# Antennas and Propagation.L04 by YAdocs

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```									  Radiation Integrals and Auxiliary Potential
Functions
Ranga Rodrigo
June 23, 2010

Lecture notes are fully based on Balanis [?]. Some diagrams and text are directly
from the books.

Contents
1 The Vector Potential for an Electric Current Source J                            4

2 Solution of the Inhomogeneous Vector Potential Equation                          6

Introduction

• In the analysis of radiation problems, the usual procedure is to specify
the sources and then require the ﬁelds radiated by the sources. This is in
contrast to the synthesis problem where the radiated ﬁelds are speciﬁed,
and we are required to determine the sources.

• It is a very common practice in the analysis procedure to introduce auxil-
iary functions, known as vector potentials, which will aid in the solution
of the problems.

• The most common vector potential functions are the A (magnetic vector
potential) and F (electric vector potential).

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Integration path 1
Sources J , M                                             Radiated ﬁelds E , H

Integration                                             Differentiation
path 2                                                  path 2

Vector potentials A, F

The Divergence
The basic deﬁnition of divergence is

S   A · dS
∇ · A = lim                             .
∆v→0              ∆v
The expansion of divergence in Cartesian coordinates is
∂A x ∂A y ∂A z
∇· A =       +    +     .
∂x   ∂y   ∂z

The Curl
The basic deﬁnition of curl is
C   A · dl
∇ × A = lim                                    i n.
∆S→0            ∆S        max

The expansion of curl in Cartesian coordinates is
ix       iy       iz
∂        ∂       ∂
∇× A =      ∂x       ∂y       ∂z   .
Ax       Ay       Az

Vector Identities

div curl = 0                                 ∇·∇× A = 0
curl grad = 0                                     ∇ × ∇φ = 0

2
Maxwell’s Equations in Integral Form

d
E · dl = −              B · d S.                                       (Faraday’s law)
C                  dt   S
d
H · dl =        J · dS +               D · d S.               (Ampère’s circuital law)
C                S                dt    S

D · dS =        ρd v.                                (Gauss’ law for the electric ﬁeld)
S               V

B · d S = 0.                                      (Gauss’ law for the magnetic ﬁeld)
S

Auxiliary equation

d
J · dS = −               ρd v. (Law of conservation of charges)
S                  dt   V

Other relations

D = εE ,
B
H =   ,
µ

Maxwell’s Equations in Differential Form

∂B
∇×E = −        .
∂t
∂D
∇×H = J +        .
∂t
∇ · D = ρ.
∇ · B = 0.

Auxiliary equation
∂ρ
∇· J = −        .
∂t

Maxwell’s Equations in Free Space

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∂B
∇×E = −       .
∂t
∂D
∇×H =        .
∂t
∇ · D = 0.
∇ · B = 0.

1 The Vector Potential for an Electric Current Source
J

• The vector potential A is useful in solving for the EM ﬁeld generated by a
given harmonic electric current J .

• The magnetic ﬂux B is always solenoidal; that is, ∇ · B = 0.

Therefore, it can be represented as the curl of another vector because it obeys
the vector identity
∇ · ∇ × A = 0.
where A is an arbitrary vector. Thus we deﬁne

B A = µH A = ∇ × A.                           (1)

We can substitute this into Maxwell’s curl equation ∇ × E A = − j ωµH A ,

∇ × E A = − j ω∇ × A.

This can also be written as

∇ × E A + j ω × A = 0.

From the vector identity ∇ × (−∇φe ) = 0, it follows that

E A + j ωA = −∇φe ,

or
E A = −∇φe − j ωA.

4
The scalar function φe represents an arbitrary electric scalar potential which is
a function of position. Taking the curl of both sides of Eq. 1 and using the vector
identity ∇ × ∇ × A = ∇(∇ · A) − ∇2 A, reduces it to

∇ × (µH A ) = ∇(∇ · A) − ∇2 A

For homogeneous medium,

µ∇ × H A = ∇(∇ · A) − ∇2 A

Using Maxwell’s equation, ∇ × H A = J + j ω E A ,

µJ + j ωµ E A = ∇(∇ · A) − ∇2 A

Substituting 11,
∇2 A + k 2 A = −µJ + ∇(∇ · A + j ωµ φe )
where k 2 = ω2 µ . In 1 the curl of A was deﬁned. Now we are at liberty to deﬁne
the divergence of A, which is independent of its curl. Let

∇ · A = − j ω µφe ,

which is known as the Lorentz condition. Substituting,

∇2 A + k 2 A = −µJ .

1
E A = −∇φe − j ωA = − j ωA − j         ∇(∇ · A).             (2)
ωµ

Once A is known, H A can be found from Eq. 1 and E A from Eq. 2. E A can just
as easily be found from Maxwell’s equation, ∇ × H A = J + j ω E A , with J = 0. It
will be shown later how to ﬁnd A in terms of the current density J . It will be a
solution to the inhomogeneous Helmholtz equation of Eq. 11.

• Although magnetic currents appear to be physically unrealizable, equiv-
alent magnetic currents arise when we use the volume or the surface
equivalence theorems.

• The ﬁelds generated by a harmonic magnetic current in a homogeneous
region, with J = 0 but M = 0, must satisfy ∇ · D = 0.

5
• Therefore, E F can be expressed as the curl of the vector potential F by
1
EF = − ∇×F .                             (3)

• By letting
∇ · F = − j ωµ φm                            (4)
we can obtain
∇2 F + k 2 F = − M.

Once F is known, E F can be found from Eq. 3 and H F with M = 0. It will be
shown later how to ﬁnd F once M is known. It will be a solution to the inhomo-
geneous Helmholtz equation of 4.

