## 2 System Model
### Phase 1: Source Broadcasts Symbol to Destination and Relay (for Relay)  
The source broadcasts to the receiver (destination) and relay. For the relay, the received time domain signal is expressed as the point-to-point SISO model as follows:  
\[\begin{array}{c}r_{r}(t)=h_{s,r}s(t-\tau_{+s,r})+\\ h_{s,d,r}e^{j\pi\
u_{s,d,r}(t-\tau_{s,d,r})}s(t-\tau_{s,d,r})\end{array} \tag{24}\]  
where \(h_{s,r}\) is the channel gain of the LoS path, the Doppler frequency (shift) of the LoS path = 0, \(\tau_{s,r}\) is the delay shift of the LoS path, \(s(t)\) is the transmitted signal from the source, \(h_{s,d,r}\) is the channel gain of the NLoS (reflected) path, \(\
u_{s,d,r}\) is the Doppler frequency of the NLoS path, and \(\tau_{s,d,r}\) is the delay of the NLoS path.  
For two propagation paths, a specific 2-tap doubly selective time-variant channel impulse response (CIR) for this transmission is given below. When \(s(t)\) passes through this channel, Equation (24) is obtained.  
\[\begin{array}{c}\mathrm{g}(t,\tau)=h_{s,r}\delta(t-\tau_{s,r})+\\ h_{s,d,r}e^{j\pi\
u_{s,d,r}(t-\tau_{s,d,r})}\delta(t-\tau_{s,d,r})\end{array} \tag{25}\]To make the mathematical expressions for the time-varying channel above easier to understand, we present below the received signal from the source to the relay in the time domain in vector and matrix forms, in the context of OTFS modulation:  
\[\overline{r}_{s,r}=H_{s,r}\overline{s}+\overline{n} \tag{26}\]  
The general mathematical expression of the channel matrix \(H_{s,r}\), effective channel matrix \(H_{eff_{s,r}}^{rect}\) and vectorized form of the received signal in the delay-Doppler domain are similarly obtained using Equations (4)-(8).  
First Path (LoS Path): Path 1 (LoS path) is the shortest path for i = 1. Since the relay is stationary, the Doppler effect for the LoS path is zero, and therefore, the value of \(v_{i}\) will be zero (very low). Therefore, \(k_{1}\) = 0 is obtained.  
Since the LoS path is the shortest path, the delay value will be minimum and \(\tau_{i}\) will be low. Therefore, \(l_{1}\) = 0 is obtained.  
Second Path (NLoS Path): Since the vehicle is moving in the V direction, there is a velocity component in the relay direction. Additionally, path 2 (NLoS path) is the longest path for i = 2. 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 direction of motion of our vehicle is in the V direction, there is a velocity component in the relay direction as mentioned 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 value of \(v_{i}\) will be greater than zero. Therefore, the value of \(k_{i}\) can be considered as 1.  
Since the NLoS path is the longest path, both the delay value and the \(\tau\)i value will be high. Therefore, the value of li can be considered as 1.  
For LoS and NLoS paths, the values of \(k_{i}\) and \(l_{i}\) are 0 for i = 1 and 1 for i = 2 since they are the same for the source-destination scenario. Therefore, the channel matrix \(H_{s,r}\) and the effective channel matrix \(H_{eff_{s,r}}^{rect}\) can be obtained from the equations written above for source-destination Equations (9)-(23).  
In vectorized form, the received signal from the source to the relay in the delay-Doppler domain can be written for rectangular waveforms:  
\[\overline{y}_{s,r}=H_{eff_{s,r}}^{rect}\overline{x}+\overline{n} \tag{27}\]