Spatial sampling restrictions

Problems in Exploration Seismology and their Solutions

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

${\begin{aligned}f<f_{\rm {N}}=1/2\Delta .\end{aligned}}$

(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 $2\Delta x$.

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Table of Contents (book)

Also in this chapter

• Spatial sampling restrictions
• Bin size in marine work
• Effect of crosscurrents
• Number of seismic sources
• Circle shooting
• Ocean-bottom cable surveys
• Vibroseis land survey
• Loop layout for a 3D survey
• Fault interpretation using time slices
• Acquisition direction for marine 3D surveys

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