Btl Anten- Vu Minh Hai DTV50

8/14/2019 Btl Anten- Vu Minh Hai DTV50

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Bi Tp Ln Mn L Thuyt V K Thut Anten

Sinh vin: V Minh Hi Lp: TV50-H 1

LI M U

Trong nhng nm gn y, khoa hc cng ngh pht trin nh v bo trn

mi lnh vc vi hng lot nhng nghin cu, pht minh mi gp phn khngnh trong vic nng cao trnh sn xut v i sng ca con ngi. Mt trong

nhng lnh vc c nh gi l c trin vng nht v c coi l th mnh ca

Vit Nam hin nay phi k n vin thng, n lm cho con ngi xch li gn

nhau hn, lm cho khong cch a l khng cn ngha na.

ng gp vo s pht trin mnh m ni trn chng ta phi ni n s pht

trin ca cc thit b thu pht v kh nng truyn lan sng in t hin nay, bi l

hu ht cc h thng truyn dn thng tin, lin lc chng u s dng phng thctruyn lan sng in t l ch yu.

Cc thit b thu pht v chuyn tip sng in t gi chung l anten. Tu

theo iu kin cng tc, mc ch s dng cng nh kt cu ca cc h thng vin

thng m ta s dng nhiu loi anten khc nhau: anten chn t, anten khe, anten

mch di, anten gng, anten xon

Do nhu cu thng tin, lin lc, truyn ti d liu ngy cng cao nn cc

bng tn di sng di, sng trung dn dn b thay th bi cc bng tn di sngngn v cc ngn. Vi li th l kh nng bc x tt cc di sng ny cng vi

kt cu tng i n gin, d dng iu chnh v kt hp vi cc loi anten khc

to thnh mt h bc x m anten chn t l la chn ti u trong hu ht cc

thit b v tuyn in.

Trong bai t p l n nay, em xin c trnh bay v c tnh phng hng ca

chn t i xng nhm tm ra c gi tr gii hn di chn t sao cho hng

tnh ca n cn t cc i hng 090 .y l mt trong nhng hng bc

x quan trng ca anten trong vic thu pht sng in t.


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Mc lc, bao gm IVphn :

I: GII THIU CHUNG V CHN T I XNG

c tnh phng hng ca anten chn t1.1. M hnh ton

1.2. Hm phng hng v th phng hng ca chn t i xng

II: T VN

III: GII QUYT VN

IV: A NH GIA , K T LUN VA H NG PHA T TRI N TI

1. anh gia , kt lun

2. H ng pha t trin tai


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I. GII THIU CHUNG V CHN T I XNG

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

(hnh tr, hnh chp, elipsoit) c kch thc ging nhau, t thng hng trongkhng gian, v gia chng c ni vi ngun dao ng cao tn.

Chn t i xng l mt trong nhng ngun bc x c s dng kh ph bin

trong k thut anten.N c th xem nh l mt anten c lp hon chnh (anten

chn t i xng ),ng thi trong nhiu trng hp n cng l phn t kt cu

nhng anten phc tp.

1. c tnh phng hng ca anten chn t

1.1: M hnh ton

Coi chn t l tp hp ca nhiu dipol(I(z)=const) v kho xt trng vngxa.

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

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


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

trn mt phng x0y v 0x

, cn gc c xc nh bi gc gia R

v 0z

, tc l

mt phng x0y c 090

+ Cng thc tnh cng trng bc x ca chn t i xng:

Ta c: E=

/ 2

/ 2

l

l

dE =2

/ 2

0

l

dE (1)

M dE=W2

i 1

Re

-ikRI(z)

dz

sin( ) (2)

Thay (2) vo (1) ta c:

E= 2/ 2

0

l

W2

i 1

Re-ikR I(z)

dz

sin( ) (3)

(4)

(5)

lk lk os( cos )-cos( )

12 2W 2 sin

ikR

c

E iIbH eR

(6)

*Trong :W: tr khngsng (W=120 ( ))Ib:bin dng in im bng sng

k: h s sng (k=2

)

l: chiu di ca chn t:bc sng

1.2.Hm phng hng v th phng hng ca chn t i xng-p dng nguyn l nhn th hm phng hng ta c:

f( ,)=fd( ,).fk( ,) (7)


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Trong :fd( ,)=sin() :hm phng hng thnh phn

fk( ,)=

2

cos cos cos

2 2sin

kl kl

:l hm phng hng t hp

*Hm phng hng dng chun ha bin c dng:

