# Antenna.Lectures.L17_Aperture

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```					LECTURE 17: Aperture Antennas – Part I
(The uniqueness theorem. The equivalence principle. The application of the
equivalence principle to aperture problem. The uniform rectangular aperture.
The tapered rectangular aperture.)

Equation Section 17
1. Introduction
Aperture antennas constitute a large class of antennas, which emit
electromagnetic waves through an opening (or aperture). These antennas have
close analogs in acoustics: the megaphone and the parabolic microphone. The
pupil of the human eye is a typical aperture receiver for optical EM radiation.
At radio and microwave frequencies, horns, waveguide apertures and reflectors
are examples of aperture antennas. Aperture antennas are of common use at
UHF and above. It is because aperture antennas have their gain increase as
∼ f 2 . For an aperture antenna to be efficient and have high directivity, it has to
have an area comparable or larger than λ 2 . Obviously, these antennas would be
impractical at low frequencies. Another positive feature of the aperture
antennas is their near-real valued input impedance and geometry compatibility
with waveguide feeds.
To facilitate the analysis of these antennas, the equivalence principle is
applied. This allows us to carry out the far-field analysis in the outer
(unbounded) region only, which is external to the radiating aperture and the
antenna. This requires the knowledge of the tangential field components at the
aperture, as it follows from the equivalence principle.

2. Uniqueness theorem
A solution is said to be unique if it is the only one possible among a given
class of solutions.

The EM field in a given region V[ S ] is uniquely defined if
- all sources are given;
- either the tangential Eτ components or the tangential Hτ components
are specified at the boundary S.

Nikolova 2004                                                                     1
The uniqueness theorem is proven by making use of the Poynting’s theorem in
integral form:
(                  )
∫∫ E × H ds + jω ∫∫∫ µ | H | −ε | E | dv + ∫∫∫ σ | E | dv =
*                   2        2                2
(                          )
S                                            V[ S ]                                    V[ S ]
(17.1)
(
− ∫∫∫ E ⋅ J + H ⋅ M dv    i*          *             i
)
V[ S ]

Poynting’s theorem states the conservation of energy law in EM systems.
We starts with the supposition that a given EM problem has two solutions
(due to the same sources and the same boundary conditions): E a , H a and                                       (   )
( E , H ) . The difference field is then formed as
b       b

δ E = Ea − Eb ,
(17.2)
δ H = H a − H b.
Since the difference field has no sources, it satisfies the source-free form of
(17.1):
∫∫ (δ E × δ H )ds + jω ∫∫∫ ( µ | δ H | −ε | δ E | ) dv + ∫∫∫ σ | δ E | dv = 0 (17.3)
*                        2          2                    2

S                                                  V[ S ]                                       V[ S ]

Since both fields satisfy the same boundary conditions at S, then δ E = 0 and
δ H = 0 over S. This leaves us with
(
jω ∫∫∫ µ | δ H |2 −ε | δ E |2 dv + ∫∫∫ σ | δ E |2 dv = 0 , (17.4)     )
V[ S ]                                                   V[ S ]

which is true only if
ω ∫∫∫ ( µ | δ H |2 −ε | δ E |2 ) dv = 0,
V[ S ]
(17.5)
∫∫∫ σ | δ E |                     dv = 0.
2

V[ S ]

If we assume some dissipation, however slight, equations (17.5) are satisfied
only if δ E = δ H = 0 everywhere in the volume V[ S ] . This implies the
uniqueness of the solution. If σ = 0 , which is a physical impossibility, but is
often used approximation, multiple solutions (δ E , δ H ) may exist in the form
of self-resonant modes of the structure under consideration. In open problems,
resonance is impossible in the whole region.

Nikolova 2004                                                                                                            2
Notice that the uniqueness theorem holds if either δ E = 0 or δ H = 0 is true
on any part of the boundary.

3. Equivalence principles
The equivalence principle follows from the uniqueness theorem. It allows us
to build simpler to solve problems. As long as the equivalent problem preserves
the boundary conditions of the original problem for the field at S, it is going to
produce the only one possible solution for the region outside V[ S ] .

( Eo , H o )                               ( Eo , H o )

V[ S ]                                          V[ S ]
J se
( Eo , H o )            ⇒                       ( Ee , H e )
S           sources                                 S no sources                       M se
ˆ
n                                              ˆ
n
(a) Original problem                            (b) General equivalent
problem    ⇒

(
J se = n × H o − H e
ˆ                       )                           ( Eo , H o )
(17.6)
(               )
V[ S ]
M se = Eo − Ee × n
ˆ
Js
no fields
S no sources
Js = n × Ho
ˆ                                                                                        Ms
(17.7)
M s = Eo × n
ˆ                                                              ˆ
n
(c) Equivalent problem
with zero fields

The zero-field formulation is often referred to as Love’s equivalence principle.

