Separation and Grouping

First, we can observe only the signal f(t), where f(t)=f₁(t)+f₂(t), f₁(t) is the desired signal and f₂(t) is a noise masker. The observed signal f(t) is decomposed into its frequency components by an auditory filterbank. Second, outputs of the k-th channel, which correspond to f₁(t) and f₂(t), are assumed to be

$$f_1(t): A_k(t)\sin(\omega_k t + \theta_{1k}(t))$$ (14)

and

$$f_2(t): B_k(t)\sin(\omega_k t + \theta_{2k}(t)).$$ (15)

Here, $\omega_k$ is the center frequency of the auditory filter and $\theta_{1k}(t)$ and $\theta_{2k}(t)$ are the input phases of f₁(t) and f₂(t), respectively. Since the output of the k-th channel X_k(t) is the sum of Eqs. () and (),

$$X_k(t)=S_k(t)\sin(\omega_k t + \phi_k(t)).$$ (16)

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

$$A_k(t)={S_k(t)\sin(\theta_{2k}(t)-\phi_k(t))}/{\sin\theta_k(t)}$$ (17)

and

$$B_k(t)={S_k(t)\sin(\phi_k(t)-\theta_{1k}(t))}/{\sin\theta_k(t)},$$ (18)

where $\theta_{k}(t)=\theta_{2k}(t)-\theta_{1k}(t)$ and $\theta_k(t)\not=n\pi, n\in {\bf{Z}}$. Since the amplitude envelope S_k(t) and the output phase $\phi_k(t)$ are observable, then if $\theta_{1k}(t)$ and $\theta_{2k}(t)$ are determined, A_k(t) and B_k(t) can be determined by the above equations. Finally, all the components are synthesized from Eqs. () and () in the grouping block. Then f₁(t) and f₂(t) can be reconstructed by the grouping block using the inverse wavelet transform. Here, $\hat{f}_{1,{\rm {A}}}(t)$ and $\hat{f}_{2,{\rm{B}}}(t)$ are the reconstructed f₁(t) and f₂(t), respectively.

In this paper, we assume that the center frequency of the auditory filter corresponds to the signal frequency. Therefore, we consider the problem of segregating f₁(t) from f(t) when $\theta_{1k}(t)=0$ and $\theta_k(t)=\theta_{2k}(t)$.