# Rateless Coded Uplink Transmission Design for Multi-User C-RAN
## 4 Decoding Performance Analysis and Degree Profile Design
### EXIT Analysis of the Decoding Process

Table 1: Annotations to symbols. As shown in Figure 4, we redraw the decoding graph of Figure 2 for a clearer illustration of the EI update procedure between each node during the decoding process. The EI update during the decoding iteration is given in the following, which corresponds to the decoding process given in Section 3.3. For the \(l\)th iteration of the first stage decoding, firstly EI is exchanged on the Raptor decoding graph of user 1, i.e., on the upper part of Figure 3:  
_Step 1:_ The LLR messages are transferred from user 1's input nodes to the LDPC check nodes. The EI of the messages is:  
\[x_{1,\;ext}^{(l-1)}=\sum\
olimits_{d=1}^{d_{\varphi}}\alpha_{1,d}\bigg{/} \bigg{(}dJ^{-1}\bigg{(}x_{1,u}^{(l-1)}\bigg{)}\bigg{)}, \tag{16}\]  
where \(x_{1,u}^{(l-1)}\) is the EI of the messages from the output nodes to the input nodes in the \((l-1)\)th iteration. _Step 2:_ EI from the LDPC check nodes to the input nodes of user 1 is:  
\[T\bigg{(}x_{1,ext}^{(l-1)}\bigg{)}=\sum\
olimits_{d=1}^{d_{\varphi}^{\prime}} \xi_{d}\bigg{/}\bigg{(}dJ^{-1}\bigg{(}1-\sum\
olimits_{j=1}^{d_{\varphi}^{ \prime}}\widetilde{\rho}_{j}\bigg{/}\bigg{(}(j-1)J^{-1}\bigg{(}1-x_{1,ext}^{(l -1)}\bigg{)}\bigg{)}\bigg{)}\bigg{)}. \tag{17}\]  
_Step 3:_ EI from user 1's input nodes to the output nodes is:  
\[x_{1,\varphi}^{l}=\sum\
olimits_{d=1}^{d_{\varphi}}\widetilde{\alpha}_{1,d} \bigg{/}\bigg{(}(d-1)J^{-1}\bigg{(}x_{u(1)}^{(l-1)}\bigg{)}+J^{-1}\bigg{(}T \bigg{(}x_{ext(1)}^{(l-1)}\bigg{)}\bigg{)}\bigg{)}. \tag{18}\]  
_Step 4:_ EI from the output nodes of user 1 to the input nodes:  
\[x_{1,u}^{(l)}=1-\sum\
olimits_{d=1}^{d_{\varphi}}\omega_{1,d}\bigg{/}\bigg{(} (d-1)J^{-1}\bigg{(}1-x_{1,\varphi}^{(l)}\bigg{)}+J^{-1}\bigg{(}1-I_{DET1} \bigg{(}I_{out2}^{(l-1)};\mathbf{H}\bigg{)}\bigg{)}\bigg{)}, \tag{19}\]  
where \(I_{DET1}\bigg{(}I_{out2}^{(l-1)};\mathbf{H}\bigg{)}\) is the EI of the messages from the MU detector. Its value is determined by the channel matrix \(\mathbf{H}\) and the EI \(I_{out2}^{(l-1)}\) corresponding to the messages from the output nodes of user 2 in the \((l-1)\)th iteration. The derivation of \(I_{DET1}\) will be discussed later. _Step 5:_ EI from LT output nodes of user 1 to the MU detector:  
\[I_{out1}^{(l)}=1-\sum\
olimits_{d=1}^{c}\Omega_{1,d}\bigg{/}\bigg{(}dJ^{-1} \bigg{(}1-x_{1,\varphi}^{(l-1)}\bigg{)}\bigg{)}. \tag{20}\]  
The EI update on the Raptor decoding graph of user 2 is similar, which is omitted here. In step 4, \(I_{DET1}(I_{out2};\mathbf{H})\) is the output EI from the detector. The input EI is \(I_{out2}\), which is the EI from the LT output node of user 2 to the detector. This function can be numerically derived by Monte Carlo simulations [44]. In detail, \(N_{mont}\) transmitted symbols \(x_{1}\) and \(x_{2}\) of user 1 and user 2 are generated, respectively. For user 1, the symbol value is fixed to be 1, while for user 2, the symbol value is randomly 1 or \(-1\). With randomly generated Gaussian noise samples, \(N_{mont}\) samples of quantized  
Figure 4: Extrinsic information (EI) transfer on the decoding graph during the decoding iteration. signals from each RRH can be generated. For each fixed \(I_{out2}\), we generate \(N_{mut}\) samples of \(LLR^{e}[c_{2}]\) from the symmetric Gaussian distribution with mean \(\tau\) and \(-\tau\), depending on the corresponding \(x_{2}\) being either \(+1\) or \(-1\), respectively, where \(\tau\) is calculated from (14). Then, \(N_{mut}\) samples of \(LLR^{a}[c_{1}]\) can be generated. \(I_{DET1}\) is evaluated using the histogram for the probability distribution of \(LLR^{a}[c_{1}]\). According to the results of Monte Carlo simulations, we find that under fixed \(\mathbf{H}\), \(I_{DET1}\) can be well approximated by a linear function:  
\[I_{DET1}=al_{out2}+b, \tag{21}\]  
where \(a\) and \(b\) are coefficients which can be determined by two end points of the line, i.e., \(I_{DET1}(0;\mathbf{H})=I(x_{1};\hat{y}_{1},\hat{y}_{2})\) and \(I_{DET1}(1;\mathbf{H})=I(x_{2};\hat{y}_{1},\hat{y}_{2}|x_{1})\).