chuong4-anten mảng
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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
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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)
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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
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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
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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
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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:
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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 ====
+=
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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
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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)
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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
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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
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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)
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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)
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(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)
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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:
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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)
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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.
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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
>=
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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)
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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.
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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
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