng sut bc x ton phn (t l vi din tch gii hn bi ng F trong vng kh kin.)

___________________________________________

26

4.2 MNG NG NHT 2 CHIU

- Phn b trong mt phg xoz - C (N+1) (M+1) phn t - Kho st mng gm cc anten phn t l cc dipole // oz c kch thch bi cc

dng in c cng bin , phn b pha c dng tng ng vi v tr (n,m) dmdjn je =- C th xem h l 1 mng ca M+1 mng mt chiu N+1 phn t => H s mng ca 2 chiu = tch ca h s mng ca M+1 anten phn b theo

trc oz vi h s mng ca mng N+1 phn t phn b theo trc ox ( )

( )( )[ ]( )

( )( )[ ]

+

+

+

+

+

+

=cos2sin

)cos(21sin

cossin2sin

)cossin(21sin

0

0

0

0

0),( dkd

ddkM

dkd

ddkN

IF

- n gin t: ,cossin0 dku = du =0

,cos0 dkv = dv =0 =>

( )[ ]{ } ( )[ ]{ }( )[ ] ( )[ ]2/sin.2/sin

)(2/1sin.)(2/1sin

00

000)( vvuu

vvMuuNIF u ++++++=

+ Hng bc x cc i chnh c xc nh t iu kin: u = u0 v v = v0.

+ Nu 0== (mng ng pha) th hung chnh mt phng mng nh hng y. Vi v thch hp c th iu khin hng chnh theo mt hng tu mng pha

+ rng gc ca bc sng chnh, trong cc mt phng xy v yz c xc nh bi cc iu kin:

=+=+ vMuN2

1;2

1

Tng t vi mng mt chiu :

( )dNxy )1(

2BW 0+= ; ( )

dMyz )1(2BW 0+=

Tng t nh mng 1 chiu, gc na cng sut c xc nh :

dNxy

)1(

65,2BW 02

1 +=

; dMyz

)1(

65,2BW 02

1 +=

+ H s nh hng c tnh gn ng:

27

02

12

1

(max) 83,8BWBW2

4

ADyzxy

=

Vi A = (N+1)(M+1)d2 : din tch ca mng 2 chiu. => H s nh hng cc i t l thun vi din tch o theo dn v bnh phng

bc sng. y l c trng chung cho tt c cc anten. + S thay i ca rng tia chnh khi tia chnh lch khi hng vung gc vi

trc ca mng: gi thit 00 cos,0 dkd == tia chnh ti gc o so vi trc x v nm trong mt phng xy. S dng khai trin Taylor c th vit :

( ) ))(sin(coscos 000000 == dkdkuu Biu thc xc nh rng tia chnh :

0

0000 sin)1(

22))(sin(2

)1(

dN

dkN +==>=+

Nhn xt : 0sin

1 ln, ( di mng chiu ln phng

vung gc vi

+ 0sin)1( dN0 )

_______________________________________________

4.3. TNG HP KIU MNG Nhn xt: - Khi cc phn t ca mng c kch thch bi cc dng c cng bin

, c th to ra cc kiu bc x vi cc bp sng hp nh phn b pha thch hp ca cc dng kch thch.

- C th dng phn b bin ca cc dng kch thch iu khin hnh dng v rng ca cc bp sng chnh cng nh v tr v ln ca cc bp sng ph c th to ra mt kiu bc x gn ging vi mt kiu bc x cho trc bi ton tng hp kiu mng, hay tng hp mng.

1) * Phng php chui Fourier : Xt chui 2N+1 phn t c kch thch bi cc ng pha, c bin Cn,

=

==>=N

Nn

ndjkneCFNNn

cos0

+ Nu chn Cn = C-n => =+=N

nn nuCCF

10 cos2

28

- Bng cch chn cc h s cn thch hp c th lm gn ng mt kiu bc x Fd(u) tu .

+ Ch : 0 - kod u kod l khong ng vi vng kh kin ; Tuy nhin Fd(u) s c tng hp trong mt chu k - u .

* Mng Chebyshep : p dng c thit k mng vi rng nh nht cho mt mc ph cho trc hoc ngc li mt mc ph nh nht vi rng cho trc . C th s dng cc a thc Chebyshev tm ra phn b dng ph hp vi mc tiu thit k. Phng php c ngh u tin bi C. L . Dolph mng Dolph Chebyshev

* Cc tnh cht c bn ca cc a thc Chebyshev - nh ngha : T1(x) = x , T2(x) = 2x2 1 , T3(x) = 4x3 3x, T4(x) = 8x4 8x2 + 1 , Tn(x) = 2x Tn-1 Tn-2- Tn(x)dao ng trong khong 1 khi x dao ng trong khong -11 v c n

nghim trong khong 1. Khi x>1, Tn(x) tng n iu - Nghim ca Tn(x) c cho bi ;

)1(0,221coscos =+== nmnmx mm (1)

- Xt hm: ubuabubaubaT 2coscos4)1)cos(2()cos( 222 +++=+ ,

C dng chui Fourier cosine hu hn n cos2u Tng qut: Tn(a+b cosu) c dng chui Fourier cosine hu hn n cosNu v c

th tng ng vi mt h s mng ca mt mng 2N+1 phn t. - H s mng ca mt mng ng pha i xng gm 2N+1 phn t c dng:

