Useful attenuation mechanisms

Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing

Series	Geophysical References Series
Title	Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing
Author	Enders A. Robinson and Sven Treitel
Chapter	14
DOI	http://dx.doi.org/10.1190/1.9781560801610
ISBN	9781560801481
Store	SEG Online Store

A traveling sine wave has the form ${\rm {\ sin\ 2}}\pi \left(ft-\ \kappa x\right)$, where f, t, $\kappa$, and x are, respectively, the temporal frequency, time, spatial frequency (or wavenumber), and distance. The period is T = 1/f, and the wavelength is $\lambda ={1/}\kappa$. During one period of time, the wave travels one wavelength of distance. Thus, the wave velocity is equal to wavelength over period. The wave velocity is specified by any of the expressions

${\begin{aligned}V={\frac {f}{\kappa }}\mathrm {\;\;or\;\;} V={\frac {\lambda }{T}}\mathrm {\;\;\;or\;\;\;} {\frac {1}{\lambda }}={\frac {f}{V}}.\end{aligned}}$

(3)

As we have seen, in addition to transmission losses and spherical spreading, a seismic wave loses amplitude because of absorption (Ecevitoglu and Costain, 1988). Absorption is the loss of elastic energy by irreversible conversion into heat. Absorption decreases the amplitude of a sinusoidal wave as it travels. The quality factor Q is

${\begin{aligned}Q={\frac {\pi f}{\alpha V}}={\frac {\pi }{\alpha \lambda }}.\end{aligned}}$

(4)

The generally accepted mechanism for attenuation is the linear model given by equation 1. If we eliminate the effect of spherical spreading (i.e., if we eliminate the $x_{0}/x$ factor) and if we let the initial amplitude $A_{0}$ be unity, then equation 1 reduces to

${\begin{aligned}A={\rm {\ exp\ }}\left[-\alpha \ x\right]={\rm {\ exp\ }}\left[-{\frac {\pi fx}{QV}}\right]={\rm {\ exp\ }}\left[-{\frac {\pi x}{Q\lambda }}\right].\end{aligned}}$

(5)

Equation 5 gives the amplitude loss that occurs because of absorption. In addition, equation 5 shows that the absorption loss is a function of the quotient $x{/}\lambda$. This quotient is the distance traveled divided by the wavelength. The mechanism is wavelength selective in that for a given travel distance, a greater loss occurs for a small wavelength than for a large wavelength. In other words, for a given travel distance, a greater loss occurs for a large frequency than for a small frequency. If x is the travel distance, then $t=x/V$ is the traveltime. Thus, the amplitude loss resulting from absorption can be written as

${\begin{aligned}A={\rm {\ exp\ }}\left[-{\frac {\pi x}{Q\lambda }}\right]={\rm {\ exp\ }}\left[-{\frac {\pi fx}{QV}}\right]={\rm {\ exp\ }}\left[-{\frac {\pi ft}{Q}}\right].\end{aligned}}$

(6)

For example, suppose that the wave velocity in the medium is 3048 m/s and that Q is 100. A plane wave of frequency 19.5 Hz has a wavelength of 3048/19.5, which is 156 m. A plane wave of twice the given frequency — that is, a plane wave of frequency 39 Hz — has a wavelength of 3048/39, which is 78 m. Suppose that both waves travel a distance of x = 1524 m. The amplitude loss for the 19.5-Hz wave is

${\begin{aligned}{\rm {\ exp\ }}\left[-{\frac {\pi }{100}}{\frac {1524}{156}}\right]={0.74},\end{aligned}}$

(7)

whereas the amplitude loss for the 39-Hz wave is

${\begin{aligned}{\rm {\ exp\ }}\left[-{\frac {\pi }{100}}{\frac {1524}{78}}\right]={0.54}.\end{aligned}}$

(8)

Thus, to restore the original amplitude, the 19.5-Hz wave should be multiplied by the boosting factor 1/0.74 = 1.35. Similarly, the 39-Hz wave should be multiplied by the boosting factor 1/0.54 = 1.85.

