Plane wave solutions for right-angled interior impedance wedges

Authors


Abstract

[1] We consider the reflection of a plane electromagnetic wave incident obliquely in a right-angled corner region with impedance walls. The surface impedances are taken in their most general tensor form. We determine the conditions under which the sum of incident, singly and doubly reflected waves provides an exact solution of the problem and report several new explicitly solvable cases.

1. Introduction

[2] The impedance boundary condition [Senior and Volakis, 1995] is a convenient tool for simulating the material properties of a surface. For most materials the surface impedance is a scalar, but there are materials whose properties are anisotropic for which a tensor impedance is required. The simplest cases are those for which the tensors are diagonal but recent work with metamaterials has made possible the creation of very general materials for which the tensor may be nondiagonal. These are the focus of the present study.

[3] The problem considered is a plane electromagnetic wave incident on the interior of a right-angled impedance wedge. This is a geometry that is relevant to the analysis of finitely conducting waveguides and resonators as well as to the propagation of radio waves inside buildings. In general the solution consists of plane waves reflected off the two faces of the wedge and a diffracted field associated with the vertex, but if the surface impedances satisfy certain restrictions the diffracted field disappears. The exact solution is then the sum of four plane waves, one of which is the incident field. We seek the restrictions on the impedances for which this is so, and follow a procedure similar to that by Senior [1978].

2. Formulation

[4] The geometry is shown in Figure 1. The tensor impedance boundary condition is

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where equation image is the normalized tensor surface impedance, equation image is the outward unit vector normal to the surface and Z is the intrinsic impedance of the free space. On the horizontal surface (y = 0) the tensor impedance is

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and the boundary conditions derived from (1) are

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Figure 1.

Geometry.

[5] Similarly, on the vertical face (x = 0),

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and

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For the boundary conditions to ensure a unique solution it is necessary that Reη1 ≥ 0, Reη4 ≥ 0 and 4Reη1Reη4 ≥ ∣η2 + η3*∣2 where the asterisk denotes the complex conjugate, with similar restrictions on the primed quantities [see Senior and Volakis, 1995, p. 43].

[6] In terms of the single component Hertz vectors

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where a time factor exp(−iωt) has been assumed and suppressed, we have

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where h = k cosβ and λ = k sinβ so that

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Substituting (5) into (3), the boundary conditions on y = 0 can be written as

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and similarly the boundary conditions on x = 0 are

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[7] The incident field is a plane wave with z components

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and direction of propagation

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for 0 ≤ απ/2. We seek a plane wave solution

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where, for brevity, c = cosα and s = sinα. The second and third terms in the right-hand side of (10) represent the single reflections of the incident plane wave (first term) off the two faces of the wedge and the fourth term is the doubly reflected wave. When (10) is substituted into the boundary conditions (6) and (7) on y = 0, the coefficients of exp(±iλxc) give

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and from the boundary conditions (8) and (9) we obtain

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[8] Equations (11) and (13) can be used to express A1 and B1 in terms of A and B. The elimination of B1 gives

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where

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with

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and similarly, by eliminating A1, we find

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We can also use (12) and (14) to express A3 and B3 in terms of A2 and B2, and since the equations differ from (11) and (13) only in the sign of h, we have

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Taking next the 4 equations derived from the boundary conditions on x = 0, (15) and (17) specify A2 and B2 in terms of A and B as

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where

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with

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and from (16) and (18)

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[9] We now have 8 equations linking the 4 pairs of coefficients A, B, A1, B1, A2, B2 and A3, B3. By substituting (23) and (24) into (21) we can express A3 in terms of A and B as

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and the substitution of (19) and (20) into (25) gives

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[10] The analogous expressions for B3 are

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[11] If the field inside the wedge is to consist of the 4 plane waves shown in (10), the above expressions for A3 must be identical, as must those for B3. We now seek the restrictions on equation image and equation image′ to make this so, and start with the simple case of normal incidence.

3. Normal Incidence

[12] If the plane wave is incident in a plane perpendicular to the edge, β = π/2 implying h = 0 and therefore λ = k. Then

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and (26) and (27) become

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[13] Similarly, from (28) and (29),

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[14] Comparing (30) and (31), the coefficients of A are the same if

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and those of B are the same if

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[15] For (32) and (33) the coefficients of A are the same if

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and (34) is sufficient to make the coefficients of B identical. Hence for a plane wave having any angle of incidence α, i.e., arbitrary c and s, the requirements for the existence of a plane wave solution are

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[16] In view of (35) the condition (36) implies (37) and vice versa, so that (35) and (36) are sufficient. We now have two relations connecting the 8 quantities η1,2,3,4 and η1,2,3,4. This allows us to freely choose 6 of them, e.g., η1,2,3,4 and η1,2, with the other two specified as

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[17] A few special cases are worthy of note. If the impedance tensors are diagonal (η2 = η3 = η2′ = η3′ = 0), the problem is easily solved using Maliuzhinets' technique [Maliuzhinets, 1958], and since the angular spectra are 2π-periodic functions, the diffracted field is clearly zero. The conditions are also satisfied by polarization-independent surfaces whose impedances are such that η2 = η3 and det equation image = 1, and if η2 = η3 = 0 with det equation image = 1, a plane wave solutions exists regardless of the impedance of the other face.

