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Phng trnh sng, nghim ca phng trnh sng

Dipole Hertz

Dipole ngn

Dipole c ti

Monopole

Anten thng

Nguyn t anten vng

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Divergence:

Curl (Rot):

Mts ton t

3

213

2

132

1

321

321

)()()(1

u

hhA

u

hhA

u

hhA

hhhAdiv

332211

321

332211

321

1

hAhAhA

uuu

ihihih

hhhArot

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nh l divergence v nh l stock

nh l divergence

VS

SdAdVAdiv

.

nh l stokes

CldASdArot

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Cc nh lut

nh lut lu s AmpereMaxwell:

Lu s ca vector cng trng ttheo ng kn Cty bng tng i scc dng in chy qua din tch baobi ng kn C

nh lut cm ng in t Faraday:

Sc in ng cm ng c gi trbng v ngc du vitc bin thin t thng gi qua din tch gii hn bivng dy

nh lut Gauss i vi trng in:

thng lng ca vector cm ng in gi qua mt kn S bngtng cc in tch t do phn b trong th tch V bao bi mt S

nh lut Gauss i vi trng t:

thng lng vector cm ng t (t thng) gi qua mt kn S ty

lun bng khng

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Trong :

mt thng lngin[C / m2 ]

mt thng lng t [T] [Tesla] [Weber /

m2 ]

mt dng in [A / m2 ]

mtin tch[C / m3 ]

Ton t Gradient , Nabla , Hamilton

= . Ton t Laplace

D

B

J

v

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jm ezyxtzyx ,,),,,(

zyx jzmzj

ymyj

xmx eEieEieEiz,y,xEt,z,y,xE

zyx jzmzj

ymyj

xmx eDieDieDiz,y,xDt,z,y,xD

zyx jzmzj

ymyj

xmx eBieBieBiz,y,xBt,z,y,xB

zyx jzmzj

ymyj

xmx eHieHieHiz,y,xHt,z,y,xH

jt/

Biu din phc ho:

Mt khc:

Suy ra:

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Gi s ta bit vector mt dng in J, ta mong mun tnhton gi tr ca vector trng E v H c sinh ra bi J=>gii h trn

HjE

EjJH

/E

0 H

BjE

DjJH

D

0 B

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0 A

Trong ton hc ngi ta chng minh nu Divergence camt vector=0 th vector lun c th c biu din nhl xoy ca mt trng khc

V (hay ) nn lun tn ti vector sao

cho

0 H

A

L th vectorA

0 B

Hoc:(Biu din theo vector B) (*)

(Biu din theo vector H)

(*) Cch biu din trong ti liu

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0 Trng vector c xoy bng khng lun c th din t nhl Gradient ca mt trng v hng

Ngoi ra, ta c(Biu din theo vector B)

0

AjE

Suy ra

BjE

0 BjE

)( AB

nx

f

x

f

x

ff ,...,,

21

Li c=>

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AB

etAE

A

e

- Th vector

- in th v hng

Trong min tn s:

eAjE AB

Tm li

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Hng ng thc:

Maxwell: EjJH

EJHBED

;;

AB

=>

iu kin Lorentz:

=> Phng trnh sng 3 chiu theo v ngun)(rA

)(rJ

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Phng trnh sng 3 chiu theo v ngun)(rA

)(rJ

Trong min thi gian:

Trong min tn s:

,

,4

V

RJ r t dV

vA r t

R

Nghim:

dVV

rP R

r

0

x

y

z

V l vng c cha ngun J,

l vn tc truyn sng1v

Jt

AA

2

22

JAA

22

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16

Nguyn t anten thng, cn gi l dipole Hertz, l on dythng,rtmnh,h hai u, mang dng inbin thin tns, di rtnh so vibc sng l

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Nghim ca phng trnh sng:

V

dVr

v

r

tJA

4

V l th tch on dy dn: ddSdV

r

dv

rti

dSv

rtJ

r

dddS

r

v

rtJ

ASS

444

r

krtI

r

vrtI

r

vrti

A mm

4

sin

4

sin

42 /k v L h s sng

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Trong min tn s mII

r

eIA

jkr

4

r

eIAAjkr

mR

cos4

cos

r

eIAA

jkr

m

sin4

sin

0A

Trong h ta cu:

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Vy ta c cc thnh phn trng bc x nh sau:

L tr sng ca mi trng

Cc thnh phn cn li( ) bng 0 E,H,HR

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vng xa: R

Ta c:

Rt nh c th b qua.

Vy ti cc im vng xa, sng bc x c dng

gnnh cc sng phng, v cng pha nhau, vung

gc vi nhau v cng vung gc viphngtruynE

H

ri

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Cc ng sc

trng

in bc x

Trc Dipole nm

thng ng vung

gc mp ngang (mp cha trc

ngang , 2, 3

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Khi chiu di anten ln hn => s bp sng bt u tng

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23

Vector Poynting trung bnh, tcmt dng cng sut

catrngbcx l:

*HERe)r(W 2

1

2

2

0

2

22

222

sinsin32

,m

WW

r

IkrW

m

i vi dipole Hertz:

W(r,q) tlnghchvikhong cch R tim

kho st n anten v cngbcxph

thuc gc q, phn bu theo hng,phn

b khng u theo hng .

