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d-faces of the bounding hyper-rectangle of sources corresponding tô u m (Z) andîandî m (Z) for a scenario with 3 sources and 3 mixtures. 

d-faces of the bounding hyper-rectangle of sources corresponding tô u m (Z) andîandî m (Z) for a scenario with 3 sources and 3 mixtures. 

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Article
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Bounded Component Analysis (BCA) is a recently introduced approach including Independent Component Analysis (ICA) as a special case under the assumption of source boundedness. In this article, we provide a stationary point analysis for the recently proposed instantaneous BCA algorithms that are capable of separating dependent, even correlated as we...

Contexts in source publication

Context 1
... an example, Figure 2 illustrates the faces F m,+ of the bounding hyper-rectangle of sources, for a scenario with 3-sources and 3-mixtures scenario and ...
Context 2
... an example, Figure 3 illustrates the subdifferential setsû setsˆsetsû m (Z) andîandî m (Z) for the example in Figure 2. Based on this figure, we can see that • ∂ ˆ u 1 (Z) and ∂ ˆ l 1 (Z) are 2-faces of the parallelepiped, and they are the images of F 1,+ (Z) and F 1,− (Z) in Figure 2 respectively. ...
Context 3
... an example, Figure 3 illustrates the subdifferential setsû setsˆsetsû m (Z) andîandî m (Z) for the example in Figure 2. Based on this figure, we can see that • ∂ ˆ u 1 (Z) and ∂ ˆ l 1 (Z) are 2-faces of the parallelepiped, and they are the images of F 1,+ (Z) and F 1,− (Z) in Figure 2 respectively. ...
Context 4
... ∂ ˆ u 2 and ∂ ˆ l 2 are singleton sets each containing one vertex, and they are the images of F 2,+ (Z) and F 2,− (Z) in Figure 2 respectively. ...
Context 5
... ∂ ˆ u 3 and ∂ ˆ l 3 are 1-faces (edges) of the parallelepiped, and they are the images of F 3,+ (Z) and F 3,− (Z) in Figure 2 respectively. We also note that ∂ ˆ u i , ∂ ˆ l i pairs are located symmetrically with respect to the center of the parallelepiped. ...
Context 6
... p, what proves the desired equality in (109). Figure 9 illustrates the orthogonality of H † b 1 to the corresponding faces F 1,+ (Z) and F 1,− (Z), for the example in Figure 2. For this example, H † b 1 = (u 1 − l 1 )e 1 and I 1,0 = {2, 3}, which confirms the orthogonality property put forward in (114). ...

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