Statistically Optimal Joint Multimicrophone MAP Estimators Under Super-Gaussian Assumption

Abstract

This paper presents two super-Gaussian-based multimicrophone maximum a posteriori (MAP) estimators which exploit both amplitude and phase of speech signal from noisy observations. It is well known that super-Gaussian distributions model the statistical properties of speech signal more accurately. Under the independent Gaussian statistical assumption for noise signals, which is usually valid in wireless acoustic sensor networks, two joint multimicrophone estimators are derived while the speech signal is modeled by super-Gaussian distribution. Since the microphones are distributed randomly and may also belong to different devices, the independency assumption of noise signals is more reasonable in these networks. The performance of the proposed estimators is compared to that of four baseline estimators; the first is the multimicrophone minimum mean square error (MMSE) estimation, where both amplitude and phase are derived assuming Gaussian properties for speech signal. The second baseline is the multimicrophone MAP-based amplitude estimator, that utilizes the super-Gaussian statistics to just obtain the amplitude of speech and keeps the phase unchanged. As the third one, we have considered a minimum variance distortion-less response filter followed by a super-Gaussian MMSE estimator. We have also compared the performance of the proposed estimators with the centralized multichannel Wiener filter. The simulation experiments demonstrate remarkable ability of the proposed estimators to enhance speech quality and intelligibility when the clean speech is degraded by a mixture of both point source interference and additive noise in reverberant environments.

Appendix A

A generalized form of (9) has been presented in Example 1 of Chapter 5 of [34], that expresses the probability distribution of random variable Y, which is a function of variable X as follows:

$$\begin{aligned} Y = aX + b, \end{aligned}$$

(A1)

where a and b represent deterministic variables. In the case of $b = 0$, this equation is simplified to our case. Although in general, the division of two random variables X and Y, i.e., Y/X yields a random variable, however, in special case like the current situation, ($a = \dfrac{Y}{X}$) represents a deterministic value. In the problem at hand

$$\begin{aligned} {\left\{ \begin{array}{ll} Y \longleftarrow A_{m} \\ X \longleftarrow A_{1} \end{array}\right. } \end{aligned}$$

(A2)

where random variables are with Rayleigh distribution, and

$$\begin{aligned} {\left\{ \begin{array}{ll} a \longleftarrow C_{m} \\ b \longleftarrow 0 \end{array}\right. } \end{aligned}$$

(A3)

so, $C_{m}$ represents a deterministic value (the ratio of two standard deviations) as explained in the manuscript.

Based on [34], the distribution function of $F_y(y)$ is computed as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} F_y(y) = P\{X\le \dfrac{y-b}{a}\}= F_x(\dfrac{y-b}{a}), \qquad &{} a > 0, \\ F_y(y)= P\{X \ge \dfrac{y-b}{a}\}= 1- F_x(\dfrac{y-b}{a}), \qquad &{} a < 0, \end{array}\right. } \end{aligned}$$

(A4)

and the PDF is computed as

$$\begin{aligned} f_y(y) =\frac{1}{\mid a \mid } f_x(\dfrac{y-b}{a}). \end{aligned}$$

(A5)

In our problem the amplitude $A_1$ has the super-Gaussian distribution

$$\begin{aligned} p(A_1) = {\left\{ \begin{array}{ll} \dfrac{\mu ^{\nu +1} A_1^\nu }{\Gamma (\nu +1) \sigma ^{\nu +1}_x(1)} \exp \left( \dfrac{-\mu A_1}{\sigma _x(1)} \right) , \qquad &{} A_1 > 0, \\ 0, \qquad &{} \text {else}, \end{array}\right. } \end{aligned}$$

(A6)

hence, the PDF of $A_{m} = C_{m}A_1$ is given by

$$\begin{aligned} p(A_{m}) =\frac{1}{C_{m}} \, p(\dfrac{A_{m}}{C_{m}}), \end{aligned}$$

(A7)

consequently:

$$\begin{aligned} p(A_{m}) = {\left\{ \begin{array}{ll} \dfrac{\mu ^{\nu +1}A_m^\nu }{ \Gamma (\nu +1)(C_m\sigma _x(1))^{\nu +1}} \exp \left( \dfrac{-\mu A_m}{C_m\sigma _x(1)} \right) , \qquad &{} A_m > 0, \\ 0, \qquad &{} \text {else}, \end{array}\right. } \end{aligned}$$

(A8)

which again represents super-Gaussian distribution with variance $\sigma ^2_x(m) = C_m^2\sigma ^2_x(1)$ as mentioned in the manuscript.