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

Transcription

1 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 H-plane Plane Corner Feed Feed Pyramidal Conical Parabolic Cassegrain Feed Typical Horn Antennas Feed (focal point) Parabolic Reflector Feed Hyperbolic Subreflector Blockage Parabola focus Typical Reflector Antennas 1

2 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);. 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 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 = jωb M (5.1) H = jωd + J (5.) B = ρ m (5.3) D = ρ (5.4)

3 where ρ m is a fictitious magnetic charge density and ρ is the usual electric charge density. Boundary Conditions In Section 1. we saw that at the interface between two media ˆn [ H1 H ] = Js (5.5) where we have used J s rather than K for the surface current density. Of course, if medium was a perfect electric conductor ˆn H = J s (5.6) where H exists outside the conductor. By analogy to this boundary condition we consider next that the surface between media 1 and can sustain a magnetic rather than electric current sheet. With a view to equation (5.1), (5.) and (5.5) we write by analogy to the latter ˆn [ E1 E ] = Ms (5.7) where M s is a fictitious magnetic surface current density. If medium is this time a perfect magnetic conductor so that E = 0, ˆn E = M s (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 ɛ F (5.9) 3

4 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 φ m with its Lorentz guage F = (compare this with equation (1.37). Noting by analogy with equation (1.3), 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 (5.1) Clearly, the solution to (5.10) has the same form as (5.1). Noting (1.40) for a surface current density, J s, as A( r) = (5.13) we similarly write for the electric vector potential F ( r) = (5.14) in which J s has been replaced by M s. 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

5 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 J s and M s, respectively, E A = (5.19) H A = (5.0) H F = (5.1) E F = (5.) Duality Before proceeding with further analysis, it is useful to observe the similarities between the equations for electric sources and those for magnetic sources. When two equations decribing the behaviour of two different quantities have the same mathematical 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

6 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. 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

7 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) 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 principle 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

8 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, J s and M s, which give rise to the same external fields. From equations (5.6) and (5.7) we have (5.3) (5.4) 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 J s shorts out. Then, on the surface we have M s = ˆn E.. Suppose V is filled with a perfect magnetic conductor so that M s shorts out. Then, on the surface we have J s = ˆn H. 8

9 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 : 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, 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 M s and J s 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

10 5.1.4 Application Rectangular Aperture on an Infinite Ground Plane 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 E a. E a exists only in the aperture so M s = 0 outside the aperture. However, at this point J s is unknown everywhere. 10

11 . Step. Replace the imaginary surface with an infinite conducting plate. J s is shorted out. M s 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 M s = ˆn E a in the aperture. Thus, we have succeeded in eliminating J s 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 E a = E 0 ŷ with E 0 being constant. The problem is to find the far-field E and H. Using the last step of the field equivalence example, M s = J s = 11

12 Rectangular Aperture on an Infinite Ground Plane P z r aperture R ' a r' ' dy' dx' y b ground plane x 1

13 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 = (5.5) For our aperture orientation (see diagram) we note: and r ˆr = (5.6) Next, Ms ( r ) = ẑ E a = ẑ (E 0 ŷ) = E 0ˆx. Since, in the far field, the ˆr-component of the potentials are zero, M s ( r ) = E 0ˆx = = (5.7) Considering equations (5.5) to (5.7) and writing (5.5) as L = L θ ˆθ + Lφ ˆφ (5.8) let s examine each component of L. From the M s stipulation, L θ = which implies L θ = (5.9) From (5.17), (5.) and (5.9) it is possible to deduce an element factor as jke jkr 4πr 13 cos θ cos φ

14 and the E θ portion of the electric field in (5.) is simply the product of the element factor and the space factor. To complete the integration in (5.9) we note that l/ l/ e jpz dz = Applying this to (5.9) readily yields for the x -integral a/ a/ E 0 ( e jkx sin θ cos φ ) dx = Similarly doing the y -integral, we get in total L θ = E 0 ab cos θ cos φ sin ( ) ka sin θ cos φ ) sin ( ) kb sin θ sin φ ) ( ka sin θ cos φ ( kb sin θ sin φ It is similarly straightforward to show that for the φ-component L φ = E 0 ab sin φ sin ( ) ka sin θ cos φ ) sin ( ) kb sin θ sin φ ) ( ka sin θ cos φ ( kb sin θ sin φ (5.30) (5.31) From equations (5.17), (5.18), (5.1), (5.), (5.30) and (5.31), on noting that A = 0 so that (5.19) and (5.0) 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 cos φ and Y = kb sin θ sin φ. From the cited equations, the following result: E r = 0 (5.3) { E θ = jωηf φ = jke jkr E 0 ab sin φ sin X } sin Y (5.33) πr X Y { E φ = jωηf θ = jke jkr E 0 ab cos θ cos φ sin X } sin Y. (5.34) πr X Y Also, for the far field where H = ˆr η E = ˆr η ( E θ ˆθ + Eφ ˆφ) 14

