论文标题

过度确定的独立矢量分析

Overdetermined independent vector analysis

论文作者

Ikeshita, Rintaro, Nakatani, Tomohiro, Araki, Shoko

论文摘要

我们解决了(过度)确定的情况的备受纪录的盲源分离问题,在该情况下,(i)非平稳目标$ k $的数量少于麦克风$ m $的数量,并且(ii)最多有不需要提取的$ m-k $ k $ sentary高斯声音。独立矢量分析(IVA)可以通过将其分为$ M $来源并在其中选择顶级$ K $高度非平稳的信号来解决问题,但是这种方法浪费了计算,尤其是当$ k \ ll m $ $。 IVA预处理中的频道减少,例如,原理分析具有删除目标信号的风险。我们在这里扩展IVA以解决这些问题。通过假设正交性约束(OC),目标和噪声信号之间的样本相关性为零,从而实现了这样的扩展。另一方面,拟议的IVA不依赖OC,而仅利用噪音的来源和平稳性之间的独立性。这使我们能够基于具有特定问题加速度的块坐标下降方法开发几种有效的算法。我们澄清说,一种这样的算法与传统的IVA与OC完全一致,并且还解释说,其他新开发的算法速度比IVA更快。实验结果表明,与常规方法相比,新算法的计算负载提高了。特别是,一种专门针对$ k = 1 $的新算法优于其他算法。

We address the convolutive blind source separation problem for the (over-)determined case where (i) the number of nonstationary target-sources $K$ is less than that of microphones $M$, and (ii) there are up to $M - K$ stationary Gaussian noises that need not to be extracted. Independent vector analysis (IVA) can solve the problem by separating into $M$ sources and selecting the top $K$ highly nonstationary signals among them, but this approach suffers from a waste of computation especially when $K \ll M$. Channel reductions in preprocessing of IVA by, e.g., principle component analysis have the risk of removing the target signals. We here extend IVA to resolve these issues. One such extension has been attained by assuming the orthogonality constraint (OC) that the sample correlation between the target and noise signals is to be zero. The proposed IVA, on the other hand, does not rely on OC and exploits only the independence between sources and the stationarity of the noises. This enables us to develop several efficient algorithms based on block coordinate descent methods with a problem specific acceleration. We clarify that one such algorithm exactly coincides with the conventional IVA with OC, and also explain that the other newly developed algorithms are faster than it. Experimental results show the improved computational load of the new algorithms compared to the conventional methods. In particular, a new algorithm specialized for $K = 1$ outperforms the others.

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