26)

Derivation of equation (

• Using the kinematics of the transformation to the angle domain (equation (10)) and the equations of the migration impulse response (equations (1) and (3)),

  $$\begin{eqnarray} \frac{\partial z_{\tilde{\gamma}}}{\partial L} &=& \frac{\par... ...a} + \frac{\sin \gamma}{\cos \alpha_x} \tan \widehat{\gamma}. \end{eqnarray}$$ (40) (41)

• Differentiating the equality L_sS_s+L_rS_r=t_D with respect to S_s,

  $$\begin{eqnarray} \frac{\partial L_s}{\partial S_s} S_s+ L_s &=& 0,\\ \frac{\partial L_s}{\partial S_s} &=& -\frac{L_s}{S_s}. \end{eqnarray}$$ (42) (43)

  Using the simple geometric relationship $L_s\cos \beta_s=z_\xi$, we eventually get

  $$\begin{eqnarray} \frac{\partial L_s}{\partial S_s} = -\frac{z_\xi}{S_s\cos \beta_s} = -\frac{z_\xi}{S_s\cos\left(\alpha_x + \gamma \right)} . \end{eqnarray}$$ (44)

  From the geometric relationship $L_r\cos \beta_r=z_\xi$, we similarly show

  $$\begin{eqnarray} \frac{\partial L_r}{\partial S_r} = -\frac{z_\xi}{S_r\cos \beta_r} = -\frac{z_\xi}{S_r\cos\left(\alpha_x - \gamma \right)}. \end{eqnarray}$$ (45)