Databases Reference
In-Depth Information
Substituting
x
=
2
t
−
k
with
dx
=
2
dt
in the right-hand side of the equation, we get
∞
h
k
√
2
∞
−∞
1
2
dx
φ(
t
)
dt
=
φ(
x
)
(66)
−∞
k
∞
1
√
2
=
h
k
φ(
x
)
dx
(67)
−∞
k
Assuming that the average value of the scaling function is not zero, we can divide both sides
by the integral to obtain
√
2
h
k
=
(68)
k
If we normalize the scaling function to have a magnitude of one, we can use the orthogonality
condition on the scaling function to get another condition on
{
h
k
}
:
h
k
√
2
h
m
√
2
2
dt
|
φ(
t
)
|
=
φ(
2
t
−
k
)
φ(
2
t
−
m
)
dt
(69)
m
k
h
k
h
m
2
=
φ(
2
t
−
k
)φ(
2
t
−
m
)
dt
(70)
m
k
h
k
h
m
=
φ(
x
−
k
)φ(
x
−
m
)
dx
(71)
m
k
=
where in the last equation we have used the substitution
x
2
t
. The integral on the right-hand
=
=
side is zero except when
k
m
. When
k
m
, the integral is unity and we obtain
h
k
=
1
(72)
k
We can actually elicit a more general property by using the orthogonality of the translates
of the scaling function
φ(
t
)φ(
t
−
m
)
dt
=
δ
m
(73)
Rewriting this using the MRA equation to substitute for
φ(
t
)
and
φ(
t
−
m
)
, we obtain
dt
h
k
√
2
h
l
√
2
φ(
2
t
−
k
)
φ(
2
t
−
2
m
−
l
)
k
l
h
k
h
l
2
=
φ(
2
t
−
k
)φ(
2
t
−
2
m
−
l
)
dt
(74)
k
l
Substituting
x
=
2
t
, we obtain
h
k
h
l
φ(
t
)φ(
t
−
m
)
dt
=
φ(
x
−
k
)φ(
x
−
2
m
−
l
)
dx
(75)
k
l
=
h
k
h
l
δ
k
−
(
2
m
+
l
)
(76)
k
l
=
h
k
h
k
−
2
m
(77)
k