Total Fields
Now, we have developed equations that can be used to ﬁnd the electric and
magnetic ﬁelds generated by an electric current source J and a magnetic cur-
rent source M. The procedure requires that the auxiliary potential functions
A and F generated, respectively, by J and M are found ﬁrst. In turn, the cor-
responding electric and magnetic ﬁelds are then determined (E A ,H A due to A
and E F ,H F due to F ). The total ﬁelds are then obtained by the superposition of
the individual ﬁelds due to A and F (J and M).

1            1            1          1
E = E A +EF     = − j ωA − j    ∇(∇ · A) − ∇ × F =       ∇×HA − ∇×F.
ωµ                        jω
1                  1             1          1
H = H A + HF        = ∇ × A − j ωF − j     ∇(∇ · F ) = ∇ × A −      ∇×EF .
µ                 ωµ             µ        j ωµ

2 Solution of the Inhomogeneous Vector Potential
Equation
The solution of the inhomogeneous vector wave equation is
µ              e − j kR
A=               J            dv .
4π     V           R
To derive it, let us assume that a source with current density J z , which in the
limit is an inﬁnitesimal source, is placed at the origin of a x, y, z coordinate
system.

6
z

(x, y, z)
θ

(x , y , z )

φ
y

x

Since the current density is directed along the z-axis (J z ), only an A z com-
ponent will exist. Thus we can write ∇2 A + k 2 A = −µJ as

∇2 A z + k 2 A z = −µJ z .

At points outside of the inﬁnitesimal source,

∇2 A z + k 2 A z = 0.                          (5)

Point source: A z is not a function of direction (θ and φ) and A z = A z (r ), where
1 ∂      ∂A z (r )
∇2 A z (r ) + k 2 A z (r ) =        2 ∂r
r2           + k 2 A z (r ) = 0.
r           ∂r

d 2 A z (r ) 2 d A z (r )
+             + k 2 A z (r ) = 0.
dr 2       r dr
This differential equation has two independent solutions

e − j kr
A z1 = C 1          :       radially outward traveling wave.
r

e + j kr
A z2 = C 2            :     radially inward traveling wave.
r

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Therefore, we choose
e − j kr
A z = A z1 = C 1      .
r
In the static case (ω = 0, k = 0), this simpliﬁes to

C1
Az =        .
r
which is the solution to the wave equation 6 when k = 0. Thus at points outside
the source, the time-varying and static solutions differ only by e − j kr , the phase
retardation factor.
In the presence of the source (J z = 0), and k = 0, the wave equation reduces
to
∇2 A z = −µJ z .                                 (6)
This equation is recognized to be Poisson’s equation, the most familiar form
being that relates the scalar electric potential φ to the electric charge density ρ.
ρ
∇2 φ = − .

The solution is
1                   ρ
φ=                          dv ,
4π              V    r
where r is the distance from any point in the charge density to the observation
point. Using a similar form

µ               Jz
Az =                        dv .
4π           V   r

This represents the static solution. The time-varying solution is

µ                    e − j kr
Az =                 Jz               dv .
4π        V              r

If we consider the current densities in x- and y-directions (J x , and J y ),

µ                e − j kr
A=                 J            dv .
4π        V          r

If the source is removed from the origin and placed at a position represented by
(x , y , z ),
µ                   e − j kR
A(x, y, z) =       J (x , y , z )          dv .
4π  V                   R

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where (x, y, z) is the observation point and R is the distance from any point on
the source to the observation point. In a similar fashion we can obtain

e − j kR
F (x, y, z) =                M(x , y , z )               dv .
4π     V                          R
If J and M represent linear densities,

µ                            e − j kR
A(x, y, z) =              J s (x , y , z )            ds .
4π     S                         R

e − j kR
F (x, y, z) =            M s (x , y , z )          ds .
4π     S                     R
For electric and magnetic current I e and I m

µ                           e − j kR
A(x, y, z) =              I e (x , y , z )            dl .
4π   C                          R

e − j kR
F (x, y, z) =              I m (x , y , z )             dl .
4π     C                          R

The far-ﬁelds radiated by antennas of ﬁnite dimensions are spherical waves.
For these radiators, a general solution to the vector wave equation is spherical
components is

A = a r A r (r, θ, φ) + a θ A θ (r, θ, φ) + a φ A φ (r, θ, φ).
ˆ                   ˆ                   ˆ

The amplitude variations of r in each component are of the form 1/r n . Neglect-
ing the higher order terms, for large r ,

e − j kr
A    a r A r (r, θ, φ) + a θ A θ (r, θ, φ) + a φ A φ (r, θ, φ)
ˆ                   ˆ                   ˆ                                  .
r
Consider
1
E A = −∇φe − j ωA = − j ωA − j                 ∇(∇ · A).
ωµ
If we evaluate the second term in spherical coordinates, we will notice that
higher powers of 1/r are produced. Neglecting them,

EA     − j ωA       for θ and φ components.

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Neglecting the higher order terms of 1/r n , the radiated E - and H -ﬁelds have
only θ and φ components. They can be expressed as

Er    0,
Eθ    − j ωA θ ,
Eφ    − j ωA φ .

Hr    0,
ω       Eθ
Hθ    +j   Aφ = − ,
η        η
ω       Eφ
Hφ    − j Aθ = + .
η       η
Radial far-ﬁeld terms exist only for higher order terms of 1/r n .
In a similar manner, the far-zone ﬁelds due to magnetic source M (potential
F )can be written as

Hr    0,
Hθ    − j ωF θ ,
Hφ    − j ωF φ .

Er    0,
Eθ    − j ωηF φ = ηHφ ,
Eφ    + j ωηF θ = −ηHθ
The corresponding far-zone E - and H -ﬁeld components are orthogonal to each
other and form TEM (to r ) mode ﬁelds.

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