# A Fourier Approach to Model Electromagnetic Fields Scattered by a Buried Rectangular Cavity

Authors: John L. Fleming; Jessica Moser
Journal: Mathematical Problems in Engineering
 (2009)

---

## Abstract

We consider the problem of a two-dimensional rectangular cavity in a PEC half plane covered by layers of material with uniform thickness. The rectangular geometry allows for an application of Fourier methods to solve the problem. The paper will also discuss how to compute the far field scattering once the solution is found. The fast mode matching approximation technique
developed by Morgan and Schwering will be applied to this layered cavity problem setting. Numerical results will show that using the Fourier solution on the layered problem combined with Morgan and Schwering's technique can produce good approximations while using much less computation time than would be required for the entire solution.


---

## Body

## 1. Introduction

The cavity problem has been extensively studied due to its importance in the computation of radar cross section. A two-dimensional cavity can be used to model long seems or cracks in metallic surfaces which can significantly contribute to the overall radar profile of large objects. The cavity problem has been solved with a variety of methods including integral equations, finite elements, and Fourier methods. These approaches are primarily applied to a problem with a material filled cavity in a PEC ground place opening into an empty half space.A neglected aspect of the problem is when the cavity is buried beneath a layered material. A two-dimensional cavity beneath a layered material can serve as model for seams and cracks in metallic surfaces which are covered by paint or materials applied during a manufacturing process. The material covering would mean that the cavities are invisible to a visual inspection. However, they may be revealed by understanding the scattering characteristics of the cavity when exposed to an electromagnetic field. The mathematical model can serve as a predictor of the scattering characteristics for use in nondestructive testing.We want to compute the fields in the finite cavity region shown in Figure1. The computation can be done with integral equation methods. The disadvantage of this approach is that a layered Green's function must be used which adds difficulty to the problem and brings up convergence issues [1]. A finite element approach can also be used but would require appropriate boundary conditions at the cavity opening [2]. A key contribution of this manuscript is to provide the understanding of the Fourier approach in the overlayer setting on which an appropriate boundary condition is based.Figure 1
Layered geometry.In addition, the Fourier approach has some unique advantages. It is the sensible and natural approach for a rectangular geometry. Further, it is the only approach which can employ Morgan and Schwering's mode matching method and improved mode matching methods to get very fast approximations for large cavities and high frequencies. Very large problems with high wave numbers will become computationally cumbersome for any technique. Using the fast mode matching approach opens the door to dealing with very large problems which could not be dealt with by other conventional approaches [3, 4].


## 2. Problem Setting

We consider a time harmonic plane wave with fixed frequencyω incident onto a ideal metal or a perfect electrical conductor (PEC) half plane which is covered by a material layer(s) of uniform thickness. Embedded in the PEC half space is a cavity with an opening to the upper half space. As indicated in Figure 1, the region above the material layer is empty space and will be denoted by region 0, the region inside the material layer will be denoted by region 1, while the cavity will be called region 2. The opening of the cavity is the interval [0,L] in the x-direction. The depth of the cavity is denoted by d and top opening of the cavity is located at y=0 while the bottom of the cavity is at y=-d. The cavity may be filled with a uniform material. The electric permittivities and magnetic permeabilities of the materials are denoted by ϵk and μk where k=0,1,2 represents the region. Multiple material layers can be considered for both the inside of the cavity as well as the material covering the PEC half space. The paper will deal exclusively with a single material in both cases. The generalization to multiple layered materials follows easily once a single material is understood.We will solve for the electric and magnetic fields denoted byE and H which satisfy Maxwell's equations [5]. In the  two-dimensional setting, Maxwell's equations can be separated into two fundamental polarizations [5]. In the transverse magnetic (TM) polarization the electric field has only a z component which will be computed. We denote the z component of E by u. The other polarization is the transverse electric (TE) where the magnetic field H has only a z component which is denoted by v. In Sections 5 and 6, we will solve for the unknown functions u and v inside the cavity region. In both cases u and v will satisfy a Helmholtz equation with appropriate boundary conditions. Once the solutions are found in region 2 we will show how to compute far field values.Note that in regions 0 and 1 the solutions can be decomposed into a superpositions of fields. We use a subscript to denote which region the field is contained. Thusuk and vk for k=0,1,2 are the solutions in each of the respective regions. In region 0, the fields u0 and v0 are expressed as a sum of an incoming plane waves u0i and v0i, outgoing plane waves u0r and v0r, and a scattered fields u0s and v0s. The incoming waves in region 0 are known and specified. The remaining fields will be computed in following sections below. In region 1, the fields u1 and v1 also consist of incoming plane waves u1i and v1i, outgoing plane waves u1r and v1r and scattered fields u1s and v0s all of which are unknown. When appropriate we will combine the incoming and outgoing plane waves into a single term ukir=uki+ukr and vkir=vki+vkr for k=0,1. Finally the fields u2 and v2 in the cavity region 2 will not be decomposed but solved as single fields.When the known incident plane wavesu0i and v0i come in from region 0 they interact with the material layer (region 1). The interaction produces reflected plane waves u0r and v0r as well as transmitted plane waves into region 1. The transmitted plane waves propagate into region 1 which produces plane waves propagating in the positive y-direction (u1r and v1r) and in the negative y-direction (u1r and v1r). While the plane waves u0i and v0i are given, the remainder of the plane waves must be computed. They are computed assuming that the cavity (region 2) is not present. Once the plane waves are completely known they are used as a source field which interacts with the cavity to produce scattered fields uks and vks with k=0,1 in regions 0 and 1 as well as transmitted fields u2 and v2 in region 2.Once the transmitted field in the cavity is computed we want to find far field values based on this solution. The difficulty here is that the cavity does not open to an empty half space. The upper half space is partially filled with material layer. The Fourier representations will allow the solution of the transmitted field in the cavity to be transferred to the upper half space and interpreted appropriately to allow far field computations.


## 3. Previous Work

There have been numerous techniques developed to account for the electromagnetic scattering from a cavity aperture in a ground plane, including Fourier transform, finite element, integral equations, cavity mode coupling and impedance boundary conditions. The solution of the cavity problem when buried underneath uniform layers of material has received much less attention.The Fourier transform technique was utilized by Park and Eom to examine the TE and TM scattering from an empty rectangular cavity embedded in an infinite ground plane [6, 7]. The method allowed for a closed form solution to be developed by approximating a series solution for the scattered field. It analyzed the effect that filling the rectangular cavity with a dielectric material has on the radar cross section. It did not account for material outside of the cavity. Other techniques have dealt with the same situation by employing Green's functions, integral equations and impedance boundary conditions [8, 9]. Since these methods tend to be computationally burdensome as the size of the cavity increases, Morgan and Schwering approached this problem by implementing a mode coupling method, and solved for a rapid approximation technique by extending the cavity walls using propagating waveguide modes [4]. Morgan and Schwering's approach was followed by an improved mode matching technique which employed an asymptotic expansion which converges as the frequency increases [3]. This paper presents a similar methodology to duplicate Morgan and Schwering's fast approximation technique for the cavity underneath the material layer.The finite element method has the advantage of being able to model arbitrary shaped cavities embedded in a ground plane. Analysis of the TM case of a material filled arbitrarily shaped cavity was analyzed by Wood using a set of scalar integral equations, where Howe's masters thesis expanded this to look at the results of the TE cases [10, 11]. An alternate solution to this problem was presented by Van and Wood which coupled the finite element method with Fourier transforms, expressing the results of both the TM and TE case [2]. While these papers provide efficient solutions for an arbitrarily shaped cavity, they are dependent on the ideal situation where the material layer is restricted to the cavity space below the half plane. Wood was able to account for this problem, and extended the research to consider the effect of over filling the cavity space with a dielectric material. This was accomplished by creating an artificial boundary condition on a semicircle inscribing the overfilled material region, and combining a hybrid finite element method with Fourier transforms to solve for the far field scattering [12]. This methodology fails when the entire surface is coated with a material since no semicircle can contain the entire material layer.The work proposed in this paper is an extension of previously published literature, where we are considering an embedded rectangular cavity, in which the entire surface is coated with a dielectric material. Understanding the problem with the restricted cavity shape allows for the boundary conditions to be appropriately defined so that future research can investigate the use of the finite element method to solve for the scattering based on any arbitrary cavity.


