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Inverse diffraction grating of Maxwell’s equations in biperiodic structures

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Abstract

Consider a time-harmonic electromagnetic plane wave incident on a perfectly conducting biperiodic surface (crossed grating). The diffraction is modeled as a boundary value problem for the three-dimensional Maxwell equation. The surface is assumed to be a small and smooth deformation of a planar surface. In this paper, a novel approach is developed to solve the inverse diffraction grating problem in the near-field regime, which is to reconstruct the surface with resolution beyond Rayleigh’s criterion. The method requires only a single incident field with one polarization, one frequency, and one incident direction, and is realized by using the fast Fourier transform. Numerical results show that the method is simple, efficient, and stable to reconstruct biperiodic surfaces with subwavelength resolution.

© 2014 Optical Society of America

1. Introduction

Consider the scattering of a time-harmonic electromagnetic plane wave by a biperiodic structure, known as crossed grating or two-dimensional grating. Scattering theory in periodic structures has many applications in micro-optics including the design and fabrication of optical elements such as corrective lenses, anti-reflective interfaces, beam splitters, and sensors. Depending on the direction and polarization of the incident plane wave, the governing mathematical model can be simplified from the full three-dimensional Maxwell equations to two fundamental polarizations: the transverse electric polarization and the transverse magnetic polarization, known as linear grating or one-dimensional grating. In both polarizations, the scalar components of electromagnetic waves satisfy the two-dimensional Helmholtz equation. We refer to the monograph [1] for a good introduction to the problems of electromagnetic diffraction.

Recently, the scattering problems in periodic structures have been studied extensively on both mathematical and numerical aspects. We refer to [2] and references therein for the mathematical studies of existence and uniqueness of the diffraction grating problems. Numerical methods can be found in [35] for either an integral equation approach or a variational approach. A comprehensive review can be found in [6] on diffractive optics technology and its mathematical modeling as well as computational methods.

We consider an inverse problem, which is to reconstruct the grating surface from a measured data field at a constant height above the surface. The mathematical questions on uniqueness and stability of the inverse problem have been studied by many researchers for the one-dimensional grating [711]. Computationally, a number of methods have been developed for the reconstruction of perfectly conducting grating surfaces in the transverse electric polarization [1219]. We refer to [2023] for related optimal design problems in diffractive optics, which are to design grating structures to obtain some specified diffraction patterns, and [2427] for general inverse surface scattering problems. These work addressed conventional far-field imaging, where the scattering data is taken at distances which are greater than the wavelength of the incident field. The role of evanescent wave components were ignored and the resolution of reconstructions was limited by Rayleigh’s criterion, approximately half of the incident wavelength, also known as the diffraction limit. We refer to [2831] for the existence, uniqueness, and numerical approximations of solutions for the direct two-dimensional grating problems. Mathematical studies can be found in [3237] on the uniqueness results for detecting biperiodic grating surfaces. In contrast, numerical results are very rare due to nonlinearity and ill-posedness of the inverse problem plus complexity of the three-dimensional Maxwell equations. A qualitative imaging method can be found in [38] for solving an inverse scattering problem from penetrable biperiodic structures. More recent reviews can be found in [39,40] on the direct and inverse scattering problems in periodic media.

In this work, we develop an efficient and stable computational method to solve the inverse problem. The grating surface is assumed to be a small and smooth deformation of a planar surface. Based on the transformed field expansion, the method reduces the boundary value problem into a successive sequence of two-point boundary value problems. For transformed field expansion and related boundary perturbation method, we refer to [4147] for solving the direct and inverse diffraction grating problems. An explicit reconstruction formula is derived for the linearized inverse problem by dropping higher order terms in the expansions. A spectral cutoff regularization is adopted to suppress the exponential growth of the noise in the evanescent wave components, which carry high spatial frequency of the scattering surface and contribute to the super resolution in the near-field regime. The method requires only a single incident field with one polarization, one frequency, and one incident direction, and is realized by using the fast Fourier transform. The numerical results are computed by using synthetic scattering data provided by an adaptive edge element method with a perfectly matched absorbing layer [29]. Two numerical examples, one smooth surface and one non-smooth surface, are presented to demonstrate the effectivenss of the proposed method. The influence is carefully investigated on the reconstructions for such parameters as surface deformation, measurement distance, and noise level. The numerical results show that the method is simple, efficient, stable to reconstruct biperiodic grating surfaces with subwavelength resolution.

This paper significantly extends our previous work on near-field imaging of one-dimensional surfaces [4850], where the two-dimensional scalar Helmholtz equation was considered, to two-dimensional grating surfaces. Apparently, the techniques differ greatly from existing work because of the complicated model problem of three-dimensional Maxwell’s equations. To the best of our knowledge, we develop the first quantitative method for solving the inverse diffraction grating problem of Maxwell’s equations in biperiodic structures and provide numerical examples of reconstruction with super resolved resolution. We point out a closely related work on the inverse surface scattering in near-field imaging [51], where the scattering surface is assumed to be a small and local perturbation of a planar surface. Other related work may be found in [52, 53] for solving an inverse medium scattering problem in near-field optical imaging and in [54, 55] for resolution and stability analysis of conductivity imaging.

2. Model problem

In this section, we define some notations and introduce a boundary value problem for the diffraction by a biperiodic structure.

2.1. Maxwell’s equations

Let us first specify the diffraction grating problem geometry. Denote (ρ, z) ∈ ℝ3, where ρ = (x, y) ∈ ℝ2. As seen in Fig. 1, the problem may be restricted to a single period of Λ = (Λ1, Λ2) in ρ due to the periodicity of the structure. Let the surface in one period be described by S = {(ρ, z) ∈ ℝ3 : z = ϕ(ρ), 0 < x < Λ1, 0 < y < Λ2}, where ϕC2(ℝ2) is a biperiodic function satisfying ϕ(ρ + Λ, z) = ϕ(ρ, z). The grating surface function ϕ is assumed to be in the form

ϕ(ρ)=εψ(ρ),
where ψC2(ℝ2) is a biperiodic function with period Λ and is called the grating profile, and ε is sufficiently small and is called the surface deformation parameter.

 figure: Fig. 1

Fig. 1 Geometry of the diffraction grating problem.

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Denote by ΩS = {(ρ, z) ∈ ℝ3 : z > ϕ(ρ), 0 < x < Λ1, 0 < y < Λ2} the space above S, which is filled with some homogeneous medium characterized by a positive constant wavenumber κ. Denote Ω = {(ρ, z) ∈ ℝ3 : ϕ(ρ) < z < h, 0 < x < Λ1, 0 < y < Λ2} be the domain bounded below by S and bounded above by the plane surface Γ = {(ρ, z) ∈ ℝ3 : z = h, 0 < x < Λ1, 0 < y < Λ2}, where h > max0 < x < Λ1,0 < y < Λ2ϕ(ρ).

