## 5 Illustrative Examples  
Throughout this section, it is assumed that the rectangular aperture is irradiated by a uniform \(E\)-polarized TD plane-wave:  
\[\mathbf{E}^{\mathrm{i}}(\mathbf{r},t)=\mathbf{\alpha}e^{\mathrm{i}}(t-\kappa_{0}x-\sigma_ {0}y+\gamma_{0}z), \tag{25}\]  
where the unit vector in the direction of polarization, \(\mathbf{\alpha}\), can be expressed via the azimuthal angle, \(\phi\), as \(\mathbf{\alpha}=\sin(\phi)i_{x}-\cos(\phi)i_{y}\). The pertaining slowness parameters are then described by \(\kappa_{0}=\cos(\phi)\sin(\theta)/c_{0}\), \(\sigma_{0}=\sin(\phi)\sin(\theta)/c_{0}\) and \(\gamma_{0}=\cos(\theta)/c_{0}\), where \(\theta\) denotes the polar angle. The (causal) plane-wave signature, \(e^{\mathrm{i}}(t)\), has the property \(e^{\mathrm{i}}(t)=0\) for all \(t<0\). With reference to the right-hand side of the TD reciprocity relation (4), the corresponding tangential components of the magnetic-field strength at the plane of aperture, \(z=0\), then follow as  
\[H_{x}^{\mathrm{i}}(x,y,0,t) =-Y_{0}e^{\mathrm{i}}(t-\kappa_{0}x-\sigma_{0}y)\cos(\phi)\cos( \theta), \tag{26}\] \[H_{y}^{\mathrm{i}}(x,y,0,t) =-Y_{0}e^{\mathrm{i}}(t-\kappa_{0}x-\sigma_{0}y)\sin(\phi)\cos( \theta). \tag{27}\]  
To determine the corresponding TD excitation array \(\mathbf{I}(t)\) (see Equation (23)), we may use either Equations (9) and (10) with (26) and (27) on the right-hand side of Equation (4) or, equivalently, evaluate the right-hand side of the transform domain reciprocity relation (13). Both ways yield the following elements of the excitation array:  
\[I_{x}(t) =\frac{Y_{0}}{\kappa_{0}}\biggl{[}\int_{\tau=0}^{t+\kappa_{0} \Delta_{x}/2}e^{\mathrm{i}}(\tau)\mathrm{d}\tau-\int_{\tau=0}^{t-\kappa_{0} \Delta_{x}/2}e^{\mathrm{i}}(\tau)\mathrm{d}\tau\biggr{]}\cos(\phi)\cos(\theta), \tag{28}\] \[I_{y}(t) =\frac{Y_{0}}{\sigma_{0}}\biggl{[}\int_{\tau=0}^{t+\sigma_{0} \Delta_{y}/2}e^{\mathrm{i}}(\tau)\mathrm{d}\tau-\int_{\tau=0}^{t-\sigma_{0} \Delta_{y}/2}e^{\mathrm{i}}(\tau)\mathrm{d}\tau\biggr{]}\sin(\phi)\cos(\theta). \tag{29}\]  
The limits pertaining to the normal incidence characterized by \(\theta=0\), implying \(\kappa_{0}=\sigma_{0}=0\), follow as  
\[\lim_{\theta\to 0}I_{x}(t) =Y_{0}\Delta_{x}e^{\mathrm{i}}(t)\cos(\phi), \tag{30}\] \[\lim_{\theta\to 0}I_{y}(t) =Y_{0}\Delta_{y}e^{\mathrm{i}}(t)\sin(\phi). \tag{31}\]  
For the presented example, we take \(2\Delta_{x}=0.10\,\mathrm{m}\) and \(\Delta_{y}=\Delta_{x}/2\) with \(\theta=0\) and \(\phi=\pi/4\). Furthermore, the plane-wave signature has the shape of a bipolar triangle, which can be described by  
\[ \begin{array}{l}e^{\mathrm{i}}(t)=(2\varepsilon_{\mathrm{m}}/t_{\mathrm{w}}) \bigl{[}t\,\mathrm{H}(t)-2\bigl{(}t-t_{\mathrm{w}}/2\bigr{)}\mathrm{H}\bigl{(} t-t_{\mathrm{w}}/2\bigr{)}\\ +2\bigl{(}t-3t_{\mathrm{w}}/2\bigr{)}\mathrm{H}\bigl{(}t-3t_{\mathrm{w}}/2 \bigr{)}-\bigl{(}t-2t_{\mathrm{w}}\bigr{)}\,\mathrm{H}\bigl{(}t-2t_{\mathrm{ w}}\bigr{)}\bigr{]},\end{array}  \tag{32}\]where we take \(e_{\rm m}=1.0\,\rm kV/m\) and \(c_{0}t_{\rm w}=1.0\,\rm m\) (see Figure 2). Consequently, \(c_{0}t_{\rm w}=10\,\max(2\Delta_{x},2\Delta_{y})\), which implies that the aperture is relatively small, as assumed in our analysis. In the first step, the marching-on-in-time solution (24) is applied to calculate the magnetic-current coefficients, \(v_{kx}\) and \(v_{ky}\), which, through Equations (20) and (21) with (5) and (6), determine the space-time distribution of the magnetic-current surface density induced throughout the aperture. Its tangential components can be expressed using the actual electric-field strengths in the aperture as \(\partial K_{x}^{\rm s}(x,y,t)=E_{y}(x,y,0,t)\) and \(\partial K_{y}^{\rm s}(x,y,t)=-E_{x}(x,y,0,t)\). The electric-field pulses at the center of the aperture as calculated with the aid of the CdH-MoM method and via the finite-integration technique (FIT) as implemented in CST Microwave Studio(r) are shown in Figure 3a,b. As can be seen, the results agree very well.