Chuong4-Anten mảng

1

CHNG 4 ANTEN MNG ....................................................................................................... 2

4.1 M U ....................................................................................................................... 2

4.2 MNG NG NHT MT CHIU ........................................................................... 4

4.3 MNG NG NHT HAI CHIU ........................................................................... 11

4.4 TNG HP KIU MNG ......................................................................................... 14

4.4.1 Phng php chui Fourier ................................................................................. 14

4.4.2 Mng Chebyshev ................................................................................................. 18

4.5 MNG CP IN CHO MNG ............................................................................... 24

4.6 MNG K SINH........................................................................................................ 27

2

CHNG 4 ANTEN MNG

4.1 M U

Trong chng trc, mt s loi Anten dipole tiu biu c gii thiu v phn tch. Cc anten u l cc anten n phn t c cu trc n gin. Tuy nhin cc anten n ny u c h s nh hng thp. Trong nhiu ng dng c bit l cc h thng thng tin Point-to-point i hi anten phi c tnh nh hng cao. Mt phng php hiu qu lm tng tnh nh hng ca anten l t hp cc anten n gin theo mt trt t nht nh to thnh mt anten mng gm nhiu anten phn t.

Trng tng ca anten mng c xc nh bng phng php cng vector ca cc trng bc x t cc anten thnh phn. y b qua tc ng tng h gia cc anten phn t. Trong mt mng gm cc anten phn t ging nhau c t nht nm yu t quan trng nh hng n kiu bc x ca anten mng:

1. Cch sp xp cc phn t ( xp xp theo ng thng, ng trn, tam gic....)

2. Khong cch gia phn t 3. Bin dng c kch thch trn cc phn t 4. Pha ca dng c kch thch trn cc phn t 5. Kiu bc x ca cc phn t

Xt mt mng gm N anten ging nhau, c t cng mt hng trong khng gian c kch thch vi bin Ci v pha i i vi anten phn t th i. V tr ca phn t i c xc nh bi vector ir .

Trng bc x ca mt anten chun t ti gc ta c kch thch vi bin 1 v pha bng 0 c dng:

refrE

rjk

4),()(

0

= (4.1)

khu xa rri, cc tia t cc anten phn t n im kho st c th xem l song song vi nhau do khong cch t phn t th i n im kho st c tnh nh sau:

iri rarR =

Trng to bi phn t th i s chm pha mt lng ir rak 0 so vi anten chun t ti gc ta . i vi bin ta s dng rRi .

Trng tng c dng:

iri rajkjN

ii

rjk

eCr

efrE +=

= 00

14),()(

(4.2)

3

Biu thc (4.2) c th c vit di dng:

refFrE

rjk

4),(),()(

0

=

iri rajkjN

iieCF

+

== 0

1),( (4.3)

Vi F(,) l hm phng hng ca mng.

H s nh hng t l vi 2 o h s nh hng ca mng c dng:

22 ),(),(),( FfD = (4.4)

Nguyn tc nhn gin phng hng: hm phng hng ca mng c xc nh bng hm phng hng ca mt anten phn t nhn vi hm phng hng c trng ca mng. y b qua tc ng tng h gia cc phn t trong mng.

Hnh 4. 1 Anten mng gm N phn t

z

x

yir ra

ra

ir

Nr

1r

2r 1R

iR

NR

1RAntenphn t

4

4.2 MNG NG NHT MT CHIU

Xt mt mng gm N +1 phn t dipole na sng t cch nhau mt khong d, c kch thch bi cc dng c cng bin C=I0 lch pha lin tip mt lng d v do n=nd.

Hnh 4. 2 Anten mng gm N+1 dipole na sng

H s mng c tnh bi cng thc (4.3):

=

+=N

i

ndjkdjneIF0

cos0

0),( (4.5)

Vi l gc gia vector v trc ca mng, trong trng hp ny l trc x. Bin i biu thc (4.5) c h s mng F (kiu ca trng bc x) c dng:

( )

( ) 2/cossin

cos2

1sin

0

0

0 dk

dkN

IF

+

+

+

= (4.6)

tin li cho vic tm hiu kiu bc x ca mng, t:

cos0dku = (4.7)

du =0 (4.8)

z

x

y

ra

r

0

12

i

N

5

H s ca mng c dng v c biu din trn hnh (4.3):

( )

( ) 2/sin2

1sin

0

0

0 uu

uuN

IF+

+

+

= (4.9)

Cc i ca tia chnh xy ra khi u= u0 Khong cch gia im cc i ca tia chnh v im cc i ca tia ph u tin:

13+

=N

u

Khi N8 th t s gia bin tia ph u tin v tia chnh c gi tr 2/(3 ) (hay 0.21). C N-1 tia ph gia hai tia chnh. Kiu mng tun hon vi chu k 2 theo bin u.

