An Improved Underdetermined Blind Source Separation Method for Insufficiently Sparse Sources

Abstract

Recovering M sources from N mixtures in underdetermined cases, i.e., M > N, is a great challenge, especially for insufficiently sparse sources in noisy cases. To solve this problem, an improved underdetermined blind source separation (UBSS) method is proposed based on single source points (SSPs) identification and l₀-norm. Firstly, we present a mixing matrix estimation method based on SSPs that is identified by directly searching the identical normalized time–frequency (TF) vectors of mixed signals. This method considers the linear representation relations among these TF vectors and therefore could achieve more accurate SSPs identification even in noisy cases. Then, we prove that a non-SSP will be misjudged as SSP with probability zero under some assumptions, which guarantees the stability and effectiveness of the proposed method. Secondly, SSPs are only searched in some optimal frequency bins so that all SSPs in these frequency bins can be identified at one time. Then, the mixing matrix is estimated using hierarchical clustering technique. Thirdly, to recover source signals with real number of active sources, a l₀-norm-based source recovery method is proposed which would be transformed to find the submatrix with the least column of the mixing matrix that can linearly represent TF vectors of mixed signals. Therefore, source signals can be recovered with the real number of active sources, which improves the estimation accuracy of source signals. Some experiments are carried out to show the effectiveness of the proposed method. The present research could improve the estimation accuracy of sources for insufficiently sparse sources with noise in underdetermined cases.

Appendices

Appendix A

Theorem 1

The nonzero normalized TF vector ${\tilde{\mathbf{x}}}\left( {u{\prime} ,v{\prime} } \right)$ is not identical with the nonzero normalized TF vector ${\tilde{\mathbf{x}}}\left( {\psi{\prime} ,\omega{\prime} } \right)$ with probability one if at least one of ${\tilde{\mathbf{x}}}\left( {u{\prime} ,v{\prime} } \right)$ and ${\tilde{\mathbf{x}}}\left( {\psi{\prime} ,\omega{\prime} } \right)$ is non-SSP.

Proof

Assume that the set of active sources at $\left( {u{\prime} ,v{\prime} } \right)$ and $\left( {\psi{\prime} ,\omega{\prime} } \right)$ are $\mathcal{H}$ and $\mathcal{T}$, respectively. The number of elements in $\mathcal{H}$, $\mathcal{T}$ and their union are $m$, $n$, and $l$, respectively. The indices of the $m$ active sources at $\left( {u{\prime} ,v{\prime} } \right)$ are denoted as $\phi_{1} ,\;\phi_{2} ,\; \ldots ,\;\phi_{m}$, i.e., $\mathcal{H} = \left\{ {s_{{\phi_{1} }} \left( {u{\prime} ,v{\prime} } \right),} \right.$$\left. {s_{{\phi_{2} }} \left( {u{\prime} ,v{\prime} } \right),\; \ldots ,\;s_{{\phi_{m} }} \left( {u{\prime} ,v{\prime} } \right)} \right\}$. According to Assumption 3, $m \le N - 1$ and $n \le N - 1$. Since ${\tilde{\mathbf{x}}}\left( {u{\prime} ,v{\prime} } \right)$ and ${\tilde{\mathbf{x}}}\left( {\psi{\prime} ,\omega{\prime} } \right)$ are nonzero vectors, then $m \ge 1$ and $n \ge 1$. Thus, $1 \le m \le N - 1$ and $1 \le n \le N - 1$ can be obtained.

Four cases are discussed as follows.

Case 1: $\mathcal{H} \ne \mathcal{T}$ and $l \le N$.

