ANTEN CHẤN TỬ ĐỐI XỨNG

7/22/2019 ANTEN CHN T I XNG

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CHNG I: KHI QUT CHUNG V ANTEN CHN T I

XNG

1. Khi nim, cu to v ng dng ca Anten chn t i xng:

Chn t i xng l loi Anten n gin nht v l mt trong nhng ngun

bc x c s dng kh ph bin. Chn t i xng c th s dng nh mt Anten

c lp hoc c th c s dng cu to cc Anten phc tp khc.

Chn t i xng l mt cu trc gm hai on vt dn c hnh dng tu

( hnh tr, hnh chp, elipsoit) c kch thc ging nhau, t thng hng trong

khng gian, v gia chng c ni vi ngun dao ng cao tn.

Hnh 1.7: Chn t i xng

Chn t c dng nh hnh v trn, vi mt dy gm hai na thng hng, chiu

di lv 2lhoc l/2 v l. Gi thit2

0.01a

vi a l bn knh dy.

Trong mt s ti liu k thut, ngi ta dng thut ng Anten dipol ( Anten

lng cc) ch cho chn t i xng.

Anten chn t i xng l vic cc di sng cc ngn, sng ngn, sng di v

sng trung. Nhng ch yu c ng dng trong di sng ngn v sng cc ngn

lm Anten thu v pht. Trong cc di sng ny Anten c th lm vic c lp hoc

lm vic phi hp. Trong di sng cc ngn chn t i xng cn c s dng

lm b chiu x cho cc Anten phc tp khc(vd: Anten gng parabon).


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2. Thc hin Anten bng kt cu c vn tc pha nh(kt cu sng chm).

3. Kt hp Anten vi mch tch cu.

Phng php dng ti in khng

Mt trong nhng yu t quan trng nh hng quyt nh n vic hnh thnh

th phng hng bc x l quy lut phn b dng in dc theo chn t.

phng php ny thay i phn b dng in trn chn t ta mc u cui ca

n ti thun khng dung tnh c dng khi kim loi hnh a hoc hnh cu.

Phn b dng in trong trng hp ny c th c xc nh theo phng

php gn ng, khi coi chn t tng ng vi mt on dy song hnh mc tiin dung u cui. Do mc ti nn tr khng u cui c gi tr hu hn, dng

in u cui s khc khng, ngha l phn b dng in s tng t trng hp

chn t c ko di thm mt on tdl . Nh vy mt chn t i xng c ti

chiu di mi nhnh2

lc th c thay th bng mt chn t i xng khng

ti vi di mi nhnh bng: 2 2L l

k

= + . Trong : Asin =1

Acos1

C

=

Hm phn b dng in:( ) sin ( )

2sin2

Ic LI z k z

kL=

Trong : Ic l dng in cui chn t(ch mc ti).

z l ta im kho st tnh t u vo chn t.Cng bc x ca chn t:

( )W

2

Riki Ib e

ER

=

Trong :( )

os os os os sin os sin os2 2 2

sin

kL kl kLc c c c c c

+ =


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Hnh v v th phng hng ca chn t i xng mc ti in dung c

0,422

l= ;

0

1502

kl = ng vi cc gi tr khc nhau ca ti (

0 0

45 ;80 = ). Trn

hnh cng v th phng hng ca chn t khng ti vi 0.52

l= (ng gch

chm).

Kho st th trn rt ra c mt s kt lun thc t, quan trng l c th

bo ton dng ca th phng hng ca chn t khi gim nh kch thc ca

chng bng cch mc ti in dung thch hp u cui chn t. Phng php nyc ng dng rng ri thit lp cc Anten sng di v sng trung, cho php

gim nh kch thc ca Anten khong 20-30%.

trn ta kho st phng php gim nh kch thc bng cch mc ti dung

tnh u cui phn b dng in. Vic ny cng c th thc hin c bng

cch mc ti cm tnh ni tip ti im gia trn hai nhnh chn t. Trong mt vi

trng c th s dng hn hp c hai cch c hiu qu cao hn.

Phng php dng ng dy sng chm

Vic gim nh kch thc ca chn t c th c thc hin bng cch s

dng cc ng dy sng chm.

