Aperture Antennas Word

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Aperture Antennas1 Introduction

Very often, we have antennas in aperture forms, for example, the antennas shown below:

Hon Tat Hui 1 Aperture Antennas

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Pyramidal horn antenna Conical horn antenna


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Paraboloidal antenna Slot antenna

2 Analysis Method for Aperture Antennas2.1 Uniqueness TheoremAn 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


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× 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])2.2 Equivalence Principle


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Actual problem Equivalent problem 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


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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.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 = Ms


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This is exactly the boundary condition specified on the tangential component of the electric field across S with an added magnetic current source. 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.


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Only equivalent Twice the magnetic current is equivalent Aperture fields required magnetic

knowncurrent radiates in free space

Ea, Ha M = Ea × nsV1

V2V1

ε2,μ2 ε1,μ12 ε2,μ2 2

(a) (b) (c)Aperture in a ground plane Equivalent problem Equivalent problem

after using image theorem


1μ,1ε

1V

Aperture

Ground planen

μ,2ε

2V

s

μ,2ε

2V

M2

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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.


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

with M s with J

∇×H = jωεE ∇×E = − jωμH ∇×E = −Ms − jωμH ∇×H = J + jωεE ∇⋅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.


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with M s with J

E F H A

∇2F + k 2F = −εMs ∇2A + k 2A = −μJε e− jkR μ e− jkR

F R( ) = 4 ∫∫∫

v Ms (R')

R dv' (A R) = 4π∫∫∫

v ' J R( ')

R dv' π '

Hence if there is only a magnetic current source, E can be calculated from the electric vector potential F, whose


==− ∇× ∇×με 11

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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.

Far-Field ApproximationsWhen the far-field of aperture radiation is interested, the following approximations can be used to simplify the factor (see ref. [1]):e− jkRR

R ≈ r − r′cosψ, for numeratorR ≈ r, for denominator


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Then,

whereHon Tat Hui 13 Aperture Antennas

y

x

z

′ ′′ φθr ,, )( )φ,θ,r(ψ

RRR'

L

RMF R s∫=

=−

′−

πε

πε ψ

re

r' e dve

jkrv

jkrjkr

4

4')(

'

cos( )

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L = ∫∫∫Ms (R' e) jkr′cosψdv'v '

= ∫∫∫ ⎡⎣aˆ xM x (R')+ aˆ yM y (R')+ aˆ zM z (R')⎤⎦e jkr

′cosψdv'v '

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

L a= ˆ rLr +aˆθLθ φ

φ+aˆ L

where


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Lθ=∫∫∫(M x cosθ φcos +M y cosθ φsin −M z sinθ)e jkr

′cosψdv'v'

Lφ=∫∫∫(−M x sinφ+M y cosφ)e jkr′cosψdv'v'

Lr =∫∫∫(M x sinθ φcos +M y sinθ φsin −M z cosθ)e jkr

′cosψdv'v'

1 1


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E = − ∇×F H, = −∇×E ε jωμ

For far fields (see ref. [3], Chapter 3),Eθ = − jωηFφ

Eφ = + jωηFθ

EφHθ = − η

EHφ = +

εe− jkr

Fθ = LθHon Tat Hui 16 Aperture Antennas

ηθ

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4π

r − jkr εeFφ = 4πr Lφ Note: there is no

need to know Fr. Hence there is no need to find Lr.

Example 1Find the far-field produced by a rectangular aperture opened on an infinitely large ground plane with the following


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SolutionsThe equivalent magnetic current density is:


aE⎩− ≤ ≤′⎨= ⎧ ≤≤− ′

bybE

a x aya 22

ˆ22

0

aperture field distribution:

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Ms = Ea ×aˆ z = aˆ y ×aˆ zE0 = aˆ xE0

⎧⎨−a2 ≤ x′ ≤ a2⎩−b2 ≤ y′ ≤ b2

M x = E0, M y = 0, M z = 0

Lθ = ∫∫M x cosθ φcos e jkr′cosψds′s′

Lφ = −∫∫M x sinφe jkr′cosψds′s′


.rLfind is no need to Actually, there

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Lr = ∫∫ M x sinθ φcos e jkr′cosψds′s′

r′cosψ= r a′⋅ ˆ r


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= (aˆ x x′+aˆ y y′) (⋅ aˆ x sinθcosφ+ aˆ y sinθ φsin + aˆ z

cosθ)= x′sinθ φcos+ y′sinθ φsin

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

b 2 a 2

Lθ = cosθ φcos ∫ ∫)dx dy′ ′−b 2 −a 2


+ ′′ θφθφMexyjkx2 sinsincossin(

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= 2abE0 ⎣⎡⎢cosθ φcos ⎛⎜⎝sin XX ⎟⎜⎠⎝⎞⎛sin YY ⎟⎠⎞⎤⎥⎦ From image theorem

ka kb

X = sinθcos , φ Y = sinθsinφ2 2

Similarly,

Lφ=−2abE0 ⎡⎢⎣sinφ⎛⎜⎝sin XX ⎞⎛⎟⎜⎠⎝sinYY ⎞⎟⎠⎤⎥⎦


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Therefore,

Fθ =ε 4eπ− jkrr Lθ =ε 2eπ− jkrr abE0 ⎢⎣⎡cosθ φcos ⎝⎜⎛sin XX ⎟⎜⎞⎛⎠⎝sin YY ⎟⎠⎞⎦⎤⎥

εe− jkr εe− jkr ⎡sin ⎛sin X ⎞⎛sinY ⎞⎤ Fφ= 4πr

Lφ=− 2πr abE0 ⎣⎢ φ⎜⎝ X ⎟⎜⎠⎝ Y ⎟⎠⎥⎦The E and H far-fields can be found to be:Er = 0 Eθ = − jωηFφ =


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Eφ = + jωηFθ

= j abkE e2π0r−

jkr ⎡⎢⎣sinφ⎛⎜⎝sinXX

⎞⎛⎟⎜⎠⎝sinYY ⎞⎟⎠⎤⎥⎦

j abkE e2π0r− jkr ⎡⎢⎣cosθ φcos⎛⎜⎝sin XX ⎞⎛⎟⎜⎠⎝sin YY ⎞⎟⎠⎤⎥⎦

Hr = 0Eφ

Hθ = −η

Eθ


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Hφ =η

The radiation patterns are plotted on next page.


