Radio Science

Diffraction of a plane wave by a right-angled penetrable wedge

Authors


Abstract

[1] The problem of electromagnetic diffraction (E-polarization) by two screens is analyzed. The screens are half-planes which form a right-angled wedge. The first half-plane is an electrically resistive sheet, and the second one is a perfectly magnetically conductive surface (ideal ferrite). The problem is formulated as a boundary-value problem for the Helmholtz equation with respect to the Ez-component of the electric field. On the conductive screen, the normal derivative of the function Ez vanishes. On the resistive half-plane, the function Ez is continuous and it is proportional to the jump of its normal derivative. The Sommerfeld integral representation is used to convert the problem to a difference equation of the second order. For a special value of the impedance parameter the problem reduces to two scalar Riemann-Hilbert (RH) problems on a segment with coefficients having a pole and a zero on the segment. The general solution to the RH problems is derived by quadratures. The RH problems are equivalent to the governing boundary-value problem when certain conditions are satisfied. These conditions are used to determine unknown meromorphic functions in the solution of the RH problems. Exact formulas for the reflected, transmitted, and diffracted waves are derived, and numerical results for the diffraction coefficient are reported.

1. Introduction

[2] The geometrical theory of diffraction (GTD) is the most widely used theory for analysis of the edge diffraction. The GTD in the context of the time harmonic theory [Keller, 1962] deals with high-frequency asymptotic expansions of electromagnetic fields scattered from structures. It requires the solution of canonical problems on scattering of radio waves in order to recover transmitted, surface, and diffracted waves or, equivalently, to evaluate transmission, surface and diffraction coefficients. High fidelity analytical methods are essential for solving the canonical problems. Solution of these problems by pure numerical schemes is not always simple, and it is often costly in computational time. Considerable insight to the problem can be obtained if the solution is carried out analytically.

[3] Reliable detection of targets, specification of antenna ground plates and radar scattering evaluation requires a further investigation into electromagnetic scattering by plates and wedges. The analysis of half-plane diffraction problems is generally carried out using the Wiener-Hopf technique or the Clemmow dual integral equations method [Senior and Volakis, 1995, chapter 3]. Model problems for the exterior of wedges with impedance faces are effectively treated by the Maliuzhinets [1958] technique. This method is based on the use of the classical Sommerfeld integral representation of the solution to the Helmholtz equation and requires the solution of certain difference equations. When the electric and magnetic fields are not coupled by the boundary conditions and the corresponding difference equations are scalar, the solution can be derived in terms of special functions (known as the Maliuzhinets functions). For penetrative wedges, when the electromagnetic field has to be recovered in the interior and the exterior of the domain, in general, the difference equations are of the second order. For an arbitrary wedge angle, an analytical technique is not available in the literature. For a special case of the second-order difference equation, by using bilinear Riemann relations for abelian differentials, a partial solution was analyzed by Senior and Legault [2000] and Legault and Senior [2002]. A multi-valued non-physical solution for some canonical diffraction problems on right-angled scatterers was proposed by Demetrescu et al. [1998a, 1998b].

[4] Recently, Antipov and Silvestrov [2004a, 2004b] developed a novel method for second-order difference equations with meromorphic periodic coefficients and applied it to a diffraction problem for a right-angled wedge. One of the sheets of the scatterer was a conductive surface, and the second one was perfectly conductive. This method is a two-step-procedure which requires first to reduce the model problem to a scalar RH problem on a Riemann surface [Antipov and Silvestrov, 2004c] (in particular cases, this results in two scalar RH problems on a complex plane). The general single-valued meromorphic solution to the RH problems is derived in terms of 2π-periodic meromorphic functions with specified zeros and poles. These functions are arbitrary for the solution of the RH problem but not free for the main difference equation. The second step of the method is to find these functions from certain extra conditions. In the case considered by Antipov and Silvestrov [2004b], the coefficients l1(t) and l2(t) of the RH problems are continuous on the contour and do not vanish. It turns out that for some boundary conditions the coefficients of the associated RH problems may have poles and zeros on the contour and therefore the functions log lj(t) are multi-valued. The procedure by Antipov and Silvestrov [2004b] if remains unchanged does not work for this multi-valued case. The main aims of the current paper are as follows: (1) to analyze the diffraction problem for a right-angled scatterer formed by an electrically resistive half-plane and a perfectly magnetically conductive half-plane; (2) to develop further the theory of second-order difference equations of diffraction theory focusing on the case when the coefficients of the associated RH problems have zeros and poles on the contour; (3) to determine the far field asymptotic expansion of the electric field.