Nikolova 2004                                                                                              3
We can apply Love’s equivalence principle in three different ways:
(a) We can assume that the boundary S is a perfect conductor. This eliminates
the surface electric currents, i.e., J s = 0 , and leaves just surface magnetic
currents M s , which radiate in the presence of a perfect electric surface.
(b) We can assume that the boundary S is a perfect magnetic conductor. This
eliminates the surface magnetic currents, i.e., M s = 0 , and leaves just
surface electric currents J s , which radiate in the presence of a perfect
magnetic surface.
(c) Make no assumptions about the materials inside S, and define both J s and
M s currents, which are radiating in free space (no fictitious conductors
behind them). It can be shown that these equivalent currents create zero
fields inside V[ S ] (Ewald-Oseen extinction theorem: A. Ishimaru,
Electromagnetic Wave Propagation, Radiation, and Scattering, Prentice
Hall, 1991, p.173)
All three approaches lead to the same field solution according to the uniqueness
theorem. The first two approaches are not very useful in the general case of
curvilinear boundary surface S. However, in the case of flat infinite planes
(walls), the image theory can be used to reduce the problem to an open one.
Image theory can be successfully applied to curved surfaces provided the
curvature’s radius is large compared to the wavelength. Here is how we can
implement Love’s equivalence principles in conjunction with image theory.

( Eo , H o )       ( Eo , H o )        no fields       ( Eo , H o )        no fields       ( Eo , H o )
V[ S ]   S                                   S                                   S
ˆ
n                                   ˆ
n                                   ˆ
n
sources                             no sources                          no sources
⇒                    Ms             ⇒                2M s
Js = 0                              Js = 0

(a) Original problem                   (b) Equivalent problem              (c) Equivalent problem
- electric wall                     - images

Nikolova 2004                                                                                             4
( Eo , H o )       ( Eo , H o )        no fields       ( Eo , H o )        no fields       ( Eo , H o )
V[ S ]   S                                   S                                   S
ˆ
n                                   ˆ
n                                    ˆ
n
sources                             no sources                          no sources
⇒                    Js             ⇒                    2J s
Ms = 0                              Js = 0

(a) Original problem                   (b) Equivalent problem              (c) Equivalent problem
- magnetic wall                     - images

The above approach is used to evaluate fields in half-space as excited by
apertures. The field behind S is assumed known. This is enough to define
equivalent surface currents. Using image theory, the open-region far-zone
solutions for the vector potentials, A (resulting from J s ) and F (resulting from
M s ), are found from
e− jβ r
∫∫ J s (r′)e ds′ ,
j β r ⋅r ′
A( P ) = µ                         ˆ
(17.8)
4π r S
e− jβ r
∫∫ M s (r′)e ds′ .
j β r ⋅r ′
F ( P) = ε                            ˆ
(17.9)
4π r S
ˆ
Here, r denotes the unit vector pointing from the origin of the coordinate
system to the point of observation P. The integration point Q is specified
through the position vector r′ . In the far zone, it is assumed that the field
propagates radially away from the antenna. It is convenient to introduce the so-
called propagation vector or wave vector:
β = βr ,ˆ                           (17.10)
which characterizes both the phase constant and the direction of propagation of
the wave. The vector potentials can then be written as
e− jβ r
A( P) = µ
4π r S  ∫∫ J s (r′)e j β ⋅r′ds′ ,      (17.11)

Nikolova 2004                                                                                             5
e− jβ r
∫∫ M s (r′)e ds′ .
j β ⋅r′
F ( P) = ε                                           (17.12)
4π r S
The relations between the far-zone fields and the vector potentials are:
E A = − jω ( Aθθˆ + Aϕϕ ) ,
far
ˆ                     (17.13)
H far = − jω ( F θˆ + F ϕ ) .
F             θ     ˆϕ                    (17.14)
Since
EFfar = η H Ffar × r ,                         (17.15)
the total far-zone electric field (due to both A and F ) is found as
E far = E A + EFfar = − jω ⎡( Aθ − η Fϕ )θˆ + ( Aϕ + η Fθ ) ϕ ⎤ .
far
⎣
ˆ
⎦     (17.16)
Equation (17.16) involves both vector potentials as arising from both types of
surface currents. Computations are reduced in half if image theory is used in
conjunction with an electric or magnetic wall assumption.