=

+=N

nnu nuCCF

10)( cos2

- Tng ng vi TN(a+b cosu) = TN(x). Cc h s a,b c chn sao khon kh kin ca u tng ng vi gi tr ca [ ]1,1 xx vi x1 > 1. Khi gi tr TN(x1)tng ng vi gi tr ln nht ca F(u) k hiu F(u)max > 1. Cc cc i ph tng ng vi

c ln =1 11 xXt trng hp

20d : khi cos

20 0dku ==> thay i t dkdk 00 0 ,

bin thay i t a + bcoskubax cos+= 0d -> a + b -> dkbadkba 00 cos)cos( +=+ . tng ng khong kh bin ca u vi 0 th gi tr xmax = a + b = x1

1cos 0min =+= dkbax hay dk

dkxa0

01

cos1cos1

+= , dk

xb0

1

cos11+= (2)

* Thit k mng c rng tia chnh cho trc: Gi im khng cui cng ca Tn(x) trc khu x ri vo on [1,x1] l xz tng ng vi gc z v 29

)coscos(cos,cos 00 zzzzz dkbaubaxku +=+== (3) Theo (1) => (4) zz ubax cos+= t h :

=>

=+=+zz xuba

dkbacos

1cos 0

+=+=

dkuxb

dkudkxua

z

z

z

zz

0

0

0

coscos1

coscoscoscos

(5)

* T s mc chnh/ mc ph R: [ ])(coshcosh)()( 11 baNRbaTxT NN +==+= (6) * Thit k: t (2) => a v b. rng tia chnh x, nh t x1 v xz cho bi (4), z

cho bi (3). Hoc tn a,b t (5) v R t (6). Phn b dng c tm t khai trin chui Fourier ca Tn(a,b cosu)

Bi tp : c thm mng Chebyshev v cc v d thit k. 2) Mng siu hng : c h s nh hng rt ln hn h s nh hng ca anten

mng ng nht mt chiu. + Xt mng c chiu di L c nh, gm 2N + 1 phn t,

NLd

2= c 2N im

khng. Nu chng u nm trong vng kh kin v cc cc i ph rt nh so vi cc i chnh (c rng rt nh) th mng s c h s nh hng rt cao.

+Thit k mng siu hng c tia chnh 2 = ng u = uo = 0

Vng kh kin : NLku

NLk

2200

Dng mng Chebyshev vi 4

0=L c 7 phn t, mc ph = 0,1 mc chnh (hay 20dB thp hn)

=> 54,1,556,74,015,73,4 1

0 ==== xbad => Phn b dng :

53

62

61

60 10.072,2,10.217,1,10.006,3,10.99,3 ==== IIII

* Mng siu hng ch c ngha ton hc v yu cu dng cung cp cho mi phn t rt ln, hiu sut bc x rt thp li rt thp.

____________________________________________

30

4.4 MNG CP IN CHO MNG (feed netword)

+ Vn thit k mng dng truyn cung cp cc dng in c bin v pha

cho trc ti mi phn t c th rt phc tp v : - Tr khng vo ca mi phn t chu nh hng ca tr khng tng h vi tt

c cc phn t khc . - Trng hp yu cu s kch thch khng ng nht cn s dng mt s dng

mch chia cng sut vi tn hao thp - S nh hng ca di tn cng tc ln s phi hp tr khng + Phng php tng qut : Chia mng thnh nhm hoc vng tu theo s i

khng chung ca mng. Cc vng tng t nhau s c nui bi cc mng nui i xng

+ V d : Mng 9 phn t c chia thnh 3 nhm, mi nhm c nui bi 1 ng truyn n. Nh vy s kch thch ca mng s c tnh i xng cao, khng ph thuc vo nh hng ca s phi hp tr khng v tr khng tng h. (hnh v)

Theo hnh v, c cc nhm c cng kch thch sau : (1,3,7,9) ; (2,8) ; (4,6) + Phng php chi tit : Mi phn t c ni vi ng truyn chnh nh mt

on ng truyn c di bc sng . Dng in trn mi phn t lin quan trc tip vi bin v pha ca in th trn ng truyn chnh.

K hiu : + Zf : Tr khng c trng ca ng truyn chnh (hnh v) + Za : Tr khng c trng ca on bc sng + Za,in :Tr khng ca phn t anten + Vf : Th ti u vo on bc sng Dng ti u vo ca on bc sng + += aaa III

)(. + +== aaaaaf IIZIZV + Ti u vo anten phn t :

faj

aj

ain VjYeIeII .22 =+= +

* V d : p dng nguyn tc trn cho 3 anten phn t + Cc dng trn cc phn t l ng pha v c bin t l vi Ya , Yb , Yc v :

- Th ti cc u vo cc im a a , b b , c c u bng nhau do khong cch 0 nh nhau trn ng truyn chnh.