V dkdkudk 000 cos = do ch c mt khong ca u vi c ngha vt l gi l vng kh kin (visible region): dkudk 00 .

Hnh 4. 3 H s mng ca mng ng nht

Thc t thng ch yu cu c mt tia chnh trong vng kh kin iu ny i hi chn khong cch nh vng kh kin ch cha mt tia chnh nh biu din trong hnh 4.3. C hai trng hp chnh di y:

1. Trng hp =0 ( mng ng pha): Khi u0=0, cc i ca tia chnh xy ra khi: u = 0 hay cos =0 v =/2. Nh vy cc i ca tia chnh xy ra ti mt phng vung gc vi trc ca mng.

Nu chn khong cch gia cc phn t d

6

rng tia chnh BW( beam width- gc gia hai im khng ca tia chnh) c xc nh t diu kin :

=+ cos2

10dk

N

dNdkN )1()1(2cos 0

0 +

=+

=

Khi N c gi tr rt ln, cos c gi tr rt nh. Lc c th xem 2

t =2

, ta c = sin)2

cos( do rng ca tia chnh:

LdNBW 00 2

)1(22 =+

== (4.10)

Vi dNL )1( += l chiu di ca mng.

Nh vy rng ca tia chnh t l nghch vi chiu di ca mng. tn s cao, c th to mng ng nht ng pha vi bp sng hp.

V d: Mt anten mng c chiu di L=200, c rng tia chnh BW =0.1 rad =6o. Nu anten hot ng tn s 30GHz (0=1cm), th chiu di ca mng l 20cm. Tuy nhin nu bc sng vo khong 1MHz (0=300m), chiu di ca mng l 6km khng kh thi.

Vic tnh chnh xc h s nh hng ca mng l rt kh. Trong trng hp ang xt l mng ng pha gm N+1 dipole na sng th phi tnh cng sut bc x ton phn bng cch tnh tch phn sau:

dddk

dkN

sin)cossin

2sin(

)cossin2

1sin(

sin

)cos2

cos(2

2

0 0 0

0

+

Cng c th tnh gn ng h s nh hng cc i ca mng bng cng thc:

HEA HPBWHPBW

D

= 44

0 (4.11)

Vi A l gc c chim bi chm tia chnh, HPBWE l gc na cng sut trong mt phng E, trng hp ny l gc na cng sut ca dipole na sng c gi tr 780 (1.36 rad). L gc na cng sut trong mt phng H; trong trng hp ny chnh l gc na cng sut xc nh bi h s mng:

7

dNdkNHPBWH

)1(

65.2)1(65.222 0

02/1 +

=+

== (4.12)

Nh vy h s nh hng cc i ca mt mng ng pha gm N+1 dipole na sng c tnh bi cng thc:

00

)1(48.536.12

4

dNHPBW

DH

+=

(4.13)

Tha s 2 mu s tnh cho 2 tia.

Nu phn t mng l anten v hng th kiu bc x s c tnh i xng trc quanh trc ca mng, khi h s nh hng cc i c tnh bi cng thc:

00

)1(37.22

4

dNHPBW

DH

+=

(4.14)

V d: Tnh h s nh hng cc i D0 ca mt mng ng nht gm 10 anen v hng c t cch nhau mt khong bng bc sng v c kch thch ng pha.

dBdND 72.7)925.5(log10925.54/1037.2)1(37.2 100

0

00 ====

+=

8

Hnh 4. 4 (a) Kiu mng ca mng ng pha (b) Kiu bc x ca dipole na sng

(c) Kiu bc x tng hp ca mng gm cc dipole na sng ng pha

2. Trng hp mng c pha ca dng in bin i theo qui lut sng chy (mng End-fire):

Trng hp chn dku 00 = , bp sng chnh t cc i khi:

0cos000 ==== dkdkuu

Hng bc x cc i chnh l hng +x. Nu chn dku 00 = , hng bc x cc i s c hng x.

iu kin ch c mt tia chnh trong vng kh kin: d

9

Kiu mng c tnh i xng quanh trc ca mng nh biu din trn hnh 4.5.

