Formulation of the problem of segregating two acoustic sources

In this paper, we define the problem of segregating two acoustic sources as ``the segregation of the mixed signal into original signal components, where mixed signal is composed of two signals generated by any two acoustic sources''. The problem of segregating two acoustic sources is formulated as follows.

Firstly, we can observe only the signal f(t):

 

f(t)=f₁(t)+f₂(t), (1)

where f₁(t) is the desired signal and f₂(t) is a noise. The observed signal f(t) is decomposed into its frequency components by an auditory filterbank. Secondly, outputs of the k-th channel, which correspond to f₁(t) and f₂(t), are assumed to be

$$f_1(t) : A_k(t)\sin(\omega_k t + \theta_{1k}(t))$$ (2)

and

$$f_2(t) : B_k(t)\sin(\omega_k t + \theta_{2k}(t)),$$ (3)

respectively. Since the output of the k-th channel X_k(t) is represented by

$$X_k(t)=S_k(t)\sin(\omega_k t + \phi_k(t)),$$ (4)

where

$$S_k(t)=\sqrt{A_k^2(t)+2A_k(t)B_k(t)\cos\theta_k(t)+B_k^2(t)}$$ (5)

and

$$\phi_k(t)=\tan^{-1}\left(\frac{A_k(t)\sin\theta_{1k}(t)+B_k(t... ...t)}{A_k(t)\cos\theta_{1k}(t)+B_k(t)\cos\theta_{2k}(t)}\right),$$ (6)

then the amplitude envelopes of the two signals A_k(t) and B_k(t) can be determined by

$$A_k(t)=\frac{S_k(t)\sin(\theta_{2k}(t)-\phi_k(t))}{\sin\theta_k(t)}$$ (7)

and

$$B_k(t)=\frac{S_k(t)\sin(\phi_k(t)-\theta_{1k}(t))}{\sin\theta_k(t)},$$ (8)

respectively, where $\theta_k(t)=\theta_{2k}(t)-\theta_{1k}(t)$ and $\theta_k(t)\not= n\pi, n\in{\bf {Z}}$. Thus, if the four parameters, S_k(t), $\phi_k(t)$, $\theta_{1k}(t)$, and $\theta_{2k}(t)$ are calculated, A_k(t) and B_k(t) can be determined by the above equations. Finally, f₁(t) and f₂(t) can be reconstructed by grouping constraints. $\hat{f}_1(t)$ and $\hat{f}_2(t)$ are the reconstructed f₁(t) and f₂(t), respectively.

In this paper, we assume $\theta_{1k}(t)=0$ and $\theta_{k}(t)=\theta_{2k}(t)$. Moreover, we consider the problem of segregating two acoustic sources in which the localized f₁(t) is added to f₂(t).