1 Introduction Very often, we have antennas in aperture forms, for example, the antennas shown below: Pyramidal horn antenna Conical horn antenna 1
Paraboloidal antenna Slot antenna
Analysis Method for.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]) 3
. Equivalence Principle Actual problem Equivalent problem 4
For the actual problem on the left-hand side, if we are interested only to find the fields (E 1, H 1 ) outside S (i.e., V ), we can replace region V 1 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 M s =E 1 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 V. In the actual problem, the tangential component of the electric field at the outside of S is E 1 n. In the equivalent problem, the tangential component of the electric field at the outside of S is also E 1 n as a magnetic current density M s =E 1 n has been specified on S already. 5
( 0) E1 n= E1 n= Ms 6 Hence by using the uniqueness theorem, the fields (E 1, H 1 ) in V in the equivalent problem will be the same as those in the actual problem. Note that the requirement for zero fields inside V 1 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 E 1 while there is also a magnetic current density M s. But just inside S, the electric field is zero. Hence on S, 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 (E 1, H 1 )in V by knowing M s on the surface of a perfect conductor. A modified case with practical interest is shown below. 7
Ground plane Aperture fields known Only equivalent magnetic current is required n Twice the equivalent magnetic current radiates in free space E a, H a M s = E a n V 1 V V 1 V V 1 M s Aperture V ε 1,μ 1 ε,μ ε 1,μ 1 ε,μ ε,μ ε,μ (a) (b) (c) Aperture in a ground plane Equivalent problem Equivalent problem after using image theorem 8
Thus the problem of an aperture in a perfectly conducting ground plane is equivalent to the finding of (i) the fields in V due to an equivalent magnetic current density of M s radiating in a half-space bounded by the ground plane, or (ii) the fields in V due to an equivalent magnetic current density of M s radiating in a free space having the properties of V. Note that for the equivalent problem in (c), the field so calculated in V 1 may not be equal to the original fields in V 1 in actual problem in (a). To find the electromagnetic field due to a magnetic current density M s, we need to construct an equation with the source M s and solve it. 9
.3 Radiation of a Magnetic Current Density Maxwell s equations with a magnetic current source with M with s H = jωεe E= jωμh E= M jωμh H = J+ jωεe B = ρ D= ρ m s D = 0 B = 0 J When there is only a magnetic current source M s, an electric vector potential F can be defined similar to the magnetic vector potential A. 10
with M with 1 1 E= F H = A ε μ F k F M A k A μj + = ε s + = FR ( ) = M( R ' ) AR = J( R) s ε 4π v' s e jkr R 11 J v' jkr μ e dv ' ( ) ' dv' 4π R 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.
Far-Field Approximations When the far-field of aperture radiation is interested, the following approximations can be used to simplify the jkr factor e R (see ref. [1]): R r r cos ψ, for numerator R r, for denominator z ( r, θ, φ ) R' ψ R R y (r,θ,φ) x 1
Then, where εe FR Ms ( R' ) e 4π r jkr jkr cosψ ( ) = ' = εe 4π r jkr v ' L dv v ' v ' ( ) L= M R' e jkr cosψ dv' ( ) ( ) ( ) jkr cosψ xm x ' ym y ' zm z ' e dv = aˆ R + aˆ R + aˆ R s ' 13
In spherical coordinates (see ref. [1]), L= aˆ L + aˆ L + aˆ L r r θ θ φ φ where ( ) x θ φ y θ φ z θ L = M + M M e dv θ v' jkr cosψ cos cos cos sin sin ' ( ) x φ y φ L = M + M e dv φ v' jkr cosψ sin cos ' ( θ φ θ φ θ) jkr cosψ r = xsin cos + ysin sin zcos ' L M M M e dv v' 14
1 1 E =, ε F H = jωμ E For far fields (see ref. [3], Chapter 3), E = jωη F θ E =+ jωη F H H φ θ φ = E φ η Eθ =+ η φ θ 15 F F θ φ εe = 4π r εe = 4π r jkr jkr L L θ φ Note: there is no need to know F r. Hence there is no need to find L r.
