Amplitude/energy of reflections and multiples

Problems in Exploration Seismology and their Solutions

Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
Chapter	3
Pages	47 - 77
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

Problem 3.8a

3.8a Assume horizontal layering (as shown in Figure 3.8a), a source just below interface $A$, and a geophone at the surface. Calculate (ignoring absorption and divergence) the relative amplitudes and energy densities of the primary reflections from $B$ and $C$ and the multiples $BSA$, $BAB$, and $BSB$ (where the letters denote the interfaces involved). Compare traveltimes, amplitudes, and energy densities of these five events for normal incidence.

Background

Multiples are events that have been reflected more than once. They are generally weak because the energy decreases at each reflection, but where the reflection coefficients are large, multiples may be strong enough to cause problems. Multiples are of two kinds as shown in Figure 3.8b: long-path multiples which arrive long enough after the primary reflection that they appear as separate events, and short-path multiples which arrive so soon after the primary wave that they add to it and change its shape. The most important short-path multiples are two in number: (i) ghosts (Figure 3.8b) where part of the energy leaving the source travels upward and is reflected downward either at the base of the LVL (see problem 4.16) or at the surface, (ii) peg-leg multiples resulting from the addition to a primary reflection of energy reflected from both the top and bottom of a thin bed, either on the way to or on the way back from the principal reflecting horizon. Short-path near-surface multiples are also called ghosts and long-path interformational multiples are also called peg-leg multiples. A notable example of the latter occurs in marine work when wave energy bounces back and forth within the water layer.

The energy density of a wave (see problem 3.7) decreases continuously as the wave progresses because of two factors: absorption and spreading or divergence. The energy density is proportional to the square of the amplitude, so both effects are usually expressed in terms of the decrease in amplitude with distance.

Figure 3.8a.  A layered model.

Figure 3.8b.  Types of multiples.

Absorption causes the amplitude to decrease exponentially, the relation being $A=$ $A_{0}e^{-\eta x}$ where the amplitude decreases from $A_{0}$ to $A$ over a distance $x$; the absorption coefficient $\eta$ is often expressed in terms of per wavelength, $\lambda .$

For a point source in an infinite constant-velocity medium, divergence causes the energy density to decrease inversely as the square of the distance from the source, the amplitude decreasing inversely as the first power of the distance from the source.

Nepers and decibels are defined in problem 2.17.

Solution

We first calculate the impedances $Z_{i}$ for each layer, the coefficients of reflection and downgoing and upgoing transmission $R,$ $T\downarrow$, $T\uparrow$ (see problem 3.6), and the reflected and transmitted energy coefficients, $E_{R}$ and $E_{T}$, for each interface. The results are shown in Table 3.8a.

Table 3.8a. Reflection and transmission coefficients.

Interface	$Z$	$R^{*}$	$T\downarrow$	$T\uparrow$	$E_{R}$	$E_{T}$
S		1.000	0.000	0.000	1.000	0.000
Layer 1	0.870
${\textit {A}}$		0.733	0.267	1.733	0.537	0.463
Layer 2	5.640
${\textit {B}}$		0.207	0.793	1.207	0.043	0.957
Layer 3	8.576
${\textit {C}}$		0.034	0.966	1.034	0.001	0.999
Layer 4	9.180

* Signs are for incidence from above.

Assuming unit amplitude and unit energy density for the downgoing wave incident on interface $B$ and neglecting absorption and divergence, we arrive at the following values:

Reflection $B$:

Amplitude of reflection $B=R_{B}T_{A}\uparrow =0.207\times 1.733=0.359.$ Energy density $=E_{RB}E_{TA}=0.043\times 0.463=0.020.$ Arrival time $t_{B}=2\times 0.600/2.400+0.010/0.600=0.517s.$

Reflection $C$

Amplitude $=T_{B}\downarrow R_{C}T_{B}\uparrow T_{A}\uparrow =0.793\times 0.034\times 1.207\times 1.733=0.056.$

Energy density $=E_{TB}^{2}E_{RC}E_{TA}=0,0004.$

Arrival time $t_{C}=t_{B}+2\times 0.800/3.20=1.017s.$

Multiple BSA

Amplitude $=R_{B}T_{A}\uparrow \left(-R_{S}\right)R_{A}=-0.207\times 1.733\times 1.000\times 0.733=-0.263.$