Suppose that there are $k$ elements ($k \le m$ and $k \le n$) in the intersection of $\mathcal{H}$ and $\mathcal{T}$. Without loss of generality, the $k$ elements can be set as $s_{{\phi_{1} }} \left( {u^{\prime},v^{\prime}} \right),\;s_{{\phi_{2} }} \left( {u^{\prime},v^{\prime}} \right),\; \ldots ,\;s_{{\phi_{k} }} \left( {u^{\prime},v^{\prime}} \right)$. Obviously, the other $n - k$ elements in $\mathcal{T}$, denoted as $s_{{\gamma_{1} }} \left( {u^{\prime},v^{\prime}} \right),\;s_{{\gamma_{2} }} \left( {u^{\prime},v^{\prime}} \right),\; \ldots ,\;s_{{\gamma_{n - k} }} \left( {u^{\prime},v^{\prime}} \right)$, come from the set $\left\{ {s_{{\phi_{m + 1} }} \left( {u^{\prime},v^{\prime}} \right),\;s_{{\phi_{m + 2} }} \left( {u^{\prime},v^{\prime}} \right),\; \ldots ,\;s_{{\phi_{M} }} \left( {u^{\prime},v^{\prime}} \right)} \right\}$, then we obtain $\mathcal{T} = \left\{ {s_{{\phi_{1} }} \left( {u^{\prime},v^{\prime}} \right),\; \ldots ,\;s_{{\phi_{k} }} \left( {u^{\prime},v^{\prime}} \right),} \right.$$\left. {s_{{\gamma_{1} }} \left( {u^{\prime},v^{\prime}} \right),\; \ldots ,\;s_{{\gamma_{n - k} }} \left( {u^{\prime},v^{\prime}} \right)} \right\}$.

Then, at $\left( {u^{\prime},v^{\prime}} \right)$ and $\left( {\psi^{\prime},\omega^{\prime}} \right)$, Eq. (3) can be rewritten as Eq. (A1) and Eq. (A2), respectively.

$$ {\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right) = \sum\limits_{i = 1}^{m} {{\mathbf{a}}_{{\phi_{i} }} s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right)} $$

(A1)

$$ {\mathbf{x}}\left( {\psi^{\prime},\omega^{\prime}} \right) = \sum\limits_{h = 1}^{k} {{\mathbf{a}}_{{\phi_{h} }} s_{{\phi_{h} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} + \sum\limits_{j = 1}^{n - k} {{\mathbf{a}}_{{\gamma_{j} }} s_{{\gamma_{j} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} $$

(A2)

Now, we assume that ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$, then ${\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right) = \alpha {\mathbf{x}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ will hold, i.e.,

$$ \sum\limits_{i = 1}^{m} {{\mathbf{a}}_{{\phi_{i} }} s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right)} = \alpha \left[ {\sum\limits_{h = 1}^{k} {{\mathbf{a}}_{{\phi_{h} }} s_{{\phi_{h} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} + \sum\limits_{j = 1}^{n - k} {{\mathbf{a}}_{{\gamma_{j} }} s_{{\gamma_{j} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} } \right] $$

(A3)

Eq. (A3) can be rewritten as

$$ \sum\limits_{i = 1}^{k} {\left[ {s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right) - \alpha s_{{\phi_{i} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} \right]{\mathbf{a}}_{{\phi_{i} }} } + \sum\limits_{i = k + 1}^{m} {s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right){\mathbf{a}}_{{\phi_{i} }} } - \sum\limits_{j = 1}^{n - k} {\alpha s_{{\gamma_{j} }} \left( {\psi^{\prime},\omega^{\prime}} \right){\mathbf{a}}_{{\gamma_{j} }} } = {\mathbf{0}} $$

(A4)

The matrix form of Eq. (A4) is

$$ {\mathbf{A^{\prime}s^{\prime}}} = {\mathbf{0}} $$

(A5)

where ${\mathbf{A^{\prime}}}{ = }\left[ {{\mathbf{a}}_{{\phi_{1} }} ,\; \ldots ,\;{\mathbf{a}}_{{\phi_{k} }} ,{\mathbf{a}}_{{\phi_{k + 1} }} ,\; \ldots ,\;{\mathbf{a}}_{{\phi_{m} }} \;{\mathbf{a}}_{{\gamma_{1} }} ,\; \ldots ,\;{\mathbf{a}}_{{\gamma_{n - k} }} } \right]$ and ${\mathbf{s^{\prime}}} = \left[ {s_{{\phi_{1} }} \left( {u^{\prime},v^{\prime}} \right) - \alpha s_{{\phi_{1} }} \left( {\psi^{\prime},\omega^{\prime}} \right),\; \ldots ,\;} \right.s_{{\phi_{k} }} \left( {u^{\prime},v^{\prime}} \right) - \alpha s_{{\phi_{k} }} \left( {\psi^{\prime},\omega^{\prime}} \right),$$s_{{\phi_{k + 1} }} \left( {u^{\prime},v^{\prime}} \right),\;\left. { \ldots ,\;s_{{\phi_{m} }} \left( {u^{\prime},v^{\prime}} \right),\; - \alpha s_{{\gamma_{1} }} \left( {\psi^{\prime},\omega^{\prime}} \right),\; \ldots ,\; - \alpha s_{{\gamma_{n - k} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} \right]^{{\text{T}}}$.