[18] The plots in Figures 2 and 3 are sample distributions of total electric and magnetic fields in a right-angled interior wedge with tensor impedances compliant with (35) and (36). The patterns in the field distributions are explained by the analytical structure of expression (10), which is a continuous function of x and y whose magnitude is periodic in the x and y directions with periods δx = λ0/(2cosα sinβ) and δy = λ0/(2sinα sinβ), respectively (λ0 is the wavelength).

Figure 2.

Total field distribution ∣Ez(x, y)∣ for an E-polarized plane wave of unit amplitude (∣Ezinc∣ = 1, ∣Hzinc∣ = 0) incident from direction α = 3π/8, β = π/2 in a sector 0 ≤ x, y ≤ 4λ0 with η1 = 2, η2 = 0.5, η3 = −0.5, η4 = 0.5, η1′ = 1, η2′ = i, η3′ = −i, η4′ = 2 − 0.5i. The field level varies between the following minimum (black) and maximum (white) values: min ∣Ez∣ = 0.222, max ∣Ez∣ = 2.169.

Figure 3.

Same as in Figure 2 but for ∣ZHz(x, y)∣, min ∣ZHz∣ = 0.003, max ∣ZHz∣ = 0.610.

4. Arbitrary Incidence

[19] This is the most general case and the analysis is more laborious, primarily because Γ1 and Γ2 differ in the sign of η2 + η3 = Δ (say), as do Γ1′ and Γ2′ in the sign of η2′ + η3′ = Δ′.

[20] If the expressions for A3 in (26) and (27) are labeled A3(1) and A3(2) respectively, we have

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where

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[21] When multiplied out this is found to be the sum of sixth power polynomials in s multiplied by s and c, and while the coefficients of the highest power turn out to be zero, the coefficients of the next (and all subsequent) powers vanish only if Δ = Δ′ = 0.

[22] This greatly simplifies the analysis. Since Γ2 = Γ1 and Γ′2 = Γ′1 the coefficients of A in the expressions for A3 are clearly equal, and from (26), (27) and (38),

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which vanishes for all angles of incidence if (36) is satisfied and

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[23] This is the standard impedance compatibility condition [Dybdal et al., 1971]. In the case of B3 the coefficients of B are equal and those of A agree if (37) and (39) are satisfied.

[24] Thus, for arbitrary angles of incidence, the requirements for a plane wave solution are

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plus (36), (37) and (39). We note that (40) implies (35) but not the reverse and that (36) in conjunction with (40) implies (37). A sufficient set of conditions is therefore (36), (39) and (40) and this allows the free choice of 4 elements of the impedance tensors. If, for example, we choose η1, η2, η4 and η2′, the remaining elements are

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[25] Either sign of the square root is permissible subject, of course, to the requirements of physical realizability [Senior and Volakis, 1995].

[26] Figures 4 and 5 present sample field distributions for a configuration with tensor impedances which are in agreement with conditions (36), (39), and (40).

Figure 4.

Same as in Figure 2 but for α = π/4, β = π/4, η1 = 2, η2 = 0.5i, η3 = −0.5i, η4 = 0.5, η1 = equation image, η2 = 0.5i, η3 = −0.5i, η4 = 1.5/equation image. The field level is such that 0.726 ≤ ∣Ez∣ ≤ 1.299.

Figure 5.

Same as in Figure 4 but for ∣ZHz(x, y)∣ with 0.107 ≤ ∣ZHz∣ ≤ 1.035.

[27] A case of special interest is a wedge having one side, e.g., the vertical, perfectly conducting (η1 = η2 = η3 = η4 = 0) with the other having a diagonal impedance tensor (η2 = η3 = 0). A plane wave solution then exists for arbitrary η1 and η4.

5. Conclusions

[28] For a plane wave incident on the interior of a right-angled wedge with impedance walls we have established the conditions for the existence of a solution consisting only of four plane waves without any diffracted field. In general, a diffracted field is necessary to compensate for the discontinuity in the plane waves doubly reflected off the vertical and horizontal and horizontal and vertical faces of the wedge. The discontinuity exists across the plane ϕ (= arctan y/x) = α, but when the conditions that we have derived are satisfied, there is no discontinuity and, hence, no diffracted field. The results are an extension of those previously found [Senior, 1978] for diagonal impedance tensors.

[29] An immediate consequence is that in a rectangular waveguide whose neighboring walls have the surface impedances (2) and (4), the conditions (36), (39) and (40) are necessary to ensure the existence of a separable modal solution. In other words, the conditions guarantee that the modal spectrum is discrete with no continuous component.

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