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H
10
(
z
)
2
z
-1
H
11
(
z
)
H
20
(
z
)
2
z
-1
H
21
(
z
)
F I GU R E 14 . 23
Alternative representation of the analysis portion of a two-band
subband coder.
By grouping the odd and even terms together, we can write this as
h
2
z
−
2
h
4
z
−
4
z
−
1
h
3
z
−
2
h
5
z
−
4
H
1
(
z
)
=
(
h
0
+
+
+···
)
+
(
h
1
+
+
+···
)
(94)
Define
h
2
z
−
1
h
4
z
−
2
H
10
(
z
)
=
h
0
+
+
+···
(95)
h
3
z
−
1
h
5
z
−
2
H
11
(
z
)
=
h
1
+
+
+···
(96)
z
2
z
−
1
H
11
(
z
2
Then
H
1
(
z
)
=
H
10
(
)
+
)
. Similarly, we can decompose the filter
H
2
(
z
)
into
components
H
20
(
z
)
and
H
21
(
z
)
, and we can represent the system of Figure
14.22
as shown in
Figure
14.23
. The filters
H
10
(
z
)
,
H
11
(
z
)
and
H
20
(
z
),
H
21
(
z
)
are called the polyphase compo-
nents of
H
1
(
.
Let's take the inverse Z-transform of the polyphase components of
H
1
(
z
)
and
H
2
(
z
)
z
)
:
h
10
(
n
)
=
h
2
n
n
=
0
,
1
,...
(97)
h
11
(
n
)
=
h
2
n
+
1
n
=
0
,
1
,...
(98)
Thus,
h
10
(
are simply the impulse response
h
n
downsampled by two. Consider
the output of the downsampler for a given input
X
n
)
and
h
11
(
n
)
(
z
)
. The input to the downsampler is
X
(
z
)
H
1
(
z
)
; thus, the output from Equation (
35
)is
2
X
z
2
H
1
z
2
2
X
z
2
H
1
z
2
1
1
Y
1
(
z
)
=
+
−
−
(99)