## 4. Focus of the Paper

In this paper we focus on modifying the standard rectangular cavity model to accommodate uniform material layers over the cavity opening. The modification of the Fourier approach has intrinsic value as a computational model of a buried rectangular cavity but it also is important since it can be applied to finite element formulation for electromagnetic scattering by nonrectangular buried cavities in future work.In addition to providing the layered modification to the standard Fourier approach the paper will also demonstrate that the fast approximation method used by Morgan and Schwering and extended by Bao and Zhang, works well with the modification [3, 4]. The fast approximations are very important for electrically large cavities which take a lot of computation time. Also, the fast approximation can be used for inverse and design problems where solutions must be computed many times over as parameters are determined in order optimize scattering characteristics.


## 5. Overview

The  two-dimensional cavity problem can be split into two fundamental polarizations. The transverse magnetic (TM) polarization whereE=(0,0,u) and the transverse electric (TE) polarization where H=(0,0,v). In both cases we find the z-component of the field in question as a solution. In general the fields satisfy Maxwell's equations, but these can easily be reduced to the scalar Helmholtz equation for both polarizations. The difference between the two polarizations is the boundary conditions which will be clarified in the following sections.Much of the calculations to follow rely on enforcing the continuity or boundary conditions. The continuity conditions for electromagnetic fields state that the tangential components of the electric and magnetic field are continuous at interfaces. Thus we have thatn×E0=n×E1 and n×H0=n×H1 where n is the normal vector. At the interfaces between regions 0 and 1 and between regions 1 and 2 we will have n=(0,1,0). Note that E0 and E1 are the electric fields in regions 0 and 1 while H0 and H1 are the magnetic fields in regions 0 and 1. The same continuity holds for the interface between regions 1 and 2 for the fields E1, E2, H1 and H2.Note that inside a PEC material we haveE=0. Therefore when enforcing the continuity conditions we get that E×n=0 at the boundary of any PEC material. For the TM polarization this means u=0 at the boundary of the PEC material. For the TE case we have by applying Maxwell's equations that (1/ϵ)(dv/dn)=0. In both cases we will solve for u2 and v2 in region 2. The appropriate boundary conditions at the walls of the cavity will be enforced along with the continuity conditions at the interfaces between regions 0 and 1 and regions 1 and 2.For the TM case the continuity conditions of the electric field simply states thatu0=u1 at y=a and u1=u2 at y=0. The magnetic continuity condition reduces to (1/μ0)(∂u0/∂y)=(1/μ1)(∂u1/∂y) at y=a and (1/μ1)(∂u1/∂y)=(1/μ2)(∂u2/∂y) at y=0. For the TE case the continuity conditions of the magnetic field simply states that v0=v1 at y=a and v1=v2 at y=0. The electric continuity condition reduces to (1/ϵ0)(∂v0/∂y)=(1/ϵ1)(∂v1/∂y) at y=a and (1/ϵ1)(∂v1/∂y)=(1/ϵ2)(∂v2/∂y) at y=0.


## 6. TM Case

The TM polarization has only az-component for the electric field E=(0,0,u). In region 0 u is denoted by u0 consists of the incident/reflected plane waves u0i and u0r as well as the scattered field u0s. Thus we have u0=u0i+u0r+u0s. In region 1, u is denoted by u1 and is again a combination of an incoming and outgoing plane wave along with a scattered field. Therefore u1=u1i+u1r+u1s. The transmitted field in region 2, is denoted by u2 but it will not be decomposed into a superposition of fields. It will be found in the cavity by appropriately specifying the incoming and outgoing plane waves in the upper half space with the material over layer (regions 0 and 1) and by using Fourier theory to represent the scattered and transmitted fields in all three regions.By enforcing continuity conditions at the interfaces along with the boundary conditions at the cavity walls, the transmitted fieldu2 will be determined. At the PEC walls we will have u2=0. At the opening to the upper half space we must enforce continuity with a representation of the scattered field us combined with the incident and reflected fields which captures the effect of the material layer in region 0. Once all of the boundary conditions are enforced we will be able to determine the solution u2.The solutionu2 satisfies the Helmholtz equation Δu2+k2u2=0. The PEC boundary condition u2=0 is enforced at x=0, x=L and y=-d. At y=0 the continuity conditions are enforced which give the final boundary condition. Note that k is known as the “wave number” and is given by k=ω2ϵμ.The Helmholtz equation and all of the boundary conditions dictate the type of Fourier representation we use for the various fields. Once we have the appropriate representation and enforce the continuity conditions we can solve for the desired solutionu2.


### 6.1. TM: Incoming and Outgoing Plane Waves

When dealing with a cavity problem without a material covering an incident plane wave is introduced along with a reflected plane wave both of which are as if no cavity is present. The reflected field is easily computed by enforcing the PEC boundary condition aty=0. The scattered field is then viewed as a perturbation of these fields when the cavity is included in the geometry. In this paper, when dealing with a material coating, the incoming and reflected waves increase in complexity because of the layer in region 1. There will be plane waves propagating in both the positive and negative y directions inside region 0 and region 1. We denote the superposition of these forward and backward propagating plane waves by u0ir and u1ir in regions 0 and 1 respectively. The incident plane wave u0i in region 0 is assumed to be given. The remaining plane waves are calculated by enforcing the continuity conditions at y=0 and y=a. In region 0 we have u0ir=u0i(x,y)+u0r(x,y) and in region 1 we have u1ir=u1i(x,y)+u1r(x,y):(6.1)u0i(x,y)=eıkx0x-ıky0y,u0r(x,y)=ATM-eıkx0x+ıky0y,u1i(x,y)=BTM+eıkx1x-ıky1y,u1r(x,y)=BTM-eıkx1x+ıky1y,
where kx0 and ky0 are related to the wavenumber of region 0 by (kx0)2+(ky0)2=k02. The same relationship holds for kx1 and ky1 in region 1 where we have (kx1)2+(ky1)2=k12.The PEC boundary condition states thatu1ir=0 at y=0. Enforcing this boundary condition allows us to eliminate one of the unknown coefficients(6.2)u1ır(x,0)=0,u1i(x,0)+u1r(x,0)=0,BTM+eıkx1x=-BTM-eıkx1x,BTM+=-BTM-.In order to solve for the remaining unknown coefficients representing the incoming and outgoing plane waves in regions 0 and 1, the boundary condition of the material interface aty=a must be enforced. Applying the electric field continuity condition yields(6.3)u0ir(x,a)=u1ir(x,a),u0i(x,a)+u0r(x,a)=u1i(x,a)+u1r(x,a),eıkx0x-iky0a+ATM-eıkx0x+ıky0a=BTM+(eıkx1x-ıky1a-eıkx1x+ıky1a).
Enforcing the magnetic field continuity condition gives(6.4)1μ0∂u0ir∂y(x,a)=1μ1∂u1ir∂y(x,a),μ1∂u0i∂y(x,a)+μ1∂u0r∂y(x,a)=μ0∂u1i∂y(x,a)+μ0∂u1r∂y(x,a),-ıky0μ1eıkx0x-ıky0a+ıky0μ1ATM-eıkx0x+ıky0a=-ıky1μ0BTM+(eıkx1x-ıky1a+eıkx1x+ıky1a).Combining the above equations gives a system of which can easily be solved by using Cramer's rule:(6.5)ATM-=e-2ıky0a(ky0μr1(e-ıky1a-eıky1a)-ky1(e-ıky1a+eıky1a))ky1(e-ıky1a+eıky1a)+ky0μr1(e-ıky1a-eıky1a),BTM+=2ky0μr1e-ıky0aky1(e-ıky1a+eıky1a)+ky0μr1(e-ıky1a-eıky1a).Note that for multiple layers the same approach works in principle but the system of equations will be larger. Each layer will have a set of unknown coefficients and the continuity conditions would be enforced at each interface.


