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:
The conjugate operator LH brings us back to time and space:
|LH m = d.||(3)|
The conjugate operator is the more straightforward of the two. If we have a point in space, the operator LH 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 LH:
|m = (LH)-1 d.||(4)|
However, the matrix LH 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 LH m = L d,||(5)|
|m = ( L LH )-1 L d.||(6)|
The matrix ( L LH )-1 L is the least-squares inverse of LH.
The least-squares slant stack is summarized in Figure .