Unit 5

Aperture Antennas

So far, we have considered three basic classes of antennas – electrically small, resonant

(narrowband) and broadband (the travelling wave antenna). There are amny other

types of broadband antennas. In this unit we introduce the aperture antenna, part

of which is an opening or aperture through which e-m waves may be transmitted or

received. These include horns and reflector antennas whose apertures may have any

one of a variety of shapes (rectangular, circular, parabolic, etc.). A few are shown

below (taken from the Balanis reference).

E-plane

Pyramidal

H-plane

Conical

Typical Horn Antennas

Feed

Plane

Feed

Corner

Feed

(focal point)

Parabolic

Feed

Cassegrain

Feed

Parabolic

Reflector

Typical Reflector Antennas

Hyperbolic

Subreflector

Parabola

focus

Blockage

1

These kinds of antennas are in common use for frequencies at and above UHF

(> 0.3 GHz) – why is this? They are most common at microwave frequencies where

wavelengths are on the order of centimetres. Their characteristics include:

1. high gain (when aperture dimensions are several wavelengths);

2. gain increasing with frequency;

3. moderate bandwidth

4. nearly real-valued input impedance.

In this unit we shall only briefly consider a method of analysis for such antennas with

a few simple examples.

5.1 Field equivalence and Related Concepts

Carefully read Balanis, Section 12.1–12.3.

5.1.1 Magnetic Current Density, Electric Vector Potential

and Duality

It is sometimes convenient to introducce a fictitious magnetic current density, ~M ,

into Maxwell’s equations. This ~M is the magnetic, but fictitious, analogy to the

electric current density, ~J , and its utility in the analysis of aperture antennas will be

addressed shortly. For analysis purposes, then, Maxwell’s equations in complex form

may be cast as

~∇× ~E = −jω ~B − ~M (5.1)

~∇× ~H = jω ~D + ~J (5.2)

~∇ · ~B = ρm (5.3)

~∇ · ~D = ρ (5.4)

2

where ρm is a fictitious magnetic charge density and ρ is the usual electric charge

density.

Boundary Conditions

In Section 1.2 we saw that at the interface between two media

n×[

~H1 − ~H2

]

= ~Js (5.5)

where we have used ~Js rather than ~K for the surface current density. Of course, if

medium 2 was a perfect electric conductor

n× ~H = ~Js (5.6)

where ~H exists outside the conductor.

By analogy to this boundary condition we consider next that the surface between

media 1 and 2 can sustain a magnetic rather than electric current sheet. With a view

to equation (5.1), (5.2) and (5.5) we write by analogy to the latter

n×[

~E1 − ~E2

]

= − ~Ms (5.7)

where ~Ms is a fictitious magnetic surface current density. If medium 2 is this time a

“perfect magnetic conductor” so that ~E2 = 0,

−n× ~E = ~Ms (5.8)

where ~E exists outside the magnetic conductor.

Electric Vector Potential and Magnetic Scalar Potential

Just as we defined a magnetic vector potential ~A with

~H =1

µ~∇× ~A

we now define an electric vector potential ~F by

~E = −1

ε~∇× ~F (5.9)

3

where we are considering a homogeneous region in which ~M exist and ~J = 0 (and

therefore ~∇ · ~D = 0 like ~∇ · ~B = 0 in the Section 1.3 case).

Carrying out the same mathematics as in Section 1.3 leads to

(5.10)

where we have used a “magnetic scalar potetntial” φm with its Lorentz guage

~∇ · ~F =

(compare this with equation (1.37). Noting by analogy with equation (1.32), the

contribution to the ~H-field due to the presence of magnetic sources is given by

~H = (5.11)

Note that in Section 1.3, equation (1.38) read for the magnetic vector potential

(5.12)

Clearly, the solution to (5.10) has the same form as (5.12). Noting (1.40) for a surface

current density, ~Js, as

~A(~r) = (5.13)

we similarly write for the electric vector potential

~F (~r) = (5.14)

in which ~Js has been replaced by ~Ms.

Considering only the far field with the usual approximations for amplitude and

phase (R ≈ r in amplitude and R ≈ r − r′ cos Ψ in phase for the geometry shown)

4

Then

~A(~r) = (5.15)

with

~N = (5.16)

By analogy we write

~F (~r) = (5.17)

with

~L = (5.18)

In these equations, ~L and ~N are referred to as radiation vectors. They will be used

later in determining fields from apertures.