Let (Einc, Hinc) be the incoming plane waves that are incident upon the grating surface from above, where

Einc=texp(iκ(αρβz)),Hinc=sexp(iκ(αρβz)).
Here α = (α1, α2), α1 = sinθ1 cosθ2, α2 = sinθ1 sinθ2, and β = cosθ1, where θ1 and θ2 are the latitudinal and longitudinal incident angles, which satisfy 0 ≤ θ1 < π/2, 0 ≤ θ2 < 2π. Denote by d = (α1, α2, −β) the unit propagation direction vector. The unit polarization vectors t = (t1, t2, t3), s = (s1, s2, s3) satisfy t · d = 0 and s = d × t, which gives explicitly
s1=α2t3+βt2,s2=(α1t3+βt1),s3=α1t2α2t1.
For normal incident, i.e., θ1 = 0, we have α1 = 0, α2 = 0, β = 1, and s1 = t2, s2 = −t1, s3 = 0. Hence we get from |t| = |s| = 1 that t12+t22=1, t3 = 0.

For the sake of simplicity, we focus on the case of normal incidence from now on since our method requires only a single incident wave for solving the inverse problem. We mention that the method works for general non-normal incidence with obvious modifications.

Denote Einc=(E1inc,E2inc,E3inc) and Hinc=(H1inc,H2inc,H3inc). Under the normal incidence, the incoming plane waves (2) reduce to

Ejinc=tjexp(iκz),Hjinc=sjexp(iκz).
It can be verified that the incident electromagnetic waves satisfy the three-dimensional time-harmonic Maxwell equation:
×EinciκHinc=0,×Hinc+iκEinc=0,in3.

The scattering of time-harmonic electromagnetic waves follows Maxwell’s equations in the space above the grating surface:

×EiκH=0,×H+iκE=0,inΩS,
where E is the total electric field and H is the total magnetic field. Due to the homogeneous medium, the electromagnetic fields satisfy the divergence free condition:
E=0andH=0inΩS.
We consider the perfect electric conductor condition:
E×νS=0onS,
where νS = (ν1, ν2, ν3) ∈ ℝ3 is the unit normal vector on S, given explicitly as
ν1=ϕx(1+ϕx2+ϕy2)1/2,ν2=ϕy(1+ϕx2+ϕy2)1/2,ν3=1(1+ϕx2+ϕy2)1/2.
Here ϕx = xϕ(x, y) and ϕy = yϕ(x, y) are the partial derivatives.

2.2. Transparent boundary condition

To reduced the diffraction grating problem from an unbounded domain ΩS into a bounded domain Ω, a transparent boundary condition needs to be imposed on Γ.

Let n = (n1, n2) ∈ 𝕑2 and denote αn = (α1n, α2n), where α1n = 2πn11 and α2n = 2πn22. For a biperiodic function u(ρ) with period Λ in ρ, it has the Fourier series expansion

u(ρ)=n2unexp(iαnρ),un=Λ11Λ210Λ10Λ2u(ρ)exp(iαnρ)dρ.
For any vector field u = (u1, u2, u3), denote its tangential component on Γ by
uΓ=νΓ×(u×νΓ)=(u1(ρ,h),u2(ρ,h),0),
where νΓ = (0, 0, 1) is the unit normal vector on Γ.

For any tangential vector u(ρ, h) = (u1(ρ, h), u2(ρ, h), 0) on Γ, where uj is a biperiodic function in ρ with period Λ, we define a boundary operator T :

Tu=(v1(ρ,h),v2(ρ,h),0),
where vj is also a biperiodic function in ρ with the same period Λ. Here uj and vj have the following Fourier expansions
uj(ρ,h)=n2ujn(h)exp(iαnρ),vj(ρ,h)=n2vjn(h)exp(iαnρ),
and the Fourier coefficients ujn and vjn satisfy
{v1n(h)=1κβn[(κ2α2n2)u1n(h)+α1nα2nu2n(h)],v2n(h)=1κβn[(κ2α1n2)u2n(h)+α1nα2nu1n(h))].

Using the boundary operator (9), we may derive the transparent boundary condition [29]:

(×E)×νΓ=iκTEΓ+f,onΓ,
where
f=iκ(Hinc×νΓTEΓinc)=(f1,f2,0).
Recalling the incident fields (3) and using the boundary operator (9), we have explicitly that
f1=2iκt1exp(iκh)andf2=2iκt2exp(iκh).

2.3. Reduced model problem

Taking curl on both sides of (5), we may eliminate the magnetic field and deduce a boundary value problem for the electric field:

{×(×E)κ2E=0inΩ,E×νS=0onS,(×E)×νΓiκTEΓ=fonΓ.

It is convenient introducing an equivalent scalar form of the problem (11) in order to apply the transformed field expansion. Denote E = (E1, E2, E3). Noting the divergence condition (6), we may reduce Maxwell’s equations to the Helmholtz equation for Ej:

ΔEj+κ2Ej=0inΩ.
The divergence free condition (6) can be explicitly written as
xE1+yE2+zE3=0inΩ.
Substituting (8) into (7) yields
E2+ϕyE3=0,E1+ϕxE3=0,ϕyE1ϕxE2=0,onz=ϕ(ρ).
The transparent boundary condition (10) becomes
zE1xE3=iκH1+f1,zE2yE3=iκH2+f2,onz=h,
where the Fourier coefficients of H1 and H2 are given by
{H1n(h)=1κβn[(κ2α2n2)E1n(h)+α1nα2nE2n(h)],H2n(h)=1κβn[(κ2α1n2)E2n(h)+α1nα2nE1n(h)].
Here E1n(h) and E2n(h) are the Fourier coefficients of E1(ρ, h) and E2(ρ, h), respectively.

Given the incident field, the direct problem is to solve the boundary value problem (12)(15) for the known surface function ϕ(ρ). The inverse problem is to reconstruct the function ϕ(ρ) from the tangential trace of the total field measured at Γ, i.e., Eδ(ρ,h)×νΓ=(E2δ(ρ,h),E1δ(ρ,h),0), where δ is the noise level. In particular, we are interested in the inverse problem in the near-field regime where the measurement distance h is much smaller than the wavelength λ = 2π/κ.

3. Transformed field expansion

In this section, we introduce a transformed field expansion to find a power series solution for the direct problem (12)(15).

3.1. Change of variables

The transformed field expansion method begins with the change of variables:

x˜=x,y˜=y,z˜=h(zϕhϕ),
which maps the domain Ω into a rectangular slab
D={(x˜,y˜,z˜)3:0<z˜<h}=2×(0,h).

Introduce a new function = (1, 2, 3) and let j(, , ) = Ej(x, y, z) under the transformation. After tedious but straightforward calculations, it can be verified from (12) that the total electric field, upon dropping the tilde, satisfies the equation

c12Ejx2+c12Ejy2+c22Ejz2c32Ejxzc42Ejyzc5Ejz+κ2c1Ej=0inD,
where
c1=(hϕ)2,c2=(ϕx2+ϕy2)(hz)2+h2,c3=2ϕx(hz)(hϕ),c4=2ϕy(hz)(hϕ),c5=(hz)[(ϕxx+ϕyy)(hϕ)+2(ϕx2+ϕy2)].
The divergence free condition (13) becomes
xE1+yE2(hzhϕ)(ϕxzE1+ϕyzE2)+(hhϕ)zE3=0inD.
The perfect electric conductor condition (14) is
E2+ϕyE3=0,E1+ϕxE3=0,ϕyE1ϕxE2=0,z=0.
The transparent boundary condition (15) on z = h reduces to
(hhϕ)zE1xE3=iκH1+f1,(hhϕ)zE2yE3=iκH2+f2.