Energy density $=0.043\times 0.463\times 1.000\times 0.537=0.011.$

Arrival time $t_{BSA}=2\times 0.600/2.40+3\times 0.010/0.600=0.550s.$

Multiple BAB

Amplitude $=(R_{B})^{2}\left(-R_{A}\right)T_{A}\uparrow =0.207^{2}\times \left(-0.733\right)\times 1.733=0.0544.$

Energy density $=0.043^{2}\times 0.537\times 0.463=0.0005.$

Arrival time $t_{BAB}=4\times 0.600/2.40+0.010/0.600=1.017s.$

Multiple BSB

Amplitude $=(R_{B})^{2}(T_{A}\uparrow )^{2}T_{A}\downarrow \left(R_{S}\right)=-0.0344.$

Energy density $=(E_{RB})^{2}(E_{TA})^{3}\left(E_{RS}\right)=0.0002.$

Arrival time $t_{BSB}=4\times 0.600/2.400+3\times 0.010/0.600=1.050.$

The results are summarized in Table 3.8b.

BSA arrives 33 ms after $B$ (one period for a 33-Hz wave) with reversed polarity and about 75% of the amplitude and 50% of the energy of $B$, so BSA will significantly alter the waveshape of B. BSA involves an extra bounce at the surface and is a type of ghost whose effect is mainly that of changing waveshape rather than showing up as a distinct event.

$C$ and BAB arrive simultaneously with opposite polarities, $C$ being slightly stronger than BAB; the multiple will obscure and significantly alter the waveshape of the primary reflection.

Table 3.8b. Amplitude/energy density of primary/multiple reflections.

Event	$t$	Amplitude	$20\log \left(A/A_{B}\right)$	Energy
$B$	$0.517s$	$0.359$	$0dB$	$0.020$
$BSA$	$0.550$	$-0.263$	$-2.7$	$0.011$
$C$	$1.017$	$0.056$	$-16.1$	$0.0004$
$BAB$	$1.017$	$-0.054$	$-16.4$	$0.0004$
${\textit {BSB}}$	$1.050$	$-0.034$	$-20.5$	$0.0002$

The surface multiple BSB is smaller than the multiple from the base of the near-surface layer BAB; on land the base of the near-surface layer is often the most important interface in generating multiples.

Problem 3.8b

3.8b Recalculate for 15- and 75-Hz waves allowing for absorption.

Solution

The absorption coefficient $\eta$ has the values 0.45, 0.30, and 0.25 ${\rm {dB}}/\lambda$ in layers SA, $AB$, and $BC$, respectively. Using $\Delta z$ for the layer thicknesses, the results are given in Table 3.8c.

Table 3.8c. Absorption for one-way travel in each layer.

			$f=15$ Hz			$f=75$ Hz
Layer	Velocity	$\Delta z$	$\lambda$	$\Delta z/\lambda$	$\eta \Delta z$	$\lambda$	$\Delta z/\lambda$	$\eta \Delta z$
SA	600 m/s	10 m	40 m	0.25	0.11 dB	8 m	1.25	0.56 dB
AB	2400	600	160	3.75	1.12	32	18.8	5.64
BC	3200	800	213	3.76	0.94	43	18.6	4.65

For 15-Hz waves, the travelpath for reflection $B$ involves two-way travel through $AB$ and one-way travel through $SA$, hence attenuation due to absorption is $2\times 1.12+0.11=2.35\ {\rm {dB}}$ the amplitude being decreased by the factor 0.763. For the multiple BSA we add attenuation for the extra two-way path through $SA$ to the attenuation for $B$, giving 2.57 $dB$, or an amplitude reduction of 0.744. For reflection $C$, we add to the attenuation of reflection $B$ the attenuation for the two-way travel through $BC$, giving 4.23 $dB$, and an amplitude ratio of 0.614. For the multiple BAB we get attenuation of 4.59 $dB$, an amplitude ratio of 0.590. For BSB, attenuation is 4.81 $dB$ and an amplitude ratio is 0.575. Attenuation for 75 Hz is 5 times that for 15 Hz because $\lambda$ is only one-fifth that for 15 Hz, hence $\Delta z/\lambda \$$ will be five times greater. Table 3.8d repeats the reflection amplitudes in Table 3.8b to compare them with the amplitudes after allowing for absorption for 15 Hz and 75 Hz.

