Suppressing multiples by NMO differences

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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}} $

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Differential moveout between primary and multiple Distinguishing horizontal/vertical discontinuities
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