Time & space sampling; aliasing

Physics prerequisites for acquisition

Learning objectives

State the Nyquist limits in time, $f_{\mathrm N} = 1/(2\Delta t)$ , and in space, $k_{\mathrm N} = 1/(2\Delta x)$
Predict the frequency $f'$ at which a signal above $f_{\mathrm N}$ is recorded
Explain why a dipping event aliases along a receiver line above $f_{\mathrm a} = 1/(2p\Delta x)$ , and where its energy goes
Choose $\Delta t$ and $\Delta x$ that keep the signal band at the steepest dip of interest

Every digital seismic record is sampled twice: in time, every $\\Delta t$ at each receiver, and in space, every $\\Delta x$ along the line. Each sampling keeps only a band. Samples every $\\Delta t$ hold frequencies below the Nyquist frequency $f\_{\\mathrm N} = 1/(2\\Delta t)$ ; traces every $\\Delta x$ hold wavenumbers below $k\_{\\mathrm N} = 1/(2\\Delta x)$ . Anything beyond folds back into the band, where its samples are identical to those of a real signal. This is aliasing, and nothing done to the data afterwards can undo it.

Temporal aliasing

A sinusoid of frequency $f$ sampled at $f\_{\\mathrm s} = 1/\\Delta t$ gives exactly the samples of one at $f' = |f - m f\_{\\mathrm s}|$ , where $m$ is the whole number nearest $f/f\_{\\mathrm s}$ , so $f'$ always lies between 0 and $f\_{\\mathrm N}$ . At $\\Delta t$ = 4 ms, $f\_{\\mathrm N}$ = 125 Hz and a 170 Hz component is recorded as 80 Hz. Because the fold cannot be undone, every recording system passes the analog signal through an anti-alias filter before its sampler, cutting at roughly 0.8 $f\_{\\mathrm N}$ . In practice, then, choosing $\\Delta t$ chooses the top of the recorded band: about 100 Hz at 4 ms and 200 Hz at 2 ms.

Spatial aliasing

In space the only filter ahead of the sampler is the receiver group itself, and it removes only part of the steep energy. A dipping event reaches each receiver a little later than the last, with time dip $p = dt/dx$ . At frequency $f$ that is a wavenumber $k = pf$ along the line, so the higher the frequency, the higher the wavenumber. For a wave reaching the surface at angle $\\theta$ through velocity $V$ , as in a shot record, $p = \\sin\\theta/V$ . For a reflector dipping at $\\theta$ in a stack or zero-offset section, where time is two-way, $p = 2\\sin\\theta/V$ . The event keeps its true wavenumber only up to the frequency where $pf$ reaches $k\_{\\mathrm N}$ :

f\_{\\mathrm a} = \\frac{1}{2p\\,\\Delta x} = \\frac{V}{4\\,\\Delta x\\sin\\theta}\\ \\text{(stack)}, \\qquad \\frac{V}{2\\,\\Delta x\\sin\\theta}\\ \\text{(shot record)}

Above $f\_{\\mathrm a}$ its energy folds to false dips; between $f\_{\\mathrm a}$ and $2f\_{\\mathrm a}$ the dip is reversed. The traces themselves can look innocent, because the eye follows an isolated event from trace to trace. The damage appears when processing treats the traces as samples of a continuous wavefield: f-k filtering, trace interpolation, DMO and migration. In the figure, drag the trace spacing and watch plate (b), which rebuilds the wavefield between the traces the way those processes do, and plate (c), where the folded energy crosses to the wrong side of $k = 0$ .

The figure opens on a 30 Hz Ricker wavelet on a 30° dip under 2000 m/s, in a stack. Its band reaches $f\_{\\max}$ = 66 Hz, but traces every 25 m keep that dip only to 40 Hz: 21 % of the wavelet’s energy folds to false dips, and (b) rebuilds the event as beads with ghosts between the traces. The largest spacing that keeps the whole band is 15 m: 27 traces on the 400 m line instead of 17.

Why this controls bin size

Turn the condition around and it becomes the design rule. To keep every frequency up to $f\_{\\max}$ at dips up to $\\theta\_{\\max}$ , the spacing must satisfy

\\Delta x \\le \\frac{V}{4 f\_{\\max}\\sin\\theta\_{\\max}} = \\frac{\\lambda\_{\\min}}{4\\sin\\theta\_{\\max}}

for a CMP bin, and $V/(2 f\_{\\max}\\sin\\theta\_{\\max})$ for receivers in a shot record, with $V$ the velocity just above the steepest reflector. It is the shortest wavelength, $\\lambda\_{\\min} = V/f\_{\\max}$ , that sets the limit, not the dominant one. The two rules agree, because CMP bins are half the station interval: 50 m stations give 25 m bins. With $V$ = 3000 m/s, $f\_{\\max}$ = 60 Hz and a 30° dip, the bin must be 25 m or less, which is where the familiar 25 m bin of Part 3 comes from. Steeper dips, slower rock or a higher $f\_{\\max}$ all shrink it, and every halving of the bin doubles the channel count along that direction.

References

Bracewell, R. N. (1999). The Fourier Transform and Its Applications (3rd ed.). McGraw-Hill.
Shannon, C. E. (1949). Communication in the presence of noise. Proceedings of the IRE, 37(1), 10–21.
Vermeer, G. J. O. (1990). Seismic Wavefield Sampling. SEG Geophysical Monograph 4.
Vermeer, G. J. O. (2002). 3-D Seismic Survey Design. SEG Geophysical References 12.
Yilmaz, Ö. (2001). Seismic Data Analysis: Processing, Inversion, and Interpretation of Seismic Data (2 vols.). SEG Investigations in Geophysics 10.