[5] It is known [Senior and Volakis, 1995, p. 53] that for an electrically resistive half-plane, the Ez-component of the electric field is continuous, and the Hx-component of the magnetic field is discontinuous:

equation image

where Re denotes the surface resistivity, and [f]+ = fy=+0fy=−0. The electromagnetic dual of an electrically resistive screen is a magnetically conductive one. The magnetically conductive sheet boundary conditions stipulate the continuity of the Hx-component and the discontinuity of the Ez-component [Senior and Volakis, 1995, p. 74]:

equation image

where Rm is the conductivity. In the case of normal incidence (the incident wave Ezi is orthogonal to the z-axis), the Ez- and Hx-components are linked by

equation image

where k0 is the wave number, and Z0 is the intrinsic impedance of the medium. Therefore, if Re = 0 (a perfectly electrically conductive sheet), then the Ez-component is continuous and vanishes on the faces of the screen while its normal derivative is discontinuous. In the case Rm = 0 (a perfectly magnetically conductive sheet), the function ∂Ez/∂y is continuous and equal to zero, while the Ez-component is discontinuous. If Re = ∞ in the case (1) or Rm = ∞ in the case (2), then the sheet ceases to exist.

[6] The paper is organized as follows. In section 2, the canonical problem of diffraction by a right-angled wedge with electrically resistive and perfectly magnetically conductive surfaces is formulated. It is reduced to a certain difference equation in section 3. In the next section, to solve this difference equation we analyze an auxiliary second-order difference equation and convert it into two RH problems. For the particular case Z0/Re = 4/equation image, the general solution to the auxiliary difference equation is derived in terms of some unknown meromorphic periodic functions. In section 5, we establish under which conditions the auxiliary equation is equivalent to the main difference equation. These conditions are used to find the unknown meromorphic functions from section 4. An asymptotic expansion of the Ez-component of the electric field far away from the junction of the screens is derived in section 6. Numerical results for the diffraction coefficient are also reported.

2. Formulation

[7] Consider diffraction of a time-harmonic plane wave of unit amplitude,

equation image

by two half-planes, W1 = {0 < r < ∞, equation image = 3π/4} and W2 = {0 < r < ∞, ϕ = 5π/4} in a medium {0 < r < ∞, −3π/4 < ϕ < 5π/4} with permittivity ɛ, permeability μ, and intrinsic impedance Z0. The incident wave is normal to the z-axis and is traveling at an angle ϕ0 (∣ϕ0∣ < 3π/4) to the plane ϕ = 0. Here k0 is the wave number and ω = k0/equation image. The time factor eiωt will be suppressed henceforth. It is assumed that the first screen W1 is an electrically resistive surface, and the second screen W2 is a perfectly magnetically conductive sheet (Figure 1).

Figure 1.

Geometry of the problem.

[8] By expressing the component Hr of the magnetic field through the component Ez of the electric field,

equation image

we can transform the electrically resistive boundary conditions (1) on the sheet W1 to the form

equation image

Here γ = Z0/(2Re) ≠ 0 and Re is the surface resistivity. It will be convenient to represent the complex parameter γ as γ = sinθ, 0 < Re θ < π/2.

[9] On the second sheet W2, a perfectly magnetically conductive surface, the parameter Rm vanishes, and the boundary conditions (2) become

equation image

Thus, the canonical diffraction problem reduces to the Helmholtz equation in the free space for the Ez-component of the electric field

equation image

which has to satisfy the boundary conditions (6) and (7) on the lines ϕ = 3π/4 and ϕ = 5π/4, respectively.

