Databases Reference
In-Depth Information
Writing out
R
(
z
)
as the Z-transform of the sequence
{
ρ(
n
)
}
we obtain
z
N
z
N
−
1
z
−
N
−
1
z
−
N
(85)
R
(
z
)
=
ρ(
N
)
+
ρ(
N
−
1
)
+···+
ρ(
0
)
+···+
ρ(
N
−
1
)
+
ρ(
N
)
Then
R
(
−
z
)
is
z
N
z
N
−
1
z
−
N
−
1
z
−
N
(86)
R
(
−
z
)
=−
ρ(
N
)
+
ρ(
N
−
1
)
−···+
ρ(
0
)
−···+
ρ(
N
−
1
)
−
ρ(
N
)
Adding
R
(
z
)
and
R
(
−
z
)
, we obtain
Q
(
z
)
as
z
N
−
1
z
N
−
3
z
−
N
−
1
Q
(
z
)
=
2
ρ(
N
−
1
)
+
2
ρ(
N
−
3
)
+···+
ρ(
0
)
+···+
2
ρ(
N
−
1
)
(87)
Notice that the terms containing the odd powers of
z
got canceled out. Thus, for
Q
(
z
)
to
be a constant all we need is that for even values of the lag
n
(except for
n
=
0),
ρ(
n
)
be zero.
In other words
N
ρ(
)
=
h
k
h
k
+
2
n
=
,
=
(88)
2
n
0
n
0
k
=
0
Writing this requirement in terms of the impulse response,
0
N
n
=
0
h
k
h
k
+
2
n
=
(89)
ρ(
0
)
n
=
0
k
=
0
If we now normalize the impulse response,
N
2
0
|
h
k
|
=
1
(90)
k
=
we obtain the perfect reconstruction requirement
N
h
k
h
k
+
2
n
=
δ
n
(91)
k
=
0
In other words, for perfect reconstruction, the impulse response of the prototype filter is
orthogonal to the twice-shifted version of itself.
14.7
M
-Band Quadrature Mirror Filter Banks
We have looked at howwe can decompose an input signal into two bands. In many applications
it is necessary to divide the input into multiple bands. We can do this by using a recursive
two-band splitting as shown in Figure
14.19
, or we can obtain banks of filters that directly split
the input into multiple bands. Given that we have good filters that provide two-band splitting,
it would seem that using a recursive splitting, as shown in Figure
14.19
, would be an efficient
way of obtaining an
M
-band split. Unfortunately, even when the spectral characteristics of the
