Geoscience Reference
In-Depth Information
Equation (6.6)
is a physical-space representation of filtering. With
Eqs. (6.2)
and
(6.4)
we can also view it from wavenumber space:
N
N
f(κ
n
)e
iκ
n
x
,
r
(x)
f(κ
n
)T (κ
n
)e
iκ
n
x
.
f(x)
=
=
(6.7)
n
=−
N
n
=−
N
G
and
T
comprise a Fourier transform pair,
∞
∞
1
2
π
e
−
iκx
T(κ)dκ,
e
iκx
G(x) dx.
=
=
G(x)
T(κ)
(6.8)
−∞
−∞
For the one-dimensional running-mean operator of
Eq. (6.1)
,
G
is
1
,
2
;
>
G(x)
=
|
x
|≤
G(x)
=
0
,
|
x
|
2
,
(6.9)
which is plotted in
Figure 6.1
.
This is the Fourier transform of
T
given in
Eq. (6.5)
(Problem 6.12)
.
6.2.3 The wave-cutoff filter
If we filter
f(x)
a second time it follows from
Eq. (6.7)
that
N
(f
r
)
r
(x)
T
2
(κ
n
) f(κ
n
)e
iκ
n
x
,
=
(6.10)
n
=−
N
so the amplitude transfer function is
T
2
. If we filter
n
times the amplitude transfer
function is
T
n
.Wesawin
Chapter 2
that subsequent applications of the ensemble-
averaging operator have no effect, which prompts the question: Is there a spatial
filter with this property?
For subsequent applications of a filter to have no effect it is necessary that
T(κ)
×
T(κ)
T(κ)
, which means that
T
can have only the values 0 and
1. This is the
wave-cutoff
filter. One that passes Fourier components of smaller
wavenumbers and rejects those of larger wavenumbers (the
low-pass
form) is
··· ×
T(κ)
=
T(κ)
=
1
,
κ
≤
κ
c
;
T(κ)
=
0
,
κ>
c
,
(6.11)
with
κ
c
the
cutoff wavenumber
.From
Eq. (6.8)
its filter function is
(Problem 6.9)
∞
κ
c
1
2
π
1
2
π
e
−
iκx
T(κ)dκ
e
−
iκx
dκ
G(x)
=
=
−∞
−
κ
c
(6.12)
κ
c
π
sin
κ
c
x
κ
c
x
sin
κ
c
x
πx
=
=
.