Attenuation and Q from acquisition view

The earth is a low-pass filter whose cut-off falls with travel time. The loss comes from intrinsic attenuation, the conversion of wave energy to heat, described by the quality factor $Q$: a wave loses a fraction of about $2\\pi/Q$ of its energy in each cycle. In real data an apparent loss from scattering by fine layering adds to it, and the two are hard to tell apart, so field values of $Q$ are usually effective values.

The law

For a wave of frequency $f$ that has travelled for time $t$ through rock of constant $Q$:

In decibels the loss is a straight line in frequency, $20\\log\_{10}e \\cdot \\pi f t/Q \\approx 8.69\\,\\pi f t/Q$ dB, so doubling $f$ or $t$ doubles the loss in decibels and squares the amplitude factor. Only the ratio $t/Q$ enters (through layered rock it is the sum of $\\Delta t/Q$ along the path, often written $t^{\*}$), so a short path through lossy rock and a long path through stiff rock look alike. A 60 Hz component after $t$ = 2 s at $Q$ = 50 keeps $e^{-7.54}$, about 0.05% of its amplitude, a loss of 65 dB; the 10 Hz component on the same path keeps $e^{-1.26}$, about 28%, a loss of 11 dB. A causal loss of this kind must also delay the low frequencies against the high ones (dispersion), so the arriving pulse is not only weaker and broader but late and lopsided.

In the figure, pick a target by its two-way time and the rock by its $Q$, and read how high in frequency the echo still stands above the noise when it gets back to the receivers. Then set the frequency your target needs, $f\_w$, and read what the source or the receivers would have to give to reach it.

Why this drives source and receiver choice

In the opening state, a target at $t$ = 2 s in rock of $Q$ = 50, the earth takes 65 dB at 60 Hz and 33 dB at 30 Hz, and a 30 Hz Ricker source against a noise floor 50 dB down leaves a usable band that ends at 43 Hz. Reaching 60 Hz needs 30 dB more signal-to-noise at 60 Hz itself. A broadband source gives back the 14 dB the Ricker lacks there; the other 16 dB must come from quieter receivers or more fold, about forty times as much against random noise. Deeper targets cost more: at $t$ = 3 s and $Q$ = 100 the loss is 3 dB by about 4 Hz, 25 dB at 30 Hz and 49 dB at 60 Hz. The receivers must also respond flat and quiet across the band the target can still return, because a sensor that rolls off inside that band throws away frequencies the earth let through.

Inverse-$Q$ filtering in processing (Section 7.3 of the Processing textbook) inverts this filter: it boosts the attenuated high frequencies. But it boosts the noise recorded at those frequencies by the same gain, so it cannot recover a frequency that arrived under the noise, and it cannot recover energy the source never put in. Plan for $Q$ at the acquisition stage: estimate $t/Q$ for the deepest target, set the frequency the target needs, and design the source output and the noise budget (receivers and fold) to meet it.

References

• Aki, K., Richards, P. G. (2002). Quantitative Seismology (2nd ed.). University Science Books.
• Kjartansson, E. (1979). Constant Q-wave propagation and attenuation. Journal of Geophysical Research 84(B9), 4737-4748.
• Yilmaz, Ö. (2001). Seismic Data Analysis: Processing, Inversion, and Interpretation of Seismic Data (2 vols.). SEG Investigations in Geophysics 10.
• Sheriff, R. E., Geldart, L. P. (1995). Exploration Seismology (2nd ed.). Cambridge University Press.