A wavelet is made up of sinusoidal waves of all frequencies from zero to the Nyquist frequency. To restore a wavelet that has traveled 1524 m, each component sinusoid should be multiplied by the appropriate boosting factor.

Equation 5 can be written as

${\begin{aligned}A={\rm {\ exp\ }}\left[-bf\right]\mathrm {\;\;with\;\;} b={\frac {\pi }{Q}}{\frac {x}{V}}.\end{aligned}}$

(9)

The expression A, considered to be a function of frequency, represents the amplitude spectrum of the earth-absorption filter, also called the Q filter. The log of the amplitude spectrum is

${\begin{aligned}{\rm {\ log\ A=}}-bf\mathrm {\;\;\;{\rm {with\;\;\;}}} b={\frac {\pi x}{QV}}.\end{aligned}}$

(10)

Thus, the log of the amplitude spectrum is a straight line with slope b. The filter is causal, and usually it is assumed to be minimum phase (Aki and Richards, 2002). The minimum-phase spectrum is the Hilbert transform of the log-amplitude spectrum (Silvia and Robinson, 1979, p. 90). The amplitude and phase spectra make up the frequency spectrum of the absorption filter (i.e., of the Q filter). The inverse Fourier transform of this spectrum gives the impulse-response function of the Q filter. This impulse-response function is called the Q pulse. Because the Q filter is minimum phase, it has a minimum-phase inverse.

As an example, let x = 1524 m, Q = 100, and V = 3048 m/s so that

$b={\frac {{1524}\pi }{{100}\cdot {3048}}}={0.0157.}$

Figure 1.  (a) Log amplitude spectra of a Q pulse and its inverse Q pulse. (b) Amplitude spectra of a Q pulse and its inverse Q pulse. (c) Phase spectra of a Q pulse and an inverse Q pulse. (d) A Q pulse and its inverse Q pulse.

The log of the amplitude spectrum of the Q pulse is the straight line with slope –b shown in Figure 1a. Figure 1a also shows the log of the amplitude spectrum of the inverse Q pulse. The log of the amplitude spectrum of the inverse Q pulse is simply the negative of the log of the amplitude spectrum of the Q pulse because the two must add to zero.

The amplitude spectrum of the Q pulse is shown in Figure 1b, as is the amplitude spectrum of the inverse Q pulse. The amplitude spectrum of the inverse Q pulse is simply the reciprocal of the amplitude spectrum of the Q pulse because the two must multiply to one.

The phase spectrum of the Q pulse and the phase spectrum of the inverse Q pulse are shown in Figure 1c. The phase spectrum of the inverse Q pulse is simply the negative of the phase spectrum of the Q pulse because the two must add to zero.

Finally, in Figure 1d, we see the Q pulse and the inverse Q pulse for x = 1524 m. The two pulses must convolve to a spike. The points on each spectrum in Figure 1 that correspond to the frequencies 19.5 Hz and 39 Hz are indicated in the figure. In particular, the points on the amplitude spectrum (Figure 1b) give the value 0.74 for 19.5 Hz and the value 0.54 for 39 Hz, which are the absorption factors calculated above. The points on the amplitude spectrum of the inverse give the value 1.35 for 19.5 Hz and the value 1.85 for 39 Hz, which are the boosting factors calculated above. Thus, to restore the original amplitude to a signal that has suffered absorption, the signal should be convolved with the inverse of the Q pulse.

