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Because the time axis is well sampled and unaliased, we can safely
Fourier transform the data between time t and frequency f:
| ![\begin{displaymath}
d (s,r,t) \equiv \int \exp( i 2 \pi f t ) \tilde d(s,r,f) df .\end{displaymath}](img7.gif) |
(3) |
Tildes will indicate Fourier transforms.
Transform the slant stack from
to its frequency fs:
| ![\begin{displaymath}
\tilde S (s,p_s,f_s) =
\int \exp( -i 2 \pi f_s \tau_s ) S (s,p_s,\tau_s) d \tau_s .\end{displaymath}](img8.gif) |
(4) |
The slant stack simplifies numerically.
Substitute the transform (3) into the slant stack (2),
then take the transform (4) of both sides of the equation:
| ![\begin{displaymath}
\tilde S (s,p_s ,f_s ) =
\int \int \int \exp ( -i 2 \pi f_...
...\pi f (\tau_s + p_s h) )
\tilde d (s,r=s+h,f) dh df d \tau_s.\end{displaymath}](img9.gif) |
(5) |
Rearranging terms, we reduce integrals
| ![\begin{eqnarray}
\tilde S (s,p_s ,f_s ) &=&
\int \int \{ \int \exp[ -i 2 \pi (f...
... &=&
\int \exp( i 2 \pi f_s p_s h ) \tilde d (s,r=s+h,f=f_s) dh .\end{eqnarray}](img10.gif) |
|
| |
| (6) |
The second step uses the Fourier transform of a delta function
.The third uses
the behavior of a delta function in an integral
.
Next: The midpoint gather
Up: NOTES FROM TIEMAN's SEMINAR
Previous: A conventional slant stack
Stanford Exploration Project
11/12/1997