F( ,)=

,

max( , )

f

f

(8)

+Vi di tng i ca chn t so vi bc sng l

1.44

Khi =900=>f(max)=(1-cos

2

kl)

Suy ra ( , ) ( , ). ( , )d k

F F F =

2

cos cos cos2 2

sin

sin 1 cos2

kl kl

kl

(9)

+ Vi di tng i ca chn t so vi bc sng l

>1.44(hm phng hng

khng cn cc i gc =900)

f( ,)=2

cos cos cos2 2

sin

kl kl

+ V hm phng hng dng chun ha bin c dng

F( ,)=

,

max( , )

f

f

* th phng hngng vi cc trng hpl


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+ Xt trong mt s trng hp:

0l

: dipol in

0.5l

: chn t na sng

1l

: chn t ton sng

l

=1.25 v

l

= 1.44 ;

l

=1.5 ;

l

=2

*Gin phng hng trong h ta cc ng vi cc gi tr ca l/ :

Hnh a:l

0 Hnh b:

l

=0.5

Hnh c:l

=1 Hnh d:

l

=1.25


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Hnh e:l

=1.44 Hnh f:

l

=1.5

Hnh g:l

=2

*Nhn xt: Hm phng hng anten cng ch ph thuc vo gc ,ngha l bc x

ca anten c hng tnh trong mt phng E v v hng trong mt phngH.

Hm phng hng ca anten chn t ng hng vi mt phng vunggc vi trc ca chn t (mt phng ).

Hm phng hng bc x bng khng ti =00hoc =1800. c tnh hng ch xc nh trong mt phng cha trc ca chn t v ph

thuc vo chiu di l.

*T cc th phng hng ta nhn thy:


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Khi anten c di l/ nh, th phng hng chun ha c dng gnging th phng hng ca dipol in (hnh b), ch khc l n c

rng hp hn (so snh a vi b).iu ny c gii thch nh sau: - V trng bc x ca dy dn ti im kho st bng tng vecto ca trng

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

l/ rt nh, dng in mi im trn anten ng pha nhau. ng thi v

c th coi khong cch t cc dipol n cc im kho st bng nhau nn

trng bc x ca cc dipol ring r s ng pha v c cng i s vi

nhau. Bin trng bc x ca dy dn cc im trong khng gian u

tng ln mt s ln ging nhau so vi cng trng bc x ca mtdipol in ring r. V vy m th phng hng ca anten khng khc

so vi th phng hng ca dipol in.

- th phng hng ca n hp hn l do sai pha khong cch gia ccdipol thnh phn.

Khi tng dn di anten (trong gii hn vn m bo ng pha dng intrn anten, ngha l l/ 1) th th phng hng s hp dn li (hnh b,

c)

- Tht vys tng di anten trong gii hn ni trn s tng ng vivic tng s dipol ng pha sp xp theo ng thng. Cng trng

khu xa theo hng vung gc vi anten s bng tng i s cng

trng ca cc dipole in ring r, v theo hng ny khng c sai pha

khong cch. Bc x c tng cng theo hng = 90. Khi dch

chuyn im kho st khi hng ny s xut hin sai pha khong cch.

Cng trng ti im kho st trong trng hp ny s nh hn

trng hng = 90. Tng vecto s gim nhanh nu im kho st

cng dch chuyn xa hng = 90.


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Khi tng di anten qa gii hn mt bc sng (l>) s xut hin khuvc dng in ngc pha. th phng hng c xu th hp li nhng

ng thi s xut hin cc cc i ph (hnh d,e,f,g). S xut hin cc cci ph l do bc x theo hng vung gc vi trc dy dn ca cc dipol

thuc khu vc dng in ngc pha s b trit tiu bi bc x ca cc dipol

thuc khu vc dng in mang du dng, v theo hng ny khng c sai

pha khong cch nhng sai pha dng in bng . Nu dch chuyn im

kho st khi hng = 90 th s xut hin gc sai pha khong cch ca

trng to bi cc dipol thuc hai khu vc ni trn.Ta c th tm c

hng m theo hng sai pha khong cch ca trng to bi haikhu vc dng in s bng . Tng sai pha ca trng s bng2 , ngha

l trng bc x to bi cc dipol thuc hai khu vc dng in ngc pha

s tr nn ng pha nhau, v hng s xut hin cc i ph.

