## 2 Formulation of the Problem  
We examine a semi-infinite parallel-plate waveguide that has a five-layer material loading and is excited by an H-polarized plane wave. The structure of the waveguide is depicted in Figure 1 and features infinitely thin plates that are uniformly conducting in the y-direction. The relative permittivity and permeability \((\varepsilon_{m},\mu_{m})\) for \(m=1,2,3,4\), and \(5\) characterize the material layers I, II, III, IV, and V, respectively. Let the total magnetic field \(\phi^{t}(x,z)[\equiv H^{t}_{y}(x,z)]\) be  
\[\phi^{t}(x,z)=\phi^{t}(x,z)+\phi(x,z) \tag{1}\]for \(-\infty<x<\infty\) and \(-\infty<z<\infty\), where \(\phi^{i}(x,z)\) is the incident field of H-polarization defined by  
\[\phi^{i}(x,z)=e^{-ik(x\sin\theta_{0}+z\cos\theta_{0})},\;0<\theta_{0}<\pi/2 \tag{2}\]  
with \(k[\equiv\omega(\varepsilon_{0}\mu_{0})^{1/2}]\) being the free-space wavenumber. The total field \(\phi^{t}(x,z)\) satisfies the two-dimensional Helmholtz equation  
\[[\partial^{2}/\partial x^{2}+\partial^{2}/\partial z^{2}+\mu(x,z)\varepsilon( x,z)k^{2}]\phi^{t}(x,z)=0, \tag{3}\]  
where  
\[\mu(x,z)=\left\{ \begin{array}{l}\mu_{1}(\text{layer I})\\ \mu_{2}(\text{layer II})\\ \mu_{3}(\text{layer III})\\ \mu_{4}(\text{layer IV})\\ \mu_{5}(\text{layer V})\\ 1\;\;(\text{otherwise})\end{array} \right.,\;\;\varepsilon(x,z)=\left\{ \begin{array} []{l}\varepsilon_{1}(\text{layer I})\\ \varepsilon_{2}(\text{layer II})\\ \varepsilon_{3}(\text{layer III})\\ \varepsilon_{4}(\text{layer IV})\\ \varepsilon_{5}(\text{layer V})\\ 1\;\;(\text{otherwise})\end{array} \right., \tag{4}\]  
Once the solution of (3) is found, we derive non-zero components of the total electromagnetic field as follows:  
\[(H_{y}^{t},E_{x}^{t},E_{z}^{t})=\left[\phi^{t},\frac{1}{i\omega\varepsilon_{0} \varepsilon(x,z)}\frac{\partial\phi^{t}}{\partial z},\frac{i}{\omega\varepsilon _{0}\varepsilon(x,z)}\frac{\partial\phi^{t}}{\partial x}\right]. \tag{5}\]  
For the convenience of analysis, we assume that the vacuum is slightly lossy as in \(k=k_{1}+ik_{2}\) with \(0<k_{2}\ll k_{1}\). Taking the limit \(k_{2}\to 0\) at the end of the analysis, we can obtain the solution for the lossless case. Taking into account the asymptotic behavior of the scattered field together with the radiation condition, we can show that  
\[\phi(x,z)=\left\{ \begin{array}{l}O(e^{k_{2}z\cos\theta_{0}})\text{ as }z \rightarrow-\infty,\\ O(e^{-k_{2}z})\text{ \ \ \ as }z\rightarrow\infty.\end{array} \right. \tag{6}\]  
Let us define the Fourier transform of the scattered field with respect to \(z\) as  
\[\Phi(x,\alpha)=(2\pi)^{-1/2}\int_{-\infty}^{\infty}\phi(x,z)e^{i\alpha z}dz, \tag{7}\]  
Figure 1: Geometry of the problem. where \(\alpha=\sigma+i\tau(\equiv{\rm Re}\alpha+i{\rm Im}\alpha)\). Using (6), we can show that \(\Phi(x,\alpha)\) is regular in the strip \(-k_{2}<\tau<k_{2}\cos\theta_{0}\) of the complex \(\alpha-\)plane. It can also be verified that as \(|x|\)\(\rightarrow\infty\), \(\Phi(x,\alpha)\) is bounded for any \(\alpha\) in the strip \(-k_{2}<\tau<k_{2}\cos\theta_{0}\). We also introduce the Fourier integrals as  
\[\left. \begin{array}{l}\Phi_{+}(x,\alpha)=(2\pi)^{-1/2}\int_{0}^{\infty}\phi( x,z)e^{i\alpha z}dz,\\ \Phi_{-}(x,\alpha)=(2\pi)^{-1/2}\int_{-\infty}^{-L_{1}}\phi^{\prime}(x,z)e^{i \alpha z}dz,\\ \Phi_{m}^{\prime}(x,\alpha)=(2\pi)^{-1/2}\int_{-L_{m}}^{-L_{m+1}}\phi^{\prime }(x,z)e^{i\alpha z}dz,\ m=1,2,3,4,\\ \Phi_{1}^{0}(x,\alpha)=(2\pi)^{-1/2}\int_{-L_{5}}^{0}\phi^{\prime}(x,z)e^{i \alpha z}dz.\end{array} \right\} \tag{8}\]  
It follows from (6) that \(\Phi_{+}(x,\alpha)\) and \(\Phi_{-}(x,\alpha)\) are regular in \(\tau>-k_{2}\) and \(\tau<k_{2}\cos\theta_{0}\) respectively, while \(\Phi_{1}^{0}(x,\alpha)\) and \(\Phi_{m}^{\prime}(x,\alpha)\) for \(m=1\), 2, 3, 4 are all entire functions. In the following analysis, These conventions will be utilized to denote the areas of regularity in the complex \(\alpha-\)plane.