Conclusions

We presented the converted-mode PS 3-D common-azimuth migration operator. The difference between this operator and the single-mode PP operator is the use of two different velocity fields. Therefore, a more careful implementation is needed to ensure the correct velocity model. We demonstrate that the subsurface area covered by the PS common-azimuth migration operator is different than that covered by the PP common-azimuth migration operator; therefore, only the area that the two surfaces share can be used for rock-properties analysis based on the two complementary images. This might have important impacts on the reservoir-characterization process. A This section derives the exact solution for the common azimuth prestack migration for a reflecting point within an homogeneous Earth. The total travel time is

$$t_D = \frac{\sqrt{z_\xi^2 + \Vert{\xi_{{\bf {xy}}}- {\bf {m}... ..._\xi^2 + \Vert{\xi_{{\bf {xy}}}- {\bf {m}}- h_D}\Vert^2}}{v_p}.$$ (8)

The following procedure shows how to go from $t_D(z_\xi,{\bf {m}},{\bf {h}})$ to $z_\xi(t_D,{\bf {m}},{\bf {h}})$: 

$$t_D v_p = \phi \sqrt{z_\xi^2 + \Vert{\xi_{{\bf {xy}}}- {\bf ... ...sqrt{z_\xi^2 + \Vert{\xi_{{\bf {xy}}}- {\bf {m}}- h_D}\Vert^2},$$ (9)

where $\phi$ represents the P-to-S velocities ratio. If we make the following definitions,

$$\begin{eqnarray} 2 A &=& t_D v_p, \nonumber\\ \alpha &=& z_\xi+ \Vert{\xi_{{\bf ... ... \beta &=& z_\xi+ \Vert{\xi_{{\bf {xy}}}- {\bf {m}}- h_D}\Vert^2, \end{eqnarray}$$ (10)

(9) becomes

$$2 A = \phi \sqrt{\alpha} + \sqrt{\beta}.$$ (11)

We square both sides to get a new equation with only one square root:

$$4 A^2 - (\phi^2 \alpha + \beta) = 2\phi \sqrt{\alpha \beta}.$$ (12)

Squaring again to eliminate the square root, and combining elements, we obtain

$$16 A^4 - 8 A^2 (\phi^2 \alpha + \beta) + (\phi^2 \alpha - \beta)^2 = 0.$$ (13)

This expression is a 4^th degree polynomial in $z_\xi$; which is:

$$\begin{eqnarray} 0 &=& 16 A^4 - 8A^2 ((\phi^2+1)z_\xi^2 + \phi^2 (\Vert{\xi_{{\b... ...Vert^2)^2 - (\Vert{\xi_{{\bf {xy}}}- {\bf {m}}- h_D}\Vert^2)^2)^2.\end{eqnarray}$$ (14)

This can also be writen as follows

$$\begin{eqnarray} 0 &=& (\phi^2-1)^2 z_\xi^4 \nonumber\\ &+& (2 \phi^2 (\Vert{\x... ...\Vert^2)^2 - (\Vert{\xi_{{\bf {xy}}}- {\bf {m}}- h_D}\Vert^2)^2)^2\end{eqnarray}$$ (15)

This polynomial equation has 4 solutions, which take the following well known form:

$$z_\xi= \pm \sqrt{\frac{-b \pm \sqrt{b^2 - 4a c}}{2 a}},$$ (16)

where

$$\begin{eqnarray} a &=& \left ( \phi^2 -1 \right )^2, \nonumber\\ b &=& 2\phi^2 (... ...Vert^2)^2 - (\Vert{\xi_{{\bf {xy}}}- {\bf {m}}- h_D}\Vert^2)^2)^2.\end{eqnarray}$$ (17)