# Apparent-velocity 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 9.25a

On a north-south line, the noise arriving from the south is mainly in the band $V_{S}\leq 6\ {\hbox{km/s}}$ and the noise arriving from the north in the band $V_{N}\leq 3\ {\hbox{km/s}}$ , where $V_{S}$ and $V_{N}$ are apparent velocities. Given that $\Delta x=50\ {\hbox{m}}$ , sketch an $f-k$ (or $\omega -\kappa$ ) plot.

### Background

Apparent velocity is defined in problem 4.2d by the equation

{\begin{aligned}V_{a}=V/\sin \alpha ,\end{aligned}} where $\alpha$ is the angle of approach. Using the relation $V=f\lambda =\omega /\kappa$ , where the angular frequency $\omega$ equals $2\pi f$ and the angular wavenumber $\kappa$ is $2\pi /\lambda$ . we obtain the relation

 {\begin{aligned}V_{a}=f\lambda _{a}=2\pi f/\kappa _{a}=\omega /\kappa _{a},\end{aligned}} (9.25a)

$\lambda _{a}$ being the apparent wavelength and $\kappa _{a}=2\pi /\lambda _{a}$ is the apparent wavenumber. Writing this equation as $\omega =V_{a}{\kappa _{a}}$ , we have

 {\begin{aligned}{\frac {{\hbox{d}}\omega }{{\hbox{d}}{\kappa _{a}}}}={\frac {{\hbox{d}}f}{{\hbox{d}}k_{a}}}=V_{a}.\end{aligned}} (9.25b)

The slope of a line on an $f-k$ plot gives the apparent velocity.

The one-dimensional Fourier transform relation was defined in equation (9.3c.d) as

{\begin{aligned}G(\omega )={\int }_{-\infty }^{+\infty }g(t)e^{-{\hbox{j}}\omega t}{\hbox{d}}t,\;g(t)=(1/2\pi ){\int }_{-\infty }^{+\infty }G(\omega )e^{{\hbox{j}}\omega t}{\hbox{d}}\omega .\end{aligned}} The two-dimensional Fourier transform is defined by the equation

 {\begin{aligned}G(\kappa ,\;\omega )=\int \int _{-\infty }^{+\infty }g(x,\;t)e^{-{\hbox{j}}(\kappa x+\omega t)}{\hbox{d}}x\ {\hbox{d}}t,\end{aligned}} (9.25c)

the inverse transform being

 {\begin{aligned}g\;(x,\;t)=(1/2\pi )^{2}\int \int _{-\infty }^{+\infty }G(\kappa ,\;\omega )\;e^{{\hbox{j}}(\kappa x+\omega t)}{\hbox{d}}{\kappa }\ {\hbox{d}}\omega .\end{aligned}} (9.25d)

One-dimensional convolution [equation (9.2a)] becomes in two dimensions

 {\begin{aligned}h(x,\;t)=g(x,\;t)*f(x,\;t)=\int {\int }_{-\infty }^{+\infty }g(\sigma ,\;\tau )f(x-\sigma ,\;t-\tau )\,\mathrm {d} \sigma \,\mathrm {d} {\tau }.\end{aligned}} (9.25e)

In digital form, equation (9.25e) becomes

 {\begin{aligned}h_{xt}=g_{xt}*f_{xt}=\mathop {\sum } \limits _{m}^{}\mathop {\sum } \limits _{n}^{}g_{mn}\ f_{x-m,t-n}.\end{aligned}} (9.25f)

In applying apparent-velocity filtering, we deal with data that are sampled in both time and space. Corresponding to the temporal Nyquist frequency $f_{N}=1/2\Delta$ in time sampling [see equation (9.4c)], spatial sampling involves a spatial Nyquist “frequency” = Nyquist wave-number = $1/(2\Delta x)$ . Using the symbols $\Delta _{t}$ and $\Delta _{x}$ for the time and spatial sampling intervals, the Nyquist frequencies are

{\begin{aligned}f_{N}=1/2\Delta _{t},\;\omega _{N}=\pi /\Delta _{t},\end{aligned}} and

{\begin{aligned}\kappa _{N}=\pi /\Delta _{x}.\end{aligned}}  Figure 9.25a.  An $f-k$ plot.

Apparent-velocity filters can be designed to remove pie-sliced portions, e.g., a filter in Figure 9.25a to remove $<6\ {\hbox{km/s}}$ would remove the noise between the 6 km/s line and the $x$ -axis.

