Evaluation of the impulse response of the transformation to DDOCIGs

The transformation to DDHOCIG of an image $I_{x_h}\left(k_z,k_{x},h_x\right)$ is defined as

$$I_0\left(k_z,k_{x},{\vec k_h}\right) = \int dh_0 I_0\left(k_... ...ec k_h}h_x\left(1+\frac{k_{x}^2}{k_z^2}\right)^{-\frac{1}{2}}}.$$ (74)

The transformation to DDHOCIG of an impulse located at $\left(\bar{z},\bar{x},\bar{x_h}\right)$is thus (after inverse Fourier transforms):  

$$\widetilde{\rm Imp}\left(z,x,h_0\right) = \int d{\vec k_h} \... ...\left(\bar{z}-z\right) + k_{x}\left(\bar{x}-x\right) \right\}}.$$ (75)

We now approximate by stationary phase the inner double integral. The phase of this integral is,  

$$\Phi\equiv {\vec k_h}\left[\bar{x_h}\left(1+\frac{k_{x}^2}{... ...ight] + k_z\left(\bar{z}-z\right) + k_{x}\left(\bar{x}-x\right)$$ (76)

The stationary path is defined by the solutions of the following system of equations:

$$\begin{eqnarray} \frac{\partial \Phi}{\partial k_z} &=& {\vec k_h}\bar{x_h} \fr... ...}}\right)^{-\frac{3}{2}}+ \left(\bar{x}-x\right) =0 , \ \nonumber\end{eqnarray}$$ (77) (78)

By moving both $\left(\bar{z}-z\right)$ and $\left(\bar{x}-x\right)$ on the right of equations () and (), and then dividing equation () by equation (), we obtain the following relationships between $\left(\bar{z}-z\right)$ and $\left(\bar{x}-x\right)$: 

$$\frac{\bar{z}-z}{\bar{x}-x}= -\frac{k_{x}}{k_z}.$$ (79)

Furthermore, by multiplying equations () by k_z and equation () by k_x, and then substituting them appropriately in the phase function (), we can evaluate the phase function along the stationary path as  

$$\Phi_{\rm stat}= {\vec k_h}\left[\bar{x_h}\left(1+\frac{k_{x}^2}{k_z^2}\right)^{-\frac{1}{2}}-h_0\right],$$ (80)

that becomes, by substituting equation (),  

$$\Phi_{\rm stat}= {\vec k_h}\left\{-\bar{x_h}\left[1+\frac{\... ...2}{\left(\bar{x}-x\right)^2}\right]^{-\frac{1}{2}}-h_0\right\}.$$ (81)

Notice that the minus sign comes from the ${\rm sign}$ function in expression (). By substituting expression () in equation () it is immediate to evaluate the kinematics of the impulse response as  

$$h_0=-h_x\left[1+\frac{\left(\bar{z}-z\right)^2}{\left(\bar{x}-x\right)^2}\right]^{-\frac{1}{2}}$$ (82)