3.2. Power series solution

Consider a formal expansion of Ej in a power series of ε:

Ej(ρ,z;ε)=k=0Ej(k)(ρ,z)εk.
Substituting ϕ = εψ into cj and inserting (20) into (16), we may derive
ΔEj(k)+κ2Ej(k)=Fj(k)inD,
where
Fj(k)=2ψh2Ej(k1)x2+2ψh2Ej(k1)y2+2(hz)ψxh2Ej(k1)xx+2(hz)ψyh2Ej(k1)yz+(hz)(ψxx+ψyy)hEj(k1)z+2κ2ψhEj(k1)ψ2h22Ej(k2)x2ψ2h22Ej(k2)y2(hz)2(ψx2+ψy2)h22Ej(k2)z22ψψx(hz)h22Ej(k2)xz2ψψy(hz)h22Ej(k2)yz+(hz)[2(ψx2+ψy2)ψ(ψxx+ψyy)]h2Ej(k2)zκ2ψ2h2Ej(k2).
Here ψx = xψ(x, y) and ψy = yψ(x, y) are the partial derivatives.

Substituting (20) into the divergence free condition (17) yields

xE1(k)+yE2(k)+zE3(k)=w(k)inD,
where
w(k)=ψh(xE1(k1)+yE2(k1))+(hzh)(ψxzE1(k1)+ψyzE2(k1)).
The perfect electric conductor boundary condition (18) can be written as
E1(k)=u1(k),E2(k)=u2(k),
where
u1(k)(ρ)=ψxE3(k1),u2(k)=ψyE3(k1).
Evaluating (22) at z = 0, we have
xE1(k)+yE2(k)+zE3(k)=u3(k),z=0,
where
u3(k)(ρ)=ψh(xE1(k1)+yE2(k1))+(ψxzE1(k1)+ψyzE2(k1)).

Inserting (20) into the transparent boundary condition (19), we get

{zE1(k)xE3(k)=iκH1(k)+v1(k),zE2(k)yE3(k)=iκH2(k)+v2(k),
where
v1(0)=f1,v1(1)=ψhzE1(0),v1(k)=ψh(xE3(k1)+iκH1(k1)),v2(0)=f2,v2(1)=ψhzE2(0),v2(k)=ψh(yE3(k1)+iκH2(k1)),
and the Fourier coefficients of H1(k)(ρ,h) and H2(k)(ρ,h) are
{H1n(k)(h)=1κβn[(κ2α2n2)E1n(k)(h)+α1nα2nE2n(k)(h)],H2n(k)(h)=1κβn[(κ2α1n2)E2n(k)(h)+α1nα2nE1n(k)(h)].
Here E1n(k)(h) and E2n(k)(h) are the Fourier coefficients of E1(k)(ρ,h) and E1(k)(ρ,h), respectively. The divergence free condition (22) on z = h reduces to
xE1(k)+yE2(k)+zE3(k)=v3(k),
where
v3(k)(ρ)=ψh(xE1(k1)+yE2(k1)).

Clearly, the problem (21)(26) for Ej(k) involves Fj(k), uj(k), vj(k), which depend only on previous two terms of Ej(k1) and Ej(k2). Thus, the transformed problem (21)(26) indeed can be solved efficiently in a recursive manner starting from k = 0.

3.3. Zeroth order term

Recalling the recurrence relation (21) and letting k = 0, we have

ΔEj(0)+κ2Ej(0)=0inD.
The divergence free condition (22) reduces to
xE1(0)+yE2(0)+zE3(0)=0inD.
The perfect electric conductor boundary condition (23) is
E1(0)(ρ,0)=0,E2(0)(ρ,0)=0.
Using (28) and (29), we have
zE3(0)(ρ,0)=xE1(0)(ρ,0)yE2(0)(ρ,0)=0.

The transparent boundary condition (25) becomes

{zE1(0)(ρ,h)xE3(0)(ρ,h)=iκH1(0)(ρ,h)+f1(ρ),zE2(0)(ρ,h)yE3(0)(ρ,h)=iκH2(0)(ρ,h)+f2(ρ).
In addition, the divergence free condition (28) gives one more boundary condition:
xE1(0)(ρ,h)+yE2(0)(ρ,h)+zE3(0)(ρ,h)=0.

Since Ej(0)(ρ,z) and fj are periodic functions of ρ, they have the Fourier expansions

Ej(0)(ρ,z)=n2Ejn(0)(z)exp(iαnρ),fj=n2fjnexp(iαnρ),
where fj0 = −2iκtj exp(−iκh) and fjn = 0 for n ≠ 0.

Substituting (33) into (27), we derive a second order ordinary differential equation for the Fourier coefficient Ejn(0)(z):

d2Ejn(0)(z)dz2+(κ2|αn|2)Ejn(0)(z)=0,0<z<h.
Similarly, we have the boundary conditions at z = 0:
E1n(0)=0,E2n(0)=0,E3n(0)=0,
and the boundary conditions at z = h:
{E1n(0)iα1nE3n(0)=iβn[(κ2α2n2)E1n(0)+α1nα2nE2n(0)]+f1n,E2n(0)iα2nE3n(0)=iβn[(κ2α1n2)E2n(0)+α1nα2nE1n(0)]+f2n,E3n(0)+iα1nE1n(0)+iα2nE2n(0)=0.
Simple calculations yield that the solution of the two-point bounary value problem (34)(36) is
Ej0(ρ,z)=tj(exp(iκz)exp(iκz)).

