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de Bruin's imaging condition

In order to retrieve the full reflectivity matrix, de Bruin et al. (1990) proposes another imaging condition (Figure [*]). Instead of deconvolving in the spatial domain, they deconvolve in the Fourier domain $(k_x,\omega)$after Fourier transforming both downgoing and upcoming waves, as follows:  
 \begin{displaymath}
X(k_x,z_n,\omega)={{{\bf g}(k_x,z_n,\omega){\bf s}^\ast(k_x,...
 ... s}(k_x,z_n,\omega){\bf s}^\ast(k_x,z_n,\omega) + \epsilon^2}}.\end{displaymath} (52)
The next step is to transform $X(k_x,z_n,\omega)$ into $X(p,z_n,\omega)$by replacing the wavenumber kx with the ray parameter $p = k_x / \omega$.Then imaging is carried out in the $p - \omega$domain along the lines of the constant p:  
 \begin{displaymath}
{\bf r}(p,z_n) = \sum_{\omega} X(p,z_n,\omega).\end{displaymath} (53)

This imaging condition retrieve the full angle-dependent reflectivity. However, since the deconvolution is performed in the Fourier domain, it implies a spatially invariant reflectivity, which is the case of horizontal reflector in constant velocity medium. Thus it has the disadvantage that is of being difficult to implement a realistic model.

 
fk-img
fk-img
Figure 5
Schematic representation of de Bruin's imaging condition
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next up previous print clean
Next: Imaging condition in plane Up: Imaging conditions Previous: Conventional prestack imaging
Stanford Exploration Project
2/5/2001