Nhn xt:

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Cng bc x

Cng bc x chun ha

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th nh hng:

rng ca th nh hng :

ooo 9045135

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L thuyt AntenDipole Hertz

WdFWrddFWr

ddRWrPr

4

0

2

2

0 0

0

2

2

0 0

2

,sin,

sin,,

Cng sut bc x:

2 2 2

0 max 2 2 232

mk I WW WR m

2k

1200

00

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2

2

0 0

2

2

0

0 0

22 2 2 2 2 2 2 2 22 3 2

2 2 2 2

0 0

, , sin

, sin

8sin

32 32 3 12

r

m m m

P R W R d d W

R W F d d

k I k I k I R d d R

R R

Cng thc cng sut bc x

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in tr bc x:22 2 2 2 2 2 2

2

2 2 222P 120 20.4 8012 6

mrr

m m

k I kRI I

Cng sut bc x:2 2 2

12

mr

k IP

in tr tn hao

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in tr tn hao dy di l , bn knh a , dn in , h st

2 2 2loss S

l lR R

a a

in tr b mt dy dn in , h s t 2

SR

in tr dy bn knh a , di 1 n v di

1 1

2 2 2SR

a a

2 f

7 7

0 4 10 ; 5,8 10H m S m Vt liu ng

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2

2P

losslossm

R I

Cong suat bc xa:2 2 2

12

mr

k IP

ien tr ton hao:

Cong suat ton hao

Hieu suat Dipole

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Goc khoi:

22

2

4 0 0

2 2

2

0 0

22 3

0 0

, 2 sin sin

2 1 cos sin

cos 1 82.2 cos 4 1

3 3 3

A F d d d

d d

d

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o li hng tnh:

5,1384

ddsin

4

ddsin,F

4

D 2

0 0

3

4

dB76,15,1log10dBD

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V du:

Cho anten la mot oan day dan l = 4 cm bc xa

tan so 75 MHz. Anten c lam bang ong va co

ban knh a = 0,4 mm. Cho bieu thc tnh Rloss nh

sau:

2 2loss lR

a

Tnh ien tr bc xa va hieu suat bc xa cua anten.

l

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Gii:

f=75 MHz m,f

c 4

1057

1037

8

1100

110

4

4 2

m

cm

Vay co the coi anten nay la nguyen to anten thang.

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=>0

24W

D

RPrad

D = 1.5, suy ra

2

22

2

2

22

4015

5,1

4

m

m

radI

R

IRP

2 2 2 2 2 2 2 2

0 max 2 2 2 2 2 2 2

120 .(2 ) 15 .

32 32

m m mk I I I

W W R R R

2

max maxmax

4 ( , ) 4 . ( , )( , )

rad rad

U R WD D

P P

Ta co: 120

0

00

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,R r 080802

2

Vat lieu Cu: mS,;mH 77

01085104

0,036

21

7

76

4

2

c

closs

108,5

1041075

1042

104f

a2

1R

%6969,0036,008,0

08,0

RR

Re

lossr

r

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Neu chieu dai cua oan day nhng khong

the coi oan day nh dipole Hertz th phan bo dong

ien tren oan day co the coi nh hnh tam giac.

z

2/

2/

20

2

21

|z|khi

;|z|khi|z|

I)z(I

m

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V

dVR

vRtJ

A4

R

dv

R

tidS

v

RtJ

R

dddS

R

v

R

tJA

SS

444

Sau khi lay tch phan va chuyen sang mien tan so tanhan c:

R

eIA

jkR

m

8

a biet:

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Bai tap:

Tm mat o cong suat bc xa, ve o th nh

hng, tm he so nh hng, ien tr bc xacua dipole ngan???

Gi y:

T the vector cua dipole ngan bang lan so vi dipole Hertz!

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Do the vect Dipole ngan = cua the vect Dipole Hertz nen Trng ien va t Dipole ngan cung = cua Dipole Hertz

Cng o bc xa, cong suat bc xa ien tr bc xa Dipole

ngan cung = cua Dipole Hertz

Nhac lai ve Dipole Hertz

r

sin /4

jkR

mjI k e

E V m

R

m/AE

H

2 2 2

12

mr

k IP

2

280rR

2 2 22

2( , ) sin

32

mk IU

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R

sin /8

jkR

mjI k eE V mR

m/A

EH

Suy ra Dipole ngan

2 2 2

48mr

k IP

2

220rR

2 2 22

2( , ) sin

128

mk IU

Cng o bc xa:

Cong suat bc xa:

ien tr bc xa:

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z

2/

2/z

2/

2/

e dong ien phan bo eu tren dipole ngan ta co the

s dung tai cam hoac dung (tai khang).

z

2/

2/

2/

mI

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20

221

2

1

2

121

zkhi

zkhi)(

zI

zkhiz)(

I

)z(I m

m

z

2/

2/

2/

mI

Bai tap:

Tm the vector, mat o dong cong suat,ve o th nh hng, tnh he so nh

hng, cong suat bc xa, ien tr bc

xa trong trng hp nay.

