## 4 LLR Calculation under Rayleigh Fading Channel  
As shown in Section 3, we assume that the instantaneous amplitude information for the fading channel is available, and a phase noncoherent channel is considered. In this section, we will relax this condition and assume that only the statistical characteristic for the fading amplitude is available. Note that the main problem for the receiver now becomes to how to exploit this statistical information, thereby optimizing the design and then coming up with a detection scheme. In this context, the statistical average is also considered to eliminate the impact on the detection, which is similar to the method we adopted in Section 3. In particular, the PDF of \(\rho\) is known as the Rayleigh distribution:  
\[p(\rho)\!=\!\frac{\rho}{\sigma_{\rho}^{2}}\exp\!\left(-\frac{\rho^{2}}{2\sigma _{\rho}^{2}}\right),\rho\geq 0 \tag{18}\]  
Figure 3: Bessel function \(I_{0}\). As shown in (22), the implementation process for the LLR is complicated, especially when the observation interval \(N\) is large. Using (10), (11), and (12), (22) can be simplified as:  
\[\zeta_{m} \approx\ \ln\frac{\sum\limits_{\mathbf{S}_{y}:c_{m}=0,l}\exp\left( \frac{\left|\sum\limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}\sigma_{p}^{2}}{2 \sigma^{2}\left(\sigma_{p}^{2}\sum\limits_{k=1}^{N}\frac{16}{\sum\limits_{k=1 }^{16}\left|\mathbf{s}_{y_{i}}\right|^{2}+\sigma^{2}}\right)}\right)}{\sum\limits _{\mathbf{S}_{y}:c_{m}=1,l}\exp\left(\frac{\left|\sum\limits_{k=1}^{N}w_{k, \delta_{l}}\right|^{2}\sigma_{p}^{2}}{2\sigma^{2}\left(\sigma_{p}^{2}\sum \limits_{k=1}^{16}\left|\mathbf{s}_{y_{i}}\right|^{2}+\sigma^{2}\right)} \right)} \tag{23}\] \[=\ln\frac{\sum\limits_{\mathbf{S}_{y}:c_{m}=0,l}\exp\left(\frac{ \left|\sum\limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}\sigma_{p}^{2}}{2\sigma ^{2}\left(\sigma_{p}^{2}\left(\sigma_{p}^{2}\left(\mathbf{S}\mathbf{S}^{*} \right)+\sigma^{2}\right)\right)}\right)}{\sum\limits_{\mathbf{S}_{y}:c_{m}=1,l}\exp\left(\frac{\left|\sum\limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}\sigma _{p}^{2}}{2\sigma^{2}\left(\sigma_{p}^{2}\left(\mathbf{S}\mathbf{S}^{*} \right)+\sigma^{2}\right)}\right)}\,,\,l\in\{1,\,2,\,3\},\,m\in\{0,\,1,\,\,\cdots,4N-1\}\]  
Obviously, the next task is how to simplify the exponential operation in (23). Directly following the fact that [33]  
\[\ln\bigl{[}\exp(\delta_{1})+\cdots+\exp\bigl{(}\delta_{l}\bigr{)} \bigr{]} \approx\ln\bigl{\{}\max\bigl{[}\exp(\delta_{1}),\cdots,\exp\bigl{(} \delta_{l}\bigr{)}\bigr{]}\bigr{\}} \tag{24}\] \[=\max\bigl{[}\ln\exp(\delta_{1}),\cdots,\ln\exp(\delta_{l})\bigr{]} \tag{25}\]  
(23) can be further simplified as:  
\[\zeta_{m} \approx\frac{\sigma_{p}^{2}}{2\sigma^{2}\Bigl{(}\sigma_{p}^{2} \sum\limits_{k=1}^{N}\sum\limits_{i=1}^{16}\left|\mathbf{s}_{y_{i}}\right|^{2 }+\sigma^{2}\Bigr{)}}\left\{\max\limits_{\mathbf{S}_{y}:c_{m}=0,l}\left|\sum \limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}-\max\limits_{\mathbf{S}_{y}:c_{m }=1,l}\left|\sum\limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}\right\} \tag{26}\] \[=\frac{\sigma_{p}^{2}}{2\sigma^{2}\Bigl{(}\sigma_{p}^{2}\bigl{(} \sigma_{p}^{2}\left(\mathbf{S},\mathbf{S}^{*}\right)+\sigma^{2}\Bigr{)}\Bigr{)} }\left\{\max\limits_{\mathbf{S}_{y}:c_{m}=0,l}\left|\sum\limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}-\max\limits_{\mathbf{S}_{y}:c_{m}=1,l}\left|\sum \limits_{k=1}^{N}w_{k,\delta_{l}}\right|^{2}\right\},\]  
where \(l\in\{1,2,3\},m\in\{0,1,\cdots,4N-1\}\).