# Advanced Technology in ATSC3.0: Boosting Data Signal Rates Using Polarization Properties and FSS +
## 2 System Model
### Xpd  
Figure 1 illustrates the configuration of the transmitter and receiver equipped with vertical and horizontal polarization antennas. Figure 2 shows the incident electric field vectors arriving at the receiver, which are used to derive the analytical model of XPD in a multipath fading environment. For clarity, we assume that the two receiving antennas lie in the y-z plane. The incident electric field at the location of the receiving antennas is represented by  
\[\mathcal{E}=\mathcal{E}_{v}+\mathcal{E}_{h}\] \[\mathcal{E}_{v}=\mathcal{E}_{v,1}\hat{d}_{v,1}+\mathcal{E}_{v,2} \hat{d}_{v,2}\] (5) \[\mathcal{E}_{h}=\mathcal{E}_{h,1}\hat{d}_{  
The elevation angles are defined as \(\gamma_{v}=\frac{\pi}{2}-\theta_{v}\) and \(\gamma_{h}=\frac{\pi}{2}-\theta_{h}\). The unit vectors \(\hat{x},\hat{y}\) and \(\hat{z}\) indicate directions along the Cartesian axes. The vectors representing the orientation of the receiving antennas are given by  
\[\hat{s}_{1} =\sin\xi\cdot\hat{y}+\cos\xi\cdot\hat{z}, \tag{10}\] \[\hat{s}_{2} =-\sin\xi\cdot\hat{y}+\cos\xi\cdot\hat{z}. \tag{11}\]  
Given that the incident electric field follows a Rayleigh distribution, let  
\[\mathcal{E}_{v,1} =\mathcal{A}_{v,1}\cos\left(\omega t+\alpha_{v,1}\right), \tag{12}\] \[\mathcal{E}_{h,1} =\mathcal{A}_{h,1}\cos\left(\omega t+\alpha_{h,1}\right),\] (13) \[\mathcal{E}_{v,2} =\mathcal{A}_{v,2}\cos\left(\omega t+\alpha_{v,2}\right),\] (14) \[\mathcal{E}_{h,2} =\mathcal{A}_{h,2}\cos\left(\omega t+\alpha_{h,2}\right), \tag{15}\]  
where the random variables \(\mathcal{A}_{v,1},\mathcal{A}_{v,2},\mathcal{A}_{h,1}\) and \(\mathcal{A}_{h,2}\) are mutually independent and follow a Rayleigh distribution. The phase variables \(\alpha_{v,1},\alpha_{h,1},\alpha_{v,2}\) and \(\alpha_{h,2}\) are independent and iden  
Figure 1: The transmitter and receiver equipped two branch polarization diversity  
Figure 2: Wave arriving at the receiver  
where \(\hat{v}_{v}=\cos\gamma_{v}\cos\xi+\sin\gamma_{v}\sin\varphi_{v}\sin\xi,\hat{v}_{h}= \cos\gamma_{h}\cos\xi+\sin\gamma_{h}\sin\varphi_{h}\sin\xi\). The amplitudes of the received vertical polarization component signals by \(R_{1}\) and \(R_{2}\) are, therefore, proportional to  
\[\Lambda_{v,1}= [(\mathcal{A}_{v,1}\mu_{v}\cos\alpha_{v,1}+\mathcal{A}_{v,2}v_{v} \cos\alpha_{v,2})^{2} \tag{18}\] \[+(\mathcal{A}_{v,1}\mu_{v}\sin\alpha_{v,1}+\mathcal{A}_{v,2}v_{v} \sin\alpha_{v,2})^{2}]^{1/2}\] \[= [\mathcal{A}_{v,1}^{2}\mu_{v}^{2}+\mathcal{A}_{v,2}^{2}\
u_{v}^{ 2}+2\mathcal{A}_{v,1}\mathcal{A}_{v,2}\mu_{v}v_{v}\] \[\cdot\cos(\alpha_{v,1}-\alpha_{v,2})]^{1/2}\] \[\Lambda_{v,2}= [\mathcal{A}_{v,1}^{2}\mu_{v}^{2}+\mathcal{A}_{v,2}^{2}\hat{v}_{ v}^{2}-2\mathcal{A}_{v,1}\mathcal{A}_{v,2}\mu_{v}\hat{v}_{v}\] \[\cdot\cos(\alpha_{v,1}-\alpha_{v,2})]^{1/2}\]  
According to Equation (18), the mean received signal strength at the receiving antennas is lower than that received by a vertically polarized antenna. The mean signal received by the vertically polarized antenna is expressed as \(\langle\Lambda_{v,0}^{2}\rangle=\langle\mathcal{A}_{v,2}^{2}\rangle\cos\gamma_ {v}\), while the mean signal level received by the polarization diversity antenna \(R_{1}\) is \(\langle\Lambda_{v,1}^{2}\rangle\). The ratio between these two signal levels is, therefore, given by:  
\[L_{v,1}= \frac{\langle\Lambda_{v,1}^{2}\rangle}{\langle\Lambda_{v,0}^{2} \rangle}=\frac{\langle\mathcal{A}_{v,1}^{2}\mu_{v}^{2}+\mathcal{A}_{v,2}^{2} \
u_{v}^{2}\rangle}{\langle\mathcal{A}_{v,2}^{2}\rangle\cos\gamma_{v}}\] (19) \[= \frac{\langle\mathcal{A}_{