Derivative and integral operators

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 9.31a

The operator $\displaystyle f_t=[-1,\; +1]$ is called the “derivative operator”; explain why.

Solution

In terms of finite differences (see problem 9.29), the derivative of a function is approximately equal to the change in values between two adjacent points divided by the distance between the two points. Convolving $\displaystyle f_{t}$ with $\displaystyle g_{t}$ gives values of the finite-difference derivative of $\displaystyle g_{t}$ for different values of $\displaystyle t$ . To show this, we find the $\displaystyle (n+1)^{th}$ term of $\displaystyle h_{t} =f_{t} *g_{t}$ . From equation (9.2b), we have

\displaystyle \begin{align} h_{n} =\mathop{\sum}\limits_{k=0}^{} f_{k} g_{n-k} =[-g_{0} ,\; (-g_{1} +g_{0}),\; .\; .\; .\; (-g_{n} +g_{n-1})]. \end{align}

Figure 9.30a.  Seismic section. (i) Unmigrated; (ii) migrated.
Figure 9.30b.  Interpreted migrated seismic section.

Each term (except the first) equals $\displaystyle -\Delta g$ at $\displaystyle =n\Delta$ . Taking $\displaystyle \Delta =1$ , we see that $\displaystyle h_{n}$ is minus the finite difference derivative at $\displaystyle t=n\Delta$ .

Problem 9.31b

What is the integral operator?

Solution

The integral, $\displaystyle \int_{g_0}^{g_n}\ g(t)\,\mathrm{d}t$ , for continuous functions becomes a summation for digital functions, the equivalent of the above integral being $\displaystyle \Delta \mathop{\sum}\limits{_{g_{0}}^{g_{n}}} g_{t}$ . If we take the operator $\displaystyle f_{t} = [1, 1, 1, \ldots.., 1]$ where both $\displaystyle f_{t}$ and $\displaystyle g_{t}$ have $\displaystyle (n+1)$ terms, we obtain the following for the terms of $\displaystyle f_{t} *g_{t}$ (omitting $\displaystyle \Delta$ ):

\displaystyle \begin{align} &[(g_{0}) , (g_{0} +g_{1}) , . . . , (g_{0} +g_{1} +\cdots +g_{r}) , (g_{0} +g_{1} +\cdots +g_{n}) ,\\ &\quad (g_{1} +g_{2} +\cdots +g_{n}) , (g_{2} +g_{3} +\cdots +g_{n}) , . . . , (g_{n-1} +g_{n}) , (g_{n}) . \end{align}

The terms are equivalent to the integrals between 0 and $\displaystyle r$ for $\displaystyle 0\le r\le n$ . After the value $\displaystyle r=n$ is reached, the upper limit remains $\displaystyle n$ while the lower limit is successively 1, 2, . . . , $\displaystyle n$ . Thus we conclude that the operator [1, 1, 1, . . . , 1] is an integral operator. By taking $\displaystyle f_{t}$ with fewer terms than $\displaystyle g_{t}$ , we can get the equivalent of an integral over selected parts of $\displaystyle g_{t}$ .