4. Application of the equivalence principle to aperture problems
The equivalence principle is widely used in the analysis of aperture
antennas. To calculate exactly the far field, the exact field distribution at the
aperture is needed. In the case of exact knowledge of the aperture field
distribution, all three approaches given above produce the same results.
However, such exact knowledge of the aperture field distribution is usually
impossible, and certain approximations are used. Then, the three equivalence-
principle approaches produce slightly different results, the consistency being
dependent on how accurate our knowledge about the aperture field is. Usually,
it is assumed that the field is to be determined in half-space, leaving the feed
and the antenna behind a infinite wall S (electric or magnetic). The aperture of
the antenna S A is this portion of S where we have an approximate knowledge of
the field distribution based on the type of the feed line or the incident wave
illuminating the aperture. This is the so-called physical optics approximation,
which certainly is more accurate than the geometrical optics approach of ray
tracing. The larger the aperture (as compared to the wavelength), the more
accurate the approximation based on the incident wave.
Let us assume that the fields at the aperture S A are known: Ea , H a , and they
are zero everywhere else at S . The equivalent current densities are:

Nikolova 2004                                                                         6
Js = n × Ha ,
ˆ
(17.17)
M s = Ea × n.   ˆ
Using (17.17) in (17.11) and (17.12) produces:
e − jβ r
A( P ) = µ          n × ∫∫ H a e jβ ⋅r′ ds′ ,
ˆ                                    (17.18)
4π r         SA

e− jβ r
F ( P) = −ε         n × ∫∫ Ea e jβ ⋅r′ ds′ .
ˆ                                 (17.19)
4π r        SA

The radiation integrals in (17.18) and (17.19) are denoted as
I H = ∫∫ H a e j β ⋅r′ds′ ,                       (17.20)
S

I = ∫∫ Ea e j β ⋅r′ds′ ,
E
(17.21)
S
for brevity.
We can find general vector expression for the far-field E vector making use
of equation (17.16) written as
E far = − jω A − jωη F × r ,ˆ                          (17.22)
where the longitudinal Ar component is to be neglected. Substituting (17.18)
and (17.19) into (17.22) yields
e− jβ r
E = − jβ
far
4π r
r × ∫∫ ⎡ n × Ea − η r × n × H a ⎤ e j β ⋅r ′ ds′ .
ˆ
⎣
ˆ          ˆ ˆ       ( ⎦         )        (17.23)
S       A

This is the full vector form of the radiated field resulting from the aperture
field, and it is referred to as the vector diffraction integral (or vector Kirchhoff
integral).
We now consider a practical case of a flat aperture lying in the x − y plane
with n ≡ z . Then,
ˆ ˆ
e− jβ r
A=µ
4π r
( −I yH x + I xH y ) ,
ˆ         ˆ                (17.24)
e− jβ r
F = −ε
4π r
( −I yE x + I xE y ) .
ˆ        ˆ                (17.25)
The integrals in the above expressions can be explicitly written for this case in
which r ′ = x′x + y′y :
ˆ     ˆ

Nikolova 2004                                                                         7
I xE = ∫∫ Eax ( x′, y′)e j β ( x′ sin θ cos ϕ + y′ sin θ sin ϕ ) dx′dy′ ,    (17.26)
SA

I y = ∫∫ Ea y ( x′, y′)e j β ( x′ sin θ cos ϕ + y′ sin θ sin ϕ ) dx′dy′ ,
E
(17.27)
SA

I xH = ∫∫ H ax ( x′, y′)e j β ( x′ sin θ cos ϕ + y′ sin θ sin ϕ ) dx′dy′ ,    (17.28)
SA

I y = ∫∫ H a y ( x′, y′)e jβ ( x′ sin θ cos ϕ + y′ sin θ sin ϕ ) dx′dy′ .
H
(17.29)
SA
Note that the above integrals are exactly the double inverse Fourier transforms
of the aperture field components.
The vector potentials in spherical terms are
e− j β r ⎡ ˆ
A=µ             θ cosθ ( I xH sin ϕ − I yH cos ϕ ) + ϕ ( I xH cos ϕ + I yH sin ϕ ) ⎤ , (17.30)
ˆ
4π r ⎣                                                                      ⎦
e − jβ r ⎡ ˆ
F = −ε           θ cosθ ( I xE sin ϕ − I yE cos ϕ ) + ϕ ( I xE cos ϕ + I yE sin ϕ ) ⎤ . (17.31)
ˆ
4π r ⎣                                                                       ⎦
By substituting the above expressions in (17.16), we obtain the far E field
components as
e − jβ r E
Eθ = j β           [I x cos ϕ + I yE sin ϕ + η cosθ (I yH cos ϕ − I xH sin ϕ )] ,      (17.32)
4π r
e − jβ r
Eϕ = j β          [-η (I xH cosϕ + I yH sin ϕ ) + cosθ (I yE cos ϕ − I xE sin ϕ )] .   (17.33)
4π r
For apertures mounted on a conducting plane, the preferred equivalent
model is the one with electric wall with magnetic current density
M s = 2 ⋅ Ea × n
ˆ          (        )(17.34)
radiating in open space. The solution, of course, is valid only for z ≥ 0 . In this
case, I H = 0 .
For apertures in open space, the dual current formulation is used. Then, a
usual assumption is that the aperture fields are related as in the TEM-wave case
1
H a = z × Ea .
ˆ                                 (17.35)
η
This implies that
1                            I yE H I xE
I   H
=        z×I
ˆ     E
or I    H
=−     ,I y =   .                  (17.36)
η                             η         η
x

