# Difference between revisions of "Suppressing multiples by NMO differences"

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.11

A primary and a multiple each arrive at 0.600 s at $\displaystyle x=0$ ; 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 $\displaystyle n$ , where $\displaystyle n=1, 2,...$ . 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 $\displaystyle (0.600\times 1800)/2=540\ {\rm m}$ and to the reflector responsible for the multiple, assuming it is simply a double bounce, is $\displaystyle (0.600\times 1500)/4=225\ {\rm m}$ . NMO is given by equation (4.1c), $\displaystyle \Delta t_{\rm NMO} =x^{2} /2V^{2} t_{0}$ . 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 $\displaystyle x$ where

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