3.4. First order term

Recalling (21) and letting k = 1, we have

ΔEj(1)+κ2Ej(1)=Fj(1)inD,
where
Fj(1)=2ψh2Ej(0)x2+2ψh2Ej(0)y2+2(hz)ψxh2Ej(0)xz+2(hz)ψyh2Ej(0)yz+(hz)(ψxx+ψyy)hEj(0)z+2κ2ψhEj(0).
It follows from (37) that we have explicitly
Fj(1)(ρ,z)=2κ2tjhψ(exp(iκz)exp(iκz))iκtj(hz)h(ψxx+ψyy)(exp(iκz)+exp(iκz)).
The divergence free condition (22) reduces to
xE1(1)+yE2(1)+zE3(1)=w(1)inD,
where
w(1)(ρ,z)=ψh(xE1(0)+yE2(0))+(hzh)(ψxzE1(0)+ψyzE2(0))=iκ(hz)h(t1ψx+t2ψy)(exp(iκz)+exp(iκz)).
The perfectly conducting boundary condition (23) on z = 0 is
E1(1)(ρ,0)=u1(1)(ρ)=ψx(ρ)E3(0)(ρ,0)=0,E2(1)(ρ,0)=u2(1)(ρ)=ψy(ρ)E3(0)(ρ,0)=0.
Evaluating (39) at z = 0 gives
zE3(1)(ρ,0)=w(1)(ρ,0)xE1(1)(ρ,0)yE2(1)(ρ,0)=2iκ(t1ψx+t2ψy).
The transparent boundary condition (25) becomes
zE1(1)xE3(1)=iκH1(1)+v1(1),zE2(1)yE3(1)=iκH2(1)+v2(1),
where
vj(1)(ρ)=ψhzEj(0)(ρ,h)=iκtjh(exp(iκh)+exp(iκh))ψ.
The divergence free condition (26) gives one more boundary condition on z = h:
xE1(1)(ρ,h)+yE2(1)(ρ,z)+zE3(1)(ρ,z)=0.
Consider the Fourier expansions for periodic functions of Ej(1)(ρ,z), Fj(1)(ρ,z), and ψ(ρ):
ψ(ρ)=n2ψnexp(iαnρ),Ej(1)(ρ,z)=n2Ejn(1)(z)exp(iαnρ),Fj(1)(ρ,z)=n2Fjn(1)(z)exp(iαnρ),
where
Fjn(1)(z)=[2κ2tjh(exp(iκz)exp(iκz))+iκtj(hz)h(α1n2+α2n2)(exp(iκz)+exp(iκz))]ψn.

Substituting the above Fourier expansions into (38), we derive an equation for Ejn(1)(z):

d2Ejn(1)(z)dz2+(κ2|αn|2)Ejn(1)(z)=Fjn(1)(z),0<z<h,
together with boundary conditions at z = 0:
E1n(1)=0,E2n(1)=0,E3n(1)(0)=2κ(α1n+α2n)ψn,
and boundary conditions at z = h:
{E1n(1)iα1nE3n(1)=iβn[(κ2α2n2)E1n(1)+α1nα2nE2n(1)]+v1n(1),E2n(1)iα2nE3n(1)=iβn[(κ2α1n2)E2n(1)+α1nα2nE1n(1)]+v2n(1),E3n(1)+iα1nE1n(1)+iα2nE2n(1)=0,
where v1n(1) and v1n(2) are the Fourier coefficients of v1(1)(ρ) and v2(1)(ρ). Explicitly, we have
vjn(1)=iκtjh(exp(iκh)+exp(iκh))ψn.

Following straightforward but tedious calculations, we may solve the two-point boundary value problem (41)(43) and obtain elegant equations

E1n(1)(h)=2iκt1exp(iβnh)ψnandE2n(1)(h)=2iκt2exp(iβnh)ψn,
which relate the Fourier coefficients of order one terms Ej(1)(ρ,h) with the Fourier coefficient of the grating profile function ψ(ρ).

4. Reconstruction formula

Assume that the noisy data takes the form

Ejδ(ρ,h)=Ej(ρ,h)+𝒪(δ),
where Ej(ρ, h), j = 1, 2 is the exact data and δ is the noise level.

Evaluating the power series (20) at z = h and replacing Ej(ρ, h) with Ejδ(ρ,h), we have

Ej(ρ,h)=Ej(0)(ρ,h)+εEj(1)(ρ,h)+𝒪(ε2)+𝒪(δ).
Rearranging (45), and dropping 𝒪(ε2) and 𝒪(δ) yield
εEj(1)(ρ,h)=Ejδ(ρ,h)Ej(0)(ρ,h)
which is the linearization of the nonlinear inverse problem and enables us to find an explicit reconstruction formula for the linearized inverse problem.

Noting ϕ = εψ and thus ϕn = εψn, where ϕn is the Fourier coefficient of ϕ. Substituting (44) into (46), we deduce that

ϕn=(2iκtj)1[Ejnδ(h)Ejn(0)(h)]exp(iβnh),
where Ejnδ(h) is the Fourier coefficient of the noisy data Ejδ(x,h) and Ejn(0)(h) is the Fourier coefficient of Ej(0)(x,h) given as
Ejn(0)(h)=tj(exp(iκh)exp(iκh))δ0n.
Here δ0n the Kronecker’s delta function.

It follows from the definition of βn and (47) that it is well-posed to reconstruct those Fourier coefficients ϕn with |αn| < κ, since the small variations of the measured data will not be amplified and lead to large errors in the reconstruction, but the resolution of the reconstructed function ϕ is restricted by the given wavenumber κ. In contrast, it is severely ill-posed to reconstruct those Fourier coefficients ϕn with |αn| > κ, since the small variations in the data will be exponentially enlarged and lead to huge errors in the reconstruction, but they contribute to the super resolution of the reconstructed function ϕ. To obtain a stable and super-resolved reconstruction, we may adopt a regularization to suppress the exponential growth of the reconstruction errors.

Following [51], we consider the spectral cut-off regularization. Define the signal-to-noise ratio (SNR) by

SNR=min{ε2,δ1}.
For fixed h, the cut-off wavenumber κc is chosen in such a way that
exp((κc2κ2)1/2h)=SNR,
which implies that the spatial frequency will be cut-off for those below the noise level. More explicitly, we have
κcκ=[1+(logSNRκh)2]1/2,
which indicates κc > κ as long as SNR > 0 and super resolution may be achieved.

Taking into account the frequency cut-off, we have a regularized reconstruction formulation

ϕn=(2iκtj)1[Ejnδ(h)Ejn(0)(h)]exp(iβnh)χn,
where the characteristic function
χn={1for|αn|κc,0for|αn|>κc.
Once ϕn are computed, the grating surface function can be approximated by
ϕ(ρ)nϕnexp(iαnρ)=|αn|κc(2iκtj)1[Ejnδ(h)Ejn(0)(h)]exp(i(αnρβnh))=|αn|κc(2iκtj)1Ejnδ(h)exp(i(αnρβnh))+(2iκ)1(1exp(2iκh)).
Hence, the method requires only two fast Fourier transforms: one is done for the data to obtain Ejnδ(h) and another is done to obtain the approximated function ϕ.

5. Numerical experiment

In this section, we discuss the algorithmic implementation for the direct and inverse problems, present two numerical examples to illustrate the effectiveness of the proposed method, and examine influence of the parameters ε, h, and δ on the reconstruction results. As seen in Fig. 2, we consider two types of grating profiles: one is a smooth function with finitely many Fourier modes and another is a non-smooth function with infinitely many Fourier modes. Although the method requires that the grating profile function ψ(ρ) is C2(ℝ2), it is still applicable to non-smooth functions numerically.

 figure: Fig. 2

Fig. 2 The exact grating profile ψ. (a) Example 1: smooth grating profile with finite Fourier modes; (b) Example 2: non-smooth grating profile with infinite Fourier modes.