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Three-dimensional field pattern of a constant field rectangular aperture opened on an infinite ground plane ( a =3 λ , b =2 λ )


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Two-dimensional field patterns of a constant field rectangular aperture opened on an infinite ground plane ( a =3 λ , b =2 λ )


H-planeE-plane

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3. Parabolic Reflector AntennasParabolic 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.


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Parabolic reflector antennas


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A typical paraboloidal antenna for satellite communication


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3.1 Front-Fed Paraboloidal Reflector Antenna


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Important geometric parameters and description of a paraboloidal reflector antenna:

θ0 = Half subtended angled = Aperture diameter f = focal length

Defining equation for the paraboloidal surface:

OP + PQ = constant = 2f Physical

area of the aperture Ap:2


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⎛ d ⎞ Ap =π⎜⎝ 2 ⎟⎠

The half subtended angle θ0 can be calculated by the following formula:

⎡ 1⎛ ⎞f ⎤θ = − ⎢ 2⎜ ⎟⎝ ⎠ d ⎥⎥ tan

1

0 ⎢ 2

⎢⎛ ⎞f 1 ⎥⎢⎣⎜ ⎟⎝ ⎠d−16⎥⎦


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Aperture Efficiency εap

Aem maximum effective area εap ==

Ap physical area2

= cot2 ⎛⎜θ20 ⎞⎟⎠ G (θ')tan⎛⎜⎝θ2'⎞⎟⎠dθ' ⎝

Directivity: maximum directivity = 4π2 Aem

=⎛⎜πd⎞⎟2εap


∫θ

f0

0

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D0 =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 Gf (θ') =⎨ ⎩0, π/ 2 ≤θ π' ≤

where n is a number chosen to match the directivity of the feed horn.


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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

εap(n= 2) = 24⎨⎧⎩sin2⎜⎛⎝θ20 ⎟⎞⎠+ln cos⎡⎢⎜⎛⎝θ20

⎟⎞⎠⎤⎥⎦⎬⎫⎭ cot2⎜⎛⎝θ20 ⎞⎠⎟⎣

2


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εap(n = 4) = 40⎧⎩⎨sin4⎛⎝⎜θ20 ⎞⎠⎟+ ln cos⎡⎢⎛⎝⎜θ20

⎞⎠⎟⎤⎥⎦⎫⎭⎬ cot2⎛⎝⎜θ20 ⎞⎠⎟⎣

εap (n = 6) =14 2⎨⎩⎧ ln cos⎡⎢⎣⎛⎜⎝θ20 ⎠⎞⎟⎤⎥⎦+[1− cos(3 θ0)]3

+ 1sin (2 θ )}2 cot2⎛⎜θ20 ⎞⎟⎠2 0 ⎝


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1− cos (4 θ0)

εap (n = 8) =18⎩⎧ 4 ⎢⎣ ⎜⎝ 2 ⎟⎠⎥⎦ ⎨− 2ln

cos⎡ ⎛θ0 ⎞⎤

−[1− cos(θ0)]3 − 12sin (2 θ0)⎬⎫cot2

⎝⎜⎛θ20 ⎞⎟⎠3 ⎭


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) apε(

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Effective Aperture (Area) Aem

With the aperture efficiency, the maximum effective aperture can be calculated as:

Aem = Apεap2

p physical area =π⎛⎜ d ⎞⎟

A =⎝ 2 ⎠


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Radiation PatternThe 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.


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An example of the radiation pattern of a paraboloidal reflector antenna with an axially symmetric feed pattern. Note that the half-power beamwidth is only about 2°.


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Example 2A 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 Gf (θ') =⎨

⎩0, π/ 2 ≤θ π' ≤(a) Find the aperture efficiency and the maximum directivity

of the antenna.


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(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?

Solution

(a) With f/d = 0.5, the half subtended angle θ0 can be calculated.

−1⎢⎡⎢⎛ ⎞f12⎛ ⎞⎜ ⎟⎝ ⎠2df− 1 ⎥⎤⎥⎥ = tan−1⎢⎢⎡⎢(0.512()0.52 −)1 ⎥⎥⎤⎥ = 53.13°


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θ0 = tan ⎢

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

εap (n = 2) = 24{sin2 (26.57° +) ln cos 26.57[ (°)]}2 cot2 (26.57°)

= 0.75 = 75%2


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maximum directivity =⎛⎜πd⎞⎟ εap

D0 =⎝ λ ⎠

2

⎛π10⎞ 0.75 = 74022.03 = 48.69 dB=⎜ ⎟

⎝ 0.1 ⎠(b) Frequency = 3 GHz, ⇒λ= 0.1 m.

2

Maximum effective area = Aem =D0λ = 58.9 m2


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4πPL ⇒ Aem = PL

Ae (θφ, )=Pavi Pav (θφ, ) Pavi

Hence, PL = A Pemavi = 58.9 10× −5 = 5.89 10× −4 WReferences:

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.


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3. R. F. Harrington, Time-Harmonic Electromagnetic Fields, McGraw-Hill, New York, 1962, pp. 100-103, 143-263, 365-367.


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