## 2 System Model
### Phase 1: Source broadcasts Symbol to Destination and Relay (for Destination)
The first Path (LoS path) is the shortest path for i = 1. There is no delay or Doppler shift in that channel. Therefore, the channel with zero delay and Doppler in the first channel case, as presented in [24], and \(l_{1}=0\), \(k_{1}=0\) are obtained. Since the movement direction of our vehicle is in the \(V\) direction, it moves perpendicularly to the source, as we stated before. For this reason, the velocity vector value along the LoS path line is zero. Since there is no velocity component in the source direction, the Doppler effect for the LoS path is zero, and therefore, the \(v_{i}\) value will be zero (very low). Therefore, the \(k_{i}\) value is taken as 0. Since the LoS path is the shortest path, the delay value will be minimum and the \(\tau_{i}\) value will be low. Therefore, the \(l_{i}\) value can be accepted as 0. Second Path (NLoS Path-Reflected Path): Since the vehicle moves in the \(V\) direction, the receiver moves perpendicular to the source. However, there is a velocity component in the relay direction. Additionally, path 2 (NLoS path) is the longest path for i = 2. There are delay and Doppler shifts in that channel. Therefore, the channel with both delay and Doppler in the fourth channel case, presented in [22], and \(l_{2}\) = 1, \(k_{2}\) = 1 are obtained. Since the movement direction of our vehicle is in the \(V\) direction, it moves perpendicularly to the source, as we stated before. For this reason, the velocity vector component occurs along the NLoS path. Since there is a velocity component in the relay direction, the Doppler effect also occurs for the NLoS path, and therefore, the \(v_{i}\) value will be greater than zero. Therefore, the \(k_{i}\) value can be accepted as 1. Since the NLoS path is the longest, both the delay value and the \(\tau\)i value will be high. Therefore, the value of \(l_{i}\) can be accepted as 1. According to our model, as expressed above, P = 2, and \(k_{i}= \begin{bmatrix}0&1\end{bmatrix} \) and \(l_{i}= \begin{bmatrix}0&1\end{bmatrix} \) is defined for Phase 1 for the destination. When M = 2 and N = 2 are taken into account, then Equation (4) is transformed to  
\[H_{s,d}=\sum_{i=1}^{2}h_{i}\Pi_{\text{MN}}^{\dagger j}\Delta_{MN}^{ki} \tag{9}\]  
When i = 1, then \(H_{1_{s,d}}=h_{1}\Pi^{0}\Delta^{0}\) is obtained, where  
\[\Gamma^{0}= \begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} ,\ \Delta^{0}= \begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}  \tag{10}\]  
Substituting \(\Pi^{0}\) and \(\Delta^{0}\) in the equation, the \(H_{1_{s,d}}\) channel matrix is obtained,  
\[H_{1_{s,d}}= \begin{bmatrix}h_{1}&0&0&0\\ 0&h_{1}&0&0\\ 0&0&h_{1}&0\\ 0&0&0&h_{1}\end{bmatrix}  \tag{11}\]  
When i = 2, \(H_{2_{s,d}}=h_{2}\Pi^{1}\Delta^{1}\) is obtained, where  
\[\Pi^{1}= \begin{bmatrix}0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\end{bmatrix} ,\ \Delta^{1}= \begin{bmatrix}1&0&0&0\\ 0&z&0&0\\ 0&0&z^{2}&0\\ 0&0&0&z^{3}\end{bmatrix}  \tag{12}\]  
\[z=i^{\frac{2\Pi}{\text{MN}}}=i^{\frac{1}{2}}=j \tag{13}\]  
\[\Delta^{1}= \begin{bmatrix}1&0&0&0\\ 0&j&0&0\\ 0&0&-1&0\\ 0&0&0&-j\end{bmatrix}  \tag{14}\]  
\[H_{2_{s,d}}= \begin{bmatrix}0&0&0&-h_{2}j\\ h_{2}&0&0&0\\ 0&h_{2}j&0&0\\ 0&0&-h_{2}&0\end{bmatrix}  \tag{15}\]  
Since \(H_{s,d}=H_{1_{s,d}}+H_{2_{s,d}}\), we sum the matrices \(H_{1_{s,d}}\) and \(H_{2_{s,d}}\), and then we obtain the \(H_{s,d}\) matrice:  
\[H_{s,d}= \begin{bmatrix}h_{1}&0&0&-h_{2}j\\ h_{2}&h_{1}&0&0\\ 0&h_{2}j&h_{1}&0\\ 0&0&-h_{2}&h_{1}\end{bmatrix}  \tag{16}\]\[H^{rect}_{eff_{s,d}}=(F_{N}\otimes I_{M})H_{s,d}\Big{(}F_{N}^{H}\otimes I_{M} \Big{)} \tag{17}\]  
\[\left[F_{N}\right]_{k,l\rightarrow(N-1)}\,=\{\frac{1}{\sqrt{N}}\Big{\rvert}^{ \frac{j\Omega H}{N}}\} \tag{18}\]  
\[F_{N}=\,\,\frac{1}{\sqrt{2}}\Big{[} \begin{matrix}1&1\\ 1&e^{\Pi}\end{matrix} \Big{]}=\frac{1}{\sqrt{2}}\Big{[} \begin{matrix}1&1\\ 1&-1\end{matrix} \Big{]} \tag{19}\]  
\[F_{N}^{H}=\frac{1}{\sqrt{2}}\Big{[} \begin{matrix}1&1\\ 1&-1\end{matrix} \Big{]}\,,\,I_{M}=\,\,\Big{[} \begin{matrix}1&0\\ 0&1\end{matrix} \Big{]} \tag{20}\]  
\[