+ Tr khng vo ca c mng :

=

inc

c

inb

b

ina

ain Z

ZZZ

ZZZ

,

2

,

2

,

2

31

_______________________________________

4.5. MNG K SINH

t ca mng u l phn t tch cc (c kch thch b

h cc v vi cc phn t th ng khc thng qua tr khng tng h gia ch

it k bng con ng thc nghim. c bit n nhiu nh a.

+Xem mng mt mng ca cp phn t 2 u. 2121111 0

IZIZIZV +==

=>

+ Khng phi tt c cc phn

i dng nui): driven element + Cc phn t th ng (nondriven element) c kch thch bi s cm ng vi

phn t tcng. + Mng thng c tht l mng Yagi Ud1) Mng 2 phn t:

11

12

2

12122211

21122

122211

2121 ,,

.ZZ

II

ZZZVZI

ZZZVZI ==

= 1211 IZV += 222

H s mng:

cos11Z12

)(01 djkdju e

ZF = => nhn c hng bc x i khi = o th: cc

= dkd o hay ( ) =d+ iu kin bc x theo hng ngc ( = ) bng khng (phn x ton phn):

ok

1;,011

12 ==+ ZZdkd o

+ Nu 1101 2Zl < : dung khng ( dn x )

1101 2Zl > : cm khng ( phn x )

+ Gc pha ca Z11 thay i theo chiu di ca phn t 1 + Tr khng tng h Z12 ph thuc vo khong cch d

tr thnh phn t dn x v hng bc x cc i t k sinh

n t k sinh - Nu dng dipole gp th Ra 80 khi c mt phn t k sinh

+ Thc t, phn x tt th: (cng thc) + Nu l1 < l2 th phn t k sinh 1 s hng v pha phn2) Mng Yagi Uda: + Nhc im ln ca mng k sinh l Ra ca phn t driven nh - Vi phn t driven l dipole na sng Ra 20 khi c mt ph

32

- Di tn cng tc rt hp (~ 32%) v phi iu chnh tht chnh xc phn t k sinh c kt qu ti u :

NNNNN

NN

NN

IZIZIZ

IZIZIZVIZIZIZIZ

+++=

+++=++=

.....0........................................................

.........0

0011

00001100

1111010111

+ Yu cu thit k:Chn di v (do Zi)sao cho dng Iil i c pha rin tha mn iu kin cng ng

pha vo trng bc x theo hng thun (hng t refector driver directors). y l bi ton rt kh gii chnh xc .

+Thng gp: mng 810 phn t, G 14dB, ph hp vi % +Cu trc n gin (u im) ____________________________________________

4.6 MNG LOG - CHU K - L anten gii rng. - C th c thit k hot ng gi tn bt k. - Yu cu chung cho anten log_chu k dng lm anten thu l c rng di tng

3-1 hoc ln hn . - Nguyn tc thit k ct yu l xy dng 1 cu hnh m kch thc ca n t iu

chnh 1 cch c chu k tu theo tng s hot ng. - Mt anten hot ng tt bc sng 1 s hot ng tt boc sng 2 nu kch

thc ca n thay i 1 lng bng 2

1 Xt mng theo hnh v tho mn iu kin :

11111 >==== ++++ n

n

n

n

n

n

n

n

aa

dd

ll

xx

Mng s oc xc nh hon ton bi 2 trong s 3 thng s :

n

nl

d2, = v

- Nu tt c cc kch thc ca mng vi th phn t n tr thnh phn t n+1, phn t n+1 n+2 Mng c cc c trng bc x nh nhau cc tng s ......,., 123121 fffff ==

33

V ln2lnln,lnln 21

3

1

2 ===ff

ff : nn mng c tn l log chu k.

- Thc nghim cho thy rng cn phi nui anten bng dy song hnh v to ra gc lch pha 180o ca dng gia cc phn t lin tip.

- Ti mt tng s nht nh dnh trong tt c cc phn t s rt nh, ngoi tr phn t s c chiu di tng ong bc sng /2(ph/ t cng hng )

- Dng in dc theo ng truyn s suy gim rt nhanh pha sau phn t cng hng .

- C th b sau cc phn t cng hng tng s thp nht mt vi phn t cng nh c th b i cc phn t ng trc phn t cng hng tng s cao nht 1 s phn t.

- c trng bc x ca mng s tong t tng s v khi fn . fn .1+ 1 . - Cc tnh ton gn ng s dng cc chng trnh tnh s ch ra khong ti u

cho 8,096,0 = v 14,018,0 = - Ch 96,0= li =12dB , 8,0= , li =8 dB ====================================================

4.7. ANTEN DY DI - Chiu di bng 1 s ln bc sng, c treo bi 1 s thp thch hp, hot ng

gii tng s t 2 30MHz; - Cc dng thng gp: Anten hnh ch V, hnh trm, (nm ngang hoc ng) v

anten dy n nm ngang. - a s anten dy di c th hot ng nh anten cng hng, khi dng trn n c

dng sng ng hnh sin. nhng anten ny thng hot ng tt 1 tn s nht nh no v cc hi bc cao hn ca n.

- Tr khng vo ca loi anten ny rt nhy vi tn s, do di tn cng tc rt hp.

