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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
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- Thng tin v tinh - Kinh t - Bo mt * Nhc im : Hiu sut thp
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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
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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
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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.
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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
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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
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- 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
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- 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)
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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
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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)
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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)
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),( 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)
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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
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=> 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 +=
___________________________________________
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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