Infinitesimal Dipole

Outline

• Maxwell’s equations– Wave equations for A and for

• Power: Poynting Vector• Dipole antenna

Maxwell Equations

Jt

DH

vD

0 B

t

BE

Ampere:

Faraday:

Gauss:

Constitution Relation

EJt

J

HB

ED

Vector Magnetic PotentialA

HAB • Applying in Faraday’s Law:

0

t

AE

0

: vectorial

Identidad

t

AE

At

At

E

is the Electric Scalar Potential

Ampere’s Law:J

t

EB

Jt

DH

2

221

t

A

tJAA

t

A

tJAA

21

AAA 2

Jt

AA

tA

2

22

Lorentz’ condition

tA

• Assuming

• The wave equation

Jt

AA

tA

2

22

Jt

AA

2

22

Gauss’ Law

vD

vE

v

t

A

2

v

t

2

22

v

t

A

v

tt

2

Wave equation

• For sinusoidal fields (harmonics):

whereJAkA

JAjA

Jt

AA

22

22

2

22

)(

22 k

Outline

• Maxwell’s equations– Wave equations for A and for

• Power: Poynting Vector• Dipole antenna

Poynting Vector *

2

1HES

*Re2

1HESave

yx EyExE ˆˆ

yx HyHxH ˆˆ

)(2

)(1

ztjy

ztjx

eEE

eEE

)(2

)(1

ztj

y

ztjx

eHH

eHH

*ˆˆˆˆRe2

1yxyxave HyHxEyExS

Average Poynting Vector S yxyxave HyHxEyExS ** ˆˆˆˆRe

2

1

zHEHES xyyxave ˆRe2

1 **

zeHEHES zave ˆcos

2

1 2*12

*21

zeZ

E

Z

ES z

ooave ˆcos

2

1 2

2

2

2

1

zEER

So

ave ˆ2

1 2

2

2

1

In free space:

Outline

• Maxwell’s equations– Wave equations for A and for

• Power: Poynting Vector• Infinitesimal Dipole antenna

Find A from Dipole with current J

• Line charge w/uniform charge density, L

Jt

AA

2

22

x

z

r

r

0

Jz

)()( yxIoo

Az zyxJAkA zzz ,,22

Assume the simplest solution Az(r):

To find….

a

sin

1aa

AA

r

AA r

F

r

F

rr

Fr

rF r

sin

1sin

sin

11 2

2

zz rAdr

d

rrA

2

22 1

)(

AA 2

Assume the simplest solution Az(r):

Homogenous Equation (J=0)

01 2

2

2

zz AkrAdr

d

r

022

2

zz rAkrAdr

d

jkrjkrz ececrA 21

Which has general solution of:

Apply B.C.• If radiated wave travels outwards from the source:

• To find C2, let’s examine what happens near the source. (in that case k tends to 0)

• So the wave equation reduces to

r

ecrA

jkr

z

2)(

r

crAz

2)(

jkrz ecrrA 2)(

)()(2 yxIAA oozz

Now we integrate the volume around the dipole:

• And using the Divergence Theorem

zIr

Arddr

r

Aoo

zz

22 4sin

zIdSA ooz

ddrdS sin2

dxdydzyxIdvA ooz )()(

Comparing both, we get:

42

zIc oo

24 r

zI

dr

dA ooz

jkrooz e

r

zIrA

4)(

22

r

c

dr

dAz

Now from A we can find E & H

• Using the Victoria IDENTITY:

• And

• Substitute:

zrzzz aarAr

aAH ˆˆ)(1

ˆ1

zzzzzz aAaAaA ˆˆˆ

zzaAAH ˆ11

)( GfGffG

sinˆˆˆ aaa zr

ˆsinˆ1

Har

AH z

ˆsinˆ1

Har

AH z

sin

1

4 2jkro e

rr

jklIH

jkrooz e

r

zIrA

4)(

The magnetic filed intensity from the dipole is:

Now the E field:

t

EH

Jt

DH

rr ArA

rrrA

r

A

r

AA

rrA

1ˆsin

1ˆsin

sin

1ˆ

ˆˆ

ˆsin

sin

1ˆ

ErEj

rHrr

HrrH

r

HEj

The electric field from infinitesimal dipole:

cossin21

4sin

1

sinsin

1

2jkro

r

err

jklI

r

Hr

Ej

cos1/

2 32jkro

r erjr

lIE

sin1

4

1

22 jkro e

rr

jkk

r

ljI

rHrr

Ej

sin

1/

4 32jkro e

rjrr

jlIE

General @Far field r> 2D2/

sin

1

4 2jkro e

rr

jklIH

cos1/

2 32jkro

r erjr

lIE

sin

1/

4 32jkro e

rjrr

jlIE

0rE

sin

4jkro e

r

jlIE

sin

4jkro e

r

jklIH

Note that the ratio of E/H is the intrinsic impedance of the medium.

Power :Hertzian Dipole

ˆˆ

2

12

1

**

*

HErHE

HES

r

ddrHErdASAdSP sin2

1ˆ 2*

2

0 0

2322

2

211/

sin32 rr

jk

rjrr

jlIS or

3

222

21

1sin8 krr

lIS or

sin1

4sin

1/

42

12

*

32jkrojkro

r err

jklIe

rjrr

jlIS

dr

krr

lIdP o sin

11sin

82

32

22

2

3

2

13 kr

jlIP o

Radiation Resistance

reactivarado PP

kr

jlIP

3

2

13

radoo

rad RIlI

P2

2

2

1

3

3

2

13

2

kr

jlZ

2

280

lRrad