next up previous print clean
Next: Controlling the angle Up: Theory review Previous: Theory review

Redefining R

The principle of causality: no response before a stimulus, is not considered in the R definition [equation (2)]. The omission of causality translates into improper behavior of the high frequencies. In order to include causality into the definition of R, let $s=-i\hat{\omega}$ be the causal, positive, discrete representation of the differentiation operator,

\begin{displaymath}
s=-i\hat{\omega} = \frac{2}{\Delta t} \frac{1-\rho Z}{1+ \rho Z},\end{displaymath} (3)

which is simplified by writting $s=-i\omega + \epsilon$.

Claerbout (1985) proposes the use of the following Muir recursion starting from R0=s:

 
 \begin{displaymath}
R_{n+1} = s + \frac{X^{2}}{s + R_{n}}.\end{displaymath} (4)

This recursion produces a continuous fraction. Studying the limit for $n \rightarrow \infty$we obtain:

   \begin{eqnarray}
R_{\infty} \ \ & = & \ \ s + \frac{X^{2}}{s + R_{\infty}}, \non...
 ...2} + X^{2}, \nonumber \\ R \ \ & = & \ \ \pm \sqrt{s^{2} + X^{2}}.\end{eqnarray}
(5)

If we let X2 = v2 kx2 in equation (5) then R is $\pm ik_{z}v$. The retarded time expression of $e^{-R\frac{z}{v}}$ [equation (6)] will downward continue, in time, the data for phase-shift migration.

 
 \begin{displaymath}
e^{-R\frac{z}{v}} = e^{-(R - s) \frac{z}{v}} e^{-s \frac{z}{v}}.\end{displaymath} (6)

The following change of variables:

\begin{displaymath}
R^{\prime} = R - s\end{displaymath} (7)

transforms the Muir recursion (4) into:

 
 \begin{displaymath}
R^{\prime}_{n+1} = \frac{v^{2} k_{x}^{2}}{2s + R^{\prime}_{n}}.\end{displaymath} (8)

Again, taking the limit for $n \rightarrow \infty$ in this recursion we will obtain

\begin{eqnarray}
R^{\prime}_{\infty} \cdot (2s + R^{\prime}_{n}) \ \ & = & \ \ v...
 ...ty}}^{2} + 2sR^{\prime}_{\infty} - v^{2} k_{x}^{2} \ \ & = & \ \ 0\end{eqnarray}
(9)

This quadratic expression yields to two square roots for $R^{\prime}$,

\begin{displaymath}
R^{\prime} = -s \pm \sqrt{s^{2} + v^{2} k_{x}^{2}}.\end{displaymath} (10)

We need to select the one that is able to handle the evanescent region, i.e, the square root that goes to zero at kx = 0, which corresponds to the positive square root.

 
 \begin{displaymath}
R^{\prime} = -s + \sqrt{s^{2} + v^{2} k_{x}^{2}},\end{displaymath} (11)
multiplying numerator and denominator by $s+\sqrt{s^{2} + v^{2} k_{x}^{2}}$, equation (11) transforms into:

 
 \begin{displaymath}
R^{\prime} = \frac{v^2 k_x^2}{\sqrt{s^{2} + v^{2} k_{x}^{2}} + s}.\end{displaymath} (12)

I already showed that the real part of $\sqrt{s^{2} + v^{2} k_{x}^{2}}$ is positive (Figure 1); therefore, $R^{\prime}$, as defined in equation (12), also has a positive real part.

Finally, we downward continue the data in the Fourier domain by multiplying by:

 
 \begin{displaymath}
e^{-R^{\prime} \frac{z}{v}} e^{-i\omega \frac{z}{v}}\end{displaymath} (13)

Expression (13) incorporates the causality and viscosity concepts in phase shift migration, controls the evanescent energy and will not allow discontinuity between evanescent and non-evanescent regions.


next up previous print clean
Next: Controlling the angle Up: Theory review Previous: Theory review
Stanford Exploration Project
4/29/2001