Suppressing multiples by NMO differences

**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	6
Pages	181 - 220
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

Problem 6.11

A primary and a multiple each arrive at 0.600 s at $x=0$ ; their stacking velocities are 1800 and 1500 m/s, respectively. Calculate the residual NMO (after NMO correction for the primary velocity) for offsets of 300 $n$ , where $n=1,2,...$ . What is the shortest offset that will give good multiple suppression for a wavelet with a 50-ms dominant period?

Solution

The distance to the primary reflector is $(0.600\times 1800)/2=540\ {\rm {m}}$ and to the reflector responsible for the multiple, assuming it is simply a double bounce, is $(0.600\times 1500)/4=225\ {\rm {m}}$ . NMO is given by equation (4.1c), $\Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}$ . We obtain the following values for the moveouts:

Offset	300 m	600 m	900 m
Primary NMO	0.023 s	0.093 s	0.208 s
Multiple NMO	0.033 s	0.133 s	0.300 s
NMO Difference	0.010 s	0.040 s	0.092 s

Multiple suppression should be maximum when the NMO difference approximates half the wavelet period so that some of the traces are out-of-phase, which is achieved at offset $x$ where

${\begin{aligned}x^{2}/1.200\times 1500^{2}-x^{2}/1.200\times \;1800^{2}\;=0.050,\;x=660\ {\rm {m}}.\end{aligned}}$

Continue reading

Previous section	Next section
Differential moveout between primary and multiple	Distinguishing horizontal/vertical discontinuities
Previous chapter	Next chapter
Geometry of seismic waves	Characteristics of seismic events
Table of Contents (book)

Also in this chapter

External links

find literature about
Suppressing multiples by NMO differences