Nikolova 2004                                                                                        8
This assumption is valid for moderate and high-gain apertures; therefore, the
apertures should be at least a couple of wavelengths in extent. The above
assumptions reduce (17.32)-(17.33) to
e − j β r (1 + cosθ ) E
Eθ = j βη                      ⎡I x cos ϕ + I yE sin ϕ ⎤ ,
⎣                       ⎦   (17.37)
4π r           2
e− j β r (1 + cosθ ) E
Eϕ = j βη                      ⎡I y cos ϕ − I xE sin ϕ ⎤ .
⎣                       ⎦   (17.38)
4π r           2

5. The uniform rectangular aperture on an infinite ground plane
A rectangular aperture is defined in the x − y plane as shown below.

y

Ly
x

Lx

If the fields are uniform in amplitude and phase across the aperture, it is
referred to as a uniform rectangular aperture. Let us assume that the aperture
field is y-polarized.
L            L
Ea = E0 y, | x |≤ x and | y |≤ y
ˆ                                       (17.39)
2            2
According to the equivalence principle, we assume an electric wall at z = 0 ,
where the equivalent magnetic current density is given by M se = E0 × n .   ˆ

Nikolova 2004                                                                 9
Applying image theory, we can find the equivalent sources radiating in open
space as
M s = 2M se = 2 E0 y × z = 2 E0 x .
ˆ ˆ          ˆ             (17.40)
The only non-zero radiation integral is
Lx / 2                                     Ly / 2

∫                                          ∫
j β x′ sin θ cos ϕ
I yE = 2 E0              e                        dx′ ⋅            e jβ y′sinθ sin ϕ dy′ =
− Lx / 2                                − Ly / 2

⎡βL                   ⎤      ⎡ β Ly             ⎤  (17.41)
sin ⎢ x sin θ cos ϕ ⎥ sin ⎢             sin θ sin ϕ ⎥
= 2 E0 Lx Ly     ⎣ 2                   ⎦⋅     ⎣ 2                ⎦.
β Lx                         β Ly
sin θ cos ϕ               sin θ sin ϕ
2                            2
It is appropriate to introduce the pattern variables:
β Lx
u=          sin θ cos ϕ ,
2
(17.42)
β Ly
v=          sin θ sin ϕ .
2
The complete radiation fields are found by substituting (17.41) in (17.32) and
(17.33):
e − jβ r                sin u sin v
Eθ = j β            E0 Lx Ly sin ϕ
2π r                      u     v
− jβ r
(17.43)
e                              sin u sin v
Eϕ = j β            E0 Lx Ly cosθ cos ϕ
2π r                             u     v
The total-field amplitude pattern is, therefore,
sin u sin v
| E |= F (θ , ϕ ) = sin 2 ϕ + cos 2 θ cos 2 ϕ ⋅                 =
u     v
(17.44)
sin u sin v
= 1 − sin 2 θ cos 2 ϕ ⋅
u     v

Nikolova 2004                                                                                          10
The principal plane patterns are:
E-plane pattern (ϕ = π / 2)
⎛ β Ly       ⎞
sin ⎜      sin θ ⎟
Eθ =     ⎝ 2          ⎠                      (17.45)
⎛ β Ly       ⎞
⎜ 2 sin θ ⎟
⎝            ⎠
H-plane pattern (ϕ = 0)
⎛ βL         ⎞
sin ⎜ x sin θ ⎟
Eϕ = cosθ     ⎝ 2          ⎠                    (17.46)
⎛ β Lx       ⎞
⎜      sin θ ⎟
⎝ 2          ⎠

Principle patterns for aperture of size: Lx = 3λ , Ly = 2λ

0

30                            30
E-plane
H-plane

60                                               60

1    0.8    0.6   0.4   0.2
90                                                          90

