Calculating the pressure gradient in the $\omega$-$\kappa_x$domain

It is easy to obtain the horizontal component of the gradient in the $\omega$-$\kappa_x$ domain  

$${\partial \over \partial x}\phi(x,z_0,\omega) \;\; \stackrel... ...ngrightarrow} \;\; i \; \kappa_x \; \phi(\kappa_x,z_0,\omega).$$ (13)

To obtain the vertical component however, it is necessary to uncouple the downgoing and upcoming components. Taking the vertical derivative of equation (5), moving to the $\omega$-$\kappa_x$ domain, and using the relations in (7), we get  

$${\partial \over \partial z}\phi(\kappa_x,z_0,\omega) \; = \;... ...0,\omega) \; + \; i \; \kappa_z \; \phi_d(\kappa_x,z_0,\omega).$$ (14)

Finally, equations (11) and (12) can be substituted into equation (14) to give  

$${\partial \over \partial z}\phi(\kappa_x,z_0,\omega) \; = \;... ...\over 1 - \cos(2 z_0 \kappa_z) } \; \phi(\kappa_x,z_0,\omega).$$ (15)

A first glance at equation (15) shows that it has a set of singular strings in the $\omega$-$\kappa_x$ plane, which are defined by  

$$\omega_n(\kappa_x) = \pm \sqrt{\left({K \over \rho}\right) \left[ \left({n \pi \over z_0}\right)^2 + \kappa_x^2 \right]},$$ (16)

and which correspond to vertical wavelengths of ${\textstyle2 \: z_0 \over n}$.For waves with these wavelengths, the downgoing and upcoming fields cancel each other at the cable depth (assuming perfect reflection at the water's surface), and the result is a zero in the recorded wavefield. Nevertheless, this cancellation does not restrict the complete recovery of the displacement field because, fortunately, the associated singularities are removable.

In the neighborhood of these strings, the vertical component of the pressure gradient is given by  

$$\lim_{\omega \rightarrow \omega_n(\kappa_x)} {\partial \over... ... {\partial \over \partial \omega} \; \phi(\kappa_x,z_0,\omega).$$ (17)