# Autocorrelation

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 9.21a

Show that the autocorrelation $\phi _{gg}(i-j)$ is given by

 {\begin{aligned}\phi _{gg}(i-j)=\mathop {\sum } \limits _{t}^{}g_{t-j}g_{t-i}.\end{aligned}} (9.21a)

### Solution

The autocorrelation of $g_{t}$ is given by equation (9.8e):

{\begin{aligned}\phi _{gg}(\tau )=\mathop {\sum } \limits _{k}^{}g_{k}\ g_{k+\tau }.\end{aligned}} If we equate $g_{t-j}$ in equation (9.21a) to $g_{k}$ in equation (9.8e), we have $k=t-j$ , so equation (9.21a) becomes

{\begin{aligned}\mathop {\sum } \limits _{k}^{}g_{k}\ g_{k+i-j}=\phi _{gg}(i-j).\end{aligned}} ## Problem 9.21b

Calculate $\phi _{ff}$ for the right triangle shown in Figure 9.21a.

{\begin{aligned}f(t)=0,\qquad t\leq 0,\\=(1-t)\;,\quad 0\leq t\leq 1,\\=0,\qquad t\geq 1.\end{aligned}} ### Solution

Let $t=OA$ , $\tau =DO=BC$ , $AB=AF=1-\tau -t$ , $AG=1-t$ :

{\begin{aligned}\phi _{ff}(\tau )={\int }_{-\infty }^{+\infty }f(t)f(t+\tau ){\rm {d}}t={\int }_{0}^{B}AG\times AF\ {\rm {d}}t\\={\int }_{}^{1-\tau }(1-t)(1-\tau -t){\rm {d}}t={\int }_{0}^{1-\tau }[(1-t)^{2}-\tau (1-t)]{\rm {d}}t\\=\left[(-1/3+t-t^{2}+t^{3}/3){\frac {\tau }{2}}-\tau t+\tau t^{2}/2\right]|_{0}^{1-\tau }\\=1/3-\tau /2+\tau ^{3}/6.\\\end{aligned}} Figure 9.21a shows that a displacement of $-\tau$ gives the same result.