Tia chnh bng 0 khi:

=+ )1(cos2

10dk

N

Khi N rt ln, ca ti im 0 ca tia chnh c gi tr rt nh do c th thc hin php tnh gn ng: 2/)(1cos 2 .

dkN 0

2

)1(2

2)(

+=

rng tia chnh:

2/10

2/10 22

1/222

=

+==

LNdBW

Vi L l chiu di ca mng.

Nh vy rng tia chnh t l nghch vi cn bc hai ca chiu di mng tnh theo bc sng.

H s nh hng cc i c tnh gn ng bng rng tia na cng sut.

2/10

2/1 )1(63.122

+

==Nd

HPBW

(4.16)

Gc c gii hn bi chm tia na cng sut:

22/12/1

2

0 0

)()cos1(2sin2/1

==

dd

H s nh hng cc i:

00

)1(73.44

dND +=

= (4.17)

V d: H s nh hng cc i ca mng end-fire gm 10 phn t t cch nhau mt khong bng bc sng:

dBdND 728.10)825.11(log10825.114/1073.4)1(73.4 100

0

00 ====

+=

10

Hnh 4. 5 H s mng v kiu bc x ca mng End-fire

iu kin Hansen-Woodyard: tng tnh nh hng ca mng

Ndkud == 00 (4.18)

Cc i ca tia chnh xy ra khi:

Ndkuu +== 00 (4.19)

1cos

cos 00

>

+==

Ndkdku

Nh vy cc i ca chm tia chnh xy ra ngoi vng kh kin do th ca |F(u)| b dch sang pha tri mt on /N. So vi mng thng thng th bp sng chnh b thu hp v cc i chnh t gi tr nh hn. Tuy nhin do cng sut bc x ton phn gim do h s nh hng tng.

Vng kh kin

uu=0

= dk0

F

000

=

=

udk

Tia ph u tin

z

y

x

(a)

(b)

11

Hnh 4. 6 H s mng ca mng end-fire vi iu kin Hansen-Woodyard

4.3 MNG NG NHT HAI CHIU

Xt mt mng gm (N +1)(M+1) phn t dipole na sng t song song vi trc z trong mt phng xOz nh biu din trong hnh v 4.7. Cc anten phn t c kch thch bi cc dng c cng bin C=I0, phn b pha c dng djmdjne + tng ng vi v tr (n,m).

C th xem h l mt mng gm (M+1) mng mt chiu c (N +1) phn t. S dng nguyn tc nhn gin phng hng, h s ca mng hai chiu bng tch ca h s mng ca mng gm (M+1) phn t vi h s mng ca mng gm (N +1) phn t.

Ta c: cossin= xr aa v cos= rz aa , do h s mng ca mng hai chiu c biu din nh sau:

( )

( ){ }

( )

( ){ }2/cossin

cos2

1sin

2/cossinsin

cossin2

1sin),(

0

0

0

0

0 dk

ddkM

dk

ddkN

IF

+

+

+

+

+

+

=

(4.20)

n gin ta t:

dvdkvdudku

==

==

00

00

,cos,cossin

Cc i

F

Vng kh kin

udk0 0=u

0)1( IN +

dk0 0u

N

12

H s ca mng c vit li thnh:

( ) ( )

( ){ } ( ){ }2/sin2/sin2

1sin2

1sin

00

00

0 vvuu

vvMuuN

IF++

+

+

+

+

=

(4.21)

Hnh 4. 7 Mng hai chiu

Hng bc x cc i chnh c xc nh t iu kin:

=

=

0

0

vvuu

Nu cc phn t ca mng c kch thch ng pha, 0== th hng bc x cc i chnh vung gc vi mt phng ca mng. Trong trng hp ny mt phng ca mng l mt phng xOz nn hng bc x cc i chnh l hng y. Bng cch chn cc gi tr v thch hp c th iu khin hng bc x cc i theo mt hng ty (mng pha).

Trong trng hp mng c kch thch ng pha, rng tia chnh trong mt phng xy v xz c xc nh bi cc iu kin sau:

=+

=+

vM

uN

21

21

z

x

yra

r

d

i

d

M+1 phn t

N+1 phn t

13

Tng t vi mng mt chiu, rng tia chnh:

+=

+=

dMBW

dNBW

yz

xy

)1(2)(

)1(2)(

0

0

(4.22)

Gc na cng sut c xc nh:

+=

+=

dMHPBW

dNHPBW

yz

xy

)1(65.2)(

)1(65.2)(

0

0

(4.23)

H s nh hng cc i c tnh gn ng nh sau:

20

20

2

0 33.8)1)(1(33.8

)()(24

AdMN

HPBWHPBWD

yzxy

=++

=

(4.24)

Vi 2)1)(1( dMNA ++= l din tch ca mng.