Table 3.8d. Illustrating the effect of absorption.

Event	$A$	$\mathop {\sum } \limits _{}^{}\eta \Delta z\left(15\right)$	Ratio(15)	$A_{a}(15)$	${\sum }\eta \Delta z\left(75\right)$	Ratio(75)	$A_{a}(75)$
$B$	0.359	2. 35 dB	0.763	0.274	11.8 dB	0.257	0.092
$BSA$	-0.263	2.57	0.744	-0.196	12.8	0.229	0.060
$C$	0.056	4.23	0.614	0.034	21.2	0.087	0.005
$BAB$	-0.054	4.59	0.590	-0.032	23.0	0.071	0.004
$BSB$	-0.034	4.81	0.575	-0.020	24.0	0.063	0.002

Problem 3.8c

3.8c Recalculate amplitudes for divergence without absorption. Normalize values by letting the divergence effect of reflection $B$ be unity.

Solution

Divergence depends upon the distance traveled, not upon the traveltime. In Table 3.8e, $L$ is the distance traveled by the event in column 1 (assuming normal incidence), $F$ is the divergence factor obtained by dividing $L_{B}$ by $L$, $A_{\rm {no~div}}$ is the reflection amplitude from Table 3.8b, $A_{\rm {div}}=F\times A_{\rm {no~div}}$, and the column headed $dB$ is $A_{\rm {div}}$ expressed in decibels.

Divergence generally affects multiples less than primaries with the same traveltime because they travel at lower velocities and therefore have not gone as far. Thus, allowing for divergence, $C$ is weaker than BAB rather than slightly stronger.

Problem 3.8d

3.8d Summarize your conclusions regarding (i) the importance of multiples and (ii) the relative importance of absorption and divergence.

Solution

The 3rd column of Table 3.8f gives the attenuation because of reflectivity only and the following columns also include the effects of reflectivity changes. The 4th column shows the changes because of absorption beginning at the source, whereas the 5th and following columns reference to reflection B.

Comparing multiples with primaries involves considering interference, noting that the three multiples all have opposite polarity to the primaries. Multiples can strongly affect the wave-shape of primaries with which they interfere as well as being confused as primaries. As noted earlier, absorption and divergence effects for multiples are different than for primaries because of differences in the distances traveled.

Table 3.8e. Effect of divergence.

Event	$L\left(m\right)$	$F$	$A_{\rm {no~div}}$	$A_{\rm {div}}$	dB
${\textit {B}}$	1210	1.000	0.359	0.359	0.0
${\textit {BSA}}$	1230	0.984	–0.263	–0.259	–2.8
${\textit {C}}$	2810	0.431	0.056	0.024	–23.8
${\textit {BAB}}$	2410	0.502	–0.054	–0.027	–23.5
${\textit {BSB}}$	2430	0.498	–0.034	–0.017	–26.6

Note: Minus sign on amplitudes indicates 180° phase shift.

Table 3.8f Effects of absorption and divergence.

Event	Time	Reflect. only	75 Hz abs, no div	75 Hz abs, no div (ref B)	Div, no abs (ref B)	Abs and Div (ref B)
B	0.517 s	0 dB	$-11.8dB$	0 dB	0 dB	0 dB
BSA	0.550	$-2.7$	$-12.8$	$-1.0$	$-2.8$	$-15.6$
C	1.017	$-16.1$	$-21.2$	$-9.4$	$-23.8$	$-45.0$
BAB	1.017	$-16.5$	$-23.0$	$-11.2$	$-22.5$	$-45.0$
BSB	1.050	$-20.5$	$-24.0$	$-12.2$	$-26.6$	$-50.6$

Divergence is more important than absorption for early arrival times, whereas the opposite is true for longer arrival times. This effect is not well illustrated by this problem.

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Also in this chapter

• General form of Snell’s law
• Reflection/refraction at a solid/solid interface and displacement of a free surface
• Reflection/refraction at a liquid/solid interface
• Zoeppritz’s equations for incident SV- and SH-waves
• Reinforcement depth in marine recording
• Complex coefficient of reflection
• Reflection and transmission coefficients
• Reflection/transmission coefficients at small angles and magnitude
• Magnitude
• AVO versus AVA and effect of velocity gradient
• Variation of reflectivity with angle (AVA)

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