Difference between revisions of "Water reverberation filter"

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  | part    =  
  | part    =  
  | chapter = 9
  | chapter = 9
  | pages  = 295 - 294
  | pages  = 295 - 366
  | author  = [[Lloyd P. Geldart]] and [[Robert E. Sheriff]]
  | author  = [[Lloyd P. Geldart]] and [[Robert E. Sheriff]]
  | doi    = 10.1190/1.9781560801733
  | doi    = 10.1190/1.9781560801733

Latest revision as of 09:14, 16 July 2019

Problem 9.7a

Verify equation (9.7d) for the inverse filter for water reverberation by convolving it with equation (9.7a) for the water reverberation.


Multiple reflections within a water layer can cause severe reverberation (ringing) when the reflection coefficient at the sea floor is sufficiently high. We take the reflection coefficients at the top and bottom of the water layer as and , and as the round-trip traveltime in the layer, being an integer and the sampling interval. Bounces can occur before or after a wave is reflected at deeper reflectors. We assign unit amplitude to a primary reflection with traveltime . A reflection from the same horizon that has suffered one bounce in the water layer has traveltime and amplitude . Since the bounce may take place either before or after reflection at depth, there are two waves arriving at , the combined amplitude being . There are three waves that suffer two bounces: two bounces before the deep reflection, two after, and one before plus one after, so the multiple that arrives at has amplitude . Reflections suffering three bounces arrive at with amplitude , and so on. The impulse response (see Sheriff and Geldart, 1955, 279) of the water layer is thus the infinite series



and so on for other values of .

A filter (see problem 7.11) that removes the effect of another filter is an inverse filter, and the inverse filter in this case is ; it satisfies the equation




[See Sheriff and Geldart, 1995, section 15.5.3 for a discussion of -transforms.]


For , , and . We wish to verify that . Using the procedure described in problem 9.2, we replace each element of by a scaled version of ; thus

1, , +, , +
+, , +,
+, , +
= 1, 0, 0, 0, 0, … .

We can proceed in the same way for other values of .

Problem 9.7b

The spectrum of the water-layer filter is shown in Figure 9.7a for ; the large peaks occur at the “singing frequency” , where is any odd number. Sketch the amplitude spectrum of the inverse filter.


In the time domain, ; in the frequency domain this becomes . Thus the amplitude spectrum of is the inverse of that of , that is, for a given frequency,

as shown in Figure 9.7b.

Figure 9.7a.  Spectrum of water-layer filter; water depth, .
Figure 9.7b.  Inverse filter spectrum (dashed).

Problem 9.7c

Verify your sketch of the water-reverberation inverse filter by transforming


and calculating the value of the spectrum for , , , and .


We start with the -transform of , that is, . This is a complex quantity and we multiply by the conjugate complex (see Sheriff and Geldart, 1995, section 15.1.5) to get the magnitude squared. Since , the conjugate complex is and ; thus,

Substituting and using Euler’s formula (Sheriff and Geldart, 1995, problem 15.12a), we get

Substituting , , and , we get the following values

The peaks and troughs in Figure 9.7a have the relative values 9, 1, 9; since these values are normalized, the agreement is exact.

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