### Solution

Taking north as the positive direction, on an $f-k$ plot, $V_{a}=6$ km/s is a straight line through the origin with slope $df/dk_{a}=6$ km/s and the noise from the south is mainly between this line and the $+\kappa _{a}$ -axis. Similarly the noise from the north lies between the $-\kappa _{a}$ -axis and a straight line with slope –3 km/s extending from the origin. The $f-k$ plot is shown in Figure 9.25a for the Nyquist wavenumber $k_{N}=(1/2\times 50){\hbox{m}}^{-1}=10/{\hbox{km}}$ , or $\kappa _{N}=\pi /50\ {\hbox{m}}^{-1}=62.8/{\hbox{km}}$ .

If Figure 9.25a were rolled into a vertical cylinder by matching $-\kappa _{N}$ with $+\kappa _{N}$ , it can be seen that the alias slopes are simply extensions of the apparent velocity lines.

## Problem 9.25b

Repeat for $\Delta x=25\ {\hbox{m}}$ .

### Solution

The only change from part (a) is that the Nyquist frequency is now double that in (a), that is, $\kappa _{N}=2\pi /2\times 25\ {\hbox{m}}^{-1}=126\ \mathrm {km} ^{-1}$ . This will move the alias lines upward in Figure 9.25a.

## Problem 9.25c

Calculate a filter that will prevent aliasing in both the wavenumber and time domains for parts (a) and (b) [see equation (9.25c)].

### Solution

We require a filter $f(x,\;t)$ whose transform is defined by the equations

{\begin{aligned}F(\kappa _{a},\;\omega )=+1,\quad |\kappa _{a}|\leq \kappa _{N}\quad \mathrm {and} \quad |\omega |\leq \omega _{N},\\=0,\quad |\kappa _{a}|\geq \kappa _{N}\quad \mathrm {and/or} \quad |\omega |\geq \omega _{N},\end{aligned}} $\kappa _{N}$ and $\omega _{N}$ being the Nyquist wavenumber and Nyquist frequency. Transforming $F(\kappa _{a},\;\omega )$ to the $x-t$ domain gives

{\begin{aligned}f\;(x,\;t)=(1/2\pi )^{2}\int _{-\kappa _{N}}^{+\kappa _{N}}\int _{-\omega _{N}}^{+\omega _{N}}e^{{\hbox{j}}(\kappa _{a}x+\omega t)}{\hbox{d}}{\kappa _{a}}{\hbox{d}}\omega .\end{aligned}} The recorded data are real, hence $f(x,\;t)$ must also be real, and therefore we can set the imaginary part in the integrand equal to zero. Using Euler’s formula (Sheriff and Geldart, 1995, problem 15.12a), we get

{\begin{aligned}f\;(x,\;t)=(1/2\pi )^{2}\int _{-\kappa _{N}}^{+\kappa _{N}}\int _{-\omega _{N}}^{+\omega _{N}}\cos(\kappa _{a}x+\omega t){\hbox{d}}{\kappa _{a}}{\hbox{d}}\omega .\end{aligned}} We integrate first with respect to $\omega$ and obtain

{\begin{aligned}f(x,\;t)=(1/2\pi )^{2}{\int }_{-\kappa _{N}}^{+\kappa _{N}}\left[{\frac {{\rm {\;sin\;}}(\kappa _{a}x+\omega t)}{t}}|_{-\omega _{N}}^{+\omega _{N}}\right]{\hbox{d}}{\kappa _{a}}\\=(1/4\pi ^{2}t){\int }_{-\kappa _{N}}^{+\kappa _{N}}[{\rm {\;sin\;}}(\kappa _{a}x+\omega _{N}t)-{\rm {\;sin\;}}(\kappa _{a}x-\omega _{N}t)]{\hbox{d}}{\kappa _{a}}\\=(1/4\pi ^{2}t)\left({\frac {-{\rm {\;cos\;}}(\kappa _{a}x+\omega _{N}t)}{x}}+{\frac {{\rm {\;cos\;}}(\kappa _{a}x-\omega _{N}t)}{x}}\right)|_{-\kappa _{N}}^{+\kappa _{N}}\\=(1/2\pi ^{2}xt)[{\rm {\;sin\;}}(\kappa _{a}x){\rm {\;sin\;}}(\omega _{N}t)|_{-\kappa _{N}}^{+\kappa _{N}}\\=(1/\pi ^{2}xt)[{\rm {\;sin\;}}(\kappa _{N}x){\rm {\;sin\;}}(\omega _{N}t)]\\=(\kappa _{N}\omega _{N}/\pi ^{2})\mathrm {sinc} \ \kappa _{N}x\ \mathrm {sinc} \ \omega _{N}t=(1/\Delta _{x}\Delta _{t})\ \mathrm {sinc} \ \kappa _{N}x\ \mathrm {sinc} \ \omega _{N}t_{j},\end{aligned}} where sinc $x={\rm {\;sin\;}}x/x$ .