Biomedical Engineering Reference
In-Depth Information
where U 0 is the eigenvector matrix, and is a diagonal matrix with corresponding
eigenvalues as its diagonal elements. The variance in the space spanned by U 0 com-
ponents can be equalized by the following whitening matrix P :
12
T
P
=
Σ
U
(8.10)
0
It can be shown that, if R L and R R are transformed into S L and S R by whitening
matrix P :
T
T
S RPS RP
L
=
=
(8.11)
L
R
R
then S L and S R will share common eigenvalues. This means, given the SVD of S L and
S R ,
T
T
SUUSU U
L
=
Σ
=
Σ
(8.12)
L
L
L
R
R
R
R
the following equation holds true:
UUU
=
=
ΣΣ
+
=
I
(8.13)
L
R
L
R
Thus,
L and
R may look like the following diagonal matrix:
m
m
m
C
L
R
Σ
=
diag
11
σσ
00
L
1
m
C
(8.14)
m
m
m
C
L
R
Σ
=
diag
00
δδ
1
11
R
m
C
Because the sum of corresponding eigenvalues in L and R is always 1, the big-
gest eigenvalue of S L corresponds to the smallest eigenvalue of S R. The eigenvectors
in L corresponding to the first m eigenvalues in L are used to form a new transform
matrix U l , which makes up the spatial filter with whitening matrix P , for extracting
the so-called source activity of left-hand imagery. The spatial filters for the left and
the right cases are constructed as follows:
T
T
FUPFUP
L
=
=
(8.15)
r
l
R
Then the source activities Y L and Y R are derived by applying the preceding spa-
tial filter on bandpass-filtered EEG data matrix X , that is,
YFXYFX
L
=⋅
=⋅
(8.16)
L
R
R
Because of the way in which the spatial filter is derived, the filtered source activ-
ities Y L and Y R are expected to be better features for discriminating these two imag-
ery tasks, compared with the original EEGs. Usually, the following inequation holds
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