V nguyn tc c th s dng bt c ng dy sng chm no m i vi n

c th p dng khi nim tr khng b mt (hay ipedang b mt), ngha l khi trn

mt ngoi ca n c cc thnh phn tip tuyn ca in trng v t trng c gi

tr khc khng. Tuy nhin vic la chn loi ng dy sng chm xut pht t cc

yu cu c lin quan n thong s cu cc kt cu y, trong hai thng s quan

trng l h s chm v h s suy gim ca kt cu. Thng s h s chm c nh

hng n kh nng rt ngn kch thc ca Anten, thng s h s suy gim nh

hng n hiu sut ca Anten.


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Cc ng dy sng chm thng gp l cc dy dn kim loi c ph lp in

mi hoc ferit, trc kim loi hnh rng lc. Cc chn t s dng loi dy sng

chm ny c gi l chn t impedang.

S dng cc ng truyn sng chm thit lp Anten chn t cho php

nhn c h s rt ngn Anten khong 2-5 ln (h s rt ngn Anten c nh

ngha bng t s gia tn s cng hng ca chn t kim loi thng c cng chiu

di v tn s cng hng ca chn t lm bng ng dy sng chm).

Cc chn t impedang c nhc im l phi s dng cc vt liu in mi t

mi gy tn hao trong cc mi trng y v do lm gim hiu sut ca Anten.

khc phc c th thay th mi trng bao quanh dy dn(in mi hay ferit) bi

ng dy xon.

Kt hp Anten vi cc phn t tch cc

Bit rng khi n thun gim nh kch thc ca Anten th di hiu dng

ca Anten cng ng thi gim i v s dn n gim sc in ng nhn c

u ra Aten khi Anten lm vic ch thu v gim cng trng bc x ca

Anten khi Anten lm vic ch pht.

m bo c tnh ca Anten khi gim nh kch thc cn c bin php b

li s gim di hiu dng ca Anten. Mt trong nhng bin php c hiu qu

khc phc nhc im khi gim nh kch thc Anten l kt hp Anten vi cc

phn t (hay mch) tch cc. Ta gi Anten l Anten tch cc.vic hp nht Anten v

mch nu thc hin tt s to ra mt cu trc hp l ci thin c tnh ca Anten,

v trong mt s trng hp cn c th to ra cho Anten mt s chc nng mi m

cc Anten thng khng c. Ngoi ra khi kt hp Anten v mch th gia Anten v

my thu hay my pht khng cn cc phn t phi hp iu chnh nh cc trng

hp thng thng, gim bt chiu di fide mc gia Anten v thit b thu pht, do

gim tn hao cao tn v gim tp m nhit ca Anten.

Cn lu rng kt hp Anten vi mch tch cc th s ci thin h s tng ch

Anten khng c lin quan n vic ci thin gin hng tnh. Trong cc trng

hp ny hm phng hng chun ha ca Anten vn ch c quyt nh bi


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di thc ca Anten v do gim nh kch thc Anten cng vn dn n gim

hng tnh, ngha l dn n m rng th phng hng.

Tuy nhin, vic kt hp Anten vi phn t hay mch tch cc trong mt s

trng hp cho php d dng s dng Aanten lm phn t ca cc h thng bc x

thit lp th phng hng theo yu cu cho trc, thit lp Anten iu

khin th phng hng bng phng php in hay h thng bc x c thc

hin bc u vic s l tn hiu.

Chng II:CC C TNH CA ANTEN CHN T IXNG

1 .PHN B DNG IN TRN ANTEN CHN T I XNG:

Mt trong nhng vn c bn khi kho st cc Anten l xc nh trng bc

x to ra trong khng gian v cc thng s ca Anten. Nh vy cn bit phn bdng in trn Anten . C th s dng l thuyt ng dy song hnh xc

nh phn b dng in trn chn t i xng da trn suy lun v s tng t gia

chn t v ng dy song hnh h mch u cui khng tn hao.

Mt ng dy song hnh h mch du cui, nu m rng hai nhnh ca ng dy

ra 1800 ta s c chn t i xng. Vic m rng ny lm mt tnh i xng ca ng

dy song hnh v lm cho sng in t bc x ra khng gian bn ngoi to thnh Anten.

l z

a) b)


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Hnh 2.1: S tng quan gia chn t i xng v ng dy song hnh

Gia hai h thng ny c nhng im khc bit, l:

- Cc thng s phn b ca ng dy khng bin i dc theo dy, cn cc

thng s phn b ca chn t th bin i ng vi cc v tr khc nhau trn chn t.

- ng dy song hnh l h thng truyn dn nng lng sng in t cn

chn t i xng l h thng bc x.

