# Difference between revisions of "The 2-D Fourier transform"

Series Investigations in Geophysics Öz Yilmaz http://dx.doi.org/10.1190/1.9781560801580 ISBN 978-1-56080-094-1 SEG Online Store

Multichannel processing operations can be loosely defined as those that must operate on several data traces, simultaneously. Multichannel processes can be useful in discriminating against noise and enhancing signal on the basis of a criterion that can be distinguished from trace to trace, such as dip or moveout. The 2-D Fourier transform is a basis for both analysis and implementation of multichannel processes.

Consider the six zero-offset sections in Figure 1.2-1. The trace spacing is 25 m with 24 traces per section. All have monochromatic events with 12-Hz frequency, but with dips that vary from 0 to 15 ms/trace. From the discussion on the 1-D Fourier transform, we know about frequency, particularly temporal frequency, or the number of cycles per unit time. This is the Fourier dual for the time variable. However, a seismic wavefield is not only a function of time, but also a function of a space variable (offset or midpoint axis). The Fourier dual for the space variable is defined as spatial frequency, which is the number of cycles per unit distance, or wavenumber. Just as the temporal frequency of a given sinusoid is determined by counting the number of peaks within a unit time, say 1 s, the wavenumber of a dipping event is determined by counting the number of peaks within a unit distance, say 1 km, along the horizontal direction. Just as the temporal Nyquist frequency is defined as in equation (1), the Nyquist wavenumber is defined as

 $k_{Nyq}={\frac {1}{2\Delta x}},$ (4)

where Δx is the spatial sampling interval. For all of the sections in Figures 1.2-1 through 1.2-6, the Nyquist wavenumber is 20 cycles/km, since the trace interval is 25 m.

To compute the wavenumber that is associated with the section corresponding, say, to the 15-ms/trace dip in Figure 1.2-1, follow a peak or trough across the section. First compute the total time difference along the selected peak or trough across the section:

$\left({23\;{\text{traces}}}\right)\left({15\;{\text{ms/trace}}}\right)=\;345\;{\text{ms}}.$ Then convert this to number of cycles by dividing by the (temporal) period:

${\frac {345\;{\text{ms}}}{\left({1000\;{\text{ms/s}}}\right)\left({12\;{\text{cycles/s}}}\right)}}=4.14\;{\text{cycles}}.$ The spatial extent of the section is 575 m; therefore, the wavenumber associated with the 15-ms/trace dip and the 12-Hz frequency is

${\frac {4.14\;{\text{cycles}}}{0.575\;{\text{km}}}}=7.2\;{\text{cycles/km}}.$ To continue this discussion, we will map these sections to the plane of temporal frequency versus spatial wavenumber, then look at two quadrants of this plane. The following convention will be used: Events with downdip to the right are assigned positive dip, while events with updip to the right are assigned negative dip. Additionally, positive dips map onto the right quadrant, which corresponds to positive wavenumbers, while negative dips map onto the left quadrant, which corresponds to negative wavenumbers.

The plane of frequency-wavenumber (the f – k plane) appears at the bottom of each section in Figure 1.2-1. The section with zero-dip events maps onto a single point on the frequency axis at 12 Hz. Zero dip is equivalent to zero wavenumber. The magnitude of the spike corresponds to the peak amplitude of the sinusoids that make up the traces in the section. Therefore, the f – k plane actually represents the 2-D amplitude spectrum of the section in the t – x domain. These data have been transferred from the time-space domain to the frequency-wavenumber domain. This process is described mathematically by the 2-D Fourier transform.

There is a practical relationship between the four variables: time-space (t – x), and their Fourier duals, frequency-wavenumber (f – k). Measure the inverse of the stepout in the 15-ms/trace section in Figure 1.2-1 by following a peak, trough, or zero crossing from trace to trace. Stepout is defined as the slope Δtx. In this case, the inverse of the stepout is

${\frac {\Delta x}{\Delta t}}={\frac {0.575\;{\text{km}}}{0.345\;{\text{s}}}}=1.67\;{\text{km/s}}.$ Now, compute the ratio:

${\frac {f}{k}}={\frac {12\;{\text{cycles/s}}}{7.2\;{\text{cycles/km}}}}=1.67\;{\text{km/s}}.$ From this, the inverse of the stepout measured in the t – x domain along a constant phase is equal to the ratio of the frequency to the wavenumber associated with the event

 ${\frac {\Delta x}{\Delta t}}={\frac {f}{k}}.$ (5)

Therefore, while retaining fixed stepout, doubling the frequency means doubling the wavenumber.

Note that all sections in Figure 1.2-1 have the same frequency component. However, from 0 to 15 ms/trace, the number of peaks increases horizontally across each section. That is, for a given frequency, higher dips are assigned to higher wavenumbers, as seen on the f – k plots.