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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


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                            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
      added magnetic current source.
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
                      known                                                           current radiates
                                                       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⎠
              Radiation Pattern
              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
              beamwidth is only about 2°.
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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