# 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 $\Delta _{x}$ can be written

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

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

### Background

A wave is a function of time and space, e.g., $g(x,t)$ (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 $\Delta$ and be recovered exactly from the sampled data provided all frequencies are less than the Nyquist frequency $f_{N}$ , that is, less than half the sampling frequency:

 {\begin{aligned}f (12.1b)

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

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

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

### Solution

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

 {\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 $D_{\max }$ is

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

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

### Solution

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

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