According to Assumption 2, ${\mathbf{A^{\prime}}}$ is of full column rank due to $l \le N$. Therefore, Eq. (A5) has a unique zero solution, meaning ${\mathbf{s^{\prime}}} = {\mathbf{0}}$, i.e.,

$$ s_{{\phi_{k + 1} }} \left( {u^{\prime},v^{\prime}} \right) = \ldots = s_{{\phi_{m} }} \left( {u^{\prime},v^{\prime}} \right) = s_{{\gamma_{1} }} \left( {\psi^{\prime},\omega^{\prime}} \right) = \ldots = s_{{\gamma_{n - k} }} \left( {\psi^{\prime},\omega^{\prime}} \right) = 0 $$

(A6)

As $\mathcal{H} \ne \mathcal{T}$, there exists at least one element in the set $\left\{ {s_{{\phi_{k + 1} }} \left( {u^{\prime},v^{\prime}} \right),} \right.\; \ldots ,\;s_{{\phi_{m} }} \left( {u^{\prime},v^{\prime}} \right),$$\left. {s_{{\gamma_{1} }} \left( {\psi^{\prime},\omega^{\prime}} \right),\; \ldots ,\;s_{{\gamma_{n - k} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} \right\}$, and without loss of generality, we suppose $s_{{\phi_{k + 1} }} \left( {u^{\prime},v^{\prime}} \right)$ exists.

However, we have $s_{{\phi_{k + 1} }} \left( {u^{\prime},v^{\prime}} \right) = 0$ from Eq. (A6), which is a contradiction with the above result.

Therefore, ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is not hold in Case 1.

Case 2: $\mathcal{H} \ne \mathcal{T}$ and $l > N$.

Similar to Case 1, we also suppose that there are also $k$ elements, i.e., $s_{{\phi_{1} }} \left( {u^{\prime},v^{\prime}} \right),\;s_{{\phi_{2} }} \left( {u^{\prime},v^{\prime}} \right),\; \ldots ,\;s_{{\phi_{k} }} \left( {u^{\prime},v^{\prime}} \right)$, in the intersection of $\mathcal{H}$ and $\mathcal{T}$. Thus, Eq. (A1) and Eq. (A2) can be also obtained. It can be seen that ${\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right)$ is in the m-dimensional subspace spanned by ${\mathbf{a}}_{{\phi_{1} }} ,\;{\mathbf{a}}_{{\phi_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\phi_{m} }}$ and ${\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is in the n-dimensional subspace spanned by ${\mathbf{a}}_{{\phi_{1} }} ,\;{\mathbf{a}}_{{\phi_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\phi_{k} }} ,\;{\mathbf{a}}_{{\gamma_{1} }} ,$$\;{\mathbf{a}}_{{\gamma_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\gamma_{n - k} }} \;$.If ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$, Eq. (A7) will hold.

$$ {\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right) = \beta {\mathbf{x}}\left( {\psi^{\prime},\omega^{\prime}} \right) $$

(A7)

where $\beta$ is a real constant. Eq. (A7) shows that ${\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right)$ is also in the n-dimensional subspace spanned by ${\mathbf{a}}_{{\phi_{1} }} ,\;{\mathbf{a}}_{{\phi_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\phi_{k} }} ,\;{\mathbf{a}}_{{\gamma_{1} }} ,\;{\mathbf{a}}_{{\gamma_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\gamma_{n - k} }}$.

Let $\mathcal{P}$ be the union of the m-dimensional subspace and the n-dimensional subspace, and $\mathcal{Q}$ be the intersections of these two subspaces. Clearly, $\mathcal{Q}$ has a measure zero in $\mathcal{P}$, indicating that any given TF vector in $\mathcal{P}$ happens to be in $\mathcal{Q}$ with probability zero. Therefore, Eq. (A7) will hold with probability zero, i.e., ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is not hold with probability one in Case 2.

Case 3: $\mathcal{H} = \mathcal{T}$ and $m \le N - 1$.