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The first-order Nédélec edge element is used for solving the direct problem and obtaining the synthetic scattering data. Uniaxial perfect matched layer (PML) boundary condition is imposed on z direction so that no artificial wave reflection occurs to ruin the wave field inside the domain. Adaptive refinement technique [29] is used to achieve the solution having a specified accuracy in an optimal fashion. Our implementation is based on parallel hierarchical grid (PHG) [56], which is a toolbox for developing parallel adaptive finite element programs on unstructured tetrahedral meshes and it is under active development at the State Key Laboratory of Scientific and Engineering Computing. The finite elements implemented in PHG are the Largrange element, hierarchical H1 and H(curl) element. In order to generate the tetrahedral mesh with a biperiodic structure, we generate a non-uniform hexahedral mesh firstly and divide each hexahedron into six tetrahedrons. The linear system resulted from finite element discretization is solve by the multifrontal massively parallel sparse direct solver [57, 58].

In the following two examples, the incident wave is taken as Einc = (1, 0, 0)exp(−iκz), i.e., t1 = 1 and t2 = t3 = 0, and only the first component of the electric field, E1(ρ, h), needs to be measured. The wavenumber is κ = 2π, which corresponds to the wavenlength λ = 1. Define by R the unit rectangular domain, i.e., R = [0, 1.0λ] × [0, 1.0λ]. The computational domain is R × [ϕ, 1.0λ] with the PML region R × [0.5λ, 1.0λ]. The scattering data E1(ρ, h) is obtained by interpolation into the uniform 256 × 256 grid points on the measurment plane z = h. In all the figures, the plots are rescaled with respect to the wavelength λ to clearly show the relative size, and the meshes are done in 32 × 32 instead of 256 × 256 grid points in order to reduce the display sizes. To test the stability of the method, some relative random noise is added to the scattering data, i.e., the scattering data takes the form

E1δ(ρ,h)=E1(ρ,h)(1+δrand),
where rand stands for uniformly distributed random numbers in [−1, 1]. The relative L2(R) error is defined by
e=ϕϕδ,ε0,Rϕ0,R,
where ϕ is the exact surface function and ϕδ,ε is the reconstructed surface function.

Example 1. This example illustrates the reconstruction results of a smooth grating profile with finitely many Fourier modes. The exact grating surface function is given by ϕ(ρ) =εψ(ρ), where the grating profile function

ψ(x,y)=0.6sin(2πx)sin(2πy)+sin(4πx)sin(4πy).

First, consider the surface deviation parameter ε. The measurement is taken at h = 0.4λ and no additional random noise is added to the scattering data, i.e., δ = 0. This test is to investigate the influence of surface deviation parameter on the reconstructions. In (46), higher order terms of ε are dropped in the power series to linearize the inverse problem and to obtain the explicit reconstruction formula. As expected, the smaller the surface deviation ε is, the more accurate is the approximation of the linearized model to the original nonlinear model problem. Table 1 shows the relative L2(R) error of the reconstructions with four different surface deformation parameter ε = 0.2λ, 0.1λ, 0.05λ, 0.025λ for fixed measurement distance h = 0.4λ. It is clear to note that the error decreases from 85.0% to 9.86% as ε decreases from 0.2λ to 0.025λ.

Tables Icon

Table 1. Example 1: Relative error of the reconstructions by using different ε with h = 0.4λ and δ = 0.0.

Next, consider the noise level δ and the measurement distance h. In practice, the scattering data always contains certain level of noise. To test the stability and super resolving capability of the method, we add an amount of 5% random noise to the scattering data. Table 2 reports the relative L2(R) error of the reconstructions with four different measurement distance h = 0.4λ, 0.3λ, 0.2λ, 0.1λ for fixed ε = 0.025λ. Comparing the results for the same ε = 0.025λ and h = 0.4λ in Tables 1 and 2, we can see that the relative error increases dramatically from 9.86% by using noise free data to 86.3% by using 5% noise data. The reason is that a smaller cut-off should be chosen to suppress the expotentially increasing noise in the data and thus the Fourier modes of the exact grating surface function can not be recovered for those higher than the cutoff frequency, which leads to a large error and poor resolution in the reconstruction. A smaller measurement distance is desirable in order to have a large cut-off frequency, which enhances the resolution and reduces the error. As can be seen in Table 2, the reconstruction error decreases from 86.3% by using h = 0.4λ to as low as 12.0% by using h = 0.1λ even for 5% noise data. Figure 3 plots the reconstructed surfaces by using h = 0.4λ, 0.3λ, 0.2λ, 0.1λ. Comparing the exact surface profile in Fig. 2(a) and the reconstructed surface in Fig. 3(d), we can see that the reconstruction is nearly perfect and the difference is really minor by carefully checking the contour plots.

Tables Icon

Table 2. Example 1: Relative error of the reconstructions by using different h with ε = 0.025λ and δ = 5%.

 figure: Fig. 3

Fig. 3 Example 1: Reconstructed grating surfaces by using different h with ε = 0.025λ and δ = 5%. (a) h = 0.4λ; (b) h = 0.3λ; (c) h = 0.2λ; (d) h = 0.1λ.

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Example 2. This example illustrates the reconstruction results of a non-smooth grating profile with infinitely many Fourier modes, as seen in Fig. 2(b). The exact grating surface function is given by ϕ(ρ) = εψ(ρ), where the grating profile function

ψ(x,y)=|sin(2πx)sin(2πy)||cos(2πx)cos(2πy)|.
Clearly, the profile function (51) is nondifferentiable and its Fourier coefficients decay slowly. Comparing with the grating profile (50), it is more challenging to obtain as good reconstructions as those in Example 1 since a much higher cutoff frequency is desirable to recover as many Fourier modes as possible for (51).

Consider the influence of ε by using noise-free data. The measurement is taken at h = 0.2λ. Table 3 presents the relative L2(R) error of the reconstructions with four different surface deformation parameter ε = 0.1λ, 0.05λ, 0.025λ, 0.0125λ. The error decreases from 72.9% to 15.2% as ε decreases from 0.1λ to 0.0125λ. Based on these results, the following observation can be made: a smaller deformation parameter ε yields a better reconstruction; smaller ε and h are required in order to obtain comparable error with that in Table 2 for Example 1 due to the non-smooth nature of the grating surface function of Example 2.

Tables Icon

Table 3. Example 2: Relative error of the reconstructions by using different ε with h = 0.2λ and δ = 0.0.

Next, consider the influence of the noise level δ and the measurement distance h. An amount of 5% random noise is added to the scattering data. Table 4 reports the relative L2(R) error of the reconstructions with four different measurement distance h = 0.2λ, 0.1λ, 0.05λ, 0.025λ for fixed ε = 0.0125λ. Comparing the results for the same ε = 0.0125λ and h = 0.2λ in Tables 3 and 4, we can see that the relative error is more than doubled from 15.2% by using noise-free data to 39.9% by using 5% noise data. Again, the reason is that a smaller cut-off is chosen to suppress the expotentially increasing noise in the data and thus higher Fourier modes of the exact grating surface function can not be recovered. A smaller measurement distance helps to enhance the resolution and reduce the error. In Table 4, the reconstruction error decreases from 39.9% by using h = 0.2λ to as low as 11.9% by using h = 0.025λ. Figure 4 shows the reconstructed surfaces by using h = 0.2λ, 0.1λ, 0.05λ, 0.025λ. Comparing the exact surface profile in Fig. 2(b) and the reconstructed surface in Fig. 4(d), we can see that a good reconstruction can still be possible when using a small measurement distance.