- a s anten dy di cng c th hot ng ch khng cng hng (dng in c dng sng chy) nu ni vo u cui ca n vi 1 in tr c gi tr bng tr khng c trng ca anten, khi coi n nh 1 ng truyn sng. Khi di tng cng tc c m rng ng k.

- Thng dng cho di sng ngn (2 30MHz) khi s dng sng phn x t tn in li. Khi gc bc x ti u t 10 300 so vi phng nm ngang

- nh hng ca mt t l rt ln.. - Bi ton thit k ch yul nhn c hng bc x nh mong mun so vi

mt t sao cho vic truyn thng tin khong cch xa s dng s phn x t tn in ly l ti u v tho mn yu cu cho trc v mt phi hp tr khng vi ng truyn.

34

1) anten cng hng.

xkII

nl

x 00)(

0

sin2

==

- Tng t tnh ton nh anten dipole na sng, s dng h to cc vi gc cc , in trng E c dng :

( ) ( )

sin

)2

(sinsin.1.

2

2cos1

200 0

= +

+

ne

rZIE n

njrjk

- Kiu bc x trong khng gian 3 chiu bao gm 1 s hnh nn( bp sng) - S hnh nn = n - Kiu bc x i xng qua mt phng v i qua im gia ca dy. hng khng( =

2 ) dy.

- n cc bp sng nhn hn, o - Khi Dl max cng vi in tr bc x. - C th cp in 1 u hoc gia (h---) - Tuy nhin do phn b khng i xng ca dng trong ng truyn, mt s bc

x s xy ra do bn thn ng truyn. khc phc, c th ni ng truyn ti tm ca mt vng in((hnh v)

- Khi dng on chuyn tip 041 th c th ni in tr bc x ca anten vi cc ng truyn song hnh c tr khng 600. tr khng c trng ca on 40 lc ny s l ac RZ 600= vi Ra: in tr bc x ca anten.

2) Anten hnh ch V (hnh v) - Trng bc x l chng chp 2 trng ca 2 nhnh. - Thng chn o trng bc x tng tho mn iu kin cho trc. - Chng hn, nu yu cu hng bc x cc i hng theo ng phn gic ca

gc to bi 2 nhnh ca hnh V th phi chn o = 21, vi 1: gc tng ng vi bc x cc i ca mi nhnh; v d :

000

0 8442223 === l 00

00 723622 === l . ( Tham kho phn 1) - Dng in trn mi nhnh ngc pha nhau E ng pha hng ca ng

phn gic.

35

- to ra hng bc x cc i lm vi phng nm ngang mt gc , c th dng anten V c 1 nhnh nm ngang, nhnh th 2 lm vi nhnh th 1 mt gc 0 < 21, v mt ca hnh V thng ng

(hnh v) - c thm v anten trm ( Phan Anh ) 3)anten dy di vi sng dng c dng sng chy; - Do c ni tr R u cui m dng in trn dy c dng sng chy

(hnh v)

xjkeII 00=

- Trng bc x theo hng :

( )( )( )

)2(sin

)2(sin.sin.sin

4 2

20cos12000 0

= +

lke

rZIjkE

lrjk

- Hng bc x cc i c xc nh bi :

lk

tg0

22 = vi )2(sin

20

lk= - Vi (hay 1>>lko oolk >> ) th 165,1= - Nu tnh ti s suy gim bin ng dng do mt mc nng lngtrn ng

truyn th bin dng ti im c to x c dng ; h s suy gim ch gy ra 1 thay i rt nh trong kiu bc x, thng c b qua.

xxjkx eII

= 00* c im chnh ca kiu bc x l ch c 1 hnh nn bc x chnh theo hng

trc x. - Nu 0l- Nu = 0nl c 2n bp sng * u im ca anten dy sng chy l tr khng vo ca n c th xem nh thun

tr, t ph thuc vo tn sdi tn cng tc rng. Hn ch ch yu l s mt i xng ca cc bp sng khi tn s thay i.

4/ Cc nh hng giao thoa ca mt t : - Trng bc x theo hng l tng ca trng bc x t anten v trng phn

t mt t. Trng phn x c gc chm pha tng ng trng bc x t anten nh di mt t.

- Nu trng bc x ca anten trong khng gian t do c dng :

( )r

efErjk

40 = (1)

Th trng tng s c dng :

( ) ( ) sin2 00

14

hjkjrjk

er

efE += (2)

* Biu thc trng (2) c dn tng ng nh ca 1 mng vi h s mng: 36

sin2)( 01 hjkjeE += (3)

* H s phn x ph thuc vo dn in ca mt t vo gc , v c tnh phn cc ca trng theo phng ngang hay phng thng ng. Thng c th coi mt t l mt dn l tng, khi : 1= v = i vi trng phn cc ngang

=> )sinsin(2 0)( hkE = 0= vi phn cc ng => )sincos(2 0)( hkE =

______________________________________________

37

CHNG 5 ANTEN MT 5.1 BC X T MT MT PHNG, PHNG PHP BIN I FOURIER + C 1 lp rt rng cc anten tin dng hn gi l cc anten mt trong bc x

c coi nh t 1 mt m: anten parabol v anten loa. + Thng c kch thc mt m (khu )ln hn vi ln boc sng c li

cao, v do c ng dng ch yu di tng s viba. + Mt phng php quan trng nghin cu anten mt l dng bin i Fourier.