Nh vy h s nh hng cc i t l thun vi din tch o theo n v bnh phng bc sng. y l c trng cho tt c cc loi anten.

Khi tia chnh lch khi trc ca mng rng ca tia chnh s thay i. Gi s 00 cos,0 dkd == , tia chnh nm trong mt phng xy v lch so vi trc x mt gc

0 . S dng php khai trin Taylor c th vit:

))(sin()cos(cos 00000 dkdk

rng tia chnh c xc nh bi biu thc:

0

00

000

0

sin)1(222)(

))(sin(2

1

)(2

1

dNBW

dkN

uuN

xy +===

=+

=++

(4.25)

14

Nh vy rng tia chnh trong mt phng xy tng t l vi 1/sin0. Ta nhn thy 0sin)1( dN + chnh l di ca mng chiu ln hng vung gc vi hng bc x

cc i, v di ca hnh chiu th bao gi cng nh hn di thc do rng tia chnh c m rng.

4.4 TNG HP KIU MNG

Trong phn trc chng ta tm hiu v mng ng nht. Mng ng nht c th to ra kiu bc x vi bp sng hp bng cch phn b phn b pha thch hp cho cc dng c kch thch trn cc phn t. Bng cch iu khin phn b pha gia cc phn t bc x c th thay i hng bc x ca tia chnh.

Nu thay i phn b ca cc dng kch thch, ta c th iu khin hnh dng v d rng ca bp sng chnh cng nh v tr v ln ca cc bp sng ph. Do c th to ra mt kiu bc x gn ging vi kiu bc x cho trc. y chnh l bi ton tng hp kiu mng hay c th gi n gin l tng hp mng.

C rt nhiu phng php khc nhau c nghin cu v pht trin tng hp kiu mng, trong phn ny ch gii hn mt s phng php tiu biu. Trong phn ny chng ta ch cp n vic tng hp kiu mng cho mng mt chiu. Tuy nhin cc phng php ny hon ton c th ng dng cho mng hai chiu da trn nguyn tc nhn gin phng hng.

4.4.1 Phng php chui Fourier Xt mt mng gm 2N+1 phn t v mt mng gm 2N phn t c kch thch ng pha nh biu din trong hnh 4.8 vi gc ta trng vi tm ca mng.

Hnh 4. 8 (a) Mng mt chiu vi 2N+1 phn t; (b) Mng mt chiu vi 2N phn t

d

z

y

x -N

N

1

-1

r

d

z

y

x -N

N

1-1

r

(a) (b)

15

(a) Trng hp mng c s phn t l, 2N+1 phn t:

=

==N

Nn

jnuneCuFF )()(

(4.26a)

(b) Trng hp mng c s phn t l s chn, 2N phn t

=

=

+ +==N

n

unjn

Nn

unjn eCeCuFF

1

]2/)12[(1

]2/)12[()()( (4.26b)

Vi cos0dku = .

Ch : V 0 , do dkudk 00 ng vi vng kh kin.

Nu mng c tnh i xng, Cn=C-n, h s mng c th vit li thnh:

(a) Mng c 2N+1 phn t:

=

=

=

+=

++===

N

nn

N

n

jnujnun

N

Nn

jnun

nuCCuF

eeCCeCuFF

10

10

cos2)(

)()()(

(4.27a)

(b) Mng c 2N phn t:

=

=

=

+

+=

+==

N

n

unjunjn

N

n

unjn

Nn

unjn

eeC

eCeCuFF

1

]2/)12[(]2/)12[(

1

]2/)12[(1

]2/)12[(

)(

)()(

=

=

N

nn u

nCuF1

)2

12cos(2)( (4.27b)

Bng cch kch thch cc phn t mng vi bin Cn thch hp c th to c mt mng c h s mng gn ng vi h s mng Fd(u) cho trc. Cn chnh l h s ca chui Fourier v c tnh bng cng thc di y:

(c) Mng c 2N+1 phn t:

NndueuFCC jnudnn ==

0,)(21

(4.28a)

(d) Mng c 2N phn t:

