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26)
- 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}](img72.gif) |
(40) |
| (41) |
-
Differentiating the equality LsSs+LrSr=tD with respect to
Ss,
| ![\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}](img73.gif) |
(42) |
| (43) |
Using the simple geometric relationship
, 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}](img75.gif) |
(44) |
From the geometric relationship
, 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}](img77.gif) |
(45) |
Next: Derivation of equation (29)
Up: Derivation of the derivative
Previous: Derivation of the derivative
Stanford Exploration Project
4/6/2006