Mu cht ca phng php l: kiu bc x ca mt chnh l nh Fourier ca trng ca mt bc x v s dng cc c trng ca cp bin i Fourier m t c im ca anten mt.

- Trn hnh 5.1l 1 anten mt c din tch Sa nh x gc to , mt phng z= 0

- Gi thit bin thnh phn tip tuyn ca in trng trn mt ca anten aEr

- Chng ta s i xc nh trng bc x trong min z>0 . - Tng tng trng b mt Sa oc hnh thnh bi 1 phn b ngun thch hp

no pha sau anten z< 0. Chng ta s khng cn bit phn b ngun ny m ch quan tm n trng trn b mt anten , bi v n s xc nh duy nht trng trong na khng gian z>0.

* Php bin i Fourier : - nh Fourier ca hm w(x) c dng :

(5.1a) dxexw xjkx

=-

x )()W(k

V khi quan h ngc c dng :

xxjk dkexw x

=-

x )W(k21)( (5.1b)

- Cc bin kx v x ng vai tr tng t nh bin thi gian t v tng s gc trong cc ph tn hiu ph thuc thi gian.

- Tng t vi hm 2 bin u(x,y) :

(5.2a) dydxeyxuyjkxjk yx +

=

-yx ),()k,U(k

(5.2b) yxyjkxjk dkdkeyxu yx

=

-yx )k,U(k),(

* Trong chng 1chng ta c quan h:

=> (5.3)

==+

0

0202

E

EkE

38

* c trng ca ton t Fourier :

( )

=

=

=

=

),(),(

),(),(

),(),(

)(

2

22

2

)(

yxukkyxyxu

yxujkx

yxu

yxujkx

yxu

jdt

tds

yxyxyx

xxx

xxx

tst

(5.4)

(5.3) c th vit li di dng :

=+

+

=

+

++

0

0),,(2022

2

2

2

2

zE

yE

xE

Ekzyx

zyx

zyx

(5.5)

* Bin i Fourier h (5.5) =>

=++

=

+

0)z,k,k()z,k,k(k)z,k,k(k

0)(

yxzyxyyyxxx

)z,k,k(222

02

2

yx

Ez

jEE

Ekkkz yx

(5.6)

* t th (5.6) => 22202

yxz kkkk =

0)z,k,k()z,k,k(

yx2

2yx

2

=+

Ekz

Ez (5.7)

=> Nghim tng qut ca (5.7) c dng :

(5.8) zjefE .kyxyx z).k,k()z,k,k( =

* Tm : Thay (5.8) vo (5.6) => )k,k( yxf

(5.9) 0. =fk

* S dng php bin i Fourier ngc s thu c biu thc ca cng in

trng : ( s dng (5.2) v (5.8)) )z,,x( yE

39

yxrkj

zyx dkdkeE

=-

yx2),,( )k,(kf41 (5.10)

* ngha ca (5.10) trong min khng gian z>0 in trng c dng ph ca cc

sng mt phng v hm l sng phng vi bin rkje)k,(kf yx

f , lan truyn theo

hng ca vector lan truyn k .

* : 0222

02 kkkkkk yxz ==>=

* Nu => hng s sng k2022 kkk xz >+ z l o cc sng phng trong vng ph ny

suy yu dn theo hng Z. Ni cch khc, ch c cac sng phng trong vng ph tnh ng vi mi ng gp vo trng khu xa. 20

22 kkk yx +* Khi z=0 ta phi c iu kin bin :

yx

yjkxjk dkdkeE yx

=-

yxt2 )k,(kf41

(5.11) T ( 5.2 ) ta c :

+ =a

yx

S

yjkxjk dydxe .y)(x,E41)k,(kf a2yxt (5.12)

* T (5.9) => 2220

..

yx

yyxx

z

ttz

kkk

fkfkk

kff ==

(5.13)

* Nu tnh c tch phn (5.10) th xc nh c Er

, nhng iu ny ch d dng thc hin khi r>> 0 hay kor>>1

Khi : )sin.sin,cos.sin(2cos

000

)( 0 kkfe

rjkE rjkr

= (5.14) * Nhn xt : - Trng bc x khu xa t l vi nh Fourier ca trng b

mt vi (cng thc) l cc thnh phn ca vectorng ca sng cu lan truyn theo hng (,).

- Theo hngZ,fz0v cos1thng bc x c tnh theo (5.14)+(5.12) v t cc i; E

rch c thnh phn Ex,Ey t l vi fx,fy .

