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```									NUS/ECE                                                           EE5308

Aperture Antennas
1 Introduction
Very often, we have antennas in aperture forms, for
example, the antennas shown below:

Pyramidal horn antenna       Conical horn antenna

Hon Tat Hui
1                 Aperture Antennas
NUS/ECE                                             EE5308

Paraboloidal antenna   Slot antenna

Hon Tat Hui
2         Aperture Antennas
NUS/ECE                                                        EE5308

2 Analysis Method for Aperture Antennas
2.1 Uniqueness Theorem
An electromagnetic field (E, H) in a lossy region is uniquely
specified by the sources (J, M) within the region plus (i) the
tangential component of the electric field over the boundary,
or (ii) the tangential component of the magnetic field over
the boundary, or (iii) the former over part of the boundary
and the latter over the rest of boundary. The case for a
lossless region is considered to be the limiting case as the
losses go to zero. Here M is the magnetic current density
assumed that it exists or its existence is derived from M = E
× n, where E is the electric field on a surface and n is the
normal vector on that surface. (For a proof, see ref. [3])
Hon Tat Hui
3                       Aperture Antennas
NUS/ECE                                                       EE5308

2.2 Equivalence Principle

Actual problem        Equivalent problem

Hon Tat Hui
4                        Aperture Antennas
NUS/ECE                                                             EE5308

For the actual problem on the left-hand side, if we are
interested only to find the fields (E1, H1) outside S (i.e.,
V2), we can replace region V1 with a perfect conductor as
on the right-hand side so that the fields inside it are zero.
We also need to place a magnetic current density
Ms=E1×n on the surface of the perfect conductor in order
to satisfy the boundary condition on S. Now for both the
actual problem and the equivalent problem, there are no
sources inside V2. In the actual problem, the tangential
component of the electric field at the outside of S is E1×n.
In the equivalent problem, the tangential component of the
electric field at the outside of S is also E1×n as a magnetic
current density Ms=E1×n has been specified on S already.
Hon Tat Hui
5                        Aperture Antennas
NUS/ECE                                                            EE5308

Hence by using the uniqueness theorem, the fields (E1,
H1) in V2 in the equivalent problem will be the same as
those in the actual problem.
Note that the requirement for zero fields inside V1 is to
satisfy the boundary condition specified on the tangential
component of the electric field across S. Because now in
the equivalent problem just outside S, the electric field is
E1 while there is also a magnetic current density Ms. But
just inside S, the electric field is zero. Hence on S,
( E1 − 0 ) × n = E1 × n = M s
This is exactly the boundary condition specified on the
tangential component of the electric field across S with an
Hon Tat Hui
6                        Aperture Antennas
NUS/ECE                                                          EE5308

The advantage of the equivalent problem is that we can
calculate (E1, H1) in V2 by knowing Ms on the surface of a
perfect conductor.
A modified case with practical interest is shown below.

Hon Tat Hui
7                       Aperture Antennas
NUS/ECE                                                                                      EE5308

Only equivalent                Twice the
magnetic current is            equivalent
Ground plane          Aperture fields                  required                       magnetic
n
in free space

Ea, Ha                           Ms = Ea × n                       2Ms
V1                                    V1
V2                                 V2                    V1
V2
Aperture

ε1,μ1             ε2,μ2             ε1,μ1         ε2,μ2                  ε2,μ2      ε2,μ2

(a)                              (b)                                  (c)
Equivalent problem
Aperture in a ground plane            Equivalent problem
after using image
theorem

Hon Tat Hui
8                                    Aperture Antennas
NUS/ECE                                                             EE5308

Thus the problem of an aperture in a perfectly conducting
ground plane is equivalent to the finding of (i) the fields in
V2 due to an equivalent magnetic current density of Ms
radiating in a half-space bounded by the ground plane, or
(ii) the fields in V2 due to an equivalent magnetic current
density of 2Ms radiating in a free space having the
properties of V2. Note that for the equivalent problem in
(c), the field so calculated in V1 may not be equal to the
original fields in V1 in actual problem in (a).

