# Spatial sampling restrictions

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 12 469 - 484 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 12.1a

Show that the maximum spatial sampling ${\displaystyle \Delta _{x}}$ can be written

 {\displaystyle {\begin{aligned}\Delta _{x}=V/(2f_{\max }\sin \alpha _{\max }),\end{aligned}}} (12.1a)

where ${\displaystyle f_{\max }}$ is the maximum frequency of interest, and ${\displaystyle \alpha _{\max }}$ is the maximum angle of approach.

### Background

A wave is a function of time and space, e.g., ${\displaystyle g(x,t)}$ (see problem 2.5); therefore it can be sampled in time at a fixed location (problem 9.4) or in space at a fixed time (see Sheriff and Geldart, 1995, section 8.3.10). In both cases the sampling theorem (see problem 9.4) states that the wave can be sampled at fixed intervals ${\displaystyle \Delta }$ and be recovered exactly from the sampled data provided all frequencies are less than the Nyquist frequency ${\displaystyle f_{N}}$, that is, less than half the sampling frequency:

 {\displaystyle {\begin{aligned}f (12.1b)

For spatial sampling, ${\displaystyle 1/\lambda _{a}}$ gives the number of waves per unit length and, hence, corresponds to frequency in the time domain. Therefore, for spatial sampling at intervals ${\displaystyle \Delta _{x}}$, the equivalent of equation (12.1b) is

 {\displaystyle {\begin{aligned}1/\lambda _{a}<1/\lambda _{\rm {N}}=1/2\Delta _{x},\end{aligned}}} (12.1c)

so ${\displaystyle \lambda _{a}>\lambda _{\rm {N}}=V_{a}/f_{\rm {N}}=2\Delta _{x},}$

where ${\displaystyle \lambda _{a}={\mbox{apparent wavelength}}=V_{a}/f=V/f\sin \alpha }$, ${\displaystyle f_{\rm {N}}=V/\lambda _{\rm {N}}=V/2\Delta _{x}}$.

### Solution

The maximum sampling interval ${\displaystyle \Delta _{x}}$ is associated with the minimum apparent wavelength. From equation (12.1c) we have

 {\displaystyle {\begin{aligned}(\Delta _{x})_{\max }={\frac {1}{2}}(\lambda _{a})_{\min }={\frac {1}{2}}(V_{a}/f)_{\min }=(V/2f_{\max }\sin \alpha _{\max }).\end{aligned}}} (12.1d)

## Problem 12.1b

Show that the maximum group spacing ${\displaystyle D_{\max }}$ is

 {\displaystyle {\begin{aligned}D_{\max }<1000/[2f_{\max }(\Delta t/\Delta x)_{\max }],\end{aligned}}} (12.1e)

where the dip moveout ${\displaystyle (\Delta t/\Delta x)_{\max }}$ is in milliseconds/unit distance.

### Solution

Assuming ${\displaystyle \alpha \approx \zeta }$, we replace ${\displaystyle \sin \alpha }$ in equation (12.1d) with ${\displaystyle \sin \zeta =(V/2)(\Delta t/\Delta x)}$ [see equation (4.2b)]:

{\displaystyle {\begin{aligned}D_{\max }=(\lambda _{a})_{\min }/2=(V_{a}/2f_{\max })=V/2f_{\max }\sin \zeta _{\max }\\=1000/f_{\max }(\Delta t/\Delta x)_{\max },\end{aligned}}}

where ${\displaystyle \Delta t={\rm {time}}}$ time difference in milliseconds between two geophones separated by a distance ${\displaystyle 2\Delta x}$.