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 x a Ea = aˆ ye0 b y b 16
Solutions The equivalent magnetic current density is: M E a a a a s = a ˆz = ˆ y ˆzE0 = ˆxE0 a x a b y b Actually, there is no need to find L r. M = E, 0 M = 0, M = 0 x y z jkr cosψ Lθ = Mx cosθcosφe ds s = jkr cosψ Lφ Mx sinφe ds s jkr cosψ Lr = Mxsinθcosφe ds s 17
r cosψ = r aˆ r ( aˆ x aˆ y ) ( aˆ sinθ cosφ aˆ sinθsinφ aˆ cosθ) = + + + x y x y z = x sinθcosφ+ y sinθsinφ 18
After using the image theorem to remove the ground plane, we have: b a 19 ( sinθcosφ+ y sinθsinφ) jk x Lθ = cosθcosφ M e dx dy b a sin X siny X Y = abe0 cosθcosφ ka kb X = sinθ cos φ, Y = sinθsinφ Similarly, L φ = 0 x sin X siny abe sinφ X Y From image theorem
Therefore, jkr jkr εe εe sin X siny Fθ = Lθ = abe0 cosθcosφ 4πr πr X Y jkr jkr εe εe sin X siny Fφ = Lφ = abe0 sinφ 4πr πr X Y The E and H far-fields can be found to be: E = 0 r jkr abke0e sin X siny Eθ = jωηfφ = j sinφ π r X Y jkr abke0e sin X siny Eφ =+ jωηfθ = j cosθ cosφ π r X Y 0
H r H H θ φ = 0 Eφ = η Eθ = η The radiation patterns are plotted on next page. 1
Three-dimensional field pattern of a constant field rectangular aperture opened on an infinite ground plane (a=3λ, b= λ)
E-plane H-plane Two-dimensional field patterns of a constant field rectangular aperture opened on an infinite ground plane (a=3λ, b= λ) 3
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.. 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. 4
Parabolic reflector antennas 5
A typical paraboloidal antenna for satellite communication 6
3.1 Front-Fed Paraboloidal Reflector Antenna 7
Important geometric parameters and description of a paraboloidal reflector antenna: θ 0 = Half subtended angle d = Aperture diameter f = focal length Defining equation for the paraboloidal surface: OP + PQ = constant = f Physical area of the aperture A p : d Ap = π 8
The half subtended angle θ 0 can be calculated by the following formula: 1 f 1 θ 0 tan d = f 1 d 16 Aperture Efficiency ε ap ε ap A A em = = p maximum effective area physical area θ 0 0 ' cot θ θ Gf ( θ ') tan dθ ' 0 = 9
Directivity: D 4π πd = maximum directivity = A em = ε ap λ λ 0 Feed Pattern G f (θ ) The feed pattern is the radiation pattern produced by the feeding horn and is given by: G f ( θ ') n ( n + 1)cos ( θ'), 0 θ' π / = 0, π / θ' π where n is a number chosen to match the directivity of the feed horn. 30
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: ε ε ap ap ( n ) 0 0 0 θ θ θ = = 4 sin ln cos cot + ( n ) 4 0 0 0 θ θ θ = 4 = 40 sin ln cos cot + 31
ε ε ap ap θ0 [1 cos( θ0)] = 6 = 14 ln cos + 3 ( n ) 4 1 cos ( θ0) θ0 = 8 = 18 ln cos 4 ( n ) } 1 sin ( ) cot θ 0 + θ0 3 [1 cos( θ0)] 1 sin ( θ0 θ0 ) cot 3 3 3
(ε ap ) 33
Effective Aperture (Area) A em With the aperture efficiency, the maximum effective aperture can be calculated as: A A em p ap p = A Radiation Pattern ε d = physical area = π 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. 34
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. 35
Example 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 ( θ '), 0 θ' π / G f ( θ ') = 0, π / 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 P avi = 10-5 W/m, what is the power P L delivered to a matched load? 36
Solution 37 (a) With f/d = 0.5, the half subtended angle θ 0 can be calculated. 1 f 1 ( 0.5 ) 1 1 0 tan d θ = = tan 53.13 f 1 1 = ( 0.5) d 16 16 From the feed pattern, n =. Hence using the aperture efficiency formula with n =, we find: ( n ) ( ) ( ) { [ ]} ( ) ε = = 4 sin 6.57 + ln cos 6.57 cot 6.57 ap = 0.75 = 75%
D 0 π d = maximum directivity = λ ε ap π10 = 0.75 = 740.03 = 48.69 db 0.1 (b) Frequency = 3 GHz, λ = 0.1 m. D λ Maximum effective area = Aem 4π PL Ae( θφ, ) = Aem P P θφ, = avi av ( ) 0 = = 58.9 m P P L avi Hence, P = A P = = L em avi 5 4 58.9 10 5.89 10 W 38
References: 1. C. A. Balanis, Antenna Theory, Analysis and Design, John Wiley & Sons, Inc., New Jersey, 005.. 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, 196, pp. 100-103, 143-63, 365-367. 39