The absorption coefficient, as given by equation 2, increases with frequency. Thus, as a wavelet travels, it loses more and more high-frequency values. As a result, a wavelet broadens with the distance it has traveled. The deeper the reflection is, the greater is the loss of the higher frequencies. How far can a wave (of frequency 39 Hz) travel before it is reduced to one-tenth of its original amplitude? From equation 2, we see that the absorption coefficient is

${\begin{aligned}\alpha ={\frac {\pi f}{QV}}={\frac {\pi \left({39}\right)}{\left({100}\right)\left({3048}\right)}}={0.000402}.\end{aligned}}$

(11)

From equation 5, we have

${\begin{aligned}{\frac {1}{10}}=e^{-{0.000402}x}.\end{aligned}}$

(12)

The solution of this equation gives x = 5729 m. Thus, merely by absorption, a plane wave’s amplitude drops by a factor of 0.1 as the wave travels to a depth of 2864 m and back up to the surface.

When it comes to absorption, a basic assumption is made. When a wave travels one wavelength, absorption reduces the wave’s amplitude by a certain fraction. The assumption is that the fraction is independent of the wavelength. The absorption can be approximated as an exponential decay of energy with propagation distance. The decay is roughly constant with each frequency cycle. A high-frequency 39-Hz wave traveling at 3048 m/s has a wavelength of 78 m. A low-frequency 19.5-Hz wave traveling at 3048 m/s has a wavelength of 156 m. The high-frequency wave suffers the same amount of absorption by traveling 78 m as the low-frequency wave does in traveling 156 m, or twice the distance. The net result is that high-frequency components suffer greater attenuation than do low-frequency components per unit distance traveled. Absorption is measured by the quality factor Q. A weathered rock at the surface can have a Q factor as low as 10 or 20, whereas a deep, less-absorptive rock can have a Q factor of 200. That is, Q increases and absorption decreases with depth.

At great enough depths, the presence of random noise generally precludes the possibility of obtaining good-quality seismic acoustic-impedance logs. The random noise level depends on agents such as wind noise (either acting directly on the geophones or indirectly by shaking the ground) and the scattering of horizontally traveling waves by random inhomogeneities. In addition, although transmission through a water layer is excellent, irregularities in the form of bottom scatterers within the water (fish, gas bubbles, or man-made artifacts) combine to give incoherent scattered energy that persists up to high frequencies. Penetration of high-frequency energy directly into the earth and the return of high-frequency energy from the target layers is limited both by irreversible absorption through solid friction and by such purely elastic phenomena as the loss of energy by reverberations and transmission through many layer boundaries.

Let us summarize. To a good approximation, amplitude loss resulting from absorption follows the exponential law given by equation 5, for which the absorption coefficient is given by equation 2. The loss, in decibels, in traveling a distance x is

${\begin{aligned}\mathrm {dBloss} ={8.686}{\frac {\pi fx}{QV}}.\end{aligned}}$

(13)

The dB loss is proportional to both f and x. The effect of doubling Q produces the same effect as does halving distance x. Experiments with shear waves show how drastic exponential decay is. Caution is required to ensure that an objective is within the realms of possibility. Often, results can be attained only in certain types of lithologic sections.

Let us look at an example. Suppose that the Pierre Shale has a thickness of 1524 m, a velocity of 2377 m/s, and an effective Q of 20. Suppose that a certain signal-to-noise ratio has been established in propagating a 15-Hz signal to a depth just below the Pierre Shale. How much higher must the signal-to-noise ratio be to propagate a 30-Hz signal to the same depth? The decibel loss at 15 Hz is

${\begin{aligned}{\rm {dB\ }}\mathrm {loss} ={8.686}{\frac {\pi {15}\left({3048}\right)}{{20}\left({237}{7}\right)}}={26.23}.\end{aligned}}$

(14)

The decibel loss at 30 Hz is

${\begin{aligned}{\rm {dB\ }}\mathrm {loss} ={8.686}{\frac {\pi {30}\left({3048}\right)}{{20}\left({2377}\right)}}={52.46}.\end{aligned}}$

(15)

Thus, a gain of 26.23 dB in the signal is necessary.

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• Attenuation
• Practical considerations
• Appendix O: Exercises

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