- Trn ng dy song hnh khng tn hao, h mch u cui, dng in ch

bin i theo quy lut sng ng thun ty, dng sin, cn i vi chn t lun c s

mt mt nng lng do bc x (mt mt hu ch). Do ni mt cch chnh xc th

phn b dng in trn chn t s khng theo quy lut sng ng hnh sin. Tuy

nhin vi cc chn t rt mnh (ng knh


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2.2.Cng sut bc x, in tr bc x v h s tnh hng:

Cng sut bc x ca chn t i xng c th c xc nh theo phng php

vec t Poyting, ging nh khi tnh ton cho dipol in.

Ta tnh tng thng lng ca vec t Poyting qua mt mt cu bao bc chn t,

khi mt cu c bn knh kh ln so vi bc sng cng tc, hnh 2.4

Hnh 2.4 Xc nh cng sut bc x ca chn t i xng


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Tng t nh i vi dipol in, ta cng c nh ngha v cng sut bc x

ca chn t i xng l i lng biu th quan h gia cng sut bc x v bnh

phng dng in trn chn t. Tuy nhin, do dng in c phn b khng ngu dc theo chn t nn khi biu th cng sut bc x qua bin dng in ti v

tr no trn chn t (v d qua dng in u vo, hay qua dng in ti im

bng sng ng) th s c in tr bc x tng ng (in tr bc x ng vi

dng in im vo, in tr bc x ng vi dng in im bng).

2.3. Tr khng sng ca chn t i xng:

Tng t nh ng dy song hnh, i vi chn t i xng cng c th

a vo khi nim tr khng sng. Theo l thuyt ng dy, tr khng sng ca

ng dy song hnh khng tn hao c xc nh theo cng thc:

11

A

LZ

C= (2.18)

Trong L1 in cm phn b ca ng dy

C1 in dung phn b ca ng dy

Mt khc ta c:

1 1

1 1v

L C= =

l vn tc sng truyn trn ng dy

Nu ng dy c t trong khng gian t do th = 0, = 0. Tr khng

sng ca ng dy c th c biu th qua thng s ca mi trng v mttrong hai thng s phn b L1, hoc C1 ca ng dy:

0 01

AZ C=

(2.19)


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i vi ng dy song hnh, C1 l i lng khng bin i theo chiu di

ca dy v c gii hn bng kch thc ca ng dy. Khi biu th tr khng

sng qua cc kch thc hnh hc ca ng dy, cng thc (2.19) s c dng:

276lgAD

Zr

=

(2.20)

D: khong cch gia hai dy dn (tnh t trc dy)

r: bn knh dy dn

i vi chn t i xng hoc cc loi anten dy khc, c th p dng cng

thc (2.20) tnh tr khng sng ca anten nhng cn ch rng in dung phn

b C1 lc ny khng phi l hng s m thay i dc theo chiu di chn t. V

vy khi tnh C1 cn ly gi tr trung bnh ca n, ngha l ly in dung tnh tng

cng ca anten chia cho chiu di ca n.

Cng thc tnh tr khng sng ca chn t i xng khi chiu di chn t

nh hn bc sng cng tc s l:

2

120 ln 1Al

Zr

=

() (2.21)

Khi tng chiu di chn t th sai s tnh theo cng thc trn s tng. Nh

vy khi chiu di chn t ln hn bc sng cng tc th tr khng sng ca chn

t s c tnh theo cng thc Kesenich:

120 lnAZ Er =

() (2.22)

Trong E = 0,577 l hng s Euler

2.4. Tr khng vo ca chn t i xng


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Nh cp trong phn e ca mc 1.2.3 chng 1 tr khng vo ca

chn t i xng bao gm c phn thc v phn khng.

vA vA vAZ R jX= +

Phn thc bao gm in tr bc x v phn in tr tn hao ca chn t. ivi chn t i xng, in tr tn hao khng ng k (c th coi bng 0) phn

cng sut thc a vo anten hu nh c chuyn thnh cng sut bc x

A bxP P

(2.23)

Nu biu th cng sut bc x theo dng in u vo Ia th cng thc

(2.23) c th vit:

2 2

0

2 2a bx a vAI R I R

(2.24)

Rbx0 l in tr bc x tnh theo dng in u vo

2

0 2 2sinb bxb bxb

bx

a

I R RR

I kl= =

Ta c: 2sinbxb

vA RR kl=

(2.25)

Phn khng ca tr khng vo ca chn t i xng chnh l tr khng ca

ng dy song hnh h mch u cui v c tnh theo cng thc:

cotvA AX iZ gkl=

(2.26)

Trong ZA l tr khng sng ca chn t i xng.

