# Examples

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Series Geophysical References Series Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing Enders A. Robinson and Sven Treitel 8 http://dx.doi.org/10.1190/1.9781560801610 9781560801481 SEG Online Store

Let us now discuss a model consisting of a surface interface overlying two buried interfaces, with the three interfaces separated by arbitrary two-way layer traveltimes. Let the reflection coefficients be given by a, b, c. Let the two-way traveltime between the surface and the first buried interface be S, and let the two-way traveltime between the first buried interface and the second buried interface be T. In other words, the surface reflection coefficient is ${\displaystyle {\varepsilon }_{0}=a}$, the reflection coefficient for the first buried interface is ${\displaystyle {\varepsilon }_{S}=b}$, and the reflection coefficient for the second buried interface is ${\displaystyle {\varepsilon }_{T+S}=c}$. The Z-transform of the reflectivity is

 {\displaystyle {\begin{aligned}a+bZ^{S}+cZ^{S+T}.\end{aligned}}} (30)

The right-hand side of the autocorrelation of the reflectivity is

 {\displaystyle {\begin{aligned}g_{0}+g_{S}Z^{S}+g_{T}Z^{T}+g_{S+T}Z^{S+T}=g_{0}+abZ^{S}+bcZ^{T}+acZ^{S+T}.\end{aligned}}} (31)

If we replace ${\displaystyle g_{0}}$ by 1, we obtain the feedback loop

 {\displaystyle {\begin{aligned}{1}+abZ^{S}+acZ^{T}+acZ^{S+T}.\end{aligned}}} (32)

There are three reverberations. The reverberation between the surface and the first buried interface contributes ${\displaystyle g_{S}=a\ b\ Z^{s}}$, the reverberation between the second and third buried interfaces contributes ${\displaystyle g_{T}=b\ c\ Z^{T}}$, and the reverberation between the surface and third buried interface contributes ${\displaystyle g_{T+S}=a\ c\ Z^{S+T}}$. The synthetic trace now is given by the feedforward-feedback filter (Figure 11):

Figure 11.  (a) The reflectivity for the case a = 0.8, b = –0.4, and c = 0.7. (b) The resulting impulsive synthetic trace.

 {\displaystyle {\begin{aligned}{\frac {bZ^{S}+cZ^{S+T}}{{1}+abZ^{S}+bcZ^{T}+acZ^{S+T}}}.\end{aligned}}} (33)

Suppose that ${\displaystyle S={2,}T={=5}}$. Then the synthetic trace is given by

 {\displaystyle {\begin{aligned}{\frac {bZ^{2}+cZ^{7}}{{1}+abZ^{2}+bcZ^{5}+acZ^{7}}},\end{aligned}}} (34)

which is

${\displaystyle {\rm {shot}},0,b,0,-\left(ab^{2}\right),{\rm {0,}}a^{2}b^{3},c-b^{2}c,-\left(a^{3}b^{4}\right),2ab\left(-{1}+b^{2}\right)c,a^{4}b^{5},}$

${\displaystyle 3a^{2}b^{2}\left(1-b^{2}\right)c{,\ }-\left(a^{5}b^{6}\right)-bc^{2}+b^{3}c^{2}{\ ,\ 4}a^{3}b^{3}\left(-{1+}b^{2}\right)c,}$

${\displaystyle a\left(a^{5}b^{7}-c^{2}+{4}b^{2}c^{2}-2b^{2}c^{2}\right){\ ,\ 5}a^{4}b^{4}\left(1-b^{2}\right)c,}$

 {\displaystyle {\begin{aligned}a^{2}b\left(-\left(a^{5}b^{7}\right)+{3}c^{2}-9b^{2}c^{2}+{6}b^{4}c^{2}\right){\ ,\ }b^{2}\left(-{1+}b^{2}\right)c\left(6a^{5}b^{3}-c^{2}\right),\dots .\end{aligned}}} (35)

The shot occurs at time 0, the first primary b occurs at time 2, the first-surface- first-interface multiple ${\displaystyle {\text{ – }}ab^{2}}$ occurs at time 4, the second-surface–first-interface multiple ${\displaystyle a^{2}b^{3}}$ occurs at time 6, the second primary occurs at time 7, the third-surface–first-interface multiple ${\displaystyle -\left(a^{3}b^{4}\right)}$ occurs at time 8, the peg-leg multiple ${\displaystyle 2ab\left(-{1\ +}b^{2}\right)c}$ occurs at time 9, and so on. Figure 11a shows the reflectivity for the case ${\displaystyle a\;=0.8;\;b=-0.4;\;c=0.7}$. The leading portion of the corresponding impulsive synthetic trace is shown in Figure 11b.