Tables Icon

Table 4. Example 2: Relative error of the reconstructions by using different h with ε = 0.0125λ and δ = 5%.

 figure: Fig. 4

Fig. 4 Example 2: Reconstructed grating surfaces by using different h with ε = 0.0125λ and δ = 5%. (a) h = 0.2λ; (b) h = 0.1λ; (c) h = 0.05λ; (d) h = 0.025λ.

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6. Conclusion

We have presented a simple, stable, and effective computational method for reconstructing biperiodic grating surfaces and achieved subwavelength resolution. Using the transformed field and Fourier expansions, we deduced a power series solution for the direct problem. By dropping higher order terms in power series, we linearized the nonlinear inverse problem and obtained an explicit reconstruction formula, which was implemented by using the fast Fourier transform. We considered two examples, one of which has finite Fourier modes and another one has infinite Fourier modes, and investigated how the parameters influence the reconstructions. The results show that super resolution may be achieved by using small measurement distance. As for future work, we will study a more complicated transmission problem where the surface is penetrable, and investigate the mathematical issues such as uniqueness, stability, and resolution.

Acknowledgments

The research of GB was supported in part by the NSF grants DMS-0968360 and DMS-1211292, the ONR grant N00014-12-1-0319, a Key Project of the Major Research Plan of NSFC (No. 91130004), and a special research grant from Zhejiang University. The research of TC was partially supported by the National Basic Research Project Grant 2011CB309700, the National High Technology Research and Development Program of China Grant 2012AA01A309, and NSFC Grants 11101417 and 11171334. The research of PL was supported in part by the NSF grant DMS-1151308.

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Figures (4)

Fig. 1
Fig. 1 Geometry of the diffraction grating problem.
Fig. 2
Fig. 2 The exact grating profile ψ. (a) Example 1: smooth grating profile with finite Fourier modes; (b) Example 2: non-smooth grating profile with infinite Fourier modes.
Fig. 3
Fig. 3 Example 1: Reconstructed grating surfaces by using different h with ε = 0.025λ and δ = 5%. (a) h = 0.4λ; (b) h = 0.3λ; (c) h = 0.2λ; (d) h = 0.1λ.
Fig. 4
Fig. 4 Example 2: Reconstructed grating surfaces by using different h with ε = 0.0125λ and δ = 5%. (a) h = 0.2λ; (b) h = 0.1λ; (c) h = 0.05λ; (d) h = 0.025λ.

Tables (4)

Tables Icon

Table 1 Example 1: Relative error of the reconstructions by using different ε with h = 0.4λ and δ = 0.0.

Tables Icon

Table 2 Example 1: Relative error of the reconstructions by using different h with ε = 0.025λ and δ = 5%.

Tables Icon

Table 3 Example 2: Relative error of the reconstructions by using different ε with h = 0.2λ and δ = 0.0.

Tables Icon

Table 4 Example 2: Relative error of the reconstructions by using different h with ε = 0.0125λ and δ = 5%.

Equations (87)

Equations on this page are rendered with MathJax. Learn more.