3. Main Difference Equation

[10] To solve the problem, we use the Sommerfeld integral representations

equation image

which satisfy the Helmholtz equation. Here ℒ is the Sommerfeld double loop contour (Figure 2). It is symmetric with respect to the origin s = 0. For large positive Im s, the upper contour asymptotically approaches the lines s = 3π/2 (from the left) and s = −π/2 (from the right), whilst the lower contour approaches the lines s = π/2 and s = −3π/2. The function &#55349;&#56494;1(s) is analytic in the strip ∣Re s∣ < 3π/4 and continuous in the strip up to the boundary ∣Re s∣ = 3π/4 apart from the geometrical optics pole at the point s = ϕ0 with the prescribed residue

equation image

The second spectral function &#55349;&#56494;2(s) is analytic in the strip ∣Re s∣ < π/4 and continuous everywhere in the strip ∣Re s∣ ≤ π/4. At infinity, as Im s → ±∞ and Re s is finite, both functions are assumed to be bounded: ∣&#55349;&#56494;j(s)∣ ≤ const, j = 1, 2.

Figure 2.

Sommerfeld double loop contour ℒ (the broken lines) and the two steepest descent paths (the solid lines).

[11] To determine the functions &#55349;&#56494;1(s) and &#55349;&#56494;2(s) and to satisfy the boundary conditions, we substitute the integral representations (9) into the boundary conditions (6) and (7). Then, taking into account the symmetry of the contour ℒ with respect to the origin, we arrive at the following system of difference equations

equation image

By using the first two equations in (11), we find

equation image

Then we transform the third and the forth equations in (11) to new ones with respect to the terms &#55349;&#56494;1(s + 3π/4), &#55349;&#56494;1(s − 9π/4), &#55349;&#56494;2(s + 3π/4), and &#55349;&#56494;2(sπ/4). It will be helpful to express the function &#55349;&#56494;2(sπ/4) through &#55349;&#56494;1(s + 3π/4) and &#55349;&#56494;1(s − 9π/4). This can be done by eliminating the function &#55349;&#56494;2(s + 3π/4) from those two new equations. After the substitution sπ/4 = s′, we have

equation image

Thus, the function &#55349;&#56494;2(s) can be excluded from (11). By doing this, we derive a single difference equation for the function &#55349;&#56494;1

equation image

By following Maliuzhinets [1958], in order to eliminate the pole at the point s = ϕ0, we split the function &#55349;&#56494;1(s) into two parts

equation image

where the function ψ(s) is free of the geometrical optics pole of the function &#55349;&#56494;1(s). Equation (14) is a difference equation of order 3. To reduce it to a new one of order 2, we introduce a function

equation image

and transform equation (14) to the form

equation image

where sin θ = γ. It follows from (15), (12), and (16) that

equation image

At infinity, the function f(s) may grow:

equation image

[12] It is clear that in order to recover the function &#55349;&#56494;1(s), it is required to express the function ψ(s) through the solution to equation (17). From (16), we find the most general form of the function ψ(s)

equation image

Here C1 and C2 are constants to be fixed, Ω = {sCs: Re s = π}, Cs is a complex plane, and s0 is an arbitrary fixed internal point in the strip Π = {sCs: −π < Re s < π} such that Re s0 ≠ 0. The choice of the kernel guarantees the convergence of the integral (20) with the density growing at infinity as in (19). The values of the function ψ(s) as Re s = ±π are recovered by the Sokhotski-Plemelj formulas for the Hilbert kernel (a periodic analogue of the Cauchy kernel)

equation image

where the integral in (21) is understood in the sense of the principal value.

4. Auxiliary Second-Order Difference Equation

[13] The main step of the method is to determine the function f(s). If this function is known, then the Ez-component of the electric field is found by formulas (9), (13), (15), and (20). The function f(s) satisfies the difference equation (17). A second-order difference equation

equation image

with meromorphic T-periodic coefficients a(s), b(s), and c(s) can be reduced to a scalar Riemann-Hilbert problem on a Riemann surface [Antipov and Silvestrov, 2004a] if h = mT, where m is an integer. In certain cases, the genus of the surface equals zero, and the equation is equivalent to two scalar Riemann-Hilbert problems on a complex plane. Clearly, in the case of equation (17), h = π, T = 2π, and the condition h = mT is not valid (m is not an integer in this case). To transform the equation to another one which meets the condition h = mT with m being an integer, replace first s by s + π and then s by sπ, sum the new equations and finally obtain