120                                              120

150                           150

180

Nikolova 2004                                                                         11
For electrically large apertures, the main beam is narrow and the
1 − sin 2 θ cos 2 ϕ in (17.44) is negligible, i.e., it is roughly equal to 1 for all
observation angles within the main beam. That is why, in the theory of large
arrays, it is assumed that the amplitude pattern of a rectangular aperture is
sin u sin v
f (u , v )             ,                     (17.47)
u     v
β Lx                         β Ly
where u =             sin θ cos ϕ and v =          sin θ sin ϕ . Here is a view of the
2                           2
| (sin u ) / u | function for Lx = 20λ and ϕ = 0 (H-plane pattern):

|sin[20*pi*sin(theta)]/[20*pi*sin(theta)]|
1

0.8

0.6

0.4

0.2

0
-1            -0.5             0              0.5             1
sin(theta)

Nikolova 2004                                                                       12
Here is a view of the | sin v / v | function for Ly = 10λ and ϕ = 90 (E-plane
pattern):

|sin(10 π sin(theta))/(10 π sin(theta))|
1

0.8

0.6

0.4

0.2

0
-1             -0.5             0            0.5            1
sin(theta)

Beamwidths

(a) first-null beamwidth
We need the locations of the first nulls in the pattern in order to calculate the
FNBW. The nulls of the E-plane pattern are determined from (17.45) as
β Ly
sin θ / θ =θ n = nπ , n = 1, 2,… ,                (17.48)
2
⎛ nλ ⎞
⇒ θ n = arcsin ⎜ ⎟ , rad.                           (17.49)
⎜ Ly ⎟
⎝ ⎠
The first null occurs at n = 1 .
Nikolova 2004                                                                     13
⎛λ ⎞
⇒ FNBWE = 2θ n = 2arcsin ⎜ ⎟ , rad.                    (17.50)
⎜ Ly ⎟
⎝ ⎠
In a similar fashion, FNBWH is determined to be
⎛λ ⎞
FNBWH = 2arcsin ⎜ ⎟ , rad.                         (17.51)
⎝ Lx ⎠

(b) half-power beamwidth
The half-power point in the E-plane occurs when
⎛ β Ly       ⎞
sin ⎜      sin θ ⎟
⎝ 2          ⎠= 1 ,                      (17.52)
⎛ β Ly       ⎞     2
⎜ 2    sin θ ⎟
⎝            ⎠
or
β Ly
sin θ / θ =θ h = 1.391,                (17.53)
2
⎛ 0.443λ ⎞
⇒ θ h = arcsin ⎜
⎜ Ly ⎟
⎟
⎝         ⎠
⎛ 0.443λ ⎞
HPBWE = 2 arcsin ⎜
⎜ Ly ⎟
.                (17.55)
⎟
⎝          ⎠
A first-order approximation is possible for very small arguments in (17.55), i.e.,
when Ly 0.443λ (large aperture):
λ
HPBWE            .
0.886                          (17.56)
Ly
The half-power beamwidth in the H-plane is analogous:
⎛ 0.443λ ⎞
HPBWH = 2arcsin ⎜           ⎟.                        (17.57)
⎝    Lx ⎠

Nikolova 2004                                                                   14
Side-lobe level
It is obvious from the properties of the | sin x / x | function that the first side
lobe has the largest maximum of all side lobes, and it is
sin 4.494
| Eθ (θ = θ s ) |=           = 0.217 = −13.26 , dB.           (17.58)
4.494
When evaluating side-lobe levels and beamwidths in the H-plane, one has to
include the cosθ factor, too. The larger the aperture, the less important this
factor is.

Directivity
In a general approach to the calculation of the directivity, the total radiated
power Π has to be calculated first using the far-field pattern expression (17.44)
4π          U
D0 =        = 4π max .                           (17.59)
Here,
1
U (θ ,ϕ ) =    ⎡| Eθ |2 + | Eϕ |2 ⎤ r 2 = U max | F (θ , ϕ ) |2 , (17.60)
2η ⎣                  ⎦
2π π / 2
ΩA =    ∫ ∫ | F (θ ,ϕ ) |    sin θ dθ dϕ .
2
(17.61)
0    0
However, in the case of an aperture illuminated by a TEM wave, we can use
a simpler approach. Generally, for all aperture antennas, the assumption of a
uniform TEM wave at the aperture ( E = yE0 ),
ˆ
E
Ha = −x 0 ,
ˆ                               (17.62)
η
is quite accurate (although η is not necessarily the intrinsic impedance of
vacuum). The far-field components in this case were already derived in (17.37)
and (17.38). They lead to the following expression for the radiation intensity,
β2
U (θ ,ϕ ) =       (1 + cosθ )2 ⎡| J xE |2 + | J y |2 ⎤ .
⎣
E
⎦       (17.63)
32π η
2