### 6.2. TM: Scattered and Transmitted Fields

We now consider the incoming and outgoing plane waves described above as a source field impinging on the rectangular cavity in region 2 embedded in an infinite PEC ground plane that has been entirely coated with a dielectric material in region 1. These plane waves already satisfy the Helmholtz equation. We can add in the scattered fields to these plane waves by the superposition principle. The scattered fields are produced by the interaction of the plane waves with the cavity opening. The interaction produces scattered fields in regions 1 and 0 and a transmitted field in region 2. The scattered and transmitted fields will satisfy the Helmholtz equation as well as the same electric and magnetic continuity conditions.Since the scattered field in region 0 is not bounded above in they-direction, it consists of only an outgoing field. Therefore, the Fourier representation is(6.6)u0s(x,y)=∫-∞∞DTM(λ)e-α0ye2πıλxdλ,
where α0=(2πλ)2-k02.Since region 1 is bounded both above and below, the scattered field can propagate in both the positive and negativey-directions. The representation of the scattered field in region 1 is given by(6.7)u1s(x,y)=∫-∞∞(GTM(λ)e-α1y+HTM(λ)eα1y)e2πıλxdλ,
where α1=(2πλ)2-k12. Note that both representations are solutions to the Helmholtz equation but that the unknown Fourier transforms DTM(λ), GTM(λ) and HTM(λ) have not been specified.The transmitted field in region 2 has compact support and must satisfy the PEC boundary conditions at the cavity walls. Thus, the Fourier representation of the transmitted field is given by(6.8)u2(x,y)=∑n=1∞AnTMsin(nπLx)sinh(γ2(y+d)),
where γ2=(nπ/L)2-k22.  The unknown Fourier coefficients An will be found approximately. In order to solve for the coefficients the unknown functions DTM(λ), GTM(λ) and HTM(λ) must be eliminated by using the continuity conditions at y=0 and y=a. We must enforce the continuity conditions at y=0 and at y=a. Once the conditions are enforced we can reduce the problem to a system of equations for the unknown Fourier coefficients An's.


### 6.3. Boundaryy=a

By first applying the electric field continuity we are able to solve for the transformDTM(λ):(6.9)u0(x,a)=u1(x,a),u0ir(x,a)+u0s(x,a)=u1ir(x,a)+u1s(x,a).
Since uir already satisfies the continuity conditions we have u0s(x,a)=u1s(x,a).Now going to the Fourier representations of the scattered field in regions 0 and 1 we have(6.10)∫-∞∞DTM(λ)e-α0ae2πλxdλ=∫-∞∞(GTM(λ)e-α1a+HTM(λ)eα1a)e2πıλxdλ,ℱ-1(DTM(λ)e-α0a)=ℱ-1(GTM(λ)e-α1a+HTM(λ)eα1a),DTM(λ)e-α0a=GTM(λ)e-α1a+HTM(λ)eα1a.
Next, applying the boundary constraints controlling for the continuity of the magnetic field at y=a also yields a solution for the coefficient DTM(λ):(6.11)μ1∂u0ir∂y(x,a)+μ1∂u0s∂y(x,a)=μ0∂u1ir∂y(x,a)+μ0∂u1s∂y(x,a).
Again, knowing that uir already satisfies the magnetic continuity conditions we have(6.12)μ1∂u0s∂y(x,a)=μ0∂u1s∂y(x,a).Returning to the Fourier representations we have that(6.13)-μ1∫-∞∞DTM(λ)α0e-α0ae2πıλxdλ=μ0∫-∞∞α1(HTM(λ)eα1a-GTM(λ)e-α1a)e2πıλxdλ,-μ1ℱ-1(α0DTM(λ)e-α0a)=μ0ℱ-1(α1(HTM(λ)eα1a-GTM(λ)e-α1a)),-μ1α0DTM(λ)e-α0a=μ0α1HTM(λ)eα1a-μ0α1GTM(λ)e-α1a.The results from the boundary condition aty=a then provide a system of equations that can be solved using Cramer's rule. In general, if there were multiple layers the system of equations would be larger and the continuity conditions would be enforced at each material interface. We solve the system to get the unknown functions GTM(λ) and HTM(λ) written in terms of DTM(λ):(6.14)GTM(λ)=DTM(λ)e-α0a(eα1a(μ0α1+μ1α0)2μ0α1),HTM(λ)=DTM(λ)e-α0a(e-α1a(μ0α1-μ1α0)2μ0α1).
We now want to eliminate DTM(λ) from the problem to be left with only the Fourier coefficients An's of the transmitted field u2 as unknowns. Further, these coefficients will be the solution to a system of equations which we will solve to get our approximate solution.


