## 2 System Model
### Phase 2: Relay Retransmits the Symbol to Destination  
The relay afterwards retransmits to the destination only if it is able to correctly decode within the SDF scheme. For the destination, the received time domain signal is expressed as a point-to-point SISO as follows:  
\[r_{d}(t)=h_{r,d}e^{i2\pi v_{r,d}(t-\tau_{r,d})}s(t-\tau_{r,d})+ \tag{28}\] \[h_{r,s,d}s(t-\tau_{r,s,d})\]  
where \(h_{r,d}\) is the channel gain of the LoS path, \(v_{r,d}\) is the Doppler frequency (shift) of the LoS path, \(s(t)\) is the transmitted signal from the source, \(\tau_{r,d}\) is the delay shift of the LoS path, \(h_{r,s,d}\) is the channel gain of the NLoS (reflected) path, and \(\tau_{r,s,d}\) is the delay shift of the NLoS path. The time-variant channel impulse response (CIR) for this transmission is given below:  
\[ \begin{split}\mathsf{g}_{(t,v)}=h_{r,d}e^{i2\pi v_{r,d}(t-v_{r,d })}\delta(t-\tau_{r,d})+\\ h_{r,s,d}\delta(t-\tau_{r,s,d})\end{split}  \tag{29}\]To make the mathematical expressions for the time-varying channel above easier to understand, we present below the received signal from the relay to the destination in the time domain in vector and matrix forms, in the context of OTFS modulation:  
\[\overline{r}_{r,d}=H_{r,d}\overline{s}+\overline{n} \tag{30}\]  
First Path (LoS Path): Since the vehicle is moving in V direction, the receiver has a velocity component in the direction of the relay. Path 1 (LoS path) is the shortest path for i = 1. Therefore, the channel with zero delay and one Doppler in the third channel case, as presented in [22], and \(l_{1}=0,k_{1}=1\) are obtained. Since the direction of motion of our vehicle is V, there is a velocity component in the direction of the relay as mentioned before. Since there is a velocity component in the relay direction, the Doppler effect for the LoS path will be greater than zero, and therefore, the value of \(v_{i}\) will be greater than zero. Therefore, the value of \(k_{i}\) is taken as 1. Since the LoS path is the shortest path, the delay value will be minimal and \(\tau_{i}\) will be low. Therefore, the value of \(l_{i}\) can be taken as 0. Second Path (NLoS Path): In this scenario, the source acts as a reflector as it reflects the signal sent by the relay, which retransmits the incoming signal with the SDF protocol. Since the vehicle moves in the V direction, the receiver moves perpendicularly to the source (relay). Path 2 (NLoS path) is the longest path for i = 2. Therefore, the channel with one delay and zero Doppler in the second channel case, as presented in [24], and \(l_{2}=1\), \(k_{2}=0\) are obtained. Since the direction of movement of our vehicle is in the V direction, it moves perpendicularly to the relay as we mentioned before. Therefore, the velocity vector component along the NLoS path is 0. Since there is no velocity component in the source direction, the Doppler effect for the NLoS path is zero, and therefore, the value of \(v_{i}\) will be zero (very low). Therefore, the value of \(k_{i}\) is taken as 0. Since the NLoS path is the longest path, the delay value will be maximum and the \(\tau_{i}\) value will be high. Therefore, the value of \(l_{i}\) can be taken as 1. According to our model, as expressed above, P = 2, and \(k_{i}= \begin{bmatrix}1&0\end{bmatrix} \) and \(l_{i}= \begin{bmatrix}0&1\end{bmatrix} \) is defined for Phase 2 for the destination. When M = 2 and N = 2 are taken into account, then Equation (4) is transformed into  
\[H_{r,d}=\sum_{i=1}^{2}h_{i}\Pi_{\text{MN}}^{\text{i}j}\Delta_{MN}^{\text{i}i} \tag{31}\]  
\[H_{r,d}=h_{1}\Pi^{0}\Delta^{1}+h_{2}\Pi^{1}\Delta^{0} \tag{32}\]  
When i = 1, then \(H_{1_{r,d}}=h_{1}\Pi^{0}\Delta^{1}\) is obtained and i = 2; moreover, \(H_{2_{r,d}}=h_{2}\Pi^{1}\Delta^{0}\) is obtained, where  
\[\Pi^{0}\]  
\[h_{1} \begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} \times \begin{bmatrix}1&0&0&0\\ 0&j&0&0\\ 0&0&-1&0\\ 0&0&0&-j\end{bmatrix} +\]  
\[\Pi^{1}\]  
\[h_{2} \begin{bmatrix}0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\end{bmatrix} \times