## 4 Problem Solution  
The problem will be solved via the CdH joint transform technique [19]. Accordingly, we combine a unilateral Laplace transformation, i.e.,  
\[\tilde{\mathbf{H}}(x,y,z,s)=\int_{t=0}^{\infty}\exp(-st)\mathbf{H}(x,y,z,t)\text{d}t, \tag{11}\]  
for \(\{s\in\mathbb{R};s>0\}\), by virtue of [PERSON]'s uniqueness theorem ([26], Appendix), with the wave slowness representation taken along the surface parallel to the PEC screen:  
\[\tilde{\mathbf{H}}(x,y,z,s)=\Big{(}\frac{s}{2\pi\text{i}}\Big{)}^{2}\int_{x=- \text{i}\infty}^{\text{i}\infty}\text{d}x\int_{\sigma=-\text{i}\infty}^{ \text{i}\infty}\exp[-s(\kappa x+\sigma y)]\tilde{\mathbf{H}}(\kappa,\sigma,z,s) \text{d}\sigma, \tag{12}\]  
where \(\kappa\) and \(\sigma\) are (imaginary-valued) slowness parameters in the \(x\)- and \(y\)-direction, respectively. As a matter of fact, Equation (12) is a two-dimensional Fourier inversion integral, where the (real-valued and positive) Laplace-transform parameter, \(s\), plays the role of a scaling parameter. This representation entails the properties \(\partial_{x}\to-\kappa s\) and \(\partial_{y}\to-\sigma\). Under the wave slowness representation, the TD reciprocity relation (4) can be written as  
\[ \begin{split}\Big{(}\frac{s}{2\pi\text{i}}\Big{)}^{2}\int_{\kappa =-\text{i}\infty}^{\text{i}\infty}\text{d}x\int_{\sigma=-\text{i}\infty}^{ \text{i}\infty}&\big{[}\tilde{H}_{x}^{\text{T}}(x,\sigma,0^{+},s )\partial\tilde{K}_{x}^{\text{s}}(-\kappa,-\sigma,s)\\ &+\tilde{H}_{y}^{\text{T}}(\kappa,\sigma,0^{+},s)\partial\tilde{K }_{y}^{\text{s}}(-\kappa,-\sigma,s)\big{]}\text{d}\sigma\\ =-\Big{(}\frac{s}{2\pi\text{i}}\Big{)}^{2}\int_{\kappa=-\text{i} \infty}^{\text{i}\infty}\text{d}x\int_{\sigma=-\text{i}\infty}^{\text{i} \infty}&\big{[}\tilde{H}_{x}^{\text{i}}(x,\sigma,0^{+},s)\partial \tilde{K}_{x}^{\text{T}}(-\kappa,-\sigma,s)\\ &+\tilde{H}_{y}^{\text{i}}(\kappa,\sigma,0^{+},s)\partial\tilde{K }_{y}^{\text{T}}(-\kappa,-\sigma,s)\big{]}\text{d}\sigma.\end{split}  \tag{13}\]  
The (tangential components of the) transform domain testing fields as generated by the equivalent magnetic-current surface density, \(\partial\tilde{\mathbf{K}}^{\text{T}}\), can be determined from the transform domain counterparts of standard source-type representations for the electric and magnetic-field strengths in a loss-free medium (cf. ([24], Equations (26.10-14) and (26.10-15))) as  
\[\bar{E}_{x}^{\mathrm{T}}(\kappa,\sigma,z,s) =\partial\bar{K}_{y}^{\mathrm{T}}(\kappa,\sigma,s)\partial_{z} \bar{G}(\kappa,\sigma,z,s), \tag{14}\] \[\bar{E}_{y}^{\mathrm{T}}(\kappa,\sigma,z,s) =-\partial\bar{K}_{x}^{\mathrm{T}}(\kappa,\sigma,s)\partial_{z} \bar{G}(\kappa,\sigma,z,s),\] (15) \[\bar{H}_{x}^{\mathrm{T}}(\kappa,\sigma,z,s) =-(s/\mu_{0})\Omega_{0}^{2}(\kappa)\delta\bar{K}_{x}^{\mathrm{T} }(\kappa,\sigma,s)\bar{G}(\kappa,\sigma,z,s)\] \[+(s/\mu_{0})\kappa\sigma\,\partial\bar{K}_{y}^{\mathrm{T}}( \kappa,\sigma,s)\bar{G}(\kappa,\sigma,z,s),\] (16) \[\bar{H}_{y}^{\mathrm{T}}(\kappa,\sigma,z,s) =-(s/\mu_{0})\Omega_{0}^{2}(\sigma)\partial\bar{K}_{y}^{\mathrm{ T}}(\kappa,\sigma,s)\bar{G}(\kappa,\sigma,z,s)\] \[+(s/\mu_{0})\sigma\kappa\,\partial\bar{K}_{x}^{\mathrm{T}}( \kappa,\sigma,s)\bar{G}(\kappa,\sigma,z,s), \tag{17}\]  
where \(\Omega_{0}^{2}(\kappa)=1/c_{0}^{2}-\kappa^{2}\), and the transform domain Green's function, \(\bar{G}\), follows as the bounded solution of  
\[\big{(}\partial_{z}^{2}-s^{2}\gamma^{2}\big{)}\bar{G}=0\text{ with }\lim_{z\downarrow 0} \partial_{z}\bar{G}=-1. \tag{18}\]  
Using the limit from Equation (18) in Equations (14) and (15), it can be easily verified that the transform domain testing fields satisfy the excitation conditions \(\bar{E}_{x}^{\mathrm{T}}(\kappa,\sigma,0^{+},s)=-\partial\bar{K}_{y}^{\mathrm{ T}}(\kappa,\sigma,s)\) and \(\bar{E}_{y}^{\mathrm{T}}(\kappa,\sigma,0^{+},s)=\partial\bar{K}_{x}^{\mathrm{ T}}(\kappa,\sigma,s)\).