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canonical correlation matrix
K
, we find
IK
KI
D0
0D
∗
IK
KI
D
H
0D
T
0
H
H
E
=
=
=
E
(3.101)
DD
H
D
∗
D
T
DKD
T
=
.
(3.102)
D
∗
KD
H
This shows that
D
is unitary and
DKD
T
=
K
. The latter can be true only if
D
ij
=
0
whenever
k
i
=
k
j
. Therefore,
D
is diagonal and unitary if all circularity coefficients are
distinct. Since
K
is real, the corresponding diagonal entries of all nonzero circularity
coefficients are actually
1. On the other hand, components with identical circularity
coefficient cannot be separated.
±
x
2
]
T
. The first component
x
1
is the signal-
space representation of a QPSK signal with amplitude 2 and phase offset
Example 3.5.
Consider a source
x
=
[
x
1
,
π/
8, i.e.,
2e
j
π/
8
2je
j
π/
8
x
1
∈{±
. The second component
x
2
, independent of
x
1
, is the signal-
space representation of a BPSK signal with amplitude 1 and phase offset
,
±
}
π/
4, i.e.,
e
j
π/
4
x
1
∈{±
}
. Hence,
40
01
00
0j
R
xx
=
R
xx
=
and
.
In order to take
x
into canonical coordinates, we use the strong uncorrelating transform
01
10
2
0
01
01
1
2
F
xx
R
−
1
/
2
A
xx
=
=
=
.
xx
0
We see that
F
xx
is a permutation matrix,
R
−
1
/
2
xx
is diagonal, and the product of the two is
monomial. The circularity coefficients are
k
1
=
0. Note that the circularity
coefficients carry no information about the amplitude or phase of the two signals.
Now consider the linear mixture
y
1 and
k
2
=
=
Mx
with
−
j
1
M
=
.
2
−
j1
+
j
and
R
yy
=
MR
xx
M
T
, we compute the SVD of the coherence
MR
xx
M
H
With
R
yy
=
R
yy
R
−
T
/
2
R
−
1
/
2
yy
UKV
H
. The unitary Takagi factor
F
yy
is then obtained
matrix
C
=
=
yy
U
(
V
H
U
∗
)
1
/
2
. The circularity coefficients of
y
are the same as those of
x
.
In order to take
y
into canonical coordinates, we use the strong uncorrelating transform
(rounded to four decimals)
as
F
yy
=
0
.
7071
−
2
.
1213j
0
.
7071
+
0
.
7071j
F
yy
R
−
1
/
2
A
yy
=
=
.
yy
−
0
.
7045
−
0
.
0605j
0
.
3825
−
0
.
3220j
We see that
0
0
.
7071
−
0
.
7071j
A
yy
M
=
0
.
3825
−
0
.
3220j
0
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