Fourier decomposition

Fourier analyzing the function f(x,t,z=0), seen on the earth's surface, requires the Fourier kernel $\exp ( -\, i \omega t\,+\,i\,k_x x )$.Moving on the earth's surface at an inverse speed of ${\partial t_0}/{\partial x} \,=\, k_x / \omega$,the phase of the Fourier kernel, hence the kernel itself, remains constant. Only those sinusoidal components that move at the same speed as the Snell wave can have a nonzero correlation with it. So if the disturbance is a single Snell wave, then all Fourier components vanish except for those that satisfy $p\,=\,k_x / \omega$.You should memorize these basic relations:

 

$$\fbox {\rule[-.4cm]{0cm}{1cm}$\displaystyle {\strut \partia... ...{.2in}=\hspace{.2in} \displaystyle {\strut k_x \over \omega}$}$$ (51)

In theoretical seismology a square-root function often appears as a result of using (51) to make a cosine.

Utilization of this Fourier domain interpretation of Snell's parameter p enables us to write the square-root equations (48), (49) and (50) in an even more useful form. But first the square-root equation must be reexpressed in the Fourier domain. This is done by replacing the $\partial / \partial t$ operator in (48), (49) and (50) by $- i \omega$.The result is  

$${\partial P( \omega , k_x ,z) \over \partial z} \quad =\quad... ...z)^2} \ \ -\ { k_x^2 \over \omega^2} } \ \ P( \omega , k_x ,z)$$ (52)

At present it is equivalent to specify either the differential equation (52) or its solution (47) with f as the complex exponential:  

$$P( \omega , k_x ,z) \quad =\quad\exp \left( \, i \omega \i... ...1 \over v(z)^2}\ -\ {k_x^2 \over \omega^2 }\ } \ \ dz \ \right)$$ (53)

Later, when we consider lateral velocity variation v(x), the solution (53) becomes wrong, whereas the differential equation (50) is a valid description of any local plane-wave behavior. But before going to lateral velocity gradients we should look more carefully at vertical velocity gradients.