# CMOS Fixed Pattern Noise Removal Based on Low Rank Sparse Variational Method
## 4 Proposed Model
\begin{array}{cc}\frac{\partial\mathrm{U}}{\partial\mathrm{x}}- \frac{R_{2}}{\omega_{2}}-\frac{\alpha_{2}}{\omega_{2}},&if\ \frac{\partial\mathrm{U}}{\partial\mathrm{x}}- \frac{R_{2}}{\omega_{2}}>\frac{\alpha_{2}}{\omega_{2}}\\ 0,&if\ \left|\frac{\partial\mathrm{U}}{\partial\mathrm{x}}-\frac{R_{2}}{\omega_{2} }\right|\leq\frac{\alpha_{2}}{\omega_{2}}\\ \frac{\partial\mathrm{U}}{\partial\mathrm{x}}-\frac{R_{2}}{\omega_{2}}+\frac{ \alpha_{2}}{\omega_{2}},&if\ \frac{\partial\mathrm{U}}{\partial\mathrm{x}}- \frac{R_{2}}{\omega_{2}}<-\frac{\alpha_{2}}{\omega_{2}}\end{array} \right. \tag{14}\]
5. Sub-questions about \(J\) \[(\mathrm{U},\mathrm{S},\mathrm{H},\mathrm{J},\mathrm{K})=\alpha_{3}\|J\|_{1}+ \frac{\omega_{3}}{2}\left\|J-\frac{\partial\mathrm{U}}{\partial\mathrm{y}}+ \frac{R_{3}}{\omega_{3}}\right\|_{2}^{2}\] (15) Solve the optimal solution of \(J\). \[\frac{\partial\mathrm{E}(U,S,H,J,K)}{\partial J}=0\]  
\[J=\left\{ \begin{array}{cc}\frac{\partial\mathrm{U}}{\partial\mathrm{y}}- \frac{R_{3}}{\omega_{3}}-\frac{\alpha_{3}}{\omega_{3}},&if\ \frac{\partial\mathrm{U}}{\partial\mathrm{y}}- \frac{R_{3}}{\omega_{3}}>\frac{\alpha_{3}}{\omega_{3}}\\ 0,&if\ \left|\frac{\partial\mathrm{U}}{\partial\mathrm{y}}-\frac{R_{3}}{\omega_{3} }\right|\leq\frac{\alpha_{3}}{\omega_{3}}\\ \frac{\partial\mathrm{U}}{\partial\mathrm{y}}-\frac{R_{3}}{\omega_{3}}+\frac{ \alpha_{3}}{\omega_{3}},&if\ \frac{\partial\mathrm{U}}{\partial\mathrm{y}}- \frac{R_{3}}{\omega_{3}}<-\frac{\alpha_{3}}{\omega_{3}}\end{array} \right. \tag{16}\]
6. Sub-questions about \(K\) \[E(\mathrm{U},\mathrm{S},\mathrm{H},\mathrm{J},\mathrm{K})=\alpha_{4}\|K\|_{1}+ \frac{\omega_{4}}{2}\left\|\mathrm{K}-\left(\frac{\partial\mathrm{Y}}{\partial \mathrm{y}}-\frac{\partial\mathrm{U}}{\partial\mathrm{y}}\right)+\frac{R_{4}}{ \omega_{4}}\right\|_{2}^{2}\] (17)Solve the optimal solution of \(K\). \[K=\left\{ \begin{array}{cc}\frac{\partial E(U,S,H,J,K)}{\partial K}=0\\ \\ \left(\frac{\partial Y}{\partial y}-\frac{\partial U}{\partial y}\right)-\frac{R_{ 4}}{\omega_{4}}-\frac{\alpha_{4}}{\omega_{4}},&if\left(\frac{\partial Y}{ \partial y}-\frac{\partial U}{\partial y}\right)-\frac{R_{4}}{\omega_{4}}> \frac{\alpha_{4}}{\omega_{4}}\\ \\ 0,&if\left(\frac{\partial Y}{\partial y}-\frac{\partial U}{\partial y}\right) \right|-\frac{R_{4}}{\omega_{4}}\leq\frac{\alpha_{4}}{\omega_{4}}\\ \\ \left(\frac{\partial Y}{\partial y}-\frac{\partial U}{\partial y}\right)-\frac{R _{4}}{\omega_{4}}+\frac{\alpha_{4}}{\omega_{4}},&if\left(\frac{\partial Y}{ \partial y}-\frac{\partial U}{\partial y}\right)-\frac{R_{4}}{\omega_{4}}<- \frac{\alpha_{4}}{\omega_{4}}\\ \end{array} \right. \tag{18}\]  
7. Updating the language multiplier \(R_{2}\)\(R_{3}\)\(R_{4}\) by the dual gradient rise method  
\[R_{2}=R_{2}+\omega_{2}\left(\mathrm{H}-\frac{\partial U}{\partial\mathbf{x}}\right) \tag{19}\]  
\[R_{3}=R_{3}+\omega_{3}\left(\mathrm{J}-\frac{\partial U}{\partial y}\right) \tag{20}\]  
\[R_{4}=R_{4}+\omega_{4}\left(\mathbf{K}-\left(\frac{\partial Y}{\partial y}- \frac{\partial U}{\partial y}\right)\right) \tag{21}\]  
where \(\omega_{2},\omega_{3},\omega_{4}\) is the iterative step length in the process of gradient rise. In order to facilitate computer computation, we need to discretize continuous operators. The discretization of partial differential is defined as follows: \(\frac{\partial A}{\partial\mathbf{x}}\) discrete operation is \(\mathrm{A}_{\mathrm{ij}+1}-\mathrm{A}_{ij}\), \(\frac{\partial A}{\partial y}\) discrete operation is \(\mathrm{A}_{\mathrm{i}+1\mathrm{j}}-\mathrm{A}_{ij}\), \(\frac{\partial^{2}A}{\partial x^{2}}\) discrete operation is \(\mathrm{A}_{\mathrm{ij}+1}-\mathrm{A}_{ij-1}-2\mathrm{A}_{ij}\). The complete calculation process is shown in Algorithm 1. The matlab code used in this article can be obtained from the download link in Supplementary Materials. ```
1. Get image Y with FPN
2. The initial matrix \(\mathbf{U=0,S=0,}\)\(R_{2}=\mathbf{0,}\)\(R_{3}=\mathbf{0,}\)\(R_{4}=\mathbf{0,}\)\(\mathbf{H=0,J=0,K=0}\)
3.