In the far-field, where orders of (1/r) > 1 may be justifiably neglected, it is possible

to show (but we won’t here) that, using A and F subscripts to indicate contributions

to the fields from ~Js and ~Ms, respectively,

~EA = (5.19)

~HA = (5.20)

~HF = (5.21)

~EF = (5.22)

Duality

Before proceeding with further analysis, it is useful to observe the similarities be-

tween the equations for electric sources and those for magnetic sources. When two

equations decribing the behaviour of two different quantities have the same mathemat-

ical form, so will their solutions. The variables in the respective equations occupying

similar positions are called dual quantities. A solution of one equation may be effected

5

by systematically interchanging the dual quatities with those in the other. This is

the duality theorem and its application to the present discussion is obvious.

Equations for Equations for

Electric Sources ( ~J); ( ~M = 0) Magnetic Sources ( ~M); ( ~J = 0)

Quantities for Quantities for

Electric Sources ( ~J); ( ~M = 0) Magnetic Sources ( ~M); ( ~J = 0)

(See the Balanis reference.)

5.1.2 More on Images

In a previous Term, we saw that when an e-m wave with parallel polarization – i.e.

the ~E-field is in the plane of incidence – strikes a perfectly conducting surface the

reflection coefficient Γ has a value of −1. Thus, removing the conductor, the analysis

in the field region may be analyzed by introducing an image field along with the

original incident field.

6

Similarly, for vertical polarization of the incident field

Assuming the existance of magnetic sources ~M , the boundary conditions would

lead to the above depictions (on the right of each diagram). If on the other hand, the

reflecting surface is changed to a fictitious “perfect magnetic conductor”, the following

images result. (This should be not too surprising based on the duality relationships

– but we’ll not “prove” it here).

5.1.3 Field Equivalence

Germane to the present analysis is the statement of Huygen’s principle which says

that “every point on a primary wavefront can be considered to be a new source of a

secondary spherical wave and that a secondary wavefront can be constructed as the

envelope of these secondary spherical waves.

A mathematical formulation of this principle, called the field equivalence princi-

ple allows for the replacement of an antenna aperture (as a source) with equivalent

currents that produce radiation fields that are equivalent to those from the antenna.

It’s derivation lies in the uniqueness theorem as it applies to Maxwell’s equations –

this theorem says that a differential equation subject to given boundary conditions

is unique (that is, a solution is the solution). The application to the present problem

may be addressed as follows:

Consider a volume V enclosing sources and in which and outside of which there

exist ~E and ~H fields. The bounding “closed” surface is S (V and S do not need to

7

be physical entities).

The equivalence principle says, for example, that the sources and fields internal

to S may be removed and replaced with appropriate surface current densities, ~Js and

~Ms, which give rise to the same external fields.

From equations (5.6) and (5.7) we have

(5.23)

(5.24)

where ~H and ~E exist outside V . This zero-internal-field formulation of the equivalence

principle is called Love’s equivalence principle. It is not the most general case, but

rather a special case, of the original theorem. Consider, next, two possible further

simplifications:

1. Suppose V is filled with a perfect electric conductor so that ~Js “shorts out”.

Then, on the surface we have

~Ms = −n× ~E .

2. Suppose V is filled with a perfect magnetic conductor so that ~Ms “shorts out”.

Then, on the surface we have

~Js = n× ~H .

8

Problem: The potentials ~A and ~F and their use in the relevant equations are for

radiation into an unbounded medium. However, with no fields inside V as suggested

above, this is no longer the case. The “solution” to this dilemma, appropriate to

aperture antennas, is to let the “closed surface” be a flat plane conductor extending

to infinity (say through the z-axis) – i.e., the surface closes at infinity. Then we may

invoke image theory.

CASE 1: S ≡ perfect magnetic ground plane

CASE 2: S ≡ perfect electric ground plane

Note: If the fields in an aperture are known and if the problem of finding the radiated

fields reduces to either Case 1 or Case 2, then either ~A or ~F will be zero, and the

evaluation of the ~E and ~H fields will be straightforward conceptually. If the case is

more general such that both ~Ms and ~Js are required to describe the aperture fields,

the mechanics of finding the far-fields is still essentially the same. As usual, the

solutions only apply to the half-space outside the conducting planes in the regions

where the source current densities “exist”.