Tip tc tng di dy dn th cng trng theo hng = 90 sgim (do nh hng bc x ca cc dipol c dng in ngc pha gy ra),

ng thi bin cc i ph s tng.

Khi l

=1.44 th cc i chnh bng cc i ph.

Khi l

=1.5 th cc i ph ln t cc i chnh.

Khi l

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

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

Nh vy rng bp sng chnh trong c tnh phng hng ca chn tph

thucvo thng s l/

II: t vn


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Gi 3(hay 21/2) l rng bp sng chnh - l gc to bi hai hng m ti

cng sut bc x suy gim mt na so vi hng bc x cc i. Theo hnh v

ta cng thy rng trong khong bin i ca l

, t rt nh ( 0 ) n gi tr gii

hn, rng bp sng chnh 3( hay 21/2) cng thay i theo. Vy quy lut bin

i ca 3 theol

nh th no?Bi ton t ra l chng ta phi tm -c mi lin

h ( tc l tm hm 3=f( l

) ).

III: GII QUYT VN

Vi bi ton ny chng ta c th gii quyt c theo ba cch sau:

Biu din di dng th (phng php ny rt trc quan, thhin r rngs ph thuc 3vo

l

).

Biu din di dng bng (ta c th tra cu cc gi tr c th ca 3ng vimi gi tr ca

l

).

S dng chng trnh trong Matlab tra cu nhanh chng gi tr 3 khinhp bt k gi tr

l

no.

Ta s s dng cng c phn mm Matlab v ng cong biu din s ph

thuc ca rng bp sng chnh trong c tnh phng hng ca anten chn

t vo di tng i ca chn t so vi bc sng(L= l

).

Ta s trnh by 2 hng gii bi ton ny:

+Hng 1:Theo hng F( )=-3dB( hm phng hng suy gim 3dB)

+Hng 2:Theo hng F()=0.7071(hm phng hng suy gim na

cng sut)

Lnh v hng 1 Lnh v hng 2


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th hm 3=f( l

) c dng nh hnh v:

clear

clc

for i=1:1:144L=0.01*i;

F=0;

teta=90;

b=1-cos(pi*L);

while F>-3.0102

teta=teta-0.01;

t=teta*pi/180;

a=(cos(pi*L*cos(t))-

cos(pi*L))./sin(t);

f=abs(a./b);

F=20*log10(f);

end

l(i)=L;

teta3(i)=2*(90-teta);

end

plot(l,teta3)

grid on

xlabel('do dai cua chan tu so voi buoc

song')

ylabel(' gia tri goc teta3')

title ('do thi cua ham teta3')

clear

clc

for i = 1:1:144L= 0.01*i;

b = 1-cos(pi*L);

F = 1;

teta = 90;

while F > 0.7071

teta=teta-0.01;

t = teta*pi/180;


cos(pi*L))./(sin(t));

F=abs(a./b);

end

l(i)=L;

teta3(i)=2*(90-teta);

end

plot(l,teta3)

grid on

xlabel('l/lamda')

ylabel(' gia tri goc teta3')

title (' do thi ham teta3 )')


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0 0.5 1 1.44 1.620

30

40

50

60

70

80

90

100

l/lamda

gia

trigoc

teta3

do thi ham teta3 )

Trong matlab guide, ta c th lp trnh yu cu matlab v th, giao din nh

sau:


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Hnh 3.1: l p trnh trong matlab guide biu din s phu thuc cua go c 3vo l/

+Biu din s ph thuc ca gc 3theol

nh trong bng sau:

di

(l/)

Gc 3 di (l/) Gc 3 di (l/) Gc 3

0.01 90.00 0.50 78.08 0.94 51.72

0.04 89.94 0.52 77.12 0.98 49.14

0.08 89.70 0.54 76.12 1.00 47.84

0.10 89.54 0.56 75.10 1.04 45.30

0.12 89.34 0.60 72.94 1.08 42.78

0.14 89.08 0.62 71.82 1.10 41.54

0.16 88.80 0.64 70.68 1.12 40.30


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0.18 88.48 0.66 69.52 1.14 39.08