NndueuFCCunj

dnn ==

1,)(21 2

12

(4.28b)

16

V d: Dng phng php bin i Fourier thit k mng i xng c 7 phn t t cch nhau d=0/2 c kiu mng nh sau:

17

Hnh 4. 9 Kiu mng cho trc v kiu mng tng hp bng phng php chui Fourier vi 7 phn t

-0.5

0

0.5

1

1.5

-4 4dk0dk0 20dk

20dk

)(uFd

uuuF 3cos32cos2

21)(

+=

u

-0.5

0

0.5

1

1.5

0 30 60 90 120 150 180

)(uFd

)(F

(a)

(b)

18

V d 2: Lp li bi ton trn vi mng c 6 phn t:

V mng c tnh i xng do :

)22

12sin()12(

22)12(

2

2122

121)(

21

2212

2212

2/

2/

212

2/

2/2

122

12

=

=

====

nnj

een

nj

eduedueuFCC

njnj

unjunjunj

dnn

Bin dng kch thch trn cc phn t:

2)

4sin(2)

22112sin(

)112(2

1 ==

=C

32)

43sin(

32)

22122sin(

)122(2

2 ==

=C

52)

45sin(

52)

22132sin(

)132(2

3 ==

=C

Kiu mng:

)25cos(

522)

23cos(

322)

21cos(22)( uuuuF

+=

)cos2

5cos(5

22)cos2

3cos(3

22)cos2

cos(22)(

+=F

4.4.2 Mng Chebyshev

Mng Chebyshev c tng hp bng nh s dng a thc Chebyshev. Phng php ny c p ng thit k mng vi rng nh nht ng vi mt mc ph cho trc hoc ngc li mt mc ph nh nht vi rng ca mng cho trc. a thc Chebyshev c s dng tm ra phn b dng ph hp vi mc tiu thit k. Phng php ny c xut u tin bi C.L. Dolph do mng loi ny cn c gi l mng Dolph-Chebyshev.

19

Cc tnh cht c bn ca a thc Chebyshev:

+nh ngha:

,1)(0 =xT xxT =)(1

12)( 22 = xxT xxxT 34)( 33 =

188)( 244 += xxxT xxxxT 52016)( 355 +=

1184832)( 2466 += xxxxT xxxxxT 75611264)( 3577 +=

132160256128)( 24688 ++= xxxxxT xxxxxxT 9120432276256)( 35799 ++=

212)( = nnn TxTxT (4.29)

)(xTn dao ng trong khong 1 khi x dao ng trong khong 1 v c n nghim trong khong 1. Khi ,1>x )(xTn tng n iu nh biu din trong hnh 4. 10. Nghim ca )(xTn c cho bi:

1...,,2,1,0,2

21coscos =+== nmn

mx mm (4.30)

Gi tr ca a thc c th c tnh theo cng thc di y:

1)](coshcosh[)(

11)](coscos[)(1

1

>=

20

Hnh 4. 11 Da thc Chebyshev

Xt mng ng pha i xng c 2N v 2N+1 nh biu din trong hnh 4. 12.

Hnh 4. 12 Mng Chebyshev vi 2N+1 phn t v 2N phn t

H s ca mng c dng (s dng 4.27):

==

==+

=

=

=+=

N

nn

N

nnN

N

nn

N

nnN

vnCunCF

nvCnuCCF

112

01012

])12cos[(2)2

12cos(2

2cos2cos22

T1T2

1

Tn(x)

-1

x-1 1

d

z

y

x

2C0

r

C1

C1

CN

CN

C1C1

CN

CN

d

z

y

x

r

(a) (b)

21

H s chun ha ca mng c dng:

=

=+

=

=

N

nnN

N

nnN

vnCF

nvCF

12

012

])12cos[(

2cos (4.32)

Vi

coscos212/

00

ddkuv === .

Nh vy h s mng ca mt mng ng pha i xng c dng chi cosin hu hn v c dng ging vi a thc Chebyshev. Phn b bin Cn ca cc phn t c th c xc nh bng cch ng dng trin khai a thc Chebyshev thch hp. Ch bc ca a thc bng s phn t ca mng tr i mt.

Cc bc thit k mng Chebyshev:

1. Xc nh dng ca h s mng da theo biu thc (4.32) 2. Khai trin h s mng da vo cng thc khai trin a thc Chebyshev(4.29) 3. Xc nh gi tr x = x1 sao cho Tm(x1) = R vi R l t s gia mc chnh v mc

ph. Hoc xc nh x1 da vo rng tia chnh.