To find the electromagnetic field due to a magnetic current
density Ms, we need to construct an equation with the
source Ms and solve it.
Hon Tat Hui
9                         Aperture Antennas
NUS/ECE                                                                  EE5308

2.3 Radiation of a Magnetic Current Density
Maxwell’s equations with a magnetic current source

with M s                    with J
∇ × H = jωε E               ∇ × E = − jωμ H
∇ × E = −M s − jωμ H        ∇ × H = J + jωε E
∇ ⋅ B = ρm                  ∇⋅D = ρ
∇⋅D = 0                     ∇⋅B = 0

When there is only a magnetic current source Ms, an
electric vector potential F can be defined similar to the
magnetic vector potential A.
Hon Tat Hui
10                       Aperture Antennas
NUS/ECE                                                           EE5308

with M s                               with J
1                            1
E = − ∇×F                             H=       ∇× A
ε                            μ
∇ 2F + k 2F = −ε M s                  ∇2A + k 2A = −μJ
ε              e − jkR         μ             e − jkR
F(R ) =    ∫∫∫ Ms ( R' ) R dv ' A(R) = 4π ∫∫∫ J ( R' ) R dv '
4π v '                             v'

Hence if there is only a magnetic current source, E can
be calculated from the electric vector potential F, whose
solution is given about. In general, when there are both
magnetic and electric current sources, E and H can be
calculated by the superposition principle.
Hon Tat Hui
11                      Aperture Antennas
NUS/ECE                                                                           EE5308

Far-Field Approximations
When the far-field of aperture radiation is interested, the
following approximations can be used to simplify the
factor e − jkR R (see ref. [1]):
R ≈ r − r ′ cosψ ,                 for numerator
R ≈ r,                              for denominator
z

( r ′,θ ′, φ ′ )    R            (r,θ,φ)
R' ψ                      R
y

x
Hon Tat Hui
12                            Aperture Antennas
NUS/ECE                                                                                 EE5308

Then,
ε e − jkr
M s ( R' ) e jkr′ cosψ dv '
4π r ∫∫∫
F(R ) =
v'

ε e − jkr
=           L
4π r
where

L = ∫∫∫ M s ( R' ) e jkr′ cosψ dv '
v'

= ∫∫∫ ⎡a x M x ( R' ) + a y M y ( R' ) + a z M z ( R' )⎤ e jkr′ cosψ dv '
⎣ˆ                ˆ                ˆ             ⎦
v'

Hon Tat Hui
13                               Aperture Antennas
NUS/ECE                                                                       EE5308

In spherical coordinates (see ref. [1]),

L = a r Lr + aθ Lθ + aφ Lφ
ˆ        ˆ       ˆ
where

Lθ = ∫∫∫ ( M x cos θ cos φ + M y cos θ sin φ − M z sin θ ) e    jkr′ cosψ
dv '
v'

Lφ = ∫∫∫ ( − M x sin φ + M y cos φ ) e   jkr ′ cosψ
dv '
v'

Lr = ∫∫∫ ( M x sin θ cos φ + M y sin θ sin φ − M z cosθ ) e jkr′ cosψ dv '
v'

Hon Tat Hui
14                            Aperture Antennas
NUS/ECE                                                                   EE5308

1                       1
E = − ∇ × F,          H=−          ∇×E
ε                      jωμ

For far fields (see ref. [3], Chapter 3),

Eθ = − jωη Fφ
Eφ = + jωη Fθ             ε e− jkr
Fθ =          Lθ
Eφ                4π r
Hθ = −                    ε e− jkr
η           Fφ =          Lφ
Eθ                4π r         Note: there is no
Hφ = +                                  need to know Fr.
η                              Hence there is no
need to find Lr.
Hon Tat Hui
15                            Aperture Antennas
NUS/ECE                                                      EE5308

Example 1
Find the far-field produced by a rectangular aperture opened
on an infinitely large ground plane with the following
aperture field distribution:
⎧− a 2 ≤ x′ ≤ a 2
Ea = a y E0 ⎨
ˆ
⎩ − b 2 ≤ y′ ≤ b 2

Hon Tat Hui
16                     Aperture Antennas
NUS/ECE                                                                         EE5308

Solutions
The equivalent magnetic current density is:
⎧− a 2 ≤ x′ ≤ a 2
M s = E a × a z = a y × a z E0 = a x E0 ⎨
ˆ     ˆ ˆ            ˆ
⎩ − b 2 ≤ y′ ≤ b 2
M x = E0 , M y = 0, M z = 0

Actually, there        Lθ = ∫∫ M x cosθ cos φ e     jkr′ cosψ
ds′
is no need to                 s′
find Lr.
Lφ = − ∫∫ M x sin φ e jkr′ cosψ ds′
s′

