Databases Reference
In-Depth Information
Therefore, we have
h
k
h
k
−
2
m
=
δ
m
(78)
k
Notice that this is the same relationship we had to satisfy for perfect reconstruction in the
previous chapter.
Using these relationships, we can generate scaling coefficients for filters of various lengths.
Example15.5.1:
For
k
=
2, we have from (
68
) and (
72
)
h
1
=
√
2
h
0
+
(79)
h
0
+
h
1
=
1
(80)
These equations are uniquely satisfied by
1
√
2
h
0
=
h
1
=
which is the Haar scaling function.
An orthogonal expansion does not exist for all lengths.
In the following example, we
consider the case of
k
=
3.
Example15.5.2:
For
k
=
3, from the three conditions (
68
), (
72
), and (
78
), we have
h
2
=
√
2
h
0
+
h
1
+
(81)
h
0
+
h
1
+
h
2
=
1
(82)
h
0
h
2
=
0
(83)
The last condition can only be satisfied if
h
0
=
0or
h
2
=
0. In either case we will be left with
the two-coefficient filter for the Haar scaling function.
In fact, we can see that for
k
odd, we will always end up with a condition that will force
one of the coefficients to zero, thus leaving an even number of coefficients. When the number
of coefficients gets larger than the number of conditions, we end up with an infinite number of
solutions.