)](cosh1cosh[ 11 Rnx =

Hoc z

z

vxx

cos1=

4. Thay

1

cosxxv = (4.33)

vo biu thc trin khai ca h s mng bc 2.

5. S dng biu thc khai trin ca h s mng sau khi thay (4.33) vo v so snh vi biu thc khai trin ca Tm(x). Vi m c gi tr nh hn s phn t mng l 1. So snh h s ca cc s hng cng bc tm h s Cn.

6. Dng cc h s tm c chun ha theo h s ca phn t ngoi cng biu din h s mng.

V d1: Thit k mng Chebyshev vi t s mc chnh ph cho trc

Hy thit k mt mng i xng ng pha vi 5 phn t t cch nhau d= 0/2 c t s gia mc chnh ph R=20dB.

22

Bi gii: 1. H s mng ca mng ng pha gm 5 phn t c biu din di dng:

=

=+ =

2

012 2cos

N

nnN nvCF

Vi cos

2cos

0

==dv

2. Bin i h s mng:

vCvCCCCCvvCvCC

vCvCCF

42

221210

242

210

2105

cos8cos)82()(

)1cos8cos8()1cos2(

4cos2cos

+++=

+++=

++=

3. Xc nh x1 101020 20/20 === RdBR

T (4.31),ta c :

293.1)]10(cosh41cosh[

10)](cosh4cosh[)(

11

11

14

==

===

x

RxxT

4. Thay 1

cosxxv = vo h s mng:

424

41

4

221

2

212105

881)(

8)82()(

xxxTxxC

xxCCCCCF

+==

+++=

ng nht cc h s tng ng, ta c h phng trnh:

=

=

=

+=

=

=

=+

=

=

7.25.48.2

431

44

1

882

88

0

1

2

21

410

21

411

412

210

21

221

1

41

2

CCC

xxCxxC

xC

CCCxC

xCxC

5. Cc h s chun ha :

=

=

=

97.061.1

1

0

1

2

CCC

6. H s mng chun ha:

cos2

4cos2cos61.197.05

=

++=

v

vvF

23

Bin dng c kch thch trn cc phn t ln lt l: 1.94I0 cho phn t trung tm ca mng, 1.61 I0 i vi phn t v tr 1, v I0 i vi hai phn t v tr 2, v tr ngoi cng ca mng.

rng tia chnh:

im 0 ca tia chnh chnh l nghim u tin ca a thc T4(x):

924.0)8/cos(21cos === n

xz

01

1

43.60054.1)/775.02(cos

775.0cos2

714.0293.1924.0)cos

2cos(cos

===

==

====

rad

v

xxv

z

zz

zzz

rng tia chnh: BW=2(90-60.43)=59.130

V d 2: Thit k mng Chebyshev vi rng tia chnh cho trc.

Hy thit k mt mng i xng ng pha vi nm phn t t cch nhau d= 0/2 c rng tia chnh bng 400.

Gii:

Cc bc 1, 2, 4, 5 v 6 lm tng t v d 1, ch khc bc th 2, xc nh x1,

3. Xc nh x1 v t s gia mc chnh v ph: im khng ca tia chnh xc nh bi:

924.0)8/cos(21cos === n

xz

859.0537.0coscos

537.0cos2

2040902/90 0

==

==

===

z

zz

z

v

v

BW

Ta c

dBxTR

x

xxxv zz

75.7)44.2(log2044.2)]075.1(cosh4cosh[)(

075.1859.0924.0

859.0924.0cos

101

14

1

11

=====

==

==

24

Cc bc cn li lm tng t v d trn ta thu c h s dng kch thch v h s chun ha nh sau:

=

=

=

=

=

=

+=

=

=

29.054.0

1

39.072.034.1

431

44

0

1

2

0

1

2

21

410

21

411

412

CCC

CCC

xxCxxC

xC

H s ca mng c dng:

cos2

4cos2cos54.1029.05

=

++=

v

vvF

)cos2cos()coscos(54.1029.0)( ++=F

4.5 MNG CP IN CHO MNG

Vn thit k mng ng truyn cung cp cc dng in c bin v pha nh trc cho mi phn t ca mng 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 dng kch thch c bin khng ng nht cn s dng mt s mch chia cng sut vi tn hao thp kch thch cc dng c bin d khc nhau trn mi phn t. Cn phi phi hp tr khng gia cc anten phn t vi ng truyn cung cp trong di tn cng tc.

