# Calculation of inverse filters

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 9.18

Assuming that the signature of an air-gun array is a unit impulse and that the recorded wavelet after transmission through the earth is ${\displaystyle [-12,\;-4,\;+3,\;+1]}$, find the inverse filter that will remove the earth filtering. How many terms should the filter include?

### Solution

The inverse filter ${\displaystyle i_{t}}$ (see problem 9.7) is a filter that will restore the source impulse, i.e.,

{\displaystyle {\begin{aligned}g_{t}*i_{t}=\delta _{t},\end{aligned}}}

or in the frequency domain where ${\displaystyle i_{t}\leftrightarrow I(z)}$,

{\displaystyle {\begin{aligned}G(z)I(z)=1.\end{aligned}}}

Thus,

 {\displaystyle {\begin{aligned}I(z)=1/G(z)=1/[-12-4z+3z^{2}+z^{3}]\\=-{\frac {1}{12}}\left[1+{\frac {z}{3}}-{\frac {z^{2}}{4}}-{\frac {z^{3}}{12}}\right]^{-1}=-{\frac {1}{12}}(1+A)^{-1},\end{aligned}}} (9.18a)

where ${\displaystyle A={\frac {z}{3}}-{\frac {z^{2}}{4}}-{\frac {z^{3}}{12}}}$.

Since ${\displaystyle z}$ has the magnitude ${\displaystyle |z|=1}$, the magnitude of ${\displaystyle A<1}$ for all values of ${\displaystyle z}$, and we can expand equation (9.18a) using the binomial theorem [see equation (4.1b)] and Sheriff and Geldart, 1995, equation (15.43):

 {\displaystyle {\begin{aligned}I(z)=-{\frac {1}{12}}(1-A+A^{2}-A^{3}+A^{4}-\ldots ).\end{aligned}}} (9.18b)

We first find ${\displaystyle (1-A+A^{2}-A^{3}+A^{4})}$ neglecting powers higher than ${\displaystyle z^{4}}$:

{\displaystyle {\begin{aligned}-A&=-0.3333z+0.2500z^{2}+0.0833z^{3},\\A^{2}&=+0.1111z^{2}-0.1667z^{3}+0.0069z^{4},\\-A^{3}&=-0.0370z^{3}+0.0833z^{4},\\A^{4}&=+0.0123z^{4}\\\mathrm {Sum} &=-0.3333z+0.3611z^{2}-0.1204z^{3}+0.1025z^{4}\\I(z)&=-{\frac {1}{12}}(1-A+A^{2}-A^{3}+A^{4})\\&=-{\frac {1}{12}}(1-0.3333z+0.3611z^{2}-0,1204z^{3}+0.1025z^{4}).\end{aligned}}}

We can verify the accuracy of ${\displaystyle I(z)}$ by multiplying ${\displaystyle G(z)}$ by ${\displaystyle I(z)}$. We have

{\displaystyle {\begin{aligned}I(z):1-0.3333z+0.3611z^{2}-0,1204z^{3}+0.1025z^{4},\\G(z):1+0.3333z-0.2500z^{2}-0.0833z^{3}\\1-0.3333z+0.3611z^{2}-0.1204z^{3}+0.1025z^{4}\\+0.3333z-0.1111z^{2}+0.1204z^{3}-0.0401z^{4}+0.0342z^{5}\\-0.2500z^{2}+0.0833z^{3}-0.0903z^{4}+0.0301z^{5}-0.0256z^{6}\\{-0.0833z^{3}+0.0278z^{4}-0.0301z^{5}+0.0100z^{6}-0.0085z^{7}}\\{1+0+0+0-0.0001z^{4}+0.0342z^{5}-0.0156z^{6}-0.0085z^{7}}.\end{aligned}}}

We see that the inverse filter is exact as far as the term ${\displaystyle z^{3}}$ and terms for higher powers are small. The overall effect is to create a small tail whose energy is 0.00149 or 0.1%.

To determine how the accuracy depends on the number of terms used in ${\displaystyle I(z)}$, we observe the effect on the product ${\displaystyle I(z)G(z)}$ as we successively drop high powers in ${\displaystyle I(z)}$. Dropping the ${\displaystyle z^{4}}$ term in ${\displaystyle I(z)}$ yields the product ${\displaystyle 1-0.1026z^{4}+0.0100z^{6}}$ and the energy of the tail is now 0.01063 or 1.1%. If we want accuracy of at least 1%, we must therefore retain the ${\displaystyle z^{4}}$ term.

If we also delete the ${\displaystyle z^{3}}$ term in ${\displaystyle I(z)}$ [but not in ${\displaystyle G(z)}$] the product becomes ${\displaystyle 1+0.1204z^{3}-0.0625z^{4}-0.0100z^{5}}$ and the energy of the tail is 0.01850 or 1.8%. If we go one step further and drop the ${\displaystyle z^{2}}$ term in ${\displaystyle I(z)}$, we get ${\displaystyle 1-0.3611z^{2}+0.0278z^{4}}$ and the energy of the tail is 0.13107 or 13.1%.