The maximum value of the function in (17.63) is easily derived after
substituting the radiation integrals from (17.26) and (17.27):

Nikolova 2004                                                                      15
2
β 2
U max =
8π 2η   ∫∫ Ea ds′       .        (17.64)
SA
The integration of the radiation intensity (17.63) over a closed sphere is in
general not easy. It can be avoided by observing that the total power reaching
the far zone must have passed through the aperture in the first place. In an
aperture, where the field obeys (17.62), this power is determined as
1
Π = ∫∫ Pav ⋅ ds =    ∫∫ | Ea | ds .
2
(17.65)
S
2η S
A

Substituting (17.64) and (17.65) in (17.59) finally yields
2
Aeff
∫∫ Ea ds′
4π      S
D0 = 2 × A               .                  (17.66)
λ ∫∫ | Ea |2 ds′
SA

In the case of a uniform rectangular aperture,
| E0 |2
Π = Lx Ly         ,                      (17.67)
2η
2
⎛ Lx Ly ⎞ | E0 |2
U max = ⎜
λ ⎟ 2η
.               (17.68)
⎝       ⎠
Thus, the directivity is found to be
U       4π           4π       4π
D0 = 4π max = 2 Lx Ly = 2 Ap = 2 Aeff .               (17.69)
Π     λ            λ        λ
The physical and the effective areas of a uniform aperture are equal.

Nikolova 2004                                                               16
6. The uniform rectangular aperture in open space
Now, we examine the same aperture when it is not mounted on a ground
plane. The field distribution is the same as in (17.39) but now the H field must
be defined, too, in order to apply the general form of the equivalence principle
with both types of surface currents,
Ea = yE0 ⎫
ˆ
⎪ − Lx / 2 ≤ x′ ≤ Lx / 2
E ⎬,                                      (17.70)
H a = − x 0 ⎪ − L y / 2 ≤ y′ ≤ L y / 2
ˆ
η ⎭
Above, again an assumption was made that there is a direct relation between the
electric and the magnetic field components.
To form the equivalent problem, an infinite surface is chosen again to
extend in the z = 0 plane. Over the entire surface, the equivalent J s and M s
must be defined. Both J s and M s are not zero outside the aperture in the z = 0
plane because the field is not zero there. Moreover, the field is not known a
priori outside the aperture. Thus, the exact equivalent problem cannot be built
in practice (at least, not by making use of the infinite plane model).
The usual assumption made is that Ea and H a are zero outside the aperture
in the z = 0 plane, and, therefore, so are the equivalent currents J s and M s ,
M s = −n × Ea = − z × y E0 ⎫
ˆ           ˆ ˆ
⎪ − L / 2 ≤ x′ ≤ L / 2
xˆ      ⎪      x             x
⎬,                             (17.71)
J s = n × H a = z × (− x) ⎪
ˆ         ˆ        ˆ
E0    − L y / 2 ≤ y′ ≤ L y / 2
η ⎪
−yˆ        ⎭
Since the equivalent currents are related via the TEM-wave assumption, only
E
the integral J y is needed for substitution in the far-field expressions derived in
(17.37) and (17.38), same as in (17.41):

Nikolova 2004                                                                      17
Lx / 2                                     Ly / 2

∫                                          ∫
j β x′ sin θ cos ϕ
I = 2 E0
E
y                   e                        dx′ ⋅            e jβ y′sinθ sin ϕ dy′ =
− Lx / 2                                − Ly / 2

⎡βL              ⎤       ⎡ β Ly              ⎤  (17.72)
sin ⎢ x sin θ cos ϕ ⎥ sin ⎢           sin θ sin ϕ ⎥
= 2 E0 Lx Ly       ⎣ 2              ⎦⋅      ⎣ 2                 ⎦.
β Lx                     β Ly
sin θ cos ϕ                sin θ sin ϕ
2                         2
Now, the far-field components are obtained by substituting in (17.37) and
(17.38):
Eθ = C sin ϕ
(1 + cosθ ) sin u sin v
2       u       v
(17.73)
Eϕ = C cos ϕ
(1 + cosθ ) sin u sin v
2        u      v
where:
e− j β r
C = j β Lx Ly E0           ;
2π r
β Lx
u=        sin θ cos ϕ ;
2
β Ly
v=        sin θ sin ϕ .
2
The far-field expressions in (17.73) are very similar to those of the aperture
mounted on a ground plane, see (17.43). For small values of θ , the patterns of
both apertures are practically identical.
An exact analytical evaluation of the directivity is difficult. However,
according to the approximations made, the directivity formula derived in
(17.66) should provide accurate enough value. According to (17.66), the
directivity is the same as in the case of the aperture mounted on a ground plane.

