Geology Reference
In-Depth Information
In the case of unequally spaced samples, the representation (2.134) of the time
sequence g
j
may be used if it is understood that the sample points
t
j
are not at
uniform intervals. If the sample points are at equally spaced intervals, we shall see
that a least squares adjustment of this representation gives the DFT (2.148).
This suggests the following approach to the analysis of unequally spaced data. If
we have a record segment of length
M
, with 2
L
+
1 samples, equally or unequally
spaced, we try to represent it in the frequency domain for
N
≤
L
by 2
N
+
1 sinu-
soids, equally spaced in frequency, as
N
1
M
g
j
=
G
k
e
i
2π
(
k
/
M
)
t
j
.
(2.178)
k
=−
N
The 2
N
1values,
G
k
, of the DFT, once calculated, give a frequency domain
representation that is equally spaced in frequency with sample interval
+
1/
M
.
The FFT algorithm can then be used to give an equivalent, equally spaced time
sequence.
To find the DFT sequence
G
k
, we first construct the error sequence
j
as the
Δ
f
=
erence of the sequence g
j
and its representation g
j
expressed by the sum of
sinusoids (2.178),
di
ff
−
g
j
.
j
=
g
j
(2.179)
If each member of the sequence g
j
has an associated standard deviation σ
j
,the
G
k
are chosen to minimise the objective function
j
∗
j
σ
L
I
=
j
,
(2.180)
2
j
=−
L
the superscript asterisk again indicating complex conjugation. Substitution for g
j
from (2.178) in expression (2.180) for the objective function gives
⎩
g
j
g
∗
j
−
g
∗
j
M
L
N
N
1
σ
g
j
M
G
∗
l
e
−
i
2π(
l
/
M
)
t
j
G
k
e
i
2π(
k
/
M
)
t
j
I
=
−
2
j
j
=−
L
l
=−
N
k
=−
N
G
k
G
∗
l
e
i
2π((
k
−
l
)/
M
)
t
j
⎭
.
N
N
1
M
2
+
(2.181)
k
=−
N
l
=−
N
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