em-ch6-lecture-01-s1-12-13

1 Trn Quang Vit BMCS Khoa in HBK Tp.HCM Tran Quang Viet Faculty of EEE HCMUT-Semester 1/12-13

Chng 6 Nguyn l bc x in t v anten

6.1. Gii thiu6.2. Nguyn t Anten thng (dipole Hetzian)6.3. p dng cho anten sng6.4. Cc thng s c trng ca anten

Trn Quang Vit BMCS Khoa in HBK Tp.HCM Tran Quang Viet Faculty of EEE HCMUT-Semester 1/12-13

6.1. Gii thiu

Hin tng bc x in t: anten sinh ra sng in t

V, v(t)

J(t)

z

xy

EM wave

EM wave

EM wave

EM wave

EM wave

EM wave

EM wave


6.1. Gii thiu

Dng thc t c bn ca anten:


6.1. Gii thiu

Mt s anten thc t:


6.1. Gii thiu

Tnh trng bc x:

Trn thc t V l dng dy v ngun xem xt l iu ha:

V

J(t-R/v)dV A(t)=

4 Rpi

anten

R-j

v

V

Je dV A =

4 R

pi

i

i

/ 2 /v pi = =Vi: (H s pha)

Rpi

ii

-j

L

I e d A =

4 R

Tnh th vect dng biu thc th chm:


6.1. Gii thiu

B Arot=i i

Tnh trng t dng nh ngha th:

Tnh trng in dng hpt Maxwell: (mi trng in mi)

rot H Ej=i i

Trn thc t ngi ta thng dng cch trn tnh chocc loi anten c bn (Hetzian dipole,) sau dng ktqu ny xp chng tnh cho cc anten phc tp hn!!!

1 H Arot

=

i i

1 E rot Hj=

i i


6.2. Nguyn t Anten thng (dipole Hetzian)

Xt nguyn t Anten thng mang dng i(t)=Imcos(t+)

6.2.1. Tnh trng in t6.2.2. Trng in t trong min gn6.2.3. Trng in t trong min xa


6.2.1. Tnh trng in t

' zd dz a= / 2

/ 2+

Tnh th vect dng:

/ 2

/ 2

'-jI e

A =4 R

Rzdz a

pi i

i

-jI

A e4

rza

r

pi

ii

Rpi

ii

-j

L

I e d A =

4 R



Biu din th vect trong h ta cu:

r

za

ra

aa

( ) ( ) ( )z r z r z za a a a a a a a a a = + + cos sinz ra a a =

( )I cos sin4 j r rA e a ar pi = i

i

-jI A e

4r

zar

pi

ii



Tnh trng t:

1H = rot A

i i 2-jr

2 2I j 1H = sin + e a4 r r

ii

Tnh trng in:

E =(1/j)rot Hi i

3-jr

r2 2 3 3

3-jr

2 2 3 3

jI j 1E =- cos + e a2 r r

jI 1 j 1 - sin - + + e a

4 r r r

ii

i


6.2.2. Trng in t trong min gn

Min gn c nh ngha l: r


6.2.2. Trng in t trong min gn

mr 3

IE = sin(t+ )(2cosa +sina )4r

Mt cng sut in t trong min gn:

2 22m

r 2 5IP=EH= sin(2t+2 ) sin a -sin2a

32 r

( )m 2I sinH = cos t+ a4r

V d ti im trong min gn c =pi/2:

r02P(r, , )=P sin(2t+2 )api

Cng sut in t lan truyn c tnh cht dao ng; mingn c gi l min cm ng ng to thnh phnkhng trong tr khng tng ng ca anten (ZA)


6.2.3. Trng in t trong min xa

Min xa c nh ngha l: r>>1 r>> /2 pi

2 2 3 31 1 1r r r

>> >>

Vi nh ngha trn ta c:

2-jr

2 2I j 1H = sin + e a4 r r

ii

-jrjI H = sine a4r

ii

mI H= sincos t- r+ + a2r 2

-jrjIH = sine a2 r

ii


6.2.3. Trng in t trong min xa3

-jrr2 2 3 3

3-jr

2 2 3 3

jI j 1E =- cos + e a2 r r

jI 1 j 1 - sin - + + e a

4 r r r

ii

i

2-j r

j I E = sine a4 r

ii

m

I E= sincos t- r+ + a2r 2

-j r

j IE = sine a2 r

ii -j r

j I

= sine a2 r

i



Mt cng sut trong min xa:

mI H = sincos t- r+ + a2r 2

m

I E= sincos t- r+ + a2r 2

( )2 2

2 2mr2 2

IP=EH= sin cos t- r+ +/2 a4 r

Cng sut in t lun truyn t ngun ra min xa vihng truyn l +r ; min xa c gi l min bc x. Min bc x ng gp vo phn thc ca tr khng tngca anten (ZA)



Mt s tnh cht i vi trng trong min bc x: Sng trong min xa l sng TEM Bin sng suy gim theo quy lut 1/r

Mt ng pha: t+r++pi/2=cost r=const sng cu(Thc t gn ng l sng phng!!!)