### 6.4. Boundaryy=0

We use the continuity conditions once again foru1 and u2 at the interface between regions 1 and 2. First the electric field continuity gives(6.15)u1(x,0)=u2(x,0),u1ir(x,0)+u1s(x,0)=u2(x,0).
Since uir=0 at y=0 we have(6.16)u1s(x,0)=u2(x,0).
Using the Fourier representations we have(6.17)∫-∞∞(GTM(λ)+HTM(λ))e2πıλxdλ=∑n=1∞AnTMsin(nπLx)sinh(γ2d),ℱ-1(GTM(λ)+HTM(λ))=∑n=1∞AnTMsin(nπLx),GTM(λ)+HTM(λ)=∑n=1∞AnTMℱ(sin(nπLx)).
Substitute in the solution for GTM(λ) and HTM(λ), into the equation to get(6.18)DTM(λ)e-α0a(𝒬ATM2μ0α1)=∑n=1∞AnTMℱ(sin(nπLx))
and solving for DTM(λ) yields:(6.19)DTM(λ)e-α0a=2μ0α1∑n=1∞AnTMℱ(sin((nπ/L)x))𝒬ATM,
where(6.20)𝒬ATM=e-α1a(μ0α1-μ1α0)+eα1a(μ0α1+μ1α0).Next, the continuity of the magnetic field across the cavity opening aty=0 gives(6.21)-1ıwμ1∂u1∂y(x,0)=-1ıwμ2∂u2∂y(x,0),μ2∂u1ir∂y(x,0)+μ2∂u1s∂y(x,0)=μ1∂u2∂y(x,0).
Using the Fourier representations we have(6.22)-2ıkky1μ2BTM+eıkx1x+μ2∫-∞∞α1(HTM(λ)-GTM(λ))e2πıλxdλ=μ1∑n=1∞AnTMγ2sin(nπLx)cosh(γ2d),-2ıky1μ2BTM+eıkx1x+μ2ℱ-1(α1(HTM(λ)-GTM(λ)))=μ1∑n=1∞AnTMγ2sin(nπLx)coth(γ2d).
Again, substitute in for GTM(λ) and HTM(λ) resulting in(6.23)-2ıky1μ2BTM+eıkx1x+μ2ℱ-1(α1(DTM(λ)e-α0a𝒬BTM2μ0α1))=μ1∑n=1∞AnTMγ2sin(nπLx)coth(γ2d),
where(6.24)𝒬BTM=e-α1a(μ0α1-μ1α0)-eα1a(μ0α1+μ1α0).Using the orthogonality of the sine terms this equation can be reduced to.(6.25)-2ıky1μ2BTM+∫0Leıkx1xsin(mπLx)dx+μ2∫0Lℱ-1(DTM(λ)e-α0a𝒬BTM2μ0)sin(mπLx)dx=μ1L2AmTMγ2mcoth(γ2md).
Then by performing integration and utilizing Parseval's Theorem, our results show. 								(6.26)-ky1μ2BTM+(eıkx1+ımπ-1ıkx1+(ımπ/L)+1-eıkx1-ımπıkx1-(ımπ/L))+μ2∫-∞∞(DTM(λ)e-α0a𝒬BTM2μ0)ℱ(sin(mπLx))¯dλ=μ1L2AmTMγ2mcoth(γ2md).
Lastly, the values for DTM(λ) can be substituted in, and rearranging the equation results in:(6.27)∑n=1∞AnTMKm,n-μ1μ2L2AmTMγ2mcoth(γ2md)=ky1BTM+(eıkx1+ımπ-1ıkx1+ımπ/L+1-eıkx1-ımπıkx1-ımπ/L),
where(6.28)Km,n=∫-∞∞α1𝒬BTM𝒬ATMℱ(sin(nπLx))ℱ(sin(mπLx))¯dλ
The Fourier coefficients, AnTM, satisfy an infinite system of linear equations. The solution can be approximated computationally by truncating the infinite sum to a finite number of terms. Typically, the number of terms should be at least large enough to include all of the “guided modes." The convergence issues around truncating such an infinite systems of equations are not well understood but a partial discussion is given in [13]. This solution of the finite system is used as an approximate set of Fourier coefficients for the representation of u2 in the cavity.


## 6.1. TM: Incoming and Outgoing Plane Waves

When dealing with a cavity problem without a material covering an incident plane wave is introduced along with a reflected plane wave both of which are as if no cavity is present. The reflected field is easily computed by enforcing the PEC boundary condition aty=0. The scattered field is then viewed as a perturbation of these fields when the cavity is included in the geometry. In this paper, when dealing with a material coating, the incoming and reflected waves increase in complexity because of the layer in region 1. There will be plane waves propagating in both the positive and negative y directions inside region 0 and region 1. We denote the superposition of these forward and backward propagating plane waves by u0ir and u1ir in regions 0 and 1 respectively. The incident plane wave u0i in region 0 is assumed to be given. The remaining plane waves are calculated by enforcing the continuity conditions at y=0 and y=a. In region 0 we have u0ir=u0i(x,y)+u0r(x,y) and in region 1 we have u1ir=u1i(x,y)+u1r(x,y):(6.1)u0i(x,y)=eıkx0x-ıky0y,u0r(x,y)=ATM-eıkx0x+ıky0y,u1i(x,y)=BTM+eıkx1x-ıky1y,u1r(x,y)=BTM-eıkx1x+ıky1y,
where kx0 and ky0 are related to the wavenumber of region 0 by (kx0)2+(ky0)2=k02. The same relationship holds for kx1 and ky1 in region 1 where we have (kx1)2+(ky1)2=k12.The PEC boundary condition states thatu1ir=0 at y=0. Enforcing this boundary condition allows us to eliminate one of the unknown coefficients(6.2)u1ır(x,0)=0,u1i(x,0)+u1r(x,0)=0,BTM+eıkx1x=-BTM-eıkx1x,BTM+=-BTM-.In order to solve for the remaining unknown coefficients representing the incoming and outgoing plane waves in regions 0 and 1, the boundary condition of the material interface aty=a must be enforced. Applying the electric field continuity condition yields(6.3)u0ir(x,a)=u1ir(x,a),u0i(x,a)+u0r(x,a)=u1i(x,a)+u1r(x,a),eıkx0x-iky0a+ATM-eıkx0x+ıky0a=BTM+(eıkx1x-ıky1a-eıkx1x+ıky1a).
Enforcing the magnetic field continuity condition gives(6.4)1μ0∂u0ir∂y(x,a)=1μ1∂u1ir∂y(x,a),μ1∂u0i∂y(x,a)+μ1∂u0r∂y(x,a)=μ0∂u1i∂y(x,a)+μ0∂u1r∂y(x,a),-ıky0μ1eıkx0x-ıky0a+ıky0μ1ATM-eıkx0x+ıky0a=-ıky1μ0BTM+(eıkx1x-ıky1a+eıkx1x+ıky1a).Combining the above equations gives a system of which can easily be solved by using Cramer's rule:(6.5)ATM-=e-2ıky0a(ky0μr1(e-ıky1a-eıky1a)-ky1(e-ıky1a+eıky1a))ky1(e-ıky1a+eıky1a)+ky0μr1(e-ıky1a-eıky1a),BTM+=2ky0μr1e-ıky0aky1(e-ıky1a+eıky1a)+ky0μr1(e-ıky1a-eıky1a).Note that for multiple layers the same approach works in principle but the system of equations will be larger. Each layer will have a set of unknown coefficients and the continuity conditions would be enforced at each interface.