equation image

Now we have T = π, h = 2π, m = 2, and the technique by Antipov and Silvestrov [2004a] may be applied to equation (23). Obviously, the main equation (17) and the new one (23) are not equivalent. Not each solution to equation (23) satisfies equation (17). However, each solution to equation (17) is a solution to equation (23). That is why it is crucial to separate the functions which do not satisfy equation (17) from the general solution to the auxiliary equation (23).

[14] First we need to construct the general solution to the auxiliary equation (23). By following Antipov and Silvestrov [2004a] we introduce two new functions

equation image

and rewrite equation (23) as a system of two difference equations of the first order

equation image

where

equation image

In general, the eigenvalues of the matrix G(s) are two-valued functions

equation image

where

equation image

In what follows we confine ourselves to considering the case when the function Δ1/2(s) does not have branch points, namely when sinequation image = 2/equation image (equation image = π/2 ± ilog equation image). In this case, the eigenvalues become single-valued

equation image

and the problem (25) can be decoupled

equation image

by using the following single-valued transformation matrix

equation image

The next step of the solution procedure is to introduce a new vector ϕ(s) = [T(s)]−1Φ(s), s ∈ Π. The components of the vector, the functions ϕ1(s) and ϕ2(s), form the general solution of the two scalar first-order difference equations

equation image

These two equations can be transformed into two scalar Riemann-Hilbert problems

equation image

by the map

equation image

Here Cz is a complex plane,

equation image

a = 2 − equation image, b = 1/a = 2 + equation image, and t = z(σ). The function z = icot equation images maps the strip Π into the z-plane cut along the segment [−1, 1]. The upper side of the cut corresponds to the contour Ω, and the lower bank z = xi0 (−1 ≤ x ≤ 1) corresponds to the left boundary Re s = −π of the strip.

[15] To factorize the function l1(t), fix its argument by the condition arg l1(t) = 0 as t > b. Because the points t = −a and t = b are poles and t = a and t = −b are zeros of the function l1(t) we have

equation image

Consequently,

equation image

Observe now that l1(t)l1(−t) = 1, t ∈ [−1, 1] and establish the following properties

equation image

In view of these relations it can be shown that

equation image

where Γ is the upper side of the cut [−1, 1], and Γ0 = {zCz: z = x + i0, 0 ≤ x ≤ 1}. Therefore, the functions

equation image

are the canonical functions to the Riemann-Hilbert problems (33). They are even and factorize the functions lj(t):

equation image

The general solution to the auxiliary difference equation (23) can be written in terms of the canonical functions in the form

equation image

where Φj0(s) = Fj0(z). The functions Pj(s) (j = 1, 2) are even 2π-periodic meromorphic functions.

5. Exact Solution to the Diffraction Problem

[16] In what follows we find the most general form of the functions P1(s) and P2(s) for which the function (42) is the general solution to the main difference equation (17). Afterwards we determine the functions &#55349;&#56494;1(s) and &#55349;&#56494;2(s) and therefore complete the solution procedure for the diffraction problem (8), (6), and (7).

5.1. Connection Between the Functions P1(s) and P2(s)

[17] Substitute the expression (42) into (17)

equation image

Let Re s = 0. The function z = icot equation images maps the imaginary axis into the two segments (−∞, −1) ∪ (1, ∞), while the points s ± π fall into the points (1/x)± = 1/x ± i0, 1/x ∈ [−1, 1]. Represent next the functions Φ10(s) and Φ10(s ± π) (Re s = 0) in the form

equation image

Here we used the Sokhotski-Plemelj formulas. The integral χ*(x) is understood in the sense of the principal value. Establish now a connection between χ1(x) and χ*(x). It is directly verified that

equation image

and since log l1(±1) = −πi we have

equation image

By making the substitution t → 1/t we find

equation image

where

equation image

Notice that log l1(t)/t is bounded at the point t = 0, and Re C = 0. By using now the Cauchy theorem