Vn tc pha bng vn tc truyn sng v c tnh gingstps trong in mi l tng

E & H cng pha tr sng c tnh ging nh stpstrong in mi l tng

Tnh nh hng: bin sng ph thuc vo lnca sng khng u theo mi hng, max khi =900, min=0khi =0 hoc 1800


Cng sut bc x: cng sut in t trung bnh qua mt cubn knh r>>; tm ti gc ta

bx rS SP P S P Sd d

= < > = 2 2

* 22 2

1P Re{ } sin2 8

mr

IE H ar

< >= =i i

2 22 2 2bx 2 20 0

P sin sin8

mI

r d dr

pi pi =

22

bx mP I3pi

=


10


in tr bc x Rbx l in tr tng ng m cng suttiu tn trn n bng cng sut bc x:

2bx bx mR =2P /I

2

bx2R3pi

=

21bx bx m2P R I=



6.3. p dng cho anten sng

Xt anten sng vi phn b ca bin dng nh h.v

~

Z

0

( )0I cos z

( )0i(t)=I cos cos for (-L/2

11


Xt trng min xa v dng kt qu ca dipole Hetzian:

-j r'0'

jI cos( ') 'sin'e a

2r'z dzd E =

i

-j r'0jI cos( ') ' sin'e a2r'

z dzd H

=

i

~

Z

0

z'

dz' 'r'

r

a

aa '

zcos

2L

2L

4

=

4

=



/ 4-j r'0

'/ 4

jI cos( ') 'sin'e a

2r'z dzE

= i

' / 2

' / 2

z L

z LE d E

=

=

= i i

/ 4-j (r-z'cos )0

/ 4

jI cos( ') 'sine a

2rz dzE

i

/ 4-j r j z'cos0

/ 4

jI sine cos( ')e 'a

2rE z dz

i

2L

2L

4

=

4

=


12


j z' -j z'/ 4-j r j z'cos0

/ 4

jI sin e +ee e 'a

2r 2E dz

i

/ 4-j r j z'(1+cos ) -j z'(1-cos )0

/ 4

jI sine [e +e ] 'a

4rE dz

i

-j r02

jI sin 4e cos cos a

4r sin 2E pi

i

( )2 -j r0

cos cosjIe a

2 r sinE

pi

pi

i

Tng t:

( )' / 2 2 -j r0' / 2

cos cosjIe a

2 r sinz L

z LH d H

pi

pi

=

=

= i i



6.4. Cc thng s c trng ca anten

6.4.1. Cng bc x v th bc x6.4.2. li nh hng v nh hng6.4.3. Hiu sut, li v HPBW (Haft Power Beamwidth)

13


6.4.1. Cng bc x v th bc x

nh ngha: cng bc x l CS in t trung bnh trnmt n v gc c (solid angle) theo hng kho st

r

y

x

z

od

dS

Hng kho st2

dSd= =sindd ( )r

Sr2

S 0 0ex: S = d =4 ( )Srpi pi pi

2rr

dSu( , )= =r

dWSr

2

bx 0 0P = u( , )d = u( , )sin d dpi pi


6.4.1. Cng bc x v th bc x

Cng bc x ca dipole Hetzian:2 2 2 2

2 2 2 2r 2 2 2u( , )=r = sin .r sin8 8

m mI I Wr Sr

=

th bc x: biu din th cho hm cng bc xtheo cc hng khc nhau (thng thng dng cng bc x chun): un(,)=u(,)/umax

14


6.4.2. li nh hng v nh hng

Cng bc x ca anten ng hng (isotropic anten):cng bc x ri u theo mi hng sao cho cng sutbc x bng vi cng sut bc x ca anten c hng angxt.

2

i 0 0

1u = u( , )sin d d

4 4bxP pi pi pi pi

=

li nh hng:

2i

0 0

u( , ) 4 u( , ) 4 u( , )D( , )=u u( , )sin d dbxP pi pi pi pi

= =

V d: dipole Hetzian:

2 22

2u( , ) sin8m

I =

2D( , )=1.5sin


6.4.2. li nh hng v nh hng

li nh hng thng c tnh theo decibel (dBi):D( , ) dBi=10log[D( , )]

nh hng: li nh hng cc i

D=max[D( , )] V d: dipole Hetzian:

2D( , )=1.5sin D=1.5

D dBi=10logD

15


SP

lossP

bxP

+-

Zn

RLoss

Rbx

jXantenE

Anten pht

6.4.3. Hiu sut, li v HPBW (Haft Power Beamwidth ) Hiu sut:

in tr tnhao nhit(dy dn)

in trbc x

(min xa)in

khng(min gn)

bx bx bx

S bx loss bx loss

P P R [%]

P P P R R= = =

+ +


6.4.3. Hiu sut, li v HPBW (Haft Power Beamwidth ) li ca anten:

S bx

4 u( , ) 4 u( , )G( , ) D( , )P P

pi pi = = =

li chun ha:G(, )( , ) (, )

Max[G(, )] ng u = =

HPBW

rng na cng sut (HPBW or 3-dB):

Dipole Hetzian;

=0 / 4pi 3 / 4pi

HPBW=/2

16


6.4.3. Hiu sut, li v HPBW (Haft Power Beamwidth )

Example: A TV station is transmitting 10kW of power with a gain of 15dB towards a particular direction. Determine the peak and rms value of the electric field at a distance of 5km from the station?

15/10G dB=15dB 10 31.62G = =

bxi SP =G.P 31.62 10 316.2kW= =

i = / 4bxiu P pi 2 2 2i1/r = / 4

2r i bxi mP u P r Epi

< > = =

3

31 1 120 316.2 10 0.87 /

2 5 10 2bxi

m

PE V mr

pipi pi

= = =

/ 2 0.62 /rms mE E V m = =

:solution

em-ch6-lecture-01-s1-12-13

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