Downward continuation with the DSR

In this section we apply the double square root (DSR) operator to extrapolate the recorded wavefield of equation (13) through the depth coordinate of the lattice. After initially Fourier transforming the wavefield into $k_{r_{\xi}}$, $k_{s_{\xi}}$, and $\omega_{\xi}$, the first application of the DSR yields a new wavefield at depth step $z_{\xi}=a_4\Delta z_{\xi}$. Mathematically, this is represented by:

$$\begin{eqnarray} W(k_{r_{\xi}},k_{s_{\xi}},\omega,z_{\xi})= & \d(z_{\xi}-a_z\Del... ...z_{\xi}+a_z\Delta z_{\xi})\d(z_{\xi}-a_z\Delta z_{\xi})\nonumber .\end{eqnarray}$$ (13)

The periodicity of the lattice over depth coordinate $z_{\xi}$ enables the Shah function index u_z to be shifted by $a_z\Delta z_{\xi}$ such that equation (14) reads,

$$W(k_{r_{\xi}},k_{s_{\xi}},\omega,z_{\xi})=\left[DSR(k_{r_{\x... ...i}},k_{s_{\xi}},\omega,z_{\xi})\d(z_{\xi}-a_4\Delta z_{\xi})\;.$$ (14)

By extension, any continuation step operating on a wavefield will take the same form. Applying an inverse Fourier transform over coordinates $k_{r_{\xi}}$ and $k_{s_{\xi}}$ yields,  

$$W(\r,s_{\xi},\omega,z_{\xi}=a_4\Delta z_{\xi})=H(\r,s_{\xi},... ...\r,s_{\xi},\omega,z_{\xi})\d(z_{\xi}-a_4\Delta z_{\xi})\right]$$ (15)

where , for convenience, H is defined by,

$$H(\r,s_{\xi},\omega,z_{\xi})={\cal F}^{-1}_{k_{r_{\xi}},k_{s... ...}\left[DSR(k_{r_{\xi}},k_{s_{\xi}},\omega,z_{\xi})f^W\right]\;,$$ (16)

where ${\cal F}$ is the Fourier transform operator. It is important to note here that the convolution of lattice ${\cal L}_{FL}$with filter H does not change the location of the sample points. Rather, it operates only on the amplitudes at the predefined locations.