15 H r = 0 (5.35) H θ = E φ η Principal Plane Patterns for this Example: See Section.3.1 for definitions. (5.36) H φ = E θ η. (5.37) In view of the definitions, it may be observed that the principal E-plane occurs in the plane φ = π/ (i.e., the y-z plane). We note the following: (1) the source field is in the y-direction; () sin φ φ=π/ = 1; sin X (3) lim = 1 when φ π/ and θ 0; X 0 X sin Y (4) lim = 1 when θ 0; Y 0 Y Under these conditions, E φ 0 since cos(π/) = 0. Therefore, the principal E-plane field is given by E θ = jke jkr E 0 ab πr [ sin kb sin θ kb sin θ ] (5.38) The principal H-plane field is perpendicular to the E-plane, in this case in the φ = 0 plane. Then, E r = E θ = 0 and the principal H-plane E-field pattern is easily seen to be E φ = jke jkr E 0 ab πr [ sin ka sin θ ka sin θ ] (5.39) 15

16 5.1.5 Application The Parabolic Reflector (One Approximate Example) The parabolic reflector is very useful and is widely used in microwave communications systems (microwave includes VHF, UHF, SHF, EHF, 100 MHz 10 1 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. 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 dimensions 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; 16

17 . 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 E a = E 0ˆx. Illustration: As before, in the far field, for amplitude considerations, R = r but for phase purposes, R = r r cos ψ. 17

18 For use in equations (5.16) and (5.18) we note that r cos ψ = r ˆr = x sin θ cos φ + y sin θ sin φ similarly as in equation (5.6). Given that x = r cos φ and y = r sin φ, it is clear that r cos ψ = sin θ [r cos φ cos φ + r sin φ sin φ] or r cos ψ = r sin θ cos(φ φ ). (5.40) The equivalent electric and magnetic surface current densities in the aperture are, from (5.3) and (5.4), Therefore, J s = (5.41) M s = (5.4) 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), (5.4) and (5.43) in equation (5.18), it is easy to verify that N = (5.44) L = (5.45) 18

19 where C = The value of this integral is C = (5.46) where J 1 ( ) is a first-order Bessel function of the first kind. Again, remember that these radiation integrals are for the far field where the ˆr components do not appear i.e. E r = 0 and H r = 0. From equations (5.15), (5.17), (5.19), (5.), (5.44) and (5.45) it is easy to verify (DO THIS) that E θ = (5.47) while E φ = (5.48) The H-field components can be also gotten using the radiation integrals and corresponding equations or since E, H, and ˆr, are mutually perpendicular in the far field (i.e. E and H are perpendicular to each other and to the direction of propagation) H θ = (5.49) while H φ = (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 Do this. U(θ, φ) = r η E. 19

20 5. 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 ˆr direction and its wave vector is defined by k = kˆr = ; k = k = π/λ. (5.51). Aspatial Fourier transform pair may be DEFINED as f(x) = 1 F(k x )e jxkx dk x (5.5) π F(k x ) = f(x)e +jxkx dx (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 F(k 4π x, k y )e j(kxx+kyy) dk x dk y (5.54) F(k x, k y ) = 0 f(x, y)e +j(kxx+kyy) dxdy (5.55)

21 The Problem: Consider as before a rectangular aperture mounted on an infinite ground plane. Let the electric field in the aperture be E a (x, y, z = 0) = and let the spatial transform in the z = 0 plane be given by E(k x, k y ) = Then, since the aperture field exists only where x a/ and y b/, we may write using (5.55) E ax (k x, k y ) = = (5.56) Note that k r ˆr = with k = k x + k y + k z. Similar expressions for E a y and E az may of course be found by using the appropriate component of E a in (5.56). 1

22 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 E r = 0. Notice that E az has been eliminated from explicit appearance in the answer. Equation (5.57) is very powerful. It indicates that if the aperture field E a is known, the the far-field E may be determinde via the (inverse) Fourier transform of E a, component by component. Example: Consider the earlier problem in which the aperture field is given by E a = { E0 ŷ, a/ x a/ 0, otherwise. Let s again determine the far-field electric field E = E θ ˆθ + Eφ ˆφ. We note that since the aperture field has only a ŷ component, (5.56) implies E ax = E az = 0. Also, k( r ˆr) = = Therefore, from (5.56), on using E ay instead of E ax, we have E ay = Clearly, these integrals look like those which led to (5.30) etc. and we write immediately that E ay = where X = ; Y =

23 Thus, from (5.57), since E ax = 0, E θ = and E φ = As before, for the far field, E r = 0. Clearly, the above solution is identical to that obtained in (5.33) and (5.34) using the field equivalence principle. 3