- V 0. = Er

v 0. =fkrr

nn thng l ng phn cc ngang TEM trong vng bc x (khu x)

40

_____________________________________________

41

5.2 BC X T MT MING CH NHT - Gi thit trng trn ming l ng nht v cho bi :

vi = xaEE 0

by

ax

= 0 vi cc gi tr khc ca x,y (5.12) tr thnh

vv

uuaabE x

sinsin4f 0t = (5.16)

=> Cng trng bc x c cho bi (5.15) :

)coscos(sinsin

24

00)( = aa

vv

uue

rabjkE rjkr (5.17)

* Nhn xt : - (5.17) c dng tng t nh mng ng pha 1 chiu - C dng tng t nh ca kiu bc x trong vng kh kin ca khng gian

u,v vi bkvaku 00 , - Cc cc i ph c ln gim dn

* Trong mt phng 0= :

( )

sinsinsin4

2 00

00

)( 0

akakabEae

rjkE rjkr

= (5.18) rng tia chnh :

aB 0W = vi 0>>a

____________________________________________

5.3 BC X T MING TRN - Gi thit trng ng nht

vi = xaEE 0 222 ayx +

= 0 vi 222 ayx >+Khi :

sin)sin(2f

0

010

2t ak

akJaEa x =

42

Trong J0(x) ,J1(x) l cc hm Bessel bc 0,1 loi 1. * J1(x) tng t nh hm sin tt dn (hm lng gic c bin gim dn )

* Vi x>> )4

sin(2)(2

1

1

xx

xJ

* th ca hm phng hngc dng tng t nh ca bc x t ming ch nht, vi s suy gim nhanh hn ca cc cc i ph

aB 0832,3W =

R = -17,6dB _________________________________________

5.4 MING VI TRNG NG NHT C PHA BIN I TUYN TNH

Xt ming hnh ch nht vi cng trng c dng :

vi yjxj

x eaEE = 0

by

ax

=> (5.21) dydxeaEa

a

b

b

ybkjxkjx

yx .f )()(0t

+ =

=> )cos.sincos())(()sin()sin(

24

00

0000)( 0

= aa

vvuuvvuue

rabEjkE rjkr (5.22)

- Kiu bc x trong h to u,v tng t nh ming ch nht ng nht vi cc i ti u=uov v=vo, tc l:

auak == 00 cos.sin

bubk == 00 cos.sin =>

=tg , ( )0

2122

sink +=

=>c th iu khin hng bc x cc i tng t mn 2 chiu. * Nu = 0 th hng bc x cc i nm trong mt phng = 0( hay mt phng

xoz) vi gc = 0 cho bi :

=

2arcsinarcsin 00k

43

- Hng khng tho mn iu kin = ouu

=> ( )0

00 cos

2

aBW ==

.5.5 MING VI TRNG C BIN GIM T TM RA BIN:

xt ming ch nht vi phn b trng c dng

)1(0 ax

aEE x =

vi by

ax

=>

2

0t)2(

)2(sinsin2f

= ak

ak

bkbk

aabEx

x

y

yx (5.23)

=> ( )( ) )cos.sincos(22

sinsin2

00)( 0

= aau

u

vve

rabEjkE rjkr (5.24)

- T s mc chnh trn mc ph R = 26dB - Cc i chnh xy ra khi u = v =0

rabEkE

00

max

= ---------------------------------------------------------------------------------------

44

CHNG 6: MING NG DN SNG ANTEN LOA. -Ming ng dn sng( hnh ch nht hoc trn) thng khng c s dng lm

anten pht v tnh nh hng km, nhng thng c s dng lm b chiu x cho anten parabol phn x

6.1 NG DN SNG CH NHT - Xt ng dn sng ch nht c kch thc tit din ngang l a x b , ming ng

nh x trong mt z = 0 - Mode truyn sng ch yu l sng TE10 (transverse electric), c Ey, Hx, v Hz

zjy eaxEE = cos0

zjWx eaxYEH = cos0 (6.1)

- Vi z = 0 => axEE y

cos0= a

xYEH Wxcos0=

* Ti gn ming ng xut hin sng phn x ca mode TE10 v cc mode bc coa

hn vi bin nh * Nu ch s dng minh ng dn sng cho b chiu x, c th b qua cc mode

bc cao v coi trng trong mt z = 0 , ch 0 trn ming ng - Theo nguyn l i ln ca

trng in t, c th coi tn ti dng t mt a

xaEJ xmscos0

=

- Tng bc x khu x c tnh bi cng thc (5.15) vi fx =0 (theo 5.12, lu

Ex =0)

( )[ ]( ) ( )[ ]( )220y 2cos2 2

sin2f

ak

akbk

bkabE

x

x

y

y

= (6.3) - Trong mt phng

2 = (yoz) , kx =0, t l vi E ( )[ ]( )[ ] sin2

sin2sin2f0

00y bk

bkabE=

- Trong mt phng Ek y ,0,0 == t l vi 45

( )[ ]

( )( )

cossin2

sin2cosf 20

2

0y ak

ak

=

- Cng sut bc x ton phn theo (6.2) 204

EYabP Wr = => H s nh hng :

