Databases Reference
In-Depth Information
V
1
(
e
j
ω
)
π/2
ω
F I GU R E 14 . 16
Output of the low-pass filter.
y
1,
n
w
1,
n
v
1,
n
u
1,
n
H
1
(
z
)
2
2
K
1
(
z
)
x
n
ˆ
x
n
y
2,
n
w
2,
n
v
2,
n
u
2,
n
H
2
(
z
)
2
2
K
2
(
z
)
F I GU R E 14 . 17
Two-channel subband decimation and interpolation.
Therefore, we need to find
V
1
(
z
)
and
V
2
(
z
)
. The sequence
v
1
,
n
is obtained by upsampling
w
1
,
n
. Therefore, from Equation (
41
),
z
2
V
1
(
z
)
=
W
1
(
)
(44)
The sequence
w
1
,
n
is obtained by downsampling
y
1
,
n
,
Y
1
(
z
)
=
X
(
z
)
H
1
(
z
)
Therefore, from Equation (
35
),
X
1
2
z
2
z
2
z
2
z
2
W
1
(
z
)
=
(
)
H
1
(
)
+
X
(
−
)
H
1
(
−
)
(45)
and
1
2
[
X
V
1
(
z
)
=
(
z
)
H
1
(
z
)
+
X
(
−
z
)
H
1
(
−
z
)
]
(46)
Similarly, we can also show that
1
2
[
X
V
2
(
z
)
=
(
z
)
H
2
(
z
)
+
X
(
−
z
)
H
2
(
−
z
)
]
(47)
Substituting the expressions for
V
1
(
z
)
and
V
2
(
z
)
into Equation (
43
) we obtain
1
2
[
H
1
(
X
(
z
)
=
z
)
K
1
(
z
)
+
H
2
(
z
)
K
2
(
z
)
]
X
(
z
)
1
2
[
H
1
(
−
+
z
)
K
1
(
z
)
+
H
2
(
−
z
)
K
2
(
z
)
]
X
(
−
z
)
(48)
