Apparent-velocity filtering

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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 $.

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