Lr = ∫∫ M x sin θ cos φ e   jkr′ cosψ
ds′
s′
Hon Tat Hui
17                            Aperture Antennas
NUS/ECE                                                                              EE5308

r ′ cosψ = r ′ ⋅ a r
ˆ
= ( a x x ′ + a y y ′) ⋅ ( a x sin θ cos φ + a y sin θ sin φ + a z cosθ )
ˆ         ˆ            ˆ                 ˆ                 ˆ
= x ′ sin θ cos φ + y ′ sin θ sin φ
Hon Tat Hui
18                              Aperture Antennas
NUS/ECE                                                                                              EE5308

After using the image theorem to remove the ground plane,
we have:            b2 a2

∫ ∫         2 M x e jk ( x sin θ cos φ + y sin θ sin φ ) dx′dy′
′             ′
Lθ = cosθ cos φ
−b 2 − a 2
From image theorem
⎡            ⎛ sin X ⎞⎛ sin Y ⎞ ⎤
= 2abE0 ⎢ cosθ cos φ ⎜       ⎟⎜       ⎟⎥
⎣            ⎝ X ⎠⎝ Y ⎠ ⎦
ka                                   kb
X = sin θ cos φ ,                    Y = sin θ sin φ
2                                    2
Similarly,
⎡      ⎛ sin X ⎞⎛ sin Y ⎞ ⎤
Lφ = −2abE0 ⎢sin φ ⎜       ⎟⎜       ⎟⎥
⎣      ⎝ X ⎠⎝ Y ⎠ ⎦
Hon Tat Hui
19                                        Aperture Antennas
NUS/ECE                                                                     EE5308

Therefore,
ε e− jkr      ε e− jkr      ⎡            ⎛ sin X ⎞⎛ sin Y ⎞ ⎤
Fθ =          Lθ =          abE0 ⎢cos θ cos φ ⎜       ⎟⎜       ⎟⎥
4π r          2π r         ⎣            ⎝ X ⎠⎝ Y ⎠ ⎦
ε e− jkr        ε e− jkr      ⎡      ⎛ sin X ⎞⎛ sin Y ⎞ ⎤
Fφ =          Lφ = −          abE0 ⎢sin φ ⎜       ⎟⎜       ⎟⎥
4π r            2π r         ⎣      ⎝ X ⎠⎝ Y ⎠ ⎦
The E and H far-fields can be found to be:
Er = 0
abkE0 e− jkr   ⎡       ⎛ sin X ⎞⎛ sin Y ⎞ ⎤
Eθ = − jωη Fφ = j                ⎢ sin φ ⎜        ⎟⎜      ⎟⎥
2π r        ⎣       ⎝ X ⎠⎝ Y ⎠ ⎦
abkE0 e− jkr   ⎡              ⎛ sin X ⎞⎛ sin Y ⎞ ⎤
Eφ = + jωη Fθ = j                ⎢ cos θ cos φ ⎜        ⎟⎜       ⎟
2π r        ⎣              ⎝   X ⎠⎝ Y ⎠ ⎥     ⎦
Hon Tat Hui
20                           Aperture Antennas
NUS/ECE                                                     EE5308

Hr = 0
Eφ
Hθ = −
η
Eθ
Hφ =
η

The radiation patterns are plotted on next page.

Hon Tat Hui
21                   Aperture Antennas
NUS/ECE                                                                            EE5308

Three-dimensional field pattern of a constant field rectangular aperture opened on
an infinite ground plane (a=3λ, b=2 λ)
Hon Tat Hui
22                               Aperture Antennas
NUS/ECE                                                                           EE5308

E-plane
H-plane

Two-dimensional field patterns of a constant field rectangular aperture opened on
an infinite ground plane (a=3λ, b=2 λ)
Hon Tat Hui
23                              Aperture Antennas
NUS/ECE                                                                 EE5308

3. Parabolic Reflector Antennas
Parabolic reflector antennas are frequently used in radar
systems. They are very high gain antennas. There are
two types of parabolic reflector antennas:
1. Parabolic right cylindrical reflector antenna
This antenna is usually fed by a line source such as a
dipole antenna and converts a cylindrical wave from
the source into a plane wave at the aperture.
2. Paraboloidal reflector antenna
This antenna is usually fed by a point source such as
a horn antenna and converts a spherical wave from
the feeding source into a plane wave at the aperture.
Hon Tat Hui
24                       Aperture Antennas
NUS/ECE                                               EE5308