Thng thng ta chia mng thnh cc nhm nh hoc vng da trn tnh i xng ca mng. Cc vng s c cp in vi cc mng cp in i xng.

V d: Mng gm 9 phn t c chia thnh 3 nhm, mi nhm c nui bi mt ng truyn n ni vi ng truyn chnh nh biu din trong hnh v 4.13. Nh tnh i xng ca mng cp in 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.

25

Hnh 4. 13 Mng gm 9 phn t c tnh i xng chia thnh 3 vng

Trong mng biu din trn hnh v, c 3 nhm c cng dng kch thch: (1,3,7,9), (2,8), (4,6).

Mi phn t c ni vi ng truyn chnh nh mt on ng truyn c di bng bc sng. Dng in kch thch trn mi anten phn t lin quan trc tip n pha v bin ca in th trn ng truyn chnh.

t: Zf : tr khng c trng ca ng truyn chnh Za : tr khng c trng ca on bc sng Za,in : tr khng vo ca phn t anten Vf : in th ti u vo ca on bc sng

Hnh 4. 14 Dipole phn t c cp in bi ng truyn bc sng

Dng v in th ti u vo ca on bc sng:

)( +

+

+=

=

aaaf

aaa

IIZVIII

(4.34)

1

2

3

4

5

6

7

8

9

fZfV

aZ

aI

+aI4

0

26

Vi +aI v aI l dng ti v dng phn x. Ti u vo ca mi anten phn t dng in

s b tr pha 900 so vi u vo ca on bc sng. Dng ti u vo ca anten phn t:

faa

f

aaj

aj

ain

VjYZjV

IIjeIeII

=

=

+== ++ )(2/2/

(4.35)

Vi aY l dn np c trng ca on bc sng.

Nh vy, dng vo ti anten phn t ch ph thuc vo in th ca ng truyn chnh v tr khng c trng ca on bc sng m khng ph thuc vo tr khng c trng ca ng truyn chnh cng nh tr khng ca anten phn t.

p dng nguyn tc trn cho ba anten phn t dng kch thch ng pha v bin t l vi aY , bY , cY biu din trong hnh. V in th ti cc u vo ca mi on bc sng ging nhau do cc u vo ny cch nhau ng bng bc sng 0 trn ng truyn chnh. Nu mun dng kch thch trn phn t b ngc pha vi dng kch thch trn hai phn t cn li th khong cch gia u vo ca mi on bc sng trn ng truyn chnh bng bc sng.

Hnh 4. 15 Mng gm 3 anten dipole phn t c cp in bi ng truyn bc sng

Tr khng vo ca c mng chnh bng tr khng tng ng ca mng c ba phn t vi tr khng c gi tr ln lt l inccinbbinaa ZZZZZZ ,

2,

2,

2 /,/,/ mc song vi nhau. Nu c s phi hp tr khng hon ton gia on bc sng v cc anten phn t th tr khng vo ca mng chnh bng tr khng tng ng ca mng c ba phn t vi tr khng bng tr khng c trng ca mi on bc sng mc song song.

aY

bY

cY4/0 in

Z

ng truync chiu di 0

27

4.6 MNG K SINH

Mng k sinh l mng m khng phi tt c cc phn t ca mng u l phn t tch cc (phn t c kch thch bi dng nui-driven element). Cc phn t th ng (nondriven element) c kch thch bi s cm ng vi phn t tch cc cng nh vi cc phn t th ng khc thng qua tr khng tng h gia chng.

Mng k sinh thng c thit k bng con ng thc nghim bi v rt kh tnh tr khng tng h, kch thc cc phn t v khong cch thch hp gia cc phn t do cc phn t tc ng tng h phc tp gia cc phn t.

Hnh 4. 16 (a)Mng k sinh vi 2 phn t (b) Mng k sinh vi 3 phn t

Xt mng k sinh n gin nht vi 2 phn t gm mt phn t tch cc v mt phn t phn x c biu din trong hnh 4.16a. Ta c th xem mng nh mt mng ca cp phn t hai u. Vi phn t phn x, do khng c dng nui nn in th bng 0.

