Anti-aliasing constraints for 3-D prestack time migration

  For 3-D prestack time migration, the reflectors' dips $p^{{\rm \xi}}_x$ and $p^{{\rm \xi}}_y$, and the wavelet-stretch factor ${dt_{D}}/{d\tau_\xi}$,can be analytically derived as functions of the input and output trace geometry. Starting from the prestack time-migration ellipsoid, expressed as a parametric function of the angles $\alpha$ and $\beta$

$$\begin{eqnarray} \tau_\xi& = & t_{N}\sin \alpha\cos \beta, \nonumber \\ x_\xi& =... ...ha, \nonumber \\ y_\xi& = & \frac{t_{N}V}{2}\sin \alpha\sin \beta,\end{eqnarray}$$ (7)

where t_D is the time of the input impulse and t_N is the time after application of NMO. We differentiate the image coordinates with respect to the angles $\alpha$ and $\beta$;that is,

$$\begin{eqnarray} d\tau_\xi& = & t_{N}\left(\cos \alpha\cos \beta\;d\alpha- \sin ... ...s \alpha\sin \beta\;d\alpha+ \sin \alpha\cos \beta\;d\beta\right).\end{eqnarray}$$ (8)

We then eliminate the differentials $d\alpha$ and $d\beta$from this set of equations by setting respectively $dy_\xi$ equal to zero when evaluating the dip $p^{{\rm \xi}}_x$ in the in-line direction, and set $dx_\xi$ equal to zero when evaluating the dip in the cross-line direction $p^{{\rm \xi}}_y$.The second step is to eliminate the angles themselves and express the image dips as a function of the image coordinates $\left(\tau_\xi,x_\xi,y_\xi\right)$,

$$\begin{eqnarray} p^{{\rm \xi}}_x & = & \frac{d\tau_\xi}{dx_\xi} = \frac{2t_{N}}{... ...xi}{dy_\xi} = \frac{2\tan \beta}{V} = \frac{4y_\xi}{V^2\tau_\xi}. \end{eqnarray}$$ (9)

The wavelet-stretch factor can be easily derived by differentiating the summation surfaces of 3-D prestack time migration expressed as the hyperboloids  

$$t_{D}= t_s + t_g = \sqrt{\frac{\tau_\xi^2}{4} + \frac{\left\... ...{\left\vert{\bf \vec\xi_{xy}}-{\bf \vec g}\right\vert^2}{V^2}},$$ (10)

where ${\bf \vec s}$ and ${\bf \vec g}$ are the source and receiver coordinates vector, and ${\bf \vec\xi_{xy}}=\left(x_\xi,y_\xi\right)$ represents the horizontal components of the image coordinates vector. After a few simple algebra steps, we obtain

$$\frac{dt_{D}}{d\tau_\xi} = \frac{\tau_\xi}{4}\left(\frac{1}{t_s}+\frac{1}{t_g}\right).$$ (11)