Parabolic reflector antennas

Hon Tat Hui
25                 Aperture Antennas
NUS/ECE                                                                  EE5308

A typical paraboloidal antenna for satellite communication
Hon Tat Hui
26                        Aperture Antennas
NUS/ECE                                                       EE5308

3.1 Front-Fed Paraboloidal Reflector Antenna

Hon Tat Hui
27                   Aperture Antennas
NUS/ECE                                                                  EE5308

Important geometric parameters and description of a
paraboloidal reflector antenna:

θ0 = Half subtended angle
d = Aperture diameter
f = focal length
Defining equation for the paraboloidal surface:
OP + PQ = constant = 2f
Physical area of the aperture Ap:
2
⎛d ⎞
Ap = π ⎜ ⎟
⎝2⎠
Hon Tat Hui
28                       Aperture Antennas
NUS/ECE                                                                          EE5308

The half subtended angle θ0 can be calculated by the
following formula:
⎡ 1⎛ f ⎞ ⎤
⎢ 2⎜ d ⎟ ⎥
θ 0 = tan −1 ⎢ ⎝2 ⎠ ⎥
⎢⎛ f ⎞ − 1 ⎥
⎢ ⎜ d ⎟ 16 ⎥
⎣⎝ ⎠       ⎦
Aperture Efficiency εap
Aem maximum effective area
ε ap   =    =
Ap      physical area
θ0                           2
⎛ θ0 ⎞              ⎛ θ ' ⎞ dθ '
= cot ⎜ ⎟ ∫ G f (θ ') tan ⎜ ⎟
2

⎝2⎠0                ⎝2⎠
Hon Tat Hui
29                          Aperture Antennas
NUS/ECE                                                                      EE5308

Directivity:
⎛πd ⎞ ε
4π
2

D0 = maximum directivity = 2 Aem = ⎜   ⎟ ap
λ        ⎝ λ ⎠
Feed Pattern Gf(θ’)
The feed pattern is the radiation pattern produced by
the feeding horn and is given by:
⎧ 2(n + 1)cos n (θ '),    0 ≤θ '≤π / 2
G f (θ ') = ⎨
⎩0,                       π / 2 ≤θ '≤π
where n is a number chosen to match the directivity of
the feed horn.
Hon Tat Hui
30                        Aperture Antennas
NUS/ECE                                                                     EE5308

The above formula for feed pattern represents the
major part of the main lobe of many practical feeding
horns. Note that this feed pattern is axially symmetric
about the z axis, independent of φ’.
With this feed pattern formula, the aperture efficiency
can be evaluated to be:
⎧ 2 ⎛ θ0 ⎞    ⎡ ⎛ θ0 ⎞ ⎤ ⎫    2 ⎛ θ0 ⎞
2

ε ap ( n = 2 ) = 24 ⎨sin ⎜ ⎟ + ln ⎢cos ⎜ ⎟ ⎥ ⎬ cot ⎜ ⎟
⎩    ⎝2⎠      ⎣ ⎝ 2 ⎠⎦ ⎭        ⎝2⎠
⎧ 4 ⎛ θ0 ⎞    ⎡ ⎛ θ0 ⎞⎤ ⎫     2 ⎛ θ0 ⎞
2

ε ap ( n = 4 ) = 40 ⎨sin ⎜ ⎟ + ln ⎢cos ⎜ ⎟ ⎥ ⎬ cot ⎜ ⎟
⎩    ⎝2⎠      ⎣ ⎝ 2 ⎠⎦ ⎭        ⎝2⎠

Hon Tat Hui
31                         Aperture Antennas
NUS/ECE                                                                          EE5308

⎧          θ 0 ⎞ ⎤ [1 − cos(θ 0 )]3
ε ap ( n = 6 ) = 14 ⎨2ln ⎡cos ⎛ ⎟ ⎥ +
⎢ ⎜ 2 ⎠⎦
⎩    ⎣ ⎝                  3

