On extending the Noisy Independent Component Analysis to Impulsive Components

Abstract

As an important factor in the fast fixed-point algorithm of independent component analysis (ICA), noise has a significant influence on the separate performance of ICA. Unfortunately, the traditional algorithm of noisy ICA did not address the influence of impulsive components. Because the sources were signals mixed with impulsive noise, the Gaussian noisy algorithm will be invalid for separating the sources. In general, those measurements that significantly deviate from the normal pattern of sensed data are considered impulses. In this paper, we introduce a non-linear function based on the S-estimator to identify the impulsive components in the observed data. This approach guarantees that the impulse noise can be detected from the observed signal. Furthermore, a threshold for the impulse components and methods to remove impulse noise and reconstruct the signal is proposed. The proposed technique improves the separate performance of the traditional algorithm for Gaussian noisy ICA. With the proposed method, the fast fixed-point algorithm of ICA is more reliable for noisy situations. The simulation results show the effectiveness of the proposed method.

Appendix

1.1 Proof Lemma 1

We first suppose the case l odd (l = 2q + 1). Using S = s(r ₁, …, r _l ) for ease of notation, we can obtain

$$\frac{{med_{i} \left| {r_{i} } \right|}}{c} \le S \le \frac{{med_{i} \left| {r_{i} } \right|}}{{\text{g}^{ - 1} \left( {\frac{{\text{g} \left( c \right)}}{l + 1}} \right)}}$$

(31)

Suppose med _i |r _i | > cS. Because $med_{i} \left| {r_{i} } \right| = \left| r \right|_{{q + 1{\kern 1pt} :l}}$, it holds that at least q + 1 of the $\frac{{\left| {r_{i} } \right|}}{S}$ is larger than c. Consequently,

$$\frac{1}{l}\sum\limits_{i = 1}^{l} {\text{g} \left( {\frac{{\left| {r_{i} } \right|}}{S}} \right)} \ge \frac{1}{l}\left( {q + 1} \right)\text{g} \left( c \right) > \frac{{\text{g} \left( c \right)}}{2} = {\rm K}$$

(32)

Therefore, $med_{i} \left| {r_{i} } \right| \le cS$.

Now, we suppose that $\text{g}^{ - 1} \left( {\frac{{\text{g} \left( c \right)}}{l + 1}} \right)S > med_{i} \left| {r_{i} } \right|$. This would imply that the first q + 1 of the $\frac{{\left| {r_{i} } \right|}}{S}$ is strictly smaller than $\text{g}^{ - 1} \left( {\frac{{\text{g} \left( c \right)}}{l + 1}} \right)$. Introducing this in $\frac{1}{l}\sum\nolimits_{i = 1}^{l} {{\text{g}}\left( {\frac{{\left| {r_{i} } \right|}}{S}} \right)}$, we find that

$$\begin{aligned} \frac{1}{l}\sum\limits_{i = 1}^{l} {\text{g} \left( {\frac{{\left| {r_{i} } \right|}}{S}} \right)} & \le \frac{1}{l}\left( {q + 1} \right)\text{g} \left( {g^{ - 1} \left( {\frac{{\text{g} \left( c \right)}}{l + 1}} \right)} \right) + \frac{{q\text{g} \left( c \right)}}{l} \\ & \le \frac{q + 1}{l}\left( {\frac{{\text{g} \left( c \right)}}{l + 1}} \right) + \frac{{q\text{g} \left( c \right)}}{l} = \frac{{\text{g} \left( c \right)}}{2} = {\rm K} \\ \end{aligned}$$

(33)

Therefore, $\text{g}^{ - 1} \left( {\frac{g\left( c \right)}{l + 1}} \right)S \le med_{i} \left| {r_{i} } \right|$.

Now, n is even (l = 2q), which proves that

$$\frac{{med_{i} \left| {r_{i} } \right|}}{c} \le S \le \frac{{med_{i} \left| {r_{i} } \right|}}{{\frac{1}{2}\text{g}^{ - 1} \left( {\frac{{2\text{g} \left( c \right)}}{l + 2}} \right)}}$$

(34)

Because $med_{i} \left| {r_{i} } \right| = \frac{1}{2}\left( {\left| r \right|_{{q{\kern 1pt} :l}} + \left| r \right|_{q + 1:l} } \right)$, we have that at least q of the $\frac{{\left| {r_{i} } \right|}}{s}$ is strictly larger than c, and

$$\frac{1}{l}\sum\limits_{i = 1}^{l} {\text{g} \left( {\frac{{\left| {r_{i} } \right|}}{S}} \right)} \ge \frac{q}{l}\text{g} \left( c \right) > \frac{{\text{g} \left( c \right)}}{2} = {\rm K}$$

(35)

Except when all other |r _i | are zero, the set of solutions for Eq. (10) is the interval $\left( {0,\frac{{2\,med_{i} \left| {r_{i} } \right|}}{c}} \right]$. Therefore, $S = \frac{{2\,med_{i} \left| {r_{i} } \right|}}{c}$. In other cases, med _i |r _i | ≤ cS.

Suppose now $med_{i} \left| {r_{i} } \right| < \frac{1}{2}\text{g}^{ - 1} \left( {\frac{{2\text{g} \left( c \right)}}{l + 2}} \right)S$. Then, $\left| r \right|_{{q + 1{\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} n}} < \text{g}^{ - 1} \left( {\frac{{2\text{g} \left( c \right)}}{l + 2}} \right)S$. Hence, the first q + 1 of the $\frac{{\left| {r_{i} } \right|}}{S}$ is less than $g^{ - 1} \left( {\frac{{2\text{g} \left( c \right)}}{l + 2}} \right)$. Then, we obtain

$$\begin{aligned} \frac{1}{l}\sum\limits_{i = 1}^{l} {\text{g} \left( {\frac{{\left| {r_{i} } \right|}}{S}} \right)} & < \frac{1}{l}\left( {l + 1} \right)\frac{2}{l + 2}\text{g} \left( c \right) + \frac{{\left( {q - 1} \right)\text{g} \left( c \right)}}{l} \\ & \le \frac{q}{l}\text{g} \left( c \right) = {\rm K} \\ \end{aligned}$$

(36)

That means $med_{i} \left| {r_{i} } \right| \ge \frac{1}{2}\text{g}^{ - 1} \left( {\frac{{2\text{g} \left( c \right)}}{l + 2}} \right)S$.