9

5.1.4 Application – Rectangular Aperture on an Infinite GroundPlane

The rectangular aperture is a common microwave antenna configuration. Initially,

consider the side view of such an aperture as fed by a waveguide mounted on an

infinite electric ground plane. A waveguide, as the name implies, is a structure used

for guiding waves in a confined cross section – rather than unbounded space – and

they vary widely in type (eg., coaxial lines, rectangular or circular metallic “tubes”,

optic fibres to name a few).

Setting up the equivalent model in terms of ~Ms:

(Qualifying note: we will assume no diffraction at the edges of the aperture – this

is a reasonable approximation if the dimensions of the structures involved are much

larger than a wavelength. This is essentially equivalent to using so-called geometrical

optics (GO) techniques. The geometrical theory of diffraction (GTD) and the method

of moments (MOM) methods are used to deal with such issues as edge diffraction.

1. Step 1. Form an imaginary “closed” surface – here a flat plane whose edges

extend to infinity. Form the current densities equivalent to ~Ea.

~Ea exists only in the aperture

so ~Ms = 0 outside the aperture. However,

at this point ~Js is unknown everywhere.

10

2. Step 2. Replace the imaginary surface with an infinite conducting plate.

~Js is “shorted out”.

~Ms exists in the space occupied

by original aperture.

3. Step 3. Remove the conducting plane and replace it with the image of ~Ms to

give ~Ms = −2n× ~Ea in the aperture.

Thus, we have succeeded in eliminating ~Js using a magnetic surface current

density of twice the original magnitude.

Starting with these ideas, let us orient the rectangular aperture as shown. The

field in the aperture is

~Ea = E0y

with E0 being constant.

The problem is to find the far-field ~E and ~H.

Using the last step of the field equivalence example,

~Ms =

~Js =

11

z

y

x

r

R

P

dy'dx'r'

a

b

'

'

aperture

ground

plane

Rectangular�Aperture�on�an�Infinite�Ground�Plane

12

From equations (5.15) and (5.16), ~A(~r) = ~0 and ~N = ~0. There remain equations

(5.17) and (5.18) to consider. Particularly, let us examine the primed quantities in

(5.18).

~L = (5.25)

For our aperture orientation (see diagram) we note:

and

~r′ · r = (5.26)

Next, ~Ms(~r′) = −2z × ~Ea = −2z × (E0y) = 2E0x.

Since, in the far field, the r-component of the potentials are zero,

~Ms(~r′) = 2E0x =

= (5.27)

Considering equations (5.25) to (5.27) and writing (5.25) as

~L = Lθθ + Lφφ (5.28)

let’s examine each component of ~L. From the ~Ms stipulation,

Lθ =

which implies

Lθ = (5.29)

From (5.17), (5.22) and (5.29) it is possible to deduce an element factor as

−jke−jkr

4πrcos θ cosφ

13

and the Eθ portion of the electric field in (5.22) is simply the product of the element

factor and the space factor. To complete the integration in (5.29) we note that

∫ `/2

−`/2ejpzdz =

Applying this to (5.29) readily yields for the x′-integral

∫ a/2

−a/22E0

(

ejkx′ sin θ cos φ)

dx′ =

Similarly doing the y′-integral, we get in total

Lθ = 2E0ab

cos θ cosφ

sin(

ka sin θ cos φ2

)

(

ka sin θ cos φ2

)

sin(

kb sin θ sinφ2

)

(

kb sin θ sinφ2

)

(5.30)

It is similarly straightforward to show that for the φ-component

Lφ = −2E0ab

sin φ

sin(

ka sin θ cos φ2

)

(

ka sin θ cos φ2

)

sin(

kb sin θ sinφ2

)

(

kb sin θ sin φ2

)

(5.31)

From equations (5.17), (5.18), (5.21), (5.22), (5.30) and (5.31), on noting that ~A =

0 so that (5.19) and (5.20) do not contribute, the far-field ~E and ~H are readily

determined. To facilitate the compactness of our “answer” we let

X =ka sin θ cosφ

2and Y =

kb sin θ sinφ

2.