} 2 ⎛ θ0 ⎞
2
1 2
+ sin (θ 0 ) cot ⎜ ⎟
2                ⎝2⎠

⎧1 − cos 4 (θ 0 )       ⎡ ⎛ θ0 ⎞⎤
ε ap ( n = 8 ) = 18 ⎨                 − 2ln ⎢cos ⎜ ⎟ ⎥
⎩      4                ⎣ ⎝ 2 ⎠⎦
[1 − cos(θ 0 )] 1 2
3
⎫ 2 ⎛ θ0 ⎞
−                − sin (θ 0 ) ⎬ cot ⎜ ⎟
3        2           ⎭     ⎝2⎠

Hon Tat Hui
32                            Aperture Antennas
NUS/ECE                             EE5308

(εap)

Hon Tat Hui
33   Aperture Antennas
NUS/ECE                                                               EE5308

Effective Aperture (Area) Aem
With the aperture efficiency, the maximum effective
aperture can be calculated as:
Aem = Apε ap
2
⎛d ⎞
Ap = physical area = π ⎜ ⎟
⎝2⎠
The radiation pattern of a paraboloidal reflector
antenna is highly directional with a narrow half-power
beamwidth. An example of a typical radiation pattern
is shown below.
Hon Tat Hui
34                    Aperture Antennas
NUS/ECE                                                                                 EE5308

An example of the radiation pattern of a paraboloidal reflector antenna
with an axially symmetric feed pattern. Note that the half-power
Hon Tat Hui
35                               Aperture Antennas
NUS/ECE                                                       EE5308

Example 2
A 10-m diameter paraboloidal reflector antenna with an f/d
ratio of 0.5, is operating at a frequency = 3 GHz. The
reflector is fed by an antenna whose feed pattern is axially
symmetric and which can be approximated by:
⎧ 6cos 2 (θ '), 0 ≤θ '≤π / 2
G f (θ ') = ⎨
⎩0,             π / 2 ≤θ '≤π
(a) Find the aperture efficiency and the maximum
directivity of the antenna.
(b) If this antenna is used for receiving an electromagnetic
wave with a power density Pavi = 10-5 W/m2, what is the
power PL delivered to a matched load?
Hon Tat Hui
36                      Aperture Antennas
NUS/ECE                                                                  EE5308

Solution
(a) With f/d = 0.5, the half subtended angle θ0 can be
calculated.
⎡ 1⎛ f ⎞ ⎤              ⎡ 1             ⎤
⎢      ⎜ ⎟ ⎥
2 ⎝ d ⎠ ⎥ = tan −1 ⎢ 2
( 0.5 ) ⎥
θ 0 = tan −1 ⎢       2               ⎢               ⎥ = 53.13°
⎢⎛ f ⎞ − 1 ⎥                          1
⎢ ( 0.5 ) − ⎥
2

⎢ ⎜ d ⎟ 16 ⎥
⎣⎝ ⎠         ⎦          ⎣            16 ⎦
From the feed pattern, n = 2. Hence using the aperture
efficiency formula with n = 2, we find:

ε ap ( n = 2 ) = 24{sin ( 26.57° ) + ln [ cos ( 26.57° )]} cot 2 ( 26.57° )
2                                  2

= 0.75 = 75%
Hon Tat Hui
37                           Aperture Antennas
NUS/ECE                                                                 EE5308

⎛πd ⎞ ε
2

D0 = maximum directivity = ⎜   ⎟ ap
⎝ λ ⎠
⎛ π 10 ⎞ 0.75 = 74022.03 = 48.69 dB
2

=⎜      ⎟
⎝ 0.1 ⎠

(b) Frequency = 3 GHz, ⇒ λ = 0.1 m.
D0λ 2
Maximum effective area = Aem =          = 58.9 m 2
4π
PL                  PL
Ae (θ ,φ ) =                  ⇒ Aem =
Pavi Pav (θ ,φ )         Pavi

Hence, PL = Aem Pavi = 58.9 × 10−5 = 5.89 × 10−4 W
Hon Tat Hui
38                          Aperture Antennas
NUS/ECE                                                           EE5308

References:
1. C. A. Balanis, Antenna Theory, Analysis and Design, John Wiley
& Sons, Inc., New Jersey, 2005.
2. W. L. Stutzman and G. A. Thiele, Antenna Theory and Design,
Wiley, New York, 1998.
3. R. F. Harrington, Time-Harmonic Electromagnetic Fields,
McGraw-Hill, New York, 1962, pp. 100-103, 143-263, 365-367.

Hon Tat Hui
39                        Aperture Antennas

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