## 3 Outage Analysis of Blind-RIS-NOMA Uplink System
### Outage Probability of Blind-RIS-AP-NOMA System for Uplink Transmission  
RIS is typically placed near the BS or receiver in real applications. In [30; 31], RIS as an AP scenario is discussed in which the RIS is deployed quite close to the BS, and both are connected via a dedicated link. As a result, the BS-RIS channel is insignificant. The transmission is possible without RF processing. Figure 2 illustrates a typical blind-RIS-AP-NOMA uplink scenario. The signals received by each RIS element are combined at AP/BS. The superposed signal received at the BS is given by [32]  
\[y_{BS}=\left(\sum_{i=1}^{N}g_{1i}\right)\sqrt{\mu_{1}\;\kappa_{s}}x_{1}+\left( \sum_{i=1}^{N}g_{2i}\right)\sqrt{\mu_{2}\;\kappa_{s}}x_{2}+n_{BS} \tag{23}\]  
Let  
\[L=\sum_{i=1}^{N}g_{1i}\in\bar{\Omega}\aleph(0,\;N) \tag{24}\]  
\[M=\sum_{i=1}^{N}g_{2i}\in\bar{\Omega}\aleph(0,\;N) \tag{25}\]  
We can obtain expressions like (14) and (22) for outages of user 1 and user 2, if we perform the analysis stated previously. **4. Optimal Power Allocation for Uplink Sum Capacity Maximization**  
The uplink sum capacity is given by  
\[C=C^{User1}+C^{User2}=\log_{2}\biggl{(}1+\frac{\mu_{1}\;\kappa\;\alpha_{1}}{\mu _{2}\;\kappa\;\alpha_{2}+1}\biggr{)}+\log_{2}(1+\mu_{2}\;\kappa\;\alpha_{2}) \tag{26}\]  
where \(C^{User1}\) and \(C^{User2}\) are the capacity (b/s/Hz) achieved by user 1 and user 2, respectively. \[C=\log_{2}\biggl{(}\frac{\mu_{2}\;\kappa\;\alpha_{2}+1+\mu_{1}\;\kappa\;\alpha _{1}}{\mu_{2}\;\kappa\;\alpha_{2}+1}\biggr{)}+\log_{2}(1+\mu_{2}\;\kappa\; \alpha_{2}) \tag{27}\]  
Simplifying (27) results in  
\[C=\log_{2}(1+\mu_{2}\;\kappa\;\alpha_{2}+\mu_{1}\;\kappa\;\alpha_{1}) \tag{28}\]  
Substituting \(\mu_{2}=1-\mu_{1}\) in (28) gives  
\[C=\log_{2}(1+\kappa\alpha_{2}+\mu_{1}\kappa(\alpha_{1}-\alpha_{2})) \tag{29}\]  
Figure 2: A sample scenario of the blind-RIS-AP-NOMA uplink. To maximize the sum capacity, the power fractions allocated to users 1 and 2 must be optimized. The users should be able to meet the expected QoS requirements without experiencing an outage, and the sum of the power fractions should equal 1. The following optimization problem is formulated in order to identify the feasible region for both users:  
\[\max_{\mu_{1},\mu_{2}}\ \{\log_{2}(1+\kappa\alpha_{2}+\mu_{1}\kappa(\alpha_{1}- \alpha_{2}))\} \tag{30}\]  
Subject to  
\[ \begin{array}{c}\log_{2}(1+v^{x_{1}})\geq\widetilde{D}_{1}\\ \log_{2}(1+v^{x_{2}})\geq\widetilde{D}_{2}\\ \mu_{1}\ +\mu_{2}\ =1\\ 0<\mu_{j}<1,\ j=1,2\end{array} \]  
From (29), it is clear that \(C\) is a monotonically increasing function of \(\mu_{1}\). When \(\mu_{1}=0\), \(C\) becomes \(1+\kappa\alpha_{2}\). For \(\mu_{1}=1\), \(C\) reaches \(1+\kappa\alpha_{1}\). For maximum uplink capacity, \(\mu_{1}=1\). This may not satisfy the QoS demand of user 2. The optimization problem formulated in (30) for optimal power allocation to user 1 and user 2 is non-convex in nature. Alternatively, an analytical method is used to identify the feasibility region. Solving (17) for \(\mu_{1}\), the feasible region for user 1 can be obtained. \[\mu_{1}\geq\frac{D_{1}(1+\kappa\alpha_{2})}{\kappa(\alpha_{1}+\alpha_{2}D_{1})} \tag{31}\]  
The detailed derivation of (31) is highlighted in Appendix F. Similarly, solving (20) for \(\mu_{1}\) gives a feasible region for user 2.