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

### Background

Apparent velocity is defined in problem 4.2d by the equation

{\displaystyle {\begin{aligned}V_{a}=V/\sin \alpha ,\end{aligned}}}

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

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

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

 {\displaystyle {\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 ${\displaystyle f-k}$ plot gives the apparent velocity.

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

{\displaystyle {\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

 {\displaystyle {\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

 {\displaystyle {\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

 {\displaystyle {\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

 {\displaystyle {\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 ${\displaystyle f_{N}=1/2\Delta }$ in time sampling [see equation (9.4c)], spatial sampling involves a spatial Nyquist “frequency” = Nyquist wave-number = ${\displaystyle 1/(2\Delta x)}$. Using the symbols ${\displaystyle \Delta _{t}}$ and ${\displaystyle \Delta _{x}}$ for the time and spatial sampling intervals, the Nyquist frequencies are

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

and

{\displaystyle {\begin{aligned}\kappa _{N}=\pi /\Delta _{x}.\end{aligned}}}

Figure 9.25a.  An ${\displaystyle f-k}$ plot.

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

### Solution

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

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

## Problem 9.25b

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

### Solution

The only change from part (a) is that the Nyquist frequency is now double that in (a), that is, ${\displaystyle \kappa _{N}=2\pi /2\times 25\ {\hbox{m}}^{-1}=126\ \mathrm {km} ^{-1}}$. 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 ${\displaystyle f(x,\;t)}$ whose transform is defined by the equations

{\displaystyle {\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}}}

${\displaystyle \kappa _{N}}$ and ${\displaystyle \omega _{N}}$ being the Nyquist wavenumber and Nyquist frequency. Transforming ${\displaystyle F(\kappa _{a},\;\omega )}$ to the ${\displaystyle x-t}$ domain gives

{\displaystyle {\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 ${\displaystyle f(x,\;t)}$ 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

{\displaystyle {\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 ${\displaystyle \omega }$ and obtain

{\displaystyle {\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 ${\displaystyle x={\rm {\;sin\;}}x/x}$.