Thay cc cng thc (2.24) v (2.25) vo cng thc (1.17) ta c cng thc

tnh tr khng vo ca chn t i xng:


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2cot

sinbxb

vA A

RZ iZ gkl

lk=

(2.27)

Cng thc ny nhn c khi tnh ton theo gi thit dng in trn chn t

phn b hnh sin. Khi di ca chn t gn bng na bc sng cng tc th

cng thc (2.26) cho kt qu hp l nhng khi chn t c di ln hn th

chnh xc ca cng thc s gim i. n khi di ca chn t bng ng bc

sng cng tc th cng thc ny khng cn ngha v lc c phn thc v phn

o ca tr khng vo u c gi tr v cng ln.

Cng thc (2.26) cho php ng dng khi im nt dng in nm cch u

vo chn t mt khong ln hn (0,1 0,15) ngha l khi t s l nm trong

khong 0 0,35 v 0,65 0,85.

Hnh 2.6 S ph thuc ca ZvA vo l

Nhn xt: Tr khng vo ca chn t i xng ph thuc vo chiu di tng

i ca chn t.

- 300

- 400

- 200- 100

0

100200300400

0 0,2 0,4 0,6 0,8 1,0 1,2

XVA

()

l/r=40l/r=20

l

l/r=60

100

-100

200300400

500600700800

0 0,2 0,4 0,6 0,7 1,0 1,2

RVA

()

l

l/r =40

l/r =20

l/r =60

0


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- Khi chiu di ca chn t (2 l) bng bi s ca /2 th lc tr khng ca

chn t c th xem gn ng ch c in tr thun. Cng hng ni tip xy ra khi

chiu di chn t bng 0,5; 1,5; 2,5;v tr khng vo RvA l thc v c gi

tr bng tr khng bc x ca anten. Trong trng hp ny s c gi tr l 73,1 i vi chn t na sng v 200 i vi chn t ton sng.

- Cng hng song song xy ra khi chiu di chn t (2 l) bng bi s ca

bc sng cng tc (, 2, 3,). Tr khng vo trong trng hp ny cng l

in tr thc v c gi tr v cng ln, c th t n 5000 . Gi tr ca tr

khng vo thc ph thuc vo t s chiu di trn bn knh chn t v tr khng

sng ca chn t.- Khi chiu di ca chn t khng phi l bi s ca /2 th lc tr khng

vo ca anten bao gm c phn thc v phn khng. Thnh phn khng ca tr

khng vo ca chn t i xng gn cc im cng hng ni tip ( l= /4)

tng t nh thnh phn khng ca mt mch cng hng ni tip: vng tn s

thp hn tn s cng hng n mang tnh dung khng v cc vng tn s cao

hn tn s cng hng n mang tnh cm khng. cc im gn im cng hng

song song ( l= /2) tr khng ca chn t tng t nh thnh phn khng ca mt

mch cng hng song song: vng tn s thp hn tn s cng hng n mang

tnh cm khng v cc vng tn s cao hn tn s cng hng n mang tnh

dung khng.

Thng thng tr khng vo ca chn t i xng c chn bng phng

php thc nghim. Bng cch thay i chiu di v ng knh ca dy chn t

ng vi tn s trung tm ca di tn cng tc, khi tr khng s thun tr .Khi bit gi tr ca tr khng vo ca anten ta c th d dng thc hin vic

phi hp tr khng gia anten v my pht hoc my thu.


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2.3.5 Chiu di hiu dng ca chn t i xng

Bc x ca anten ph thuc vo s phn b dng in trn anten. xem

xt quan h gia phn b dng in v chiu di chn t, ngi ta a vo khi

nim chiu di hiu dng, hdl .

Chiu di hiu dng l chiu di ca mt chn t tng ng vi chn t

tht, c dng in phn b ng u trn chn t v bng dng in u vo ca

chn t tht, vi din tch phn b dng in trn chn t tht bng din tch phn

b dng in trn chn t tng ng.

Cu trc ca anten cng tt nu nh chiu di hiu dng ca anten c gi tr

gn bng chiu di thc ca anten.

Chiu di hiu dng ca chn t i xng c xc nh theo biu thc:

t2hdkl

l g

= (2.28)

Vi chn t na sng c chiu di 22

l=

, do 1tgkl= v chiu di hiu

dng ca chn t s l hdl =

. Nu l chn t ton sng c chiu di2l=

, th

chiu di hiu dng ca chn t s l2

hdl =

.