200

20

2 642

4 ab

PEYr

Dr

== * nh gi h s nh hng: chng hn cho di X (812 GHz),

6.2 ANTEN LOA H - nhn c trng bc x c tnh nh hng cao khi so vi ming ng dn

sng, c th m rng cc ming ng dn sngthnh cc anten loa. - Nu ming ng dn sng ch nht c m rng trong mpanten loa H (hnh v) - Trng bc x t pha ming ODS v pha ming loa c dng mt sng tr trn

(hnh v) - trng ming loa gn ng pha th gc m phi nh. - li v kiu bc x s ging vi ming bc xng pha, nu lng sai khc v

pha ra ming loa v tm loa 4 hay

'444)( 0120 a

tgRRk Vy c ming loa rng thf gc m nh hn ch phm vi s dng( v loa

di). - Nu b qua sai khc v pha v coi phn b trng ming loa tng t nh

trng ming ng dn sng TE10 th :

'cos0 a

xaEE ya = vi

2'2

'

by

ax

- Trng bc x c tnh tng t trng hp ng dn sng ch nht vi a a

v hng s truyn sng :

( ) 021220 ' kak = 46

- H s nh hng : 20

'2,10 baD =

- li G D - Vi cng 1 chiu di ca loa th li s tng nu tng gc m . Tuy nhin khi

sai pha trn ming loa tng theo gim li - Cc tnh ton l thuyt vi cng chiu di loa th li cc i nhn c do

tng rng ming loa a cho n khi sai pha 0,75 .

6.3 MING NG DN SNG HNH TRN

- Mode TE11 phn b in trng trn tit din thng (s dng h to cc ( , ))

)84,1(sin2

1 aJE

= (5.3.1)

d

adJaE

)84,1(

84,1cos2 1= (5.3.2)

(hnh v) - Trong h to Decarte :

cossincoscos

EEEEEE

y

x

==

(5.3.3)

(hnh v) - S dng tnh cht ca hm Bessel

2sin)84,1(2 aJE x = (5.3.4) 2cos)84,1()84,1( 20 aJaJE y = (5.3.5) - S dng cng thc tch phn Lommel :

uuJJaae

rjkE rjk )(

84,1)84,1(2sin2 11220 0 = (5.3.7)

=

du

udJu

Jaer

jkE rjk )()84,1(

)84,1(84,1cos2 122120 0 (5.3.8)

* Nhn xt

47

- trong mp 2 = (mt E), kiu bc x tng t nh kiu bc x ca ming bc x

ng nht hnh trn (chnh 4_) - trong mp = 0(mt H) kiu bc x hon ton tng t kiu bc x ca ming

ng ch nht. -h s nh hng c tm theo cch tng t nh vi ng ch nht

20

2

0

66

aD = (5.3.9)

6.4 LOA H. - nhn c trng bc xc tnh nh hng cao hn so vi ming ng dn

sng, c th m rng(hay lm loe ra) cc ming ng dn sng thnh cc anten loa. - Nu ming ng dn sng ch nhtc m rng trong mp H , ta c anten loa H, - Trng bc x vo loa t pha ming ng s c dng sng tr vi cc mt ng

pha dng mt tr trn (hnh v) - trng ming loa gn ng pha th gc m phi nh. - li v kiu bc x s rt ging vi ming bc x ng pha, nu lng sai

khc v pha ra ca loa v tm loa 4 hay

'444)( 0120 a

tgRRk * Nhn xt : c ming loa rng th gc m phi nh loa di gii hn

phm vi ng dng. - Nu b qua s sai khc v pha ca trng ming loa th c th coi phn b

trng ming loa tng t nh trng ming ng dn sng ng vi mode TE10

,'

cos0 axaEE ya = vi

2

2'

by

ax

- Trng bc x dc tnh tnh tnh trng hp ming ng ch nht 5.2 vi aa, vi hng s truyn 0k

- H s nh hng : 20

'2,10 baD =

- li (cng thc) - Vi cng 1 chiu di ca loa, li s tng nu tng gc m. tuy nhin khi sai

pha trn ming loa tng v lm gim li. cc tnh ton l thuyt ch ra rng: vi cng 1 chiu di loa th li cc i nhn c do tng rng ming loa a s t c cho n khi sai pha 0,75.

---------------------------------------------------------------------

48

49

CHNG 7 : ANTEN PARABOL

7.1 CU TO V NGUYN L HOT NG - L anten c tnh nh hng tng i cao, s dng ch yu di sng cc ngn

(thng tin di ng v v tinh) - Phng trnh mt parabol trong h to cc:

cos12+=

fr (7.1.1)

- c trng b mt parabol l tt c cc tia bc x xut pht t t tiu im (ni t loa chiu x ) sau khi phn x u song song vi trc parabol c th p dng cc nguyn l quang hnh tm trng trn ming parabol.