Ta c:

2121110 IZIZ += (4.36a)

2221122 IZIZV += (4.36b)

Vi Z11: tr khng ring ca phn t phn x

Z22: l tr khngphn t tch cc

Z12: l tr khng tng h gia hai phn t

T (4.36) tnh dng trn cc phn t thu c:

22211

2121

12ZZZ

VZI

= (4.37a)

d

Phn tphn x

Phn ttch cc

(a) (b)

Phn tphn x

Phn ttch cc

Phn tdn x

Hng catia chnh

28

22211

2112

12ZZZ

VZI

= (4.37b)

11

12

2

1

ZZ

II = (4.37c)

Biu din t s gia dng I1 v dng I1 di dng: djeZZ

ZZ

II

11

12

11

12

2

1 =

= th h s ca

mng c dng:

cos

11

12 01 djkdjeZZF = (4.38)

iu kin c hng bc x cc i ti =0 : = dkd 0 hay

=0k

d

iu kin hng = c bc x bng khng:

=

=

1

2,0

11

12

0

ZZ

dkd

Tuy nhin rt kh c 11

12

ZZ

=1 o ch c th lm cho bc x theo hng = c gi tr nh nht ch khng th lm cho bc x theo hng ny b trit tiu.

Gc pha ca Z11 c th thay i bng cch thay i chiu di ca phn t:

Nu :2/ 1101 Zl < dung khng

:2/ 1101 Zl > cm khng

Tr khng tng h Z12 ph thuc vo khong cch d gia hai phn t.

Phn t phn x phi c chiu di ln hn na bc sng v khong cch d phi c gi tr vo khong 0.150.

iu kin l tng c phn x ton phn:

29

=

=

=

1

2/4/

11

12

0

ZZ

dd

(4.39)

Nu phn t k sinh c chiu di nh hn phn t tch cc th phn t k sinh s l phn t dn x v hng bc x cc i s hng v pha phn t dn x. tng tnh nh hng ca mng k sinh ta s dng kt hp c phn t phn x v phn t dn x nh biu din trong hnh 4.16b. y l loi anten Yagi-Uda n gin nht vi 3 phn t.

Mng Yagi-Uda

Nhc im ln ca mng k sinh l in tr bc x nh. Vi phn t tch cc l dipole na sng in tr bc x c gi tr vo khong 20 khi c mt phn t k sinh.

Nu dng mt dipole gp th in tr bc x tng ln khong 4 ln v c gi tr vo khong 80.

Di tn cng tc rt hp ch vo khong t 2~3 % do phi thit lp phn t k sinh tht chnh xc c kt qu ti u.

Mng Yagi-Uda l mt mng end-fire vi mt phn t tch cc, mt phn t phn x v nhiu phn t dn x nh biu din trong hnh v di y.

Hnh 4. 17 Mng Yagi-Uda

t Zii: tr khng ring ca phn t th i

Zij: l tr khng tng h gia hai phn t th i v j

Ch ngoi phn t tch cc c in th u vo l V0, cc phn t cn li c in th u vo bng 0, ta c h phng trnh sau cho mng Uda-yagi vi N phn t:

Phn tphn x

Phn ttch cc

Phn tdn x

Hng catia chnh

l1

d1

V0

-1 0 1 2 N

30

NNNNNN

NN

NN

IZIZIZIZ

IZIZIZIZVIZIZIZIZ

++++=

++++=

++++=

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

......0

110011

01010001100

1111010111

(4.40)

Nu c th xc nh c tr khng tng h Zij v dng Ii trn mi phn t th c th tnh c trng bc x to bi anten mng.

Yu cu thit k l chn khong cch gia cc phn t v chiu di ca cc phn t thch hp dng trn cc phn t c pha thch hp tha mn iu kin cng ng pha v trng bc x theo hng thun (Hng t phn t phn x n phn t tch cc v phn t dn x). Tuy nhin cc thng s nh hng ln nhau nn rt kh gii bi ton mt cch chnh xc. Do vic thit k mng Yagi-Uda thng da trn thc nghim.

Mng Yagi-Uda thng gp c t 8 n 10 phn t vi li vo khong 14dB. Mng c di tn hot ng hp ch khong vi phn trm. Tuy nhin mng c cu trc n gin nn c s dng rt ph bin.

CHNG 4 ANTEN MNG4.1 M U4.2 MNG NG NHT MT CHIU4.3 MNG NG NHT HAI CHIU4.4 TNG HP KIU MNG4.4.1 Phng php chui Fourier4.4.2 Mng Chebyshev

4.5 MNG CP IN CHO MNG4.6 MNG K SINH

Documents

Chuong4-Anten mảng