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0z

2ksinI

z

2

ksinI

zI

m

m

z2

-vi

2

z0vi

2z

Mt Dipole co chieu dai hu han la mot day co chieu

dai l (so sanh c vi bc song) c kch thch tai

iem gia . Bien o dong trong oan day phu

thuoc vao toa o z nh sau

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0z

2

ksinI

z2

ksinI

zI

m

m

z

2

-vi

2z0vi

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sin

2coscos

2cos

2

kk

r

eI

jE

jkr

m

sin

2coscos

2cos

2

kk

r

eIj

EH

jkr

m

Ta co the tm c cac thanh phan trng bc xa nh sau [1]:

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Ta co the the hien vector trng theo mot dang khac:

sin

2

kcoscos

2

kcos

R

eI60jE

jkR

m

Mat o cong suat trung bnh:

2

2

22

2215

2

sin

kcoscoskcos

R

IEW m

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5.0l l 5.1l

o th nh hng cua mot so anten thang

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Xet vong day hnh tron co ban knh a rat nho (a

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Chn 1 cp phn t dng in vi xng qua mt P cha im ang xtv Oz

'4

' ldr

eIAd

jkr

m

''

4'' ld

r

eIAd

jkrm

'ld

''ld

2 vector vi phn th cng li

ti thnh 1 vector vung gcmt phng P chnh lvector

i

'lid

''lid

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Khi o: iAA

'd'cosar

eI

'd'cosa

r

eIAdAdA

2

0

jkrm

2

0

jkrm

4

4

vung xa a

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1.2. aak 'cossinjkae 'cossinjka 1

'd'cossinjkacosR4

eaIA 2'

jkRm

2

0

sin

4sin

4

2

R

eSkIj

R

ekIajA

jkR

m

jkR

m

Vi S la dien tch hnh tron

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ie

rjrr

jISjH jkRm

.sin.

1

.

1

4

.

32

ie

rr

jkISjE jkRm

.sin.

1

4

.2

Loai bo cac thanh phan bac cao:

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Trong mien thi gian:

Song ien t bc xa bi nguyen to anten vong ch phu thuoc

vao . Phng cua vect E , H cua anten vong khac phng cuavect E , H cua dipole (hoan v)

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Mat o cong suat cua trng bc xa:

RR i,RWiE

*HEReW

2

2

1

2

1

4

2 2 2max2

1, sin sin2

maW R I W

R

2sin,F

Vay o th nh hng cua nguyen to anten vong cunggiong nh cua DIPOLE HERTZ .

4

2

max 2

1

2

m

aW I

R

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2

2

0 0

42 2

2 2 2 3

max2

0 0 0 0

4 4 4

2 2 2 2

2

, , sin

1, sin sin

2

1 8 2 2

2 3 4.3 12

r

m

m m m

P R W R d d W

aR W F d d R I d d

R

a a aR I I I

R

Cong suat bc xa:

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Cong suat bc xa:4

2 22 1

12 2rad m rad m

aP I R I

ien tr bc xa:4

2

6rad

aR

Trong khong kh:

1200

00

4

2 2210rad maP I

4

2 220radaR

So sanh vi cong suat bc xa cua anten thang?

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Anten monopole la cac anten n cc.

(VD: noi vo cap cua cap ong truc en mat phang at vadung vat dan ben trong keo dai nh la mot anten)

L

z

x L

2V

L

L

a) b) c)

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Tr khang vao cua monopole:

zdipoleHertA

monopoleA ZI

VVZ

2

12

2

1

Tr khang vao cua monopole phan t song

/4 bang motna tr khang vao cua dipole na song /2, neu bo qua matmat.

Tong cong suat c bc xa bi dipole gap oi monopole

dipolerad

monopolerad PP 2

1

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o nh hng cua monopole:

),(DP

),(U

P

),(U),(D

dipole

dipole

R

dipole

monopole

R

monopolemonopole

22

1

44

Ta a biet o nh hng cua dipole /2 la 1.64.

o nh hng cua monopole /4 la:

D = 21.64 = 3.28

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Hieu suat bc xa:

lossrad

rad

RRRe

V chieu dai dipole gap oi monopole nen ta co:

dipoleloss

monopoleloss RR

2

1

Ngoai ra: dipole

rad

monopole

rad PP 2

1

dipolerad

monopolerad RR

2

1

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62

Xac nh hng ma tai o cng o bc xa cc ai, tnh gockhi a, he so nh hng, o rong theo mc 3 dB cua anten

bc xa ch tren na cau tren va co cng o bc xa chuan

hoa la . 2cos,F

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V hng bc xa ch na cau tren nen ta co the viet:

laiconiemcactai0

20

20cos

F,F

2

Gii:

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2

0

2

0

2

0

3

4

2

0

2

0

2

32

31

3cos

sincos,

dd

dddFA

6

2

34

4

A

D dB78,76log10dBD

o rong theo mc 3 dB:

50cosF 21221 oo 4545

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Anten3