Digital Signal Processing Reference
In-Depth Information
To determine
θ
opt
, another step is possible. In effect, from [9.5] we immediately
obtain:
σ
S
=cos
2
(
θ
)
σ
2
X
1
+2cos(
θ
)sin(
θ
)
ρ
(0)
σ
X
1
σ
X
2
+sin
2
(
θ
)
σ
2
X
2
σ
S
=sin
2
(
θ
)
σ
2
X
1
−
2cos(
θ
)sin(
θ
)
ρ
(0)
σ
X
1
σ
X
2
+cos
2
(
θ
)
σ
2
X
2
or:
c
+
1
c
+
y
(
θ
)
2
σ
S
σ
X
1
σ
X
2
=
c
cos
2
(
θ
)+2
ρ
(0) cos(
θ
)sin(
θ
)+
1
c
sin
2
(
θ
)=
1
σ
S
=
2
c
+
1
c
σ
S
σ
X
1
σ
X
2
2
ρ
(0) cos(
θ
)sin(
θ
)+
1
c
cos
2
(
θ
)=
1
y
(
θ
)
2
σ
S
=
=
c
sin
2
(
θ
)
−
−
2
with:
y
(
θ
)=(
c
−
1
/c
)cos(2
θ
)+2
ρ
(0) sin(2
θ
)
Since:
∂y
(
θ
)
∂θ
=
−
2(
c
−
1
/c
)sin(2
θ
)+4
ρ
(0) cos(2
θ
)
the function
y
(
θ
) has an extremum when:
θ
=
1
2
arctan
2
ρ
(0)
c
−
1
/c
1
/c
, which leads to taking
θ
modulo
π/
2 as in equation [9.6]. Knowing
θ
opt
, we can deduce
s
1
(
n
) and
s
2
(
n
).
The four possible cases follow the sign of
ρ
(0) and of
c
−
9.3.2.
In the frequency domain
The procedure shown earlier for the time domain can be realized for each tile
in the time/frequency decomposition. The angles
θ
opt
(
b
) for
b
=1
···
B
must be
determined. These depend only on:
c
(
b
)=
||
X
1
(
b
)
||
||
X
2
(
b
)
||
and:
=
X
t
(
b
)
X
2
(
b
)
X
t
1
(
b
)
X
2
(
b
)+
X
t
(
b
)
X
1
(
b
)
ρ
(0
,b
)=
1
2
1
2
||
X
1
(
b
)
|| ||
X
2
(
b
)
||
||
X
1
(
b
)
|| ||
X
2
(
b
)
||
X
t
1
(
b
)
X
2
(
b
)
|
|
ρ
(0
,b
)=
cos[ICPD(
b
)]
||
X
1
(
b
)
|| ||
X
2
(
b
)
||
ρ
(0
,b
) = ICC(
b
)cos[ICPD(
b
)]
The three inter-channel indices given in [9.4] are therefore sufficient to calculate
the monophonic signal
s
(
n
).