ϕ ( ρ ) = ε ψ ( ρ ) ,
E inc = t exp ( i κ ( α ρ β z ) ) , H inc = s exp ( i κ ( α ρ β z ) ) .
s 1 = α 2 t 3 + β t 2 , s 2 = ( α 1 t 3 + β t 1 ) , s 3 = α 1 t 2 α 2 t 1 .
E j inc = t j exp ( i κ z ) , H j inc = s j exp ( i κ z ) .
× E inc i κ H inc = 0 , × H inc + i κ E inc = 0 , in 3 .
× E i κ H = 0 , × H + i κ E = 0 , in Ω S ,
E = 0 and H = 0 in Ω S .
E × ν S = 0 on S ,
ν 1 = ϕ x ( 1 + ϕ x 2 + ϕ y 2 ) 1 / 2 , ν 2 = ϕ y ( 1 + ϕ x 2 + ϕ y 2 ) 1 / 2 , ν 3 = 1 ( 1 + ϕ x 2 + ϕ y 2 ) 1 / 2 .
u ( ρ ) = n 2 u n exp ( i α n ρ ) , u n = Λ 1 1 Λ 2 1 0 Λ 1 0 Λ 2 u ( ρ ) exp ( i α n ρ ) d ρ .
u Γ = ν Γ × ( u × ν Γ ) = ( u 1 ( ρ , h ) , u 2 ( ρ , h ) , 0 ) ,
T u = ( v 1 ( ρ , h ) , v 2 ( ρ , h ) , 0 ) ,
u j ( ρ , h ) = n 2 u j n ( h ) exp ( i α n ρ ) , v j ( ρ , h ) = n 2 v j n ( h ) exp ( i α n ρ ) ,
{ v 1 n ( h ) = 1 κ β n [ ( κ 2 α 2 n 2 ) u 1 n ( h ) + α 1 n α 2 n u 2 n ( h ) ] , v 2 n ( h ) = 1 κ β n [ ( κ 2 α 1 n 2 ) u 2 n ( h ) + α 1 n α 2 n u 1 n ( h ) ) ] .
( × E ) × ν Γ = i κ T E Γ + f , on Γ ,
f = i κ ( H inc × ν Γ T E Γ inc ) = ( f 1 , f 2 , 0 ) .
f 1 = 2 i κ t 1 exp ( i κ h ) and f 2 = 2 i κ t 2 exp ( i κ h ) .
{ × ( × E ) κ 2 E = 0 in Ω , E × ν S = 0 on S , ( × E ) × ν Γ i κ T E Γ = f on Γ .
Δ E j + κ 2 E j = 0 in Ω .
x E 1 + y E 2 + z E 3 = 0 in Ω .
E 2 + ϕ y E 3 = 0 , E 1 + ϕ x E 3 = 0 , ϕ y E 1 ϕ x E 2 = 0 , on z = ϕ ( ρ ) .
z E 1 x E 3 = i κ H 1 + f 1 , z E 2 y E 3 = i κ H 2 + f 2 , on z = h ,
{ H 1 n ( h ) = 1 κ β n [ ( κ 2 α 2 n 2 ) E 1 n ( h ) + α 1 n α 2 n E 2 n ( h ) ] , H 2 n ( h ) = 1 κ β n [ ( κ 2 α 1 n 2 ) E 2 n ( h ) + α 1 n α 2 n E 1 n ( h ) ] .
x ˜ = x , y ˜ = y , z ˜ = h ( z ϕ h ϕ ) ,
D = { ( x ˜ , y ˜ , z ˜ ) 3 : 0 < z ˜ < h } = 2 × ( 0 , h ) .
c 1 2 E j x 2 + c 1 2 E j y 2 + c 2 2 E j z 2 c 3 2 E j x z c 4 2 E j y z c 5 E j z + κ 2 c 1 E j = 0 in D ,
c 1 = ( h ϕ ) 2 , c 2 = ( ϕ x 2 + ϕ y 2 ) ( h z ) 2 + h 2 , c 3 = 2 ϕ x ( h z ) ( h ϕ ) , c 4 = 2 ϕ y ( h z ) ( h ϕ ) , c 5 = ( h z ) [ ( ϕ x x + ϕ y y ) ( h ϕ ) + 2 ( ϕ x 2 + ϕ y 2 ) ] .
x E 1 + y E 2 ( h z h ϕ ) ( ϕ x z E 1 + ϕ y z E 2 ) + ( h h ϕ ) z E 3 = 0 in D .
E 2 + ϕ y E 3 = 0 , E 1 + ϕ x E 3 = 0 , ϕ y E 1 ϕ x E 2 = 0 , z = 0 .
( h h ϕ ) z E 1 x E 3 = i κ H 1 + f 1 , ( h h ϕ ) z E 2 y E 3 = i κ H 2 + f 2 .
E j ( ρ , z ; ε ) = k = 0 E j ( k ) ( ρ , z ) ε k .
Δ E j ( k ) + κ 2 E j ( k ) = F j ( k ) in D ,
F j ( k ) = 2 ψ h 2 E j ( k 1 ) x 2 + 2 ψ h 2 E j ( k 1 ) y 2 + 2 ( h z ) ψ x h 2 E j ( k 1 ) x x + 2 ( h z ) ψ y h 2 E j ( k 1 ) y z + ( h z ) ( ψ x x + ψ y y ) h E j ( k 1 ) z + 2 κ 2 ψ h E j ( k 1 ) ψ 2 h 2 2 E j ( k 2 ) x 2 ψ 2 h 2 2 E j ( k 2 ) y 2 ( h z ) 2 ( ψ x 2 + ψ y 2 ) h 2 2 E j ( k 2 ) z 2 2 ψ ψ x ( h z ) h 2 2 E j ( k 2 ) x z 2 ψ ψ y ( h z ) h 2 2 E j ( k 2 ) y z + ( h z ) [ 2 ( ψ x 2 + ψ y 2 ) ψ ( ψ x x + ψ y y ) ] h 2 E j ( k 2 ) z κ 2 ψ 2 h 2 E j ( k 2 ) .
x E 1 ( k ) + y E 2 ( k ) + z E 3 ( k ) = w ( k ) in D ,
w ( k ) = ψ h ( x E 1 ( k 1 ) + y E 2 ( k 1 ) ) + ( h z h ) ( ψ x z E 1 ( k 1 ) + ψ y z E 2 ( k 1 ) ) .
E 1 ( k ) = u 1 ( k ) , E 2 ( k ) = u 2 ( k ) ,
u 1 ( k ) ( ρ ) = ψ x E 3 ( k 1 ) , u 2 ( k ) = ψ y E 3 ( k 1 ) .
x E 1 ( k ) + y E 2 ( k ) + z E 3 ( k ) = u 3 ( k ) , z = 0 ,
u 3 ( k ) ( ρ ) = ψ h ( x E 1 ( k 1 ) + y E 2 ( k 1 ) ) + ( ψ x z E 1 ( k 1 ) + ψ y z E 2 ( k 1 ) ) .
{ z E 1 ( k ) x E 3 ( k ) = i κ H 1 ( k ) + v 1 ( k ) , z E 2 ( k ) y E 3 ( k ) = i κ H 2 ( k ) + v 2 ( k ) ,
v 1 ( 0 ) = f 1 , v 1 ( 1 ) = ψ h z E 1 ( 0 ) , v 1 ( k ) = ψ h ( x E 3 ( k 1 ) + i κ H 1 ( k 1 ) ) , v 2 ( 0 ) = f 2 , v 2 ( 1 ) = ψ h z E 2 ( 0 ) , v 2 ( k ) = ψ h ( y E 3 ( k 1 ) + i κ H 2 ( k 1 ) ) ,
{ H 1 n ( k ) ( h ) = 1 κ β n [ ( κ 2 α 2 n 2 ) E 1 n ( k ) ( h ) + α 1 n α 2 n E 2 n ( k ) ( h ) ] , H 2 n ( k ) ( h ) = 1 κ β n [ ( κ 2 α 1 n 2 ) E 2 n ( k ) ( h ) + α 1 n α 2 n E 1 n ( k ) ( h ) ] .
x E 1 ( k ) + y E 2 ( k ) + z E 3 ( k ) = v 3 ( k ) ,
v 3 ( k ) ( ρ ) = ψ h ( x E 1 ( k 1 ) + y E 2 ( k 1 ) ) .
Δ E j ( 0 ) + κ 2 E j ( 0 ) = 0 in D .
x E 1 ( 0 ) + y E 2 ( 0 ) + z E 3 ( 0 ) = 0 in D .
E 1 ( 0 ) ( ρ , 0 ) = 0 , E 2 ( 0 ) ( ρ , 0 ) = 0 .
z E 3 ( 0 ) ( ρ , 0 ) = x E 1 ( 0 ) ( ρ , 0 ) y E 2 ( 0 ) ( ρ , 0 ) = 0 .
