An Algorithm for Mixing Matrix Estimation in Underdetermined Blind Source Separation

Abstract

In this paper, we propose a noble algorithm for mixing matrix estimation in Underdetermined blind source separation. A concept of confidence measure and that of being able to be single source points at all the points in time frequency plane are introduced in the proposed algorithm. At first, we can detect the single source points from real parts and imaginary parts of time frequency coefficients of mixture signals and calculate principal vectors and its confidence measures through principal component analysis at the single source points using this algorithm. Finally, mixing matrix is obtained by clustering principal vectors according to its confidence measure. Experimental results show that the proposed algorithm is very suitable for actual situation of UBSS.

Appendix: Statistical Analysis

From assumption that the mixture has some SSPs in TF plane and at every SSP, in neighborhood of this point only one signal is dominant, all columns of matrix $\mathbf{X} _{\varOmega _{t,f}}$ must be same column of mixing matrix $\mathbf{A}$. However, the absolute direction of matrix $\mathbf{X} _{\varOmega _{t,f}}$ are different. Hence, at SSP (t, f), we can choose the principal vector $\hat{\theta }(t,f)$ with a unit norm as the vector that is collinear with the column of mixing matrix $\mathbf{A}$. Now we model the STFT coefficients of the dominant source in the TF region $\varOmega _{t,f}$ with a centered normal distribution of large variance $\sigma ^2_1$, and the rest of the sources as centered normal distribution with variances $\sigma ^2_2\ge \cdots \ge \sigma ^2_M$, respectively. Suppose that noise $\mathbf{n}$ has centered normal distribution with variance $\sigma ^2_{noise}$, i.e., $\mathbf{n} \sim N(0,\sigma ^2_{noise}{} \mathbf{I} _M)$. For point $(t',f') \in \varOmega _{t,f}$, we have the following expression:

$$\begin{aligned} \mathbf{X} (t',f')=\sum ^{M}_{i=1}s_i(t',f')\mathbf{a} _i+\mathbf{n} (t',f'), \end{aligned}$$

where $s_i(t',f') \sim N(0,\sigma ^2_i)$ and $\mathbf{a} _i$ is column of mixing matrix $\mathbf{A}$ corresponding to i-th strong source of mixture. We’d like to stress that $\mathbf{a} _i$ may be not i-th column of $\mathbf{A}$. Therefore, $\mathbf{X} (t',f') \sim N(0,\varSigma ), ~\varSigma =\sum ^{M}_{i=1}\sigma ^2_i \mathbf{a} _i \mathbf{a} _i^T+\sigma ^2_{noise}{} \mathbf{I} _M$. Let the eigenvalues of the covariance matrix $\varSigma$ be $\lambda _1\ge \cdots \ge \lambda _M$ and a unit eigenvectors corresponding with $\lambda _i$ be $\mathbf{u} _i$. If $\sigma ^2_2= \cdots \sigma ^2_M=\sigma ^2_{noise}=0$, then $\mathbf{a} _1$ is collinear with $\mathbf{u} _1$. At this point, we can regard the ratio $\frac{\lambda _1}{\sum ^M_{i=1}\lambda _i}$ as a measure of how accurately the first principal component represents a space spanned by columns of the mixing matrix $\mathbf{A}$. Since this is relatively measured, we can replace the ratio $\frac{\lambda _1}{\sum ^M_{i=1}\lambda _i}$ with the ratio $T=\frac{\lambda _1}{\sum ^M_{i=2}\lambda _i}$. On the other hand, according to [9, Theorem 5.7], the matrix $\mathbf{X} _{\varOmega _{t,f}}{} \mathbf{X} _{\varOmega _{t,f}}^T$ has a Wishart distribution $W_M(\varSigma ,card(\varOmega _{t,f})-1)$ and its expectation is $(card(\varOmega _{t,f})-1)\varSigma$, i.e.,

$$\begin{aligned}&\mathbf{X} _{\varOmega _{t,f}}{} \mathbf{X} _{\varOmega _{t,f}}^T \sim W_M(\varSigma ,(card(\varOmega _{t,f})-1)),\\&E(\mathbf{X} _{\varOmega _{t,f}}{} \mathbf{X} _{\varOmega _{t,f}}^T)=(card(\varOmega _{t,f})-1)\varSigma . \end{aligned}$$

Hence, $\hat{T}(t,f)$ in Eq. (7) is an estimator of the ratio $\frac{\lambda _1}{\sum ^M_{i=2}\lambda _i}$. Then, from [9, Theorem 9.4], we can conclude that $\sqrt{card(\varOmega _{t,f})-1}(\hat{\theta }(t,f)-\theta (t,f))$ has asymptotically normal distribution

$$\begin{aligned} \sqrt{card(\varOmega _{t,f})-1}(\hat{\theta }(t,f)-\theta (t,f)) \sim N_M(0,\mathbf{V} _1) \end{aligned}$$

and $\sqrt{card(\varOmega _{t,f})-1}(\hat{\lambda }_1-\lambda _1)$ has asymptotically normal distribution

$$\begin{aligned} \sqrt{card(\varOmega _{t,f})-1}(\hat{\lambda }_1-\lambda _1) \sim N{}(0,2\lambda ^2_1), \end{aligned}$$

where $\mathbf{V} _1=\lambda _1\sum ^M_{k=2}\lambda _k\mathbf{u} _k\mathbf{u} _k^T/(\lambda _k-\lambda _1)^2, ~\theta (t,f)=\mathbf{u} _1$ and $\hat{\lambda }_1$ is an estimator of $\lambda _1$. Now taking $f(\lambda )=\lambda _1/\sum ^M_{i=2}\lambda _i=T$, according to [9, Theorem 4.11]

$$\begin{aligned} \sqrt{card(\varOmega _{t,f})-1}(f(\hat{\lambda })-f(\lambda )) \sim N(0,\mathbf{D} ^T \varSigma \mathbf{D} ), ~card(\varOmega _{t,f}) \rightarrow \infty , \end{aligned}$$

where $\mathbf{D} =(\partial f/\partial t_i)|_{t=\lambda }$ is the M-dimensional vector. From this we have Eq. (8). In Eqs. (9) and (11), we have $q_1$ and $q_2$ as $q_1=q_2=\sqrt{2}\times erfinv(0.98)=2.3263 \approx 2.33$, where erfinv is an inverse error function in MATLAB.