Nikolova 2004                                                                                       18
7. The tapered rectangular aperture on a ground plane
The uniform rectangular aperture has the maximum possible effective area
(for an aperture-type antenna) equal to its physical area. This also implies that it
has the highest possible directivity for all constant-phase excitations of a
rectangular aperture. However, the directivity is not the only important factor in
the design of an antenna. A factor that frequently comes into a conflict with the
directivity is the side-lobe level (SLL). The uniform distribution excitation
produces the highest SLL of all constant-phase excitations of a rectangular
aperture. It will be shown that a reduction of the SLL can be achieved by
tapering the equivalent sources distribution from a maximum at the aperture’s
center to zero values at its edges.
One practical aperture of tapered source distribution is the open rectangular
waveguide. The dominant TE10 mode has the following distribution:
⎛ π ⎞ ⎧− Lx / 2 ≤ x′ ≤ Lx / 2
Ea = yE0 cos ⎜ x′ ⎟ , ⎨
ˆ                                                   (17.74)
⎝ Lx ⎠ ⎩− Ly / 2 ≤ y′ ≤ Ly / 2

y

Ey                             x

The general procedure for the far-field analysis is the same as before (in
Section 5). The only difference is in the field distribution. Again, only the
E
integral J y is to be evaluated:

Nikolova 2004                                                                    19
⎛ π ⎞ jβ x′ sinθ cosϕ
Lx / 2                                  L /2
y

I = 2 E0 ∫ cos ⎜ x′ ⎟ e
E
y                                        dx′ ⋅ ∫ e j β y′sinθ sin ϕ dy′ .   (17.75)
− Lx / 2 ⎝ Lx ⎠                       − Ly / 2

The integral of the y′ variable was already encountered in (17.41):
⎡βL                    ⎤
Ly / 2                       sin ⎢ y sin θ sin ϕ ⎥
I ( y ) = ∫ e j β y′sinθ sin ϕ dy′ = Ly     ⎣ 2                    ⎦.         (17.76)
β Ly
− Ly / 2                                sin θ sin ϕ
2
The integral of the x′ variable is also easily solved:
⎛π ⎞
Lx / 2

I ( x) = ∫ cos ⎜ x′ ⎟ e j β x′sinθ cosϕ dx′ =
− Lx / 2     ⎝ Lx ⎠
⎛π ⎞
Lx / 2

=            cos ⎜ x′ ⎟ ⎡cos ( β x′ sin θ cos ϕ ) + j sin ( β x′ sin θ cos ϕ ) ⎤ dx′ =
∫ ⎝ Lx ⎠ ⎣                                                              ⎦
− Lx / 2

1 x ⎧ ⎡⎛ π                          ⎞ ⎤         ⎡⎛ π               ⎞ ⎤⎫
L /2
⎪                                                             ⎪
2 − L∫ / 2 ⎪ ⎣⎝ Lx
=            ⎨ cos ⎢⎜ − β sin θ cos ϕ ⎟ x′⎥ + cos ⎢⎜ + β sin θ cos ϕ ⎟ x′⎥ ⎬dx′ +
x     ⎩                        ⎠ ⎦         ⎣⎝ Lx              ⎠ ⎦⎪  ⎭
j x ⎧ ⎡⎛                     π ⎞ ⎤        ⎡⎛             π ⎞ ⎤⎫
L /2
⎪                                                    ⎪
+ ∫ ⎨sin ⎢⎜ β sin θ cos ϕ − ⎟ x′⎥ + cos ⎢⎜ β sin θ cos ϕ + ⎟ x′⎥ ⎬dx′
2 − Lx / 2 ⎪ ⎣⎝
⎩                 Lx ⎠ ⎦       ⎣⎝             Lx ⎠ ⎦ ⎪
⎭
⎛ βL             ⎞
cos ⎜ x sin θ cos ϕ ⎟
π Lx      ⎝ 2              ⎠
⇒ I ( x) =
2 ⎛ π ⎞ 2 β Lx
⎜ ⎟ −        sin θ cos ϕ
⎝ 2⎠      2

⎛ βL            ⎞       ⎡βL                ⎤
cos ⎜ x sin θ cos ϕ ⎟   sin ⎢ y sin θ sin ϕ ⎥
⇒Jy
E
= π E0 Lx Ly         ⎝ 2             ⎠ ×     ⎣ 2                ⎦ (17.77)
⎡                       ⎤     ⎛ β Ly             ⎞
2
⎢⎛ π ⎞ β Lx             ⎥     ⎜      sin θ sin ϕ ⎟
⎜ ⎟ −
⎢⎝ 2 ⎠      sin θ cos ϕ ⎥     ⎝ 2                ⎠
2
⎢                       ⎥
⎣                       ⎦
v
u

