# Amplitude/energy of reflections and multiples

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

## Problem 3.8a

Assume horizontal layering (as shown in Figure 3.8a), a source just below interface ${\displaystyle A}$, and a geophone at the surface. Calculate (ignoring absorption and divergence) the relative amplitudes and energy densities of the primary reflections from ${\displaystyle B}$ and ${\displaystyle C}$ and the multiples ${\displaystyle BSA}$, ${\displaystyle BAB}$, and ${\displaystyle BSB}$ (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 ${\displaystyle A=}$ ${\displaystyle A_{0}e^{-\eta x}}$ where the amplitude decreases from ${\displaystyle A_{0}}$ to ${\displaystyle A}$ over a distance ${\displaystyle x}$; the absorption coefficient ${\displaystyle \eta }$ is often expressed in terms of per wavelength, ${\displaystyle \lambda .}$

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

Table 3.8a. Reflection and transmission coefficients.
Interface ${\displaystyle Z}$ ${\displaystyle R^{*}}$ ${\displaystyle T\downarrow }$ ${\displaystyle T\uparrow }$ ${\displaystyle E_{R}}$ ${\displaystyle E_{T}}$
S 1.000 0.000 0.000 1.000 0.000
Layer 1 0.870
${\displaystyle {\textit {A}}}$ 0.733 0.267 1.733 0.537 0.463
Layer 2 5.640
${\displaystyle {\textit {B}}}$ 0.207 0.793 1.207 0.043 0.957
Layer 3 8.576
${\displaystyle {\textit {C}}}$ 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 ${\displaystyle B}$ and neglecting absorption and divergence, we arrive at the following values:

Reflection ${\displaystyle B}$:

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

Reflection ${\displaystyle C}$

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

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

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

#### Multiple BSA

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

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

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

#### Multiple BAB

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

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

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

#### Multiple BSB

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

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

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

The results are summarized in Table 3.8b.

BSA arrives 33 ms after ${\displaystyle B}$ (one period for a 33-Hz wave) with reversed polarity and about 75% of the amplitude and 50% of the energy of ${\displaystyle B}$, 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.

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

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

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

### Solution

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

Table 3.8c. Absorption for one-way travel in each layer.
${\displaystyle f=15}$ Hz ${\displaystyle f=75}$ Hz
Layer Velocity ${\displaystyle \Delta z}$ ${\displaystyle \lambda }$ ${\displaystyle \Delta z/\lambda }$ ${\displaystyle \eta \Delta z}$ ${\displaystyle \lambda }$ ${\displaystyle \Delta z/\lambda }$ ${\displaystyle \eta \Delta z}$
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 ${\displaystyle B}$ involves two-way travel through ${\displaystyle AB}$ and one-way travel through ${\displaystyle SA}$, hence attenuation due to absorption is ${\displaystyle 2\times 1.12+0.11=2.35\ {\rm {dB}}}$ the amplitude being decreased by the factor 0.763. For the multiple BSA we add attenuation for the extra two-way path through ${\displaystyle SA}$ to the attenuation for ${\displaystyle B}$, giving 2.57 ${\displaystyle dB}$, or an amplitude reduction of 0.744. For reflection ${\displaystyle C}$, we add to the attenuation of reflection ${\displaystyle B}$ the attenuation for the two-way travel through ${\displaystyle BC}$, giving 4.23 ${\displaystyle dB}$, and an amplitude ratio of 0.614. For the multiple BAB we get attenuation of 4.59 ${\displaystyle dB}$, an amplitude ratio of 0.590. For BSB, attenuation is 4.81 ${\displaystyle dB}$ and an amplitude ratio is 0.575. Attenuation for 75 Hz is 5 times that for 15 Hz because ${\displaystyle \lambda }$ is only one-fifth that for 15 Hz, hence ${\displaystyle \Delta z/\lambda \}$ 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 ${\displaystyle A}$ ${\displaystyle \mathop {\sum } \limits _{}^{}\eta \Delta z\left(15\right)}$ Ratio(15) ${\displaystyle A_{a}(15)}$ ${\displaystyle {\sum }\eta \Delta z\left(75\right)}$ Ratio(75) ${\displaystyle A_{a}(75)}$
${\displaystyle B}$ 0.359 2. 35 dB 0.763 0.274 11.8 dB 0.257 0.092
${\displaystyle BSA}$ -0.263 2.57 0.744 -0.196 12.8 0.229 0.060
${\displaystyle C}$ 0.056 4.23 0.614 0.034 21.2 0.087 0.005
${\displaystyle BAB}$ -0.054 4.59 0.590 -0.032 23.0 0.071 0.004
${\displaystyle BSB}$ -0.034 4.81 0.575 -0.020 24.0 0.063 0.002

## Problem 3.8c

Recalculate amplitudes for divergence without absorption. Normalize values by letting the divergence effect of reflection ${\displaystyle B}$ be unity.

### Solution

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

## Problem 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 ${\displaystyle L\left(m\right)}$ ${\displaystyle F}$ ${\displaystyle A_{\rm {no~div}}}$ ${\displaystyle A_{\rm {div}}}$ dB
${\displaystyle {\textit {B}}}$ 1210 1.000 0.359 0.359 0.0
${\displaystyle {\textit {BSA}}}$ 1230 0.984 –0.263 –0.259 –2.8
${\displaystyle {\textit {C}}}$ 2810 0.431 0.056 0.024 –23.8
${\displaystyle {\textit {BAB}}}$ 2410 0.502 –0.054 –0.027 –23.5
${\displaystyle {\textit {BSB}}}$ 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 ${\displaystyle -11.8dB}$ 0 dB 0 dB 0 dB
BSA 0.550 ${\displaystyle -2.7}$ ${\displaystyle -12.8}$ ${\displaystyle -1.0}$ ${\displaystyle -2.8}$ ${\displaystyle -15.6}$
C 1.017 ${\displaystyle -16.1}$ ${\displaystyle -21.2}$ ${\displaystyle -9.4}$ ${\displaystyle -23.8}$ ${\displaystyle -45.0}$
BAB 1.017 ${\displaystyle -16.5}$ ${\displaystyle -23.0}$ ${\displaystyle -11.2}$ ${\displaystyle -22.5}$ ${\displaystyle -45.0}$
BSB 1.050 ${\displaystyle -20.5}$ ${\displaystyle -24.0}$ ${\displaystyle -12.2}$ ${\displaystyle -26.6}$ ${\displaystyle -50.6}$

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.