## 6.2. TM: Scattered and Transmitted Fields

We now consider the incoming and outgoing plane waves described above as a source field impinging on the rectangular cavity in region 2 embedded in an infinite PEC ground plane that has been entirely coated with a dielectric material in region 1. These plane waves already satisfy the Helmholtz equation. We can add in the scattered fields to these plane waves by the superposition principle. The scattered fields are produced by the interaction of the plane waves with the cavity opening. The interaction produces scattered fields in regions 1 and 0 and a transmitted field in region 2. The scattered and transmitted fields will satisfy the Helmholtz equation as well as the same electric and magnetic continuity conditions.Since the scattered field in region 0 is not bounded above in they-direction, it consists of only an outgoing field. Therefore, the Fourier representation is(6.6)u0s(x,y)=∫-∞∞DTM(λ)e-α0ye2πıλxdλ,
where α0=(2πλ)2-k02.Since region 1 is bounded both above and below, the scattered field can propagate in both the positive and negativey-directions. The representation of the scattered field in region 1 is given by(6.7)u1s(x,y)=∫-∞∞(GTM(λ)e-α1y+HTM(λ)eα1y)e2πıλxdλ,
where α1=(2πλ)2-k12. Note that both representations are solutions to the Helmholtz equation but that the unknown Fourier transforms DTM(λ), GTM(λ) and HTM(λ) have not been specified.The transmitted field in region 2 has compact support and must satisfy the PEC boundary conditions at the cavity walls. Thus, the Fourier representation of the transmitted field is given by(6.8)u2(x,y)=∑n=1∞AnTMsin(nπLx)sinh(γ2(y+d)),
where γ2=(nπ/L)2-k22.  The unknown Fourier coefficients An will be found approximately. In order to solve for the coefficients the unknown functions DTM(λ), GTM(λ) and HTM(λ) must be eliminated by using the continuity conditions at y=0 and y=a. We must enforce the continuity conditions at y=0 and at y=a. Once the conditions are enforced we can reduce the problem to a system of equations for the unknown Fourier coefficients An's.


## 6.3. Boundaryy=a

By first applying the electric field continuity we are able to solve for the transformDTM(λ):(6.9)u0(x,a)=u1(x,a),u0ir(x,a)+u0s(x,a)=u1ir(x,a)+u1s(x,a).
Since uir already satisfies the continuity conditions we have u0s(x,a)=u1s(x,a).Now going to the Fourier representations of the scattered field in regions 0 and 1 we have(6.10)∫-∞∞DTM(λ)e-α0ae2πλxdλ=∫-∞∞(GTM(λ)e-α1a+HTM(λ)eα1a)e2πıλxdλ,ℱ-1(DTM(λ)e-α0a)=ℱ-1(GTM(λ)e-α1a+HTM(λ)eα1a),DTM(λ)e-α0a=GTM(λ)e-α1a+HTM(λ)eα1a.
Next, applying the boundary constraints controlling for the continuity of the magnetic field at y=a also yields a solution for the coefficient DTM(λ):(6.11)μ1∂u0ir∂y(x,a)+μ1∂u0s∂y(x,a)=μ0∂u1ir∂y(x,a)+μ0∂u1s∂y(x,a).
Again, knowing that uir already satisfies the magnetic continuity conditions we have(6.12)μ1∂u0s∂y(x,a)=μ0∂u1s∂y(x,a).Returning to the Fourier representations we have that(6.13)-μ1∫-∞∞DTM(λ)α0e-α0ae2πıλxdλ=μ0∫-∞∞α1(HTM(λ)eα1a-GTM(λ)e-α1a)e2πıλxdλ,-μ1ℱ-1(α0DTM(λ)e-α0a)=μ0ℱ-1(α1(HTM(λ)eα1a-GTM(λ)e-α1a)),-μ1α0DTM(λ)e-α0a=μ0α1HTM(λ)eα1a-μ0α1GTM(λ)e-α1a.The results from the boundary condition aty=a then provide a system of equations that can be solved using Cramer's rule. In general, if there were multiple layers the system of equations would be larger and the continuity conditions would be enforced at each material interface. We solve the system to get the unknown functions GTM(λ) and HTM(λ) written in terms of DTM(λ):(6.14)GTM(λ)=DTM(λ)e-α0a(eα1a(μ0α1+μ1α0)2μ0α1),HTM(λ)=DTM(λ)e-α0a(e-α1a(μ0α1-μ1α0)2μ0α1).
We now want to eliminate DTM(λ) from the problem to be left with only the Fourier coefficients An's of the transmitted field u2 as unknowns. Further, these coefficients will be the solution to a system of equations which we will solve to get our approximate solution.


## 6.4. Boundaryy=0

We use the continuity conditions once again foru1 and u2 at the interface between regions 1 and 2. First the electric field continuity gives(6.15)u1(x,0)=u2(x,0),u1ir(x,0)+u1s(x,0)=u2(x,0).
Since uir=0 at y=0 we have(6.16)u1s(x,0)=u2(x,0).
Using the Fourier representations we have(6.17)∫-∞∞(GTM(λ)+HTM(λ))e2πıλxdλ=∑n=1∞AnTMsin(nπLx)sinh(γ2d),ℱ-1(GTM(λ)+HTM(λ))=∑n=1∞AnTMsin(nπLx),GTM(λ)+HTM(λ)=∑n=1∞AnTMℱ(sin(nπLx)).
Substitute in the solution for GTM(λ) and HTM(λ), into the equation to get(6.18)DTM(λ)e-α0a(𝒬ATM2μ0α1)=∑n=1∞AnTMℱ(sin(nπLx))
and solving for DTM(λ) yields:(6.19)DTM(λ)e-α0a=2μ0α1∑n=1∞AnTMℱ(sin((nπ/L)x))𝒬ATM,
where(6.20)𝒬ATM=e-α1a(μ0α1-μ1α0)+eα1a(μ0α1+μ1α0).Next, the continuity of the magnetic field across the cavity opening aty=0 gives(6.21)-1ıwμ1∂u1∂y(x,0)=-1ıwμ2∂u2∂y(x,0),μ2∂u1ir∂y(x,0)+μ2∂u1s∂y(x,0)=μ1∂u2∂y(x,0).
Using the Fourier representations we have(6.22)-2ıkky1μ2BTM+eıkx1x+μ2∫-∞∞α1(HTM(λ)-GTM(λ))e2πıλxdλ=μ1∑n=1∞AnTMγ2sin(nπLx)cosh(γ2d),-2ıky1μ2BTM+eıkx1x+μ2ℱ-1(α1(HTM(λ)-GTM(λ)))=μ1∑n=1∞AnTMγ2sin(nπLx)coth(γ2d).
Again, substitute in for GTM(λ) and HTM(λ) resulting in(6.23)-2ıky1μ2BTM+eıkx1x+μ2ℱ-1(α1(DTM(λ)e-α0a𝒬BTM2μ0α1))=μ1∑n=1∞AnTMγ2sin(nπLx)coth(γ2d),
where(6.24)𝒬BTM=e-α1a(μ0α1-μ1α0)-eα1a(μ0α1+μ1α0).Using the orthogonality of the sine terms this equation can be reduced to.(6.25)-2ıky1μ2BTM+∫0Leıkx1xsin(mπLx)dx+μ2∫0Lℱ-1(DTM(λ)e-α0a𝒬BTM2μ0)sin(mπLx)dx=μ1L2AmTMγ2mcoth(γ2md).
Then by performing integration and utilizing Parseval's Theorem, our results show. 								(6.26)-ky1μ2BTM+(eıkx1+ımπ-1ıkx1+(ımπ/L)+1-eıkx1-ımπıkx1-(ımπ/L))+μ2∫-∞∞(DTM(λ)e-α0a𝒬BTM2μ0)ℱ(sin(mπLx))¯dλ=μ1L2AmTMγ2mcoth(γ2md).
Lastly, the values for DTM(λ) can be substituted in, and rearranging the equation results in:(6.27)∑n=1∞AnTMKm,n-μ1μ2L2AmTMγ2mcoth(γ2md)=ky1BTM+(eıkx1+ımπ-1ıkx1+ımπ/L+1-eıkx1-ımπıkx1-ımπ/L),
where(6.28)Km,n=∫-∞∞α1𝒬BTM𝒬ATMℱ(sin(nπLx))ℱ(sin(mπLx))¯dλ
The Fourier coefficients, AnTM, satisfy an infinite system of linear equations. The solution can be approximated computationally by truncating the infinite sum to a finite number of terms. Typically, the number of terms should be at least large enough to include all of the “guided modes." The convergence issues around truncating such an infinite systems of equations are not well understood but a partial discussion is given in [13]. This solution of the finite system is used as an approximate set of Fourier coefficients for the representation of u2 in the cavity.


