NUS/ECE EE5308 Aperture Antennas 1 Introduction Very often, we have antennas in aperture forms, for example, the antennas shown below: Pyramidal horn antenna Conical horn antenna Hon Tat Hui 1 Aperture Antennas NUS/ECE EE5308 Paraboloidal antenna Slot antenna Hon Tat Hui 2 Aperture Antennas NUS/ECE EE5308 2 Analysis Method for Aperture Antennas 2.1 Uniqueness Theorem An electromagnetic field (E, H) in a lossy region is uniquely specified by the sources (J, M) within the region plus (i) the tangential component of the electric field over the boundary, or (ii) the tangential component of the magnetic field over the boundary, or (iii) the former over part of the boundary and the latter over the rest of boundary. The case for a lossless region is considered to be the limiting case as the losses go to zero. Here M is the magnetic current density assumed that it exists or its existence is derived from M = E × n, where E is the electric field on a surface and n is the normal vector on that surface. (For a proof, see ref. [3]) Hon Tat Hui 3 Aperture Antennas NUS/ECE EE5308 2.2 Equivalence Principle Actual problem Equivalent problem Hon Tat Hui 4 Aperture Antennas NUS/ECE EE5308 For the actual problem on the left-hand side, if we are interested only to find the fields (E1, H1) outside S (i.e., V2), we can replace region V1 with a perfect conductor as on the right-hand side so that the fields inside it are zero. We also need to place a magnetic current density Ms=E1×n on the surface of the perfect conductor in order to satisfy the boundary condition on S. Now for both the actual problem and the equivalent problem, there are no sources inside V2. In the actual problem, the tangential component of the electric field at the outside of S is E1×n. In the equivalent problem, the tangential component of the electric field at the outside of S is also E1×n as a magnetic current density Ms=E1×n has been specified on S already. Hon Tat Hui 5 Aperture Antennas NUS/ECE EE5308 Hence by using the uniqueness theorem, the fields (E1, H1) in V2 in the equivalent problem will be the same as those in the actual problem. Note that the requirement for zero fields inside V1 is to satisfy the boundary condition specified on the tangential component of the electric field across S. Because now in the equivalent problem just outside S, the electric field is E1 while there is also a magnetic current density Ms. But just inside S, the electric field is zero. Hence on S, ( E1 − 0 ) × n = E1 × n = M s This is exactly the boundary condition specified on the tangential component of the electric field across S with an 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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