# 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 $\displaystyle V_{S} \le 6\ {\hbox{km/s}}$ and the noise arriving from the north in the band $\displaystyle V_{N} \le 3\ {\hbox{km/s}}$ , 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{align} V_{a} =V/\sin\alpha , \end{align}

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{align} V_{a} =f\lambda _{a} =2\pi f/\kappa_{a} =\omega /\kappa_{a}, \end{align} (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{align} \frac{{\hbox{d}}\omega}{{\hbox{d}}{\kappa_{a}}} =\frac{{\hbox{d}}f}{{\hbox{d}}k_{a}} =V_{a}. \end{align} (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{align} 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{align}

The two-dimensional Fourier transform is defined by the equation

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

the inverse transform being

 \displaystyle \begin{align} 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{align} (9.25d)

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

 \displaystyle \begin{align} 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{align} (9.25e)

In digital form, equation (9.25e) becomes

 \displaystyle \begin{align} h_{xt} =g_{xt} *f_{xt} =\mathop{\sum}\limits_{m}^{} \mathop{\sum}\limits_{n}^{} g_{mn} \ f_{x-m,t-n}. \end{align} (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{align} f_{N} =1/2\Delta _{t} ,\; \omega _{N} =\pi /\Delta _{t}, \end{align}

and

\displaystyle \begin{align} \kappa_{N} =\pi /\Delta _{x}. \end{align}

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{align} F (\kappa_{a} ,\; \omega)=+1 ,\quad |\kappa_{a} |\le \kappa_{N}\quad \mathrm{and}\quad |\omega |\le \omega _{N} ,\\ =0,\quad |\kappa_{a} |\ge \kappa_{N}\quad \mathrm{and/or}\quad |\omega |\ge \omega _{N} , \end{align}

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