## 7. TE Case

We now look at the transverse electric case where assumeH=(0,0,v). The approach is much the same as the TM case. The unknown function will satisfy the Helmholtz equation Δv+k2v=0. The primary difference is the enforcement of the PEC boundary condition. Since we are now solving for the z component of the magnetic field we must account for how the PEC boundary condition is represented. If we have H=(0,0,v) we can apply Maxwell's equations to produce E. We can then compute the tangential component of E and set it equal to zero. Doing so for the TE case gives us that (1/ϵ)(dv/dn)=0. That is the normal derivative of v will be equal to zero. This condition must be enforced in region 1 when we compute the incident and reflected field and also in region 2 when we represent the transmitted field in the cavity.


### 7.1. TE: Incoming and Outgoing Plane Waves

For the TE case we are assuming that we have a TE-polarized incident magnetic field. The reflections and transmissions at the material interface and at the PEC boundary will produce forward and backward propagating plane waves in both regions. We can solve for the these fieldsv0ir and v1ir in regions 0 and 1 just as we did in the TM case. The only difference in the PEC boundary condition at y=0 now is applied to the normal derivative of v instead of v itself. We have the following:(7.1)vi,0(x,y)=eıkx0x-ıky0y,vr,0(x,y)=ATE-eıkx0x+ıky0y,vi,1(x,y)=BTE+eıkx1x-ıky1y,vr,1(x,y)=BTE-eıkx1x+ıky1y.
These results provide us a system of equations in two unknowns that can be solved using Cramer's rule to give(7.2)ATE-=e-2ıky0a(-ky1(eıky1a-e-ıky1a)-ky0εr1(e-ıky1a+eıky1a))ky1(eıky1a-e-ıky1a)-ky0εr1(e-ıky1a+eıky1a),BTE+=-2ky0εr1e-ıky0aky1(eıky1a-e-ıky1a)-ky0εr1(e-ıky1a+eıky1a).
As in the TM case, these incoming and outgoing plane waves in regions 0 and 1 are used as a source field. The plane waves impinge on the cavity in region 2 and produce scattered fields in regions 0 and 1 along with a transmitted field inside region 2.


### 7.2. TE: Scattered and Transmitted Fields

We now consider the TE case when an incident plane wave impinges on a rectangular cavity embedded in an infinite ground plane that has been entirely coated with a dielectric material. The incoming and outgoing plane waves of regions 0 and 1 were found above. Now the embedded cavity requires representations for the transmitted and scattered fields.  The representation of the scattered field in region 0 is(7.3)v0s(x,y)=∫-∞∞DTE(λ)e-α0ye2πıλxdλ.
Since region 1 is bounded both above and below, and the scattered filed propagates in both the positive and negative y direction, we must account for this using the following Fourier representation:(7.4)v1s(x,y)=∫-∞∞(GTE(λ)e-α1y+HTE(λ)eα1y)e2πıλxdλ.
Lastly, the representation of the transmitted field is(7.5)v2(x,y)=∑n=0∞AnTEcos(nπLx)cosh(γ2(y+d)).
Note that this Fourier representation is chosen to satisfy the PEC boundary condition when it is enforced on the normal derivative.The continuity conditions will again be used aty=a to solve for the unknown functions GTE(λ) and HTE(λ) in terms of DTE(λ). Once this is done the continuity conditions are enforced at y=0 to eliminate DTE(λ) from the problem and to be left with only the unknown Fourier coefficients for v2. Using orthogonality and Parseval's theorem these coefficients are shown to satisfy an infinite system of equations:(7.6)BTE+(eıkx1L+ımπ-1ıkx1+ımπ/L+eıkx1L-ımπ-1ıkx1-ımπ/L)=-ε1ε2∑n=0∞AnTEγ2Lm,n+{AmTELcoth(γ2md)ifm=n=0,AmTEL2coth(γ2md)ifm=n>0,
where(7.7)Lm,n=∫-∞∞1α1ℱ(cos(nπLx))ℱ(cos(mπLx))¯𝒬BTE𝒬ATEdλ,𝒬ATE=e-α1a(ε0α1-ε1α0)-eα1a(ε0α1+ε1α0),𝒬BTE=e-α1a(ε0α1-ε1α0)+eα1a(ε0α1+ε1α0).
The Fourier coefficients, AnTE, form a system of equations that can be solved for computationally by truncating the infinite system to a suitably large finite system.


## 7.1. TE: Incoming and Outgoing Plane Waves

For the TE case we are assuming that we have a TE-polarized incident magnetic field. The reflections and transmissions at the material interface and at the PEC boundary will produce forward and backward propagating plane waves in both regions. We can solve for the these fieldsv0ir and v1ir in regions 0 and 1 just as we did in the TM case. The only difference in the PEC boundary condition at y=0 now is applied to the normal derivative of v instead of v itself. We have the following:(7.1)vi,0(x,y)=eıkx0x-ıky0y,vr,0(x,y)=ATE-eıkx0x+ıky0y,vi,1(x,y)=BTE+eıkx1x-ıky1y,vr,1(x,y)=BTE-eıkx1x+ıky1y.
These results provide us a system of equations in two unknowns that can be solved using Cramer's rule to give(7.2)ATE-=e-2ıky0a(-ky1(eıky1a-e-ıky1a)-ky0εr1(e-ıky1a+eıky1a))ky1(eıky1a-e-ıky1a)-ky0εr1(e-ıky1a+eıky1a),BTE+=-2ky0εr1e-ıky0aky1(eıky1a-e-ıky1a)-ky0εr1(e-ıky1a+eıky1a).
As in the TM case, these incoming and outgoing plane waves in regions 0 and 1 are used as a source field. The plane waves impinge on the cavity in region 2 and produce scattered fields in regions 0 and 1 along with a transmitted field inside region 2.