equation image

and comparing (44), (47), and (49) we express the function χ*(x) through the function χ1(x)

equation image

On using (44) and (50), we observe that the boundary values Φj0(s ± π), j = 1, 2, have the following representations

equation image

which are required in the sequel. Therefore,

equation image

Substitute now these expressions into equation (43)

equation image

Because of the linear independence of the functions Φ10(s) and Φ20(s), on one hand,

equation image

On the other hand,

equation image

Replace s by s + π. Then x will be replaced by 1/x, and we obtain

equation image

Since x = icot equation images, we have the relations

equation image

and it follows that B(s) = A(s) as Re s = 0. By principle of analytical continuation, this is valid everywhere in the strip Π, and therefore, the general solution (42) becomes

equation image

5.2. Function P1(s)

[18] The determination of the function P1(s) requires a further study into the properties of the functions Φ10(s) and f(s). It will be helpful to rewrite formulas (51) for Φ10(s ± π) as follows

equation image

where x = icot equation images, x is real, x > 1, and Re s = 0. Clearly, if s = ±ilog equation image, then x = ±b. Therefore, the function Φ(σ) has simple zeros at the points σ = π + ilog equation image and σ = −πilog equation image. At all other finite points of the strip Π including the points σ = πilog equation image and σ = −π + ilog equation image, it is bounded and nonzero. Now, because of the factor 2 + equation image cos s, the second term in (58) is bounded and nonzero everywhere in the strip Π apart from the points s = πilog equation image and s = −π + ilog equation image which are simple zeros of the function (2 + equation image cos s)/Φ10(s).

[19] Show next that the 2π-periodic meromorphic function P1(s) may grow as exp{∣Im s∣} when Im s → ±∞. Indeed, since log l1(±1 + i0) = −log l1(±1 − i0) = −π, by analyzing the Cauchy integral (39) at the points z = ±1, we find F10(z) = O((1 ∓ z)−1/2) as z → ±1. Correspondingly, at the infinite points of the strip,

equation image

Check now the asymptotics of the function equation image(s) at infinity. By using the connection between the Cauchy and Hilbert kernels

equation image

we have

equation image

where equation image(t) = f(ilog{(1 + t)/(1 − t)}). Since equation image(t) = O((t ± 1)−3/2), t → ∓1, the function equation image(s) grows as e3/2∣Im s as ∣Im s∣ → ∞. This means that the function ψ(s) given by (20) grows at infinity as e2∣Im s as it must be. Therefore, the most general form of the even 2π-periodic meromorphic function P1(s) which grows at infinity as e∣Im s and which has simple poles at the zeros of the function Φ10(s) is given by

equation image

and from (58) we have

equation image

Clearly, for arbitrary constants E0, E1, and E2, the function f(s) has inadmissible poles at the points s = ±(πilog equation image) (the first term) and s = ±ilog equation image (the second term). To remove them, we require

equation image

This condition reduces formula (63) to the form P1(s) = D0 + D1 cos s, where D0 = E1/equation image − 4E2/3 and D1 = 2E2/equation image are free constants. This simplifies the expression for the function f(s)

equation image

where

equation image

Formula (66) represents the general solution to the main difference equation (17). It is expressed through one quadrature χ1(z) given by (44) (C = −χ1(0)). The function ψ(s) and therefore the spectral functions &#55349;&#56494;1(s) and &#55349;&#56494;2(s) have 4 unknown constants C1, C2, D0, and D1. They will be fixed next.

