## 3 Proposed Method
### Separation of Non-Sparse Components
\right] \tag{16}\]  
where \(\mu_{\mathcal{I}_{1}}\), \(\mu_{\mathcal{I}_{2}}\), \(\mu_{\mathcal{I}_{3}}\) denote the estimated DOA of the active source by applying the method in [21]. Equation (16) can be rewritten as a vector form as:  
\[\mathbf{\mathfrak{K}}\mathbf{S_{\mathcal{I}}}=\mathbf{S} \tag{17}\]  
where \(\mathbf{S_{\mathcal{I}}}=[S_{\mathcal{I}_{1}(n,l)},S_{\mathcal{I}_{2}(n,l)},S_{ \mathcal{I}_{3}(n,l)}]^{T}\) and \(\mathbf{S}=[S_{W}(n,l),S_{X}(n,l),S_{Y}(n,l)]^{T}\).The separation problem is converted to solve the linear equations by regarding the TF coefficient of each active source as an independent variable. It should be mentioned that the process is based on the hypothesis that the mixing matrix of Equation (16) is a column full rank matrix. To jointly consider the case \(||\mathbf{\mathfrak{D}}||_{0}>3\), the aim is converted to find a vector \(\mathbf{S_{\mathcal{I}}}^{N}=[S_{\mathcal{I}_{1}}^{N}(n,l),S_{\mathcal{I}_{2}}^{N }(n,l),S_{\mathcal{I}_{3}}^{N}(n,l)]\) to minimize \(||\mathbf{S}-\mathbf{\mathfrak{K}}\mathbf{S_{\mathcal{I}}}||_{F}^{2}\), i.e.,  
\[\mathbf{S_{\mathcal{I}}}^{N}=\arg\min_{\mathbf{S_{\mathcal{I}}}}||\mathbf{S}-\mathbf{ \mathfrak{K}}\mathbf{S_{\mathcal{I}}}||_{F}^{2} \tag{18}\]  
where \(||\cdot||_{F}\) represents the Frobenius norm. \(\mathbf{S_{\mathcal{I}}}^{N}\) represents the separated TF coefficients of the active sources at \((n,l)\). For the case \(||\mathbf{\mathfrak{D}}||_{0}=2\), i.e., \(\mathbf{\mathcal{I}}=[\mathcal{I}_{1},\mathcal{I}_{2}]\), we can omit the first equation in Equation (16) i.e.,  
\[\left[ \begin{array}{cc}\cos\mu_{\mathcal{I}_{1}}&\cos\mu_{\mathcal{I}_{2}}\\ \sin\mu_{\mathcal{I}_{1}}&\sin\mu_{\mathcal{I}_{2}}\end{array} \right]\left[ \begin{array}{c}S_{\mathcal{I}_{1}}(n,l)\\ S_{\mathcal{I}_{2}}(n,l)\end{array} \right]=\left[ \begin{array}{c}S_{X}(n,l )\\ S_{Y}(n,l)\end{array} \right] \tag{19}\]  
We can still get the solution by solving Equation (18) for this case. Then, we can get the actual TF coefficients of \(S_{\mathcal{I}_{1}}(n,l)\) and \(S_{\mathcal{I}_{2}}(n,l)\) by:  
\[\left\{ \begin{array}{c}S_{\mathcal{I}_{1}}^{N}(n,l)=\frac{sin(\mu_{\mathcal{ I}_{2}})\cdot S_{X}(n,l)-cos(\mu_{\mathcal{I}_{2}})\cdot S_{Y}(n,l)}{sin(\mu_{ \mathcal{I}_{2}}-\mu_{\mathcal{I}_{1}})}\\ S_{\mathcal{I}_{2}}^{N}(n,l)=\frac{sin(\mu_{\mathcal{I}_{1}})\cdot S_{X}(n,l)- cos(\mu_{\mathcal{I}_{2}})\cdot S_{Y}(n,l)}{sin(\mu_{\mathcal{I}_{1}}-\mu_{ \mathcal{I}_{2}})}\end{array} \right. \tag{20}\]  
Finally, the final recovered signal can be obtained by a synthesis as:  
\[\hat{S_{i}}(n,l)=S_{i}^{C}(n,l)+S_{i}^{N}(n,l) \tag{21}\]  
Figure (a)a illustrates an example of the TF component extraction of the proposed method. Figure (b)b is an example of the recovered signals from the mixture signal. Figure 5: **(a)** Example of a TF component extraction of the proposed framework. Both the frame length and number of STFT points are 2048. Each square represents a TF instant. Blue-shadowed squares denote the sparse component, while the squares in other colors denote the non-sparse component. The respective TF instants for each source (six sources for example) are indicated by a ball; **(b)** Example of six recovered signals from the mixture signal in the frequency domain.