From Eq. (3), at $\left( {u^{\prime},v^{\prime}} \right)$, we have

$$ {\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right) = \sum\limits_{i = 1}^{m} {{\mathbf{a}}_{{\phi_{i} }} s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right)} $$

(A8)

Since $\mathcal{T} = \mathcal{H}$, then we have

$$ {\mathbf{x}}\left( {\psi^{\prime},\omega^{\prime}} \right) = \sum\limits_{j = 1}^{m} {{\mathbf{a}}_{{\phi_{j} }} s_{{\phi_{j} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} $$

(A9)

If ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$, then ${\mathbf{x}}\left( {u^{\prime},v^{\prime}} \right) = \kappa {\mathbf{x}}\left( {\psi^{\prime},\omega^{\prime}} \right)$, i.e.,

$$ \sum\limits_{i = 1}^{m} {{\mathbf{a}}_{{\phi_{i} }} s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right)} = \kappa \left[ {\sum\limits_{j = 1}^{m} {{\mathbf{a}}_{{\phi_{j} }} s_{{\phi_{j} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} } \right] $$

(A10)

Thus, we can have

$$ \sum\limits_{i = 1}^{m} {\left[ {s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right) - \kappa s_{{\phi_{i} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} \right]{\mathbf{a}}_{{\phi_{i} }} } = {\mathbf{0}} $$

(A11)

Eq. (A11) can be rewritten in matrix form as

$$ {\mathbf{A^{\prime}s^{\prime}}} = {\mathbf{0}} $$

(A12)

where ${\mathbf{A^{\prime}}}{ = }\left[ {{\mathbf{a}}_{{\phi_{1} }} ,\; \ldots ,\;{\mathbf{a}}_{{\phi_{m} }} } \right]$ and ${\mathbf{s^{\prime}}} = \left[ {s_{{\phi_{1} }} \left( {u^{\prime},v^{\prime}} \right) - \kappa s_{{\phi_{1} }} \left( {\psi^{\prime},\omega^{\prime}} \right),\;} \right. \ldots ,\;\left. {s_{{\phi_{m} }} \left( {u^{\prime},v^{\prime}} \right) - \kappa s_{{\phi_{m} }} \left( {\psi^{\prime},\omega^{\prime}} \right)} \right]^{{\text{T}}}$. As $m < N$,${\mathbf{A^{\prime}}}$ is of full column rank, indicating that Eq. (A12) has unique zero solution, that is,

$$ \begin{array}{*{20}c} {s_{{\phi_{i} }} \left( {u^{\prime},v^{\prime}} \right) - \kappa s_{{\phi_{i} }} \left( {\psi^{\prime},\omega^{\prime}} \right) = 0,} & {i = 1,\;2,\; \ldots ,\;m} \\ \end{array} $$

(A13)

As at least one of ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right)$ and ${\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is non-SSP, if $\mathcal{H} = \mathcal{T}$, we get $m \ge 2$. Thus, the following equation can be obtained.

$$ \frac{{s_{{\phi_{1} }} \left( {u^{\prime},v^{\prime}} \right)}}{{s_{{\phi_{1} }} \left( {\psi^{\prime},\omega^{\prime}} \right)}} = \frac{{s_{{\phi_{2} }} \left( {u^{\prime},v^{\prime}} \right)}}{{s_{{\phi_{2} }} \left( {\psi^{\prime},\omega^{\prime}} \right)}} = \ldots = \frac{{s_{{\phi_{m} }} \left( {u^{\prime},v^{\prime}} \right)}}{{s_{{\phi_{m} }} \left( {\psi^{\prime},\omega^{\prime}} \right)}} = \kappa $$

(A14)

From Assumption 4, source signals are mutually independent, which implies that Eq. (A14) will hold with probability zero. Therefore, ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is not hold with probability one in Case 3.

Case 4: $\mathcal{H} = \mathcal{T}$ and $m \ge N - 1$.

Case 4 is a contradiction with Assumption 3 due to $m \ge N - 1$.

From the above analysis of Cases 1–4, we can conclude that ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right) = {\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is not hold with probability one if at least one of ${\tilde{\mathbf{x}}}\left( {u^{\prime},v^{\prime}} \right)$ and ${\tilde{\mathbf{x}}}\left( {\psi^{\prime},\omega^{\prime}} \right)$ is non-SSP.

