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Evaluation of the impulse response of the transformation to GOCIGs

The transformation to GOCIG of an image $I_{x_h}\left(k_z,k_{x},x_h\right)$ is defined as
\begin{displaymath}
I_0\left(k_z,k_{x},k_{h}\right) =
\int d\widetilde{h_0}
I_0\...
 ...{ik_{h}x_h\left(1+\frac{k_{x}^2}{k_z^2}\right)^{-\frac{1}{2}}}.\end{displaymath} (26)
The transformation to GOCIG of an impulse located at $\left(\bar{z},\bar{x},\bar{x_h}\right)$is thus (after inverse Fourier transforms):  
 \begin{displaymath}
\widetilde{\rm Imp}\left(z,x,\widetilde{h_0}\right) =
\int d...
 ...\left(\bar{z}-z\right) +
k_{x}\left(\bar{x}-x\right)
\right\}}.\end{displaymath} (27)

We now approximate by stationary phase the inner double integral. The phase of this integral is:  
 \begin{displaymath}
\Phi\equiv 
k_{h}\left[\bar{x_h}\left(1+\frac{k_{x}^2}{k_z^2...
 ...ht] +
k_z\left(\bar{z}-z\right) +
k_{x}\left(\bar{x}-x\right)
.\end{displaymath} (28)
The stationary path is defined by the solutions of the following system of equations:
      \begin{eqnarray}
\frac{\partial \Phi}{\partial k_z} &=& 
k_{h}\bar{x_h}
\frac{k_...
 ...2}}\right)^{-\frac{3}{2}}+ \left(\bar{x}-x\right) =0.
\\ \nonumber\end{eqnarray} (29)
(30)
By moving both $\left(\bar{z}-z\right)$ and $\left(\bar{x}-x\right)$ to the right of equations (29) and (30), and then dividing equation (29) by equation (30), we obtain the following relationship between $\left(\bar{z}-z\right)$ and $\left(\bar{x}-x\right)$: 
 \begin{displaymath}
\frac{\bar{z}-z}{\bar{x}-x}=
-\frac{k_{x}}{k_z}.\end{displaymath} (31)
Furthermore, by multiplying equation (29) by kz and equation (30) by kx, and then substituting them appropriately in the phase function (28), we can evaluate the phase function along the stationary path as follows:  
 \begin{displaymath}
\Phi_{\rm stat}= 
k_{h}\left[\bar{x_h}\left(1+\frac{k_{x}^2}{k_z^2}\right)^{-\frac{1}{2}}-\widetilde{h_0}\right],\end{displaymath} (32)
which becomes, by substituting equation (31),  
 \begin{displaymath}
\Phi_{\rm stat}= 
k_{h}\left\{-\bar{x_h}\left[1+\frac{\left(...
 ...r{x}-x\right)^2}\right]^{-\frac{1}{2}}-\widetilde{h_0}\right\}.\end{displaymath} (33)
Notice that the minus sign comes from the ${\rm sign}$ function in expression (23). By substituting expression (33) in equation (27) it is immediate to evaluate the kinematics of the impulse response as follows:  
 \begin{displaymath}
\widetilde{h_0}=-x_h\left[1+\frac{\left(\bar{z}-z\right)^2}{\left(\bar{x}-x\right)^2}\right]^{-\frac{1}{2}}.\end{displaymath} (34)


next up previous print clean
Next: Evaluation of the image Up: REFERENCES Previous: REFERENCES
Stanford Exploration Project
7/8/2003