# Attenuation calculations

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 2.18

A refraction seismic wavelet assumed to be essentially harmonic with a frequency of 40 Hz is found to have amplitudes of 5.00 and 4.57 mm on traces 2500 and 3000 m from the source. Assuming a velocity of 3200 m/s, constant subsurface conditions, and ideal recording conditions, what is the ratio of the amplitudes on a given trace of the first and fourth cycles? What percentage of the energy is lost over three cycles? What is the value of the damping factor $h$ ?

### Background

As a wave travels through a medium, the energy of the wave is gradually absorbed by the medium. This results in attenuation of the wave, the decrease in amplitude being approximately exponential with both distance and time. For a fixed time $t$ , we have

 {\begin{aligned}A_{1}=A_{0}e^{-\eta x},\end{aligned}} (2.18a)

where the initial amplitude $A_{0}$ has decreased to $A_{1}$ after the wave travels a distance $x;\eta$ is the absorption coefficient. On the other hand, at a fixed location, the amplitude varies with time according to the equation

 {\begin{aligned}A_{1}=A_{0}e^{-ht},\end{aligned}} (2.18b)

$h$ being the damping factor. During a period $T$ , the wave travels a distance $\lambda$ , hence equations (2.18a) and (2.18b) show that

 {\begin{aligned}hT=\eta \lambda .\end{aligned}} (2.18c)

A damped harmonic wave can be written

{\begin{aligned}A_{1}=A_{0}e^{-ht}\cos \omega t.\end{aligned}} The logarithmic decrement (“log dec”) $\delta$ is defined as

 {\begin{aligned}\delta =\ln \left({\frac {\mathrm {amplitude} }{\hbox{amplitude one cycle later}}}\right).\end{aligned}} (2.18d)

If $T$ is the period, equations (2.18b) and (2.18d) show that

 {\begin{aligned}\delta =hT=h/f=2\pi h/\omega =\eta \lambda .\end{aligned}} (2.18e)

The quality factor $Q$ is another attenuation constant; it is defined by the relation

{\begin{aligned}Q=2\pi /\left(\Delta E/E\right),\end{aligned}} where $\Delta E/E$ is the fractional wave energy loss/cycle. Because $E$ is proportional, to $A^{2}$ , $E=E_{0}e^{-2ht}$ . Since $\Delta E={}$ energy loss in one period $T$ ,

 {\begin{aligned}\Delta E/E=\Delta \left(e^{-2ht}\right)/e^{-2ht}=2h\left(\Delta t\right)=2hT=2\delta .\end{aligned}} (2.18f)

($\Delta E$ is the loss per cycle, so we have dropped the minus sign and set $\Delta t=T$ ). Thus, $Q$ becomes

 {\begin{aligned}Q=\pi /hT=\pi /\delta .\end{aligned}} (2.18g)

### Solution

The wavelength is $\lambda =3200/40=80$ m. From equation (2.18a) we get

{\begin{aligned}\eta =\left(1/x\right)\ln \left(A_{0}/A_{1}\right)=\left(1/0.50\right)\ln \left(5.00/4.57\right)=0.180\ \mathrm {km} ^{-1}.\end{aligned}} From equation (2.18e),

{\begin{aligned}\log \mathrm {dec} =\delta =\eta \lambda =0.180\times 0.080=0.0144.\end{aligned}} Equation (2.18b) shows that the amplitude ratio decreases by the same fraction over each interval $T$ , hence the decrease in the ratio from the first to the fourth cycle is

{\begin{aligned}\ln \left(A_{1}/A_{4}\right)=3hT=38=0.0432,\end{aligned}} so

{\begin{aligned}A_{1}/A_{4}=e^{3\delta }=e^{0.0432}=1.044,\end{aligned}} {\begin{aligned}A_{4}=0.958\ A_{1}.\end{aligned}} The fraction of the energy lost per cycle, $\Delta E/E$ , is equal to $2\delta$ from equation (2.18f). For 3 cycles, the fractional energy loss is $6\delta =0.0864=8.64\%$ .

From equation (2.18e), $h=f\delta =40\times 0.0144=0.576$ s$^{-1}$ .