Nikolova 2004                                                                                      20
To derive the far-field components, (17.77) is substituted in (17.32) and (17.33)
π           cos u       sin v
Eθ = − C sin ϕ                ×
2        ⎡ 2 ⎛ π ⎞2 ⎤     v
⎢u − ⎜ ⎟ ⎥
⎢
⎣    ⎝2⎠ ⎥ ⎦
(17.78)
π                 cos u        sin v
Eϕ = − C cosθ cos ϕ                  ×
2            ⎡ 2 ⎛π ⎞   2⎤      v
⎢ u −⎜ ⎟ ⎥
⎢
⎣     ⎝2⎠ ⎥  ⎦
where
e− j β r
C = j β Lx Ly E0           ;
2π r
β Lx
u=        sin θ cos ϕ ;
2
β Ly
v=        sin θ sin ϕ .
2

Principle plane patterns
In the E-plane, the aperture is not tapered. As expected, the E-plane principal
pattern is the same as that of a uniform aperture.

E-plane ( ϕ = 90 ):
⎛ β Ly       ⎞
sin ⎜      sin θ ⎟
Eθ =     ⎝ 2          ⎠                       (17.79)
⎛ β Ly       ⎞
⎜ 2    sin θ ⎟
⎝            ⎠
H-plane ( ϕ = 0 ):
⎛ β Lx       ⎞
cos ⎜      sin θ ⎟
Eϕ = cosθ        ⎝ 2          ⎠                    (17.80)
⎛ β Lx       ⎞ ⎛π ⎞
2       2

⎜      sin θ ⎟ − ⎜ ⎟
⎝  2         ⎠ ⎝ 2⎠

Nikolova 2004                                                                  21
H-PLANE PATTERN – UNIFORM VS. TAPERED ILLUMINATION ( Lx = 3λ ):

0
uniform
30                            30                         tapered

60                                                    60

1     0.8   0.6   0.4   0.2                                 90
90

120                                                   120

150                            150

180

The lower SLL of the tapered-source far field is obvious. It is better seen in the
rectangular plot given below. The price to pay for the lower SLL is the decrease
in directivity (the beamwidth of the major lobe increases).

Nikolova 2004                                                                    22
tapered
1                                                              uniform
0.9

0.8
H-plane amplitude pattern

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

-1   -0.8   -0.6   -0.4   -0.2        0     0.2   0.4   0.6    0.8      1
sin(theta)

The above example of Lx = 3λ illustrates well the effect of source
distribution on the far-field pattern. However, a more practical example is the
rectangular-waveguide open-end aperture, where the waveguide operates in a
dominant mode, i.e. λ0 / 2 < Lx < λ0 . Here, λ0 is the wavelength in open space
( λ0 = c / f 0 ). Consider the case Lx = 0.75λ0 . The principal-plane patterns for an
aperture on a ground plane look like this:

Nikolova 2004                                                                                                  23
0                    H-plane
30                   30               E-plane

60                                  60

90    1    0.8 0.6   0.4 0.2
90

120                                 240

150                  210
180

In the above example, a practical X-band waveguide was considered whose
cross-section has the following sizes: Lx = 2.286 cm, Ly = 1.016 cm.
( λ0 = 3.048 cm and f 0 = 9.84 GHz)
The case of a dominant-mode open-end waveguide radiating in free space
can be analyzed following the approaches outlined in this Section and in
Section 6.
The calculation of beamwidths and directivity is analogous to previous
cases. Only the final results will be given here for the case of the x-tapered
(cosine taper) aperture on a ground plane:
8 ⎛ 4π      ⎞
Directivity: D0 = 2 ⎜ 2 Lx Ly ⎟                                       (17.81)
π ⎝λ          ⎠
8
Effective area: Aeff = 2 Lx Ly = 0.81Ap                               (17.82)
π
Note the decrease in the effective area.

Nikolova 2004                                                               24
Half-power beamwidths:
50.6
HPBWE =           , deg. (= HPBWE of the uniform aperture)            (17.83)
Ly / λ
68.8
HPBWH =           , deg. (> HPBWH of the uniform aperture)            (17.84)
Lx / λ
The above results are approximate. Better results would be obtained if the
following factors were taken into account:
• the phase constant of the waveguide β g and its wave impedance Z g are
not equal to the free-space phase constant β 0 = ω µ 0ε 0 and intrinsic
impedance Z 0 = µ0 / ε 0 ; they are dispersive;
• the abrupt termination at the waveguide open end introduces reflection,
which affects the field at the aperture;
• there are strong fringe currents at the waveguide walls, which contribute

Nikolova 2004                                                               25

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