## 4 Problem Formulate
\begin{bmatrix}\mathbf{g}_{1}\mathbf{W}\mathbf{W}^{T}\mathbf{g }_{1}^{T}+n_{b}^{2}+n_{b}&0&\cdots&0\\ 0&\mathbf{g}_{2}\mathbf{W}\mathbf{W}^{T}\mathbf{g}_{2}^{T}+n_{b}^{2}+n_{b}& \cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\mathbf{g}_{N_{s}}\mathbf{W}\mathbf{W}^{T}\mathbf{g}_{N_{r}}^{T}+ n_{b}^{2}+n_{b}\end{bmatrix}  \tag{15}\] \[=\text{diag}\Big{(}\text{diag}\Big{(}\mathbf{G}\mathbf{W} \mathbf{W}^{T}\mathbf{G}^{T}+n_{b}^{2}+n_{b}\Big{)}\Big{)}.\]  
Next, \(\mathbf{R}_{\mathbf{ys}}\) is given by  
\[\mathbf{R}_{\mathbf{ys}}=\mathbb{E} \begin{pmatrix}y_{1}s_{1}&y_{1}s_{2}&\cdots &y_{1}s_{N_{s}}\\ y_{2}s_{1}&y_{2}s_{2}&\cdots&y_{2}s_{N_{s}}\\ \vdots&\vdots&\ddots&\vdots\\ y_{N_{r}}s_{1}&y_{N_{r}}s_{2}&\cdots&y_{N_{r}}s_{N_{s}}\end{pmatrix} . \tag{16}\]  
For \(1\leqslant i\leqslant N_{r},1\leqslant j\leqslant N_{s}\), we have  
\[\mathbb{E}\big{(}y_{i}s_{j}\big{)} =\mathbb{E}\big{(}y_{i}s_{j}|s_{j}=-1\big{)}\mathrm{Pr}\big{(}s_{ j}=-1\big{)}+\mathbb{E}\big{(}y_{i}s_{j}|s_{j}=1\big{)}\mathrm{Pr}\big{(}s_{j}=1 \big{)}\] \[=-\frac{1}{2}\mathbb{E}\big{(}y_{i}|s_{j}=-1\big{)}+\frac{1}{2} \mathbb{E}\big{(}y_{i}|s_{j}=1\big{)} \tag{17}\] \[=\mathbf{g}_{i}\mathbf{W}\mathbf{e}_{j}.\]  
Note that we use the conclusion that \(\mathbb{E}\big{(}y_{i}|s_{j}=1\big{)}=\mathbf{g}_{i}\mathbf{W}\mathbf{e}_{j}+n _{b}\) and \(\mathbb{E}\big{(}y_{i}|s_{j}=-1\big{)}=-\mathbf{g}_{i}\mathbf{W}\mathbf{e}_{j} +n_{b}\), and the proof is given in Appendix A. According to (17), we have that  
\[\mathbf{R}_{\mathbf{ys}}= \begin{pmatrix}\mathbf{g}_{1}\mathbf{W}\mathbf{e}_{ 1}&\mathbf{g}_{1}\mathbf{W}\mathbf{e}_{2}&\cdots&\mathbf{g}_{1}\mathbf{W} \mathbf{e}_{N_{s}}\\ \mathbf{g}_{2}\mathbf{W}\mathbf{e}_{1}&\mathbf{g}_{2}\mathbf{W}\mathbf{e}_{ 2}&\cdots&\mathbf{g}_{2}\mathbf{W}\mathbf{e}_{N_{s}}\\ \vdots&\vdots&\ddots&\vdots\\ \mathbf{g}_{N_{r}}\mathbf{W}\mathbf{e}_{1}&\mathbf{g}_{N_{r}}\mathbf{W} \mathbf{e}_{2}&\cdots&\mathbf{g}_{N_{r}}\mathbf{W}\mathbf{e}_{N_{s}}\end{pmatrix}  =\mathbf{G}\mathbf{W}. \tag{18}\]  
According to (8), (15), and (18), the closed form of the MSE is  
\[J_{\text{MSE}}=\mathrm{tr}\Big{\{}\mathbf{F}^{T}\text{diag}\Big{(}\text{diag} \Big{(}\mathbf{g}\mathbf{W}\mathbf{W}^{T}\mathbf{g}^{T}+n_{b}^{2}+n_{b}\Big{)} \Big{)}\mathbf{F}-2\mathbf{F}^{T}\mathbf{G}\mathbf{W}\Big{\}}+N_{s}. \tag{19}\]  
In order to make the LED work within a linear dynamic range, the transmit signal satisfies  
\[I_{L}\leqslant e_{i}^{T}\mathbf{W}\mathbf{s}+b_{i}\leqslant I_{H}, \tag{20}\]  
where \(I_{L}\) and \(I_{H}\) represent the minimum and maximum currents corresponding to the linear dynamic range, respectively. Since  
\[-||e_{i}^{T}\mathbf{W}||_{1}\leqslant e_{i}^{T}\mathbf{W}\mathbf{s}\leqslant|| e_{i}^{T}\mathbf{W}||_{1}, \tag{21}\]  
adding DC bias to (21) yields  
\[-||e_{i}^{T}\mathbf{W}||_{1}+b_{i}\leqslant e_{i}^{T}\mathbf{W}\mathbf{s}+b_{ i}\leqslant||e_{i}^{T}\mathbf{W}||_{1}+b_{i}. \tag{22}\]To satisfy the constraint in (20), we have  
\[ \begin{cases}-||e_{i}^{T}\mathbf{W}||_{1}+b_{i}\geqslant I_{L}\\ ||e_{i}^{T}\mathbf{W}||_{1}+b_{i}\leqslant I_{H},\end{cases}  \tag{23}\]  
(23) can be further expressed as  
\[||e_{i}^{T}\mathbf{W}||_{1}\leqslant D,\forall i=1,\cdots,N_{T}, \tag{24}\]  
where \(D\) is given by  
\[D=\min(I_{H}-b_{i},b_{i}-I_{L}). \tag{25}\]  
The optimization problem can thus be formulated as  
\[ \begin{split}\underset{\mathbf{W},\mathbf{F}^{T}}{\text{arg}}& \text{min}\ \ \text{tr}\Big{\{}\mathbf{F}^{T}\text{diag}\Big{(}\text{diag}\Big{(}g \mathbf{W}\mathbf{W}^{T}\mathbf{g}^{T}+n_{b}^{2}+n_{b}\Big{)}\Big{)}\mathbf{F} -2\mathbf{F}^{T}\mathbf{GW}\Big{\}}\\ &\text{s.t.}\ ||e_{i}^{T}\mathbf{W}||_{1}\leqslant D,\forall i=1, \cdots,N_{t}\\ &||\mathbf{e}_{i}^{T}\mathbf{W}||_{1}+b_{i}\leqslant\frac{\eta P _{\text{max}}T}{h\upsilon},\forall i=1,\cdots,N_{t},\end{split}  \tag{26}\]  
where \(P_{\text{max}}\) is the maximum linear optical power of the LEDs. For optical communications, the transmit signals must be non-negative.