This completes the proof of Theorem 1.

Appendix B

Theorem 2

For any given vector ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right)$ with $r$ active sources, $r < N$, there does not exist a vector ${\mathbf{s^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)$ that $\left\| {{\mathbf{s^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)} \right\|_{0} < r$ and ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right) = {\mathbf{As^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)$ with probability one.

Proof

Assume that the indices of $r$ active sources are $\chi_{1} ,\;\chi_{2} ,\; \ldots ,\;\chi_{r}$, then, Eq. (21) can be obtained with $\left\| {{\mathbf{s}}_{{\chi_{1} ,\;\chi_{2} ,\; \ldots ,\;\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right)} \right\|_{0} = r$, which indicates that ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right)$ belongs to the r-dimensional subspace spanned by the column of ${\mathcal{A}}_{\;r}^{ * }$.

Assume that there exists a vector ${\mathbf{s^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)$ that ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right) = {\mathbf{As^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)$ and $\left\| {{\mathbf{s^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)} \right\|_{0} = k$, $k < r$. The indices of $k$ nonzero elements of ${\mathbf{s^{\prime}}}\left( {t_{0} ,\;k_{0} } \right)$ are denoted as $\sigma_{1} ,\;\sigma_{2} ,\; \ldots ,\;\sigma_{k}$, then

$$ {\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right) = {\mathcal{A}^{\prime}}_{\;k} {\mathbf{s^{\prime}}}_{{\sigma_{1} ,\;\sigma_{2} ,\; \ldots ,\;\sigma_{k} }} \left( {t_{0} ,\;k_{0} } \right) $$

(B1)

where ${\mathcal{A}^{\prime}}_{\;k} = \left[ {{\mathbf{a}}_{{\sigma_{1} }} ,\;{\mathbf{a}}_{{\sigma_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\sigma_{k} }} } \right] \in {\mathcal{A}}_{\;k}$ and ${\mathbf{s^{\prime}}}_{{\sigma_{1} ,\;\sigma_{2} ,\; \ldots ,\;\sigma_{k} }} \left( {t_{0} ,\;k_{0} } \right) = \left[ {s^{\prime}_{{\sigma_{1} }} \left( {t_{0} ,\;k_{0} } \right),\;s^{\prime}_{{\sigma_{2} }} \left( {t_{0} ,\;k_{0} } \right),\; \ldots ,\;s^{\prime}_{{\sigma_{k} }} \left( {t_{0} ,\;k_{0} } \right)} \right]^{{\text{T}}}$.

Two cases are considered:

Case 1: ${\mathcal{A}^{\prime}}_{\;k}$ is a submatrix of ${\mathcal{A}}_{\;r}^{ * }$.

Without loss of generality, suppose that ${\mathcal{A}^{\prime}}_{\;k}$ is the first $k$ columns of ${\mathcal{A}}_{\;r}^{ * }$, i.e., ${\mathcal{A}^{\prime}}_{\;i} = \left[ {{\mathbf{a}}_{{\chi_{1} }} ,\;{\mathbf{a}}_{{\chi_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\chi_{k} }} } \right]$. Then we has ${\mathcal{A}}_{\;r}^{ * } {\mathbf{s}}_{{\chi_{1} ,\;\chi_{2} ,\; \ldots ,\;\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right) = {\mathcal{A}^{\prime}}_{\;k} {\mathbf{s^{\prime}}}_{{\sigma_{1} ,\;\sigma_{2} ,\; \ldots ,\;\sigma_{k} }} \left( {t_{0} ,\;k_{0} } \right)$, that is,

$$ \begin{gathered} \left[ {{\mathbf{a}}_{{\chi_{1} }} ,\;{\mathbf{a}}_{{\chi_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\chi_{r} }} } \right]\left[ {s_{{\chi_{1} }} \left( {t_{0} ,\;k_{0} } \right),\;s_{{\chi_{2} }} \left( {t_{0} ,\;k_{0} } \right),\; \ldots ,\;s_{{\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right)} \right]^{{\text{T}}} \hfill \\ = \left[ {{\mathbf{a}}_{{\chi_{1} }} ,\;{\mathbf{a}}_{{\chi_{2} }} ,\; \ldots ,\;{\mathbf{a}}_{{\chi_{k} }} } \right]\left[ {s^{\prime}_{{\chi_{1} }} \left( {t_{0} ,\;k_{0} } \right),\;s^{\prime}_{{\chi_{2} }} \left( {t_{0} ,\;k_{0} } \right),\; \ldots ,\;s^{\prime}_{{\chi_{k} }} \left( {t_{0} ,\;k_{0} } \right)} \right]^{{\text{T}}} \hfill \\ \end{gathered} $$

