The conventional slant-stack transform takes us from the recorded data d(x,t) to the slant stack domain :
L d = m. | (1) |
In the frequency domain, a separate system of equations can be constructed for each frequency. The matrix L contains time shifts, expressed as complex exponentials:
(2) |
The conjugate operator L^{H} brings us back to time and space:
L^{H} m = d. | (3) |
The conjugate operator is the more straightforward of the two. If we have a point in space, the operator L^{H} does a good job of turning it into a line in (x,t) space, regardless of the number of traces in x; there is no aperture effect. The forward transform, however, is affected by the aperture. If we transform a single dipping event in (x,t) to there will be artifacts and a loss of resolution, caused by the limited aperture.
We would do better to transform from (x,t) to using the inverse of the matrix L^{H}:
m = (L^{H})^{-1} d. | (4) |
However, the matrix L^{H} is typically not square; the problem is either overdetermined or underdetermined. In this case, we can find the best least-squares solution by multiplying both sides of the conjugate transform by L:
L L^{H} m = L d, | (5) |
giving:
m = ( L L^{H} )^{-1} L d. | (6) |
The matrix ( L L^{H} )^{-1} L is the least-squares inverse of L^{H}.
The least-squares slant stack is summarized in Figure .