0.20 88.12 0.68 68.32 1.16 37.88

0.24 87.28 0.70 67.12 1.18 36.680.26 86.80 0.72 65.90 1.20 35.50

0.28 86.28 0.74 64.64 1.24 33.18

0.30 85.74 0.76 63.38 1.25 32.62

0.32 85.14 0.78 62.12 1.28 30.92

0.34 84.50 0.80 60.84 1.27 31.48

0.36 83.84 0.82 59.54 1.28 30.92

0.37 83.48 0.84 58.24 1.30 29.82

0.40 82.38 0.88 55.64 1.34 27.64

0.44 80.78 0.90 54.34 1.36 26.58

0.46 79.92 0.91 53.68 1.40 24.50

0.48 79.02 0.92 53.02 1.44 22.48

Bng 1: Cc gi tr ca 3ng vil

thay i t 01.44

+nh gi:

Gi tr l/ Gc 3theo l thuyt Gc 3theo thc t

0.5 800 78.08

0

1.0 44 47.84

1.25 31 32.62

Bng 2: So snh kt qu gia thc t v l thuyt gc 3

+Lnh trong Matlab tra cu nhanh chng gi tr 3khi nhp bt k gi tr l

no.


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Lnh v hng 1 Lnh v hng 2

clear

clcdisp('day la bai toan tim goc teta3 theo

huong suy giam 3 dB')

L=input('nhap L=')

F=0;

teta=90;

b=1-cos(pi*L);

while F>-3.0102

teta=teta-0.01;

t=teta*pi/180;


cos(pi*L))./sin(t);

f=abs(a./b);

F=20*log10(f);

end

disp('goc teta3 can tim la:')

disp(2*(90-teta))

clear

clcdisp('nhap vao gia tri do dai chan tu

theo buoc song :');

L= input('nhap gia tri L=');

b= 1-cos(pi*L);

F=1;

teta=90;

while F> 0.7071

teta= teta-0.01;

t=teta*pi/180;

a= (cos(pi*L*cos(t))- cos(pi*L))./sin(t);

F= abs(a/b);

end

disp('goc teta 3 can tim la=');

disp(2*(90-teta));

Tuy nhin, nu s dung matlab thng thng , mi khi cn xac i nh 1 gi tr 3ta

phi copy dng lnh v paste vo ca s comman , ho c phai n F 5 chy li m-

file o 1 ln. iu o khng c tin l i khi ta c n tra c u nhanh cac gia tricu a

gc 3.Matlab guide acho ta l p trnh giao din tnh toan 1 cch d dng hn.


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Hnh 3.2: l p trnh trong matlab guide tnh goc 3khi nh p vao l/


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IV: a nh gia , kt lun va h ng pha t trin ta i:

1. anh gia , kt lun:

T th biu din s ph thuc 3= lf

ta c th nhn thy mt cch

trc quan rng khi gi trl

tng th rng bp sng chnh gim.

Kt qu cngtng t khi ta biu din bng bng , hay mt hm . Vi bngtra c u lp c vaham l p trnh trong matlab guide ta c th tra cu

mt cch d dng cc gi tr ca 3khil

thay i t 0 n gi tr ti hn.

So snh vi cc s liu c vi kt qu tnh ton c ta nhn thy giachng c s sai lch. R rng l kt qu tnh ton c trn c

chnh xc cao hn. Mc d kt qu a ra trong cc ti liu c sai s l

khng ng k.

Vi vic tm ra gi tr ti hn ( l

)gh c tnh hng ca chn t vn t

cc i gc = 90 th vic thit k ra cc anten chn t c kh nng bc

x tt s tr nn thun tin, d dng hn.

Vic phi ghp anten chn t vi nhau s cho hiu qu cao nht khita bitc hng tnh ca mi anten, ta s to ra mt h thng thu pht sng in

t p ng yu cu ngy cng cao ca k thut anten.

2. H ng pha t trin bi tp ln:

- Vi ti ny chng ta c th xy dng c tp hp cc anten chn t vi

di thch hp, sao cho n c kh nng bc x tt nht sng in t.

- V thi gian lm bi tp ln c hn nn em ch tm hiu c 1phn rt nho

trong l p trnh matlab guide nn ban l p trnh matlab guide con rt th s ,cha

thhin c ht tnh trc quan ca bi tp ln . hoan thin hn v mang


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l p trnh , h ng pha t trin cu a bai t p l n nay lal p trnh bng matlab guide

ho c bng cc ngn ng khc nh Visual basic , C+. v i hoa e p hn, trc

quan hn, kt h p thc hin c nhiu vi c m t lu c nh v a xac i nh cgc 3khi nh p vao L t ban phm, v a vera vachra c goc 3 o trong toa

cc cung nh trong khng gian 3D.

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