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The Fourier version

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} (3)
Tildes will indicate Fourier transforms. Transform the slant stack from $\tau_s$ 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} (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} (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}
(6)
The second step uses the Fourier transform of a delta function $\delta(f) = \int \exp ( -i 2 \pi f t ) dt$.The third uses the behavior of a delta function in an integral $ \int g(f) \delta (f-f_0 ) df = g(f_0)$.


previous up next print clean
Next: The midpoint gather Up: NOTES FROM TIEMAN's SEMINAR Previous: A conventional slant stack
Stanford Exploration Project
11/12/1997