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Radiation integrals and auxiliarypotential functions

There are several possible ways to analyze radiation from

harmonic current sources. A usual approach introduces

auxiliary vector potentials to simplify the mathematical

treatment of radiation problems.

Contents

1. Introduction: Analysis of radiation problems

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2. Vector potential A ξ

for electric current source

J ξ

3. Vector potential F ξ

for magnetic current source

M ξ

4. Solution of the inhomogeneous vector potential

equation

5. Far-field approximation

6. Example: The infinitesimal dipole

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1. Analysis of radiation problemsThe general problem is to find the fields radiated by a

current source

J ξ

( x ′ , y ′ , z

′ ) Exyz ξ

( , , ) ?

z

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Maxwell’s equationsThe physics of the fields radiated by an antenna are

described by Maxwell’s equations. For harmonic variations

of the fields , we can write

(1)

(2)

(3)

0 (4)

( e j w t )    H

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ξ = j we

E ξ +

J ξ

    E ξ = -

 j wm

H ξ

with

ξ

ì1ï ïï í ï= ïïî 

0

0

in source region

elsewhere

⋅ D ξ = e ⋅ E ξ

=

r ⋅ B ξ = m

⋅ H ξ

=

 J 

E, H: Electric and magnetic field D: Dielectric displacement B: Magnetic flux

(induction) J: Electric source current density r: Charge density

ξ ξ div

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curl ξ

ξ

grad

⋅ (    V )

o 0 The

continuity equation

can be derived from (1) and (3).

    V =

V ⋅ V =

V f =

f 

Note that by using the vector identity ξ

, we

can show that (4) follows from (2).

⋅ J ξ

= -

 jwr (5)

3

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Vector potentialsTo analyze the fields radiated by sources, it is common

practice to introduce auxiliary functions known as vector

potentials, which will aid in the solution of Maxwell’s

equations.

Sources ,J ξ M ξIntegration Path 1

Radiated ,E ξ H ξ

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fields

Path 2 Integration

 ,A F 

Path 2 Vector ξ potentials

ξ

Differentiation

• The two-step procedure usually involves simpler

integrations than the direct path.

• All field functions can be retrieved from the vector

potentials through differentiations.

• Other auxiliary functions may be used (Hertz potentials):

P ξ e ,

P ξ

h .

• The use of vector potentials basically permits to reduce

the number of unknowns.

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Useful identities from ector analysisDuring the following mathematical treatment, we will make

use of the following relations:

div curl = 0 r 

⋅ (    V ξ

) = 0 curl grad = 0 r 

  ( f ) = 0

Reciprocally, it can then be demonstrated that

1) If the divergence of a vector field equals zero, then

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there exists a potential

vector field so that the curl of the potential field equals the

vector field

 V ξ = 0 there exists U ξ so that V ξ = ⋅

! ξ

(*)

2) If the curl of a vector field equals zero, then this vector

field can be written as

the gradient of a potential function

  V ξ = 0 there exists f so that V ξ

=

f (**)

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! ". Vector potentialA

for electric current source

J ξ

Since B =

0 , the magnetic flux can be represented as

the curl of another vector

or

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ξ

" ξ

# ξ

# ξ

 A

=  

 A ξ ( ⋅ B ξ

= ⋅ (    A ξ ) =

0 according to (*)

)

 H ξ

 A

= m 1

  

" ξ ( ξ " ξ

 A

)

The vector A

is called magnetic vector potential. B = Hm ξ 

The definition of A ξ

is put into Faraday’s law (Maxwell equation (2))

    E ξ A = - j wm H ξ A

= - j w  

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 A ξ    ( E ξ A

+ j Aw ξ )

=0

We can now apply the identity (**) from the previous page

to define the electric Since scalar ( E ξ A

+ potential j Aw ξ )

=0 f , e

there exists a scalar function f

e

 A e

(    ( - f e

) =

)0

so that

E ξ

 A + j Aw ξ = -

f e$ ξ

= - f -

 j w A ξ 

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f The scalar function function ofposition.

e

represents an arbitrary electric scalar potential which is a

Relations#ip between A !

and

f 

e

We now consider Ampere’s law (Maxwell equation (1))

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Introducing the defined vector and scalar potentials, we

can write

1( %

)

    H ξ A = j we E ξ A

+

J ξ

m

       A ξ = j we - f - j w

 A ξ       +

J ξ A ξ = ( ⋅ A ξ )

-

2 Aξ ( ⋅ A ξ )

- 2 A ξ = - j wem f e

+ w 2 

em A ξ +

m J ξ The vector potential A ξ

is defined through its curl.

According to the Helmholtz theorem, this leaves us one

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degree of freedom to define its divergence. We can then

use the Lorenz condition

⋅ A ξ= -

 jwemf e

 2 A ξ + w 2 em A ξ = -

m J ξ Inhomogeneous equation for A

ξwave

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A solution for this inhomogeneous wave equation will bederived later

In addition, using the Lorenz condition, once is known, the

fields can be determined

H ξ

 A

= m 1

  

 A ξ E ξ A = - f e

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- j w A ξ = - j w

" ξ & j 1

wme

 ( ⋅

" ξ )

ξ

Thus the electromagnetic field can be calculated from a

single vector potential .

 A = - jwemf A ξ ⋅

H ξ " ,

$ ξ

 A

ξ " = 4 '

 p

VJ ξ

e

-

 R

 jkR

dv

ξ " ξ

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! $. Vector potential % !

for magnetic current source M

Maxwell’s equation for magnetic current source

M: Magnetic source current density due (2)

to magnetic charge density

(3)

Note: Magnetic sources are not physically

(4)

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realizable. However, equivalent magnetic currents arise when we apply the

field equivalence theorem (see section on aperture antennas).

We proceed as before. Since , the dielectric displacement

can be represented as the curl of another vector

or

    H ξ =

 j we

$ ξ

(1)    E ξ = - j w

m

H ξ

-

 M ξ 

r 

m  ⋅ D ξ = e

⋅ E ξ

=

0 ⋅ B ξ

= m ⋅ H ξ

=

r 

m

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⋅ D ξ

= 0 D ξ ! = -  

ξ

ξ

* ξ

( ⋅ D ξ = ⋅ ( -    ! ξ

) =

0 according to (*) )

$ ξ = - 1 e

  

ξ

( ξ ξ

 F 

)The vector F

is called electric vector potential. D = Ee ξ 

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ξ The definition of Fis inserted into Ampere’s law (Maxwell equation (1))

    H ξ ! = j we E ξ !

= - j w  

ξ

    ( H ξ !+ j Fw ξ 

)=

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0

We can now apply the identity (**) to define the magnetic

scalar potentialf 

m

Since ( H ξ F

+ j Fw ξ 

)=0

, there exists a scalar function f

m

 F m

(    ( - f m ) =

)0

so that

H ξ F + j Fw ξ

= -

f m

H ξ = - f -

 j w F ξ 

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wave

10