Digital Signal Processing Reference
In-Depth Information
section 8.7,z raised to a negative power gives a negative value for an odd power,
but returns a positive value for an even one.
−P (−z)
0
Input
Output
a, b, c, d
d, c, b, a
P (z)
0
Figure 9.21: Designing a lter bank with four taps each.
That is, we can determine the lters' responses and see how the terms cancel
each other out. Then we can divide by two, which is the same eect that down/up-
samplers have.
Without down/up-samplers:
1
+ 2(aa + bb + cc + dd)z
3
+ (ac + bd)z
5
= 2z
L
:
P
0
(z)P
0
(z) = 2(ac + bd)z
With down/up-samplers:
1
+ (aa + bb + cc + dd)z
3
+ (ac + bd)z
5
= z
L
:
P
0
(z)P
0
(z) = (ac + bd)z
1
,
We are left with the output in terms of three delayed versions of the input: z
5
. Two of these must be eliminated, which can be accomplished by
setting ac + bd = 0. Also, we do not want the result to be a scaled version of the
input, so aa+bb+cc+dd = 1. It is also desirable that the sum of the highpass lter
coecients be zero [3], i.e., dc + ba = 0, for this to be a wavelet transform.
The question now becomes, can we nd values for a;b;c; and d to satisfy these
equations?
3
, and z
z
ac + bd =
0
a
2
+ b
2
+ c
2
+ d
2
=
1
dc + ba =
0
One possibility is to use the Daubechies wavelet lter coecients, since they
satisfy these criteria.
