Suppressing multiples by NMO differences

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Problem

A primary and a multiple each arrive at 0.600 s at ; 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 , where . 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 and to the reflector responsible for the multiple, assuming it is simply a double bounce, is . NMO is given by equation (4.1c), . 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 where

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Differential moveout between primary and multiple Distinguishing horizontal/vertical discontinuities
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