5.3. Definition of the Constants

[20] In general, the spectral functions may have some inadmissible poles which should be eliminated by fixing the free constants. The function &#55349;&#56494;1(s) must be analytic everywhere in the strip −3π/4 ≤ Re s ≤ 3π/4 apart from the point s = ϕ0, where it has to have a simple pole to reproduce the incident field (4). The function &#55349;&#56494;2(s) must be free of poles everywhere in the strip −π/4 ≤ Re sπ/4. To analyze the functions &#55349;&#56494;1(s) and &#55349;&#56494;2(s), express them through the functions f(s) and ψ(s). To do this we continue analytically the function ψ(s) into the strips −2π ≤ Re s ≤ −π and π ≤ Re s ≤ 2π

equation image

and find then

equation image

Notice that the point s = −π/4 is a removable singularity of the function &#55349;&#56494;2(s). This follows from (18). Analysis of formulas (69) [see also Antipov and Silvestrov, 2004b] shows that the function &#55349;&#56494;1(s) has two simple poles at the points s = α1 and s = α2, and the function &#55349;&#56494;2(s) has one pole at the point s = α3:

equation image

All these poles are inadmissible singularities and should be removed. Thus, we have four additional conditions

equation image

By evaluating these residues we rewrite the conditions (71) in terms of the functions f(s) and ψ(s)

equation image

where

equation image

The requirements (72) when satisfied give the following values for the free constants:

equation image

where

equation image

ν = 1 for −3π/4 ≤ ϕ0π/4, and ν = 2 for π/4 ≤ ϕ0 ≤ 3π/4.

6. Reflected, Transmitted, Surface, and Diffracted Waves

[21] In this section we will derive the asymptotic expansion for the far field. For large k0r, the Ez-component of the electric field can be represented as follows

equation image

where Ezi, Ezr, Ezt, Ezs, and Ezd are the incident, reflected, transmitted, surface, and diffracted waves, respectively. By using the method of steepest descent we deform the contour ℒ into another one consisting of two steepest descent paths (Figure 2). The right-hand path is given by Re s = π + gd(Im s)sgn Im s, where gd x = arccos(1/cosh x) is the Gudermann function. This curve goes from the infinite point s = π/2 − i∞, crosses the real axis at the point s = π and then travels to the upper infinite point s = 3π/2 + i∞. The lines Re s = π/2 and Re s = 3π/2 are the asymptotes for the lower and the upper part of the path, respectively. The second path is symmetric to the first one with respect to the origin. To derive the expansion (76), we need to continue the function &#55349;&#56494;1(s) into the strip 5π/4 ≤ Re s ≤ 9π/4 and the function &#55349;&#56494;2(s) into the strips −7π/4 ≤ Re s ≤ −3π/4 and π/4 ≤ Re s ≤ 5π/4

equation image

Let

equation image

In view of formulas (69) and (77) we deduce the following result. Let first ∣ϕ∣ < 3π/4. If −3π/4 < ϕ0 < −π/4, then

equation image

In the case −π/4 < ϕ0 < π/4 the incident, transmitted and reflected waves are

equation image

In the last possible case π/4 < ϕ0 < 3π/4, the waves have the form

equation image

The reflected and transmitted waves in the sector 3π/4 < ϕ < 5π/4 are defined by the corresponding poles of the function &#55349;&#56494;2(s). The reflected and transmitted waves can be written straightforwardly (Ezi = 0)

equation image

where

equation image

[22] Check next if there are surface waves for −equation image < equation image < equation image. They may be produced by the zeros sm of the function sin(s + equation imageπ) − sinequation image, equation image = equation imageπ ± ilog equation image which satisfy the condition

equation image

where g0 = gd log equation image = equation image. Analysis of formulas (69) and (77) shows that all possible zeros (surface poles) of the function sin(s + equation imageπ) − sinθ in the strip (84) give ϕ which is outside the range (−equation image, equation image) that is impossible. Therefore, no surface waves are observed for ∣ϕ∣ < equation image. The use of a similar argument yields the same result for the second wedge ϕ ∈ (equation image, equation image), and we conclude that Ezs = 0 everywhere in the medium for the impedance parameter γ = 2/equation image.

