previous up next print clean
Next: About this document ... Up: Popovici : DMO basics Previous: MZO by Fourier transform

References

APPENDIX A

The Fourier transform of a finite segment differs only in the amplitude term from the Fourier transform of an infinitely long segment. This is to be expected since we actually multiply the infinite segment by a boxcar filter which in Fourier domain means convolution with a sync function.

A two dimensional function representing a segment of constant amplitude in (t,y) space can be described by

\begin{displaymath}
H(y) \delta ({{t-t_0} \over p} - y)\end{displaymath}

where

\begin{displaymath}
\left \{ \begin{array}
{ll}
H(y)=1; & y \in [a,b]
\\ \\ H(y)=0; & y \in (-\infty,a) \cup (b,\infty)\end{array} \right.\end{displaymath}

where t and y are the coordinates, t0 is the intersection point on the t axis and the value p is the tangent of the slope. The function $\delta ({{t-t_0} \over p}-y)$ has unitary amplitude when the argument is zero or t=t0+py. The 2-D Fourier transform is

\begin{displaymath}
\begin{array}
{lcl}
S(\omega,k_y) & = & \displaystyle{
\int ...
 ...omega p-k_y)}+i \int_a^b {\sin y(\omega p-k_y)} 
] }\end{array}\end{displaymath}

and using the well known trigonometric transformations

\begin{displaymath}
\begin{array}
{ccc}
{\sin \alpha - \sin \beta} & = & {2 \sin...
 ...pha+ \beta} \over 2 }
\sin {{\alpha-\beta} \over 2}}\end{array}\end{displaymath}

we obtain
\begin{displaymath}
\begin{array}
{lcl}
S(\omega,k_y) & = & \displaystyle{
{{e^{...
 ...(\omega p -k_y)
e^{{{b+a} \over 2}(\omega p -k_y)}}.\end{array}\end{displaymath} (29)


previous up next print clean
Next: About this document ... Up: Popovici : DMO basics Previous: MZO by Fourier transform
Stanford Exploration Project
11/17/1997