From the cited equations, the following result:

Er = 0 (5.32)

Eθ = −jωηFφ =jke−jkrE0ab

2πr

{

sin φsinX

X

sinY

Y

}

(5.33)

Eφ = jωηFθ =jke−jkrE0ab

2πr

{

cos θ cosφsinX

X

sinY

Y

}

. (5.34)

Also, for the far field where

~H =r

η× ~E =

r

η×

(

Eθθ + Eφφ)

14

Hr = 0 (5.35)

Hθ = −Eφ

η(5.36)

Hφ =Eθ

η. (5.37)

Principal Plane Patterns for this Example:

See Section 2.3.1 for definitions.

In view of the definitions, it may be observed that the principal E-plane occurs

in the plane φ = π/2 (i.e., the y-z plane). We note the following:

(1) the source field is in the y-direction;

(2) sinφ|φ=π/2 = 1;

(3) limX→0

sinX

X= 1 when φ→ π/2 and θ → 0;

(4) limY →0

sin Y

Y= 1 when θ → 0;

Under these conditions, Eφ → 0 since cos(π/2) = 0. Therefore, the principal E-plane

field is given by

Eθ =jke−jkrE0ab

2πr

[

sin kb sin θ2

kb sin θ2

]

(5.38)

The principal H-plane field is perpendicular to the E-plane, in this case in the

φ = 0 plane. Then, Er = Eθ = 0 and the principal H-plane E-field pattern is easily

seen to be

Eφ =jke−jkrE0ab

2πr

[

sin ka sin θ2

ka sin θ2

]

(5.39)

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5.1.5 Application – The Parabolic Reflector (One Approxi-mate Example)

The parabolic reflector is very useful and is widely used in microwave communica-

tions systems (microwave includes VHF, UHF, SHF, EHF, ∼100 MHz →∼1012 Hz,

spanning, for example, the TV, FM and satellite communications channels). The

feed elements for such antennas may be dipoles, horns, or waveguides, depending on

the surface geometry of the reflector – for example, parabolic cylindrical reflectors

are most generally fed with a linear dipole, linear array or slotted waveguide. The

arrangements of the feed and the reflecting surface depend on particular applications

and design constraints (see, for example, Figures 15.1 and 15.2 of the text).

Illustration:

Recall, that for a parabolic reflector, rays from the focus are reflected parallel to

the principal axis – we may talk in terms of “rays” if the reflectors have large dimen-

sions relative to the wavelength since then diffraction effects at the edges and other

finite dimensional effects are reduced. That is, we are again considering geometrical

optics (ray tracing) results. If more detailed knowledge of operational characteristics

is required, more sophisticated analyses has to be done (again, using GTD or MOM

methods). In the example which follows it is considered that:

1. the induced current density is zero on the shadow side of the reflector;

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2. the discontinuity of the current density at the reflector “edges” is ignored;

3. direct radiation of the feed into the space where “observations” are being made

are ignored; and

4. aperture blockage due to placement of the feed and its associated equipment

are neglected.

For electrically large apertures (i.e. large relative to wavelength) these approximations

still give good results for the mainbeam and nearby minor lobes.

Example – Using Field Equivalence

Suppose that a particular feed produced in an aperture of a parabolic reflector a

portion of a plane wave whose E-field is given by

~Ea = E0x .

Illustration:

As before, in the far field, for amplitude considerations, | ~R| = |~r| but for phase

purposes,

R = r − r′ cosψ .

17

For use in equations (5.16) and (5.18) we note that

r′ cosψ = ~r ′ · r

= x′ sin θ cosφ+ y′ sin θ sin φ

similarly as in equation (5.26). Given that x′ = r′ cosφ′ and y′ = r ′ sinφ′, it is clear

that

r′ cosψ = sin θ [r′ cosφ′ cos φ+ r′ sin φ′ sin φ]

or

r′ cosψ = r′ sin θ cos(φ− φ′) . (5.40)

The equivalent electric and magnetic surface current densities in the aperture are,

from (5.23) and (5.24),

Therefore,

~Js = (5.41)

~Ms = (5.42)

The aperture surface element is

dS ′ = (5.43)

We are now in a position to determine the far-field ~E and ~H fields via the potentials

and radiation integrals used earlier. Using (5.40), (5.41) and (5.43) in equation (5.16)

and (5.40), (5.42) and (5.43) in equation (5.18), it is easy to verify that

~N = (5.44)

