# Water reverberation filter

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## 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.

### Background

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 ${\displaystyle -1}$ and ${\displaystyle R}$, and ${\displaystyle n\Delta }$ as the round-trip traveltime in the layer, ${\displaystyle n}$ being an integer and ${\displaystyle \Delta }$ 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 ${\displaystyle t}$. A reflection from the same horizon that has suffered one bounce in the water layer has traveltime ${\displaystyle t+n\Delta }$ and amplitude ${\displaystyle -R}$. Since the bounce may take place either before or after reflection at depth, there are two waves arriving at ${\displaystyle t+n\Delta }$, the combined amplitude being ${\displaystyle -2R}$. 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 ${\displaystyle t+2n\Delta }$ has amplitude ${\displaystyle +3R^{2}}$. Reflections suffering three bounces arrive at ${\displaystyle t+3n\Delta }$ with amplitude ${\displaystyle -4R^{3}}$, and so on. The impulse response (see Sheriff and Geldart, 1955, 279) of the water layer is thus the infinite series

 {\displaystyle {\begin{aligned}w_{t}=[1,\;-2R,\;3R^{2},\;-4R^{3},\;5R^{4},\ldots \ldots .]\quad \mathrm {when} \ n=1;\end{aligned}}} (9.7a)

 {\displaystyle {\begin{aligned}{\mbox{or}}\qquad \qquad w_{t}=[1,\;0,\;-2R,\;0,\;3R^{2},\;0,\;-4R^{3},\ldots .]\qquad \mathrm {when} \ n=2;\end{aligned}}} (9.7b)

and so on for other values of ${\displaystyle n}$.

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 ${\displaystyle i_{t}}$; it satisfies the equation

 {\displaystyle {\begin{aligned}w_{t}*i_{t}=\delta _{t};\end{aligned}}} (9.7c)

 {\displaystyle {\begin{aligned}i_{t}&=[1,\;2R,\;R^{2},\;0,\;0,\;.\;.\;.]\qquad \mathrm {when} \ n=1\end{aligned}}} (9.7d)

 {\displaystyle {\begin{aligned}&=[1,\;0,\;2R,\;0,\;R^{2},\;0,\;.\;.\;.]\quad \mathrm {when} \ n=2.\end{aligned}}} (9.7e)

[See Sheriff and Geldart, 1995, section 15.5.3 for a discussion of ${\displaystyle z}$-transforms.]

### Solution

For ${\displaystyle n=1}$, ${\displaystyle i_{t}=\left[1,\;2R,\;R^{2}\right]}$, and ${\displaystyle w_{t}=[1,-2R,3R^{2},-4R^{3},5R^{4},\ldots ]}$. We wish to verify that ${\displaystyle w_{t}*i_{t}=\delta _{t}}$. Using the procedure described in problem 9.2, we replace each element of ${\displaystyle i_{t}}$ by a scaled version of ${\displaystyle w_{t}}$; thus

 1, –${\displaystyle 2R}$, +${\displaystyle 3R^{2}}$, –${\displaystyle 4R^{3}}$, +${\displaystyle 5R^{4},\ldots }$ +${\displaystyle 2R}$, –${\displaystyle 4R^{2}}$, +${\displaystyle 6R^{3}}$, –${\displaystyle 8R^{4},\ldots }$ +${\displaystyle R^{2}}$, –${\displaystyle 2R^{3}}$, +${\displaystyle 3R^{4},\ldots }$ = 1, 0, 0, 0, 0, … ${\displaystyle =\delta _{t}}$.

We can proceed in the same way for other values of ${\displaystyle n}$.

## Problem 9.7b

The spectrum of the water-layer filter is shown in Figure 9.7a for ${\displaystyle n=1}$; the large peaks occur at the “singing frequency” ${\displaystyle nf_{1}}$, where ${\displaystyle n}$ is any odd number. Sketch the amplitude spectrum of the inverse filter.

### Solution

In the time domain, ${\displaystyle w_{i}*i_{t}=\delta _{t}}$; in the frequency domain this becomes ${\displaystyle W(f)\times I(f)=1}$. Thus the amplitude spectrum of ${\displaystyle I(f)}$ is the inverse of that of ${\displaystyle W(f)}$, that is, for a given frequency,

{\displaystyle {\begin{aligned}|I(f)|=1/|W(f)|,\end{aligned}}}

as shown in Figure 9.7b.

Figure 9.7a.  Spectrum of water-layer filter; ${\displaystyle f_{1}=V_{W}/4h,h=}$ water depth, ${\displaystyle R=0.5}$.
Figure 9.7b.  Inverse filter spectrum (dashed).

## Problem 9.7c

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

 {\displaystyle {\begin{aligned}\left[1,\;2R,\;R^{2}\right]\leftrightarrow [1+2Re^{-{\hbox{j}}2\pi f{\Delta }}+R^{2}e^{-{\hbox{j}}4\pi f{\Delta }},\end{aligned}}} (9.7f)

and calculating the value of the spectrum for ${\displaystyle R=0.5}$, ${\displaystyle f=0}$, ${\displaystyle 1/2\Delta }$, and ${\displaystyle 1/\Delta }$.

### Solution

We start with the ${\displaystyle z}$-transform of ${\displaystyle i_{t}}$, that is, ${\displaystyle I(f)=(1+2Rz+R^{2}z^{2})}$. 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 ${\displaystyle z=e^{{\hbox{j}}2\pi \Delta }}$, the conjugate complex is ${\displaystyle e^{+{\hbox{j}}2\pi \Delta }}$ and ${\displaystyle zz^{-1}=1}$; thus,

{\displaystyle {\begin{aligned}|I(f)|^{2}&=(1+2Rz+R^{2}z^{2})(1+2Rz^{-1}+R^{2}z^{-2})\\&=(1+Rz)^{2}(1+Rz^{-1})^{2}=[(1+Rz)(1+Rz^{-1})]^{2},\end{aligned}}} {\displaystyle {\begin{aligned}{\mbox{so}}\qquad \qquad |I(f)|=(1+Rz)(1+Rz^{-1})=[1+R^{2}+R(e^{+{\hbox{j}}2\pi f\Delta }+e^{-{\hbox{j}}2\pi f\Delta })].\end{aligned}}}

Substituting ${\displaystyle R=0.5}$ and using Euler’s formula (Sheriff and Geldart, 1995, problem 15.12a), we get

{\displaystyle {\begin{aligned}|I(f)|=[1.25+\cos \,(2\pi f\Delta )].\end{aligned}}}

Substituting ${\displaystyle f=0}$, ${\displaystyle 1/2\Delta }$, and ${\displaystyle 1/\Delta }$, we get the following values

{\displaystyle {\begin{aligned}|I(f)|=2.25,\ 0.25,\ 2.25,\ \quad \mathrm {that\ is} ,9/4,\ 1/4,\ 9/4.\end{aligned}}}

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