[23] Finally, define the diffracted field. By applying the steepest descent method for k0r ≫ 1 we obtain

equation image

where

equation image

Numerical computations are implemented for the diffraction coefficient in the case γ = 2/equation image. The numerical algorithm is very simple and efficient. To evaluate the diffraction coefficient D, one needs to compute just two integral F10 and ℱ given by (40) and (62). Notice that these two integrals are convergent as improper ones. The density of the former integral has the logarithmic singularity at the point x = a, while the density for the integral ℱ has the square root singularities at the ends z = ±1. For integration, we used the Simpson formula for F10 and the Gauss formula for ℱ. Sample plots of the coefficients d1 and d2 against the angle of observation ϕ for the incident angles ϕ0 = π/3 and ϕ0 = −π/3 are given in Figures 3 and 4, respectively. We also present two graphs which show how the coefficients Re d1 and Re d2 depend on the angle of incidence. For Figure 5, the angle of diffraction is π/8 (the corresponding diffraction coefficient is d1) which is in the larger wedge, and for Figure 6, the point of observation is in the wedge 3π/4 < ϕ < 5π/4 (ϕ0 = 7π/8 and therefore the diffraction coefficient is d2). Notice that in the cases considered, the imaginary parts of the coefficients d1 and d2 if plotted together with Re d1 and Re d2 are negligible to be visible.

Figure 3.

The dependence of the diffraction coefficients d1 and d2 on the angle of observation ϕ ∈ (−3π/4, 5π/4) for ϕ0 = π/3.

Figure 4.

The dependence of the diffraction coefficients d1 and d2 on the angle of observation ϕ ∈ (−3π/4, 5π/4) for ϕ0 = −π/3.

Figure 5.

The dependence of the diffraction coefficient d1 on the angle of incident ϕ0 ∈ (−3π/4, 3π/4) for ϕ = π/8.

Figure 6.

The dependence of the diffraction coefficient d1 on the angle of incident ϕ0 ∈ (−3π/4, 3π/4) for ϕ = 7π/8.

7. Conclusion

[24] Extension of the Sommerfeld-Maliuzhinets theory for canonical problems of diffraction for penetrable wedges is an important and not trivial task. If a wedge is of an arbitrary aperture, this is still an open problem. The main contribution of this paper is the development of an exact method for a right-angled penetrable wedge with one of its walls being electrically resistive and the second one being perfectly magnetically conductive. In comparison with the previous work by Antipov and Silvestrov [2004b] dealing with another type of penetrable boundary conditions on a right-angled wedge, the main difficulty of the problem in this paper was poles and zeros of the coefficients of the associated Riemann-Hilbert problems. When the penetrable wall is magnetically conductive [Antipov and Silvestrov, 2004b], the corresponding coefficients are continuous and do not vanish on the contour of discontinuity of the Riemann-Hilbert problems. The presence of zeros and poles makes the level of mathematical difficulties high even in the particular case of the impedance parameter (γ = 2/equation image) when the main second-order difference equation can be reduced to a pair of first-order difference equations, and the transformation matrix is single-valued.

[25] In this paper, we have derived an exact solution to the diffraction problem for the case γ = 2/equation image. The solution has been presented in a simple form convenient for computation. To evaluate the diffracted waves in the far field expansion of the electric field, it is required to compute only two integrals. To find the numerical values of the Ez-component of the electric field, one needs to evaluate three integrals. It has been shown that no surface wave can be observed in the medium −3π/4 < ϕ < 5π/4. It has also been found that the diffraction coefficient is continuous through the electrically resistive wall but it is discontinuous and bounded through the perfectly magnetically conductive surface.

[26] In the general case of the impedance parameter, γ ≠ 2/equation image, the transformation matrix is two-valued, and the use of Riemann surfaces is unavoidable. Because of the zeros and poles of the coefficient, the technique by Antipov and Silvestrov [2004b] has to be modified accordingly. It turns out that the first step of the procedure, the solution of the auxiliary second-order difference equation can be implemented without essential difficulties. What is hard is to satisfy the conditions under which the auxiliary equation is equivalent to the main second-order difference equation and to find unknown periodic meromorphic functions on the Riemann surface. We hope to report a closed-form solution to the general case in the nearest future. The solution presented in this paper for a particular value of the impedance parameter provides a reliable test for the validation of future analytical solutions for the general case and also for pure numerical schemes for more general diffraction problems.

Acknowledgments

[27] This work was partly funded by Louisiana Board of Regents grant LEQSF(2005-07)-ENH-TR-09 and Russian Foundation for Basic Research through contract 07-01-00038. The authors are thankful to the referees for their comments.

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