Im

Im

2 = /2

Hnh 2.7. Chiu di thc v chiu di hiu dng ca chn ti xng


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CHNG III: C TNH PHNG HNG CA

ANTEN CHN T I XNG TRNH BY BNG

MATLAB

3.1: C TNH PHNG HNG CA ANTEN CHN T

3.1.1: M hnh ton

Gi s chn t c di l, c t dc theo trc 0z, tm pha trng vi gc

ta . Ta im kho st l (R, , ). Khi kho st trng vng xa ta lun c

R ? .

Hnh 3.1: Biu din chn t i xng trong ta cu

Theo quy c nh hnh v trn, gc c xc nh bi hnh chiu ca

R trn mt phng x0y v 0x , cn gc c xc nh bi gc gia R v 0z , tc

l mt phng x0y c 090 = .

T hnh v ta thy rng:


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Hm phng hng ca anten chn t ng hng vi mt phng vung

gc vi trc ca chn t (mt phng ).

Hm phng hng bc x bng khng ti =00 hoc =1800.

c tnh hng ch xc nh trong mt phng cha trc ca chn t.

Vy ta c hm phng hng ca anten chn t c xc nh bi :

kl kl

os( cos )-cos2 2( , )

sin

cf

=

: l bc sng.

2k

= : l tr khng sng.

Xt hm tnh hng v th phng hng ca chn t c chiu di tng

i khc nhau:

Chn t ngn ( l


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( )( )

22 os cosos cos 1 2

sin sin

cc

f

+ = =

(3.5)

Khi hm tnh hng bin chun ha l:

( )

2os cos2

sin

cF

=

(3.6)

Chn t c chiu di ln hn

Trong trng hp ny do trn mi nhnh chn t xut hin dng in ngc

pha nn hng vung gc khng c sai pha v ng i ca cc on dz nhng

v dng in c on ngc pha, do cng in trng tng hng ny s

gim xung, ng thi xut hin cc bp ph cc hng c sai pha ng i b

ht cho sai pha dng in. Nu on dng in ngc pha ln dn, ngha l l tin

dn ti , bp ph s ln dn, bp chnh nh dn. Khi l = , on ngc pha trn

mi nhnh chn t l bng nhau, bc x hng chnh (tc hng vung gc vi

trc chn t) s bng 0, bn bp ph tr thnh bn bp chnh.

Hnh 3.2 th phng hngca chn t i xng trong mt phng E

90o

0o180o

a) 0,1

90o

0o180o

b) 0,25

90o

0o180o

c) 0,5

90o

0o180o

d) 0,75

90o

0o180o

e)


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T th phng hng ta c nhn xt: Tnh hng ca chn t i xng

ph thuc vo chiu di in l. V vy khi nghin cu v anten chn t ngi ta

thng s dng di tng i so vi bc sng. K hiu:l

Xt trong mt s trng hp:

0l

: dipol in

0.5l

= : chn t na sng

1l

= : chn t ton sng

3.1.2: th phng hng trnh by bng Matlab

Phn code lnh trng trnh v th phng hng ca Anten chn t i

xng bng Matlab:

Lnh v th phng hng Anten chn t i xng trong khng gian 2

chiu (2D):

L=input('\n nhapl /lamda='); % Nhp gi tr

teta=0:0.01*pi:2*pi;tuso=(cos(pi.*L.*cos(teta))-cos(pi.*L));

ifL==2

mauso=sin(teta)*2.5;

else

mauso=sin(teta).*(1-cos(pi.*L))*1.5;

end

f=abs(tuso./mauso);polar(teta,f); % V th


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Lnh v th phng hng Anten chn t i xng trong khng gian 3

chiu (3D):

[teta,phi]=meshgrid(0:2.*pi./180:pi, 0:4.*pi./180:1.5.*pi);

L=input('\n nhap l/lamda=');tuso=(cos(pi.*L.*cos(teta))-cos(pi.*L));

ifL==2

mauso=sin(teta)*2.5;

else

mauso=sin(teta).*(1-cos(pi.*L))*1.5;

end

f=abs(tuso./mauso);x=f.*sin(teta).*cos(phi);

y=f.*sin(teta).*sin(phi);

z=f.*cos(teta);

surfl(x,y,z);

Phn kt qu chy trng trnh v th phng hng ca Anten chn t

i xng ng vi cc trng hp kho st:


