1
Source Separation and Clustering of Phase-Locked Subspaces: Derivations and Proofs
arXiv:1106.2474v1 [stat.ML] 13 Jun 2011
Miguel Almeida, Jan-Hendrik Schleimer, Jos´e Bioucas-Dias, Ricardo Vig´ario
Abstract—Due to space limitations, our submission “Source Separation and Clustering of Phase-Locked Subspaces”, accepted for publication on the IEEE Transactions on Neural Networks in 2011, presented some results without proof. Those proofs are provided in this paper. Index Terms—phase-locking, synchrony, source separation, clustering, subspaces
A PPENDIX A 2 G RADIENT OF |̺| IN RPA In this section we derive that the gradient of |̺|2 is given by Eq. 6 of [1], where |̺| is defined as in Eq. 5 of [1]. Recall that ∆φ(t) = φ(t)− ψ(t), where φ(t) is the phase of the estimated source y(t) = wT x(t) and ψ(t) is the phase of the reference u(t). Further, define ̺ ≡ |̺|eiΦ . We begin by noting that |̺|2 = (|̺| cos(Φ))2 +(|̺| sin(Φ))2 , so that ∇|̺|2 = ∇(|̺| cos(Φ))2 + ∇(|̺| sin(Φ))2 = 2|̺| [cos(Φ)∇(|̺| cos(Φ)) + sin(Φ)∇(|̺| sin(Φ))] . P Note that we have T1 Tt=1 cos(∆φ(t))) = |̺| cos(Φ) and PT 1 t=1 sin(∆φ(t))) = |̺| sin(Φ), so we get T # T X 1 cos(∆φ(t))) + ∇|̺|2 = 2|̺| cos(Φ)∇ T t=1 " #) T 1X + sin(Φ)∇ sin(∆φ(t))) T t=1 (
=
T 2|̺| X h
T
t=1
"
i sin(Φ) cos(∆φ(t)) − cos(Φ) sin(∆φ(t)) ×
T
2|̺| X sin[Φ − ∆φ(t)]∇∆φ(t). T t=1
Because of we can say, if wT x(t) 6= 0, that ∇φ(t) = this T h (t) ∇ arctan wwTxx(t) . On the other hand, since ∆φ(t) = φ(t)− ψ(t) and ψ(t) does not depend on w, we have (we will omit the time dependence for the sake of clarity): T w xh ∇∆φ = ∇φ − ∇ψ = ∇φ = ∇ arctan wT x T T xh · w x − x · w xh Γ x (t) · w = T 2 2 = Y 2 (t) , 1 + wwTxxh · wT x
2 2 where Y 2 (t) = wT x(t) + wT xh (t) is the squared magnitude of the estimated source, and Γ x (t) = xh (t)xT (t)− x(t)xh T (t), thus Γ xij (t) = Xi (t)Xj (t) sin(φi (t) − φj (t)). We can now replace ∇∆φ(t) in (1) to obtain " T # 2|̺| X sin[Φ − ∆φ(t)] 2 ∇|̺| = Γ x (t) w T t=1 Y 2 (t) sin[Φ − ∆φ(t)] = 2|̺| Γ x (t) w. (2) Y 2 (t) A PPENDIX B G RADIENT OF Jl IN IPA
× ∇∆φ(t)
=
Let’s now take a closer look on ∇∆φ(t). Note that T w xh (t) or φ(t) = arctan wT x(t) T w xh (t) φ(t) = arctan + π. wT x(t)
(1)
Miguel Almeida is with the Institute of Telecommunications, Superior Technical Institute, Portugal. Email:
[email protected] Jan-Hendrik Schleimer is with the Bernstein Center for Computational Neuroscience, Humboldt University, Berlin. Email:
[email protected] Jos´e Bioucas-Dias is with the Institute of Telecommunications, Superior Technical Institute, Portugal. Email:
[email protected] Ricardo Vig´ario is with the Adaptive Informatics Research Centre, Aalto University, Finland. Email:
[email protected]
In this section we show that the gradient of Jl in Eq. 7 of [1] is given by Eq. 8 of [1]. Throughout this whole section, we will omit the dependence on the subspace l, for the sake of clarity. In other words, we are assuming (with no loss of generality) that only one subspace was found. Whenever we write W, y, ym or z, we will be referring to Wl , yl , (yl )m or zl . The derivative of log |det W| is W−T . We will therefore focus on the gradient of the first term of Eq. 7 of [1], which we will denote by P : P ≡
1−λ X |̺mn |2 . N 2 m,n
P 2 Let’s rewrite P as P = 1−λ m,n pmn with pmn = |̺mn | . N2 Define ∆φmn = φm − φn . Omitting the time dependency, we
2
have i∆φ e mn
∇wj pmn = 2|̺mn |∇wj
2
2 −1/2 = |̺mn | × cos(∆φmn ) + i sin(∆φmn ) × h
× 2 cos(∆φmn ) ∇wj cos(∆φmn ) +
i
+ 2 sin(∆φmn ) ∇wj sin(∆φmn )
−1 = 2|̺mn | ei∆φmn × E h
D × − cos(∆φmn ) sin(∆φmn )∇wj ∆φmn + Ei
D (3) + sin(∆φmn ) cos(∆φmn )∇wj ∆φmn ,
where we have interchanged the partial derivative and the time average operators, and used
2
2 1/2 i∆φmn . cos(∆φmn ) + i sin(∆φmn ) = e
Since φm is the phase of the m-th measurement, its derivative with respect to any wj is zero unless m = j or n = j. In the former case, a reasoning similar to Appendix A shows that ∇wj ∆φjk ≡ ∇wj φj − ∇wj φk = ∇wj φj = [zh · z − z · zh ] · wj Γ z · wj = = , 2 Yj Yj2
(4)
where Γ z (t) = zh (t)zT (t) − z(t)zh T (t). It is easy to see that ∇wj ∆φjk = −∇wj ∆φkj . Furthermore, pmm = 1 by definition, hence ∇wj pmm = 0 for all m and j. From these considerations, the only nonzero terms in the derivative of P are of the form D E −1 ∇wj pjk = ∇wj pkj = 2|̺jk | ei(φj −φk ) × + * "
Γ z · wj + × − cos(∆φjk ) sin(∆φjk ) Yj2 +# *
Γ z · wj . (5) + sin(∆φjk ) cos(∆φjk ) Yj2 We now define Ψjk ≡ hφj − φk i = h∆φjk i. Plugging in this definition into Eq. (5) we obtain ∇wj pjk = 2|̺jk |× "
*
+ Γ z · wj × − cos(Ψjk ) sin(∆φjk ) + Yj2 +# * Γ z · wj + sin(Ψjk ) cos(∆φjk ) Yj2 + "* Γ z · wj + = 2|̺jk | − cos Ψjk sin ∆φjk Yj2 +# * Γ z · wj + sin Ψjk cos ∆φjk Yj2 * + Γz = 2|̺jk | sin (Ψjk − ∆φjk ) 2 · wj , Yj
where we again used sin(a − b) = sin a cos b − cos a sin b in the last step. Finally, 1−λX 1−λ X ∇wj P = p = 2 ∇ ∇wj pmn = mn w j N 2 m,n N 2 m