## 7.2. TE: Scattered and Transmitted Fields

We now consider the TE case when an incident plane wave impinges on a rectangular cavity embedded in an infinite ground plane that has been entirely coated with a dielectric material. The incoming and outgoing plane waves of regions 0 and 1 were found above. Now the embedded cavity requires representations for the transmitted and scattered fields.  The representation of the scattered field in region 0 is(7.3)v0s(x,y)=∫-∞∞DTE(λ)e-α0ye2πıλxdλ.
Since region 1 is bounded both above and below, and the scattered filed propagates in both the positive and negative y direction, we must account for this using the following Fourier representation:(7.4)v1s(x,y)=∫-∞∞(GTE(λ)e-α1y+HTE(λ)eα1y)e2πıλxdλ.
Lastly, the representation of the transmitted field is(7.5)v2(x,y)=∑n=0∞AnTEcos(nπLx)cosh(γ2(y+d)).
Note that this Fourier representation is chosen to satisfy the PEC boundary condition when it is enforced on the normal derivative.The continuity conditions will again be used aty=a to solve for the unknown functions GTE(λ) and HTE(λ) in terms of DTE(λ). Once this is done the continuity conditions are enforced at y=0 to eliminate DTE(λ) from the problem and to be left with only the unknown Fourier coefficients for v2. Using orthogonality and Parseval's theorem these coefficients are shown to satisfy an infinite system of equations:(7.6)BTE+(eıkx1L+ımπ-1ıkx1+ımπ/L+eıkx1L-ımπ-1ıkx1-ımπ/L)=-ε1ε2∑n=0∞AnTEγ2Lm,n+{AmTELcoth(γ2md)ifm=n=0,AmTEL2coth(γ2md)ifm=n>0,
where(7.7)Lm,n=∫-∞∞1α1ℱ(cos(nπLx))ℱ(cos(mπLx))¯𝒬BTE𝒬ATEdλ,𝒬ATE=e-α1a(ε0α1-ε1α0)-eα1a(ε0α1+ε1α0),𝒬BTE=e-α1a(ε0α1-ε1α0)+eα1a(ε0α1+ε1α0).
The Fourier coefficients, AnTE, form a system of equations that can be solved for computationally by truncating the infinite system to a suitably large finite system.


## 8. Use in Finite Element Methods

The Fourier approach in the previous sections clearly only works when dealing with rectangular cavities. For cavities with arbitrary shape other methods such as the finite element method must be employed [2]. Application of the finite element method finds a solution within a space of functions for the weak form of the Helmholtz equation. Typically the cavity region is approximated by triangular grid and linear tetrahedral functions defined on the grid for a basis for a space of piecewise linear functions. Boundary conditions must be provided in order to solve the problem on a finite region. Therefore, a transparent boundary condition at the cavity opening must be used to properly represent the transmitted field's continuity with the total fields in the regions above it.Equations (6.27) and (7.6) can be used to represent such a boundary condition for each of the respective polarizations. Equation (6.27) from the TM case can be viewed as a solution of(8.1)∂u2∂y=μ2μ1ℱ-1(𝒬BTM𝒬ATMℱ(u2))+μ2μ1∂uir∂y
at y=0. Such an expression is known as a Dirichlet to Neumann map. It connects the derivative values of u2 with the function values. In the Fourier case presented here, the equation is solved by substituting in the Fourier representation of u2 and using orthogonality to produce a system of equations. However, (8.1) can be plugged directly into the boundary term of the weak form of the Helmholtz equation. On the other hand, (7.6) from the TE case can be written as(8.2)v2=ℱ(𝒬BTE𝒬ATEℱ-1(ϵ2ϵ1∂v2∂y))+vir
at y=0. In this case, the expression is a Neumann to Dirichlet map. Again, in the Fourier case, (8.2) is solved by inserting the Fourier representation of v2 and using orthogonality to produce a system of equations. For finite elements in the TE case (8.2) cannot be plugged directly into the weak form of the Helmholtz equation. However, it can be used as auxiliary equation for a mixed finite element method approach.


## 9. Far Field Computation

Computation of far field or radar cross section (RCS) quantities is a postprocessing step. We will use the computed transmitted fieldsu2 and v2 above to compute the strength of the scattered fields at distance far from the cavity opening. The trick is that to use the standard formula for far field values, we need to have the scattered field at the interface with free space. That is we need to know the values of u0s and v0s at y=a. The far field formulas are [14](9.1)σTM(ϕ,ϕi)=k02|∫-∞∞u0s(x′,a)sinϕeık0(x′cosϕ+asinϕ)dx′|2,σTE(ϕ,ϕi)=k02|∫-∞∞iηϵω∂v0s∂y(x′,a)eık0(x′cosϕ+asinϕ)dx′|2.The computation of these integrals can potentially be difficult since they are over the entire real line. However, if we relate the scattered fields back to the transmitted fieldsu2 and v2 in the cavity we can avoid evaluating such an improper integral. If we rewrite above we see that(9.2)σTM(ϕ,ϕi)=k02|sinϕeık0(asinϕ)∫-∞∞u0s(x′,a)eık0(x′cosϕ)dx′|2,σTE(ϕ,ϕi)=k02|iηϵωeık0(asinϕ)∫-∞∞∂v0s∂y(x′,a)eık0(x′cosϕ)dx′|2.We can now see that the integrals are actually Fourier transforms ofu0s and ∂v0s/∂y. In Sections 5 and 6, these Fourier transforms are also expressed as DTM(λ)e-α0a and -α0DTE(λ)e-α0a:(9.3)σTM(ϕ,ϕi)=k02|sinϕeık0(asinϕ)DTM(λ)e-α0a|2,σTE(ϕ,ϕi)=k02|iηϵωeık0(asinϕ)α0DTE(λ)e-α0a|2.
Recall from (6.19) that  we have from the TM case that(9.4)DTM(λ)e-α0a=2μ0α1∑n=1∞AnTMℱ(sin((nπ/L)x))𝒬ATM
and a similar calculation in the TE case will give(9.5)α0DTE(λ)e-α0a=2ϵ0ϵ1∑n=1∞γ2AnTEℱ(sin((nπ/L)x))𝒬ATEThese equations can be substituted into (9.3). Hence, the RCS can be evaluated using the computed solutions u2 and v2 from inside the cavity.


## 10. Fast RCS Approximation

Morgan and Schwering presented a simple method for rapidly producing approximations of the RCS for filled rectangular cavities. A mode matching technique replaces the open half space above the cavity with a vertical channel containing outgoing waveguide modes. The coefficients are computed directly without building and solving a linear system by matching the cavity modes with the outgoing waveguide modes [4]. The approach yields surprisingly good agreement with the true solutions when used for RCS computations. Bao and Zhang attempted to refine the mode matching approach by improving the approximation using an asymptotic argument [3]. The “impoved” mode matching solution converges to the exact solution as the frequency increases.The fast mode matching approximation of Morgan and Schwering extends the cavity walls into the upper half space and uses a mode matching solution as an approximation [4]. If the same mode matching approach is applied to an overlayer problem, it would be performed at the interface from the cavity material to the overlayer material and then again from the overlayer material to free space parameters. It is the author's observation based on comparison with [4] that this is essentially equivalent to using only the diagonal terms of (6.27) and (7.6) in the full Fourier derivation and solving the corresponding diagonal system. Although this equivalence is not formally investigated here, it can be seen below that the solution of the diagonal system yields solid results. It is possible that this equivalence is the path to a future formal Mathematical analysis of the convergence properties of the mode matching approach.


## 11. Numerical Results

The following results demonstrate that the derivations are sound and show the efficacy of the method. As was done in [4], the parameters are chosen arbitrarily for demonstration purposes.