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a: l/ 0 (Trng hp dipol in).

b:l/=0.5 (Trng hp chn t na sng)


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c:l/=1 (Trng hp chn t ton sng)


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d:l/=1.25

e: l/=1.5

f:l/=2

Hnh 3.3: c tnh phng hng ca anten chn t i xng


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3.1.3: Nhn xt v kt lun

T cc th phng hng ta nhn thy: Khi anten c di l/ nh, th phng hng c dng gn ging

th phng hng ca dipol in (hnh 3.3.b), ch khc l n c rng hp hn

(so snh 3.3.a vi 3.3.b).iu ny c gii thch nh sau:

- V trng bc x ca dy dn ti im kho st bng tng vecto ca trng

to bi cc dipol thnh phn (anten l tp hp ca cc dipole). Khi di l/ rt

nh, dng in mi im trn anten ng pha nhau. ng thi v c th coi

khong cch t cc dipol n cc im kho st bng nhau nn trng bc x cacc dipol ring r s ng pha v c cng i s vi nhau. Bin trng bc

x ca dy dn cc im trong khng gian u tng ln mt s ln ging nhau so

vi cng trng bc x ca mt dipol in ring r. V vy m th phng

hng ca anten khng khc so vi th phng hng ca dipol in.

- th phng hng ca n hp hn l do sai pha khong cch gia cc

dipol thnh phn.

Khi tng dn di anten (trong gii hn vn m bo ng pha dng

in trn anten, ngha l l/ 1) th th phng hng s hp dn li (hnh

3.3.b, 3.3.c)

- Tht vy, s tng di anten trong gii hn ni trn s tng ng vi

vic tng s dipol ng pha sp xp theo ng thng . Cng trng khu xa

theo hng vung gc vi anten s bng tng i s cng trng ca cc

dipole in ring r, v theo hng ny khng c sai pha khong cch. Bc x

c tng cng theo hng = 90. Khi dch chuyn im kho st khi

hng ny s xut hin sai pha khong cch. Cng trng ti im kho st

trong trng hp ny s nh hn trng hng = 90. Tng vecto s gim

nhanh nu im kho st cng dch chuyn xa hng = 90.


24/25

Khi tng di anten qa gii hn mt bc sng (l>) s xut hin khu

vc dng in ngc pha. th phng hng c xu th hp li nhng ng thi

s xut hin cc cc i ph (hnh 3.3.d, 3.3.e). S xut hin cc cc i ph l do

bc x theo hng vung gc vi trc dy dn ca cc dipol thuc khu vc dngin ngc pha s b trit tiu bi bc x ca cc dipol thuc khu vc dng in

mang du dng, v theo hng ny khng c sai pha khong cch nhng sai pha

dng in bng . Nu dch chuyn im kho st khi hng = 90 th s

xut hin gc sai pha khong cch ca trng to bi cc dipol thuc hai khu vc

ni trn.Ta c th tm c hng m theo hng sai pha khong cch ca

trng to bi hai khu vc dng in s bng . Tng sai pha ca trng s bng

2, ngha l trng bc x to bi cc dipol thuc hai khu vc dng in ngcpha s tr nn ng pha nhau, v hng s xut hin cc i ph.

Tip tc tng di dy dn th cng trng theo hng = 90 s

gim (do nh hng bc x ca cc dipol c dng in ngc pha gy ra), ng

thi bin cc i ph s tng.

Khil

=2 trng bc x theo hng = 90 s hon ton b trit tiu, cn

cc cc i ph s tr thnh cc cc i chnh ca anten.(hnh 3.3.f)

Nh vy th phng hng ca anten chn t s thay i khi thay i

di ca ca n. C th nh sau:

Khil

bin i, tng t gi tr rt nh (tng ng vi mt dipol in)

n mt gii hn nht nh th anten chn t cng tng nh hng gc

090 = tc l hm phng hng t cc i 090 = , v gi tr 3 cng nh( 3

l l gc to bi hai hng m ti cng sut bc x suy gim mt na so vi

hng bc x cc i) cng thay i theo. ), bp sng chnh cng hp (hnh 3.3.a,

3.3.b, 3.3.c v 3.3.d).


25/25

Khi gi trl

kh ln (

l

=1.5 hnh 3.3.e) hm phng hng khng cn

t cc i gc 090 = .

Khi l =2 ( hnh 3.3.f) th cc i chnh theo hng090 = cn b trit

tiu hon ton.

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