## 4 Proposed Methodology of Hybrid Channel Modeling
### Electromagnetic Field Analysis in the Mine Gallery  
The entire distribution field on the excitation plane is obtained by summing up the contribution of the source and total reflected images. In a rectangular-shaped mine gallery, this geometry exhibits the protocol that vertical walls reflect the image \(I_{pq}|p|\) times and horizontal ceilings or floors reflect \(|q|\) times. Assume that the incident angle for the horizontal floor or ceiling is \(\alpha\) and the incident angle for vertical walls is \(\beta\), as shown in Figure 4 with the basic design of the mine. If the transmitter is set at coordinates \((x_{o},y_{o},0)\), and the observant node is located at \((x,y,z)\), then the net field on an observant node can be considered to be the summation of all rays, and the field at the transmitter node \(E_{o}\) can be calculated as \[E^{Rx}(x,y,z)=E_{o}\sum_{p=-\infty}^{\infty}\sum_{q=-\infty}^{\infty}\left[\frac{ exp(-jkr_{p,q})}{r_{p,q}}\right]S(\overline{k_{p}})^{|p|}R(\overline{k_{h}})^{|q|} \tag{9}\]  
where \(R(\overline{k_{h}})\) and \(S(\overline{k_{v}})\) are reflection coefficients for horizontal and vertical parameters, respectively, \(r_{p,q}\) denotes the distance between the receiver and image \(I_{p,q}\), as described in Equations (10) and (11). The \(r_{p,q}\) is given in Equation (14), where a +ve sign represents the scenario when \(p\) or \(q\) is even and a -ve sign for \(p\) or \(q\) represents the odd. \[R(\overline{k_{h}})=-exp\left(\frac{-2 sin\alpha}{\sqrt{\overline{k_{h}}-1}}\right) \tag{10}\]  
\[S(\overline{k_{v}})=-exp\left(\frac{-2 sin\beta}{\sqrt{\overline{k_{v}}-1}}\right) \tag{11}\]  
The parameters \(R(\overline{k_{h}})\) and \(S(\overline{k_{v}})\) can be transformed into simplified equations, using the following  
\[R(\overline{k_{h}})=-exp\left(\frac{-2}{\sqrt{\overline{k_{h}}-1}}\frac{|2 qb \pm y_{o}-y|}{r_{p,q}}\right) \tag{12}\]  
\[S(\overline{k_{v}})=-exp\left(\frac{-2\overline{k_{v}}}{\sqrt{\overline{k_{v} }-1}}\frac{|2 pa\pm x_{o}-x|}{r_{p,q}}\right) \tag{13}\]  
\[r_{(}p,q)=\sqrt{((2 pa\pm x_{o}-x)^{2}+(2 qb\pm y_{o}-y)^{2}+z^{2})} \tag{14}\]  
The ray sum is reorganized from Equation (9) and split into four parts, as given below  
\[ \begin{split} E^{Rx}&(x,y,z)=\sum_{p,q=-\infty}^{ \infty}f(4 qa+x_{o}-x,4 pb+y_{o}-y)\\ &+\sum_{p,q=-\infty}^{\infty}f(4 qa+x_{o}-x,4 pb+2b-y_{o}-y)\\ &+\sum_{p,q=-\infty}^{\infty}f(4 qa+2a-x_{o}-x,4 pb+y_{o}-y)\\ &+\sum_{p,q=-\infty}^{\infty}f(4 qa+2a-x_{o}-x,4 pb+2b-y_{o}-y) \end{split}  \tag{15}\]  
And the function \(f(u,v)\) is calculated using  
\[ \begin{split} f(u,v)=E_{o}.\frac{exp(-jk\sqrt{u^{2}+v^{2}+z^{2}} )}{\sqrt{u^{2}+v^{2}+z^{2}}}(-1)^{p(v)+q(u)}\\.exp\left[\frac{-2}{\sqrt{u^{2}+v^{2}+z^{2}}}\left(\frac{|v|p(v) }{\sqrt{k_{h}}-1}+\frac{|u|\overline{k_{v}}q(u)}{\sqrt{\overline{k_{v}}-1}} \right)\right]\end{split}  \tag{16}\]  
The parameters \(p(v)\) and \(q(u)\) need to be transformed from discontinuous to continuous functions for mode-matching purpose, as given below  
\[p(v)=\frac{|v|}{2b}\qquad;\qquad q(u)=\frac{|u|}{2a} \tag{17}\]  
It is noteworthy to point out that every segment of Equation (15) reflects the periodic function of \(4a\) and \(4b\). The first part is entertained first, and the sum is transformed into the following expression (18) under a two-dimensional Poisson summation expression [34]\[ \begin{split}\sum_{p,q=-\infty}^{\infty}f(4 qa+x_{o}-x,4 pb+y_{o}-y)=\\ \frac{1}{4a}\frac{1}{4b}\sum_{m=-\infty}^{\infty}\sum_{n=-\infty} ^{\infty}F_{1}(m,n)e^{j\frac{m\pi}{2a}x}e^{j\frac{m\pi}{2b}y}\end{split}  \tag{18}\]  
The coefficient \(F_{1}(m,n)\) denotes Fourier transform of the function \(f(x_{o}-x,y_{o}-y)\) for the first part given in (15):  
\[F_{1}(m,n)=\int\int_{-\infty}^{\infty}f(x_{o}-x,y_{o}-y)e^{-j\frac{m\pi}{2b}x} e^{-j\frac{m\pi}{2b}y}dxdy \tag{19}\]  
The Saddle-point methodology is used to obtain the result of integration, as the study is focused on the field \(E^{Rx}(x,y,z)\) on the excitation plane with \(z=0\).