### 11.1. Basic Computations

Figures2, 3, 4, 5, 6, 7, 8 and 9 show the solutions obtained by solving (6.27) for the TM case and 19 for the TE case. Since there are no results to compare with directly, the output is validated by physical interpretation. First, a cavity underneath a very thin layer of dielectric material is compared to the same cavity with no dielectric overlayer. Note that the layered problem results in a small perturbation of the solution of the nonlayered problem as one would expect. The cavity geometry is given by L=1 and d=10. The material parameters are given by ϵ1=2, μ1=3, ϵ2=2 and μ2=1 and the frequency is 14π. The layer has thicknes a=.1. The computation of a monostatic radar cross section is given for both the TE and TM cases.Figure 2
Thin layer perturbation for the TM case.Figure 3
Thin layer perturbation for the TE case.Figure 4
Thick layer perturbation for the TM case.Figure 5
Thick layer perturbation for the TE case.Figure 6
Increasing layer of conducting material.Figure 7
Increasing layer of conducting material.Figure 8
Comparison of full solution with Morgan and Schwering mode matching method for the TM case.Figure 9
Comparison of full solution with Morgan and Schwering mode matching method for the TE case.Note that for angles near 90 degrees we may expect more prominent disagreement. The deviation is due to the fact that the fields will travel through much more material at angles near grazing incidence. The deviation is much more prominent in the TE example for Figure3 due to the presence of surface waves traveling in the overlayer dielectric medium. In the TM case, far fields tend to zero near grazing incidence which explains the better agreement for Figure 2 [15].A second example deals with an thicker overlayer but with a only a small change in the dielectric materials. Again, as one would expect only a small perturbation of the solution is seen. A similar explanation as given above regarding Figure3 explains the deviations found near 90 degrees. The cavity geometry is the same as above but the material parameters are given by ϵ1=1.01, μ1=1.01, ϵ2=2, and μ2=1. The thickness of the overlayer material is a=1. The frequency is also the same as above. Note that the material parameters in the layer only slightly differ from that of free space. Again the monostatic radar cross section is computed for both the TE and TM cases.Finally, a computation of bistatic radar cross sections is performed for which the thickness of the layer is increased. The layer material is a conductor which means that as the thickness of the layer increases there is a corresponding decrease in the strength of the far field. In this example the geometry is given byL=1.25 and d=.0625. The material parameters are ϵ1=16-5ı, μ1=4-1.25ı, ϵ2=1 and μ2=1. The frequency is 2π. Again the results agree with the expected physical interpretation.


### 11.2. Fast Approximations

Here the mode matching approximation of Morgan and Schwering as applied to the layered problem is demonstrated. The comparisons between the mode matching approximation and the full solution agree well. Although it has not been explained mathematically, the numerical experiments show that the mode matching method appears to work effectively with electrically large deep cavities. Some mathematical explanation is given by Bao and Zhang [3]. The geometry used here is again L=1 and d=10. The material parameters are ϵ1=2, μ1=3, ϵ2=2 and μ2=1. The frequency is 14π.The agreement between the full and mode matching solutions is remarkable given the time savings. The full problem requires a fullN×N matrix to be built and the corresponding system to be solved. The mode matching approximation used only the diagonal terms of the system and hence needs only to compute N matrix entries and solve a diagonal system. The time saved is obviously quite dramatic. For example, the computation of the results in Figures 8 and 9 took nearly 3000 seconds with the full solution while the mode matching approximation took less than 40 seconds.


## 11.1. Basic Computations

Figures2, 3, 4, 5, 6, 7, 8 and 9 show the solutions obtained by solving (6.27) for the TM case and 19 for the TE case. Since there are no results to compare with directly, the output is validated by physical interpretation. First, a cavity underneath a very thin layer of dielectric material is compared to the same cavity with no dielectric overlayer. Note that the layered problem results in a small perturbation of the solution of the nonlayered problem as one would expect. The cavity geometry is given by L=1 and d=10. The material parameters are given by ϵ1=2, μ1=3, ϵ2=2 and μ2=1 and the frequency is 14π. The layer has thicknes a=.1. The computation of a monostatic radar cross section is given for both the TE and TM cases.Figure 2
Thin layer perturbation for the TM case.Figure 3
Thin layer perturbation for the TE case.Figure 4
Thick layer perturbation for the TM case.Figure 5
Thick layer perturbation for the TE case.Figure 6
Increasing layer of conducting material.Figure 7
Increasing layer of conducting material.Figure 8
Comparison of full solution with Morgan and Schwering mode matching method for the TM case.Figure 9
Comparison of full solution with Morgan and Schwering mode matching method for the TE case.Note that for angles near 90 degrees we may expect more prominent disagreement. The deviation is due to the fact that the fields will travel through much more material at angles near grazing incidence. The deviation is much more prominent in the TE example for Figure3 due to the presence of surface waves traveling in the overlayer dielectric medium. In the TM case, far fields tend to zero near grazing incidence which explains the better agreement for Figure 2 [15].A second example deals with an thicker overlayer but with a only a small change in the dielectric materials. Again, as one would expect only a small perturbation of the solution is seen. A similar explanation as given above regarding Figure3 explains the deviations found near 90 degrees. The cavity geometry is the same as above but the material parameters are given by ϵ1=1.01, μ1=1.01, ϵ2=2, and μ2=1. The thickness of the overlayer material is a=1. The frequency is also the same as above. Note that the material parameters in the layer only slightly differ from that of free space. Again the monostatic radar cross section is computed for both the TE and TM cases.Finally, a computation of bistatic radar cross sections is performed for which the thickness of the layer is increased. The layer material is a conductor which means that as the thickness of the layer increases there is a corresponding decrease in the strength of the far field. In this example the geometry is given byL=1.25 and d=.0625. The material parameters are ϵ1=16-5ı, μ1=4-1.25ı, ϵ2=1 and μ2=1. The frequency is 2π. Again the results agree with the expected physical interpretation.


## 11.2. Fast Approximations

Here the mode matching approximation of Morgan and Schwering as applied to the layered problem is demonstrated. The comparisons between the mode matching approximation and the full solution agree well. Although it has not been explained mathematically, the numerical experiments show that the mode matching method appears to work effectively with electrically large deep cavities. Some mathematical explanation is given by Bao and Zhang [3]. The geometry used here is again L=1 and d=10. The material parameters are ϵ1=2, μ1=3, ϵ2=2 and μ2=1. The frequency is 14π.The agreement between the full and mode matching solutions is remarkable given the time savings. The full problem requires a fullN×N matrix to be built and the corresponding system to be solved. The mode matching approximation used only the diagonal terms of the system and hence needs only to compute N matrix entries and solve a diagonal system. The time saved is obviously quite dramatic. For example, the computation of the results in Figures 8 and 9 took nearly 3000 seconds with the full solution while the mode matching approximation took less than 40 seconds.


## 12. Conclusion

A complete Fourier solution has been provided for the problem of electromagnetic scattering by a recatangular buried cavity beneath a uniform material layer. It has been demonstrated that the method gives solutions that agree with physical interpretations. Also, the mode matching method described by Morgan and Schwering has been extended to the buried cavity problem. The mode matching technique works very well in this context and provides enormous speed up in terms of computation.The technique can be extended to rectangular cavities buried under multiple uniform material layers. It will also be applied in finite element techniques as a transparent boundary condition for similar problems involving nonrectangular cavities. In the future, it would be very useful to extend the approach to three dimensional cavities. Also, a thorough mathematical investigation of the convergence properties of the mode matching technique would be desirable.



---

*Source: 937327-2010-01-24.xml*