Geology Reference
In-Depth Information
In the Fourier integral (2.249) and Fourier transform (2.250) pair,
f
and
t
play
symmetrical roles. The Fourier transform of the product of the functions of time,
g(
t
)
h
(
t
)
=
k
(
t
), is
∞
g(
t
)
h
(
t
)
e
−
i
2π
ft
dt
−∞
∞
−∞
·
∞
∞
G
(
f
1
)
e
i
2π
f
1
t
df
1
H
(
f
2
)
e
i
2π
f
2
t
df
2
e
−
i
2π
ft
dt
=
·
·
−∞
−∞
∞
G
(
f
1
)
∞
−∞
H
(
f
2
)
∞
−∞
e
i
2π(
f
1
+
f
2
−
f
)
t
dt d f
2
df
1
.
=
(2.270)
−∞
If we replace
t
by
f
and
t
0
by
f
0
in the Fourier integral representation of the Dirac
delta function (2.262), we have
∞
e
i
2π
t
(
f
−
f
0
)
dt
,
δ(
f
−
f
0
)
=
(2.271)
−∞
or
∞
e
i
2π(
f
0
−
f
)
t
dt
,
δ(
f
0
−
f
)
=
(2.272)
−∞
and
∞
e
i
2π(
f
1
+
f
2
−
f
)
t
dt
.
δ(
f
1
+
f
2
−
f
)
=
(2.273)
−∞
Expression (2.270) for the Fourier transform of the product of the two functions of
time,
k
(
t
)
=
g(
t
)
h
(
t
), becomes
∞
g(
t
)
h
(
t
)
e
−
i
2π
ft
dt
=
K
(
f
)
−∞
∞
G
(
f
1
)
∞
−∞
=
H
(
f
2
)δ(
f
1
+
f
2
−
f
)
df
2
df
1
−∞
∞
=
G
(
f
1
)
H
(
f
−
f
1
)
df
1
.
(2.274)
−∞
Thus, the Fourier transform of the product of two functions of time is the convolu-
tion, in the frequency domain, of their Fourier transforms.
2.4.2 The e
ff
ect of finite record length
Any real, physical record of a function of time, g(
t
), is bound to be of finite
length
T
. Yet, to calculate its Fourier transform (2.250), we require a record extend-
ing indefinitely in both directions in time. Instead, we have only g(
t
)for
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