~L = (5.45)

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where

C =

The value of this integral is

C = (5.46)

where J1(·) is a first-order Bessel function of the first kind. Again, remember that

these radiation integrals are for the far field where the r components do not appear

– i.e. Er = 0 and Hr = 0. From equations (5.15), (5.17), (5.19), (5.22), (5.44) and

(5.45) it is easy to verify (DO THIS) that

Eθ = (5.47)

while

Eφ = (5.48)

The H-field components can be also gotten using the radiation integrals and corre-

sponding equations or since ~E, ~H, and r, are mutually perpendicular in the far field

(i.e. ~E and ~H are perpendicular to each other and to the direction of propagation)

Hθ = (5.49)

while

Hφ = (5.50)

The Bessel functions may be evaluated numerically or determined from tables.

Finally, note that the radiation intensity for the aperture may be determined as

usual from

U(θ, φ) =r2

2η| ~E|2 .

Do this.

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5.2 Fourier Transforms and Aperture Antennas –

A Brief Summary

As in many other areas of engineering and applied science, aperture antenna analysis

carried out in the temporal (or spatial) domain is often more complex than when the

same problems are approached in the frequency (or spectral) domain. To give a very

brief overview of the latter and a result that has many applications, we seek a Fourier

transform approach to the problem of finding the far-field of an aperture. We will

focus on the important results rather than all of the intervening details.

Preliminaries:

1. As we have noted, in the vicinity of a particular far-field point the field itself

may be treated as a plane wave. At any such position ~r, we shall consider that

the wave is travelling in the r direction and its wave vector is defined by

~k = kr = ; k = |~k| = 2π/λ . (5.51)

2. Aspatial Fourier transform pair may be DEFINED as

f(x) =1

2π

∫

∞

−∞

F(kx)e−jxkxdkx (5.52)

F(kx) =∫

∞

−∞

f(x)e+jxkxdx (5.53)

Note that as long as the transform pair is defined, which of that pair is referred

to as the direct and inverse transform is not really critical.

Extending the transform to include two-dimensions (or more if required) we

write

f(x, y) =1

4π2

∫

∞

−∞

∫

∞

−∞

F(kx, ky)e−j(kxx+kyy)dkxdky (5.54)

F(kx, ky) =∫

∞

−∞

∫

∞

−∞

f(x, y)e+j(kxx+kyy)dxdy (5.55)

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The Problem:

Consider as before a rectangular aperture mounted on an infinite ground plane.

Let the electric field in the aperture be

~Ea(x′, y′, z′ = 0) =

and let the spatial transform in the z = 0 plane be given by

~E(kx, ky) =

Then, since the aperture field exists only where |x| ≤ a/2 and |y| ≤ b/2, we may

write using (5.55)

Eax(kx, ky) =

= (5.56)

Note that

k~r ′ · r =

with k =√

k2x + k2

y + k2z . Similar expressions for Eay

and Eazmay of course be found

by using the appropriate component of ~Ea in (5.56).

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After a substantial amount of analytical detail (stationary phase analysis), it may

be deduced that in the far field of the aperture

~E(r, θ, φ) = (5.57)

Recall that for the far field Er = 0. Notice that Eazhas been eliminated from explicit

appearance in the “answer”. Equation (5.57) is very powerful. It indicates that if the

aperture field ~Ea is known, the the far-field ~E may be determinde via the (inverse)

Fourier transform of ~Ea, component by component.

Example: Consider the earlier problem in which the aperture field is given by

~Ea =

{

E0y , −a/2 ≤ x′ ≤ a/20 , otherwise.

Let’s again determine the far-field electric field ~E = Eθθ + Eφφ.

We note that since the aperture field has only a y component, (5.56) implies

Eax= Eaz

= 0 .

Also,

k(~r ′ · r) =

=

Therefore, from (5.56), on using Eayinstead of Eax

, we have

Eay=

Clearly, these integrals look like those which led to (5.30) etc. and we write immedi-

ately that

Eay=

where

X = ; Y =

22

Thus, from (5.57), since Eax= 0,

Eθ =

and

Eφ =

As before, for the far field, Er = 0. Clearly, the above solution is identical to that

obtained in (5.33) and (5.34) using the field equivalence principle.

23