Discretizing the output space

To clarify all this talk of operators which push and pull, here are a few samples of code to show exactly what pushing and pulling mean from an operator design perspective. Below is the computational part of the subroutine nmo1 from Jon's book Basic Earth Imaging.

do iz= 1, n {z = t0 + dt*(iz-1)
             xs = x * slow(iz)
             t = sqrt ( z * z + xs * xs ) + 1.e-20
             wt = z/t * (1./sqrt(t))    #weighting function
             it = 1 + .5 + (t - t0) / dt
             if( it <= n )
               if( conj == 0 )
                   tt(it) = tt(it) + zz(iz) * wt
               else 
                   zz(iz) = zz(iz) + tt(it) * wt
            }

One can see that in both the forward (conj == 0 ) and adjoint operation, a loop over z (which represents vertical travel time $\tau$) is performed. This means that each sample along the zero offset trace is visited once. Time t is calculated and nearest-neighbor interpolation is performed to find the correct sample along the non-zero offset trace. A given sample along the non-zero offset trace may be visited multiple times or not at all.

An alternative is to use two subroutines. The first performs the adjoint operation exactly as above (I will avoid reprinting it and just say that the differences are that there is no test for conj, and the weighting function is different). The second subroutine performs the forward operation. In the forward operation, discretizing in the output space means looping over time t.

do it= 2, n {   t = t0 + dt*(it-1)
                xs = x * slow(it)  
                z = sqrt ( t * t - xs * xs ) 
                iz = 1 + .5 + (z - t0) / dt
                if( iz >=1 )
                        tt(it) = tt(it) + zz(iz) * z/t
            }

Here the non-zero offset trace is sampled evenly, just as the zero-offset trace is above. There is, however, one serious complication to the computation shown directly above. Calculating z requires knowing the velocity as a function of travel time t rather than vertical travel time $\tau$. The equation for moveout in a vertically varying velocity field doesn't permit an explicit solution for v(t). This is an important future project. For now I stick to constant velocity.