{ z E 1 ( 0 ) ( ρ , h ) x E 3 ( 0 ) ( ρ , h ) = i κ H 1 ( 0 ) ( ρ , h ) + f 1 ( ρ ) , z E 2 ( 0 ) ( ρ , h ) y E 3 ( 0 ) ( ρ , h ) = i κ H 2 ( 0 ) ( ρ , h ) + f 2 ( ρ ) .
x E 1 ( 0 ) ( ρ , h ) + y E 2 ( 0 ) ( ρ , h ) + z E 3 ( 0 ) ( ρ , h ) = 0 .
E j ( 0 ) ( ρ , z ) = n 2 E j n ( 0 ) ( z ) exp ( i α n ρ ) , f j = n 2 f j n exp ( i α n ρ ) ,
d 2 E j n ( 0 ) ( z ) d z 2 + ( κ 2 | α n | 2 ) E j n ( 0 ) ( z ) = 0 , 0 < z < h .
E 1 n ( 0 ) = 0 , E 2 n ( 0 ) = 0 , E 3 n ( 0 ) = 0 ,
{ E 1 n ( 0 ) i α 1 n E 3 n ( 0 ) = i β n [ ( κ 2 α 2 n 2 ) E 1 n ( 0 ) + α 1 n α 2 n E 2 n ( 0 ) ] + f 1 n , E 2 n ( 0 ) i α 2 n E 3 n ( 0 ) = i β n [ ( κ 2 α 1 n 2 ) E 2 n ( 0 ) + α 1 n α 2 n E 1 n ( 0 ) ] + f 2 n , E 3 n ( 0 ) + i α 1 n E 1 n ( 0 ) + i α 2 n E 2 n ( 0 ) = 0 .
E j 0 ( ρ , z ) = t j ( exp ( i κ z ) exp ( i κ z ) ) .
Δ E j ( 1 ) + κ 2 E j ( 1 ) = F j ( 1 ) in D ,
F j ( 1 ) = 2 ψ h 2 E j ( 0 ) x 2 + 2 ψ h 2 E j ( 0 ) y 2 + 2 ( h z ) ψ x h 2 E j ( 0 ) x z + 2 ( h z ) ψ y h 2 E j ( 0 ) y z + ( h z ) ( ψ x x + ψ y y ) h E j ( 0 ) z + 2 κ 2 ψ h E j ( 0 ) .
F j ( 1 ) ( ρ , z ) = 2 κ 2 t j h ψ ( exp ( i κ z ) exp ( i κ z ) ) i κ t j ( h z ) h ( ψ x x + ψ y y ) ( exp ( i κ z ) + exp ( i κ z ) ) .
x E 1 ( 1 ) + y E 2 ( 1 ) + z E 3 ( 1 ) = w ( 1 ) in D ,
w ( 1 ) ( ρ , z ) = ψ h ( x E 1 ( 0 ) + y E 2 ( 0 ) ) + ( h z h ) ( ψ x z E 1 ( 0 ) + ψ y z E 2 ( 0 ) ) = i κ ( h z ) h ( t 1 ψ x + t 2 ψ y ) ( exp ( i κ z ) + exp ( i κ z ) ) .
E 1 ( 1 ) ( ρ , 0 ) = u 1 ( 1 ) ( ρ ) = ψ x ( ρ ) E 3 ( 0 ) ( ρ , 0 ) = 0 , E 2 ( 1 ) ( ρ , 0 ) = u 2 ( 1 ) ( ρ ) = ψ y ( ρ ) E 3 ( 0 ) ( ρ , 0 ) = 0 .
z E 3 ( 1 ) ( ρ , 0 ) = w ( 1 ) ( ρ , 0 ) x E 1 ( 1 ) ( ρ , 0 ) y E 2 ( 1 ) ( ρ , 0 ) = 2 i κ ( t 1 ψ x + t 2 ψ y ) .
z E 1 ( 1 ) x E 3 ( 1 ) = i κ H 1 ( 1 ) + v 1 ( 1 ) , z E 2 ( 1 ) y E 3 ( 1 ) = i κ H 2 ( 1 ) + v 2 ( 1 ) ,
v j ( 1 ) ( ρ ) = ψ h z E j ( 0 ) ( ρ , h ) = i κ t j h ( exp ( i κ h ) + exp ( i κ h ) ) ψ .
x E 1 ( 1 ) ( ρ , h ) + y E 2 ( 1 ) ( ρ , z ) + z E 3 ( 1 ) ( ρ , z ) = 0 .
ψ ( ρ ) = n 2 ψ n exp ( i α n ρ ) , E j ( 1 ) ( ρ , z ) = n 2 E j n ( 1 ) ( z ) exp ( i α n ρ ) , F j ( 1 ) ( ρ , z ) = n 2 F j n ( 1 ) ( z ) exp ( i α n ρ ) ,
F j n ( 1 ) ( z ) = [ 2 κ 2 t j h ( exp ( i κ z ) exp ( i κ z ) ) + i κ t j ( h z ) h ( α 1 n 2 + α 2 n 2 ) ( exp ( i κ z ) + exp ( i κ z ) ) ] ψ n .
d 2 E j n ( 1 ) ( z ) d z 2 + ( κ 2 | α n | 2 ) E j n ( 1 ) ( z ) = F j n ( 1 ) ( z ) , 0 < z < h ,
E 1 n ( 1 ) = 0 , E 2 n ( 1 ) = 0 , E 3 n ( 1 ) ( 0 ) = 2 κ ( α 1 n + α 2 n ) ψ n ,
{ E 1 n ( 1 ) i α 1 n E 3 n ( 1 ) = i β n [ ( κ 2 α 2 n 2 ) E 1 n ( 1 ) + α 1 n α 2 n E 2 n ( 1 ) ] + v 1 n ( 1 ) , E 2 n ( 1 ) i α 2 n E 3 n ( 1 ) = i β n [ ( κ 2 α 1 n 2 ) E 2 n ( 1 ) + α 1 n α 2 n E 1 n ( 1 ) ] + v 2 n ( 1 ) , E 3 n ( 1 ) + i α 1 n E 1 n ( 1 ) + i α 2 n E 2 n ( 1 ) = 0 ,
v j n ( 1 ) = i κ t j h ( exp ( i κ h ) + exp ( i κ h ) ) ψ n .
E 1 n ( 1 ) ( h ) = 2 i κ t 1 exp ( i β n h ) ψ n and E 2 n ( 1 ) ( h ) = 2 i κ t 2 exp ( i β n h ) ψ n ,
E j δ ( ρ , h ) = E j ( ρ , h ) + 𝒪 ( δ ) ,
E j ( ρ , h ) = E j ( 0 ) ( ρ , h ) + ε E j ( 1 ) ( ρ , h ) + 𝒪 ( ε 2 ) + 𝒪 ( δ ) .
ε E j ( 1 ) ( ρ , h ) = E j δ ( ρ , h ) E j ( 0 ) ( ρ , h )
ϕ n = ( 2 i κ t j ) 1 [ E j n δ ( h ) E j n ( 0 ) ( h ) ] exp ( i β n h ) ,
E j n ( 0 ) ( h ) = t j ( exp ( i κ h ) exp ( i κ h ) ) δ 0 n .
SNR = min { ε 2 , δ 1 } .
exp ( ( κ c 2 κ 2 ) 1 / 2 h ) = SNR ,
κ c κ = [ 1 + ( log SNR κ h ) 2 ] 1 / 2 ,
ϕ n = ( 2 i κ t j ) 1 [ E j n δ ( h ) E j n ( 0 ) ( h ) ] exp ( i β n h ) χ n ,
χ n = { 1 for | α n | κ c , 0 for | α n | > κ c .
ϕ ( ρ ) n ϕ n exp ( i α n ρ ) = | α n | κ c ( 2 i κ t j ) 1 [ E j n δ ( h ) E j n ( 0 ) ( h ) ] exp ( i ( α n ρ β n h ) ) = | α n | κ c ( 2 i κ t j ) 1 E j n δ ( h ) exp ( i ( α n ρ β n h ) ) + ( 2 i κ ) 1 ( 1 exp ( 2 i κ h ) ) .
E 1 δ ( ρ , h ) = E 1 ( ρ , h ) ( 1 + δ rand ) ,
e = ϕ ϕ δ , ε 0 , R ϕ 0 , R ,
ψ ( x , y ) = 0.6 sin ( 2 π x ) sin ( 2 π y ) + sin ( 4 π x ) sin ( 4 π y ) .
ψ ( x , y ) = | sin ( 2 π x ) sin ( 2 π y ) | | cos ( 2 π x ) cos ( 2 π y ) | .
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