- Trong cc ng dng cho nghin cu bc x v tr, trnh nh hng ca nhiu t mt t (c th lm gim nhy), ngi ta thng dng h thng chiu x th cp. H parabol lc ny c gi l anten Cassegrain

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7.2 Hiu sut ming bc x 1) Mt cng sut bc x trn ming parabol: - K hiu g (,): cng sut bc x/ n v gc t theo hng (,) ca ngun

chiu x. => cng sut n mt phn x trong gc sindd l :

ddgPi ..sin).,(),( = (7.2.1) - Mt cng sut tng t phi xut hin trong mt trn ming rsin d ds (cng thc) 2) Tn hao trn: c trng bi hiu sut trn: t s cng sut c gng ph x

tr li/tng cng sut bc x ca b chiu x ddgddrP ..sin).,(..sin),( = (7.2.2) Hay

dd

rgP 1),(),( = (7.2.3)

- S dng (7.1.1) =>

2222

44cos

ffff

+= (7.2.4) ( )

222

2

2

2

)4(16),(

4cos1),(),(

fffg

fgP +=

+= (7.2.5)

50

Nu constg =),( th ( )2cos4),( 4 =P * H s nh hng ca b chiu x :

=ch

f PgD )0,0(4 (7.2.7)

3) Hiu sut ming: m t tt c cc dng tn hao (phn b bin , pha v c tnh phn cc)

- Trng trn ming:

+== yyxxa aEaEE ),(),(),(

- Cng sut bc x ton phn t ming: gi thit sng l sng phng

= 20 0

..),(a

a ddPP

- Mt cng sut bc x trn n v gc c :

22

0 020

2022

0 ..),(821 =

a

a ddEYkErY (*)

Nu cng sut bc x ton phn Pa t ming parabol c phn b ng nht vi mt Pa / a2 th :

212

02

= Ya

PE aa (**) Khi trng trn ming l phn cc thng v ng pha. Trng ny to ra mt

mt cng sut trn n v gc di dc theo trc z l : 222

0

4 aPak

- Thng chiu thnh phn phn cc c s dng, chn thnh phn Ey, kh : t s cng sut bc x do thnh phn Ey Gi l hiu sut ming mt cng sut bc x trn n v gc c

+=

2

0 0

222

22

0 0

..)(

..),(

a

yx

a

a

ddEEa

ddE

(7.2.9)

+ Hiu sut ming c th c biu din bng tch ca 3 s hng, bao gm tn hao do chiu x khng ng nht (1-i), tn hao do s khng ng pha ca trng ming (1- p) v tn hao phn cc ngang (1- x), tc l:

xpiA = 51

+ li trc : inin PI

PIg

4

4

==

Vi I: Cng bc x ca anten theo hng trc (watts/ n v gc c) ng vi dng phn cc thng cho trc, Pin: cng sut t vo b chiu x.

+ Cng sut bc x bi b chiu x : infT PP .= vi f : l hiu sut ca b chiu x

+ Cng sut n ming parabol l:

aTS PP = =>

afS P

IG 4..=

+ Tng qut :

= 2

0

4.... aIG ipxfS * H s nh hng: Cng bc x t ming ng nht phn cc thng:

== aae

rEjkEE rjk 200 0

2 => Mt cng sut gc c :

40

20

2022

0 821 aYEkErY =

+ Cng sut bc x ton phn

200

2

21 EYaPa =

=> )(4 2

20

aD = (7.2.10)

+ Tng qut : ASaD .)(4 220

= (7.2.11)

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52

CHNG 8 : ANTEN THU

8.1 DIN TCH HIU DNG + Trong a s trng hp, c th dng nguyn l thun nghch kho st cc c

trng ca anten thu: Cc c trng ca 1 anten s dng thu sng in t rt gn vi cc c trng tng ng ca anten khi n c s dng bc x sng in t. Nu 1 anten c li G theo 1 hng cho trc khi bc x s c cng li

nh vy khi nhn bc x t cng mt hng khi sng ti c cng dng phn cc + tin kho st c trng nhn ca anten thang ta s dng khi nim din tch

hiu dng Ae sao cho nng lng nhn c bi anten bng mt nng lng n trn n v din tch nhn vi Ae.

+ Khi nu iu kin v dng phn cc ca sng n v iu kin tr khng c tho mn th:

G4

A20

e = Vi anten ming Ae din tch thc ca ming V Ae ~ din tch thc ca ming + c trng phn cc ca 1 anten c th c m t bi vic s dng thng s

chiu di hiu dng phc hr

. = ic EhV .0 trong iEr cng in trng sng ti

Voc: Th h mch thu c. Vi anten dipole hr

~ chiu di ca anten nhng chiu di ca anten do phn b dng bt ng nht.

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8.2 CNG THC FRIIS

- Xt h R T antennas Anten pht c li Gt (t t) gi Pin:cng sut t vo T anten

t : H s phn x ca ng truyn nui T anten => cng sut bc x ton phn l: int P)1(

2 (8.1) + Mt cng sut trn n v din tch theo hng Ranten, khong cch r s l:

22

4),(.)1( r

GPP tttintinc = (8.2)

53

=> cng sut tn hiu thu c s l:

),(4)1)(,()1( 2

22ttt

intrrrrrec Gr

PGP = (8.3) Cng thc (8.3) l cng thc Friss + Nu iu kin v tnh phn cc khng tho mn th

),(),(4)1)(1( 2

2022

tttrrrin

trrec GGrPpP = (8.4)

Vi : 22

2

.

=i

i

Eh

Ehp (8.5)

54