Inverse filter to remove ghosting and recursive filtering

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

Problem 9.19a

A source is located 8 m below the base of the LVL. Given that the velocity below the LVL is ${\displaystyle V_{H}=2000}$ m/s, that ${\displaystyle V_{W}=300}$ m/s in the LVL, the densities are ${\displaystyle \rho _{H}=2.3\ {\hbox{g/cm}}^{3}}$, ${\displaystyle \rho _{W}=1.8g/{\hbox{cm}}^{3}}$, the sampling interval ${\displaystyle \Delta =4}$ ms, and that the reflected signal is ${\displaystyle [6,\;-7,\;-2.8,\;5.6,\;-1.6]}$, find the original wavelet using the inverse filter ${\displaystyle 1/(1+Rz^{n})}$ [see the discussion of equation (9.17a)]

Background

The LVL is discussed in problem 4.16, inverse filters in problem 9.7, ghosting as a filter in problem 9.17.

Following the discussion in problem 9.17, we write

{\displaystyle {\begin{aligned}H(z)=G(z)(1+Rz^{n})=G(z)F(z),\end{aligned}}}

where ${\displaystyle G(z)}$ is the original (unghosted) signal, ${\displaystyle H(z)}$ is the ghosted signal, the filter ${\displaystyle F(z)}$ is ${\displaystyle (1+Rz^{n})}$ where ${\displaystyle R}$ is the reflection coefficient at the base of the LVL, and ${\displaystyle z^{n}}$ accounts for the delay of the ghost relative to ${\displaystyle G(z)}$.

The inverse filter ${\displaystyle I(z)}$ is

 {\displaystyle {\begin{aligned}I(z)=1/F(z)=(1+Rz^{n})^{-1}.\end{aligned}}} (9.19a)

The original signal ${\displaystyle G(z)}$ is given by

{\displaystyle {\begin{aligned}G(z)=H(z)/(1+Rz^{n}),\\G(z)(1+Rz^{n})=H(z),\\G(z)=H(z)-Rz^{n}G(z).\\\end{aligned}}}

In the time domain this becomes

 {\displaystyle {\begin{aligned}g_{t}=h_{t}-Rg_{t-n}.\end{aligned}}} (9.19b)

This process is called recursive filtering or feedback filtering.

Solution

The reflection coefficient involved in ghosting at the base of the LVL by a wave approaching from below is given by equation (3.6a), the value being

{\displaystyle {\begin{aligned}R={\frac {\rho _{W}V_{W}-\rho _{H}V_{H}}{\rho _{W}V_{W}+\rho _{H}V_{H}}}={\frac {1.8\times 0.3-2.3\times 2.0}{1.8\times 0.3+2.3\times 2.0}}=-0.79.\end{aligned}}}

The filter term is ${\displaystyle (1+Rz^{n})}$ which now becomes ${\displaystyle (1-0.79z^{n})}$. To get ${\displaystyle n}$, we have for the two-way traveltime from the source to the base of the LVL ${\displaystyle 2\times 8/2000=8}$ ms ${\displaystyle =2\Delta }$ if ${\displaystyle \Delta =4}$ ms. Thus,

{\displaystyle {\begin{aligned}F(z)=(1-0.79z^{2}),\end{aligned}}}

and the inverse filter of equation (9.19a) is

{\displaystyle {\begin{aligned}I(z)=(1-0.79z^{2})^{-1}.\end{aligned}}}

Since ${\displaystyle |0.79z^{2}|<1}$, we can expand using equation (4.1b) and get

{\displaystyle {\begin{aligned}I(z)&=1+(0.79z^{2})+(0.79z^{2})^{2}+(0.79z^{2})^{3}+\ldots \\&=1+0.79z^{2}+0.62z^{4}+0.39z^{6}+\ldots .\end{aligned}}}

Omitting inverse filter terms higher than ${\displaystyle z^{4}}$, we get

{\displaystyle {\begin{aligned}G(z)=H(z)I(z)\\=(6-7z-2.8z^{2}+5.6z^{3}+1.6z^{4})(1+0.79z^{2}+0.62z^{4}).\end{aligned}}}

We get ${\displaystyle G(z)}$ by multiplication as follows

{\displaystyle {\begin{aligned}6-7z-2.8z^{2}+5.6z^{3}-1.6z^{4}\\4.7z^{2}-5.5z^{3}-2.2z^{4}+4.4z^{5}-1.3z^{6}\\+3.7z^{4}-4.3z^{5}-1.7z^{6}+3.5z^{7}-1.0z^{8}\\G(z)=6-7z+1.9z^{2}+0.1z^{3}-0.1z^{4}+0.1z^{5}-3.0z^{6}+3.5z^{7}-1.0z^{8}.\end{aligned}}}

Assuming the original wavelet was ${\displaystyle g_{t}=[6,\;-7,\;2]}$, the third term is close but not exact, the next three terms are quite small, but the final three terms are sizable and will cause the highly undesirable result of creating the appearance of a fictitious event.

We can verify this answer by adding ${\displaystyle g_{t}}$ and ${\displaystyle -0.79g_{t}}$ as follows:

{\displaystyle {\begin{aligned}6,\;-7,\;2.\\-4.7,5.5,-1.6\\g_{t}=6,-7,-2.7,5.5,-1.6.\\\end{aligned}}}

Problem 9.19b

Find the original wavelet by recursive filtering.

Solution

Equation (9.19b) gives

{\displaystyle {\begin{aligned}g_{t}=h_{t}-Rg_{t-2}=h_{t}+0.79g_{t-2},\end{aligned}}}

that is, each element of ${\displaystyle g_{t}}$ is equal to the corresponding element of ${\displaystyle h_{t}}$ plus 0.79 times the element of ${\displaystyle g_{t}}$ that ocurred two time units earlier. Since ${\displaystyle h_{t}=0}$, ${\displaystyle t<0}$, equation (9.17a) shows that ${\displaystyle g_{t}}$ must also be causal, that is, ${\displaystyle g_{t}=0}$, ${\displaystyle t<0}$. Thus,

{\displaystyle {\begin{aligned}g_{0}=h_{0}=6,\\g_{1}=h_{1}=7,\\g_{2}=h_{2}+0.79g_{0}=-2.8+4.7=1.9,\\g_{3}=h_{3}+0.79g_{1}=5.6-5.5=0,\;1,\\g_{4}=h_{4}+0.79g_{2}=-1.6+1.5=-0,\;1,\\{\mbox{so}}\qquad \qquad g_{t}\approx [6,\;-7,\;1.9,\;0.1,\;-0.1].\\\end{aligned}}}

This result seems to be better than that obtained in part (a) only because the calculation terminated at ${\displaystyle g_{4}}$. However, if we apply the rule to ${\displaystyle g_{5}}$ and subsequent elements, we would see a small tail added.