(B2)

Thus we can obtain

$$ {\mathcal{A}}_{\;r}^{ * } \left[ {s_{{\chi_{1} }} \left( {t_{0} ,\;k_{0} } \right) - s^{\prime}_{{\sigma_{1} }} \left( {t_{0} ,\;k_{0} } \right),\;\; \ldots ,\;s_{{\chi_{k} }} \left( {t_{0} ,\;k_{0} } \right) - s^{\prime}_{{\sigma_{k} }} \left( {t_{0} ,\;k_{0} } \right),\;s_{{\chi_{k + 1} }} \left( {t_{0} ,\;k_{0} } \right),\;\; \ldots ,\;s_{{\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right)} \right]^{{\text{T}}} = {\mathbf{0}} $$

(B3)

According to Assumption 2, ${\mathcal{A}}_{\;r}^{ * }$ is of full column rank, therefore $\left[ {s_{{\chi_{1} }} \left( {t_{0} ,\;k_{0} } \right) - s^{\prime}_{{\sigma_{1} }} \left( {t_{0} ,\;k_{0} } \right),\;\; \ldots ,} \right.$$\left. {s_{{\chi_{k} }} \left( {t_{0} ,\;k_{0} } \right) - s^{\prime}_{{\sigma_{k} }} \left( {t_{0} ,\;k_{0} } \right),\;s_{{\chi_{k + 1} }} \left( {t_{0} ,\;k_{0} } \right),\;\; \ldots ,\;s_{{\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right)} \right]^{{\text{T}}} = {\mathbf{0}}$, that is, $s_{{\chi_{k + 1} }} \left( {t_{0} ,\;k_{0} } \right) = \;\; \ldots = \;s_{{\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right) = 0$, which is a contradiction $\left\| {{\mathbf{s}}_{{\chi_{1} ,\;\chi_{2} ,\; \ldots ,\;\chi_{r} }} \left( {t_{0} ,\;k_{0} } \right)} \right\|_{0} = r$.

Therefore, Eq (B1) does not hold in Case 1, i.e., the Theorem 2 is true in Case 1.

Case 2: ${\mathcal{A}^{\prime}}_{\;k}$ is not a submatrix of ${\mathcal{A}}_{\;r}^{ * }$.

If ${\mathcal{A}^{\prime}}_{\;k}$ is not a submatrix of ${\mathcal{A}}_{\;r}^{ * }$, then it must be a submatrix of one of other elements in set ${\mathcal{A}}_{\;r}$ because $k < r$. Therefore, suppose that ${\mathcal{A}^{\prime}}_{\;k}$ is a submatrix of ${\mathcal{A}^{\prime}}_{\;r}$, where ${\mathcal{A}^{\prime}}_{\;r} \in {\mathcal{A}}_{\;r}$ and ${\mathcal{A}^{\prime}}_{\;r} \ne {\mathcal{A}}_{\;r}^{ * }$. From Eq. (B1), ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right)$ belongs to the k-dimensional subspace spanned by the column of ${\mathcal{A}^{\prime}}_{\;k}$, therefore ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right)$ must belong to the r-dimensional subspace spanned by the column of ${\mathcal{A}^{\prime}}_{\;r}$.

However, for any given TF vector with $r$ active sources, it belongs to only one r-dimensional subspace spanned by the columns of the element in ${\mathcal{A}}_{\;r}$ with probability one, which indicates that ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right)$ belongs to two different r-dimensional subspaces spanned by the column of ${\mathcal{A}}_{\;r}^{ * }$ and ${\mathcal{A}^{\prime}}_{\;r}$, respectively, at the same time with probability zero. As a result, ${\mathbf{x}}\left( {t_{0} ,\;k_{0} } \right)$ belongs to the k-dimensional subspace spanned by the column of ${\mathcal{A}^{\prime}}_{\;k}$ with probability zero, i.e., the Theorem 2 is true in Case 2.

This completes the proof of Theorem 2.