Uncoupling the downgoing and upcoming components of the recorded field

Most of the treatment that follows is carried out either in the $\tau$-p or in the $\omega$-$\kappa_x$ domain. To simplify the analysis, avoiding the unnecessary intrusion of convolutions along the $\kappa_x$ axis, the discussion focuses on the points in space where the wavefield is measured, that is, at the points (x,z = z₀) where z₀ is the cable depth. At these points the bulk moduli and density are constant and known: K(x,z₀)=K, and $\rho(x,z_0)=\rho.$

The pressure field $\phi$ at any position of the space can be represented as the superposition of an upward propagating wavefield $\phi_u$ with a downward propagating wavefield $\phi_d$: 

$$\phi(x,z_0,\omega) \; = \; \phi_u(x,z_0,\omega) \; + \; \phi_d(x,z_0,\omega).$$ (5)

Continuation of these wavefields in the frequency-horizontal wavenumber domain ($\omega$-$\kappa_x$) is controlled by the following equations:

$$\begin{eqnarray} {\partial \over \partial z} \phi_u(\kappa_x,z_0,\omega) & = & ... ...x,z_0,\omega) & = & i \; \kappa_z \; \phi_d(\kappa_x,z_0,\omega),\end{eqnarray}$$ (6) (7)

where $\kappa_z$ is the vertical wavenumber, which relates to the horizontal slowness p through the dispersion relation  

$$\kappa_z \;\; = \;\; \omega \sqrt{{\rho \over K} - p^2} \;\;... ...qrt{{\rho \over K} - \left( {\kappa_x \over \omega} \right)^2}.$$ (8)

To separate these two wavefields, we must make the following assumptions:

• The water-air reflection coefficient is $\; -1.$
  
• The water surface is nearly horizontal.
  
• The cable depth is a smooth function of the receiver position.

Under these assumptions, the upcoming and downgoing fields can be related by a simple time shift equation in the $\tau$-p domain:  

$$\phi_d(p,z_0,\tau) \; = \; - \; \phi_u(p,z_0,\tau - 2 z_0 q(p)),$$ (9)

where q is the vertical slowness. When the source is located below the cable equation (9) can be applied without restrictions, and the only event for which it doesn't hold perfectly true, for cases in which the source is above the cable, is the direct wave.

Substituting equation (9) into equation (5), both in the $\omega$-p domain, we obtain  

$$\phi(p,z_0,\omega) \; = \; \phi_u(p,z_0,\omega) \; [ 1 - \exp{(i 2 z_0 \omega \sqrt{\rho/K - p^2})} ];$$ (10)

which results in the following equations for separation of the two wavefields:  

$$\phi_u(p,z_0,\omega) \; = \; {\phi(p,z_0,\omega) \over 1 - \exp (i 2 z_0 \omega \sqrt{\rho/K - p^2})},$$ (11)

 

$$\phi_d(p,z_0,\omega) \; = \; {\phi(p,z_0,\omega) \over 1 - \exp (-i 2 z_0 \omega \sqrt{\rho/K - p^2})}.$$ (12)